HKDSE Mathematics Extended Part Module 1 Mock Paper (8 questions, 100 marks)

2026-10-31

HKDSE Mathematics Module 1

HKDSE Mathematics Extended Part Module 1 mock paper: 8 questions, 100 marks, 2 hours 30 minutes, covering the binomial theorem, exponential and logarithmic equations, the product and chain rules, definite integrals and areas, the trapezoidal rule, binomial, Poisson and normal distributions, with full answers and marking notes.

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Paper details　Level: HKDSE | Subject: Mathematics (Extended Part Module 1) | Time: 2 hours 30 minutes | Marks: 100 | Questions: 8

Section A: binomial theorem, exponentials and logarithms, differentiation and integration (50 marks)

Structured questions: show full working, substitutions and units; for probability state the distribution and its parameters first. 8 questions, 100 marks, 2 hours 30 minutes, calculator allowed.

#	Question	Answer and marking notes
1	(a) Expand (1 + 2x)⁷ up to and including the term in x³. (b) Find the coefficient of x⁴ in the expansion of (1 − 3x)⁹.	Answer: (a) By the binomial theorem the coefficients are C(7,k)·2^k: 1, 14, 84, 280, so (1 + 2x)^7 = 1 + 14x + 84x² + 280x³ + …. (b) The general term is C(9,r)(-3x)^r; with r = 4: C(9,4) x 81 = 126 x 81 = 10206.
2	(a) Solve 2^x+1 = 5^2x, giving the answer correct to 3 significant figures. (b) Solve log₂x + log₂(x − 3) = 2 and explain why one root is rejected.	Answer: (a) Taking natural logarithms, (x + 1)ln2 = 2x·ln5, so x(2ln5 – ln2) = ln2 and x = ln2 / (2ln5 – ln2) = 0.2744 ≈ 0.274. (b) log₂[x(x – 3)] = 2 gives x² – 3x – 4 = 0, i.e. (x – 4)(x + 1) = 0, so x = 4 or x = -1; since the logarithm requires x > 3, x = -1 is rejected and x = 4.
3	Let y = x²e^3x. (a) Find dy/dx, showing the use of the product rule and the chain rule. (b) Find the equation of the tangent to the curve at x = 1, in terms of e.	Answer: (a) With u = x² and v = e^3x, u’ = 2x and v’ = 3e^3x, so dy/dx = 2x·e^3x + 3x²·e^3x = e^3x(2x + 3x²). (b) At x = 1, y = e³ and the gradient is e³(2 + 3) = 5e³ = 100.4; the tangent is y – e³ = 5e³(x – 1), i.e. y = 5e³x – 4e³.
4	The curve y = x² – x – 2 cuts the x-axis at A and B. (a) Find the coordinates of A and B. (b) Find the area of the region bounded by the curve and the x-axis, giving the answer as a fraction.	Answer: (a) x² – x – 2 = 0 gives (x – 2)(x + 1) = 0, so A(-1, 0) and B(2, 0). (b) Area = \|∫(from -1 to 2) (x² – x – 2) dx\| with primitive x³/3 – x²/2 – 2x; at x = 2 the value is -3.3333 and at x = -1 it is 1.1667, so the area is 4.5000 = 9/2 square units.

Section B: numerical integration, discrete and continuous probability distributions (50 marks)

Structured questions: show full working, substitutions and units; for probability state the distribution and its parameters first. 8 questions, 100 marks, 2 hours 30 minutes, calculator allowed.

#	Question	Answer and marking notes
5	The table gives f(x) at five points: at x = 0, 0.5, 1.0, 1.5 and 2.0, f(x) equals 1.000, 1.284, 1.649, 2.117 and 2.718 respectively. (a) Use the trapezoidal rule (5 ordinates, 4 sub-intervals) to estimate ∫₀²f(x)dx, correct to 3 decimal places. (b) State how the estimate generally changes when the number of sub-intervals is doubled, and comment on the direction of the error.	Answer: (a) With h = 0.5, the trapezoidal rule gives h/2 x [f₀ + f₄ + 2(f₁ + f₂ + f₃)] = 0.25 x 13.818 = 3.4545 ≈ 3.454. (b) Doubling the sub-intervals (halving h) reduces the error by roughly a factor of 4, so the estimate approaches the true value; for a curve that is concave up the rule over-estimates and for one concave down it under-estimates.
6	A random variable X follows a binomial distribution B(20, 0.3). (a) Find P(X = 6), correct to 4 decimal places. (b) Find P(X ≥ 2), and state E(X) and Var(X).	Answer: (a) P(X = 6) = C(20,6)(0.3)^6(0.7)^14 = 38760 x 0.000729 x 0.0067822 ≈ 0.1916. (b) P(X ≥ 2) = 1 – P(X = 0) – P(X = 1) = 1 – 0.0007979 – 0.0068393 ≈ 0.9924; E(X) = np = 20 x 0.3 = 6 and Var(X) = np(1 – p) = 4.2.
7	Vehicles arrive at a junction according to a Poisson distribution with a mean of 4 vehicles per 30 seconds; let X be the number of arrivals in 30 seconds. (a) Find P(X = 3), correct to 4 decimal places. (b) Find P(X ≤ 1), and find the probability that no vehicle arrives in 15 seconds.	Answer: (a) With X ~ Po(4), P(X = 3) = e^-4 x 4³ / 3! = 0.0183156 x 64 / 6 ≈ 0.1954. (b) P(X ≤ 1) = e^-4(1 + 4) ≈ 0.0916; in 15 seconds the mean is 2, so the probability of no arrival is e^-2 = 0.1353.
8	Examination scores X are normally distributed with mean 50 and standard deviation 10. (a) Find P(X > 62) and P(45 < X < 65), each correct to 4 decimal places. (b) If only the top 10% obtain grade A, find the minimum grade-A mark k (correct to 1 decimal place) and explain the standard score used.	Answer: (a) With z = (x – 50)/10, P(X > 62) = P(Z > 1.2) ≈ 0.1151 and P(45 < X < 65) = P(-0.5 < Z < 1.5) ≈ 0.6247. (b) We need P(Z < z₀) = 0.9, so z₀ ≈ 1.2816 and k = 50 + 1.2816 x 10 = 62.8.

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