FREE HKDSE PRACTICE PAPER

HKDSE Maths M1 Mock Paper: Calculus and Statistics

This Mathematics Extended Part Module 1 sample contains 14 questions over two 50-mark sections. Allow 2 hours 30 minutes. It covers binomial expansion, differentiation, integration, probability distributions, sampling and confidence intervals. Worked answers are grouped by question and include marking allocations.

14 questions100 marks2 hr 30 min

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An original AI-generated practice paper from Studify. This is an independent sample, not an official HKEAA paper.

Paper structure

Section A10 questions · 50 marks
Section B4 questions · 50 marks

Topics in this sample

Binomial expansion
Exponential and logarithmic functions
Derivative of a function
Differentiation of a function
Second derivative
Applications of differentiation
Indefinite integration and its applications
Definite integration and its applications
Approximation of definite integrals using the trapezoidal rule
Conditional probability and Bayes' theorem
Discrete random variables
Probability distribution, expectation and variance
The binomial distribution
The Poisson distribution
Applications of the binomial and the Poisson distributions
Basic definition and properties of normal distribution
Standardisation of a normal variable and the standard normal table
Applications of the normal distribution
Sampling distribution and point estimates
Confidence interval for a population mean

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Hong Kong Diploma of Secondary Education Examination Mock Paper

Mathematics M1

Paper: Calculus and Statistics (2027 onwards)

Time Allowed: 2 hr 30 minTotal Marks: 100

INSTRUCTIONS TO CANDIDATES:

This paper consists of multiple sections.
Answer ALL questions in this paper.
Write your answers in the spaces provided.
Unless otherwise specified, all working must be clearly shown.
Unless otherwise specified, numerical answers should be either exact or correct to 3 significant figures.

Generated by studify.life • 16/10/2026

Section A

Question 1

Answer all parts.

(a)

Find the coefficient of

x^3

in the expansion of

(1+2x)^5

(2 marks)

(b)

Find the first three terms, in ascending powers of

x

, in the expansion of

(1-3x)^{-2}

(3 marks)

Question 2

Answer all parts.

(a)

Solve

e^{2x}=7

for

x

, giving your answer to 3 significant figures.

(2 marks)

(b)

Solve

\log_2(x-1)+\log_2(x+1)=3

, stating any restriction on

x

(3 marks)

Question 3

Let

f(x)=x^3-4x^2+5x+2

(a)

Find

f'(x)

(2 marks)

Page2

Question 3 (continued)

(b)

Find

f''(x)

and hence find the value of

f''(2)

(3 marks)

Question 4

The height

h

metres of a particle above a horizontal surface after

t

seconds is

h(t)=12+8t-2t^2

, for

0\leq t\leq 4

(a)

Find the time at which the particle reaches its maximum height.

(3 marks)

(b)

Find this maximum height.

(2 marks)

Question 5

Answer all parts.

(a)

Find

\int (6x^2-4x+3)\,dx

(3 marks)

(b)

A particle has velocity

v(t)=6t-4

m s

^{-1}

. Given that its displacement from a fixed origin is 5 m when

t=0

, find its displacement at time

t

(2 marks)

Page3

Question 6

The curve

y=2x+1

is considered for

0\leq x\leq 4

(a)

Find the exact area under the curve and above the

x

-axis from

x=0

x=4

(2 marks)

(b)

Estimate the same area using the trapezoidal rule with 4 equal strips.

(3 marks)

Question 7

A factory uses machines

A

and

B

to produce 60% and 40% of its components respectively. The probability that a component from

A

is defective is 0.02, while that from

B

is 0.05.

(a)

Find the probability that a randomly selected component is defective.

(2 marks)

(b)

Given that a selected component is defective, find the probability that it was produced by machine

B

(3 marks)

Question 8

A discrete random variable

X

has probability distribution

P(X=x)=kx

for

x=1,2,3

(a)

Find

k

(2 marks)

Page4

Question 8 (continued)

(b)

Find

E(X)

(2 marks)

(c)

Find

\operatorname{Var}(X)

(1 mark)

Question 9

Answer all parts.

(a)

A fair coin is tossed 5 times. Find the probability of obtaining exactly 3 heads.

(3 marks)

(b)

The number of calls received by a switchboard in a minute follows a Poisson distribution with mean 2. Find the probability of receiving no calls in a particular minute.

(2 marks)

Question 10

The random variable

X

is normally distributed with mean 50 and standard deviation 8.

Standard normal distribution values

$z$	$P(Z<z)$
-1.00	0.1587
1.00	0.8413
1.50	0.9332

(a)

Find

P(X>62)

(3 marks)

Page5

Question 10 (continued)

(b)

Find

c

such that

P(X<c)=0.1587

(2 marks)

Page6

Section B

Question 11

A rectangular garden is to be built beside a straight wall, so fencing is required for only three sides. The total length of fencing available is 24 m. Let

x

m be the width perpendicular to the wall and

y

m be the length parallel to the wall.

(a)

Express

y

in terms of

x

(2 marks)

(b)

Find the value of

x

that gives the maximum area of the garden.

(4 marks)

(c)

Find the maximum area.

(2 marks)

(d)

For this value of

x

, find the area of the garden by evaluating a definite integral of its width profile

\displaystyle w(u)=x-\frac{u}{6}

for

0\leq u\leq 6

. Give the value in square metres.

(4 marks)

Page7

Question 12

A quality-control test for a component has probability 0.9 of giving a positive result when the component is defective, and probability 0.08 of giving a positive result when it is not defective. A production line makes defective components with probability 0.04. Components are tested independently. Let

X

be the number of defective components among 20 randomly selected components.

(a)

Find the probability that a randomly selected component gives a positive test result.

(4 marks)

(b)

Given that a component gives a positive result, find the probability that it is defective.

(4 marks)

(c)

State the distribution of

X

and find

P(X\geq 2)

(4 marks)

Question 13

The lifetimes, in hours, of a certain type of battery are normally distributed with mean

\mu

hours and known standard deviation 12 hours. A random sample of 64 batteries has sample mean lifetime 72 hours.

(a)

State the sampling distribution of the sample mean

\bar X

in terms of

\mu

, and find its standard error.

(3 marks)

Page8

Question 13 (continued)

(b)

Using the sample, find a 95% confidence interval for

\mu

. Use

z_{0.975}=1.96

(4 marks)

(c)

Find the probability that, for a sample of 64 batteries from this population, the sample mean exceeds 75 hours if

\mu=72

(3 marks)

(d)

How large a sample is required so that the 95% confidence interval for

\mu

has margin of error at most 2 hours? Use

z_{0.975}=1.96

(3 marks)

Question 14

A manufacturer models the number of minor faults found in a batch of products by a Poisson distribution with mean

\lambda

faults per batch. In a sample of 50 batches, the observed numbers of faults were recorded as follows.

Observed number of faults in 50 batches

Number of faults, $x$	0	1	2	3	4 or more
Number of batches	12	18	12	6	2

(a)

Use the midpoint values

0,1,2,3,4

for the respective categories to estimate

\lambda

from the sample.

(4 marks)

Page9

Question 14 (continued)

(b)

Using your estimate of

\lambda

, find the probability that a future batch has no faults.

(4 marks)

(c)

Using your estimate of

\lambda

, find the probability that at least 3 faults occur in a combined production run of 2 independent batches. State the distribution used and give your answer to 4 decimal places.

(5 marks)

Page10

Model answers

Choose a question to reveal the saved solution and marking guidance. Try it yourself first.

Question 15 marks

(a) The

x^3

coefficient is

\binom{5}{3}2^3=10\times8=80

. (M1: identify the binomial term; A1: 80.) (b) Using

(1-u)^{-2}=1+2u+3u^2+\cdots

with

u=3x

, the first three terms are

1+6x+27x^2

. (M1: use the expansion; A1: substitute correctly; A1: all three terms.)

Question 25 marks

(a) Taking natural logarithms gives

2x=\ln 7

, so

\displaystyle x=\frac{\ln 7}{2}=0.973

(3 s.f.). (M1: take logarithms; A1: answer.) (b) The logarithms require

x>1

. Combining them gives

\log_2((x-1)(x+1))=3

, hence

x^2-1=8

and

x^2=9

. Thus

x=3

-3

, but only

x=3

satisfies

x>1

. (M1: state restriction/combine logs; M1: solve quadratic; A1: valid solution.)

Question 35 marks

(a) Differentiating term by term,

f'(x)=3x^2-8x+5

. (M1: differentiate correctly; A1: expression.) (b) Differentiating again,

f''(x)=6x-8

. Therefore

f''(2)=12-8=4

. (M1: second differentiation; A1: expression; A1: value.)

Question 45 marks

(a)

h'(t)=8-4t

. At a stationary point,

8-4t=0

, giving

t=2

. Since

h''(t)=-4<0

, this is a maximum. (M1: derivative; M1: stationary time; A1: verify maximum.) (b)

h(2)=12+16-8=20

m. (M1: substitute; A1: 20 m.)

Question 55 marks

(a) Integrating term by term gives

2x^3-2x^2+3x+C

. (M1: integrate powers; A1: correct terms; A1: include constant.) (b) Since displacement

s

satisfies

s'(t)=v(t)

s=3t^2-4t+C

. From

s(0)=5

C=5

, so

s(t)=3t^2-4t+5

m. (M1: integrate and use initial condition; A1: result.)

Question 65 marks

(a)

\int_0^4(2x+1)\,dx=[x^2+x]_0^4=20

. (M1: definite integral; A1: 20.) (b) The strip width is

h=1

and the ordinates are

1,3,5,7,9

. The estimate is

\displaystyle \frac{h}{2}[y_0+y_4+2(y_1+y_2+y_3)]=\frac12[1+9+2(3+5+7)]=20

. (M1: width and ordinates; M1: trapezoidal formula; A1: 20.)

Question 75 marks

(a) By total probability,

P(D)=0.6(0.02)+0.4(0.05)=0.032

. (M1: total-probability expression; A1: 0.032.) (b) By Bayes' theorem,

\displaystyle P(B\mid D)=\frac{P(B)P(D\mid B)}{P(D)}=\frac{0.4(0.05)}{0.032}=0.625

. (M1: Bayes' expression; M1: substitute correctly; A1: 0.625.)

Question 85 marks

(a)

k+2k+3k=1

, so

k=\frac16

. (M1: normalisation; A1: value.) (b)

E(X)=1\cdot\frac16+2\cdot\frac13+3\cdot\frac12=\frac73

. (M1: expectation calculation; A1: value.) (c)

E(X^2)=\frac16+4(\frac13)+9(\frac12)=6

, hence

\operatorname{Var}(X)=6-(\frac73)^2=\frac59

. (A1: variance.)

Question 95 marks

(a) Let

X\sim B(5,\frac12)

. Then

\displaystyle P(X=3)=\binom53(\frac12)^3(\frac12)^2=\frac{10}{32}=0.3125

. (M1: binomial model; M1: substitute; A1: answer.) (b) For

Y\sim\operatorname{Poisson}(2)

\displaystyle P(Y=0)=e^{-2}\frac{2^0}{0!}=e^{-2}\approx0.1353

. (M1: Poisson probability; A1: answer.)

Question 105 marks

(a) Standardising,

\displaystyle z=\frac{62-50}{8}=1.5

. Thus

P(X>62)=P(Z>1.5)=1-0.9332=0.0668

. (M1: standardise; M1: complement; A1: answer.) (b) The table gives

P(Z<-1)=0.1587

, so

c=50+8(-1)=42

. (M1: identify standard score; A1: value.)

Question 1112 marks

11 / (a) The fencing constraint is

2x+y=24

, hence

y=24-2x

. (M1: constraint; A1: expression.) (b) The area is

A=x(24-2x)=24x-2x^2

. Thus

A'=24-4x=0

gives

x=6

. Also

A''=-4<0

, so this is a maximum. (M1: area function; M1: stationary condition and solution; M1: maximum test; A1:

x=6

m.) (c)

A(6)=6(12)=72

^2

. (M1: substitute; A1: answer.) (d) Using

x=6

\displaystyle \int_0^6(6-\frac{u}{6})\,du=[6u-\frac{u^2}{12}]_0^6=36-3=33

^2

. (M1: substitute; M1: integrate; M1: limits; A1: 33 m

^2

Question 1212 marks

12 / (a) Let

D

denote defective and

+

positive.

P(+)=P(D)P(+\mid D)+P(D^c)P(+\mid D^c)=0.04(0.9)+0.96(0.08)=0.1128

. (M1: total probability; M1: correct substitution; A1: sum; A1: answer.) (b)

\displaystyle P(D\mid +)=\frac{0.04(0.9)}{0.1128}=\frac{0.036}{0.1128}\approx0.3191

. (M1: Bayes' theorem; M1: numerator and denominator; A1: calculation; A1: answer.) (c) Each selection is an independent Bernoulli trial with defect probability 0.04, so

X\sim B(20,0.04)

P(X\geq2)=1-P(X=0)-P(X=1)=1-(0.96)^{20}-20(0.04)(0.96)^{19}\approx0.1917

. (M1: distribution; M1: complement; M1: probabilities; A1: answer.)

Question 1313 marks

13 / (a) Since the population is normal,

\bar X\sim N(\mu,12^2/64)

, so its standard deviation (standard error) is

12/8=1.5

hours. (M1: sampling distribution; M1: variance/standard deviation; A1: 1.5.) (b) The point estimate is

\bar x=72

. The margin is

1.96(12/\sqrt{64})=1.96(1.5)=2.94

. Hence the interval is

(72-2.94,72+2.94)=(69.06,74.94)

hours. (M1: point estimate and standard error; M1: margin; M1: interval; A1: endpoints.) (c) Under

\mu=72

\bar X\sim N(72,1.5^2)

. Thus

P(\bar X>75)=P(Z>2)=1-0.9772=0.0228

. (M1: standardise; M1: tail probability; A1: answer.) (d) Require

1.96(12/\sqrt n)\leq2

, so

\sqrt n\geq11.76

and

n\geq138.2976

. The smallest integer sample size is 139. (M1: margin inequality; M1: solve; A1: round up correctly.)

Question 1413 marks

14 / (a) The estimated mean is the sample mean:

\displaystyle \hat\lambda=\frac{0(12)+1(18)+2(12)+3(6)+4(2)}{50}=\frac{68}{50}=1.36

. (M1: weighted total; M1: frequency total; M1: quotient; A1: estimate.) (b) For

Y\sim\operatorname{Poisson}(1.36)

P(Y=0)=e^{-1.36}\approx0.2567

. (M1: Poisson model; M1: formula; A1: answer; A1: suitable accuracy.) (c) Independent Poisson counts add, so the combined count

T\sim\operatorname{Poisson}(2(1.36))=\operatorname{Poisson}(2.72)

\displaystyle P(T\geq3)=1-[P(T=0)+P(T=1)+P(T=2)] =1-e^{-2.72}(1+2.72+\frac{2.72^2}{2})\approx0.5079

. (M1: combined mean/distribution; M1: complement; M1: probabilities; M1: calculation; A1: answer to 4 d.p.)