FREE HKDSE PRACTICE PAPER

HKDSE Maths M2 Mock Paper: Algebra and Calculus

This 100-mark sample follows the HKEAA M2 format: a page of trigonometric formulas for reference, then Section A with 9 short questions of 4 to 7 marks, each with one mark total, and Section B with 4 long questions of 12 or 13 marks, 50 marks per section. Allow 2 hours 30 minutes. It opens with differentiation from first principles and covers the binomial theorem, mathematical induction, trigonometric identities, implicit differentiation, integration by parts, matrices, vectors, systems of linear equations, curve sketching, definite integrals and volumes of revolution. Model answers use HKEAA 1M/1A marking with exact answers.

13 questions100 marks2 hr 30 min

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An original practice paper written by Studify in the HKEAA format. This is an independent sample, not an official HKEAA paper.

Paper structure

Section A9 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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2027-DSEMATH EPM2

STUDIFY MOCK EXAMINATION

HONG KONG DIPLOMA OF SECONDARY EDUCATION EXAMINATION 2027 · PRACTICE PAPER

MATHEMATICS Extended PartModule 2 (Algebra and Calculus)

Question-Answer Book

8.30 am – 11.00 am (2½ hours)
This paper must be answered in English

INSTRUCTIONS

(1)After the announcement of the start of the examination, you should first write your Candidate Number in the space provided on Page 1 and stick barcode labels in the spaces provided.
(2)This paper consists of TWO sections, A and B.
(3)Attempt ALL questions in this paper. Write your answers in the spaces provided in this Question-Answer Book. Do not write in the margins. Answers written in the margins will not be marked.
(4)Graph paper and supplementary answer sheets will be supplied on request. Write your Candidate Number, mark the question number box and stick a barcode label on each sheet, and fasten them with string INSIDE this book.
(5)Unless otherwise specified, all working must be clearly shown.
(6)Unless otherwise specified, numerical answers must be exact.
(7)The diagrams in this paper are not necessarily drawn to scale.
(8)No extra time will be given to candidates for sticking on the barcode labels or filling in the question number boxes after the 'Time is up' announcement.

Studify practice paper. Not an official HKEAA examination paper.

Please stick the barcode label here.

Candidate Number

FORMULAS FOR REFERENCE

\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B

\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B

\displaystyle \tan(A\pm B)=\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}

2\sin A\cos B=\sin(A+B)+\sin(A-B)

2\cos A\cos B=\cos(A+B)+\cos(A-B)

2\sin A\sin B=\cos(A-B)-\cos(A+B)

\displaystyle \sin A+\sin B=2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2}

\displaystyle \sin A-\sin B=2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2}

\displaystyle \cos A+\cos B=2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2}

\displaystyle \cos A-\cos B=-2\sin\dfrac{A+B}{2}\sin\dfrac{A-B}{2}

Answers written in the margins will not be marked.

SECTION A (50 marks)

Find

\displaystyle \dfrac{d}{dx}\left(\dfrac{1}{3x+2}\right)

from first principles.(4 marks)

Answers written in the margins will not be marked.

In the expansion of

(1+2x)^{n}

, the coefficients of

x

and

x^{2}

are 16 and 112 respectively, where

n

is a positive integer. (a) Find

n

. (b) Find the constant term in the expansion of

\displaystyle (1+2x)^{n}\left(1-\dfrac{1}{x}\right)^{2}

.(5 marks)

Answers written in the margins will not be marked.

Prove, by mathematical induction, that

\displaystyle 1\times 2+2\times 3+\cdots+n(n+1)=\dfrac{n(n+1)(n+2)}{3}

for all positive integers

n

.(4 marks)

Answers written in the margins will not be marked.

(a) Prove that

\sin 3\theta=3\sin\theta-4\sin^{3}\theta

. (b) Solve

\sin 3\theta=2\sin\theta

, where

0\leq\theta\leq\pi

.(6 marks)

Answers written in the margins will not be marked.

Consider the curve

C: x^{2}+xy+y^{3}=3

. (a) Find

\displaystyle \dfrac{dy}{dx}

in terms of

x

and

y

. (b) Find the equation of the tangent to

C

at the point

(1,\,1)

.(6 marks)

Answers written in the margins will not be marked.

(a) Using integration by parts, find

\displaystyle\int xe^{2x}\,dx

. (b) Find the exact value of the area of the region bounded by the curve

y=xe^{2x}

, the

x

-axis and the straight line

x=1

.(7 marks)

Answers written in the margins will not be marked.

Let

A=\begin{pmatrix}2&1\\3&2\end{pmatrix}

. (a) Find

A^{-1}

. (b) Using (a), solve the system of linear equations

\begin{cases}2x+y=5\\3x+2y=8\end{cases}

.(5 marks)

Answers written in the margins will not be marked.

Let

O

be the origin. It is given that

\overrightarrow{OA}=\mathbf{i}+2\mathbf{j}-\mathbf{k}

and

\overrightarrow{OB}=3\mathbf{i}-\mathbf{j}+2\mathbf{k}

. (a) Find

\overrightarrow{OA}\cdot\overrightarrow{OB}

and

\overrightarrow{OA}\times\overrightarrow{OB}

. (b) Is

\angle AOB

an obtuse angle? Explain your answer. (c) Find the area of

\triangle OAB

.(6 marks)

Answers written in the margins will not be marked.

Consider the system of linear equations in real variables

x

y

and

z

(E):\ \begin{cases}x+y+z=1\\x+2y+az=2\\2x+3y+4z=b\end{cases}

, where

a

and

b

are real constants. (a) Find the range of values of

a

such that

(E)

has a unique solution. (b) Suppose that

a=3

. If

(E)

is consistent, find

b

and solve

(E)

.(7 marks)

Answers written in the margins will not be marked.

SECTION B (50 marks)

10.

Let

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

, where

x\neq 1

(a)

Find

f'(x)

.(2 marks)

(b)

Find the maximum point(s) and the minimum point(s) of the graph of

y=f(x)

.(3 marks)

(c)

Find the asymptote(s) of the graph of

y=f(x)

.(3 marks)

(d)

Find the area of the region bounded by the graph of

y=f(x)

, the oblique asymptote found in (c) and the straight lines

x=2

and

x=5

.(4 marks)

Answers written in the margins will not be marked.

11.

Let

f(x)

be a continuous function defined on the interval

[0,\,\pi]

(a)

Prove that

\displaystyle \displaystyle\int_{0}^{\pi}\sin^{2}x\,dx=\dfrac{\pi}{2}

.(2 marks)

(b)

Find the volume of the solid of revolution generated by revolving the region bounded by the curve

y=\sin x

(0\leq x\leq\pi)

and the

x

-axis about the

x

-axis.(2 marks)

(c)

Find

\displaystyle\int x\sin x\,dx

.(3 marks)

(d)

Prove that

\displaystyle \displaystyle\int_{0}^{\pi}xf(\sin x)\,dx=\dfrac{\pi}{2}\int_{0}^{\pi}f(\sin x)\,dx

.(3 marks)

(e)

Using (a) and (d), evaluate

\displaystyle\int_{0}^{\pi}x\sin^{2}x\,dx

.(3 marks)

Answers written in the margins will not be marked.

12.

Let

M=\begin{pmatrix}3&1\\-2&0\end{pmatrix}

and let

I

be the

2\times 2

identity matrix.

(a)

Prove that

M^{2}=3M-2I

.(2 marks)

(b)

Prove, by mathematical induction, that

M^{n}=(2^{n}-1)M-(2^{n}-2)I

for all positive integers

n

.(4 marks)

(c)

Find

M^{10}

.(2 marks)

(d)

Let

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

. Find

P^{-1}

and

P^{-1}MP

.(2 marks)

(e)

Using (d), verify the entry in the first row and first column of

M^{n}

obtained in (b).(2 marks)

Answers written in the margins will not be marked.

13.

Let

O

be the origin. The position vectors of the points

A

B

and

C

are

\mathbf{i}+2\mathbf{k}

3\mathbf{i}+\mathbf{j}

and

2\mathbf{j}+\mathbf{k}

respectively. Let

\Pi

be the plane which contains

A

B

and

C

(a)

Find

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

.(2 marks)

(b)

Find the area of

\triangle ABC

.(2 marks)

(c)

Find the distance from

O

\Pi

.(3 marks)

(d)

Let

D

be the foot of the perpendicular from

O

\Pi

. Find

\overrightarrow{OD}

.(3 marks)

(e)

Someone claims that the angle between the straight line

OA

and

\Pi

is greater than

45^\circ

. Do you agree? Explain your answer.(3 marks)

Answers written in the margins will not be marked.

END OF PAPER

Model answers

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

Question 14 marks

\displaystyle \dfrac{d}{dx}\left(\dfrac{1}{3x+2}\right)=\lim\limits_{h\to 0}\dfrac{1}{h}\left[\dfrac{1}{3(x+h)+2}-\dfrac{1}{3x+2}\right]

\displaystyle =\lim\limits_{h\to 0}\dfrac{-3h}{h(3x+3h+2)(3x+2)}

\displaystyle =\lim\limits_{h\to 0}\dfrac{-3}{(3x+3h+2)(3x+2)}

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

1A (4)

Question 25 marks

(a)

2n=16

, so

n=8

. 1A (Check:

C^{8}_{2}(2)^{2}=112

.) (b)

(1+2x)^{8}=1+16x+112x^{2}+\cdots

and

\displaystyle \left(1-\dfrac{1}{x}\right)^{2}=1-\dfrac{2}{x}+\dfrac{1}{x^{2}}

1M Constant term

=1(1)+16(-2)+112(1)

1M+1M

=81

1A (5)

Question 34 marks

When

n=1

, L.H.S.

=2

and R.H.S.

\displaystyle =\dfrac{1(2)(3)}{3}=2

, so the statement is true for

n=1

. 1 Assume

\displaystyle 1\times 2+\cdots+k(k+1)=\dfrac{k(k+1)(k+2)}{3}

for some positive integer

k

. 1M Then

\displaystyle 1\times 2+\cdots+k(k+1)+(k+1)(k+2)=\dfrac{k(k+1)(k+2)}{3}+(k+1)(k+2)=\dfrac{(k+1)(k+2)(k+3)}{3}

1M So the statement is true for

n=k+1

if it is true for

n=k

. By the principle of mathematical induction, it is true for all positive integers

n

. 1 (4)

Question 46 marks

(a)

\sin 3\theta=\sin 2\theta\cos\theta+\cos 2\theta\sin\theta

=2\sin\theta\cos^{2}\theta+(1-2\sin^{2}\theta)\sin\theta

=2\sin\theta(1-\sin^{2}\theta)+\sin\theta-2\sin^{3}\theta=3\sin\theta-4\sin^{3}\theta

1 (b)

3\sin\theta-4\sin^{3}\theta=2\sin\theta

, so

\sin\theta(1-4\sin^{2}\theta)=0

\sin\theta=0

\displaystyle \sin\theta=\pm\dfrac{1}{2}

\displaystyle \theta=0,\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6}

\pi

1A (6)

Question 56 marks

(a)

\displaystyle 2x+y+x\dfrac{dy}{dx}+3y^{2}\dfrac{dy}{dx}=0

1M+1A

\displaystyle \dfrac{dy}{dx}=-\dfrac{2x+y}{x+3y^{2}}

1A (b) At

(1,\,1)

\displaystyle \dfrac{dy}{dx}=-\dfrac{3}{4}

\displaystyle y-1=-\dfrac{3}{4}(x-1)

3x+4y-7=0

1A (6)

Question 67 marks

(a)

\displaystyle \displaystyle\int xe^{2x}\,dx=\dfrac{x}{2}e^{2x}-\int\dfrac{1}{2}e^{2x}\,dx

1M+1A

\displaystyle =\dfrac{x}{2}e^{2x}-\dfrac{1}{4}e^{2x}+C

1A (b) The curve meets the

x

-axis at

x=0

and

xe^{2x}\geq 0

for

0\leq x\leq 1

. 1M Area

\displaystyle =\displaystyle\int_{0}^{1}xe^{2x}\,dx=\left[\dfrac{x}{2}e^{2x}-\dfrac{1}{4}e^{2x}\right]_{0}^{1}

\displaystyle =\dfrac{e^{2}}{4}+\dfrac{1}{4}

\displaystyle =\dfrac{e^{2}+1}{4}

1A (7)

Question 75 marks

(a)

\det A=4-3=1

A^{-1}=\begin{pmatrix}2&-1\\-3&2\end{pmatrix}

1A (b)

\begin{pmatrix}x\\y\end{pmatrix}=A^{-1}\begin{pmatrix}5\\8\end{pmatrix}

=\begin{pmatrix}2\\1\end{pmatrix}

1M So

x=2

and

y=1

. 1A (5)

Question 86 marks

(a)

\overrightarrow{OA}\cdot\overrightarrow{OB}=3-2-2=-1

\overrightarrow{OA}\times\overrightarrow{OB}=3\mathbf{i}-5\mathbf{j}-7\mathbf{k}

1M+1A (b)

\displaystyle \cos\angle AOB=\dfrac{-1}{|\overrightarrow{OA}||\overrightarrow{OB}|}<0

, so

\angle AOB

is obtuse. 1A f.t. (c) Area

\displaystyle =\dfrac{1}{2}\left|\overrightarrow{OA}\times\overrightarrow{OB}\right|=\dfrac{1}{2}\sqrt{9+25+49}

\displaystyle =\dfrac{\sqrt{83}}{2}

1A (6)

Question 97 marks

(a)

\begin{vmatrix}1&1&1\\1&2&a\\2&3&4\end{vmatrix}=(8-3a)-(4-2a)+(3-4)=3-a

1M+1A

(E)

has a unique solution if and only if

a\neq 3

. 1A (b) Subtracting:

y+2z=1

and

y+2z=b-2

1M Consistency requires

b-2=1

, so

b=3

. 1A Let

z=t

. Then

y=1-2t

and

x=1-y-z=t

. 1M The solution is

(x,\,y,\,z)=(t,\,1-2t,\,t)

, where

t

is any real number. 1A (7)

Question 1012 marks

(a)

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

1M+1A (b)

f'(x)=0

gives

x=-1

x=3

. 1M

(x-1)^{2}>0

, so

f'(x)

has the sign of

(x-3)(x+1)

: it changes from

+

-

x=-1

and from

-

+

x=3

. 1M Maximum point

(-1,\,-2)

; minimum point

(3,\,6)

. 1A (c) Vertical asymptote:

x=1

. 1A

\displaystyle f(x)=x+1+\dfrac{4}{x-1}

, so the oblique asymptote is

y=x+1

. 1M+1A (d) Area

\displaystyle =\displaystyle\int_{2}^{5}\left[f(x)-(x+1)\right]dx=\int_{2}^{5}\dfrac{4}{x-1}\,dx

1M+1M

=\big[4\ln(x-1)\big]_{2}^{5}

=8\ln 2

1A (12)

Question 1113 marks

(a)

\displaystyle \displaystyle\int_{0}^{\pi}\sin^{2}x\,dx=\int_{0}^{\pi}\dfrac{1-\cos 2x}{2}\,dx=\left[\dfrac{x}{2}-\dfrac{\sin 2x}{4}\right]_{0}^{\pi}=\dfrac{\pi}{2}

1M+1 (b) Volume

=\pi\displaystyle\int_{0}^{\pi}\sin^{2}x\,dx

\displaystyle =\dfrac{\pi^{2}}{2}

1A (c)

\displaystyle\int x\sin x\,dx=-x\cos x+\int\cos x\,dx

1M+1A

=-x\cos x+\sin x+C

1A (d) Let

u=\pi-x

. Then

dx=-du

and

\sin x=\sin u

. 1M

\displaystyle\int_{0}^{\pi}xf(\sin x)\,dx=\int_{0}^{\pi}(\pi-u)f(\sin u)\,du=\pi\int_{0}^{\pi}f(\sin u)\,du-\int_{0}^{\pi}uf(\sin u)\,du

1M So

2\displaystyle\int_{0}^{\pi}xf(\sin x)\,dx=\pi\int_{0}^{\pi}f(\sin x)\,dx

. 1 (e) Taking

f(t)=t^{2}

\displaystyle \displaystyle\int_{0}^{\pi}x\sin^{2}x\,dx=\dfrac{\pi}{2}\int_{0}^{\pi}\sin^{2}x\,dx

\displaystyle =\dfrac{\pi}{2}\cdot\dfrac{\pi}{2}

\displaystyle =\dfrac{\pi^{2}}{4}

1A (13)

Question 1212 marks

(a)

M^{2}=\begin{pmatrix}7&3\\-6&-2\end{pmatrix}=3\begin{pmatrix}3&1\\-2&0\end{pmatrix}-2\begin{pmatrix}1&0\\0&1\end{pmatrix}

1M+1 (b) For

n=1

(2-1)M-(2-2)I=M

, true. 1 Assume

M^{k}=(2^{k}-1)M-(2^{k}-2)I

. Then

M^{k+1}=(2^{k}-1)M^{2}-(2^{k}-2)M

=(2^{k}-1)(3M-2I)-(2^{k}-2)M

=(2^{k+1}-1)M-(2^{k+1}-2)I

, so the statement is true for all positive integers

n

. 1 (c)

M^{10}=1023M-1022I

=\begin{pmatrix}2047&1023\\-2046&-1022\end{pmatrix}

1A (d)

P^{-1}=\begin{pmatrix}2&1\\-1&-1\end{pmatrix}

P^{-1}MP=\begin{pmatrix}2&0\\0&1\end{pmatrix}

1A (e)

(P^{-1}MP)^{n}=P^{-1}M^{n}P

, so

M^{n}=P\begin{pmatrix}2^{n}&0\\0&1\end{pmatrix}P^{-1}

1M The entry in the first row and first column is

2^{n}(2)+1(-1)=2^{n+1}-1

, which agrees with (b):

3(2^{n}-1)-(2^{n}-2)=2^{n+1}-1

. 1 (12)

Question 1313 marks

(a)

\overrightarrow{AB}=2\mathbf{i}+\mathbf{j}-2\mathbf{k}

and

\overrightarrow{AC}=-\mathbf{i}+2\mathbf{j}-\mathbf{k}

\overrightarrow{AB}\times\overrightarrow{AC}=3\mathbf{i}+4\mathbf{j}+5\mathbf{k}

1A (b) Area

\displaystyle =\dfrac{1}{2}\sqrt{3^{2}+4^{2}+5^{2}}

\displaystyle =\dfrac{5\sqrt{2}}{2}

1A (c) Let

\mathbf{n}=3\mathbf{i}+4\mathbf{j}+5\mathbf{k}

, a normal vector to

\Pi

. 1M Distance

\displaystyle =\dfrac{|\overrightarrow{OA}\cdot\mathbf{n}|}{|\mathbf{n}|}=\dfrac{3+10}{5\sqrt{2}}

\displaystyle =\dfrac{13\sqrt{2}}{10}

1A (d)

\displaystyle \overrightarrow{OD}=\dfrac{\overrightarrow{OA}\cdot\mathbf{n}}{|\mathbf{n}|^{2}}\mathbf{n}=\dfrac{13}{50}(3\mathbf{i}+4\mathbf{j}+5\mathbf{k})

1M+1M

\displaystyle =\dfrac{39}{50}\mathbf{i}+\dfrac{26}{25}\mathbf{j}+\dfrac{13}{10}\mathbf{k}

1A (e) Let

\phi

be the angle between

OA

and

\Pi

\displaystyle \sin\phi=\dfrac{OD}{OA}=\dfrac{13\sqrt{2}/10}{\sqrt{5}}=\dfrac{13\sqrt{10}}{50}

\displaystyle \sin 45^\circ=\dfrac{\sqrt{2}}{2}

and

\displaystyle \left(\dfrac{13\sqrt{10}}{50}\right)^{2}=\dfrac{169}{250}>\dfrac{1}{2}

1M So

\phi>45^\circ

and the claim is agreed. 1A f.t. (13)