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HKDSE 數學延伸部分 M2 模擬試卷：代數與微積分

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13 題100 分2 小時 30 分鐘

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Studify 按考評局格式編寫嘅原創練習卷，屬獨立樣本，並非考評局官方試卷。

試卷結構

甲部9 題 · 50 分
乙部4 題 · 50 分

呢份試卷涵蓋嘅課題

奇函數和偶函數
數學歸納法
二項式定理
續三角函數
e的簡介
極限
求導法
不定積分法及其應用
定積分法
定積分法的應用
行列式
矩陣
線性方程組
向量的簡介
純量積與向量積
求導法的應用
向量的應用

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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 分)

(b)

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

y=f(x)

.(3 分)

(c)

Find the asymptote(s) of the graph of

y=f(x)

.(3 分)

(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 分)

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 分)

(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 分)

(c)

Find

\displaystyle\int x\sin x\,dx

.(3 分)

(d)

Prove that

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

.(3 分)

(e)

Using (a) and (d), evaluate

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

.(3 分)

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 分)

(b)

Prove, by mathematical induction, that

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

for all positive integers

n

.(4 分)

(c)

Find

M^{10}

.(2 分)

(d)

Let

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

. Find

P^{-1}

and

P^{-1}MP

.(2 分)

(e)

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

M^{n}

obtained in (b).(2 分)

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 分)

(b)

Find the area of

\triangle ABC

.(2 分)

(c)

Find the distance from

O

\Pi

.(3 分)

(d)

Let

D

be the foot of the perpendicular from

O

\Pi

. Find

\overrightarrow{OD}

.(3 分)

(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 分)

Answers written in the margins will not be marked.

END OF PAPER

參考答案

揀一條題目就會顯示答案同評分指引。建議先自己試做。

第 1 題4 分

\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)

第 2 題5 分

(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)

第 3 題4 分

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)

第 4 題6 分

(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)

第 5 題6 分

(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)

第 6 題7 分

(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)

第 7 題5 分

(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)

第 8 題6 分

(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)

第 9 題7 分

(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)

第 10 題12 分

(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)

第 11 題13 分

(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)

第 12 題12 分

(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)

第 13 題13 分

(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)