Cambridge AS & A Level Mathematics 9709/12 is the May/June 2026 Paper 1 (Pure Mathematics 1), variant 2. It is the AS Level pure paper that every 9709 candidate sits, lasts 1 hour 50 minutes and is worth 75 marks. It had ten compulsory questions, answered in an answer booklet. The first six questions were worth 3 to 7 marks each; the big marks came at the end, with the progressions in Question 7 (10 marks), the circle in Question 8 (11 marks), the area and normal in Question 9 (11 marks) and the quadratic function in Question 10 (9 marks). 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.

9709/12 at a glance

Question	Topic	Marks
1	An equation in x² and 1/x², solved as a quadratic in x²	3
2	Graph transformations from a diagram, then g(x) in the form af(x + b) + c	7
3	Perpendicular bisector meeting a given line	6
4	Binomial expansions and the coefficient of x² in their product	7
5	A curve from its second derivative: the third stationary point	7
6	An inverse cosine equation leading to a quadratic in sin x	4
7	Arithmetic progression sum; geometric progression with negative k and its sum to infinity	10
8	Circle from a diameter; exact length of a chord	11
9	Area under y = 3√(2x + 1); the normal at R and a triangle of equal area	11
10	Quadratic function: repeated roots, domain for an inverse, the inverse	9
	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 and DB 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, and warns that unsupported calculator answers get no marks.

Question 1: a quadratic in x² (3 marks)

The equation had terms in x², a constant and 16/x².

• Answer: x = ±2.
• Working: multiply through by x² to get 3x⁴ − 8x² − 16 = 0, which factorises as (3x² + 4)(x² − 4) = 0. So x² = 4 or x² = −4/3, and only x² = 4 gives real solutions.
• Marking: M1 for a three-term quadratic in x² (a substitution such as u = x² is fine, or factorising the original equation directly), A1 for x² = 4, A1 for x = ±2 only. Complex solutions were ignored. If you use a substitution, define it, although a correct final answer implied it.

Question 2: graph transformations (7 marks)

The diagram showed y = f(x) as solid line segments through (−1, −10), (2, −4), (4, −6) and (6, −6), and y = g(x) as broken segments through (1, 0), (3, 3), (5, 2) and (7, 2).

• (a) Describe the transformations (5 marks): g is f moved 1 to the right and then squashed vertically by a factor of a half, with a vertical move. Two orders were accepted:
  • Translation by the vector (1, 10), then a stretch with factor 0.5 parallel to the y-axis, or
  • a stretch with factor 0.5 parallel to the y-axis, then a translation by (1, 5). B2 for the three parts of the translation (B1 for two), B2 for the three parts of the stretch (type, factor 0.5, direction), and B1 for the correct order. The vertical translation is 10 if it comes before the stretch and 5 if it comes after. Words like “factor” for a translation, or just “right” or “up”, did not earn the translation marks; a column vector was enough on its own.
• (b) g(x) in the form af(x + b) + c (2 marks): g(x) = 0.5f(x − 1) + 5, so a = ½, b = −1, c = 5. B2 for all three, B1 for two. B1 was also given for 0.5{f(x − 1) + k}.

Question 3: perpendicular bisector (6 marks)

The bisector of the line joining (4, 3) and (−2, 6) meets x − 4y − 4 = 0 at P.

• Answer: P = (−2, −3/2).
• Working: midpoint (1, 9/2); gradient of the line (6 − 3)/(−2 − 4) = −½, so the bisector has gradient 2 and equation y = 2x + 5/2. From the given line x = 4y + 4, so y = 2(4y + 4) + 5/2, giving y = −3/2 and x = −2.
• Marking: M1* for the midpoint, M1* for the gradient and M1* for the negative reciprocal (one sign slip condoned in each), A1 for the bisector’s equation, DM1 for forming an equation in one variable, A1 for P. An alternative route equating the distances from (x, y) to the two points, then expanding so the squared terms cancel, earned the same marks.

Question 4: binomial expansions (7 marks)

• (a)(i) First three terms of (3 − ½x)⁴ (2 marks): 81 − 54x + (27/2)x². B1 for the second or third term correct and simplified, B1 for all three. Equivalent fractions or decimals for 27/2 were allowed.
• (a)(ii) First three terms of (2 + ax)³ (1 mark): 8 + 12ax + 6a²x².
• (b) Find a (4 marks): the coefficient of x² in the product of the two expansions is −108.
  • Answer: a = 2/3.
  • Working: the x² terms come from three products: 81 × 6a², (−54) × 12a and (27/2) × 8. So 486a² − 648a + 108 = −108, which simplifies to 9a² − 12a + 4 = 0, that is (3a − 2)² = 0.
  • Marking: M1* for selecting the three right products, DM1 for adding them and equating to −108, A1 for the quadratic, A1 for a = 2/3 only.

Question 5: a curve from its second derivative (7 marks)

You were given d²y/dx² = 6x² + p and two minimum points, (2, 3) and (−2, 3).

• Answer: the other stationary point is (0, 11).
• Working: integrate once: dy/dx = 2x³ + px + q. Both minima are stationary points, so 16 + 2p + q = 0 and −16 − 2p + q = 0, giving p = −8 and q = 0. Then dy/dx = 2x³ − 8x = 2x(x² − 4), which is also zero at x = 0. Integrate again: y = ½x⁴ − 4x² + r, and (2, 3) gives 3 = 8 − 16 + r, so r = 11.
• Marking: B1 for 2x³ + px + q, M1* for substituting x = 2 and x = −2 and setting both to zero, A1 for p = −8 and q = 0, A1 for x = 0 (only if the first three marks were scored), DM1 for integrating again, DM1 for substituting (2, 3) or (−2, 3) with the constant present, A1 for (0, 11). Every mark after the B1 depended on the M1*, so leaving out the constant q or not using both minimum points cost almost the whole question.

Question 6: an inverse cosine equation (4 marks)

The equation set cos⁻¹ of (sin²x + (3/2) sin x) equal to (2/3)π, for 0 ≤ x ≤ 2π.

• Answer: x = 7π/6, 3π/2, 11π/6.
• Working: take the cosine of both sides: sin²x + (3/2) sin x = cos(2π/3) = −½. Multiply by 2: 2sin²x + 3 sin x + 1 = 0, so (2 sin x + 1)(sin x + 1) = 0, giving sin x = −½ or sin x = −1.
• Marking: B1 for sin²x + (3/2) sin x = cos(2π/3) or better, B1* for sin x = −½ and −1, DB1 for any one correct angle and DB1 for all three with no others in the range. Decimals that round to 3.67, 4.71 and 5.76 were accepted. Answers in degrees (210°, 270°, 330°) earned only a special case mark (SC), and correct final answers on their own did not imply full marks.

Question 7: progressions (10 marks)

• (a) Arithmetic progression (4 marks): the 2nd, 4th and last terms are 60, 52 and −28.
  • Answer: 432.
  • Working: two steps of d take 60 to 52, so d = −4 and a = 64. Then 64 − 4(n − 1) = −28 gives n = 24, and the sum is (24/2)(64 + (−28)) = 12 × 36 = 432.
  • Marking: B1 for d = −4, M1* for finding n with the nth-term formula, DM1 for a correct sum formula, A1 for 432. If your equation for n didn’t give a positive whole number, only the first M1 was available.
• (b)(i) Find k (3 marks): the 2nd, 3rd and 4th terms of a geometric progression are 9, k + 4 and k², with k negative.
  • Answer: k = −1.
  • Working: the common ratio gives (k + 4)/9 = k²/(k + 4), so (k + 4)² = 9k² and k + 4 = ±3k. That gives k = 2 or k = −1, and k must be negative.
  • Marking: B1 for an equation in k only, B1 for a correct quadratic such as 8k² − 8k − 16 = 0 (or k + 4 = ±3k, with the ±), B1 for k = −1. The last mark was withheld if k = 2 was also given as an answer.
• (b)(ii) Sum to infinity (3 marks): with k = −1 the terms are 9, 3, 1, so r = ⅓ and the first term is 27. S∞ = 27 ÷ (1 − ⅓) = 81/2 (40.5). B1 for r = ⅓ (or a = 27), M1 for using the sum to infinity formula with your values (only allowed when |r| < 1), A1 for 40.5.

Question 8: circle and chord (11 marks)

A diameter has end-points (−2, 4) and (8, 8).

• (a) Equation of the circle (4 marks): centre at the midpoint (3, 6); radius² = 5² + 2² = 29. (x − 3)² + (y − 6)² = 29 (or x² + y² − 6x − 12y + 16 = 0). B1 for the centre, B1 for the radius, M1A1 for the equation. Writing the right-hand side as (√29)² lost the A1.
• (b) Exact length PQ (7 marks): the line y = x + 6 cuts the circle at P and Q, and the answer had to be in the form a√b.
  • Answer: PQ = 7√2.
  • Working: substitute: (x − 3)² + x² = 29, so 2x² − 6x − 20 = 0, or x² − 3x − 10 = 0, giving x = −2 and 5. The points are (−2, 4) and (5, 11), and PQ² = 7² + 7² = 98.
  • Marking: M1 for a quadratic in one variable, M1 for simplifying to three terms, A1 for 2x² − 6x − 20 = 0 (or 2y² − 30y + 88 = 0), A1 for the x-values, A1 for the y-values, M1 for Pythagoras with your points, A1 for 7√2 (a = 7, b = 2 only).

Question 9: area and the normal (11 marks)

The diagram showed y = 3√(2x + 1), the normal at R(4, 9) and a horizontal line y = a. The shaded region lay under the curve between the y-axis and x = 4.

• (a) Area of the shaded region (4 marks): ∫₀⁴ 3(2x + 1)^½ dx = [(2x + 1)^(3/2)] from 0 to 4 = 27 − 1 = 26.
  • B2 for integrating to (2x + 1)^(3/2) (B1 for the right power with the wrong coefficient), M1 for using the limits 0 and 4, A1 for 26. Using “27 − 0” for the limits was M0.
  • An alternative was to integrate with respect to y: the 4 × 9 rectangle (36) minus ∫₃⁹ (y²/18 − ½) dy, which also gives 26.
  • An answer with no working scored 0 out of 4, and working out a different area afterwards was not ignored.
• (b) Exact value of a (7 marks): the triangle between the y-axis, the normal and y = a has area 26.
  • Answer: a = 13 − 2√13 (or 13 − √52).
  • Working: dy/dx = 3(2x + 1)^(−½), which is 1 at x = 4. The normal has gradient −1, so y = −x + 13. The triangle has vertices (0, 13), (0, a) and (13 − a, a), so its area is ½(13 − a)². Setting this to 26 gives (13 − a)² = 52, so a = 13 ± √52. The line y = a in the diagram is below the y-intercept 13, so a = 13 − √52.
  • Marking: B2* for the derivative 3(2x + 1)^(−½) (B1 for k(2x + 1)^(−½)), DB1 for the gradient 1, M1 for the normal (or the y-intercept 13), B1 for the triangle’s area in terms of a, M1 for equating it to your answer from (a) (this M1 did not depend on the earlier marks), A1 for 13 − √52. Giving 13 + √52 as well lost the A1.

Question 10: a quadratic function and its inverse (9 marks)

f(x) = px² − 2x + p + 4 for all real x; g is the same expression with domain x ≥ a, where p and a are positive.

• (a) Repeated roots (3 marks): b² − 4ac = 0 gives 4 − 4p(p + 4) = 0, so p² + 4p − 1 = 0 and (p + 2)² = 5. p = −2 ± √5. B1* for the discriminant (or completing the square), M1 for solving the quadratic in p, DB1 for the exact values.
• (b)(i) Show that the least a is 1/p (2 marks, AG): g′(x) = 2px − 2 = 0 at x = 1/p, which is the minimum point because p > 0, so g is one-one only from x = 1/p onwards. B1* for 2px − 2 or the completed square p(x − 1/p)² + …, DB1 for reaching x = 1/p with no errors. Stating that the vertex is at (1/p, …) or a sketch with the minimum labelled was also accepted.
• (b)(ii) g⁻¹(x) when p = 2 (4 marks): g(x) = 2x² − 2x + 6 = 2(x − ½)² + 11/2. Set y equal to this and rearrange: (x − ½)² = (y − 11/2)/2, so x = ½ + √((y − 11/2)/2). The positive root is taken because the domain is x ≥ ½.
  • Answer: g⁻¹(x) = ½ + √((2x − 11)/4), or equivalently (1 + √(2x − 11))/2.
  • Marking: M1* for completing the square (or setting up the quadratic formula), DM1* for solving as far as (x − ½)² = …, DM1 for an expression for x (± allowed at this stage), A1 for the final answer. The final answer had to be written as g⁻¹(x) = …; an answer left as “x = …” or “g(x) = …” lost the A1.

Where the marks are on 9709/12

• Functions and quadratics carried 19 of the 75 marks: the quadratic in x² (Question 1, 3), graph transformations (Question 2, 7) and the quadratic function and inverse (Question 10, 9).
• Calculus carried 18 marks: the curve from its second derivative (Question 5, 7) and the area and normal (Question 9, 11).
• Coordinate geometry carried 17 marks: the perpendicular bisector (Question 3, 6) and the circle and chord (Question 8, 11).
• Series carried 17 marks: binomial expansions (Question 4, 7) and arithmetic and geometric progressions (Question 7, 10).
• Trigonometry carried 4 marks (Question 6).

The last four questions were worth 41 of the 75 marks. Questions 8 and 9 alone carried 22.

How to self-mark 9709/12 with the mark scheme

1. Mark the method before the answer. Each part is built from M, A and B marks, so a wrong final answer can still earn most of the marks. In Question 8(b), forming and simplifying the quadratic and using Pythagoras are three M marks even if your coordinates are wrong.
2. Check the starred marks first. If you missed the M1* in Question 5 (substituting both x = 2 and x = −2 into your first integral and setting it to zero), only the opening B1 can count. In Question 4(b), the DM1 needs the M1* for picking the right three products.
3. Give no marks for unsupported answers. Question 9(a) with no working scores 0 out of 4, and correct final angles alone in Question 6 do not give full marks.
4. Treat “show that” parts strictly. Question 10(b)(i) is AG: you must reach x = 1/p from a correct derivative or completed square, with no errors.
5. Check for extra answers. k = 2 in Question 7(b)(i), 13 + √52 in Question 9(b) and anything outside 0 ≤ x ≤ 2π in Question 6 each cost the final mark if left in.
6. Check the form of the answer. Question 6 wanted radians (degrees earned only a special case mark), Question 8(b) wanted a√b with a and b greater than 1, Question 9(b) and Question 10(a) wanted exact values, and Question 10(b)(ii) had to be written as g⁻¹(x) = ….
7. Log each lost mark as a missing method, an algebra slip, a missed condition (negative k, positive a, domain) or an answer in the wrong form, then revise in that order.

Sit 9709/12 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 3	16	21 minutes
Questions 4 to 6	18	25 minutes
Question 7 (progressions)	10	14 minutes
Question 8 (circle and chord)	11	16 minutes
Question 9 (area and normal)	11	16 minutes
Question 10 (quadratic and inverse)	9	12 minutes
Check, especially exact forms and extra roots	–	6 minutes

Question 9 needs the most care per mark: part (b) has several linked steps (derivative, normal, triangle area), and it uses your answer to part (a).

Other May/June 2026 9709 papers

• 9709 Paper 22 (Pure Mathematics 2)
• 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 9709/12 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

Need help with a 9709 topic? Book a free trial lesson with an A Level Maths tutor →

Frequently asked questions

What topics were on 9709 Paper 12 May/June 2026? +

Ten questions: an equation in x² and 1/x², graph transformations, a perpendicular bisector, binomial expansions, a curve found from its second derivative, an inverse cosine equation, arithmetic and geometric progressions, a circle and a chord, an area under a curve with a normal, and a quadratic function with its inverse.

How many marks is 9709/12 and how long is it? +

It is out of 75 marks and lasts 1 hour 50 minutes. Questions 8 (the circle) and 9 (the area and the normal) were the biggest at 11 marks each, followed by Question 7 (progressions) at 10 marks.

What was the answer to the circle question in 9709/12? +

The circle is (x − 3)² + (y − 6)² = 29, with centre (3, 6) at the midpoint of the diameter. The line y = x + 6 meets it at (−2, 4) and (5, 11), so PQ = √98 = 7√2.

How do you find the other stationary point in 9709/12 Question 5? +

Integrate 6x² + p to get dy/dx = 2x³ + px + q, then set it to zero at x = 2 and x = −2 to get p = −8 and q = 0. So dy/dx = 2x³ − 8x, which is also zero at x = 0. Integrate again and use (2, 3) to find the constant: the other stationary point is (0, 11).

Is 9709 Paper 1 an AS or A Level paper? +

Paper 1 (Pure Mathematics 1) is an AS Level component, and every 9709 candidate sits it. AS candidates pair it with Paper 2, 4 or 5; A Level candidates add Paper 3 and two of the mechanics and statistics papers.

Where can I download 9709/12 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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