· PDF file4 1. Give your answer in the form ax2 + bx + c = 0, where a, b and c are constants....

3472/1 [Lihat sebelah

SULIT

JABATAN PELAJARAN WILAYAH PERSEKUTUAN

KUALA LUMPUR

UJIAN INTERVENSI

TINGKATAN LIMA

2010

Kertas soalan ini mengandungi 15 halaman bercetak

MATEMATIK TAMBAHAN

Kertas 1

Dua jam

JANGAN BUKA KERTAS SOALAN INI

SEHINGGA DIBERITAHU 1. This question paper consists of 25

questions.

2. Answer all questions.

3. Give only one answer for each

question.

4. Write your answers clearly in the

spaces provided in the question paper.

5. Show your working. It may help you to

get marks.

6. If you wish to change your answer,

cross out the work that you have done.

Then write down the new answer.

7. The diagrams in the questions provided

are not drawn to scale unless stated.

8. The marks allocated for each question

are shown in brackets.

9. A list of formulae is provided on pages

2 and 3.

10. You may use a non-programmable

scientific calculator.

11. This question paper must be handed in

at the end of the examination.

.

.

.

For examiner’s use only

Question Total Marks Marks

Obtained

1 2

2 2

3 4

4 4

5 2

6 3

7 3

8 3

9 3

10 3

11 4

12 3

13 4

14 3

15 3

16 4

17 3

18 4

19 2

20 3

21 3

22 4

23 4

24 4

25 3

TOTAL 80

Name : ………………..……………

Form : ………………………..……

3472/1

Matematik Tambahan

Kertas 1

Januari 2010

2 Jam

http://tutormansor.wordpress.com/

SULIT 3472/1

3472/1 [ Lihat sebelah

SULIT

2

BLANK PAGE

HALAMAN KOSONG


SULIT 3472/1

3472/1 2010 Hak Cipta JPWPKL [ Lihat

sebelah

SULIT

3

3

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

commonly used.

ALGEBRA

1

2 4

2

b b acx

a

2 a

m a

n = a

m + n

3 am a

n = a

m - n

4 (am)

n = a

mn

5 loga mn = log am + loga n

6 loga n

m = log am – loga n

7 log a mn = n log a m

8 logab = a

b

c

c

log

log

CALCULUS

1 y = uv , dx

duv

dx

dvu

dx

dy

2 v

uy ,

2v

dx

dvu

dx

duv

dy

dx

,

3 dx

du

du

dy

dx

dy

3 A point dividing a segment of a line

( x,y) = ,21

nm

mxnx

nm

myny 21

4 Area of triangle

= )()(2

1312312133221 1

yxyxyxyxyxyx

1 Distance = 2

21

2

21 )()( yyxx

2 Midpoint

(x , y) =

2

21 xx ,

2

21 yy

GEOMETRY

2


SULIT 3472/1

3472/1 2010 Hak Cipta JPWPKL

SULIT

4

STATISTICS

1 Arc length, s = r

2 Area of sector , L = 21

2r

TRIGONOMETRY

3 C

c

B

b

A

a

sinsinsin

4 a2 = b

2 + c

2 - 2bc cos A

5 Area of triangle = Cabsin2

1

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

FN

Lm

2

1

6 1

0

100Q

IQ

7 iw

iIiwI

3


SULIT 3472/1

5


[Lihat sebelah

SULIT

Answer all questions.

1 Given that A ={ – 2, –1, 1, 2 } and B ={ 1, 4, 9 }. The relation between sets A and B is

represented by the set of ordered pairs { (–2, 4), (–1, 1 ), (1, 1), (2, 4) }.

Diberi bahawa A = { – 2, –1, 1, 2 } dan B = { 1, 4, 9 }. Hubungan antara set A dan B

diwakili sebagai set pasangan bertertib { (–2, 4), (–1, 1 ), (1, 1), (2, 4) }.

State

Nyatakan

(a) the object of 4,

objek bagi 4,

(b) the image of –1 .

imej bagi –1 . [2 marks / markah]

Answer / Jawapan : (a) ……………………..

(b) ……………………...

2 Diagram 1 represents the relation between set R and set S.

Rajah 1 mewakili hubungan antara set R dan set S.

State

Nyatakan

(a) the range,

julat,

(b) the type of relation.

jenis hubungan. [2 marks / markah]

For

examiner’s

use only

2

3

1

2

2

S

R 1 2 3

2

4

6

8

10

Diagram 1 / Rajah 1

Answer / Jawapan: (a) ……………………..

(b) ……………………...

4


SULIT 3472/1


SULIT

6

3.

The information given above refers to the functions f and g .

Maklumat yang diberi di atas merujuk kepada fungsi f dan g .

(a) State the value of h.

Nyatakan nilai h.

(b) Find the value of )3(1fg .

Nyatakan nilai bagi )3(1fg .

[4 marks / markah]

Answer / Jawapan: (a) …………………….

(b) ..............................

4 Given the function xxf 31)( and the composite function 234)( xxfg , find

Diberi fungsi xxf 31)( dan fungsi gubahan 234)( xxfg , cari

(a) )(xg ,

)(xg ,

(b) the value of x when 5)(1 xf .

nilai x apabila .5)(1 xf

[4 marks / markah]

Answer / Jawapan : (a) ...………………………......

(b)...............................................

For

examiner’s

use only

hxx

xxf

,

3

4:

xxg 21:

4

3

4

4

5


SULIT 3472/1

7


[Lihat sebelah

SULIT

5 Form the quadratic equation which has the roots 2 and 4

1. Give your answer in

the form ax2 + bx + c = 0, where a, b and c are constants.

Bentukkan persamaan kuadratik yang mempunyai punca-punca 2 dan 4

1. Berikan

jawapan anda dalam bentuk ax2 + bx + c = 0, dengan keadaan a, b, dan c adalah

pemalar.

[2 marks / markah]

Answer / Jawapan: .........…………………

___________________________________________________________________________

6 One of the roots of the quadratic equation 0242 2 nxx is three times the value

of the other root. Find the value of n , if n > 0 .

Satu daripada punca persamaan kuadratik 0242 2 nxx adalah tiga kali ganda

nilai punca yang satu lagi. Cari nilai n jika n > 0.

[3 marks / markah]

Answer / Jawapan: .........…………………

7 Given the roots of 3x2 – 4x + 5 = 0 are and . Form a quadratic equation with roots

13 and 3 +1.

Diberi punca-punca persamaan bagi 3x2 – 4x + 5 = 0 ialah dan . Bentukkan

persamaan kuadratik dengan punca-punca 13 dan 3 +1.

[3 marks / markah]

Answer / Jawapan: ……............................

3

6

For

examiner’s

use only

2

5

3

7

6


SULIT 3472/1


SULIT

8

8. Find the range of values of x if 2 4y x and 2 11y x .

Cari julat nilai x jika 2 4y x dan 2 11y x .

[3 marks / markah]

Answer / Jawapan: .........…………………

9 Diagram 2 shows the graph of the quadratic function y = (x p)2 –5, where p is a

constant.

Rajah 2 menunjukkan graf kuadratik y = (x p)2 – 5, dengan keadaan p ialah pemalar.

Find

Cari

(a) the value of p,

nilai p,

(b) the equation of the axis of symmetry, persamaan paksi simetri,

(c) the coordinates of the minimum point.

koordinat bagi titik minimum.

[3 marks / markah]

Answer/Jawapan : (a) …….............................

(b) ....................................

(c) ....................................

For

examiner’s

use only

3

8

3

9

x

4

O

Diagram 2 / Rajah 2

y

7


SULIT 3472/1

9


[Lihat sebelah

SULIT

10 Express 12 n 3 ( 2n ) + 7( 12 n ) in the simplest form.

Ungkapkan 12 n 3 ( 2n ) + 7( 12 n ) dalam bentuk termudah.

[3 marks / markah]

Answer / Jawapan : ……………...……….....

11 Solve the equation )4(log21)2(log 42 xx .

Selesaikan persamaan )4(log21)2(log 42 xx .

[4 marks / markah]

Answer/Jawapan : ……………...……….....

12 Given xp 3log and yp 2log , express 12log8 in terms of x and y.

Diberi xp 3log dan yp 2log , ungkapkan 12log8 dalam sebutan x dan y.

[3 marks / markah]

Answer / Jawapan : ..................................

3

10

4

11

3

12

For

examiner’s

use only

8


SULIT 3472/1


SULIT

10

13 A point P moves such that its distance from two fixed points, A(3, 2) and B ( 1, 4),

are in the ratio of PA : PB = 4 : 3. Find the equation of locus P.

Titik P bergerak dengan keadaan jarak dari dua titik tetap A(3, 2) dan B ( 1, 4) adalah

dalam nisbah PA : PB = 4 : 3. Cari persamaan lokus P.

[4 marks / markah]

Answer /Jawapan: .....................................

14 Given points P(5, 6), Q(10, 3), R(7, n) and S(6, 1) are the vertices of a kite. Find the

value of n.

Diberi titik-titik P(5, 6), Q(10, 3), R(7, n) dan S(6, 1) adalah bucu-bucu bagi suatu

lelayang. Cari nilai n.

[3 marks / markah]

Answer / Jawapan: n =…………...………....

For

examiner’s

use only

4

3

13

3

14

9


SULIT 3472/1

11


[Lihat sebelah

SULIT

15 For a set of four numbers, it is given that the sum of all the numbers, x , is 28 and the

standard deviation is 2 3 . Find the value of 2x .

Satu set yang terdiri daripada empat nombor, diberi hasil tambah bagi semua nombor-

nombor itu x , ialah 28 dan sisihan piawainya ialah 2 3 . Cari nilai bagi 2x .

[3 marks / markah]

Answer / Jawapan: …...…………..……..…...

16 A set of positive integers 1, m – 1, 6 and 8 is in ascending order.

Set nombor integer positif 1, m – 1, 6 dan 8 adalah dalam tertib menaik.

Find the value of m if

Cari nilai m jika

(a) the mode is 1,

mod ialah 1,

(b) the mean is 4,

min ialah 4,

(c) the median is 5.

median ialah 5.

[4 marks / markah]

Answer /Jawapan: (a) ...................................

(b)....................................

(c) .....................................

3

15

For

examiner’s

use only

4

16

10


SULIT 3472/1


SULIT

12

17 Diagram 3 shows a circle with centre O.

Rajah 3 menunjukkan suatu bulatan berpusat O.

The length of the minor arc AB is 3.9275 cm and the angle of the major sector AOB is

315o. Using 142.3 , find

Panjang lengkok minor AB ialah 3.9275 cm dan sudut sektor major AOB ialah

315o. Dengan menggunakan 142.3 , cari

(a) the value of , in radians,

(Give your answer correct to four significant figures.)

nilai bagi , dalam radian,

(Berikan jawapan anda betul kepada empat angka bererti.)

(b) the length, in cm, of the radius of the circle.

panjang, dalam cm, jejari bulatan tersebut.

[3 marks / markah]

Answer/ Jawapan: (a) ……..…….…………...

(b) ………………………..

For examiner’s

use only

θ O

A

B

Diagram 3 / Rajah 3

3

17

11


SULIT 3472/1

13


[Lihat sebelah

SULIT

18 Diagram 4 shows sector OAB with centre O and radius of 5p.

Rajah 4 menunjukkan sector OAB berpusat O dan berjejari 5p.

Point C is on OB such that OC : CB = 3 : 2. Given that the perimeter of the shaded

region is 34 cm, calculate the value of p.

(Use = 3.142)

Titik C terletak pada OB dengan keadaan OC: CB = 3 : 2. Diberi bahawa perimeter

rantau berlorek ialah 34 cm, hitungkan nilai p.

(Guna = 3.142)

[4 marks / markah]

Answer / Jawapan :…...…………..……..…...

19 Given that 2)31(

1)(

xxf

, find f (x).

Diberi bahawa 2)31(

1)(

xxf

, cari f (x).

[2 marks / markah]

Answer / Jawapan : …...…………..……..…...

For examiner’s

use only

4

18

2

19

O

A

B C

Diagram 4 / Rajah 4

12


SULIT 3472/1


SULIT

14

20 Given that 22 1

( )3 4

xf x

x

, find )1(' f .

Diberi bahawa 22 1

( )3 4

xf x

x

, cari )1(' f .

[3 marks / markah]

Answer / Jawapan: …...…………..……..…...

___________________________________________________________________________

21 Find the equation of the tangent to the curve y = 3x 2 – 2x + 1 that passes through

point (1, 2).

Cari persamaan garis tangen kepada lengkung y = 3x 2 – 2x + 1 yang melalui titik

(1, 2).

[3 marks / markah]

Answer / Jawapan: …...…………..……..…...

3

8

For

examiner’s

use only

3

20

3

21

13


SULIT 3472/1

15


[Lihat sebelah

SULIT

22 Given the curve22

1

xxy .

Diberi suatu lengkung 22

1

xxy .

(a) Find the coordinates of the turning point.

Cari koordinat bagi titik pusingan.

(b) Hence, determine whether it is a maximum or a minimum point.

Seterusnya, tentukan sama ada ianya titik maksimum atau titik minimum.

[4 marks / markah]

Answer / Jawapan: (a)…...…………..….......

(b)....................................

23 It is given that y = u – u 2 and x = 2u 3. Find

dx

dy in terms of x.

Diberi bahawa y = u – u 2 dan x = 2u 3. Cari

dx

dy dalam sebutan x.

[4 marks / markah]

Answer / Jawapan :…...…………..……..…...

For

examiner’s

use only

4

22

4

23

14


SULIT 3472/1


SULIT

16

24 A cylinder has a fixed height of 10 cm and a radius of 5 cm. If the radius decreases

by 0.05 cm, find (in terms of )

Sebuah silinder mempunyai ketinggian tetap 10 cm dan jejari 5 cm. Jika jejari

berkurang sebanyak 0.05 cm, cari (dalam sebutan )

(a) the approximate change in the volume of the cylinder,

perubahan hampir dalam isipadu silinder tersebut,

(b) the final volume of the cylinder.

isipadu akhir silinder itu.

[4 marks / markah]

Answer / Jawapan: (a)……………………………

(b)……………………………

25 Given that 4)23( xy , and x decreases at the rate of 4 unit per second, find the rate

at which y changes when x =3

1.

Diberi bahawa 4)23( xy , dan x menyusut pada kadar 4 unit per saat, cari kadar

perubahan y apabila x =3

1

.

[3 marks / markah]

Answer / Jawapan: …...…………..……..…...

END OF QUESTION PAPER

4

24

For

examiner’s

use only

3

25

15


3472/1 SULIT

SULIT

3472/1

Additional

Mathematics

Paper 1

January

2010

JABATAN PELAJARAN WILAYAH PERSEKUTUAN

KUALA LUMPUR

UJIAN INTERVENSI

TINGKATAN 5

2010

ADDITIONAL MATHEMATICS

Paper 1

MARK SCHEME

This mark scheme consists of 7 printed pages


2

Number Solution and mark scheme Sub

Marks

Full

Marks

1

(a) –2, 2

(b) 1

1

1

2

2

(a) { 2, 4, 6, 10 }

(b) many to many

1

1

2

3

(a) h = –3

(b) –2

31

)1(4

1)3(gor2

1)(1 1

xxg

1

3

B2

B1

4

4

(a) g(x) = x 2 + 1

21 3 ( ) 4 3g x x

(b) x = 16

1 1( ) or 1 3(5)

3

xf x x

2

B1

2

B1

4

5

4x2 +7x –2 = 0

(x + 2)(4x –1) = 0 OR SOR = 4

7 & POR = –

2

1

2

B1

2


3


Marks

Full

Marks

6

n = 16

SOR = 4 = 2

n and POR = 23 12

SOR = 4 = 2

n or POR = 23 12

3

B2

B1

3

7

02062 xx

613()13 or 201313 )(

3

4 or

3

5

3

B2

B1

3

8

31 ,

2x x

(2 3)( 1) 0x x

2(4 + x2) x + 11

3

B2

B1

3

9

(a) p = 3

(b) x = 3

(c) ( 3 , –5 )

1

1

1

3

10

5 ( 2 n – 1

)

2n

2

732

2n (2

1) – 3 ( 2

n ) + 7 ( 2

n × 2

- 1)

3

B2

B1

3


4


Marks

Full

Marks

11

x = 10

3

22

4

x

x

log 2 2

4

x

x

= 1

log 2 ( x – 2 ) = 1 + 2 2

2

log (4 )

log 4

x

4

B3

B2

B1

4

12

y

yx

3

2

2log3

2log23plogor

2log3

2log2log3log

p

p

p

ppp

32log

)223(plogor

32log

)223(log

pp

p

3

B2

B1

3

13

7x2 + 7y

2 – 86x – 164y + 155 = 0

9 [ x2

+ 6x + 9 + y2 – 4y + 4 ] = 16 [ x

2 – 2x + 1 + y

2 + 8y + 16 ]

3 22 23 )y()x( = 4 22 41 )y()x(

3PA = 4PB

4

B3

B2

B1

4

14

n = –2

3 1 61

10 6 7 5

n

mQS = 3 1

10 6

or mPR =

6

7 5

n

3

B2

B1

3


5


Marks

Full

Marks

15

2 244x

2

2(7) 2 34

x

77

28x

3

B2

B1

3

16

(a) 2

(b) 2

44

8611

)m(

(c) 5

1

2

B1

1

4

17

(a) 0.7855

(b) r = 5

3.9275 = r (0.7855)

1

2

B1

3

18

p = 3.196

2p + 4p + 0.9274(5p) = 34

Arc AB = 0.9274(5p) = 4.637p

AOB = 53.13 or 0.9274 rad

4

B3

B2

B1

4

19

3)31(

6)(

xxf

)3(3)31(2)( xxf

2

B1

2


6


Marks

Full

Marks

20

7

2

2

4 1 3 1 4 3 2 1 1

[3 1 4]

2

2

4 3 4 3 2 1

[3 4]

x x x

x

3

B2

B1

3

21

y = 4x – 2

2 = 4 (1) + c OR 2

41

y

x

OR y –2 = 4(x –1)

dy

dx = 6x – 2 or 4

dy

dx

3

B2

B1

3

22

(a)

2

3,1

01 3 x

(b) Minimum point

4

2

2

3 xdx

yd

03

2

2

dx

yd

2

B1

2

B1

4

or


7


Marks

Full

Marks

23

2

2 x

dx

dy

2

2 x

dx

dy

dx

dy

2

2

321

x

)2

1()21( u

dx

dy

udu

dy21 or 2

du

dx or u =

2

3x

4

B3

B2

B1

4

24

(a) 5

)5(20dr

dV or 100 )05.0)(5(20 V

(b) 245

5)5(10 2 newV

2

B1

2

B1

4

25

48dy

dt

dy

dt

3

112 3 2 ( 4)

3

dy

dx = 4 (3x – 2)

3 (3)

3

B2

B1

3

or

or


· PDF file4 1. Give your answer in the form ax2 + bx + c = 0, where a, b and c are constants....

Documents

Transcript of · PDF file4 1. Give your answer in the form ax2 + bx + c = 0, where a, b and c are constants....