Spm Format Practice

CHAPTER 9 DIFFERENTIATION FORM 4

49

PAPER 1

1. Given that )6(15 xxy , calculate

(a) the value of x when y is maximum,(b) the maximum value of y. [3 marks]

2. Given that 24 xxy , use differentiation to find the small change in y when x increases

from 5 to 5.01. [3 marks]

3. Differentiate 42 )53(2 xx with respect to x. [3 marks]

4. Two variables, x and y are related by the equation .5

4x

xy Given that y increases at a

constant rate of 3 units per second, find the rate of change of x when x 5. [3 marks]

5. Given that2)42(

1)(

xxg , evaluate g”(1). [4 marks]

6. The volume of water, V cm3, in a container is given by hhV 63

2 3 , where h is the height in

cm, of the water in the container. Water is poured into the container at the rate of 5 cm3 s1.Find the rate of change of the height of water, in cm s1, at the instant when its height is 3 cm.

[3 marks]

7. The point R lies on the curve 2)3( xy . It is given that the gradient of the normal at R is

6

1 . Find the coordinates of R. [3 marks]

8. It is given that 3

5

4uy , where .25 xu Find

dx

dyin terms of x. [3 marks]

9. Given that ,432 2 xxy

(a) find the value ofdx

dywhen x 3,

(b) express the approximate change in y, in terms of m, when x changes from1 to 1 + m, where m is a small value. [4 marks]

10. The curve y = f(x) is such that 25 pxdx

dy, where p is a constant. The gradient of the curve

at x = 2 is 10. Find the value of p. [2 marks]

11. The curve y = x2 – 28x + 52 has a minimum point at x = k, where k is a constant. Find thevalue of k. [3 marks]


50

ANSWERS PAPER 1Marks

1. a) y = 15x (6 – x)= 90x – 15x2

xdx

dy3090 1

b)03090,0 x

dx

dy

30x = 90x = 3 # 1

y = 15(3)(6 – 3)= 135 # 1

2. y = 24 xx

dx

dy= 4 + 2x 1

x = 5,dx

dy= 4 + 2(5)

= 14

y =dx

dy x

= 14 0.01= 0.14 #

11

3. Let y = 42 )53(2 xx

dx

dy= )4()53(]3)53(4[2 432 xxxx 1

= 432 )53(4)53(24 xxxx

= )]53(6[)53(4 3 xxxx1

= )59()53(4 3 xxx # 1

4.y =

xx

54 = 154 xx

dx

dy=

2

54

x 1

dt

dy=

dt

dx

dx

dy

3 = (2

54

x )

dt

dx

3 = (25

54 )

dt

dx1

3 =dt

dx

5

14

dt

dx=

7

5unit per second # 1


51

5. g(x) = 2)42( x

g (x) = )2()42(2 3 x

= 3)42(4 x 1

g(x) = )2()42(12 4x

= 4)42(24 x1

g(1) = 4)2(24

=16

24

=2

3# 1

6.V = hh 6

3

2 3

dt

dV= 62 2h 1

dt

dV=

dh

dV

dt

dh

5 = ( 62 2h ) dt

dh

dt

dh=

6)3(2

52

1

= 11 2083.0@24

5 cmscms #1

7. y = 2)3( x

dx

dy= 2(x – 3) 1

Given gradient of normal =6

1

2(x – 3) = 6x = 6 1

y = 2)36(

= 9R (6, 9) # 1

8.y = 3

5

4u

= 3)25(5

4x 1

dx

dy= 5)25(

5

12 2x1

= 2)25(12 x # 1

9. a) y = 432 2 xx

1


52

dx

dy= 4x + 3

When x = 3,dx

dy= 4(3) + 3

= 15 # 1

b) x = m

x

y

dx

dy

x = 1,m

y 4(1) + 3 = 7 1

y 7m # 1

10.

dx

dy= 5px + 2

5p(2) + 2 = 10 1

10p = 8

p =5

4# 1

11. y = 52282 xx

dx

dy= 2x – 28 1

Minimum point at x = k :2(k) – 28 = 0 1

2k = 28k = 14 # 1


53

PAPER 2

1. Diagram shows a conical container of diameter 0.8 m and height 0.6 m.Water is poured into the container at a constant rate of 0.3 m3 s1.

Calculate the rate of change of the height of the water level at the instant when the height of

the water level is 0.4 m. (Use 3.142; Volume of a cone = hr 2

3

1 ) [4 marks]

2. Diagram shows part of the curve2)32(

4

xy which passes through A(2, 5).

Find the equation of the tangent to the curve at the point A. [4 marks]

2)32(

4

xy

y

xO

A(2, 5)

0.8 m

0.6 mwater


54

3. In the diagram, the straight line PQ is a normal to the curve 32

2

x

y at A(2, 5).

Find the value of k. [3 marks]

4. Diagram shows part of the curve y = k(x – 2)3, where k is a constant. The curve intersects thestraight line x = 4 at point A.

At point A, .36dx

dyFind the value of k. [3 marks]

y

x

P

A(2, 5)

O Q(k, 0)

y = k(x – 2)3

y

x

A

O

x = 4


55

5. Diagram shows the curve y = x2 + 3 and the tangent to the curve at the point P(1, 5).Calculate the equation of the tangent at P. [3 marks]

y

x

O

y = x2 + 3

P(1, 5)


56

ANSWERS PAPER 2

1.

h

r=

6.0

4.0

r = hh3

2

6

4 1

V = hr 2

3

1

= hh 2)3

2(

3

1

= 3

27

4h 1

dh

dV=

9

4 2h

dh

dV=

dt

dV

dh

dt

2

9

4h = 0.3

dh

dt1

2)4.0)(142.3(9

4= 0.3

dh

dt

dt

dh= 1.343 1ms # 1

2. y = 2)32(4 x

dx

dy= )2()32(8 3 x

=3)32(

16

x1

At A (2, 5) ,dx

dy=

3]3)2(2[

16

= –161

y – 5 = – 16( x 2 )= – 16x + 32

1

y = –16x + 37 # 13.

y = 32

2

x

dx

dy= x

At point A(2 , 5),dx

dy= 2

1

0.6 m

h

r

0.4 m


57

Gradient of normal,2

12 m

2

1

2

50

k1

– 10 = –k + 2k = 12 # 1

4. y = 3)2( xk

dx

dy= 2)2(3 xk 1

When x = 4,dx

dy= 36 :

2)24(3 k = 36 1

3k = 9k = 3 # 1

5. y = 32 x

dx

dy= 2x

At P (1,5),dx

dy= 2(1)

= 2 1

y – 5 = 2 (x – 1)= 2x – 2

y = 2x + 3 #

1

1

CHAPTER 10 SOLUTION OF TRIANGLES FORM 4

58

6510cm

12cm

Q

R

P

PAPER 21.

Diagram 1 shows a quadrilateral ABCD such that ABC is acute.

a ) Calculate

i ) ABC ii ) ADC

iii ) the area, in cm2, of the quadrilateral ABCD [ 8 marks]

b ) A triangle A’B’C’ has the same measurements as those given for triangle ABC but it isdifferent in shape to triangle ABC. Sketch the triangle A’B’C’ [ 2 marks ]

2.

º

Diagram 2

The Diagram 2 shows a triangle PQR.

(a) Calculate the length, in cm, of PR [ 2 marks ]

(b) A quadrilateral PQRS is then formed so that PR is a diagonal, PRS = 30 and PS = 8.2 cm.

Diagram 1


59

Calculate the two possible values of PSR. [2 marks ]

(c) Using the acute PSR from ( b ) , calculate [ 6 marks]i ) the length, in cm, of RSii ) the area, in cm2, of quadrilateral PQRS.

3.

Diagram 3

Diagram 3 shows a quadrilateral ABCD.

The area of ∆BCD is 20cm2 and BCD is acute. Calculate

(a) BCD [ 2 marks](b) the length in cm, of BD. [ 2 marks](c) ADB [ 3 marks](d) the area, in cm2 , of quadrilateral ABCD [ 3 marks]

4.

80 48'

12010cm

5cm5cm D

C

B

A

Diagram 4

Diagram 4 shows quadrilateral ABCD.

(a) Calculate ( 4 marks )

53º

10cm

7cm

6cm

D

A

BC


60

R P

Q

S

(i ) the length, in cm. of AC,(ii) ACB

(b) Point A’ lies on AC such that A’B = AB ( 6 marks )(i ) Sketch ∆A’BC(ii) Calculate the area, in cm2, of ∆A’BC

5.

Diagram 1

Diagram 1 shows a pyramid PQRS with triangle PQR as the horizontal base. S is the vertex of thepyramid and the angle between the inclined plane QRS and the base PQR is 50º. Given that PQand PR =5.6 cm and SQ = SR = 4.2 cm, calculate

(a) the length of RQ if the area of the base PQR is 12.4 cm2 [ 3 marks ](b) the length of SP, [ 3 marks ](c) the area of the triangle PQS. [ 4 marks ]


61

ANSWER(PAPER2)

1.( a ) ( i )

8.5

sin 42.5=

10.8

sin ABC

ABC =59.14 º

(ii) 10.82=6.52 +5.22- 2(6.5)(5.2) cosADC

ADC = 134.46 º

(iii) Total area

=1

2(10.8)(8.5)sin(180º- 42.5º -59.14 º) +

1

2(6.5)(5.2)sin134.46 º

= 57.02 cm2

( b )

1

1

1

1

31

242.5°

C’

A’

B’10.8 cm

8.5 cm


62

2.( a ) PR2 =122 +102- 2(12)(10)cos65º

= 142.6PR = 11.94cm

( b )

sin

11.94

PSR=

sin 30

8.2

PSR = 46.72 º or 180 º - 46.72 º=133.28 º

( c) ( i ) RPS = 180 º - 30º - 46.72 º= 103.28 º

sin103.28

RS

=

8.2

sin 30

RS = 15.96cm

( ii) Area =1

2(10)(12)(sin65 º ) +

1

2(11.94)(15.96)(sin30 º)

=102.02cm2

1

1

2

1

1

1

2

1

30

S'

S

65

10cm12cm

Q

R

P


63

3.

( a )1

2(6)(7)sinBCD = 20

BCD =72 º15’

( b ) BD2 = 6 2 + 7 2 – 2(6)(7)cos 72 º15’=59.39

BD = 7.706

(c )sin 53

10

=

706.7

sin BAD

BAD =37 º59’

ADB =180º- 37º59’-53 º=89 º 1’

(d) Area =1

2(7)(6)(sin72 º15’ ) +

1

2(7.706)(10)(sin89 º1’)

=58.52cm2

1

1

1

1

1

1

1

21

4. ( a ) ( i ) AC2 = 5 2 + 10 2 – 2(5)(10)cos 80 º48’AC = 10.44

( ii )sin

5

ACB=

sin120

10.44

ACB = 24 º30’

( b) ( i)

11

1

1

2


64

5.( a ) Area of PQR = 12.4

1

2(5.6) (5.6)sinQPR = 12.4

sinQPR = 0.7908

QPR=52.26

RQ2 = 5.6 2 + 5.6 2 – 2(5.6)(5.6)cos 52.26=24.33

RQ = 4.933cm

1

1

1

( ii ) BAC = 180 º -120 º - 24 º30’=35 º 30’

A’BA = 180 º - 2 (35 º 30’)=109 º

Area of ∆ABC =1

2(5)(10.4408)(sin35 º30’ )

=15.1575

Area of ∆ABA’ =1

2(5)(5)(sin109 º )

=11.8190

Area of ∆A’BC =15.1575 - 11.8190=3.3385 cm2

1

1

11


65

(b) QM =1

2QR where M is the midpoint of RQ

=1

2(4.933)

= 2.466

SM2 = 4.22- 2.4662

=11.56SM =3.4 cmPM2 = 5.62- 2.4662

PM = 5.028SP2 = 3.42 + 5.0282 – 2(3.4)( 5.028) cos 50ºSP =3.855 cm

(c) cos SQP =2 2 25.6 4.2 3.855

2(5.6)(4.2)

=0.7257

SQP = 43.47 º

Area of PQS

=1

2(5.6) (4.2)sin 43.47 º

=8.091 cm2

111

1

1

1

1

CHAPTER 11 INDEX NUMBER FORM 4

66

N

ML

K

80

100120

PAPER 2

1 Table 1 shows the price indices and percentage of usage of four items, P, Q, R and S, whichare the main ingredients in the making of a type of cake.

ItemsPrice index for the year 2007 based

on the year 2004Percentage of usage

P 125 40Q x 20R 110 10S 130 30

Table 1(a) Calculate

(i) the price of S in the year 2004 if its price in year 2007 is RM 44.85,(ii) the price index of P in the year 2007 based on the year 2000 if its price index in the

year 2004 based on the year 2000 is 120. [5 marks](b) The composite index number of the cost in making the cake for the year 2007 based on the

year 2004 is 125.Calculate(i) the value of x,(ii) the price of the cake in the year 2004 if the corresponding price in the year 2007 is

RM 40. [5 marks]

2 Table 2 shows the prices and the price indices for the four ingredients K, L, M, and N, used inmaking a particular kind of cake. Diagram 1 is a pie chart which represents the relative amountof the ingredients K, L, M and N, used in making the cake.

Table 2Diagram 1

(a) Find the value of p, q and r . [3 marks]

(b) (i) Calculate the composite index for the cost of making this cake in the year 2006 basedon the year 2003.

(ii) Hence, calculate the corresponding cost of making this cake in the year 2003 if thecost in the year 2006 was RM 40. [5 marks]

(c) The cost of making this cake is expected to increase by 40% from the year 2006 to theyear 2010. Find the expected composite index for the year 2010 based on the year 2003

[2 marks]

IngredientsPrice per Kg

(RM)Price index for theyear 2006 basedon the year 2003Year 2003 Year 2006

K 1.60 2.00 pL 4.00 q 120M 0.80 1.20 150N r 1.60 80


67

T

S36

Q

P

R

9072

144

3 A particular type of muffin is made by using four ingredients, P, Q, R and S. Table 3 shows theprices of the ingredients.

IngredientsPrice per kilogram (RM)

Year 2005 Year 2008P 4.00 xQ 2.50 3.00R y zS 2.00 2.20

Table 3

(a) The index number of ingredient P in the year 2008 based on the year 2005 is 125.Calculate the value of x. [2 marks]

(b) The index number of ingredient R in the year 2008 based on the year 2005 is 140. Theprice per kilogram of ingredient R in the year 2008 is RM2.00 more than its correspondingprice in the year 2005.Calculate the value of y and z. [3 marks]

(c) The composite index for the cost of making the muffin in the year 2008 based on the year2005 is 126Calculate

(i) the price of the muffin in the year 2005 if its corresponding price in the year 2008 isRM 6.30

(ii) the value of r if the quantities of ingredients P, Q, R and S used are in the ratio of6 : 3 : r : 2 [5 marks]

4 Table 4 shows the prices and the price indices of five components, P, Q, R, S and T, used toproduce a type of toy car.Diagram 2 shows a pie chart which represents the relative quantity of components used.

Table 4 Diagram 2

(a) Find the value of m and of n. [3 marks](b) Calculate the composite index for the production cost of the toy car in the year 2007 based

on the year 2005. [3 marks](c) The price of each component increases by 25% from the year 2007 to the year 2009.

Given that the production cost of one toy car in the year 2005 is RM60, calculate thecorresponding cost in the year 2009. [4 marks]

ComponentPrice (RM) for the year Price index for the

year 2007 based onthe year 2005

2005 2007

P m 4.40 110Q 4.00 5.60 140R 2.40 3.00 125S 6.00 5.40 nT 8.00 12.00 150


68

S

R

QP

25%

40%

20%

15%

5 Table 5 shows the prices and the price indices of four ingredients P, Q, R and S, to make adish.Diagram 3 shows a pie chart which represents the relative quantity of the ingredients used.

IngredientsPrice (RM) per kg

for the yearPrice index for theYear 2008 based

2006 2008 on the year 2006P 2.25 2.70 xQ 4.50 6.75 150R y 1.35 112.5S 2 2.10 105

Table 5

Diagram 3

(a) Find the value of x and of y. [3 marks](b) Calculate the composite index for the cost of making this dish in the year 2008 based on

the year 2006. [3 marks](c) The composite index for the cost of making this dish increases by 20% from the year 2008

to the year 2009.Calculate

(i) the composite index for the cost of making this dish in the year 2009 based on theyear 2006

(ii) the price of a bowl of this dish in the year 2009 if its corresponding price in the year2006 is RM 25 [4 marks]


69

ANSWERS (PAPER 2)

1 (a) (i)85.44

130

1002004

RMp 1

= RM 34.5 1

(ii) 50.34120

1002000

RMP or 75.282000

RMP 1

10075.28

85.442000/2007

RM

RMI 1

= 156 1

(b) (i) 125100

)30130()10110()20()40125(

x1

20x + 10000 = 12500 1

x = 125 1

(ii) 40125

1002004

RMP 1

= RM 32 1

2 (a)

12510060.1

00.2

RM

RMp 1

80.400.4100

120RMRMq 1

60.180

100RMr 1

(b) (i)360

)8080()120150()100120()60125(2003/2006

I 1

=360

439001

= 121.9 1

(ii) 409.121

1002003

RMP 1

= RM 32.81 1

(c) 5640100

1402010

RMRMP 1

7.17010081.32

00.562003/2010

RM

RMI 1


70

3 (a)4

100

125RMx 1

= RM 5.00 1

(b)

140

100

2

y

y1

y = RM 5.00 1

z = RM 7.00 1

(c) (i)30.6

126

1002005

RMP 1

= RM 5.00 1

(ii)126

11

)2110()140()3120()6125(

r

r1

1330 + 140r = 126r + 1386 1

r = 4 1

4 (a)40.4

110

100RMm

1

= RM4.00 1

n = 901006

40.5

RM

RM1

(b)360

)72150()1890()144125()36140()90110( 1

=360

453601

= 126 1

(c) 60100

1262007

RMP 1

= RM 75.60 1

60.75100

1252009

RMP 1

= RM 94.50 1


71

5 (a)120100

25.2

70.2

RM

RMx

1

y = 35.15.112

100RM 1

= RM 1.20 1

(b)100

)20105()405.112()25150()15120( 1

=100

121501

= 121.5 1

(c) (i)100

1202006/20082006/2009

II

= 121.5 100

1201

= 145.8 1

(ii) 25100

8.1452009

RMP 1

= RM 36.45 1

CHAPTER 1 FUNCTIONS FORM 4

1

Diagram 1

Set BSet A

p

q

r 8

6

4

2

g(x)x

0246

-2

k0

4

PAPER 1

1. Diagram 1 shows the relation between set A and set B.

State(a) the range of the relation,(b) the type of the relation.

[2 marks]2.

Based on the above information, the relation between R and S is defined by the set ofordered pairs ),(),,(),,(),,( hbfbdaba .

State(a) the images of a(b) the object of b

[2 marks]3. Diagram 2 shows the linear function .g

(a) State the value of k.(b) Using the function notation, express g in terms of x.

[2 marks]

4.Diagram 3 shows the function 0,:

x

x

kxxg where k is a constant.

Find the value of k.

2

1

3

x

kx x

Diagram 2

Diagram 3

jhfdbS

cbaR

,,,,

,,


2

[2 marks]5. Given the function 1: xxg , find the value of x such that 2)( xg .

[2 marks]

6. Diagram 5 shows the graph of the function 62)( xxf for domain 40 x .

State(a) the value of t,(b) the range of f(x) corresponding to the given domain.

[3 marks]

7. Given the function 12)( xxf and kxxg 3)( , find

(a) )2(f

(b) the value of k such that 7)2( gf

[3 marks]

8. The following information is about the function g and the composite function 2g .

Find the value of a and b.[3 marks]

9. Given the function 0,2

1)( x

xxf and the composite function xxfg 4)( .

Find(a) )(xg

(b) the value of x when 2)( xgf

[4 marks]

10.The function h is defined as 3,

3

7)(

x

xxh .

Find

(a) )(1 xh

(b) )2(1h

x4t0

6

f(x)

Diagram 5

bxaxg : , where a and b are constant and b > 0

89:2 xxg


3

[3 marks]

11. Diagram 9 shows the function f maps x to y and the function g maps y to z.

Determine

(a) )1(1f

(b) )5(gf

[2 marks]

12. The following information refers to the function f and g.

Find )(1 xgf .

[3 marks]

13.Given the function hxxg 3: and

2

1:1 kxxg , where h and k are constants. Find

the value of h and of k.[3marks]

14.Given the function 13)( xxh and

3)(

xxg . Find

(a) )7(1h

(b) )(1 xgh

[4 marks]

15. Given the function 23: xxf and 32: 2 xxg .

Find

(a) )4(1f

(b) )(xgf

[4 marks]

gfzyx

4

1

5

3:

15:

xxg

xxf


4

ANSWER (PAPER 1)

1 (a) 8,4 1

(b) many-to-one 1

2 (a) b , d 1

(b) a 1

3 (a) 2k 1

(b) 2)( xxg 1

4

3

2

12

1

2

1

k

g 1

1k 1

5 21 x or 2)1( x 1

1x 3x 1

6 (a) When 0)( xf , 062 x 1

3x

3 t 1

(b) Range : 6)(0 xf1

7 (a) (a) 5)2( f 1

(b) (b) 7)5( g

7)5(3 k 1

2k 1

8 )()(2 bxabaxg 1

xbaba 2

92 b and 8 aba 1

3b 4a 1

9 (a) xxg

4)(2

1

1

0,8

1)( x

xxg

1


5

(b)2

2

18

1

x

1

8

1x 1

10 (a)3

7

yx

1

37

)(1

xxh , 0x

1

(b)

2

1)2(1 h

1

11 (a) 5 1

(b) 4 1

12

5

1

yx

1

5

1)3()(1 x

xgf1

5

4

x 1

13.

3

hyx

1

3

1k

1

2

3h

1

14.(a) 3

1

yx

1

23

17)7(1

h

1

(b)

33

1

)(1

x

xgh 1

9

1

x 1

15 (a)

3

2

yx

1

2)4(1 f 1

(b) 3)23(2)( 2 xxgf 1

52418 2 xx 1

CHAPTER 2 QUADRATIC EQUATIONS FORM 4

6

PAPER 1

1. The quadratic equation 2x 2422 xpx , where p is a constant, has no real roots. Find the

range of the values of p. [3 marks]

2. The quadratic equation m 542 mxx ,where m 0, has real and different roots. Find therange of values of m. [4 marks]

3. Find the values of n for which the curve y = n + 8x x 2 intersect the straight line y = 3 at a point.[4 marks]

4. Solve the quadratic equation 5(2x – 1) = (3x + 1)(x – 3) . Give your answer correct to foursignificant figures [3 marks]

5. The straight line y + x = 4 intersects the curve y = wxx 72 at two points . Find the range ofvalues of w. [4 marks]

6. Form the quadratic equation which has the roots 7 and3

2. Give your answer in the form

02 cbxax , where a , b and c are constants.[2 marks]

7. Given the roots of the quadratic equation 24 8 0kx hx are equal. Express k in terms of h.[2 marks]

8. The straight line y = 9 4px is a tangent to the curve y = 21 .p x Find the possible values of p.

[5 marks]

9. The straight line y =2x1 does not intersect the curve y = 2 2 .x x p Find the range of values of

p. [5 marks]

10. The quadratic equation 23 0x kx h has roots 4 and 3. Find the values of k and h.[3 marks]


7

ANSWERS (PAPER 1)1 2(2 ) 4 8 0p x x 1

2

4 4 2 8 0p

48 32 0p

1

p <3

2

1

2 0542 mxmx 1

mm 5442 > 0 1

41 mm > 0 1

m < 1 , m > 4 1

3 n + 8x -x 2 = 3 1

0382 nxx 1

13482

n = 0 1

n = 13 1

4 3x 2 18x + 2 = 0 1

2( 18) ( 18) 4(3)(2)

2(3)

1

x = 0.1132, 5.887 1

5 4 – x = wxx 72 12 8 4 0x x w 1

64 – 4(w – 4 ) > 0 120w 1

6 014193

02372

xx

xx 11

7 2 4 4 8 0h k 1

2

128

hk

1

8 29 4 1px p x 1

21 4 9 0p x px 1

2

4 4 9 1 0p p 1


8

3 4 3 0p p 1

p = 3 and p =4

3 1

9 2x – 1 = 2x -2x + p 12 4 1 0x x p 1

16 – 4(1)(p + 1) < 0 112 – 4p <0 1

p > 3 1

10 (x + 4)(x – 3) = 0 123 3 36 0x x 1

k = -3 and h = -36 1

CHAPTER 3 QUADRATIC FUNCTIONS FORM 4

9

Paper 1

1. The quadratic function f(x) = a(x+p)2 + q, where a, p and q are constants, has a maximum valueof 5. The equation of the axis of symmetry is x=3.State(a) the range of values of a,(b) the value of p(c) the value of q [3 marks]

2. Find the range of values of x for which (x 4)2 < 12 3x[3 marks]

3. Find the range of values of x for which 2x2 3 5x. [3 marks]

4. The quadratic function f(x) = x2 6x + 5 can be expressed in the formf(x) = (x + m)2 + n where m and n are constants.Find the value of m and of n. [3 marks]

5. The following diagram shows the graph of quadratic function y = g(x).The straight line y = 9 is a tangent to the curve y = g(x).

(a) Write the equation of the axis of symmetry of the curve

(b) Express g(x) in the form (x + b)2 + c where b and c are constant. [3 marks]

6. The following diagram shows the graph of a quadratic function 2)(25)( pxxf , where p

is a constant.

x

y

O 15

y = 9

A(1, q)

.

0x

y


10

The curve )(xfy has a maximum point at A(1, q), where q is a constant. State

(a) the value of p,(b) the value of q,(c) the equation of the axis of symmetry.

[3 marks ]

7. Find the range of values of x for which x(2 x) 15. [3 marks ]

8. The quadratic equation x(p x ) = x + 4 has no real roots. Find the range of values of p[3 marks ]

Paper 2

1. Diagram below shows the curve of a quadratic function 62

1)( 2 kxxxf .

A is the point of intersection of the quadratic graph and y-axis. The x-intercepts are 6 and 2.

(a) State the value of r and of p. [2 marks]

(b) The function can be expressed in the form qpxxf 2)(2

1)( , find the value of q and

of k. [4 marks]

(c) Determine the range of values of x if f(x)< 6. [2 marks]

x

y

O 26

A(0, r)

(p, q)


11

Answers ( Paper 1)

Q Solution Marks1 (a) a<0 1

(b) p=3 1

(c) q=5 12 x2 5x + 4 < 0 1

(x 1)(x 4) < 0 1

1<x<4 13 2x2 + 5x. 3 0 1

(2x 1)(x + 3) 0 1

3 x ½ 1

4 f(x) = x2 6x + 32 32 + 5= (x 3)2 4

1

m = 3 n = 4 1,1

5 (a) x = 2 1

(b) g(x) = (x + 2)2 9

1,1

6 (a) p = 1 1(b) q =5 1(c) x=1 1

7 x2 2x 15 0 1

(x + 3)(x 5) 0 1

x3, x5 1

8 x2 + (1 p)x + 4 =0(1 p)2 4(1)(4) < 0 1

p2 2p 15 < 0(p+3)(p 5) <0 1

3 < p < 5 1

Answer(Paper 2)1 (a) r = 6, 1

p = 2 1

(b)6

2

1

2

1

2

1 222 kxxqppxx(Expand or completing the square) 1

62

1 2 qpor k=p

2)2(2

16 q

1

q = 8 1

k =2 1(c) x2 + 4x<0

x(x + 4)<0

1

4 < x < 0 1

4 0x

CHAPTER 4 SIMULTANEOUS EQUATIONS FORM 4

12

PAPER 2

1. Solve the simultaneous equations 82 xy and x2 - 3x – y = 2

[5 marks]

2.Solve the simultaneous equations j – k = 2 and j2 + 2k = 8. Give your answers correct to threedecimal places

[5 marks]

3. Solve the simultaneous equations x + 2y = 1 and y2 - 10 = 2x.[5 marks]

4. Solve the simultaneous equations 2x + y = 1 and x2 + y2 + xy = 7.[5 marks]

5. Solve the following simultaneous equations.82 yx

374 22 yx

Give your answers correct to three decimal places.[5 marks]

CHAPTER 4 SIMULTANEOUS EQUATIONS FORM 4

13

ANSWERS (PAPER 2)1. y – 2x = -8

y = - 8 + 2x ___________(1)x2 - 3x – y = 2___________(2)x2 - 3x – (-8 + 2x) = 2

1

1(x – 3)(x – 2) = 0x = 2, 3

11

y = - 8 + 2(2) y = - 8 + 2(3)y = -4, y = - 2 1

2. k = j – 2 _____________(1)j2 + 2k = 8 _____________(2)j2 + 2(j – 2) = 8

1

12( 4 )

2

b b acj

a

)1(2

)12)(1(422 2

1

j = 2.606 j = - 4.606 1k = (2.606) – 2 k = (- 4.606) - 2

= 0.606 = - 6.606 1

3. x = 1 – 2y __________(1)y2 - 10 = 2x __________(2)y2 - 10 = 2(1 – 2y)

1

1(y + 6)(y – 2) = 0y = 2, - 6

11

x = 1 – 2(2), x = 1 – 2(- 6)= -3, = 13 1

4. y = 1 – 2x __________(1)x2 + y2 + xy = 7 __________(2)x2 + (1 – 2x)2 + x(1 - 2x) = 7

1

10)2)(1( xx

x = - 1, 2

11

y= 1 – 2(- 1), y = 1 – 2(2)= 3, = - 3 1

5. x = 8 + 2y __________(1)x2 + 4y2 = 37 __________(2)(8 + 2y)2 + 4y2 = 37

1

1

a

acbby

2

42

232 32 4(8)(27)

2(8)

1

y ≈ - 1.209, -2.791 1x = 8 + 2(- 1.209), x = 8 + 2(-2.791)

= 5.582, = 2.418 1

CHAPTER 5 INDICES AND LOGARITHM FORM 4

14

PAPER 1

1. Solve the equation 3 13(5 ) 36x [3marks]

2. Solve the equation 3 7 432 4x x [3marks]

3. Solve the equation 2 13 5x x [4marks]

4. Solve the equation 5 9 616 4x x [3marks]

5. Solve the equation 3 3

3

127

9

x

x

[3marks]

6. Solve the equation 2 12 13(2 ) 24 0x x [4marks]

7. Given that 3 9log log 2K L , express K in terms of L. [4marks]

8.Given that log 3p r and log 7p s , express

49log

27p

p

in terms

of r and s.

[4marks]

9.Given that 5log 2 q and 5log 9 p , express 5log 8.1 in terms of q

and p.[4marks)

10. Solve the equation 3 33 log (2 1) logx x . [3marks]

11.Given that log 2x p and log 7x q , express

2

56log x

x

in terms

of p and q.

[4marks]

12. Given that 3 3 3log 3 2log logmn m n express m in terms of n. [4marks]

13. Given that 9 3log log 18y , find the value of y [3marks]

14.Given that 3log m v and 3log n w , express 9

81log

m

n

in

terms of v and w.

[4marks]

15. Solve the equation 2)14(loglog 33 xx . [3marks]

16. Solve the equation 1 45 5

25x x . [3marks]


15

ANSWERS (PAPER 1)

1.3 15 12x

3 1log 5 log12x 1

(3 1) log5 log12x 1

log123 1

log 5x

6990.0

0792.113 x

54392.113 x

54392.03 x

0.1813x 1

2. 3 7 45 22 2

x x

15 14 82 2x x 1

15 14 8x x 1

8x 1

3.2 1log 3 log5x x 1

2 1 log 3 log5x x 1

2 log3 log3 log 5x x

2 log3 log5 log3x x

2log3 log5 log 3x

log 3

2log3 log 5x

1

1.8691x 1


16

4. 5 9 64 22 2

x x 1

20 18 122 2x x

20 18 12x x 1

6x 1

5. 1

3 3 3 227 9x x

1

3 3 3 23 23 3x x

1

1

9 9 2 6 23 3x x

9 9 3x x 1

10 6x

3

5x 1

6.2 12 13(2 ) 24 0x x

2 12 .2 13(2 ) 24 0x x

2

2 .2 13(2 ) 24 0x x 1

Let u = 2x

22 13 24 0u u

2 3 8 0u u

3or 8

2u 1

But 2x must be positive, so 2x = 8

32 2x

x =3 1


17

7.3

3

3

loglog 2

log 9

LK 1

33

loglog 2

2

LK

3 32log log 4K L

2

3log 4K

L 1

243

K

L 1

9K L 1

8.2 3log 7 log log 3p p pp 1,1

2log 7 log 3log 3p p pp 1

2 1 3s r 1

9. 5log 8.1 = 5

81log

101

= 5 5log 81 log 10 1

= 25 5log 9 log 2 5 1

= 5log2log9log2 555

= 12 qp 1

10.3 3 33log 3 log (2 1) logx x

33 3 3log 3 log (2 1) logx x

3 3log 27 2 1 logx x 1

27 2 1x x 1

54 27x x

27

53x 1


18

11. 2

56log x

x

= 2log 56 logx x x 1

= log (7 8) 2 logx x x

= log 7 log 8 2logx x x x 1

= 3log 7 log 2 2logx x x x

= log 7 3log 2 2logx x x x 1

= 3 2q p 1

12.3 2

3 3 3 3log log 3 log logmn m n 1

2

3 3

27log log

mmn

n 1

227mmn

n 1

27

2nm 1

13.3

3

3

loglog 18

log 9

y 1

33

3

loglog 18

log 9

y

33

loglog 18

2

y

3 3log 2log 18y

23 3log log 18y 1

324y 1

14.9

81log

m

n

=3

3

81log

log 9

m

n

1

= 3 3 3

3

log 81 log log

2log 3

m n 1

= 43 3 3

1log 3 log log

2m n 1

= 1

42

v w 1


19

15. 3 3log log (4 1) 2 0x x

3 3 3log log (4 1) 2 log 3 0x x

3

(9)log 0

4 1

x

x

1

093

4 1

x

x

1

9 4 1x x

5 1x

1

5x 1

16.1

2

45 .5 5

5x x

2

1 45 1

5 5x

1

2

4 45

5 5x

2

4 55

5 4x

15 5x 1

1x 1

CHAPTER 6 COORDINATE GEOMETRY FORM 4

20

PAPER 1

1. A point T divides the line segment joining the points A(1, -2) and B(-5, 4) internally in theratio 2 : 1. Find the coordinates of point T.

[2 marks]

2. Diagram below shows a straight line PQ with the equation3

x+

5

y= 1. The point Q lies

on the x-axis and the point P lies on the y-axis.

Find the equation of the straight line perpendicular to PQ and passing through the point Q.[3 marks]

3. The line 8x + 4hy - 6 = 0 is perpendicular to the line 3x + y = 16. Find the value of h.[3 marks]

4. Diagram below shows the straight line AB which is perpendicular to the straight line CB atthe point B.

The equation of the straight line CB is y = 3x 4. Find the coordinates of B.

[3 marks]

5. The straight line14

x+

m

y= 1 has a y-intercept of 3 and is parallel to the straight line

y + nx = 0. Determine the value of m and of n.

x

P

Q

y

0

A(0,6) B

x

y

C

0


21

[3 marks]

6. Diagram below shows a straight line passing through A(2, 0) and B (0, 6).

a) Write down the equation of the straight line AB in the forma

x+

b

y= 1.

[1 mark]

b) A point P(x, y) moves such that PA = PB. Find the equation of the locus of P.[2 marks]

x

B(0, 6)

A(2, 0)

y

0


22

PAPER 2

1. Solutions to this question by scale drawing will not be accepted.

Diagram shows a straight line CD which meets a straight line AB at the point D. The point Clies on the y-axis.

0

a) Write down the equation of AB in the form of intercepts. [1 mark ]

b) Given that 2AD = DB, find the coordinates of D. [2 marks]

c) Given that CD is perpendicular to AB , find the y-intercept of CD. [3 marks]


In the diagram the straight line BC has an equation of 3y + x + 6 = 0 and is perpendicular tostraight line AB at point B.

(a) Findi) the equation of the straight line ABii) the coordinates of B. [5 marks]

(b) The straight line AB is extended to a point D such that AB : BD = 2 : 3. Find thecoordinates of D. [2 marks]

(c) A point P moves such that its distance from point A is always 5 units.Find the equation of the locus of P. [3 marks]

0x

y

DA(0 , -3)

C

B (12, 0)

A(-6, 5)

B

C

3y + x + 6 = 0

x

y

0


23


Diagram shows the triangle AOB where O is the origin. Point C lies on the straight line AB.

(a) Calculate the area, in unit2, of triangle AOB. [2 marks]

(b) Given that AC : CB = 3 : 2, find the coordinates of C. [2 marks]

(c) A point P moves such that its distance from point A is always twice its distance from pointB.(i) Find the equation of the locus of P.(ii) Hence, determine whether or not this locus intercepts the y-axis. [6 marks]

4. In the diagram, the straight line PQ has an equation of y + 3x + 9 = 0. PQ intersects thex-axis at point P and the y-axis at point Q.

Point R lies on PQ such that PR : RQ = 1 : 2. Find(a) the coordinates of R, [3 marks]

(b) the equation of the straight line that passes through R and perpendicular to PQ.[3 marks]

y + 3x + 9 = 0

y

x

A(-2, 5)

B(5, -1)

0

C

y

xP

Q

0

R


24


Diagram shows the triangle OPQ. Point S lies on the line PQ.

a) A point W moves such that its distance from point S is always 22

1units.

Find the equation of the locus of W. [3 marks]

b) It is given that point P and point Q lie on the locus of W.Calculatei) the value of k,ii) the coordinates of Q.

[5 marks]c) Hence, find the area , in unit2 , of triangle OPQ.

[2 marks]

0x

y P(3 , k)

S (5, 1)

Q


25

ANSWERS ( PAPER 1 )

1.T (

3

)2)(5()1)(1( ,

3

)2)(4()1)(2( )

2

= T( -3 , 2 ) 1

2.Gradient of PQ , m1 = -

3

5and the coordinates of Q (3 , 0)

1Let the gradient of straight line perpendicular to PQ and passing through Q= m2 . Then m1 m2 = -1.

m2 =5

3

The equation of straight line is3

0

x

y=

5

3

5y = 3(x – 3)

1

5y = 3x – 9 1

3. Given 8x + 4hy – 6 = 04hy = -8x + 6

y = -h4

8x +

h4

6

y = -h

2x +

h2

3

Gradient , m1 = -h

2

3x + y = 16y = -3x + 16

Gradient , m2 = -31

Since the straight lines are perpendicular to each other , then m1 m2 = -1.

(-h

2)(-3) = -1

1

6 = -hh = -6 1

4. Gradient of CB , m1 = 3Since AB is perpendicular to CB, then m1 m2 = 1

Gradient of AB, m2 = 3

1 1

The equation of AB is y = -3

1x + 6

B is the point of intersection.y = 3x 4 ……………(1)

y = 3

1x + 6 ……………(2)

3x 4 = 3

1x + 6

1


26

3

10x = 10

x = 3

y = 3(3) 4= 5

The coordinates of B are (3, 5). 1

5.

14

x+

m

y= 1

y-intercept = m = 3

1

From14

x+

3

y= 1, the gradient m1 = -

14

3

From y = -nx , the gradient m2 = -n .Since the two straight lines are parallel , then m1 = m2

-14

3= -n 1

n =14

31

6. a) From the graph given, x- intercept = 2 and y-intercept = 6.

The equation of AB is2

x+

6

y= 1 . 1

b) Let the coordinates of P = (x , y) and since PA = PB22 )0()2( yx = 22 )6()0( yx

(x – 2)2 + y2 = x2 + (y – 6)2

x2 – 4x + 4 + y2 = x2 + y2 – 12y + 36

1

12y – 4x -32 = 03y – x - 8 = 0 1


27

ANSWERS ( PAPER 2 )

1a)

12

x-

3

y= 1

1

b) Given 2AD = DB , soDB

AD=

2

1

D = (3

)1(12)2(0 ,

3

)1(0)2(3 ) 1

= ( 4 , -2 ) 1

c) Gradient of AB, mAB = -(12

3)

=4

11

Since AB is perpendicular to CD, then mAB mCD = 1. Gradient of CD, mCD = - 4

Let, coordinates of C = (0 , h) ,

mCD =40

)2(

h

- 4 =4

2

h

16 = h + 2h = 14

1

y-intercept of CD = 14 1

2 a) i) Given equation of BC, 3y + x + 6 = 0

y = -3

1x – 2

Gradient of BC = -3

11

Since AB is perpendicular to BC , then mAB mBC = 1.Gradient of AB, mAB = 3

The equation of AB ,)6(

5

x

y= 3

y – 5 = 3x + 18

1

y = 3x + 23 1

ii) B is the point of intersection.Equation of AB , y = 3x + 23 …………. (1)Equation of BC , 3y + x + 6 = 0 ………….(2)

Substitute (1) into (2), 3(3x + 23) + x + 6 = 01


28

9x + 69 + x + 6 = 0

x = -2

15

Substitute value of x into (1), y = 3(-2

15) + 23

y =2

1

The coordinates of B are ( -2

15,

2

1)

1

b) Let D (h, k)

B( -2

15,

2

1) = (

5

)18(2 h,

5

152 k) 1

-2

15=

5

)18(2 h,

-75 = 4h – 36

h =4

39

2

1=

5

152 k

5 = 4k + 30

k =4

25

1 The coordinates of D are (4

39,

4

25 )

c) Given PA = 522 )5())6(( yx = 5 1

( x + 6)2 + ( y – 5)2 = 25 1

x2 + 12x + 36 + y2 -10y + 25 = 25x2 + y2 + 12x -10y + 36 = 0 1

3 .)

a) Area =2

1

0510

0250

=2

1)2()25(

1

=2

23unit2 1

b) C = (5

)2(2)5(3 ,

5

)5(2)1(3 1

= (5

11,

5

7)

1

c) i) Since PA = 2PB22 )5()2( yx = 2 22 )1()5( yx 1

x2 + 4x + 4 + y2 10y + 25 = 4 (x2 10x + 25 + y2 +2y + 1) 1


29

x2 + y2 + 4x 10y + 29 = 4x2 + 4y2 40x + 8y + 1043x2 + 3y2 44x + 18y + 75 = 0 1

(ii) When it intersects the y-axis, x = 0. 3y2 +1 8y + 75 = 0 1

Use b2 4ac= (18)2 4(3)(75) 1

= 576b2 4ac < 0

It does not cut the y-axis since there is no real root. 1

4. a) y + 3x + 9 = 0When y = 0, 0 + 3x + 9 = 0

x = –3 P(–3, 0)

When x = 0, y + 0 + 9 = 0y = –9

Q(0, –9)

1

R(x, y) = (3

)3(2)0(1 ,

3

)0(2)9(1 )

1

= (-2 , -3 ) 1

b) y + 3x + 9 = 0y = -3x - 9

Gradient of PQ , m1 = –3

1

Since PQ is perpendicular to the straight line, then m1 m2 = 1

Thus,3

12 m

The equation of straight line that passes through R(-2, -3) andperpendicular to PQ is

2

3

x

y=

3

1

1

3y = x - 7 1

5. a) Equation of the locus of W,

22 )1()5( yx =2

51

(x – 5)2 + ( y – 1)2 = (2

5)2 1

x2 -10x +25 + y2 – 2y + 1 =4

25

4 x2 + 4y2 – 40x - 8y + 79 = 0 1

b) i) P(3 , k) lies on the locus of W,substitute x =3 and y = k into the equation of the locus of W.4(3)2 + 4(k)2 – 40(3) – 8(k) + 79 = 0 1


30

4k2 - 8k -5 = 0(2k + 1)(2k – 5) = 0

k = -2

1, k =

2

5

Since k > 0, k =2

5

1

1

ii) Since S is the centre of the locus of W, then S is themid-point of PQ.

S(5 , 1) = (2

3x,

22

5y

) 1

5 =2

3x, 1 =

22

5y

x = 7 , y = -2

1

Hence, the coordinates of Q are ( 7 , -2

1). 1

c) Area of triangle OPQ =2

1

02

5

2

10

0370

=2

1[ (7)(

2

5) – (-

2

3) ]

1

=2

19unit2 1

CHAPTER 7 STATISTICS FORM 4

31

PAPER 1

1. A set of eight numbers has a mean of 11.

(a) Find x .

(b) When a number k is added to this set, the new mean is 10. Find the value of k.[3 marks]

2. A set of six numbers has a mean of 9.5.

(a) Find x .

(b) When a number p is added to this set, the new mean is 9. Find the value of p.[3 marks]

3. Table 3 shows the distribution of the lengths of 30 fish caught by Syukri in a day.

Length (cm) 20 - 24 25 - 29 30 - 34 35 - 39 40 -44Number of fish 5 3 9 7 6

Table 3Find the mean length of the fish. [3 marks]

4. Table 4 shows the distribution of marks obtained by 56 students in a test.

Marks 20 - 29 30 - 39 40 - 49 50 - 59 60 – 69Number of Students 4 20 16 10 6

Table 4Find,

(a) the midpoint of the modal class,(b) the mean marks of the distribution.

[4 marks]

5. The mean of the set of numbers 3, 2n + 1, 4n, 14, 17, 19 which are arranged in ascendingorder is q. If the median for the set of numbers is 13, find the value of

(a) n,(b) q.

[4 marks]

6. A set of data consists of six numbers. The sum of the numbers is 72 and the sum of thesquares of the numbers is 960.Find, for the six numbers

(a) the mean,(b) the standard deviation.

[3 marks]

7. A set of positive integers consists of 2, 5 and q. The variance for this set of integers is 14.Find the value of q.


32

[3 marks]

8. The mean of eight numbers is p. The sum of the squares of the numbers is 200 and thestandard deviation is 4m. Express p in terms of m.

[3 marks]

9. Given a set of numbers 5, 12, 17, 19 and 20. Find the standard deviation of the numbers.[3 marks]

10. The sum of 8 numbers is 120 and the sum of the squares of these eight numbers is 2 200.(a) Find the variance of the eight numbers.(b) Two numbers with the values 6 and 12 are added to the eight numbers. Find the new

mean of the 10 numbers.[4 marks]

PAPER 2

1. A set of data consists of 11 numbers. The sum of the numbers is 176 and the sum of thesquares of the numbers is 3212.

(a) Find the mean and variance of the 11 numbers[3 marks]

(b) Another number is added to the set of data and the mean is increased by 2.Find

(i) the value of this number,(ii) the standard deviation of the set of 12 numbers. [4 marks]

2. Diagram 2 is a histogram which represents the distribution of the marks obtained by 60students in a test.

Diagram 2(a) Without using an ogive, calculate the median mark.

[3 marks]

0

2

4

6

8

10

12

14

16

18

20

1

Marks

Nu

mb

er

of

stu

de

nts

0.5 10.5 20.5 30.5 40.5 50.5 60.5

0

2

4

6

8

10

12

14

16

18

20

1

Marks

Nu

mb

er

of

stu

de

nts

0.5 10.5 20.5 30.5 40.5 50.5 60.5


33

(b) Calculate the standard deviation of the distribution.[4 marks]

3. Table 3 shows the frequency distribution of the scores of a group of students in a game.

Score Number of students20 - 29 130 – 39 240 - 49 850 - 59 1260 - 69 k70 - 79 1

Table 3

(a) It is given that the median score of the distribution is 52.Calculate the value of k. [3 marks]

(b) Use the graph paper to answer the question.Using a scale of 2 cm to 10 scores on the horizontal axis and 2 cm to 2 students on thevertical axis, draw a histogram to represent the frequency distribution of the scores.Find the mode score.

[4 marks]

(c) What is the mode score if the score of each pupil is increased by 4?[1 mark]

4. Table 4 shows the cumulative frequency distribution for the scores of 40 students in aMathematics Quiz.

Score < 20 <30 < 40 < 50 <60Number of students 6 16 28 36 40

Table 4

(a) Based on Table 4, copy and complete Table 4.

Score 10 - 19 20 - 29 30 - 39 40 – 49 50 - 59Number of students

(b) Without drawing an ogive, find the interquartile range of the distribution.[5 marks]


34

ANSWER

PAPER 1

No. Solution Marks

1(a) ∑x = 88 1

(b)x + k

9= 10

1

k = 2 12. (a)

∑x = 57 1

(b)x + p

7= 9

1

p = 6 1

3. x =(22 x 5) + ( 27 x 3) + (37 x 7) + (42 x 6 )

30 1,1

=990

30

= 33 14.(a) = 34.5 1

(b)

x =(24.5 x 4 ) + ( 34.5 x 20 ) + ( 44.5 x 16 ) + ( 54.5 x 10 ) + ( 64.5 x 6)

561,1

=2432

56

= 43.43 1


35

5.(a) 4n + 14

2= 13 1

n* = 3 1

(b)

3 + 19 + 2n* + 1 + 4n* + 14 + 17

6= q

1

q = 12 1

6.(a) 12 1

(b)

=960

6- (12)2

1

= 4 1

7.14 =

22 + 52 + q2

3-

2 + 5 + q

3 2 1

(q + 4 )(q – 11) = 0 1

q = 11 1

8. x2 = 200 or x = 8

1

22 168

200mp

1

p = 25 - 16m2 1

9.x = 14.6 1

Standard deviation =1219

5- (14.6)2

1

= 5.5353 1


36

10.(a)

Variance =2200

8- (15) 2 1

= 50 1

(b)

x =120 + 6 + 12

10

1

= 13.8 1


37

PAPER 2

No. Solution Marks

1(a) Mean = 16 1

Variance =3212

11- (16)2

1

= 36 1(b)(i)

18 =176 + k

121

k = 40 1(ii)

Standard deviation =3212 + 402

12- (18)2

1

= 8.775 12(a). 30.5 or ½ (60) or 30 or median class 20.5 – 30.5 1

20.5 + 1020

20)60(2

1

1

= 25.5 1(b)

x = 26.67 1

fx2 = 54285 1

Standard deviation =54285

60-

1600

60 2

1

= 13.92 1

3.(a) 49.5 or ½ (24 + k) or 11 1


38

52 = 49.5 +

1

2(24 + k) - 11

12 10

1

k = 4 1(b)

4

(c) = 57 14.(a) Score 10 - 19 20 - 29 30 – 39 40 - 49 50 - 59

Number ofstudents

6 10 12 8 4 1

(b) 19.5 or ¼ (40) or 6 139.5 or ¾ (40) or 28 1Interquartile range = 42* - 23.5* 1

= 18.5 1

Frequency

19.5 29.5 39.5 49.5 59.5 69.5 79.5

Score

0

14

12

10

8

6

4

2

Frequency

19.5 29.5 39.5 49.5 59.5 69.5 79.5

Score

0

14

12

10

8

6

4

2

19.5 29.5 39.5 49.5 59.5 69.5 79.5

Score

0

14

12

10

8

6

4

2

Mode score = 53

CHAPTER 8 CIRCULAR MEASURE FORM 4

39

PAPER 1

1. Diagram 1 shows a sector AOB with centre O .

The length of the arc AB is 7.5 cm and the perimeter of the sector AOB is 25 cm. Find thevalue of , in radian. [ 3 marks ]

2. Diagram 2 shows a circle with centre O .

Given that the length of the major arc RS is 45 cm , find the length , in cm , of the radius.( Use = 3.142 ) [ 3 marks ]

O

A

B

DIAGRAM 1

O

R

S

0.35 rad

DIAGRAM 2


40

3. Diagram 3 shows a circle with centre O .

The length of the minor arc is 15 cm and the angle of the major sector POR is 280o .Using = 3.142 , find

(a) the value of , in radians.( Give your answer correct to four significant figures )

(b) the length, in cm, of the radius of the circle . [ 3 marks ]

4. Diagram 4 shows sector OPQ with centre O and sector PXY with centre P .

Given that OQ = 20 cm , PY = 8 cm , XPY = 1.1 radians and the length of arc PQ = 14cm ,calculate

( a) the value of , in radian ,

( b) the area, in cm2, of the shaded region . [ 4 marks ]

O

P

R

DIAGRAM 3

P

Y

QO

DIAGRAM 4

X


41

5. Diagram 5 shows a sector QOR of a circle with centre O. It is given that PS = 10 cm and QP =PO = OS = SR = 6 cm.Find(a) the length, in cm, of the arc QR,(b) the area, in cm2, of the shaded region.

[4 marks]

S

R

P

Q

1.97 rad

0

DIAGRAM 5

6. Diagram 6 shows a circle with centre O and radius 12 cm. Given that A, B and C are pointssuch that OA = AB and OAC = 90o, find[Use = 3.142](a) BOC, in radians,(b) the area, in cm2, of the coloured region

[4 marks]

C

B

A

O

DIAGRAM 6


42

PAPER 2

1. Diagram 1 shows the sectors AOB, centre O with radius 15 cm. The point C on OA is suchthat OC : OA = 3 : 5 .

Calculate

(a) the value of , in radian, [ 3 marks ]

(b) the area of the shaded region, in cm2 . [ 4 marks ]

2. Diagram 2 shows a circle PQRT , centre O and radius 10 cm. AQB is a tangent to the circleat Q. The straight lines, AO and BO, intersect the circle at P and R respectively. OPQR is arhombus. ACB is an arc of a circle, centre O.

Calculate(a) the angle , in terms of , [ 2 marks ]

(b) the length, in cm, of the arc ACB, [ 4 marks ]

(c) the area, in cm2, of the shaded region. [ 4 marks ]

O

C

B

DIAGRAM 1

A

O

rad

Q

RP

A

C

B

T

DIAGRAM 2


43

3. Diagram 3 shows a sector AOB of a circle , centre O. The point P lies on OA , the point Q

lies on OB and PQ is perpendicular to OB. The length of OP is 9 cm and AOB = .6

rad

It is given that OP: OA= 3 : 5 .( Using = 3.142 )

Calculate(a) the length, in cm, of PA, [ 1 mark ]

(b) the perimeter, in cm, of the shaded region, [ 5 marks ]

( c) the area, in cm2, of the shaded region. [ 4 marks ]

4. In Diagram 4, PBQ is a semicircle with centre O and has a radiusof 10 m. RAQ is a sector of a circle with centre A and has a radius of 16 m .

It is given that AB = 10 m and BOQ = 1.876 radians. [ Use = 3.142 ]

O

PA

BQ

rad6

DIAGRAM 3

P A O Q

R

B

DIAGRAM 4


44

Calculate(a) the area , in m2 , of sector BOQ [ 2 marks ]

(b) the perimeter, in m , of the shaded region , [ 4 marks ]

(c ) the area , in m2 , of the shaded region . [ 4 marks ]

5. Diagram 5 shows a circle, centre O and radius 20 cm inscribed in a sector PAQ of a circle,centre A. The straight lines, AP and AQ, are tangents to the circle at point B and point C,respectively.[Use = 3.142]Calculate(a) the length, in cm, of the arc PQ, [5 marks](b) the area, in cm2, of the shaded region. [5 marks]

DIAGRAM 5


45

6. Diagram 6 shows two circles. The larger circle has centre P and radius 24 cm. The smaller circlehas centre Q and radius 16 cm. The circles touch at point R. The straight line XY is a commontangent to the circles at point X and Y.[Use = 3.142]

Given that XPQ = radians,

(a) show that = 1.37 ( to two decimals places), [2 marks](b) calculate the length, in cm, of the minor arc YR, [3 marks](c) calculate the area, in cm2, of the coloured region [5 marks]

DIAGRAM 6


46

ANSWERS (PAPER 1)

1 2

5.725 OA 1

7.5 = 8.75 18571.0 1

2

35.02 1)35.02(45 r 1

r = 7.583 1

3a 396.1 1

3b 396.1

15r 1

r = 10.74 1

4a rad7.0 1

4b

)7.0()20(2

1 2 or )1.1()8(2

1 2 1

)1.1()8(2

1)7.0()20(

2

1 22 1

104.8 1

5a 23.64 cm 1

5b

97.1)12(2

1 2 or rad97.1sin)6(2

1 2 1

rad97.1sin)6(2

197.1)12(

2

1 22 1

125.26 1

6a 1.047 rad or rad3

1

6b

047.1)12(2

1 2 or 6)612(2

1 22 1

6)612(2

1047.1)12(

2

1 222 1

44.21 1


47

ANSWERS (PAPER 2)

1a 15

9cos 1

9273.0 rad. 2

1b

BC = 12

)9)(12(2

1)9273.0(15

2

1 2 3

50.32 1

2a

120O 1

3

2rad 1

2b

o

OB60cos

10 1

OB = 20 cm 1

arc ACB = 20

3

21

41.89 1

2co120sin)20(

2

1

3

2)20(

2

1 22 3

245.72 1

3a PA = 6 cm 1

3b9sin30o + (15 9cos30o ) + 15 )

6(

+ 6 4

25.56 1

3c

PQ = 4.5 cm , OQ = 9cos 30o

)30cos9)(5.4(2

1

6)15(

2

1 2 o

3

41.38 1

4a876.1)10(

2

1 2 1

93.8 1

4b10(1.876) + 16 )876.1( + 6 3

45.016 1

4c)91)(6(

2

1876.1)10(

2

1)876.1()16(

2

1 22 3

39.63 1


48

5a

o

OA30sin

20 1

OA = 40 cm 1PA = 60 cm 1

arc PQ = 60

3

1

62.84 1

5b

AB = 22 2040 1

]20)2040(2

1

3

2)20(

2

1

6)60(

2

1[2 2222

3

354.513 1

6acos =

40

81

1.37 rad 1

6b

YQP = 369.1 1

arc YR = 16( 369.1 ) 128.37 1

6c

22 840 XY= 39.19

1

1.369)-()16(2

1)24)(369.1(

2

119.39)2416(

2

1 22 3

162.58 1

CHAPTER 1 PROGRESSIONS FORM 5

72

PAPER 1

1. Three consecutive terms of an arithmetic progression are pp 3,9,22 . Find the

common difference of the progression.[3 marks]

2. The first three terms of an arithmetic progression are ......,7,,1 x

Find(a) the common difference of the progression(b) the sum of the first 10 terms after the 3rd term.

[4 marks]

3. Given an arithmetic progression .......4,1,2 , state three consecutive terms in this

progression which sum up to 84 .[3 marks]

4. The sum of the first n terms of the geometric progression 5, 15, 45,….. is 5465.Find(a) the common ratio of the progression,(b) the value of n.

[4 marks]

5. The first three terms of a geometric progression are 48, 12, 3.Find the sum to infinity of the geometric progression.

[3 marks]

6. In a geometric progression, the first term is 27 and the fourth term is 1 .Calculate(a) the common ratio(b) the sum to infinity of the geometric progression.

[4 marks]

7. Express the recurring decimal 0.121212…… as a fraction in its simplest form.[4 marks]


73

PAPER 2

1. Ali and Borhan start to collect stamps at the same time.(a) Ali collects p stamps in the first month and his collection increase constantly by q

stamps every subsequent month. He collects 220 stamps in the 7th month and the totalcollection for the first 12 month are 2520 stamps. Find the value of p and q.

[5 marks]

(b) Borhan collects 60 stamps in the first month and his collection increase constantly by 25stamps every subsequent month.If both of them collect the same number of stamps in the nth month, find the value of n.

[2 marks]

2. Diagram 2 shows the arrangement of the first three of an infinite series of similar rectangles.The first rectangle has a base of x cm and a height of y cm. The measurements of the baseand height of each subsequent rectangle are half of the measurements of its previous one.

(a) Show that the areas of the rectangles form a geometric progression and state thecommon ratio.

[3 marks](b) Given that 20x cm and 80y cm,

(i) determine which rectangle has an area of16

91 cm2

(ii) find the sum to infinity of the areas, in cm2, of the rectangles.[5 marks]

y cm

x cm

Diagram 2


74

3. Diagram 3 shows part of an arrangement of circle of equal size.

The number of circles in the lowest row is 80. For each of the other rows, the number ofcircles is 3 less than in the row below. The diameter of each circle is 10 cm . The number ofcircles in the highest row is 5.Calculate(a) the height, in cm, of the arrangement of circles

[3 marks]

(b) the total length of the circumference of circles, in terms of cm.[3marks]

10 cm

Diagram 3


75

ANSWER (PAPER 1)

1 93)22(9 pp 1

4p 1

369 differencecommon 1

2 (a) xx 7)1(

3x 1

4 differencecommon 1

(b) the sum of the first 10 terms after the 3rd term:

9)4(12)1(22

13313 SS

1

290 1

3 84)3()2()1( dnadnadna 1

12n 1

Three consecutive terms:25,28,31 101112 TTT 1

4 (a)3

5

15r 1

(b)

13

)13(55465

n

1

n337 1

7n 1

5

4

1

48

12r 1

4

11

48

S 1

64 1

6 (a) 1)(27 14 r 1

3

1r

1

(b)

3

11

27S 1


76

4

120

1

7. 0.121212…… = 0.12 + 0.0012 + 0.000012 + ……..

12.0a 1

01.012.0

0012.0r

1

01.01

12.0

S 1

33

4

1

ANSWER (PAPER 2)

1 (a) 220)17( qp …….eq(1) 1

2520)112(22

12 qp ……..eq(2)

1

eq (1) x 2 : 440122 qp ……. eq(3) 1

From eq (2) : 420112 qp ……. eq(4)

(3) – (4) : 20q 1

100p 1

(b) Ali : 20)1(100 nTn

n2080

Borhan : 25)1(60 nTn

n2535

nn 25352080 1

9n 1

2 (a)Area : ..............,

16,

4,

xyxyxy 1

4

141

xy

xy

r4

1

4

162

xy

xy

r both 1

either


77

21 rr , the areas of the rectangles form a geometric progression with

the common ratio =4

1

1

(b)(i) 1600)80(20 a 1

4

1r

16

91

4

11600

1

n

1

51

4

1

4

1

n

6n 1

(ii)

4

11

1600

S1

3

12133

1

3 (a) 80, 77, 74, …………, 55nT , 3,80 da

5)3)(1(80 n 1

26n 1

Height of the arrangement = 2601026 cm 1

(b) )3)(126()80(22

2626 S

1

1105 1

The total length of circumference of circles = )5(21105

11050 cm 1

CHAPTER 2 LINEAR LAW FORM 5

78

PAPER 1

1.

Diagram 1 shows part of a straight line graph drawn to represent10

1.yx

Find the values of k

and n. [4 marks]

2.

Diagram 2 shows part of a straight line graph drawn to represent y = 10kxn, where k and n areconstants. Find the values of k and n.

[4 marks]

n •

• ( 8, k )

0x

xy

• ( 7,1)

( 3,9 )•

0 10log x

10log y

Diagram 1

Diagram 2


79

3.

Diagram 3 shows that the variables x and y are related in such away that wheny

xis plotted

against1

x, a straight line that passes through the points (12 , 7 ) and (2 , - 3 ) is obtained .

Express y in terms of x. [3 marks]

4.

Diagram 4 shows part of the graph of log10 y against x. The variables x and y are related by the

equationx

ay

b where a and b are constants. Find the values of a and b.

[4 marks ]

0

•(2 , - 3 )

1

x

y

x

• 12,7

Diagram 3

log10 y

0

•( 5 , -7 )

3 •

x

Diagram 4


80

5.

Diagram 5 shows part of the graph of 10log y against x. The variables x and y are related by the

equation y = a (10bx

) where a and b are constants. Find the values of a and b. [3 marks]

6.

Diagram 6 shows part of a straight line graph when 10log y against 10log x is plotted. Express y

in terms of x. [4 marks]

4

2

0

10log y

x

•-4

• 2

0

10log y

10log x

Diagram 5

Diagram 6


81

7. The variable x and y are related by equation xpky 3 , where k and p are constant.

Diagram 7 shows the straight line obtained by plotting y10log against x .

a) Reduce the equation xpky 3 to linear form Y = mX + c.

b) Find the value of,

i) p10log ,

ii) k. [4 marks]

y10log

xO

( 0, 8 )

( 2 , 2 )

Diagram 7


82

PAPER 2

1. Use graph paper to answer this question.

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

and y are related by the equation 22 ,n

y rx xr

where r and n are constants.

x 2 3 4 5 6 7y 8 13.2 20 27.5 36.6 45.5

(a). Ploty

xagainst x , using a scale of 2 cm to 1 unit on both axes.

Hence, draw the line of best fit. [5 marks]

(b). Use your graph in (a), to find the value of(i). n,(ii). r,(iii). y when x = 1.5. [5 marks]



and y are related by the equation ,h

y kxx

where h and k are constants.

x 1 2 3 4 5 6y 5.1 6.9 9.7 12.5 15.4 18.3

(a). Plot xy against 2x , using a scale of 2 cm to 5 units on the 2x -axis and

2 cm to 10 units on the xy-axis.Hence, draw the line of best fit. [5 marks]

(b). Use your graph in (a), to find the value of(i). h,(ii). k,(iii). y when x = 2.5. [5 marks]

Table 1

Table 2


83



and y are related by the equation ,w

ny

x where n and w are constants.

x 3 4 5 6 7 8y 103 87 76 68 62 57.4

(a). Plot 10log y against 10log x , using a scale of 2 cm to 0.1 unit on the 10log x -axis and

2 cm to 0.2 units on the 10log y -axis. Hence, draw the line of best fit.

[5 marks]

(b). Use your graph in (a), to find the value of(i). n,(ii). w,(iii). y when x = 2. [5 marks]


Table 4 shows the values of two variables, x and y obtained from an experiment. Variables x

and y are related by the equation ,b

y a xx

where a and b are constants.

x 0.2 0.4 0.6 0.8 1.2 1.4y 12.40 8.50 6.74 5.66 4.90 3.87

(a). Plot y x against x , using a scale of 2 cm to 0.2 unit on the x -axis and 2 cm to 0.2

units on the y x -axis. Hence, draw the line of best fit.

[5 marks]

(b). Use your graph in (a), to find the value of(i). a,(ii). b,(iii). y when x = 0.9. [5 marks]

Table 3

Table 4


84


Table 5 shows the values of two variables, x and y obtained from an experiment. Variables x and

y are related by the equation ,xy pm where m and p are constants.

x 1.5 3.0 4.5 6.0 7.5 9.0y 2.51 3.24 4.37 5.75 7.76 10.00

(a). Plot 10log y against x , using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 0.1

units on the 10log y -axis. Hence, draw the line of best fit.

[5 marks]

(b). Use your graph in (a), to find the value of(i). m,(ii). p,(iii). x when y = 4.8. [5 marks]


Table 6 shows the values of two variables, x and y obtained from an experiment. Variables x and

y are related by the equation ,xy hk where h and k are constants.

x 3 4 5 6 7 8y 10.2 16.4 26.2 42 67.1 107.4

(a). Plot 10log y against x , using a scale of 2 cm to 1 unit on the x -axis and 4 cm to 0.5

units on the 10log y -axis. Hence, draw the line of best fit.

[5 marks](b). Use your graph in (a), to find the value of

(i). h,(ii). k,(iii). x when y = 35.6. [5 marks]

Table 5

Table 6


85

ANSWER

PAPER 1No. Solution Marks1. 10xy x 1

Calculate gradient from graph 110n 12k 1

2.10 10log 2log 15y x 1

Calculate gradient from graph 12n 1

15k 1

3. c = -5 1

15

y

x x

1

1 5y x 1

4. log10 y = - log10 b + log10 a 1Gradient = -2 1

b = 100 1a = 1000 1

5. log10 y = bx + log10 a or log10 y = bxlog10 10 + log10 a 1a = 100 1b = - ½ 1

6. Gradient = ½ 1log10 y = ½ log10 x + 2 1

log10 x1/2 or log10 102 or log10 x1/2.102 1

100y x 1

7(a) 3log3 10 k 1

(b)(i)10 10 10log ( 3log ) logy k x p 1

8log10 p 1

(ii) 10k 1


86

PAPER 2

No. Solution Marks1(a) 1

Uniform scale for x-axis and y-axis 16 points plotted correctly 1Draw the line of best fit 1

Reduce non-linear to linear equation

2y n

rxx r

1

(b) Calculate gradient from graph 1(i) n = 0.77 1(ii) r = 0.275 1(iii)

From graph , 3.6y

x 1

y = 5.4 1

2.(a) 1


Reduce non-linear to linear equation2xy kx h

1

(b) Calculate gradient from graph 1(i) 2h 1(ii) 3k 1(iii) From graph, xy = 20 1

y = 8 1

x 2 3 4 5 6 7y

x

4 4.4 5 5.5 6.1 6.5

2x 1 4 9 16 25 36xy 5.1 13.8 29.1 50 77 109.8


87

3.(a) 1



10 10 10log log logy w x n

1

(b) Calculate gradient from graph 1(i) 2.310 199.526n 1

(ii) 0.6w 1(iii) From graph, 10log 2.12y 1

2.1210 131.825y 1

4.(a) 1



y x ax b

1

(b) Calculate gradient from graph 1(i) a - 0.8 1(ii) 5.7b 1(iii) From graph 4.9y x 1

y = 5.165 1

x 0.2 0.4 0.6 0.8 1.2 1.4

y x 5.55 5.38 5.22 5.06 4.9 4.74

10log x 0.48 0.60 0.70 0.78 0.85 0.90

10log y 2.01 1.94 1.88 1.83 1.79 1.76


88

5.(a) 1



10 10 10log log logy x m p

1

Calculate gradient from graph 1(i) 0.08410m or 1.2134m 1

(ii) 0.2610 1.8197p or 1

(iii)10log 4.8 0.68 1

x = 5 1

6.(a) All values of 10log y are correct 1



10 10 10log log logy x k h

1

Calculate gradient from graph 1(i) 0.210k or 1.588k 1

(ii) 0.42510h or 2.66h 1

(iii)10log 35.6 1.55 1

x = 5.7 1

x 3 4 5 6 7 8

10log y 1.0 1.21 1.42 1.62 1.82 2.03

x 1.5 3.0 4.5 6 7.5 9

10log y 0.4 0.51 0.64 0.76 0.89 1

Chapter 3 Intergration Form 5

89

PAPER 1

1. Given that cxkxdxx 2)25( 43 , where k and c are constants.

Find (a) the value of k

(b) the value of c if 3(5 2 ) 3 0x d x when x=2. [3 marks]

2. Evaluate

4

3

3)2(

1dx

x[3 marks]

3. Given that 5

,632p

dxx where p < 5 , find the possible values of p.

[4 marks]

4. Given that 8)(4

2

dxxf and 8)(24

2

dxxfkx , find the value of k. [4 marks]

5.

Given that the area of the shaded region under the curve y=x(a x) is2

14 unit2, find the value

of a. [ 4 marks]

6.

Diagram above shows the curve y = f(x). Given that the area of the shaded region is 5 unit2, find

the value of .2)(24

0

dxxf [3 marks]

a0

y=x(a x)

x

y

40

y = f(x)

x

y


90

PAPER 2

1. A curve is such that22

11

xdx

dy . Given that the curve passes through the point

2

1,1 , find the

equation of the curve. [ 4 marks]

(b) Diagram above shows part of the curve14

1

xy which passes through A(0,1). A region is

bounded by the curve, the x-axis, straight line x = 2 and y-axis. Find the area of the region.[4 marks]

2. Diagram below shows part of the curve 2)4(2

1 xy which passes through A(2, 2).

(a) Find the equation of the tangent to the curve at the point A [ 4 marks]

(b) The shaded region is bounded by the curve, the x-axis and the straight line x = 2.(i) Find the area of the shaded region(ii) The region is revolved through 360º about the x-axis. Find the volume generated, in terms

of . [ 6 marks]

20x

y

A(0, 1)

14

1

xy

0x

y

A(2, 2)


91

3. In the diagram, the straight line PQ is a tangent to the curve 12

1 2 yx at Q(3, 2)

(a) Show that the value of k is2

1 , [ 3 marks]

(b) Find the area of the shaded region. [ 4 marks](c) Find the volume generated, in terms of , when the region bound by the curve, the x-axis,

and the straight line x=3 is revolved through 360º about the x-axis.

4. The diagram shows a curve such that 62 xdx

dy . The minimum point of the curve is A(3,1). AB

is a straight line passing through A and B where B is the point of intersection between thecurve and y-axis.

(a) Find the equation of the curve. [3 marks]

(b) Calculate the area of the shaded region. [4 marks]

(c) Calculate the volume of revolution, in term of , when the region bounded by the line AB, y-axis and the line x=3 is rotated through 360o about the x-axis. [3 marks]

0x

y

Q(3, 2)

P(0,k)

x

y

O

A

B


92

5. Diagram shows the straight line y=3x intersecting the curve y = 4 x2 at point P.

Find(a) the coordinates of P, [3 marks]

(b) the area of region R which is bounded by the line y = 3x, the curve y = 4 x2 and the x-axis.[4 marks]

(c) the volume generated by region bounded by the curve, straight line y = 4, x-axis, andy-axis is revolve 360o about the y-axis. [3 marks]

y=3x

y = 4 x 2

0

y

x

P

R


93

PAPER 1 (Answer)

Q Solution Marks1 ( a )

k =5

4

1

( b ) 5

4( 2 )4 + 2( 2 ) + c = 30

1

c = 6 12 4

3( 2)x

-3 dx

=

42

3

( 2)

2

x

1

=1

2

1 1

4 1

1

=3

8

1

3 52 3p

x x = 61

[52-3(5) ] – (p2-3p) = 6 1(p + 1) (p – 4 ) =0 1

p = 1, 4 1

4

k

424

22

2 ( ) 82

xf x dx

1,1

6k-2(8)=8 1k = 4 1

52

0

9

2

a

ax x dx 1

2 3

0

9

2 3 2

a

ax x

1

3 3 9

2 3 2

a a

1

a = 3 16

24 4

0 0( ) 2f x dx dx

= 2 5 + [2x] 40

1

= 18 1


94

PAPER 2 (Answer)

Q Solution Marks

1 (a)y =

2

11

2dx

x

= 211

2x dx 1

= x 1

2c

x

1

Substitute1

1, , 12

c

1

y = x 1

12x

1

(b)Area=

2

0

1

4 1dx

x correct limit1

=2

0

24 1

4x

1

=1

( 9 12

1

= 1 12 (a) dy

dx= x 4

1

x = 2,dy

dx= 24 = 2

1

y 2 = 2(x2) 1

y = 2x+6 1

(b)(i) 42

2

1( 4)

2x dx

1

=

43

2

( 4)

2(3)

x

1

= 24

3unit

1

(ii)4

4

2

1( 4)

4x dx

1

=

45

2

1 ( 4)

4 5

x

1

=8

5 unit3 1

3 (a)y = (2x2)

1

2

dy

dx=

1

2 2x 1


95

x=3,1 1

22(3) 2

dy

dx

1

2 1 1,

3 0 2 2

kk

1

(b) 2

2

0

1 1 3( 1) 32 2 2

y dy

1, 1

=2

3

0

1 9

6 4y y

=10 9

3 4

1

=13

12unit 2 1

(c) 3

1(2 2)x dx

1

=32

12x x

1

=4 unit3 14 (a) y = (2 6)x dx

y = x2-6x + c

1

Substitute x = 3, y = 1 in the equation, c = 10 1y = x2-6x + 10 1

(b) Equation of AB is y = 3x + 103

2

0( 3 10) ( 6 10)x x x dx

=3

2

03x x dx

1

=

32 3

0

3

2 3

x x

1

=9

2unit2 1

(c)Volume=

32

0( 3 10)x dx

=3

2

0(9 60 100)x x dx

1

=33 2

03 30 100x x x

1

=111 unit3 15 (a) x2 +3x 4 = 0 1

(x1) (x+4)=0 1

x= 1, y =3(1)=3P(1, 3)

1

(b)Area of triangle=

1

2(1)(3) = 1.5 unit2 1

232

2

11

4 43

xx dx x

1


96

=12

3unit2 1

Area of R = 31

6

1

(c) 4

0(4 )y dy

1

=

42

0

42

yy

1

=8 unit3 1

CHAPTER 4 VECTORS FORM 5

97

PAPER 1

1. Diagram below shows two vectors, OP and QO

Q(-8,4)

P(5,3)

Express

(a) OP in the form ,

y

x

(b) QO in the form x i + y j

[2 marks]

2. Use the above information to find the values of h and k when r = 3p – 2q.[3 marks]

3. Diagram below shows a parallelogram ABCD with BED as a straight line.

D C

EA B

Given that AB = 6p , AD = 4q and DE = 2EB, express, in terms of p and q

(a) BD

(b) EC[4 marks]

O

p = 2a + 3bq = 4a – br = ha + ( h – k ) b, where h and k are constants


98

4. Given that O(0,0), A(-3,4) and B(2, 16), find in terms of the unit vectors, i and j,

(a) AB

(b) the unit vector in the direction of AB[4 marks]

5. Given that A(-2, 6), B(4, 2) and C(m, p), find the value of m and of p such that

AB + 2 BC = 10i – 12j.[4 marks]

6. Diagram below shows vector OA drawn on a Cartesian plane.y

A

0 2 4 6 8 10 12 x

(a) Express OA in the form

y

x

(b) Find the unit vector in the direction of OA[3 marks]

7. Diagram below shows a parallelogram, OPQR, drawn on a Cartesian plane.

y

Q

R P

O x

It is given that OP = 6i + 4j and PQ = - 4i + 5j. Find PR .

[3 marks]

6

4

2


99

8. Diagram below shows two vectors, OA and AB .

yA(4,3)

O x

-5

Express

(a) OA in the form

y

x

(b) AB in the form xi + yj[2 marks]

9. The points P, Q and R are collinear. It is given that PQ = 4a – 2b and

bkaQR )1(3 , where k is a constant. Find

(a) the value of k

(b) the unit vector in the direction of PQ

[4 marks]

10. Given that jia 76 and ,2 jpib find the possible value (or values) of p for following

cases:-a) ba and are parallel

b) ba

[5 marks]

B


100

PAPER 2

1. Give that

3

2,

7

5OBAB and

5

kCD , find

(a) the coordinates of A, [2 marks]

(b) the unit vector in the direction of OA , [2 marks]

(c) the value of k, if CD is parallel to AB [2 marks]

2. Diagram below shows triangle OAB. The straight line AP intersects the straight line OQ at R. It

is given that OP = 1/3 OB, AQ = ¼ AB, xOP 6 and .2yOA

A

R

O P B

(a) Express in terms of x and/or y:

(i) AP

(ii) OQ [4 marks]

(b) (i) Given that ,APhAR state AR in terms of h, x and y.

(ii) Given that ,OQkRQ state RQ in terms of k, x and y.

[2 marks]

(c) Using AR and RQ from (b), find the value of h and of k.

[4 marks]

Q


101

3. In diagram below, ABCD is a quadrilateral. AED and EFC are straight lines.

D

E F C

A B

It is given that AB 20x, AE 8y, DC = 25x – 24y, AE = ¼ AD

and EF =5

3EC.

(a) Express in terms of x and/or y:

(i) BD

(ii) EC [3 marks]

(b) Show that the points B, F and D are collinear. [3 marks]

(c) If | x | = 2 and | y | = 3, find | BD |. [2 marks]

4. Diagram below shows a trapezium ABCD.

B CF•

•A E D

It is given that A B

=2y, AD = 6x, AE =3

2AD and BC =

6

5AD

(a) Express AC in terms of x and y [2 marks]

(b) Point F lies inside the trapezium ABCD such that 2 EF = m AB , and m is aconstant.

(i) Express AF in terms of m , x and y(j) Hence, if the points A, F and C are collinear, find the value of m.

[5 marks]


102

ANSWERS (PAPER 1)

1. a)

3

51

b) 8i – 4j 1

2. r = - 2a + 11br = ha + (h – k)b

1

h = -2 1(h – k) = 11

k = −13 1

3 a) BD = −6p + 4q 1

b) DB = − BD= 6p −4q

1

EB =3

DB

42

3p q 1

BCEBEC 8

23

p q 1

4. a) jiAB )416())3(2( 1

= 5i + 12j 1

b)u )125(

125

1

22ji

1

)125(13

1ji

1

5. ))2(2)4((2)46(2 jpimjiBCAB 1

= (-2+2m)i + (-8+2p)j 1m = 6 1p = -2 1

6. a)

5

12OA

1

b))512(

512

1

22jiu

1

)512(13

1ji 1


103

7. OPPO ji 46 1

PQOR

ji 54

1

PR PO OR

ji 101

8. a)

3

4OA 1

b) jiAB 84 1

9. a) PQmQR

2

4

1

3m

k1

3 = m(4)

4

3m

1

1+ k = m(-2)

2

5k

1

b)

2 24 ( 2)

PQu

1(4 2 )

20a b 1

10 a) bka

)2(76 jpikji 1

2

7k 1

7

12p 1

b) ba

2 2 2 26 7 2p 1

9p 1


104

ANSWERS (PAPER 2)

1 (a)

4

3AO

1

4

3OA

A = (-3,-4) 1(b)

OA

OAu

OA22 43

1

1

jiu 435

1

1

(c) ABmCD

7

5

5m

k1

7

5m

7

25k 1

2 (a) (i) 6 2AP x y

1

(ii) 18 2AB x y

1

9 1

2 2AQ x y 1

9 3

2 2OQ x y 1

(b) (i) hyhxAR 26 1

9 3

2 2RQ kx ky 1

(c) AQRQAR

yxykhxkh2

1

2

9

2

32

2

96

1

3

1k

1

2

1h

1


105

3 (a) (i) yxBD 3220 1

(ii)ADED

4

3

= 24y 1

xEC 25 1

(b) xFC 10

DCBDBC yx 85

1

CFBCBF yx 85 1

yxBD 3220

)85(4 yx

)(4 BF 1

(c) 223220 yxBD

22 332)220( 1

= 104 1

4 (a)ADBC

6

5

= 5x 1

BCABAC = 5x + 2y 1

(b) (i) ABmEF 2

myEF 1

ADAE3

2

= 4x 1

EFAEAF = 4x + my 1

(ii) yxAC 25

yxAC 255

4

5

4

yxAC5

84

5

4

1

Assume A, F, C collinear,

AFAC 5

4

= 4x + my

5

8m 1

CHAPTER 5 TRIGONOMETRIC FUNCTIONS FORM 5

106

PAPER 1

1. Given is an acute angle and sin p . Express each of the following in

terms of p.[3 marks]

a) tan

b) cos 90oec

2. Given cos p and 270 360o o .Express each of the following in

terms of p.[3 marks]

a) sec

b) cot 90o

3. Given tan r , where r is a constant and 180 270o o . Find interms of r. [3 marks]

a) cot

b) tan 2

4. Solve the equation 26cos 13cot 0ec x x for 0 360o ox [4 marks]

5. Solve the equation 2 22sin cos sin 1A A A for 0 360o oA [4 marks]

6. Solve the equation 2sin7cos2 2 yy for 0 360o oy [4 marks]

7. Solve the equation 2 015cos cos 4cos 60x x for 0 360o oA [4 marks]

8. Solve the equation 3cot 2sin 0x x for 0 360o ox [4 marks]


107

PAPER 2

1. (a) Sketch the graph of sin 2y x for 0 180o ox [4 marks]

(b) (b) Hence, by drawing suitable straight line on the same axes, find the

number of solution to the equation1

sin cos2 360o

xx x for

0 180o ox

[3 marks]

2. (a) Sketch the graph of 3cosy x for 0 2x [4 marks]

(b) (b) Hence, by using the same axes, sketch a suitable graph to find the

number of solution to the equation2

3cos 0xx

for 0 2x .

State number of solutions.

[3 marks]

3. (a) Sketch the graph of 3sin 2y x for 0 2x [4 marks]

(b) Hence, by using the same axes, sketch a suitable straight line to find

the number of solution to the equation 2 3sin 22

xx

for

0 2x

[3 marks]

4.(a) Prove that

cot tancos 2

2

x xec x

[2 marks]

(b) (i) Sketch the graph of3

2sin2

y x for 0 2x [6 marks]

(ii) Find the equation of a suitable straight line to solve the

equation3 3 1

sin2 2 2

x x

.

Hence, on the same axes, sketch the straight line and state the

number of solutions to the equation3 3 1

sin2 2 2

x x

for

0 2x .5. (a) Prove that 2 2 2sec 2cos tan cos 2x x x x [2 marks]

(b) (i) Sketch the graph of cos 2y x for 0 2x [6 marks]

(ii) Hence, using the same axes, draw a suitable straight line to find

the number of solutions to the equation 22cos 1x

x

for

0 2x .


108

6. (a) Prove that 2 22 2sin 2cosx x [2 marks]

(b) Sketch the graph of 12tan xy for 20 x . By using the same

axes, draw the straight line9

32

y x

and state the number of solution to

equation9

tan 2 22

x x

for 20 x

[6 marks]

7. (a) Prove that 2 2 2 2cot cos tan secx ec x x x [2 marks]

(b) Sketch the graph3

cos2

y x and 2siny x for 0 2x . State

the number of solution to equation1 3

sin cos2 2

x x

for

0 2x

[6 marks]


109

ANSWERS (PAPER 1)

1 a)2

tan1

p

p

1

b) cos 90oec =

1

sin 90o

1

= 21 p 1

2 a)sec =

1

cos

=1

p

1

b) cot 90o = tan 1

=21 p

p

1

3 a)cot =

1

tan

=1

r

1

b)tan 2 =

2

2 tan

1 tan

1

=2

2

1

r

r

1

4 26 1 cot 13cot 0x x 1

26cot 13cot 6 0x x

3cot 2 2cot 3 0x x 1

3cot 2 0x OR 2cot 3 0x

3tan

2x OR

2tan

3x

' 056 19 or 56.31ox and '33 41 or 33.69o ox 1

' o ' ' o '56 19 , 236 19 33 41 , 213 41o ox

Or 56.31 , 236.31 ,33.69 , 213.69o o o o

1

5 2 22sin 1 sin sin 1A A A 1

2 22sin 1 sin sin 1A A A

23sin sin 2 0A A

3sin 2 sin 1 0A A 1


110

3sin 2 0A OR sin 1 0A

2sin

3A OR sin 1A

90 and 41.81o oA 1

90 , 221.81 , 318.19o o oA 1

6 22cos 7sin 2 0y y

22(1 sin ) 7sin 2 0y y 1

22sin 7sin 4 0y y

2sin 1 sin 4 0y y 1

1sin

2y

sin 4y (not accepted)

30oy 1

30 , 150o oy 1

7 2 015cos cos 4cos 60x x

2 015cos cos 4cos 60 0x x

215cos cos 4(0.5) 0x x

215cos cos 2 0x x 1

5cos 2 3cos 1 0x x 1

5cos 2 0x OR 3cos 1 0x

2cos

5x OR

1cos

3x

42.66x or '2566 and 53.70 or '3170 1

47.289,53.70,58.293,42.66 x' ' ' o '66 25 , 70 31 , 289 28 , 293 35o o ox

1

8 cos3 2sin 0

sin

xx

x

23cos 2sin 0x x

23cos 2 1 cos 0x x 1

23cos 2 2cos 0x x

22cos 3cos 2 0x x


111

45o 90o 135o

x

y

1180o

xy

180o

sin 2y x

(ANSWERS)PAPER 2

2cos 1 cos 2 0x x 1

2cos 1 0x OR cos 2 0x

1cos

2x OR cos 2x (unaccepted)

120ox 1

o120 , 240ox 1

1 1 ( shape)1(max/min)1(one

period)1(complete

from 0 to180o)

1 (straightline)

1sin cos

2 360o

xx x

1

2 2 360o

y x

1180o

xy 1

Number of solutions= 3

1


112

x

y

2 a) 1 ( shape)1(max/min)1(one

period)1(complete

from 0 to2 or360 o)

1(for line2

yx

b) 23cos 0x

x

20y

x

(b)2

yx

1

Number of solution =21

3

x

y

1 ( shape)1(max/min)1(one

period)1(complete

from 0o

to 2 )

1( for thestraightline)

2 3sin 22

xx

22

xy

22

xy

1

Numbers of solutions= 81

2y

x

3cosy x


113

4(a) Prove that

cot tancos 2

2

x xec x

LHS

1 cos sin

2 sin cos

x x

x x

2 21 cos sin

2 sin cos

x x

x x

1

1 1

2 sin cosx x

1

2sin cosx x

1

sin 2x

cos 2ec x 1

b) 1(shape)1(max/min)1(one

period)

1(for thestraightline)

3 3 1sin

2 2 2x x

3 12

2 2y x

31y x

1

Number of solution = 11

5(a) LHS 2 2 2sec 2cos tanx x x

2 2 21 tan 2cos tanx x x 1

21 2cos x

cos 2 (proved)x 1

2

3

2

2

x

y

31y x

32sin

2y x


114

xy

2

3

2

-

x

y

cos 2y x

b) 1(shape)1(max/min)1(one

period)

1(for thestraightline )

22cos 1x

x

xy

1

Number of solutions = 2 1

6 22 2sin x

22 1 sin x 1

22cos x (proved) 1

b)

Number of solution = 3

1(shape)1(max/min)1(one

period)

1(completecyclefrom 0 to2 )

1(for thestraightline)

1

2

3

2

2

2

x

y

tan 2 1y x

93

2y x

-2


115

3cos

2y x

7 (a) 2 2 2cos tan secRHS ec x x x

2 2 21 cot tan secx x x 1

2 2 2cot 1 tan secx x x

2 2 2cot sec secx x x

2cot (proved)x 1

Number of solutions = 3

1,1(shapes)1(max/min)1(one

period)1(complete

cyclefrom 0 to2 )

1

y=2sinx

2

3

2

2

x

y

CHAPTER 6 PERMUTATIONS AND COMBINATIONS FORM 5

116

PAPER 1

1. A committee of 3 boys and 3 girls are to be formed from 10 boys and 11 girls.In how many ways can the committee be formed?

[2 marks]

2. How many 4-letter codes can be formed using the letters in the word 'GRACIOUS' withoutrepetition such that the first letter is a vowel?

[2 marks]

3. Find the number of ways of choosing 6 letters including the letter G from the word'GRACIOUS'.

[2 marks]

4. How many 3-digit numbers that are greater than 400 can be formed using the digits 1, 2, 3, 4,and 5 without repetition?

[2 marks]

5. How many 4-digit even numbers can be formed using the digits 1, 2, 3, 4, and 5 withoutrepetition?

[2 marks]

6. Diagram shows 4 letters and 4 digits.

A code is to be formed using those letters and digits. The code must consists of 3 lettersfollowed by 2 digits. How many codes can be formed if no letter or digit is repeated in eachcode ?

[3 marks]

7. A badminton team consists of 8 students. The team will be chosen from a group of 8 boys and5 girls. Find the number of teams that can be formed such that each team consists of

a) 5 boys,b) not more than 2 girls.

[4 marks]

8. Diagram shows five cards of different letters.

a) Find the number of possible arrangements, in a row, of all the cards.b) Find the number of these arrangements in which the letters A and N are side by side.

[4 marks]

A B C D 5 6 7 8

R A J I N


117

9. A debating team consists of 6 students. These 6 students are chosen from 2 monitors, 3assistant monitors and 5 prefects.

a) there is no restriction,b) the team contains only 1 monitor and exactly 3 prefects.

[4 marks]

10. Diagram shows seven letter cards.

A five-letter code is to be formed using five of these cards. Finda) the number of different five-letter codes that can be formed,b) the number of different five-letter codes which end with a consonant.

[4 marks]

11. How many 5-digit numbers that are greater than 50000 can be formed using the digits 1, 2, 3, 4,5, 6, 7, 8, and 9 without repetition?

[4 marks]

12. How many 4-digit even numbers can be formed using the digits 1, 2, 3, 4, 5 and 6 without anydigit being repeated?

[4 marks]

13. A coach wants to choose 9 players consisting of 6 boys and 3 girls to form a squash team.These 9 players are chosen from a group of 8 boys and 6 girls. Find(a) the number of ways the team can be formed,(b) the number of ways the team members can be arranged in a row for a group photograph,

if the 6 boys sit next to each other. [4 marks]

14. 2 girls and 8 boys are to be seated in a row of 5 chairs. Find the number of ways they can beseated if no two persons of the same sex are next to each other.

[3 marks]

15. Diagram shows six numbered cards.

A four-digit number is to be formed by using four of these cards.

How manya) different numbers can be formed?b) different odd numbers can be formed?

[4 marks]

ROFINU M

987541


118

ANSWERS ( PAPER 1 )

1.3

10C x 311C 1

= 19800 1

2.1

4 p x 37 p 1

= 840 1

3. 1 x 57C 1

= 21 1

4.1

2 p x 24 p 1

= 24 1

5.3

4 p x 12 p 1

= 48 1

6.3

4 p x 24 p 2

= 288 1

7. a) 58C x 3

5C = 560 1

b) If the team consists of 8 boys and 0 girl 88C x 0

5C = 1

If the team consists of 7 boys and 1 girl 78C x 1

5C = 40

If the team consists of 6 boys and 2 girl 68C x 2

5C = 280

1

The number of teams that can be formed = 1 + 40 + 280 1= 321 1

8. a) 5! = 120 1

b) 4! x 2! 2

= 481

9. a) 610C = 210 2

b) 12C x 3

5C x 23C = 60 2

10. a) 57 p = 2520 2

b) 46 p x 1

4 p 1

= 14401

11.1

5 p x 48 p 2

= 84001

12.3

5 p x 13 p = 180 2

= 1801


119

13. a) 68C 3

6C 1

= 560 1

b) 6! x 4! 1

= 17280 1

14.3

8 P x 22 P 2

= 672 1

15. a) 6P4 = 360 1

b) 5P3 x 4P1 2= 240 1

CHAPTER 7 PROBABILITY FORM 5

120

PAPER 1

1. A box contains 8 blue marbles and f white marbles. If a marble is picked randomly from the

box, the probability of getting a white marble is3

5.

Find the value of f.[3 marks]

2. A bag contains 4 blue pens and 6 black pens. Two pens are drawn at random from the bag oneafter another without replacement. Find the probability that the two pens drawn are ofdifferent colours.

[3 marks]3. Table 2 shows the number of coloured cards in a box.

Colour Number of CardsYellow 6Green 4Blue 2

Table 2

Two cards are drawn at random from the box.Find the probability that both cards are of the same colour.

[3 marks]

4. The probability that Amir qualifies for the final of a track event is1

5while the probability

that Rajes qualifies is2

3.

Find the probability that,(a) both of them qualify for the final,(b) only one of them qualifies for the final. [3 marks]

5. A box contains 4 cards with the digits 1, 2, 3 and 4. Two numbers are picked randomly fromthe box. Find the probability that both the numbers are prime numbers.

[3 marks]

6. Team A will play against Team B and Team C in a sepak takraw competition. The

probabilities that Team A will beat Team B and Team C are1

3and

2

5respectively. Find the

probability that Team A will beat at least one of the teams.

[3 marks]


121

7. The probability of a particular netball player scoring a goal in a netball match is1

5. Find the

probability that this player scores only one goal in three matches.[3 marks]

8. The probabilities that Hamid and Lisa are selected for a Science Quiz are2

5and

1

3respectively. Find the probability that at least one of them are selected.

[3 marks]

9. Table 9 shows the number of coloured marbles in a box.

Colour Number of MarblesBlack 2Red 5

Green 3Yellow 8

Table 9

Two marbles are drawn at random from the box.Find the probability that both marbles are of the same colour.

[3 marks]

10. The probability of Zainal scoring a goal in a football match is2

5. Find the probability that

Zainal scores at least one goal in two matches.[3 marks]


122

ANSWER (PAPER 1)

No. Solution Marks

1. 8 + f 1

f

8 + f=

3

5

1

f = 12 1

2.=

4

10x

6

9 @6

10x

4

9 1

=4

10x

6

9 +6

10x

4

9 1

=8

151

3.=

6

12x

5

11 or4

12x

3

11 or2

12x

1

11 1

=6

12x

5

11 +4

12x

3

11 +2

12x

1

11 1

=1

3

1

4(a)=

2

15

1

(b)=

1

5x

1

3 +2

3x

4

5 1


123

=3

5

1

5.=

2

4or

1

3

1

=2

4x

1

3

1

=1

61

6.=

1

3x

3

5 o r2

3x

2

5 or1

3x

2

5 1

=1

3x

3

5 +2

3x

2

5 +1

3x

2

5 1

=3

51

7. =1

5x

4

5x

4

5 1

=1

5x

4

5x

4

5 +4

5x

1

5x

4

5 +4

5x

4

5x

1

5 1

=48

1251

8. =2

5x

2

3 or1

3x

3

5 or1

3x

2

5 1


124

=2

5x

2

3 +1

3x

3

5 +1

3x

2

5 1

=3

51

9.=

2

18x

1

17 or5

18x

4

17 or3

18x

2

17 or8

18x

7

17 1

=2

18x

1

17 +5

18x

4

17 +3

18x

2

17 +8

18x

7

17 1

=14

51

1

10.=

2

5x

2

5 or2

5x

3

5 or3

5x

2

5 1

=2

5x

2

5 +2

5x

3

5 +3

5x

2

5 1

=16

25

1

CHAPTER 8 PROBABILITY DISTRIBUTION FORM 5

125

PAPER 1

1. Diagram 1 shows a standard normal distribution graph.

Diagram 1

If P(0 < z < k) = 0.3125, find P(z > k). [2 marks]

2. In an examination, 85% of the students passed. If a sample of 12 students is randomly selected,find the probability that 10 students from the sample passed the examination. [3 marks]

3. X is a random variable of a normal distribution with a mean of 12.5 and a variance of 2.25.Find(a) the Z score if X= 14.75(b) P(12.5 X 14.75) [4 marks]

4. The mass of students in a school has a normal distribution with a mean of 55 kg and a standarddeviation of 10 kg. Find(a) the mass of the students which give a standard score of 0.5,(b) the percentage of students with mass greater than 48 kg. [4 marks]

5. Diagram 2 below shows a standard normal distribution graph.

Diagram 2

The probability represented by the area of the shaded region is 0.3264 .(a) Find the value of k.(b) X is a continuous random variable which is normally distributed with a mean of 180 and a

standard deviation of 5.5.Find the value of X when z-score is k. [4 marks]

f(z)

0 k z

f(z)

0 k z

0.3264


126

6 X is a continuous random variable of a normal distribution with a mean of 52 and a standarddeviation of 10. Find(a) the z-score when X = 67(b) the value of k when P(z < k) = 0.8643

[4 marks]

7 The masses of a group of students in a school have a normal distribution with a mean of 45 kgand a standard deviation of 5 kg.Calculate the probability that a student chosen at random from this group has a mass of(a) more than 50.6 kg,(b) between 40.5 and 52.1 kg [4 marks]

PAPER 2

1. (a) Senior citizens make up 15% of the population of a settlement.(i) If 8 people are randomly selected from the settlement, find the probability that at least two

of them are senior citizens.(ii) If the variance of the senior citizens is 165.75, what is the population of the settlement?

[5 marks]

(b) The mass of the workers in a factory is normally distributed with a mean of 65.34 kg and avariance of 56.25 kg2. 321 of the workers in the factory weigh between 48 kg and 72 kg.Find the total number of workers in the factory. [5 marks]

2. (a) A club organizes a practice session for trainees on scoring goals from penalty kicks. Eachtrainee takes 6 penalty kicks. The probability that a trainee scores a goal from a penalty kickis p. After the session, it is found that the mean number of goals for a trainee is 4.5(i) Find the value of p.(ii) If a trainee is chosen at random, find the probability that he scores at least one goal.

[5 marks]

(b) A survey on body-mass is done on a group of students. The mass of a student has a normaldistribution with a mean of 65 kg and a standard deviation of 12 kg.(i) If a student is chosen at random, calculate the probability that his mass is less than 59 kg.(ii) Given that 15.5% of the students have a mass of more than m kg, find the value of m.

[5 marks]


127

3. For this question, give your answer correct to three significant figures.(a) The result of a study shows that 18% of pupils in a city cycle to school.

If 10 pupils from the city are chosen at random, calculate the probability that(i) exactly 2 of them cycle to school,(ii) less than 3 of them cycle to school. [4 marks]

(b) The mass of water-melons produced from an orchard follows a normal distribution with amean of 4.8 kg and a standard deviation of 0.6 kg.Find(i) the probability that a water-melon chosen randomly from the orchard has a mass of not

more than 5.7 kg,(ii) the value of m if 80% of the water-melons from the orchard has a mass of more than

m kg. [6 marks]

4. An orchard produces oranges.Only oranges with diameter, x greater than k cm are graded and marketed.Table below shows the grades of the oranges based on their diameters.

Grade A B CDiameter, x (cm) x > 6.5 6.5 x > 4.5 4.5 x k

It is given that the diameter of oranges has a normal distribution with a mean of 5.3 cm and astandard deviation of 0.75 cm.(a) If an orange is picked at random, calculate the probability that it is of grade A. [2 marks](b) In a basket of 312 oranges, estimate the number of grade B oranges. [4 marks](c) If 78.76% of the oranges is marketed, find the value of k. [4 marks]

5 (a) In a survey carried out in a school, it is found that 3 out of 5 students have handphones. If8 students from the school are chosen at random, calculate the probability that(i) exactly 2 students have handphones.(ii) more than 2 students have handphones.

[5 marks]

(b) A group of teachers are given medical check up. The blood pressure of a teacher has anormal distribution with a mean of 128 mmHg and a standard deviation of 10 mmHg.Blood pressure that is more than 140 mmHg is classified as “high blood pressure”.(i) A teacher is chosen at random from the group.

Find the probability that the teacher has a pressure between 110 mmHg and 140mmHg.

(ii) It is found that 16 teachers have “high blood pressure”. Find the total number ofteachers in the group.

[5 marks]


128

6 The masses of mangoes from an orchard has a normal distribution with a mean of 285 g and astandard deviation of 75 g.(a) Find the probability that a mango chosen randomly from this orchard has a mass of more

than 191.25 g. [3 marks](b) A random sample of 520 mangoes is chosen.

(i) Calculte the number of mangoes from this sample that have a mass of more than191.25.

(ii) Given that 416 mangoes from this sample have a mass of more than m g, find thevalue of m. [7 marks]

ANSWERS (PAPER 1)

10.5 - 0.3125 10.1875 1

2

21010

12 )15.0()85.0(C 1

0.2924 1p = 0.85, q = 0.15 1

3a 5.1

5.1275.14 1

1.5 1

3b 5.1

5.125.12 or

5.1

5.1275.14 1

0.4332 1

4a 10

555.0

X1

60 1

4b 10

5548 1

75.8% 1

5a0.5 - 0.3264 10.94 1

5b94.0

5.5

180

X1

X = 185.17 1

6a 10

5267 z 1

1.5 1

6b1 - 0.8643 1k=1.1 1


129

7a 5

456.50 1

0.1314 1

7b 5

455.40 or

5

451.52 1

0.7381 1

ANSWERS (PAPER 2)

1(a)(i)

p = 0.15, q = 0.85 17

1880

08 )85.0)(15.0()85.0()15.0(1)2( CCXP

1 - 0.2725 - 0.3847 10.3428 1

1(a)(ii)165.75 = n(0.15)(0.85) 1n = 1300 1

1(b)

34.65 , 5.7 1

)7248( xP

)5.7

34.6572

5.7

34.6548(

zP 1

P(-2.312 < z < 0.888)1 - 0.0103 - 0.1872 1400 1

8025.0321

xx = 400 1

2(a)(i)

mean = np

6

5.4p 1

0.75 1

2(a)(ii)

)1( XP

1 - P(X <1) 11 - P(X=0)

600

6 )25.0()75.0(1 C 1

0.9998 1

2(b)(i))

12

6559()59(

zPxP 1

P(z < - 0.5)0.3085 1

2(b)(ii)015.1

12

65

m2

m = 77.18 1


130

3(a)(i)

p = 0.18, q = 0.82

P(x = 2) = 822

10 )82.0()18.0(C 1

0.298 1

3(a)(ii)

P(X < 3) = P(x = 0) + P(x = 1) + P(x = 2)82

2109

110100

010 )82.0()15.0()82.0)(18.0()82.0()18.0( CCC 1

0.1374 + 0.3017 + 0.29800.7371 1

3(b)(i)P(

6.0

8.47.5 z ) 1

1 - 0.0668 10.9332 1

3(b)(ii)842.0

6.0

8.4

m2

m = 48 1

4(a)P(

75.0

3.55.6 z ) 1

0.0548 1

4(b)

P(75.0

3.55.6

75.0

3.55.4

z ) 1

1 - 0.1430 - 0.05481

312 x 0.8022 1250 1

4(c)

1- 0.7876 1

798.075.0

3.5

k2

k = 4.7015 1


131

5(a)(i)P(x = 2) = 62

28 )4.0()6.0(C 1

0.04129 1

5(a)(ii)P(X > 2) = 62

287

1880

08 )4.0()6.0()4.0)(6.0()4.0()6.0(1 CCC 1

1 – 0.0006554 – 0.007864 – 0.04129 10.9502 1

5(b)(i) 10

128140

10

128110(

zP ) 1

1 - 0.0359 - 0.11510.849 1

5(b)(ii)

P( )10

128140 z 1

0.1151np = 16

n =1151.0

16

1139 1

6(a)P(

75

28525.191 z ) 1

1 - 0.1056 10.8944 1

6(b)(i)0.8944 x 520 1465 1

6(b)(ii)

P(X > m ) =520

4161

= 0.8 1

842.075

285

m2

m =221.85 1

CHAPTER 9 MOTION ALONG A STRAIGHT LINE FORM 5

132

PAPER 2

1. A particle moves in a straight line and passes through a fixed point O, with a velocity of

10 m s 1 . Its acceleration, a m s 2 , t seconds after passing through O is given by .48 ta The particle stops after k seconds.

(a) Find(i) the maximum velocity of the particle,(ii) the value of k. [6 marks]

(b) Sketch a velocity-time graph for kt 0 .Hence, or otherwise, calculate the total distance travelled during that period.

[4 marks]

2. A particle moves along a straight line from a fixed point R. Its velocity, V m s 1 , is given by2220 ttV , where t is the time, in seconds, after leaving the point R.

(Assume motion to the right is positive)

Find(a) the maximum velocity of the particle, [3 marks](b) the distance travelled during the 4th second, [3 marks](c) the value of t when the particle passes the points R again, [2 marks](d) the time between leaving R and when the particle reverses its direction of motion.

[2 marks]

3. The following diagram shows the positions and directions of motion of two objects, A and B,moving in a straight line passing two fixed points, P and Q, respectively. Object A passes thefixed point P and object B passes the fixed point Q simultaneously. The distance PQ is 90 m.

The velocity of A, AV m s 1 , is given 22810 ttVA , where t is the time, in seconds, after

it passes P while B travels with a constant velocity of - 3 m s 1 . Object A stops instantaneouslyat point M.

(Assume that the positive direction of motion is towards the right.)

Find

(a) the maximum velocity , in, m s 1 , of A, [3 marks](b) the distance, in m, of M from P, [4 marks](c) the distance, in m, between A and B when A is at the points M. [3 marks]

A B

P M Q

90 m


133

4. A particle moves in a straight line and passes through a fixed point O.

Its velocity, v ms 1 , is given by 342 ttv , where t is the time, in seconds, after leaving O .[Assume motion to the right is positive.]

(a) Find(i) the initial velocity of the particle,(ii) the time interval during which the particle moves towards the left,(iii) the time interval during which the acceleration of the particle is positive.

[5 marks](b) Sketch the velocity-time graph of the motion of the particle for 30 t .

[2 marks](c) Calculate the total distance travelled during the first 3 seconds after leaving O.

[3 marks]

5. A particle moves along a straight line and passes through a fixed point O. Its velocity, v ms 1 , is

given by 1582 ttv , where t is the time, in seconds, after passing through O .[Assume motion to the right is positive.]

Find

(a) the initial velocity, in ms 1 , [1 mark]

(b) the minimum velocity, in ms 1 , [3 marks](c) the range of values of t during which the particle moves to the left, [2 marks](d) the total distance, in m, travelled by the particle in the first 5 seconds. [4 marks]

6. A particle moves along a straight line and passes through a fixed point O. Its velocity, v ms 1 ,

is given by 228 ttv , where t is the time, in seconds, after passing through O . Theparticle stops instantaneously at point M.[Assume motion to the right is positive.]

Find

(a) the acceleration, in ms 2 , of the particle at M, [3 marks]

(b) the maximum velocity, in ms 1 , of the particle [3 marks](c) the total distance, in m, travelled by the particle in the first 10 seconds, after passing

through O. [4 marks]


134

7. A particle moves along a straight line and passes through a fixed point O, with velocity of

20 ms 1 . Its acceleration, a ms 2 , is given by a = – 2t + 8 where t is the time, in seconds, afterpassing through point O. The particle stops after k s.

(a) Find:(i) the maximum velocity of the particle,(ii) the value of k [6 marks]

(b) Sketch a velocity-time graph of the motion of the particle for kt 0 .Hence, or otherwise, calculate the total distance travelled during that period.

[4 marks]

8. A particle moves along a straight line. Its velocity, v ms- 1 , from a fixed point, O, is given byv = t2 – 10t + 24 where t is the time, in seconds, after passing through point O.[Assume motion to the right is positive.]

Find(a) the initial velocity of the particle, [1 marks](b) the minimum velocity of the particle, [3 marks](c) the range of values of t during which the particle moves to the left, [2 marks](d) the total distance travelled by the particle in the first 6 seconds. [4 marks]


135

ANSWERS PAPER 2

1. a = 8 – 4tV = ( 8-4t ) dt

= ct

t 2

48

2

= ctt 228 1V = 10, t = 0.

10 = c 2)0(2)0(8

10 = c

V = 1028 2 tt1

a) (i) maximum velocity, a= 08 – 4t = 08 = 4t

t = 2 s 1

imumVmax = 10)2(2)2(8 2

= 18 ms-1 1

(ii) particle stops: V = 0

1028 2 tt = 0 1

1082 2 tt = 0

1042 tt = 0(t + 1)(t – 5) = 0

t = 5 k = 5 1

b) V = 1028 2 tt

t 0 2 5V 10 18 0

2

Total Distance = 5

0

dtV

= dttt 5

0

2 1028

=5

0

32 103

24

ttt 1

V

10

18

0 2 5t


136

= 0)5(10)5(3

2)5(4 32

= m3

266 1

2. a) 220 ttV

tdt

dVa 420 1

Maximum velocity, a = 020 – 4t = 0

20 = 4tt = 5 1

Vmax = 20(5) – 2(52)= 50 ms- 1 1

b) dtvs

dttt 2220

ctt 32

3

210

s = 0, t = 0 c = 0

32

3

210 tts 1

mst 72)3(3

2)3(10;3 32

mst3

1117)4(

3

2)4(10;4 32 1

Distance travelled during the fourth second = 723

1117

m3

145 1

c) At point R again, s = 0

03

210 32 tt

0)3

210(2 tt 1

03

210 t

103

2t

15t1


137

d) Maximum displacement, V = 0

0220 2 tt 1

0)10(2 tt

10t Time = 10 seconds 1

3. a) 22810 ttVA

taA 48 1

AV maximum when 0Aa

8 – 4t = 08 = 4tt = 2 1

)2(2)2(810max 2AV

= 18 ms- 1 1

b) Object A at M when 0AV

02810 2 tt

01082 2 tt

0542 tt 1

051 tt

t = 5 1

Distance M from P 5

0

dtv

5

0

22810 dttt

5

0

32

3

2410

ttt 1

0)5(3

2)5(4)5(10 32

m3

266 1

c) When t = 5, distance B from Q = v t= 3 5= 15 m 1

Distance between A and B =

15

3

26690 1

=3

18 m # 1


138

4. a) i) 342 ttv

t = 0, 3)0(402 v

= 3 ms-1 #1

ii) particle moves to the left, v < 0

342 tt < 0 1

31 tt < 0

1 < t < 3 # 1

iii) 342 ttv

42 tdt

dva 1

a is positive, a > 02t – 4 > 0

2t > 0t > 2 # 1

b) 342 ttv

t 0 1 3v 3 0 0

2

c)Total distance =

3

1

1

0

dtvdtv

= 1

0

3

1

22 )34(34 dtttdttt

=

3

1

231

0

23

323

323

tt

ttt

t1

=

32

3

1918932

3

11

=3

11

3

11

= 23

2m # 1

5. a) 1582 ttv12 1515)0(80,0 msvt # 1

v

3

0 1 3t

1 3t


139

b) 1582 ttv

82 tdt

dva 1

When ,0dt

dv

2t – 8 = 02t = 8

t = 4 1

When t = 4, vmin = 42 – 8(4) + 15 = - 1 ms-1 # 1

c) Particle moves to the left, v < 0

01582 tt

053 tt 1

3 < t < 5 1d)

Total distance = 5

3

3

0

dtvdtv

= 3

0

5

3

22 )158(158 dtttdttt 1

=

5

3

233

0

23

1543

1543

tt

ttt

t1

=

4536975100

3

12545369 1

= 183

21618

= 193

1m # 1

6. a) v = 228 tt

When v = 0, 228 tt = 0

822 tt = 0(t + 2) (t – 4) = 0

t = 41

a =dt

dv= 2 – 2t 1

At M, a = 2 – 2(4)

= – 6 ms 2 # 1

3 5t


140

b) Maximum velocity, a = 02 – 2t = 0 1

2 = 2tt = 1 1

When t = 1 ; v imummax = 8 + 2(1) – 1

= 9 ms 1 # 1

c) Total distance,s = v dt

= 228 tt dt

= ct

tt 3

83

2

1

When t = 0, s = 0 c = 0

s =3

83

2 ttt 1

s 4t =3

44)4(8

32 =

3

226 m

s 10t =3

1010)10(8

32 =

3

1153 m 1

Total distance in the first 10 seconds

=3

1153)

3

226(2

=3

2206 m # 1

7. a) i) a = – 2t + 8V = – 2t + 8 dt

= ctt 82 1

When t = 0, V = 20 :

20 = ctt 82

20 = 0

V = 2082 tt 1

When maximum velocity, a = 0- 2t + 8= 0

2t = 8t = 4 1

V imummax = 20)4(8)4( 2 1

t = 0

03

1153 m

3

226 m

t = 4

t = 10


141

= 36 ms 1 #

(ii) Particle stops, V = 0

2082 tt = 0

2082 tt = 0(t + 2)(t – 10) = 0

t = 10k = 10 1

b) V = 2082 tt

t 0 4 10V 20 36 0

2

Distance = ( 2082 tt ) dt

= 100

23

]2043

[ ttt

1

= )10(20)10(43

10 23

– 0

=3

2266 m # 1

8. a) v = 24102 ttWhen t = 0, initial velocity,

v = 24)0(1002

= 24 ms 1 #1

b)a =

dt

dv

= 2t – 10 1

When minimum velocity, a = 02t -10 = 0

t = 5 1

v imummax = 24)5(1052

= – 1 ms 1 #1

V36

20

0 4 10 t


142

c) When moves to the left, v 0

24102 tt 0( t – 4 )( t – 6 ) 0

1

4 t 6 1

d) Distance, s = ( 24102 tt ) dt

= cttt 2453

1 23

1

When t = 0, s = 0 c = 0

s = ttt 2453

1 23 1

t = 0, s = 0

t = 4, s = )4(24)4(5)4(3

1 23 =3

137 m

t = 6, s = )6(24)6(5)6(3

1 23 = 36 m 1

Total distance travelled during the first 6 seconds

=3

137 +

3

137 – 36

=3

238 m # 1

4 6t

t = 0

363

137 m

t = 4

0

t = 6

CHAPTER 10 LINEAR PROGRAMMING FORM 5

143

PAPER 2

1. Amirah has an allocation of RM200 to buy x workbooks and y reference books. The totalnumber of books is not less than 20. The number of workbooks is at most twice the number ofthe references. The price of a workbook is RM10 and that of a reference is RM5.

a ) Write down three inequalities, other than x0 and y0, which satisfy all the aboveconstraints.

( 3 marks)b ) Hence, using a scale of 2cm to 5 books on the x-axis and 2cm to 5 books on the y-axis,

construct and shade the region R that satisfies all the above constraints. ( 4 marks )

c ) If Amirah buys 15 reference books, find the maximum amount of money that is left.( 3 marks )

2. A university wants to organise a course for x medical undergraduates and y dentistryundergraduates. The method in which the number of medical undergraduates and dentistryundergraduates are chosen are as follows.

I : The total number of participants is at least 30.II : The number of medical undergraduates is not more than three times the number of dentistry

undergraduates.III : The maximum allocation for the course is RM6 000 with RM100 for a medical

undergraduates and RM80 for a dentistry undergraduates.

a) Write down three inequalities, other than x0 and y0, which satisfy all the aboveconstraints.

( 3 marks )

b) Hence, by using a scale of 2cm to 10 participants on both axes, construct and shade the regionR that satisfies all the above constraints. ( 3 marks )

c) Using your graph from (b), find( i) The maximum and minimum number of dentistry undergraduates , if the number of

medical undergraduates that participate in the course is 20.(ii) The minimum expenditure to run the course in this case. ( 4 marks )

3. A tuition centre offers two different subjects, science, S, and mathematics, M, for Form 4students. The number of students for S is x and for M is y. The intake of the students is basedon the following constraints.

I : The total number of students is not more than 90.II : The number of students for subject S is at most twice the number of students for subjectM.III : The number of students for subject M must exceed the number of students for subject S

by at most 10


144

a) Write down three inequalities, other than x0 and y0, which satisfy all the aboveconditions.

( 3 marks)

b) Hence, by using a scale of 2 cm to represent 10 students on both axes, construct and shade theregion R that satisfies all the above conditions. ( 3 marks )

c) Using the graph from ( b ), find ( 4 marks )( i ) the range of the number of students for subject M if the number of students for subjects S

is 20(ii) the maximum total fees per month that can be collected if the fees per month for subject S

and M are RM12 and RM10 respectively.

4 A bakery shop produces two types of bread, L and M. The production of the bread involves twoprocesses, mixing the ingredients and baking the breads. Table 1 shows the time taken to makebread L and M respectively.

Type of bread Time taken ( minutes )Mixing the ingredients Baking the breads

L 30 40M 30 30

Table 1

The shop produces x breads of type L and y breads of type M per day. The production of breadsper day are based on the following constraints:

I : The maximum total time used for mixing ingredients for both breads is not more than 540minutes.

II : The total time for baking both breads is at least 480 minutes.III : The ratio of the number of breads for type L to the number of breads for type M is not less

than 1 : 2

a ) Write down three inequalities, other than x0 and y0, which satisfy all the aboveconstraints. ( 3 marks )

b) Using a scale of 2 cm to represent 2 breads on both axes, construct and shade the region Rthat satisfies all the above constraints. ( 3 marks )

c) By using your graph from 4(b), find

( i ) the maximum number of bread L if 10 breads of type M breads are produced per day.

(ii ) the minimum total profit per day if the profit from one bread of type L is RM2.00 and


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from one bread of type M is RM1.00 . ( 4 marks )

5. A factory produced x toys of model A and y toys of model B. The profit from the sales of anumber of model A is RM 15 per unit and a number of model B is RM 12 per unit.The production of the models per day is based on the following conditions:-

I : The total number of models produced is not more than 500.II : The number of model A produced is at most three times the number of model B.III : The minimum total profit for model A and model B is RM4200.

a) Write down three inequalities, other than x0 and y0, which satisfy all the aboveconditions.

( 3 marks)

b) Hence, by using a scale of 2 cm to represent 50 models on both axes, construct and shade theregion R that satisfies all the above conditions. ( 3 marks )

c) Based on ypur graph, find ( 4 marks )( i ) the minimum number of model B if the number of model A produced on a particular

day is 100.

(ii) the maximum total profit per day


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1. ( a ) I : x + y ≥ 20II : x ≤ 2yIII : 10 x + 5y ≤ 200

2x + y ≤ 40

( b )

( c ) Draw the line y = 15From the graph, the minimum value occurs at ( 5, 15 )Hence, minimum expenditure = 10 (5) + 5 (15)

= RM125Therefore, the maximum amount of money left = RM200 –RM125

=RM 75

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4

11

1

40

35

30

25

20

15

10

5

10 20 30 40 50 60

R y =15

x + y = 20

2x + y= 40

x =2y


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2.( a ) I : x + y ≥ 30

II : x ≤ 3yIII : 100 x + 80y ≤ 6000

5x + 4y ≤ 300

( b )

( c ) ( i ) Draw the line x = 20,From the graph, the minimum number of dentistry undergraduates is 10 and themaximum number of dentistry undergraduates is 50.

(ii) For the minumum expenditure, there are 20 medical undergraduates and 10 dentistryundergraduates.Thus , minimum expenditure = 100 (20) + 80(10)

=RM2800

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11

11

90

80

70

60

50

40

30

20

10

-10

-20 20 40 60 80 100 120 140

x =3y

R

5x + 4y= 300

x =20

x + y=30


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3. ( a ) I : x + y ≤ 90

II : x ≤ 2yIII : y - x ≤ 10

( b )90

80

70

60

50

40

30

20

10

-10

20 40 60 80 100 120 140

12x + 10y = 600

x + y = 90

x =2y

R

y - x = 10

( c ) ( i ) From the graph, when x = 20, the range of y is 10 ≤ y ≤ 30

( ii ) The maximum value occurs at ( 60, 30)

Thus, maximum total fee = 12 (60) +10 (30)= RM1 020

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1

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4( a ) I : 3x + 3y ≤ 54

II : 4x + 3y ≥ 48III : 2x ≥ y

( b )22

20

18

16

14

12

10

8

6

4

2

-2

5 10 15 20 25 30 35

R

2x + y = 2

4x + 3y = 4 8

3x + 3y =54

y =2x

( c ) (i ) 8(ii ) (5 x RM2.00) + ( 10 x RM1.00)

= RM 20.00

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121


150

5 (a ) I : x + y ≤ 500II : x ≤ 3yIII : 15x + 12y ≥ 4200

5x + 4y ≥ 1400

( b )

( c ) ( i ) When x = 100, the minimum number of model B is 225.

(ii ) Maximum point (375, 125)The maximum total profit per day= 15(375) +12(125)= RM7125

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