Ｌ ﹁︲． ︵ ︲¨ ︲し 一 Ｌ ︹ ︶﹁ ｌｔ ︻ ｒ︲ ﹁︱． ︹︲﹁ ︱●´ ■︱． 一 ¨ ﹃ ︵、 一︲ 一 ︻ 一 ︻ 一 ︲ ・ ／
￨_ ￨
薦 A Level Preliminary Examination Paper
H2MATHEMTICS Hwa Chong Junior College Anderson Junior Nttional Junior St.Andrew's Junior Serangoon Junior Pioneer Junior
College College College College College
Yishull Junior College
Rames Junior College Victoria JuniOr College
Tampines Junior Collcge Nanya7ng Junior College
Mc五 dian Jllnior College Ca■ olic JuniOr College Anglo― Chinese Junior Colicge
lnnova Junior College Millennia lnstimte
．
Ｌ
，
一
︱
一
ヽ
一
一 ￨
一
一
一
︲
一
十
一
一 、 、 ．︼ 一 一 一 ．
HWA CHONG:NST:TUT:ON
JC2 PRELiMiNARY EXAMiNAT:ON 2007
MATHEMAT:CS
9740ノ 01
Higher 2 Paper l
Wednesday Addiliona:mateda:s:
12 September 2007
3 hours
Answer paper Cover Page List of Fomula(MF15)
READ THESE INSTRUCTIONS FIRST yoqr 1a.me and CTdass on all the wod( you hand in, induding the Cover page. Write in dark blue or black pen on both siles of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or coneciion fluid. Do not write anythlng on the List of Formula (MFlS). w_1'!e ヽ一
し
Answer all the questions. Begin each answer on a fresh page of paper. Give nonexacl numerkxl answers conect to 3 significant figures, or 1 decimal place in the.case of angles h degrees, unless a differeni level of aco.rracy is specified in the question. You are oxpeded to use a graphic calculator. . Unsupported alstErs from a graphic catcutator are allowed unless a question specifrcally states otherube. where unsupported an$v6rs from a graphic calculator are not allowed in a question, you are requlrcd tg pesent thefiatheinali, steps using mathemathat notations and not calculator cornmands. You are reguired of the need for dear presentation in your answers.
The number given in brad0, find
J" lr"'l*,givingyour
answer in termsof
(rln.r)2 dx, giving your answer in an exact form.
●7…
ア
a.
t31
[4〕
3 ‐
4.
The
points'l andB
the planes
fll
and
/r: r= where tr, 一一 ´
l.
have position vectoo 3l +
[I2
(2\ tt': x+22=3' '
I,
and
r'l r0) "tt=^[l,l[l,,'
f 1R.
・
It is given that the planes fI, ard n2
(l)
and 3i + 3J respcctively. Thc line
have equations as follows:
".1',,l
and
j
ht
rscct in thc linc
{
.
Find a vecor cquation of thc line /, and show that tlrc linc I, is parallel to the line Hence, find thc shortGt distancs betwecn drc
lin6
4r
/,
ad ,r.
.
[5]
(1)The pOints C and D arc on慟 c lines t andち 峰 pectiveけ suCh血 乙 OC=90° and ′BDgヽ a pardLIogram Find thc pOsmiOn v∝ tors ofdle poins C and D. 141
(1)1■ e plane n3お
Π2 andお equidおtantto both point Иand ofthe plane Π3お given by ro(1j+k)=1
parallei to the plane
plane n2・ Sh° W thtt thc cqm● otl
ule
Find the posidon vector ofthe foot ofthe perpendicular from the point/to dlc planc Π3
[5】
Sectlon B: Statistics [60 martst Tlree single men, two single women and two families take their places at Each ofthe two families consists of nro parents and one child. Find the number ofpossible se*ing arrangements
(i) (iD
members
of the
separated,
if
same family are seated together and the two single women are t31
sis between thcir
the seats are numbered and each child
parents.
lo orange‐ ■avollret i4 strawbq‐ aav4Dured and 16 chc町 ‐
navoured sw∝ ts which ate ofidentical shapcs and sizes. Ben■ y sciects a sweet at random ttam■ E bag.Ifit is not cheny‐ navoured he and selects another sweet at mndom.Hc repcats thc process unti:hc oЫ 」ns
navOured sw∝ t Caiculate hc probability that
(i) (ii)
l2l
¨ ヽ ¨
A bag∞ n面ns
a round table.
“
y‐
the first swe€t selected is strawberryflavoured and the fourth sweet is orangeflavoured,
tzl
he selects an even number ofswee6.
t3I
OHOこ MATH 2007
97…
7
ffum over
.4 7.
A random vaiable ,Y is rrcrmatly dffibuted with meot p
and variance 2.5.
tt
is given
that P(x>13)=0.2s.
(i)
Find thc value
of p, giving pur arswer to 3 signilicant
Another random variable
I/
figures.
is.also normally disributed widr known mean and
I3I
vriance'
ftid thc fbability thc 3 indcpcndent obscrvlioos of X difFer ftbm 3 timcs m obscrvairn of I by at lcast l. State an nsrmptim dnt the strdcnt hr to make to solvc ltc $estion. .
60
A stdc[lt
was rsked to
Part of ttlc studcnt's solutiqr to thh
3x
3r  rv(3E(x)
quction is givan
3E (r), evar(x)
r
follous:
evr(r))
Plχ 3y≧ 1)=・ … 131
Point out 3 mistakes that the student has made.
restaurant colmts th€ nurnber of banquets received by the reStaurant at the ofeach week. State a condition under which d Poisson distibution model would be
The owner
ind
ofa
suitable.
lU
disfibution, thc pmbabitity that tlre rcstaurant rcceives at least two 'in a randonrly chosen week is 0.3. Find tlre mean number of banq,as received
Assuming a PoiSson
L*quit
perweek.
A week is considered busy if

t31
tlre restauranl receives at
leat two wedding banquets in
the
week.
(D
Calculate the probability that in a period of 4 comecutive weeks, thcrc is at most one busy
(ii)
weeli.
l2l
Hence calcutate the probability that in a pcriod of 6 consecutive week5, the 66 week four^is the second busy week, given that thcrc is at most one busy wcek in thc firs
wecks.
O HC:こ MATH 2007
131
977
いn clbtt ulluT圏 2007hれ 鳳咄
(‖
2 hlls町 む
リ
5 The deparhrcnt managers of a company are to study thc number of days of sick tearre taken by ernployees under droir charge. The oompany coosists of 2 deparhents namely thc Sales Departsnent and ttrc Finarx* Dcpaftnent The nunber of employec in the 2 d€partrnents and tho rcspectivc probabilities that thc employecs would take sick leave in a montlr are givcn in thc table below:
Department
Numbcr
Probab‖ ity
Sales
55 45
0.2
FinalB
(a)
0.3
Using a suitable approximuioq furd the probdility thd therc are at lcast 15 pcople in ltc company taking sic& leave in a pcriod ofolrc non0r. 13]
ne nulnber ofdays ofsict icave per mondl taken by an accomts clerk fo:lows a Poisson dお tributioo with nlean 12.
(り USing a suitabic approximation,■ nd the maximuln nulnber oF days″ ,h apc● od
of oncぃ、thc
aCCOunts dcrk can take so that the probabi:ity matthe number of
days oFsick lcave is icss than″
wi::not excecd O.7.
14]
10. Bcn,a fmit sc:lcr,oniy sells tnangoes from‖ o ormrds,A and B.In cach oFChard ule weigh“ of dle mangocs are norlnJ:ydお tributed with a standard deviatlon 39g in A and an unknowi standard dcviatlon in B.The mcan weigllも are 200g:n A and160g m B, (■
)OnCぬ
L And■ a business parmcr of BcL brought h a largc batch of II angocs
mm Orchardん 10 mangocs were randolnly sddcd“ m mis batch and the
wcigh● (n9ofulc ntangOcs weκ
175
182
167
160
as fo■ ows:
169
189
137
164
χ 207
1f a燿 掟at 2 5%sittd6cance icvel indicates thcrcお a dccreasc in dle mean wcight of dle mangocs from orchardへ flnd ule g“ 山 valuc of x,∞ rect to l dccimal place
“
141
(b)On mOther day,Bcn exolnlned 20 mangoes lom orchd B which have nun= wcight 140g and a standard devi面 。口453 He was worried dlat tt mean wei』 t of manpes in orchard B might havc changed.Andy helped him by“ 膚ng out a statヽ 通cal testtO flnd out」 mdl “ “ (1)
State the nui:and altemat市 c hypoulescs in ca_ing out such a test
〔 l〕
OD Andy has to rcicctthe nd‖ hypoth、 おat l%icvel oFsipincance.0」 ctllⅢ the minirllm valuc of l,dcnoted byち ,00mctto 2 dccimJ p:"鴫 in order to arrive at this dccision,
[2J
(ili) Ifa mangoい m orchard B which wcitts 140gお added to the samplc of20. Would this afFect Andy's∞ nclusion for thc test at t%ieVe1 0f 雨gnilcance?
O HC:こ MATH 2007
12]
974●‐
7ヽ
レ07
ffum over
■ュ ■ュ
Considq thc
de
x:
set and its omesponding scatrer diagram as output from a GC: L3
v
e
sEE'
2
.
E',
lrtl
521,,
,lr.0z' .
I
,E
Give a rcron why it might bc unwise to use cither of the regression lines or r on y to estimatc tlr valrrc of y whcn
(b)
Thc cquation of thc rcgression linc of y oo : is /=l'069x+l'0E3' Find thc obscrvedvalrrcofywtrco,=.S2,andhenccdeducethelinear.(productmoment) t31 conclatioo cdefficien( r, bctwecn r 8nd /. Adding more points (a y) to Ot autu may chango the value of r ' Give an example
on .r
r=1.5.
ofa pair ofpoinB that would not carso such
(c)
12.
y
(r)
a
tU
Ill
change'
to Find rhe equation of the rcgression line of (xt.108)' on y. Use this equation I31 estimare dre vatu{s) of x wlren Can we use thc same cquation to estimate the value of y when x =
y=3.
3?
tll
th3 Si1eaq.o1 In order to find out how much ouriss spcnd during tlrcir stay in SinBSnore,
TourismBoard(STB)interviewedeverytenthtouristleavingtheairport,simmlgratlon departure checlqoint to obtain a sample of250 responses'
tzt
(a)
Give two reasors why the sample obtained may not be representative'
(b)
checkpoint' obtain a.De.scribe how STB may, ar the airport's,immigration departure tU sampre of zso tourlss via a luoa ia*pling
mihoa.
random, samplc of 250 tourists surnmarised as follows:
A
ws
intcrviewed and their expendiure' in
f,(r160)=l2so md f,(r165f
(i)
$x
'
is
=2813?0
Calculate the unbiased cstimdcs of the population mean
p
and variance
o''
I2l
STB bcliwes ttur eaclr tourist spads $ 170 on average
will have a I2l
(ii)
Estimatc the probability .tlrat another random samPle of 250 tourists sampte mean ixpendinnc exoecding that found in
(iii) ' '
Find the lcct numbcr of additional touriss sTB should sample so that the sample expcndiore of lhc enlargcd sarnple excceds the mean found in (i) with a
(i).
;;
p.U$ifi''tyhi$"tthan0.99.
O HCl&MATH 2007

07,(Xr2nvo7
t3l
2007 HCI Prgliminary Examination, H2 Mathomatics Paper
l.
Let P, be the statement "r,
4:
I
YneZ"' ={IJI n RHS=9=l=LHs
LHS=″ :=l
I
Thereforc,Plお tuc.
Assume P, is tsuc, thal is,
,,
=
E#

Jor some k e
Z'
ЪSbW 4ギ 崎 J〓 出 LHS=″ 1.=面 崎 c市en)
1) .● =二 ■+1 1
…
1) ==● 1+1 ■ !
← +1) i
Plも mc,PI tne=,P.J also me,by mattcmaticalinductioL ι おmc
∀″cZ・
2(b) The 8raph of y = t1 (.r)
Sketch ttrc graph
ot
2007 HCI H2 Mathcoratics Paper I
,
=
is shown below.
6
3〕 【
5 aG
+x=
(2G)'
+t > O for anY.r> 0
x(5aG+x) (x a\(x b) =O Since
5
4rE+.r
>
0,
←¨α)(エ ーb)
≦0
xS0 or a
l)
￨―
2007‖ CI‖ 2山 憮 N薇
嚇
町ぎ
2 Sol● 6鮨
￨
Some possiblc conditbns(any Onc ofbelow for answering qucstioo:
8.
● Mcan number ofbanquets propottonatc to thc numbcr ofwceks ・ Banqucts at ulc rcstaurant occur at random ● Each banquct is indcpcndcnt oFcach other o Receiving banquc"iS a rarc event
P(X≧ 2)=0.3⇒ l,P(X")― P(X=1)=03⇒ P(X却 )+P(X=1)=07
ハ ● ⇒ √ (1+え )=07 ― ) ⇒ え=1.10(anS) ― By GC:Graphわ 」 (mustShOW graphsD≦丞 Equalion So:verlgiVe kcystrokcs】 【 8(i)
Lct、 一B(■ 0.3)L ule humbcr ofbusy w∞ bo■ of"wecも
,
P(L≦ 1)=binOmcdf“ ,0.3,1)=0.652(a幅 ) 8(i
D
P
(6th weck is 2nd brsy week I at most I brsy wcek in tirst 4 weeks) P(t weck busy in lst 5 week & 6th wcek busy
Pは nlpSt
l WCek busy h■ は4 wce魅
=P(Ys=ll(Q31=5(0.3)(07)4(03)=a166.cans) 06517 0 ALIER}IATTVE SOLUTION
P(6th week is 2nd busy week I at most I busy week in first 4 weeks) _ P(wk l 4 rct busy, wk 5 & 6 busy)+P(t wk busy on first 4 wks, busy on wk6) P(at most t wk tusy in first 4

P(Y. = 0x0.3)'z+.1(Y. = lxo.7xo 3) 0.65 r7
wk)

= 0. 166 . (ans)
9(a) Lct χ∼β(55,Oa and γ∼B(45,0.3) Thcn χ ∼〃(11,8.D apprOXimatcly and r∼ Ⅳ(135,9.45)apprOXimatcly Thus χ +r― ″(24.5,1825)apprOXimatciy. Required probab‖ ity=Rχ +γ ≧lつ =0987(to 3 sFw/o ccp(ans)処 P(χ +γ ≧1つ → P(χ +γ >14つ =0.990(t03 sfw∝ )(邸 ) ALTERNATIVE SOLUT10N La r∼
重 β 評型1)Then ttproximatc r∼ ″ (55+45,二 需
(245,18.4975)
Requircd probabnity=P(r≧ 15)→ Pc>14.5)=0:"O Ct0 3 so(aぃ )(MUST CCヽ χb)bt″ ∼ 鳥 (14.4〉 dcIIOtC dle no.ofdays ofsick:cavc per month the derk w‖ i takc. Sinccえ >10,7〃 (14.4,14.4)apprOXimately Then P(″ ≦″)≦ 07→ P(″ く″+05)≦ 07(do cc fOr using nollllal disttibution) From CC,P(″ く16.38996)=07 Thercfore″ +0.5≦ 1638996 =o ″≦:588996 二 largcst vdue of″ Is 15(alls)
( tsso+xr*) o'ozs l0(a) To indicate decrease in mean weight, ,l , '*1;5' [.r
I
'

1550+ x
have l0
And we

r'o
39/Jl0 Solving for x: tO.06772 and thus (o = 6'78 ' (2 dp) (ans)
(iii)
Addition of the special mango with weight l40g means New sample mean remains unchanged '
(D
(lD
Newu.e
sr*
=#E(xJ
=452
tos) = normatcdf(16s, rEs9,
= I130. (ars)
Ｑ
／ｆ ｌｌ ヽ
¨
Ｎ
Sincc n=25O larBe, so by CLT,
r(x l2(iii)
o,
+ roo
一 Ｘ
l2(ii)
p,, ; = I(::teo)
響
=452)apprOX
tlo, Jlsz)= o.s9r. (ans)
Let the sample size needed be n . Since 0.999 > 0.991, so z > 250, and by CLT,
x  N(ruo, !!19) unp,o*. set P(X > 155) > 0.99e
=
f
)
{,.iii'21..
一 一
Ji rZO.'tB + = We should increase
(.) = ff.,n*orm(o.oot)=r.oe62
n>431.6*432
tfte sample size by at least 4322SO
Note: GC method to solve
=lBZ (arc)
(r)
(l)
Graphical: Plot a glaph such as Yr = normalcd(165, lE99, tlO, X ana Yz= 0.999 and find intersection point bctween thc two graphs, thcn deduce the inequality. OR
(ll)
Use Equation Solver: 0 = normatcdf 65, I 899, I 7 O, \tt :^]lOI X)  O.gg9 . 0
Jitly
l
一 一
ヽ
￨
―
Anderson Junior College Preliminary Exam inati,on 2007 H2 illathem atics Paper 1 Expand
^1* Q,f
in ascending powers
ofx
up to and including the term in
of/
ofx for which the expansion is valid. Find the coefticient simplifuing your answer. range
G“ nthat
y=cCOSヨ
l,
stating the
in dre expansion, 14]
2,show dlat← 2)(者 χ _ ⇒ )2=4ノ
図
O By価 面 Cr difFeに ndttbn ofぬ c above"suL■ nd the Madaudn'ssel,9fyup tO and including the tcnln in x2. [31 (D Dcducc dle equation ofttc tangent to thc curvc ofy=(∞ s lX)2 atthC point where
X=0
[11
3(→ Πnd ttCCtttVducof[Jn χ卜nJと (b)Find j:ミ
二 可 『 dX
7:事
13]
by using the substitution
χ =上
[3〕
″
′ハ
ヽ ご
itF｀
0
The points A and B have coordinates
/,
passes through
the point
?=T''='
I
)
二 し ♂ t'を う
(2,l
,3) and (4,13,_3) respectively. Thc linc and is parallel to the line /, with cartesian oquatiOns
'{'11u, "2:Zi"
tl)'(i")+i1g; tlI the vector equation
of t41
r\ Y . cos\r\
ftO
5
The functions fand g are defined by
f :xr+coslrll, xeil I +.r
gtrr+
cos''
1= 1
X 
1
lqEl1tr
!'(r)'rtslx{r
l2
l
(111)
Let
n=2, LHS= z, =1 RHS= 20=l
:
Therefore the result is true for n 2. Assume that the result is true for n = 19 k t_l
i.e. u, =\u, For
>z
=2'1, k>2 ,
k_t
i=l
,=l
n=,t+1, ur*r=Zu, =lu,
+ u,
_)L2 L1e_2 _ 1
aL_2
 "rl .r(r.lF2
Therefore the result is true for n = k
+l
.
Hence by induction, the result is true for
12a)
a) :[+
dl
ne
Z, n>
2
〔1+(・ ―ッア ]∞ s2x=sin2 x
Using ν=χ ―ノ,
多=1多 ⇒多=l一 分 卜分+[1+ツ 2 1cos∼ =Sh2x = [1+ッ 2][cos'χ ]+1_sin2 x 分
21 II‐ cos2.[2+ッ
σ)
.
Solutaon
∫ 嘉
d' = :∫
ll+CoS2ェ )a"
+tan・ +=:(X+卜 tan 12予
7テ
≠
n2χ
)+C
=:(工 +:Sin 2r)十 C
y=″ ―V2tan(13(X+ :Sin 2x)+V5C) 12b)
ap面 ve conmnt 争=R― を ヽ ,た
２ 一３
=R―
争
:RI=0
〓 ｋ
Thus,
=0 =)R―
券
⇒
At,=15R,
:X(ShOWn)
り 坤″ は 法 ― 力
inlR―
+ι :χ l=′
inl R ―
:χ l=― :′
R―
:χ
=И
X=百 At′
(R―
=θ ,χ =θ
っ χ=サ
´
: C
]′
/
:′
ι
)
, 0=(R― /)iC/=R
(1` ÷ )
‖)As′ → ∞
,` :`→ 0
x→ ユR⇒ ie regardless
13a)
―
α =ユ R
oftime,
the amount of drug in the patient's body
2+(18x)2
ι=ぜ ノ
o
+(lEr)2
′ 乙一 雄
=χ
4+(18χ
ι ２
=4χ
)2
3+2(18χ
x1)
will n"u".
"*"."d
1R' 2
Solution
=0
At nlin pt券
_4x3=362χ 2x'+χ 18=0 From GC,χ =2 is the only solution Thcrcforc the pohtお (2,4)
χ=2+,望 生 >0
x=2 ,#く 0∴ Mh pOhL 13b)
i)
2=χ ⇒x=6 4 6+χ
2=壼二⇒′=2+上 ″
∠4=子
(″
+6)2
激
滋
ο ″
グッ a″
ル
180
=轟 (20)
″力 `″
手
力=3ο ″
,
=デ
音卜
=券
6+力
ι″/S
π(力 +6)2
3
Anderson Junior College Preliminary Examanation 2007 H2 Mathematics Paper 2 Section A: Pure Mathematics (40 marks)
la)
r. b) The curve Chas parametric equations ,=?+l', t Differentiarc tan(ln
r')
with resp€ct to
tzl /=12r+l,whererisanonzero
parameter.
O
Show that the gradient
of the curve at any point (ry)
satisfies the equation
″ (2r1)r2 =2'2・ 冨
(iD
The
linex:p
E21
is a tangent to the curve
valueofp.
C. By
using the result in (i), find the exact
I2l
2 Express(2r+3)h thc fOrln 2(′ +1)+/′ +3(r1),WhCrc/and B are∞ Ushg thc method ofdifFeκn∝ ,■ nd
nstanも
[5]
Σ(2″ +3)2′ h“ nns of″ r‐ [
2"
Hcncc,or othepise,■ nd an cxPに ssbn br
3.
[n an Argand diagram, the point 0く arg(a)く
I
・ Σ(2″ +3)2′
h“ rlns
of晟
repres€nts the fixed complex number
[2]
a,
where l and
,
ThC COmplcx numbcrs z and w arc such that lz2ial=lα ial. Sketch, in a single diagram, the loci of the points representing z aod w frvf = fw+
[31
Find
a) b)
in terms of lal,
the minimum value
of
lzwl
the range orvatues
of
are(j)
in
t"r.,
org1o1.
Page 2
[1]
[31
_
4a)
By completing the square, or otherwise, describe the geom€trical transformation by which the curve 12 y2
4y5=0
can be obtained from the curve
12y2
=1. Pl
The diagram below shows the graph of y = f(r) with asymptotes y = 2 and x = 0. The curve has turning points at (2, 2) and (3, 2).
b)
On scparatc diagrallns,sketch the graphs Of
2=f(_χ
(i) ッ
3] 〔
)
(li) y=f'(X)
[3]
Show all interceptq asymptotes and turning poins clearly on your diagrams be found.
can i
_io, 5.
ifthey
j
/6)
l"l lin" /, 2  t*J
Relative to an origin O, the point,{ has position vector
 , th"
has equation
,i'
'=[l].^[joJ',.0 andthepranerl hasCartesianequation 'ir.,trr#' (i)
Find the shortest distance from z{ to ihe plane the
origin O
are on the same side
The plane
fI,
ifthe planes
has Cartesian
fl,, fI,
and
equation
flr
and determine whether
oron opposite sides
(ii) tI, is the plane that passes through point I angle between the planes fI, and fI, . (iii)
fI,
of fI,.
and contains rhe line
r+8y+az =6
t3l
/,.
Find the acute t4I
. Find the values
intersect along a common
Page 3
I and //'
line.
of a and D t5I
o ,1
\\
^
,
、 一
Section B:Statttics c60 marks)
〆 FЮ m me Ю に■cぃ 。fthe word“ LEG「IMSFl■ nd↑ 口 ↑で ↑ て,XttIIτ し [°
c(T'1
ア ,r
are nextto caCh OthcL
MS
Z/1n a∽ 由 in∞ unり ,16%ofthe total popuhtbn are aged 60 years or more.and 18%of
面 b“ “ l獄 :盤l踏 ぎ
r ヽ
ヽklYキ ｀ 3c2メ て 、 '4is‖ 3
,― II
・C、
/LSS than 60 years o同
′γ
`
｀31=.(｀
fi
" (i.iif 一／
平 ｀ グ タ■ i
lミ
け
thc total population have myoph.Furthermore,65%ofthosc aged 60 years or more havc
t( :
■ ￨■ f′ ●′
:￨
￨￨:￨111[l:￨:l盤 llI[111:￨」 l彗‖灘l:鷲
∬
1・
,電
l
n at ran“
lTマ
讐響,lTttpЮ
hЫ
and has myopit
12]
aged 60 yearsor more, given that the person does not have myopiq
[3]
either aged 60 years or more, or has myopiq or both.
[2]
A manufacturer produces pens in two colours, red and blue On average, 35% ofthe pens
、 ぃヽ 、
produced are red pens.
Ｔ ´／
14 (b)
fi"a
the probability that in a sample
of l0 p€ns, more than half of them are .*
*tr,
The manufacturer offers a special price on red pens by selling them in packs ofsix. Pens are selected at random to fill each pack, Given that five red pens have been selected, find the least value ofn such that the probability that at most n more pens need to be selected to complete a pack of 6 red pens is greater than t4]
0.98.
9.
An IT services company has two helplines which receive calls independently and at random times. The helpline manned by Alice receives calls at a rate of I call per hour and the helpJine manned by Brenda receives calls at a rate of 3.6 calls per hour.
(a)
Find the probabilitythat both helplines receive a total of less than 5 calls in each 3 successive 30minute periods.
(b)
Find the probability that Alice receives at most call in a randomly chosen 30minute period, given that both helplines receive a total ofless than 5 t31
(c)
I
of 2】 〔
calls.
The helpJines are open for 8 hours a day. By using a suitable approximation, find the probability that, in a random sample of60 days, Alice receives less than 6 calls on at least ten but less than twenty days. 14]
Pagc 4
10.
A manufacturer claims that his slimming diet helps people lose weight. A random sample of 20 people whose original weights were 80kg took the diet for a month and their new weights, r kg were summarized as follows:
)t,aoy =r 4.7, I(rso)'? (a)
= t02.5
Assume that the population variance is not known. Test whether or not the manufacturer's claim is justified ar the 57o level significance. You should state any assumptions that you need to
(i)
of
make.
(ii)
t6I
Explain what is meant by the expression..at the 5% level of significance. in the context of this question.
llI
(b)
Suppose the population variance is now known
i
to be l0 kg. Determine the set of will lead to
values of the mean weight, kg of a random sample of 20 people which the rejection ofthe null hypothesis 5yo level
^tthe
ofsignificance.
t}l
If the test is now conducted at the 87o level of significance, explain briefly what can be concluded about the manufacturer's claim if a random sample of 20 people yielded a sample mean weight of 78 kg. tl I
I
l.
Large beer cans contain a volume of beer which is normally distributed with mean 500 ml and standard deviation 3.3 ml, while small beer cans contain a volume of beer which is also normally distributed with mean 340 ml and standard deviation 2.4. The volume of beer in any can is independent ofthe volume of boer in any other can.
(a) lf 5% of
oft
the small cans produced contain more than
/c
ml of beer each, find the value
lzl
(b) Find the probability that the volume of beer in two large cans differ by not more than l0 ml. ti1
(c) Beer is also sold in bottles, each of which contains four times the volume of a large can. Find the probability that a crate of six bottles contain in total more than t2M0 ml of beer. 131
A
sample of n large cans and a sample of zsm all cans are selected. If drere is a probability ofat most 0.01 that the average volume ofbeer in a small can exceeds halfthe average volume of beer in a large can by more than 92 ml, find the least value ofn t4]
Page 5
12(a)The mゴb
Of"vo short qulzzes in Mathematics,,and y.of 10 students are shown in
dlc tablc:
(D
1
2
3
4
5
6
7
8
9
10
Mttb for ist quiz(1)
6
5
8
8
9
6
10
4
9
6
MJb for2ndquИ 力
8
7
8
10
l
8
10
5
8
7
Find the productmoment correlation coefticient between x and on dre relationship between r and
y.
(iD (iiD (b)
Studetrt
Plot the data on a scatter diagram
ofy
against
y and comment I21
r.
121
By using (ii), state with a reason whether or not your interpretation in (i) should beamended. Justi$ your answer. l2'1
Drums of hair shampoo are kept in storage for a number of week before being rebottled for retail sale. To investigate the relationship between the number of weeks(.r), 0 0.e8 o.3s(r
(0.6s)')
> o.e8
 0.65 (0.6s)'+
(0.6s)
ln
z > 9.08 least z 9a
: l0 : Let A number of calls Alice receives in 30 minutes Let B : number of calls Brenda receives in 30 minutes AD Po(o.s) and B0 Po(l.E)
:.
A+ BD Po(2.3)
required
[r(;+a< s)]' : lr(e* o 3x2.
Hcnce,soivel号
3.
[51
A cubic curve is dcfined by the equation y=d
+a+d'
where a' b' c and d 4 The are real comtants. The cflrve cuts the laxis at Jr=8 and the 'r'otis at 'x = line y = 82r intersecb the curve at r = 3 and is parallel to ttre tangent of the curve at x =2.
(D (ii)
4
1>3c".
+ bxz
'
[1]
Write down the value of d
Form a system of linear equations involving a, D and c, and find the equation t5] ofthe curvc. Show your workings
clearly.
欧 四
dtth¨
昴
g"“
叩 Ю鋼
mh′
耐 面 昭
“ “
…
[3]
(D Uing the■ st ω
ttc釧
Ⅲ
tto terrns ofttc cxPanSb■ Of
onof“
Cq田 」On台
=ψ +エ
ψ +x,° Ⅳ
ル
bttn an approximttion
e p町
Explain why thiS approximation is a valid Onc.
5.
Byconsidering
D
W∝
な a frac“ o■
[3】
#,showthat
+静 “ 基 嵩静可
are constants to be determined.
,
ｐ Ｆ
where a and
#
椰
Prove lhe above result by Mathematical Induction
[Turn over
36 ッ=f(χ )
Sketch, on separate diagrams, the fottowing graphs, indicating cteuly any asymptotes, axial inrcrceps and tuming poins, where possible. ッ=― f(χ )1:
r2l
ノ=f(― ‖ );
I2t
ノ =f(χ )
t3l
＞
一 一
(1) り
7
The functions fand g are given by
f :r
t+
(x 2)2
3,re R,r S 2
g:rHae',reR (i)
” ． 一
8
Show that f r exists and express
clearly.
ft in similar form,
stating the domain t3I
exists.
(iD
Determine the largest intcger value a such that fg
(iii)
For the largest value ofa obtained in part (ii), dctermine tlre domain and range of fg. r2t
Giventhat
r= . il
121
,.detcrmine the cxact value ofthe modulus of zand show
(J'.r''
Str . t3] 12 (a) Find the fourth roots of the complex numbcr z. Give yow answer in the form reid , where d is in radians. I2l (b) Find the exact values ofa and.D which suisfy the equation z'=eo*iD, where aeR,and r o.
t2t l,rlt A and B be the minimum and maximum tuming points of y= f(x) rtspectively. Show that coordinarcs of A md B ue (a+2,a(a+l)) and .
(a,0) respecively. (iii) Skach the graph of v=f(t)
t3l
, indicating clearly the 121 asy{nptotes, Eming points and axial intErccpts.
(iv)
wheo
00.
wc
h,f,vc
in order ror dte 6shes notto b● ● o●
"exttnct, ノ
'0,V′ y。 20563)0 y.>20563
Thus.mlrLッ 。‐20563 ie the min iiti」 steady∞ nstant population k 2057
仰2鴎
肛К 2001 ttn 珈
卜 Ｌ一 “
H(→
‐ sec′ ⇒ ∝s′
PQF I品 1嶋
:
=―
χ
=Scc`tan′
よ Secrtan′ ∫ 丁器可 =∫ 冨冨浩万可 “ ′ m′ =∫ 函尭雨摯 “ =∫
喜 山 い′ “ F∫
=sin′ +C
=亘
・
°
IG雨
+c
X
=響
僣 百赫
By∞ ver‐ up
ml●
両
■
C=:
,2=(ル +8)(χ 1)+:(2ェ 2+ウ 一 ︼
WLcn
χ=0,β ←!)+:=0⇒ β=:
Wllcll
χ=1,(― ′+:)(2)+:(3)=l=o/=:
Thuヽ
=:(1議
流 一 ︻
∫ 品
aF‐ :∫
+7≒ 二≒ )
」警告+デ≒&
=:∫
:(ぅ
メ≒1)+手 テ 号 ア む ￨=ィ 可￨+:≒
ヨ 喜り +十 m・ (a)+hト ー 11+D :[:h← ・(a)+D =告 hい 2+→ +:hト ー 1+金 軸
一 ¨ 一
● ●
∈ Ｌ ＜ １ ． ↓ ︱叫リ ー叫ｌ ︲覇 ︲覇
‖_(0
(d,"*n)
d,=*(lc) (2,) cos
(2,) sio
c"
J;
10
コ●■ :3.‐ :)
:2(a)
′ フ ′
缶
ｎ
州=
Ⅲ
:r=￨:￨+え
G∵ 動
えCR
￨キ ￨,
a7V=￨:iiilForsOIIEえ CR 3・
Thuヽ
￨:li:llllll=4 7+え 2(182ス )「 :+′ =43●
え =8
0N=llil:i)￨=￨￨￨(ShOW→
〓 ｕ■■■■■■■ョぅ〓■４
２ ︲ ︲ ・ ′︰１１１１ヽ
〓 ヽ１ １ １ １ １ ４ １ ７ ８ ・ ・ ︲ ′Ｉ ︰ ︱ ︱ ︱ ヽ 一
2=4.
Thercfore point,{ liesin
︲
[l;.
(Vcri6cd)
﹃ 鍋 レ い ︲ ︲ ︲ ︰ ′ ヽ
〃
３ ７ ４ １ ・ ︲ ・ ′︰︱︱ ︱︱ヽ
〓
ヽ１１１１１１′ ４ ３ １ ７ ・ ︰ ′︰︱︱︱︰ヽ
＋
ヽ１︰︰︱︱ノ ６ １ ８ ８ ¨ ′︱︱︱︱︱ヽ
一
ヽ︱︱︰︱︱′ ２ ０ ２ ／ ︰ ︰ ︰ ︰ ヽ
〓
７ 博 １ ・ ・
′︰︰︱︱︰Ｉヽ
一 ＝
￨
十
〓
２ ・
ИB′ =′ β+
:2(c)
8Vl+4+1=8ぜ 7
︲
′︰ＩＩＩＩＩヽ ヽ１ １ １ ︰ １ １ ノ ２ ０ ２ ′︰︱ｌｌｌｌヽ
:2o)
１ ２ ７
=PI=1面 ― 硼
７ｕ口＝＝ト 〓
Pcrpcndicular distancc
ァフ=￨:￨lil=￨::￨=311)
TWO VC口 Ors″ tO n,arelilmdill・
￨￨:IX1121=liil=21il VeCtOr perpendiculaF tO n,, I12=lil
Also,pOint′
:iCS ilt n,
一 一 ・
TherefOre,riil=lil・
lil=4 E4uttiOn oF n2:『
{￨)'4 12.“ xi)I Dinnce′
●● mn.
1司 =p― 可
‐
￨￨￨1:￨=￨:￨
=Vl+4+25=√ 0 l2(dxii) I Acutc angle
betwccn I and II1
=“
肥=m・ 器… ・嘉
=“ 6・ (01d∝
〕 口
NATTONAL JUNIOR COLLEGE 2OO7 PRELIMINARY EXAMINATIONS
MATHEMATTCS
974OIO2
Higher 2 Paper 2
Paper
3 hours
Additional Matcrials: Writing List of Formulae (MFl5) Cover Sheet 196 Septcmbcr 2007,
Wcdncsday
1300

l6lXl hn
INSTRUCTIONS TO CANDIDATES Write your narne, registration number, subject tutorial group, on all the work you hand inWrite in dark blue or black pen on both sides ofthe paper. You may use a sofl pencil for diagrams or graphs. Do not use paper clips, highlighters, glue or conection fluid.
Answer
ALL
the questions.
Give nonexact numerical answers corrert to 3 signi{icant hgures, or t decimal place in the case
of
angles in degrees, unless a different level ofaccuracyis specified in the question.
You arc expected to use a graphic
calculator.
I
Unsupported answers from a graphic calculator are allowe.d unless a question specifically states
ofherwise. Where unsupporled answers from a graphic colculator are not allowed in
I
question, you
are required to present the mathematical steps using mathemalical notations 8nd not calculstor
commands.
You are reminded of the oeed for clear presentation io your answes.
Ilp to 3 maths may be deductedlor giterol poot ptaenlatioa in The number of marks is given in the brackes [.]
u
atry
Pa ol lhts pape?'
lhc end ofeach question or part question
At $e end of lhe examination, fasten all your wort securely togedrcr.
This question prper consists of 8 printed pagcs' including I btank pege'
[Turn over
2Scction A: Purc Mrthcmatics [40 Mrrksl
The curve C is defined by the parametric equations 工 =′ ― ＞ Ｄ ０ ０
Show that器 =ヂ
ツ =′
:,
+:,
′ cRヽ {0}
暑・
[2】
Detcrminc thc coordinates of the poins where the grdient of the tangent line is parallel to thc.raxiS.
(iii)
[2]
Show 0rat nonc of the tangent lines of this curve will pass through thc origin. [3]
Given that y=vl_χ
2'Sh°
ｉ
① ←2)器 =ッ
W alat
;
i) =2=O when,=0
0btain thc Maciaurin's expallsion dfy up to and including the term in
Hence find the Maclaurin's expansion
of y = 5int x
１
・
３
_χ
2 χ
up to and including the term in
x'.
tzl The sequcnce of nonzero numbers at,a2,ay... forms a convergcnt geometric progtession. Given that progression
d
apal utd a,
are three consecutive terms in an
arithmetic
tr, =(tr,,)',
(D
find thc common ratio of the gcometric
(iD
Show that a, =
b(Zt  l)
progression.
whorc 6 is a real constant to be determined
t3l
.
t41
[Turu over
3
4.
Sketch on an Argand diagram the set ofpoints representing all complex numbers z
satisfoing both of the following inequalities:
3tr q s uc(z
) 25
Under H。 ,wc haveア
_N(25,::)by CLT
Since o2 is unknown, we set out to calcutate s,:
",
=
i9l
I rrzsrs)fl27o)'l=
4elso''l.50/J
We conduct
a
660 4e
z test at 5% significance level.
Using GC, we obtain pvatue = O.220 We reject Hu
ud
zvalue = 0.7?l
ifpvalue < 005.
Hence we do not reject
H,
and conctude at 5% sig. level
that there is insufficient evidence to say that
p
> 25
.
)
2007 JC). I lP4 NJC Preliminary Examinations Paper 2
I
l.(ii)
Ho
:
lt =25 vs H,: P *25
Nowpvalue = 0.22O x 2 = O.44O We will reject Ho if;r'value < 0.05. Hence we do not
rejed
fL
and corrclude at 5%
sig level
that B€tty's claim is justified. I
l.(iii)
Ho
:
lt =25 vs Htl. P *
Under Hn, we have X
25
 N[rs,{. ( '10l
We conduct a ,' test at 5% significance level since d2 is unknown and samPle size is small. We need to assumc that Xfollows a normal distribution Using GC, we obtain pvalue = 0.M27 and
t'valrc:2'159
We reject Ho ifPvalue < 0.05. Hence we reject Hu and conclude at 5oZ sig level that there is sufficient evidence to say that * 25 '
I
12.(i) Since v = I.oo3lx 4 !018,then we have 6 = I 0031 and Y6.r= 4'30t8' Hence
y
1.0031(52.75) = 4.301
8= v = 48'611725
From the table we have
f,l
=
338.7+t = 48.61 l?2s(8)
=+
t
= 50
2
(ans)
12.(D FЮ m GC,we haVe r=0863(ans) 12.(iii) Model C is suitable since it is observed from the scatter plot diagram that as r increases, thc rate at.which y increases is decreasing Since b>0,MOdel I)iS not
12.(iv) L, +.;r
Lz+
!
L,
ln(L1)
+
︼ 一
I I .(iv) This is not advisable since the 2 data seb might not be. independent, e.g. the same interviewee is in both samples'
￨
￨ _
2007 JC2′ IP4 NJC Preliminary Ex3minatiolls Paper 2
Using GC,LinReg L,,L2yiclds y=5372!nχ Hencc α=163 1ttb=5372,′ =0905(a“ )
12.(v)
80 = 53.72n.x

163.14
+
x = 92.39 kg
―16314
(urs)
Since y = 39 is out ofthe rangg the value ofx obtaincd may not be accuratr.
13.(i)
L€t
Xand
f
ot"D24" uld.Xo26" durians
be the uaeighs
respectively.
then
x  N(I.s,0.t')
y.I6yy ge1 1,7
=
X,
+"'+
Therefore P(.4 <
r  N(l.8p.2,).
and
X.j ( +"'+ f, (;,#) 
M sl.1)=0 730
(ans)
bOmttJ〔 4口 ,:,原 13.o)La r=2γ ―(χ :+ろ +為 )― N(09,019) P(14 > 0.s) =
p(r [t
> o.s)+
p(r < o.s) =0.s21 1ro;
 normatcdf (0.5,0 s,{.9,J0.19)]
13.(iii) Let C =30(X,+ X,)+35y Since we have P(C > a)=

N(1s3,67)
6.13 n(C < a) = 9 9,
we get o = t63.49 [invnorm (O.e,rsl,.67)l
13(v)Req面 red
PЮ baЫ :lty=[P(χ
t*rl
>i3)]2P(χ くi3)x]
=00652(ans)
)
SA:NT ANDREW'S JUN10R COLLEGE PRELiMiNARY EXAMiNAT10N
MATHEMATICS Higher 2
9740r01
Paper l
Thursday
30 August 2007
Additionat materials
:
3 hours
Answer paper List of Formdae(MF15) Cover Sheet
REAO THESE INSTRUCTIONS FIRST Write your name, civics group and index number on all the work you hand in. Write in dark blue or black pen on both skles of the paper. You may use a soft pencil for any diagrams or graphs.
¨ 一
Answer all the questions. Give nonexact numerical answers conect to 3 significant tigures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expecled to use a graphic calculator. Unsupported ans\ /ers from a graphic cdculator are allowed unless a question speciftcally state otherwise. \y'Vhere unsupported ansu,ers from a graphic caldllator are not allowed in a question, you are required to present the mathematic steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of eacfi question or part question. At the end of the examination, fasten all your work securely together.
This document consists of 6 printed pages including this page.
2
Sketch, on a single diagram, the graphs
of f=2x+9
and
y=l(r;12;
, showing clearly the
point(s) of intersection by giving the xcoordinates in exact values. Sotve the
2'*9
ineoualitv
[31
lxl+2 =o.
The functions f, and g are defined by
f :xt+2x2,
xeO, xSa
g:xr+ln(3x), :eD, r0, v>0. region bounded by the curve y2 +2ry+2x7 =l,the line y:r
The diagram shows the curve with the equation y2 R is the shaded and the yaxis.
y'+
2nJ +
2*=
1
一 一
Find the volume of the solid formed when R is rotated through .xaxis. Give your answer
corect to 3 decimal
End of Paper
places.
2t
radians about the t31
I
[Turn over
2007 WL lUaths Prelim Exam(Paper 1) Solutions
“
dinatcs of画 mo■ 血
For需
r画 Ons:,
≦ 4,
_ the solution set is
2(1)
１ 一６ 一
1■
0=0 f:χ 卜→ 2χ
y=2χ
2,
(χ :χ
≦―
xc□
:or
χ≧ :}‐
,x≦ 0
2
x=― V2y sinccx≦ 0 f 1(χ )=― V2χ ,ェ ≦ 2 ShCe烏 =(―‐,21⊆ (→,3)=2,gfexぉ も .
3(1)
ズ
βχ+C
=TTT+1+χ
2
2=/(1+χ 2)+(βχ+C)(1+χ ) Subts x=_1,/=I Comparing cocfflcients Of
2:B=_1
χ
:C=1 И =1,β =1,c=1
x°
°
0=論 I lr l+x l+x'
.1_
+r,f' = (l ,+,'r'+...)+( r{l r, *...) _ I x +x2 r' +. .. +  x x2 +r' +.. = (r
+,)'
= 2 2x
+ (r
+...
..{r
(Stnwn)
４０
― 樋 朧 +)=h(挙 Z郎 =h〔 一 歩 )=h:, hι
Statcmemを h〔
出=h(場 )=h(:)・ ∴ お 鳥 rme n■ TIsmeぉ n(サ
―
ih〔 +)ヨ
Ve htegcr l≧
)WhC∝
た2
2,Lc
)
Wantto prove that 4J is true based on 4 istrue,
に Ⅲ (鵜 Σ― 郎=2h〔 J ι =h(¥)Jn〔 マ キ =h(等 ― 詩 =h(午 )卜 出■ =h(等 半 h〔
)
)
)〔
)
)
)(霧
=h(#≒
)
)=郡
‖ ″ ≧ mQ 43mに お λ A撫 罰4J、 meb6d m民 ｀ 穏・ “ by Mathcmatical lnduction
(D ilh〔 _+) =']in(1_￨)1:in(1￨)
Jn(需 )刊
―
n(警
封 n(7)脇
=
=in(J:卜
_
=in(:子
5
)
)
:卜 ) 1:)
tet r
be the amounr
ofmoney invested in ttre structu.ed deposiiiccoiit
ッ be the amount Ofrnoncy invcstcd in bOnds z be dle alnount ofrnoncy invcsted in thc mutual ttnd
+Z=25000 006x+007ノ +008z=1580 ノ ーz=6600 χ+ノ
a
_ _
x=$17,467,y=S7,067,z=シ 67 6(D Lety=《 1)
l
χ=― C iS the vcrtical asymptote So ο =1
Bガ 。 喝 Jm,y=α +讐 d宙
エー
_
ッ=α is thc hodzontal asymptote
Sinccノ ヽthc
rencc」 。n of Й aboutthe lnc y=● midpoint of И'and/.
―
α=5+← 1)=2 2
_
2χ
y=χ
_1
Whcn
―
+b
χ=o,y=1_
1=上 1
b=1
ThepOht lo,α )ヽ the
(■
)TO getgo“ mo l)Shiッ
dOWn 2 units along y‐ axis
2)Rencct about x‐ axis 3)Sht up 2 ulllts along y― axis g(χ )=―
(f← )2)+2=2,5
Altemative soll■ lon l:
(1) y=f(.)=2x+1 g(χ )=― f(→
+4
2x5 エー1
Al“ mative∞ lution
(li)
Since」 F氏 →is
2・
a“ tangular hyperbol■
y=2 and χ=l
are its asymptotes
bt g(X)=響 When x=0,ッ =5
5=二 1
グ =5
7(D
ツ=
2+2 αχ χ_1 =等
多
Since C has no stationary points, there are no real roots for
(2a)'4a(2)099
IZ≦ 辟
>2.326 辟 ″>2164
Lcast″
=217
∼ ア Ⅳ ル o80,器
)
P(χ 20>178)=08271 Lct y bc“ no Ofsamples with mcan>
y∼ B(100,0.8271) E(り =lClll X O.8271∼ 83
178 out
of
100 sample"
P(a raCer wih a dynalno bcionging to a male)=09x06x001
=0_0054 P(a raCCr without a dynamO)=(09X06+07x04)0_99
=0.81i8
P(it be10ngs to a girl ifit is not a racco
=P(be10ngS to femaleヽ not a racer) =P(
)
0.3x04 0.lxO.6+0.3xO_4
=0.667
ロ
r = 0.932 The scatter diagram and the value of r suggests a strong positive linear correlation between and y.
(ii)
Regression line ofy on x
:
y
:
O.348 + 0.271x
:
When r 58,y = 9.34t * 0.273(58) So moisture content is 16.2%.
(iii)
When ;r
=
10,
y
:
:
16.2
0.348 + 0.273( l0) = 308. So moisture content is 3.08o%.
The estimation may be unreliable
(i")
as
r = t0 is outside
No. r will be the same because the value
ofr
the range
ofthe given data.
is not affected by a linear transformation on y.
９ ０
Letジ 【 be the numbcr ofca‖ s madcin a one― hour period
χ ∼Pο θ) P(χ >9)=l― P(χ ≦9)=0413(to 3 sO(ShOWn) Let y bc the numbcr ofperiods whcre morc than 9 calls are made out of力 pcriods
y∼ β(1,0413) y∼ Ⅳ (0.4131,0.24241)apprOXimately Givcn that P(y≧ 20)渕 9
P(y≧ 195)>0.9 P(γ く19.5)く
01
r
Pし
く
llil号
:;::生
)畑
く‐ 郎 2 諄
19.50413た
く1.282ぜ 0.2424■
■47_2>1.53√ Lt rbc thc total nulnbcr ofcalk madc in one day.
r∼
P。 (9o)
Si― え>10,r∼ Ⅳ(9oメ Ю)appЮ Ximately ′Oo≦ rく loo)=PC89.5≦ rく 99.5)aner continuity∞ rrection Probabi‖ けrequircd=∞ rdd【 (89.5,99.5,";輌 ) =0.363 cto 3 sつ 1∝
mdL L“ →ど 覧 ミよ ●TFlttal:ヂ 2P(>960)P(L (d)
less than 710 g in a box of 20,'.
(30,624)
l0):
0.78g
P(E7ls) > 0.9
e) \z.7tst'\
=t.2sz 7265
Least value of the mean mass
ofa small
packet = 727 g
H● )ItiS nOtnecew to,褻
ntral
nlcorem(sinCe salnple sizcお largo lズ え,`bc
bulon Limit “
urllc that tt amolmt spcnt per customer fonOws a mmal d誌
bccause dlc salnPling di面 bution ofmearls is approximatcly nonnal by Cた
dlc alnt sPcnt pCr customcr.
Test ff。 :μ
=215 vs名
>215
:μ
Under ffO,ア ∼Ц 215,器
=Ⅳ (215J∞ )
At 5%levcl,the cttical x― valuc is invNorm(095,215,lo=231.45
ヽreJ∝ ted when I>231.45
So ff。
Alternative method: At5%levcl,the ci血 」 z‐ value is invNormal(0_95)=1.645. If I>231_45, χ 215
23145215
10
10
>
z>1645
So〃 。ls ll(り
reJected at 5%lcvcl
()unbiaSed estimatc ofPopuhtlon mcan=I=■
畳+10=86.875%869
8
Unbiascd estimate ofpopulation variancc
=￨)(47343≦
:ll)
=9.268
%927 =88 vs〃 1:μ く88 GC:T― にst,Sttts,為 =88,I=86875,義 =● 268=3044,
Test ff。 :μ
″=8,μ くμo Calculatc_ Sinceρ =0165>0.1 ●rt=1045>invTo■ ,7)=141) Do notrq∝ tI。 .There b insufrlclcnt evidence at 10%levd ofsignincance that thc icssons had made an in口 provcmcnt in his gol[
Assumption:■ Ю Юre “
in a round follows a Nollllal diS輛 bution
(D h COnclusion would be thc salncCe.〃 。Win not be reioCted) 珈
SISbCCan the md面
並
」 鳴
=芳
岸 械
l'
1s
X嗣
湘 ¨
∫
bigger than r(the test statistic value found in part i)) utren s'>927 , hence it
still not lie in the critical region OR(comparing pvalues): P(T て
n∝
燐b・
2】 【
can be w‖ tten
in thc fonn ,テ
=:万 Hence, or otherwise,
find
)
色
耳 T― 漸
１ ３
Show that
131
2″ ― : (″
2_3″ +1)(″ 2+″ _1)
Usc the abovc resultto show thatthc sum to inflniけ
井=… み+井 +井 す 121
is less than
I
.
Sketch the locus
where c =
of
arg(l 2
z)L
inan Argand diagram Find the least value of lz+cl
li.
141
number z, that satisfies both
lz+cl=.k
/re0*
such
and arg(l22) =
． 血 ４ π一
Find also the range of vatues of&, where
there is exactly one comPlex
pl
,3
4
(a)
2
A sequence {.r,} ofpositive numbers is defined
{rr} exact value of 2. The sequence
(り
converges to a number
2
u ,*r=##forzfJl*
as n tends to
infinity. Find the 121
A sequen∝ (%}iS deined as lrl=3 and“ ″ +1=(″
+3)% for″
Πl+.Π nd」 に
first four terms of the sequence and show that these terms can be written as
",
=
#*
for
n:
1,2,3 and 4
With the values of a induction for z >
5
(a)
Find `」
and b
where a and 6 are constants to be determined.
found, prove that
l.
",
=
ffiAV
*e
method
of {51
ヒif
一 一
片品
,
0月n(網
121
21 【
)
(b) Givcn that ﹁ 一
′
・nd者 h“ ‖
us ofF andノ
ツ
=(■ 5y)3,
pl
.
[Turn Over
・ ︼
(a)
the ends of the diameter of a circle with centre O Uisa point such that it lies on dre circumference ofthe circle. Taking the centre 0 as the origin, the position vectors ofpoints S, i'and Uare s, t and u respectively. The points Sand
lare
Write out the vectors SU and lines SUand
(b)
IU
are
TU in terms of s, t and u and
perpendicular.
Given that OAPB is a parallelogram and at
X. The position
＞ り ６ ０
)
hence show that the
vectors
ofl
p
I3l
is the mid point of
AP.
OP meets BO
and B are a and b respectively.
)
Find OP and OQ in terms of a and b.
If R is the point
on BQ produced such that BQ
produced and that OR  2OA.
121
=
QiR, show that R is on OA 131
5
(a)
The diagram below shows &e graph ofy = clearly, the larbeled graph of y = f (.r) .
(:).
On a separate diagram, sketch a
13】
(b)
The functions fand g arc delined as follows:
f:rr+(t+2r)r, r χ9
″+2 o+2xェ ー4) X+l
(x+2xχ 4)
1l 20
4
or x>4
2_9 2+1> χ 4(x4_χ 2_2) 2χ
l
2
(2x2)_9
2)+2> (2χ
(2,2)2 212,12
s
Thus, ⇒ 2く 2x2く _1
or
2x2
>4
(司 ∝tCd)
=12>2 = ,, J) orr. J7
[Turn Over
2
Show th江
2_3″ (″
Hence, or otherwise,
+り (″
2+″ _1)Can be wdttcn h thc fonn″ 2_3″ +1 ″ 2+″ _1・ 2″
find )
籠二(″
1
2_3″ +ix″ 2+″ _1)
l l 1222+万
l l +7+4252+ 121
l.
一 一 ¨
infinity
Use the above result to show that the sum to
is less than
lll
n2
2_3″ +1 ′
2+″ 二 ″
+nl(nz 3n+1)
2_3″ +1)(″ 2+″ _1) (′
l
Wn_ FlIT, Sh° 二 ]:::li;;)耳 二 「三
=マ
F戸
“祠 ″で ， (′
2″ 2
1
3″ +1)(″
2+″ _
=:葛 詰 (漸 ― り
)
ｒｌｌｌｌＬ ｌ 一２
l
＋ ＋
1 1
/11
/ 1
5
ノ
/
/19
l
29 41

l)2
+(Nt)l
１ 一２
〓
Ⅳ ６ 一５
I
一 ︼
(Ⅳ
,t+N{ Ⅳ
2_Ⅳ _1
Ⅳ
缶
2+Ⅳ _1
]
¨ 一
=[:―
] 2
For″ ≧ 3, ″ >″
凧 Σ 蒜
2_3″
and(″ +1)2: >″ 2+″ _1
+l
1
くΣ 品
_
″ →
⇒ arg(2xz―
arg(12z)=千
:)=争
⇒ a曜 (2)+arg(Z― ⇒ arg(Z―
:)=争
:)=1:
―π =―
子
←l拌 1)。 、
ヽ
1
Locus of
arg(12z)=争
From the Argand diagram,
To have exacJy one∞ m口 cx number z thtt satisfy both,レ +J=■ and arg(12z)=1:,from thc Argand diagram,■ =d:or■ ≧d2
Now d2=√
≧ 「=写・rL8o Loた =:√ λ 頭 手 Or
I3l
￨―
￨
￨
5
4
(a)
A sequence
{:,}
The sequence value
of positive numbers is defined
{r,}
converges to a number
2
as
* ,ru=*r^*3for z ll lt
.
z tends to infinity. Find the
of 2 exactly.
121
Solution>
If {.r,} converges to 2 then
rn+r:r.
Therefore
=,
as a
2=X *z
2).+3
Solving wegetz
as
z tends to infinity,
to infinity.
.
:{a
U"n"redsincer, rOy o,
fr1
[Turn Over
6
(D
f° r″ Πl+.Find the A sequen∝ (%)お de6ned as″ l=3 and%.l=″ +2+2″ ′
flrst four tcrrns ofthe sequen∝ and show thatthcse tc.llis Can be wr詭 n as
玲 =2″
f° r″
=1,2,3 and 4 wherc α and b are∞ nttn"to be decrrnhed
+ら
With the vJues ofα and b found,prclvc that% induction for PI≧
″+α
by the method of 151
1.
くSolutio■ >
=れ Jみ =: 竹ち =■
From thc above resuLs,″
" 2″ ―l iC_α
=2 andら =1
br″ Π l+" “ ":籍キ “ κ =器 一 h“ お 町均
Ld L“ ttt
Statcme離
Plls trに
Assum L讀
眠4=括
To show P卜 l ls truc.lc〃 々tl=21+1
Thereforc Pa.l is true_ Pた
is truc l P卜 l
is truc.
Since PI is truc,by Mathemadcalinducjon P″
is truc for a!!″
≧1,″ Π
l+
︼ 二
7
(a)
Find tt r dF
1) く
月網
y=
n〔
ψ一 壷
ｄ ｎ ■
) G市 en that y`ノ =(ェ ー5ノ )3,
121121 )・
in t― s Ofx andッ
.
131
“
Solution>
=√ Ox+1ド :
(aXi) y=
」 と―― 誓
(2x+1) :(2)
√ (2x+り
3 2
2+χ
y=!n(儒
+1)― in(1+COS X)
)=ln(χ
+轟
=島
⇒ 者
(C)
y`y=(エ ー5ガ 3 ιツ+ッ
⇒
者
`ツ
:F=3(χ
5y)2(1_5者 )
ツ
一
ξ′
χ
〓
３
■︱ Ｉ Ｊ
ψ一 む
の 研
＋
５ ︲ ● ３
ツ ′
ツ
＋ 〓
隔ド ウ 一
dF (1+y)`ツ +15(■ 5ッ )2
[Turn Over
8
The points S and I are the ends of the diameter of a circle with centre O. U is a point such that it lies on the circumference of the circle. Taking the c€ntr€ O as the origin, the position vectors of points S, I and U are s, t and u respectively.
>
Write out the vectors SU and
+
fU
in terms
ofs, t and u and hence show that the lines t3l SIt and TIJ arc perpendicular.
+
)
SU=As_ and TU=UL
+
)
SU TU = 4
A
t' L =lul'  a' L  ' 1 +[l' l{cos(r 80' ) U.
L t'
4+
=lal2  a' L* u' t+lul'
l4(
=0 IU
Thus SU and
are perPendicular.
Alternatively,
+
)
surU =(us_)
[email protected] L) = (g s)(q +s) =日
2_ぽ
=0
r)
9
(b)
oflp.
Given that OAPB is a parallelogram and p is the mid point atx. The position vectors of,{ and are a and b respectively.
I
(i) (ii)
+ ) Op
Find OP and
in terms of a and
Op
meets
Be
b.
tzl
If R is the point on ^Bp produced such that rO = OR, show rhat R is on OA produced and that OR:2OA. pl
くSolution>
o
(D
+ oP q+Q
+r
o(2 =
)l1q.
+
4')*
4 = llzq * ay ll /) ) \
(ii) Now. OO = OB+oR  2\ )
I
)
_+
__>
+OR =2 O2OB
)
=
on =2.!(za +b)h =2s
> =2 +OR
) OA
Since O, R, A are collinear
=
R is on OA produced
Altematively, Bp produced such that Bg = OR
+ + ) + +BR =2 88.....+OR =2OA
.
ffurn Over
10
(a)
The diagram below shows the graph clearly, the labeled graph of y = f(r)
of7 = (.r).
On a separate diagram, sketch a
.
31 〔
ll (b)
The functions fand g are defined as follows:
r'r*(r+zr)3,,.j; g,,;!,
t31)
O ftt fft
/'
Showthat
,t+
Given that
P(x > s)=6.3a13, find the valueof&.
Given that
XyX2
Find P(x,
+Xr+Xr>2o).
and
12】
X3 ar€ three independent observations of X.
41 【
21 〔
Three boys Alex, Benjamin and Charles agreed to meet at the theatre. AII three boys forgot which of the theatres, Palace, Queen's or Royalty to meet at. Alex tosses a fair coin to decide between the Palace and the Queen's. Benjamin tosses a fair coin to decide between the Queen's dnd the Royalty. charles tosses a fair coin and if it is'head, he goes to the Palace, otherwise, he tosses again to decide between the Queen's and the Royalty. Find the probability that
4f
Alex and Benjamin meet,
tU
b> li*,
Benjamin and Charles meet,
l2l
all thre€ boys meet,
121
alt three boys go to different places,
12t
at least two boys meet.
l2l
. !9 ltl
１１ｌＪ
■、
f
The table below gives the observed values
,A
of
bivariates x and y.
χ
20
30
34
35
36
40
42
ノ
32
25
α
22
26
18
19
It is given that the equation of the regression line y on
石／
Find the value
of
a correct to the ne€rest
: is
y = 43.5

O6fl2L
integer.
l3l
Write down the equation of the regression line r on ), and the value of product moment correlation coeflicient between x and y. I2l
く「 ゃ OTll(『
:)
The fO‖
Owing are the summary Ofsix pairs ofvalues ofthe variablcs x andッ :
lx=tt,Zy=ztq, lx2 =rue2, andly2 =9274. It is given further that when
r=9
Find the equation of the regression
4
the estimated_value
tin",
on
r.
\s
of
yis37.\
tr
S.\
o*X Isl
r'!":s+>'*.;
The amount ofdonation collected by each student in a particular college is $x. Data is collected from a random sample of50 students and the results are summarized
by
Ix g
Ix2
= 360250.
Find the exact value of the unbiased estimates of the population mean and the population variarce ofthe amount ofdonation collected byihe
pl
state clearly the distribution of the mean amount coilected by the students, giving rts mean, vanance and any assumptions
I2l
studenS.
／〆
/ W 
= 4O00 and
made.
Find the probability that the mean amount collected exceeds $g5.
Find the smallest possible sample size exceeds $85 is less than
0.03.
if
121
the probability that the mean amount
l4l
一 ヽ
[Turn Over
￨
/g
_Oft
a
student's homework solution on hypothesis testing for a random variable Xwas
found to be in a bad condition as shown below. ¨ “ ︼
The test conducted used a sample size of 50.
メ メ ´ Ｘ 乙
一 ． 一
i.c
ruSecl
ll.
...rt'en
Ztu+ 7
Z+.+ < Gぃ duヽ 10n
Sincc ZⅢ だャ
=2‐ :21 ife 10 J.._a‖
瑯
i∬ (∬ Trぶ
dr崎 おo,
畷
ノ⑭
Find the sample mean, leaving your answer correct to 3 significant figures' [21
Find also the least value of the significance level such that the result will t3l be in accordance to the student's conclusion.
I
,{
9{
fn
principal ofa private college claims that graduates from his school have an
average sta(ing salary of $2050. A private body checks the claim by interviewing a random sample of60 graduates from the school.
The data obtained is summarized bclow, where
r
denotes the monthly salary per
person.
l{,
zooo) = zt+o ana
f,
{.r

zooo)2 = 162001
Carry out an appropdate test at the 3% significance level to determine whether the principal is overestimating his claim. t5l
10
(a)
According to the school rules, a student who arrives at school after 0730 hours is considered late.
During the school term, a boy cycles to school on 5 days each week. On any given day, the probability ihat he arrives after 0730 hours is 0.1. For a period of4 weeks, calculate the probability that he is late on at least one day but not more than 3
り
days.
121
When the boy has cycled to school for the nth time, the probability that he arrived at school late at least once is greater than 0.99. Calculate the least value ofn
l2l
A girl travels to school by bus on 5 days a week. Overa long period of time, the variance of the number of days per week on which she is late is 0.8. Given that p denotes the probability that she arrives late and that p < 0.5, evaluate p.
l3l
[Turn Over
・ ﹄
The Royal Sports School runs an intensive training programme to groom sports talents. The school believes that when the students participate in the annual
national sports heats, 807o
of them will qualiS for their events.
Among those who qualifo, 97Yo of them will eventually be in the top three positions and the rest will receive consolation prizes.
／／
100 such sfirdents attempted the sports heats. Using an approoriale*approximation, find
the probability that at least 5
will quali$
and receive consolation prizes. 一 ¨ 一
End of Paper
ｐ ｐ
the pmbability that the number that qualifies is at most 90 but no fewer than 70.
2OO7
I
JC2H2MATH PRELIM PAPER 2 The curve C is defined parametrically by
x:
asrre
and
y =s1a16, where a is a fixed
constant. It is known that the point P with coordinates (asecd, a tan d) lies on C. Show that the equation of dre tangent at P is given by rysin9=acosd and the equation normal at P is given by.rsin d + y = 2atanq 
15】
The tangent and the normal at P cut the.raxis at I. and ly' respectively. Prove that OT
ON :2^a2,\Nfue 0 is the origin.
t3t
C: x=asq,0, y=atanQ
!=on de
et^ne
and
!Z=a*:e d0
' d.r asecorano=.*r=(*) t ^^.o dyasec2o trnal11fr=a=*""'0 ( cos d./
At
P( P(asecd, atan9) ,
t
Equation of taneent:
+
ysin0  asin9
aulne = l=(:
tan 0 =
asocd)
r  asq?
 = xysinl =aseclasinlane =o(\cosdrinB.tind) cos? ) =_3=(r_.i"ra)
+
o =cos d *"2e
ysin d = acos d (shown)  atm? = sine(x  awe) =xsin0+atan0 xsin9 +y =)s tand (shown) .x
Equation of normal: y
) Aty:0, r = acnsO
(from eqn
T: (acosd,O) *r(freqnof
oftgt)
rsinl =2atanl
i.e.
normal)i.e.N=1?a tcosd ,g; =, =cosd ' 2o:=2o2 oTxoN =acosd . (rho*n) cos
d
[Turn Over
●) FInd the moddus and argument of the∞ mpに x number z,where z=(1、
１
giving your answers in exact form.
２
5)4'
Hence, using de Moivre's theorem, evaluate z6

くSolutions>
=静 濡→ い¨ 羽 arg(Z)=argl絲
j)4arg(1√j) ]=2argKl― =2(― :矛 )4(―
争)
(Can use g_c)
=子 Thus z=:(coSi多 +fSini「 )=。 Z6=十
二 )z6=」
(り
(coS」
F+iSini計 )6
:(cOS5π
+isin 5π )=―
Show that For any compiex numbera:θ , 光
: 」
=:―
￨:(COti:)′
12〕
Write down the solution to the equation z5_1=O ieaving your answers in
thc form″ 'θ where r>O and― πくθ≦π
121
Hence, find all possible values of the complex number w that satisfies
(*l
=l,leavingvouranswersintheform
:+vi where'r,veD'
I3l
︻ 一
″ ι
`″
(` ″ 1)
θ1=● ″1)・ ′ (′
β
1`θ
1)= 1(′ I

(cosA
θ
+ι
β
)+l
+isind)
22cosθ I . sin θ =― ― ′ 2 2(1coSθ )
÷′ 鵡 =:―
:(COt:)J (ShOWn) 2た π
z5_l=0⇒ z5=l=♂ (21π )⇒ z=」
川ト
′ ，
′︰ ︱︱ ヽ
ｂ
“ Ｗ一 祠 ・
■2`与
5 ,′
5 ,た
=0,± 1,2
±生 5
Lct Z=―
―― ″ l
=OW=― z1
2κ π
Th¨ =舟 =巻
=::(C∝ 詳 等 5 _1
た=0,± l,2
)ら
ι ⇒
″ =05± 0688′ ,0_5± 0.162′
[Turn Over
4
A differential equation is given Ay
(i)
+ . {= 4t7y QG)y"
P(r) = 4 and
[email protected]) = 4 , find the general solution of the differential equation when z = 3. lYou need not expras y as the subjed in your final answer.l For
(iD
For P(.t1 =
1 x
and Q1x1 =
J, l+
x'
show, using the substitution y = ,tx
differerrtial equation can be rcduced to Given that
y:
0 when .r
q=f= dx l+ xz
answef.
::=4ッ ー4ッ
3=4ッ
(1ノ Xl+y)
Xl+ッ )=:テ
11三
′=Lβ =:,C― :
⇒ら 浩万デ場ψ =4ト +函
― +列 =義 :h‖ J― :h卜 ⇒2hlJ― hト ー列―h卜 +冽 =8χ ⇒hッ 2_h卜 _冽 ttC2 ⇒h回
+q
+Q
11+冽 =8χ
in詢
。 ):f=:ツ +丁 Whenッ =zr⇒
dy
ッ(1ッ Xl+y)と
=4
+I.y and by cover‐ up mettod, ・ ラ
Let, y(1ッ
デ
131
terms of .r in your 121
Solutions> (1)
, the
when n = 0.
= l, solve the DE givingy explicitly in
15】
=8χ
fiテ
者
+C2
¨ ¨ ¨ ¨ C)
=″ +χ
告
5
￨
￨ _
From(1), :ノ +静
=″ +χ
:静
︻ 一
d″ ⇒ l ″+百χ +工 T・ 7=″ 百 生 =百 ⇒
7 (Sh°
︻ 一
muS,ld″ =∫ When庁 1,ノЮ
Wn)
⇒ ″ =tan 1■
か ⇒
+ε
IIl(Ы n∝ y=″ )
⇒0=tan 11+`⇒ c=― 争
⇒孝 =tanlχ チ ⇒y=χ
tan lχ
―チχ
[Turn Over 一 ︻
4
The line /r and the plane zr have equations r =2i+2j+ pk+2(i +gk) and r.(i+ j+k)=2
[email protected] Given that /1 lies in the plane rr, show that p =2 and find the value ofq. Verifo that the point A(1,2,1) lieson 11.
I3l
rl.
l3l
Find the equation ofthe plane z2 which contains the origin and intersects z1 at the line Given that the point
r
I
=
has
position vector
(2i+ j+3k)+s(i
Iind the length of projection of
i
+ 3j + 2k and plane z3 has the equarion
k)+(3i  j+2k)
)
lil
wheres,r e D,
I3l
23.
onto the plane
By finding the solution ofthe following equations,
x+y+z:2, xl z:O
and
x+ 5y+z=10,
or otherwise, give a geometrical explanation on the results you have
〓
２
ヽ︱ ︱ ︱ ︱ ︲ ︲ ︱ ノ
２ ２ ρ
l:
″ 一
SinCe rlileS On Π l 二)lil
Π
／ｆ ｌｌ ｌｌ ︰︱ ヽ
[=￨:￨+'￨￨l and
lll=2二 )′ =2 (shown)
obtained.
I3l
ヽ︱ １ １ １１ １ １ ノ ー ０ １ イ ー ー ー ー ヽ
〓
つ
ヽ︱ ︱ ︱ ︱︱ ︱ ︱ ノ
２０つ
〓
′︱︱︱︱︱ヽ
ヽ︱ ︱ ︱ ︱︱ ︱ ︱ ノ ︲ ー ０ 一 ／１１︰︱︱ヽ Ｘ ヽ１ ︰ ︱ ︱ ︱ ︱ ︰ ノ
２ ２ つ
〓
Π 2:
is
／ｉ ｌ ｌ ｌ ｌ ｌ ｌ ヽ
ThuS
of fI2
め
Normal
1・
111=0
→
Thuslength OFproJcctiOn Of И3 0n Π 3=
痛
x酬
到 √
一 一 一 一 〓 〓
[;= イ
χ+ノ +z=2, x+z=O and X+5y+z=lo,intcrsccts at rl Ottettisc mettOd
[Turn Over
円
Π ｎ ０
“ ２
ヽ１ １ １ １１ １ ︰ ノ ３ 一
１
・
〓
︲ ヽ︱︱︱︱︱ ノ ０ ２ ０
１ ０ ０ ０ １ ０
ー
１ ｌ
０
１ ｋ
５
１
１ ｒ
＝
１ ０ ０ イ ー ー ー ー ヽ
＝ ュ １
り
５
ヽ︱︱︱︱︲︲︱ノ ー ー ー ０
(' atl intersects on the line
fI, and
+fll,fl2
３
ほ︺ ２
１
︲Ｌ ︲ ゞ ⇒ ド ん ﹂ Ｈ ︱ ＩＩ
p=z,a=P Let
x + z  0.
y:2, Thtts
嗣 叫 ３ Π
ル
ρｂ Π ︲ Π 価 陥 ⇒
Solvine method
I Section B: Statistics [60 marks]
5
The random variable X has a normal distribution with mean
and variance
4.
n(r > rr).
Showthat
,=1!f
Given lhat
P(x > s) = g.8a 13, find the value of*.
口 ＝
◎ ／／
Given that e(x < s) =
I
.
Given that X1,X2 and X3 are three independent observations Find P(X, + X, + X, > 20) .
ofX. 121
(i)
Since P(x 3k'),
p
(iD
is the midpoint of 5 and
3t, ;. p =5 *=3k z
e(x>s)=s.3413
p(x 0.03, we do not reject Ho
evidence at
30%
.
Here we have insuflicient
significance level that the principal is overestimating his
claim.
一 一 ¨ ′ 一 一 ¨ ¨ 一
[Turn Over
一 ヽ
18
10
(a)
According to the school rules, a student who arrives at school after 0730 hours is considered late.
()
During the school term, a boy cycles to school on 5 days each week. On any given day, the probability that he arrives after 0730 hours is 0. l. For a period of4 weeks, calculate the probability that he is late on at least one day but not more than 3 days. When the boy has cycled to school for the zth time, the probability that he arrived at school late at least once is greater than 0.99. Calculate rhe least value
(ii)
ofn.
eYahnte p.
k.
y=E(x)
(D
Sketch, on separate diagrams, the graphs
(ii) (iii)
State the maximal domain of g for which its inverse function
(iv)
of
y=f (x)
ana
Find the range ofvalues ofa for which the equation h(x) roots, giving your answer in exact
form.
Find the least value offt for which the frmction Ih
l2l
exists.
 g(x) :0
has
tU two real 13]
exists.
t31
1l The quantities,andノ arc related by the difFcrential oquation
i″ ッdi
l=二
.
ッ
By using the substitutionッ =z ',Show that the general solution of hc equation is
Y=lx+Ae', where
I
is a
constant.
Hence sketch the two solution curves passing tkough the poins
respectively.
I5l
(t,0) ana (O,Z) t4l
lTurn ovcr
6
In a single Argand diagram, sketch the set of points representing all complex numbers z satisfuing both of the following lnequalities:
+3+到 ≦ぬ レ 卜 1鋼
0く argCZ+3+勤
≦ 子
Hcnce 5nd the range of argcz+1+■ )and the maximum vJue ofレ
13 Sketch the graph ofノ
=響
[6]
.
+1+31.
[5]
,ShOWing c!early on your graph the exact cOordinates Of
χ
its tuming points and the equations of the asymptotes.
(D
Find the equation ofthe tangent to the curve at .r =
(ii)
Explain why the tangent to the curve at
all xe0
once.
ts1
3.
\{l}
t31
will cut the graph exactly
I2t
ぬ dn"画 け ‐ 0鋼 “ 1響 ￨>χ
End of Paper
[3]
4雁 却‐メ
21X17J2Prelim Paper l― sol■ d面 l iψ +z=1i― ―― (1) 2z― (1+1)ソ FЮ m(1),Z=
ll al→
=6+2i― ― (幼
i″ 1i― ― ―
(3)
Sub O)intO(2) 21″
22i― フーiv=6+2i
3iソ ーw=8+2i
8+2 =―
―
―
１
フ
―
"=2+21: z=1+i
2
∴ +:可 厩だ [号 」 …
Using sine rule'
sind sln. 1r 6
十二χ =1+ユ エ 2 8 2
〓
+1‐ 0
︲
]―
θ
)・ =lcm and b=3cm,
1+
レト
:
２ ︲ 斗 一 計 ︲ ハ ︶ ５ ５
手
彩
２ 卜 ６
Substitute●
はェ =争
hOm)
=2ba〔
／１ ト ー ヽ ＋ ３ 一２ ３ 一 一６
="(,+)
阪勝Й風ほ√
a = 2bsin9
f(告)
“ θ =253,0 167 or 2 36 rad
Sincc θ can■ ot bc ncgativc,rg∝ tθ
=253
ltls assumod that θ ls ttal:,reJ∝ :θ =236 0nly l possiblc answer For θ,ic θ → _167
Thcrefore,
座
e arls“
はI=事
2
3
√=髪 ・ nC"釧 Ы 1・
5∼
[″
:(:)+醤 (告 )
÷=:+書 +轟
=22 
n∝ eぉ
+32
に 鳴√=3■ 需=器 ぃ mttr a¨
.1"r1 z2"( + n
rU(
=
*
\)La
1
(,  z))r(, 
I
)
 r"lr  f
=ll,zlzr,.rt
]
ヽｌｉｌノ
″ ハ ２ ⇒
zu(
*1^21
wc→
身七 げ 蒙
+42
_
o
′ l1:)力 断 能 に 面 証 n3争 /=π
2ヵ
=π
2
hll{n6\ " '=0 Sumequal ,2 tozrro = n=lotrn=6 Since a> 2+ z = 6
2_子
(α
#=0⇒
_4;)・ hOWn)
(α
=π
(r lXz6)
2ヵ
=″
(α
)
″
(a2_半 )=0
⇒た台 ・
(―
iatteCt“
)
一２
エ
く
13′
Substiauting back into ,/,
(ii) y'
=
f(1)..2
く O When力 弊一普π 千こ mttmum“ 力 n“ =台 eFore,/お
`
えcD
︲ ︱ ， ｏ ⋮ ４ ︱ 〓︲ 一 ︲え ︱〓 鍬
(iii)y=F。 (χ )
′︱ＩＩＩＩＩ、
E.
Using GC in pcramctric s€tting
Changc WIN∞ W
setting to
鵡m=:,T―
=号
Graph wi:l bc thc`ヽ op''halfofan cllipsc with horiЮ ntal dialnctcr 4,
(11)′ B=夕 :+2k
° G市 en acute anglc bctween 4 and ち｀ω
vcdi∽ l dlarnetcr 2,ccnter at(1,0)
,
When x=0,θ =― cos 6Oo
者
:WhCn,=2,θ =を
Arca und""ph=fッ
=
と ,
法
=[(∞ Sθ )(2ぃ θ)“
JI,IV *4 =2q
7.
i.
y
=2f(x+t)
 q =fl
=二 (2∞ s2θ )“ =[(い
"+ツ
=[」L;22+θ ]:二
θ
=2:]+:; unitS2
グθ
=2∞ sθ
=:∫
checking graphs with GC;
NT
"r{r,
iv)Orexissif&94=nLargat ( =(o,0) Solve h(x) :0,
″
∫ 万寿圭島τと一 夢薫七 と
k=s
=τ
‐
:∫
3
激 万≒圭告τ激― :∫ F再 き 可乙
‐ 叫
＋ セ ︱＞
一 一
一２
＋ ＋
１
＋ ＋
薇 行
一２
一 〓 一
宏 ”
︱
車小 プ ︲
肺
激
憫躊
=か
ｇ
nt _"=r{,,
l(ff)c*',r* ″=:∫蒜
∫ 轟
●
["'  l,)
=[(*.4(t"(*"))] Jsinra = [1...,)(r" (.,))] +cosl+c
0
０
=2=l
d
[(*.,)(tn (*',))]
“
=
x)
x ln(cos
v=sinr
f*=oos,
cosr
１ 一３ ／ １ １ ヽ
J sin
Suppose graphs intersect at
一３
d,
〓 エ
dz _ sin x
１
q t") I sinrln(cosr) dt trt r=lo(cosx);
dv dz . y=?r=,1l ll. dxdr Substrtutrng
.rn. ldv = y t,
=
t(4,\,=,, (:r)\dx (zx\ )
x2 y 
(dz \  l j+r=r2 l ) \dx dz _=z_l dx
l t a,=fia, !
Jl lnlz
r+s, lz  rl= e"' z
y
+
r
ll =
r',/hgre c is a constant
= Ae' +l ,where A =
= Ae' + I
=
y
=I
r+,4e'
When curve passes through
(ii) Maximal domain = (
[email protected],01 o. (iii) g(r)h(x) = 0= s(r)= h(r) Sketch the graphs
ofh(r) ard g(t)
For ho)tO be deFlncd, ェ>■
:W晰
.v
=
[0,co)
When curve passes through
(Notice thatj/ = I
on the same diagram. 】=:,h(χ )=2
t(r)
y=lr+e')
r
!e' (shown)
A=0 = y =ly (0,2),,{ = I = } =   r +e, (1,0),
is an oblique asymptote
for
12
lz
+:+ ril
<
0く arg(z―
lr
+
Jzl = l, (::i)
(3■ )≦
<
0
J:
烏 争
鮨― ■ッ =3‖ tギ￨=:,
=1「
=:
多
ず
≒
ァ
Im`zヽ
Eqn: ッー (―
ー :)=:(エ (3))(or uSingy="r+ε
7 ノ =T・
)
1
+I
J tt AsI→ ,奇 → 多 … ∝ 4;11r...6o."'dx Thcrefore, any tangant to the curve at the points whe.e
x>l+Ji and r
(iii)
data.
0.
For the appropriate model, calculate the values moment correlation coefficient.
I2l
ofa
and 6, and find the prodrrct
l2l
Obtain an estimate of the probability of a 19 yearold woman giving birth to a Down's Syndrome chjld. tII Comment on the reliability of your answer. ttl
[Turn over

9
l0
The average number of errors transmitted over an intemet data trammission system is known to be p per day, where p > I . It is fourd that the probability ofthree errors being transmitted in one day is equal to the probability of twelve errors being transmitted in twelve days. Calculate the value of p. t4I Company ,tl uses the above internet data transmission syst€m, but Company I uses an altemative system which is known to have an average of I error per day. Random samples of 60 days and 50 days are taken from companies I and I respectively. Find the probability that the differenc€ between the mean number of errors per day for the two companies exceeds 0.5. t6l
ll A film distributor I I 85 tickets a
claimed that the tickets sales for a particular movie hit an average of sales, x tickets, for
day. A random sample of 56 days was taken and the
each day were recorded. lt was found that )(;IOOO1=9SZO and
)1,rooo)' =182043e. (i) Find, correct to 2 decimal
places, the unbiased estimate of the population mean and
variance ofthe ticket sales.
(ii) (iii)
t2l
Test at 2% significance level whether the distributor was overstating his
claim.
t51
The ticket sales for a threeJay sneak preview were recorded instead. State, with reasorL what test you would use in this case. tU
It is known that the movie was screened over a period of time and that there was a difference between the ticket sales during weekends (Saturdays and Sundays) and the rest of the week. Suggest an appropriate sampling method that can be used to obtain a random sample of56 days to assess the ticket sales, and describe briefly how this can be done. l2'l
12
A
Physics examination consists of two parts: a written paper, and laboratorybased practical assessment. The marks obtained by the students taking the examination are normally distributed, with means and standard deviations as shown in the following table.
written paper laboratorybased practical assessment
Standard deviation
Mean 48
23
61
8
Three candidates are chosen at random. Find the probability that exactly one of these candidates has a written paper score ofat least 53 marks and exactly two of the candidates have laboratorybased practical scores ofnot more than 60 marks. l4l
[Turn over
l0 The point entry requirement ofa Physics course in a local university is calculated based on the sum of 70% of the written paper mark and 30% of the laboratorybased practical ass€ssment mark. To qualifu for the course, a candidate must attain at least 70 points. Find the probability that a randomly chosen candidate can qualiS for the Physics course in the local t4l
university.
State the assumption you have made assumption may not be valid in
in your calculations. Explain briefly why
practice.
this
l2l
13 The time taken for a certain brand of solid fuel to bum completely may be assumed to follow a normal distribution with a standard deviation of42 seconds. It is found ttprt l2o/o ofthe solid firels took longer than 379 seconds to bum completely.
O (ii)
Find the mean time taken for this brand answer to the nearest
second.
ofsolid fuel to burn completely, giving your t31
An adventure company buys 120 of this brand of solid fuel. By using a suitable approximation, find the probability that more than l8 of this brand of solid fuel will last longer than 379 seconds. t4l
A shopkeeper repackaged and sold this brand of solid fuel in boxes of 12. Each box needs to be exchanged if more than 2 ofthe solid fuels do not last longer than 2E0 seconds. Find the probability that a randomly chosen box ofsolid fuel needs to be exchanged. t3I
A boy scout and a girl
guide buy
l0
boxes and 5 boxes
of
such solid fuel from the
shopkeeper respectively.
(iii)
Find the probability that the shopkeeper needs to exchange 3 boxes of solid fuel with the boy scout. 121
(iv)
Find the probability that the shopkeeper needs to exchange a total of6 boxes ofsolid
fuel.
t2l End of Paper
ITurn over
21X17J2Premm Paper 2sollltion t r6 ■ =■ r3 ⇒
5α J=α
α′
3
〕 ″ ― αr2
ソ4=つ √ 2 4e r
子)for″ =0,1,23 ソ=矛 」 ― (等
「
⇒ (″ 2)(ノ 1)=(″ 2)(「 1)
Sub″ =0,1,2,3 into abovc,
Since α,「 ≠ 0;
Obtaln w=輌L“ 考
3_2′
),vЪ 70)=0.1331●
Щ
133 ︼
Assumption ls that″ and ι arc indcpendent
・
Not va‖ d because the pcrFonllan∝ oFthc nrst paper
would afFectthc pcrForrnance ofthc sccond papcr 一 ・
i3 i
P(r>379)=012 P crく
379=088
379μ μ
=InvNorrn(0_88,0,1)
=3297=330(nearCSt S―
nd)
1/No oFsolid ttels that last:ongcr than 379 s∝ onds,out oF 120 /∼ B(120,012)
“
since″ =120 is large,4ρ ‐ 144>5,べ /∼ N(!44,12672)apprOX
1‐
2)‐
1056>5,
P(Y>18)=P(Y>185)cc =nomdf(185,1099,144,V12672) =01247=0125 P
(X< 280) = normcar(to', ZtO,ra.l,+Z)2)=卜 КM≦ 2)=01616=0:62
S― No oFboxcs that needs to bc cxchanged,out oF 10 S― B(!0,0:616)
P(S=3)=0147 ofboxcs that needs to bc cxchan脚 ,Out Of
し ，・
S+G∼ No 15
S+G∼ B(:5,01616) P(S+G=6)=00182
し
し
YtsHutt JutttoR Cotleee
PRELIM:NARY EXAMINAT10N 2007
H:GHER 2 MATHEMATICS
9740′ 1
PAPER l
17 AUGUST 2007
FRIDAY 0800h1100h Addttbna:matena:s:
Anttr paper Graph paper LIst of Fomulae(MF15) り″JUNЮR60“ σε rrytlI´ υttOR COLLaCε /rsI● I´ υvρ R 60ι ECε 6″υ″′u″ ρRcο ′′ E ИS″ uV′υ″oR Cat` υⅣOR 60“ ECε Z6″ Oε Z6″ 勧 V′ ι″OR 60LI ッ″′uvoR σOLCF rrSrfυ “ 切 Vyt/vroR COコ “ "′ ′υχЮ RCOttCε ИS″ υ″′ι″70R COZLaCF ИS″ uv′ υNoRGOι り″JUNЮR60“ aCε ′ u、 “
=駆 ワ″JUNЮRCOttCε Z6″ υ″ノυNρ R COLEECa りV′ 勧VOR σOELECF ИS″ υ″′υ″OR CaZaFcε
И S″ υ″JUNЮRCO“ ε 77S″ 勧Vyt17VrOR COL`` Z6″切 V力ヽ■ORCO′ ′FcF i■ttUN′ υMOR60ι
″S″切 Vノ υ″り OR COコに Cε ″S″ uvコ ,wOR GO“` Z6〃切 V」 07VroR caLι FcF rrsrr● N′ α″ゎ R COι υ VOR CO“ εσε ″S″ υVノ uVρR 60at“ E ИS″ υⅣ ヽЮ RCO′ ′鰺 E″S″ uvノ υ OR 60LII` :ノ ″′ υ″OR COLL“ ε rrsrrt/1 `も juヽЮ RCOttσ ε ИS″ uv Jυド 勁VJONЮRCO“ ECε “ =OR COι /1SI●Iノ υ Vゆ R ″S″ 助 Vノ ιMり R COι ιFCε 4S″ 勧V ЛttWOR COι ` 勁 V′OW´ OR COム ιECε `ISHONノ ′ Fcε V′ Vノ V′ /1slaVノ aVOマ 60ι ι ε /1SI“ υ″ OR CO“ ε σε /1SIt/7V′ 切 αR COι `ι 勁 0い OR CO′ ヽ ′ひψ ″s″ avsu、森 cOLLε σε ″S″ひV′ V′OR COι ι う “OR COLaε Oε tts″ ひvノ υMOtt CO“ “ ε 7/1 JfttVρ R CO“ εCF=団 ヽ力V′ ′ 力 V“ つ′rr2ι ιattε r団tヵ V″ ′ ″70R COι ιεOε y=SH(力 V′ “ V彙 つR COι [ ″ρ R COι ιε “GF risH■ /1′ 〔
劣鵬
雛 翻脇洲器鍛震:↓
"笠
`力
T:ME
3 hours
READ THESE INSTRUCTIONS FIRST Answer all the questions. Give nonexact numerical answers correct to 3 signmcant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers ftom a graphic calqJlator are allowed unless a question specilically states othenivise. V1/here unsupported answers from a graphic calaJlator are not allowed in a question, you are
required to present the mathematical step6 using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackels I I at the end of each question or part question. At the end of the examination, fasten all your work securely together.
This question paper consists of 5 printed pages including this cover page.
1′ υ
2007 YJC Pre:iminary Examination Higher 2 Mathematics 9740 Paper l
l・
L'
The eqtation of a curve
* o'+=#where x>0,y>0 and t
is a positive
constant,
$Y
pina the coordinates of the stationary point on the curve, leaving your answer
exactly in terms
oft.
I4l
[You need not determine the nature ofthe stationary point]
,.(iU
For the case when
&
:
I
, find the equation of the normal to the curve where the
curve p― dlough tt polnt←
h exactお
that satisfu
l
χ
χ
―
Show that″
2_″ +2
=;戸
罵百三￨「 「百
QFindancxprcssionintcrmsofノ
ど
三:争}三百万
V forthc sum of SN wherc
・ 丁 赤 石 覇
Dcducc that sⅣ
12]
ご
ミ く ｀ ミ
[2〕
l・
︼ 一 ︼
=,::戸
+4≦
ナ‐
●Ｊ
SⅣ
工
、 く 、
一 一
2x+4
く ――
＋ヽ
fina ,n" range ofvalues ofr
３
Without using a graphic calculator, show that x2 + 2.r + 4 is always positive.
Deduce thc range of valucs ofχ that satisfy 2ο
/
鳴 m図
、 一 一
@
hg yOW昴
１
{ fi
希)脚
︼
The first three terms in the series expansion
aeE
and 6> 0, in
争+手 ‐
︼
asccndhg"wers ofF arc:―
of (o+r)b, where
・
r.
+…
[5]
理 糧 輩 メ
胤
年
xb晰
巾 膿 xPanSiO由 軸 よ
[1〕
Page 2 of 5
ヽ
｀
￨
2∞ 7
YJC Pre‖
mh口 呼 Fャ コmination
Higher 2 Matheantics 9740 Paper l
Ｐ
ガ
●ヽ ５
Sketch the邸 叩hs of y2+4χ
;liif
2_8χ
=O and y=三 生(2  x) on a single diagram.[3]
On your sketch,sIЮ w that tllcぃ o graphs intersect at
χ=0 5 and χ=2.
By using thesubstitution sin z = r  l, n"a frfi  1, 
rf ar.
‖ね E4]
2_8χ
=O and y=π 12χ )is
＋ π 一３
“ メ ｉｌ
。f y2+4χ
√丁
tt the exacta暉ュ ofthe region bounded by dle x‐ axis and thc graphs
夕りShOW
14]
rl Sotve z4 = S(JS ;), f.rlring your answer in exact exponenrial form. .6 ' ,,i{ oru"nt}rat (a + i)a = *(J,  r, show that one of the roots of a can be CXprCSSelas(c。 (Sin(」 :)+9)),WherC′
and 9 are∞ nstantS tO bC ｐ
(;:)―
｀ ヽ■
14]
determined.
The set of points in the tocuslz + I  24
Find the exact maximum value
arglz・
≦ り 子
ｐ
vCn by:≦ 〃 SbW deJy on a Argand dagralnthe bc厖 」
. , li". in the regionl s arCQ+t)sl.
ofr.
t3l
ヽ
The equations of the line /1 and plane
.
t1,
flt:.
(if {iif
If
.
rI
I are as
follows:
=[t,l.J],],**.2isaparameter;
[*.1
l.oJ
xa+ z = 5a+4, where a is a positive constant.
the angle between /1 and
Verify that /1 and
fI
I
fI I is1, show
inrersect at the point
that d =
[3]
l(5,1,4).
(/T) civen that C(7,3,4), find the tength of projection
{fi
l.
oflCon fI
[2]
firra ,n" position vector of N, the foot of perpendicular of C on fI
 (v) Point C
is obtained by reflecting C about
the line that passes through
fI
1.
[3]
1.
I
.
[3]
Determine the vector equation
of
lC
t3l Page 3
of
5
h葺
′
ヽ
イ ′
2007 YJC Pre:iminary Fvハ minat10n Higher 2 Mathematics,740 Paper l
￨
h… dttm血 子 +子 ギ ー ー 品ヨ 満 制κ相 q
讚
forΣ]デ
The cunre shown in the diagram has equation y =
f(r). It intersects the axes atx
= 3,
一９
５
︱ ︱ ノ ヽ︱
ー
３ ／︰ ︱ ︱ ヽ
ェ=4,y=L2and has a maximum point at
[3]
:T‐
﹂ ﹂ 、
9
イ
Usc your rcsult to flnd an expression in tenlls of″
ノ
y=1
ヽ
χ=2
χ=5
On s€pamte diagrams, sketch the graphs of: ３ ３
G) /=fftl), V(ii) y2
[email protected],
３
I .tr(rr) /=7G+DIn each case, indicate clearly the asymptotes and the coordinates of the intersection
of
the axes and tuming points whenever possible. Page 4
of 5
１ 、
︱Ｉ ＪＪ
ヽ
l0
2007 YJC Pre‖ m:naw Examination
Higher 2 Mathematics,740 Paper l
In a certain farm, the growth ofthe population of prawrs is studied. The population
of
prawns at time t is denoted by x (in thousands).
It was found that the rate of birth per day is twice ofr and the rate of death per day is proportional to x2.
(i)
Given that there is no change in the population of prawns when its popularion hits
*=rr*
'dt5
l0000,strowthar
(ii)
tzl
Its owner decides to sell away 1800 prawrs daily.
Modift the differential
6q'ati61 above and show that the resulting differential equation can be written as
&lloYrl [21
Given that the initial population of prawns is 13000, solve this modified in terms of t.
r
４
diflerential equation, exprcssing
１
Deduce the long term implications on the population of prawns in the farm.
The functions f, g and h are defined by
ixい
学 ,xく
0
g:r r. J2r, x lnx, 0 13.4
14.
+...=ett' *
"'ru'
*
"'"'
*...
,= f,Tr_t
elgu" etBu
/ "\ \u^lI =e tcl
Since
r is a constant, the series is a geometric progression.
= elg"
'ttlrt*l.t l < elc' 0‐ 75
12P(7"く 80)>075 P17"く 80)く
等
― ■ α り ″ ′ ο ・ 重>l15035 UJngル ッ つ平く バ PIZく
￨「 )く
0・
125
,一
″>64_84
Least″
10a
=65
15035⇒
7
Solu6on
Qロ
Hr: P*6 Under I1o, the test statistic is Z
=I+l{(0, l) where p=g,6=Q.g,
/J;
l,evel of sitrificancn: a = 5Yo We reject Ho tf 14>1.96 and do not reject the claim
i.e.
1.96
5.5(X
i.e. 10b
.Ir'.
if
n=le
lzl ι 05 )=0・
we haveア
⇒ P(χ
ly.
57く ―ο)+P(χ 57>ι )=005
⇒ P(χ く57c)+P(χ =〉 2P(´ Yく
>57+c)=0.05
57ο )=0_05 (by symmetly)
⇒ P(ア く57ο
)=■
望 =0‐ 025
⇒ 57ο =5467576863(ushg hvNonn(0025,57,Nり =,ο
25/160))
=57_54.67576863==2324231373
Hence,ο =2.32.(3s_1)
If the scores had not been known to be normalry distribute4 we can still apply the central Limit Theorem to estabrish the normarity of F since z: 160 > 50 is large. Hence the calculations in (ii) would still be valii.
2∞
L踊 」 鰊
郷31電
:り
:締
曖
12(1)
(→
Normal model inappropriate as weighs
of
babies bom belong
to two different
distributions.
OR Normal model approoriate as the weights of babies belong to normal distributions with means and standard deviations such that therc is a huge overlap of the probability density functions as shown in the graph. The required distribution would be unimodal and symmetrical. A normal model would thus be appropriate for the weight of a randomly chosen baby.
12(b)
Normal
as the sum
of independent normal distributions is normal
Let C and β denotc thc random variableS rcprcsenting the weight ofa baby girl and boy rcspcctivcly.
Thcn C∼ N(2.6,016)and B∼ N(29,036)
3_q+G2∼
N(0■ 036+キ
に卜∵ 型鳴Qり ≧ 05)=0.382 P(3≦ Liデ
13(D
χ=13
:)
ダ
g二
=),Ex=13(8)=104 ⇒ 10+15+11+α +13+14+b+17=104 ‐ ¨¨・ ・―(1) =>α +b=24. …_..・・・ ¨¨¨
2∞ 7 RJC JC2 Preliminary Examinalcn Paper 2
H2M」 Кmatlcs
9740/02Pagc 8 oF10
s2=21 =)'EX2_:(DEχ
7
81
=〉
)2=54
7
Σ ],2_:(】 EX)2=54
二 ){102+112+1524α 2+132+142+ら 2+1,2}_(104)2=54 =)α =)α
2+ら 2+11001352=54 2+ら 2=306.._¨
¨ ・ ・・ ・・ ・ ・ ・ ・ ・ ・ ・ ・¨ ・ ・¨¨ ・ ・ ・(2)
Subsdtuting(1)int。 (2)yieldS:
(24b)2+ら
2=306
b2_24b+135=0 ⇒ o15xb9)=0 ⇒
⇒ b=15 or b=9 Since α>ら ,thus we havc α=15,b=9
Since the population width is normally distributed, we carD/ out a two+ailed rtest at 5o/o level of significance.
Let X be the random variable representing the width, in mm, of a beetle of 'genus prometheus' species.
: p 11.5 Hr: p*11.5 H,
Since o2 is unknown, and z is small, we estimate o2 by Significance level is a = 0.05 .
Under
,'.
flr, , = !.r+  rα fOr a∝ 寵
血
of″ then
颯
,
“
O WritedOwn fdは
e‐ 8と
131
.
11〕
(lib ShoW that uに dittmm● Ol.quatiOn メ 3ノ tan x=2smχ e嘔 may be reduced by mealls ofthe substimonノ
=.s“ r
to
=2 slnェ ∞ sx e喘 お
器
.
Hcnce ttd the gend sOl面 。n OF the direredal equatiOn9 81Vlllg yOur
answerin a fOllu cXpreSsingッ in te― Ofχ .
[6]
■ ０ ＜
８
FindJ器 こ
‖
■ ectwe Cis demed P_icallyり 2
X=(1+′ )3,
ッ =In(r2),
′≦ _1.
OhOWthatthe● ●m ofthe“ JOn enCIoed by c缶
Can腱
句寝d
… be daermincd.
1饉 ″=Qx=l面 ′ andら 氣 )d′ ,who α l i
htheお m■
Hcncc cvaluate this area
the
"cOnstmts b t4l
ttl [Turu over
ll
,
10
(D
Givcnthrr(r)
(lD
Henccfindtbc sum toatermsofthcseri*
(tD
Dacrmirc thc smr to iofinity of thc scrics. 
(tv)
Showtiat r
=rh,showttst f(r l) 
***f;+...
Thc position vcr.l) ".111 F'.s
e(,
p(x1tl: o.qlr]l P (I € g'l . o.Sttrrq
"' t,rl ro'[
si X.
iflt',,
1:
$
1"t1
t
50) is uken from the population and thc sample mcan is denoted by X Find the least value ofn such that P(17 ,4 < o.s1>
o.es.
.
t4l
︱ ″
” ・
8
con mption of a brand of car callcd Missan is known to have a mean of rr krn pcr Iitsc. A ncw device has bccn intoduced in the manufacturing process to The rde of fuel
improvc thc car's fucl consumptiot rate. A sample of 100'improved'cars selccted gave the following
datr:

コけ17)=180
ρ 裾留識瀧躙
ボN Xttσ
ぃ9 、
[email protected] =25e
準載 鯉T:Ld
閾いtandけ 品eXP―
:誦 酬
6 AMg淵
ond
and“
und ulat“ de階 お お
iOn`at 50/● sign」 Rcance
level'in dle oontext [ll
譜 鳳癬辮躍畿霊鳳電 拙認
e“
もib■ n"ebJL 」
A piayer chooses 4 distinct digits randomly from the digits{1,2,3,4,5,6,7)
Thee digi徳
“
then manod ageinstttt digits On“ ,balL drawn '山
shw m ttpmbめ 市け J血
play∝
h宙 ng蜘 」y3matchingdJ埜
A and B are independent.Itお
P(A)=6.25
jI
P(A
[31
that
^B')=0.15.
ShowthatP(B)=0.S00
Thc errcnts B also
ガ
and
」ven
t31
ari C 8re muEally e4clqive
occurs.
Findヽへ B)State
.閣
■nd the pobabnity Of him ha宙 ng
Given that the player has atlost i matching digi、 exacdy 3 matching digiも
ヵrmheeven、
唱
i '.
and wlrencver event C occurs, event A
C"
tt largest possiЫ e charu that event C wi‖ occur.
t3I
―
ヽ
ー
:,
T?Jc aool ?"tm tk
(trz ttnttr P"f* :)
￨
￨
￨
,
The number of people applying for the course in Biomedical Science in a particular university in each of six sernesters is given below. Semester (.r)
1
No ofapp:icants(ハ
2 29
3
4
5
6
68
138
215
5“
The admissions office believes that the number of applicants (y), and the semester (.r) are related by the equation ,y = ,413f , iryh€(r and are conslrnts.
I
,,if .v 好
Using a suitable trarsformation invotving Y = lqg y, givc a sketch of the scafier diagrarn. Explain wtrctlrcr Ore scaflcr diagram provides evidence that the relation is a reasonable modcl. 13] ind ttc"uatiOn of ule ttmated l鵬 of regr6sim of r on x,and tte least squatcs estimate
y'( e*6rin
'
I
ofl.
121
whether the correlation coefficient supports tlre reasonability
model.
of
the 121
The'admission office realised that they had accidentally left out the data for a'tpecial semestef when the course was also open to applications.
奮て 9.r
\$
， ， ． ヽ ・ ・
一
66O* inserting the data for 'special sanester", it is found that the equation of  regession line of I/ on.r found in part (ii) remains the same. The equation of lar,'r.ntl\ regression line of x on I' is given by x =
[email protected]  0.297 . Both the line of \vr0 lt\ \\t{ regression of .x on I ard the line of regression of I on : pass through the same
J
:''fl\a
point (i,7). Show that 7 = 1.98, corrcct !o 3 significant figures. Hence, find the number of applicants in the 'tpecial semester", correct to the nearest whole number. tsl
[l}q**tr'' t
( i. ', f.7
ヽ ヽ
k
A piza @mpany receives an average of t.2 phone orders for pi"z^ delivery per minute. the probability of receiving.more J Calculate interval.
tlran 8 phone orders in a five minute I3l
jlzThe 
compary's wo*ing hours is from 9am to 9pnr. Using a gula&lg.epprorilEalis0. find the probability tlut th€re are 8t least 20 fivo minute intervals in a day that have t4l more than I plpnc
orders.
The pizza. company sells two different types of pizza with the ctroice of additional ue lgrmdly opping The urcight of the pizas and additional lopping per di$ributed with ttre following means and stardard dcviations:
pia
仙 一
Man(mms) Pレ
"A P:― ・ B Additionalわ PPing
475 50
Standard deviation
20 30 6
per p:77●
the probability that lhe lotal weight of 3 randomly chosen Pi:za B, all with chosen Pizza_ A
' iiilFind additional topping  differs from four times the weight ofa randomly by at least
5Og
t61
ヽ“
TPJC Solutions for H2 nthematics 9740/01 Prelininarv Examination 2∞
g O CscL C"面
=轟 =30 “Φmd¨ 画、
Case 2:Conぃ nS Φ and one triple、
Qse■ Does tt contah
=]=20
Φ=轟 =Ю
Total no ofdifFerent codes=30+2Cl+10=ω
O N● '
Q2.
y=
FE2
oFcod●
=]+1=4
+qx+r
y=′ (X+1)2+7(x+1)+″ =′ 2+2,+l)+9(″ +:)+r 。
2+(2′ +9>+(′ +9+′ ) 2+(2′ +9),+(′ +9+r)] ノ=2[メ 事 =メ第
=2pF2+(4P+27),十 (2′ +2,+2r) By companng― 価 cients Ofx,
4P+″
=3′ 2
nstant 濡詰 ン庶話∞ 2P=9′ 2P+29+2r=l
ing灘 :第
ユ1
①
=8
∴′ 子:9==,″ =器 =―
Papc i Rellm
TPIC 2007 H2陥 “
1
7
︼ 一
Q3,
ul=",u2=‰ u3=93,Щ =90
i) 1)
u:=1023(1) ●2=1023c2)
=10230) u4=1023o)
uっ
.・
iiO
.un=1023● where【 →=3■
I.€t P' be the st8tr.leot rh = t0Z3n When n = I, LHS = 99' RI{S = 99 .'. Po is hrc' l(}2  3k Assnrc h, is truo i.c.
rt:
whenn=k+1,
u.*r=q3=1023k3
= 1023(k+ l) Pl.r is true Herrce bY irductiorL % is Eue' Pr is tnr +
剖
／１１１１１、
５ ９
９ ９ ５ ０ 一 一２ ・ ２
９ ８ 一２ ３ ９ 一２ ０
２
5 y う
１ ０
万
2
０
∴X十
リ 哺︲ ０９ ︲
by rrefusing G.C
is reduced to
白 Ｆ い ｌｏ ｔ
︲ ３ ¨ ︺ロ Ｈ 日 Ｍ ・３︹ ・５ ︲
ιｉ ⇒ 口︲﹄︲ ヽ り︵ ｌｌＶ ⇒
α
TheaugmentedmatrlXII:: 12
―
95 Z=扇 38
OZ=う 百 Let z=29λ
x=万 2λ , y=ラ +5λ Since the plancs intersect in a linc equatiOns rcprcscnt sheafofPlanes,
TPJC
2m?
and all planes have different normal,
t{2l,lr6t P.pcl I Ptdim Errm
１ ０
０ １
０
０ ０
０
１
Augmented matrix is r€duccd to
９ ５ 一９ ０ ２ 一２ ・ ２
When b becomes 7,
The plures have diffcrrnt nonnal and thOy fonn“ a triangtllar pttm.“ nce equCiors have no solution.
Q5 o OraphiCal Medlod:
Xく
T2 ° ″ χ>4
(bXi) 1 2■ 1 >―
2
―― ■2
轟― ∴,0 >0 品
(2xlxx2x3ェ )'0 ∴
ェくO
o′
1く χく2 2
(il)
e'5,nq=144o.84724)=122>5,
npq=18637 C∼
N(22032,18637)apprO対
mady
P(C≧ 20)=P(C≧ 19.5)=Q721 ―へ Pi"n B with additional toppings. Lctへ B and T be the mass ofPi″ ″
A N(400,201
,B
N(
475. 30,
),
T
(50, 62)
N
Let W be thc total weighr of 3 randomly chosen piz,a B with toppings.
W=Bl+82+83+Tt+T2+T3 W
N(3 x 475 +3 x 50, wN(1575,2808)
wP(lw
4A

3
x302+
3
 N (r57s  4 (400),2808 N (45, e208)
x
52)
+
4?0))
4Al> so)= p (w
4A> s0) + p (w 4 A o
Thustt mud mm鯰 =2ン l
Heno′ ■0.2.
1lQ¨ +0. oN中 野 "が ),dlllsT>等 ゥυ マ卿ガー Using GC,″
6
=12
o Disagrcc Onc口 c∝ oFdodl
can have more than i deL∝
(iD ThC OCCurrcn∝ s oFthc dcFec happen r"domly The number of dc急
ガ ve
spots on a
腱淋 2stt 織 罵 嵐鮒篤瀾臨ザ』 富賓‰ll驚盟上 Thus χ ∼Po(1) Thus P(χ ≧1)=1Rχ =0)=0・ 221
Thi221%ofi m2s■
“
dm
11‐ pOiSSOnpd《 0.25,0]
cbdt wil have deFective spots
Lct r bc dle number ofi m2stt cioth out 6f 100 pに os ulat havc tO be d`eded■ hus y∼ B(100,0221)Sin∝ =22.1>5,4=779>5,thus r∼
"=100 is iarge,4ρ
N(22117227)apprOX
prob rcq電
=P(y=20) 20.5) →′ (19_5く γく
nonlndcdt19.5205,221,V17227)] 〔
∼00844 7
(→
no oFwayF 6!x2'x10=576∞ 3
2
5
(
０
(ii) P(getting a matched prh in 3 socb)

５ 一７
〓
０
=1
ヽ１１︰ノ ３ １ ′︰︰︱ヽ
ヽ︰︱︱ノ ２ １ ′︰ ︲ ︱ ︱ ヽ
=l P(all 3 sodis are of diffcrcnt coloun)
(iii) PBd sock docs not givc matcly'll 2 not matdreo Page 3 of5
18
一 ・
+Cl ゴ p=器
8
Lt χ be tt nulttber oftotint out of 10 who wili buy ati“ Ot Onc`Special Discount'疏 nus χ,B(lo,D・ ・ IbinOmdfoO,1ノ 3,7)1 四 b・ req電 =P(χ ≦η=0'リ ワ
S=4+… S∼
礼
・
Stt
Цり
=Pa∞ くSく
,V《 η
Ltt By Cenm Lh■
. “ 2803=0.341 lnomalCdfo吼
N愕 響)卿面4・
prob.roq電
響
1肺 軋
28Q800′ 3,■ 600′ 9)]
血 most7 who boun ttlettone`S画 」 Lt y be me numberOf口 四 眺 ∞unt'itcm Thus r― Bc80,0.997D. "out Of80 dlat誡
Sincc"=80 is iargc.″ =79.76>5,"9=0.24く l pЮ b
requ=P(γ =7つ =К 80y=3)=0∞ 181
〔 PoiSSOnpdf0
24,3)]
during the month of June at the particular
@
9
5,80r― Poco.24)● pprOXimateiy.
Ho:p=2!.9, Ht:P>23.9. lrqr X & the mirlday tcrnpcraErc &ring the month of Junc at tlrc patticolar plt(x'.
placc.
Level of significucc = 0. I 0. Assume thet X follmrs r normrl distribution.
Under恥 ,Z=
χ239
∼N(0,1)
R● ect rfo if「 v」 ucく 0.10 Perfom Z‐ Test OtoOSe Dattん
=239,σ =23,Lお L
Ll,′ ‐ valuc=0."8
Since,valucく Q10,thereお sumcient e宙 dencc at 100/O signinmce levet to rciect鍋 and wc cOndude衝 饉 ute mean mid‐day tcmperature during dle month of Junc at uに
s…
particular plg"l“ (i11)Assume山 ばχfollc●
Under謁 ,7=
.
a no―l dbHbution■ ith unown vttancc
χ23,
Rci醸 あ らrp‐Vmく 0.Ю 11‐Var stats Ll] Tct OoosC D江 ヘ プ PJorln T‐ 喝 =23,,List:Ll,P‐valuc=0.2ω .
s■ 4.53529 Sincc P‐
vduc>a10,6¨
b insurldel■ cviden∝
at 100/●
扇b md We conde dlat he mem mid‐ day tem口 ture the partida口 att haS nOt changed Samplc standard devialion=4.42055 Sin∝
the sampic…
gniflcance icvel to tted
dШ ing the mondl of Jullc at
“
11‐ Var S'n,C Lll
deViation`dmost"vi∝ the nccumcd valuc of 2.3 for dle
standard dc宙 ation of χ dtere is cviden∝ For」 に e∞ :ogist to question that the mld‐ day b:c. tcmpcraturC in Junc has be● ome more va面 ほ
Page 4
of
5
NJtrc 2oor 10
rreln
Exom
(Hz lrklhs &per
o卜 邸 ラ=Ψ
2
so{rl
=0.197x+0184
(il)ッ
ッ=0197χ
+0_184
等 抑 いη仰
″=2994 ″=09287
￨
￨
because
(iV)
(v)
r
is a measure
of thc
ofscaner and this is
lt is reasonable because Jr = 3.4 lies within the range of the given r = 0.9287 *
I
data and
,the regression linc y on .r is almost identicsl to .r on
Lctχ bc the amount oficmonade ddivcred (, prOb・ rcqu=P(χ >275)=00668 00 Pcχ く25o=0.158655
/.
χ∼N(2α L102) nomalCdf(275,"%260,lの 】 〔
Lct y bc the nulnbcr of cups out of Flve dlat oontam icss than 250 m:of iclnonade r∼ B(5,0.158655)
Prob.req'd=К γ≦り=0.819. (iD Lct μ be tt v」 腱 ofぬ c¨ ′
(Zく
21キ 生 テ )≦
[bhOmOdf15,0.158655,1)] nlealL We nCOd′ (χ
くわ
0≦ 0.05.Thus
005.
ShCe КZ≦ 1645)=005,
5l; / s f .oas
=
p >26.45 .Thus a suitabte vatrrc for
血 mean is 2“ .45.
(iv) Let lilbc thc total amount oflemonade dispensed for z orps. Thus ″ ∼N(20",lIXl→

p,(z
>zJi)
Whcn z is large, the required probability is approximaely 0. Pagc 5 of5
.
一 一
MttCL 2
An arittmetic progression C has flrsttcrm α and a non zcЮ common difFcrcnce a G市 en that the sum oftllc flrst 10 tcnlls in G is"宙 ce the 22日 te....ofthe scries and the 6・ term is 37,flnd thc values ofα and a 3]
〔
A new series
11
is formed by selecting every third term of G. Find the sum of the first 50
terms of 1L
Express
f(r)
[3]
=
G+
in partial fractions.
+2) Hence find the expansion of f(x), in :r, where lxl < l.
in ascending powers ofx, up to and including the term 16]
●＝”＝” ０
lS a ー
６ ´一
＋
＋ ″
ｎι
４ 一３ ｒｌ ｌ ｌｌ Ｌ ｌ 一３
[2〕
A child
was blowing bubbles and noticed that if he blows too hard, the bubble would burst immediately but if he were to blow gently, the bubble increases in size till its optimal volume and detaches itself offfte blowing stick.
鰐 ぶ F叫 卜聾 じ 鰭:鸞鷲滋 ソ ・ ´
d′
ｎ＝Ⅲ＝ン ｏ
、 子
cm3,■ nd hc dmctakcnお ra buめ
l
団
に o“ tach itserorthc bb面 電
い
slck‐
]
3] [VOlume Ofsphcre=:π ″
t_ ヽ
ヽ
ヽ
ヽ
MJα2007 JC2 Preliminary Examina● ●‐ 740/01
ITurn OVer
rYlご
′ iミ iantc
,,..
4
16)
Using the substitution x=secd,showthat
と =√ h(暑 属 二 ) whereO0)
く
: (Sin∝
0)0
For a到 ,綺
満
=:
For″
L =2, ″ 2=二
For″
=3,
For″
=4,″ 4=告
″3=士
一７
Σ4
４ 一９ 〓
３
5'
〓 ４
2
３︶ “Ｈ ≫
∴ =:,二 ″ ′ ′ 喜″
′■:
″
り
一
■Ｉ Ｊ
+
＋ ル
+l
２
2た
ｒｌＬ
=―
１
た =+ 2k+t (2k+3)(2k+t)
k(2k+3)+t
_
(2k+t[2k+3)
2k'z+3k+l
[email protected]+n!k,
_ (*+t)(z*+r) (2k +t)(zk +3)
(t+t)
=:t
':
:RHS
l2k +3)
Mlcl2oo? lCZ Prelim Exam Paper I
suggested Marking Schemer'tD Maths (g74oyMarhs
Dept
page 7
of
13
Thus
Since
P* is true
I
=
P.., is true
.
is tme, and P* is tme
=
Induction, P, is true for all n e
P**, is true , by Mathematical
Z'
.
(x3)(ェ ■2)く 0
κ C X"‖ ■ brttd“ 口 ⇒‖くつ いAQ卜￨>0∀ χCi) MJα 2KX17 JC2 Prelim Exaln Paper l Suggested Marklng Schemtt Maths(9740yMaths Dcpt
Pagc 8 of13
r °
2く 2 H)・
H く3
・
Xく
2
o r F)
‖ く3 3く χく3
AND
2
.'.3″ +05)く 03 ∴″>1972(from GC) =,lcast″ is 20 (1う
Gt I/be th" no. of days whereby no one will enter and leave the antique shop in a onehour interval out
of7 days.
W
B(7,0'0273)
P(w >2)
=1 P(w ln 3
,'.r2
A=2r ul
=叩
and笙 と=4 ″=2
= $ 26(l .04" [Bl]
From
04xlttH
uo
=15+u,
r'u,
= 15 11r,
(ii)

=15
)
Lく
⇒
為
[Bl]
く '″
+1
l)x
'$ 2611.0+'o l)x= x=
10(al X=2 sin0
$10 000
$800
[Al]
ar=2 cos θ′θ IMl〕
(c)()As n→ 3u,
⇒
[Bll
[Ml]
,1
=4 *
;2x"1>0
一 ﹁ ５
Since
5χ
5xl>2x,+l
= $[ L04x(l + I .04 +... + I .04'r) ]
acceptable
9(a)
'J
[Bl]
lAll Note: hg(x) = In(ln(e*2)), x> ln 3 is
t J6
∞,x"→ λand χ″ +′ → λ
∴∫2sin θ+1 2cosθ ′θ V44sh2 θ [MI〕
ur=J
た=―
[BI]
[BI]
512_2た
1=0
∫ 器
2cos
12shθ +D“
θ ′ θ
:
=2cos θ+θ +C [All
=2年
=清 卸 ザ■ザジに⇒ [Mll
十Sin 1(:)+C
2)%=: χ け 券 グ
2 +sln l(:)+13 = ―V4χ
:ll―
,つ %← 2xl=―
yG
JElny
0∫鶏
・
ー =器
7「
:)密
=l競
ylny dy dx ,lzxlytanyJ?:)
教 ∫ 詩
― ふ
)″ =∫ (_χ
∫ 赫
hd
ra =L a =n(tL\ \ a)
2)(vl_χ
2)教
=_χ 2v卜 x2_∫ vl_X2(_2x)教 [Ml]
教 [MI〕
hP+2x+1岩
一ノ√7_:卜 √7中 国 一ノ√7_釦 "ジ て
[A2〕
国
tBrl
Volume of Cylinder,
4=″ 2′ =π
2_ti)
[ν η
(″
Volume ofCone,/2=:″ ―■1)
」生 =滋
ね ♂(ザ )+C
IarJ
(b)
[Mll
教―∫ 満
ltanノ ￨テ
(11)
凌
[И ]
= 1n1s6r;
+(f)窪 =mツ 0讐 l 癬ν 」
II「 [Al〕
= cosy
2ヵ
[ν ]
(2″
1:i)=0
π力
[ν l]
(2″
2
″ =― α
3
[И l]
ぬ冷 力 ″ =::が う I:ll→ 〔 ;ノ ￨:4 ￨サ ]‐
12.
(i) Using long division, we have
,q
/(x)
=
/.  o\2
Gffi
=t
g0 lgx +GiX;,
624 '(r3)G+,
[A3]
(ii) Vertical asymptotes ar€ x = 3 and x : 3. Horizontal asymptotes: y = I The curve intersects x  axis at ( 9,0) and yaxis at (0, e1.
[A31
(iil)
(a)
=―
―J 一― ・
[2 marks for correct shape; I mark for indicating minimum(9,0)and maximum point(1,8) clearly]
(li) ヽ
I
mark for conect shape, point and asymptotesl
I
mark for correct intercepts with axes, I mark for correct maximum
(iii)
i2
(')'
Qx\\'z v=sgvs1= = (x}(r+D ' Qx4)8x+4 n
mark for correct asymptotes;
poinsl
I
mark for correct shape and
I mark for correct stationary
一
・
一
﹄
一
︼
．
︼
CATHOLIC JUN10R COLLEGE PREL!MINARY EXAM!NAT10NS 2007
MATHEMATiCS
9740′ 02
Higher 2
Paper 2
29 AUGuS丁 2007 3 hours
Addttonai matttals:Answer Paper Graph Paper ust of Foonulae“ F15)
READ THESE INSTRUCTIONS FTRST
Question
Marks
1
Write your name and HT group on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a sofr pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or corectbn
2
fluid.
3
Anslrer all the questions. Give nonenct numerir=l answers correct to 3 significant figures, or 1 decimal place in the case ofangles in degrees,
4 5
unless a different level of accuracy is specified in the question. You are expected to use a graphic c€lculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwiseWhere unsupported answers from a graphic calculator are not allowed in a question, you are required lo present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for cbar presenlation in youa
6 7 8
answefs. The number of marks is given in bractets I I at the end of edt
9
question or part question.
10
At the end of the examination, faslen all your work securely together.
Totai
At the end of the paper, detach the coyer page and attach it to the answer script with lte coyer page on top. Name:
Calculator Model:
HT: This document crnsists of 6 p{inted pages.
[Turn over
Section A: Pure Mathematics [40 marksl The variables.r and z are rclated by
,2" ＞
4as" =ra &.
rn
０
By means of tlre substitution y= xe' , obtz'in a diffelential equation relatingy and x. Hence show that the general rclution of(f) is
' (iii)
Sketch the solution curve for
t
= =

r"(z' * t) , r l,
121
is an arbitrarv constant'
121
21 【
stating clearly any asymptotes and axial interc€pts.
2.(a)
(D (ii)
Find the fourth roots ofthe complex number
re'dwher€r>0andr r t41 . .*iS' ,.0. mean I .65165 m and standard deviation 0.09220 m, correct to 5 decimal places. X .no\ ' i'.? O50 people from country I are chosen randomly. Using a pilqgjplloxim{q, find the t31 probability that 6 people have heights exceeding l5 m. In country B, the heights of the population follow a normal distribution with mean 1.55 m and standard deviation o' 15
[Turn over
m' 4
trt
Find tle probability that the total hcight of 3 randomly chosen people from country ,{ is more than 3 times the height of a rardomly chosen person from country B.
t41
Шm ttein h山 “ d酬 面 ぃ 維 IⅢ憮ふ “ y7f .
Tlte popularion of Singaporc (population 4 492 150) are divided into the following races: Chinese
76.7%,Malay 14.00/0,Indian 7.9‰
面 け 1.4%.A sarnple of200 Singaporealls is tO bc chOsen for
a survey
だア 篇』 ぃ 蝋 l搬 還 w路■m蝋ぶ苫T“
O blayS,50 1ndi椰
Statc one disadvantage ofdlも method
ヽ
Describe how you would carry out stratified sampling. State one disadvantage of this method.
The
lifespan,l,
in years,
ofa randomly chosen mobile
[5]
phone produced by a certain manufacturer
may be taken to have a normal distribution with mean p and standard deviation 0.5. A random sample of z such phones are purchased. The probability that the sample mean lexceeds p by at most 0.1 is more than 0.9. Find the least value of z.
./i
e
4] 〔
particular machine has been calibrared to produc€ golf balls of 5 cm diameters.
A random sample of6 golfballs prodrred by that machine were measued and the diameters in. cm were found to be 5.00, 5.02, 5.05, 5.05, 5.t0,
5.il.
Find unbiased estimates for the population mean and Perform an appmpriate test at the
2olo level
variance.
of significanc€ to check if the machine is not
producing golfballs of the correct size. State one assumption that is used in the test. [5〕
,!s\ *
r\ NSA '
{rrt\'$$On A random sample of 50 packets of the snack'Yummi Cnnch' is weighed and the mass.x in grams is recorded. The results are summarised as follows:
Σo150)=■ 0,XX150)2=25CXl.
A tt ws carnCd outatthe 5%slgnincallce level with the followingけ t't \rl0 ,1b1lb :Ψ T '
K、
/
"hes:
Hr : the population mean mass of lhe snack is < p6
ド弩 Given that
4,,;+)
Ho is rejected
in favour of H1, find the set of possible values of
15. Assume that the
distribution of the mass of the snrck is normal.
Iil.
It is believed lhat
ttre
t61
probabilityp of a randomly chosen pregnant woman giving birth to a Down
Syndrome child is related to the woman's age x, in years, by the relation p = a and 6 arc constants. The table gives observed values
ofp for 5 differcnt
χ
25
30
35
40
45
′
000067
0.00125
000333
0 0 111Cltl
003330
a6'
values
, 25
( r 3 45, where
ofr.
nd fi"a linear regression lines and their product moment correlation coefficients
of lnp on x and
lnp on ln x.
繊 ⑩
I41
Which regression line is better in establishing a model for the data? Give a reason for your 2〕 〔
n Plot on the same graph thedari ana tne regression line ofyour choice in part
(ii)
[2]
By using your choice of the rcgression line in (ii), estimate the expected number of Down Syndrome children (to the ne$est child) that
will
be bom to 50q) randomly chosen pregnant
women of age 53. Explain why this is not a good estimate.
↓
くれ い ヽ′
｀
,｀
｀^4 クめ
ヽお
ミヽ
｀く
つ
ヽ
'ゞ
E″ イグ
6
Papar
[21
Sch eme
多
=〆
l①
″ ″ lり
+χ
Vertical asymptotes are x=O and
x=士
″ θ 券
Shading the circle
揚
[Al]
Min dist=PB‐
2(a)
024_1+輌 [Al]
ッ=2χ
Max dist=PA=12
=2ノ 等 =2θ
[MI]
`121″ ′ スト 午♂ =2σ η
3oロ
+(』 ル =(ム )b=(幻
1)
2`″
z=2:′
κl'31):,k=0,_1,1,‐
2
[A2]
02)+2(2カ +1)2(2o‐ 9々
・=2x+`三 ο
z8+
χ
arbitrtt constam
2
=― l士
V3,[Ml]
l I i
[l mark for correct shape] Curve cuts xaxis at (l/2,0) and (1,0).
and z =2tr e
k=0,‐
1,1,‐
2[Al]
バJ● 0:
k=0,‐
1,1,‐
2[Al]
[Bl〕
Ⅳ い
e
6)1崩 =2ヾ 1+4+4=6
］□
=2i
[Ml]
Al] 〔
＜ 〓 う‘ ●
Therefore z
…圃 卜 ⇒
l
18
力=2
一刊
゛●0:
I
=18⇒
赫 瑚 一
(ii)24_
2z4+4=0 2士 V416
[Al]
ν
Ml〕 〔
=(ti)
[Mll
χ′'=2x2+c
"〈 "+:)a_n
8
[MI〕
=4x
″=∫ 4χ ″ ∫ 2+c
PC=、 J64+36‐ 10
ｐｐ Ⅳいい
(b)Circle centre(2,2),rad‐ 2
Prellm 07 H2 Maths Peper 2 Marking
引
︱
１
〓
Ⅳ □
=
３
(b)(D TOtal area of n rectangles
(1,3)
ル嬌 みレ =31088+2 0794
Bl] 〔
=上 ①+1(2:)+上 (2:)+.¨ +上 (2早 )
剛
=￨ll+2:+2:+.¨ +2ギ
″
″
"
″
1MI]
￨
[Mll
‐519
Al] 〔
Ⅳ
も xあ ‐ そ (l)x3(1〕
2and l intersectat χ
Ｊ
︱
the graphsマ ′ 10χ
刊
１ ３ ・ ２ ・ ・ ロ ロ ︱・ ︲ み 〓毛
＞ “ ｖ ↓ ｉ ＜
4(→ ぐ )USing enher Cc OrnOn‐ CC method,
(iり
We rOtate the curve
2+(χ _マ 幅 )2=10 abom血 ey・ α鬱 ッ
Therefore equation of plane is
instead.
[Al]
藤働 剛 ① ¨ …卜 proJection ofline BD on the plane=
V22+卜 √
IBl,Al]
Th“ ,x=V75lo_ッ
2∵ χくV75
Bll 〔
(iり
0 1
Volume required=
π∫,2の =″ ∫(V而 ― V10ッ 0
Liml ofA=∫ 2X″ =1.44 or
2)2″
in 2
[Ml,Al〕
0
=952
[Mll [Al]
5(i) P(all 4 players arc Singaporcans)
=指 =田 ρ
IMl,Al]
(ii) P(exactly I striker is chosen)
='ciiic, La
=o.tsz
[Mr,Ar]
=norlnalcdf(‐ E99,9.5,
‐Events occdr uniformly ・ Events occurring in one time interval are independent Of those occuring in any other mutua‖ y exclusive
12,「 ) =0.235
lAll
7)Let X=height ofpeople in country A
tlme intervall
(iii) P(3 Singaporeans lexactly I striker) _ P(3 Singaporeans 11  sti ker) P(lstri ker)
X∼
tn
30^
=
0.452
in one hour,X∼
= 0.46
defendcr (or both))  P( I goalkccper) + P( I defendcr) P(l goalkceper n ldefender)
_tcr^"c, toco _'cr*"c, _
'oco
Either one ofthe

(3)
Кzく 1」:=L汁 17‐ μ
…
periods with exactly 2 car tanks topped up with Diesel
5c,x!c,r'' c,
P(Y≧
o
lBl
4)=1P(Y≦
lMll
(10,0224,3)
For 4‐
hour period,X∼
I
IAll P。
X∼ N(12,12) approximately
"
‐ =invNom(0.7)¨ ‐ ……
恒 li
(l)
zく К し>∞ 1」
→ Ⅳつ 一回 対 ― (l)OV‐ =世
5
ぃ
hvNom(0つ
11
invNorrn(005)
(12)
Since λ‐ 12> 10,
"
Noml∞
3)
=0.167
‐‐
価瑚
P(Xく 15)‐ 005
:Bll
=l― binomcdf
[Ml,Al]
Evcnts occur randomly in a continuous given time interval.  Evcnts have a low probability of occunence at any givcn
lAll
= 0224
Y∼ B(10,0224)
following:
IBll
Let Y=no orone_hour
I

'oc
RZ>守 汁
(3,2)
lMl,Arl
‐0671
P。
P(X=2)=poiSSOnpdf
r
(iv) P(exactly I goalkeeper or exactly
P(X>17)=03
Let X‐ number ofcartanks topped up whh Diesel
'tc, x'c,
N(μ ,o2)
IBll
Then c=009220 and 165165
μ=
lBll
instant.
P(X≦ 9)=P(X≦ 95)using CC.
tlsrll
Comnent: [MlJ marla
Commcnb For ld.ndrying cor.ci mil
'awarded oen for '
(i)
inconect
st and ard iza t i on
:
d
used instead
ofo
(ii) incorrect use of invNorm function: did not change the sign to 'r2 lz.rtl ll > x r I I l+2x
Hencc, solve
山C
151
in cxact polar form.
Find the set ofvalues
宅
Md.rl.b r1o:
2m7
rc
2
Patc
Ptttmhrv
,
0{6
Erlhilrlb PrF
I
′ ／品 ７
l
ｒ場
the subsdtution
工=― ツ
雄 浩
, or otherwise,
ind the valuc of
″
for which i61
:・
The funaion
f
is defirBd
b,
f
,r*l#t,
xeR, x*21, tr>O,
shown below. The lirps
t = 2tr' and y = 
r
2.
intercept is given bY .r =
and the graPh is
are asymptotes of the curve, and the
v
y=l
″=え
The function f has an inverse itits domain is restricted to a < x O for posidve gradに nt oftangent to any point on the curvc) 10■
】■=亀 ―■. =6ぅ
FT (6ぅ ;￨)
6
γ
″ ∞ mmmon面 Q =争 6 =二ど 6
L=上 2
Hen∝ ,山 e
se‖
estagcome∬
れ
gession
s_=6 Page 4
of
8
一 一
は 10b〕
劉
にln臥 輌
(‖
2 hlhs Ⅲじリ
︻ 一
銑+咋 詰 +轟 =轟 :〓
i)
Or浩 蒸 =:蒸 げ 出 +轟 =:言 ← “ り [出 =:[(1+:)― =;[f‐
]
・
(:+:)+(:+:)― (:+:)ギ
― (7JLT+7ギ [T)]
ラ 弔 1「 7]
ｆ 一イ
$,Jr)
警 I Bl(2,0)
諭 ー
/
¨ 一
Page 5
of8
4=z(r+r\
12o)
*=*
*=,
Equation of normal to the curve at P: y zQ+ t)=(o+ t)(x p(p+z))
t  2( p + r) =  (p + t\ x + p (p + t)( p + 2) y + (p + t) x = (p + \(p2 + z p + z)
AtG, y =O= x= P2 +2P+2
y, y= p(p+Z)=
p2 +2p
:. NG=2
12b
2+siny = 12  vy
= *, tfr= z,(.*. r) dY
= When y
=2x,
2x Y
A= t*t

the tangent to C is parallel to thc;raxis'
...Z+sii2x= x2 x(Zx)
=
5i,r2r= sin"e
(Z+r')
has no solution,
(z+12)s2 vxeR.
Thus, any taogent to C cannot be paratlel to the,axis' (shown)
the}raxis'
[email protected]! ' 2+ sin y =l sin2 y + ycosy
For C to have a tangent parallel to .'. 2 + sin
138
y = 6s52 y +
[email protected] y
NormaiVCCirizl=(:)X〔
= =
sin2y+siny
=
YcosY I (shown)
il)=3(1)
nOrlnal t。 ″2=(ユ
) Sincc the nomals ofthe two p:anes are paralicl,thc plancs arc paral!C!
b)i)
POS:tiOn vcCtOr ofβ
=(」
Page 6
of 8
一 一
￨'4,)=(1)
一 一
Method i: Let tte foot ofperpendicular hm/to plane■ be F
EquatiOn oFlinc/Fr=(1)+/(1) PointFis ule intcrsectionoflineИ Fand Plane′
(17:;)(1)=6 =,/=: POSitiOn veCtOr ofF= :(1:)
一
Lengh Ofp"jection of/β onto″ 2=BF=浮
．
=7.86
、 一
Method 2:
oflE
Length ofprojection
14
f(工
)=x3_5,2+8χ
9,2_25x+20 (χ
9χ
1)(χ
4=器
/
2)2 =両
2_25χ
!
=7.86
onto
β
+で
11「
c
+(.2)2
】 ラ
+20=′ (χ 2)2+3(χ _2)(■ 1)+C(χ 1)
Solving,_
′ =4,
3=5,C=6
￨
J+ 6 f(r)=1+ x I x 2 (x 2),
︻ ”
=
a(r l)' + s(.t2)' +6(x2)'?
=
a(t  r)'  s(z x)' + 6(2
=
a(r,)''
=
4(r+r+.r'. )J(r.1.
,[,(,;)]',.
x)'z
r[,(,;[ "...).f(,.,.i,....)
x 3l x' +... =53342
¨ 一
The range ofvalues
{xeR.'1 83)二
9り
← ・ ∫ と ∫ ) ・
°
lldndtycorIむ n
`[Fど ・
ロ
０
●
T∼ Ⅳ(■ ,え )apprOX
There｀ 8"gatiVC∞
lation bemeen血 2 data s■ Wid1 0ne dear
“ Oudi∝ wLcn′ =54,3=58. iD
The doctor shoild ignore tt patr of山 雌 pointS
When/=54 md
:
lli→ ilib)
in Page 4 of6
'
fl,..U. 10o
z*Z
ttclq tson
(HZ lapo
f"f* Z)
To obtrain a strdiEcd surple of 50 gradu4cs based on lhcir faaliに ftom thi agcgroup with ゝ wc d― ndom s2,in dlc tlrc sizc ofeach faculty Enginecring Accountancy Veterinary MaOC Comm
r
Science ●50∼ 4
42%● 50■21 219%● 50∼ 11
860/●
27.4%● 50∼ 14
good rcpresentativc sample of the populatioq or givs more accuratc estimatcs than random sampling Disadvantage: samplc not evenly sprcad, or timc consuming or MorB difficult to conduct and analyse resulb dran
Advantage: gives
a
random sampling bi)
Unbiascd csrimue for popukition mean,
; = E'
I
=
#
=
3
I.Zf
Unbiased cstimate for the populuioo variancc:
,
‐
bli)
Ｈ Ｄ Ｄ
By CLT,χ ―Ⅳ(31_28,讐 ) ′(χ >32)=0.0896 Lct
χ and y be¨ mass Ofa shltakc and an abalone mushFOOm
rcspectively.
χ ∼Ⅳ(5.02o.832)r― Ⅳ(7.12,1052)
fげ
舅Jケ
0・
8⇒ aχ
≦ →=o2
″ (6x502,6(083)2) 『:晨務 島∼ ∼ 〃 2却 Ю つ り [ル を ∫ 統11ふ GX・ 22x4B34+&勁 I尤(1,:ち ∫ ♂ :痣
lltt」
P(2r>め =P(2r― S>の =0742 'F二
″ =80,7照 7.:2,署 詳 )
２ つ ｍ ︲
P(γ く7)=0153
Case
I : all 3 distinct colours: 2: 2 samc & I different
No.
efwap =1
No. of ways = tcrr2=6 Case 3: all 3 cards same colour: No. of wsys = Total = 9 Case
l+l
Case I :
without I of the red .ard
Case 2 : witfiout
I l! = 5l4t2t
I of the blue card 
Il!
3t612l Page 5 of6
凛3:宙 d10ut血 ガb Card=詰 ・ ・ ・1+豊・ ・ lⅥ R2BIP“ 詢 =金・ Ц :=艦 =0176 奇 書 岳 書 :・
0R
bli)
К2Ы
“
=会 +士
〓 fi=Q口
6 一 ・ ︼
畿■+号
l bu sOOre O=靱 =編 0、
1色
書 llL÷
=彗
0・
1758=0.827
It means that ulm is a o.
footlcngth is not l0crn whcn it ac$ally is l0 cm'
The ttest should be used. size is small, and population varianc€ is not known
[*ontt.rpt"
Ho: P=19 Hr: p*10 at 5% level of significance Under Ho, 7 follows ttre tdistribution & degree of freedom rest statisric t
d
=i JP = 2.678 7.t;
Since p =0.0316 < 5 %,
at 57o IIo and conclude that there is significant evidence is not l0cm' level thar the average length of a toddler's feet
W"
r"i.t"
follows a normal Assumption is 0ra the length of toddler's feet
一 一
︺ ”
Page 6
of6
￨ _
lnnovo.
JC
2 一 一
The curvc with equationッ
=ar+レ 十c PaSSeS ttЮ ugh
(5,99)FInd the equatlon oF tt curve Gven that多
=0‐ 4ズ
10"andッ =5
whenェ
tlle points(1,7),(3,日 )and 41 〔
=― l, furd an exprcssion
for
telllls ofェ
y
in
t61
1
Let y=0e3χ )] (1)
ShOW that
y2立 _ノ +9=0 (il)
t2l
By futter difFercnthiOn oF thls resuL。 『 OtttMse,ind Maclaurin's series forッ up to and including the term in
2
χ
t4l I
Deduce the equation of rhe tangenl to th€ curve
x=0.
y=(g e3\1 at the point
lrl
Sketch, on an Argand diagram, the locus
wtrcrelz2+2tl=2.
of p
representing the complex number z
el
The point O representing complex number w is the point on the locus of p such that
l: + 4il
is maximum. Find
(D (iD
the exact value
the value
of
lw+4i1,
[2]
ofw in the fom r + iy, giving
the exact values ofx and y.
3] 〔
The posliOn v∝ tOrs oFthe points′ ,B,and C are given by i+j+`4i+3j+2 k and
7i

k respectively. ｐ
Prove that the points,{,8 and C are nor collinear.
Find a vector which is perpendicular to the plane,{rCDeduce the exact length of projection o f FQon te plane ABC, given thar OP
=Zi + qi + 7k and OA =ai
+
974071′ 2007
4i + 6k.
ｐ ロ
(i) (iD (iiD
2j
ffum ovel
lnn。 」
3
Let
xt,x2,
13,... be a
of positive integers, where \=2
sequenc€
and
xn+t=x,2 xn+l forallz= 1,2,3, ...
s,=i I
for alt
rr
1,2,3, ...,
prove
by the method of
mathematical
i=l '' ６
.S,,
=I
€ +, state tlle
value
of Iim
１
for atl n
.
z
Civen that r, >
!ra+l  I
induction thar
﹄ ︼
rf
.S" .
Show that
t43 r r+l
f5r
[2]
r(r + t)(r+3)'
r+3
Herrce find
n
\'lor(r
0
g● )=′ _G+λ =(χ 3)2+λ _9, Rg=(λ 9,∞ )
GC command:sum(u(1,20)
r €. ,r65
To test″ .:μ
Pcrform a l tailed(み tesう at
α%Slg‖
1
Urder H`χ ∼Ⅳ (65,::)by CLT SInce tt ιnotreJected,P― vahe=P(ア >664)>α
%
004∞ 6>α %
o%く 401%
OR Enter into CC:μ
‐65,c=、 52,I=664,″ =50 。
Ch∞ se μ >μ。 We obtain p=004∞ 6 004Ю 6>α %
Since Ho is not reJectcd,
oO/0く 41010/0
No. Sirr.. the sample size 50 is large, the mean marks of 50 students follow a normal distribution approximately (according to CLT). 1lc
To test氏
:μ
=5
Against″ [:μ ≠5 Perforrn a 2面 ied(2‐ teSt)at 50/O Jg lvL
Since比
軋
`reJ釦 >● 96
И
Z>1 96 or Zく
―L96(na)
>1_96
">1892384
Least n‐ 190
pg 6
of6
pqet 2
Admission No. Candidate Name:
M:d¨ Year Exarllinations 2007
Preuniversity 3 MATHEMATICS 9233 PAPER 1
Monday
9233t1 1O
September 2007
3 hours
Additional materials: Answer paper List of Foimulae (MF'l 1)
INSTRUCTIONS TO CANDIDATES
write your name, class and admission number in the spaces at the top of this page and on all the work you hand in. Write in dark blue or black pen on both sides of the paper_ You may use a soft pencil for any diagrarirs or graphs. Do not use staples, pape( clips, highlighters, glue or corection fluid. Answer all the questions.
Give nonexact numerical answers corred to 3 significant figures, or 1 decimal place in the c€se of angles in degrees, unless a different level of accurary is specified in the question. At the end of the examination, fasten all your work securely together. iNFORHATION FOR CANDIDATES The number of marks is given in brackets
I lat the end of each question or part question.
The use of an electronic calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers.
This question paper consasts of S printeA pages.
ffurn over
■釧“詭h"酬 w允 ユ
陥 n―"詰 ≦ 鈴
15]
2
Express i(2「 3+5r)hね mSd″ 「 =2
3
Express
X2+5 m pa両
J ttd10n,
Hence or othettserxPand(1_χ )(χ
State the values of
4.
Prove
i51
_
χ fbr which the expansloo is va:id
171
by induction tnat' ;!+;!+...+;" (3n2)(3n , for all ^: t 1)=\ 3n+ 1x4 4x7
integers n. Hence, deduce the vatue of
posrtrve
1

5.
upto he tem m x3
2+2)h ascendlng powett ofχ
.
=trirhrE
Vl
(a) The complex numbers z and w are such that
w=1+ai, .
z
2i b,
where a and b are real and positive. Given that wz and
=
7i, find lhe exact values of a
b.
l3l
(b) The complex numbers p and q are such that
lpl=z a's(P)=;, lql=z' (i)
Express p in the form
(ii) Find the modulus
6.
a6li6;r=;'
a+,b.
l2l
,.5 I.
ana argument ot I
l.pj
t3l
Given that x, 6, x+9, are the third, fourth ard fifth terms of a geometric progression and that the sum to infinity of the series exits, find
(i)
x,
(ii) the sum to
l4l
inlinity.
t3l
2
.
7_
O The graph J y=flxl t Shom騰 ￨",meに o,か and← 10aに
∞ OdnaLs d
points A and B respeciveiy_Sketch,on separate diagrams,the graphs of ２ ２
(i) j/2=′ (x),
一 一
① y=「 (ゆ showing in each case, the
asymptotes and
the coordinates of the
points
conesponding to A atd B.
(り
SketCh the curve of y=J+(x_2)2'Stalng the equatlons of any asym口
otes,
ooordinates of any stationary points and coordinates of any intersections wlth the
axes
15]
On a single Argand diagram, sketch the following loci.
Й 図 ＝
(i)
lz+43,1 X
=:￨
::F:[∵ Flい 
11
53 1‐
&t c* L y'hu F6 D* ^<
"fnrl ,
,{}1
=
;<
+ M\Z
}'tY2+y?)
lr/ V
︱
ょ ￨.`欄 ■ `り ・
︱
S
一 ¨ ぃ っ 一 一
“
￨イ
I
一 靡 ¨
lnJ '(A
T"
∴り :HTz=76°
'
lf?i",0,
: x+Jil7,
x>3.
Find g(X) Exp:ain whythe composlte fundOn f g l exists and ind the function
(ili) Find the range of f g 1
[31
141 [21
3Section D: Probability and Statistics 6
The random variable X has a normal distribution with mean100 and standard deviation 5. 50 independent observations of X are taken. Find the probability that the sum of the lirst 30 observations exceeds the sum of the remaining 20 observations by more than 1050.
t4t 7
ln a la€e company, 1 in every 12 employees is a smoker. (D Find the pobability that, of 6 employees chosen at random, exactly two of them are
(iD
smokers.
l2l
N people are cfiosen at randorn. Find the least value of N so that the probability that at least one will be a smoker exceeds
8
O'9.
manufacturer claims that 6506 of hb products are of Grade A A sample of 1240 products was taken. 776 of them were found to be of Grade A. Test, at 5% level of significance whether he has overstated his claim. 16l
A
State what you understand by the elgression'al of this question.
(a)
5o/o
level of significance' in the context
I1l
１
9
I4l
一３ ２ ３ 一４
Ｃ Ｃ
一５
__________ c, The tree diagram shows hovrr three events A, B and C are related to each other and the various probabilities associated with them. tl l srate P(B I A)
. (ii) Find P(A'B). (i.i) Fino e (Auc)']. 0
(b)
121
t3l
(Monte Hall Problem) in the popular game show "Let's Make A Deal" which was hosted by Monte Hall, the stage was set with lhree doors. Behind each door was a prize. One prize was very d&irable and valuable. The rernaining two prizes were undesirable. After the contestant selected a door, another door was opened to show an undesirable prize (the host knew where the desirable prize was) and the contestant was given the choice between the already selected door or lhe other door that had not been l2l opened. Should the contestant switch doors? Why?
410
The demand for orp cakes in a cafeteria may be taken to follow a Poisson distribution with mean 0.6 per hour. The cafeteria opens for 5 days in a week and operates for 10 hours per day. (D Find the pIobatility that the demand for cup cakes is at least 5 in a day. (ii) Find the probaulity that there is at most 1 day with zero demand for cup cakes in13]a week. (Give your answer conecl to 4 decimal places.) I3l (iii) Given that the demand for cup cakes is exactly 6 on a particular day, find the probability that this occuned within the first half ofthe day. t3l
The (1fi)  cr)% confidence interval for a population mean p is (ab). Explain, in terms of probability, what q96 means. I1l
A teacher suspects that the students in her school are not having enough steep and hence face the dffiorlty of staying awake du,irg lessons. She surveyed 6g of them ard found out the number of sleeping hours each of them has on an average school day. The data gathered was as follows:
No. of sleeping hours 3
4 5
6 7
8
No. of students 0 4 22 30 8 4
Ｄ り
ｅ回図 同 ｈ ｔ
＞
Find the unbiased estimates of the population mean p and variance d of number of sleeping hours the students have. Find a 90% conlidence interval for p. A symmetric (100  G)o/o confidence interval is found to be (5.Sg , 6.00). Find o.
12. ln a
country, 80% of the population weigh more than.Sokg. Assuming a normal distribution, show that the mean p and the standard deviation o of the disiribution are related by p=50+0.&42o. pl
Given also that the probability of a person from the population weighing less than Tokg is 0.65, find the values of U and o.
let
'A random sample of 76 people is taken from the population. Find the probability that more than haff of them weigh more than
70kg.
l5l
5
13.
Ansnter only one of the following two aftematives. EITHER
The continuous random variable X is uniformly distributed on the interval [ 1 , 5 t. Find the cumulative distribution function of X. 121 Two independent. observations, Xr and Xz, are made of X and the smaller of the two is denoted by Y. (i) Use the fact that Y > x if and oniy if X1 > x and X2 > x to show that the qrmulative distribution function of Yfor 1