Cambridge AS & A Level Mathematics 9709/22 is the May/June 2026 Paper 2 (Pure Mathematics 2), variant 2. It is an AS Level paper, lasts 1 hour 15 minutes, is worth 50 marks and is answered on the question paper. It had eight compulsory questions. The first three were 4 or 5 marks each; the parametric curve in Question 8 was the biggest at 9 marks, followed by the quartic and iteration in Question 6 (8 marks) and the modulus inequality and exact area in Questions 5 and 7 (7 marks each). This guide goes through the paper question by question with the final answers and what the official mark scheme gave marks for. The paper and its mark scheme are free on the Tutopiya Learning Portal.

For an overview of all six 9709 papers from this series, see A Level Maths 9709 May/June 2026 past papers: what came up and how it was marked.

9709/22 at a glance

Question	Topic	Marks
1	Integrating (1 + e⁻ˣ)² between a and 2a	4
2	Equation in cosec² θ and cot² θ, for −180° < θ < 180°	4
3	y = kaˣ⁺² from a straight-line graph of ln y against x	5
4	Normal to y = e²ˣ/(eˣ + 2) where it crosses the y-axis	6
5	Modulus inequality, then the same inequality in ln(½N)	7
6	Quartic with factor (x + 2): find a, show an iterative form, iterate	8
7	Exact area under y = 8 cos 2x − 1/cos² 2x	7
8	Parametric curve: gradient at t = 0 and exact y-coordinate of the maximum	9
	Total	50

Question-by-question walkthrough

The mark scheme uses three main types of mark. M marks are for a valid method applied to this question (quoting a formula is not enough; you have to substitute into it). A marks are for accurate answers and can’t be given without the M mark before them. B marks are independent marks for a correct result or statement. A star (*) on a mark means later marks depend on it: DM marks are lost if the starred mark wasn’t earned. AG (answer given) marks a “show that” where the result is printed, so every step must be shown. The cover asks for non-exact answers to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise; Questions 3 and 6 did say otherwise.

Question 1: integrating an exponential (4 marks)

The integral of (1 + e⁻ˣ)² from a to 2a, in terms of the positive constant a.

• Answer: a + 2e⁻ᵃ − (3/2)e⁻²ᵃ − ½e⁻⁴ᵃ.
• Working: expand to 1 + 2e⁻ˣ + e⁻²ˣ and integrate to x − 2e⁻ˣ − ½e⁻²ˣ. Substitute 2a and subtract the value at a: (2a − 2e⁻²ᵃ − ½e⁻⁴ᵃ) − (a − 2e⁻ᵃ − ½e⁻²ᵃ).
• Marking: B1 for the expansion, M1 for integrating to the form x + k₁e⁻ˣ + k₂e⁻²ˣ (at least two terms), A1 for the correct integral, A1 for applying the limits correctly. The expansion needs the middle term 2e⁻ˣ; squaring each term separately loses the B1.

Question 2: a cosec and cot equation (4 marks)

The equation set 3 cosec² θ equal to 7 + cot² θ, for −180° < θ < 180°.

• Answer: θ = −144.7°, −35.3°, 35.3°, 144.7°.
• Working: use cosec² θ = 1 + cot² θ: 3 + 3cot² θ = 7 + cot² θ, so cot² θ = 2, tan² θ = ½ (or sin² θ = ⅓). tan θ = ±1/√2 gives 35.3° and 144.7° and their negatives.
• Marking: M1 for using correct identities to get an equation in one trigonometric function, A1 for sin² θ = ⅓ (or the equivalent in cos, tan, cot or cosec), A1 for at least two of the angles, A1 for all four and no others in the range. Remember the negative angles: the range runs from −180° to 180°.

Question 3: logarithms and a straight-line graph (5 marks)

y = kaˣ⁺², and the graph of ln y against x is a straight line through (1.5, 3.22) and (4.0, 4.69). k and a were wanted to 2 significant figures.

• Answer: a = 1.8, k = 3.2.
• Working: take logs: ln y = ln k + (x + 2) ln a, so the gradient of the line is ln a = (4.69 − 3.22) ÷ 2.5 = 0.588, and a = e⁰·⁵⁸⁸ = 1.8. Then 3.22 = ln k + 3.5 × 0.588, so ln k = 1.162 and k = 3.2.
• Marking: B1 for the equation ln y = ln k + (x + 2) ln a, M1 for equating the gradient to ln a, A1 for a = 1.8, M1 for substituting to find ln k, A1 for k = 3.2. Forming two simultaneous equations from the two points, or eliminating with e³·²² = ka³·⁵ and e⁴·⁶⁹ = ka⁶, earned the same marks. Watch the (x + 2): the intercept is not ln k, because the line passes through ln k + 2 ln a at x = 0.

Question 4: normal to an exponential curve (6 marks)

y = e²ˣ/(eˣ + 2) crosses the y-axis at P; the normal was wanted in the form ax + by + c = 0 with integer coefficients.

• Answer: 27x + 15y − 5 = 0.
• Working: at x = 0, y = 1/3. The quotient rule gives dy/dx = (2e²ˣ(eˣ + 2) − e²ˣ · eˣ)/(eˣ + 2)² = (e³ˣ + 4e²ˣ)/(eˣ + 2)², which is 5/9 at x = 0. The normal has gradient −9/5, so y − ⅓ = −(9/5)x; multiplying by 15 gives the answer.
• Marking: M1* for the quotient rule, A1 for the derivative, DM1 for substituting x = 0, A1 for −9/5 (from a correct derivative), M1 for the equation of the normal through (0, ⅓), A1 for the integer form equated to zero, from correct work only.

Question 5: modulus inequality (7 marks)

• (a) Solve |4x − 7| ≤ |2x + 3| (4 marks): 2/3 ≤ x ≤ 5.
  • Working: the critical values come from 4x − 7 = 2x + 3 (x = 5) and 4x − 7 = −(2x + 3) (x = 2/3). Alternatively, square both sides: (4x − 7)² = (2x + 3)² gives 12x² − 68x + 40 = 0, that is (3x − 2)(x − 5) = 0.
  • Marking: B1 for x = 5 (or B1 for the squared equation), M1 for solving the linear equation with opposite signs (or the three-term quadratic), A1 for 2/3, A1 for the final inequality with ≤ signs.
• (b) Smallest and largest integers N (3 marks): “hence” means you replace x by ln(½N), so 2/3 ≤ ln(½N) ≤ 5. That gives 2e^(2/3) ≤ N ≤ 2e⁵, about 3.90 ≤ N ≤ 296.8, so the smallest integer is 4 and the largest is 296.
  • Marking: M1 for solving ln(½N) equal to a critical value, A1 for 4, A1 for 296. The M1 allowed a non-integer N at that stage; the A marks were allowed if seen inside an inequality.

Question 6: quartic and iteration (8 marks)

f(x) = ax⁴ + 4x³ − 3ax² − 21x − 18, with (x + 2) a factor. The diagram showed one positive root.

• (a) Find a (2 marks): f(−2) = 16a − 32 − 12a + 42 − 18 = 4a − 8 = 0, so a = 2. M1 for setting f(−2) = 0, A1.
• (b) Show the positive root satisfies x = ∛(3x + 4.5) (3 marks, AG): divide f(x) = 2x⁴ + 4x³ − 6x² − 21x − 18 by (x + 2) to get 2x³ − 6x − 9. Setting this to zero, 2x³ = 6x + 9, so x³ = 3x + 4.5 and x = ∛(3x + 4.5). M1 for the division at least as far as the x term (inspection was allowed), A1 for the cubic, A1 for the rearrangement. The scheme said “necessary detail needed”, because the result is printed.
• (c) Iterate to 4 significant figures (3 marks): x = 2.238. From x₁ = 2, the iterates to 6 significant figures are 2.18976, 2.22864, 2.23644, 2.23800, 2.23831, 2.23837, 2.23838, so the root is 2.238 to 4 significant figures.
  • Marking: M1 for using the formula correctly at least once, A1 for 2.238, A1 for enough iterates at 6 significant figures to justify it (or a sign change shown in the interval 2.2375 to 2.2385). The answer had to be given to exactly four significant figures.

Question 7: exact area (7 marks)

The diagram showed y = 8 cos 2x − 1/cos² 2x, with the region between the curve and the x-axis shaded, symmetrical about the y-axis.

• Answer: 3√3.
• Working: the curve meets the x-axis where 8 cos 2x = 1/cos² 2x, so cos³ 2x = ⅛, cos 2x = ½ and x = ±π/6. Write 1/cos² 2x as sec² 2x, so the integral is 4 sin 2x − ½ tan 2x. From −π/6 to π/6 (or twice the value from 0 to π/6): 2 × (4 × (√3/2) − ½ × √3) = 2 × (3√3/2) = 3√3.
• Marking: M1 for solving 8 cos 2x − 1/cos² 2x = 0, A1 for cos 2x = ½ and x = π/6, B1 for recognising sec² 2x, M1* for integrating to the form k₁ sin 2x + k₂ tan 2x, A1 for 4 sin 2x − ½ tan 2x, DM1 for using the limits correctly to find an exact value, A1 for 3√3. Using 0 and π/6 and doubling was accepted, but not two positive limits.

Question 8: parametric curve (9 marks)

x = 6t − sin 2t and y = 3 cos 2t + 5 sin t, for 0 ≤ t < ½π. The curve has a maximum point M.

• (a) Gradient at t = 0 (4 marks): dx/dt = 6 − 2 cos 2t and dy/dt = −6 sin 2t + 5 cos t. At t = 0 these are 4 and 5, so dy/dx = 5/4. B1 for each derivative, M1 for dividing at t = 0, A1.
• (b) Exact y-coordinate of M (5 marks): set dy/dt = 0 and use sin 2t = 2 sin t cos t: −12 sin t cos t + 5 cos t = 0, so cos t (5 − 12 sin t) = 0. cos t = 0 is outside the range, so sin t = 5/12. Then y = 3(1 − 2 sin² t) + 5 sin t = 3 − 6(25/144) + 25/12 = 97/24.
  • Marking: M1 for setting the derivative to zero and using the double-angle identity, A1 for the factorised form and sin t = 5/12, M1 for writing y in terms of sin t (or sin t and cos t, with cos t = √119/12), A1 for 3 − 6 sin² t + 5 sin t or equivalent, A1 for 97/24 or an exact equivalent.

Where the marks are on 9709/22

• Calculus carried 26 of the 50 marks: integrating exponentials (Question 1, 4), the normal via the quotient rule (Question 4, 6), the exact trigonometric area (Question 7, 7) and parametric differentiation (Question 8, 9).
• Algebra carried 12 marks: the modulus inequality (Question 5, 7) and the factor theorem and division in Question 6(a) and (b) (5).
• Logarithms and exponentials carried 5 marks (Question 3).
• Trigonometry carried 4 marks (Question 2), with more trigonometry inside Questions 7 and 8.
• Numerical solution of equations carried 3 marks (Question 6(c)).

The last four questions were worth 31 of the 50 marks, and two of them (Questions 7 and 8) needed exact answers.

How to self-mark 9709/22 with the mark scheme

1. Mark the method before the answer. Each part is built from M, A and B marks. In Question 4, the quotient rule, substituting x = 0 and forming the normal are three M marks even if a slip spoils the final equation.
2. Check the starred marks. In Question 4, the DM1 for substituting x = 0 needs the M1* for a correct quotient rule; in Question 7, the DM1 for the limits needs the M1* for integrating to sin 2x and tan 2x.
3. Check the accuracy the question asked for. Question 3 wanted k and a to 2 significant figures; Question 6(c) wanted exactly 4 significant figures with iterates to 6. Questions 7 and 8(b) wanted exact values, so a decimal earns no final A mark.
4. Count your solutions. Question 2 needs four angles (two of them negative) and no others; Question 5(b) needs both the smallest and the largest integer.
5. Treat “show that” parts strictly. Question 6(b) is AG: you must show the division (or inspection) giving 2x³ − 6x − 9 and the rearrangement, not just state the result.
6. Log each lost mark as a missing method, an identity error, an accuracy or rounding problem, or a missed condition (the range of θ or t), and revise in that order.

Sit 9709/22 as a timed mock

The paper gives 75 minutes for 50 marks, exactly 1.5 minutes per mark. A workable plan:

Block	Marks	Time
Questions 1 to 3	13	18 minutes
Question 4 (normal)	6	9 minutes
Question 5 (modulus inequality)	7	10 minutes
Question 6 (quartic and iteration)	8	12 minutes
Question 7 (exact area)	7	11 minutes
Question 8 (parametric curve)	9	12 minutes
Check, especially exact forms and accuracy	–	3 minutes

Question 7 takes longer than its marks suggest: you need the limits from the equation before you can integrate, and the sec² 2x step is easy to miss under time pressure.

Other May/June 2026 9709 papers

• 9709 Paper 12 (Pure Mathematics 1)
• 9709 Paper 32 (Pure Mathematics 3)
• 9709 Paper 42 (Mechanics)
• 9709 Paper 52 (Probability & Statistics 1)
• 9709 Paper 62 (Probability & Statistics 2)

Practise and download

• Download 9709/22 and its mark scheme on the Learning Portal →
• All 9709 May/June 2026 papers: overview and marking patterns
• Try a free A Level Maths quiz

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Frequently asked questions

What topics were on 9709 Paper 22 May/June 2026? +

Eight questions: integrating (1 + e⁻ˣ)², a cosec² and cot² equation, finding k and a from a graph of ln y against x, a normal to an exponential curve, a modulus inequality with logarithms, a quartic with an iterative root, an exact area involving cos 2x and sec² 2x, and a parametric curve's gradient and maximum.

How many marks is 9709/22 and how long is it? +

It is out of 50 marks and lasts 1 hour 15 minutes, answered on the question paper. Question 8 (the parametric curve) was the biggest at 9 marks, followed by Question 6 (the quartic and iteration) at 8 marks.

What was the answer to the area question in 9709/22? +

3√3. The curve meets the x-axis where cos 2x = ½, at x = ±π/6. Writing 1/cos² 2x as sec² 2x, the integral is 4 sin 2x − ½ tan 2x, and evaluating from −π/6 to π/6 gives 3√3.

What was the root in 9709/22 Question 6? +

2.238, correct to 4 significant figures. Starting from x = 2, the iterates of x = ∛(3x + 4.5) are 2.18976, 2.22864, 2.23644, 2.23800, 2.23831, 2.23837 and 2.23838, so the root is 2.238 to 4 significant figures.

Who sits 9709 Paper 2? +

Paper 2 (Pure Mathematics 2) is an AS Level paper sat with Paper 1 on the AS-only route. That route can't be carried forward to the full A Level, which uses Paper 3 instead.

Where can I download 9709/22 May/June 2026 and its mark scheme? +

Both are free to download and print on the Tutopiya Learning Portal, on the Cambridge AS & A Level Mathematics 9709 past papers page.

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