FREE HKDSE PRACTICE PAPER

HKDSE Maths M2 Mock Paper: Algebra and Calculus

This Mathematics Extended Part Module 2 sample contains 14 questions across two 50-mark sections, with a time allowance of 2 hours 30 minutes. Topics include mathematical induction, trigonometric functions, limits, calculus, matrices and vectors. Model answers show the working and marking allocations for each question.

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

Odd and even functions
Mathematical induction
The binomial theorem
More about trigonometric functions
Introduction to e
Limits
Differentiation
Indefinite integration and its applications
Definite integration
Applications of definite integration
Determinants
Matrices
Systems of linear equations
Introduction to vectors
Scalar product and vector product
Applications of differentiation
Applications of vectors

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Hong Kong Diploma of Secondary Education Examination Mock Paper

Mathematics M2

Paper: Algebra and Calculus (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

A

Question 1

Let

\displaystyle f(x)=\dfrac{x^3}{x^2+1}+\cos x

(a)

Determine whether

\displaystyle g(x)=\dfrac{x^3}{x^2+1}

is odd, even or neither. Give a reason.

(2 marks)

(b)

Determine whether

h(x)=\cos x

is odd, even or neither.

(1 mark)

(c)

Find

f(2)+f(-2)

(2 marks)

Question 2

Use mathematical induction to prove that, for every positive integer

n

1+3+5+\cdots +(2n-1)=n^2

.(5 marks)

Question 3

Find the coefficient of

x^3

in the expansion of

(1+2x)^5

.(5 marks)

Question 4

You may use

\sin 2x=2\sin x\cos x

and

\displaystyle \cos x=\sin\left(\dfrac{\pi}{2}-x\right)

. Solve

\sin 2x=\cos x

for

0\leq x<2\pi

. Give your answers in radians.(5 marks)

Question 5

Using

\displaystyle e=\lim_{n\to\infty}\left(1+\dfrac1n\right)^n

, evaluate

\displaystyle \lim_{x\to\infty}\left(1+\dfrac3x\right)^{2x}

.(5 marks)

Question 6

The curve

C

has equation

y=(x^2+1)e^x

(a)

Find

\displaystyle \dfrac{dy}{dx}

(3 marks)

Page2

Question 6 (continued)

(b)

Find the gradient of

C

x=0

(2 marks)

Question 7

A function

G

satisfies

G'(x)=3x^2-4x+2

and

G(0)=1

(a)

Find

G(x)

(3 marks)

(b)

Hence find

G(2)

(2 marks)

Question 8

Find the area enclosed by the curve

y=x^2-1

and the

x

-axis between

x=-1

and

x=2

.(5 marks)

Question 9

Consider the simultaneous equations

2x+y=5

and

x-y=1

(a)

Find the determinant of the coefficient matrix.

(2 marks)

(b)

Solve the simultaneous equations.

(3 marks)

Page3

Question 10

Let

\mathbf{a}=(1,2,0)

and

\mathbf{b}=(2,-1,2)

(a)

Find

\mathbf{a}\cdot\mathbf{b}

(2 marks)

(b)

Find

\mathbf{a}\times\mathbf{b}

(3 marks)

Page4

B

Question 11

Consider

f(x)=xe^{-x}

for

x\geq 0

(a)

Evaluate

\displaystyle \displaystyle \lim_{x\to0}\dfrac{e^{2x}-1}{x}

(3 marks)

(b)

Find the stationary point of

f

and determine its nature.

(5 marks)

(c)

Find the equation of the tangent to the graph of

y=f(x)

x=2

(4 marks)

Page5

Question 12

Consider the curves

y=2x

and

y=x^2

for

0\leq x\leq2

(a)

Find

\displaystyle \int(2x-x^2)\,dx

(4 marks)

(b)

Find the area enclosed by the two curves.

(5 marks)

(c)

Find the value

a

0<a<2

such that the area between the curves from

x=0

x=a

is half the enclosed area.

(4 marks)

Page6

Question 13

For a real parameter

\lambda

, consider the system

\lambda x+y=1

x+\lambda y=1

, whose coefficient matrix is

A=\begin{pmatrix}\lambda&1\\1&\lambda\end{pmatrix}

(a)

Find

\det A

(2 marks)

(b)

State the values of

\lambda

for which the system has a unique solution.

(2 marks)

(c)

Solve the system when

\lambda=2

(4 marks)

(d)

Discuss the system when

\lambda=1

and when

\lambda=-1

(4 marks)

Page7

Question 14

In three-dimensional space, let

A=(0,0,0)

B=(4,0,0)

and

C=(1,3,0)

. The diagram shows the triangle

ABC

in the

xy

-plane.

(a)

Find

\overrightarrow{AB}

and

\overrightarrow{AC}

(2 marks)

(b)

Find the angle

BAC

, giving your answer to the nearest degree.

(4 marks)

(c)

Find

\overrightarrow{AB}\times\overrightarrow{AC}

and hence find the area of triangle

ABC

(4 marks)

Page8

Question 14 (continued)

(d)

Find a unit vector perpendicular to both

\overrightarrow{AB}

and

\overrightarrow{AC}

(3 marks)

Page9

Model answers

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

Question 15 marks

1 / (a):

\displaystyle g(-x)=\frac{-x^3}{x^2+1}=-g(x)

, so

g

is odd. (M1, A1) 1 / (b):

\cos(-x)=\cos x

, so

h

is even. (A1) 1 / (c): The odd terms cancel and the even terms double:

f(2)+f(-2)=2\cos 2

. (M1, A1)

Question 25 marks

2: When

n=1

, the left-hand side is

1=1^2

, so the statement holds. (B1) Assume it holds for

n=k

, so

1+3+\cdots +(2k-1)=k^2

. (M1) For

n=k+1

, add the next term

2(k+1)-1=2k+1

: the sum is

k^2+2k+1=(k+1)^2

. (M1, A1) Therefore, by mathematical induction, the identity holds for every positive integer

n

. (C1)

Question 35 marks

3: By the binomial theorem, the term containing

x^3

\binom{5}{3}(1)^2(2x)^3

. (M1) Its coefficient is

\binom{5}{3}\times 2^3=10\times 8=80

. (M1, A1) Answer: 80. (A1, B1)

Question 45 marks

2\sin x\cos x=\cos x

, so

\cos x(2\sin x-1)=0

. (M1) Thus either

\cos x=0

, giving

\displaystyle x=\frac{\pi}{2},\frac{3\pi}{2}

, or

\sin x=\frac12

, giving

\displaystyle x=\frac{\pi}{6},\frac{5\pi}{6}

in the stated interval. (M1, A1, A1) Hence

\displaystyle x=\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{3\pi}{2}

. (A1)

Question 55 marks

5: Rewrite the expression as

\left[\left(1+\frac3x\right)^{x/3}\right]^6

. (M1) Let

n=x/3

; as

x\to\infty

n\to\infty

and the expression in square brackets tends to

e

. (M1, A1) Therefore the required limit is

e^6

. (A1, B1)

Question 65 marks

6 / (a): By the product rule,

\displaystyle \frac{dy}{dx}=2xe^x+(x^2+1)e^x=e^x(x^2+2x+1)=e^x(x+1)^2

. (M1, A1, A1) 6 / (b): At

x=0

\displaystyle \frac{dy}{dx}=1

. The gradient is 1. (M1, A1)

Question 75 marks

7 / (a): Integrating,

G(x)=x^3-2x^2+2x+C

. (M1, A1) Using

G(0)=1

gives

C=1

, so

G(x)=x^3-2x^2+2x+1

. (M1, A1) 7 / (b):

G(2)=8-8+4+1=5

. (A1)

Question 85 marks

8: The curve crosses the

x

-axis at

x=-1

and

x=1

. It is below the axis on

[-1,1]

and above it on

[1,2]

. (M1, A1) Area

=\int_{-1}^{1}(1-x^2)\,dx+\int_{1}^{2}(x^2-1)\,dx

. (M1) The first integral is

\frac43

and the second is

\frac43

. (M1) Total area

=\frac83

square units. (A1)

Question 95 marks

9 / (a): The coefficient matrix is

\begin{pmatrix}2&1\\1&-1\end{pmatrix}

. Its determinant is

2(-1)-1(1)=-3

. (M1, A1) 9 / (b): Adding the equations gives

3x=6

, hence

x=2

. (M1, A1) Substitution into

x-y=1

gives

y=1

. The solution is

(x,y)=(2,1)

. (M1)

Question 105 marks

10 / (a):

\mathbf{a}\cdot\mathbf{b}=1(2)+2(-1)+0(2)=0

. (M1, A1) 10 / (b):

\mathbf{a}\times\mathbf{b}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&2&0\\2&-1&2\end{vmatrix}=(4,-2,-5)

. (M1, A1, A1)

Question 1112 marks

11 / (a): Put

u=2x

. Then

\displaystyle \frac{e^{2x}-1}{x}=2\frac{e^u-1}{u}

. (M1) Using

\displaystyle \lim_{u\to0}\frac{e^u-1}{u}=1

, the limit is 2. (M1, A1) 11 / (b):

f'(x)=e^{-x}(1-x)

. (M1, A1) The stationary point satisfies

1-x=0

, so

x=1

and

f(1)=1/e

. (M1, A1) For

0\leq x<1

f'(x)>0

, and for

x>1

f'(x)<0

, so this is a maximum. (M1) 11 / (c): At

x=2

f(2)=2e^{-2}

and

f'(2)=-e^{-2}

. (M1, A1) The tangent is

y-2e^{-2}=-e^{-2}(x-2)

, or

y=(4-x)e^{-2}

. (M1, A1)

Question 1213 marks

12 / (a):

\displaystyle \int(2x-x^2)\,dx=x^2-\frac{x^3}{3}+C

. (M1, A1, A1, A1) 12 / (b): The curves intersect where

2x=x^2

, so

x=0,2

. On this interval

2x\geq x^2

. (M1, A1) Area

\displaystyle =\int_0^2(2x-x^2)\,dx=\left[x^2-\frac{x^3}{3}\right]_0^2=\frac43

square units. (M1, A1, A1) 12 / (c): Half the enclosed area is

2/3

. Hence

\displaystyle a^2-\frac{a^3}{3}=\frac23

. (M1, A1) Rearranging gives

a^3-3a^2+2=0=(a-1)(a^2-2a-2)

. The roots are

1,1\pm\sqrt3

; only

a=1

lies in

(0,2)

. (M1, A1)

Question 1312 marks

13 / (a):

\det A=\lambda^2-1=(\lambda-1)(\lambda+1)

. (M1, A1) 13 / (b): A unique solution exists when

\det A\ne0

, hence when

\lambda\ne1,-1

. (M1, A1) 13 / (c): For

\lambda=2

, the equations are

2x+y=1

and

x+2y=1

. Subtracting gives

x-y=0

, so

x=y

. (M1, A1) Then

3x=1

, so

x=y=\frac13

. (M1, A1) 13 / (d): If

\lambda=1

, both equations reduce to

x+y=1

, so there are infinitely many solutions. (M1, A1) If

\lambda=-1

, the equations are

-x+y=1

and

x-y=1

. The first implies

x-y=-1

, contradicting the second; there is no solution. (M1, A1)

Question 1413 marks

14 / (a):

\overrightarrow{AB}=B-A=(4,0,0)

and

\overrightarrow{AC}=C-A=(1,3,0)

. (M1, A1) 14 / (b):

\overrightarrow{AB}\cdot\overrightarrow{AC}=4

and

|\overrightarrow{AB}|=4

|\overrightarrow{AC}|=\sqrt{10}

. (M1, A1) Thus

\displaystyle \cos\angle BAC=\frac{4}{4\sqrt{10}}=\frac1{\sqrt{10}}

, so

\angle BAC\approx72^\circ

. (M1, A1) 14 / (c):

\overrightarrow{AB}\times\overrightarrow{AC}=(0,0,12)

. (M1, A1, A1) The area of triangle

ABC

is half the magnitude of this vector:

\frac12(12)=6

square units. (A1) 14 / (d): A normal vector is

(0,0,12)

, whose magnitude is 12. (M1, A1) A unit vector perpendicular to both is

(0,0,1)

; its negative is also acceptable. (A1)