Cambridge AS & A Level Further Mathematics 9231/22 is the May/June 2026 Paper 2 (Further Pure Mathematics 2), variant 2. It lasts 2 hours, is worth 75 marks and is sat by full A Level candidates. There were eight compulsory questions, answered on the question paper with the MF19 list of formulae. The biggest was Question 8 on matrices (14 marks), followed by Question 3 on a hyperbolic integral and a differential equation (11 marks); Question 6, a second-order differential equation, was 10 marks with no parts. 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.

9231/22 at a glance

Question	Topic	Marks
1	Maclaurin series for sinh⁻¹(x + 1) with exact constants	5
2	Reduction formula for ∫(1 + t)ⁿ sin πt dt, then I₂	7
3	∫sinh³ x dx by substitution; first-order differential equation with coth x	11
4	Implicit differentiation: first and second derivatives at (−1, −2)	8
5	Upper and lower rectangle bounds for an integral, and their limit	10
6	Second-order differential equation with initial conditions	10
7	Geometric series in z, roots of 1 + z + … + z⁶ = 0, de Moivre’s theorem	10
8	Systems of equations, characteristic equation, diagonalisation for A⁷	14
	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, CWO means correct working only, OE means “or equivalent”, and AG (answer given) marks “show that” results, where every step must be shown because the answer is printed. The cover asks for non-exact answers to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise. Most answers on this paper were exact.

Question 1: Maclaurin series for sinh⁻¹(x + 1) (5 marks)

The answer had to be in the form ln a + bx + cx² with exact constants.

• Answer: sinh⁻¹(x + 1) = ln(1 + √2) + x/√2 − x²/(4√2).
• Working: f′(x) = (1 + (x + 1)²)^(−½) and f″(x) = −(1 + (x + 1)²)^(−3/2)(x + 1). At x = 0, f(0) = sinh⁻¹ 1 = ln(1 + √2), f′(0) = 1/√2 and f″(0) = −1/(2√2), which is halved for the x² term.
• Marking: B1 for f′(x), B1 for f″(x) (unsimplified or implicit forms, such as y′ cosh y = 1, were allowed), B1 for sinh⁻¹ 1 = ln(1 + √2), M1 for evaluating both derivatives at zero, and A1 (CWO) for the series.

Question 2: a reduction formula (7 marks)

Iₙ is the integral of (1 + t)ⁿ sin πt from 0 to 1.

• (a) Show that π²Iₙ = π(2ⁿ + 1) − n(n − 1)Iₙ₋₂ for n ≥ 2 (4 marks): integrate by parts, integrating sin πt and differentiating (1 + t)ⁿ. The boundary term [−(1/π)(1 + t)ⁿ cos πt] from 0 to 1 is (2ⁿ + 1)/π, leaving (n/π)∫(1 + t)ⁿ⁻¹ cos πt dt. Integrating by parts again, the new boundary term is zero because sin π = sin 0 = 0, leaving −(n(n − 1)/π²)Iₙ₋₂. Multiply through by π². *M1 A1 for the first integration by parts, DM1 for the second (both need the correct trigonometric functions and powers of (1 + t)), and A1 (AG, CWO), which needed the limits and correct brackets.
• (b) Exact value of I₂ (3 marks): I₀ = ∫ sin πt dt from 0 to 1 = 2/π (B1). Then π²I₂ = π(2² + 1) − 2 × 1 × I₀ = 5π − 4/π (M1), so I₂ = 5/π − 4/π³ (A1).

Question 3: a hyperbolic integral and a differential equation (11 marks)

• (a) ∫sinh³ x dx using u = cosh x (4 marks): du = sinh x dx (B1), and sinh² x = cosh² x − 1 = u² − 1, so the integral becomes ∫(u² − 1) du (M1 A1). The answer is ⅓cosh³ x − cosh x (+ C) (A1).
• (b) Solve dy/dx + y coth x = sinh² x with y = 0 when x = ln ⅓ (7 marks):
  • The integrating factor is e^(∫coth x dx) = e^(ln sinh x) = sinh x (M1 A1).
  • So d/dx(y sinh x) = sinh³ x (M1 for the correct left-hand side and an attempt to integrate the right).
  • Using part (a), y sinh x = ⅓cosh³ x − cosh x + C (*M1 A1). The M1 needed your answer to (a) to be of the form a cosh³ x + b cosh x with a and b non-zero.
  • At x = ln ⅓, cosh x = 5/3, so 0 = ⅓(5/3)³ − 5/3 + C, giving C = 10/81 (DM1). This was M0 if the expression had been wrongly rearranged before substituting, for example by not dividing every term by sinh x.
  • Answer: y = cosech x (⅓cosh³ x − cosh x + 10/81), or equivalent (A1). It had to be given as y = f(x).

Question 4: implicit differentiation (8 marks)

C has equation (x − y)⁵ + ln(xy) = 1 + ln 2, and (−1, −2) lies on it.

• (a) Show that dy/dx = 8/11 at (−1, −2) (3 marks): differentiating gives 5(x − y)⁴(1 − y′) (B1) and y′/y + 1/x for ln(xy) (B1). At (−1, −2), x − y = 1, so 5(1 − y′) − ½y′ − 1 = 0, giving y′ = 8/11 (B1, AG, CWO).
• (b) d²y/dx² at (−1, −2) (5 marks): differentiate again. The first term gives −5(x − y)⁴y″ + 20(1 − y′)²(x − y)³ (B1 B1), and y′y⁻¹ + x⁻¹ gives −(y′)²y⁻² + y⁻¹y″ − x⁻² (B1). Substituting (−1, −2) and y′ = 8/11 (M1) gives −(11/2)y″ + 43/121 = 0, so d²y/dx² = 86/1331 (A1). Two alternatives, differentiating a rearranged equation for y′ or using the quotient rule on y′ as a fraction, were marked the same way and reach the same value.

Question 5: rectangle bounds for an integral (10 marks)

The diagram showed y = (11/10)x³ + 1/10 for 0 ≤ x ≤ 1 with n rectangles of width 1/n. The curve is increasing, so rectangles whose heights are taken at the right-hand ends lie above the curve, and those using the left-hand ends lie below it.

• (a) Show the upper bound Uₙ = (11/40)(1 + 1/n)² + 1/10 (4 marks): the sum of the rectangle areas is (1/n)Σ((11/10)(r/n)³ + 1/10) for r = 1 to n, which is (11/(10n⁴))Σr³ + 1/10. Using the MF19 result for Σr³ gives 11n²(n + 1)²/(40n⁴) + 1/10, the printed answer. M1 A1 for forming the sum of n complete terms (A1 for at least three correct terms including the first and last), M1 for applying Σr³, A1 (AG, CWO).
• (b) A lower bound Lₙ (4 marks): use r = 0 to n − 1. The first rectangle has area 1/(10n), and the sum is 1/10 + (11/(10n⁴))Σr³ for r = 1 to n − 1, so Lₙ = (11/40)(1 − 1/n)² + 1/10, or an equivalent form such as (11/40)(1 + 1/n)² + 1/10 − 11/(10n). Marked M1 A1, M1, A1 as in (a).
• (c) Show that Uₙ − Lₙ → 0 (2 marks): Uₙ − Lₙ = (11/40)((1 + 1/n)² − (1 − 1/n)²) = 11/(10n), which tends to 0 as n → ∞. M1 for a difference that simplifies to c/n; A1 (AG) needed a correct Lₙ.

Question 6: a second-order differential equation (10 marks)

Solve 5y″ + 4y′ + y = x² + 7x + 1, with y = 0 and dy/dx = 0 when x = 0.

• Answer: y = e^(−2x/5)(5 cos(x/5) + 15 sin(x/5)) + x² − x − 5.
• Auxiliary equation (M1): 5m² + 4m + 1 = 0, so m = −2/5 ± i/5. M1 for complex roots of the form a ± bi.
• Complementary function (A1): e^(−2x/5)(A cos(x/5) + B sin(x/5)). Leaving out “y =” was allowed here.
• Particular integral (B1, M1, A1): try y = px² + qx + r, with y′ = 2px + q and y″ = 2p (B1). Equating coefficients gives p = 1, 8p + q = 7 and 10p + 4q + r = 1 (M1), so q = −1 and r = −5 (A1).
• General solution (A1): the complementary function plus x² − x − 5, FT on your complementary function. It must start “y =”.
• Initial conditions (M1, DM1, A1, A1): differentiate the general solution with the product rule (M1). y(0) = 0 gives A − 5 = 0, and y′(0) = 0 gives B/5 − 2A/5 − 1 = 0, so A = 5 and B = 15 (DM1, then A1 for both correct). The final A1 (CWO) needs the full solution with “y =”.

Question 7: series and roots with complex numbers (10 marks)

• (a) Sum of z + z² + … + zⁿ (1 mark): z(zⁿ − 1)/(z − 1), or an equivalent such as (z − zⁿ⁺¹)/(1 − z) (B1).
• (b) Roots of 1 + z + … + z⁶ = 0 (2 marks): multiplying by (z − 1) gives z⁷ = 1 with z ≠ 1, so the roots are e^(2kπi/7) for k = 1, 2, …, 6, that is q = 2/7, 4/7, 6/7, 8/7, 10/7 and 12/7. B1 for one root, B1 for the other five. Repeats or wrong roots lost the second mark, and z = 1 had to be left out.
• (c) Show that the sum of e⁻ᵐ cos mθ from m = 1 to ∞ is (e cos θ − 1)/(e² − 2e cos θ + 1) (7 marks): with z = e^(−1 + iθ), de Moivre’s theorem gives zᵐ = e⁻ᵐ(cos mθ + i sin mθ), so the required sum is the real part of Σzᵐ. Since |z| = e⁻¹ < 1, the sum to infinity is z/(1 − z) = e^(iθ)/(e − e^(iθ)) (*M1 A1). Rationalise the denominator by multiplying by the conjugate (DM1). Using e^(iθ) + e^(−iθ) = 2 cos θ, or sin² θ + cos² θ = 1 in the trigonometric version (DM1 A1), gives (e cos θ − 1 + ie sin θ)/(e² − 2e cos θ + 1). Taking the real part gives the printed result (DM1 A1, AG).

Question 8: matrices (14 marks)

• (a) No unique solution (3 marks): the system has no unique solution when the determinant of the coefficient matrix is zero. That determinant is 4a + ab − b − 2 (M1 A1), so a = (b + 2)/(b + 4) (A1).
• (b) A⁴ = pA² + qI (4 marks): A = (−1 −2 1; 0 1 −1; 0 0 2) is upper triangular, so the characteristic equation is (λ + 1)(λ − 1)(λ − 2) = λ³ − 2λ² − λ + 2 = 0 (B1). Replace λ with A and multiply by A: A⁴ − 2A³ − A² + 2A = 0 (M1). Substitute A³ = 2A² + A − 2I to get A⁴ = 5A² − 4I, so p = 5 and q = −4 (M1, A1 CWO).
• (c) P and D with A⁷ = PDP⁻¹ (7 marks): the eigenvalues are −1, 1 and 2, read from the diagonal or the characteristic equation (B1). Using vector products (or equations), the eigenvectors are (1, 0, 0) for λ = −1, (1, −1, 0) for λ = 1 and (1, −1, 1) for λ = 2 (M1, then A1 for each). So P = (1 1 1; 0 −1 −1; 0 0 1) and D = diag(−1, 1, 128), since (−1)⁷ = −1, 1⁷ = 1 and 2⁷ = 128 (M1 A1). Columns could be in any order as long as D matched P; a column of zeros or a repeated column in P scored M0.

Where the marks are on 9231/22

• Integration: 21 marks: the reduction formula in Question 2 (7), the hyperbolic integral in Question 3(a) (4) and the rectangle bounds in Question 5 (10).
• Differential equations: 17 marks: the first-order equation in Question 3(b) (7) and the second-order equation in Question 6 (10).
• Matrices: 14 marks (Question 8).
• Complex numbers: 10 marks (Question 7).
• Differentiation: 8 marks (implicit differentiation in Question 4).
• Maclaurin series and inverse hyperbolic functions: 5 marks (Question 1).

That is 75 marks in total. “Show that” results with an AG mark appeared in Questions 2(a), 4(a), 5(a), 5(c) and 7(c).

How to self-mark 9231/22 with the mark scheme

1. Mark every M, A and B mark separately. The long questions spread their marks across stages: in Question 6 there are marks for the auxiliary equation, the complementary function, the particular integral, the general solution and the initial conditions, so one early slip leaves most marks available.
2. Follow the dependency chains. In Question 2(a), the DM1 for the second integration by parts needs the starred first one. In Question 3(b), substituting the initial condition (DM1) needs the starred integration mark, and is lost if you divided only some terms by sinh x first. In Question 7(c), every later mark depends on the sum to infinity being set up correctly.
3. Check “y =”. In Question 6, the general solution and the final answer both had to start “y =”, and Question 3(b) asked for y = f(x).
4. Treat AG parts strictly. In Questions 2(a), 4(a), 5(a), 5(c) and 7(c), give yourself the final mark only if every step is shown: the limits and brackets in 2(a), the summation formula in 5(a), and the real part taken explicitly in 7(c).
5. Use follow through only where the scheme allows it. Question 3(b) uses your answer to 3(a) if it has the right form; Question 6 follows through from your complementary function. Question 5(c) needs a correct Lₙ.
6. Keep answers exact. Questions 1, 2(b), 3(b), 4(b) and 8 all have exact answers: ln(1 + √2), 5/π − 4/π³, 10/81, 86/1331 and 128.
7. Log each lost mark as a missing method, an algebra or sign slip, a missing step in a “show that”, or a misread condition, and revise in that order.

Sit 9231/22 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 (Maclaurin series)	5	8 minutes
Question 2 (reduction formula)	7	11 minutes
Question 3 (integral and first-order equation)	11	17 minutes
Question 4 (implicit differentiation)	8	13 minutes
Question 5 (rectangle bounds)	10	15 minutes
Question 6 (second-order equation)	10	15 minutes
Question 7 (complex numbers)	10	15 minutes
Question 8 (matrices)	14	20 minutes
Check, especially AG steps and signs	–	6 minutes

Question 4(b) is slow for its marks because the second derivative has many terms before you substitute, and Question 8(c) needs three eigenvectors before P and D can be written down.

Other May/June 2026 9231 papers

• 9231 Paper 12 (Further Pure Mathematics 1)
• 9231 Paper 32 (Further Mechanics)
• 9231 Paper 42 (Further Probability & Statistics)

Practise and download

• Download 9231/22 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

Need help with a 9231 topic? Book a free trial lesson with an A Level Further Maths tutor →

Frequently asked questions

What topics were on 9231 Paper 22 May/June 2026? +

Eight questions: a Maclaurin series for sinh⁻¹(x + 1), a reduction formula, a hyperbolic integral and a first-order differential equation, implicit differentiation, upper and lower rectangle bounds for an integral, a second-order differential equation, complex numbers with de Moivre's theorem, and matrices including diagonalisation.

How many marks is 9231/22 and how long is it? +

It is out of 75 marks and lasts 2 hours, with every question compulsory. Question 8 on matrices is the biggest at 14 marks, and Question 3 is worth 11.

What was the answer to the differential equation in 9231/22 Question 6? +

y = e^(−2x/5)(5 cos(x/5) + 15 sin(x/5)) + x² − x − 5. The auxiliary equation 5m² + 4m + 1 = 0 has roots −2/5 ± i/5, the particular integral is x² − x − 5, and the conditions y = 0 and dy/dx = 0 at x = 0 give A = 5 and B = 15.

What was A⁴ in terms of A² in 9231/22 Question 8(b)? +

A⁴ = 5A² − 4I, so p = 5 and q = −4. The characteristic equation is λ³ − 2λ² − λ + 2 = 0. Replacing λ with A, multiplying by A, and substituting A³ = 2A² + A − 2I gives the result.

Is 9231 Paper 2 an AS or A Level paper? +

It is an A Level paper. Full A Level Further Mathematics candidates sit all four papers; AS candidates sit Paper 1 with Paper 3 or Paper 4, not Paper 2.

Where can I download 9231/22 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 Further Mathematics 9231 past papers page.

Ready to Excel in Your Studies?

Get personalised help from Tutopiya's expert tutors. Whether it's IGCSE, IB, A-Levels, or any other curriculum — we match you with the perfect tutor and your first session is free.

Book Your Free Trial