2000 Mathematics

7/27/2019 2000 Mathematics

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Hong Kong Examinations Authority

All Rights Reserved 2000

2000-AS-M & S1

HONG KONG EXAMINATIONS AUTHORITY

HONG KONG ADVANCED LEVEL EXAMINATION 2000

MATHEMATICS AND STATISTICS AS-LEVEL

8.30 am 11.30 am (3 hours)

This paper must be answered in English

1. This paper consists of Section A and Section B.

2. Answer ALL questions in Section A, using the AL(E) answer book.

3. Answer any FOUR questions in Section B, using the AL(C)2 answer book.

4. Unless otherwise specified, numerical answers should either be exact or given

to 4 decimal places.

2000-ASL

M & S


2/28

2000-AS-M & S2 1

SECTION A (40 marks)

Answer ALL questions in this section.Write your answers in the AL(E) answer book.

1. Lety

xxy =)(ln where x, y > 0 . Show that

2

2

d

d

xxy

yxy

x

y

+

= .

(4 marks)

2. Let2

1


3/28

2000-AS-M & S3 2 Go on to the next page

4. A polynomial f(x) has the following properties:

x < 0 x= 0 0 < x < 1 x= 1 1 < x < 3 x= 3 x > 3

f(x) 27 16 0

f(x) < 0 0 > 0 > 0 > 0 0 > 0

f(x) > 0 > 0 > 0 0 < 0 0 > 0

(a) Find the range of values ofx for which the graph of f(x) is concave

downward (i.e. convex upward).

(b) Write down all the points of inflexion of the graph of f(x) .

(c) Sketch the graph of f(x) .

(4 marks)

5. A fitness centre advertised a programme specifically designed for women

weighing 70 kg or more, and claimed that their individual weights could be

reduced by at least 20 kg on completion of the programme. Twenty-one

women joined the programme and their weights in kg when they started are

shown below:

Stem (tens) Leaf (units)

7 0 0 2 3 5 5 7

8 1 1 4 5 6 6 7 8

9 0 2 5 8 9 9

(a) Find the median and the interquartile range of these weights.

(b) On completion of the programme, the median, lower quartile and upperquartile of the weights of these women are 73 kg , 68 kg and 77 kg

respectively. The lightest and heaviest women weigh 60 kg and 82 kg

respectively. Draw two box-and-whisker diagrams in your answer book

comparing the weights of these women before and after the programme.

(c) Referring to the box-and-whisker diagram in (b), someone claimed that

none of these women had reduced their individual weights by 20 kg or

more on completion of the programme. Determine whether this claim is

correct or not. Explain your answer briefly.

(6 marks)



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2000-AS-M & S4 3

6. Mr. Chan has 6 cups of ice-cream in his refrigerator. There are 5 different

flavours as listed:

1 cup of chocolate,

1 cup of mango,

1 cup of peach,

1 cup of strawberry and

2 cups of vanilla.

Mr. Chan randomly chooses 3 cups of the ice-cream. Find the probability that

(a) there is no vanilla flavour ice-cream,

(b) there is exactly 1 cup of vanilla flavour ice-cream.

(5 marks)

7. The number of people killed in a traffic accident follows a Poisson distribution

with mean 0.1 . There are 5 traffic accidents on a given day, find the

probability that there is at most 1 accident in which some people are killed.

(5 marks)

8. A department store uses a machine to offer prizes for customers by playing

games A or B . The probability of a customer winning a prize in game A is

9

5and that in game B is

6

5. Suppose each time the machine randomly

generates either game A or game B with probabilities 0.3 and 0.7

respectively.

(a) Find the probability of a customer winning a prize in 1 trial.

(b) The department store wants to adjust the probabilities of generating

game A and game B so that the probability of a customer winning a

prize in 1 trial is3

2. Find the probabilities of generating game A and

game B respectively.

(6 marks)



5/28


SECTION B (60 marks)

Answer any FOUR questions in this section. Each question carries 15 marks.Write your answers in the AL(C)2 answer book.

9. A department store has two promotion plans, F and G , designed to increase

its profit, from which only one will be chosen. A marketing agent forecasts that

ifx hundred thousand dollars is spent on a promotion plan, the respective rates

of change of its profit with respect to x can be modelled by

f(x) = xxe 25.0416 + and g(x) =x

x

81

616

++ .

(a) Suppose that promotion plan F is adopted.

(i) Show that f(x) f(4) forx > 0 .

(ii) If six hundred thousand dollars is spent on the plan, use the

trapezoidal rule with 6 sub-intervals to estimate the expected

increase in profit to the nearest hundred thousand dollars.

(6 marks)

(b) Suppose that promotion plan G is adopted.

(i) Show that g(x) is strictly increasing forx > 0 .

As x tends to infinity, what value would g(x) tend to?

(ii) If six hundred thousand dollars is spent on the plan, use the

substitution xu 81+= , or otherwise, to find the expected

increase in profit to the nearest hundred thousand dollars.(7 marks)

(c) The manager of the department store notices that if six hundred

thousand dollars is spent on promotion, plan F will result in a bigger

profit than G . Determine which plan will eventually result in a bigger

profit if the amount spent on promotion increases indefinitely. Explain

your answer briefly.

(2 marks)



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2000-AS-M & S6 5

10. A researcher studied the growth of a certain kind of bacteria. 100 000 such

bacteria were put into a beaker for cultivation. Let t be the number of dayselapsed after the cultivation has started and r(t) , in thousands per day, be the

growth rate of the bacteria. The researcher obtained the following data:

t 1 2 3 4

r(t) 7.9 12.3 15.3 17.5

(a) The researcher suggested that r(t) can be modelled by batt =)r( ,

where a and b are positive constants.

(i) Express )r(ln t as a linear function of tln .

(ii) Using the graph paper on Page 6, estimate graphically the value

of r(5) to 1 decimal place without finding the values of a and

b .

(5 marks)

(b) The researcher later observed that r(5) was 18.5 and considered the

model in (a) unsuitable. After reviewing some literature, he used the

model qtpet= 20)r( , where p and q are positive constants.

(i) Express )]r(20ln[ t as a linear function of t.

(ii) Using the graph paper on Page 7, estimate graphically the values

ofp and q to 3 significant figures.

(iii) Estimate the total number of bacteria, to the nearest thousand,

after 15 days of cultivation.

(10 marks)



7/28


Page Total

10. (Contd) If you attempt Question 10, fill in the details in the first three boxes

above and tie this sheet INSIDE your answer book.

Graph paper for part (a)(ii), Question 10

Candidate Number Centre Number Seat Number



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2000-AS-M & S8 7

Page Total

Graph paper for part (b)(ii), Question 10



9/28


11. A researcher studied the commercial fishing situation in a certain fishing zone.

Denoting the total catch of coral fish in that zone in t years time from January1, 1992 by N(t) (in thousand tonnes) , he obtained the following data:

t 2 4

N(t) 55 98

The researcher modelled N(t) by kteat = 1)N(ln where a and k are

constants.

(a) Show that 055

98ln

124 =+ e

ee kk .

Hence find, to 2 decimal places, two sets of values of a and k.

(4 marks)

(b) The researcher later found out that N(7) = 170 . Determine which set ofvalues of a and k obtained in (a) will make the model fit for the known

data.

Hence estimate, to the nearest thousand tonnes, the total possible catch

of coral fish in that zone since January 1, 1992.

(4 marks)

(c) The rate of change of the total catch of coral fish in that zone since

January 1, 1992 at time t is given byt

t

d

)N(d.

(i) Show that ktetkt

t = 1)N(d

)N(d.

(ii) Using the values of a and k chosen in (b), determine in which

year the maximum rate of change occurred.

Hence find, to the nearest integer, the volume of fish caught in

that year.

(7 marks)



10/28

2000-AS-M & S10 9

12. The milk produced by Farm A has been contaminated by dioxin. The amount

of dioxin presented in each bottle of milk follows a normal distribution withmean 20 ng (1 ng = 610 g) and standard deviation 5 ng . Bottles whichcontain more than 12 ng of dioxin are classified as risky, and those which

contain more than 27 ng are hazardous.

(a) Suppose a bottle of milk from Farm A is randomly chosen.

(i) Find the probability that it is risky but not hazardous.

(ii) If it is risky, find the probability that it is hazardous.

(6 marks)

(b) A distributor purchases bottles of milk from both Farm A and Farm B

and sells them under the same brand name Healthy. It is known that

60% of the milk is from Farm A and the rest from Farm B . A bottle ofmilk from Farm B has a probability of 0.058 of being risky and 0.004

of being hazardous.

(i) If a randomly chosen bottle of Healthy milk is risky, find the

probability that it is from Farm B .

(ii) If a randomly chosen bottle of Healthy milk is risky, find the

probability that it is a hazardous bottle from Farm B .

(iii) The Health Department inspects 5 randomly chosen bottles of

Healthy milk. If 2 or more bottles of milk in the batch are

risky, the distributors license will be suspended immediately.

Find the probability that the license will be suspended.

(9 marks)



11/28


13. Boys B1 , B2 and girls G1 , G2 are students who have qualified to represent

their school in a singing contest. One boy and one girl will form one team. Theteam formed by Bi and Gj is denoted by BiGj , where i= 1, 2 and j= 1, 2 .A team can enter the second round of the contest if both team members do not

make any mistakes during their performance. Suppose that a student making

mistakes in a performance is an independent event, and the probabilities that

B1 , B2 , G1 and G2 do not make any mistakes in a performance are 0.9 , 0.7 ,

0.8 and 0.6 respectively.

(a) List all the possible teams that can be formed.

(1 mark)

(b) Find the probability that B1G1 can enter the second round of the

contest.

(1 mark)

(c) If a team is selected randomly to represent the school, find the

probability that the team can enter the second round of the contest.

(2 marks)

(d) If two teams B1G1 and B2G2 are formed to represent the school, find

the probability that

(i) exactly one team can enter the second round of the contest,

(ii) at least one team can enter the second round of the contest.

(5 marks)

(e) Suppose that two teams are allowed to represent the school and each

student can only join one team.

(i) If the two teams are formed randomly, find the probability that

exactly one team can enter the second round of the contest.

(ii) How should the teams be formed so that the school has a better

chance of having at least one team that can enter the second

round of the contest?

(6 marks)



12/28

2000-AS-M & S12 11

14. The company which sells boxes of Energy Chips wanted to increase its sales

volume by issuing stamps which could be exchanged for toys. Stamps were putinto some of the boxes and a collection of 5 stamps could be exchanged for one

toy. Some students conducted a survey to study the chance of getting stamps.

They randomly selected 100 boxes of the Chips and counted the number of

stamps in each box. The result is shown in the first two columns of Table 1.

(a) The students tried to fit the data by a binomial (n= 6) and a Poissondistribution. They took 1 as the mean of each of the distributions. Fill

in the missing values in Table 1.

(4 marks)

(b) Suppose the absolute values of the differences between observed and

expected frequencies are regarded as errors. The distribution with a

smaller maximum error will fit the data better. Which distribution in (a)

is better?(1 mark)

(c) Assume the distribution chosen in (b) is adopted.

(i) Find the probability of getting at least 1 stamp in a box of the

Chips.

(ii) Mrs. Wong keeps on buying the Chips until she gets at least 1

stamp. Find the probability that she needs not buy more than 3

boxes.

(iii) Suppose Mr. Cheung buys 2 boxes of the Chips. If he gets

stamps in both boxes, find the probability that he gets just

enough stamps to exchange for a toy. (10 marks)



13/28

2000-AS-M & S13 12

Page Total

14. (Contd) If you attempt Question 14, fill in the details in the first three boxes

above and tie this sheet INSIDE your answer book.

Table 1 Observed and expected frequencies of the number of boxes

by the number of stamps in 100 boxes of Energy Chips

Expected frequency *Number of

stamps

Observed

frequency Poisson Binomial

0 34 36.79 33.49

1 39 40.19

2 21 18.39

3 5 5.36

4 1 1.53

5 0 0.31 0.06

6 0 0.05 0.00

Total 100

* Correct to 2 decimal places.

END OF PAPER

Candidate Number Centre Number Seat Number



14/28

2000-AS-M & S14 13

This is a blank page.



15/28

2000-AS-M & S15 14

Table: Area under the Standard Normal Curve

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

0.0

0.1

0.2

0.3

0.4

.0000

.0398

.0793

.1179

.1554

.0040

.0438

.0832

.1217

.1591

.0080

.0478

.0871

.1255

.1628

.0120

.0517

.0910

.1293

.1664

.0160

.0557

.0948

.1331

.1700

.0199

.0596

.0987

.1368

.1736

.0239

.0636

.1026

.1406

.1772

.0279

.0675

.1064

.1443

.1808

.0319

.0714

.1103

.1480

.1844

.0359

.0753

.1141

.1517

.1879

0.5

0.6

0.70.8

0.9

.1915

.2257

.2580

.2881

.3159

.1950

.2291

.2611

.2910

.3186

.1985

.2324

.2642

.2939

.3212

.2019

.2357

.2673

.2967

.3238

.2054

.2389

.2704

.2995

.3264

.2088

.2422

.2734

.3023

.3289

.2123

.2454

.2764

.3051

.3315

.2157

.2486

.2794

.3078

.3340

.2190

.2517

.2823

.3106

.3365

.2224

.2549

.2852

.3133

.3389

1.0

1.1

1.21.3

1.4

.3413

.3643

.3849

.4032

.4192

.3438

.3665

.3869

.4049

.4207

.3461

.3686

.3888

.4066

.4222

.3485

.3708

.3907

.4082

.4236

.3508

.3729

.3925

.4099

.4251

.3531

.3749

.3944

.4115

.4265

.3554

.3770

.3962

.4131

.4279

.3577

.3790

.3980

.4147

.4292

.3599

.3810

.3997

.4162

.4306

.3621

.3830

.4015

.4177

.4319

1.51.6

1.7

1.8

1.9

.4332

.4452

.4554

.4641

.4713

.4345

.4463

.4564

.4649

.4719

.4357

.4474

.4573

.4656

.4726

.4370

.4484

.4582

.4664

.4732

.4382

.4495

.4591

.4671

.4738

.4394

.4505

.4599

.4678

.4744

.4406

.4515

.4608

.4686

.4750

.4418

.4525

.4616

.4693

.4756

.4429

.4535

.4625

.4699

.4761

.4441

.4545

.4633

.4706

.4767

2.0

2.12.2

2.3

2.4

.4772

.4821

.4861

.4893

.4918

.4778

.4826

.4864

.4896

.4920

.4783

.4830

.4868

.4898

.4922

.4788

.4834

.4871

.4901

.4925

.4793

.4838

.4875

.4904

.4927

.4798

.4842

.4878

.4906

.4929

.4803

.4846

.4881

.4909

.4931

.4808

.4850

.4884

.4911

.4932

.4812

.4854

.4887

.4913

.4934

.4817

.4857

.4890

.4916

.4936

2.5

2.6

2.7

2.82.9

.4938

.4953

.4965

.4974

.4981

.4940

.4955

.4966

.4975

.4982

.4941

.4956

.4967

.4976

.4982

.4943

.4957

.4968

.4977

.4983

.4945

.4959

.4969

.4977

.4984

.4946

.4960

.4970

.4978

.4984

.4948

.4961

.4971

.4979

.4985

.4949

.4962

.4972

.4979

.4985

.4951

.4963

.4973

.4980

.4986

.4952

.4964

.4974

.4981

.4986

3.0

3.13.2

3.33.4

.4987

.4990.4993

.4995

.4997

.4987

.4991.4993

.4995

.4997

.4987

.4991.4994

.4995

.4997

.4988

.4991.4994

.4996

.4997

.4988

.4992.4994

.4996

.4997

.4989

.4992.4994

.4996

.4997

.4989

.4992.4994

.4996

.4997

.4989

.4992.4995

.4996

.4997

.4990

.4993.4995

.4996

.4997

.4990

.4993.4995

.4997

.4998

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

Note: An entry in the table is the proportion of the area under the entire curve which is between z= 0and a positive value ofz. Areas for negative values ofz are obtained by symmetry.

z0

A(z)

xezAz

x

d2

1)(

0

2

2

=



16/28

2000

Section A

2. (a) 21

)21( x+ = ...+++32

2

1

2

11 xxx

2

1

3 )81(

+x = ...+3

41 x

(b) 21

2 )421(

+ xx = ...++32

2

7

2

11 xxx

3. 6.2537

4. (a) 1 < x < 3

(b) (1, 16) and (3, 0)

(c)

(1, 16)

x

y

O

(0, 27)

(3, 0)



17/28

5. (a) Median = 85 kg

Interquartile range = 15 kg

(b)

(c) No. It is possible that the woman weighed 60 kg on completion of

the programme is the woman weighed 99 kg when the programme

started.

100

95

90

85

80

75

70

65

60

Weight(kg)

Old

weights

New

weights



18/28

6. (a)5

1

(b)5

3

7. 0.9255

8. (a) 0.75

(b) The probabilities of generating game A and game B are5

3and

5

2

respectively.



19/28

2000

Section B

9. (a) (i) )25.01(4)(f25.0

xex x =

>x ,

x

xxx

x81

81286816

)(g+

++

= =2

3

)81(

)41(6

x

x

+

+> 0

g(x) is strictly increasing forx > 0 .

+

+=

++

8

1

616lim

81

616lim

x

x

x

x

xx

)(g x as x



20/28

(ii) Let xu 81+= , then xu 812 += , xuu d8d2 =

xx

xxx d

81

616d)(g

6

0

6

0

++=

= uuu

ud

4

1

8

)1(616

7

1

2

+

= uuu d163

416

37

1

2

+

=7

1

23

16

32

16

1

+ uuu

116 The expected increase in profit is 116 hundred thousand dollars.

(c) From (a)(i), )4(f)(f x for 0>x .i.e. )(f x is bounded above by )4(f .

From (b)(i), )(g x increases to infinity as x increases to infinity.

f(x) > 0 and g(x) > 0 for x > 0 ,the area under the graph of g(x) will be greater than that of f(x) as

x increases indefinitely. Plan G will eventually result in a bigger profit.



21/28

10. (a) (i) batt =)r(tbat lnln)r(ln +=

(ii) ln t 0 0.69 1.10 1.39

ln r(t) 2.07 2.51 2.73 2.86

When t= 5 , ln t 1.61 ln r(5) 3 from the graph.

r(5) 20.1

0 1 22

2.5

3

ln t

ln

r(t)

1.61



22/28

(b) (i) qtpet = 20)r(qtpt = ln)]r(20ln[

(ii) t 1 2 3 4 5

)]r(20ln[ t 2.49 2.04 1.55 0.92 0.41

From the graph, ln p 3.05p 21.1

q5

05.341.0 0.528

q 0.528

0 2 4t

6

3

ln

[20r(t)]

2

1



23/28

(iii) The total number, in thousands, of bacteria after 15 days of

cultivation

= 15

0d]20[ tpe qt + 100

= 1002015

0

+

+ qteq

pt

= 10030015

++

q

p

eq

p q

260 + 100 360



24/28

11. (a)

=

=

k

k

ea

ea

41

21

98ln

55ln

Eliminating a , we have

055ln98ln2141 =+ kk ee

055

98ln

124 =+ e

ee kk

055

98ln

1)(

222 =+ e

eekk

ke

2 =2

55

98ln

411

e

0.30635 or 0.69365 0.306 or 0.694

8401.4

5915.0

a

kor

8929.5

1829.0

a

k

84.4

59.0

a

kor

89.5

18.0

a

k(2 d.p.)

(b) Using

84.4

59.0

a

k, 80.4)7lnN( , N(7) 121 .

Using

89.5

18.0

a

k, 12.5)7lnN( , N(7) 167 .

89.5

18.0

a

kwill make the model fit for the known data.

t

et eet18.01

89.5)N(ln)N(

=

361)N( 89.5 et as t

The total possible catch of coral fish in that area since January 1, 1992 is361 thousand tonnes.



25/28

(c) (i) kteat = 1)N(ln

ktket

t = 1

)N(

)(N

ktetkt

= 1)N()(N

(ii) N(t)= ])N()(N[ 11 ktkt etketk

= )1()N( 112 ktkt eetk

> 1.6) 0.4452 + 0.5 0.9452

P(hazardous | A) = P(X> 27)= P(Z> 1.4) 0.5 0.4192 0.0808

P (a risky bottle is hazardous | A) 9452.0

0808.0 0.0855

(b) (i) P(risky) = 0.6 P(risky | A) + 0.4 P(risky | B) 0.6(0.9452) + 0.4(0.058) 0.59032 0.5903

P(B and risky | risky) =P(risky)

B)P(B)|P(risky

59032.0

)4.0)(058.0(

0.0393

(ii) P(B and hazardous | risky) =P(risky)

B)P(B)|sP(hazardou

59032.0

)4.0)(004.0(

0.00271 0.0027

(iii) P(license suspended) 45 )59032.01)(59032.0(5)59032.01(1 0.9053



27/28

13. (a) Possible teams: B1G1 , B1G2 , B2G1 and B2G2 .

(b) The probability that B1G1 can enter the second round of the contest

= 0.9 0.8= 0.72

(c) Probability required = )6.07.08.07.06.09.08.09.0(4

1+++

= 0.56

(d) Suppose B1G1 and B2G2 are formed to represent the school.

(i) The probability that exactly one team can enter the second round

= ( )( ) ( )( )8.09.016.07.06.07.018.09.0 += 0.5352

(ii) The probability that at least one team can enter the second round

= 0.5352 + 0.9 0.8 0.7 0.6= 0.8376

(e) (i) If the two teams are formed randomly, the probability that

exactly one team can enter the second round

= ( )( ) ( )( )[ ]6.09.018.07.08.07.016.09.02

15352.0

2

1++

= )4952.05352.0(2

1+

= 0.5152

(ii) If B1G2 and B2G1 are formed to represent the school, the

probability that at least one team can enter the second round

= 0.4952 + 0.9 0.8 0.7 0.6= 0.7976

From (d)(ii), the combination B1G1 and B2G2 will have a better

chance of having at least one team that can enter the second

round of the contest.



28/28

14. (a) Poisson distribution Binomial distribution

frequency:!

1001

x

e

xx

xC

6

6

6

5

6

1100

missing values: 36.79 20.09

6.13 0.80

(b) The maximum error for Po(1) is 2.79 while that for B(6,

6

1) is 1.19.

The binomial distribution is better.

(c) (i) The probability of getting at least 1 stamp in a box of the Chips

=6

61

11

0.665102 0.6651

(ii) The probability of getting at least 1 stamp in buying

not more than 3 boxes

])665102.01()665102.01(1)[665102.0( 2++ 0.9624

(iii) The probability of getting exactly 5 stamps in 2 boxes with stamps

=

+

3363

4262

2464

561

6

5

6

1

6

5

6

1

6

5

6

1

6

5

6

12 CCCC

=

+

7563

62

7564

61

6

5

6

1

6

5

6

12 CCCC

=12

7

6

)2015156(52 +

0.027994

The probability of getting stamps in both boxes in buying two boxes

(0.665102)2 0.442361

The required conditional probability =3

2

p

p

442361.0

027994.0

0.0633


2000 Mathematics

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