HKDSE Mathematics Extended Part Module 2 (Algebra and Calculus) Mock Paper (10 questions, 100 marks, 2 hours 30 minutes)

2026-10-31

HKDSE Maths M2

HKDSE Mathematics Extended Part Module 2 mock paper: 10 questions, 100 marks, 2 hours 30 minutes, covering mathematical induction, the binomial theorem, limits and differentiation, integration by substitution, trigonometric equations, matrices and inverses, a 3 by 3 linear system, vectors and cross products, curve sketching with area, and integration by parts, each with worked answers and marking notes.

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Paper details　Level: HKDSE S6 | Subject: Mathematics Extended Part Module 2 | Time: 2 hours 30 minutes | Marks: 100 | Questions: 10

Section A (Questions 1 to 6, 50 marks)

Answer ALL questions. Show your working, give numerical answers to three significant figures unless told otherwise, and state units in every final answer. All data are examples and this is not a past paper of any examination board.

#	Question	Answer and marking notes
1	Let S(n) be the statement “for every positive integer n, the sum of r2^r from r = 1 to n equals (n – 1)2^(n+1) + 2”.\n(a) Prove S(n) by mathematical induction for all positive integers n.\n(b) Hence find the sum of r*2^r from r = 1 to 5.	(a) When n = 1: LHS = 1 x 2 = 2 and RHS = (1 – 1) x 4 + 2 = 2, so the statement holds.\nAssume it holds for some positive integer k, that is the sum from r = 1 to k equals (k – 1)2^(k+1) + 2.\nFor n = k + 1 the sum is (k – 1)2^(k+1) + 2 + (k + 1)2^(k+1) = (2k)2^(k+1) + 2 = k2^(k+2) + 2 = [(k + 1) – 1]2^((k+1)+1) + 2, so the statement also holds for n = k + 1.\nBy the principle of mathematical induction, S(n) is true for all positive integers n.\n(b) Put n = 5: the sum is 4 x 64 + 2 = 258.\nMarking: (a) 6 marks (base case 1, induction assumption 1, substitution 2, simplification 2); (b) 2 marks.
2	In the expansion of (2 – x/2)^8,\n(a) find the constant term;\n(b) find the coefficient of x^3;\n(c) find the coefficient of x^4.	(a) Constant term = C(8,0) x 2^8 = 256.\n(b) The term in x^3 is C(8,3) x 2^5 x (-1/2)^3 = 56 x 32 x (-1/8) = -224, so the coefficient is -224.\n(c) The term in x^4 is C(8,4) x 2^4 x (-1/2)^4 = 70 x 16 x (1/16) = 70, so the coefficient is 70.\nMarking: (a) 2 marks; (b) 4 marks (binomial coefficient, power of 2, sign, final answer); (c) 2 marks.
3	(a) Find the limit of (x^2 – 5x + 6)/(x – 2) as x tends to 2.\n(b) Let h(x) = x^2 ln x for x greater than 0. Find h'(1).\n(c) Find the equation of the tangent to y = h(x) at x = 1.	(a) Since x^2 – 5x + 6 = (x – 2)(x – 3), the quotient is x – 3 for x not equal to 2, so the limit is 2 – 3 = -1.\n(b) h'(x) = 2x ln x + x^2 x (1/x) = 2x ln x + x, hence h'(1) = 0 + 1 = 1.\n(c) h(1) = 0, so the tangent is y = 1 x (x – 1), that is y = x – 1.\nMarking: (a) 3 marks (factorisation 1, cancellation 1, answer 1); (b) 3 marks; (c) 2 marks.
4	(a) Use the substitution u = x^2 + 3 to find the integral of xsqrt(x^2 + 3) with respect to x.\n(b) Hence evaluate the definite integral of xsqrt(x^2 + 3) from x = 1 to x = 3, giving the answer to three significant figures.	(a) Let u = x^2 + 3, then du = 2x dx, so x dx = du/2.\nThe integral becomes (1/2)*integral of u^(1/2) du = (1/2)(2/3)u^(3/2) + C = (1/3)(x^2 + 3)^(3/2) + C.\n(b) Value = (1/3)(12^(3/2) – 4^(3/2)) = (1/3)(41.569 – 8) = 11.190, that is about 11.2.\nMarking: (a) 5 marks (substitution 2, antiderivative 2, constant 1); (b) 4 marks (limits 2, value 2).
5	(a) Solve 2sin^2(theta) – sin(theta) – 1 = 0 for 0 <= theta <= 360 degrees.\n(b) Hence solve 2sin^2(2x) - sin(2x) - 1 = 0 for 0 <= x <= 180 degrees.	(a) 2sin^2(theta) – sin(theta) – 1 = (2sin(theta) + 1)(sin(theta) – 1) = 0.\nIf sin(theta) = 1 then theta = 90 degrees; if sin(theta) = -0.5 then theta = 210 or 330 degrees.\nSo theta = 90, 210 or 330 degrees.\n(b) Taking 2x equal to each value in (a) with 0 <= 2x <= 360 degrees gives 2x = 90, 210 or 330 degrees, so x = 45, 105 or 165 degrees.\nMarking: (a) 5 marks (factorisation 2, three roots 3); (b) 4 marks (range 1, three roots 3).
6	Let A be the matrix with rows (2, 1) and (3, 4).\n(a) Find det A and A^(-1).\n(b) Using the inverse matrix, solve 2x + y = 5 and 3x + 4y = 6.	(a) det A = 2×4 – 1×3 = 5; A^(-1) = (1/5) x rows (4, -1) and (-3, 2), that is rows (0.8, -0.2) and (-0.6, 0.4).\n(b) (x; y) = A^(-1) (5; 6) = (1/5) x (14; -3) = (2.8; -0.6), so x = 2.8 and y = -0.6.\nMarking: (a) 4 marks (determinant 1, cofactors 2, inverse 1); (b) 4 marks (matrix product 2, two values 2).

Section B (Questions 7 to 10, 50 marks)

#	Question	Answer and marking notes
7	Consider the system x + y + z = 6, 2x – y + z = 3 and x + 2y – kz = 3, where k is a real number.\n(a) Express the determinant of the coefficient matrix in terms of k.\n(b) Find the value of k for which the system does not have a unique solution.\n(c) When k = 1, solve the system by Cramer’s rule.	(a) Determinant = 1[(k) – 2] – 1[-2k – 1] + 1[4 + 1] = 3k + 4.\n(b) No unique solution means the determinant is zero, so 3k + 4 = 0 and k = -4/3.\n(c) When k = 1 the determinant is 7; by Cramer’s rule x = 9/7, y = 15/7 and z = 18/7, which satisfy all three equations on substitution.\nMarking: (a) 4 marks; (b) 3 marks; (c) 5 marks (three determinants 3, three values 2).
8	Let a = i + 2j – 2k and b = 2i – j + k.\n(a) Find \|a\|, \|b\| and the scalar product a.b.\n(b) Find the angle between a and b, giving the answer to three significant figures.\n(c) Find the vector product a x b and \|a x b\|.\n(d) Find the scalar component of b along a.	(a) \|a\| = 3, \|b\| = sqrt(6) = 2.45 and a.b = 2 – 2 – 2 = -2.\n(b) cos(theta) = -2 / (3 sqrt(6)) = -0.2722, so theta = 106 degrees (three significant figures).\n(c) a x b = 0i – 5j – 5k and \|a x b\| = sqrt(50) = 7.07.\n(d) Scalar component = a.b / \|a\| = -2/3 = -0.667; the negative sign shows the opposite direction to a.\nMarking: (a) 3 marks; (b) 3 marks (cosine 2, angle 1); (c) 4 marks (cross product 3, magnitude 1); (d) 2 marks.
9	The curve C has equation y = x^3 – 6x^2 + 9x + 1.\n(a) Find the stationary points of C and determine their nature.\n(b) Find the point of inflexion of C.\n(c) Find the area of the region bounded by C, the lines x = 0, x = 3 and the x-axis.	(a) y’ = 3x^2 – 12x + 9 = 3(x – 1)(x – 3), so stationary points at x = 1 and x = 3. Since y” = 6x – 12 is negative at x = 1 and positive at x = 3, (1, 5) is a maximum point and (3, 1) is a minimum point.\n(b) y” = 0 at x = 2 and changes sign there, so (2, 3) is a point of inflexion.\n(c) Area = 81/4 – 54 + 81/2 + 3 = 39/4 = 9.75 square units.\nMarking: (a) 6 marks (derivative 1, stationary points 2, second derivative test 3); (b) 3 marks; (c) 4 marks (antiderivative 2, substitution 2).
10	(a) Use integration by parts to find the integral of x^2 e^(2x) with respect to x.\n(b) Hence evaluate the definite integral of x^2 e^(2x) from x = 0 to x = 1, giving the answer to three significant figures.\n(c) Explain why the substitution u = e^(2x) cannot be used directly for part (a).	(a) Take u = x^2 and dv = e^(2x) dx, so du = 2x dx and v = e^(2x)/2. Then the integral equals (x^2 e^(2x))/2 – integral of x e^(2x) dx. Applying parts again gives e^(2x)(x^2/2 – x/2 + 1/4) + C.\n(b) Substituting the limits gives 0.25 e^2 – 0.25 = 1.597, that is about 1.60.\n(c) With u = e^(2x) we get du = 2 e^(2x) dx, and the factor x^2 cannot be written in terms of u alone without logarithms, which makes the integral harder, so the substitution is not useful.\nMarking: (a) 7 marks (first application 3, second application 3, constant 1); (b) 4 marks; (c) 2 marks.

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