Special Features

Special Features

HKDSE Examination Preparation Guide

provides information about the assessment format and useful guidelines for answering questions in the HKDSE public examination.

HKDSE Examination Format

is fully adopted in all the Mock Exam Papers to help

candidates to familiarize with the public examination format.

c54h_Front matter.indd 3 2014/5/23 10:48:09 AM

Solution Guide

clearly shows the steps and the marking scheme in answering each question.

Questions similar to those in 2013–2014 HKDSE

are provided and marked clearly to help candidates to familiarize with the latest question types and thus enhance the effectiveness of their preparations for the public examination.

Guidelines

suggest useful thinking strategy in answering similar type of questions and also remind candidates about important knowledge and formulas.

Common Mistakes

remind candidates about some common misconceptions or careless mistakes.

c54h_Front matter.indd 4 2014/5/23 10:48:10 AM

Content

HKDSE Examination Preparation Guide

Mock Exam Papers

Mock Exam 1 Mock Exam 2 Mock Exam 3 Mock Exam 4 Mock Exam 5 Mock Exam 6

Solution Guide

c54h_Front matter.indd 5 2014/5/23 10:48:10 AM

In order to prepare for the examination effectively, candidates are advised to read the instruction of the assessment carefully.

A. Public Assessment Format

There is only one examination paper consisting conventional questions alone in the HKDSE Mathematics (Extended Part) Module 1 examination. The following table lists the details of the Module 1 (Calculus and Statistics) examination:

Module 1 (Calculus and Statistics)

Component Weighting Duration

Conventional questions 100% 2 ½ hours

The examination paper consists of two sections A and B, in which ALL questions are to be attempted.

Section A (50 marks) consists of shorter questions related to the whole Module 1 curriculum. Answers to questions in Section A should be written in the spaces provided in the Question-Answer Book. Section B (50 marks) consists of longer and harder questions related to the whole Module 1 curriculum. Note that the content to be examined includes knowledge of the subject matter in the Mathematics (Compulsory Part) curriculum.

B. Standard Referencing and Reporting of Results

In the HKDSE, standards-referenced reporting will be adopted to report candidates’ results. Candidates’ levels of performance will be reported with reference to a set of standards as defined by cut scores on the variable or scale for a given subject (see the figure below).

Cut scores

Variable/scale 1

U 2 3 4 5

HKDSE Examination Preparation Guide

c54h_Front matter.indd 6 2014/5/23 10:48:11 AM

There will be five cut scores in each subject to distinguish five levels of performance (1–5), with 5 being the highest. A performance below the threshold cut score for Level 1 will be labelled as “Unclassified” (U).

Level 5 candidates with the best performance will have their results annotated with the symbols ‘**’ and the next top group with the symbol ‘*’.

For each of the five levels, a set of written descriptors will be developed that describe what the typical candidate performing at this level is able to do. These descriptors will necessarily represent “on-average”

statements and may not apply precisely to individuals, whose performance within a subject may be variable and span two or more levels.

Note that the levels awarded to candidates in the Extended Part will be reported separately from the Compulsory Part.

C. Time Allocation

Section Number of questions Time spent per question

Section A 9 to 10 Questions ~ 7 minutes

Section B 4 to 5 Questions ~ 18 minutes

Time left for checking: 5 – 10 minutes

c54h_Front matter.indd 7 2014/5/23 10:48:11 AM

1

© 樂思教育出版有限公司    保留版權 Pan Lloyds Publishers Ltd

All Rights Reserved 2014

Pan Lloyds Publishers Ltd

MATHEMATICS Extended Part Module 1 (Calculus and Statistics)

Mock Exam 1 Question-Answer Book

(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 on Pages 1, 3, 5, 7, 9 and 11.

2. Answer 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.

3. 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.

4. Unless otherwise specified, all working must be clearly shown.

5. Unless otherwise specified, numerical answers should be either exact or given to 4 decimal places.

6. For definite integrals, answers obtained by using numerical integration functions in calculators are not accepted.

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.

Candidate Number

Marker’s

Use Only Examiner’s Use Only

Question

No. Marks Marks

1 2 3 4 5 6 7 8 9 10 11 12 13

Total

Please stick the barcode label here.

c54h_M1_Mock1.indd 1 2014/5/23 10:49:33 AM

Answers written in the margins will not be marked. Answers written in the margins will not be marked.

Answers written in the margins will not be marked.

© Pan Lloyds Publishers Ltd

2

HKDSE Exam Series — Mathematics (Extended Part) Mock Exam Papers (Module 1) (2015 Edition)

Section A (50 marks)

1. Peter

Snowball

Peter is rolling a spherical snowball. He keeps increasing his effort so that the radius of the snowball is increasing at a constant rate of 0.1 cm per second. Find the rate of change of the volume of the snowball when the radius is 50 cm.

(3 marks) 2014

c54h_M1_Mock1.indd 2 2014/5/23 10:49:33 AM

Answers written in the margins will not be marked. Answers written in the margins will not be marked.

Answers written in the margins will not be marked.

© Pan Lloyds Publishers Ltd

12

HKDSE Exam Series — Mathematics (Extended Part) Mock Exam Papers (Module 1) (2015 Edition)

Section B (50 marks)

10. (a) (i) Find ( ln ) d

d

t t² t .

(ii) Using (a)(i), or otherwise, find ∫ tln tdt.

(3 marks) (b)

y = xe^2x x = 1

x y

0

The above figure shows the curve y = xe^2x. The shaded region is bounded by the curve, the line x = 1 and the x-axis. Using a suitable substitution and the result of (a), show that the area of the shaded region is e

4

2 1

+ . (5 marks)

(c) (i) Let I be the estimated area of the shaded region by using the trapezoidal rule with 10 sub- intervals. Show that

I = e 20

2 + 20

1 [(0.2)e^0.2 + (0.4)e^0.4 + (0.6)e^0.6 + … + (1.8)e^1.8].

(ii) Florence claims that (0.2)e^0.2 + (0.4)e^0.4 + (0.6)e^0.6 + … + (1.8)e^1.8 > 4e² + 5. Is she correct?

Explain your answer.

(6 marks) 2014

c54h_M1_Mock1.indd 12 2014/5/23 10:49:38 AM

© Pan Lloyds Publishers Ltd

24

HKDSE Exam Series — Mathematics (Extended Part) Mock Exam Papers (Module 1) (2015 Edition)

Standard Normal Distribution Table

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

0.0 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .0359

0.1 .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0753

0.2 .0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141

0.3 .1179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 .1517

0.4 .1554 .1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .1879

0.5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224

0.6 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2517 .2549

0.7 .2580 .2611 .2642 .2673 .2704 .2734 .2764 .2794 .2823 .2852

0.8 .2881 .2910 .2939 .2967 .2995 .3023 .3051 .3078 .3106 .3133

0.9 .3159 .3186 .3212 .3238 .3264 .3289 .3315 .3340 .3365 .3389

1.0 .3413 .3438 .3461 .3485 .3508 .3531 .3554 .3577 .3599 .3621

1.1 .3643 .3665 .3686 .3708 .3729 .3749 .3770 .3790 .3810 .3830

1.2 .3849 .3869 .3888 .3907 .3925 .3944 .3962 .3980 .3997 .4015

1.3 .4032 .4049 .4066 .4082 .4099 .4115 .4131 .4147 .4162 .4177

1.4 .4192 .4207 .4222 .4236 .4251 .4265 .4279 .4292 .4306 .4319

1.5 .4332 .4345 .4357 .4370 .4382 .4394 .4406 .4418 .4429 .4441

1.6 .4452 .4463 .4474 .4484 .4495 .4505 .4515 .4525 .4535 .4545

1.7 .4554 .4564 .4573 .4582 .4591 .4599 .4608 .4616 .4625 .4633

1.8 .4641 .4649 .4656 .4664 .4671 .4678 .4686 .4693 .4699 .4706

1.9 .4713 .4719 .4726 .4732 .4738 .4744 .4750 .4756 .4761 .4767

2.0 .4772 .4778 .4783 .4788 .4793 .4798 .4803 .4808 .4812 .4817

2.1 .4821 .4826 .4830 .4834 .4838 .4842 .4846 .4850 .4854 .4857

2.2 .4861 .4864 .4868 .4871 .4875 .4878 .4881 .4884 .4887 .4890

2.3 .4893 .4896 .4898 .4901 .4904 .4906 .4909 .4911 .4913 .4916

2.4 .4918 .4920 .4922 .4925 .4927 .4929 .4931 .4932 .4934 .4936

2.5 .4938 .4940 .4941 .4943 .4945 .4946 .4948 .4949 .4951 .4952

2.6 .4953 .4955 .4956 .4957 .4959 .4960 .4961 .4962 .4963 .4964

2.7 .4965 .4966 .4967 .4968 .4969 .4970 .4971 .4972 .4973 .4974

2.8 .4974 .4975 .4976 .4977 .4977 .4978 .4979 .4979 .4980 .4981

2.9 .4981 .4982 .4982 .4983 .4984 .4984 .4985 .4985 .4986 .4986

3.0 .4987 .4987 .4987 .4988 .4988 .4989 .4989 .4989 .4990 .4990

3.1 .4990 .4991 .4991 .4991 .4992 .4992 .4992 .4992 .4993 .4993

3.2 .4993 .4993 .4994 .4994 .4994 .4994 .4994 .4995 .4995 .4995

3.3 .4995 .4995 .4995 .4996 .4996 .4996 .4996 .4996 .4996 .4997

3.4 .4997 .4997 .4997 .4997 .4997 .4997 .4997 .4997 .4997 .4998

3.5 .4998 .4998 .4998 .4998 .4998 .4998 .4998 .4998 .4998 .4998

Note: An entry in the table is the area under the standard normal curve between x = 0 and x=z (z ≥ 0).

Areas for negative values of z can be obtained by symmetry.

( ) r

A z ^z 21 e ^x dx

2 0

2

=

#

^-

0 z

A(z)

x

c54h_M1_Mock1.indd 24 2014/5/23 10:49:42 AM

27 Mock Exam 5

Mock Exam 5

Section A

1. Let x cm and V cm³ be the side length and the volume of the cube respectively.

V = x³ d d t

V = d x d

t

3 ² x 1M

When x = 15 and d d

t V = –50,

–50 = 3(15)²d d t

x 1M

d d t x = 27

-2 1A

The rate of change of the side length is 27

- 2 cm per

second and it is decreasing. 1A

2. (a) e^t= y^y3-y

t= (y³-y) ln y 1A

ddy

t = (3y²- 1) ln y+ y y³-y

ddy

t = (3y²- 1) ln y+y²- 1 1A

Common Mistakes

Candidates may mistakenly differentiate (y³-y) ln y without using the product rule which would give a wrong result of y

y 3 ²-1

.

(b) x(t- 1) =x⁴- 2 t- 1 =x³-

x 2

t=x³- x

2 + 1 1A

dd x

t = 3x²+ x

2

2 1A

(c) ddy x

= dd x t ' ddy

t 1M

= x 3y 1 lny y 1 x

3 2

2 2 2

4

- + -

^ h+

6 @ 1A

3. (a) Slant height of the cone

= r²+r²

= 2r

A= 2rr(r) + 2rr 2r

= (2 + 2 2)rr² 1A (b) ddA

r = ddr[(2 + 2 2)rr²]

= (4 + 4 2)rr ddr

t = 5 2

ddt e ^10t

2

- ^-

^ h

=-2 10 e ₁₀^t2 2t

- c- m

= t 2 e

5 ¹⁰

t²

- 1A

dt dA

= ddA r · ddr

t

= (4 + 4 2)rr · t 2 e

5 ¹⁰

t²

- 1M

When t= 3, r= 5 - 2e^-109 and dt

dA

= (4 + 4 2)r(5 - 2e^-109)2 3 5 e ¹⁰

-9

^ h 1M

= 61.9713 (cor. to 4 d.p.) 1A Guidelines

Apply the chain rule to find ddAt .

(c) When t  ∞, r= 5 5- 22e ^10t

2

- ^-

^ h  5

A= (2 + 2 2)r(5)²

= 50(1 + 2)r 1A

4. (a) y= e dx 2

2

#

x ^1M

y= d 2 e₂^x _{c m}x

#

y=e^x2 +C, where C is a constant

∵ The y-intercept of R is 1.

∴ 1 =e⁰+C C= 0

The equation of R is:

y=e^x2 1A

c54h_M1_Mock5_sol.indd 27 2014/5/23 10:57:57 AM

28

HKDSE Exam Series — Mathematics (Extended Part) Mock Exam Papers (Module 1) (2015 Edition) Soulution Guide (b) Slope of the line 2x-y+ 1 = 0 is 2.

1e 2

x 2 = 2 x 2 = ln 4

x= 2 ln 4

When x= 2 ln 4, y=e^{21^2 4lnh}= 4. 1A The equation of T is:

4 ln y x 2 4

-

- = 2

y- 4 = 2x- 4 ln 4

y= 2x+ 4(1 - ln 4) 1A (c) The required area

= ^ln e^2x 2x 4 1 ln4 dx

0

2 4" -₆ + _^ - _h@,

#

1M

=62e^2x-x²-4 1x_^ -ln4_h@₀^{2 4ln} 1A

= 2e^{ln 4}- (2 ln 4)²- 4(2 ln 4)(1 - ln 4) - 2

= 2.5969 (cor. to 4 d.p.) 1A

Guidelines

Sketch the graphs of the curve and the line.

y

R T

x (2 ln 4, 4) 1

0

5. (a) Let X_

be the mean of the sample.

P(X_ < 33)

=P

25 4 33 35

<

Z -

f p ^1M

=P(Z < -2.5)

= 0.5 - 0.4938

= 0.0062 1A

(b) The required confidence interval

= 1.645

25 25 12 1

25 12

1.645

25 , 25

12

25 25 12 1

25 12 12 -

- -

+

c c

f ^{m m}p

1.645

25 25 12 1

25 12

1.645

25 ,25

12

25 25 12 1

25 12 12-

- -

+

c c

f ^{m m}p _{1M + 1A}

= (0.3156, 0.6444) (cor. to 4 d.p.) 1A

6. (a) P(Bill wins)

= P(Bill wins in the 1st round) + P(Bill wins in the 2nd round)

= P(only the 2nd outcome is ‘4’) + P(only the 5th outcome is ‘4’)

= 4

3 3

4 4

4

1 ⁴ 1

# +_{c m} # 1M

= 1024

273 1A

(b) P(Bill wins)

= 4

3 3

4 4 4

3 4 1 4

1 ⁴ 1 ⁷#

# +_{c m} # +_{c m} +g

= 4 3

4 1 1 4

3 ³ -_c

c m

m 1M

= 12

37 1A

(c) P(Win when throws more than twice | Bill wins)

= 37 12

1024 273 37 12

- 1M

= 4096

729 1A

Guidelines

It can be observed that the sum of the probabilities forms a geometric series with common ratio as 4

3 ³ c m.

7.

Possible

outcome YB YR BR YY BB RR

Scores 15 12 7 26 16 10

Prob. C

C C13 1 4

2

12 C

C C13 1 2 12

5

C C C1 1

2 12 4 5

C C

2 12

2 3

C C

2 12

2 4

C C

2 12

2 5

1A for all probabilities correct Expected score

= 15 # C C C13

1 4

2

12 + 12 # C C C13

1 2 12

5

+ … + 10 # C

C

2 12

2 5 1M + 1A

= 66 774

= 11

129 1A

c54h_M1_Mock5_sol.indd 28 2014/5/23 10:58:01 AM