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International GCSE Maths · Pearson Edexcel 4MA1 · Unit 3

Sequences, functions and graphs

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

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+ Add to test✓ In your 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.

(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 test11IGME-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 test12IGME-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.

(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 test13IGME-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.

(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 test14IGME-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 test15IGME-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 test16IGME-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

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 test17IGME-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 test18IGME-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 test19IGME-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 test20IGME-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 test21IGME-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 test22IGME-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 test23IGME-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 test24IGME-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 test25IGME-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 test26IGME-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 test27IGME-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 test28IGME-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 test29IGME-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 test30IGME-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 test31IGME-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 test32IGME-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 test33IGME-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 test34IGME-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 test35IGME-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 test36IGME-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 test37IGME-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 test38IGME-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).

(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 test39IGME-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 test40IGME-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 test41IGME-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 test42IGME-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 test43IGME-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 test44IGME-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 test45IGME-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 test46IGME-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 test47IGME-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.

(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 test48IGME-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 test49IGME-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.

(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.