HKDSE Mathematics Extended Part Module 2 (M2) Practice Paper | 8 questions, 100 marks, 2 hours 30 minutes

2026-10-31

HKDSE Mathematics Extended Part Module 2 (M2)

HKDSE Mathematics Extended Part Module 2 (M2) practice paper: 8 questions, 100 marks, 2 hours 30 minutes, covering induction, binomial theorem, trigonometry, limits, differentiation, integration, matrices, vectors, determinants, with worked answers.

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Paper details　Level: S6 (HKDSE) | Subject: Mathematics Extended Part Module 2 | Time: 2 hours 30 minutes | Marks: 100 | Questions: 8

Section A: short questions (50 marks)

Total 100 marks, time allowed 2 hours 30 minutes. Show all working; give units where appropriate. Unless otherwise stated, numerical answers should be given to 3 significant figures.

#	Question	Answer and marking notes
1	(Mathematical induction) (a) Prove by mathematical induction that, for all positive integers n, 1² + 2² + … + n² = n(n + 1)(2n + 1) ÷ 6. (6 marks) (b) Hence find the value of 1² + 2² + … + 10². (3 marks) (c) Hence find the value of 11² + 12² + … + 20². (3 marks)	Answer: (a) Let P(n) be the statement. For n = 1, LHS = 1² = 1 and RHS = 1×2×3 ÷ 6 = 1, so P(1) is true. Assume P(k) is true, i.e. 1² + … + k² = k(k + 1)(2k + 1) ÷ 6. Then for n = k + 1, LHS = k(k + 1)(2k + 1) ÷ 6 + (k + 1)² = (k + 1)[k(2k + 1) + 6(k + 1)] ÷ 6 = (k + 1)(2k² + 7k + 6) ÷ 6 = (k + 1)(k + 2)(2k + 3) ÷ 6, so P(k + 1) is true. By the principle of mathematical induction, P(n) is true for all positive integers n. [6 marks] (b) Put n = 10: 385. [3 marks] (c) 11² + 12² + … + 20² = 2870 − 385 = 2485. [3 marks]
2	(Binomial theorem) (a) Find the coefficient of x³ in the expansion of (2 + x)⁸. (4 marks) (b) Find the coefficient of x² in the expansion of (1 + 3x)⁶. (4 marks) (c) Find the sum of all the coefficients in the expansion of (2 + x)⁸. (4 marks)	Answer: (a) The general term is C(8, r)·2^(8−r)·x^r; putting r = 3 gives C(8, 3)·2⁵ = 56 × 32 = 1792. [4 marks] (b) The general term is C(6, r)·(3x)^r; putting r = 2 gives C(6, 2)·3² = 15 × 9 = 135. [4 marks] (c) Put x = 1: the sum of the coefficients is (2 + 1)⁸ = 3⁸ = 6561. [4 marks]
3	(Trigonometry) A and B are acute angles with sin A = 3/5 and cos B = 5/13. (a) Find the value of sin(A + B). (4 marks) (b) Find the value of cos 2A. (4 marks) (c) Find the value of tan(A − B). (4 marks)	Answer: (a) Since A is acute, cos A = 4/5; since B is acute, sin B = 12/13. sin(A + B) = sin A cos B + cos A sin B = (3/5)(5/13) + (4/5)(12/13) = 15/65 + 48/65 = 63/65. [4 marks] (b) cos 2A = 1 − 2sin²A = 1 − 2(9/25) = 7/25. [4 marks] (c) tan A = 3/4 and tan B = 12/5, so tan(A − B) = (tan A − tan B) ÷ (1 + tan A tan B) = (−33/20) ÷ (56/20) = −33/56. [4 marks]
4	(Limits and differentiation) (a) Find lim(x→0) sin 3x ÷ (2x). (3 marks) (b) Find lim(x→0) (1 − cos x) ÷ x². (3 marks) (c) Let y = x³ − 3x² + 4x − 5. Find dy/dx and the equation of the tangent to the curve at x = 2. (4 marks) (d) Find d/dx [x² sin x]. (4 marks)	Answer: (a) sin 3x ÷ (2x) = (3/2)·(sin 3x ÷ 3x) and sin 3x ÷ 3x → 1 as x→0, so the limit is 3/2. [3 marks] (b) 1 − cos x = 2sin²(x/2), so (1 − cos x) ÷ x² = 2sin²(x/2) ÷ x² = (1/2)·[sin(x/2) ÷ (x/2)]² → 1/2. [3 marks] (c) dy/dx = 3x² − 6x + 4; at x = 2, dy/dx = 4 and y = -1, so the tangent is y = 4x -9. [4 marks] (d) By the product rule, d/dx [x² sin x] = 2x sin x + x² cos x. [4 marks]

Section B: long questions (50 marks)

Total 100 marks, time allowed 2 hours 30 minutes. Show all working; give units where appropriate. Unless otherwise stated, numerical answers should be given to 3 significant figures.

#	Question	Answer and marking notes
5	(Integration) (a) Evaluate ∫₀¹ x(x² + 1)³ dx. (6 marks) (b) Find ∫ x sin x dx. (6 marks)	Answer: (a) Let u = x² + 1, so du = 2x dx; when x = 0, u = 1 and when x = 1, u = 2. ∫₀¹ x(x² + 1)³ dx = (1/2)∫₁² u³ du = (1/2)[u⁴/4]₁² = (1/8)(16 − 1) = 15/8. [6 marks] (b) Integrate by parts with u = x and dv = sin x dx, so du = dx and v = −cos x: ∫ x sin x dx = −x cos x + ∫ cos x dx = −x cos x + sin x + C. [6 marks]
6	(Matrices) Let A = [[2, 1], [3, 4]]. (a) Find det A. (2 marks) (b) Find A⁻¹. (4 marks) (c) Hence solve the system 2x + y = 7, 3x + 4y = 18. (6 marks)	Answer: (a) det A = (2)(4) − (1)(3) = 5. [2 marks] (b) A⁻¹ = (1/5)·[[4, −1], [−3, 2]]. [4 marks] (c) The system is A·[x, y]ᵀ = [7, 18]ᵀ, so [x, y]ᵀ = A⁻¹·[7, 18]ᵀ; x = (4×7 − 1×18) ÷ 5 = 10/5 = 2 and y = (−3×7 + 2×18) ÷ 5 = 15/5 = 3. [6 marks]
7	(Vectors) Let a = (1, 2, −1) and b = (3, −1, 2). (a) Find a·b. (3 marks) (b) Find a × b. (5 marks) (c) Find the angle θ between a and b, correct to 0.1°. (5 marks)	Answer: (a) a·b = (1)(3) + (2)(−1) + (−1)(2) = 3 − 2 − 2 = −1. [3 marks] (b) a × b = ((2)(2) − (−1)(−1), (−1)(3) − (1)(2), (1)(−1) − (2)(3)) = (4 − 1, −3 − 2, −1 − 6) = (3, -5, -7). [5 marks] (c) \|a\| = √6 and \|b\| = √14, so cos θ = a·b ÷ (\|a\|\|b\|) = −1 ÷ √84 ≈ −0.1091 and θ ≈ 96.3°. [5 marks]
8	(Determinants and systems) Consider x + 2y + 3z = 14, 2x − y + z = 3 and 3x + y − 2z = −1. (a) Find the determinant D of the coefficient matrix. (5 marks) (b) Use Cramer rule to find x, y and z. (8 marks)	Answer: (a) D = 1[(−1)(−2) − (1)(1)] − 2[(2)(−2) − (1)(3)] + 3[(2)(1) − (−1)(3)] = 1(1) − 2(−7) + 3(5) = 1 + 14 + 15 = 30. [5 marks] (b) Dx = 30, Dy = 60 and Dz = 90, so x = 30/30 = 1, y = 60/30 = 2 and z = 90/30 = 3. [8 marks]

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