2007-ASLM&S

HONG KONG EXAMINATIONS AND ASSESSMENT AUTHORITY

HONG KONG ADVANCED LEVEL EXAMINATION 2OO7

MATHEMATICS AND STATISTICS AS.LEVEL

8.30 am -

11.30 am (3 hours)

This paper must be answered in English

L 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) 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 decimalplaces.

2007-AS-M & S*l I4

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.

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(sryuu 0,) y NOIIJSS

4. Albert conducted a survey on the time spent (in hours) on watching television by 16 students'Thedatarecorded are z.i , 1.2,2.1, 5.1 ,2.1 , 4'7 , l'g ,2'4,2'4,2'9 ,3'6 ' 2'3 ' 3'9 ' 2'2 '1.8 and ft, where t isthemissingdatum'

(a) Albert assumes that the range of these data is 5'3 hours'(i) Find the value of t '(ii) Construct a stem and leafdiagram for these data'(iii) Find the mean and the median of these data'

(b) Alben finds that the assumption in (a) is incorrect and he can only assume that the range ofthese data is greater than 5.3 hours. Describe the change in the mean and the change inthe median oith.s" data due to the revision of Albert's assumption' (7 marks)

5. Let A and 'B betwoeventswith P(A)=a and P(B)=b ' where 0

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El

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= (r]a

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ot _

xs = (r)J auua0

'

L

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( .mquo$ fi,NoxJo s

8. A financial analyst, Mary, models the rates of change of profit (in billion dollars) made by

companies A and B resPectivelY bY

f(/) = tn(et +2)+3 and c@=# ,

where / is the time measured in months'Assume that the two models are valid for 0 < t < 6 '

Using the trapezoidal rule with 6 sub-intervals, estimate the total profit made bycompany A from t=0 to t=6 '

Find *P and hence

derermine whether the estimate in (aXi) is an over-

estimate or an under-estimate' (7 marks)

(D

(ii)

(a)

(b)I(i) Expand i;:7 in ascending powers of I as far as

the term in /a

(ii) Using the result of (bxi), find the expansion "f # in ascending powers of Ias far as the term in /a

(iiD Using the result of (b)(ii), estimate the total profit madet=O to t=6 '

by companY B from(6 marks)

(c) Mary claims that the total profit made by company A from I = 0that of company B ' Do you agree? Explain your answer'

to l=6 islessthan(2 marks)

l82007-AS-M & S-5

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' tl Jo enle^ aql purC (e)

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(c)

(q)

l0 The manager, Teresa, of a superstore launches a promotion plan to increase the sales volume. Thenumber of customers shopping at the superstore in a minute can be modelled by a Poissondistribution with a mean of 2.4 customers per minute. The expenses of customers in thesuperstore are assumed to be independent and follow a normal distribution with a mean of $ 375and a standard deviation of $ 125 . A customer who spends more than $ 30O but less than $ 600in the superstore can enter lucky draw X in which the probability of winning a gift is 0.25 . Acustomer who spends $ 600 or more in the superstore can enter lucky draw I/ in which theprobabiliry of winning a gift is 0.8 . Assume that eaeh customer enters at most one lucky draw foreach visit.

Find the probability that there are more than 2 customers shopping at the superstore in acertain minute.

(3 marks)

(a)

(b)

(e)

(0

Find the probabiliry that a randomly selected customer shopping at theenter lucky draw X.

superstore can

(2 marks)

(c)

(d)

Find the probability that a randomly selected customergift.

shopping at the superstore wins a

(2 marks)

Find the probability that there are exactly 3 customers shopping at the superstore in acertain minute and each of them wins a gift. (2 marks)

Given that there are more than 2 customers shopping at the superstore in a certain minute,find the probability that there are fewer than 5 customers shopping at the superstorein this minute and each of them wins a gift. (3 marks)

IfTeresa wants to revise the least expense ofa customer for entering lucky draw X so that33%o of the customers shopping at the superstore could enter lucky draw )', what shouldbe the revised least exPense? (3 marks)

2007-AS-M & S-7 20

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12. ln game I , two players take turns to draw a ball randomly' with replacement' from a bag.ori"i"ing a' greentails and 1 red ball. The first player who draws the red ball wins the game'Ctrristlne"ana ionald play the game until one of them wins. Christine draws a ball first'

(a) Find the probability that Donald wins game I before his 4th draw'

Find the probability that Donald wins game I '

(2 marks)

(b)

(c) Given that Donald wins gamebefore his 4th draw'

Donald wins game ,4

(3 marks)

I , find the probability that Donald does not win game '4(3 marks)

(7 marks)

END OF PAPER

(d) Aftergamel, ChristineandDonaldplaygame B' Ingame B' therearebox X andb;; f. eox i contains 2 cardswhicharenumbered 4 and 8 respectivelywhilebox 1,contains 7 cards which are numbered 1 ,2 , "' ' 7 respectively'. A player randomlydraws one card from each box without replacement. If the number drawn from box x isgr""i". ifr", that from box I, then the piayer wins game 'B ' Christine and Donald takeiurns to draw cards until one of them wins game B ' Donald draws cards first'

(i) Find the probability that Donald wins game B in his l st draw'(ii) Find the probability that Christine wins game B '(iii)GiventhatChristineandDonaldwinonegameeach,findtheprobabilirythat

2007-AS-M & S-9 22

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Solution

(a) (i) [, *l)'\ o)

r r(r -l).x.,

=t+-x+ 'ri '?)'+. ., rx r(r-l) .)

=r*;*-.--a-*- *.'.r 2r r(r-l) -lSo. we have -

=-- and E: lg1-

urd , ='Solving, we have a =:2 2

The binomial expansion is valid a, l* I . ,lJlThus, the range of values of x i, a. , .1

( ii)

lM for any two terms correct

lA for both correct

1M can be absorbed

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elnr lcnpord Jo oJnr tuerlonb loJ h1

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,e2+l*ta-l

{p

=t (u) 'r

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OS + ,12'9= 'c

002

eq Illll\ ?rjolcEq Jo roqtunu eqJ'9=/ aABI{e,i\{'oS

- 8 o^El{a.t\ '9=N uot{A(q)

F

I+I r? z_=0_ ^

'0=r( uaq^\ n- ,lpz(rsa

+ I) r8a + I;7**- .fv-=

(*ru*, ,ta-r),ra+1 rpI rsrg ,rrt-)rvu-l {p

t8a*l _ rr'-l =,8a8 ,fl - {P t

( ,ra ;- 1)u1 - ( rta - 1)u1 = ,{ u1

*ga+1

-=A

a -r xb

t = (0)N Sursn rogn

_

JO{002-

uorlnlrlsqns Joc IAi I

0s + -tz-n*

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ost n) ttor*r,rloor )-

,ft osr no' l=' 002 ss+J7J

.(og+ -rz) t^o?J=Ytrul ' 6p^l = n p eAEq 0,{\ '{ueq.l.

' 09+ z^a=n

s{iEI^uorlnlos

Solution

(b) (i)

(ii)

( iii)

Since (22 +l)e3' =ed+Fx, wehave ln(22 +1) +32=a+ 0x

Since the graph of the linear function passes through the origin and theslope of the graph is 2, wehave a = 0 and B =2 '

ln( z2 + l) + 3z =2x2'

-a -, d'

z2+l' 'dzdxl 3

Therefore, we have *1,=r=, .

Notethat x=0 when z=0dvl

Af so note that -' | = '2 .dxllr=Uayldrltz=U

( a,l )(a,l )=[*1.=,Jlal,=,J

= (-2(1)"\2 )

=-3

1A

1A for both correct

--------(7)

tM for quotient rule or product rule

IA

| + (22 +l)o nt" - 6(z2 +7)2 e6'| - 122 + 112 e6' 2(22 + l)o

""=

+ (22 +l)o rt"

-2(22 +1)(22)e6')+ 4(22 +1)3 (22)et2'

28

5--

6Z

(srnoq

(srnoq

(r)----------

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0090't ldeccu) y1

(tUXe) ut pourutqo urpolu eql sB ous eql sr uerporu pesr^or oqJ'(ttt)(e) ur paurslqo ueotu eql uuql releer8 sr uoru posr^ol oq1 (q)

srnoq , z =u3rpou eqJ

srnoq 90'[ =ueou orlJ (ltt)

VI + htrI

s

IL

6 L96T'f,ZTI

682G-qrrEi)fEET

9s

nC

ZICsr-Tn=ImrS (lt)

'S9=ry o^Er4am'snq1'

z'I- 1 - {.'S o^Eq e,u '0cu0H0

5. (a)

(b)

= P(B I A',)P(A'): 0.3(l

- a)

P(A'aB)=P(A'I B) P(B)= 0.6b

Hence, we have 0.6b = 0.3(1- a) .Thus, we have a +2b =l

P(A n B')= P(B'l A)P(A)=0.ia

P(Aw B')= 1- P(l'n B)=l-0.66Note that P(Aw B') =P(A)+ P(B') -P(Aa B')Hence, we have 1

-0.6b = a + (l - b) -0.7 aSo, we have 3a=4bSolving a*2b-l and 3a=4b , wehave a=0'4 and b=03

Marks

II

---------------lIIII

either oneIIIIIIIII

-------t

1M for complementary events

1M

lA for both correct

M ------------lIIIIIII

P(A a B')= P(B'l A)P(A)=0.1a

P(Aw B')=l

-P(A' a B)=1-0.3(l-a)= 0.7 + 0.3a

Note that P(Aw B') = P(l) + P(B') -P(A r: B')Hence, we have 0.7 + 0.3a = a +7 - b - 0.1 aSo, we have b = 0.3 .By (a), we have a +2(0.3) =1 .Thus, we have a =0.4 .

IIIIII

1M

1M for complementary events

both correct

(c) Since P(l^B)= P(A)-P(AaB',) , P(A) =0.4 and P(AetB',)=0.28 ,we have P(AaB) =0.12 = (0'4X0.3) =P(A)P(B) 'Thus,,4 and B areindePendentevents.

lM for relating P(A a B) and P(l)P(B)lA f.t.

Since P(l) =a , wehave P(lrlB)=P(A)-P(AaB')=a-0'7a=0'3aWith the help of P(B) = 0.3 , we have P(l ^ B) = P(I)P(B) 'Thus, I and B are independent events.

1M for relating P(A n B) and P(l)P(B)1A f.t.

Since P(l'lB)=0.6, we have P(AlB)=l-P(A'lB)=t-0'6=0'4With the help of P(A) = 0'4 , we have P(l I B) = P('4) 'Thus, I and B areindependentevents.

lM for relating P(A1B) and P(l)1A f.t.----------(7)

30

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roJhtrI + (s + t + b + d) ToJI II

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sv_L

Etolc oe z^ t^sJsJ t

,(lpqeqord permber eq1

9SI'0 '1'r toJ l-D

YI(e) Bursn

roJNI + (s * .t + b + d) ro31atr1

9S9l'0 =9S99S99S l'0 =

9n_L

. 8 .6 .01. 09 (-x-x- ) r s's t.{rlpqeqord purnber eq1 (q)

/10'0 '1'r roJ I-D

VI

IIB roJ YI +JolBurruousp Jo JolBrolllnu roJ v I

l9l0'0 =Lggggggl}'o x

09_I

t, otdz,G= vd rd

,Qrpqeqord permbar eq1

,10'0 '1'r toJ l-D

YI

IIB roJ vl + lalclq puz roJ vl

1910'0 *

[9*999't'o''ol*09I

( f;)(,J;)Ifc.][ i: ]- ,Qrpquqo-rd parrnber aq1

lI0'0 '1'r roJ [-t2

VI

lle roJ vl + lalcelq lsl roJ vl

19l0'0 =4999999t,0.0.r,

09_I

(J;.)( ,i^l )I r][iclt) .fiqrqeqord pertnber eq1

l,10'0 '1'r JoJ I-r7

VI

IIB roJ VI +Joleuruouop Jo Jolerel'unu JoJ YI

$lreIAtr

L9t0'0 x

499999910'0 =09_I

.8 6..0i (-X-X-) = 't ilI ,Qrlrqeqord pertnber aq1

uoqnlos

'9 (u)

#

Solution Marks

The roquired probabilitY1 PsP:

-

-

L ---!---!-60' Pr'

745

*,0J55555556s 0.1556

1M for (p+q) + lM for using (a)

IA

a-l for r.t. 0.156

The required probability

= rfrrrlrtfr.tfrrrfxfi=

'7_

45J0,,r5ffi85i:56=

0.1556

lMfor(p+q+r+s)2A

a-l far r.t. 0.156

The required probabilitYclcf , ciclci

=

c;roci - t;ot'1

45

^* $,15555,55.5,5

= 0.1556

lMfor(p+q+r+s)2A

a-l for r.t. 0.156

The required probabilityp! nu , PrzPt4 P:=7r-

"r'7

45qse.$555:S$56* 0.1556

lMfor(p+q+r+s)2A

a-l for r.t. 0.156-:---*(6)

32

(x)g = ,{

Ip ;o edeqs olll roJ \i I13 yo sldeo:alur oqi ll roJ VI

13 3o saloldtuXse aqt ll? roJ VI

t- = x s!

g=,{ sr t)

'01- sl 13 goldacrelur-{aq1'

E sl t3 3o ldecralur-f, oqJIJ o1 eloldurXse lecrga,r eq1;o uorlunbe eql

*l-+t _l-?x

co- = (r)g ull pue co+ = (r)3 ullol eloldru(se lEluozrror4 eqtSo uorlenbe eql

r

-tr y'I o+-r' ,-X @;+)8: x IIIII =- IIIII

--a ' o'-x8

0,

uorlnlos

(u)

Solution

(x+4)2(x+5)(b) (i) e(x) =g'(x) =

8

3(x + 4)(x -2)

Since g'(-4) = 0 and g"(-4) -lthe coordinates of the point of inflexion are

lM forjustification

_)1ct,;)(ii) Ct:!=f(x) , where f@=#

Cz'.1 = g(x) , where g(x) = (x++)2(x-5)

Note that f(x) = g(x)8x-40 (x+4)21x-5;

a

x+4 8

-!(r)------

196'tt'l'r roJ I-,

VI

uorler8elur lcerJoc JoJ VI

uorsr^rp JoJ htrI

ldacce yq1

(s)------

uorlcesrolur Jo slurod oql IIe roJ VIuorxegurJo lurod oq1 pue$urod suerxo sql IIe roJ VI

z3 go eduqs sql roJ yI

8996'e EEnnLLq,,sftiwY

(z\ zt'';/l*'-

"::=lay"- ss6z= ol(

z 8 ze) -l

,L['o'- ,*-7.7 f ('+x)urzt-r8.]=.,[[,,-"-**i) (+-')] J =

( s l+x)01 re IG-r)z(r+r)

-

or-rtJ rJ =

rp((r)s-frl:)'[ =care perrnbor ag1 (c)

,p ((r): - trltfl

0r-\,*

(o 'r-)(x)1 = {

8=/(r)B = r(

2-

uopnlos

--.{

8. (a)

Solution

The total profit made by company I: I r(r)ar

JO

I

= 1( r(0)+ f(6)+ 2(f(r)+f(2)+f(3) +f(+)+ f(s)))

x 37.487053'41x 37 .4871 billion dollars

fQ) = ln (et +2)+3df(t)

-

et

dt e' +2a

2r(r)dt2

(et+2)et-e'(e'): ------------;---(e'+2)"

= 2u'

lA withhotd 1A for omitting this step

1M for trapezoidal rule

lA a-l forr.t.37.481

1M

lA f.t.--------(7)

I A pp - I for omitting ' --- '

lM for any four terms correct

1A for correct integration

lA a -1 for r.t. 41.822

-------(6)1A

lA f.t.-------(2)

?

-

(ii)

+

(e'+2)2sn". {10>0, f(l) isconcaveupwardfor 0

LE

t{:

serleu rrqnc uorllltu 9s?'9 l'r roJ I -, vl

VI + Sullsol roJ IAtrl

L)oZ = / roJVI

(V = t t> L Jl

L--1 JlL>1>V Jl

0J

1=7 to (r1>l

0>v+ls- zt0

Solution

lM for testing + 1A

1A a -

I for r.t. 6.456 million cubic metres

Note that A(r) = 1-r2 + 5t - 4) e' + 1

^ dA(r) ( ,' s, ,) ^,1"o, -;;- = [7 -, - ' )"

d dA0) = o when t =2 (rejected sincedt

a'eo)

t>4) or t=7

a2errll s +fore.wehave #l =Le 2 >0 'dt' | 2I t=7ote,that,there iS only one lboalrrmildlfium after,thd ad,$Wtepeiiod.

So, A(t) attains its least value when I = 7The least amount of water stored

= A(7)x 6.456447098=

6.4564 million cubic metres

d2L(l)(iii) ---idt'

=( -,' *r, -7),+ .( ,-2)r1[ + 4 2) \ 2)=(*.+-'.l,;[4 4 )

dA(r) t ( ,,

-r t, -?1\"+ > o for to

6E

a:

( e )-------YIVI

VI

( s)-------,00'0

'1'I roJ I-, YI(e) Sursn roleurruouep roJ IAtr I +

(p) pue (c) Sursn rolererunu roJ IAII

(z)------200'0

'1'r roJ I-, vIit

,\c) , * JoJhtrl vz- t? L

k)-*--:- I0z'0 'l'r roJ I-p Yl9'O>d >0 'd8'0+(q)SZ'g roJhtrI

k)----- 069'0 '1'r roJ I-r2 YIsz I 97.1 ( ( --. > z>:\d ldeccu ) y11 ' 'Ett -009

' " sl-f -00t'

( e )-------0er'0

'l'r roJ I-D vI

,(1rJrquqord uossrod roJ htr I +slue,re &sluueuelduroc roJ I trI

G)'ns

(e)

(p)

0t, $ sl esuodxe lsuel posr^er eq1'snq1' \ti = f e^eq e,tr

.ecue11

SZI ' nn'0= ,Lt_x oAEq o^{'eloJoJst{J

. ;Zt ' tt'0 = ( ^- 7 7) 4 er,eq en 'og \LL- X

' te

'0 : (r < X) a eleq errr 'ueq1sr asuedxe lseel posr^or ar{l leql asoddng

vl00'0 =re6Tffir00'0 ^,

b9zr6z0*'0

o0 t toz'o) + t9 ro/I oo'o&lllqeqord pertnber eq1

1.I00'0 Ec9I0tt00'0 ^,

it ,lLtl}z'}) _ = L'-- -- - - ,r-a P'Z

$rlrqeqord pe:rnbe.l eq1

ZIOZ'O X

ttr0r0'+fivsv'o- 9'0)(s 0) + (eess'oxsz'o) =

,(1rpqeqord perrnber eq1

8689'0 =wgv'a+ L9zz'0=

(g'r> z > 9'o-)a =. gzt SZt (s/i-oo9 > Z>

ur_ *)d= (oog>x>ooe)d=,(1rlrqeqord perrnbar aq1

' Qszt

'szt)N - x 'ueqJ'Jeruolsnc e3o esuedxe eql aq X g le.I

e0fr'O ^,wzisaosrr&.t'

( iz ir io ) |

-t-

lrr-r rp'z '

,.r-a f'z '

,r-a ov'z ) t -

firpqeqord perrnba.r aq1

(c)

!

(q)

uorlnlos

(e) 'o t

II

i

I

Solution

Ler X g be the netweight ofa can ofbrand D coffee beans'Then, X

- N(300 ,7.5\ .

(a) The required ProbabilitY=P(X 316'5)='p(2.283.5 -3oo or Z >3t6.1:3oo)- \- - 7.5 7.5:P(Z 2.2)= 2(0.013e)= 0.0278

(b) (i) The required ProbabilitY= (l - 0.0278;trlO.OZZt;x0.AA387152o 0.0204

(iD The required ProbabilitY= ciu(t -o.o27g2e Q.0278)as 0.3i58195889p 0.3682

(iiD The required ProbabilitY=

(l -

0.0278)30 + 0.368195889**,.J914045'.1'5N 0.7974

(c) (i) The required ProbabilitY=

l(0.:os rsssag)2'* 0.184097944s 0.1841

(ii) The required ProbabilitY0.184097944

= 0.7g7404575* 0.230871443x 0.2309

rM(accept ttz.4X#9 o, zt{f!9ll

1A a-l for r.t. 0.028------(2)lM for (l-p)" p ---------------l

iI

1A a-l for r.t. 0.020 ii

I

lM ror cio(r- d,, p-----]I"-%X;ji'I

lA a-l for r.t. 0.368 iI

iI

tu for ftr- p)to *q) + lM for s = (bxii)--l

lA a-l forr.t, 0.797-------(8)

I1M for

-((bxiil)2" " "lA a-l for r.t. 0.184

1M for numerator using (c)(i)+ 1M for denominator using (bXiii)

1,A' a-l for r.t. 0.231------(s)

40

ti

( e )------

Z9Z'0 't'r roJ l-,

VI

(q) Susn rolsurluouop JoJ I II +(u)-(q) = rolBJelunu roJ htrI

( r)------tft'\'1'l roJ I-r2

ecuenbes ctrleuoe8 Sutruuns roJ htrI

suJal r lsBol lu 0A3I{ puesolJes olruuur ol?slpul lsnur IAJI

k)"---- gzt'0 'l'r toJ l-oVI

,{1111qeqo:d ctrleruoe8 roJ htr t

lz9z'0 x

WIZSZ'0=92991 _960n

6

i w_q=nztg i

,!rlrqeqord pe:tnber aq1 (c)

tttt'0 >wvw?rvtv.0'A,

6=n

(s\ (s\ (s) (s\ (s)(s) . [iJ,[;J. [ , .J,[;J. l. ' .]t ;l=

&lpqeqord Pertnbe: aq1 (ql

6lzt',0 *

9t6itt:o,n9299r _YZIS

/s\ /s\ /s) (s\ (s)(s)[i],[;,. [rJ,[;J. [rJ[;J

=

&111qeqord Perrnbar eq1 (e) 'Zl

I

IZ9Z'O =bnlzgz'0 =

SZ9SI960n

9n ,_ SAst- t-

ru9,grlrqeqord partnber aq1

z9z'0 'l'l roJ I-r2

VI

(q) Sursn roleutuoueP roJ IAII +,gr 1 rqeqord,ftelueuelduoc loJ IAI I

uormlos

d

(d) (i)Solution

The required probabilitYt?l

=(;)(;)+{7)0)5

7&.h'{&ffiffi.*4q 0.7143

Marks

1M for either case

1A

a-l for r.t. 0;714The required probability

t4= I - (:X;)tt=l

7

ajw+affiffi,* 0.7143

lM for somplementary probabilitY

1A

a-l for r.t. 0.714

(ii) The required probabilitY=l-:

7

=27

}fux A.2857

lM for 1-(dxi)IA

a-1 for r:t. 0.286

The required probabilitY1A: (:X;)0Xl)

27

= 02857

lM for denominator = (2)(1)

1A

a-l for r.t. 0,286

(iii) The required probabilitya. ')(rx:)'9" 7-=,T$0]*1

_8J-t

e*0.2424

lM for Pqpq+(r- p)(t-q)I p=(b) [p=(dXii)+lMfor{ or{'lq=(d)(ii) |. q=(u)

1A

a-1 for r.l. 0.242------(7)

\.

42

Candidates' Performance

Section A (Compulsory)

QuestionNumber

Performance in General

Good. Many candidates were familiar with binomial expansion with rational index. Somecould not work out the range of values of x for which the binomial expansion is valid.

2 Good. Many candidates could handle indefinite integration, but some forgot the constant ofintegration.

l Good. Most candidates could handle quotient rule and product rule. It is more efficient to applydv dv dx:L =:/ . I but some candidates went through the tedious way by expressing the function ofdz dx dzy in terms of z .

4 Very good. Many candidates label the stem and leaf diagram perfectly well

5 Fair. Some candidates could not apply the laws of probability especially when complementaryevents are also involved.

6 Poor. Many candidates had difflrculty in identifoing the different outcomes.

4

Section B (A choice of4 out of6 questions)

QuestionNumber Popularity

Performance in General

7 (a)

(b) (i)

( ii)

(c)

76% Good. Many candidates could sketch the curves'

Fair.

Unsatisfactory. A number of candidates could not sketch the curve.

Poor. Many candidates could not reach this part. Some had difficulty in workingout the integration.

8(a) (i)

( ii)

(b) (i)

( ii)

(iii)

(c)

6\Yo Good. Most candidates could apply the trapezoidal rule.

Fair.

Good.

Fair. Some candidates could not expand the exponential function.

Fair. Failing to get the correct result was mainly due to the performance of theprevious parts.

Poor. Many candidates did not attempt this part and others could not make use ofthe concept of over- and under-estimate of a mathematical model to explain theunderlying meaning.

I

f

uJ

46

LV

'UEd srql peldueu seleprpu?o ^\oJ {JaA 'Jood

'(t)p"lo luane ,fteluaureldruoc orll sl slql osrleor tou prp soleprpuec,(uu61 ';leg

'sJueAe

luenoleJ Jo requnu eql Surlunoo ur ft1ncg3:tp peralunocuo soleprpuec otuos 'poo3

',&rpquqord ,fte1uaue1duoo eql lno Iro^ lou plnoc soleprpuec auos 'lrC

'seues culoluoe8 elrugur aql uns ol olqe lou oJe,^A soleplpuec ouos poog

'poo9%8t

(rl)(rr)

(l) (p)

(c)

(q)

(e)zt'.ftolcegsr1eg

= (v ls)aG)a u.{AouI

9I

pw (1)q tuog sr:e1n.r uorlecrldrlynu

'esec relncrped srql uIlle/$ eql ,(1dde 1ou plnoc

@ls)a@)a=@wv)a seleprpuec,(ue141'rood'pooc

'poo8 f;e1%98

(r) (c)

(q)

(e)tt

(rr)

'en1u,r pelnbar eql roJ o^[ostou plnoc oruos puu &11enbeur eLIl qsrlqetso lou plnoc seteplpuucJo requnu V 'rluC

'lq8u rolereunu eqt te8 pup sluo^o punoduoc eql .lrluepr ol elq touoJo,u euos lnq sertrlrquqord leuorlrpuoJ olpueq 01 alqe ero,4A soluptpuuc Xuetr41 ';rug

'gr8 e surm sJoruolsncaorql 3qi Jo qrDa ]eql uorlrpuoc ua,rr3 3q1 po{oolJ3^o soleplpuec ouos 'pooD

'pooc

'poo8,fuen

'poo8 Xren%18

(])

(e)

(p)

(c)

(q)

(e)or

'poued eurl eqt,$rceds lou plp .(ueu puu G),V Jo Jnorleqeqeql uleldxe ot (l), V Jo onle^ eqt Jo esn eleu lou plnoc seleptpuuc .{ueyq '.roog

ecuoq pue olqeue^ eql Jo senlun go e8ue:

'pooD

'uounlos eql olelduroc lou plnoceqt pelce13eu soleptpuec {ue61

'rlEC

'pooD

'pooc

'poo3,ten%ti

(nr)

(r rr)

(rr)

(r) (c)

(q)

(e) o

IEJouoc ur ocuuruJoJJod,grrc1ndo6Joqr.unNuorlsenI

t

General comments and recommendations

Candidates performed better in the relatively more familiar type of problems. Candidates are reminded that

1. attention shoutd be paid to the range of values of the variable for a mathematical model to stay valid. Suchrange may affect the calculation, interpretation and testing of various mathematical properties;

in addition to mathematical manipulation, candidates are expected to understand the concepts and be abl,e togive explanation by making use of the calculated results;

3. propr notations in integration, normal distribution, probability, equation and inequality, etc. should beadopted as far as possible.

48