HKDSE Mathematics (Extended Module 1) Practice Paper: Calculus And Statistics

HKDSE Mathematics (Extended Module 1) Practice Paper: Calculus and Statistics

2026-11-04

HKDSE Mathematics Module 1: Calculus and Statistics

HKDSE Mathematics Module 1 practice paper: 12 original questions, 100 marks, 75 minutes on differentiation, integration and statistics.

Download　Question paper PDF　|　Answer key PDF

Paper details　Level: S6 | Subject: Mathematics (Extended Module 1) | Time: 75 minutes | Marks: 100 | Questions: 12

Section A (Q1-Q4): Differentiation and Its Applications (32 marks)

Differentiation: polynomials, logarithmic and exponential functions; tangents, rates of change and optimisation.

#	Question	Answer and marking notes
1	(Differentiation, chain rule) Let y = (3x^2 + 1)^4. (a) Find dy/dx. (b) Find the equation of the tangent to the curve at x = 1, giving it in the form y = mx + c.	Answer: (a) Let u = 3x^2 + 1, so y = u^4, dy/du = 4u^3 and du/dx = 6x; hence dy/dx = 24x(3x^2 + 1)^3. [4 marks] (b) At x = 1, y = 4^4 = 256 and the slope is 24(1)(4)^3 = 1536, so y – 256 = 1536(x – 1), i.e. y = 1536x – 1280. [4 marks]
2	(Differentiation, logarithmic function) Let y = x^2 ln x, where x is greater than 0. (a) Find dy/dx and simplify it to x(2 ln x + 1). (b) Find the slope of the tangent at x = e, in terms of e and correct to 3 decimal places.	Answer: (a) By the product rule, dy/dx = 2x ln x + x^2(1/x) = 2x ln x + x = x(2 ln x + 1). [4 marks] (b) At x = e, y = e^2 which is about 7.389, and the slope is e(2 + 1) = 3e, about 8.155. [4 marks]
3	(Rate of change) A spherical balloon is inflated so that its volume V grows at 50 cm^3 per second, with V = (4/3)pi r^3. (a) Find dr/dt when r = 5 cm. (b) Find the rate of increase of the surface area A = 4 pi r^2 at that instant. (c) State, with a reason, whether dr/dt increases or decreases as the balloon keeps expanding.	Answer: (a) dV/dr = 4 pi r^2, which is 100 pi when r = 5, so dr/dt = 50/(100 pi) which is about 0.159 cm per second. [3 marks] (b) dA/dt = 8 pi r (dr/dt) = 8 pi (5)(0.159) = 20 cm^2 per second. [2 marks] (c) Since dr/dt = 50/(4 pi r^2), a larger r makes the denominator larger, so dr/dt decreases. [3 marks]
4	(Optimisation) An open-top rectangular box has a square base and a fixed volume of 32 m^3. Let the side of the base be x m and the height be h m. (a) Express h in terms of x. (b) Show that the surface area is S = x^2 + 128/x, and find the x that minimises S. (c) Find the minimum surface area.	Answer: (a) x^2 h = 32, so h = 32/x^2. [3 marks] (b) S = x^2 + 4xh = x^2 + 128/x; setting dS/dx = 2x – 128/x^2 = 0 gives x^3 = 64, so x = 4; d2S/dx2 = 2 + 256/x^3 equals 6 at x = 4, which is positive, so S is a minimum. [3 marks] (c) Then h = 2 and S = 16 + 32 = 48 m^2. [2 marks]

Section B (Q5-Q8): Integration, Areas and the Trapezoidal Rule (36 marks)

Integration: antiderivatives, definite integrals, areas under curves and the trapezoidal rule.

#	Question	Answer and marking notes
5	(Definite integral) Evaluate the integral of (3x^2 – 4x + 5) with respect to x from 1 to 3, stating the antiderivative used.	Answer: The antiderivative is F(x) = x^3 – 2x^2 + 5x. [4 marks] F(3) = 27 – 18 + 15 = 24 and F(1) = 1 – 2 + 5 = 4, so the integral is 24 – 4 = 20. [5 marks]
6	(Definite integral, logarithm) Evaluate the integral of (1/x + 2) with respect to x from 1 to e. (a) State the antiderivative first. (b) Give the value in terms of e and correct to 3 decimal places.	Answer: (a) The antiderivative is ln x + 2x. [4 marks] (b) (ln e + 2e) – (0 + 2) = 1 + 2e – 2 = 2e – 1, about 4.437. [5 marks]
7	(Area under a curve) The curve y = x^2 – 4x + 3 meets the x-axis at A and B. (a) Find the x-coordinates of A and B. (b) Find the area of the region bounded by the curve and the x-axis. (c) Explain why evaluating the integral of (x^2 – 4x + 3) from 1 to 3 directly gives a negative value.	Answer: (a) x^2 – 4x + 3 = 0 gives (x – 1)(x – 3) = 0, so x = 1 and x = 3. [3 marks] (b) With antiderivative x^3/3 – 2x^2 + 3x, the value is (9 – 18 + 9) – (1/3 – 2 + 3) = -4/3, so the area is 4/3, about 1.333 square units. [4 marks] (c) The curve lies below the x-axis there, so the definite integral is negative; an area must be taken as a positive quantity. [2 marks]
8	(Trapezoidal rule) Let f(x) = e^x. Use the trapezoidal rule with 4 equal sub-intervals to estimate the integral of e^x from 0 to 1. (a) State h and the grid points x0 to x4. (b) Compute the estimate to 5 decimal places. (c) Compare it with the exact value e – 1, find the absolute error, and say whether the estimate is too high or too low.	Answer: (a) h = 0.25 with grid points 0, 0.25, 0.5, 0.75 and 1. [3 marks] (b) T = (0.25/2)[1 + e + 2(e^0.25 + e^0.5 + e^0.75)], about 1.72722. [3 marks] (c) The exact value is 1.71828, so the absolute error is about 0.00894; the estimate is too high because e^x is convex and each chord lies above the curve. [3 marks]

Section C (Q9-Q12): Probability Distributions and Statistical Inference (32 marks)

Statistics: binomial, normal and Poisson distributions, standard error and confidence intervals.

#	Question	Answer and marking notes
9	(Binomial distribution) Let X ~ B(10, 0.3). (a) Find E(X) and Var(X). (b) Find P(X = 3) to 4 decimal places. (c) Find P(X >= 1) to 4 decimal places.	Answer: (a) E(X) = np = 3 and Var(X) = np(1 – p) = 2.1. [3 marks] (b) P(X = 3) = C(10,3)(0.3)^3(0.7)^7 = 120 x 0.027 x 0.0823543, about 0.2668. [3 marks] (c) P(X >= 1) = 1 – 0.7^10 = 1 – 0.0282475, about 0.9718. [2 marks]
10	(Normal distribution) Let X ~ N(50, 8^2). (a) Find P(X greater than 62). (b) Find k such that P(X less than k) = 0.9, to 3 decimal places. (c) Explain why P(42 <= X <= 58) is about 0.68.	Answer: (a) z = (62 – 50)/8 = 1.5, so P(X greater than 62) = 1 – Phi(1.5), about 0.0668. [3 marks] (b) Phi(z) = 0.9 gives z about 1.2816, so k = 50 + 8(1.2816), about 60.252. [3 marks] (c) 42 and 58 are each one standard deviation from the mean, and a normal variable lies within one standard deviation of its mean about 68.27% of the time. [2 marks]
11	(Poisson approximation) Each item from a factory is defective with probability 1%, independently. A sample of 200 items is taken. Using a Poisson approximation, find (a) the parameter lambda; (b) the probability that no item is defective; (c) the probability that at least two items are defective. Give probabilities to 4 decimal places.	Answer: (a) lambda = 200 x 0.01 = 2. [2 marks] (b) P(X = 0) = e^(-2), about 0.1353. [3 marks] (c) P(X >= 2) = 1 – e^(-2)(1 + 2) = 1 – 0.4060, about 0.5940. [3 marks]
12	(Confidence interval) The net weight of a brand of coffee is normally distributed with standard deviation 12 g. A random sample of 64 packets has mean weight 42 g. (a) Find the standard error of the sample mean. (b) Find a 95% confidence interval for the population mean weight. (c) If the sample size rises to 144, by what factor does the width of the interval become?	Answer: (a) Standard error = 12/sqrt(64) = 1.5 g. [2 marks] (b) 42 +/- 1.96(1.5) = 42 +/- 2.94, i.e. (39.06, 44.94) g. [3 marks] (c) The new standard error is 12/12 = 1, so the width becomes 1/1.5 = 2/3 of the original. [3 marks]

Find a tutor — free Tutor Registration →

A TutorZone practice paper; original questions and passages, not publisher material.