HKDSE Practice Paper 2012 Mathematics_Module 1

8/3/2019 HKDSE Practice Paper 2012 Mathematics_Module 1

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HKD SE Practice Paper 2012 Mathematics (Module 1)

1

Section A (50 marks)

1. (a) Expand3

(2 1)x

(b) Expand axe in ascending powers ofx as far as the term in 2 , where a is a constant.

(c) If the coefficient of 2 in the expansion of3(2 1)

ax

x

e

is 4 , find the value(s) ofa.

(5 marks)

2. It is given that1

3 22 1t y y

and2

1t xe x .

(a) Finddt

dy .

(b) By expressing tin terms ofx, finddt

dx.

(c) Finddy

dxin terms ofx andy.

(5 marks)

3. A glass container is in the shape of a vertically

inverted right circular cone of base radius 15 cm

and height 20 cm. Initially, the container is full of

water. Suppose the water is running out from it at a

constant rate of 2 cm3/s. Let h cm be the depth

of water remaining in the container, rcm be the

radius of water surface (see Figure 1), Vcm3 be the volume of the water, andA cm2 be the

area of the wet surface of the container. It is given that 21

3V r h and 2 2 A r r h .

(a) Express VandA in terms ofronly.

(b) When r= 3,

(i) find the rate of change of the radius of the water surface;

(ii) find the rate of change of the area of the wet surface of the container.

(6 marks)


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2

4. Consider the curve C:1

2(2 1) y x x , where1

2x .

(a) Finddy

dx.

(b) Using (a), find the equation of the two tangents to the curve Cwhich are parallel to the

straight line 2 0x y .

(6 marks)

5. Consider the two curves 1 : 1 xe

C ye

and 2 :xC y e e .

(a) Find thex-coordinates of all the points of intersection of 1C and 2C .

(b) Find the area of region bounded by 1C and 2C .

(5 marks)

6. A random sample of size 10 is drawn from a normal population with mean and variance 8.

Let X be the mean of the sample.

(a) CalculateVar(2 7)X .

(b) Suppose the mean of the sample is 50. Construct a 97% confidence interval for .

(5 marks)

7. In a game, there are two bags, A and B, each containing 5 balls. Bag A contains 3 red and 2

blue balls, while bag B contains 4 red and 1 blue balls. A player first chooses a bag at random

and then draws a ball randomly from the bag. The player will be rewarded if the ball drawn is

blue. The ball is then replaced for the next players turn.

(a) Find the probability that a player is rewarded in a particular game.

(b) Two players participate in the game. Given that at least one of them is rewarded, find the

probability that both of them are rewarded.

(c) If 60 players are rewarded, find the expected number of players among them having

drawn a blue from bag A.

(5 marks)


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8. Eggs from a farm are packed in boxes of 30. The probability that a randomly selected egg is

rotten is 0.04.

(a) Find the probability that a box contains more than 1 rotten egg.

(b) Boxes of eggs are inspected one by one.

(i) Find the probability that the 1st

box containing more than 1 rotten egg is the 6th

box inspected.

(ii) What is the expected number of boxes inspected until a box containing more than

1 rotten egg is found?

(7 marks)

9. SupposeA andB are two events. Let 'A and 'B be the complementary events ofA andB

respectively. It is given that ( | ') 0.6 P A B , ( ) 0.12P A B and ( ')P A B k , where

0k .

(a) Find ( )P A , ( )P B and ( )P A B in terms ofk.

(b) IfA andB are independent, find the value ofk.

(6 marks)


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4

Section B (50 marks)

10. An engineer models the rates of production of an alloy in the first 10 weeks by two new

machines A and B respectively by

5

2

61

( 1)

dx t

dtt

and215ln( 100)

16

dy t

dt

for 0 10t ,

wherex (in million kg) and y (in million kg) are the amount of the alloy produced by

machine A and B respectively, and t(in weeks) is the time elapsed since the beginning of the

production.

(a) Using the substitution 1u t , find the amount of the alloy produced by machine A in

the first 10 weeks.

(4 marks)

(b) Using the trapezoidal rule with 5 sub-intervals, estimate the amount of alloy produced

by machine B in the first 10 weeks.

(2 marks)

(c) The engineer use the result of (a) and (b) to claim that machine B is more productive

than machine A in the first 10 weeks. Do you agree? Explain your answer.

(4 marks)


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5

11. A researcher models the rate of change of the population size of a kind of inserts in a forest

by 20'( )a

t

P t kte , where ( )P t , in thousand, is the population size, t ( 0) is the time

measured in weeks since the start of the research, and a, kare integers.

The following table shows some values oftand '( )P t .

(a) Express'( )

lnP t

tas a linear function oft.

(1 marks)

(b) By plotting a suitable straight line on the graph paper, estimate the integera and k.

(5 marks)

(c) Suppose that (0) 30P . Using the estimates in (b),

(i) find the value oftsuch that the rate of change of the population size of the

insect is the greatest;

(ii) find 20a

tdte

dt

and hence, or otherwise, find ( )P t ;

(iii) estimate the population size after a very long time.

[Hint: You may use the fact that lim 0mtt

te

for any positive constant m.]

(9 marks)

t 1 2 3 4

'( )P t 22.83 43.43 61.97 78.60


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11. (cont)


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12. A staff of a school studies the school sick room utilization. The number of visits to the sick

room per day on 100 randomly selected school days recorded as follows:

(a) Find an unbiased estimate of the mean number of visits per day.

(1 marks)

(b) (i) Find the sample proportion of school days with less than 4 visits per day.

(ii) Construct an approximate 95% confidence interval for the proportion of school

days with less than 4 visits per day.

(3 marks)

(c) Suppose then number of visits per day follows a Poisson distribution with mean .

Assume that the unbiased estimate obtained in (a) is used for. The sick room is said

to be crowdedon a particular day if there are more than 3 visits on that day.

(i) Find the probability that the sick room is crowdedon a particular day.

(ii) In a certain week of 5 school days, give that the sick room is crowdedon at least 2

days, find the probability that the sick room is crowded on alternate days in the

week.

(6 marks)

Number of visits per day 0 1 2 3 4 5 6 7

Frequency 6 12 18 21 20 12 7 4


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13. In a supermarket, there are two cashier counters: a regular counter and an express counter. It

is known that 88% of customers pay at the regular counter. It is found that the waiting time

for a customer to pay at the regular counter follows the normal distribution with mean 6.6

minutes and standard deviation 1.2 minutes.

(a) Find the probability that the waiting time for a customer to pay at the regular counter is

more than 6 minutes.

(2 marks)

(b) Suppose 12 customers who pay at the regular counter are randomly selected. Assume

that their waiting times are independent.

(i) Find the probability that there are more than 10 of the 12 customers each having

a waiting time of more than 6 minutes.

(ii) Find the probability that the average waiting time of the 12 customers is more

than 6 minutes

(5 marks)

(c) It is found that the waiting time for a customer to pay at the express counter follows the

normal distribution with mean minutes and standard deviation 0.8 minutes.

It is known that exactly 21.19% of the customers at the regular counter wait less than k

minutes, while exactly 3.95% of the customers at the express counter wait more than k

minutes.

(i) Find kand .

(ii) Two customers are randomly selected. Assume that their waiting times are

independent. Given that both of them wait more than minutes to pay, find

the probability that exactly one of them pays at the regular counter.

(8 marks)

End of Paper


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9

Solution

1. (a)3 3 2 2 3 3 2

(2 1) (2 ) 3(2 ) (1) 3(2 )(1) (1) 8 12 6 1 x x x x x x x

(b)

2 22( )

1 ( ) ... 1 ...2! 2

ax ax a

e ax ax x

(c) 3 2

3 2 2(2 1) (2 1) 1 6 12 ... 1 ...2

ax

ax

x a x e x x ax x

e

Coefficient of2

26 12 4

2

ax a 2 12 32 0a a 8a or 4a

2. (a)

3

223

dty ydy

(b)2 1

t xe x

2

1ln xt x 2( 1) lnt x x 2

12 ln

dt xx x

dx x

(c)

23

2 22

3 7

2 2 2

12 ln

1 ( 2 ln 1)

3 (3 1)

xx x

dy dy dt dt y x x xxdtdx dt dx dx

y y x ydy

3. (a)15 20

r h (corr. sides, s)

4

3h r

2 31 4 4

3 3 9V r r r

2

2 2 24 25 5

3 9 3 A r r r r r r

(b) (i) 34

9

V r 24

3

dV dr r

dt dt

Put 2dV

dt and r= 3, we have 2

3

42 (3)

3 r

dr

dt

3

1

6r

dr

dt cm/s

The rate of change of the radius of the water surface is1

6 cm/s

(ii) 25

3A r

10

3

dA dr r

dt dt

3

10 1 5(3)

3 6 3r

dA

dt

cm2/s

The rate of change of the area of the wet surface of the container is5

3

cm2/s


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4. (a) 1 1

2 21 3 1

2 1 (2) 2 12 2 1

dy x x x x

dx x

(b) Let (a, b) be the point of contact,

Slope of tangent = 2 =3 1

2 1

a

a

29 6 14

2 1

a a

a

29 14 5 0a a

1a or5

9a

Hence the point of contact (a, b) = (1, 1) and5 5

,9 27

The equation of tangent at (1, 1) is1

21

y

2 1 0x y

The equation of tangent at5 5

,9 27

is

5

27 25

9

y

x

54 27 25 0x y

5. (a) 1xx

ee e

e 2 ( 1) 0x xe e e e

21 ( 1) 4

2

x e e e

e

1 1

2

x e ee

or1 1

2

x e ee

xe e or 1xe

1x or 0x

The point of intersection are (1, 0) and (0,1 e ).

(b) Area = 11

0 01 x x x

x

ee e dx x e e e ex

e

1 1 0 1 0 3e e e e sq. units


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11

6. (a)Var( ) 8 4

Var( )10 5

XX

n

2 4 16Var(2 7) 2 Var( ) 45 5

X X

(b) 97% confidence interval = 8 8

50 2.17 ,50 2.17 48.0591,51.940910 10

7. (a) P(rewarded)1 2 1 1 3

0.32 5 2 5 10

(b) P(both rewarded| at least one rewarded)

2

2

(both rewarded) 0.3 3 0.17651 (both no rewarded) 1 0.7 17

P

P

(c) P(from A| rewarded) =

1 2

22 51 2 1 1 3

2 5 2 5

The expected number of players =2

60 403

8. (a) LetXbe the number of rotten eggs, then ~ (30,0.04)X B

( 1) 1 ( 0) ( 1) P X P X P X

30 30 29

11 (1 0.04) (1 0.04) (0.04)

0.338820302 0.3388

C

(b) Let Ybe the number of box inspected until 1st box containing more than 1 rotten egg is

found, then ~ (0.338820302)Y Geo

5( 6) (1 0.338820302) (0.338820302) 0.042812143 0.0428P Y

(c)1

( ) 2.951470020.338820302

E Y

The required expected number of boxes is 2.9515


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9. (a) ( ) ( ) ( ')P A P A B P A B

( ) 0.12 P A k

( ') ( | ') ( ')P A B P A B P B

( ')

1 ( )( | ')

P A BP B

P A B

5( ) 1

3

kP B

( ) ( ) ( ) ( )P A B P A P B P A B

5 2( ) 0.12 1 0.12 1

3 3

k k P A B k

9. (b) ( ) ( ) ( )P A P B P A B [for independent]

5

0.12 1 0.123

kk

0.48k ork= 0 (rejected as kis positive)

Alternatively, asA andB are independent, ( | ') ( )P A B P A

0.12 0.6k 0.48k

10. (a) Let 1u t , du dt

3 510 11 11

2 25 50 1 12 2

61 161 61

( 1)

t u x dt du u u du

t u

111 3

2 2

1

122122

3 x u u

1 3

2 2122 122

122(11) (11) 1223 3

x

3904 24445.6636

333 11

The amount produced by machine A in the first 10 weeks is 45.6636 million kg

(shaded region = 'A B )


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10. (b)10

2

0

15ln( 100)

16 y t dt

15 2

ln(100) 2 ln(104) ln(116) ln(136) ln(164) ln(200)

16 2

15

ln(20000) 2 ln 26907545616

45.6792

The amount produced by machine B in the first 10 weeks is 45.6792 million kg

10. (c) 22

2ln( 100) 0

100

d tt

dt t

,

so the graph is increasing in 0 10t (1)

2 22

2 2 2 2 2

2( 100) (2 )(2 ) 2( 10)( 10)ln( 100)

( 100) ( 100)

d t t t t t t

dt t t

Since2

2

2ln( 100) 0

dt

dt for 0 10t ,

the curve of2

ln( 100)y t is concave downward for 0 10t ,..(2)

Also, at t= 0, 2ln( 100)y t attains minimum point and

it is (0,ln100) , where ln100 is positive.(3)

Combining (1), (2), and (3),

the graph of2

ln( 100)y t is above the t-axis and concave

downward for 0 10t .

Therefore the estimation of10 2

0ln( 100)t is underestimated.

Since the value ofy ( 45.6792) is underestimated

but still greater than that ofx( 45.6636),

I agree with the engineer that machine B is more productive than machine A in the

first 10 weeks


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11. (a) 20'( )

atP t

ket

20'( )

ln lna

tP tke

t

20'( )

ln ln lna

tP tk e

t

'( )

ln ln20

P t ak t

t

(b)

From the graph, ln 3.18k k 24

3.128 2.978

20 1 4

a

1a

t 1 2 3 4

'( )P t 22.83 43.43 61.97 78.60

'( )ln

P t

t 3.128 3.087 3.028 2.978


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11. (c) (i) 20( )

atdP t

ktedt

2

20 20 202

( )1

20 20

a a at t td P t a a

k te e ke t dt

12

202( ) 124 1

20td P t e t

dt

Set2

2

( )0

d P t

dt

11 0

20t

20t

t 0 20t 20t 20t 2

2

( )d P t

dt + 0 -

The rate of change of the population size of the insect is the greatest ift= 20

(ii)1 1 1

20 20 201

20

t t tdte te e

dt

1 1 1

20 20 201

20

t t t

te te dt e dt

1 1 1

20 20 20

120

240240 24

t t t

te te dt e

1 1

20 20240 '( ) 4800t t

te P t dt e

1 1

20 20( ) 240 4800t t

P t te e C

Consider (0) 30P , 30 4800 C 4830C

Hence,1 1

20 20( ) 240 4800 4830t t

P t te e

(iii) For very long time,

1 1

20 20

lim ( ) 240lim 4800lim 4830

t t

t t t P t te e

240(0) 4800(0) 4830

4830


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12. (a) mean = 3.21

(b) (i) proportion = 0.57

(ii) 95% confidence interval

=(0.57)(1 0.57) (0.57)(1 0.57)

0.57 1.96 ,0.57 1.96100 100

0.4730,0.6670

(c) (i) Let ~ (3.21) X Po be the number of days such that there are more than 3 visits

on that day

P(crowded) ( 3)P X

2 33.21

1 ( 0) ( 1) ( 2) ( 3)

3.21 3.211 1 3.21

2 3!

0.39970529 0.3997

P X P X P X P X

e

(ii) Let ~ (5,0.399705729)Y B be the number ofcrowdeddays in 5 days,

P( 2Y ) 1 ( 0) ( 1)P Y P Y

5 5 4

11 (1 0.399705729) (1 0.399705729) (0.399705729)

0.662531312

C

Let Cbe the event ofcrowded,

P(crowdedon alternate days in the week | 2Y )

= ( ' ' ) ( ' ' ')( 2)

P C C C C C P C C C C C P Y

=2 3 3 2

(1 0.39970529) (0.39970529) (1 0.39970529) (0.39970529)

0.662531312

=2 2

(1 0.39970529) (0.39970529)

0.662531312

=0.086896526 0.0869


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13. (a) Let2

~ (6.6,1.2 )X N be the waiting time (in unit of minutes) at the regular counter.

6 6.6( 6) ( 0.5) 0.5 0.1915 0.6915

1.2

P X P Z P Z

(b) (i) Let ~ (12,0.6915)M B be the number of customers (out of 12 customers) having

a waiting time of more than 6 minutes

( 10) ( 11) ( 12)P M P M P M

12 11 1 12 12 0

11 12(0.6915) (1 0.6915) (0.6915) (1 0.6915)C C

12 11 1 12 12 0

11 12(0.6915) (0.6915) (0.6915) (0.6915)C C

0.075949454 0.0759

(ii) Let21.2

~ 6.6,12

X N

be the average waiting time (in unit of minutes) of the

12 customers at the regular counter.

6 6.66 ( 1.732)1.212

P X P Z P Z p

0.4591 0.4582 0.4582

1.74 1.73 1.732 1.73

p

0.45838p

Hence the required probability is 0.4584


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13. (c) (i) Let2

~ ( ,0.8 )Y N be the waiting time (in unit of minutes) at the express

counter.

( ) 0.2119P X k 6.6

0.21191.2

kP Z

6.6

0.81.2

k

5.64k

5.64( ) ( 5.64) 0.0359

0.8P Y k P Y P Z

5.641.80.8

4.2

(ii)4.2 6.6

( 4.2) ( 2) 0.5 0.4772 0.97721.2

P X P Z P Z

Required probability

2 2

2(0.88) ( 4.2)(0.12) ( 4.2)

(0.88) ( 4.2) 2 (0.88)( 4.2)(0.12) ( 4.2) (0.12) ( 4.2)

P X P Y

P X P X P Y P Y

=

2

2(0.88)(0.12) 0.9772 0.5

0.88 0.9772 0.12 0.5

0.121936057

0.1219

End of Solution

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HKDSE Practice Paper 2012 Mathematics_Module 1