jabatan pelajaran negeri johor peperiksaan percubaan spm 2008 · pdf filepeperiksaan percubaan...

SULlT

Additional Mathematics Kertas 2 Sept. 2008

1 2 - jam

2

JABATAN PELAJARAN NEGERI JOHOR PEPERIKSAAN PERCUBAAN SPM 2008

ADDITIONAL MATHEMATICS Kertas 2

Dua jam tiga puluh minit

JANGAN BUKA KERTAS SOALAN IN1 SEHINGGA DIBERITAHU

1. Kertas soalan ini adalah dalam rhvibahasa.

2. Soalan dalam bahasa Inggeris mendahului soalan yang sepadan dalarn bahasa Meluyu.

3. C'alon dikehendaki membaca maklumat di halaman belakang kertas soalan ini.

4. Calon dikehendaki menceraikan halaman 19 dun ikat sebagai muka hadapan bersama-sama dengan jawapan anda..

Kertas soalan ini ~nengandungi 19 haIaman bercetak dan 2 halaman kosong.

347212 I3ak Cipta JPNJ 2008 SULlT

MOZ@C

SULIT

Section A Bahagian A

Answer all questions in this section . Jawab semua soalan.

I Solve thc following simultaneous equations : Scie.suikari persatnuan serentak berikui:

~ , - . r = 2 and y 2 + 2 y + s 2 = 9

Give your answer correct to three decimal places.

Berikan julvapun anda betul kepuda 3 iempai perpuluhan.

[5 marks]

[5 markah]

2 '[he gradicnt function of a curve which passes through the point P(1,2) is 2x3 - 4x.

Fungsi kecerunrrn bagi suaiu lengkzrng yang melalui titik P(1 ,2) ialah 2x3 - 4x.

Find

Cari

(a) the equation of the curve, /~cr.satnaan Iengkzing itu,

[3 marks] [3 markah]

(b) the equation of the normal to the curve at point M. pcrsamuan normal kepuda Iengkung di titik M


3 Given a set of numbers :

Diberi scrtzc set nornbor:

3 , 5 , 6 , 8 , 9 , 1 0 , 1 l , 1 2 .

(a) Find the mean and standard deviation of the set of numbers. [4 marks]

Carikan min dun sisihan pialvai bagi set nombor tersebut. [4 markah]

(b) Given that each of the number increases by 2 and then divide by 4.

Diberi bahaiva setiup notnbor tersebut meningkat sebanyak 2 dan kemudian dibahagi dengan 4.

Find the new standard deviation. C 'trri .ci.sihan pitnvui yang baru.


MOZ@C

SULlT 5

4 Diagram 1 shows part of arrangement of bricks of equal size.

Rajah 1 menunjukkan sebahagian daripada susunan baru bara yang sama saiz.

Diagram 1 Rajah 1

The number of bricks in the lowest row is 4096 and the height of each brick is 4 cm. Edham built a wall by arranging bricks such that the number of bricks is half the number of its previous one until the number of bricks in the top most row is 1.

Bilangan baru bara di baris paling bawah adalah 4096 dun tinggi seriap baru bara ialah 4 cm. Edhan? menyusun batu bata untuk membina sebuah tembok dengan keadaan bilangan baru bara adalah separuh daripada bilangan pada baris sebelumnya sehingga bilangan batu bara di haris paling atas adalah 1.

Calculate

Hitungkan

(a) the number of bricks in the 71h row,

hilangan baru bara pada baris kerujuh,

(b) the height in cm, of the wall.

iinggi rembok dalam cm.

[3 marh]

[3 markah]

[3 marh]

[3 markah]

[ Lihat sebelah SULIT

MOZ@C

SULIT

5 [n Diagram 2, the straight line joining PQ meets the x-axis at R. The point S lies on PR such thatPS:SQ=3:2.

Dalatn Rujah 2, garis llrrus PQ menyilangpaksi-x di titik R. Titik S terletak di atas garis PR dengun keadilun PS : SQ = 3 : 2.

Diagram 2 Rajah 2

A straight line passes through S meets the x-axis at Tand L RST = 90'.

Satu garis lurus melalui tirik S menyilang pabi-x di titik T dun L RST = 90'.

Find

Cari

(a) the coordinates ofS,

koordinaf bagi titik S

(b) the equation of ST.

persunlaan garis lurus ST

[2 marks]

[2 markah]

[4 marks]

[4 markah]

I Lihat sebelah SULIT -

MOZ@C

6 (a) Prove that tan 0 + cot 0 = sec 0 cosec 0. [2 marks]

Bzrktikan identiti tan 0 + cot 0 = sec 0 cosec 0. [2 markah]

(b) Sketch the graph of y = I sin x 1 + 1 for 0 s x 1 271. Hence, using the same axes, draw a

2x suitable straight line and state the number of solutions to the equation 1 sinx I= - -1 x

for 01xs2i' l . [6 marks]

Lakarkan graf bagi y = I sin x 1 + 1 untuk 0 5 x 5 2 x . Seterusnya, dengan menggunakan paki yang sama, lukiskan garis lurus yang sesuai dan nyatakan bilangan penyelesian

2x bagipersamaan I sin x 1 = - - 1 untuk 0 x s 2 x . [6 markah]

i'l

[ Lihat sebelah SULIT - MOZ@C

Section B

Bahagian B

[40 marks]

[40 markah]

Answer four questions from this section.

Jawab empat soalan dar&ada bahagian ini.

7 Use graph paper to answer this question.

G'zrr~akun kcrtus graf untlrk tnenjawab soalan ini.

Table 1 shows the values of two variables, x and y, obtained from an experiment.

Variables x and y are related by the equation y = a b X L , where a and b are constants.

Jadlitrl I menunjukkan nilai-nilai duapembolehubah x dun y , yang diperoleh daripada satu 2

eksperimcn. Pembolchubah x drrn y dihubungkun olch persamaan y = ah' , di mana a dun b ariuluh petnalar.

Table 1 Jadual 1

(a) Plot loglo y against x h y using a scale of 2 cm to 1 unit on the x-axis and 2 cm to

0.2 units on they -axis. Hence, draw the line of bcst fit. [ 5 tnarks]

Plotkan log,, y melawan x2 dengan mcngpnakan skula 2 cm kepada 1 unit padapaksi-x dan 2 cm kepada 0.2 unit pada paksi-y. Seterusnya, lukiskun garis lurus penyuaian terhaik. [5 markuh]

(b) Use your graph from 7(a) to find the value of

(;unukrrn grrif undu di 7(u) untuk mencari nilai

6) u

(ii) h [5 marks] [ 5 markuh]


MOZ@C

SULIT 9

8 Diagram 3 shows a parallelogram OPQR.

Rqjah 3 menunjukkan sebuah segiempat selari OPQR.

Diagram 3 Rajah 3

Given that = 6 j + rn j , ?% = 4j + 3 j and

Diberi bahawa = 6i + rn j , OR = 4i + 3 j dan

(a) Find

Cari

(i) the positive value o f m ,

Nilui pusitifbagi m.

2 - d

(b) Given E= - RQ and ?% = LG. Find H. 3 3 7,

Diberi = - RQ dan = !OR. Cari E. 3 3

(c) Given that T is a point such that = 5 j + 9 j -

Diberi bahawa T ioloh satu tilik dengan kcadaan KT = 51 + 9 j .

( i ) Find E,

Cari >?.

(ii) Show that the points 0, P and Tare collinear.

fitqukkan buha~va titik-titik 0, P dun Tadalah segaris

[3 markr 1 [4 tnarktrh]

[ Lihat sebelah SULI'T

MOZ@C

0 I)i,~g:.a~n 1 al~ous a semicircle ABC , centre 0 with radius 5 cm. EBD is a sector of a circle , centre E with radius I0 cm.

lic~jrrii 4 t)ii.nrii~jlikkirrr .sehuuh .semihulaian ABC, herpusai 0 dengun jejari 5 cm 1:ljl) icrltrli .sch/rtih .sek~or hagi hulatan, herpusat E dengan jejari I0 cm.

Diagram 4 Rajah 4

( ~ i v e n that 110 is perpendicular to AE.

1)iheri BO hcr.vc~cnjnng dengun AE.

(a) i B E 0 in radian,

L' 1~1<0 (Iol(1tt1 rudian,

( b j rile perimeter in cm, of the shaded region,

/x2rittictcr, c-kllum cm, hagi ka~vasan herlorek,

(c) Llie area in cm'. o l ' ~ h e shaded region.

111~1.s. ~lultini L ,~ I I , hugi k u ~ t ' a ~ u n herlorek.

[2 marks]

[2 markah]

[4 marks]

[4 markah]

[4 marks]

[4 markah]

[ Lihat sebelah SULlT

MOZ@C

10 (a) Diagram 5 shows part of the curve y = 9 - x2 . The straight line intersects the y-axis

at (0, k ) and intersects the x-axis at (6,O).

Kujuh 5 menlrnjlrkkan sehahagian daripada lengkung y = 9 - x 2 . Garis lurus hcr5ilung rior~gan paksi-y di (ilik (0, k) dun bersilang denganpaksi-x di (ilik (6, 0).

Rajah 5

Given that the area of the shaded region is 12 unit2 Find the value of k . Dibcri luas ka~rasan hcrlorck ialah 12 unit ('uri nilui k,

[ 5 marks]

[ 5 markah]

(b) Diagram 6 shows part of the curve x = y 2 . The curve intersects the straight line x = 4 at point A.

Rnjalr 6 rnenunjukkan sebahagian daripada lengklrng x = y '. Lengkung bersilang tiengan gnris lurus x = 4 di (irik A.

Diagram 6 ~ a j a h 6

Calculate the volume generated, in terms of .rr , when the shaded region is revolved through 360" about the y-axis. [ 5 marks]

Hitungkan isipndu yang dijanakan , dalam sebutan x apabila kawasan berlorek dikisnrkon rnelallri 36U0pada pnksi-y. [ 5 rnarkah]


MOZ@C

SULIT

2 11 (a) In an archery competition, the probability that Amir strikes the target is -.

3 [ f he tries six attempts, calculate the probability that,

Dalam satu pertandingan memanah, kebarangkulian Amir ferkena sasaran ialah 5 . 3

Jika dia rnelukukan enam kali cubaan, cari kebarangkalian bahawa

(i) he miss the target twice,

rliu tiduk terkena sasaran dua kali,

(ii) hc hits the target at least twice.

dra terhnu sasaran sektrrang-kurangnya dua kali.

[5 mark]

[5 markah]

(b) ?'he mass of students of a school are normally distributed with a mean of 45 kg and

the variance is 25 kg2.

Jisim bagi sekumpulan murid dalam sebuah sekolah adalah mengikut taburan normal

dengan min 45 kg don varians 25 kg2.

( i ) Find the percentage of students with weigh more than 50 kg,

Carikan pera/uspelajar yang mempunyai jisim melebihi 50 kg.

( i i ) Given that 35.2% of the students have a mass less than m kg, find

the value of m .

Diberi bahawa 35.2% daripadapelajar itu mempunyai jisim tidak melebihi m kg,

cur; nilai rn.


SULIT MOZ@C

Section C

Bahagian C

[20 marks]

[20 nlarkah]

Answer two questions from this section. Ja~vah dua soalan duripuda bahagian in;.

12 The displacement of a particle Q that moves in a straight line from a fixed point 0, is given by s = t3 -6t2 + 91 .

Su.sutwrr h~rgi surllzr zuruh Llyung hergeruk di sepanjang suatzr garis lurus clan rnclalui .crrirr titik letup 0, diberi oleh s = 1' - 6r2 + 91 .

Find

( b r i

( u ) the initial velocity of particle Q,

halqjzi ultlul zarah Q,

[ 2 marks]

[2 markuh]

( h ) the values of t when the particle Q change its direction of the motion, [ 2 marks]

nilai-nilai f apabila zuruh Q herrukar arah gerakari, [2 markah]

(c) the acceleratio~l of particle Q after 4 seconds.

pccrrtan zaruh Q selepas 4 saar.

(4 [he maximum velocity of particle Q. [3 marks]

huluju mu~itnum zarah Q. [3 nrurkuh]


MOZ@C

SULlT 15

14 Ilse graph paper to answer this question. Gur~ukan ker-tas gruf untuk menjuwub soalam ini.

A cake shop produces two types of cake, A and B . Each type of cake is made by using two ingredients, flour and butter.

Sehzruh kedui kek menghasilkun duo jenis kek, A dun B. Setiap jenis kek diperbuat dari dua ,jenis huhan. tc7pung dun ntcntega.

Table 3 shows the masses of the ingredients to make the two types of cakes. Joduul mcnunjukkun,jisim kand14ngan untuk membuat kedua-dua jenis kek.

1 Cake Mass in (g) f i k Jisim dalam ( )

Flour Butter E Table 3

Jadual3

Tepung

A i 100

The cake shop produces x cakes A and y cakes B. The making of cakes per day is based on the following constraints:

Keriai tersebut rnenghasilkan x biji kek '4 dun y biji kek B. Penghasilan bilangan biji kek dolam sehnri adalah berdasarkan kekangan berikut:

Menlega 100

I : The minirnum mass of flour used is 8 kg.

h!inimum jisim tepung yang digunakan ialah 8 kg.

I I : The maximum mass of butter used is 9 kg.

Mnk~imum jisim rnentega yang digunakan ialah 9 kg.

111 : The number of cake A produced is not more than the number of cake B.

Dilungun biji kek A adalah tidak melebihi bilangan biji kek B.

( ( I ) Write three inequalities, other than x 2 0 and y 2 0 , which satisfy all the above constraints. [3 marks]

fii1i.s tiga ketuksamaan, selain x 2 0 dun y 2 0 , yang memenuhi sentua kcktrngun tli ata.s. [3 markah]


MOZ@C

SULIT

( h ) Using a scale of 2 cm to 10 cakes on both axes, construct and shade the region R which satisfies all of the above constraints. [ 3 mark]

Dengan menggunakan skala 2 cm kepada 10 kekpada kedua-duapaksi, bina dun lorek ranfau R yang memenuhi semua kekangan di atas.

[3 markah]

(c ) Use your graph in 14(b), to find

Glrnakan graf anda di 14(b), untuk mencari

(i) the minimum number of cakes B, if 15 of cakes A are produced per day,

hilangan minimum kek B, jika 15 biji kek A dihasilkan dalam sehari,

( i i ) the maximum profit per day ifthe profit of each cake A and B is RM12 and RM9 respectively. [4 marks]

Jumlah keuntungan maksimum dalam sehari jika keuntungan bagi sebiji kek A dun kek B adalah musing-masing RM12 dun RM9.

[4 markah]

[ Lihat sebelah SULIT -

MOZ@C

SULIT 17 347212

15 Diagram 7 shows quadrilateral PQRST. Given QRS is a straight line, LPRQ is an obtuse angle and the area of triangle PST is 25 cm2.

Rajah 7 menunjukkan sisiempat PQRST. Diberi bahawa QRS ialah garis lurus, LPRQ ialah sudut cakah dun luas bagi segitiga PST ialah 25 cm2.

~ i a ~ r a m 7 Rajah 7

Calculate

Hitungkan

(a) the length, in cm, of PS,

panjang, dalam cm, bagi PS,

(b) LSPT .

(c) the area of APQS.

Luas APQS

[5 marks]

[ 5 markah]


[3 marks]

[3 markah]

END OF QUESTION PAPER

[ Lihat sebelah SULIT MOZ@C

1-3 l ,~hli. 2 xho~z i llic prices and tht: heightage for the lbur ingredients, A , B, C and I ) in 2007

~ n d 1008. L I ~ in nlahing a particular chocolate.

.Jiriirra/ 2 nic~nliiij~ikkor~ h a r p dmpernherut hag; Jjenis bahun, A, 5, C dun Dpudu tuhun 2007 ,lii~i 3008. \ ang d~,yrrnak(~n lrntuk rriembttut .sejenis coklar.

( ' 1 ) C',ilci~lnle thc index numbers of each ingredients, A , 5, C and L) in 2008 with '3(07 '15 the b ~ s e yeas.

11iillil~yk~111 irleick.~ hui.gil bog; sctiup hal~url/i, B, C t b n Dpudu ~ullun 2008 dcngan itrc~/g~~iuc/kii/i 2007 .sehu,yrri tuliun cisus.

1 I ~ n c e , cnlculdte ~ h c cornposite index for the cost of rnaking the chocolate in 2008 with 2007 as thc basc year.

[ 5 morks] . S c ' ~ ~ ~ / . i t \ r i \ ( I . Iiifzin~kuti riomhor hideks gt~bairan bugi kov rnembzrat coklat pada 2008 / I ~ ' I , ~ I \ ~ I \ kun ttriircn 1007.

[ 5 markuh]

( h I I I' the cl~uculall: is sold at RM 150 per pack in 2007, calculate the selling 1'1 icc per pack in 1008. ( 2 rnarks]

(c.) I I IC ~ ~ I C C 111j ice~ of a commodity in 2007 and 2008 based on 2006 are I25 and 120 rcspect~vely. If the price of the commodity in 2007 is Rkl.1150, crrlsulate its price in 2008. [3 tnarks]

I~itii~li.~ 11~11.gtr hogi surlc kor?lorlitipada 2007 dur~ 2008 bcrususkan tuhun 2000 ~~i~t.\.irl,y-rnrr.siti~q utiuluh 125 iiun 120. ,lib 11urgu hagi konloditi tersebur pada ?0117 ict/u/l Rk1 1 00, Ilif~irigk~iri /i~~qyc~n?'o padu 2008.

[3 murkah]

[ Lihat sebelah SULIT -

MOZ@C

1449/2 Matematics Kertas 2 Sept. 2008

.JABATAN PELAJARAN NEGERI JOHOR

PEPERIKSAAN PERCUBAAN SPM TAIFL. 2008

SKEMA PEMARKAHAN

P v] MATEMATICS

Kertas 2

Duajam tiga puluh minit

144912 0 Hak Cipta .lPNJ 2008

SULIT

MOZ@C

SKEMA PERMARKAHAN MATEMATIK TAMBAIUN KERTAS 2 PEPERIKSAAN PERCURAAN SPM TAHUN 2008

Number

I

Full Marks Solution and mark scheme

x = y - 2

y 2 + 2 y + ( y - 2 y = 9

2y2 - 2 y - 5 = 0

-(2) + J-40(-5) Y =

2(2)

5

6

5

Sub Marks P 1

K1

KI

N1 N1

K I

K l

N1

P 1 P 1

N1

NI

P 1

N1

K1

N1

x =0.158,-3.158

(b)

3(a)

(b)

= J(2x3 - dx)dr

1 =-x4 -2x2 +C

2

1 4 2 7 v=-x -2x +- - 2 2

m, = 2(113 -4(1) = -2

1 M , = -

2 2 y = x + 3

- 64 x = - - = $ 8 ,

G x 2 = 580

o = /? -- = 2.9155

- 2.9 1 55 on, - ---- 4

= 0.7289

MOZ@C

Number of solutions = 3

6(a)

N1

N1 for Sin shape

tan2 8 + 1

tan B sec2 B tan B

sec B cos ec8

N1 for 1 periodic and max = 2

N1 for modulus

N1 for

K1 for straight line

MOZ@C

Number

7

8(a>(i>

(ii)

(b)

(c>(i>

( i i )

Solution and mark scheme

logy = log a + x2 log h

log a = 0.19(+0.0 1)

a = 1.55

log b = 0.889 - 0.493

- = 0.0792 9 - 4

Graph: K I for uniform scale on both axes N 1 for a1 l correct points N I for best fit line

JG=Io m = 8

L

= --(10!+l l j ) 3 -

0, P and T are collinear.

Sub Marks

NI NI

P 1

Kl N1

KI NI

KI NI

K1 N1

K1

N1

K1 N1

K1

N1

Full Marks

10

I

10 MOZ@C

Perimeter of shaded region =AB+BD+AD

= 5j1.572) + 1 O(0.5237) + 3.6602

= 1 6 . 7 5 2 2 ~ ~

Number

9ja)

Area of the shaded region = ABEO + sector AOB - sector BED


sin LBEO = 0.5

Sub Marks K1 N1 :. LBEO = 30° = 0.5237rads

K1 for any correct formula used

N1 for any correct answer

K 1 for all correct operation

Full Marks 1

MOZ@C

-- 1 1 (a)

- - Solution and mark scheme Full Marks

--

Sub -7 Marks

K1

N1

K1 for

I ncrprqn-r

K1 for operation N1

K1 for x-P z=-

LT

N l

K1 for x-P z=--

LT

KI

NI

MOZ@C

Number -..-----.pp--


v,,,, = 3(2)' - 12(2) + 9

= -3 m/s

2 1 I,, =--x 100 = 140

15

--- Sub Marks K1

K1 for formula I

Full Marks

MOZ@C

---

Number Solution and mark scheme -

sin LQRP - sin 30' -- 9.5 5.8 LQRP = 125'1' LPRS = 180'-125'11= 54'59'

1 - (6.322)(1 1.2) sin LSPT = 25 2

SPT = 44'55'

LQPR = 24'59' Area APQS

1 1 = -(9.5)(5.8) sin 24'59'+-(5.9)(7.5)sin 54O59'

2 2 = 11.6359+18.12

= 29.7559cm2

Sub Marks - K1

NI NI

Kl N I

KI N1

P 1

K1

N1

Full Marks

MOZ@C

MO

Z@C


I :100x+400y28000

II : 1 OOx + 200y I 9000

III : x < y

refer to the graph:

K 1 for uniform scale on both axes K I for any one correct straight line drawn N 1 for correct region R

(i) 16

( i i ) 1 2 x 4 9 ~ = 180 ,(30,30) Maximum profit = 30(RM12) + 30(RM9)

= RM630

Sub Marks

K1 N1

K1

Kl

N1

N1 Nl N1

K1 K1 N1

PI

N1 K1 for k=ax+by N1

Full Marks

10

10

~ MOZ@C

jabatan pelajaran negeri johor peperiksaan percubaan spm 2008 · pdf filepeperiksaan percubaan...

Documents