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SULIT 2 3472/1

The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used

-b±~b2 -4ac x=

2a

2 a"' X tf=am+n

3 a"' + an=am- n

4 (a"') n =a mn

5 log a mn = log am + log a n

6 m

log a - = log am - log a n n

7 log a mn = n log a m

dy dv du y=uv, -=u-+v-

dx dx dx

du dv

2 u dy __ vd;-ud;;

y=-v, dx vz

3 dy dy du -=-x-dx du dx

2 Midpoint

(X, y) = (XI ; Xz

3 ld=Jxz + yz

4

3472/1

ALGEBRA

8 l b- loge b 0~- --

loge a

9 Tn =a+ (n-l)d

n 10 Sn = -[2a + (n -l)d]

2 II Tn = ar,_ 1

12 Sn = a( r" -1) = a(l- r")

r-1 1-r a lri<I 13 s --.,-1-r'

CALCULUS

4 Area under a curve I>

= Jy dx or a

b

= Jx dy a

5 Volume generated b

= r~2 dx or

a

(I

GEOMETRY

, (r :t: 1)

5 A point dividing a segment of a line

( x,y) = (nx, +mxz, nyt +myz) m+n m+n

6 Area of triangle

=±l(x,yz +XzY3 +X3Yt1 )-(XzYt +X3Y2 +x,y3)1

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SULIT

x= 2> N

r _!_N-F] 5 m = L+l2 fm C

Arc length, s = re

l 2 Area of sector, A= -r29

2 3 sin 2A +cos 2A = 1

4 sec2A = 1 + tan2A

5 cosec2 A = l + cor A

6 sin 2A = 2 sinA cosA

7 cos 2A = cos2A- sin2 A = 2 cos2A- I = l- 2 sin2A

8 tan 2A = 2tanA 1-tan 2 A

3472/l

3 3472/1

STATISTICS

8 "P = n! r (n-r)!

9 "C = n! r (n- r)!r!

10 P(AvB) = P(A) + P(B)- P(AnB)

12 Meanp = np

13

14 X -f.! Z=--

cr

TRIGONOMETRY

9

10

11

12

13

14

sin (A ±B) = sinA cosB ± cosA sinB

cos (A ± B) = cosA cosB + sinA sinB

tan (A ±B)= tan A± tan B I+tanAtanB

a b c --=--=--sin A sinB sinC

a2 = b2 + c2 - 2bc cosA

= .!.absinC Area of triangle 2

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1

'SULIT 4

1 y

Answer all questions. Jawab semua soalan.

3._---4----~----~--~

2J----__..---+-----l-----i

0 2 3

The graph above shows the relation Qetween x andy. State

(a) the type of relation,

(b) the objects of3.

X

Graf diatas menunjukkan hubungan antara x dan y. Nyatakan

(a) jenis hubungan itu,

(b) objek-objek bagi 3.

Answer/Jawapan:

(a)

(b)

3472/1

3472/1

[2 marks]

[2 markah]

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SULIT s 3472/l

2 Given that function g(x) = 2x + m and g 2 (x) = nx + 15 where mand n are

constant$. Find the value of m and .n .

3

Diberifungsi g(x) = 2x + m dan g 2 (x) = nx + 15 dengan lceadaan mdan n

ialah pemalar. Carl nilai m dan n.

Answer/Jawapan:

Given the function h( x) = x + 5

, find 2

Diberifungsi h(x) = x;S, cari

(a) h _, (2),

(b) the value of p when h-1 (p) = 7.

nilai bagi p apabila h-' (p) = 7.

Answer/Jawapan:

(a)

(b)

[3 marks] [3 markahJ

[4 marks]

[4 markah]

For examiner's

11se tlfll)i

3

3472/1

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For SULIT 6 3472/1

exomtrrer's

use only 4 Given -2 and 3 are the roots of the quadratic equation 2x 2 +hx-3k = 0. Find the

~

value of h and of k. [3 marks]

Diberi -2 dan 3 ialah punca-punca persamaan kuadratik 2 x 2 + hx-3 k = 0. Cari nilai h dan nilai k .

[3 markah]

Answer!Jawapan:

·----------------------------------------------------------------5 The graph of the quadratic function f( x) = a(x- 2)2 + q has a minimum point

(p, - 8) where a, p and q are constants.

Grafbagifungsi kuadratik J(x} = a(x-2)2 + q mempunyai titik minimum(p, -8)

dengan keadaan a, p dan q ialah pemalar.

State Nyatakan

(a) the value of p, nilai bagi p,

(b) the value of q, ni/ai bagi q,

(c) the range of values of a. julat nilai-nilai a.

Answer/Jawapan:

(a)

(b)

(c)

[3 mark~·] [3 markah]

[ s

(=m 8 347211 SULIT

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SULIT 7

6 4x Find the range of values of x for 3x -1 ~ -.

x+2

4x Cari julat nilai-nilai x bagi 3x -1 ~ --.

x+2

Answer/Jawapan:

7 Solve the equation : Selesaikan persamaan :

5 Jx X gx - 100

Answer/Jawapan:

3472/1

3472/1

[3 marks] ·

[3 markah]

[3 marks] [3 markah]

For examiner's

11se only

7

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For SULIT 8 3472/1

e:xominer 's

use only 8 Given Slog ... 2 + log ... y - 2log ... 4 = 3 , express yin terms of x. [4 mark~]

Diberi Slog ... 2 +log ... y- 2log ... 4 = 3, ungkapkan y dalam sebutan x. [4 markah]

Answer!Jawapan:

9 If 27, 3k and 48 are three consecutive terms in a geometric progression. Find the

values of k. [3 mark~J

Jika 27. 3k dan 48 ialah tiga sebutan berturut-turut bagijanjang geometri. Cari

nilai-nilai bagi k. [3 markah]

Answer/Jawapan:

9

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SULIT 9 3472/l For

10 51 , 58 , 65 , ... 191 are the first n terms of an arithmetic progression. Find the common different of the progression and the value of n.

51 , 58 , 65 , ... 191 ialah n sebutan pertama bagi suatu janjang aritmetik. Cari beza

sepunya bagi janjang terse but dan nilai bagi n.

Answer!Jawapan:

[3 markvJ [3 markah]

5 11 Given the geometric progression 1 0 , - 5 , - , ... . Find the sum to infinity of the

2 progression.

5 Diberi suatu janjang geometri 10 , - 5 , - , ... . Cari hasil tambah hingga sebutan

2 ketaklerhinggaan bagi janjang terse but.

Answer!Jawapan:

3472/l

[3 marks] [3 markah]

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10

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SULIT 10 3472/l For

examiner's useon/y . 12 y

Diagram 12 shows the straight line graph obtained by plotting 2 against x.

12

w 0

X

Rajah 12 menunjukkan graf garis lurus yang diperoleh dengan memplot ~ X

me/awan x.

(m, 13)

(2,1) ~-------------------------------.x

Diagram 12 Rajah 12

The variables x and y are related by the equation y = 3x3 + kx

2 where k is a

constant. Find the value of k and of m.

Pembolehubah x dan y dihubungkan oleh persamaan y = 3x3 + kx2

, dengan

keadaan k ialah pemalar. Cari nilai k dan nilai m.

Answer/Jawapan:

3472/1

[4 marks] [4 markah]

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SULIT 11 3472/1 For e:«Jminer 's

use on~v

13

3472/1

y~, u

D £

\' ~ \

~ ............

\ _4

~ T

~ \

](

_, 0

"-.... ...........

i'-..

~

1 ...........

~

L

Dtagram 13 Rajah 13

(

-...........

' 0

~

Diagram 13 shows a triangle PQR drawn on a Cartesian plane. Find

(a) the equation of the straight line PQ,

(b) the area of the triangle PQR.

rt• ~ I

[4 marks]

Rajah 13 menunjukkan sebuah segitiga PQR dilukis di atas sarah Cartesan. Cari

(a) persamaan garis lurus PQ,

(b) luas bagi segitiga PQR. [4 markah]

Answer/Jawapan:

13

ca [ Lihat halaman sebelah 0

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SULIT 12 3472/1 For

examiner's

useonly 14 Giventhatequationofastraightline 2y=(k-I)x+7 withgradient -3 andpasses

through point (1 • t) . Find the value of k and of t. [3 marks]

Diberi suatupersamaan garis lurus 2y= (k-1)x+7 dengan kecerunan -3 dan

melalui titik (I, t). Cari nilai k dan nilai t. [3 markah]

Answer/Jawapan:

14

w 15 Given the curve y = x(x-2)

2 has two turning points. Find the values of x for those

turning points. [3 marks]

Diberi lengkung y = x( X- 2 r mempunyai dua titik pusingan. Cari nilai-nilai X

bagi kedua-dua titikpusingan itu. [3 markah]

Answer/Jawapan:

15

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SULIT 13 3472/1

16 Diagram 16 shows a trapezium ABCD. Given that BC is parallel to AD, BC = 5[ + 2.f.,

CJ5==3i-7j and AD=2BC.

Rajah 16 menunjukkan satu trapezium ABCD. Diberi BC selari dengan AD,

BC=5{+2.f.. CD=3£-7.f. dan AD=2BC.

Find AC . Cari AC.

A

Answer/Jawapan:

3472/1

Diagram 16 Rajah 16

c

D

[3 marks] [3 markah]

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use only •

16

w 0

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For SULIT 14

examiner's useon(v 17

17

w 0

t' .- ~ ~

.A. '"t"

~ ...... .,........

"'\ .... v .,., ~

11 ..,....,. .I.YJ.

~ v ~ /

-2 0

y

/

> ..

........ / v .., v

,

L

Diagram 17 Rajah 17

(

v

Diagram 17 shows vector MN drawn on a Cartesian plane.

(a) Express MN in the form of (;) .

(b) Find the unit vector in the direction of MN .

~

-;::;'

Rajah J 7 menunjukkan vektor MN dilukis di atas satah Cartesan.

(a) Ungkapkan MN dalam bentuk (;}

(b) Cari vektor unit dalam arah MN.

Answer/Jawapan:

(a)

(b)

3472/1

3472/1

]\[ 14 ~

.lit.

1 )' . .~

[3 markY]

[3 markah]

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SULIT 15 3472/1 For

18 x1,x2 ,x3 ,x4 ,x5 and x6 is a group of data with standard deviation and variance as 2 and 4 respectively. When each data above is multiplied by 5, find

x1,x2 ,x3,x4 ,x5 dan x6 ialah satu kumpulan data dengan sisihan piawai dan varian."i

masing-masing ialah 2 dan 4. Apabila setiap data tersebut didarab dengan 5, cari

(a) the new standard deviation, sisihan piawai baru,

(b) the new variance. varians baru.

Answer/Jawapan:

(a)

(b)

[2 marks] [2 markah]

19 Solve the equation sinx-cos2x = 0 for o· ~X~ 360".

Selesaikan persamaan sin X- cos 2x = 0 bagi o· ~ X ~ 360".

Answer/Jawapan:

3472/1

[4 mark"i] [4 markah]

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18 em

19

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20

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SULIT 16 3472/1

20

A

0

Diagram 20 Rajah20

B

Diagram 20 shows a sector OAB of a circle with centre 0 . Find the area of the shaded

segment. [4 marks]

Rqjah 20 menuryuklwn sebuah sektor OAB bagi sebuah bulatan berpusat 0. Cari luas

bagi tembereng berlorek.

[4 markah]

Answer/Jawapan:

3472/1 SULIT

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SULIT 17 3472/1

21 Given y = ( 4x- 7)2 and the rate of change of x is 3 units per second. Find the rate

of change of y when x = 2 .

Dibert y = ( 4x- 7)2 dan kadar perubahan x ialah 3 unit per saat. Cari kadar

perubahan y apabila x = 2 .

Answer/Jawapan:

22 Given that f2J(x)dx = 6

(a) r 2 5f(x)dx

and J1 g(x)dx=9. Findthevalueof

-2

(b) h if f2[3h-2f(x)+g(x)]dx=l5.

Diberi f2J(x)dx = 6 dan f

2g(x)dx =9. Cari nilai bagi

(a) r2 5 f(x)dx

(b) hjika f2[3h-2f(x)+g(x)]dx=l5.

Answer!Jawapan:

(a)

(b)

[3 marks] [3 markah]

[3 marks]

[3 markah]

For examiner's

use on~~·

21

[EJ

22

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For examiner's

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23

SULIT 18 3472/1

23 A teacher wants to select 3 students to represent school in an essay-writing competition. These 3 students are selected from a group of 5 boys and 2 girls. Find the number of possible selections if (a) there are no condition for the selection, (b) the selection must consist boy and girl. [4 marh]

Seorang guru ingin memilih 3 orang murid untuk mewakili sekolah dalam pertandingan menu/is karangan. 3 orang murid ini dipilih daripada sekumpulan 5 orang murid lelaki dan 2 orang murid perempuan. Cari bilangan pilihan yang mungkin jika (a) tiada syarat bagi pemilihan itu, (b) pemilihan itu mesti mempunyai murid lelaki dan murid perempuan.

[4 markah] Answer/Jawapan:

(a)

(b)

[E] _____ _

24 w 0

24 5 The probability that Chin Loong will get distinction in a Mathematics quiz is - and

6

the probability that he will get distinction in a Chemistry quiz is 2

Find the 3

probability that Chin Loong gets distinction in (a) both quiz, (b) at least one quiz. [3 marh]

Kebarangkalian Chin Loong mendapat cemerlang da/am kuiz Matematik ia/ah ~ dan 6

kebarangkalian mendapat cemerlang dalam kuiz Kimia ialah ~. Cari kebarangkalian 3

Chin Loong mendapat cemerlang dalam (a) kedua-dua quiz, (b) sekurang-kurang satu quiz.

Answer/Jawapan:

(a)

(b)

347211

(3 markah]

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SULIT 19

25 Diagram 25 shows the graph of a binomial distribution of X.

Rajah 25 menunjukkan graf suatu tabu ran binomial bagi X

P(X =x) ... 0·36- -----------------

n------------0·21- ----------- ----- ------

0·15 -------

3472/1

----- ------~-----1 ~----~-----+------+-----~------4------.x

0·03- ------ -----

0

Find

Cari

(a) the value of n ,

nilai n,

(b) P(X~2).

Answer/Jawapan:

(a)

(b)

3472/1

2 3

Diagram 25 Rajah25

END OF QUESTION PAPER KERTAS SOALAN TAMAT

I

4

[3 marks] [3 markah]

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SULIT 20

INFORMATION FOR CANDIDATES MAKLUMAT UNTUK CALON

1. This question paper consists of25 questions. Kertas soa/an ini mengandungi 25 soalan.

2. Answer aU questions. Jawab semua soalan.

3. Write your answers in the spaces provided in the question paper. Tulis jawapan anda da/am ruang yang disediakan dalam kertas soa/an.

4. Show your working. It may help you to get marks.

3472/1

Tunjukkan langkah-langkah penting dalam kerja mengira anda. lni boleh membantu anda untuk mendapatkan markah.

5. If you wish to change your answer, cross out the answer that you have done. Then write down the new answer. Sekiranya anda hendak menu/car jawapan, batalkan jawapan yang telah dibuat. Kemudian tulis jawapan yang baru.

6. The diagrams in the questions provided are not drawn to scale unless stated. Rajah yang mengiringi soa/an tidak dilukis mengikut skala kecuali dinyatakan.

7. The marks allocated for each question are shown in brackets. Markah yang diperuntukkan bagi setiap soalan diturifukkan dalam kurungan.

8. A list of formulae is provided on pages 2 and 3. Satu senarai rumus disediakan di halaman 2 dan 3.

9. A booklet of four-figure mathematical tables is provided. Sebuah buku si.fir matematik empat angka disediakan.

I 0. You may use a non-programmable scientific calculator. Anda dibenarkan menggunakan kalkulator sainti.fik yang tidak boleh diprogram.

11. Hand in this question paper to the invigilator at the end ofthe examination. Serahkan kertas soalan ini kepada pengawas peperiksaan di akhir peperiksaan.

3472/1 SULIT

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SULIT 3472/1

3472/1 Additional Mathematics Paper 1SULIT

2PROGRAM PENINGKATAN PRESTASI AKADEMIK SPM 2012

Marking SchemeAdditional Mathematics Paper 1

Question Solution/ Marking Scheme Answer Marks1 (a) many-to-one

(b) 0 and 2

1

1

2 B2 : 5m = or 4n =

B1: 2(2 )x m m+ +5m = and 4n = 3

3(a) B1: 5 2

2a +

=

(b) B1: (7)h p= or 14a – 15

(a) 1-

(b) 6p =

2

2

4 B2 : 2h = - or 4k =

B1: 2 32h

- + =- or ( )( ) 32 32

k- = - or

( )( )2 3 0x x+ - =

2h = - and 4k = 3

5 (a) 2

(b) –8

(c) 0a >

1

11

6 B2 :

or 23

x = , 1x = -

B1: (3 2)( 1)x x- +

2 ,3

x ³ 1x £ - 3

7B2 : 1000 100x = or 3 210 10x =

B1 : 125 8 100x x´ = or 3 35 2 100x x´ =

23

x =3

1- 23

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SULIT 3472/1

3472/1 Additional Mathematics Paper 1SULIT

3Question Solution/ Marking Scheme Answer Marks

8B3 :

83

22

4y x´=

B2 :8

22log 3

4xy´=

B1 : 8log 2x or 2log 4x

3

16xy =

4

9 B2: k 2 = 144

B1: 48 33 27

kk=

k = 12 and -12 3

10 B2 : 7d = or 21n =

B1 : 51 ( 1) 191n+ - =7d = and 21n = 3

11B2 : 10

112

S¥ =æ ö- -ç ÷è ø

B1 : 12

r = -

203

3

12 B3 : k = –5 or m = 6

B2: 1 = 3(2) + k or 13 = 3m + k

B1 : 2 3y x kx

= +

k = –5 and m = 6 4

13(a) B1: 1

2PQm =-

(b) B1: 1 8(7) ( 2)(1) 8(1) ( 2)(2)2

A = + - - - -

OR equivalent

(a) 1 62

y x= - +

(b) 25

2

2

14B2 : 5k = - or 1

2t =

B1 : 1 32

k -= - or 2 ( 1)(1) 7t k= - +

5k = - and 12

t = 3

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SULIT 3472/1

3472/1 Additional Mathematics Paper 1SULIT

4Question Solution/ Marking Scheme Answer Marks

15B2 : 2x = or 2

3x =

B1: 21( 2) 2( 2)x x x- + - OR 23 8 4x x- +2x = and 2

3x =

3

16 B2 : 2AC BC CD= -uuur uuur uuur

OR 2AD BC CD= -uuur uuur uuur

B1 : AC AD DC= +uuur uuur uuur

or104

ADæ ö

= ç ÷è ø

uuur

OR AC AB BC= +uuur uuur uuur

or29

ABæ ö

= ç ÷è ø

uuur

7 11i j+% %

or

711æ öç ÷è ø

3

17 (a) B1 : MN ON OM= -uuuur uuuur uuuur

(a)113

æ öç ÷è ø

(b)1113130

æ öç ÷è ø

2

1

18 (a) 10

(b) 100

1

1

19 B3 : 30 , 150 , 270o o o

(any 2 correct answer)

B2:1sin sin 12

x x= = -or

B1: 2sin (1 2sin ) 0x x- - =

30 , 150 , 270o o o 4

20B3 : ( ) ( ) ( )2 21 110 1 22 10 sin 70

2 2o× -

B2: ( ) ( )21 10 1 222

× or ( ) 21 10 sin 702

o

B1: 70 1 22180

o

o radp´ = ×

14 02× 4

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3472/1 Additional Mathematics Paper 1SULIT

5

END OF MARKING SCHEME

Question Solution/ Marking Scheme Answer Marks21

B2: 8 3dydt

= ´

B1: dy dy dxdt dx dt

= ´ or 2(2 7) 4dy xdx

= - ·

24 3

22

(b) B1 : 123 [ ]h x - or ( )2 6 or 9

(a) 30-

(b) 2h =

1

2

23 (a) B1 : 73C

(b) B1 : 5 22 1C C´ or 5 2

1 2C C´

(a) 35

(b) 25

2

2

24

(b) B1 : 56

1æ ö´ç ÷3è øor 1

62æ ö´ç ÷3è ø

(a) 59

(b) 1718

1

2

25(a) B1 : 0 15 0 36 0 21 0 03 1n× + + × + × + × = (a) 0 25n = ×

(b) 0 × 6

2

1

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