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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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乙部4 題 · 50 分

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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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香港中學文憑考試模擬試卷

數學延伸 M2

Paper: Algebra and Calculus (2027 onwards)

作答時間： 2 hr 30 min總分： 100

考生須知：

本試卷分為多個部分。
本試卷所有題目均須作答。
請在指定位置作答。
除另有說明外，須清楚列出所有計算步驟。
除另有說明外，數值答案須為準確值或準確至三位有效數字。

由 studify.life 生成 • 16/10/2026

A

題目 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 分)

(b)

Determine whether

h(x)=\cos x

is odd, even or neither.

(1 分)

(c)

Find

f(2)+f(-2)

(2 分)

題目 2

Use mathematical induction to prove that, for every positive integer

n

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

.(5 分)

題目 3

Find the coefficient of

x^3

in the expansion of

(1+2x)^5

.(5 分)

題目 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 分)

題目 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 分)

題目 6

The curve

C

has equation

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

(a)

Find

\displaystyle \dfrac{dy}{dx}

(3 分)

頁次：2

題目 6 （續）

(b)

Find the gradient of

C

x=0

(2 分)

題目 7

A function

G

satisfies

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

and

G(0)=1

(a)

Find

G(x)

(3 分)

(b)

Hence find

G(2)

(2 分)

題目 8

Find the area enclosed by the curve

y=x^2-1

and the

x

-axis between

x=-1

and

x=2

.(5 分)

題目 9

Consider the simultaneous equations

2x+y=5

and

x-y=1

(a)

Find the determinant of the coefficient matrix.

(2 分)

(b)

Solve the simultaneous equations.

(3 分)

頁次：3

題目 10

Let

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

and

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

(a)

Find

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

(2 分)

(b)

Find

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

(3 分)

頁次：4

B

題目 11

Consider

f(x)=xe^{-x}

for

x\geq 0

(a)

Evaluate

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

(3 分)

(b)

Find the stationary point of

f

and determine its nature.

(5 分)

(c)

Find the equation of the tangent to the graph of

y=f(x)

x=2

(4 分)

頁次：5

題目 12

Consider the curves

y=2x

and

y=x^2

for

0\leq x\leq2

(a)

Find

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

(4 分)

(b)

Find the area enclosed by the two curves.

(5 分)

(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 分)

頁次：6

題目 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 分)

(b)

State the values of

\lambda

for which the system has a unique solution.

(2 分)

(c)

Solve the system when

\lambda=2

(4 分)

(d)

Discuss the system when

\lambda=1

and when

\lambda=-1

(4 分)

頁次：7

題目 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 分)

(b)

Find the angle

BAC

, giving your answer to the nearest degree.

(4 分)

(c)

Find

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

and hence find the area of triangle

ABC

(4 分)

頁次：8

題目 14 （續）

(d)

Find a unit vector perpendicular to both

\overrightarrow{AB}

and

\overrightarrow{AC}

(3 分)

頁次：9

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5 分

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)

第 2 題5 分

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)

第 3 題5 分

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)

第 4 題5 分

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)

第 5 題5 分

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)

第 6 題5 分

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)

第 7 題5 分

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)

第 8 題5 分

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)

第 9 題5 分

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)

第 10 題5 分

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)

第 11 題12 分

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)

第 12 題13 分

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)

第 13 題12 分

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)

第 14 題13 分

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)