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International GCSE Maths · Pearson Edexcel 4MA1 · Unit 3

Sequences, functions and graphs

316 free, high-quality past-paper-style questions on Sequences, functions and graphs for Edexcel IGCSE Maths, each with a full mark scheme. Covers: Graphs, gradients and real-life graphs; Sequences and nth terms; Function notation, composite and inverse functions; and more. For when you have finished the past papers, or want to practise one topic at a time.Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.2,063 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked

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+ Add to test✓ In your test1IGME-060-01-01Graphs, gradients and real-life graphs Paper FEasy6 marks

The point A has coordinates (2, 5) and the point B has coordinates (6, 17).
A straight line passes through A and B.

Diagram not to scale

(a)

Work out the gradient of the line AB.

(2)

(b)

Find an equation of the line AB. Give your answer in the form y = mx + c

(2)

(c)

Find the coordinates of the point where the line AB crosses the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{17-5}{6-2}\) or \(\frac{12}{4}\)	✓ 1
A1 gradient = 3	✓ 1
Part (b)
M1 for substituting (2, 5) or (6, 17) into y = 3x + c, e.g. 5 = 3 × 2 + c	✓ 1
A1 y = 3x − 1	✓ 1
Part (c)
M1 for setting 3x − 1 = 0	✓ 1
A1 \(\left(\frac{1}{3}, 0\right)\)	✓ 1

Exam tip: Gradient is change in y divided by change in x. Subtract the coordinates in the same order on the top and the bottom to keep the sign correct.

+ Add to test✓ In your test2IGME-060-01-02Graphs, gradients and real-life graphs Paper FEasy7 marks

A taxi company uses the formula
C = 2.5d + 4
to work out the cost, £C, of a journey of d kilometres.
The graph of C against d is a straight line.

(a)

Work out the cost of a journey of 8 km.

(2)

(b)

Write down the gradient of the graph and state what it represents in this context.

(2)

(c)

State what the number 4 represents in this context.

(1)

(d)

A customer pays £31.50 for a journey.
Work out the length of the journey.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 2.5 × 8 + 4	✓ 1
A1 £24	✓ 1
Part (b)
B1 gradient = 2.5	✓ 1
B1 the cost per kilometre is £2.50	✓ 1
Part (c)
B1 a fixed (starting) charge of £4 paid before any distance is travelled	✓ 1
Part (d)
M1 for 2.5d + 4 = 31.5 or (31.5 − 4) ÷ 2.5	✓ 1
A1 11 km	✓ 1

Exam tip: When interpreting a gradient in a real-life graph, give the rate with its units, here 'pounds per kilometre', rather than just the number.

+ Add to test✓ In your test3IGME-060-01-05Graphs, gradients and real-life graphs Paper FEasy7 marks

Here is an incomplete table of values for y = x² − 2x − 3
x: −2, −1, 0, 1, 2, 3, 4
y: ___, 0, ___, −4, −3, ___, 5

(a)

Complete the table of values.

(3)

(b)

Use the table to write down the solutions of x² − 2x − 3 = 0

(2)

(c)

The graph of y = x² − 2x − 3 is symmetrical.
Write down the coordinates of the minimum point of the graph.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 x = −2 gives y = 5	✓ 1
B1 x = 0 gives y = −3	✓ 1
B1 x = 3 gives y = 0	✓ 1
Part (b)
B1 x = −1	✓ 1
B1 x = 3	✓ 1
Part (c)
B1 x-coordinate 1 (from symmetry between −1 and 3)	✓ 1
B1 y-coordinate −4, giving (1, −4)	✓ 1

Common error: When substituting a negative value, (−2)² = +4. Typing −2² into a calculator without brackets gives −4, which is wrong.

+ Add to test✓ In your test4IGME-060-00-02Sequences and nth terms Paper FEasy4 marks

The first term of a sequence is 3.
The term-to-term rule is: multiply by 2, then subtract 1.

(a)

Write down the next three terms of the sequence.

(2)

(b)

Find the first term of the sequence that is greater than 100.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 5 and 9	✓ 1
B1: 17	✓ 1
Part (b)
M1: continues the sequence correctly to 33 and 65	✓ 1
A1: 129	✓ 1

Common error: Apply the operations in the order given. Subtracting 1 first and then multiplying by 2 gives 4, 6, 10, … which is wrong.

+ Add to test✓ In your test5IGME-060-00-03Sequences and nth terms Paper FEasy5 marks

The nth term of a sequence is 5n − 2.

(a)

Write down the first three terms of the sequence.

(2)

(b)

Work out the 40th term of the sequence.

(1)

(c)

Which term of the sequence is equal to 248?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 3 and 8	✓ 1
B1: 13	✓ 1
Part (b)
B1: 198	✓ 1
Part (c)
M1: 5n − 2 = 248 or (248 + 2) ÷ 5	✓ 1
A1: 50th term	✓ 1

Exam tip: For the 40th term, substitute n = 40. Do not count on 40 terms from the last term you wrote down.

+ Add to test✓ In your test6IGME-060-00-04Sequences and nth terms Paper FEasy4 marks

Here are the first four terms of an arithmetic sequence.

50 46 42 38

(a)

Find an expression, in terms of n, for the nth term of the sequence.

(2)

(b)

Find the first term of the sequence that is negative.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: −4n + k for any constant k	✓ 1
A1: 54 − 4n	✓ 1
Part (b)
M1: 54 − 4n < 0 giving n > 13.5, or continues the sequence as far as 2	✓ 1
A1: −2	✓ 1

Common error: A decreasing sequence has a negative common difference. Writing 4n + 46 ignores the sign of the difference.

+ Add to test✓ In your test7IGME-060-00-06Sequences and nth terms Paper FEasy4 marks

In a sequence, each term after the second is found by adding together the two terms before it.

(a)

The first two terms of one such sequence are 4 and 7.
Work out the 5th term.

(2)

(b)

A different sequence follows the same rule. Its 3rd term is 20 and its 4th term is 32.
Work out its 1st term.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 11 and 18 found	✓ 1
A1: 29	✓ 1
Part (b)
M1: 2nd term = 32 − 20 = 12	✓ 1
A1: 1st term = 20 − 12 = 8	✓ 1

Exam tip: To go backwards in a sequence built by adding, subtract. Each term is the term after it minus the term before it.

+ Add to test✓ In your test8IGME-060-00-01Sequences and nth terms Paper FEasy5 marks

Here are the first four terms of an arithmetic sequence.

7 11 15 19

(a)

Write down the next term of the sequence.

(1)

(b)

Find an expression, in terms of n, for the nth term of the sequence.

(2)

(c)

Is 150 a term of this sequence? Give a reason for your answer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 23	✓ 1
Part (b)
M1: 4n + k for any constant k (including 4n alone), using the common difference 4	✓ 1
A1: 4n + 3	✓ 1
Part (c)
M1: a valid method, for example: solving 4n + 3 = 150 to get n = 36.75; or finding (150 − 3) ÷ 4 = 36.75; or finding the consecutive terms 147 and 151; or stating that every term is odd	✓ 1
A1: No, with a correct reason. Any one of these reasons is acceptable: n = 36.75 is not a whole number; 150 lies between the consecutive terms 147 and 151; all the terms are odd but 150 is even	✓ 1

Exam tip: To test whether a number is a term, set the nth term equal to it and solve for n. The number is a term only if n is a positive whole number. Always write a clear 'yes' or 'no' with your reason.

+ Add to test✓ In your test9IGME-060-01-03Graphs, gradients and real-life graphs Paper FEasy7 marks

Sam leaves home at 09:00 and cycles at a constant speed to a lake 12 km away. The journey takes 40 minutes.
Sam rests at the lake for 20 minutes.
Sam then cycles straight back home along the same route at a constant speed, taking 30 minutes.

(a)

Work out Sam's speed, in km/h, on the way to the lake.

(2)

(b)

Work out the time at which Sam arrives home.

(1)

(c)

Work out Sam's speed, in km/h, on the way home.

(2)

(d)

Work out Sam's average speed, in km/h, for the whole time from leaving home to arriving back home.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 12 ÷ \(\frac{40}{60}\) or equivalent	✓ 1
A1 18 km/h	✓ 1
Part (b)
B1 10:30	✓ 1
Part (c)
M1 for 12 ÷ 0.5 or equivalent	✓ 1
A1 24 km/h	✓ 1
Part (d)
M1 for total distance 24 km ÷ total time 1.5 h (including the rest)	✓ 1
A1 16 km/h	✓ 1

Common error: Dividing kilometres by minutes does not give km/h. Convert the time to hours first, e.g. 40 minutes = 40/60 hours. For average speed, include the time spent resting.

+ Add to test✓ In your test10IGME-060-01-04Graphs, gradients and real-life graphs Paper FEasy4 marks

A conversion graph changes miles (x) into kilometres (y).
The graph is a straight line passing through (0, 0) and (50, 80).

(a)

Work out the gradient of the conversion graph.

(1)

(b)

Use your value of the gradient to convert 35 miles into kilometres.

(1)

(c)

Convert 200 kilometres into miles.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 1.6	✓ 1
Part (b)
B1 56 km	✓ 1
Part (c)
M1 for 200 ÷ 1.6 or 200 × \(\frac{50}{80}\)	✓ 1
A1 125 miles	✓ 1

Common error: Converting km to miles needs division by 1.6, not multiplication. Check that your answer in miles is smaller than the number of kilometres.

+ Add to test✓ In your test11IGME3-3-09d-01Sequences and nth terms Paper FEasy4 marks

A runner follows a training plan. In weeks 1, 2, 3 and 4 she runs 4.5 km, 6 km, 7.5 km and 9 km in total, and the pattern continues in the same way.

(a)

Find an expression, in terms of n, for the total distance, in km, in week n.

(2)

(b)

In which week will she first run more than 30 km?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 common difference of 1.5 identified	✓ 1
A1 1.5n + 3	✓ 1
Part (b)
M1 1.5n + 3 > 30 or 1.5n > 27	✓ 1
A1 week 19 (week 18 gives exactly 30 km, week 19 gives 31.5 km)	✓ 1

Common error: Writing the nth term as 4.5n or as n + 1.5 only copies the first term or the difference; test your rule on the first two terms.

+ Add to test✓ In your test12IGME3-3-09d-02Graphs, gradients and real-life graphs Paper FEasy8 marks

A cyclist rides 15 km in the first 40 minutes of a journey at a constant speed. She then rests for 20 minutes. She finishes by riding the remaining 9 km in 30 minutes at a different constant speed.

(a)

Work out her speed in the first 40 minutes, in km/h.

(2)

(b)

Work out her speed in the last 30 minutes, in km/h.

(2)

(c)

On a distance–time graph of the journey, state the gradient of the graph during the rest.

(1)

(d)

Work out her average speed for the whole journey, including the rest.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 15 ÷ (40/60) or 15 ÷ ²/₃	✓ 1
A1 22.5 km/h	✓ 1
Part (b)
M1 9 ÷ 0.5	✓ 1
A1 18 km/h	✓ 1
Part (c)
B1 0 (zero) as the distance does not change	✓ 1
Part (d)
M1 total distance 24 km	✓ 1
M1 total time 90 minutes = 1.5 hours	✓ 1
A1 16 km/h	✓ 1

Exam tip: For average speed use the total distance divided by the total time, and include any rest in the time. Convert minutes to hours first.

+ Add to test✓ In your test13IGME3-3-09d-03Graphs, gradients and real-life graphs Paper FEasy6 marks

On a map grid with units of kilometres, a straight pipeline passes through the points A(0, −3) and B(4, 5).

Diagram not to scale

(a)

Work out the gradient of the pipeline.

(2)

(b)

Find the equation of the line through A and B.

(2)

(c)

The point (k, 11) lies on the pipeline. Find the value of k.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (5 − (−3)) ÷ (4 − 0)	✓ 1
A1 2	✓ 1
Part (b)
M1 use of y = mx + c with m = 2 (ECF from (a))	✓ 1
A1 y = 2x − 3	✓ 1
Part (c)
M1 2k − 3 = 11	✓ 1
A1 k = 7	✓ 1

Common error: When subtracting coordinates to find a gradient, keep the order the same on top and bottom; mixing the order changes the sign.

+ Add to test✓ In your test14IGME3-3-09d-04Sequences and nth terms Paper FEasy6 marks

A biologist counts the bacteria in a sample. The counts after 1, 2, 3 and 4 hours are 5, 10, 20 and 40 (in thousands). The pattern continues.

(a)

Write down the next two terms and describe the rule for getting from one term to the next.

(2)

(b)

Work out the count after 9 hours.

(2)

(c)

The count after n hours is 5 × 2ⁿ⁻¹. After how many hours does the count first exceed 10 000?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 80 and 160	✓ 1
B1 multiply by 2 each time	✓ 1
Part (b)
M1 5 × 2⁸ or doubling 8 more times from 5	✓ 1
A1 1280 (thousand)	✓ 1
Part (c)
M1 count after 11 hours is 5120, which is below 10 000	✓ 1
A1 count after 12 hours is 10 240, so n = 12	✓ 1

Exam tip: For a doubling sequence the power is one less than the term number, because term 1 has not yet been doubled.

+ Add to test✓ In your test15IGME3-3-09d-05Graphs, gradients and real-life graphs Paper FEasy5 marks

Here is a table of values for y = x² − 2x − 3.
x: −2, −1, 0, 1, 2, 3, 4
y: a, 0, −3, b, −3, 0, 5

(a)

Work out the values of a and b.

(2)

(b)

Write down the equation of the line of symmetry of the graph.

(1)

(c)

Use the table to write down the positive solution of x² − 2x − 3 = 5.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitute x = −2 or x = 1 into the equation	✓ 1
A1 a = 5 and b = −4	✓ 1
Part (b)
B1 x = 1	✓ 1
Part (c)
M1 look for y = 5 in the table	✓ 1
A1 x = 4 (the other solution is x = −2)	✓ 1

Exam tip: The line of symmetry of a quadratic graph sits halfway between two points with the same y-value, such as x = 0 and x = 2 here.

+ Add to test✓ In your test16IGME3-3-12r6-01Graphs, gradients and real-life graphs Paper FEasy5 marks

A water company records the volume of water, V litres, in a storage tank t minutes after filling begins. The graph of V against t is a straight line passing through the points (0, 150) and (12, 510).

Diagram not to scale

(a)

Work out the gradient of the line.

(2)

(b)

State what the gradient represents in this context.

(1)

(c)

Work out the time at which the tank contains 750 litres.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (510 − 150) ÷ (12 − 0)	✓ 1
A1 30	✓ 1
Part (b)
B1 the rate at which water enters the tank, 30 litres per minute	✓ 1
Part (c)
M1 (750 − 150) ÷ 30 or 150 + 30t = 750	✓ 1
A1 20 minutes	✓ 1

Common error: The gradient of a real-life graph carries units: the unit of the vertical axis per unit of the horizontal axis.

+ Add to test✓ In your test17IGME3-3-12r6-02Sequences and nth terms Paper FEasy6 marks

In a video game, the bonus points for completing level n follow this sequence: 14, 19, 24, 29, ...

(a)

Find an expression, in terms of n, for the nth term of the sequence.

(2)

(b)

Explain why 300 cannot be the number of bonus points for any level.

(2)

(c)

A level gives 154 bonus points. Work out which level it is.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 common difference 5 (5n + ...)	✓ 1
A1 5n + 9	✓ 1
Part (b)
M1 5n + 9 = 300 gives n = 58.2	✓ 1
A1 n is not a whole number, so 300 is not in the sequence	✓ 1
Part (c)
M1 5n + 9 = 154	✓ 1
A1 level 29	✓ 1

Exam tip: Test your nth term by substituting n = 1 and n = 2 to see that it reproduces the first two terms.

+ Add to test✓ In your test18IGME3-3-12r6-03Function notation, composite and inverse functions Paper FEasy3 marks

A pharmacist uses the function D(w) = 0.6w + 12 to find the dose, D mg, of a medicine for a child of mass w kg.

(a)

Work out D(25).

(1)

(b)

A child is prescribed a dose of 36 mg. Work out the mass of the child.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 27 mg	✓ 1
Part (b)
M1 0.6w + 12 = 36 (or 36 − 12 = 24)	✓ 1
A1 w = 40 kg	✓ 1

Exam tip: D(25) means replace w by 25; it does not mean D multiplied by 25.

+ Add to test✓ In your test19IGME3-3-12r6-05Graphs, gradients and real-life graphs Paper FEasy4 marks

A water tanker leaves a depot and drives along a straight road to a village 45 km away. The journey takes 1 hour 30 minutes. The tanker stops for 20 minutes and then returns to the depot along the same road in 1 hour 15 minutes.

(a)

Calculate the average speed of the tanker on the journey to the village.

(2)

(b)

Calculate the average speed of the tanker on the journey back to the depot.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 45 ÷ 1.5	✓ 1
A1 30 km/h	✓ 1
Part (b)
M1 1 hour 15 minutes = 1.25 hours	✓ 1
A1 45 ÷ 1.25 = 36 km/h	✓ 1

Common error: Convert minutes to a fraction of an hour before dividing; 1 hour 15 minutes is 1.25 hours, not 1.15.

+ Add to test✓ In your test20IGME3-3-12r11-01Sequences and nth terms Paper FEasy5 marks

A swimmer increases the distance she swims each week. The numbers of lengths swum in the first four weeks are 8, 13, 18, 23.

(a)

Find an expression, in terms of n, for the number of lengths swum in week n.

(2)

(b)

Find the first week in which she swims more than 100 lengths, and the number of lengths in that week.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 common difference 5	✓ 1
A1 \(5n + 3\)	✓ 1
Part (b)
M1 \(5n + 3 > 100\)	✓ 1
A1 \(n > 19.4\) so week 20	✓ 1
A1 103 lengths	✓ 1

Exam tip: For 'more than' questions, solve the inequality and then round up to the next whole term.

+ Add to test✓ In your test21IGME3-3-12r11-02Graphs, gradients and real-life graphs Paper FEasy7 marks

A maintenance van leaves a wind farm depot at 09:00. It travels 18 km in 24 minutes at a constant speed. It then stays at a turbine for 20 minutes. It returns to the depot, 18 km, in 36 minutes at a constant speed.

(a)

Work out the speed of the van, in km/h, on the outward journey.

(2)

(b)

Work out the speed of the van, in km/h, on the return journey.

(2)

(c)

Work out the average speed of the van for the whole trip, including the stop.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(18 \div \frac{24}{60}\)	✓ 1
A1 45 km/h	✓ 1
Part (b)
M1 \(18 \div \frac{36}{60}\)	✓ 1
A1 30 km/h	✓ 1
Part (c)
M1 total distance 36 km and total time 80 minutes	✓ 1
M1 \(36 \div \frac{80}{60}\)	✓ 1
A1 27 km/h	✓ 1

Common error: Convert minutes to hours before dividing, and include the time stopped in the total time for average speed.

+ Add to test✓ In your test22IGME3-3-12r11-08Sequences and nth terms Paper FEasy4 marks

A library emails a notice to 3 members on day 1. On each following day, the number of members who receive the notice is double the number of the previous day.

(a)

Write down the numbers of members receiving the notice on days 5 and 6.

(2)

(b)

Find the number of members receiving the notice on the first day on which more than 1000 members receive it.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 48	✓ 1
B1 96	✓ 1
Part (b)
M1 continue doubling: 192, 384, 768, ...	✓ 1
A1 1536 (on day 10)	✓ 1

Exam tip: For a geometric sequence, write out the terms in a list; it is quick and shows your method.

+ Add to test✓ In your test23IGME3-3-12r12-01Sequences and nth terms Paper FEasy5 marks

A library counts the books returned on the first four days of a campaign: 14, 19, 24, 29. The numbers continue as an arithmetic sequence.

(a)

Find an expression for the number of books returned on day n.

(2)

(b)

Explain why exactly 100 books cannot be returned on any day.

(2)

(c)

Work out the number of books returned on day 30.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Common difference 5 so 5n	✓ 1
5n + 9	✓ 1
Part (b)
5n + 9 = 100 gives n = 18.2	✓ 1
n must be a whole number, so this is impossible	✓ 1
Part (c)
5 × 30 + 9 = 159	✓ 1

Exam tip: For a linear sequence the coefficient of n is the common difference; then adjust with a constant.

+ Add to test✓ In your test24IGME3-3-12r12-02Graphs, gradients and real-life graphs Paper FEasy6 marks

A swimmer swims one length of a 50 m pool in 40 seconds at a constant speed, rests at the wall for 20 seconds, then swims back in 45 seconds at a constant speed. Consider a distance-time graph for the whole 105 seconds.

(a)

State the time interval for which the graph is horizontal.

(1)

(b)

Work out the swimmer's speed during the first 40 seconds.

(2)

(c)

Work out the swimmer's average speed for the whole 105 seconds. Give your answer to 2 decimal places.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
From 40 s to 60 s	✓ 1
Part (b)
50 ÷ 40	✓ 1
1.25 m/s	✓ 1
Part (c)
Total distance = 100 m	✓ 1
100 ÷ 105	✓ 1
0.95 m/s	✓ 1

Exam tip: Average speed is total distance divided by total time, and the rest period counts in the time.

+ Add to test✓ In your test25IGME3-3-12r12-03Graphs, gradients and real-life graphs Paper FEasy6 marks

Over wind speeds from 2 m/s to 6 m/s, the power output P kW of a small turbine rises in a straight line with wind speed v m/s. At 2 m/s the output is 7 kW and at 6 m/s it is 19 kW.

Diagram not to scale

(a)

Work out the gradient of the line.

(2)

(b)

Find the equation of the line in the form P = mv + c.

(2)

(c)

Explain what the gradient means in this context.

(1)

(d)

Use your equation to find the output at 5 m/s.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(19 − 7) ÷ (6 − 2)	✓ 1
3	✓ 1
Part (b)
Substitute a point, e.g. 7 = 3 × 2 + c	✓ 1
P = 3v + 1	✓ 1
Part (c)
Output increases by 3 kW for every 1 m/s increase in wind speed	✓ 1
Part (d)
3 × 5 + 1 = 16 kW	✓ 1

Exam tip: The gradient of a real-life graph is the change in the vertical quantity per one unit of the horizontal quantity.

+ Add to test✓ In your test26IGME3-3-11r14-01Sequences and nth terms Paper FEasy5 marks

A shop stacks tins in rows for a display. The numbers of tins in the first four rows form a sequence: 7, 11, 15, 19.

(a)

Write down the number of tins in the next row.

(1)

(b)

Find an expression, in terms of n, for the number of tins in row n.

(2)

(c)

Show that one of the rows contains exactly 95 tins and state which row it is.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
23	✓ 1
Part (b)
Common difference of 4 / 4n seen	✓ 1
4n + 3	✓ 1
Part (c)
4n + 3 = 95 so n = (95 − 3) ÷ 4	✓ 1
n = 23, a whole number, so row 23 has 95 tins	✓ 1

Exam tip: To find the nth term of a linear sequence, the number in front of n is the common difference; then adjust with the constant.

+ Add to test✓ In your test27IGME3-3-11r14-02Sequences and nth terms Paper FEasy6 marks

During an outbreak, the numbers of new cases reported on days 1, 2 and 3 are 12, 19 and 26. The numbers of new cases on successive days form an arithmetic sequence.

(a)

Find an expression, in terms of n, for the number of new cases on day n.

(2)

(b)

On day d there are 103 new cases. Work out the value of d.

(2)

(c)

A health officer says that there will be more than 150 new cases on day 20. Use the sequence to decide whether she is correct. You must show your working.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
7n seen (common difference 7)	✓ 1
7n + 5	✓ 1
Part (b)
7d + 5 = 103	✓ 1
d = 14	✓ 1
Part (c)
7 × 20 + 5 = 145	✓ 1
145 < 150, so she is not correct	✓ 1

Common error: Writing the nth term as n + 7 by using the common difference as an addition rather than a multiplier of n.

+ Add to test✓ In your test28IGME3-3-11r14-03Graphs, gradients and real-life graphs Paper FEasy5 marks

A home battery is charged by solar panels at a constant rate. The graph of the energy stored (kWh) against time is a straight line. At 09:00 the battery stores 2.4 kWh and at 13:00 it stores 8.8 kWh. The battery stores a maximum of 12.0 kWh.

Diagram not to scale

(a)

Work out the gradient of the line.

(2)

(b)

State what the gradient represents in this situation.

(1)

(c)

Assuming the same rate continues after 13:00, work out how many hours after 13:00 the battery is full.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(8.8 − 2.4) ÷ 4	✓ 1
1.6 (kWh per hour)	✓ 1
Part (b)
The rate of charging (energy stored per hour)	✓ 1
Part (c)
(12.0 − 8.8) ÷ 1.6	✓ 1
2 hours (full at 15:00)	✓ 1

Exam tip: For a real-life graph, the gradient is the change in the vertical quantity per one unit of the horizontal quantity; quote its units.

+ Add to test✓ In your test29IGME3-3-11r14-04Graphs, gradients and real-life graphs Paper FEasy4 marks

A company hires out portable wind turbines. The cost, £C, of hiring a turbine for d days is given by C(d) = 45d + 120.

(a)

Work out C(6).

(1)

(b)

A customer pays £615. Work out the number of days she hired the turbine for.

(2)

(c)

Write down what the 120 represents in this situation.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
390	✓ 1
Part (b)
45d + 120 = 615 / 615 − 120 = 495	✓ 1
d = 11	✓ 1
Part (c)
A fixed charge (£120) paid however many days the turbine is hired	✓ 1

Exam tip: Substitute into a function as you would into a formula: C(6) means replace d with 6.

+ Add to test✓ In your test30IGME3-3-11r15-01Graphs, gradients and real-life graphs Paper FEasy5 marks

A file is downloading to a phone. The amount of data still to download, r MB, is a straight-line function of the time, t seconds, since the download began. At t = 0, r = 900. At t = 6, r = 420.

Diagram not to scale

(a)

Work out the gradient of the line.

(2)

(b)

State what the gradient tells you about the download.

(1)

(c)

The equation of the line is r = 900 − 80t. Work out the time at which the download is complete.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (420 − 900) ÷ 6 or 480 ÷ 6	✓ 1
A1 −80	✓ 1
Part (b)
A1 Data is downloaded at 80 MB per second (the amount remaining falls by 80 MB each second)	✓ 1
Part (c)
M1 900 − 80t = 0 or 900 ÷ 80	✓ 1
A1 11.25 seconds	✓ 1

Common error: When the quantity is falling, the gradient is negative. Keep the minus sign and then explain it as a rate of decrease.

+ Add to test✓ In your test31IGME3-3-11r15-02Sequences and nth terms Paper FEasy5 marks

A food scientist records the sugar content, in grams, of a drink after each stage of a recipe. The first four values are 12, 17, 22, 27, ...

(a)

Write down the next term.

(1)

(b)

Find an expression, in terms of n, for the nth term.

(2)

(c)

Is 100 a term of this sequence? You must show your working.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 32	✓ 1
Part (b)
M1 common difference 5 or 5n seen	✓ 1
A1 5n + 7	✓ 1
Part (c)
M1 5n + 7 = 100 or (100 − 7) ÷ 5 (ECF from (b))	✓ 1
A1 n = 18.6 is not a whole number, so 100 is not a term	✓ 1

Exam tip: To test whether a number is in a sequence, set the nth term equal to it. The value of n must be a positive whole number.

+ Add to test✓ In your test32IGME3-3-11r15-04Graphs, gradients and real-life graphs Paper FEasy5 marks

The recommended cooking time, T minutes, for a joint of beef of mass w kg is given by T(w) = 40w + 15.

(a)

Work out T(2.5).

(2)

(b)

A joint has a cooking time of 175 minutes. Work out its mass.

(2)

(c)

State what the number 40 represents in this formula.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40 × 2.5 + 15	✓ 1
A1 115	✓ 1
Part (b)
M1 40w + 15 = 175 or (175 − 15) ÷ 40	✓ 1
A1 4 kg	✓ 1
Part (c)
A1 Each extra kilogram adds 40 minutes to the cooking time	✓ 1

Exam tip: T(w) is read as 'T of w'. Substitute the number given for w, or set the whole expression equal to the number given as the output.

+ Add to test✓ In your test33IGME3-3-11r15-10Graphs, gradients and real-life graphs Paper FEasy5 marks

A walker's journey is shown on a distance–time graph made of three straight-line segments. The first segment goes from (0 minutes, 0 km) to (40 minutes, 2 km). The second is horizontal from 40 minutes to 60 minutes. The third goes from (60 minutes, 2 km) to (120 minutes, 5 km).

Diagram not to scale

(a)

Work out the walker's speed in km/h during the first 40 minutes.

(2)

(b)

State what the horizontal segment shows.

(1)

(c)

Work out the walker's average speed for the whole 120 minutes, in km/h.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2 ÷ (40/60) or 2 ÷ 40 × 60	✓ 1
A1 3 km/h	✓ 1
Part (b)
A1 The walker is stationary (resting) for 20 minutes	✓ 1
Part (c)
M1 total distance ÷ total time = 5 ÷ 2	✓ 1
A1 2.5 km/h	✓ 1

Common error: Convert minutes to hours before finding a speed in km/h. Average speed is total distance divided by total time, not the mean of the speeds.

+ Add to test✓ In your test34IGME3-3-11r15-12Graphs, gradients and real-life graphs Paper FEasy5 marks

The graph of y = x² − 4x + 1 is drawn for values of x from −1 to 5.

(a)

Work out the value of y when x = −1, and the value of y when x = 4.

(2)

(b)

Write down the equation of the line of symmetry of the graph.

(1)

(c)

Work out the minimum value of y.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 6	✓ 1
B1 1	✓ 1
Part (b)
B1 x = 2 (the mid-point of 0 and 4, where y = 1 for both)	✓ 1
Part (c)
M1 substitutes x = 2: 2² − 4 × 2 + 1 (ECF from (b))	✓ 1
A1 −3	✓ 1

Exam tip: The turning point of a quadratic lies on its line of symmetry. Substitute the x-value of the line of symmetry to find the y-value.

+ Add to test✓ In your test35IGME3-3-12r15-01Sequences and nth terms Paper FEasy6 marks

The numbers of books left on a returns shelf at the end of the first four days of a clear-out are 95, 88, 81, 74, ... and the pattern continues.

(a)

Write down the next term.

(1)

(b)

Find an expression, in terms of n, for the number of books left at the end of day n.

(2)

(c)

Find the number of books left at the end of the first day on which fewer than 10 books are left.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 67	✓ 1
Part (b)
M1 common difference −7 or −7n seen	✓ 1
A1 102 − 7n	✓ 1
Part (c)
M1 102 − 7n < 10 (ECF from (b))	✓ 1
A1 n > 13.14..., so day 14	✓ 1
A1 4 books	✓ 1

Common error: In a decreasing sequence the nth term has a negative coefficient of n. Check by substituting n = 1.

+ Add to test✓ In your test36IGME3-3-12r15-02Graphs, gradients and real-life graphs Paper FEasy6 marks

A swimming pool is filled by a pump. The depth of water, d cm, is shown on a graph against time, t minutes, made of three straight-line segments. The first goes from (0, 0) to (15, 30). The second goes from (15, 30) to (45, 45). The third is horizontal at d = 45 once the pool is full.

Diagram not to scale

(a)

Work out the rate at which the depth increases during the first 15 minutes.

(2)

(b)

Work out the rate at which the depth increases between 15 and 45 minutes.

(2)

(c)

Work out the depth of water after 25 minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 30 ÷ 15	✓ 1
A1 2 cm per minute	✓ 1
Part (b)
M1 (45 − 30) ÷ (45 − 15)	✓ 1
A1 0.5 cm per minute	✓ 1
Part (c)
M1 30 + 0.5 × (25 − 15) (ECF from (b))	✓ 1
A1 35 cm	✓ 1

Exam tip: A gradient on a depth–time graph is a rate of change. A smaller gradient means a slower rate of filling.

+ Add to test✓ In your test37IGME3-3-12r15-09Sequences and nth terms Paper FEasy5 marks

A library charges no fine for the first 3 days that a book is overdue. After that it charges 15p for every extra day.

(a)

Work out the fine, in pounds, for a book that is 10 days overdue.

(2)

(b)

A fine is £2.40. For how many days was the book overdue?

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (10 − 3) × 0.15	✓ 1
A1 £1.05	✓ 1
Part (b)
M1 2.40 ÷ 0.15 = 16 extra days	✓ 1
M1 16 + 3	✓ 1
A1 19 days	✓ 1

Common error: The first 3 days are free, so the number of charged days is 3 fewer than the days overdue. Add them back when working in reverse.

+ Add to test✓ In your test38IGME3-3-12r15-13Graphs, gradients and real-life graphs Paper FEasy5 marks

On a site plan of a wind farm, a cable run lies along a straight line through P(0, 4) and Q(3, 13).

Diagram not to scale

(a)

Work out the gradient of the line PQ.

(2)

(b)

Write down the equation of the line PQ.

(1)

(c)

A junction box lies on the line where y = 25. Work out its x-coordinate.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (13 − 4) ÷ (3 − 0)	✓ 1
A1 3	✓ 1
Part (b)
A1 y = 3x + 4 (ECF from (a))	✓ 1
Part (c)
M1 3x + 4 = 25 (ECF from (b))	✓ 1
A1 x = 7	✓ 1

Exam tip: The y-intercept is the y-value where the line crosses the y-axis. Here P is on the y-axis, so c = 4.

+ Add to test✓ In your test39IGME3-3-11r16-01Sequences and nth terms Paper FEasy4 marks

An archaeologist counts the pottery sherds found in four successive layers of a dig. The numbers are 7, 12, 17, 22. The pattern continues with the same constant difference.

(a)

Write down the number of sherds in layer 5.

(1)

(b)

Find an expression, in terms of n, for the number of sherds in layer n.

(2)

(c)

Work out the number of sherds in layer 40.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 27	✓ 1
Part (b)
M1 common difference 5, or 5n seen	✓ 1
A1 5n + 2	✓ 1
Part (c)
B1 202 (ECF from (b))	✓ 1

Common error: The number added to 5n is the first term minus the common difference, here 7 − 5. Check the rule on the first two terms.

+ Add to test✓ In your test40IGME3-3-11r16-02Sequences and nth terms Paper FEasy4 marks

The height of the floor of a trench above a fixed survey marker is measured at the end of each day of digging. The heights, in cm, are 45, 41, 37, 33, ... and continue to fall by 4 cm each day.

(a)

Find an expression, in terms of n, for the height at the end of day n.

(2)

(b)

On which day is the floor of the trench 15 cm below the survey marker (a height of −15 cm)?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 −4n seen	✓ 1
A1 49 − 4n oe	✓ 1
Part (b)
M1 49 − 4n = −15 (ECF from (a))	✓ 1
A1 day 16	✓ 1

Exam tip: Write the equation for the term you are asked about, then solve it. Check by substituting your answer back into the nth term.

+ Add to test✓ In your test41IGME3-3-11r16-04Sequences and nth terms Paper FEasy4 marks

A data centre adds cooling units each month. The numbers of units in the first four months are 6, 11, 21, 41. Each term after the first is found from the previous term using the same rule.

(a)

Describe the rule for finding the next term.

(1)

(b)

Work out the number of units in month 6.

(2)

(c)

Explain why every term after the first term is an odd number.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 multiply by 2 and subtract 1 oe	✓ 1
Part (b)
M1 attempt to apply the rule to 41 (gives 81)	✓ 1
A1 161	✓ 1
Part (c)
B1 doubling gives an even number and subtracting 1 from an even number gives an odd number	✓ 1

Exam tip: Test your rule on every pair of terms you are given, not just the first pair.

+ Add to test✓ In your test42IGME3-3-11r16-10Graphs, gradients and real-life graphs Paper FEasy5 marks

A trench cross-section is modelled by \(y = x^2 - 3x\), where \(x\) metres is the distance across the trench and \(y\) metres is the height of the ground relative to the trench edge.

(a)

Work out the value of \(y\) when \(x = -1\), when \(x = 2\) and when \(x = 4\).

(2)

(b)

The graph of \(y = x^2 - 3x\) is symmetrical. Write down the equation of its line of symmetry.

(1)

(c)

Work out the lowest value of \(y\).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 any two of 4, −2, 4 correct	✓ 1
B1 all three correct	✓ 1
Part (b)
B1 x = 1.5 oe	✓ 1
Part (c)
M1 substituting x = 1.5 (ECF from (b))	✓ 1
A1 −2.25	✓ 1

Exam tip: The turning point of a quadratic graph lies on its line of symmetry, halfway between the two points where it crosses the x-axis.

+ Add to test✓ In your test43IGME3-3-11r19-01Sequences and nth terms Paper FEasy4 marks

A team of archaeologists numbers the layers of a trench. The numbers of pottery fragments found in the first four layers are 7, 12, 17, 22 and the pattern continues as a linear sequence.

(a)

Find an expression, in terms of n, for the nth term of the sequence.

(2)

(b)

Layer 30 is to be checked. Is 152 a term of the sequence? You must show your working.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 common difference of 5 identified, or 5n seen	✓ 1
A1 5n + 2	✓ 1
Part (b)
M1 sets 5n + 2 = 152, ECF from (a)	✓ 1
A1 n = 30, a whole number, so yes: 152 is the 30th term	✓ 1

Exam tip: To test whether a number is in a sequence, set the nth term equal to it and check that n is a positive whole number.

+ Add to test✓ In your test44IGME3-3-11r19-02Graphs, gradients and real-life graphs Paper FEasy7 marks

A cyclist rides along a straight road in three stages.
Stage 1: 12 km at a constant speed in 30 minutes.
Stage 2: stopped for 15 minutes.
Stage 3: 15 km at a constant speed in 45 minutes.

(a)

Work out the speed in stage 1, in km/h.

(2)

(b)

Work out the speed in stage 3, in km/h.

(2)

(c)

Work out the average speed for the whole journey, in km/h.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 ÷ 0.5	✓ 1
A1 24 km/h	✓ 1
Part (b)
M1 15 ÷ 0.75	✓ 1
A1 20 km/h	✓ 1
Part (c)
M1 total distance 27 km and total time 1.5 hours (including the stop)	✓ 1
M1 27 ÷ 1.5	✓ 1
A1 18 km/h	✓ 1

Common error: Average speed uses the total time, including any time spent stopped, not just the time spent moving.

+ Add to test✓ In your test45IGME3-3-11r19-03Graphs, gradients and real-life graphs Paper FEasy5 marks

In a data centre, the power y kilowatts drawn by x racks of servers lies on a straight line that passes through the points (2, 14) and (6, 30).

Diagram not to scale

(a)

Work out the gradient of the line.

(2)

(b)

Find the equation of the line.

(2)

(c)

Use your equation to work out the power drawn by 10 racks.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (30 − 14) ÷ (6 − 2)	✓ 1
A1 4	✓ 1
Part (b)
M1 y = 4x + c with a point substituted, e.g. 14 = 8 + c	✓ 1
A1 y = 4x + 6	✓ 1
Part (c)
A1 46 kilowatts (ECF from (b))	✓ 1

Exam tip: After finding an equation, check it by substituting the second point.

+ Add to test✓ In your test46IGME3-3-12r22-01Sequences and nth terms Paper FEasy5 marks

A space probe sends back images in batches. The numbers of images in the first four batches are 7, 11, 15, 19.

(a)

Write down the next two terms of the sequence.

(1)

(b)

Find an expression, in terms of n, for the nth term of the sequence.

(2)

(c)

Show that 150 is not a term of this sequence.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 23 and 27	✓ 1
Part (b)
M1 for a common difference of 4	✓ 1
A1 for 4n + 3	✓ 1
Part (c)
M1 for 4n + 3 = 150, so 4n = 147	✓ 1
A1 for n = 36.75, which is not an integer, so 150 is not a term	✓ 1

Common error: To test whether a number is in a sequence, set the nth term equal to it; n must be a positive whole number.

+ Add to test✓ In your test47IGME3-3-12r22-02Graphs, gradients and real-life graphs Paper FEasy5 marks

The temperature of coffee beans in a roaster rises at a constant rate. The temperature is 20 °C at the start (t = 0) and 200 °C after 12 minutes. This is shown by a straight line on a graph of temperature against time.

(a)

Work out the gradient of the line.

(2)

(b)

State what the gradient represents.

(1)

(c)

Work out the temperature after 7 minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for (200 − 20) ÷ 12	✓ 1
A1 for 15	✓ 1
Part (b)
B1 for the rate of increase of temperature, 15 °C per minute	✓ 1
Part (c)
M1 for 20 + 15 × 7	✓ 1
A1 for 125 °C	✓ 1

Exam tip: On a real-life graph, the gradient is the change in the vertical quantity per unit of the horizontal quantity. Always include the units.

+ Add to test✓ In your test48IGME3-3-12r22-03Function notation, composite and inverse functions Paper FEasy4 marks

The function f is defined by \(f(x) = 5x - 2\) and the function g is defined by \(g(x) = x^2 + 1\).

(a)

Find \(f(3)\).

(1)

(b)

Solve \(f(x) = 48\).

(2)

(c)

Find \(g(-4)\).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 13	✓ 1
Part (b)
M1 for 5x − 2 = 48	✓ 1
A1 for x = 10	✓ 1
Part (c)
B1 for 17	✓ 1

Common error: In g(−4), square the whole of −4 to get +16 before adding 1. Use brackets.

+ Add to test✓ In your test49IGME3-3-12r22-04Graphs, gradients and real-life graphs Paper FEasy7 marks

A rover's distance from its base is shown on a distance–time graph made of straight line segments. The rover moves away from the base at constant speed and is 120 m from the base after 8 minutes. It then stays still for 6 minutes. Finally it returns to the base at constant speed, taking 6 minutes.

Diagram not to scale

(a)

Work out the speed of the rover during the first 8 minutes.

(2)

(b)

Work out the speed of the rover on its return to the base.

(2)

(c)

Work out the average speed of the rover for the whole journey.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 120 ÷ 8	✓ 1
A1 for 15 m/min	✓ 1
Part (b)
M1 for 120 ÷ 6	✓ 1
A1 for 20 m/min	✓ 1
Part (c)
M1 for total distance 120 + 120 = 240 m	✓ 1
M1 for total time 8 + 6 + 6 = 20 minutes	✓ 1
A1 for 240 ÷ 20 = 12 m/min	✓ 1

Exam tip: Average speed is total distance divided by total time, and the time spent stationary must be included.

+ Add to test✓ In your test50IGME3-3-12r24-01Sequences and nth terms Paper FEasy5 marks

A beekeeper builds hive boxes with frames. The numbers of frames in the first four boxes are 7, 11, 15, 19.

(a)

Find an expression, in terms of n, for the nth term of this sequence.

(2)

(b)

Is 150 a term of this sequence? You must show how you decide.

(2)

(c)

Work out the 30th term of the sequence.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 common difference 4 or 4n seen	✓ 1
A1 4n + 3	✓ 1
Part (b)
M1 4n + 3 = 150, so 4n = 147	✓ 1
A1 n = 36.75 is not a whole number, so No	✓ 1
Part (c)
A1 123	✓ 1

Common error: Test whether a number is in a sequence by solving for n; it is a term only if n is a positive whole number.

+ Add to test✓ In your test51IGME3-3-12r24-02Graphs, gradients and real-life graphs Paper FEasy6 marks

A hospital drip bag contains 500 ml of fluid at the start. After 40 minutes there are 340 ml left. The volume V ml in the bag decreases at a constant rate with time t minutes.

(a)

Work out the rate at which fluid leaves the bag, in ml per minute.

(2)

(b)

Write down an equation for V in terms of t.

(2)

(c)

Work out the time at which the bag is empty.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (500 − 340) ÷ 40	✓ 1
A1 4 ml/min	✓ 1
Part (b)
M1 gradient −4 or intercept 500 used	✓ 1
A1 V = 500 − 4t	✓ 1
Part (c)
M1 500 − 4t = 0 (ECF from (b))	✓ 1
A1 125 minutes	✓ 1

Exam tip: On a straight-line graph of a real situation, the gradient is the rate of change and the intercept is the starting value.

+ Add to test✓ In your test52IGME3-3-12r24-03Function notation, composite and inverse functions Paper FEasy4 marks

A hospital calculates the dose, D mg, of a medicine for a child of weight w kg using the function D = f(w) = 5w + 20.

(a)

Work out f(24).

(1)

(b)

State the increase in dose for each extra kilogram of weight.

(1)

(c)

A dose of 345 mg is prescribed. Work out the weight of the child.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 140	✓ 1
Part (b)
A1 5 mg	✓ 1
Part (c)
M1 5w + 20 = 345	✓ 1
A1 65 kg	✓ 1

Exam tip: For f(w) = number, replace f(w) by that number and solve the equation for w.

+ Add to test✓ In your test53IGME3-3-13r24-01Sequences and nth terms Paper FEasy5 marks

A wind farm company adds turbines to a new site each month. The numbers of turbines in service at the end of the first four months are
7, 12, 17, 22

(a)

Write down the next term of the sequence.

(1)

(b)

Find an expression, in terms of n, for the nth term of the sequence.

(2)

(c)

Show that 152 is a term of the sequence. State which term it is.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 27	✓ 1
Part (b)
M1 common difference 5 used, e.g. 5n + ... or 5n	✓ 1
A1 5n + 2	✓ 1
Part (c)
M1 5n + 2 = 152 or (152 − 2) ÷ 5	✓ 1
A1 n = 30, a whole number so 152 is the 30th term	✓ 1

Exam tip: For a linear sequence, the number in front of n is the common difference; then compare with the first term to find the constant.

+ Add to test✓ In your test54IGME3-3-13r24-02Graphs, gradients and real-life graphs Paper FEasy5 marks

A diving pool is being filled with water at a constant rate. At the start (time 0) the depth of water is 0.4 m. After 30 minutes the depth is 1.6 m. The graph of depth against time is a straight line.

Diagram not to scale

(a)

Calculate the gradient of the graph, in metres per minute.

(2)

(b)

State what the gradient represents in this situation.

(1)

(c)

Calculate the time, in minutes, at which the depth is 2.2 m.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (1.6 − 0.4) ÷ 30	✓ 1
A1 0.04	✓ 1
Part (b)
B1 the depth of the water increases by 0.04 m every minute (rate at which the depth rises)	✓ 1
Part (c)
M1 (2.2 − 0.4) ÷ 0.04 or 0.4 + 0.04t = 2.2	✓ 1
A1 45	✓ 1

Common error: Remember the graph starts at a depth of 0.4 m, not 0, so subtract this before dividing by the gradient.

+ Add to test✓ In your test55IGME3-3-13r24-03Graphs, gradients and real-life graphs Paper FEasy6 marks

A library charges a fixed fee plus a daily amount for a late book. The total charge, C pounds, for a book that is d days late is given by a straight-line graph. The charge for 4 days is £1.50 and the charge for 10 days is £3.00.

Diagram not to scale

(a)

Work out the gradient of the line.

(2)

(b)

Find the equation of the line, in the form C = md + c.

(2)

(c)

State what the value 0.50 represents.

(1)

(d)

Work out the charge for a book that is 18 days late. Give your answer in pounds.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (3.00 − 1.50) ÷ (10 − 4)	✓ 1
A1 0.25	✓ 1
Part (b)
M1 substitute a point, e.g. 1.50 = 0.25 × 4 + c	✓ 1
A1 C = 0.25d + 0.50	✓ 1
Part (c)
B1 the fixed charge of £0.50 (charge when the book is 0 days late)	✓ 1
Part (d)
A1 5.00 (ECF from (b))	✓ 1

Exam tip: Check your equation by substituting the other given point; both points must fit.

+ Add to test✓ In your test56IGME3-3-13r24-04Sequences and nth terms Paper FEasy5 marks

The nth term of a sequence is given by n² + 3. The sequence is used to number the shelves in a library annex.

(a)

Write down the first three terms of the sequence.

(2)

(b)

Work out the 10th term.

(1)

(c)

147 is a term of this sequence. Work out which term it is.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 two terms correct	✓ 1
B1 4, 7, 12	✓ 1
Part (b)
B1 103	✓ 1
Part (c)
M1 n² + 3 = 147 or n² = 144	✓ 1
A1 n = 12 (12th term)	✓ 1

Common error: Square n before adding 3: 10² + 3 is 103, not (10 + 3)².

+ Add to test✓ In your test57IGME3-3-12r26-02Sequences and nth terms Paper FEasy6 marks

In an outbreak, the number of reported cases, N, is modelled by a straight line against the number of days, d, since monitoring began. The line passes through the points (0, 12) and (10, 62).

Diagram not to scale

(a)

Work out the gradient of the line.

(2)

(b)

State what the gradient means in this situation.

(1)

(c)

Write down an equation for N in terms of d.

(1)

(d)

Use your equation to work out the number of days when the model gives 112 cases.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (62 − 12)/(10 − 0)	✓ 1
A1 5	✓ 1
Part (b)
B1 5 more reported cases each day	✓ 1
Part (c)
B1 N = 5d + 12 oe	✓ 1
Part (d)
M1 5d + 12 = 112 (ECF from (c))	✓ 1
A1 20	✓ 1

Common error: The gradient is change in y divided by change in x, with both differences taken in the same order.

+ Add to test✓ In your test58IGME3-3-12r26-03Graphs, gradients and real-life graphs Paper FEasy8 marks

A battery store is charged by a wind turbine. The energy stored, E kWh, is described by a graph against time, t hours. From t = 0 to t = 4 the graph is a straight line from (0, 20) to (4, 60). From t = 4 to t = 6 it is horizontal. From t = 6 to t = 15 it is a straight line down to (15, 15).

(a)

Work out the rate of charging between t = 0 and t = 4.

(2)

(b)

Write down the energy stored when t = 5.

(1)

(c)

Work out the rate at which energy is used between t = 6 and t = 15.

(2)

(d)

Work out the time at which 30 kWh is stored during the discharge.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (60 − 20) ÷ 4	✓ 1
A1 10 kWh per hour	✓ 1
Part (b)
B1 60 kWh	✓ 1
Part (c)
M1 (60 − 15) ÷ (15 − 6)	✓ 1
A1 5 kWh per hour	✓ 1
Part (d)
M1 energy lost = 60 − 30 (= 30)	✓ 1
M1 30 ÷ 5 (= 6 hours after t = 6)	✓ 1
A1 t = 12	✓ 1

Common error: A flat section of a graph means no change in the quantity, not zero quantity; read the value from the vertical axis.

+ Add to test✓ In your test59IGME3-3-12r26-05Sequences and nth terms Paper FEasy5 marks

The number of new cases reported on four consecutive days of an outbreak model is 14, 19, 24, 29, and the pattern continues.

(a)

Write down the next two terms of the sequence.

(1)

(b)

Find an expression for the nth term.

(2)

(c)

On which day does the model give 109 new cases?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 34, 39	✓ 1
Part (b)
M1 5n (common difference 5)	✓ 1
A1 5n + 9	✓ 1
Part (c)
M1 5n + 9 = 109 (ECF from (b))	✓ 1
A1 day 20	✓ 1

Common error: The nth term starts with the common difference times n, then you add or subtract whatever makes the first term correct.

+ Add to test✓ In your test60IGME3-3-12r26-07Sequences and nth terms Paper FEasy5 marks

A retailer displays tins in rows. The first four rows hold 12, 19, 26 and 33 tins.

(a)

Find an expression for the number of tins in row n.

(2)

(b)

Show that there is a row containing exactly 250 tins and state which row it is.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 7n (common difference 7)	✓ 1
A1 7n + 5	✓ 1
Part (b)
M1 7n + 5 = 250	✓ 1
M1 n = 245 ÷ 7	✓ 1
A1 n = 35, a whole number, so row 35	✓ 1

Exam tip: To check whether a number is in a linear sequence, solve for n and check that it is a positive whole number.

+ Add to test✓ In your test61IGME3-3-14r28-01Sequences and nth terms Paper FEasy4 marks

A beekeeper sells jars of honey. In the first four weeks of the season the numbers of jars sold are 12, 17, 22, 27 and the pattern continues.

(a)

Find an expression, in terms of n, for the number of jars sold in week n.

(2)

(b)

Show that 100 is not a term of this sequence.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 common difference 5 (5n seen)	✓ 1
A1 5n + 7	✓ 1
Part (b)
M1 5n + 7 = 100	✓ 1
A1 n = 18.6, which is not an integer, so 100 is not a term	✓ 1

Exam tip: Always check your nth term by substituting n = 1 and n = 2.

+ Add to test✓ In your test62IGME3-3-14r28-02Graphs, gradients and real-life graphs Paper FEasy5 marks

A hospital drip bag contains 500 ml of fluid. The volume, V ml, remaining after t minutes is given by V = 500 − 2.5t while the drip is running.

(a)

Work out the volume remaining after 40 minutes.

(1)

(b)

State the gradient of the graph of V against t and explain what it represents.

(2)

(c)

Work out how long the bag takes to empty.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 400 ml	✓ 1
Part (b)
B1 −2.5	✓ 1
B1 the volume decreases at a rate of 2.5 ml per minute	✓ 1
Part (c)
M1 500 ÷ 2.5 or 500 − 2.5t = 0	✓ 1
A1 200 minutes	✓ 1

Exam tip: In a real-life graph, the gradient is the rate of change: say what quantity changes per unit of the other.

+ Add to test✓ In your test63IGME3-3-14r28-03Graphs, gradients and real-life graphs Paper FEasy4 marks

A Formula 1 car accelerates at a constant rate along a pit exit lane. Its speed is 20 m/s after 1 second and 44 m/s after 4 seconds.

(a)

Work out the acceleration of the car.

(2)

(b)

Work out the speed of the car at time 0.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (44 − 20) ÷ (4 − 1)	✓ 1
A1 8 m/s²	✓ 1
Part (b)
M1 20 − 8 × 1 (ECF from (a))	✓ 1
A1 12 m/s	✓ 1

Common error: Acceleration is change in speed divided by change in time, not speed divided by time.

+ Add to test✓ In your test64IGME3-3-14r28-11Graphs, gradients and real-life graphs Paper FEasy7 marks

A hospital courier leaves ward A, walks 600 m to ward B in 5 minutes at a constant speed, waits 3 minutes, then returns to ward A in 4 minutes at a constant speed.

(a)

Work out the speed on the outward journey in metres per minute.

(2)

(b)

Work out the speed on the return journey in km/h.

(2)

(c)

Work out the average speed for the whole 12 minutes in km/h.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 600 ÷ 5	✓ 1
A1 120 m/min	✓ 1
Part (b)
M1 600 ÷ 4 = 150 m/min and × 60	✓ 1
A1 9 km/h	✓ 1
Part (c)
M1 total distance 1200 m	✓ 1
M1 1200 ÷ 12 = 100 m/min, then × 60 ÷ 1000	✓ 1
A1 6 km/h	✓ 1

Common error: Average speed is total distance divided by total time, and the waiting time must be included.

+ Add to test✓ In your test65IGME3-3-14r32-01Sequences and nth terms Paper FEasy6 marks

A space probe sends images back to Earth. The numbers of images received on the first four days are 17, 24, 31 and 38. The pattern continues in the same way.

(a)

Find an expression, in terms of n, for the number of images received on day n.

(2)

(b)

Work out the number of images received on day 40.

(1)

(c)

Work out the first day on which the probe sends more than 500 images, and state how many it sends on that day.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 common difference 7 (or 7n seen)	✓ 1
B1 7n + 10	✓ 1
Part (b)
B1 290	✓ 1
Part (c)
M1 7n + 10 > 500	✓ 1
M1 n > 70, so n = 71	✓ 1
A1 507 images (on day 71)	✓ 1

Common error: Writing the term-to-term rule 'add 7' is not an nth term; the expression must contain n.

+ Add to test✓ In your test66IGME3-3-14r32-02Graphs, gradients and real-life graphs Paper FEasy7 marks

During roasting, the temperature T °C of a batch of coffee beans rises steadily. A graph of T against time t minutes is a straight line segment from (0, 25) to (6, 205).

Diagram not to scale

(a)

Work out the gradient of the line.

(2)

(b)

State what the gradient represents.

(1)

(c)

Write down an equation for T in terms of t.

(2)

(d)

Work out the time when the temperature reaches 190 °C.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (205 − 25) ÷ (6 − 0)	✓ 1
A1 30	✓ 1
Part (b)
B1 the temperature rises by 30 °C each minute (the rate of increase of temperature)	✓ 1
Part (c)
B1 gradient 30 used and intercept 25 seen	✓ 1
B1 T = 30t + 25	✓ 1
Part (d)
M1 30t + 25 = 190 (or (190 − 25) ÷ 30)	✓ 1
A1 5.5 minutes	✓ 1

Exam tip: The gradient of a real-life graph has units 'units of y per unit of x', which tells you what it measures.

+ Add to test✓ In your test67IGME3-3-14r32-03Graphs, gradients and real-life graphs Paper FEasy7 marks

A bridge has a straight support cable. Taking the deck as the horizontal axis, the cable passes through the points A(0, 4) and B(12, 22), where distances are in metres.

Diagram not to scale

(a)

Find an equation of the line through A and B.

(3)

(b)

A second cable is parallel to the first and passes through (0, 10). Write down an equation of the second cable.

(2)

(c)

Find the value of x at the point on the first cable where y = 34.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (22 − 4) ÷ (12 − 0)	✓ 1
A1 gradient 1.5	✓ 1
A1 y = 1.5x + 4	✓ 1
Part (b)
B1 gradient 1.5	✓ 1
B1 y = 1.5x + 10	✓ 1
Part (c)
M1 1.5x + 4 = 34 (ECF from (a))	✓ 1
A1 x = 20	✓ 1

Common error: Parallel lines have the same gradient but a different y-intercept; do not change the gradient.

+ Add to test✓ In your test68IGME3-3-14r32-04Sequences and nth terms Paper FEasy6 marks

The number of users of a coffee shop's loyalty app doubles each week. In weeks 1 to 4 the numbers of users are 3, 6, 12 and 24.

(a)

Write down the numbers of users in week 5 and week 6.

(2)

(b)

Write down the term-to-term rule for this sequence.

(1)

(c)

Work out the number of users in the first week in which the number of users is greater than 1000.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 48	✓ 1
B1 96	✓ 1
Part (b)
B1 multiply by 2 (each term is double the previous term)	✓ 1
Part (c)
M1 continue doubling: 192, 384, 768	✓ 1
B1 768 < 1000 (week 9 is not enough)	✓ 1
A1 1536	✓ 1

Exam tip: When a sequence grows by a constant multiplier, keep multiplying and check each term against the target rather than guessing.

+ Add to test✓ In your test69IGME3-3-13r33-01Sequences and nth terms Paper FEasy7 marks

Archaeologists excavating a trench record the depth, in cm, of the bottom of each successive layer of soil. The first four depths are
12, 19, 26, 33, ...
The depths form an arithmetic sequence.

(a)

Write down the next two depths in the sequence.

(2)

(b)

Find an expression, in terms of n, for the nth depth.

(2)

(c)

Show that 306 cm is a depth in the sequence and find which layer it belongs to.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 40	✓ 1
B1 47	✓ 1
Part (b)
M1 for 7n (common difference 7)	✓ 1
A1 7n + 5	✓ 1
Part (c)
M1 7n + 5 = 306	✓ 1
M1 7n = 301 so n = 301 ÷ 7	✓ 1
A1 n = 43, a whole number, so 306 is the bottom of layer 43	✓ 1

Common error: Remember that the nth term is the common difference times n plus a correction, not the common difference plus the first term.

+ Add to test✓ In your test70IGME3-3-13r33-02Graphs, gradients and real-life graphs Paper FEasy8 marks

A cyclist rides along a straight road. For the first 1.5 hours she cycles at a constant speed of 24 km/h. She then stops for 20 minutes. She then cycles a further 12 km at a constant speed in 40 minutes, finishing her ride.

(a)

Calculate the distance she cycles in the first 1.5 hours.

(2)

(b)

Calculate her speed, in km/h, during the final 40 minutes of the ride.

(2)

(c)

Calculate her average speed, in km/h, for the whole ride, including the stop.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 24 × 1.5	✓ 1
A1 36 km	✓ 1
Part (b)
M1 12 ÷ (40/60) or 12 ÷ (2/3)	✓ 1
A1 18 km/h	✓ 1
Part (c)
M1 total distance = 36 + 12 = 48 km	✓ 1
M1 total time = 1.5 + 1/3 + 2/3 = 2.5 hours	✓ 1
M1 48 ÷ 2.5	✓ 1
A1 19.2 km/h	✓ 1

Exam tip: Convert minutes to hours before using speed = distance ÷ time with speeds in km/h.

+ Add to test✓ In your test71IGME3-3-13r33-03Sequences and nth terms Paper FEasy6 marks

The cooling power, P kW, needed in a small data centre depends on the number of server racks, n, installed. The relationship is linear. When n = 2, P = 35 and when n = 10, P = 83.

(a)

Work out the gradient of the line.

(2)

(b)

Find an equation for P in terms of n.

(2)

(c)

Use your equation to find the cooling power needed for 25 racks.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (83 − 35) ÷ (10 − 2)	✓ 1
A1 6	✓ 1
Part (b)
M1 35 = 6 × 2 + c or 83 = 6 × 10 + c	✓ 1
A1 P = 6n + 23	✓ 1
Part (c)
M1 6 × 25 + 23 (ECF from (b))	✓ 1
A1 173 kW	✓ 1

Common error: Subtract the coordinates in the same order on the top and bottom of the gradient fraction.

+ Add to test✓ In your test72IGME3-3-13r33-04Function notation, composite and inverse functions Paper FEasy6 marks

The function f is given by f(x) = 3x − 7 and the function g is given by g(x) = x² + 1.

(a)

Work out f(5).

(1)

(b)

Work out g(−4).

(2)

(c)

Solve f(x) = 20.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 8	✓ 1
Part (b)
M1 (−4)² + 1	✓ 1
A1 17	✓ 1
Part (c)
M1 3x − 7 = 20	✓ 1
M1 3x = 27	✓ 1
A1 x = 9	✓ 1

Common error: Square a negative number inside brackets: (−4)² is 16, not −16.

+ Add to test✓ In your test73IGME3-3-12r35-01Sequences and nth terms Paper FEasy5 marks

A space probe reports its distance from Earth, in thousand kilometres, at the end of each day. The first four distances are 12, 19, 26, 33.

(a)

Write down the next two terms of the sequence.

(1)

(b)

Find an expression, in terms of n, for the nth term of the sequence.

(2)

(c)

The probe sends a signal when it is 201 thousand kilometres from Earth. On which day does this happen?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 40 and 47	✓ 1
Part (b)
M1 common difference of 7 (or 7n seen)	✓ 1
A1 7n + 5	✓ 1
Part (c)
M1 7n + 5 = 201 (ECF from (b))	✓ 1
A1 day 28	✓ 1

Exam tip: Check an nth term by substituting n = 1: it must give the first term of the sequence.

+ Add to test✓ In your test74IGME3-3-12r35-02Graphs, gradients and real-life graphs Paper FEasy5 marks

In a coffee roaster the temperature starts at 20 °C and then rises at a constant rate of 12 °C every minute for the first 15 minutes of roasting.

(a)

Work out the temperature after 6 minutes.

(2)

(b)

The temperature, T °C, is plotted against the time, t minutes, as a straight line. State what the gradient of this line represents.

(1)

(c)

Work out the time taken for the temperature to reach 128 °C.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 20 + 12 × 6	✓ 1
A1 92 °C	✓ 1
Part (b)
B1 the rate of increase of temperature (12 °C per minute)	✓ 1
Part (c)
M1 (128 − 20) ÷ 12	✓ 1
A1 9 minutes	✓ 1

Common error: The starting value is the intercept, not the gradient: do not multiply the starting temperature by the time.

+ Add to test✓ In your test75IGME3-3-12r35-03Graphs, gradients and real-life graphs Paper FEasy5 marks

The straight deck of a bridge rises steadily from one end. When the horizontal distance from the start of the deck is x metres, the height of the deck above the water is y metres. The deck is 4.0 m above the water when x = 20 and 6.0 m above the water when x = 60.

Diagram not to scale

(a)

Work out the gradient of the deck.

(2)

(b)

Find the equation of the line in the form y = mx + c.

(2)

(c)

Work out the height of the deck above the water when x = 90.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (6 − 4) ÷ (60 − 20)	✓ 1
A1 0.05	✓ 1
Part (b)
M1 y = 0.05x + c with a correct substitution, e.g. 4 = 1 + c (ECF from (a))	✓ 1
A1 y = 0.05x + 3	✓ 1
Part (c)
B1 7.5 m (ECF from (b))	✓ 1

Exam tip: Always check your equation by substituting the other given point into it.

+ Add to test✓ In your test76IGME3-3-12r35-04Sequences and nth terms Paper FEasy5 marks

The power output of a space probe, in watts, is recorded at the end of each week. The first four readings are 40, 34, 28, 22.

(a)

Write down the next two terms of the sequence.

(1)

(b)

Find an expression, in terms of n, for the nth term of the sequence.

(2)

(c)

In one week the power output is 4 watts. Work out which week this is.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 16 and 10	✓ 1
Part (b)
M1 common difference of −6 (or −6n seen)	✓ 1
A1 46 − 6n	✓ 1
Part (c)
M1 46 − 6n = 4 (ECF from (b))	✓ 1
A1 week 7	✓ 1

Common error: In a decreasing sequence the nth term starts with a negative multiple of n; writing 6n + 34 gives a sequence that goes up.

+ Add to test✓ In your test77IGME3-3-12r37-01Sequences and nth terms Paper FEasy5 marks

In a game, the experience points (XP) needed to reach levels 1, 2, 3, 4, … are
40, 55, 70, 85, …

(a)

Write down the next term of the sequence.

(1)

(b)

Find an expression, in terms of n, for the nth term of the sequence.

(2)

(c)

Level N needs exactly 460 XP. Find N.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 100	✓ 1
Part (b)
M1 common difference 15, so 15n	✓ 1
A1 15n + 25	✓ 1
Part (c)
M1 15N + 25 = 460	✓ 1
A1 N = 29	✓ 1

Exam tip: To check an nth term, substitute n = 1 and n = 2 and compare with the first two terms.

+ Add to test✓ In your test78IGME3-3-12r37-02Graphs, gradients and real-life graphs Paper FEasy5 marks

A water tank holds 800 litres at the start. Water drains out at a constant rate. A graph of the volume of water in the tank against time is a straight line from (0, 800) to (40, 0), where time is in minutes.

Diagram not to scale

(a)

Calculate the rate at which water drains out, in litres per minute.

(2)

(b)

The gradient of the graph is negative. State what this tells you about the volume of water.

(1)

(c)

Find the time at which 250 litres remain in the tank.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 800 ÷ 40	✓ 1
A1 20 litres per minute	✓ 1
Part (b)
B1 the volume is decreasing	✓ 1
Part (c)
M1 (800 − 250) ÷ 20 (or 800 − 20t = 250)	✓ 1
A1 27.5 minutes	✓ 1

Common error: The gradient of a graph of volume against time is a rate; its units are litres per minute, not litres.

+ Add to test✓ In your test79IGME3-3-12r37-04Graphs, gradients and real-life graphs Paper FEasy7 marks

A cloud gaming service charges a cost of €C for h hours of play. The graph of C against h is a straight line passing through the points (2, 7) and (6, 19).

Diagram not to scale

(a)

Calculate the gradient of the line.

(2)

(b)

Find the equation of the line in the form C = mh + c.

(2)

(c)

State what the gradient represents in this situation.

(1)

(d)

Find the number of hours of play that cost €25.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (19 − 7) ÷ (6 − 2)	✓ 1
A1 3	✓ 1
Part (b)
M1 7 = 3 × 2 + c	✓ 1
A1 C = 3h + 1	✓ 1
Part (c)
B1 the cost per hour of play (€3 per hour)	✓ 1
Part (d)
M1 3h + 1 = 25	✓ 1
A1 8 hours	✓ 1

Exam tip: Check your equation by substituting the coordinates of both given points.

+ Add to test✓ In your test80IGME3-3-12r37-07Graphs, gradients and real-life graphs Paper FEasy6 marks

In a game, the height h metres of a ball above the ground, t seconds after it is thrown, is given by h = 20t − 5t² for 0 ≤ t ≤ 4. The graph of h against t is symmetrical about a vertical line.

(a)

Calculate the height of the ball when t = 2.5.

(2)

(b)

When t = 1, the height is 15 m. Use symmetry to find another time when the height is 15 m.

(2)

(c)

Use your answers to find the maximum height of the ball.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 20 × 2.5 − 5 × 2.5²	✓ 1
A1 18.75 m	✓ 1
Part (b)
M1 the ball is also at height 15 m when t = 4 − 1	✓ 1
A1 3 seconds	✓ 1
Part (c)
M1 maximum is halfway between t = 1 and t = 3, so t = 2	✓ 1
A1 20 × 2 − 5 × 2² = 20 m	✓ 1

Exam tip: A quadratic graph is symmetrical, so its turning point is halfway between two points with the same height.

+ Add to test✓ In your test81IGME3-3-13r44-01Sequences and nth terms Paper FEasy6 marks

A beekeeper adds new hives to an apiary each spring. The numbers of hives at the end of the first four springs are
12, 17, 22, 27
The pattern continues in the same way.

(a)

Write down the next two terms of the sequence.

(2)

(b)

Find an expression, in terms of n, for the nth term of the sequence.

(2)

(c)

The beekeeper aims to have exactly 102 hives. Work out in which spring this happens.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 32	✓ 1
B1 37	✓ 1
Part (b)
M1 common difference 5 or 5n seen	✓ 1
A1 5n + 7	✓ 1
Part (c)
M1 5n + 7 = 102 (or (102 − 7) ÷ 5) (ECF from (b))	✓ 1
A1 n = 19 (19th spring)	✓ 1

Exam tip: To find the nth term of a linear sequence, the number in front of n is the common difference. Check your rule by substituting n = 1 and n = 2.

+ Add to test✓ In your test82IGME3-3-13r44-02Graphs, gradients and real-life graphs Paper FEasy8 marks

The speed–time graph for a Formula 1 car on a test run consists of three straight-line sections. The car accelerates uniformly from rest to 60 m/s in 8 seconds. It then travels at a constant 60 m/s for 5 seconds. Finally it decelerates uniformly to rest in 12 seconds.

Diagram not to scale

(a)

Work out the acceleration of the car during the first 8 seconds.

(2)

(b)

Work out the total distance travelled by the car.

(4)

(c)

Work out the average speed of the car over the whole run.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 60 ÷ 8	✓ 1
A1 7.5 m/s²	✓ 1
Part (b)
M1 area of first triangle ½ × 8 × 60 or 240	✓ 1
M1 area of rectangle 60 × 5 or 300	✓ 1
M1 area of last triangle ½ × 12 × 60 or 360	✓ 1
A1 900 m	✓ 1
Part (c)
M1 total distance ÷ total time = 900 ÷ 25 (ECF from (b))	✓ 1
A1 36 m/s	✓ 1

Common error: Acceleration is the gradient of a speed–time graph, but distance is the area under it. Do not use the gradient when the question asks for distance.

+ Add to test✓ In your test83IGME3-3-13r44-03Sequences and nth terms Paper FEasy6 marks

A hospital pharmacy uses the rule d(w) = 4w + 10 to work out the dose, d mg, of a medicine for a child of mass w kg.

(a)

Work out d(15).

(2)

(b)

Show that the dose for a child is always 12 mg more than the dose for a child who is 3 kg lighter.

(2)

(c)

A child is given a dose of 94 mg. Work out the mass of this child.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4 × 15 + 10	✓ 1
A1 70 (mg)	✓ 1
Part (b)
M1 d(w) − d(w − 3) = (4w + 10) − (4(w − 3) + 10) or 4w − 2 expression for lighter child	✓ 1
A1 4w + 10 − 4w + 12 − 10 = 12 shown	✓ 1
Part (c)
M1 4w + 10 = 94 or (94 − 10) ÷ 4	✓ 1
A1 21 (kg)	✓ 1

Exam tip: d(15) means replace w by 15. When a value of d is given instead, you must solve an equation.

+ Add to test✓ In your test84IGME3-3-13r44-04Graphs, gradients and real-life graphs Paper FEasy7 marks

A beekeeper records the mass, M kg, of honey stored in a hive d days after a nectar flow begins. The graph of M against d is a straight line that passes through the points (4, 3.5) and (10, 6.5).

Diagram not to scale

(a)

Work out the gradient of the line.

(2)

(b)

Write down what the gradient represents in this situation.

(1)

(c)

Find the equation of the line, giving M in terms of d.

(2)

(d)

Use your equation to work out the mass of honey after 22 days.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (6.5 − 3.5) ÷ (10 − 4)	✓ 1
A1 0.5	✓ 1
Part (b)
B1 the mass of honey increases by 0.5 kg each day	✓ 1
Part (c)
M1 M = 0.5d + c with a correct point substituted, e.g. 3.5 = 0.5 × 4 + c	✓ 1
A1 M = 0.5d + 1.5	✓ 1
Part (d)
M1 0.5 × 22 + 1.5 (ECF from (c))	✓ 1
A1 12.5 (kg)	✓ 1

Exam tip: Always give the units of a gradient in words: it is the change in the vertical quantity for each one unit of the horizontal quantity.

+ Add to test✓ In your test85IGME3-3-12r48-02Graphs, gradients and real-life graphs Paper FEasy7 marks

A cyclist leaves home at 09:00. On a distance–time graph with time in hours and distance in km, the first stage is a straight line from (0, 0) to (0.5, 9). The cyclist then rests for 15 minutes (a horizontal line). The final stage is a straight line from the end of the rest to the point (1.25, 21).

Diagram not to scale

(a)

Work out the cyclist's speed during the first stage.

(2)

(b)

Write down the time at which the rest ends.

(1)

(c)

Work out the cyclist's speed during the final stage.

(2)

(d)

Work out the average speed for the whole ride, including the rest.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 9 ÷ 0.5	✓ 1
A1 18 km/h	✓ 1
Part (b)
B1 09:45	✓ 1
Part (c)
M1 (21 − 9) ÷ (1.25 − 0.75)	✓ 1
A1 24 km/h	✓ 1
Part (d)
M1 21 ÷ 1.25	✓ 1
A1 16.8 km/h	✓ 1

Common error: Average speed uses total distance divided by total time, including any rest; it is not the mean of the stage speeds.

+ Add to test✓ In your test86IGME3-3-12r48-03Graphs, gradients and real-life graphs Paper FEasy7 marks

The power, P kilowatts, used by a data-centre room depends on the number of servers, n, that are switched on. The relationship is a straight line passing through the points (10, 12) and (50, 28).

Diagram not to scale

(a)

Work out the gradient of the line.

(2)

(b)

Explain what the gradient represents in this situation.

(1)

(c)

Find the equation of the line, in terms of P and n.

(2)

(d)

Work out the power used when 120 servers are switched on.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (28 − 12) ÷ (50 − 10)	✓ 1
A1 0.4	✓ 1
Part (b)
B1 each extra server switched on uses an extra 0.4 kW	✓ 1
Part (c)
M1 12 = 0.4 × 10 + c (ECF from (a))	✓ 1
A1 P = 0.4n + 8	✓ 1
Part (d)
M1 0.4 × 120 + 8 (ECF from (c))	✓ 1
A1 56 kW	✓ 1

Exam tip: Substitute a point you were given into y = mx + c to find c, then check with the second point.

+ Add to test✓ In your test87IGME3-3-12r48-06Sequences and nth terms Paper FEasy5 marks

Archaeologists record the number of pottery fragments found at a series of sites. After the first two sites, the number at each site is the sum of the numbers at the previous two sites. Site 3 has 11 fragments and site 4 has 18 fragments.

(a)

Work out the number of fragments at site 2.

(1)

(b)

Work out the number of fragments at site 1.

(1)

(c)

Work out the number of fragments at site 5.

(1)

(d)

In a second sequence the nth term is n² + 1. Work out the 10th term.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 7 (18 − 11)	✓ 1
Part (b)
B1 4 (11 − 7)	✓ 1
Part (c)
B1 29 (11 + 18)	✓ 1
Part (d)
M1 10² + 1	✓ 1
A1 101	✓ 1

Common error: Work backwards by subtraction: the term before is the later term minus the term in between.

+ Add to test✓ In your test88IGME3-3-12r48-09Sequences and nth terms Paper FEasy6 marks

A data-centre cooling controller uses a two-step rule. The input, n, is multiplied by 3 and then 4 is subtracted to give the output.

(a)

Work out the output when the input is 6.

(1)

(b)

The output is 29. Work out the input.

(2)

(c)

Write down an expression for the output in terms of n.

(1)

(d)

Find the input for which the output is equal to the input.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 14	✓ 1
Part (b)
M1 (29 + 4) ÷ 3 or works backwards through the rule	✓ 1
A1 11	✓ 1
Part (c)
B1 3n − 4	✓ 1
Part (d)
M1 3n − 4 = n	✓ 1
A1 2	✓ 1

Common error: To reverse a function machine, undo the steps in the opposite order using inverse operations.

+ Add to test✓ In your test89IGME-060-01-06Graphs, gradients and real-life graphs Paper FMedium6 marks

The line L₁ has equation 2y = 6x − 5

(a)

Write down the gradient of L₁ and the coordinates of the point where L₁ crosses the y-axis.

(2)

(b)

The line L₂ is parallel to L₁ and passes through the point (1, 7).
Find an equation of L₂.

(2)

(c)

Find the x-coordinate of the point on L₂ where y = 10

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 gradient = 3	✓ 1
B1 (0, −2.5)	✓ 1
Part (b)
M1 for 7 = 3 × 1 + c	✓ 1
A1 y = 3x + 4	✓ 1
Part (c)
M1 for 3x + 4 = 10	✓ 1
A1 x = 2	✓ 1

Common error: The gradient is not 6. Rearrange to the form y = mx + c by dividing every term by 2 before reading off m and c.

+ Add to test✓ In your test90IGME-060-01-12Graphs, gradients and real-life graphs Paper FMedium6 marks

A wheelchair ramp rises a vertical height of 0.45 m over a horizontal distance of 3.6 m.

Not drawn to accurate scale

(a)

Work out the gradient of the ramp.

(2)

(b)

A safety guideline says the gradient of a ramp must not be greater than \(\frac{1}{12}\)
Does this ramp meet the guideline? Give a reason for your answer.

(2)

(c)

Work out the minimum horizontal distance needed for a ramp rising 0.45 m to meet the guideline.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 0.45 ÷ 3.6	✓ 1
A1 0.125	✓ 1
Part (b)
M1 for converting to a comparable form, e.g. \(\frac{1}{12}\) = 0.0833… or 0.125 = \(\frac{1}{8}\)	✓ 1
A1 No, because 0.125 > 0.0833	✓ 1
Part (c)
M1 for 0.45 ÷ \(\frac{1}{12}\) or 0.45 × 12	✓ 1
A1 5.4 m	✓ 1

Exam tip: A larger gradient means a steeper slope. To compare a gradient with a fraction, write both as decimals.

+ Add to test✓ In your test91IGME-060-00-05Sequences and nth terms Paper FMedium6 marks

Jo makes patterns of hexagons in a row using matchsticks. Hexagons that are next to each other share one matchstick.
Pattern 1 uses 6 matchsticks, Pattern 2 uses 11 matchsticks and Pattern 3 uses 16 matchsticks.
The numbers of matchsticks form an arithmetic sequence.

The first three patterns

(a)

How many matchsticks are needed for Pattern 6?

(1)

(b)

Find an expression, in terms of n, for the number of matchsticks in Pattern n.

(2)

(c)

Jo has 200 matchsticks. She makes the largest pattern she can.
Find the pattern number of this pattern and the number of matchsticks she has left over.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 31	✓ 1
Part (b)
M1: 5n + k for any constant k	✓ 1
A1: 5n + 1	✓ 1
Part (c)
M1: 5n + 1 ≤ 200 or (200 − 1) ÷ 5 = 39.8	✓ 1
A1: Pattern 39	✓ 1
A1: 4 matchsticks left over	✓ 1

Exam tip: When n has to be a whole number and the answer must not exceed a limit, round down. Do not round 39.8 to the nearest whole number.

+ Add to test✓ In your test92IGME-060-00-07Sequences and nth terms Paper FMedium6 marks

Sequence A has nth term 3n + 4.
Sequence B has nth term 40 − n.

(a)

For one value of n, the nth term of sequence A is equal to the nth term of sequence B.
Find this value of n and the value of the equal terms.

(3)

(b)

Find the smallest value of n for which the nth term of sequence A is more than 20 greater than the nth term of sequence B.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 3n + 4 = 40 − n	✓ 1
A1: n = 9	✓ 1
A1: 31	✓ 1
Part (b)
M1: (3n + 4) − (40 − n) > 20	✓ 1
M1: 4n − 36 > 20, so n > 14	✓ 1
A1: 15	✓ 1

Common error: When subtracting the expression 40 − n, keep it in brackets so that the sign of −n changes. Without brackets the n terms do not combine to 4n.

+ Add to test✓ In your test93IGME-060-01-10Graphs, gradients and real-life graphs Paper FMedium8 marks

Water drains from tank A at a constant rate.
The volume, V litres, of water in tank A after t minutes is given by
V = 600 − 15t

(a)

Write down the volume of water in tank A at the start.

(1)

(b)

Write down the gradient of the graph of V against t and explain what it represents.

(2)

(c)

Work out how long it takes for tank A to empty.

(2)

(d)

At the same time as tank A starts to drain, tank B starts with 120 litres and fills at a constant rate of 25 litres per minute.
Work out how many minutes after the start the two tanks contain the same volume of water.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 600 litres	✓ 1
Part (b)
B1 gradient = −15	✓ 1
B1 the volume decreases by 15 litres every minute; accept 'water drains at 15 litres per minute' or equivalent	✓ 1
Part (c)
M1 for 600 − 15t = 0 or 600 ÷ 15	✓ 1
A1 40 minutes	✓ 1
Part (d)
B1 V = 120 + 25t for tank B, or initial difference 600 − 120 = 480 litres, or combined rate 15 + 25 = 40 litres per minute	✓ 1
M1 for 600 − 15t = 120 + 25t or 480 ÷ 40	✓ 1
A1 12 minutes	✓ 1

Exam tip: The point where two lines on a real-life graph cross is where the two quantities are equal, so set the two formulas equal and solve.

+ Add to test✓ In your test94IGME-060-01-08Graphs, gradients and real-life graphs Paper FMedium7 marks

A train starts from rest and accelerates uniformly to a speed of 24 m/s in 30 seconds.
It then travels at 24 m/s for 90 seconds.
It then decelerates uniformly and comes to rest after a further 40 seconds.
You may find it helpful to sketch a speed–time graph of the journey.

(a)

Work out the acceleration of the train during the first 30 seconds.

(2)

(b)

Work out the total distance travelled by the train.

(3)

(c)

Work out the average speed of the train for the whole journey.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 24 ÷ 30	✓ 1
A1 0.8 m/s²	✓ 1
Part (b)
M1 for any one correct area, e.g. ½ × 30 × 24 = 360 or 90 × 24 = 2160 or ½ × 40 × 24 = 480 (or a trapezium method, e.g. ½ × (90 + 160) × 24)	✓ 1
M1 for a complete method for the total area, e.g. 360 + 2160 + 480	✓ 1
A1 3000 m	✓ 1
Part (c)
M1 for 3000 ÷ 160, or their total distance ÷ 160	✓ 1
A1 18.75 m/s (accept 18.8)	✓ 1

Model sketch

Model sketch (not to scale): speed–time graph; the area under it is the distance.

Exam tip: On a speed–time graph, the gradient gives the acceleration and the area under the graph gives the distance. Split the area into triangles and a rectangle, or treat it as one trapezium.

+ Add to test✓ In your test95IGME3-3-09d-06Graphs, gradients and real-life graphs Paper FMedium9 marks

A taxi firm, Company A, charges a fixed fee plus a rate per kilometre. The cost, €C, of a journey of d km is found from a straight-line graph that passes through the points (0, 3.50) and (10, 15.50).

(a)

Find the equation of the graph in the form C = md + c.

(2)

(b)

Explain what the gradient and the C-intercept represent in this situation.

(2)

(c)

A journey costs €21.50. Work out its length.

(2)

(d)

Company B charges €0.50 plus €1.50 per kilometre. For what journey length do both companies charge the same?

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 gradient (15.50 − 3.50) ÷ 10 = 1.2	✓ 1
B1 C = 1.2d + 3.5	✓ 1
Part (b)
B1 gradient: cost per kilometre is €1.20	✓ 1
B1 intercept: fixed charge of €3.50	✓ 1
Part (c)
M1 (21.5 − 3.5) ÷ 1.2	✓ 1
A1 15 km	✓ 1
Part (d)
M1 1.2d + 3.5 = 1.5d + 0.5	✓ 1
M1 3 = 0.3d or equivalent rearrangement	✓ 1
A1 d = 10 km	✓ 1

Exam tip: To compare two linear costs, set the two expressions equal and solve for the shared variable.

+ Add to test✓ In your test96IGME3-3-09d-07Graphs, gradients and real-life graphs Paper FMedium5 marks

On a plan of a roof, one solar-panel rail lies along the line y = 4x − 5. A second rail is parallel to the first and passes through the point (3, 2). Lengths are in metres.

(a)

Find the equation of the second rail.

(3)

(b)

Find the x-coordinate of the point where the second rail meets the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient of 4 used because the lines are parallel	✓ 1
M1 substitute (3, 2) into y = 4x + c, so 2 = 12 + c	✓ 1
A1 y = 4x − 10	✓ 1
Part (b)
M1 0 = 4x − 10 (ECF from (a))	✓ 1
A1 2.5	✓ 1

Common error: Parallel lines share the same gradient, not the same intercept. Do not copy the −5 from the first equation.

+ Add to test✓ In your test97IGME3-3-09d-08Sequences and nth terms Paper FMedium6 marks

The first row of seats in a small theatre has 14 seats. Each row has 3 more seats than the row in front of it.

(a)

Show that the number of seats in row n is 3n + 11.

(2)

(b)

Explain why no row has exactly 100 seats.

(2)

(c)

The last row has 98 seats. How many rows are there?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 common difference 3 so the expression starts 3n	✓ 1
B1 14 − 3 = 11 hence 3n + 11	✓ 1
Part (b)
M1 3n + 11 = 100 gives n = 29.67	✓ 1
A1 not a whole number, so no such row	✓ 1
Part (c)
M1 3n + 11 = 98	✓ 1
A1 29 rows	✓ 1

Exam tip: To test whether a number is in a sequence, set the nth term equal to it and check that n is a positive whole number.

+ Add to test✓ In your test98IGME3-3-12r6-11Differentiation and stationary points Paper FMedium6 marks

In a video game, a straight path on the map passes through the points (−2, 1) and (4, 10).

Diagram not to scale

(a)

Work out the gradient of the path.

(2)

(b)

Find the equation of the path in the form y = mx + c.

(2)

(c)

The point (k, 16) is on the path. Work out the value of k.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (10 − 1) ÷ (4 − (−2))	✓ 1
A1 1.5	✓ 1
Part (b)
M1 y = 1.5x + c with a point substituted (ECF from (a))	✓ 1
A1 y = 1.5x + 4	✓ 1
Part (c)
M1 1.5k + 4 = 16	✓ 1
A1 k = 8	✓ 1

Common error: When subtracting negative coordinates, 4 − (−2) is 6, not 2.

+ Add to test✓ In your test99IGME3-3-12r6-13Differentiation and stationary points Paper FMedium7 marks

In a game, the height profile of a valley is modelled by y = x² − 4x − 5, where y is the height in metres above the map's base level at horizontal position x metres.

(a)

Work out the value of y when x = −2.

(1)

(b)

Solve x² − 4x − 5 = 0.

(3)

(c)

Work out the lowest height of the valley.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 7	✓ 1
Part (b)
M1 (x − 5)(x + 1) = 0	✓ 1
A1 x = 5	✓ 1
A1 x = −1	✓ 1
Part (c)
M1 lowest point is midway between the roots: x = (−1 + 5) ÷ 2 = 2 (ECF from (b))	✓ 1
M1 substitute x = 2: 4 − 8 − 5	✓ 1
A1 −9 m	✓ 1

Exam tip: A quadratic graph is symmetrical, so its turning point is midway between its roots.

+ Add to test✓ In your test100IGME3-3-12r6-14Graphs, gradients and real-life graphs Paper FMedium5 marks

Two water suppliers charge by the volume of water used. Supplier A charges a fixed fee of £12 plus £1.80 for every m³. Supplier B charges a fixed fee of £30 plus £0.90 for every m³.

(a)

Work out the cost with Supplier A for 15 m³ of water.

(2)

(b)

Work out the volume of water for which the two suppliers charge the same amount.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 + 1.80 × 15	✓ 1
A1 £39.00	✓ 1
Part (b)
M1 12 + 1.8v = 30 + 0.9v	✓ 1
M1 0.9v = 18	✓ 1
A1 20 m³	✓ 1

Common error: Use the fixed fee once only; the per-unit charge is multiplied by the volume.

+ Add to test✓ In your test101IGME3-3-12r11-03Graphs, gradients and real-life graphs Paper FMedium6 marks

The cost, £C, of hosting a library event for n people is shown by a straight line graph. The line passes through the points \((10, 85)\) and \((30, 165)\).

Diagram not to scale

(a)

Work out the gradient of the line.

(2)

(b)

Find the equation of the line in the form \(C = an + b\).

(2)

(c)

Use your equation to work out the cost for 50 people.

(1)

(d)

Explain what the gradient represents in this context.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\frac{165 - 85}{30 - 10}\)	✓ 1
A1 4	✓ 1
Part (b)
M1 \(85 = 4 \times 10 + b\) or equivalent	✓ 1
A1 \(C = 4n + 45\)	✓ 1
Part (c)
A1 £245 (ECF from (b))	✓ 1
Part (d)
B1 the cost per extra person is £4	✓ 1

Common error: The gradient is the change in cost divided by the change in people; do not divide in the opposite order.

+ Add to test✓ In your test102IGME3-3-12r11-04Graphs, gradients and real-life graphs Paper FMedium5 marks

On a map of a wind farm, a cable runs along the line \(y = 3x - 2\), with distances in kilometres. A second cable runs parallel to the first and passes through the point \((2, 9)\).

(a)

Find the equation of the second cable.

(2)

(b)

Find the x-coordinate of the point where the second cable crosses the x-axis.

(2)

(c)

Show that the point \((10, 33)\) lies on the second cable.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient 3 used with point \((2, 9)\), e.g. \(9 = 3 \times 2 + c\)	✓ 1
A1 \(y = 3x + 3\)	✓ 1
Part (b)
M1 \(3x + 3 = 0\)	✓ 1
A1 \(x = -1\)	✓ 1
Part (c)
B1 \(3 \times 10 + 3 = 33\)	✓ 1

Common error: Parallel lines have the same gradient, but not the same intercept; do not copy the \(-2\).

+ Add to test✓ In your test103IGME3-3-12r11-11Graphs, gradients and real-life graphs Paper FMedium7 marks

A swimmer's speed is shown on a speed-time graph. It rises uniformly from 0 m/s to 2.4 m/s in 3 seconds, stays constant at 2.4 m/s for 40 seconds, then falls uniformly to 0 m/s in 6 seconds.

Diagram not to scale

(a)

Work out the acceleration during the first 3 seconds.

(2)

(b)

Work out the total distance swum.

(3)

(c)

Work out the average speed for the whole 49 seconds. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\frac{2.4}{3}\)	✓ 1
A1 0.8 m/s²	✓ 1
Part (b)
M1 area of a triangle \(\frac{1}{2} \times 3 \times 2.4 = 3.6\)	✓ 1
M1 rectangle \(40 \times 2.4 = 96\) and triangle \(\frac{1}{2} \times 6 \times 2.4 = 7.2\)	✓ 1
A1 106.8 m	✓ 1
Part (c)
M1 \(106.8 \div 49\)	✓ 1
A1 2.18 m/s	✓ 1

Common error: The distance is the area under a speed-time graph, so a triangle needs the factor \(\frac{1}{2}\).

+ Add to test✓ In your test104IGME3-3-12r11-12Graphs, gradients and real-life graphs Paper FMedium5 marks

A water jet at a swimming centre follows the path \(y = -x^2 + 4x - 3\), where x is the horizontal distance in metres from a fixed point and y is the height in metres above the water surface, for values where \(y \ge 0\).

(a)

Find the two values of x at which the jet meets the water surface.

(2)

(b)

Write down the equation of the line of symmetry of the path.

(1)

(c)

Work out the maximum height of the jet above the water surface.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(-x^2 + 4x - 3 = 0\) and factorise, \((x - 1)(x - 3) = 0\)	✓ 1
A1 \(x = 1\) and \(x = 3\)	✓ 1
Part (b)
B1 \(x = 2\)	✓ 1
Part (c)
M1 substitute \(x = 2\)	✓ 1
A1 1 m	✓ 1

Exam tip: The turning point of a parabola lies on its line of symmetry, halfway between the two roots.

+ Add to test✓ In your test105IGME3-3-12r12-04Sequences and nth terms Paper FMedium6 marks

A swimming pool is a square of side n tiles. It is surrounded by a border one tile wide. Pool size 1 has 8 border tiles, size 2 has 12 border tiles and size 3 has 16 border tiles.

Shaded: the pool. Unshaded squares: border tiles.

(a)

Find an expression, in terms of n, for the number of border tiles.

(2)

(b)

Find the size of pool that has 100 border tiles.

(2)

(c)

Each tile costs £2.35. Work out the cost of the border tiles for the pool in part (b).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Differences of 4 so 4n	✓ 1
4n + 4	✓ 1
Part (b)
4n + 4 = 100	✓ 1
n = 24	✓ 1
Part (c)
100 × 2.35	✓ 1
£235	✓ 1

Exam tip: Check your nth term against the first two or three terms of the pattern.

+ Add to test✓ In your test106IGME3-3-12r12-05Sequences and nth terms Paper FMedium7 marks

Two library branches track their memberships. In month n, Branch A has 18n + 240 members at the start of the month. Branch B has 150, 180, 210, ... members in months 1, 2, 3, ..., increasing in an arithmetic sequence.

(a)

Find an expression for the number of members of Branch B in month n.

(2)

(b)

Find the month in which both branches have the same number of members.

(3)

(c)

Work out the number of members of each branch in that month.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Common difference 30	✓ 1
30n + 120	✓ 1
Part (b)
18n + 240 = 30n + 120	✓ 1
12n = 120	✓ 1
n = 10	✓ 1
Part (c)
18 × 10 + 240 = 420 (or 30 × 10 + 120 = 420)	✓ 1
Each branch has 420 members	✓ 1

Exam tip: To find when two sequences are equal, set their nth terms equal and solve.

+ Add to test✓ In your test107IGME3-3-12r12-06Sequences and nth terms Paper FMedium6 marks

A swimmer's training plan has distances that form an arithmetic sequence. Session 5 is 1550 m and session 12 is 3300 m.

(a)

Work out the common difference.

(2)

(b)

Find an expression for the distance in session n.

(2)

(c)

In which session is the distance 3800 m?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(3300 − 1550) ÷ (12 − 5)	✓ 1
250 m	✓ 1
Part (b)
250n	✓ 1
250n + 300 (check: 5 × 250 + 300 = 1550)	✓ 1
Part (c)
250n + 300 = 3800	✓ 1
n = 14	✓ 1

Exam tip: The difference between terms 5 places apart is 5 times the common difference; divide the change in value by the gap in positions.

+ Add to test✓ In your test108IGME3-3-12r12-07Graphs, gradients and real-life graphs Paper FMedium7 marks

A library charges for printing in two ways. Plan A: £2 fixed charge plus £0.05 per page. Plan B: £0.15 per page with no fixed charge. The cost of printing p pages is £C.

(a)

Write an equation for C in terms of p for each plan.

(2)

(b)

Find the number of pages for which both plans cost the same, and that cost.

(3)

(c)

For 50 pages, work out how much cheaper Plan A is than Plan B.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Plan A: C = 0.05p + 2	✓ 1
Plan B: C = 0.15p	✓ 1
Part (b)
0.05p + 2 = 0.15p	✓ 1
p = 20	✓ 1
£3.00	✓ 1
Part (c)
Plan A: 2 + 0.05 × 50 = 4.50 and Plan B: 0.15 × 50 = 7.50	✓ 1
£3.00	✓ 1

Exam tip: The point where two cost lines cross is where the plans cost the same; check on either side to see which is cheaper.

+ Add to test✓ In your test109IGME3-3-11r14-05Graphs, gradients and real-life graphs Paper FMedium5 marks

A mobile testing van leaves a depot, drives to a village, waits there and returns. A graph of its distance from the depot (km) against time (hours) is made of three straight lines: from (0, 0) to (0.5, 9), then horizontal to (0.75, 9), then a straight line to (1.25, 0).

Diagram not to scale

(a)

Work out the speed of the van on its outward journey.

(2)

(b)

Work out how many minutes the van waits in the village.

(1)

(c)

Work out the average speed of the van for the whole journey.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
9 ÷ 0.5	✓ 1
18 km/h	✓ 1
Part (b)
(0.75 − 0.5) × 60 = 15 minutes	✓ 1
Part (c)
Total distance = 18 km, total time = 1.25 h	✓ 1
18 ÷ 1.25 = 14.4 km/h	✓ 1

Common error: Average speed is total distance divided by total time, including any time spent stationary; it is not the mean of the two travelling speeds.

+ Add to test✓ In your test110IGME3-3-11r14-06Sequences and nth terms Paper FMedium5 marks

A solar farm is built in rows of panels. The number of panels in each row increases by the same amount from one row to the next. Row 4 has 19 panels and row 10 has 49 panels.

(a)

Find an expression, in terms of n, for the number of panels in row n.

(3)

(b)

A row has 94 panels. Work out its row number.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Difference per row = (49 − 19) ÷ 6	✓ 1
5n seen	✓ 1
5n − 1	✓ 1
Part (b)
5n − 1 = 94 (ECF from (a))	✓ 1
n = 19	✓ 1

Exam tip: Check your nth term by substituting a known row number: 5 × 4 − 1 = 19 as given.

+ Add to test✓ In your test111IGME3-3-11r14-07Sequences and nth terms Paper FMedium5 marks

A model predicts the number of new cases each week, in hundreds. The first term is 4 and each later term is found by the rule: multiply the previous term by 2, then add 3.

(a)

Work out the 2nd term and the 3rd term.

(2)

(b)

Show that the 4th term is 53.

(1)

(c)

A different sequence uses the same rule but starts with another first term. Its 2nd term is 31. Work out its first term.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
2 × 4 + 3 = 11	✓ 1
2 × 11 + 3 = 25	✓ 1
Part (b)
2 × 25 + 3 = 53	✓ 1
Part (c)
2x + 3 = 31	✓ 1
x = 14	✓ 1

Exam tip: To go backwards through a term-to-term rule, undo the operations in reverse order: subtract 3, then divide by 2.

+ Add to test✓ In your test112IGME3-3-11r14-12Graphs, gradients and real-life graphs Paper FMedium7 marks

A courier charges a fixed fee plus an amount for each kilometre travelled. The graph of the charge, £C, against the distance, d km, is a straight line. A 2 km delivery costs £11 and an 8 km delivery costs £23.

Diagram not to scale

(a)

Work out the gradient of the line.

(2)

(b)

Find the equation of the line in the form C = md + c.

(2)

(c)

Write down what the 7 represents.

(1)

(d)

A delivery costs £31. Work out the distance.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(23 − 11) ÷ (8 − 2)	✓ 1
2	✓ 1
Part (b)
C = 2d + c with a point substituted	✓ 1
C = 2d + 7	✓ 1
Part (c)
The fixed fee, £7, charged for any delivery	✓ 1
Part (d)
2d + 7 = 31 (ECF from (b))	✓ 1
12 km	✓ 1

Exam tip: Check an equation by substituting both original points: 2 × 8 + 7 = 23 and 2 × 2 + 7 = 11.

+ Add to test✓ In your test113IGME3-3-11r15-03Sequences and nth terms Paper FMedium7 marks

A café buys flour from one of two suppliers. Supplier A charges a delivery fee of €12 plus €0.80 per kilogram. Supplier B has no delivery fee and charges €1.10 per kilogram.

(a)

Write down an expression, in terms of k, for the total cost in euros of buying k kg from each supplier.

(2)

(b)

Find the mass for which both suppliers charge the same.

(3)

(c)

The café needs 50 kg of flour. Work out how much more the dearer supplier charges. Give your answer in euros.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 Supplier A: 12 + 0.8k	✓ 1
B1 Supplier B: 1.1k	✓ 1
Part (b)
M1 12 + 0.8k = 1.1k (ECF from (a))	✓ 1
M1 12 = 0.3k	✓ 1
A1 40 kg	✓ 1
Part (c)
M1 A: 12 + 0.8 × 50 = 52 and B: 1.1 × 50 = 55	✓ 1
A1 €3 (Supplier B is dearer)	✓ 1

Exam tip: A common method for two linear costs is to write both as expressions and set them equal. Check the answer by substituting back into both.

+ Add to test✓ In your test114IGME3-3-11r15-08Sequences and nth terms Paper FMedium5 marks

A telecoms engineer lays cable along a trench. On day 1 she lays 35 m of cable. On each following day she lays 8 m more than on the previous day, so the lengths laid on days 1, 2, 3, ... form an arithmetic sequence.

(a)

Find an expression, in terms of n, for the length of cable laid on day n.

(2)

(b)

On which day does she first lay more than 150 m of cable?

(2)

(c)

How much cable does she lay on that day?

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 common difference 8 or 8n seen	✓ 1
A1 8n + 27	✓ 1
Part (b)
M1 8n + 27 > 150 (ECF from (a))	✓ 1
A1 day 16 (since n > 15.375)	✓ 1
Part (c)
A1 155 m (ECF from (b))	✓ 1

Exam tip: When solving an inequality in n, round up to the next whole number if you want the first term above the target.

+ Add to test✓ In your test115IGME3-3-11r15-11Graphs, gradients and real-life graphs Paper FMedium5 marks

On a map of a ski area, 1 unit represents 100 m. One cable-car line has equation y = 3x − 2. A second line, B, is parallel to the first and passes through the point (2, 9).

(a)

Write down the gradient of line B.

(1)

(b)

Find the equation of line B.

(2)

(c)

Find the x-coordinate of the point where line B crosses the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 3	✓ 1
Part (b)
M1 y − 9 = 3(x − 2) or 9 = 3 × 2 + c	✓ 1
A1 y = 3x + 3	✓ 1
Part (c)
M1 0 = 3x + 3 (ECF from (b))	✓ 1
A1 x = −1	✓ 1

Exam tip: Parallel lines have equal gradients. A line crosses the x-axis where y = 0.

+ Add to test✓ In your test116IGME3-3-12r15-11Function notation, composite and inverse functions Paper FMedium6 marks

A function is defined by f(x) = 2x² − 5.

(a)

Work out f(3).

(2)

(b)

Work out f(−2).

(1)

(c)

Solve f(x) = 45.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2 × 3² − 5	✓ 1
A1 13	✓ 1
Part (b)
A1 3	✓ 1
Part (c)
M1 2x² − 5 = 45	✓ 1
M1 x² = 25	✓ 1
A1 x = 5 and x = −5	✓ 1

Common error: In f(−2) square the negative number as a whole: (−2)² = 4, not −4. When solving x² = 25, remember both the positive and negative roots.

+ Add to test✓ In your test117IGME3-3-12r15-14Differentiation and stationary points Paper FMedium6 marks

The graph of y = x² − 2x − 8 is a smooth U-shaped curve.

(a)

Find the coordinates of the points where the graph crosses the x-axis.

(2)

(b)

Write down the coordinates of the point where the graph crosses the y-axis.

(1)

(c)

Find the coordinates of the turning point of the graph.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 x² − 2x − 8 = 0 and factorises as (x − 4)(x + 2)	✓ 1
A1 (−2, 0) and (4, 0)	✓ 1
Part (b)
A1 (0, −8)	✓ 1
Part (c)
M1 x-coordinate is half-way between the roots: (−2 + 4) ÷ 2 (ECF from (a))	✓ 1
A1 x = 1	✓ 1
A1 y = 1 − 2 − 8 = −9, so (1, −9)	✓ 1

Exam tip: A quadratic graph is symmetrical, so its turning point has the x-value half-way between the two x-intercepts.

+ Add to test✓ In your test118IGME3-3-11r16-05Graphs, gradients and real-life graphs Paper FMedium7 marks

A cyclist leaves home at 09:00 and rides directly away along a straight road at a constant speed. She reaches a café 24 km from home at 10:30. She stays at the café for 30 minutes. She then rides straight back home at a constant speed and arrives at 14:00.

(a)

Work out her speed on the outward journey.

(2)

(b)

Work out her speed on the journey home.

(2)

(c)

Work out her average speed for the whole time from 09:00 to 14:00. Give your answer in km/h.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 24 ÷ 1.5	✓ 1
A1 16 km/h	✓ 1
Part (b)
M1 24 ÷ 3 (journey home takes 3 hours)	✓ 1
A1 8 km/h	✓ 1
Part (c)
M1 total distance 48 km	✓ 1
M1 total time 5 hours	✓ 1
A1 9.6 km/h	✓ 1

Common error: Average speed is total distance divided by total time, including any rest. It is not the mean of the two travelling speeds.

+ Add to test✓ In your test119IGME3-3-11r16-06Graphs, gradients and real-life graphs Paper FMedium6 marks

In a test at a data centre, 2 racks of servers draw 7 kW of power and 6 racks draw 19 kW. The power P kW drawn by n racks follows a straight-line relationship.

Diagram not to scale

(a)

Work out the gradient of the line.

(2)

(b)

Find the equation of the line, giving P in terms of n.

(2)

(c)

State what the number 1 in your equation represents in this context.

(1)

(d)

Use the equation to work out the power drawn by 25 racks.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (19 − 7) ÷ (6 − 2)	✓ 1
A1 3	✓ 1
Part (b)
M1 7 = 3 × 2 + c or P − 7 = 3(n − 2) (ECF from (a))	✓ 1
A1 P = 3n + 1	✓ 1
Part (c)
B1 the power drawn with no racks, e.g. by lighting and cooling	✓ 1
Part (d)
B1 76 kW (ECF from (b))	✓ 1

Exam tip: Check your equation by substituting the coordinates of the second point you were given. Both points must fit.

+ Add to test✓ In your test120IGME3-3-11r16-09Sequences and nth terms Paper FMedium5 marks

A data centre compares two plans for adding cooling units. Under Plan A the number of units after n months is \(3n + 8\). Under Plan B the number of units after n months is \(5n - 6\).

(a)

Work out the number of units under Plan A after 12 months.

(1)

(b)

Find the value of n for which both plans give the same number of units.

(3)

(c)

Work out the number of units at this time.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 44	✓ 1
Part (b)
M1 3n + 8 = 5n − 6	✓ 1
M1 2n = 14 oe	✓ 1
A1 7	✓ 1
Part (c)
B1 29 (ECF from (b))	✓ 1

Exam tip: When two rules are equal, set the expressions equal to each other, then substitute the answer back into one of them.

+ Add to test✓ In your test121IGME3-3-11r16-14Graphs, gradients and real-life graphs Paper FMedium7 marks

The speed-time graph of a cyclist is made of three straight sections. She accelerates uniformly from rest to 10 m/s in 5 s. She then cycles at a constant 10 m/s for 20 s. She then decelerates uniformly to rest in 10 s.

Diagram not to scale

(a)

Work out her acceleration during the first 5 s.

(2)

(b)

Work out the total distance she travels.

(3)

(c)

Work out her average speed for the whole journey. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 10 ÷ 5	✓ 1
A1 2 m/s²	✓ 1
Part (b)
M1 attempt at one area, e.g. ½ × 5 × 10, 20 × 10 or ½ × 10 × 10	✓ 1
M1 sum of all three areas, or trapezium ½(20 + 35) × 10	✓ 1
A1 275 m	✓ 1
Part (c)
M1 275 ÷ 35 (ECF from (b))	✓ 1
A1 7.86 m/s	✓ 1

Exam tip: Distance is the area under a speed-time graph. Split it into triangles and rectangles, or use the trapezium rule, and add.

+ Add to test✓ In your test122IGME3-3-11r19-05Sequences and nth terms Paper FMedium7 marks

The numbers of server racks installed in the first four weeks of a project are 85, 79, 73, 67 and this linear sequence continues.

(a)

Find an expression, in terms of n, for the nth term of the sequence.

(2)

(b)

Show that 100 cannot be a term of this sequence.

(2)

(c)

Work out the first term of the sequence that is less than 20.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 common difference of −6, or −6n seen	✓ 1
A1 91 − 6n	✓ 1
Part (b)
M1 sets 91 − 6n = 100, ECF from (a)	✓ 1
A1 n = −1.5, which is not a positive whole number	✓ 1
Part (c)
M1 91 − 6n < 20, ECF from (a)	✓ 1
M1 n > 11.8 (to 1 d.p. or better) so n = 12	✓ 1
A1 19	✓ 1

Common error: When a question asks for the first term below a value, find the smallest whole n that works and then give the term itself, not n.

+ Add to test✓ In your test123IGME3-3-11r19-06Graphs, gradients and real-life graphs Paper FMedium7 marks

In a data centre the battery backup is fully charged (100%) at time t = 0. It discharges at a constant rate, and after 5 hours the charge is 40%. The charge, y %, at time t hours can be shown as a straight-line graph.

(a)

Work out the gradient of the line.

(2)

(b)

State what the gradient represents in this situation.

(1)

(c)

Find the equation of the line.

(2)

(d)

Work out how many hours it takes for the charge to reach 10%.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (40 − 100) ÷ 5	✓ 1
A1 −12	✓ 1
Part (b)
A1 the charge falls by 12 percentage points every hour	✓ 1
Part (c)
M1 y = mt + c with c = 100 or a point substituted	✓ 1
A1 y = 100 − 12t	✓ 1
Part (d)
M1 100 − 12t = 10, ECF from (c)	✓ 1
A1 7.5 hours	✓ 1

Exam tip: A negative gradient on a real-life graph means the quantity is decreasing; say what falls and by how much per unit of the other quantity.

+ Add to test✓ In your test124IGME3-3-11r19-07Graphs, gradients and real-life graphs Paper FMedium5 marks

On a survey grid, line L₁ has equation y = 3x − 4. Line L₂ is parallel to L₁ and passes through the point (2, 9).

(a)

Find the equation of L₂.

(3)

(b)

Work out the x-coordinate of the point where L₂ crosses the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient of L₂ is 3	✓ 1
M1 substitutes (2, 9) into y = 3x + c, so 9 = 6 + c	✓ 1
A1 y = 3x + 3	✓ 1
Part (b)
M1 sets 3x + 3 = 0, ECF from (a)	✓ 1
A1 x = −1	✓ 1

Exam tip: Parallel lines have equal gradients; the intercept is the only thing that changes.

+ Add to test✓ In your test125IGME3-3-11r19-12Graphs, gradients and real-life graphs Paper FMedium7 marks

A cyclist rides a straight track. Her speed increases uniformly from 0 to 12 m/s in 8 seconds. She then rides at a constant 12 m/s for 20 seconds. Finally she slows uniformly to rest in 4 seconds.

Diagram not to scale

(a)

Work out her acceleration in the first 8 seconds.

(2)

(b)

Work out the total distance she travels.

(3)

(c)

Work out her average speed for the whole 32 seconds.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 ÷ 8	✓ 1
A1 1.5 m/s²	✓ 1
Part (b)
M1 area of at least one section, e.g. ½ × 8 × 12 = 48	✓ 1
M1 areas 48 + 240 + 24 (accept any complete valid method)	✓ 1
A1 312 m	✓ 1
Part (c)
M1 312 ÷ 32, ECF from (b)	✓ 1
A1 9.75 m/s	✓ 1

Common error: Average speed is total distance divided by total time; it is not the mean of the highest and lowest speeds.

+ Add to test✓ In your test126IGME3-3-12r22-05Sequences and nth terms Paper FMedium5 marks

The lengths of the spans of a footbridge, in metres, form an arithmetic sequence. The 4th span is 17 m long and the 9th span is 37 m long.

(a)

Find an expression, in terms of n, for the length of the nth span.

(3)

(b)

Find the length of the first span that is longer than 100 m.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for a common difference (37 − 17) ÷ (9 − 4) = 4	✓ 1
M1 for 4n + c with 4 × 4 + c = 17	✓ 1
A1 for 4n + 1	✓ 1
Part (b)
M1 for 4n + 1 > 100, so n > 24.75	✓ 1
A1 for n = 25, length 101 m	✓ 1

Exam tip: Use two known terms to find the common difference first, then substitute one term to find the constant.

+ Add to test✓ In your test127IGME3-3-12r22-06Graphs, gradients and real-life graphs Paper FMedium7 marks

During roasting, the temperature of coffee beans is recorded. The temperature–time graph is made of two straight line segments: from (0, 25) to (9, 160), then from (9, 160) to (15, 205). Time is in minutes and temperature is in °C.

Diagram not to scale

(a)

Work out the gradient of the first segment.

(2)

(b)

Work out the gradient of the second segment.

(2)

(c)

Work out the average rate of temperature increase over the whole 15 minutes.

(2)

(d)

Explain why your answer to (c) is not the mean of the answers to (a) and (b).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for (160 − 25) ÷ 9	✓ 1
A1 for 15 °C/min	✓ 1
Part (b)
M1 for (205 − 160) ÷ (15 − 9)	✓ 1
A1 for 7.5 °C/min	✓ 1
Part (c)
M1 for (205 − 25) ÷ 15	✓ 1
A1 for 12 °C/min	✓ 1
Part (d)
B1 for the two segments last for different times (9 and 6 minutes), so the rates must be weighted	✓ 1

Common error: Do not average two gradients when the time intervals differ. Use total change divided by total time.

+ Add to test✓ In your test128IGME3-3-12r22-07Graphs, gradients and real-life graphs Paper FMedium6 marks

A space probe follows a straight path. On a map with units of 1000 km, the path passes through A(2, 7) and B(6, 19).

Diagram not to scale

(a)

Work out the gradient of the line AB.

(2)

(b)

Find the equation of the line AB.

(2)

(c)

A second probe follows a path parallel to AB through the point (1, −5). Find the equation of this path.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for (19 − 7) ÷ (6 − 2)	✓ 1
A1 for 3	✓ 1
Part (b)
M1 for y = 3x + c with 7 = 3 × 2 + c	✓ 1
A1 for y = 3x + 1	✓ 1
Part (c)
M1 for gradient 3 and −5 = 3 × 1 + c	✓ 1
A1 for y = 3x − 8	✓ 1

Exam tip: Parallel lines have equal gradients; only the y-intercept changes.

+ Add to test✓ In your test129IGME3-3-12r24-05Graphs, gradients and real-life graphs Paper FMedium7 marks

A Formula 1 car accelerates uniformly from rest to 60 m/s in 6 seconds. It then travels at a constant 60 m/s for 15 seconds. It then brakes uniformly, slowing to 20 m/s in the next 4 seconds. Its speed–time graph is made of straight lines.

Diagram not to scale

(a)

Work out the acceleration of the car during the first 6 seconds.

(2)

(b)

Work out the total distance travelled in the 25 seconds.

(3)

(c)

Work out the average speed of the car over the 25 seconds.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 60 ÷ 6	✓ 1
A1 10 m/s²	✓ 1
Part (b)
M1 first stage ½ × 6 × 60 = 180 m and second stage 15 × 60 = 900 m	✓ 1
M1 third stage ½ × (60 + 20) × 4 = 160 m	✓ 1
A1 1240 m	✓ 1
Part (c)
M1 1240 ÷ 25 (ECF from (b))	✓ 1
A1 49.6 m/s	✓ 1

Exam tip: Distance is the area under a speed–time graph; split the shape into triangles, rectangles and trapezia.

+ Add to test✓ In your test130IGME3-3-12r24-11Graphs, gradients and real-life graphs Paper FMedium7 marks

On a plan of a pit lane, with coordinates in metres, a painted line L passes through the points (0, −3) and (4, 5).

Diagram not to scale

(a)

Find the equation of L.

(3)

(b)

Find the equation of the line parallel to L that passes through (1, 7).

(2)

(c)

Find the x coordinate of the point where the line from part (b) crosses the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient = (5 − (−3)) ÷ (4 − 0)	✓ 1
A1 gradient = 2	✓ 1
A1 y = 2x − 3	✓ 1
Part (b)
M1 gradient 2 and 7 = 2 × 1 + c	✓ 1
A1 y = 2x + 5	✓ 1
Part (c)
M1 0 = 2x + 5	✓ 1
A1 x = −2.5	✓ 1

Exam tip: Parallel lines have equal gradients, so reuse the gradient and find the new intercept from a point on the line.

+ Add to test✓ In your test131IGME3-3-13r24-05Sequences and nth terms Paper FMedium7 marks

The number of visitors to an early-morning swimming session increases by the same amount each day. On day 4 there are 19 visitors. On day 9 there are 44 visitors.

(a)

Work out the increase in visitors each day.

(2)

(b)

Work out the number of visitors on day 1.

(1)

(c)

Find an expression, in terms of n, for the number of visitors on day n.

(2)

(d)

On one later day there are 154 visitors. Work out which day this is.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (44 − 19) ÷ (9 − 4)	✓ 1
A1 5	✓ 1
Part (b)
B1 4 (19 − 3 × 5)	✓ 1
Part (c)
M1 5n + ...	✓ 1
A1 5n − 1	✓ 1
Part (d)
M1 5n − 1 = 154 (ECF from (c))	✓ 1
A1 31	✓ 1

Exam tip: Two terms of a linear sequence give the common difference by dividing the change in value by the change in term number.

+ Add to test✓ In your test132IGME3-3-13r24-07Graphs, gradients and real-life graphs Paper FMedium7 marks

An open-water swimmer swims 400 m away from the shore at a constant speed in 8 minutes. She rests for 3 minutes. She then swims 400 m straight back to the shore at a constant speed, taking 10 minutes.

(a)

Work out her speed on the way out, in metres per minute.

(2)

(b)

Work out her speed on the way back, in metres per second. Give your answer correct to 3 significant figures.

(2)

(c)

Work out her average speed for the whole swim, including the rest, in metres per minute. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 400 ÷ 8	✓ 1
A1 50	✓ 1
Part (b)
M1 400 ÷ (10 × 60) or 400 ÷ 600	✓ 1
A1 0.667	✓ 1
Part (c)
M1 total distance 800 m	✓ 1
M1 total time 8 + 3 + 10 = 21 minutes and 800 ÷ 21	✓ 1
A1 38.1	✓ 1

Common error: Average speed uses the total distance and the total time, and the rest period counts as part of the time.

+ Add to test✓ In your test133IGME3-3-13r24-08Graphs, gradients and real-life graphs Paper FMedium6 marks

On a map of a wind farm site, a cable runs in a straight line from A (−2, 7) to B (4, −5). All distances are in kilometres.

Diagram not to scale

(a)

Work out the gradient of the line through A and B.

(2)

(b)

Find the equation of the line through A and B.

(2)

(c)

The cable crosses the x-axis at the point P. Work out the x-coordinate of P.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (−5 − 7) ÷ (4 − (−2))	✓ 1
A1 −2	✓ 1
Part (b)
M1 y = −2x + c with a point substituted	✓ 1
A1 y = −2x + 3	✓ 1
Part (c)
M1 0 = −2x + 3 (ECF from (b))	✓ 1
A1 1.5	✓ 1

Exam tip: A line crosses the x-axis where y = 0, so substitute 0 for y in its equation.

+ Add to test✓ In your test134IGME3-3-12r26-06Sequences and nth terms Paper FMedium8 marks

A shop sells two phone-charging plans. Plan A costs a fixed £12 plus £0.40 per charge. Plan B costs £0.70 per charge with no fixed fee. For n charges the costs are £C_A and £C_B.

(a)

Write down an equation for C_A and an equation for C_B in terms of n.

(2)

(b)

Work out the number of charges for which both plans cost the same.

(3)

(c)

A customer makes 55 charges. Work out how much cheaper the cheaper plan is.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 C_A = 12 + 0.4n oe	✓ 1
B1 C_B = 0.7n oe	✓ 1
Part (b)
M1 12 + 0.4n = 0.7n	✓ 1
M1 12 = 0.3n	✓ 1
A1 40	✓ 1
Part (c)
M1 12 + 0.4 × 55 (= 34)	✓ 1
M1 0.7 × 55 (= 38.50)	✓ 1
A1 £4.50 (Plan A cheaper)	✓ 1

Common error: Where two lines cross the values are equal, so set the two expressions equal and solve; do not subtract the gradients first.

+ Add to test✓ In your test135IGME3-3-12r26-08Differentiation and stationary points Paper FMedium7 marks

The number of new cases, C, d days after the start of an outbreak is modelled by C = d(12 − d) for 0 ≤ d ≤ 12. The graph of C against d is a smooth curve with a ∩ shape.

(a)

Work out the number of new cases when d = 3.

(2)

(b)

Write down the values of d for which C = 0.

(2)

(c)

Use symmetry to find the maximum number of new cases predicted by the model.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3 × (12 − 3)	✓ 1
A1 27	✓ 1
Part (b)
B1 d = 0	✓ 1
B1 d = 12	✓ 1
Part (c)
M1 maximum halfway between the roots: (0 + 12)/2	✓ 1
A1 d = 6	✓ 1
A1 C = 6 × 6 = 36	✓ 1

Common error: A ∩-shaped quadratic is symmetrical about the vertical line halfway between its two roots, and its maximum lies on that line.

+ Add to test✓ In your test136IGME3-3-12r26-13Graphs, gradients and real-life graphs Paper FMedium8 marks

An electric bus is charged by solar panels. Its speed-time graph for one journey has three straight sections. It accelerates uniformly from rest to 12 m/s in 8 s, then travels at a constant 12 m/s for 20 s, then decelerates uniformly to rest in 6 s.

Diagram not to scale

(a)

Work out the acceleration of the bus in the first 8 seconds.

(2)

(b)

Work out the total distance travelled.

(4)

(c)

Work out the average speed for the whole journey. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 ÷ 8	✓ 1
A1 1.5 m/s²	✓ 1
Part (b)
M1 ½ × 8 × 12 (= 48)	✓ 1
M1 12 × 20 (= 240)	✓ 1
M1 ½ × 6 × 12 (= 36)	✓ 1
A1 324 m	✓ 1
Part (c)
M1 324 ÷ (8 + 20 + 6) (ECF from (b))	✓ 1
A1 9.53 m/s	✓ 1

Common error: Distance is the area under a speed-time graph; a triangle needs the half, but the rectangle in the constant-speed section does not.

+ Add to test✓ In your test137IGME3-3-14r28-04Graphs, gradients and real-life graphs Paper FMedium5 marks

On a coordinate plan of a hospital wing (units in metres), line L passes through the points A(0, 3) and B(4, 11).

Diagram not to scale

(a)

Work out the gradient of L.

(2)

(b)

Line M is parallel to L and passes through the point (1, −4). Find the equation of M.

(2)

(c)

Find the x coordinate of the point where M crosses the x-axis.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (11 − 3) ÷ (4 − 0)	✓ 1
A1 2	✓ 1
Part (b)
M1 uses gradient 2 and the point (1, −4), e.g. −4 = 2 × 1 + c	✓ 1
A1 y = 2x − 6	✓ 1
Part (c)
B1 x = 6 ÷ 2 = 3 (ECF from (b))	✓ 1

Exam tip: Parallel lines have equal gradients; use the given point to find the intercept.

+ Add to test✓ In your test138IGME3-3-14r28-06Sequences and nth terms Paper FMedium4 marks

In a sequence, each term after the second is the sum of the two terms before it. The 3rd term is 10 and the 5th term is 26.

(a)

Work out the 4th term.

(2)

(b)

Work out the 1st term.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5th term = 3rd term + 4th term, so 26 − 10	✓ 1
A1 16	✓ 1
Part (b)
M1 2nd term = 16 − 10 = 6 (ECF from (a))	✓ 1
A1 1st term = 10 − 6 = 4	✓ 1

Exam tip: Work backwards using the rule: each earlier term is a difference of two later terms.

+ Add to test✓ In your test139IGME3-3-14r28-10Graphs, gradients and real-life graphs Paper FMedium6 marks

Two hospital cleaning companies charge for a job as follows. Company A charges a call-out fee of €40 plus €12 per hour. Company B charges a call-out fee of €25 plus €15 per hour.

(a)

Work out the cost for Company A for a 6-hour job.

(1)

(b)

Find the number of hours for which the two companies charge the same.

(3)

(c)

For a 9-hour job, work out how much cheaper Company A is than Company B.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 €112	✓ 1
Part (b)
M1 40 + 12h = 25 + 15h	✓ 1
M1 15 = 3h	✓ 1
A1 5 hours	✓ 1
Part (c)
M1 40 + 12 × 9 = 148 and 25 + 15 × 9 = 160	✓ 1
A1 €12	✓ 1

Exam tip: Set the two cost expressions equal to find the point where the graphs of cost against time cross.

+ Add to test✓ In your test140IGME3-3-14r28-12Sequences and nth terms Paper FMedium5 marks

A hospital pharmacy reduces a patient's dose each day. The doses on the first four days are 95 mg, 88 mg, 81 mg, 74 mg and the pattern continues.

(a)

Find an expression for the dose on day n.

(2)

(b)

Find the first dose that is below 20 mg.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 common difference −7	✓ 1
A1 102 − 7n	✓ 1
Part (b)
M1 102 − 7n < 20	✓ 1
A1 n > 11.7… so n = 12	✓ 1
A1 18 mg	✓ 1

Common error: With a decreasing sequence the inequality sign flips when you divide by a negative number, so check by testing values of n.

+ Add to test✓ In your test141IGME3-3-14r32-05Sequences and nth terms Paper FMedium5 marks

The rows of a bridge deck mesh contain 95, 88, 81, ... rivets, decreasing by the same amount each row.

(a)

Show that the number of rivets in row n is 102 − 7n.

(2)

(b)

The pattern stops at the last row that has at least one rivet. Work out the number of rivets in this last row.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 common difference −7 (or −7n seen)	✓ 1
B1 −7 + c = 95 gives c = 102 (or checks 102 − 7 = 95 and 102 − 14 = 88)	✓ 1
Part (b)
M1 102 − 7n > 0	✓ 1
M1 n < 14.57, so n = 14	✓ 1
A1 4	✓ 1

Common error: For a decreasing sequence the nth term starts with the constant and subtracts a multiple of n; do not write +7n.

+ Add to test✓ In your test142IGME3-3-14r32-06Graphs, gradients and real-life graphs Paper FMedium7 marks

A maintenance drone flies along a bridge deck. Its distance–time graph is made of three straight-line sections. First it moves away from its base at 12 m/s for 20 s. Then it stays still for 15 s. Finally it returns to its base at a constant speed, taking 40 s.

(a)

Work out the distance of the drone from its base after the first 20 s.

(2)

(b)

Work out the speed of the drone on its return.

(2)

(c)

Work out the average speed of the drone for the whole journey.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 × 20	✓ 1
A1 240 m	✓ 1
Part (b)
M1 240 ÷ 40 (ECF from (a))	✓ 1
A1 6 m/s	✓ 1
Part (c)
M1 total distance 480 m (240 + 240)	✓ 1
M1 total time 75 s (20 + 15 + 40)	✓ 1
A1 6.4 m/s	✓ 1

Common error: Average speed uses total distance travelled, not the final distance from the start; here the drone ends back at its base.

+ Add to test✓ In your test143IGME3-3-14r32-07Graphs, gradients and real-life graphs Paper FMedium7 marks

Two coffee-roasting services charge for roasting m kg of beans. Service A charges £40 plus £2.50 per kg. Service B charges £10 plus £4 per kg.

(a)

Write down a formula for the cost C pounds for each service.

(2)

(b)

Work out the mass for which both services cost the same, and the cost.

(3)

(c)

A customer has 50 kg of beans roasted. Work out how much less the cheaper service charges.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 C = 40 + 2.5m (Service A)	✓ 1
B1 C = 10 + 4m (Service B)	✓ 1
Part (b)
M1 40 + 2.5m = 10 + 4m	✓ 1
A1 m = 20 kg	✓ 1
A1 £90	✓ 1
Part (c)
M1 A: 40 + 2.5 × 50 = £165 and B: 10 + 4 × 50 = £210	✓ 1
A1 £45 (A is cheaper)	✓ 1

Exam tip: The graphs of two cost formulas cross where the costs are equal, so set the two expressions equal and solve.

+ Add to test✓ In your test144IGME3-3-14r32-08Sequences and nth terms Paper FMedium7 marks

The nth term of sequence A is 6n − 1. The first four terms of sequence B are 7, 11, 15, 19.

(a)

Show that 300 is not a term of sequence A.

(2)

(b)

Find the nth term of sequence B.

(2)

(c)

Find the smallest number greater than 100 that is a term in both sequences.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6n − 1 = 300	✓ 1
A1 n = 50.1… is not a whole number, so 300 is not a term (or 300 + 1 = 301 is not a multiple of 6)	✓ 1
Part (b)
B1 common difference 4	✓ 1
B1 4n + 3	✓ 1
Part (c)
M1 list or generate terms of both sequences, or find common terms 11, 23, 35, ... (or identify that common terms differ by 12)	✓ 1
M1 common terms 11 + 12k, or common terms listed up to 95	✓ 1
A1 107	✓ 1

Exam tip: Numbers that appear in two linear sequences differ by the lowest common multiple of the two common differences.

+ Add to test✓ In your test145IGME3-3-13r33-08Graphs, gradients and real-life graphs Paper FMedium7 marks

A cyclist accelerates uniformly from rest to 12 m/s in 8 seconds. She then cycles at a constant 12 m/s for 20 seconds. She then decelerates uniformly to rest in 6 seconds.

Diagram not to scale

(a)

Calculate her acceleration during the first 8 seconds.

(2)

(b)

Calculate the total distance she travels.

(3)

(c)

Calculate her average speed for the whole 34 seconds. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 ÷ 8	✓ 1
A1 1.5 m/s²	✓ 1
Part (b)
M1 area of speed–time graph split into a triangle, rectangle and triangle	✓ 1
M1 ½ × 8 × 12 = 48 and 20 × 12 = 240 and ½ × 6 × 12 = 36	✓ 1
A1 324 m	✓ 1
Part (c)
M1 324 ÷ 34 (ECF from (b))	✓ 1
A1 9.53 m/s	✓ 1

Exam tip: Distance on a speed–time graph is the area under it; split a trapezium-like shape into simple pieces.

+ Add to test✓ In your test146IGME3-3-13r33-09Sequences and nth terms Paper FMedium8 marks

The seats in a lecture theatre are arranged in rows. Row 1 has 14 seats and each row has 3 more seats than the row in front of it.

(a)

Find an expression, in terms of n, for the number of seats in row n.

(2)

(b)

Work out the number of seats in row 20.

(2)

(c)

One row has 98 seats. Find the number of that row.

(2)

(d)

Explain why no row has exactly 100 seats.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3n	✓ 1
A1 3n + 11	✓ 1
Part (b)
M1 3 × 20 + 11	✓ 1
A1 71	✓ 1
Part (c)
M1 3n + 11 = 98	✓ 1
A1 n = 29	✓ 1
Part (d)
M1 3n + 11 = 100 gives n = 29.67... (or 89 ÷ 3)	✓ 1
A1 n is not a whole number so there is no such row	✓ 1

Common error: Check an nth term by substituting n = 1; here 3 × 1 + 11 must equal 14.

+ Add to test✓ In your test147IGME3-3-13r33-12Graphs, gradients and real-life graphs Paper FMedium7 marks

The graph of y = x² − 6x + 5 is drawn for values of x from −1 to 7.

(a)

Work out the value of y when x = −1.

(2)

(b)

Solve x² − 6x + 5 = 0, so finding where the graph crosses the x-axis.

(2)

(c)

Write down the equation of the line of symmetry of the graph.

(1)

(d)

Find the coordinates of the lowest point of the graph.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (−1)² − 6(−1) + 5	✓ 1
A1 12	✓ 1
Part (b)
M1 (x − 1)(x − 5)	✓ 1
A1 x = 1 and x = 5	✓ 1
Part (c)
B1 x = 3 (midway between the roots)	✓ 1
Part (d)
M1 substitute x = 3: 9 − 18 + 5	✓ 1
A1 (3, −4)	✓ 1

Exam tip: The line of symmetry of a quadratic graph passes midway between its two roots.

+ Add to test✓ In your test148IGME3-3-12r35-05Graphs, gradients and real-life graphs Paper FMedium7 marks

During a coffee roast, the temperature of the beans rises steadily from 20 °C to 180 °C in the first 4 minutes. It then rises steadily from 180 °C to 204 °C over the next 6 minutes.

(a)

Work out the rate of temperature rise, in °C per minute, during the first 4 minutes.

(2)

(b)

Work out the rate of temperature rise, in °C per minute, during the next 6 minutes.

(2)

(c)

Describe how the graph of temperature against time changes at 4 minutes.

(1)

(d)

Work out the average rate of temperature rise over the whole 10 minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (180 − 20) ÷ 4	✓ 1
A1 40	✓ 1
Part (b)
M1 (204 − 180) ÷ 6	✓ 1
A1 4	✓ 1
Part (c)
B1 the line becomes less steep (the gradient decreases from 40 to 4)	✓ 1
Part (d)
M1 (204 − 20) ÷ 10	✓ 1
A1 18.4 °C per minute	✓ 1

Common error: The average rate over the whole roast is total change divided by total time; it is not the mean of the two gradients.

+ Add to test✓ In your test149IGME3-3-12r35-06Sequences and nth terms Paper FMedium5 marks

The support cables of a footbridge have lengths that increase in a regular pattern. The three shortest cables are 4.2 m, 5.7 m and 7.2 m long.

(a)

Find an expression, in terms of n, for the length in metres of the nth cable.

(2)

(b)

Work out the length of the 12th cable.

(1)

(c)

The 12th cable costs £18.50 per metre. Work out the cost of the 12th cable.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 common difference 1.5 (or 1.5n seen)	✓ 1
A1 1.5n + 2.7	✓ 1
Part (b)
B1 20.7 m (ECF from (a))	✓ 1
Part (c)
M1 20.7 × 18.50 (ECF from (b))	✓ 1
A1 £382.95	✓ 1

Exam tip: Give money answers to the nearest penny, and use the exact length from part (b), not a rounded one.

+ Add to test✓ In your test150IGME3-3-12r35-08Graphs, gradients and real-life graphs Paper FMedium7 marks

A truck crosses a long bridge. Its speed increases steadily from 0 to 12 m/s in 8 seconds. It then travels at a constant 12 m/s for 20 seconds. It then slows steadily to rest in 6 seconds.

Diagram not to scale

(a)

Work out the acceleration of the truck during the first 8 seconds.

(2)

(b)

Work out the distance travelled at constant speed.

(2)

(c)

Work out the total distance the truck travels.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 ÷ 8	✓ 1
A1 1.5 m/s²	✓ 1
Part (b)
M1 12 × 20	✓ 1
A1 240 m	✓ 1
Part (c)
M1 1/2 × 8 × 12 (= 48)	✓ 1
M1 1/2 × 6 × 12 (= 36)	✓ 1
A1 48 + 240 + 36 = 324 m (ECF from (b))	✓ 1

Exam tip: On a speed–time graph, distance is the area under the line; split the area into triangles and rectangles.

+ Add to test✓ In your test151IGME3-3-12r37-03Sequences and nth terms Paper FMedium7 marks

A patient is given a drug in increasing daily doses. The dose on day 1 is 5 mg. The dose increases by 2.5 mg each day, so the doses are 5 mg, 7.5 mg, 10 mg, …

(a)

Show that the dose on day n is (2.5n + 2.5) mg.

(2)

(b)

The dose must not exceed 20 mg. Find the first day on which the dose is more than 20 mg.

(3)

(c)

Work out the total amount of the drug given over the first 4 days.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 common difference 2.5, so 2.5n + constant	✓ 1
A1 constant = 5 − 2.5 = 2.5, giving 2.5n + 2.5	✓ 1
Part (b)
M1 2.5n + 2.5 > 20	✓ 1
A1 n > 7	✓ 1
A1 day 8	✓ 1
Part (c)
M1 5 + 7.5 + 10 + 12.5	✓ 1
A1 35 mg	✓ 1

Common error: A dose of exactly 20 mg does not exceed 20 mg, so day 7 is not the answer.

+ Add to test✓ In your test152IGME3-3-12r37-06Sequences and nth terms Paper FMedium5 marks

A drug is removed from the body so that the amount halves every hour. A patient has 640 mg at time 0. The amounts at the start of each hour are 640 mg, 320 mg, 160 mg, …

(a)

Write down the next two terms of the sequence.

(1)

(b)

Write down the common ratio of the sequence.

(1)

(c)

The drug becomes ineffective when the amount is less than 10 mg. Find the first whole number of hours after which this happens.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 80 and 40	✓ 1
Part (b)
B1 0.5 (or ½)	✓ 1
Part (c)
M1 640 ÷ 2ⁿ < 10 (or continuing the sequence 20, 10, 5)	✓ 1
A1 amount is exactly 10 mg after 6 hours, which is not less than 10	✓ 1
A1 7 hours	✓ 1

Common error: Check whether the boundary value satisfies the condition: exactly 10 mg is not less than 10 mg.

+ Add to test✓ In your test153IGME3-3-13r44-06Sequences and nth terms Paper FMedium7 marks

Ward A in a hospital admits 6, 10, 14, … patients on days 1, 2, 3, …. The numbers of admissions form an arithmetic sequence. Ward B admits 34, 31, 28, … patients on days 1, 2, 3, … and its numbers also form an arithmetic sequence. Both patterns continue.

(a)

Find an expression, in terms of n, for the number of patients admitted by Ward A on day n.

(2)

(b)

Find an expression, in terms of n, for the number of patients admitted by Ward B on day n.

(2)

(c)

On one day both wards admit the same number of patients. Work out the value of n for this day.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4n seen	✓ 1
A1 4n + 2	✓ 1
Part (b)
M1 −3n seen	✓ 1
A1 37 − 3n	✓ 1
Part (c)
M1 4n + 2 = 37 − 3n (ECF from (a) and (b))	✓ 1
M1 7n = 35 or equivalent rearrangement	✓ 1
A1 n = 5	✓ 1

Exam tip: When a sequence decreases, the number in front of n is negative. Test your rule on the first term before moving on.

+ Add to test✓ In your test154IGME3-3-13r44-07Sequences and nth terms Paper FMedium7 marks

The mass of fuel in a Formula 1 car at the start of laps 1, 2, 3, … is 110 kg, 108.4 kg, 106.8 kg, …. The masses form an arithmetic sequence and the pattern continues.

(a)

Find an expression for the mass of fuel, in kg, at the start of lap n.

(2)

(b)

Work out the number of the first lap for which the mass of fuel at the start of the lap is less than 20 kg.

(3)

(c)

Work out the mass of fuel, in kg, at the start of lap 45.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 −1.6n seen	✓ 1
A1 111.6 − 1.6n	✓ 1
Part (b)
M1 111.6 − 1.6n < 20 (or = 20) (ECF from (a))	✓ 1
M1 n > 57.25 (or n = 57.25)	✓ 1
A1 58	✓ 1
Part (c)
M1 111.6 − 1.6 × 45 (ECF from (a))	✓ 1
A1 39.6 (kg)	✓ 1

Common error: Because lap numbers are whole numbers, round an inequality answer to the next whole lap that satisfies it, not to the nearest whole number.

+ Add to test✓ In your test155IGME3-3-13r44-08Graphs, gradients and real-life graphs Paper FMedium7 marks

An ambulance leaves a hospital and travels 24 km at a constant speed to an accident. The journey takes 20 minutes. The ambulance stays at the accident for 10 minutes. It then returns to the hospital by the same road at a constant speed. The return journey takes 45 minutes.

(a)

Work out the speed of the ambulance on the journey to the accident. Give your answer in km/h.

(2)

(b)

Work out the speed of the ambulance on the return journey. Give your answer in km/h.

(2)

(c)

Work out the average speed of the ambulance for the whole time from leaving to returning to the hospital. Give your answer in km/h.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 24 ÷ (20 ÷ 60) or 24 ÷ ¹/₃	✓ 1
A1 72 km/h	✓ 1
Part (b)
M1 24 ÷ (45 ÷ 60) or 24 ÷ 0.75	✓ 1
A1 32 km/h	✓ 1
Part (c)
M1 total distance 48 km and total time 75 minutes	✓ 1
M1 48 ÷ (75 ÷ 60) or 48 ÷ 1.25	✓ 1
A1 38.4 km/h	✓ 1

Common error: Average speed uses the total time, including any time spent stopped, and the total distance travelled. It is not the mean of the two speeds.

+ Add to test✓ In your test156IGME3-3-13r44-11Graphs, gradients and real-life graphs Paper FMedium6 marks

Two hospital infusion pumps deliver a drug. The graph of the volume, y mL, delivered by pump A against the time, x minutes, is a straight line through the points (0, 4) and (6, 16). The graph for pump B is a straight line parallel to the line for pump A and passes through the point (5, 7). Use the model for pump B for x ≥ 2.

(a)

Work out the gradient of the line for pump A.

(2)

(b)

Find the equation of the line for pump B.

(2)

(c)

Pump B has delivered 33 mL. Work out the time that has passed.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (16 − 4) ÷ (6 − 0)	✓ 1
A1 2	✓ 1
Part (b)
M1 y = 2x + c with (5, 7) substituted (gradient 2 or ECF from (a))	✓ 1
A1 y = 2x − 3	✓ 1
Part (c)
M1 33 = 2x − 3 (ECF from (b))	✓ 1
A1 18 minutes	✓ 1

Exam tip: Parallel lines have the same gradient. Use that gradient and one point on the new line to find its equation.

+ Add to test✓ In your test157IGME3-3-12r48-01Sequences and nth terms Paper FMedium7 marks

Archaeologists dig a trench in layers. The depths, in cm, of the bottoms of layers 1, 2, 3 and 4 are 15, 23, 31 and 39. The pattern continues.

(a)

Write down the depth of the bottom of layer 5.

(1)

(b)

Find an expression for the depth of the bottom of layer n.

(2)

(c)

Show that 200 cm is not the depth of the bottom of any layer.

(2)

(d)

Find the first depth in the sequence that is greater than 100 cm.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 47	✓ 1
Part (b)
M1 8n + ...	✓ 1
A1 8n + 7	✓ 1
Part (c)
M1 8n + 7 = 200, n = 24.125	✓ 1
C1 n is not a whole number, so 200 is not a term	✓ 1
Part (d)
M1 8 × 11 + 7 = 95 and 8 × 12 + 7 or solves 8n + 7 > 100	✓ 1
A1 103	✓ 1

Exam tip: To show a number is not in a sequence, solve for n and state clearly that it is not a whole number.

+ Add to test✓ In your test158IGME3-3-12r48-04Graphs, gradients and real-life graphs Paper FMedium7 marks

A cyclist rides along a straight road. For the first 8 s the speed increases uniformly from 0 to 12 m/s. The cyclist then rides at a constant 12 m/s for 20 s. Finally the cyclist slows down uniformly to rest in 4 s.

Diagram not to scale

(a)

Work out the acceleration during the first 8 s.

(2)

(b)

Work out the total distance travelled.

(3)

(c)

Work out the average speed for the whole journey.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 ÷ 8	✓ 1
A1 1.5 m/s²	✓ 1
Part (b)
M1 uses area under the speed–time graph	✓ 1
M1 areas ½ × 8 × 12 = 48, 20 × 12 = 240 and ½ × 4 × 12 = 24	✓ 1
A1 312 m	✓ 1
Part (c)
M1 312 ÷ 32 (ECF from (b))	✓ 1
A1 9.75 m/s	✓ 1

Common error: The total time is 8 + 20 + 4 = 32 s; forgetting one of the three stages gives the wrong average speed.

+ Add to test✓ In your test159IGME3-3-12r48-12Graphs, gradients and real-life graphs Paper FMedium7 marks

On a site grid, with units in metres, an archaeologist records two finds at A(−2, 1) and B(4, 10).

Diagram not to scale

(a)

Find the coordinates of the midpoint of AB.

(2)

(b)

Work out the gradient of the line AB.

(2)

(c)

Work out the distance AB. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ((−2 + 4) ÷ 2, (1 + 10) ÷ 2)	✓ 1
A1 (1, 5.5)	✓ 1
Part (b)
M1 (10 − 1) ÷ (4 − (−2))	✓ 1
A1 1.5	✓ 1
Part (c)
M1 6² + 9² = 117	✓ 1
M1 square root of 117	✓ 1
A1 10.8 m	✓ 1

Common error: Take care with negative coordinates: 4 − (−2) = 6, not 2.

+ Add to test✓ In your test160IGME3-3-12r48-14Sequences and nth terms Paper FMedium7 marks

A data centre stores 6 terabytes (TB) of data at the start of year 1. The amount stored at the start of each later year is double the amount stored at the start of the previous year.

(a)

Work out the amount stored at the start of year 2 and at the start of year 3.

(2)

(b)

Work out the amount stored at the start of year 5.

(2)

(c)

Show that the amount stored first exceeds 1000 TB at the start of year 9.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 12 TB	✓ 1
B1 24 TB	✓ 1
Part (b)
M1 continues doubling, 48 then 96, or 6 × 2⁴	✓ 1
A1 96 TB	✓ 1
Part (c)
B1 year 8: 768 TB, which is less than 1000	✓ 1
B1 year 9: 1536 TB (6 × 2⁸)	✓ 1
C1 1536 > 1000 and 768 < 1000, so year 9 is the first	✓ 1

Exam tip: In "show that" questions, give the value on both sides of the target to prove it is the first time it is passed.

+ Add to test✓ In your test161IGME3-3-12r6-12Sequences and nth terms Paper FHard6 marks

In a game, the number of bonus points awarded for completing level n follows a linear sequence with nth term an + b, where a and b are constants. Level 3 awards 17 points and level 8 awards 42 points.

(a)

Find the values of a and b.

(3)

(b)

The points for level n and level n + 1 add up to 199. Work out the value of n.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (42 − 17) ÷ (8 − 3) or 3a + b = 17 and 8a + b = 42	✓ 1
A1 a = 5	✓ 1
A1 b = 2	✓ 1
Part (b)
M1 (5n + 2) + (5(n + 1) + 2) = 199 (ECF from (a))	✓ 1
M1 10n + 9 = 199	✓ 1
A1 n = 19	✓ 1

Exam tip: Subtracting two simultaneous equations removes the constant and leaves just the multiple of a.

+ Add to test✓ In your test162IGME3-3-12r12-08Graphs, gradients and real-life graphs Paper FHard7 marks

A service boat travels to a wind farm. Its speed-time graph has three straight sections: the speed increases uniformly from 0 to 6 m/s in 10 s, stays at 6 m/s for 40 s, then decreases uniformly to 0 in 15 s.

Diagram not to scale

(a)

Work out the acceleration during the first 10 seconds.

(2)

(b)

Work out the total distance travelled.

(3)

(c)

Work out the average speed for the whole journey. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
6 ÷ 10	✓ 1
0.6 m/s²	✓ 1
Part (b)
Section 1: ½ × 10 × 6 = 30	✓ 1
Section 2: 6 × 40 = 240 and section 3: ½ × 15 × 6 = 45	✓ 1
315 m	✓ 1
Part (c)
Total time = 65 s	✓ 1
315 ÷ 65 = 4.85 m/s	✓ 1

Exam tip: Distance is the area under a speed-time graph; split it into triangles and rectangles.

+ Add to test✓ In your test163IGME3-3-12r24-14Sequences and nth terms Paper FHard10 marks

Two beekeepers expand their apiaries each year. Aiko has 3 hives in year 1, 8 hives in year 2, 13 hives in year 3, and so on, increasing by the same number each year. Bram has 13 hives in year 1, 16 hives in year 2, 19 hives in year 3, and so on.

(a)

Find an expression for the number of hives Aiko has in year n.

(2)

(b)

Find an expression for the number of hives Bram has in year n.

(2)

(c)

In one year they have the same number of hives. Find this year and the number of hives each has.

(3)

(d)

Find the first year in which Aiko has at least 20 more hives than Bram.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 common difference 5 or 5n seen	✓ 1
A1 5n − 2	✓ 1
Part (b)
M1 common difference 3 or 3n seen	✓ 1
A1 3n + 10	✓ 1
Part (c)
M1 5n − 2 = 3n + 10	✓ 1
A1 year 6	✓ 1
A1 28 hives (ECF from (a) and (b))	✓ 1
Part (d)
M1 (5n − 2) − (3n + 10) ≥ 20	✓ 1
M1 2n − 12 ≥ 20, so n ≥ 16	✓ 1
A1 year 16	✓ 1

Exam tip: To compare two sequences, set their nth terms equal for a common term, and use an inequality for the first time one exceeds the other.

+ Add to test✓ In your test164IGME3-3-13r24-14Graphs, gradients and real-life graphs Paper FHard11 marks

The power output, P megawatts (MW), of a wind turbine depends on the wind speed, v m/s. Below 4 m/s the output is 0. From 4 m/s to 12 m/s the graph of P against v is a straight line rising from 0 MW to 2.4 MW. From 12 m/s to 25 m/s the output stays constant at 2.4 MW. Above 25 m/s the turbine is switched off.

Diagram not to scale

(a)

Work out the gradient of the straight-line section, in MW per m/s.

(2)

(b)

Work out the power output when the wind speed is 9 m/s.

(2)

(c)

Work out the wind speed at which the output is 1.8 MW.

(2)

(d)

Write down the range of wind speeds for which the output is at least 1.8 MW.

(2)

(e)

A wind farm has 12 of these turbines, all in a steady 9 m/s wind for 6 hours. Work out the total energy produced, in megawatt-hours (MWh).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (2.4 − 0) ÷ (12 − 4)	✓ 1
A1 0.3	✓ 1
Part (b)
M1 0.3 × (9 − 4)	✓ 1
A1 1.5 MW	✓ 1
Part (c)
M1 1.8 ÷ 0.3 = 6 above 4 m/s	✓ 1
A1 10 m/s	✓ 1
Part (d)
B1 lower limit 10 m/s (ECF from (c))	✓ 1
B1 upper limit 25 m/s	✓ 1
Part (e)
M1 12 × 1.5 = 18 MW (ECF from (b))	✓ 1
M1 18 × 6	✓ 1
A1 108	✓ 1

Common error: The straight line starts at 4 m/s, not at 0 m/s, so measure the wind speed from 4 when using the gradient.

+ Add to test✓ In your test165IGME3-3-12r26-14Sequences and nth terms Paper FHard8 marks

In a model of disease spread, each term of a sequence after the first two is the sum of the two terms before it. The first term is a and the second term is b.

(a)

Given that a = 3 and b = 5, work out the 6th term.

(2)

(b)

Write down expressions in terms of a and b for the 3rd, 4th and 5th terms.

(3)

(c)

A different sequence of this type has 3rd term 11 and 5th term 29. Find the value of b and the value of a.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 terms 3, 5, 8, 13, 21 (at least three further terms correct)	✓ 1
A1 34	✓ 1
Part (b)
B1 3rd term a + b	✓ 1
B1 4th term a + 2b	✓ 1
B1 5th term 2a + 3b	✓ 1
Part (c)
M1 a + b = 11 and 2a + 3b = 29	✓ 1
M1 eliminate one variable, e.g. 29 − 2 × 11 = b	✓ 1
A1 b = 7 and a = 4	✓ 1

Exam tip: Write each term in terms of a and b one step at a time, then use the two given terms to form simultaneous equations.

+ Add to test✓ In your test166IGME3-3-12r35-14Sequences and nth terms Paper FHard8 marks

Two space probes store scientific data. On day n probe A stores a number of gigabytes (GB) that follows a linear sequence. On day 3 it stores 14 GB and on day 7 it stores 26 GB. Probe B has 36 GB stored on day 1, then 32 GB on day 2, then 28 GB on day 3, and the pattern continues as it sends data back to Earth.

(a)

Find an expression, in terms of n, for the amount in GB stored by probe A on day n.

(3)

(b)

Find an expression, in terms of n, for the amount in GB stored by probe B on day n.

(2)

(c)

On one day the two probes store the same amount of data. Work out this amount.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (26 − 14) ÷ (7 − 3) = 3	✓ 1
M1 3n + c with correct substitution, e.g. 14 = 9 + c	✓ 1
A1 3n + 5	✓ 1
Part (b)
B1 −4n seen	✓ 1
A1 40 − 4n	✓ 1
Part (c)
M1 3n + 5 = 40 − 4n (ECF from (a) and (b))	✓ 1
A1 n = 5	✓ 1
A1 20 GB	✓ 1

Common error: Equal terms occur at the same value of n; solve for n first, then substitute to find the amount.

+ Add to test✓ In your test167IGME3-3-12r37-14Graphs, gradients and real-life graphs Paper FHard7 marks

A drone in a game moves in a straight line. Its speed-time graph is made of straight lines: speed increases steadily from 0 to 12 m/s in the first 4 s, stays at 12 m/s for the next 6 s, then decreases steadily to 0 in the final 3 s.

Diagram not to scale

(a)

Calculate the acceleration during the first 4 s.

(2)

(b)

Calculate the total distance travelled by the drone.

(3)

(c)

Calculate the average speed over the whole 13 s, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 ÷ 4	✓ 1
A1 3 m/s²	✓ 1
Part (b)
M1 area of any one section of the graph (24, 72 or 18)	✓ 1
M1 24 + 72 + 18 (sum of all three areas)	✓ 1
A1 114 m	✓ 1
Part (c)
M1 114 ÷ 13 (ECF from (b))	✓ 1
A1 8.77 m/s	✓ 1

Common error: The distance travelled is the area under a speed-time graph, not the final speed multiplied by the total time.

+ Add to test✓ In your test168IGME-060-02-01Function notation, composite and inverse functions Paper HEasy3 marks

The function f is defined as f(x) = 3x − 7

(a)

Find f(5)

(1)

(b)

Solve f(x) = 11

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: f(5) = 3 × 5 − 7 = 8	✓ 1
Part (b)
M1: 3x − 7 = 11 so 3x = 18	✓ 1
A1: x = 6	✓ 1

Exam tip: f(x) = 11 means the output is 11. Set the expression equal to 11 and solve. Do not substitute 11 for x.

+ Add to test✓ In your test169IGME-060-02-02Function notation, composite and inverse functions Paper HEasy3 marks

The function g is defined as g(x) = x² + 4x

(a)

Find g(2a), giving your answer in terms of a. Simplify your answer.

(2)

(b)

Find g(−3)

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: (2a)² + 4(2a)	✓ 1
A1: 4a² + 8a	✓ 1
Part (b)
B1: (−3)² + 4(−3) = 9 − 12 = −3	✓ 1

Common error: (2a)² is 4a², not 2a². Put brackets around the whole input before you square it.

+ Add to test✓ In your test170IGME-060-02-03Function notation, composite and inverse functions Paper HEasy3 marks

The functions f and g are defined as
f(x) = 2x + 5
g(x) = x²

(a)

Find gf(x)

(1)

(b)

Find fg(3)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: gf(x) = (2x + 5)² or 4x² + 20x + 25	✓ 1
Part (b)
M1: g(3) = 9	✓ 1
A1: f(9) = 2 × 9 + 5 = 23	✓ 1

Exam tip: In fg(3) the function nearest the bracket acts first. Work out g(3), then apply f to that result.

+ Add to test✓ In your test171IGME-060-02-04Function notation, composite and inverse functions Paper HEasy3 marks

The function f is defined as \(f(x) = \frac{x+4}{3}\)

(a)

Find f⁻¹(x)

(2)

(b)

Find f⁻¹(5)

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: y = (x + 4)/3 so x = 3y − 4, or swaps x and y and rearranges correctly	✓ 1
A1: f⁻¹(x) = 3x − 4	✓ 1
Part (b)
B1: 3 × 5 − 4 = 11	✓ 1

Common error: f⁻¹(x) is the inverse function, not 1/f(x). Undo the operations in reverse order.

+ Add to test✓ In your test172IGME-060-02-05Function notation, composite and inverse functions Paper HEasy4 marks

The function h is defined as \(h(x) = \frac{5}{x-2}\)

(a)

Find h(7)

(1)

(b)

State which value of x must be excluded from the domain of h.

(1)

(c)

Solve h(x) = −10

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 5/(7 − 2) = 1	✓ 1
Part (b)
B1: x = 2	✓ 1
Part (c)
M1: x − 2 = 5/(−10) = −0.5	✓ 1
A1: x = 1.5	✓ 1

Exam tip: A value is excluded when it makes the denominator zero, because division by zero is undefined.

+ Add to test✓ In your test173IGME-060-03-01Differentiation and stationary points Paper HEasy5 marks

A curve has equation y = 4x³ − 5x² + 7x − 2

(a)

Find \(\frac{dy}{dx}\)

(3)

(b)

Find the gradient of the curve at the point where x = 2

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 12x²	✓ 1
B1 for −10x	✓ 1
B1 for +7, with the constant −2 removed	✓ 1
dy/dx = 12x² − 10x + 7	—
Part (b)
M1 substitutes x = 2 into their dy/dx: 12(2)² − 10(2) + 7	✓ 1
A1 35	✓ 1

Common error: The gradient at a point comes from substituting into dy/dx, not into the equation of the curve. Substituting x = 2 into y gives a y-coordinate, not a gradient.

+ Add to test✓ In your test174IGME-060-03-02Differentiation and stationary points Paper HEasy5 marks

A curve has equation y = x² − 8x + 3

(a)

Find \(\frac{dy}{dx}\)

(2)

(b)

Use your answer to part (a) to find the coordinates of the turning point of the curve.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 2x	✓ 1
B1 for −8	✓ 1
dy/dx = 2x − 8	—
Part (b)
M1 sets their dy/dx equal to 0: 2x − 8 = 0	✓ 1
A1 x = 4	✓ 1
A1 y = 4² − 8(4) + 3 = −13, so the turning point is (4, −13)	✓ 1

Exam tip: A turning point is where the gradient is zero. Set dy/dx = 0, solve for x, then substitute back into the original equation to get y.

+ Add to test✓ In your test175IGME-060-03-05Differentiation and stationary points Paper HEasy5 marks

A curve has equation y = 2x² − 12x + 7

(a)

Write 2x² − 12x + 7 in the form a(x + b)² + c, where a, b and c are integers.

(3)

(b)

Using your answer to part (a), write down the coordinates of the turning point of the curve. State whether it is a maximum or a minimum point.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Factorises the x terms: 2(x² − 6x) + 7	✓ 1
Completes the square inside the bracket: 2[(x − 3)² − 9] + 7	✓ 1
Final answer: 2(x − 3)² − 11	✓ 1
Part (b)
Turning point at (3, −11). Follow through from the form in (a) is accepted.	✓ 1
States that it is a minimum. The reason is that a = 2 > 0, or that the squared term is never negative.	✓ 1

Common error: After taking out the factor 2, the −9 inside the bracket must also be multiplied by 2. Writing 2(x − 3)² − 9 + 7 gives the wrong constant.

+ Add to test✓ In your test176IGME3-3-12r12-09Function notation, composite and inverse functions Paper HEasy5 marks

A swimming club charges £f(n) for n lessons, where f(n) = 8 + 3n.

(a)

Work out f(6).

(1)

(b)

Find the number of lessons when the charge is £44.

(2)

(c)

Show that the charge increases by the same amount for every extra lesson, and state this amount.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
8 + 18 = 26	✓ 1
Part (b)
8 + 3n = 44	✓ 1
n = 12	✓ 1
Part (c)
(8 + 3(n + 1)) − (8 + 3n) = 3	✓ 1
£3 per extra lesson	✓ 1

Exam tip: For f(n) = value, solve the resulting equation for n; substituting is for evaluating.

+ Add to test✓ In your test177IGME3-3-11r19-04Function notation, composite and inverse functions Paper HEasy5 marks

The functions f and g are defined as f(x) = 4x − 9 and g(x) = x² + 2.

(a)

Work out f(6).

(1)

(b)

Solve f(x) = −1.

(2)

(c)

Work out gf(3).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 15	✓ 1
Part (b)
M1 4x − 9 = −1	✓ 1
A1 x = 2	✓ 1
Part (c)
M1 f(3) = 3 found first	✓ 1
A1 g(3) = 11	✓ 1

Common error: In gf(x) the function f is applied first, then g.

+ Add to test✓ In your test178IGME-060-02-06Function notation, composite and inverse functions Paper HMedium4 marks

The functions f and g are defined as
f(x) = 4x − 1
\(g(x) = \frac{x}{2} + 3\)

(a)

Show that fg(x) = 2x + 11

(2)

(b)

Solve fg(x) = f(x)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 4(x/2 + 3) − 1	✓ 1
A1: 2x + 12 − 1 = 2x + 11 shown clearly	✓ 1
Part (b)
M1: 2x + 11 = 4x − 1	✓ 1
A1: x = 6	✓ 1

Exam tip: In a 'show that' part, write every line of working, including the expanded bracket, before the given answer.

+ Add to test✓ In your test179IGME-060-02-10Function notation, composite and inverse functions Paper HMedium6 marks

The functions f and g are defined as
\(f(x) = \sqrt{x+5}\), x ≥ −5
g(x) = x² − 1

(a)

Find fg(2), giving your answer as a surd in its simplest form.

(2)

(b)

Find gf(x), simplifying your answer.

(2)

(c)

Solve f(x) = 4

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: g(2) = 3 so f(3) = √8	✓ 1
A1: 2√2	✓ 1
Part (b)
M1: (√(x + 5))² − 1	✓ 1
A1: x + 4	✓ 1
Part (c)
M1: x + 5 = 16	✓ 1
A1: x = 11	✓ 1

Exam tip: Squaring a square root cancels it, so (√(x + 5))² = x + 5. This simplification is worth looking for in composite functions.

+ Add to test✓ In your test180IGME-060-02-11Function notation, composite and inverse functions Paper HMedium6 marks

The functions h and k are defined as
\(h(x) = \frac{1}{x+3}\), x ≠ −3
k(x) = 2x − 1

(a)

Find hk(x), simplifying your answer.

(2)

(b)

Solve \(hk(x) = \frac{1}{8}\)

(2)

(c)

Find h⁻¹(x)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 1/((2x − 1) + 3)	✓ 1
A1: 1/(2x + 2)	✓ 1
Part (b)
M1: 2x + 2 = 8	✓ 1
A1: x = 3	✓ 1
Part (c)
M1: x + 3 = 1/y, or swaps x and y to get y + 3 = 1/x	✓ 1
A1: h⁻¹(x) = 1/x − 3 or (1 − 3x)/x	✓ 1

Common error: For hk(x), the whole of k(x) replaces x inside h. Do not write h(x) × k(x).

+ Add to test✓ In your test181IGME-060-03-08Differentiation and stationary points Paper HMedium6 marks

A curve has equation y = x³ − 3x² − 7x + 4

(a)

Find \(\frac{dy}{dx}\)

(2)

(b)

Find the coordinates of the points on the curve where the gradient is 2

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 at least two terms correct	✓ 1
A1 3x² − 6x − 7	✓ 1
Part (b)
M1 sets 3x² − 6x − 7 = 2	✓ 1
M1 rearranges and factorises: x² − 2x − 3 = 0, (x − 3)(x + 1) = 0	✓ 1
A1 x = 3 and x = −1	✓ 1
A1 (3, −17) and (−1, 7)	✓ 1

Common error: Setting dy/dx = 0 here finds the stationary points, not the points where the gradient is 2. Set the derivative equal to the gradient you are told.

+ Add to test✓ In your test182IGME-060-03-09Differentiation and stationary points Paper HMedium5 marks

The point P lies on the curve y = 2x² − 5x + 1
The x-coordinate of P is 3

(a)

Find the gradient of the curve at P.

(2)

(b)

Find the equation of the tangent to the curve at P. Give your answer in the form y = mx + c

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 dy/dx = 4x − 5	✓ 1
A1 4(3) − 5 = 7	✓ 1
Part (b)
B1 y-coordinate of P: 18 − 15 + 1 = 4	✓ 1
M1 y − 4 = 7(x − 3) or 4 = 7(3) + c	✓ 1
A1 y = 7x − 17	✓ 1

Exam tip: A tangent touches the curve at P, so it has the same gradient as the curve there and passes through P. Find both the gradient and the y-coordinate before writing the line.

+ Add to test✓ In your test183IGME-060-00-09Sequences and nth terms Paper HMedium5 marks

The 5th term of an arithmetic series is 23.
The 12th term of the series is 51.

(a)

Find the first term and the common difference of the series.

(3)

(b)

Find the sum of the first 40 terms of the series.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: a + 4d = 23 and a + 11d = 51	✓ 1
A1: d = 4	✓ 1
A1: a = 7	✓ 1
Part (b)
M1: \(\frac{40}{2}[2 \times 7 + 39 \times 4]\)	✓ 1
A1: 3400	✓ 1

Common error: From the 5th term to the 12th term there are 7 steps, not 8. So 7d = 28.

+ Add to test✓ In your test184IGME-060-00-11Sequences and nth terms Paper HMedium6 marks

The first three terms of an arithmetic sequence are

2k + 1 4k − 1 5k + 3

where k is a constant.

(a)

Find the value of k.

(3)

(b)

Find an expression, in terms of n, for the nth term of the sequence.

(2)

(c)

Which term of the sequence is 503?

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: (4k − 1) − (2k + 1) = (5k + 3) − (4k − 1)	✓ 1
M1: 2k − 2 = k + 4	✓ 1
A1: k = 6	✓ 1
Part (b)
M1: terms 13, 23, 33 or d = 10	✓ 1
A1: 10n + 3	✓ 1
Part (c)
B1: 50th term	✓ 1

Exam tip: In an arithmetic sequence the gap between the 1st and 2nd terms equals the gap between the 2nd and 3rd terms. Writing this as an equation lets you find k.

+ Add to test✓ In your test185IGME-060-02-08Function notation, composite and inverse functions Paper HMedium5 marks

The function g is defined for all real values of x as g(x) = x² − 6x + 11.

(a)

Write g(x) in the form (x − a)² + b, where a and b are integers.

(2)

(b)

Hence state the range of g.

(1)

(c)

Given that g(x) = 18 and x > 0, find the value of x.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: (x − 3)² − 9 + 11, or a = 3 seen	✓ 1
A1: (x − 3)² + 2	✓ 1
Part (b)
B1: g(x) ≥ 2	✓ 1
Part (c)
M1: (x − 3)² = 16 so x − 3 = ±4, or x² − 6x − 7 = 0 factorised as (x − 7)(x + 1) = 0	✓ 1
A1: x = 7 only, since x = −1 is rejected because x > 0	✓ 1

Exam tip: A squared term is never negative, so the smallest value of (x − a)² + b is b. This gives the range directly.

+ Add to test✓ In your test186IGME-060-03-07Differentiation and stationary points Paper HMedium4 marks

A curve has equation y = ax² + bx, where a and b are constants.
The curve passes through the point (2, 4).
The curve also passes through the point (−1, 10).
Find the value of a and the value of b.

(a)

Find the value of a and the value of b.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Substitutes (2, 4) to get 4a + 2b = 4 (or 2a + b = 2)	✓ 1
Substitutes (−1, 10) to get a − b = 10	✓ 1
Uses a correct method to eliminate one unknown, e.g. adds 2a + b = 2 and a − b = 10 to get 3a = 12	✓ 1
a = 4 and b = −6	✓ 1

Common error: When substituting x = −1, (−1)² = 1, so the a term stays positive while the b term becomes −b. Check your answers in both original equations.

+ Add to test✓ In your test187IGME-060-02-09Function notation, composite and inverse functions Paper HMedium6 marks

The function f is defined as f(x) = 3x + k, where k is a constant.
Given that f(4) = 10:

(a)

Find the value of k.

(2)

(b)

Find f⁻¹(x).

(2)

(c)

Solve f(x) = f⁻¹(x).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 3 × 4 + k = 10, i.e. 12 + k = 10	✓ 1
A1: k = −2	✓ 1
Part (b)
M1: y = 3x − 2 rearranged to x = (y + 2)/3, or swapping x and y to give x = 3y − 2 and rearranging	✓ 1
A1: f⁻¹(x) = (x + 2)/3, written in terms of x	✓ 1
Part (c)
M1: 3x − 2 = (x + 2)/3 leading to 9x − 6 = x + 2. Alternatively, use the fact that f is linear and increasing, so the graphs meet on y = x, and solve 3x − 2 = x	✓ 1
A1: x = 1	✓ 1

Common error: When you multiply both sides by 3, multiply every term on the left. This gives 9x − 6, not 9x − 2.

+ Add to test✓ In your test188IGME-060-02-07Function notation, composite and inverse functions Paper HMedium4 marks

The function f is defined as \(f(x) = \frac{2x+3}{x-1}\), x ≠ 1.

(a)

Express the inverse function f⁻¹ in the form f⁻¹(x) = …

(3)

(b)

State the value of x that cannot be included in the domain of f⁻¹.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: writes y = (2x + 3)/(x − 1) and multiplies out, e.g. xy − y = 2x + 3. Equally valid: swap x and y first and write xy − x = 2y + 3	✓ 1
M1: collects the terms in the new subject on one side and factorises, e.g. x(y − 2) = y + 3 or y(x − 2) = x + 3	✓ 1
A1: f⁻¹(x) = (x + 3)/(x − 2)	✓ 1
Part (b)
B1: x = 2	✓ 1

Exam tip: After multiplying out, move every term containing the variable you are solving for to one side, then factorise. You cannot make it the subject without the factorising step.

+ Add to test✓ In your test189IGME-060-00-12Sequences and nth terms Paper HMedium5 marks

The nth term of a sequence is 6n − 3.

(a)

Write down the first three terms of the sequence.

(1)

(b)

Prove that the sum of any two consecutive terms of the sequence is a multiple of 12.

(3)

(c)

Explain why 2024 cannot be a term of the sequence.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 3, 9, 15	✓ 1
Part (b)
M1: writes two consecutive terms as 6n − 3 and 6(n + 1) − 3 = 6n + 3	✓ 1
A1: (6n − 3) + (6n + 3) = 12n	✓ 1
A1: concludes that 12n is a multiple of 12 because n is an integer	✓ 1
Part (c)
B1: any one valid reason, e.g. 6n is even, so 6n − 3 is always odd, but 2024 is even; OR 6n − 3 = 2024 gives n = 2027/6 = 337.83…, which is not an integer; OR every term is 3(2n − 1), a multiple of 3, but 2024 is not a multiple of 3	✓ 1

Common error: Checking a few numerical examples does not prove a result. Use general terms written in terms of n.

+ Add to test✓ In your test190IGME-060-03-06Differentiation and stationary points Paper HMedium8 marks

A curve has equation y = 2x³ − 9x² + 12x + 1

(a)

Find \(\frac{dy}{dx}\)

(2)

(b)

Find the x-coordinates of the stationary points of the curve.

(2)

(c)

Determine the nature of each stationary point. Give reasons for your answers.

(3)

(d)

Find the y-coordinate of the maximum point.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 at least two terms correct	✓ 1
A1 6x² − 18x + 12	✓ 1
Part (b)
M1 sets 6x² − 18x + 12 = 0 and simplifies or factorises, e.g. 6(x − 1)(x − 2) = 0	✓ 1
A1 x = 1 and x = 2	✓ 1
Part (c)
M1 finds d²y/dx² = 12x − 18, or evaluates dy/dx on both sides of each stationary point	✓ 1
A1 x = 1 is a maximum, with reason: d²y/dx² = −6 < 0, or dy/dx changes from positive to negative, e.g. dy/dx = 12 at x = 0 and −1.5 at x = 1.5	✓ 1
A1 x = 2 is a minimum, with reason: d²y/dx² = 6 > 0, or dy/dx changes from negative to positive, e.g. dy/dx = −1.5 at x = 1.5 and 12 at x = 3	✓ 1
Part (d)
B1 y = 2 − 9 + 12 + 1 = 6	✓ 1

Exam tip: State each conclusion with its reason. Say that the second derivative is negative (maximum) or positive (minimum), or say how the sign of the gradient changes either side of the point. A bare 'max' with no reason may not earn the mark.

+ Add to test✓ In your test191IGME-060-01-07Graphs, gradients and real-life graphs Paper HMedium7 marks

The point A has coordinates (−1, 4) and the point B has coordinates (5, 1).

Diagram not to scale

(a)

Work out the gradient of AB.

(2)

(b)

Find an equation of the perpendicular bisector of AB.
Give your answer in the form ax + by = c where a, b and c are integers.

(4)

(c)

Write down the x-coordinate of the point where the perpendicular bisector crosses the x-axis.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{1-4}{5-(-1)}\) or \(\frac{-3}{6}\)	✓ 1
A1 \(-\frac{1}{2}\)	✓ 1
Part (b)
M1 for midpoint (2, 2.5)	✓ 1
M1 for perpendicular gradient 2	✓ 1
M1 for y − 2.5 = 2(x − 2) or 2.5 = 2 × 2 + c	✓ 1
A1 4x − 2y = 3 (or any integer multiple)	✓ 1
Part (c)
B1 \(\frac{3}{4}\) or 0.75 (ft from their equation in (b))	✓ 1

Exam tip: A perpendicular bisector needs two things: the midpoint (for 'bisector') and the negative reciprocal gradient (for 'perpendicular'). If you miss either one, the method marks that depend on it are lost as well as the final answer.

+ Add to test✓ In your test192IGME3-3-09d-09Differentiation and stationary points Paper HMedium6 marks

A signal passes through two processing stages. Stage 1 changes an input x to f(x) = 2x + 3. Stage 2 then changes this to g(x) = x² − 5x.

(a)

Show that gf(x) = 4x² + 2x − 6.

(3)

(b)

Find the positive input for which the final output is 0.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 g(2x + 3) = (2x + 3)² − 5(2x + 3)	✓ 1
M1 (2x + 3)² = 4x² + 12x + 9	✓ 1
A1 4x² + 12x + 9 − 10x − 15 = 4x² + 2x − 6	✓ 1
Part (b)
M1 4x² + 2x − 6 = 0, or 2x² + x − 3 = 0	✓ 1
M1 factorise (2x + 3)(x − 1) = 0	✓ 1
A1 x = 1 (the other solution is −1.5)	✓ 1

Common error: In gf(x) the function nearest x acts first. Here that is f, so substitute f(x) into g.

+ Add to test✓ In your test193IGME3-3-09d-10Function notation, composite and inverse functions Paper HMedium6 marks

A paediatric dose, D mg, of a medicine for a child of mass m kg is given by the function f(m) = 4m + 15.

(a)

Find f(18).

(1)

(b)

Find f⁻¹(x) and say what f⁻¹ tells you in this context.

(3)

(c)

A child is prescribed a dose of 123 mg. Find the child's mass.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 87	✓ 1
Part (b)
M1 x = 4m + 15 and rearrange for m	✓ 1
A1 f⁻¹(x) = (x − 15) ÷ 4	✓ 1
B1 it gives the child's mass for a given dose	✓ 1
Part (c)
M1 (123 − 15) ÷ 4 or f⁻¹(123)	✓ 1
A1 27 kg	✓ 1

Exam tip: To find an inverse, write y = f(x), swap the roles of x and y if you like, then make the new subject.

+ Add to test✓ In your test194IGME3-3-09d-11Graphs, gradients and real-life graphs Paper HMedium7 marks

A farmer uses 60 m of fencing to make a rectangular pen. One side of the pen is a straight river bank and needs no fence. The two sides at right angles to the river each have length x metres, and the area of the pen is A m².

Red lines show the fencing. Diagram not to scale

(a)

Show that A = 60x − 2x².

(2)

(b)

Use differentiation to find the value of x that gives the largest area.

(3)

(c)

Work out the largest area.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 side parallel to the river is 60 − 2x	✓ 1
M1 A = x(60 − 2x) then expand to 60x − 2x²	✓ 1
Part (b)
M1 dA/dx = 60 − 4x	✓ 1
M1 set 60 − 4x = 0	✓ 1
A1 x = 15	✓ 1
Part (c)
M1 substitute x = 15 (ECF from (b))	✓ 1
A1 450 m²	✓ 1

Exam tip: A stationary point is where the gradient is zero. Always substitute back into the original function to find the maximum value, not into the derivative.

+ Add to test✓ In your test195IGME3-3-12r6-04Differentiation and stationary points Paper HMedium6 marks

The curve C has equation y = x² − 5x. The point P lies on C and has x-coordinate 3.

(a)

Find dy/dx.

(2)

(b)

Work out the gradient of C at P.

(1)

(c)

Find an equation of the tangent to C at P. Give your answer in the form y = mx + c.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 2x	✓ 1
B1 −5	✓ 1
Part (b)
B1 2 × 3 − 5 = 1	✓ 1
Part (c)
M1 y-coordinate of P: 3² − 5 × 3 = −6	✓ 1
M1 y − (−6) = 1(x − 3) (ECF from (b))	✓ 1
A1 y = x − 9	✓ 1

Common error: Find the y-coordinate of the point from the curve, not from the tangent, before writing the tangent equation.

+ Add to test✓ In your test196IGME3-3-12r6-06Sequences and nth terms Paper HMedium6 marks

The number of tiles in the nth pattern of a game level is given by the sequence 5, 12, 23, 38, ...

(a)

Find an expression, in terms of n, for the nth term of this sequence.

(3)

(b)

One pattern has 173 tiles. Work out the value of n for this pattern.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 second difference is 4, so the nth term starts 2n²	✓ 1
M1 subtract 2n² from the terms to leave 3, 4, 5, 6	✓ 1
A1 2n² + n + 2	✓ 1
Part (b)
M1 2n² + n + 2 = 173, so 2n² + n − 171 = 0	✓ 1
M1 (2n + 19)(n − 9) = 0 or quadratic formula	✓ 1
A1 n = 9 (reject n = −9.5)	✓ 1

Exam tip: For a quadratic sequence, the n² coefficient is half the constant second difference.

+ Add to test✓ In your test197IGME3-3-12r6-07Function notation, composite and inverse functions Paper HMedium5 marks

The functions f and g are defined by f(x) = 3x − 2 and g(x) = x² + 4, for all real values of x.

(a)

Find fg(x). Give your answer in the form ax² + b.

(2)

(b)

Find the positive value of x for which fg(x) = 58.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3(x² + 4) − 2	✓ 1
A1 3x² + 10	✓ 1
Part (b)
M1 3x² + 10 = 58 (ECF from (a))	✓ 1
M1 x² = 16	✓ 1
A1 x = 4	✓ 1

Common error: In fg(x), the function g is applied first, then f.

+ Add to test✓ In your test198IGME3-3-12r6-09Differentiation and stationary points Paper HMedium7 marks

The curve y = x³ − 6x² + 5 has two stationary points.

(a)

Find dy/dx.

(2)

(b)

Find the x-coordinates of the two stationary points.

(2)

(c)

Show that the stationary point at x = 4 is a minimum and find its y-coordinate.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3x²	✓ 1
B1 −12x	✓ 1
Part (b)
M1 3x² − 12x = 0, so 3x(x − 4) = 0	✓ 1
A1 x = 0 and x = 4	✓ 1
Part (c)
M1 d²y/dx² = 6x − 12	✓ 1
R1 at x = 4 this is 12, which is positive, so a minimum	✓ 1
A1 y = 4³ − 6 × 4² + 5 = −27	✓ 1

Common error: A positive second derivative means a minimum; do not confuse this with the sign of the first derivative.

+ Add to test✓ In your test199IGME3-3-12r11-06Function notation, composite and inverse functions Paper HMedium7 marks

The power output of a wind turbine is modelled using two functions, \(f(x) = 2x + 3\) and \(g(x) = x^2 - 1\).

(a)

Find \(fg(x)\). Give your answer in its simplest form.

(2)

(b)

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

(2)

(c)

Solve \(gf(x) = 35\). Give your positive solution.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(f(x^2 - 1) = 2(x^2 - 1) + 3\)	✓ 1
A1 \(2x^2 + 1\)	✓ 1
Part (b)
M1 \(y = 2x + 3\) then rearrange for x	✓ 1
A1 \(f^{-1}(x) = \frac{x - 3}{2}\)	✓ 1
Part (c)
M1 \((2x + 3)^2 - 1 = 35\), so \((2x + 3)^2 = 36\)	✓ 1
M1 \(2x + 3 = \pm 6\)	✓ 1
A1 \(x = 1.5\) (other solution \(x = -4.5\))	✓ 1

Common error: In \(gf(x)\) the function \(f\) is applied first; reversing the order gives \(fg(x)\), which is a different function.

+ Add to test✓ In your test200IGME3-3-12r11-07Sequences and nth terms Paper HMedium6 marks

The number of panels in the nth stage of a wind farm display is given by the sequence 5, 11, 19, 29, ...

(a)

Work out the next term of the sequence.

(1)

(b)

Find an expression for the nth term.

(3)

(c)

Show that 109 is a term of the sequence and state which term it is.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 differences 6, 8, 10, so next difference is 12 and the term is 41	✓ 1
Part (b)
M1 second difference 2 so \(n^2\) term	✓ 1
M1 subtract \(n^2\) from terms: \(4, 7, 10, 13\) so \(3n + 1\)	✓ 1
A1 \(n^2 + 3n + 1\)	✓ 1
Part (c)
M1 \(n^2 + 3n + 1 = 109\) so \(n^2 + 3n - 108 = 0\) and \((n + 12)(n - 9) = 0\)	✓ 1
A1 \(n = 9\), the 9th term (reject \(n = -12\))	✓ 1

Common error: A constant second difference of 2 means the nth term starts with \(n^2\), not \(2n^2\).

+ Add to test✓ In your test201IGME3-3-12r12-10Function notation, composite and inverse functions Paper HMedium6 marks

The functions f and g are defined by \(f(x) = 3x + 7\) and \(g(x) = x^2\).

(a)

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

(2)

(b)

Find \(fg(x)\).

(2)

(c)

Find the value of \(x\) for which \(f^{-1}(x) = g(2)\).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Let y = 3x + 7 and rearrange to x = (y − 7)/3	✓ 1
f⁻¹(x) = (x − 7)/3	✓ 1
Part (b)
Replace x in f by x²	✓ 1
3x² + 7	✓ 1
Part (c)
g(2) = 4	✓ 1
x = 3 × 4 + 7 = 19	✓ 1

Exam tip: For fg(x), apply g first and then f. Do not swap the order.

+ Add to test✓ In your test202IGME3-3-12r12-12Differentiation and stationary points Paper HMedium6 marks

The gradient of a curve is found for \(y = 2x^3 - 5x + 1\) at the point where \(x = 2\).

(a)

Find \(\frac{dy}{dx}\).

(2)

(b)

Find the gradient of the curve at the point where \(x = 2\).

(1)

(c)

The tangent to the curve at \(x = 2\) crosses the y-axis at the point \((0, k)\). Find \(k\).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Differentiates a term correctly	✓ 1
6x² − 5	✓ 1
Part (b)
6 × 4 − 5 = 19	✓ 1
Part (c)
y = 16 − 10 + 1 = 7 when x = 2	✓ 1
y − 7 = 19(x − 2)	✓ 1
k = −31	✓ 1

Exam tip: Find the y-coordinate of the point before writing the tangent equation.

+ Add to test✓ In your test203IGME3-3-12r12-13Differentiation and stationary points Paper HMedium6 marks

A rope 40 m long marks out three sides of a rectangular swimming area of width x metres. The fourth side is the shoreline and needs no rope. The area enclosed is A m².

Red lines show the rope. Diagram not to scale

(a)

Show that \(A = 40x - 2x^2\).

(2)

(b)

Find the value of x that gives the maximum area and justify that it is a maximum.

(3)

(c)

Work out the maximum area.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Length parallel to shore = 40 − 2x	✓ 1
A = x(40 − 2x) = 40x − 2x²	✓ 1
Part (b)
dA/dx = 40 − 4x	✓ 1
40 − 4x = 0 gives x = 10	✓ 1
d²A/dx² = −4 which is negative so a maximum	✓ 1
Part (c)
40 × 10 − 2 × 100 = 200 m²	✓ 1

Exam tip: After finding a stationary point, test it with the second derivative or the gradient either side.

+ Add to test✓ In your test204IGME3-3-11r14-08Function notation, composite and inverse functions Paper HMedium5 marks

In a sale, a shop offers 20% off every item, and a customer also holds a voucher worth £15 off. The price, £x, of an item is changed by the functions
f(x) = 0.8x (the 20% discount) and g(x) = x − 15 (the voucher).

(a)

Find gf(x), giving your answer in its simplest form.

(2)

(b)

A coat has an original price of £120. Work out how much less the customer pays if the discount is applied first and then the voucher, compared with the voucher first and then the discount.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
g(0.8x) substituted	✓ 1
0.8x − 15	✓ 1
Part (b)
gf(120) = 0.8 × 120 − 15 = 81	✓ 1
fg(120) = 0.8 × (120 − 15) = 84	✓ 1
Difference = £3	✓ 1

Common error: In gf(x), the function f is applied first. Reading the letters left to right reverses the order.

+ Add to test✓ In your test205IGME3-3-11r14-10Differentiation and stationary points Paper HMedium7 marks

A shop's weekly profit, £P, from selling a product at a price of £x is modelled by P = −2x² + 48x − 100.

(a)

Find dP/dx.

(2)

(b)

Find the price that gives the maximum weekly profit.

(2)

(c)

Work out the maximum weekly profit.

(2)

(d)

Explain how you know this stationary point is a maximum.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Attempt to differentiate (at least one term correct)	✓ 1
−4x + 48	✓ 1
Part (b)
−4x + 48 = 0 (ECF from (a))	✓ 1
x = 12	✓ 1
Part (c)
Substitute x = 12: −2 × 144 + 48 × 12 − 100	✓ 1
£188	✓ 1
Part (d)
The coefficient of x² is negative, so the graph is ∩-shaped (or d²P/dx² = −4 < 0)	✓ 1

Exam tip: A stationary point is where the gradient is zero: set the derivative equal to zero, then substitute back into the original function for the value.

+ Add to test✓ In your test206IGME3-3-11r14-14Graphs, gradients and real-life graphs Paper HMedium6 marks

The volume of water, V thousand m³, in a reservoir used for hydroelectric power is modelled by V = 90 + 24t − 3t², where t is the time in hours after 06:00 and 0 ≤ t ≤ 8.

(a)

Find dV/dt.

(2)

(b)

Work out the rate at which the volume is changing when t = 3. Include the units.

(1)

(c)

Find the greatest volume of water in the reservoir during this period.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Attempt to differentiate	✓ 1
24 − 6t	✓ 1
Part (b)
6 thousand m³ per hour	✓ 1
Part (c)
24 − 6t = 0	✓ 1
t = 4	✓ 1
V = 90 + 96 − 48 = 138 thousand m³	✓ 1

Exam tip: The greatest value of a smooth function on an interval is at a stationary point or an end point; here V = 90 at both ends, so 138 is the greatest.

+ Add to test✓ In your test207IGME3-3-11r15-05Function notation, composite and inverse functions Paper HMedium9 marks

The functions f and g are defined by f(x) = 3x − 2 and g(x) = x² + 1.

(a)

Work out f(4).

(1)

(b)

Show that gf(x) = 9x² − 12x + 5.

(3)

(c)

Find f⁻¹(x).

(2)

(d)

Find the value of x for which f(x) = f⁻¹(x).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 10	✓ 1
Part (b)
M1 gf(x) = (3x − 2)² + 1	✓ 1
M1 expands (3x − 2)² = 9x² − 12x + 4	✓ 1
A1 adds 1 to reach 9x² − 12x + 5 with no errors	✓ 1
Part (c)
M1 y = 3x − 2 rearranged to x = (y + 2) ÷ 3	✓ 1
A1 f⁻¹(x) = (x + 2) ÷ 3	✓ 1
Part (d)
M1 3x − 2 = (x + 2) ÷ 3 (ECF from (c))	✓ 1
M1 9x − 6 = x + 2	✓ 1
A1 x = 1	✓ 1

Common error: In gf(x), the function f is applied first and g second. Writing the functions in the wrong order gives a different expression.

+ Add to test✓ In your test208IGME3-3-11r15-06Graphs, gradients and real-life graphs Paper HMedium8 marks

A mountain path rises and falls between a hut and a ridge. Its height, h metres above the hut, at a horizontal distance x km from the hut is modelled by h = 10x³ − 60x² + 90x + 20 for 0 ≤ x ≤ 4.

(a)

Find dh/dx.

(2)

(b)

Find the values of x at which the path is level (the gradient is zero).

(3)

(c)

Show that the path has a minimum height at x = 3, and find this minimum height.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 reduces the power of at least one term	✓ 1
A1 30x² − 120x + 90	✓ 1
Part (b)
M1 30x² − 120x + 90 = 0 (ECF from (a))	✓ 1
M1 attempt to solve, e.g. 30(x − 1)(x − 3) = 0	✓ 1
A1 x = 1 and x = 3	✓ 1
Part (c)
M1 d²h/dx² = 60x − 120 evaluated at x = 3	✓ 1
A1 60 > 0, so a minimum	✓ 1
A1 h = 270 − 540 + 270 + 20 = 20 m	✓ 1

Exam tip: A stationary point is a minimum when the second derivative is positive and a maximum when it is negative. State the sign you found.

+ Add to test✓ In your test209IGME3-3-11r15-09Sequences and nth terms Paper HMedium6 marks

The first four terms of a sequence, giving the number of ice screws in successive anchor displays, are 3, 9, 19, 33, ...

(a)

Find the next term.

(1)

(b)

Find an expression, in terms of n, for the nth term.

(3)

(c)

One display uses 289 ice screws. Work out which term of the sequence this is.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 51	✓ 1
Part (b)
M1 first differences 6, 10, 14 and second difference 4	✓ 1
M1 2n² identified and subtracted from the terms, leaving 1 each time	✓ 1
A1 2n² + 1	✓ 1
Part (c)
M1 2n² + 1 = 289 so n² = 144 (ECF from (b))	✓ 1
A1 n = 12	✓ 1

Common error: For a quadratic sequence the second difference is constant, and the coefficient of n² is half of it. Do not use the first difference as the coefficient.

+ Add to test✓ In your test210IGME3-3-12r15-03Graphs, gradients and real-life graphs Paper HMedium6 marks

On a site plan of a wind farm, 1 unit represents 100 m. Points A(1, 3) and B(5, 11) lie on a straight service road. A cable runs along line L, which is perpendicular to AB and passes through B.

Diagram not to scale

(a)

Work out the gradient of AB.

(2)

(b)

Write down the gradient of L.

(1)

(c)

Find the equation of L.

(2)

(d)

Find the y-coordinate of the point where L crosses the y-axis.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (11 − 3) ÷ (5 − 1)	✓ 1
A1 2	✓ 1
Part (b)
A1 −1/2 (ECF from (a))	✓ 1
Part (c)
M1 y − 11 = −1/2(x − 5) (ECF from (b))	✓ 1
A1 y = −1/2x + 13.5 (or equivalent)	✓ 1
Part (d)
A1 13.5	✓ 1

Common error: The gradients of perpendicular lines multiply to give −1. Do not just change the sign of the gradient.

+ Add to test✓ In your test211IGME3-3-12r15-04Function notation, composite and inverse functions Paper HMedium7 marks

The functions f and g are defined by f(x) = 2x − 7 and g(x) = x² + 3x.

(a)

Work out g(−2).

(1)

(b)

Find f⁻¹(x).

(2)

(c)

Find fg(x).

(2)

(d)

Work out f⁻¹(g(2)).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 −2	✓ 1
Part (b)
M1 x = (y + 7) ÷ 2 from y = 2x − 7	✓ 1
A1 f⁻¹(x) = (x + 7) ÷ 2	✓ 1
Part (c)
M1 2(x² + 3x) − 7	✓ 1
A1 2x² + 6x − 7	✓ 1
Part (d)
M1 g(2) = 10	✓ 1
A1 f⁻¹(10) = 8.5 (ECF from (b))	✓ 1

Exam tip: For f⁻¹(g(2)), work from the inside outwards: evaluate g(2) first, then apply the inverse function to that result.

+ Add to test✓ In your test212IGME3-3-12r15-05Differentiation and stationary points Paper HMedium7 marks

The power output, P kW, of a small wind turbine at wind speed v m/s is modelled by P = 12v² − v³ for 0 ≤ v ≤ 12.

(a)

Find dP/dv.

(2)

(b)

Find the non-zero value of v at which dP/dv = 0.

(2)

(c)

Show that this value of v gives a maximum power output, and find the maximum power.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 reduces the power of at least one term	✓ 1
A1 24v − 3v²	✓ 1
Part (b)
M1 24v − 3v² = 0 or 3v(8 − v) = 0 (ECF from (a))	✓ 1
A1 v = 8	✓ 1
Part (c)
M1 d²P/dv² = 24 − 6v evaluated at v = 8	✓ 1
A1 −24 < 0, so a maximum	✓ 1
A1 P = 768 − 512 = 256 kW	✓ 1

Exam tip: Once you have a stationary value, substitute it back into the original formula, not the derivative, to find the maximum or minimum value.

+ Add to test✓ In your test213IGME3-3-12r15-08Sequences and nth terms Paper HMedium6 marks

A swimming pool is being drained. At the start of each hour the volume of water, in cubic metres, is recorded: 1024, 768, 576, 432, ... Each term is a fixed multiple of the term before it.

(a)

Write down the common ratio.

(1)

(b)

Work out the sixth term.

(2)

(c)

The recording at the start of hour n is the nth term. Find the first term that is below 100 m³. Give this term to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3/4 or 0.75	✓ 1
Part (b)
M1 432 × 0.75 × 0.75 or 1024 × 0.75⁵ (ECF from (a))	✓ 1
A1 243	✓ 1
Part (c)
M1 continues the sequence or uses 1024 × 0.75^{(n − 1)}	✓ 1
A1 ninth term 102.5... (not below 100)	✓ 1
A1 tenth term 76.9 m³	✓ 1

Exam tip: For a geometric sequence, multiply by the common ratio each time. Keep unrounded values in your working and round only the final answer.

+ Add to test✓ In your test214IGME3-3-11r16-03Sequences and nth terms Paper HMedium7 marks

A data centre installs servers in a pattern. The total number of servers installed after weeks 1, 2, 3 and 4 is 3, 10, 21, 36. This is a quadratic sequence.

(a)

Work out the total number of servers installed after week 5.

(1)

(b)

Show that the second difference is constant and find an expression, in terms of n, for the total after week n.

(3)

(c)

In which week does the total reach exactly 300 servers?

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 55 (differences 7, 11, 15, 19)	✓ 1
Part (b)
M1 first differences 7, 11, 15 and second difference 4	✓ 1
M1 2n² subtracted from the sequence (leaving 1, 2, 3, 4)	✓ 1
A1 2n² + n	✓ 1
Part (c)
M1 2n² + n = 300 (ECF from (b))	✓ 1
M1 correct method to solve, e.g. (2n + 25)(n − 12) = 0 or the quadratic formula	✓ 1
A1 n = 12 (negative root rejected)	✓ 1

Common error: For a quadratic sequence the coefficient of n² is half of the second difference, not the second difference itself.

+ Add to test✓ In your test215IGME3-3-11r16-08Function notation, composite and inverse functions Paper HMedium6 marks

The functions \(f\) and \(g\) are given by \(f(x) = 3x - 4\) and \(g(x) = x^2 + 2\).

(a)

Work out \(g(-3)\).

(1)

(b)

Find \(fg(x)\). Give your answer in the form \(ax^2 + b\).

(2)

(c)

Given that \(fg(x) = 50\), find the positive value of \(x\).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 11	✓ 1
Part (b)
M1 3(x² + 2) − 4	✓ 1
A1 3x² + 2	✓ 1
Part (c)
M1 3x² + 2 = 50 (ECF from (b))	✓ 1
M1 x² = 16	✓ 1
A1 x = 4	✓ 1

Common error: In \(fg(x)\) the function \(g\) is applied first. Applying \(f\) first gives \(gf(x)\), a different function.

+ Add to test✓ In your test216IGME3-3-11r16-13Differentiation and stationary points Paper HMedium7 marks

The profile of a buried wall in an excavation is modelled by the curve \(y = x^3 - 6x^2 + 18\), where \(x\) and \(y\) are in metres. The point A on the curve has \(x = 3\).

(a)

Find \(\frac{dy}{dx}\).

(2)

(b)

Find the equation of the tangent to the curve at A.

(3)

(c)

The tangent meets the x-axis at the point B. Find the x-coordinate of B.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 at least one term differentiated correctly	✓ 1
A1 3x² − 12x	✓ 1
Part (b)
B1 gradient −9 and y-coordinate of A is −9 (ECF from (a))	✓ 1
M1 y + 9 = −9(x − 3) oe	✓ 1
A1 y = −9x + 18	✓ 1
Part (c)
M1 0 = −9x + 18 (ECF from (b))	✓ 1
A1 2	✓ 1

Common error: The gradient from the derivative is the gradient of the tangent, but you still need the y-coordinate of the point from the original curve.

+ Add to test✓ In your test217IGME3-3-11r19-08Function notation, composite and inverse functions Paper HMedium8 marks

The functions f and g are defined as f(x) = 5x − 2 and g(x) = x² + 3.

(a)

Find f⁻¹(x).

(2)

(b)

Show that gf(x) = 25x² − 20x + 7.

(3)

(c)

Solve fg(x) = 93.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sets y = 5x − 2 and rearranges to make x the subject	✓ 1
A1 f⁻¹(x) = (x + 2) ÷ 5	✓ 1
Part (b)
M1 g(5x − 2) = (5x − 2)² + 3	✓ 1
M1 (5x − 2)² = 25x² − 20x + 4	✓ 1
A1 25x² − 20x + 7	✓ 1
Part (c)
M1 fg(x) = 5(x² + 3) − 2	✓ 1
M1 5x² + 13 = 93, so x² = 16	✓ 1
A1 x = 4 or x = −4	✓ 1

Common error: fg(x) means f(g(x)), so g is applied first. Do not forget the negative square root when solving x² = 16.

+ Add to test✓ In your test218IGME3-3-11r19-09Differentiation and stationary points Paper HMedium9 marks

A curve has equation y = x³ − 9x² + 24x + 3.

(a)

Find dy/dx.

(2)

(b)

Find the coordinates of the two stationary points of the curve.

(4)

(c)

Determine the nature of each stationary point.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate, at least two terms correct	✓ 1
A1 3x² − 18x + 24	✓ 1
Part (b)
M1 sets dy/dx = 0, ECF from (a)	✓ 1
M1 solves 3(x − 2)(x − 4) = 0 (or equivalent)	✓ 1
A1 x = 2 and x = 4	✓ 1
A1 (2, 23) and (4, 19)	✓ 1
Part (c)
M1 d²y/dx² = 6x − 18 (or the sign of the gradient either side of each point)	✓ 1
A1 (2, 23) is a maximum, since the second derivative is −6	✓ 1
A1 (4, 19) is a minimum, since the second derivative is 6	✓ 1

Exam tip: Show the sign of d²y/dx² or the gradient either side of the point as evidence for maximum or minimum.

+ Add to test✓ In your test219IGME3-3-12r22-08Function notation, composite and inverse functions Paper HMedium7 marks

A signal-processing unit in a probe uses the functions \(f(x) = 2x + 3\) and \(g(x) = x^2 - 1\).

(a)

Find \(fg(x)\). Give your answer in the form \(ax^2 + b\).

(2)

(b)

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

(2)

(c)

Solve \(fg(x) = 19\).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for f(x² − 1) = 2(x² − 1) + 3	✓ 1
A1 for 2x² + 1	✓ 1
Part (b)
M1 for y = 2x + 3 and rearranging to x = (y − 3)/2	✓ 1
A1 for f⁻¹(x) = (x − 3)/2	✓ 1
Part (c)
M1 for 2x² + 1 = 19 (ECF from (a))	✓ 1
M1 for x² = 9	✓ 1
A1 for x = 3 and x = −3	✓ 1

Common error: For fg(x), apply g first and then f. fg(x) and gf(x) are generally different.

+ Add to test✓ In your test220IGME3-3-12r22-09Differentiation and stationary points Paper HMedium9 marks

A designer models the height y metres of a footbridge deck above a river, at a horizontal distance x metres from one end of the bridge, by \(y = x^3 - 9x^2 + 24x + 10\) for \(0 \leq x \leq 6\).

(a)

Find \(\frac{dy}{dx}\).

(2)

(b)

Find the values of x at which the deck has a stationary point.

(3)

(c)

Find \(\frac{d^2y}{dx^2}\) and use it to determine the nature of each stationary point.

(2)

(d)

Find the height of the deck at the minimum point.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for differentiating at least two terms correctly	✓ 1
A1 for 3x² − 18x + 24	✓ 1
Part (b)
M1 for setting dy/dx = 0	✓ 1
M1 for factorising: 3(x − 2)(x − 4) = 0	✓ 1
A1 for x = 2 and x = 4	✓ 1
Part (c)
M1 for d²y/dx² = 6x − 18	✓ 1
A1 for x = 2: −6 < 0 so a maximum; x = 4: 6 > 0 so a minimum	✓ 1
Part (d)
M1 for substituting x = 4 into y	✓ 1
A1 for 26 m	✓ 1

Exam tip: A negative second derivative means a maximum and a positive one means a minimum. Give the nature of every stationary point you find.

+ Add to test✓ In your test221IGME3-3-12r22-10Differentiation and stationary points Paper HMedium8 marks

The cost £C of running a coffee roaster for one hour at a throughput of x kg of beans per hour is modelled by \(C = 4x + \frac{9}{x}\), where \(x > 0\).

(a)

Find \(\frac{dC}{dx}\).

(2)

(b)

Find the value of x for which C is a minimum. Show that it is a minimum.

(4)

(c)

Find the minimum cost.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for differentiating 9/x as −9x⁻² (or one term correct)	✓ 1
A1 for 4 − 9/x²	✓ 1
Part (b)
M1 for setting 4 − 9/x² = 0	✓ 1
A1 for x² = 9/4, so x = 1.5 (reject −1.5 as x > 0)	✓ 1
M1 for d²C/dx² = 18/x³	✓ 1
A1 for 18/x³ > 0 when x = 1.5, so a minimum	✓ 1
Part (c)
M1 for 4 × 1.5 + 9 ÷ 1.5	✓ 1
A1 for £12	✓ 1

Common error: Rewrite 9/x as 9x⁻¹ before differentiating, and remember to reject roots that are outside the allowed range of x.

+ Add to test✓ In your test222IGME3-3-12r24-06Sequences and nth terms Paper HMedium7 marks

The first five patterns in a sequence of floor tilings in a hospital corridor use 3, 8, 15, 24 and 35 tiles.

(a)

Write down the number of tiles in the 6th pattern.

(1)

(b)

Find an expression, in terms of n, for the number of tiles in the nth pattern.

(3)

(c)

Show that 120 tiles are used in one of the patterns, and state which pattern.

(2)

(d)

Work out the number of tiles in the 20th pattern.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 48	✓ 1
Part (b)
M1 first differences 5, 7, 9, 11 and second difference 2, so n² seen	✓ 1
M1 subtract n² from the terms to leave 2, 4, 6, 8 (or equivalent)	✓ 1
A1 n² + 2n	✓ 1
Part (c)
M1 n² + 2n = 120 and (n + 12)(n − 10) = 0, or 10² + 2 × 10 evaluated	✓ 1
A1 n = 10, the 10th pattern (n = −12 rejected)	✓ 1
Part (d)
A1 440 (ECF from (b))	✓ 1

Exam tip: If the second difference of a sequence is 2, the nth term starts with n²; subtract it and what is left is linear.

+ Add to test✓ In your test223IGME3-3-12r24-07Function notation, composite and inverse functions Paper HMedium8 marks

The functions f and g are defined by f(x) = 3x − 7 and g(x) = x² + 1.

(a)

Work out f(5).

(1)

(b)

Find f⁻¹(x).

(2)

(c)

Find fg(x). Give your answer in its simplest form.

(2)

(d)

Find the value of x, with x > 0, for which gf(x) = 50.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 8	✓ 1
Part (b)
M1 y = 3x − 7 and rearranging to make x the subject	✓ 1
A1 f⁻¹(x) = (x + 7) ÷ 3	✓ 1
Part (c)
M1 3(x² + 1) − 7	✓ 1
A1 3x² − 4	✓ 1
Part (d)
M1 (3x − 7)² + 1 = 50	✓ 1
M1 3x − 7 = ±7	✓ 1
A1 x = 14/3 (the solution x = 0 is rejected)	✓ 1

Common error: In gf(x) the function f is applied first, so substitute f(x) into g, not the other way round.

+ Add to test✓ In your test224IGME3-3-12r24-09Differentiation and stationary points Paper HMedium8 marks

A beekeeper models the yearly profit P thousand euros with x hundred hives by P = 2x³ − 9x² + 12x, for x > 0.

(a)

Find dP/dx.

(2)

(b)

Find the coordinates of the two stationary points of the graph of P against x.

(4)

(c)

Determine the nature of each stationary point.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 one term differentiated correctly	✓ 1
A1 6x² − 18x + 12	✓ 1
Part (b)
M1 6x² − 18x + 12 = 0 and factorising 6(x − 1)(x − 2)	✓ 1
A1 x = 1 and x = 2	✓ 1
M1 substituting a value of x into P	✓ 1
A1 (1, 5) and (2, 4)	✓ 1
Part (c)
M1 d²P/dx² = 12x − 18 evaluated at 1 and 2, giving −6 and 6	✓ 1
A1 (1, 5) is a maximum and (2, 4) is a minimum	✓ 1

Exam tip: Setting the derivative equal to zero finds where stationary points are; the second derivative, or the sign of the gradient either side, tells you which type each is.

+ Add to test✓ In your test225IGME3-3-12r24-10Differentiation and stationary points Paper HMedium8 marks

A section of a Formula 1 circuit is modelled by the curve y = x³ − 3x² + 4x, where x and y are in hundreds of metres.

(a)

Find dy/dx.

(2)

(b)

Find the equation of the tangent to the curve at the point where x = 2.

(3)

(c)

The gradient of the curve is also 4 at another point. Find the coordinates of this point.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 at least two terms differentiated correctly	✓ 1
A1 3x² − 6x + 4	✓ 1
Part (b)
B1 y = 4 when x = 2 and gradient = 12 − 12 + 4 = 4	✓ 1
M1 y − 4 = 4(x − 2)	✓ 1
A1 y = 4x − 4	✓ 1
Part (c)
M1 3x² − 6x + 4 = 4	✓ 1
A1 3x(x − 2) = 0, so x = 0 (the other root is the point from (b))	✓ 1
A1 (0, 0)	✓ 1

Common error: Substitute x into the derivative to get the gradient, but into the original equation to get y.

+ Add to test✓ In your test226IGME3-3-13r24-09Function notation, composite and inverse functions Paper HMedium8 marks

The functions f and g are defined by f(x) = 3x − 5 and g(x) = x² + 1.

(a)

Find gf(x). Give your answer in the form ax² + bx + c.

(3)

(b)

Find f⁻¹(x).

(2)

(c)

Find the positive value of x for which fg(x) = 25.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 g(3x − 5) = (3x − 5)² + 1	✓ 1
M1 (3x − 5)² = 9x² − 30x + 25	✓ 1
A1 9x² − 30x + 26	✓ 1
Part (b)
M1 y = 3x − 5 and rearrange for x, or flow-chart method	✓ 1
A1 (x + 5) ÷ 3	✓ 1
Part (c)
M1 fg(x) = 3(x² + 1) − 5 = 3x² − 2	✓ 1
M1 3x² − 2 = 25 so x² = 9	✓ 1
A1 x = 3	✓ 1

Common error: gf(x) means apply f first and then g; swapping the order gives a different function.

+ Add to test✓ In your test227IGME3-3-13r24-11Differentiation and stationary points Paper HMedium8 marks

A curve has equation y = x³ − 3x² − 24x + 5.

(a)

Find dy/dx.

(2)

(b)

Find the x-coordinates of the stationary points of the curve.

(3)

(c)

Find the coordinates of the minimum point of the curve. Show how you know it is a minimum.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3x²	✓ 1
B1 −6x − 24 (no extra terms)	✓ 1
Part (b)
M1 3x² − 6x − 24 = 0 (ECF from (a))	✓ 1
M1 3(x − 4)(x + 2) = 0 or use of the quadratic formula	✓ 1
A1 x = 4 and x = −2	✓ 1
Part (c)
M1 d²y/dx² = 6x − 6	✓ 1
A1 at x = 4 the value is 18, which is positive, so it is a minimum	✓ 1
A1 y = −75, so the minimum point is (4, −75)	✓ 1

Common error: A stationary point is where the gradient is zero; do not set the original equation y equal to zero.

+ Add to test✓ In your test228IGME3-3-13r24-12Differentiation and stationary points Paper HMedium9 marks

A curve has equation y = 4x³ − 5x² + 2x − 7.

(a)

Find dy/dx.

(3)

(b)

Work out the gradient of the curve at the point where x = 2.

(2)

(c)

Find the equation of the tangent to the curve at the point where x = 2. Give your answer in the form y = mx + c.

(3)

(d)

Write down the y-coordinate of the point where this tangent crosses the y-axis.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 12x²	✓ 1
B1 −10x	✓ 1
B1 +2 (constant term disappears)	✓ 1
Part (b)
M1 12 × 4 − 10 × 2 + 2	✓ 1
A1 30	✓ 1
Part (c)
B1 y = 32 − 20 + 4 − 7 = 9 when x = 2	✓ 1
M1 y − 9 = 30(x − 2) (ECF from (b))	✓ 1
A1 y = 30x − 51	✓ 1
Part (d)
A1 −51 (ECF from (c))	✓ 1

Exam tip: The tangent passes through the point on the curve, so find its y-coordinate from the original equation, not from the gradient function.

+ Add to test✓ In your test229IGME3-3-12r26-01Function notation, composite and inverse functions Paper HMedium7 marks

An online shop gives a discount and then adds a delivery charge. The sale price in pounds for a list price of £x is given by f(x) = 0.85x. Adding delivery of £4 is given by g(x) = x + 4.

(a)

Work out gf(60).

(2)

(b)

Find (gf)⁻¹(x).

(3)

(c)

A customer pays a total of £89. Use your answer to (b) to work out the list price.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 f(60) = 51	✓ 1
A1 55	✓ 1
Part (b)
M1 y = 0.85x + 4 (gf(x) found)	✓ 1
M1 attempt to rearrange for x	✓ 1
A1 (x − 4)/0.85 oe	✓ 1
Part (c)
M1 (89 − 4)/0.85 (ECF from (b))	✓ 1
A1 £100	✓ 1

Common error: In gf(x), apply f first and then g; reversing the order gives 0.85(x + 4), which is a different function.

+ Add to test✓ In your test230IGME3-3-12r26-04Function notation, composite and inverse functions Paper HMedium8 marks

The functions f and g are given by f(x) = 2x + 1 and g(x) = x² − 3.

(a)

Work out g(4).

(1)

(b)

Find f⁻¹(x).

(2)

(c)

Show that fg(x) = 2x² − 5.

(2)

(d)

Find the positive value of x for which fg(x) = 45.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 13	✓ 1
Part (b)
M1 y = 2x + 1 and attempt to rearrange for x	✓ 1
A1 (x − 1)/2 oe	✓ 1
Part (c)
M1 2(x² − 3) + 1	✓ 1
A1 correct simplification to 2x² − 5 shown	✓ 1
Part (d)
M1 2x² − 5 = 45	✓ 1
M1 x² = 25	✓ 1
A1 5	✓ 1

Common error: Substitute the whole inner function into the outer one, and remember to square the bracket correctly when the outer function contains x².

+ Add to test✓ In your test231IGME3-3-12r26-09Function notation, composite and inverse functions Paper HMedium7 marks

The function f is given by f(x) = (2x + 1)/(x − 3), where x ≠ 3.

(a)

Work out f(5).

(1)

(b)

Find f⁻¹(x).

(4)

(c)

Work out f⁻¹(4).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 11/2 (5.5)	✓ 1
Part (b)
M1 y(x − 3) = 2x + 1	✓ 1
M1 xy − 3y = 2x + 1, collecting x terms: x(y − 2) = 3y + 1	✓ 1
A1 x = (3y + 1)/(y − 2)	✓ 1
A1 f⁻¹(x) = (3x + 1)/(x − 2)	✓ 1
Part (c)
M1 (3 × 4 + 1)/(4 − 2) (ECF from (b))	✓ 1
A1 6.5 oe	✓ 1

Exam tip: When the unknown appears twice in the rearrangement, collect all the x terms on one side and factorise before dividing.

+ Add to test✓ In your test232IGME3-3-12r26-11Function notation, composite and inverse functions Paper HMedium8 marks

The functions f and g are given by f(x) = 3x − 2 and g(x) = x² + 1.

(a)

Find fg(x).

(2)

(b)

Find gf(x) and write it in the form ax² + bx + c.

(2)

(c)

Solve fg(x) = gf(x). Give the larger solution to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3(x² + 1) − 2	✓ 1
A1 3x² + 1	✓ 1
Part (b)
M1 (3x − 2)² + 1	✓ 1
A1 9x² − 12x + 5	✓ 1
Part (c)
M1 3x² + 1 = 9x² − 12x + 5	✓ 1
A1 3x² − 6x + 2 = 0 oe	✓ 1
M1 correct use of the quadratic formula (discriminant 12)	✓ 1
A1 1.58	✓ 1

Common error: Expand (3x − 2)² as three terms, not 9x² + 4; the middle term −12x is the one most often lost.

+ Add to test✓ In your test233IGME3-3-14r28-05Sequences and nth terms Paper HMedium5 marks

The first four terms of a sequence are 4, 11, 22, 37, which models the number of cells in a honeycomb pattern.

(a)

Find an expression for the nth term of the sequence.

(3)

(b)

Show that 301 is a term of the sequence and state its position.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 first differences 7, 11, 15 and second difference 4	✓ 1
M1 subtract 2n² (2, 8, 18, 32) to leave 2, 3, 4, 5	✓ 1
A1 2n² + n + 1	✓ 1
Part (b)
M1 2n² + n + 1 = 301 or 2n² + n − 300 = 0	✓ 1
A1 n = 12 (from factorising or the formula) and check 2 × 144 + 12 + 1 = 301	✓ 1

Exam tip: For a quadratic sequence, half the second difference gives the coefficient of n².

+ Add to test✓ In your test234IGME3-3-14r28-07Function notation, composite and inverse functions Paper HMedium7 marks

The functions f and g are defined as f(x) = 3x − 4 and g(x) = x² + 1, and model the effect of two successive tyre-pressure adjustments.

(a)

Find fg(2).

(2)

(b)

Find f⁻¹(x).

(2)

(c)

Solve gf(x) = 26. Give the positive solution.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 g(2) = 5	✓ 1
A1 f(5) = 11	✓ 1
Part (b)
M1 y = 3x − 4 rearranged to make x the subject	✓ 1
A1 f⁻¹(x) = (x + 4)/3	✓ 1
Part (c)
M1 gf(x) = (3x − 4)² + 1	✓ 1
M1 (3x − 4)² = 25 so 3x − 4 = ±5	✓ 1
A1 x = 3 (the other solution is −¹⁄₃)	✓ 1

Common error: In fg(x) the function g is applied first; reversing the order gives gf instead.

+ Add to test✓ In your test235IGME3-3-14r28-08Differentiation and stationary points Paper HMedium7 marks

A beekeeper models the height, y metres, of a hillside path by y = x³ − 6x² + 9x + 2, where x is the horizontal distance in hundreds of metres, 0 ≤ x ≤ 4.

(a)

Find ^dy⁄_dx.

(2)

(b)

Find the coordinates of the two stationary points.

(3)

(c)

Determine the nature of the stationary point at x = 3.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiates at least two terms correctly	✓ 1
A1 3x² − 12x + 9	✓ 1
Part (b)
M1 sets 3x² − 12x + 9 = 0	✓ 1
A1 x = 1 and x = 3	✓ 1
A1 (1, 6) and (3, 2)	✓ 1
Part (c)
M1 second derivative 6x − 12 = 6 > 0 (or gradient test either side)	✓ 1
A1 minimum	✓ 1

Exam tip: A positive second derivative means a minimum and a negative one means a maximum.

+ Add to test✓ In your test236IGME3-3-14r32-09Function notation, composite and inverse functions Paper HMedium8 marks

Two stages of an instrument calibration on a space probe are modelled by the functions f(x) = 3x + 8 and g(x) = x² − 5.

(a)

Work out f(4).

(1)

(b)

Find fg(x). Give your answer in its simplest form.

(2)

(c)

Find f⁻¹(x).

(2)

(d)

Solve fg(x) = 41, given that x > 0.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 20	✓ 1
Part (b)
M1 3(x² − 5) + 8	✓ 1
A1 3x² − 7	✓ 1
Part (c)
M1 y = 3x + 8 and rearrange for x (or reverse the operations)	✓ 1
A1 (x − 8) ÷ 3	✓ 1
Part (d)
M1 3x² − 7 = 41 (ECF from (b))	✓ 1
M1 x² = 16	✓ 1
A1 x = 4	✓ 1

Common error: In fg(x) the function g is applied first; writing g into f, not f into g, gives the wrong composite.

+ Add to test✓ In your test237IGME3-3-14r32-10Differentiation and stationary points Paper HMedium9 marks

The height y metres of the road on a bridge approach is modelled by y = x³ − 6x² + 9x + 2 for 0 ≤ x ≤ 4, where x is the horizontal distance in hundreds of metres.

(a)

Find ^dy⁄_dx.

(2)

(b)

Show that the road has stationary points at x = 1 and x = 3.

(3)

(c)

Determine the nature of the stationary point at x = 3.

(2)

(d)

Work out the height of the road at the stationary point where x = 3.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 at least one term differentiated correctly	✓ 1
A1 3x² − 12x + 9	✓ 1
Part (b)
M1 3x² − 12x + 9 = 0	✓ 1
M1 3(x − 1)(x − 3) = 0 (or correct use of the formula)	✓ 1
A1 x = 1 and x = 3	✓ 1
Part (c)
M1 ^d²y⁄_dx² = 6x − 12 = 6 at x = 3 (or a gradient test either side)	✓ 1
A1 positive, so a minimum	✓ 1
Part (d)
M1 3³ − 6 × 3² + 9 × 3 + 2	✓ 1
A1 2 m	✓ 1

Exam tip: To classify a stationary point, find the second derivative and substitute the x-value: positive means a minimum, negative means a maximum.

+ Add to test✓ In your test238IGME3-3-14r32-11Differentiation and stationary points Paper HMedium8 marks

A suspension cable follows the curve y = x³ − 4x + 1, where x and y are measured in metres.

(a)

Find ^dy⁄_dx.

(2)

(b)

Work out the gradient of the curve at the point where x = 2.

(1)

(c)

Find an equation of the tangent to the curve at the point where x = 2.

(3)

(d)

Work out the value of x at the point where this tangent crosses the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 at least one term differentiated correctly	✓ 1
A1 3x² − 4	✓ 1
Part (b)
B1 8	✓ 1
Part (c)
B1 y = 1 when x = 2	✓ 1
M1 y − 1 = 8(x − 2) (using their gradient)	✓ 1
A1 y = 8x − 15	✓ 1
Part (d)
M1 8x − 15 = 0	✓ 1
A1 x = 1.875 (or ¹⁵⁄₈)	✓ 1

Common error: The tangent needs the y-value of the point as well as the gradient; substitute x into the original curve, not into the derivative.

+ Add to test✓ In your test239IGME3-3-13r33-05Function notation, composite and inverse functions Paper HMedium8 marks

The functions f and g are defined as f(x) = 2x + 3 and g(x) = x² − 1.

(a)

Find fg(x). Give your answer in its simplest form.

(2)

(b)

Find f⁻¹(x).

(2)

(c)

Find the positive value of x for which gf(x) = 15.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2(x² − 1) + 3	✓ 1
A1 2x² + 1	✓ 1
Part (b)
M1 y = 2x + 3 and rearrange to make x the subject (or reverse the operations)	✓ 1
A1 f⁻¹(x) = (x − 3)/2	✓ 1
Part (c)
M1 gf(x) = (2x + 3)² − 1	✓ 1
M1 (2x + 3)² = 16 so 2x + 3 = ±4	✓ 1
A1 x = 0.5 or x = −3.5	✓ 1
A1 positive value is 0.5	✓ 1

Common error: fg(x) means apply g first, then f; the order is not interchangeable.

+ Add to test✓ In your test240IGME3-3-13r33-06Differentiation and stationary points Paper HMedium7 marks

On a cycling route, the height of the road above sea level is modelled by h = x³ − 6x² + 9x + 2 for 0 ≤ x ≤ 4, where h is in metres and x is the horizontal distance in km from the start.

(a)

Find dh/dx.

(2)

(b)

Find the coordinates of the two stationary points of the model.

(3)

(c)

Determine whether each stationary point is a maximum or a minimum. Give a reason for each.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 one correct term differentiated	✓ 1
A1 3x² − 12x + 9	✓ 1
Part (b)
M1 set 3x² − 12x + 9 = 0	✓ 1
M1 3(x − 1)(x − 3) = 0 so x = 1 and x = 3	✓ 1
A1 (1, 6) and (3, 2) with both h values correct	✓ 1
Part (c)
M1 d²h/dx² = 6x − 12 and evaluates at one point	✓ 1
A1 at x = 1 second derivative is −6 < 0, so maximum; at x = 3 it is +6 > 0, so minimum (minimum height 2 m)	✓ 1

Exam tip: A sign test on the second derivative (or on the gradient either side) is needed to justify maximum or minimum.

+ Add to test✓ In your test241IGME3-3-13r33-10Differentiation and stationary points Paper HMedium6 marks

A curve has equation y = 2x³ − 5x + 1.

(a)

Find dy/dx.

(2)

(b)

Find the equation of the tangent to the curve at the point where x = 2. Give your answer in the form y = mx + c.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 one term correct	✓ 1
A1 6x² − 5	✓ 1
Part (b)
B1 y = 16 − 10 + 1 = 7 at x = 2	✓ 1
M1 gradient = 6 × 4 − 5 (ECF from (a))	✓ 1
M1 y − 7 = 19(x − 2)	✓ 1
A1 y = 19x − 31	✓ 1

Exam tip: Substitute into the derivative, not the original equation, to find the gradient of the tangent.

+ Add to test✓ In your test242IGME3-3-12r35-07Sequences and nth terms Paper HMedium7 marks

Solar cells are added to a space probe in stages. The numbers of cells after stages 1, 2, 3 and 4 are 3, 8, 15 and 24.

(a)

Work out the number of cells after stage 5.

(1)

(b)

Show that the nth term of the sequence is n² + 2n.

(3)

(c)

After which stage does the probe have exactly 195 cells?

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 35 (differences 5, 7, 9, 11)	✓ 1
Part (b)
B1 second difference is 2, so the nth term starts with n²	✓ 1
M1 subtracts n² from each term to leave 2, 4, 6, 8	✓ 1
A1 remaining sequence is 2n, so the nth term is n² + 2n	✓ 1
Part (c)
M1 n² + 2n = 195 and rearranges to n² + 2n − 195 = 0	✓ 1
M1 (n + 15)(n − 13) = 0 or correct use of the quadratic formula	✓ 1
A1 stage 13 (reject −15)	✓ 1

Common error: After finding a root of a quadratic in context, check it makes sense: a stage number cannot be negative.

+ Add to test✓ In your test243IGME3-3-12r35-09Function notation, composite and inverse functions Paper HMedium6 marks

The functions f and g are defined by f(x) = 4x − 7 and g(x) = x².

(a)

Find f(5).

(1)

(b)

Find fg(x).

(2)

(c)

Find f⁻¹(x).

(2)

(d)

Hence find f⁻¹(33).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 13	✓ 1
Part (b)
M1 substitutes g(x) into f, e.g. 4(x²) − 7	✓ 1
A1 4x² − 7	✓ 1
Part (c)
M1 y = 4x − 7 with x and y swapped, or rearranges x = 4y − 7	✓ 1
A1 f⁻¹(x) = (x + 7)/4	✓ 1
Part (d)
B1 10 (ECF from (c))	✓ 1

Common error: fg(x) means apply g first, then f. Applying f first gives (4x − 7)², which is gf(x).

+ Add to test✓ In your test244IGME3-3-12r35-10Graphs, gradients and real-life graphs Paper HMedium6 marks

A test probe is fired vertically upwards from a platform. Its height, h metres above the ground, at time t seconds after firing is given by h = 3 + 40t − 5t², for the time while the probe is in the air.

(a)

Find dh/dt.

(2)

(b)

Find the time at which the probe reaches its greatest height.

(2)

(c)

Find the greatest height of the probe.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 40	✓ 1
B1 −10t	✓ 1
Part (b)
M1 sets dh/dt = 0, e.g. 40 − 10t = 0 (ECF from (a))	✓ 1
A1 t = 4 s	✓ 1
Part (c)
M1 substitutes t = 4, i.e. 3 + 160 − 80 (ECF from (b))	✓ 1
A1 83 m	✓ 1

Exam tip: At the greatest height the gradient of the height–time graph is zero, so set the derivative equal to zero.

+ Add to test✓ In your test245IGME3-3-12r35-11Differentiation and stationary points Paper HMedium8 marks

A coffee roaster works out the cost, £C per kilogram of coffee, of roasting a batch of x kg. It uses C = 0.5x² − 6x + 25, where x > 0.

(a)

Find dC/dx.

(2)

(b)

Find the batch size for which the cost per kilogram has a stationary value.

(2)

(c)

Show that this stationary value is a minimum.

(2)

(d)

Work out the minimum cost per kilogram.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 x	✓ 1
B1 −6	✓ 1
Part (b)
M1 sets dC/dx = 0 (ECF from (a))	✓ 1
A1 x = 6 kg	✓ 1
Part (c)
M1 gradient evaluated either side of x = 6, e.g. −1 at x = 5 and 1 at x = 7, or d²C/dx² = 1	✓ 1
A1 gradient changes from negative to positive (or second derivative is positive), so it is a minimum	✓ 1
Part (d)
M1 0.5 × 6² − 6 × 6 + 25	✓ 1
A1 £7	✓ 1

Exam tip: To justify a minimum, say what the gradient does either side of the stationary point, or give the sign of the second derivative.

+ Add to test✓ In your test246IGME3-3-12r37-08Function notation, composite and inverse functions Paper HMedium6 marks

In a game, a base score x is converted using the functions f(x) = 3x + 4 and g(x) = x² − 2.

(a)

Find f(5).

(1)

(b)

Find fg(x). Give your answer in its simplest form.

(2)

(c)

Solve gf(x) = 62.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 19	✓ 1
Part (b)
M1 3(x² − 2) + 4	✓ 1
A1 3x² − 2	✓ 1
Part (c)
M1 (3x + 4)² − 2 = 62, so (3x + 4)² = 64	✓ 1
M1 3x + 4 = 8 or 3x + 4 = −8	✓ 1
A1 x = 4/3 or x = −4	✓ 1

Common error: In fg(x), g is applied first; do not apply the functions in reading order from the left.

+ Add to test✓ In your test247IGME3-3-12r37-09Function notation, composite and inverse functions Paper HMedium8 marks

A pharmacist uses the function f(w) = 4w + 10 to find a dose D mg of a drug for a child of weight w kg. Another function is h(x) = 5/(x − 1), where x ≠ 1.

(a)

Find f⁻¹(x).

(2)

(b)

A child is given a dose of 90 mg. Use f⁻¹ to find the child's weight.

(2)

(c)

Find h⁻¹(x).

(3)

(d)

Find h⁻¹(4).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 y = 4w + 10 and rearrange for w	✓ 1
A1 f⁻¹(x) = (x − 10) ÷ 4	✓ 1
Part (b)
M1 (90 − 10) ÷ 4	✓ 1
A1 20 kg	✓ 1
Part (c)
M1 y = 5/(x − 1), so x − 1 = 5/y	✓ 1
M1 x = 5/y + 1, then swap x and y	✓ 1
A1 h⁻¹(x) = 5/x + 1	✓ 1
Part (d)
B1 2.25 (ECF from (c))	✓ 1

Exam tip: To find an inverse function, write y = f(x), make x the subject, then swap the letters.

+ Add to test✓ In your test248IGME3-3-12r37-10Differentiation and stationary points Paper HMedium7 marks

The volume of water in a reservoir, V thousand m³, t weeks after a dry spell begins, is modelled by V = 90 + 12t − t² for 0 ≤ t ≤ 12.

(a)

Find dV/dt.

(2)

(b)

Find the value of t at which the volume is a maximum.

(2)

(c)

Find the maximum volume of water.

(2)

(d)

Explain how you know this stationary point is a maximum.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiate at least one term correctly	✓ 1
A1 12 − 2t	✓ 1
Part (b)
M1 12 − 2t = 0	✓ 1
A1 t = 6	✓ 1
Part (c)
M1 90 + 12 × 6 − 6²	✓ 1
A1 126 thousand m³	✓ 1
Part (d)
B1 the gradient changes from positive to negative (or d²V/dt² = −2 < 0)	✓ 1

Exam tip: At a stationary point set the gradient function equal to zero, then substitute back into the original equation for the value.

+ Add to test✓ In your test249IGME3-3-12r37-13Sequences and nth terms Paper HMedium7 marks

The number of tokens needed for levels 1, 2, 3, 4, 5 in a game is
5, 12, 23, 38, 57

(a)

Find the number of tokens needed for level 6.

(1)

(b)

Find an expression, in terms of n, for the nth term of the sequence.

(3)

(c)

Find the first level that needs more than 500 tokens, and the number of tokens it needs.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 80 (second differences are constant at 4, first differences 7, 11, 15, 19, 23)	✓ 1
Part (b)
M1 second difference 4, so the n² coefficient is 2	✓ 1
M1 subtract 2n² from the terms: 3, 4, 5, … so remainder is n + 2	✓ 1
A1 2n² + n + 2	✓ 1
Part (c)
M1 test values or solve 2n² + n + 2 > 500 (level 15 gives 467)	✓ 1
A1 level 16	✓ 1
A1 530 tokens	✓ 1

Exam tip: For a quadratic sequence, the coefficient of n² is half the constant second difference.

+ Add to test✓ In your test250IGME3-3-13r44-05Sequences and nth terms Paper HMedium6 marks

In a new bee colony the number of bees, P thousand, t weeks after the colony is moved to a new site is modelled by
P = 2t² − 12t + 25 for 0 ≤ t ≤ 8

(a)

Find dP/dt.

(2)

(b)

Find the value of t at which the number of bees is stationary.

(2)

(c)

Work out the smallest number of bees predicted by the model. Give your answer in thousands.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 one term differentiated correctly	✓ 1
A1 4t − 12	✓ 1
Part (b)
M1 4t − 12 = 0 (ECF from (a))	✓ 1
A1 t = 3	✓ 1
Part (c)
M1 2 × 3² − 12 × 3 + 25 (ECF from (b))	✓ 1
A1 7 (thousand)	✓ 1

Common error: To find the value of the stationary point, substitute t back into the original expression for P, not into the derivative.

+ Add to test✓ In your test251IGME3-3-13r44-10Sequences and nth terms Paper HMedium8 marks

A Formula 1 team models the lap time, L seconds, of a car by
L = 0.5p² − 16p + 217.5 for 12 ≤ p ≤ 20
where p psi is the tyre pressure.

(a)

Find dL/dp.

(2)

(b)

Find the value of p for which dL/dp = 0.

(2)

(c)

Show that this value of p gives a minimum lap time.

(2)

(d)

Work out the minimum lap time.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 one term differentiated correctly	✓ 1
A1 p − 16	✓ 1
Part (b)
M1 p − 16 = 0 (ECF from (a))	✓ 1
A1 p = 16	✓ 1
Part (c)
B1 gradient is negative for p < 16 and positive for p > 16 (a value either side tested), or d²L/dp² = 1 > 0	✓ 1
B1 so the stationary point is a minimum	✓ 1
Part (d)
M1 0.5 × 16² − 16 × 16 + 217.5 (ECF from (b))	✓ 1
A1 89.5 s	✓ 1

Exam tip: Show how you know a stationary point is a minimum by checking the sign of the gradient on either side, or by finding the sign of the second derivative.

+ Add to test✓ In your test252IGME3-3-12r48-08Function notation, composite and inverse functions Paper HMedium7 marks

The functions f and g are given by f(x) = 3x − 5 and g(x) = x² + 2.

(a)

Find fg(x). Give your answer in its simplest form.

(2)

(b)

Find f⁻¹(x).

(2)

(c)

Solve gf(x) = 18. Give both values of x.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3(x² + 2) − 5	✓ 1
A1 3x² + 1	✓ 1
Part (b)
M1 y = 3x − 5 and rearranges to make x the subject	✓ 1
A1 f⁻¹(x) = (x + 5) ÷ 3	✓ 1
Part (c)
M1 (3x − 5)² + 2 = 18	✓ 1
M1 3x − 5 = 4 or 3x − 5 = −4	✓ 1
A1 x = 3 and x = 1/3	✓ 1

Exam tip: fg(x) means apply g first, then f; gf(x) reverses this and gives a different function.

+ Add to test✓ In your test253IGME3-3-12r48-10Graphs, gradients and real-life graphs Paper HMedium8 marks

The height, h metres, of a cycle route above sea level after x kilometres is modelled by h = 5x³ − 30x² + 45x + 20 for 0 ≤ x ≤ 4.

(a)

Find dh/dx.

(2)

(b)

Find the values of x at the stationary points of the route.

(3)

(c)

Show that the stationary point at x = 3 is a local minimum and work out the height of the route there.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 at least one term differentiated correctly	✓ 1
A1 15x² − 60x + 45	✓ 1
Part (b)
M1 15x² − 60x + 45 = 0 (ECF from (a))	✓ 1
M1 15(x − 1)(x − 3) = 0 or equivalent solving	✓ 1
A1 x = 1 and x = 3	✓ 1
Part (c)
M1 d²h/dx² = 30x − 60	✓ 1
A1 at x = 3 the value is 30 > 0, so a local minimum	✓ 1
A1 h = 135 − 270 + 135 + 20 = 20 m	✓ 1

Common error: After finding the x-values of stationary points, substitute into the original equation, not the derivative, to find the height.

+ Add to test✓ In your test254IGME3-3-12r48-13Function notation, composite and inverse functions Paper HMedium7 marks

The function f is defined by f(x) = \(\frac{x+2}{x-1}\) for x ≠ 1.

(a)

Work out f(5).

(1)

(b)

Find f⁻¹(x).

(3)

(c)

Show that ff(x) = x.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 7/4 or 1.75	✓ 1
Part (b)
M1 y(x − 1) = x + 2 after swapping or rearranging	✓ 1
M1 collects x terms and factorises, x(y − 1) = y + 2	✓ 1
A1 f⁻¹(x) = (x + 2) ÷ (x − 1)	✓ 1
Part (c)
M1 ff(x) = ((x + 2) ÷ (x − 1) + 2) ÷ ((x + 2) ÷ (x − 1) − 1)	✓ 1
M1 multiplies through by (x − 1) to get (x + 2 + 2(x − 1)) ÷ (x + 2 − (x − 1))	✓ 1
A1 3x ÷ 3 = x shown	✓ 1

Exam tip: A function that equals its own inverse satisfies ff(x) = x; use this to check a result.

+ Add to test✓ In your test255IGME-060-02-12Function notation, composite and inverse functions Paper HHard7 marks

The functions f and g are defined as
f(x) = x² + 2x, x ≥ −1
g(x) = x − 3

(a)

Show that f(x) can be written as (x + 1)² − 1

(1)

(b)

Find f⁻¹(x)

(3)

(c)

Given that fg(x) is defined only when g(x) ≥ −1, solve fg(x) = 35

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: (x + 1)² − 1 = x² + 2x + 1 − 1 = x² + 2x shown	✓ 1
Part (b)
M1: y = (x + 1)² − 1 so (x + 1)² = y + 1	✓ 1
M1: x + 1 = √(y + 1), using the positive root since x ≥ −1	✓ 1
A1: f⁻¹(x) = √(x + 1) − 1	✓ 1
Part (c)
M1: fg(x) = (x − 3 + 1)² − 1 = (x − 2)² − 1	✓ 1
M1: (x − 2)² = 36 so x = 8 or x = −4	✓ 1
A1: x = 8 only, since x = −4 gives g(x) = −7 < −1	✓ 1

Exam tip: Check each solution against the domain. A restricted domain often means one root of a quadratic has to be rejected.

+ Add to test✓ In your test256IGME-060-02-13Function notation, composite and inverse functions Paper HHard6 marks

The function f is defined as f(x) = ax + b, where a and b are constants.
Given that f(2) = 11 and ff(0) = 20

(a)

Show that 2a² − 9a + 9 = 0

(3)

(b)

Given that a is an integer, find the values of a and b.

(2)

(c)

Find f⁻¹(26)

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 2a + b = 11	✓ 1
M1: f(0) = b so ff(0) = ab + b = 20	✓ 1
A1: substitutes b = 11 − 2a to get (11 − 2a)(a + 1) = 20, leading to 2a² − 9a + 9 = 0	✓ 1
Part (b)
M1: (2a − 3)(a − 3) = 0	✓ 1
A1: a = 3, b = 5	✓ 1
Part (c)
B1: (26 − 5)/3 = 7	✓ 1

Exam tip: To find f⁻¹(26) you can solve f(x) = 26. There is no need to find the full inverse function first.

+ Add to test✓ In your test257IGME-060-02-14Function notation, composite and inverse functions Paper HHard7 marks

The function f is defined as \(f(x) = \frac{x+5}{2x-1}\), x ≠ 0.5

(a)

Show that ff(x) = x

(3)

(b)

Hence write down f⁻¹(x)

(1)

(c)

Solve f(x) = x, giving your answers correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: ff(x) = [(x + 5)/(2x − 1) + 5] ÷ [2(x + 5)/(2x − 1) − 1]	✓ 1
M1: numerator simplifies to 11x/(2x − 1) or denominator simplifies to 11/(2x − 1)	✓ 1
A1: both simplified and divided to give x	✓ 1
Part (b)
B1: f⁻¹(x) = (x + 5)/(2x − 1)	✓ 1
Part (c)
M1: x + 5 = x(2x − 1) leading to 2x² − 2x − 5 = 0	✓ 1
M1: x = [2 ± √(4 + 40)]/4	✓ 1
A1: x = 2.16 or x = −1.16	✓ 1

Exam tip: If ff(x) = x, then f is its own inverse. You can state f⁻¹(x) straight away without rearranging.

+ Add to test✓ In your test258IGME-060-01-14Graphs, gradients and real-life graphs Paper HHard8 marks

The line M has equation 3x + 4y = 12
The line L passes through the point P(3, 7) and is perpendicular to M.

(a)

Find an equation of L. Give your answer in the form y = mx + c

(3)

(b)

Show that L and M intersect at the point (0, 3).

(2)

(c)

The lines L and M and the x-axis form a triangle.
Work out the area of this triangle.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for gradient of M = \(-\frac{3}{4}\)	✓ 1
M1 for gradient of L = \(\frac{4}{3}\) and substituting (3, 7)	✓ 1
A1 y = \(\frac{4}{3}\)x + 3	✓ 1
Part (b)
M1 for substituting y = \(\frac{4}{3}\)x + 3 into 3x + 4y = 12, giving \(\frac{25}{3}\)x + 12 = 12	✓ 1
A1 x = 0, y = 3 shown	✓ 1
Part (c)
M1 for x-intercepts: M at x = 4 and L at x = \(-\frac{9}{4}\)	✓ 1
M1 for ½ × (4 + \(\frac{9}{4}\)) × 3	✓ 1
A1 \(\frac{75}{8}\) or 9.375	✓ 1

Exam tip: For the area of a triangle bounded by lines and an axis, use the distance between the axis intercepts as the base and the y-coordinate of the intersection point as the height.

+ Add to test✓ In your test259IGME-060-03-14Differentiation and stationary points Paper HHard8 marks

A curve has equation y = x³ + kx² − 9x + 2, where k is a constant.
The curve has a stationary point where x = 3

(a)

Find the value of k.

(3)

(b)

Find the x-coordinate of the other stationary point.

(2)

(c)

Determine the nature of each stationary point.

(2)

(d)

Find the y-coordinate of the maximum point.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 dy/dx = 3x² + 2kx − 9	✓ 1
M1 substitutes x = 3 and sets equal to 0: 27 + 6k − 9 = 0	✓ 1
A1 k = −3	✓ 1
Part (b)
M1 3x² − 6x − 9 = 0, so (x − 3)(x + 1) = 0	✓ 1
A1 x = −1	✓ 1
Part (c)
M1 d²y/dx² = 6x − 6, evaluated at both points, or a correct sign test	✓ 1
A1 x = 3: 12 > 0, minimum; x = −1: −12 < 0, maximum	✓ 1
Part (d)
B1 (−1)³ − 3(−1)² − 9(−1) + 2 = 7	✓ 1

Common error: The condition 'stationary point at x = 3' means dy/dx = 0 at x = 3. Substituting x = 3 into y and setting y = 0 is a different condition and gives the wrong k.

+ Add to test✓ In your test260IGME-060-00-08Sequences and nth terms Paper HHard7 marks

Priya does press-ups every day.
On day 1 she does 20 press-ups. Each day after that, she does 3 more press-ups than on the day before.

(a)

How many press-ups does she do on day 15?

(2)

(b)

Find the first day on which her total number of press-ups since day 1 is more than 1000.

(3)

(c)

Work out the total number of press-ups she does over the first 30 days.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 20 + 14 × 3	✓ 1
A1: 62	✓ 1
Part (b)
M1: \(\frac{n}{2}[2 \times 20 + (n-1) \times 3] > 1000\), or cumulative totals found	✓ 1
M1: total after 20 days = 970, or solves 3n² + 37n − 2000 = 0 to get n = 20.4	✓ 1
A1: day 21, since the total after 21 days is 1050	✓ 1
Part (c)
M1: \(\frac{30}{2}[2 \times 20 + 29 \times 3]\)	✓ 1
A1: 1905	✓ 1

Exam tip: Day n is the nth term, so the number of steps of 3 is n − 1, not n. Use S = ½n[2a + (n − 1)d] for the running total.

+ Add to test✓ In your test261IGME-060-01-11Graphs, gradients and real-life graphs Paper HHard7 marks

The curve C has equation y = x² − 3x
The line L has equation y = 2x − 4
The line L intersects the curve C at the points P and Q.

(a)

Show that the x-coordinates of P and Q satisfy the equation x² − 5x + 4 = 0

(2)

(b)

Find the coordinates of P and Q.

(2)

(c)

Find the exact length of PQ. Give your answer as a surd in its simplest form.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for equating the two equations: x² − 3x = 2x − 4	✓ 1
A1 for rearranging correctly to x² − 5x + 4 = 0	✓ 1
Part (b)
M1 for a correct method to solve x² − 5x + 4 = 0, e.g. (x − 1)(x − 4) = 0 or the quadratic formula, giving x = 1 and x = 4	✓ 1
A1 for (1, −2) and (4, 4)	✓ 1
Part (c)
M1 for differences in coordinates of 3 and 6	✓ 1
M1 for √(3² + 6²) = √45	✓ 1
A1 for 3√5	✓ 1

Common error: Leaving the answer as √45 does not earn the final mark when the simplest surd form is asked for: √45 = √9 × √5 = 3√5.

+ Add to test✓ In your test262IGME-060-00-14Sequences and nth terms Paper HHard7 marks

Here is an arithmetic series.

3 + 7 + 11 + 15 + …

(a)

Show that the sum of the first n terms of the series is 2n² + n.

(2)

(b)

The sum of the first n terms is 1275. Find the value of n.

(3)

(c)

Work out the sum of the 26th term to the 50th term inclusive.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: \(\frac{n}{2}[2 \times 3 + (n-1) \times 4]\)	✓ 1
A1: \(\frac{n}{2}(4n+2) = 2n^2 + n\) with no errors seen	✓ 1
Part (b)
M1: 2n² + n − 1275 = 0	✓ 1
M1: (2n + 51)(n − 25) = 0, or correct use of the quadratic formula	✓ 1
A1: n = 25 only, with n = −25.5 rejected or ignored	✓ 1
Part (c)
M1: S₅₀ − S₂₅ = 5050 − 1275, or \(\frac{25}{2}(103 + 199)\) using 26th term 103 and 50th term 199	✓ 1
A1: 3775	✓ 1

Common error: The terms from the 26th to the 50th give S₅₀ − S₂₅. Subtracting S₂₆ would leave out the 26th term.

+ Add to test✓ In your test263IGME-060-03-12Differentiation and stationary points Paper HHard8 marks

An open box is made from a square sheet of card of side 30 cm. A square of side x cm is cut from each corner, and the sides are folded up to form a box with a square base and no lid.
The volume of the box is V cm³.

Shaded squares are cut away. Diagram not to scale

(a)

Show that V = 4x³ − 120x² + 900x

(2)

(b)

Explain why x must satisfy 0 < x < 15

(1)

(c)

Work out the value of V when x = 4 and when x = 6

(2)

(d)

Ali says, "V is bigger when x = 6 than when x = 4, so the maximum volume must happen for a value of x between 5 and 6."
By working out V when x = 5 and when x = 5.5, show that Ali is wrong.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
States that the base has side (30 − 2x) and the height is x, so V = x(30 − 2x)²	✓ 1
Expands (30 − 2x)² = 900 − 120x + 4x² and multiplies by x to reach the given result	✓ 1
Part (b)
x > 0 because a square must be cut from each corner, and 30 − 2x > 0 (the base side must be positive), so x < 15	✓ 1
Part (c)
x = 4: V = 4 × 22² = 1936	✓ 1
x = 6: V = 6 × 18² = 1944	✓ 1
Part (d)
x = 5: V = 5 × 20² = 2000	✓ 1
x = 5.5: V = 5.5 × 19² = 1985.5	✓ 1
Conclusion: V(5) = 2000 is greater than V(5.5) and V(6), so the volume falls as x goes from 5 to 6. The maximum is at or close to x = 5, not between 5 and 6, so Ali is wrong	✓ 1

Exam tip: Substitute into the factorised form V = x(30 − 2x)². The arithmetic is quicker and less error-prone than using the expanded cubic, and you can use the cubic to check your answer.

+ Add to test✓ In your test264IGME-060-00-13Sequences and nth terms Paper HHard6 marks

The sum of the first n terms of an arithmetic series is given by

S_n = 2n² + 3n

(a)

Write down the first term of the series.

(1)

(b)

Find the 10th term of the series.

(2)

(c)

Show that the nth term of the series is 4n + 1.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 5	✓ 1
Part (b)
M1: S₁₀ = 230 and S₉ = 189 seen, OR common difference 4 found (e.g. S₂ = 14 so 2nd term = 9, d = 9 − 5 = 4) and 5 + 9 × 4 attempted	✓ 1
A1: 41	✓ 1
Part (c)
M1: S_n − S_n−1 = 2n² + 3n − [2(n − 1)² + 3(n − 1)], OR first term 5 and common difference 4 both stated with justification from values of S_n	✓ 1
M1: correct expansion, 2n² + 3n − 2n² + 4n − 2 − 3n + 3, OR nth term written as 5 + 4(n − 1)	✓ 1
A1: = 4n + 1 with no errors seen	✓ 1

Exam tip: The nth term of a series equals Sn − Sn−₁. Substituting n = 10 into Sn gives the sum of the first ten terms, not the 10th term.

+ Add to test✓ In your test265IGME-060-03-13Differentiation and stationary points Paper HHard8 marks

A closed metal can is in the shape of a cylinder of radius r cm and height h cm.
The volume of the can is 250π cm³.

Diagram not to scale

(a)

Show that the total surface area, A cm², of the can is given by \(A = 2\pi r^2 + \frac{500\pi}{r}\)

(3)

(b)

The total surface area of the can is 500 cm².
Show that r satisfies the equation \(r^3 - \frac{250}{\pi}r + 250 = 0\)

(2)

(c)

Use the iteration formula \(r_{n+1} = \sqrt[3]{\frac{250}{\pi}r_n - 250}\) with \(r_0 = 6\) to find the values of \(r_1\), \(r_2\) and \(r_3\). Give each value correct to 3 decimal places.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
πr²h = 250π, so h = 250/r²	✓ 1
A = 2πr² + 2πrh, i.e. two circular ends plus the curved surface	✓ 1
Substitutes for h to obtain A = 2πr² + 500π/r	✓ 1
Part (b)
Sets 2πr² + 500π/r = 500 and multiplies by r to get 2πr³ + 500π = 500r	✓ 1
Divides by 2π and rearranges to r³ − (250/π)r + 250 = 0	✓ 1
Part (c)
r₁ = 6.104	✓ 1
r₂ = 6.178	✓ 1
r₃ = 6.228	✓ 1

Exam tip: Type r₀ into your calculator, then enter the formula using the ANS key. Each press of = gives the next term at full accuracy, so rounding errors do not build up.

+ Add to test✓ In your test266IGME3-3-09d-12Differentiation and stationary points Paper HHard12 marks

A box is made from a square sheet of card of side 24 cm. A square of side x cm is cut from each corner and the sides are folded up to make an open box of volume V cm³, where 0 < x < 12.

Shaded squares are cut away. Diagram not to scale

(a)

Show that V = 4x³ − 96x² + 576x.

(3)

(b)

Find dV/dx.

(2)

(c)

Find the values of x for which dV/dx = 0.

(3)

(d)

Show that x = 4 gives a maximum value of V.

(2)

(e)

Find the maximum volume of the box.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 base is a square of side 24 − 2x	✓ 1
M1 V = x(24 − 2x)²	✓ 1
A1 expand (24 − 2x)² = 576 − 96x + 4x² to reach the given expression	✓ 1
Part (b)
M1 at least two terms differentiated correctly	✓ 1
A1 12x² − 192x + 576	✓ 1
Part (c)
M1 set 12x² − 192x + 576 = 0	✓ 1
M1 divide by 12 and factorise (x − 4)(x − 12) = 0	✓ 1
A1 x = 4 and x = 12	✓ 1
Part (d)
M1 d²V/dx² = 24x − 192	✓ 1
A1 at x = 4 this is −96, which is negative, so a maximum	✓ 1
Part (e)
M1 substitute x = 4 into 4 × (24 − 2x)² or the cubic	✓ 1
A1 1024 cm³	✓ 1

Common error: Both stationary values satisfy dV/dx = 0, but only one gives a maximum. Check the sign of the second derivative and the allowed range of x.

+ Add to test✓ In your test267IGME3-3-09d-13Function notation, composite and inverse functions Paper HHard9 marks

Two functions are defined by f(x) = 2x + 1 and g(x) = x² + 6.

(a)

Find fg(x), giving your answer in its simplest form.

(2)

(b)

Find gf(x), giving your answer in its simplest form.

(2)

(c)

Solve fg(x) = gf(x).

(3)

(d)

Work out f⁻¹(g(2)).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2(x² + 6) + 1	✓ 1
A1 2x² + 13	✓ 1
Part (b)
M1 (2x + 1)² + 6	✓ 1
A1 4x² + 4x + 7	✓ 1
Part (c)
M1 2x² + 13 = 4x² + 4x + 7 (ECF from (a), (b))	✓ 1
M1 rearrange to x² + 2x − 3 = 0 and factorise (x + 3)(x − 1)	✓ 1
A1 x = 1 and x = −3	✓ 1
Part (d)
M1 g(2) = 10 and f⁻¹(x) = (x − 1) ÷ 2	✓ 1
A1 4.5	✓ 1

Exam tip: fg(x) and gf(x) are usually different functions. Write out each substitution line by line to avoid swapping the order.

+ Add to test✓ In your test268IGME3-3-09d-14Sequences and nth terms Paper HHard11 marks

The height, h metres, of a drone above the ground t seconds after take-off is modelled by h = t³ − 9t² + 24t for 0 ≤ t ≤ 5.

(a)

Find dh/dt.

(2)

(b)

Find the times at which the drone is momentarily not rising or falling.

(3)

(c)

Find the height of the drone at each of these times.

(2)

(d)

Show that the drone is descending for all times between t = 2 and t = 4.

(2)

(e)

Find the vertical speed of the drone at t = 1.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 at least two terms correct	✓ 1
A1 3t² − 18t + 24	✓ 1
Part (b)
M1 set 3t² − 18t + 24 = 0	✓ 1
M1 factorise 3(t − 2)(t − 4)	✓ 1
A1 t = 2 and t = 4	✓ 1
Part (c)
M1 substitute t = 2 or t = 4 into h	✓ 1
A1 20 m and 16 m	✓ 1
Part (d)
M1 dh/dt at t = 3 is 27 − 54 + 24	✓ 1
A1 −3, which is negative, and there are no stationary points between 2 and 4, so the drone is descending throughout	✓ 1
Part (e)
M1 substitute t = 1 into dh/dt	✓ 1
A1 9 m/s upwards	✓ 1

Common error: Differentiating gives the rate of change, not the height. Use h for heights and dh/dt for speeds.

+ Add to test✓ In your test269IGME3-3-12r6-08Function notation, composite and inverse functions Paper HHard7 marks

A function is defined by f(x) = (4x − 3)/5, for all real values of x.

(a)

Find f⁻¹(x).

(3)

(b)

Work out f⁻¹(7).

(1)

(c)

Solve f(x) = f⁻¹(x).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 y = (4x − 3)/5, then swap x and y or make x the subject	✓ 1
M1 5y = 4x − 3 rearranged correctly	✓ 1
A1 f⁻¹(x) = (5x + 3)/4	✓ 1
Part (b)
B1 38/4 = 9.5	✓ 1
Part (c)
M1 (4x − 3)/5 = (5x + 3)/4	✓ 1
M1 4(4x − 3) = 5(5x + 3), so 16x − 12 = 25x + 15	✓ 1
A1 x = −3	✓ 1

Exam tip: Check the final answer: f(−3) and f⁻¹(−3) should both equal −3.

+ Add to test✓ In your test270IGME3-3-12r6-10Graphs, gradients and real-life graphs Paper HHard8 marks

After an injection, the concentration of a drug in a patient's blood is modelled by C = 9t² − t³, where C is in mg per litre and t is the time in hours after the injection, for 0 ≤ t ≤ 9.

(a)

Find dC/dt.

(2)

(b)

Work out the rate of change of concentration when t = 2. Include units.

(1)

(c)

Find the time at which the concentration is greatest.

(3)

(d)

Work out the greatest concentration of the drug.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 18t	✓ 1
B1 −3t²	✓ 1
Part (b)
B1 18 × 2 − 3 × 2² = 24 mg per litre per hour	✓ 1
Part (c)
M1 18t − 3t² = 0	✓ 1
M1 3t(6 − t) = 0	✓ 1
A1 t = 6 hours (t = 0 is the start and gives the least concentration)	✓ 1
Part (d)
M1 9 × 6² − 6³ (ECF from (c))	✓ 1
A1 108 mg per litre	✓ 1

Exam tip: After finding the time of a maximum, substitute it into the original function, not the derivative, to get the maximum value.

+ Add to test✓ In your test271IGME3-3-12r11-13Differentiation and stationary points Paper HHard6 marks

The profile of a ramp in a wind farm service building is modelled by \(y = 2x^3 - 5x + 1\), where x and y are in metres.

(a)

Find \(\frac{dy}{dx}\).

(2)

(b)

Work out the gradient of the ramp when \(x = 2\).

(1)

(c)

The tangent to the curve at \(x = 2\) crosses the y-axis. Find the y-coordinate of this point.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 one term correctly differentiated	✓ 1
A1 \(6x^2 - 5\)	✓ 1
Part (b)
A1 19 (ECF from (a))	✓ 1
Part (c)
M1 \(y = 2(8) - 10 + 1 = 7\) when \(x = 2\)	✓ 1
M1 tangent \(y - 7 = 19(x - 2)\) (ECF from (b))	✓ 1
A1 \(y = -31\)	✓ 1

Common error: Use the gradient from the derivative at the given x, and the y value from the original curve, when forming a tangent.

+ Add to test✓ In your test272IGME3-3-12r11-14Function notation, composite and inverse functions Paper HHard7 marks

In a wind farm model, \(f(x) = \frac{x + 3}{x - 2}\) for \(x \ne 2\).

(a)

Work out \(f(7)\).

(1)

(b)

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

(4)

(c)

Work out \(f^{-1}(5)\).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 2	✓ 1
Part (b)
M1 \(y(x - 2) = x + 3\)	✓ 1
M1 \(xy - 2y = x + 3\) and collect x terms, \(x(y - 1) = 2y + 3\)	✓ 1
M1 \(x = \frac{2y + 3}{y - 1}\)	✓ 1
A1 \(f^{-1}(x) = \frac{2x + 3}{x - 1}\)	✓ 1
Part (c)
M1 \(\frac{2 \times 5 + 3}{5 - 1}\)	✓ 1
A1 \(\frac{13}{4}\) (3.25)	✓ 1

Common error: To find an inverse, swap the roles of x and y and rearrange; factorise out x when it appears twice.

+ Add to test✓ In your test273IGME3-3-12r12-11Function notation, composite and inverse functions Paper HHard8 marks

The function f is defined by \(f(x) = \frac{5}{x-2}\) for \(x \neq 2\).

(a)

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

(3)

(b)

Show that solving \(f(x) = f^{-1}(x)\) leads to \(x^2 - 2x - 5 = 0\).

(3)

(c)

Solve \(x^2 - 2x - 5 = 0\) for \(x > 0\). Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
y = 5/(x − 2) so x − 2 = 5/y	✓ 1
f⁻¹(x) = 2 + 5/x	✓ 1
x ≠ 0 stated or implied	✓ 1
Part (b)
5/(x − 2) = 2 + 5/x	✓ 1
5x = (2x + 5)(x − 2)	✓ 1
2x² − 4x − 10 = 0 so x² − 2x − 5 = 0	✓ 1
Part (c)
x = (2 ± √24)/2 = 1 ± √6	✓ 1
x = 3.45	✓ 1

Exam tip: To find where a function equals its inverse, set the two expressions equal and clear the fractions carefully.

+ Add to test✓ In your test274IGME3-3-12r12-14Differentiation and stationary points Paper HHard8 marks

A curve has equation \(y = x^3 + ax^2 + bx\), where \(a\) and \(b\) are constants. The curve has stationary points at \(x = 1\) and \(x = 3\).

(a)

Show that \(3 + 2a + b = 0\) and \(27 + 6a + b = 0\).

(2)

(b)

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

(3)

(c)

Determine the nature of the stationary point at \(x = 1\) and find its y-coordinate.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
dy/dx = 3x² + 2ax + b	✓ 1
Substituting x = 1 and x = 3 and setting equal to zero gives both equations	✓ 1
Part (b)
Subtracting: 24 + 4a = 0	✓ 1
a = −6	✓ 1
b = 9	✓ 1
Part (c)
d²y/dx² = 6x − 12 = −6 which is negative so a maximum	✓ 1
y = 1 − 6 + 9	✓ 1
y = 4	✓ 1

Exam tip: Use each stationary condition to form an equation in the unknown constants, then solve them together.

+ Add to test✓ In your test275IGME3-3-11r14-09Function notation, composite and inverse functions Paper HHard6 marks

A sensor converts a raw reading x into a voltage using the function f(x) = (x + 9) ÷ 4.

(a)

Find f⁻¹(x).

(2)

(b)

Work out f⁻¹(5).

(1)

(c)

Find the value of x for which f(x) = f⁻¹(x).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Let y = (x + 9) ÷ 4 and attempt to make x the subject	✓ 1
f⁻¹(x) = 4x − 9	✓ 1
Part (b)
11 (ECF from (a))	✓ 1
Part (c)
(x + 9) ÷ 4 = 4x − 9	✓ 1
x + 9 = 16x − 36 / 45 = 15x	✓ 1
x = 3	✓ 1

Exam tip: Check an inverse by testing a value: f(11) = 20 ÷ 4 = 5, which confirms f⁻¹(5) = 11.

+ Add to test✓ In your test276IGME3-3-11r14-11Differentiation and stationary points Paper HHard9 marks

A model for the number of new daily cases of an illness, N hundred, on day t is N = t³ − 6t² + 9t + 2 for 0 ≤ t ≤ 4.

(a)

Find dN/dt.

(2)

(b)

Find the values of t at which dN/dt = 0.

(3)

(c)

Determine the nature of each stationary point.

(2)

(d)

Work out the value of N at the local minimum point.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Attempt to differentiate (at least two terms correct)	✓ 1
3t² − 12t + 9	✓ 1
Part (b)
3t² − 12t + 9 = 0	✓ 1
3(t − 1)(t − 3) = 0 or correct use of the formula	✓ 1
t = 1 and t = 3	✓ 1
Part (c)
d²N/dt² = 6t − 12 evaluated (or gradients either side tested)	✓ 1
t = 1 is a maximum (−6 < 0) and t = 3 is a minimum (+6 > 0)	✓ 1
Part (d)
Substitute t = 3: 27 − 54 + 27 + 2	✓ 1
N = 2	✓ 1

Common error: Forgetting to substitute the stationary t-values back into N; the derivative gives where the turning points are, not their heights.

+ Add to test✓ In your test277IGME3-3-11r14-13Sequences and nth terms Paper HHard7 marks

The first four terms of a sequence of LED panel counts are 5, 12, 23, 38.

(a)

Work out the next term.

(1)

(b)

Find an expression, in terms of n, for the nth term.

(3)

(c)

One term of the sequence is 212. Work out its position in the sequence.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
57	✓ 1
Part (b)
Second difference is 4 so the n² term is 2n²	✓ 1
Subtract 2n² (2, 8, 18, 32) to leave 3, 4, 5, 6	✓ 1
2n² + n + 2	✓ 1
Part (c)
2n² + n + 2 = 212 (ECF from (b))	✓ 1
(2n + 21)(n − 10) = 0 or correct use of the formula	✓ 1
n = 10	✓ 1

Common error: In a quadratic sequence, halve the second difference to get the coefficient of n²; using the second difference itself gives 4n².

+ Add to test✓ In your test278IGME3-3-11r15-07Differentiation and stationary points Paper HHard9 marks

A food scientist designs an open-topped box (no lid) for freezing sauce. The base is a square of side x cm and the height is h cm. The volume of the box is 500 cm³. The total outer surface area, S cm², is the base plus the four sides.

Diagram not to scale

(a)

Show that S = x² + 2000/x.

(2)

(b)

Find dS/dx.

(2)

(c)

Find the value of x that gives a stationary value of S.

(2)

(d)

Show that this value of x gives a minimum surface area, and find the minimum surface area.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 x²h = 500 so h = 500/x²	✓ 1
A1 S = x² + 4x × 500/x² = x² + 2000/x	✓ 1
Part (b)
M1 differentiates one term correctly, e.g. 2000x⁻¹ becomes −2000x⁻²	✓ 1
A1 2x − 2000/x²	✓ 1
Part (c)
M1 2x − 2000/x² = 0 so 2x³ = 2000 (ECF from (b))	✓ 1
A1 x = 10	✓ 1
Part (d)
M1 d²S/dx² = 2 + 4000/x³ evaluated at x = 10	✓ 1
A1 = 6 > 0 so a minimum	✓ 1
A1 S = 100 + 200 = 300 cm²	✓ 1

Common error: Always write the formula for the quantity in one variable before differentiating. Do not forget to state which test shows a minimum.

+ Add to test✓ In your test279IGME3-3-11r15-13Function notation, composite and inverse functions Paper HHard7 marks

The function f is defined by f(x) = \(\frac{4}{x-1}\) for x ≠ 1.

(a)

Find f(5).

(1)

(b)

Find f⁻¹(x).

(3)

(c)

State the value of x that must be excluded from the domain of f⁻¹.

(1)

(d)

Work out ff(3).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 1	✓ 1
Part (b)
M1 y(x − 1) = 4	✓ 1
M1 x = 4/y + 1	✓ 1
A1 f⁻¹(x) = 4/x + 1 (or (x + 4)/x)	✓ 1
Part (c)
B1 x = 0	✓ 1
Part (d)
M1 f(3) = 2	✓ 1
A1 f(2) = 4	✓ 1

Common error: To find an inverse, swap the roles of x and y and then make y the subject. Check your inverse by testing a number: f(5) = 1 so f⁻¹(1) should be 5.

+ Add to test✓ In your test280IGME3-3-11r15-14Function notation, composite and inverse functions Paper HHard8 marks

The functions f and g are defined by f(x) = 2x + 3 and g(x) = x² − 4.

(a)

Find fg(x).

(2)

(b)

Find gf(x).

(2)

(c)

Solve fg(x) = gf(x).

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2(x² − 4) + 3	✓ 1
A1 2x² − 5	✓ 1
Part (b)
M1 (2x + 3)² − 4	✓ 1
A1 4x² + 12x + 5	✓ 1
Part (c)
M1 2x² − 5 = 4x² + 12x + 5 (ECF from (a) and (b))	✓ 1
A1 x² + 6x + 5 = 0 (or equivalent)	✓ 1
M1 attempt to factorise or use the quadratic formula	✓ 1
A1 x = −1 and x = −5	✓ 1

Exam tip: Check a solution of a composite equation by substituting it into both sides. Here x = −1 gives −3 in each case.

+ Add to test✓ In your test281IGME3-3-12r15-06Differentiation and stationary points Paper HHard8 marks

The bed of a swimming lake is modelled by y = x⁴ − 8x² for −3 ≤ x ≤ 3, where x is the horizontal distance in tens of metres from the centre line and y is the height of the bed in metres relative to the water surface level of the shore.

(a)

Find dy/dx.

(2)

(b)

Find the x-coordinates of all the stationary points of the bed.

(3)

(c)

Show that the stationary point at x = 2 is a minimum, and find the minimum value of y.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 reduces the power of at least one term	✓ 1
A1 4x³ − 16x	✓ 1
Part (b)
M1 4x³ − 16x = 0 (ECF from (a))	✓ 1
M1 4x(x² − 4) = 0 or equivalent factorising	✓ 1
A1 x = 0, x = 2 and x = −2	✓ 1
Part (c)
M1 d²y/dx² = 12x² − 16, which is 32 at x = 2	✓ 1
A1 32 > 0, so a minimum	✓ 1
A1 y = 16 − 32 = −16 m	✓ 1

Common error: A zero of the derivative can have more than one solution. Solving by dividing both sides by x loses the solution x = 0, so factorise instead.

+ Add to test✓ In your test282IGME3-3-12r15-12Function notation, composite and inverse functions Paper HHard7 marks

The function f is defined by f(x) = \(\frac{x-2}{5}\).

(a)

Work out f(17).

(1)

(b)

Find f⁻¹(x).

(2)

(c)

Show that ff(x) = \(\frac{x-12}{25}\).

(2)

(d)

Solve ff(x) = 2.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 3	✓ 1
Part (b)
M1 x = 5y + 2 from y = (x − 2) ÷ 5	✓ 1
A1 f⁻¹(x) = 5x + 2	✓ 1
Part (c)
M1 ((x − 2)/5 − 2) ÷ 5	✓ 1
A1 (x − 2 − 10) ÷ 25 = (x − 12)/25	✓ 1
Part (d)
M1 (x − 12) ÷ 25 = 2 so x − 12 = 50	✓ 1
A1 x = 62	✓ 1

Exam tip: Check by applying f twice to your answer: f(62) = 12 and f(12) = 2.

+ Add to test✓ In your test283IGME3-3-11r16-07Graphs, gradients and real-life graphs Paper HHard6 marks

On a map grid with units of km, Road A has equation \(y = 2x + 1\). Road B is perpendicular to Road A and passes through the point (8, 3). A cyclist wants to know where the two roads meet.

(a)

Find the equation of Road B.

(3)

(b)

Find the coordinates of the point where the roads meet.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 gradient of Road B is −½	✓ 1
M1 y − 3 = −½(x − 8) or 3 = −½ × 8 + c	✓ 1
A1 y = −½x + 7 oe	✓ 1
Part (b)
M1 2x + 1 = −½x + 7 (ECF from (a))	✓ 1
A1 x = 2.4	✓ 1
A1 y = 5.8, so (2.4, 5.8)	✓ 1

Common error: The gradient of a perpendicular line is the negative reciprocal. Changing only the sign, or only inverting, will not give it.

+ Add to test✓ In your test284IGME3-3-11r16-11Function notation, composite and inverse functions Paper HHard6 marks

The function \(f\) is defined by \(f(x) = \frac{3x + 1}{x - 2}\) for \(x \neq 2\).

(a)

Work out \(f(4)\).

(1)

(b)

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

(4)

(c)

Work out \(f^{-1}(5)\).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 6.5 oe	✓ 1
Part (b)
M1 y = (3x + 1)/(x − 2) and attempt to make x the subject, or swap x and y	✓ 1
M1 multiply out: y(x − 2) = 3x + 1 and expand	✓ 1
M1 collect x terms on one side and factorise	✓ 1
A1 \(f^{-1}(x) = \frac{2x + 1}{x - 3}\) oe	✓ 1
Part (c)
B1 5.5 oe (ECF from (b))	✓ 1

Exam tip: Check an inverse by testing one value: if \(f(4) = 6.5\) then \(f^{-1}(6.5)\) must come out as 4.

+ Add to test✓ In your test285IGME3-3-11r16-12Differentiation and stationary points Paper HHard9 marks

The height \(h\) metres of a cycle-track surface is modelled by \(h = 2x^3 - 9x^2 + 12x + 1\) for \(0 \leq x \leq 3\), where \(x\) is the distance in hundreds of metres along the track.

(a)

Find \(\frac{dh}{dx}\).

(2)

(b)

Find the values of \(x\) at which the track surface has a stationary point.

(3)

(c)

Determine the nature of each stationary point.

(2)

(d)

Work out the difference in height between the local maximum and the local minimum.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 at least one term differentiated correctly	✓ 1
A1 6x² − 18x + 12	✓ 1
Part (b)
M1 6x² − 18x + 12 = 0 (ECF from (a))	✓ 1
M1 correct method to solve, e.g. 6(x − 1)(x − 2) = 0	✓ 1
A1 x = 1 and x = 2	✓ 1
Part (c)
M1 second derivative 12x − 18 evaluated at both values, or gradient test either side	✓ 1
A1 maximum at x = 1 and minimum at x = 2	✓ 1
Part (d)
M1 substituting both x values to get 6 and 5	✓ 1
A1 1 m	✓ 1

Exam tip: After finding the stationary points, always substitute back into the original function for the height. Substituting into the derivative gives zero by definition.

+ Add to test✓ In your test286IGME3-3-11r19-11Graphs, gradients and real-life graphs Paper HHard9 marks

Archaeologists will fence a rectangular trench of area 200 m². A cliff forms one side, so fencing is needed on only three sides: two sides of length x metres and one side of length y metres parallel to the cliff. The total length of fencing is L metres.

Red lines show the fencing. Diagram not to scale

(a)

Show that L = 2x + 200/x.

(2)

(b)

Find dL/dx.

(2)

(c)

Find the value of x that gives a stationary value of L.

(2)

(d)

Show that this value of x gives the minimum length of fencing, and state that minimum length.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 xy = 200 so y = 200/x	✓ 1
A1 L = 2x + y = 2x + 200/x	✓ 1
Part (b)
M1 attempt to differentiate 200x⁻¹	✓ 1
A1 2 − 200/x²	✓ 1
Part (c)
M1 sets 2 − 200/x² = 0	✓ 1
A1 x = 10 (the value −10 is rejected as a length)	✓ 1
Part (d)
M1 d²L/dx² = 400/x³	✓ 1
A1 positive (0.4) when x = 10, so a minimum	✓ 1
A1 minimum length 40 m	✓ 1

Exam tip: In optimisation questions, always justify that your stationary point is a minimum or maximum and then answer the question that was asked (here, the length).

+ Add to test✓ In your test287IGME3-3-11r19-13Function notation, composite and inverse functions Paper HHard6 marks

The function f is defined as f(x) = (x + 4)/(x − 2) for x ≠ 2.

(a)

Find f⁻¹(x).

(3)

(b)

State the value of x for which f⁻¹(x) is not defined.

(1)

(c)

Solve f(x) = 5.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 y(x − 2) = x + 4	✓ 1
M1 collects terms in x: x(y − 1) = 2y + 4	✓ 1
A1 f⁻¹(x) = (2x + 4)/(x − 1)	✓ 1
Part (b)
A1 x = 1	✓ 1
Part (c)
M1 x + 4 = 5(x − 2)	✓ 1
A1 x = 3.5	✓ 1

Common error: When rearranging for the inverse, gather every term containing x on one side before factorising.

+ Add to test✓ In your test288IGME3-3-11r19-14Sequences and nth terms Paper HHard7 marks

The numbers of servers in the first five cabinet rows of a data centre are 5, 12, 23, 38, 57. The sequence is quadratic.

(a)

Find an expression, in terms of n, for the nth term.

(3)

(b)

Work out the 20th term.

(1)

(c)

Work out the first term of the sequence that is greater than 1000.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 first differences 7, 11, 15, 19 and a constant second difference of 4, so 2n²	✓ 1
M1 subtracts 2n² (2, 8, 18, 32, 50) leaving 3, 4, 5, 6, 7	✓ 1
A1 2n² + n + 2	✓ 1
Part (b)
A1 822 (ECF from (a))	✓ 1
Part (c)
M1 attempt to solve or test 2n² + n + 2 > 1000	✓ 1
M1 n = 22 gives 992 and n = 23 gives 1083	✓ 1
A1 1083 (the 23rd term)	✓ 1

Exam tip: For a quadratic sequence, half the second difference is the coefficient of n².

+ Add to test✓ In your test289IGME3-3-12r22-12Sequences and nth terms Paper HHard8 marks

The numbers of steel rivets used in the first four sections of a bridge are 5, 14, 27 and 44. The number of rivets in the nth section follows a quadratic sequence.

(a)

Show that the second differences of the sequence are constant.

(2)

(b)

Find an expression, in terms of n, for the number of rivets in the nth section.

(3)

(c)

One section uses 152 rivets. Work out which section this is.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for first differences 9, 13, 17	✓ 1
A1 for second differences 4 and 4, which are constant	✓ 1
Part (b)
M1 for 2n² (half the second difference)	✓ 1
M1 for subtracting 2n² from each term: 3, 6, 9, 12	✓ 1
A1 for 2n² + 3n	✓ 1
Part (c)
M1 for 2n² + 3n = 152, i.e. 2n² + 3n − 152 = 0	✓ 1
M1 for factorising (2n + 19)(n − 8) = 0, or using the quadratic formula	✓ 1
A1 for n = 8 (the root −9.5 is rejected)	✓ 1

Exam tip: For a quadratic sequence the coefficient of n² is half of the constant second difference. Subtract that part, then fit the remaining linear sequence.

+ Add to test✓ In your test290IGME3-3-12r22-13Function notation, composite and inverse functions Paper HHard8 marks

A calibration function for a bridge sensor is \(f(x) = \frac{3x + 1}{x - 2}\), where \(x \neq 2\). Another function is \(g(x) = 2x - 1\).

(a)

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

(4)

(b)

Find \(fg(x)\). Give your answer as a single fraction in its simplest form.

(2)

(c)

Solve \(fg(x) = 4\).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for y(x − 2) = 3x + 1	✓ 1
M1 for collecting x terms: xy − 3x = 1 + 2y	✓ 1
M1 for factorising: x(y − 3) = 2y + 1	✓ 1
A1 for f⁻¹(x) = (2x + 1)/(x − 3)	✓ 1
Part (b)
M1 for (3(2x − 1) + 1)/((2x − 1) − 2)	✓ 1
A1 for (6x − 2)/(2x − 3)	✓ 1
Part (c)
M1 for 6x − 2 = 4(2x − 3) (ECF from (b))	✓ 1
A1 for x = 5	✓ 1

Common error: When finding an inverse, swap the roles of x and y carefully, and collect every term containing the subject on one side before factorising.

+ Add to test✓ In your test291IGME3-3-12r22-14Differentiation and stationary points Paper HHard10 marks

A roaster makes an open cooling tray from a rectangular metal sheet 30 cm by 20 cm. A square of side x cm is cut from each corner and the sides are folded up to make a tray of height x cm with no lid.

Shaded squares are cut away. Diagram not to scale

(a)

Show that the volume of the tray is \(V = 4x^3 - 100x^2 + 600x\) cm³.

(2)

(b)

Find \(\frac{dV}{dx}\).

(2)

(c)

Find the value of x, to 3 significant figures, that gives the maximum volume. Show that your value gives a maximum.

(4)

(d)

Find the maximum volume of the tray, to the nearest cm³. Use the unrounded value of x.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for length 30 − 2x and width 20 − 2x	✓ 1
A1 for expanding x(30 − 2x)(20 − 2x) to 4x³ − 100x² + 600x	✓ 1
Part (b)
M1 for differentiating at least two terms correctly	✓ 1
A1 for 12x² − 200x + 600	✓ 1
Part (c)
M1 for setting 12x² − 200x + 600 = 0, i.e. 3x² − 50x + 150 = 0	✓ 1
M1 for using the quadratic formula	✓ 1
A1 for x = 3.92 (reject x = 12.7 since x must be less than 10)	✓ 1
B1 for d²V/dx² = 24x − 200 < 0 at x = 3.92, so a maximum	✓ 1
Part (d)
M1 for substituting x = (50 − √700)/6 into V	✓ 1
A1 for 1056 cm³	✓ 1

Exam tip: In an optimisation question, state the range of x that makes physical sense and use it to reject roots that fall outside it.

+ Add to test✓ In your test292IGME3-3-12r24-12Differentiation and stationary points Paper HHard10 marks

A hospital stores sterile instruments in an open-topped box with a square base of side x cm and height h cm. The volume of the box is 4000 cm³. The outside surface area of the box, S cm², is the base plus the four sides.

Diagram not to scale

(a)

Show that S = x² + 16 000 ÷ x.

(3)

(b)

Find dS/dx.

(2)

(c)

Find the value of x for which S is a minimum.

(2)

(d)

Find the minimum value of S and show that it is a minimum.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 h = 4000 ÷ x²	✓ 1
M1 S = x² + 4xh	✓ 1
A1 S = x² + 4x(4000 ÷ x²) = x² + 16 000 ÷ x	✓ 1
Part (b)
M1 one term differentiated correctly, e.g. 16 000x⁻¹ to −16 000x⁻²	✓ 1
A1 2x − 16 000 ÷ x²	✓ 1
Part (c)
M1 2x − 16 000 ÷ x² = 0, so x³ = 8000	✓ 1
A1 x = 20	✓ 1
Part (d)
M1 substituting x = 20 into S	✓ 1
A1 400 + 800 = 1200 cm²	✓ 1
R1 d²S/dx² = 2 + 32 000 ÷ x³ = 6 > 0, so a minimum	✓ 1

Exam tip: In an optimisation question, first reduce the quantity to a single variable using the fixed constraint, then differentiate.

+ Add to test✓ In your test293IGME3-3-12r24-13Function notation, composite and inverse functions Paper HHard9 marks

The function f is defined by f(x) = (x + 3) ÷ (x − 1), for x ≠ 1.

(a)

Work out f(5).

(1)

(b)

Find f⁻¹(x).

(3)

(c)

Use your answer to part (b) to write down ff(x), and hence the value of ff(10).

(2)

(d)

Find the positive value of x for which f(x) = x.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 2	✓ 1
Part (b)
M1 y(x − 1) = x + 3	✓ 1
M1 xy − x = y + 3, so x(y − 1) = y + 3	✓ 1
A1 f⁻¹(x) = (x + 3) ÷ (x − 1)	✓ 1
Part (c)
A1 ff(x) = x because f is its own inverse	✓ 1
A1 ff(10) = 10	✓ 1
Part (d)
M1 x + 3 = x(x − 1)	✓ 1
M1 x² − 2x − 3 = 0 and (x − 3)(x + 1) = 0	✓ 1
A1 x = 3	✓ 1

Common error: When rearranging for an inverse, collect all the x terms on one side before factorising out x.

+ Add to test✓ In your test294IGME3-3-13r24-10Function notation, composite and inverse functions Paper HHard8 marks

The functions f and g are defined by f(x) = 2x + 3 and g(x) = x² − 7.

(a)

Show that fg(x) = 2x² − 11.

(2)

(b)

Show that gf(x) = 4x² + 12x + 2.

(2)

(c)

Solve fg(x) = gf(x). Find the greater solution, correct to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 f(x² − 7) = 2(x² − 7) + 3	✓ 1
A1 2x² − 14 + 3 = 2x² − 11	✓ 1
Part (b)
M1 g(2x + 3) = (2x + 3)² − 7	✓ 1
A1 4x² + 12x + 9 − 7 = 4x² + 12x + 2	✓ 1
Part (c)
M1 2x² − 11 = 4x² + 12x + 2	✓ 1
M1 rearrange to 2x² + 12x + 13 = 0	✓ 1
M1 use of the quadratic formula, e.g. x = (−12 ± √40) ÷ 4	✓ 1
A1 −1.42 (the other solution is −4.58)	✓ 1

Exam tip: When asked to solve for the greater solution, find both roots from the formula and compare them before choosing.

+ Add to test✓ In your test295IGME3-3-13r24-13Differentiation and stationary points Paper HHard11 marks

A library makes open-top storage boxes for returned books. Each box has a square base of side x cm and height h cm, and a volume of 500 cm³. The total outside surface area is A cm².

Diagram not to scale

(a)

Show that A = x² + 2000 ÷ x.

(3)

(b)

Find dA/dx.

(2)

(c)

Find the value of x for which A is a minimum.

(3)

(d)

Work out the minimum surface area. Show that your value is a minimum.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 x²h = 500 so h = 500 ÷ x²	✓ 1
M1 A = x² + 4xh (base and four sides)	✓ 1
A1 4x × 500 ÷ x² = 2000 ÷ x, giving the result	✓ 1
Part (b)
B1 2x	✓ 1
B1 −2000x⁻² or −2000 ÷ x²	✓ 1
Part (c)
M1 2x − 2000 ÷ x² = 0 (ECF from (b))	✓ 1
M1 x³ = 1000	✓ 1
A1 x = 10	✓ 1
Part (d)
M1 substitute x = 10 into A	✓ 1
A1 100 + 200 = 300	✓ 1
B1 d²A/dx² = 2 + 4000 ÷ x³ = 6, which is positive, so a minimum	✓ 1

Exam tip: In an optimisation problem, eliminate one variable first so that the quantity to be optimised depends on a single letter.

+ Add to test✓ In your test296IGME3-3-12r26-10Differentiation and stationary points Paper HHard8 marks

A battery's stored energy, E MWh, is modelled by E = t³ − 9t² + 24t + 10 for 0 ≤ t ≤ 6, where t is the time in hours.

(a)

Find dE/dt.

(2)

(b)

Find the values of t at which E has a stationary point.

(3)

(c)

Determine the nature of the stationary point at t = 4 and find the value of E there.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 at least two terms differentiated correctly	✓ 1
A1 3t² − 18t + 24	✓ 1
Part (b)
M1 3t² − 18t + 24 = 0 (ECF from (a))	✓ 1
M1 3(t − 2)(t − 4) = 0 or correct use of the formula	✓ 1
A1 t = 2 and t = 4	✓ 1
Part (c)
M1 d²E/dt² = 6t − 18, which is 6 at t = 4	✓ 1
A1 positive so a minimum	✓ 1
A1 E = 64 − 144 + 96 + 10 = 26 MWh	✓ 1

Common error: A stationary point is where the gradient is zero; check its nature with the second derivative (or gradients on either side) before calling it a maximum or minimum.

+ Add to test✓ In your test297IGME3-3-12r26-12Differentiation and stationary points Paper HHard10 marks

A retailer makes an open-top box from a square sheet of card of side 30 cm by cutting a square of side x cm from each corner and folding up the sides. The volume of the box is V cm³, where 0 < x < 15.

Shaded squares are cut away. Diagram not to scale

(a)

Show that V = 4x³ − 120x² + 900x.

(2)

(b)

Find dV/dx.

(2)

(c)

Find the value of x that gives a stationary value of V inside the given range.

(3)

(d)

Show that this value of x gives a maximum and find the maximum volume.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 base side 30 − 2x and height x, so V = x(30 − 2x)²	✓ 1
A1 expansion (30 − 2x)² = 900 − 120x + 4x² leading to the given result	✓ 1
Part (b)
M1 at least two terms differentiated correctly	✓ 1
A1 12x² − 240x + 900	✓ 1
Part (c)
M1 12x² − 240x + 900 = 0	✓ 1
A1 x = 5 and x = 15 (from 12(x − 5)(x − 15))	✓ 1
A1 x = 5, since 15 is outside the range	✓ 1
Part (d)
M1 d²V/dx² = 24x − 240	✓ 1
A1 −120 at x = 5, which is negative, so a maximum	✓ 1
A1 V = 2000 cm³	✓ 1

Exam tip: For a practical problem, always check that each stationary point lies within the allowed range of the variable before using it.

+ Add to test✓ In your test298IGME3-3-14r28-09Sequences and nth terms Paper HHard8 marks

The concentration of a drug in a patient's blood is C mg per litre, t hours after an injection, where C = 0.6t² − 0.05t³ for 0 ≤ t ≤ 12.

(a)

Find ^dC⁄_dt.

(2)

(b)

Find the value of t at which the concentration is greatest (other than t = 0).

(2)

(c)

Show that this value of t gives a maximum.

(2)

(d)

Work out the greatest concentration.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiates at least one term correctly	✓ 1
A1 1.2t − 0.15t²	✓ 1
Part (b)
M1 1.2t − 0.15t² = 0 or 0.15t(8 − t) = 0	✓ 1
A1 t = 8	✓ 1
Part (c)
M1 second derivative 1.2 − 0.3t evaluated at t = 8	✓ 1
A1 −1.2 < 0, so a maximum	✓ 1
Part (d)
M1 0.6 × 8² − 0.05 × 8³	✓ 1
A1 12.8 mg per litre	✓ 1

Common error: Setting the derivative to zero finds stationary points only; you must show which one is a maximum.

+ Add to test✓ In your test299IGME3-3-14r28-13Function notation, composite and inverse functions Paper HHard6 marks

A Formula 1 team models tyre grip with the function f(x) = (2x + 1)/(x − 3), where x ≠ 3.

(a)

Find f(5).

(1)

(b)

Find f⁻¹(x).

(3)

(c)

Find f⁻¹(7).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 ¹¹⁄₂ or 5.5	✓ 1
Part (b)
M1 y(x − 3) = 2x + 1	✓ 1
M1 collects x terms: x(y − 2) = 3y + 1	✓ 1
A1 f⁻¹(x) = (3x + 1)/(x − 2)	✓ 1
Part (c)
M1 (3 × 7 + 1)/(7 − 2) (ECF from (b))	✓ 1
A1 4.4	✓ 1

Exam tip: To find an inverse, swap to y, clear the fraction and collect all x terms on one side before factorising.

+ Add to test✓ In your test300IGME3-3-14r28-14Differentiation and stationary points Paper HHard9 marks

A hospital sterile container is a closed cuboid with a square base of side x cm and a volume of 2000 cm³. Its total surface area is S cm².

Diagram not to scale

(a)

Show that S = 2x² + ⁸⁰⁰⁰⁄_x.

(3)

(b)

Find ^dS⁄_dx.

(2)

(c)

Find the value of x that gives the minimum surface area. Give your answer to 3 significant figures.

(2)

(d)

Work out the minimum surface area. Give your answer to 3 significant figures. Use the unrounded value of x.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 height = 2000 ÷ x²	✓ 1
M1 S = 2x² + 4xh	✓ 1
A1 4x × 2000/x² = 8000/x, so the result is shown	✓ 1
Part (b)
M1 differentiates, writing 8000x⁻¹	✓ 1
A1 4x − 8000x⁻²	✓ 1
Part (c)
M1 4x − 8000x⁻² = 0 so x³ = 2000	✓ 1
A1 12.6 cm	✓ 1
Part (d)
M1 substitutes x = 2000^1/3 = 12.599… into S	✓ 1
A1 952 cm²	✓ 1

Common error: Rounding x to 12.6 before the final calculation gives 952.56, which rounds to 953; keep the unrounded value in your calculator.

+ Add to test✓ In your test301IGME3-3-14r32-12Function notation, composite and inverse functions Paper HHard9 marks

The functions f and g are defined as f(x) = (2x + 3)/(x − 1), for x ≠ 1, and g(x) = 3x − 1. They model two stages in the signal processing of a space probe.

(a)

Work out f(5).

(1)

(b)

Find f⁻¹(x).

(4)

(c)

Solve fg(x) = 4.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 ¹³⁄₄ (or 3.25)	✓ 1
Part (b)
M1 y(x − 1) = 2x + 3 (starting from y = f(x), then clearing the fraction)	✓ 1
M1 collect x terms: xy − 2x = y + 3	✓ 1
M1 factorise: x(y − 2) = y + 3	✓ 1
A1 f⁻¹(x) = (x + 3)/(x − 2)	✓ 1
Part (c)
M1 f(3x − 1) = (2(3x − 1) + 3)/((3x − 1) − 1)	✓ 1
A1 (6x + 1)/(3x − 2)	✓ 1
M1 6x + 1 = 4(3x − 2)	✓ 1
A1 x = 1.5	✓ 1

Exam tip: When finding an inverse, swap the roles of x and y by writing y = f(x), then make x the subject; collect the x terms on one side before factorising.

+ Add to test✓ In your test302IGME3-3-14r32-13Differentiation and stationary points Paper HHard11 marks

A coffee company designs an open-topped box with a square base of side x cm and height h cm. The box is made from 300 cm² of card with no waste, so the base and the four vertical sides use all of the card. The volume of the box is V cm³, and x > 0.

Diagram not to scale

(a)

Show that V = 75x − ¹⁄₄x³.

(3)

(b)

Find the value of x for which V is stationary.

(4)

(c)

Work out the maximum volume, and show that this value is a maximum.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 x² + 4xh = 300	✓ 1
M1 h = (300 − x²)/(4x) and V = x²h	✓ 1
A1 V = x(300 − x²)/4 = 75x − ¹⁄₄x³ (shown with no errors)	✓ 1
Part (b)
M1 differentiate: at least one term correct	✓ 1
A1 ^dV⁄_dx = 75 − ³⁄₄x²	✓ 1
M1 75 − ³⁄₄x² = 0, giving x² = 100	✓ 1
A1 x = 10 (reject −10)	✓ 1
Part (c)
M1 75 × 10 − ¹⁄₄ × 10³	✓ 1
A1 500 cm³	✓ 1
M1 ^d²V⁄_dx² = −³⁄₂x	✓ 1
A1 −15 at x = 10, which is negative, so a maximum	✓ 1

Exam tip: In an optimisation question, first write the quantity to maximise in terms of one variable, then differentiate, set to zero and justify the type of stationary point.

+ Add to test✓ In your test303IGME3-3-14r32-14Sequences and nth terms Paper HHard9 marks

A bridge truss pattern uses 4, 11, 22 and 37 beams in patterns 1, 2, 3 and 4. The numbers of beams form a quadratic sequence.

(a)

Find an expression for the number of beams in pattern n.

(4)

(b)

Pattern n uses 466 beams. Find n.

(3)

(c)

Show that no pattern uses exactly 500 beams.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 first differences 7, 11, 15 and second differences 4	✓ 1
M1 coefficient of n² is 2; subtract 2n² from the terms to leave 2, 3, 4, 5	✓ 1
M1 remaining linear sequence n + 1	✓ 1
A1 2n² + n + 1	✓ 1
Part (b)
M1 2n² + n + 1 = 466 (ECF from (a)), i.e. 2n² + n − 465 = 0	✓ 1
M1 (2n + 31)(n − 15) = 0 or correct use of the quadratic formula	✓ 1
A1 n = 15 (reject −15.5)	✓ 1
Part (c)
M1 pattern 15 uses 466 beams and pattern 16 uses 2 × 256 + 16 + 1 = 529 beams	✓ 1
A1 466 < 500 < 529 and the sequence is increasing, so 500 is not a term	✓ 1

Common error: For a quadratic sequence, the second difference is 2a, so halve it to get the coefficient of n².

+ Add to test✓ In your test304IGME3-3-13r33-07Sequences and nth terms Paper HHard8 marks

In a data centre, the number of cable connections, C, needed when n racks are linked in a network follows a quadratic pattern. For n = 1, 2, 3, 4, 5 the values of C are
5, 12, 23, 38, 57

(a)

Show that the second differences of the sequence are constant.

(2)

(b)

Find an expression, in terms of n, for C.

(3)

(c)

Show that 1277 is a value in the sequence and find the number of racks n that gives it.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 first differences 7, 11, 15, 19	✓ 1
A1 second differences 4, 4, 4 (constant)	✓ 1
Part (b)
M1 2n² (half of the second difference as the coefficient)	✓ 1
M1 subtract 2n² from the terms to get 3, 4, 5, 6, 7 (or equivalent)	✓ 1
A1 2n² + n + 2	✓ 1
Part (c)
M1 2n² + n + 2 = 1277 leading to 2n² + n − 1275 = 0	✓ 1
M1 use of quadratic formula or factorising: (2n + 51)(n − 25) = 0	✓ 1
A1 n = 25 (the negative root −25.5 is rejected as n must be positive)	✓ 1

Exam tip: For a quadratic nth term the coefficient of n² is half the constant second difference.

+ Add to test✓ In your test305IGME3-3-13r33-11Function notation, composite and inverse functions Paper HHard8 marks

The function f is defined by f(x) = (3x + 1)/(x − 2) for x ≠ 2.

(a)

Find the value of f(4).

(1)

(b)

Find f⁻¹(x).

(3)

(c)

Solve f(x) = f⁻¹(x). Give the larger solution to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 6.5 (or 13/2)	✓ 1
Part (b)
M1 y(x − 2) = 3x + 1 (rearranging with x and y swapped or equivalent)	✓ 1
M1 collect x terms: x(y − 3) = 2y + 1	✓ 1
A1 f⁻¹(x) = (2x + 1)/(x − 3)	✓ 1
Part (c)
M1 (3x + 1)(x − 3) = (2x + 1)(x − 2) (ECF from (b))	✓ 1
M1 3x² − 8x − 3 = 2x² − 3x − 2	✓ 1
M1 x² − 5x − 1 = 0 and use of the quadratic formula	✓ 1
A1 x = (5 + √29)/2 = 5.19	✓ 1

Common error: Cross-multiply only after both functions are written as single fractions, and expand each bracket carefully.

+ Add to test✓ In your test306IGME3-3-13r33-14Differentiation and stationary points Paper HHard8 marks

An archaeological team uses 60 m of fencing to enclose a rectangular area for a dig. One side of the area is a long straight wall, so fencing is needed on the other three sides only. The sides perpendicular to the wall each have length x metres.

Red lines show the fencing. Diagram not to scale

(a)

Show that the area, A m², enclosed is given by A = 60x − 2x².

(2)

(b)

Find the value of x for which A is stationary.

(3)

(c)

Find the maximum area and show that this stationary value is a maximum.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 side parallel to the wall is 60 − 2x	✓ 1
A1 A = x(60 − 2x) = 60x − 2x²	✓ 1
Part (b)
M1 dA/dx = 60 − 4x	✓ 1
M1 60 − 4x = 0	✓ 1
A1 x = 15	✓ 1
Part (c)
M1 A = 60 × 15 − 2 × 15²	✓ 1
A1 450 m²	✓ 1
R1 d²A/dx² = −4 < 0 so it is a maximum	✓ 1

Exam tip: In optimisation problems, write the quantity as a function of one variable before differentiating.

+ Add to test✓ In your test307IGME3-3-12r35-12Function notation, composite and inverse functions Paper HHard7 marks

The function f is defined by f(x) = 5/(x − 2), for x ≠ 2.

(a)

Find f⁻¹(x).

(3)

(b)

There are two values of x for which f(x) = x. Find the positive value, correct to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 x = 5/(y − 2) after swapping, or y = 5/(x − 2) rearranged for x	✓ 1
M1 y − 2 = 5/x	✓ 1
A1 f⁻¹(x) = 5/x + 2 (or equivalent)	✓ 1
Part (b)
M1 5/(x − 2) = x	✓ 1
M1 rearranges to x² − 2x − 5 = 0	✓ 1
M1 correct use of the quadratic formula, e.g. (2 ± √24) ÷ 2	✓ 1
A1 3.45	✓ 1

Common error: When the unknown is on the bottom of a fraction, multiply both sides by the denominator before collecting terms.

+ Add to test✓ In your test308IGME3-3-12r35-13Differentiation and stationary points Paper HHard10 marks

A sensor housing for a space probe is a closed box with a square base of side x cm and height h cm. Its volume is 500 cm³. The surface area of the box is S cm².

Diagram not to scale

(a)

Show that S = 2x² + 2000/x.

(3)

(b)

Find dS/dx.

(2)

(c)

Find the value of x for which S is a minimum. Give your answer correct to 3 significant figures.

(3)

(d)

Work out the minimum surface area, correct to 3 significant figures. Either the exact value of x or 7.94 may be used.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 h = 500/x²	✓ 1
M1 S = 2x² + 4xh	✓ 1
A1 substitutes h and simplifies 4x × 500/x² = 2000/x	✓ 1
Part (b)
M1 differentiates, writing 2000/x as 2000x⁻¹	✓ 1
A1 4x − 2000/x²	✓ 1
Part (c)
M1 sets dS/dx = 0 (ECF from (b))	✓ 1
M1 x³ = 500	✓ 1
A1 x = 7.94	✓ 1
Part (d)
M1 substitutes x into 2x² + 2000/x (ECF from (c))	✓ 1
A1 378 cm²	✓ 1

Exam tip: In a 'show that' question every step must be written down, ending exactly on the given result.

+ Add to test✓ In your test309IGME3-3-12r37-11Differentiation and stationary points Paper HHard9 marks

In a game, the height y of a track above the ground at horizontal distance x is given by y = x³ − 6x² + 9x + 2 for 0 ≤ x ≤ 4.

(a)

Find dy/dx.

(2)

(b)

Find the x-coordinates of the two stationary points of the track.

(3)

(c)

Determine which stationary point is a local minimum, and find its y-coordinate.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiate at least two terms correctly	✓ 1
A1 3x² − 12x + 9	✓ 1
Part (b)
M1 3x² − 12x + 9 = 0	✓ 1
M1 3(x − 1)(x − 3) = 0 (or correct use of the quadratic formula)	✓ 1
A1 x = 1 and x = 3	✓ 1
Part (c)
M1 d²y/dx² = 6x − 12	✓ 1
M1 at x = 3 the value is 6, which is positive	✓ 1
A1 local minimum at x = 3	✓ 1
A1 y = 27 − 54 + 27 + 2 = 2	✓ 1

Common error: A positive second derivative means a minimum and a negative one means a maximum; do not mix these up.

+ Add to test✓ In your test310IGME3-3-12r37-12Function notation, composite and inverse functions Paper HHard9 marks

The function f is defined by f(x) = (x + 3)/(x − 1), where x ≠ 1.

(a)

Find f(5).

(1)

(b)

Show that ff(x) = x.

(4)

(c)

Use the result in (b) to write down f⁻¹(x).

(1)

(d)

Solve f(x) = x. Give the positive solution as your final answer.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 2	✓ 1
Part (b)
M1 ff(x) = ((x + 3)/(x − 1) + 3) ÷ ((x + 3)/(x − 1) − 1)	✓ 1
M1 write numerator and denominator over the common denominator x − 1	✓ 1
A1 numerator (4x) and denominator 4 correct	✓ 1
A1 ff(x) = 4x ÷ 4 = x	✓ 1
Part (c)
B1 f⁻¹(x) = (x + 3)/(x − 1)	✓ 1
Part (d)
M1 x + 3 = x(x − 1)	✓ 1
M1 x² − 2x − 3 = 0 and factorise or use the formula	✓ 1
A1 x = 3 (and x = −1)	✓ 1

Exam tip: If ff(x) = x then f is its own inverse, so f⁻¹ and f are the same function.

+ Add to test✓ In your test311IGME3-3-13r44-12Function notation, composite and inverse functions Paper HHard8 marks

The function f is defined as
f(x) = (4x + 3) ÷ (x + 2) for x ≠ −2

(a)

Find f⁻¹(x).

(4)

(b)

State the value of x that must be excluded from the domain of f⁻¹.

(1)

(c)

Given that x > 0, solve f(x) = x.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 y(x + 2) = 4x + 3 (letting y = f(x) and clearing the fraction)	✓ 1
M1 xy + 2y = 4x + 3 and collecting terms in x on one side	✓ 1
M1 x(y − 4) = 3 − 2y factorised	✓ 1
A1 f⁻¹(x) = (3 − 2x) ÷ (x − 4)	✓ 1
Part (b)
B1 x = 4	✓ 1
Part (c)
M1 4x + 3 = x(x + 2)	✓ 1
M1 x² − 2x − 3 = 0 solved, e.g. (x − 3)(x + 1) = 0	✓ 1
A1 x = 3 (reject x = −1)	✓ 1

Common error: When finding an inverse function, swap to make x the subject only after collecting every x term on one side and factorising.

+ Add to test✓ In your test312IGME3-3-13r44-13Differentiation and stationary points Paper HHard12 marks

A beekeeper makes an open-topped feeding tray in the shape of a cuboid with a square base of side x cm and height h cm. The tray must hold a volume of 4000 cm³. The beekeeper wants to use as little material as possible. The total outer surface area of the tray is S cm².

Diagram not to scale

(a)

Show that S = x² + 16000x⁻¹.

(3)

(b)

Find dS/dx.

(2)

(c)

Find the value of x for which S has a stationary value.

(3)

(d)

Show that this value of x gives a minimum value of S.

(2)

(e)

Work out the minimum surface area.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 x²h = 4000 so h = 4000 ÷ x²	✓ 1
M1 S = x² + 4xh (base plus four sides)	✓ 1
A1 4x × 4000 ÷ x² = 16000 ÷ x leading to the given expression	✓ 1
Part (b)
M1 one term differentiated correctly	✓ 1
A1 2x − 16000x⁻²	✓ 1
Part (c)
M1 2x − 16000x⁻² = 0 (ECF from (b))	✓ 1
M1 x³ = 8000	✓ 1
A1 x = 20	✓ 1
Part (d)
M1 d²S/dx² = 2 + 32000x⁻³ (or gradient tested either side of x = 20)	✓ 1
A1 value at x = 20 is 6 > 0 so a minimum (or gradient changes from negative to positive)	✓ 1
Part (e)
M1 20² + 16000 ÷ 20	✓ 1
A1 1200 cm²	✓ 1

Exam tip: In a ‘show that’ question, write every step of the algebra. Use the volume to eliminate one variable before differentiating.

+ Add to test✓ In your test313IGME3-3-13r44-14Sequences and nth terms Paper HHard7 marks

The number of data channels recorded by a Formula 1 team’s simulator after each of the first three software updates is
7, 16, 29
These numbers are the first three terms of a quadratic sequence.

(a)

Find an expression, in terms of n, for the nth term of the sequence.

(4)

(b)

One update produces exactly 121 channels. Work out the value of n.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 first differences 9 and 13, second difference 4	✓ 1
M1 coefficient of n² is 2 (half of second difference)	✓ 1
M1 subtracting 2n² from the terms gives 5, 8, 11 (or equivalent method for b and c)	✓ 1
A1 2n² + 3n + 2	✓ 1
Part (b)
M1 2n² + 3n + 2 = 121 rearranged to 2n² + 3n − 119 = 0 (ECF from (a))	✓ 1
M1 factorising (2n + 17)(n − 7) = 0 or correct use of the formula	✓ 1
A1 n = 7 (reject −8.5)	✓ 1

Exam tip: For a quadratic sequence, the second difference is twice the coefficient of n². Subtract this part from the sequence to leave a linear sequence.

+ Add to test✓ In your test314IGME3-3-12r48-05Sequences and nth terms Paper HHard8 marks

A data centre is built in stages. The numbers of cables needed at stages 1, 2, 3 and 4 are 4, 11, 22 and 37. These numbers form a quadratic sequence.

(a)

Work out the number of cables needed at stage 5.

(2)

(b)

Find an expression for the number of cables at stage n.

(3)

(c)

At which stage are 211 cables needed?

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 first differences 7, 11, 15 and next difference 19	✓ 1
A1 56	✓ 1
Part (b)
M1 second difference 4, so 2n²	✓ 1
M1 subtracts 2n² leaving 2, 3, 4, 5	✓ 1
A1 2n² + n + 1	✓ 1
Part (c)
M1 2n² + n + 1 = 211, so 2n² + n − 210 = 0 (ECF from (b))	✓ 1
M1 (2n + 21)(n − 10) = 0 or the quadratic formula	✓ 1
A1 stage 10, rejecting n = −10.5	✓ 1

Exam tip: For a quadratic sequence, the second difference divided by 2 gives the coefficient of n².

+ Add to test✓ In your test315IGME3-3-12r48-07Graphs, gradients and real-life graphs Paper HHard9 marks

On a plan of a data centre, one cable run follows the line L₁ with equation y = 2x + 3. A second cable run, L₂, is perpendicular to L₁ and passes through the point A(6, −1). All lengths are in metres.

(a)

Find an equation of L₂.

(3)

(b)

Find the coordinates of the point where the two cable runs meet.

(3)

(c)

Work out the distance from A to the point where the cable runs meet. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient of L₂ = −½	✓ 1
M1 −1 = −½ × 6 + c or y + 1 = −½(x − 6)	✓ 1
A1 y = −½x + 2	✓ 1
Part (b)
M1 2x + 3 = −½x + 2 (ECF from (a))	✓ 1
A1 x = −0.4	✓ 1
A1 y = 2.2	✓ 1
Part (c)
M1 (6 − (−0.4))² + (−1 − 2.2)² = 40.96 + 10.24	✓ 1
M1 square root of 51.2	✓ 1
A1 7.16 m	✓ 1

Common error: Perpendicular gradients multiply to −1: turn the gradient upside down and change the sign.

+ Add to test✓ In your test316IGME3-3-12r48-11Graphs, gradients and real-life graphs Paper HHard9 marks

A data-centre compound is a rectangle of area 144 m². The width of the rectangle is x metres. A fence goes around the whole compound. The length of the fence is P metres.

Diagram not to scale

(a)

Show that P = 2x + 288 ÷ x.

(2)

(b)

Find dP/dx.

(2)

(c)

Find the value of x for which the fence is shortest, and work out the shortest length of fence.

(3)

(d)

Show that this value of x gives a minimum.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 length = 144 ÷ x	✓ 1
A1 P = 2(x + 144 ÷ x) = 2x + 288 ÷ x shown	✓ 1
Part (b)
M1 288x⁻¹ differentiated to −288x⁻²	✓ 1
A1 2 − 288 ÷ x²	✓ 1
Part (c)
M1 2 − 288 ÷ x² = 0 (ECF from (b))	✓ 1
A1 x = 12 (rejecting x = −12)	✓ 1
A1 P = 48 m	✓ 1
Part (d)
M1 d²P/dx² = 576 ÷ x³	✓ 1
A1 at x = 12 this is positive, so a minimum	✓ 1

Exam tip: In an optimisation question, always justify the minimum or maximum with the second derivative or a sign check.