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PEPERIKSAAN PERCUBAAN SPM

TAHUN 2013

MATEMATIK TAMBAHANKertas 1Dua Jam

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

Nama :

Tingkatan:

3472/1Matematik TambahanKertas 1Sept 20132 jam

1. This question paper consists of 25 questionsKertas soalan ini mengandungi 25 soalan.

2. Answer all questions.Jawab semua soalan.

3. Give only one answer for each questionBagi setiap soalan berikan SATU jawapan sahaja.

4. Write the answers clearly in the space provided in the question paper.Jawapan hendaklah ditulis pada ruang yang disediakan dalam kertas soalan.

5. Show your working. It may help you to get marks.Tunjukkan langkah-langkah penting dalam kerja mengira anda. Ini bolehmembantu anda untuk mendapatkan markah.

6. If you wish to change your answer, cross out the work thatyou have done. Then write down the new answer.Sekiranya anda hendak menukar jawapan, batalkan kerja mengira yang telahdibuat. Kemudian tulis jawapan yang baru.

7 The diagram in the questions provided are not drawn to scale unlessstated.

Rajah yang mengiringi soalan ini tidak dilukiskan mengikut skala kecuali dinyatakan.

8. The marks allocated for each question and sub-part of a question areshown in brackets.Markah yang diperuntukkan bagi setiap soalan atau ceraian soalan ditunjukkandalam kurungan.

9. A list of formulae is provided on page 2 to 3Satu senarai rumus disediakan di halaman 2 hingga 3

10. You may use a non-programmable scientific calculator.Anda dibenasrkan menggunakan kalkulator sainstifik yang tidak boleh diprogram.

11 This question paper must be handed in at the end of the examination.Kertas soalan ini hendaklah diserahkan pada akhirpeperiksaan .

KodPemeriksa

SoalanMarkahPenuh

MarkahDiperoleh

1 32 33 24 35 36 37 48 39 4

10 311 312 313 214 315 316 317 418 419 420 321 322 423 424 325 3

Jumlah80

Kertas soalan ini mengandungi 23 halaman bercetak

http://www.chngtuition.blogspot.comwww.tutormansor.com

[ Lihat sebelah3472/1

The following formulae may be helpful in answering the questions. The symbols given are the ones commonlyused.Rumus-rumus berikut boleh digunakan untuk membantu anda menjawab soalan. . Simbol-simbol yang diberi adalah yang biasadigunakan.

ALGEBRA

12 4

2

b b acx

a

2 am an = a m + n

3 am an = a m - n

4 (am) n = a nm

5 loga mn = log am + loga n

6 logan

m= log am - loga n

7 log a mn = n log a m

8 logab =a

b

c

c

log

log

9 Tn = a + (n -1)d

10 Sn = ])1(2[2

dnan

11 Tn = ar n-1

12 Sn =r

ra

r

ra nn

1

)1(

1

)1(, (r 1)

13r

aS

1, r <1

CALCULUS( KALKULUS)

1 y = uv ,dx

duv

dx

dvu

dx

dy

2v

uy ,

2vdx

dvu

dx

duv

dy

dx

,

3dx

du

du

dy

dx

dy

4 Area under a curve ( Luas dibawah lengkung)

= b

a

y dx or

= b

a

x dy

5 Volume generated ( Isipadu Janaan)

= b

a

y 2 dx or

= b

a

x 2 dy

5 A point dividing a segment of a lineTitik yang membahagi suatu tembereng garis

( x,y) = ,21

nm

mxnx

nm

myny 21

6 Area of triangle ( Luas Segitiga )

)()(2

1312312133221 1

yxyxyxyxyxyx

1 Distance (Jarak) = 221

221 )()( yyxx

2 Midpoint ( Titik Tengah )

(x , y) =

221 xx

,

221 yy

3 22 yxr

42 2

ˆxi yj

rx y

GEOMETRY

2


3472/1 [ Lihat sebelah

3

1 Arc length, s = r( Panjang lengkok) s = j

2 Area of sector , L = 21

2r

( Luas sektor L = 2

2

1j )

3 sin 2A + cos 2A = 1

4 sec2A = 1 + tan2A

5 cosec2 A = 1 + cot2 A

6 sin 2A = 2 sinA cosA

7 cos 2A = cos2A – sin2 A= 2 cos2A - 1= 1 - 2 sin2A

8 tan 2A =A

A2tan1

tan2

TRIGONOMETRY

9 sin (A B) = sinA cosB cosA sinB

10 cos (A B) = cosA cosB sinA sinB

11 tan (A B) =BtanAtan

BtanAtan

1

12C

c

B

b

A

a

sinsinsin

13 a2 = b2 + c2 - 2bc cosA

14 Area of triangle = Cab sin2

1

( Luas Segitiga )

1 x =N

x

2 x =

f

fx

3 =N

xx 2)(=

2_2

xN

x

4 =

f

xxf 2)(=

22

xf

fx

5 m = Cf

FNL

m

2

1

6 1

0

100Q

IQ

71

11

w

IwI

8)!(

!

rn

nPr

n

9!)!(

!

rrn

nCr

n

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

11 P (X = r) = rnrr

n qpC , p +q = 1

12 Mean µ = np

13 npq

14 z =

x

STATISTICS



Answer all questions.Jawab semua soalan

1 Diagram 1 shows the linear function f.Rajah 1 menunjukkan fungsi linear f.

Diagram 1Rajah 1

State:Nyatakan:

(a) value of mnilai m

(b) value of nnilai n

(c) image of 10imej bagi 10

[3 marks][3 markah]

Answer / Jawapan :

(a)

(b)

(c)

Forexaminer’s

use only

3

1

1 2

m 4

3 n

4 8

x f(x)

4



2 Given function g and h as g(x) = 2x – 3 and h(x) = 2x2 + 2. Find hg(x).Fungsi g dan h diberi sebagai g(x) = 2x – 3 dan h(x) = 2x2 + 2. Cari hg(x).

[3 marks][3 markah]

Answer / Jawapan :

3 Find f – 1 (x) given f(x) = –2 x +10.Cari f – 1 (x) diberi f(x) = –2 x +10.

[2 marks][2 markah]

Answer/ Jawapan :

Forexaminer’s

use only

3

2

2

3

5



4 Find the values of k if x2 + 2kx + 2 – k = 0 has two equal roots.Cari nilai-nilai k jika x2 + 2kx + 2 – k = 0 mempunyai dua punca sama.

[3 marks ][3 markah ]

Answer /Jawapan:

5 Find the range of values of x for (x – 4)2 < 6 – x.Cari julat nilai x bagi (x – 4)2 < 6 – x.

[3 marks][3 markah]

Answer/Jawapan:

Forexaminer’s

use only

3

4

3

5

6



6 Solve the equation √27 = .

Selesaikan persamaan √27 = .

[3 marks][3 markah]

Answer / Jawapan:

7 Given that log3 p = x and log3 q = y, express log9 in terms of x and y.

Diberi log3 p = x dan log3 q = y, ungkapkan log9 dalam bentuk x dan y.

[4 marks][4 markah]

Answer / Jawapan :

3

6

Forexaminer’s

use only

4

7

7



8. Diagram 8 shows the graph of the quadratic function y = f(x). The straight line y = –14is a tangent to the curve y = f(x).Rajah 8 menunjukkan fungsi kuadratik y = f(x). Garis lurus y = –14 ialah tangen kepadalengkung y = f(x).

Diagram 8Rajah 8

(a) State the equation of the axis of symmetry of the curve.Nyatakan persamaan paksi simetri bagi lengkung itu.

(b) Express f(x) in the form of (x + p)2 + q, where p and q are constants.Ungkapkan f(x) dalam bentuk (x + p)2 + q, di mana p dan q adalah pemalar.

(c) Express f(x) in general form.Ungkapkan f(x) dalam bentuk am.

[3 marks][3 markah]

Answer / Jawapan :

(a)

(b)

(c)

5–1

y

y = –14

x0

Forexaminer’s

use only

3

8

8



9 The sum of first n terms of an arithmetic progression is given by S = 2 (9 − ).Hasil tambah n sebutan pertama bagi suatu janjang aritmetik diberi oleh = 2 (9 − ).

FindCari

(a) the common difference,beza sepunya,

(b) the sum of the first 6 terms.hasil tambah 6 sebutan pertama.

[4 marks][4 markah]

Answer / Jawapan :

(a)

(b)

Forexaminer’s

use only

4

9

9



10 Given the geometric progression –10, ,− …….., find the sum to infinity of the

progression.

Diberi janjang geometri –10, ,− …….., cari hasil tambah hingga ketakterhinggaan

janjang itu.[3 marks]

[3 markah]

Answer / Jawapan:

11 The vertices of a parallelogram are ( 9, 8 ), ( 9, 2 ), ( 2, p ) and ( 4, 7 ). Given that itsarea is 40 unit2, find the values of p.Bucu-bucu sebuah segi empat selari ialah ( 9, 8 ), ( 9, 2 ), ( 2, p ) dan ( 4, 7 ). Diberiluasnya ialah 40 unit2. Cari nilai-nilai p.

[3 marks][3markah]

Answer /Jawa pan:

Forexaminer’s

use only

3

10

3

111

10



12.

Diagram 12Rajah 12

Diagram 12 shows a straight line ST. Point P lies on ST such that SP : PT = 2 : 3.Find the coordinates of P.Rajah 12 menunjukkan garis lurus ST. Titik P terletak di atas ST dengan keadaanSP : PT = 2 : 3. Cari koordinat P.

[3 marks][3 markah]

Answer/ Jawapan:

3

12

Forexaminer’s

use only

P(h, k)

y

xS(–2, 0)

T(8,10 )

11



13 Diagram 13 shows the vectors OA

= a and OB

= b

Rajah 13 menunjukkan vektor OA

= a dan OB

= b

Diagram 13Rajah 13

If C is the midpoint of AB, find OC

Jika C adalah titik tengah AB, cari OC

.

[2 marks][2 markah]

Answer / Jawapan:

2

13

Forexaminer’s

use only

O

A

B

12



14 Given that AB

=5

m

and CD

=

2

k

, find

Diberi AB

=5

m

dan CD

=

2

k

, cari

(a) the value of m, where m >0 , if the unit vector in the direction of AB

is5 12

13 13i j

nilai m, dengan keadaan m > 0, jika vektor unit dalam arah AB

ialah5 12

13 13i j

(b) the value of k, if AB

is parallel to CD

.

nilai k, jika AB

selari denganCD

.

[3 marks][ 3 markah]

Answer/ Jawapan:

(a)

(b)

3

14

Forexaminer’s

use only

13



15 Diagram 15 shows a straight line graph ofy

xagainst

1

x. Express y in terms of x.

Rajah 15 menunjukkan graf garis lurusy

xmelawan

1

x. Ungkapkan y dalam sebutan x.

Diagram 15Rajah 15

[3 marks][ 3 markah ]

Answer / Jawapan:

O

(4, 16)

(– 2, 4)

x

y

2

1

x

Forexaminer’s

use only

3

151

14



16 Given that sin A =5

3 and tan B =

12

5such that angle A and B lie on the same

quadrant. Evaluate cos (A + B). [3 marks]

Diberi bahawa sin A =5

3 dan tan B =

12

5di mana sudut A dan sudut B berada

pada sukuan yang sama. Nilaikan cos (A + B).[3 markah]

Answer / Jawapan:

17 Solve the equation 6 sec2 x 13 tan x = 0, for 0 x 360.

Selesaikan persamaan 6 sec2 x 13 tan x = 0, bagi oo x 3600 .

[4 marks][4 markah]

Answer / Jawapan:

4

17

Forexaminer’s

use only

3

16

15



18 Diagram 18 shows two sectors OAB and OCD of circles with centre O.Rajah 18 menunjukkan dua sektor OAB dan OCD sebuah bulatan berpusat di O.

C

A

rad.

O B D

Diagram 18Rajah 18

Given AOB = radian, arc length AB is twice the length of radius OB and radiusOD = 6 cm.Diberi AOB = radian, panjang lengkok AB dua kali panjang jejari OB dan OD = 6 cm.

Find,Cari

(a) the value of ,nilai ,

(b) perimeter of the shaded region.perimeter kawasan berlorek.

[4 marks][4 markah]

Answer / Jawapan:

(a)

(b)

4

18

Forexaminer’s

use only

16



19 Given the equation of the curve, y = 2x2 16 x + 25.Diberi persamaan lengkung, y = 2x2 16 x + 25.

Find,Cari

(a) the coordinate – x of turning point.koordinat – x bagi titik pertukaran.

(b) the equation of tangent at point A ( 3 , 4).persamaan tangen pada titik A (3,4)

[ 4 marks][4 markah]

Answer / Jawapan:

(a)

(b)

Forexaminer’s

use only

4

19

17



20 Given that f x( )0

4

dx = 5 and g x( )1

3

dx = 6.

Di beri bahawa f x( )0

4

dx = 5 dan g x( )1

3

dx = 6.

Find the value of,Cari nilai bagi,

(a) 23

1

0

4

f x dx g x dx( ) ( ) ,

(b) k if [ ( ) ]g x kx dx13

=14. [3 marks]

[3 markah]Answer / Jawapan:

(a)

(b)

21 A set of data consists of 10 numbers. The sum of the numbers is 150.Satu set data mempunyai 10 nombor. Hasil tambah nombor-nombor tersebut ialah 150.

(a) Find the meanCari min.

(b) A number p is added to the set of data, the new mean is increased by 1.Find, the value of p.Satu nombor p ditambahkan kepada set data ini, min baru bertambah sebanyak 1.

Cari nilai p.

[3 marks][3 markah]

Answer / Jawapan:

(a)

(b)

3

20

Forexaminer’s

use only

3

21

18



22. Diagram 22 shows nine letter cards.Rajah 22 menunjukkan sembilan keping kad huruf .

T U M B D R I V E

Diagram 22Rajah 22

A five -letter code is to be formed by using five of these cards.Suatu kod lima huruf hendak dibentuk dengan menggunakan lima daripada kad-kad ini.

Find,Cari,

(a) the number of different five -letter codes that can be formed,bilangan kod lima huruf yang berlainan yang dapat dibentuk.

(b) the number of different five -letter codes which begin with a vowel and endWith a consonant.bilangan kod lima huruf yang berlainan yang bermula dengan huruf vokal danberakhir dengan konsonan.

[4 marks][4 markah]

Answer / Jawapan:

(a)

(b)

Forexaminer’s

use only

4

22

19



23 Ah Chong and Abu compete in a game which has two outcomes, win or lose.The game will end when any of the players has won two sets. The probability that

Ah Chong will win any one set is5

3.

Ah Chong dan Abu bertanding dalam satu permainan yang mempunyai dua kesudahan,menang atau kalah. Permainan akan berakhir apabila salah seorang menang dua set

permainan. Kebarangkalian Ah Chong menang salah satu permainan ialah5

3

.Calculate the probability thatKira kebarangkalian bahawa.

(a) the game will end in only two set,permainan akan berakhir dalam dua set sahaja.

(b) Ah Chong will win the competition after playing 3 sets.Ah Chong akan menang permainan itu selepas bermain 3 set.

[4 marks][4 markah]

Answer / Jawapan:

(a)

(b)

4

23

Forexaminer’s

use only

20



24 The probability of a student having laptop in a particular class is p. Given that themean and the standard deviation of the number of students having laptops are 18

and5

36respectively. Find the value of p.

Kebarangkalian seorang pelajar mempunyai komputer riba dalam suatu kelas ialah p.Diberi bahawa min dan sisihan piawai bilangan pelajar yang mempunyai komputer riba

ialah 18 dan5

36masing –masing. Cari nilai p

[3 marks][3 markah]

Answer / Jawapan:

3

24

Forexaminer’s

use only

21



25 The length of the pen produced by a factory are normally distributed witha mean of 14 cm and a standard deviation of 0.1 cm. Diagram 25 shows thenormal distribution graph for the length of the pen , X cm.Panjang bagi sebatang pen dari sebuah kilang adalah tertabur secara normal denganmin 14 cm dan sisihan piawai 0.1 cm. Rajah 25 menunjukkan graf taburan normal untukpanjang pen, X cm.

It is given that the area of the shaded region is 0.4483. Find the value of h.Diberi luas kawasan berlorek ialah 0.4483. Cari nilai h

[3 marks][3 markah]

Answer / Jawapan:

END OF THE QUESTION PAPERKERTAS SOALAN TAMAT

3

25

f (X )

Forexaminer’s

use only

Diagram 25Rajah 25

X cm = 14 h

22



PAPER1 MARKING SCEMEQUESTIONNUMBERS

WORKING MARKS FULL MARKS

1 a) m = 2

b) n = 6

c) 20

111 3

2 8x2 – 24x + 202(4x2-12x + 9) + 22(2x – 3)2 + 2

3B2B1 3

3 f-1(x) =

10 – y = 2x

2B1 2

4 k = -2, k = 1(both)(k-1)(k+2)=0(2k)2 – 4(1)(2-k)=0

3B2B1 3

5 2< x < 5(x – 2)(x – 5) < 0 orx2-7x + 10 < 0

3B2B1 3

6 x = 10

x + 3 = -2 + 2x

33(x+2) or 32(1-x)

3B2B1

3

7

81

9

4B3B2

B1 4

8 a) x = 2

b) (x – 2)2 – 14

c) x2 – 4x – 10

111 3

9 (a) – 4

28 – 2(16) or S2 –2 S1

T1 = 16(b) 36

3B2B11 4

10 –6 3

52



QUESTIONNUMBERS


( )

r = –

B2

B1 3

11 p = – 1 and p = 315p –75 = 80 or 5p – 75 = –80

9 98 2

2 47

98

= 40

3B2

B1 312 P( 2, 4)

h= =

h =( ) ( ) ( ) ( )

3B2B1 3

13OC

= ba2

1

2

1

OC

= OA +2

1AB OR OC

= OB +2

1AB

2

B1

214 (a) m = 12

(b)5

24k

25

km

12

B13

15x

xy 8

2

cxx

y )

1(2

2

16 = 2(4) + C or 4 = 2(-2) + C or Gradient = 2

3

B2B1

3

16

65

33

13

5

5

3

13

12

5

4

Cos A =5

4 or Cos B =

13

12 or sin B =

13

5

3

B2

B1

317 x = 56o 19’ , 236 o 19’ , 33o 41’, 213 o 41’

or x = 56.31o 236.31o 33.69o 213.69o

x = 56o 19’ 33o 41’(2 tan x – 3) ( 3 tan x – 2 ) = 06( 1 + tan2x ) – 13 tan x = 0

4

B3B2B1 4



QUESTIONNUMBERS


18 (a) = 2 radianr = 2 r

(b) 24 cm12 – 2OB + 2OB +12

2B12B1

4

19 (a) 4

164 xdx

dy

y = - 4x + 16

416)3(4)( dx

dyb

2B12B1

4

20 (a) 4(b) k = –2

6 –

3

1

2

2

kx

12

B1

321 (a) 15

(b) 26

1611

150

x

12

B13

22 (a) 15120

59 p

(b) 3780

3 7 6 5 6 or 63 37 p

2B12B1

4

23a)

25

13

5

2

5

2

5

3

5

3

b)125

36

5

3

5

3

5

2

5

3

5

2

5

3

2

B1

2B1

424

p =5

3 3

B2



QUESTIONNUMBERS


5

3618 q

npq =5

36

B1

3

25 14.013

13.01.0

14

h

P( z > )1.0

14h= 0.4483

3

B2

B1

3

END OF MARKING SCHEME


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