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IB Maths AA · Concept clusters

IB Maths AA concept clusters

Original, past-paper-style synoptic questions. Each one is built around a single setup and needs ideas from several subchapters at once, so you practise connecting topics the way long exam questions expect. 64 questions · 1,609 marks · two per subchapter or merged group · every part with a worked solution · All courses

Topic

Subchapter

Level

Skill

Topic 1 · Number and algebra

● home subchapter · ○ linked subchapter · last column: links to other topics

Setup	1.1	1.2	1.3	1.4	1.5	1.6	1.7	1.8	1.9	1.10	1.11	1.12	1.13	1.14	1.15	1.16	Other
A new arena's tiered seating: rows, capacity and ticket income	○	●	○														2.1
A data centre's storage demand: geometric growth in standard form	○	○	●		○												—
A first car: a growing loan against a falling resale value			○	●	○		○										—
Earthquake magnitudes and the energy they release	○				○		●										2.9
Repeated medicine doses: the long-term level in the blood			○		○	○		●									—
A striker's penalty kicks and the binomial expansion						○			●								4.8 4.5
How secure is a password? Counting arrangements	○						○		○	●							4.5
An audio filter's gain: partial fractions and series										○	●						5.15 2.13
Impedance in an AC circuit: complex numbers in polar and Euler form												○	●				3.5 3.10
A drone light show: roots of a complex number												○	○	●			2.12 3.2
Cabling a server room: induction, contradiction and counterexample		○				○				○					●		—
A smoothie bar's recipe: three ingredients, three targets																●	3.17 3.18 2.1

1.1 Standard form · 1.2 Arithmetic sequences and series · 1.3 Geometric sequences and seriesmerged group

+ Add to test✓ In your testCC-AA-11-12-13-122 marks~25 min

A new arena's tiered seating: rows, capacity and ticket income

1.2 Arithmetic sequences and series1.1 Standard form2.1 Equations of straight lines, parallel and perpendicular lines1.3 Geometric sequences and seriesSL

The lower tier of seating in a new indoor arena has 40 rows. Row 1, nearest the court, has 64 seats, and every row after that has 6 more seats than the row in front of it.

Let \(u_n\) be the number of seats in row \(n\) of the lower tier. Figure 1 shows \(u_n\) for the first 10 rows.

Figure 1: the number of seats in each of the first 10 rows of the lower tier

(a)

Find the number of seats in row 25.

No calculatorROUTINENotes

Show solution

\(u_{25}=64+(25-1)\times6\)	M1	Substitution into \(u_n=u_1+(n-1)d\) with \(u_1=64,\ d=6\).
208	A1

Answer: 208 seats

(2)

(b)

Find the total number of seats in the lower tier.

No calculatorROUTINENotes

Show solution

\(S_{40}=\frac{40}{2}\left(2\times64+39\times6\right)\)	M1	Or \(u_{40}=298\) and \(S_{40}=\frac{40}{2}(64+298)\).
\(7240\)	A1

Answer: 7240 seats

(2)

(c)

Show that the total number of seats in the first \(n\) rows is \(3n^{2}+61n\).

No calculatorPROOFNotes

Show solution

\(S_n=\frac{n}{2}\left(2\times64+(n-1)\times6\right)\)	M1	Correct substitution into the sum formula.
\(=\frac{n}{2}(6n+122)\) or \(=64n+3n^{2}-3n\)	A1	A correct intermediate simplified form must be seen.
\(S_n=3n^{2}+61n\)	AG

(2)

(d)

For a concert, 6000 tickets are sold for the lower tier. The spectators are seated from row 1 backwards, and each row is filled completely before the next row is used.

Find the number of rows that are needed to seat all 6000 spectators.

MULTINotes

Show solution

\(3n^{2}+61n\ge6000\)	M1	Accept an equation \(3n^{2}+61n=6000\), or a table of values of \(S_n\).
\(n=35.6957\ldots\) or \(S_{35}=5810\) and \(S_{36}=6084\)	A1	Ignore the negative root \(n=-56.03\).
36 rows	A1	Must be a whole number. 35 or 35.7 scores A0.

Answer: 36 rows

(3)

(e)(i)

The points in Figure 1 lie on a straight line with equation \(y=mx+c\).

Find the value of \(m\) and the value of \(c\).

No calculatorGRAPHNotes

Show solution

gradient from two points, e.g. \(m=\frac{118-64}{10-1}\), or \(u_n=64+6(n-1)\)	M1
\(m=6,\ c=58\)	A1	Accept \(y=6x+58\).

Answer: \(m=6,\ c=58\)

(2)

(e)(ii)

Interpret, in context, the value of \(m\). Explain why the point \((0,\ c)\) is not part of the graph of \(u_n\).

No calculatorINTERPRETNotes

Show solution

\(m\) is the number of extra seats in each row compared with the row in front of it	R1	Accept "the common difference: each row has 6 more seats".
there is no row 0: \(n\) must be a positive integer, so 58 is not the number of seats in any row	R1	Accept a reason based on the graph being separate points with \(n\ge1\).

(2)

(f)

Every seat in the lower tier is sold for each of the 38 home matches of the season. The mean price of a lower-tier ticket is €46.50.

Find the total income from lower-tier tickets for the season. Give your answer in the form \(a\times10^{k}\), where \(1\le a\lt10\) and \(k\in\mathbb{Z}\), to three significant figures.

ROUTINENotes

Show solution

\(38\times7240\times46.50\ (=12\,793\,080)\)	M1	Allow ECF from (b).
€\(1.28\times10^{7}\)	A1	Award A0 for 12 800 000 or 1.28E7 written in calculator style.

Answer: €\(1.28\times10^{7}\)

(2)

(g)(i)

The upper tier will have 50 rows, and row 1 of the upper tier will also have 64 seats. Two patterns are being compared.

Pattern A: each row has 6 more seats than the row in front, so row \(n\) has \(u_n\) seats.
Pattern B: each row has 4% more seats than the row in front, so row \(n\) has \(v_n=64\times1.04^{\,n-1}\) seats, rounded to the nearest whole number.

Find the number of seats in row 25 for pattern B.

ROUTINENotes

Show solution

\(64\times1.04^{24}\ (=164.05\ldots)\)	M1	Exponent \(25\) (giving 170.6) scores M0.
164	A1

Answer: 164 seats

(2)

(g)(ii)

Find the first row in which pattern B has more seats than pattern A.

MULTINotes

Show solution

attempt to compare \(64\times1.04^{\,n-1}\) with \(6n+58\), e.g. a table of values, graphs or an equation solver	M1	Rounded or unrounded values of \(v_n\) may be used.
\(v_{40}=295\lt u_{40}=298\) and \(v_{41}=307\gt u_{41}=304\), or intersection at \(n=40.4\ldots\)	A1	Both rows 40 and 41 must be seen if a table is used.
row 41	A1

Answer: row 41

(3)

(g)(iii)

In rows 2 to 40, pattern A has more seats than pattern B. Explain why pattern B grows faster than pattern A for large values of \(n\).

No calculatorINTERPRETNotes

Show solution

in pattern A the increase from one row to the next is always the same, 6 seats	R1
in pattern B the increase is 4% of the previous row, which gets larger as the rows get longer; it is more than 6 once a row has more than 150 seats, so B eventually overtakes A	R1	Accept: geometric (exponential) growth with \(r\gt1\) eventually exceeds any arithmetic (linear) growth, with a reason.

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(u_{25}=64+(25-1)\times6\)	M1	Substitution into \(u_n=u_1+(n-1)d\) with \(u_1=64,\ d=6\).
208	A1
Part (b)
\(S_{40}=\frac{40}{2}\left(2\times64+39\times6\right)\)	M1	Or \(u_{40}=298\) and \(S_{40}=\frac{40}{2}(64+298)\).
\(7240\)	A1
Part (c)
\(S_n=\frac{n}{2}\left(2\times64+(n-1)\times6\right)\)	M1	Correct substitution into the sum formula.
\(=\frac{n}{2}(6n+122)\) or \(=64n+3n^{2}-3n\)	A1	A correct intermediate simplified form must be seen.
\(S_n=3n^{2}+61n\)	AG
Part (d)
\(3n^{2}+61n\ge6000\)	M1	Accept an equation \(3n^{2}+61n=6000\), or a table of values of \(S_n\).
\(n=35.6957\ldots\) or \(S_{35}=5810\) and \(S_{36}=6084\)	A1	Ignore the negative root \(n=-56.03\).
36 rows	A1	Must be a whole number. 35 or 35.7 scores A0.
Part (e)(i)
gradient from two points, e.g. \(m=\frac{118-64}{10-1}\), or \(u_n=64+6(n-1)\)	M1
\(m=6,\ c=58\)	A1	Accept \(y=6x+58\).
Part (e)(ii)
\(m\) is the number of extra seats in each row compared with the row in front of it	R1	Accept "the common difference: each row has 6 more seats".
there is no row 0: \(n\) must be a positive integer, so 58 is not the number of seats in any row	R1	Accept a reason based on the graph being separate points with \(n\ge1\).
Part (f)
\(38\times7240\times46.50\ (=12\,793\,080)\)	M1	Allow ECF from (b).
€\(1.28\times10^{7}\)	A1	Award A0 for 12 800 000 or 1.28E7 written in calculator style.
Part (g)(i)
\(64\times1.04^{24}\ (=164.05\ldots)\)	M1	Exponent \(25\) (giving 170.6) scores M0.
164	A1
Part (g)(ii)
attempt to compare \(64\times1.04^{\,n-1}\) with \(6n+58\), e.g. a table of values, graphs or an equation solver	M1	Rounded or unrounded values of \(v_n\) may be used.
\(v_{40}=295\lt u_{40}=298\) and \(v_{41}=307\gt u_{41}=304\), or intersection at \(n=40.4\ldots\)	A1	Both rows 40 and 41 must be seen if a table is used.
row 41	A1
Part (g)(iii)
in pattern A the increase from one row to the next is always the same, 6 seats	R1
in pattern B the increase is 4% of the previous row, which gets larger as the rows get longer; it is more than 6 once a row has more than 150 seats, so B eventually overtakes A	R1	Accept: geometric (exponential) growth with \(r\gt1\) eventually exceeds any arithmetic (linear) growth, with a reason.

Answers: (a) 208 seats · (b) 7240 seats · (d) 36 rows · (e)(i) \(m=6,\ c=58\) · (f) €\(1.28\times10^{7}\) · (g)(i) 164 seats · (g)(ii) row 41

Common error (d) Rounding \(n=35.7\) down to 35 gives only 5810 seats, which is not enough. When a total must reach a target, take the next whole number up and check it with the sum formula.

Exam tip (g)(iii) Comparing an arithmetic and a geometric sequence, look at the step between terms: \(d\) is fixed, but \(u_{n+1}-u_n=(r-1)u_n\) grows with \(u_n\).

More 1.2 questions in the question bank

+ Add to test✓ In your testCC-AA-11-12-13-221 marks~25 min

A data centre's storage demand: geometric growth in standard form

1.3 Geometric sequences and series1.1 Standard form1.2 Arithmetic sequences and series1.5 Exponents and introduction to logarithmsSL

A data centre records the amount of new data written to its storage in each quarter (three-month period). In quarter 1, \(2.4\times10^{15}\) bytes of new data were written.

A model assumes that the amount of new data written in each quarter is 9% more than in the quarter before. Let \(D_n\) be the amount of new data, in bytes, written in quarter \(n\).

(a)

Find \(D_5\). Give your answer in the form \(a\times10^{k}\), where \(1\le a\lt10\) and \(k\in\mathbb{Z}\), to three significant figures.

ROUTINENotes

Show solution

\(2.4\times10^{15}\times1.09^{4}\)	M1	Ratio \(r=1.09\) and exponent \(n-1=4\).
\(3.39\times10^{15}\) bytes	A1

Answer: \(3.39\times10^{15}\) bytes

(2)

(b)

Find the total amount of new data written in the first 12 quarters.

ROUTINENotes

Show solution

\(S_{12}=\frac{2.4\times10^{15}\left(1.09^{12}-1\right)}{1.09-1}\)	M1
\(4.83\times10^{16}\) bytes	A1	Accept \(4.8338\ldots\times10^{16}\).

Answer: \(4.83\times10^{16}\) bytes

(2)

(c)

The model can also be written as \(D_n=2.4\times10^{15}\,e^{k(n-1)}\), where \(e^{k}=1.09\).

Show that \(D_n\gt10^{16}\) when \(n-1\gt\dfrac{\ln\left(\frac{25}{6}\right)}{\ln1.09}\).

No calculatorPROOFNotes

Show solution

\(2.4\times10^{15}\,e^{k(n-1)}\gt10^{16}\)	M1	Or \(2.4\times10^{15}\times1.09^{\,n-1}\gt10^{16}\).
\(e^{k(n-1)}\gt\frac{10^{16}}{2.4\times10^{15}}=\frac{10}{2.4}=\frac{25}{6}\)	A1	The step \(\frac{10}{2.4}=\frac{25}{6}\) must be seen.
\(k(n-1)\gt\ln\frac{25}{6}\), with \(k=\ln1.09\) because \(e^{k}=1.09\)	M1	Taking \(\ln\) of both sides and using \(k=\ln1.09\). Accept \((n-1)\ln1.09\gt\ln\frac{25}{6}\) from the laws of logarithms.
\(n-1\gt\frac{\ln\left(\frac{25}{6}\right)}{\ln1.09}\) (since \(\ln1.09\gt0\))	AG

(3)

(d)

Hence find the first quarter in which more than \(10^{16}\) bytes of new data are written.

ROUTINENotes

Show solution

\(n-1\gt16.56\ldots\) or \(n\gt17.56\ldots\)	(A1)	Or \(D_{17}=9.53\times10^{15}\) and \(D_{18}=1.04\times10^{16}\).
quarter 18	A1	Answer 17 scores A0.

Answer: quarter 18

(2)

(e)(i)

When the data centre was built, its owners planned the storage capacity for new data, \(C_n\) bytes, in quarter \(n\). The plan allows for \(4.0\times10^{15}\) bytes in quarter 1, increasing by \(3.0\times10^{14}\) bytes each quarter.

Show that \(C_n=(0.3n+3.7)\times10^{15}\).

No calculatorPROOFNotes

Show solution

\(C_n=4.0\times10^{15}+(n-1)\times3.0\times10^{14}\)	M1
\(=4.0\times10^{15}+0.3(n-1)\times10^{15}\)	A1	Writing \(3.0\times10^{14}=0.3\times10^{15}\) must be seen.
\(C_n=(0.3n+3.7)\times10^{15}\)	AG

(2)

(e)(ii)

On the axes in Figure 1, sketch curves through the values of \(D_n\) and of \(C_n\) for \(1\le n\le20\). Label each curve.

Figure 1: axes for the sketch in (e)(ii)

GRAPHNotes

Show solution

\(D_n\): increasing curve, gradient increasing, from about \((1,\ 2.4)\) to about \((20,\ 12.3)\)	A1	Points or a smooth curve through them are both acceptable.
\(C_n\): straight line from about \((1,\ 4.0)\) to about \((20,\ 9.7)\)	A1
curves cross between \(n=15\) and \(n=16\), with \(D_n\) above \(C_n\) after the crossing; both curves labelled	A1

(3)

(e)(iii)

Find the first quarter in which the new data written is greater than the planned capacity. Interpret your answer in context.

INTERPRETNotes

Show solution

attempt to compare \(D_n\) with \(C_n\), e.g. \(D_{15}=8.02\times10^{15}\lt C_{15}=8.2\times10^{15}\)	M1
quarter 16 (\(D_{16}=8.74\times10^{15}\gt C_{16}=8.5\times10^{15}\))	A1
from quarter 16 onwards the model predicts more new data than the plan can store, so extra capacity is needed before then	R1

Answer: quarter 16

(3)

(f)(i)

Newer figures suggest that the amount of new data increases by only 6% each quarter, starting from \(2.4\times10^{15}\) bytes in quarter 1. The capacity plan does not change.

Find the first quarter in which the new data written is greater than the planned capacity, using this new model.

MODELNotes

Show solution

\(2.4\times1.06^{\,n-1}\gt0.3n+3.7\) (in units of \(10^{15}\) bytes), or a table of values	M1	Values \(12.27\lt12.4\) for \(n=29\) and \(13.00\gt12.7\) for \(n=30\) may be seen.
quarter 30	A1

Answer: quarter 30

(2)

(f)(ii)

Comment on the reliability of using either model to predict the amount of new data written in quarter 40, the last quarter of the tenth year.

No calculatorINTERPRETNotes

Show solution

the predictions are very sensitive to the growth rate: the crossover moves from quarter 16 to quarter 30, and \(D_{40}\) is about \(6.9\times10^{16}\) at 9% but about \(2.3\times10^{16}\) at 6%	R1	Any valid numerical comparison of the two models.
quarter 40 is a long way beyond the data the models are based on (a long extrapolation), and a constant percentage growth is unlikely to continue for ten years, so the predictions are not reliable	R1

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(2.4\times10^{15}\times1.09^{4}\)	M1	Ratio \(r=1.09\) and exponent \(n-1=4\).
\(3.39\times10^{15}\) bytes	A1
Part (b)
\(S_{12}=\frac{2.4\times10^{15}\left(1.09^{12}-1\right)}{1.09-1}\)	M1
\(4.83\times10^{16}\) bytes	A1	Accept \(4.8338\ldots\times10^{16}\).
Part (c)
\(2.4\times10^{15}\,e^{k(n-1)}\gt10^{16}\)	M1	Or \(2.4\times10^{15}\times1.09^{\,n-1}\gt10^{16}\).
\(e^{k(n-1)}\gt\frac{10^{16}}{2.4\times10^{15}}=\frac{10}{2.4}=\frac{25}{6}\)	A1	The step \(\frac{10}{2.4}=\frac{25}{6}\) must be seen.
\(k(n-1)\gt\ln\frac{25}{6}\), with \(k=\ln1.09\) because \(e^{k}=1.09\)	M1	Taking \(\ln\) of both sides and using \(k=\ln1.09\). Accept \((n-1)\ln1.09\gt\ln\frac{25}{6}\) from the laws of logarithms.
\(n-1\gt\frac{\ln\left(\frac{25}{6}\right)}{\ln1.09}\) (since \(\ln1.09\gt0\))	AG
Part (d)
\(n-1\gt16.56\ldots\) or \(n\gt17.56\ldots\)	(A1)	Or \(D_{17}=9.53\times10^{15}\) and \(D_{18}=1.04\times10^{16}\).
quarter 18	A1	Answer 17 scores A0.
Part (e)(i)
\(C_n=4.0\times10^{15}+(n-1)\times3.0\times10^{14}\)	M1
\(=4.0\times10^{15}+0.3(n-1)\times10^{15}\)	A1	Writing \(3.0\times10^{14}=0.3\times10^{15}\) must be seen.
\(C_n=(0.3n+3.7)\times10^{15}\)	AG
Part (e)(ii)
\(D_n\): increasing curve, gradient increasing, from about \((1,\ 2.4)\) to about \((20,\ 12.3)\)	A1	Points or a smooth curve through them are both acceptable.
\(C_n\): straight line from about \((1,\ 4.0)\) to about \((20,\ 9.7)\)	A1
curves cross between \(n=15\) and \(n=16\), with \(D_n\) above \(C_n\) after the crossing; both curves labelled	A1
Part (e)(iii)
attempt to compare \(D_n\) with \(C_n\), e.g. \(D_{15}=8.02\times10^{15}\lt C_{15}=8.2\times10^{15}\)	M1
quarter 16 (\(D_{16}=8.74\times10^{15}\gt C_{16}=8.5\times10^{15}\))	A1
from quarter 16 onwards the model predicts more new data than the plan can store, so extra capacity is needed before then	R1
Part (f)(i)
\(2.4\times1.06^{\,n-1}\gt0.3n+3.7\) (in units of \(10^{15}\) bytes), or a table of values	M1	Values \(12.27\lt12.4\) for \(n=29\) and \(13.00\gt12.7\) for \(n=30\) may be seen.
quarter 30	A1
Part (f)(ii)
the predictions are very sensitive to the growth rate: the crossover moves from quarter 16 to quarter 30, and \(D_{40}\) is about \(6.9\times10^{16}\) at 9% but about \(2.3\times10^{16}\) at 6%	R1	Any valid numerical comparison of the two models.
quarter 40 is a long way beyond the data the models are based on (a long extrapolation), and a constant percentage growth is unlikely to continue for ten years, so the predictions are not reliable	R1

Answers: (a) \(3.39\times10^{15}\) bytes · (b) \(4.83\times10^{16}\) bytes · (d) quarter 18 · (e)(iii) quarter 16 · (f)(i) quarter 30

Common error (a) Using \(1.09^{5}\) gives the amount in quarter 6. Quarter 1 has had no increases, so quarter 5 has had four.

More 1.3 questions in the question bank

1.4 Financial applications · 1.5 Exponents and introduction to logarithms · 1.7 Rational exponents and laws of logarithmsmerged group

+ Add to test✓ In your testCC-AA-14-15-17-124 marks~30 min

A first car: a growing loan against a falling resale value

1.4 Financial applications1.3 Geometric sequences and series1.5 Exponents and introduction to logarithms1.7 Rational exponents and laws of logarithmsSL

Nikos buys a new car for €18 500. He pays a deposit of 10% of the price and borrows the rest from his aunt. She charges interest at a nominal annual rate of 6.2%, compounded monthly. Nikos makes no payments during the loan: he repays everything he owes in a single payment at the end of 5 years.

(a)

Find the amount that Nikos must repay at the end of the 5 years.

ROUTINENotes

Show solution

amount borrowed \(=18\,500\times0.9=16\,650\)	(A1)
\(FV=16\,650\left(1+\frac{6.2}{1200}\right)^{60}\)	M1	Or finance solver: \(N=60,\ I\%=6.2,\ PV=-16\,650,\ PMT=0,\ P/Y=C/Y=12\). Exponent or \(N\) of 5 scores M0.
€22 682.92	A1	Accept 22 700.

Answer: €22 682.92

(3)

(b)

Find the total interest that Nikos pays on the loan.

ROUTINENotes

Show solution

\(22\,682.92-16\,650\)	M1	Allow ECF from (a).
€6032.92	A1	Accept 6030.

Answer: €6032.92

(2)

(c)(i)

The car loses 20% of its value each year, so its value after \(t\) years is modelled by \(V=18\,500\times0.8^{t}\) euros.

Show that the value of the car is less than €9000 when \(t\gt\dfrac{\log37-\log18}{1-3\log2}\).

No calculatorPROOFNotes

Show solution

\(18\,500\times0.8^{t}\lt9000\Rightarrow0.8^{t}\lt\frac{9000}{18\,500}=\frac{18}{37}\)	M1
\(t\log0.8\lt\log18-\log37\)	M1	Taking logarithms of both sides and using the power and quotient laws.
\(\log0.8=\log8-\log10=3\log2-1\)	A1
\(3\log2-1\lt0\), so the inequality reverses when dividing	R1	The reason for the change of direction must be stated.
\(t\gt\frac{\log18-\log37}{3\log2-1}=\frac{\log37-\log18}{1-3\log2}\)	AG

(4)

(c)(ii)

Hence find the number of complete years after which the value of the car is first less than €9000.

ROUTINENotes

Show solution

\(t\gt3.229\ldots\)	(A1)	Or \(V(3)=9472\) and \(V(4)=7577.60\).
4 years	A1

Answer: 4 years

(2)

(d)(i)

Figure 1 shows the amount Nikos owes and the modelled value of the car during the 5 years of the loan. Nikos is in negative equity when the amount he owes is greater than the value of the car.

Use Figure 1 to estimate the time at which Nikos first owes more than the car is worth. Hence write down the period during which he is in negative equity.

Figure 1: the amount owed (blue) and the modelled value of the car (red) during the 5 years of the loan

GRAPHNotes

Show solution

\(t\approx0.4\) years	A1	Accept 0.3 to 0.45 (about 4 to 5 months).
from about 0.4 years until the loan is repaid at \(t=5\)	A1	Allow ECF from their crossing time.

Answer: first at about 0.4 years; negative equity from about 0.4 years to 5 years

(2)

(d)(ii)

Explain why, once Nikos is in negative equity, he stays in negative equity for the rest of the loan.

No calculatorINTERPRETNotes

Show solution

the amount owed is multiplied by \(1+\frac{6.2}{1200}\gt1\) each month, so it keeps increasing	R1	Accept: geometric growth, ratio greater than 1.
the value of the car is multiplied by \(0.8\lt1\) each year, so it keeps decreasing; the gap between them can only widen	R1	Accept a valid reference to the shapes of the two curves in Figure 1.

(2)

(e)(i)

At the end of the first 2 years, the nominal annual interest rate rises to 7.5%, compounded monthly, for the remaining 3 years. Nikos still makes no payments until the end of the 5 years.

Find the amount Nikos owes at the end of the first 2 years.

ROUTINENotes

Show solution

\(16\,650\left(1+\frac{6.2}{1200}\right)^{24}\)	M1	Or finance solver with \(N=24,\ I\%=6.2,\ PV=-16\,650,\ PMT=0,\ P/Y=C/Y=12\).
€18 842.05	A1	Accept 18 800.

Answer: €18 842.05

(2)

(e)(ii)

Find the amount Nikos must now repay at the end of the 5 years.

MULTINotes

Show solution

\(18\,842.05\left(1+\frac{7.5}{1200}\right)^{36}\)	M1	Allow ECF from (e)(i). Or finance solver with \(N=36,\ I\%=7.5,\ PV=-18\,842.05,\ PMT=0,\ P/Y=C/Y=12\).
€23 579.81	A1	Accept 23 600.

Answer: €23 579.81

(2)

(e)(iii)

Find how much more Nikos pays in total because of the rise in the interest rate.

ROUTINENotes

Show solution

\(23\,579.81-22\,682.92\)	M1	Allow ECF from (a) and (e)(ii).
€896.89	A1	Accept 897.

Answer: €896.89

(2)

(f)

The interest rate rises as in part (e). During the 5 years of the loan, prices rise by 3% each year.

Find the value, in today's money, of Nikos's final repayment. Hence explain how inflation affects the real cost of the loan to Nikos.

INTERPRETNotes

Show solution

\(\frac{23\,579.81}{1.03^{5}}\)	M1	Allow ECF from (e)(ii).
€20 340.15	A1	Accept 20 300.
in today's money he repays about €20 340, so the real cost of borrowing is about €3690, much less than the €6930 of interest he pays in euros: inflation reduces the real cost of the loan	R1	Accept: inflation reduces the real value of the repayment, which helps the borrower.

Answer: €20 340.15

(3)

Show full solution

Working	Mark	Notes
Part (a)
amount borrowed \(=18\,500\times0.9=16\,650\)	(A1)
\(FV=16\,650\left(1+\frac{6.2}{1200}\right)^{60}\)	M1	Or finance solver: \(N=60,\ I\%=6.2,\ PV=-16\,650,\ PMT=0,\ P/Y=C/Y=12\). Exponent or \(N\) of 5 scores M0.
€22 682.92	A1	Accept 22 700.
Part (b)
\(22\,682.92-16\,650\)	M1	Allow ECF from (a).
€6032.92	A1	Accept 6030.
Part (c)(i)
\(18\,500\times0.8^{t}\lt9000\Rightarrow0.8^{t}\lt\frac{9000}{18\,500}=\frac{18}{37}\)	M1
\(t\log0.8\lt\log18-\log37\)	M1	Taking logarithms of both sides and using the power and quotient laws.
\(\log0.8=\log8-\log10=3\log2-1\)	A1
\(3\log2-1\lt0\), so the inequality reverses when dividing	R1	The reason for the change of direction must be stated.
\(t\gt\frac{\log18-\log37}{3\log2-1}=\frac{\log37-\log18}{1-3\log2}\)	AG
Part (c)(ii)
\(t\gt3.229\ldots\)	(A1)	Or \(V(3)=9472\) and \(V(4)=7577.60\).
4 years	A1
Part (d)(i)
\(t\approx0.4\) years	A1	Accept 0.3 to 0.45 (about 4 to 5 months).
from about 0.4 years until the loan is repaid at \(t=5\)	A1	Allow ECF from their crossing time.
Part (d)(ii)
the amount owed is multiplied by \(1+\frac{6.2}{1200}\gt1\) each month, so it keeps increasing	R1	Accept: geometric growth, ratio greater than 1.
the value of the car is multiplied by \(0.8\lt1\) each year, so it keeps decreasing; the gap between them can only widen	R1	Accept a valid reference to the shapes of the two curves in Figure 1.
Part (e)(i)
\(16\,650\left(1+\frac{6.2}{1200}\right)^{24}\)	M1	Or finance solver with \(N=24,\ I\%=6.2,\ PV=-16\,650,\ PMT=0,\ P/Y=C/Y=12\).
€18 842.05	A1	Accept 18 800.
Part (e)(ii)
\(18\,842.05\left(1+\frac{7.5}{1200}\right)^{36}\)	M1	Allow ECF from (e)(i). Or finance solver with \(N=36,\ I\%=7.5,\ PV=-18\,842.05,\ PMT=0,\ P/Y=C/Y=12\).
€23 579.81	A1	Accept 23 600.
Part (e)(iii)
\(23\,579.81-22\,682.92\)	M1	Allow ECF from (a) and (e)(ii).
€896.89	A1	Accept 897.
Part (f)
\(\frac{23\,579.81}{1.03^{5}}\)	M1	Allow ECF from (e)(ii).
€20 340.15	A1	Accept 20 300.
in today's money he repays about €20 340, so the real cost of borrowing is about €3690, much less than the €6930 of interest he pays in euros: inflation reduces the real cost of the loan	R1	Accept: inflation reduces the real value of the repayment, which helps the borrower.

Answers: (a) €22 682.92 · (b) €6032.92 · (c)(ii) 4 years · (d)(i) first at about 0.4 years; negative equity from about 0.4 years to 5 years · (e)(i) €18 842.05 · (e)(ii) €23 579.81 · (e)(iii) €896.89 · (f) €20 340.15

Exam tip (a) With monthly compounding, both the rate and the number of periods change: the rate per month is \(\frac{6.2}{12}\%\) and there are \(12\times5=60\) periods. If you use the finance solver, write down every value you enter.

Common error (c)(i) Dividing by \(\log0.8\) without reversing the inequality. The logarithm of a number between 0 and 1 is negative.

More 1.4 questions in the question bank

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Earthquake magnitudes and the energy they release

1.7 Rational exponents and laws of logarithms1.5 Exponents and introduction to logarithms1.1 Standard form2.9 Exponential and logarithmic functionsSL

The energy \(E\) joules released by an earthquake of magnitude \(M\) can be modelled by

\[\log E=4.8+1.5M,\]

where \(\log\) is the logarithm to base 10. Figure 1 shows the graph of \(\log E\) against \(M\).

Figure 1: the graph of log E against M for 3 ≤ M ≤ 9

(a)

Calculate the energy released by an earthquake of magnitude 6.0. Give your answer in the form \(a\times10^{k}\), where \(1\le a\lt10\) and \(k\in\mathbb{Z}\), to three significant figures.

ROUTINENotes

Show solution

\(E=10^{4.8+1.5\times6.0}=10^{13.8}\)	M1
\(6.31\times10^{13}\ \text{J}\)	A1	Award A0 for calculator notation such as 6.31E13.

Answer: \(6.31\times10^{13}\) J

(2)

(b)

Find the magnitude of an earthquake that releases \(4.5\times10^{13}\) J.

ROUTINENotes

Show solution

\(4.8+1.5M=\log\left(4.5\times10^{13}\right)\ (=13.653\ldots)\)	M1
\(M=5.90\)	A1	Accept 5.9 or 5.902.

Answer: \(M=5.90\)

(2)

(c)

Show that \(E=10^{4.8}\times1000^{\frac{M}{2}}\).

No calculatorPROOFNotes

Show solution

\(E=10^{4.8+1.5M}=10^{4.8}\times10^{1.5M}\)	M1
\(10^{1.5M}=10^{3\times\frac{M}{2}}=\left(10^{3}\right)^{\frac{M}{2}}\)	A1
\(E=10^{4.8}\times1000^{\frac{M}{2}}\)	AG

(2)

(d)

Hence show that when the magnitude increases by 2, the energy released is multiplied by 1000.

No calculatorPROOFNotes

Show solution

\(\frac{10^{4.8}\times1000^{\frac{M+2}{2}}}{10^{4.8}\times1000^{\frac{M}{2}}}\)	M1
\(=1000^{\frac{M+2}{2}-\frac{M}{2}}=1000^{1}\)	A1
the energy is multiplied by 1000	AG

(2)

(e)

Show that an earthquake of magnitude 6.0 releases \(10\sqrt{10}\) times as much energy as an earthquake of magnitude 5.0.

No calculatorPROOFNotes

Show solution

\(\frac{E_6}{E_5}=10^{1.5\times6.0-1.5\times5.0}=10^{1.5}\)	M1	Or \(1000^{\frac12}\) from (c).
\(10^{\frac32}=10^{1}\times10^{\frac12}=10\sqrt{10}\)	A1
\(10\sqrt{10}\)	AG

(2)

(f)

Earthquake B releases 8 times as much energy as earthquake A. Their magnitudes are \(M_B\) and \(M_A\).

Show that \(M_B-M_A=2\log2\).

No calculatorPROOFNotes

Show solution

\(\log E_B=\log(8E_A)=\log8+\log E_A\)	M1	Product law.
\(4.8+1.5M_B=3\log2+4.8+1.5M_A\)	M1	Power law \(\log8=3\log2\) and substitution of the model.
\(1.5(M_B-M_A)=3\log2\)	A1
\(M_B-M_A=2\log2\)	AG

(3)

(g)(i)

On the axes in Figure 2, sketch the graph of \(E\) against \(M\) for \(6\le M\le7\). Label the end points of your graph with their coordinates.

Figure 2: axes for the sketch in (g)(i)

GRAPHNotes

Show solution

increasing curve with increasing gradient	A1	A straight line scores A0.
end point \(\left(6,\ 6.31\times10^{13}\right)\), plotted at about 0.06 on the vertical scale	A1	Allow ECF from (a).
end point \(\left(7,\ 2.00\times10^{15}\right)\)	A1	Accept \(10^{15.3}\).

(3)

(g)(ii)

With reference to Figures 1 and 2, explain why earthquakes are usually compared using magnitudes, or \(\log E\), rather than the energy \(E\).

No calculatorINTERPRETNotes

Show solution

the values of \(E\) cover many powers of ten, so on a linear scale smaller earthquakes are hard to tell apart (the point for \(M=6\) is almost on the axis in Figure 2)	R1
\(\log E\) is a linear function of \(M\), so each increase of 1 in \(M\) adds the same amount, 1.5, to \(\log E\), and values are easy to read and compare (Figure 1)	R1

(2)

(h)(i)

For shallow earthquakes, a researcher proposes the modified model \(\log E=4.8+1.4M\).

Find the factor by which the energy is multiplied when the magnitude increases by 2, using the modified model.

MODELNotes

Show solution

\(10^{1.4\times2}=10^{2.8}\)	M1
631	A1	Accept 630.96.

Answer: 631

(2)

(h)(ii)

Comment on how sensitive energy comparisons are to the value of the coefficient of \(M\).

No calculatorINTERPRETNotes

Show solution

the coefficient changes by only 0.1 (about 7%), but the energy factor falls from 1000 to 631 (about 37%)	R1	Allow ECF from (h)(i).
the coefficient is in the exponent (the factor is \(10^{2b}\)), so a small change in it causes a large change in energy ratios: comparisons are very sensitive to it	R1

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(E=10^{4.8+1.5\times6.0}=10^{13.8}\)	M1
\(6.31\times10^{13}\ \text{J}\)	A1	Award A0 for calculator notation such as 6.31E13.
Part (b)
\(4.8+1.5M=\log\left(4.5\times10^{13}\right)\ (=13.653\ldots)\)	M1
\(M=5.90\)	A1	Accept 5.9 or 5.902.
Part (c)
\(E=10^{4.8+1.5M}=10^{4.8}\times10^{1.5M}\)	M1
\(10^{1.5M}=10^{3\times\frac{M}{2}}=\left(10^{3}\right)^{\frac{M}{2}}\)	A1
\(E=10^{4.8}\times1000^{\frac{M}{2}}\)	AG
Part (d)
\(\frac{10^{4.8}\times1000^{\frac{M+2}{2}}}{10^{4.8}\times1000^{\frac{M}{2}}}\)	M1
\(=1000^{\frac{M+2}{2}-\frac{M}{2}}=1000^{1}\)	A1
the energy is multiplied by 1000	AG
Part (e)
\(\frac{E_6}{E_5}=10^{1.5\times6.0-1.5\times5.0}=10^{1.5}\)	M1	Or \(1000^{\frac12}\) from (c).
\(10^{\frac32}=10^{1}\times10^{\frac12}=10\sqrt{10}\)	A1
\(10\sqrt{10}\)	AG
Part (f)
\(\log E_B=\log(8E_A)=\log8+\log E_A\)	M1	Product law.
\(4.8+1.5M_B=3\log2+4.8+1.5M_A\)	M1	Power law \(\log8=3\log2\) and substitution of the model.
\(1.5(M_B-M_A)=3\log2\)	A1
\(M_B-M_A=2\log2\)	AG
Part (g)(i)
increasing curve with increasing gradient	A1	A straight line scores A0.
end point \(\left(6,\ 6.31\times10^{13}\right)\), plotted at about 0.06 on the vertical scale	A1	Allow ECF from (a).
end point \(\left(7,\ 2.00\times10^{15}\right)\)	A1	Accept \(10^{15.3}\).
Part (g)(ii)
the values of \(E\) cover many powers of ten, so on a linear scale smaller earthquakes are hard to tell apart (the point for \(M=6\) is almost on the axis in Figure 2)	R1
\(\log E\) is a linear function of \(M\), so each increase of 1 in \(M\) adds the same amount, 1.5, to \(\log E\), and values are easy to read and compare (Figure 1)	R1
Part (h)(i)
\(10^{1.4\times2}=10^{2.8}\)	M1
631	A1	Accept 630.96.
Part (h)(ii)
the coefficient changes by only 0.1 (about 7%), but the energy factor falls from 1000 to 631 (about 37%)	R1	Allow ECF from (h)(i).
the coefficient is in the exponent (the factor is \(10^{2b}\)), so a small change in it causes a large change in energy ratios: comparisons are very sensitive to it	R1

Answers: (a) \(6.31\times10^{13}\) J · (b) \(M=5.90\) · (h)(i) 631

Common error (f) Writing \(\log(8E_A)=8\log E_A\). The product law gives \(\log8+\log E_A\); the power law only applies to \(\log\left(E_A^{8}\right)\).

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1.6 Deductive proof · 1.8 Infinite geometric series · 1.9 The binomial theoremmerged group

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Repeated medicine doses: the long-term level in the blood

1.8 Infinite geometric series1.3 Geometric sequences and series1.6 Deductive proof1.5 Exponents and introduction to logarithmsSL

A patient is prescribed tablets that each contain \(200\) mg of a drug. A tablet is absorbed into the body as soon as it is taken, and the amounts from different tablets add together.

Between tablets, the amount of the drug in the body, \(M\) mg, decreases according to \(M=M_0e^{-kt}\), where \(M_0\) is the amount just after the most recent tablet, \(t\) is the time in hours since that tablet and \(k\) is a positive constant.

Tests show that \(36\%\) of the drug present in the body at any moment is still present \(8\) hours later.

(a)

Find the value of \(k\), giving your answer correct to three significant figures.

ROUTINENotes

Show solution

\(e^{-8k}=0.36\)	M1	Or \(200e^{-8k}=72\).
\(k=-\frac{\ln0.36}{8}=0.128\)	A1	Accept \(0.127706\ldots\). Accept \(\frac18\ln\frac{25}{9}\) followed by \(0.128\).

Answer: \(k=0.128\) (3 s.f.)

(2)

(b)(i)

The patient takes one tablet every \(8\) hours, starting at \(t=0\). Let \(A_n\) mg be the amount of the drug in the body just after the \(n\)th tablet is taken, so \(A_1=200\).

Show that \(A_3=297.92\).

No calculatorROUTINENotes

Show solution

\(A_2=200+0.36\times200=272\)	M1	Award for multiplying by \(0.36\) and adding \(200\), or for \(200+200(0.36)+200(0.36)^2\).
\(A_3=200+0.36\times272=200+97.92\)	A1	Or \(200+72+25.92\).
\(A_3=297.92\)	AG

Answer: \(297.92\)

(2)

(b)(ii)

Explain why \(A_n\) is the sum of the first \(n\) terms of a geometric series, and hence find an expression for \(A_n\) in terms of \(n\).

No calculatorALGEBRANotes

Show solution

\(A_n=200+200(0.36)+\dots+200(0.36)^{n-1}\): each earlier tablet has been reduced by a further factor of \(0.36\) for every \(8\) hours since it was taken	R1	Must identify \(u_1=200\) and \(r=0.36\) (or show the terms).
\(A_n=\frac{200(1-0.36^n)}{1-0.36}\)	A1	Accept \(312.5(1-0.36^n)\) or \(\frac{625}{2}(1-0.36^n)\).

Answer: \(A_n=\dfrac{200(1-0.36^n)}{0.64}=312.5(1-0.36^n)\)

(2)

(c)

Show that, in the long term, the amount of the drug just after a tablet is taken approaches \(312.5\) mg.

No calculatorCONCEPTNotes

Show solution

\(\|r\|=0.36\lt1\), so the series converges (\(0.36^n\to0\))	R1	The condition must be stated.
\(S_\infty=\frac{200}{1-0.36}=\frac{200}{0.64}\)	M1
\(=312.5\)	AG

Answer: \(312.5\) mg

(2)

(d)(i)

Write down the long-term amount of the drug in the body just before a tablet is taken.

No calculatorMODELNotes

Show solution

\(0.36\times312.5=112.5\) mg

Allow ECF from (c).

Answer: \(112.5\) mg

(1)

(d)(ii)

On the axes in Figure 1, sketch the graph of the amount of the drug in the body for \(0\le t\le48\). Show the long-term level found in part (c) as a horizontal line.

Figure 1: axes for the amount of the drug in the body during the first 48 hours

GRAPHNotes

Show solution

vertical rises of \(200\) mg at \(t=0,\,8,\,16,\,24,\,32,\,40\) (six tablets)	A1	Peaks at approximately \(200,\ 272,\ 298,\ 307,\ 311,\ 312\). A seventh rise at \(t=48\) (to about \(312\)) may be shown or omitted.
decreasing curves between tablets, concave up (exponential decay), falling to \(72\) at \(t=8\) and to about \(98\) at \(t=16\)	A1	Straight-line segments score A0.
peaks increasing towards, but staying below, the line \(312.5\) (drawn); troughs approaching \(112.5\)	A1	The line at \(312.5\) must be drawn and labelled.

Answer: saw-tooth graph with peaks rising towards \(312.5\) mg

(3)

(e)

Consider any geometric series with first term \(u_1\) and common ratio \(r\), where \(|r|\lt1\). Let \(S_n\) be the sum of its first \(n\) terms and \(S_\infty\) its sum to infinity.

Prove that \(S_\infty-S_n=\dfrac{u_1r^n}{1-r}\).

No calculatorPROOFNotes

Show solution

\(S_\infty-S_n=\frac{u_1}{1-r}-\frac{u_1(1-r^n)}{1-r}\)	M1	Both formulae must be quoted correctly.
\(=\frac{u_1-u_1+u_1r^n}{1-r}\)	A1	Must show the numerator expanded over the common denominator.
\(=\frac{u_1r^n}{1-r}\), true for every such series	AG	Starting from the result and working backwards scores M0.

(2)

(f)

Hence find the least number of tablets that must be taken for the amount just after a tablet to be at least \(99\%\) of the long-term level.

MULTINotes

Show solution

\(312.5-A_n=312.5(0.36)^n\) and \(312.5(0.36)^n\le0.01\times312.5\)	M1	Or \(312.5(1-0.36^n)\ge309.375\). Allow ECF from (b)(ii) and (c).
\(0.36^n\le0.01\Rightarrow n\ge\frac{\ln0.01}{\ln0.36}=4.51\)	(A1)	Or GDC table: \(A_4=307.25\lt309.375\), \(A_5=310.61\ge309.375\).
\(n=5\) tablets	A1	Award A0 for \(4.51\) or for \(n=4\).

Answer: \(5\) tablets

(3)

(g)(i)

The doctor changes the schedule: the patient now takes one \(200\) mg tablet every \(4\) hours.

Show that \(60\%\) of the drug present in the body at any moment is still present \(4\) hours later.

No calculatorALGEBRANotes

Show solution

\(e^{-4k}=\left(e^{-8k}\right)^{\frac12}\)	M1	Or \(e^{-4k}\) with \(k=-\frac{\ln0.36}{8}\) gives \(e^{\frac12\ln0.36}\).
\(=0.36^{\frac12}=0.6\)	A1
so \(60\%\) remains	AG	Substituting \(k=0.128\) to get \(0.599\ldots\approx0.6\) scores M1 A0.

Answer: \(0.6\)

(2)

(g)(ii)

Find the long-term amount of the drug in the body just after a tablet is taken on this new schedule.

No calculatorROUTINENotes

Show solution

\(\frac{200}{1-0.6}\)	M1	Award only if \(r=0.6\) is used.
\(500\) mg	A1

Answer: \(500\) mg

(2)

(g)(iii)

The maximum safe amount of the drug in the body is \(450\) mg. On the new schedule, find which tablet first makes the amount in the body exceed \(450\) mg.

MODELNotes

Show solution

\(500(1-0.6^n)\gt450\Rightarrow0.6^n\lt0.1\Rightarrow n\gt4.51\)	(M1)	Or listing \(A_4=435.2\) and \(A_5=461.12\). Allow ECF from (g)(ii).
the \(5\)th tablet	A1	Accept: the tablet taken at \(t=16\).

Answer: the 5th tablet (\(A_5=461.12\) mg)

(2)

(g)(iv)

The doctor keeps the \(4\)-hour schedule but reduces the amount of drug in each tablet to \(D\) mg. Find the largest value of \(D\) for which the amount in the body never exceeds \(450\) mg. Justify your answer.

No calculatorINTERPRETNotes

Show solution

the amounts just after each tablet increase and approach \(\frac{D}{1-0.6}\) but never exceed it, so the condition is \(\frac{D}{0.4}\le450\)	R1	Must explain that the highest amounts are the peaks and that the peaks never pass the sum to infinity.
\(D=180\)	A1

Answer: \(D=180\)

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(e^{-8k}=0.36\)	M1	Or \(200e^{-8k}=72\).
\(k=-\frac{\ln0.36}{8}=0.128\)	A1	Accept \(0.127706\ldots\). Accept \(\frac18\ln\frac{25}{9}\) followed by \(0.128\).
Part (b)(i)
\(A_2=200+0.36\times200=272\)	M1	Award for multiplying by \(0.36\) and adding \(200\), or for \(200+200(0.36)+200(0.36)^2\).
\(A_3=200+0.36\times272=200+97.92\)	A1	Or \(200+72+25.92\).
\(A_3=297.92\)	AG
Part (b)(ii)
\(A_n=200+200(0.36)+\dots+200(0.36)^{n-1}\): each earlier tablet has been reduced by a further factor of \(0.36\) for every \(8\) hours since it was taken	R1	Must identify \(u_1=200\) and \(r=0.36\) (or show the terms).
\(A_n=\frac{200(1-0.36^n)}{1-0.36}\)	A1	Accept \(312.5(1-0.36^n)\) or \(\frac{625}{2}(1-0.36^n)\).
Part (c)
\(\|r\|=0.36\lt1\), so the series converges (\(0.36^n\to0\))	R1	The condition must be stated.
\(S_\infty=\frac{200}{1-0.36}=\frac{200}{0.64}\)	M1
\(=312.5\)	AG
Part (d)(i)
\(0.36\times312.5=112.5\) mg	A1	Allow ECF from (c).
Part (d)(ii)
vertical rises of \(200\) mg at \(t=0,\,8,\,16,\,24,\,32,\,40\) (six tablets)	A1	Peaks at approximately \(200,\ 272,\ 298,\ 307,\ 311,\ 312\). A seventh rise at \(t=48\) (to about \(312\)) may be shown or omitted.
decreasing curves between tablets, concave up (exponential decay), falling to \(72\) at \(t=8\) and to about \(98\) at \(t=16\)	A1	Straight-line segments score A0.
peaks increasing towards, but staying below, the line \(312.5\) (drawn); troughs approaching \(112.5\)	A1	The line at \(312.5\) must be drawn and labelled.
Part (e)
\(S_\infty-S_n=\frac{u_1}{1-r}-\frac{u_1(1-r^n)}{1-r}\)	M1	Both formulae must be quoted correctly.
\(=\frac{u_1-u_1+u_1r^n}{1-r}\)	A1	Must show the numerator expanded over the common denominator.
\(=\frac{u_1r^n}{1-r}\), true for every such series	AG	Starting from the result and working backwards scores M0.
Part (f)
\(312.5-A_n=312.5(0.36)^n\) and \(312.5(0.36)^n\le0.01\times312.5\)	M1	Or \(312.5(1-0.36^n)\ge309.375\). Allow ECF from (b)(ii) and (c).
\(0.36^n\le0.01\Rightarrow n\ge\frac{\ln0.01}{\ln0.36}=4.51\)	(A1)	Or GDC table: \(A_4=307.25\lt309.375\), \(A_5=310.61\ge309.375\).
\(n=5\) tablets	A1	Award A0 for \(4.51\) or for \(n=4\).
Part (g)(i)
\(e^{-4k}=\left(e^{-8k}\right)^{\frac12}\)	M1	Or \(e^{-4k}\) with \(k=-\frac{\ln0.36}{8}\) gives \(e^{\frac12\ln0.36}\).
\(=0.36^{\frac12}=0.6\)	A1
so \(60\%\) remains	AG	Substituting \(k=0.128\) to get \(0.599\ldots\approx0.6\) scores M1 A0.
Part (g)(ii)
\(\frac{200}{1-0.6}\)	M1	Award only if \(r=0.6\) is used.
\(500\) mg	A1
Part (g)(iii)
\(500(1-0.6^n)\gt450\Rightarrow0.6^n\lt0.1\Rightarrow n\gt4.51\)	(M1)	Or listing \(A_4=435.2\) and \(A_5=461.12\). Allow ECF from (g)(ii).
the \(5\)th tablet	A1	Accept: the tablet taken at \(t=16\).
Part (g)(iv)
the amounts just after each tablet increase and approach \(\frac{D}{1-0.6}\) but never exceed it, so the condition is \(\frac{D}{0.4}\le450\)	R1	Must explain that the highest amounts are the peaks and that the peaks never pass the sum to infinity.
\(D=180\)	A1

Answers: (a) \(k=0.128\) (3 s.f.) · (b)(i) \(297.92\) · (b)(ii) \(A_n=\dfrac{200(1-0.36^n)}{0.64}=312.5(1-0.36^n)\) · (c) \(312.5\) mg · (d)(i) \(112.5\) mg · (d)(ii) saw-tooth graph with peaks rising towards \(312.5\) mg · (f) \(5\) tablets · (g)(i) \(0.6\) · (g)(ii) \(500\) mg · (g)(iii) the 5th tablet (\(A_5=461.12\) mg) · (g)(iv) \(D=180\)

Exam tip (d)(ii) Find the first few peaks and troughs before you sketch: \(200\to72\), \(272\to97.9\), \(297.9\to107.3\). Plot those points first, then join each peak to the next trough with a decreasing curve.

Common error (f) When you divide by \(\ln0.36\), which is negative, the inequality sign reverses. The answer must also be a whole number of tablets, so round up.

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A striker's penalty kicks and the binomial expansion

1.9 The binomial theorem1.6 Deductive proof4.8 Binomial distribution4.5 Sample spaces and probabilitySL

In a training session a striker takes \(6\) penalty kicks. She scores with each kick with probability \(0.8\), independently of the other kicks. Let \(X\) be the number of goals she scores, so that \(X\sim B(6,0.8)\).

(a)

Use Pascal's triangle to write down the expansion of \((q+p)^6\) in ascending powers of \(p\).

No calculatorROUTINENotes

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coefficients \(1,\ 6,\ 15,\ 20,\ 15,\ 6,\ 1\)	A1	Seen as row \(6\) of Pascal's triangle or in the expansion.
\(q^6+6q^5p+15q^4p^2+20q^3p^3+15q^2p^4+6qp^5+p^6\)	A1	Powers must be correct in every term.

Answer: \(q^6+6q^5p+15q^4p^2+20q^3p^3+15q^2p^4+6qp^5+p^6\)

(2)

(b)(i)

The result of each kick is recorded as G (goal) or M (miss), so that one possible record of the session is GGMGMG.

Write down the number of different possible records.

No calculatorROUTINENotes

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\(2^6=64\)

Answer: \(64\)

(1)

(b)(ii)

Explain why exactly \(15\) of these records contain exactly \(4\) goals, and find the probability of any one of these records.

No calculatorCONCEPTNotes

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a record with \(4\) goals is fixed by choosing which \(4\) of the \(6\) kicks are goals, in \(\binom64=15\) ways	R1	Accept a reference to the \(15\) in row \(6\) of Pascal's triangle with this reason.
\(0.8^4\times0.2^2=0.016384\)	A1	Accept \(0.0164\).

Answer: \(0.016384\)

(2)

(c)

Hence identify the term in the expansion of \((0.2+0.8)^6\) that equals \(\mathrm{P}(X=4)\), and find \(\mathrm{P}(X=4)\).

MULTINotes

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the term \(15q^2p^4\) with \(q=0.2,\ p=0.8\): \(15(0.2)^2(0.8)^4\)	M1	Or \(15\times\) their answer to (b)(ii).
\(0.24576\)	A1	Accept \(0.246\).

Answer: \(\mathrm{P}(X=4)=0.246\) (3 s.f.)

(2)

(d)(i)

Show that the coefficient of \(x^3\) in the expansion of \((1+4x)^6\) is \(1280\).

No calculatorALGEBRANotes

Show solution

\(\binom63(4x)^3\)	M1	Award for the correct term in any form.
\(20\times64x^3\)	A1	Must see \((4x)^3=64x^3\), not \(4x^3\).
\(1280\)	AG

Answer: \(1280\)

(2)

(d)(ii)

Explain why the coefficient of \(x^r\) in the expansion of \((0.2+0.8x)^6\) is equal to \(\mathrm{P}(X=r)\).

By noting that \(0.2+0.8x=\dfrac{1+4x}{5}\), find the exact value of \(\mathrm{P}(X=3)\).

No calculatorMULTINotes

Show solution

general term \(\binom6r(0.2)^{6-r}(0.8x)^r\) has coefficient \(\binom6r(0.8)^r(0.2)^{6-r}=\mathrm{P}(X=r)\)	R1
\((0.2+0.8x)^6=\frac{(1+4x)^6}{5^6}\), so \(\mathrm{P}(X=3)=\frac{1280}{5^6}\)	M1	Allow ECF from (d)(i).
\(\frac{1280}{15625}=\frac{256}{3125}\ (=0.08192)\)	A1	Accept any equivalent exact form.

Answer: \(\mathrm{P}(X=3)=\dfrac{256}{3125}=0.08192\)

(3)

(e)(i)

Prove that \(\dbinom n0+\dbinom n1+\dbinom n2+\dots+\dbinom nn=2^n\) for all \(n\in\mathbb{Z}^+\).

No calculatorPROOFNotes

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use \((a+b)^n=\sum_{r=0}^{n}\binom nr a^{n-r}b^r\) with \(a=b=1\)	M1
\((1+1)^n=\sum_{r=0}^{n}\binom nr 1^{n-r}1^r=\binom n0+\binom n1+\dots+\binom nn\), so the sum equals \(2^n\) for every \(n\in\mathbb{Z}^+\)	A1	Checking particular values of \(n\) only scores M0 A0.

(2)

(e)(ii)

Interpret the result of part (e)(i) for \(n=6\) in terms of the records described in part (b).

No calculatorINTERPRETNotes

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\(\binom6r\) counts the records with exactly \(r\) goals; every record has some number of goals from \(0\) to \(6\), so adding over \(r\) counts all \(2^6=64\) records

Must link each coefficient to a number of goals.

(1)

(f)(i)

Figure 1 shows the probability distribution of \(X\), with the bars for \(x=3\) and \(x=4\) missing.

Complete the bar chart.

Figure 1: the distribution of X, with the bars for x = 3 and x = 4 not yet drawn

No calculatorGRAPHNotes

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bar at \(x=3\) of height \(0.082\)	A1	Allow ECF from (d)(ii). Height within half a small square.
bar at \(x=4\) of height \(0.246\)	A1	Allow ECF from (c).

Answer: bars of heights \(0.082\) at \(x=3\) and \(0.246\) at \(x=4\)

(2)

(f)(ii)

Describe the shape of the distribution and explain it in terms of the probability of scoring.

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negatively skewed (tail to the left), with mode \(x=5\)	A1	Accept "skewed to the left" or "tail to the left". Stating the mode is not required for this mark.
because \(p=0.8\gt0.5\), scoring is more likely than missing, so the probability is concentrated at high numbers of goals	R1

Answer: negatively skewed, mode \(5\)

(2)

(g)(i)

After extra practice the striker's probability of scoring with each kick rises to \(0.98\). In a later session she takes \(8\) kicks.

Write down an expression for the probability that she scores with all \(8\) kicks, in the form \((1-a)^8\). Use the first three terms of a binomial expansion to estimate this probability.

No calculatorMODELNotes

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\((1-0.02)^8\)	A1
\(1+8(-0.02)+\binom82(-0.02)^2\)	M1	Signs must alternate; \(\binom82=28\).
\(1-0.16+0.0112=0.8512\)	A1

Answer: \(0.8512\)

(3)

(g)(ii)

The exact value of the probability is \(0.850\,763\), correct to six significant figures. Find the percentage error in your estimate and explain why the estimate is so close.

INTERPRETNotes

Show solution

\(\frac{0.8512-0.850763}{0.850763}\times100=0.0514\%\)	A1	Accept \(0.05\%\). Allow ECF from (g)(i).
\(0.02\) is small, so each later term is much smaller than the one before; the next term is \(-\binom83(0.02)^3=-0.000448\), which accounts for almost all of the error	R1	Must refer to the size of the omitted terms.

Answer: \(0.0514\%\)

(2)

Show full solution

Working	Mark	Notes
Part (a)
coefficients \(1,\ 6,\ 15,\ 20,\ 15,\ 6,\ 1\)	A1	Seen as row \(6\) of Pascal's triangle or in the expansion.
\(q^6+6q^5p+15q^4p^2+20q^3p^3+15q^2p^4+6qp^5+p^6\)	A1	Powers must be correct in every term.
Part (b)(i)
\(2^6=64\)	A1
Part (b)(ii)
a record with \(4\) goals is fixed by choosing which \(4\) of the \(6\) kicks are goals, in \(\binom64=15\) ways	R1	Accept a reference to the \(15\) in row \(6\) of Pascal's triangle with this reason.
\(0.8^4\times0.2^2=0.016384\)	A1	Accept \(0.0164\).
Part (c)
the term \(15q^2p^4\) with \(q=0.2,\ p=0.8\): \(15(0.2)^2(0.8)^4\)	M1	Or \(15\times\) their answer to (b)(ii).
\(0.24576\)	A1	Accept \(0.246\).
Part (d)(i)
\(\binom63(4x)^3\)	M1	Award for the correct term in any form.
\(20\times64x^3\)	A1	Must see \((4x)^3=64x^3\), not \(4x^3\).
\(1280\)	AG
Part (d)(ii)
general term \(\binom6r(0.2)^{6-r}(0.8x)^r\) has coefficient \(\binom6r(0.8)^r(0.2)^{6-r}=\mathrm{P}(X=r)\)	R1
\((0.2+0.8x)^6=\frac{(1+4x)^6}{5^6}\), so \(\mathrm{P}(X=3)=\frac{1280}{5^6}\)	M1	Allow ECF from (d)(i).
\(\frac{1280}{15625}=\frac{256}{3125}\ (=0.08192)\)	A1	Accept any equivalent exact form.
Part (e)(i)
use \((a+b)^n=\sum_{r=0}^{n}\binom nr a^{n-r}b^r\) with \(a=b=1\)	M1
\((1+1)^n=\sum_{r=0}^{n}\binom nr 1^{n-r}1^r=\binom n0+\binom n1+\dots+\binom nn\), so the sum equals \(2^n\) for every \(n\in\mathbb{Z}^+\)	A1	Checking particular values of \(n\) only scores M0 A0.
Part (e)(ii)
\(\binom6r\) counts the records with exactly \(r\) goals; every record has some number of goals from \(0\) to \(6\), so adding over \(r\) counts all \(2^6=64\) records	R1	Must link each coefficient to a number of goals.
Part (f)(i)
bar at \(x=3\) of height \(0.082\)	A1	Allow ECF from (d)(ii). Height within half a small square.
bar at \(x=4\) of height \(0.246\)	A1	Allow ECF from (c).
Part (f)(ii)
negatively skewed (tail to the left), with mode \(x=5\)	A1	Accept "skewed to the left" or "tail to the left". Stating the mode is not required for this mark.
because \(p=0.8\gt0.5\), scoring is more likely than missing, so the probability is concentrated at high numbers of goals	R1
Part (g)(i)
\((1-0.02)^8\)	A1
\(1+8(-0.02)+\binom82(-0.02)^2\)	M1	Signs must alternate; \(\binom82=28\).
\(1-0.16+0.0112=0.8512\)	A1
Part (g)(ii)
\(\frac{0.8512-0.850763}{0.850763}\times100=0.0514\%\)	A1	Accept \(0.05\%\). Allow ECF from (g)(i).
\(0.02\) is small, so each later term is much smaller than the one before; the next term is \(-\binom83(0.02)^3=-0.000448\), which accounts for almost all of the error	R1	Must refer to the size of the omitted terms.

Answers: (a) \(q^6+6q^5p+15q^4p^2+20q^3p^3+15q^2p^4+6qp^5+p^6\) · (b)(i) \(64\) · (b)(ii) \(0.016384\) · (c) \(\mathrm{P}(X=4)=0.246\) (3 s.f.) · (d)(i) \(1280\) · (d)(ii) \(\mathrm{P}(X=3)=\dfrac{256}{3125}=0.08192\) · (f)(i) bars of heights \(0.082\) at \(x=3\) and \(0.246\) at \(x=4\) · (f)(ii) negatively skewed, mode \(5\) · (g)(i) \(0.8512\) · (g)(ii) \(0.0514\%\)

Common error (g)(ii) In \((1-0.02)^8\) the second term is \(8(-0.02)\), not \(8(0.02)\). Write the bracket as \((1+(-0.02))^8\) so the signs take care of themselves.

More 1.9 questions in the question bank

1.10 Counting principles and the extended binomial theorem · 1.11 Partial fractionsmerged group

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How secure is a password? Counting arrangements

1.10 Counting principles and the extended binomial theorem1.1 Standard form4.5 Sample spaces and probability1.9 The binomial theorem1.7 Rational exponents and laws of logarithmsHL

An online service requires every password to have exactly \(8\) characters. Each character is one of the \(26\) lower-case letters a to z or one of the \(10\) digits 0 to 9. Characters may be repeated unless stated otherwise.

(a)

Show that the number of possible passwords is \(2.82\times10^{12}\), correct to three significant figures.

HL onlyROUTINENotes

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\(36^8\)	M1	Award for \(36\) choices for each of \(8\) positions multiplied.
\(2\,821\,109\,907\,456\)	A1	Must see more than 3 s.f. or the exact value.
\(2.82\times10^{12}\)	AG

Answer: \(2.82\times10^{12}\)

(2)

(b)

Find the number of passwords in which no character is repeated. Give your answer in the form \(a\times10^k\), where \(1\le a\lt10\) and \(k\in\mathbb{Z}\), with \(a\) correct to three significant figures.

HL onlyROUTINENotes

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\({}^{36}P_8=36\times35\times34\times\dots\times29\)	M1	Accept \(\frac{36!}{28!}\).
\(1.22\times10^{12}\)	A1	Exact value \(1\,220\,096\,908\,800\). Calculator notation such as 1.22E12 scores A0.

Answer: \(1.22\times10^{12}\)

(2)

(c)(i)

Find the number of passwords that contain exactly two digits.

HL onlyMULTINotes

Show solution

\(\binom82=28\) ways to choose the positions of the digits	(M1)
\(10^2\times26^6\) ways to fill the positions	(M1)
\(28\times100\times26^6=8.65\times10^{11}\)	A1	Exact value \(864\,964\,172\,800\).

Answer: \(8.65\times10^{11}\)

(3)

(c)(ii)

Find the number of passwords that contain exactly two digits, where the two digits are not next to each other.

HL onlyMULTINotes

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pairs of non-adjacent positions: \(28-7=21\)	M1	Or \(\binom72=21\) (gaps method).
\(21\times100\times26^6=6.49\times10^{11}\)	A1	Exact value \(648\,723\,129\,600\). Allow ECF from (c)(i): \(\frac34\) of their answer.

Answer: \(6.49\times10^{11}\)

(2)

(d)

Explain why \(\displaystyle\sum_{k=0}^{8}\binom8k10^k\,26^{8-k}=36^8\), with reference to the number of passwords and to the binomial theorem.

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\(\binom8k10^k26^{8-k}\) is the number of passwords with exactly \(k\) digits, and every password has exactly one value of \(k\) from \(0\) to \(8\), so the sum counts all \(36^8\) passwords	R1
the sum is the binomial expansion of \((26+10)^8=36^8\)	R1

(2)

(e)

A password is created by choosing each of its \(8\) characters at random from the \(36\) available characters. Find the probability that the password contains at least one digit.

ROUTINENotes

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\(1-\left(\frac{26}{36}\right)^8\)	M1	Or \(\frac{36^8-26^8}{36^8}\).
\(0.926\)	A1	Accept \(0.925977\ldots\).

Answer: \(0.926\)

(2)

(f)

An attacker's computer can test \(4.0\times10^{10}\) passwords per second. Find the time, in seconds, that it needs to test every possible password.

ROUTINENotes

Show solution

\(\frac{2.821\ldots\times10^{12}}{4.0\times10^{10}}\)	M1
\(70.5\) s	A1	Accept \(71\) s.

Answer: \(70.5\) s

(2)

(g)(i)

The service decides to allow longer passwords. Figure 1 shows \(\log_{10}N\) against the password length \(L\), where \(N\) is the number of possible passwords of length \(L\), for the present set of \(36\) characters and for a larger set of \(46\) characters.

Use Figure 1 to find the smallest length \(L\) of a \(36\)-character password for which \(N\gt10^{15}\).

Figure 1: \(\log_{10}N\) against password length \(L\), where \(N\) is the number of possible passwords, for character sets of 36 and 46 characters

GRAPHNotes

Show solution

\(L=10\)

At \(L=9\), \(\log_{10}N\approx14.0\); at \(L=10\), \(\log_{10}N\approx15.6\).

Answer: \(L=10\)

(1)

(g)(ii)

Explain why each point for the \(36\)-character set is about \(1.56\) units higher than the point to its left.

No calculatorINTERPRETNotes

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one more character multiplies \(N\) by \(36\), which adds \(\log_{10}36=1.556\ldots\) to \(\log_{10}N\)

Must link multiplying by \(36\) to adding a constant amount to the logarithm.

(1)

(g)(iii)

The service wants a complete search by the attacker's computer to take at least one year (\(365\) days). Find the smallest password length, using the \(36\)-character set, that achieves this.

MULTINotes

Show solution

number tested in a year \(=4.0\times10^{10}\times365\times24\times3600=1.26\times10^{18}\)	M1
\(36^L\ge1.26\times10^{18}\), e.g. \(36^{11}=1.32\times10^{17}\) and \(36^{12}=4.74\times10^{18}\)	(M1)	Or \(L\log_{10}36\ge18.1\Rightarrow L\ge11.6\), or from Figure 1.
\(L=12\)	A1

Answer: \(L=12\)

(3)

(h)

Instead of making passwords longer, the service could keep the length at \(8\) and add \(10\) special characters, giving the \(46\)-character set. Compare the time for a complete search under this rule with the time for \(36\)-character passwords of length \(12\), and comment on which change is more effective.

INTERPRETNotes

Show solution

\(\frac{46^8}{4.0\times10^{10}}=501\) s (about \(8\) minutes)	A1	\(46^8=2.00\times10^{13}\).
\(\frac{36^{12}}{4.0\times10^{10}}=1.18\times10^8\) s (about \(3.76\) years)	A1	Allow any consistent unit.
increasing the length is far more effective: the length is the exponent, so each extra character multiplies \(N\) by \(36\), while enlarging the set only changes the base (the \(46\)-character points in Figure 1 are only slightly steeper)	R1	Must compare the two effects, not just state the times.

Answer: about \(501\) s against about \(3.76\) years: longer passwords win

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(36^8\)	M1	Award for \(36\) choices for each of \(8\) positions multiplied.
\(2\,821\,109\,907\,456\)	A1	Must see more than 3 s.f. or the exact value.
\(2.82\times10^{12}\)	AG
Part (b)
\({}^{36}P_8=36\times35\times34\times\dots\times29\)	M1	Accept \(\frac{36!}{28!}\).
\(1.22\times10^{12}\)	A1	Exact value \(1\,220\,096\,908\,800\). Calculator notation such as 1.22E12 scores A0.
Part (c)(i)
\(\binom82=28\) ways to choose the positions of the digits	(M1)
\(10^2\times26^6\) ways to fill the positions	(M1)
\(28\times100\times26^6=8.65\times10^{11}\)	A1	Exact value \(864\,964\,172\,800\).
Part (c)(ii)
pairs of non-adjacent positions: \(28-7=21\)	M1	Or \(\binom72=21\) (gaps method).
\(21\times100\times26^6=6.49\times10^{11}\)	A1	Exact value \(648\,723\,129\,600\). Allow ECF from (c)(i): \(\frac34\) of their answer.
Part (d)
\(\binom8k10^k26^{8-k}\) is the number of passwords with exactly \(k\) digits, and every password has exactly one value of \(k\) from \(0\) to \(8\), so the sum counts all \(36^8\) passwords	R1
the sum is the binomial expansion of \((26+10)^8=36^8\)	R1
Part (e)
\(1-\left(\frac{26}{36}\right)^8\)	M1	Or \(\frac{36^8-26^8}{36^8}\).
\(0.926\)	A1	Accept \(0.925977\ldots\).
Part (f)
\(\frac{2.821\ldots\times10^{12}}{4.0\times10^{10}}\)	M1
\(70.5\) s	A1	Accept \(71\) s.
Part (g)(i)
\(L=10\)	A1	At \(L=9\), \(\log_{10}N\approx14.0\); at \(L=10\), \(\log_{10}N\approx15.6\).
Part (g)(ii)
one more character multiplies \(N\) by \(36\), which adds \(\log_{10}36=1.556\ldots\) to \(\log_{10}N\)	R1	Must link multiplying by \(36\) to adding a constant amount to the logarithm.
Part (g)(iii)
number tested in a year \(=4.0\times10^{10}\times365\times24\times3600=1.26\times10^{18}\)	M1
\(36^L\ge1.26\times10^{18}\), e.g. \(36^{11}=1.32\times10^{17}\) and \(36^{12}=4.74\times10^{18}\)	(M1)	Or \(L\log_{10}36\ge18.1\Rightarrow L\ge11.6\), or from Figure 1.
\(L=12\)	A1
Part (h)
\(\frac{46^8}{4.0\times10^{10}}=501\) s (about \(8\) minutes)	A1	\(46^8=2.00\times10^{13}\).
\(\frac{36^{12}}{4.0\times10^{10}}=1.18\times10^8\) s (about \(3.76\) years)	A1	Allow any consistent unit.
increasing the length is far more effective: the length is the exponent, so each extra character multiplies \(N\) by \(36\), while enlarging the set only changes the base (the \(46\)-character points in Figure 1 are only slightly steeper)	R1	Must compare the two effects, not just state the times.

Answers: (a) \(2.82\times10^{12}\) · (b) \(1.22\times10^{12}\) · (c)(i) \(8.65\times10^{11}\) · (c)(ii) \(6.49\times10^{11}\) · (e) \(0.926\) · (f) \(70.5\) s · (g)(i) \(L=10\) · (g)(iii) \(L=12\) · (h) about \(501\) s against about \(3.76\) years: longer passwords win

Exam tip (h) Very large counts are easiest to compare through their logarithms: \(\log_{10}(46^8)\approx13.3\) while \(\log_{10}(36^{12})\approx18.7\), a difference of more than five orders of magnitude.

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An audio filter's gain: partial fractions and series

1.11 Partial fractions1.10 Counting principles and the extended binomial theorem5.15 Further derivatives and integrals; partial fractions2.13 Rational functions with oblique and horizontal asymptotesHL

An engineer models the gain of an audio filter by

\[G(x)=\frac{2+5x}{(1+x)(1-2x)},\]

where \(x\) is a normalised frequency and \(0\le x\lt0.5\) for the filter. In parts (d) and (e), \(G\) is considered for all real \(x\) with \(x\ne-1\) and \(x\ne0.5\).

(a)

Express \(G(x)\) in partial fractions.

HL onlyNo calculatorALGEBRANotes

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\(\frac{A}{1+x}+\frac{B}{1-2x}\), so \(2+5x\equiv A(1-2x)+B(1+x)\)	M1
\(x=-1\): \(-3=3A\Rightarrow A=-1\)	A1
\(x=\frac12\): \(\frac92=\frac32B\Rightarrow B=3\)	A1

Answer: \(G(x)=\dfrac{3}{1-2x}-\dfrac{1}{1+x}\)

(3)

(b)(i)

Hence use the extended binomial theorem to show that \(G(x)=2+7x+11x^2+25x^3+\dots\)

HL onlyNo calculatorMULTINotes

Show solution

\((1-2x)^{-1}=1+2x+4x^2+8x^3+\dots\)	M1	Award for an attempt at the expansion with \(n=-1\) and \(u=-2x\).
\((1+x)^{-1}=1-x+x^2-x^3+\dots\)	M1
\(3(1+2x+4x^2+8x^3)-(1-x+x^2-x^3)\)	A1	Both expansions correct and combined with the constants from (a).
\(=3+6x+12x^2+24x^3-1+x-x^2+x^3\)	A1	Must see terms collected.
\(=2+7x+11x^2+25x^3+\dots\)	AG

Answer: \(2+7x+11x^2+25x^3\)

(4)

(b)(ii)

State the values of \(x\) for which this expansion is valid. Give a reason for your answer.

HL onlyNo calculatorCONCEPTNotes

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\((1+x)^{-1}\) needs \(\|x\|\lt1\) and \((1-2x)^{-1}\) needs \(\|2x\|\lt1\); both must hold	R1
\(\|x\|\lt\frac12\)	A1	Accept \(-\frac12\lt x\lt\frac12\).

Answer: \(|x|\lt\tfrac12\)

(2)

(c)(i)

Use the four terms of the expansion to estimate \(G(0.05)\).

HL onlyROUTINENotes

Show solution

\(2+0.35+0.0275+0.003125=2.380625\)

Accept \(2.3806\) or \(2.381\).

Answer: \(2.380625\)

(1)

(c)(ii)

Find the exact value of \(G(0.05)\) as a fraction, and hence find the percentage error of your estimate.

ROUTINENotes

Show solution

\(G(0.05)=\frac{2.25}{1.05\times0.9}=\frac{50}{21}\)	A1
\(\frac{2.380952-2.380625}{2.380952}\times100=0.0138\%\)	A1	Accept \(0.0137\%\) to \(0.0138\%\). Allow ECF from (c)(i).

Answer: \(\dfrac{50}{21}\); \(0.0138\%\)

(2)

(d)

On the axes in Figure 1, sketch the graph of \(y=G(x)\) for \(-3\le x\le3\). Show the equations of all asymptotes and the coordinates of the points where the graph meets the axes.

Figure 1: axes for the graph of \(y=G(x)\)

HL onlyGRAPHNotes

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vertical asymptotes \(x=-1\) and \(x=\frac12\)	A1
horizontal asymptote \(y=0\)	A1	Degree of numerator less than degree of denominator.
intercepts \((-0.4,\,0)\) and \((0,\,2)\)	A1
three branches, each increasing: for \(x\lt-1\) rising from just above \(y=0\) towards \(+\infty\); for \(-1\lt x\lt\frac12\) rising from \(-\infty\) to \(+\infty\); for \(x\gt\frac12\) rising from \(-\infty\) towards \(y=0\) from below	A1	Correct behaviour on both sides of each asymptote.

Answer: asymptotes \(x=-1\), \(x=\tfrac12\), \(y=0\); intercepts \((-0.4,0)\), \((0,2)\)

(4)

(e)

The cubic \(2+7x+11x^2+25x^3\) gives \(9.66\) at \(x=0.45\), but \(G(0.45)=29.3\). Explain why the estimate is so poor at this value of \(x\), with reference to your graph and to part (b)(ii).

HL onlyNo calculatorINTERPRETNotes

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\(x=0.45\) is close to the limit \(\|x\|\lt\frac12\): the terms of \((1-2x)^{-1}\) shrink only by the factor \(2x=0.9\) each time, so the omitted terms are far from negligible	R1
near \(x=\frac12\) the graph rises towards a vertical asymptote, \(G(x)\to\infty\), while a cubic stays finite there and cannot follow it	R1

(2)

(f)

Show that \(\displaystyle\int_0^{\frac14}G(x)\,dx=\ln\left(\dfrac{8\sqrt2}{5}\right)\).

HL onlyNo calculatorMULTINotes

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\(\int\left(\frac{3}{1-2x}-\frac{1}{1+x}\right)dx=-\frac32\ln\|1-2x\|-\ln\|1+x\|\)	M1	Award for integrating both partial fractions to logarithms.
coefficient \(-\frac32\) correct	A1	A common error is \(3\ln\|1-2x\|\) or \(-6\ln\|1-2x\|\).
\(\left(-\frac32\ln\frac12-\ln\frac54\right)-(0-0)\)	M1	Substituting both limits.
\(=\frac32\ln2+2\ln2-\ln5=\frac72\ln2-\ln5=\ln\frac{2^{7/2}}{5}\)	A1	Must see \(2^{7/2}=8\sqrt2\) or equivalent.
\(=\ln\left(\frac{8\sqrt2}{5}\right)\)	AG

Answer: \(\ln\left(\dfrac{8\sqrt2}{5}\right)\approx0.817\)

(4)

(g)(i)

The filter is redesigned so that its gain is

\[H(x)=\frac{4+3x-4x^2}{(1+x)(1-2x)}.\]

Show that \(H(x)=2+G(x)\).

HL onlyNo calculatorALGEBRANotes

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\((1+x)(1-2x)=1-x-2x^2\) and \(4+3x-4x^2=2(1-x-2x^2)+(2+5x)\)	M1	Or long division giving quotient \(2\).
remainder \(2+5x\), so \(H(x)=2+\frac{2+5x}{(1+x)(1-2x)}\)	A1
\(=2+G(x)\)	AG

Answer: \(H(x)=2+G(x)\)

(2)

(g)(ii)

Write down the equation of the horizontal asymptote of \(y=H(x)\) and the first three terms of the expansion of \(H(x)\) in ascending powers of \(x\).

HL onlyNo calculatorROUTINENotes

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\(y=2\)	A1
\(4+7x+11x^2\)	A1	Allow ECF from (b)(i).

Answer: \(y=2\); \(4+7x+11x^2\)

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(\frac{A}{1+x}+\frac{B}{1-2x}\), so \(2+5x\equiv A(1-2x)+B(1+x)\)	M1
\(x=-1\): \(-3=3A\Rightarrow A=-1\)	A1
\(x=\frac12\): \(\frac92=\frac32B\Rightarrow B=3\)	A1
Part (b)(i)
\((1-2x)^{-1}=1+2x+4x^2+8x^3+\dots\)	M1	Award for an attempt at the expansion with \(n=-1\) and \(u=-2x\).
\((1+x)^{-1}=1-x+x^2-x^3+\dots\)	M1
\(3(1+2x+4x^2+8x^3)-(1-x+x^2-x^3)\)	A1	Both expansions correct and combined with the constants from (a).
\(=3+6x+12x^2+24x^3-1+x-x^2+x^3\)	A1	Must see terms collected.
\(=2+7x+11x^2+25x^3+\dots\)	AG
Part (b)(ii)
\((1+x)^{-1}\) needs \(\|x\|\lt1\) and \((1-2x)^{-1}\) needs \(\|2x\|\lt1\); both must hold	R1
\(\|x\|\lt\frac12\)	A1	Accept \(-\frac12\lt x\lt\frac12\).
Part (c)(i)
\(2+0.35+0.0275+0.003125=2.380625\)	A1	Accept \(2.3806\) or \(2.381\).
Part (c)(ii)
\(G(0.05)=\frac{2.25}{1.05\times0.9}=\frac{50}{21}\)	A1
\(\frac{2.380952-2.380625}{2.380952}\times100=0.0138\%\)	A1	Accept \(0.0137\%\) to \(0.0138\%\). Allow ECF from (c)(i).
Part (d)
vertical asymptotes \(x=-1\) and \(x=\frac12\)	A1
horizontal asymptote \(y=0\)	A1	Degree of numerator less than degree of denominator.
intercepts \((-0.4,\,0)\) and \((0,\,2)\)	A1
three branches, each increasing: for \(x\lt-1\) rising from just above \(y=0\) towards \(+\infty\); for \(-1\lt x\lt\frac12\) rising from \(-\infty\) to \(+\infty\); for \(x\gt\frac12\) rising from \(-\infty\) towards \(y=0\) from below	A1	Correct behaviour on both sides of each asymptote.
Part (e)
\(x=0.45\) is close to the limit \(\|x\|\lt\frac12\): the terms of \((1-2x)^{-1}\) shrink only by the factor \(2x=0.9\) each time, so the omitted terms are far from negligible	R1
near \(x=\frac12\) the graph rises towards a vertical asymptote, \(G(x)\to\infty\), while a cubic stays finite there and cannot follow it	R1
Part (f)
\(\int\left(\frac{3}{1-2x}-\frac{1}{1+x}\right)dx=-\frac32\ln\|1-2x\|-\ln\|1+x\|\)	M1	Award for integrating both partial fractions to logarithms.
coefficient \(-\frac32\) correct	A1	A common error is \(3\ln\|1-2x\|\) or \(-6\ln\|1-2x\|\).
\(\left(-\frac32\ln\frac12-\ln\frac54\right)-(0-0)\)	M1	Substituting both limits.
\(=\frac32\ln2+2\ln2-\ln5=\frac72\ln2-\ln5=\ln\frac{2^{7/2}}{5}\)	A1	Must see \(2^{7/2}=8\sqrt2\) or equivalent.
\(=\ln\left(\frac{8\sqrt2}{5}\right)\)	AG
Part (g)(i)
\((1+x)(1-2x)=1-x-2x^2\) and \(4+3x-4x^2=2(1-x-2x^2)+(2+5x)\)	M1	Or long division giving quotient \(2\).
remainder \(2+5x\), so \(H(x)=2+\frac{2+5x}{(1+x)(1-2x)}\)	A1
\(=2+G(x)\)	AG
Part (g)(ii)
\(y=2\)	A1
\(4+7x+11x^2\)	A1	Allow ECF from (b)(i).

Answers: (a) \(G(x)=\dfrac{3}{1-2x}-\dfrac{1}{1+x}\) · (b)(i) \(2+7x+11x^2+25x^3\) · (b)(ii) \(|x|\lt\tfrac12\) · (c)(i) \(2.380625\) · (c)(ii) \(\dfrac{50}{21}\); \(0.0138\%\) · (d) asymptotes \(x=-1\), \(x=\tfrac12\), \(y=0\); intercepts \((-0.4,0)\), \((0,2)\) · (f) \(\ln\left(\dfrac{8\sqrt2}{5}\right)\approx0.817\) · (g)(i) \(H(x)=2+G(x)\) · (g)(ii) \(y=2\); \(4+7x+11x^2\)

Common error (f) \(\int\frac{3}{1-2x}\,dx\) is \(-\frac32\ln|1-2x|\), not \(3\ln|1-2x|\): by the reverse chain rule you divide by the coefficient of \(x\), which is \(-2\).

More 1.11 questions in the question bank

1.12 Complex numbers in Cartesian form · 1.13 Polar and Euler forms · 1.14 Conjugate roots, De Moivre’s theorem, powers and rootsmerged group

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Impedance in an AC circuit: complex numbers in polar and Euler form

1.13 Polar and Euler forms1.12 Complex numbers in Cartesian form3.5 The unit circle, exact values and the ambiguous case3.10 Compound and double angle identitiesHL

In alternating-current (AC) circuits, each part of a circuit is described by a complex number \(Z\), its impedance, measured in ohms (Ω). The real part of \(Z\) is the resistance and the imaginary part is the reactance. The argument of \(Z\) is called its phase angle. In this question \(i^2=-1\) and angles are in radians.

Figure 1 shows two branches connected in parallel to an AC supply. Branch A is a 30 Ω resistor in series with a coil of reactance 40 Ω, so its impedance is \(Z_A=30+40i\). Branch B is a 40 Ω resistor in series with a capacitor of reactance 30 Ω. A capacitor gives a negative imaginary part, so the impedance of branch B is \(Z_B=40-30i\).

Two impedances \(Z_1\) and \(Z_2\) in series have total impedance \(Z_1+Z_2\). Two impedances in parallel have total impedance \(Z\), where \(\dfrac{1}{Z}=\dfrac{1}{Z_1}+\dfrac{1}{Z_2}\). When a current \(I\) flows through an impedance \(Z\), the voltage across it is \(V=IZ\), where \(I\) and \(V\) are also complex numbers.

Figure 1: two branches in parallel with an AC supply (not to scale)

(a)

Find the modulus and the argument of \(Z_A\). Hence write \(Z_A\) in Euler form, giving the argument correct to three significant figures.

ROUTINENotes

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\(\|Z_A\|=\sqrt{30^2+40^2}=50\)	A1
\(\arg Z_A=\arctan\frac{40}{30}\)	(M1)	The point is in the first quadrant, so the calculator value is the principal argument.
\(Z_A=50e^{0.927i}\)	A1	\(0.927295\ldots\) Accept \(50e^{i\,0.927}\). Award A0 for an argument in degrees (\(53.1^\circ\)).

Answer: \(|Z_A|=50\), \(\arg Z_A=0.927\), \(Z_A=50e^{0.927i}\)

(3)

(b)(i)

Show that \(Z_A=iZ_B\).

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\(iZ_B=i(40-30i)=40i-30i^2\)	M1	Expanding the bracket.
\(=40i+30\)	A1	Uses \(i^2=-1\).
\(=30+40i=Z_A\)	AG

(2)

(b)(ii)

Hence describe fully the single geometric transformation that maps the point representing \(Z_B\) onto the point representing \(Z_A\) on an Argand diagram.

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rotation through \(\frac{\pi}{2}\) anticlockwise about the origin

Must state angle, direction and centre. Accept \(90^\circ\). Reason: \(i=e^{i\pi/2}\) has modulus 1.

Answer: rotation through \(\frac{\pi}{2}\) anticlockwise about \(O\)

(1)

(c)

If the two branches were connected in series instead, the total impedance would be \(Z_A+Z_B\).

On the Argand diagram in Figure 2, plot and label the points \(P\), \(Q\) and \(S\) representing \(Z_A\), \(Z_B\) and \(Z_A+Z_B\) respectively.

Figure 2: Argand diagram for part (c)

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\(P\) at \((30,\,40)\) and \(Q\) at \((40,\,-30)\)	A1	Both points correct and labelled.
\(S\) at \((70,\,10)\)	A1	\(Z_A+Z_B=70+10i\).

Answer: \(P(30,40)\), \(Q(40,-30)\), \(S(70,10)\)

(2)

(d)

Explain why \(OPSQ\) is a square, where \(O\) is the origin.

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\(\overrightarrow{OS}=\overrightarrow{OP}+\overrightarrow{OQ}\), so \(OPSQ\) is a parallelogram	R1	Vector addition of complex numbers (parallelogram rule).
\(OP=OQ=50\) and \(\angle POQ=\frac{\pi}{2}\) by (b), so the parallelogram is a square	R1	Need both equal adjacent sides and the right angle. Accept \(\|Z_A\|=\|Z_B\|=50\) with a reference to the rotation in (b)(ii).

(2)

(e)

Let \(\beta\) be the acute angle with \(\tan\beta=\dfrac34\), so that \(\arg Z_B=-\beta\).

Use the square \(OPSQ\) to explain why \(\arg(Z_A+Z_B)=\dfrac{\pi}{4}-\beta\). Hence, without using a calculator, show that \(\tan\bigl(\arg(Z_A+Z_B)\bigr)=\dfrac17\).

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the diagonal \(OS\) bisects the right angle \(POQ\), so \(\angle QOS=\frac{\pi}{4}\) and \(\arg(Z_A+Z_B)=-\beta+\frac{\pi}{4}\)	R1
\(\tan\left(\frac{\pi}{4}-\beta\right)=\frac{\tan\frac{\pi}{4}-\tan\beta}{1+\tan\frac{\pi}{4}\tan\beta}\)	M1	Correct compound angle identity for tangent.
\(=\frac{1-\frac34}{1+\frac34}=\frac{\frac14}{\frac74}\)	A1	Uses \(\tan\frac{\pi}{4}=1\).
\(=\frac17\)	AG	Reading \(\frac{10}{70}\) from \(70+10i\) is a valid check but does not earn the M1 or A1.

(3)

(f)

The branches are connected in parallel, as in Figure 1. Use the result of (b)(i) to show that the total impedance of the circuit is \(Z=35+5i\).

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\(\frac{1}{Z}=\frac{1}{iZ_B}+\frac{1}{Z_B}=\frac{1+i}{iZ_B}\)	M1	Or \(\frac{1}{Z}=\frac{-i+1}{Z_B}\), using \(\frac1i=-i\).
\(Z=\frac{iZ_B}{1+i}\) or \(Z=\frac{Z_B}{1-i}\)	A1
\(Z=\frac{(40-30i)(1+i)}{(1-i)(1+i)}\)	M1	Multiplying numerator and denominator by the conjugate of the denominator.
\(=\frac{70+10i}{2}\)	A1	Or from \(\frac{i(40-30i)}{1+i}=\frac{(30+40i)(1-i)}{2}\).
\(=35+5i\)	AG	A direct method with \(\frac{1}{Z_A}+\frac{1}{Z_B}\) that does not use (b)(i) can score M1A1M1A1 if fully correct.

(4)

(g)(i)

A current \(I=2e^{i\pi/6}\) flows through branch A, so the voltage across branch A is \(V_A=IZ_A\).

Find \(V_A\) in Euler form, and describe the geometric effect of multiplying \(I\) by \(Z_A\).

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\(V_A=2\times50\,e^{i(\pi/6+0.927)}\)	M1	Multiplies the moduli and adds the arguments. Allow ECF from (a).
\(V_A=100e^{1.45i}\)	A1	\(1.45089\ldots\)
enlargement with scale factor 50 and rotation anticlockwise through \(0.927\) (\(=\arg Z_A\)) about the origin	A1	Both the scale factor and the rotation are needed.

Answer: \(V_A=100e^{1.45i}\)

(3)

(g)(ii)

Let \(\alpha=\arg Z_A\). Write down the exact values of \(\cos\alpha\) and \(\sin\alpha\). Hence use a compound angle identity to show that \(\mathrm{Re}(V_A)=30\sqrt3-40\).

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\(\cos\alpha=\frac35\), \(\sin\alpha=\frac45\)	A1	From the right-angled triangle with sides 30, 40, 50.
\(\mathrm{Re}(V_A)=100\cos\left(\frac{\pi}{6}+\alpha\right)=100\left(\cos\frac{\pi}{6}\cos\alpha-\sin\frac{\pi}{6}\sin\alpha\right)\)	M1	Correct expansion of \(\cos(A+B)\).
\(=100\left(\frac{\sqrt3}{2}\times\frac35-\frac12\times\frac45\right)\)	A1	Exact values substituted.
\(=30\sqrt3-40\)	AG

(3)

(g)(iii)

Verify the result of (g)(ii) by writing \(I\) in the form \(a+bi\) and multiplying in Cartesian form. Write down \(\mathrm{Im}(V_A)\).

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\(I=2\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)=\sqrt3+i\) and \((\sqrt3+i)(30+40i)\)	M1
\(=(30\sqrt3-40)+(40\sqrt3+30)i\), so \(\mathrm{Im}(V_A)=40\sqrt3+30\)	A1	Accept \(30+40\sqrt3\).

Answer: \(\mathrm{Im}(V_A)=40\sqrt3+30\)

(2)

(h)

The frequency of the supply is now doubled. This doubles the reactance of the coil, so the impedance of branch A becomes \(30+80i\).

A student claims that the phase angle of branch A also doubles, from \(\alpha\) to \(2\alpha\). By finding the exact value of \(\tan2\alpha\), show that the student's claim is false.

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\(\tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}=\frac{2\times\frac43}{1-\frac{16}{9}}\)	M1	Uses \(\tan\alpha=\frac43\).
\(\tan2\alpha=-\frac{24}{7}\)	A1
\(\tan2\alpha\lt0\), so \(2\alpha\) is obtuse (\(\frac{\pi}{2}\lt2\alpha\lt\pi\)); but \(30+80i\) is in the first quadrant, so the new phase angle is acute (\(\tan=\frac83\gt0\)). Hence the new phase angle is not \(2\alpha\)	R1	Accept: the new phase angle has tangent \(\frac{80}{30}=\frac83\ne-\frac{24}{7}\), with both angles in \((0,\pi)\). Do not award R1 for decimal angles alone.

Answer: \(\tan2\alpha=-\frac{24}{7}\); claim false

(3)

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Working	Mark	Notes
Part (a)
\(\|Z_A\|=\sqrt{30^2+40^2}=50\)	A1
\(\arg Z_A=\arctan\frac{40}{30}\)	(M1)	The point is in the first quadrant, so the calculator value is the principal argument.
\(Z_A=50e^{0.927i}\)	A1	\(0.927295\ldots\) Accept \(50e^{i\,0.927}\). Award A0 for an argument in degrees (\(53.1^\circ\)).
Part (b)(i)
\(iZ_B=i(40-30i)=40i-30i^2\)	M1	Expanding the bracket.
\(=40i+30\)	A1	Uses \(i^2=-1\).
\(=30+40i=Z_A\)	AG
Part (b)(ii)
rotation through \(\frac{\pi}{2}\) anticlockwise about the origin	A1	Must state angle, direction and centre. Accept \(90^\circ\). Reason: \(i=e^{i\pi/2}\) has modulus 1.
Part (c)
\(P\) at \((30,\,40)\) and \(Q\) at \((40,\,-30)\)	A1	Both points correct and labelled.
\(S\) at \((70,\,10)\)	A1	\(Z_A+Z_B=70+10i\).
Part (d)
\(\overrightarrow{OS}=\overrightarrow{OP}+\overrightarrow{OQ}\), so \(OPSQ\) is a parallelogram	R1	Vector addition of complex numbers (parallelogram rule).
\(OP=OQ=50\) and \(\angle POQ=\frac{\pi}{2}\) by (b), so the parallelogram is a square	R1	Need both equal adjacent sides and the right angle. Accept \(\|Z_A\|=\|Z_B\|=50\) with a reference to the rotation in (b)(ii).
Part (e)
the diagonal \(OS\) bisects the right angle \(POQ\), so \(\angle QOS=\frac{\pi}{4}\) and \(\arg(Z_A+Z_B)=-\beta+\frac{\pi}{4}\)	R1
\(\tan\left(\frac{\pi}{4}-\beta\right)=\frac{\tan\frac{\pi}{4}-\tan\beta}{1+\tan\frac{\pi}{4}\tan\beta}\)	M1	Correct compound angle identity for tangent.
\(=\frac{1-\frac34}{1+\frac34}=\frac{\frac14}{\frac74}\)	A1	Uses \(\tan\frac{\pi}{4}=1\).
\(=\frac17\)	AG	Reading \(\frac{10}{70}\) from \(70+10i\) is a valid check but does not earn the M1 or A1.
Part (f)
\(\frac{1}{Z}=\frac{1}{iZ_B}+\frac{1}{Z_B}=\frac{1+i}{iZ_B}\)	M1	Or \(\frac{1}{Z}=\frac{-i+1}{Z_B}\), using \(\frac1i=-i\).
\(Z=\frac{iZ_B}{1+i}\) or \(Z=\frac{Z_B}{1-i}\)	A1
\(Z=\frac{(40-30i)(1+i)}{(1-i)(1+i)}\)	M1	Multiplying numerator and denominator by the conjugate of the denominator.
\(=\frac{70+10i}{2}\)	A1	Or from \(\frac{i(40-30i)}{1+i}=\frac{(30+40i)(1-i)}{2}\).
\(=35+5i\)	AG	A direct method with \(\frac{1}{Z_A}+\frac{1}{Z_B}\) that does not use (b)(i) can score M1A1M1A1 if fully correct.
Part (g)(i)
\(V_A=2\times50\,e^{i(\pi/6+0.927)}\)	M1	Multiplies the moduli and adds the arguments. Allow ECF from (a).
\(V_A=100e^{1.45i}\)	A1	\(1.45089\ldots\)
enlargement with scale factor 50 and rotation anticlockwise through \(0.927\) (\(=\arg Z_A\)) about the origin	A1	Both the scale factor and the rotation are needed.
Part (g)(ii)
\(\cos\alpha=\frac35\), \(\sin\alpha=\frac45\)	A1	From the right-angled triangle with sides 30, 40, 50.
\(\mathrm{Re}(V_A)=100\cos\left(\frac{\pi}{6}+\alpha\right)=100\left(\cos\frac{\pi}{6}\cos\alpha-\sin\frac{\pi}{6}\sin\alpha\right)\)	M1	Correct expansion of \(\cos(A+B)\).
\(=100\left(\frac{\sqrt3}{2}\times\frac35-\frac12\times\frac45\right)\)	A1	Exact values substituted.
\(=30\sqrt3-40\)	AG
Part (g)(iii)
\(I=2\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)=\sqrt3+i\) and \((\sqrt3+i)(30+40i)\)	M1
\(=(30\sqrt3-40)+(40\sqrt3+30)i\), so \(\mathrm{Im}(V_A)=40\sqrt3+30\)	A1	Accept \(30+40\sqrt3\).
Part (h)
\(\tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}=\frac{2\times\frac43}{1-\frac{16}{9}}\)	M1	Uses \(\tan\alpha=\frac43\).
\(\tan2\alpha=-\frac{24}{7}\)	A1
\(\tan2\alpha\lt0\), so \(2\alpha\) is obtuse (\(\frac{\pi}{2}\lt2\alpha\lt\pi\)); but \(30+80i\) is in the first quadrant, so the new phase angle is acute (\(\tan=\frac83\gt0\)). Hence the new phase angle is not \(2\alpha\)	R1	Accept: the new phase angle has tangent \(\frac{80}{30}=\frac83\ne-\frac{24}{7}\), with both angles in \((0,\pi)\). Do not award R1 for decimal angles alone.

Answers: (a) \(|Z_A|=50\), \(\arg Z_A=0.927\), \(Z_A=50e^{0.927i}\) · (b)(ii) rotation through \(\frac{\pi}{2}\) anticlockwise about \(O\) · (c) \(P(30,40)\), \(Q(40,-30)\), \(S(70,10)\) · (g)(i) \(V_A=100e^{1.45i}\) · (g)(iii) \(\mathrm{Im}(V_A)=40\sqrt3+30\) · (h) \(\tan2\alpha=-\frac{24}{7}\); claim false

Exam tip (a) Sketch the point before you take an inverse tangent. Here the point is in the first quadrant, so the calculator value is already the principal argument; in the second or third quadrant it would not be.

Common error (h) Doubling the imaginary part doubles the tangent of the phase angle, not the angle itself: \(\tan\) is not a linear function.

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A drone light show: roots of a complex number

1.14 Conjugate roots, De Moivre’s theorem, powers and roots1.12 Complex numbers in Cartesian form1.13 Polar and Euler forms2.12 Polynomial functions, factor and remainder theorems, sum and product of roots3.2 Right-angled and non-right-angled trianglesHL

At an evening event, six drones hover at the same height above a stage. Their positions in a horizontal plane are represented by complex numbers on an Argand diagram. The origin \(O\) is directly above the centre of the stage, the positive real axis points east, the positive imaginary axis points north, and distances are in metres.

The positions of the six drones are the six solutions of the equation \(z^6=64i\).

(a)

Find the six solutions of \(z^6=64i\). Give your answers in the form \(re^{i\theta}\), where \(r\gt0\) and \(-\pi\lt\theta\le\pi\).

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\(64i=64e^{i\pi/2}\)	(A1)
\(r=64^{1/6}=2\)	A1
\(\theta=\frac{\frac{\pi}{2}+2\pi k}{6}=\frac{\pi}{12}+\frac{k\pi}{3}\), \(k=0,1,\dots,5\)	M1	Or any equivalent \(2\pi k\) method, e.g. rotating one root by \(\frac{\pi}{3}\).
\(2e^{i\pi/12},\ 2e^{i5\pi/12},\ 2e^{i3\pi/4},\ 2e^{-i11\pi/12},\ 2e^{-i7\pi/12},\ 2e^{-i\pi/4}\)	A1	All six, with principal arguments. Arguments \(\frac{13\pi}{12},\frac{17\pi}{12},\frac{7\pi}{4}\) not reduced to \((-\pi,\pi]\): A0 for this row.

Answer: \(2e^{i\pi/12},\ 2e^{i5\pi/12},\ 2e^{i3\pi/4},\ 2e^{-i11\pi/12},\ 2e^{-i7\pi/12},\ 2e^{-i\pi/4}\)

(4)

(b)

The drones are labelled \(D_1,D_2,\dots,D_6\) in anticlockwise order, where \(D_1\) is at \(2e^{i\pi/12}\).

On the Argand diagram in Figure 1, plot and label the positions of the six drones.

Figure 1: Argand diagram for part (b) (1 unit = 1 m)

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six points on the circle of radius 2 centred at \(O\), spaced \(\frac{\pi}{3}\) apart	A1	Allow a small tolerance in plotting.
correct positions and labels, with \(D_1\) at argument \(\frac{\pi}{12}\) (just above east) and \(D_3\) at \(\frac{3\pi}{4}\) (north-west)	A1	Allow ECF from (a) for labels if the points are consistent with their answers.

(2)

(c)

Explain why the six drones are the vertices of a regular hexagon, and find the exact area of this hexagon.

No calculatorMULTINotes

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all six drones are 2 m from \(O\) and consecutive arguments differ by \(\frac{\pi}{3}\), so the six sides and six angles are equal	R1	Accept: the hexagon splits into six congruent isosceles triangles with apex angle \(\frac{\pi}{3}\) at \(O\).
area \(=6\times\frac12\times2\times2\times\sin\frac{\pi}{3}\)	M1	Uses \(\frac12ab\sin C\) for one triangle and multiplies by 6.
\(=6\sqrt3\ \mathrm{m}^2\)	A1

Answer: \(6\sqrt3\) m²

(3)

(d)

By considering the sum of the roots of the polynomial equation \(z^6-64i=0\), show that the sum of the six drone positions is zero. Interpret this result in context.

No calculatorINTERPRETNotes

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sum of the roots \(=-\frac{\text{coefficient of }z^5}{\text{coefficient of }z^6}=-\frac{0}{1}\)	M1	The coefficient of \(z^5\) in \(z^6-64i\) is 0. Accept a symmetry argument: the roots form three pairs \(z,\,-z\).
\(=0\)	AG
the mean position of the six drones is \(O\): the formation is centred directly above the centre of the stage	R1

(2)

(e)(i)

Write the positions of \(D_3\) and \(D_6\) in the form \(a+bi\).

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\(D_3=2\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)\)	M1	Uses \(a=r\cos\theta\), \(b=r\sin\theta\) for either drone.
\(D_3=-\sqrt2+\sqrt2\,i\), \(D_6=\sqrt2-\sqrt2\,i\)	A1	Both. Accept \(D_6=-D_3\) shown.

Answer: \(D_3=-\sqrt2+\sqrt2\,i\), \(D_6=\sqrt2-\sqrt2\,i\)

(2)

(e)(ii)

A polynomial \(P(z)\) has real coefficients. The positions of \(D_3\) and \(D_6\) are both roots of \(P(z)=0\).

Show that \(z^4+16\) is a factor of \(P(z)\).

No calculatorALGEBRANotes

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real coefficients, so the conjugates \(-\sqrt2-\sqrt2\,i\) and \(\sqrt2+\sqrt2\,i\) are also roots	R1	Must refer to real coefficients.
pair \(-\sqrt2\pm\sqrt2\,i\): sum \(-2\sqrt2\), product \(4\), giving \(z^2+2\sqrt2z+4\)	M1	Using sum and product, or expanding \((z-w)(z-w^*)\), for either pair.
\(z^2+2\sqrt2z+4\) and \(z^2-2\sqrt2z+4\)	A1	Both quadratic factors.
\((z^2+4+2\sqrt2z)(z^2+4-2\sqrt2z)=(z^2+4)^2-8z^2\)	M1	Any correct method of multiplying the two quadratics.
\(=z^4+16\)	AG	The four roots are distinct, so the product of the four linear factors divides \(P(z)\).

(4)

(f)(i)

Use De Moivre's theorem to show that \(\cos3\theta=4\cos^3\theta-3\cos\theta\).

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\(\cos3\theta+i\sin3\theta=(\cos\theta+i\sin\theta)^3=\cos^3\theta+3i\cos^2\theta\sin\theta-3\cos\theta\sin^2\theta-i\sin^3\theta\)	M1	De Moivre and an attempt at the binomial expansion.
real parts: \(\cos3\theta=\cos^3\theta-3\cos\theta\sin^2\theta\)	A1
\(=\cos^3\theta-3\cos\theta(1-\cos^2\theta)\)	A1	Uses \(\sin^2\theta=1-\cos^2\theta\).
\(=4\cos^3\theta-3\cos\theta\)	AG

(3)

(f)(ii)

The east coordinate of drone \(D_1\) is \(x_1=2\cos\dfrac{\pi}{12}\). Show that \(x_1\) is a root of the equation \(x^3-3x-\sqrt2=0\).

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\(\theta=\frac{\pi}{12}\): \(\cos\frac{\pi}{4}=4\cos^3\frac{\pi}{12}-3\cos\frac{\pi}{12}\), with \(\cos\frac{\pi}{12}=\frac{x_1}{2}\)	M1
\(\frac{\sqrt2}{2}=\frac{x_1^3}{2}-\frac{3x_1}{2}\)	A1
\(x_1^3-3x_1-\sqrt2=0\)	AG

(2)

(f)(iii)

Explain why the east coordinates of drones \(D_3\) and \(D_5\) are also roots of \(x^3-3x-\sqrt2=0\).

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\(\arg D_3=\frac{\pi}{12}+\frac{2\pi}{3}=\frac{3\pi}{4}\) and \(\arg D_5=\frac{\pi}{12}-\frac{2\pi}{3}=-\frac{7\pi}{12}\)	R1	The arguments differ from \(\frac{\pi}{12}\) by \(\pm\frac{2\pi}{3}\).
so \(3\theta=\frac{\pi}{4}\pm2\pi\) and \(\cos3\theta=\frac{\sqrt2}{2}\) again; hence \(x=2\cos\theta\) satisfies the same equation	R1	Accept verifying \(x=2\cos\frac{3\pi}{4}=-\sqrt2\): \(-2\sqrt2+3\sqrt2-\sqrt2=0\), for the first R1 only.

(2)

(g)(i)

For the finale, every drone moves from its position \(z\) to the position \(w=\dfrac32e^{i\pi/12}z\).

Describe this change geometrically, and show that the new positions satisfy \(w^6=-729\).

No calculatorMULTINotes

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enlargement scale factor \(\frac32\), centre \(O\), and rotation \(\frac{\pi}{12}\) anticlockwise about \(O\)	A1
\(w^6=\left(\frac32\right)^6e^{i\pi/2}z^6=\frac{729}{64}\times i\times64i\)	M1	De Moivre applied to \(\frac32e^{i\pi/12}\), and \(z^6=64i\) substituted.
\(=729i^2\)	A1
\(=-729\)	AG

(3)

(g)(ii)

Write down the exact area of the new hexagon.

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\(\frac{27\sqrt3}{2}\ \mathrm{m}^2\)

\(\left(\frac32\right)^2\times6\sqrt3\). Allow ECF from (c).

Answer: \(\frac{27\sqrt3}{2}\) m²

(1)

Show full solution

Working	Mark	Notes
Part (a)
\(64i=64e^{i\pi/2}\)	(A1)
\(r=64^{1/6}=2\)	A1
\(\theta=\frac{\frac{\pi}{2}+2\pi k}{6}=\frac{\pi}{12}+\frac{k\pi}{3}\), \(k=0,1,\dots,5\)	M1	Or any equivalent \(2\pi k\) method, e.g. rotating one root by \(\frac{\pi}{3}\).
\(2e^{i\pi/12},\ 2e^{i5\pi/12},\ 2e^{i3\pi/4},\ 2e^{-i11\pi/12},\ 2e^{-i7\pi/12},\ 2e^{-i\pi/4}\)	A1	All six, with principal arguments. Arguments \(\frac{13\pi}{12},\frac{17\pi}{12},\frac{7\pi}{4}\) not reduced to \((-\pi,\pi]\): A0 for this row.
Part (b)
six points on the circle of radius 2 centred at \(O\), spaced \(\frac{\pi}{3}\) apart	A1	Allow a small tolerance in plotting.
correct positions and labels, with \(D_1\) at argument \(\frac{\pi}{12}\) (just above east) and \(D_3\) at \(\frac{3\pi}{4}\) (north-west)	A1	Allow ECF from (a) for labels if the points are consistent with their answers.
Part (c)
all six drones are 2 m from \(O\) and consecutive arguments differ by \(\frac{\pi}{3}\), so the six sides and six angles are equal	R1	Accept: the hexagon splits into six congruent isosceles triangles with apex angle \(\frac{\pi}{3}\) at \(O\).
area \(=6\times\frac12\times2\times2\times\sin\frac{\pi}{3}\)	M1	Uses \(\frac12ab\sin C\) for one triangle and multiplies by 6.
\(=6\sqrt3\ \mathrm{m}^2\)	A1
Part (d)
sum of the roots \(=-\frac{\text{coefficient of }z^5}{\text{coefficient of }z^6}=-\frac{0}{1}\)	M1	The coefficient of \(z^5\) in \(z^6-64i\) is 0. Accept a symmetry argument: the roots form three pairs \(z,\,-z\).
\(=0\)	AG
the mean position of the six drones is \(O\): the formation is centred directly above the centre of the stage	R1
Part (e)(i)
\(D_3=2\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)\)	M1	Uses \(a=r\cos\theta\), \(b=r\sin\theta\) for either drone.
\(D_3=-\sqrt2+\sqrt2\,i\), \(D_6=\sqrt2-\sqrt2\,i\)	A1	Both. Accept \(D_6=-D_3\) shown.
Part (e)(ii)
real coefficients, so the conjugates \(-\sqrt2-\sqrt2\,i\) and \(\sqrt2+\sqrt2\,i\) are also roots	R1	Must refer to real coefficients.
pair \(-\sqrt2\pm\sqrt2\,i\): sum \(-2\sqrt2\), product \(4\), giving \(z^2+2\sqrt2z+4\)	M1	Using sum and product, or expanding \((z-w)(z-w^*)\), for either pair.
\(z^2+2\sqrt2z+4\) and \(z^2-2\sqrt2z+4\)	A1	Both quadratic factors.
\((z^2+4+2\sqrt2z)(z^2+4-2\sqrt2z)=(z^2+4)^2-8z^2\)	M1	Any correct method of multiplying the two quadratics.
\(=z^4+16\)	AG	The four roots are distinct, so the product of the four linear factors divides \(P(z)\).
Part (f)(i)
\(\cos3\theta+i\sin3\theta=(\cos\theta+i\sin\theta)^3=\cos^3\theta+3i\cos^2\theta\sin\theta-3\cos\theta\sin^2\theta-i\sin^3\theta\)	M1	De Moivre and an attempt at the binomial expansion.
real parts: \(\cos3\theta=\cos^3\theta-3\cos\theta\sin^2\theta\)	A1
\(=\cos^3\theta-3\cos\theta(1-\cos^2\theta)\)	A1	Uses \(\sin^2\theta=1-\cos^2\theta\).
\(=4\cos^3\theta-3\cos\theta\)	AG
Part (f)(ii)
\(\theta=\frac{\pi}{12}\): \(\cos\frac{\pi}{4}=4\cos^3\frac{\pi}{12}-3\cos\frac{\pi}{12}\), with \(\cos\frac{\pi}{12}=\frac{x_1}{2}\)	M1
\(\frac{\sqrt2}{2}=\frac{x_1^3}{2}-\frac{3x_1}{2}\)	A1
\(x_1^3-3x_1-\sqrt2=0\)	AG
Part (f)(iii)
\(\arg D_3=\frac{\pi}{12}+\frac{2\pi}{3}=\frac{3\pi}{4}\) and \(\arg D_5=\frac{\pi}{12}-\frac{2\pi}{3}=-\frac{7\pi}{12}\)	R1	The arguments differ from \(\frac{\pi}{12}\) by \(\pm\frac{2\pi}{3}\).
so \(3\theta=\frac{\pi}{4}\pm2\pi\) and \(\cos3\theta=\frac{\sqrt2}{2}\) again; hence \(x=2\cos\theta\) satisfies the same equation	R1	Accept verifying \(x=2\cos\frac{3\pi}{4}=-\sqrt2\): \(-2\sqrt2+3\sqrt2-\sqrt2=0\), for the first R1 only.
Part (g)(i)
enlargement scale factor \(\frac32\), centre \(O\), and rotation \(\frac{\pi}{12}\) anticlockwise about \(O\)	A1
\(w^6=\left(\frac32\right)^6e^{i\pi/2}z^6=\frac{729}{64}\times i\times64i\)	M1	De Moivre applied to \(\frac32e^{i\pi/12}\), and \(z^6=64i\) substituted.
\(=729i^2\)	A1
\(=-729\)	AG
Part (g)(ii)
\(\frac{27\sqrt3}{2}\ \mathrm{m}^2\)	A1	\(\left(\frac32\right)^2\times6\sqrt3\). Allow ECF from (c).

Answers: (a) \(2e^{i\pi/12},\ 2e^{i5\pi/12},\ 2e^{i3\pi/4},\ 2e^{-i11\pi/12},\ 2e^{-i7\pi/12},\ 2e^{-i\pi/4}\) · (c) \(6\sqrt3\) m² · (e)(i) \(D_3=-\sqrt2+\sqrt2\,i\), \(D_6=\sqrt2-\sqrt2\,i\) · (g)(ii) \(\frac{27\sqrt3}{2}\) m²

Common error (a) Dividing only \(\frac{\pi}{2}\) by 6 gives a single drone. Write \(64i=64e^{i(\pi/2+2\pi k)}\) before taking the sixth root, so that all six positions appear.

More 1.14 questions in the question bank

1.15 Proof by induction, contradiction and counterexample · 1.16 Systems of linear equationsmerged group

+ Add to test✓ In your testCC-AA-115-116-127 marks~30 min

Cabling a server room: induction, contradiction and counterexample

1.15 Proof by induction, contradiction and counterexample1.2 Arithmetic sequences and series1.10 Counting principles and the extended binomial theorem1.6 Deductive proofHL

A data centre has \(n\) servers, where \(n\ge2\). In a full mesh network every pair of servers is joined by exactly one cable. Let \(C(n)\) be the number of cables in a full mesh of \(n\) servers.

The servers stand in one straight row and are numbered \(S_1,S_2,\dots,S_n\) from one end, with adjacent servers 1 m apart. The cables run along a tray beside the row, so the cable joining \(S_i\) and \(S_j\) has length \(|i-j|\) metres. Figure 1 shows the full mesh for \(n=5\).

Figure 1: the full mesh of 5 servers in a row; each arc is one cable

(a)

Explain why \(C(n)=\dbinom{n}{2}\), and hence show that \(C(n)=\dfrac{n(n-1)}{2}\).

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each cable corresponds to choosing 2 of the \(n\) servers, and the order does not matter (\(S_i\) to \(S_j\) is the same cable as \(S_j\) to \(S_i\))	R1
\(\binom{n}{2}=\frac{n!}{2!\,(n-2)!}=\frac{n(n-1)(n-2)!}{2\,(n-2)!}\)	M1	Uses the formula for \(\binom{n}{r}\) and cancels \((n-2)!\).
\(=\frac{n(n-1)}{2}\)	AG

(2)

(b)

The servers are installed one at a time, in order. When server \(S_k\) is installed (\(k\ge2\)), it is cabled to each of \(S_1,S_2,\dots,S_{k-1}\).

Show that the total length of the new cables needed for \(S_k\) is \(\dfrac{k(k-1)}{2}\) metres.

No calculatorROUTINENotes

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the new cables have lengths \(1,2,\dots,k-1\) m: an arithmetic series with \(k-1\) terms, \(u_1=1\), \(d=1\)	R1	Accept the lengths listed in either order.
sum of the \(k-1\) terms \(=\frac{k-1}{2}\bigl(1+(k-1)\bigr)\)	M1	Or \(\frac{k-1}{2}\bigl(2(1)+(k-2)(1)\bigr)\).
\(=\frac{k(k-1)}{2}\)	AG

(2)

(c)

Let \(L(n)\) be the total length, in metres, of all the cables in the full mesh, so that \(L(n)=\displaystyle\sum_{r=2}^{n}\frac{r(r-1)}{2}\).

Prove by mathematical induction that \(L(n)=\dfrac{(n-1)n(n+1)}{6}\) for all integers \(n\ge2\).

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\(n=2\): \(\mathrm{LHS}=\frac{2\times1}{2}=1\), \(\mathrm{RHS}=\frac{1\times2\times3}{6}=1\), so true for \(n=2\)	A1	Both sides must be evaluated.
assume true for \(n=k\), for some integer \(k\ge2\): \(\sum_{r=2}^{k}\frac{r(r-1)}{2}=\frac{(k-1)k(k+1)}{6}\)	M1	Do not accept "let \(n=k\)" with no statement of the assumption.
\(\sum_{r=2}^{k+1}\frac{r(r-1)}{2}=\frac{(k-1)k(k+1)}{6}+\frac{(k+1)k}{2}\)	M1	Adds the \((k+1)\)th term to the assumed sum.
\(=\frac{k(k+1)\bigl[(k-1)+3\bigr]}{6}\)	A1	Correct factorising or full expansion.
\(=\frac{k(k+1)(k+2)}{6}\), which is the formula with \(n=k+1\)	A1
true for \(n=2\), and true for \(n=k\) implies true for \(n=k+1\); so true for all integers \(n\ge2\) by mathematical induction	R1	Award R1 only if the A1 for \(n=2\) and both M1 marks are awarded.

(6)

(d)

Without using induction, prove that \((n-1)n(n+1)\) is a multiple of 6 for every integer \(n\ge2\).

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\(n-1,\ n,\ n+1\) are three consecutive integers, so at least one of them is even: the product is a multiple of 2	R1
one of them is a multiple of 3 (the remainders on division by 3 are \(0,1,2\) in some order): the product is a multiple of 3	R1
2 and 3 have no common factor (other than 1), so the product is a multiple of \(2\times3=6\)	R1	Must justify combining the two facts.

(3)

(e)

Figure 2 shows \(C(n)\) for \(2\le n\le8\). A technician joins the points for \(n=2\) and \(n=8\) with a straight line and uses this line to estimate the number of cables needed for 5 servers.

Use Figure 2 to write down the technician's estimate, and find the true value of \(C(5)\). By considering the differences \(C(n+1)-C(n)\), explain why the straight line overestimates the number of cables.

Figure 2: the number of cables in a full mesh, with the technician's straight line (dashed)

GRAPHNotes

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estimate \(=14.5\)	B1	Accept 14 to 15 read from the graph.
\(C(5)=10\)	B1	From \(\frac{5\times4}{2}\) or read from Figure 2.
\(C(n+1)-C(n)=n\), so the increases \(2,3,4,\dots\) grow (an arithmetic sequence): the points curve upwards, so between the end points they lie below the straight line	R1	Accept: the graph of \(C(n)\) is quadratic (gradient increasing), not linear.

Answer: estimate 14.5; \(C(5)=10\)

(3)

(f)(i)

In a different design, not every pair of servers needs to be joined, but no two servers are joined by more than one cable. The degree of a server is the number of cables attached to it.

Suppose, for a contradiction, that in such a network of \(n\ge2\) servers all \(n\) servers have different degrees. Explain why the degrees must then be exactly \(0,1,2,\dots,n-1\).

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a server can be joined to at most the \(n-1\) other servers, so each degree is one of the \(n\) integers \(0,1,\dots,n-1\)	R1
\(n\) different degrees chosen from only \(n\) possible values, so each value occurs exactly once	R1

(2)

(f)(ii)

Hence complete the proof that every such network has at least two servers with the same degree.

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the server of degree \(n-1\) is joined to every other server, including the server of degree 0	R1	Since \(n\ge2\), these are two different servers.
this contradicts the server of degree 0 having no cables; so the assumption is false and at least two servers have the same degree	R1	A clear concluding statement is required.

(2)

(g)

A technician claims that, for \(n\ge3\), the number of cables \(C(n)\) is never a perfect square. Show that the claim is false.

PROOFNotes

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\(n=9\): \(C(9)=\frac{9\times8}{2}=36\)	A1	Accept any valid counterexample, e.g. \(C(50)=1225=35^2\) or \(C(289)=41\,616=204^2\).
\(36=6^2\) is a perfect square, so the claim is false	R1	A conclusion is required. Award R0 for \(n=2\) (\(C(2)=1\)), which is outside the claim.

Answer: e.g. \(C(9)=36=6^2\)

(2)

(h)(i)

The 12 servers of a small data centre are split between two racks: 7 in rack 1 and 5 in rack 2. To save cable, every server in rack 1 is joined to every server in rack 2, and there are no cables between servers in the same rack.

Find the number of cables used, and the number saved compared with a full mesh of 12 servers.

ROUTINENotes

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\(7\times5=35\) cables	A1	Multiplication principle.
\(C(12)-35=66-35=31\) saved	A1	Allow ECF from their 35.

Answer: 35 cables; 31 saved

(2)

(h)(ii)

Now \(n=2m\) servers are split equally, with \(m\) servers in each rack and cables only between the racks. Show that the number of cables saved, compared with a full mesh of \(2m\) servers, is \(m(m-1)\).

No calculatorALGEBRANotes

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\(C(2m)-m^2=\frac{2m(2m-1)}{2}-m^2\)	M1	\(m\times m=m^2\) cross-rack cables.
\(=2m^2-m-m^2\)	A1
\(=m^2-m=m(m-1)\)	AG

(2)

(h)(iii)

Interpret the expression \(m(m-1)\) in terms of the cables in a full mesh.

No calculatorINTERPRETNotes

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\(m(m-1)=2\times\frac{m(m-1)}{2}=2C(m)\): the saved cables are exactly the cables a full mesh would have inside each of the two racks

Accept a check with (h)(i): \(31=C(7)+C(5)=21+10\).

(1)

Show full solution

Working	Mark	Notes
Part (a)
each cable corresponds to choosing 2 of the \(n\) servers, and the order does not matter (\(S_i\) to \(S_j\) is the same cable as \(S_j\) to \(S_i\))	R1
\(\binom{n}{2}=\frac{n!}{2!\,(n-2)!}=\frac{n(n-1)(n-2)!}{2\,(n-2)!}\)	M1	Uses the formula for \(\binom{n}{r}\) and cancels \((n-2)!\).
\(=\frac{n(n-1)}{2}\)	AG
Part (b)
the new cables have lengths \(1,2,\dots,k-1\) m: an arithmetic series with \(k-1\) terms, \(u_1=1\), \(d=1\)	R1	Accept the lengths listed in either order.
sum of the \(k-1\) terms \(=\frac{k-1}{2}\bigl(1+(k-1)\bigr)\)	M1	Or \(\frac{k-1}{2}\bigl(2(1)+(k-2)(1)\bigr)\).
\(=\frac{k(k-1)}{2}\)	AG
Part (c)
\(n=2\): \(\mathrm{LHS}=\frac{2\times1}{2}=1\), \(\mathrm{RHS}=\frac{1\times2\times3}{6}=1\), so true for \(n=2\)	A1	Both sides must be evaluated.
assume true for \(n=k\), for some integer \(k\ge2\): \(\sum_{r=2}^{k}\frac{r(r-1)}{2}=\frac{(k-1)k(k+1)}{6}\)	M1	Do not accept "let \(n=k\)" with no statement of the assumption.
\(\sum_{r=2}^{k+1}\frac{r(r-1)}{2}=\frac{(k-1)k(k+1)}{6}+\frac{(k+1)k}{2}\)	M1	Adds the \((k+1)\)th term to the assumed sum.
\(=\frac{k(k+1)\bigl[(k-1)+3\bigr]}{6}\)	A1	Correct factorising or full expansion.
\(=\frac{k(k+1)(k+2)}{6}\), which is the formula with \(n=k+1\)	A1
true for \(n=2\), and true for \(n=k\) implies true for \(n=k+1\); so true for all integers \(n\ge2\) by mathematical induction	R1	Award R1 only if the A1 for \(n=2\) and both M1 marks are awarded.
Part (d)
\(n-1,\ n,\ n+1\) are three consecutive integers, so at least one of them is even: the product is a multiple of 2	R1
one of them is a multiple of 3 (the remainders on division by 3 are \(0,1,2\) in some order): the product is a multiple of 3	R1
2 and 3 have no common factor (other than 1), so the product is a multiple of \(2\times3=6\)	R1	Must justify combining the two facts.
Part (e)
estimate \(=14.5\)	B1	Accept 14 to 15 read from the graph.
\(C(5)=10\)	B1	From \(\frac{5\times4}{2}\) or read from Figure 2.
\(C(n+1)-C(n)=n\), so the increases \(2,3,4,\dots\) grow (an arithmetic sequence): the points curve upwards, so between the end points they lie below the straight line	R1	Accept: the graph of \(C(n)\) is quadratic (gradient increasing), not linear.
Part (f)(i)
a server can be joined to at most the \(n-1\) other servers, so each degree is one of the \(n\) integers \(0,1,\dots,n-1\)	R1
\(n\) different degrees chosen from only \(n\) possible values, so each value occurs exactly once	R1
Part (f)(ii)
the server of degree \(n-1\) is joined to every other server, including the server of degree 0	R1	Since \(n\ge2\), these are two different servers.
this contradicts the server of degree 0 having no cables; so the assumption is false and at least two servers have the same degree	R1	A clear concluding statement is required.
Part (g)
\(n=9\): \(C(9)=\frac{9\times8}{2}=36\)	A1	Accept any valid counterexample, e.g. \(C(50)=1225=35^2\) or \(C(289)=41\,616=204^2\).
\(36=6^2\) is a perfect square, so the claim is false	R1	A conclusion is required. Award R0 for \(n=2\) (\(C(2)=1\)), which is outside the claim.
Part (h)(i)
\(7\times5=35\) cables	A1	Multiplication principle.
\(C(12)-35=66-35=31\) saved	A1	Allow ECF from their 35.
Part (h)(ii)
\(C(2m)-m^2=\frac{2m(2m-1)}{2}-m^2\)	M1	\(m\times m=m^2\) cross-rack cables.
\(=2m^2-m-m^2\)	A1
\(=m^2-m=m(m-1)\)	AG
Part (h)(iii)
\(m(m-1)=2\times\frac{m(m-1)}{2}=2C(m)\): the saved cables are exactly the cables a full mesh would have inside each of the two racks	R1	Accept a check with (h)(i): \(31=C(7)+C(5)=21+10\).

Answers: (e) estimate 14.5; \(C(5)=10\) · (g) e.g. \(C(9)=36=6^2\) · (h)(i) 35 cables; 31 saved

Exam tip (c) Before you start the inductive step, write down the target: \(\frac{k(k+1)(k+2)}{6}\). Taking out the common factor \(k(k+1)\) at once then reaches it in two lines.

Common error (g) A counterexample must satisfy the conditions of the claim. \(C(2)=1=1^2\), but \(n=2\) is excluded by \(n\ge3\).

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A smoothie bar's recipe: three ingredients, three targets

1.16 Systems of linear equations3.17 Equations of a plane3.18 Intersections and angles of lines and planes2.1 Equations of straight lines, parallel and perpendicular linesHL

A smoothie bar makes a jug of smoothie for two people from natural yoghurt, oats and mixed berries. The jug contains \(x\) hundred grams of yoghurt, \(y\) hundred grams of oats and \(z\) hundred grams of berries. The table shows the mass of protein, fibre and sugar in 100 g of each ingredient.

per 100 g	protein (g)	fibre (g)	sugar (g)
yoghurt	5	0	5
oats	12	10	1
berries	1	5	7

The jug must contain exactly 29 g of protein, 20 g of fibre and 30 g of sugar.

(a)

Write down a system of three linear equations in \(x\), \(y\) and \(z\) that represents these requirements.

No calculatorMODELNotes

Show solution

one correct equation formed from a column of the table	M1
\(5x+12y+z=29\), \(10y+5z=20\), \(5x+y+7z=30\)	A1	All three correct. Accept equivalent forms, e.g. \(2y+z=4\).

Answer: \(5x+12y+z=29\), \(10y+5z=20\), \(5x+y+7z=30\)

(2)

(b)

Solve the system by elimination, showing all your working. Hence state the mass of each ingredient in the jug.

No calculatorALGEBRANotes

Show solution

eliminating \(x\): (3) \(-\) (1)	M1	Or row operations on the augmented array. The fibre equation simplifies to \(2y+z=4\).
\(-11y+6z=1\) and \(2y+z=4\)	A1
\(z=4-2y\Rightarrow-11y+24-12y=1\), so \(y=1\)	M1	Solving their two equations in two unknowns.
\(x=3,\ y=1,\ z=2\): 300 g yoghurt, 100 g oats, 200 g berries	A1	Answers from a GDC with no working score 0.

Answer: \(x=3\), \(y=1\), \(z=2\): 300 g yoghurt, 100 g oats, 200 g berries

(4)

(c)(i)

Another recipe card fixes the berries at 200 g, so \(z=2\). Show that the protein and sugar requirements become \(5x+12y=27\) and \(5x+y=16\). Write down the gradient of each line when it is drawn with \(x\) on the horizontal axis and \(y\) on the vertical axis.

No calculatorALGEBRANotes

Show solution

\(5x+12y+2=29\) and \(5x+y+14=30\)	M1	Substituting \(z=2\) into both equations.
\(5x+12y=27\) and \(5x+y=16\)	AG
gradients \(-\frac{5}{12}\) and \(-5\)	A1	Both.

Answer: gradients \(-\frac{5}{12}\) and \(-5\)

(2)

(c)(ii)

On the axes in Figure 1, draw the lines \(5x+12y=27\) and \(5x+y=16\). Write down the coordinates of their point of intersection, and interpret this point in context.

Figure 1: axes for part (c)(ii)

No calculatorGRAPHNotes

Show solution

\(5x+12y=27\) drawn through \((0,\,2.25)\) and \((5.4,\,0)\)	A1
\(5x+y=16\) drawn through \((3.2,\,0)\) and \((2.4,\,4)\)	A1
\((3,\,1)\): with 200 g of berries, 300 g of yoghurt and 100 g of oats meet the protein and sugar targets (the recipe of (b), which also meets the fibre target)	R1	Coordinates and an interpretation in context are both needed.

Answer: \((3,1)\)

(3)

(d)(i)

The oats are replaced by a sweetened granola. Per 100 g it has the same protein and fibre as the oats, but contains \(k\) grams of sugar. The three targets are unchanged.

Show that the system has a unique solution if and only if \(k\ne24\).

No calculatorALGEBRANotes

Show solution

new (3) \(-\) (1): \((k-12)y+6z=1\), and substitute \(z=4-2y\)	M1
\((k-24)y=-23\)	A1
if \(k\ne24\), \(y=\frac{23}{24-k}\) is unique, and then \(z\) and \(x\) are unique; if \(k=24\) the equation becomes \(0=-23\), so there is no unique solution	R1	Both directions are needed.
unique solution \(\iff k\ne24\)	AG

(3)

(d)(ii)

Given that \(k\lt24\), express \(x\) and \(z\) in terms of \(y\). Hence find the largest value of \(k\) for which all three targets can be met with non-negative amounts of every ingredient, and state the recipe for this value of \(k\).

MODELNotes

Show solution

\(z=4-2y\) and \(5x=29-12y-(4-2y)\)	M1	Using the fibre and protein equations, which do not involve \(k\).
\(x=5-2y\), \(z=4-2y\)	A1
\(z\ge0\iff y\le2\) (this is stronger than \(x\ge0\iff y\le2.5\)), so \(\frac{23}{24-k}\le2\)	M1	Allow ECF from (d)(i).
\(k=12.5\); recipe: 100 g yoghurt, 200 g granola, no berries	A1	\(y=2\), \(x=1\), \(z=0\).

Answer: \(x=5-2y\), \(z=4-2y\); \(k=12.5\): 100 g yoghurt, 200 g granola, 0 g berries

(4)

(e)(i)

Now let \(k=24\). Each equation of the system represents a plane.

Explain why the system has no solution, and describe how the three planes are arranged. Justify your answer.

No calculatorINTERPRETNotes

Show solution

\((k-24)y=-23\) becomes \(0=-23\), which is impossible: the system is inconsistent	R1
normals \((5,12,1)\), \((0,10,5)\), \((5,24,7)\): no one is a multiple of another, so no two planes are parallel	R1	Normals read from the coefficients.
each pair of planes meets in a line, and the three lines are parallel: the planes form a triangular prism	A1	Accept "triangular tube".

Answer: no solution; the planes form a triangular prism

(3)

(e)(ii)

With \(k=24\), the sugar target is changed to \(m\) grams. Find the value of \(m\) for which the system has infinitely many solutions. For this value of \(m\), find the general solution in terms of a parameter \(\lambda\).

No calculatorALGEBRANotes

Show solution

(3) \(-\) (1): \(12y+6z=m-29\), and \(12y+6z=6(2y+z)=24\)	M1	Comparing with the fibre equation.
\(m=53\)	A1
let \(z=2\lambda\), then \(y=2-\lambda\) and \(x=\frac{29-12(2-\lambda)-2\lambda}{5}\)	M1	Any choice of parameter.
\(x=1+2\lambda,\ y=2-\lambda,\ z=2\lambda\), \(\lambda\in\mathbb{R}\)	A1	Or equivalent, e.g. \(y=\lambda\): \(x=5-2\lambda,\ z=4-2\lambda\).

Answer: \(m=53\); \(x=1+2\lambda,\ y=2-\lambda,\ z=2\lambda\)

(4)

(e)(iii)

Find the set of values of \(\lambda\) for which the general solution in (e)(ii) gives a possible recipe.

No calculatorINTERPRETNotes

Show solution

\(1+2\lambda\ge0\), \(2-\lambda\ge0\) and \(2\lambda\ge0\)	R1	All three amounts non-negative.
\(0\le\lambda\le2\)	A1	Follow through from their parametrisation (e.g. \(0\le\lambda\le2\) for \(y=\lambda\)).

Answer: \(0\le\lambda\le2\)

(2)

Show full solution

Working	Mark	Notes
Part (a)
one correct equation formed from a column of the table	M1
\(5x+12y+z=29\), \(10y+5z=20\), \(5x+y+7z=30\)	A1	All three correct. Accept equivalent forms, e.g. \(2y+z=4\).
Part (b)
eliminating \(x\): (3) \(-\) (1)	M1	Or row operations on the augmented array. The fibre equation simplifies to \(2y+z=4\).
\(-11y+6z=1\) and \(2y+z=4\)	A1
\(z=4-2y\Rightarrow-11y+24-12y=1\), so \(y=1\)	M1	Solving their two equations in two unknowns.
\(x=3,\ y=1,\ z=2\): 300 g yoghurt, 100 g oats, 200 g berries	A1	Answers from a GDC with no working score 0.
Part (c)(i)
\(5x+12y+2=29\) and \(5x+y+14=30\)	M1	Substituting \(z=2\) into both equations.
\(5x+12y=27\) and \(5x+y=16\)	AG
gradients \(-\frac{5}{12}\) and \(-5\)	A1	Both.
Part (c)(ii)
\(5x+12y=27\) drawn through \((0,\,2.25)\) and \((5.4,\,0)\)	A1
\(5x+y=16\) drawn through \((3.2,\,0)\) and \((2.4,\,4)\)	A1
\((3,\,1)\): with 200 g of berries, 300 g of yoghurt and 100 g of oats meet the protein and sugar targets (the recipe of (b), which also meets the fibre target)	R1	Coordinates and an interpretation in context are both needed.
Part (d)(i)
new (3) \(-\) (1): \((k-12)y+6z=1\), and substitute \(z=4-2y\)	M1
\((k-24)y=-23\)	A1
if \(k\ne24\), \(y=\frac{23}{24-k}\) is unique, and then \(z\) and \(x\) are unique; if \(k=24\) the equation becomes \(0=-23\), so there is no unique solution	R1	Both directions are needed.
unique solution \(\iff k\ne24\)	AG
Part (d)(ii)
\(z=4-2y\) and \(5x=29-12y-(4-2y)\)	M1	Using the fibre and protein equations, which do not involve \(k\).
\(x=5-2y\), \(z=4-2y\)	A1
\(z\ge0\iff y\le2\) (this is stronger than \(x\ge0\iff y\le2.5\)), so \(\frac{23}{24-k}\le2\)	M1	Allow ECF from (d)(i).
\(k=12.5\); recipe: 100 g yoghurt, 200 g granola, no berries	A1	\(y=2\), \(x=1\), \(z=0\).
Part (e)(i)
\((k-24)y=-23\) becomes \(0=-23\), which is impossible: the system is inconsistent	R1
normals \((5,12,1)\), \((0,10,5)\), \((5,24,7)\): no one is a multiple of another, so no two planes are parallel	R1	Normals read from the coefficients.
each pair of planes meets in a line, and the three lines are parallel: the planes form a triangular prism	A1	Accept "triangular tube".
Part (e)(ii)
(3) \(-\) (1): \(12y+6z=m-29\), and \(12y+6z=6(2y+z)=24\)	M1	Comparing with the fibre equation.
\(m=53\)	A1
let \(z=2\lambda\), then \(y=2-\lambda\) and \(x=\frac{29-12(2-\lambda)-2\lambda}{5}\)	M1	Any choice of parameter.
\(x=1+2\lambda,\ y=2-\lambda,\ z=2\lambda\), \(\lambda\in\mathbb{R}\)	A1	Or equivalent, e.g. \(y=\lambda\): \(x=5-2\lambda,\ z=4-2\lambda\).
Part (e)(iii)
\(1+2\lambda\ge0\), \(2-\lambda\ge0\) and \(2\lambda\ge0\)	R1	All three amounts non-negative.
\(0\le\lambda\le2\)	A1	Follow through from their parametrisation (e.g. \(0\le\lambda\le2\) for \(y=\lambda\)).

Answers: (a) \(5x+12y+z=29\), \(10y+5z=20\), \(5x+y+7z=30\) · (b) \(x=3\), \(y=1\), \(z=2\): 300 g yoghurt, 100 g oats, 200 g berries · (c)(i) gradients \(-\frac{5}{12}\) and \(-5\) · (c)(ii) \((3,1)\) · (d)(ii) \(x=5-2y\), \(z=4-2y\); \(k=12.5\): 100 g yoghurt, 200 g granola, 0 g berries · (e)(i) no solution; the planes form a triangular prism · (e)(ii) \(m=53\); \(x=1+2\lambda,\ y=2-\lambda,\ z=2\lambda\) · (e)(iii) \(0\le\lambda\le2\)

Common error (e)(ii) A row \(0=0\) means one equation is redundant, not that there is no solution. When a question says "find the general solution", give \(x\), \(y\) and \(z\) in terms of a parameter.

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Topic 2 · Functions

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Planning a new tram line on a city grid	●	○	○														3.2
A phone battery’s charge: modelling with functions	○	●	○		○												—
A basketball shot: is the ball going in?				○		●	○				○						—
Changing money on a school trip: composite and inverse functions	○	○		○	●												—
Designing a smartphone screen: quadratic constraints				○		○	●										3.2
A 3D printing workshop: a cost model, its inverse and printing times		○					○	●			○						—
A cup of coffee cooling in a room									●	○	○						1.7
Streaming services racing for subscribers				○					○	●	○						—
Designing a section of rollercoaster track with a cubic				○			○					●					5.8
A delivery truck's fuel use at different speeds												○	●		○		5.8
A full-wave rectifier: turning AC into DC														○		●	3.7 5.11
Keeping a greenhouse in the safe temperature band												○		○	●	○	—

2.1 Equations of straight lines, parallel and perpendicular lines · 2.2 Functions, domain, range and inverse functions · 2.3 Graphs of functions and sketching from a contextmerged group

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Planning a new tram line on a city grid

2.1 Equations of straight lines, parallel and perpendicular lines2.2 Functions, domain, range and inverse functions2.3 Graphs of functions and sketching from a context3.2 Right-angled and non-right-angled trianglesSL

A city is planning a new tram line. On a map of the city the \(x\)-axis points east and the \(y\)-axis points north. One unit on each axis represents 100 metres.

The tram line is straight and passes through station \(\mathrm{A}(2,\,3)\) and station \(\mathrm{B}(14,\,11)\). A school is at \(\mathrm{S}(12,\,1)\) and a bus depot is at \(\mathrm{C}(4,\,12)\). Harbour Road runs due east–west through \(\mathrm{A}\).

This information is shown in Figure 1.

Figure 1: the city map; one unit represents 100 m

(a)

Show that the equation of the tram line can be written as \(2x-3y+5=0\).

No calculatorPROOFNotes

Show solution

\(m=\frac{11-3}{14-2}\)	M1
\(=\frac{8}{12}=\frac{2}{3}\)	A1
\(y-3=\frac{2}{3}(x-2)\) or \(3=\frac{2}{3}(2)+c\Rightarrow c=\frac{5}{3}\)	M1	Substituting \(\mathrm{A}\) or \(\mathrm{B}\) into a line equation with their gradient.
\(3y-9=2x-4\), so \(2x-3y+5=0\)	AG	A correct intermediate line, with fractions cleared, must be seen.

Answer: \(2x-3y+5=0\)

(3)

(b)

A bus lane is to be built parallel to the tram line, passing through \(\mathrm{C}\).

Find the equation of the bus lane. Give your answer in the form \(y=mx+c\).

No calculatorROUTINENotes

Show solution

gradient \(\frac{2}{3}\) and \(y-12=\frac{2}{3}(x-4)\)	(M1)	Same gradient as the tram line, with \(\mathrm{C}\) substituted.
\(y=\frac{2}{3}x+\frac{28}{3}\)	A1	Accept \(y=0.667x+9.33\).

Answer: \(y=\frac{2}{3}x+\frac{28}{3}\)

(2)

(c)(i)

A straight footpath is to be built from the school \(\mathrm{S}\) to the tram line, perpendicular to the tram line.

Find the equation of the line containing the footpath. Give your answer in the form \(ax+by+d=0\), where \(a,b,d\in\mathbb{Z}\).

No calculatorROUTINENotes

Show solution

gradient \(-\frac{3}{2}\) and \(y-1=-\frac{3}{2}(x-12)\)	(M1)	Negative reciprocal of their gradient from (a), with \(\mathrm{S}\) substituted.
\(3x+2y-38=0\)	A1	Accept any integer multiple.

Answer: \(3x+2y-38=0\)

(2)

(c)(ii)

The footpath meets the tram line at the point \(\mathrm{F}\).

Find the coordinates of \(\mathrm{F}\).

No calculatorALGEBRANotes

Show solution

attempt to solve \(2x-3y+5=0\) and \(3x+2y-38=0\) simultaneously, e.g. \(4x-6y=-10\) and \(9x+6y=114\)	M1	Allow ECF from (c)(i).
\(13x=104\), so \(x=8\)	A1
\(y=7\)	A1	\((8,\,7)\) stated with no working, or only read from Figure 1, scores 1 mark at most.

Answer: \(\mathrm{F}(8,\,7)\)

(3)

(d)

Find the length of the footpath, in metres. Give your answer to the nearest metre.

MULTINotes

Show solution

\(\mathrm{SF}=\sqrt{(12-8)^2+(1-7)^2}=\sqrt{52}\ (=7.2111\ldots)\)	M1	Allow ECF from their \(\mathrm{F}\).
\(721\) m	A1	Answer \(7.21\) with no conversion to metres scores A0.

Answer: 721 m

(2)

(e)

Find the acute angle between the tram line and Harbour Road.

ROUTINENotes

Show solution

\(\tan\theta=\frac{2}{3}\) (gradient = \(\frac{\text{rise}}{\text{run}}\))	M1	Or a right-angled triangle with sides 8 and 12 drawn from \(\mathrm{A}\) and \(\mathrm{B}\).
\(\theta=33.7^\circ\)	A1	Accept \(0.588\) radians.

Answer: \(33.7^\circ\)

(2)

(f)(i)

The city’s tram fare depends on the distance travelled, \(d\) km. For any journey of up to 2 km the fare is €1.50. For journeys longer than 2 km, the fare increases at a constant rate of €0.25 per km until it reaches a maximum fare of €3.50. The fare stays at €3.50 for all longer journeys. The longest possible journey on the network is 14 km.

The fare, in euros, is given by the function \(F(d)\).

Find the distance at which the fare first reaches €3.50.

No calculatorMODELNotes

Show solution

\(1.50+0.25(d-2)=3.50\) or \(\frac{3.50-1.50}{0.25}=8\)	M1	A linear model for the middle section, or the extra distance found.
\(d=10\) (km)	A1

Answer: 10 km

(2)

(f)(ii)

On Figure 2, sketch the graph of \(F\). Show clearly the fare and the distance at each point where the graph changes.

Figure 2: axes for the graph of the fare F against the distance d

No calculatorGRAPHNotes

Show solution

horizontal segment at \(F=1.50\) from \(d=0\) to \(d=2\)	A1	Accept an open circle or no circle at \(d=0\).
straight segment from \((2,\,1.50)\) to \((10,\,3.50)\)	A1	Allow ECF from (f)(i).
horizontal segment at \(F=3.50\) from \(d=10\) to \(d=14\), ending at \(d=14\)	A1	The graph must not continue beyond \(d=14\).

Answer: Three joined segments: \((0,\,1.50)\) to \((2,\,1.50)\), to \((10,\,3.50)\), to \((14,\,3.50)\)

(3)

(f)(iii)

Write down the domain and the range of \(F\).

No calculatorCONCEPTNotes

Show solution

domain \(0<d\le14\)	A1	Accept \(0\le d\le14\).
range \(1.50\le F(d)\le3.50\)	A1	Do not accept \(1.50<F(d)\le3.50\).

Answer: Domain \(0<d\le14\); range \(1.50\le F(d)\le3.50\)

(2)

(f)(iv)

Explain why \(F\) does not have an inverse function.

No calculatorCONCEPTNotes

Show solution

\(F\) is not one-to-one: for example every \(d\) with \(0<d\le2\) gives the same fare €1.50 (a horizontal line meets the graph more than once)

Any correct reason that refers to one output coming from more than one input.

(1)

(g)(i)

To reduce costs, the planners consider a different route. The tram line would instead run from \(\mathrm{A}\) due east along Harbour Road to a new terminus at \((16,\,3)\).

Write down the equation of the shortest footpath from \(\mathrm{S}\) to this new route, and the coordinates of the point \(\mathrm{G}\) where it meets the route.

No calculatorCONCEPTNotes

Show solution

\(x=12\)	A1	The route is horizontal (\(y=3\)), so the perpendicular is vertical.
\(\mathrm{G}(12,\,3)\)	A1

Answer: \(x=12\); \(\mathrm{G}(12,\,3)\)

(2)

(g)(ii)

Determine which of the two routes gives the shorter walk from the school to the tram line, and by how many metres.

INTERPRETNotes

Show solution

\(\mathrm{SG}=2\) units \(=200\) m, compared with \(721\) m from (d)	M1	Allow ECF from (d).
the route along Harbour Road, shorter by \(521\) m	A1	Both the route and the difference are needed.

Answer: The Harbour Road route, shorter by 521 m

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(m=\frac{11-3}{14-2}\)	M1
\(=\frac{8}{12}=\frac{2}{3}\)	A1
\(y-3=\frac{2}{3}(x-2)\) or \(3=\frac{2}{3}(2)+c\Rightarrow c=\frac{5}{3}\)	M1	Substituting \(\mathrm{A}\) or \(\mathrm{B}\) into a line equation with their gradient.
\(3y-9=2x-4\), so \(2x-3y+5=0\)	AG	A correct intermediate line, with fractions cleared, must be seen.
Part (b)
gradient \(\frac{2}{3}\) and \(y-12=\frac{2}{3}(x-4)\)	(M1)	Same gradient as the tram line, with \(\mathrm{C}\) substituted.
\(y=\frac{2}{3}x+\frac{28}{3}\)	A1	Accept \(y=0.667x+9.33\).
Part (c)(i)
gradient \(-\frac{3}{2}\) and \(y-1=-\frac{3}{2}(x-12)\)	(M1)	Negative reciprocal of their gradient from (a), with \(\mathrm{S}\) substituted.
\(3x+2y-38=0\)	A1	Accept any integer multiple.
Part (c)(ii)
attempt to solve \(2x-3y+5=0\) and \(3x+2y-38=0\) simultaneously, e.g. \(4x-6y=-10\) and \(9x+6y=114\)	M1	Allow ECF from (c)(i).
\(13x=104\), so \(x=8\)	A1
\(y=7\)	A1	\((8,\,7)\) stated with no working, or only read from Figure 1, scores 1 mark at most.
Part (d)
\(\mathrm{SF}=\sqrt{(12-8)^2+(1-7)^2}=\sqrt{52}\ (=7.2111\ldots)\)	M1	Allow ECF from their \(\mathrm{F}\).
\(721\) m	A1	Answer \(7.21\) with no conversion to metres scores A0.
Part (e)
\(\tan\theta=\frac{2}{3}\) (gradient = \(\frac{\text{rise}}{\text{run}}\))	M1	Or a right-angled triangle with sides 8 and 12 drawn from \(\mathrm{A}\) and \(\mathrm{B}\).
\(\theta=33.7^\circ\)	A1	Accept \(0.588\) radians.
Part (f)(i)
\(1.50+0.25(d-2)=3.50\) or \(\frac{3.50-1.50}{0.25}=8\)	M1	A linear model for the middle section, or the extra distance found.
\(d=10\) (km)	A1
Part (f)(ii)
horizontal segment at \(F=1.50\) from \(d=0\) to \(d=2\)	A1	Accept an open circle or no circle at \(d=0\).
straight segment from \((2,\,1.50)\) to \((10,\,3.50)\)	A1	Allow ECF from (f)(i).
horizontal segment at \(F=3.50\) from \(d=10\) to \(d=14\), ending at \(d=14\)	A1	The graph must not continue beyond \(d=14\).
Part (f)(iii)
domain \(0<d\le14\)	A1	Accept \(0\le d\le14\).
range \(1.50\le F(d)\le3.50\)	A1	Do not accept \(1.50<F(d)\le3.50\).
Part (f)(iv)
\(F\) is not one-to-one: for example every \(d\) with \(0<d\le2\) gives the same fare €1.50 (a horizontal line meets the graph more than once)	R1	Any correct reason that refers to one output coming from more than one input.
Part (g)(i)
\(x=12\)	A1	The route is horizontal (\(y=3\)), so the perpendicular is vertical.
\(\mathrm{G}(12,\,3)\)	A1
Part (g)(ii)
\(\mathrm{SG}=2\) units \(=200\) m, compared with \(721\) m from (d)	M1	Allow ECF from (d).
the route along Harbour Road, shorter by \(521\) m	A1	Both the route and the difference are needed.

Answers: (a) \(2x-3y+5=0\) · (b) \(y=\frac{2}{3}x+\frac{28}{3}\) · (c)(i) \(3x+2y-38=0\) · (c)(ii) \(\mathrm{F}(8,\,7)\) · (d) 721 m · (e) \(33.7^\circ\) · (f)(i) 10 km · (f)(ii) Three joined segments: \((0,\,1.50)\) to \((2,\,1.50)\), to \((10,\,3.50)\), to \((14,\,3.50)\) · (f)(iii) Domain \(0<d\le14\); range \(1.50\le F(d)\le3.50\) · (g)(i) \(x=12\); \(\mathrm{G}(12,\,3)\) · (g)(ii) The Harbour Road route, shorter by 521 m

Exam tip (a) In a “show that” the final line is given, so every step that leads to it must be written down. Checking that \(\mathrm{B}\) also satisfies the equation is a good habit, but on its own it does not show where the equation comes from.

Common error (e) Using \(\tan\theta=\frac{3}{2}\) gives the angle between the tram line and a north–south street, not Harbour Road. Draw the right-angled triangle first: the run is east–west and the rise is north–south.

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A phone battery’s charge: modelling with functions

2.2 Functions, domain, range and inverse functions2.1 Equations of straight lines, parallel and perpendicular lines2.3 Graphs of functions and sketching from a context2.5 Composite functions and inverse functionsSL

Lena’s phone is fully charged (100%) at 08:00. She then streams music on it, and the battery charge falls at a constant rate of 8 percentage points per hour.

Let \(B(t)\) be the battery charge, in %, \(t\) hours after 08:00, while the phone is streaming.

(a)

Write down \(B(t)\) in the form \(B(t)=mt+c\), and interpret the value of \(m\) in context.

No calculatorMODELNotes

Show solution

\(B(t)=100-8t\)	A1	Accept \(B(t)=-8t+100\).
\(m=-8\): the charge falls by 8 percentage points for each hour of streaming	R1	Must refer to the rate per hour. “8% per hour” is accepted.

Answer: \(B(t)=100-8t\)

(2)

(b)

The model is valid from \(t=0\) until the battery is empty.

Find the domain and the range of \(B\).

No calculatorCONCEPTNotes

Show solution

domain \(0\le t\le12.5\)	A1	From \(100-8t=0\).
range \(0\le B(t)\le100\)	A1

Answer: Domain \(0\le t\le12.5\); range \(0\le B(t)\le100\)

(2)

(c)(i)

Find \(B^{-1}(x)\), and state its domain.

No calculatorALGEBRANotes

Show solution

\(x=100-8y\) or \(t=\frac{100-B}{8}\)	M1	Swapping the variables or rearranging.
\(B^{-1}(x)=\frac{100-x}{8}\)	A1	Accept \(12.5-\frac{x}{8}\).
domain \(0\le x\le100\)	A1	Allow ECF from the range in (b).

Answer: \(B^{-1}(x)=\dfrac{100-x}{8}\), \(0\le x\le100\)

(3)

(c)(ii)

Find \(B^{-1}(40)\) and interpret your answer in context.

No calculatorINTERPRETNotes

Show solution

\(B^{-1}(40)=\frac{60}{8}=7.5\)	A1
the battery charge is 40% at 7.5 hours after 08:00, that is at 15:30	R1	Must state that 7.5 is a time at which the charge is 40%.

Answer: \(7.5\): the charge is 40% at 15:30

(2)

(d)(i)

On one day Lena streams from 08:00 until the battery charge reaches 20%. She then switches the phone off for exactly one hour, and the charge stays constant. She then charges the phone, while it is switched off, at a constant rate of 25 percentage points per hour until it is full.

Show that the total time that the phone spends streaming or charging is 13.2 hours.

No calculatorPROOFNotes

Show solution

streaming: \(\frac{100-20}{8}=10\) hours	M1	Or \(B^{-1}(20)=10\).
charging: \(\frac{100-20}{25}=3.2\) hours	A1
\(10+3.2=13.2\) hours	AG

Answer: 13.2 hours

(2)

(d)(ii)

On Figure 1, sketch the graph of the battery charge against \(t\) for this day, from 08:00 until the phone is full. Label the coordinates of each point where the graph changes.

Figure 1: axes for the battery charge B against the time t

No calculatorGRAPHNotes

Show solution

straight line from \((0,\,100)\) to \((10,\,20)\)	A1
horizontal segment from \((10,\,20)\) to \((11,\,20)\)	A1
straight line from \((11,\,20)\) to \((14.2,\,100)\), with the graph ending there	A1	Coordinates must be labelled for full marks; penalise once.

Answer: Segments \((0,\,100)\to(10,\,20)\to(11,\,20)\to(14.2,\,100)\)

(3)

(d)(iii)

Find the equation of the charging section of the graph, in the form \(B=mt+c\).

No calculatorROUTINENotes

Show solution

\(B-20=25(t-11)\) or gradient \(\frac{100-20}{14.2-11}=25\) with a point substituted	M1	Allow ECF from their graph in (d)(ii).
\(B=25t-255\)	A1	Accept with a domain \(11\le t\le14.2\); the domain is not required.

Answer: \(B=25t-255\)

(2)

(e)(i)

On another day Lena uses battery-saver mode, which halves the rate at which the charge falls while streaming. Starting at 100%, the charge after \(t\) hours is \(S(t)=100-4t\).

The function \(h(n)=\dfrac{n}{60}\) converts a time of \(n\) minutes into hours.

Find \((S\circ h)(n)\) and state what it represents.

No calculatorALGEBRANotes

Show solution

\((S\circ h)(n)=S\!\left(\frac{n}{60}\right)=100-\frac{4n}{60}=100-\frac{n}{15}\)	A1	Accept unsimplified \(100-4\left(\frac{n}{60}\right)\).
the battery charge (in %) after \(n\) minutes of streaming in battery-saver mode	R1

Answer: \(100-\dfrac{n}{15}\): the charge after \(n\) minutes

(2)

(e)(ii)

Hence find how many more minutes it takes for the charge to fall from 100% to 20% in battery-saver mode than in normal mode.

No calculatorMULTINotes

Show solution

\(100-\frac{n}{15}=20\Rightarrow n=1200\)	M1	Allow ECF from (e)(i).
\(1200-600=600\) minutes more	A1	Normal mode takes 10 hours = 600 minutes, from (d)(i).

Answer: 600 minutes

(2)

(e)(iii)

Suggest one reason why a linear model may not describe the battery charge well when it is close to 0%.

No calculatorINTERPRETNotes

Show solution

e.g. the phone switches itself off (or into a low-power mode) before reaching 0%; the charge does not fall at a constant rate when the battery is nearly empty

Any sensible reason in context.

(1)

Show full solution

Working	Mark	Notes
Part (a)
\(B(t)=100-8t\)	A1	Accept \(B(t)=-8t+100\).
\(m=-8\): the charge falls by 8 percentage points for each hour of streaming	R1	Must refer to the rate per hour. “8% per hour” is accepted.
Part (b)
domain \(0\le t\le12.5\)	A1	From \(100-8t=0\).
range \(0\le B(t)\le100\)	A1
Part (c)(i)
\(x=100-8y\) or \(t=\frac{100-B}{8}\)	M1	Swapping the variables or rearranging.
\(B^{-1}(x)=\frac{100-x}{8}\)	A1	Accept \(12.5-\frac{x}{8}\).
domain \(0\le x\le100\)	A1	Allow ECF from the range in (b).
Part (c)(ii)
\(B^{-1}(40)=\frac{60}{8}=7.5\)	A1
the battery charge is 40% at 7.5 hours after 08:00, that is at 15:30	R1	Must state that 7.5 is a time at which the charge is 40%.
Part (d)(i)
streaming: \(\frac{100-20}{8}=10\) hours	M1	Or \(B^{-1}(20)=10\).
charging: \(\frac{100-20}{25}=3.2\) hours	A1
\(10+3.2=13.2\) hours	AG
Part (d)(ii)
straight line from \((0,\,100)\) to \((10,\,20)\)	A1
horizontal segment from \((10,\,20)\) to \((11,\,20)\)	A1
straight line from \((11,\,20)\) to \((14.2,\,100)\), with the graph ending there	A1	Coordinates must be labelled for full marks; penalise once.
Part (d)(iii)
\(B-20=25(t-11)\) or gradient \(\frac{100-20}{14.2-11}=25\) with a point substituted	M1	Allow ECF from their graph in (d)(ii).
\(B=25t-255\)	A1	Accept with a domain \(11\le t\le14.2\); the domain is not required.
Part (e)(i)
\((S\circ h)(n)=S\!\left(\frac{n}{60}\right)=100-\frac{4n}{60}=100-\frac{n}{15}\)	A1	Accept unsimplified \(100-4\left(\frac{n}{60}\right)\).
the battery charge (in %) after \(n\) minutes of streaming in battery-saver mode	R1
Part (e)(ii)
\(100-\frac{n}{15}=20\Rightarrow n=1200\)	M1	Allow ECF from (e)(i).
\(1200-600=600\) minutes more	A1	Normal mode takes 10 hours = 600 minutes, from (d)(i).
Part (e)(iii)
e.g. the phone switches itself off (or into a low-power mode) before reaching 0%; the charge does not fall at a constant rate when the battery is nearly empty	R1	Any sensible reason in context.

Answers: (a) \(B(t)=100-8t\) · (b) Domain \(0\le t\le12.5\); range \(0\le B(t)\le100\) · (c)(i) \(B^{-1}(x)=\dfrac{100-x}{8}\), \(0\le x\le100\) · (c)(ii) \(7.5\): the charge is 40% at 15:30 · (d)(i) 13.2 hours · (d)(ii) Segments \((0,\,100)\to(10,\,20)\to(11,\,20)\to(14.2,\,100)\) · (d)(iii) \(B=25t-255\) · (e)(i) \(100-\dfrac{n}{15}\): the charge after \(n\) minutes · (e)(ii) 600 minutes

Exam tip (c)(i) The domain of an inverse function is the range of the original function, so the answer to (b) gives the domain here without any more work.

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2.4 Key features of graphs and intersections using technology · 2.5 Composite functions and inverse functions · 2.6 The quadratic function and its formsmerged group

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A basketball shot: is the ball going in?

2.6 The quadratic function and its forms2.4 Key features of graphs and intersections using technology2.7 Quadratic equations, inequalities and the discriminant2.11 Transformations of graphsSL

A basketball player takes a jump shot. The path of the ball is modelled in a vertical plane, where \(x\) is the horizontal distance in metres from the point of release and \(h\) is the height of the ball above the floor in metres.

The ball is released at \((0,\,2.5)\) and reaches its maximum height of 4.3 m when \(x=3\). The path is modelled by

\(h(x)=a(x-3)^2+4.3\), for \(x\ge0\).

The centre of the hoop is at \((5.5,\,3.05)\). This information is shown in Figure 1.

Figure 1: the path of the ball up to the hoop

(a)

Find the value of \(a\).

No calculatorROUTINENotes

Show solution

\(2.5=a(0-3)^2+4.3\)	M1	Substituting \((0,\,2.5)\).
\(a=-0.2\)	A1	Accept \(-\frac{1}{5}\).

Answer: \(a=-0.2\)

(2)

(b)

Write \(h(x)\) in the form \(px^2+qx+r\).

No calculatorALGEBRANotes

Show solution

\(-0.2(x^2-6x+9)+4.3\)	M1	Correct expansion of the bracket with their \(a\).
\(h(x)=-0.2x^2+1.2x+2.5\)	A1	The constant term must match the release height 2.5.

Answer: \(h(x)=-0.2x^2+1.2x+2.5\)

(2)

(c)

Show that, according to the model, the ball passes through the centre of the hoop.

No calculatorPROOFNotes

Show solution

\(h(5.5)=-0.2(5.5-3)^2+4.3\) or \(-0.2(5.5)^2+1.2(5.5)+2.5\)	M1
\(=-0.2(6.25)+4.3=-1.25+4.3\)	A1	Or \(-6.05+6.6+2.5\).
\(=3.05\), so the ball passes through \((5.5,\,3.05)\)	AG

Answer: \(h(5.5)=3.05\)

(2)

(d)

Without using a calculator, solve the equation \(h(x)=3.05\). Use the axis of symmetry of the graph to explain the relationship between your two solutions.

No calculatorCONCEPTNotes

Show solution

\(-0.2(x-3)^2+4.3=3.05\Rightarrow(x-3)^2=6.25\)	M1	Or rearranging to \(x^2-6x+2.75=0\) and factorising \((x-0.5)(x-5.5)=0\).
\(x=0.5\) or \(x=5.5\)	A1
both solutions are 2.5 m from the axis of symmetry \(x=3\) (the point \((0.5,\,3.05)\) is the mirror image of the hoop centre): the ball is at hoop height once while rising and once while falling	R1	Must refer to the axis \(x=3\) or to equal distances either side of it.

Answer: \(x=0.5\) or \(x=5.5\)

(3)

(e)(i)

A defender stands between the player and the hoop and jumps with one arm raised. In Figure 2 the arm is modelled by the line segment from the shoulder at \((1.6,\,2.0)\) to the fingertips at \((1.0,\,3.2)\). This segment lies on the line \(y=-2x+5.2\).

Use your GDC to find the coordinates of the point, with \(0\le x\le2\), where the graph of \(h\) meets the line \(y=-2x+5.2\).

Figure 2: the start of the ball’s path and the defender’s arm

GRAPHNotes

Show solution

\(x=0.894\)	A1	\(0.893664\ldots\)
\(y=3.41\)	A1	Accept \((0.894,\,3.41)\).

Answer: \((0.894,\,3.41)\)

(2)

(e)(ii)

Hence determine whether the ball touches the defender’s arm. Justify your answer.

INTERPRETNotes

Show solution

the arm only occupies \(1.0\le x\le1.6\), and \(0.894\) is outside this interval	R1	Or: at \(x=1.0\), \(h=3.5>3.2\), so the ball is above the fingertips, and the path stays above the line for \(1.0\le x\le1.6\).
the ball does not touch the arm	A1	Do not award A1 without the R1.

Answer: No: the meeting point is not on the arm

(2)

(f)

Find the value of \(x\) for which \(h(x)=0\), and state what it would represent if there were no hoop or backboard.

ROUTINENotes

Show solution

\(-0.2x^2+1.2x+2.5=0\) or \((x-3)^2=21.5\); \(x=7.64\) (3 s.f.)	A1	The negative root \(-1.64\) must be rejected or not given.
the horizontal distance from the point of release at which the ball would hit the floor	A1

Answer: \(x=7.64\): where the ball would land

(2)

(g)

On Figure 3, sketch the graph of \(h\) for \(0\le x\le7.64\). Label the point of release, the maximum point, the centre of the hoop and the \(x\)-intercept with their coordinates.

Figure 3: axes for the sketch of the graph of h

No calculatorGRAPHNotes

Show solution

concave-down parabola from \((0,\,2.5)\) to the \(x\)-axis, ending there	A1
maximum \((3,\,4.3)\) labelled, with the curve symmetric about \(x=3\)	A1
\((5.5,\,3.05)\) and \((7.64,\,0)\) labelled	A1	Allow ECF from (f).

Answer: Parabola through (0, 2.5), (3, 4.3), (5.5, 3.05) and (7.64, 0)

(3)

(h)(i)

The player now takes the same shot from a point 0.5 m further back from the hoop. The ball is released at the same height, speed and angle, so the path has the same shape. The point of release of the original shot is still the origin, and the hoop has not moved.

The new path is the graph of \(y=g(x)\). Describe fully the transformation that maps the graph of \(h\) onto the graph of \(g\), and write down \(g(x)\) in the form \(a(x-p)^2+q\).

CONCEPTNotes

Show solution

translation by \(\begin{pmatrix}-0.5\\0\end{pmatrix}\)	A1	Accept “translation 0.5 units in the negative \(x\)-direction”. “Shift left” alone is A0.
\(g(x)=-0.2(x-2.5)^2+4.3\)	A1

Answer: Translation by \(\begin{pmatrix}-0.5\\0\end{pmatrix}\); \(g(x)=-0.2(x-2.5)^2+4.3\)

(2)

(h)(ii)

Determine whether the new shot passes through the centre of the hoop. Justify your answer.

INTERPRETNotes

Show solution

\(g(5.5)=-0.2(3)^2+4.3\)	M1	Allow ECF from (h)(i).
\(=2.5\)	A1
\(2.5<3.05\): the ball is 0.55 m below the hoop when \(x=5.5\) (it falls to hoop height at \(x=5\), 0.5 m short), so it does not pass through the centre	R1	The conclusion must follow from a comparison with 3.05.

Answer: No: the ball is at 2.5 m when it reaches the hoop

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(2.5=a(0-3)^2+4.3\)	M1	Substituting \((0,\,2.5)\).
\(a=-0.2\)	A1	Accept \(-\frac{1}{5}\).
Part (b)
\(-0.2(x^2-6x+9)+4.3\)	M1	Correct expansion of the bracket with their \(a\).
\(h(x)=-0.2x^2+1.2x+2.5\)	A1	The constant term must match the release height 2.5.
Part (c)
\(h(5.5)=-0.2(5.5-3)^2+4.3\) or \(-0.2(5.5)^2+1.2(5.5)+2.5\)	M1
\(=-0.2(6.25)+4.3=-1.25+4.3\)	A1	Or \(-6.05+6.6+2.5\).
\(=3.05\), so the ball passes through \((5.5,\,3.05)\)	AG
Part (d)
\(-0.2(x-3)^2+4.3=3.05\Rightarrow(x-3)^2=6.25\)	M1	Or rearranging to \(x^2-6x+2.75=0\) and factorising \((x-0.5)(x-5.5)=0\).
\(x=0.5\) or \(x=5.5\)	A1
both solutions are 2.5 m from the axis of symmetry \(x=3\) (the point \((0.5,\,3.05)\) is the mirror image of the hoop centre): the ball is at hoop height once while rising and once while falling	R1	Must refer to the axis \(x=3\) or to equal distances either side of it.
Part (e)(i)
\(x=0.894\)	A1	\(0.893664\ldots\)
\(y=3.41\)	A1	Accept \((0.894,\,3.41)\).
Part (e)(ii)
the arm only occupies \(1.0\le x\le1.6\), and \(0.894\) is outside this interval	R1	Or: at \(x=1.0\), \(h=3.5>3.2\), so the ball is above the fingertips, and the path stays above the line for \(1.0\le x\le1.6\).
the ball does not touch the arm	A1	Do not award A1 without the R1.
Part (f)
\(-0.2x^2+1.2x+2.5=0\) or \((x-3)^2=21.5\); \(x=7.64\) (3 s.f.)	A1	The negative root \(-1.64\) must be rejected or not given.
the horizontal distance from the point of release at which the ball would hit the floor	A1
Part (g)
concave-down parabola from \((0,\,2.5)\) to the \(x\)-axis, ending there	A1
maximum \((3,\,4.3)\) labelled, with the curve symmetric about \(x=3\)	A1
\((5.5,\,3.05)\) and \((7.64,\,0)\) labelled	A1	Allow ECF from (f).
Part (h)(i)
translation by \(\begin{pmatrix}-0.5\\0\end{pmatrix}\)	A1	Accept “translation 0.5 units in the negative \(x\)-direction”. “Shift left” alone is A0.
\(g(x)=-0.2(x-2.5)^2+4.3\)	A1
Part (h)(ii)
\(g(5.5)=-0.2(3)^2+4.3\)	M1	Allow ECF from (h)(i).
\(=2.5\)	A1
\(2.5<3.05\): the ball is 0.55 m below the hoop when \(x=5.5\) (it falls to hoop height at \(x=5\), 0.5 m short), so it does not pass through the centre	R1	The conclusion must follow from a comparison with 3.05.

Answers: (a) \(a=-0.2\) · (b) \(h(x)=-0.2x^2+1.2x+2.5\) · (c) \(h(5.5)=3.05\) · (d) \(x=0.5\) or \(x=5.5\) · (e)(i) \((0.894,\,3.41)\) · (e)(ii) No: the meeting point is not on the arm · (f) \(x=7.64\): where the ball would land · (g) Parabola through (0, 2.5), (3, 4.3), (5.5, 3.05) and (7.64, 0) · (h)(i) Translation by \(\begin{pmatrix}-0.5\\0\end{pmatrix}\); \(g(x)=-0.2(x-2.5)^2+4.3\) · (h)(ii) No: the ball is at 2.5 m when it reaches the hoop

Common error (e)(ii) The line \(y=-2x+5.2\) goes on for ever, but the arm does not. An intersection of the graphs only matters if it lies inside the interval of \(x\)-values that the model describes.

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Changing money on a school trip: composite and inverse functions

2.5 Composite functions and inverse functions2.1 Equations of straight lines, parallel and perpendicular lines2.2 Functions, domain, range and inverse functions2.4 Key features of graphs and intersections using technologySL

On a school trip, a student changes euros (EUR) into US dollars (USD) at an airport kiosk, and later changes the dollars into Japanese yen (JPY) at a bank in Tokyo.

The kiosk converts \(x\) euros at a rate of 1.08 USD per euro and then deducts a fixed fee of 3 USD. The number of dollars received is \(f(x)=1.08x-3\). The kiosk accepts amounts from €20 to €1000 inclusive.

The bank converts \(x\) dollars into yen with no fee. The number of yen received is \(g(x)=148x\).

The process is shown in Figure 1.

Figure 1: the two stages of the exchange

(a)

Find \((g\circ f)(x)\), giving your answer in the form \(mx+c\).

ALGEBRANotes

Show solution

\(g(1.08x-3)=148(1.08x-3)\)	M1	Correct order: \(f\) is applied first.
\((g\circ f)(x)=159.84x-444\)	A1

Answer: \(159.84x-444\)

(2)

(b)

Find the number of yen the student receives for €250.

ROUTINENotes

Show solution

\(159.84(250)-444\) or \(f(250)=267\), then \(g(267)\)	M1	Allow ECF from (a).
\(39\,516\) yen	A1

Answer: \(39\,516\) yen

(2)

(c)

Interpret the number 444 in your answer to (a) in context.

INTERPRETNotes

Show solution

the 3 USD fee, expressed in yen (\(3\times148=444\)), which is lost from every exchange

(1)

(d)

Find the range of \(g\circ f\) for the amounts that the kiosk accepts.

CONCEPTNotes

Show solution

\((g\circ f)(20)=2752.8\) and \((g\circ f)(1000)=159\,396\)	A1	Both end values; \(g\circ f\) is increasing, so the end points give the least and greatest values.
\(2752.8\le(g\circ f)(x)\le159\,396\)	A1	Correct inequality signs needed. Allow ECF from (a).

Answer: \(2752.8\le(g\circ f)(x)\le159\,396\)

(2)

(e)(i)

Find \(f^{-1}(x)\) and \(g^{-1}(x)\).

ALGEBRANotes

Show solution

\(x=1.08y-3\) or equivalent rearrangement	M1
\(f^{-1}(x)=\frac{x+3}{1.08}\)	A1
\(g^{-1}(x)=\frac{x}{148}\)	A1

Answer: \(f^{-1}(x)=\dfrac{x+3}{1.08}\), \(g^{-1}(x)=\dfrac{x}{148}\)

(3)

(e)(ii)

Show that \(\left(f^{-1}\circ g^{-1}\right)(x)=\dfrac{x+444}{159.84}\).

PROOFNotes

Show solution

\(f^{-1}\!\left(\frac{x}{148}\right)=\frac{\frac{x}{148}+3}{1.08}\)	M1	Correct order of composition.
multiplying numerator and denominator by 148: \(\frac{x+444}{148\times1.08}\)	A1
\(=\frac{x+444}{159.84}\)	AG

Answer: \(\dfrac{x+444}{159.84}\)

(2)

(e)(iii)

Hence find the amount in euros that the student must change to receive 30 000 yen. Give your answer to the nearest cent.

MULTINotes

Show solution

\((g\circ f)^{-1}(30\,000)=\frac{30\,000+444}{159.84}\)	M1	Recognising that \(f^{-1}\circ g^{-1}=(g\circ f)^{-1}\), or solving \(159.84x-444=30\,000\).
€190.47	A1	\(190.4654\ldots\); accept €190.47 only.

Answer: €190.47

(2)

(f)(i)

A second kiosk at the airport charges no fee but uses a lower rate. It gives \(r(x)=1.06x\) dollars for \(x\) euros, and it accepts the same amounts as the first kiosk.

The function \(D(x)=f(x)-r(x)\) gives the number of extra dollars that the first kiosk gives compared with the second.

On Figure 2, sketch the graph of \(y=D(x)\) for \(20\le x\le300\).

Figure 2: axes for the graph of D, the extra dollars given by the first kiosk

GRAPHNotes

Show solution

\(D(x)=1.08x-3-1.06x=0.02x-3\)	(M1)	May be implied by a correct graph.
straight line from \((20,\,-2.6)\) to \((300,\,3)\), crossing the \(x\)-axis between \(x=100\) and \(x=200\)	A1	Drawn over the stated domain only, with both end points approximately correct.

Answer: Straight line from \((20,\,-2.6)\) to \((300,\,3)\), crossing the \(x\)-axis at \(x=150\)

(2)

(f)(ii)

Use your GDC to find the amount of euros for which both kiosks give the same number of dollars, and write down this number of dollars.

ROUTINENotes

Show solution

\(x=150\) (euros)	A1	From the zero of \(D\) or the intersection of \(y=f(x)\) and \(y=r(x)\).
\(159\) (dollars)	A1	\(f(150)=r(150)=159\).

Answer: €150, giving 159 dollars at both kiosks

(2)

(f)(iii)

The student wants to change €100 on one trip and €500 on another. Explain which kiosk the student should use each time.

INTERPRETNotes

Show solution

€100: second kiosk, as \(r(100)=106>f(100)=105\) (\(D(x)<0\) for \(x<150\))	R1	Reasoning from values or from the graphs, either side of \(x=150\). Allow ECF from (f)(ii).
€500: first kiosk, as \(f(500)=537>r(500)=530\) (\(D(x)>0\) for \(x>150\))	R1

Answer: €100: second kiosk; €500: first kiosk

(2)

(g)

The first kiosk changes its rate to 1.05 USD per euro, still deducting the 3 USD fee. Determine whether there is now any amount that the kiosks accept for which the first kiosk gives more dollars than the second. Justify your answer.

INTERPRETNotes

Show solution

\(1.05x-3=1.06x\Rightarrow x=-300\), or \(r(x)-(1.05x-3)=0.01x+3\)	M1
the lines meet only at \(x=-300\), outside \(20\le x\le1000\) (or \(0.01x+3>0\) for all \(x>0\)), so the second kiosk always gives more: there is no such amount	R1

Answer: No: the second kiosk now gives more for every accepted amount

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(g(1.08x-3)=148(1.08x-3)\)	M1	Correct order: \(f\) is applied first.
\((g\circ f)(x)=159.84x-444\)	A1
Part (b)
\(159.84(250)-444\) or \(f(250)=267\), then \(g(267)\)	M1	Allow ECF from (a).
\(39\,516\) yen	A1
Part (c)
the 3 USD fee, expressed in yen (\(3\times148=444\)), which is lost from every exchange	R1
Part (d)
\((g\circ f)(20)=2752.8\) and \((g\circ f)(1000)=159\,396\)	A1	Both end values; \(g\circ f\) is increasing, so the end points give the least and greatest values.
\(2752.8\le(g\circ f)(x)\le159\,396\)	A1	Correct inequality signs needed. Allow ECF from (a).
Part (e)(i)
\(x=1.08y-3\) or equivalent rearrangement	M1
\(f^{-1}(x)=\frac{x+3}{1.08}\)	A1
\(g^{-1}(x)=\frac{x}{148}\)	A1
Part (e)(ii)
\(f^{-1}\!\left(\frac{x}{148}\right)=\frac{\frac{x}{148}+3}{1.08}\)	M1	Correct order of composition.
multiplying numerator and denominator by 148: \(\frac{x+444}{148\times1.08}\)	A1
\(=\frac{x+444}{159.84}\)	AG
Part (e)(iii)
\((g\circ f)^{-1}(30\,000)=\frac{30\,000+444}{159.84}\)	M1	Recognising that \(f^{-1}\circ g^{-1}=(g\circ f)^{-1}\), or solving \(159.84x-444=30\,000\).
€190.47	A1	\(190.4654\ldots\); accept €190.47 only.
Part (f)(i)
\(D(x)=1.08x-3-1.06x=0.02x-3\)	(M1)	May be implied by a correct graph.
straight line from \((20,\,-2.6)\) to \((300,\,3)\), crossing the \(x\)-axis between \(x=100\) and \(x=200\)	A1	Drawn over the stated domain only, with both end points approximately correct.
Part (f)(ii)
\(x=150\) (euros)	A1	From the zero of \(D\) or the intersection of \(y=f(x)\) and \(y=r(x)\).
\(159\) (dollars)	A1	\(f(150)=r(150)=159\).
Part (f)(iii)
€100: second kiosk, as \(r(100)=106>f(100)=105\) (\(D(x)<0\) for \(x<150\))	R1	Reasoning from values or from the graphs, either side of \(x=150\). Allow ECF from (f)(ii).
€500: first kiosk, as \(f(500)=537>r(500)=530\) (\(D(x)>0\) for \(x>150\))	R1
Part (g)
\(1.05x-3=1.06x\Rightarrow x=-300\), or \(r(x)-(1.05x-3)=0.01x+3\)	M1
the lines meet only at \(x=-300\), outside \(20\le x\le1000\) (or \(0.01x+3>0\) for all \(x>0\)), so the second kiosk always gives more: there is no such amount	R1

Answers: (a) \(159.84x-444\) · (b) \(39\,516\) yen · (d) \(2752.8\le(g\circ f)(x)\le159\,396\) · (e)(i) \(f^{-1}(x)=\dfrac{x+3}{1.08}\), \(g^{-1}(x)=\dfrac{x}{148}\) · (e)(ii) \(\dfrac{x+444}{159.84}\) · (e)(iii) €190.47 · (f)(i) Straight line from \((20,\,-2.6)\) to \((300,\,3)\), crossing the \(x\)-axis at \(x=150\) · (f)(ii) €150, giving 159 dollars at both kiosks · (f)(iii) €100: second kiosk; €500: first kiosk · (g) No: the second kiosk now gives more for every accepted amount

Common error (a) \(f(g(x))=1.08(148x)-3\) converts dollars to yen first, which is the wrong order. The function applied first, here \(f\), is the one written nearest to \(x\).

More 2.5 questions in the question bank

2.7 Quadratic equations, inequalities and the discriminant · 2.8 The reciprocal function and rational functionsmerged group

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Designing a smartphone screen: quadratic constraints

2.7 Quadratic equations, inequalities and the discriminant2.6 The quadratic function and its forms3.2 Right-angled and non-right-angled triangles2.4 Key features of graphs and intersections using technologySL

A designer models the screen of a smartphone as a rectangle of width \(x\) cm and height \((x+7)\) cm, as shown in Figure 1.

Figure 1: the screen (not to scale)

(a)

The diagonal of the screen is 17 cm long.

Show that \(x^2+7x-120=0\).

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Show solution

\(x^2+(x+7)^2=17^2\)	M1	Pythagoras with the given sides; accept \(=289\).
\(2x^2+14x+49=289\) or \(2x^2+14x-240=0\)	A1	Correct expansion seen before the given answer.
\(x^2+7x-120=0\)	AG

Answer: \(x^2+7x-120=0\)

(2)

(b)

Hence find the width and the height of the screen.

No calculatorROUTINENotes

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\((x+15)(x-8)=0\)	M1	Or a correct substitution into the quadratic formula: \(x=\frac{-7\pm\sqrt{529}}{2}\).
\(x=8\), rejecting \(x=-15\) because a width cannot be negative	A1	A reason must be given for rejecting \(-15\). Do not award if \(x=-15\) is simply ignored.
width 8 cm, height 15 cm	A1

Answer: 8 cm by 15 cm

(3)

(c)(i)

The designer also wants the screen area to be at least 60 cm², so that \(x(x+7)\ge 60\).

Let \(f(x)=x^2+7x-60\). Write \(f(x)\) in the form \((x-h)^2+k\).

No calculatorALGEBRANotes

Show solution

\(\left(x+\frac{7}{2}\right)^2-\frac{49}{4}-60\)	(M1)	Attempt to complete the square, with \(h=-3.5\) seen.
\(f(x)=\left(x+\frac{7}{2}\right)^2-\frac{289}{4}\)	A1	Accept \((x+3.5)^2-72.25\), or \(h=-3.5\), \(k=-72.25\).

Answer: \(\left(x+\tfrac{7}{2}\right)^2-\tfrac{289}{4}\)

(2)

(c)(ii)

On the axes in Figure 2, sketch the graph of \(y=f(x)\) for \(-14\le x\le 8\). Label the vertex and the points where the graph meets the axes with their coordinates.

Figure 2: axes for the graph of \(y=f(x)\)

No calculatorGRAPHNotes

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upward parabola, symmetric about \(x=-3.5\), drawn over the full domain	A1	Endpoints approximately \((-14,\,38)\) and \((8,\,60)\).
vertex \((-3.5,\,-72.25)\) labelled	A1	Allow ECF from (c)(i).
intercepts \((-12,\,0)\), \((5,\,0)\) and \((0,\,-60)\) labelled	A1	All three needed.

Answer: parabola, vertex \((-3.5,-72.25)\), intercepts \((-12,0)\), \((5,0)\), \((0,-60)\)

(3)

(c)(iii)

The diagonal of the screen must be at most 17 cm. Using your answers to parts (a) and (c)(ii), find the set of possible values of \(x\).

No calculatorINTERPRETNotes

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area condition: \(x\ge 5\), since \(x\le -12\) gives a negative width	R1	Accept a reading from their graph in (c)(ii). A reason for rejecting \(x\le-12\) is required.
diagonal condition \(x^2+7x-120\le 0\) gives \(x\le 8\); so \(5\le x\le 8\)	A1	Accept \(x\in[5,\,8]\). Do not accept strict inequalities.

Answer: \(5\le x\le 8\)

(2)

(d)(i)

A second design keeps the perimeter equal to that of the screen in part (b), but the height no longer has to be 7 cm more than the width. The width is \(w\) cm and the area is \(A\) cm².

Show that \(w^2-23w+A=0\).

No calculatorMODELNotes

Show solution

perimeter \(=2(8+15)=46\), so height \(=23-w\)	M1	Allow ECF from (b).
\(A=w(23-w)=23w-w^2\)	A1
\(w^2-23w+A=0\)	AG

Answer: \(w^2-23w+A=0\)

(2)

(d)(ii)

By considering the discriminant of this equation, find the greatest possible value of \(A\).

No calculatorCONCEPTNotes

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real \(w\) requires \(\Delta\ge 0\)	M1	The condition must be stated (accept \(\Delta\ge 0\) or "real roots").
\((-23)^2-4(1)(A)\ge 0\), that is \(529-4A\ge 0\)	A1
\(A\le 132.25\), so the greatest area is 132.25 cm²	A1	Accept \(\frac{529}{4}\).

Answer: 132.25 cm²

(3)

(d)(iii)

Find the dimensions of the screen with this greatest area, and comment on whether this shape is suitable for a smartphone.

No calculatorINTERPRETNotes

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\(\Delta=0\) gives a repeated root \(w=\frac{23}{2}=11.5\); the screen is 11.5 cm by 11.5 cm	A1
it is a square, but a phone screen needs to be long and narrow to be held in one hand, so this shape is not suitable	R1	Any sensible comment based on the square shape.

Answer: 11.5 cm by 11.5 cm (a square)

(2)

(e)(i)

Return to the first design, with height \((x+7)\) cm. The phone case adds a border 0.4 cm wide to every edge of the screen.

Write down expressions for the width and the height of the front face of the case.

MODELNotes

Show solution

width \((x+0.8)\) cm and height \((x+7.8)\) cm

Both needed.

Answer: \((x+0.8)\) cm and \((x+7.8)\) cm

(1)

(e)(ii)

The front face of the case must have an area of at most 144 cm². Ignoring the conditions on the diagonal and on the area of the screen, find the greatest possible width of the screen.

MULTINotes

Show solution

\((x+0.8)(x+7.8)\le 144\) or \(x^2+8.6x-137.76=0\)	M1	Allow ECF from (e)(i). Accept a GDC sketch of \(y=(x+0.8)(x+7.8)\) and \(y=144\).
\(x=8.2\) or \(x=-16.8\)	A1
\(x=-16.8\) rejected (negative width); greatest width 8.2 cm	R1	The case is then 9 cm by 16 cm.

Answer: 8.2 cm

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(x^2+(x+7)^2=17^2\)	M1	Pythagoras with the given sides; accept \(=289\).
\(2x^2+14x+49=289\) or \(2x^2+14x-240=0\)	A1	Correct expansion seen before the given answer.
\(x^2+7x-120=0\)	AG
Part (b)
\((x+15)(x-8)=0\)	M1	Or a correct substitution into the quadratic formula: \(x=\frac{-7\pm\sqrt{529}}{2}\).
\(x=8\), rejecting \(x=-15\) because a width cannot be negative	A1	A reason must be given for rejecting \(-15\). Do not award if \(x=-15\) is simply ignored.
width 8 cm, height 15 cm	A1
Part (c)(i)
\(\left(x+\frac{7}{2}\right)^2-\frac{49}{4}-60\)	(M1)	Attempt to complete the square, with \(h=-3.5\) seen.
\(f(x)=\left(x+\frac{7}{2}\right)^2-\frac{289}{4}\)	A1	Accept \((x+3.5)^2-72.25\), or \(h=-3.5\), \(k=-72.25\).
Part (c)(ii)
upward parabola, symmetric about \(x=-3.5\), drawn over the full domain	A1	Endpoints approximately \((-14,\,38)\) and \((8,\,60)\).
vertex \((-3.5,\,-72.25)\) labelled	A1	Allow ECF from (c)(i).
intercepts \((-12,\,0)\), \((5,\,0)\) and \((0,\,-60)\) labelled	A1	All three needed.
Part (c)(iii)
area condition: \(x\ge 5\), since \(x\le -12\) gives a negative width	R1	Accept a reading from their graph in (c)(ii). A reason for rejecting \(x\le-12\) is required.
diagonal condition \(x^2+7x-120\le 0\) gives \(x\le 8\); so \(5\le x\le 8\)	A1	Accept \(x\in[5,\,8]\). Do not accept strict inequalities.
Part (d)(i)
perimeter \(=2(8+15)=46\), so height \(=23-w\)	M1	Allow ECF from (b).
\(A=w(23-w)=23w-w^2\)	A1
\(w^2-23w+A=0\)	AG
Part (d)(ii)
real \(w\) requires \(\Delta\ge 0\)	M1	The condition must be stated (accept \(\Delta\ge 0\) or "real roots").
\((-23)^2-4(1)(A)\ge 0\), that is \(529-4A\ge 0\)	A1
\(A\le 132.25\), so the greatest area is 132.25 cm²	A1	Accept \(\frac{529}{4}\).
Part (d)(iii)
\(\Delta=0\) gives a repeated root \(w=\frac{23}{2}=11.5\); the screen is 11.5 cm by 11.5 cm	A1
it is a square, but a phone screen needs to be long and narrow to be held in one hand, so this shape is not suitable	R1	Any sensible comment based on the square shape.
Part (e)(i)
width \((x+0.8)\) cm and height \((x+7.8)\) cm	A1	Both needed.
Part (e)(ii)
\((x+0.8)(x+7.8)\le 144\) or \(x^2+8.6x-137.76=0\)	M1	Allow ECF from (e)(i). Accept a GDC sketch of \(y=(x+0.8)(x+7.8)\) and \(y=144\).
\(x=8.2\) or \(x=-16.8\)	A1
\(x=-16.8\) rejected (negative width); greatest width 8.2 cm	R1	The case is then 9 cm by 16 cm.

Answers: (a) \(x^2+7x-120=0\) · (b) 8 cm by 15 cm · (c)(i) \(\left(x+\tfrac{7}{2}\right)^2-\tfrac{289}{4}\) · (c)(ii) parabola, vertex \((-3.5,-72.25)\), intercepts \((-12,0)\), \((5,0)\), \((0,-60)\) · (c)(iii) \(5\le x\le 8\) · (d)(i) \(w^2-23w+A=0\) · (d)(ii) 132.25 cm² · (d)(iii) 11.5 cm by 11.5 cm (a square) · (e)(i) \((x+0.8)\) cm and \((x+7.8)\) cm · (e)(ii) 8.2 cm

Common error (b) Leaving out \(x=-15\) without comment. In a context, write down both roots and state why one is rejected.

Exam tip (c)(iii) Sketch every quadratic inequality. A parabola that opens upward is above the axis outside its roots and below it between them.

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A 3D printing workshop: a cost model, its inverse and printing times

2.8 The reciprocal function and rational functions2.7 Quadratic equations, inequalities and the discriminant2.11 Transformations of graphs2.2 Functions, domain, range and inverse functionsSL

A small workshop makes phone stands on 3D printers. The average cost, in euros, of one stand when \(n\) stands are made in a week is modelled by

\(C(n)=\dfrac{an+b}{n},\quad n\gt 0,\)

where \(a\) is the cost of the plastic and electricity for one stand and \(b\) is the fixed weekly cost (rent and machine leasing), both in euros. In a week when 60 stands are made the average cost is €21.00, and in a week when 150 stands are made it is €12.60.

(a)(i)

Find the value of \(a\) and the value of \(b\).

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Show solution

\(60a+b=1260\) and \(150a+b=1890\)	M1	Substituting both data points and multiplying by \(n\); accept \(a+\frac{b}{60}=21\) and \(a+\frac{b}{150}=12.6\).
\(90a=630\), so \(a=7\)	A1
\(b=840\)	A1

Answer: \(a=7\), \(b=840\)

(3)

(a)(ii)

By first writing \(C(n)\) in the form \(p+\dfrac{q}{n}\), describe a sequence of two transformations that maps the graph of \(y=\dfrac{1}{n}\) onto the graph of \(y=C(n)\).

No calculatorCONCEPTNotes

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\(C(n)=7+\frac{840}{n}\): vertical stretch, scale factor 840 (parallel to the \(y\)-axis)	A1	Allow ECF from (a)(i).
followed by a translation by \(\begin{pmatrix}0\\7\end{pmatrix}\)	A1	Accept "translation 7 units up". If the translation is done first it must be by \(\begin{pmatrix}0\\ \frac{1}{120}\end{pmatrix}\).

Answer: vertical stretch, scale factor 840, then translation by \(\begin{pmatrix}0\\7\end{pmatrix}\)

(2)

(b)(i)

Find an expression for \(C^{-1}(x)\).

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Show solution

\(x=7+\frac{840}{y}\), so \(x-7=\frac{840}{y}\)	M1	Interchanging \(x\) and \(y\) (at any stage) and an attempt to make \(y\) the subject. Allow ECF from (a).
\(C^{-1}(x)=\frac{840}{x-7}\)	A1	Accept \(y=\frac{840}{x-7}\).

Answer: \(C^{-1}(x)=\dfrac{840}{x-7}\)

(2)

(b)(ii)

Find \(C^{-1}(10)\) and interpret your answer in context.

No calculatorINTERPRETNotes

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\(C^{-1}(10)=\frac{840}{3}=280\)	A1	Allow ECF from (b)(i).
280 stands must be made in a week for the average cost of a stand to be €10	R1	The interpretation must refer to the number of stands and the average cost.

Answer: 280

(2)

(c)

The workshop moves to a larger unit. The fixed weekly cost rises to €1050 and the cost of the plastic and electricity for one stand rises to €9.

Find how many more stands than 150 must now be made in a week for the average cost of a stand to be no more than €12.60.

MULTINotes

Show solution

\(\frac{9n+1050}{n}\le 12.6\) or \(9+\frac{1050}{n}\le 12.6\)	M1	Accept an equation with \(=12.6\), or a GDC intersection with \(y=12.6\).
\(n\ge 291.66\ldots\)	A1	Accept \(n=291.7\) from an equation.
292 stands, so 142 more stands	A1	291 stands gives an average cost of €12.61, which is too high.

Answer: 142 more stands (292 in total)

(3)

(d)(i)

The workshop has two printers, P and Q. Working together, they print a large batch of stands in 6 hours. Working alone, P would take \(t\) hours to print the batch and Q would take 5 hours longer than P.

By considering the fraction of the batch printed by each printer in one hour, show that \(\dfrac{1}{t}+\dfrac{1}{t+5}=\dfrac{1}{6}\).

No calculatorMODELNotes

Show solution

in one hour P prints \(\frac{1}{t}\) of the batch and Q prints \(\frac{1}{t+5}\)	M1
together they print \(\frac{1}{6}\) of the batch in one hour	A1
\(\frac{1}{t}+\frac{1}{t+5}=\frac{1}{6}\)	AG

Answer: \(\dfrac{1}{t}+\dfrac{1}{t+5}=\dfrac{1}{6}\)

(2)

(d)(ii)

Hence find the time that each printer takes to print the batch alone.

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Show solution

\(6(t+5)+6t=t(t+5)\), giving \(t^2-7t-30=0\)	M1	Multiplying by \(6t(t+5)\) and rearranging to a quadratic equal to zero.
\((t-10)(t+3)=0\), \(t=10\) (\(t=-3\) rejected as a time cannot be negative)	A1	Reason for rejection required.
P takes 10 hours and Q takes 15 hours	A1

Answer: P: 10 hours, Q: 15 hours

(3)

(e)(i)

Printer Q is replaced by a printer R, which would take \(x\) hours to print the batch alone. Printer P is unchanged. The time, in hours, for P and R together to print the batch is \(T(x)\).

Show that \(T(x)=\dfrac{10x}{x+10}\).

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Show solution

\(\frac{1}{T}=\frac{1}{10}+\frac{1}{x}\)	M1	Allow ECF from (d)(ii) for the 10.
\(\frac{1}{T}=\frac{x+10}{10x}\)	A1
\(T(x)=\frac{10x}{x+10}\)	AG

Answer: \(T(x)=\dfrac{10x}{x+10}\)

(2)

(e)(ii)

On the axes in Figure 1, sketch the graph of \(y=T(x)\) for \(-40\le x\le 30\), \(x\ne -10\). Show the asymptotes with their equations and any points where the graph meets the axes.

Figure 1: axes for the graph of \(y=T(x)\)

GRAPHNotes

Show solution

asymptotes \(x=-10\) and \(y=10\) drawn and labelled	A1
two branches in the correct regions: one above \(y=10\) for \(x\lt -10\), one below \(y=10\) for \(x\gt -10\)	A1	Branches must approach the asymptotes without crossing them.
right branch passes through the origin \((0,\,0)\)	A1

Answer: asymptotes \(x=-10\), \(y=10\); passes through \((0,0)\)

(3)

(e)(iii)

State the domain and the range of \(T\) in this context, and explain what the range tells the workshop.

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domain \(x\gt 0\); range \(0\lt T(x)\lt 10\)	A1	Both needed.
however slow printer R is, P and R together always finish faster than P alone (less than 10 hours)	R1

Answer: \(x\gt 0\); \(0\lt T(x)\lt 10\)

(2)

(e)(iv)

The workshop wants the batch printed in 4 hours. Find the greatest time that R can take to print the batch alone.

ROUTINENotes

Show solution

\(\frac{10x}{x+10}=4\), giving \(6x=40\)	M1	Accept a GDC intersection of \(y=T(x)\) and \(y=4\).
\(x=\frac{20}{3}\) hours (6 hours 40 minutes)	A1	Accept 6.67 hours.

Answer: \(\dfrac{20}{3}\) hours

(2)

Show full solution

Working	Mark	Notes
Part (a)(i)
\(60a+b=1260\) and \(150a+b=1890\)	M1	Substituting both data points and multiplying by \(n\); accept \(a+\frac{b}{60}=21\) and \(a+\frac{b}{150}=12.6\).
\(90a=630\), so \(a=7\)	A1
\(b=840\)	A1
Part (a)(ii)
\(C(n)=7+\frac{840}{n}\): vertical stretch, scale factor 840 (parallel to the \(y\)-axis)	A1	Allow ECF from (a)(i).
followed by a translation by \(\begin{pmatrix}0\\7\end{pmatrix}\)	A1	Accept "translation 7 units up". If the translation is done first it must be by \(\begin{pmatrix}0\\ \frac{1}{120}\end{pmatrix}\).
Part (b)(i)
\(x=7+\frac{840}{y}\), so \(x-7=\frac{840}{y}\)	M1	Interchanging \(x\) and \(y\) (at any stage) and an attempt to make \(y\) the subject. Allow ECF from (a).
\(C^{-1}(x)=\frac{840}{x-7}\)	A1	Accept \(y=\frac{840}{x-7}\).
Part (b)(ii)
\(C^{-1}(10)=\frac{840}{3}=280\)	A1	Allow ECF from (b)(i).
280 stands must be made in a week for the average cost of a stand to be €10	R1	The interpretation must refer to the number of stands and the average cost.
Part (c)
\(\frac{9n+1050}{n}\le 12.6\) or \(9+\frac{1050}{n}\le 12.6\)	M1	Accept an equation with \(=12.6\), or a GDC intersection with \(y=12.6\).
\(n\ge 291.66\ldots\)	A1	Accept \(n=291.7\) from an equation.
292 stands, so 142 more stands	A1	291 stands gives an average cost of €12.61, which is too high.
Part (d)(i)
in one hour P prints \(\frac{1}{t}\) of the batch and Q prints \(\frac{1}{t+5}\)	M1
together they print \(\frac{1}{6}\) of the batch in one hour	A1
\(\frac{1}{t}+\frac{1}{t+5}=\frac{1}{6}\)	AG
Part (d)(ii)
\(6(t+5)+6t=t(t+5)\), giving \(t^2-7t-30=0\)	M1	Multiplying by \(6t(t+5)\) and rearranging to a quadratic equal to zero.
\((t-10)(t+3)=0\), \(t=10\) (\(t=-3\) rejected as a time cannot be negative)	A1	Reason for rejection required.
P takes 10 hours and Q takes 15 hours	A1
Part (e)(i)
\(\frac{1}{T}=\frac{1}{10}+\frac{1}{x}\)	M1	Allow ECF from (d)(ii) for the 10.
\(\frac{1}{T}=\frac{x+10}{10x}\)	A1
\(T(x)=\frac{10x}{x+10}\)	AG
Part (e)(ii)
asymptotes \(x=-10\) and \(y=10\) drawn and labelled	A1
two branches in the correct regions: one above \(y=10\) for \(x\lt -10\), one below \(y=10\) for \(x\gt -10\)	A1	Branches must approach the asymptotes without crossing them.
right branch passes through the origin \((0,\,0)\)	A1
Part (e)(iii)
domain \(x\gt 0\); range \(0\lt T(x)\lt 10\)	A1	Both needed.
however slow printer R is, P and R together always finish faster than P alone (less than 10 hours)	R1
Part (e)(iv)
\(\frac{10x}{x+10}=4\), giving \(6x=40\)	M1	Accept a GDC intersection of \(y=T(x)\) and \(y=4\).
\(x=\frac{20}{3}\) hours (6 hours 40 minutes)	A1	Accept 6.67 hours.

Answers: (a)(i) \(a=7\), \(b=840\) · (a)(ii) vertical stretch, scale factor 840, then translation by \(\begin{pmatrix}0\\7\end{pmatrix}\) · (b)(i) \(C^{-1}(x)=\dfrac{840}{x-7}\) · (b)(ii) 280 · (c) 142 more stands (292 in total) · (d)(i) \(\dfrac{1}{t}+\dfrac{1}{t+5}=\dfrac{1}{6}\) · (d)(ii) P: 10 hours, Q: 15 hours · (e)(i) \(T(x)=\dfrac{10x}{x+10}\) · (e)(ii) asymptotes \(x=-10\), \(y=10\); passes through \((0,0)\) · (e)(iii) \(x\gt 0\); \(0\lt T(x)\lt 10\) · (e)(iv) \(\dfrac{20}{3}\) hours

Common error (b)(ii) Reading \(C^{-1}(10)\) as a cost. The inverse takes an average cost as its input and gives the number of stands as its output.

More 2.8 questions in the question bank

2.9 Exponential and logarithmic functions · 2.10 Solving equations graphically and analytically · 2.11 Transformations of graphsmerged group

+ Add to test✓ In your testCC-AA-29-210-211-123 marks~30 min

A cup of coffee cooling in a room

2.9 Exponential and logarithmic functions2.11 Transformations of graphs2.10 Solving equations graphically and analytically1.7 Rational exponents and laws of logarithmsSL

A cup of freshly made coffee is left on a desk in a room where the air temperature is a constant 20 °C. The temperature of the coffee, \(T\) °C, \(t\) minutes after it is made, is modelled by

\(T(t)=20+70e^{-kt},\quad t\ge 0,\)

where \(k\) is a positive constant. After 10 minutes the temperature of the coffee is 62 °C.

(a)

Write down the temperature of the coffee when it is made, and the equation of the horizontal asymptote of the graph of \(T\). Interpret the asymptote in context.

No calculatorINTERPRETNotes

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\(T(0)=90\) °C and asymptote \(T=20\)	A1	Both needed.
the coffee cools towards the room temperature of 20 °C but, according to the model, never quite reaches it	R1

Answer: 90 °C; \(T=20\)

(2)

(b)

Show that \(k=\dfrac{1}{10}\ln\dfrac{5}{3}\).

No calculatorALGEBRANotes

Show solution

\(20+70e^{-10k}=62\)	M1	Substituting \(t=10\), \(T=62\).
\(e^{-10k}=\frac{42}{70}=\frac{3}{5}\), so \(-10k=\ln\frac{3}{5}\)	A1	Taking natural logarithms correctly.
\(k=-\frac{1}{10}\ln\frac{3}{5}=\frac{1}{10}\ln\left(\frac{3}{5}\right)^{-1}\)	A1	Use of a log law to remove the minus sign must be seen; accept \(\frac{1}{10}(\ln 5-\ln 3)\) with the step shown.
\(k=\frac{1}{10}\ln\frac{5}{3}\)	AG

Answer: \(k=\dfrac{1}{10}\ln\dfrac{5}{3}\)

(3)

(c)

Find the time it takes for the coffee to cool to 60 °C.

ROUTINENotes

Show solution

\(20+70e^{-kt}=60\), so \(e^{-kt}=\frac{4}{7}\)	M1	Or a GDC intersection of \(y=T(t)\) with \(y=60\).
\(t=\frac{1}{k}\ln\frac{7}{4}=11.0\) minutes	A1	Accept 10.96 (10.9551…).

Answer: 11.0 minutes

(2)

(d)(i)

Show that the time taken for the coffee to cool from 60 °C to 40 °C is \(\dfrac{\ln 2}{k}\) minutes.

No calculatorPROOFNotes

Show solution

\(e^{-kt_1}=\frac{40}{70}\) and \(e^{-kt_2}=\frac{20}{70}\)	M1	Setting up both times; accept \(t_1=\frac{1}{k}\ln\frac{7}{4}\), \(t_2=\frac{1}{k}\ln\frac{7}{2}\).
\(t_2-t_1=\frac{1}{k}\left(\ln\frac{7}{2}-\ln\frac{7}{4}\right)=\frac{1}{k}\ln\frac{7/2}{7/4}\)	A1	Or \(e^{-k(t_2-t_1)}=\frac{20}{40}=\frac{1}{2}\).
\(t_2-t_1=\frac{\ln 2}{k}\)	AG

Answer: \(\dfrac{\ln 2}{k}\)

(2)

(d)(ii)

Hence find this time.

ROUTINENotes

Show solution

\(\frac{10\ln 2}{\ln(5/3)}=13.6\) minutes

Accept 13.57.

Answer: 13.6 minutes

(1)

(e)(i)

The graph of \(y=T(t)\) can be obtained from the graph of \(y=e^{t}\) by a sequence of transformations. Describe a suitable sequence.

No calculatorCONCEPTNotes

Show solution

reflection in the \(y\)-axis, then horizontal stretch with scale factor \(\frac{1}{k}\) (or \(\frac{10}{\ln(5/3)}\))	A1	These two may be in either order.
vertical stretch, scale factor 70	A1
translation by \(\begin{pmatrix}0\\20\end{pmatrix}\), applied after the vertical stretch	A1	If the translation comes before the stretch it must be by \(\begin{pmatrix}0\\ \frac{2}{7}\end{pmatrix}\).

(3)

(e)(ii)

On the axes in Figure 1, sketch the graph of \(T\) for \(0\le t\le 60\). Show the asymptote with its equation and the point where the graph meets the vertical axis.

Figure 1: axes for the graph of \(T\)

GRAPHNotes

Show solution

decreasing curve, concave up, starting at \((0,\,90)\)	A1	The point \((0,\,90)\) must be labelled.
asymptote \(T=20\) drawn dashed and labelled; curve approaches it without crossing	A1
curve passes approximately through \((10,\,62)\) and \((60,\,23)\)	A1	Allow ±2 on the vertical scale.

Answer: decreasing curve from \((0,90)\) towards \(T=20\)

(3)

(f)(i)

A second cup of the same coffee, also at 90 °C when made, is carried outside where the air is at 5 °C. Because of the wind, its temperature \(U\) °C is modelled by \(U(t)=5+85e^{-2kt}\), \(t\ge 0\), with the same value of \(k\).

Let \(u=e^{-kt}\). Show that the equation \(T(t)-U(t)=25\) can be written as \(17u^2-14u+2=0\).

No calculatorALGEBRANotes

Show solution

\(e^{-2kt}=\left(e^{-kt}\right)^2=u^2\)	B1
\((20+70u)-(5+85u^2)=25\)	M1
\(85u^2-70u+10=0\), then divide by 5	A1
\(17u^2-14u+2=0\)	AG

Answer: \(17u^2-14u+2=0\)

(3)

(f)(ii)

Hence find the two times at which the coffee indoors is 25 °C hotter than the coffee outdoors.

MULTINotes

Show solution

\(u=\frac{7\pm\sqrt{15}}{17}\), that is \(u=0.6396\) or \(u=0.1839\)	A1	Accept either form.
both values lie in \(0\lt u\le 1\), so both are valid; \(t=-\frac{1}{k}\ln u\)	M1	Attempt to convert at least one value of \(u\) back to \(t\).
\(t=8.75\) minutes and \(t=33.1\) minutes	A1	Both needed. Accept a GDC intersection as a check, not as the method.

Answer: 8.75 minutes and 33.1 minutes

(3)

(f)(iii)

Figure 2 shows the graph of \(y=T(t)-U(t)\). With reference to the graph, explain why there are two answers to part (f)(ii).

Figure 2: the graph of \(y=T(t)-U(t)\) for \(0\le t\le 60\), with the line \(y=25\)

No calculatorINTERPRETNotes

Show solution

the difference starts at 0, rises to a maximum (about 29 °C) above 25 and then falls, so the line \(y=25\) meets the graph twice

Reference to the maximum lying above 25 is needed.

(1)

Show full solution

Working	Mark	Notes
Part (a)
\(T(0)=90\) °C and asymptote \(T=20\)	A1	Both needed.
the coffee cools towards the room temperature of 20 °C but, according to the model, never quite reaches it	R1
Part (b)
\(20+70e^{-10k}=62\)	M1	Substituting \(t=10\), \(T=62\).
\(e^{-10k}=\frac{42}{70}=\frac{3}{5}\), so \(-10k=\ln\frac{3}{5}\)	A1	Taking natural logarithms correctly.
\(k=-\frac{1}{10}\ln\frac{3}{5}=\frac{1}{10}\ln\left(\frac{3}{5}\right)^{-1}\)	A1	Use of a log law to remove the minus sign must be seen; accept \(\frac{1}{10}(\ln 5-\ln 3)\) with the step shown.
\(k=\frac{1}{10}\ln\frac{5}{3}\)	AG
Part (c)
\(20+70e^{-kt}=60\), so \(e^{-kt}=\frac{4}{7}\)	M1	Or a GDC intersection of \(y=T(t)\) with \(y=60\).
\(t=\frac{1}{k}\ln\frac{7}{4}=11.0\) minutes	A1	Accept 10.96 (10.9551…).
Part (d)(i)
\(e^{-kt_1}=\frac{40}{70}\) and \(e^{-kt_2}=\frac{20}{70}\)	M1	Setting up both times; accept \(t_1=\frac{1}{k}\ln\frac{7}{4}\), \(t_2=\frac{1}{k}\ln\frac{7}{2}\).
\(t_2-t_1=\frac{1}{k}\left(\ln\frac{7}{2}-\ln\frac{7}{4}\right)=\frac{1}{k}\ln\frac{7/2}{7/4}\)	A1	Or \(e^{-k(t_2-t_1)}=\frac{20}{40}=\frac{1}{2}\).
\(t_2-t_1=\frac{\ln 2}{k}\)	AG
Part (d)(ii)
\(\frac{10\ln 2}{\ln(5/3)}=13.6\) minutes	A1	Accept 13.57.
Part (e)(i)
reflection in the \(y\)-axis, then horizontal stretch with scale factor \(\frac{1}{k}\) (or \(\frac{10}{\ln(5/3)}\))	A1	These two may be in either order.
vertical stretch, scale factor 70	A1
translation by \(\begin{pmatrix}0\\20\end{pmatrix}\), applied after the vertical stretch	A1	If the translation comes before the stretch it must be by \(\begin{pmatrix}0\\ \frac{2}{7}\end{pmatrix}\).
Part (e)(ii)
decreasing curve, concave up, starting at \((0,\,90)\)	A1	The point \((0,\,90)\) must be labelled.
asymptote \(T=20\) drawn dashed and labelled; curve approaches it without crossing	A1
curve passes approximately through \((10,\,62)\) and \((60,\,23)\)	A1	Allow ±2 on the vertical scale.
Part (f)(i)
\(e^{-2kt}=\left(e^{-kt}\right)^2=u^2\)	B1
\((20+70u)-(5+85u^2)=25\)	M1
\(85u^2-70u+10=0\), then divide by 5	A1
\(17u^2-14u+2=0\)	AG
Part (f)(ii)
\(u=\frac{7\pm\sqrt{15}}{17}\), that is \(u=0.6396\) or \(u=0.1839\)	A1	Accept either form.
both values lie in \(0\lt u\le 1\), so both are valid; \(t=-\frac{1}{k}\ln u\)	M1	Attempt to convert at least one value of \(u\) back to \(t\).
\(t=8.75\) minutes and \(t=33.1\) minutes	A1	Both needed. Accept a GDC intersection as a check, not as the method.
Part (f)(iii)
the difference starts at 0, rises to a maximum (about 29 °C) above 25 and then falls, so the line \(y=25\) meets the graph twice	R1	Reference to the maximum lying above 25 is needed.

Answers: (a) 90 °C; \(T=20\) · (b) \(k=\dfrac{1}{10}\ln\dfrac{5}{3}\) · (c) 11.0 minutes · (d)(i) \(\dfrac{\ln 2}{k}\) · (d)(ii) 13.6 minutes · (e)(ii) decreasing curve from \((0,90)\) towards \(T=20\) · (f)(i) \(17u^2-14u+2=0\) · (f)(ii) 8.75 minutes and 33.1 minutes

Exam tip (d)(ii) Store \(k\) in your GDC rather than typing a rounded value such as 0.05; early rounding changes the third significant figure.

Common error (f)(ii) Stopping at the values of \(u\). After a substitution, convert back to \(t\) and check each \(u\) is possible: here \(u=e^{-kt}\) must lie between 0 and 1.

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Streaming services racing for subscribers

2.10 Solving equations graphically and analytically2.9 Exponential and logarithmic functions2.11 Transformations of graphs2.4 Key features of graphs and intersections using technologySL

Two video streaming services, A and B, launch at the same time. The number of subscribers, in millions, \(t\) years after launch is modelled by

\(S_A(t)=2+1.5\ln(t+1)\quad\) and \(\quad S_B(t)=0.6e^{0.25t},\quad t\ge 0.\)

(a)

Find the exact time at which service A reaches 5 million subscribers.

No calculatorROUTINENotes

Show solution

\(1.5\ln(t+1)=3\), so \(\ln(t+1)=2\)	M1
\(t=e^{2}-1\) years	A1	Exact answer required.

Answer: \(e^2-1\) years

(2)

(b)

Find the time at which service B reaches 4.8 million subscribers. Give your answer in the form \(a\ln 2\), where \(a\in\mathbb{Z}\).

No calculatorROUTINENotes

Show solution

\(e^{0.25t}=8\), so \(0.25t=\ln 8\)	M1
\(t=4\ln 8=12\ln 2\) years	A1	Use of \(\ln 8=3\ln 2\) needed for this mark.

Answer: \(12\ln 2\) years

(2)

(c)

Describe a sequence of transformations that maps the graph of \(y=\ln t\) onto the graph of \(y=S_A(t)\).

No calculatorCONCEPTNotes

Show solution

translation by \(\begin{pmatrix}-1\\0\end{pmatrix}\)	A1	Accept "1 unit to the left".
vertical stretch, scale factor 1.5	A1
translation by \(\begin{pmatrix}0\\2\end{pmatrix}\), after the stretch	A1	Accept a single translation by \(\begin{pmatrix}-1\\2\end{pmatrix}\) after the stretch for the first and third marks.

(3)

(d)

On the axes in Figure 1, sketch the graphs of \(y=S_A(t)\) and \(y=S_B(t)\) for \(0\le t\le 10\). Label the points where each graph meets the vertical axis.

Figure 1: axes for the graphs of \(S_A\) and \(S_B\)

GRAPHNotes

Show solution

\(S_A\): increasing, concave down curve from \((0,\,2)\), reaching about 5.6 at \(t=10\)	A1
\(S_B\): increasing, concave up curve from \((0,\,0.6)\), reaching about 7.3 at \(t=10\)	A1
a single crossing at about \(t=8.8\), with B above A after it	A1

Answer: two increasing curves crossing once near \(t=8.8\)

(3)

(e)

Use your graphic display calculator to find when service B overtakes service A, and the number of subscribers each service has at that time.

ROUTINENotes

Show solution

\(t=8.81\) years	A1	Accept 8.807…
5.42 million subscribers	A1

Answer: \(t=8.81\) years; 5.42 million

(2)

(f)

If both models are drawn for \(t\gt -1\), the equation \(S_A(t)=S_B(t)\) has a second solution. Write down this solution and explain why it is not relevant.

INTERPRETNotes

Show solution

\(t=-0.629\)	A1
it is a negative time, before the services launched, and the models are defined only for \(t\ge 0\)	R1

Answer: \(t=-0.629\)

(2)

(g)(i)

A third service, C, launched at the same time, has \(S_C(t)=0.1e^{0.5t}+0.8\) million subscribers.

Let \(u=e^{0.25t}\). Show that the equation \(S_B(t)=S_C(t)\) can be written as \(u^2-6u+8=0\).

No calculatorALGEBRANotes

Show solution

\(e^{0.5t}=\left(e^{0.25t}\right)^2=u^2\)	B1
\(0.6u=0.1u^2+0.8\), so \(0.1u^2-0.6u+0.8=0\), then multiply by 10	A1
\(u^2-6u+8=0\)	AG

Answer: \(u^2-6u+8=0\)

(2)

(g)(ii)

Hence find the exact times at which services B and C have the same number of subscribers.

No calculatorALGEBRANotes

Show solution

\((u-2)(u-4)=0\)	M1	Or a correct substitution into the quadratic formula.
\(u=2\) or \(u=4\)	A1
\(e^{0.25t}=2\Rightarrow t=4\ln 2\); \(e^{0.25t}=4\Rightarrow t=4\ln 4=8\ln 2\) years	A1	Both needed. Accept \(4\ln 4\). Allow ECF from their values of \(u\) if both are positive.

Answer: \(4\ln 2\) and \(8\ln 2\) years

(3)

(g)(iii)

Write down the interval of time during which service B has more subscribers than service C.

No calculatorINTERPRETNotes

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\(4\ln 2\lt t\lt 8\ln 2\)

Accept \(2.77\lt t\lt 5.55\). Allow ECF from (g)(ii).

Answer: \(4\ln 2\lt t\lt 8\ln 2\)

(1)

(h)

A rival lowers the growth of service B, so that its model becomes \(S_B(t)=0.6e^{0.2t}\). Service A's model is unchanged.

Find the new time at which B overtakes A, and comment on whether either model is likely to remain valid for very large values of \(t\).

MODELNotes

Show solution

\(2+1.5\ln(t+1)=0.6e^{0.2t}\)	(M1)	Implied by a correct answer.
\(t=11.3\) years	A1	Accept 11.31.
neither model can stay valid: both grow without limit, but the number of possible subscribers is limited (B's exponential growth is the less realistic)	R1	Any comment based on the limited size of the market.

Answer: 11.3 years

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(1.5\ln(t+1)=3\), so \(\ln(t+1)=2\)	M1
\(t=e^{2}-1\) years	A1	Exact answer required.
Part (b)
\(e^{0.25t}=8\), so \(0.25t=\ln 8\)	M1
\(t=4\ln 8=12\ln 2\) years	A1	Use of \(\ln 8=3\ln 2\) needed for this mark.
Part (c)
translation by \(\begin{pmatrix}-1\\0\end{pmatrix}\)	A1	Accept "1 unit to the left".
vertical stretch, scale factor 1.5	A1
translation by \(\begin{pmatrix}0\\2\end{pmatrix}\), after the stretch	A1	Accept a single translation by \(\begin{pmatrix}-1\\2\end{pmatrix}\) after the stretch for the first and third marks.
Part (d)
\(S_A\): increasing, concave down curve from \((0,\,2)\), reaching about 5.6 at \(t=10\)	A1
\(S_B\): increasing, concave up curve from \((0,\,0.6)\), reaching about 7.3 at \(t=10\)	A1
a single crossing at about \(t=8.8\), with B above A after it	A1
Part (e)
\(t=8.81\) years	A1	Accept 8.807…
5.42 million subscribers	A1
Part (f)
\(t=-0.629\)	A1
it is a negative time, before the services launched, and the models are defined only for \(t\ge 0\)	R1
Part (g)(i)
\(e^{0.5t}=\left(e^{0.25t}\right)^2=u^2\)	B1
\(0.6u=0.1u^2+0.8\), so \(0.1u^2-0.6u+0.8=0\), then multiply by 10	A1
\(u^2-6u+8=0\)	AG
Part (g)(ii)
\((u-2)(u-4)=0\)	M1	Or a correct substitution into the quadratic formula.
\(u=2\) or \(u=4\)	A1
\(e^{0.25t}=2\Rightarrow t=4\ln 2\); \(e^{0.25t}=4\Rightarrow t=4\ln 4=8\ln 2\) years	A1	Both needed. Accept \(4\ln 4\). Allow ECF from their values of \(u\) if both are positive.
Part (g)(iii)
\(4\ln 2\lt t\lt 8\ln 2\)	A1	Accept \(2.77\lt t\lt 5.55\). Allow ECF from (g)(ii).
Part (h)
\(2+1.5\ln(t+1)=0.6e^{0.2t}\)	(M1)	Implied by a correct answer.
\(t=11.3\) years	A1	Accept 11.31.
neither model can stay valid: both grow without limit, but the number of possible subscribers is limited (B's exponential growth is the less realistic)	R1	Any comment based on the limited size of the market.

Answers: (a) \(e^2-1\) years · (b) \(12\ln 2\) years · (d) two increasing curves crossing once near \(t=8.8\) · (e) \(t=8.81\) years; 5.42 million · (f) \(t=-0.629\) · (g)(i) \(u^2-6u+8=0\) · (g)(ii) \(4\ln 2\) and \(8\ln 2\) years · (g)(iii) \(4\ln 2\lt t\lt 8\ln 2\) · (h) 11.3 years

More 2.10 questions in the question bank

2.12 Polynomial functions, factor and remainder theorems, sum and product of roots · 2.13 Rational functions with oblique and horizontal asymptotesmerged group

+ Add to test✓ In your testCC-AA-212-213-128 marks~35 min

Designing a section of rollercoaster track with a cubic

2.12 Polynomial functions, factor and remainder theorems, sum and product of roots2.4 Key features of graphs and intersections using technology5.8 Stationary points, inflexion and optimization2.7 Quadratic equations, inequalities and the discriminantHL

An engineer designs a section of rollercoaster track that leaves the ground, passes over a hill and returns to the ground. Horizontal distances are measured from a support post at O. At a horizontal distance of \(x\) metres from O, the height of the track above level ground is \(h(x)\) metres, where

\(h(x)=0.001\left(x^3-125x^2+ax+b\right),\quad 5\le x\le40,\)

and \(a\) and \(b\) are constants. Let \(P(x)=x^3-125x^2+ax+b\), so that \(h(x)=0.001P(x)\).

The track meets the ground at A, where \(x=5\), and at B, where \(x=40\), as shown in Figure 1.

Figure 1: the track section between A and B (no vertical scale shown)

(a)

Use the factor theorem to show that \(a=3800\) and \(b=-16\,000\).

No calculatorPROOFNotes

Show solution

\(P(5)=0\) or \(P(40)=0\) used, e.g. \(125-3125+5a+b=0\)	M1	Substituting a root of the divisor and equating to zero.
\(5a+b=3000\) and \(40a+b=136\,000\)	A1	Both equations, in any equivalent form.
eliminating \(b\), e.g. \(35a=133\,000\)	A1	Or eliminating \(a\).
\(a=3800\), \(b=3000-19\,000=-16\,000\)	AG	Checking that the given values satisfy one equation only scores M1 at most.

Answer: \(a=3800\), \(b=-16\,000\)

(3)

(b)

If the model is extended beyond B, the curve \(y=P(x)\) crosses the \(x\)-axis at a third point C.

Use the sum of the roots of \(P(x)=0\) to find the \(x\)-coordinate of C. Verify your answer using the product of the roots, and hence write \(h(x)\) in fully factorised form.

No calculatorALGEBRANotes

Show solution

sum of roots \(=-\frac{-125}{1}=125\)	M1
\(5+40+x_{\mathrm{C}}=125\), so \(x_{\mathrm{C}}=80\)	A1
product of roots \(=-\frac{-16\,000}{1}=16\,000\) and \(5\times40\times80=16\,000\)	A1	Both values needed for the verification.
\(h(x)=0.001(x-5)(x-40)(x-80)\)	A1	Accept \(\frac{1}{1000}(x-5)(x-40)(x-80)\).

Answer: \(x=80\); \(h(x)=0.001(x-5)(x-40)(x-80)\)

(4)

(c)

Find the remainder when \(P(x)\) is divided by \((x-25)\). Hence write down the height of the track at \(x=25\).

No calculatorROUTINENotes

Show solution

remainder \(=P(25)=15\,625-78\,125+95\,000-16\,000\) \((=16\,500)\)	M1	Or \(P(25)=20\times(-15)\times(-55)\) from (b).
height \(=16.5\) m	A1	Remainder 16 500 must be seen or implied.

Answer: remainder 16 500; height 16.5 m

(2)

(d)

Use your GDC to find the coordinates of the highest point of this section of track.

ROUTINENotes

Show solution

\(x=20\)	A1
\(y=18\)	A1	Answer as coordinates \((20,\,18)\).

Answer: \((20,\,18)\)

(2)

(e)

Show that \(h'(20)=0\), and that the stationary point at \(x=20\) is a local maximum.

No calculatorPROOFNotes

Show solution

\(h'(x)=0.001\left(3x^2-250x+3800\right)\)	M1	Differentiating term by term.
\(h\'(20)=0.001(1200-5000+3800)=0\)	A1	Substitution must be seen.
\(h''(x)=0.001(6x-250)\), \(h''(20)=-0.13<0\), so a local maximum	R1	Or a sign change of \(h'\): \(h'(19)=0.133>0\), \(h'(21)=-0.127<0\).
\(h'(20)=0\) and the point is a local maximum	AG

(3)

(f)(i)

A safety net must be fitted beneath every part of the track that is at least 12.5 m above the ground.

Show that the values of \(x\) at which \(h(x)=12.5\) satisfy \(Q(x)=0\), where \(Q(x)=x^3-125x^2+3800x-28\,500\), and show that \((x-30)\) is a factor of \(Q(x)\).

No calculatorPROOFNotes

Show solution

\(0.001P(x)=12.5\Rightarrow P(x)-12\,500=0\), which is \(Q(x)=0\)	M1
\(Q(30)=27\,000-112\,500+114\,000-28\,500=0\)	A1	Each term must be seen.
so \((x-30)\) is a factor, by the factor theorem	AG

(2)

(f)(ii)

Hence find the exact value of the other \(x\)-coordinate in the interval \(5\le x\le40\) at which \(h(x)=12.5\).

No calculatorALGEBRANotes

Show solution

attempt to divide \(Q(x)\) by \((x-30)\): \(Q(x)=(x-30)\left(x^2-95x+950\right)\)	M1	By division or by comparing coefficients.
\(x=\frac{95\pm\sqrt{95^2-4(950)}}{2}=\frac{95\pm\sqrt{5225}}{2}\)	M1	Quadratic formula (or completing the square) applied to their quadratic.
\(x=\frac{95-5\sqrt{209}}{2}\)	A1	Accept \(\frac{95-\sqrt{5225}}{2}\). The root \(\frac{95+5\sqrt{209}}{2}\approx83.6\) must be rejected (outside the section).

Answer: \(x=\frac{95-5\sqrt{209}}{2}\ (\approx11.4)\)

(3)

(f)(iii)

Find the horizontal length of track that needs the net. Justify your answer.

MODELNotes

Show solution

the track is above 12.5 m between the two solutions, since it rises to 18 m at \(x=20\) and is below 12.5 m near A and B, so \(\frac{95-5\sqrt{209}}{2}\le x\le30\)	R1	Accept a sign argument, a test value (e.g. \(h(20)=18\ge12.5\)) or a reference to the graph.
\(30-11.357\ldots=18.6\) m	A1	Allow ECF from (f)(ii).

Answer: 18.6 m

(2)

(g)

On the axes in Figure 2, sketch the graph of \(y=h(x)\) for \(5\le x\le40\) and the line \(y=12.5\). Label the end points and the maximum point with their coordinates, and indicate the part of the track that needs the net.

Figure 2: axes for part (g)

GRAPHNotes

Show solution

smooth single hump from \((5,\,0)\) to \((40,\,0)\), end points labelled	A1	No turning point other than the maximum in \(5\lt x\lt 40\).
maximum labelled \((20,\,18)\), clearly to the left of the middle of the interval	A1
horizontal line \(y=12.5\) meeting the curve at \(x\approx11.4\) and \(x=30\), with the part between them indicated	A1	Allow ECF from (f).

(3)

(h)(i)

For a smaller version of the ride, the engineer keeps A and B and uses the model \(h_k(x)=k(x-5)(x-40)(x-80)\), \(5\le x\le40\), where \(k>0\). A safety rule says that this track must never be more than 9 m above the ground.

Find the greatest possible value of \(k\).

No calculatorMODELNotes

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highest point still at \(x=20\) (a positive multiple of \(P\) has the same stationary points), so \(18\,000k\le9\)	M1	Accept \(\frac{h_k(20)}{h(20)}=\frac{9}{18}\), i.e. a vertical scale factor of \(\frac12\).
\(k=0.0005\)	A1	Accept \(\frac{1}{2000}\) or \(5\times10^{-4}\).

Answer: \(k=0.0005\)

(2)

(h)(ii)

With this value of \(k\), a net is needed beneath every part of the smaller track that is at least 6.25 m above the ground.

Explain why the horizontal length of track that needs the net is the same as your answer to (f)(iii).

No calculatorINTERPRETNotes

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\(0.0005P(x)\ge6.25\iff P(x)\ge12\,500\)	R1
this is the same inequality as \(0.001P(x)\ge12.5\), so it has the same solution interval and the same length (18.6 m)	R1	Alternative: \(h_k\) is a vertical stretch of \(h\) with factor \(\frac12\) and \(6.25=\frac12\times12.5\), so the intersection points have the same \(x\)-coordinates.

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(P(5)=0\) or \(P(40)=0\) used, e.g. \(125-3125+5a+b=0\)	M1	Substituting a root of the divisor and equating to zero.
\(5a+b=3000\) and \(40a+b=136\,000\)	A1	Both equations, in any equivalent form.
eliminating \(b\), e.g. \(35a=133\,000\)	A1	Or eliminating \(a\).
\(a=3800\), \(b=3000-19\,000=-16\,000\)	AG	Checking that the given values satisfy one equation only scores M1 at most.
Part (b)
sum of roots \(=-\frac{-125}{1}=125\)	M1
\(5+40+x_{\mathrm{C}}=125\), so \(x_{\mathrm{C}}=80\)	A1
product of roots \(=-\frac{-16\,000}{1}=16\,000\) and \(5\times40\times80=16\,000\)	A1	Both values needed for the verification.
\(h(x)=0.001(x-5)(x-40)(x-80)\)	A1	Accept \(\frac{1}{1000}(x-5)(x-40)(x-80)\).
Part (c)
remainder \(=P(25)=15\,625-78\,125+95\,000-16\,000\) \((=16\,500)\)	M1	Or \(P(25)=20\times(-15)\times(-55)\) from (b).
height \(=16.5\) m	A1	Remainder 16 500 must be seen or implied.
Part (d)
\(x=20\)	A1
\(y=18\)	A1	Answer as coordinates \((20,\,18)\).
Part (e)
\(h'(x)=0.001\left(3x^2-250x+3800\right)\)	M1	Differentiating term by term.
\(h\'(20)=0.001(1200-5000+3800)=0\)	A1	Substitution must be seen.
\(h''(x)=0.001(6x-250)\), \(h''(20)=-0.13<0\), so a local maximum	R1	Or a sign change of \(h'\): \(h'(19)=0.133>0\), \(h'(21)=-0.127<0\).
\(h'(20)=0\) and the point is a local maximum	AG
Part (f)(i)
\(0.001P(x)=12.5\Rightarrow P(x)-12\,500=0\), which is \(Q(x)=0\)	M1
\(Q(30)=27\,000-112\,500+114\,000-28\,500=0\)	A1	Each term must be seen.
so \((x-30)\) is a factor, by the factor theorem	AG
Part (f)(ii)
attempt to divide \(Q(x)\) by \((x-30)\): \(Q(x)=(x-30)\left(x^2-95x+950\right)\)	M1	By division or by comparing coefficients.
\(x=\frac{95\pm\sqrt{95^2-4(950)}}{2}=\frac{95\pm\sqrt{5225}}{2}\)	M1	Quadratic formula (or completing the square) applied to their quadratic.
\(x=\frac{95-5\sqrt{209}}{2}\)	A1	Accept \(\frac{95-\sqrt{5225}}{2}\). The root \(\frac{95+5\sqrt{209}}{2}\approx83.6\) must be rejected (outside the section).
Part (f)(iii)
the track is above 12.5 m between the two solutions, since it rises to 18 m at \(x=20\) and is below 12.5 m near A and B, so \(\frac{95-5\sqrt{209}}{2}\le x\le30\)	R1	Accept a sign argument, a test value (e.g. \(h(20)=18\ge12.5\)) or a reference to the graph.
\(30-11.357\ldots=18.6\) m	A1	Allow ECF from (f)(ii).
Part (g)
smooth single hump from \((5,\,0)\) to \((40,\,0)\), end points labelled	A1	No turning point other than the maximum in \(5\lt x\lt 40\).
maximum labelled \((20,\,18)\), clearly to the left of the middle of the interval	A1
horizontal line \(y=12.5\) meeting the curve at \(x\approx11.4\) and \(x=30\), with the part between them indicated	A1	Allow ECF from (f).
Part (h)(i)
highest point still at \(x=20\) (a positive multiple of \(P\) has the same stationary points), so \(18\,000k\le9\)	M1	Accept \(\frac{h_k(20)}{h(20)}=\frac{9}{18}\), i.e. a vertical scale factor of \(\frac12\).
\(k=0.0005\)	A1	Accept \(\frac{1}{2000}\) or \(5\times10^{-4}\).
Part (h)(ii)
\(0.0005P(x)\ge6.25\iff P(x)\ge12\,500\)	R1
this is the same inequality as \(0.001P(x)\ge12.5\), so it has the same solution interval and the same length (18.6 m)	R1	Alternative: \(h_k\) is a vertical stretch of \(h\) with factor \(\frac12\) and \(6.25=\frac12\times12.5\), so the intersection points have the same \(x\)-coordinates.

Answers: (a) \(a=3800\), \(b=-16\,000\) · (b) \(x=80\); \(h(x)=0.001(x-5)(x-40)(x-80)\) · (c) remainder 16 500; height 16.5 m · (d) \((20,\,18)\) · (f)(ii) \(x=\frac{95-5\sqrt{209}}{2}\ (\approx11.4)\) · (f)(iii) 18.6 m · (h)(i) \(k=0.0005\)

Common error (b) For \(x^3-125x^2+\dots=0\) the sum of the roots is \(+125\), not \(-125\): the sum is \(-\frac{b}{a}\), and for a cubic the product is \(-\frac{d}{a}\).

Exam tip (e) A value from the GDC does not prove a maximum. Write \(h''(20)\) with its sign, or a sign table for \(h'\), and state the conclusion.

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A delivery truck's fuel use at different speeds

2.13 Rational functions with oblique and horizontal asymptotes2.12 Polynomial functions, factor and remainder theorems, sum and product of roots5.8 Stationary points, inflexion and optimization2.15 Solving inequalities g(x) ≥ f(x)HL

A logistics company tests one of its delivery trucks on a flat test track. When the truck travels at a constant speed of \(v\) km h⁻¹, its fuel consumption is \(E(v)\) litres per 100 km, where

\(E(v)=\dfrac{v^2+20v+3700}{5(v+10)},\quad 0\lt v\le120.\)

(a)

Use the remainder theorem to find the remainder when \(v^2+20v+3700\) is divided by \((v+10)\).

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\((-10)^2+20(-10)+3700\)	M1	Substituting \(v=-10\). Long division instead of the remainder theorem: award M1 A0 for a correct remainder of 3600.
\(3600\)	A1

Answer: 3600

(2)

(b)

Hence show that \(E(v)=\dfrac{v+10}{5}+\dfrac{720}{v+10}\).

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\(v^2+20v+3700=(v+10)(v+10)+3600\)	M1	Quotient \(v+10\) by inspection or division, with the remainder from (a).
\(E(v)=\frac{(v+10)^2+3600}{5(v+10)}=\frac{v+10}{5}+\frac{3600}{5(v+10)}\)	A1
\(=\frac{v+10}{5}+\frac{720}{v+10}\)	AG

(2)

(c)

The formula for \(E\) is defined for all real \(v\ne-10\). Write down the equations of the asymptotes of the graph of \(E\).

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\(v=-10\)	A1
\(E=\frac{v}{5}+2\)	A1	Accept \(E=\frac{v+10}{5}\) or \(y=0.2v+2\). Answers that are not equations score A0.

Answer: \(v=-10\) and \(E=\frac{v}{5}+2\)

(2)

(d)

Interpret the gradient of the oblique asymptote in the context of the truck travelling at high speeds.

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for large \(v\), \(\frac{720}{v+10}\) is small, so \(E(v)\approx\frac{v}{5}+2\)	R1
each extra 1 km h⁻¹ uses about 0.2 more litres per 100 km (about 1 litre more for each extra 5 km h⁻¹)	R1	Units or context required.

(2)

(e)(i)

Use calculus to show that the fuel consumption is least when \(v=50\).

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\(E'(v)=\frac15-\frac{720}{(v+10)^2}\)	M1	Differentiating the form from (b); or a correct quotient-rule derivative of the original fraction.
\(E'(v)=0\Rightarrow(v+10)^2=3600\)	M1
\(v+10=\pm60\), and \(v=-70\) is rejected because \(v>0\)	A1
\(E''(v)=\frac{1440}{(v+10)^3}>0\) at \(v=50\), so a minimum	R1	Or a sign change of \(E'\) from negative to positive.
\(v=50\)	AG

(4)

(e)(ii)

Find the least fuel consumption.

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\(E(50)=12+12=24\) litres per 100 km

Answer: 24 litres per 100 km

(1)

(f)

On the axes in Figure 1, sketch the graph of \(E\) for \(0\lt v\le120\). Show the oblique asymptote, and label the minimum point and the end point at \(v=120\) with their coordinates.

Figure 1: axes for part (f)

GRAPHNotes

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curve decreasing from near \((0,\,74)\) to a single minimum, then increasing	A1	Open circle or no point at \(v=0\) (\(E\to74\)).
minimum \((50,\,24)\) and end point \((120,\,31.5)\) labelled	A1	Accept \(\left(120,\,\frac{410}{13}\right)\).
line \(E=\frac{v}{5}+2\) drawn, with the curve above it and approaching it as \(v\) increases	A1

(3)

(g)

The company wants the truck to use no more than 26 litres per 100 km. Solve the inequality \(E(v)\le26\) algebraically, for \(0\lt v\le120\).

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\(\frac{v^2+20v+3700-130(v+10)}{5(v+10)}\le0\)	M1	Rearranging to a single fraction. Multiplying both sides by \(5(v+10)\) is accepted only with a statement that \(5(v+10)>0\) for \(v>0\).
\(\frac{v^2-110v+2400}{5(v+10)}\le0\)	A1
\(\frac{(v-30)(v-80)}{5(v+10)}\le0\), critical values 30 and 80 (denominator positive for \(v>0\))	M1	A sign diagram or a sketch may be seen.
\(30\le v\le80\)	A1	Strict inequalities score A0.

Answer: \(30\le v\le80\)

(4)

(h)

When the truck is fully loaded, its fuel consumption is modelled by \(E_{\mathrm{L}}(v)=\dfrac{v^2+20v+c}{5(v+10)}\), where \(c>100\). Tests show that the fully loaded truck is most economical at \(v=60\).

Find the value of \(c\).

No calculatorMULTINotes

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remainder \(c-100\), so \(E_{\mathrm{L}}(v)=\frac{v+10}{5}+\frac{c-100}{5(v+10)}\)	M1	Or quotient rule on the original fraction.
\(E_{\mathrm{L}}'(60)=\frac15-\frac{c-100}{5(70)^2}=0\)	M1
\(c=5000\)	A1

Answer: \(c=5000\)

(3)

(i)

For the unloaded truck, the manager also considers the distance travelled per litre of fuel, \(K(v)=\dfrac{100}{E(v)}\) km per litre.

Write down the equation of the horizontal asymptote of the graph of \(K\), and explain why \(K\) has a horizontal asymptote although \(E\) does not.

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\(K=0\)	A1	Accept \(y=0\).
\(E(v)\) grows without limit (it follows the oblique asymptote), so \(\frac{100}{E(v)}\to0\): at higher and higher speeds the truck travels less and less far per litre	R1	Accept: \(K(v)=\frac{500(v+10)}{v^2+20v+3700}\) is linear over quadratic, so it tends to 0.

(2)

Show full solution

Working	Mark	Notes
Part (a)
\((-10)^2+20(-10)+3700\)	M1	Substituting \(v=-10\). Long division instead of the remainder theorem: award M1 A0 for a correct remainder of 3600.
\(3600\)	A1
Part (b)
\(v^2+20v+3700=(v+10)(v+10)+3600\)	M1	Quotient \(v+10\) by inspection or division, with the remainder from (a).
\(E(v)=\frac{(v+10)^2+3600}{5(v+10)}=\frac{v+10}{5}+\frac{3600}{5(v+10)}\)	A1
\(=\frac{v+10}{5}+\frac{720}{v+10}\)	AG
Part (c)
\(v=-10\)	A1
\(E=\frac{v}{5}+2\)	A1	Accept \(E=\frac{v+10}{5}\) or \(y=0.2v+2\). Answers that are not equations score A0.
Part (d)
for large \(v\), \(\frac{720}{v+10}\) is small, so \(E(v)\approx\frac{v}{5}+2\)	R1
each extra 1 km h⁻¹ uses about 0.2 more litres per 100 km (about 1 litre more for each extra 5 km h⁻¹)	R1	Units or context required.
Part (e)(i)
\(E'(v)=\frac15-\frac{720}{(v+10)^2}\)	M1	Differentiating the form from (b); or a correct quotient-rule derivative of the original fraction.
\(E'(v)=0\Rightarrow(v+10)^2=3600\)	M1
\(v+10=\pm60\), and \(v=-70\) is rejected because \(v>0\)	A1
\(E''(v)=\frac{1440}{(v+10)^3}>0\) at \(v=50\), so a minimum	R1	Or a sign change of \(E'\) from negative to positive.
\(v=50\)	AG
Part (e)(ii)
\(E(50)=12+12=24\) litres per 100 km	A1
Part (f)
curve decreasing from near \((0,\,74)\) to a single minimum, then increasing	A1	Open circle or no point at \(v=0\) (\(E\to74\)).
minimum \((50,\,24)\) and end point \((120,\,31.5)\) labelled	A1	Accept \(\left(120,\,\frac{410}{13}\right)\).
line \(E=\frac{v}{5}+2\) drawn, with the curve above it and approaching it as \(v\) increases	A1
Part (g)
\(\frac{v^2+20v+3700-130(v+10)}{5(v+10)}\le0\)	M1	Rearranging to a single fraction. Multiplying both sides by \(5(v+10)\) is accepted only with a statement that \(5(v+10)>0\) for \(v>0\).
\(\frac{v^2-110v+2400}{5(v+10)}\le0\)	A1
\(\frac{(v-30)(v-80)}{5(v+10)}\le0\), critical values 30 and 80 (denominator positive for \(v>0\))	M1	A sign diagram or a sketch may be seen.
\(30\le v\le80\)	A1	Strict inequalities score A0.
Part (h)
remainder \(c-100\), so \(E_{\mathrm{L}}(v)=\frac{v+10}{5}+\frac{c-100}{5(v+10)}\)	M1	Or quotient rule on the original fraction.
\(E_{\mathrm{L}}'(60)=\frac15-\frac{c-100}{5(70)^2}=0\)	M1
\(c=5000\)	A1
Part (i)
\(K=0\)	A1	Accept \(y=0\).
\(E(v)\) grows without limit (it follows the oblique asymptote), so \(\frac{100}{E(v)}\to0\): at higher and higher speeds the truck travels less and less far per litre	R1	Accept: \(K(v)=\frac{500(v+10)}{v^2+20v+3700}\) is linear over quadratic, so it tends to 0.

Answers: (a) 3600 · (c) \(v=-10\) and \(E=\frac{v}{5}+2\) · (e)(ii) 24 litres per 100 km · (g) \(30\le v\le80\) · (h) \(c=5000\)

Exam tip (g) Before multiplying an inequality by an expression in \(v\), check its sign on the domain. Here \(5(v+10)>0\) for \(v>0\), so the sign is kept; if the sign is unknown, combine into one fraction and use a sign diagram.

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2.14 Odd and even functions, inverse and self-inverse functions · 2.15 Solving inequalities g(x) ≥ f(x) · 2.16 Modulus, reciprocal and other graph transformationsmerged group

+ Add to test✓ In your testCC-AA-214-215-216-128 marks~35 min

A full-wave rectifier: turning AC into DC

2.16 Modulus, reciprocal and other graph transformations2.14 Odd and even functions, inverse and self-inverse functions3.7 Circular functions and their transformations5.11 Definite integrals and areasHL

An alternating voltage supply gives a voltage of \(V(t)\) volts at time \(t\) seconds, where

\(V(t)=12\sin(100\pi t).\)

A full-wave rectifier circuit turns every negative voltage into the positive voltage of the same size, so its output is \(|V(t)|\) volts.

Figure 1 shows the graph of \(y=V(t)\) for \(-0.02\le t\le0.02\).

Figure 1: the graph of \(y=V(t)\) for \(-0.02\le t\le0.02\)

(a)

Write down the amplitude of \(V\) and find its period.

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amplitude 12 (volts)	A1
period \(\frac{2\pi}{100\pi}=0.02\) s	A1	Accept \(\frac{1}{50}\) s.

Answer: amplitude 12 V; period 0.02 s

(2)

(b)

Show that \(V\) is an odd function.

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\(V(-t)=12\sin(-100\pi t)\)	M1	A general \(t\) is needed; numerical checks score M0.
\(=-12\sin(100\pi t)\), since \(\sin(-\theta)=-\sin\theta\)	A1
\(=-V(t)\), so \(V\) is odd	AG

(2)

(c)

On Figure 1, sketch the graph of \(y=|V(t)|\) for \(-0.02\le t\le0.02\).

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parts below the \(t\)-axis reflected in the \(t\)-axis, parts above unchanged; nothing below the axis	A1
four equal arches of height 12, touching the axis with sharp corners at \(t=0,\ \pm0.01,\ \pm0.02\)	A1

(2)

(d)

Write down the period and the range of \(|V|\).

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period 0.01 s	A1
\(0\le\|V(t)\|\le12\)	A1	Accept \([0,\,12]\).

Answer: period 0.01 s; range \(0\le|V(t)|\le12\)

(2)

(e)(i)

Show that \(|V|\) is an even function.

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\(\|V(-t)\|=\left\|-V(t)\right\|\)	M1	Using the result of (b).
\(=\|V(t)\|\), because \(\left\|-a\right\|=\|a\|\)	A1
so \(\|V\|\) is even	AG

(2)

(e)(ii)

Hence explain why the graph of \(y=|V(|t|)|\) is the same as the graph of \(y=|V(t)|\).

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for \(t\ge0\), \(|V(|t|)|=|V(t)|\); for \(t<0\), \(|V(|t|)|=|V(-t)|=|V(t)|\) by (e)(i)

Accept: \(y=f(|t|)\) keeps the right half and reflects it in the \(y\)-axis, and the graph of the even function \(|V|\) is already symmetric in the \(y\)-axis.

(1)

(f)

Solve the equation \(|V(t)|=6\) for \(0\le t\le0.02\), giving your answers as exact fractions.

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\(\sin(100\pi t)=\frac12\) or \(\sin(100\pi t)=-\frac12\)	M1	Both signs considered.
\(100\pi t=\frac{\pi}{6},\ \frac{5\pi}{6}\)	A1
\(100\pi t=\frac{7\pi}{6},\ \frac{11\pi}{6}\)	A1
\(t=\frac{1}{600},\ \frac{1}{120},\ \frac{7}{600},\ \frac{11}{600}\)	A1	Extra values in the interval: A0 for this mark.

Answer: \(t=\frac{1}{600},\ \frac{1}{120},\ \frac{7}{600},\ \frac{11}{600}\)

(4)

(g)

The mean output voltage of the rectifier is the mean value of \(|V(t)|\) over one of its periods, \(\dfrac{1}{0.01}\displaystyle\int_0^{0.01}|V(t)|\,\mathrm{d}t\).

Show that the mean output voltage is \(\dfrac{24}{\pi}\) volts.

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\(V(t)\ge0\) for \(0\le t\le0.01\), so \(\|V(t)\|=V(t)\)	R1
\(\int12\sin(100\pi t)\,\mathrm{d}t=-\frac{12}{100\pi}\cos(100\pi t)\)	M1	Integrating; the factor \(\frac{1}{100\pi}\) must be present.
\(-\frac{12}{100\pi}\left(\cos\pi-\cos0\right)\)	M1	Substituting both limits, upper minus lower.
\(=\frac{24}{100\pi}\)	A1
mean \(=100\times\frac{24}{100\pi}=\frac{24}{\pi}\)	AG

(4)

(h)

Consider the graph of \(y=\dfrac{1}{|V(t)|}\) for \(0\lt t\lt 0.02\), \(t\ne0.01\).

Write down the equation of the vertical asymptote that lies inside this interval, and the coordinates of the minimum points of the graph.

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\(t=0.01\)	A1	Condone also stating \(t=0\) and \(t=0.02\).
\(\left(0.005,\ \frac{1}{12}\right)\) and \(\left(0.015,\ \frac{1}{12}\right)\)	A1	Both points needed.

Answer: \(t=0.01\); \(\left(0.005,\frac{1}{12}\right)\), \(\left(0.015,\frac{1}{12}\right)\)

(2)

(i)(i)

The circuit is changed so that a steady 3 volts is added to the supply voltage before it is rectified. The output is now \(W(t)=|V(t)+3|\) volts.

Find the greatest value of \(W(t)\), and the value of \(W(t)\) at its other local maximum points.

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greatest value \(12+3=15\)	A1
other local maxima \(\left\|-12+3\right\|=9\)	A1

Answer: 15 V; 9 V

(2)

(i)(ii)

State the period of \(W\), and explain why it is not the same as the period of \(|V|\).

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0.02 s	A1
the arches now alternate between heights 15 and 9 (the reflected negative half-cycles are lower than the positive ones), so the pattern repeats only after a full period of \(V\)	R1

Answer: 0.02 s

(2)

(i)(iii)

Solve the equation \(W(t)=6\) for \(0\le t\le0.02\).

ROUTINENotes

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\(V(t)+3=6\) or \(V(t)+3=-6\), i.e. \(\sin(100\pi t)=\frac14\) or \(\sin(100\pi t)=-\frac34\)	M1	Or graphs of \(y=\|12\sin(100\pi t)+3\|\) and \(y=6\) on the GDC.
\(t=0.000804,\ 0.00920\)	A1
\(t=0.0127,\ 0.0173\)	A1

Answer: \(t=0.000804,\ 0.00920,\ 0.0127,\ 0.0173\)

(3)

Show full solution

Working	Mark	Notes
Part (a)
amplitude 12 (volts)	A1
period \(\frac{2\pi}{100\pi}=0.02\) s	A1	Accept \(\frac{1}{50}\) s.
Part (b)
\(V(-t)=12\sin(-100\pi t)\)	M1	A general \(t\) is needed; numerical checks score M0.
\(=-12\sin(100\pi t)\), since \(\sin(-\theta)=-\sin\theta\)	A1
\(=-V(t)\), so \(V\) is odd	AG
Part (c)
parts below the \(t\)-axis reflected in the \(t\)-axis, parts above unchanged; nothing below the axis	A1
four equal arches of height 12, touching the axis with sharp corners at \(t=0,\ \pm0.01,\ \pm0.02\)	A1
Part (d)
period 0.01 s	A1
\(0\le\|V(t)\|\le12\)	A1	Accept \([0,\,12]\).
Part (e)(i)
\(\|V(-t)\|=\left\|-V(t)\right\|\)	M1	Using the result of (b).
\(=\|V(t)\|\), because \(\left\|-a\right\|=\|a\|\)	A1
so \(\|V\|\) is even	AG
Part (e)(ii)
for \(t\ge0\), \(\|V(\|t\|)\|=\|V(t)\|\); for \(t<0\), \(\|V(\|t\|)\|=\|V(-t)\|=\|V(t)\|\) by (e)(i)	R1	Accept: \(y=f(\|t\|)\) keeps the right half and reflects it in the \(y\)-axis, and the graph of the even function \(\|V\|\) is already symmetric in the \(y\)-axis.
Part (f)
\(\sin(100\pi t)=\frac12\) or \(\sin(100\pi t)=-\frac12\)	M1	Both signs considered.
\(100\pi t=\frac{\pi}{6},\ \frac{5\pi}{6}\)	A1
\(100\pi t=\frac{7\pi}{6},\ \frac{11\pi}{6}\)	A1
\(t=\frac{1}{600},\ \frac{1}{120},\ \frac{7}{600},\ \frac{11}{600}\)	A1	Extra values in the interval: A0 for this mark.
Part (g)
\(V(t)\ge0\) for \(0\le t\le0.01\), so \(\|V(t)\|=V(t)\)	R1
\(\int12\sin(100\pi t)\,\mathrm{d}t=-\frac{12}{100\pi}\cos(100\pi t)\)	M1	Integrating; the factor \(\frac{1}{100\pi}\) must be present.
\(-\frac{12}{100\pi}\left(\cos\pi-\cos0\right)\)	M1	Substituting both limits, upper minus lower.
\(=\frac{24}{100\pi}\)	A1
mean \(=100\times\frac{24}{100\pi}=\frac{24}{\pi}\)	AG
Part (h)
\(t=0.01\)	A1	Condone also stating \(t=0\) and \(t=0.02\).
\(\left(0.005,\ \frac{1}{12}\right)\) and \(\left(0.015,\ \frac{1}{12}\right)\)	A1	Both points needed.
Part (i)(i)
greatest value \(12+3=15\)	A1
other local maxima \(\left\|-12+3\right\|=9\)	A1
Part (i)(ii)
0.02 s	A1
the arches now alternate between heights 15 and 9 (the reflected negative half-cycles are lower than the positive ones), so the pattern repeats only after a full period of \(V\)	R1
Part (i)(iii)
\(V(t)+3=6\) or \(V(t)+3=-6\), i.e. \(\sin(100\pi t)=\frac14\) or \(\sin(100\pi t)=-\frac34\)	M1	Or graphs of \(y=\|12\sin(100\pi t)+3\|\) and \(y=6\) on the GDC.
\(t=0.000804,\ 0.00920\)	A1
\(t=0.0127,\ 0.0173\)	A1

Answers: (a) amplitude 12 V; period 0.02 s · (d) period 0.01 s; range \(0\le|V(t)|\le12\) · (f) \(t=\frac{1}{600},\ \frac{1}{120},\ \frac{7}{600},\ \frac{11}{600}\) · (h) \(t=0.01\); \(\left(0.005,\frac{1}{12}\right)\), \(\left(0.015,\frac{1}{12}\right)\) · (i)(i) 15 V; 9 V · (i)(ii) 0.02 s · (i)(iii) \(t=0.000804,\ 0.00920,\ 0.0127,\ 0.0173\)

Common error (f) Solving only \(\sin(100\pi t)=\frac12\) loses half the solutions: \(|V(t)|=6\) means \(V(t)=6\) or \(V(t)=-6\).

More 2.16 questions in the question bank

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Keeping a greenhouse in the safe temperature band

2.15 Solving inequalities g(x) ≥ f(x)2.12 Polynomial functions, factor and remainder theorems, sum and product of roots2.16 Modulus, reciprocal and other graph transformations2.14 Odd and even functions, inverse and self-inverse functionsHL

A grower models the air temperature inside a greenhouse on a spring day. Without any heating, the temperature \(t\) hours after midnight would be \(T(t)\) °C, where

\(T(t)=-0.02t^3+0.54t^2-3t+18,\quad 0\le t\le20.\)

Figure 1 shows the graph of \(y=T(t)\). An electric heater switches on whenever this temperature would be below 14 °C.

Figure 1: the graph of \(y=T(t)\) for \(0\le t\le20\) (the vertical axis starts at 8)

(a)

Show that the inequality \(T(t)\ge14\) is equivalent to \(p(t)\le0\), where \(p(t)=t^3-27t^2+150t-200\).

No calculatorPROOFNotes

Show solution

\(-0.02t^3+0.54t^2-3t+4\ge0\)	M1	Rearranging so that one side is zero.
multiplying by \(-50\) and reversing the inequality sign	A1	The reversal must be stated or clearly shown.
\(t^3-27t^2+150t-200\le0\)	AG

(2)

(b)

Use the factor theorem to factorise \(p(t)\) completely.

No calculatorALGEBRANotes

Show solution

\(p(2)=8-108+300-200=0\), so \((t-2)\) is a factor	M1	Or \(p(5)=0\) or \(p(20)=0\) shown.
\(p(t)=(t-2)\left(t^2-25t+100\right)\)	A1	Correct quadratic factor for their linear factor.
\(p(t)=(t-2)(t-5)(t-20)\)	A1

Answer: \(p(t)=(t-2)(t-5)(t-20)\)

(3)

(c)

Hence solve \(T(t)\ge14\) for \(0\le t\le20\), and state the times of day between which the heater is on.

No calculatorALGEBRANotes

Show solution

sign diagram or test values for \((t-2)(t-5)(t-20)\le0\), critical values 2, 5, 20	M1
\(0\le t\le2\) or \(5\le t\le20\)	A1
heater on for \(2\lt t\lt 5\), i.e. from 02:00 to 05:00	A1	Allow ECF from their solution.

Answer: \(0\le t\le2\) or \(5\le t\le20\); heater on 02:00–05:00

(3)

(d)(i)

To protect a more delicate crop, the grower changes the setting so that the heater switches on whenever the temperature would be below 18 °C.

Solve the inequality \(T(t)\ge18\) for \(0\le t\le20\), Give the complete solution set, with the boundary values in exact form.

No calculatorALGEBRANotes

Show solution

\(-0.02t^3+0.54t^2-3t\ge0\Rightarrow t\left(t^2-27t+150\right)\le0\)	M1	Factor \(t\) taken out; the sign reversal must be correct.
\(t=\frac{27\pm\sqrt{27^2-4(150)}}{2}=\frac{27\pm\sqrt{129}}{2}\)	A1	Quadratic formula or completing the square. Decimals only (7.82, 19.2): A0.
\(\frac{27-\sqrt{129}}{2}\le t\le\frac{27+\sqrt{129}}{2}\)	A1	From a sign diagram for \(t\left(t^2-27t+150\right)\): the product is \(\le0\) for \(t\le0\) and between the two roots. Strict inequalities: A0.
\(t=0\) also satisfies the inequality, since \(T(0)=18\)	A1	The isolated value \(t=0\) (the factor \(t\) is zero at the end of the domain) must be included in the final answer, e.g. \(t=0\) or \(\frac{27-\sqrt{129}}{2}\le t\le\frac{27+\sqrt{129}}{2}\). This mark is independent of the previous A1.

Answer: \(t=0\) or \(\frac{27-\sqrt{129}}{2}\le t\le\frac{27+\sqrt{129}}{2}\)

(4)

(d)(ii)

Find the total time for which the heater is on with this new setting.

MODELNotes

Show solution

\(\frac{27-\sqrt{129}}{2}+\left(20-\frac{27+\sqrt{129}}{2}\right)\)	M1	Or \(7.82+0.821\). Allow ECF from (d)(i).
\(20-\sqrt{129}=8.64\) hours	A1	Accept 8 hours 38 minutes.

Answer: \(20-\sqrt{129}\approx8.64\) hours

(2)

(e)(i)

The plants grow best when \(|T(t)-22|\le4\).

On the axes in Figure 2, sketch the graph of \(y=|T(t)-22|\) for \(6\le t\le18\), together with the line \(y=4\).

Figure 2: axes for part (e)(i)

GRAPHNotes

Show solution

starts at \((6,\,6.88)\), decreases to a sharp corner on the \(t\)-axis at \((10,\,0)\)	A1	No part of the graph below the \(t\)-axis.
rises to a local maximum at about \((14.6,\,5.06)\), then falls to \((18,\,0.32)\)	A1
line \(y=4\) drawn, meeting the curve three times (near \(t=7.8,\ 12.7,\ 16.3\))	A1

(3)

(e)(ii)

Hence, using your answer to (d)(i), solve \(|T(t)-22|\le4\) for \(6\le t\le18\).

MULTINotes

Show solution

\(\|T(t)-22\|\le4\iff18\le T(t)\le26\)	M1
\(T(t)=26\) at \(t=12.67\ldots\) and \(t=16.26\ldots\)	A1	From the GDC.
\(7.82\le t\le12.7\) or \(16.3\le t\le18\)	A1	Accept \(\frac{27-\sqrt{129}}{2}\) as the first boundary. Allow ECF from (d)(i).

Answer: \(7.82\le t\le12.7\) or \(16.3\le t\le18\)

(3)

(f)(i)

A sensor sends the temperature to the heater's controller as a voltage signal. A temperature of \(x\) °C, where \(10\le x\le40\), is sent as a signal of \(s\) volts, where \(s=f(x)\) and

\(f(x)=\dfrac{2x+20}{x-2},\quad x\in\mathbb{R},\ x\ne2.\)

Show that \(f(f(x))=x\) for all \(x\ne2\).

No calculatorPROOFNotes

Show solution

\(f(f(x))=\frac{2\left(\frac{2x+20}{x-2}\right)+20}{\frac{2x+20}{x-2}-2}\)	M1
numerator \(\times(x-2)\): \(2(2x+20)+20(x-2)=24x\)	A1
denominator \(\times(x-2)\): \((2x+20)-2(x-2)=24\)	A1
\(f(f(x))=\frac{24x}{24}=x\)	AG

(3)

(f)(ii)

The controller receives a signal of 3.6 volts. Find the temperature in the greenhouse, and explain why the controller can use the same formula to convert signals back into temperatures.

No calculatorINTERPRETNotes

Show solution

\(\frac{2(3.6)+20}{3.6-2}=\frac{27.2}{1.6}=17\) °C	A1
\(f(f(x))=x\) means \(f\) is self-inverse, so \(f^{-1}(s)=\frac{2s+20}{s-2}\): the formula that turns temperatures into signals also turns signals back into temperatures	R1

Answer: 17 °C

(2)

(f)(iii)

Write the condition for the plants' best growth, \(18\le T\le26\), as a condition on the signal \(s\) in the form \(|s-c|\le d\).

No calculatorMULTINotes

Show solution

\(f(18)=3.5\) and \(f(26)=3\), so \(3\le s\le3.5\) (\(f\) is decreasing)	M1
\(\|s-3.25\|\le0.25\)	A1

Answer: \(|s-3.25|\le0.25\)

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(-0.02t^3+0.54t^2-3t+4\ge0\)	M1	Rearranging so that one side is zero.
multiplying by \(-50\) and reversing the inequality sign	A1	The reversal must be stated or clearly shown.
\(t^3-27t^2+150t-200\le0\)	AG
Part (b)
\(p(2)=8-108+300-200=0\), so \((t-2)\) is a factor	M1	Or \(p(5)=0\) or \(p(20)=0\) shown.
\(p(t)=(t-2)\left(t^2-25t+100\right)\)	A1	Correct quadratic factor for their linear factor.
\(p(t)=(t-2)(t-5)(t-20)\)	A1
Part (c)
sign diagram or test values for \((t-2)(t-5)(t-20)\le0\), critical values 2, 5, 20	M1
\(0\le t\le2\) or \(5\le t\le20\)	A1
heater on for \(2\lt t\lt 5\), i.e. from 02:00 to 05:00	A1	Allow ECF from their solution.
Part (d)(i)
\(-0.02t^3+0.54t^2-3t\ge0\Rightarrow t\left(t^2-27t+150\right)\le0\)	M1	Factor \(t\) taken out; the sign reversal must be correct.
\(t=\frac{27\pm\sqrt{27^2-4(150)}}{2}=\frac{27\pm\sqrt{129}}{2}\)	A1	Quadratic formula or completing the square. Decimals only (7.82, 19.2): A0.
\(\frac{27-\sqrt{129}}{2}\le t\le\frac{27+\sqrt{129}}{2}\)	A1	From a sign diagram for \(t\left(t^2-27t+150\right)\): the product is \(\le0\) for \(t\le0\) and between the two roots. Strict inequalities: A0.
\(t=0\) also satisfies the inequality, since \(T(0)=18\)	A1	The isolated value \(t=0\) (the factor \(t\) is zero at the end of the domain) must be included in the final answer, e.g. \(t=0\) or \(\frac{27-\sqrt{129}}{2}\le t\le\frac{27+\sqrt{129}}{2}\). This mark is independent of the previous A1.
Part (d)(ii)
\(\frac{27-\sqrt{129}}{2}+\left(20-\frac{27+\sqrt{129}}{2}\right)\)	M1	Or \(7.82+0.821\). Allow ECF from (d)(i).
\(20-\sqrt{129}=8.64\) hours	A1	Accept 8 hours 38 minutes.
Part (e)(i)
starts at \((6,\,6.88)\), decreases to a sharp corner on the \(t\)-axis at \((10,\,0)\)	A1	No part of the graph below the \(t\)-axis.
rises to a local maximum at about \((14.6,\,5.06)\), then falls to \((18,\,0.32)\)	A1
line \(y=4\) drawn, meeting the curve three times (near \(t=7.8,\ 12.7,\ 16.3\))	A1
Part (e)(ii)
\(\|T(t)-22\|\le4\iff18\le T(t)\le26\)	M1
\(T(t)=26\) at \(t=12.67\ldots\) and \(t=16.26\ldots\)	A1	From the GDC.
\(7.82\le t\le12.7\) or \(16.3\le t\le18\)	A1	Accept \(\frac{27-\sqrt{129}}{2}\) as the first boundary. Allow ECF from (d)(i).
Part (f)(i)
\(f(f(x))=\frac{2\left(\frac{2x+20}{x-2}\right)+20}{\frac{2x+20}{x-2}-2}\)	M1
numerator \(\times(x-2)\): \(2(2x+20)+20(x-2)=24x\)	A1
denominator \(\times(x-2)\): \((2x+20)-2(x-2)=24\)	A1
\(f(f(x))=\frac{24x}{24}=x\)	AG
Part (f)(ii)
\(\frac{2(3.6)+20}{3.6-2}=\frac{27.2}{1.6}=17\) °C	A1
\(f(f(x))=x\) means \(f\) is self-inverse, so \(f^{-1}(s)=\frac{2s+20}{s-2}\): the formula that turns temperatures into signals also turns signals back into temperatures	R1
Part (f)(iii)
\(f(18)=3.5\) and \(f(26)=3\), so \(3\le s\le3.5\) (\(f\) is decreasing)	M1
\(\|s-3.25\|\le0.25\)	A1

Answers: (b) \(p(t)=(t-2)(t-5)(t-20)\) · (c) \(0\le t\le2\) or \(5\le t\le20\); heater on 02:00–05:00 · (d)(i) \(t=0\) or \(\frac{27-\sqrt{129}}{2}\le t\le\frac{27+\sqrt{129}}{2}\) · (d)(ii) \(20-\sqrt{129}\approx8.64\) hours · (e)(ii) \(7.82\le t\le12.7\) or \(16.3\le t\le18\) · (f)(ii) 17 °C · (f)(iii) \(|s-3.25|\le0.25\)

Common error (a) Multiplying an inequality by a negative number reverses it. Multiplying by \(-50\) turns \(\ge\) into \(\le\).

More 2.15 questions in the question bank

Topic 3 · Geometry and trigonometry

● home subchapter · ○ linked subchapter · last column: links to other topics

Setup	3.1	3.2	3.3	3.4	3.5	3.6	3.7	3.8	3.9	3.10	3.11	3.12	3.13	3.14	3.15	3.16	3.17	3.18	Other
Support cables in a glass atrium	●	○	○																1.1
A coastguard rescue: bearings and angles of depression	○	○	●		○														—
A car's windscreen wiper: arcs, sectors and cleaned area		○		●	○	○													—
A two-link robot arm reaching a target		○			○	●													2.10
When can a ship enter the harbour? Modelling the tide				○			●	○											2.11
A ride on a giant observation wheel						○	○	●											5.6
Noise-cancelling headphones: combining sound waves							○	○		●	○								—
The best seat in the cinema: maximising the viewing angle									●	○									5.15 5.8
Air traffic control: how close do two aircraft get?												○	○	●					5.8
Angling a solar panel towards the sun							○					○	●	○					—
Two tunnels under a city: do they meet?													○	○	●	○			—
Roof planes and a surveying laser														○		○	○	●	—

3.1 Three-dimensional geometry: distance, solids and angles · 3.2 Right-angled and non-right-angled triangles · 3.3 Applications of trigonometry: elevation, depression and bearingsmerged group

+ Add to test✓ In your testCC-AA-31-32-33-126 marks~30 min

Support cables in a glass atrium

3.1 Three-dimensional geometry: distance, solids and angles3.2 Right-angled and non-right-angled triangles3.3 Applications of trigonometry: elevation, depression and bearings1.1 Standard formSL

A glass atrium has the shape of a cuboid \(ABCDEFGH\). The horizontal floor \(ABCD\) is a rectangle with \(AB=24\) m and \(BC=18\) m. The vertical edges \([AE]\), \([BF]\), \([CG]\) and \([DH]\) are each 20 m long, and the roof \(EFGH\) is horizontal.

Coordinates are taken with \(A\) at the origin, the \(x\)-axis along \([AB]\), the \(y\)-axis along \([AD]\) and the \(z\)-axis vertically upwards, with units in metres. So \(B\) is \((24,0,0)\), \(D\) is \((0,18,0)\) and \(E\) is \((0,0,20)\).

A light fitting \(L\) hangs from four straight cables, one from each of the top corners \(E\), \(F\), \(G\) and \(H\). The fitting is vertically above the centre \(M\) of the floor, at a height of 12 m above the floor, as shown in Figure 1.

Figure 1: the atrium and the four cables (not to scale)

(a)

Find the exact length of the diagonal \([AG]\).

No calculatorROUTINENotes

Show solution

\(AG=\sqrt{24^2+18^2+20^2}\)	(M1)	Or \(AC=30\) then \(\sqrt{30^2+20^2}\).
\(\sqrt{1300}=10\sqrt{13}\) m	A1	Accept \(\sqrt{1300}\). A decimal answer (36.1) alone scores A0.

Answer: \(10\sqrt{13}\) m (\(\approx 36.1\) m)

(2)

(b)(i)

Write down the coordinates of \(M\) and of \(L\).

No calculatorROUTINENotes

Show solution

\(M(12,\,9,\,0)\) and \(L(12,\,9,\,12)\)

Both required.

Answer: \(M(12,9,0)\), \(L(12,9,12)\)

(1)

(b)(ii)

Show that each cable is 17 m long.

No calculatorPROOFNotes

Show solution

\(EL=\sqrt{(12-0)^2+(9-0)^2+(12-20)^2}\)	M1	FT their \(L\) only if it leads to 17. Any one of the four cables.
\(=\sqrt{144+81+64}=\sqrt{289}\)	A1
\(=17\) m, and by symmetry the four cables are equal	AG	The symmetry remark is not required for the marks.

Answer: 17 m

(2)

(c)

Find the angle between the cable \([EL]\) and the roof \(EFGH\).

ROUTINENotes

Show solution

\(N(12,9,20)\) above \(L\): \(EN=15\), \(LN=8\); \(\tan\theta=\frac{8}{15}\) or \(\sin\theta=\frac{8}{17}\) or \(\cos\theta=\frac{15}{17}\)	(M1)	Correct right-angled triangle with the angle at \(E\).
\(28.1^\circ\)	A1	\(28.0724\ldots\). \(61.9^\circ\) (angle with the vertical) scores M0A0.

Answer: \(28.1^\circ\)

(2)

(d)

Find \(\angle ELH\), the angle between the cables \([EL]\) and \([HL]\).

MULTINotes

Show solution

\(EH=18\)	(A1)
\(\cos\angle ELH=\frac{17^2+17^2-18^2}{2(17)(17)}\ \left(=\frac{127}{289}\right)\)	M1	Or \(\sin\frac{1}{2}\angle ELH=\frac{9}{17}\) in the isosceles triangle.
\(63.9^\circ\)	A1	\(63.9314\ldots\)

Answer: \(63.9^\circ\)

(3)

(e)(i)

Find the volume of the atrium. Give your answer in the form \(a\times10^{k}\), where \(1\le a<10\) and \(k\in\mathbb{Z}\).

ROUTINENotes

Show solution

\(24\times18\times20=8640\); \(8.64\times10^{3}\text{ m}^3\)

\(86.4\times10^{2}\) or 8640 scores A0.

Answer: \(8.64\times10^{3}\text{ m}^3\)

(1)

(e)(ii)

The roof \(EFGH\) is a single sheet of glass of uniform thickness. The sheet has a mass of \(1.30\times10^{4}\) kg, and \(1\text{ m}^3\) of this glass has a mass of \(2.50\times10^{3}\) kg.

Find the thickness of the glass, in millimetres.

MULTINotes

Show solution

volume of glass \(=\frac{1.30\times10^{4}}{2.50\times10^{3}}=5.2\ \text{m}^3\)	(M1)
thickness \(=\frac{5.2}{24\times18}=0.01203\ldots\) m	M1	Dividing their volume by the roof area \(432\text{ m}^2\).
\(12.0\) mm	A1	Accept 12 mm. An answer left in metres scores A0.

Answer: 12.0 mm

(3)

(f)(i)

The height of the fitting can be adjusted. When the fitting is \(h\) metres above the floor, still vertically above \(M\), each cable has length \(l\) metres.

Show that \(l=\sqrt{225+(20-h)^2}\), for \(0\le h\le20\).

No calculatorPROOFNotes

Show solution

\(L(12,\,9,\,h)\): \(l^2=12^2+9^2+(20-h)^2\)	M1	Or horizontal distance \(\frac12 AC=15\) and vertical distance \(20-h\).
\(12^2+9^2=225\)	A1
\(l=\sqrt{225+(20-h)^2}\)	AG

(2)

(f)(ii)

The graph of \(l\) against \(h\) is shown in Figure 2.

Use Figure 2 to estimate the height of the fitting when each cable is 19 m long.

Figure 2: cable length \(l\) against the height \(h\) of the fitting

GRAPHNotes

Show solution

horizontal line at \(l=19\) meeting the curve, or \(225+(20-h)^2=361\)	(M1)	Evidence of reading from the graph, or solving the equation.
\(h\approx8.3\) m	A1	Accept 8.2 to 8.5. The exact value is \(20-\sqrt{136}=8.34\ldots\)

Answer: \(h\approx8.3\) m

(2)

(f)(iii)

Explain, with reference to the shape of the atrium, why \(l\ge15\) for all values of \(h\), and state the value of \(h\) for which \(l=15\).

INTERPRETNotes

Show solution

each top corner is always 15 m horizontally from the vertical line through \(M\) (half the floor diagonal), and the cable is the hypotenuse of a right-angled triangle with this side, so \(l\ge15\)	R1	Or \((20-h)^2\ge0\) so \(l\ge\sqrt{225}=15\), linked to the 15 m horizontal distance.
\(l=15\) when \(h=20\) (the fitting at roof level)	A1

Answer: \(h=20\)

(2)

(g)

For safety, each cable must make an angle of at least \(30^\circ\) with the horizontal.

Determine whether the design with the fitting 12 m above the floor meets this requirement, and find the greatest height at which the fitting can hang so that it does.

MODELNotes

Show solution

\(28.1^\circ<30^\circ\), so the design does not meet the requirement	R1	FT from (c).
\(\tan30^\circ=\frac{20-h}{15}\)	M1	Or \(20-h=15\tan30^\circ=8.66\ldots\)
\(h=20-5\sqrt3=11.3\) m	A1	\(11.339\ldots\). Accept \(h\le11.3\).

Answer: No; greatest height \(11.3\) m

(3)

(h)

The design is changed. A glass skylight in the shape of a right pyramid is built on the roof. Its base is the rectangle \(EFGH\) and its apex \(V\) is vertically above \(M\).

A visitor standing at the corner \(A\) of the floor looks up at \(V\). His eye level is 1.5 m above the floor, and the line of sight to \(V\) makes an angle of elevation of \(55^\circ\).

Find the height of the skylight.

MULTINotes

Show solution

horizontal distance from \(A\) to the point below \(V\) \(=AM=\frac{30}{2}=15\)	(A1)	FT from their \(AC\).
\(15\tan55^\circ\ (=21.42\ldots)\)	M1
\(21.42+1.5-20=2.92\) m	A1	\(2.922\ldots\). 21.4 or 22.9 scores A0.

Answer: \(2.92\) m

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(AG=\sqrt{24^2+18^2+20^2}\)	(M1)	Or \(AC=30\) then \(\sqrt{30^2+20^2}\).
\(\sqrt{1300}=10\sqrt{13}\) m	A1	Accept \(\sqrt{1300}\). A decimal answer (36.1) alone scores A0.
Part (b)(i)
\(M(12,\,9,\,0)\) and \(L(12,\,9,\,12)\)	A1	Both required.
Part (b)(ii)
\(EL=\sqrt{(12-0)^2+(9-0)^2+(12-20)^2}\)	M1	FT their \(L\) only if it leads to 17. Any one of the four cables.
\(=\sqrt{144+81+64}=\sqrt{289}\)	A1
\(=17\) m, and by symmetry the four cables are equal	AG	The symmetry remark is not required for the marks.
Part (c)
\(N(12,9,20)\) above \(L\): \(EN=15\), \(LN=8\); \(\tan\theta=\frac{8}{15}\) or \(\sin\theta=\frac{8}{17}\) or \(\cos\theta=\frac{15}{17}\)	(M1)	Correct right-angled triangle with the angle at \(E\).
\(28.1^\circ\)	A1	\(28.0724\ldots\). \(61.9^\circ\) (angle with the vertical) scores M0A0.
Part (d)
\(EH=18\)	(A1)
\(\cos\angle ELH=\frac{17^2+17^2-18^2}{2(17)(17)}\ \left(=\frac{127}{289}\right)\)	M1	Or \(\sin\frac{1}{2}\angle ELH=\frac{9}{17}\) in the isosceles triangle.
\(63.9^\circ\)	A1	\(63.9314\ldots\)
Part (e)(i)
\(24\times18\times20=8640\); \(8.64\times10^{3}\text{ m}^3\)	A1	\(86.4\times10^{2}\) or 8640 scores A0.
Part (e)(ii)
volume of glass \(=\frac{1.30\times10^{4}}{2.50\times10^{3}}=5.2\ \text{m}^3\)	(M1)
thickness \(=\frac{5.2}{24\times18}=0.01203\ldots\) m	M1	Dividing their volume by the roof area \(432\text{ m}^2\).
\(12.0\) mm	A1	Accept 12 mm. An answer left in metres scores A0.
Part (f)(i)
\(L(12,\,9,\,h)\): \(l^2=12^2+9^2+(20-h)^2\)	M1	Or horizontal distance \(\frac12 AC=15\) and vertical distance \(20-h\).
\(12^2+9^2=225\)	A1
\(l=\sqrt{225+(20-h)^2}\)	AG
Part (f)(ii)
horizontal line at \(l=19\) meeting the curve, or \(225+(20-h)^2=361\)	(M1)	Evidence of reading from the graph, or solving the equation.
\(h\approx8.3\) m	A1	Accept 8.2 to 8.5. The exact value is \(20-\sqrt{136}=8.34\ldots\)
Part (f)(iii)
each top corner is always 15 m horizontally from the vertical line through \(M\) (half the floor diagonal), and the cable is the hypotenuse of a right-angled triangle with this side, so \(l\ge15\)	R1	Or \((20-h)^2\ge0\) so \(l\ge\sqrt{225}=15\), linked to the 15 m horizontal distance.
\(l=15\) when \(h=20\) (the fitting at roof level)	A1
Part (g)
\(28.1^\circ<30^\circ\), so the design does not meet the requirement	R1	FT from (c).
\(\tan30^\circ=\frac{20-h}{15}\)	M1	Or \(20-h=15\tan30^\circ=8.66\ldots\)
\(h=20-5\sqrt3=11.3\) m	A1	\(11.339\ldots\). Accept \(h\le11.3\).
Part (h)
horizontal distance from \(A\) to the point below \(V\) \(=AM=\frac{30}{2}=15\)	(A1)	FT from their \(AC\).
\(15\tan55^\circ\ (=21.42\ldots)\)	M1
\(21.42+1.5-20=2.92\) m	A1	\(2.922\ldots\). 21.4 or 22.9 scores A0.

Answers: (a) \(10\sqrt{13}\) m (\(\approx 36.1\) m) · (b)(i) \(M(12,9,0)\), \(L(12,9,12)\) · (b)(ii) 17 m · (c) \(28.1^\circ\) · (d) \(63.9^\circ\) · (e)(i) \(8.64\times10^{3}\text{ m}^3\) · (e)(ii) 12.0 mm · (f)(ii) \(h\approx8.3\) m · (f)(iii) \(h=20\) · (g) No; greatest height \(11.3\) m · (h) \(2.92\) m

Common error (c) The angle between a line and a plane is the angle between the line and its shadow on the plane. Here the shadow of \([EL]\) on the roof runs from \(E\) to the point directly above \(L\), so the right angle is at that point, not at \(E\).

Exam tip (h) An angle of elevation is measured from the horizontal at eye level. Add the eye height to the height found with trigonometry, then subtract the height of the roof to get the height of the skylight itself.

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A coastguard rescue: bearings and angles of depression

3.3 Applications of trigonometry: elevation, depression and bearings3.2 Right-angled and non-right-angled triangles3.5 The unit circle, exact values and the ambiguous case3.1 Three-dimensional geometry: distance, solids and anglesSL

A coastguard station \(S\) is at the top of a vertical cliff, 85 m above sea level. The point \(O\) is at sea level, vertically below \(S\). The sea is modelled as a horizontal plane, and all boats are treated as points on it.

(a)

From \(S\), the angle of depression of a yacht \(Y\) is \(4^\circ\).

Find the horizontal distance \(OY\).

ROUTINENotes

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\(\tan4^\circ=\frac{85}{OY}\)	(M1)	Angle of elevation of \(S\) from \(Y\) is \(4^\circ\).
\(OY=1220\) m	A1	\(1215.56\ldots\) Accept 1.22 km.

Answer: \(1220\) m (\(1215.6\) m)

(2)

(b)

A dinghy \(D\) is seen from \(S\) at an angle of depression of \(30^\circ\). The yacht is on a bearing of \(130^\circ\) from \(O\) and the dinghy is on a bearing of \(230^\circ\) from \(O\), as shown in Figure 1.

Find \(\angle YSD\), the angle between the lines of sight from \(S\) to the yacht and to the dinghy.

Figure 1: the lines of sight from \(S\) to the yacht \(Y\) and the dinghy \(D\) (not to scale)

MULTINotes

Show solution

\(SY=\frac{85}{\sin4^\circ}=1218.5\ldots\), \(SD=\frac{85}{\sin30^\circ}=170\), \(OD=\frac{85}{\tan30^\circ}=147.2\ldots\)	(M1)	Attempt at the lengths needed; Pythagoras is fine. FT from (a).
\(\angle YOD=100^\circ\); \(YD^2=1215.6^2+147.2^2-2(1215.6)(147.2)\cos100^\circ\), \(YD=1249.6\ldots\)	M1
\(\cos\angle YSD=\frac{1218.5^2+170^2-1249.6^2}{2(1218.5)(170)}\)	M1	Cosine rule in triangle \(YSD\) with their three lengths.
\(96.6^\circ\)	A1	\(96.611\ldots\). Giving \(100^\circ\) (the horizontal angle) scores M1M1M0A0 at most.

Answer: \(96.6^\circ\)

(4)

(c)(i)

A lifeboat leaves the harbour \(H\) and sails 6.4 km on a bearing of \(062^\circ\) to a point \(P\). It then sails 3.8 km on a bearing of \(145^\circ\) to a point \(Q\), as shown in Figure 2.

Show that \(\angle HPQ=97^\circ\).

Figure 2: the route of the lifeboat (not to scale)

No calculatorPROOFNotes

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bearing of \(H\) from \(P\) is \(062^\circ+180^\circ=242^\circ\)	M1	Or co-interior angles: \(180^\circ-62^\circ=118^\circ\) between \(PH\) and north at \(P\), measured anticlockwise.
\(\angle HPQ=242^\circ-145^\circ\)	A1	Or \(62^\circ+(180^\circ-145^\circ)\) using the south line at \(P\).
\(=97^\circ\)	AG

Answer: \(97^\circ\)

(2)

(c)(ii)

Find the distance \(HQ\).

ROUTINENotes

Show solution

attempt to use the cosine rule	M1
\(HQ^2=6.4^2+3.8^2-2(6.4)(3.8)\cos97^\circ\ (=61.33\ldots)\)	A1
\(HQ=7.83\) km	A1	\(7.8312\ldots\)

Answer: \(7.83\) km

(3)

(c)(iii)

Find the bearing of \(Q\) from \(H\).

MULTINotes

Show solution

\(\frac{\sin\angle PHQ}{3.8}=\frac{\sin97^\circ}{7.831}\)	M1	FT from (c)(ii). Or the cosine rule for \(\angle PHQ\).
\(\angle PHQ=28.8^\circ\)	A1	\(28.79\ldots\)
bearing \(=062^\circ+28.8^\circ=091^\circ\)	A1	Accept \(090.8^\circ\). FT from their angle.

Answer: \(091^\circ\)

(3)

(d)(i)

The lifeboat travels at a constant speed for the whole journey. Figure 3 shows the graph of its distance \(d\) km from \(H\) against the time \(t\) minutes after it leaves \(H\).

Use Figure 3 to find the speed of the lifeboat, in km per hour.

Figure 3: the distance \(d\) of the lifeboat from \(H\) against time \(t\)

GRAPHNotes

Show solution

the graph changes at \(t=16\), \(d=6.4\): \(6.4\) km in 16 minutes	(M1)	Or any correct reading on the straight section, e.g. 4 km in 10 minutes.
\(24\) km per hour	A1

Answer: 24 km per hour

(2)

(d)(ii)

Explain why the graph is a straight line for \(0\le t\le16\) but is not a straight line for \(16\le t\le25.5\).

INTERPRETNotes

Show solution

for \(0\le t\le16\) the lifeboat sails directly away from \(H\), so \(d\) increases at the constant speed (\(d=0.4t\))	R1
after turning at \(P\) it no longer moves along the line through \(H\), so \(d\) increases by less than the distance travelled and the rate of change of \(d\) varies (\(d\) comes from the cosine rule)	R1	Accept: the gradient is smaller after the turn because the direction of travel is at an angle to \(HQ\).

(2)

(e)(i)

A fishing boat \(F\) sends a distress call. A radio fix shows that \(F\) is on a bearing of \(100^\circ\) from \(H\) and is 4.6 km from \(P\).

Find the two possible values of \(\angle HFP\).

MULTINotes

Show solution

\(\angle PHF=100^\circ-62^\circ=38^\circ\)	(A1)
\(\frac{\sin\angle HFP}{6.4}=\frac{\sin38^\circ}{4.6}\ (=0.8565\ldots)\)	M1
\(58.9^\circ\) or \(121.1^\circ\)	A1	Both required. \(58.93\ldots\), \(121.07\ldots\)

Answer: \(58.9^\circ\) or \(121.1^\circ\)

(3)

(e)(ii)

Find the two possible distances \(HF\).

MULTINotes

Show solution

\(\angle HPF=83.1^\circ\) or \(20.9^\circ\)	(A1)	FT from (e)(i).
\(HF=\frac{4.6\sin83.07^\circ}{\sin38^\circ}\) or \(\frac{4.6\sin20.93^\circ}{\sin38^\circ}\)	M1	Or \(4.6^2=6.4^2+x^2-2(6.4)x\cos38^\circ\) solved for \(x\).
\(7.42\) km or \(2.67\) km	A1	Both required. \(7.4169\ldots\), \(2.6695\ldots\)

Answer: \(7.42\) km or \(2.67\) km

(3)

(e)(iii)

A second radio fix shows that \(F\) is still on a bearing of \(100^\circ\) from \(H\), but is now 7.2 km from \(P\).

Explain why there is now only one possible position for \(F\).

INTERPRETNotes

Show solution

\(\sin\angle HFP=\frac{6.4\sin38^\circ}{7.2}\Rightarrow\angle HFP=33.2^\circ\) or \(146.8^\circ\)	M1	Or: \(PF=7.2>HP=6.4\), so \(\angle HFP<\angle PHF=38^\circ\).
\(146.8^\circ+38^\circ>180^\circ\), so the obtuse angle is impossible and only one triangle exists	R1	Do not award R1 without a reason.

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(\tan4^\circ=\frac{85}{OY}\)	(M1)	Angle of elevation of \(S\) from \(Y\) is \(4^\circ\).
\(OY=1220\) m	A1	\(1215.56\ldots\) Accept 1.22 km.
Part (b)
\(SY=\frac{85}{\sin4^\circ}=1218.5\ldots\), \(SD=\frac{85}{\sin30^\circ}=170\), \(OD=\frac{85}{\tan30^\circ}=147.2\ldots\)	(M1)	Attempt at the lengths needed; Pythagoras is fine. FT from (a).
\(\angle YOD=100^\circ\); \(YD^2=1215.6^2+147.2^2-2(1215.6)(147.2)\cos100^\circ\), \(YD=1249.6\ldots\)	M1
\(\cos\angle YSD=\frac{1218.5^2+170^2-1249.6^2}{2(1218.5)(170)}\)	M1	Cosine rule in triangle \(YSD\) with their three lengths.
\(96.6^\circ\)	A1	\(96.611\ldots\). Giving \(100^\circ\) (the horizontal angle) scores M1M1M0A0 at most.
Part (c)(i)
bearing of \(H\) from \(P\) is \(062^\circ+180^\circ=242^\circ\)	M1	Or co-interior angles: \(180^\circ-62^\circ=118^\circ\) between \(PH\) and north at \(P\), measured anticlockwise.
\(\angle HPQ=242^\circ-145^\circ\)	A1	Or \(62^\circ+(180^\circ-145^\circ)\) using the south line at \(P\).
\(=97^\circ\)	AG
Part (c)(ii)
attempt to use the cosine rule	M1
\(HQ^2=6.4^2+3.8^2-2(6.4)(3.8)\cos97^\circ\ (=61.33\ldots)\)	A1
\(HQ=7.83\) km	A1	\(7.8312\ldots\)
Part (c)(iii)
\(\frac{\sin\angle PHQ}{3.8}=\frac{\sin97^\circ}{7.831}\)	M1	FT from (c)(ii). Or the cosine rule for \(\angle PHQ\).
\(\angle PHQ=28.8^\circ\)	A1	\(28.79\ldots\)
bearing \(=062^\circ+28.8^\circ=091^\circ\)	A1	Accept \(090.8^\circ\). FT from their angle.
Part (d)(i)
the graph changes at \(t=16\), \(d=6.4\): \(6.4\) km in 16 minutes	(M1)	Or any correct reading on the straight section, e.g. 4 km in 10 minutes.
\(24\) km per hour	A1
Part (d)(ii)
for \(0\le t\le16\) the lifeboat sails directly away from \(H\), so \(d\) increases at the constant speed (\(d=0.4t\))	R1
after turning at \(P\) it no longer moves along the line through \(H\), so \(d\) increases by less than the distance travelled and the rate of change of \(d\) varies (\(d\) comes from the cosine rule)	R1	Accept: the gradient is smaller after the turn because the direction of travel is at an angle to \(HQ\).
Part (e)(i)
\(\angle PHF=100^\circ-62^\circ=38^\circ\)	(A1)
\(\frac{\sin\angle HFP}{6.4}=\frac{\sin38^\circ}{4.6}\ (=0.8565\ldots)\)	M1
\(58.9^\circ\) or \(121.1^\circ\)	A1	Both required. \(58.93\ldots\), \(121.07\ldots\)
Part (e)(ii)
\(\angle HPF=83.1^\circ\) or \(20.9^\circ\)	(A1)	FT from (e)(i).
\(HF=\frac{4.6\sin83.07^\circ}{\sin38^\circ}\) or \(\frac{4.6\sin20.93^\circ}{\sin38^\circ}\)	M1	Or \(4.6^2=6.4^2+x^2-2(6.4)x\cos38^\circ\) solved for \(x\).
\(7.42\) km or \(2.67\) km	A1	Both required. \(7.4169\ldots\), \(2.6695\ldots\)
Part (e)(iii)
\(\sin\angle HFP=\frac{6.4\sin38^\circ}{7.2}\Rightarrow\angle HFP=33.2^\circ\) or \(146.8^\circ\)	M1	Or: \(PF=7.2>HP=6.4\), so \(\angle HFP<\angle PHF=38^\circ\).
\(146.8^\circ+38^\circ>180^\circ\), so the obtuse angle is impossible and only one triangle exists	R1	Do not award R1 without a reason.

Answers: (a) \(1220\) m (\(1215.6\) m) · (b) \(96.6^\circ\) · (c)(i) \(97^\circ\) · (c)(ii) \(7.83\) km · (c)(iii) \(091^\circ\) · (d)(i) 24 km per hour · (e)(i) \(58.9^\circ\) or \(121.1^\circ\) · (e)(ii) \(7.42\) km or \(2.67\) km

Common error (b) The angle between the two lines of sight is not the angle \(\angle YOD\) on the sea. The lines of sight slope downwards from \(S\), so find the three sides of triangle \(YSD\) and use the cosine rule.

More 3.3 questions in the question bank

3.4 Radians, arc length and sector area · 3.5 The unit circle, exact values and the ambiguous case · 3.6 Pythagorean and double angle identitiesmerged group

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A car's windscreen wiper: arcs, sectors and cleaned area

3.4 Radians, arc length and sector area3.5 The unit circle, exact values and the ambiguous case3.6 Pythagorean and double angle identities3.2 Right-angled and non-right-angled trianglesSL

A windscreen wiper consists of a straight arm that turns about a fixed pivot \(O\). The rubber blade is fixed along the arm and covers the part of the arm from 15 cm to 55 cm from \(O\). In one sweep the arm turns through an angle of \(\dfrac{2\pi}{3}\) radians, and the tip of the blade moves from \(A\) to \(B\). The blade cleans the shaded region shown in Figure 1.

Figure 1: one sweep of the wiper; the blade cleans the shaded region (not to scale)

(a)

Write down the angle of the sweep in degrees.

No calculatorROUTINENotes

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\(120^\circ\)

Answer: \(120^\circ\)

(1)

(b)

Find the exact length of the arc traced by the tip of the blade in one sweep.

No calculatorROUTINENotes

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\(l=55\times\frac{2\pi}{3}\)	(M1)
\(\frac{110\pi}{3}\) cm	A1	Accept \(36\frac{2}{3}\pi\). 115 cm alone scores A0 (exact answer required).

Answer: \(\dfrac{110\pi}{3}\) cm

(2)

(c)

Show that the area cleaned in one sweep is \(\dfrac{2800\pi}{3}\ \text{cm}^2\).

No calculatorPROOFNotes

Show solution

use of \(\frac12r^2\theta\) with \(r=55\) or \(r=15\) and \(\theta=\frac{2\pi}{3}\)	M1
\(\frac12(55)^2\left(\frac{2\pi}{3}\right)=\frac{3025\pi}{3}\) and \(\frac12(15)^2\left(\frac{2\pi}{3}\right)=75\pi\)	A1
\(\frac{3025\pi}{3}-\frac{225\pi}{3}\)	A1	Or \(\frac12(55^2-15^2)\left(\frac{2\pi}{3}\right)=\frac12(2800)\left(\frac{2\pi}{3}\right)\).
\(=\frac{2800\pi}{3}\)	AG

(3)

(d)

Find the exact length of the chord \([AB]\).

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Show solution

\(AB^2=55^2+55^2-2(55)(55)\cos\frac{2\pi}{3}\)	M1	Or \(AB=2\times55\sin\frac{\pi}{3}\).
\(\cos\frac{2\pi}{3}=-\frac12\)	A1	Or \(\sin\frac{\pi}{3}=\frac{\sqrt3}{2}\).
\(AB^2=3(55^2)\), \(AB=55\sqrt3\) cm	A1

Answer: \(55\sqrt3\) cm

(3)

(e)

Find the area of the region between the chord \([AB]\) and the arc traced by the tip.

ROUTINENotes

Show solution

\(\frac12(55)^2\left(\frac{2\pi}{3}-\sin\frac{2\pi}{3}\right)\)	(M1)	Or sector \(\frac{3025\pi}{3}\) minus triangle \(\frac{3025\sqrt3}{4}\).
\(1860\ \text{cm}^2\)	A1	\(1857.9\ldots\). Accept \(\frac{3025\pi}{3}-\frac{3025\sqrt3}{4}\).

Answer: \(1860\ \text{cm}^2\)

(2)

(f)(i)

The wiper can be set to sweep through any angle \(\theta\) radians, where \(0\le\theta\le\pi\). The area cleaned in one sweep is \(A\ \text{cm}^2\).

On the axes in Figure 2, sketch the graph of \(A\) against \(\theta\).

Figure 2: axes for the graph of \(A\) against \(\theta\)

GRAPHNotes

Show solution

straight line through the origin	A1	Award for \(A=1400\theta\) drawn as a line from \((0,0)\).
ending at \((\pi,\,1400\pi)\approx(\pi,\,4400)\), or passing through \(\left(\frac{2\pi}{3},\,2930\right)\)	A1	Accept the end point within one small square.

Answer: \(A=1400\theta\), a straight line from \((0,0)\) to \((\pi,4400)\)

(2)

(f)(ii)

Explain why the graph is a straight line, and state what its gradient represents.

INTERPRETNotes

Show solution

\(A=\frac12(55^2-15^2)\theta=1400\theta\): the radii are fixed, so \(A\) is directly proportional to \(\theta\)	R1
the gradient, \(1400\), is the area cleaned (in \(\text{cm}^2\)) for each radian of sweep	R1

Answer: gradient 1400 cm² per radian

(2)

(g)

To clean a larger area, a designer considers two changes to the original wiper.

Option 1: increase the sweep angle to \(\dfrac{3\pi}{4}\), with the same blade.
Option 2: keep the sweep angle \(\dfrac{2\pi}{3}\), but use a longer arm and blade, so that the blade covers the part of the arm from 15 cm to 59 cm from \(O\).

Determine which option cleans the larger area in one sweep.

MODELNotes

Show solution

Option 1: \(1400\times\frac{3\pi}{4}=1050\pi\ (=3299)\)	M1	Or using the graph in (f)(i) for option 1.
Option 2: \(\frac12(59^2-15^2)\left(\frac{2\pi}{3}\right)=\frac{3256\pi}{3}\ (=3410)\)	A1	Both areas correct.
Option 2 cleans the larger area	R1	Must follow from two correctly compared areas. Accept a comparison of factors: \(\frac{9}{8}\) against \(\frac{3256}{2800}=1.16\).

Answer: Option 2 (3410 cm² against 3300 cm²)

(3)

(h)(i)

The rear window of the car has a different wiper. Its arm \([OT]\) has length 40 cm and sweeps through an angle \(\varphi\) radians, symmetrically about the vertical line through \(O\). At each end of the sweep, \(T_1\) and \(T_2\), the tip is 24 cm above the horizontal line through \(O\), as shown in Figure 3.

Show that \(\cos\varphi=-\dfrac{7}{25}\).

Figure 3: the rear wiper at the two ends of its sweep (not to scale)

No calculatorPROOFNotes

Show solution

\(\cos\frac{\varphi}{2}=\frac{24}{40}=\frac35\)	(A1)	Angle between \([OT_2]\) and the vertical is \(\frac{\varphi}{2}\).
\(\cos\varphi=2\cos^2\frac{\varphi}{2}-1=2\left(\frac{9}{25}\right)-1\)	M1	Correct double angle identity with their value substituted. Or the cosine rule in triangle \(OT_1T_2\) with \(T_1T_2=2\times32=64\): \(\cos\varphi=\frac{40^2+40^2-64^2}{2(40)(40)}=-\frac{896}{3200}\).
\(=-\frac{7}{25}\)	AG

(2)

(h)(ii)

Find the exact value of \(\sin\varphi\).

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\(\sin^2\varphi=1-\left(-\frac{7}{25}\right)^2=\frac{576}{625}\)	M1	Or \(\sin\varphi=2\sin\frac{\varphi}{2}\cos\frac{\varphi}{2}=2\left(\frac45\right)\left(\frac35\right)\).
\(\sin\varphi=\frac{24}{25}\) (positive, as \(0<\varphi<\pi\))	A1	\(\pm\frac{24}{25}\) scores A0.

Answer: \(\dfrac{24}{25}\)

(2)

(h)(iii)

Hence find the exact area of triangle \(OT_1T_2\).

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\(\frac12(40)(40)\left(\frac{24}{25}\right)\)	M1	FT from (h)(ii). Or \(\frac12\times64\times24\).
\(768\ \text{cm}^2\)	A1

Answer: \(768\ \text{cm}^2\)

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(120^\circ\)	A1
Part (b)
\(l=55\times\frac{2\pi}{3}\)	(M1)
\(\frac{110\pi}{3}\) cm	A1	Accept \(36\frac{2}{3}\pi\). 115 cm alone scores A0 (exact answer required).
Part (c)
use of \(\frac12r^2\theta\) with \(r=55\) or \(r=15\) and \(\theta=\frac{2\pi}{3}\)	M1
\(\frac12(55)^2\left(\frac{2\pi}{3}\right)=\frac{3025\pi}{3}\) and \(\frac12(15)^2\left(\frac{2\pi}{3}\right)=75\pi\)	A1
\(\frac{3025\pi}{3}-\frac{225\pi}{3}\)	A1	Or \(\frac12(55^2-15^2)\left(\frac{2\pi}{3}\right)=\frac12(2800)\left(\frac{2\pi}{3}\right)\).
\(=\frac{2800\pi}{3}\)	AG
Part (d)
\(AB^2=55^2+55^2-2(55)(55)\cos\frac{2\pi}{3}\)	M1	Or \(AB=2\times55\sin\frac{\pi}{3}\).
\(\cos\frac{2\pi}{3}=-\frac12\)	A1	Or \(\sin\frac{\pi}{3}=\frac{\sqrt3}{2}\).
\(AB^2=3(55^2)\), \(AB=55\sqrt3\) cm	A1
Part (e)
\(\frac12(55)^2\left(\frac{2\pi}{3}-\sin\frac{2\pi}{3}\right)\)	(M1)	Or sector \(\frac{3025\pi}{3}\) minus triangle \(\frac{3025\sqrt3}{4}\).
\(1860\ \text{cm}^2\)	A1	\(1857.9\ldots\). Accept \(\frac{3025\pi}{3}-\frac{3025\sqrt3}{4}\).
Part (f)(i)
straight line through the origin	A1	Award for \(A=1400\theta\) drawn as a line from \((0,0)\).
ending at \((\pi,\,1400\pi)\approx(\pi,\,4400)\), or passing through \(\left(\frac{2\pi}{3},\,2930\right)\)	A1	Accept the end point within one small square.
Part (f)(ii)
\(A=\frac12(55^2-15^2)\theta=1400\theta\): the radii are fixed, so \(A\) is directly proportional to \(\theta\)	R1
the gradient, \(1400\), is the area cleaned (in \(\text{cm}^2\)) for each radian of sweep	R1
Part (g)
Option 1: \(1400\times\frac{3\pi}{4}=1050\pi\ (=3299)\)	M1	Or using the graph in (f)(i) for option 1.
Option 2: \(\frac12(59^2-15^2)\left(\frac{2\pi}{3}\right)=\frac{3256\pi}{3}\ (=3410)\)	A1	Both areas correct.
Option 2 cleans the larger area	R1	Must follow from two correctly compared areas. Accept a comparison of factors: \(\frac{9}{8}\) against \(\frac{3256}{2800}=1.16\).
Part (h)(i)
\(\cos\frac{\varphi}{2}=\frac{24}{40}=\frac35\)	(A1)	Angle between \([OT_2]\) and the vertical is \(\frac{\varphi}{2}\).
\(\cos\varphi=2\cos^2\frac{\varphi}{2}-1=2\left(\frac{9}{25}\right)-1\)	M1	Correct double angle identity with their value substituted. Or the cosine rule in triangle \(OT_1T_2\) with \(T_1T_2=2\times32=64\): \(\cos\varphi=\frac{40^2+40^2-64^2}{2(40)(40)}=-\frac{896}{3200}\).
\(=-\frac{7}{25}\)	AG
Part (h)(ii)
\(\sin^2\varphi=1-\left(-\frac{7}{25}\right)^2=\frac{576}{625}\)	M1	Or \(\sin\varphi=2\sin\frac{\varphi}{2}\cos\frac{\varphi}{2}=2\left(\frac45\right)\left(\frac35\right)\).
\(\sin\varphi=\frac{24}{25}\) (positive, as \(0<\varphi<\pi\))	A1	\(\pm\frac{24}{25}\) scores A0.
Part (h)(iii)
\(\frac12(40)(40)\left(\frac{24}{25}\right)\)	M1	FT from (h)(ii). Or \(\frac12\times64\times24\).
\(768\ \text{cm}^2\)	A1

Answers: (a) \(120^\circ\) · (b) \(\dfrac{110\pi}{3}\) cm · (d) \(55\sqrt3\) cm · (e) \(1860\ \text{cm}^2\) · (f)(i) \(A=1400\theta\), a straight line from \((0,0)\) to \((\pi,4400)\) · (f)(ii) gradient 1400 cm² per radian · (g) Option 2 (3410 cm² against 3300 cm²) · (h)(ii) \(\dfrac{24}{25}\) · (h)(iii) \(768\ \text{cm}^2\)

Common error (d) \(\cos\frac{2\pi}{3}\) is negative because \(\frac{2\pi}{3}\) is in the second quadrant. Using \(+\frac12\) gives \(AB=55\), which is wrong: the chord must be longer than the radius when the angle is more than \(\frac{\pi}{3}\).

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+ Add to test✓ In your testCC-AA-34-35-36-223 marks~30 min

A two-link robot arm reaching a target

3.6 Pythagorean and double angle identities3.5 The unit circle, exact values and the ambiguous case3.2 Right-angled and non-right-angled triangles2.10 Solving equations graphically and analyticallySL

A robot arm moves in a plane. It has a fixed shoulder joint \(O\), an elbow joint \(E\) and a hand \(H\). The links \([OE]\) and \([EH]\) are each 30 cm long. The elbow angle is \(\angle OEH=\theta\) radians, where \(0\le\theta\le\pi\), and the distance from the shoulder to the hand is \(OH=d\) cm, as shown in Figure 1.

Figure 1: the robot arm (not to scale)

(a)

Show that \(d^2=1800-1800\cos\theta\).

No calculatorPROOFNotes

Show solution

\(d^2=30^2+30^2-2(30)(30)\cos\theta\)	M1	Cosine rule in triangle \(OEH\).
\(=900+900-1800\cos\theta\)	A1
\(d^2=1800-1800\cos\theta\)	AG

(2)

(b)

Hence show that \(d=60\sin\dfrac{\theta}{2}\).

No calculatorPROOFNotes

Show solution

\(\cos\theta=1-2\sin^2\frac{\theta}{2}\)	M1	Double angle identity with angle \(\frac{\theta}{2}\).
\(d^2=1800-1800\left(1-2\sin^2\frac{\theta}{2}\right)=3600\sin^2\frac{\theta}{2}\)	A1
\(0\le\frac{\theta}{2}\le\frac{\pi}{2}\), so \(\sin\frac{\theta}{2}\ge0\), and \(d\ge0\): take the positive root	R1	Must justify the positive square root.
\(d=60\sin\frac{\theta}{2}\)	AG

(3)

(c)

Find the exact value of \(d\) when \(\theta=\dfrac{2\pi}{3}\).

No calculatorROUTINENotes

Show solution

\(60\sin\frac{\pi}{3}\) or \(d^2=1800-1800\left(-\frac12\right)=2700\)	(M1)
\(d=30\sqrt3\)	A1	Accept \(\sqrt{2700}\).

Answer: \(30\sqrt3\) cm

(2)

(d)(i)

In a different position, \(\sin\theta=\dfrac{12}{13}\) and the elbow angle is obtuse.

Find the exact value of \(\cos\theta\).

No calculatorALGEBRANotes

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\(\cos^2\theta=1-\frac{144}{169}=\frac{25}{169}\)	M1
\(\cos\theta=-\frac{5}{13}\) (negative, since \(\theta\) is obtuse)	A1	\(\pm\frac{5}{13}\) or \(\frac{5}{13}\) scores A0.

Answer: \(-\dfrac{5}{13}\)

(2)

(d)(ii)

Hence find the exact value of \(d\) in this position.

No calculatorALGEBRANotes

Show solution

\(d^2=1800-1800\left(-\frac{5}{13}\right)=\frac{32400}{13}\)	M1	FT from (d)(i). Or \(\sin^2\frac{\theta}{2}=\frac{1-\cos\theta}{2}=\frac{9}{13}\).
\(d=\frac{180}{\sqrt{13}}=\frac{180\sqrt{13}}{13}\)	A1	Either form. \(49.9\) alone scores A0.

Answer: \(\dfrac{180\sqrt{13}}{13}\) cm

(2)

(e)(i)

On the axes in Figure 2, sketch the graph of \(d=60\sin\dfrac{\theta}{2}\) for \(0\le\theta\le\pi\).

Figure 2: axes for the graph of \(d\) against \(\theta\)

GRAPHNotes

Show solution

increasing curve from \((0,0)\) to \((\pi,60)\)	A1	End points labelled or clearly placed.
concave down throughout (gradient decreasing), approximately through \(\left(\frac{\pi}{2},\,42.4\right)\)	A1	Award A0 for a straight line.

(2)

(e)(ii)

The hand must reach a point 50 cm from \(O\). Find the value of \(\theta\).

ROUTINENotes

Show solution

\(\sin\frac{\theta}{2}=\frac56\), or intersection of \(Y_1=60\sin\frac{x}{2}\) and \(Y_2=50\)	(M1)
\(\theta=1.97\)	A1	\(1.9702\ldots\) Accept \(112.9^\circ\). Do not accept other solutions outside \([0,\pi]\).

Answer: \(\theta=1.97\)

(2)

(e)(iii)

Explain why each distance \(d\) with \(0\le d\le60\) is reached for exactly one value of \(\theta\), and why the hand cannot reach a point 65 cm from \(O\).

INTERPRETNotes

Show solution

\(\frac{\theta}{2}\) runs from 0 to \(\frac{\pi}{2}\), where sine is increasing, so \(d\) is increasing and each horizontal line \(d=k\) meets the graph once	R1
the greatest value of \(d\) is 60 (arm straight, \(\theta=\pi\)), and \(65>60\)	R1	Accept: \(\sin\frac{\theta}{2}=\frac{65}{60}>1\) has no solution.

(2)

(f)(i)

The lower link is replaced by a link of length 24 cm, so \(EH=24\) cm and \(OE=30\) cm. A target \(T\) is placed so that \(\angle EOT=30^\circ\), and the hand \(H\) has to lie on the line \([OT]\).

Find the two possible values of \(\angle OHE\).

MULTINotes

Show solution

\(\frac{\sin\angle OHE}{30}=\frac{\sin30^\circ}{24}\)	M1
\(\sin\angle OHE=\frac{5}{8}\)	A1	Using \(\sin30^\circ=\frac12\).
\(38.7^\circ\) or \(141.3^\circ\)	A1	Both required. \(38.682\ldots\), \(141.317\ldots\)

Answer: \(38.7^\circ\) or \(141.3^\circ\)

(3)

(f)(ii)

Find the two possible distances \(OH\).

MULTINotes

Show solution

\(\angle OEH=111.3^\circ\) or \(8.68^\circ\)	(A1)	FT from (f)(i).
\(OH=\frac{24\sin111.3^\circ}{\sin30^\circ}\) or \(\frac{24\sin8.68^\circ}{\sin30^\circ}\)	M1	Or \(24^2=30^2+x^2-2(30)x\cos30^\circ\), giving \(x^2-30\sqrt3\,x+324=0\) and \(x=15\sqrt3\pm3\sqrt{39}\).
\(44.7\) cm or \(7.25\) cm	A1	Both required. \(44.715\ldots\), \(7.2457\ldots\)

Answer: \(44.7\) cm or \(7.25\) cm

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(d^2=30^2+30^2-2(30)(30)\cos\theta\)	M1	Cosine rule in triangle \(OEH\).
\(=900+900-1800\cos\theta\)	A1
\(d^2=1800-1800\cos\theta\)	AG
Part (b)
\(\cos\theta=1-2\sin^2\frac{\theta}{2}\)	M1	Double angle identity with angle \(\frac{\theta}{2}\).
\(d^2=1800-1800\left(1-2\sin^2\frac{\theta}{2}\right)=3600\sin^2\frac{\theta}{2}\)	A1
\(0\le\frac{\theta}{2}\le\frac{\pi}{2}\), so \(\sin\frac{\theta}{2}\ge0\), and \(d\ge0\): take the positive root	R1	Must justify the positive square root.
\(d=60\sin\frac{\theta}{2}\)	AG
Part (c)
\(60\sin\frac{\pi}{3}\) or \(d^2=1800-1800\left(-\frac12\right)=2700\)	(M1)
\(d=30\sqrt3\)	A1	Accept \(\sqrt{2700}\).
Part (d)(i)
\(\cos^2\theta=1-\frac{144}{169}=\frac{25}{169}\)	M1
\(\cos\theta=-\frac{5}{13}\) (negative, since \(\theta\) is obtuse)	A1	\(\pm\frac{5}{13}\) or \(\frac{5}{13}\) scores A0.
Part (d)(ii)
\(d^2=1800-1800\left(-\frac{5}{13}\right)=\frac{32400}{13}\)	M1	FT from (d)(i). Or \(\sin^2\frac{\theta}{2}=\frac{1-\cos\theta}{2}=\frac{9}{13}\).
\(d=\frac{180}{\sqrt{13}}=\frac{180\sqrt{13}}{13}\)	A1	Either form. \(49.9\) alone scores A0.
Part (e)(i)
increasing curve from \((0,0)\) to \((\pi,60)\)	A1	End points labelled or clearly placed.
concave down throughout (gradient decreasing), approximately through \(\left(\frac{\pi}{2},\,42.4\right)\)	A1	Award A0 for a straight line.
Part (e)(ii)
\(\sin\frac{\theta}{2}=\frac56\), or intersection of \(Y_1=60\sin\frac{x}{2}\) and \(Y_2=50\)	(M1)
\(\theta=1.97\)	A1	\(1.9702\ldots\) Accept \(112.9^\circ\). Do not accept other solutions outside \([0,\pi]\).
Part (e)(iii)
\(\frac{\theta}{2}\) runs from 0 to \(\frac{\pi}{2}\), where sine is increasing, so \(d\) is increasing and each horizontal line \(d=k\) meets the graph once	R1
the greatest value of \(d\) is 60 (arm straight, \(\theta=\pi\)), and \(65>60\)	R1	Accept: \(\sin\frac{\theta}{2}=\frac{65}{60}>1\) has no solution.
Part (f)(i)
\(\frac{\sin\angle OHE}{30}=\frac{\sin30^\circ}{24}\)	M1
\(\sin\angle OHE=\frac{5}{8}\)	A1	Using \(\sin30^\circ=\frac12\).
\(38.7^\circ\) or \(141.3^\circ\)	A1	Both required. \(38.682\ldots\), \(141.317\ldots\)
Part (f)(ii)
\(\angle OEH=111.3^\circ\) or \(8.68^\circ\)	(A1)	FT from (f)(i).
\(OH=\frac{24\sin111.3^\circ}{\sin30^\circ}\) or \(\frac{24\sin8.68^\circ}{\sin30^\circ}\)	M1	Or \(24^2=30^2+x^2-2(30)x\cos30^\circ\), giving \(x^2-30\sqrt3\,x+324=0\) and \(x=15\sqrt3\pm3\sqrt{39}\).
\(44.7\) cm or \(7.25\) cm	A1	Both required. \(44.715\ldots\), \(7.2457\ldots\)

Answers: (c) \(30\sqrt3\) cm · (d)(i) \(-\dfrac{5}{13}\) · (d)(ii) \(\dfrac{180\sqrt{13}}{13}\) cm · (e)(ii) \(\theta=1.97\) · (f)(i) \(38.7^\circ\) or \(141.3^\circ\) · (f)(ii) \(44.7\) cm or \(7.25\) cm

Exam tip (b) When a "show that" needs a square root, say why the positive root is the right one. Here both \(d\) and \(\sin\frac{\theta}{2}\) are non-negative for \(0\le\theta\le\pi\).

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3.7 Circular functions and their transformations · 3.8 Solving trigonometric equationsmerged group

+ Add to test✓ In your testCC-AA-37-38-127 marks~30 min

When can a ship enter the harbour? Modelling the tide

3.7 Circular functions and their transformations3.8 Solving trigonometric equations3.4 Radians, arc length and sector area2.11 Transformations of graphsSL

The depth of water at the entrance to a harbour is modelled by

\(D(t)=a\cos\big(b(t-c)\big)+d\), \(0\le t\le24\),

where \(D\) is the depth in metres, \(t\) is the time in hours after midnight on a particular day, and \(a\), \(b\), \(c\) and \(d\) are positive constants with \(0<c<12\).

On this day the first high tide is at 03:30, when the depth is 8.0 m. The next low tide is at 09:42, when the depth is 2.4 m.

(a)

Find the values of \(a\), \(c\) and \(d\).

No calculatorROUTINENotes

Show solution

\(a=\frac{8.0-2.4}{2}=2.8\)	A1
\(d=\frac{8.0+2.4}{2}=5.2\)	A1
\(c=3.5\)	A1	A cosine has its maximum where the bracket is zero, and the high tide is at 03:30 = 3.5 hours. \(c=3.3\) (reading 03:30 as 3.3) is A0.

Answer: \(a=2.8\), \(c=3.5\), \(d=5.2\)

(3)

(b)

Show that \(b=0.507\), correct to three significant figures.

ROUTINENotes

Show solution

period \(=2\times(9.7-3.5)=12.4\) hours	M1	High tide to the next low tide is half a period.
\(b=\frac{2\pi}{12.4}=0.50670\ldots\)	A1	Must see \(\frac{2\pi}{12.4}\) or a value to at least 4 s.f.
\(b=0.507\)	AG

Answer: \(b=\frac{2\pi}{12.4}=0.507\)

(2)

(c)

In the rest of the question, use \(D(t)=2.8\cos\big(0.507(t-3.5)\big)+5.2\).

Describe a sequence of transformations that maps the graph of \(y=\cos t\) onto the graph of \(y=D(t)\).

No calculatorCONCEPTNotes

Show solution

horizontal stretch (parallel to the \(t\)-axis) with scale factor \(\frac{1}{0.507}\) (\(=1.97\))	A1	Accept \(\frac{12.4}{2\pi}\). "Compression by factor 0.507" is A0.
vertical stretch (parallel to the \(D\)-axis) with scale factor 2.8	A1
translation by \(\begin{pmatrix}3.5\\5.2\end{pmatrix}\)	A1	Accept 3.5 to the right and 5.2 up, as two separate translations. Award this mark only if each translation comes after the stretch in the same direction (or the values are adjusted correctly for another order).

Answer: horizontal stretch, s.f. \(\frac{1}{0.507}\); vertical stretch, s.f. 2.8; then translation \(\begin{pmatrix}3.5\\5.2\end{pmatrix}\)

(3)

(d)

On the axes in Figure 1, sketch the graph of \(D\) for \(0\le t\le24\). Show the coordinates of all maximum and minimum points and of the end points of the graph.

Figure 1: axes for the graph of \(D\)

GRAPHNotes

Show solution

smooth cosine-shaped curve between \(D=2.4\) and \(D=8.0\), with almost two complete cycles	A1
maxima at \((3.5,\,8.0)\) and \((15.9,\,8.0)\)	A1
minima at \((9.7,\,2.4)\) and \((22.1,\,2.4)\)	A1
end points \((0,\,4.63)\) and \((24,\,3.61)\)	A1	Accept \(D(0)\) from 4.6 to 4.64 and \(D(24)\) from 3.58 to 3.62 (the exact value of \(b\) gives 4.64 and 3.60).

Answer: max \((3.5,8.0)\), \((15.9,8.0)\); min \((9.7,2.4)\), \((22.1,2.4)\); ends \((0,4.63)\), \((24,3.61)\)

(4)

(e)(i)

A cargo ship needs a depth of at least 4.0 m to pass safely through the harbour entrance.

Find the values of \(t\), \(0\le t\le24\), for which \(D(t)=4.0\).

MULTINotes

Show solution

\(\cos\big(0.507(t-3.5)\big)=-\frac{3}{7}\), or intersection of \(y=D(t)\) with \(y=4\) on the GDC	(M1)
\(t=7.47\)	A1
\(t=11.9\) and \(t=19.9\)	A1	Both needed. Accept 11.92 and 19.86 (or 11.93 and 19.87 from the exact \(b\)). Check by symmetry: \(2(9.7)-7.47=11.93\) and \(7.47+12.4=19.87\).

Answer: \(t=7.47,\ 11.9,\ 19.9\)

(3)

(e)(ii)

Hence find the total time during this day for which the ship can pass safely through the entrance. Give your answer in hours and minutes, to the nearest minute.

INTERPRETNotes

Show solution

\(D\ge4\) for \(0\le t\le7.47\) and \(11.92\le t\le19.86\): total \(7.47+7.94\)	(M1)	Allow ECF from (e)(i).
\(15.4\) hours \(=15\) h 25 min	A1	Accept 15 h 24 min to 15 h 28 min (the upper value comes from using the 3 s.f. answers).

Answer: 15 h 25 min

(2)

(f)

The harbour office has a tide clock with a single hand of length 15 cm. The hand turns clockwise at a constant rate, completing one full turn every 12.4 hours, and it points vertically upwards at each high tide (Figure 2).

Find the length of the arc traced by the tip of the hand between the high tide at 03:30 and the first time after that at which the depth is 4.0 m.

Figure 2: the tide clock at a high tide (not to scale)

ROUTINENotes

Show solution

angle turned \(=0.507\times(7.47-3.5)=2.01\) radians	(M1)	Or \(\frac{3.97}{12.4}\times2\pi\). This angle is the argument of the cosine, \(\arccos\left(-\frac37\right)\). Allow ECF from (e)(i).
\(l=15\times2.01=30.2\) cm	A1	Accept 30.1 to 30.3. Degrees used in \(l=r\theta\) is A0.

Answer: 30.2 cm

(2)

(g)(i)

On days of a neap tide, the high tides are lower and the low tides are higher. A model for the depth on a neap-tide day is

\(N(t)=2.4\cos\big(b(t-3.5)\big)+5.2\), where \(b=\dfrac{2\pi}{12.4}\).

Without using a calculator, show that in each tidal cycle the depth is below 4.0 m for exactly one third of the cycle.

No calculatorPROOFNotes

Show solution

\(2.4\cos\theta+5.2<4\Rightarrow\cos\theta<-\frac12\), where \(\theta=b(t-3.5)\)	M1	Accept the equation \(\cos\theta=-\frac12\).
\(\cos\theta=-\frac12\) at \(\theta=\frac{2\pi}{3}\) and \(\theta=\frac{4\pi}{3}\) in one cycle \(0\le\theta\le2\pi\)	A1
\(\cos\theta<-\frac12\) for \(\frac{2\pi}{3}<\theta<\frac{4\pi}{3}\), an interval of length \(\frac{2\pi}{3}\), which is \(\frac13\) of \(2\pi\)	R1
so the depth is below 4.0 m for one third of each cycle	AG

(3)

(g)(ii)

Hence find, in hours and minutes, the length of time in each tidal cycle during which the ship cannot pass through the entrance on a neap-tide day.

No calculatorROUTINENotes

Show solution

\(\frac13\times12.4=4.1\dot{3}\) hours	(M1)
4 h 8 min	A1

Answer: 4 h 8 min

(2)

(g)(iii)

Use your answers to parts (e)(i) and (g)(ii) to compare the time in each tidal cycle for which the ship cannot pass under the two models. Explain the difference with reference to the depth at low tide.

INTERPRETNotes

Show solution

first model: \(11.92-7.47=4.45\) hours (about 4 h 27 min), which is longer than 4 h 8 min	A1	Allow ECF from (e)(i).
at a neap tide the low-tide depth is \(5.2-2.4=2.8\) m, more than 2.4 m, so the depth spends less of each cycle below 4.0 m	R1	Accept: the smaller amplitude keeps the curve closer to the midline 5.2 m.

Answer: 4 h 27 min compared with 4 h 8 min

(2)

(h)

Suggest one reason why the model \(D\) should not be used to predict the depth at the harbour entrance one week later.

INTERPRETNotes

Show solution

for example: the amplitude changes between spring and neap tides over about two weeks (as in part (g)), so \(a\) and \(d\) do not stay fixed; or wind and air pressure change the depth; or the period is not exactly 12.4 hours, so errors in the times build up

Any one sensible reason linked to the model.

(1)

Show full solution

Working	Mark	Notes
Part (a)
\(a=\frac{8.0-2.4}{2}=2.8\)	A1
\(d=\frac{8.0+2.4}{2}=5.2\)	A1
\(c=3.5\)	A1	A cosine has its maximum where the bracket is zero, and the high tide is at 03:30 = 3.5 hours. \(c=3.3\) (reading 03:30 as 3.3) is A0.
Part (b)
period \(=2\times(9.7-3.5)=12.4\) hours	M1	High tide to the next low tide is half a period.
\(b=\frac{2\pi}{12.4}=0.50670\ldots\)	A1	Must see \(\frac{2\pi}{12.4}\) or a value to at least 4 s.f.
\(b=0.507\)	AG
Part (c)
horizontal stretch (parallel to the \(t\)-axis) with scale factor \(\frac{1}{0.507}\) (\(=1.97\))	A1	Accept \(\frac{12.4}{2\pi}\). "Compression by factor 0.507" is A0.
vertical stretch (parallel to the \(D\)-axis) with scale factor 2.8	A1
translation by \(\begin{pmatrix}3.5\\5.2\end{pmatrix}\)	A1	Accept 3.5 to the right and 5.2 up, as two separate translations. Award this mark only if each translation comes after the stretch in the same direction (or the values are adjusted correctly for another order).
Part (d)
smooth cosine-shaped curve between \(D=2.4\) and \(D=8.0\), with almost two complete cycles	A1
maxima at \((3.5,\,8.0)\) and \((15.9,\,8.0)\)	A1
minima at \((9.7,\,2.4)\) and \((22.1,\,2.4)\)	A1
end points \((0,\,4.63)\) and \((24,\,3.61)\)	A1	Accept \(D(0)\) from 4.6 to 4.64 and \(D(24)\) from 3.58 to 3.62 (the exact value of \(b\) gives 4.64 and 3.60).
Part (e)(i)
\(\cos\big(0.507(t-3.5)\big)=-\frac{3}{7}\), or intersection of \(y=D(t)\) with \(y=4\) on the GDC	(M1)
\(t=7.47\)	A1
\(t=11.9\) and \(t=19.9\)	A1	Both needed. Accept 11.92 and 19.86 (or 11.93 and 19.87 from the exact \(b\)). Check by symmetry: \(2(9.7)-7.47=11.93\) and \(7.47+12.4=19.87\).
Part (e)(ii)
\(D\ge4\) for \(0\le t\le7.47\) and \(11.92\le t\le19.86\): total \(7.47+7.94\)	(M1)	Allow ECF from (e)(i).
\(15.4\) hours \(=15\) h 25 min	A1	Accept 15 h 24 min to 15 h 28 min (the upper value comes from using the 3 s.f. answers).
Part (f)
angle turned \(=0.507\times(7.47-3.5)=2.01\) radians	(M1)	Or \(\frac{3.97}{12.4}\times2\pi\). This angle is the argument of the cosine, \(\arccos\left(-\frac37\right)\). Allow ECF from (e)(i).
\(l=15\times2.01=30.2\) cm	A1	Accept 30.1 to 30.3. Degrees used in \(l=r\theta\) is A0.
Part (g)(i)
\(2.4\cos\theta+5.2<4\Rightarrow\cos\theta<-\frac12\), where \(\theta=b(t-3.5)\)	M1	Accept the equation \(\cos\theta=-\frac12\).
\(\cos\theta=-\frac12\) at \(\theta=\frac{2\pi}{3}\) and \(\theta=\frac{4\pi}{3}\) in one cycle \(0\le\theta\le2\pi\)	A1
\(\cos\theta<-\frac12\) for \(\frac{2\pi}{3}<\theta<\frac{4\pi}{3}\), an interval of length \(\frac{2\pi}{3}\), which is \(\frac13\) of \(2\pi\)	R1
so the depth is below 4.0 m for one third of each cycle	AG
Part (g)(ii)
\(\frac13\times12.4=4.1\dot{3}\) hours	(M1)
4 h 8 min	A1
Part (g)(iii)
first model: \(11.92-7.47=4.45\) hours (about 4 h 27 min), which is longer than 4 h 8 min	A1	Allow ECF from (e)(i).
at a neap tide the low-tide depth is \(5.2-2.4=2.8\) m, more than 2.4 m, so the depth spends less of each cycle below 4.0 m	R1	Accept: the smaller amplitude keeps the curve closer to the midline 5.2 m.
Part (h)
for example: the amplitude changes between spring and neap tides over about two weeks (as in part (g)), so \(a\) and \(d\) do not stay fixed; or wind and air pressure change the depth; or the period is not exactly 12.4 hours, so errors in the times build up	R1	Any one sensible reason linked to the model.

Answers: (a) \(a=2.8\), \(c=3.5\), \(d=5.2\) · (b) \(b=\frac{2\pi}{12.4}=0.507\) · (c) horizontal stretch, s.f. \(\frac{1}{0.507}\); vertical stretch, s.f. 2.8; then translation \(\begin{pmatrix}3.5\\5.2\end{pmatrix}\) · (d) max \((3.5,8.0)\), \((15.9,8.0)\); min \((9.7,2.4)\), \((22.1,2.4)\); ends \((0,4.63)\), \((24,3.61)\) · (e)(i) \(t=7.47,\ 11.9,\ 19.9\) · (e)(ii) 15 h 25 min · (f) 30.2 cm · (g)(ii) 4 h 8 min · (g)(iii) 4 h 27 min compared with 4 h 8 min

Common error (a) Clock times are not decimals: 03:30 is \(t=3.5\) and 09:42 is \(t=9.7\), since 42 minutes is 0.7 of an hour.

Exam tip (e)(i) Count the crossings on your sketch first: the line \(D=4\) meets the curve three times between \(t=0\) and \(t=24\), because the curve starts above 4 and ends below 4.

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A ride on a giant observation wheel

3.8 Solving trigonometric equations3.7 Circular functions and their transformations3.6 Pythagorean and double angle identities5.6 Differentiation rulesSL

A giant observation wheel turns at a constant speed. A passenger boards a capsule at the lowest point of the wheel at time \(t=0\). The height \(h\) of the capsule above the ground, in metres, \(t\) minutes after boarding is modelled by

\(h(t)=68-64\cos\left(\dfrac{\pi t}{18}\right)\), \(t\ge0\).

(a)

Find the diameter of the wheel and the time taken for one complete revolution.

No calculatorROUTINENotes

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greatest height \(68+64=132\), least height \(68-64=4\), so diameter \(=128\) m	A1	Accept \(2\times64\).
period \(=\frac{2\pi}{\pi/18}=36\) minutes	A1

Answer: 128 m; 36 minutes

(2)

(b)(i)

Without using a calculator, solve \(h(t)=100\) for \(0\le t\le36\).

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\(-64\cos\left(\frac{\pi t}{18}\right)=32\Rightarrow\cos\left(\frac{\pi t}{18}\right)=-\frac12\)	M1
\(\frac{\pi t}{18}=\frac{2\pi}{3}\) or \(\frac{\pi t}{18}=\frac{4\pi}{3}\)	A1	With \(0\le\frac{\pi t}{18}\le2\pi\). Both needed.
\(t=12\) or \(t=24\)	A1

Answer: \(t=12\) or \(t=24\)

(3)

(b)(ii)

Hence write down the length of time in each revolution for which the capsule is more than 100 m above the ground.

No calculatorINTERPRETNotes

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\(24-12=12\) minutes

Allow ECF from (b)(i).

Answer: 12 minutes

(1)

(c)

Find the two values of \(t\) in the first revolution at which the capsule is 50 m above the ground.

MULTINotes

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\(\cos\left(\frac{\pi t}{18}\right)=\frac{18}{64}\ \left(=\frac{9}{32}\right)\), or intersection of \(y=h(t)\) with \(y=50\)	(M1)
\(t=7.37\)	A1	\(\frac{18}{\pi}\arccos\frac{9}{32}=7.3665\ldots\)
\(t=28.6\)	A1	By symmetry about the maximum at \(t=18\): \(36-7.37\).

Answer: \(t=7.37\) and \(t=28.6\) minutes

(3)

(d)

On the axes in Figure 1, sketch the graph of \(h\) for \(0\le t\le36\), showing the coordinates of the maximum and minimum points. On the same axes, draw the line \(h=100\) and mark the time interval found in part (b)(ii).

Figure 1: axes for the graph of \(h\)

GRAPHNotes

Show solution

one complete cycle of a cosine-shaped curve, starting at the minimum \((0,\,4)\) and ending at \((36,\,4)\)	A1
maximum at \((18,\,132)\)	A1
line \(h=100\) meeting the curve at \(t=12\) and \(t=24\), with the interval between them marked	A1	Allow ECF from (b)(i).

Answer: min \((0,4)\) and \((36,4)\); max \((18,132)\)

(3)

(e)(i)

Show that \(h'(t)=\dfrac{32\pi}{9}\sin\left(\dfrac{\pi t}{18}\right)\).

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\(\frac{d}{dt}\cos\left(\frac{\pi t}{18}\right)=-\frac{\pi}{18}\sin\left(\frac{\pi t}{18}\right)\)	M1	Chain rule: the factor \(\frac{\pi}{18}\) must be seen.
\(h'(t)=-64\times\left(-\frac{\pi}{18}\right)\sin\left(\frac{\pi t}{18}\right)\)	A1
\(=\frac{32\pi}{9}\sin\left(\frac{\pi t}{18}\right)\)	AG

(2)

(e)(ii)

Find the greatest rate at which the capsule rises, and the value of \(t\) in the first revolution at which this occurs. Describe the position of the capsule on the wheel at this moment.

INTERPRETNotes

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greatest value of the sine is 1, so the greatest rate is \(\frac{32\pi}{9}=11.2\) metres per minute	A1	Accept \(\frac{32\pi}{9}\).
\(\frac{\pi t}{18}=\frac{\pi}{2}\Rightarrow t=9\)	A1
the capsule is level with the centre of the wheel (68 m high), where it is moving vertically upwards	R1	Accept: it is halfway up, on the rising side; this is where the graph in (d) is steepest.

Answer: 11.2 m per minute at \(t=9\); level with the centre

(3)

(f)(i)

Let \(O\) be the centre of the wheel and \(P\) the lowest point, where the passenger boarded. After \(t\) minutes the capsule \(C\) has turned through an angle \(\dfrac{\pi t}{18}\) about \(O\), and the straight-line distance from \(P\) to \(C\) is \(d\) metres (Figure 2).

Show that \(d=128\sin\left(\dfrac{\pi t}{36}\right)\) for \(0\le t\le36\).

Figure 2: the wheel after \(t\) minutes (not to scale)

No calculatorPROOFNotes

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cosine rule in triangle \(OPC\): \(d^2=64^2+64^2-2(64)(64)\cos\left(\frac{\pi t}{18}\right)\)	M1	For \(t>18\) the angle is reflex, but \(\cos(2\pi-\phi)=\cos\phi\), so the same expression holds.
\(d^2=8192\left(1-\cos\frac{\pi t}{18}\right)\)	A1
\(\cos2A=1-2\sin^2A\) with \(A=\frac{\pi t}{36}\): \(1-\cos\frac{\pi t}{18}=2\sin^2\frac{\pi t}{36}\), so \(d^2=16384\sin^2\frac{\pi t}{36}\)	M1
\(0\le\frac{\pi t}{36}\le\pi\), so \(\sin\frac{\pi t}{36}\ge0\) and the positive root is taken	R1
\(d=128\sin\left(\frac{\pi t}{36}\right)\)	AG

(4)

(f)(ii)

Without using a calculator, find the values of \(t\), \(0\le t\le36\), for which the capsule is 64 m from \(P\).

No calculatorROUTINENotes

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\(\sin\left(\frac{\pi t}{36}\right)=\frac12\), with \(0\le\frac{\pi t}{36}\le\pi\)	M1
\(\frac{\pi t}{36}=\frac{\pi}{6}\) or \(\frac{5\pi}{6}\), so \(t=6\) or \(t=30\)	A1	Check: at \(t=6\) the angle turned is \(\frac{\pi}{3}\), so triangle \(OPC\) is equilateral and \(d=64\).

Answer: \(t=6\) or \(t=30\)

(2)

(g)(i)

The operator increases the speed of the wheel so that one revolution now takes 24 minutes. The lowest and highest points of the capsule are unchanged.

Write down the new expression for \(h(t)\).

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\(h(t)=68-64\cos\left(\frac{\pi t}{12}\right)\)

Accept \(\frac{2\pi t}{24}\).

Answer: \(h(t)=68-64\cos\left(\frac{\pi t}{12}\right)\)

(1)

(g)(ii)

Without solving any further equations, find the new length of time in each revolution for which the capsule is more than 100 m above the ground, and the new greatest rate of climb. Justify your answers.

INTERPRETNotes

Show solution

the graph is stretched horizontally by scale factor \(\frac{24}{36}=\frac23\), so the capsule is still above 100 m for \(\frac13\) of a revolution	R1	Accept: every time in the ride is multiplied by \(\frac23\).
\(\frac13\times24=8\) minutes	A1	Or \(\frac23\times12\).
greatest rate \(=64\times\frac{\pi}{12}=\frac{16\pi}{3}=16.8\) metres per minute	A1	Or \(\frac{32\pi}{9}\times\frac{36}{24}\). Allow ECF from (e)(ii).

Answer: 8 minutes; \(\frac{16\pi}{3}=16.8\) m per minute

(3)

Show full solution

Working	Mark	Notes
Part (a)
greatest height \(68+64=132\), least height \(68-64=4\), so diameter \(=128\) m	A1	Accept \(2\times64\).
period \(=\frac{2\pi}{\pi/18}=36\) minutes	A1
Part (b)(i)
\(-64\cos\left(\frac{\pi t}{18}\right)=32\Rightarrow\cos\left(\frac{\pi t}{18}\right)=-\frac12\)	M1
\(\frac{\pi t}{18}=\frac{2\pi}{3}\) or \(\frac{\pi t}{18}=\frac{4\pi}{3}\)	A1	With \(0\le\frac{\pi t}{18}\le2\pi\). Both needed.
\(t=12\) or \(t=24\)	A1
Part (b)(ii)
\(24-12=12\) minutes	A1	Allow ECF from (b)(i).
Part (c)
\(\cos\left(\frac{\pi t}{18}\right)=\frac{18}{64}\ \left(=\frac{9}{32}\right)\), or intersection of \(y=h(t)\) with \(y=50\)	(M1)
\(t=7.37\)	A1	\(\frac{18}{\pi}\arccos\frac{9}{32}=7.3665\ldots\)
\(t=28.6\)	A1	By symmetry about the maximum at \(t=18\): \(36-7.37\).
Part (d)
one complete cycle of a cosine-shaped curve, starting at the minimum \((0,\,4)\) and ending at \((36,\,4)\)	A1
maximum at \((18,\,132)\)	A1
line \(h=100\) meeting the curve at \(t=12\) and \(t=24\), with the interval between them marked	A1	Allow ECF from (b)(i).
Part (e)(i)
\(\frac{d}{dt}\cos\left(\frac{\pi t}{18}\right)=-\frac{\pi}{18}\sin\left(\frac{\pi t}{18}\right)\)	M1	Chain rule: the factor \(\frac{\pi}{18}\) must be seen.
\(h'(t)=-64\times\left(-\frac{\pi}{18}\right)\sin\left(\frac{\pi t}{18}\right)\)	A1
\(=\frac{32\pi}{9}\sin\left(\frac{\pi t}{18}\right)\)	AG
Part (e)(ii)
greatest value of the sine is 1, so the greatest rate is \(\frac{32\pi}{9}=11.2\) metres per minute	A1	Accept \(\frac{32\pi}{9}\).
\(\frac{\pi t}{18}=\frac{\pi}{2}\Rightarrow t=9\)	A1
the capsule is level with the centre of the wheel (68 m high), where it is moving vertically upwards	R1	Accept: it is halfway up, on the rising side; this is where the graph in (d) is steepest.
Part (f)(i)
cosine rule in triangle \(OPC\): \(d^2=64^2+64^2-2(64)(64)\cos\left(\frac{\pi t}{18}\right)\)	M1	For \(t>18\) the angle is reflex, but \(\cos(2\pi-\phi)=\cos\phi\), so the same expression holds.
\(d^2=8192\left(1-\cos\frac{\pi t}{18}\right)\)	A1
\(\cos2A=1-2\sin^2A\) with \(A=\frac{\pi t}{36}\): \(1-\cos\frac{\pi t}{18}=2\sin^2\frac{\pi t}{36}\), so \(d^2=16384\sin^2\frac{\pi t}{36}\)	M1
\(0\le\frac{\pi t}{36}\le\pi\), so \(\sin\frac{\pi t}{36}\ge0\) and the positive root is taken	R1
\(d=128\sin\left(\frac{\pi t}{36}\right)\)	AG
Part (f)(ii)
\(\sin\left(\frac{\pi t}{36}\right)=\frac12\), with \(0\le\frac{\pi t}{36}\le\pi\)	M1
\(\frac{\pi t}{36}=\frac{\pi}{6}\) or \(\frac{5\pi}{6}\), so \(t=6\) or \(t=30\)	A1	Check: at \(t=6\) the angle turned is \(\frac{\pi}{3}\), so triangle \(OPC\) is equilateral and \(d=64\).
Part (g)(i)
\(h(t)=68-64\cos\left(\frac{\pi t}{12}\right)\)	A1	Accept \(\frac{2\pi t}{24}\).
Part (g)(ii)
the graph is stretched horizontally by scale factor \(\frac{24}{36}=\frac23\), so the capsule is still above 100 m for \(\frac13\) of a revolution	R1	Accept: every time in the ride is multiplied by \(\frac23\).
\(\frac13\times24=8\) minutes	A1	Or \(\frac23\times12\).
greatest rate \(=64\times\frac{\pi}{12}=\frac{16\pi}{3}=16.8\) metres per minute	A1	Or \(\frac{32\pi}{9}\times\frac{36}{24}\). Allow ECF from (e)(ii).

Answers: (a) 128 m; 36 minutes · (b)(i) \(t=12\) or \(t=24\) · (b)(ii) 12 minutes · (c) \(t=7.37\) and \(t=28.6\) minutes · (d) min \((0,4)\) and \((36,4)\); max \((18,132)\) · (e)(ii) 11.2 m per minute at \(t=9\); level with the centre · (f)(ii) \(t=6\) or \(t=30\) · (g)(i) \(h(t)=68-64\cos\left(\frac{\pi t}{12}\right)\) · (g)(ii) 8 minutes; \(\frac{16\pi}{3}=16.8\) m per minute

Exam tip (g)(ii) Changing only the period is a horizontal stretch: heights do not change, all times scale by the same factor, and every rate of change scales by the reciprocal factor.

More 3.8 questions in the question bank

3.9 Reciprocal and inverse trigonometric functions · 3.10 Compound and double angle identities · 3.11 Symmetry properties of trigonometric graphsmerged group

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Noise-cancelling headphones: combining sound waves

3.10 Compound and double angle identities3.11 Symmetry properties of trigonometric graphs3.8 Solving trigonometric equations3.7 Circular functions and their transformationsHL

In a test of a pair of noise-cancelling headphones, two loudspeakers each produce a pure tone of the same frequency. At a microphone inside the ear cup, the pressure variations caused by the two tones are modelled by

\(p_1=3\sin x\) and \(p_2=4\cos x\),

where \(x\) is a phase angle in radians that increases at a constant rate with time, and pressures are in arbitrary units. The combined pressure variation is \(p=p_1+p_2\). Figure 1 shows \(p_1\) and \(p_2\) for \(0\le x\le2\pi\).

Figure 1: the two tones \(p_1\) and \(p_2\) for \(0\le x\le2\pi\)

(a)

Show that \(p\) can be written as \(p=5\sin(x+\alpha)\), where \(\tan\alpha=\dfrac43\) and \(0<\alpha<\dfrac{\pi}{2}\).

HL onlyNo calculatorPROOFNotes

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\(R\sin(x+\alpha)=R\sin x\cos\alpha+R\cos x\sin\alpha\)	M1	Compound angle identity.
comparing with \(3\sin x+4\cos x\): \(R\cos\alpha=3\) and \(R\sin\alpha=4\)	A1
\(R^2=3^2+4^2=25\Rightarrow R=5\); \(\tan\alpha=\frac{R\sin\alpha}{R\cos\alpha}=\frac43\), with \(\alpha\) acute as \(\sin\alpha,\cos\alpha>0\)	A1
\(p=5\sin(x+\alpha)\)	AG

(3)

(b)

Write down the greatest value of \(p\). Show that the smallest positive value of \(x\) at which it occurs satisfies \(\tan x=\dfrac34\).

HL onlyNo calculatorCONCEPTNotes

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greatest value 5	A1
\(x+\alpha=\frac{\pi}{2}\Rightarrow x=\frac{\pi}{2}-\alpha\)	M1
\(\tan\left(\frac{\pi}{2}-\alpha\right)=\frac{1}{\tan\alpha}=\frac34\)	R1	Complementary angle identity; or \(\frac{\sin(\pi/2-\alpha)}{\cos(\pi/2-\alpha)}=\frac{\cos\alpha}{\sin\alpha}\).
\(\tan x=\frac34\)	AG

Answer: 5

(3)

(c)

On the axes in Figure 2, sketch the graph of \(p\) for \(0\le x\le2\pi\). Show the coordinates of the maximum and minimum points and of the points where the graph meets the axes, either exactly in terms of \(\alpha\) or correct to three significant figures.

Figure 2: axes for the graph of \(p\)

GRAPHNotes

Show solution

sine-shaped curve of amplitude 5, one complete cycle, starting at \((0,\,4)\) increasing and ending at \((2\pi,\,4)\)	A1
maximum \(\left(\frac{\pi}{2}-\alpha,\,5\right)=(0.644,\,5)\); minimum \(\left(\frac{3\pi}{2}-\alpha,\,-5\right)=(3.79,\,-5)\)	A1
intercepts \((0,\,4)\), \((\pi-\alpha,\,0)=(2.21,\,0)\) and \((2\pi-\alpha,\,0)=(5.36,\,0)\)	A1	\(\alpha=0.927\).

Answer: max \((0.644,5)\), min \((3.79,-5)\), zeros \(2.21\), \(5.36\), \(y\)-intercept 4

(3)

(d)

Without using a calculator, solve \(p=2.5\) for \(0\le x\le2\pi\). Give your answers in the form \(x=k\pi-\alpha\), where \(k\in\mathbb{Q}\).

No calculatorMULTINotes

Show solution

\(\sin(x+\alpha)=\frac12\), with \(\alpha\le x+\alpha\le2\pi+\alpha\)	M1	The interval for \(x+\alpha\) must be considered.
\(x+\alpha=\frac{5\pi}{6}\) or \(\frac{13\pi}{6}\)	A1	\(\frac{\pi}{6}\) is rejected because \(\frac{\pi}{6}<\alpha\) (as \(\tan\alpha=\frac43>1\), \(\alpha>\frac{\pi}{4}\)).
\(x=\frac{5\pi}{6}-\alpha\) or \(x=\frac{13\pi}{6}-\alpha\)	A1	\(=1.69\) or \(5.88\).

Answer: \(x=\frac{5\pi}{6}-\alpha\) or \(x=\frac{13\pi}{6}-\alpha\)

(3)

(e)(i)

Let \(x=\beta\) be any solution of the equation \(p=k\), where \(k\) is a constant. Use a symmetry property of the sine function to show that \(x=\pi-2\alpha-\beta\) is also a solution of \(p=k\).

HL onlyNo calculatorPROOFNotes

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\(5\sin\big((\pi-2\alpha-\beta)+\alpha\big)=5\sin\big(\pi-(\beta+\alpha)\big)\)	M1
\(=5\sin(\beta+\alpha)\) since \(\sin(\pi-\theta)=\sin\theta\), and \(5\sin(\beta+\alpha)=k\)	R1
so \(x=\pi-2\alpha-\beta\) is a solution	AG

(2)

(e)(ii)

The solutions found in part (d) are \(x_1\) and \(x_2\). Show that \(\tan(x_1+x_2)=\dfrac{24}{7}\).

HL onlyNo calculatorALGEBRANotes

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\(x_1+x_2=3\pi-2\alpha\)	M1	Or from (e)(i): \(x_2=(\pi-2\alpha-x_1)+2\pi\).
\(\tan(3\pi-2\alpha)=\tan(-2\alpha)=-\tan2\alpha\)	M1	Tangent has period \(\pi\) and is odd.
\(\tan2\alpha=\frac{2\left(\frac43\right)}{1-\frac{16}{9}}=\frac{8/3}{-7/9}=-\frac{24}{7}\)	A1
\(\tan(x_1+x_2)=\frac{24}{7}\)	AG

(3)

(e)(iii)

Use the identity \(\sin(\theta+\pi)=-\sin\theta\) to write down the solutions of \(p=-2.5\) for \(0\le x\le2\pi\), in the same form as in part (d).

HL onlyNo calculatorCONCEPTNotes

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\(x=\frac{11\pi}{6}-\alpha\)	A1	From \(x_1+\pi\).
\(x=\frac{7\pi}{6}-\alpha\)	A1	From \(x_2-\pi\). Allow ECF from (d).

Answer: \(x=\frac{7\pi}{6}-\alpha\) or \(\frac{11\pi}{6}-\alpha\)

(2)

(f)

The headphones cancel the combined tone by producing a third tone \(p_3=5\sin(x+\gamma)\), where \(0\le\gamma<2\pi\), so that \(p+p_3=0\) for all \(x\). Find \(\gamma\) in terms of \(\alpha\), and justify your answer.

HL onlyNo calculatorCONCEPTNotes

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\(\gamma=\pi+\alpha\)	A1	Accept 4.07.
\(5\sin(x+\alpha+\pi)=-5\sin(x+\alpha)=-p\) for all \(x\)	R1	The cancelling tone is half a period out of phase with \(p\).

Answer: \(\gamma=\pi+\alpha\)

(2)

(g)(i)

The second loudspeaker is moved, so that now \(p_2=4\cos\left(x-\dfrac{\pi}{6}\right)\), while \(p_1\) is unchanged.

Show that the new combined pressure is \(p=5\sin x+2\sqrt3\cos x\).

HL onlyNo calculatorPROOFNotes

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\(4\cos\left(x-\frac{\pi}{6}\right)=4\left(\cos x\cos\frac{\pi}{6}+\sin x\sin\frac{\pi}{6}\right)\)	M1
\(=2\sqrt3\cos x+2\sin x\)	A1
\(p=3\sin x+2\sin x+2\sqrt3\cos x=5\sin x+2\sqrt3\cos x\)	AG

(2)

(g)(ii)

Hence find the exact greatest value of the new combined pressure.

HL onlyNo calculatorALGEBRANotes

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\(R^2=5^2+\left(2\sqrt3\right)^2=37\)	M1	Method as in (a).
greatest value \(\sqrt{37}\)	A1	Not 6.08 alone, since an exact value is required.

Answer: \(\sqrt{37}\)

(2)

(g)(iii)

By considering where the maxima of \(p_1\) and \(p_2\) occur, explain why the greatest value of \(p\) is now larger than in part (b).

HL onlyNo calculatorINTERPRETNotes

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\(p_1\) has a maximum at \(x=\frac{\pi}{2}\); \(p_2\) now has its maximum at \(x=\frac{\pi}{6}\) instead of \(x=0\), so the phase difference falls from \(\frac{\pi}{2}\) to \(\frac{\pi}{3}\)	R1
the peaks of the two tones are closer together (more nearly in phase), so they reinforce each other more and the combined amplitude increases	R1	Accept a comment that it is still less than \(3+4=7\) because the tones are not exactly in phase.

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(R\sin(x+\alpha)=R\sin x\cos\alpha+R\cos x\sin\alpha\)	M1	Compound angle identity.
comparing with \(3\sin x+4\cos x\): \(R\cos\alpha=3\) and \(R\sin\alpha=4\)	A1
\(R^2=3^2+4^2=25\Rightarrow R=5\); \(\tan\alpha=\frac{R\sin\alpha}{R\cos\alpha}=\frac43\), with \(\alpha\) acute as \(\sin\alpha,\cos\alpha>0\)	A1
\(p=5\sin(x+\alpha)\)	AG
Part (b)
greatest value 5	A1
\(x+\alpha=\frac{\pi}{2}\Rightarrow x=\frac{\pi}{2}-\alpha\)	M1
\(\tan\left(\frac{\pi}{2}-\alpha\right)=\frac{1}{\tan\alpha}=\frac34\)	R1	Complementary angle identity; or \(\frac{\sin(\pi/2-\alpha)}{\cos(\pi/2-\alpha)}=\frac{\cos\alpha}{\sin\alpha}\).
\(\tan x=\frac34\)	AG
Part (c)
sine-shaped curve of amplitude 5, one complete cycle, starting at \((0,\,4)\) increasing and ending at \((2\pi,\,4)\)	A1
maximum \(\left(\frac{\pi}{2}-\alpha,\,5\right)=(0.644,\,5)\); minimum \(\left(\frac{3\pi}{2}-\alpha,\,-5\right)=(3.79,\,-5)\)	A1
intercepts \((0,\,4)\), \((\pi-\alpha,\,0)=(2.21,\,0)\) and \((2\pi-\alpha,\,0)=(5.36,\,0)\)	A1	\(\alpha=0.927\).
Part (d)
\(\sin(x+\alpha)=\frac12\), with \(\alpha\le x+\alpha\le2\pi+\alpha\)	M1	The interval for \(x+\alpha\) must be considered.
\(x+\alpha=\frac{5\pi}{6}\) or \(\frac{13\pi}{6}\)	A1	\(\frac{\pi}{6}\) is rejected because \(\frac{\pi}{6}<\alpha\) (as \(\tan\alpha=\frac43>1\), \(\alpha>\frac{\pi}{4}\)).
\(x=\frac{5\pi}{6}-\alpha\) or \(x=\frac{13\pi}{6}-\alpha\)	A1	\(=1.69\) or \(5.88\).
Part (e)(i)
\(5\sin\big((\pi-2\alpha-\beta)+\alpha\big)=5\sin\big(\pi-(\beta+\alpha)\big)\)	M1
\(=5\sin(\beta+\alpha)\) since \(\sin(\pi-\theta)=\sin\theta\), and \(5\sin(\beta+\alpha)=k\)	R1
so \(x=\pi-2\alpha-\beta\) is a solution	AG
Part (e)(ii)
\(x_1+x_2=3\pi-2\alpha\)	M1	Or from (e)(i): \(x_2=(\pi-2\alpha-x_1)+2\pi\).
\(\tan(3\pi-2\alpha)=\tan(-2\alpha)=-\tan2\alpha\)	M1	Tangent has period \(\pi\) and is odd.
\(\tan2\alpha=\frac{2\left(\frac43\right)}{1-\frac{16}{9}}=\frac{8/3}{-7/9}=-\frac{24}{7}\)	A1
\(\tan(x_1+x_2)=\frac{24}{7}\)	AG
Part (e)(iii)
\(x=\frac{11\pi}{6}-\alpha\)	A1	From \(x_1+\pi\).
\(x=\frac{7\pi}{6}-\alpha\)	A1	From \(x_2-\pi\). Allow ECF from (d).
Part (f)
\(\gamma=\pi+\alpha\)	A1	Accept 4.07.
\(5\sin(x+\alpha+\pi)=-5\sin(x+\alpha)=-p\) for all \(x\)	R1	The cancelling tone is half a period out of phase with \(p\).
Part (g)(i)
\(4\cos\left(x-\frac{\pi}{6}\right)=4\left(\cos x\cos\frac{\pi}{6}+\sin x\sin\frac{\pi}{6}\right)\)	M1
\(=2\sqrt3\cos x+2\sin x\)	A1
\(p=3\sin x+2\sin x+2\sqrt3\cos x=5\sin x+2\sqrt3\cos x\)	AG
Part (g)(ii)
\(R^2=5^2+\left(2\sqrt3\right)^2=37\)	M1	Method as in (a).
greatest value \(\sqrt{37}\)	A1	Not 6.08 alone, since an exact value is required.
Part (g)(iii)
\(p_1\) has a maximum at \(x=\frac{\pi}{2}\); \(p_2\) now has its maximum at \(x=\frac{\pi}{6}\) instead of \(x=0\), so the phase difference falls from \(\frac{\pi}{2}\) to \(\frac{\pi}{3}\)	R1
the peaks of the two tones are closer together (more nearly in phase), so they reinforce each other more and the combined amplitude increases	R1	Accept a comment that it is still less than \(3+4=7\) because the tones are not exactly in phase.

Answers: (b) 5 · (c) max \((0.644,5)\), min \((3.79,-5)\), zeros \(2.21\), \(5.36\), \(y\)-intercept 4 · (d) \(x=\frac{5\pi}{6}-\alpha\) or \(x=\frac{13\pi}{6}-\alpha\) · (e)(iii) \(x=\frac{7\pi}{6}-\alpha\) or \(\frac{11\pi}{6}-\alpha\) · (f) \(\gamma=\pi+\alpha\) · (g)(ii) \(\sqrt{37}\)

Common error (d) Solving \(\sin\theta=\frac12\) on \([0,2\pi]\) gives \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\), but here \(\theta=x+\alpha\) runs from \(\alpha\) to \(2\pi+\alpha\). The value \(\frac{\pi}{6}\) falls outside, and its replacement \(\frac{13\pi}{6}\) is easily missed.

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The best seat in the cinema: maximising the viewing angle

3.9 Reciprocal and inverse trigonometric functions3.10 Compound and double angle identities5.15 Further derivatives and integrals; partial fractions5.8 Stationary points, inflexion and optimizationHL

In a cinema, a viewer's eye is at a horizontal distance \(x\) metres from the wall on which the screen is fixed, where \(x>0\). The lower edge of the screen is 3 m above the viewer's eye level and the upper edge is 11 m above eye level, as shown in Figure 1. The angle subtended at the viewer's eye by the screen is \(\theta\) radians, where

\(\theta=\arctan\left(\dfrac{11}{x}\right)-\arctan\left(\dfrac{3}{x}\right)\).

Figure 1: side view of the viewer and the screen (not to scale)

(a)

Find the value of \(\theta\) when \(x=4\).

HL onlyROUTINENotes

Show solution

\(\theta=\arctan(2.75)-\arctan(0.75)=1.2220-0.6435\)	(M1)
\(\theta=0.579\)	A1	Accept 0.5785. An answer of 33.1 (degrees) is A0.

Answer: 0.579

(2)

(b)

Show that \(\tan\theta=\dfrac{8x}{x^2+33}\).

HL onlyNo calculatorPROOFNotes

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\(\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}\) with \(\tan A=\frac{11}{x}\), \(\tan B=\frac{3}{x}\)	M1	Uses \(\tan(\arctan u)=u\).
\(\tan\theta=\frac{\frac{11}{x}-\frac{3}{x}}{1+\frac{33}{x^2}}\)	A1
multiplying numerator and denominator by \(x^2\): \(\frac{11x-3x}{x^2+33}\)	A1
\(\tan\theta=\frac{8x}{x^2+33}\)	AG

(3)

(c)

State the range of the function \(y=\arctan u\), \(u\in\mathbb{R}\). Hence explain why \(\theta=\arctan\left(\dfrac{8x}{x^2+33}\right)\) for all \(x>0\).

HL onlyNo calculatorCONCEPTNotes

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\(-\frac{\pi}{2}<y<\frac{\pi}{2}\)	A1
for \(x>0\), \(0<\arctan\frac{3}{x}<\arctan\frac{11}{x}<\frac{\pi}{2}\), so \(0<\theta<\frac{\pi}{2}\): \(\theta\) lies in the range of arctan, so it is the value of arctan of its own tangent	R1	Must refer to \(\theta\) lying in the range of arctan.

Answer: \(-\frac{\pi}{2}<y<\frac{\pi}{2}\)

(2)

(d)(i)

On the axes in Figure 2, sketch the graph of \(\theta\) against \(x\) for \(0<x\le20\). Show the coordinates of the maximum point.

Figure 2: axes for the graph of \(\theta\) against \(x\)

HL onlyGRAPHNotes

Show solution

curve rising from the origin to a single maximum, then decreasing slowly, with \(\theta(20)\approx0.354\)	A1	The curve must not reach the \(x\)-axis for \(x>0\).
maximum at \((5.74,\,0.608)\)	A1

Answer: maximum \((5.74,\,0.608)\)

(2)

(d)(ii)

Explain, in the context of the cinema, why \(\theta\) is small both for very small and for very large values of \(x\).

HL onlyINTERPRETNotes

Show solution

very close to the wall, both edges of the screen are almost vertically above the eye, so both lines of sight are close to vertical (both arctan values close to \(\frac{\pi}{2}\)) and the angle between them is small	R1
far from the wall, both lines of sight are almost horizontal (both arctan values close to 0), so the screen subtends a small angle	R1

(2)

(e)

Show that \(\dfrac{d\theta}{dx}=\dfrac{8(33-x^2)}{(x^2+9)(x^2+121)}\).

HL onlyNo calculatorPROOFNotes

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\(\frac{d}{dx}\arctan\left(\frac{11}{x}\right)=\frac{1}{1+\left(\frac{11}{x}\right)^2}\times\left(-\frac{11}{x^2}\right)\)	M1	Chain rule with the derivative of arctan.
\(=-\frac{11}{x^2+121}\)	A1
\(\frac{d}{dx}\arctan\left(\frac{3}{x}\right)=-\frac{3}{x^2+9}\)	A1
\(\frac{d\theta}{dx}=\frac{3}{x^2+9}-\frac{11}{x^2+121}=\frac{3(x^2+121)-11(x^2+9)}{(x^2+9)(x^2+121)}\)	M1	Combining over a common denominator.
\(=\frac{264-8x^2}{(x^2+9)(x^2+121)}=\frac{8(33-x^2)}{(x^2+9)(x^2+121)}\)	AG	Alternative: differentiate \(\tan\theta=\frac{8x}{x^2+33}\) implicitly.

(4)

(f)

Hence find the exact value of \(x\) for which \(\theta\) is greatest, justifying that it gives a maximum.

HL onlyNo calculatorMULTINotes

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\(\frac{d\theta}{dx}=0\Rightarrow33-x^2=0\)	M1
\(x=\sqrt{33}\) (since \(x>0\))	A1
the denominator is positive, so \(\frac{d\theta}{dx}>0\) for \(0<x<\sqrt{33}\) and \(\frac{d\theta}{dx}<0\) for \(x>\sqrt{33}\): a maximum	R1	Accept a reference to the graph in (d)(i) only together with a sign argument.

Answer: \(x=\sqrt{33}\) m

(3)

(g)

Find the exact value of \(\sec\theta\) at this maximum.

HL onlyNo calculatorALGEBRANotes

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\(\tan\theta=\frac{8\sqrt{33}}{33+33}=\frac{4}{\sqrt{33}}\)	A1	Allow ECF from (f).
\(\sec^2\theta=1+\tan^2\theta=1+\frac{16}{33}=\frac{49}{33}\)	M1
\(\sec\theta=\frac{7}{\sqrt{33}}=\frac{7\sqrt{33}}{33}\)	A1	Positive root, because \(\theta\) is acute.

Answer: \(\sec\theta=\frac{7\sqrt{33}}{33}\)

(3)

(h)(i)

The cinema is refurbished with a sloping floor, which raises the viewer's eye level by 1 m. The screen is unchanged.

Find the new distance from the wall that gives the greatest viewing angle, and the new greatest angle.

HL onlyMULTINotes

Show solution

edges now 2 m and 10 m above eye level: \(\tan\theta=\frac{\frac{10}{x}-\frac{2}{x}}{1+\frac{20}{x^2}}=\frac{8x}{x^2+20}\)	M1	Or \(\theta=\arctan\frac{10}{x}-\arctan\frac{2}{x}\) maximised on the GDC.
\(x=\sqrt{20}=2\sqrt5=4.47\) m	A1	By the method of (e) and (f), the derivative has the factor \(20-x^2\).
\(\theta=\arctan\frac{2}{\sqrt5}=0.730\)	A1

Answer: \(x=2\sqrt5=4.47\) m; \(\theta=0.730\)

(3)

(h)(ii)

Comment on the effect of the sloping floor for a viewer seated at the best distance.

HL onlyINTERPRETNotes

Show solution

the greatest viewing angle increases (from 0.608 to 0.730) and the best seat moves closer to the screen (from 5.74 m to 4.47 m)

Both features needed. Allow ECF.

(1)

Show full solution

Working	Mark	Notes
Part (a)
\(\theta=\arctan(2.75)-\arctan(0.75)=1.2220-0.6435\)	(M1)
\(\theta=0.579\)	A1	Accept 0.5785. An answer of 33.1 (degrees) is A0.
Part (b)
\(\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}\) with \(\tan A=\frac{11}{x}\), \(\tan B=\frac{3}{x}\)	M1	Uses \(\tan(\arctan u)=u\).
\(\tan\theta=\frac{\frac{11}{x}-\frac{3}{x}}{1+\frac{33}{x^2}}\)	A1
multiplying numerator and denominator by \(x^2\): \(\frac{11x-3x}{x^2+33}\)	A1
\(\tan\theta=\frac{8x}{x^2+33}\)	AG
Part (c)
\(-\frac{\pi}{2}<y<\frac{\pi}{2}\)	A1
for \(x>0\), \(0<\arctan\frac{3}{x}<\arctan\frac{11}{x}<\frac{\pi}{2}\), so \(0<\theta<\frac{\pi}{2}\): \(\theta\) lies in the range of arctan, so it is the value of arctan of its own tangent	R1	Must refer to \(\theta\) lying in the range of arctan.
Part (d)(i)
curve rising from the origin to a single maximum, then decreasing slowly, with \(\theta(20)\approx0.354\)	A1	The curve must not reach the \(x\)-axis for \(x>0\).
maximum at \((5.74,\,0.608)\)	A1
Part (d)(ii)
very close to the wall, both edges of the screen are almost vertically above the eye, so both lines of sight are close to vertical (both arctan values close to \(\frac{\pi}{2}\)) and the angle between them is small	R1
far from the wall, both lines of sight are almost horizontal (both arctan values close to 0), so the screen subtends a small angle	R1
Part (e)
\(\frac{d}{dx}\arctan\left(\frac{11}{x}\right)=\frac{1}{1+\left(\frac{11}{x}\right)^2}\times\left(-\frac{11}{x^2}\right)\)	M1	Chain rule with the derivative of arctan.
\(=-\frac{11}{x^2+121}\)	A1
\(\frac{d}{dx}\arctan\left(\frac{3}{x}\right)=-\frac{3}{x^2+9}\)	A1
\(\frac{d\theta}{dx}=\frac{3}{x^2+9}-\frac{11}{x^2+121}=\frac{3(x^2+121)-11(x^2+9)}{(x^2+9)(x^2+121)}\)	M1	Combining over a common denominator.
\(=\frac{264-8x^2}{(x^2+9)(x^2+121)}=\frac{8(33-x^2)}{(x^2+9)(x^2+121)}\)	AG	Alternative: differentiate \(\tan\theta=\frac{8x}{x^2+33}\) implicitly.
Part (f)
\(\frac{d\theta}{dx}=0\Rightarrow33-x^2=0\)	M1
\(x=\sqrt{33}\) (since \(x>0\))	A1
the denominator is positive, so \(\frac{d\theta}{dx}>0\) for \(0<x<\sqrt{33}\) and \(\frac{d\theta}{dx}<0\) for \(x>\sqrt{33}\): a maximum	R1	Accept a reference to the graph in (d)(i) only together with a sign argument.
Part (g)
\(\tan\theta=\frac{8\sqrt{33}}{33+33}=\frac{4}{\sqrt{33}}\)	A1	Allow ECF from (f).
\(\sec^2\theta=1+\tan^2\theta=1+\frac{16}{33}=\frac{49}{33}\)	M1
\(\sec\theta=\frac{7}{\sqrt{33}}=\frac{7\sqrt{33}}{33}\)	A1	Positive root, because \(\theta\) is acute.
Part (h)(i)
edges now 2 m and 10 m above eye level: \(\tan\theta=\frac{\frac{10}{x}-\frac{2}{x}}{1+\frac{20}{x^2}}=\frac{8x}{x^2+20}\)	M1	Or \(\theta=\arctan\frac{10}{x}-\arctan\frac{2}{x}\) maximised on the GDC.
\(x=\sqrt{20}=2\sqrt5=4.47\) m	A1	By the method of (e) and (f), the derivative has the factor \(20-x^2\).
\(\theta=\arctan\frac{2}{\sqrt5}=0.730\)	A1
Part (h)(ii)
the greatest viewing angle increases (from 0.608 to 0.730) and the best seat moves closer to the screen (from 5.74 m to 4.47 m)	R1	Both features needed. Allow ECF.

Answers: (a) 0.579 · (c) \(-\frac{\pi}{2}<y<\frac{\pi}{2}\) · (d)(i) maximum \((5.74,\,0.608)\) · (f) \(x=\sqrt{33}\) m · (g) \(\sec\theta=\frac{7\sqrt{33}}{33}\) · (h)(i) \(x=2\sqrt5=4.47\) m; \(\theta=0.730\)

Exam tip (g) To move between \(\tan\) and \(\sec\) use \(1+\tan^2\theta=\sec^2\theta\) directly: there is no need to find \(\theta\) itself. Take the sign from the quadrant of the angle.

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3.12 Vectors: concepts and algebra · 3.13 Scalar product · 3.14 Vector equation of a linemerged group

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Air traffic control: how close do two aircraft get?

3.14 Vector equation of a line3.12 Vectors: concepts and algebra3.13 Scalar product5.8 Stationary points, inflexion and optimizationHL

An air traffic controller tracks two aircraft, A and B. The origin O is a radar station at sea level, the \(x\)-, \(y\)- and \(z\)-axes point east, north and vertically upwards, distances are in kilometres and \(t\) is the time in minutes after tracking starts. The position vectors of the aircraft at time \(t\) are

\(\mathbf{r}_A=\begin{pmatrix}-45\\-60\\10\end{pmatrix}+t\begin{pmatrix}9\\12\\0\end{pmatrix}\) and \(\mathbf{r}_B=\begin{pmatrix}35\\-56\\11\end{pmatrix}+t\begin{pmatrix}-7\\12\\-0.2\end{pmatrix}\).

Aircraft A flies at a constant altitude of 10 km and aircraft B is descending. Figure 1 shows the paths of the two aircraft seen from above for \(0\le t\le 10\).

In this airspace, two aircraft must never be less than 5 km apart.

Figure 1: plan view of the paths of aircraft A and B for 0 ≤ t ≤ 10

(a)

Find the speed of aircraft A, giving your answer in km h⁻¹.

HL onlyNo calculatorROUTINENotes

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\(\sqrt{9^2+12^2+0^2}\,\left(=15\text{ km per minute}\right)\)	M1	Magnitude of the velocity vector.
\(900\text{ km h}^{-1}\)	A1	Answer 15 with no conversion scores M1A0.

Answer: \(900\text{ km h}^{-1}\)

(2)

(b)

Find the acute angle between the flight paths of the two aircraft.

HL onlyROUTINENotes

Show solution

\(\begin{pmatrix}9\\12\\0\end{pmatrix}\cdot\begin{pmatrix}-7\\12\\-0.2\end{pmatrix}=-63+144+0=81\)	M1	Attempt at the scalar product of the two direction vectors.
\(\cos\theta=\frac{81}{15\sqrt{193.04}}\;(=0.3887\ldots)\)	A1	\(\sqrt{193.04}=13.89\ldots\)
\(\theta=67.1^\circ\)	A1	Accept 1.17 radians.

Answer: \(67.1^\circ\)

(3)

(c)

Show that, at time \(t\), the displacement vector from A to B is \(\overrightarrow{AB}=\begin{pmatrix}80-16t\\4\\1-0.2t\end{pmatrix}\).

HL onlyNo calculatorALGEBRANotes

Show solution

\(\overrightarrow{AB}=\mathbf{r}_B-\mathbf{r}_A\)	M1	Must be \(\mathbf{r}_B-\mathbf{r}_A\), not \(\mathbf{r}_A-\mathbf{r}_B\).
\(\begin{pmatrix}35-(-45)\\-56-(-60)\\11-10\end{pmatrix}+t\begin{pmatrix}-7-9\\12-12\\-0.2-0\end{pmatrix}=\begin{pmatrix}80\\4\\1\end{pmatrix}+t\begin{pmatrix}-16\\0\\-0.2\end{pmatrix}\)	A1
\(\overrightarrow{AB}=\begin{pmatrix}80-16t\\4\\1-0.2t\end{pmatrix}\)	AG

(2)

(d)

Let \(d\) km be the distance between the aircraft at time \(t\). Show that \(d^2=256.04t^2-2560.4t+6417\).

HL onlyNo calculatorALGEBRANotes

Show solution

\(d^2=\left\|\overrightarrow{AB}\right\|^2=(80-16t)^2+4^2+(1-0.2t)^2\)	M1	Or \(\overrightarrow{AB}\cdot\overrightarrow{AB}\).
\(\left(6400-2560t+256t^2\right)+16+\left(1-0.4t+0.04t^2\right)\)	A1	All three squares expanded correctly.
\(d^2=256.04t^2-2560.4t+6417\)	AG

(2)

(e)

Find the time at which the aircraft are closest together and the minimum distance between them. Justify that the distance you have found is a minimum.

MULTINotes

Show solution

\(\frac{\mathrm{d}}{\mathrm{d}t}\left(d^2\right)=512.08t-2560.4=0\)	M1	Setting the derivative of \(d^2\) (or of \(d\)) equal to zero. Completing the square or the vertex formula \(t=-\frac{b}{2a}\) are equally valid.
\(t=5\) (minutes)	A1
\(d^2=256.04(25)-2560.4(5)+6417\;(=16)\)	M1	Substituting their \(t\) into \(d^2\) or \(d\).
\(d=4\) km	A1
\(\frac{\mathrm{d}^2}{\mathrm{d}t^2}\left(d^2\right)=512.08>0\), so a minimum	R1	Accept: the graph of \(d^2\) is a parabola with positive leading coefficient, or a sign change of the derivative from negative to positive. Since \(d>0\), \(d\) is least when \(d^2\) is least.

Answer: \(t=5\) minutes; \(d=4\) km

(5)

(f)

Verify that, at the time found in (e), the displacement vector \(\overrightarrow{AB}\) is perpendicular to the relative velocity \(\mathbf{v}_B-\mathbf{v}_A=\begin{pmatrix}-16\\0\\-0.2\end{pmatrix}\).

HL onlyNo calculatorCONCEPTNotes

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\(\overrightarrow{AB}=\begin{pmatrix}0\\4\\0\end{pmatrix}\) at \(t=5\)	M1	Substituting \(t=5\) from (e) into \(\overrightarrow{AB}\).
\(\begin{pmatrix}0\\4\\0\end{pmatrix}\cdot\begin{pmatrix}-16\\0\\-0.2\end{pmatrix}=0+0+0=0\), so perpendicular	A1	Must show the scalar product equals zero and conclude.

Answer: scalar product \(=0\)

(2)

(g)

Find the length of time, in seconds, for which the aircraft are less than 5 km apart.

MULTINotes

Show solution

\(256.04t^2-2560.4t+6417=25\)	M1	Or \(d=5\) solved with a GDC, or \(256.04(t-5)^2+16=25\).
\(t=4.8125\ldots\) and \(t=5.1874\ldots\)	A1	Both values; accept 4.81 and 5.19.
\(0.37497\ldots\times60=22.5\) s	A1	Using 3 s.f. values 5.19 and 4.81 gives 22.8 s: accept answers in the range 22 to 23 s.

Answer: \(22.5\) s

(3)

(h)(i)

Figure 2 shows the graph of \(d\) against \(t\) for \(0\le t\le10\). Use Figure 2 to estimate the gradient of the graph for \(8\le t\le10\).

Figure 2: the distance d km between the aircraft against time t minutes

HL onlyGRAPHNotes

Show solution

readings such as \(d\approx48\) at \(t=8\) and \(d\approx80\) at \(t=10\)	M1	Two readings from the graph used in a gradient calculation.
gradient \(\approx16\)	A1	Accept values from 15 to 17.

Answer: \(\approx16\) km per minute

(2)

(h)(ii)

Interpret this gradient in the context of the question, and explain why the graph is very close to a straight line when \(t\) is far from 5.

HL onlyINTERPRETNotes

Show solution

the rate at which the distance between the aircraft increases, about 16 km per minute, equal to the relative speed \(\left\|\mathbf{v}_B-\mathbf{v}_A\right\|=\sqrt{256.04}\approx16.0\) km per minute	R1	Must refer to the rate of separation or the relative speed, with units.
\(d=\sqrt{256.04(t-5)^2+16}\); when \(\|t-5\|\) is large the 16 is negligible, so \(d\approx16.0\,\|t-5\|\), which is linear in \(t\)	R1	Accept an equivalent argument: far from the closest approach almost all the relative velocity is directed along \(\overrightarrow{AB}\).

(2)

(i)

At \(t=0\) the controller instructs aircraft B to climb instead of descending, so that its velocity becomes \(\begin{pmatrix}-7\\12\\0.6\end{pmatrix}\) and A is unchanged. Find the new minimum distance between the aircraft and determine whether the separation rule is now satisfied.

HL onlyMULTINotes

Show solution

\(\overrightarrow{AB}=\begin{pmatrix}80-16t\\4\\1+0.6t\end{pmatrix}\)	M1	New relative position vector.
\(d^2=(80-16t)^2+16+(1+0.6t)^2=256.36t^2-2558.8t+6417\), minimised	M1	Any valid method of minimising, including GDC. Least value at \(t=4.99\) minutes.
\(d_{\min}=5.65\) km	A1	Accept 5.66 or \(\sqrt{31.98}\).
\(5.65>5\), so the rule is now satisfied	R1	Conclusion must follow from their value; R1 can be awarded after A0 if consistent with a minimum found by a valid method.

Answer: \(5.65\) km; the rule is satisfied

(4)

Show full solution

Working	Mark	Notes
Part (a)
\(\sqrt{9^2+12^2+0^2}\,\left(=15\text{ km per minute}\right)\)	M1	Magnitude of the velocity vector.
\(900\text{ km h}^{-1}\)	A1	Answer 15 with no conversion scores M1A0.
Part (b)
\(\begin{pmatrix}9\\12\\0\end{pmatrix}\cdot\begin{pmatrix}-7\\12\\-0.2\end{pmatrix}=-63+144+0=81\)	M1	Attempt at the scalar product of the two direction vectors.
\(\cos\theta=\frac{81}{15\sqrt{193.04}}\;(=0.3887\ldots)\)	A1	\(\sqrt{193.04}=13.89\ldots\)
\(\theta=67.1^\circ\)	A1	Accept 1.17 radians.
Part (c)
\(\overrightarrow{AB}=\mathbf{r}_B-\mathbf{r}_A\)	M1	Must be \(\mathbf{r}_B-\mathbf{r}_A\), not \(\mathbf{r}_A-\mathbf{r}_B\).
\(\begin{pmatrix}35-(-45)\\-56-(-60)\\11-10\end{pmatrix}+t\begin{pmatrix}-7-9\\12-12\\-0.2-0\end{pmatrix}=\begin{pmatrix}80\\4\\1\end{pmatrix}+t\begin{pmatrix}-16\\0\\-0.2\end{pmatrix}\)	A1
\(\overrightarrow{AB}=\begin{pmatrix}80-16t\\4\\1-0.2t\end{pmatrix}\)	AG
Part (d)
\(d^2=\left\|\overrightarrow{AB}\right\|^2=(80-16t)^2+4^2+(1-0.2t)^2\)	M1	Or \(\overrightarrow{AB}\cdot\overrightarrow{AB}\).
\(\left(6400-2560t+256t^2\right)+16+\left(1-0.4t+0.04t^2\right)\)	A1	All three squares expanded correctly.
\(d^2=256.04t^2-2560.4t+6417\)	AG
Part (e)
\(\frac{\mathrm{d}}{\mathrm{d}t}\left(d^2\right)=512.08t-2560.4=0\)	M1	Setting the derivative of \(d^2\) (or of \(d\)) equal to zero. Completing the square or the vertex formula \(t=-\frac{b}{2a}\) are equally valid.
\(t=5\) (minutes)	A1
\(d^2=256.04(25)-2560.4(5)+6417\;(=16)\)	M1	Substituting their \(t\) into \(d^2\) or \(d\).
\(d=4\) km	A1
\(\frac{\mathrm{d}^2}{\mathrm{d}t^2}\left(d^2\right)=512.08>0\), so a minimum	R1	Accept: the graph of \(d^2\) is a parabola with positive leading coefficient, or a sign change of the derivative from negative to positive. Since \(d>0\), \(d\) is least when \(d^2\) is least.
Part (f)
\(\overrightarrow{AB}=\begin{pmatrix}0\\4\\0\end{pmatrix}\) at \(t=5\)	M1	Substituting \(t=5\) from (e) into \(\overrightarrow{AB}\).
\(\begin{pmatrix}0\\4\\0\end{pmatrix}\cdot\begin{pmatrix}-16\\0\\-0.2\end{pmatrix}=0+0+0=0\), so perpendicular	A1	Must show the scalar product equals zero and conclude.
Part (g)
\(256.04t^2-2560.4t+6417=25\)	M1	Or \(d=5\) solved with a GDC, or \(256.04(t-5)^2+16=25\).
\(t=4.8125\ldots\) and \(t=5.1874\ldots\)	A1	Both values; accept 4.81 and 5.19.
\(0.37497\ldots\times60=22.5\) s	A1	Using 3 s.f. values 5.19 and 4.81 gives 22.8 s: accept answers in the range 22 to 23 s.
Part (h)(i)
readings such as \(d\approx48\) at \(t=8\) and \(d\approx80\) at \(t=10\)	M1	Two readings from the graph used in a gradient calculation.
gradient \(\approx16\)	A1	Accept values from 15 to 17.
Part (h)(ii)
the rate at which the distance between the aircraft increases, about 16 km per minute, equal to the relative speed \(\left\|\mathbf{v}_B-\mathbf{v}_A\right\|=\sqrt{256.04}\approx16.0\) km per minute	R1	Must refer to the rate of separation or the relative speed, with units.
\(d=\sqrt{256.04(t-5)^2+16}\); when \(\|t-5\|\) is large the 16 is negligible, so \(d\approx16.0\,\|t-5\|\), which is linear in \(t\)	R1	Accept an equivalent argument: far from the closest approach almost all the relative velocity is directed along \(\overrightarrow{AB}\).
Part (i)
\(\overrightarrow{AB}=\begin{pmatrix}80-16t\\4\\1+0.6t\end{pmatrix}\)	M1	New relative position vector.
\(d^2=(80-16t)^2+16+(1+0.6t)^2=256.36t^2-2558.8t+6417\), minimised	M1	Any valid method of minimising, including GDC. Least value at \(t=4.99\) minutes.
\(d_{\min}=5.65\) km	A1	Accept 5.66 or \(\sqrt{31.98}\).
\(5.65>5\), so the rule is now satisfied	R1	Conclusion must follow from their value; R1 can be awarded after A0 if consistent with a minimum found by a valid method.

Answers: (a) \(900\text{ km h}^{-1}\) · (b) \(67.1^\circ\) · (e) \(t=5\) minutes; \(d=4\) km · (f) scalar product \(=0\) · (g) \(22.5\) s · (h)(i) \(\approx16\) km per minute · (i) \(5.65\) km; the rule is satisfied

Exam tip (e) Minimise \(d^2\) rather than \(d\): it avoids differentiating a square root, and because \(d\ge0\) both are least at the same value of \(t\).

Common error (g) Leaving the answer in minutes. Here \(t\) is measured in minutes, so the difference of the two roots must be multiplied by 60.

More 3.14 questions in the question bank

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Angling a solar panel towards the sun

3.13 Scalar product3.12 Vectors: concepts and algebra3.14 Vector equation of a line3.7 Circular functions and their transformationsHL

A rectangular solar panel \(ABCD\) is fixed to the south-facing face of a house roof, as shown in Figure 1. The origin O is at ground level, the \(x\)-, \(y\)- and \(z\)-axes point east, north and vertically upwards, and units are metres. Three corners of the panel are \(A(2,\,5,\,6)\), \(B(3.6,\,5,\,6)\) and \(D(2,\,4.2,\,5.4)\).

The power output of the panel is \(P=400\cos\theta\) watts, where \(\theta\) is the angle between a vector \(\mathbf{n}\), normal to the panel and pointing away from the roof, and a vector \(\mathbf{s}\) pointing from the panel towards the sun, with \(0^\circ\le\theta\le90^\circ\).

Figure 1: the house, the solar panel ABCD and the flagpole (not to scale)

(a)

Find \(\overrightarrow{AD}\) and show that the edge \([AD]\) has length 1 m.

HL onlyNo calculatorROUTINENotes

Show solution

\(\overrightarrow{AD}=\begin{pmatrix}0\\-0.8\\-0.6\end{pmatrix}\)	A1
\(\left\|\overrightarrow{AD}\right\|=\sqrt{0^2+0.8^2+0.6^2}=\sqrt{0.64+0.36}\)	M1	Must see the sum of squares.
\(=\sqrt{1}=1\) m	AG

Answer: \(\overrightarrow{AD}=\begin{pmatrix}0\\-0.8\\-0.6\end{pmatrix}\)

(2)

(b)

The vector \(\mathbf{n}=\begin{pmatrix}0\\-3\\k\end{pmatrix}\) is perpendicular to the panel. Find the value of \(k\), and verify that \(\mathbf{n}\) is also perpendicular to \(\overrightarrow{AB}\).

HL onlyNo calculatorALGEBRANotes

Show solution

\(\mathbf{n}\cdot\overrightarrow{AD}=0+2.4-0.6k=0\)	M1	Scalar product with \(\overrightarrow{AD}\) set equal to zero. Allow ECF from (a).
\(k=4\)	A1
\(\overrightarrow{AB}=\begin{pmatrix}1.6\\0\\0\end{pmatrix}\), \(\mathbf{n}\cdot\overrightarrow{AB}=0+0+0=0\)	A1	Must show \(\overrightarrow{AB}\) and the zero scalar product.

Answer: \(k=4\)

(3)

(c)

Find the unit vector in the direction of \(\mathbf{n}\).

HL onlyNo calculatorROUTINENotes

Show solution

\(\frac{1}{5}\begin{pmatrix}0\\-3\\4\end{pmatrix}=\begin{pmatrix}0\\-0.6\\0.8\end{pmatrix}\)

Allow ECF from their \(k\).

Answer: \(\begin{pmatrix}0\\-0.6\\0.8\end{pmatrix}\)

(1)

(d)(i)

At one moment on a spring morning, \(\mathbf{s}=\begin{pmatrix}2\\-6\\9\end{pmatrix}\). Show that \(\cos\theta=\dfrac{54}{55}\).

HL onlyNo calculatorROUTINENotes

Show solution

\(\mathbf{n}\cdot\mathbf{s}=0+18+36=54\) and \(\|\mathbf{s}\|=\sqrt{4+36+81}=11\)	M1	Both the scalar product and \(\|\mathbf{s}\|\) seen.
\(\cos\theta=\frac{54}{5\times11}\)	A1	\(\|\mathbf{n}\|=5\) must be used.
\(\cos\theta=\frac{54}{55}\)	AG

(2)

(d)(ii)

Find the power output of the panel at this moment.

HL onlyROUTINENotes

Show solution

\(400\times\frac{54}{55}=393\) W

Accept 392.7 W.

Answer: \(393\) W

(1)

(e)

A vertical flagpole stands on horizontal ground, the plane \(z=0\). The top of the flagpole is \(T(-3,\,-10,\,9)\). At the same moment, sunlight travels in the direction of \(-\mathbf{s}\).

Write down a vector equation of the ray of sunlight through \(T\), and hence find the coordinates of the tip of the flagpole's shadow on the ground.

HL onlyNo calculatorROUTINENotes

Show solution

\(\mathbf{r}=\begin{pmatrix}-3\\-10\\9\end{pmatrix}+\lambda\begin{pmatrix}-2\\6\\-9\end{pmatrix}\)	A1	Accept direction \(\mathbf{s}\) (then \(\lambda=-1\) below). Must include \(\mathbf{r}=\).
\(9-9\lambda=0\Rightarrow\lambda=1\)	M1	Setting the \(z\)-component equal to zero.
\((-5,\,-4,\,0)\)	A1

Answer: \((-5,\,-4,\,0)\)

(3)

(f)(i)

The panel is now mounted on a frame that can turn about its top edge \([AB]\). After turning, a normal to the panel is \(\mathbf{n}'=\begin{pmatrix}0\\-2\\3\end{pmatrix}\). Find the new power output at the moment described in (d), and comment on the size of the change.

HL onlyMULTINotes

Show solution

\(\cos\theta'=\frac{0+12+27}{\sqrt{13}\times11}=\frac{39}{11\sqrt{13}}\;(=0.9833\ldots)\)	M1
\(P=393.3\) W, an increase of only about 0.6 W (0.15%), which is negligible	A1	Both the value and a comment that the gain is very small. Accept 393 W with comment that the output is almost unchanged.

Answer: \(393.3\) W; increase of about 0.6 W, negligible

(2)

(f)(ii)

Explain why, however the frame is turned, the sun's rays at this moment can never be perpendicular to the panel.

HL onlyNo calculatorINTERPRETNotes

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turning about \([AB]\), which is parallel to the \(x\)-axis, keeps every normal perpendicular to \(\overrightarrow{AB}\), so every possible normal has the form \(\begin{pmatrix}0\\p\\q\end{pmatrix}\)	R1	Accept: every possible normal has zero \(x\)-component.
the rays are perpendicular to the panel only if \(\mathbf{s}\) is parallel to a normal, but \(\mathbf{s}\) has \(x\)-component \(2\ne0\), so \(\mathbf{s}\) is never a multiple of such a normal	R1

(2)

(g)(i)

On a clear day in December the panel is back in its original position, with normal \(\mathbf{n}\). Figure 2 shows the power output \(P\) watts, \(t\) hours after midnight, for \(7.5\le t\le17.5\). It is modelled by \(P(t)=a\cos\left(b(t-12.5)\right)\), where \(a,b>0\).

Write down the value of \(a\) and find the value of \(b\).

Figure 2: the power output P watts of the panel, t hours after midnight, on a clear day in December

GRAPHNotes

Show solution

\(a=360\)	A1	Maximum value read from the graph.
\(b(17.5-12.5)=\frac{\pi}{2}\) or period \(=4\times5=20\) hours	M1	Using the zero at \(t=7.5\) or \(t=17.5\), or the period.
\(b=\frac{\pi}{10}\;(=0.314)\)	A1

Answer: \(a=360\), \(b=\frac{\pi}{10}\)

(3)

(g)(ii)

Find the smallest angle between \(\mathbf{n}\) and \(\mathbf{s}\) on this day.

HL onlyMODELNotes

Show solution

\(400\cos\theta=360\)	M1	The greatest power corresponds to the smallest angle. Allow ECF from their \(a\).
\(\theta=25.8^\circ\)	A1

Answer: \(25.8^\circ\)

(2)

(g)(iii)

Find the length of time for which the power output is more than 300 W.

MODELNotes

Show solution

\(360\cos\left(\frac{\pi}{10}(t-12.5)\right)=300\)	M1	Allow ECF from (g)(i). Using the graph alone to estimate scores at most M1.
\(t=10.64\ldots\) and \(t=14.36\ldots\)	A1	Both values.
\(3.73\) hours	A1	Accept 3 hours 44 minutes.

Answer: \(3.73\) hours

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(\overrightarrow{AD}=\begin{pmatrix}0\\-0.8\\-0.6\end{pmatrix}\)	A1
\(\left\|\overrightarrow{AD}\right\|=\sqrt{0^2+0.8^2+0.6^2}=\sqrt{0.64+0.36}\)	M1	Must see the sum of squares.
\(=\sqrt{1}=1\) m	AG
Part (b)
\(\mathbf{n}\cdot\overrightarrow{AD}=0+2.4-0.6k=0\)	M1	Scalar product with \(\overrightarrow{AD}\) set equal to zero. Allow ECF from (a).
\(k=4\)	A1
\(\overrightarrow{AB}=\begin{pmatrix}1.6\\0\\0\end{pmatrix}\), \(\mathbf{n}\cdot\overrightarrow{AB}=0+0+0=0\)	A1	Must show \(\overrightarrow{AB}\) and the zero scalar product.
Part (c)
\(\frac{1}{5}\begin{pmatrix}0\\-3\\4\end{pmatrix}=\begin{pmatrix}0\\-0.6\\0.8\end{pmatrix}\)	A1	Allow ECF from their \(k\).
Part (d)(i)
\(\mathbf{n}\cdot\mathbf{s}=0+18+36=54\) and \(\|\mathbf{s}\|=\sqrt{4+36+81}=11\)	M1	Both the scalar product and \(\|\mathbf{s}\|\) seen.
\(\cos\theta=\frac{54}{5\times11}\)	A1	\(\|\mathbf{n}\|=5\) must be used.
\(\cos\theta=\frac{54}{55}\)	AG
Part (d)(ii)
\(400\times\frac{54}{55}=393\) W	A1	Accept 392.7 W.
Part (e)
\(\mathbf{r}=\begin{pmatrix}-3\\-10\\9\end{pmatrix}+\lambda\begin{pmatrix}-2\\6\\-9\end{pmatrix}\)	A1	Accept direction \(\mathbf{s}\) (then \(\lambda=-1\) below). Must include \(\mathbf{r}=\).
\(9-9\lambda=0\Rightarrow\lambda=1\)	M1	Setting the \(z\)-component equal to zero.
\((-5,\,-4,\,0)\)	A1
Part (f)(i)
\(\cos\theta'=\frac{0+12+27}{\sqrt{13}\times11}=\frac{39}{11\sqrt{13}}\;(=0.9833\ldots)\)	M1
\(P=393.3\) W, an increase of only about 0.6 W (0.15%), which is negligible	A1	Both the value and a comment that the gain is very small. Accept 393 W with comment that the output is almost unchanged.
Part (f)(ii)
turning about \([AB]\), which is parallel to the \(x\)-axis, keeps every normal perpendicular to \(\overrightarrow{AB}\), so every possible normal has the form \(\begin{pmatrix}0\\p\\q\end{pmatrix}\)	R1	Accept: every possible normal has zero \(x\)-component.
the rays are perpendicular to the panel only if \(\mathbf{s}\) is parallel to a normal, but \(\mathbf{s}\) has \(x\)-component \(2\ne0\), so \(\mathbf{s}\) is never a multiple of such a normal	R1
Part (g)(i)
\(a=360\)	A1	Maximum value read from the graph.
\(b(17.5-12.5)=\frac{\pi}{2}\) or period \(=4\times5=20\) hours	M1	Using the zero at \(t=7.5\) or \(t=17.5\), or the period.
\(b=\frac{\pi}{10}\;(=0.314)\)	A1
Part (g)(ii)
\(400\cos\theta=360\)	M1	The greatest power corresponds to the smallest angle. Allow ECF from their \(a\).
\(\theta=25.8^\circ\)	A1
Part (g)(iii)
\(360\cos\left(\frac{\pi}{10}(t-12.5)\right)=300\)	M1	Allow ECF from (g)(i). Using the graph alone to estimate scores at most M1.
\(t=10.64\ldots\) and \(t=14.36\ldots\)	A1	Both values.
\(3.73\) hours	A1	Accept 3 hours 44 minutes.

Answers: (a) \(\overrightarrow{AD}=\begin{pmatrix}0\\-0.8\\-0.6\end{pmatrix}\) · (b) \(k=4\) · (c) \(\begin{pmatrix}0\\-0.6\\0.8\end{pmatrix}\) · (d)(ii) \(393\) W · (e) \((-5,\,-4,\,0)\) · (f)(i) \(393.3\) W; increase of about 0.6 W, negligible · (g)(i) \(a=360\), \(b=\frac{\pi}{10}\) · (g)(ii) \(25.8^\circ\) · (g)(iii) \(3.73\) hours

Common error (g)(iii) Working in degrees: the model uses radians, so \(b=\frac{\pi}{10}\) and the GDC must be in radian mode when solving.

More 3.13 questions in the question bank

3.15 Intersecting, parallel and skew lines · 3.16 Vector product · 3.17 Equations of a plane · 3.18 Intersections and angles of lines and planesmerged group

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Two tunnels under a city: do they meet?

3.15 Intersecting, parallel and skew lines3.16 Vector product3.14 Vector equation of a line3.13 Scalar productHL

Engineers are planning two tunnels beneath a city. The origin is a point on the ground surface, which is the plane \(z=0\); the \(x\)- and \(y\)-axes are horizontal (east and north) and the \(z\)-axis points vertically upwards. Units are metres.

The centre line of a horizontal metro tunnel is \(L_1:\ \mathbf{r}=\begin{pmatrix}0\\0\\-20\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\0\end{pmatrix}\), and the centre line of an inclined access tunnel is \(L_2:\ \mathbf{r}=\begin{pmatrix}2\\-6\\-18\end{pmatrix}+\mu\begin{pmatrix}4\\3\\1\end{pmatrix}\).

Figure 1 shows the two centre lines seen from above.

Figure 1: plan view of the centre lines L₁ and L₂

(a)

Show that \(L_1\) and \(L_2\) are skew.

HL onlyNo calculatorPROOFNotes

Show solution

\(\begin{pmatrix}2\\1\\0\end{pmatrix}\ne k\begin{pmatrix}4\\3\\1\end{pmatrix}\) for any \(k\) (the \(z\)-components force \(k=0\)), so the lines are not parallel	R1	Any valid reason: e.g. ratios \(\frac{2}{4}\ne\frac{1}{3}\).
\(2\lambda=2+4\mu,\quad\lambda=-6+3\mu,\quad-20=-18+\mu\)	M1	Equating the components, with different parameters.
\(\mu=-2,\ \lambda=-12\)	A1	From the second and third equations. Alternatively the first two give \(\lambda=15,\ \mu=7\).
first equation: \(-24\ne2-8=-6\) (or third: \(-20\ne-11\)), so there is no common point; not parallel and not intersecting, so skew	R1	Both facts (not parallel, no intersection) needed for the conclusion.

(4)

(b)

Find the acute angle between the two tunnels.

HL onlyROUTINENotes

Show solution

\(\cos\theta=\frac{\|8+3+0\|}{\sqrt5\sqrt{26}}=\frac{11}{\sqrt{130}}\)	(M1)	Scalar product of the direction vectors and both magnitudes.
\(\theta=15.3^\circ\)	A1	Accept 0.266 radians.

Answer: \(15.3^\circ\)

(2)

(c)

Find \(\begin{pmatrix}2\\1\\0\end{pmatrix}\times\begin{pmatrix}4\\3\\1\end{pmatrix}\).

HL onlyNo calculatorROUTINENotes

Show solution

\(\begin{pmatrix}1\times1-0\times3\\0\times4-2\times1\\2\times3-1\times4\end{pmatrix}\)	M1	At least two components correct, or the correct method seen.
\(\begin{pmatrix}1\\-2\\2\end{pmatrix}\)	A1

Answer: \(\begin{pmatrix}1\\-2\\2\end{pmatrix}\)

(2)

(d)(i)

The point \(P\) lies on \(L_1\) with parameter \(\lambda\) and the point \(Q\) lies on \(L_2\) with parameter \(\mu\). Show that \(\overrightarrow{PQ}=\begin{pmatrix}2+4\mu-2\lambda\\-6+3\mu-\lambda\\2+\mu\end{pmatrix}\).

HL onlyNo calculatorALGEBRANotes

Show solution

\(\overrightarrow{PQ}=\mathbf{q}-\mathbf{p}=\begin{pmatrix}2+4\mu\\-6+3\mu\\-18+\mu\end{pmatrix}-\begin{pmatrix}2\lambda\\\lambda\\-20\end{pmatrix}\)	M1	Position vectors of \(Q\) and \(P\) subtracted in the correct order.
\(z\)-component: \(-18+\mu+20=2+\mu\)	A1	All components clearly shown.
\(\overrightarrow{PQ}=\begin{pmatrix}2+4\mu-2\lambda\\-6+3\mu-\lambda\\2+\mu\end{pmatrix}\)	AG

(2)

(d)(ii)

The shortest distance between two skew lines is the length of the line segment that joins them and is perpendicular to both lines. Let \([PQ]\) be this segment, so that \([PQ]\) is perpendicular to both \(L_1\) and \(L_2\).

Use your answer to (c) to find the values of \(\lambda\) and \(\mu\).

HL onlyNo calculatorMULTINotes

Show solution

\(\overrightarrow{PQ}=k\begin{pmatrix}1\\-2\\2\end{pmatrix}\)	M1	Or \(\overrightarrow{PQ}\cdot\begin{pmatrix}2\\1\\0\end{pmatrix}=0\) and \(\overrightarrow{PQ}\cdot\begin{pmatrix}4\\3\\1\end{pmatrix}=0\), giving \(5\lambda-11\mu=-2\) and \(11\lambda-26\mu=-8\).
\(2+\mu=2k,\quad-6+3\mu-\lambda=-2k,\quad2+4\mu-2\lambda=k\)	A1	Correct system in any form.
\(\lambda=4,\ \mu=2\)	A1	\(k=2\) need not be seen.

Answer: \(\lambda=4\), \(\mu=2\)

(3)

(d)(iii)

Hence find the coordinates of \(P\) and \(Q\), and show that the shortest distance between the two centre lines is 6 m.

HL onlyNo calculatorROUTINENotes

Show solution

\(P(8,\,4,\,-20)\) and \(Q(10,\,0,\,-16)\)	A1	Allow ECF from their \(\lambda\) and \(\mu\) for the coordinates only.
\(\left\|\overrightarrow{PQ}\right\|=\left\|\begin{pmatrix}2\\-4\\4\end{pmatrix}\right\|=\sqrt{4+16+16}=\sqrt{36}\)	A1
\(=6\) m	AG

Answer: \(P(8,4,-20)\), \(Q(10,0,-16)\)

(2)

(e)

Seen from above, the centre lines appear to cross at the point with \(x=30\) and \(y=15\) (Figure 1). Find the vertical distance between the two centre lines at this point, and explain why it is greater than the shortest distance found in (d)(iii).

HL onlyNo calculatorINTERPRETNotes

Show solution

\(\lambda=15\) on \(L_1\) gives \(z=-20\); \(\mu=7\) on \(L_2\) gives \(z=-18+7=-11\)	M1	Both parameter values used to find the two heights.
vertical distance \(=9\) m	A1
the shortest distance is along the common perpendicular \([PQ]\), which is parallel to \(\begin{pmatrix}1\\-2\\2\end{pmatrix}\) and so is not vertical; any other segment joining the lines, including the vertical one, is longer	R1	Accept: the vertical segment is not perpendicular to both lines, so it is not the shortest.

Answer: \(9\) m

(3)

(f)(i)

A tunnel-boring machine moves along \(L_1\). When it is at the point \(R\) with parameter \(\lambda\), its distance from the point \(Q\) found in (d)(iii) is \(D\) metres. Figure 2 shows the graph of \(D\) against \(\lambda\) for \(0\le\lambda\le8\).

Write down the coordinates of the minimum point of the graph, and explain why its \(D\)-coordinate equals the answer to (d)(iii).

Figure 2: the distance D metres from the point R on L₁ to the point Q, against the parameter λ of R

HL onlyNo calculatorGRAPHNotes

Show solution

\((4,\,6)\)	A1
at \(\lambda=4\), \(R\) is the point \(P\); \([PQ]\) is perpendicular to \(L_1\), so \(P\) is the point of \(L_1\) closest to \(Q\)	R1	Must link \(\lambda=4\) to \(P\) and to perpendicularity.

Answer: \((4,\,6)\)

(2)

(f)(ii)

While the machine is less than 9 m from \(Q\), the access tunnel must be closed. Use Figure 2 to find the values of \(\lambda\) between which this applies, and hence find the distance the machine travels during this time.

HL onlyGRAPHNotes

Show solution

\(\lambda=1\) and \(\lambda=7\)	A1	Read from the graph; may be confirmed from \(D^2=5\lambda^2-40\lambda+116=81\).
distance \(=(7-1)\left\|\begin{pmatrix}2\\1\\0\end{pmatrix}\right\|=6\sqrt5\)	M1	Change in \(\lambda\) multiplied by the length of the direction vector, or the distance between the points \((2,1,-20)\) and \((14,7,-20)\).
\(13.4\) m	A1

Answer: \(13.4\) m

(3)

(g)

A vertical ventilation shaft is drilled down from the point \(G(18,\,9,\,0)\) on the surface. Show that the shaft meets \(L_1\), and find the length of the shaft from \(G\) to the centre line of \(L_1\).

HL onlyNo calculatorROUTINENotes

Show solution

shaft: \(\mathbf{r}=\begin{pmatrix}18\\9\\0\end{pmatrix}+\nu\begin{pmatrix}0\\0\\-1\end{pmatrix}\); \(2\lambda=18\) and \(\lambda=9\) are consistent, so the shaft meets \(L_1\) at \((18,\,9,\,-20)\)	M1	Both horizontal components must give the same \(\lambda\).
length \(=20\) m	A1

Answer: meets at \((18,9,-20)\); \(20\) m

(2)

(h)

The access tunnel is redesigned so that its centre line is \(\mathbf{r}=\begin{pmatrix}2\\-6\\c\end{pmatrix}+\mu\begin{pmatrix}4\\3\\1\end{pmatrix}\). Find the value of \(c\) for which the two centre lines meet, and the coordinates of the junction.

HL onlyNo calculatorMULTINotes

Show solution

\(2\lambda=2+4\mu\) and \(\lambda=-6+3\mu\) give \(\lambda=15,\ \mu=7\)	M1	Using the two equations that do not contain \(c\). Allow values carried from (e).
\(-20=c+7\)	M1
\(c=-27\); junction \((30,\,15,\,-20)\)	A1	Both required.

Answer: \(c=-27\); \((30,\,15,\,-20)\)

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(\begin{pmatrix}2\\1\\0\end{pmatrix}\ne k\begin{pmatrix}4\\3\\1\end{pmatrix}\) for any \(k\) (the \(z\)-components force \(k=0\)), so the lines are not parallel	R1	Any valid reason: e.g. ratios \(\frac{2}{4}\ne\frac{1}{3}\).
\(2\lambda=2+4\mu,\quad\lambda=-6+3\mu,\quad-20=-18+\mu\)	M1	Equating the components, with different parameters.
\(\mu=-2,\ \lambda=-12\)	A1	From the second and third equations. Alternatively the first two give \(\lambda=15,\ \mu=7\).
first equation: \(-24\ne2-8=-6\) (or third: \(-20\ne-11\)), so there is no common point; not parallel and not intersecting, so skew	R1	Both facts (not parallel, no intersection) needed for the conclusion.
Part (b)
\(\cos\theta=\frac{\|8+3+0\|}{\sqrt5\sqrt{26}}=\frac{11}{\sqrt{130}}\)	(M1)	Scalar product of the direction vectors and both magnitudes.
\(\theta=15.3^\circ\)	A1	Accept 0.266 radians.
Part (c)
\(\begin{pmatrix}1\times1-0\times3\\0\times4-2\times1\\2\times3-1\times4\end{pmatrix}\)	M1	At least two components correct, or the correct method seen.
\(\begin{pmatrix}1\\-2\\2\end{pmatrix}\)	A1
Part (d)(i)
\(\overrightarrow{PQ}=\mathbf{q}-\mathbf{p}=\begin{pmatrix}2+4\mu\\-6+3\mu\\-18+\mu\end{pmatrix}-\begin{pmatrix}2\lambda\\\lambda\\-20\end{pmatrix}\)	M1	Position vectors of \(Q\) and \(P\) subtracted in the correct order.
\(z\)-component: \(-18+\mu+20=2+\mu\)	A1	All components clearly shown.
\(\overrightarrow{PQ}=\begin{pmatrix}2+4\mu-2\lambda\\-6+3\mu-\lambda\\2+\mu\end{pmatrix}\)	AG
Part (d)(ii)
\(\overrightarrow{PQ}=k\begin{pmatrix}1\\-2\\2\end{pmatrix}\)	M1	Or \(\overrightarrow{PQ}\cdot\begin{pmatrix}2\\1\\0\end{pmatrix}=0\) and \(\overrightarrow{PQ}\cdot\begin{pmatrix}4\\3\\1\end{pmatrix}=0\), giving \(5\lambda-11\mu=-2\) and \(11\lambda-26\mu=-8\).
\(2+\mu=2k,\quad-6+3\mu-\lambda=-2k,\quad2+4\mu-2\lambda=k\)	A1	Correct system in any form.
\(\lambda=4,\ \mu=2\)	A1	\(k=2\) need not be seen.
Part (d)(iii)
\(P(8,\,4,\,-20)\) and \(Q(10,\,0,\,-16)\)	A1	Allow ECF from their \(\lambda\) and \(\mu\) for the coordinates only.
\(\left\|\overrightarrow{PQ}\right\|=\left\|\begin{pmatrix}2\\-4\\4\end{pmatrix}\right\|=\sqrt{4+16+16}=\sqrt{36}\)	A1
\(=6\) m	AG
Part (e)
\(\lambda=15\) on \(L_1\) gives \(z=-20\); \(\mu=7\) on \(L_2\) gives \(z=-18+7=-11\)	M1	Both parameter values used to find the two heights.
vertical distance \(=9\) m	A1
the shortest distance is along the common perpendicular \([PQ]\), which is parallel to \(\begin{pmatrix}1\\-2\\2\end{pmatrix}\) and so is not vertical; any other segment joining the lines, including the vertical one, is longer	R1	Accept: the vertical segment is not perpendicular to both lines, so it is not the shortest.
Part (f)(i)
\((4,\,6)\)	A1
at \(\lambda=4\), \(R\) is the point \(P\); \([PQ]\) is perpendicular to \(L_1\), so \(P\) is the point of \(L_1\) closest to \(Q\)	R1	Must link \(\lambda=4\) to \(P\) and to perpendicularity.
Part (f)(ii)
\(\lambda=1\) and \(\lambda=7\)	A1	Read from the graph; may be confirmed from \(D^2=5\lambda^2-40\lambda+116=81\).
distance \(=(7-1)\left\|\begin{pmatrix}2\\1\\0\end{pmatrix}\right\|=6\sqrt5\)	M1	Change in \(\lambda\) multiplied by the length of the direction vector, or the distance between the points \((2,1,-20)\) and \((14,7,-20)\).
\(13.4\) m	A1
Part (g)
shaft: \(\mathbf{r}=\begin{pmatrix}18\\9\\0\end{pmatrix}+\nu\begin{pmatrix}0\\0\\-1\end{pmatrix}\); \(2\lambda=18\) and \(\lambda=9\) are consistent, so the shaft meets \(L_1\) at \((18,\,9,\,-20)\)	M1	Both horizontal components must give the same \(\lambda\).
length \(=20\) m	A1
Part (h)
\(2\lambda=2+4\mu\) and \(\lambda=-6+3\mu\) give \(\lambda=15,\ \mu=7\)	M1	Using the two equations that do not contain \(c\). Allow values carried from (e).
\(-20=c+7\)	M1
\(c=-27\); junction \((30,\,15,\,-20)\)	A1	Both required.

Answers: (b) \(15.3^\circ\) · (c) \(\begin{pmatrix}1\\-2\\2\end{pmatrix}\) · (d)(ii) \(\lambda=4\), \(\mu=2\) · (d)(iii) \(P(8,4,-20)\), \(Q(10,0,-16)\) · (e) \(9\) m · (f)(i) \((4,\,6)\) · (f)(ii) \(13.4\) m · (g) meets at \((18,9,-20)\); \(20\) m · (h) \(c=-27\); \((30,\,15,\,-20)\)

Common error (a) Concluding that the lines are skew only because the equations are inconsistent. Parallel lines also have no common point, so you must show that the directions are not parallel as well.

Common error (f)(ii) Giving 6 m. The parameter \(\lambda\) is not a distance: one unit of \(\lambda\) moves the machine \(\sqrt5\) m along the tunnel.

More 3.15 questions in the question bank

+ Add to test✓ In your testCC-AA-315-316-317-318-227 marks~30 min

Roof planes and a surveying laser

3.18 Intersections and angles of lines and planes3.17 Equations of a plane3.16 Vector product3.14 Vector equation of a lineHL

A building stands on level ground. Its roof is made of two plane faces that meet along a sloping ridge \([RS]\), as shown in Figure 1. The origin O is at ground level at one corner of the building, the \(x\)- and \(y\)-axes are horizontal, pointing east and north, the \(z\)-axis points vertically upwards and units are metres.

The south face of the roof lies in the plane \(\Pi_1\) and contains the points \(A(0,\,0,\,6)\), \(B(16,\,0,\,4)\) and \(R(0,\,5,\,8.5)\). The north face lies in the plane \(\Pi_2\) with equation \(x+4y+8z=88\).

Figure 1: the building, its roof faces and the laser at L (not to scale)

(a)

Find \(\overrightarrow{AB}\times\overrightarrow{AR}\).

HL onlyNo calculatorROUTINENotes

Show solution

\(\overrightarrow{AB}=\begin{pmatrix}16\\0\\-2\end{pmatrix}\), \(\overrightarrow{AR}=\begin{pmatrix}0\\5\\2.5\end{pmatrix}\)	M1	Both vectors, and an attempt at the vector product.
\(\begin{pmatrix}10\\-40\\80\end{pmatrix}\)	A1	Reversing the order gives \(\begin{pmatrix}-10\\40\\-80\end{pmatrix}\): M1A0.

Answer: \(\begin{pmatrix}10\\-40\\80\end{pmatrix}\)

(2)

(b)

Hence show that \(\Pi_1\) has Cartesian equation \(x-4y+8z=48\).

HL onlyNo calculatorROUTINENotes

Show solution

normal \(\mathbf{n}_1=\begin{pmatrix}1\\-4\\8\end{pmatrix}\) (dividing by 10)	A1	Any non-zero multiple of the answer to (a).
\(\mathbf{r}\cdot\mathbf{n}_1=\mathbf{a}\cdot\mathbf{n}_1=0-0+48\)	M1	Substituting a point of the plane, e.g. \(A\); \(B\) gives \(16+32\), \(R\) gives \(-20+68\).
\(x-4y+8z=48\)	AG

(2)

(c)

Find the acute angle between the planes \(\Pi_1\) and \(\Pi_2\). Hence write down the angle between the two faces of the roof measured inside the roof space, under the ridge.

HL onlyMULTINotes

Show solution

\(\cos\theta=\frac{\|1-16+64\|}{9\times9}\)	M1	Scalar product of the normals \(\begin{pmatrix}1\\-4\\8\end{pmatrix}\) and \(\begin{pmatrix}1\\4\\8\end{pmatrix}\), and their magnitudes.
\(\theta=52.8^\circ\)	A1
\(180^\circ-52.8^\circ=127.2^\circ\)	A1	Accept 127°. Allow ECF from their acute angle.

Answer: \(52.8^\circ\); \(127.2^\circ\)

(3)

(d)

Find a vector equation of the ridge line \(RS\).

HL onlyNo calculatorMULTINotes

Show solution

\(\begin{pmatrix}1\\-4\\8\end{pmatrix}\times\begin{pmatrix}1\\4\\8\end{pmatrix}\)	M1	Vector product of the two normals.
\(=\begin{pmatrix}-64\\0\\8\end{pmatrix}\), so direction \(\begin{pmatrix}8\\0\\-1\end{pmatrix}\)	A1	Any non-zero multiple.
a point on both planes, e.g. \(R(0,\,5,\,8.5)\): \(0-20+68=48\) and \(0+20+68=88\)	M1	Or setting \(x=0\) and solving \(-4y+8z=48,\ 4y+8z=88\).
\(\mathbf{r}=\begin{pmatrix}0\\5\\8.5\end{pmatrix}+s\begin{pmatrix}8\\0\\-1\end{pmatrix}\)	A1	Must include \(\mathbf{r}=\). Accept any correct equivalent form.

Answer: \(\mathbf{r}=\begin{pmatrix}0\\5\\8.5\end{pmatrix}+s\begin{pmatrix}8\\0\\-1\end{pmatrix}\)

(4)

(e)

A surveying laser is mounted at \(L(-4,\,-2,\,14)\) on top of a mast. Its beam follows the line \(\mathbf{r}=\begin{pmatrix}-4\\-2\\14\end{pmatrix}+t\begin{pmatrix}3\\1\\-2\end{pmatrix}\). Find the coordinates of the point \(I\) where the beam meets the south face of the roof.

HL onlyNo calculatorROUTINENotes

Show solution

\((-4+3t)-4(-2+t)+8(14-2t)=48\)	M1	Substituting the parametric form into \(\Pi_1\).
\(116-17t=48\Rightarrow t=4\)	A1
\(I(8,\,2,\,6)\)	A1

Answer: \(I(8,\,2,\,6)\)

(3)

(f)

Find the acute angle between the laser beam and the south face of the roof.

HL onlyROUTINENotes

Show solution

\(\begin{pmatrix}3\\1\\-2\end{pmatrix}\cdot\begin{pmatrix}1\\-4\\8\end{pmatrix}=3-4-16=-17\)	M1	Scalar product of the direction of the beam with the normal of \(\Pi_1\).
\(\sin\theta=\frac{17}{\sqrt{14}\times9}\;(=0.504\ldots)\)	A1	Using sine (or \(\cos(90^\circ-\theta)\)).
\(\theta=30.3^\circ\)	A1	\(59.7^\circ\) (the angle with the normal) scores M1A0A0.

Answer: \(30.3^\circ\)

(3)

(g)(i)

Figure 2 shows the cross-section of the building in the vertical plane \(x=8\). Use Figure 2 to write down the height of the ridge above the ground at \(x=8\), and verify your answer using the equation found in (d).

Figure 2: the cross-section of the building in the vertical plane x = 8

HL onlyNo calculatorGRAPHNotes

Show solution

\(7.5\) m	A1
\(s=1\) gives \((8,\,5,\,7.5)\), so the ridge is at height 7.5 m above \(x=8\)	A1	Allow ECF from their line in (d) if it passes through a point with \(x=8\).

Answer: \(7.5\) m

(2)

(g)(ii)

In Figure 2 the edge of the south face makes an angle of \(\arctan\dfrac12\approx26.6^\circ\) with the horizontal. Find the angle between \(\Pi_1\) and the horizontal ground, and explain why it is greater than \(26.6^\circ\).

HL onlyINTERPRETNotes

Show solution

\(\cos\alpha=\frac{\begin{pmatrix}1\\-4\\8\end{pmatrix}\cdot\begin{pmatrix}0\\0\\1\end{pmatrix}}{9\times1}=\frac{8}{9}\)	M1	Normal of the ground \(\begin{pmatrix}0\\0\\1\end{pmatrix}\), or the plane \(z=0\).
\(\alpha=27.3^\circ\)	A1
the steepest line in \(\Pi_1\) runs in the direction of the horizontal part of \(\mathbf{n}_1\), \((1,-4)\), which is not in the plane \(x=8\) because \(\mathbf{n}_1\) has a non-zero \(x\)-component (the ridge slopes); the section line in Figure 2 is not the steepest line, so it is less steep	R1	Accept: the angle between two planes is measured along a line perpendicular to their line of intersection, and the plane \(x=8\) does not cut \(\Pi_1\) along such a line.

Answer: \(27.3^\circ\)

(3)

(h)(i)

A solar canopy is to be fixed in the plane \(\Pi_3:\ 3x+4y+24z=k\). Show that a normal to \(\Pi_3\) is \(\mathbf{n}_1+2\mathbf{n}_2\), where \(\mathbf{n}_1=\begin{pmatrix}1\\-4\\8\end{pmatrix}\) and \(\mathbf{n}_2=\begin{pmatrix}1\\4\\8\end{pmatrix}\). Hence find the value of \(k\) for which the three planes \(\Pi_1\), \(\Pi_2\) and \(\Pi_3\) have a common line.

HL onlyNo calculatorALGEBRANotes

Show solution

\(\begin{pmatrix}1\\-4\\8\end{pmatrix}+2\begin{pmatrix}1\\4\\8\end{pmatrix}=\begin{pmatrix}3\\4\\24\end{pmatrix}\)	B1
equation of \(\Pi_1\) \(+\,2\times\) equation of \(\Pi_2\): \(3x+4y+24z=48+2(88)\)	M1	Or: \(\Pi_3\) must contain the ridge line, so substitute \(R(0,\,5,\,8.5)\): \(0+20+204\).
\(k=224\)	A1

Answer: \(k=224\)

(3)

(h)(ii)

The canopy is installed with \(k=230\) instead. Describe the geometrical arrangement of the three planes, justifying your answer.

HL onlyNo calculatorINTERPRETNotes

Show solution

every point on both \(\Pi_1\) and \(\Pi_2\) satisfies \(3x+4y+24z=224\ne230\), so no point lies on all three planes	R1
no two of the normals are parallel, so no two planes are parallel: each pair meets in a line and the three lines are parallel, forming a triangular prism	R1	"Parallel planes" scores R0 for this row.

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(\overrightarrow{AB}=\begin{pmatrix}16\\0\\-2\end{pmatrix}\), \(\overrightarrow{AR}=\begin{pmatrix}0\\5\\2.5\end{pmatrix}\)	M1	Both vectors, and an attempt at the vector product.
\(\begin{pmatrix}10\\-40\\80\end{pmatrix}\)	A1	Reversing the order gives \(\begin{pmatrix}-10\\40\\-80\end{pmatrix}\): M1A0.
Part (b)
normal \(\mathbf{n}_1=\begin{pmatrix}1\\-4\\8\end{pmatrix}\) (dividing by 10)	A1	Any non-zero multiple of the answer to (a).
\(\mathbf{r}\cdot\mathbf{n}_1=\mathbf{a}\cdot\mathbf{n}_1=0-0+48\)	M1	Substituting a point of the plane, e.g. \(A\); \(B\) gives \(16+32\), \(R\) gives \(-20+68\).
\(x-4y+8z=48\)	AG
Part (c)
\(\cos\theta=\frac{\|1-16+64\|}{9\times9}\)	M1	Scalar product of the normals \(\begin{pmatrix}1\\-4\\8\end{pmatrix}\) and \(\begin{pmatrix}1\\4\\8\end{pmatrix}\), and their magnitudes.
\(\theta=52.8^\circ\)	A1
\(180^\circ-52.8^\circ=127.2^\circ\)	A1	Accept 127°. Allow ECF from their acute angle.
Part (d)
\(\begin{pmatrix}1\\-4\\8\end{pmatrix}\times\begin{pmatrix}1\\4\\8\end{pmatrix}\)	M1	Vector product of the two normals.
\(=\begin{pmatrix}-64\\0\\8\end{pmatrix}\), so direction \(\begin{pmatrix}8\\0\\-1\end{pmatrix}\)	A1	Any non-zero multiple.
a point on both planes, e.g. \(R(0,\,5,\,8.5)\): \(0-20+68=48\) and \(0+20+68=88\)	M1	Or setting \(x=0\) and solving \(-4y+8z=48,\ 4y+8z=88\).
\(\mathbf{r}=\begin{pmatrix}0\\5\\8.5\end{pmatrix}+s\begin{pmatrix}8\\0\\-1\end{pmatrix}\)	A1	Must include \(\mathbf{r}=\). Accept any correct equivalent form.
Part (e)
\((-4+3t)-4(-2+t)+8(14-2t)=48\)	M1	Substituting the parametric form into \(\Pi_1\).
\(116-17t=48\Rightarrow t=4\)	A1
\(I(8,\,2,\,6)\)	A1
Part (f)
\(\begin{pmatrix}3\\1\\-2\end{pmatrix}\cdot\begin{pmatrix}1\\-4\\8\end{pmatrix}=3-4-16=-17\)	M1	Scalar product of the direction of the beam with the normal of \(\Pi_1\).
\(\sin\theta=\frac{17}{\sqrt{14}\times9}\;(=0.504\ldots)\)	A1	Using sine (or \(\cos(90^\circ-\theta)\)).
\(\theta=30.3^\circ\)	A1	\(59.7^\circ\) (the angle with the normal) scores M1A0A0.
Part (g)(i)
\(7.5\) m	A1
\(s=1\) gives \((8,\,5,\,7.5)\), so the ridge is at height 7.5 m above \(x=8\)	A1	Allow ECF from their line in (d) if it passes through a point with \(x=8\).
Part (g)(ii)
\(\cos\alpha=\frac{\begin{pmatrix}1\\-4\\8\end{pmatrix}\cdot\begin{pmatrix}0\\0\\1\end{pmatrix}}{9\times1}=\frac{8}{9}\)	M1	Normal of the ground \(\begin{pmatrix}0\\0\\1\end{pmatrix}\), or the plane \(z=0\).
\(\alpha=27.3^\circ\)	A1
the steepest line in \(\Pi_1\) runs in the direction of the horizontal part of \(\mathbf{n}_1\), \((1,-4)\), which is not in the plane \(x=8\) because \(\mathbf{n}_1\) has a non-zero \(x\)-component (the ridge slopes); the section line in Figure 2 is not the steepest line, so it is less steep	R1	Accept: the angle between two planes is measured along a line perpendicular to their line of intersection, and the plane \(x=8\) does not cut \(\Pi_1\) along such a line.
Part (h)(i)
\(\begin{pmatrix}1\\-4\\8\end{pmatrix}+2\begin{pmatrix}1\\4\\8\end{pmatrix}=\begin{pmatrix}3\\4\\24\end{pmatrix}\)	B1
equation of \(\Pi_1\) \(+\,2\times\) equation of \(\Pi_2\): \(3x+4y+24z=48+2(88)\)	M1	Or: \(\Pi_3\) must contain the ridge line, so substitute \(R(0,\,5,\,8.5)\): \(0+20+204\).
\(k=224\)	A1
Part (h)(ii)
every point on both \(\Pi_1\) and \(\Pi_2\) satisfies \(3x+4y+24z=224\ne230\), so no point lies on all three planes	R1
no two of the normals are parallel, so no two planes are parallel: each pair meets in a line and the three lines are parallel, forming a triangular prism	R1	"Parallel planes" scores R0 for this row.

Answers: (a) \(\begin{pmatrix}10\\-40\\80\end{pmatrix}\) · (c) \(52.8^\circ\); \(127.2^\circ\) · (d) \(\mathbf{r}=\begin{pmatrix}0\\5\\8.5\end{pmatrix}+s\begin{pmatrix}8\\0\\-1\end{pmatrix}\) · (e) \(I(8,\,2,\,6)\) · (f) \(30.3^\circ\) · (g)(i) \(7.5\) m · (g)(ii) \(27.3^\circ\) · (h)(i) \(k=224\)

Exam tip (d) Check your line of intersection by substituting a second point, for example \(s=1\) giving \((8,\,5,\,7.5)\), into both plane equations.

Common error (f) Using cosine with the normal gives the angle between the beam and the normal (59.7°), not the angle between the beam and the plane.

More 3.18 questions in the question bank

Topic 4 · Statistics and probability

● home subchapter · ○ linked subchapter · last column: links to other topics

Setup	4.1	4.2	4.3	4.4	4.5	4.6	4.7	4.8	4.9	4.10	4.11	4.12	4.13	4.14	Other
Finish times in a city marathon	○	○	●												—
Placing air-quality sensors across a city	●	○	○												—
Cooking at altitude: boiling point and height above sea level			○	●						○					2.1
Forestry survey: tree height and trunk diameter			○	○						●					2.1
First serve, second serve: probability on the tennis court					○	●	○				○				—
Cycling or taking the bus to school: a student survey	○				○	○					●				—
A dice game at a charity fair: is it fair?					○	○	●	○							—
Quality control on an LED bulb production line						○	○	●							1.5
How full is the bottle? A juice bottling line						○		○	●			○			—
Airline checked baggage: finding the mean and spread								○	○		○	●			—
A screening test: what does a positive result mean?						○		○			○		●		—
Waiting for the bus: a continuous random variable							○							●	5.11 5.10

4.1 Sampling, bias and outliers · 4.2 Presentation of data · 4.3 Measures of central tendency and dispersionmerged group

+ Add to test✓ In your testCC-AA-41-42-43-127 marks~30 min

Finish times in a city marathon

4.3 Measures of central tendency and dispersion4.1 Sampling, bias and outliers4.2 Presentation of dataSL

The finishing times, \(t\) minutes, of the 400 runners who completed a city marathon are summarised in the table.

Finishing time \(t\) (minutes)	Number of runners
\(120\le t<150\)	20
\(150\le t<180\)	60
\(180\le t<210\)	120
\(210\le t<240\)	80
\(240\le t<270\)	60
\(270\le t<300\)	40
\(300\le t<330\)	12
\(330\le t<360\)	8

The winner finished in 2 hours 8 minutes and the slowest of the 400 runners finished in 5 hours 56 minutes.

(a)

Write down the modal class.

ROUTINENotes

Show solution

\(180\le t<210\)

Answer: \(180\le t<210\)

(1)

(b)

Use the mid-interval values to find an estimate of the mean finishing time and an estimate of the standard deviation of the finishing times.

ROUTINENotes

Show solution

mid-interval values \(135,\,165,\,195,\,\ldots,\,345\) used with the frequencies	(M1)	May be implied by a correct answer. Use of class boundaries scores M0.
mean \(=\frac{87\,240}{400}=218.1\) minutes	A1	Accept 218.
standard deviation \(=46.8\) minutes	A1	\(46.8336\ldots\) Accept 46.8 or 46.83. The value \(s_x=46.89\ldots\) (sample formula) scores A0.

Answer: mean \(\approx 218\) min, \(\sigma\approx 46.8\) min

(3)

(c)

On the grid in Figure 1, draw a cumulative frequency graph for the finishing times.

Figure 1: grid for the cumulative frequency graph

GRAPHNotes

Show solution

cumulative frequencies \(20,\,80,\,200,\,280,\,340,\,380,\,392,\,400\)	(M1)	Running totals of the frequencies; may be seen on the graph.
points plotted at the upper class boundaries, starting at \((120,\,0)\): \((150,20),\,(180,80),\,(210,200),\,\ldots,\,(360,400)\)	A1	Award A0 if the points are plotted at the mid-interval values. Allow \(\pm\) half a small square.
smooth increasing curve, or straight segments, through all the points	A1

Answer: graph through \((120,0),(150,20),(180,80),(210,200),(240,280),(270,340),(300,380),(330,392),(360,400)\)

(3)

(d)

Use your graph to estimate

the median finishing time;
the interquartile range of the finishing times.

GRAPHNotes

Show solution

median \(\approx 210\) minutes	A1	Accept 205 to 215. Reading at position \(\frac12\times400=200\).
readings at 100 and 300: \(Q_1\approx185\), \(Q_3\approx250\)	(M1)	Accept \(Q_1\) from 180 to 190 and \(Q_3\) from 245 to 255.
\(\text{IQR}\approx 250-185=65\) minutes	A1	Accept 60 to 70. FT from their graph.

Answer: median \(\approx 210\) min; IQR \(\approx 65\) min

(3)

(e)

The fastest \(10\%\) of the 400 runners are offered a place in the elite start group of next year's race. Use your graph to estimate the longest finishing time that earns a place.

INTERPRETNotes

Show solution

reading at cumulative frequency \(0.1\times400=40\)	(M1)	Reading at 360 (the slowest 10%) scores M0.
\(t\approx160\) minutes	A1	Accept 155 to 165. FT from their graph.

Answer: \(\approx160\) min (2 h 40 min)

(2)

(f)(i)

Using your answers to part (d), determine whether the winner's time and the slowest runner's time are outliers.

MULTINotes

Show solution

\(185-1.5\times65\) or \(250+1.5\times65\)	M1	Either fence calculation, with their quartiles.
lower fence \(87.5\), upper fence \(347.5\)	A1	FT from their quartiles.
winner: \(128>87.5\), so not an outlier; slowest: \(356>347.5\), so an outlier	R1	Both conclusions with the comparisons. Times must be in minutes (2 h 8 min \(=128\), 5 h 56 min \(=356\)). FT from their fences.

Answer: fences 87.5 and 347.5 min: 128 is not an outlier, 356 is an outlier

(3)

(f)(ii)

Use your graph to estimate the number of runners whose finishing times are outliers.

GRAPHNotes

Show solution

cumulative frequency read at \(t=347.5\): about 397	(M1)	FT their upper fence.
\(400-397\approx3\) runners	A1	Accept any whole number from 2 to 5 from a correct reading of their graph.

Answer: about 3 runners

(2)

(g)

In fact exactly three runners had finishing times that are outliers: 349, 352 and 356 minutes. The longest finishing time that is not an outlier was 344 minutes.

Using your answers to part (d), draw a box-and-whisker diagram of the finishing times on the scale in Figure 2.

Figure 2: scale for the box-and-whisker diagram

GRAPHNotes

Show solution

box from \(Q_1=185\) to \(Q_3=250\) with the median line at 210	A1	FT their values from (d).
whiskers from 128 to 344	A1	A whisker drawn to 356 scores A0.
the three outliers 349, 352 and 356 marked separately (for example as crosses)	A1

Answer: box 185 to 250, median 210; whiskers 128 to 344; crosses at 349, 352, 356

(3)

(h)

The results are published in hours, so every finishing time \(t\) minutes is converted to \(h=\dfrac{t}{60}\) hours.

Let \(\sigma\) be the standard deviation of the times in minutes. Show that the standard deviation of the times in hours is \(\dfrac{\sigma}{60}\).

No calculatorPROOFNotes

Show solution

\(\bar h=\frac{\bar t}{60}\), so each deviation is \(h-\bar h=\frac{t-\bar t}{60}\)	M1	Or: multiplying every value by \(k\) multiplies the standard deviation by \(\|k\|\), with \(k=\frac1{60}\).
variance in hours \(=\frac{\sum f(t-\bar t)^2}{3600\,n}=\frac{\sigma^2}{3600}\)	A1	Or \(\|k\|\sigma=\frac{1}{60}\sigma\) clearly stated.
standard deviation \(=\sqrt{\frac{\sigma^2}{3600}}=\frac{\sigma}{60}\)	AG

Answer: \(\dfrac{\sigma}{60}\) (about 0.781 hours)

(2)

(i)(i)

It is later found that 20 runners who finished after the 6-hour cut-off were left out of the table. All 20 had times in the class \(360\le t<420\). They are now included, so there are 420 runners.

Find a new estimate of the mean finishing time and a new estimate of the median finishing time.

MULTINotes

Show solution

\(\frac{87\,240+20\times390}{420}\)	M1	Or the GDC with the extra mid-interval value 390 and frequency 20.
mean \(\approx 226\) minutes	A1	\(226.285\ldots\) Accept 226.3.
median at position 210, in \(210\le t<240\): \(210+\frac{210-200}{80}\times30\approx 214\) minutes	A1	\(213.75\). Accept 213 to 215, including a reading from an extended graph.

Answer: mean \(\approx 226\) min; median \(\approx 214\) min

(3)

(i)(ii)

Explain why the estimate of the mean increases by more than the estimate of the median.

INTERPRETNotes

Show solution

the mean uses the size of every value, and the 20 added times are all far above the mean, so they pull it up strongly; the median depends only on position, which moves just 10 places within a crowded class

Must refer to both measures.

(1)

(j)

A journalist suggests removing the slowest runner's time from the original data because it is an outlier. Comment on this suggestion.

INTERPRETNotes

Show solution

the time is a genuine result (a real runner who finished), not a recording error, so it should be kept; removing it would misrepresent the field of runners

Accept an answer that says it should be removed only if it is shown to be an error. "It is an outlier so remove it" scores R0.

(1)

Show full solution

Working	Mark	Notes
Part (a)
\(180\le t<210\)	A1
Part (b)
mid-interval values \(135,\,165,\,195,\,\ldots,\,345\) used with the frequencies	(M1)	May be implied by a correct answer. Use of class boundaries scores M0.
mean \(=\frac{87\,240}{400}=218.1\) minutes	A1	Accept 218.
standard deviation \(=46.8\) minutes	A1	\(46.8336\ldots\) Accept 46.8 or 46.83. The value \(s_x=46.89\ldots\) (sample formula) scores A0.
Part (c)
cumulative frequencies \(20,\,80,\,200,\,280,\,340,\,380,\,392,\,400\)	(M1)	Running totals of the frequencies; may be seen on the graph.
points plotted at the upper class boundaries, starting at \((120,\,0)\): \((150,20),\,(180,80),\,(210,200),\,\ldots,\,(360,400)\)	A1	Award A0 if the points are plotted at the mid-interval values. Allow \(\pm\) half a small square.
smooth increasing curve, or straight segments, through all the points	A1
Part (d)
median \(\approx 210\) minutes	A1	Accept 205 to 215. Reading at position \(\frac12\times400=200\).
readings at 100 and 300: \(Q_1\approx185\), \(Q_3\approx250\)	(M1)	Accept \(Q_1\) from 180 to 190 and \(Q_3\) from 245 to 255.
\(\text{IQR}\approx 250-185=65\) minutes	A1	Accept 60 to 70. FT from their graph.
Part (e)
reading at cumulative frequency \(0.1\times400=40\)	(M1)	Reading at 360 (the slowest 10%) scores M0.
\(t\approx160\) minutes	A1	Accept 155 to 165. FT from their graph.
Part (f)(i)
\(185-1.5\times65\) or \(250+1.5\times65\)	M1	Either fence calculation, with their quartiles.
lower fence \(87.5\), upper fence \(347.5\)	A1	FT from their quartiles.
winner: \(128>87.5\), so not an outlier; slowest: \(356>347.5\), so an outlier	R1	Both conclusions with the comparisons. Times must be in minutes (2 h 8 min \(=128\), 5 h 56 min \(=356\)). FT from their fences.
Part (f)(ii)
cumulative frequency read at \(t=347.5\): about 397	(M1)	FT their upper fence.
\(400-397\approx3\) runners	A1	Accept any whole number from 2 to 5 from a correct reading of their graph.
Part (g)
box from \(Q_1=185\) to \(Q_3=250\) with the median line at 210	A1	FT their values from (d).
whiskers from 128 to 344	A1	A whisker drawn to 356 scores A0.
the three outliers 349, 352 and 356 marked separately (for example as crosses)	A1
Part (h)
\(\bar h=\frac{\bar t}{60}\), so each deviation is \(h-\bar h=\frac{t-\bar t}{60}\)	M1	Or: multiplying every value by \(k\) multiplies the standard deviation by \(\|k\|\), with \(k=\frac1{60}\).
variance in hours \(=\frac{\sum f(t-\bar t)^2}{3600\,n}=\frac{\sigma^2}{3600}\)	A1	Or \(\|k\|\sigma=\frac{1}{60}\sigma\) clearly stated.
standard deviation \(=\sqrt{\frac{\sigma^2}{3600}}=\frac{\sigma}{60}\)	AG
Part (i)(i)
\(\frac{87\,240+20\times390}{420}\)	M1	Or the GDC with the extra mid-interval value 390 and frequency 20.
mean \(\approx 226\) minutes	A1	\(226.285\ldots\) Accept 226.3.
median at position 210, in \(210\le t<240\): \(210+\frac{210-200}{80}\times30\approx 214\) minutes	A1	\(213.75\). Accept 213 to 215, including a reading from an extended graph.
Part (i)(ii)
the mean uses the size of every value, and the 20 added times are all far above the mean, so they pull it up strongly; the median depends only on position, which moves just 10 places within a crowded class	R1	Must refer to both measures.
Part (j)
the time is a genuine result (a real runner who finished), not a recording error, so it should be kept; removing it would misrepresent the field of runners	R1	Accept an answer that says it should be removed only if it is shown to be an error. "It is an outlier so remove it" scores R0.

Answers: (a) \(180\le t<210\) · (b) mean \(\approx 218\) min, \(\sigma\approx 46.8\) min · (c) graph through \((120,0),(150,20),(180,80),(210,200),(240,280),(270,340),(300,380),(330,392),(360,400)\) · (d) median \(\approx 210\) min; IQR \(\approx 65\) min · (e) \(\approx160\) min (2 h 40 min) · (f)(i) fences 87.5 and 347.5 min: 128 is not an outlier, 356 is an outlier · (f)(ii) about 3 runners · (g) box 185 to 250, median 210; whiskers 128 to 344; crosses at 349, 352, 356 · (h) \(\dfrac{\sigma}{60}\) (about 0.781 hours) · (i)(i) mean \(\approx 226\) min; median \(\approx 214\) min

Exam tip (b) Anything calculated from mid-interval values is an estimate. Check that \(n=400\) on the GDC screen before you copy the results.

Common error (f)(i) Convert both times to minutes before comparing with the fences: 2 hours 8 minutes is 128 minutes, not 208 or 2.08.

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Placing air-quality sensors across a city

4.1 Sampling, bias and outliers4.2 Presentation of data4.3 Measures of central tendency and dispersionSL

A city council will install air-quality sensors at 30 of its 600 bus stops. Each sensor records the daily mean concentration of fine particles (PM2.5) in micrograms per cubic metre (µg/m³). The bus stops lie in three districts: Riverside (252 stops), Old Town (186 stops) and Hillside (162 stops). Three plans are proposed.

Plan A: number the 600 stops from 1 to 600 in the order of the council's list, choose a random number from 1 to 20 as the first stop, and then take every 20th stop after it.
Plan B: use the 30 stops nearest to the council's maintenance depot, which is in the centre of Riverside, beside its busiest road.
Plan C: in each district, number the stops and use random numbers to choose a number of stops proportional to the number of stops in that district.

(a)

Name the sampling method used in Plan A and in Plan B.

CONCEPTNotes

Show solution

Plan A: systematic sampling	A1
Plan B: convenience sampling	A1

Answer: A: systematic; B: convenience

(2)

(b)

Explain why Plan B is likely to give a biased picture of the air quality at bus stops across the city.

INTERPRETNotes

Show solution

all 30 stops are close together in one part of Riverside, so most stops in the city (including all of Old Town and Hillside) have no chance of being chosen	R1
stops beside the busiest road are likely to have higher PM2.5 readings than typical stops, so the results would overestimate the city-wide level	R1	Must refer to the context (traffic, location). "Not random" alone scores R0.

(2)

(c)

Plan C is chosen. Find the number of sensors placed in each district.

ROUTINENotes

Show solution

\(\frac{252}{600}\times30\;(=12.6)\), \(\frac{186}{600}\times30\;(=9.3)\), \(\frac{162}{600}\times30\;(=8.1)\)	(M1)	At least one correct calculation.
Riverside 13, Old Town 9, Hillside 8	A1	All three, with total 30.

Answer: Riverside 13, Old Town 9, Hillside 8

(2)

(d)

On one day, the 13 sensors in Riverside recorded the following readings, in µg/m³.

360

Find

the mean of the readings;
the median of the readings;
the interquartile range of the readings.

ROUTINENotes

Show solution

mean \(=\frac{650}{13}=50\) µg/m³	A1
median \(=24\) µg/m³	A1
\(Q_1=20\), \(Q_3=30\), so \(\text{IQR}=10\) µg/m³	A1	Quartiles may be seen in (e).

Answer: mean 50; median 24; IQR 10 (µg/m³)

(3)

(e)

Show that the reading of 360 is an outlier.

No calculatorPROOFNotes

Show solution

upper fence \(=Q_3+1.5\times\text{IQR}=30+1.5\times10\)	M1	Must use the quartiles. A comparison with the mean scores M0.
\(=45\)	A1
\(360>45\), so 360 is an outlier	AG

Answer: upper fence 45; \(360>45\)

(2)

(f)(i)

The council checks the sensor's own log and finds that the reading was 36, and that 360 was a typing error. The value is corrected to 36.

Find the mean and the standard deviation of the corrected Riverside readings.

ROUTINENotes

Show solution

mean \(=\frac{326}{13}=25.1\) µg/m³	A1	\(25.0769\ldots\)
standard deviation \(=5.92\) µg/m³	A1	\(5.9155\ldots\) The sample value \(6.157\ldots\) scores A0.

Answer: mean 25.1 µg/m³; \(\sigma=5.92\) µg/m³

(2)

(f)(ii)

The median and the interquartile range of the Riverside readings are not changed by the correction. Explain why the median and interquartile range are unchanged, while the mean is roughly halved.

INTERPRETNotes

Show solution

the median and quartiles depend only on the positions of the values in order; 36 is still the largest reading, so every value keeps its position and the median, \(Q_1\) and \(Q_3\) stay the same	R1
the mean uses the size of every value: the total falls by \(360-36=324\), so the mean falls by \(\frac{324}{13}\approx24.9\)	R1	Accept a clear statement that the mean (unlike the median) is strongly affected by one extreme value.

(2)

(g)

On the same day, the 8 sensors in Hillside recorded the following readings, in µg/m³.

On Figure 1, draw box-and-whisker diagrams for the corrected Riverside readings and for the Hillside readings.

Figure 1: scale for the two box-and-whisker diagrams

GRAPHNotes

Show solution

Riverside box from 20 to 30 with median 24, whiskers from 16 to 36	A1	Values from (d) with 36 in place of 360. Neither data set has an outlier (fences 5 and 45; 4.75 and 22.75).
Hillside: minimum 9, \(Q_1=11.5\), median 14.5, \(Q_3=16\), maximum 19	(M1)	Five-number summary found; may be seen on the diagram.
Hillside box from 11.5 to 16 with median 14.5, whiskers from 9 to 19	A1	FT their summary values.

Answer: Riverside 16, 20, 24, 30, 36; Hillside 9, 11.5, 14.5, 16, 19

(3)

(h)

Use your diagrams to compare the PM2.5 readings in the two districts.

INTERPRETNotes

Show solution

centre: the median for Hillside (14.5) is lower than for Riverside (24), so the air at Hillside stops is typically cleaner	R1	Must include the comparison in context.
spread: the IQR for Hillside (4.5) is smaller than for Riverside (10), so Hillside readings are more consistent from stop to stop	R1	Accept the range (10 compared with 20).

(2)

(i)

All the readings were taken on a single weekday in winter. Suggest one reason why they may not be reliable as a measure of typical air quality at these bus stops.

INTERPRETNotes

Show solution

PM2.5 levels vary with weather, season and traffic (for example weekends or summer), so one day may not represent typical conditions

Must give a specific reason.

(1)

Show full solution

Working	Mark	Notes
Part (a)
Plan A: systematic sampling	A1
Plan B: convenience sampling	A1
Part (b)
all 30 stops are close together in one part of Riverside, so most stops in the city (including all of Old Town and Hillside) have no chance of being chosen	R1
stops beside the busiest road are likely to have higher PM2.5 readings than typical stops, so the results would overestimate the city-wide level	R1	Must refer to the context (traffic, location). "Not random" alone scores R0.
Part (c)
\(\frac{252}{600}\times30\;(=12.6)\), \(\frac{186}{600}\times30\;(=9.3)\), \(\frac{162}{600}\times30\;(=8.1)\)	(M1)	At least one correct calculation.
Riverside 13, Old Town 9, Hillside 8	A1	All three, with total 30.
Part (d)
mean \(=\frac{650}{13}=50\) µg/m³	A1
median \(=24\) µg/m³	A1
\(Q_1=20\), \(Q_3=30\), so \(\text{IQR}=10\) µg/m³	A1	Quartiles may be seen in (e).
Part (e)
upper fence \(=Q_3+1.5\times\text{IQR}=30+1.5\times10\)	M1	Must use the quartiles. A comparison with the mean scores M0.
\(=45\)	A1
\(360>45\), so 360 is an outlier	AG
Part (f)(i)
mean \(=\frac{326}{13}=25.1\) µg/m³	A1	\(25.0769\ldots\)
standard deviation \(=5.92\) µg/m³	A1	\(5.9155\ldots\) The sample value \(6.157\ldots\) scores A0.
Part (f)(ii)
the median and quartiles depend only on the positions of the values in order; 36 is still the largest reading, so every value keeps its position and the median, \(Q_1\) and \(Q_3\) stay the same	R1
the mean uses the size of every value: the total falls by \(360-36=324\), so the mean falls by \(\frac{324}{13}\approx24.9\)	R1	Accept a clear statement that the mean (unlike the median) is strongly affected by one extreme value.
Part (g)
Riverside box from 20 to 30 with median 24, whiskers from 16 to 36	A1	Values from (d) with 36 in place of 360. Neither data set has an outlier (fences 5 and 45; 4.75 and 22.75).
Hillside: minimum 9, \(Q_1=11.5\), median 14.5, \(Q_3=16\), maximum 19	(M1)	Five-number summary found; may be seen on the diagram.
Hillside box from 11.5 to 16 with median 14.5, whiskers from 9 to 19	A1	FT their summary values.
Part (h)
centre: the median for Hillside (14.5) is lower than for Riverside (24), so the air at Hillside stops is typically cleaner	R1	Must include the comparison in context.
spread: the IQR for Hillside (4.5) is smaller than for Riverside (10), so Hillside readings are more consistent from stop to stop	R1	Accept the range (10 compared with 20).
Part (i)
PM2.5 levels vary with weather, season and traffic (for example weekends or summer), so one day may not represent typical conditions	R1	Must give a specific reason.

Answers: (a) A: systematic; B: convenience · (c) Riverside 13, Old Town 9, Hillside 8 · (d) mean 50; median 24; IQR 10 (µg/m³) · (e) upper fence 45; \(360>45\) · (f)(i) mean 25.1 µg/m³; \(\sigma=5.92\) µg/m³ · (g) Riverside 16, 20, 24, 30, 36; Hillside 9, 11.5, 14.5, 16, 19

Common error (c) After rounding the stratum sizes, add them up. Here \(13+9+8=30\), so no adjustment is needed; if the total were 29 or 31 you would change the stratum with the largest rounding remainder.

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4.4 Correlation and regression of y on x · 4.10 Regression line of x on ymerged group

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Cooking at altitude: boiling point and height above sea level

4.4 Correlation and regression of y on x4.10 Regression line of x on y4.3 Measures of central tendency and dispersion2.1 Equations of straight lines, parallel and perpendicular linesSL

Water boils at a lower temperature at high altitude, which makes cooking slower in the mountains. Members of a hiking club measured the temperature, \(y\) °C, at which water boils at eight campsites at different altitudes, \(x\) km above sea level.

Altitude \(x\) (km)	0.3	0.8	1.5	2.0	2.6	3.1	3.7	4.4
Boiling point \(y\) (°C)	99.1	97.0	95.3	95.8	91.7	90.2	87.6	85.7

(a)

On the grid in Figure 1, draw a scatter diagram of the data.

Figure 1: grid for the scatter diagram

GRAPHNotes

Show solution

all eight points plotted correctly

A1 for six or seven correct points. Allow \(\pm\) half a small square.

Answer: eight points plotted

(2)

(b)

Find the value of Pearson's product-moment correlation coefficient, \(r\), and describe the correlation in context.

ROUTINENotes

Show solution

\(r=-0.984\)	A1	\(-0.98375\ldots\)
strong negative linear correlation: the higher the campsite, the lower the boiling point tends to be	A1	Need strength, direction and the context.

Answer: \(r=-0.984\); strong negative linear correlation

(2)

(c)(i)

Find the equation of the regression line of \(y\) on \(x\), in the form \(y=ax+b\).

ROUTINENotes

Show solution

\(a=-3.30\)	A1	\(-3.29545\ldots\)
\(b=100.4\)	A1	\(100.3795\ldots\) Accept 100 (3 s.f.).

Answer: \(y=-3.30x+100.4\)

(2)

(c)(ii)

Interpret, in context, the gradient and the \(y\)-intercept of this line.

INTERPRETNotes

Show solution

gradient: the boiling point is predicted to fall by about 3.3 °C for each extra kilometre of altitude	R1	"Decreases by 3.3" without the rate per km scores R0.
intercept: the predicted boiling point at sea level (\(x=0\)) is about 100.4 °C	R1	FT their \(b\). Accept a remark that \(x=0\) is just outside the data, so this is a small extrapolation.

(2)

(d)(i)

Use your regression line to estimate the boiling point of water at a campsite 2.4 km above sea level.

ROUTINENotes

Show solution

\(y=-3.29545\ldots(2.4)+100.3795\ldots\)	(M1)
\(92.5\) °C	A1	\(92.470\ldots\) Using \(-3.30\) and 100.4 gives 92.48, which also rounds to 92.5. FT from (c)(i).

Answer: \(92.5\) °C

(2)

(d)(ii)

A climber uses the same line to predict the boiling point of water at the top of a mountain 8.8 km above sea level. Find this prediction and comment on its reliability.

INTERPRETNotes

Show solution

\(71.4\) °C	A1	\(71.379\ldots\) FT from (c)(i).
unreliable: 8.8 km is far outside the measured altitudes (0.3 to 4.4 km), so this is extrapolation and the linear pattern may not continue	R1	A strong \(r\) does not justify extrapolation; an answer that calls it reliable because \(\|r\|\) is large scores R0.

Answer: \(71.4\) °C, unreliable (extrapolation)

(2)

(e)(i)

At another campsite, a hiker finds that water boils at 88.0 °C. The hiker wants to estimate the altitude of the campsite.

Explain why the regression line found in part (c)(i) should not be used for this estimate.

CONCEPTNotes

Show solution

the line of \(y\) on \(x\) minimises the vertical distances, so it is designed to predict \(y\) from \(x\); here \(x\) is estimated from a known \(y\), which needs the line of \(x\) on \(y\)

(1)

(e)(ii)

Find the equation of the regression line of \(x\) on \(y\), in the form \(x=cy+d\), and use it to estimate the altitude of this campsite.

ROUTINENotes

Show solution

\(x=-0.294y+29.6\)	A1	\(c=-0.293670\ldots\), \(d=29.5526\ldots\) Both values required.
\(x=-0.293670\ldots(88.0)+29.5526\ldots\)	(M1)	Rearranging the line of \(y\) on \(x\) (giving 3.76) scores M0A0.
\(3.71\) km	A1	\(3.7096\ldots\) Accept 3.73 from the 3 s.f. coefficients. FT their line.

Answer: \(x=-0.294y+29.6\); altitude \(\approx 3.71\) km

(3)

(f)

Show that both regression lines pass through the point \((\bar x,\bar y)\).

PROOFNotes

Show solution

\(\bar x=\frac{18.4}{8}=2.3\) and \(\bar y=\frac{742.4}{8}=92.8\)	B1	Both means.
substituting: \(-3.29545\ldots(2.3)+100.3795\ldots\) and \(-0.293670\ldots(92.8)+29.5526\ldots\)	M1	Substitution into both lines. Unrounded coefficients needed for an exact match; with 3 s.f. values accept answers that round to 92.8 and 2.3.
\(=92.8=\bar y\) and \(=2.3=\bar x\)	A1
so \((2.3,\,92.8)\) lies on both lines	AG

Answer: \((\bar x,\bar y)=(2.3,\,92.8)\) satisfies both equations

(3)

(g)

The club's report gives each boiling point in degrees Fahrenheit, \(F\), where \(F=1.8y+32\).

Write down

the value of the correlation coefficient between \(x\) and \(F\);
the gradient of the regression line of \(F\) on \(x\).

CONCEPTNotes

Show solution

\(r=-0.984\) (unchanged)	A1	A linear change \(F=1.8y+32\) with a positive multiplier leaves \(r\) unchanged.
gradient \(=1.8\times(-3.29545\ldots)=-5.93\) °F per km	A1	\(-5.9318\ldots\) Accept recalculation with the GDC. FT from (c)(i).

Answer: (i) \(-0.984\); (ii) \(-5.93\)

(2)

(h)

The club later discovers that the thermometer used at the campsite 2.0 km above sea level was faulty, and that reading is removed.

Find the new value of \(r\) and explain why it has changed in this way.

INTERPRETNotes

Show solution

\(r=-0.999\)	A1	\(-0.99853\ldots\)
\(\|r\|\) increases (closer to \(-1\)) because the faulty reading (95.8 °C) lay well above the line formed by the other seven points, so without it the remaining points are much closer to a straight line	R1	Accept reference to the scatter diagram.

Answer: \(r=-0.999\); the correlation becomes stronger

(2)

Show full solution

Working	Mark	Notes
Part (a)
all eight points plotted correctly	A2	A1 for six or seven correct points. Allow \(\pm\) half a small square.
Part (b)
\(r=-0.984\)	A1	\(-0.98375\ldots\)
strong negative linear correlation: the higher the campsite, the lower the boiling point tends to be	A1	Need strength, direction and the context.
Part (c)(i)
\(a=-3.30\)	A1	\(-3.29545\ldots\)
\(b=100.4\)	A1	\(100.3795\ldots\) Accept 100 (3 s.f.).
Part (c)(ii)
gradient: the boiling point is predicted to fall by about 3.3 °C for each extra kilometre of altitude	R1	"Decreases by 3.3" without the rate per km scores R0.
intercept: the predicted boiling point at sea level (\(x=0\)) is about 100.4 °C	R1	FT their \(b\). Accept a remark that \(x=0\) is just outside the data, so this is a small extrapolation.
Part (d)(i)
\(y=-3.29545\ldots(2.4)+100.3795\ldots\)	(M1)
\(92.5\) °C	A1	\(92.470\ldots\) Using \(-3.30\) and 100.4 gives 92.48, which also rounds to 92.5. FT from (c)(i).
Part (d)(ii)
\(71.4\) °C	A1	\(71.379\ldots\) FT from (c)(i).
unreliable: 8.8 km is far outside the measured altitudes (0.3 to 4.4 km), so this is extrapolation and the linear pattern may not continue	R1	A strong \(r\) does not justify extrapolation; an answer that calls it reliable because \(\|r\|\) is large scores R0.
Part (e)(i)
the line of \(y\) on \(x\) minimises the vertical distances, so it is designed to predict \(y\) from \(x\); here \(x\) is estimated from a known \(y\), which needs the line of \(x\) on \(y\)	R1
Part (e)(ii)
\(x=-0.294y+29.6\)	A1	\(c=-0.293670\ldots\), \(d=29.5526\ldots\) Both values required.
\(x=-0.293670\ldots(88.0)+29.5526\ldots\)	(M1)	Rearranging the line of \(y\) on \(x\) (giving 3.76) scores M0A0.
\(3.71\) km	A1	\(3.7096\ldots\) Accept 3.73 from the 3 s.f. coefficients. FT their line.
Part (f)
\(\bar x=\frac{18.4}{8}=2.3\) and \(\bar y=\frac{742.4}{8}=92.8\)	B1	Both means.
substituting: \(-3.29545\ldots(2.3)+100.3795\ldots\) and \(-0.293670\ldots(92.8)+29.5526\ldots\)	M1	Substitution into both lines. Unrounded coefficients needed for an exact match; with 3 s.f. values accept answers that round to 92.8 and 2.3.
\(=92.8=\bar y\) and \(=2.3=\bar x\)	A1
so \((2.3,\,92.8)\) lies on both lines	AG
Part (g)
\(r=-0.984\) (unchanged)	A1	A linear change \(F=1.8y+32\) with a positive multiplier leaves \(r\) unchanged.
gradient \(=1.8\times(-3.29545\ldots)=-5.93\) °F per km	A1	\(-5.9318\ldots\) Accept recalculation with the GDC. FT from (c)(i).
Part (h)
\(r=-0.999\)	A1	\(-0.99853\ldots\)
\(\|r\|\) increases (closer to \(-1\)) because the faulty reading (95.8 °C) lay well above the line formed by the other seven points, so without it the remaining points are much closer to a straight line	R1	Accept reference to the scatter diagram.

Answers: (a) eight points plotted · (b) \(r=-0.984\); strong negative linear correlation · (c)(i) \(y=-3.30x+100.4\) · (d)(i) \(92.5\) °C · (d)(ii) \(71.4\) °C, unreliable (extrapolation) · (e)(ii) \(x=-0.294y+29.6\); altitude \(\approx 3.71\) km · (f) \((\bar x,\bar y)=(2.3,\,92.8)\) satisfies both equations · (g) (i) \(-0.984\); (ii) \(-5.93\) · (h) \(r=-0.999\); the correlation becomes stronger

Common error (e)(ii) Solving \(88=-3.30x+100.4\) for \(x\) uses the wrong line. Put the variable you are estimating on the left: \(x=cy+d\).

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Forestry survey: tree height and trunk diameter

4.10 Regression line of x on y4.4 Correlation and regression of y on x4.3 Measures of central tendency and dispersion2.1 Equations of straight lines, parallel and perpendicular linesSL

A forester measures the trunk diameter at chest height, \(x\) cm, and the height, \(y\) m, of 10 spruce trees in a mature plantation. The results are shown in the table and in Figure 1.

Diameter \(x\) (cm)	18	22	25	27	29	31	33	35	38	42
Height \(y\) (m)	14.6	15.9	18.1	17.0	19.8	21.4	20.2	22.1	23.0	27.9

Figure 1: scatter diagram for the ten mature trees

(a)

Find the mean and the standard deviation of the diameters, and the mean and the standard deviation of the heights.

ROUTINENotes

Show solution

\(\bar x=30\) cm and \(\bar y=20\) m	A1	Both means.
\(\sigma_x=6.97\) cm and \(\sigma_y=3.69\) m	A1	\(6.9713\ldots\), \(3.6937\ldots\) Sample values (7.35, 3.89) score A0.

Answer: \(\bar x=30\) cm, \(\sigma_x=6.97\) cm; \(\bar y=20\) m, \(\sigma_y=3.69\) m

(2)

(b)(i)

Find the value of the correlation coefficient, \(r\).

ROUTINENotes

Show solution

\(r=0.962\)

\(0.96152\ldots\)

Answer: \(r=0.962\)

(1)

(b)(ii)

Find the equation of the regression line of \(y\) on \(x\), in the form \(y=mx+k\).

ROUTINENotes

Show solution

\(m=0.509\)	A1	\(0.509465\ldots\)
\(k=4.72\)	A1	\(4.71604\ldots\)

Answer: \(y=0.509x+4.72\)

(2)

(c)(i)

Another tree in the plantation is measured as 18.0 m tall, but its diameter is not recorded. The forester wants to estimate its diameter.

Explain why the line found in part (b)(ii) should not be used for this estimate.

CONCEPTNotes

Show solution

the line of \(y\) on \(x\) minimises vertical distances and is the best line for predicting height from diameter; to estimate the diameter from a known height the line of \(x\) on \(y\) (which minimises horizontal distances) is needed

(1)

(c)(ii)

Find the equation of the regression line of \(x\) on \(y\), in the form \(x=cy+d\). Use it to estimate the diameter of this tree and comment on the reliability of your estimate.

MULTINotes

Show solution

\(x=1.81y-6.29\)	A1	\(c=1.81471\ldots\), \(d=-6.29434\ldots\) Both values required.
\(x=1.81471\ldots(18.0)-6.29434\ldots\)	(M1)	Rearranging the line of \(y\) on \(x\) (giving 26.1) scores M0A0.
\(26.4\) cm	A1	\(26.3705\ldots\) FT their line.
reliable: 18.0 m lies within the measured heights (14.6 m to 27.9 m), so this is interpolation, and \(r=0.962\) shows strong correlation	R1	Both reasons needed.

Answer: \(x=1.81y-6.29\); diameter \(\approx 26.4\) cm; reliable

(4)

(d)

Show that the line of \(x\) on \(y\) can be written as \(y=0.551x+3.47\), with the coefficients given correct to 3 significant figures.

PROOFNotes

Show solution

\(y=\frac{x+6.29434\ldots}{1.81471\ldots}\)	M1	Correct rearrangement to make \(y\) the subject.
\(y=0.55105\ldots x+3.46849\ldots\)	A1	Values to at least 4 s.f. must be seen. Using the 3 s.f. values 1.81 and \(-6.29\) gives \(0.552x+3.48\), which does not establish the result: A0.
\(y=0.551x+3.47\)	AG

Answer: \(y=0.551x+3.47\)

(2)

(e)(i)

On Figure 1, draw the regression line of \(y\) on \(x\) and the regression line of \(x\) on \(y\). Label each line.

GRAPHNotes

Show solution

line of \(y\) on \(x\) drawn through, for example, \((15,\,12.4)\) and \((45,\,27.6)\)	A1	Allow \(\pm\) half a small square. FT from (b)(ii).
line of \(x\) on \(y\) drawn through, for example, \((15,\,11.7)\) and \((45,\,28.3)\)	A1	FT from (d).
both lines pass through \((30,\,20)\) and are correctly labelled	A1

Answer: two lines crossing at \((30,\,20)\); the line of \(x\) on \(y\) is the steeper

(3)

(e)(ii)

Write down the coordinates of the point where the two lines meet. Explain why the angle between the two lines is small.

INTERPRETNotes

Show solution

\((30,\,20)\), the mean point \((\bar x,\bar y)\)	A1	FT from (a).
\(\|r\|=0.962\) is close to 1, so the points lie close to a straight line and the two regression lines nearly coincide (they would be the same line if \(r=1\))	R1	Accept \(m\times c=r^2=0.925\), close to 1.

Answer: \((30,\,20)\); \(r\) is close to 1

(2)

(f)(i)

The forester also measures five trees in a younger plantation nearby.

Diameter \(x\) (cm)	12	14	15	17	19
Height \(y\) (m)	13.8	15.1	15.4	16.6	17.5

Figure 2 shows all fifteen trees.

Find the value of \(r\) for all fifteen trees together.

Figure 2: all fifteen trees

ROUTINENotes

Show solution

\(r=0.940\)

\(0.93993\ldots\)

Answer: \(r=0.940\)

(1)

(f)(ii)

A colleague says that, because \(r\) is still close to 1, a single regression line for all fifteen trees should be used to estimate the height of any tree in either plantation. Discuss this claim.

INTERPRETNotes

Show solution

the younger trees all lie about 3 m above the line of \(y\) on \(x\) for the mature trees (from (b)(ii) or (e)(i)), so the two plantations follow different relationships; a single line would underestimate the heights of young trees and distort the estimates for mature trees	R1	Must refer to Figure 2 or to the position of the young trees. For example, at \(x=15\) the mature line gives 12.4 m but the young tree is 15.4 m.
so separate regression lines should be used for each plantation (the high \(r\) partly comes from the wider range of diameters, not from one model fitting both groups)	R1	Accept the point that the young trees' diameters (12 to 19 cm) are mostly below the mature range (18 to 42 cm), so using the mature line for them would be extrapolation.

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(\bar x=30\) cm and \(\bar y=20\) m	A1	Both means.
\(\sigma_x=6.97\) cm and \(\sigma_y=3.69\) m	A1	\(6.9713\ldots\), \(3.6937\ldots\) Sample values (7.35, 3.89) score A0.
Part (b)(i)
\(r=0.962\)	A1	\(0.96152\ldots\)
Part (b)(ii)
\(m=0.509\)	A1	\(0.509465\ldots\)
\(k=4.72\)	A1	\(4.71604\ldots\)
Part (c)(i)
the line of \(y\) on \(x\) minimises vertical distances and is the best line for predicting height from diameter; to estimate the diameter from a known height the line of \(x\) on \(y\) (which minimises horizontal distances) is needed	R1
Part (c)(ii)
\(x=1.81y-6.29\)	A1	\(c=1.81471\ldots\), \(d=-6.29434\ldots\) Both values required.
\(x=1.81471\ldots(18.0)-6.29434\ldots\)	(M1)	Rearranging the line of \(y\) on \(x\) (giving 26.1) scores M0A0.
\(26.4\) cm	A1	\(26.3705\ldots\) FT their line.
reliable: 18.0 m lies within the measured heights (14.6 m to 27.9 m), so this is interpolation, and \(r=0.962\) shows strong correlation	R1	Both reasons needed.
Part (d)
\(y=\frac{x+6.29434\ldots}{1.81471\ldots}\)	M1	Correct rearrangement to make \(y\) the subject.
\(y=0.55105\ldots x+3.46849\ldots\)	A1	Values to at least 4 s.f. must be seen. Using the 3 s.f. values 1.81 and \(-6.29\) gives \(0.552x+3.48\), which does not establish the result: A0.
\(y=0.551x+3.47\)	AG
Part (e)(i)
line of \(y\) on \(x\) drawn through, for example, \((15,\,12.4)\) and \((45,\,27.6)\)	A1	Allow \(\pm\) half a small square. FT from (b)(ii).
line of \(x\) on \(y\) drawn through, for example, \((15,\,11.7)\) and \((45,\,28.3)\)	A1	FT from (d).
both lines pass through \((30,\,20)\) and are correctly labelled	A1
Part (e)(ii)
\((30,\,20)\), the mean point \((\bar x,\bar y)\)	A1	FT from (a).
\(\|r\|=0.962\) is close to 1, so the points lie close to a straight line and the two regression lines nearly coincide (they would be the same line if \(r=1\))	R1	Accept \(m\times c=r^2=0.925\), close to 1.
Part (f)(i)
\(r=0.940\)	A1	\(0.93993\ldots\)
Part (f)(ii)
the younger trees all lie about 3 m above the line of \(y\) on \(x\) for the mature trees (from (b)(ii) or (e)(i)), so the two plantations follow different relationships; a single line would underestimate the heights of young trees and distort the estimates for mature trees	R1	Must refer to Figure 2 or to the position of the young trees. For example, at \(x=15\) the mature line gives 12.4 m but the young tree is 15.4 m.
so separate regression lines should be used for each plantation (the high \(r\) partly comes from the wider range of diameters, not from one model fitting both groups)	R1	Accept the point that the young trees' diameters (12 to 19 cm) are mostly below the mature range (18 to 42 cm), so using the mature line for them would be extrapolation.

Answers: (a) \(\bar x=30\) cm, \(\sigma_x=6.97\) cm; \(\bar y=20\) m, \(\sigma_y=3.69\) m · (b)(i) \(r=0.962\) · (b)(ii) \(y=0.509x+4.72\) · (c)(ii) \(x=1.81y-6.29\); diameter \(\approx 26.4\) cm; reliable · (d) \(y=0.551x+3.47\) · (e)(i) two lines crossing at \((30,\,20)\); the line of \(x\) on \(y\) is the steeper · (e)(ii) \((30,\,20)\); \(r\) is close to 1 · (f)(i) \(r=0.940\)

Exam tip (f)(ii) A large \(|r|\) measures how close the points are to one straight line overall; it does not show that every group in the data follows the same line. Always look at the scatter diagram.

More 4.10 questions in the question bank

4.5 Sample spaces and probability · 4.6 Combined, conditional and independent events · 4.11 Conditional probability and independencemerged group

+ Add to test✓ In your testCC-AA-45-46-411-127 marks~30 min

First serve, second serve: probability on the tennis court

4.6 Combined, conditional and independent events4.5 Sample spaces and probability4.11 Conditional probability and independence4.7 Discrete random variablesSL

In tennis, the player who serves has two attempts to start each point. If the first serve lands in, the point is played. If the first serve is out, the player takes a second serve. If the second serve is also out, this is a double fault and the server loses the point.

Data from Elena's recent matches give the following model for each point she serves.

Her first serve is in with probability 0.62. When her first serve is in, she wins the point with probability 0.74.
When her first serve is out, her second serve is in with probability 0.91. When her second serve is in, she wins the point with probability 0.52.

Assume that the outcomes of different points are independent.

Figure 1: tree diagram for one point served by Elena

(a)

Copy and complete the tree diagram in Figure 1 by writing down the four missing probabilities.

No calculatorGRAPHNotes

Show solution

\(0.38\) on the first-serve “out” branch	A1
\(0.26\) and \(0.48\) on the two “point lost” branches	A1	Both required, each on the correct branch.
\(0.09\) on the second-serve “out” branch	A1

Answer: first serve out \(0.38\); lost after first serve in \(0.26\); second serve out \(0.09\); lost after second serve in \(0.48\)

(3)

(b)

Find the probability that Elena serves a double fault on a randomly chosen point.

ROUTINENotes

Show solution

\(0.38\times0.09\)	M1	Allow ECF from (a).
\(=0.0342\)	A1

Answer: \(0.0342\)

(2)

(c)

Show that the probability that Elena wins a randomly chosen point that she serves is \(0.639\), correct to three significant figures.

MULTINotes

Show solution

\(0.62\times0.74\) or \(0.38\times0.91\times0.52\)	M1	Multiplying along one path that ends in “point won”.
\(0.62\times0.74+0.38\times0.91\times0.52\)	M1	Adding the two “point won” paths.
\(=0.4588+0.179816=0.638616\)	A1	Must see a value to at least 4 s.f. (e.g. 0.6386) before the given answer.
\(\approx0.639\)	AG

Answer: \(0.638616\approx0.639\)

(3)

(d)

Given that Elena wins a point she serves, find the probability that her first serve was in.

ROUTINENotes

Show solution

\(\frac{0.4588}{0.638616}\)	M1	Numerator must be the single path; denominator the answer to (c). Allow 0.639 as denominator.
\(=0.718\)	A1	Accept 0.718 from 0.4588/0.639 (0.71800…).

Answer: \(0.718\)

(2)

(e)

Let \(F\) be the event “Elena's first serve is in” and \(W\) the event “Elena wins the point”. Determine whether \(F\) and \(W\) are independent. Justify your answer.

CONCEPTNotes

Show solution

\(P(F\mid W)=0.718\) but \(P(F)=0.62\)	R1	Or compare \(P(F\cap W)=0.4588\) with \(P(F)P(W)=0.62\times0.639=0.396\), or \(P(W\mid F)=0.74\) with \(P(W)=0.639\). Allow ECF from (d).
these are not equal, so \(F\) and \(W\) are not independent	A1	Do not award A1R0. A conclusion with no numerical comparison scores 0.

Answer: not independent, since \(P(F\mid W)\ne P(F)\)

(2)

(f)

In a long match Elena serves 80 points. Find the expected number of double faults she serves.

ROUTINENotes

Show solution

\(80\times0.0342\)	M1	Allow ECF from (b).
\(=2.74\)	A1	Accept 2.736. Do not penalise an answer that is not a whole number.

Answer: \(2.74\)

(2)

(g)

In one service game Elena serves 6 points. Find the probability that she serves at least one double fault in this game.

MULTINotes

Show solution

\(P(\text{no double fault on a point})=1-0.0342=0.9658\)	(M1)	Allow ECF from (b).
\(1-0.9658^{6}\)	M1	Use of the complement of “none in six points”.
\(=0.188\)	A1	\(0.18843\ldots\)

Answer: \(0.188\)

(3)

(h)(i)

Let \(X\) be the number of points Elena wins out of the next two points that she serves.

Copy and complete the table to show the probability distribution of \(X\). Give the probabilities correct to three significant figures.

\(x\)	0	1	2
\(P(X=x)\)

MULTINotes

Show solution

\(P(X=0)=(1-0.638616)^{2}=0.131\)	A1	Accept 0.130 (from 0.361 squared). Allow ECF from (c).
\(P(X=1)=2\times0.638616\times0.361384=0.462\)	A1	The factor 2 (two orders: won-lost and lost-won) is required.
\(P(X=2)=0.638616^{2}=0.408\)	A1	Award A1 for any one correct value found by subtraction from 1 after the other two are correct.

Answer: \(0.131,\ 0.462,\ 0.408\)

(3)

(h)(ii)

Find \(E(X)\).

ROUTINENotes

Show solution

\(0\times0.131+1\times0.462+2\times0.408\)	M1	Allow ECF from (h)(i).
\(=1.28\)	A1	\(1.277\ldots\) Accept 1.278 from the rounded table values.

Answer: \(1.28\)

(2)

(i)(i)

Elena's coach suggests a faster first serve. With the faster serve, her first serve is in with probability 0.55, and when it is in she wins the point with probability 0.80. Her second serve does not change.

Find the new value of \(E(X)\).

MULTINotes

Show solution

\(P(\text{win})=0.55\times0.80+0.45\times0.91\times0.52\)	M1	Correct structure with the new first-stage values and the unchanged second serve.
\(=0.65294\)	A1	Accept 0.653.
\(E(X)=2\times0.65294=1.31\)	A1	Accept a new distribution table (0.120, 0.453, 0.426) followed by the sum of \(xP(X=x)\).

Answer: \(1.31\)

(3)

(i)(ii)

With the faster first serve, Elena serves more double faults. Discuss whether she should follow her coach's advice.

INTERPRETNotes

Show solution

double-fault probability rises from \(0.0342\) to \(0.45\times0.09=0.0405\), but the expected number of points won out of two rises from \(1.28\) to \(1.31\) (\(P(\text{win})\) from \(0.639\) to \(0.653\))	R1	Both comparisons, with values, are needed. Allow ECF.
so yes, she should follow the advice: more double faults, but the gain on the points where the first serve is in outweighs them	R1	Accept a sensible caveat (e.g. the 0.55 and 0.80 are estimates, or the model may differ against stronger opponents) in addition to a conclusion.

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(0.38\) on the first-serve “out” branch	A1
\(0.26\) and \(0.48\) on the two “point lost” branches	A1	Both required, each on the correct branch.
\(0.09\) on the second-serve “out” branch	A1
Part (b)
\(0.38\times0.09\)	M1	Allow ECF from (a).
\(=0.0342\)	A1
Part (c)
\(0.62\times0.74\) or \(0.38\times0.91\times0.52\)	M1	Multiplying along one path that ends in “point won”.
\(0.62\times0.74+0.38\times0.91\times0.52\)	M1	Adding the two “point won” paths.
\(=0.4588+0.179816=0.638616\)	A1	Must see a value to at least 4 s.f. (e.g. 0.6386) before the given answer.
\(\approx0.639\)	AG
Part (d)
\(\frac{0.4588}{0.638616}\)	M1	Numerator must be the single path; denominator the answer to (c). Allow 0.639 as denominator.
\(=0.718\)	A1	Accept 0.718 from 0.4588/0.639 (0.71800…).
Part (e)
\(P(F\mid W)=0.718\) but \(P(F)=0.62\)	R1	Or compare \(P(F\cap W)=0.4588\) with \(P(F)P(W)=0.62\times0.639=0.396\), or \(P(W\mid F)=0.74\) with \(P(W)=0.639\). Allow ECF from (d).
these are not equal, so \(F\) and \(W\) are not independent	A1	Do not award A1R0. A conclusion with no numerical comparison scores 0.
Part (f)
\(80\times0.0342\)	M1	Allow ECF from (b).
\(=2.74\)	A1	Accept 2.736. Do not penalise an answer that is not a whole number.
Part (g)
\(P(\text{no double fault on a point})=1-0.0342=0.9658\)	(M1)	Allow ECF from (b).
\(1-0.9658^{6}\)	M1	Use of the complement of “none in six points”.
\(=0.188\)	A1	\(0.18843\ldots\)
Part (h)(i)
\(P(X=0)=(1-0.638616)^{2}=0.131\)	A1	Accept 0.130 (from 0.361 squared). Allow ECF from (c).
\(P(X=1)=2\times0.638616\times0.361384=0.462\)	A1	The factor 2 (two orders: won-lost and lost-won) is required.
\(P(X=2)=0.638616^{2}=0.408\)	A1	Award A1 for any one correct value found by subtraction from 1 after the other two are correct.
Part (h)(ii)
\(0\times0.131+1\times0.462+2\times0.408\)	M1	Allow ECF from (h)(i).
\(=1.28\)	A1	\(1.277\ldots\) Accept 1.278 from the rounded table values.
Part (i)(i)
\(P(\text{win})=0.55\times0.80+0.45\times0.91\times0.52\)	M1	Correct structure with the new first-stage values and the unchanged second serve.
\(=0.65294\)	A1	Accept 0.653.
\(E(X)=2\times0.65294=1.31\)	A1	Accept a new distribution table (0.120, 0.453, 0.426) followed by the sum of \(xP(X=x)\).
Part (i)(ii)
double-fault probability rises from \(0.0342\) to \(0.45\times0.09=0.0405\), but the expected number of points won out of two rises from \(1.28\) to \(1.31\) (\(P(\text{win})\) from \(0.639\) to \(0.653\))	R1	Both comparisons, with values, are needed. Allow ECF.
so yes, she should follow the advice: more double faults, but the gain on the points where the first serve is in outweighs them	R1	Accept a sensible caveat (e.g. the 0.55 and 0.80 are estimates, or the model may differ against stronger opponents) in addition to a conclusion.

Answers: (a) first serve out \(0.38\); lost after first serve in \(0.26\); second serve out \(0.09\); lost after second serve in \(0.48\) · (b) \(0.0342\) · (c) \(0.638616\approx0.639\) · (d) \(0.718\) · (e) not independent, since \(P(F\mid W)\ne P(F)\) · (f) \(2.74\) · (g) \(0.188\) · (h)(i) \(0.131,\ 0.462,\ 0.408\) · (h)(ii) \(1.28\) · (i)(i) \(1.31\)

Common error (c) There are two separate paths that end in “point won”. Multiply along each path and then add the paths; do not multiply all the branch probabilities together.

Exam tip (i)(ii) When asked to discuss a decision, compare the quantity that actually matters (here points won) and quote numbers for both options before giving your conclusion.

More 4.6 questions in the question bank

+ Add to test✓ In your testCC-AA-45-46-411-224 marks~30 min

Cycling or taking the bus to school: a student survey

4.11 Conditional probability and independence4.5 Sample spaces and probability4.6 Combined, conditional and independent events4.1 Sampling, bias and outliersSL

A student council wants to know how students travel to school. One morning, a council member asks the first 120 students who arrive at the school gate two questions: “Do you sometimes cycle to school?” and “Do you sometimes take the bus to school?”

Of these 120 students, 48 sometimes cycle, 60 sometimes take the bus and 32 do neither.

A student is chosen at random from the 120. Let \(C\) be the event “the student sometimes cycles to school” and \(B\) the event “the student sometimes takes the bus to school”.

(a)

Copy and complete the Venn diagram in Figure 1 to show the number of students in each region.

Figure 1: Venn diagram for the 120 students

No calculatorGRAPHNotes

Show solution

\(48+60+32-120\)	(M1)	Or \(120-32=88\) in at least one circle, then \(48+60-88\).
\(20\) in \(C\cap B\)	A1
\(28\) in \(C\) only, \(40\) in \(B\) only and \(32\) outside both circles	A1	All three required. Allow ECF from an incorrect intersection if the four regions total 120.

Answer: \(C\) only 28, both 20, \(B\) only 40, neither 32

(3)

(b)(i)

Find \(P(C\cup B)\).

No calculatorROUTINENotes

Show solution

\(\frac{28+20+40}{120}\) or \(1-\frac{32}{120}\) or \(\frac{48}{120}+\frac{60}{120}-\frac{20}{120}\)	(M1)	Allow ECF from (a).
\(\frac{88}{120}=\frac{11}{15}\)	A1	Accept 0.733.

Answer: \(\frac{11}{15}\)

(2)

(b)(ii)

Find the probability that the student uses exactly one of these two ways of travelling.

No calculatorROUTINENotes

Show solution

\(\frac{28+40}{120}\)	(M1)	Allow ECF from (a).
\(=\frac{68}{120}=\frac{17}{30}\)	A1	Accept 0.567.

Answer: \(\frac{17}{30}\)

(2)

(c)

Find \(P(C\mid B)\).

No calculatorROUTINENotes

Show solution

\(\frac{20}{60}\)	(M1)	Denominator must be the 60 bus users. Allow ECF from (a).
\(=\frac13\)	A1

Answer: \(\frac13\)

(2)

(d)

Determine whether the events \(C\) and \(B\) are independent. Justify your answer.

No calculatorCONCEPTNotes

Show solution

\(P(C)=\frac{48}{120}=\frac25\)	A1
\(P(C\mid B)=\frac13\ne\frac25=P(C)\)	R1	Or \(P(C)P(B)=\frac25\times\frac12=\frac15\ne\frac16=P(C\cap B)\). Allow ECF from (c).
so \(C\) and \(B\) are not independent	A1	Do not award A1R0.

Answer: not independent

(3)

(e)(i)

Two of the 120 students are chosen at random, without replacement.

Find the probability that both students use exactly one of the two ways of travelling.

MULTINotes

Show solution

\(\frac{68}{120}\)	B1	Probability that the first student uses exactly one way. Allow ECF from (b)(ii).
\(\frac{68}{120}\times\frac{67}{119}\)	M1	Second factor must be out of 119 (without replacement).
\(=\frac{4556}{14280}=\frac{67}{210}=0.319\)	A1	Accept \(\frac{67}{210}\). Using \(\frac{68}{120}\times\frac{68}{120}\) (with replacement) gives 0.321: award B1M0A0.

Answer: \(0.319\)

(3)

(e)(ii)

Find the probability that at least one of the two students uses neither way of travelling.

MULTINotes

Show solution

\(P(\text{neither student is in the neither group})=\frac{88}{120}\times\frac{87}{119}\)	M1	Or add the three paths \(\frac{32}{120}\times\frac{88}{119}+\frac{88}{120}\times\frac{32}{119}+\frac{32}{120}\times\frac{31}{119}\).
\(1-\frac{7656}{14280}\)	M1	Complement of the correct product.
\(=0.464\)	A1	\(\frac{276}{595}=0.46386\ldots\)

Answer: \(0.464\)

(3)

(f)(i)

The school has 960 students.

Use the survey to estimate the number of students at the school who neither cycle nor take the bus.

No calculatorROUTINENotes

Show solution

\(\frac{32}{120}\times960=256\)

Answer: \(256\)

(1)

(f)(ii)

Comment on the reliability of your estimate in (f)(i), with reference to how the sample was chosen.

No calculatorINTERPRETNotes

Show solution

the sample is a convenience sample (the first 120 to arrive on one morning), not a random sample of all 960 students	R1
early arrivals are likely to travel in similar ways (e.g. many arrive together on the same bus, or live close by), so the sample may not be representative and the estimate may be unreliable	R1	Must refer to the context. Accept a reason based on one morning only (e.g. weather that day affects cycling).

(2)

(g)

The council member then surveys 15 more students, all of whom sometimes cycle and sometimes take the bus. These are added to the original 120 students, and a student is chosen at random from the 135.

Show that the events \(C\) and \(B\) are now independent.

No calculatorPROOFNotes

Show solution

\(P(C)=\frac{63}{135}=\frac{7}{15}\) and \(P(B)=\frac{75}{135}=\frac59\)	A1	Both required.
\(P(C\cap B)=\frac{35}{135}=\frac{7}{27}\)	A1
\(P(C)P(B)=\frac{7}{15}\times\frac59=\frac{35}{135}=\frac{7}{27}=P(C\cap B)\)	R1	Or \(P(C\mid B)=\frac{35}{75}=\frac{7}{15}=P(C)\). Must compare exact values (or decimals to at least 3 s.f., 0.259 = 0.259) and conclude.
so \(C\) and \(B\) are independent	AG

Answer: \(P(C)P(B)=\frac7{27}=P(C\cap B)\)

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(48+60+32-120\)	(M1)	Or \(120-32=88\) in at least one circle, then \(48+60-88\).
\(20\) in \(C\cap B\)	A1
\(28\) in \(C\) only, \(40\) in \(B\) only and \(32\) outside both circles	A1	All three required. Allow ECF from an incorrect intersection if the four regions total 120.
Part (b)(i)
\(\frac{28+20+40}{120}\) or \(1-\frac{32}{120}\) or \(\frac{48}{120}+\frac{60}{120}-\frac{20}{120}\)	(M1)	Allow ECF from (a).
\(\frac{88}{120}=\frac{11}{15}\)	A1	Accept 0.733.
Part (b)(ii)
\(\frac{28+40}{120}\)	(M1)	Allow ECF from (a).
\(=\frac{68}{120}=\frac{17}{30}\)	A1	Accept 0.567.
Part (c)
\(\frac{20}{60}\)	(M1)	Denominator must be the 60 bus users. Allow ECF from (a).
\(=\frac13\)	A1
Part (d)
\(P(C)=\frac{48}{120}=\frac25\)	A1
\(P(C\mid B)=\frac13\ne\frac25=P(C)\)	R1	Or \(P(C)P(B)=\frac25\times\frac12=\frac15\ne\frac16=P(C\cap B)\). Allow ECF from (c).
so \(C\) and \(B\) are not independent	A1	Do not award A1R0.
Part (e)(i)
\(\frac{68}{120}\)	B1	Probability that the first student uses exactly one way. Allow ECF from (b)(ii).
\(\frac{68}{120}\times\frac{67}{119}\)	M1	Second factor must be out of 119 (without replacement).
\(=\frac{4556}{14280}=\frac{67}{210}=0.319\)	A1	Accept \(\frac{67}{210}\). Using \(\frac{68}{120}\times\frac{68}{120}\) (with replacement) gives 0.321: award B1M0A0.
Part (e)(ii)
\(P(\text{neither student is in the neither group})=\frac{88}{120}\times\frac{87}{119}\)	M1	Or add the three paths \(\frac{32}{120}\times\frac{88}{119}+\frac{88}{120}\times\frac{32}{119}+\frac{32}{120}\times\frac{31}{119}\).
\(1-\frac{7656}{14280}\)	M1	Complement of the correct product.
\(=0.464\)	A1	\(\frac{276}{595}=0.46386\ldots\)
Part (f)(i)
\(\frac{32}{120}\times960=256\)	A1
Part (f)(ii)
the sample is a convenience sample (the first 120 to arrive on one morning), not a random sample of all 960 students	R1
early arrivals are likely to travel in similar ways (e.g. many arrive together on the same bus, or live close by), so the sample may not be representative and the estimate may be unreliable	R1	Must refer to the context. Accept a reason based on one morning only (e.g. weather that day affects cycling).
Part (g)
\(P(C)=\frac{63}{135}=\frac{7}{15}\) and \(P(B)=\frac{75}{135}=\frac59\)	A1	Both required.
\(P(C\cap B)=\frac{35}{135}=\frac{7}{27}\)	A1
\(P(C)P(B)=\frac{7}{15}\times\frac59=\frac{35}{135}=\frac{7}{27}=P(C\cap B)\)	R1	Or \(P(C\mid B)=\frac{35}{75}=\frac{7}{15}=P(C)\). Must compare exact values (or decimals to at least 3 s.f., 0.259 = 0.259) and conclude.
so \(C\) and \(B\) are independent	AG

Answers: (a) \(C\) only 28, both 20, \(B\) only 40, neither 32 · (b)(i) \(\frac{11}{15}\) · (b)(ii) \(\frac{17}{30}\) · (c) \(\frac13\) · (d) not independent · (e)(i) \(0.319\) · (e)(ii) \(0.464\) · (f)(i) \(256\) · (g) \(P(C)P(B)=\frac7{27}=P(C\cap B)\)

Common error (c) \(P(C\mid B)\) restricts the sample space to the bus users, so the denominator is 60, not 120 and not 48.

More 4.11 questions in the question bank

4.7 Discrete random variables · 4.8 Binomial distributionmerged group

+ Add to test✓ In your testCC-AA-47-48-126 marks~30 min

A dice game at a charity fair: is it fair?

4.7 Discrete random variables4.5 Sample spaces and probability4.6 Combined, conditional and independent events4.8 Binomial distributionSL

At a charity fair, a player pays €2 to play a game. The player rolls two fair six-sided dice and receives a prize according to the scores shown.

Result	Prize
a double six	€12
any other double (both dice show the same score)	€6
not a double, and the two scores add up to 7	€2
anything else	nothing

Let \(X\) be the player's profit, in euros, from one game (the prize minus the €2 paid).

(a)

Write down the number of equally likely outcomes in the sample space when two dice are rolled, and hence find the probability that the two scores add up to 7.

No calculatorROUTINENotes

Show solution

\(36\) outcomes	A1
\(\frac{6}{36}=\frac16\)	A1	From (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).

Answer: \(36\); \(\frac16\)

(2)

(b)(i)

Copy and complete the table to show the probability distribution of \(X\).

\(x\)	\(-2\)	0	4	10
\(P(X=x)\)

No calculatorROUTINENotes

Show solution

\(P(X=10)=\frac1{36}\) and \(P(X=4)=\frac5{36}\)	A1	Both required.
\(P(X=0)=\frac6{36}\)	A1	Allow ECF from (a).
\(P(X=-2)=\frac{24}{36}\)	A1	Accept equivalent fractions (\(\frac16\), \(\frac23\)) or decimals to 3 s.f. Award for \(1-\) the sum of the others.

Answer: \(\frac{24}{36},\ \frac6{36},\ \frac5{36},\ \frac1{36}\)

(3)

(b)(ii)

On the axes in Figure 1, draw a bar chart to show the probability distribution of \(X\).

Figure 1: axes for the probability distribution of X

No calculatorGRAPHNotes

Show solution

four vertical bars (or lines) at \(x=-2,\,0,\,4,\,10\) only	A1
heights \(\frac{24}{36},\,\frac6{36},\,\frac5{36},\,\frac1{36}\) respectively	A1	Allow ECF from (b)(i). Heights within half a small square.

(2)

(c)

Show that the charity expects to make €0.50 per game.

No calculatorMULTINotes

Show solution

\(E(X)=-2\times\frac{24}{36}+0\times\frac6{36}+4\times\frac5{36}+10\times\frac1{36}\)	M1	Sum of \(xP(X=x)\) over all four values.
\(=\frac{-48+0+20+10}{36}=-\frac{18}{36}=-0.5\)	A1	Or expected prize \(=\frac{12+30+12}{36}=\frac{54}{36}=1.5\), then \(2-1.5\).
the player expects to lose €0.50 per game, so the charity expects to gain €0.50 per game	R1	Must link the negative expected profit of the player to the charity's gain.
€0.50	AG

Answer: \(E(X)=-0.5\)

(3)

(d)

The organiser considers changing only the prize for a double six. Find the prize for a double six that would make the game fair.

No calculatorALGEBRANotes

Show solution

\((w-2)\times\frac1{36}+4\times\frac5{36}-2\times\frac{24}{36}=0\)	M1	Or the expected prize must equal €2: \(\frac{w+30+12}{36}=2\).
\(w=30\) (€30)	A1

Answer: €30

(2)

(e)

A player rolls the two dice and tells you that at least one of the dice shows a six. Find the probability that this player receives a prize.

No calculatorMULTINotes

Show solution

\(11\) outcomes contain at least one six	B1	Or \(P(\text{at least one six})=\frac{11}{36}\).
\(\frac{P(\text{prize}\cap\text{at least one six})}{P(\text{at least one six})}\) or counting the prize outcomes among these 11: \((6,6),\,(1,6),\,(6,1)\)	M1	Restricting the sample space to the outcomes with at least one six. \(\frac{3}{36}\) on its own is M0.
\(\frac{3}{11}\)	A1	Accept \(\frac{3/36}{11/36}\) simplified.

Answer: \(\frac3{11}\)

(3)

(f)(i)

Nikos plays the game 10 times. Let \(D\) be the number of games in which he rolls a double.

Explain why \(D\) can be modelled by a binomial distribution, and state its parameters.

CONCEPTNotes

Show solution

a fixed number of games (10), each with two outcomes (double or not), the rolls are independent and \(P(\text{double})\) is the same each time	R1	At least two conditions, in context.
\(D\sim B\left(10,\frac16\right)\)	A1	\(p=\frac{6}{36}\).

Answer: \(D\sim B\left(10,\frac16\right)\)

(2)

(f)(ii)

Find the probability that Nikos rolls a double in at least 3 of the 10 games.

ROUTINENotes

Show solution

\(1-P(D\le2)\)	(M1)	Using \(P(D\le3)\) or \(P(D=3)\) is M0.
\(=0.225\)	A1	\(0.22477\ldots\)

Answer: \(0.225\)

(2)

(g)(i)

The charity also tries a three-dice version of the game. The player pays €2 and rolls three fair dice. If exactly two of the dice show the same score, the prize is €2. If all three dice show the same score, the prize is €24. Otherwise there is no prize.

Show that the probability that exactly two of the three dice show the same score is \(\dfrac{5}{12}\).

No calculatorPROOFNotes

Show solution

\(P(\text{all different})=\frac{6\times5\times4}{216}=\frac{120}{216}\) and \(P(\text{all same})=\frac{6}{216}\)	M1	Or count directly: 6 choices for the repeated score, 5 for the other score and 3 positions for the odd die.
\(1-\frac{120}{216}-\frac{6}{216}\) or \(6\times5\times3=90\) outcomes	A1
\(\frac{90}{216}=\frac{5}{12}\)	A1	Must see \(\frac{90}{216}\) or equivalent before the given answer.
\(\frac5{12}\)	AG

Answer: \(\frac{90}{216}=\frac5{12}\)

(3)

(g)(ii)

Find the charity's expected profit per game for the three-dice version.

No calculatorMULTINotes

Show solution

expected prize \(=2\times\frac5{12}+24\times\frac{6}{216}\;\left(=\frac{90}{216}\times2+\frac{144}{216}=1.5\right)\)	M1	\(P(\text{all same})=\frac1{36}\). Or \(E(\text{player's profit})=0\times\frac5{12}+22\times\frac1{36}-2\times\frac{120}{216}\).
charity's expected profit \(=2-1.5=\) €0.50	A1

Answer: €0.50

(2)

(g)(iii)

Suggest, with a reason, which version of the game the charity should run.

INTERPRETNotes

Show solution

both versions give the charity the same expected profit, €0.50 per game	R1	Allow ECF from (c) and (g)(ii).
so the choice depends on something else, e.g. the three-dice version gives a prize more often (\(\frac49\) of games compared with \(\frac13\)), which may attract more players and raise more money overall; however its largest prize (€24) is bigger, so the charity's takings vary more over a day	R1	Accept either version with a reason that goes beyond the expected value. A choice based only on the expected values scores R1R0.

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(36\) outcomes	A1
\(\frac{6}{36}=\frac16\)	A1	From (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).
Part (b)(i)
\(P(X=10)=\frac1{36}\) and \(P(X=4)=\frac5{36}\)	A1	Both required.
\(P(X=0)=\frac6{36}\)	A1	Allow ECF from (a).
\(P(X=-2)=\frac{24}{36}\)	A1	Accept equivalent fractions (\(\frac16\), \(\frac23\)) or decimals to 3 s.f. Award for \(1-\) the sum of the others.
Part (b)(ii)
four vertical bars (or lines) at \(x=-2,\,0,\,4,\,10\) only	A1
heights \(\frac{24}{36},\,\frac6{36},\,\frac5{36},\,\frac1{36}\) respectively	A1	Allow ECF from (b)(i). Heights within half a small square.
Part (c)
\(E(X)=-2\times\frac{24}{36}+0\times\frac6{36}+4\times\frac5{36}+10\times\frac1{36}\)	M1	Sum of \(xP(X=x)\) over all four values.
\(=\frac{-48+0+20+10}{36}=-\frac{18}{36}=-0.5\)	A1	Or expected prize \(=\frac{12+30+12}{36}=\frac{54}{36}=1.5\), then \(2-1.5\).
the player expects to lose €0.50 per game, so the charity expects to gain €0.50 per game	R1	Must link the negative expected profit of the player to the charity's gain.
€0.50	AG
Part (d)
\((w-2)\times\frac1{36}+4\times\frac5{36}-2\times\frac{24}{36}=0\)	M1	Or the expected prize must equal €2: \(\frac{w+30+12}{36}=2\).
\(w=30\) (€30)	A1
Part (e)
\(11\) outcomes contain at least one six	B1	Or \(P(\text{at least one six})=\frac{11}{36}\).
\(\frac{P(\text{prize}\cap\text{at least one six})}{P(\text{at least one six})}\) or counting the prize outcomes among these 11: \((6,6),\,(1,6),\,(6,1)\)	M1	Restricting the sample space to the outcomes with at least one six. \(\frac{3}{36}\) on its own is M0.
\(\frac{3}{11}\)	A1	Accept \(\frac{3/36}{11/36}\) simplified.
Part (f)(i)
a fixed number of games (10), each with two outcomes (double or not), the rolls are independent and \(P(\text{double})\) is the same each time	R1	At least two conditions, in context.
\(D\sim B\left(10,\frac16\right)\)	A1	\(p=\frac{6}{36}\).
Part (f)(ii)
\(1-P(D\le2)\)	(M1)	Using \(P(D\le3)\) or \(P(D=3)\) is M0.
\(=0.225\)	A1	\(0.22477\ldots\)
Part (g)(i)
\(P(\text{all different})=\frac{6\times5\times4}{216}=\frac{120}{216}\) and \(P(\text{all same})=\frac{6}{216}\)	M1	Or count directly: 6 choices for the repeated score, 5 for the other score and 3 positions for the odd die.
\(1-\frac{120}{216}-\frac{6}{216}\) or \(6\times5\times3=90\) outcomes	A1
\(\frac{90}{216}=\frac{5}{12}\)	A1	Must see \(\frac{90}{216}\) or equivalent before the given answer.
\(\frac5{12}\)	AG
Part (g)(ii)
expected prize \(=2\times\frac5{12}+24\times\frac{6}{216}\;\left(=\frac{90}{216}\times2+\frac{144}{216}=1.5\right)\)	M1	\(P(\text{all same})=\frac1{36}\). Or \(E(\text{player's profit})=0\times\frac5{12}+22\times\frac1{36}-2\times\frac{120}{216}\).
charity's expected profit \(=2-1.5=\) €0.50	A1
Part (g)(iii)
both versions give the charity the same expected profit, €0.50 per game	R1	Allow ECF from (c) and (g)(ii).
so the choice depends on something else, e.g. the three-dice version gives a prize more often (\(\frac49\) of games compared with \(\frac13\)), which may attract more players and raise more money overall; however its largest prize (€24) is bigger, so the charity's takings vary more over a day	R1	Accept either version with a reason that goes beyond the expected value. A choice based only on the expected values scores R1R0.

Answers: (a) \(36\); \(\frac16\) · (b)(i) \(\frac{24}{36},\ \frac6{36},\ \frac5{36},\ \frac1{36}\) · (c) \(E(X)=-0.5\) · (d) €30 · (e) \(\frac3{11}\) · (f)(i) \(D\sim B\left(10,\frac16\right)\) · (f)(ii) \(0.225\) · (g)(i) \(\frac{90}{216}=\frac5{12}\) · (g)(ii) €0.50

Common error (b)(i) A sum of 7 can never be a double, so the “sum of 7” outcomes do not overlap with the doubles. The four prize groups are mutually exclusive, which is why their probabilities must add to 1.

Exam tip (f)(ii) Write “at least 3” as \(1-P(D\le2)\) before you use the binomial cdf: the cdf adds from 0 up to and including the value you enter.

More 4.7 questions in the question bank

+ Add to test✓ In your testCC-AA-47-48-227 marks~30 min

Quality control on an LED bulb production line

4.8 Binomial distribution4.7 Discrete random variables4.6 Combined, conditional and independent events1.5 Exponents and introduction to logarithmsSL

A factory makes LED bulbs. Each bulb is faulty with probability 0.04, independently of all other bulbs. The bulbs are packed in boxes of 20.

Let \(X\) be the number of faulty bulbs in a randomly chosen box.

(a)

State two assumptions, in context, that are needed for \(X\) to be modelled by \(B(20,\,0.04)\). For one of your assumptions, suggest a reason why it might not hold on a real production line.

INTERPRETNotes

Show solution

each bulb is faulty or not independently of the others (e.g. faults are not caused by one problem affecting a whole batch)	R1
the probability that a bulb is faulty is the same (0.04) for every bulb in the box	R1	Accept “the number of bulbs (20) is fixed” or “each bulb is either faulty or not faulty” as one of the two. Statements not in context score at most R1R0.
a sensible reason, in context, why one assumption may fail, e.g. a machine fault can make several bulbs in a row faulty, so faults in the same box are not independent; or the fault rate may differ between machines or shifts, so \(p\) is not constant	R1	The reason must match one of the assumptions stated.

(3)

(b)(i)

Find the probability that a box contains exactly one faulty bulb.

ROUTINENotes

Show solution

\(\binom{20}{1}(0.04)(0.96)^{19}\) or binomial pdf with \(n=20,\ p=0.04,\ r=1\)	(M1)
\(=0.368\)	A1	\(0.36833\ldots\)

Answer: \(0.368\)

(2)

(b)(ii)

Find \(P(X\le2)\).

ROUTINENotes

Show solution

binomial cdf with \(n=20,\ p=0.04,\ r=2\), or \(P(X=0)+P(X=1)+P(X=2)\)	(M1)	\(0.442+0.368+0.146\).
\(=0.956\)	A1	\(0.95613\ldots\)

Answer: \(0.956\)

(2)

(c)

Find the mean and the variance of \(X\).

ROUTINENotes

Show solution

\(E(X)=20\times0.04=0.8\)	A1
\(\mathrm{Var}(X)=20\times0.04\times0.96=0.768\)	A1	Do not accept the standard deviation 0.876 as the variance.

Answer: mean \(0.8\); variance \(0.768\)

(2)

(d)

Figure 1 shows the probability distribution of \(X\) for \(x\le6\). (The probabilities for \(x\ge7\) are too small to show.)

Describe the shape of the distribution, and use Figure 1 to explain why the most likely number of faulty bulbs is less than the mean.

Figure 1: the distribution of X, where X ~ B(20, 0.04)

GRAPHNotes

Show solution

positively skewed (tail to the right; the bars decrease from \(x=0\))	A1	Accept “skewed to the right”.
the tallest bar (mode) is at \(x=0\), but the long tail of small probabilities at \(x=2,3,4,\ldots\) pulls the mean up to \(0.8\)	R1	Must refer to the mode at 0 and to the tail (or to \(p\) being small). Allow ECF from (c).

Answer: positively skewed; mode 0 < mean 0.8

(2)

(e)(i)

A box is rejected if it contains more than 2 faulty bulbs.

Show that the probability that a box is rejected is \(0.0439\), correct to three significant figures.

ROUTINENotes

Show solution

\(P(X\gt2)=1-P(X\le2)\)	M1	Allow ECF from (b)(ii) only if the given answer is reached.
\(=1-0.956137\ldots=0.043862\ldots\)	A1	Must see a value to at least 4 s.f. \(1-0.956=0.044\) alone is A0.
\(\approx0.0439\)	AG

Answer: \(0.0439\)

(2)

(e)(ii)

A shipment contains 5 boxes. Find the probability that at least one box in the shipment is rejected.

MULTINotes

Show solution

\(P(\text{a box is not rejected})=1-0.0439=0.9561\)	(M1)
\(1-0.9561^{5}\)	M1	Complement of “no box rejected”, assuming boxes are independent.
\(=0.201\)	A1	\(0.2009\ldots\) Accept 0.200 to 0.201.

Answer: \(0.201\)

(3)

(f)(i)

An inspector tests bulbs one at a time, straight from the production line. She wants the probability that she finds at least one faulty bulb among the first \(n\) bulbs tested to be greater than 0.95.

Show that \(n\gt\dfrac{\ln0.05}{\ln0.96}\).

PROOFNotes

Show solution

\(P(\text{at least one faulty})=1-P(\text{none faulty})=1-0.96^{n}\)	M1	Use of the complement of “no faulty bulbs in \(n\)”.
\(1-0.96^{n}\gt0.95\Rightarrow0.96^{n}\lt0.05\)	A1
\(n\ln0.96\lt\ln0.05\), and dividing by \(\ln0.96\lt0\) reverses the inequality	R1	The reason for reversing the inequality must be seen.
\(n\gt\frac{\ln0.05}{\ln0.96}\)	AG

Answer: \(n\gt\frac{\ln0.05}{\ln0.96}\)

(3)

(f)(ii)

Hence find the smallest value of \(n\).

ROUTINENotes

Show solution

\(\frac{\ln0.05}{\ln0.96}=73.4\ldots\), so \(n=74\)

Answer 73 is A0.

Answer: \(74\)

(1)

(g)(i)

Each box is checked before it leaves the factory. The cost to the factory, \(C\) euros, depends on the number of faulty bulbs found, as shown below.

Faulty bulbs in the box	0	1	2	more than 2 (box rejected)
Cost \(C\) (€)	0	3	6	40

Find \(E(C)\).

MULTINotes

Show solution

\(E(C)=\sum c\,P(C=c)\) over the four cases, with \(P(C=40)=P(X\gt2)\)	M1	Allow ECF from (b) and (e).
\(0\times0.442+3\times0.368+6\times0.146+40\times0.0439\)	A1	\(P(X=2)=0.146\), \(P(X=1)=0.368\) from (b)(i) and \(0.0439\) from (e)(i).
\(=3.73\) (€)	A1	\(3.734\ldots\) Accept 3.73 or 3.74.

Answer: €3.73

(3)

(g)(ii)

A new process reduces the probability that a bulb is faulty to 0.015. The new process adds €2.50 to the cost of producing each box.

By finding the new value of \(E(C)\), determine whether the factory should use the new process.

MODELNotes

Show solution

\(Y\sim B(20,0.015)\): \(P(Y=0)=0.739,\ P(Y=1)=0.225,\ P(Y=2)=0.0326,\ P(Y\gt2)=0.00318\)	(M1)	At least two correct values seen, or implied by the next line.
\(E(C)=3\times0.225+6\times0.0326+40\times0.00318=0.998\) (€)	A1	Accept 0.997 to 1.00.
expected saving \(=3.73-1.00=2.74\) (€) per box	M1	Allow ECF from (g)(i).
\(2.74\gt2.50\), so the new process lowers the expected cost per box (by about €0.24) and should be used	R1	Accept a conclusion that notes the margin is small, provided the comparison is made. Do not award R1 without the numerical comparison.

Answer: new \(E(C)=€1.00\); yes, saving €2.74 > €2.50

(4)

Show full solution

Working	Mark	Notes
Part (a)
each bulb is faulty or not independently of the others (e.g. faults are not caused by one problem affecting a whole batch)	R1
the probability that a bulb is faulty is the same (0.04) for every bulb in the box	R1	Accept “the number of bulbs (20) is fixed” or “each bulb is either faulty or not faulty” as one of the two. Statements not in context score at most R1R0.
a sensible reason, in context, why one assumption may fail, e.g. a machine fault can make several bulbs in a row faulty, so faults in the same box are not independent; or the fault rate may differ between machines or shifts, so \(p\) is not constant	R1	The reason must match one of the assumptions stated.
Part (b)(i)
\(\binom{20}{1}(0.04)(0.96)^{19}\) or binomial pdf with \(n=20,\ p=0.04,\ r=1\)	(M1)
\(=0.368\)	A1	\(0.36833\ldots\)
Part (b)(ii)
binomial cdf with \(n=20,\ p=0.04,\ r=2\), or \(P(X=0)+P(X=1)+P(X=2)\)	(M1)	\(0.442+0.368+0.146\).
\(=0.956\)	A1	\(0.95613\ldots\)
Part (c)
\(E(X)=20\times0.04=0.8\)	A1
\(\mathrm{Var}(X)=20\times0.04\times0.96=0.768\)	A1	Do not accept the standard deviation 0.876 as the variance.
Part (d)
positively skewed (tail to the right; the bars decrease from \(x=0\))	A1	Accept “skewed to the right”.
the tallest bar (mode) is at \(x=0\), but the long tail of small probabilities at \(x=2,3,4,\ldots\) pulls the mean up to \(0.8\)	R1	Must refer to the mode at 0 and to the tail (or to \(p\) being small). Allow ECF from (c).
Part (e)(i)
\(P(X\gt2)=1-P(X\le2)\)	M1	Allow ECF from (b)(ii) only if the given answer is reached.
\(=1-0.956137\ldots=0.043862\ldots\)	A1	Must see a value to at least 4 s.f. \(1-0.956=0.044\) alone is A0.
\(\approx0.0439\)	AG
Part (e)(ii)
\(P(\text{a box is not rejected})=1-0.0439=0.9561\)	(M1)
\(1-0.9561^{5}\)	M1	Complement of “no box rejected”, assuming boxes are independent.
\(=0.201\)	A1	\(0.2009\ldots\) Accept 0.200 to 0.201.
Part (f)(i)
\(P(\text{at least one faulty})=1-P(\text{none faulty})=1-0.96^{n}\)	M1	Use of the complement of “no faulty bulbs in \(n\)”.
\(1-0.96^{n}\gt0.95\Rightarrow0.96^{n}\lt0.05\)	A1
\(n\ln0.96\lt\ln0.05\), and dividing by \(\ln0.96\lt0\) reverses the inequality	R1	The reason for reversing the inequality must be seen.
\(n\gt\frac{\ln0.05}{\ln0.96}\)	AG
Part (f)(ii)
\(\frac{\ln0.05}{\ln0.96}=73.4\ldots\), so \(n=74\)	A1	Answer 73 is A0.
Part (g)(i)
\(E(C)=\sum c\,P(C=c)\) over the four cases, with \(P(C=40)=P(X\gt2)\)	M1	Allow ECF from (b) and (e).
\(0\times0.442+3\times0.368+6\times0.146+40\times0.0439\)	A1	\(P(X=2)=0.146\), \(P(X=1)=0.368\) from (b)(i) and \(0.0439\) from (e)(i).
\(=3.73\) (€)	A1	\(3.734\ldots\) Accept 3.73 or 3.74.
Part (g)(ii)
\(Y\sim B(20,0.015)\): \(P(Y=0)=0.739,\ P(Y=1)=0.225,\ P(Y=2)=0.0326,\ P(Y\gt2)=0.00318\)	(M1)	At least two correct values seen, or implied by the next line.
\(E(C)=3\times0.225+6\times0.0326+40\times0.00318=0.998\) (€)	A1	Accept 0.997 to 1.00.
expected saving \(=3.73-1.00=2.74\) (€) per box	M1	Allow ECF from (g)(i).
\(2.74\gt2.50\), so the new process lowers the expected cost per box (by about €0.24) and should be used	R1	Accept a conclusion that notes the margin is small, provided the comparison is made. Do not award R1 without the numerical comparison.

Answers: (b)(i) \(0.368\) · (b)(ii) \(0.956\) · (c) mean \(0.8\); variance \(0.768\) · (d) positively skewed; mode 0 < mean 0.8 · (e)(i) \(0.0439\) · (e)(ii) \(0.201\) · (f)(i) \(n\gt\frac{\ln0.05}{\ln0.96}\) · (f)(ii) \(74\) · (g)(i) €3.73 · (g)(ii) new \(E(C)=€1.00\); yes, saving €2.74 > €2.50

Common error (f)(i) \(\ln0.96\) is negative, because 0.96 is less than 1. Dividing both sides of an inequality by a negative number reverses the inequality sign.

More 4.8 questions in the question bank

4.9 The normal distribution · 4.12 Standardization and unknown normal parametersmerged group

+ Add to test✓ In your testCC-AA-49-412-126 marks~30 min

How full is the bottle? A juice bottling line

4.9 The normal distribution4.8 Binomial distribution4.6 Combined, conditional and independent events4.12 Standardization and unknown normal parametersSL

A machine fills bottles with orange juice. Each bottle is labelled 500 ml and holds at most 512 ml, so a bottle that receives more than 512 ml overflows. A bottle that contains less than 500 ml is underfilled.

The volume of juice, \(X\) ml, that the machine puts into a bottle is modelled by \(X\sim N(505,3^2)\). The volumes in different bottles are independent.

(a)

Find the probability that a bottle overflows.

ROUTINENotes

Show solution

\(P(X\gt 512)\)	(M1)	For the correct region: a sketch, or normal cdf with lower bound 512, upper bound \(10^{99}\), \(\sigma=3\), \(\mu=505\).
\(0.00982\)	A1	Accept \(0.0098153\ldots\) or \(9.82\times10^{-3}\). Entering the variance 9 for \(\sigma\) gives 0.218: A0 (the M1 only if the correct region is seen).

Answer: \(0.00982\)

(2)

(b)(i)

Find the probability that a bottle is underfilled.

ROUTINENotes

Show solution

\(P(X\lt 500)=0.0478\)

Accept \(0.0477903\ldots\)

Answer: \(0.0478\)

(1)

(b)(ii)

Hence, without using a calculator, write down \(P(X\gt 510)\) and find \(P(500\lt X\lt 510)\).

No calculatorCONCEPTNotes

Show solution

510 is 5 ml above the mean and 500 is 5 ml below it, so by symmetry \(P(X\gt 510)=P(X\lt 500)=0.0478\)	A1	Allow ECF from (b)(i).
\(P(500\lt X\lt 510)=1-2(0.0478)=0.904\)	A1	Accept 0.9044. Allow ECF from (b)(i).

Answer: \(0.0478\); \(0.904\)

(2)

(c)

On the axes in Figure 1, sketch the curve of the distribution of \(X\). Indicate the mean, mark 512 on the horizontal axis, and shade the regions that represent a bottle that is underfilled and a bottle that overflows.

Figure 1: axes for the distribution of \(X\)

GRAPHNotes

Show solution

bell-shaped curve, symmetric about a peak at 505, approaching the axis at both ends	A1	The curve must not cross or end on the axis inside the window.
500 and 512 in the correct positions, 512 further from 505 than 500 is	A1
region to the left of 500 and region to the right of 512 shaded, the left region visibly larger	A1

(3)

(d)

Ninety per cent of bottles contain more than \(v\) ml of juice. Find \(v\).

ROUTINENotes

Show solution

\(P(X\lt v)=0.1\)	(M1)	For converting to a left area of 0.1, seen or implied by inverse normal with area 0.1. Using 0.9 gives 508.8: M0A0.
\(v=501\) (501.155…)	A1	Accept 501.2.

Answer: \(v=501\) ml (501.2)

(2)

(e)

The juice is sold in packs of 12 bottles. Show that the probability that a pack contains at least one underfilled bottle is 0.444, correct to three significant figures.

MULTINotes

Show solution

number of underfilled bottles \(Y\sim B(12,\,0.0477903\ldots)\)	M1	Independent bottles, each underfilled with the probability from (b)(i).
\(P(Y\ge1)=1-P(Y=0)=1-(1-0.0477903\ldots)^{12}\)	M1
\(=1-0.55564\ldots=0.44436\ldots\)	A1	Must see a value to at least 4 s.f. or \(1-0.5556\).
\(\approx0.444\)	AG

Answer: \(0.444\)

(3)

(f)

A bottle is chosen at random and found to contain more than 508 ml. Find the probability that it overflows.

MULTINotes

Show solution

\(P(X\gt512\mid X\gt508)=\frac{P(X\gt512)}{P(X\gt508)}\)	M1	The event "more than 512 and more than 508" is "more than 512".
\(P(X\gt508)=0.158655\ldots\)	A1	508 is one standard deviation above the mean.
\(\frac{0.0098153}{0.158655}=0.0619\)	A1	Allow ECF from (a).

Answer: \(0.0619\)

(3)

(g)(i)

A new regulation requires that at most 2% of bottles are underfilled. The mean of the machine is changed to \(\mu\) ml and the standard deviation stays at 3 ml.

Find the least value of \(\mu\) that meets the regulation. Give your answer correct to one decimal place.

ALGEBRANotes

Show solution

\(\frac{500-\mu}{3}=z\) where \(P(Z\lt z)=0.02\)	M1	Or \(500=\mu+3z\). Standardising 500 with the unknown mean and setting the left tail equal to 0.02.
\(z=-2.05374\ldots\)	(A1)	Negative, because 500 must be below the mean.
\(\mu=500+2.05374\ldots\times3=506.161\ldots\approx506.2\) ml	A1	506.1 is A0: it gives \(P(X\lt500)=0.021\), which does not meet the regulation.

Answer: \(\mu=506.2\) ml

(3)

(g)(ii)

Find the probability that a bottle overflows when the mean is the value found in (g)(i).

ROUTINENotes

Show solution

\(P(X\gt512)\) with \(\mu=506.16\ldots\), \(\sigma=3\)	(M1)
\(0.0258\)	A1	Accept 0.0266 (from \(\mu=506.2\)). Allow ECF from (g)(i).

Answer: \(0.0258\)

(2)

(h)(i)

The manager would like both the proportion of underfilled bottles and the proportion of bottles that overflow to be at most 2%. Show that this cannot be done by changing the mean alone while the standard deviation is 3 ml.

INTERPRETNotes

Show solution

with the mean at the midpoint 506 of 500 and 512, the two tails are equal	M1	Or: the conditions need \(\mu-2.054\sigma\ge500\) and \(\mu+2.054\sigma\le512\).
each tail is \(P(Z\gt2)=0.0228\), which is more than 0.02	A1	Or: these need \(4.108\sigma\le12\), but \(4.108\times3=12.3\gt12\).
moving the mean away from 506 makes one of the two tails larger still, so both proportions cannot be at most 2%	R1	Not required with the inequality method, where the R1 is for the conclusion that no \(\mu\) satisfies both.
so it cannot be done	AG

(3)

(h)(ii)

Find the largest standard deviation for which it can be done.

ALGEBRANotes

Show solution

\(\frac{512-506}{\sigma}\ge2.05374\ldots\)	M1	Or \(2\times2.05374\sigma\le12\). Allow "=".
\(\sigma\le2.92\) ml	A1	Accept 2.9215. Answer 2.92 alone (from correct working) is fine.

Answer: \(\sigma=2.92\) ml

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(P(X\gt 512)\)	(M1)	For the correct region: a sketch, or normal cdf with lower bound 512, upper bound \(10^{99}\), \(\sigma=3\), \(\mu=505\).
\(0.00982\)	A1	Accept \(0.0098153\ldots\) or \(9.82\times10^{-3}\). Entering the variance 9 for \(\sigma\) gives 0.218: A0 (the M1 only if the correct region is seen).
Part (b)(i)
\(P(X\lt 500)=0.0478\)	A1	Accept \(0.0477903\ldots\)
Part (b)(ii)
510 is 5 ml above the mean and 500 is 5 ml below it, so by symmetry \(P(X\gt 510)=P(X\lt 500)=0.0478\)	A1	Allow ECF from (b)(i).
\(P(500\lt X\lt 510)=1-2(0.0478)=0.904\)	A1	Accept 0.9044. Allow ECF from (b)(i).
Part (c)
bell-shaped curve, symmetric about a peak at 505, approaching the axis at both ends	A1	The curve must not cross or end on the axis inside the window.
500 and 512 in the correct positions, 512 further from 505 than 500 is	A1
region to the left of 500 and region to the right of 512 shaded, the left region visibly larger	A1
Part (d)
\(P(X\lt v)=0.1\)	(M1)	For converting to a left area of 0.1, seen or implied by inverse normal with area 0.1. Using 0.9 gives 508.8: M0A0.
\(v=501\) (501.155…)	A1	Accept 501.2.
Part (e)
number of underfilled bottles \(Y\sim B(12,\,0.0477903\ldots)\)	M1	Independent bottles, each underfilled with the probability from (b)(i).
\(P(Y\ge1)=1-P(Y=0)=1-(1-0.0477903\ldots)^{12}\)	M1
\(=1-0.55564\ldots=0.44436\ldots\)	A1	Must see a value to at least 4 s.f. or \(1-0.5556\).
\(\approx0.444\)	AG
Part (f)
\(P(X\gt512\mid X\gt508)=\frac{P(X\gt512)}{P(X\gt508)}\)	M1	The event "more than 512 and more than 508" is "more than 512".
\(P(X\gt508)=0.158655\ldots\)	A1	508 is one standard deviation above the mean.
\(\frac{0.0098153}{0.158655}=0.0619\)	A1	Allow ECF from (a).
Part (g)(i)
\(\frac{500-\mu}{3}=z\) where \(P(Z\lt z)=0.02\)	M1	Or \(500=\mu+3z\). Standardising 500 with the unknown mean and setting the left tail equal to 0.02.
\(z=-2.05374\ldots\)	(A1)	Negative, because 500 must be below the mean.
\(\mu=500+2.05374\ldots\times3=506.161\ldots\approx506.2\) ml	A1	506.1 is A0: it gives \(P(X\lt500)=0.021\), which does not meet the regulation.
Part (g)(ii)
\(P(X\gt512)\) with \(\mu=506.16\ldots\), \(\sigma=3\)	(M1)
\(0.0258\)	A1	Accept 0.0266 (from \(\mu=506.2\)). Allow ECF from (g)(i).
Part (h)(i)
with the mean at the midpoint 506 of 500 and 512, the two tails are equal	M1	Or: the conditions need \(\mu-2.054\sigma\ge500\) and \(\mu+2.054\sigma\le512\).
each tail is \(P(Z\gt2)=0.0228\), which is more than 0.02	A1	Or: these need \(4.108\sigma\le12\), but \(4.108\times3=12.3\gt12\).
moving the mean away from 506 makes one of the two tails larger still, so both proportions cannot be at most 2%	R1	Not required with the inequality method, where the R1 is for the conclusion that no \(\mu\) satisfies both.
so it cannot be done	AG
Part (h)(ii)
\(\frac{512-506}{\sigma}\ge2.05374\ldots\)	M1	Or \(2\times2.05374\sigma\le12\). Allow "=".
\(\sigma\le2.92\) ml	A1	Accept 2.9215. Answer 2.92 alone (from correct working) is fine.

Answers: (a) \(0.00982\) · (b)(i) \(0.0478\) · (b)(ii) \(0.0478\); \(0.904\) · (d) \(v=501\) ml (501.2) · (e) \(0.444\) · (f) \(0.0619\) · (g)(i) \(\mu=506.2\) ml · (g)(ii) \(0.0258\) · (h)(ii) \(\sigma=2.92\) ml

Exam tip (d) "90% contain more than \(v\)" puts 90% to the right of \(v\), so the area entered into inverse normal is 0.1. The answer must be below the mean.

Common error (f) Writing \(P(X\gt512\cap X\gt508)=P(X\gt512)\times P(X\gt508)\). The events are not independent: every bottle with more than 512 ml already has more than 508 ml.

More 4.9 questions in the question bank

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Airline checked baggage: finding the mean and spread

4.12 Standardization and unknown normal parameters4.9 The normal distribution4.11 Conditional probability and independence4.8 Binomial distributionSL

An airline records the mass of every checked bag. The mass of a checked bag, \(X\) kg, is modelled by a normal distribution with unknown mean \(\mu\) and unknown standard deviation \(\sigma\). A bag with mass more than 23 kg is overweight, and the passenger pays a fee.

The records show that 8% of checked bags are overweight and 15% have mass less than 12 kg.

(a)

On the axes in Figure 1, sketch the distribution of \(X\). Mark 12 and 23 on the horizontal axis, and shade and label the regions whose probabilities are 0.15 and 0.08.

Figure 1: axes for the distribution of \(X\)

GRAPHNotes

Show solution

bell-shaped curve with 12 in the left tail and 23 in the right tail, peak between them	A1
region left of 12 labelled 0.15 and region right of 23 labelled 0.08, the left region visibly larger	A1

(2)

(b)(i)

Write down the \(z\)-values that correspond to 12 kg and to 23 kg.

ROUTINENotes

Show solution

\(z_1=-1.03643\ldots\)	A1	Inverse normal for \(N(0,1)\) with area 0.15. Accept \(-1.036\).
\(z_2=1.40507\ldots\)	A1	Area to the left is 0.92. Accept 1.405. Using 0.08 (giving \(-1.405\)) is A0.

Answer: \(-1.04\) (\(-1.0364\)) and \(1.41\) (\(1.4051\))

(2)

(b)(ii)

Hence find \(\sigma\) and show that \(\mu=16.7\), correct to three significant figures.

ALGEBRANotes

Show solution

\(12=\mu-1.03643\sigma\) and \(23=\mu+1.40507\sigma\)	M1	Or \(\frac{12-\mu}{\sigma}=-1.03643\) and \(\frac{23-\mu}{\sigma}=1.40507\). Allow ECF from (b)(i).
subtracting: \(11=2.44150\sigma\)	M1	Or solving the pair with the GDC.
\(\sigma=4.51\) (4.50542…)	A1
\(\mu=23-1.40507\times4.50542=16.6696\ldots\)	A1	Must see a value to at least 4 s.f.
\(\mu\approx16.7\)	AG

Answer: \(\sigma=4.51\) kg, \(\mu=16.7\) kg

(4)

(c)

In the rest of the question, use \(\mu=16.67\) and \(\sigma=4.505\).

Find the probability that a bag has mass between 15 kg and 20 kg.

ROUTINENotes

Show solution

\(P(15\lt X\lt20)\)	(M1)	Normal cdf with bounds 15 and 20, or by standardising.
\(0.415\)	A1	Accept 0.4147 or 0.4146.

Answer: \(0.415\)

(2)

(d)

A bag is chosen at random and found to have mass more than 20 kg. Find the probability that it is overweight.

MULTINotes

Show solution

\(P(X\gt23\mid X\gt20)=\frac{P(X\gt23)}{P(X\gt20)}\)	M1	Every bag over 23 kg is also over 20 kg.
\(P(X\gt20)=0.229899\ldots\)	A1
\(\frac{0.08}{0.229899}=0.348\)	A1	Accept 0.3480 from the model value of \(P(X\gt23)\).

Answer: \(0.348\)

(3)

(e)(i)

A small aircraft carries 30 checked bags. The masses of the bags are independent.

Write down the expected number of overweight bags.

ROUTINENotes

Show solution

\(30\times0.08=2.4\)

Answer: \(2.4\)

(1)

(e)(ii)

Find the probability that more than 3 of the 30 bags are overweight.

ROUTINENotes

Show solution

\(Y\sim B(30,\,0.08)\)	M1	Recognising the binomial model, seen or implied.
\(P(Y\gt3)=1-P(Y\le3)\)	M1	\(1-P(Y\le4)=0.0874\) is M0.
\(=0.216\)	A1	Accept 0.2158.

Answer: \(0.216\)

(3)

(f)(i)

The airline raises the limit for an overweight bag from 23 kg to 25 kg. Assume that the distribution of masses does not change.

Find the \(z\)-value of 25 kg and hence the percentage of bags that are now overweight.

ROUTINENotes

Show solution

\(z=\frac{25-16.67}{4.505}=1.849\)	A1
\(P(Z\gt1.849)=0.0322\), so 3.22%	A1	Accept 3.2%.

Answer: \(z=1.85\); 3.22%

(2)

(f)(ii)

The fee is 65 euros for each overweight bag. On one route the airline carries 180 checked bags on each flight. Find the expected reduction in the fee income per flight caused by the change of limit.

MULTINotes

Show solution

expected number overweight: \(180\times0.08=14.4\) before, \(180\times0.03222=5.80\) after	M1	Either expected number.
\(65\times(14.4-5.80)\)	M1
\(559\) euros	A1	Accept 558 to 560. Allow ECF from (f)(i).

Answer: 559 euros

(3)

(f)(iii)

Suggest why the actual reduction in fee income may be smaller than your answer to (f)(ii).

INTERPRETNotes

Show solution

the calculation assumes the distribution of masses stays the same after the change	R1
passengers who packed to just under 23 kg may now pack up to 25 kg, so the mean increases and more bags exceed 25 kg than predicted	R1	Accept any reasoned change in packing behaviour that increases the proportion over 25 kg.

(2)

Show full solution

Working	Mark	Notes
Part (a)
bell-shaped curve with 12 in the left tail and 23 in the right tail, peak between them	A1
region left of 12 labelled 0.15 and region right of 23 labelled 0.08, the left region visibly larger	A1
Part (b)(i)
\(z_1=-1.03643\ldots\)	A1	Inverse normal for \(N(0,1)\) with area 0.15. Accept \(-1.036\).
\(z_2=1.40507\ldots\)	A1	Area to the left is 0.92. Accept 1.405. Using 0.08 (giving \(-1.405\)) is A0.
Part (b)(ii)
\(12=\mu-1.03643\sigma\) and \(23=\mu+1.40507\sigma\)	M1	Or \(\frac{12-\mu}{\sigma}=-1.03643\) and \(\frac{23-\mu}{\sigma}=1.40507\). Allow ECF from (b)(i).
subtracting: \(11=2.44150\sigma\)	M1	Or solving the pair with the GDC.
\(\sigma=4.51\) (4.50542…)	A1
\(\mu=23-1.40507\times4.50542=16.6696\ldots\)	A1	Must see a value to at least 4 s.f.
\(\mu\approx16.7\)	AG
Part (c)
\(P(15\lt X\lt20)\)	(M1)	Normal cdf with bounds 15 and 20, or by standardising.
\(0.415\)	A1	Accept 0.4147 or 0.4146.
Part (d)
\(P(X\gt23\mid X\gt20)=\frac{P(X\gt23)}{P(X\gt20)}\)	M1	Every bag over 23 kg is also over 20 kg.
\(P(X\gt20)=0.229899\ldots\)	A1
\(\frac{0.08}{0.229899}=0.348\)	A1	Accept 0.3480 from the model value of \(P(X\gt23)\).
Part (e)(i)
\(30\times0.08=2.4\)	A1
Part (e)(ii)
\(Y\sim B(30,\,0.08)\)	M1	Recognising the binomial model, seen or implied.
\(P(Y\gt3)=1-P(Y\le3)\)	M1	\(1-P(Y\le4)=0.0874\) is M0.
\(=0.216\)	A1	Accept 0.2158.
Part (f)(i)
\(z=\frac{25-16.67}{4.505}=1.849\)	A1
\(P(Z\gt1.849)=0.0322\), so 3.22%	A1	Accept 3.2%.
Part (f)(ii)
expected number overweight: \(180\times0.08=14.4\) before, \(180\times0.03222=5.80\) after	M1	Either expected number.
\(65\times(14.4-5.80)\)	M1
\(559\) euros	A1	Accept 558 to 560. Allow ECF from (f)(i).
Part (f)(iii)
the calculation assumes the distribution of masses stays the same after the change	R1
passengers who packed to just under 23 kg may now pack up to 25 kg, so the mean increases and more bags exceed 25 kg than predicted	R1	Accept any reasoned change in packing behaviour that increases the proportion over 25 kg.

Answers: (b)(i) \(-1.04\) (\(-1.0364\)) and \(1.41\) (\(1.4051\)) · (b)(ii) \(\sigma=4.51\) kg, \(\mu=16.7\) kg · (c) \(0.415\) · (d) \(0.348\) · (e)(i) \(2.4\) · (e)(ii) \(0.216\) · (f)(i) \(z=1.85\); 3.22% · (f)(ii) 559 euros

Common error (b)(ii) Rounding the \(z\)-values to 2 d.p. before solving. Using \(-1.04\) and \(1.41\) gives \(\sigma=4.49\), which is not correct to 3 s.f.

More 4.12 questions in the question bank

4.13 Bayes’ theorem · 4.14 Variance, continuous random variables and linear transformationsmerged group

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A screening test: what does a positive result mean?

4.13 Bayes’ theorem4.6 Combined, conditional and independent events4.11 Conditional probability and independence4.8 Binomial distributionHL

A regional health service screens adults for an eye condition, \(G\). Everyone screened belongs to one of three age groups. The table shows the percentage of the people screened who are in each group, and the percentage of each group who have \(G\).

Age group	A: 40 to 59	B: 60 to 74	C: 75 and over
Percentage of people screened	50%	35%	15%
Percentage of the group who have \(G\)	1%	4%	10%

The screening test gives a positive result for 88% of people who have \(G\) and for 5% of people who do not have \(G\), whatever their age group. Let \(T\) be the event that a person tests positive.

(a)

A person screened is chosen at random. Show that the probability that they have \(G\) is 0.034.

No calculatorROUTINENotes

Show solution

\(P(G)=0.5(0.01)+0.35(0.04)+0.15(0.10)\)	M1	Sum of the three products. The groups form a partition.
\(=0.005+0.014+0.015\)	A1
\(=0.034\)	AG

Answer: \(0.034\)

(2)

(b)

Copy and complete the tree diagram in Figure 1 by writing the missing probabilities.

Figure 1: tree diagram for a randomly chosen person screened

No calculatorGRAPHNotes

Show solution

\(P(G')=0.966\)	A1
\(P(T'\mid G)=0.12\) and \(P(T'\mid G')=0.95\)	A1	Both needed.

Answer: \(0.966\); \(0.12\); \(0.95\)

(2)

(c)

Find \(P(T)\).

ROUTINENotes

Show solution

\(0.034\times0.88+0.966\times0.05\)	(M1)	Two routes that end in \(T\). Allow ECF from (b).
\(=0.02992+0.0483=0.0782\)	A1	Accept 0.07822.

Answer: \(0.0782\)

(2)

(d)(i)

A person tests positive. Find the probability that they have \(G\).

ROUTINENotes

Show solution

\(P(G\mid T)=\frac{0.02992}{0.07822}\)	M1	Allow ECF from (c).
\(=0.383\)	A1	Accept 0.3825.

Answer: \(0.383\)

(2)

(d)(ii)

The test gives a positive result for 88% of people who have \(G\). Explain why your answer to (d)(i) is so much lower than 0.88.

INTERPRETNotes

Show solution

\(G\) is rare (3.4%), so the 5% false positives from the large group without \(G\) (0.0483) outnumber the true positives from the small group with \(G\) (0.0299)

Must refer to the low proportion with \(G\), or compare the two groups of positive results. "The test is not accurate" alone is R0.

(1)

(e)(i)

A person tests positive. Find the probability that they are in age group C.

MULTINotes

Show solution

\(P(T\mid C)=0.10\times0.88+0.90\times0.05=0.133\)	M1	Within group C the prevalence is 10%.
\(P(C\mid T)=\frac{0.15\times0.133}{0.07822}\)	M1	Allow ECF from (c).
\(=0.255\)	A1	Accept 0.2550.

Answer: \(0.255\)

(3)

(e)(ii)

Determine whether the events \(T\) and \(C\) are independent. Justify your answer.

CONCEPTNotes

Show solution

\(P(T\mid C)=0.133\) and \(P(T)=0.0782\)	A1	Or \(P(C\mid T)=0.255\) and \(P(C)=0.15\); or \(P(T\cap C)=0.01995\) and \(P(T)P(C)=0.0117\).
these are not equal, so \(T\) and \(C\) are not independent	R1	Do not award R1 without the numerical comparison. Allow ECF.

Answer: not independent

(2)

(f)(i)

On one day, 25 people screened receive a positive result. Assume that whether each of them has \(G\) is independent of the others. Let \(X\) be the number of these 25 people who have \(G\).

Find \(E(X)\).

ROUTINENotes

Show solution

\(25\times0.38251=9.56\)

Allow ECF from (d)(i).

Answer: \(9.56\)

(1)

(f)(ii)

Find the probability that fewer than 6 of these 25 people have \(G\).

ROUTINENotes

Show solution

\(X\sim B(25,\,0.38251)\)	M1
\(P(X\lt6)=P(X\le5)\)	M1	\(P(X\le6)=0.101\) is M0.
\(=0.0431\)	A1	Accept 0.04307. Allow ECF from (d)(i).

Answer: \(0.0431\)

(3)

(g)(i)

To reduce the number of false positive results, the health service considers screening only people in group C.

Show that, for a person in group C who tests positive, the probability that they have \(G\) is 0.662, correct to three significant figures.

ROUTINENotes

Show solution

\(\frac{0.10\times0.88}{0.10\times0.88+0.90\times0.05}=\frac{0.088}{0.133}\)	M1
\(=0.66165\ldots\)	A1	Must see a value to at least 4 s.f. or \(\frac{88}{133}\).
\(\approx0.662\)	AG

Answer: \(0.662\)

(2)

(g)(ii)

Find the probability that a person screened who has \(G\) is in group C. Hence discuss one advantage and one disadvantage of screening only group C.

INTERPRETNotes

Show solution

\(P(C\mid G)=\frac{0.015}{0.034}=0.441\)	A1	Accept \(\frac{15}{34}\).
advantage: a positive result is far more reliable (0.662 compared with 0.383), so fewer people without \(G\) are sent for further tests	R1
disadvantage: \(1-0.441=0.559\), so more than half of the people with \(G\) are in groups A and B and would not be detected	R1	Allow ECF from their \(P(C\mid G)\).

Answer: \(0.441\)

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(P(G)=0.5(0.01)+0.35(0.04)+0.15(0.10)\)	M1	Sum of the three products. The groups form a partition.
\(=0.005+0.014+0.015\)	A1
\(=0.034\)	AG
Part (b)
\(P(G')=0.966\)	A1
\(P(T'\mid G)=0.12\) and \(P(T'\mid G')=0.95\)	A1	Both needed.
Part (c)
\(0.034\times0.88+0.966\times0.05\)	(M1)	Two routes that end in \(T\). Allow ECF from (b).
\(=0.02992+0.0483=0.0782\)	A1	Accept 0.07822.
Part (d)(i)
\(P(G\mid T)=\frac{0.02992}{0.07822}\)	M1	Allow ECF from (c).
\(=0.383\)	A1	Accept 0.3825.
Part (d)(ii)
\(G\) is rare (3.4%), so the 5% false positives from the large group without \(G\) (0.0483) outnumber the true positives from the small group with \(G\) (0.0299)	R1	Must refer to the low proportion with \(G\), or compare the two groups of positive results. "The test is not accurate" alone is R0.
Part (e)(i)
\(P(T\mid C)=0.10\times0.88+0.90\times0.05=0.133\)	M1	Within group C the prevalence is 10%.
\(P(C\mid T)=\frac{0.15\times0.133}{0.07822}\)	M1	Allow ECF from (c).
\(=0.255\)	A1	Accept 0.2550.
Part (e)(ii)
\(P(T\mid C)=0.133\) and \(P(T)=0.0782\)	A1	Or \(P(C\mid T)=0.255\) and \(P(C)=0.15\); or \(P(T\cap C)=0.01995\) and \(P(T)P(C)=0.0117\).
these are not equal, so \(T\) and \(C\) are not independent	R1	Do not award R1 without the numerical comparison. Allow ECF.
Part (f)(i)
\(25\times0.38251=9.56\)	A1	Allow ECF from (d)(i).
Part (f)(ii)
\(X\sim B(25,\,0.38251)\)	M1
\(P(X\lt6)=P(X\le5)\)	M1	\(P(X\le6)=0.101\) is M0.
\(=0.0431\)	A1	Accept 0.04307. Allow ECF from (d)(i).
Part (g)(i)
\(\frac{0.10\times0.88}{0.10\times0.88+0.90\times0.05}=\frac{0.088}{0.133}\)	M1
\(=0.66165\ldots\)	A1	Must see a value to at least 4 s.f. or \(\frac{88}{133}\).
\(\approx0.662\)	AG
Part (g)(ii)
\(P(C\mid G)=\frac{0.015}{0.034}=0.441\)	A1	Accept \(\frac{15}{34}\).
advantage: a positive result is far more reliable (0.662 compared with 0.383), so fewer people without \(G\) are sent for further tests	R1
disadvantage: \(1-0.441=0.559\), so more than half of the people with \(G\) are in groups A and B and would not be detected	R1	Allow ECF from their \(P(C\mid G)\).

Answers: (a) \(0.034\) · (b) \(0.966\); \(0.12\); \(0.95\) · (c) \(0.0782\) · (d)(i) \(0.383\) · (e)(i) \(0.255\) · (e)(ii) not independent · (f)(i) \(9.56\) · (f)(ii) \(0.0431\) · (g)(i) \(0.662\) · (g)(ii) \(0.441\)

Exam tip (d)(ii) \(P(T\mid G)\) and \(P(G\mid T)\) answer different questions. Write each probability in words before you calculate, and check which event is given.

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+ Add to test✓ In your testCC-AA-413-414-228 marks~35 min

Waiting for the bus: a continuous random variable

4.14 Variance, continuous random variables and linear transformations5.11 Definite integrals and areas5.10 Indefinite integrals and reverse chain rule4.7 Discrete random variablesHL

Buses are timetabled to leave a stop every 10 minutes, but traffic makes them irregular. The time that a passenger waits for a bus, \(T\) minutes, is modelled by the probability density function

\(f(t)=\begin{cases}kt(10-t)^2, & 0\le t\le10\\ 0, & \text{otherwise}\end{cases}\)

where \(k\) is a positive constant.

(a)

Show that \(k=\dfrac{3}{2500}\).

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\(\int_0^{10}kt(10-t)^2\,dt=1\)	M1	Setting the total area equal to 1.
\(k\left[50t^2-\frac{20t^3}{3}+\frac{t^4}{4}\right]_0^{10}\)	A1	From \(t(10-t)^2=100t-20t^2+t^3\).
\(k\left(5000-\frac{20000}{3}+2500\right)=\frac{2500}{3}k=1\)	A1
\(k=\frac{3}{2500}\)	AG

Answer: \(k=\frac{3}{2500}\)

(3)

(b)(i)

Find the mode of \(T\).

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\(f'(t)=k\left[(10-t)^2-2t(10-t)\right]=k(10-t)(10-3t)\)	M1	Product rule, or differentiating \(k(100t-20t^2+t^3)\) to get \(k(100-40t+3t^2)\). Setting \(f'(t)=0\).
mode \(=\frac{10}{3}\) (3.33) minutes	A1	\(t=10\) is a minimum (\(f(10)=0\)) and must be rejected.

Answer: \(\frac{10}{3}\) minutes

(2)

(b)(ii)

On the axes in Figure 1, sketch the graph of \(y=f(t)\). Label the maximum point with its coordinates.

Figure 1: axes for the graph of \(y=f(t)\)

GRAPHNotes

Show solution

curve starting at the origin, single maximum, falling to touch the \(t\)-axis at \(t=10\), skewed with the peak left of centre	A1	The curve must meet the axis at 10 tangentially (repeated factor), not at a steep angle.
maximum labelled \(\left(\frac{10}{3},\,\frac{8}{45}\right)\) or \((3.33,\,0.178)\)	A1	Allow ECF from (b)(i).

Answer: maximum \(\left(\frac{10}{3},\frac{8}{45}\right)\approx(3.33,\,0.178)\)

(2)

(c)

By using the substitution \(u=10-t\), show that \(P(T\gt7)=0.0837\).

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\(du=-dt\); when \(t=7\), \(u=3\) and when \(t=10\), \(u=0\)	M1	Both the differential and the new limits.
\(P(T\gt7)=k\int_3^0(10-u)u^2(-du)=k\int_0^3(10u^2-u^3)\,du\)	A1
\(k\left[\frac{10u^3}{3}-\frac{u^4}{4}\right]_0^3=k(90-20.25)=69.75k\)	M1	Integrating in \(u\) and substituting the limits.
\(=\frac{3}{2500}\times69.75=0.0837\)	A1	Or \(\frac{837}{10000}\).
\(P(T\gt7)=0.0837\)	AG

Answer: \(0.0837\)

(4)

(d)

Show that \(E(T)=4\).

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\(E(T)=k\int_0^{10}(100t^2-20t^3+t^4)\,dt=k\left[\frac{100t^3}{3}-5t^4+\frac{t^5}{5}\right]_0^{10}\)	M1
\(=\frac{3}{2500}\left(\frac{100000}{3}-50000+20000\right)=\frac{3}{2500}\times\frac{10000}{3}\)	A1
\(=4\)	AG

Answer: \(4\)

(2)

(e)

Find \(\mathrm{Var}(T)\).

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\(E(T^2)=k\int_0^{10}(100t^3-20t^4+t^5)\,dt=k\left[25t^4-4t^5+\frac{t^6}{6}\right]_0^{10}\)	M1
\(=\frac{3}{2500}\times\frac{50000}{3}=20\)	A1
\(\mathrm{Var}(T)=20-4^2=4\)	A1

Answer: \(4\)

(3)

(f)(i)

The median waiting time is \(m\) minutes. Show that \(3m^4-80m^3+600m^2-5000=0\).

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\(\int_0^m kt(10-t)^2\,dt=\frac{3}{2500}\left(50m^2-\frac{20m^3}{3}+\frac{m^4}{4}\right)=\frac{1}{2}\)	M1	Setting the area up to \(m\) equal to 0.5.
multiplying by 10000: \(600m^2-80m^3+3m^4=5000\)	A1
\(3m^4-80m^3+600m^2-5000=0\)	AG

(2)

(f)(ii)

Hence find the median waiting time.

ROUTINENotes

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\(m=3.86\) minutes

Accept 3.857. The only root in \(0\le m\le10\) (the others are \(-2.47\) and two complex roots).

Answer: \(3.86\) minutes

(1)

(g)

Each journey costs a passenger a fare of 1.50 euros. A transport study values waiting time at 0.30 euros per minute, so the total cost of a journey, in euros, is \(C=1.5+0.3T\).

Find \(E(C)\) and the standard deviation of \(C\).

ROUTINENotes

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\(E(C)=1.5+0.3\times4=2.70\) euros	A1
\(\mathrm{Var}(C)=0.3^2\times4=0.36\), so the standard deviation is 0.60 euros	A1	\(0.3\times4+1.5\) for the variance, or a standard deviation of \(\sqrt{2.7}\), is A0.

Answer: \(E(C)=2.70\) euros; standard deviation 0.60 euros

(2)

(h)

A transport app uses a simpler model, in which the waiting time \(W\) minutes takes only the values in the table.

\(w\)	1	3	5	7	9
\(P(W=w)\)	0.18	0.34	\(p\)	0.15	0.03

Find \(p\) and \(E(W)\), and comment on how well \(E(W)\) agrees with \(E(T)\).

MODELNotes

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\(0.18+0.34+p+0.15+0.03=1\), so \(p=0.30\)	A1
\(E(W)=1(0.18)+3(0.34)+5(0.30)+7(0.15)+9(0.03)\)	M1	Allow ECF from their \(p\).
\(E(W)=4.02\)	A1
very close to \(E(T)=4\) (only 0.02 minutes, about 1 second, more), so the simpler model gives almost the same mean wait	R1	Accept a comment that each value stands for a 2-minute interval, so a small difference is expected.

Answer: \(p=0.30\); \(E(W)=4.02\)

(4)

(i)

The timetable changes so that buses leave every 12 minutes. The waiting time, \(S\) minutes, is now modelled by the pdf \(g(s)=cs(12-s)^2\) for \(0\le s\le12\), and \(g(s)=0\) otherwise, where \(c\) is a positive constant.

Without further integration, explain why \(E(S)=4.8\) and find \(\mathrm{Var}(S)\).

INTERPRETNotes

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\(g\) has the same shape as \(f\), stretched horizontally by factor 1.2 (\(12=1.2\times10\)), so \(S\) has the same distribution as \(1.2T\)	R1	Accept: every waiting time is scaled up by the factor \(\frac{12}{10}\). Need a reason, not only \(1.2\times4\).
\(E(S)=1.2\,E(T)=1.2\times4=4.8\)	AG
\(\mathrm{Var}(S)=1.2^2\,\mathrm{Var}(T)\)	(M1)	Using \(\mathrm{Var}(aT)=a^2\,\mathrm{Var}(T)\).
\(=1.44\times4=5.76\)	A1	\(1.2\times4=4.8\) for the variance is M0A0. Allow ECF from (e).

Answer: \(\mathrm{Var}(S)=5.76\)

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(\int_0^{10}kt(10-t)^2\,dt=1\)	M1	Setting the total area equal to 1.
\(k\left[50t^2-\frac{20t^3}{3}+\frac{t^4}{4}\right]_0^{10}\)	A1	From \(t(10-t)^2=100t-20t^2+t^3\).
\(k\left(5000-\frac{20000}{3}+2500\right)=\frac{2500}{3}k=1\)	A1
\(k=\frac{3}{2500}\)	AG
Part (b)(i)
\(f'(t)=k\left[(10-t)^2-2t(10-t)\right]=k(10-t)(10-3t)\)	M1	Product rule, or differentiating \(k(100t-20t^2+t^3)\) to get \(k(100-40t+3t^2)\). Setting \(f'(t)=0\).
mode \(=\frac{10}{3}\) (3.33) minutes	A1	\(t=10\) is a minimum (\(f(10)=0\)) and must be rejected.
Part (b)(ii)
curve starting at the origin, single maximum, falling to touch the \(t\)-axis at \(t=10\), skewed with the peak left of centre	A1	The curve must meet the axis at 10 tangentially (repeated factor), not at a steep angle.
maximum labelled \(\left(\frac{10}{3},\,\frac{8}{45}\right)\) or \((3.33,\,0.178)\)	A1	Allow ECF from (b)(i).
Part (c)
\(du=-dt\); when \(t=7\), \(u=3\) and when \(t=10\), \(u=0\)	M1	Both the differential and the new limits.
\(P(T\gt7)=k\int_3^0(10-u)u^2(-du)=k\int_0^3(10u^2-u^3)\,du\)	A1
\(k\left[\frac{10u^3}{3}-\frac{u^4}{4}\right]_0^3=k(90-20.25)=69.75k\)	M1	Integrating in \(u\) and substituting the limits.
\(=\frac{3}{2500}\times69.75=0.0837\)	A1	Or \(\frac{837}{10000}\).
\(P(T\gt7)=0.0837\)	AG
Part (d)
\(E(T)=k\int_0^{10}(100t^2-20t^3+t^4)\,dt=k\left[\frac{100t^3}{3}-5t^4+\frac{t^5}{5}\right]_0^{10}\)	M1
\(=\frac{3}{2500}\left(\frac{100000}{3}-50000+20000\right)=\frac{3}{2500}\times\frac{10000}{3}\)	A1
\(=4\)	AG
Part (e)
\(E(T^2)=k\int_0^{10}(100t^3-20t^4+t^5)\,dt=k\left[25t^4-4t^5+\frac{t^6}{6}\right]_0^{10}\)	M1
\(=\frac{3}{2500}\times\frac{50000}{3}=20\)	A1
\(\mathrm{Var}(T)=20-4^2=4\)	A1
Part (f)(i)
\(\int_0^m kt(10-t)^2\,dt=\frac{3}{2500}\left(50m^2-\frac{20m^3}{3}+\frac{m^4}{4}\right)=\frac{1}{2}\)	M1	Setting the area up to \(m\) equal to 0.5.
multiplying by 10000: \(600m^2-80m^3+3m^4=5000\)	A1
\(3m^4-80m^3+600m^2-5000=0\)	AG
Part (f)(ii)
\(m=3.86\) minutes	A1	Accept 3.857. The only root in \(0\le m\le10\) (the others are \(-2.47\) and two complex roots).
Part (g)
\(E(C)=1.5+0.3\times4=2.70\) euros	A1
\(\mathrm{Var}(C)=0.3^2\times4=0.36\), so the standard deviation is 0.60 euros	A1	\(0.3\times4+1.5\) for the variance, or a standard deviation of \(\sqrt{2.7}\), is A0.
Part (h)
\(0.18+0.34+p+0.15+0.03=1\), so \(p=0.30\)	A1
\(E(W)=1(0.18)+3(0.34)+5(0.30)+7(0.15)+9(0.03)\)	M1	Allow ECF from their \(p\).
\(E(W)=4.02\)	A1
very close to \(E(T)=4\) (only 0.02 minutes, about 1 second, more), so the simpler model gives almost the same mean wait	R1	Accept a comment that each value stands for a 2-minute interval, so a small difference is expected.
Part (i)
\(g\) has the same shape as \(f\), stretched horizontally by factor 1.2 (\(12=1.2\times10\)), so \(S\) has the same distribution as \(1.2T\)	R1	Accept: every waiting time is scaled up by the factor \(\frac{12}{10}\). Need a reason, not only \(1.2\times4\).
\(E(S)=1.2\,E(T)=1.2\times4=4.8\)	AG
\(\mathrm{Var}(S)=1.2^2\,\mathrm{Var}(T)\)	(M1)	Using \(\mathrm{Var}(aT)=a^2\,\mathrm{Var}(T)\).
\(=1.44\times4=5.76\)	A1	\(1.2\times4=4.8\) for the variance is M0A0. Allow ECF from (e).

Answers: (a) \(k=\frac{3}{2500}\) · (b)(i) \(\frac{10}{3}\) minutes · (b)(ii) maximum \(\left(\frac{10}{3},\frac{8}{45}\right)\approx(3.33,\,0.178)\) · (c) \(0.0837\) · (d) \(4\) · (e) \(4\) · (f)(ii) \(3.86\) minutes · (g) \(E(C)=2.70\) euros; standard deviation 0.60 euros · (h) \(p=0.30\); \(E(W)=4.02\) · (i) \(\mathrm{Var}(S)=5.76\)

Common error (c) Changing the variable but keeping the limits 7 and 10. With \(u=10-t\), the limits become 3 and 0, and the minus sign from \(du=-dt\) puts them back in increasing order.

More 4.14 questions in the question bank

Topic 5 · Calculus

● home subchapter · ○ linked subchapter · last column: links to other topics

Setup	5.1	5.2	5.3	5.4	5.5	5.6	5.7	5.8	5.9	5.10	5.11	5.12	5.13	5.14	5.15	5.16	5.17	5.18	5.19	Other
A bakery's daily profit: rates of change	○	○	●																	2.6
A melting ice cap: average and instantaneous rates of change	●	○	○																	2.4
A curved road and its straight slip road				●		○	○													2.1
When do an app's users change fastest? Second derivatives						○	●	○												2.9
How much water flows down a river channel?					○					○	●									2.4
Fast-charging an electric car: energy from power					○					●	○									2.9
Designing an athletics track around a playing field							○	●												3.4 2.6
A lift to an observation deck: velocity, acceleration and distance								○	●	○	○									—
From monthly to continuous compounding: a limit												○	●							1.4 2.9
Smooth rides on a metro train: higher derivatives						○			○			●	○							—
A rotating lighthouse beam sweeping the shore														●	○					3.3 3.9
Two masts and the phone signal along a road											○			○	●					1.11
The shape and volume of a rocket nose cone											○					○	●			2.11
A car's suspension settling after a bump						○					○					●				3.7
A skydiver approaching terminal velocity									○									●	○	2.9
The sag of a suspension cable: series approximations											○		○						●	2.9

5.1 Limits and the derivative · 5.2 Increasing and decreasing functions · 5.3 Differentiating polynomials and integer powersmerged group

+ Add to test✓ In your testCC-AA-51-52-53-126 marks~30 min

A bakery's daily profit: rates of change

5.3 Differentiating polynomials and integer powers5.1 Limits and the derivative5.2 Increasing and decreasing functions2.6 The quadratic function and its formsSL

A bakery bakes \(x\) hundred loaves of bread each day, where \(0\le x\le8\). Its daily profit, \(P\) euros, is modelled by

\(P(x)=-2x^3+15x^2+36x-40\).

Figure 1 shows the graph of \(P\).

Figure 1: the graph of \(P\) for \(0\le x\le8\)

(a)

The gradient of the chord joining the points on the graph of \(P\) where \(x=2\) and \(x=2+h\) is

\(g(h)=\dfrac{P(2+h)-P(2)}{h}\), \(h\ne0\).

Complete the following table, giving the values of \(g(h)\) exactly.

\(h\)	\(0.5\)	\(0.1\)	\(0.01\)
\(g(h)\)	\(73\)

ROUTINENotes

Show solution

\(P(2)=-16+60+72-40=76\) and \(P(2.1)=83.228\), so \(g(0.1)=\frac{83.228-76}{0.1}\)	M1	Award for substituting correctly into the chord formula for either \(h=0.1\) or \(h=0.01\).
\(g(0.1)=72.28\)	A1
\(g(0.01)=72.0298\)	A1	Exact value required; \(72.03\) is A0.

Answer: \(72.28\) and \(72.0298\)

(3)

(b)

Write down the value that \(g(h)\) appears to approach as \(h\to0\), and state what this value represents on the graph of \(P\).

CONCEPTNotes

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\(72\)	A1
the gradient of the tangent to the graph of \(P\) at the point where \(x=2\)	R1	Accept "the gradient of the curve at \(x=2\)" or "\(P'(2)\)". "The gradient of the chord" is R0.

Answer: \(72\); the gradient of the tangent at \(x=2\)

(2)

(c)

Show that \(P'(x)=-6(x-6)(x+1)\).

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differentiates term by term, with at least three terms correct	M1
\(P'(x)=-6x^2+30x+36\)	A1
\(=-6(x^2-5x-6)=-6(x-6)(x+1)\)	AG	Expanding \(-6(x-6)(x+1)\) to reach \(-6x^2+30x+36\) is equally acceptable.

Answer: \(P'(x)=-6x^2+30x+36=-6(x-6)(x+1)\)

(2)

(d)

Hence find \(P'(2)\), and interpret this value in the context of the bakery.

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\(P'(2)=-6(-4)(3)=72\)	A1	Agrees with (b).
when 200 loaves are baked, the daily profit is increasing at 72 euros per extra hundred loaves	R1	Must mention the rate and the units (euros per hundred loaves). Accept "about 0.72 euros per extra loaf".

Answer: \(72\) euros per hundred loaves

(2)

(e)

Find the values of \(x\), for \(0\le x\le8\), for which \(P\) is increasing. Justify your answer.

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\(P'(x)=0\) when \(x=6\) or \(x=-1\)	M1	Award for setting \(P'(x)=0\) or \(P'(x)>0\) and finding the roots.
\(P'\) is a quadratic with negative leading coefficient, so \(P'(x)>0\) between its roots (or test values, e.g. \(P'(0)=36>0\), \(P'(7)=-48<0\))	R1
\(0\le x<6\)	A1	Accept \(0\le x\le6\) or \(0<x<6\). Do not award A1 for \(-1<x<6\) (outside the domain).

Answer: \(0\le x<6\)

(3)

(f)

On the axes in Figure 2, sketch the graph of \(y=P'(x)\) for \(0\le x\le8\). Show clearly the coordinates of the \(y\)-intercept, the \(x\)-intercept, the vertex and the end point of the graph.

Figure 2: axes for the graph of \(y=P'(x)\)

No calculatorGRAPHNotes

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concave-down parabola drawn only for \(0\le x\le8\)	A1
\(y\)-intercept \((0,\,36)\) and \(x\)-intercept \((6,\,0)\)	A1
vertex \((2.5,\,73.5)\)	A1	The axis of symmetry is halfway between the zeros \(-1\) and \(6\).
end point \((8,\,-108)\)	A1

Answer: parabola through \((0,36)\), vertex \((2.5,73.5)\), \((6,0)\), \((8,-108)\)

(4)

(g)

Using your graph in part (f), explain why the bakery should not bake more than 600 loaves a day. Find the greatest possible daily profit.

No calculatorINTERPRETNotes

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for \(6<x\le8\) the graph of \(P'\) is below the \(x\)-axis, so \(P\) is decreasing: baking more than 600 loaves reduces the profit	R1	Must refer to the sign of \(P'\) (or the graph of \(P'\) below the axis).
\(P(6)=-432+540+216-40=284\) euros	A1

Answer: \(284\) euros (at \(x=6\))

(2)

(h)

The bakery's rent increases by 25 euros per day, and its new daily profit is \(Q(x)=P(x)-25\).

Explain why \(Q\) is increasing for exactly the same values of \(x\) as \(P\).

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\(Q'(x)=P'(x)\), because the derivative of the constant \(-25\) is 0	R1
so \(Q'\) has the same sign as \(P'\) for every \(x\) (the graph of \(Q\) is the graph of \(P\) translated 25 units down, which does not change where it rises or falls)	R1

(2)

(i)(i)

The average profit per hundred loaves is \(A(x)=\dfrac{Q(x)}{x}\), for \(0<x\le8\).

Find \(A'(x)\).

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\(A(x)=\frac{-2x^3+15x^2+36x-65}{x}=-2x^2+15x+36-65x^{-1}\)	M1	Award for dividing each term by \(x\). Differentiating the numerator and the denominator separately is M0.
\(A'(x)=-4x+15\ldots\)	A1
\(\ldots+65x^{-2}\)	A1	Accept \(+\frac{65}{x^2}\).

Answer: \(A'(x)=-4x+15+\dfrac{65}{x^2}\)

(3)

(i)(ii)

Find \(A'(4)\) and \(A'(5)\). Hence explain why the average profit per hundred loaves is greatest for a value of \(x\) between 4 and 5.

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\(A'(4)=-16+15+\frac{65}{16}=\frac{49}{16}\ (=3.0625)\)	A1
\(A'(5)=-20+15+\frac{65}{25}=-2.4\)	A1
\(A'\) changes from positive to negative, so \(A\) increases and then decreases: its greatest value is between \(x=4\) and \(x=5\)	R1	Allow ECF from (i)(i) if the signs of the two values differ. \(A'\) is continuous for \(x>0\), so it equals zero between 4 and 5. (Not required: \(A''(x)=-4-\frac{130}{x^3}<0\), so \(A'\) is decreasing and this is the only stationary point.)

Answer: \(A'(4)=\frac{49}{16}\), \(A'(5)=-2.4\)

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(P(2)=-16+60+72-40=76\) and \(P(2.1)=83.228\), so \(g(0.1)=\frac{83.228-76}{0.1}\)	M1	Award for substituting correctly into the chord formula for either \(h=0.1\) or \(h=0.01\).
\(g(0.1)=72.28\)	A1
\(g(0.01)=72.0298\)	A1	Exact value required; \(72.03\) is A0.
Part (b)
\(72\)	A1
the gradient of the tangent to the graph of \(P\) at the point where \(x=2\)	R1	Accept "the gradient of the curve at \(x=2\)" or "\(P'(2)\)". "The gradient of the chord" is R0.
Part (c)
differentiates term by term, with at least three terms correct	M1
\(P'(x)=-6x^2+30x+36\)	A1
\(=-6(x^2-5x-6)=-6(x-6)(x+1)\)	AG	Expanding \(-6(x-6)(x+1)\) to reach \(-6x^2+30x+36\) is equally acceptable.
Part (d)
\(P'(2)=-6(-4)(3)=72\)	A1	Agrees with (b).
when 200 loaves are baked, the daily profit is increasing at 72 euros per extra hundred loaves	R1	Must mention the rate and the units (euros per hundred loaves). Accept "about 0.72 euros per extra loaf".
Part (e)
\(P'(x)=0\) when \(x=6\) or \(x=-1\)	M1	Award for setting \(P'(x)=0\) or \(P'(x)>0\) and finding the roots.
\(P'\) is a quadratic with negative leading coefficient, so \(P'(x)>0\) between its roots (or test values, e.g. \(P'(0)=36>0\), \(P'(7)=-48<0\))	R1
\(0\le x<6\)	A1	Accept \(0\le x\le6\) or \(0<x<6\). Do not award A1 for \(-1<x<6\) (outside the domain).
Part (f)
concave-down parabola drawn only for \(0\le x\le8\)	A1
\(y\)-intercept \((0,\,36)\) and \(x\)-intercept \((6,\,0)\)	A1
vertex \((2.5,\,73.5)\)	A1	The axis of symmetry is halfway between the zeros \(-1\) and \(6\).
end point \((8,\,-108)\)	A1
Part (g)
for \(6<x\le8\) the graph of \(P'\) is below the \(x\)-axis, so \(P\) is decreasing: baking more than 600 loaves reduces the profit	R1	Must refer to the sign of \(P'\) (or the graph of \(P'\) below the axis).
\(P(6)=-432+540+216-40=284\) euros	A1
Part (h)
\(Q'(x)=P'(x)\), because the derivative of the constant \(-25\) is 0	R1
so \(Q'\) has the same sign as \(P'\) for every \(x\) (the graph of \(Q\) is the graph of \(P\) translated 25 units down, which does not change where it rises or falls)	R1
Part (i)(i)
\(A(x)=\frac{-2x^3+15x^2+36x-65}{x}=-2x^2+15x+36-65x^{-1}\)	M1	Award for dividing each term by \(x\). Differentiating the numerator and the denominator separately is M0.
\(A'(x)=-4x+15\ldots\)	A1
\(\ldots+65x^{-2}\)	A1	Accept \(+\frac{65}{x^2}\).
Part (i)(ii)
\(A'(4)=-16+15+\frac{65}{16}=\frac{49}{16}\ (=3.0625)\)	A1
\(A'(5)=-20+15+\frac{65}{25}=-2.4\)	A1
\(A'\) changes from positive to negative, so \(A\) increases and then decreases: its greatest value is between \(x=4\) and \(x=5\)	R1	Allow ECF from (i)(i) if the signs of the two values differ. \(A'\) is continuous for \(x>0\), so it equals zero between 4 and 5. (Not required: \(A''(x)=-4-\frac{130}{x^3}<0\), so \(A'\) is decreasing and this is the only stationary point.)

Answers: (a) \(72.28\) and \(72.0298\) · (b) \(72\); the gradient of the tangent at \(x=2\) · (c) \(P'(x)=-6x^2+30x+36=-6(x-6)(x+1)\) · (d) \(72\) euros per hundred loaves · (e) \(0\le x<6\) · (f) parabola through \((0,36)\), vertex \((2.5,73.5)\), \((6,0)\), \((8,-108)\) · (g) \(284\) euros (at \(x=6\)) · (i)(i) \(A'(x)=-4x+15+\dfrac{65}{x^2}\) · (i)(ii) \(A'(4)=\frac{49}{16}\), \(A'(5)=-2.4\)

Exam tip (e) Intervals of increase are statements about \(x\). Solve \(P'(x)>0\), then keep only the part of the solution that lies in the domain \(0\le x\le8\).

Common error (i)(i) A common error is to write \(-65x^{-2}\) for the derivative of \(-65x^{-1}\). Multiplying by the power \(-1\) changes the sign: \((-65)(-1)x^{-2}=+65x^{-2}\).

More 5.3 questions in the question bank

+ Add to test✓ In your testCC-AA-51-52-53-223 marks~30 min

A melting ice cap: average and instantaneous rates of change

5.1 Limits and the derivative5.2 Increasing and decreasing functions5.3 Differentiating polynomials and integer powers2.4 Key features of graphs and intersections using technologySL

The volume of ice in a mountain ice cap is modelled by

\(V(t)=120-0.9t-0.02t^2\), \(t\ge0\),

where \(V\) is the volume in km³ and \(t\) is the time in years after the start of 2000. Figure 1 shows the graph of \(V\) for \(0\le t\le45\).

Figure 1: the graph of \(V\) for \(0\le t\le45\)

(a)

Complete the following table, which shows the average rate of change of \(V\) over three intervals.

Interval	\(10\le t\le12\)	\(10\le t\le10.1\)	\(10\le t\le10.01\)
Average rate of change (km³ per year)

ROUTINENotes

Show solution

\(\frac{V(12)-V(10)}{12-10}=\frac{106.32-109}{2}\)	M1	Award for a correct quotient for any one interval.
\(-1.34\)	A1
\(-1.302\) and \(-1.3002\)	A1	Both needed.

Answer: \(-1.34\), \(-1.302\), \(-1.3002\)

(3)

(b)

Write down the limit of the average rate of change of \(V\) over the interval \(10\le t\le10+h\) as \(h\to0\). State what this limit represents.

No calculatorCONCEPTNotes

Show solution

\(-1.3\)	A1
the instantaneous rate of change of \(V\) at \(t=10\) (the gradient of the tangent to the graph at \(t=10\), or \(V'(10)\))	R1

Answer: \(-1.3\)

(2)

(c)

Find \(V'(t)\) and hence verify your answer to part (b).

No calculatorROUTINENotes

Show solution

\(V'(t)=-0.9-0.04t\)	A1
\(V'(10)=-0.9-0.4=-1.3\)	A1	Must see the substitution.

Answer: \(V'(t)=-0.9-0.04t\); \(V'(10)=-1.3\)

(2)

(d)

Interpret the value of \(V'(10)\) in context. Include its units.

No calculatorINTERPRETNotes

Show solution

km³ per year (km³ year⁻¹)	A1
at the start of 2010 the volume of ice is decreasing at a rate of 1.3 km³ per year	R1	Must say "decreasing" (or "being lost") and refer to the time \(t=10\). "The volume is \(-1.3\)" is R0.

Answer: decreasing at 1.3 km³ per year at the start of 2010

(2)

(e)

Show that, according to the model, \(V\) is decreasing for all \(t\ge0\).

No calculatorPROOFNotes

Show solution

considers the sign of \(V'(t)=-0.9-0.04t\)	M1
for \(t\ge0\), \(0.04t\ge0\), so \(V'(t)\le-0.9<0\)	R1	Accept: \(V'(t)=0\) only at \(t=-22.5\), and \(V'\) is a decreasing linear function, so \(V'(t)<0\) for \(t\ge0\).
so \(V\) is decreasing for all \(t\ge0\)	AG

(2)

(f)(i)

Find the equation of the tangent to the graph of \(V\) at \(t=10\). Give your answer in the form \(V=mt+c\).

No calculatorROUTINENotes

Show solution

\(V-109=-1.3(t-10)\)	M1	Award for using the point \((10,\,109)\) and gradient \(-1.3\).
\(V=-1.3t+122\)	A1

Answer: \(V=-1.3t+122\)

(2)

(f)(ii)

Draw this tangent on Figure 1.

No calculatorGRAPHNotes

Show solution

straight line touching the curve at \((10,\,109)\)	A1
line passing through \((0,\,122)\) and, for example, \((40,\,70)\)	A1	Allow \(\pm2\) km³ in the readings. Allow ECF from (f)(i).

Answer: line through \((0,122)\), \((10,109)\), \((40,70)\)

(2)

(g)(i)

Use your graphic display calculator to find the year in which the model predicts the volume of ice will fall to 60 km³.

ROUTINENotes

Show solution

\(t=36.7\) (\(36.7135\ldots\))	A1	Award for the intersection of \(V=V(t)\) and \(V=60\), or for solving \(0.02t^2+0.9t-60=0\). Reject the negative root.
in 2036	A1	Do not accept 2037.

Answer: \(t=36.7\), so in 2036

(2)

(g)(ii)

Find the value of \(t\) at which your tangent from part (f)(i) reaches 60 km³. Explain why this value is later than your answer to part (g)(i).

INTERPRETNotes

Show solution

\(-1.3t+122=60\)	M1	Allow ECF from (f)(i).
\(t=47.7\) (\(47.69\ldots\))	A1
\(V'(t)=-0.9-0.04t\) becomes more negative as \(t\) increases (the curve bends down below the tangent for \(t>10\)), so the ice is lost faster than the constant 1.3 km³ per year that the tangent assumes: the tangent reaches 60 km³ later	R1	Must refer to the rate of loss increasing (or to the curve lying below the tangent).

Answer: \(t=47.7\)

(3)

(h)

A revised model, which allows for heavier snowfall in some years, is

\(W(t)=120-1.2t+0.03t^2-0.0005t^3\), \(t\ge0\).

Determine whether the revised model predicts that the volume of ice will increase at any time.

MULTINotes

Show solution

\(W'(t)=-1.2+0.06t-0.0015t^2\)	M1	Award for differentiating \(W\) with at least two terms correct and considering the sign of \(W'\).
greatest value of \(W'(t)\) is at \(t=\frac{0.06}{2\times0.0015}=20\), where \(W'(20)=-0.6\)	A1	Accept the maximum point \((20,\,-0.6)\) of the graph of \(W'\) from a GDC, or a discriminant argument: \(0.06^2-4(0.0015)(1.2)=-0.0036<0\) with a negative leading coefficient.
so \(W'(t)\le-0.6<0\) for all \(t\): the revised model never predicts an increase in the volume of ice	R1	Must include a conclusion.

Answer: No: \(W'(t)\le-0.6<0\) for all \(t\)

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(\frac{V(12)-V(10)}{12-10}=\frac{106.32-109}{2}\)	M1	Award for a correct quotient for any one interval.
\(-1.34\)	A1
\(-1.302\) and \(-1.3002\)	A1	Both needed.
Part (b)
\(-1.3\)	A1
the instantaneous rate of change of \(V\) at \(t=10\) (the gradient of the tangent to the graph at \(t=10\), or \(V'(10)\))	R1
Part (c)
\(V'(t)=-0.9-0.04t\)	A1
\(V'(10)=-0.9-0.4=-1.3\)	A1	Must see the substitution.
Part (d)
km³ per year (km³ year⁻¹)	A1
at the start of 2010 the volume of ice is decreasing at a rate of 1.3 km³ per year	R1	Must say "decreasing" (or "being lost") and refer to the time \(t=10\). "The volume is \(-1.3\)" is R0.
Part (e)
considers the sign of \(V'(t)=-0.9-0.04t\)	M1
for \(t\ge0\), \(0.04t\ge0\), so \(V'(t)\le-0.9<0\)	R1	Accept: \(V'(t)=0\) only at \(t=-22.5\), and \(V'\) is a decreasing linear function, so \(V'(t)<0\) for \(t\ge0\).
so \(V\) is decreasing for all \(t\ge0\)	AG
Part (f)(i)
\(V-109=-1.3(t-10)\)	M1	Award for using the point \((10,\,109)\) and gradient \(-1.3\).
\(V=-1.3t+122\)	A1
Part (f)(ii)
straight line touching the curve at \((10,\,109)\)	A1
line passing through \((0,\,122)\) and, for example, \((40,\,70)\)	A1	Allow \(\pm2\) km³ in the readings. Allow ECF from (f)(i).
Part (g)(i)
\(t=36.7\) (\(36.7135\ldots\))	A1	Award for the intersection of \(V=V(t)\) and \(V=60\), or for solving \(0.02t^2+0.9t-60=0\). Reject the negative root.
in 2036	A1	Do not accept 2037.
Part (g)(ii)
\(-1.3t+122=60\)	M1	Allow ECF from (f)(i).
\(t=47.7\) (\(47.69\ldots\))	A1
\(V'(t)=-0.9-0.04t\) becomes more negative as \(t\) increases (the curve bends down below the tangent for \(t>10\)), so the ice is lost faster than the constant 1.3 km³ per year that the tangent assumes: the tangent reaches 60 km³ later	R1	Must refer to the rate of loss increasing (or to the curve lying below the tangent).
Part (h)
\(W'(t)=-1.2+0.06t-0.0015t^2\)	M1	Award for differentiating \(W\) with at least two terms correct and considering the sign of \(W'\).
greatest value of \(W'(t)\) is at \(t=\frac{0.06}{2\times0.0015}=20\), where \(W'(20)=-0.6\)	A1	Accept the maximum point \((20,\,-0.6)\) of the graph of \(W'\) from a GDC, or a discriminant argument: \(0.06^2-4(0.0015)(1.2)=-0.0036<0\) with a negative leading coefficient.
so \(W'(t)\le-0.6<0\) for all \(t\): the revised model never predicts an increase in the volume of ice	R1	Must include a conclusion.

Answers: (a) \(-1.34\), \(-1.302\), \(-1.3002\) · (b) \(-1.3\) · (c) \(V'(t)=-0.9-0.04t\); \(V'(10)=-1.3\) · (d) decreasing at 1.3 km³ per year at the start of 2010 · (f)(i) \(V=-1.3t+122\) · (f)(ii) line through \((0,122)\), \((10,109)\), \((40,70)\) · (g)(i) \(t=36.7\), so in 2036 · (g)(ii) \(t=47.7\) · (h) No: \(W'(t)\le-0.6<0\) for all \(t\)

Common error (g)(i) \(t\) counts years after the start of 2000, so \(t=36.7\) is during 2036, not 2037. Rounding \(t\) up before converting to a year gives the wrong year.

More 5.1 questions in the question bank

5.4 Tangents and normals · 5.6 Differentiation rules · 5.7 The second derivative and graphsmerged group

+ Add to test✓ In your testCC-AA-54-56-57-125 marks~30 min

A curved road and its straight slip road

5.4 Tangents and normals5.6 Differentiation rules2.1 Equations of straight lines, parallel and perpendicular lines5.7 The second derivative and graphsSL

In this question, distances are measured in hundreds of metres.

On a map, a straight main road lies along the \(x\)-axis and a straight footpath lies along the \(y\)-axis. A curved road leaves the main road at the junction \(O(0,\,0)\) and follows the curve

\(y=12xe^{-0.2x}\), \(0\le x\le15\).

The point \(P\) on the curved road has \(x\)-coordinate 2. Figure 1 shows the roads and the footpath.

Figure 1: the curved road, the main road (x-axis) and the footpath (y-axis); equal scales, distances in hundreds of metres

(a)

Show that \(\dfrac{dy}{dx}=12e^{-0.2x}(1-0.2x)\).

No calculatorPROOFNotes

Show solution

product rule with \(u=12x\), \(v=e^{-0.2x}\)	M1
\(\frac{du}{dx}=12\), \(\frac{dv}{dx}=-0.2e^{-0.2x}\)	A1	Award for the derivative of \(e^{-0.2x}\) with the factor \(-0.2\) from the chain rule.
\(\frac{dy}{dx}=12x(-0.2e^{-0.2x})+12e^{-0.2x}=12e^{-0.2x}-2.4xe^{-0.2x}\)	A1
\(=12e^{-0.2x}(1-0.2x)\)	AG

(3)

(b)

Find the \(y\)-coordinate of \(P\) and the gradient of the curved road at \(P\).

ROUTINENotes

Show solution

\(y=24e^{-0.4}=16.1\) (\(16.0876\ldots\))	A1
\(\frac{dy}{dx}=12e^{-0.4}(0.6)=7.2e^{-0.4}=4.83\) (\(4.82630\ldots\))	A1

Answer: \(y=16.1\); gradient \(4.83\)

(2)

(c)(i)

A straight slip road leaves the curved road at \(P\) along the tangent to the curve at \(P\).

Find the equation of this tangent, giving your answer in the form \(y=mx+c\).

ROUTINENotes

Show solution

\(y-16.0876=4.8263(x-2)\)	M1	Allow ECF from (b).
\(y=4.83x+6.44\)	A1	Accept \(c=6.435\) or the exact \(c=9.6e^{-0.4}\).

Answer: \(y=4.83x+6.44\)

(2)

(c)(ii)

The slip road meets the main road at \(Q\). Without using decimal approximations, show that the \(x\)-coordinate of \(Q\) is \(-\dfrac{4}{3}\).

No calculatorPROOFNotes

Show solution

at \(Q\), \(y=0\): \(0-24e^{-0.4}=7.2e^{-0.4}(x-2)\)	M1	Award for using the exact point and exact gradient with \(y=0\).
\(x-2=-\frac{24e^{-0.4}}{7.2e^{-0.4}}=-\frac{10}{3}\)	A1	Must see the factor \(e^{-0.4}\) cancel.
\(x=2-\frac{10}{3}=-\frac{4}{3}\)	AG

(2)

(d)

A lamp post is placed at the point \(R\), where the normal to the curve at \(P\) meets the footpath.

Find the coordinates of \(R\).

MULTINotes

Show solution

gradient of normal \(=-\frac{1}{4.8263}=-0.207\) (\(-0.20719\ldots\))	(M1)
substitutes \(x=0\) into \(y-16.0876=-0.20720(x-2)\)	M1	Allow ECF from (b).
\(R(0,\,16.5)\)	A1	Accept \(y=16.50\). Do not accept \(16.1\) (the height of \(P\)).

Answer: \((0,\,16.5)\)

(3)

(e)

On Figure 1, draw the slip road from \(P\) to \(Q\) and the normal from \(P\) to \(R\). Label the points \(Q\) and \(R\).

GRAPHNotes

Show solution

straight line from \(P\) to \((-1.33,\,0)\) that touches the curve at \(P\)	A1
straight line from \(P\) to \((0,\,16.5)\), at right angles to the slip road	A1	The axes in Figure 1 have equal scales, so the right angle should be visible.
\(Q\) and \(R\) labelled in the correct positions	A1	Allow ECF from (c)(i) and (d).

Answer: tangent through \(Q(-1.33,0)\); normal through \(R(0,16.5)\)

(3)

(f)

A cycle path also starts at \(O\). It follows the curve \(y=\dfrac{60x}{x+5}\), \(x\ge0\).

Show that the cycle path and the curved road leave \(O\) in the same direction.

No calculatorPROOFNotes

Show solution

quotient rule with \(u=60x\), \(v=x+5\): \(\frac{dy}{dx}=\frac{(x+5)(60)-60x(1)}{(x+5)^2}\)	M1	Award for the correct order in the numerator.
\(=\frac{300}{(x+5)^2}\)	A1
at \(x=0\): cycle path gradient \(\frac{300}{25}=12\); road gradient \(12e^{0}(1-0)=12\)	A1	Both gradients must be seen.
equal gradients at \(O\), so both have the same tangent \(y=12x\) at \(O\)	AG

Answer: both gradients equal \(12\) at \(O\)

(3)

(g)(i)

Show that \(\dfrac{d^2y}{dx^2}=0.48e^{-0.2x}(x-10)\).

No calculatorPROOFNotes

Show solution

product rule on \(12e^{-0.2x}(1-0.2x)\) with \(u=12e^{-0.2x}\), \(v=1-0.2x\)	M1
\(\frac{d^2y}{dx^2}=12\left[-0.2e^{-0.2x}(1-0.2x)+e^{-0.2x}(-0.2)\right]\)	A1	Any correct unsimplified expression.
\(=12e^{-0.2x}(0.04x-0.4)=0.48e^{-0.2x}(x-10)\)	AG

(2)

(g)(ii)

Hence explain why the curved road is concave down for \(0\le x<10\). State what this tells you about the position of the curved road relative to the slip road close to \(P\).

No calculatorINTERPRETNotes

Show solution

\(e^{-0.2x}>0\) for all \(x\) and \(x-10<0\) for \(x<10\), so \(\frac{d^2y}{dx^2}<0\): concave down	R1
a concave-down curve lies below its tangents, so near \(P\) the curved road lies below the slip road (the two separate on both sides of \(P\))	R1

(2)

(h)

A second straight road is to leave the curved road at the point \(S\), where the curved road is parallel to the main road.

Find the coordinates of \(S\) and write down the equation of this second road.

MULTINotes

Show solution

\(\frac{dy}{dx}=0\Rightarrow 12e^{-0.2x}(1-0.2x)=0\)	M1
\(e^{-0.2x}\ne0\), so \(x=5\)	A1
\(S(5,\,60e^{-1})=(5,\,22.1)\); road \(y=22.1\)	A1	Accept \(y=60e^{-1}\) or \(y=22.07\).

Answer: \(S(5,\,22.1)\); \(y=22.1\)

(3)

Show full solution

Working	Mark	Notes
Part (a)
product rule with \(u=12x\), \(v=e^{-0.2x}\)	M1
\(\frac{du}{dx}=12\), \(\frac{dv}{dx}=-0.2e^{-0.2x}\)	A1	Award for the derivative of \(e^{-0.2x}\) with the factor \(-0.2\) from the chain rule.
\(\frac{dy}{dx}=12x(-0.2e^{-0.2x})+12e^{-0.2x}=12e^{-0.2x}-2.4xe^{-0.2x}\)	A1
\(=12e^{-0.2x}(1-0.2x)\)	AG
Part (b)
\(y=24e^{-0.4}=16.1\) (\(16.0876\ldots\))	A1
\(\frac{dy}{dx}=12e^{-0.4}(0.6)=7.2e^{-0.4}=4.83\) (\(4.82630\ldots\))	A1
Part (c)(i)
\(y-16.0876=4.8263(x-2)\)	M1	Allow ECF from (b).
\(y=4.83x+6.44\)	A1	Accept \(c=6.435\) or the exact \(c=9.6e^{-0.4}\).
Part (c)(ii)
at \(Q\), \(y=0\): \(0-24e^{-0.4}=7.2e^{-0.4}(x-2)\)	M1	Award for using the exact point and exact gradient with \(y=0\).
\(x-2=-\frac{24e^{-0.4}}{7.2e^{-0.4}}=-\frac{10}{3}\)	A1	Must see the factor \(e^{-0.4}\) cancel.
\(x=2-\frac{10}{3}=-\frac{4}{3}\)	AG
Part (d)
gradient of normal \(=-\frac{1}{4.8263}=-0.207\) (\(-0.20719\ldots\))	(M1)
substitutes \(x=0\) into \(y-16.0876=-0.20720(x-2)\)	M1	Allow ECF from (b).
\(R(0,\,16.5)\)	A1	Accept \(y=16.50\). Do not accept \(16.1\) (the height of \(P\)).
Part (e)
straight line from \(P\) to \((-1.33,\,0)\) that touches the curve at \(P\)	A1
straight line from \(P\) to \((0,\,16.5)\), at right angles to the slip road	A1	The axes in Figure 1 have equal scales, so the right angle should be visible.
\(Q\) and \(R\) labelled in the correct positions	A1	Allow ECF from (c)(i) and (d).
Part (f)
quotient rule with \(u=60x\), \(v=x+5\): \(\frac{dy}{dx}=\frac{(x+5)(60)-60x(1)}{(x+5)^2}\)	M1	Award for the correct order in the numerator.
\(=\frac{300}{(x+5)^2}\)	A1
at \(x=0\): cycle path gradient \(\frac{300}{25}=12\); road gradient \(12e^{0}(1-0)=12\)	A1	Both gradients must be seen.
equal gradients at \(O\), so both have the same tangent \(y=12x\) at \(O\)	AG
Part (g)(i)
product rule on \(12e^{-0.2x}(1-0.2x)\) with \(u=12e^{-0.2x}\), \(v=1-0.2x\)	M1
\(\frac{d^2y}{dx^2}=12\left[-0.2e^{-0.2x}(1-0.2x)+e^{-0.2x}(-0.2)\right]\)	A1	Any correct unsimplified expression.
\(=12e^{-0.2x}(0.04x-0.4)=0.48e^{-0.2x}(x-10)\)	AG
Part (g)(ii)
\(e^{-0.2x}>0\) for all \(x\) and \(x-10<0\) for \(x<10\), so \(\frac{d^2y}{dx^2}<0\): concave down	R1
a concave-down curve lies below its tangents, so near \(P\) the curved road lies below the slip road (the two separate on both sides of \(P\))	R1
Part (h)
\(\frac{dy}{dx}=0\Rightarrow 12e^{-0.2x}(1-0.2x)=0\)	M1
\(e^{-0.2x}\ne0\), so \(x=5\)	A1
\(S(5,\,60e^{-1})=(5,\,22.1)\); road \(y=22.1\)	A1	Accept \(y=60e^{-1}\) or \(y=22.07\).

Answers: (b) \(y=16.1\); gradient \(4.83\) · (c)(i) \(y=4.83x+6.44\) · (d) \((0,\,16.5)\) · (e) tangent through \(Q(-1.33,0)\); normal through \(R(0,16.5)\) · (f) both gradients equal \(12\) at \(O\) · (h) \(S(5,\,22.1)\); \(y=22.1\)

Common error (f) A common error is to reverse the numerator of the quotient rule. It is \(v\dfrac{du}{dx}-u\dfrac{dv}{dx}\): bottom times derivative of top, minus top times derivative of bottom.

More 5.4 questions in the question bank

+ Add to test✓ In your testCC-AA-54-56-57-225 marks~30 min

When do an app's users change fastest? Second derivatives

5.7 The second derivative and graphs5.6 Differentiation rules5.8 Stationary points, inflexion and optimization2.9 Exponential and logarithmic functionsSL

A company launches a new fitness app. The number of active users, \(N\) thousand, \(t\) days after the launch is modelled by

\(N(t)=50t^2e^{-0.25t}\), \(0\le t\le30\).

Figure 1 shows the graph of \(y=N'(t)\). The points \(A\) and \(B\) are the maximum point and the minimum point of this graph.

Figure 1: the graph of \(y=N'(t)\) for \(0\le t\le30\)

(a)

Show that \(N'(t)=12.5te^{-0.25t}(8-t)\).

No calculatorPROOFNotes

Show solution

product rule with \(u=50t^2\), \(v=e^{-0.25t}\)	M1
\(\frac{dv}{dt}=-0.25e^{-0.25t}\)	A1
\(N'(t)=100te^{-0.25t}-12.5t^2e^{-0.25t}\)	A1
\(=12.5te^{-0.25t}(8-t)\)	AG

(3)

(b)

Find the greatest number of active users predicted by the model. Justify that your answer is a maximum.

MULTINotes

Show solution

\(N'(t)=0\)	M1
\(e^{-0.25t}>0\) for all \(t\), so \(t=0\) or \(t=8\); the maximum is at \(t=8\)	A1	Must give a reason for discarding the exponential factor (\(t=0\) gives \(N=0\)).
\(N'\) changes from positive to negative at \(t=8\) (e.g. \(N'(7)=15.2>0\), \(N'(9)=-11.9<0\), or from Figure 1)	R1	Accept a second derivative argument: \(N''(8)=-13.5<0\).
\(N(8)=3200e^{-2}=433\) thousand (\(433.072\ldots\))	A1	Accept 433 000 users.

Answer: \(433\) thousand, at \(t=8\)

(4)

(c)

Show that \(N''(t)=3.125e^{-0.25t}(t^2-16t+32)\).

No calculatorPROOFNotes

Show solution

product rule on \(N'(t)=12.5e^{-0.25t}(8t-t^2)\)	M1
\(N''(t)=12.5\left[e^{-0.25t}(8-2t)-0.25e^{-0.25t}(8t-t^2)\right]\)	A1
\(=12.5e^{-0.25t}(0.25t^2-4t+8)\)	A1
\(=3.125e^{-0.25t}(t^2-16t+32)\)	AG

(3)

(d)

Find the exact values of \(t\) at the points of inflexion on the graph of \(N\). Justify that they are points of inflexion.

No calculatorALGEBRANotes

Show solution

\(t^2-16t+32=0\Rightarrow t=\frac{16\pm\sqrt{256-128}}{2}\)	M1	Accept completing the square: \((t-8)^2=32\).
\(t=8\pm4\sqrt2\)	A1	Both values, exact. Both lie in \(0\le t\le30\).
\(3.125e^{-0.25t}>0\) and the quadratic has two distinct roots, so \(N''\) changes sign at each value	R1

Answer: \(t=8-4\sqrt2\) and \(t=8+4\sqrt2\)

(3)

(e)

The points \(A\) and \(B\) in Figure 1 correspond to the points of inflexion of \(N\). Explain what \(A\) and \(B\) tell the company about its active users.

INTERPRETNotes

Show solution

at \(A\) (\(t=8-4\sqrt2\approx2.34\) days) \(N'\) is greatest: the number of active users is increasing most rapidly	R1
at \(B\) (\(t=8+4\sqrt2\approx13.7\) days) \(N'\) is least (most negative): the number of active users is decreasing most rapidly	R1	Must refer to rates of change of the number of users, not to the number itself.

(2)

(f)

On the axes in Figure 2, sketch the graph of \(N\) for \(0\le t\le30\). Show clearly the coordinates of the maximum point, the two points of inflexion and the end point of the graph.

Figure 2: axes for the graph of \(N\)

GRAPHNotes

Show solution

curve from the origin rising to a single maximum and then falling, staying above the \(t\)-axis	A1
maximum \((8,\,433)\) and end point \((30,\,24.9)\)	A1	Allow ECF from (b).
points of inflexion \((2.34,\,153)\) and \((13.7,\,307)\)	A1
concave up for \(0<t<2.34\), concave down between the inflexions, concave up for \(t>13.7\)	A1

Answer: max \((8,433)\); inflexions \((2.34,153)\), \((13.7,307)\); end \((30,24.9)\)

(4)

(g)

Explain why the model predicts that there are always some active users for \(0<t\le30\). Describe what the model predicts for the number of active users if it is used for values of \(t\) greater than 30.

CONCEPTNotes

Show solution

\(t^2>0\) and \(e^{-0.25t}>0\) for all \(t\), so \(N(t)>0\)	R1
\(N\) continues to decrease and approaches 0 (the \(t\)-axis is an asymptote), but never reaches 0	R1	Accept a description from a GDC graph of \(N\) on a wider window.

(2)

(h)(i)

After a marketing campaign, the company revises its model to \(M(t)=50t^2e^{-0.2t}\), \(0\le t\le30\). It is given that

\(M''(t)=2e^{-0.2t}(t^2-20t+50)\).

Find the exact values of \(t\) at the points of inflexion on the graph of \(M\).

No calculatorROUTINENotes

Show solution

\(t^2-20t+50=0\Rightarrow t=\frac{20\pm\sqrt{200}}{2}\)	M1
\(t=10\pm5\sqrt2\) (\(2.93\) and \(17.1\))	A1

Answer: \(t=10\pm5\sqrt2\)

(2)

(h)(ii)

By comparing your answers to parts (d) and (h)(i), describe the effect of the campaign on when the number of active users grows fastest and when it falls fastest.

INTERPRETNotes

Show solution

both inflexions are later: fastest growth at about 2.93 days instead of 2.34 days, and fastest decline at about 17.1 days instead of 13.7 days	R1
they are further apart (\(10\sqrt2\approx14.1\) days instead of \(8\sqrt2\approx11.3\) days), so the period of rising and then falling use is stretched out	R1	Accept: every time is multiplied by 1.25 (\(=\frac{0.25}{0.2}\)).

(2)

Show full solution

Working	Mark	Notes
Part (a)
product rule with \(u=50t^2\), \(v=e^{-0.25t}\)	M1
\(\frac{dv}{dt}=-0.25e^{-0.25t}\)	A1
\(N'(t)=100te^{-0.25t}-12.5t^2e^{-0.25t}\)	A1
\(=12.5te^{-0.25t}(8-t)\)	AG
Part (b)
\(N'(t)=0\)	M1
\(e^{-0.25t}>0\) for all \(t\), so \(t=0\) or \(t=8\); the maximum is at \(t=8\)	A1	Must give a reason for discarding the exponential factor (\(t=0\) gives \(N=0\)).
\(N'\) changes from positive to negative at \(t=8\) (e.g. \(N'(7)=15.2>0\), \(N'(9)=-11.9<0\), or from Figure 1)	R1	Accept a second derivative argument: \(N''(8)=-13.5<0\).
\(N(8)=3200e^{-2}=433\) thousand (\(433.072\ldots\))	A1	Accept 433 000 users.
Part (c)
product rule on \(N'(t)=12.5e^{-0.25t}(8t-t^2)\)	M1
\(N''(t)=12.5\left[e^{-0.25t}(8-2t)-0.25e^{-0.25t}(8t-t^2)\right]\)	A1
\(=12.5e^{-0.25t}(0.25t^2-4t+8)\)	A1
\(=3.125e^{-0.25t}(t^2-16t+32)\)	AG
Part (d)
\(t^2-16t+32=0\Rightarrow t=\frac{16\pm\sqrt{256-128}}{2}\)	M1	Accept completing the square: \((t-8)^2=32\).
\(t=8\pm4\sqrt2\)	A1	Both values, exact. Both lie in \(0\le t\le30\).
\(3.125e^{-0.25t}>0\) and the quadratic has two distinct roots, so \(N''\) changes sign at each value	R1
Part (e)
at \(A\) (\(t=8-4\sqrt2\approx2.34\) days) \(N'\) is greatest: the number of active users is increasing most rapidly	R1
at \(B\) (\(t=8+4\sqrt2\approx13.7\) days) \(N'\) is least (most negative): the number of active users is decreasing most rapidly	R1	Must refer to rates of change of the number of users, not to the number itself.
Part (f)
curve from the origin rising to a single maximum and then falling, staying above the \(t\)-axis	A1
maximum \((8,\,433)\) and end point \((30,\,24.9)\)	A1	Allow ECF from (b).
points of inflexion \((2.34,\,153)\) and \((13.7,\,307)\)	A1
concave up for \(0<t<2.34\), concave down between the inflexions, concave up for \(t>13.7\)	A1
Part (g)
\(t^2>0\) and \(e^{-0.25t}>0\) for all \(t\), so \(N(t)>0\)	R1
\(N\) continues to decrease and approaches 0 (the \(t\)-axis is an asymptote), but never reaches 0	R1	Accept a description from a GDC graph of \(N\) on a wider window.
Part (h)(i)
\(t^2-20t+50=0\Rightarrow t=\frac{20\pm\sqrt{200}}{2}\)	M1
\(t=10\pm5\sqrt2\) (\(2.93\) and \(17.1\))	A1
Part (h)(ii)
both inflexions are later: fastest growth at about 2.93 days instead of 2.34 days, and fastest decline at about 17.1 days instead of 13.7 days	R1
they are further apart (\(10\sqrt2\approx14.1\) days instead of \(8\sqrt2\approx11.3\) days), so the period of rising and then falling use is stretched out	R1	Accept: every time is multiplied by 1.25 (\(=\frac{0.25}{0.2}\)).

Answers: (b) \(433\) thousand, at \(t=8\) · (d) \(t=8-4\sqrt2\) and \(t=8+4\sqrt2\) · (f) max \((8,433)\); inflexions \((2.34,153)\), \((13.7,307)\); end \((30,24.9)\) · (h)(i) \(t=10\pm5\sqrt2\)

More 5.7 questions in the question bank

5.5 Introduction to integration · 5.10 Indefinite integrals and reverse chain rule · 5.11 Definite integrals and areasmerged group

+ Add to test✓ In your testCC-AA-55-510-511-125 marks~30 min

How much water flows down a river channel?

5.11 Definite integrals and areas5.10 Indefinite integrals and reverse chain rule5.5 Introduction to integration2.4 Key features of graphs and intersections using technologySL

Figure 1 shows the cross-section of a straight river channel. At normal level the water surface lies along the line \(y=0\), from the left bank at \(x=0\) to the right bank at \(x=20\). The river bed is modelled by the curve \(y=f(x)\), where

\(f(x)=0.03x^2-0.6x\), \(0\le x\le20\),

and \(x\) and \(y\) are measured in metres.

Figure 1: cross-section of the river channel at normal water level

(a)

Find \(\displaystyle\int f(x)\,dx\).

No calculatorROUTINENotes

Show solution

\(0.01x^3\)	A1
\(-0.3x^2+C\)	A1	Missing \(+C\): award A1A0 (penalise once in the question).

Answer: \(0.01x^3-0.3x^2+C\)

(2)

(b)

Hence show that the area of the cross-section of the water is \(40\) m².

No calculatorPROOFNotes

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\(\int_0^{20}f(x)\,dx=\left[0.01x^3-0.3x^2\right]_0^{20}\)	M1	Correct limits needed. Allow ECF from (a).
\(=(80-120)-0=-40\)	A1
the bed lies below the \(x\)-axis, so the integral is negative and the area is \(\|-40\|\)	R1	Accept \(-\int_0^{20}f(x)\,dx\) or \(\int_0^{20}\big(0-f(x)\big)\,dx\) written at the start.
area \(=40\) m²	AG

Answer: \(40\) m²

(3)

(c)

At normal level, the water flows through this cross-section with a mean speed of \(1.2\) m s⁻¹.

Find the volume of water that flows through the cross-section in one hour.

ROUTINENotes

Show solution

\(40\times1.2=48\) m³ each second	(M1)	Award for \(40\times1.2\times3600\) seen.
\(172\,800\) m³ \((1.73\times10^5\) m³\()\)	A1

Answer: \(172\,800\) m³

(2)

(d)(i)

After a storm, a layer of sediment settles on the river bed near the left bank. The vertical thickness of the layer is modelled by

\(h(x)=0.03x\sqrt{100-x^2}\) metres, \(0\le x\le10\).

There is no sediment for \(10<x\le20\).

Show that the area of the cross-section of the sediment is \(10\) m².

No calculatorPROOFNotes

Show solution

area \(=\int_0^{10}0.03x\sqrt{100-x^2}\,dx\)	M1
recognises the reverse chain rule, e.g. \(u=100-x^2\), \(du=-2x\,dx\), or a multiple of \((100-x^2)^{3/2}\)	M1
\(\left[-0.01(100-x^2)^{3/2}\right]_0^{10}\)	A1	By substitution: \(-0.015\int_{100}^{0}u^{1/2}\,du=0.01\left[u^{3/2}\right]_0^{100}\).
\(=0-\left(-0.01\times100^{3/2}\right)=0.01\times1000\)	A1	Must see \(100^{3/2}=1000\) or equivalent.
area \(=10\) m²	AG

Answer: \(10\) m²

(4)

(d)(ii)

The upper surface of the sediment is the curve \(y=g(x)\), where \(g(x)=f(x)+h(x)\), \(0\le x\le10\).

On Figure 1, sketch the curve \(y=g(x)\) and shade the region that represents the sediment.

GRAPHNotes

Show solution

curve from \((0,\,0)\), decreasing, lying between the line \(y=0\) and the bed	A1
passes close to \((5,\,-0.95)\) and \((9,\,-1.79)\), becomes very steep near \(x=10\) and meets the bed at \((10,\,-3)\)	A1	Award if the general position is right (within one small square).
region between \(y=g(x)\) and \(y=f(x)\) for \(0\le x\le10\) shaded	A1

(3)

(e)

After the storm, the river carries the same volume of water per second as in part (c), and the water surface stays at normal level.

Assuming the water has the same speed across the whole cross-section, find the new mean speed of the water.

MULTINotes

Show solution

new area \(=40-10=30\) m²	(M1)
\(\frac{48}{30}=1.6\) m s⁻¹	A1	Allow ECF from (c).

Answer: \(1.6\) m s⁻¹

(2)

(f)

In a dry summer, with the sediment still in place, the water surface falls to the line \(y=-1.5\).

Find the width of the water surface.

MULTINotes

Show solution

intersections of \(y=-1.5\) with \(y=g(x)\) and with \(y=f(x)\)	(M1)	May be seen on a GDC sketch.
\(x=8.218\) (on the sediment) and \(x=17.07\) (on the bed)	A1	Both needed. \(17.07=10+5\sqrt2\) may also be found by solving \(f(x)=-1.5\) algebraically.
width \(=17.07-8.218=8.85\) m	A1	Accept 8.85 to 8.88 (8.88 comes from 3 s.f. intersections). \(17.07-2.93=14.1\) (using \(f\) on both sides) is A0.

Answer: \(8.85\) m

(3)

(g)

The sediment is later removed and the channel is dredged, so that the bed becomes the curve \(y=-kx(20-x)\), \(0\le x\le20\), where \(k\) is a positive constant.

At normal level the new cross-sectional area is \(25\%\) greater than the original area of \(40\) m².

Find the value of \(k\).

No calculatorALGEBRANotes

Show solution

area \(=\int_0^{20}kx(20-x)\,dx=k\left[10x^2-\frac{x^3}{3}\right]_0^{20}\)	M1	Accept \(-\int_0^{20}-kx(20-x)\,dx\).
\(=k\left(4000-\frac{8000}{3}\right)=\frac{4000k}{3}\)	A1
\(\frac{4000k}{3}=1.25\times40=50\)	M1
\(k=0.0375\) \(\left(=\frac{3}{80}\right)\)	A1

Answer: \(k=0.0375\)

(4)

(h)

After dredging, the river again carries \(48\) m³ of water per second at normal level.

Find the new mean speed of the water, and suggest one consequence of this change in speed for the sediment carried by the river.

INTERPRETNotes

Show solution

\(\frac{48}{50}=0.96\) m s⁻¹	A1	Allow ECF from (g) only if the area used is 50.
the water is slower than before (0.96 < 1.2), so it carries sediment less easily and more is likely to settle on the bed (the channel may need dredging again)	R1	Accept any sensible consequence that is linked to the lower speed.

Answer: \(0.96\) m s⁻¹

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(0.01x^3\)	A1
\(-0.3x^2+C\)	A1	Missing \(+C\): award A1A0 (penalise once in the question).
Part (b)
\(\int_0^{20}f(x)\,dx=\left[0.01x^3-0.3x^2\right]_0^{20}\)	M1	Correct limits needed. Allow ECF from (a).
\(=(80-120)-0=-40\)	A1
the bed lies below the \(x\)-axis, so the integral is negative and the area is \(\|-40\|\)	R1	Accept \(-\int_0^{20}f(x)\,dx\) or \(\int_0^{20}\big(0-f(x)\big)\,dx\) written at the start.
area \(=40\) m²	AG
Part (c)
\(40\times1.2=48\) m³ each second	(M1)	Award for \(40\times1.2\times3600\) seen.
\(172\,800\) m³ \((1.73\times10^5\) m³\()\)	A1
Part (d)(i)
area \(=\int_0^{10}0.03x\sqrt{100-x^2}\,dx\)	M1
recognises the reverse chain rule, e.g. \(u=100-x^2\), \(du=-2x\,dx\), or a multiple of \((100-x^2)^{3/2}\)	M1
\(\left[-0.01(100-x^2)^{3/2}\right]_0^{10}\)	A1	By substitution: \(-0.015\int_{100}^{0}u^{1/2}\,du=0.01\left[u^{3/2}\right]_0^{100}\).
\(=0-\left(-0.01\times100^{3/2}\right)=0.01\times1000\)	A1	Must see \(100^{3/2}=1000\) or equivalent.
area \(=10\) m²	AG
Part (d)(ii)
curve from \((0,\,0)\), decreasing, lying between the line \(y=0\) and the bed	A1
passes close to \((5,\,-0.95)\) and \((9,\,-1.79)\), becomes very steep near \(x=10\) and meets the bed at \((10,\,-3)\)	A1	Award if the general position is right (within one small square).
region between \(y=g(x)\) and \(y=f(x)\) for \(0\le x\le10\) shaded	A1
Part (e)
new area \(=40-10=30\) m²	(M1)
\(\frac{48}{30}=1.6\) m s⁻¹	A1	Allow ECF from (c).
Part (f)
intersections of \(y=-1.5\) with \(y=g(x)\) and with \(y=f(x)\)	(M1)	May be seen on a GDC sketch.
\(x=8.218\) (on the sediment) and \(x=17.07\) (on the bed)	A1	Both needed. \(17.07=10+5\sqrt2\) may also be found by solving \(f(x)=-1.5\) algebraically.
width \(=17.07-8.218=8.85\) m	A1	Accept 8.85 to 8.88 (8.88 comes from 3 s.f. intersections). \(17.07-2.93=14.1\) (using \(f\) on both sides) is A0.
Part (g)
area \(=\int_0^{20}kx(20-x)\,dx=k\left[10x^2-\frac{x^3}{3}\right]_0^{20}\)	M1	Accept \(-\int_0^{20}-kx(20-x)\,dx\).
\(=k\left(4000-\frac{8000}{3}\right)=\frac{4000k}{3}\)	A1
\(\frac{4000k}{3}=1.25\times40=50\)	M1
\(k=0.0375\) \(\left(=\frac{3}{80}\right)\)	A1
Part (h)
\(\frac{48}{50}=0.96\) m s⁻¹	A1	Allow ECF from (g) only if the area used is 50.
the water is slower than before (0.96 < 1.2), so it carries sediment less easily and more is likely to settle on the bed (the channel may need dredging again)	R1	Accept any sensible consequence that is linked to the lower speed.

Answers: (a) \(0.01x^3-0.3x^2+C\) · (b) \(40\) m² · (c) \(172\,800\) m³ · (d)(i) \(10\) m² · (e) \(1.6\) m s⁻¹ · (f) \(8.85\) m · (g) \(k=0.0375\) · (h) \(0.96\) m s⁻¹

Common error (d)(i) Integrating \(x\) and \(\sqrt{100-x^2}\) separately and multiplying the results: there is no product rule for integrals. The factor \(x\) is a multiple of the derivative of \(100-x^2\), which is what makes the reverse chain rule work.

Exam tip (f) Before using the GDC, look at your sketch: at this height the left edge of the water is on the sediment surface \(y=g(x)\), and the right edge is on the bed \(y=f(x)\).

More 5.11 questions in the question bank

+ Add to test✓ In your testCC-AA-55-510-511-224 marks~30 min

Fast-charging an electric car: energy from power

5.10 Indefinite integrals and reverse chain rule5.5 Introduction to integration5.11 Definite integrals and areas2.9 Exponential and logarithmic functionsSL

An electric car has a battery with a capacity of \(80\) kWh. It is connected to a fast charger. The charging power, \(P\) kilowatts (kW), \(t\) minutes after the car is connected, is modelled by

\(P(t)=37.5\sqrt{3t+1}\) for \(0\le t\le5\), and \(P(t)=150e^{-0.04(t-5)}\) for \(t>5\).

The energy delivered to the battery, in kilowatt-hours (kWh), between \(t=a\) and \(t=b\) is \(\dfrac{1}{60}\displaystyle\int_a^b P(t)\,dt\). Assume that all the energy delivered is stored in the battery.

(a)

Find \(\displaystyle\int37.5\sqrt{3t+1}\,dt\).

No calculatorROUTINENotes

Show solution

attempt at the reverse chain rule, obtaining a multiple of \((3t+1)^{3/2}\)	M1
\(\frac{25}{3}(3t+1)^{3/2}+C\)	A1	Accept \(8.33(3t+1)^{3/2}+C\). Missing \(+C\): A0.

Answer: \(\dfrac{25}{3}(3t+1)^{3/2}+C\)

(2)

(b)

Hence show that \(8.75\) kWh of energy is delivered during the first \(5\) minutes.

No calculatorPROOFNotes

Show solution

\(\left[\frac{25}{3}(3t+1)^{3/2}\right]_0^5=\frac{25}{3}\left(16^{3/2}-1^{3/2}\right)\)	M1	Allow ECF from (a) for the method.
\(=\frac{25}{3}(64-1)=525\)	A1
\(\frac{525}{60}\)	M1	Dividing by 60 must be seen.
\(=8.75\) kWh	AG

Answer: \(8.75\) kWh

(3)

(c)

When \(t=5\), the battery holds \(14.0\) kWh.

Find the energy held by the battery when the car was connected, as a percentage of the battery's capacity.

ROUTINENotes

Show solution

\(14-8.75=5.25\) kWh	(M1)
\(\frac{5.25}{80}\times100=6.56\%\)	A1	Accept 6.5625%.

Answer: \(6.56\%\)

(2)

(d)

Find \(\displaystyle\int150e^{-0.04(t-5)}\,dt\).

No calculatorROUTINENotes

Show solution

a multiple of \(e^{-0.04(t-5)}\)	M1
\(-3750e^{-0.04(t-5)}+C\)	A1	Missing \(+C\): A0 if not already penalised in (a).

Answer: \(-3750e^{-0.04(t-5)}+C\)

(2)

(e)

Let \(E(t)\) be the energy held by the battery, in kWh, at time \(t\).

Show that, for \(t\ge5\), \(E(t)=76.5-62.5e^{-0.04(t-5)}\).

No calculatorPROOFNotes

Show solution

\(E(t)=\frac{1}{60}\int P(t)\,dt=-62.5e^{-0.04(t-5)}+C\)	M1	Allow ECF from (d). Accept \(E(t)=14+\frac{1}{60}\int_5^tP\,dt\) as an alternative method.
\(E(5)=14\): \(14=-62.5e^{0}+C\)	M1
\(C=76.5\)	A1	Alternative method: \(14+62.5\left(1-e^{-0.04(t-5)}\right)\).
\(E(t)=76.5-62.5e^{-0.04(t-5)}\)	AG

(3)

(f)

Show that the battery holds \(80\%\) of its capacity when \(t=5+25\ln5\).

No calculatorPROOFNotes

Show solution

\(76.5-62.5e^{-0.04(t-5)}=0.8\times80=64\)	M1
\(e^{-0.04(t-5)}=\frac{12.5}{62.5}=\frac{1}{5}\)	A1
\(-0.04(t-5)=\ln\frac{1}{5}=-\ln5\)	M1	Taking natural logarithms of both sides.
\(t-5=\frac{\ln5}{0.04}=25\ln5\), so \(t=5+25\ln5\)	AG

(3)

(g)

On the axes in Figure 1, sketch the graph of \(P\) for \(0\le t\le60\). On your sketch, shade the region whose area, divided by \(60\), is the energy delivered from \(t=5\) until the battery holds \(80\%\) of its capacity.

Figure 1: axes for the graph of the charging power

GRAPHNotes

Show solution

increasing curve, concave down, from \((0,\,37.5)\) to \((5,\,150)\)	A1
decreasing curve for \(t>5\), concave up, approaching the \(t\)-axis without reaching it, ending near \((60,\,16.6)\)	A1
region under the curve between \(t=5\) and \(t=45.2\) shaded	A1	\(5+25\ln5=45.2\). Allow ECF from their curve.

(3)

(h)

Write down the equation of the horizontal asymptote of the graph of \(y=E(t)\), \(t\ge5\). Interpret this asymptote in context.

INTERPRETNotes

Show solution

\(E=76.5\)	A1
the stored energy approaches \(76.5\) kWh (\(95.6\%\) of capacity) but never reaches it, so under this model the battery is never fully charged	R1	Must refer to the battery not reaching full charge, or to the energy levelling off below 80 kWh.

Answer: \(E=76.5\)

(2)

(i)

In cold weather the charger reduces its output, and the model becomes

\(P(t)=30\sqrt{3t+1}\) for \(0\le t\le5\), and \(P(t)=120e^{-0.05(t-5)}\) for \(t>5\).

The car is connected with the battery holding the same energy as in part (c). Determine whether, under this model, the battery ever holds \(80\%\) of its capacity. Justify your answer.

MULTINotes

Show solution

first 5 minutes: \(\frac{30}{37.5}\times8.75=7\) kWh	M1	Or \(\frac{1}{60}\times\frac{20}{3}(64-1)=7\).
for \(t>5\): \(\frac{1}{60}\int_5^t120e^{-0.05(u-5)}\,du=40\left(1-e^{-0.05(t-5)}\right)\), which approaches \(40\) kWh	M1	Accept \(\frac{120}{60\times0.05}=40\).
greatest possible energy \(=5.25+7+40=52.25\) kWh	A1	Allow ECF from (c).
\(52.25<64\) (it is only \(65.3\%\) of capacity), so the battery never holds \(80\%\) of its capacity	R1	Do not award R1 without a value compared with 64 kWh or 80%.

Answer: No: the energy approaches \(52.25\) kWh, less than \(64\) kWh

(4)

Show full solution

Working	Mark	Notes
Part (a)
attempt at the reverse chain rule, obtaining a multiple of \((3t+1)^{3/2}\)	M1
\(\frac{25}{3}(3t+1)^{3/2}+C\)	A1	Accept \(8.33(3t+1)^{3/2}+C\). Missing \(+C\): A0.
Part (b)
\(\left[\frac{25}{3}(3t+1)^{3/2}\right]_0^5=\frac{25}{3}\left(16^{3/2}-1^{3/2}\right)\)	M1	Allow ECF from (a) for the method.
\(=\frac{25}{3}(64-1)=525\)	A1
\(\frac{525}{60}\)	M1	Dividing by 60 must be seen.
\(=8.75\) kWh	AG
Part (c)
\(14-8.75=5.25\) kWh	(M1)
\(\frac{5.25}{80}\times100=6.56\%\)	A1	Accept 6.5625%.
Part (d)
a multiple of \(e^{-0.04(t-5)}\)	M1
\(-3750e^{-0.04(t-5)}+C\)	A1	Missing \(+C\): A0 if not already penalised in (a).
Part (e)
\(E(t)=\frac{1}{60}\int P(t)\,dt=-62.5e^{-0.04(t-5)}+C\)	M1	Allow ECF from (d). Accept \(E(t)=14+\frac{1}{60}\int_5^tP\,dt\) as an alternative method.
\(E(5)=14\): \(14=-62.5e^{0}+C\)	M1
\(C=76.5\)	A1	Alternative method: \(14+62.5\left(1-e^{-0.04(t-5)}\right)\).
\(E(t)=76.5-62.5e^{-0.04(t-5)}\)	AG
Part (f)
\(76.5-62.5e^{-0.04(t-5)}=0.8\times80=64\)	M1
\(e^{-0.04(t-5)}=\frac{12.5}{62.5}=\frac{1}{5}\)	A1
\(-0.04(t-5)=\ln\frac{1}{5}=-\ln5\)	M1	Taking natural logarithms of both sides.
\(t-5=\frac{\ln5}{0.04}=25\ln5\), so \(t=5+25\ln5\)	AG
Part (g)
increasing curve, concave down, from \((0,\,37.5)\) to \((5,\,150)\)	A1
decreasing curve for \(t>5\), concave up, approaching the \(t\)-axis without reaching it, ending near \((60,\,16.6)\)	A1
region under the curve between \(t=5\) and \(t=45.2\) shaded	A1	\(5+25\ln5=45.2\). Allow ECF from their curve.
Part (h)
\(E=76.5\)	A1
the stored energy approaches \(76.5\) kWh (\(95.6\%\) of capacity) but never reaches it, so under this model the battery is never fully charged	R1	Must refer to the battery not reaching full charge, or to the energy levelling off below 80 kWh.
Part (i)
first 5 minutes: \(\frac{30}{37.5}\times8.75=7\) kWh	M1	Or \(\frac{1}{60}\times\frac{20}{3}(64-1)=7\).
for \(t>5\): \(\frac{1}{60}\int_5^t120e^{-0.05(u-5)}\,du=40\left(1-e^{-0.05(t-5)}\right)\), which approaches \(40\) kWh	M1	Accept \(\frac{120}{60\times0.05}=40\).
greatest possible energy \(=5.25+7+40=52.25\) kWh	A1	Allow ECF from (c).
\(52.25<64\) (it is only \(65.3\%\) of capacity), so the battery never holds \(80\%\) of its capacity	R1	Do not award R1 without a value compared with 64 kWh or 80%.

Answers: (a) \(\dfrac{25}{3}(3t+1)^{3/2}+C\) · (b) \(8.75\) kWh · (c) \(6.56\%\) · (d) \(-3750e^{-0.04(t-5)}+C\) · (h) \(E=76.5\) · (i) No: the energy approaches \(52.25\) kWh, less than \(64\) kWh

Common error (a) Forgetting to divide by the 3 from the inside of the bracket: \(\int(3t+1)^{1/2}\,dt=\frac{1}{3}\times\frac{2}{3}(3t+1)^{3/2}+C\), not \(\frac{2}{3}(3t+1)^{3/2}+C\). Differentiate your answer to check.

Exam tip (f) Make the exponential the subject before you take logarithms. \(\ln\left(76.5-62.5e^{-0.04(t-5)}\right)\) cannot be split into separate logarithms.

More 5.10 questions in the question bank

5.8 Stationary points, inflexion and optimization · 5.9 Kinematicsmerged group

+ Add to test✓ In your testCC-AA-58-59-123 marks~30 min

Designing an athletics track around a playing field

5.8 Stationary points, inflexion and optimization3.4 Radians, arc length and sector area5.7 The second derivative and graphs2.6 The quadratic function and its formsSL

The inside edge of the innermost lane of an athletics track is \(400\) m long. It is made of two parallel straights, each of length \(x\) metres, joined at each end by a semicircle of radius \(r\) metres, as shown in Figure 1.

The rectangular field between the two straights is used for football.

Figure 1: inside edge of the innermost lane, with the field between the straights (not to scale)

(a)

Show that \(x=200-\pi r\).

No calculatorPROOFNotes

Show solution

each semicircle has arc length \(l=r\theta=\pi r\) (\(\theta=\pi\) radians)	M1	Accept \(\frac12(2\pi r)\).
\(2x+2\pi r=400\)	A1
\(x=200-\pi r\)	AG

(2)

(b)(i)

Show that the area of the field, \(A\) m², is given by \(A=400r-2\pi r^2\).

No calculatorALGEBRANotes

Show solution

\(A=2r\times x=2r(200-\pi r)\)	A1	The width of the field is the diameter \(2r\).
\(A=400r-2\pi r^2\)	AG

(1)

(b)(ii)

Write down the set of possible values of \(r\) for this design.

No calculatorCONCEPTNotes

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\(0<r<\frac{200}{\pi}\)

From \(r>0\) and \(x>0\). Accept \(0<r<63.7\), and \(\le\) at either end.

Answer: \(0<r<\dfrac{200}{\pi}\)

(1)

(c)

Use differentiation to find the exact value of \(r\) for which \(A\) is greatest. Hence find the greatest area of the field.

MULTINotes

Show solution

\(\frac{dA}{dr}=400-4\pi r=0\)	M1	Attempt to differentiate and set the derivative equal to zero.
\(r=\frac{100}{\pi}\)	A1	Must be exact for this mark; \(31.8\) alone is A0.
\(A=400\times\frac{100}{\pi}-2\pi\left(\frac{100}{\pi}\right)^2\)	M1	Substituting their \(r\) into \(A\).
\(A=\frac{20000}{\pi}=6370\) m²	A1	Accept 6366. Then \(x=200-100=100\) m.

Answer: \(r=\dfrac{100}{\pi}\), \(A=6370\) m²

(4)

(d)

Use the second derivative to justify that this value of \(r\) gives a maximum.

No calculatorCONCEPTNotes

Show solution

\(\frac{d^2A}{dr^2}=-4\pi\)	A1
\(-4\pi<0\), so the stationary point is a maximum	R1	Do not award R1 without a correct second derivative.

(2)

(e)

On the axes in Figure 2, sketch the graph of \(A\) against \(r\) for the values of \(r\) in part (b)(ii). Label the maximum point and the points where the graph meets the \(r\)-axis.

Figure 2: axes for the graph of A against r

GRAPHNotes

Show solution

concave-down parabola, symmetric about \(r=31.8\)	A1
meets the \(r\)-axis at \((0,\,0)\) and \((63.7,\,0)\)	A1	From the factorised form \(A=2r(200-\pi r)\). Open or closed end points are both accepted.
maximum point \((31.8,\,6370)\) labelled	A1	Allow ECF from (c).

(3)

(f)

The football pitch must be at least \(68\) m wide, so \(2r\ge68\).

Find the greatest possible area of the field under this condition, and the corresponding length of each straight. Justify your answer.

MULTINotes

Show solution

\(\frac{100}{\pi}=31.8<34\) and \(A\) is decreasing for \(r>\frac{100}{\pi}\), so the greatest area is at the end point \(r=34\)	R1	Accept a reference to the graph in (e), or \(\frac{dA}{dr}<0\) for \(r\ge34\).
\(A=68(200-34\pi)=6340\) m²	A1	Accept 6337.
\(x=200-34\pi=93.2\) m	A1

Answer: \(6340\) m², \(x=93.2\) m

(3)

(g)(i)

The region enclosed by the inside edge of the lane consists of the field together with the two semicircular regions at its ends. Let the total area of this region be \(T\) m².

Show that \(T=400r-\pi r^2\).

No calculatorPROOFNotes

Show solution

the two semicircles together have area \(\pi r^2\)	M1	Accept \(2\times\frac12r^2\pi\) (sector area with \(\theta=\pi\)).
\(T=400r-2\pi r^2+\pi r^2\)	A1
\(T=400r-\pi r^2\)	AG

(2)

(g)(ii)

Find the value of \(r\) for which \(T\) is greatest, and the corresponding value of \(x\).

No calculatorROUTINENotes

Show solution

\(\frac{dT}{dr}=400-2\pi r=0\)	M1	Or the axis of symmetry \(r=-\frac{400}{2(-\pi)}\) of the quadratic.
\(r=\frac{200}{\pi}\)	A1	Accept 63.7.
\(x=200-\pi\times\frac{200}{\pi}=0\)	A1

Answer: \(r=\dfrac{200}{\pi}\), \(x=0\)

(3)

(h)

Interpret your answer to part (g)(ii) geometrically, and explain why this design would not be used for an athletics track.

INTERPRETNotes

Show solution

with \(x=0\) the lane is a complete circle of circumference \(400\) m: for a fixed perimeter, the circle encloses the greatest area	R1
a track with no straights cannot be used for races that need a straight (for example the 100 m) and leaves no rectangular field for a pitch	R1	Accept any sensible practical reason.

(2)

Show full solution

Working	Mark	Notes
Part (a)
each semicircle has arc length \(l=r\theta=\pi r\) (\(\theta=\pi\) radians)	M1	Accept \(\frac12(2\pi r)\).
\(2x+2\pi r=400\)	A1
\(x=200-\pi r\)	AG
Part (b)(i)
\(A=2r\times x=2r(200-\pi r)\)	A1	The width of the field is the diameter \(2r\).
\(A=400r-2\pi r^2\)	AG
Part (b)(ii)
\(0<r<\frac{200}{\pi}\)	A1	From \(r>0\) and \(x>0\). Accept \(0<r<63.7\), and \(\le\) at either end.
Part (c)
\(\frac{dA}{dr}=400-4\pi r=0\)	M1	Attempt to differentiate and set the derivative equal to zero.
\(r=\frac{100}{\pi}\)	A1	Must be exact for this mark; \(31.8\) alone is A0.
\(A=400\times\frac{100}{\pi}-2\pi\left(\frac{100}{\pi}\right)^2\)	M1	Substituting their \(r\) into \(A\).
\(A=\frac{20000}{\pi}=6370\) m²	A1	Accept 6366. Then \(x=200-100=100\) m.
Part (d)
\(\frac{d^2A}{dr^2}=-4\pi\)	A1
\(-4\pi<0\), so the stationary point is a maximum	R1	Do not award R1 without a correct second derivative.
Part (e)
concave-down parabola, symmetric about \(r=31.8\)	A1
meets the \(r\)-axis at \((0,\,0)\) and \((63.7,\,0)\)	A1	From the factorised form \(A=2r(200-\pi r)\). Open or closed end points are both accepted.
maximum point \((31.8,\,6370)\) labelled	A1	Allow ECF from (c).
Part (f)
\(\frac{100}{\pi}=31.8<34\) and \(A\) is decreasing for \(r>\frac{100}{\pi}\), so the greatest area is at the end point \(r=34\)	R1	Accept a reference to the graph in (e), or \(\frac{dA}{dr}<0\) for \(r\ge34\).
\(A=68(200-34\pi)=6340\) m²	A1	Accept 6337.
\(x=200-34\pi=93.2\) m	A1
Part (g)(i)
the two semicircles together have area \(\pi r^2\)	M1	Accept \(2\times\frac12r^2\pi\) (sector area with \(\theta=\pi\)).
\(T=400r-2\pi r^2+\pi r^2\)	A1
\(T=400r-\pi r^2\)	AG
Part (g)(ii)
\(\frac{dT}{dr}=400-2\pi r=0\)	M1	Or the axis of symmetry \(r=-\frac{400}{2(-\pi)}\) of the quadratic.
\(r=\frac{200}{\pi}\)	A1	Accept 63.7.
\(x=200-\pi\times\frac{200}{\pi}=0\)	A1
Part (h)
with \(x=0\) the lane is a complete circle of circumference \(400\) m: for a fixed perimeter, the circle encloses the greatest area	R1
a track with no straights cannot be used for races that need a straight (for example the 100 m) and leaves no rectangular field for a pitch	R1	Accept any sensible practical reason.

Answers: (b)(ii) \(0<r<\dfrac{200}{\pi}\) · (c) \(r=\dfrac{100}{\pi}\), \(A=6370\) m² · (f) \(6340\) m², \(x=93.2\) m · (g)(ii) \(r=\dfrac{200}{\pi}\), \(x=0\)

Common error (a) Using \(2\pi r\) for each semicircle. The two semicircles together make one full circle, so their total length is \(2\pi r\).

Exam tip (f) When the stationary point lies outside the allowed values, the greatest value is at an end point of the interval. Check the sign of the derivative or the graph to choose the end.

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+ Add to test✓ In your testCC-AA-58-59-225 marks~30 min

A lift to an observation deck: velocity, acceleration and distance

5.9 Kinematics5.8 Stationary points, inflexion and optimization5.11 Definite integrals and areas5.10 Indefinite integrals and reverse chain ruleSL

A lift carries visitors from the ground floor of a tower up to an observation deck. The velocity of the lift, \(v\) m s⁻¹, \(t\) seconds after it starts moving upwards, is modelled by

\(v(t)=0.06t(20-t)\), \(0\le t\le20\).

The lift is at rest at the deck when \(t=20\). Figure 1 shows the graph of \(v\) against \(t\).

Figure 1: velocity of the lift during the ascent

(a)

Find an expression for the acceleration, \(a\) m s⁻², of the lift in terms of \(t\). Hence find the acceleration when \(t=5\).

No calculatorROUTINENotes

Show solution

\(a=1.2-0.12t\)	A1	From \(v=1.2t-0.06t^2\).
\(a(5)=0.6\) m s⁻²	A1	Allow ECF from their \(a\).

Answer: \(a=1.2-0.12t\); \(0.6\) m s⁻²

(2)

(b)

Use differentiation to find the greatest speed of the lift, and justify that it is a maximum.

No calculatorROUTINENotes

Show solution

\(1.2-0.12t=0\Rightarrow t=10\)	M1
\(v(10)=0.06\times10\times10=6\) m s⁻¹	A1
\(a\) changes from positive to negative at \(t=10\) (or \(\frac{da}{dt}=-0.12<0\)), so this is a maximum	R1	Comparing with \(v(0)=v(20)=0\) at the end points is also accepted.

Answer: \(6\) m s⁻¹

(3)

(c)

Show that the observation deck is \(80\) m above the point where the lift starts.

No calculatorPROOFNotes

Show solution

\(\int_0^{20}\left(1.2t-0.06t^2\right)dt\)	M1	\(v\ge0\) throughout, so the displacement is the height gained.
\(\left[0.6t^2-0.02t^3\right]_0^{20}\)	A1
\(=240-160\)	A1	Must see both values.
\(=80\) m	AG

Answer: \(80\) m

(3)

(d)

On the axes in Figure 2, sketch the graph of \(a\) against \(t\) for \(0\le t\le20\). State the feature of your graph that corresponds to the highest point of the graph in Figure 1.

Figure 2: axes for the graph of acceleration against time

GRAPHNotes

Show solution

straight line with negative gradient	A1
from \((0,\,1.2)\) to \((20,\,-1.2)\), crossing the \(t\)-axis at \(t=10\)	A1	Allow ECF from (a).
\(a=0\) (the graph meets the \(t\)-axis) at \(t=10\): the gradient of the \(v\)–\(t\) graph is zero where \(v\) is greatest	R1

(3)

(e)

A restaurant floor is \(20\) m above the point where the lift starts.

Find the time at which the lift passes the restaurant floor on its way up, and the speed of the lift at that moment.

MULTINotes

Show solution

\(0.6t^2-0.02t^3=20\) or \(\int_0^{T}v\,dt=20\)	M1
\(t=6.53\) s	A1	Reject \(-5.32\) and \(28.8\), which lie outside \(0\le t\le20\).
\(v(6.527)=5.28\) m s⁻¹	A1	Accept 5.27 to 5.28.

Answer: \(t=6.53\) s, \(5.28\) m s⁻¹

(3)

(f)

The lift then returns to the ground floor. On the way down its velocity, \(w\) m s⁻¹, is modelled by

\(w(t)=-k\sin\left(\dfrac{\pi t}{20}\right)\), \(0\le t\le20\),

where \(t\) is now the time in seconds since the lift started moving down, and \(k\) is a positive constant.

Show that \(k=2\pi\).

No calculatorPROOFNotes

Show solution

\(\int_0^{20}-k\sin\left(\frac{\pi t}{20}\right)dt=-80\)	M1	Displacement of \(-80\) m (or a distance of 80 m) must be used.
\(\left[\frac{20k}{\pi}\cos\left(\frac{\pi t}{20}\right)\right]_0^{20}\)	A1
\(=\frac{20k}{\pi}(-1-1)=-\frac{40k}{\pi}\)	A1
\(-\frac{40k}{\pi}=-80\Rightarrow k=2\pi\)	AG

(3)

(g)

For the complete round trip, up and down, write down the displacement of the lift and the total distance it travels. Explain why these two values are different.

INTERPRETNotes

Show solution

displacement \(0\) m and distance \(160\) m	A1	Both needed.
the velocity is negative on the way down, so the integral of velocity adds \(+80\) and \(-80\); distance adds the speed (\(\|v\|\)), so the two 80 m journeys add	R1

Answer: \(0\) m; \(160\) m

(2)

(h)

For passenger comfort, the magnitude of the acceleration of the lift should never exceed \(1.0\) m s⁻².

Determine whether the ascent and the descent each meet this condition.

MULTINotes

Show solution

ascent: greatest \(\|a\|=1.2\) m s⁻² (at \(t=0\) and \(t=20\)), which exceeds 1.0, so the ascent does not meet the condition	A1	Allow ECF from (a) or (d).
descent: \(\frac{dw}{dt}=-2\pi\times\frac{\pi}{20}\cos\left(\frac{\pi t}{20}\right)=-\frac{\pi^2}{10}\cos\left(\frac{\pi t}{20}\right)\)	M1
greatest \(\|a\|=\frac{\pi^2}{10}=0.987\) m s⁻² \(\le1.0\), so the descent meets the condition	A1

Answer: ascent: no (\(1.2\)); descent: yes (\(0.987\))

(3)

(i)

The ascent is redesigned so that the velocity is \(v(t)=ct(T-t)\), \(0\le t\le T\), where \(c\) and \(T\) are positive constants. The lift must still rise \(80\) m, and the greatest magnitude of its acceleration must be exactly \(1.0\) m s⁻².

Find the value of \(T\).

MODELNotes

Show solution

\(\int_0^{T}ct(T-t)\,dt=\frac{cT^3}{6}=80\)	M1
\(a=c(T-2t)\), greatest \(\|a\|=cT\) (at \(t=0\) and \(t=T\)), so \(cT=1\)	M1
\(\frac{T^2}{6}=80\Rightarrow T=\sqrt{480}=21.9\) s	A1

Answer: \(T=21.9\) s

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(a=1.2-0.12t\)	A1	From \(v=1.2t-0.06t^2\).
\(a(5)=0.6\) m s⁻²	A1	Allow ECF from their \(a\).
Part (b)
\(1.2-0.12t=0\Rightarrow t=10\)	M1
\(v(10)=0.06\times10\times10=6\) m s⁻¹	A1
\(a\) changes from positive to negative at \(t=10\) (or \(\frac{da}{dt}=-0.12<0\)), so this is a maximum	R1	Comparing with \(v(0)=v(20)=0\) at the end points is also accepted.
Part (c)
\(\int_0^{20}\left(1.2t-0.06t^2\right)dt\)	M1	\(v\ge0\) throughout, so the displacement is the height gained.
\(\left[0.6t^2-0.02t^3\right]_0^{20}\)	A1
\(=240-160\)	A1	Must see both values.
\(=80\) m	AG
Part (d)
straight line with negative gradient	A1
from \((0,\,1.2)\) to \((20,\,-1.2)\), crossing the \(t\)-axis at \(t=10\)	A1	Allow ECF from (a).
\(a=0\) (the graph meets the \(t\)-axis) at \(t=10\): the gradient of the \(v\)–\(t\) graph is zero where \(v\) is greatest	R1
Part (e)
\(0.6t^2-0.02t^3=20\) or \(\int_0^{T}v\,dt=20\)	M1
\(t=6.53\) s	A1	Reject \(-5.32\) and \(28.8\), which lie outside \(0\le t\le20\).
\(v(6.527)=5.28\) m s⁻¹	A1	Accept 5.27 to 5.28.
Part (f)
\(\int_0^{20}-k\sin\left(\frac{\pi t}{20}\right)dt=-80\)	M1	Displacement of \(-80\) m (or a distance of 80 m) must be used.
\(\left[\frac{20k}{\pi}\cos\left(\frac{\pi t}{20}\right)\right]_0^{20}\)	A1
\(=\frac{20k}{\pi}(-1-1)=-\frac{40k}{\pi}\)	A1
\(-\frac{40k}{\pi}=-80\Rightarrow k=2\pi\)	AG
Part (g)
displacement \(0\) m and distance \(160\) m	A1	Both needed.
the velocity is negative on the way down, so the integral of velocity adds \(+80\) and \(-80\); distance adds the speed (\(\|v\|\)), so the two 80 m journeys add	R1
Part (h)
ascent: greatest \(\|a\|=1.2\) m s⁻² (at \(t=0\) and \(t=20\)), which exceeds 1.0, so the ascent does not meet the condition	A1	Allow ECF from (a) or (d).
descent: \(\frac{dw}{dt}=-2\pi\times\frac{\pi}{20}\cos\left(\frac{\pi t}{20}\right)=-\frac{\pi^2}{10}\cos\left(\frac{\pi t}{20}\right)\)	M1
greatest \(\|a\|=\frac{\pi^2}{10}=0.987\) m s⁻² \(\le1.0\), so the descent meets the condition	A1
Part (i)
\(\int_0^{T}ct(T-t)\,dt=\frac{cT^3}{6}=80\)	M1
\(a=c(T-2t)\), greatest \(\|a\|=cT\) (at \(t=0\) and \(t=T\)), so \(cT=1\)	M1
\(\frac{T^2}{6}=80\Rightarrow T=\sqrt{480}=21.9\) s	A1

Answers: (a) \(a=1.2-0.12t\); \(0.6\) m s⁻² · (b) \(6\) m s⁻¹ · (c) \(80\) m · (e) \(t=6.53\) s, \(5.28\) m s⁻¹ · (g) \(0\) m; \(160\) m · (h) ascent: no (\(1.2\)); descent: yes (\(0.987\)) · (i) \(T=21.9\) s

Common error (h) Checking the acceleration only at the time of greatest speed, where it is zero. For the ascent, \(a\) is linear in \(t\), so its greatest size is at an end point of the interval.

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5.12 Limits, first principles and higher derivatives · 5.13 Limits and l’Hôpital’s rulemerged group

+ Add to test✓ In your testCC-AA-512-513-127 marks~30 min

From monthly to continuous compounding: a limit

5.13 Limits and l’Hôpital’s rule5.12 Limits, first principles and higher derivatives1.4 Financial applications2.9 Exponential and logarithmic functionsHL

Eleni deposits €6000 in a savings account for one year. The account pays a nominal annual interest rate of 4.8%, compounded \(k\) times during the year at equal intervals, where \(k\in\mathbb{Z}^+\). The value of the account at the end of the year, in euros, is

\(V(k)=6000\left(1+\dfrac{0.048}{k}\right)^{k}.\)

(a)

Find the value of the account at the end of the year when the interest is compounded annually, monthly and daily (take \(k=365\)). Give your answers to the nearest cent.

ROUTINENotes

Show solution

annually: €6288.00	A1	\(6000\times1.048\).
monthly: \(6000\times1.004^{12}=\)€6294.42; daily: \(6000\left(1+\frac{0.048}{365}\right)^{365}=\)€6295.00	A1	Both values needed. Accept 6295 for the daily value.

Answer: €6288.00, €6294.42, €6295.00

(2)

(b)

Show that \(\ln V(k)=\ln6000+k\ln\left(1+\dfrac{0.048}{k}\right)\).

No calculatorALGEBRANotes

Show solution

\(\ln V(k)=\ln6000+\ln\left(1+\frac{0.048}{k}\right)^{k}\)	M1	Use of \(\ln(ab)=\ln a+\ln b\).
\(\ln\left(1+\frac{0.048}{k}\right)^{k}=k\ln\left(1+\frac{0.048}{k}\right)\)	A1	Use of \(\ln(u^k)=k\ln u\).
\(\ln V(k)=\ln6000+k\ln\left(1+\frac{0.048}{k}\right)\)	AG

(2)

(c)

Let \(x=\dfrac1k\). By writing \(k\ln\left(1+\dfrac{0.048}{k}\right)\) in terms of \(x\), use l’Hôpital’s rule to show that

\(\displaystyle\lim_{k\to\infty}k\ln\left(1+\dfrac{0.048}{k}\right)=0.048.\)

No calculatorPROOFNotes

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\(k\ln\left(1+\frac{0.048}{k}\right)=\frac{\ln(1+0.048x)}{x}\), and \(x\to0^+\) as \(k\to\infty\)	M1
at \(x=0\) the numerator is \(\ln1=0\) and the denominator is 0, so the form is \(\frac00\)	R1	The form must be stated to justify the rule.
differentiating numerator and denominator separately: \(\frac{\frac{0.048}{1+0.048x}}{1}\)	M1	Not the quotient rule.
\(\to\frac{0.048}{1}=0.048\) as \(x\to0\)	A1
\(\displaystyle\lim_{k\to\infty}k\ln\left(1+\frac{0.048}{k}\right)=0.048\)	AG

(4)

(d)

Hence find the exact value of \(\displaystyle\lim_{k\to\infty}V(k)\), and give this value to the nearest cent.

MULTINotes

Show solution

\(\ln V(k)\to\ln6000+0.048\), so \(V(k)\to e^{\ln6000+0.048}\)	(M1)	Or \(V(k)=6000\,e^{k\ln(1+0.048/k)}\).
\(6000e^{0.048}=\)€6295.02	A1	Both the exact form and the value are needed.

Answer: \(6000e^{0.048}\approx\)€6295.02

(2)

(e)(i)

Let \(f(x)=6000\left(1+\dfrac{0.048}{x}\right)^{x}\) for \(x\ge1\), so that \(f(k)=V(k)\) for every positive integer \(k\).

On the axes in Figure 1, sketch the graph of \(y=f(x)\) for \(1\le x\le40\). Show the horizontal asymptote, with its equation, and the coordinates of the end point where \(x=1\).

Figure 1: axes for the graph of \(y=f(x)\)

GRAPHNotes

Show solution

curve starting at \((1,\,6288)\)	A1
increasing curve that rises steeply at first and then flattens, staying below the asymptote	A1	Must not cross or touch the asymptote.
horizontal asymptote \(y=6000e^{0.048}\) (\(\approx6295.02\)) drawn and labelled	A1	Allow ECF from (d).

(3)

(e)(ii)

Interpret the horizontal asymptote in the context of Eleni’s account.

INTERPRETNotes

Show solution

however often the interest is compounded, the value after one year cannot exceed €6295.02; it approaches this value (continuous compounding) as \(k\) increases

Must refer to the value at the end of the year and to more frequent compounding.

(1)

(f)

With interest compounded continuously, the value of the account after \(t\) years is \(A(t)=6000e^{0.048t}\).

Write down a limit, from first principles, that is equal to \(A'(0)\). Use l’Hôpital’s rule to evaluate this limit, and interpret your answer in context.

No calculatorCONCEPTNotes

Show solution

\(A'(0)=\displaystyle\lim_{h\to0}\frac{6000e^{0.048h}-6000}{h}\)	A1
form \(\frac00\); differentiating top and bottom with respect to \(h\): \(\frac{288e^{0.048h}}{1}\)	M1
\(A'(0)=288\)	A1
at the start the account is growing at €288 per year (the same as the annual interest under simple interest at 4.8%)	R1	Units (euros per year) needed.

Answer: \(A'(0)=288\) euros per year

(4)

(g)

Use l’Hôpital’s rule to show that \(\displaystyle\lim_{x\to0}\frac{\ln(1+0.048x)-0.048x}{x^2}=-\frac{0.048^2}{2}\).

No calculatorPROOFNotes

Show solution

form \(\frac00\); differentiating numerator and denominator separately	M1	The form must be named.
\(\frac{\frac{0.048}{1+0.048x}-0.048}{2x}\), which is again \(\frac00\) at \(x=0\)	A1	Correct expression and the form both needed.
second application: \(\frac{-\frac{0.048^2}{(1+0.048x)^2}}{2}\)	M1	Accept \(-0.002304(1+0.048x)^{-2}\) for the numerator.
at \(x=0\): \(\frac{-0.048^2}{2}\)	A1
\(\displaystyle\lim_{x\to0}\frac{\ln(1+0.048x)-0.048x}{x^2}=-\frac{0.048^2}{2}\)	AG

(4)

(h)

Let \(L\) be the limit found in part (d). It can be deduced from part (g) that, for large values of \(k\),

\(V(k)\approx L\,e^{-0.001152/k}.\)

Use this approximation to estimate \(L-V(365)\). Compare your estimate with your answers to parts (a) and (d).

MULTINotes

Show solution

\(L-V(365)\approx6000e^{0.048}\left(1-e^{-0.001152/365}\right)\)	(M1)
\(\approx0.0199\) euros (about 2 cents), which agrees with \(6295.02-6295.00=0.02\)	A1	Accept 0.02. Comparison needed.

Answer: about €0.02

(2)

(i)

The bank changes the nominal annual interest rate to 7%. All other conditions are unchanged.

Find the value of the account after one year with daily compounding and with continuous compounding. Hence comment on whether, for this deposit, it matters in practice if interest is compounded daily or continuously.

INTERPRETNotes

Show solution

daily: \(6000\left(1+\frac{0.07}{365}\right)^{365}=\)€6435.01	A1
continuous: \(6000e^{0.07}=\)€6435.05	A1
the difference is only about 4 cents on €6000, so in practice it does not matter; the change of rate (about €140 more) matters far more than the compounding frequency	R1	Any sensible comment based on the size of the difference.

Answer: €6435.01 and €6435.05

(3)

Show full solution

Working	Mark	Notes
Part (a)
annually: €6288.00	A1	\(6000\times1.048\).
monthly: \(6000\times1.004^{12}=\)€6294.42; daily: \(6000\left(1+\frac{0.048}{365}\right)^{365}=\)€6295.00	A1	Both values needed. Accept 6295 for the daily value.
Part (b)
\(\ln V(k)=\ln6000+\ln\left(1+\frac{0.048}{k}\right)^{k}\)	M1	Use of \(\ln(ab)=\ln a+\ln b\).
\(\ln\left(1+\frac{0.048}{k}\right)^{k}=k\ln\left(1+\frac{0.048}{k}\right)\)	A1	Use of \(\ln(u^k)=k\ln u\).
\(\ln V(k)=\ln6000+k\ln\left(1+\frac{0.048}{k}\right)\)	AG
Part (c)
\(k\ln\left(1+\frac{0.048}{k}\right)=\frac{\ln(1+0.048x)}{x}\), and \(x\to0^+\) as \(k\to\infty\)	M1
at \(x=0\) the numerator is \(\ln1=0\) and the denominator is 0, so the form is \(\frac00\)	R1	The form must be stated to justify the rule.
differentiating numerator and denominator separately: \(\frac{\frac{0.048}{1+0.048x}}{1}\)	M1	Not the quotient rule.
\(\to\frac{0.048}{1}=0.048\) as \(x\to0\)	A1
\(\displaystyle\lim_{k\to\infty}k\ln\left(1+\frac{0.048}{k}\right)=0.048\)	AG
Part (d)
\(\ln V(k)\to\ln6000+0.048\), so \(V(k)\to e^{\ln6000+0.048}\)	(M1)	Or \(V(k)=6000\,e^{k\ln(1+0.048/k)}\).
\(6000e^{0.048}=\)€6295.02	A1	Both the exact form and the value are needed.
Part (e)(i)
curve starting at \((1,\,6288)\)	A1
increasing curve that rises steeply at first and then flattens, staying below the asymptote	A1	Must not cross or touch the asymptote.
horizontal asymptote \(y=6000e^{0.048}\) (\(\approx6295.02\)) drawn and labelled	A1	Allow ECF from (d).
Part (e)(ii)
however often the interest is compounded, the value after one year cannot exceed €6295.02; it approaches this value (continuous compounding) as \(k\) increases	R1	Must refer to the value at the end of the year and to more frequent compounding.
Part (f)
\(A'(0)=\displaystyle\lim_{h\to0}\frac{6000e^{0.048h}-6000}{h}\)	A1
form \(\frac00\); differentiating top and bottom with respect to \(h\): \(\frac{288e^{0.048h}}{1}\)	M1
\(A'(0)=288\)	A1
at the start the account is growing at €288 per year (the same as the annual interest under simple interest at 4.8%)	R1	Units (euros per year) needed.
Part (g)
form \(\frac00\); differentiating numerator and denominator separately	M1	The form must be named.
\(\frac{\frac{0.048}{1+0.048x}-0.048}{2x}\), which is again \(\frac00\) at \(x=0\)	A1	Correct expression and the form both needed.
second application: \(\frac{-\frac{0.048^2}{(1+0.048x)^2}}{2}\)	M1	Accept \(-0.002304(1+0.048x)^{-2}\) for the numerator.
at \(x=0\): \(\frac{-0.048^2}{2}\)	A1
\(\displaystyle\lim_{x\to0}\frac{\ln(1+0.048x)-0.048x}{x^2}=-\frac{0.048^2}{2}\)	AG
Part (h)
\(L-V(365)\approx6000e^{0.048}\left(1-e^{-0.001152/365}\right)\)	(M1)
\(\approx0.0199\) euros (about 2 cents), which agrees with \(6295.02-6295.00=0.02\)	A1	Accept 0.02. Comparison needed.
Part (i)
daily: \(6000\left(1+\frac{0.07}{365}\right)^{365}=\)€6435.01	A1
continuous: \(6000e^{0.07}=\)€6435.05	A1
the difference is only about 4 cents on €6000, so in practice it does not matter; the change of rate (about €140 more) matters far more than the compounding frequency	R1	Any sensible comment based on the size of the difference.

Answers: (a) €6288.00, €6294.42, €6295.00 · (d) \(6000e^{0.048}\approx\)€6295.02 · (f) \(A'(0)=288\) euros per year · (h) about €0.02 · (i) €6435.01 and €6435.05

Common error (c) In l’Hôpital’s rule the numerator and the denominator are differentiated separately. Using the quotient rule on \(\frac{\ln(1+0.048x)}{x}\) gives a different expression and a wrong limit.

Common error (d) Since \(k\ln\left(1+\frac{0.048}{k}\right)\to0.048\) is the limit of a logarithm, the value tends to \(6000e^{0.048}\), not to \(6000\times0.048\).

Exam tip (g) Write the form, \(\frac00\), after every substitution. It justifies each use of the rule and tells you when to stop.

More 5.13 questions in the question bank

+ Add to test✓ In your testCC-AA-512-513-227 marks~30 min

Smooth rides on a metro train: higher derivatives

5.12 Limits, first principles and higher derivatives5.13 Limits and l’Hôpital’s rule5.9 Kinematics5.6 Differentiation rulesHL

A metro train leaves a station from rest at time \(t=0\), where \(t\) is in seconds. During its acceleration phase, \(0\le t\le20\), its displacement from the station, \(s\) metres, is modelled by

\(s(t)=0.04t^3-0.001t^4.\)

At \(t=20\) the train reaches its cruising velocity of \(16\ \text{m s}^{-1}\). It travels at this velocity until \(t=60\) and then brakes, coming to rest at the next station at \(t=80\). Figure 1 shows the velocity–time graph of the journey.

The rate of change of acceleration is called the jerk, \(j=\dfrac{d^3s}{dt^3}\), measured in \(\text{m s}^{-3}\). A design guideline for passenger comfort requires that, at all times, the magnitude of the acceleration is at most \(1.3\ \text{m s}^{-2}\) and the magnitude of the jerk is at most \(0.3\ \text{m s}^{-3}\).

Figure 1: velocity–time graph of the journey between two stations

(a)(i)

Show that the average velocity of the train over the interval \(10\le t\le10+h\), where \(0\lt h\le10\), is \(\left(8+0.6h-0.001h^3\right)\ \text{m s}^{-1}\).

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Show solution

average velocity \(=\frac{s(10+h)-s(10)}{h}\), with \(s(10)=40-10=30\)	M1
\(0.04(10+h)^3=40+12h+1.2h^2+0.04h^3\) and \(0.001(10+h)^4=10+4h+0.6h^2+0.04h^3+0.001h^4\)	A1	Both expansions.
\(s(10+h)-s(10)=8h+0.6h^2-0.001h^4\)	A1
dividing by \(h\): \(8+0.6h-0.001h^3\)	AG

(3)

(a)(ii)

Hence find the velocity of the train when \(t=10\), giving a reason for your answer.

No calculatorCONCEPTNotes

Show solution

\(v(10)=\displaystyle\lim_{h\to0}\left(8+0.6h-0.001h^3\right)=8\ \text{m s}^{-1}\)

Reference to the limit as \(h\to0\) needed.

Answer: \(8\ \text{m s}^{-1}\)

(1)

(b)

Find expressions for the velocity \(v(t)\) and the acceleration \(a(t)\) during the acceleration phase. Verify that at \(t=20\) the velocity is \(16\ \text{m s}^{-1}\) and the acceleration is zero.

No calculatorROUTINENotes

Show solution

\(v(t)=0.12t^2-0.004t^3\)	A1
\(a(t)=0.24t-0.012t^2\)	A1	Allow ECF from \(v\).
\(v(20)=48-32=16\) and \(a(20)=4.8-4.8=0\)	A1	Both substitutions must be seen.

Answer: \(v=0.12t^2-0.004t^3\), \(a=0.24t-0.012t^2\)

(3)

(c)

Find \(j(t)\). Hence find the maximum acceleration during the acceleration phase, justifying that it is a maximum.

No calculatorMULTINotes

Show solution

\(j(t)=0.24-0.024t\)	A1	Allow ECF from (b).
\(j=0\) when \(t=10\), and \(j\) changes from positive to negative there (or \(\frac{d^4s}{dt^4}=-0.024<0\)), so \(a\) is a maximum	R1	Also accept \(a(0)=a(20)=0\) compared with \(a(10)\).
maximum acceleration \(a(10)=2.4-1.2=1.2\ \text{m s}^{-2}\)	A1

Answer: \(j=0.24-0.024t\); \(1.2\ \text{m s}^{-2}\) at \(t=10\)

(3)

(d)

On the axes in Figure 2, sketch the graphs of \(a\) and \(j\) against \(t\) for \(0\le t\le20\). Label each graph and show the coordinates of its end points and of any maximum point or axis intercept.

Figure 2: axes for the graphs of \(a\) and \(j\) against \(t\)

GRAPHNotes

Show solution

\(a\): concave-down parabola from \((0,0)\) to \((20,0)\) with maximum \((10,\,1.2)\)	A1
\(j\): straight line from \((0,\,0.24)\) to \((20,\,-0.24)\)	A1
\(j\) crosses the \(t\)-axis at \(t=10\), at the same time as the maximum of \(a\)	A1

(3)

(e)

Determine whether the acceleration phase meets the design guideline. Justify your answer.

No calculatorINTERPRETNotes

Show solution

\(0\le a\le1.2\) on \(0\le t\le20\), and \(1.2\le1.3\)	R1	Allow ECF from (c).
\(j\) is linear, so its greatest magnitude is at an end point: \(\|j(0)\|=\|j(20)\|=0.24\le0.3\); so the guideline is met	R1	Conclusion needed.

Answer: yes, the guideline is met

(2)

(f)(i)

The engineers consider changing the model to \(s(t)=0.04t^3-ct^4\), where \(c\) is a positive constant chosen so that the maximum acceleration still occurs during the acceleration phase.

Show that the maximum acceleration is \(\dfrac{0.0012}{c}\ \text{m s}^{-2}\).

No calculatorPROOFNotes

Show solution

\(a=0.24t-12ct^2\)	M1	Differentiating twice.
\(j=0.24-24ct=0\Rightarrow t=\frac{0.01}{c}\)	A1
\(a\left(\frac{0.01}{c}\right)=\frac{0.0024}{c}-\frac{0.0012}{c}\)	M1	Substitution must be seen.
\(=\frac{0.0012}{c}\)	AG

(3)

(f)(ii)

Find the least value of \(c\) for which the acceleration guideline is met. Explain why changing \(c\) has no effect on the jerk at the moment the train starts.

CONCEPTNotes

Show solution

\(\frac{0.0012}{c}\le1.3\Rightarrow c\ge0.000923\)	A1	Accept \(\frac{3}{3250}\) or \(9.23\times10^{-4}\).
\(j(0)=0.24-24c(0)=0.24\) for every value of \(c\), because the term in \(c\) contains a factor \(t\)	R1

Answer: \(c\approx0.000923\)

(2)

(g)

During braking, the velocity of the train \(\tau\) seconds after braking begins is modelled by

\(u(\tau)=8\left(1+\cos\dfrac{\pi\tau}{20}\right),\quad0\le\tau\le20.\)

Find \(u'(\tau)\) and \(u''(\tau)\). Hence suggest an expression for \(u^{(n)}(\tau)\), where \(n\in\mathbb{Z}^+\).

No calculatorALGEBRANotes

Show solution

\(u'(\tau)=-\frac{2\pi}{5}\sin\frac{\pi\tau}{20}\)	A1	Chain-rule factor \(\frac{\pi}{20}\) needed.
\(u''(\tau)=-\frac{\pi^2}{50}\cos\frac{\pi\tau}{20}\)	A1
\(u^{(n)}(\tau)=8\left(\frac{\pi}{20}\right)^n\cos\left(\frac{\pi\tau}{20}+\frac{n\pi}{2}\right)\)	A1	Accept any correct equivalent, e.g. four cases according to \(n\) modulo 4.

Answer: \(u^{(n)}(\tau)=8\left(\frac{\pi}{20}\right)^n\cos\left(\frac{\pi\tau}{20}+\frac{n\pi}{2}\right)\)

(3)

(h)

Use l’Hôpital’s rule to find the exact value of \(\displaystyle\lim_{\tau\to20}\frac{u(\tau)}{(20-\tau)^2}\).

No calculatorMULTINotes

Show solution

at \(\tau=20\): \(u(20)=8(1+\cos\pi)=0\) and \((20-20)^2=0\), form \(\frac00\); first application: \(\frac{u'(\tau)}{-2(20-\tau)}\)	M1	Allow ECF from (g).
\(\frac{-\frac{2\pi}{5}\sin\frac{\pi\tau}{20}}{-2(20-\tau)}\), again \(\frac00\) at \(\tau=20\)	A1
second application: \(\frac{u''(\tau)}{2}=\frac{-\frac{\pi^2}{50}\cos\frac{\pi\tau}{20}}{2}\)	M1
\(\frac{-\frac{\pi^2}{50}\cos\pi}{2}=\frac{\pi^2}{100}\)	A1	\(\approx0.0987\).

Answer: \(\frac{\pi^2}{100}\)

(4)

Show full solution

Working	Mark	Notes
Part (a)(i)
average velocity \(=\frac{s(10+h)-s(10)}{h}\), with \(s(10)=40-10=30\)	M1
\(0.04(10+h)^3=40+12h+1.2h^2+0.04h^3\) and \(0.001(10+h)^4=10+4h+0.6h^2+0.04h^3+0.001h^4\)	A1	Both expansions.
\(s(10+h)-s(10)=8h+0.6h^2-0.001h^4\)	A1
dividing by \(h\): \(8+0.6h-0.001h^3\)	AG
Part (a)(ii)
\(v(10)=\displaystyle\lim_{h\to0}\left(8+0.6h-0.001h^3\right)=8\ \text{m s}^{-1}\)	A1	Reference to the limit as \(h\to0\) needed.
Part (b)
\(v(t)=0.12t^2-0.004t^3\)	A1
\(a(t)=0.24t-0.012t^2\)	A1	Allow ECF from \(v\).
\(v(20)=48-32=16\) and \(a(20)=4.8-4.8=0\)	A1	Both substitutions must be seen.
Part (c)
\(j(t)=0.24-0.024t\)	A1	Allow ECF from (b).
\(j=0\) when \(t=10\), and \(j\) changes from positive to negative there (or \(\frac{d^4s}{dt^4}=-0.024<0\)), so \(a\) is a maximum	R1	Also accept \(a(0)=a(20)=0\) compared with \(a(10)\).
maximum acceleration \(a(10)=2.4-1.2=1.2\ \text{m s}^{-2}\)	A1
Part (d)
\(a\): concave-down parabola from \((0,0)\) to \((20,0)\) with maximum \((10,\,1.2)\)	A1
\(j\): straight line from \((0,\,0.24)\) to \((20,\,-0.24)\)	A1
\(j\) crosses the \(t\)-axis at \(t=10\), at the same time as the maximum of \(a\)	A1
Part (e)
\(0\le a\le1.2\) on \(0\le t\le20\), and \(1.2\le1.3\)	R1	Allow ECF from (c).
\(j\) is linear, so its greatest magnitude is at an end point: \(\|j(0)\|=\|j(20)\|=0.24\le0.3\); so the guideline is met	R1	Conclusion needed.
Part (f)(i)
\(a=0.24t-12ct^2\)	M1	Differentiating twice.
\(j=0.24-24ct=0\Rightarrow t=\frac{0.01}{c}\)	A1
\(a\left(\frac{0.01}{c}\right)=\frac{0.0024}{c}-\frac{0.0012}{c}\)	M1	Substitution must be seen.
\(=\frac{0.0012}{c}\)	AG
Part (f)(ii)
\(\frac{0.0012}{c}\le1.3\Rightarrow c\ge0.000923\)	A1	Accept \(\frac{3}{3250}\) or \(9.23\times10^{-4}\).
\(j(0)=0.24-24c(0)=0.24\) for every value of \(c\), because the term in \(c\) contains a factor \(t\)	R1
Part (g)
\(u'(\tau)=-\frac{2\pi}{5}\sin\frac{\pi\tau}{20}\)	A1	Chain-rule factor \(\frac{\pi}{20}\) needed.
\(u''(\tau)=-\frac{\pi^2}{50}\cos\frac{\pi\tau}{20}\)	A1
\(u^{(n)}(\tau)=8\left(\frac{\pi}{20}\right)^n\cos\left(\frac{\pi\tau}{20}+\frac{n\pi}{2}\right)\)	A1	Accept any correct equivalent, e.g. four cases according to \(n\) modulo 4.
Part (h)
at \(\tau=20\): \(u(20)=8(1+\cos\pi)=0\) and \((20-20)^2=0\), form \(\frac00\); first application: \(\frac{u'(\tau)}{-2(20-\tau)}\)	M1	Allow ECF from (g).
\(\frac{-\frac{2\pi}{5}\sin\frac{\pi\tau}{20}}{-2(20-\tau)}\), again \(\frac00\) at \(\tau=20\)	A1
second application: \(\frac{u''(\tau)}{2}=\frac{-\frac{\pi^2}{50}\cos\frac{\pi\tau}{20}}{2}\)	M1
\(\frac{-\frac{\pi^2}{50}\cos\pi}{2}=\frac{\pi^2}{100}\)	A1	\(\approx0.0987\).

Answers: (a)(ii) \(8\ \text{m s}^{-1}\) · (b) \(v=0.12t^2-0.004t^3\), \(a=0.24t-0.012t^2\) · (c) \(j=0.24-0.024t\); \(1.2\ \text{m s}^{-2}\) at \(t=10\) · (e) yes, the guideline is met · (f)(ii) \(c\approx0.000923\) · (g) \(u^{(n)}(\tau)=8\left(\frac{\pi}{20}\right)^n\cos\left(\frac{\pi\tau}{20}+\frac{n\pi}{2}\right)\) · (h) \(\frac{\pi^2}{100}\)

Common error (a)(i) \((10+h)^4\) has five terms, \(10^4+4\cdot10^3h+6\cdot10^2h^2+4\cdot10h^3+h^4\). Dropping the middle terms makes the average velocity, and so its limit, wrong.

Exam tip (h) Check the form before each application. After the second application the fraction is no longer indeterminate, so stop and substitute.

More 5.12 questions in the question bank

5.14 Implicit differentiation, related rates and optimisation · 5.15 Further derivatives and integrals; partial fractionsmerged group

+ Add to test✓ In your testCC-AA-514-515-123 marks~30 min

A rotating lighthouse beam sweeping the shore

5.14 Implicit differentiation, related rates and optimisation5.15 Further derivatives and integrals; partial fractions3.3 Applications of trigonometry: elevation, depression and bearings3.9 Reciprocal and inverse trigonometric functionsHL

A lighthouse stands at point L, 0.8 km due west of a point O on a straight shoreline that runs north–south. Its lamp sends out a narrow horizontal beam that rotates anticlockwise, when viewed from above, at a constant rate of 4 revolutions per minute.

At time \(t\) minutes the beam meets the shoreline at the point P, which is \(x\) km north of O. The angle OLP is \(\theta\) radians, where \(-\dfrac{\pi}{2}\lt\theta\lt\dfrac{\pi}{2}\); when P is south of O, both \(x\) and \(\theta\) are negative. The situation is shown in Figure 1.

Figure 1: plan view of the lighthouse and the shoreline (not to scale)

(a)

When P is 0.6 km north of O, find the bearing of P from L.

ROUTINENotes

Show solution

\(\tan\theta=\frac{0.6}{0.8}\), so \(\theta=36.87^\circ\)	M1
bearing \(=090^\circ-36.87^\circ=053.1^\circ\)	A1	Accept \(053^\circ\).

Answer: \(053.1^\circ\)

(2)

(b)

Show that \(\dfrac{dx}{dt}=6.4\pi\sec^2\theta\ \text{km min}^{-1}\).

No calculatorPROOFNotes

Show solution

\(\frac{d\theta}{dt}=4\times2\pi=8\pi\) radians per minute	B1
\(x=0.8\tan\theta\), so \(\frac{dx}{d\theta}=0.8\sec^2\theta\)	M1
\(\frac{dx}{dt}=\frac{dx}{d\theta}\times\frac{d\theta}{dt}=0.8\sec^2\theta\times8\pi\)	M1	Chain rule.
\(\frac{dx}{dt}=6.4\pi\sec^2\theta\)	AG

(3)

(c)

Find the exact speed of P when \(x=0.6\).

No calculatorROUTINENotes

Show solution

\(\sec^2\theta=1+\tan^2\theta=1+0.75^2=1.5625\)	M1	Or from \(LP=1\): \(\sec\theta=\frac{1}{0.8}\).
\(\frac{dx}{dt}=6.4\pi\times1.5625=10\pi\ \text{km min}^{-1}\)	A1	\(\approx31.4\).

Answer: \(10\pi\ \text{km min}^{-1}\)

(2)

(d)

Let \(r\) km be the length LP of the beam between the lamp and the shoreline. Find the rate at which \(r\) is changing when \(x=0.6\).

No calculatorMULTINotes

Show solution

\(r^2=x^2+0.64\), so \(2r\frac{dr}{dt}=2x\frac{dx}{dt}\)	M1	Implicit differentiation with respect to \(t\).
when \(x=0.6\), \(r=\sqrt{0.36+0.64}=1\)	A1
\(\frac{dr}{dt}=\frac{0.6\times10\pi}{1}=6\pi\ \text{km min}^{-1}\) (increasing)	A1	\(\approx18.8\). Allow ECF from (c).

Answer: \(6\pi\approx18.8\ \text{km min}^{-1}\)

(3)

(e)(i)

On the axes in Figure 2, sketch the graph of \(\dfrac{dx}{dt}\) against \(\theta\) for \(-\dfrac{\pi}{2}\lt\theta\lt\dfrac{\pi}{2}\). Show the coordinates of the minimum point and the equations of any asymptotes.

Figure 2: axes for the graph of \(\dfrac{dx}{dt}\) against \(\theta\)

GRAPHNotes

Show solution

U-shaped curve, symmetric about \(\theta=0\), with minimum \((0,\,6.4\pi)\)	A1	Accept \((0,\,20.1)\).
vertical asymptotes \(\theta=-\frac{\pi}{2}\) and \(\theta=\frac{\pi}{2}\)	A1
curve above the horizontal axis throughout, rising without bound towards both asymptotes	A1

(3)

(e)(ii)

Interpret the vertical asymptotes in the context of the lighthouse.

INTERPRETNotes

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as \(\theta\to\pm\frac{\pi}{2}\) the beam becomes parallel to the shoreline, P moves further and further from O and its speed increases without limit	R1
for \(\theta\) beyond \(\pm\frac{\pi}{2}\) the beam points out to sea and does not meet the shoreline, so P does not exist	R1

(2)

(f)

By writing \(\theta\) in terms of \(x\) and differentiating with respect to \(t\), show that \(\dfrac{dx}{dt}=10\pi\left(x^2+0.64\right)\).

No calculatorPROOFNotes

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\(\theta=\arctan\frac{x}{0.8}\)	A1
\(\frac{d\theta}{dt}=\frac{1}{1+\left(\frac{x}{0.8}\right)^2}\times\frac{1}{0.8}\times\frac{dx}{dt}\)	M1	Chain rule with the derivative of arctan.
\(8\pi=\frac{0.8}{0.64+x^2}\frac{dx}{dt}\), so \(\frac{dx}{dt}=\frac{8\pi}{0.8}\left(x^2+0.64\right)\)	A1
\(\frac{dx}{dt}=10\pi\left(x^2+0.64\right)\)	AG

(3)

(g)

The lamp is now slowed to 3 revolutions per minute. A camera on the shoreline can record the moving spot of light only while its speed is at most \(30\ \text{km min}^{-1}\).

Find the length of shoreline over which the camera can record the spot.

MODELNotes

Show solution

\(\frac{d\theta}{dt}=6\pi\), so \(\frac{dx}{dt}=\frac{6\pi}{0.8}\left(x^2+0.64\right)=7.5\pi\left(x^2+0.64\right)\)	M1	Or \(4.8\pi\sec^2\theta\).
\(7.5\pi\left(x^2+0.64\right)\le30\)	M1	Accept an equation.
\(x^2\le\frac{4}{\pi}-0.64\), so \(-0.796\le x\le0.796\)	A1
length \(=2\times0.7957\ldots=1.59\) km	A1

Answer: 1.59 km

(4)

(h)

Without calculating any further positions, explain whether this length would be greater or smaller if the lamp rotated at 4 revolutions per minute.

No calculatorINTERPRETNotes

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smaller: at every point the speed of P is proportional to the rate of rotation, so at the faster rate the speed limit is reached closer to O

Answer with a reason.

Answer: smaller

(1)

Show full solution

Working	Mark	Notes
Part (a)
\(\tan\theta=\frac{0.6}{0.8}\), so \(\theta=36.87^\circ\)	M1
bearing \(=090^\circ-36.87^\circ=053.1^\circ\)	A1	Accept \(053^\circ\).
Part (b)
\(\frac{d\theta}{dt}=4\times2\pi=8\pi\) radians per minute	B1
\(x=0.8\tan\theta\), so \(\frac{dx}{d\theta}=0.8\sec^2\theta\)	M1
\(\frac{dx}{dt}=\frac{dx}{d\theta}\times\frac{d\theta}{dt}=0.8\sec^2\theta\times8\pi\)	M1	Chain rule.
\(\frac{dx}{dt}=6.4\pi\sec^2\theta\)	AG
Part (c)
\(\sec^2\theta=1+\tan^2\theta=1+0.75^2=1.5625\)	M1	Or from \(LP=1\): \(\sec\theta=\frac{1}{0.8}\).
\(\frac{dx}{dt}=6.4\pi\times1.5625=10\pi\ \text{km min}^{-1}\)	A1	\(\approx31.4\).
Part (d)
\(r^2=x^2+0.64\), so \(2r\frac{dr}{dt}=2x\frac{dx}{dt}\)	M1	Implicit differentiation with respect to \(t\).
when \(x=0.6\), \(r=\sqrt{0.36+0.64}=1\)	A1
\(\frac{dr}{dt}=\frac{0.6\times10\pi}{1}=6\pi\ \text{km min}^{-1}\) (increasing)	A1	\(\approx18.8\). Allow ECF from (c).
Part (e)(i)
U-shaped curve, symmetric about \(\theta=0\), with minimum \((0,\,6.4\pi)\)	A1	Accept \((0,\,20.1)\).
vertical asymptotes \(\theta=-\frac{\pi}{2}\) and \(\theta=\frac{\pi}{2}\)	A1
curve above the horizontal axis throughout, rising without bound towards both asymptotes	A1
Part (e)(ii)
as \(\theta\to\pm\frac{\pi}{2}\) the beam becomes parallel to the shoreline, P moves further and further from O and its speed increases without limit	R1
for \(\theta\) beyond \(\pm\frac{\pi}{2}\) the beam points out to sea and does not meet the shoreline, so P does not exist	R1
Part (f)
\(\theta=\arctan\frac{x}{0.8}\)	A1
\(\frac{d\theta}{dt}=\frac{1}{1+\left(\frac{x}{0.8}\right)^2}\times\frac{1}{0.8}\times\frac{dx}{dt}\)	M1	Chain rule with the derivative of arctan.
\(8\pi=\frac{0.8}{0.64+x^2}\frac{dx}{dt}\), so \(\frac{dx}{dt}=\frac{8\pi}{0.8}\left(x^2+0.64\right)\)	A1
\(\frac{dx}{dt}=10\pi\left(x^2+0.64\right)\)	AG
Part (g)
\(\frac{d\theta}{dt}=6\pi\), so \(\frac{dx}{dt}=\frac{6\pi}{0.8}\left(x^2+0.64\right)=7.5\pi\left(x^2+0.64\right)\)	M1	Or \(4.8\pi\sec^2\theta\).
\(7.5\pi\left(x^2+0.64\right)\le30\)	M1	Accept an equation.
\(x^2\le\frac{4}{\pi}-0.64\), so \(-0.796\le x\le0.796\)	A1
length \(=2\times0.7957\ldots=1.59\) km	A1
Part (h)
smaller: at every point the speed of P is proportional to the rate of rotation, so at the faster rate the speed limit is reached closer to O	R1	Answer with a reason.

Answers: (a) \(053.1^\circ\) · (c) \(10\pi\ \text{km min}^{-1}\) · (d) \(6\pi\approx18.8\ \text{km min}^{-1}\) · (g) 1.59 km · (h) smaller

Common error (b) The rate of rotation must be in radians per minute before it is used with \(\sec^2\theta\): 4 revolutions per minute is \(8\pi\), not 4.

Exam tip (d) Differentiate \(r^2=x^2+0.64\) while \(x\) and \(r\) are still variables; substitute \(x=0.6\) and \(r=1\) only afterwards.

More 5.14 questions in the question bank

+ Add to test✓ In your testCC-AA-514-515-227 marks~30 min

Two masts and the phone signal along a road

5.15 Further derivatives and integrals; partial fractions1.11 Partial fractions5.14 Implicit differentiation, related rates and optimisation5.11 Definite integrals and areasHL

A straight road leaves a tunnel at the point T. Two masts, A and B, provide a mobile-phone signal along the road. At a point on the road \(x\) km from T, the signal index from mast A is modelled by

\(S(x)=\dfrac{6x}{(x+1)(x+4)},\quad x\ge0.\)

Mast B stands 3 km from the road, on the line through T perpendicular to the road. Its signal index follows an inverse-square law and is modelled by

\(R(x)=\dfrac{6}{x^2+9},\quad x\ge0.\)

A signal index has no units; a larger value means a stronger signal. Figure 1 shows the graph of \(y=R(x)\).

Figure 1: the graph of \(y=R(x)\) for \(0\le x\le10\)

(a)

Express \(S(x)\) in partial fractions.

No calculatorALGEBRANotes

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\(\frac{6x}{(x+1)(x+4)}=\frac{A}{x+1}+\frac{B}{x+4}\Rightarrow6x=A(x+4)+B(x+1)\)	M1
\(x=-1\): \(-6=3A\), so \(A=-2\)	A1
\(x=-4\): \(-24=-3B\), so \(B=8\)	A1

Answer: \(S(x)=\dfrac{8}{x+4}-\dfrac{2}{x+1}\)

(3)

(b)

Hence show that \(\displaystyle\int_0^5S(x)\,dx=14\ln3-18\ln2\).

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\(\Big[8\ln(x+4)-2\ln(x+1)\Big]_0^5\)	M1	Integrating both terms to logarithms. Allow ECF from (a).
\(=(8\ln9-2\ln6)-(8\ln4-2\ln1)\)	A1
\(=16\ln3-2\ln2-2\ln3-16\ln2\)	A1	Use of \(\ln9=2\ln3\), \(\ln4=2\ln2\), \(\ln6=\ln2+\ln3\), \(\ln1=0\).
\(=14\ln3-18\ln2\)	AG

(3)

(c)

The mean signal index over the stretch \(a\le x\le b\) is defined as \(\dfrac{1}{b-a}\displaystyle\int_a^bS(x)\,dx\).

Find the mean signal index from mast A over \(0\le x\le5\).

ROUTINENotes

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\(\frac{14\ln3-18\ln2}{5}=0.581\)

\(0.58078\ldots\)

Answer: 0.581

(1)

(d)

Use your answer to part (a) to find \(S'(x)\). Hence find the coordinates of the point on the graph of \(y=S(x)\) where the signal from mast A is strongest, justifying that it is a maximum.

No calculatorMULTINotes

Show solution

\(S'(x)=\frac{2}{(x+1)^2}-\frac{8}{(x+4)^2}\)	A1	Allow ECF from (a).
\(S'(x)=0\Rightarrow(x+4)^2=4(x+1)^2\Rightarrow x+4=2(x+1)\)	M1	The root \(x=-2\) is outside the domain.
\(x=2\), \(S(2)=\frac{8}{6}-\frac{2}{3}=\frac23\), giving \(\left(2,\,\frac23\right)\)	A1
\(S'(x)\gt0\) for \(0\le x\lt2\) and \(S'(x)\lt0\) for \(x\gt2\) (e.g. \(S'(1)=0.18\), \(S'(3)=-0.0383\)), so a maximum	R1	Or \(S'(x)=\frac{6(4-x^2)}{(x+1)^2(x+4)^2}\), whose sign is that of \(4-x^2\).

Answer: \(\left(2,\,\frac23\right)\)

(4)

(e)

On Figure 1, sketch the graph of \(y=S(x)\) for \(0\le x\le10\).

GRAPHNotes

Show solution

curve from the origin rising to a maximum at \(\left(2,\,\frac23\right)\)	A1	Allow ECF from (d).
then decreasing gradually towards the \(x\)-axis, crossing \(y=R(x)\) once and staying above it for larger \(x\)	A1

(2)

(f)

Show that the graphs of \(y=S(x)\) and \(y=R(x)\) meet only where \(x=1\).

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\(\frac{6x}{(x+1)(x+4)}=\frac{6}{x^2+9}\Rightarrow x\left(x^2+9\right)=(x+1)(x+4)\)	M1
\(x^3-x^2+4x-4=0\)	A1
\((x-1)\left(x^2+4\right)=0\), and \(x^2+4\gt0\) for all \(x\), so \(x=1\) is the only solution	R1	Factorisation and the reason for no other root are both needed.
the graphs meet only where \(x=1\)	AG

(3)

(g)

Show that the mean signal index from mast B over \(0\le x\le3\) is \(\dfrac{\pi}{6}\).

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\(\int\frac{6}{x^2+9}\,dx=6\times\frac13\arctan\frac{x}{3}\)	M1	Use of \(\int\frac{dx}{x^2+a^2}=\frac1a\arctan\frac xa\) with \(a=3\).
\(\Big[2\arctan\frac{x}{3}\Big]_0^3=2\times\frac{\pi}{4}=\frac{\pi}{2}\)	A1
mean \(=\frac13\times\frac{\pi}{2}\)	A1
\(=\frac{\pi}{6}\)	AG

(3)

(h)

A car drives away from T along the road at a constant speed of 90 km h⁻¹.

Find the rate of change, per minute, of each signal index as the car passes the point where \(x=1\). State which signal is getting stronger.

No calculatorMULTINotes

Show solution

\(\frac{dS}{dt}=S'(x)\frac{dx}{dt}\) with \(\frac{dx}{dt}=1.5\) km per minute	M1
\(S'(1)=\frac{2}{4}-\frac{8}{25}=0.18\), so \(\frac{dS}{dt}=0.27\) per minute	A1	Allow ECF from (d).
\(R'(x)=-\frac{12x}{\left(x^2+9\right)^2}\), \(R'(1)=-0.12\), so \(\frac{dR}{dt}=-0.18\) per minute; the signal from A is getting stronger	A1

Answer: \(\frac{dS}{dt}=0.27\), \(\frac{dR}{dt}=-0.18\) per minute; A stronger

(3)

(i)

For \(x\ge2\), the operator replaces the model for mast A by \(S_2(x)=\dfrac23(0.8)^{x-2}\).

Find \(S_2'(x)\). Hence find the rate of change, per minute, of the signal index from mast A as the car in part (h) passes the point where \(x=4\), according to each of the models \(S\) and \(S_2\).

MULTINotes

Show solution

\(S_2'(x)=\frac23(0.8)^{x-2}\ln0.8\)	A1
multiplying a derivative at \(x=4\) by \(\frac{dx}{dt}=1.5\)	M1
\(S_2\): \(\frac23(0.64)(\ln0.8)(1.5)=-0.143\) per minute; \(S\): \(S'(4)=-0.045\), giving \(-0.0675\) per minute	A1	Both values. Accept \(-0.068\).

Answer: \(-0.143\) (model \(S_2\)) and \(-0.0675\) (model \(S\)) per minute

(3)

(j)

Mast A’s signal index is now modelled by \(S(x)\) for \(0\le x\le2\) and by \(S_2(x)\) for \(x\gt2\). Determine whether the graph of this combined model is smooth at \(x=2\). Justify your answer.

INTERPRETNotes

Show solution

\(S(2)=S_2(2)=\frac23\), so the model is continuous at \(x=2\)	R1
\(S'(2)=0\) but \(S_2'(2)=\frac23\ln0.8\approx-0.149\ne0\), so the gradient changes suddenly and the graph has a corner: it is not smooth	R1	Conclusion needed.

Answer: not smooth (continuous, but the gradients differ)

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(\frac{6x}{(x+1)(x+4)}=\frac{A}{x+1}+\frac{B}{x+4}\Rightarrow6x=A(x+4)+B(x+1)\)	M1
\(x=-1\): \(-6=3A\), so \(A=-2\)	A1
\(x=-4\): \(-24=-3B\), so \(B=8\)	A1
Part (b)
\(\Big[8\ln(x+4)-2\ln(x+1)\Big]_0^5\)	M1	Integrating both terms to logarithms. Allow ECF from (a).
\(=(8\ln9-2\ln6)-(8\ln4-2\ln1)\)	A1
\(=16\ln3-2\ln2-2\ln3-16\ln2\)	A1	Use of \(\ln9=2\ln3\), \(\ln4=2\ln2\), \(\ln6=\ln2+\ln3\), \(\ln1=0\).
\(=14\ln3-18\ln2\)	AG
Part (c)
\(\frac{14\ln3-18\ln2}{5}=0.581\)	A1	\(0.58078\ldots\)
Part (d)
\(S'(x)=\frac{2}{(x+1)^2}-\frac{8}{(x+4)^2}\)	A1	Allow ECF from (a).
\(S'(x)=0\Rightarrow(x+4)^2=4(x+1)^2\Rightarrow x+4=2(x+1)\)	M1	The root \(x=-2\) is outside the domain.
\(x=2\), \(S(2)=\frac{8}{6}-\frac{2}{3}=\frac23\), giving \(\left(2,\,\frac23\right)\)	A1
\(S'(x)\gt0\) for \(0\le x\lt2\) and \(S'(x)\lt0\) for \(x\gt2\) (e.g. \(S'(1)=0.18\), \(S'(3)=-0.0383\)), so a maximum	R1	Or \(S'(x)=\frac{6(4-x^2)}{(x+1)^2(x+4)^2}\), whose sign is that of \(4-x^2\).
Part (e)
curve from the origin rising to a maximum at \(\left(2,\,\frac23\right)\)	A1	Allow ECF from (d).
then decreasing gradually towards the \(x\)-axis, crossing \(y=R(x)\) once and staying above it for larger \(x\)	A1
Part (f)
\(\frac{6x}{(x+1)(x+4)}=\frac{6}{x^2+9}\Rightarrow x\left(x^2+9\right)=(x+1)(x+4)\)	M1
\(x^3-x^2+4x-4=0\)	A1
\((x-1)\left(x^2+4\right)=0\), and \(x^2+4\gt0\) for all \(x\), so \(x=1\) is the only solution	R1	Factorisation and the reason for no other root are both needed.
the graphs meet only where \(x=1\)	AG
Part (g)
\(\int\frac{6}{x^2+9}\,dx=6\times\frac13\arctan\frac{x}{3}\)	M1	Use of \(\int\frac{dx}{x^2+a^2}=\frac1a\arctan\frac xa\) with \(a=3\).
\(\Big[2\arctan\frac{x}{3}\Big]_0^3=2\times\frac{\pi}{4}=\frac{\pi}{2}\)	A1
mean \(=\frac13\times\frac{\pi}{2}\)	A1
\(=\frac{\pi}{6}\)	AG
Part (h)
\(\frac{dS}{dt}=S'(x)\frac{dx}{dt}\) with \(\frac{dx}{dt}=1.5\) km per minute	M1
\(S'(1)=\frac{2}{4}-\frac{8}{25}=0.18\), so \(\frac{dS}{dt}=0.27\) per minute	A1	Allow ECF from (d).
\(R'(x)=-\frac{12x}{\left(x^2+9\right)^2}\), \(R'(1)=-0.12\), so \(\frac{dR}{dt}=-0.18\) per minute; the signal from A is getting stronger	A1
Part (i)
\(S_2'(x)=\frac23(0.8)^{x-2}\ln0.8\)	A1
multiplying a derivative at \(x=4\) by \(\frac{dx}{dt}=1.5\)	M1
\(S_2\): \(\frac23(0.64)(\ln0.8)(1.5)=-0.143\) per minute; \(S\): \(S'(4)=-0.045\), giving \(-0.0675\) per minute	A1	Both values. Accept \(-0.068\).
Part (j)
\(S(2)=S_2(2)=\frac23\), so the model is continuous at \(x=2\)	R1
\(S'(2)=0\) but \(S_2'(2)=\frac23\ln0.8\approx-0.149\ne0\), so the gradient changes suddenly and the graph has a corner: it is not smooth	R1	Conclusion needed.

Answers: (a) \(S(x)=\dfrac{8}{x+4}-\dfrac{2}{x+1}\) · (c) 0.581 · (d) \(\left(2,\,\frac23\right)\) · (h) \(\frac{dS}{dt}=0.27\), \(\frac{dR}{dt}=-0.18\) per minute; A stronger · (i) \(-0.143\) (model \(S_2\)) and \(-0.0675\) (model \(S\)) per minute · (j) not smooth (continuous, but the gradients differ)

Common error (g) The factor \(\frac1a\) in \(\int\frac{dx}{x^2+a^2}=\frac1a\arctan\frac xa+C\) is often left out. Differentiate your answer to check it.

More 5.15 questions in the question bank

5.16 Integration by substitution and by parts · 5.17 Areas with the y-axis and volumes of revolutionmerged group

+ Add to test✓ In your testCC-AA-516-517-128 marks~35 min

The shape and volume of a rocket nose cone

5.17 Areas with the y-axis and volumes of revolution5.16 Integration by substitution and by parts5.11 Definite integrals and areas2.11 Transformations of graphsHL

The nose cone of a small rocket is a solid of revolution. In a side view, the \(x\)-axis lies along the axis of the rocket, with the origin \(O\) at the tip of the nose cone, and \(y\) is the radius of the nose cone at a distance \(x\) from the tip. All lengths are in metres.

The nose cone is \(2\) m long and the radius of its base, at \(x=2\), is \(0.5\) m. Two profiles are being considered:

profile P: \(y=\dfrac{\sqrt{2x}}{4}\), \(0\le x\le2\)
profile Q: \(y=\dfrac{x}{4}e^{1-\frac{x}{2}}\), \(0\le x\le2\).

For each profile, the nose cone is the solid formed when the region between the profile, the \(x\)-axis and the line \(x=2\) is rotated through \(360^\circ\) about the \(x\)-axis. Profile P is shown in Figure 1.

Figure 1: profile P, with the region R between P, the y-axis and the line y = 0.5 shaded

(a)

Find the exact volume of the nose cone with profile P.

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\(V=\pi\int_0^2\left(\frac{\sqrt{2x}}{4}\right)^2dx\)	M1	Award for \(\pi\int y^2\,dx\) with the correct limits. Missing \(\pi\) or \(y\) not squared: M0.
\(=\pi\int_0^2\frac{x}{8}\,dx=\pi\left[\frac{x^2}{16}\right]_0^2\)	A1
\(V=\frac{\pi}{4}\) m\(^3\)	A1	Units not required.

Answer: \(\dfrac{\pi}{4}\) m\(^3\) \((\approx0.785\) m\(^3)\)

(3)

(b)

The region \(R\) is bounded by profile P, the \(y\)-axis and the line \(y=0.5\), as shown in Figure 1.

By first writing \(x\) in terms of \(y\), find the area of \(R\).

No calculatorROUTINENotes

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\(y^2=\frac{2x}{16}\Rightarrow x=8y^2\)	B1	Independent of the method mark.
\(A=\int_0^{0.5}8y^2\,dy\)	M1	Integral of their \(x\) with respect to \(y\), with limits \(0\) and \(0.5\). Limits \(0\) and \(2\): M0.
\(=\left[\frac{8y^3}{3}\right]_0^{0.5}=\frac13\) m\(^2\)	A1	A rectangle-minus-area method \(\left(1-\frac23\right)\) does not follow the instruction: M0A0 (B1 can still be awarded if \(x=8y^2\) is seen).

Answer: \(\dfrac13\) m\(^2\)

(3)

(c)

For transport, the nose cone with profile P is packed in a closed cylindrical case of radius \(0.5\) m and length \(2\) m, whose axis is the \(x\)-axis. The space between the nose cone and the case is filled with foam. This space is the solid formed when \(R\) is rotated through \(360^\circ\) about the \(x\)-axis.

Find the volume of foam needed.

No calculatorMODELNotes

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\(\pi(0.5)^2(2)-\frac{\pi}{4}\) or \(\pi\int_0^2\left(0.5^2-\frac{x}{8}\right)dx\)	M1	Cylinder minus their (a), or outer radius squared minus inner radius squared. Allow ECF from (a).
\(\frac{\pi}{4}\) m\(^3\)	A1

Answer: \(\dfrac{\pi}{4}\) m\(^3\) \((\approx0.785\) m\(^3)\)

(2)

(d)(i)

Let \(g(x)=xe^{-x}\).

Show that the equation of profile Q can be written as \(y=\dfrac{e}{2}\,g\!\left(\dfrac{x}{2}\right)\).

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\(\frac{e}{2}\,g\!\left(\frac{x}{2}\right)=\frac{e}{2}\cdot\frac{x}{2}\,e^{-\frac{x}{2}}\)	M1	Replaces \(x\) by \(\frac x2\) in \(g\).
\(=\frac{x}{4}\,e^{1}e^{-\frac{x}{2}}\)	A1	Uses \(e\cdot e^{-\frac x2}=e^{1-\frac x2}\).
\(=\frac{x}{4}e^{1-\frac{x}{2}}\)	AG

(2)

(d)(ii)

Hence describe fully two transformations that map the graph of \(y=g(x)\) onto the graph of profile Q.

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horizontal stretch (parallel to the \(x\)-axis) with scale factor \(2\)	A1	Scale factor \(\frac12\): A0.
vertical stretch (parallel to the \(y\)-axis) with scale factor \(\frac{e}{2}\)	A1	Either order. “Stretch” with a direction and a scale factor is needed for each mark.

(2)

(e)

Figure 2 shows the graph of \(y=g(x)\) for \(0\le x\le4\). Its maximum point is \(\left(1,\,e^{-1}\right)\).

On Figure 2, sketch the graph of profile Q for \(0\le x\le2\). Write down the coordinates of the image of the maximum point of \(g\), and explain why profile Q joins smoothly onto the cylindrical body of the rocket, which has radius \(0.5\) m and begins at \(x=2\).

Figure 2: the graph of y = g(x) for 0 ≤ x ≤ 4

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curve from \(O\), increasing throughout, ending at \(x=2\) with a horizontal tangent	A1	Q lies slightly below \(g\) for \(0\lt x\lt0.77\) and above it after; this detail is not required.
\(\left(2,\ 0.5\right)\)	A1	From \(\left(1\times2,\ e^{-1}\times\frac e2\right)\). Allow ECF from (d)(ii).
the maximum of Q is at \(x=2\), so \(\frac{dy}{dx}=0\) there: the profile meets the body at the same radius, \(0.5\) m, and with the same (zero) gradient as the edge \(y=0.5\) of the cylinder, so there is no corner	R1	Both the matching radius and the zero gradient needed.

Answer: \((2,\ 0.5)\)

(3)

(f)

The area of the side view of the nose cone is \(2\displaystyle\int_0^2y\,dx\).

Show that, for profile Q, the area of the side view is \(2(e-2)\) m\(^2\).

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by parts with \(u=x\), \(\frac{dv}{dx}=e^{1-\frac x2}\): \(\int xe^{1-\frac x2}dx=-2xe^{1-\frac x2}+\int2e^{1-\frac x2}dx\)	M1	Correct choice of \(u\) and attempt at \(uv-\int v\frac{du}{dx}\,dx\).
\(=-2xe^{1-\frac x2}-4e^{1-\frac x2}\;(+C)\)	A1
\(\int_0^2\frac{x}{4}e^{1-\frac x2}dx=\frac14\Big[-2xe^{1-\frac x2}-4e^{1-\frac x2}\Big]_0^2=\frac14\big((-4-4)-(0-4e)\big)=e-2\)	A1	Correct substitution of both limits seen.
area of side view \(=2(e-2)\) m\(^2\)	AG

(3)

(g)(i)

Show that the volume of the nose cone with profile Q is \(\dfrac{\pi\left(e^2-5\right)}{8}\) m\(^3\).

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\(V=\pi\int_0^2\frac{x^2}{16}e^{2-x}dx\)	M1	Correct squaring of \(y\) needed: \(\left(e^{1-\frac x2}\right)^2=e^{2-x}\).
\(\int x^2e^{2-x}dx=-x^2e^{2-x}+\int2xe^{2-x}dx\)	M1	First application of parts with \(u=x^2\).
\(\int2xe^{2-x}dx=-2xe^{2-x}-2e^{2-x}\)	A1	Correct second application of parts, giving a fully correct antiderivative.
\(V=\frac{\pi}{16}\Big[-e^{2-x}\left(x^2+2x+2\right)\Big]_0^2=\frac{\pi}{16}\left(-10+2e^2\right)\)	A1	Correct substitution of both limits seen.
\(=\frac{\pi\left(e^2-5\right)}{8}\)	AG

(4)

(g)(ii)

The space available for instruments is proportional to the volume of the nose cone.

Find the percentage by which the volume of the nose cone with profile Q exceeds the volume with profile P.

ROUTINENotes

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\(\frac{0.938180\ldots-0.785398\ldots}{0.785398\ldots}\times100\)	M1	Or \(\left(\frac{e^2-5}{2}-1\right)\times100\). Allow ECF from (a).
\(19.5\%\)	A1	\(19.4528\ldots\)

Answer: \(19.5\%\)

(2)

(h)(i)

The designers decide to lengthen the nose cone to \(2.5\) m, keeping the radius of its base as \(0.5\) m. Each profile is stretched horizontally, parallel to the \(x\)-axis, with scale factor \(1.25\).

Show that the stretched profile P satisfies \(y^2=\dfrac{x}{10}\), and find the exact volume of the new nose cone with this profile.

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replacing \(x\) by \(\frac{x}{1.25}\): \(y=\frac14\sqrt{\frac{2x}{1.25}}\Rightarrow y^2=\frac{1.6x}{16}=\frac{x}{10}\)	M1	Replacing \(x\) by \(1.25x\): M0.
\(V=\pi\int_0^{2.5}\frac{x}{10}\,dx=\pi\left[\frac{x^2}{20}\right]_0^{2.5}=\frac{5\pi}{16}\) m\(^3\)	A1	Upper limit \(2.5\) required.

Answer: \(\dfrac{5\pi}{16}\) m\(^3\) \((\approx0.982\) m\(^3)\)

(2)

(h)(ii)

Explain why stretching a profile horizontally with scale factor \(k\) multiplies the volume of the nose cone by \(k\). Hence state, with a reason, whether the percentage found in part (g)(ii) changes.

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each disc keeps its radius but its thickness is multiplied by \(k\) (the solid is \(k\) times as long); or, with \(u=\frac xk\): \(\pi\int_0^{2k}\left(f\!\left(\frac xk\right)\right)^2dx=k\,\pi\int_0^2\left(f(u)\right)^2du\)	R1
both volumes are multiplied by \(1.25\), so their ratio, and hence the percentage \((19.5\%)\), is unchanged	R1	Reason required.

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(V=\pi\int_0^2\left(\frac{\sqrt{2x}}{4}\right)^2dx\)	M1	Award for \(\pi\int y^2\,dx\) with the correct limits. Missing \(\pi\) or \(y\) not squared: M0.
\(=\pi\int_0^2\frac{x}{8}\,dx=\pi\left[\frac{x^2}{16}\right]_0^2\)	A1
\(V=\frac{\pi}{4}\) m\(^3\)	A1	Units not required.
Part (b)
\(y^2=\frac{2x}{16}\Rightarrow x=8y^2\)	B1	Independent of the method mark.
\(A=\int_0^{0.5}8y^2\,dy\)	M1	Integral of their \(x\) with respect to \(y\), with limits \(0\) and \(0.5\). Limits \(0\) and \(2\): M0.
\(=\left[\frac{8y^3}{3}\right]_0^{0.5}=\frac13\) m\(^2\)	A1	A rectangle-minus-area method \(\left(1-\frac23\right)\) does not follow the instruction: M0A0 (B1 can still be awarded if \(x=8y^2\) is seen).
Part (c)
\(\pi(0.5)^2(2)-\frac{\pi}{4}\) or \(\pi\int_0^2\left(0.5^2-\frac{x}{8}\right)dx\)	M1	Cylinder minus their (a), or outer radius squared minus inner radius squared. Allow ECF from (a).
\(\frac{\pi}{4}\) m\(^3\)	A1
Part (d)(i)
\(\frac{e}{2}\,g\!\left(\frac{x}{2}\right)=\frac{e}{2}\cdot\frac{x}{2}\,e^{-\frac{x}{2}}\)	M1	Replaces \(x\) by \(\frac x2\) in \(g\).
\(=\frac{x}{4}\,e^{1}e^{-\frac{x}{2}}\)	A1	Uses \(e\cdot e^{-\frac x2}=e^{1-\frac x2}\).
\(=\frac{x}{4}e^{1-\frac{x}{2}}\)	AG
Part (d)(ii)
horizontal stretch (parallel to the \(x\)-axis) with scale factor \(2\)	A1	Scale factor \(\frac12\): A0.
vertical stretch (parallel to the \(y\)-axis) with scale factor \(\frac{e}{2}\)	A1	Either order. “Stretch” with a direction and a scale factor is needed for each mark.
Part (e)
curve from \(O\), increasing throughout, ending at \(x=2\) with a horizontal tangent	A1	Q lies slightly below \(g\) for \(0\lt x\lt0.77\) and above it after; this detail is not required.
\(\left(2,\ 0.5\right)\)	A1	From \(\left(1\times2,\ e^{-1}\times\frac e2\right)\). Allow ECF from (d)(ii).
the maximum of Q is at \(x=2\), so \(\frac{dy}{dx}=0\) there: the profile meets the body at the same radius, \(0.5\) m, and with the same (zero) gradient as the edge \(y=0.5\) of the cylinder, so there is no corner	R1	Both the matching radius and the zero gradient needed.
Part (f)
by parts with \(u=x\), \(\frac{dv}{dx}=e^{1-\frac x2}\): \(\int xe^{1-\frac x2}dx=-2xe^{1-\frac x2}+\int2e^{1-\frac x2}dx\)	M1	Correct choice of \(u\) and attempt at \(uv-\int v\frac{du}{dx}\,dx\).
\(=-2xe^{1-\frac x2}-4e^{1-\frac x2}\;(+C)\)	A1
\(\int_0^2\frac{x}{4}e^{1-\frac x2}dx=\frac14\Big[-2xe^{1-\frac x2}-4e^{1-\frac x2}\Big]_0^2=\frac14\big((-4-4)-(0-4e)\big)=e-2\)	A1	Correct substitution of both limits seen.
area of side view \(=2(e-2)\) m\(^2\)	AG
Part (g)(i)
\(V=\pi\int_0^2\frac{x^2}{16}e^{2-x}dx\)	M1	Correct squaring of \(y\) needed: \(\left(e^{1-\frac x2}\right)^2=e^{2-x}\).
\(\int x^2e^{2-x}dx=-x^2e^{2-x}+\int2xe^{2-x}dx\)	M1	First application of parts with \(u=x^2\).
\(\int2xe^{2-x}dx=-2xe^{2-x}-2e^{2-x}\)	A1	Correct second application of parts, giving a fully correct antiderivative.
\(V=\frac{\pi}{16}\Big[-e^{2-x}\left(x^2+2x+2\right)\Big]_0^2=\frac{\pi}{16}\left(-10+2e^2\right)\)	A1	Correct substitution of both limits seen.
\(=\frac{\pi\left(e^2-5\right)}{8}\)	AG
Part (g)(ii)
\(\frac{0.938180\ldots-0.785398\ldots}{0.785398\ldots}\times100\)	M1	Or \(\left(\frac{e^2-5}{2}-1\right)\times100\). Allow ECF from (a).
\(19.5\%\)	A1	\(19.4528\ldots\)
Part (h)(i)
replacing \(x\) by \(\frac{x}{1.25}\): \(y=\frac14\sqrt{\frac{2x}{1.25}}\Rightarrow y^2=\frac{1.6x}{16}=\frac{x}{10}\)	M1	Replacing \(x\) by \(1.25x\): M0.
\(V=\pi\int_0^{2.5}\frac{x}{10}\,dx=\pi\left[\frac{x^2}{20}\right]_0^{2.5}=\frac{5\pi}{16}\) m\(^3\)	A1	Upper limit \(2.5\) required.
Part (h)(ii)
each disc keeps its radius but its thickness is multiplied by \(k\) (the solid is \(k\) times as long); or, with \(u=\frac xk\): \(\pi\int_0^{2k}\left(f\!\left(\frac xk\right)\right)^2dx=k\,\pi\int_0^2\left(f(u)\right)^2du\)	R1
both volumes are multiplied by \(1.25\), so their ratio, and hence the percentage \((19.5\%)\), is unchanged	R1	Reason required.

Answers: (a) \(\dfrac{\pi}{4}\) m\(^3\) \((\approx0.785\) m\(^3)\) · (b) \(\dfrac13\) m\(^2\) · (c) \(\dfrac{\pi}{4}\) m\(^3\) \((\approx0.785\) m\(^3)\) · (e) \((2,\ 0.5)\) · (g)(ii) \(19.5\%\) · (h)(i) \(\dfrac{5\pi}{16}\) m\(^3\) \((\approx0.982\) m\(^3)\)

Exam tip (f) The exponent \(1-\frac x2\) gives an antiderivative \(-2e^{1-\frac x2}\): the factor \(-2\) is the reciprocal of the coefficient of \(x\). Check it by differentiating before you use it in parts.

Common error (g)(i) Squaring \(\frac x4e^{1-\frac x2}\) gives \(\frac{x^2}{16}e^{2-x}\). Writing \(e^{(1-\frac x2)^2}\) squares the exponent instead of the power.

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A car's suspension settling after a bump

5.16 Integration by substitution and by parts5.11 Definite integrals and areas3.7 Circular functions and their transformations5.6 Differentiation rulesHL

When the front wheel of a car goes over a bump, the wheel moves up and down relative to the body of the car before it settles. The displacement of the wheel from its equilibrium position, \(x\) cm, \(t\) seconds after the bump is modelled by

\[x(t)=4e^{-t}\sin2t,\qquad t\ge0,\]

where upward displacement is positive.

(a)

Write down the period of \(\sin2t\). Hence find the values of \(t\), for \(0\le t\le\pi\), at which the wheel is at its equilibrium position.

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period \(=\pi\)	A1
\(t=0,\ \frac{\pi}{2},\ \pi\)	A1	All three required. \(e^{-t}\ne0\), so \(x=0\) only when \(\sin2t=0\).

Answer: \(\pi\); \(t=0,\ \dfrac{\pi}{2},\ \pi\)

(2)

(b)

Show that \(x'(t)=4e^{-t}\left(2\cos2t-\sin2t\right)\).

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product rule: \(x'(t)=4\left(-e^{-t}\right)\sin2t+4e^{-t}\left(2\cos2t\right)\)	M1	Award for a sum of two products with at least one term correct.
both terms correct, including the chain-rule factors \(-1\) and \(2\)	A1
\(x'(t)=4e^{-t}\left(2\cos2t-\sin2t\right)\)	AG

(2)

(c)

Find the time at which the displacement first reaches a maximum, and find this maximum displacement.

MULTINotes

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\(2\cos2t-\sin2t=0\Rightarrow\tan2t=2\)	M1	Award for setting \(x'(t)=0\) and reaching an equation in one trigonometric function, or for a GDC maximum.
\(t=\frac12\arctan2=0.554\) s	A1	\(0.553574\ldots\)
\(x=2.06\) cm	A1	\(2.05679\ldots\)

Answer: \(t=0.554\) s, \(x=2.06\) cm

(3)

(d)

On Figure 1, sketch the graph of \(x(t)\) for \(0\le t\le\pi\). Show clearly the points where your graph meets the curves \(x=\pm4e^{-t}\).

Figure 1: the curves x = 4e⁻ᵗ and x = −4e⁻ᵗ for 0 ≤ t ≤ π

GRAPHNotes

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starts at the origin; crosses the \(t\)-axis at \(t=\frac{\pi}{2}\) and ends at \(t=\pi\) on the axis; positive then negative	A1
maximum at about \((0.55,\ 2.06)\) and minimum at about \((2.12,\ -0.43)\)	A1	Allow ECF from (c). The minimum is \(\frac{\pi}{2}\) later than the maximum.
touches \(x=4e^{-t}\) at \(t=\frac{\pi}{4}\) and \(x=-4e^{-t}\) at \(t=\frac{3\pi}{4}\), without crossing them	A1	These are where \(\sin2t=\pm1\); they are not the turning points.

(3)

(e)

Let \(I=\displaystyle\int e^{-t}\sin2t\,dt\).

Use integration by parts twice to show that \(I=-\dfrac15e^{-t}\left(\sin2t+2\cos2t\right)+C\).

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\(I=-e^{-t}\sin2t+\int2e^{-t}\cos2t\,dt\)	M1	First application of parts, e.g. \(u=\sin2t\), \(\frac{dv}{dt}=e^{-t}\). Either choice of \(u\) is acceptable if kept consistently.
\(\int2e^{-t}\cos2t\,dt=-2e^{-t}\cos2t-\int4e^{-t}\sin2t\,dt\)	M1	Second application of parts with the same type of choice.
\(I=-e^{-t}\sin2t-2e^{-t}\cos2t-4I\)	A1	Recognising that the original integral has returned.
\(5I=-e^{-t}\left(\sin2t+2\cos2t\right)\) (+ constant)	A1
\(I=-\frac15e^{-t}\left(\sin2t+2\cos2t\right)+C\)	AG

(4)

(f)

Hence show that \(\displaystyle\int_0^{\pi/2}x(t)\,dt=\dfrac85\left(1+e^{-\pi/2}\right)\).

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\(4\left[-\frac15e^{-t}\left(\sin2t+2\cos2t\right)\right]_0^{\pi/2}\)	M1	Uses \(4I\) with both limits.
\(=4\left(-\frac15e^{-\pi/2}(0-2)+\frac15(0+2)\right)=\frac85e^{-\pi/2}+\frac85\)	A1	Correct values of \(\sin\) and \(\cos\) at both limits seen.
\(=\frac85\left(1+e^{-\pi/2}\right)\)	AG

(2)

(g)(i)

An engineer measures the discomfort of the ride by \(D=\displaystyle\int_0^{\pi}\left|x(t)\right|dt\), the total area between the graph of \(x\) and the \(t\)-axis for \(0\le t\le\pi\).

Show that \(x\!\left(t+\dfrac{\pi}{2}\right)=-e^{-\pi/2}\,x(t)\).

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\(x\!\left(t+\frac{\pi}{2}\right)=4e^{-t-\frac{\pi}{2}}\sin(2t+\pi)\)	M1
\(=4e^{-\pi/2}e^{-t}\left(-\sin2t\right)\)	A1	Both \(\sin(2t+\pi)=-\sin2t\) and \(e^{-t-\frac\pi2}=e^{-\pi/2}e^{-t}\) seen.
\(=-e^{-\pi/2}\,x(t)\)	AG

(2)

(g)(ii)

By using the substitution \(u=t-\dfrac{\pi}{2}\), or otherwise, show that \(D=\dfrac85\left(1+e^{-\pi/2}\right)^2\).

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\(x\le0\) for \(\frac\pi2\le t\le\pi\), so \(D=\int_0^{\pi/2}x\,dt-\int_{\pi/2}^{\pi}x\,dt\)	M1	Splits at \(t=\frac{\pi}{2}\) and changes the sign of the negative part.
\(-\int_{\pi/2}^{\pi}x(t)\,dt=-\int_0^{\pi/2}x\!\left(u+\frac{\pi}{2}\right)du=e^{-\pi/2}\int_0^{\pi/2}x(u)\,du\)	M1	Substitution with new limits \(0\) and \(\frac{\pi}{2}\), and use of (g)(i).
\(D=\frac85\left(1+e^{-\pi/2}\right)+e^{-\pi/2}\cdot\frac85\left(1+e^{-\pi/2}\right)\)	A1	Uses (f) twice.
\(=\frac85\left(1+e^{-\pi/2}\right)^2\)	AG	Direct integration with the result of (e) is acceptable for full marks.

(3)

(h)(i)

The car is fitted with stiffer springs. The displacement of the wheel after the same bump is now modelled by \(x_s(t)=4e^{-1.5t}\sin3t\).

Show that \(x_s(t)=x(1.5t)\). Hence write down the time at which \(x_s\) first reaches a maximum, and the value of this maximum.

CONCEPTNotes

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\(x(1.5t)=4e^{-1.5t}\sin(2\times1.5t)=4e^{-1.5t}\sin3t=x_s(t)\)	A1
\(t=\frac{0.554}{1.5}=0.369\) s and maximum \(2.06\) cm (unchanged)	A1	Both values needed. Allow ECF from (c).

Answer: \(t=0.369\) s; \(2.06\) cm

(2)

(h)(ii)

Use the substitution \(u=1.5t\) to express \(D_s=\displaystyle\int_0^{2\pi/3}\left|x_s(t)\right|dt\) in terms of \(D\), and find the value of \(D_s\).

MULTINotes

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\(dt=\frac{du}{1.5}\), limits \(0\) and \(\pi\): \(D_s=\frac{1}{1.5}\int_0^{\pi}\left\|x(u)\right\|du\)	M1	Both the factor and the new limits needed.
\(D_s=\frac23D=1.56\)	A1	\(1.55623\ldots\) Allow ECF from their \(D=2.33435\ldots\)

Answer: \(D_s=\dfrac23D\approx1.56\)

(2)

(h)(iii)

With reference to your answers to parts (h)(i) and (h)(ii), discuss the effect of the stiffer springs on the comfort of the passengers.

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the greatest displacement is the same (\(2.06\) cm) but the motion dies away more quickly: \(D\) falls by one third	R1
but the period falls from \(\pi\approx3.14\) s to \(\frac{2\pi}{3}\approx2.09\) s, so the wheel moves up and down faster and the ride may feel harsher; a reasoned conclusion	R1	Accept any conclusion supported by both effects.

(2)

Show full solution

Working	Mark	Notes
Part (a)
period \(=\pi\)	A1
\(t=0,\ \frac{\pi}{2},\ \pi\)	A1	All three required. \(e^{-t}\ne0\), so \(x=0\) only when \(\sin2t=0\).
Part (b)
product rule: \(x'(t)=4\left(-e^{-t}\right)\sin2t+4e^{-t}\left(2\cos2t\right)\)	M1	Award for a sum of two products with at least one term correct.
both terms correct, including the chain-rule factors \(-1\) and \(2\)	A1
\(x'(t)=4e^{-t}\left(2\cos2t-\sin2t\right)\)	AG
Part (c)
\(2\cos2t-\sin2t=0\Rightarrow\tan2t=2\)	M1	Award for setting \(x'(t)=0\) and reaching an equation in one trigonometric function, or for a GDC maximum.
\(t=\frac12\arctan2=0.554\) s	A1	\(0.553574\ldots\)
\(x=2.06\) cm	A1	\(2.05679\ldots\)
Part (d)
starts at the origin; crosses the \(t\)-axis at \(t=\frac{\pi}{2}\) and ends at \(t=\pi\) on the axis; positive then negative	A1
maximum at about \((0.55,\ 2.06)\) and minimum at about \((2.12,\ -0.43)\)	A1	Allow ECF from (c). The minimum is \(\frac{\pi}{2}\) later than the maximum.
touches \(x=4e^{-t}\) at \(t=\frac{\pi}{4}\) and \(x=-4e^{-t}\) at \(t=\frac{3\pi}{4}\), without crossing them	A1	These are where \(\sin2t=\pm1\); they are not the turning points.
Part (e)
\(I=-e^{-t}\sin2t+\int2e^{-t}\cos2t\,dt\)	M1	First application of parts, e.g. \(u=\sin2t\), \(\frac{dv}{dt}=e^{-t}\). Either choice of \(u\) is acceptable if kept consistently.
\(\int2e^{-t}\cos2t\,dt=-2e^{-t}\cos2t-\int4e^{-t}\sin2t\,dt\)	M1	Second application of parts with the same type of choice.
\(I=-e^{-t}\sin2t-2e^{-t}\cos2t-4I\)	A1	Recognising that the original integral has returned.
\(5I=-e^{-t}\left(\sin2t+2\cos2t\right)\) (+ constant)	A1
\(I=-\frac15e^{-t}\left(\sin2t+2\cos2t\right)+C\)	AG
Part (f)
\(4\left[-\frac15e^{-t}\left(\sin2t+2\cos2t\right)\right]_0^{\pi/2}\)	M1	Uses \(4I\) with both limits.
\(=4\left(-\frac15e^{-\pi/2}(0-2)+\frac15(0+2)\right)=\frac85e^{-\pi/2}+\frac85\)	A1	Correct values of \(\sin\) and \(\cos\) at both limits seen.
\(=\frac85\left(1+e^{-\pi/2}\right)\)	AG
Part (g)(i)
\(x\!\left(t+\frac{\pi}{2}\right)=4e^{-t-\frac{\pi}{2}}\sin(2t+\pi)\)	M1
\(=4e^{-\pi/2}e^{-t}\left(-\sin2t\right)\)	A1	Both \(\sin(2t+\pi)=-\sin2t\) and \(e^{-t-\frac\pi2}=e^{-\pi/2}e^{-t}\) seen.
\(=-e^{-\pi/2}\,x(t)\)	AG
Part (g)(ii)
\(x\le0\) for \(\frac\pi2\le t\le\pi\), so \(D=\int_0^{\pi/2}x\,dt-\int_{\pi/2}^{\pi}x\,dt\)	M1	Splits at \(t=\frac{\pi}{2}\) and changes the sign of the negative part.
\(-\int_{\pi/2}^{\pi}x(t)\,dt=-\int_0^{\pi/2}x\!\left(u+\frac{\pi}{2}\right)du=e^{-\pi/2}\int_0^{\pi/2}x(u)\,du\)	M1	Substitution with new limits \(0\) and \(\frac{\pi}{2}\), and use of (g)(i).
\(D=\frac85\left(1+e^{-\pi/2}\right)+e^{-\pi/2}\cdot\frac85\left(1+e^{-\pi/2}\right)\)	A1	Uses (f) twice.
\(=\frac85\left(1+e^{-\pi/2}\right)^2\)	AG	Direct integration with the result of (e) is acceptable for full marks.
Part (h)(i)
\(x(1.5t)=4e^{-1.5t}\sin(2\times1.5t)=4e^{-1.5t}\sin3t=x_s(t)\)	A1
\(t=\frac{0.554}{1.5}=0.369\) s and maximum \(2.06\) cm (unchanged)	A1	Both values needed. Allow ECF from (c).
Part (h)(ii)
\(dt=\frac{du}{1.5}\), limits \(0\) and \(\pi\): \(D_s=\frac{1}{1.5}\int_0^{\pi}\left\|x(u)\right\|du\)	M1	Both the factor and the new limits needed.
\(D_s=\frac23D=1.56\)	A1	\(1.55623\ldots\) Allow ECF from their \(D=2.33435\ldots\)
Part (h)(iii)
the greatest displacement is the same (\(2.06\) cm) but the motion dies away more quickly: \(D\) falls by one third	R1
but the period falls from \(\pi\approx3.14\) s to \(\frac{2\pi}{3}\approx2.09\) s, so the wheel moves up and down faster and the ride may feel harsher; a reasoned conclusion	R1	Accept any conclusion supported by both effects.

Answers: (a) \(\pi\); \(t=0,\ \dfrac{\pi}{2},\ \pi\) · (c) \(t=0.554\) s, \(x=2.06\) cm · (h)(i) \(t=0.369\) s; \(2.06\) cm · (h)(ii) \(D_s=\dfrac23D\approx1.56\)

Exam tip (d) The graph touches the curve \(x=4e^{-t}\) where \(\sin2t=1\), but its maximum comes earlier, where \(x'(t)=0\). At \(t=\frac{\pi}{4}\) the decreasing factor \(e^{-t}\) already makes the graph fall.

Common error (e) Changing the choice of \(u\) in the second application (for example \(u=e^{-t}\) after \(u=\sin2t\)) undoes the first step and gives \(I=I\). Keep the trigonometric factor as \(u\) both times.

More 5.16 questions in the question bank

5.18 First order differential equations · 5.19 Maclaurin seriesmerged group

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A skydiver approaching terminal velocity

5.18 First order differential equations5.19 Maclaurin series2.9 Exponential and logarithmic functions5.9 KinematicsHL

A skydiver steps out of a hovering helicopter and falls vertically. Her velocity \(t\) seconds after leaving the helicopter is \(v\) m s\(^{-1}\), measured downwards. Until her parachute opens, her motion is modelled by the differential equation

\[\frac{dv}{dt}=9.8-0.2v,\qquad v=0\ \text{when}\ t=0.\]

(a)

Use Euler's method with a step length of \(1\) second to estimate the velocity of the skydiver when \(t=3\).

ROUTINENotes

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\(v_{n+1}=v_n+1\times\left(9.8-0.2v_n\right)\)	M1	Award for one correct step from \(v_0=0\).
\(v_1=9.8,\ v_2=17.64\)	A1
\(v_3=23.9\) m s\(^{-1}\)	A1	\(23.912\)

Answer: \(23.9\) m s\(^{-1}\)

(3)

(b)

By solving the differential equation, show that \(v=49\left(1-e^{-0.2t}\right)\).

No calculatorALGEBRANotes

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\(\int\frac{1}{9.8-0.2v}\,dv=\int1\,dt\)	M1	Correct separation. An integrating factor \(e^{0.2t}\) is an acceptable alternative: \(\frac{d}{dt}\left(ve^{0.2t}\right)=9.8e^{0.2t}\).
\(-5\ln\left\|9.8-0.2v\right\|=t+c\)	A1	IF method: \(ve^{0.2t}=49e^{0.2t}+c\).
\(t=0,\ v=0\Rightarrow c=-5\ln9.8\)	M1	IF method: \(c=-49\).
\(\ln\frac{9.8-0.2v}{9.8}=-0.2t\Rightarrow1-\frac{v}{49}=e^{-0.2t}\)	A1
\(v=49\left(1-e^{-0.2t}\right)\)	AG

(4)

(c)

Find the value of \(v\) when \(t=3\) given by this solution. Explain why the estimate found in part (a) is greater than this value.

MULTINotes

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\(v=22.1\) m s\(^{-1}\)	A1	\(22.1082\ldots\)
\(\frac{dv}{dt}=9.8-0.2v\) decreases as \(v\) increases, so the graph of \(v\) is concave down; each Euler step follows the tangent at the start of the step, which lies above the curve	R1	Accept: the gradient used in each step is the largest gradient on that step.

Answer: \(22.1\) m s\(^{-1}\)

(2)

(d)

Write down the terminal velocity of the skydiver according to this model. Find the time taken for her to reach \(95\%\) of this velocity.

ROUTINENotes

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\(49\) m s\(^{-1}\)	A1	As \(t\to\infty\), \(e^{-0.2t}\to0\).
\(49\left(1-e^{-0.2t}\right)=0.95\times49\Rightarrow e^{-0.2t}=0.05\)	M1
\(t=5\ln20=15.0\) s	A1	\(14.9786\ldots\)

Answer: \(49\) m s\(^{-1}\); \(15.0\) s

(3)

(e)

Figure 1 shows the slope field of the differential equation for \(0\le t\le30\).

On Figure 1, sketch the solution curve that passes through \((0,\,0)\), and draw its horizontal asymptote. Explain what the slope field shows about the motion of a skydiver whose velocity is greater than the terminal velocity.

Figure 1: the slope field of dv/dt = 9.8 − 0.2v

GRAPHNotes

Show solution

curve from the origin, increasing and concave down, following the direction of the segments	A1	Passes close to \((5,\ 31)\) and \((15,\ 47)\).
asymptote \(v=49\) drawn; curve approaches it from below without crossing it	A1	Allow ECF from (d).
above \(v=49\) the segments slope downwards, so \(\frac{dv}{dt}\lt0\): such a skydiver slows down towards \(49\) m s\(^{-1}\) (the segments on \(v=49\) are horizontal)	R1

(3)

(f)

Show that the distance fallen by the skydiver in the first \(10\) seconds is \(245\left(1+e^{-2}\right)\) metres.

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distance \(=\int_0^{10}49\left(1-e^{-0.2t}\right)dt\)	M1	\(v\ge0\), so the distance is the integral of \(v\).
\(=\Big[49t+245e^{-0.2t}\Big]_0^{10}\)	A1
\(=\left(490+245e^{-2}\right)-245\)	A1	Correct substitution of both limits seen.
\(=245\left(1+e^{-2}\right)\)	AG

(3)

(g)(i)

By differentiating the differential equation, find the value of \(\dfrac{d^2v}{dt^2}\) when \(t=0\). Hence find the Maclaurin series for \(v\) up to and including the term in \(t^2\).

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\(\frac{d^2v}{dt^2}=-0.2\frac{dv}{dt}\)	M1
at \(t=0\): \(\frac{dv}{dt}=9.8\) and \(\frac{d^2v}{dt^2}=-1.96\)	A1
\(v\approx0+9.8t-\frac{1.96}{2!}t^2=9.8t-0.98t^2\)	A1	Expanding \(49\left(1-e^{-0.2t}\right)\) with the series for \(e^x\) gives the same result and is acceptable.

Answer: \(v\approx9.8t-0.98t^2\)

(3)

(g)(ii)

Use your answer to part (g)(i) to estimate the velocity when \(t=10\). Comment on your result.

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\(9.8(10)-0.98(100)=0\)	A1
not reasonable: the skydiver is still falling fast (the model gives \(49\left(1-e^{-2}\right)\approx42\) m s\(^{-1}\)); the truncated series is accurate only for \(t\) close to \(0\), and the neglected terms are not small at \(t=10\)	R1

(2)

(h)(i)

Later in the jump, the skydiver is falling at a steady \(49\) m s\(^{-1}\) when she opens her parachute. Let \(\tau\) be the time in seconds after the parachute opens. Her motion is now modelled by

\[\frac{dv}{d\tau}=9.8-1.4v,\qquad v=49\ \text{when}\ \tau=0.\]

Solve this differential equation to find \(v\) in terms of \(\tau\).

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\(\int\frac{1}{9.8-1.4v}\,dv=\int1\,d\tau\) or integrating factor \(e^{1.4\tau}\)	M1
\(v=7+Ae^{-1.4\tau}\)	A1	Or \(-\frac{1}{1.4}\ln\left\|9.8-1.4v\right\|=\tau+c\).
\(v=7+42e^{-1.4\tau}\)	A1

Answer: \(v=7+42e^{-1.4\tau}\)

(3)

(h)(ii)

Find the acceleration of the skydiver immediately after the parachute opens. Describe her motion after that, and interpret your answers in context.

INTERPRETNotes

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\(9.8-1.4(49)=-58.8\) m s\(^{-2}\)	A1	Accept a deceleration of \(58.8\) m s\(^{-2}\) (about six times \(9.8\)).
she slows down sharply at first, then more and more gently, and her velocity approaches a new terminal velocity of \(7\) m s\(^{-1}\) from above: she keeps falling, but at a speed that is safe for landing	R1	Must refer to the limit \(7\) m s\(^{-1}\). Allow ECF from (h)(i).

Answer: \(-58.8\) m s\(^{-2}\)

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(v_{n+1}=v_n+1\times\left(9.8-0.2v_n\right)\)	M1	Award for one correct step from \(v_0=0\).
\(v_1=9.8,\ v_2=17.64\)	A1
\(v_3=23.9\) m s\(^{-1}\)	A1	\(23.912\)
Part (b)
\(\int\frac{1}{9.8-0.2v}\,dv=\int1\,dt\)	M1	Correct separation. An integrating factor \(e^{0.2t}\) is an acceptable alternative: \(\frac{d}{dt}\left(ve^{0.2t}\right)=9.8e^{0.2t}\).
\(-5\ln\left\|9.8-0.2v\right\|=t+c\)	A1	IF method: \(ve^{0.2t}=49e^{0.2t}+c\).
\(t=0,\ v=0\Rightarrow c=-5\ln9.8\)	M1	IF method: \(c=-49\).
\(\ln\frac{9.8-0.2v}{9.8}=-0.2t\Rightarrow1-\frac{v}{49}=e^{-0.2t}\)	A1
\(v=49\left(1-e^{-0.2t}\right)\)	AG
Part (c)
\(v=22.1\) m s\(^{-1}\)	A1	\(22.1082\ldots\)
\(\frac{dv}{dt}=9.8-0.2v\) decreases as \(v\) increases, so the graph of \(v\) is concave down; each Euler step follows the tangent at the start of the step, which lies above the curve	R1	Accept: the gradient used in each step is the largest gradient on that step.
Part (d)
\(49\) m s\(^{-1}\)	A1	As \(t\to\infty\), \(e^{-0.2t}\to0\).
\(49\left(1-e^{-0.2t}\right)=0.95\times49\Rightarrow e^{-0.2t}=0.05\)	M1
\(t=5\ln20=15.0\) s	A1	\(14.9786\ldots\)
Part (e)
curve from the origin, increasing and concave down, following the direction of the segments	A1	Passes close to \((5,\ 31)\) and \((15,\ 47)\).
asymptote \(v=49\) drawn; curve approaches it from below without crossing it	A1	Allow ECF from (d).
above \(v=49\) the segments slope downwards, so \(\frac{dv}{dt}\lt0\): such a skydiver slows down towards \(49\) m s\(^{-1}\) (the segments on \(v=49\) are horizontal)	R1
Part (f)
distance \(=\int_0^{10}49\left(1-e^{-0.2t}\right)dt\)	M1	\(v\ge0\), so the distance is the integral of \(v\).
\(=\Big[49t+245e^{-0.2t}\Big]_0^{10}\)	A1
\(=\left(490+245e^{-2}\right)-245\)	A1	Correct substitution of both limits seen.
\(=245\left(1+e^{-2}\right)\)	AG
Part (g)(i)
\(\frac{d^2v}{dt^2}=-0.2\frac{dv}{dt}\)	M1
at \(t=0\): \(\frac{dv}{dt}=9.8\) and \(\frac{d^2v}{dt^2}=-1.96\)	A1
\(v\approx0+9.8t-\frac{1.96}{2!}t^2=9.8t-0.98t^2\)	A1	Expanding \(49\left(1-e^{-0.2t}\right)\) with the series for \(e^x\) gives the same result and is acceptable.
Part (g)(ii)
\(9.8(10)-0.98(100)=0\)	A1
not reasonable: the skydiver is still falling fast (the model gives \(49\left(1-e^{-2}\right)\approx42\) m s\(^{-1}\)); the truncated series is accurate only for \(t\) close to \(0\), and the neglected terms are not small at \(t=10\)	R1
Part (h)(i)
\(\int\frac{1}{9.8-1.4v}\,dv=\int1\,d\tau\) or integrating factor \(e^{1.4\tau}\)	M1
\(v=7+Ae^{-1.4\tau}\)	A1	Or \(-\frac{1}{1.4}\ln\left\|9.8-1.4v\right\|=\tau+c\).
\(v=7+42e^{-1.4\tau}\)	A1
Part (h)(ii)
\(9.8-1.4(49)=-58.8\) m s\(^{-2}\)	A1	Accept a deceleration of \(58.8\) m s\(^{-2}\) (about six times \(9.8\)).
she slows down sharply at first, then more and more gently, and her velocity approaches a new terminal velocity of \(7\) m s\(^{-1}\) from above: she keeps falling, but at a speed that is safe for landing	R1	Must refer to the limit \(7\) m s\(^{-1}\). Allow ECF from (h)(i).

Answers: (a) \(23.9\) m s\(^{-1}\) · (c) \(22.1\) m s\(^{-1}\) · (d) \(49\) m s\(^{-1}\); \(15.0\) s · (g)(i) \(v\approx9.8t-0.98t^2\) · (h)(i) \(v=7+42e^{-1.4\tau}\) · (h)(ii) \(-58.8\) m s\(^{-2}\)

Common error (f) The lower limit does not give zero: \(245e^{0}=245\) must be subtracted. Forgetting it gives \(490+245e^{-2}\), which is more than \(49\times10\), impossible when \(v\lt49\) throughout.

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The sag of a suspension cable: series approximations

5.19 Maclaurin series2.9 Exponential and logarithmic functions5.11 Definite integrals and areas5.13 Limits and l’Hôpital’s ruleHL

During the construction of a suspension footbridge, the main cable hangs freely between the tops of two towers that are \(120\) m apart. The origin \(O\) is at the lowest point of the cable, the \(x\)-axis is horizontal and the \(y\)-axis is vertical, with lengths in metres. The cable is modelled by

\[y=100\left(e^{\frac{x}{200}}+e^{-\frac{x}{200}}\right)-200,\qquad -60\le x\le60.\]

The deck of the bridge will be horizontal and \(2\) m below \(O\), as shown in Figure 1.

Figure 1: the cable, the towers and the deck (not to scale; only some of the hangers are shown)

(a)(i)

Write down the Maclaurin series for \(e^u\) up to and including the term in \(u^4\). Hence show that \(e^u+e^{-u}=2+u^2+\dfrac{u^4}{12}+\cdots\).

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\(e^u=1+u+\frac{u^2}{2!}+\frac{u^3}{3!}+\frac{u^4}{4!}+\cdots\)	A1
\(e^{-u}=1-u+\frac{u^2}{2!}-\frac{u^3}{3!}+\frac{u^4}{4!}-\cdots\), so the odd powers cancel when the series are added	A1
\(e^u+e^{-u}=2+u^2+\frac{2u^4}{24}+\cdots=2+u^2+\frac{u^4}{12}+\cdots\)	AG

(2)

(a)(ii)

Hence show that, for the cable, \(y=\dfrac{x^2}{400}+\dfrac{x^4}{k}+\cdots\), where \(k\) is a positive integer to be found.

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\(u=\frac{x}{200}\): \(y=100\left(2+\frac{x^2}{200^2}+\frac{x^4}{12\times200^4}+\cdots\right)-200\)	M1
\(=\frac{100x^2}{40000}+\frac{100x^4}{12\times1.6\times10^{9}}+\cdots=\frac{x^2}{400}+\cdots\)	A1	The constant terms cancel.
\(k=192\,000\,000\;\left(=1.92\times10^{8}\right)\)	A1

Answer: \(k=192\,000\,000\)

(3)

(b)(i)

Use l’Hôpital’s rule to find \(\displaystyle\lim_{x\to0}\frac{y}{x^2}\).

No calculatorROUTINENotes

Show solution

form \(\frac00\); \(\lim_{x\to0}\frac{\frac12\left(e^{\frac{x}{200}}-e^{-\frac{x}{200}}\right)}{2x}\)	M1	Differentiates numerator and denominator separately.
still \(\frac00\); \(\lim_{x\to0}\frac{\frac{1}{400}\left(e^{\frac{x}{200}}+e^{-\frac{x}{200}}\right)}{2}\)	M1	Second application, with the form checked.
\(=\frac{\frac{2}{400}}{2}=\frac{1}{400}\)	A1	A method using only the series of (a)(ii) scores M0M0A0.

Answer: \(\dfrac{1}{400}\)

(3)

(b)(ii)

Write down the value of \(\displaystyle\lim_{x\to0}\frac{y-\frac{x^2}{400}}{x^4}\).

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\(\frac{1}{192\,000\,000}\)

\(\frac1k\). Allow ECF from (a)(ii).

Answer: \(\dfrac{1}{192\,000\,000}\)

(1)

(c)

Find the exact values of \(x\) at which the cable is \(5\) m above \(O\).

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\(100\left(e^{u}+e^{-u}\right)-200=5\) with \(u=\frac{x}{200}\), so \(e^u+e^{-u}=2.05\)	M1
\(e^{2u}-2.05e^u+1=0\) or \(20e^{2u}-41e^u+20=0\)	M1	Multiplies by \(e^u\) to form a quadratic in \(e^u\).
\(\left(4e^u-5\right)\left(5e^u-4\right)=0\Rightarrow e^u=\frac54\) or \(\frac45\)	A1
\(x=\pm200\ln\frac54\)	A1	Accept \(200\ln\frac54\) and \(200\ln\frac45\). Decimals only \((\pm44.6)\): A0.

Answer: \(x=\pm200\ln\dfrac54\;(\approx\pm44.6)\)

(4)

(d)

Find the height of the tops of the towers above \(O\). Find the percentage error in this height when it is estimated using \(y\approx\dfrac{x^2}{400}\).

ROUTINENotes

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\(9.07\) m	A1	\(9.06770\ldots\)
\(\frac{9.0677\ldots-9}{9.0677\ldots}\times100=0.747\%\)	A1	Accept \(0.746\%\) to \(0.75\%\).

Answer: \(9.07\) m; \(0.747\%\)

(2)

(e)(i)

Vertical hangers will join the cable to the deck at intervals of \(1\) m. The total length of the hangers, in metres, is estimated by \(H=\displaystyle\int_{-60}^{60}(y+2)\,dx\).

Show that \(H=40000\left(e^{0.3}-e^{-0.3}\right)-23760\).

No calculatorROUTINENotes

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\(\int\left(100e^{\frac{x}{200}}+100e^{-\frac{x}{200}}-198\right)dx\)	M1	Attempt to integrate term by term.
\(=20000e^{\frac{x}{200}}-20000e^{-\frac{x}{200}}-198x\)	A1
\(\left(20000e^{0.3}-20000e^{-0.3}-11880\right)-\left(20000e^{-0.3}-20000e^{0.3}+11880\right)\)	A1	Or uses symmetry: \(2\int_0^{60}\ldots\,dx\).
\(=40000\left(e^{0.3}-e^{-0.3}\right)-23760\)	AG

(3)

(e)(ii)

Find the value of \(H\). Find also the estimate of \(H\) obtained when \(y\) is replaced by \(\dfrac{x^2}{400}\).

ROUTINENotes

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\(H=602\) m	A1	\(601.623\ldots\)
\(\int_{-60}^{60}\left(\frac{x^2}{400}+2\right)dx=360+240=600\) m	A1

Answer: \(602\) m; \(600\) m

(2)

(f)(i)

A second footbridge uses a cable with the same equation, but its towers are \(160\) m apart, so that \(-80\le x\le80\).

Find the percentage error in the height of its towers when \(y\approx\dfrac{x^2}{400}\) is used.

ROUTINENotes

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exact height \(16.214\ldots\) m; estimate \(\frac{80^2}{400}=16\) m	M1
\(1.32\%\)	A1	\(1.32273\ldots\)

Answer: \(1.32\%\)

(2)

(f)(ii)

Figure 2 shows the graph of \(y=100\left(e^{\frac{x}{200}}+e^{-\frac{x}{200}}\right)-200\) for \(0\le x\le300\).

On Figure 2, sketch the graph of \(y=\dfrac{x^2}{400}\) for \(0\le x\le300\).

Figure 2: the graph of the cable model for 0 ≤ x ≤ 300

GRAPHNotes

Show solution

curve from \(O\), almost indistinguishable from the given graph for small \(x\) (up to about \(x=100\))	A1
below the given graph for \(x\gt0\), with the gap increasing as \(x\) increases	A1
passes through about \((200,\ 100)\) and ends at about \((300,\ 225)\), compared with \((300,\ 270)\) on the given graph	A1

(3)

(g)

With reference to the series in part (a)(ii), explain why the approximation \(y\approx\dfrac{x^2}{400}\) becomes less accurate as the distance between the towers increases.

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the neglected terms contain only even powers of \(x\) with positive coefficients, so the approximation is always too small, by an amount that grows with \(x\)	R1
relative to \(\frac{x^2}{400}\), the first neglected term is \(\frac{x^4}{k}\div\frac{x^2}{400}=\frac{x^2}{480\,000}\), which increases with \(x\) (with the square of the half-span: from about \(0.75\%\) at \(x=60\) to about \(1.3\%\) at \(x=80\))	R1	Accept a correct argument that the percentage error grows roughly with \(x^2\).

(2)

Show full solution

Working	Mark	Notes
Part (a)(i)
\(e^u=1+u+\frac{u^2}{2!}+\frac{u^3}{3!}+\frac{u^4}{4!}+\cdots\)	A1
\(e^{-u}=1-u+\frac{u^2}{2!}-\frac{u^3}{3!}+\frac{u^4}{4!}-\cdots\), so the odd powers cancel when the series are added	A1
\(e^u+e^{-u}=2+u^2+\frac{2u^4}{24}+\cdots=2+u^2+\frac{u^4}{12}+\cdots\)	AG
Part (a)(ii)
\(u=\frac{x}{200}\): \(y=100\left(2+\frac{x^2}{200^2}+\frac{x^4}{12\times200^4}+\cdots\right)-200\)	M1
\(=\frac{100x^2}{40000}+\frac{100x^4}{12\times1.6\times10^{9}}+\cdots=\frac{x^2}{400}+\cdots\)	A1	The constant terms cancel.
\(k=192\,000\,000\;\left(=1.92\times10^{8}\right)\)	A1
Part (b)(i)
form \(\frac00\); \(\lim_{x\to0}\frac{\frac12\left(e^{\frac{x}{200}}-e^{-\frac{x}{200}}\right)}{2x}\)	M1	Differentiates numerator and denominator separately.
still \(\frac00\); \(\lim_{x\to0}\frac{\frac{1}{400}\left(e^{\frac{x}{200}}+e^{-\frac{x}{200}}\right)}{2}\)	M1	Second application, with the form checked.
\(=\frac{\frac{2}{400}}{2}=\frac{1}{400}\)	A1	A method using only the series of (a)(ii) scores M0M0A0.
Part (b)(ii)
\(\frac{1}{192\,000\,000}\)	A1	\(\frac1k\). Allow ECF from (a)(ii).
Part (c)
\(100\left(e^{u}+e^{-u}\right)-200=5\) with \(u=\frac{x}{200}\), so \(e^u+e^{-u}=2.05\)	M1
\(e^{2u}-2.05e^u+1=0\) or \(20e^{2u}-41e^u+20=0\)	M1	Multiplies by \(e^u\) to form a quadratic in \(e^u\).
\(\left(4e^u-5\right)\left(5e^u-4\right)=0\Rightarrow e^u=\frac54\) or \(\frac45\)	A1
\(x=\pm200\ln\frac54\)	A1	Accept \(200\ln\frac54\) and \(200\ln\frac45\). Decimals only \((\pm44.6)\): A0.
Part (d)
\(9.07\) m	A1	\(9.06770\ldots\)
\(\frac{9.0677\ldots-9}{9.0677\ldots}\times100=0.747\%\)	A1	Accept \(0.746\%\) to \(0.75\%\).
Part (e)(i)
\(\int\left(100e^{\frac{x}{200}}+100e^{-\frac{x}{200}}-198\right)dx\)	M1	Attempt to integrate term by term.
\(=20000e^{\frac{x}{200}}-20000e^{-\frac{x}{200}}-198x\)	A1
\(\left(20000e^{0.3}-20000e^{-0.3}-11880\right)-\left(20000e^{-0.3}-20000e^{0.3}+11880\right)\)	A1	Or uses symmetry: \(2\int_0^{60}\ldots\,dx\).
\(=40000\left(e^{0.3}-e^{-0.3}\right)-23760\)	AG
Part (e)(ii)
\(H=602\) m	A1	\(601.623\ldots\)
\(\int_{-60}^{60}\left(\frac{x^2}{400}+2\right)dx=360+240=600\) m	A1
Part (f)(i)
exact height \(16.214\ldots\) m; estimate \(\frac{80^2}{400}=16\) m	M1
\(1.32\%\)	A1	\(1.32273\ldots\)
Part (f)(ii)
curve from \(O\), almost indistinguishable from the given graph for small \(x\) (up to about \(x=100\))	A1
below the given graph for \(x\gt0\), with the gap increasing as \(x\) increases	A1
passes through about \((200,\ 100)\) and ends at about \((300,\ 225)\), compared with \((300,\ 270)\) on the given graph	A1
Part (g)
the neglected terms contain only even powers of \(x\) with positive coefficients, so the approximation is always too small, by an amount that grows with \(x\)	R1
relative to \(\frac{x^2}{400}\), the first neglected term is \(\frac{x^4}{k}\div\frac{x^2}{400}=\frac{x^2}{480\,000}\), which increases with \(x\) (with the square of the half-span: from about \(0.75\%\) at \(x=60\) to about \(1.3\%\) at \(x=80\))	R1	Accept a correct argument that the percentage error grows roughly with \(x^2\).

Answers: (a)(ii) \(k=192\,000\,000\) · (b)(i) \(\dfrac{1}{400}\) · (b)(ii) \(\dfrac{1}{192\,000\,000}\) · (c) \(x=\pm200\ln\dfrac54\;(\approx\pm44.6)\) · (d) \(9.07\) m; \(0.747\%\) · (e)(ii) \(602\) m; \(600\) m · (f)(i) \(1.32\%\)

Exam tip (c) An equation containing both \(e^u\) and \(e^{-u}\) becomes a quadratic in \(e^u\) when you multiply every term by \(e^u\). Check that each root for \(e^u\) is positive before taking logarithms.

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