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IB Maths Applications & Interpretation · SL and HL · Topic 2

Functions

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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 15 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →

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+ Add to test✓ In your test1IBMI-031-00-01Linear models and piecewise models Paper SLEasy5 marks

A taxi company charges a fixed fee plus a charge for each kilometre travelled. The cost, C dollars, of a journey of d kilometres is modelled by
$C = 3.50 + 1.85d$.

(a)

Find the cost of a 12 km journey.

(2)

(b)

Interpret the value 1.85 in the context of the model.

(1)

(c)

Priya has 40.00 dollars. Find the greatest distance she can travel.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substituting d = 12: 3.50 + 1.85 × 12	✓ 1
A1 25.70 dollars	✓ 1
Part (b)
R1 each additional kilometre travelled adds 1.85 dollars to the cost (cost per kilometre)	✓ 1
Part (c)
M1 forming 3.50 + 1.85d = 40 (or 3.50 + 1.85d ≤ 40)	✓ 1
A1 d = 19.7 km (accept 19.7297…; do not accept 19.8, because that would cost more than 40 dollars)	✓ 1

Exam tip: When asked to interpret a gradient, give its meaning with units and context, for example 'dollars per kilometre', not just 'the slope'.

+ Add to test✓ In your test2IBMI-031-00-03Linear models and piecewise models Paper SLEasy5 marks

A mobile phone contract charges a monthly cost, C dollars, for x gigabytes (GB) of data used, given by
$C(x)=\begin{cases}15, & 0\le x\le 10\\ 15+4.5(x-10), & x>10\end{cases}$

(a)

Write down the cost for a month in which 7 GB of data is used.

(1)

(b)

Find the cost for a month in which 16.4 GB of data is used.

(2)

(c)

In one month the bill was $60.45. Find the amount of data used.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $15.00	✓ 1
Part (b)
M1 15 + 4.5(16.4 − 10)	✓ 1
A1 $43.80	✓ 1
Part (c)
M1 15 + 4.5(x − 10) = 60.45, using the x > 10 rule because the bill exceeds $15	✓ 1
A1 x = 20.1 GB	✓ 1

Exam tip: Check which piece of a piecewise function applies before substituting: the domain condition tells you which rule to use.

+ Add to test✓ In your test3IBMI-031-00-04Linear models and piecewise models Paper SLEasy6 marks

A water tank initially contains 4800 litres. A valve is opened and water drains out at a constant rate of 65 litres per minute.
The volume of water remaining in the tank after t minutes is V litres.

(a)

Write down an equation for V in terms of t.

(1)

(b)

Find the volume remaining after 40 minutes.

(2)

(c)

Find the time taken for the tank to empty.

(2)

(d)

Hence state a suitable domain for the model.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 V = 4800 − 65t	✓ 1
Part (b)
M1 Substituting t = 40: 4800 − 65 × 40	✓ 1
A1 2200 litres	✓ 1
Part (c)
M1 Setting 4800 − 65t = 0	✓ 1
A1 t = 73.8 minutes (73.846…, exact value 960/13)	✓ 1
Part (d)
A1 0 ≤ t ≤ 73.8. Accept equivalent endpoints such as 73.846…, 960/13 or 4800/65.	✓ 1

Exam tip: A model's domain should stop where the context stops making sense. A negative volume is impossible, so the domain ends at the moment the tank is empty.

+ Add to test✓ In your test4IBMI-031-07-01Exponential and sinusoidal models Paper SLEasy6 marks

A cup of coffee is left to cool in a room. Its temperature, T °C, t minutes after it is poured, is modelled by
$T(t) = 20 + 70(0.92)^t$, $t \ge 0$.

(a)

Write down the temperature of the coffee when it is poured.

(1)

(b)

Find the temperature of the coffee 10 minutes after it is poured.

(2)

(c)

Find the time taken for the coffee to cool to 40 °C.

(2)

(d)

The graph of T has a horizontal asymptote at T = 20. Interpret this value in context.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: $T(0) = 20 + 70 = 90$ °C	✓ 1
Part (b)
M1: substituting $t = 10$: $20 + 70(0.92)^{10}$	✓ 1
A1: 50.4 °C	✓ 1
Part (c)
M1: setting $20 + 70(0.92)^t = 40$ or using a graph / solver	✓ 1
A1: $t = 15.0$ minutes	✓ 1
Part (d)
R1: 20 °C is the room temperature; the coffee approaches this temperature over time but never cools below it	✓ 1

Exam tip: For 'find the time' questions, a GDC graph intersection or solver is fully acceptable, but write down the equation you are solving to earn the method mark.

+ Add to test✓ In your test5IBMI-031-07-02Exponential and sinusoidal models Paper SLEasy6 marks

The height, h metres, of a passenger capsule on an observation wheel above the ground is modelled by
$h(t) = 26 - 24\cos(12t)$,
where t is the time in minutes after the passenger boards and the angle is measured in degrees.

(a)

Write down the maximum and minimum heights of the capsule.

(2)

(b)

Find the time taken for the wheel to complete one full revolution.

(2)

(c)

Find the height of the capsule 8 minutes after the passenger boards.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: maximum height $26 + 24 = 50$ m	✓ 1
A1: minimum height $26 - 24 = 2$ m	✓ 1
Part (b)
M1: period $= \frac{360}{12}$	✓ 1
A1: 30 minutes	✓ 1
Part (c)
M1: $h(8) = 26 - 24\cos(96°)$	✓ 1
A1: 28.5 m	✓ 1

Common error: Check your GDC is in degree mode when the model uses degrees; in radian mode h(8) gives a completely different (and wrong) height.

+ Add to test✓ In your test6IBMI-031-07-03Exponential and sinusoidal models Paper SLEasy5 marks

Mei invests 5000 US dollars (USD) in a savings account. The value of the investment, V USD, after t years is modelled by
$V = 5000(1.035)^t$.

(a)

Find the number of complete years after which the value of the investment first exceeds 8000 USD.

(3)

(b)

Find the value of the investment after 6 years. Give your answer correct to two decimal places.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $5000(1.035)^t = 8000$ or $(1.035)^t = 1.6$	✓ 1
A1: $t = 13.66\ldots$	✓ 1
A1: 14 complete years	✓ 1
Part (b)
M1: $5000(1.035)^6$	✓ 1
A1: 6146.28 USD	✓ 1

Exam tip: When the question asks for complete years, round the time up to the next whole number, since the value only exceeds the target after that point.

+ Add to test✓ In your test7IBMI-031-07-04Exponential and sinusoidal models Paper SLEasy6 marks

The depth of water in a harbour varies between a minimum of 2.4 m and a maximum of 8.8 m. The time between consecutive high tides is 12.4 hours. The depth, d metres, t hours after midnight is modelled by
$d(t) = a\sin(bt) + c$,
where a, b, c > 0 and the angle is in degrees.

(a)

Find the value of a.

(1)

(b)

Find the value of c.

(1)

(c)

Find the value of b.

(2)

(d)

Find the depth of water at 05:00.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: $a = \frac{8.8 - 2.4}{2} = 3.2$	✓ 1
Part (b)
A1: $c = \frac{8.8 + 2.4}{2} = 5.6$	✓ 1
Part (c)
M1: $b = \frac{360}{12.4}$	✓ 1
A1: $b = 29.0$ (29.032…)	✓ 1
Part (d)
M1: $d(5) = 3.2\sin(29.032\ldots \times 5) + 5.6$	✓ 1
A1: 7.43 m (7.428…)	✓ 1

Common error: Using the rounded value b = 29.0 in part (d) gives 7.44 m instead of 7.43 m. Keep the full GDC value of b in memory when you substitute.

+ Add to test✓ In your test8IBMI-031-03-r1-01Choosing and evaluating models Paper SLEasy5 marks

A taxi company charges a fixed amount plus a constant rate per kilometre travelled. A 4 km journey costs $9.80 and a 10 km journey costs $17.60. The cost, $C$ dollars, of a journey of $k$ km is modelled by $C = mk + c$.

(a)

Find the value of m and the value of c.

(3)

(b)

Interpret the value of c in the context of the question.

(1)

(c)

Use the model to find the cost of a 23 km journey.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to find gradient, e.g. (17.60 − 9.80)/(10 − 4)	✓ 1
A1 m = 1.30	✓ 1
A1 c = 4.60	✓ 1
Part (b)
R1 c = 4.60 is the fixed charge in dollars (the cost before any distance is travelled)	✓ 1
Part (c)
A1 C = 1.30 × 23 + 4.60 = $34.50	✓ 1

Exam tip: When interpreting a parameter, give its value, its units and its meaning in the context. Saying only 'the y-intercept' does not earn the mark.

+ Add to test✓ In your test9IBMI-031-02-r1-01Exponential and sinusoidal models Paper SLEasy5 marks

The mass of caffeine, C mg, remaining in a person's bloodstream t hours after drinking a large coffee is modelled by
$C(t) = 180(0.87)^t$, $t \ge 0$.

(a)

Write down the mass of caffeine in the bloodstream immediately after the coffee is drunk.

(1)

(b)

Find the mass of caffeine remaining after 5 hours.

(2)

(c)

The person can sleep comfortably once the mass of caffeine falls below 20 mg. Find the time, to the nearest tenth of an hour, after which the person can sleep comfortably.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
180 mg	✓ 1
Part (b)
Substitutes t = 5: 180(0.87)⁵ (M1)	✓ 1
= 89.7 mg (A1)	✓ 1
Part (c)
Sets up 180(0.87)^t = 20, e.g. using a graph intersection, a solver or logarithms (M1)	✓ 1
t = 15.8 hours (A1)	✓ 1

Exam tip: The multiplier 0.87 means the mass falls by 13% each hour. Equations such as $180(0.87)^t = 20$ can be solved with the GDC's solver or by intersecting graphs; you do not need logarithms, but you must show the equation you solved.

+ Add to test✓ In your test10IBMI-031-02-r1-02Exponential and sinusoidal models Paper SLEasy6 marks

The depth of water, D metres, at the entrance to a small fishing harbour is modelled by
$D(t) = 1.6\sin\left(\frac{\pi t}{6.2}\right) + 5.0$,
where t is the number of hours after midnight, $0 \le t \le 24$.

(a)

Write down the maximum depth and the minimum depth of water predicted by the model.

(2)

(b)

Find the period of the model, and interpret this value in context.

(2)

(c)

Find the depth of water at 03:00.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Maximum depth = 6.6 m (A1)	✓ 1
Minimum depth = 3.4 m (A1)	✓ 1
Part (b)
Period = $\frac{2\pi}{\pi/6.2}$ = 12.4 hours (A1)	✓ 1
The time between two successive high tides (or low tides) is 12.4 hours (R1)	✓ 1
Part (c)
Substitutes t = 3: 1.6 sin(3π/6.2) + 5.0 (M1)	✓ 1
= 6.60 m (A1)	✓ 1

Common error: This model uses radians. Evaluating $\sin\left(\frac{3\pi}{6.2}\right)$ with the calculator in degree mode gives about 5.04 m instead of 6.60 m – check the mode before substituting.

+ Add to test✓ In your test11IBMI-031-01-r2-01Quadratic and cubic models Paper SLEasy5 marks

A netball is thrown upwards from a player's hand. Its height above the ground, h metres, t seconds after it is released is modelled by
$h(t) = -4.9t^2 + 14.7t + 1.8$.

(a)

Write down the height of the ball at the moment it is released.

(1)

(b)

Find the maximum height reached by the ball.

(2)

(c)

The ball is not caught and lands on the ground. Find the time taken for the ball to reach the ground.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
1.8 m	✓ 1
Part (b)
Uses vertex, e.g. $t = \frac{14.7}{2 \times 4.9} = 1.5$, or the GDC maximum (M1)	✓ 1
Maximum height = 12.825, so 12.8 m (A1)	✓ 1
Part (c)
Sets $-4.9t^2 + 14.7t + 1.8 = 0$ and solves, e.g. with the GDC (M1)	✓ 1
Rejects the negative root; t = 3.12 s (A1)	✓ 1

Exam tip: A quadratic model usually gives two roots. In context only one makes sense here, so state which root you reject and why: time cannot be negative.

+ Add to test✓ In your test12IBMI-031-01-r2-02Quadratic and cubic models Paper SLEasy6 marks

An open-topped tray is made from a rectangular sheet of card measuring 30 cm by 20 cm. A square of side x cm is cut from each corner. The sides are then folded up to form the tray.

(a)

Show that the volume of the tray, V cm³, is given by $V = 4x^3 - 100x^2 + 600x$.

(2)

(b)

State the set of possible values of x in this context.

(1)

(c)

Find the value of x that gives the maximum volume, and state this maximum volume.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Base dimensions $(30-2x)$ and $(20-2x)$, height $x$, so $V = x(30-2x)(20-2x)$ (M1)	✓ 1
Expands correctly: $x(600 - 100x + 4x^2) = 4x^3 - 100x^2 + 600x$ (A1) (AG)	✓ 1
Part (b)
0 < x < 10, because the 20 cm side requires 20 − 2x > 0 (A1)	✓ 1
Part (c)
Uses the GDC maximum on the graph of V, or solves V′(x) = 12x² − 200x + 600 = 0 (M1)	✓ 1
x = 3.92 cm (A1)	✓ 1
V = 1060 cm³ (1056.3…) (A1)	✓ 1

Common error: The cubic has a second turning point at x ≈ 12.7. It lies outside the domain 0 < x < 10, so it does not give a real tray. Always check turning points against the domain.

+ Add to test✓ In your test13IBMI-031-03-r2-01Choosing and evaluating models Paper SLEasy5 marks

A taxi company charges a fixed booking fee plus a constant rate per kilometre. A journey of 4 km costs $9.80 and a journey of 10 km costs $18.20. The cost, $C$ dollars, of a journey of $d$ km is modelled by $C = md + c$.

(a)

Find the value of $m$ and the value of $c$.

(3)

(b)

Interpret the value of $c$ in the context of the question.

(1)

(c)

Use the model to find the cost of a 25 km journey.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at gradient: $m = \frac{18.20-9.80}{10-4}$	✓ 1
A1 $m = 1.40$	✓ 1
A1 $c = 9.80 - 4(1.40) = 4.20$	✓ 1
Part (b)
R1 $c$ is the fixed booking fee of $4.20, charged before any distance is travelled	✓ 1
Part (c)
A1 $C = 1.40(25) + 4.20 = $39.20	✓ 1

Exam tip: Interpretation marks need the context: say what the number means for the taxi fare, not just that it is the y-intercept.

+ Add to test✓ In your test14IBMI-031-05-r1-02Linear models and piecewise models Paper SLEasy6 marks

A decorative candle burns at a constant rate. Its height, h cm, t hours after it was lit, is modelled by $h=mt+c$. After 2 hours the height is 18.4 cm, and after 5 hours the height is 13.9 cm.

(a)

Find the value of m.

(2)

(b)

Find the value of c.

(1)

(c)

Interpret the value of m in context.

(1)

(d)

Find how long after it was lit the candle burns down completely.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$m=\frac{13.9-18.4}{5-2}$ M1 ; $m=-1.5$ A1	✓ 2
Part (b)
$18.4=-1.5(2)+c \Rightarrow c=21.4$ A1	✓ 1
Part (c)
The height of the candle decreases by 1.5 cm every hour R1	✓ 1
Part (d)
Sets $-1.5t+21.4=0$ M1 ; $t=14.3$ hours (14.266…) A1	✓ 2

Common error: A negative gradient must be interpreted as a decrease. Saying 'the height changes by −1.5 cm' without stating units per hour does not fully interpret the value in context.

+ Add to test✓ In your test15IBMI-031-06-r2-01Quadratic and cubic models Paper SLEasy5 marks

At a school science fair, a small catapult on a raised platform launches a ball upwards. The height of the ball above the ground, h metres, t seconds after launch is modelled by
$h(t) = -4.9t^2 + 14.7t + 1.8$, for $t \geq 0$.

(a)

Write down the height of the platform from which the ball is launched.

(1)

(b)

Find the maximum height reached by the ball.

(2)

(c)

Find the time taken for the ball to reach the ground.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
h(0) = 1.8 m A1	✓ 1
Part (b)
Attempt to find the vertex, e.g. $t = \frac{-14.7}{2(-4.9)} = 1.5$ or use of GDC maximum M1	✓ 1
Maximum height = 12.8 m (12.825) A1	✓ 1
Part (c)
Setting h(t) = 0, or use of GDC to find the positive zero M1	✓ 1
t = 3.12 s (3.1178…), rejecting the negative root A1	✓ 1

Exam tip: When a quadratic model gives two roots, only one usually makes sense in context. State clearly which root you reject and why, e.g. time cannot be negative.

+ Add to test✓ In your test16IBMI-031-06-r2-02Quadratic and cubic models Paper SLEasy6 marks

A footbridge arch spans a river. The arch is modelled by $y = ax^2 + bx + c$, where x metres is the horizontal distance from the left base of the arch and y metres is the height of the arch above the base. The arch meets the ground at (0, 0) and (40, 0), and passes through the point (10, 9).

(a)

Write down the value of c.

(1)

(b)

Find the values of a and b.

(3)

(c)

Find the maximum height of the arch.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
c = 0 A1	✓ 1
Part (b)
Substituting both points to form equations, e.g. 1600a + 40b = 0 and 100a + 10b = 9 M1	✓ 1
a = −0.03 A1	✓ 1
b = 1.2 A1	✓ 1
Part (c)
Use of axis of symmetry x = 20, or GDC maximum M1	✓ 1
Maximum height = 12 m (12.0 m) A1	✓ 1

Common error: A common mistake is to substitute the coordinates the wrong way round, e.g. putting 9 in place of x. Write each point as (x, y) before you substitute.

+ Add to test✓ In your test17IBMI-031-05-r2-01Linear models and piecewise models Paper SLEasy5 marks

A taxi company in a city charges fares using the linear model $C = 3.50 + 1.85d$, where $C$ is the fare in dollars and $d$ is the distance travelled in kilometres.

(a)

Find the fare for a journey of 12.4 km.

(2)

(b)

Interpret the value 1.85 in the context of the model.

(1)

(c)

A passenger pays $41.30 for a journey. Find the distance travelled.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: substitutes $d = 12.4$: $C = 3.50 + 1.85 \times 12.4$	✓ 1
A1: $C = \$26.44$	✓ 1
Part (b)
A1: the fare increases by $1.85 for each additional kilometre travelled (cost per km)	✓ 1
Part (c)
M1: sets up $3.50 + 1.85d = 41.30$	✓ 1
A1: $d = 20.4$ km (20.432…)	✓ 1

Exam tip: When interpreting a gradient, always state the units of both variables: 'dollars per kilometre', not just 'the rate'.

+ Add to test✓ In your test18IBMI-031-01-r1-02Quadratic and cubic models Paper SLEasy7 marks

A small bakery sells a speciality loaf at a price of x dollars. Its daily profit from these loaves, P dollars, is modelled by $P(x) = ax^2 + bx + c$.
When the price is \$2 the daily profit is \$60, when the price is \$4 the daily profit is \$140, and when the price is \$6 the daily profit is \$156.

(a)

Find the values of a, b and c.

(3)

(b)

Find the price that maximises the daily profit and state this maximum profit.

(2)

(c)

Find the range of prices for which the bakery makes a profit on these loaves.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: forms three equations, e.g. 4a + 2b + c = 60, 16a + 4b + c = 140, 36a + 6b + c = 156	✓ 1
A1: a = −8, b = 88	✓ 1
A1: c = −84	✓ 1
Part (b)
A1: price = \$5.50	✓ 1
A1: maximum daily profit = \$158.00	✓ 1
Part (c)
A1: solves −8x² + 88x − 84 = 0 to obtain x = 1.06 and x = 9.94	✓ 1
A1: profit made for \$1.06 < x < \$9.94	✓ 1

Exam tip: Money answers should be given to 2 decimal places, so write \$5.50 and \$158.00 rather than 5.5 and 158.

+ Add to test✓ In your test19IBMI-031-02-r2-01Exponential and sinusoidal models Paper SLEasy5 marks

A used car is bought for $24 500. Its value, $V$ dollars, $t$ years after purchase is modelled by $V(t) = 24500 \times 0.86^{t}$.

(a)

Find the value of the car 3 years after purchase. Give your answer to 2 decimal places.

(2)

(b)

Find the value of $t$ at which the value of the car first falls below $8000, and hence state in which year after purchase this happens.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Substitutes t = 3: 24500 × 0.86³ (M1)	✓ 1
$15 583.37 (A1)	✓ 1
Part (b)
Sets up 24500 × 0.86^t = 8000 (or uses a graph/solver with V = 8000) (M1)	✓ 1
t = 7.42 years (A1)	✓ 1
The value first falls below $8000 during the 8th year after purchase (A1)	✓ 1

Exam tip: When a question asks 'in which year', a time of 7.42 means the event happens during the 8th year, not the 7th: year 1 runs from t = 0 to t = 1.

+ Add to test✓ In your test20IBMI-031-02-r2-02Exponential and sinusoidal models Paper SLEasy7 marks

The height, $h$ metres, of a passenger capsule on an observation wheel above the ground is modelled by
$h(t) = 21 - 18\cos(12t)$, $t \ge 0$,
where $t$ is the time in minutes after the passenger boards and the angle is measured in degrees.

(a)

Write down the maximum and minimum heights of the capsule.

(2)

(b)

Find the time taken for the wheel to complete one full rotation.

(1)

(c)

Find the height of the capsule 7 minutes after boarding.

(2)

(d)

Find the first time after boarding at which the capsule is 30 m above the ground.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Maximum height 39 m (A1)	✓ 1
Minimum height 3 m (A1)	✓ 1
Part (b)
360 ÷ 12 = 30 minutes (A1)	✓ 1
Part (c)
Substitutes t = 7: 21 − 18cos(84°) (M1)	✓ 1
h = 19.1 m (A1)	✓ 1
Part (d)
Solves 21 − 18cos(12t) = 30, e.g. cos(12t) = −0.5 or uses graph intersection (M1)	✓ 1
t = 10 minutes (A1)	✓ 1

Common error: Check your calculator is in degree mode. In radian mode cos(84) gives a completely different height, and the answer will not match the physical situation.

+ Add to test✓ In your test21IBMI-031-04-r2-02Transformations of graphs and logarithmic scales Paper SLEasy5 marks

The sound level $L$, in decibels (dB), of a sound with intensity $I$ W m⁻² is given by
$L=10\log_{10}\left(\frac{I}{I_0}\right)$, where $I_0=1.0\times10^{-12}$ W m⁻².
The sound from a busy café has an intensity of $3.2\times10^{-5}$ W m⁻².

(a)

Find the sound level of the café.

(2)

(b)

A leaf blower has a sound level of 92 dB. Find its intensity.

(2)

(c)

Find how many times greater the intensity of the leaf blower is than the intensity of the café.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $10\log_{10}\left(\frac{3.2\times10^{-5}}{1.0\times10^{-12}}\right)$	✓ 1
A1: 75.1 dB (75.05...)	✓ 1
Part (b)
M1: $\frac{I}{I_0}=10^{9.2}$	✓ 1
A1: $I=1.58\times10^{-3}$ W m⁻² (1.5849...×10⁻³)	✓ 1
Part (c)
A1: $\frac{1.585\times10^{-3}}{3.2\times10^{-5}}=49.5$ times. Also accept $10^{(92-75.05)/10}\approx49.5$. Accept 49.0 to 49.6 where it follows from rounded earlier answers.	✓ 1

Common error: A difference of about 17 dB does not mean 17 times the intensity. Every 10 dB multiplies the intensity by 10, so compare the intensities, not the decibel values. Keep full calculator values between parts to avoid rounding drift.

+ Add to test✓ In your test22IBMI-031-06-r1-01Quadratic and cubic models Paper SLEasy6 marks

A netball is thrown upwards from a player's hand. Its height above the ground, h metres, t seconds after release is modelled by
$h(t) = -4.9t^2 + 14.7t + 1.8$, for $t \ge 0$, until the ball reaches the ground.

(a)

Find the maximum height reached by the ball.

(3)

(b)

Find the time at which the ball hits the ground, assuming nothing stops it.

(2)

(c)

Interpret the value 1.8 in the context of the model.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses a vertex method, e.g. $t = -\frac{14.7}{2(-4.9)}$, or the GDC maximum	✓ 1
$t = 1.5$ s	✓ 1
Maximum height = 12.8 m (12.825)	✓ 1
Part (b)
Sets $h(t) = 0$ and solves, e.g. by GDC or the quadratic formula	✓ 1
$t = 3.12$ s, rejecting the negative root	✓ 1
Part (c)
The ball is 1.8 m above the ground at the moment it is released (at $t = 0$)	✓ 1

Exam tip: When solving $h(t)=0$, state clearly that the negative root is rejected because time cannot be negative in this model.

+ Add to test✓ In your test23IBMI-031-05-r1-01Linear models and piecewise models Paper SLEasy5 marks

A taxi company in a city charges a fixed fee of USD 3.50 for every journey plus USD 1.85 per kilometre travelled. The cost, in USD, of a journey of d kilometres is $C(d)$.

(a)

Write down an expression for $C(d)$ in terms of d.

(2)

(b)

Find the cost of a journey of 12.4 km.

(1)

(c)

Amira has USD 40.00 to spend on a single taxi journey. Find the greatest distance she can travel.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$C(d)=1.85d+3.50$: coefficient of d is 1.85 A1 ; constant term is 3.50 A1	✓ 2
Part (b)
$C(12.4)=3.50+1.85\times12.4=26.44$, so the cost is USD 26.44 A1	✓ 1
Part (c)
Sets $1.85d+3.50=40$ (or $\le 40$) M1 ; $d=19.7$ km (19.729…), also accept 19.72 km, but not a value rounded up to 19.73 km A1	✓ 2

Exam tip: In a linear cost model, the gradient is the rate (cost per km) and the vertical intercept is the fixed charge. Identify each before writing the equation. When a budget is a maximum, round the distance down so the cost stays within budget.

+ Add to test✓ In your test24IBMI-031-08-r2-02Choosing and evaluating models Paper SLEasy5 marks

A food truck sells fruit smoothies. The owner models the daily profit, P dollars, when each smoothie is sold at a price of x dollars by
$P(x)=-40x^2+560x-1000$.
Give all monetary answers correct to the nearest cent.

(a)

Find the price that maximises the daily profit and the corresponding maximum profit.

(2)

(b)

Find the two prices at which the food truck makes zero profit.

(2)

(c)

Suggest one reason why this model may not be realistic for prices well above 12 dollars.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: price = 7.00 dollars (from $x=\frac{-560}{2(-40)}$ or using GDC)	✓ 1
A1: maximum profit = 960.00 dollars	✓ 1
Part (b)
M1: attempt to solve $-40x^2+560x-1000=0$ (e.g. GDC or quadratic formula)	✓ 1
A1: $x=2.10$ dollars and $x=11.90$ dollars	✓ 1
Part (c)
R1: e.g. the model predicts an ever-increasing loss, whereas in reality almost no smoothies would be sold, so the loss would be limited to roughly the fixed costs (or the owner would stop selling); OR such prices are likely outside the range of data used to build the model	✓ 1

Exam tip: When a question asks for answers to the nearest cent, give exactly 2 decimal places, e.g. 2.10 rather than 2.1, and do not round until the final step.

+ Add to test✓ In your test25IBMI-031-03-r2-02Choosing and evaluating models Paper SLEasy6 marks

The mass of caffeine, $A$ mg, remaining in a person's bloodstream $t$ hours after drinking a large coffee is modelled by $A(t) = 200(0.87)^t$, $t \ge 0$.

(a)

Write down the initial mass of caffeine in the bloodstream.

(1)

(b)

Write down the percentage of caffeine removed from the bloodstream each hour.

(1)

(c)

Find the mass of caffeine remaining after 5 hours.

(2)

(d)

The person wants to sleep once the mass of caffeine has fallen to 25 mg. Find the time after drinking the coffee at which the mass of caffeine first equals 25 mg.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 200 mg	✓ 1
Part (b)
A1 13%	✓ 1
Part (c)
M1 substituting $t = 5$: $200(0.87)^5$	✓ 1
A1 99.7 mg	✓ 1
Part (d)
M1 setting $200(0.87)^t = 25$ or equivalent, e.g. graph intersection, a table of values or logarithms	✓ 1
A1 $t = 14.9$ hours (accept 14.93; 15 alone does not earn this mark)	✓ 1

Common error: A base of 0.87 means 13% is lost each hour, not 87%. The base is the fraction that remains.

+ Add to test✓ In your test26IBMI-031-08-r2-01Choosing and evaluating models Paper SLEasy6 marks

A candle is lit and its height is recorded. After 10 minutes the candle is 23.5 cm tall and after 40 minutes it is 17.5 cm tall. The height, $h$ cm, of the candle $t$ minutes after it was lit is modelled by a linear function $h(t)=mt+c$.

(a)

Find the value of $m$ and the value of $c$.

(3)

(b)

Interpret the value of $m$ in the context of the question.

(1)

(c)

Use the model to estimate how long after being lit the candle will have burnt down completely.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: attempt at gradient using both points, $m=\frac{17.5-23.5}{40-10}$	✓ 1
A1: $m=-0.2$	✓ 1
A1: $c=25.5$ (from $23.5=-0.2(10)+c$)	✓ 1
Part (b)
R1: the candle's height decreases by 0.2 cm every minute (it burns down at 0.2 cm per minute)	✓ 1
Part (c)
M1: setting $25.5-0.2t=0$	✓ 1
A1: $t=127.5$ minutes (accept 128 minutes)	✓ 1

Exam tip: When interpreting a gradient, always give its units and say what changes per unit of the input — a bare number like 'the gradient is −0.2' earns no interpretation mark.

+ Add to test✓ In your test27IBMI-031-03-r1-02Choosing and evaluating models Paper SLEasy6 marks

The value, $V$ dollars, of a car $t$ years after it is bought new is modelled by
$V(t) = 24000(0.85)^t$, $t \ge 0$.

(a)

State the annual percentage decrease in the value of the car.

(1)

(b)

Find the value of the car after 3 years.

(2)

(c)

Find the time taken for the value of the car to fall to 8000 dollars.

(2)

(d)

With reference to the model, explain why it may not be suitable for predicting the value of the car many years after purchase.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 15%	✓ 1
Part (b)
M1 substituting t = 3: 24000(0.85)³	✓ 1
A1 14 739 dollars	✓ 1
Part (c)
M1 setting up 24000(0.85)^t = 8000, solved by GDC or logarithms	✓ 1
A1 t = 6.76 years	✓ 1
Part (d)
R1 as t increases, V approaches 0 (horizontal asymptote V = 0), giving ever-smaller values, whereas a real car keeps a minimum value, e.g. its scrap value; or the model ignores factors such as condition, mileage or demand (e.g. classic cars may rise in value)	✓ 1

Common error: Reading 0.85 as an 85% decrease. The multiplier 0.85 means 85% of the value is kept each year, so the decrease is 15%.

+ Add to test✓ In your test28IBMI-031-04-r1-01Transformations of graphs and logarithmic scales Paper SLEasy6 marks

The sound level L, in decibels (dB), of a sound of intensity I W m⁻² is given by $L = 10\log_{10}\left(\frac{I}{10^{-12}}\right)$.
A leaf blower produces sound of intensity 3.2×10⁻⁵ W m⁻² at the position of its operator.

(a)

Find the sound level experienced by the operator.

(2)

(b)

A motorbike engine produces a sound level of 92 dB. Find the intensity of this sound.

(2)

(c)

The sound level at a concert increases by 15 dB. Find the factor by which the sound intensity increases.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: substitutes correctly, $L = 10\log_{10}(3.2\times10^{-5}/10^{-12}) = 10\log_{10}(3.2\times10^{7})$	✓ 1
A1: L = 75.1 dB (accept 75.05)	✓ 1
Part (b)
M1: rearranges to $I = 10^{-12}\times10^{9.2}$	✓ 1
A1: I = 1.58×10⁻³ W m⁻² (accept 1.585×10⁻³)	✓ 1
Part (c)
M1: recognises that an increase of 15 dB multiplies the intensity by $10^{15/10}$	✓ 1
A1: factor = 31.6	✓ 1

Common error: An increase of 15 dB does not mean the intensity increases 15 times or 1.5 times. Each 10 dB step multiplies the intensity by 10, so 15 dB multiplies it by 10¹.⁵.

+ Add to test✓ In your test29IBMI-031-08-r1-02Choosing and evaluating models Paper SLEasy6 marks

The value $V$, in dollars, of a new electric car $t$ years after purchase is modelled by
$V(t) = 32000 \times 0.85^{t}$, $t \geq 0$.

(a)

Find the value of the car after 4 years. Give your answer in dollars to two decimal places.

(2)

(b)

Interpret the number 0.85 in this context.

(1)

(c)

Find the time, in years, at which the model predicts the value of the car falls to 5000 dollars. Give your answer to three significant figures.

(2)

(d)

State one reason why this model may not be appropriate for predicting the car's value after 30 years.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Substitutes $t = 4$: $32000 \times 0.85^4$ M1	✓ 1
16704.20 dollars A1	✓ 1
Part (b)
Each year the car keeps 85% of its value from the previous year, so it loses 15% of its value per year. R1	✓ 1
Part (c)
Sets up $32000 \times 0.85^{t} = 5000$ and solves by GDC or logarithms M1	✓ 1
$t = 11.4$ years A1	✓ 1
Part (d)
Any valid reason, e.g. a very old car may be scrapped, so its value is only its scrap value; or a rare model may gain value as a classic; or the rate of depreciation is unlikely to stay at 15% per year for 30 years. R1	✓ 1

Common error: Reading 0.85 as 'loses 85% per year' confuses the fraction of value kept with the percentage lost.

+ Add to test✓ In your test30IBMI-031-06-r1-02Quadratic and cubic models Paper SLEasy6 marks

An open-topped tray is made from a rectangular sheet of card measuring 30 cm by 20 cm. A square of side x cm is cut from each corner and the sides are folded up.

(a)

Show that the volume of the tray, V cm³, is given by $V = 4x^3 - 100x^2 + 600x$.

(2)

(b)

State the set of possible values of x for which the model is valid.

(1)

(c)

Find the value of x that maximises the volume, and the maximum volume.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Base dimensions $(30-2x)$ and $(20-2x)$ with height $x$, so $V = x(30-2x)(20-2x)$	✓ 1
Expands correctly: $x(600 - 100x + 4x^2) = 4x^3 - 100x^2 + 600x$	✓ 1
Part (b)
$0 < x < 10$	✓ 1
Part (c)
Valid method: either a GDC to locate the local maximum of V in \(0	✓ 1
$x = 3.92$ cm	✓ 1
$V_{max} = 1060$ cm³ (3 s.f.; 1056.3… also accepted)	✓ 1

Common error: Do not give the other turning point, x ≈ 12.7, as an answer. It lies outside 0 < x < 10 because the 20 cm side would then have negative length. Always check a turning point against the domain.

+ Add to test✓ In your test31IBMI-031-01-r1-01Quadratic and cubic models Paper SLEasy5 marks

A footballer kicks a ball from a raised platform. The height of the ball above the ground, h metres, t seconds after it is kicked is modelled by
$h(t) = -4.9t^2 + 12t + 0.8$, for $t \ge 0$.

(a)

Find the maximum height reached by the ball.

(2)

(b)

Find the time taken for the ball to reach the ground. Justify your choice of solution.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: uses the GDC maximum, or finds the vertex $t = \frac{-12}{2(-4.9)} = 1.22$ s and substitutes it into h	✓ 1
A1: maximum height = 8.15 m (8.14694…)	✓ 1
Part (b)
M1: sets $-4.9t^2 + 12t + 0.8 = 0$, using the graph meeting the t-axis or the quadratic formula	✓ 1
A1: t = 2.51 s (2.51393…)	✓ 1
R1: rejects t = −0.0649 (−0.06494…) because time cannot be negative in this context	✓ 1

Exam tip: When a quadratic gives two roots, state which one you are rejecting and explain why in terms of the context. The reasoning mark is only earned if the reason refers to the situation, which here is that time cannot be negative.

+ Add to test✓ In your test32IBMI-031-00-05Linear models and piecewise models Paper SLMedium12 marks

In a certain country, annual income tax is calculated as follows:
• no tax is paid on the first $12 000 of income;
• 20% tax is paid on income between $12 000 and $45 000;
• 35% tax is paid on income above $45 000.
Let T(x) be the tax, in dollars, paid on an annual income of x dollars.

(a)

Find the tax paid on an income of $30 000.

(2)

(b)

Show that the tax paid on an income of $45 000 is $6600.

(1)

(c)

Write down a piecewise expression for T(x), stating the domain of each piece.

(4)

(d)

Amara paid $11 230 in tax. Find her annual income.

(3)

(e)

The effective tax rate is the total tax paid as a percentage of total income. Find the effective tax rate for an income of $80 000.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.20 × (30000 − 12000)	✓ 1
A1 $3600.00	✓ 1
Part (b)
M1 0.20 × (45000 − 12000) = 0.20 × 33000 = 6600	✓ 1
Part (c)
A1 T(x) = 0 for first piece	✓ 1
A1 T(x) = 0.2(x − 12000) for second piece	✓ 1
A1 T(x) = 6600 + 0.35(x − 45000) for third piece	✓ 1
A1 correct domains 0 ≤ x ≤ 12000, 12000 < x ≤ 45000, x > 45000	✓ 1
Part (d)
R1 11230 > 6600 so third piece applies	✓ 1
M1 6600 + 0.35(x − 45000) = 11230	✓ 1
A1 x = $58 228.57	✓ 1
Part (e)
M1 T(80000) = 6600 + 0.35 × 35000 = 18850 and dividing by 80000	✓ 1
A1 23.6%	✓ 1

Common error: Applying the top rate to the whole income overstates the tax; the higher rate only applies to the portion of income above each threshold.

+ Add to test✓ In your test33IBMI-031-00-06Linear models and piecewise models Paper SLMedium7 marks

A small business makes and sells handmade candles. It has fixed monthly costs of $2400 and each candle costs $3.75 to produce. Each candle is sold for $9.50.
Let n be the number of candles made and sold in a month.

(a)

Find the least number of candles that must be sold in a month for the business to make a profit.

(3)

(b)

Find the profit made in a month when 600 candles are sold.

(2)

(c)

The business wants to break even at exactly 300 candles per month, with costs unchanged. Find the selling price per candle needed.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 equating cost and revenue: 2400 + 3.75n = 9.50n	✓ 1
A1 n = 417.39…	✓ 1
A1 418 candles	✓ 1
Part (b)
M1 9.50 × 600 − (2400 + 3.75 × 600)	✓ 1
A1 $1050.00	✓ 1
Part (c)
M1 300p = 2400 + 3.75 × 300	✓ 1
A1 p = $11.75	✓ 1

Exam tip: When a count must make a profit, round the break-even value up to the next whole number, even if the decimal part is small.

+ Add to test✓ In your test34IBMI-031-00-07Linear models and piecewise models Paper SLMedium11 marks

A courier charges C dollars to deliver a parcel of mass m kg, where
\(C(m)=\begin{cases}6+2.5m, & 0The function C is continuous at m = 4.

(a)

Find the value of a.

(2)

(b)

Find the cost of delivering a 13 kg parcel.

(2)

(c)

A parcel costs $24.10 to deliver. Find its mass.

(2)

(d)

Find the range of C.

(2)

(e)

Jonas has two parcels, each of mass 3 kg. Find how much he saves by packing them together and sending them as one 6 kg parcel rather than sending them separately.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6 + 2.5(4) = a + 1.8(4)	✓ 1
A1 a = 8.8	✓ 1
Part (b)
M1 8.8 + 1.8 × 13	✓ 1
A1 $32.20	✓ 1
Part (c)
M1 8.8 + 1.8m = 24.10 (second piece since 24.10 > 16)	✓ 1
A1 m = 8.5 kg	✓ 1
Part (d)
A1 lower bound 6 < C (6 not included)	✓ 1
A1 C ≤ 44.80	✓ 1
Part (e)
M1 separately: 2 × (6 + 2.5 × 3) = $27.00	✓ 1
M1 combined: 8.8 + 1.8 × 6 = $19.60	✓ 1
A1 saving $7.40	✓ 1

Exam tip: 'Continuous at m = 4' means both pieces give the same value there; set the two expressions equal at that point to find the unknown.

+ Add to test✓ In your test35IBMI-031-00-08Linear models and piecewise models Paper SLMedium7 marks

Two candles are lit at the same time. Candle A is initially 24 cm tall and its height decreases at a constant 1.6 cm per hour. Candle B is initially 18 cm tall and its height decreases at a constant 0.9 cm per hour.
Let t be the time in hours after the candles are lit.

(a)

Write down expressions for the heights h_A and h_B, in cm, in terms of t.

(2)

(b)

Find the time at which the candles have the same height.

(2)

(c)

Find this common height.

(1)

(d)

Find how much longer candle B burns than candle A.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 h_A = 24 − 1.6t	✓ 1
A1 h_B = 18 − 0.9t	✓ 1
Part (b)
M1 24 − 1.6t = 18 − 0.9t	✓ 1
A1 t = 8.57 hours (8.571…)	✓ 1
Part (c)
A1 10.3 cm (10.285…)	✓ 1
Part (d)
M1 burn-out times 24 ÷ 1.6 = 15 h and 18 ÷ 0.9 = 20 h	✓ 1
A1 5 hours	✓ 1

Common error: Using a rounded time such as 8.6 h to find the common height introduces error; use the full calculator value.

+ Add to test✓ In your test36IBMI-031-00-10Linear models and piecewise models Paper SLMedium5 marks

The speed of sound in air, v m s⁻¹, at temperature T °C is modelled by $v = 331 + 0.6T$.

(a)

Find the speed of sound when the temperature is 25 °C.

(1)

(b)

Find the inverse function, giving T in terms of v, and state what it allows you to calculate.

(3)

(c)

A sound is measured to travel at 340 m s⁻¹. Find the air temperature.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 346 m s⁻¹	✓ 1
Part (b)
M1 attempt to rearrange for T	✓ 1
A1 T = (v − 331)/0.6	✓ 1
R1 it gives the air temperature from a measured speed of sound	✓ 1
Part (c)
A1 15.0 °C	✓ 1

Exam tip: In context, an inverse function swaps the roles of input and output; say in words what the new input and output represent.

+ Add to test✓ In your test37IBMI-031-07-07Exponential and sinusoidal models Paper SLMedium10 marks

In a northern city, the number of hours of daylight varies through the year. The maximum of 16.4 hours occurs on day 172 and the minimum of 8.0 hours occurs half a year later. The number of hours of daylight, D, on day t is modelled by
$D(t) = a\cos\big(b(t - 172)\big) + d$,
where a, b, d > 0, a year has 365 days and the angle is in degrees.

(a)

Find the values of a and d.

(2)

(b)

Find the value of b.

(1)

(c)

Find the number of hours of daylight on day 60.

(2)

(d)

Find the number of days in the year on which there are more than 14 hours of daylight.

(4)

(e)

Suggest one reason why the model might not give exactly the right number of daylight hours on a particular day.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: $a = \frac{16.4 - 8.0}{2} = 4.2$	✓ 1
A1: $d = \frac{16.4 + 8.0}{2} = 12.2$	✓ 1
Part (b)
A1: $b = \frac{360}{365} = 0.986$ (0.98630…)	✓ 1
Part (c)
M1: $4.2\cos\big(0.98630\ldots(60 - 172)\big) + 12.2$	✓ 1
A1: 10.7 hours	✓ 1
Part (d)
M1: setting $D(t) = 14$, i.e. $\cos\big(b(t-172)\big) = \frac{1.8}{4.2}$, or sketching $D$ with the line $y = 14$	✓ 1
A1: $t = 106.5$ (106.48…)	✓ 1
A1: $t = 237.5$ (237.52…)	✓ 1
A1: $237.5 - 106.5 \approx 131$ days (accept 131.0; counting days 107 to 237 inclusive also gives 131)	✓ 1
Part (e)
R1: a valid reason, e.g. the real variation is not exactly sinusoidal, the year is not exactly 365 days (leap years), or terrain/atmospheric effects	✓ 1

Common error: In part (d), a frequent error is giving one of the intersection days instead of the length of the interval between them. Read what the question is asking for.

+ Add to test✓ In your test38IBMI-031-07-08Exponential and sinusoidal models Paper SLMedium7 marks

Ravi buys a car for 28 500 euros (EUR). Its value decreases by 14% each year.

(a)

Write down a model for the value, V EUR, of Ravi's car t years after he buys it.

(1)

(b)

Find the value of Ravi's car after 4 years. Give your answer to two decimal places.

(2)

(c)

At the same time, Sana buys a car for 32 000 EUR. The value of her car t years later is modelled by $W = 32000(0.81)^t$. Find the time at which the two cars have the same value.

(3)

(d)

Describe how the values of the two cars compare after this time.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: $V = 28500(0.86)^t$	✓ 1
Part (b)
M1: $28500(0.86)^4$	✓ 1
A1: 15 589.73 EUR	✓ 1
Part (c)
M1: $28500(0.86)^t = 32000(0.81)^t$	✓ 1
M1: using GDC intersection, or $\left(\frac{0.86}{0.81}\right)^t = \frac{32000}{28500}$	✓ 1
A1: $t = 1.93$ years	✓ 1
Part (d)
R1: after 1.93 years Sana's car is worth less than Ravi's car, because her car loses value at a faster rate (19% compared with 14% per year)	✓ 1

Common error: A 14% decrease gives a multiplier of 0.86, not 0.14. Using 0.14 would wipe out 86% of the value each year.

+ Add to test✓ In your test39IBMI-031-07-10Exponential and sinusoidal models Paper SLMedium10 marks

On a summer day in a desert town, the maximum temperature of 31 °C occurs at 15:00 and the minimum temperature of 19 °C occurs at 03:00. The temperature, T °C, t hours after midnight is modelled by
$T(t) = a\sin\big(b(t - c)\big) + d$,
where a, b, d > 0, 0 ≤ c < 24 and the angle is in degrees. Assume the pattern repeats every 24 hours.

(a)

Find the values of a and d.

(2)

(b)

Find the value of b.

(2)

(c)

Find the value of c.

(2)

(d)

Find the first time after midnight at which the temperature reaches 28 °C.

(2)

(e)

Find the number of hours during the day for which the temperature is above 28 °C.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: $a = 6$	✓ 1
A1: $d = 25$	✓ 1
Part (b)
M1: period = 24 hours, so $b = \frac{360}{24}$	✓ 1
A1: $b = 15$	✓ 1
Part (c)
M1: the temperature is 25 °C and rising halfway between 03:00 and 15:00	✓ 1
A1: $c = 9$	✓ 1
Part (d)
M1: $6\sin\big(15(t - 9)\big) + 25 = 28$ or graph intersection	✓ 1
A1: $t = 11$, i.e. 11:00	✓ 1
Part (e)
M1: second solution $t = 19$, or use of symmetry about $t = 15$	✓ 1
A1: $19 - 11 = 8$ hours	✓ 1

Exam tip: For a sine model, the horizontal shift c is the time when the curve crosses the mean value going upwards, which is halfway between a minimum and the next maximum.

+ Add to test✓ In your test40IBMI-031-03-r1-03Choosing and evaluating models Paper SLMedium7 marks

A footbridge over a canal has a parabolic arch. The height of the arch above the water, $h$ metres, at a horizontal distance $x$ metres from one end is modelled by $h(x) = ax^2 + bx + c$.
The arch meets the water at $x = 0$ and at $x = 40$. At $x = 10$ the arch is 7.5 m above the water.

(a)

Find the values of a, b and c.

(3)

(b)

Find the maximum height of the arch above the water.

(2)

(c)

A barge carrying a load that is 8.4 m high must pass under the arch. Find the width of the part of the canal in which the arch is at least 8.4 m above the water.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 forming equations from the three points, e.g. h = ax(x − 40) with 7.5 = a(10)(−30), or solving a system on GDC	✓ 1
A1 a = −0.025	✓ 1
A1 b = 1, c = 0	✓ 1
Part (b)
M1 using the axis of symmetry x = 20 or GDC maximum	✓ 1
A1 10 m	✓ 1
Part (c)
M1 solving −0.025x² + x = 8.4 to give x = 12 and x = 28	✓ 1
A1 width = 16 m	✓ 1

Exam tip: If you know where a quadratic crosses the axis, starting from the factorised form h = ax(x − 40) leaves only one unknown to find.

+ Add to test✓ In your test41IBMI-031-02-r1-03Exponential and sinusoidal models Paper SLMedium12 marks

On a particular summer day in a coastal city, the air temperature reached a maximum of 27 °C at 15:00 and a minimum of 13 °C at 03:00.
The temperature, T °C, is modelled by
$T(t) = a\cos\big(b(t - c)\big) + d$,
where t is the number of hours after midnight and a, b, c, d are positive constants, with $0 < c < 24$. Assume the temperature pattern repeats every 24 hours.

(a)

Find the value of a and the value of d.

(2)

(b)

Find the value of b.

(2)

(c)

Write down the value of c.

(1)

(d)

Use the model to estimate the temperature at 10:00.

(2)

(e)

Outdoor workers must take extra breaks whenever the temperature is above 24 °C. Find the total length of time during this day for which the temperature is above 24 °C.

(4)

(f)

State one reason why this model may not give reliable predictions of the temperature one week later.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
a = (27 − 13)/2 = 7 (A1)	✓ 1
d = (27 + 13)/2 = 20 (A1)	✓ 1
Part (b)
Uses period = 24, so b = 2π/24 (M1)	✓ 1
b = π/12 (= 0.262) (A1)	✓ 1
Part (c)
c = 15 (since the cosine curve has its maximum when t − c = 0) (A1)	✓ 1
Part (d)
Substitutes t = 10: 7cos(π(10 − 15)/12) + 20 (M1)	✓ 1
= 21.8 °C (A1)	✓ 1
Part (e)
Sets up 7cos(π(t − 15)/12) + 20 = 24, e.g. graph intersection (M1)	✓ 1
t = 11.3 (A1)	✓ 1
t = 18.7 (A1)	✓ 1
Total time = 18.677 − 11.323 = 7.35 hours (A1)	✓ 1
Part (f)
Valid reason in context, e.g. the model assumes every day has the same maximum and minimum, but weather conditions (cloud, wind, rain) change from day to day (R1)	✓ 1

Exam tip: For a cosine model, the horizontal shift c is the time of a maximum; for a sine model it would be the time the curve crosses its principal axis going up. Choose whichever matches the information given to avoid extra working.

+ Add to test✓ In your test42IBMI-031-02-r1-04Exponential and sinusoidal models Paper SLMedium7 marks

A new electric car is bought for $28 500. Three years later its value is $17 200.
The value of the car, V dollars, t years after it was bought is modelled by $V(t) = k\,r^t$, where k and r are constants.

(a)

Find the value of k and the value of r.

(3)

(b)

Find the predicted value of the car 7 years after it was bought. Give your answer to the nearest dollar.

(2)

(c)

The owner plans to sell the car when its value first falls below $5000. Find the time after purchase at which this occurs.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
k = 28 500 (A1)	✓ 1
Sets up 28 500 r³ = 17 200 (M1)	✓ 1
r = 0.845 (0.845073…) (A1)	✓ 1
Part (b)
Substitutes t = 7 using unrounded r: 28 500(0.845073…)⁷ (M1)	✓ 1
$8772 (A1)	✓ 1
Part (c)
Sets up 28 500(0.845073…)^t = 5000 (M1)	✓ 1
t = 10.3 years (A1)	✓ 1

Common error: Using the rounded value r = 0.845 in later parts gives $8763 in part (b) instead of $8772. Store the unrounded value of r in your calculator and use it throughout.

+ Add to test✓ In your test43IBMI-031-01-r2-04Quadratic and cubic models Paper SLMedium7 marks

A museum opens a temporary exhibition. The weekly number of visitors, N thousand, t weeks after opening is modelled by
$N(t) = -0.02t^3 + 0.6t^2 + 1.5t + 4$, for 0 ≤ t ≤ 25.

(a)

Write down the number of visitors in the week the exhibition opens.

(1)

(b)

Find the value of t when the weekly number of visitors first reaches 50 000.

(2)

(c)

Find the maximum weekly number of visitors predicted by the model and the value of t at which it occurs.

(3)

(d)

Explain why this model would not be suitable for predicting visitor numbers long after week 25.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
4000 visitors (A1)	✓ 1
Part (b)
Solves N(t) = 50, e.g. finds the intersection with y = 50 on the GDC (M1)	✓ 1
t = 8.80 weeks (A1)	✓ 1
Part (c)
Uses the GDC maximum, or solves −0.06t² + 1.2t + 1.5 = 0 (M1)	✓ 1
t = 21.2 weeks (A1)	✓ 1
Maximum ≈ 115 thousand, i.e. about 115 000 visitors (A1)	✓ 1
Part (d)
The leading term −0.02t³ dominates, so N decreases without limit and eventually becomes negative, which is impossible for visitor numbers (R1)	✓ 1

Common error: N is measured in thousands. Give 115 as '115 thousand' or '115 000 visitors', not '115 visitors'. Credit for interpretation depends on stating the units in context.

+ Add to test✓ In your test44IBMI-031-05-r1-03Linear models and piecewise models Paper SLMedium7 marks

An electricity supplier uses the following monthly tariff.
• A standing charge of $12.00 per month.
• $0.18 per kWh for the first 200 kWh used.
• $0.25 per kWh for every kWh used above 200 kWh.
The total monthly bill, in dollars, for x kWh of electricity is $C(x)$.

(a)

Find the bill for a month in which 150 kWh is used.

(1)

(b)

Write down a piecewise function for $C(x)$, stating the domain of each piece.

(3)

(c)

One household receives a bill of $103.50 for a month. Find the amount of electricity it used.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$12+0.18\times150=$ $39.00 A1	✓ 1
Part (b)
$C(x)=12+0.18x$ for $0\le x\le200$ A1 ; recognises bill at 200 kWh is $48 and adds $0.25(x-200)$ M1 ; $C(x)=0.25x-2$ (or $48+0.25(x-200)$) for $x>200$ A1	✓ 3
Part (c)
Since $103.50>48$, usage exceeds 200 kWh so the second piece applies R1 ; $0.25x-2=103.5$ M1 ; $x=422$ kWh A1	✓ 3

Exam tip: Before solving with a piecewise model, check which piece applies by evaluating the function at the boundary, here $C(200)=48$.

+ Add to test✓ In your test45IBMI-031-05-r1-06Linear models and piecewise models Paper SLMedium9 marks

A cyclist rides along a straight route. Her distance from the start, d km, t minutes after setting off is modelled by
\(d(t)=\begin{cases}0.35t, & 0\le t\le 20\\ 7, & 20The route is 15.4 km long and ends at time T.

(a)

Find the distance travelled after 12 minutes.

(1)

(b)

Interpret the second piece of the function in context.

(2)

(c)

Find the value of T.

(2)

(d)

Find the average speed for the whole journey, in km h⁻¹.

(2)

(e)

Find the time at which the cyclist is 11 km from the start.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$d(12)=0.35\times12=4.2$ km A1	✓ 1
Part (b)
Distance stays constant at 7 km R1 ; so the cyclist stops (rests) for 10 minutes, from $t=20$ to $t=30$ R1	✓ 2
Part (c)
$7+0.28(T-30)=15.4$ M1 ; $T=60$ A1	✓ 2
Part (d)
$\frac{15.4}{60}$ km min⁻¹ or 15.4 km in 1 hour M1 ; 15.4 km h⁻¹ A1	✓ 2
Part (e)
Uses third piece: $7+0.28(t-30)=11$ M1 ; $t=44.3$ minutes (44.285…) A1	✓ 2

Exam tip: Decide which piece applies by comparing the target value with the values at the boundaries: 11 km is more than 7 km, so only the third piece can give it.

+ Add to test✓ In your test46IBMI-031-06-r2-03Quadratic and cubic models Paper SLMedium7 marks

An open-topped tray is made from a rectangular sheet of card measuring 30 cm by 20 cm. A square of side x cm is cut from each corner and the sides are folded up. The volume of the tray is V cm³.

(a)

Show that $V = 4x^3 - 100x^2 + 600x$.

(2)

(b)

State the possible values of x in this context.

(1)

(c)

The designer wants the tray to have a volume of 800 cm³. Find the possible values of x.

(2)

(d)

Find the maximum possible volume of the tray.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Base dimensions (30 − 2x) and (20 − 2x), so V = x(30 − 2x)(20 − 2x) M1	✓ 1
Correct expansion, e.g. x(600 − 100x + 4x²), leading to the given answer A1	✓ 1
Part (b)
0 < x < 10 A1	✓ 1
Part (c)
Solving 4x³ − 100x² + 600x = 800 with GDC (graph or solver) M1	✓ 1
x = 1.88 cm or x = 6.36 cm (16.8 rejected, as it is outside 0 < x < 10) A1	✓ 1
Part (d)
Use of GDC maximum, or V′ = 12x² − 200x + 600 = 0 giving x = 3.92 M1	✓ 1
Maximum volume = 1060 cm³ (1056.3…) A1	✓ 1

Exam tip: A cubic model can give roots that lie outside the valid domain. Check every GDC solution against the restriction you stated earlier in the question.

+ Add to test✓ In your test47IBMI-031-06-r2-04Quadratic and cubic models Paper SLMedium13 marks

The temperature inside a greenhouse, T °C, t hours after 06:00 is modelled by
$T(t) = at^3 + bt^2 + ct + d$, for $0 \leq t \leq 16$.
The following readings were taken: T(0) = 12, T(2) = 12.8, T(6) = 26.4 and T(10) = 40.

(a)

Write down the value of d, and write down three equations in a, b and c.

(3)

(b)

Find the values of a, b and c.

(2)

(c)

Find the maximum temperature in the greenhouse and the time of day at which it occurs.

(3)

(d)

Find T(20) and comment on the use of the model at this time.

(2)

(e)

Find the length of time for which the temperature is above 30 °C.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
d = 12 A1	✓ 1
Substituting at least one data point into the cubic M1	✓ 1
8a + 4b + 2c = 0.8, 216a + 36b + 6c = 14.4, 1000a + 100b + 10c = 28 (or equivalent forms) A1	✓ 1
Part (b)
a = −0.05 and b = 0.9 A1	✓ 1
c = −1.2 A1	✓ 1
Part (c)
Use of GDC maximum, or T′(t) = −0.15t² + 1.8t − 1.2 = 0 M1	✓ 1
Maximum temperature = 41.2 °C A1	✓ 1
t = 11.3 hours, so at about 17:17 (accept 17:18) A1	✓ 1
Part (d)
T(20) = −52 °C A1	✓ 1
This is unrealistic, since t = 20 lies outside the domain of the model, so the model should not be used to extrapolate R1	✓ 1
Part (e)
Setting T(t) = 30, or using the intersection with y = 30 on a GDC M1	✓ 1
t = 6.86 and t = 14.7 A1	✓ 1
Length of time = 7.84 hours A1	✓ 1

Exam tip: For time-of-day answers, convert the decimal part of the hours into minutes (0.29 h ≈ 17 min) and add the result to the starting time.

+ Add to test✓ In your test48IBMI-031-05-r2-04Linear models and piecewise models Paper SLMedium8 marks

A small bakery sells celebration cakes. Its monthly costs are $2400 in fixed costs plus $6.50 for the ingredients of each cake. Each cake sells for $14. Let $n$ be the number of cakes made and sold in a month.

(a)

Write down an expression for the monthly profit, $P(n)$, in dollars.

(2)

(b)

Find the number of cakes that must be sold for the bakery to break even.

(2)

(c)

Find the minimum number of cakes that must be sold for a monthly profit of at least $1750.

(2)

(d)

The bakery changes the selling price to $p$ dollars per cake so that it breaks even at 250 cakes. Assuming costs are unchanged, find $p$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: revenue minus cost: $14n - (2400 + 6.5n)$	✓ 1
A1: $P(n) = 7.5n - 2400$	✓ 1
Part (b)
M1: $7.5n - 2400 = 0$	✓ 1
A1: $n = 320$ cakes	✓ 1
Part (c)
M1: $7.5n - 2400 \ge 1750$, giving $n \ge 553.3\ldots$	✓ 1
A1: 554 cakes (must be a whole number, rounded up)	✓ 1
Part (d)
M1: $250p = 2400 + 6.5 \times 250$	✓ 1
A1: $p = \$16.10$	✓ 1

Common error: Rounding 553.3 down to 553 gives a profit just below the target; in context the number of cakes must be rounded up.

+ Add to test✓ In your test49IBMI-031-05-r2-05Linear models and piecewise models Paper SLMedium12 marks

Amira travels from home to work. Her distance from home, $d$ km, $t$ minutes after leaving is modelled by
$d(t) = \begin{cases} 0.08t, & 0 \le t \le 10 \\ 0.8, & 10 < t \le 16 \\ 0.8 + 0.5(t - 16), & 16 < t \le 36 \end{cases}$
She walks to a bus stop, waits, and then travels by bus along a straight road.

(a)

(i) Interpret the value 0.08 in context and convert it to km h⁻¹.
(ii) Explain why $d$ is constant for $10 < t \le 16$.

(3)

(b)

Find Amira's distance from home 30 minutes after leaving.

(2)

(c)

Find the time at which Amira is 5 km from home.

(2)

(d)

Her brother leaves the same house 8 minutes after Amira and cycles along the same route at a constant 0.3 km per minute, so his distance from home is $f(t) = 0.3(t - 8)$. Find the time at which he passes the bus stop, and state what Amira is doing at that moment.

(2)

(e)

Find the time at which Amira's bus catches up with her brother, and how far from home this happens.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: (i) Amira's walking speed is 0.08 km per minute	✓ 1
A1: $0.08 \times 60 = 4.8$ km h⁻¹	✓ 1
R1: (ii) she is waiting at the bus stop, so her distance from home does not change	✓ 1
Part (b)
M1: uses third piece: $0.8 + 0.5(30 - 16)$	✓ 1
A1: $d = 7.8$ km	✓ 1
Part (c)
M1: $0.8 + 0.5(t - 16) = 5$	✓ 1
A1: $t = 24.4$ minutes	✓ 1
Part (d)
A1: $0.3(t - 8) = 0.8 \Rightarrow t = 10.7$ minutes (10.666…)	✓ 1
R1: $10 < 10.7 \le 16$, so Amira is waiting at the bus stop	✓ 1
Part (e)
M1: $0.3(t - 8) = 0.8 + 0.5(t - 16)$	✓ 1
A1: $t = 24$ minutes (within $16 < t \le 36$)	✓ 1
A1: $d = 0.3(24 - 8) = 4.8$ km	✓ 1

Exam tip: After solving for an intersection, confirm the solution lies in the domain of the piece you used; otherwise it is not valid in context.

+ Add to test✓ In your test50IBMI-031-08-r2-04Choosing and evaluating models Paper SLMedium11 marks

The depth of water at a harbour entrance varies with the tides. On a particular day, high tide occurs at 03:00 with a depth of 8.6 m, and the next low tide occurs 6.2 hours later with a depth of 2.2 m.
The depth, $D$ metres, $t$ hours after midnight is modelled by
$D(t)=a\cos\big(b(t-3)\big)+d$, where $a,b,d>0$ and $b$ is measured in degrees per hour.

(a)

Find the value of $a$ and the value of $d$.

(2)

(b)

Find the value of $b$.

(2)

(c)

Use the model to find the depth of water at 14:00.

(2)

(d)

A fishing boat needs a depth of at least 4.0 m to pass through the entrance. Find the length of time between $t=3$ and $t=15.4$ for which the boat cannot pass.

(4)

(e)

Suggest one reason why the harbour master should not rely on this model for planning several weeks ahead.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: $a=\frac{8.6-2.2}{2}=3.2$	✓ 1
A1: $d=\frac{8.6+2.2}{2}=5.4$	✓ 1
Part (b)
M1: period $=2\times6.2=12.4$ hours	✓ 1
A1: $b=\frac{360}{12.4}=29.0$ (29.032…)	✓ 1
Part (c)
M1: substituting $t=14$, $3.2\cos(29.032\times11)+5.4$	✓ 1
A1: $D=7.83$ m	✓ 1
Part (d)
M1: setting $3.2\cos(b(t-3))+5.4=4$ or sketching on GDC with line $D=4$	✓ 1
A1: $t=6.99$	✓ 1
A1: $t=11.4$ (11.406…)	✓ 1
A1: duration $=4.41$ hours	✓ 1
Part (e)
R1: e.g. tidal heights vary over the month (spring/neap tides) so the amplitude is not constant; weather such as storm surges alters depth; small errors in the period accumulate over many cycles	✓ 1

Common error: Check your GDC is in degree mode when $b$ is in degrees per hour — using radian mode gives a wildly wrong depth with no warning.

+ Add to test✓ In your test51IBMI-031-08-r1-03Choosing and evaluating models Paper SLMedium12 marks

A jet of water in a garden fountain leaves a nozzle and its height $h$ metres above ground level, $t$ seconds after leaving the nozzle, is modelled by
$h(t) = at^2 + bt + c$.
Measurements show that $h = 1.2$ when $t = 0$, $h = 4.45$ when $t = 0.5$ and $h = 5.2$ when $t = 1$.

(a)

Write down the value of $c$ and two further equations in $a$ and $b$.

(3)

(b)

Find the value of $a$ and the value of $b$.

(2)

(c)

Find the maximum height of the water and the time at which it occurs.

(3)

(d)

Find the time at which the model predicts the water reaches ground level.

(2)

(e)

Video footage shows the water actually reaches ground level after 2.05 seconds. Calculate the percentage error in the model's prediction.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$c = 1.2$ A1	✓ 1
$0.25a + 0.5b + 1.2 = 4.45$ A1	✓ 1
$a + b + 1.2 = 5.2$ A1	✓ 1
Part (b)
Attempt to solve simultaneously (e.g. GDC) M1	✓ 1
$a = -5$, $b = 9$ A1	✓ 1
Part (c)
Uses $t = -\frac{b}{2a}$ or GDC maximum M1	✓ 1
$t = 0.9$ s A1	✓ 1
$h = 5.25$ m A1	✓ 1
Part (d)
Solves $-5t^2 + 9t + 1.2 = 0$ M1	✓ 1
$t = 1.92$ s (1.9247…; reject negative root) A1	✓ 1
Part (e)
$\frac{\|1.9247 - 2.05\|}{2.05} \times 100$ M1	✓ 1
6.11% A1	✓ 1

Exam tip: Carry the unrounded root 1.9247… into the percentage error; using 1.92 gives 6.34%, which may lose the final accuracy mark.

+ Add to test✓ In your test52IBMI-031-01-r1-04Quadratic and cubic models Paper SLMedium7 marks

The underside of a stone arch over a canal is modelled by $h(x) = a(x - 30)^2 + 18$, where h is the height in metres above the water and x is the horizontal distance in metres from one end of the arch. The arch meets the water at x = 0 and x = 60.

(a)

Show that a = −0.02.

(2)

(b)

A lamp is fixed to the underside of the arch at a horizontal distance of 12 m from one end. Find the height of the lamp above the water.

(2)

(c)

A barge carries a load whose top is 10 m above the water. Find the width of the part of the canal in which the top of the load can pass beneath the arch.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: substitutes (0, 0): 0 = a(0 − 30)² + 18, i.e. 900a + 18 = 0	✓ 1
A1: a = −18/900 = −0.02	✓ 1
Part (b)
M1: h(12) = −0.02(12 − 30)² + 18	✓ 1
A1: 11.5 m (11.52)	✓ 1
Part (c)
M1: sets −0.02(x − 30)² + 18 = 10	✓ 1
A1: x = 10 and x = 50	✓ 1
A1: width = 40.0 m	✓ 1

Exam tip: Vertex form a(x − h)² + k gives the maximum point directly; use the symmetry about x = 30 to check that your two solutions are equally spaced from the centre.

+ Add to test✓ In your test53IBMI-031-01-r1-05Quadratic and cubic models Paper SLMedium10 marks

A workshop makes x handmade lamps per week, where 0 ≤ x ≤ 70. The weekly cost, in dollars, of making x lamps is modelled by
$C(x) = 0.02x^3 - 1.5x^2 + 60x + 400$.
Each lamp is sold for \$45 and every lamp made is sold.

(a)

Interpret the value 400 in this context.

(1)

(b)

Show that the weekly profit, P dollars, is $P(x) = -0.02x^3 + 1.5x^2 - 15x - 400$.

(2)

(c)

Find the values of x for which the workshop breaks even.

(2)

(d)

Hence state the numbers of lamps the workshop should make each week in order to make a profit.

(1)

(e)

Find the number of lamps that gives the maximum weekly profit, and the value of this maximum profit.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: the workshop has fixed weekly costs of \$400, paid even if no lamps are made	✓ 1
Part (b)
M1: uses P = revenue − cost: 45x − (0.02x³ − 1.5x² + 60x + 400)	✓ 1
A1: simplifies correctly to the given expression	✓ 1
Part (c)
A1: x = 32.0 (31.97)	✓ 1
A1: x = 54.5 (54.51)	✓ 1
Part (d)
A1: 32 ≤ x ≤ 54, with x a whole number	✓ 1
Part (e)
M1: uses GDC maximum or sets $P'(x) = -0.06x^2 + 3x - 15 = 0$	✓ 1
A1: x = 44.4 (44.36)	✓ 1
R1: lamps must be a whole number, so compares P(44) = 140.32 and P(45) = 140.00	✓ 1
A1: 44 lamps, maximum profit \$140.32	✓ 1

Common error: Rounding 44.36 to 44 without checking is not enough: for a discrete quantity, compare the profit at the whole numbers either side of the turning point.

+ Add to test✓ In your test54IBMI-031-02-r2-04Exponential and sinusoidal models Paper SLMedium8 marks

In a northern city, the number of hours of daylight, $D$, on day $t$ of the year is modelled by
$D(t) = a\cos\big(b(t - 172)\big) + d$,
where angles are in degrees and $t = 1$ is 1 January. The longest day, day 172, has 16.4 hours of daylight and the shortest day, day 355, has 8.0 hours of daylight. The model has a period of 365 days.

(a)

Find the value of $a$ and the value of $d$.

(2)

(b)

Find the value of $b$.

(1)

(c)

Use the model to estimate the number of hours of daylight on day 100.

(2)

(d)

Find the number of days in the year on which the model predicts more than 14 hours of daylight.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
a = (16.4 − 8.0) ÷ 2 = 4.2 (A1)	✓ 1
d = (16.4 + 8.0) ÷ 2 = 12.2 (A1)	✓ 1
Part (b)
b = 360 ÷ 365 = 0.986 (0.98630…) (A1)	✓ 1
Part (c)
Substitutes t = 100: 4.2cos(0.98630 × (100 − 172)) + 12.2 (M1)	✓ 1
D = 13.6 hours (A1)	✓ 1
Part (d)
Sets up D(t) = 14 or uses graph intersections with D = 14 (M1)	✓ 1
t = 106.5 and t = 237.5 (106.479…, 237.521…) (A1)	✓ 1
Approximately 131 days (A1)	✓ 1

Common error: The amplitude is half the difference between the maximum and minimum, not the maximum value itself. Using a = 16.4 gives a model that predicts negative daylight hours.

+ Add to test✓ In your test55IBMI-031-01-r1-03Quadratic and cubic models Paper SLMedium10 marks

An open-topped box is made from a rectangular sheet of card measuring 40 cm by 25 cm. A square of side x cm is cut from each corner and the sides are folded up.
The volume of the box, V cm³, is given by $V(x) = x(40 - 2x)(25 - 2x)$.

(a)

State the set of possible values of x for this model, giving a reason for the upper limit.

(2)

(b)

Find the volume of the box when x = 3.

(2)

(c)

The box must hold at least 2000 cm³. Find the range of values of x for which this is possible.

(3)

(d)

Find the value of x that maximises the volume, and the maximum volume.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: x > 0	✓ 1
A1: x < 12.5, because the width 25 − 2x must be positive (this is more restrictive than 40 − 2x > 0)	✓ 1
Part (b)
M1: 3 × 34 × 19	✓ 1
A1: V = 1938 cm³ (1940 cm³ to 3 s.f.)	✓ 1
Part (c)
M1: solves V(x) = 2000, e.g. intersection of graphs on GDC	✓ 1
A1: x = 3.20	✓ 1
A1: x = 7.01, so 3.20 ≤ x ≤ 7.01	✓ 1
Part (d)
M1: uses GDC maximum, or sets $V'(x) = 12x^2 - 260x + 1000 = 0$	✓ 1
A1: x = 5 (cm)	✓ 1
A1: maximum volume = 2250 cm³	✓ 1

Common error: Setting V′(x) = 0 gives x = 5 and x ≈ 16.7. The second turning point lies outside the domain 0 < x < 12.5 and has no physical meaning. Always check your solutions against the domain of the model.

+ Add to test✓ In your test56IBMI-031-06-r1-05Quadratic and cubic models Paper SLMedium6 marks

The entrance to a road tunnel is a parabolic arch. Take the origin at one foot of the arch on the level road. The arch meets the road again 10 m away, and its highest point is 6 m above the road. The height of the arch above the road, y metres, at a horizontal distance x metres from the origin is modelled by $y = ax^2 + bx$.

(a)

Find the values of a and b.

(3)

(b)

A lorry carries a rectangular load whose top is 4.2 m above the road. The lorry drives through the tunnel centred on the arch's line of symmetry. Find the maximum possible width of the load, giving your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Forms two equations, using $(10, 0)$ and the vertex $(5, 6)$: $100a + 10b = 0$ and $25a + 5b = 6$	✓ 1
$a = -0.24$	✓ 1
$b = 2.4$	✓ 1
Part (b)
Sets $-0.24x^2 + 2.4x = 4.2$	✓ 1
$x = 2.26$ and $x = 7.74$ (to 3 s.f.)	✓ 1
Maximum width $= 7.7386 - 2.2614 = 5.48$ m	✓ 1

Exam tip: A symmetric arch has its vertex halfway between its feet. The point (5, 6) therefore gives a second equation without needing the vertex formula.

+ Add to test✓ In your test57IBMI-031-06-r2-05Quadratic and cubic models Paper SLMedium10 marks

A concert hall manager models the number of tickets sold, n, when the ticket price is p dollars by $n = 1200 - 40p$. The hall has fixed costs of 3500 dollars per concert, plus a cost of 4 dollars for each ticket sold.

(a)

Write down an expression for the revenue, R dollars, in terms of p.

(1)

(b)

Find the ticket price that maximises revenue, and the maximum revenue.

(2)

(c)

Show that the profit, P dollars, is given by $P = -40p^2 + 1360p - 8300$.

(2)

(d)

Find the range of ticket prices, to the nearest cent, for which the concert makes a profit.

(3)

(e)

Find the maximum profit and the ticket price at which it occurs.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
R = p(1200 − 40p) or R = 1200p − 40p² A1	✓ 1
Part (b)
p = 15 dollars A1	✓ 1
Maximum revenue = 9000 dollars A1	✓ 1
Part (c)
P = 1200p − 40p² − 3500 − 4(1200 − 40p) M1	✓ 1
Correct expansion, e.g. −40p² + 1200p + 160p − 3500 − 4800, leading to the given answer A1	✓ 1
Part (d)
Setting P = 0, or P > 0, and solving, e.g. with a GDC M1	✓ 1
p = 7.972… and p = 26.027… A1	✓ 1
7.98 ≤ p ≤ 26.02 (dollars), with the prices rounded towards the inside of the interval so that the profit stays positive; accept the equivalent form 7.97 < p < 26.03 R1	✓ 1
Part (e)
p = 17 dollars A1	✓ 1
Maximum profit = 3260 dollars A1	✓ 1

Common error: Rounding a break-even price to the nearest cent in the usual way can give a price outside the profitable range. Round towards the inside of the interval.

+ Add to test✓ In your test58IBMI-031-08-r2-03Choosing and evaluating models Paper SLMedium8 marks

A community podcast records its number of subscribers, $N$, at the start of month $t$. The data are:
$t=0$: 500 subscribers
$t=2$: 720 subscribers
$t=4$: 1040 subscribers
Two models are proposed, each fitted using only the data for $t=0$ and $t=4$.
Model A: $N=a+bt$
Model B: $N=500r^t$

(a)

Find the values of $a$ and $b$ for Model A.

(2)

(b)

Find the value of $r$ for Model B.

(2)

(c)

By considering the data value at $t=2$, determine which model is more appropriate. Justify your answer.

(2)

(d)

Use your chosen model to predict the number of subscribers at the start of month 10. Give your answer to the nearest whole number, using the unrounded value of any parameter you found earlier.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: $a=500$	✓ 1
A1: $b=\frac{1040-500}{4}=135$	✓ 1
Part (b)
M1: $500r^4=1040$ so $r^4=2.08$	✓ 1
A1: $r=2.08^{1/4}=1.20$ (1.20093…)	✓ 1
Part (c)
A1: Model A predicts $500+135(2)=770$ and Model B predicts $500(1.20093…)^2\approx 721$	✓ 1
R1: Model B is more appropriate because its prediction is much closer to the recorded value of 720	✓ 1
Part (d)
M1: Correct substitution into their chosen model, e.g. $500(1.20093…)^{10}$ or $500(2.08)^{10/4}$	✓ 1
A1: 3120 subscribers (3119.8…)	✓ 1
Follow-through: if Model A was chosen in part (c), $500+135(10)=1850$ earns M1A1.	—

Common error: Rounding $r$ to 1.20 before raising it to the power 10 gives about 3096, which is noticeably wrong. Keep the full calculator value of $r$ (or use $2.08^{10/4}$ directly) until the final step.

+ Add to test✓ In your test59IBMI-031-03-r2-03Choosing and evaluating models Paper SLMedium12 marks

A new online video channel records its number of subscribers, $S$, $t$ months after launch.

$t$ (months)	0	4	8
$S$	1500	2400	3900

Two models are proposed, each passing through the points for $t = 0$ and $t = 8$:
Model A: $L(t) = pt + q$
Model B: $E(t) = 1500b^t$

(a)

Find the equation of Model A.

(2)

(b)

Show that $b = 1.13$, correct to 3 significant figures.

(2)

(c)

By evaluating both models at $t = 4$, determine which model fits the data better. Justify your answer.

(3)

(d)

At $t = 12$ the channel has 6100 subscribers. Use Model B to predict the number of subscribers at $t = 12$, and find the percentage error in this prediction.

(3)

(e)

Explain why Model B is unlikely to be suitable for predicting the number of subscribers several years after launch.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient $\frac{3900-1500}{8} = 300$	✓ 1
A1 $L(t) = 300t + 1500$	✓ 1
Part (b)
M1 $1500b^8 = 3900$ so $b^8 = 2.6$	✓ 1
A1 $b = 2.6^{1/8} = 1.12686...$, which is 1.13 to 3 s.f.	✓ 1
Part (c)
A1 $L(4) = 2700$	✓ 1
A1 $E(4) = 2420$ (accept 2418.7, or 2450 from using $b = 1.13$)	✓ 1
R1 Model B is better because its value is much closer to the observed 2400 than 2700 is	✓ 1
Part (d)
A1 $E(12) = 6290$ (accept 6288.6; using $b = 1.13$ gives 6502, which earns this mark only as follow-through)	✓ 1
M1 $\frac{\|E(12) - 6100\|}{6100} \times 100$ with their $E(12)$	✓ 1
A1 3.09% (accept answers from 3.08% to 3.11% from a rounded $E(12)$; follow-through 6.59% from $b = 1.13$)	✓ 1
Part (e)
R1 the exponential model increases without limit	✓ 1
R1 the number of potential subscribers is limited, so growth is likely to slow down (level off) as the market saturates	✓ 1

Common error: Using the rounded value b = 1.13 for t = 12 gives about 6500 instead of 6290. Store the unrounded value of b (2.6¹/⁸) and use it in later parts.

+ Add to test✓ In your test60IBMI-031-05-r2-03Linear models and piecewise models Paper SLMedium11 marks

A water company charges households for each quarter according to the volume of water used, $x$ m³. The charge, $C$ dollars, is modelled by
$C(x) = \begin{cases} 0.95x, & 0 \le x \le 20 \\ 19 + 1.60(x - 20), & 20 < x \le 50 \\ 67 + 2.40(x - 50), & x > 50 \end{cases}$

(a)

Find the charge for a household that uses 14 m³.

(2)

(b)

Find the charge for a household that uses 63 m³.

(2)

(c)

Show that the expressions for the second and third pieces give the same value when $x = 50$.

(2)

(d)

A household receives a bill of 54.36 dollars. Find the volume of water it used.

(3)

(e)

The company considers replacing this tariff with a single rate of $k$ dollars per m³. The rate is chosen so that a household using 63 m³ pays the same as under the current model. Find $k$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: uses first piece: $0.95 \times 14$	✓ 1
A1: 13.30 dollars	✓ 1
Part (b)
M1: uses third piece: $67 + 2.40(63 - 50)$	✓ 1
A1: 98.20 dollars	✓ 1
Part (c)
M1: second expression at $x = 50$: $19 + 1.60 \times 30 = 67$	✓ 1
A1: third expression at $x = 50$: $67 + 2.40 \times 0 = 67$, so both give 67	✓ 1
Part (d)
R1: $19 < 54.36 \le 67$, so the second piece applies	✓ 1
M1: $19 + 1.60(x - 20) = 54.36$	✓ 1
A1: $x = 42.1$ m³	✓ 1
Part (e)
M1: $63k = 98.20$	✓ 1
A1: $k = 1.56$ dollars per m³ (3 s.f.)	✓ 1

Exam tip: Before solving a piecewise equation, evaluate the function at the boundaries to find which interval the given output lies in.

+ Add to test✓ In your test61IBMI-031-01-r2-03Quadratic and cubic models Paper SLMedium13 marks

The fuel consumption of a car, F litres per 100 km, when it travels at a constant speed of v km h⁻¹ is modelled by
$F(v) = av^2 + bv + c$, for 30 ≤ v ≤ 130.
Tests give the following results: at 40 km h⁻¹ the consumption is 7.6 litres per 100 km; at 80 km h⁻¹ it is 6.0 litres per 100 km; at 120 km h⁻¹ it is 8.0 litres per 100 km.

(a)

Write down a system of three equations in a, b and c.

(2)

(b)

Find the values of a, b and c.

(2)

(c)

Find the speed at which the fuel consumption is lowest, and state this lowest consumption.

(3)

(d)

Find the range of speeds for which the fuel consumption is at most 7 litres per 100 km.

(3)

(e)

A driver makes a 250 km journey at a constant 100 km h⁻¹. Fuel costs 1.62 USD per litre. Find the cost of the fuel used, in USD, giving your answer to two decimal places.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Substitutes at least one data pair into the model (M1)	✓ 1
1600a + 40b + c = 7.6; 6400a + 80b + c = 6.0; 14400a + 120b + c = 8.0 (A1)	✓ 1
Part (b)
a = 0.001125 (A1)	✓ 1
b = −0.175 and c = 12.8 (A1)	✓ 1
Part (c)
Uses $v = -\frac{b}{2a}$ or the GDC minimum (M1)	✓ 1
v = 77.8 km h⁻¹ (A1)	✓ 1
F = 5.99 litres per 100 km (A1)	✓ 1
Part (d)
Sets F(v) = 7, e.g. finds where the graph meets y = 7 (M1)	✓ 1
v = 47.9 and v = 108 (accept 107.7) (A1)	✓ 1
47.9 ≤ v ≤ 108 km h⁻¹ (A1)	✓ 1
Part (e)
F(100) = 6.55 litres per 100 km (M1)	✓ 1
Fuel used = 6.55 × 2.5 = 16.375 litres (A1)	✓ 1
Cost = 16.375 × 1.62 = 26.53 USD (A1)	✓ 1

Exam tip: Keep the exact coefficients (0.001125, −0.175, 12.8) stored in your GDC for every later part. Rounding a coefficient early can change the final cost in the second decimal place.

+ Add to test✓ In your test62IBMI-031-03-r1-04Choosing and evaluating models Paper SLMedium12 marks

The depth of water, $D$ metres, at a harbour entrance is recorded on a particular day. High tide gives a maximum depth of 7.8 m at 04:00. The next low tide gives a minimum depth of 1.4 m at 10:12.
The depth is modelled by $D(t) = a\cos(b(t - c)) + d$, where $t$ is the time in hours after midnight and $a, b, c > 0$.

(a)

Find the value of a and the value of d.

(2)

(b)

Find the value of b.

(2)

(c)

Write down a possible value of c.

(1)

(d)

Use the model to find the depth of water at 13:00.

(2)

(e)

A ferry can only use the harbour entrance when the depth is at least 5.5 m. Find the total length of time between 04:00 and 16:24 on this day during which the ferry can use the entrance.

(4)

(f)

Suggest one reason why this model may not predict the depth accurately on later days.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 a = (7.8 − 1.4)/2 = 3.2	✓ 1
A1 d = (7.8 + 1.4)/2 = 4.6	✓ 1
Part (b)
M1 period = 2 × 6.2 = 12.4 hours	✓ 1
A1 b = 2π/12.4 = 0.507 (0.506707…)	✓ 1
Part (c)
A1 c = 4, the time of the maximum (accept any equivalent value 4 + 12.4k, e.g. 16.4)	✓ 1
Part (d)
M1 substituting t = 13 into their model	✓ 1
A1 D = 4.12 m	✓ 1
Part (e)
M1 setting up D(t) = 5.5 or D(t) ≥ 5.5	✓ 1
A1 t = 6.537… and t = 13.86…	✓ 1
M1 adding the intervals at each end, e.g. (6.537 − 4) + (16.4 − 13.863)	✓ 1
A1 5.07 hours (accept 5.08 from rounded intermediate values)	✓ 1
Part (f)
R1 e.g. the tidal period and range vary over the month (spring and neap tides), or weather such as wind or storms changes the water level	✓ 1

Common error: Using a period of 6.2 hours. The time from a maximum to the next minimum is only half a period.

+ Add to test✓ In your test63IBMI-031-06-r1-03Quadratic and cubic models Paper SLMedium11 marks

A small business sells scented candles. The weekly profit, $P, when each candle is sold at a price of $x is modelled by $P(x) = ax^2 + bx + c$.
Market research gives the following information:
• at a price of $5, the weekly profit is $1200;
• at a price of $10, the weekly profit is $2700;
• at a price of $20, the weekly profit is $1200.

(a)

Write down a system of three equations in a, b and c.

(2)

(b)

Find the values of a, b and c.

(2)

(c)

Interpret the value of c in context.

(1)

(d)

Find the price that maximises weekly profit and the maximum weekly profit.

(3)

(e)

Find the range of prices for which the business makes a positive profit.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Any two correct equations from $25a + 5b + c = 1200$, $100a + 10b + c = 2700$, $400a + 20b + c = 1200$	✓ 1
All three equations correct	✓ 1
Part (b)
Solves the system using a GDC or by elimination, e.g. $b = -25a$	✓ 1
$a = -30$, $b = 750$, $c = -1800$	✓ 1
Part (c)
At a price of $0 (candles given away) the business makes a weekly loss of $1800, i.e. its fixed weekly costs are $1800	✓ 1
Part (d)
Uses the symmetry of the data, $x = -\frac{b}{2a}$ or a GDC	✓ 1
Price = $12.50	✓ 1
Maximum weekly profit = $2887.50	✓ 1
Part (e)
Solves $-30x^2 + 750x - 1800 = 0$ or uses the graph's x-intercepts	✓ 1
$x = 2.69$	✓ 1
$x = 22.31$, so $2.69 < x < 22.31$	✓ 1

Exam tip: Two of the data points share the same profit, so the axis of symmetry lies midway between their prices. This gives a quick check on your vertex.

+ Add to test✓ In your test64IBMI-031-06-r1-04Quadratic and cubic models Paper SLMedium7 marks

The volume of water in a reservoir, V million litres, t months after monitoring began is modelled by
$V(t) = 0.02t^3 - 0.9t^2 + 10.8t + 50$, for $0 \le t \le 24$.

(a)

Write down the volume of water when monitoring began.

(1)

(b)

Find the maximum volume of water in the reservoir during the 24 months, and the time at which it occurs.

(2)

(c)

The reservoir is classed as low when it holds less than 66 million litres. Find the total length of time during the 24 months for which the reservoir is classed as low.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$V(0) = 50$ million litres	✓ 1
Part (b)
$t = 8.29$ months (accept $15-\sqrt{45}$)	✓ 1
$V = 89.1$ million litres	✓ 1
(Note: $V(0)=50$ and $V(24)=67.28$ are both less, so the local maximum is the overall maximum.)	—
Part (c)
Attempts to solve $V(t) = 66$, e.g. by GDC or graph intersection	✓ 1
$t = 1.718$	✓ 1
$t = 20$ and $t = 23.28$	✓ 1
Low for $0 \le t < 1.718$ and $20 < t < 23.28$, so total time $= 1.718 + 3.282 = 5.00$ months (accept 5 or 5.0)	✓ 1

Common error: Forgetting the interval at the very start: the reservoir begins below 66 million litres, so the time from t = 0 to the first intersection also counts. Keep unrounded roots from your GDC when subtracting, or the total drifts to 5.02.

+ Add to test✓ In your test65IBMI-031-02-r1-05Exponential and sinusoidal models Paper SLMedium9 marks

A cup of tea is placed on a desk in a room. The temperature of the tea, H °C, t minutes after it is placed on the desk, is modelled by
$H(t) = 20 + 75e^{-kt}$, $t \ge 0$,
where k is a positive constant. After 4 minutes the temperature of the tea is 70 °C.

(a)

Write down the initial temperature of the tea.

(1)

(b)

Find the value of k.

(2)

(c)

Find the temperature of the tea after 10 minutes.

(2)

(d)

Write down the temperature that the tea approaches in the long term, and interpret this value in context.

(2)

(e)

The tea is considered too cool to enjoy once its temperature drops below 60 °C. Find the time after which this happens.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
95 °C (A1)	✓ 1
Part (b)
Sets up 20 + 75e^−4k = 70, or equivalent, e.g. e^−4k = 2/3 (M1)	✓ 1
k = 0.101 (0.101366…; exact form ln(1.5)/4) (A1)	✓ 1
Part (c)
Substitutes t = 10 with their k: 20 + 75e^−10k (M1)	✓ 1
= 47.2 °C (unrounded k). Accept 47.3 °C from k = 0.101 (A1)	✓ 1
Part (d)
20 °C (A1)	✓ 1
This is the temperature of the room; the tea cools towards room temperature (R1)	✓ 1
Part (e)
Sets up 20 + 75e^−kt = 60 with their k (or the inequality H < 60) (M1)	✓ 1
t = 6.20 minutes (unrounded k). Accept 6.22 minutes from k = 0.101 (A1)	✓ 1

Exam tip: Keep the unrounded value of k in your calculator memory for later parts. Rounding k early can change the third significant figure of later answers.

+ Add to test✓ In your test66IBMI-031-08-r1-04Choosing and evaluating models Paper SLMedium10 marks

At a harbour entrance on one particular day, high tide gives a water depth of 8.6 m at 03:00 and the following low tide gives a depth of 2.2 m at 09:15.
The depth $D$ metres, $t$ hours after midnight, is modelled by
$D(t) = p\cos\big(q(t - r)\big) + s$, where $p > 0$, $q$ is measured in degrees per hour and $0 \le r < 12$.

(a)

Find the values of $p$, $q$, $r$ and $s$.

(5)

(b)

Use the model to estimate the depth of water at 12:00.

(2)

(c)

A cargo boat can only pass the entrance when the depth is at least 4 m. Find the length of time, around the 09:15 low tide, during which the boat cannot pass.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$p = \frac{8.6 - 2.2}{2} = 3.2$ A1	✓ 1
$s = \frac{8.6 + 2.2}{2} = 5.4$ A1	✓ 1
Period $= 2 \times 6.25 = 12.5$ hours M1	✓ 1
$q = \frac{360}{12.5} = 28.8$ A1	✓ 1
$r = 3$ (maximum of the cosine at the 03:00 high tide) A1	✓ 1
Part (b)
Substitutes $t = 12$: $3.2\cos(28.8 \times 9) + 5.4$ M1	✓ 1
$D = 4.80$ m A1	✓ 1
Part (c)
Solves $3.2\cos(28.8(t-3)) + 5.4 = 4$ (e.g. by GDC intersection) M1	✓ 1
$t = 7.03$ and $t = 11.5$ (7.0257…, 11.474…) A1	✓ 1
Duration $= 4.45$ hours A1	✓ 1

Common error: Leaving the calculator in radian mode when $q$ is in degrees gives a completely wrong depth, so check the mode before substituting.

+ Add to test✓ In your test67IBMI-031-02-r2-03Exponential and sinusoidal models Paper SLMedium11 marks

A cup of coffee is left to cool in a room. Its temperature, $T$ °C, $t$ minutes after it is poured is modelled by
$T(t) = 20 + 70e^{-kt}$, $t \ge 0$,
where $k$ is a positive constant. After 5 minutes the temperature of the coffee is 62 °C.

(a)

Write down the temperature of the coffee when it is poured.

(1)

(b)

Find the value of $k$.

(2)

(c)

Find the temperature of the coffee 12 minutes after it is poured.

(2)

(d)

The coffee is considered pleasant to drink once its temperature is 45 °C or lower. Find the time after pouring at which the coffee first becomes pleasant to drink.

(2)

(e)

Write down the equation of the horizontal asymptote of the graph of $T$ and interpret it in context.

(2)

(f)

A student claims that, according to the model, the coffee has reached exactly 20 °C after one hour. Use the model to comment on this claim.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
90 °C (A1)	✓ 1
Part (b)
Sets up 20 + 70e^−5k = 62 (M1)	✓ 1
k = 0.102 (0.102165…) (A1)	✓ 1
Part (c)
Substitutes t = 12 with their k: 20 + 70e^−12k (M1)	✓ 1
T = 40.5 °C (40.54…); accept 40.6 °C from k = 0.102 (A1)	✓ 1
Part (d)
Sets up 20 + 70e^−kt = 45 (M1)	✓ 1
t = 10.1 minutes (10.08…) (A1)	✓ 1
Part (e)
T = 20 (A1)	✓ 1
The coffee cools towards room temperature, 20 °C, but never falls below it (R1)	✓ 1
Part (f)
T(60) = 20.15 °C (20.152…) (A1)	✓ 1
The temperature is very close to 20 °C. However, e^−kt > 0 for all t, so the model never reaches exactly 20 °C and the claim is incorrect (R1)	✓ 1

Exam tip: Carry the unrounded value of k (0.102165…) into later parts. Using 0.102 throughout can shift answers in the third significant figure.

+ Add to test✓ In your test68IBMI-031-03-r2-04Choosing and evaluating models Paper SLMedium7 marks

At a harbour, the depth of water reaches a maximum of 8.6 m at 03:00 and the next minimum of 2.2 m at 09:15. The depth, $D$ metres, is modelled by
$D(t) = a\cos(b(t - c)) + d$,
where $t$ is the number of hours after midnight, $a, b > 0$ and $0 \le c < 6$.

(a)

Find the value of $a$ and the value of $d$.

(2)

(b)

Find the value of $b$.

(2)

(c)

Write down the value of $c$.

(1)

(d)

A fishing boat in the harbour needs a depth of at least 4.0 m to leave. It is ready to leave at 03:00. Find the latest time at which it can leave before the depth first drops below 4.0 m. Give your answer to the nearest minute.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $a = \frac{8.6 - 2.2}{2} = 3.2$	✓ 1
A1 $d = \frac{8.6 + 2.2}{2} = 5.4$	✓ 1
Part (b)
M1 period $= 2 \times 6.25 = 12.5$ hours	✓ 1
A1 $b = \frac{2\pi}{12.5} = \frac{4\pi}{25} = 0.503$ (3 s.f.)	✓ 1
Part (c)
A1 $c = 3$	✓ 1
Part (d)
M1 solving $3.2\cos(0.503(t - 3)) + 5.4 = 4.0$ and choosing the first solution after $t = 3$, e.g. by graph intersection	✓ 1
A1 $t = 7.03$ hours, i.e. 07:02 (accept 07:01 from using $b = 0.503$)	✓ 1

Exam tip: With a cosine model, the horizontal shift is the time of a maximum, so c can be read straight from the data. Keep b unrounded in your calculator when solving for t.

+ Add to test✓ In your test69IBMI-031-00-11Linear models and piecewise models Paper SLHard14 marks

A water company offers two pricing plans for x cubic metres (m³) of water used per quarter.
Plan A: a fixed charge of $18, plus $1.40 per m³ for the first 20 m³, $2.10 per m³ for the next 30 m³, and $3.20 per m³ for all water above 50 m³.
Plan B: a fixed charge of $45, plus $2.05 per m³ for all water used.

(a)

Find the Plan A cost for 35 m³.

(2)

(b)

Find the Plan A cost, A(x), as a piecewise linear function, giving each piece in the form mx + c with its domain.

(4)

(c)

Write down the Plan B cost, B(x).

(1)

(d)

Find the set of values of x for which Plan B is cheaper than Plan A.

(5)

(e)

A household uses 120 m³ in one quarter. Find how much it saves by choosing Plan B.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 18 + 20 × 1.40 + 15 × 2.10	✓ 1
A1 $77.50	✓ 1
Part (b)
A1 A(x) = 1.4x + 18 for 0 ≤ x ≤ 20	✓ 1
M1 A(20) = 46 used to form 46 + 2.1(x − 20), or equivalent	✓ 1
A1 A(x) = 2.1x + 4 for 20 < x ≤ 50	✓ 1
A1 A(x) = 3.2x − 51 for x > 50	✓ 1
Part (c)
A1 B(x) = 45 + 2.05x	✓ 1
Part (d)
M1 comparing B with each piece of A	✓ 1
R1 for 0 ≤ x ≤ 50, A(x) < B(x) (e.g. B − A = 27 + 0.65x > 0 and B − A = 41 − 0.05x > 0 on these domains)	✓ 1
M1 3.2x − 51 = 45 + 2.05x	✓ 1
A1 x = 83.478… (= 96/1.15)	✓ 1
A1 Plan B cheaper for x > 83.478… m³ (accept x > 83.5 to 3 s.f.)	✓ 1
Part (e)
M1 A(120) = 333 and B(120) = 291	✓ 1
A1 $42.00	✓ 1

Common error: Solving A(x) = B(x) using only one piece can give a solution outside that piece's domain. Always check that the solution lies in the interval you used.

+ Add to test✓ In your test70IBMI-031-07-11Exponential and sinusoidal models Paper SLHard10 marks

After a filter is installed at a factory, the concentration of a pollutant in a nearby river is monitored. The concentration, C mg L⁻¹, t weeks after installation is modelled by
$C(t) = ab^t + c$,
where a > 0 and 0 < b < 1. The measurements are C(0) = 85, C(2) = 61 and C(4) = 45.

(a)

Show that $b^2 = \frac{2}{3}$.

(3)

(b)

Hence find the values of a, b and c.

(3)

(c)

The river is considered safe for fishing when the concentration falls below 20 mg L⁻¹. Find the time taken for the river to become safe.

(3)

(d)

Interpret the value of c in context.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: forming two equations from differences, $a(1 - b^2) = 24$ and $ab^2(1 - b^2) = 16$	✓ 1
M1: dividing to eliminate a: $b^2 = \frac{16}{24}$	✓ 1
A1: $b^2 = \frac{2}{3}$ shown	✓ 1
Part (b)
A1: $b = 0.816$ (0.81649…)	✓ 1
A1: $a = \frac{24}{1/3} = 72$	✓ 1
A1: $c = 85 - 72 = 13$	✓ 1
Part (c)
M1: $72(0.81649\ldots)^t + 13 = 20$	✓ 1
M1: $(0.81649\ldots)^t = \frac{7}{72}$ or GDC intersection	✓ 1
A1: 11.5 weeks	✓ 1
Part (d)
R1: 13 mg L⁻¹ is the long-term (background) concentration that the pollutant level approaches; the model predicts it will never fall below this	✓ 1

Exam tip: With equally spaced data for a model ab^t + c, subtracting consecutive values removes c, and dividing the differences removes a.

+ Add to test✓ In your test71IBMI-031-07-12Exponential and sinusoidal models Paper SLHard12 marks

A conservation team introduces 400 deer into a nature reserve that can support at most 12 000 deer. The number of deer, P, t years after introduction is modelled by
$P(t) = \frac{L}{1 + Ce^{-kt}}$,
where L, C and k are positive constants and L = 12 000. After 3 years there are 1500 deer.

(a)

Find the value of C.

(2)

(b)

Find the value of k.

(3)

(c)

The population grows fastest when it reaches half of its carrying capacity. Find the time at which this occurs.

(3)

(d)

Find the time taken for the population to reach 90% of the carrying capacity.

(3)

(e)

Interpret the value of L in context.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $\frac{12000}{1 + C} = 400$	✓ 1
A1: $C = 29$	✓ 1
Part (b)
M1: $\frac{12000}{1 + 29e^{-3k}} = 1500$	✓ 1
M1: $e^{-3k} = \frac{7}{29}$ or GDC solver	✓ 1
A1: $k = 0.474$ (0.47379…)	✓ 1
Part (c)
M1: $P = 6000$	✓ 1
M1: $29e^{-kt} = 1$ or GDC solver with their k	✓ 1
A1: 7.11 years	✓ 1
Part (d)
M1: $P = 10800$	✓ 1
M1: $29e^{-kt} = \frac{1}{9}$ or GDC solver	✓ 1
A1: 11.7 years	✓ 1
Part (e)
R1: 12 000 is the maximum number of deer the reserve can support, or the value the population approaches over time; accept either statement	✓ 1

Common error: Substituting t = 0 gives P = L/(1 + C), not L/C. Forgetting the 1 in the denominator gives C = 30, which spoils every later part.

+ Add to test✓ In your test72IBMI-031-03-r2-06Choosing and evaluating models Paper SLHard13 marks

A cup of coffee is placed in a room kept at 21 °C. Its temperature is measured at three times.

$t$ (minutes)	0	5	10
Temperature (°C)	88	70	57

Model 1 for the temperature is $T(t) = 21 + Be^{-kt}$, where $B, k > 0$.

(a)

Find the value of $B$, and interpret the value 21 in the context of the model.

(2)

(b)

Show that $k = 0.0626$, correct to 3 significant figures.

(3)

(c)

Model 2 is a quadratic, $Q(t) = pt^2 + qt + 88$, that passes through all three data points. Find the value of $p$ and the value of $q$.

(2)

(d)

Find the minimum value of $Q(t)$. Hence explain why Model 2 is unsuitable for large values of $t$.

(3)

(e)

The coffee is considered cold when its temperature reaches 45 °C. Use Model 1 to find the time at which this occurs, and explain why Model 2 cannot answer this question.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $88 = 21 + B$, so $B = 67$	✓ 1
R1 21 °C is the temperature the coffee approaches in the long term, which is room temperature	✓ 1
Part (b)
M1 $70 = 21 + 67e^{-5k}$	✓ 1
M1 $e^{-5k} = \frac{49}{67}$, so $k = \frac{1}{5}\ln\frac{67}{49}$	✓ 1
A1 $k = 0.062574...$, which is 0.0626 to 3 s.f.	✓ 1
Part (c)
A1 $p = 0.1$	✓ 1
A1 $q = -4.1$	✓ 1
Part (d)
M1 minimum at $t = \frac{4.1}{0.2} = 20.5$	✓ 1
A1 minimum value 46.0 °C	✓ 1
R1 after 20.5 minutes Model 2 predicts the coffee warming up again, and it never approaches room temperature, which is unrealistic	✓ 1
Part (e)
M1 solving $21 + 67e^{-0.06257t} = 45$	✓ 1
A1 $t = 16.4$ minutes	✓ 1
R1 the minimum of $Q$ is 46.0 °C, which is above 45 °C, so Model 2 never reaches 45 °C	✓ 1

Exam tip: A model can fit every data point exactly and still be a poor choice. Check its long-term behaviour against the physical situation.

+ Add to test✓ In your test73IBMI-031-08-r2-06Choosing and evaluating models Paper SLHard13 marks

A cup of coffee is poured at time $t=0$ minutes in a room kept at a constant 21 °C. Its temperature is 90 °C when poured and 62 °C after 10 minutes. A later measurement at $t=30$ minutes gives 32 °C.

(a)

A student fits a linear model through the readings at $t=0$ and $t=10$. Use this model to predict the temperature at $t=30$, and explain why the linear model is unsuitable for longer times.

(3)

(b)

A second model is $T=21+69e^{-kt}$. Using the reading at $t=10$, show that $k=0.0521$ to 3 significant figures.

(3)

(c)

Use the exponential model to predict the temperature at $t=30$ and find the percentage error compared with the measured value.

(3)

(d)

State the equation of the horizontal asymptote of the exponential model and interpret it in context.

(2)

(e)

Use the exponential model to find the time taken for the coffee to cool to 50 °C.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: gradient $=\frac{62-90}{10}=-2.8$, so $T=90-2.8t$	✓ 1
A1: $T(30)=6$ °C	✓ 1
R1: this is below room temperature (21 °C), which is impossible for coffee cooling in the room / the model eventually predicts negative temperatures	✓ 1
Part (b)
M1: substituting $62=21+69e^{-10k}$	✓ 1
M1: $e^{-10k}=\frac{41}{69}$ and taking logs, $k=\frac{1}{10}\ln\frac{69}{41}$	✓ 1
A1: $k=0.052053…=0.0521$	✓ 1
Part (c)
A1: $T(30)=35.5$ °C (35.476…)	✓ 1
M1: $\frac{\|35.476-32\|}{32}\times100$	✓ 1
A1: 10.9%	✓ 1
Part (d)
A1: $T=21$	✓ 1
R1: the coffee's temperature approaches (but never falls below) room temperature as time increases	✓ 1
Part (e)
M1: solving $21+69e^{-0.05205t}=50$, e.g. $t=\frac{\ln(69/29)}{k}$	✓ 1
A1: $t=16.7$ minutes (accept 16.6 from using $k=0.0521$)	✓ 1

Exam tip: When evaluating a model, check its long-term behaviour against the context: a model with the right asymptote can be judged more appropriate even if it is not perfect at every data point.

+ Add to test✓ In your test74IBMI-031-03-r1-06Choosing and evaluating models Paper SLHard13 marks

The population, $P$ thousand, of a town $t$ years after the start of 2000 is given in the table.

t: 0, 5, 10
P: 12.0, 15.6, 20.4

Two models are proposed.
Model A: $P = 12r^t$
Model B: $P = at^2 + bt + 12$

(a)

Using the data for t = 10, find the value of r for Model A. Hence show that Model A gives a population that rounds to 15.6 thousand when t = 5.

(3)

(b)

Using the data for t = 5 and t = 10, find the values of a and b for Model B.

(3)

(c)

In 2020 (t = 20) the population was found to be 33.1 thousand. Determine which model fits this value better, justifying your answer.

(4)

(d)

Use Model B to find the value of t when the population first reaches 50 thousand.

(2)

(e)

Give one reason why Model B might not be suitable for predicting the population far into the future.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12r¹⁰ = 20.4, so r¹⁰ = 1.7	✓ 1
A1 r = 1.7^0.1 = 1.0545 (1.05 to 3 s.f.)	✓ 1
A1 using the unrounded r: 12 × 1.0545⁵ = 12√1.7 = 15.646, which rounds to 15.6	✓ 1
Part (b)
M1 forming two equations: 25a + 5b = 3.6 and 100a + 10b = 8.4	✓ 1
A1 a = 0.024	✓ 1
A1 b = 0.6	✓ 1
Part (c)
M1 substituting t = 20 into both models	✓ 1
A1 Model A: 34.7 thousand (34.68)	✓ 1
A1 Model B: 33.6 thousand	✓ 1
R1 Model B is better, as its prediction is closer to 33.1 (error 0.5 thousand compared with about 1.6 thousand)	✓ 1
Part (d)
M1 solving 0.024t² + 0.6t + 12 = 50	✓ 1
A1 t = 29.2 years (positive root)	✓ 1
Part (e)
R1 e.g. the quadratic increases without limit, but limited housing, land or resources would restrict growth; or the model would give unrealistic values outside the data range	✓ 1

Common error: In a 'show that' step, keep the unrounded value of r in your calculator. Using r = 1.05 gives 15.3 and fails to show the result.

+ Add to test✓ In your test75IBMI-031-05-r2-06Linear models and piecewise models Paper SLHard13 marks

A traveller exchanges pounds for euros at an airport kiosk and later uses a card that converts euros into US dollars. The kiosk converts $x$ pounds into euros according to $f(x) = 1.16(x - 3)$, $x > 3$, which includes a fixed £3 fee. The card converts $y$ euros into US dollars according to $g(y) = 1.08y - 2$, which includes a fixed fee of 2 US dollars.

(a)

Find $(g \circ f)(x)$, giving your answer in the form $ax + b$.

(3)

(b)

Find the number of US dollars the traveller obtains from £250, and interpret what $(g \circ f)(x)$ represents.

(2)

(c)

Find $(g \circ f)^{-1}(x)$ and hence find the number of pounds needed to obtain 500 US dollars.

(4)

(d)

A second kiosk converts $x$ pounds into euros according to $k(x) = 1.20x - 9$. Find the amount of pounds for which both kiosks give the same number of euros, and state the amounts for which the second kiosk gives more euros.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $g(f(x)) = 1.08 \times 1.16(x - 3) - 2$	✓ 1
A1: $a = 1.2528$	✓ 1
A1: $b = -5.7584$, so $(g \circ f)(x) = 1.2528x - 5.7584$	✓ 1
Part (b)
A1: $(g \circ f)(250) = 307.44$ US dollars	✓ 1
R1: $(g \circ f)(x)$ gives the number of US dollars received from $x$ pounds after both conversions and both fees	✓ 1
Part (c)
M1: rearranges $y = 1.2528x - 5.7584$ for $x$ (or swaps variables first)	✓ 1
A1: $(g \circ f)^{-1}(x) = \frac{x + 5.7584}{1.2528}$	✓ 1
M1: substitutes 500	✓ 1
A1: £403.70	✓ 1
Part (d)
M1: sets $1.16(x - 3) = 1.20x - 9$	✓ 1
M1: simplifies to $0.04x = 5.52$	✓ 1
A1: $x = 138$ pounds	✓ 1
R1: the second kiosk gives more euros when more than £138 is exchanged ($x > 138$), because its gradient is larger	✓ 1

Common error: $(g \circ f)(x)$ means apply $f$ first, then $g$. Reversing the order applies the dollar conversion to pounds and gives a different function.

+ Add to test✓ In your test76IBMI-031-01-r2-06Quadratic and cubic models Paper SLHard12 marks

A small company makes handmade guitars. When it makes and sells x guitars per week, the selling price of each guitar is (80 − 0.4x) hundred dollars. The weekly cost, in hundreds of dollars, is modelled by
$C(x) = 0.002x^3 - 0.3x^2 + 25x + 600$, where 600 represents fixed costs.

(a)

Show that the weekly profit, P hundred dollars, is given by $P(x) = -0.002x^3 - 0.1x^2 + 55x - 600$.

(2)

(b)

Find the number of guitars the company should make each week to maximise its profit. Justify your answer.

(3)

(c)

Write down the maximum weekly profit in dollars.

(1)

(d)

Find the selling price of each guitar at this level of production.

(1)

(e)

Find the set of whole-number values of x for which the company makes a profit.

(3)

(f)

The fixed costs increase from 600 to F hundred dollars. All other parts of the model are unchanged. Find the largest value of F, to the nearest whole number, for which the company can still avoid making a loss.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Revenue = x(80 − 0.4x) = 80x − 0.4x², and uses P = R − C (M1)	✓ 1
Simplifies correctly to −0.002x³ − 0.1x² + 55x − 600 (A1) (AG)	✓ 1
Part (b)
GDC maximum, or solves P′(x) = 0, giving x = 80.5 (80.516…) (M1)	✓ 1
Because x must be a whole number, compares P(80) = 2136 with P(81) = 2136.018 (R1)	✓ 1
81 guitars (A1)	✓ 1
Part (c)
2136.018 hundred dollars, so 213 601.80 dollars (accept 213 602 dollars) (A1)	✓ 1
Part (d)
80 − 0.4 × 81 = 47.6 hundred dollars, so 4760 dollars (A1)	✓ 1
Part (e)
Solves P(x) = 0 (M1)	✓ 1
x = 11.2 and x = 136.0 (135.99…) (A1)	✓ 1
12 ≤ x ≤ 135, where x is an integer (A1)	✓ 1
Part (f)
Recognises that a change in fixed costs lowers the whole profit curve by the same amount, so the maximum still occurs at x = 81, and that the company needs F − 600 ≤ 2136.018 (M1)	✓ 1
F = 2736 (hundred dollars) (A1)	✓ 1

Exam tip: When the variable counts whole items, a turning point at x = 80.5 is not the final answer. Evaluate the model at the integers on either side, and give the one with the larger profit as your justification.

+ Add to test✓ In your test77IBMI-031-04-r2-01Transformations of graphs and logarithmic scales Paper HLEasy5 marks

The profile of a water slide at a theme park is modelled by $f(x)=0.04x^2-1.6x+20$, for $0\le x\le 20$. Here $x$ is the horizontal distance in metres from the top of the slide, and $f(x)$ is the height in metres above the ground. The slide ends on a raised platform beside a pool.
At a second park the same slide is rebuilt. Its profile is modelled by $g(x)=f(x-4)+2$, for $4\le x\le 24$.

(a)

Describe fully the single transformation that maps the graph of $f$ onto the graph of $g$.

(2)

(b)

Find the coordinates of the vertex of the graph of $f$.

(2)

(c)

Hence write down the height above the ground of the bottom of the new slide.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: translation	✓ 1
A1: by vector $\begin{pmatrix}4\\2\end{pmatrix}$, i.e. 4 m right (horizontally) and 2 m up (vertically)	✓ 1
Part (b)
M1: $x=-\frac{-1.6}{2(0.04)}=20$, or uses the GDC minimum	✓ 1
A1: $(20,\,4)$	✓ 1
Part (c)
A1: the vertex of $g$ is $(24,\,6)$, so the bottom is 6 m above the ground	✓ 1

Exam tip: For $f(x-h)+k$ the graph moves $h$ to the right, not the left. Apply the same translation to any key point, such as a vertex, instead of recalculating it.

+ Add to test✓ In your test78IBMI-031-04-r1-02Transformations of graphs and logarithmic scales Paper HLEasy5 marks

A small company models its weekly profit, P thousand dollars, from producing x hundred items by
$P(x) = -0.5x^2 + 12x - 40$.
After moving to a larger factory, the company's new profit model is $Q(x) = P(x-2) + 5$.

(a)

Describe fully the two transformations that map the graph of P onto the graph of Q.

(2)

(b)

Find the coordinates of the vertex of the graph of Q.

(2)

(c)

Interpret your answer to the previous part in context.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: horizontal translation 2 units to the right (or translation by vector (2, 0))	✓ 1
A1: vertical translation 5 units up (or translation by vector (0, 5))	✓ 1
Part (b)
M1: finds the vertex of P at (12, 32), e.g. using $x = -\frac{12}{2(-0.5)}$ or a GDC, or finds the vertex of Q directly	✓ 1
A1: vertex of Q is (14, 37)	✓ 1
Part (c)
R1: the maximum weekly profit under the new model is $37 000, achieved when 1400 items are produced	✓ 1

Exam tip: Apply the transformations to the vertex of the original graph rather than expanding Q(x). Replacing x with x − 2 moves points right by 2, not left.

+ Add to test✓ In your test79IBMI-031-00-09Linear models and piecewise models Paper HLMedium8 marks

A print shop charges P(x) dollars to print x posters, where
\(P(x)=\begin{cases}40+2x, & 0The following year all prices increase by 8% and then a fixed $5 administration fee is added to every order. The new cost is given by $Q(x)=1.08P(x)+5$.

(a)

Find the new cost of printing 30 posters.

(2)

(b)

Describe fully the sequence of transformations that maps the graph of P onto the graph of Q.

(2)

(c)

Write down Q(x) as a piecewise function.

(2)

(d)

A customer's bill under the new prices was $221.00. Find the number of posters printed.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 P(30) = 100 and 1.08 × 100 + 5	✓ 1
A1 $113.00	✓ 1
Part (b)
A1 vertical stretch, scale factor 1.08	✓ 1
A1 followed by a vertical translation of 5 units up	✓ 1
Part (c)
A1 Q(x) = 48.2 + 2.16x for 0 < x ≤ 50	✓ 1
A1 Q(x) = 102.2 + 1.08x for 50 < x ≤ 200	✓ 1
Part (d)
M1 102.2 + 1.08x = 221 (second piece as Q(50) = 156.2 < 221)	✓ 1
A1 110 posters	✓ 1

Common error: The order matters: stretching then translating gives 1.08P(x) + 5, whereas translating first would give 1.08(P(x) + 5), a different function.

+ Add to test✓ In your test80IBMI-031-07-06Exponential and sinusoidal models Paper HLMedium8 marks

A rumour spreads through a town of 4000 people. The number of people, N, who have heard the rumour t days after it starts is modelled by
$N(t) = \frac{4000}{1 + 79e^{-0.6t}}$.

(a)

Find the number of people who knew the rumour at the start.

(1)

(b)

Find the number of people who have heard the rumour after 5 days.

(2)

(c)

Find the time taken for 3000 people to have heard the rumour.

(3)

(d)

Write down the long-term number of people who will have heard the rumour according to this model, and interpret this value in context.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: $N(0) = \frac{4000}{80} = 50$ people	✓ 1
Part (b)
M1: $\frac{4000}{1 + 79e^{-3}}$	✓ 1
A1: 811 people	✓ 1
Part (c)
M1: $\frac{4000}{1 + 79e^{-0.6t}} = 3000$	✓ 1
M1: rearranging to $e^{-0.6t} = \frac{1}{237}$ or using GDC solver/graph intersection	✓ 1
A1: 9.11 days	✓ 1
Part (d)
A1: 4000	✓ 1
R1: eventually (almost) everyone in the town hears the rumour, but the number never exceeds the town population	✓ 1

Exam tip: For a logistic model L/(1 + Ce^{-kt}), the numerator L is the carrying capacity; recognise it directly rather than recalculating.

+ Add to test✓ In your test81IBMI-031-07-09Exponential and sinusoidal models Paper HLMedium7 marks

A pharmacologist measures the mass, M mg, of a drug remaining in a patient's bloodstream t hours after it is given. A graph of ln M against t is approximately a straight line with equation
$\ln M = 4.38 - 0.0275t$.

(a)

Show that M can be written in the form $M = Ae^{kt}$, and find the values of A and k.

(3)

(b)

Find the time taken for the mass of drug in the bloodstream to halve.

(2)

(c)

Find the mass of drug remaining 48 hours after it is given.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $M = e^{4.38 - 0.0275t} = e^{4.38}\times e^{-0.0275t}$	✓ 1
A1: $A = 79.8$ (79.838…)	✓ 1
A1: $k = -0.0275$	✓ 1
Part (b)
M1: $e^{-0.0275t} = 0.5$	✓ 1
A1: 25.2 hours	✓ 1
Part (c)
M1: $79.838\ldots \times e^{-0.0275\times 48}$ or $e^{4.38 - 1.32}$	✓ 1
A1: 21.3 mg	✓ 1

Exam tip: A straight line on a graph of ln y against x signals an exponential model: the gradient is k and the intercept is ln A.

+ Add to test✓ In your test82IBMI-031-03-r2-05Choosing and evaluating models Paper HLMedium10 marks

A biologist measures the resting metabolic rate, $P$ watts, and the body mass, $m$ kg, of mammals ranging from 0.02 kg to 70 kg. She plots $\log_{10} P$ against $\log_{10} m$. The points lie close to the straight line
$\log_{10} P = 0.75\log_{10} m + 0.55$.

(a)

Show that the relationship can be written in the form $P = a m^b$. Find the value of $a$ and the value of $b$.

(3)

(b)

Use the model to estimate the metabolic rate of a 60 kg mammal.

(2)

(c)

A colleague suggests an exponential model, $P = ke^{rm}$. Explain why the straight line on the log–log graph supports the power model rather than the exponential model.

(2)

(d)

An elephant has a mass of 5000 kg. Use the model to estimate its metabolic rate, and comment on the reliability of this estimate.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $P = 10^{0.75\log_{10} m + 0.55} = 10^{0.55} \times m^{0.75}$	✓ 1
A1 $b = 0.75$	✓ 1
A1 $a = 10^{0.55} = 3.55$	✓ 1
Part (b)
M1 $3.548 \times 60^{0.75}$	✓ 1
A1 76.5 W	✓ 1
Part (c)
R1 a power model gives a straight line when $\log P$ is plotted against $\log m$	✓ 1
R1 an exponential model would give a straight line when $\log P$ is plotted against $m$ itself, so a linear log–log graph indicates a power relationship	✓ 1
Part (d)
M1 $3.548 \times 5000^{0.75}$	✓ 1
A1 2110 W	✓ 1
R1 unreliable because it is an extrapolation far beyond the data range of 0.02 kg to 70 kg	✓ 1

Common error: The intercept 0.55 is log₁₀ a, not a itself. You must calculate 10^0.55 to find a.

+ Add to test✓ In your test83IBMI-031-04-r1-04Transformations of graphs and logarithmic scales Paper HLMedium8 marks

An ecologist studies the resting metabolic rate, R watts, of mammals with body mass M kg, for masses between 0.02 kg and 4000 kg. A graph of log₁₀ R against log₁₀ M is approximately linear with equation
$\log_{10} R = 0.74\log_{10} M + 0.53$.

(a)

Show that R can be written in the form $R = aM^{b}$, stating the values of a and b.

(3)

(b)

Estimate the resting metabolic rate of a 65 kg mammal.

(2)

(c)

Find the factor by which the body mass of a mammal must increase for its resting metabolic rate to double.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: writes $R = 10^{0.74\log_{10}M + 0.53} = 10^{0.53}\times10^{0.74\log_{10}M}$	✓ 1
M1: uses $10^{0.74\log_{10}M} = M^{0.74}$	✓ 1
A1: a = 3.39 (3.388…), b = 0.74	✓ 1
Part (b)
M1: substitutes, $R = 3.388\times65^{0.74}$	✓ 1
A1: R = 74.4 W	✓ 1
Part (c)
M1: sets up $a(kM)^{0.74} = 2aM^{0.74}$	✓ 1
M1: simplifies to $k^{0.74} = 2$	✓ 1
A1: k = 2.55	✓ 1

Common error: Doubling the rate does not mean doubling the mass. Because R is proportional to M⁰.⁷⁴, the mass must be multiplied by 2¹/⁰.⁷⁴.

+ Add to test✓ In your test84IBMI-031-08-r1-05Choosing and evaluating models Paper HLMedium7 marks

A biologist studies the resting metabolic rate $P$ watts of mammals of mass $M$ kg, for masses between 0.02 kg and 4000 kg. Plotting $\log_{10}P$ against $\log_{10}M$ gives points lying close to the straight line
$\log_{10}P = 0.75\log_{10}M + 0.55$.

(a)

Show that the relationship can be written in the form $P = kM^{n}$, stating the values of $n$ and $k$.

(3)

(b)

Use the model to estimate the resting metabolic rate of a 70 kg mammal.

(2)

(c)

A student uses the model to estimate the metabolic rate of a 0.002 kg insect. Give two reasons why this estimate may be unreliable.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses laws of logs: $\log_{10}P = \log_{10}(M^{0.75}) + \log_{10}(10^{0.55})$, so $P = 10^{0.55}M^{0.75}$ M1	✓ 1
$n = 0.75$ A1	✓ 1
$k = 10^{0.55} = 3.55$ A1	✓ 1
Part (b)
Substitutes $M = 70$: $3.548 \times 70^{0.75}$ M1	✓ 1
$P = 85.9$ W A1	✓ 1
Part (c)
Extrapolation: 0.002 kg lies outside the data range 0.02–4000 kg R1	✓ 1
Insects are not mammals, so the relationship found for mammals may not apply R1	✓ 1

Exam tip: On a log–log graph the gradient gives the power $n$ and the intercept gives $\log_{10}k$, not $k$ itself.

+ Add to test✓ In your test85IBMI-031-04-r2-03Transformations of graphs and logarithmic scales Paper HLMedium13 marks

A biologist studies the resting metabolic rate $P$ watts of mammals of mass $M$ kg, for masses between 3 kg and 320 kg. She proposes a model of the form $P=aM^{b}$, where $a$ and $b$ are constants.

(a)

Show that $\log_{10}P=b\log_{10}M+\log_{10}a$.

(2)

(b)

She plots $\log_{10}P$ against $\log_{10}M$. Her line of best fit passes through the points $(0.5,\,0.919)$ and $(2.5,\,2.403)$. Find the gradient and the vertical-axis intercept of this line.

(3)

(c)

Hence find the value of $a$ and write down the value of $b$.

(2)

(d)

Use the model to estimate the metabolic rate of a 60 kg mammal.

(2)

(e)

Find the mass of a mammal whose metabolic rate is predicted to be 250 W.

(3)

(f)

Comment on the reliability of using this model to predict the metabolic rate of a 4000 kg elephant.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: takes $\log_{10}$ of both sides: $\log_{10}P=\log_{10}(aM^{b})$	✓ 1
A1: uses product and power laws to get $\log_{10}a+b\log_{10}M$, AG	✓ 1
Part (b)
M1: gradient $=\frac{2.403-0.919}{2.5-0.5}$	✓ 1
A1: gradient $=0.742$	✓ 1
A1: intercept $=0.919-0.742(0.5)=0.548$	✓ 1
Part (c)
M1: $a=10^{0.548}$	✓ 1
A1: $a=3.53$ and $b=0.742$	✓ 1
Part (d)
M1: $3.532\times60^{0.742}$	✓ 1
A1: 73.7 W	✓ 1
Part (e)
M1: $250=3.532M^{0.742}$	✓ 1
M1: $M=\left(\frac{250}{3.532}\right)^{1/0.742}$ or solves with GDC	✓ 1
A1: 311 kg	✓ 1
Part (f)
R1: unreliable, because 4000 kg is far outside the 3–320 kg data range, so it would be extrapolation	✓ 1

Exam tip: On a log–log graph the gradient is the power $b$ directly, but the intercept is $\log_{10}a$. Raise 10 to the intercept to find $a$, and keep unrounded values for later parts.

+ Add to test✓ In your test86IBMI-031-04-r2-04Transformations of graphs and logarithmic scales Paper HLMedium6 marks

The depth of water at a harbour entrance is modelled by $f(t)=2.4\sin(30t)+5$, where $t$ is the time in hours after midnight and $f(t)$ is the depth in metres. The argument of sine is in degrees.
At a second harbour the depth is modelled by $g(t)=1.25\,f(t-1.5)$.

(a)

Describe fully the two transformations that map the graph of $f$ onto the graph of $g$.

(2)

(b)

Write $g(t)$ in the form $a\sin(b(t-c))+d$.

(2)

(c)

Find the maximum depth at the second harbour and the first time after midnight at which it occurs.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: horizontal translation 1.5 units to the right, i.e. 1.5 hours later	✓ 1
A1: vertical stretch with scale factor 1.25	✓ 1
Part (b)
M1: $1.25\times2.4$ and $1.25\times5$ seen	✓ 1
A1: $g(t)=3\sin(30(t-1.5))+6.25$	✓ 1
Part (c)
A1: maximum depth $3+6.25=9.25$ m	✓ 1
A1: $30(t-1.5)=90$, so $t=4.5$, i.e. 04:30	✓ 1

Common error: A vertical stretch multiplies the principal axis $d$ as well as the amplitude. Only the amplitude changing would give 8 m as the maximum, which is wrong.

+ Add to test✓ In your test87IBMI-031-04-r2-05Transformations of graphs and logarithmic scales Paper HLMedium10 marks

The magnitude $M$ of an earthquake is defined by $M=\log_{10}\left(\frac{A}{A_0}\right)$, where $A$ is the amplitude recorded by a seismograph and $A_0$ is a fixed reference amplitude.
The energy $E$ joules released by an earthquake of magnitude $M$ is modelled by $\log_{10}E=4.8+1.5M$.

(a)

Two earthquakes have magnitudes 6.3 and 4.1. Find how many times larger the recorded amplitude of the first is than the amplitude of the second.

(2)

(b)

Find the energy released by the magnitude 6.3 earthquake.

(2)

(c)

Show that an increase of 1 in magnitude multiplies the energy released by approximately 31.6.

(2)

(d)

An earthquake releases $5.0\times10^{15}$ J of energy. Find its magnitude.

(2)

(e)

A poster shows energies on a logarithmic axis. The axis starts at $10^{10}$ J, and each power of 10 is 2 cm long. Find the distance from the start of the axis at which the earthquake releasing $5.0\times10^{15}$ J should be plotted.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $\frac{A_1}{A_2}=10^{6.3-4.1}=10^{2.2}$	✓ 1
A1: 158 (accept 158.5)	✓ 1
Part (b)
M1: $E=10^{4.8+1.5(6.3)}=10^{14.25}$	✓ 1
A1: $1.78\times10^{14}$ J	✓ 1
Part (c)
M1: $\frac{10^{4.8+1.5(M+1)}}{10^{4.8+1.5M}}$, or subtracting the logs to get a difference of 1.5	✓ 1
A1: $=10^{1.5}=31.62\ldots\approx31.6$, AG	✓ 1
Part (d)
M1: $1.5M=\log_{10}(5.0\times10^{15})-4.8$	✓ 1
A1: $M=7.27$	✓ 1
Part (e)
M1: $(\log_{10}(5.0\times10^{15})-10)\times2$	✓ 1
A1: 11.4 cm	✓ 1

Exam tip: On a logarithmic axis, distance is proportional to the logarithm of the value. Find how many powers of ten the value lies above the start, then multiply by the length of one power.

+ Add to test✓ In your test88IBMI-031-05-r1-05Linear models and piecewise models Paper HLMedium8 marks

A biologist studies the resting metabolic rate, P watts, of mammals with mass m kg. For mammals with masses between 0.02 kg and 100 kg, a plot of $\ln P$ against $\ln m$ is a straight line with equation
$\ln P = 0.75\ln m + 1.2$.

(a)

Show that the model can be written as $P=am^n$, and find the values of a and n.

(3)

(b)

Use the model to predict the metabolic rate of a mammal of mass 65 kg.

(2)

(c)

Find the mass of a mammal predicted to have a metabolic rate of 500 W.

(2)

(d)

Comment on the reliability of the mass you found for a metabolic rate of 500 W.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Exponentiates both sides, e.g. $P=e^{0.75\ln m+1.2}=e^{1.2}m^{0.75}$ M1 ; $n=0.75$ A1 ; $a=e^{1.2}=3.32$ A1	✓ 3
Part (b)
$P=3.3201\times65^{0.75}$ (or $\ln P=0.75\ln65+1.2$) M1 ; $P=76.0$ W A1	✓ 2
Part (c)
$\ln m=\frac{\ln500-1.2}{0.75}$ (or $500=3.3201m^{0.75}$) M1 ; $m=801$ kg A1	✓ 2
Part (d)
Unreliable, as 801 kg lies outside the 0.02–100 kg range of the data (extrapolation) R1	✓ 1

Common error: The intercept of the log–log line is $\ln a$, not $a$. Leaving $a=1.2$ gives predictions that are wrong by a factor of about 3.

+ Add to test✓ In your test89IBMI-031-04-r1-05Transformations of graphs and logarithmic scales Paper HLMedium8 marks

The depth of water, D metres, at the entrance to a harbour t hours after midnight is modelled by a function of the form $D(t) = a\sin\left(b(t-c)\right) + d$, where angles are measured in degrees.
The graph of D is obtained from the graph of $y = \sin t$ by a vertical stretch with scale factor 1.8, a horizontal stretch with scale factor $\frac{1}{30}$, a horizontal translation of 2 units to the right and a vertical translation of 5.5 units up.

(a)

Write down an expression for D(t).

(3)

(b)

Find the maximum depth of water and the first time after midnight at which it occurs.

(2)

(c)

A cargo boat can only enter the harbour when the depth of water is at least 6.5 m. Find the length of time between 02:00 and 14:00 for which the boat can enter. Give your answer in hours to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: a = 1.8 and d = 5.5	✓ 1
A1: b = 30	✓ 1
A1: c = 2, giving $D(t) = 1.8\sin(30(t-2)) + 5.5$	✓ 1
Part (b)
A1: maximum depth 7.3 m	✓ 1
A1: occurs when 30(t − 2) = 90, i.e. t = 5 (05:00)	✓ 1
Part (c)
M1: sets $1.8\sin(30(t-2)) + 5.5 = 6.5$ or uses GDC intersection	✓ 1
A1: t = 3.12 and t = 6.88 (3.12497… and 6.87503…)	✓ 1
A1: duration = 6.87503… − 3.12497… = 3.75 hours . Accept 3.76 if it comes from subtracting the rounded values 6.88 and 3.12.	✓ 1

Common error: Subtracting rounded intersection times (6.88 − 3.12) gives 3.76. Keep the full GDC values until the final step to get 3.75 hours.

+ Add to test✓ In your test90IBMI-031-03-r1-05Choosing and evaluating models Paper HLMedium8 marks

A biologist records the number of bacteria, $N$, in a culture every hour for $0 \le t \le 8$, where $t$ is the time in hours. She plots $\ln N$ against $t$ and finds that the points lie close to a straight line. The regression line is
$\ln N = 0.231t + 4.62$.
She wants to model the population with $N = Ae^{kt}$.

(a)

Find the value of A and the value of k.

(3)

(b)

Use the model to estimate the number of bacteria when t = 12.

(2)

(c)

Find the time it takes for the number of bacteria to double, according to the model.

(2)

(d)

Comment on the reliability of your estimate in part (b).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 recognising ln N = ln A + kt, so ln A = 4.62	✓ 1
A1 A = e^4.62 = 101	✓ 1
A1 k = 0.231	✓ 1
Part (b)
M1 substituting t = 12, e.g. e^{0.231×12 + 4.62} or 101e^0.231×12	✓ 1
A1 N = 1620 (accept answers from 1610 to 1630, e.g. 1615 from using A = 101)	✓ 1
Part (c)
M1 setting up e^0.231t = 2 or t = ln 2 / 0.231	✓ 1
A1 3.00 hours	✓ 1
Part (d)
R1 unreliable, with either of these reasons: t = 12 is outside the data range 0 ≤ t ≤ 8 (extrapolation), or the growth may not stay exponential because nutrients or space run out	✓ 1

Exam tip: A straight line on a graph of ln N against t shows exponential growth: the gradient gives k and the intercept gives ln A, not A.

+ Add to test✓ In your test91IBMI-031-04-r1-03Transformations of graphs and logarithmic scales Paper HLMedium8 marks

A biologist records the number of bacteria, N, in a culture every hour for the first 12 hours of an experiment. She plots ln N against time t hours and finds that the points lie close to the straight line
$\ln N = 0.231t + 4.62$.

(a)

Show that this relationship can be written in the form $N = a\mathrm{e}^{kt}$, and state the values of a and k.

(3)

(b)

Use the model to find the time at which the number of bacteria reaches 1000.

(2)

(c)

Find the time taken for the number of bacteria to double.

(2)

(d)

Comment on the reliability of using the model to predict the number of bacteria after 60 hours.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: exponentiates both sides, $N = \mathrm{e}^{0.231t + 4.62}$	✓ 1
M1: uses laws of indices, $N = \mathrm{e}^{4.62}\times\mathrm{e}^{0.231t}$	✓ 1
A1: a = 101 (101.49…), k = 0.231	✓ 1
Part (b)
M1: sets up $0.231t + 4.62 = \ln 1000$ or $101.49\mathrm{e}^{0.231t} = 1000$	✓ 1
A1: t = 9.90 hours (9.904…)	✓ 1
Part (c)
M1: uses $\mathrm{e}^{0.231T} = 2$ or equivalent, e.g. $T = \frac{\ln 2}{0.231}$	✓ 1
A1: T = 3.00 hours (3.0006…)	✓ 1
Part (d)
R1: states that the prediction is unreliable, with a valid reason: either 60 hours is far outside the 12-hour data range (extrapolation), or limited nutrients and space would slow growth so exponential growth cannot continue	✓ 1

Exam tip: The vertical intercept of a ln N against t line is ln a, not a. Remember to exponentiate it to find the initial value.

+ Add to test✓ In your test92IBMI-031-08-r2-05Choosing and evaluating models Paper HLMedium8 marks

A biologist studies the resting metabolic rate, $P$ watts, of mammals with mass $m$ kg. Data were collected for mammals with masses between 0.02 kg and 60 kg. When $\log_{10}P$ is plotted against $\log_{10}m$ the points lie close to a straight line, and the regression line is
$\log_{10}P=0.75\log_{10}m+0.55$.

(a)

Show that the relationship can be written in the form $P=km^n$, and find the values of $k$ and $n$.

(3)

(b)

According to the model, find the factor by which the metabolic rate increases when the mass of a mammal doubles.

(2)

(c)

Use the model to estimate the metabolic rate of a 250 kg mammal.

(2)

(d)

Comment on the reliability of your estimate for the 250 kg mammal.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $P=10^{0.75\log_{10}m+0.55}=10^{0.55}\times(10^{\log_{10}m})^{0.75}=10^{0.55}\times m^{0.75}$	✓ 1
A1: $n=0.75$	✓ 1
A1: $k=10^{0.55}=3.55$ (3.548…)	✓ 1
Part (b)
M1: $\frac{k(2m)^{0.75}}{km^{0.75}}=2^{0.75}$	✓ 1
A1: 1.68	✓ 1
Part (c)
M1: $3.548\times250^{0.75}$	✓ 1
A1: $P=223$ W	✓ 1
Part (d)
R1: unreliable because 250 kg is outside the data range (0.02–60 kg), so this is extrapolation	✓ 1

Exam tip: A straight line on a log–log graph means a power model $y=kx^n$; a straight line on a semi-log graph (log y against x) means an exponential model. Identify which before rearranging.

+ Add to test✓ In your test93IBMI-031-02-r2-05Exponential and sinusoidal models Paper HLMedium8 marks

A biologist measures the number of bacteria, $N$, in a culture at regular intervals over the first 10 hours of an experiment. She plots $\ln N$ against time $t$ hours and finds that the points lie close to the straight line
$\ln N = 0.231t + 6.12$.

(a)

Show that the data can be modelled by $N = Ae^{kt}$, and find the values of $A$ and $k$.

(3)

(b)

Find the time taken for the number of bacteria to double.

(2)

(c)

Use the model to estimate the time at which the number of bacteria reaches 1.0 × 10⁶.

(2)

(d)

Comment on the reliability of your estimate of the time at which the number of bacteria reaches 1.0 × 10⁶.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Exponentiates both sides: N = e^{0.231t + 6.12} = e^6.12 × e^0.231t, which is of the form Ae^kt	✓ 1
A = e^6.12 = 455 (454.86…)	✓ 1
k = 0.231	✓ 1
Part (b)
Solves e^0.231t = 2, i.e. t = ln 2 ÷ 0.231	✓ 1
t = 3.00 hours	✓ 1
Part (c)
Sets up 0.231t + 6.12 = ln(10⁶) or 455e^0.231t = 10⁶	✓ 1
t = 33.3 hours	✓ 1
Part (d)
Unreliable, with either valid reason: 33.3 h is far outside the 0–10 h data range (extrapolation), OR growth is likely to slow as nutrients or space run out	✓ 1

Exam tip: The gradient of a ln N against t graph gives k directly, but the intercept is ln A, not A. Exponentiate the intercept before writing down A.

+ Add to test✓ In your test94IBMI-031-01-r2-05Quadratic and cubic models Paper HLMedium12 marks

The cross-section of a footbridge arch is modelled by $f(x) = -0.05x^2 + 2x$, for 0 ≤ x ≤ 40, where x and f(x) are measured in metres from a point at ground level.
A larger arch for a second bridge is modelled by
$g(x) = 0.75\,f\left(\frac{x-5}{1.5}\right)$.

(a)

Find the coordinates of the highest point of the arch modelled by f.

(2)

(b)

Describe fully a sequence of transformations that maps the graph of f onto the graph of g.

(3)

(c)

Hence find the coordinates of the highest point of the arch modelled by g.

(2)

(d)

Express g(x) in the form $ax^2 + bx + c$.

(3)

(e)

A walkway is placed horizontally inside the larger arch, 9 m above the ground, with each end touching the arch. Find the length of the walkway.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$x = -\frac{2}{2(-0.05)} = 20$, or use of a GDC (M1)	✓ 1
(20, 20) (A1)	✓ 1
Part (b)
Horizontal stretch, scale factor 1.5 (A1)	✓ 1
Vertical stretch, scale factor 0.75 (A1)	✓ 1
Then a translation of 5 units in the positive x-direction, i.e. by the vector (5, 0), applied after the horizontal stretch (A1)	✓ 1
Accept any correct equivalent sequence, e.g. a translation by $\binom{10/3}{0}$ followed by a horizontal stretch of scale factor 1.5 about the y-axis, with the vertical stretch at any stage.	—
Part (c)
Applies the transformations to (20, 20): x = 1.5 × 20 + 5, y = 0.75 × 20 (M1)	✓ 1
(35, 15) (A1)	✓ 1
Part (d)
Substitutes: $0.75\left[-0.05\frac{(x-5)^2}{2.25} + 2\cdot\frac{x-5}{1.5}\right]$ (M1)	✓ 1
Simplifies to $-\frac{1}{60}(x-5)^2 + (x-5)$ and expands (M1)	✓ 1
$g(x) = -\frac{1}{60}x^2 + \frac{7}{6}x - \frac{65}{12}$ (A1)	✓ 1
Part (e)
Solves g(x) = 9 (e.g. by GDC), giving x = 16.03 and x = 53.97, i.e. $x = 35 \pm \sqrt{360}$ (M1)	✓ 1
Length = $2\sqrt{360}$ = 37.9 m (A1)	✓ 1
Use unrounded roots: subtracting the 3 s.f. values 16.0 and 54.0 gives 38.0, which is inaccurate.	—

Common error: Replacing x by x/1.5 stretches the graph horizontally by factor 1.5, not 1/1.5. Inside the bracket, the effect on the graph is the inverse of the operation applied to x.

+ Add to test✓ In your test95IBMI-031-00-12Linear models and piecewise models Paper HLHard12 marks

The battery level, B %, of a phone t hours after it is unplugged is modelled by
$B(t)=\begin{cases}100-12t, & 0\le t\le 5\\ 40+30(t-5), & 5where the phone is in use for the first 5 hours and then placed on charge.
A newer model of phone uses power more efficiently, so that each stage takes 25% longer. Its battery level is modelled by \(N(t)=B\left(\frac{t}{1.25}\right)$.

(a)

Find B(3).

(1)

(b)

Find all the times at which the battery level of the original phone is 70%.

(3)

(c)

Describe the transformation that maps the graph of B onto the graph of N.

(2)

(d)

Write down N(t) as a piecewise function and state its domain.

(4)

(e)

Find how much later the newer phone first reaches a battery level of 52% than the original phone.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 64 (%)	✓ 1
Part (b)
M1 equating either piece to 70	✓ 1
A1 t = 2.5 hours	✓ 1
A1 t = 6 hours	✓ 1
Part (c)
A1 horizontal stretch (parallel to the t-axis)	✓ 1
A1 scale factor 1.25	✓ 1
Part (d)
A1 N(t) = 100 − 9.6t	✓ 1
A1 N(t) = 40 + 24(t − 6.25) or 24t − 110	✓ 1
A1 pieces valid for 0 ≤ t ≤ 6.25 and 6.25 < t ≤ 8.75	✓ 1
A1 domain 0 ≤ t ≤ 8.75	✓ 1
Part (e)
M1 100 − 12t = 52 gives t = 4 and 100 − 9.6t = 52 gives t = 5	✓ 1
A1 1 hour later	✓ 1

Common error: Replacing t with t/1.25 stretches the graph by factor 1.25, not 1/1.25; the newer phone takes longer, so its graph is wider.

+ Add to test✓ In your test96IBMI-031-02-r1-06Exponential and sinusoidal models Paper HLHard12 marks

An ecologist records the number of grey seals, N, on an island, where t is the number of years after 1 January 2000. A plot of ln N against t is approximately linear, and the regression line is
$\ln N = 0.0815t + 6.42$.

(a)

Show that the model can be written in the form $N = a e^{bt}$, and find the values of a and b.

(3)

(b)

Find the time taken for the seal population to double, according to this model.

(2)

(c)

A report requires the model to use s, the number of years after 1 January 2010.
(i) Write the model in the form $N = c\,e^{0.0815s}$, giving the value of c.
(ii) Describe the single transformation that maps the graph of N against t onto the graph of N against s.

(3)

(d)

A second species, harbour seals, has population M on the same island, modelled by $\ln M = -0.046t + 8.10$. Find the value of t when the two populations are predicted to be equal.

(3)

(e)

Give one reason why the model for N is unlikely to remain valid for many decades.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Exponentiates both sides: N = e^{0.0815t + 6.42} = e^6.42 × e^0.0815t (M1)	✓ 1
a = e^6.42 = 614 (A1)	✓ 1
b = 0.0815 (A1)	✓ 1
Part (b)
Sets up e^0.0815t = 2, or equivalent (M1)	✓ 1
t = ln 2 / 0.0815 = 8.50 years (A1)	✓ 1
Part (c)
(i) Substitutes t = s + 10: N = 614e^{0.0815(s + 10)} = 614e^0.815e^0.0815s (M1)	✓ 1
c = 1390 (1387.14…) (A1)	✓ 1
(ii) Horizontal translation of 10 units in the negative direction (to the left), or equivalently a vertical stretch with scale factor e^0.815 ≈ 2.26 (A1)	✓ 1
Part (d)
Equal populations means ln N = ln M (M1)	✓ 1
0.0815t + 6.42 = −0.046t + 8.10, so 0.1275t = 1.68 (A1)	✓ 1
t = 13.2 years (A1)	✓ 1
Part (e)
Valid reason in context, e.g. exponential growth is unbounded, but the island has limited space and food, so growth must slow down (R1)	✓ 1

Exam tip: If ln N is linear in t, then N is exponential in t: the gradient gives b and the intercept gives ln a. Comparing two populations is easier on the log scale, because two straight lines can be equated directly.

+ Add to test✓ In your test97IBMI-031-04-r1-06Transformations of graphs and logarithmic scales Paper HLHard16 marks

A cup of coffee is left to cool in a room. Its temperature, T °C, t minutes after it is poured is modelled by
$T(t) = 70\mathrm{e}^{-0.05t} + 20$, $t \geq 0$.

(a)

Describe a sequence of transformations that maps the graph of $y = \mathrm{e}^{x}$ onto the graph of T.

(3)

(b)

A café displays temperatures in degrees Fahrenheit, using $F = 1.8T + 32$. Find an expression for the temperature of the coffee, F(t) °F, and describe the transformations that map the graph of T onto the graph of F.

(3)

(c)

The café recommends that coffee is drunk once it has cooled to 140 °F. Find the time taken for the coffee to reach this temperature.

(3)

(d)

Show that a graph of $\ln(T - 20)$ against t is a straight line, and state its gradient and vertical intercept.

(3)

(e)

A second cup of coffee in the same room gives data for which the line of best fit is $\ln(T - 20) = 4.38 - 0.062t$.
(i) Find the initial temperature of the second cup.
(ii) Find the time taken for the second cup to cool to 40 °C.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: reflection in the vertical axis and horizontal stretch with scale factor 20	✓ 1
A1: vertical stretch with scale factor 70	✓ 1
A1: followed by a vertical translation of 20 units up (stretch before translation)	✓ 1
Part (b)
M1: substitutes, $F(t) = 1.8(70\mathrm{e}^{-0.05t} + 20) + 32$	✓ 1
A1: $F(t) = 126\mathrm{e}^{-0.05t} + 68$	✓ 1
A1: vertical stretch scale factor 1.8 followed by vertical translation 32 units up	✓ 1
Part (c)
M1: sets $126\mathrm{e}^{-0.05t} + 68 = 140$ (or equivalent 60 °C in T)	✓ 1
M1: rearranges to $\mathrm{e}^{-0.05t} = \frac{4}{7}$ or uses GDC	✓ 1
A1: t = 11.2 minutes	✓ 1
Part (d)
M1: $T - 20 = 70\mathrm{e}^{-0.05t}$ so $\ln(T-20) = \ln 70 - 0.05t$, which is linear in t	✓ 1
A1: gradient −0.05	✓ 1
A1: vertical intercept ln 70 = 4.25	✓ 1
Part (e)
(i) M1: at t = 0, $T = \mathrm{e}^{4.38} + 20$	✓ 1
A1: T = 99.8 °C	✓ 1
(ii) M1: sets $4.38 - 0.062t = \ln 20$	✓ 1
A1: t = 22.3 minutes	✓ 1

Common error: When linearising, subtract the room temperature before taking logarithms: ln T is not linear in t for this model, but ln(T − 20) is.

+ Add to test✓ In your test98IBMI-031-01-r1-06Quadratic and cubic models Paper HLHard13 marks

The profile of a children's slide is modelled by $f(x) = ax^3 + bx^2 + cx + d$ for 0 ≤ x ≤ 5, where x is the horizontal distance in metres from the top of the slide and f(x) is the height in metres above the ground.
The slide starts horizontally at a height of 4 m at x = 0, and finishes horizontally at ground level at x = 5.

(a)

Find the values of a, b, c and d.

(4)

(b)

Find the horizontal distance from the top at which the slide is steepest, and the angle the slide makes with the horizontal at this point.

(4)

(c)

A redesigned slide has profile $g(x) = f\left(\frac{x}{1.5}\right)$. Describe fully the single transformation that maps the graph of f onto the graph of g.

(2)

(d)

A safety rule states that a slide must make an angle of no more than 40° with the horizontal at any point. A slide is designed with profile $h(x) = f\left(\frac{x}{k}\right)$, where k > 1. Find the least value of k for which this slide meets the safety rule.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: f(0) = 4 gives d = 4, and f '(0) = 0 gives c = 0	✓ 1
M1: differentiates: f '(x) = 3ax² + 2bx + c	✓ 1
M1: forms 125a + 25b + 4 = 0 and 75a + 10b = 0	✓ 1
A1: a = 0.064, b = −0.48	✓ 1
Part (b)
M1: sets f ''(x) = 0.384x − 0.96 = 0, or finds the minimum of f '(x) on GDC	✓ 1
A1: x = 2.5 m	✓ 1
M1: f '(2.5) = 0.192(2.5)² − 0.96(2.5) = −1.2	✓ 1
A1: angle = tan⁻¹(1.2) = 50.2°	✓ 1
Part (c)
A1: horizontal stretch (parallel to the x-axis)	✓ 1
A1: scale factor 1.5	✓ 1
Part (d)
M1: recognises the steepest gradient of h is $\frac{1}{k}f'(2.5) = -\frac{1.2}{k}$	✓ 1
M1: sets $\frac{1.2}{k} \le \tan 40°$	✓ 1
A1: k = 1.43 (1.4301…)	✓ 1

Exam tip: A horizontal stretch by scale factor k leaves heights unchanged but divides every gradient by k, so the steepest point moves to x = 2.5k while its gradient becomes f′(2.5)/k.

+ Add to test✓ In your test99IBMI-031-02-r2-06Exponential and sinusoidal models Paper HLHard13 marks

The depth of water, $d$ metres, at the entrance to a harbour is modelled by
$d(t) = 2.6\cos(29t) + 7.4$,
where $t$ is the time in hours after midnight and angles are in degrees. A cargo ship can only pass through the entrance when the depth of water is at least 9.0 m.

(a)

Find the period of the model, giving your answer in hours and minutes to the nearest minute.

(2)

(b)

Find the total length of time, in each tidal cycle, for which the ship can pass through the entrance.

(4)

(c)

A storm surge raises the water level by 0.8 m at all times and delays each high tide by 1.5 hours. Write down the new model, $d_2(t)$, for the depth of water.

(2)

(d)

Using $d_2$, find the first time after midnight at which the depth falls to 9.0 m. Give your answer as a time of day to the nearest minute.

(3)

(e)

Using $d_2$, find the percentage of each tidal cycle during which the ship can pass through the entrance.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Period = 360 ÷ 29 = 12.414 hours (M1)	✓ 1
12 hours 25 minutes (A1)	✓ 1
Part (b)
Sets up 2.6cos(29t) + 7.4 = 9.0, i.e. cos(29t) = 0.615… (M1)	✓ 1
t = 1.79 h (1.7938…) (A1)	✓ 1
t = 10.6 h (10.620…) (A1)	✓ 1
Total time = 2 × 1.7938 = 3.59 hours (A1)	✓ 1
Part (c)
Horizontal translation: cos(29(t − 1.5)) (A1)	✓ 1
d₂(t) = 2.6cos(29(t − 1.5)) + 8.2 (A1)	✓ 1
Part (d)
Sets up 2.6cos(29(t − 1.5)) + 8.2 = 9.0 (M1)	✓ 1
t = 3.99 hours (3.9855…) (A1)	✓ 1
03:59 (A1)	✓ 1
Part (e)
Time per cycle with d ≥ 9.0: 2 × 2.4855 = 4.971 h, divided by period 12.414 h (M1)	✓ 1
40.0% (A1)	✓ 1

Common error: A delay of 1.5 hours is a translation in the positive t-direction, so replace t with (t − 1.5), not (t + 1.5). Check by confirming that the new maximum occurs at t = 1.5.

+ Add to test✓ In your test100IBMI-031-08-r1-06Choosing and evaluating models Paper HLHard13 marks

A conservation group introduces 400 fish into a lake. Three years later a survey estimates there are 1100 fish. Two models are proposed for the number of fish $N$, $t$ years after introduction.
Model A: $N = 400e^{kt}$
Model B: $N = \dfrac{6000}{1 + 14e^{-rt}}$
Both models are fitted to the survey data at $t = 0$ and $t = 3$.

(a)

Find the value of $k$.

(2)

(b)

Verify that Model B gives $N = 400$ when $t = 0$, and show that $r = 0.382$ to 3 significant figures.

(3)

(c)

Find the number of fish predicted by each model when $t = 8$.

(3)

(d)

A survey at $t = 8$ estimates 3900 fish. State, with a reason, which model is more appropriate.

(2)

(e)

Interpret the value 6000 in Model B in the context of the question.

(1)

(f)

Using Model B, find the time at which the fish population reaches 5000.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$400e^{3k} = 1100$ M1	✓ 1
$k = \frac{1}{3}\ln 2.75 = 0.337$ A1	✓ 1
Part (b)
$\frac{6000}{1+14} = 400$ A1	✓ 1
Sets $\frac{6000}{1 + 14e^{-3r}} = 1100$, giving $e^{-3r} = \frac{7}{22}$ M1	✓ 1
$r = \frac{1}{3}\ln\frac{22}{7} = 0.3817… = 0.382$ A1	✓ 1
Part (c)
Substitutes $t = 8$ into either model M1	✓ 1
Model A: $N = 5940$ (5937.6…) A1	✓ 1
Model B: $N = 3610$ (3613.2… using unrounded $r$; accept 3620 from using $r = 0.382$) A1	✓ 1
Part (d)
Compares both predictions with 3900, e.g. differences of about 2040 and 290 M1	✓ 1
Model B, because its prediction is much closer to the observed value R1	✓ 1
Part (e)
6000 is the carrying capacity: the maximum number of fish the lake can sustain in the long term R1	✓ 1
Part (f)
Solves $\frac{6000}{1 + 14e^{-rt}} = 5000$, giving $e^{-rt} = \frac{1}{70}$ M1	✓ 1
$t = \frac{\ln 70}{r} = 11.1$ years (unrounded $r$ or $r = 0.382$) A1	✓ 1

Exam tip: When choosing a model, quote both predictions next to the observed value and then state the conclusion. A conclusion with no comparison usually earns no reasoning mark. Keep unrounded parameter values in your calculator to avoid rounding drift.

+ Add to test✓ In your test101IBMI-031-06-r2-06Quadratic and cubic models Paper HLHard12 marks

The population of an insect species in Region A, in thousands, is modelled by
$f(x) = 0.5x^3 - 6x^2 + 18x + 40$, for $0 \leq x \leq 8$,
where x is the time in months after monitoring began.
In Region B the population, in thousands, follows a slower cycle and is modelled by
$g(x) = 1.5\,f(0.5x) - 12$, for $0 \leq x \leq 16$.

(a)

Find the coordinates of the local maximum and the local minimum of the graph of f.

(3)

(b)

Describe fully a sequence of transformations that maps the graph of f onto the graph of g.

(3)

(c)

Hence write down the coordinates of the local maximum and the local minimum of the graph of g.

(2)

(d)

Find the total length of time, within the 16 months, for which the population of Region B exceeds 60 thousand.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Use of GDC, or f′(x) = 1.5x² − 12x + 18 = 0 M1	✓ 1
Local maximum (2, 56) A1	✓ 1
Local minimum (6, 40) A1	✓ 1
Part (b)
Horizontal stretch with scale factor 2 (parallel to the x-axis), in any position in the sequence A1	✓ 1
Vertical stretch with scale factor 1.5 (parallel to the y-axis) A1	✓ 1
Vertical translation by the vector (0, −12) applied after the vertical stretch, OR a vertical translation by (0, −8) applied before the vertical stretch A1	✓ 1
Part (c)
Local maximum (4, 72) A1	✓ 1
Local minimum (12, 48) A1	✓ 1
Part (d)
g(x) > 60 ⇔ f(0.5x) > 48, or solving g(x) = 60 directly with GDC M1	✓ 1
x = 1.07, x = 8 and x = 14.9 (exact values 8 − 4√3, 8, 8 + 4√3) A1	✓ 1
g(x) > 60 for 1.07 < x < 8 and 14.9 < x ≤ 16, using the shape of the graph R1	✓ 1
Total time = 6.93 + 1.07 = 8.00 months A1	✓ 1

Exam tip: In y = a f(bx) + d, the horizontal scale factor is 1/b, not b. The order of the vertical stretch and the vertical translation matters. If you stretch first, translate by d. If you translate first, translate by d/a. State the order clearly.

+ Add to test✓ In your test102IBMI-031-04-r2-06Transformations of graphs and logarithmic scales Paper HLHard15 marks

The temperature of a cup of coffee $t$ minutes after it is poured is modelled by $f(t)=70e^{-0.05t}+20$, in °C.
A second cup, in an insulated mug in a warmer room, is modelled by $g(t)=f(0.8t)+4$.

(a)

Write down the equation of the horizontal asymptote of the graph of $f$ and interpret it in context.

(2)

(b)

Describe fully the transformations that map the graph of $f$ onto the graph of $g$.

(3)

(c)

Write $g(t)$ in the form $ae^{-kt}+c$.

(2)

(d)

Find the time taken for the second cup to cool to 60 °C, and hence find how much longer the second cup takes than the first cup to cool to 60 °C.

(5)

(e)

Find the maximum value of $g(t)-f(t)$ for $t\ge0$, and the time at which it occurs.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: $y=20$ (accept $T=20$ or $f(t)=20$)	✓ 1
R1: the coffee cools towards room temperature, 20 °C, but never falls below it	✓ 1
Part (b)
A1: horizontal stretch (parallel to the $t$-axis)	✓ 1
A1: scale factor $\frac{1}{0.8}=1.25$	✓ 1
A1: vertical translation 4 units up	✓ 1
Part (c)
M1: $70e^{-0.05(0.8t)}+20+4$	✓ 1
A1: $g(t)=70e^{-0.04t}+24$	✓ 1
Part (d)
M1: $70e^{-0.04t}+24=60$	✓ 1
A1: $t=\frac{\ln(70/36)}{0.04}=16.6$ minutes	✓ 1
M1: $70e^{-0.05t}+20=60$	✓ 1
A1: $t=\frac{\ln 1.75}{0.05}=11.2$ minutes	✓ 1
A1: $16.62-11.19=5.43$ minutes	✓ 1
Part (e)
M1: forms $70e^{-0.04t}-70e^{-0.05t}+4$ and uses a GDC maximum or sets the derivative to zero	✓ 1
A1: $t=100\ln1.25=22.3$ minutes	✓ 1
A1: maximum difference 9.73 °C	✓ 1

Common error: $f(0.8t)$ is a horizontal stretch with scale factor $\frac{1}{0.8}=1.25$, not 0.8. A factor of 0.8 would mean the second cup cools faster, which contradicts the insulated mug.

+ Add to test✓ In your test103IBMI-031-06-r1-06Quadratic and cubic models Paper HLHard15 marks

The profile of a section of a children's roller-coaster track is modelled by
$f(x) = -0.004x^3 + 0.12x^2$, for $0 \le x \le 30$,
where $f(x)$ is the height in metres above the ground at a horizontal distance x metres from the start.
A larger version of the ride is designed with profile $g(x) = p\,f(qx)$, where $p, q > 0$. The new track is the image of the whole original section. It must reach its maximum height of 20 m at a horizontal distance of 30 m from the start.

(a)

Find the maximum height of the original track and the horizontal distance at which it occurs.

(2)

(b)

Find the value of p and the value of q.

(3)

(c)

Describe fully the two transformations that map the graph of f onto the graph of g.

(2)

(d)

Find the total horizontal length of the new track section.

(2)

(e)

Show that $g(x) = -\frac{1}{675}x^3 + \frac{1}{15}x^2$.

(3)

(f)

A horizontal safety walkway is to be fixed between the two points on the new track that are 12 m above the ground. Find the length of the walkway.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$x = 20$ m (e.g. from $f'(x) = -0.012x^2 + 0.24x = 0$ or GDC)	✓ 1
Maximum height = 16 m	✓ 1
Part (b)
Recognises the maximum of g occurs where $qx = 20$ and equals $16p$	✓ 1
$p = 1.25$	✓ 1
$q = \frac{2}{3}$	✓ 1
Part (c)
Vertical stretch (parallel to the y-axis), scale factor 1.25	✓ 1
Horizontal stretch (parallel to the x-axis), scale factor 1.5	✓ 1
Part (d)
Original track ends at $x = 30$, so $\frac{2}{3}x = 30$ (or $30 \times 1.5$)	✓ 1
Length = 45 m	✓ 1
Part (e)
Substitutes: $g(x) = 1.25\left[-0.004\left(\frac{2x}{3}\right)^3 + 0.12\left(\frac{2x}{3}\right)^2\right]$	✓ 1
$x^3$ coefficient: $1.25 \times (-0.004) \times \frac{8}{27} = -\frac{1}{675}$	✓ 1
$x^2$ coefficient: $1.25 \times 0.12 \times \frac{4}{9} = \frac{1}{15}$	✓ 1
Part (f)
Solves $g(x) = 12$ by GDC, or solves $f(x) = 9.6$ and scales the solutions by 1.5	✓ 1
$x = 17.0$ and $x = 39.9$	✓ 1
Length = 22.9 m	✓ 1

Common error: Replacing x by qx stretches the graph horizontally by scale factor 1/q, not q. Here q = 2/3 gives a stretch of 1.5, which is why the track lengthens from 30 m to 45 m.

Marking point	Mark	Notes
Part (a)
A1: \(T(0) = 20 + 70 = 90\) °C	✓ 1
Part (b)
M1: substituting \(t = 10\): \(20 + 70(0.92)^{10}\)	✓ 1
A1: 50.4 °C	✓ 1
Part (c)
M1: setting \(20 + 70(0.92)^t = 40\) or using a graph / solver	✓ 1
A1: \(t = 15.0\) minutes	✓ 1
Part (d)
R1: 20 °C is the room temperature; the coffee approaches this temperature over time but never cools below it	✓ 1

Marking point	Mark	Notes
Part (a)
A1: maximum height \(26 + 24 = 50\) m	✓ 1
A1: minimum height \(26 - 24 = 2\) m	✓ 1
Part (b)
M1: period \(= \frac{360}{12}\)	✓ 1
A1: 30 minutes	✓ 1
Part (c)
M1: \(h(8) = 26 - 24\cos(96°)\)	✓ 1
A1: 28.5 m	✓ 1

Marking point	Mark	Notes
Part (a)
M1: \(5000(1.035)^t = 8000\) or \((1.035)^t = 1.6\)	✓ 1
A1: \(t = 13.66\ldots\)	✓ 1
A1: 14 complete years	✓ 1
Part (b)
M1: \(5000(1.035)^6\)	✓ 1
A1: 6146.28 USD	✓ 1

Marking point	Mark	Notes
Part (a)
A1: \(a = \frac{8.8 - 2.4}{2} = 3.2\)	✓ 1
Part (b)
A1: \(c = \frac{8.8 + 2.4}{2} = 5.6\)	✓ 1
Part (c)
M1: \(b = \frac{360}{12.4}\)	✓ 1
A1: \(b = 29.0\) (29.032…)	✓ 1
Part (d)
M1: \(d(5) = 3.2\sin(29.032\ldots \times 5) + 5.6\)	✓ 1
A1: 7.43 m (7.428…)	✓ 1

Marking point	Mark	Notes
Part (a)
Base dimensions \((30-2x)\) and \((20-2x)\), height \(x\), so \(V = x(30-2x)(20-2x)\) (M1)	✓ 1
Expands correctly: \(x(600 - 100x + 4x^2) = 4x^3 - 100x^2 + 600x\) (A1) (AG)	✓ 1
Part (b)
0 < x < 10, because the 20 cm side requires 20 − 2x > 0 (A1)	✓ 1
Part (c)
Uses the GDC maximum on the graph of V, or solves V′(x) = 12x² − 200x + 600 = 0 (M1)	✓ 1
x = 3.92 cm (A1)	✓ 1
V = 1060 cm³ (1056.3…) (A1)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 attempt at gradient: \(m = \frac{18.20-9.80}{10-4}\)	✓ 1
A1 \(m = 1.40\)	✓ 1
A1 \(c = 9.80 - 4(1.40) = 4.20\)	✓ 1
Part (b)
R1 \(c\) is the fixed booking fee of $4.20, charged before any distance is travelled	✓ 1
Part (c)
A1 $C = 1.40(25) + 4.20 = $39.20	✓ 1

Marking point	Mark	Notes
Part (a)
\(m=\frac{13.9-18.4}{5-2}\) M1 ; \(m=-1.5\) A1	✓ 2
Part (b)
\(18.4=-1.5(2)+c \Rightarrow c=21.4\) A1	✓ 1
Part (c)
The height of the candle decreases by 1.5 cm every hour R1	✓ 1
Part (d)
Sets \(-1.5t+21.4=0\) M1 ; \(t=14.3\) hours (14.266…) A1	✓ 2

Marking point	Mark	Notes
Part (a)
M1: substitutes \(d = 12.4\): \(C = 3.50 + 1.85 \times 12.4\)	✓ 1
A1: \(C = \\)26.44$	✓ 1
Part (b)
A1: the fare increases by $1.85 for each additional kilometre travelled (cost per km)	✓ 1
Part (c)
M1: sets up \(3.50 + 1.85d = 41.30\)	✓ 1
A1: \(d = 20.4\) km (20.432…)	✓ 1

Marking point	Mark	Notes
Part (a)
M1: \(10\log_{10}\left(\frac{3.2\times10^{-5}}{1.0\times10^{-12}}\right)\)	✓ 1
A1: 75.1 dB (75.05...)	✓ 1
Part (b)
M1: \(\frac{I}{I_0}=10^{9.2}\)	✓ 1
A1: \(I=1.58\times10^{-3}\) W m⁻² (1.5849...×10⁻³)	✓ 1
Part (c)
A1: \(\frac{1.585\times10^{-3}}{3.2\times10^{-5}}=49.5\) times. Also accept \(10^{(92-75.05)/10}\approx49.5\). Accept 49.0 to 49.6 where it follows from rounded earlier answers.	✓ 1

Marking point	Mark	Notes
Part (a)
Uses a vertex method, e.g. \(t = -\frac{14.7}{2(-4.9)}\), or the GDC maximum	✓ 1
\(t = 1.5\) s	✓ 1
Maximum height = 12.8 m (12.825)	✓ 1
Part (b)
Sets \(h(t) = 0\) and solves, e.g. by GDC or the quadratic formula	✓ 1
\(t = 3.12\) s, rejecting the negative root	✓ 1
Part (c)
The ball is 1.8 m above the ground at the moment it is released (at \(t = 0\))	✓ 1

Marking point	Mark	Notes
Part (a)
\(C(d)=1.85d+3.50\): coefficient of d is 1.85 A1 ; constant term is 3.50 A1	✓ 2
Part (b)
\(C(12.4)=3.50+1.85\times12.4=26.44\), so the cost is USD 26.44 A1	✓ 1
Part (c)
Sets \(1.85d+3.50=40\) (or \(\le 40\)) M1 ; \(d=19.7\) km (19.729…), also accept 19.72 km, but not a value rounded up to 19.73 km A1	✓ 2

Marking point	Mark	Notes
Part (a)
A1: price = 7.00 dollars (from \(x=\frac{-560}{2(-40)}\) or using GDC)	✓ 1
A1: maximum profit = 960.00 dollars	✓ 1
Part (b)
M1: attempt to solve \(-40x^2+560x-1000=0\) (e.g. GDC or quadratic formula)	✓ 1
A1: \(x=2.10\) dollars and \(x=11.90\) dollars	✓ 1
Part (c)
R1: e.g. the model predicts an ever-increasing loss, whereas in reality almost no smoothies would be sold, so the loss would be limited to roughly the fixed costs (or the owner would stop selling); OR such prices are likely outside the range of data used to build the model	✓ 1

Marking point	Mark	Notes
Part (a)
M1: substitutes correctly, \(L = 10\log_{10}(3.2\times10^{-5}/10^{-12}) = 10\log_{10}(3.2\times10^{7})\)	✓ 1
A1: L = 75.1 dB (accept 75.05)	✓ 1
Part (b)
M1: rearranges to \(I = 10^{-12}\times10^{9.2}\)	✓ 1
A1: I = 1.58×10⁻³ W m⁻² (accept 1.585×10⁻³)	✓ 1
Part (c)
M1: recognises that an increase of 15 dB multiplies the intensity by \(10^{15/10}\)	✓ 1
A1: factor = 31.6	✓ 1

Marking point	Mark	Notes
Part (a)
Substitutes \(t = 4\): \(32000 \times 0.85^4\) M1	✓ 1
16704.20 dollars A1	✓ 1
Part (b)
Each year the car keeps 85% of its value from the previous year, so it loses 15% of its value per year. R1	✓ 1
Part (c)
Sets up \(32000 \times 0.85^{t} = 5000\) and solves by GDC or logarithms M1	✓ 1
\(t = 11.4\) years A1	✓ 1
Part (d)
Any valid reason, e.g. a very old car may be scrapped, so its value is only its scrap value; or a rare model may gain value as a classic; or the rate of depreciation is unlikely to stay at 15% per year for 30 years. R1	✓ 1

Marking point	Mark	Notes
Part (a)
A1: \(a = \frac{16.4 - 8.0}{2} = 4.2\)	✓ 1
A1: \(d = \frac{16.4 + 8.0}{2} = 12.2\)	✓ 1
Part (b)
A1: \(b = \frac{360}{365} = 0.986\) (0.98630…)	✓ 1
Part (c)
M1: \(4.2\cos\big(0.98630\ldots(60 - 172)\big) + 12.2\)	✓ 1
A1: 10.7 hours	✓ 1
Part (d)
M1: setting \(D(t) = 14\), i.e. \(\cos\big(b(t-172)\big) = \frac{1.8}{4.2}\), or sketching \(D\) with the line \(y = 14\)	✓ 1
A1: \(t = 106.5\) (106.48…)	✓ 1
A1: \(t = 237.5\) (237.52…)	✓ 1
A1: \(237.5 - 106.5 \approx 131\) days (accept 131.0; counting days 107 to 237 inclusive also gives 131)	✓ 1
Part (e)
R1: a valid reason, e.g. the real variation is not exactly sinusoidal, the year is not exactly 365 days (leap years), or terrain/atmospheric effects	✓ 1

Marking point	Mark	Notes
Part (a)
A1: \(V = 28500(0.86)^t\)	✓ 1
Part (b)
M1: \(28500(0.86)^4\)	✓ 1
A1: 15 589.73 EUR	✓ 1
Part (c)
M1: \(28500(0.86)^t = 32000(0.81)^t\)	✓ 1
M1: using GDC intersection, or \(\left(\frac{0.86}{0.81}\right)^t = \frac{32000}{28500}\)	✓ 1
A1: \(t = 1.93\) years	✓ 1
Part (d)
R1: after 1.93 years Sana's car is worth less than Ravi's car, because her car loses value at a faster rate (19% compared with 14% per year)	✓ 1

Marking point	Mark	Notes
Part (a)
A1: \(a = 6\)	✓ 1
A1: \(d = 25\)	✓ 1
Part (b)
M1: period = 24 hours, so \(b = \frac{360}{24}\)	✓ 1
A1: \(b = 15\)	✓ 1
Part (c)
M1: the temperature is 25 °C and rising halfway between 03:00 and 15:00	✓ 1
A1: \(c = 9\)	✓ 1
Part (d)
M1: \(6\sin\big(15(t - 9)\big) + 25 = 28\) or graph intersection	✓ 1
A1: \(t = 11\), i.e. 11:00	✓ 1
Part (e)
M1: second solution \(t = 19\), or use of symmetry about \(t = 15\)	✓ 1
A1: \(19 - 11 = 8\) hours	✓ 1

Marking point	Mark	Notes
Part (a)
\(12+0.18\times150=\) $39.00 A1	✓ 1
Part (b)
\(C(x)=12+0.18x\) for \(0\le x\le200\) A1 ; recognises bill at 200 kWh is $48 and adds $0.25(x-200)$ M1 ; $C(x)=0.25x-2$ (or $48+0.25(x-200)$) for $x>200$ A1	✓ 3
Part (c)
Since \(103.50>48\), usage exceeds 200 kWh so the second piece applies R1 ; \(0.25x-2=103.5\) M1 ; \(x=422\) kWh A1	✓ 3

Marking point	Mark	Notes
Part (a)
\(d(12)=0.35\times12=4.2\) km A1	✓ 1
Part (b)
Distance stays constant at 7 km R1 ; so the cyclist stops (rests) for 10 minutes, from \(t=20\) to \(t=30\) R1	✓ 2
Part (c)
\(7+0.28(T-30)=15.4\) M1 ; \(T=60\) A1	✓ 2
Part (d)
\(\frac{15.4}{60}\) km min⁻¹ or 15.4 km in 1 hour M1 ; 15.4 km h⁻¹ A1	✓ 2
Part (e)
Uses third piece: \(7+0.28(t-30)=11\) M1 ; \(t=44.3\) minutes (44.285…) A1	✓ 2

Marking point	Mark	Notes
Part (a)
M1: revenue minus cost: \(14n - (2400 + 6.5n)\)	✓ 1
A1: \(P(n) = 7.5n - 2400\)	✓ 1
Part (b)
M1: \(7.5n - 2400 = 0\)	✓ 1
A1: \(n = 320\) cakes	✓ 1
Part (c)
M1: \(7.5n - 2400 \ge 1750\), giving \(n \ge 553.3\ldots\)	✓ 1
A1: 554 cakes (must be a whole number, rounded up)	✓ 1
Part (d)
M1: \(250p = 2400 + 6.5 \times 250\)	✓ 1
A1: \(p = \\)16.10$	✓ 1

Marking point	Mark	Notes
Part (a)
A1: \(a=\frac{8.6-2.2}{2}=3.2\)	✓ 1
A1: \(d=\frac{8.6+2.2}{2}=5.4\)	✓ 1
Part (b)
M1: period \(=2\times6.2=12.4\) hours	✓ 1
A1: \(b=\frac{360}{12.4}=29.0\) (29.032…)	✓ 1
Part (c)
M1: substituting \(t=14\), \(3.2\cos(29.032\times11)+5.4\)	✓ 1
A1: \(D=7.83\) m	✓ 1
Part (d)
M1: setting \(3.2\cos(b(t-3))+5.4=4\) or sketching on GDC with line \(D=4\)	✓ 1
A1: \(t=6.99\)	✓ 1
A1: \(t=11.4\) (11.406…)	✓ 1
A1: duration \(=4.41\) hours	✓ 1
Part (e)
R1: e.g. tidal heights vary over the month (spring/neap tides) so the amplitude is not constant; weather such as storm surges alters depth; small errors in the period accumulate over many cycles	✓ 1

Marking point	Mark	Notes
Part (a)
\(c = 1.2\) A1	✓ 1
\(0.25a + 0.5b + 1.2 = 4.45\) A1	✓ 1
\(a + b + 1.2 = 5.2\) A1	✓ 1
Part (b)
Attempt to solve simultaneously (e.g. GDC) M1	✓ 1
\(a = -5\), \(b = 9\) A1	✓ 1
Part (c)
Uses \(t = -\frac{b}{2a}\) or GDC maximum M1	✓ 1
\(t = 0.9\) s A1	✓ 1
\(h = 5.25\) m A1	✓ 1
Part (d)
Solves \(-5t^2 + 9t + 1.2 = 0\) M1	✓ 1
\(t = 1.92\) s (1.9247…; reject negative root) A1	✓ 1
Part (e)
\(\frac{\|1.9247 - 2.05\|}{2.05} \times 100\) M1	✓ 1
6.11% A1	✓ 1

Marking point	Mark	Notes
Part (a)
Forms two equations, using \((10, 0)\) and the vertex \((5, 6)\): \(100a + 10b = 0\) and \(25a + 5b = 6\)	✓ 1
\(a = -0.24\)	✓ 1
\(b = 2.4\)	✓ 1
Part (b)
Sets \(-0.24x^2 + 2.4x = 4.2\)	✓ 1
\(x = 2.26\) and \(x = 7.74\) (to 3 s.f.)	✓ 1
Maximum width \(= 7.7386 - 2.2614 = 5.48\) m	✓ 1

Marking point	Mark	Notes
Part (a)
A1: \(a=500\)	✓ 1
A1: \(b=\frac{1040-500}{4}=135\)	✓ 1
Part (b)
M1: \(500r^4=1040\) so \(r^4=2.08\)	✓ 1
A1: \(r=2.08^{1/4}=1.20\) (1.20093…)	✓ 1
Part (c)
A1: Model A predicts \(500+135(2)=770\) and Model B predicts \(500(1.20093…)^2\approx 721\)	✓ 1
R1: Model B is more appropriate because its prediction is much closer to the recorded value of 720	✓ 1
Part (d)
M1: Correct substitution into their chosen model, e.g. \(500(1.20093…)^{10}\) or \(500(2.08)^{10/4}\)	✓ 1
A1: 3120 subscribers (3119.8…)	✓ 1
Follow-through: if Model A was chosen in part (c), \(500+135(10)=1850\) earns M1A1.	—

Marking point	Mark	Notes
Part (a)
M1 gradient \(\frac{3900-1500}{8} = 300\)	✓ 1
A1 \(L(t) = 300t + 1500\)	✓ 1
Part (b)
M1 \(1500b^8 = 3900\) so \(b^8 = 2.6\)	✓ 1
A1 \(b = 2.6^{1/8} = 1.12686...\), which is 1.13 to 3 s.f.	✓ 1
Part (c)
A1 \(L(4) = 2700\)	✓ 1
A1 \(E(4) = 2420\) (accept 2418.7, or 2450 from using \(b = 1.13\))	✓ 1
R1 Model B is better because its value is much closer to the observed 2400 than 2700 is	✓ 1
Part (d)
A1 \(E(12) = 6290\) (accept 6288.6; using \(b = 1.13\) gives 6502, which earns this mark only as follow-through)	✓ 1
M1 \(\frac{\|E(12) - 6100\|}{6100} \times 100\) with their \(E(12)\)	✓ 1
A1 3.09% (accept answers from 3.08% to 3.11% from a rounded \(E(12)\); follow-through 6.59% from \(b = 1.13\))	✓ 1
Part (e)
R1 the exponential model increases without limit	✓ 1
R1 the number of potential subscribers is limited, so growth is likely to slow down (level off) as the market saturates	✓ 1

Marking point	Mark	Notes
Part (a)
M1: uses first piece: \(0.95 \times 14\)	✓ 1
A1: 13.30 dollars	✓ 1
Part (b)
M1: uses third piece: \(67 + 2.40(63 - 50)\)	✓ 1
A1: 98.20 dollars	✓ 1
Part (c)
M1: second expression at \(x = 50\): \(19 + 1.60 \times 30 = 67\)	✓ 1
A1: third expression at \(x = 50\): \(67 + 2.40 \times 0 = 67\), so both give 67	✓ 1
Part (d)
R1: \(19 < 54.36 \le 67\), so the second piece applies	✓ 1
M1: \(19 + 1.60(x - 20) = 54.36\)	✓ 1
A1: \(x = 42.1\) m³	✓ 1
Part (e)
M1: \(63k = 98.20\)	✓ 1
A1: \(k = 1.56\) dollars per m³ (3 s.f.)	✓ 1

Marking point	Mark	Notes
Part (a)
Any two correct equations from \(25a + 5b + c = 1200\), \(100a + 10b + c = 2700\), \(400a + 20b + c = 1200\)	✓ 1
All three equations correct	✓ 1
Part (b)
Solves the system using a GDC or by elimination, e.g. \(b = -25a\)	✓ 1
\(a = -30\), \(b = 750\), \(c = -1800\)	✓ 1
Part (c)
At a price of $0 (candles given away) the business makes a weekly loss of $1800, i.e. its fixed weekly costs are $1800	✓ 1
Part (d)
Uses the symmetry of the data, \(x = -\frac{b}{2a}\) or a GDC	✓ 1
Price = $12.50	✓ 1
Maximum weekly profit = $2887.50	✓ 1
Part (e)
Solves \(-30x^2 + 750x - 1800 = 0\) or uses the graph's x-intercepts	✓ 1
\(x = 2.69\)	✓ 1
\(x = 22.31\), so \(2.69 < x < 22.31\)	✓ 1

Marking point	Mark	Notes
Part (a)
\(V(0) = 50\) million litres	✓ 1
Part (b)
\(t = 8.29\) months (accept \(15-\sqrt{45}\))	✓ 1
\(V = 89.1\) million litres	✓ 1
(Note: \(V(0)=50\) and \(V(24)=67.28\) are both less, so the local maximum is the overall maximum.)	—
Part (c)
Attempts to solve \(V(t) = 66\), e.g. by GDC or graph intersection	✓ 1
\(t = 1.718\)	✓ 1
\(t = 20\) and \(t = 23.28\)	✓ 1
Low for \(0 \le t < 1.718\) and \(20 < t < 23.28\), so total time \(= 1.718 + 3.282 = 5.00\) months (accept 5 or 5.0)	✓ 1

Marking point	Mark	Notes
Part (a)
\(p = \frac{8.6 - 2.2}{2} = 3.2\) A1	✓ 1
\(s = \frac{8.6 + 2.2}{2} = 5.4\) A1	✓ 1
Period \(= 2 \times 6.25 = 12.5\) hours M1	✓ 1
\(q = \frac{360}{12.5} = 28.8\) A1	✓ 1
\(r = 3\) (maximum of the cosine at the 03:00 high tide) A1	✓ 1
Part (b)
Substitutes \(t = 12\): \(3.2\cos(28.8 \times 9) + 5.4\) M1	✓ 1
\(D = 4.80\) m A1	✓ 1
Part (c)
Solves \(3.2\cos(28.8(t-3)) + 5.4 = 4\) (e.g. by GDC intersection) M1	✓ 1
\(t = 7.03\) and \(t = 11.5\) (7.0257…, 11.474…) A1	✓ 1
Duration \(= 4.45\) hours A1	✓ 1

Marking point	Mark	Notes
Part (a)
A1 \(a = \frac{8.6 - 2.2}{2} = 3.2\)	✓ 1
A1 \(d = \frac{8.6 + 2.2}{2} = 5.4\)	✓ 1
Part (b)
M1 period \(= 2 \times 6.25 = 12.5\) hours	✓ 1
A1 \(b = \frac{2\pi}{12.5} = \frac{4\pi}{25} = 0.503\) (3 s.f.)	✓ 1
Part (c)
A1 \(c = 3\)	✓ 1
Part (d)
M1 solving \(3.2\cos(0.503(t - 3)) + 5.4 = 4.0\) and choosing the first solution after \(t = 3\), e.g. by graph intersection	✓ 1
A1 \(t = 7.03\) hours, i.e. 07:02 (accept 07:01 from using \(b = 0.503\))	✓ 1