u7212

u7a2 ADDITIONAL MATHEMATICS Paper 2

Aug / Sept 2014

2lzhours

PEPERIKSAAN PERCUBAAN SPII TAHUN 2014

ADDITIONAL MATHEMATICS Tingkatan 5 Kertas 2 Dua jam

tigapuluh minit

JANGAI\I BUKA KERTAS SOALAI{ INISEHINGGA DIBERITAHU

L

Kertas soalan ini adalah dalam dwibahasa.

2.

Soalan dalam Bahasa Inggeris mendahului soalan yang sepadan dalam Bohasa Melayu.

3.

Calon dikehendaki membaca maHumat di halaman belakang kertas soalan ini.

Kertas soalan ini mengandrrngi 20 halaman bercetak.

[Lihat halaman sebelah

u7u2 given are the The following formulae may be helpful in answering the questions. The symbols used. ones commonly Rurus-runrusberifut boleh membottu oda menjau'ab soalor. Simbol-simbol yary diboi adalahyotg biasa digunakn .

ALGEBRA

-2a

8

-bxJb? -4ac

log,b =

log" 6

log"a

2 {xd:a'*"

9 \=a+(n-l)d

3 e +d:a'-'

ro. E

4 ({f =a'n

ll

T, = atn-l

t2

""=+?=s,,l

log,mn=lo& m+logon

log"L= n

7

log

on{

logo

m-logon

13

=|[2"+{n-l)d]

E =*,1'l.r

: nlogom

CALCULUS KALKALAS

I Y="v'

dv dv+ V-du --.!- ay-

&&&

z^udy Y=-r'-l= vdx

4

Areaunderacurve Luas di bowahlenghrng

bb

= I,

du v--udx v2

3

-a-2Ly-

du

or(atau) l*

*

dv dx

Volume of revolution Isi padu kisuan b

dv dv

e

= lx

y'&

b

or(dau\ =

[r x'dl

&du&

[Uhat halaman sebelah

3

utaz

STATISTICS STATISTIK

,

Lx

l/ t--

I-

ZW,

g ,P-- nl ' (n- r\l

,. fx

2f

, "=F6.I):=,F-o

7_EWJi

I

g 'C-= 'l ' (n r)t'7t' -

2ft2 _) ---:- - X-

o=-@-= lJ E"f

Ef

[l 5 m=t*|2 "- ')lc

[,1

l0

P(AwB) = P(A) + P(B)

- P(AnB)

11 P(X=r) = oc,p" {1"-', p+q=l

)

t2

Mean/mi4

l3 o

6 I=-4x100

=

z t4 L=

%

p=

np

"lnPq

x-P 6 -

GEOMETRY GEOMETNI

I

Distance

I

juak

= JG,-x,)'+ (v,-

v,Y

Mid point / Titik tengah

(*,y)=(ry,ry)

4 fu€a of a tiangle I Luas segitiga = I" * xzlt + ,ryr)-(tryr r 4tz +:,.yr)l ,l6tn

, ld=,{z;7

A point dividing a segment of a line Titik ymg membalwgi suatu tembereng goris

*'

+

^', \ m+n

(r, y)=(

mlz\ m+n )

t'h +

^ ri+yj 6 r= -:

-

J*'*v'

[Lihat halaman sebetah

y72t2

4

TRIGONOMETRY TNIGONOMETRI

I 2

t sin(ltB) = sinl cos.B+ cos lsin B

Arc length, s : r0 Panjanglengkok"

s:jO

I = !r'0 2 Luas sebor, t = ! j'e 2 Areaofasector,

3 sin2A+cos'l=l sinzA+k66'zA=l

kas

cos(,et8) =

cosBT sin

g

kos(.1* B)

=

"otl

tanzA

=

6

13

?

M = 2 sinl cosl.

cos2A

kos2A

= : :

kosl

l-2sin2

L

_2bccosA

14 Area of tiangle/ Luas segitiga

:

A

= kos2l -sn2 : ZkoszA - | = t -2sn2 A

a2 =bz +c2

a, =b2 +c2 -2bc tas A

cos2A-sin2A Z cos2A -

B

#h

12 a = b =mc ra, rt,

= 2 sinl

lsin

kosl,tos.B:F sin ,4sin B

5 cosec2A=l+gotzA kosek2A =l + kotz A sin2A

kos Asn B

tuAttanB l0 .*n(A+Bl_ .ITtanAtanB

^ Y-'oi==lr**Yn'nu rr

sin

Bt

sin(Ztf) = sinl

labsinC

)

rq,

[Lihathalaman sebelah

u7u2

5

TTIE UPPER TAIL PROBABILITY €(z) TOR TITF NORMAL DISTRIBUTION il(O' BAGI TABUNAN NOR]IOI,L Nrc.I

l)

KEBA.NANGKALAN HAJANG ATAS 0 u.u

U.9III.|

0.1

0.$m

0.2

0ta7

o3

0.302r

0.1

0.3+16

0.5

0.fl85

0.6

0.?13

0.7

0.2{a)

0.8

02fi9

0.9

2

3

u.r$rou u.4u1(,

0.$n 0..n68 U1A 0.1s62

4

5

6

7

u.rtdu,

u./w{, u.adol

u.4/bt

0.{83

0.1143

0.{01

0.$Ar

0,4721 0,{681 0.1325 0.12S

0.052 0,,r0t3 0.xt71

0.395 0.397

0.314 0.37'15 0.3m7 0.3{dt 03372 0.33$

0.$s

0.32A

0.3551 0.3520 0.319 0.3156

0.050 0.3015 02981 0.2709 0.26?6 0.2613 0.238S 0.2350 0.A/1

0.29$ 0.n12 o.nn

0.a43 0.2810

0.2611 0.2578

0.244 0.226 0.21n

0.181r

0.4$ 0.m1 0.n33 0.t8r{ 0.1788 0.1762

o.m(b 0,t9r, 0.17S 0.1711

0.1605

1.0

0.rscr

0.1562 0.1530

0.1515

0.il92 0.1i1$

0.11it6

t.t

0.r357

0.1f;t5 0.13t1

0.12&

1.2

0.1151

0.10s

1.3

0.@68

0.1131 0.fl12 0.G51 0.0931

0.1271 0.'t25t 0.1075 01066

0.0918

t.1

0.m(I

0.07!13

0,0761

0.0rt8

0.{09

0.:132 0.330 0.326{

0.w

0.59r

0.2518

0.2268 o.nfi 0.19{9

I

7

0.i1611

4

0.1217

1

8

2

3

1

03859

1

0.3{e}

1

0312r

I

Et2 812 812 r11 711

16 20 16 20 t519n 1519n 15lEn 1417n 13 16 12 15 ilt416 10 13

0.?n6

3

7

10

0.2151

3

I

t0

0.21,18

3

6

9

0.1887

3

5

I

0.t611

3

5

I

0.1{t1

0.tf,rg

2

5

I

0.11m

0.1

170

2

4

6

I I

0.2511

0.$n 01891 0.1660 0.r$5

n3135 26il31 EAn 21n3l ?326A 2121n $n6 't8 20 B

19 18

t5

0

't6 1,f 13 tl t0

0.1423 0.1210

0.1038

0.r0n

0.1003

0.0985

2

4

6

7

0.@0r 0.0885

0,669

0.04$

0.m38

0.6A

2

3

5

6

0.07,19

0.0x5

0.02r

0.076

0.0691

0.6Er

I

3

{

12

1.5

0.(E6E

0.$6

0.(59a

0.0582 0.051.1

0.(559

I

2

1

5

6

T

8

0.$1E

00655 0.$43 0.6n 0.063t 0.052t 0.s16

0.0618

1.6

0.056

0.0495

0.0185

0..0475 0.0185

00455

1

2

3

4

5

0

7

1.7

0.&16

0.0{5 o.un

0.0392

0.0367

1

2

3

1

1

5

6

1.8

00359

0.0291

1

2

3

I

I

5

t.9

0.w7

0@33

1

I

2

2

3

1

1

20

o@

o.lltlI}

0

I

I

z1

0.0179

0J1,6

0

I

I

0.0110

0

t

1

0

1

1

3

5

t

5

7

223 222 122 It2 r0 t3 s1214

2

4

6

E

fi

13

2

4

0

1

I

1l

0.m1t0

2

3

5

a

E

I

0.mxr7

'|

2

3

5

6

7

u u

0.0336

0.0109 0.(X0t 0,0329 0.0322

0@68

0.&2 o.affi

0.u9

0.0381 0.fi175 0.0307 0.m01 0-@,44 0@39

uwl

0@12

rulllT c@

0.019

o.dsa 0.0rt0

0.ur7

0_0fi8

0.013e

0.0174 0.t'1?0 0.0ts o0il0 0.0132 ofln

0.01o

00104

0.0162 0.0r5t 0.0125 00122

0,0311

00rt1 0M10

q0150 0.01{6 0.0110 00113

0.0102

0_mtc)

000901 0.@39

o.m$a

0.0@ 0.@s0 2.1

o06a

0.6r$

0.m7r6

0.mrc5

0.olt4 0m 15

q62t

a0

0.s{t0

0.ffi0t 0.6sut o.rsrr 00ast 0.m.r0 0.@$1

zt

0.m3/l7

u

0.0336 0-m326

0.mac

2e

t.0

0.6nt

,|

0.0612

dw3a 0.trEft

la 3t 23 22 rta 'tc

rt

t9 16 15 13 tl

3 2

B 16

r5 t? t3 t5

0.@ 0.M2

0m264

0.m&

3

a

5

0

0m212

0.m103

I I

2

I

2

I

4

ootsa

0.qx{0 0.@t{ oo13

3

oFru

0@r9 0016{ 0.o1s

0

'l

t

2

2

3

o.mr35

0.ol3t omt26 ofiln

oollt oolr{

oofir

I

2

2

2

3

Example / Contoh:

r(z)=#4-:r)

0.q'1m

0

Jika X

Q@\ =

lr
t

f z) IfX-.lV(0, l),

then

- N(0, l), maha

P(x> k)= g(k)

k

4X>2.1):

11 13

I

oal1t 0.@{ oMr3 001il o.ortt 06ie

omloa

17

1

tt t2 9010 7Eg 56t ta4

0.mi0e

10

f8 66 55

0.m$8 0.(niF{ 0.mtr9 0.qF60

0.m220

21

t0 lt E9

000651 0.m$0 0m5a 0.m415 0.0402 0.m301 osr)T 0_ma8 0m@

0.m3r7

9

nns nn36

21 21

t2 11 10 9 1l 810 78

0.12X)

0.0351 0.0311 o028r 0.0211

I

7

6

5

MiNs / folal

QQ.t) = 0.0179

d t0

1f la

a

u72t2

Section A Bahogian A

140 marksl [40 narkah]

Answer allquestions. Jawab semula soalan.

1"

Solvethesimultaneousequations x+ 2y Selesaikan persamcun serentak x +

2.

lt is given

that

2y

4 374, x

: I - xy = l.

: t - xl :

,

486

,

I

.

[5 marks ] [5

narkah]

is part of a geometric

progression with positive terms and the sum of the first four terms is 19 440.

Diberi bahnrya ...... , 4 374, x , 486, .... ....ialah sebahagian daripada suatu janjang geometri dengan sebutan-sebutan positif dan hasil tanbah empat sebutan pertamajanjang itu ialah 19 440. Find,

Cari,

(a)

the common ratio, nisbah sepunya,

(b)

the first term, sebutan pertama,

(c)

12

mad
12 markahl

12 marksl

p

markahl

the smallest value of n such that the nm term is less than 0.01 . l3 nilai n yang terkecil supaya sebutan ke-n adalah hnang dmipada 0.01.

ma*g

[3 morkoh]

[Lihat halaman sebelah

u72J2

(a)

Prove

that **-

BuHikan

(b)

(i)

=

cosec2x.

baluwa #

=

[2 marks]

kosekzx

[2 markahl

.

of y = 1 - cos 2x for O s x s 2n. 13 marl
(ii) Hence, use the same axes, sketch a suitable straight line to find the numberof solutionsfortheequation

- z- *,0=x

#

State the number of solutions.

32tr.

[3 marks ]

Seterusnya dengan nenggwrakan paksi yotg sama, lakar satu garis hnts yang sesuai tmtuk mencmi bilangan penyelesaiot bagi Tiersamacm

ffi

2

x t = - 2- *,

Nyatalrm bilangan penye

o3x<2rles aian

itu

(a) Find the equation of tangent to the cUils y

[3

nokahl

- ) ^tthe point (- 1, 4). Cari persamaot tangen kepada lmghmg y = 3xz | *at fifik ( -1, 4). =

3x2

[3 marks]

13nskahl

(b) A piece of wire of length 360 cm is used to make a frame in the form of a cuboid. The base of the cuboid has sides measuring r cm by 2x cm. The height is ft crn. Seutas dmtai dengan pmjang 360cm digwakm tmtuk membuat sebuah bingkai berbentuk kuboid. Tapak kuboid itu beruburot x cm dan 2x cm. Tingginya ialah h cm.

(i)

Show that the volume of the cuboid, in cm3, Twtjukkan bahawa isipdu kuboid

(ii)

v:

18M

-

in,

is

Z: IB0l -

6x3,

datam cmi , diberi oleh

61.

Find the maximum volume of the cuboid.

Cari isipadu maksimum htboid itu. [4 marks]

[Uhathalaman sebelah

3/-7212 14 markahl

5.

for the scores obtained by 32 Table 5 shows the cumulative frequency distribution students in a test. bagi skor32 orory murid dalam satu Jadual5 menmiuklmn taburan kekcropan longgokan ujian.

Table 5 Jodual

5

(a) Based on Table 5, copy and complete table 5'1' Berdasmlrsn Jadual5, salin dan tengkapknn Jadual

S'l'

Table 5.1 Jadual5-l

(b)

Find the mean score of the students' Cui skor min mwid.

(c)

Use graph paper to ansrler this part of question' Guakan kertas graf mtuk menjawab balagian ini'

11 markl narkah]

ll

12 marksl 12 narkahl

2 qnlo 2 Using a scale of 2 cm to 20 scores on the horizontal axis and students on the vertical axis, draw a histogram to represent the frequency distribution of the scores. Find the mode s@re. mengufuk dan 2 cm Dengan menggtmakan skalaT cm kepada 20 skor pada paksi tmtuk kepada 2 orang murid pada paksi mencancang, lukis sebwh histogram mewakili tabwan frehtensi skor itu'

coi

skor

md'

[ 3 marks ]

[Lihat halaman sebelah

u7a2 [3 markahl ln Diagram 6, ABCD is a quadrilateral. The diagonals AC and BD intersect at point R. Point P lies on AD. Dalam Rajah 6, ABCD ialah sebuoh sisiempat. Perpenjtru-perpenjunr AC dsn BD bersilang di titik R Titik P terletakpada AD.

Diagram 6 Rajah 6

that AP = lAD, BR= 1BD,

ffi=x Diberi bahawa ef =l,lo, BR:;BD ,m-L It is given

(a)

Express in terms of x andlor y : Ungkapkan dalam sebutqn x dan/atau y

(i) (ii)

TF=y dn dF =2. and

:

m va

[ 3 marks ]

[3 markah]

(b)

Given that

Dd = hx -

/

and.4F =

k-ft

,where h andf are constants, find

the value of h and of k. Diberi bahawa dd = hL- y dan Art = kVe ,dengan keadaan hdan kadalah pemalm, cari nilai h dan nilai k.

14 marksl

lUhat halaman sebelah

u7a2

l0

[4 mukah)

Section B BahagianB 140 marksl 140 mmkahl Answer any four questions from this section' Jnsab mcma-mans empat soalan daripada bahagian ini. 7.

Use graph paper to answer this question. Gunakfrn kertas graf untukmeniawab soalan ini'

Table 7 shows the values of two variables, x and y, which are related by the equation y = hk-x where h and k are constants. Jadual T menunjukkon nilai-nilai dua pembolehubah x dan y, yang dilwbungkot oleh persamaan

y = hk-x,

dengan keadaan h dan k ialah pemalm'

x

2

3

4

5

o

7

v

7.80

6.00

4.61

3.55

2.73

2.10

Table 7 Jadual 7

(a)

Plot/ogroyagainstx, by using a scale of 2cmto0.1 unit on the logny axis and 2 crn to 1 unit on x-axis . Hence, draw the line of best fit.

Plot

logrc

palrsi-log

y melm'an x , dengon menggtmakan skala 2 cm kepada 0.1 mit rc y dm 2 cm kepada I mit pada paksi'x.

Seterusnya, lukis garis lunts penyuaian terbaik.

5

pada

ma*sl

[5narlnh] (b)

Use the graph from 7(a) to find the value of Gtna grof di 7(a) mtuk mencari nilai (i)

h,

(ii)

k,

(iii)

v when x = 4.5 v apabila x : 4.5

5 marks l 5

mukahl

[Lihat halaman sebelah

ll 8.

347212

Solution by scale drawing will not be accepted. Penyelesaian sectra lukisan berskala tidak akan diterimo.

Diagram 8 shows a quadrilateral PQRS. Pos and equation of PS is y: 2x and ZSPQ = 9Oo.

pre

are straight lines. The

RajahB mentmjukkan sebuah sisi empat PQRS Pos dan PTe adatah garis lurus. Persamaan PS ialah y 2x dan ZSPQ = 900.

:

r (r0, 0)

R(2,4) Diagram I RaiahS (a) Find / Cari

(i)

the value of k, nilai k

(ii) the coordinates of

P,

koordinat titik P,

(b) It is given that PT: TQ

Diberi PT : TQ

= 2:

= 2 : 1.

14 marksl [4 markahl

I

Find / Cari

(i) (ii)

the coordinates of Q, kaordinat Q, Area of quadrilateral PQRS. luas sisiempat PQRS

14 marksl

(c)

A point E moves such that its distance from the origin o is arwaj:ffi:.t Find the equataon of the locus of E. Satu titik E bergerak dengat kcadaan juafuya dti asalot O odalah sentiasa 3 wtil. Cui persam(utn lohts bagi E.

lZmarksl

12 narkahl

[Lihat halaman se0e/aD

t2 9.

Diagram 9 shows two circles. The larger circle has centre P and radius 10cm. The smaller circle has centre Q and radius 6cm. The circles touch at T. The straight line RS is a common tangent to the circles at point R and point S. Rojah 9 menwfukkan dua bulatan. Bulatan yang lebih besar berpusat P dan beriejari I0 cm. Bulatan yang lebih kecil berpusat Q dor berjejari 6 cm. Kedua4ua bulatan bersentuh di T. Gmis lurus RS adalah tangen sepwlya kcpada ke&n-&n bulatan itu di

titik R dan titik

S.

Pr*__

T

Diagram 9 Raiah9

[UselGuna r =3.1421 Given that ZQPR = 0 radians. Diberi bohmva zQPR : 0 radiwl

a) b)

[2 marksl

Show that 0 = 1.318 radians. hmjukltan bahawa 0 = 1.3|8 radian.

[

2 markahJ

Find

Cori (i)

the perimeter of the shaded region, perimeter rantau berlorek,

14 marksl [4 markah]

(ii)

the area, in cm2, of the shaded region. luas , dalam cm2 , bagi rontau berlorek.

[ 4 marks ] 14 nnkahl

[Li hat h al aman seberat,

13 10

u72J2

(a) lt is bund that 9O% of graduates from UniversitiGemilang are employed. Didopati baha+'a 90% daripada graduan Universili Gemilang mendapat pekerjaan.

lf

10 graduates from the universtty are chosen at random,

Jikn 10 orang murid daripada universiti itu dipilih secaro rawak, find the probability that at least 9 of them are employed

(i)

,

cari lreborangkalian bahowa selarang-kurangnya 9 orang mendapot pekerjaa4

determine the standard deviation of the number of graduates being employed.

(ii)

tentukan sisihan piawai bilangan groduot y@E mendapal

[4 marks] la nnkahl

(b) ln a mathematics test taken by 300 stdents of SMK Gemilang, it is found that the marks obtained follow a normal distribution with a mean of 60 marks and a standard deviation of 15 marks. Dalam suatu qiior matematik yotg diambil oleh 300 orang pelajar SMK Gemilang didapati markah yang diperoleh mengihrt taburan normal dengon min 60 markah dan sislhon piawai 15 markah.

(D

Find the number of students who passed the test if the passing mark is

45, Cari bilangan pelajar yang lulus ujian tersebut jika msrkah lulus iolah 45.

(ii)

ll

of the students passed the test with grade A+, determine the minimum mark to obtain grade A+. 12o/o

Jika I2% pelajar lulus dengan gted A+, tentukan morkoh minimum untuk memperoleh gred A+. [ 6 marks ] [6 narkah]

[Lihat halaman sebelah

t4 11.

Diagram 11 shows part of the graph of the function y = flxl wtrich has a gradient function of -Jr and passes through 8(6, 3). The straight line AB is parallelto the x-axis. Rajah ll menwjul*an sebahagian dtipada Srdfun4si y =.f(x) yang memptmyai fungsi kecenman -!x dan memalui titik 8(6, 3). Garis lurus AB adalah selni dengan paksi-x.

Find

Csri

(a) the equation of curve, persamcum lengkang itu,

[3 marks / marlmhf

(b) area of shaded region, Iuas rsntau berlorek,

[4 marks / narkah]

(c) the volume of revolution, in terms of z, when the shaded region is rotated through 3600 about the y-axis. isi

pada kisaran, dalam sebutan n, apabila rontqu berlorek diputarkan melatui

pdapaksi-y.

36f

[3 marks I narkahl

[Lihat halaman sebelah

y7u2

15

Section C Bahagian C [ 20 Marks ]

120 Marlmhl Answer any two questions from this section. Jawab mcme-mcma dug soalsn duipada batagian ini.

12.

A particle moves along a straight line such that its displacement, s m, is given by s -2J* + 18t, where f is the time, in seconds, after it passes through

:f

a fixed point O. Sdu zmah bergerak di sepof otg suatu guis lurus dengan kcadasn sescy(mnyq s m, diberi oleh s = t3 * LBt dengan keadaan t iarah masa, dalam saat, selepas

?r'

melalui satu titik tetap O.

(a)

Find

Cori

(i) (ii)

the values of f when the particle stops instantaneously, [ 3 marks ] nilai-nilai t apabila zuah itu berhenti seketika, t-3 narkah j the distance between the two points where the particles stops instantaneously, jarak di antoa titik-titik apabila zarah itu dalam keadaan berlenti seketika,

[2

marks]

[2

marksl

[2 nnkah]

(iii) (b)

the minimum velocity of the lulaju minimum zuah itu,

particle,

tZ narkah

Sketch a velocity-time graph for 0 S f < 7. Lakarkan sebuah gral halaju-masa

mtuk|
j

13 marksl [3 narknh]

[Uhat halaman sebelalr

l6 13.

3/72J2

Solution by scale drawing will not be accepted. Penyelesaian secora lukisan berslcala tiilsk akan diterima. Diagram 13 shovvs a quadrilateralABCD.

Pajah

13 menunjukkan sisi empat ABCD.

Diagram 13 Rajah 13 Given AB = 8cm, AD= 25 cm, CD = I cm and Diberi AB = 8cm, AD : 25 cm, CD : 9 cm dan

I

ABC =108o.

/ABC :1082

(a) Find Cui

(i)

the length, in cm, ol AC, panjan AC dalam cm,

(ii) t (b)

AcB,

Point A' lies on AC such that A'B = AB. Titik

(i) (ii)

A'

terletak

pda

AC dengan keadaon A'B

14 marksl [4 norkah]

:

AB.

Sketcfr triangle A'8C. Lakar segi tiga A'BC.

Calculate the area, in cm2, of triangle A'BC. Hitungkan luas segttiga A'BC dalam cm2.

[6 marks] [ 6 narkahJ

[Uhat halaman sebelah

l7

14.

347A2

Table 14 shows the prices of four ingredients p, e, R and S used for making a particular kind of cake. Jadual 14 menwjukkan harga bagi empat balnn p,e, R dan S yang digunalcon untuk membuat sejenis kek.

Ingredient Bahsn

Price per kg (RM) Hargq sekilogram (RM) Year 2012 Year 2013 Tahun2012 Tahun2Ol3

P

E.00

'l'9

0

3.00

4.80

R

x

^S

5.00

v 4.40

Table 14lJoduol t4

(a)

The index number of ingredient P in the year 2013 based on the year 2012 is 125. [2 marks] Nombor indeks bagi bahan P dalam tahtm 2013 berasaskan tahrn 2012 ialah I25. Carikan w. Find the value of

ur.

f2markah]

(b)

The index number of ingredient R in the year 2013 based on the year ZO12 is 150. The price per kiligram of ingredient R in the year 2013 is RM5.00 more than its conesponding price in the year 2012. Calculate the value of x and of y. Indeks nombor bahan Rdalam tatrun 2013 berasaskan tahun 2012 ialah lS0. Huga sekilogram baltu R pada talrun 2013 adalah RM 5.00 tebih mahat daripada har{anya yong sepadan dalam tahun 2012. Hitungfran nilosi x dan nilai y.

[3 marks]

[3nukah)

(c) The composite index for the cost of making the cake in the year 2013 based on the year2O12 is 131.5.

Indeks gubahan kos membuat kek itu dalam

tahm

2013 berasaskant tahun 2012 iatah I3g.

Calcrilate

Hitwg

(i) (i,

the price in the year 2013 if its conesponding price in the year -ol1-"{e 2012is RM 50.00. the value of m if the quantities of ingredients P, Q, R and S used are in the

ratio

nilai

8:4:

m

m:

3.

jika laontiti bohan P, Q, R don S ymg digunakan adolah mengikut nisbah

8:4:m:3.

[5 ma*s] lSmarkahl

[Uhat halaman sebelah

l8 15. Use graph paper to answer this question. Gunakm kertas grof untuk menjawab soalan ini.

ln an effort to strengthen Bahasa Melayu and to improve the standard of Bahasa lnggeris, a training centre offers two courses, BM and Bl. The number of participants for the BM course is r and the number of participants for Bl course is y. The intake of participants is based on the following constraints : Dalam usaha memperlasalcan Bahasa Melayu dan memperh*uhknt Balnsa Inggeris, sebuah pusat latihan menawarknt dtn hrrsus, BM du BI. Bilongan peserta kursus BM ialah x dan biloryan peserta htrsus BI ialah y. Pengambilan peserta dalah berdasarkanr kekangan berikut:

I :

The number of BM participants is more than 20 Bilangan peserta BM

dalah melebihi

20,

ll :

The number of Bl participants is at least 10, Bilangan peserto BI adoloh sekwang-kurangnya lO,

lll :

The maximum number of participants is 80, Jumlah maksimum bilangan peserta ialah 8A,

lV :

The ratio of the number of BM particapants to the number of Bl participants is not more than 3 : 1. Nisbah bilangan peserta BM kepada biloryan peserta BI adalah tidak melebihi 3 : L

(a)

Write down four inequalities, other than x > 0, y > 0, which satisfy all the above constraints. t 4 marksl Tulis empat ketaksamaan, selain x 2 0, y > 0, yang memenuhi semua kekangan di atas. 14 narkahl

(b)

Using a scale of 2 cm to 10 participants on both axes, construct and shade the region which satisfy allthe constraints. [ 3 marks ] Menggmakor skala2 cm kepada 10 peserta pado kedua-dua paksi, bina dan loreh rantau Ryang memenuhi semua kekangut di atas. [3 marlmhl

(c)

Using the graph constructed in 14(b), find Dengan menggunakan

(i) (ii)

gof yong dibina di l4(b\, cari

the range of the number of BM participants if the number of Bl participants is 20, julat bilmgan peserta BMjikn biloryan peserta BI ielah 20, the maximum totalfees that can be collected if the fees for BM and Bl courses are RM 200 and RM 400 respectively. Jumlah maksimum kutipan yuran yang diperoleh jika yuror bagi hnsus BM dan BI ialah RM 200 danRNl400 masing-masing. 13 marksl

[3 mukah]

END OF QUESTION PAPER KERTAS SOAL,AN TAMAT

[Lihat halaman seDerah

ADDtt

r

o

NrPrL

KEITffi t)r r cst i, rn

M ft-T f+e M4Tr qs

-

2

PEpcL.r e

A-erN/ 8Pm >A+

Workinr

Marks

x+2y:1, x--xy- lor "rrui"ale'it--_\-l

or

'\,

(t-2y)'-(I-2y)y= y (6y--5) : o v:u. 6 X: .2 I.

x

2(a)

x:

'):l x-r -x('.(1-r)z (3x+2)(x- t;=g

or or

x- -? . x:

or

X: -:

(1-x)

I

J

5-

!= ., v:0

or

+46

I

x 1458 1

I

(b)

s5:

19440 I

'f=q) \t-=/ a: (c)

B

D44o 1

13 122

122

G)"

<

I

o.or

I

Taking log both sides + simplifo..

n:14 I

3(a)

7

LHS: r-Il-2, sinz. x7

I

1

sinz

x

,

I

cosec- x

3(b) (i)

t

;ffi I

"t

'

=[ , \, 'f:,n':2o

,

\r,

o.

-r{ I

I

Shape

3(b)

(ii)

I I

ofcurve correct

2 periods rvith correct magnitud,: All correct

OR vertical shift

l unrt

1

i-cos2.t = 2-'2n -f c)secz"x Straisht

linev: 2-'

correctlvsketchetl

1

1

lji li I

tl -T--

l

lol I

)1.

Ll

4(a)

#= ur*;

&

substitute x = -L

1

--5

(b)

y-4=-5(x+1)

1

Y=-5x-1

1

v = 2lh

= 2x2 (90 180x2 - 6x3

= dv

350x dt =

x=

V-",

s(a)

(b)

I I

=

-

-

L

3x)

1

18x2 1

2O 1

24 000 cm3

7

score

| 0_L9 | Noofstudents | 0 |

29.5

x5+

49.5

x

16

*69.5 x

Zo_gs

s

B

I +o_sg I eozg I 8 | | 16 I

so_gs

+ 89.5 x 3

6(aXi) (ii

)

1

Refer to Graph in page 8

DB: -4y+ AF

3

5

L

1

: -x *;ED. or 1+

Art.

:

4y

*:Od *

*foilow through

3 = -4- x* y

(b)

3 L

55.1_3

(c)

L

Using DA

1

+ dZ =

De

-4y* i(io*Z)= hz-y h=a or -4 * i:-t

n:?

L

1

1.

k:l

1+1 7

7(a) X

2

3

4

Log ro y

0.89

0.78

0.66

Graph -

-

correct axes with uniform scales All points correctly plotted Line of best fit

5

5

7

0.5s o.M 0.32 (at least two decimal places)

1

1 L L

3

20b

tb)

Refei'to graph on page Use Y- intercept

=

9

logle h

gradient= -logrok Gradient = -0.114t0.01 k=1.3010.05 h = 13.18 10.10

y = 4.05+0.05

8(aXi) (ii)

Gradient ot

pe = -

I

Findingequationof PQ: y=

-1x *5

any method to find intersection point between line SP and QP P12,4)

(b) (i)

t(z)+z(a) ,O _

1(4)+2(b)

Q(14, -2)

(ii)

1

=

(c)

e(a)

Menggunakan formula dengan betul

60

(z-s;z +(y-0)2 x'+y' - 9

=

1.318 rad

=

16 sin

RS

=

= 3 or equivalent

=.or-t{1) '16'

zOPR B

(b)(i)

. I

Area: t |

0 or {Td,

--F

15.492

Arc length RT = s = (10)

(1.318)

0R

arc length TS = 6{1.824)

1 1

perimeter= 15.492 + 10.944 + 13.18 39.62 cm

(ii)

Areaof trapezittm = l(to + 6) (15.492)) /trectufspcloi : 1iIo)z(1.318) -' oR -" I(6)2(LB?+)

1

123.936 55.90C

1

'

25-20 cm2

z

2t

32.832

i

10(aXi)

p=0.9 orq=0.1 P(X > 9) = 106,(0.9)e(0.1)1

1

1oc,o (0.9)10(0.1)o

'J.

o.7361,

(ii)

1.

o =./ro(o.s)(o.r)

1

o.9487

(bxi)

P(x>4s)

=r(' r #)

L

= p(z>-1) = 0.8413 1

No. of students who passed = 0.8413 x 300

(ii)

=

252 orang

1

P(X>k)=0.1200 1

k-60 - 15 = 77.63

k=

7.175

=

78 rnarks

1

1

10 1L(a)(i)

(ii)

rglorfl= -:, v

=

[ -lxdx 7-

Y= - 7x'*

L

c

y- -!x2+s (b)

1

1

a, = ff Ydx

=

f(-

lx2 +s1ax

1

= [-* * n'].

=42

1

Shaded area

= 42-18

= (c)

v,

24 unit2

= Jl rcx2d.t r[54y 108n

- 3y']Z

L L

il 10

f,o

3(t-1Xt-6)=0

1i 1i

t=1, t =6

1l

v=

(ii)

3t'-21t

ttf itrf

+18

+1s(1)

tsf

-?tsf

I

i

+18(6)

62.5

(iii)

6t -21=0 -18.75 m/s

(b)

13(a)(i)

Correct shape Correct minimum point and the roots or y-intercept

1

All correct

1

AC' = 25' + 9' - 2(25)(9) cos AC = 20.41cm

20-411080

(ii)

:

sin

I

ACB

=

1 1

B

sin

C

21.890

A

tb) ti)

{ii}

1

I

BA'C

=

129-B9o

Finding BC = 16.46cm (U.sing Area

7 Sine

ntle or any relevont methotl to.fincl a second.sirle)

of aA'BC = ]{e){ro.r5) sin 129.89"

oR

31.15

] {e){ia_+e; sinzB.z2" = 31-.13

1 L 1_

l

I I

I

r

ut(

A;eaof

AABC *'

67.C'4

area

af AAA'B

33..49

= 31.15

I

-l1jD

1a(a)

- r25 w=8x100 =

tb)

10.00

1

y=x+5

I x 100= x Solving,

(cXi)

1

50

*any one eqn correct

150

x=

10

1

Y=15

1

x t3t's 100

1

RM 65.7s (ii)

131,.5

1

1

tz s (B) + t 60 (4)+ =

1s

o(m)

i80(3)

1.

(1s+m)

Solving...

1

m=5

1

10 1s(a)

(a)

x >20 y >ro

1 1

x+y < 80 xs3y (or y>lx) (b)

(b)

(c)

1_

1

Axes correct and one *straight line correct Draw correctly all 4 straight lines Region R is correctly shaded and labelled. (Pleose refer to appendix)

{i)

20
(ii)

2oo

(2Il + a00

rujukgraf

L 1

1

(5e)

The maximum fees collected is RM 27

Sila

l-

1_

8OO

1

(Lampiran 1)

10

)o7

Appendix

/

Lampiran

1

Graph for Question 15 IUNEAR PROGRAMMING]

tr

100

x+f=$Q

.

P=3S

\ 60

70

80 it

>i

"--"€ 'j a

r$ tc t4

tt t9 tti

Use either class boundaries or midpoints

/

x-axis and y-axis drawn correctly

LM

Histogram drawn correctly

tM

Mode = 50.51 1-0

1M

class marks]

)(-\

log

to

!

t,!

I

i

i I I

i

o.z

:-=i:;r:.:-u:*r;

-) 1i'l