2001 Mathematics

7/27/2019 2001 Mathematics

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Hong Kong Examinations Authority

All Rights Reserved 2001

2001-AS-M & S1

HONG KONG EXAMINATIONS AUTHORITY

HONG KONG ADVANCED LEVEL EXAMINATION 2001

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, all working must be clearly shown.

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

to 4 decimal places.

2001-ASL

M & S


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2001-AS-M & S2 1

SECTION A (40 marks)

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

1. A and B are two independent events. If 4.0)P( =A and 7.0)P( = BA ,

find P(B) .

(4 marks)

2. Let xeu2= and u

uu

y2

1

d

d= .

(a) Express

x

u

d

dand

x

y

d

din terms ofx .

(b) It is known that y= 1 when x= 0 . Express y in terms ofx .(5 marks)

3. The ages of 35 members of a golf club are shown below:

Stem (tens) Leaf (units)

1 a 8 8 9 9

2 0 1 2 3 3 4 7 8

3 1 2 2 5 b 9 9

4 0 2 5 5 6

5 2 2 5 5 8 8

6 0 1 c 6

It is known that the median and the range of the ages are 36 and 48

respectively, and the ages of the two eldest members differ by 1 .

(a) Find the unknown digits a , b and c .

(b) The three members whose ages correspond to the three unknown digits

a , b and c are replaced with three new members with ages 12 , 38

and 68 respectively. Draw two box-and-whisker diagrams in your

answer book comparing the age distributions of the members before and

after replacement.

(6 marks)


3/27

2001-AS-M & S3 2 Go on to the next page

4. The binomial expansion of nax

1

)1(

+ in ascending powers of x is

++ 29

32

3

41 xx , where a is a constant and n is a positive integer.

(a) Find the values of a and n .

(b) State the range of values ofx for which the expansion is valid.

(6 marks)

5. Suppose the rate of change of the accumulated bonus, R thousand dollars per

month, for a group of salesmen can be modelled by

150

20012

+

=

t

R (0 t 6) ,

where t is the time in months since January 1, 2001.

(a) Use the trapezoidal rule with 4 sub-intervals to estimate the total bonus

for the first 6 months in 2001.

(b) Find 2

2

d

d

t

R.

Hence or otherwise, state with reasons whether the approximation in (a)

is an overestimate or an underestimate.

(6 marks)

6. 3 students are randomly selected from 10 students of different weights. Findthe probability that

(a) the heaviest student is in the selection,

(b) the heaviest one out of the 3 selected students is the 4 th heaviest

among the 10 students,

(c) the 2 heaviest students are not both selected.

(7 marks)


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2001-AS-M & S

4

3

7. In the election of the Legislative Council, 48% of the voters support Party A ,

39% Party B and 13% Party C . Suppose on the polling day, 65% , 58%and 50% of the supporting voters of Parties A , B and C respectively cast

their votes.

(a) A voter votes on the polling day. Find the probability that the voter

supports Party B .

(b) Find the probability that exactly 2 out of 5 voting voters support Party

B .

(6 marks)


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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.

8. A chemical factory continually discharges a constant amount of biochemical

waste into a river. The microorganisms in the waste material flow down the

river and remove dissolved oxygen from the water during biodegradation. The

concentration of dissolved oxygen (CDO) of the river is given by

G(x) = kxkx eaea 2)12(122 ++ ,

where G(x) mg/L is the CDO of the river at position x km downstream from

the location of discharge of the waste, and a , k are positive constants.

At the location of the discharge of waste (i.e. x= 0) , the CDO of the river is

9 mg/L .

(a) (i) Show that a= 3 .

(ii) Find the minimum CDO of the river.

(5 marks)

(b) Figure 1 shows a sketch of the graph of G(x) against x . It is found that

downstream from the location of the discharge of waste, a stretch of

2.85 km of the river has a CDO of 4.5 mg/L or below.

(i) Find the value of k correct to 1 decimal place.

(ii) Find G(x) .

Hence determine the position of the river, to the nearest 0.1 km ,

where the rate of change of the CDO is greatest.

(iii) A river is said to be healthy if the CDO of the river is 5.5 mg/L

or above. Will the river in this case become healthy? If yes, find

the position of the river, to the nearest 0.1 km , where it

becomes healthy again.

(10 marks)

0

9

4.5

2.85

G(x)

x

Figure 1


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2001-AS-M & S

6

5

9. The spread of an epidemic in a town can be measured by the value of PPI (the

proportion of population infected). The value of PPI will increase when theepidemic breaks out and will stabilize when it dies out.

The spread of the epidemic in town A last year could be modelled by the

equationa

aket

kt

=

1

04.0)(P , where a, k > 0 and P(t) was the PPI t days

after the outbreak of the epidemic. Figure 2 shows the graph of )(Pln t

against t , which was plotted based on some observed data obtained last year.

The initial value of PPI is 0.09 (i.e. P(0) = 0.09).

(a) (i) Express )(Pln t as a linear function of t and use Figure 2 to

estimate the values of a and k correct to 2 decimal places.

Hence find P(t) .

(ii) Let be the PPI 3 days after the outbreak of the epidemic.

Find .

(iii) Find the stabilized PPI.

(8 marks)

(b) In another town B , the health department took precautions so as to

reduce the PPI of the epidemic. It is predicted that the rate of spread of

the epidemic will follow the equation 23

)43)(05.0(6)(Q

+= tbt ,

where Q(t) is the PPI t days after the outbreak of the epidemic in town

B and b is the initial value of PPI.

(i) Suppose b= 0.09 .

(I) Determine whether the PPI in town B will reach the

value in (a)(ii).

(II) How much is the stabilized PPI reduced in town B as

compared with that in town A ?

(ii) Find the range of possible values of b if the epidemic breaks out

in town B . Explain your answer briefly.

(7 marks)


7/27


t

ln P(t)

0 2 4 6 8 10 12 14 16

1

2

3

4

5

6

7

The graph of )(Pln t against t

Figure 2


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2001-AS-M & S

8

7

10. Let f(x) = 3

455

+

+

x

xfor 3x and g(x) =

x

ka 9

1

where k and a are

positive constants. Figure 3 on Page 8 shows a sketch of y= g(x) . It is known

that f(0) = g(0) and f(9) = g(9) .

(a) Determine the values ofk and a .

(2 marks)

(b) Find the equations of the horizontal and vertical asymptotes ofy= f(x) .

(2 marks)

(c) Sketch y= f(x) and its asymptotes on Figure 3. Indicate the points

where the curve cuts the axes and y= g(x).(3 marks)

(d) Let A be the area bounded by the curve y= f(x) , the x-axis, the y-axis

and the line x= 9 .

(i) Find the value of A .

(ii) If the area bounded by the curve y= g(x) , the x-axis, the lines

x= and x=+ 9 is also A , find the value of .

(8 marks)


9/27


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.

Candidate Number Centre Number Seat Number

Figure 3

O x

yy= g(x)

(0, 15)


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2001-AS-M & S10 9

This is a blank page.


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2001-AS-M & S

11

10 Go on to the next page

11. A building has only two entrances A and B . Within a 15-minute period, the

numbers of persons who entered the building by using entrances A and Bfollow the Poisson distributions with means 3.2 and 2.7 respectively.

(a) Find the probability that, on a given 15-minute period,

(i) no one entered the building by using entrance A ;

(ii) no one entered the building by using entrance B ;

(iii) at least one person entered the building;

(iv) exactly two persons entered the building.

(7 marks)

(b) Let X be the number of persons who entered the building within a 15-

minute period. Suppose X follows a Poisson distribution with mean

and k is the most probable number of persons who entered the buildingwithin a 15-minute period.

(i) By considering P(X=k 1) , P(X=k) and P(X=k+ 1) , show

that k1 .

(ii) Suppose = 5.9 . For any 5 successive 15-minute periods, find

the probability that the third time that exactly k persons entered

the building within a 15-minute period will occur during the fifth

15-minute period.

(8 marks)


12/27

2001-AS-M & S12 11

12. Table 1 gives the probability distributions of the lifetimes of two brands of

compact fluorescent lamps (CFLs). The lifetime of a Brand X CFL follows anormal distribution with mean hours and standard deviation 400 hours.The lifetime of a Brand Y CFL follows another normal distribution with mean

8 800 hours and standard deviation hours.

Table 1 Probability distributions of the lifetimes of

brands X and Y CFLs

Probability *Lifetime of a CFL

(in hours)BrandX: N(, 4002) Brand Y: N(8 800,2)

Under 8 200 0.0808 0.1587

8 200 to 8 600 0.2638 1b

8 600 to 9 000 1a 2b

9 000 to 9 400 0.2195 3b

Over 9 400 2a 0.1587

* Correct to 4 decimal places.

(a) Using the probabilities provided in Table 1, find and .

Hence find the values of 1a , 2a , 1b , 2b and 3b in Table 1.

(5 marks)

(b) Based on the results of (a), which brand of CFL would you choose to

buy? Explain.

(1 mark)

(c) Figure 4 shows a lighting system formed by three lamps. The system

will work only if lamp a works and either lamp b or lamp c works.

Lamp a

Lamp b

Lamp c

Figure 4


13/27

2001-AS-M & S

13

12

Go on to the next page

12. (continued)

(i) Suppose all the lamps in the system are brand X CFLs.

(I) Find the probability that the lifetime of the lighting

system is more than 8 200 hours.

(II) It is known that the lifetime of the lighting system is less

than 8 200 hours. Find the probability that only the

lifetime of lamp a is less than 8 200 hours.

(ii) Suppose the lighting system is formed by 2 brand X and 1

brand Y CFLs. In order for the system to have a better chance of

having a lifetime of more than 8 200 hours, where would you

put the brand Y CFL in the system? Explain.

(9 marks)


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2001-AS-M & S

14

13

13. You may use the probabilities listed in Table 2 to answer this question.

A salesman is promoting a new fertilizer which will improve the growth of

potatoes. He claims that using the fertilizer, farmers will produce 65% of

Grade A and 35% of Grade B potatoes (referred as the claim below). A

farmer uses the fertilizer on his potatoes. In order to test the effectiveness of the

fertilizer, he randomly selects 8 potatoes as a sample for testing.

(a) If the claim is valid, find the probability that there is at most 1 Grade A

potato in the sample.

(2 marks)

(b) The farmer will reject the claim if there are not more than 3 Grade A

potatoes in the sample.

(i) If the claim is valid, find the probability that the farmer will

reject the claim.

(ii) If the fertilizer can only produce 20% Grade A and 80%

Grade B potatoes, find the probability that the farmer will not

reject the claim.

(5 marks)

(c) The farmers wife takes 3 independent samples of 8 potatoes each to

checkthe claim. She will reject the claim if not more than 3 Grade A

potatoes are found in 2 or more of the 3 samples. Ifthe claim is valid,

find the probability that the farmers wife will reject the claim.

(4 marks)

(d) Suppose the claim is valid. By comparing the methods described in (b)

and (c), determine who, the farmer or his wife, will have a bigger chanceof rejecting the claim wrongly.

(1 mark)

(e) The farmers son will reject the claim if there are not more than k Grade

A potatoes in a sample of 8 potatoes. Find the greatest value of k such

that the probability of rejecting the claim is less than 0.05 given that the

claim is valid.

(3 marks)


15/27

2001-AS-M & S15 14

Table 2 Probabilities of two binomial distributions

Probability *Number of success

B(8, 0.65) B(8, 0.2)

0 0.0002 0.1678

1 0.0033 0.3355

2 0.0217 0.2936

3 0.0808 0.1468

4 0.1875 0.0459

5 0.2786 0.0092

6 0.2587 0.0011

7 0.1373 0.0001

8 0.0319 0.0000

* Correct to 4 decimal places.

END OF PAPER


16/27

2001-AS-M & S16 15

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

0.7

0.80.9

.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

=


17/27

2001

Section A

1. 0.5

2. (a)x

u

d

d=

xe

22

xy

dd = xe442

(b) 22 4 +=xexy

3. (a) a= 8 , b= 6 , c= 5

(b)

Ages

70

60

50

40

30

20

10

Beforereplacement

Afterreplacement


18/27

4. (a) 4=a , 3=n

(b)4

1

4

1


19/27

2001

Section B

8. (a) (i) Since G(0) = 9 , 9)12(122 =++ aa

a= 3

(ii) G(x) = kxkx ee 215126 +G(x)= kxkx keke 23012

= )52(6 kxkx eke

G(x) = 0 when5

2=kxe or

2

5ln

1

kx =

and G(x)

>>


20/27

(ii) G(x) = xx ee 156 5.0

G(x) = xx ee + 153 5.0

= )15(3 5.05.0 xx ee

G(x) = 0 when5

1ln

5.0

1=x ( 3.2)

and G(x)

>+t

3.29ti.e. the PPI will reach the value of .

(II) Stabilized PPI in townB= 17.0)Q(lim =

tt

The stabilized PPI will be reduced by 0.04 .

(ii) 0.05 < b (< 1).

Otherwise, Q(t) 0 and the PPI will not increase.It follows that the epidemic will not break out.


22/27

10. (a) f(0) = g(0) 153

45==k

f(9) = g(9) 12

9015=

a 2=a

(b) Since )f(lim3

xx

=

3

455lim

3 ++

x

x

x= and )f(lim

3x

x+

=3

455lim

3 ++

+ x

x

x=+ ,

3=x is a vertical asymptote.

Since )f(lim xx

=

x

x

x 31

455

lim

+

+

= 5 ,

5=y is a horizontal asymptote.

(c)

y= 5

O x

yy= g(x) x=3

(9, 7.5)

(0, 15)

(9, 0)

y= f(x)


23/27

(d) (i) A = xx

xd

3

4559

0 ++

= xx

d3

305

9

0

++

= [ ]90)3ln(305 ++ xx

= 4ln3045 +

(ii) Letx

u 91

2

= , then xu9

2lnln = and u

ux d

1

2ln

9d = .

xx

d2159

9

1

+

= uu

u d1

2ln

915

1)9/(

9/

2

2

=

1)9/(

9/

2

22ln

135

u

=

199 22

2ln

135

= 922ln2

135

If 4ln304522ln2

1359 +=

, then

( )

+=

135

2ln24ln3045ln2ln

9

5253.1


24/27

11. Let AX and BX be the numbers of persons entered the building using

entrances A and B respectively within a 15-minute period.

(a) (i) )0P( =AX =!0

)2.3(2.30

e= 2.3e 0.0408

(ii) )0P( =BX =!0

)7.2(7.20

e= 7.2e 0.0672

(iii) )1P( + BA XX = )0and0P(1 == BA XX= )0P()0P(1 == BA XX

= 7.22.31 ee= 9.51 e 0.9973

(iv) )2P( =+ BA XX= )2P()0P()1P()1P()0P()2P( ==+==+== BABABA XXXXXX

=!2

)7.2(

!1

7.2

!1

2.3

!2

)2.3(7.22

2.37.22.3

7.22.32

++e

eee

ee

= 9.5405.17 e 0.0477

(b) (i) Since k is the most probable number of persons entered

the building within a 15-minute period,

P(X=k 1) P(X=k) and P(X=k+ 1) P(X=k)

Hence!)!1(

1

k

e

k

e kk

k

and!)!1(

1

k

e

k

e kk +

+

1+ k k1

(ii) From (b)(i), k= 5 .The probability required

= )][P()]P(1[)][P( 2242 kXkXkXC ===

=

!5

)9.5(

!5

)9.5(1

!5

)9.5(9.55

29.55

29.55

42

eeeC

0.0183


25/27

12. Let XE and YE be the lifetimes of brand X and brand Y CFLs respectively.

(a) 1151.0)8200P( => cba XXX

= 2)]8200[P(1)]8200P(1[ pX 0.0459 + 0.0092 + 0.0011 + 0.0001 + 0.0000 0.0563

(c) The required probability

= 33323

2 )1( qCqqC +

33323

2 )1060.0()1060.01()1060.0( CC + 0.0313

(d) The probability that the farmer will wrongly reject the claim is 0.1060

whereas the probability that his wife will wrongly reject the claim is 0.0313 .

Therefore the farmer will have a bigger chance of rejecting the claim wrongly.

(e) )65.0|2P( = pX 0.0252)65.0|3P( = pX 0.08080.0252 + 0.1060

Since )65.0|2P( = pX < 0.05 < )65.0|3P( = pX k= 2 .

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2001 Mathematics