A Level Mathematics 9709 Paper 31, May/June 2026

Question paper and mark scheme75 marks11 questions

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Question paper

9709/31 · May/June 2026

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1

Solve the inequality 3∣x+4∣>∣2x−1∣.

[4]
2

The variables x and y satisfy the equation y3=Aekx, where A and k are constants. The graph of ln⁡y against x is a straight line passing through the points (−1.32,5.09) and (0.25,1.16). Find the values of A and k.

[5]
3

The equation of a curve is y=2x2+7x3x2−1. Find an equation of the normal to the curve at the point where x=1. Give your answer in the form ax+by+c=0, where a, b and c are integers.

[6]
4

The square roots of −71−(203)i can be expressed in the Cartesian form x+iy, where x and y are real and exact. By first forming a quartic equation, find the square roots of −71−(203)i. Give your answer in exact Cartesian form.

[5]
5
a

Show that (cos⁡x+sin⁡x)(cosec⁡x−sec⁡x)≡2cot⁡2x.

[4]
b

Hence, solve the equation (cos⁡x+sin⁡x)(cosec⁡x−sec⁡x)=3 for 0⩽x⩽π.

[3]
6

A curve has the equation y=e(x2+4x)cos⁡2x, where cos⁡x≠0.

a

Show that the x-coordinates of the stationary points of this curve satisfy the equation tan⁡x=x+2.

[3]
b

By sketching a suitable pair of graphs, show that the equation tan⁡x=x+2 has only one root in the interval −π<x<π.

[2]
c

Show by calculation that this root lies between 1.2 and 1.3.

[2]
d

Use the iterative formula xn+1=tan⁡−1(xn+2) to calculate this root correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

[3]
7

The shaded region on the Argand diagram shows points representing the complex numbers, z, defined by two inequalities. The shaded region is bounded by the circle and the line parallel to the imaginary axis shown in the diagram and includes its boundaries.

Figure 1 of 1 for question 7
a

Find two inequalities in terms of z that define the shaded region.

[3]
b

Calculate the greatest value of ∣z∣ for points in the shaded region.

[3]
8

Find the exact value of ∫013x2e3x dx.

[6]
9

The points A and B have position vectors (2−53) and (−641) respectively. The line t passes through the points A and B.

a

Find a vector equation for t.

[2]
b

The line s has equation r=(134)+μ(−12−1). Determine whether s and t are parallel, intersect or are skew.

[5]
10

The polynomial ax3−11x2+bx+4, where a and b are constants, is denoted by p(x). It is given that (x−2) is a factor of p(x), and that when p′(x) is divided by (x−2) the remainder is 24.

a

Find the value of a and the value of b.

[5]
b(i)

When a and b have these values, factorise p(x) completely.

[3]
b(ii)

Hence, solve the inequality p(x)⩾0.

[2]
11
a

By differentiating cos⁡4xsin⁡4x, show that if y=cot⁡4x, then dydx=−4cosec⁡24x.

[3]
b

The variables x and y satisfy the differential equation e(3x+2)dydx=sin⁡24y. It is given that y=116π when x=−23. Solve the differential equation, and find the value of y in the range −18π⩽y⩽0 when x=1.

[6]

Mark scheme

9709/31 · May/June 2026

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1 · 4 marks

  • M1

    State or imply non-modular inequality 32(x+4)2>(2x−1)2, or corresponding quadratic equation, or pair of linear equations

  • M1

    Make a reasonable attempt at solving a 3-term quadratic, or solve two linear equations for x

  • A1

    Obtain critical values x=−13 and x=−115

  • A1

    State final answer x<−13, x>−115

2 · 5 marks

  • B1

    State or imply 3ln⁡y=ln⁡A+kx

  • M1

    Substitute values of ln⁡y and x in equation of correct form and solve for k

  • A1

    Obtain k=−7.51

  • M1

    Solve for A

  • A1

    Obtain A=212

3 · 6 marks

  • M1

    Use the correct product or quotient rule

  • A1

    Obtain the correct derivative in any form, e.g (3x2−1)(4x+7)−(2x2+7x)(6x)(3x2−1)2

  • M1

    Substitute x=1 into their dydx

  • M1

    Find gradient of normal =−1÷their dydx

  • M1

    Substitute their gradient of normal, x=1 and their y into y=mx+c

  • A1

    x−8y+35=0

4 · 5 marks

  • M1

    Square x+iy and equate real and imaginary parts to −71 and −203 respectively

  • A1

    Obtain equations x2−y2=−71 and 2xy=−203

  • 1 mark

    Eliminate one variable and find an equation in the other

  • A1

    Obtain x4+71x2−300=0 or y4−71y2−300=0, or three-term equivalents

  • A1

    Obtain answers ±(2−53i)

5a · 4 marks

  • M1

    Expand the brackets to obtain cos⁡xcosec⁡x−1+1−sin⁡xsec⁡x or equivalent

  • A1

    Simplify to cot⁡x−tan⁡x or 1tan⁡x−tan⁡x

  • M1

    Use double angle formula

  • A1

    Obtain 2cot⁡2x following full and correct working

5b · 3 marks

  • M1

    Find a value of x in the given interval

  • A1

    Obtain one answer, e.g. 0.294

  • A1

    Obtain all answers 0.294, 1.86 and no others in the interval

6a · 3 marks

  • M1

    Use the correct product rule

  • A1

    Obtain the correct derivative in any form

  • A1

    Equate dydx to 0 and obtain tan⁡x=x+2 with no omissions and without wrong working

6b · 2 marks

  • M1

    Sketch y=tan⁡x for −π<x<π

  • A1

    Sketch y=x+2 for −π<x<π and justify the given statement

6c · 2 marks

  • M1

    Calculate the values of a relevant expression or pair of expressions at x=1.2 and x=1.3

  • A1

    Complete the argument correctly with correct calculated values

6d · 3 marks

  • M1

    Use the iterative process at least once

  • A1

    Obtain final answer 1.274

  • A1

    Show sufficient iterations to 5 d.p. to justify 1.274 to 3 d.p. or show there is a sign change in the interval (1.2735, 1.2745)

7a · 3 marks

  • B1

    Obtain Re⁡z⩾−2 Or ∣z+3∣⩾∣z+1∣ o.e.

  • M1

    Obtain answer of the form ∣z−a∣≤b

  • A1

    Obtain answer ∣z+3+i∣≤2

7b · 3 marks

  • M1

    Attempt to solve (x+3)2+(y+1)2=22 with x=−2

  • 1 mark

    Carry out a correct method for finding the greatest value of ∣z∣

  • A1

    Obtain answer 3.39 or 8+23

8 · 6 marks

  • M1

    Commence integration by parts and reach Ax2e3x−∫Bxe3x dx

  • A1

    Obtain x2e3x−∫2xe3x dx

  • M1

    Complete integration by parts and reach Ax2e3x−Bxe3x+Ce3x

  • A1

    Obtain x2e3x−23xe3x+29e3x

  • 1 mark

    Substitute limits correctly in an expression of the form Ax2e3x−Bxe3x+Ce3x

  • A1

    Obtain answer 59e3−29

9a · 2 marks

  • M1

    Use a correct method to form an equation for line t

  • A1

    Obtain r=(2−53)+λ(−89−2)

9b · 5 marks

  • B1

    Justify lines are not parallel, e.g. (−12−1)≠d(−89−2)

  • B1

    Express s or t in component form e.g. (1−μ,3+2μ,4−μ) or (2−8λ1,−5+9λ1,3−2λ1) or (−6−8λ2,4+9λ2,1−2λ2) or (−2−8λ3,−12+9λ3,2−2λ3)

  • M1

    Equate two pairs of components of general points on s and *their* t and solve simultaneously for λ or for μ

  • A1

    Obtain correct answer for λ or μ, e.g. μ=−557, λ1=−67

  • A1

    Determine that all three equations are not satisfied and the lines fail to intersect, and conclude the lines are skew Conclusion needs to follow correct working

10a · 5 marks

  • M1

    Substitute x=2 or −2 into p(x) and equate to 0

  • A1

    Obtain 8a−44+2b+4=0

  • M1

    Substitute x=2 into p′(x) and equate to 24

  • A1

    Obtain 12a−44+b=24

  • A1

    Obtain a=6 and b=−4

10b(i) · 3 marks

  • M1

    Equate (x−2)(6x2+Ax+B) to 6x3−11x2−4x+4 and obtain equations to solve for A and B or divide 6x3−11x2−4x+4 by x−2 and reach ax2±kx

  • A1

    (x−2)(6x2+x−2)

  • A1

    Obtain (x−2)(3x+2)(2x−1)

10b(ii) · 2 marks

  • B1

    Obtain one correct region −23≤x≤12 or x≥2

  • B1

    Obtain both regions −23≤x≤12, x≥2

11a · 3 marks

  • M1

    Use correct quotient rule to obtain expression of the form sin⁡4xddx(cos⁡4x)−cos⁡4xddx(sin⁡4x)(sin⁡4x)2

  • A1

    Obtain sin⁡4x(−4sin⁡4x)−cos⁡4x(4cos⁡4x)sin⁡24x

  • A1

    Obtain dydx=−4cosec⁡24x correctly

11b · 6 marks

  • B1

    Separate variables correctly and reach acosec⁡24y or be−3x−2

  • B1

    Obtain term −13e−3x−2

  • B1

    Obtain term −14cot⁡4y

  • M1

    Use x=−23, y=116π to evaluate a constant or as limits in a solution containing terms of the form acot⁡by and ce±(3x+2)

  • A1

    Obtain correct answer in any form

  • A1

    Obtain y=−0.314

Questions

  1. 1
    Question 14 marks
    Mark scheme ›
  2. 2
    Question 25 marks
    Mark scheme ›
  3. 3
    Question 36 marks
    Mark scheme ›
  4. 4
    Question 45 marks
    Mark scheme ›
  5. 5
    Question 57 marks
    Mark scheme ›
  6. 6
    Question 610 marks
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  7. 7
    Question 76 marks
    Mark scheme ›
  8. 8
    Question 86 marks
    Mark scheme ›
  9. 9
    Question 97 marks
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  10. 10
    Question 1010 marks
    Mark scheme ›
  11. 11
    Question 119 marks
    Mark scheme ›

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