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HKDSE 數學延伸部分 M1 模擬試卷：微積分與統計

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試卷結構

甲部10 題 · 50 分
乙部4 題 · 50 分

呢份試卷涵蓋嘅課題

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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香港中學文憑考試模擬試卷

數學延伸 M1

Paper: Calculus and Statistics (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

Section A

題目 1

Answer all parts.

(a)

Find the coefficient of

x^3

in the expansion of

(1+2x)^5

(2 分)

(b)

Find the first three terms, in ascending powers of

x

, in the expansion of

(1-3x)^{-2}

(3 分)

題目 2

Answer all parts.

(a)

Solve

e^{2x}=7

for

x

, giving your answer to 3 significant figures.

(2 分)

(b)

Solve

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

, stating any restriction on

x

(3 分)

題目 3

Let

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

(a)

Find

f'(x)

(2 分)

頁次：2

題目 3 （續）

(b)

Find

f''(x)

and hence find the value of

f''(2)

(3 分)

題目 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 分)

(b)

Find this maximum height.

(2 分)

題目 5

Answer all parts.

(a)

Find

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

(3 分)

(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 分)

頁次：3

題目 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 分)

(b)

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

(3 分)

題目 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 分)

(b)

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

B

(3 分)

題目 8

A discrete random variable

X

has probability distribution

P(X=x)=kx

for

x=1,2,3

(a)

Find

k

(2 分)

頁次：4

題目 8 （續）

(b)

Find

E(X)

(2 分)

(c)

Find

\operatorname{Var}(X)

(1 分)

題目 9

Answer all parts.

(a)

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

(3 分)

(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 分)

題目 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 分)

頁次：5

題目 10 （續）

(b)

Find

c

such that

P(X<c)=0.1587

(2 分)

頁次：6

Section B

題目 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 分)

(b)

Find the value of

x

that gives the maximum area of the garden.

(4 分)

(c)

Find the maximum area.

(2 分)

(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 分)

頁次：7

題目 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 分)

(b)

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

(4 分)

(c)

State the distribution of

X

and find

P(X\geq 2)

(4 分)

題目 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 分)

頁次：8

題目 13 （續）

(b)

Using the sample, find a 95% confidence interval for

\mu

. Use

z_{0.975}=1.96

(4 分)

(c)

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

\mu=72

(3 分)

(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 分)

題目 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 分)

頁次：9

題目 14 （續）

(b)

Using your estimate of

\lambda

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

(4 分)

(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 分)

頁次：10

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

(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.)

第 2 題5 分

(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.)

第 3 題5 分

(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.)

第 4 題5 分

(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.)

第 5 題5 分

(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.)

第 6 題5 分

(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.)

第 7 題5 分

(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.)

第 8 題5 分

(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.)

第 9 題5 分

(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.)

第 10 題5 分

(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.)

第 11 題12 分

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

第 12 題12 分

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.)

第 13 題13 分

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.)

第 14 題13 分

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.)