Special Features

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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

Sample

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Solution Guide

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

Questions similar to those in 2014–2015 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.

Sample

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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

Sample

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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 8 to 12 Questions ~ 7 minutes

Section B 3 to 5 Questions ~ 18 minutes

Time left for checking: 5 – 10 minutes

Sample

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1

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. 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 should be either exact or given to 4 decimal places.

7. 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.

Sample

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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.

HKDSE Exam Series — Mathematics (Extended Part) Mock Exam Papers (Module 1) (2016/17 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

Sample

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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.

HKDSE Exam Series — Mathematics (Extended Part) Mock Exam Papers (Module 1) (2016/17 Edition)

Section B (50 marks)

10. (a) (i) Find (d ln ) 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 = e20

2 + 201 [(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

Sample

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Mock Exam 2

Section A

1. (a) (1 + ax)ⁿ

= 1 + C1nax + C2na²x² + C3na³x³ + … 1A (b) From (a),

an : a n n 1 2

2 ^ - h

: a n n 1 n 2 6

3 ^ - h^ - h

= 74 : 1 : 1 an : a n n 1

2

2 ^ - h

= 4 : 1 and 7 a n n 1

2

2 ^ - h

: a n n 1 n 2 6

3 ^ - h^ - h

= 1 : 1 1M 7an = a n n 1

2 4 ² ^ - h

and

a n n 1 2

2 ^ - h

= a n n 1 n 2 6

3 ^ - h^ - h

7 = 2a(n - 1) and 3 = a(n - 2) ∴ 37 = ( )

n n

2

2 1

- - 7n - 14 = 6n - 6

n = 8 1A

Guidelines

By using the given ratio, find the relationships between a and n.

2. dx

dt = 4 (t + 4)² H = ln x^x = x ln x

dH

dt = (ln x + 1) dx

dt 1M

=

(

^ln

(

^{15 –}t + 4⁴

)

^{+ 1}

)

(t + 4)⁴ ² 1A When t = 7,

d

dtH =

(

^{ln 14}^{11 + 1}⁷

)

4 4 7+ ²

^ h 1M

= 0.1218 (cor. to 4 d.p.) 1A

∴ The rate of change of the amount of water produced is 0.1218 mL/min.

3. (a) y = e^x - x + 1 ddy

x = e^x - 1 1A

When ddy x = 0, e^x - 1 = 0 x = 0

When x = 0, y = e⁰ - 0 + 1 = 2.

x x < 0 x = 0 x > 0 d

dy

x - 0 +

The extreme point P is a minimum point.

The minimum point P is (0, 2). 1A The equation of the tangent L is y = 2. 1A (b) When x = ln 2,

y = e^{ln 2} - ln 2 + 1 = 3 - ln 2 ddy

x = e^{ln 2} - 1 = 1

Slope of the normal N = -1 Equation of N is:

3 ln2 ln y

x 2

- - -

^ h

= -1 1M y - 3 + ln 2 = -x + ln 2

x + y - 3 = 0 1A

(c) Consider the following diagram.

y

x C

N L 0

2

When y = 2, x + 2 - 3 = 0

x = 1

The coordinates of the point of intersection of N and L are (1, 2).

(ln 2, 3 - ln 2)

Sample

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8

HKDSE Exam Series — Mathematics (Extended Part) Mock Exam Papers (Module 1) (2016/17 Edition) Soulution Guide Required area

= 1 2 1 2 3 2 2

d ln ln

e x x

2

ln x 0

2 - + - - - -

^ +^ ^

h h h

#

1M

= 2

2 ln ln

e x x 1 2 2

2

2 ln 2

x

0 2

- - - +

+ ^{^} ^h

; E 1M

=

2

2 2

ln ln ln

e ln e 1 2 2

2 2

2

0

2

ln2- - - +

- +

^ h ^ ^

h h

; E

=

23 - 2 ln 2 1A

Guidelines

Divide the region into 2 parts: the region bounded by the curve C and L from x = 0 to x = ln 2 and the triangle bounded by the line N and L from x = ln 2 to x = 1.

4. (a) (i) u = ae^-bt ln u = ln (ae^-bt)

= ln a - bt 1A

(ii) When t  3, e^-bt  0.

( )

3 [3 4 (0)]

a a 1+ 0

- + k = 15

k = 6 1A

N = 3 4

u u 1 3 -

+

^ h

+ 6

N - 6 = u

u 1 9 12

+ -

N - 6 + Nu - 6u = 9 - 12u

u = N

N 6 15

+

- 1A

(b) From (a),

ln N

N 6 15

+

- = ln a - bt

t 1 3 5 7

ln N N 6 15

+

- 0.09 -0.49 -0.97 -1.51

The graph is as follows:

ln u

O t

-0.5 0.5

-1.0

-1.5

-2.0

1 2 3 4 5 6 7

1A

From the graph, ln a . 0.35

a = 1.4 (cor. to 1 d.p.) 1A –b . . .

5 0 1 0 0 35

- - -

b = 0.3 (cor. to 1 d.p.) 1A

5. (a) f(x) = e^x x x√ + 5

f'(x) = (e^x)²

e^x

(

^x√ ^{+ 5 +}

)

– x x + 5 e√ ^x

√ x 2 x + 5

1M +1A

=

= 2(x + 5) – 2x(x + 5) + x 2e^x√ x + 5

= 10 – 7x – 2x²

2e^x√ x + 5 1A (b) For x > 2, 2e^x√x + 5 > 0.

10 – 7x – 2x² = –(2x² + 7x – 10)

For x > 2, –(2x² + 7x – 10) < 0 1A ∴ f'(x) = 10 – 7x – 2x²

2e^x√x + 5 < 0 for x > 2 1A f(x) is decreasing for x > 2.

e^x

√ x + 5 – x x + 5 +

√ √

x 2 x + 5

Special Features