Cambridge AS & A Level Mathematics 9709/32 is the May/June 2026 Paper 3 (Pure Mathematics 3), variant 2. It is the A Level pure paper, lasts 1 hour 50 minutes and is worth 75 marks. It had ten compulsory questions, answered in an answer booklet. The first four were short (3 to 5 marks); the big marks came from the vectors question (Question 9, 12 marks), the partial fractions and differential equation (Question 10, 11 marks) and the parametric and inverse tan questions (Questions 5 and 6, 10 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.

9709/32 at a glance

Question	Topic	Marks
1	Quotient and remainder when a cubic is divided by 2x − 1	3
2	Exponential equation, answer as ln a / ln b	4
3	Complex numbers: st in the form re^(iθ), and the geometric effect of multiplying by t	4
4	Square roots of 3 − 6√2 i via a quartic	5
5	Parametric curve: dy/dx in terms of tan 2θ, the point where the gradient is −1	10
6	y = x tan⁻¹ 3x: exact gradient, exact integral by parts	10
7	Quotient rule to a quadratic in sin 3a, then all solutions in radians	7
8	cosec(½x) = ½(x + 3): sketch, sign change, rearrangement, iteration	9
9	Vectors: line through A and B, skew lines, a point P with angle PAB = 60°	12
10	Partial fractions, then a separable differential equation with ln(2y)	11
	Total	75

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. FT (follow through) lets a mark follow from your own earlier wrong answer, and 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, and warns that unsupported calculator answers get no marks.

Question 1: polynomial division (3 marks)

4x³ − 4x² + 7x + 2 divided by 2x − 1.

• Answer: quotient 2x² − x + 3, remainder 5.
• Working: long division, or write the cubic as (2x − 1)(Ax² + Bx + C) + D and compare coefficients: A = 2, B = −1, C = 3, D = 5.
• Marking: M1 for reaching a partial quotient of the form 2x² + kx, A1 for the quotient, A1 for the remainder. Swapping quotient and remainder in your conclusion cost a mark, and writing “the remainder is 5/(2x − 1)” did not earn the last A1. Using the remainder theorem, f(½) = 5, earned a special case B1 for the remainder.

Question 2: exponential equation (4 marks)

2 to the power (5 − 2x) equals 4 to the power (x + 1), divided by 3. The answer was wanted as ln a / ln b with integers a and b.

• Answer: x = ln 24 / ln 16 (equivalent forms such as ln 576 / ln 256 accepted).
• Working: write 4^(x + 1) as 2^(2x + 2) and multiply by 3: 3 × 2^(5 − 2x) = 2^(2x + 2), so 3 = 2^(4x − 3). Taking logs, ln 3 = (4x − 3) ln 2, so 4x ln 2 = ln 3 + 3 ln 2 = ln 24, and x = ln 24 / (4 ln 2) = ln 24 / ln 16.
• Marking: M1 for a law of logs of a product or quotient, M1 for a correct use of the power law, A1 for a correct linear equation, A1 for the final form. Working with a common base instead earned B1, B1, M1, A1. No working scored no marks, and later working that spoiled the ln a / ln b form was not ignored.

Question 3: complex numbers in exponential form (4 marks)

s = 3(cos 2 + i sin 2) and t = 2e⁻ⁱ.

• (a) st in the form re^(iθ) (2 marks): s = 3e²ⁱ, so st = 6e^(2i − i) = 6eⁱ: r = 6, θ = 1. B1 for each. 6(cos 1 + i sin 1) also scored both marks.
• (b) The two geometric effects of multiplying by t (2 marks): |t| = 2 and arg t = −1, so OB is OA enlarged by scale factor 2 and rotated clockwise through 1 radian (about O). B1 for the enlargement with its factor (“stretch”, “dilation” or “the length is doubled” were allowed; “reduction” or “stretched along the y-axis” were not), B1 for the rotation with its direction and amount (an anticlockwise rotation of 2π − 1 was also accepted). If neither scored, a special case B1 was given for naming both an enlargement and a rotation.

Question 4: square roots of a complex number (5 marks)

The square roots of 3 − 6√2 i, in exact Cartesian form, by first forming a quartic.

• Answer: ±(√6 − √3 i), that is √6 − √3 i and −√6 + √3 i.
• Working: (x + iy)² = 3 − 6√2 i gives x² − y² = 3 and 2xy = −6√2, so y = −3√2/x. Substituting: x² − 18/x² = 3, so x⁴ − 3x² − 18 = 0, which factorises as (x² − 6)(x² + 3) = 0. x is real, so x² = 6, x = ±√6 and y = ∓√3.
• Marking: M1* for squaring and equating real and imaginary parts, A1 for the two equations, DM1 for eliminating a variable, A1 for the quartic (x⁴ − 3x² − 18 = 0 or y⁴ + 3y² − 18 = 0), A1 for both roots in the form x + iy. Giving x and y separately rather than as complex numbers lost the last mark.

Question 5: parametric curve with tan 2θ and ln(cos 2θ) (10 marks)

x = 2θ − tan 2θ and y = 6θ + 2 ln(cos 2θ), for 0 ≤ θ ≤ ¼π.

• (a) dy/dx in terms of tan 2θ (6 marks):
  • Answer: dy/dx = (2 tan 2θ − 3)/tan² 2θ (or (6 − 4 tan 2θ)/(−2 tan² 2θ)).
  • Working: dx/dθ = 2 − 2 sec² 2θ = −2 tan² 2θ. By the chain rule, the derivative of ln(cos 2θ) is −2 sin 2θ / cos 2θ, so dy/dθ = 6 − 4 tan 2θ. Divide dy/dθ by dx/dθ.
  • Marking: B1 for dx/dθ, M1 for the chain rule on ln(cos 2θ), A1 for dy/dθ, M1 for dividing, M1 for using sec² 2θ = 1 + tan² 2θ and sin 2θ/cos 2θ = tan 2θ, A1 for the answer.
• (b) Exact coordinates where the gradient is −1 (4 marks):
  • Answer: (π/4 − 1, 3π/4 − ln 2).
  • Working: (2T − 3)/T² = −1 with T = tan 2θ gives T² + 2T − 3 = 0, so (T + 3)(T − 1) = 0. In the range, tan 2θ ≥ 0, so tan 2θ = 1, 2θ = π/4 and θ = π/8. Then x = π/4 − tan(π/4) = π/4 − 1 and y = 3π/4 + 2 ln(1/√2) = 3π/4 − ln 2.
  • Marking: M1* for a three-term quadratic in tan 2θ, DM1 for solving it, A1 for θ = π/8 only, A1 for both exact coordinates. The trigonometry had to be evaluated: coordinates left in terms of tan⁻¹ 1 or 45° lost the last mark.

Question 6: x tan⁻¹ 3x (10 marks)

y = x tan⁻¹ 3x.

• (a) Exact gradient at x = ⅓ (4 marks): by the product rule, dy/dx = tan⁻¹ 3x + 3x/(1 + 9x²). At x = ⅓: π/4 + ½. M1 for the product rule, B1 for differentiating tan⁻¹ 3x to the form p/(1 + qx²), A1 for the correct derivative, A1 for the exact value (with tan⁻¹ 1 evaluated as π/4).
• (b) Exact value of the integral of y from 0 to ⅓ (6 marks):
  • Answer: π/36 − 1/18 (or (π − 2)/36).
  • Working: by parts, integrating x and differentiating tan⁻¹ 3x: ½x² tan⁻¹ 3x − (3/2)∫x²/(1 + 9x²) dx. Split x²/(1 + 9x²) = 1/9 − (1/9)/(1 + 9x²), which integrates to give ½x² tan⁻¹ 3x − x/6 + (1/18) tan⁻¹ 3x. At ⅓: π/72 − 1/18 + π/72; at 0 everything is zero.
  • Marking: M1* for starting integration by parts in the right direction, A1 for the correct first stage, DM1 for reducing the fraction to a constant plus a multiple of 1/(1 + 9x²), A1 for the complete integral, DM1 for substituting the limits (with the trigonometry evaluated), A1 for the exact two-term answer.

Question 7: quotient rule and a trig equation (7 marks)

y = (3 + sin 3x)/(2 − cos 3x) for 0 ≤ x ≤ π; the gradient is −3/2 when x = a.

• (a) Show that sin² 3a + 6 sin 3a − 3 = 0 (4 marks, AG): the quotient rule gives a numerator of 3 cos 3x(2 − cos 3x) − 3 sin 3x(3 + sin 3x) = 6 cos 3x − 9 sin 3x − 3, using sin² + cos² = 1. Setting the derivative equal to −3/2 and expanding (2 − cos 3a)² leads to cos² 3a = 6 sin 3a − 2; replacing cos² 3a by 1 − sin² 3a gives the result.
  • Marking: M1* for the quotient (or product) rule with an attempt at the derivatives, A1 for the correct derivative, DM1 for equating to −3/2 and reaching a three-term quadratic in sin 3a with a correct identity, A1 for the printed equation with full, correct working. An error not corrected in the next line lost the A1.
• (b) The values of a (3 marks): sin 3a = −3 + √12 = 0.4641…. For 0 ≤ 3a ≤ 3π there are four solutions: a = 0.161, 0.886, 2.26, 2.98 (radians).
  • Marking: M1 for finding one value (with an intermediate step such as sin 3a = −3 + √12), A1 for one correct answer, A1 for all four and no others. Each value needed at least 3 significant figures in radians. Answers in degrees were treated as a misread, costing one A mark, and a correct answer with no working scored 0.

Question 8: iteration with cosec (9 marks)

The equation cosec(½x) = ½(x + 3), for 0 ≤ x ≤ π.

• (a) Show there is exactly one root (2 marks): sketch y = cosec(½x), which falls from a vertical asymptote at x = 0 to its minimum (π, 1), and the straight line y = ½(x + 3), which rises from 1.5. They cross once. M1 for the cosec curve with the right shape and key points, A1 for the line with a consistent scale and the single intersection clearly marked.
• (b) Root between 1 and 1.2 (2 marks): with f(x) = cosec(½x) − ½(x + 3), f(1) = 2.0858… − 2 = 0.0858… > 0 and f(1.2) = 1.771… − 2.1 = −0.328… < 0, so there is a sign change. M1 for both values (one must be correct), A1 for the correct conclusion. Working in degrees was M0.
• (c) Show the iteration converges to the root (2 marks, AG): if the sequence converges to x, then x = 2 sin⁻¹(2/(x + 3)), so sin(½x) = 2/(x + 3), and taking reciprocals gives cosec(½x) = ½(x + 3). M1 for writing the limit equation, A1 for the full rearrangement (working in either direction was accepted).
• (d) The root to 3 decimal places (3 marks): x = 1.037. From x₁ = 1 the iterates to 5 decimal places are 1.04720, 1.03376, 1.03755, 1.03648, 1.03678, 1.03669. M1 for using the formula correctly at least once, A1 for 1.037, A1 for enough iterates to 5 decimal places to justify it (or a sign change in the interval 1.0365 to 1.0375). Working in degrees was M0.

Question 9: vectors and skew lines (12 marks)

A and B have position vectors (3, 4, 1) and (4, 3, 1); line l is r = (6, −1, −3) + μ(1, −1, −2).

• (a) Vector equation of m through A and B (2 marks): AB = (1, −1, 0), so r = (3, 4, 1) + λ(1, −1, 0). M1 for a correct method, A1 with “r =” included (writing “m = …” or “AB = …” lost the A1).
• (b) Parallel, intersecting or skew (4 marks): the lines are skew.
  • Working: general points (6 + μ, −1 − μ, −3 − 2μ) and (3 + λ, 4 − λ, 1). The k-components give μ = −2, and then the i-components give λ = 1, but the j-components are 1 and 3, which don’t match. The direction vectors (1, −1, −2) and (1, −1, 0) are not multiples of each other, so the lines are not parallel.
  • Marking: B1 for a line in component form, M1 for equating pairs of components to find a parameter, A1 for showing all three equations can’t be satisfied, A1 for showing the lines are not parallel and concluding they are skew (this needs the three earlier marks). Saying only “the direction vectors are different” was not enough.
• (c) Position vector of P (6 marks): P is on l and angle PAB = 60°.
  • Answer: (3, 2, 3).
  • Working: AP = (3 + μ, −5 − μ, −4 − 2μ) and AP · AB = 8 + 2μ. Using cos 60° = ½: 8 + 2μ = ½ × √2 × √(6μ² + 32μ + 50). Squaring and simplifying gives μ² + 16μ + 39 = 0, so μ = −3 or μ = −13. μ = −13 makes the scalar product negative, so the angle would be 120°, not 60°; μ = −3 gives P = (3, 2, 3).
  • Marking: B1 for AP or PA in component form with the parameter, M1 for the scalar product with AB, M1 for using the moduli and cos 60° = ½, M1* for an equation without square roots, DM1 for solving the quadratic and finding a position vector, A1 for (3, 2, 3) only. The position vector from μ = −13 had to be rejected.

Question 10: partial fractions and a differential equation (11 marks)

• (a) Partial fractions of (x + 3)/((x − 1)(2x − 6)) (4 marks): −1/(x − 1) + 3/(2x − 6). Put x = 1 to get A = 4/(−4) = −1, and x = 3 to get B = 6/2 = 3. B1 for the form A/(x − 1) + B/(2x − 6), M1 for a correct method for a constant, A1 for one, A1 for both.
• (b) Solve the differential equation (7 marks): (x − 1)(2x − 6) dy/dx = (x + 3)/ln(2y), with y = 1 when x = 4.
  • Answer: y ln 2y − y = −ln(x − 1) + (3/2) ln(2x − 6) + ln 3 − ½ ln 2 − 1 (the constant is about −0.2479). Writing ln(x − 3) in place of ln(2x − 6) changes the constant to ln 6 − 1.
  • Working: separate: ∫ln(2y) dy = ∫(x + 3)/((x − 1)(2x − 6)) dx. Use part (a) on the right: −ln(x − 1) + (3/2) ln(2x − 6). Integrate ln(2y) by parts (treat it as 1 × ln 2y): y ln 2y − y. Substituting x = 4, y = 1: ln 2 − 1 = −ln 3 + (3/2) ln 2 + c, so c = ln 3 − ½ ln 2 − 1.
  • Marking: B1 for separating correctly, B1FT for −ln(x − 1) and B1FT for (3/2) ln(2x − 6), both following through from your constants in (a), M1 for starting integration by parts on ln 2y, A1 for y ln 2y − y, M1 for using x = 4, y = 1 to find the constant, A1 for a correct equation in any form.

Where the marks are on 9709/32

• Calculus carried 31 of the 75 marks: parametric differentiation (Question 5, 10), the product rule and integration by parts (Question 6, 10), the quotient rule in Question 7(a) (4) and the differential equation in Question 10(b) (7).
• Vectors carried 12 marks (Question 9).
• Algebra carried 11 marks: polynomial division (Question 1, 3), the exponential equation (Question 2, 4) and partial fractions (Question 10(a), 4).
• Complex numbers carried 9 marks (Questions 3 and 4).
• Numerical solution of equations carried 9 marks (Question 8).
• Trigonometric equations carried 3 marks (Question 7(b)), with more trigonometry inside Questions 5, 6 and 8.

Questions 5 to 10 were worth 59 of the 75 marks, and Questions 4, 5(b) and 6 all asked for exact answers.

How to self-mark 9709/32 with the mark scheme

1. Mark the method before the answer. Each part is built from M, A and B marks. In Question 9(c), the scalar product, the use of cos 60° and the squared equation are three M marks even if an arithmetic slip spoils P.
2. Check the starred marks. In Question 6(b), the two DM1 marks need the M1* for integration by parts the right way round; in Question 7(a), the DM1 for the quadratic in sin 3a needs a correct quotient rule.
3. Apply follow through where the scheme allows it. In Question 10(b), the log terms are B1FT from your partial fractions in (a), so a wrong A or B there costs only the (a) marks.
4. Give no marks for unsupported answers. Question 2 with no working scores 0, and Question 7(b) gives nothing for correct values without working.
5. Treat “show that” parts strictly. Questions 7(a) and 8(c) are AG: an error left uncorrected in 7(a) loses the final A1 even if you reach the printed equation.
6. Check radians and accuracy. Question 7(b) needs radians to at least 3 significant figures (degrees cost one A mark), Questions 8(b) and 8(d) are M0 in degrees, and 8(d) needs iterates to 5 decimal places and the root to 3. Questions 4, 5(b), 6(a) and 6(b) need exact answers with the trigonometry evaluated.
7. Reject the extra solution. Question 5(b) needs only θ = π/8, and Question 9(c) needs only (3, 2, 3). Leaving the other value in costs the final mark.

Sit 9709/32 as a timed mock

The paper gives 110 minutes for 75 marks, just under 1.5 minutes per mark. A workable plan:

Block	Marks	Time
Questions 1 to 4	16	22 minutes
Question 5 (parametric curve)	10	14 minutes
Question 6 (x tan⁻¹ 3x)	10	14 minutes
Question 7 (quotient rule and trig equation)	7	10 minutes
Question 8 (iteration)	9	13 minutes
Question 9 (vectors)	12	17 minutes
Question 10 (partial fractions and differential equation)	11	15 minutes
Check, especially radians, exact forms and rejected roots	–	5 minutes

Questions 9 and 10 come last and carry 23 marks between them, so don’t let the early questions run over. Question 9(c) in particular needs time for the algebra after squaring.

Other May/June 2026 9709 papers

• 9709 Paper 12 (Pure Mathematics 1)
• 9709 Paper 22 (Pure Mathematics 2)
• 9709 Paper 42 (Mechanics)
• 9709 Paper 52 (Probability & Statistics 1)
• 9709 Paper 62 (Probability & Statistics 2)

Practise and download

• Download 9709/32 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 32 May/June 2026? +

Ten questions: polynomial division, an exponential equation, complex numbers in exponential form, square roots of a complex number, a parametric curve with tan 2θ and ln(cos 2θ), differentiating and integrating x tan⁻¹ 3x, a quotient-rule trig equation, iteration with cosec, vectors with skew lines, and a differential equation with partial fractions.

How many marks is 9709/32 and how long is it? +

It is out of 75 marks and lasts 1 hour 50 minutes. Question 9 (vectors) was the biggest at 12 marks, followed by Question 10 (partial fractions and a differential equation) at 11 marks and Questions 5 and 6 at 10 marks each.

What were the square roots of 3 − 6√2 i in 9709/32? +

√6 − √3 i and −√6 + √3 i. Squaring x + iy gives x² − y² = 3 and 2xy = −6√2. Eliminating y gives the quartic x⁴ − 3x² − 18 = 0, so x² = 6. The answer had to be in the form x + iy.

Were the lines skew in 9709/32 Question 9? +

Yes. Equating the k-components gives μ = −2 and the i-components then give λ = 1, but the j-components don't match (1 and 3). The direction vectors (1, −1, −2) and (1, −1, 0) are not multiples of each other, so the lines are not parallel and must be skew.

Is 9709 Paper 3 an AS or A Level paper? +

Paper 3 (Pure Mathematics 3) is an A Level paper. A Level candidates sit it with Paper 1 and either Papers 4 and 5 or Papers 5 and 6.

Where can I download 9709/32 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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