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CHAPTER 9
DIFFERENTIATION
FORM 4
PAPER 1 1.
Given that y 15 x(6 x ) , calculate (a) the value of x when y is maximum, (b) the maximum value of y.
[3 marks]
2.
Given that y 4 x x 2 , use differentiation to find the small change in y when x increases from 5 to 5.01. [3 marks]
3.
Differentiate 2 x 2 (3 x 5) 4 with respect to x.
4.
[3 marks]
5 Two variables, x and y are related by the equation y 4 x . Given that y increases at a x constant rate of 3 units per second, find the rate of change of x when x 5. [3 marks] 1 , evaluate g”(1). ( 2 x 4) 2
5.
Given that g ( x )
6.
The volume of water, V cm3, in a container is given by V
7.
The point R lies on the curve y ( x 3) 2 . It is given that the gradient of the normal at R is 1 [3 marks] . Find the coordinates of R. 6
8.
It is given that y
9.
Given that y 2 x 2 3 x 4, dy (a) find the value of when x 3, dx (b) express the approximate change in y, in terms of m, when x changes from 1 to 1 + m, where m is a small value. [4 marks]
10.
The curve y = f(x) is such that
11.
The curve y = x2 – 28x + 52 has a minimum point at x = k, where k is a constant. Find the value of k. [3 marks]
[4 marks]
2 3 h 6h , where h is the height in 3 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]
4 3 dy in terms of x. u , where u 5 x 2. Find 5 dx
[3 marks]
dy 5 px 2 , where p is a constant. The gradient of the curve dx at x = 2 is 10. Find the value of p. [2 marks]
49
CHAPTER 9
DIFFERENTIATION
FORM 4
ANSWERS PAPER 1 Marks 1.
a)
b)
2.
3.
y = 15x (6 – x) = 90x – 15x2 dy 90 30 x dx dy 0, 90 30 x 0 dx 30x = 90 x=3 # y = 15(3)(6 – 3) = 135 # y = 4x x2 dy = 4 + 2x dx dy x = 5, = 4 + 2(5) dx = 14 dy = x y dx = 14 0.01 = 0.14 # 2 Let y = 2 x (3x 5) 4 dy = 2 x 2 [4(3 x 5) 3 3] (3 x 5) 4 (4 x) dx = 24 x 2 (3 x 5) 3 4 x(3x 5) 4
4.
y
dy dx dy dt
1
1 1
1
1 1
1
=
4 x(3 x 5) 3 [6 x (3x 5)]
1
=
4 x(3 x 5) 3 (9 x 5) # 5 4 x = 4 x 5 x 1 x 5 4 2 x dy dx dx dt 5 dx (4 2 ) x dt 5 dx (4 2 ) 5 dt 1 dx 4 5 dt 5 unit per second # 7
1
= = =
3
=
3
=
3
=
dx dt
=
1
1
1 50
CHAPTER 9
DIFFERENTIATION
5.
g(x)
=
g (x) = = g(x) = = g(1) = = = 6.
V dV dt dV dt
=
7.
(2 x 4) 2 2(2 x 4) 3 (2) 4(2 x 4) 3 12(2 x 4) (2) 24(2 x 4) 4 24(2) 4 24 16 3 # 2 2 3 h 6h 3 2 h 2 6
=
dV dh dh dt
5
=
( 2 h 2 6 )
dh dt
=
y dy dx
8.
y
dy dx
9.
a)
y
1
dh dt
= =
2(x – 3)
2
Given gradient of normal =
1 6 6 6 (6 3) 2 9
= = = = R (6, 9) # 4 3 = u 5 4 = (5 x 2) 3 5 12 = (5 x 2) 2 5 5 = 12(5 x 2) 2 # =
1
1
5 2(3 ) 6 5 cms 1 @ 0.2083 cms 1 # 24 ( x 3) 2
2(x – 3) x y
1
4
=
=
FORM 4
1 1
1
1
1
1 1 1
2 x 2 3x 4 1 51
CHAPTER 9
DIFFERENTIATION dy dx
=
4x + 3
When x = 3, b)
x y x y x = 1, m
=
=
4(3) + 3
=
15 #
4(1) + 3
dy dx 5p(2) + 2 10p p
11.
dy dx
m dy dx
y 10.
FORM 4
= 7
1 1
7m # =
5px + 2
= =
10 8 4 # 5 x 2 28 x 52
=
1
y = dy = 2x – 28 dx Minimum point at x = k : 2(k) – 28 = 0 2k = 28 k = 14 #
1
1
1 1 1
52
CHAPTER 9
DIFFERENTIATION
FORM 4
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. 0.8 m
0.6 m
water
Calculate the rate of change of the height of the water level at the instant when the height of 1 the water level is 0.4 m. (Use 3.142; Volume of a cone = r 2 h ) [4 marks] 3
2.
4 which passes through A(2, 5). (2 x 3) 2 Find the equation of the tangent to the curve at the point A.
Diagram shows part of the curve y
[4 marks]
y
A(2, 5)
y
O
4 (2 x 3) 2
x
53
CHAPTER 9 3.
DIFFERENTIATION
FORM 4
In the diagram, the straight line PQ is a normal to the curve y
x2 3 at A(2, 5). 2
y
P A(2, 5)
O
Q(k, 0)
Find the value of k.
4.
x
[3 marks]
Diagram shows part of the curve y = k(x – 2)3, where k is a constant. The curve intersects the straight line x = 4 at point A. dy At point A, [3 marks] 36. Find the value of k. dx
y
y = k(x – 2)3 A
x
O x=4
54
CHAPTER 9
5.
DIFFERENTIATION
FORM 4
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 2
y=x +3
P(1, 5)
x O
55
CHAPTER 9
DIFFERENTIATION
FORM 4
ANSWERS PAPER 2 1.
r h
=
r
=
V
=
0.4 m
r 0.6 m
=
h
= dV dh dV dh 4 2 h 9
2.
y dy dx
dy dx
=
1 2 r h 3 1 2 2 ( h) h 3 3 4 h 3 27 4 2 h 9 dV dt dt dh dt 0.3 dh dt 0.3 dh
1.343 ms 1 #
1
1
1
1
8(2 x 3) 3 (2)
=
= = y–5 y
3.
=
4 (3.142)(0.4) 2 = 9 dh = dt = 4(2 x 3) 2
= At A (2, 5) ,
=
0. 4 0. 6 4 2 h h 6 3
16 (2 x 3) 3 16 [2(2) 3]3 –16 = – 16( x 2 ) = – 16x + 32 = –16x + 37 #
1
1 1 1
2
y = dy = dx
x 3 2
x
At point A(2 , 5),
dy =2 dx
1 56
CHAPTER 9
DIFFERENTIATION Gradient of normal, m2
4.
=
dy dx 3k (4 2) 2 3k k y dy dx dy At P (1,5), dx
When x = 4,
5.
1 2
05 1 k 2 2 – 10 = –k + 2 k = 12 # = k ( x 2) 3
y dy dx
y–5 y
FORM 4
3k ( x 2) 2
1 1
1
= 36 : = 36 = 9 = 3 # = x2 3 =
2x
=
2(1)
= = = =
2 2 (x – 1) 2x – 2 2x + 3 #
1 1
1 1 1
57
CHAPTER 10
SOLUTION OF TRIANGLES
FORM 4
PAPER 2 1.
Diagram 1 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 is different in shape to triangle ABC. Sketch the triangle A’B’C’ [ 2 marks ] 2. P R
12cm 65 º
10cm
Diagram 2 Q
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. 58
CHAPTER 10
SOLUTION OF TRIANGLES
FORM 4
Calculate the two possible values of PSR.
[2 marks ]
(c) Using the acute PSR from ( b ) , calculate i ) the length, in cm, of RS ii ) the area, in cm2, of quadrilateral PQRS.
[ 6 marks]
3. B
6cm
C
53º 7cm A 10cm
D
Diagram 3 Diagram 3 shows a quadrilateral ABCD. The area of ∆BCD is 20cm2 and BCD is acute. Calculate (a) BCD (b) the length in cm, of BD. (c) ADB (d) the area, in cm2 , of quadrilateral ABCD
[ 2 marks] [ 2 marks] [ 3 marks] [ 3 marks]
A
4.
5cm 5cm
B
80 48' 120
D
10cm
C
Diagram 4
Diagram 4 shows quadrilateral ABCD. (a) Calculate
( 4 marks ) 59
CHAPTER 10
SOLUTION OF TRIANGLES
FORM 4
(i ) the length, in cm. of AC, (ii) ACB (b) Point A’ lies on AC such that A’B = AB (i ) Sketch ∆A’BC (ii) Calculate the area, in cm2, of ∆A’BC
5.
( 6 marks )
S
R
P
Q
Diagram 1 Diagram 1 shows a pyramid PQRS with triangle PQR as the horizontal base. S is the vertex of the pyramid and the angle between the inclined plane QRS and the base PQR is 50º. Given that PQ and 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 (b) the length of SP, (c) the area of the triangle PQS.
[ 3 marks ] [ 3 marks ] [ 4 marks ]
60
CHAPTER 10
SOLUTION OF TRIANGLES
FORM 4
ANSWER(PAPER2) 1.
(a) (i)
1
8.5 10.8 = sin 42.5 sin ABC ABC =59.14 º
1
(ii) 10.82=6.52 +5.22- 2(6.5)(5.2) cos ADC
1
ADC = 134.46 º
1
(iii) Total area 1 1 = (10.8)(8.5)sin(180º- 42.5º -59.14 º) + (6.5)(5.2)sin134.46 º 2 2 = 57.02 cm2
3 1
(b) C’
42.5°
B’
10.8 cm
2
8.5 cm A’
61
CHAPTER 10
SOLUTION OF TRIANGLES
FORM 4
2. ( a ) PR2 =122 +102- 2(12)(10)cos65º = 142.6 PR = 11.94cm (b)
1 1
S
S' P
30
12 cm
R
10 cm 65
Q
sin PSR sin 30 = 11.94 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
2
1
RS 8.2 = sin103.28 sin 30
1
RS = 15.96cm
1
1 1 (10)(12)(sin65 º ) + (11.94)(15.96)(sin30 º) 2 2 =102.02cm2
( ii) Area =
2 1
62
CHAPTER 10
SOLUTION OF TRIANGLES
FORM 4
3. 1 (6)(7)sin BCD = 20 2 BCD =72 º15’
1
( b ) BD2 = 6 2 + 7 2 – 2(6)(7)cos 72 º15’ =59.39 BD = 7.706
1
(a)
(c )
1
1
sin 53 sin BAD = 10 7.706
1
BAD =37 º59’
1
ADB =180º- 37º59’-53 º
1
=89 º 1’
1 1 (7)(6)(sin72 º15’ ) + (7.706)(10)(sin89 º1’) 2 2 =58.52cm2
4.
(d) Area =
2 1
( a ) ( i ) AC2 = 5 2 + 10 2 – 2(5)(10)cos 80 º48’ AC = 10.44
1 1
( ii )
sin ACB sin120 = 5 10.44
1
ACB = 24 º30’
1
( b) ( i) 2
63
CHAPTER 10
SOLUTION OF TRIANGLES
FORM 4
( 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 (5)(10.4408)(sin35 º30’ ) 2
1
=15.1575 Area of ∆ABA’ =
1 (5)(5)(sin109 º ) 2
=11.8190 Area of ∆A’BC =15.1575 - 11.8190 =3.3385 cm2
5. ( a ) Area of
1
1 1
PQR = 12.4
1 (5.6) (5.6)sin QPR = 12.4 2
1
sin QPR = 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
64
CHAPTER 10
SOLUTION OF TRIANGLES
FORM 4
1 QR where M is the midpoint of RQ 2 1 = (4.933) 2 = 2.466
(b) QM =
SM2 = 4.22- 2.4662 =11.56 SM =3.4 cm PM2 = 5.62- 2.4662 PM = 5.028 SP2 = 3.42 + 5.0282 – 2(3.4)( 5.028) cos 50º SP =3.855 cm (c) cos
SQP
=
1 1 1
1
5.62 4.22 3.8552 2(5.6)(4.2) 1
=0.7257
SQP
1 = 43.47 º
1
Area of PQS 1 = (5.6) (4.2)sin 43.47 º 2 =8.091 cm2
65
CHAPTER 11
INDEX NUMBER
FORM 4
PAPER 2 1 Table 1 shows the price indices and percentage of usage of four items, P, Q, R and S, which are the main ingredients in the making of a type of cake. Items P Q R S
Price index for the year 2007 based on the year 2004 125 x 110 130
Percentage of usage 40 20 10 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 in making a particular kind of cake. Diagram 1 is a pie chart which represents the relative amount of the ingredients K, L, M and N, used in making the cake.
Ingredients K L M N
Price per Kg (RM) Year 2003 Year 2006 1.60 2.00 4.00 q 0.80 1.20 r 1.60
Price index for the year 2006 based on the year 2003 p 120 150 80
K
N 80 120
100 L
M
Table 2 Diagram 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 based on the year 2003. (ii) Hence, calculate the corresponding cost of making this cake in the year 2003 if the cost 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 the year 2010. Find the expected composite index for the year 2010 based on the year 2003 [2 marks] 66
CHAPTER 11
INDEX NUMBER
FORM 4
3 A particular type of muffin is made by using four ingredients, P, Q, R and S. Table 3 shows the prices of the ingredients.
Ingredients P Q R S
Price per kilogram (RM) Year 2005 Year 2008 4.00 x 2.50 3.00 y z 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. The price per kilogram of ingredient R in the year 2008 is RM2.00 more than its corresponding price 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 year 2005 is 126 Calculate (i) the price of the muffin in the year 2005 if its corresponding price in the year 2008 is RM 6.30 (ii) the value of r if the quantities of ingredients P, Q, R and S used are in the ratio of 6:3 :r :2 [5 marks]
4 Table 4 shows the prices and the price indices of five components, P, Q, R, S and T, used to produce a type of toy car. Diagram 2 shows a pie chart which represents the relative quantity of components used.
Price (RM) for the year Component 2005 2007 P Q R S T
m 4.00 2.40 6.00 8.00
4.40 5.60 3.00 5.40 12.00 Table 4
Price index for the year 2007 based on the year 2005 110 140 125 n 150
R 144 36 S 72
Q
90
T
P
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 the corresponding cost in the year 2009. [4 marks] 67
CHAPTER 11
INDEX NUMBER
FORM 4
5 Table 5 shows the prices and the price indices of four ingredients P, Q, R and S, to make a dish. Diagram 3 shows a pie chart which represents the relative quantity of the ingredients used.
Ingredients P Q R S
Price (RM) per kg for the year 2006 2008 2.25 2.70 4.50 6.75 y 1.35 2 2.10
Price index for the Year 2008 based on the year 2006 x 150 112.5 105
P Q 25%
15% S 20%
R 40%
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 the year 2006 (ii) the price of a bowl of this dish in the year 2009 if its corresponding price in the year 2006 is RM 25 [4 marks]
68
CHAPTER 11
INDEX NUMBER
FORM 4
ANSWERS (PAPER 2) 1 (a) (i)
p
2004
100 RM 44.85 130
1
= RM 34.5 (ii)
100 RM 34.50 or 120 RM 44.85 I 2007 / 2000 RM 28.75 100
P
2000
1
P
2000
RM 28.75
= 156 (b) (i)
(ii)
(ii)
1 1
20x + 10000 = 12500
1
P
2004
x = 125
1
100 RM 40 125
1
RM 2.00 100 125 RM 1.60 120 q RM 4.00 RM 4.80 100 100 r RM 1.60 80 (125 60) (120 100) (150 120) (80 80) I 2006 / 2003 360 43900 = 360 p
(b) (i)
1
(125 40) (20 x ) (110 10) (130 30) 125 100
= RM 32
2 (a)
1
P
2003
P I
2010
1 1 1 1 1
100 RM 40 121.9
1
2010 / 2003
1
= 121.9
= RM 32.81 (c)
1
140 RM 40 RM 56 100 RM 56.00 100 170.7 RM 32.81
1 1 1
69
CHAPTER 11 3 (a)
INDEX NUMBER
x
125 RM 4 100
= RM 5.00 (b)
y 100 y2 140
(c) (i)
1
P
2005
100 RM 6.30 126
(125 6) (120 3) (140 r ) (110 2) 126 11 r
m
100 RM 4.40 110
= RM4.00 RM 5.40 100 90 RM 6 (110 90) (140 36) (125 144) (90 18) (150 72) 360 45360 = 360
n = (b)
= 126
P
2007
126 RM 60 100
= RM 75.60
P
1
z = RM 7.00
r = 4
(c)
1
1
1330 + 140r = 126r + 1386
4 (a)
1
y = RM 5.00
= RM 5.00 (ii)
FORM 4
2009
125 RM 75.60 100
= RM 94.50
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
70
CHAPTER 11
5 (a)
INDEX NUMBER
RM 2.70 100 120 RM 2.25 100 y = RM 1.35 112.5
x
= RM 1.20
= 121.5 (c) (i)
I
2009 / 2006
I
P
2009
1 1
= 145.8 (ii)
1
1
120 2008 / 2006 100 120 = 121.5 100
1
1
(120 15) (150 25) (112.5 40) (105 20) 100 12150 = 100
(b)
FORM 4
145.8 RM 25 100
= RM 36.45
1 1 1 1
71
CHAPTER 1
FUNCTIONS
FORM 4
PAPER 1 1.
Diagram 1 shows the relation between set A and set B. 2
p
4 6 8
q r Set A
Set B
Diagram 1 State (a) the range of the relation, (b) the type of the relation. [2 marks] 2.
R a, b, c
S b, d , f , h, j
3.
Based on the above information, the relation between R and S is defined by the set of ordered pairs (a, b), (a, d ), (b, f ), (b, h) . State (a) the images of a (b) the object of b [2 marks] Diagram 2 shows the linear function g . x
g(x)
0 2 4 6
-2 0 k 4
Diagram 2 (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 g : x x
xk , x 0 where k is a constant. x xk x
3 1 2
Diagram 3 Find the value of k. 1
CHAPTER 1
FUNCTIONS
FORM 4 [2 marks]
5.
Given the function g : x x 1 , find the value of x such that g ( x) 2 . [2 marks]
6.
Diagram 5 shows the graph of the function f ( x) 2 x 6 for domain 0 x 4 . f(x) 6
0
t
4
x
Diagram 5 State (a) the value of t, (b) the range of f(x) corresponding to the given domain. [3 marks] 7.
Given the function f ( x) 2 x 1 and g ( x) 3 kx , find (a) f (2) (b) the value of k such that gf (2) 7 [3 marks]
8.
The following information is about the function g and the composite function g 2 . g : x a bx , where a and b are constant and b > 0
g 2 : x 9x 8
Find the value of a and b. [3 marks]
9.
Given the function f ( x)
1 , x 0 and the composite function fg ( x) 4 x . 2x
Find (a) g (x) (b) the value of x when gf ( x) 2 [4 marks] 10.
The function h is defined as h( x)
7 , x 3 . 3 x
Find (a) h 1 ( x) (b) h 1 (2) 2
CHAPTER 1
FUNCTIONS
FORM 4 [3 marks]
11. Diagram 9 shows the function f maps x to y and the function g maps y to z. x
f
y
g
z
5 4 1
Determine (a) f 1 (1) (b) gf (5)
[2 marks] 12. The following information refers to the function f and g. f : x 5x 1 g : x x3
Find f
1
g ( x) . [3 marks]
13.
Given the function g : x 3 x h and g 1 : x kx
1 , where h and k are constants. Find 2
the value of h and of k. [3marks] 14.
Given the function h( x) 3 x 1 and g ( x)
x . Find 3
(a) h 1 (7) (b) gh 1 ( x) [4 marks] 15. Given the function f : x 3 x 2 and g : x 2 x 2 3 . Find (a) f 1 (4) (b) gf (x) [4 marks]
3
CHAPTER 1
FUNCTIONS
FORM 4
ANSWER (PAPER 1) 1 (a) (b) 2 (a) (b) 3 (a) (b) 4
4, 8
1
many-to-one
1
b , d
1
a
1
k2
1
g ( x) x 2
1
1 k 1 g 2 3 1 2 2
1
k 1
1
x 1 2
5
or ( x 1) 2
x 1
6 (a)
When f ( x) 0 ,
1
x 3
1
2x 6 0
1
x3
t 3
(b)
Range :
7 (a)
(a)
(b)
(b)
0 f ( x) 6
f ( 2) 5
1 1
g (5) 7 3 k (5) 7
1
k2
8
1
1
g 2 ( x) a b(a bx)
1
a ab b 2 x b2 9
and
a ab 8
b3
9 (a)
1 4x 2 g ( x) 1 g ( x) , x0 8x
a 4
1 1 1 1 4
CHAPTER 1 (b)
FUNCTIONS
FORM 4
1 2 1 8 x 2 1 x 8 7 x 3 y 7 h 1 ( x) 3 , x 0 x 1 h 1 (2) 2
1
11 (a)
5
1
(b)
4
1
10 (a)
(b)
12
x
y 1 5
( x 3) 1 5 x4 5 yh x 3 1 k 3 3 h 2 y 1 x 3 7 1 h 1 (7) 2 3 x 1 gh 1 ( x) 3 3 x 1 9 y2 x 3 1 f (4) 2 f 1 g ( x)
13.
14. (a)
(b)
15 (a)
(b)
1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1
gf ( x) 2(3 x 2) 2 3
1
18 x 2 24 x 5
1
5
CHAPTER 2
QUADRATIC EQUATIONS
FORM 4
PAPER 1 1. The quadratic equation 2x 2 px 2 4x 2 , where p is a constant, has no real roots. Find the range of the values of p. [3 marks] 2. The quadratic equation m x 2 4 x m 5 ,where m 0, has real and different roots. Find the range 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 four significant figures [3 marks] 5. The straight line y + x = 4 intersects the curve y = x 2 7 x w at two points . Find the range of values of w. [4 marks] 6. Form the quadratic equation which has the roots 7 and
2 . Give your answer in the form 3
ax 2 bx c 0 , where a , b and c are constants.
[2 marks] 7. Given the roots of the quadratic equation 4kx 2 hx 8 0 are equal. Express k in terms of h. [2 marks] 8. The straight line y = 9 4px is a tangent to the curve y = p 1 x 2 . Find the possible values of p. [5 marks] 9. The straight line y =2x1 does not intersect the curve y = x 2 2 x p. Find the range of values of p. [5 marks] 10. The quadratic equation 3 x 2 kx h 0 has roots 4 and 3. Find the values of k and h. [3 marks]
6
CHAPTER 2
QUADRATIC EQUATIONS
FORM 4
ANSWERS (PAPER 1) 1
2
3
4
5
(2 p ) x 4 x 8 0
1
4
1
2
2
4 2 p 8 0
48 32 p 0 3 p < 2 2 mx 4 x m 5 0 4 2 45 m m > 0 m 1m 4 > 0 m<1 , m > 4
1 1 1 1
n + 8x -x 2 = 3 x2 8x n 3 0 82 43 n 1 = 0 n = 13
1 1 1
3x 2 18x + 2 = 0
1
(18) (18)2 4(3)(2) 2(3)
1
x = 0.1132,
5.887
1
4 – x = x2 7x w
1 1 1 1
x 8x w 4 0 2
64 – 4(w – 4 ) > 0 w 20
x 7 3x 2 0
6
3x 2 19 x 14 0
7
k 8
1
1
1 1
h2 4 4k 8 0
1
h2 128
1
9 4 px p 1 x 2
1
p 1 x 2 4 px 9 0 2 4 p 4 9 p 1 0
1 1 7
CHAPTER 2
QUADRATIC EQUATIONS
p 3 4 p 3 0
10
1
3 4 2 2x – 1 = x -2x + p x2 4x p 1 0 16 – 4(1)(p + 1) < 0 12 – 4p <0 p > 3
1
(x + 4)(x – 3) = 0 3 x 2 3 x 36 0 k = -3 and h = -36
1 1 1
p = 3 and p = 9
FORM 4
1 1 1 1 1
8
CHAPTER 3
QUADRATIC FUNCTIONS
FORM 4
Paper 1 1. The quadratic function f(x) = a(x+p)2 + q, where a, p and q are constants, has a maximum value of 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 form f(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). y
x
5
O
1 y = 9
(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 f ( x) 5 2( x p ) 2 , where p is a constant. y A(1, q)
. 0
x
9
CHAPTER 3
QUADRATIC FUNCTIONS
FORM 4
The curve y f (x) 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.
1 2 x kx 6 . 2 A is the point of intersection of the quadratic graph and y-axis. The x-intercepts are 6 and 2.
Diagram below shows the curve of a quadratic function
f ( x)
y
x
6
O
2
A(0, r)
(p, q)
(a) State the value of r and of p. (b) The function can be expressed in the form f ( x ) of k. (c) Determine the range of values of x if f(x)< 6.
[2 marks]
1 ( x p ) 2 q , find the value of q and 2 [4 marks] [2 marks]
10
CHAPTER 3
QUADRATIC FUNCTIONS
FORM 4
Answers ( Paper 1) Q 1 (a) (b) (c) 2
3
4
5 (a) (b) 6 (a) (b) (c) 7
8
Solution a<0 p=3 q=5 x2 5x + 4 < 0 (x 1)(x 4) < 0 1<x<4 2x2 + 5x. 3 0 (2x 1)(x + 3) 0 3 x ½ f(x) = x2 6x + 32 32 + 5 = (x 3)2 4 m = 3 n = 4 x = 2 g(x) = (x + 2)2 9 p=1 q =5 x=1 x2 2x 15 0 (x + 3)(x 5) 0 x3, x5 2 x + (1 p)x + 4 =0 (1 p)2 4(1)(4) < 0 p2 2p 15 < 0 (p+3)(p 5) <0 3 < p < 5
Marks 1 1 1 1 1 1 1 1 1 1 1,1 1 1,1 1 1 1 1 1 1 1 1 1 Answer(Paper 2)
1 (a) (b)
r = 6, p = 2 (Expand or completing the square) 1 2 1 1 x px p 2 q x 2 kx 6 2 2 2 or k=p 1 2 2
1 1 1
p q 6
1 q 6 ( 2) 2 2
1 1 1
q = 8 k =2 (c) x2 + 4x<0 4
x(x + 4)<0 4 < x < 0
0
x
1
1
11
CHAPTER 4
SIMULTANEOUS EQUATIONS
FORM 4
PAPER 2 2 1. Solve the simultaneous equations y 2 x 8 and x - 3x – y = 2
[5 marks]
2.
Solve the simultaneous equations j – k = 2 and j2 + 2k = 8. Give your answers correct to three decimal 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. x 2y 8 2 x 4 y 2 37 Give your answers correct to three decimal places. [5 marks]
12
CHAPTER 4
SIMULTANEOUS EQUATIONS
FORM 4
ANSWERS (PAPER 2) 1.
2.
y – 2x = -8 y = - 8 + 2x ___________(1) x2 - 3x – y = 2___________(2) x2 - 3x – (-8 + 2x) = 2 (x – 3)(x – 2) = 0 x = 2, 3 y = - 8 + 2(2) y = - 8 + 2(3) y = -4, y=-2 k=j–2 _____________(1) 2 j + 2k = 8 _____________(2) j2 + 2(j – 2) = 8 j
4.
5.
1 1 1 1 1 1
b (b 2 4ac) 2a
2 2 2 4(1)(12) 2(1) j = 2.606 j = - 4.606 k = (2.606) – 2 k = (- 4.606) - 2 = 0.606 = - 6.606
1
x = 1 – 2y __________(1) y2 - 10 = 2x __________(2) y2 - 10 = 2(1 – 2y) (y + 6)(y – 2) = 0 y = 2, - 6 x = 1 – 2(2), x = 1 – 2(- 6) = -3, = 13
1
y = 1 – 2x __________(1) x2 + y2 + xy = 7 __________(2) x2 + (1 – 2x)2 + x(1 - 2x) = 7 ( x 1)( x 2) 0 x = - 1, 2 y= 1 – 2(- 1), y = 1 – 2(2) = 3, =-3
1
x = 8 + 2y __________(1) 2 2 x + 4y = 37 __________(2) (8 + 2y)2 + 4y2 = 37
1
3.
1
y
1
1 1 1 1
1 1 1 1
1
b b 2 4ac 2a
32 322 4(8)(27) 2(8) y ≈ - 1.209, -2.791 x = 8 + 2(- 1.209), x = 8 + 2(-2.791) = 5.582, = 2.418
1
1 1 1 13
CHAPTER 5
INDICES AND LOGARITHM
FORM 4
PAPER 1
1.
Solve the equation 3(53 x1 ) 36
[3marks]
2.
Solve the equation 323 x 47 x 4
[3marks]
3.
Solve the equation 32 x 1 5 x
[4marks]
4.
Solve the equation 165 x 49 x 6
[3marks]
5.
Solve the equation 273 x 3
6.
Solve the equation 22 x 1 13(2 x ) 24 0
[4marks]
7.
Given that log 3 K log 9 L 2 , express K in terms of L.
[4marks]
8.
9.
1
[3marks]
9 x 3
49 p Given that log p 3 r and log p 7 s , express log p in terms 27 of r and s. Given that log 5 2 q and log 5 9 p , express log 5 8.1 in terms of q and p.
[4marks]
[4marks)
10.
Solve the equation
3 log 3 (2 x 1) log 3 x .
[3marks]
11.
56 Given that log x 2 p and log x 7 q , express log x 2 in terms x of p and q.
[4marks]
12.
Given that log 3 mn 3 2 log 3 m log3 n express m in terms of n.
[4marks]
13.
Given that log 9 y log 3 18 , find the value of y
[3marks]
14.
81m Given that log 3 m v and log 3 n w , express log 9 in n terms of v and w.
[4marks]
15.
Solve the equation
log 3 x log 3 (4 x 1) 2 .
16.
Solve the equation
5 x 1 5x
4 . 25
[3marks] [3marks]
14
CHAPTER 5
INDICES AND LOGARITHM
FORM 4
ANSWERS (PAPER 1)
1.
53 x1 12
log 53 x1 log12
1
(3 x 1) log 5 log12
1
3x 1
log12 log 5
3x 1
1.0792 0.6990
3 x 1 1.54392 3 x 0.54392 x 0.1813
2.
2
5 3x
22
1 7 x 4
215 x 214 x8
1
15 x 14 x 8
3.
1
x 8
1
log 32 x 1 log 5x
1
2 x 1 log 3
1
x log 5
2 x log 3 log 3 x log 5 2 x log 3 x log 5 log 3
x 2 log 3 log 5 log 3 x
log 3 2 log 3 log 5
x 1.8691
1 1
15
CHAPTER 5
4.
INDICES AND LOGARITHM
2
4 5x
22
9 x 6
FORM 4 1
2 20 x 218 x12 20 x 18 x 12
1
x 6
5.
1
273 x 3 9 x 3
3
3 3 x 3
1 2
1 2 2 x 3
3
39 x 9 32 x 6
1
1 2
9x 9 x 3
1
10 x 6
x
6.
3 5
1
22 x 1 13(2 x ) 24 0 22 x.21 13(2 x ) 24 0
2
x 2
.2 13(2 x ) 24 0
1
Let u = 2x 2u 2 13u 24 0
2u 3 u 8 0 u
3 or 8 2
1
But 2x must be positive, so 2x = 8 2 x 23 x =3
1
16
CHAPTER 5
7.
INDICES AND LOGARITHM
log 3 K
log 3 L 2 log 3 9
log 3 K
log 3 L 2 2
FORM 4 1
2 log 3 K log3 L 4 log 3
8.
9.
10.
K2 4 L
1
K2 34 L
1
K 9 L
1
log p 7 2 log p p log p 33
1,1
2 log p 7 log p p 3log p 3
1
2 s 1 3r
1
log 5 8.1 =
log 5
81 10
1
=
log 5 81 log 5 10
1
=
log 5 92 log5 2 5
1
=
2 log 5 9 log 5 2 log 5 5
=
2 p q 1
1
3log 3 3 log3 (2 x 1) log 3 x log 3 33 log 3 (2 x 1) log 3 x
log 3 27 2 x 1 log 3 x
1
27 2 x 1 x
1
54 x x 27
x
27 53
1
17
CHAPTER 5
11.
12.
INDICES AND LOGARITHM
56 log x 2 = x
log x 56 log x x 2
1
=
log x (7 8) 2 log x x
=
log x 7 log x 8 2 log x x
=
log x 7 log x 23 2 log x x
=
log x 7 3log x 2 2 log x x
1
=
q 3p 2
1
1
log 3 mn log 3 33 log 3 m 2 log 3 n
1
27 m 2 n
1
log 3 mn log 3
27m 2 n 2 n m 27 log 3 y log 3 18 log 3 9 log 3 y log 3 18 log 3 9 mn
13.
FORM 4
1 1 1
log 3 y log 3 18 2
log 3 y 2 log 3 18
14.
log 3 y log 3 182
1
y 324
1
81m log 3 81m n = log 9 log3 9 n log 3 81 log 3 m log 3 n = 2 log 3 3 1 log3 34 log3 m log3 n 2 1 = 4 v w 2
=
1
1 1 1 18
CHAPTER 5
15.
INDICES AND LOGARITHM
FORM 4
log 3 x log3 (4 x 1) 2 0 log 3 x log3 (4 x 1) 2 log3 3 0 log 3
x(9) 0 4x 1
9x 30 4x 1
1 1
9x 4x 1 5x 1
x
16.
1 5
4 52 4 1 5 x 1 2 5 5 4 4 5x 2 5 5 4 5 5x 2 5 4
1
5 x.51 5x
1
5 x 5 1
1
x 1
1
19
CHAPTER 6
COORDINATE GEOMETRY
FORM 4
PAPER 1 1.
A point T divides the line segment joining the points A(1, -2) and B(-5, 4) internally in the ratio 2 : 1. Find the coordinates of point T. [2 marks]
2.
Diagram below shows a straight line PQ with the equation
x y + = 1. The point Q lies 3 5
on the x-axis and the point P lies on the y-axis. y P
0
Q
x
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 at the point B. y
B
A(0,6)
x
0 C
The equation of the straight line CB is y = 3x 4.
Find the coordinates of B. [3 marks]
5.
x y + = 1 has a y-intercept of 3 and is parallel to the straight line 14 m y + nx = 0. Determine the value of m and of n.
The straight line
20
CHAPTER 6
COORDINATE GEOMETRY
FORM 4 [3 marks]
6.
Diagram below shows a straight line passing through A(2, 0) and B (0, 6). y B(0, 6)
0
A(2, 0)
x
x y + = 1. a b [1 mark]
a)
Write down the equation of the straight line AB in the form
b)
A point P(x, y) moves such that PA = PB. Find the equation of the locus of P. [2 marks]
21
CHAPTER 6
COORDINATE GEOMETRY
FORM 4
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 C lies on the y-axis. y C
0
x 0
B (12, 0) D
A(0 , -3)
2.
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]
Solutions to this question by scale drawing will not be accepted.
In the diagram the straight line BC has an equation of 3y + x + 6 = 0 and is perpendicular to straight line AB at point B. y A(-6, 5)
B x
0 3y + x + 6 = 0 C (a) Find i) the equation of the straight line AB ii) the coordinates of B.
[5 marks]
(b) The straight line AB is extended to a point D such that AB : BD = 2 : 3. Find the coordinates 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] 22
CHAPTER 6
COORDINATE GEOMETRY
FORM 4
3. Solutions to this question by scale drawing will not be accepted. Diagram shows the triangle AOB where O is the origin. Point C lies on the straight line AB. y A(-2, 5)
C x 0 B(5, -1)
(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 point B. (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 the x-axis at point P and the y-axis at point Q. y x P
0 R
Q
y + 3x + 9 = 0
Point R lies on PQ such that PR : RQ = 1 : 2. Find (a) the coordinates of R, (b)
[3 marks]
the equation of the straight line that passes through R and perpendicular to PQ. [3 marks] 23
CHAPTER 6
5.
COORDINATE GEOMETRY
FORM 4
Solutions to this question by scale drawing will not be accepted. Diagram shows the triangle OPQ. Point S lies on the line PQ. P(3 , k)
y
S (5, 1) x
0 Q a)
A point W moves such that its distance from point S is always 2 Find the equation of the locus of W.
b)
1 units. 2 [3 marks]
It is given that point P and point Q lie on the locus of W. Calculate i) the value of k, ii) the coordinates of Q. [5 marks]
c)
2
Hence, find the area , in unit , of triangle OPQ. [2 marks]
24
CHAPTER 6
COORDINATE GEOMETRY
FORM 4
ANSWERS ( PAPER 1 ) 1.
2.
(1)(1) (5)(2) (2)(1) (4)(2) , ) 3 3 = T( -3 , 2 )
T(
Gradient of PQ , m1 = -
5 and the coordinates of Q (3 , 0) 3
Let the gradient of straight line perpendicular to PQ and passing through Q = m2 . Then m1 m2 = -1. 3 m2 = 5 y0 3 The equation of straight line is = x3 5 5y = 3(x – 3) 5y = 3x – 9 3.
4.
8x + 4hy – 6 = 0 3x + y = 16 4hy = -8x + 6 y = -3x + 16 8 6 y = x+ 4h 4h 2 3 y = - x + h 2h 2 Gradient , m2 = -3 Gradient , m1 = h Since the straight lines are perpendicular to each other , then m1 m2 = -1. 2 (- )(-3) = -1 h 6 = -h h = -6
2 1
1
1
1
Given
Gradient of CB , m1 = 3 Since AB is perpendicular to CB, then m1 m2 = 1 1 Gradient of AB, m2 = 3 1 The equation of AB is y=- x+6 3 B is the point of intersection. y = 3x 4 ……………(1) 1 y = x + 6 ……………(2) 3 1 3x 4 = x + 6 3
1
1
1
1
1
25
CHAPTER 6
COORDINATE GEOMETRY
FORM 4
10 x = 10 3 x=3
y = 3(3) 4 = 5 The coordinates of B are (3, 5). 5.
x y + 14 m
1
= 1
y-intercept = m = 3 1 x y 3 From + = 1, the gradient m1 = 14 3 14 From y = -nx , the gradient m2 = -n . Since the two straight lines are parallel , then m1 = m2 3 = -n 14 3 n = 14
6.
a) From the graph given, x- intercept = 2 and y-intercept = 6. x y The equation of AB is + =1. 2 6
1 1
1
b) Let the coordinates of P = (x , y) and since PA = PB ( x 2) 2 ( y 0) 2 = (x – 2)2 + y2 = x2 – 4x + 4 + y2 =
12y – 4x -32 = 0 3y – x - 8 = 0
( x 0) 2 ( y 6) 2 x2 + (y – 6)2 x2 + y2 – 12y + 36
1
1
26
CHAPTER 6
COORDINATE GEOMETRY
FORM 4
ANSWERS ( PAPER 2 ) 1
a) b)
c)
1
x y =1 12 3 AD 1 = DB 2 0(2) 12(1) 3(2) 0(1) D=( , ) 3 3 = ( 4 , -2 ) 3 Gradient of AB, mAB = -( ) 12 1 = 4
Given 2AD = DB , so
1 1
1
Since AB is perpendicular to CD, then mAB mCD = 1. Gradient of CD, mCD = - 4 Let, coordinates of C = (0 , h) , h ( 2) mCD = 04 h2 -4 = 4 16 = h + 2 h = 14 y-intercept of CD = 14 2
a) i)
1
1
Given equation of BC, 3y + x + 6 = 0 y =Gradient of BC = -
1 x–2 3
1 3
1
Since AB is perpendicular to BC , then mAB mBC = 1. Gradient of AB, mAB = 3 y 5 The equation of AB , =3 x ( 6) y – 5 = 3x + 18 y = 3x + 23 ii)
B is the point of intersection. Equation of AB , y = 3x + 23 Equation of BC , 3y + x + 6 = 0
1 1
…………. (1) ………….(2) 1
Substitute (1) into (2), 3(3x + 23) + x + 6 = 0
27
CHAPTER 6
COORDINATE GEOMETRY
FORM 4
9x + 69 + x + 6 = 0 x =Substitute value of x into (1),
y = 3(-
y =
15 2
15 ) + 23 2
1 2
The coordinates of B are ( -
15 1 , ) 2 2
1
b) Let D (h, k)
15 1 2h (18) 2k 15 , )= ( , ) 5 2 2 5 15 1 2k 15 2h (18) = , = 5 2 2 5 -75 = 4h – 36 5 = 4k + 30 39 25 h= k= 4 4 39 25 The coordinates of D are ( , ) 4 4 c) Given PA = 5
B( -
1
= 25
1
x2 + 12x + 36 + y2 -10y + 25 = 25 x2 + y2 + 12x -10y + 36 = 0
1
( x + 6)2 + ( y – 5)2
)
a)
b)
1
= 5
( x (6)) 2 ( y 5) 2
3.
1
Area
C= (
=
1 2
0 5 2 0 0 1 5 0 1 = (25) (2) 2 23 = unit2 2
3(5) 2(2) 3(1) 2(5) , 5 5 11 7 = ( , ) 5 5
1
1
1 1
c) i) Since PA = 2PB ( x 2) 2 ( y 5) 2 = 2 ( x 5) 2 ( y 1) 2
x2 + 4x + 4 + y2 10y + 25 = 4 (x2 10x + 25 + y2 +2y + 1)
1 1
28
CHAPTER 6
COORDINATE GEOMETRY
x2 + y2 + 4x 10y + 29 = 4x2 + 4y2 40x + 8y + 104 3x2 + 3y2 44x + 18y + 75 = 0 (ii)
When it intersects the y-axis, x = 0. 3y2 +1 8y + 75 = 0 Use b2 4ac = (18)2 4(3)(75) = 576 b2 4ac < 0
a)
y + 3x + 9 = 0 When y = 0,
0 + 3x + 9 x
1
1
= 0 = –3
P(–3, 0) When x = 0,
1 1
It does not cut the y-axis since there is no real root. 4.
FORM 4
1 y+0+9 y
= 0 = –9
Q(0, –9) 1(0) 2(3) 1(9) 2(0) , ) 3 3 = (-2 , -3 )
R(x, y) = (
b)
y + 3x + 9 = 0 y = -3x - 9
Gradient of PQ , m1 = –3 Since PQ is perpendicular to the straight line, then m1 m2 = 1 1 Thus, m2 3 The equation of straight line that passes through R(-2, -3) and perpendicular to PQ is y3 1 = x2 3 3y = x - 7 5.
a) Equation of the locus of W, 5 ( x 5) 2 ( y 1) 2 = 2 5 (x – 5)2 + ( y – 1)2 = ( )2 2
1 1
1
1
1 1
1
25 4 4 x2 + 4y2 – 40x - 8y + 79 = 0
1
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
x2 -10x +25 + y2 – 2y + 1 =
b) i)
29
CHAPTER 6
ii)
c)
COORDINATE GEOMETRY
4k2 - 8k -5 = 0 (2k + 1)(2k – 5) = 0 1 5 k=- , k= 2 2 5 Since k > 0, k = 2 Since S is the centre of the locus of W, then S is the mid-point of PQ. 5 y x3 2 ) S(5 , 1) = ( , 2 2 5 y x3 2 5= , 1= 2 2 1 x=7 , y =2 1 Hence, the coordinates of Q are ( 7 , ). 2 0 7 3 0 1 Area of triangle OPQ = 1 5 0 2 0 2 2 1 5 3 = [ (7)( ) – (- ) ] 2 2 2 19 = unit2 2
FORM 4
1
1
1
1
1
1
30
CHAPTER 7
STATISTICS
FORM 4
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) Number of fish
20 - 24 5
25 - 29 3
30 - 34 9 Table 3
35 - 39 7
40 -44 6
Find the mean length of the fish.
[3 marks]
4. Table 4 shows the distribution of marks obtained by 56 students in a test. Marks Number of Students
20 - 29 4
30 - 39 20
40 - 49 16
50 - 59 10
60 – 69 6
Table 4 Find, (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 ascending order 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 the squares 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. 31
CHAPTER 7
STATISTICS
FORM 4 [3 marks]
8. The mean of eight numbers is p. The sum of the squares of the numbers is 200 and the standard 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 the squares 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 60 students in a test. 20
Num ber of students
18 16 14 12 10 8 6 4 2 0
0.5 1 10.5
20.5
30.5
40.5
50.5
60.5
Marks
Diagram 2 (a) Without using an ogive, calculate the median mark. [3 marks] 32
CHAPTER 7
STATISTICS
FORM 4
(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 20 - 29 30 – 39 40 - 49 50 - 59 60 - 69 70 - 79
Number of students 1 2 8 12 k 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 the vertical 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 a Mathematics Quiz. Score Number of students
< 20 6
<30 16
< 40 28
< 50 36
<60 40
40 – 49
50 - 59
Table 4 (a) Based on Table 4, copy and complete Table 4. Score Number of students
10 - 19
20 - 29
30 - 39
(b) Without drawing an ogive, find the interquartile range of the distribution. [5 marks]
33
CHAPTER 7
STATISTICS
FORM 4
ANSWER PAPER 1 No. 1(a)
Solution
Marks
∑x = 88
(b)
1
x + k 9
1 = 10
k=2
1
∑x = 57
1
2. (a)
(b)
1
x + p
=9
7
p=6
1 (22 x 5) + ( 27 x 3) + (37 x 7) + (42 x 6 )
3.
x =
1,1
30
990 =
4.(a) (b)
30
= 33 = 34.5
1 1 (24.5 x 4 ) + ( 34.5 x 20 ) + ( 44.5 x 16 ) + ( 54.5 x 10 ) + ( 64.5 x 6)
x =
1,1
56
2432 =
56
= 43.43
1
34
CHAPTER 7
STATISTICS
4n + 14
5.(a)
1
= 13
2
FORM 4
n* = 3
1
3 + 19 + 2n* + 1 + 4n* + 14 + 17
(b)
=q
6
q = 12 6.(a)
1 1
12
1
(b) 960 =
6
1
- (12) 2
=4
1 22 + 52 + q2
7.
8.
14 =
3
-
2 +5+ q 3
2
1
(q + 4 )(q – 11) = 0
1
q = 11
1
x2 = 200 or x = 8
1 1
200 p 2 16m 2 8 p=
25 - 16m2
1
9. x = 14.6
1 1219
Standard deviation =
= 5.5353
5
- (14.6) 2
1
1 35
CHAPTER 7
STATISTICS
10.(a)
2200 Variance =
8
= 50 (b)
= 13.8
2
1
1
120 + 6 + 12 x =
- (15)
FORM 4
1
10
1
36
CHAPTER 7
STATISTICS
FORM 4
PAPER 2 No. 1(a)
Solution
Marks
Mean = 16
1 3212
Variance =
11
1
- (16) 2
= 36 (b)(i)
1
176 + k 18 =
1
12
k = 40
1
(ii)
3212 + 402
Standard deviation =
2(a).
12
1
- (18)2
= 8.775 30.5 or ½ (60) or 30 or median class 20.5 – 30.5 1 (60) 20 10 20.5 + 2 20 = 25.5
1 1
1 1
(b) 1
x = 26.67 fx 2 = 54285
1 54285
Standard deviation =
3.(a)
60
1600
-
2
1
60
= 13.92
1
49.5 or ½ (24 + k) or 11
1
37
CHAPTER 7
STATISTICS
52 = 49.5 +
1 2
(24 + k) - 11 12
FORM 4
1
10
k=4
1
(b) Frequency
4
14 12
10
8 6 4
2
0 19.5
29.5
39.5
49.5
59.5
69.5
Mode score = 53
(c) 4.(a)
79.5 Score
= 57
1 Score
10 - 19
20 - 29
Number of 6 10 students (b) 19.5 or ¼ (40) or 6 39.5 or ¾ (40) or 28 Interquartile range = 42* - 23.5* = 18.5
30 – 39
40 - 49
50 - 59
12
8
4
1 1 1 1 1
38
CHAPTER 8
CIRCULAR MEASURE
FORM 4
PAPER 1 1.
Diagram 1 shows a sector AOB with centre O . A
B
O DIAGRAM 1
The length of the arc AB is 7.5 cm and the perimeter of the sector AOB is 25 cm. Find the value of , in radian. [ 3 marks ]
2.
Diagram 2 shows a circle with centre O . R
O
0.35 rad S
DIAGRAM 2 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 ]
39
CHAPTER 8
3.
CIRCULAR MEASURE
FORM 4
Diagram 3 shows a circle with centre O . P
O
R
DIAGRAM 3 The length of the minor arc is 15 cm and the angle of the major sector POR is 280o . Using = 3.142 , find (a) (b)
4.
the value of , in radians. ( Give your answer correct to four significant figures ) the length, in cm, of the radius of the circle .
[ 3 marks ]
Diagram 4 shows sector OPQ with centre O and sector PXY with centre P . P
Y X
Q
O
DIAGRAM 4 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 ]
40
CHAPTER 8
5.
CIRCULAR MEASURE
FORM 4
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]
Q
P
R
1.97 rad
S
0
DIAGRAM
6.
5
Diagram 6 shows a circle with centre O and radius 12 cm. Given that A, B and C are points such 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]
O A
C
B
DIAGRAM 6
41
CHAPTER 8
CIRCULAR MEASURE
FORM 4
PAPER 2 1.
Diagram 1 shows the sectors AOB, centre O with radius 15 cm. The point C on OA is such that OC : OA = 3 : 5 . A C B
O DIAGRAM 1 Calculate
2.
(a)
the value of , in radian,
[ 3 marks ]
(b)
the area of the shaded region, in cm2 .
[ 4 marks ]
Diagram 2 shows a circle PQRT , centre O and radius 10 cm. AQB is a tangent to the circle at Q. The straight lines, AO and BO, intersect the circle at P and R respectively. OPQR is a rhombus. ACB is an arc of a circle, centre O. C
A
B
Q P
rad
R
O T DIAGRAM 2
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 ] 42
CHAPTER 8
3.
CIRCULAR MEASURE
FORM 4
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 = rad . 6 A P
O
rad 6 Q
B
DIAGRAM 3 It is given that OP: OA= 3 : 5 . ( Using = 3.142 )
4.
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 ]
In Diagram 4, PBQ is a semicircle with centre O and has a radius of 10 m. RAQ is a sector of a circle with centre A and has a radius of 16 m . R
B
P
A
O
Q
DIAGRAM 4 It is given that AB = 10 m and BOQ = 1.876 radians. [ Use = 3.142 ]
43
CHAPTER 8
5.
CIRCULAR MEASURE
FORM 4
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 ]
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
44
CHAPTER 8 6.
CIRCULAR MEASURE
FORM 4
Diagram 6 shows two circles. The larger circle has centre P and radius 24 cm. The smaller circle has centre Q and radius 16 cm. The circles touch at point R. The straight line XY is a common tangent to the circles at point X and Y. [Use = 3.142] Given that XPQ = radians, (a) show that = 1.37 ( to two decimals places), (b) calculate the length, in cm, of the minor arc YR, (c) calculate the area, in cm2, of the coloured region
[2 marks] [3 marks] [5 marks]
DIAGRAM 6
45
CHAPTER 8
CIRCULAR MEASURE
FORM 4
ANSWERS (PAPER 1)
1
25 7.5 2 7.5 = 8.75 0.8571
2
2 0.35 45 r (2 0.35) r = 7.583
1 1 1
1.396 15 r 1.396 r = 10.74
1
0.7 rad 1 1 2 (20) 2 (0.7) or (8) (1.1) 2 2 1 1 (20) 2 (0.7) (8) 2 (1.1) 2 2 104.8
1
23.64 cm 1 1 (12) 2 1.97 or (6) 2 sin 1.97 rad 2 2 1 1 (12) 2 1.97 (6) 2 sin 1.97 rad 2 2 125.26
1
OA
3a 3b
4a
4b
5a
5b
6a
6b
rad 3 1 1 (12) 2 1.047 or ( 12 2 6 2 )6 2 2 1 1 (12) 2 1.047 ( 12 2 6 2 )6 2 2 44.21 1.047 rad or
1 1 1
1 1
1 1 1
1 1 1 1 1 1 1
46
CHAPTER 8
CIRCULAR MEASURE
FORM 4
ANSWERS (PAPER 2)
1a
1b
2a
2b
2c
3a 3b
3c
4a 4b 4c
9 15 0.9273 rad. BC = 12 1 2 1 15 (0.9273) (12)(9) 2 2 50.32 cos
120O 2 rad 3 10 cos 60 o OB OB = 20 cm
2 arc ACB = 20 3 41.89 1 2 1 (20) 2 (20) 2 sin 120 o 2 3 2 245.72 PA = 6 cm
9sin30o + (15 9cos30o ) + 15 ( ) + 6 6 25.56 PQ = 4.5 cm , OQ = 9cos 30o 1 1 (15) 2 (4.5)(9 cos 30 o ) 2 6 2 41.38 1 (10) 2 1.876 2 93.8 10(1.876) + 16 ( 1.876) + 6 45.016 1 1 1 (16) 2 ( 1.876) (10) 2 1.876 (6)( 91) 2 2 2 39.63
1 2 3 1 1 1 1 1 1 1 3 1 1 4 1 3 1 1 1 3 1 3 1
47
CHAPTER 8
5a
5b
6a
6b
6c
CIRCULAR MEASURE
20 sin 30 o OA OA = 40 cm PA = 60 cm arc PQ = 60 3 62.84
AB = 40 2 20 2 1 1 2 1 2[ (60) 2 (20) 2 ( 40 2 20 2 )20] 2 6 2 3 2 354.513 cos =
8 40
1.37 rad YQP = 1.369 arc YR = 16( 1.369 ) 28.37
XY 40 2 8 2 = 39.19 1 1 1 (16 24)39.19 (1.369)(24) 2 (16) 2 ( - 1.369) 2 2 2 162.58
FORM 4
1 1 1 1 1 1 3 1 1 1 1 1 1 1
3 1
48
CHAPTER 1
PROGRESSIONS
FORM 5
PAPER 1 1.
Three consecutive terms of an arithmetic progression are 2 p 2, 9, 3 p . Find the common difference of the progression. [3 marks]
2.
The first three terms of an arithmetic progression are 1, x, 7, ...... 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 2, 1, 4....... , 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]
72
CHAPTER 1
PROGRESSIONS
FORM 5
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 total collection 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 25 stamps 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 base and height of each subsequent rectangle are half of the measurements of its previous one.
y cm
x cm Diagram 2 (a) Show that the areas of the rectangles form a geometric progression and state the common ratio. [3 marks] (b) Given that x 20 cm and y 80 cm, 9 (i) determine which rectangle has an area of 1 cm2 16 (ii) find the sum to infinity of the areas, in cm2, of the rectangles. [5 marks]
73
CHAPTER 1
3.
PROGRESSIONS
FORM 5
Diagram 3 shows part of an arrangement of circle of equal size.
10 cm Diagram 3 The number of circles in the lowest row is 80. For each of the other rows, the number of circles is 3 less than in the row below. The diameter of each circle is 10 cm . The number of circles 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]
74
CHAPTER 1
PROGRESSIONS
FORM 5
ANSWER (PAPER 1) 9 ( 2 p 2) 3 p 9
1
1
p4
1
common difference 9 6 3
2 (a)
1
x (1) 7 x x3
1
common difference 4
1
(b) the sum of the first 10 terms after the 3rd term: 13 S13 S 3 2(1) 12(4) 9 2
1
290
1
a (n 1)d a (n 2)d
3
a (n 3)d
84 n 12
1 1
Three consecutive terms: T12 31, T11 28, T10 25 4 (a) (b)
5
15 3 5 5(3 n 1) 5465 3 1 r
1 1
n7
1
12 1 48 4
1
48 1 1 4 64
S
1 1
27(r ) 41 1
1
1 3
1
r
(b)
1
37 3 n
r
6 (a)
1
S
27 1 1 3
1
75
CHAPTER 1 20
7.
PROGRESSIONS
FORM 5 1
1 4
0.121212…… = 0.12 + 0.0012 + 0.000012 + …….. a 0.12
r
1 1
0.0012 0.01 0.12
S
0.12 1 0.01
1 1
4 33
ANSWER (PAPER 2) 1 (a)
p (7 1)q 220 …….eq(1)
1
12 2 p (12 1)q 2520 ……..eq(2) 2
1
eq (1) x 2
: 2 p 12q 440 ……. eq(3) : 2 p 11q 420 ……. eq(4)
From eq (2)
(3) – (4) :
(b)
2 (a)
1
either
q 20
1
p 100
1
Tn 100 (n 1)20 80 20n Borhan : Tn 60 (n 1)25 35 25n Ali :
80 20n 35 25n
1
n9
1
Area : xy 1 r1 4 xy 4
xy,
xy xy , , .............. 4 16
xy 1 r2 16 xy 4 4
1
both
1
76
CHAPTER 1
(b)(i)
3 (a)
1
a 20(80) 1600
1
1 4
1 1600 4
n 1
1 4
n 1
S
1
9 16
1 4 n6
1 5
1 1
1600 1 1 4
2133
1 3 80, 77, 74, …………, 5 Tn 5 , a 80, d 3
1
80 (n 1)(3) 5
1
n 26
1
Height of the arrangement = 26 10 260 cm (b)
FORM 5
r1 r2 , the areas of the rectangles form a geometric progression with 1 the common ratio = 4
r
(ii)
PROGRESSIONS
S 26
26 2(80) (26 1)(3) 2
1 1
1105
1
The total length of circumference of circles = 1105 2 (5) 11050 cm
1
77
CHAPTER 2
LINEAR LAW
FORM 5
PAPER 1 xy
1.
n •
• ( 8, k ) x
0 Diagram 1 Diagram 1 shows part of a straight line graph drawn to represent y and n.
2.
10 1. Find the values of k x [4 marks]
log10 y ( 3,9 ) •
• ( 7,1) log10 x
0 Diagram 2
Diagram 2 shows part of a straight line graph drawn to represent y = 10kxn, where k and n are constants. Find the values of k and n. [4 marks]
78
CHAPTER 2
LINEAR LAW
FORM 5
y x
3.
• 12, 7
1 x
0 • (2 , - 3 ) Diagram 3
Diagram 3 shows that the variables x and y are related in such away that when
y is plotted x
1 , a straight line that passes through the points (12 , 7 ) and (2 , - 3 ) is obtained . x Express y in terms of x. [3 marks]
against
4.
log10 y
3 •
x
0
• ( 5 , -7 ) Diagram 4 Diagram 4 shows part of the graph of log10 y against x. The variables x and y are related by the a equation y x where a and b are constants. Find the values of a and b. b [4 marks ]
79
CHAPTER 2
LINEAR LAW
FORM 5
log10 y 5. 2
0
x
4 Diagram 5
Diagram 5 shows part of the graph of log10 y against x. The variables x and y are related by the bx
equation y = a (10 ) where a and b are constants. Find the values of a and b.
[3 marks]
6. log10 y
• 2
• -4
0
log10 x
Diagram 6
Diagram 6 shows part of a straight line graph when log10 y against log10 x is plotted. Express y in terms of x. [4 marks]
80
CHAPTER 2
LINEAR LAW
FORM 5
7. The variable x and y are related by equation y pk 3 x , where k and p are constant. Diagram 7 shows the straight line obtained by plotting log10 y against x . log10 y ( 0, 8 )
(2,2) x
O Diagram 7
a) Reduce the equation y pk 3 x to linear form Y = mX + c. b) Find the value of, i) ii)
log10 p , k.
[4 marks]
81
CHAPTER 2
LINEAR LAW
FORM 5
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 n and y are related by the equation y 2rx 2 x, where r and n are constants. r x y
2 8
3 13.2
4 20
5 27.5
6 36.6
7 45.5
Table 1 (a).
(b).
y against x , using a scale of 2 cm to 1 unit on both axes. x Hence, draw the line of best fit.
[5 marks]
Use your graph in (a), to find the value of (i). n, (ii). r, (iii). y when x = 1.5.
[5 marks]
Plot
2. Use graph paper to answer this question. Table 2 shows the values of two variables , x and y obtained from an experiment. Variables x h and y are related by the equation y kx, where h and k are constants. x x y
1 5.1
2 6.9
3 9.7
4 12.5
5 15.4
6 18.3
Table 2 (a).
(b).
Plot xy against x 2 , using a scale of 2 cm to 5 units on the x 2 -axis and 2 cm to 10 units on the xy-axis. Hence, draw the line of best fit.
[5 marks]
Use your graph in (a), to find the value of (i). h, (ii). k, (iii). y when x = 2.5.
[5 marks]
82
CHAPTER 2
LINEAR LAW
FORM 5
3. Use graph paper to answer this question. Table 3 shows the values of two variables , x and y obtained from an experiment. Variables x n and y are related by the equation y w , where n and w are constants. x x y
3 103
4 87
5 76
6 68
7 62
8 57.4
Table 3 (a).
Plot log10 y against log10 x , using a scale of 2 cm to 0.1 unit on the log10 x -axis and 2 cm to 0.2 units on the log10 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]
4. Use graph paper to answer this question. Table 4 shows the values of two variables, x and y obtained from an experiment. Variables x b and y are related by the equation y a x , where a and b are constants. x x y
0.2 12.40
0.4 8.50
0.6 6.74
0.8 5.66
1.2 4.90
1.4 3.87
Table 4 (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]
83
CHAPTER 2
LINEAR LAW
FORM 5
5. Use graph paper to answer this question. Table 5 shows the values of two variables, x and y obtained from an experiment. Variables x and y are related by the equation y pm x , where m and p are constants. x y
1.5 2.51
3.0 3.24
4.5 4.37
6.0 5.75
7.5 7.76
9.0 10.00
Table 5 (a).
Plot log10 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 log10 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]
6. Use graph paper to answer this question. Table 6 shows the values of two variables, x and y obtained from an experiment. Variables x and y are related by the equation y hk x , where h and k are constants. x y
3 10.2
4 16.4
5 26.2
6 42
7 67.1
8 107.4
Table 6 (a).
Plot log10 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 log10 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]
84
CHAPTER 2
LINEAR LAW
FORM 5
ANSWER
No. 1.
PAPER 1 Solution xy x 10 Calculate gradient from graph n 10 k 2
Marks 1 1 1 1
log10 y 2 log10 x 15 Calculate gradient from graph n 2 k 15
1
c = -5 y 1 5 x x y 1 5x
1 1
4.
log10 y = - log10 b + log10 a Gradient = -2 b = 100 a = 1000
1 1 1 1
5.
log10 y = bx + log10 a or log10 y = bxlog10 10 + log10 a a = 100 b=-½
1 1 1
2.
3.
6. log10 x1/2
Gradient = ½ log10 y = ½ log10 x + 2 or log10 102 or log10 x1/2.102 y 100 x
1 1 1
1
1 1 1 1
7(a)
3 log10 k 3
1
(b)(i)
log10 y (3log10 k ) x log10 p
1
log10 p 8 k 10
1
(ii)
1 85
CHAPTER 2
LINEAR LAW
FORM 5
PAPER 2 No. 1(a)
Solution x y x
(b) (i) (ii) (iii)
3 4.4
4 5
5 5.5
6 6.1
7 6.5
Uniform scale for x-axis and y-axis 6 points plotted correctly Draw the line of best fit Reduce non-linear to linear equation y n 2rx x r Calculate gradient from graph n = 0.77 r = 0.275 y From graph , 3.6 x y = 5.4
(b) (i) (ii) (iii)
2.(a)
2 4
x2 xy
1 4 5.1 13.8
9 29.1
16 50
25 77
Uniform scale for x-axis and y-axis 6 points plotted correctly Draw the line of best fit Reduce non-linear to linear equation xy kx 2 h Calculate gradient from graph h 2 k 3 From graph, xy = 20 y= 8
Marks 1
1 1 1 1
1 1 1 1 1 36 109.8
1
1 1 1 1 1 1 1 1 1
86
CHAPTER 2
3.(a)
LINEAR LAW
log10 x
0.48 0.60
0.70
0.78
0.85
0.90
log10 y
2.01 1.94
1.88
1.83
1.79
1.76
(b) (i) (ii) (iii)
4.(a)
(b) (i) (ii) (iii)
FORM 5
x y x
1
Uniform scale for x-axis and y-axis 6 points plotted correctly Draw the line of best fit Reduce non-linear to linear equation log10 y w log10 x log10 n Calculate gradient from graph n 102.3 199.526 w 0.6 From graph, log10 y 2.12
1 1 1 1
y 102.12 131.825
1
0.2 0.4 5.55 5.38
0.6 5.22
0.8 5.06
1.2 4.9
Uniform scale for x-axis and y-axis 6 points plotted correctly Draw the line of best fit Reduce non-linear to linear equation y x ax b Calculate gradient from graph a - 0.8 b 5.7 From graph y x 4.9 y = 5.165
1 1 1 1
1.4 4.74
1
1 1 1 1 1 1 1 1 1
87
CHAPTER 2
5.(a)
(i) (ii) (iii)
LINEAR LAW
x log10 y
1.5 3.0 0.4 0.51
6 0.76
7.5 0.89
1
9 1
Uniform scale for x-axis and y-axis 6 points plotted correctly Draw the line of best fit Reduce non-linear to linear equation log10 y x log10 m log10 p Calculate gradient from graph m 100.084 or m 1.2134 p 100.26 or1.8197 log10 4.8 0.68 x= 5
1 1 1 1
All values of log10 y are correct
1
6.(a) x log10 y
(i) (ii) (iii)
4.5 0.64
FORM 5
3 4 1.0 1.21
5 1.42
6 1.62
7 1.82
Uniform scale for x-axis and y-axis 6 points plotted correctly Draw the line of best fit Reduce non-linear to linear equation log10 y x log10 k log10 h Calculate gradient from graph k 100.2 or k 1.588 h 100.425 or h 2.66 log10 35.6 1.55 x = 5.7
1 1 1 1 1
8 2.03 1 1 1 1 1 1 1 1 1
88
Chapter 3
Intergration
Form 5
PAPER 1 1. Given that ( 5 x 3 2 ) dx kx 4 2 x c , where k and c are constants. Find (a) the value of k (b) the value of c if ( 5 x 3 2 ) d x 3 0 when x=2. [3 marks] 4
2. Evaluate
1
( x 2)
3
[3 marks]
dx
3
5
3. Given that 2x 3dx 6, where p < 5 , find the possible values of p. p
4. Given that
4
4
2
2
[4 marks] [4 marks]
f ( x)dx 8 and kx 2 f ( x)dx 8 , find the value of k.
5.
y y=x(a x)
0
a
x
Given that the area of the shaded region under the curve y=x(a x) is of a.
4
1 2
unit2, find the value [ 4 marks]
6. y y = f(x)
0
4
x
Diagram above shows the curve y = f(x). Given that the area of the shaded region is 5 unit2, find 4
the value of 2 f ( x) 2dx.
[3 marks]
0
89
Chapter 3
Intergration
Form 5
PAPER 2
1. A curve is such that dy 1 1 2 . Given that the curve passes through the point dx
2x
equation of the curve.
1 1, , 2
find the
[ 4 marks] y A(0, 1) y
0
1 4x 1
x 2 1 which passes through A(0,1). A region is 4x 1
(b) Diagram above shows part of the curve y
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 y 1 ( x 4) 2 which passes through A(2, 2). 2
y A(2, 2)
0
x
(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]
90
Chapter 3
Intergration
Form 5
3. In the diagram, the straight line PQ is a tangent to the curve x 1 y 2 1 at Q(3, 2) 2
y
Q(3, 2) P(0,k)
x 0
(a) Show that the value of k is
1, 2
[ 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
dy 2x 6 . dx
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 the curve and y-axis. y
B
A
x
O
(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, yaxis and the line x=3 is rotated through 360o about the x-axis. [3 marks]
91
Chapter 3
Intergration
Form 5
5. Diagram shows the straight line y=3x intersecting the curve y = 4 x2 at point P. y P
R
y=3x
y = 4 x 2 x
0
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, and y-axis is revolve 360o about the y-axis. [3 marks]
92
Chapter 3
Intergration
Form 5
PAPER 1 (Answer)
Q 1(a) (b)
2
Solution 5 k= 4 5 ( 2 )4 + 2( 2 ) + c = 30 4 c=6
4
( x 2) -3 dx
3
Marks 1 1 1 1
4
( x 2) 2 = 2 3 1 1 1 = 2 4 1 3 = 8
3
4
1 1
5
x 2 3x = 6 p 2 [5 -3(5) ] – (p2-3p) = 6 (p + 1) (p – 4 ) =0 p = 1, 4
1 1 1 1 1,1
4
4 x2 k 2 f ( x)dx 8 2 2 2 6k-2(8)=8 k=4
5
a
0
ax x 2 dx
9 2
a
ax 2 x3 9 3 0 2 2 a 3 a3 9 2 3 2 a=3
6
4
4
0
0 4 0
1 1 1 1 1 1
2 f ( x)dx 2dx = 2 5 + [2x] = 18
1 1
93
Chapter 3
Intergration
Form 5
PAPER 2 (Answer) Q
Solution
1 (a)
y = 1
(b)
Marks 1 dx 2 x2
1 = 1 x 2 dx 2 1 =x c 2x 1 Substitute 1, , c 1 2 1 y=x 1 2x 2 1 Area= dx correct limit 0 4x 1 2
2 (a)
(b)(i)
2 = 4 x 1 4 0 1 = ( 9 1 2 =1 dy = x 4 dx dy x = 2, = 24 = 2 dx y 2 = 2(x2) y = 2x+6 41 2 2 2 ( x 4) dx 4
(ii)
( x 4)3 = 2(3) 2 4 = unit 2 3 41 ( x 4) 4 dx 2 4
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 ( x 4)5 = 4 5 2
1
8 = unit3 5
1
4
3 (a)
1
1
y = (2x2) 2 dy 1 = dx 2x 2
1
94
Chapter 3
Intergration
1
dy 1 1 dx 2(3) 2 2 2k 1 1 ,k 3 0 2 2 2 1 1 3 2 0 ( 2 y 1)dy 2 2 3
x=3,
(b)
1 1, 1
1
2
1 9 = y3 y 6 0 4 10 9 = 3 4 13 = unit 2 12 (c)
Form 5
1 1
3
(2 x 2)dx 1
= x 2 2 x
1
3
1
4 (a)
(b)
=4 unit3 y = (2 x 6)dx
1 1
y = x2-6x + c Substitute x = 3, y = 1 in the equation, c = 10 y = x2-6x + 10 Equation of AB is y = 3x + 10
3
0
(3 x 10) ( x 2 6 x 10)dx
1 1 1
3
= 3x x 2 dx 0
1
3
3x 2 x3 = 3 0 2 9 = unit2 2
(c)
1 1
3
Volume= (3 x 10) 2 dx 0
3
= (9 x 2 60 x 100)dx 0
3
= 3x 3 30 x 2 100 x
1
0
5 (a)
(b)
=111 unit3 x2 +3x 4 = 0 (x1) (x+4)=0 x= 1, y =3(1)=3 P(1, 3) 1 Area of triangle= (1)(3) = 1.5 unit2 2 2
2
1
x3 4 x dx 4 x 3 1
1 1 1 1 1 1
2
95
Chapter 3
Intergration =1
2 unit2 3
Form 5 1
1 1 Area of R = 3 6
(c)
1
4
(4 y )dy 0
4
y2 = 4 y 2 0 =8 unit3
1 1
96
CHAPTER 4
VECTORS
FORM 5
PAPER 1
1. Diagram below shows two vectors, OP and QO
Q(-8,4) P(5,3) O Express x (a) OP in the form , y (b) QO in the form x i + y j [2 marks]
p = 2a + 3b q = 4a – b r = ha + ( h – k ) b, where h and k are constants
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
E A
B
Given that AB = 6p , AD = 4q and DE = 2EB, express, in terms of p and q (a) BD (b) EC [4 marks]
97
CHAPTER 4
VECTORS
FORM 5
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 6 A 4 2
0
2
4
6
8
10
12
x
x y (b) Find the unit vector in the direction of OA (a) Express OA in the form
[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] 98
CHAPTER 4
VECTORS
FORM 5
8. Diagram below shows two vectors, OA and AB .
y A(4,3)
O
-5
x
B
Express (a)
x in the form y AB in the form xi + yj
OA
(b)
[2 marks] 9. The points P, Q and R are collinear. It is given that PQ = 4a – 2b and
QR 3a (1 k )b , 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 a 6i 7 j and b pi 2 j , find the possible value (or values) of p for following cases:a) a and b are parallel b) a b [5 marks]
99
CHAPTER 4
VECTORS
FORM 5
PAPER 2
1.
5 2 k Give that AB , OB and CD , find 7 3 5 (a) the coordinates of A, (b) the unit vector in the direction of OA , (c) the value of k, if CD is parallel to AB
2.
[2 marks] [2 marks] [2 marks]
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, OP 6 x and OA 2 y. A Q R
O
P
(a) Express in terms of x and/or y: (i) AP (ii) OQ (b)
B
[4 marks]
(i) Given that AR h AP, state AR in terms of h, x and y. (ii) Given that RQ k OQ, 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]
100
CHAPTER 4 3.
VECTORS
FORM 5
In diagram below, ABCD is a quadrilateral. AED and EFC are straight lines. D
E
F
A
C
B
It is given that AB 20x, AE 8y, DC = 25x – 24y, AE = ¼ AD 3 and EF = EC. 5 (a) Express in terms of x and/or y: (i) BD (ii) EC
4.
[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]
Diagram below shows a trapezium ABCD. B
C F •
A
• E
D
2 5 It is given that A B =2y, AD = 6x, AE = AD and BC = AD 3 6 (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 a constant. (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] 101
CHAPTER 4
VECTORS
FORM 5
ANSWERS (PAPER 1) 1. a) b) 2.
3 a) b)
5 3
1
8i – 4j
1
r = - 2a + 11b r = ha + (h – k)b
1
h = -2 (h – k) = 11 k = −13
1
BD = −6p + 4q DB = − BD = 6p −4q
1
EB =
EC EB BC 8 2p q 3
b)
AB (2 (3))i (16 4) j = 5i + 12j 1 u (5i 12 j ) 5 2 12 2
5.
6. a) b)
1 (5i 12 j ) 13
AB 2 BC (6i 4 j ) 2((m 4)i 2( p 2) j ) = (-2+2m)i + (-8+2p)j m=6 p = -2 12 OA 5 1 u (12i 5 j ) 12 2 5 2
1
DB 3
4 2p q 3
4. a)
1
1 (12i 5 j ) 13
1
1 1 1 1 1
1 1 1 1 1 1
1
102
CHAPTER 4 7.
VECTORS
PO OP 6i 4 j
OR PQ 4i 5 j PR PO OR 10i j
8. a) b)
9. a)
b)
b)
1 1
1
4 OA 3
1
AB 4i 8 j
1
QR m PQ 3 4 m 1 k 2 3 = m(4) 3 m 4 1+ k = m(-2) 5 k 2 PQ u 2 4 (2) 2
10 a)
FORM 5
1 (4a 2b) 20
a kb 6i 7 j k ( pi 2 j ) 7 k 2 12 p 7 a b 62 72 p 9
p 2 22
1
1
1
1
1 1 1 1 1
103
CHAPTER 4
VECTORS
FORM 5
ANSWERS (PAPER 2) 3 AO 4 3 OA 4 A = (-3,-4)
1 (a)
(b)
u
1
1
OA OA
u
(c)
32 4 2
1 OA
1 3i 4 j 5
CD m AB k 5 m 5 7 5 m 7 25 k 7
2 (a) (i) (ii)
(b) (i)
(c)
1
AP 6 x 2 y AB 18 x 2 y 9 1 AQ x y 2 2 9 3 OQ x y 2 2
AR 6hx 2hy 9 3 RQ kx ky 2 2 AR RQ AQ 9 3 9 1 6h k x 2h k y x y 2 2 2 2 1 k 3 1 h 2
1
1
1
1 1 1 1 1 1
1 1 1
104
CHAPTER 4 3 (a) (i) (ii)
VECTORS 1
BD 20 x 32 y 3 ED AD 4 = 24y
1 1
EC 25 x (b)
FORM 5
FC 10 x BC BD DC 5x 8 y
1
BF BC CF 5 x 8 y
1
BD 20 x 32 y 4( 5 x 8 y ) 4( BF ) (c)
BD
1 2
20 x 32 y
2
(20 2) 2 32 3 = 104
4 (a)
(b) (i)
1 1
5 AD 6 = 5x
1
AC AB BC = 5x + 2y
1
BC
2 EF m AB EF my 2 AE AD 3 = 4x
AF AE EF = 4x + my (ii)
2
AC 5 x 2 y 4 4 AC 5 x 2 y 5 5 4 8 AC 4 x y 5 5 Assume A, F, C collinear, 4 AC AF 5 = 4x + my 8 m 5
1
1 1
1
1
105
CHAPTER 5
TRIGONOMETRIC FUNCTIONS
FORM 5
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 ec 90o 2.
Given cos p and 270o 360o .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 180o 270o . Find in terms of r.
[3 marks]
a) cot b) tan 2 4.
Solve the equation 6 cos ec 2 x 13cot x 0 for 0o x 360o
5.
Solve the equation 2 sin 2 A cos 2 A sin A 1 for 0o A 360o
6.
Solve the equation 2 cos 2 y 7 sin y 2 for 0o y 360o
7.
Solve the equation 15 cos 2 x cos x 4 cos 600 for 0o A 360o
8.
Solve the equation 3cot x 2 sin x 0 for 0o x 360o
[4 marks] [4 marks] [4 marks] [4 marks] [4 marks]
106
CHAPTER 5
TRIGONOMETRIC FUNCTIONS
FORM 5
PAPER 2 1.
(a) Sketch the graph of y sin 2 x for 0o x 180o
[4 marks]
(b)
(b) Hence, by drawing suitable straight line on the same axes, find the 1 x number of solution to the equation sin x cos x for 2 360o 0o x 180o
[3 marks]
(a) Sketch the graph of y 3cos x for 0 x 2
[4 marks]
2.
(b) (b) Hence, by using the same axes, sketch a suitable graph to find the 2 number of solution to the equation 3cos x 0 for 0 x 2 . x State number of solutions. 3. (a) Sketch the graph of y 3sin 2 x for 0 x 2
[3 marks]
(b) Hence, by using the same axes, sketch a suitable straight line to find x the number of solution to the equation 2 3sin 2 x for 2 0 x 2 cot x tan x (a) Prove that cos ec 2 x 2 3 (b) (i) Sketch the graph of y 2sin x for 0 x 2 2 (ii) Find the equation of a suitable straight line to solve the 3 3 1 equation sin x x . 2 2 2 Hence, on the same axes, sketch the straight line and state the 3 3 1 number of solutions to the equation sin x x for 2 2 2 0 x 2 . (a) Prove that sec 2 x 2 cos 2 x tan 2 x cos 2 x
[3 marks]
(b) (i) Sketch the graph of y cos 2 x for 0 x 2
[6 marks]
4.
5.
[4 marks]
[2 marks] [6 marks]
[2 marks]
(ii) Hence, using the same axes, draw a suitable straight line to find x the number of solutions to the equation 2 cos 2 x 1 for 0 x 2 .
107
CHAPTER 5 6.
7.
TRIGONOMETRIC FUNCTIONS
FORM 5
(a) Prove that 2 2sin 2 x 2 cos 2 x
[2 marks]
(b) Sketch the graph of y tan 2 x 1 for 0 x 2 . By using the same 9 axes, draw the straight line y 3 x and state the number of solution to 2 9 equation tan 2 x x 2 for 0 x 2 2
[6 marks]
(a) Prove that cot 2 x cos ec 2 x tan 2 x sec 2 x
[2 marks]
3 (b) Sketch the graph y cos x and y 2sin x for 0 x 2 . State 2 1 3 the number of solution to equation sin x cos x for 2 2 0 x 2
[6 marks]
108
CHAPTER 5
TRIGONOMETRIC FUNCTIONS
FORM 5
ANSWERS (PAPER 1)
1 a) b)
tan
1
p 1 p2
cos ec 90o =
1 sin 90o
= 2 a)
b)
1 cos 1 = p
b)
4
1
sec =
1
cot 90o = tan
= 3 a)
1 p2
1
1 1
1 p2 p
1 tan 1 = r 2 tan tan 2 = 1 tan 2 2r = 1 r2 6 1 cot 2 x 13cot x 0 cot =
1 1 1 1
6 cot 2 x 13cot x 6 0
3cot x 2 2 cot x 3 0 3cot x 2 0 OR
1
2 cot x 3 0
3 2 OR tan x 2 3 o ' 0 x 56 19 or 56.31 and x 33o 41' or 33.69o
tan x
x 56o19' , 236o 19' 33o 41' , 213o 41' Or 56.31o , 236.31o ,33.69o , 213.69o
5
2 sin 2 A 1 sin 2 A sin A 1
1 1 1
2 sin 2 A 1 sin 2 A sin A 1 3sin 2 A sin A 2 0
3sin A 2 sin A 1 0
1 109
CHAPTER 5
TRIGONOMETRIC FUNCTIONS
3sin A 2 0
sin A 1 0
OR
2 OR 3 A 90o and 41.81o sin A
A 90o , 221.81o , 6
FORM 5
sin A 1
1
318.19o
1
2 cos 2 y 7 sin y 2 0 2(1 sin 2 y ) 7 sin y 2 0
1
2 sin 2 y 7 sin y 4 0
2sin y 1 sin y 4 0
1
1 2 sin y 4 (not accepted) sin y
7
y 30o
1
y 30o , 150o
1
15 cos 2 x cos x 4 cos 600 15 cos 2 x cos x 4 cos 600 0
15cos 2 x cos x 4(0.5) 0 15 cos 2 x cos x 2 0
1
5cos x 2 3cos x 1 0
1
5cos x 2 0 OR
cos x
8
2 5
3cos x 1 0
OR cos x
1 3
x 66.42 or 66 25' and 70.53 or 70 31'
1
x 66.42 ,293.58 ,70.53 ,289.47 x 66o 25' , 70o31' , 289o 28' , 293o 35' cos x 3 2sin x 0 sin x 3cos x 2 sin 2 x 0
1
3cos x 2 1 cos 2 x 0
1
3cos x 2 2 cos 2 x 0 2 cos 2 x 3cos x 2 0
110
CHAPTER 5
TRIGONOMETRIC FUNCTIONS
FORM 5
2 cos x 1 cos x 2 0 2 cos x 1 0 cos x
1 2
1
cos x 2 0
OR
OR cos x 2 (unaccepted)
x 120o
1
x 120o , 240o
1
(ANSWERS)PAPER 2
1
y
y 1
x 180o
x
45o
90o
135o
180o
1 ( shape) 1(max/min) 1(one period) 1(complete from 0 to 180o)
1 (straight line)
y sin 2 x
sin x cos x
1 x 2 360o
y 1 x 2 2 360o x y 1 180o Number of solutions= 3
1
1
111
CHAPTER 5
TRIGONOMETRIC FUNCTIONS
FORM 5
2 a)
y
y
2 x
x
1 ( shape) 1(max/min) 1(one period) 1(complete from 0 to 2 or 360 o)
1(for line 2 y x
y 3cos x
b)
(b)
2 3cos x 0 x 2 y0 x 2 y x
1 1
Number of solution =2 3
y
x
1 ( shape) 1(max/min) 1(one period) 1(complete from 0o to 2 )
1( for the straight line)
2 3sin 2 x
x 2
x 2 x y 2 2
2 y
1 1
Numbers of solutions= 8 112
CHAPTER 5 4
TRIGONOMETRIC FUNCTIONS
(a) Prove that
FORM 5
cot x tan x cos ec 2 x 2
LHS 1 cos x sin x 2 sin x cos x 1 cos 2 x sin 2 x 2 sin x cos x
1
1 1 2 sin x cos x 1 2sin x cos x 1 sin 2x cos ec 2 x
1
b)
1(shape) 1(max/min) 1(one period)
y
y
3 x1
1(for the straight 2 line)
x
2
3 2
3 y 2sin x 2
sin
3 3 1 x x 2 2 2
1 3 y 2 x 2 2 3 y x 1
1
Number of solution = 1 5
1
(a) LHS sec 2 x 2 cos 2 x tan 2 x 1 tan 2 x 2 cos 2 x tan 2 x
1
1 2 cos 2 x cos 2 x
(proved)
1 113
CHAPTER 5
TRIGONOMETRIC FUNCTIONS
FORM 5
b)
1(shape) 1(max/min) 1(one period)
y
y cos 2 x
1(for the straight line ) 2
x
y
3 2
x
-
2 cos 2 x 1
x
1
x Number of solutions = 2 y
6
1
2 2 sin 2 x 2 1 sin 2 x
1
2 cos 2 x
1
(proved)
b)
1(shape) 1(max/min) 1(one period)
y
y tan 2 x 1
1(complete cycle from 0 to 2 )
2
2
x
3 2
2
-2
9 y 3 x 2
1(for the straight line) 1
Number of solution = 3
114
CHAPTER 5 7
(a)
TRIGONOMETRIC FUNCTIONS
FORM 5
RHS cos ec 2 x tan 2 x sec 2 x
1 cot 2 x tan 2 x sec 2 x
1
cot 2 x 1 tan 2 x sec 2 x
cot 2 x sec 2 x sec 2 x
cot 2 x
1
(proved)
y
3 y cos x 2
x
2
3 2
2
1,1(shapes) 1(max/min) 1(one period) 1(complete cycle from 0 to 2 )
Number of solutions = 3
y=2sinx
1
115
CHAPTER 6
PERMUTATIONS AND COMBINATIONS
FORM 5
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' without repetition 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 without repetition? [2 marks]
6.
Diagram shows 4 letters and 4 digits. A B C D
5 6 7 8
A code is to be formed using those letters and digits. The code must consists of 3 letters followed by 2 digits. How many codes can be formed if no letter or digit is repeated in each code ? [3 marks] 7.
A badminton team consists of 8 students. The team will be chosen from a group of 8 boys and 5 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. R
A
J
I
N
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]
116
CHAPTER 6 9.
PERMUTATIONS AND COMBINATIONS
FORM 5
A debating team consists of 6 students. These 6 students are chosen from 2 monitors, 3 assistant 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. U
N
F
I
O
M
R
A five-letter code is to be formed using five of these cards. Find a) 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 any digit 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 be seated if no two persons of the same sex are next to each other. [3 marks]
15.
Diagram shows six numbered cards. 1
4
5
7
8
9
A four-digit number is to be formed by using four of these cards. How many a) different numbers can be formed? b) different odd numbers can be formed? [4 marks]
117
CHAPTER 6
PERMUTATIONS AND COMBINATIONS
FORM 5
ANSWERS ( PAPER 1 ) 1.
2.
3.
4.
5.
6.
7.
C 3 x 11C 3 = 19800
1
p1 x 7 p 3 = 840
1
10
1
4
1 1
1 x 7C5 = 21 2
1 4
p1 x = 24
1
p2
1
4
2
p3 x = 48
1
p1
1 2
4
p3 x 4 p2 = 288
1
C5 x 5 C3
1
a)
8
b)
If the team consists of 8 boys and 0 girl
8
If the team consists of 7 boys and 1 girl
8
= 560 C8 x 5 C 0
=1
5
C 7 x C1 = 40
If the team consists of 6 boys and 2 girl 8 C 6 x 5 C 2 = 280 The number of teams that can be formed = 1 + 40 + 280 = 321 8.
9.
10.
11.
12.
a)
5!
b)
4! x 2! = 48
a)
10
b)
2
a)
7
b)
6
C6
= 120
1
1
3
2
C1 x C3 x C 2 = 60
2
p 5 = 2520
2
p 4 x 4 p1 = 1440
5
p1 x 8 p 4 = 8400
5
1 1
2
= 210 5
1
p 3 x 3 p1 = 180 = 180
1 1 2 1 2 1 118
CHAPTER 6 13.
8
a) b)
14.
15.
PERMUTATIONS AND COMBINATIONS
8
FORM 5
C6 6C3
1
= 560 6! x 4!
1 1
= 17280
1 2
P3 x 2 P2 = 672
a)
6
b)
5
P4
P3 x 4P1 = 240
1 = 360
1 2 1
119
CHAPTER 7
PROBABILITY
FORM 5
PAPER 1 1. A box contains 8 blue marbles and f white marbles. If a marble is picked randomly from the 3 box, the probability of getting a white marble is . 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 one after another without replacement. Find the probability that the two pens drawn are of different colours. [3 marks] 3. Table 2 shows the number of coloured cards in a box.
Colour Yellow Green Blue
Number of Cards 6 4 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 is
2 . 3 Find the probability that, (a) both of them qualify for the final, (b) only one of them qualifies for the final.
1 while the probability 5
that Rajes qualifies is
[3 marks]
5. A box contains 4 cards with the digits 1, 2, 3 and 4. Two numbers are picked randomly from the 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 1 2 probabilities that Team A will beat Team B and Team C are and respectively. Find the 3 5 probability that Team A will beat at least one of the teams. [3 marks]
120
CHAPTER 7
PROBABILITY
FORM 5
7. The probability of a particular netball player scoring a goal in a netball match is
1 . Find the 5
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 are
2 1 and 5 3
respectively. 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 Black Red Green Yellow
Number of Marbles 2 5 3 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 is
2 . Find the probability that 5
Zainal scores at least one goal in two matches. [3 marks]
121
CHAPTER 7
PROBABILITY
FORM 5
ANSWER (PAPER 1)
No. 1.
Solution
Marks
8+f
1 f
3 =
8+f
1
5
f = 12 2.
1
=
=
4 10
4 10
6 x
9
6 x
9
6
@
+
10
6
4 x
10
4 x
9
9
1
1
8 =
3.
1
15
=
=
6 12
5 x
6 12
11 5
x
11
or
+
4
3 x
12 4
11 3
12
x
11
1 =
4(a)
(b)
=
+
2 12
2 12
1 x
11
1 x
11
1
1
1
3
2 =
or
1
15
1 5
1 x
3
+
2 3
4 x
5
1
122
CHAPTER 7
PROBABILITY
FORM 5
3 =
1
5
2
5.
=
1 or
4 2
=
1
3 1
x
4
1
3
1 =
1
6
6.
1
=
1
=
3
3
3 x
5
3 x
5
or
+
2
2 x
3
2 3
5
2 x
5
or
1 3
2 x
1
+
3
2 x
5
5
1
1
3 =
7.
1
5
1
=
1
=
5
5
4 x
5
4 x
4 x
5
5
4 x
5
1
+
4 5
1 x
5
4 x
5
+
48 =
8.
=
4 5
4 x
5
1 x
5
1
1
125
2 5
2 x
3
or
1 3
3 x
5
or
1 3
2 x
5
1
123
CHAPTER 7
=
PROBABILITY
2 5
2 x
3
+
1 3
3 x
+
5
1
2 x
3
5
FORM 5
1
3 =
9.
1
5
=
=
2 18
1 x
17
2 18
1 x
17
5
or
+
4 x
18 5 18
17 4
x
17
or
+
3
2 x
18
3
2 x
18
17
14 =
10.
or
+
8 18
8 18
7 x
7 x
17
17
1
1
1
51
2
=
2
=
16 =
17
5
5
2 x
5
2 x
5
or
+
2 5
3 x
2 5
5
3 x
5
or
+
3 5
3 5
1
1
2 x
5
2 x
5
1
25
124
CHAPTER 8
PROBABILITY DISTRIBUTION
FORM 5
PAPER 1 1. Diagram 1 shows a standard normal distribution graph. f(z)
0 k Diagram 1
z
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 standard deviation 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. f(z) 0.3264
0
k
z
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]
125
CHAPTER 8
PROBABILITY DISTRIBUTION
FORM 5
6
X is a continuous random variable of a normal distribution with a mean of 52 and a standard deviation 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 kg and 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 a variance 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. Each trainee takes 6 penalty kicks. The probability that a trainee scores a goal from a penalty kick is 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 normal distribution 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]
126
CHAPTER 8
PROBABILITY DISTRIBUTION
FORM 5
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 a mean 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 Diameter, x (cm)
A x > 6.5
B 6.5 x > 4.5
C 4.5 x k
It is given that the diameter of oranges has a normal distribution with a mean of 5.3 cm and a standard 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. If 8 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 a normal 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 140 mmHg. (ii) It is found that 16 teachers have “high blood pressure”. Find the total number of teachers in the group. [5 marks]
127
CHAPTER 8 6
PROBABILITY DISTRIBUTION
FORM 5
The masses of mangoes from an orchard has a normal distribution with a mean of 285 g and a standard 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 than 191.25. (ii) Given that 416 mangoes from this sample have a mass of more than m g, find the value of m. [7 marks]
ANSWERS (PAPER 1) 1
2
3a
3b
4a
4b
5a 5b
6a 6b
0.5 - 0.3125 0.1875
1 1
12
1 1 1
C10 (0.85)10 (0.15) 2 0.2924 p = 0.85, q = 0.15
14.75 12.5 1 .5 1.5 12.5 12.5 14.75 12.5 or 1 .5 1 .5 0.4332 0 .5
X 55 10
1 1 1 1 1
60 48 55 10 75.8%
1
0.5 - 0.3264 0.94 X 180 0.94 5 .5 X = 185.17
1 1
z
67 52 10
1.5 1 - 0.8643 k=1.1
1 1
1 1 1 1 1 1
128
CHAPTER 8
7a
7b
PROBABILITY DISTRIBUTION
50.6 45 5 0.1314 52.1 45 40.5 45 or 5 5 0.7381
FORM 5
1 1 1 1
ANSWERS (PAPER 2)
1(a)(i)
1(a)(ii)
1(b)
2(a)(i)
2(a)(ii)
2(b)(i) 2(b)(ii)
p = 0.15, q = 0.85 P ( X 2) 1 8C 0 (0.15) 0 (0.85) 8 8 C1 (0.15)(0.85) 7 1 - 0.2725 - 0.3847 0.3428 165.75 = n(0.15)(0.85) n = 1300 65.34 , 7.5 P (48 x 72) 48 65.34 72 65.34 P( z ) 7 .5 7 .5 P(-2.312 < z < 0.888) 1 - 0.0103 - 0.1872 400 321 0.8025 x x = 400 mean = np 4 .5 p 6 0.75 P ( X 1) 1 - P(X <1) 1 - P(X=0) 1 6 C 0 (0.75) 0 (0.25) 6 0.9998 59 65 P ( x 59) P ( z ) 12 P(z < - 0.5) 0.3085 m 65 1.015 12 m = 77.18
1 1 1 1 1 1 1 1 1
1
1 1 1 1 1 1 1 2 1
129
CHAPTER 8
3(a)(i)
3(a)(ii)
3(b)(i)
3(b)(ii)
4(a)
4(b)
4(c)
PROBABILITY DISTRIBUTION
p = 0.18, q = 0.82 P(x = 2) = 10 C 2 (0.18) 2 (0.82) 8 0.298 P(X < 3) = P(x = 0) + P(x = 1) + P(x = 2) 10 C 0 (0.18) 0 (0.82)10 10C1 (0.18)(0.82) 9 10C 2 (0.15) 2 (0.82) 8 0.1374 + 0.3017 + 0.2980 0.7371 5 .7 4 .8 P( z ) 0 .6 1 - 0.0668 0.9332 m 4 .8 0.842 0 .6 m = 48 6 .5 5 .3 ) 0.75 0.0548 4 . 5 5 .3 6 .5 5 .3 P( z ) 0.75 0.75 1 - 0.1430 - 0.0548
P( z
312 x 0.8022 250 1- 0.7876 k 5 .3 0.798 0.75 k = 4.7015
FORM 5
1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 2 1
130
CHAPTER 8
5(a)(i) 5(a)(ii)
5(b)(i)
5(b)(ii)
P(x = 2) = 8 C 2 (0.6) 2 (0.4) 6 0.04129 P(X > 2) = 1 8 C 0 (0.6) 0 (0.4) 8 8 C1 (0.6)(0.4) 7 8 C 2 (0.6) 2 (0.4) 6 1 – 0.0006554 – 0.007864 – 0.04129 0.9502 110 128 140 128 P( z ) 10 10 1 - 0.0359 - 0.1151 0.849 140 128 P( z ) 10 0.1151 np = 16 16 n= 0.1151 139 191.25 285 ) 75 1 - 0.1056 0.8944 0.8944 x 520 465 416 P(X > m ) = 520 = 0.8 m 285 0.842 75 m =221.85
P( z 6(a)
6(b)(i)
6(b)(ii)
PROBABILITY DISTRIBUTION
FORM 5
1 1 1 1 1 1 1 1
1 1
1 1 1 1 1 1 1 2 1
131
CHAPTER 9
MOTION ALONG A STRAIGHT LINE
FORM 5
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 a 8 4t. 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 0 t k . 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 by V 20t 2t 2 , where t is the time, in seconds, after leaving the point R. (Assume motion to the right is positive)
(a) (b) (c) (d)
Find the maximum velocity of the particle, [3 marks] the distance travelled during the 4th second, [3 marks] the value of t when the particle passes the points R again, [2 marks] 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 the fixed point P and object B passes the fixed point Q simultaneously. The distance PQ is 90 m.
A
B
P
M
Q
90 m
The velocity of A, V A m s 1 , is given V A 10 8t 2t 2 , 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 instantaneously at point M. (Assume that the positive direction of motion is towards the right.) Find (a) the maximum velocity , in, m s 1 , of A, (b) the distance, in m, of M from P, (c) the distance, in m, between A and B when A is at the points M.
[3 marks] [4 marks] [3 marks] 132
CHAPTER 9
4.
MOTION ALONG A STRAIGHT LINE
FORM 5
A particle moves in a straight line and passes through a fixed point O. Its velocity, v ms 1 , is given by v t 2 4t 3 , 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. (b) Sketch the velocity-time graph of the motion of the particle for 0 t 3 .
[5 marks] [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 v t 2 8t 15 , where t is the time, in seconds, after passing through O . [Assume motion to the right is positive.] Find (a) (b) (c) (d)
6.
the initial velocity, in ms 1 , the minimum velocity, in ms 1 , the range of values of t during which the particle moves to the left, the total distance, in m, travelled by the particle in the first 5 seconds.
[1 mark] [3 marks] [2 marks] [4 marks]
A particle moves along a straight line and passes through a fixed point O. Its velocity, v ms 1 , is given by v 8 2t t 2 , where t is the time, in seconds, after passing through O . The particle stops instantaneously at point M. [Assume motion to the right is positive.] Find (a) (b) (c)
the acceleration, in ms 2 , of the particle at M, [3 marks] the maximum velocity, in ms 1 , of the particle [3 marks] the total distance, in m, travelled by the particle in the first 10 seconds, after passing through O. [4 marks]
133
CHAPTER 9 7.
FORM 5
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, after passing through point O. The particle stops after k s. (a)
(b)
8.
MOTION ALONG A STRAIGHT LINE
Find: (i) the maximum velocity of the particle, (ii) the value of k [6 marks] Sketch a velocity-time graph of the motion of the particle for 0 t k . Hence, or otherwise, calculate the total distance travelled during that period. [4 marks]
A particle moves along a straight line. Its velocity, v ms- 1 , from a fixed point, O, is given by v = t2 – 10t + 24 where t is the time, in seconds, after passing through point O. [Assume motion to the right is positive.] Find (a) (b) (c) (d)
the initial velocity of the particle, the minimum velocity of the particle, the range of values of t during which the particle moves to the left, the total distance travelled by the particle in the first 6 seconds.
[1 marks] [3 marks] [2 marks] [4 marks]
134
CHAPTER 9
MOTION ALONG A STRAIGHT LINE
FORM 5
ANSWERS PAPER 2 1.
a V
= =
8 – 4t ( 8-4t ) dt 4t 2 8t c 2 8t 2t 2 c
= =
a)
(i)
(ii)
b)
V = t V
1
V = 10, t = 0. 10 = 8(0) 2(0) 2 c 10 = c V = 8t 2t 2 10 maximum velocity, a= 0 8 – 4t = 0 8 = 4t t = 2s Vmax imum = 8(2) 2(2) 2 10 = 18 ms-1 particle stops: V = 0 0 8t 2t 2 10 = 2 0 2t 8t 10 = 0 t 2 4t 10 = (t + 1)(t – 5) = 0 t = 5 k = 5 2 8t 2t 10 V 0 10
2 18
5 0
1
1 1 1
1
18 10
5
2
0
t
2
5
Total Distance
=
V dt
=
8t 2t
0 5
2
10 dt
0
5
=
2 2 3 4t 3 t 10t 0
1
135
CHAPTER 9
MOTION ALONG A STRAIGHT LINE = =
2.
a)
b)
2 4(5 2 ) (5 3 ) 10(5) 0 3 2 66 m 3
V 20t t 2 dV a 20 4t dt Maximum velocity, a = 0 20 – 4t = 0 20 = 4t t=5 Vmax = 20(5) – 2(52) = 50 ms- 1 s v dt
FORM 5
1
1
1 1
20t 2t 2 dt
2 10t 2 t 3 c 3 s = 0, t = 0 c = 0 2 s 10t 2 t 3 3 2 t 3; s 10(3 2 ) (33 ) 72m 3 2 1 t 4; s 10(4 2 ) (4 3 ) 117 m 3 3
Distance travelled during the fourth second
c)
1
1 1 = 117 72 3 1 45 m 3
1
At point R again, s = 0 2 10t 2 t 3 0 3 2 t 2 (10 t ) 0 3 2 10 t 0 3 2 t 10 3 t 15
1
1
136
CHAPTER 9 d)
3.
MOTION ALONG A STRAIGHT LINE Maximum displacement, V = 0 20t 2t 2 0 2t (10 t ) 0 t 10 Time = 10 seconds 2 V A 10 8t 2t a A 8 4t V A maximum when a A 0 8 – 4t = 0 8 = 4t t=2 V A max 10 8(2) 2(2 2 ) = 18 ms- 1 Object A at M when V A 0
a)
b)
10 8t 2t 2 0 2t 2 8t 10 0 t 2 4t 5 0
FORM 5
1
1 1
1 1
1
t 1t 5 0
1
t=5 5
Distance M from P v dt 0
5
10 8t 2t 2 dt 0 5
c)
2 10t 4t 2 t 3 3 0 2 10(5) 4(52 ) (53 ) 0 3 2 66 m 3 When t = 5, distance B from Q = v t =35 = 15 m 2 Distance between A and B = 90 66 15 3 1 = 8 m # 3
1
1
1 1 1
137
CHAPTER 9 4.
a)
MOTION ALONG A STRAIGHT LINE
v t 2 4t 3 t = 0, v 0 2 4(0) 3 = 3 ms-1 # particle moves to the left, v < 0 t 2 4t 3 < 0 t 1t 3 < 0 1 < t < 3 # v t 2 4t 3 dv a 2t 4 dt a is positive, a > 0 2t – 4 > 0 2t > 0 t >2 # 2 v t 4t 3 v
i)
ii)
iii)
b) t v
0 3
1 0
3 0
1 1 1
1
1
3
1
Total distance
t
3
1
0
c)
FORM 5
v dt
v dt
0
1
=
t
2
t
3
2
3
=
1
1
3
4t 3 dt
0
(t
2
4t 3)dt
1
1
3
=
t 3 t 3 2 2 2 t 3 t 2t 3t 3 0 3 1
=
1 3 2 3
1 9 18 9 2 3 3
1 1 1 1 3 3 2 = 2 m # 3 2 v t 8t 15 t 0, v 0 2 8(0) 15 15ms 1 #
1
1
=
5.
a)
1 1 138
CHAPTER 9 b)
MOTION ALONG A STRAIGHT LINE
FORM 5
v t 2 8t 15 dv a 2t 8 dt
1
dv 0, dt 2t – 8 = 0 2t = 8 t=4
When
1 42 – 8(4) + 15 = - 1 ms-1
When t = 4, vmin = c)
Particle moves to the left, v < 0 t 2 8t 15 0 t 3t 5 0
#
3
1
5
3 < t < 5 d) Total distance =
3
5
0
3
1
1
v dt v dt
t 3
=
t
2
5
(t
8t 15 dt
0
2
8t 15) dt
1
3
3
5
=
t 3 t 3 2 2 4 t 15 t 4t 15t 3 0 3 3
=
9 36 45
125 100 75 9 36 45 3
2 18 16 18 3 1 = 19 m # 3 v = 2 When v = 0, 8 2t t = t 2 2t 8 = (t + 2) (t – 4) = t = 4 dv a = = 2 – 2t dt At M, a = 2 – 2(4) = – 6 ms 2
1
1
=
6.
a)
1 8 2t t 2 0 0 0
1 1
#
1
139
CHAPTER 9 b)
MOTION ALONG A STRAIGHT LINE Maximum velocity, a = 2 – 2t = 2 = t = When t = 1 ; v max imum =
0 0 2t 1 8 + 2(1) – 1
c)
Total distance, s = =
1 1 1
9 ms 1 #
= v dt 8 2t t 2 dt t3 8t t 2 c 3
=
FORM 5
1
When t = 0, s = 0 c = 0 s =
8t t 2
t3 3
1
43 3 10 3 8(10) 10 2 3 8(4) 4 2
s t 4 = s t 10 =
=
t=0 153
1m 3
2 m 3 1 153 m 3 26
=
1
t=4
0
26
2m 3
t = 10
7.
a)
i)
Total distance in the first 10 seconds 2 1 = 2(26 ) 153 3 3 2 = 206 m # 3 a = – 2t + 8 V = – 2t + 8 dt = t 2 8t c When t = 0, V = 20 : 20 = t 2 8t c 20 = 0 V = t 2 8t 20 When maximum velocity, a = 0 - 2t + 8= 0 2t = 8 t= 4 V max imum
= (4) 2 8(4) 20
1
1
1
1 1 140
CHAPTER 9
MOTION ALONG A STRAIGHT LINE = 36 ms
(ii)
b)
Particle stops, V t 2 8t 20 t 2 8t 20 (t + 2)(t – 10) t = 10 k = 10 V
= = = =0
= t 2 8t 20
1
FORM 5
#
0 0 0 V 36
1
20
t 0 4 10 V 20 36 0
2 0
Distance
= = = =
8.
a)
( t 2 8t 20 ) dt 3 t [ 4t 2 20t ]10 0 3 3 10 4(10 2 ) 20(10) – 0 3 2 266 m # 3
= t 2 10t 24 When t = 0, initial velocity, v = 0 2 10(0) 24 = 24 ms 1 #
v
b)
a
= =
dv dt 2t – 10
max imum
10
t
1
1
1
1
When minimum velocity, a = 0 2t -10 = 0 t = 5 v
4
1
=
5 2 10(5) 24
=
– 1 ms 1
#
1 141
CHAPTER 9 c)
d)
MOTION ALONG A STRAIGHT LINE When moves to the left, v 0 t 2 10t 24 ( t – 4 )( t – 6 )
0 0
4
FORM 5
6
4 t 6 Distance, s = ( t 2 10t 24 ) dt 1 3 = t 5t 2 24t c 3 When t = 0, s = 0 c = 0 1 3 s = t 5t 2 24t 3 t = 0, s = 0 1 1 t = 4, s = (4) 3 5(4) 2 24(4) = 37 m 3 3 1 t = 6, s = (6) 3 5(6) 2 24(6) = 36 m 3
t
1
1
1
1
1
t=6 t=4 0
36
t=0 Total distance travelled during the first 6 seconds 1 1 = 37 + 37 – 36 3 3 2 = m # 38 3
37
1m 3
1
142
CHAPTER 10
LINEAR PROGRAMMING
FORM 5
PAPER 2 1.
Amirah has an allocation of RM200 to buy x workbooks and y reference books. The total number of books is not less than 20. The number of workbooks is at most twice the number of the references. The price of a workbook is RM10 and that of a reference is RM5. a ) Write down three inequalities, other than x 0 and y 0, which satisfy all the above constraints. ( 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 dentistry undergraduates. The method in which the number of medical undergraduates and dentistry undergraduates 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 x 0 and y 0, which satisfy all the above constraints. ( 3 marks ) b) Hence, by using a scale of 2cm to 10 participants on both axes, construct and shade the region R 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 4 students. The number of students for S is x and for M is y. The intake of the students is based on 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 subject M. III : The number of students for subject M must exceed the number of students for subject S by at most 10 143
CHAPTER 10
LINEAR PROGRAMMING
FORM 5
a) Write down three inequalities, other than x 0 and y 0, which satisfy all the above conditions. ( 3 marks) b) Hence, by using a scale of 2 cm to represent 10 students on both axes, construct and shade the region 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 two processes, mixing the ingredients and baking the breads. Table 1 shows the time taken to make bread L and M respectively.
Type of bread L M
Time taken ( minutes ) Mixing the ingredients Baking the breads 30 40 30 30 Table 1
The shop produces x breads of type L and y breads of type M per day. The production of breads per day are based on the following constraints: I : The maximum total time used for mixing ingredients for both breads is not more than 540 minutes. 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 x 0 and y 0, which satisfy all the above constraints. ( 3 marks ) b) Using a scale of 2 cm to represent 2 breads on both axes, construct and shade the region R that 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 144
CHAPTER 10
LINEAR PROGRAMMING
from one bread of type M is RM1.00 . 5.
FORM 5 ( 4 marks )
A factory produced x toys of model A and y toys of model B. The profit from the sales of a number 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 x 0 and y 0, which satisfy all the above conditions. ( 3 marks) b) Hence, by using a scale of 2 cm to represent 50 models on both axes, construct and shade the region 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
145
CHAPTER 10
1.
LINEAR PROGRAMMING
FORM 5
( a ) I : x + y ≥ 20 II : x ≤ 2y III : 10 x + 5y ≤ 200 2x + y ≤ 40
1 1 1
(b)
4 40
35
30
25
20
2x + y= 40
R
y = 15
15
10
x + y = 20
x =2y
5
10
20
30
40
( c ) Draw the line y = 15 From the graph, the minimum value occurs at ( 5, 15 ) Hence, minimum expenditure = 10 (5) + 5 (15) = RM125 Therefore, the maximum amount of money left = RM200 –RM125 =RM 75
50
60
1 1
1
146
CHAPTER 10
LINEAR PROGRAMMING
FORM 5
2. ( a ) I : x + y ≥ 30 II : x ≤ 3y III : 100 x + 80y ≤ 6000 5x + 4y ≤ 300
1 1 1
(b)
3 90
80
5x + 4y= 300 70
x =20 60
50
40
R
30
20
x =3y x + y=30
10
-20
20
40
60
80
100
120
140
-10
( c ) ( i ) Draw the line x = 20, From the graph, the minimum number of dentistry undergraduates is 10 and the maximum number of dentistry undergraduates is 50.
1 1
(ii) For the minumum expenditure, there are 20 medical undergraduates and 10 dentistry undergraduates. Thus , minimum expenditure = 100 (20) + 80(10) =RM2800
1 1
147
CHAPTER 10
LINEAR PROGRAMMING
FORM 5
3 . ( a ) I : x + y ≤ 90 II : x ≤ 2y III : y - x ≤ 10
1 1 1
(b)
3 90
80
x + y = 90
70
y - x = 10
60
50
40
R x =2y
30
20
12x + 10y = 600 10
20
40
60
80
100
120
140
-10
( c ) ( i ) From the graph, when x = 20, the range of y is 10 ≤ y ≤ 30
1
( ii ) The maximum value occurs at ( 60, 30)
1
Thus, maximum total fee = 12 (60) +10 (30) = RM1 020
1 1
148
CHAPTER 10
LINEAR PROGRAMMING
FORM 5
4 ( a ) I : 3x + 3y ≤ 54 II : 4x + 3y ≥ 48 III : 2x ≥ y
1 1 1
(b)
3 22
20
18
16
y =2x 14
12
10
3x + 3y =54 8
6
R 4x + 3y = 4 8
4
2
2x + y = 2
5
10
15
20
25
30
35
-2
1 2 1
( c ) (i ) 8 (ii ) (5 x RM2.00) + ( 10 x RM1.00) = RM 20.00
149
CHAPTER 10 5
LINEAR PROGRAMMING
(a ) I : x + y ≤ 500 II : x ≤ 3y III : 15x + 12y ≥ 4200 5x + 4y ≥ 1400
FORM 5 1 1 1
(b)
3
( c ) ( i ) When x = 100, the minimum number of model B is 225. 1 (ii ) Maximum point (375, 125) The maximum total profit per day = 15(375) +12(125) = RM7125 1 1
150
DIFFERENTIATION
FORM 4
PAPER 1 1.
Given that y 15 x(6 x ) , calculate (a) the value of x when y is maximum, (b) the maximum value of y.
[3 marks]
2.
Given that y 4 x x 2 , use differentiation to find the small change in y when x increases from 5 to 5.01. [3 marks]
3.
Differentiate 2 x 2 (3 x 5) 4 with respect to x.
4.
[3 marks]
5 Two variables, x and y are related by the equation y 4 x . Given that y increases at a x constant rate of 3 units per second, find the rate of change of x when x 5. [3 marks] 1 , evaluate g”(1). ( 2 x 4) 2
5.
Given that g ( x )
6.
The volume of water, V cm3, in a container is given by V
7.
The point R lies on the curve y ( x 3) 2 . It is given that the gradient of the normal at R is 1 [3 marks] . Find the coordinates of R. 6
8.
It is given that y
9.
Given that y 2 x 2 3 x 4, dy (a) find the value of when x 3, dx (b) express the approximate change in y, in terms of m, when x changes from 1 to 1 + m, where m is a small value. [4 marks]
10.
The curve y = f(x) is such that
11.
The curve y = x2 – 28x + 52 has a minimum point at x = k, where k is a constant. Find the value of k. [3 marks]
[4 marks]
2 3 h 6h , where h is the height in 3 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]
4 3 dy in terms of x. u , where u 5 x 2. Find 5 dx
[3 marks]
dy 5 px 2 , where p is a constant. The gradient of the curve dx at x = 2 is 10. Find the value of p. [2 marks]
49
CHAPTER 9
DIFFERENTIATION
FORM 4
ANSWERS PAPER 1 Marks 1.
a)
b)
2.
3.
y = 15x (6 – x) = 90x – 15x2 dy 90 30 x dx dy 0, 90 30 x 0 dx 30x = 90 x=3 # y = 15(3)(6 – 3) = 135 # y = 4x x2 dy = 4 + 2x dx dy x = 5, = 4 + 2(5) dx = 14 dy = x y dx = 14 0.01 = 0.14 # 2 Let y = 2 x (3x 5) 4 dy = 2 x 2 [4(3 x 5) 3 3] (3 x 5) 4 (4 x) dx = 24 x 2 (3 x 5) 3 4 x(3x 5) 4
4.
y
dy dx dy dt
1
1 1
1
1 1
1
=
4 x(3 x 5) 3 [6 x (3x 5)]
1
=
4 x(3 x 5) 3 (9 x 5) # 5 4 x = 4 x 5 x 1 x 5 4 2 x dy dx dx dt 5 dx (4 2 ) x dt 5 dx (4 2 ) 5 dt 1 dx 4 5 dt 5 unit per second # 7
1
= = =
3
=
3
=
3
=
dx dt
=
1
1
1 50
CHAPTER 9
DIFFERENTIATION
5.
g(x)
=
g (x) = = g(x) = = g(1) = = = 6.
V dV dt dV dt
=
7.
(2 x 4) 2 2(2 x 4) 3 (2) 4(2 x 4) 3 12(2 x 4) (2) 24(2 x 4) 4 24(2) 4 24 16 3 # 2 2 3 h 6h 3 2 h 2 6
=
dV dh dh dt
5
=
( 2 h 2 6 )
dh dt
=
y dy dx
8.
y
dy dx
9.
a)
y
1
dh dt
= =
2(x – 3)
2
Given gradient of normal =
1 6 6 6 (6 3) 2 9
= = = = R (6, 9) # 4 3 = u 5 4 = (5 x 2) 3 5 12 = (5 x 2) 2 5 5 = 12(5 x 2) 2 # =
1
1
5 2(3 ) 6 5 cms 1 @ 0.2083 cms 1 # 24 ( x 3) 2
2(x – 3) x y
1
4
=
=
FORM 4
1 1
1
1
1
1 1 1
2 x 2 3x 4 1 51
CHAPTER 9
DIFFERENTIATION dy dx
=
4x + 3
When x = 3, b)
x y x y x = 1, m
=
=
4(3) + 3
=
15 #
4(1) + 3
dy dx 5p(2) + 2 10p p
11.
dy dx
m dy dx
y 10.
FORM 4
= 7
1 1
7m # =
5px + 2
= =
10 8 4 # 5 x 2 28 x 52
=
1
y = dy = 2x – 28 dx Minimum point at x = k : 2(k) – 28 = 0 2k = 28 k = 14 #
1
1
1 1 1
52
CHAPTER 9
DIFFERENTIATION
FORM 4
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. 0.8 m
0.6 m
water
Calculate the rate of change of the height of the water level at the instant when the height of 1 the water level is 0.4 m. (Use 3.142; Volume of a cone = r 2 h ) [4 marks] 3
2.
4 which passes through A(2, 5). (2 x 3) 2 Find the equation of the tangent to the curve at the point A.
Diagram shows part of the curve y
[4 marks]
y
A(2, 5)
y
O
4 (2 x 3) 2
x
53
CHAPTER 9 3.
DIFFERENTIATION
FORM 4
In the diagram, the straight line PQ is a normal to the curve y
x2 3 at A(2, 5). 2
y
P A(2, 5)
O
Q(k, 0)
Find the value of k.
4.
x
[3 marks]
Diagram shows part of the curve y = k(x – 2)3, where k is a constant. The curve intersects the straight line x = 4 at point A. dy At point A, [3 marks] 36. Find the value of k. dx
y
y = k(x – 2)3 A
x
O x=4
54
CHAPTER 9
5.
DIFFERENTIATION
FORM 4
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 2
y=x +3
P(1, 5)
x O
55
CHAPTER 9
DIFFERENTIATION
FORM 4
ANSWERS PAPER 2 1.
r h
=
r
=
V
=
0.4 m
r 0.6 m
=
h
= dV dh dV dh 4 2 h 9
2.
y dy dx
dy dx
=
1 2 r h 3 1 2 2 ( h) h 3 3 4 h 3 27 4 2 h 9 dV dt dt dh dt 0.3 dh dt 0.3 dh
1.343 ms 1 #
1
1
1
1
8(2 x 3) 3 (2)
=
= = y–5 y
3.
=
4 (3.142)(0.4) 2 = 9 dh = dt = 4(2 x 3) 2
= At A (2, 5) ,
=
0. 4 0. 6 4 2 h h 6 3
16 (2 x 3) 3 16 [2(2) 3]3 –16 = – 16( x 2 ) = – 16x + 32 = –16x + 37 #
1
1 1 1
2
y = dy = dx
x 3 2
x
At point A(2 , 5),
dy =2 dx
1 56
CHAPTER 9
DIFFERENTIATION Gradient of normal, m2
4.
=
dy dx 3k (4 2) 2 3k k y dy dx dy At P (1,5), dx
When x = 4,
5.
1 2
05 1 k 2 2 – 10 = –k + 2 k = 12 # = k ( x 2) 3
y dy dx
y–5 y
FORM 4
3k ( x 2) 2
1 1
1
= 36 : = 36 = 9 = 3 # = x2 3 =
2x
=
2(1)
= = = =
2 2 (x – 1) 2x – 2 2x + 3 #
1 1
1 1 1
57
CHAPTER 10
SOLUTION OF TRIANGLES
FORM 4
PAPER 2 1.
Diagram 1 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 is different in shape to triangle ABC. Sketch the triangle A’B’C’ [ 2 marks ] 2. P R
12cm 65 º
10cm
Diagram 2 Q
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. 58
CHAPTER 10
SOLUTION OF TRIANGLES
FORM 4
Calculate the two possible values of PSR.
[2 marks ]
(c) Using the acute PSR from ( b ) , calculate i ) the length, in cm, of RS ii ) the area, in cm2, of quadrilateral PQRS.
[ 6 marks]
3. B
6cm
C
53º 7cm A 10cm
D
Diagram 3 Diagram 3 shows a quadrilateral ABCD. The area of ∆BCD is 20cm2 and BCD is acute. Calculate (a) BCD (b) the length in cm, of BD. (c) ADB (d) the area, in cm2 , of quadrilateral ABCD
[ 2 marks] [ 2 marks] [ 3 marks] [ 3 marks]
A
4.
5cm 5cm
B
80 48' 120
D
10cm
C
Diagram 4
Diagram 4 shows quadrilateral ABCD. (a) Calculate
( 4 marks ) 59
CHAPTER 10
SOLUTION OF TRIANGLES
FORM 4
(i ) the length, in cm. of AC, (ii) ACB (b) Point A’ lies on AC such that A’B = AB (i ) Sketch ∆A’BC (ii) Calculate the area, in cm2, of ∆A’BC
5.
( 6 marks )
S
R
P
Q
Diagram 1 Diagram 1 shows a pyramid PQRS with triangle PQR as the horizontal base. S is the vertex of the pyramid and the angle between the inclined plane QRS and the base PQR is 50º. Given that PQ and 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 (b) the length of SP, (c) the area of the triangle PQS.
[ 3 marks ] [ 3 marks ] [ 4 marks ]
60
CHAPTER 10
SOLUTION OF TRIANGLES
FORM 4
ANSWER(PAPER2) 1.
(a) (i)
1
8.5 10.8 = sin 42.5 sin ABC ABC =59.14 º
1
(ii) 10.82=6.52 +5.22- 2(6.5)(5.2) cos ADC
1
ADC = 134.46 º
1
(iii) Total area 1 1 = (10.8)(8.5)sin(180º- 42.5º -59.14 º) + (6.5)(5.2)sin134.46 º 2 2 = 57.02 cm2
3 1
(b) C’
42.5°
B’
10.8 cm
2
8.5 cm A’
61
CHAPTER 10
SOLUTION OF TRIANGLES
FORM 4
2. ( a ) PR2 =122 +102- 2(12)(10)cos65º = 142.6 PR = 11.94cm (b)
1 1
S
S' P
30
12 cm
R
10 cm 65
Q
sin PSR sin 30 = 11.94 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
2
1
RS 8.2 = sin103.28 sin 30
1
RS = 15.96cm
1
1 1 (10)(12)(sin65 º ) + (11.94)(15.96)(sin30 º) 2 2 =102.02cm2
( ii) Area =
2 1
62
CHAPTER 10
SOLUTION OF TRIANGLES
FORM 4
3. 1 (6)(7)sin BCD = 20 2 BCD =72 º15’
1
( b ) BD2 = 6 2 + 7 2 – 2(6)(7)cos 72 º15’ =59.39 BD = 7.706
1
(a)
(c )
1
1
sin 53 sin BAD = 10 7.706
1
BAD =37 º59’
1
ADB =180º- 37º59’-53 º
1
=89 º 1’
1 1 (7)(6)(sin72 º15’ ) + (7.706)(10)(sin89 º1’) 2 2 =58.52cm2
4.
(d) Area =
2 1
( a ) ( i ) AC2 = 5 2 + 10 2 – 2(5)(10)cos 80 º48’ AC = 10.44
1 1
( ii )
sin ACB sin120 = 5 10.44
1
ACB = 24 º30’
1
( b) ( i) 2
63
CHAPTER 10
SOLUTION OF TRIANGLES
FORM 4
( 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 (5)(10.4408)(sin35 º30’ ) 2
1
=15.1575 Area of ∆ABA’ =
1 (5)(5)(sin109 º ) 2
=11.8190 Area of ∆A’BC =15.1575 - 11.8190 =3.3385 cm2
5. ( a ) Area of
1
1 1
PQR = 12.4
1 (5.6) (5.6)sin QPR = 12.4 2
1
sin QPR = 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
64
CHAPTER 10
SOLUTION OF TRIANGLES
FORM 4
1 QR where M is the midpoint of RQ 2 1 = (4.933) 2 = 2.466
(b) QM =
SM2 = 4.22- 2.4662 =11.56 SM =3.4 cm PM2 = 5.62- 2.4662 PM = 5.028 SP2 = 3.42 + 5.0282 – 2(3.4)( 5.028) cos 50º SP =3.855 cm (c) cos
SQP
=
1 1 1
1
5.62 4.22 3.8552 2(5.6)(4.2) 1
=0.7257
SQP
1 = 43.47 º
1
Area of PQS 1 = (5.6) (4.2)sin 43.47 º 2 =8.091 cm2
65
CHAPTER 11
INDEX NUMBER
FORM 4
PAPER 2 1 Table 1 shows the price indices and percentage of usage of four items, P, Q, R and S, which are the main ingredients in the making of a type of cake. Items P Q R S
Price index for the year 2007 based on the year 2004 125 x 110 130
Percentage of usage 40 20 10 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 in making a particular kind of cake. Diagram 1 is a pie chart which represents the relative amount of the ingredients K, L, M and N, used in making the cake.
Ingredients K L M N
Price per Kg (RM) Year 2003 Year 2006 1.60 2.00 4.00 q 0.80 1.20 r 1.60
Price index for the year 2006 based on the year 2003 p 120 150 80
K
N 80 120
100 L
M
Table 2 Diagram 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 based on the year 2003. (ii) Hence, calculate the corresponding cost of making this cake in the year 2003 if the cost 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 the year 2010. Find the expected composite index for the year 2010 based on the year 2003 [2 marks] 66
CHAPTER 11
INDEX NUMBER
FORM 4
3 A particular type of muffin is made by using four ingredients, P, Q, R and S. Table 3 shows the prices of the ingredients.
Ingredients P Q R S
Price per kilogram (RM) Year 2005 Year 2008 4.00 x 2.50 3.00 y z 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. The price per kilogram of ingredient R in the year 2008 is RM2.00 more than its corresponding price 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 year 2005 is 126 Calculate (i) the price of the muffin in the year 2005 if its corresponding price in the year 2008 is RM 6.30 (ii) the value of r if the quantities of ingredients P, Q, R and S used are in the ratio of 6:3 :r :2 [5 marks]
4 Table 4 shows the prices and the price indices of five components, P, Q, R, S and T, used to produce a type of toy car. Diagram 2 shows a pie chart which represents the relative quantity of components used.
Price (RM) for the year Component 2005 2007 P Q R S T
m 4.00 2.40 6.00 8.00
4.40 5.60 3.00 5.40 12.00 Table 4
Price index for the year 2007 based on the year 2005 110 140 125 n 150
R 144 36 S 72
Q
90
T
P
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 the corresponding cost in the year 2009. [4 marks] 67
CHAPTER 11
INDEX NUMBER
FORM 4
5 Table 5 shows the prices and the price indices of four ingredients P, Q, R and S, to make a dish. Diagram 3 shows a pie chart which represents the relative quantity of the ingredients used.
Ingredients P Q R S
Price (RM) per kg for the year 2006 2008 2.25 2.70 4.50 6.75 y 1.35 2 2.10
Price index for the Year 2008 based on the year 2006 x 150 112.5 105
P Q 25%
15% S 20%
R 40%
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 the year 2006 (ii) the price of a bowl of this dish in the year 2009 if its corresponding price in the year 2006 is RM 25 [4 marks]
68
CHAPTER 11
INDEX NUMBER
FORM 4
ANSWERS (PAPER 2) 1 (a) (i)
p
2004
100 RM 44.85 130
1
= RM 34.5 (ii)
100 RM 34.50 or 120 RM 44.85 I 2007 / 2000 RM 28.75 100
P
2000
1
P
2000
RM 28.75
= 156 (b) (i)
(ii)
(ii)
1 1
20x + 10000 = 12500
1
P
2004
x = 125
1
100 RM 40 125
1
RM 2.00 100 125 RM 1.60 120 q RM 4.00 RM 4.80 100 100 r RM 1.60 80 (125 60) (120 100) (150 120) (80 80) I 2006 / 2003 360 43900 = 360 p
(b) (i)
1
(125 40) (20 x ) (110 10) (130 30) 125 100
= RM 32
2 (a)
1
P
2003
P I
2010
1 1 1 1 1
100 RM 40 121.9
1
2010 / 2003
1
= 121.9
= RM 32.81 (c)
1
140 RM 40 RM 56 100 RM 56.00 100 170.7 RM 32.81
1 1 1
69
CHAPTER 11 3 (a)
INDEX NUMBER
x
125 RM 4 100
= RM 5.00 (b)
y 100 y2 140
(c) (i)
1
P
2005
100 RM 6.30 126
(125 6) (120 3) (140 r ) (110 2) 126 11 r
m
100 RM 4.40 110
= RM4.00 RM 5.40 100 90 RM 6 (110 90) (140 36) (125 144) (90 18) (150 72) 360 45360 = 360
n = (b)
= 126
P
2007
126 RM 60 100
= RM 75.60
P
1
z = RM 7.00
r = 4
(c)
1
1
1330 + 140r = 126r + 1386
4 (a)
1
y = RM 5.00
= RM 5.00 (ii)
FORM 4
2009
125 RM 75.60 100
= RM 94.50
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
70
CHAPTER 11
5 (a)
INDEX NUMBER
RM 2.70 100 120 RM 2.25 100 y = RM 1.35 112.5
x
= RM 1.20
= 121.5 (c) (i)
I
2009 / 2006
I
P
2009
1 1
= 145.8 (ii)
1
1
120 2008 / 2006 100 120 = 121.5 100
1
1
(120 15) (150 25) (112.5 40) (105 20) 100 12150 = 100
(b)
FORM 4
145.8 RM 25 100
= RM 36.45
1 1 1 1
71
CHAPTER 1
FUNCTIONS
FORM 4
PAPER 1 1.
Diagram 1 shows the relation between set A and set B. 2
p
4 6 8
q r Set A
Set B
Diagram 1 State (a) the range of the relation, (b) the type of the relation. [2 marks] 2.
R a, b, c
S b, d , f , h, j
3.
Based on the above information, the relation between R and S is defined by the set of ordered pairs (a, b), (a, d ), (b, f ), (b, h) . State (a) the images of a (b) the object of b [2 marks] Diagram 2 shows the linear function g . x
g(x)
0 2 4 6
-2 0 k 4
Diagram 2 (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 g : x x
xk , x 0 where k is a constant. x xk x
3 1 2
Diagram 3 Find the value of k. 1
CHAPTER 1
FUNCTIONS
FORM 4 [2 marks]
5.
Given the function g : x x 1 , find the value of x such that g ( x) 2 . [2 marks]
6.
Diagram 5 shows the graph of the function f ( x) 2 x 6 for domain 0 x 4 . f(x) 6
0
t
4
x
Diagram 5 State (a) the value of t, (b) the range of f(x) corresponding to the given domain. [3 marks] 7.
Given the function f ( x) 2 x 1 and g ( x) 3 kx , find (a) f (2) (b) the value of k such that gf (2) 7 [3 marks]
8.
The following information is about the function g and the composite function g 2 . g : x a bx , where a and b are constant and b > 0
g 2 : x 9x 8
Find the value of a and b. [3 marks]
9.
Given the function f ( x)
1 , x 0 and the composite function fg ( x) 4 x . 2x
Find (a) g (x) (b) the value of x when gf ( x) 2 [4 marks] 10.
The function h is defined as h( x)
7 , x 3 . 3 x
Find (a) h 1 ( x) (b) h 1 (2) 2
CHAPTER 1
FUNCTIONS
FORM 4 [3 marks]
11. Diagram 9 shows the function f maps x to y and the function g maps y to z. x
f
y
g
z
5 4 1
Determine (a) f 1 (1) (b) gf (5)
[2 marks] 12. The following information refers to the function f and g. f : x 5x 1 g : x x3
Find f
1
g ( x) . [3 marks]
13.
Given the function g : x 3 x h and g 1 : x kx
1 , where h and k are constants. Find 2
the value of h and of k. [3marks] 14.
Given the function h( x) 3 x 1 and g ( x)
x . Find 3
(a) h 1 (7) (b) gh 1 ( x) [4 marks] 15. Given the function f : x 3 x 2 and g : x 2 x 2 3 . Find (a) f 1 (4) (b) gf (x) [4 marks]
3
CHAPTER 1
FUNCTIONS
FORM 4
ANSWER (PAPER 1) 1 (a) (b) 2 (a) (b) 3 (a) (b) 4
4, 8
1
many-to-one
1
b , d
1
a
1
k2
1
g ( x) x 2
1
1 k 1 g 2 3 1 2 2
1
k 1
1
x 1 2
5
or ( x 1) 2
x 1
6 (a)
When f ( x) 0 ,
1
x 3
1
2x 6 0
1
x3
t 3
(b)
Range :
7 (a)
(a)
(b)
(b)
0 f ( x) 6
f ( 2) 5
1 1
g (5) 7 3 k (5) 7
1
k2
8
1
1
g 2 ( x) a b(a bx)
1
a ab b 2 x b2 9
and
a ab 8
b3
9 (a)
1 4x 2 g ( x) 1 g ( x) , x0 8x
a 4
1 1 1 1 4
CHAPTER 1 (b)
FUNCTIONS
FORM 4
1 2 1 8 x 2 1 x 8 7 x 3 y 7 h 1 ( x) 3 , x 0 x 1 h 1 (2) 2
1
11 (a)
5
1
(b)
4
1
10 (a)
(b)
12
x
y 1 5
( x 3) 1 5 x4 5 yh x 3 1 k 3 3 h 2 y 1 x 3 7 1 h 1 (7) 2 3 x 1 gh 1 ( x) 3 3 x 1 9 y2 x 3 1 f (4) 2 f 1 g ( x)
13.
14. (a)
(b)
15 (a)
(b)
1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1
gf ( x) 2(3 x 2) 2 3
1
18 x 2 24 x 5
1
5
CHAPTER 2
QUADRATIC EQUATIONS
FORM 4
PAPER 1 1. The quadratic equation 2x 2 px 2 4x 2 , where p is a constant, has no real roots. Find the range of the values of p. [3 marks] 2. The quadratic equation m x 2 4 x m 5 ,where m 0, has real and different roots. Find the range 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 four significant figures [3 marks] 5. The straight line y + x = 4 intersects the curve y = x 2 7 x w at two points . Find the range of values of w. [4 marks] 6. Form the quadratic equation which has the roots 7 and
2 . Give your answer in the form 3
ax 2 bx c 0 , where a , b and c are constants.
[2 marks] 7. Given the roots of the quadratic equation 4kx 2 hx 8 0 are equal. Express k in terms of h. [2 marks] 8. The straight line y = 9 4px is a tangent to the curve y = p 1 x 2 . Find the possible values of p. [5 marks] 9. The straight line y =2x1 does not intersect the curve y = x 2 2 x p. Find the range of values of p. [5 marks] 10. The quadratic equation 3 x 2 kx h 0 has roots 4 and 3. Find the values of k and h. [3 marks]
6
CHAPTER 2
QUADRATIC EQUATIONS
FORM 4
ANSWERS (PAPER 1) 1
2
3
4
5
(2 p ) x 4 x 8 0
1
4
1
2
2
4 2 p 8 0
48 32 p 0 3 p < 2 2 mx 4 x m 5 0 4 2 45 m m > 0 m 1m 4 > 0 m<1 , m > 4
1 1 1 1
n + 8x -x 2 = 3 x2 8x n 3 0 82 43 n 1 = 0 n = 13
1 1 1
3x 2 18x + 2 = 0
1
(18) (18)2 4(3)(2) 2(3)
1
x = 0.1132,
5.887
1
4 – x = x2 7x w
1 1 1 1
x 8x w 4 0 2
64 – 4(w – 4 ) > 0 w 20
x 7 3x 2 0
6
3x 2 19 x 14 0
7
k 8
1
1
1 1
h2 4 4k 8 0
1
h2 128
1
9 4 px p 1 x 2
1
p 1 x 2 4 px 9 0 2 4 p 4 9 p 1 0
1 1 7
CHAPTER 2
QUADRATIC EQUATIONS
p 3 4 p 3 0
10
1
3 4 2 2x – 1 = x -2x + p x2 4x p 1 0 16 – 4(1)(p + 1) < 0 12 – 4p <0 p > 3
1
(x + 4)(x – 3) = 0 3 x 2 3 x 36 0 k = -3 and h = -36
1 1 1
p = 3 and p = 9
FORM 4
1 1 1 1 1
8
CHAPTER 3
QUADRATIC FUNCTIONS
FORM 4
Paper 1 1. The quadratic function f(x) = a(x+p)2 + q, where a, p and q are constants, has a maximum value of 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 form f(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). y
x
5
O
1 y = 9
(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 f ( x) 5 2( x p ) 2 , where p is a constant. y A(1, q)
. 0
x
9
CHAPTER 3
QUADRATIC FUNCTIONS
FORM 4
The curve y f (x) 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.
1 2 x kx 6 . 2 A is the point of intersection of the quadratic graph and y-axis. The x-intercepts are 6 and 2.
Diagram below shows the curve of a quadratic function
f ( x)
y
x
6
O
2
A(0, r)
(p, q)
(a) State the value of r and of p. (b) The function can be expressed in the form f ( x ) of k. (c) Determine the range of values of x if f(x)< 6.
[2 marks]
1 ( x p ) 2 q , find the value of q and 2 [4 marks] [2 marks]
10
CHAPTER 3
QUADRATIC FUNCTIONS
FORM 4
Answers ( Paper 1) Q 1 (a) (b) (c) 2
3
4
5 (a) (b) 6 (a) (b) (c) 7
8
Solution a<0 p=3 q=5 x2 5x + 4 < 0 (x 1)(x 4) < 0 1<x<4 2x2 + 5x. 3 0 (2x 1)(x + 3) 0 3 x ½ f(x) = x2 6x + 32 32 + 5 = (x 3)2 4 m = 3 n = 4 x = 2 g(x) = (x + 2)2 9 p=1 q =5 x=1 x2 2x 15 0 (x + 3)(x 5) 0 x3, x5 2 x + (1 p)x + 4 =0 (1 p)2 4(1)(4) < 0 p2 2p 15 < 0 (p+3)(p 5) <0 3 < p < 5
Marks 1 1 1 1 1 1 1 1 1 1 1,1 1 1,1 1 1 1 1 1 1 1 1 1 Answer(Paper 2)
1 (a) (b)
r = 6, p = 2 (Expand or completing the square) 1 2 1 1 x px p 2 q x 2 kx 6 2 2 2 or k=p 1 2 2
1 1 1
p q 6
1 q 6 ( 2) 2 2
1 1 1
q = 8 k =2 (c) x2 + 4x<0 4
x(x + 4)<0 4 < x < 0
0
x
1
1
11
CHAPTER 4
SIMULTANEOUS EQUATIONS
FORM 4
PAPER 2 2 1. Solve the simultaneous equations y 2 x 8 and x - 3x – y = 2
[5 marks]
2.
Solve the simultaneous equations j – k = 2 and j2 + 2k = 8. Give your answers correct to three decimal 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. x 2y 8 2 x 4 y 2 37 Give your answers correct to three decimal places. [5 marks]
12
CHAPTER 4
SIMULTANEOUS EQUATIONS
FORM 4
ANSWERS (PAPER 2) 1.
2.
y – 2x = -8 y = - 8 + 2x ___________(1) x2 - 3x – y = 2___________(2) x2 - 3x – (-8 + 2x) = 2 (x – 3)(x – 2) = 0 x = 2, 3 y = - 8 + 2(2) y = - 8 + 2(3) y = -4, y=-2 k=j–2 _____________(1) 2 j + 2k = 8 _____________(2) j2 + 2(j – 2) = 8 j
4.
5.
1 1 1 1 1 1
b (b 2 4ac) 2a
2 2 2 4(1)(12) 2(1) j = 2.606 j = - 4.606 k = (2.606) – 2 k = (- 4.606) - 2 = 0.606 = - 6.606
1
x = 1 – 2y __________(1) y2 - 10 = 2x __________(2) y2 - 10 = 2(1 – 2y) (y + 6)(y – 2) = 0 y = 2, - 6 x = 1 – 2(2), x = 1 – 2(- 6) = -3, = 13
1
y = 1 – 2x __________(1) x2 + y2 + xy = 7 __________(2) x2 + (1 – 2x)2 + x(1 - 2x) = 7 ( x 1)( x 2) 0 x = - 1, 2 y= 1 – 2(- 1), y = 1 – 2(2) = 3, =-3
1
x = 8 + 2y __________(1) 2 2 x + 4y = 37 __________(2) (8 + 2y)2 + 4y2 = 37
1
3.
1
y
1
1 1 1 1
1 1 1 1
1
b b 2 4ac 2a
32 322 4(8)(27) 2(8) y ≈ - 1.209, -2.791 x = 8 + 2(- 1.209), x = 8 + 2(-2.791) = 5.582, = 2.418
1
1 1 1 13
CHAPTER 5
INDICES AND LOGARITHM
FORM 4
PAPER 1
1.
Solve the equation 3(53 x1 ) 36
[3marks]
2.
Solve the equation 323 x 47 x 4
[3marks]
3.
Solve the equation 32 x 1 5 x
[4marks]
4.
Solve the equation 165 x 49 x 6
[3marks]
5.
Solve the equation 273 x 3
6.
Solve the equation 22 x 1 13(2 x ) 24 0
[4marks]
7.
Given that log 3 K log 9 L 2 , express K in terms of L.
[4marks]
8.
9.
1
[3marks]
9 x 3
49 p Given that log p 3 r and log p 7 s , express log p in terms 27 of r and s. Given that log 5 2 q and log 5 9 p , express log 5 8.1 in terms of q and p.
[4marks]
[4marks)
10.
Solve the equation
3 log 3 (2 x 1) log 3 x .
[3marks]
11.
56 Given that log x 2 p and log x 7 q , express log x 2 in terms x of p and q.
[4marks]
12.
Given that log 3 mn 3 2 log 3 m log3 n express m in terms of n.
[4marks]
13.
Given that log 9 y log 3 18 , find the value of y
[3marks]
14.
81m Given that log 3 m v and log 3 n w , express log 9 in n terms of v and w.
[4marks]
15.
Solve the equation
log 3 x log 3 (4 x 1) 2 .
16.
Solve the equation
5 x 1 5x
4 . 25
[3marks] [3marks]
14
CHAPTER 5
INDICES AND LOGARITHM
FORM 4
ANSWERS (PAPER 1)
1.
53 x1 12
log 53 x1 log12
1
(3 x 1) log 5 log12
1
3x 1
log12 log 5
3x 1
1.0792 0.6990
3 x 1 1.54392 3 x 0.54392 x 0.1813
2.
2
5 3x
22
1 7 x 4
215 x 214 x8
1
15 x 14 x 8
3.
1
x 8
1
log 32 x 1 log 5x
1
2 x 1 log 3
1
x log 5
2 x log 3 log 3 x log 5 2 x log 3 x log 5 log 3
x 2 log 3 log 5 log 3 x
log 3 2 log 3 log 5
x 1.8691
1 1
15
CHAPTER 5
4.
INDICES AND LOGARITHM
2
4 5x
22
9 x 6
FORM 4 1
2 20 x 218 x12 20 x 18 x 12
1
x 6
5.
1
273 x 3 9 x 3
3
3 3 x 3
1 2
1 2 2 x 3
3
39 x 9 32 x 6
1
1 2
9x 9 x 3
1
10 x 6
x
6.
3 5
1
22 x 1 13(2 x ) 24 0 22 x.21 13(2 x ) 24 0
2
x 2
.2 13(2 x ) 24 0
1
Let u = 2x 2u 2 13u 24 0
2u 3 u 8 0 u
3 or 8 2
1
But 2x must be positive, so 2x = 8 2 x 23 x =3
1
16
CHAPTER 5
7.
INDICES AND LOGARITHM
log 3 K
log 3 L 2 log 3 9
log 3 K
log 3 L 2 2
FORM 4 1
2 log 3 K log3 L 4 log 3
8.
9.
10.
K2 4 L
1
K2 34 L
1
K 9 L
1
log p 7 2 log p p log p 33
1,1
2 log p 7 log p p 3log p 3
1
2 s 1 3r
1
log 5 8.1 =
log 5
81 10
1
=
log 5 81 log 5 10
1
=
log 5 92 log5 2 5
1
=
2 log 5 9 log 5 2 log 5 5
=
2 p q 1
1
3log 3 3 log3 (2 x 1) log 3 x log 3 33 log 3 (2 x 1) log 3 x
log 3 27 2 x 1 log 3 x
1
27 2 x 1 x
1
54 x x 27
x
27 53
1
17
CHAPTER 5
11.
12.
INDICES AND LOGARITHM
56 log x 2 = x
log x 56 log x x 2
1
=
log x (7 8) 2 log x x
=
log x 7 log x 8 2 log x x
=
log x 7 log x 23 2 log x x
=
log x 7 3log x 2 2 log x x
1
=
q 3p 2
1
1
log 3 mn log 3 33 log 3 m 2 log 3 n
1
27 m 2 n
1
log 3 mn log 3
27m 2 n 2 n m 27 log 3 y log 3 18 log 3 9 log 3 y log 3 18 log 3 9 mn
13.
FORM 4
1 1 1
log 3 y log 3 18 2
log 3 y 2 log 3 18
14.
log 3 y log 3 182
1
y 324
1
81m log 3 81m n = log 9 log3 9 n log 3 81 log 3 m log 3 n = 2 log 3 3 1 log3 34 log3 m log3 n 2 1 = 4 v w 2
=
1
1 1 1 18
CHAPTER 5
15.
INDICES AND LOGARITHM
FORM 4
log 3 x log3 (4 x 1) 2 0 log 3 x log3 (4 x 1) 2 log3 3 0 log 3
x(9) 0 4x 1
9x 30 4x 1
1 1
9x 4x 1 5x 1
x
16.
1 5
4 52 4 1 5 x 1 2 5 5 4 4 5x 2 5 5 4 5 5x 2 5 4
1
5 x.51 5x
1
5 x 5 1
1
x 1
1
19
CHAPTER 6
COORDINATE GEOMETRY
FORM 4
PAPER 1 1.
A point T divides the line segment joining the points A(1, -2) and B(-5, 4) internally in the ratio 2 : 1. Find the coordinates of point T. [2 marks]
2.
Diagram below shows a straight line PQ with the equation
x y + = 1. The point Q lies 3 5
on the x-axis and the point P lies on the y-axis. y P
0
Q
x
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 at the point B. y
B
A(0,6)
x
0 C
The equation of the straight line CB is y = 3x 4.
Find the coordinates of B. [3 marks]
5.
x y + = 1 has a y-intercept of 3 and is parallel to the straight line 14 m y + nx = 0. Determine the value of m and of n.
The straight line
20
CHAPTER 6
COORDINATE GEOMETRY
FORM 4 [3 marks]
6.
Diagram below shows a straight line passing through A(2, 0) and B (0, 6). y B(0, 6)
0
A(2, 0)
x
x y + = 1. a b [1 mark]
a)
Write down the equation of the straight line AB in the form
b)
A point P(x, y) moves such that PA = PB. Find the equation of the locus of P. [2 marks]
21
CHAPTER 6
COORDINATE GEOMETRY
FORM 4
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 C lies on the y-axis. y C
0
x 0
B (12, 0) D
A(0 , -3)
2.
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]
Solutions to this question by scale drawing will not be accepted.
In the diagram the straight line BC has an equation of 3y + x + 6 = 0 and is perpendicular to straight line AB at point B. y A(-6, 5)
B x
0 3y + x + 6 = 0 C (a) Find i) the equation of the straight line AB ii) the coordinates of B.
[5 marks]
(b) The straight line AB is extended to a point D such that AB : BD = 2 : 3. Find the coordinates 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] 22
CHAPTER 6
COORDINATE GEOMETRY
FORM 4
3. Solutions to this question by scale drawing will not be accepted. Diagram shows the triangle AOB where O is the origin. Point C lies on the straight line AB. y A(-2, 5)
C x 0 B(5, -1)
(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 point B. (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 the x-axis at point P and the y-axis at point Q. y x P
0 R
Q
y + 3x + 9 = 0
Point R lies on PQ such that PR : RQ = 1 : 2. Find (a) the coordinates of R, (b)
[3 marks]
the equation of the straight line that passes through R and perpendicular to PQ. [3 marks] 23
CHAPTER 6
5.
COORDINATE GEOMETRY
FORM 4
Solutions to this question by scale drawing will not be accepted. Diagram shows the triangle OPQ. Point S lies on the line PQ. P(3 , k)
y
S (5, 1) x
0 Q a)
A point W moves such that its distance from point S is always 2 Find the equation of the locus of W.
b)
1 units. 2 [3 marks]
It is given that point P and point Q lie on the locus of W. Calculate i) the value of k, ii) the coordinates of Q. [5 marks]
c)
2
Hence, find the area , in unit , of triangle OPQ. [2 marks]
24
CHAPTER 6
COORDINATE GEOMETRY
FORM 4
ANSWERS ( PAPER 1 ) 1.
2.
(1)(1) (5)(2) (2)(1) (4)(2) , ) 3 3 = T( -3 , 2 )
T(
Gradient of PQ , m1 = -
5 and the coordinates of Q (3 , 0) 3
Let the gradient of straight line perpendicular to PQ and passing through Q = m2 . Then m1 m2 = -1. 3 m2 = 5 y0 3 The equation of straight line is = x3 5 5y = 3(x – 3) 5y = 3x – 9 3.
4.
8x + 4hy – 6 = 0 3x + y = 16 4hy = -8x + 6 y = -3x + 16 8 6 y = x+ 4h 4h 2 3 y = - x + h 2h 2 Gradient , m2 = -3 Gradient , m1 = h Since the straight lines are perpendicular to each other , then m1 m2 = -1. 2 (- )(-3) = -1 h 6 = -h h = -6
2 1
1
1
1
Given
Gradient of CB , m1 = 3 Since AB is perpendicular to CB, then m1 m2 = 1 1 Gradient of AB, m2 = 3 1 The equation of AB is y=- x+6 3 B is the point of intersection. y = 3x 4 ……………(1) 1 y = x + 6 ……………(2) 3 1 3x 4 = x + 6 3
1
1
1
1
1
25
CHAPTER 6
COORDINATE GEOMETRY
FORM 4
10 x = 10 3 x=3
y = 3(3) 4 = 5 The coordinates of B are (3, 5). 5.
x y + 14 m
1
= 1
y-intercept = m = 3 1 x y 3 From + = 1, the gradient m1 = 14 3 14 From y = -nx , the gradient m2 = -n . Since the two straight lines are parallel , then m1 = m2 3 = -n 14 3 n = 14
6.
a) From the graph given, x- intercept = 2 and y-intercept = 6. x y The equation of AB is + =1. 2 6
1 1
1
b) Let the coordinates of P = (x , y) and since PA = PB ( x 2) 2 ( y 0) 2 = (x – 2)2 + y2 = x2 – 4x + 4 + y2 =
12y – 4x -32 = 0 3y – x - 8 = 0
( x 0) 2 ( y 6) 2 x2 + (y – 6)2 x2 + y2 – 12y + 36
1
1
26
CHAPTER 6
COORDINATE GEOMETRY
FORM 4
ANSWERS ( PAPER 2 ) 1
a) b)
c)
1
x y =1 12 3 AD 1 = DB 2 0(2) 12(1) 3(2) 0(1) D=( , ) 3 3 = ( 4 , -2 ) 3 Gradient of AB, mAB = -( ) 12 1 = 4
Given 2AD = DB , so
1 1
1
Since AB is perpendicular to CD, then mAB mCD = 1. Gradient of CD, mCD = - 4 Let, coordinates of C = (0 , h) , h ( 2) mCD = 04 h2 -4 = 4 16 = h + 2 h = 14 y-intercept of CD = 14 2
a) i)
1
1
Given equation of BC, 3y + x + 6 = 0 y =Gradient of BC = -
1 x–2 3
1 3
1
Since AB is perpendicular to BC , then mAB mBC = 1. Gradient of AB, mAB = 3 y 5 The equation of AB , =3 x ( 6) y – 5 = 3x + 18 y = 3x + 23 ii)
B is the point of intersection. Equation of AB , y = 3x + 23 Equation of BC , 3y + x + 6 = 0
1 1
…………. (1) ………….(2) 1
Substitute (1) into (2), 3(3x + 23) + x + 6 = 0
27
CHAPTER 6
COORDINATE GEOMETRY
FORM 4
9x + 69 + x + 6 = 0 x =Substitute value of x into (1),
y = 3(-
y =
15 2
15 ) + 23 2
1 2
The coordinates of B are ( -
15 1 , ) 2 2
1
b) Let D (h, k)
15 1 2h (18) 2k 15 , )= ( , ) 5 2 2 5 15 1 2k 15 2h (18) = , = 5 2 2 5 -75 = 4h – 36 5 = 4k + 30 39 25 h= k= 4 4 39 25 The coordinates of D are ( , ) 4 4 c) Given PA = 5
B( -
1
= 25
1
x2 + 12x + 36 + y2 -10y + 25 = 25 x2 + y2 + 12x -10y + 36 = 0
1
( x + 6)2 + ( y – 5)2
)
a)
b)
1
= 5
( x (6)) 2 ( y 5) 2
3.
1
Area
C= (
=
1 2
0 5 2 0 0 1 5 0 1 = (25) (2) 2 23 = unit2 2
3(5) 2(2) 3(1) 2(5) , 5 5 11 7 = ( , ) 5 5
1
1
1 1
c) i) Since PA = 2PB ( x 2) 2 ( y 5) 2 = 2 ( x 5) 2 ( y 1) 2
x2 + 4x + 4 + y2 10y + 25 = 4 (x2 10x + 25 + y2 +2y + 1)
1 1
28
CHAPTER 6
COORDINATE GEOMETRY
x2 + y2 + 4x 10y + 29 = 4x2 + 4y2 40x + 8y + 104 3x2 + 3y2 44x + 18y + 75 = 0 (ii)
When it intersects the y-axis, x = 0. 3y2 +1 8y + 75 = 0 Use b2 4ac = (18)2 4(3)(75) = 576 b2 4ac < 0
a)
y + 3x + 9 = 0 When y = 0,
0 + 3x + 9 x
1
1
= 0 = –3
P(–3, 0) When x = 0,
1 1
It does not cut the y-axis since there is no real root. 4.
FORM 4
1 y+0+9 y
= 0 = –9
Q(0, –9) 1(0) 2(3) 1(9) 2(0) , ) 3 3 = (-2 , -3 )
R(x, y) = (
b)
y + 3x + 9 = 0 y = -3x - 9
Gradient of PQ , m1 = –3 Since PQ is perpendicular to the straight line, then m1 m2 = 1 1 Thus, m2 3 The equation of straight line that passes through R(-2, -3) and perpendicular to PQ is y3 1 = x2 3 3y = x - 7 5.
a) Equation of the locus of W, 5 ( x 5) 2 ( y 1) 2 = 2 5 (x – 5)2 + ( y – 1)2 = ( )2 2
1 1
1
1
1 1
1
25 4 4 x2 + 4y2 – 40x - 8y + 79 = 0
1
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
x2 -10x +25 + y2 – 2y + 1 =
b) i)
29
CHAPTER 6
ii)
c)
COORDINATE GEOMETRY
4k2 - 8k -5 = 0 (2k + 1)(2k – 5) = 0 1 5 k=- , k= 2 2 5 Since k > 0, k = 2 Since S is the centre of the locus of W, then S is the mid-point of PQ. 5 y x3 2 ) S(5 , 1) = ( , 2 2 5 y x3 2 5= , 1= 2 2 1 x=7 , y =2 1 Hence, the coordinates of Q are ( 7 , ). 2 0 7 3 0 1 Area of triangle OPQ = 1 5 0 2 0 2 2 1 5 3 = [ (7)( ) – (- ) ] 2 2 2 19 = unit2 2
FORM 4
1
1
1
1
1
1
30
CHAPTER 7
STATISTICS
FORM 4
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) Number of fish
20 - 24 5
25 - 29 3
30 - 34 9 Table 3
35 - 39 7
40 -44 6
Find the mean length of the fish.
[3 marks]
4. Table 4 shows the distribution of marks obtained by 56 students in a test. Marks Number of Students
20 - 29 4
30 - 39 20
40 - 49 16
50 - 59 10
60 – 69 6
Table 4 Find, (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 ascending order 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 the squares 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. 31
CHAPTER 7
STATISTICS
FORM 4 [3 marks]
8. The mean of eight numbers is p. The sum of the squares of the numbers is 200 and the standard 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 the squares 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 60 students in a test. 20
Num ber of students
18 16 14 12 10 8 6 4 2 0
0.5 1 10.5
20.5
30.5
40.5
50.5
60.5
Marks
Diagram 2 (a) Without using an ogive, calculate the median mark. [3 marks] 32
CHAPTER 7
STATISTICS
FORM 4
(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 20 - 29 30 – 39 40 - 49 50 - 59 60 - 69 70 - 79
Number of students 1 2 8 12 k 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 the vertical 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 a Mathematics Quiz. Score Number of students
< 20 6
<30 16
< 40 28
< 50 36
<60 40
40 – 49
50 - 59
Table 4 (a) Based on Table 4, copy and complete Table 4. Score Number of students
10 - 19
20 - 29
30 - 39
(b) Without drawing an ogive, find the interquartile range of the distribution. [5 marks]
33
CHAPTER 7
STATISTICS
FORM 4
ANSWER PAPER 1 No. 1(a)
Solution
Marks
∑x = 88
(b)
1
x + k 9
1 = 10
k=2
1
∑x = 57
1
2. (a)
(b)
1
x + p
=9
7
p=6
1 (22 x 5) + ( 27 x 3) + (37 x 7) + (42 x 6 )
3.
x =
1,1
30
990 =
4.(a) (b)
30
= 33 = 34.5
1 1 (24.5 x 4 ) + ( 34.5 x 20 ) + ( 44.5 x 16 ) + ( 54.5 x 10 ) + ( 64.5 x 6)
x =
1,1
56
2432 =
56
= 43.43
1
34
CHAPTER 7
STATISTICS
4n + 14
5.(a)
1
= 13
2
FORM 4
n* = 3
1
3 + 19 + 2n* + 1 + 4n* + 14 + 17
(b)
=q
6
q = 12 6.(a)
1 1
12
1
(b) 960 =
6
1
- (12) 2
=4
1 22 + 52 + q2
7.
8.
14 =
3
-
2 +5+ q 3
2
1
(q + 4 )(q – 11) = 0
1
q = 11
1
x2 = 200 or x = 8
1 1
200 p 2 16m 2 8 p=
25 - 16m2
1
9. x = 14.6
1 1219
Standard deviation =
= 5.5353
5
- (14.6) 2
1
1 35
CHAPTER 7
STATISTICS
10.(a)
2200 Variance =
8
= 50 (b)
= 13.8
2
1
1
120 + 6 + 12 x =
- (15)
FORM 4
1
10
1
36
CHAPTER 7
STATISTICS
FORM 4
PAPER 2 No. 1(a)
Solution
Marks
Mean = 16
1 3212
Variance =
11
1
- (16) 2
= 36 (b)(i)
1
176 + k 18 =
1
12
k = 40
1
(ii)
3212 + 402
Standard deviation =
2(a).
12
1
- (18)2
= 8.775 30.5 or ½ (60) or 30 or median class 20.5 – 30.5 1 (60) 20 10 20.5 + 2 20 = 25.5
1 1
1 1
(b) 1
x = 26.67 fx 2 = 54285
1 54285
Standard deviation =
3.(a)
60
1600
-
2
1
60
= 13.92
1
49.5 or ½ (24 + k) or 11
1
37
CHAPTER 7
STATISTICS
52 = 49.5 +
1 2
(24 + k) - 11 12
FORM 4
1
10
k=4
1
(b) Frequency
4
14 12
10
8 6 4
2
0 19.5
29.5
39.5
49.5
59.5
69.5
Mode score = 53
(c) 4.(a)
79.5 Score
= 57
1 Score
10 - 19
20 - 29
Number of 6 10 students (b) 19.5 or ¼ (40) or 6 39.5 or ¾ (40) or 28 Interquartile range = 42* - 23.5* = 18.5
30 – 39
40 - 49
50 - 59
12
8
4
1 1 1 1 1
38
CHAPTER 8
CIRCULAR MEASURE
FORM 4
PAPER 1 1.
Diagram 1 shows a sector AOB with centre O . A
B
O DIAGRAM 1
The length of the arc AB is 7.5 cm and the perimeter of the sector AOB is 25 cm. Find the value of , in radian. [ 3 marks ]
2.
Diagram 2 shows a circle with centre O . R
O
0.35 rad S
DIAGRAM 2 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 ]
39
CHAPTER 8
3.
CIRCULAR MEASURE
FORM 4
Diagram 3 shows a circle with centre O . P
O
R
DIAGRAM 3 The length of the minor arc is 15 cm and the angle of the major sector POR is 280o . Using = 3.142 , find (a) (b)
4.
the value of , in radians. ( Give your answer correct to four significant figures ) the length, in cm, of the radius of the circle .
[ 3 marks ]
Diagram 4 shows sector OPQ with centre O and sector PXY with centre P . P
Y X
Q
O
DIAGRAM 4 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 ]
40
CHAPTER 8
5.
CIRCULAR MEASURE
FORM 4
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]
Q
P
R
1.97 rad
S
0
DIAGRAM
6.
5
Diagram 6 shows a circle with centre O and radius 12 cm. Given that A, B and C are points such 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]
O A
C
B
DIAGRAM 6
41
CHAPTER 8
CIRCULAR MEASURE
FORM 4
PAPER 2 1.
Diagram 1 shows the sectors AOB, centre O with radius 15 cm. The point C on OA is such that OC : OA = 3 : 5 . A C B
O DIAGRAM 1 Calculate
2.
(a)
the value of , in radian,
[ 3 marks ]
(b)
the area of the shaded region, in cm2 .
[ 4 marks ]
Diagram 2 shows a circle PQRT , centre O and radius 10 cm. AQB is a tangent to the circle at Q. The straight lines, AO and BO, intersect the circle at P and R respectively. OPQR is a rhombus. ACB is an arc of a circle, centre O. C
A
B
Q P
rad
R
O T DIAGRAM 2
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 ] 42
CHAPTER 8
3.
CIRCULAR MEASURE
FORM 4
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 = rad . 6 A P
O
rad 6 Q
B
DIAGRAM 3 It is given that OP: OA= 3 : 5 . ( Using = 3.142 )
4.
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 ]
In Diagram 4, PBQ is a semicircle with centre O and has a radius of 10 m. RAQ is a sector of a circle with centre A and has a radius of 16 m . R
B
P
A
O
Q
DIAGRAM 4 It is given that AB = 10 m and BOQ = 1.876 radians. [ Use = 3.142 ]
43
CHAPTER 8
5.
CIRCULAR MEASURE
FORM 4
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 ]
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
44
CHAPTER 8 6.
CIRCULAR MEASURE
FORM 4
Diagram 6 shows two circles. The larger circle has centre P and radius 24 cm. The smaller circle has centre Q and radius 16 cm. The circles touch at point R. The straight line XY is a common tangent to the circles at point X and Y. [Use = 3.142] Given that XPQ = radians, (a) show that = 1.37 ( to two decimals places), (b) calculate the length, in cm, of the minor arc YR, (c) calculate the area, in cm2, of the coloured region
[2 marks] [3 marks] [5 marks]
DIAGRAM 6
45
CHAPTER 8
CIRCULAR MEASURE
FORM 4
ANSWERS (PAPER 1)
1
25 7.5 2 7.5 = 8.75 0.8571
2
2 0.35 45 r (2 0.35) r = 7.583
1 1 1
1.396 15 r 1.396 r = 10.74
1
0.7 rad 1 1 2 (20) 2 (0.7) or (8) (1.1) 2 2 1 1 (20) 2 (0.7) (8) 2 (1.1) 2 2 104.8
1
23.64 cm 1 1 (12) 2 1.97 or (6) 2 sin 1.97 rad 2 2 1 1 (12) 2 1.97 (6) 2 sin 1.97 rad 2 2 125.26
1
OA
3a 3b
4a
4b
5a
5b
6a
6b
rad 3 1 1 (12) 2 1.047 or ( 12 2 6 2 )6 2 2 1 1 (12) 2 1.047 ( 12 2 6 2 )6 2 2 44.21 1.047 rad or
1 1 1
1 1
1 1 1
1 1 1 1 1 1 1
46
CHAPTER 8
CIRCULAR MEASURE
FORM 4
ANSWERS (PAPER 2)
1a
1b
2a
2b
2c
3a 3b
3c
4a 4b 4c
9 15 0.9273 rad. BC = 12 1 2 1 15 (0.9273) (12)(9) 2 2 50.32 cos
120O 2 rad 3 10 cos 60 o OB OB = 20 cm
2 arc ACB = 20 3 41.89 1 2 1 (20) 2 (20) 2 sin 120 o 2 3 2 245.72 PA = 6 cm
9sin30o + (15 9cos30o ) + 15 ( ) + 6 6 25.56 PQ = 4.5 cm , OQ = 9cos 30o 1 1 (15) 2 (4.5)(9 cos 30 o ) 2 6 2 41.38 1 (10) 2 1.876 2 93.8 10(1.876) + 16 ( 1.876) + 6 45.016 1 1 1 (16) 2 ( 1.876) (10) 2 1.876 (6)( 91) 2 2 2 39.63
1 2 3 1 1 1 1 1 1 1 3 1 1 4 1 3 1 1 1 3 1 3 1
47
CHAPTER 8
5a
5b
6a
6b
6c
CIRCULAR MEASURE
20 sin 30 o OA OA = 40 cm PA = 60 cm arc PQ = 60 3 62.84
AB = 40 2 20 2 1 1 2 1 2[ (60) 2 (20) 2 ( 40 2 20 2 )20] 2 6 2 3 2 354.513 cos =
8 40
1.37 rad YQP = 1.369 arc YR = 16( 1.369 ) 28.37
XY 40 2 8 2 = 39.19 1 1 1 (16 24)39.19 (1.369)(24) 2 (16) 2 ( - 1.369) 2 2 2 162.58
FORM 4
1 1 1 1 1 1 3 1 1 1 1 1 1 1
3 1
48
CHAPTER 1
PROGRESSIONS
FORM 5
PAPER 1 1.
Three consecutive terms of an arithmetic progression are 2 p 2, 9, 3 p . Find the common difference of the progression. [3 marks]
2.
The first three terms of an arithmetic progression are 1, x, 7, ...... 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 2, 1, 4....... , 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]
72
CHAPTER 1
PROGRESSIONS
FORM 5
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 total collection 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 25 stamps 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 base and height of each subsequent rectangle are half of the measurements of its previous one.
y cm
x cm Diagram 2 (a) Show that the areas of the rectangles form a geometric progression and state the common ratio. [3 marks] (b) Given that x 20 cm and y 80 cm, 9 (i) determine which rectangle has an area of 1 cm2 16 (ii) find the sum to infinity of the areas, in cm2, of the rectangles. [5 marks]
73
CHAPTER 1
3.
PROGRESSIONS
FORM 5
Diagram 3 shows part of an arrangement of circle of equal size.
10 cm Diagram 3 The number of circles in the lowest row is 80. For each of the other rows, the number of circles is 3 less than in the row below. The diameter of each circle is 10 cm . The number of circles 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]
74
CHAPTER 1
PROGRESSIONS
FORM 5
ANSWER (PAPER 1) 9 ( 2 p 2) 3 p 9
1
1
p4
1
common difference 9 6 3
2 (a)
1
x (1) 7 x x3
1
common difference 4
1
(b) the sum of the first 10 terms after the 3rd term: 13 S13 S 3 2(1) 12(4) 9 2
1
290
1
a (n 1)d a (n 2)d
3
a (n 3)d
84 n 12
1 1
Three consecutive terms: T12 31, T11 28, T10 25 4 (a) (b)
5
15 3 5 5(3 n 1) 5465 3 1 r
1 1
n7
1
12 1 48 4
1
48 1 1 4 64
S
1 1
27(r ) 41 1
1
1 3
1
r
(b)
1
37 3 n
r
6 (a)
1
S
27 1 1 3
1
75
CHAPTER 1 20
7.
PROGRESSIONS
FORM 5 1
1 4
0.121212…… = 0.12 + 0.0012 + 0.000012 + …….. a 0.12
r
1 1
0.0012 0.01 0.12
S
0.12 1 0.01
1 1
4 33
ANSWER (PAPER 2) 1 (a)
p (7 1)q 220 …….eq(1)
1
12 2 p (12 1)q 2520 ……..eq(2) 2
1
eq (1) x 2
: 2 p 12q 440 ……. eq(3) : 2 p 11q 420 ……. eq(4)
From eq (2)
(3) – (4) :
(b)
2 (a)
1
either
q 20
1
p 100
1
Tn 100 (n 1)20 80 20n Borhan : Tn 60 (n 1)25 35 25n Ali :
80 20n 35 25n
1
n9
1
Area : xy 1 r1 4 xy 4
xy,
xy xy , , .............. 4 16
xy 1 r2 16 xy 4 4
1
both
1
76
CHAPTER 1
(b)(i)
3 (a)
1
a 20(80) 1600
1
1 4
1 1600 4
n 1
1 4
n 1
S
1
9 16
1 4 n6
1 5
1 1
1600 1 1 4
2133
1 3 80, 77, 74, …………, 5 Tn 5 , a 80, d 3
1
80 (n 1)(3) 5
1
n 26
1
Height of the arrangement = 26 10 260 cm (b)
FORM 5
r1 r2 , the areas of the rectangles form a geometric progression with 1 the common ratio = 4
r
(ii)
PROGRESSIONS
S 26
26 2(80) (26 1)(3) 2
1 1
1105
1
The total length of circumference of circles = 1105 2 (5) 11050 cm
1
77
CHAPTER 2
LINEAR LAW
FORM 5
PAPER 1 xy
1.
n •
• ( 8, k ) x
0 Diagram 1 Diagram 1 shows part of a straight line graph drawn to represent y and n.
2.
10 1. Find the values of k x [4 marks]
log10 y ( 3,9 ) •
• ( 7,1) log10 x
0 Diagram 2
Diagram 2 shows part of a straight line graph drawn to represent y = 10kxn, where k and n are constants. Find the values of k and n. [4 marks]
78
CHAPTER 2
LINEAR LAW
FORM 5
y x
3.
• 12, 7
1 x
0 • (2 , - 3 ) Diagram 3
Diagram 3 shows that the variables x and y are related in such away that when
y is plotted x
1 , a straight line that passes through the points (12 , 7 ) and (2 , - 3 ) is obtained . x Express y in terms of x. [3 marks]
against
4.
log10 y
3 •
x
0
• ( 5 , -7 ) Diagram 4 Diagram 4 shows part of the graph of log10 y against x. The variables x and y are related by the a equation y x where a and b are constants. Find the values of a and b. b [4 marks ]
79
CHAPTER 2
LINEAR LAW
FORM 5
log10 y 5. 2
0
x
4 Diagram 5
Diagram 5 shows part of the graph of log10 y against x. The variables x and y are related by the bx
equation y = a (10 ) where a and b are constants. Find the values of a and b.
[3 marks]
6. log10 y
• 2
• -4
0
log10 x
Diagram 6
Diagram 6 shows part of a straight line graph when log10 y against log10 x is plotted. Express y in terms of x. [4 marks]
80
CHAPTER 2
LINEAR LAW
FORM 5
7. The variable x and y are related by equation y pk 3 x , where k and p are constant. Diagram 7 shows the straight line obtained by plotting log10 y against x . log10 y ( 0, 8 )
(2,2) x
O Diagram 7
a) Reduce the equation y pk 3 x to linear form Y = mX + c. b) Find the value of, i) ii)
log10 p , k.
[4 marks]
81
CHAPTER 2
LINEAR LAW
FORM 5
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 n and y are related by the equation y 2rx 2 x, where r and n are constants. r x y
2 8
3 13.2
4 20
5 27.5
6 36.6
7 45.5
Table 1 (a).
(b).
y against x , using a scale of 2 cm to 1 unit on both axes. x Hence, draw the line of best fit.
[5 marks]
Use your graph in (a), to find the value of (i). n, (ii). r, (iii). y when x = 1.5.
[5 marks]
Plot
2. Use graph paper to answer this question. Table 2 shows the values of two variables , x and y obtained from an experiment. Variables x h and y are related by the equation y kx, where h and k are constants. x x y
1 5.1
2 6.9
3 9.7
4 12.5
5 15.4
6 18.3
Table 2 (a).
(b).
Plot xy against x 2 , using a scale of 2 cm to 5 units on the x 2 -axis and 2 cm to 10 units on the xy-axis. Hence, draw the line of best fit.
[5 marks]
Use your graph in (a), to find the value of (i). h, (ii). k, (iii). y when x = 2.5.
[5 marks]
82
CHAPTER 2
LINEAR LAW
FORM 5
3. Use graph paper to answer this question. Table 3 shows the values of two variables , x and y obtained from an experiment. Variables x n and y are related by the equation y w , where n and w are constants. x x y
3 103
4 87
5 76
6 68
7 62
8 57.4
Table 3 (a).
Plot log10 y against log10 x , using a scale of 2 cm to 0.1 unit on the log10 x -axis and 2 cm to 0.2 units on the log10 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]
4. Use graph paper to answer this question. Table 4 shows the values of two variables, x and y obtained from an experiment. Variables x b and y are related by the equation y a x , where a and b are constants. x x y
0.2 12.40
0.4 8.50
0.6 6.74
0.8 5.66
1.2 4.90
1.4 3.87
Table 4 (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]
83
CHAPTER 2
LINEAR LAW
FORM 5
5. Use graph paper to answer this question. Table 5 shows the values of two variables, x and y obtained from an experiment. Variables x and y are related by the equation y pm x , where m and p are constants. x y
1.5 2.51
3.0 3.24
4.5 4.37
6.0 5.75
7.5 7.76
9.0 10.00
Table 5 (a).
Plot log10 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 log10 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]
6. Use graph paper to answer this question. Table 6 shows the values of two variables, x and y obtained from an experiment. Variables x and y are related by the equation y hk x , where h and k are constants. x y
3 10.2
4 16.4
5 26.2
6 42
7 67.1
8 107.4
Table 6 (a).
Plot log10 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 log10 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]
84
CHAPTER 2
LINEAR LAW
FORM 5
ANSWER
No. 1.
PAPER 1 Solution xy x 10 Calculate gradient from graph n 10 k 2
Marks 1 1 1 1
log10 y 2 log10 x 15 Calculate gradient from graph n 2 k 15
1
c = -5 y 1 5 x x y 1 5x
1 1
4.
log10 y = - log10 b + log10 a Gradient = -2 b = 100 a = 1000
1 1 1 1
5.
log10 y = bx + log10 a or log10 y = bxlog10 10 + log10 a a = 100 b=-½
1 1 1
2.
3.
6. log10 x1/2
Gradient = ½ log10 y = ½ log10 x + 2 or log10 102 or log10 x1/2.102 y 100 x
1 1 1
1
1 1 1 1
7(a)
3 log10 k 3
1
(b)(i)
log10 y (3log10 k ) x log10 p
1
log10 p 8 k 10
1
(ii)
1 85
CHAPTER 2
LINEAR LAW
FORM 5
PAPER 2 No. 1(a)
Solution x y x
(b) (i) (ii) (iii)
3 4.4
4 5
5 5.5
6 6.1
7 6.5
Uniform scale for x-axis and y-axis 6 points plotted correctly Draw the line of best fit Reduce non-linear to linear equation y n 2rx x r Calculate gradient from graph n = 0.77 r = 0.275 y From graph , 3.6 x y = 5.4
(b) (i) (ii) (iii)
2.(a)
2 4
x2 xy
1 4 5.1 13.8
9 29.1
16 50
25 77
Uniform scale for x-axis and y-axis 6 points plotted correctly Draw the line of best fit Reduce non-linear to linear equation xy kx 2 h Calculate gradient from graph h 2 k 3 From graph, xy = 20 y= 8
Marks 1
1 1 1 1
1 1 1 1 1 36 109.8
1
1 1 1 1 1 1 1 1 1
86
CHAPTER 2
3.(a)
LINEAR LAW
log10 x
0.48 0.60
0.70
0.78
0.85
0.90
log10 y
2.01 1.94
1.88
1.83
1.79
1.76
(b) (i) (ii) (iii)
4.(a)
(b) (i) (ii) (iii)
FORM 5
x y x
1
Uniform scale for x-axis and y-axis 6 points plotted correctly Draw the line of best fit Reduce non-linear to linear equation log10 y w log10 x log10 n Calculate gradient from graph n 102.3 199.526 w 0.6 From graph, log10 y 2.12
1 1 1 1
y 102.12 131.825
1
0.2 0.4 5.55 5.38
0.6 5.22
0.8 5.06
1.2 4.9
Uniform scale for x-axis and y-axis 6 points plotted correctly Draw the line of best fit Reduce non-linear to linear equation y x ax b Calculate gradient from graph a - 0.8 b 5.7 From graph y x 4.9 y = 5.165
1 1 1 1
1.4 4.74
1
1 1 1 1 1 1 1 1 1
87
CHAPTER 2
5.(a)
(i) (ii) (iii)
LINEAR LAW
x log10 y
1.5 3.0 0.4 0.51
6 0.76
7.5 0.89
1
9 1
Uniform scale for x-axis and y-axis 6 points plotted correctly Draw the line of best fit Reduce non-linear to linear equation log10 y x log10 m log10 p Calculate gradient from graph m 100.084 or m 1.2134 p 100.26 or1.8197 log10 4.8 0.68 x= 5
1 1 1 1
All values of log10 y are correct
1
6.(a) x log10 y
(i) (ii) (iii)
4.5 0.64
FORM 5
3 4 1.0 1.21
5 1.42
6 1.62
7 1.82
Uniform scale for x-axis and y-axis 6 points plotted correctly Draw the line of best fit Reduce non-linear to linear equation log10 y x log10 k log10 h Calculate gradient from graph k 100.2 or k 1.588 h 100.425 or h 2.66 log10 35.6 1.55 x = 5.7
1 1 1 1 1
8 2.03 1 1 1 1 1 1 1 1 1
88
Chapter 3
Intergration
Form 5
PAPER 1 1. Given that ( 5 x 3 2 ) dx kx 4 2 x c , where k and c are constants. Find (a) the value of k (b) the value of c if ( 5 x 3 2 ) d x 3 0 when x=2. [3 marks] 4
2. Evaluate
1
( x 2)
3
[3 marks]
dx
3
5
3. Given that 2x 3dx 6, where p < 5 , find the possible values of p. p
4. Given that
4
4
2
2
[4 marks] [4 marks]
f ( x)dx 8 and kx 2 f ( x)dx 8 , find the value of k.
5.
y y=x(a x)
0
a
x
Given that the area of the shaded region under the curve y=x(a x) is of a.
4
1 2
unit2, find the value [ 4 marks]
6. y y = f(x)
0
4
x
Diagram above shows the curve y = f(x). Given that the area of the shaded region is 5 unit2, find 4
the value of 2 f ( x) 2dx.
[3 marks]
0
89
Chapter 3
Intergration
Form 5
PAPER 2
1. A curve is such that dy 1 1 2 . Given that the curve passes through the point dx
2x
equation of the curve.
1 1, , 2
find the
[ 4 marks] y A(0, 1) y
0
1 4x 1
x 2 1 which passes through A(0,1). A region is 4x 1
(b) Diagram above shows part of the curve y
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 y 1 ( x 4) 2 which passes through A(2, 2). 2
y A(2, 2)
0
x
(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]
90
Chapter 3
Intergration
Form 5
3. In the diagram, the straight line PQ is a tangent to the curve x 1 y 2 1 at Q(3, 2) 2
y
Q(3, 2) P(0,k)
x 0
(a) Show that the value of k is
1, 2
[ 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
dy 2x 6 . dx
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 the curve and y-axis. y
B
A
x
O
(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, yaxis and the line x=3 is rotated through 360o about the x-axis. [3 marks]
91
Chapter 3
Intergration
Form 5
5. Diagram shows the straight line y=3x intersecting the curve y = 4 x2 at point P. y P
R
y=3x
y = 4 x 2 x
0
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, and y-axis is revolve 360o about the y-axis. [3 marks]
92
Chapter 3
Intergration
Form 5
PAPER 1 (Answer)
Q 1(a) (b)
2
Solution 5 k= 4 5 ( 2 )4 + 2( 2 ) + c = 30 4 c=6
4
( x 2) -3 dx
3
Marks 1 1 1 1
4
( x 2) 2 = 2 3 1 1 1 = 2 4 1 3 = 8
3
4
1 1
5
x 2 3x = 6 p 2 [5 -3(5) ] – (p2-3p) = 6 (p + 1) (p – 4 ) =0 p = 1, 4
1 1 1 1 1,1
4
4 x2 k 2 f ( x)dx 8 2 2 2 6k-2(8)=8 k=4
5
a
0
ax x 2 dx
9 2
a
ax 2 x3 9 3 0 2 2 a 3 a3 9 2 3 2 a=3
6
4
4
0
0 4 0
1 1 1 1 1 1
2 f ( x)dx 2dx = 2 5 + [2x] = 18
1 1
93
Chapter 3
Intergration
Form 5
PAPER 2 (Answer) Q
Solution
1 (a)
y = 1
(b)
Marks 1 dx 2 x2
1 = 1 x 2 dx 2 1 =x c 2x 1 Substitute 1, , c 1 2 1 y=x 1 2x 2 1 Area= dx correct limit 0 4x 1 2
2 (a)
(b)(i)
2 = 4 x 1 4 0 1 = ( 9 1 2 =1 dy = x 4 dx dy x = 2, = 24 = 2 dx y 2 = 2(x2) y = 2x+6 41 2 2 2 ( x 4) dx 4
(ii)
( x 4)3 = 2(3) 2 4 = unit 2 3 41 ( x 4) 4 dx 2 4
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 ( x 4)5 = 4 5 2
1
8 = unit3 5
1
4
3 (a)
1
1
y = (2x2) 2 dy 1 = dx 2x 2
1
94
Chapter 3
Intergration
1
dy 1 1 dx 2(3) 2 2 2k 1 1 ,k 3 0 2 2 2 1 1 3 2 0 ( 2 y 1)dy 2 2 3
x=3,
(b)
1 1, 1
1
2
1 9 = y3 y 6 0 4 10 9 = 3 4 13 = unit 2 12 (c)
Form 5
1 1
3
(2 x 2)dx 1
= x 2 2 x
1
3
1
4 (a)
(b)
=4 unit3 y = (2 x 6)dx
1 1
y = x2-6x + c Substitute x = 3, y = 1 in the equation, c = 10 y = x2-6x + 10 Equation of AB is y = 3x + 10
3
0
(3 x 10) ( x 2 6 x 10)dx
1 1 1
3
= 3x x 2 dx 0
1
3
3x 2 x3 = 3 0 2 9 = unit2 2
(c)
1 1
3
Volume= (3 x 10) 2 dx 0
3
= (9 x 2 60 x 100)dx 0
3
= 3x 3 30 x 2 100 x
1
0
5 (a)
(b)
=111 unit3 x2 +3x 4 = 0 (x1) (x+4)=0 x= 1, y =3(1)=3 P(1, 3) 1 Area of triangle= (1)(3) = 1.5 unit2 2 2
2
1
x3 4 x dx 4 x 3 1
1 1 1 1 1 1
2
95
Chapter 3
Intergration =1
2 unit2 3
Form 5 1
1 1 Area of R = 3 6
(c)
1
4
(4 y )dy 0
4
y2 = 4 y 2 0 =8 unit3
1 1
96
CHAPTER 4
VECTORS
FORM 5
PAPER 1
1. Diagram below shows two vectors, OP and QO
Q(-8,4) P(5,3) O Express x (a) OP in the form , y (b) QO in the form x i + y j [2 marks]
p = 2a + 3b q = 4a – b r = ha + ( h – k ) b, where h and k are constants
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
E A
B
Given that AB = 6p , AD = 4q and DE = 2EB, express, in terms of p and q (a) BD (b) EC [4 marks]
97
CHAPTER 4
VECTORS
FORM 5
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 6 A 4 2
0
2
4
6
8
10
12
x
x y (b) Find the unit vector in the direction of OA (a) Express OA in the form
[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] 98
CHAPTER 4
VECTORS
FORM 5
8. Diagram below shows two vectors, OA and AB .
y A(4,3)
O
-5
x
B
Express (a)
x in the form y AB in the form xi + yj
OA
(b)
[2 marks] 9. The points P, Q and R are collinear. It is given that PQ = 4a – 2b and
QR 3a (1 k )b , 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 a 6i 7 j and b pi 2 j , find the possible value (or values) of p for following cases:a) a and b are parallel b) a b [5 marks]
99
CHAPTER 4
VECTORS
FORM 5
PAPER 2
1.
5 2 k Give that AB , OB and CD , find 7 3 5 (a) the coordinates of A, (b) the unit vector in the direction of OA , (c) the value of k, if CD is parallel to AB
2.
[2 marks] [2 marks] [2 marks]
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, OP 6 x and OA 2 y. A Q R
O
P
(a) Express in terms of x and/or y: (i) AP (ii) OQ (b)
B
[4 marks]
(i) Given that AR h AP, state AR in terms of h, x and y. (ii) Given that RQ k OQ, 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]
100
CHAPTER 4 3.
VECTORS
FORM 5
In diagram below, ABCD is a quadrilateral. AED and EFC are straight lines. D
E
F
A
C
B
It is given that AB 20x, AE 8y, DC = 25x – 24y, AE = ¼ AD 3 and EF = EC. 5 (a) Express in terms of x and/or y: (i) BD (ii) EC
4.
[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]
Diagram below shows a trapezium ABCD. B
C F •
A
• E
D
2 5 It is given that A B =2y, AD = 6x, AE = AD and BC = AD 3 6 (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 a constant. (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] 101
CHAPTER 4
VECTORS
FORM 5
ANSWERS (PAPER 1) 1. a) b) 2.
3 a) b)
5 3
1
8i – 4j
1
r = - 2a + 11b r = ha + (h – k)b
1
h = -2 (h – k) = 11 k = −13
1
BD = −6p + 4q DB = − BD = 6p −4q
1
EB =
EC EB BC 8 2p q 3
b)
AB (2 (3))i (16 4) j = 5i + 12j 1 u (5i 12 j ) 5 2 12 2
5.
6. a) b)
1 (5i 12 j ) 13
AB 2 BC (6i 4 j ) 2((m 4)i 2( p 2) j ) = (-2+2m)i + (-8+2p)j m=6 p = -2 12 OA 5 1 u (12i 5 j ) 12 2 5 2
1
DB 3
4 2p q 3
4. a)
1
1 (12i 5 j ) 13
1
1 1 1 1 1
1 1 1 1 1 1
1
102
CHAPTER 4 7.
VECTORS
PO OP 6i 4 j
OR PQ 4i 5 j PR PO OR 10i j
8. a) b)
9. a)
b)
b)
1 1
1
4 OA 3
1
AB 4i 8 j
1
QR m PQ 3 4 m 1 k 2 3 = m(4) 3 m 4 1+ k = m(-2) 5 k 2 PQ u 2 4 (2) 2
10 a)
FORM 5
1 (4a 2b) 20
a kb 6i 7 j k ( pi 2 j ) 7 k 2 12 p 7 a b 62 72 p 9
p 2 22
1
1
1
1
1 1 1 1 1
103
CHAPTER 4
VECTORS
FORM 5
ANSWERS (PAPER 2) 3 AO 4 3 OA 4 A = (-3,-4)
1 (a)
(b)
u
1
1
OA OA
u
(c)
32 4 2
1 OA
1 3i 4 j 5
CD m AB k 5 m 5 7 5 m 7 25 k 7
2 (a) (i) (ii)
(b) (i)
(c)
1
AP 6 x 2 y AB 18 x 2 y 9 1 AQ x y 2 2 9 3 OQ x y 2 2
AR 6hx 2hy 9 3 RQ kx ky 2 2 AR RQ AQ 9 3 9 1 6h k x 2h k y x y 2 2 2 2 1 k 3 1 h 2
1
1
1
1 1 1 1 1 1
1 1 1
104
CHAPTER 4 3 (a) (i) (ii)
VECTORS 1
BD 20 x 32 y 3 ED AD 4 = 24y
1 1
EC 25 x (b)
FORM 5
FC 10 x BC BD DC 5x 8 y
1
BF BC CF 5 x 8 y
1
BD 20 x 32 y 4( 5 x 8 y ) 4( BF ) (c)
BD
1 2
20 x 32 y
2
(20 2) 2 32 3 = 104
4 (a)
(b) (i)
1 1
5 AD 6 = 5x
1
AC AB BC = 5x + 2y
1
BC
2 EF m AB EF my 2 AE AD 3 = 4x
AF AE EF = 4x + my (ii)
2
AC 5 x 2 y 4 4 AC 5 x 2 y 5 5 4 8 AC 4 x y 5 5 Assume A, F, C collinear, 4 AC AF 5 = 4x + my 8 m 5
1
1 1
1
1
105
CHAPTER 5
TRIGONOMETRIC FUNCTIONS
FORM 5
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 ec 90o 2.
Given cos p and 270o 360o .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 180o 270o . Find in terms of r.
[3 marks]
a) cot b) tan 2 4.
Solve the equation 6 cos ec 2 x 13cot x 0 for 0o x 360o
5.
Solve the equation 2 sin 2 A cos 2 A sin A 1 for 0o A 360o
6.
Solve the equation 2 cos 2 y 7 sin y 2 for 0o y 360o
7.
Solve the equation 15 cos 2 x cos x 4 cos 600 for 0o A 360o
8.
Solve the equation 3cot x 2 sin x 0 for 0o x 360o
[4 marks] [4 marks] [4 marks] [4 marks] [4 marks]
106
CHAPTER 5
TRIGONOMETRIC FUNCTIONS
FORM 5
PAPER 2 1.
(a) Sketch the graph of y sin 2 x for 0o x 180o
[4 marks]
(b)
(b) Hence, by drawing suitable straight line on the same axes, find the 1 x number of solution to the equation sin x cos x for 2 360o 0o x 180o
[3 marks]
(a) Sketch the graph of y 3cos x for 0 x 2
[4 marks]
2.
(b) (b) Hence, by using the same axes, sketch a suitable graph to find the 2 number of solution to the equation 3cos x 0 for 0 x 2 . x State number of solutions. 3. (a) Sketch the graph of y 3sin 2 x for 0 x 2
[3 marks]
(b) Hence, by using the same axes, sketch a suitable straight line to find x the number of solution to the equation 2 3sin 2 x for 2 0 x 2 cot x tan x (a) Prove that cos ec 2 x 2 3 (b) (i) Sketch the graph of y 2sin x for 0 x 2 2 (ii) Find the equation of a suitable straight line to solve the 3 3 1 equation sin x x . 2 2 2 Hence, on the same axes, sketch the straight line and state the 3 3 1 number of solutions to the equation sin x x for 2 2 2 0 x 2 . (a) Prove that sec 2 x 2 cos 2 x tan 2 x cos 2 x
[3 marks]
(b) (i) Sketch the graph of y cos 2 x for 0 x 2
[6 marks]
4.
5.
[4 marks]
[2 marks] [6 marks]
[2 marks]
(ii) Hence, using the same axes, draw a suitable straight line to find x the number of solutions to the equation 2 cos 2 x 1 for 0 x 2 .
107
CHAPTER 5 6.
7.
TRIGONOMETRIC FUNCTIONS
FORM 5
(a) Prove that 2 2sin 2 x 2 cos 2 x
[2 marks]
(b) Sketch the graph of y tan 2 x 1 for 0 x 2 . By using the same 9 axes, draw the straight line y 3 x and state the number of solution to 2 9 equation tan 2 x x 2 for 0 x 2 2
[6 marks]
(a) Prove that cot 2 x cos ec 2 x tan 2 x sec 2 x
[2 marks]
3 (b) Sketch the graph y cos x and y 2sin x for 0 x 2 . State 2 1 3 the number of solution to equation sin x cos x for 2 2 0 x 2
[6 marks]
108
CHAPTER 5
TRIGONOMETRIC FUNCTIONS
FORM 5
ANSWERS (PAPER 1)
1 a) b)
tan
1
p 1 p2
cos ec 90o =
1 sin 90o
= 2 a)
b)
1 cos 1 = p
b)
4
1
sec =
1
cot 90o = tan
= 3 a)
1 p2
1
1 1
1 p2 p
1 tan 1 = r 2 tan tan 2 = 1 tan 2 2r = 1 r2 6 1 cot 2 x 13cot x 0 cot =
1 1 1 1
6 cot 2 x 13cot x 6 0
3cot x 2 2 cot x 3 0 3cot x 2 0 OR
1
2 cot x 3 0
3 2 OR tan x 2 3 o ' 0 x 56 19 or 56.31 and x 33o 41' or 33.69o
tan x
x 56o19' , 236o 19' 33o 41' , 213o 41' Or 56.31o , 236.31o ,33.69o , 213.69o
5
2 sin 2 A 1 sin 2 A sin A 1
1 1 1
2 sin 2 A 1 sin 2 A sin A 1 3sin 2 A sin A 2 0
3sin A 2 sin A 1 0
1 109
CHAPTER 5
TRIGONOMETRIC FUNCTIONS
3sin A 2 0
sin A 1 0
OR
2 OR 3 A 90o and 41.81o sin A
A 90o , 221.81o , 6
FORM 5
sin A 1
1
318.19o
1
2 cos 2 y 7 sin y 2 0 2(1 sin 2 y ) 7 sin y 2 0
1
2 sin 2 y 7 sin y 4 0
2sin y 1 sin y 4 0
1
1 2 sin y 4 (not accepted) sin y
7
y 30o
1
y 30o , 150o
1
15 cos 2 x cos x 4 cos 600 15 cos 2 x cos x 4 cos 600 0
15cos 2 x cos x 4(0.5) 0 15 cos 2 x cos x 2 0
1
5cos x 2 3cos x 1 0
1
5cos x 2 0 OR
cos x
8
2 5
3cos x 1 0
OR cos x
1 3
x 66.42 or 66 25' and 70.53 or 70 31'
1
x 66.42 ,293.58 ,70.53 ,289.47 x 66o 25' , 70o31' , 289o 28' , 293o 35' cos x 3 2sin x 0 sin x 3cos x 2 sin 2 x 0
1
3cos x 2 1 cos 2 x 0
1
3cos x 2 2 cos 2 x 0 2 cos 2 x 3cos x 2 0
110
CHAPTER 5
TRIGONOMETRIC FUNCTIONS
FORM 5
2 cos x 1 cos x 2 0 2 cos x 1 0 cos x
1 2
1
cos x 2 0
OR
OR cos x 2 (unaccepted)
x 120o
1
x 120o , 240o
1
(ANSWERS)PAPER 2
1
y
y 1
x 180o
x
45o
90o
135o
180o
1 ( shape) 1(max/min) 1(one period) 1(complete from 0 to 180o)
1 (straight line)
y sin 2 x
sin x cos x
1 x 2 360o
y 1 x 2 2 360o x y 1 180o Number of solutions= 3
1
1
111
CHAPTER 5
TRIGONOMETRIC FUNCTIONS
FORM 5
2 a)
y
y
2 x
x
1 ( shape) 1(max/min) 1(one period) 1(complete from 0 to 2 or 360 o)
1(for line 2 y x
y 3cos x
b)
(b)
2 3cos x 0 x 2 y0 x 2 y x
1 1
Number of solution =2 3
y
x
1 ( shape) 1(max/min) 1(one period) 1(complete from 0o to 2 )
1( for the straight line)
2 3sin 2 x
x 2
x 2 x y 2 2
2 y
1 1
Numbers of solutions= 8 112
CHAPTER 5 4
TRIGONOMETRIC FUNCTIONS
(a) Prove that
FORM 5
cot x tan x cos ec 2 x 2
LHS 1 cos x sin x 2 sin x cos x 1 cos 2 x sin 2 x 2 sin x cos x
1
1 1 2 sin x cos x 1 2sin x cos x 1 sin 2x cos ec 2 x
1
b)
1(shape) 1(max/min) 1(one period)
y
y
3 x1
1(for the straight 2 line)
x
2
3 2
3 y 2sin x 2
sin
3 3 1 x x 2 2 2
1 3 y 2 x 2 2 3 y x 1
1
Number of solution = 1 5
1
(a) LHS sec 2 x 2 cos 2 x tan 2 x 1 tan 2 x 2 cos 2 x tan 2 x
1
1 2 cos 2 x cos 2 x
(proved)
1 113
CHAPTER 5
TRIGONOMETRIC FUNCTIONS
FORM 5
b)
1(shape) 1(max/min) 1(one period)
y
y cos 2 x
1(for the straight line ) 2
x
y
3 2
x
-
2 cos 2 x 1
x
1
x Number of solutions = 2 y
6
1
2 2 sin 2 x 2 1 sin 2 x
1
2 cos 2 x
1
(proved)
b)
1(shape) 1(max/min) 1(one period)
y
y tan 2 x 1
1(complete cycle from 0 to 2 )
2
2
x
3 2
2
-2
9 y 3 x 2
1(for the straight line) 1
Number of solution = 3
114
CHAPTER 5 7
(a)
TRIGONOMETRIC FUNCTIONS
FORM 5
RHS cos ec 2 x tan 2 x sec 2 x
1 cot 2 x tan 2 x sec 2 x
1
cot 2 x 1 tan 2 x sec 2 x
cot 2 x sec 2 x sec 2 x
cot 2 x
1
(proved)
y
3 y cos x 2
x
2
3 2
2
1,1(shapes) 1(max/min) 1(one period) 1(complete cycle from 0 to 2 )
Number of solutions = 3
y=2sinx
1
115
CHAPTER 6
PERMUTATIONS AND COMBINATIONS
FORM 5
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' without repetition 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 without repetition? [2 marks]
6.
Diagram shows 4 letters and 4 digits. A B C D
5 6 7 8
A code is to be formed using those letters and digits. The code must consists of 3 letters followed by 2 digits. How many codes can be formed if no letter or digit is repeated in each code ? [3 marks] 7.
A badminton team consists of 8 students. The team will be chosen from a group of 8 boys and 5 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. R
A
J
I
N
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]
116
CHAPTER 6 9.
PERMUTATIONS AND COMBINATIONS
FORM 5
A debating team consists of 6 students. These 6 students are chosen from 2 monitors, 3 assistant 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. U
N
F
I
O
M
R
A five-letter code is to be formed using five of these cards. Find a) 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 any digit 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 be seated if no two persons of the same sex are next to each other. [3 marks]
15.
Diagram shows six numbered cards. 1
4
5
7
8
9
A four-digit number is to be formed by using four of these cards. How many a) different numbers can be formed? b) different odd numbers can be formed? [4 marks]
117
CHAPTER 6
PERMUTATIONS AND COMBINATIONS
FORM 5
ANSWERS ( PAPER 1 ) 1.
2.
3.
4.
5.
6.
7.
C 3 x 11C 3 = 19800
1
p1 x 7 p 3 = 840
1
10
1
4
1 1
1 x 7C5 = 21 2
1 4
p1 x = 24
1
p2
1
4
2
p3 x = 48
1
p1
1 2
4
p3 x 4 p2 = 288
1
C5 x 5 C3
1
a)
8
b)
If the team consists of 8 boys and 0 girl
8
If the team consists of 7 boys and 1 girl
8
= 560 C8 x 5 C 0
=1
5
C 7 x C1 = 40
If the team consists of 6 boys and 2 girl 8 C 6 x 5 C 2 = 280 The number of teams that can be formed = 1 + 40 + 280 = 321 8.
9.
10.
11.
12.
a)
5!
b)
4! x 2! = 48
a)
10
b)
2
a)
7
b)
6
C6
= 120
1
1
3
2
C1 x C3 x C 2 = 60
2
p 5 = 2520
2
p 4 x 4 p1 = 1440
5
p1 x 8 p 4 = 8400
5
1 1
2
= 210 5
1
p 3 x 3 p1 = 180 = 180
1 1 2 1 2 1 118
CHAPTER 6 13.
8
a) b)
14.
15.
PERMUTATIONS AND COMBINATIONS
8
FORM 5
C6 6C3
1
= 560 6! x 4!
1 1
= 17280
1 2
P3 x 2 P2 = 672
a)
6
b)
5
P4
P3 x 4P1 = 240
1 = 360
1 2 1
119
CHAPTER 7
PROBABILITY
FORM 5
PAPER 1 1. A box contains 8 blue marbles and f white marbles. If a marble is picked randomly from the 3 box, the probability of getting a white marble is . 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 one after another without replacement. Find the probability that the two pens drawn are of different colours. [3 marks] 3. Table 2 shows the number of coloured cards in a box.
Colour Yellow Green Blue
Number of Cards 6 4 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 is
2 . 3 Find the probability that, (a) both of them qualify for the final, (b) only one of them qualifies for the final.
1 while the probability 5
that Rajes qualifies is
[3 marks]
5. A box contains 4 cards with the digits 1, 2, 3 and 4. Two numbers are picked randomly from the 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 1 2 probabilities that Team A will beat Team B and Team C are and respectively. Find the 3 5 probability that Team A will beat at least one of the teams. [3 marks]
120
CHAPTER 7
PROBABILITY
FORM 5
7. The probability of a particular netball player scoring a goal in a netball match is
1 . Find the 5
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 are
2 1 and 5 3
respectively. 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 Black Red Green Yellow
Number of Marbles 2 5 3 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 is
2 . Find the probability that 5
Zainal scores at least one goal in two matches. [3 marks]
121
CHAPTER 7
PROBABILITY
FORM 5
ANSWER (PAPER 1)
No. 1.
Solution
Marks
8+f
1 f
3 =
8+f
1
5
f = 12 2.
1
=
=
4 10
4 10
6 x
9
6 x
9
6
@
+
10
6
4 x
10
4 x
9
9
1
1
8 =
3.
1
15
=
=
6 12
5 x
6 12
11 5
x
11
or
+
4
3 x
12 4
11 3
12
x
11
1 =
4(a)
(b)
=
+
2 12
2 12
1 x
11
1 x
11
1
1
1
3
2 =
or
1
15
1 5
1 x
3
+
2 3
4 x
5
1
122
CHAPTER 7
PROBABILITY
FORM 5
3 =
1
5
2
5.
=
1 or
4 2
=
1
3 1
x
4
1
3
1 =
1
6
6.
1
=
1
=
3
3
3 x
5
3 x
5
or
+
2
2 x
3
2 3
5
2 x
5
or
1 3
2 x
1
+
3
2 x
5
5
1
1
3 =
7.
1
5
1
=
1
=
5
5
4 x
5
4 x
4 x
5
5
4 x
5
1
+
4 5
1 x
5
4 x
5
+
48 =
8.
=
4 5
4 x
5
1 x
5
1
1
125
2 5
2 x
3
or
1 3
3 x
5
or
1 3
2 x
5
1
123
CHAPTER 7
=
PROBABILITY
2 5
2 x
3
+
1 3
3 x
+
5
1
2 x
3
5
FORM 5
1
3 =
9.
1
5
=
=
2 18
1 x
17
2 18
1 x
17
5
or
+
4 x
18 5 18
17 4
x
17
or
+
3
2 x
18
3
2 x
18
17
14 =
10.
or
+
8 18
8 18
7 x
7 x
17
17
1
1
1
51
2
=
2
=
16 =
17
5
5
2 x
5
2 x
5
or
+
2 5
3 x
2 5
5
3 x
5
or
+
3 5
3 5
1
1
2 x
5
2 x
5
1
25
124
CHAPTER 8
PROBABILITY DISTRIBUTION
FORM 5
PAPER 1 1. Diagram 1 shows a standard normal distribution graph. f(z)
0 k Diagram 1
z
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 standard deviation 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. f(z) 0.3264
0
k
z
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]
125
CHAPTER 8
PROBABILITY DISTRIBUTION
FORM 5
6
X is a continuous random variable of a normal distribution with a mean of 52 and a standard deviation 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 kg and 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 a variance 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. Each trainee takes 6 penalty kicks. The probability that a trainee scores a goal from a penalty kick is 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 normal distribution 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]
126
CHAPTER 8
PROBABILITY DISTRIBUTION
FORM 5
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 a mean 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 Diameter, x (cm)
A x > 6.5
B 6.5 x > 4.5
C 4.5 x k
It is given that the diameter of oranges has a normal distribution with a mean of 5.3 cm and a standard 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. If 8 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 a normal 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 140 mmHg. (ii) It is found that 16 teachers have “high blood pressure”. Find the total number of teachers in the group. [5 marks]
127
CHAPTER 8 6
PROBABILITY DISTRIBUTION
FORM 5
The masses of mangoes from an orchard has a normal distribution with a mean of 285 g and a standard 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 than 191.25. (ii) Given that 416 mangoes from this sample have a mass of more than m g, find the value of m. [7 marks]
ANSWERS (PAPER 1) 1
2
3a
3b
4a
4b
5a 5b
6a 6b
0.5 - 0.3125 0.1875
1 1
12
1 1 1
C10 (0.85)10 (0.15) 2 0.2924 p = 0.85, q = 0.15
14.75 12.5 1 .5 1.5 12.5 12.5 14.75 12.5 or 1 .5 1 .5 0.4332 0 .5
X 55 10
1 1 1 1 1
60 48 55 10 75.8%
1
0.5 - 0.3264 0.94 X 180 0.94 5 .5 X = 185.17
1 1
z
67 52 10
1.5 1 - 0.8643 k=1.1
1 1
1 1 1 1 1 1
128
CHAPTER 8
7a
7b
PROBABILITY DISTRIBUTION
50.6 45 5 0.1314 52.1 45 40.5 45 or 5 5 0.7381
FORM 5
1 1 1 1
ANSWERS (PAPER 2)
1(a)(i)
1(a)(ii)
1(b)
2(a)(i)
2(a)(ii)
2(b)(i) 2(b)(ii)
p = 0.15, q = 0.85 P ( X 2) 1 8C 0 (0.15) 0 (0.85) 8 8 C1 (0.15)(0.85) 7 1 - 0.2725 - 0.3847 0.3428 165.75 = n(0.15)(0.85) n = 1300 65.34 , 7.5 P (48 x 72) 48 65.34 72 65.34 P( z ) 7 .5 7 .5 P(-2.312 < z < 0.888) 1 - 0.0103 - 0.1872 400 321 0.8025 x x = 400 mean = np 4 .5 p 6 0.75 P ( X 1) 1 - P(X <1) 1 - P(X=0) 1 6 C 0 (0.75) 0 (0.25) 6 0.9998 59 65 P ( x 59) P ( z ) 12 P(z < - 0.5) 0.3085 m 65 1.015 12 m = 77.18
1 1 1 1 1 1 1 1 1
1
1 1 1 1 1 1 1 2 1
129
CHAPTER 8
3(a)(i)
3(a)(ii)
3(b)(i)
3(b)(ii)
4(a)
4(b)
4(c)
PROBABILITY DISTRIBUTION
p = 0.18, q = 0.82 P(x = 2) = 10 C 2 (0.18) 2 (0.82) 8 0.298 P(X < 3) = P(x = 0) + P(x = 1) + P(x = 2) 10 C 0 (0.18) 0 (0.82)10 10C1 (0.18)(0.82) 9 10C 2 (0.15) 2 (0.82) 8 0.1374 + 0.3017 + 0.2980 0.7371 5 .7 4 .8 P( z ) 0 .6 1 - 0.0668 0.9332 m 4 .8 0.842 0 .6 m = 48 6 .5 5 .3 ) 0.75 0.0548 4 . 5 5 .3 6 .5 5 .3 P( z ) 0.75 0.75 1 - 0.1430 - 0.0548
P( z
312 x 0.8022 250 1- 0.7876 k 5 .3 0.798 0.75 k = 4.7015
FORM 5
1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 2 1
130
CHAPTER 8
5(a)(i) 5(a)(ii)
5(b)(i)
5(b)(ii)
P(x = 2) = 8 C 2 (0.6) 2 (0.4) 6 0.04129 P(X > 2) = 1 8 C 0 (0.6) 0 (0.4) 8 8 C1 (0.6)(0.4) 7 8 C 2 (0.6) 2 (0.4) 6 1 – 0.0006554 – 0.007864 – 0.04129 0.9502 110 128 140 128 P( z ) 10 10 1 - 0.0359 - 0.1151 0.849 140 128 P( z ) 10 0.1151 np = 16 16 n= 0.1151 139 191.25 285 ) 75 1 - 0.1056 0.8944 0.8944 x 520 465 416 P(X > m ) = 520 = 0.8 m 285 0.842 75 m =221.85
P( z 6(a)
6(b)(i)
6(b)(ii)
PROBABILITY DISTRIBUTION
FORM 5
1 1 1 1 1 1 1 1
1 1
1 1 1 1 1 1 1 2 1
131
CHAPTER 9
MOTION ALONG A STRAIGHT LINE
FORM 5
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 a 8 4t. 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 0 t k . 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 by V 20t 2t 2 , where t is the time, in seconds, after leaving the point R. (Assume motion to the right is positive)
(a) (b) (c) (d)
Find the maximum velocity of the particle, [3 marks] the distance travelled during the 4th second, [3 marks] the value of t when the particle passes the points R again, [2 marks] 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 the fixed point P and object B passes the fixed point Q simultaneously. The distance PQ is 90 m.
A
B
P
M
Q
90 m
The velocity of A, V A m s 1 , is given V A 10 8t 2t 2 , 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 instantaneously at point M. (Assume that the positive direction of motion is towards the right.) Find (a) the maximum velocity , in, m s 1 , of A, (b) the distance, in m, of M from P, (c) the distance, in m, between A and B when A is at the points M.
[3 marks] [4 marks] [3 marks] 132
CHAPTER 9
4.
MOTION ALONG A STRAIGHT LINE
FORM 5
A particle moves in a straight line and passes through a fixed point O. Its velocity, v ms 1 , is given by v t 2 4t 3 , 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. (b) Sketch the velocity-time graph of the motion of the particle for 0 t 3 .
[5 marks] [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 v t 2 8t 15 , where t is the time, in seconds, after passing through O . [Assume motion to the right is positive.] Find (a) (b) (c) (d)
6.
the initial velocity, in ms 1 , the minimum velocity, in ms 1 , the range of values of t during which the particle moves to the left, the total distance, in m, travelled by the particle in the first 5 seconds.
[1 mark] [3 marks] [2 marks] [4 marks]
A particle moves along a straight line and passes through a fixed point O. Its velocity, v ms 1 , is given by v 8 2t t 2 , where t is the time, in seconds, after passing through O . The particle stops instantaneously at point M. [Assume motion to the right is positive.] Find (a) (b) (c)
the acceleration, in ms 2 , of the particle at M, [3 marks] the maximum velocity, in ms 1 , of the particle [3 marks] the total distance, in m, travelled by the particle in the first 10 seconds, after passing through O. [4 marks]
133
CHAPTER 9 7.
FORM 5
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, after passing through point O. The particle stops after k s. (a)
(b)
8.
MOTION ALONG A STRAIGHT LINE
Find: (i) the maximum velocity of the particle, (ii) the value of k [6 marks] Sketch a velocity-time graph of the motion of the particle for 0 t k . Hence, or otherwise, calculate the total distance travelled during that period. [4 marks]
A particle moves along a straight line. Its velocity, v ms- 1 , from a fixed point, O, is given by v = t2 – 10t + 24 where t is the time, in seconds, after passing through point O. [Assume motion to the right is positive.] Find (a) (b) (c) (d)
the initial velocity of the particle, the minimum velocity of the particle, the range of values of t during which the particle moves to the left, the total distance travelled by the particle in the first 6 seconds.
[1 marks] [3 marks] [2 marks] [4 marks]
134
CHAPTER 9
MOTION ALONG A STRAIGHT LINE
FORM 5
ANSWERS PAPER 2 1.
a V
= =
8 – 4t ( 8-4t ) dt 4t 2 8t c 2 8t 2t 2 c
= =
a)
(i)
(ii)
b)
V = t V
1
V = 10, t = 0. 10 = 8(0) 2(0) 2 c 10 = c V = 8t 2t 2 10 maximum velocity, a= 0 8 – 4t = 0 8 = 4t t = 2s Vmax imum = 8(2) 2(2) 2 10 = 18 ms-1 particle stops: V = 0 0 8t 2t 2 10 = 2 0 2t 8t 10 = 0 t 2 4t 10 = (t + 1)(t – 5) = 0 t = 5 k = 5 2 8t 2t 10 V 0 10
2 18
5 0
1
1 1 1
1
18 10
5
2
0
t
2
5
Total Distance
=
V dt
=
8t 2t
0 5
2
10 dt
0
5
=
2 2 3 4t 3 t 10t 0
1
135
CHAPTER 9
MOTION ALONG A STRAIGHT LINE = =
2.
a)
b)
2 4(5 2 ) (5 3 ) 10(5) 0 3 2 66 m 3
V 20t t 2 dV a 20 4t dt Maximum velocity, a = 0 20 – 4t = 0 20 = 4t t=5 Vmax = 20(5) – 2(52) = 50 ms- 1 s v dt
FORM 5
1
1
1 1
20t 2t 2 dt
2 10t 2 t 3 c 3 s = 0, t = 0 c = 0 2 s 10t 2 t 3 3 2 t 3; s 10(3 2 ) (33 ) 72m 3 2 1 t 4; s 10(4 2 ) (4 3 ) 117 m 3 3
Distance travelled during the fourth second
c)
1
1 1 = 117 72 3 1 45 m 3
1
At point R again, s = 0 2 10t 2 t 3 0 3 2 t 2 (10 t ) 0 3 2 10 t 0 3 2 t 10 3 t 15
1
1
136
CHAPTER 9 d)
3.
MOTION ALONG A STRAIGHT LINE Maximum displacement, V = 0 20t 2t 2 0 2t (10 t ) 0 t 10 Time = 10 seconds 2 V A 10 8t 2t a A 8 4t V A maximum when a A 0 8 – 4t = 0 8 = 4t t=2 V A max 10 8(2) 2(2 2 ) = 18 ms- 1 Object A at M when V A 0
a)
b)
10 8t 2t 2 0 2t 2 8t 10 0 t 2 4t 5 0
FORM 5
1
1 1
1 1
1
t 1t 5 0
1
t=5 5
Distance M from P v dt 0
5
10 8t 2t 2 dt 0 5
c)
2 10t 4t 2 t 3 3 0 2 10(5) 4(52 ) (53 ) 0 3 2 66 m 3 When t = 5, distance B from Q = v t =35 = 15 m 2 Distance between A and B = 90 66 15 3 1 = 8 m # 3
1
1
1 1 1
137
CHAPTER 9 4.
a)
MOTION ALONG A STRAIGHT LINE
v t 2 4t 3 t = 0, v 0 2 4(0) 3 = 3 ms-1 # particle moves to the left, v < 0 t 2 4t 3 < 0 t 1t 3 < 0 1 < t < 3 # v t 2 4t 3 dv a 2t 4 dt a is positive, a > 0 2t – 4 > 0 2t > 0 t >2 # 2 v t 4t 3 v
i)
ii)
iii)
b) t v
0 3
1 0
3 0
1 1 1
1
1
3
1
Total distance
t
3
1
0
c)
FORM 5
v dt
v dt
0
1
=
t
2
t
3
2
3
=
1
1
3
4t 3 dt
0
(t
2
4t 3)dt
1
1
3
=
t 3 t 3 2 2 2 t 3 t 2t 3t 3 0 3 1
=
1 3 2 3
1 9 18 9 2 3 3
1 1 1 1 3 3 2 = 2 m # 3 2 v t 8t 15 t 0, v 0 2 8(0) 15 15ms 1 #
1
1
=
5.
a)
1 1 138
CHAPTER 9 b)
MOTION ALONG A STRAIGHT LINE
FORM 5
v t 2 8t 15 dv a 2t 8 dt
1
dv 0, dt 2t – 8 = 0 2t = 8 t=4
When
1 42 – 8(4) + 15 = - 1 ms-1
When t = 4, vmin = c)
Particle moves to the left, v < 0 t 2 8t 15 0 t 3t 5 0
#
3
1
5
3 < t < 5 d) Total distance =
3
5
0
3
1
1
v dt v dt
t 3
=
t
2
5
(t
8t 15 dt
0
2
8t 15) dt
1
3
3
5
=
t 3 t 3 2 2 4 t 15 t 4t 15t 3 0 3 3
=
9 36 45
125 100 75 9 36 45 3
2 18 16 18 3 1 = 19 m # 3 v = 2 When v = 0, 8 2t t = t 2 2t 8 = (t + 2) (t – 4) = t = 4 dv a = = 2 – 2t dt At M, a = 2 – 2(4) = – 6 ms 2
1
1
=
6.
a)
1 8 2t t 2 0 0 0
1 1
#
1
139
CHAPTER 9 b)
MOTION ALONG A STRAIGHT LINE Maximum velocity, a = 2 – 2t = 2 = t = When t = 1 ; v max imum =
0 0 2t 1 8 + 2(1) – 1
c)
Total distance, s = =
1 1 1
9 ms 1 #
= v dt 8 2t t 2 dt t3 8t t 2 c 3
=
FORM 5
1
When t = 0, s = 0 c = 0 s =
8t t 2
t3 3
1
43 3 10 3 8(10) 10 2 3 8(4) 4 2
s t 4 = s t 10 =
=
t=0 153
1m 3
2 m 3 1 153 m 3 26
=
1
t=4
0
26
2m 3
t = 10
7.
a)
i)
Total distance in the first 10 seconds 2 1 = 2(26 ) 153 3 3 2 = 206 m # 3 a = – 2t + 8 V = – 2t + 8 dt = t 2 8t c When t = 0, V = 20 : 20 = t 2 8t c 20 = 0 V = t 2 8t 20 When maximum velocity, a = 0 - 2t + 8= 0 2t = 8 t= 4 V max imum
= (4) 2 8(4) 20
1
1
1
1 1 140
CHAPTER 9
MOTION ALONG A STRAIGHT LINE = 36 ms
(ii)
b)
Particle stops, V t 2 8t 20 t 2 8t 20 (t + 2)(t – 10) t = 10 k = 10 V
= = = =0
= t 2 8t 20
1
FORM 5
#
0 0 0 V 36
1
20
t 0 4 10 V 20 36 0
2 0
Distance
= = = =
8.
a)
( t 2 8t 20 ) dt 3 t [ 4t 2 20t ]10 0 3 3 10 4(10 2 ) 20(10) – 0 3 2 266 m # 3
= t 2 10t 24 When t = 0, initial velocity, v = 0 2 10(0) 24 = 24 ms 1 #
v
b)
a
= =
dv dt 2t – 10
max imum
10
t
1
1
1
1
When minimum velocity, a = 0 2t -10 = 0 t = 5 v
4
1
=
5 2 10(5) 24
=
– 1 ms 1
#
1 141
CHAPTER 9 c)
d)
MOTION ALONG A STRAIGHT LINE When moves to the left, v 0 t 2 10t 24 ( t – 4 )( t – 6 )
0 0
4
FORM 5
6
4 t 6 Distance, s = ( t 2 10t 24 ) dt 1 3 = t 5t 2 24t c 3 When t = 0, s = 0 c = 0 1 3 s = t 5t 2 24t 3 t = 0, s = 0 1 1 t = 4, s = (4) 3 5(4) 2 24(4) = 37 m 3 3 1 t = 6, s = (6) 3 5(6) 2 24(6) = 36 m 3
t
1
1
1
1
1
t=6 t=4 0
36
t=0 Total distance travelled during the first 6 seconds 1 1 = 37 + 37 – 36 3 3 2 = m # 38 3
37
1m 3
1
142
CHAPTER 10
LINEAR PROGRAMMING
FORM 5
PAPER 2 1.
Amirah has an allocation of RM200 to buy x workbooks and y reference books. The total number of books is not less than 20. The number of workbooks is at most twice the number of the references. The price of a workbook is RM10 and that of a reference is RM5. a ) Write down three inequalities, other than x 0 and y 0, which satisfy all the above constraints. ( 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 dentistry undergraduates. The method in which the number of medical undergraduates and dentistry undergraduates 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 x 0 and y 0, which satisfy all the above constraints. ( 3 marks ) b) Hence, by using a scale of 2cm to 10 participants on both axes, construct and shade the region R 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 4 students. The number of students for S is x and for M is y. The intake of the students is based on 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 subject M. III : The number of students for subject M must exceed the number of students for subject S by at most 10 143
CHAPTER 10
LINEAR PROGRAMMING
FORM 5
a) Write down three inequalities, other than x 0 and y 0, which satisfy all the above conditions. ( 3 marks) b) Hence, by using a scale of 2 cm to represent 10 students on both axes, construct and shade the region 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 two processes, mixing the ingredients and baking the breads. Table 1 shows the time taken to make bread L and M respectively.
Type of bread L M
Time taken ( minutes ) Mixing the ingredients Baking the breads 30 40 30 30 Table 1
The shop produces x breads of type L and y breads of type M per day. The production of breads per day are based on the following constraints: I : The maximum total time used for mixing ingredients for both breads is not more than 540 minutes. 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 x 0 and y 0, which satisfy all the above constraints. ( 3 marks ) b) Using a scale of 2 cm to represent 2 breads on both axes, construct and shade the region R that 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 144
CHAPTER 10
LINEAR PROGRAMMING
from one bread of type M is RM1.00 . 5.
FORM 5 ( 4 marks )
A factory produced x toys of model A and y toys of model B. The profit from the sales of a number 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 x 0 and y 0, which satisfy all the above conditions. ( 3 marks) b) Hence, by using a scale of 2 cm to represent 50 models on both axes, construct and shade the region 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
145
CHAPTER 10
1.
LINEAR PROGRAMMING
FORM 5
( a ) I : x + y ≥ 20 II : x ≤ 2y III : 10 x + 5y ≤ 200 2x + y ≤ 40
1 1 1
(b)
4 40
35
30
25
20
2x + y= 40
R
y = 15
15
10
x + y = 20
x =2y
5
10
20
30
40
( c ) Draw the line y = 15 From the graph, the minimum value occurs at ( 5, 15 ) Hence, minimum expenditure = 10 (5) + 5 (15) = RM125 Therefore, the maximum amount of money left = RM200 –RM125 =RM 75
50
60
1 1
1
146
CHAPTER 10
LINEAR PROGRAMMING
FORM 5
2. ( a ) I : x + y ≥ 30 II : x ≤ 3y III : 100 x + 80y ≤ 6000 5x + 4y ≤ 300
1 1 1
(b)
3 90
80
5x + 4y= 300 70
x =20 60
50
40
R
30
20
x =3y x + y=30
10
-20
20
40
60
80
100
120
140
-10
( c ) ( i ) Draw the line x = 20, From the graph, the minimum number of dentistry undergraduates is 10 and the maximum number of dentistry undergraduates is 50.
1 1
(ii) For the minumum expenditure, there are 20 medical undergraduates and 10 dentistry undergraduates. Thus , minimum expenditure = 100 (20) + 80(10) =RM2800
1 1
147
CHAPTER 10
LINEAR PROGRAMMING
FORM 5
3 . ( a ) I : x + y ≤ 90 II : x ≤ 2y III : y - x ≤ 10
1 1 1
(b)
3 90
80
x + y = 90
70
y - x = 10
60
50
40
R x =2y
30
20
12x + 10y = 600 10
20
40
60
80
100
120
140
-10
( c ) ( i ) From the graph, when x = 20, the range of y is 10 ≤ y ≤ 30
1
( ii ) The maximum value occurs at ( 60, 30)
1
Thus, maximum total fee = 12 (60) +10 (30) = RM1 020
1 1
148
CHAPTER 10
LINEAR PROGRAMMING
FORM 5
4 ( a ) I : 3x + 3y ≤ 54 II : 4x + 3y ≥ 48 III : 2x ≥ y
1 1 1
(b)
3 22
20
18
16
y =2x 14
12
10
3x + 3y =54 8
6
R 4x + 3y = 4 8
4
2
2x + y = 2
5
10
15
20
25
30
35
-2
1 2 1
( c ) (i ) 8 (ii ) (5 x RM2.00) + ( 10 x RM1.00) = RM 20.00
149
CHAPTER 10 5
LINEAR PROGRAMMING
(a ) I : x + y ≤ 500 II : x ≤ 3y III : 15x + 12y ≥ 4200 5x + 4y ≥ 1400
FORM 5 1 1 1
(b)
3
( c ) ( i ) When x = 100, the minimum number of model B is 225. 1 (ii ) Maximum point (375, 125) The maximum total profit per day = 15(375) +12(125) = RM7125 1 1
150
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