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函數及其圖像
圓的基本性質

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DSEMATH CPPAPER 1

STUDIFY MOCK EXAMINATION

HONG KONG DIPLOMA OF SECONDARY EDUCATION EXAMINATION · PRACTICE PAPER

MATHEMATICS Compulsory PartPAPER 1

Question-Answer Book

8.30 am – 10.45 am (2¼ 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.
(2)This paper consists of THREE sections, A(1), A(2) 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 correct to 3 significant figures.
(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.

Studify practice paper. Not an official HKEAA examination paper.

Please stick the barcode label here.

Candidate Number

Answers written in the margins will not be marked.

SECTION A(1) (35 marks)

Simplify

\displaystyle \dfrac{(x^{-2}y^{3})^{4}}{x^{5}y^{-1}}

and express your answer with positive indices.(3 marks)

Answers written in the margins will not be marked.

Make

b

the subject of the formula

\displaystyle \dfrac{a+3b}{b-2}=5a

.(3 marks)

Answers written in the margins will not be marked.

Simplify

\displaystyle \dfrac{4}{3x-2}-\dfrac{5}{2x+1}

.(3 marks)

Answers written in the margins will not be marked.

Factorize (a)

4m^{2}-12mn+9n^{2}

, (b)

4m^{2}-12mn+9n^{2}-6m+9n

.(4 marks)

Answers written in the margins will not be marked.

The marked price of a bag is 40% above its cost. The bag is sold at a discount of 15% on its marked price and the profit is $57. Find the cost of the bag.(4 marks)

Answers written in the margins will not be marked.

Consider the compound inequality

\displaystyle \dfrac{5x-3}{2}\geq 4x+1

and

7-x>9

\quad\cdots\cdots(*)

. (a) Solve (*). (b) Find the greatest integer satisfying (*).(4 marks)

Answers written in the margins will not be marked.

The coordinates of the points

A

and

B

are

(-4,\,7)

and

(5,\,-2)

respectively.

A

is rotated clockwise about the origin

O

through

90^\circ

A'

B'

is the reflection image of

B

with respect to the

y

-axis. (a) Write down the coordinates of

A'

and

B'

. (b) Find the slope of

A'B'

.(4 marks)

Answers written in the margins will not be marked.

In Figure 1,

ABCD

is a parallelogram.

E

and

F

are points lying on

AD

and

BC

respectively such that

AE=CF

. (a) Prove that

\triangle ABE\cong\triangle CDF

. (b) If

\angle BAD=110^\circ

and

\angle ABE=32^\circ

, find

\angle CFD

.(5 marks)

Figure 1

Answers written in the margins will not be marked.

The table below shows the distribution of the numbers of books read last month by a group of students, where

k

is a positive integer. The mean of the distribution is 1.8. (a) Find

k

. (b) Write down the median of the distribution. (c) If a student is randomly selected from the group, find the probability that the number of books read by the selected student is greater than the mean of the distribution.(5 marks)

Number of books	0	1	2	3	4
Number of students	5	8	k	6	3

Answers written in the margins will not be marked.

SECTION A(2) (35 marks)

10.

It is given that

f(x)

partly varies as

x^{2}

and partly varies as

x

. Suppose that

f(2)=-4

and

f(-3)=21

(a)

Find

f(x)

.(3 分)

(b)

The graph of

y=g(x)

is obtained by reflecting the graph of

y=f(x)

with respect to the

x

-axis and then translating the resulting graph leftwards by 2 units. Find

g(x)

and write down the coordinates of the vertex of the graph of

y=g(x)

.(3 分)

Answers written in the margins will not be marked.

11.

The stem-and-leaf diagram below shows the distribution of the weights (in kg) of the 15 members of a club, where

a

and

b

are non-negative integers less than 10. The mean of the distribution is 58.2 kg and the inter-quartile range of the distribution is 14 kg.

Stem (tens)	Leaf (units)
4	5 7 8
5	0 2 a 6 9
6	1 3 3 b 8
7	0 4

(a)

Find

a

and

b

.(3 分)

(b)

Two new members join the club. Their weights are 58 kg and

x

kg, where

x

is an integer. (i) If the median of the weights of the 17 members is equal to the median of the weights of the original 15 members, find the least possible value of

x

. (ii) Someone claims that the range of the weights of the 17 members must be greater than that of the original 15 members. Do you agree? Explain your answer.(4 分)

Answers written in the margins will not be marked.

12.

A solid right circular cone has base radius 9 cm and height 12 cm.

(a)

Find the curved surface area of the cone in terms of

\pi

.(2 分)

(b)

The cone is cut into two parts by a plane parallel to its base. The upper part is a small right circular cone of height 4 cm and the lower part is a frustum. (i) Find the volume of the frustum in terms of

\pi

. (ii) The frustum is melted and recast into a solid sphere. Find the radius of the sphere.(5 分)

Answers written in the margins will not be marked.

13.

Let

p(x)=2x^{3}+ax^{2}+bx+6

, where

a

and

b

are constants. It is given that

x+2

is a factor of

p(x)

. When

p(x)

is divided by

x-1

, the remainder is 12.

(a)

Find

a

and

b

.(3 分)

(b)

Factorize

p(x)

.(2 分)

(c)

How many real roots does the equation

p(x)=0

have? Explain your answer.(2 分)

Answers written in the margins will not be marked.

14.

The coordinates of the points

A

and

B

are

(-2,\,5)

and

(6,\,-1)

respectively. Let

P

be a moving point in the rectangular coordinate plane such that

P

is equidistant from

A

and

B

. Denote the locus of

P

\Gamma

(a)

Describe the geometric relationship between

\Gamma

and the line segment

AB

. Hence, find the equation of

\Gamma

.(3 分)

(b)

The circle

C

passes through

A

and

B

, and its centre lies on the

x

-axis. Find the equation of

C

.(3 分)

(c)

Is the origin

O

inside

C

? Explain your answer.(2 分)

Answers written in the margins will not be marked.

SECTION B (35 marks)

15.

A box contains 6 red balls and 5 white balls. If 3 balls are randomly drawn from the box at the same time, find the probability that

(a)

all the balls drawn are red;(2 分)

(b)

both colours appear among the balls drawn.(2 分)

Answers written in the margins will not be marked.

16.

Let

T(n)

be the

n

th term of a geometric sequence. It is given that

T(2)=48

and

T(5)=6

(a)

Find

T(1)

and the common ratio of the sequence.(2 分)

(b)

Find the least value of

n

such that the sum of the first

n

terms of the sequence is greater than 191.99.(3 分)

Answers written in the margins will not be marked.

17.

Figure 2 shows a triangular paper card

ABC

with

AB=26

cm,

BC=24

cm and

\angle ABC=72^\circ

Figure 2

(a)

Find (i) the length of

AC

, (ii)

\angle BAC

.(4 分)

(b)

The card is held so that

BC

lies on the horizontal ground and the angle between the card and the horizontal ground is

50^\circ

. A student claims that the angle between

AB

and the horizontal ground exceeds

45^\circ

. Is the claim correct? Explain your answer.(3 分)

Answers written in the margins will not be marked.

18.

Let

f(x)=x^{2}-6kx+10k^{2}+4

, where

k

is a non-zero real constant.

(a)

Using the method of completing the square, express, in terms of

k

, the coordinates of the vertex of the graph of

y=f(x)

.(2 分)

(b)

The graph of

y=g(x)

is obtained by reflecting the graph of

y=f(x)

with respect to the

y

-axis and then translating the resulting graph upwards by 2 units. Denote the vertices of the graphs of

y=f(x)

and

y=g(x)

A

and

B

respectively. (i) Express

g(x)

in terms of

k

. (ii) If the area of

\triangle OAB

is 18, where

O

is the origin, find

k

.(5 分)

Answers written in the margins will not be marked.

19.

The equation of the circle

C

x^{2}+y^{2}-8x-6y=0

. Denote the centre of

C

G

. Let

L

be the tangent to

C

at the origin

O

(a)

(i) Write down the coordinates of

G

and the radius of

C

. (ii) Find the equation of

L

.(3 分)

(b)

The point

A(-3,\,4)

lies on

L

. Another tangent from

A

C

touches

C

at the point

T

, where

T

is different from

O

. (i) Find the equation of the straight line passing through

A

and

G

. (ii) Find the coordinates of

T

.(4 分)

(c)

Find the equation of the circle passing through

A

O

and

T

.(2 分)

(d)

Someone claims that the in-centre of

\triangle AOT

lies on

C

. Do you agree? Explain your answer.(3 分)

Answers written in the margins will not be marked.

END OF PAPER

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3 分

\displaystyle \dfrac{(x^{-2}y^{3})^{4}}{x^{5}y^{-1}}=\dfrac{x^{-8}y^{12}}{x^{5}y^{-1}}

1M (for

(ab)^n=a^nb^n

(a^m)^n=a^{mn}

)

=x^{-13}y^{13}

1M (for

\displaystyle \dfrac{a^m}{a^n}=a^{m-n}

)

\displaystyle =\dfrac{y^{13}}{x^{13}}

1A (3)

第 2 題3 分

a+3b=5a(b-2)

a+3b=5ab-10a

3b-5ab=-11a

1M (for putting

b

on one side)

\displaystyle b=\dfrac{11a}{5a-3}

1A (3)

第 3 題3 分

\displaystyle \dfrac{4}{3x-2}-\dfrac{5}{2x+1}=\dfrac{4(2x+1)-5(3x-2)}{(3x-2)(2x+1)}

\displaystyle =\dfrac{8x+4-15x+10}{(3x-2)(2x+1)}

\displaystyle =\dfrac{7(2-x)}{(3x-2)(2x+1)}

\displaystyle \dfrac{14-7x}{(3x-2)(2x+1)}

1A (3)

第 4 題4 分

(a)

4m^{2}-12mn+9n^{2}=(2m-3n)^{2}

1A (b)

4m^{2}-12mn+9n^{2}-6m+9n=(2m-3n)^{2}-3(2m-3n)

1M (for using (a))

=(2m-3n)[(2m-3n)-3]

1M (for taking out the common factor)

=(2m-3n)(2m-3n-3)

1A (4)

第 5 題4 分

Let

C

be the cost of the bag. Selling price

=C(1+40\%)(1-15\%)=1.19C

1M+1M

1.19C-C=57

C=300

, so the cost of the bag is $300. 1A (4)

第 6 題4 分

(a)

\displaystyle \dfrac{5x-3}{2}\geq 4x+1 \Rightarrow 5x-3\geq 8x+2 \Rightarrow x\leq -\dfrac{5}{3}

7-x>9 \Rightarrow x<-2

1A So the solution of (*) is

x<-2

. 1A (b) The greatest integer satisfying (*) is

-3

. 1A (4)

第 7 題4 分

(a)

A'=(7,\,4)

B'=(-5,\,-2)

1A (b) Slope of

\displaystyle A'B'=\dfrac{-2-4}{-5-7}

\displaystyle =\dfrac{1}{2}

1A (4)

第 8 題5 分

(a)

AB=CD

(opp. sides of // gram)

\angle BAE=\angle DCF

(opp.

\angle

s of // gram)

AE=CF

(given)

\therefore\ \triangle ABE\cong\triangle CDF

(SAS) Marking: 2 for a correct proof with correct reasons; 1 for a correct proof without reasons. (b)

\angle AEB=180^\circ-110^\circ-32^\circ=38^\circ

(

\angle

sum of

\triangle

) 1M

\angle CFD=\angle AEB

(corr.

\angle

\cong\triangle

s) 1M

=38^\circ

1A (5)

第 9 題5 分

(a)

\displaystyle \dfrac{0(5)+1(8)+2k+3(6)+4(3)}{5+8+k+6+3}=1.8

38+2k=39.6+1.8k

, so

k=8

1A (b) There are 30 students. The 15th and 16th data are both 2, so the median is 2. 1A (c) Required probability

\displaystyle =\dfrac{8+6+3}{30}

\displaystyle =\dfrac{17}{30}

1A (5)

第 10 題6 分

(a) Let

f(x)=ax^{2}+bx

, where

a

and

b

are non-zero constants. 1A

4a+2b=-4

and

9a-3b=21

1M Solving,

a=1

and

b=-4

, so

f(x)=x^{2}-4x

. 1A (b) Reflecting the graph with respect to the

x

-axis:

y=-(x^{2}-4x)

. 1M Translating leftwards by 2 units:

g(x)=-[(x+2)^{2}-4(x+2)]=-x^{2}+4

1A The coordinates of the vertex of the graph of

y=g(x)

are

(0,\,4)

. 1A (6)

第 11 題7 分

(a) The lower quartile is 50 kg and the upper quartile is

(60+b)

kg.

60+b-50=14

, so

b=4

. 1M+1A

\displaystyle \dfrac{870+a}{15}=58.2

, so

a=3

. 1A (b) (i) The median of the 15 weights is 59 kg. For the median of the 17 weights to be 59 kg, the 9th datum must be 59, so

x\geq 59

. 1M The least possible value of

x

is 59. 1A (ii) The range of the 15 weights is

74-45=29

kg. If, for example,

x=60

, the range of the 17 weights is still 29 kg. 1M So the claim is incorrect. 1A f.t. (7)

第 12 題7 分

(a) Slant height

=\sqrt{9^{2}+12^{2}}=15

cm 1M Curved surface area

=\pi(9)(15)=135\pi

^{2}

1A (b) (i) Volume of the original cone

\displaystyle =\dfrac{1}{3}\pi(9)^{2}(12)=324\pi

^{3}

1M Volume of the small cone

\displaystyle =324\pi\left(\dfrac{4}{12}\right)^{3}=12\pi

^{3}

1M Volume of the frustum

=312\pi

^{3}

1A (ii)

\displaystyle \dfrac{4}{3}\pi r^{3}=312\pi

r=\sqrt[3]{234}\approx 6.16

cm 1A (7)

第 13 題7 分

(a)

p(-2)=0

-16+4a-2b+6=0

, i.e.

2a-b=5

p(1)=12

2+a+b+6=12

, i.e.

a+b=4

1M Solving,

a=3

and

b=1

. 1A (b)

p(x)=2x^{3}+3x^{2}+x+6=(x+2)(2x^{2}-x+3)

1M+1A (c) For

2x^{2}-x+3=0

\Delta=(-1)^{2}-4(2)(3)=-23<0

, so it has no real roots. 1M Hence

p(x)=0

has only one real root (

x=-2

). 1A (7)

第 14 題8 分

(a)

\Gamma

is the perpendicular bisector of

AB

. 1M

(x+2)^{2}+(y-5)^{2}=(x-6)^{2}+(y+1)^{2}

4x-3y-2=0

1A (b) The centre of

C

lies on

\Gamma

and on the

x

-axis. Put

y=0

\displaystyle x=\dfrac{1}{2}

. 1M Radius

\displaystyle ^{2}=\left(-2-\dfrac{1}{2}\right)^{2}+5^{2}=\dfrac{125}{4}

1M The equation of

C

\displaystyle \left(x-\dfrac{1}{2}\right)^{2}+y^{2}=\dfrac{125}{4}

, i.e.

x^{2}+y^{2}-x-31=0

. 1A (c)

0^{2}+0^{2}-0-31=-31<0

1M So

O

lies inside

C

. 1A f.t. (8)

第 15 題4 分

(a) Required probability

\displaystyle =\dfrac{C^{6}_{3}}{C^{11}_{3}}

\displaystyle =\dfrac{4}{33}

1A (b) Required probability

\displaystyle =1-\dfrac{C^{6}_{3}}{C^{11}_{3}}-\dfrac{C^{5}_{3}}{C^{11}_{3}}

\displaystyle =\dfrac{9}{11}

1A (4)

第 16 題5 分

(a) Let

a

be the first term and

r

the common ratio.

ar=48

and

ar^{4}=6

, so

\displaystyle r^{3}=\dfrac{1}{8}

. 1M

\displaystyle r=\dfrac{1}{2}

and

T(1)=96

1A (b)

\displaystyle \dfrac{96\left[1-\left(\frac{1}{2}\right)^{n}\right]}{1-\frac{1}{2}}>191.99

\displaystyle \left(\dfrac{1}{2}\right)^{n}<\dfrac{0.01}{192}

, so

\displaystyle n\log\dfrac{1}{2}<\log\dfrac{0.01}{192}

n>14.2

, so the least value of

n

is 15. 1A (5)

第 17 題7 分

(a) (i)

AC^{2}=26^{2}+24^{2}-2(26)(24)\cos 72^\circ

AC\approx 29.4

cm 1A (ii)

\displaystyle \dfrac{\sin\angle BAC}{24}=\dfrac{\sin 72^\circ}{29.434}

\angle BAC\approx 50.8^\circ

1A (b) Let

N

be the foot of the perpendicular from

A

BC

AN=26\sin 72^\circ\approx 24.727

cm. Height of

A

above the ground

=AN\sin 50^\circ\approx 18.942

cm 1M Let

\theta

be the angle between

AB

and the ground.

\displaystyle \sin\theta=\dfrac{18.942}{26}

, so

\theta\approx 46.8^\circ

1M Since

46.8^\circ>45^\circ

, the claim is correct. 1A f.t. (7)

第 18 題7 分

(a)

f(x)=(x^{2}-6kx+9k^{2})+k^{2}+4=(x-3k)^{2}+k^{2}+4

1M The coordinates of the vertex are

(3k,\,k^{2}+4)

. 1A (b) (i)

g(x)=f(-x)+2=x^{2}+6kx+10k^{2}+6

1M+1A (ii) The vertex of the graph of

y=g(x)

B(-3k,\,k^{2}+6)

. Area of

\displaystyle \triangle OAB=\dfrac{1}{2}\left|3k(k^{2}+6)-(-3k)(k^{2}+4)\right|=3|k|(k^{2}+5)

3|k|(k^{2}+5)=18

, i.e.

|k|^{3}+5|k|-6=0

, so

(|k|-1)(|k|^{2}+|k|+6)=0

k=1

k=-1

1A (7)

第 19 題12 分

(a) (i)

G=(4,\,3)

and the radius of

C

\sqrt{4^{2}+3^{2}}=5

. 1A (ii) Slope of

\displaystyle OG=\dfrac{3}{4}

, so the slope of

L

\displaystyle -\dfrac{4}{3}

(tangent

\perp

radius). 1M The equation of

L

4x+3y=0

. 1A (b) (i) Slope of

\displaystyle AG=\dfrac{3-4}{4-(-3)}=-\dfrac{1}{7}

, so

AG

x+7y-25=0

1M+1A (ii) By tangent properties,

AO=AT

and

GO=GT

, so

AG

is the perpendicular bisector of

OT

and

T

is the reflection of

O

AG

. 1M

OT

y=7x

. Solving with

AG

, the mid-point of

OT

\displaystyle \left(\dfrac{1}{2},\,\dfrac{7}{2}\right)

, so

T=(1,\,7)

. 1A (c)

\angle AOG=\angle ATG=90^\circ

, so

A

O

G

and

T

are concyclic with

AG

as a diameter. 1M Centre

\displaystyle \left(\dfrac{1}{2},\,\dfrac{7}{2}\right)

, radius

\displaystyle ^{2}=\dfrac{50}{4}

x^{2}+y^{2}-x-7y=0

1A (d) Let

I

be the point where

AG

meets

C

between

A

and

G

. Since

AG\perp OT

through

G

I

is the mid-point of arc

OT

, so

\angle IOT=\angle ITO

(eq. chords, eq.

\angle

s).

\angle AOI=\angle ITO

(

\angle

in alt. segment), so

\angle AOI=\angle IOT

and

OI

bisects

\angle AOT

. 1M By symmetry about

AG

AI

bisects

\angle OAT

. 1M So

I

is the in-centre of

\triangle AOT

and it lies on

C

. The claim is agreed. 1A f.t. (12)