Cambridge AS & A Level Further Mathematics 9231/12 is the May/June 2026 Paper 1 (Further Pure Mathematics 1), variant 2. It lasts 2 hours, is worth 75 marks and is sat by both AS and full A Level candidates. There were seven compulsory questions, answered on the question paper with the MF19 list of formulae. The last two questions carried 31 of the 75 marks: the rational curve in Question 6 (14 marks) and the polar curve in Question 7 (17 marks, 9 of them on the area in part (c)). 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 four 9231 papers from this series, see A Level Further Maths 9231 May/June 2026 past papers: what came up and how it was marked.

9231/12 at a glance

Question	Topic	Marks
1	Series: standard results, method of differences, sum to infinity	8
2	Proof by induction: 3⁶ⁿ + 92ⁿ + 89 divisible by 91	6
3	Matrices: stretch then shear, line of invariant points, area and determinant	9
4	Roots of 5x³ − 2x + 1 = 0: new cubic, sums of powers of the roots	10
5	Vectors: shortest distance between skew lines, angle between two planes	11
6	Rational curve: asymptotes, stationary points, sketches, modulus inequality	14
7	Polar curve: furthest point from the initial line, sketch, area by substitution	17
	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), A marks are for accurate answers and need the M mark before them, and B marks are independent marks for a correct result or statement. A DM mark depends on an earlier starred mark (written *M1), so losing the starred mark loses the DM too. FT (follow through) lets you score using your own earlier answer, CAO means correct answer only, and AG (answer given) marks “show that” results, where every step must be shown because the answer is printed. AWRT means “answer which rounds to”. The cover asks for non-exact answers to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Question 1: series and the method of differences (8 marks)

The general term is (4r − 3)(12r + 3).

• (a) Σ(4r − 3)(12r + 3) as an³ + bn² + cn (3 marks): expand to 48r² − 24r − 9, then use the MF19 results for Σr² and Σr: 48 × ⅙n(n + 1)(2n + 1) − 24 × ½n(n + 1) − 9n = 16n³ + 12n² − 13n. B1 for the expansion, M1 for substituting both formulae, and A1 (CAO) for the answer in the required form.
• (b) Σ 1/((4r − 3)(12r + 3)) (4 marks): since 12r + 3 = 3(4r + 1), the partial fractions are (1/12)(1/(4r − 3) − 1/(4r + 1)). Writing out the terms, (1/12)(1/1 − 1/5), (1/12)(1/5 − 1/9), … , (1/12)(1/(4n − 3) − 1/(4n + 1)), everything cancels except the first and last fractions: 1/12 − 1/(12(4n + 1)), in any equivalent form. B1 for the partial fractions; M1 for writing enough terms to show what is left plus one pair that cancels; A1 for all the terms needed to show the cancellation (two at the start and one at the end, or one at the start and two at the end); B1 for the answer.
• (c) Sum to infinity (1 mark): as n → ∞ the second term vanishes, so the sum is 1/12. B1 FT from an answer to (b) of a similar form.

Question 2: proof by induction (6 marks)

Prove that 3⁶ⁿ + 92ⁿ + 89 is divisible by 91 for every positive integer n.

• Base case (B1): n = 1 gives 729 + 92 + 89 = 910 = 10 × 91.
• Hypothesis (B1): assume the result is true for some positive integer k. Assuming it for all values of k scored B0.
• Inductive step (M1, A1, A1): write 3⁶⁽ᵏ⁺¹⁾ + 92ᵏ⁺¹ + 89 as (728 + 1)3⁶ᵏ + (91 + 1)92ᵏ + 89 (M1), then separate it as 728 × 3⁶ᵏ + 91 × 92ᵏ + (3⁶ᵏ + 92ᵏ + 89) (A1). Both 728 = 8 × 91 and 91 are multiples of 91, and the bracket is divisible by 91 by the hypothesis, so the whole expression is (A1 for linking this to the hypothesis).
• Conclusion (A1): state the result in full for every positive integer n. This mark needed the previous three marks.
• Alternative: subtracting consecutive terms gives 728 × 3⁶ᵏ + 91 × 92ᵏ = 91(8 × 3⁶ᵏ + 92ᵏ), which was marked the same way.

Question 3: a stretch followed by a shear (9 marks)

M represents a stretch parallel to the x-axis with scale factor k (k ≠ 0, 1), followed by a shear with the x-axis fixed and (0, 1) mapped to (2, 1).

• (a) Show that M = (k 2; 0 1) (4 marks): the stretch is (k 0; 0 1) (B1) and the shear is (1 2; 0 1) (B1). The second transformation goes on the left, so M = (1 2; 0 1)(k 0; 0 1) (M1 for the correct order). AG: the A1 needed an intermediate step, with at least one element shown before its final form.
• (b) Line of invariant points (3 marks): M(x, y) = (kx + 2y, y) (B1, with the matrix applied to both coordinates). For a fixed point, kx + 2y = x (M1), so y = ½(1 − k)x (A1).
• (c) Values of k (2 marks): area of P = |det M| × area of S, and det M = k. Since S is half of P, |k| = 2, so k = ±2. M1 for using det M = k with P and S the right way round; A1 CAO.

Question 4: roots of a cubic (10 marks)

5x³ − 2x + 1 = 0 has roots α, β, γ.

• (a) Cubic with roots 1/α², 1/β², 1/γ² (3 marks): substitute y = 1/x², so x = y^(−½). This gives 5y^(−3/2) − 2y^(−½) + 1 = 0, or y^(3/2) = 2y − 5. Squaring to remove the fractional power gives y³ − 4y² + 20y − 25 = 0. B1 for the substitution, M1 for squaring, A1 for an equation with like terms collected. An alternative was to square (5x² − 2)x = −1 first.
• (b) Sum of 1/α⁴ (3 marks): from the new cubic, the sum of the 1/α² terms is 4 and the sum of their pairwise products is 20 (B1 FT from your cubic, provided it has positive powers and like terms collected). Then Σ1/α⁴ = 4² − 2 × 20 = −24 (M1, A1 FT).
• (c) Sums of 1/α⁶ and 1/α⁸ (4 marks): each root of the new cubic satisfies y³ = 4y² − 20y + 25, so S₋₆ = 4S₋₄ − 20S₋₂ + 75 (using S₀ = 3) = −96 − 80 + 75 = −101. Multiplying by y again, S₋₈ = 4S₋₆ − 20S₋₄ + 25S₋₂ = −404 + 480 + 100 = 176. M1 and A1 (CAO) for each value. The constant term 75 comes from 25 × 3, so forgetting S₀ = 3 is the trap.

Question 5: skew lines and planes (11 marks)

l₁ passes through (−2, 4, 4) with direction 3i − 4j + 2k; l₂ passes through (−3, −2, 4) with direction 2i + 4j + 3k.

• (a) Shortest distance (5 marks): the vector between the two given points is (−1, −6, 0) (B1). The common perpendicular is the cross product of the directions, (−20, −5, 20), or (−4, −1, 4) (M1, A1). The distance is |(−1, −6, 0) · (−4, −1, 4)| ÷ √33 = 10/√33 (about 1.74, also accepted). M1 for the formula with your own vectors; the scalar product and magnitude had to be worked out, not just written down. An alternative method found the points on each line where the joining vector is perpendicular to both lines.
• (b) Acute angle between the planes (6 marks): P₁ contains l₁ and P(1, −1, 1), so its normal is the cross product of (−3, 5, 3) and (3, −4, 2): (22, 15, −3). P₂ contains l₂ and the origin, so its normal is (−3, −2, 4) × (2, 4, 3) = (−22, 17, −8). The scalar product is −205, so cos θ = −205 ÷ √(718 × 837), giving 105.3°, and the acute angle is 74.7° (or 1.30 radians). *M1 A1 for each normal, DM1 for the scalar product with the magnitudes evaluated, A1 for the final answer (AWRT 74.7°).

Question 6: a rational curve and a modulus inequality (14 marks)

C has equation y = (x² − x + 2)/(4x + 4).

• (a) Asymptotes (3 marks): x = −1 (B1). Writing the numerator as (x + 1)(x − 2) + 4 gives y = ¼x − ½ + 1/(x + 1), so the oblique asymptote is y = ¼x − ½ (M1 for an attempt that finds the constant term, A1 for a linear equation).
• (b) Stationary points (3 marks): set dy/dx = 0. The numerator of the derivative is 4x² + 8x − 12, so x = 1 or x = −3, giving (1, ¼) and (−3, −7/4). M1 for setting the derivative to zero, A1 for both x-coordinates, A1 for the matching y-coordinates (or A1 for one completely correct point).
• (c) Sketch C (3 marks): B1 FT for the axes and both asymptotes labelled (follow through applies to this mark only), B1 for the left branch with the right approaches to the asymptotes, and B1 for the right branch with a smooth minimum in the first quadrant.
• (d) Sketch y = |(x² − x + 2)/(4x + 4)| (1 mark): reflect the left branch, which lies below the x-axis, in the x-axis. B1 needed the sketch fully correct, including the vertical asymptote.
• (e) Solve |(x² − x + 2)/(4x + 4)| < 2 in exact form (4 marks): set the fraction equal to −2 and to 2. These give x² + 7x + 10 = 0, so x = −5 or −2, and x² − 9x − 6 = 0, so x = ½(9 ± √105). M2 (M1 for each equation), A1 for all four exact critical values, then A1 (CAO) for −5 < x < −2 or ½(9 − √105) < x < ½(9 + √105). The sketch in (d) shows which intervals to take.

Question 7: a polar curve (17 marks)

C has polar equation r² = a/(1 + tan 2θ) for 0 ≤ θ < ¼π, where a is a positive constant.

• (a) Point furthest from the initial line, and a sign change (6 marks): the distance from the initial line is y = r sin θ = √a sin θ (1 + tan 2θ)^(−½) (B1). Differentiating with the product or quotient rule (*M1, A1) and setting dy/dθ = 0 (DM1) gives cos θ(1 + tan 2θ) − sin θ sec² 2θ = 0. Since sec² 2θ ÷ (1 + tan 2θ) simplifies to 1/(cos 2θ(sin 2θ + cos 2θ)), this is the printed equation (A1, AG, with intermediate steps shown). For the root, the left-hand side is 0.235 at θ = 0.5 and −0.379 at θ = 0.6, so it changes sign (B1, values to at least one decimal place). An alternative was to differentiate y² instead of y.
• (b) Sketch C (2 marks): the curve starts at (√a, 0) on the initial line and returns to the pole O as θ approaches ¼π, staying in the first quadrant. B1 for the initial line, the pole and a first-quadrant curve through (√a, 0) and O; B1 for the shape, roughly symmetrical, with its highest point between 40% and 60% of the way across the arc.
• (c) Area by the substitution v = tan 2θ (9 marks):
  • The area is ½∫r² dθ from 0 to ¼π, which is (a/2)∫ 1/(1 + tan 2θ) dθ (B1 for the formula and limits).
  • dv/dθ = 2 sec² 2θ (M1), and sec² 2θ = 1 + v² (M1). The limits become 0 and ∞, so the area is (a/4)∫₀^∞ 1/((1 + v)(1 + v²)) dv (A1, AG: some working had to be shown, including the change of limits).
  • Partial fractions: 1/((1 + v)(1 + v²)) = ½(1/(v + 1) − (v − 1)/(v² + 1)) (*M1 for the correct form), separated as ½(1/(v + 1) − v/(v² + 1) + 1/(v² + 1)) (A1).
  • Integrating with the three MF19 results gives (a/8)[ln(v + 1) − ½ ln(v² + 1) + tan⁻¹ v] from 0 to ∞ (DM1, A1).
  • Combining the logarithms as ln((v + 1)/√(v² + 1)), which tends to 0, the area is πa/16 (A1). This last mark needed the four previous marks, and the logarithms had to be combined, because taken separately neither has a finite limit as v → ∞.

Where the marks are on 9231/12

• Polar coordinates: 17 marks (Question 7), including 9 for the area.
• Rational functions and graphs: 14 marks (Question 6): asymptotes, stationary points, two sketches and a modulus inequality.
• Vectors: 11 marks (Question 5).
• Roots of polynomial equations: 10 marks (Question 4).
• Matrices: 9 marks (Question 3).
• Series: 8 marks (Question 1).
• Proof by induction: 6 marks (Question 2).

That is 75 marks in total. “Show that” results with an AG mark appeared in Questions 3(a), 7(a) and 7(c), and sketches carried 6 marks across Questions 6(c), 6(d) and 7(b).

How to self-mark 9231/12 with the mark scheme

1. Mark each part line by line, not just the final answer. Every question breaks into M, A and B marks, so a wrong final answer can still be worth most of the part. In Question 5(b), finding the two normals alone is worth 4 marks.
2. Check the dependent marks. In Question 7(c), the DM1 for integrating depends on the starred M1 for partial fractions of the correct form, and the final A1 needs all four marks before it. In Question 7(a), setting the derivative to zero (DM1) needs the differentiation mark.
3. Be strict with AG parts. Questions 3(a), 7(a) and 7(c) print the result. Give yourself the A mark only if your working shows an intermediate step: in Question 3(a), the product of the matrices before simplifying; in Question 7(c), the change of limits.
4. Apply follow through where the scheme allows it. Question 4(b) follows through from your cubic in 4(a), but Question 4(c) is CAO. In Question 6(c), only the asymptote mark follows through.
5. Check the induction wording. In Question 2, “assume true for all k” loses the hypothesis mark, and the final A1 needs the result stated in full.
6. Check exact and rounded answers. Questions 6(e) and 7(c) need exact answers (√105 and πa/16). Question 5(a) accepts 10/√33 or 1.74, and Question 5(b) accepts 74.7° or 1.30 radians.
7. Log each lost mark as a missing method, an algebra slip, a missing step in a “show that”, or a sketch detail, and revise in that order.

Sit 9231/12 as a timed mock

The paper gives 120 minutes for 75 marks, about 1.6 minutes per mark. A workable plan:

Block	Marks	Time
Question 1 (series)	8	12 minutes
Question 2 (induction)	6	9 minutes
Question 3 (matrices)	9	13 minutes
Question 4 (roots)	10	14 minutes
Question 5 (vectors)	11	16 minutes
Question 6 (rational curve)	14	22 minutes
Question 7 (polar curve)	17	28 minutes
Check, especially AG steps and exact answers	–	6 minutes

Question 7 deserves the most time per mark: part (c) has several stages after the substitution, and the final mark depends on all of them.

Other May/June 2026 9231 papers

• 9231 Paper 22 (Further Pure Mathematics 2)
• 9231 Paper 32 (Further Mechanics)
• 9231 Paper 42 (Further Probability & Statistics)

Practise and download

• Download 9231/12 and its mark scheme on the Learning Portal →
• All 9231 May/June 2026 papers: overview and marking patterns
• Try a free A Level Further Maths quiz

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Frequently asked questions

What topics were on 9231 Paper 12 May/June 2026? +

Seven questions: a series using standard results and the method of differences, proof by induction of divisibility by 91, a stretch and shear matrix, roots of a cubic, skew lines and planes, a rational curve with a modulus inequality, and a polar curve with an area found by substitution.

How many marks is 9231/12 and how long is it? +

It is out of 75 marks and lasts 2 hours, with every question compulsory. Question 7 on the polar curve is the biggest at 17 marks, followed by Question 6 on the rational curve at 14 marks.

What was the area in 9231/12 Question 7(c)? +

πa/16. After the substitution v = tan 2θ, the area is (a/4) times the integral of 1/((1 + v)(1 + v²)) from 0 to infinity. Partial fractions and the MF19 results give (a/8)[ln((v + 1)/√(v² + 1)) + tan⁻¹ v] from 0 to infinity, which is πa/16.

What was the answer to the modulus inequality in 9231/12 Question 6(e)? +

−5 < x < −2 or ½(9 − √105) < x < ½(9 + √105). The critical values come from setting the fraction equal to −2 and to 2, which give x² + 7x + 10 = 0 and x² − 9x − 6 = 0. The answers had to be exact.

Is 9231 Paper 1 an AS or A Level paper? +

Both. Paper 1 Further Pure Mathematics 1 is sat by AS candidates, together with Paper 3 or Paper 4, and by full A Level candidates, who sit all four papers.

Where can I download 9231/12 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 Further Mathematics 9231 past papers page.

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