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.

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.

Solution Guide

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

c54i_Front matter(edited).indd 4 2014/5/29 3:42:29 PM

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

In order to prepare for the examination effectively, students 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 2 examination. The following table lists the details of the Module 2 (Algebra and Calculus) examination:

Module 2 (Algebra and Calculus)

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 2 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 2 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

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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 9 to 10 Questions ~ 7 minutes

Section B 4 to 5 Questions ~ 18 minutes

Time left for checking: 5 – 10 minutes

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 2 (Algebra and Calculus)

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 must be exact.

6. In this paper, vectors may be represented by bold- type letters such as u, but candidates are expected to use appropriate symbols such as u^" in their working.

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 14

Total

Please stick the barcode label here.

c54i_M2_Mock1.indd 1 2014/5/29 3:41:46 PM

HKDSE Exam Series — Mathematics (Extended Part) Mock Exam Papers (Module 2) (2015 Edition)

Answers written in the margins will not be marked.

Answers written in the margins will not be marked.

Section A (50 marks)

1. Consider y = 2x³ – 3x². (a) Find

d d x

y from first principle.

(b) Find the range of x where y is increasing.

(5 marks) 2014

FORMULAS FOR REFERENCE

sin (A!B) = sin A cos B! cos A sin B cos (A!B) = cos A cos B" sin A sin B tan (A ! B) = tan tan

tan tan

A B

A B 1

!

"

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)

sin A+ sin B= 2 sin A B 2

+ cos A B 2 -

sin A- sin B= 2 cos A B 2

+ sin A B 2 -

cos A+ cos B= 2 cos A B 2

+ cos A B 2 -

cos A- cos B=-2 sin A B 2

+ sin A B 2 -

Mock Exam 1

© Pan Lloyds Publishers Ltd

13

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.

Go on to the next page

Section B (50 marks) 11. Let M = f p k + 1 –kf p

1 0 and A = f p f pp 1

1 1 , where k and p are real numbers and p ≠ 1.

(a) (i) Find A^–1 in terms of p.

(ii) Show that A^–1MA = f pk f p0 p – k 1 .

(iii) Suppose p = k. Using (ii), find Mⁿ in terms of k and n, where n is a positive integer.

(5 marks)

(b) Let B = f p f p 3 –2

1 0 . For any positive integer n, find B + B³ + B⁵ + … + B^{2n – 1}. (4 marks) (c) A sequence is defined by

x₁ = 1, x₂ = 3 and xn = 3xn – 1 – 2xn – 2 for n = 3, 4, 5, …

It is known that this sequence can be expressed in the form of x x 1

n

e n- o = f p f p 3 –2 1 0

x x 2

n n -1

e - o.

Using the result of (a)(iii), express xn in terms of n.

(3 marks) 2014

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Mock Exam 2

Mock Exam 2

Section A

1. f l(-1) = lim

h h

1 2

1

1 2 1

h 0

- + + - - +

"

= 1

lim h h h

h h 1

1

1 1 1

h 0 $1

+ - +

+ + + +

" 1M

=

( )

( )

limh h h

h

1 1 1

1 1

h 0 + + +

- +

"

=

( )

lim 1 h 1 1 h 1

h 0 + + +

-

" 1M

=

( )

1 1 1

1

0 0

+ + +

-

=

2

-1 1A

2. (1 +x)²(1 - 3x)ⁿ= (1 + 2x + x²)(1 - 3C₁ⁿx+ 9C₂ⁿx² - 27C₃ⁿx³+ …)

-3C₁ⁿ+ 2 = –19 1M

-3n+ 2 = –19 1M

n= 7 1A

The coefficient of x³

=-27C₃⁷+ 18C₂⁷ – 3C₁⁷ 1M

=-588 1A

Common Mistakes

Some candidates may miss the negative signs out.

3. (a)   x + y + z = 10 ... (1)   x^* + 5y + 10z = 40 ... (2) (2) – (1):

4y + 9z = 30 1M

Let z = t, where t is any real number. 1M Then y = 30 9t

4 - . Substituting y = 30 9t

4

- and z = t into (1),

x + 30 9t 4

- + t = 10

x = t 4 10+5

∴ The solutions are ( t 4 10+5

, 30 9t 4 - , t),

(b) a, b and c satisfy the following equations,   a + b + c = 10 ... (1)

  a^* + 5b + 10c = 40 ... (2) By (a),

a = t 4 10+5

, b = 30 9t 4

- and c = t for some value(s) of t such that a, b and c are

non-negative integers. 1M

Since b = 30 9t 4

- and c = t are non-negative, we have 0 ≤ t ≤ 3

10.

For t = 2, a = 5, b = 3 and c = 2 which is a set of solution.

For t = 0, 1 and 3, a and b are not 1M Z [

\ ] ] ] ] ] ] non-negative integers.

∴ There is only one set of combination

of a, b and c. 1A

Billy is correct.

4. (a) y=

1 x 3

1

3

- 2

^ - h

#

^dx

=

1 x 3

1

3

- 2

^ h

#

^{d(1 -x)}

=_^1-x_h³¹ +C, where C is a real constant.

1A ∵ The curve passes through (2, 1).

∴ 1 =_^1-2_h³¹ +C 1 =-1 +C C= 2

∴ The equation of the curve is:

y=_^1-x_h³¹ + 2 1A

(b) When x= 0, y=_^1-0_h³¹ + 2 = 3

∴ The coordinates of A are (0, 3).

dd x y

x=0=-

1 0 1 3_^ - _h³²

1M

=-1

3

∴ The equation of the tangant at A is:

y=-1

3x+ 3 1A

10

HKDSE Exam Series — Mathematics (Extended Part) Mock Exam Papers (Module 2) (2015 Edition) Solution Guide 5. (a)

#

^{x cos x dx}

=

#

^{xd (sin x)}

=x sin x-

#

^{sin xdx 1M}

=x sin x+ cos x+C 1A

(where C is an arbitrary constant) (b) *^{y= 2x cos x}

y=x

2x cos x=x       x(2 cos x- 1) = 0

x= 0 or cos x= 1

2

x= 0 or 3

r 1A

The required area =

0 3

#

r^{(2x cos x – x)dx 1M}

= xsinx cosx x

2 2

2

2 0

-

< + F³

r

= 3

3 r – 18

r2 – 1 1A

6. Let S(n) be the statement

‘1 + 1 # 1! + 2 # 2! + … +n # n! = (n+ 1)!’.

For n= 1,

L.H.S. = 1 + 1 # 1! = 2 R.H.S. = (1 + 1)! = 2

∵ L.H.S. = R.H.S.

∴ S(1) is true. 1

Assume S(k) is true, i.e.,

1 + 1 # 1! + 2 # 2! + … +k # k! = (k+ 1)!,

where k is a positive integer. 1

For n=k+ 1,

1 + 1 # 1! + 2 # 2! + … +k#k! + (k+ 1) # (k+ 1)!

= (k+ 1)! + (k+ 1) # (k+ 1)! 1

= (k+ 1)! [1 + (k+ 1)]

= (k+ 2)!

∴ S(k+ 1) is true.

By the principle of mathematical induction, S(n) is true for all positive integers n. 1

7. (a) A+B =

sin cos sin

i i

i

#

+ ^{di +}

#

_sini^cos₊ⁱ_cosi^di

=

sin cos sin cos

i i

i i

+

#

+ ^di

= i +C 1A

(where C is an arbitrary constant) A-B

=

sin cos sin cos

i i

i i

+

#

- ^di

=-

sin cos 1 i+ i

#

d(cos i + sin i)

=-ln |sin i + cos i| +Cl 1A (where Cl is an arbitrary constant)

(b) Let x= sin i.

Then dx= cos i di.

1 x x

1 - 2

#

+ ^dx

=

1 sin sin

cos

2i i

i -

#

+ ^{di 1M}

=

cos sin

cos i i

i

#

+ ^di

= A B A B

2

+ - -

^ h ^ h 1M

= 1 2i + 1

2ln |sin i + cos i| +Cm = 1

2sin^-1x+ 1

2ln |x+ 1-x²| +Cm 1A (where Cm is an arbitrary constant)

Guidelines

This substitution is made as we need to simplify x

1- ² by using the identity sin² i + cos² i ≡ 1.

Candidates can try to use the other substitution x= cos i. The result

-1

2cos^-1x+ 1

2ln |x+ 1-x²| +Cm would follow, which differs from the answer given above by a constant 2

r.

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