Additional Mathematics Module

1 When a number is added or subtracted from each value in the data, the measures of dispersion do not change.

2. When each value in the data is multiplied by a number k, the measures of dispersion will be change.

new range = k  original range
new interquartile range = = k  original interquartile range
new variance = k2  original variance
new standard deviation = k  original standard deviation

3. When an extreme value exists in a set of data,

i) the range becomes vary big and will not give accurate information.

ii) the interquartile range is not affected because the extremes values is not taken into account.

iii) the variance becomes much bigger .

Example/ Exercise 12 Solve the following problems [ ans a) 24 ; 32 ]

a) The range and the variance of a set of data are 12 a) The range and the variance of a set of data is 30,

and 13 respectively. If each value in the set is multiplied interquartile range of 5 and standard deviation of 2.5

by 2 and then subtracted by 5, find Each value in the set is divided by 4 and then added by 5

(a) the new range find

(b) the new variance (a) the new range

(c) the new standard deviation (b) the new variance

(c) the new standard deviation

Find the range, the interquartile and the variance of a set of data 3, 5, 6, 8,11, 13. From the result, find the range,

interquartile range and variance of each set of data below.

a) 4, 6, 7, 9,12, 14 b) 6,10,12,16,22,26

c) 0, 2, 3, 5,8 , 10 d) 3 , 5 , 6 , 8 ,11, 13
2 2 22 2 2

 a) Given the value of x and x 2 of a set of 6 b) A set of data x1, x2, x3, x4, x5 has a mean of 10 and a
variance of 4. When a value x6 is added to the set of data,
numbers are 39 and 271 respectively. If a number 5 is the mean remains unchanged find
taken out from the set, find the new mean and standard
deviation a) the value of x6 [ ans 10]
b) the variance of the new set of data [ 3.333]

96

c) A set of data consists of 6 numbers. The sum of the d) A set of data has seven numbers. Its mean is 9. If a
numbers is 39 and the sum of the squares is 271. number p is added to the set, the new mean is 12.
(a) Find the mean and variance of the set of data. What is the possible value of p?
(b) If a number 5 is taken out from the set of data,
find the new mean and standard deviation of the
new data.

SPM QUESTION

PAPER 1

YEAR 2005 [3 marks]

1. The mean of four numbers is m . The sum of the squares of the numbers is 100 and the

standard deviation is 3k. Express m in terms of k.

YEAR 2006
2. The positive integers consists of 2, 5 and m. The variance for this set of integers is 14. Find the value of m.

[4 marks]

Year 2007

3. A set of data consists of five numbers. The sum of the numbers is 60 and the sum of the squares of the numbers is 800.

Find, for the five numbers

(a) the mean,

(b) the standard deviation. [3 marks]

PAPER 2

YEAR 2003
1. A set of examination marks x1, x2, x3, x4, x5, x6 has a mean of 5 and a standard deviation of 1.5.

(a) Find

(i) the sum of the marks,  x ,

(ii) the sum of the squares of the marks,  x 2 . [3 marks]

(b) Each mark is multiplied by 2 and then 3 is added to it. [4 marks]
Find, for the new set of marks,
(i) the mean,
(ii) the variance.

YEAR 2004
2. A set of data consists of 10 numbers. The sum of the numbers is 150 and the sum of the

squares of the numbers is 2472.

a. Find the mean and variance of the 10 numbers,

[3 marks]

b. Another number is added to the set of data and the mean is increased by 1.
Find
(i) the value of this number,
(ii) the standard deviation of the set of 11 numbers.

97

[4 marks]
YEAR 2006
3. Table below shows the frequency distribution of the scores of a group of pupils in a game.

Score Number of pupils
10 – 19 1
20 – 29 2
30 – 39 8
40 – 49 12
50 – 59 k
60 – 69 1

(a) It is given that the median score of the distribution is 42.
Calculate the value of k.

[3 marks]

YEAR 2005
4. Diagram below shows a histogram which represents the distribution of the marks obtained by 40 pupils in a test.

14 a. Without using an ogive, calculate
the median mark.

[3 marks]

9

7 b. Calculate the standard deviation of
6 the distribution.

[4 marks]

4

0.5 10.5 20.5 30.5 40.5 50.5

(b) Use the graph paper to answer this question

Using a scale of 2 cm to 10 scores on the horizontal axis and 2 cm to 2 pupils 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 5

[1 marks]

98

5. YEAR 2007

Table 1 shows the cumulative frequency distribution for the scores of 32 students in a competition.

Score < 10 < 20 < 30 < 40 < 50
32
Number of 4 10 20 28
students

Table 1

(a) Based on Table 1, copy and complete Table 2.

Score 0–9 10 – 20 – 30 – 40 –
19 29 39 49
Number of
students

Table 2

(b) Without drawing an ogive, find the interquartile range of the distribution.

[5 marks] (SPM2007/Section A/Paper 2

ANSWER (STATISTICS)

PAPER 1
1. m = 25 – 9k2
2. m = - 4 or 11
3. (a) mean = 12 (b) 4

PAPER 2

1. (a)  x = 30 ,  x2. =163.5

(b) mean = 13, variance = 9

2. (a) mean = 15, variance = 22.2
(b) k = 26, standard deviation = 5.494

4. (a) median = 24.07 (b) standard deviation= 11.74

3. (a) k = 4 (b) mode = 43 (c) mode score = 48

5. Interquartile range = 43.5 - 16.17
= 18.33

99

CHAPTER 8 – CIRCULAR MEASURES

1. Radian

Converting measurements in radians to degrees and vice versa.

a) Angles can be measure in degrees ( o ) or radians ( rad )

rA

o 1rad r
r

B

b) i) The angle subtended at the centre of a circle by an arc with the same length as the radius of the circle is
define as 1 radian

ii) If the length of the arc AB = 2r , then the angle subtended at the centre of the circle is 2 radians and so on

c) When the arc length = 2  r (circumference of the circle) then the subtended angle = 2  radian
Since the angle at the centre of a circle = 360o then 2  rad = 360o
 rad = 180o

1800 = 57.3 0

Thus 1 rad =


I 0 =  rad = 0.01745 rad
180

Example 1/ Exercise 1 Convert the following angles in radian to degree and minutes

a) 0.45 rad Ans( 810 ) e) 2  rad Ans(1200 ) c) 2.57 rad a) 3.75 rad

3 Ans ( 462036' @ 462.60 ) Ans ( 214050' @ 214.830 )

Example 2 Convert the following angles in degree and minutes to radians

a) 750 ( 1.309) b) 50.30 (0.878) c) 180018' (3.147) f) 256040' (4.48)

Homework Skill Practice 8.1 texts book page 181

Students will be able to:
2. Understand and use the concept of length of arc of a circle to solve problems.
2.1 Determine:
a) length of arc; b) radius; and c) angle subtended at the centre of a circle
based on given information
2.2 Find perimeter of segments of circles.
2.3 Solve problems involving lengths of arcs.

100

r A Length of arc AB, s = r  where r is the radius of
the circle and  is the angle in radians
o s
r
B

Example 3 Find the length of the arcs in each of the circles below

a) [Ans 10.8 cm ] b) ans [ 7.333] c) Given that the length of the major arc
AB is 62 cm , find the length of the
A 6cm A radius [r = 12.25cm ]
o 70
4cm A
B o 70
o 2.7 rad
B
B

Exercise 3. [Ans a) 14.4 cm b) 32.47 c)  114035' ]

a) Find the length of AB b) Find the length of ADB c) Given that the length of arc AB is
15 cm and its radius is 7.5 cm
8cm A 6cm A
o 1.8 rad D o 50 find  in degree and minutes

B B A

o


B

Homework Text Book exercise 8.2.1 page 183 s

Complete the table below by finding the values of  , r or s .
r

1.5 rad 9 cm

Ans:2.1429 14 cm Ans:13.5
540 30 cm
Ans:15.002
2.828 cm
101

2.2 Finding perimeter of segments of circles.

r A (a) Perimeter of the segment AB
B
oθ = 2 r sin   r
2
r
(b) For  , is expressed in degrees
sin
2

Line segment AB b) Find the perimeter of the shaded segment

Example 3 [ ans a) 19.1323 cm b) 20.96 cm P
a) Find the perimeter of the shaded segments below
7 cm
A
O 66.5º
5cm

O 120º

BQ

Exercise 3 [ Ans 17.275 cm b) 7.218 cm ]

a) Find the perimeter of the shaded segments below b) The diagram shows OAB is a sector of a circle

O with centre O and radius 5 cm. BC is 1 cm
5 cm
Find the perimeter of the shaded segments below
1.855 rad
B
AB
C

O A

5 cm

Homework Text Book Exercise 8.8.2 page 184

102

Solving problems involving lengths of arc

Example 4 [ Answer i) 0.75 rad ii) 14.5 cm ] b) 13.09 cm )

a) The diagram shows B The Diagram shows , O is a
P centre of a circle APC and ABC
Q two arc QS and PR , of b) is a semicircle where AC is a
C diameter. Find the perimeter of
P two circles with centre A 60 the shaded region.

O and with radii OS 5 cm

and OR . Given the O

O ratio
OS : OR = 4 : 3 and

RS = 2 cm and QS = 6

cm find

R i)  in rad

S ii) perimeter of the

shaded region

Exercise 4 [ Answer a) 60o 9' b) 8.4 cm 2 The diagrams shows
OQR and PQS are
a) The diagram shows two sectors of circles with
circle with centre O . centers O and P
Given that OD = 8cm , respectively. PS is
DC = 2 cm and the perpendicular to OR
length of arc BC = 10.5 and the length of the
cm .Find circular arc QR is 8π.
i) the angle of AOD in Calculate
degree and minutes
ii) the length of arc AD a) the length of PS,

(b)the perimeter of

the shaded region,

Homework Text Book Exercise 8.2.3 page 185 and Skill practice 8.2 page 185

103

Students will be able to:

3. Understand and use the concept of area of sector of a circle to solve problems.

3.1 Determine the: a)area of sector; b) radius; and c)angle subtended at the centre of a circle
based on given information.

3.2 Find the area of segments of circles.
3.2 Solve problems involving areas of sectors.

r A Area of sector AOB is Area of segment AB is
given by the formula,
o given by the formula,
r A = 1 r 2 - 1 bh
A = 1 r 2 where  is 22
2
where b = base
B an angle in radians h = heights

of triangle AOB  is

an angle in radians

Example 5 Jb a) 65.46 cm2 b 165.888cm2 c) 8  cm2
b)

Find the area of the shaded Find the area of the shaded region
sector
Find the area of the shaded sector

Exercise 5 [ Ans 42.4cm2 b 244.38 cm2 c 26 2  cm2 ]
3

a) Find the area of the shaded b) Find the area of the shaded

sector sector

Find the are
shaded reg
is 8 cm and
4 cm.

Homework Text Book Exercise 8.3.1 page 187

104

2.2 Finding the area of areas of circles. b) Find the area of the shaded segment
Example 6 [ ans a) 15.3547 cm2 b) ( 5.96 cm2 )
b) Find the area of the shaded segments below P

A 7 cm

5cm O 66.5º

O 120º Q

B

Exercise 6 [ Ans (11.19 cm2 ) b) ]

a) Find the area of the shaded segments below b) The diagram shows OAB is a sector of a circle
with centre O and radius 5 cm. BC is 1 cm. Find the
O area of the shaded segments below
5 cm
B
1.855 rad
C
AB

O 5 cm A

Homework Text Book Exercise 8.3.2 page 189

105

Solving problems involving area of a circles Diagram 2 shows a circle
AKBP centred at O, with radius
The diagram shows an isosceles triangle ABC in which j cm and a sector APBH centred
BC = AC= 20 cm, and angle BAC = 0.7
radians. DC is an arc of a circle, centre A. Find, correct at P with radius 15 cm.
to 1 decimal place, Given that the ratio of the arc
(i) the area of the shaded region, [4]
AHB to the arc AKB is 6:7 and 

AOB = 1.2 radian.
Calculate
(a) the value of j [3 marks]
(b) the area of shaded region.

[4 marks]

Homework Text Book Exercise 8.3.3 page 190 and skill practice 8.3 pg 191

SPM QUESTION

PAPER 1
YEAR 2003
1. Diagram 1 shows a sector ROS with centre O .

R

 S
O

DIAGRAM 1

The length of the arc RS is 7.24 cm and the perimeter of the sector ROS is 25 cm. Find the

value of  , in radian. [ 3 marks ]

106

YEAR 2004
2. Diagram 2 shows a circle with centre O .

A Given that the length of the major arc AB
is 45.51 cm , find the length , in cm , of the
O 0.354 rad radius.
B
( Use  = 3.142 )

[ 3 marks ]

DIAGRAM 2

YEAR 2005
3. Diagram 3 shows a circle with centre O .

A

O

B

DIAGRAM 3

The length of the minor arc is 16 cm and the angle of the major sector AOB is 290o .
Using  = 3.142 , find

(a) the value of  , in radians. [ 3 marks ]
( Give your answer correct to four significant figures )

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

107

YEAR 2006
4. Diagram 3 shows sector OAB with centre O and sector AXY with centre A .

A

Y
X

O
B

DIAGRAM 4

Given that OB = 10 cm , AY = 4 cm ,  XAY = 1.1 radians and the length of arc AB = 7cm ,
calculate

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

Year 2007
5. Diagram 4 shows a sector BOC of a circle with centre O.

B

A C
O 1.85 rad

D
O

It is given that AD = 8 cm and BA = AO = OD = DC = 5 cm. [ 4 marks]

Find

a) the length, in cm, of the arc BC,
b) the area, in cm2, of the shaded region.

108

YEAR 2008

6. Diagram 18 shows a circle with centre O and radius 10 cm.

O

10 cm

=
PR

=
Q
Diagram 18
Given that P, Q and R are points such that OP = PQ and OPR = 90o, find
(a) QOR, in radians,
(b) the area, in cm2, of the coloured region.

[ 4 marks]

PAPER 2

YEAR 2003

1. Diagram 1 shows the sectors POQ, centre O with radius 10 cm. The point R on OP is such

that OR : OP = 3 : 5 . Calculate
P

(a) the value of  , in radian .

R [ 3 marks ]

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

O
DIAGRAM 1

109

YEAR 2004
2. Diagram 2 shows a circle PQRT , centre O and radius 5 cm. JQK is a tangent to the circle

at Q . The straight lines , JO and KO , intersect the circle at P andR respectively. OPQR is a
rhombus . JLK is an arc of a circle , centre O .

L

J Q K Calculate the angle  , in terms of  ,
P  rad R (a) [ 2 marks ]

O (b) the length , in cm , of the arc JLK ,
[ 4 marks ]
(c)
the area , in cm2 , of the shaded

T region. [ 4 marks ]
DIAGRAM 2

YEAR 2005

3. Diagram 3 shows a sector POQ of a circle , centre O. The point A lies on OP , the point B

lies on OQ and AB is perpendicular to OQ.

P It is given that OA: OP= 4 : 7 .
A ( Using  = 3.142 )

8 cm Q Calculate
 (a) the length , in cm , of AP ,

rad [ 1 mark ]
6
OB (b) the perimeter , in cm , of the shaded
region ,
DIAGRAM 3 [ 5 marks ]

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

110

YEAR 2006
4. Diagram 4 shows the plan of a garden. PCQ is a semicircle with centre O and has a radius

of 8 m. RAQ is a sector of a circle with centre A and has a radius of 14 m .

R Sector COQ is a lawn . The shaded region is a flower
C bed and has to be fenced . It is given that AC = 8 m and
 COQ = 1.956 radians .

[ use  = 3.142 ]

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

PA O Q (b) the length , in m , of the fence required
DIAGRAM 4 for fencing the flower bed , [ 4 marks ]

(c ) the area , in m2 , of the flower bed .
[ 4 marks ]

YEAR 2007

5. Diagram 4 shows a circle, centre O and radius 10 cm inscribed in a sector APB of a

circle , centre P. The straight lines, AP and BP, are tangents to the circle at point Q and

point R, respectively. [Use  = 3.142]

Calculate

A O B (a) the length, in cm, of the arc AB,

10 R [5 marks]
(b) the area, in cm2, of the shaded region.
Q cm
[5 marks]

60

o

Diagram 4 P
P

O

111

YEAR 2008

6. Diagram 9 shows two circles. The larger circle has centre X and radius 12 cm.
The smaller circle has centre Y and radius 8 cm. The circles touch at point R.
The straight line PQ is common tangent to the circles at point P and point Q.

Q  8 cm [use  = 3.142 ]
Y
Given that  PXR =  radians,
 (a) show that  = 1.37 (to two decimal places),
R
P  [2 marks]
12 cm X (b) calculate the length, in cm, of the minor arc QR,

[3 marks]
(c) calculate the area, in cm2, of the coloured region.

[5 marks]

Diagram 9

PAPER 1 ANSWERS (CIRCULAR MEASURE)

1.  = 0.8153 rad. (b) Area of the shaded region = 22.37
2. (a)  POR = 2 
2. r  7.675
3
3. (a)  = 1.222 rad (b) The length of arc JLK = 20.94
(c) Area of the shaded region = 61.40
(b) r  13.09
4. (a)   7  0.7 3. (a) AP = 6
(b) Perimeter of ehe shaded region = 24.40
10 (c) Area of the shaded region = 37.46
(b) A = 26.2
4. (a) Area of COQ = 62.59
5. (a) 18.5 (b) 80.48 (b) The perimeter = 38.25
6. (a) 1.047 (b) 30.70 (c) Area of the shaded region = 31.3

PAPER 5. (a) 31.42 (b) 88.63cm2
1. (a)   0.9273
6. (a) 1.37 rad (b) 14.18 cm (c) 40.66cm

112

113

Chapter 9 – differentiation

Students will be able to:

1. Understand and use the concept of gradients of curve and differentiation.

1.1 Determine the value of a function when its variable approaches a certain value.
1.2 Find the gradient of a chord joining two points on a curve.
1.3 Find the first derivative of a function y = f(x), as the gradient of tangent to its graph.
1.4 Find the first derivative of polynomials using the first principles.
1.5 Deduce the formula for first derivative of the function y = f(x) by induction.

1.1 Determine the value of a function when its variable approaches a certain value

. limit f(x) is the value of f(x) when x approaches the value of a
xa

Example 1

Find the limit for each question below

(a) lim (0.2) x = (b) lim (0.2)x = c) lim (0.2) x = d) lim (0.2) x =
x0 x x2 x2

e) lim 3 1  n  (f) lim (1)n = (g) lim 4x = h) lim 1  n 
n2 n 3 x 1 x2 n3 3  n2
n0

Exercises 1 : Find the limit for each question below

a) lim 1  2n b) lim 1  2n c) lim 1  2n
n0 2  n n 2  n n2 2  n

n2  4 e) lim n2  4 (SPM97) n2  4
d) lim f ) lim
n2 n  2
x0 n  2 x n  2

2x2 4n3 I) lim x2  x  6
g) lim h) lim
x3 x3
x 3x2  4x 1 x 43n  5n3

Homework : Text Book Exercise 9.1.1page 198

113

1.2 Find the gradient of a chord joining two points on a curve

Example 2 : Determine the gradient of the cord AB as shown below

(a) y (b) y (c) y
y=2x2 y=x2+1
y  x2  4
32- ----------- B--. 10--------------- -F
5------------------ B

8 -------A 2---------- E x 03
02 13
-4 x
A
4
Gredient of cord AB =
Gredient of cord EF =

Gredient of cord AB =

1.3 Finding the first derivative of a function y = f(x), as the gradient of tangent to its graph.

y y = x2
q

Q (3.05 , 9.30)
q

Tangent at P

P(3 9)

q

Complete the table below and hence determine the first derivative of the function y = x2

Gradient PQ

Point Q (x2, y2) x2 -x1 y2 - y1
x2 y2

3.05 9.30

3.01

3.001

3.0001

When Q  P , gradient PQ  6, then the gradient at point P = 6 , so , the gradient of the tangent at P

= lim y  dy = 6 at x = 3
x0 dx dx

1.4 Find the first derivative of polynomials using the first principles

The first derivative, dy of the polynomial y = axn + bx n-1 + ……….+ c where a b c and n are constants can be
dx

determined as shown below.

a) Let  x be a small change in x and  y be a small change in y.

b) Substitute x and y in the equation y = f(x) with x +  x and y +  y respectively

c) Express  y in terms of x and  x

d) Find x and then, lim  y  dy .
y  x dx
 x0

114

Example 1 y = x2 + 4 by using 1
a) Determine the first derivative of
the first principle x b) Determine the first derivative of y = by using first
Solution
x

principle x
Solution :

Exercise 1 : 2. Determine the first derivative of y = 4  3 by using
x
1. Given y = 3x2 + 5,find dy by using the first principle
dx first principle SPM (97)
Solution :
(SPM 94)
Solution

Homework : Text Book Exercise 9.1.4 page 201
115

1.5 Deduce the formula for first derivative of the function y = f(x) by induction
Complete the table below by using the first principle.

Function Derivative

3x
x2
4x2

From the pattern, a formula of the first derivative of axn can be deduced which is d (axn ) = nax n -1
dx

Students will be able to:

2. Understand and use the concept of first derivative of polynomial functions to solve problems

2.1 Determine the first derivative of the function y = axn using formula.
2.2 Determine value of the first derivative of the function y = axn for a given value of x.
2.3 Determine first derivative of a function involving:

a) addition, or b) subtraction of algebraic terms.
2.4 Determine the first derivative of a product of two polynomials.
2.5 Determine the first derivative of a quotient of two polynomials.
2.6 Determine the first derivative of composite function using chain rule.
2.7 Determine the gradient of tangent at a point on a curve.
2.8 Determine the equation of tangent at a point on a curve.
2.9 Determine the equation of normal at a point on a curve.

2.1 Determining the first derivative of the function y = ax n using formula

1. If y = k, where k is a constant then or dy = d (k ) = 0
dx dx

2. If y = axn where k is a constant and n is positive and negative integer

then dy = d (axn ) = nax n-1
dx dx

3. If f(x) = axn, then f (x)  anxn1 . Notation f (x) is read as f prime x

Example 2: find dy for each of the following.

dx

(a) y = 2x (b) y = x (c) y = 3x2 d ) y = - 4x e) y = 3
x4

Exercise 2: find dy for each of the following.

dx

a) y = -5 b) y = 3 x3 c) y = -6x3 1 e) y = 3x3
d) y = 12
5
2x2

Homework : Text Book Exercise 9.2.1 page 204
116

2.2 Determining value of the first derivative of the function y = axn for a given value of x.

Example3: dy at each of the given value of x
Find the value of
dx
(a) y = 15x3 when x = -1 (d) f(t) = 3t3, find f(-3)
(b)f(x)= 1 , find f’(2) (c) y=4 when x=0.5
2x2 x3

Exercise 3: Find the value of dy at each of the given value of x
dx
(a) y = -2x3 when x = -1 (d) f(t) = 5t4, find f(-3)
(b)f(x)= 3 , find f’(2) (c) y = 2 when x=0.5
2x3 x4

Homework : Text Book Exercise 9.2.2 page 205

2.3 Determining first derivative of a function involving:
a) addition, or b) subtraction of algebraic terms.

If f(x) = p(x) + q(x) then dd [ p(x) ] + d [ q(x)] or f '(x) = p ' (x) + q ' (x)
[ f(x)] =
dx dx dx

Example4 : Find dy for each of the following

dx
(a) y = 3x4 + 5x 5 + 1 (b) y = ( 2x -3)(x2 + 5)
6x3  7x  4 d)y = 2 x5  3 x4
54
(c) f(x) =

x

Exercise4. Find dy for each of the following

dx
a) y = 5x2 + 3x3+ 2x -1 d) 7s7 +5s3-8s
b) t + 3t3 - 6t2 + 7 43

c) t 5  t 2  5t 2

Homework : Text Book Exercise 9.2.3 page 206
2.4 Determining the first derivative of a product of two polynomials

dy dv du
Let y , u and v be the functions of x. If y = u v then =u + v or y ' = u v ' + v u' .
dx dx dx

It is called the product rule. 117

Example 5 : Given that y = x3 (3 - 5x 3), find dy Example 6 : Differentiate ( 4x 3 + 3 )(x2 - 2x) with respect
dx Solution

to x
Solution

Example 6 : Find dy for each of the following function.

dx
a) x4 (1 + x7 + 6x4) (SPM 96) b) y = (x5 + 3x4 + 2x)(x7 + 7x2+ 6)
c) x ( 4x - 6x3 + 3)

Exercise 6 : Find dy for each of the following function

dx
a) y = (2x+1) (x3- x2 -2x)
(b) (x3 - 4x )(x2 - 1 c) (1 - 1 ) ( 2x + x2 - 1 )
x x2
)
x

d )( x6 -1) (3x2 - 5x3) e) x3 ( x2 + 4x -1) 11

f) ( - 2x)( - 4x + 3)

x x3

Homework : Text Book Exercise 9.2.4 page 208
2.5 Determining the first derivative of a quotient of two polynomials

u then dy  v du  u dv = or dy = vu'uv'
dx dx dx v2
Let y, u and v be the function of x y=
dx v 2
v

It is called the quotient rule

118

Example 7 : Find dy for each of the following functions

dx

x4  x 2. y = t  2t 1

1. y = 3

4  x3 t2

3.Given that f(x) = 1 2x2 find f ' (SPM 93) 4.Given that f(x) = 1 2x3 find f '(x) (SPM 95)

4x  3 x 1

Exercise 7

1. I f y = (1  x)(1  x) find dy 1 2x3

(x  3)(3  x) dx 2. Given that F(x) = x2 find F ' (x)

119

3.Differentiate 2x3  3 with respect to x 2t

x2 4. Differentiate t2  4 with respect to t

Homework : Text Book Exercise 9.2.5 page 210
2.6 Determining the first derivative of composite function using chain rule

If a composite function is in the form y = u n where u = f(x) and n is an
integer, then by the chain rule

dy  dy  du .
dx du dx

Example 8 3
1. Differentiate (4x2 - 3)5 with respect to x

2. Differentiate (t 2  3t)2 with respect to t

Exercise 8 2.Find y ' if y = 4 SPM 98

1. Find d ( 1 ) SPM94 3x  2
dx 2x 1

120

3. Differentiate x4(1+3x)7 with respect to x SPM 96 4 Given that f(x) = 4x (2x - 1)5 find f '

x2 (t 2  2)3
5. Differentiate (2x  1)2 with respect to x 6. Differentiate (4t 2  5)2 with respect to t

Homework : Text Book Exercise 9.2.6 page 212
2.7 Determining the gradient of tangent at a point on a curve

Notes:

dy

a) represents the gradient of the tangent of the curve y = f(x)

dx

b) The gradient of the tangent at a point A (p,q) on a curve y = f(x) can be determined by

dy

substituting x = p into

dx

Example 9 : Find the gradient of a tangent at the given point for each of the following curves

(a) y = 4x2 -6x + 1, (2,5) (b) y = x2 – 2x, ( -1,1) (c ) y = 6 – 4x , (-2,7)

x

Chek the answer by using calculator function d/dx

121

Exercise 9 : Find the gradient of a tangent at the given point for each of the following curves

(a) y = 3x2 -7x + 3, (3,5) (b) y = 2x3 – 3x, ( -2 ,1) (c ) y = 6 – 3x , (-1,7)
x2

Homework : Text Book Exercise 9.2.7 page 213
2.8 Determining the equation of tangent at a point on a curve

Notes: The equation of the tangent at point P(x1,y1 ) on the curve y =f(x) can be determined by

(a) Finding the gradient, m, of the tangent at point P
(b) Use the formula y – y1 = m (x – x1) to find the equation of the tangent.

Example 10 : Find the equation of the tangent for each of following equations and the corresponding points.

(a) y = 2x2 – 3x +4 at point (2,3) (b) y = x3 + 3x2 at point (-1, 2)

Example 10 : Find the equation of the tangent for each of following equations and the corresponding points.

(a) y = 3x2 – 2x - 4 at point (3, 17 ) (b) y = 3x3 + 2x2 at point (-1, -1)

Homework : Text Book Exercise 9.2.8 page 214

122

2.9 Determining the equation of normal at a point on a curve
Notes

The equation of the normal to the curve y = f(x) at point P( x1 ,y1), can be determined by

(a) Finding the gradient ,m1,of the tangent at point P,
(b) Finding the gradient, m2, of the normal at point P for which m1m2 = -1
(c) using the formula y – y1 = m2 (x – x1) to find the equation of the normal

Example 11 : Find the equation of the normal for each of following equations and the corresponding points.

(a) y = x2- 4x + 1 at point (3,-2) 8

(b) y = 2x + at point (1,10)

x

Exercise 11 : Find the equation of the normal for each of following equations and the corresponding points.

(a) y = x2- 2x -4 at point (3,-1) x

(b ) y = x5 at x = 7

Homework : Text Book Exercise 9.2.9 page 215
123

Example 12 : [ Ans 1 a) 14 b ) y = 14x -23 , 14y = -x + 72 ]

1. Given y = (x2 + 1)(x -1)2 find 2.Given f(x) = x4 find the value of x if
(a) the gradient of the curve for which x = 2
(b) the equation of the tangent and the normal at x =2 x 1

Solution 5
(a)
f ' (x) = - and hence , find the equation of the

4

tangent and normal at that point..

b)

Exercise 12 : 2.The gradient of the normal on the curve y = (2x -3 )2
1. Find the equation of the tangent on a curve
1
y = (x2 + 1)(2x + 5) at point (-1,6)
is . Find the x-coordinates of the point.

2

124

k7 4. Find the gradient of the curve y = 4 at point

3. The curve y = h x + has the gradient of 2 at (-1, - ) 3x  2

x2 2 (-2, -1). Hence , find the equation of the normal .

Find the values h and k . (SPM 98)
(SPM 96)

5 Find the equation of the tangent on a curve 1
y = 3x2 - 4x +2 which is parallel to the line y = 2x + 5
6. Given that f(x) = x - . Find the positive value of x

x

for which f'(x) = 2 , Hence, find the equation of the
tangent and normal.

Homework : Skill Practice 9.2 Page 216

Students will be able to:

6. Understand and use the concept of second derivative to solve problems

6.1 Determine the second derivative of function y = f (x).
6.2 Determine whether a turning point is maximum or minimum point of a curve using the

second derivative

125

6.1 Determine the second derivative of function y = f (x).

dy d dy d 2 y
If : y = f(x), then = f '(x) and = = f " (x) is called the second derivative of y = f(x)
dx dx dx dx2

Example 13:

y = 4x3 + 2x - 5 dy d2y 2 . Given that f(x) = (1-2x2)4 .Find f " (x) and f " (0)
dx2
1. Given that find and
x dx

Exercise 13 : d2y 1 2 dy d2y
1. Given that y = (1+x2)4 find dy and dx2 2. Given that y = + find and
x x2 dx dx2
dx

3.Find f " and f" (2) for f(x) = 2x4 - x3 +15x +1 4. Given f(x) = (1 - x2)3 Find f ' and f "

126

d 2 y dy 6. Given that f(x) = (2x - 3)5 Find f " (SPM 97)

5 .Given that y = x(3-x), Express y - x +12
dx2 dx

d 2 y dy
in term x. then solve y - x = 12 (SPM 95)
dx2 dx

Homework : Text Book Exercise 9.6.1 page 231

Students will be able to:

3. Understand and use the concept of maximum and minimum values to solve problems

3.1 Determine coordinates of turning points of a curve.
3.2 Determine whether a turning point is a maximum or a minimum point.
3.3 Solve problems involving maximum or minimum values.

3.1 Determining coordinates of turning points of a curve

Notes :

dy

A curve y = f(x) has a turning point or stationary point where = 0. At this point the tangent is parallel

dx

to the x- axis

Example 14. Find the coordinates of the turning point of the following curve

a) y = 8 –x2 b) y = 3x –x3

4 4x2  9
d) y =
c) y = x + -2
x
x

Homework : Text Book Exercise 9.3.1 page 219

127

3.2 Determining whether a turning point is maximum or minimum point of a curve using the second derivative

3.3 Determining whether a turning point is a maximum or a minimum point

Notes :

The types of turning points of a curve y = f(x), can be tested as follows

a) find dy dy
of the curve y = f (x) . then find the value of x for which = 0
dx dx

b) Find the corresponding value of y by substituting the value of x into y = f(x)

c) Determine whether a turning point is maximum or minimum point by substituting the values of x

into d2y d2y d2y < 0 the point is maximum point.
. If > 0 the point is minimum point or if
dx2 dx2 dx2

Example 15 . 2. Show that the curve y = 5x2 - 5x +1 has only one
1. Find the coordinates of turning point of the curve turning point. Hence determine whether the turning
point is maximum point or minimum point
y = 12x - 3x2.Hence determine whether the turning
point is maximum point or minimum point.

Exercise 15 2. Find the coordinates of turning point of the curve

1 y = x3  x2  6x . Hence determine whether the
1. . Show that the curve y = 8x + 2x2 has only one 32

turning point. Hence determine whether the turning point turning point is maximum point or minimum point.
is maximum point or minimum point [Ans :(1/2,6)min] [ Ans (3,-27/2)min, (-2,22/3)max]

Homework : Text Book Exercise 9.3.2 page 221

128

3.3 Solve problems involving maximum or minimum values.

Example 16 ; 2. A solid cylinder with a x radius has a volume of 800cm3
1. A piece of wire of length 60 cm is bent to form a a) Show that the total surface area, A cm2 . of the

rectangle. Find the dimensions of the rectangle for

which the area is a maximum. [ Ans 15 cm] cylinder is given by A  2x2  1600
x

b) Hence, find the value of x which makes the surface

area a minimum [ Answer x = 5

Exercise 16 2. TSheeudtiagsradmawshaowi s a
1. Given a rectangle of area 12 cm2 and length x cm. pppeearninmtajgeatoennrgAonfBtyChaeDEp1e.0nIf0tathcgemon
x x
express the perimeter, P, of the rectangle in terms of
x . Hence find the minimum value of P . ids i1b8ecnmg, kfinodkthkeevpalaudesa

[Ans 2x +24/x, 8 3 ] iobsfeaxnmatnuadxkiymywuahmenngthe area

[dxi=tu4n.2j2u, ky=a2n.6d7 a] lam
y y rajah disebelah

Tunjukkan bahawa
x luas bentuk itu Lcm2

ialah
300x - 60x2. Cari

nilai x dan nilai y

supaya

L adalah maksimum

Homework : Text Book Exercise 9.3.3 page 223

129

Students will be able to:

4. Understand and use the concept of rates of change to solve problems.

4.1 Determine rates of change for related quantities

4.1 Determining rates of change for related quantities

Notes :

If y is a function of x and x is a function of time, t , then the link between y and t can be determine by

chain rule . dy  dy  dx . If the rate of change is positive, it means increment and if the rate of change
dt dx dt

is negative, it means decrement.

Example 16 . 2. The area of a circle is decreasing at the rate 2 cm2s -1
a)The radius, r cm of a circle is 20 cm and it is
increasing at the rate of 3cms-1. At what rate is the area How fast is the radius decreasing when the area is 32 
of the circle increasing.
cm2

Exercise 16 . 2. Two variable, x and y are related by equation y = 3x  5
a)The radius, r cm of a circle is 10 cm and it is x
increasing at the rate of 4 cms-1. At what rate is the
area of the circle increasing. If y is changing at a rate of 11.5 units per seconds , find
the rate of change of x when x = -2

Homework : Text Book Skill Practice 9.4 Pg 226
130

Students will be able to:
5. Understand and use the concept of small changes and approximations to solve problems.
5.1 Determine small changes in quantities.
5.2 Determine approximate values using differentiation

5.1 Determining small changes in quantities

Notes : 1. Given a function y = f(x), then as x makes a small change, y will also change by a small quantity

2. The small change in x and y are denoted by x and y respectively

3. If the value of x and y are positive its means a small increment in x and y respectively while If

the value of x and y are negative its means a small decrement in x and y respectively.

Example 16:

1. If y = 3x2 + 5x + 2, find the small change in y when 2.Find the approximate change in volume of a sphere if

x change from 3 to 3.02 its radius increase from 5 cm to 5.02

Exercise 16 2. Given that y = 6x3 +x2, find the small change in y
1 Given that y = 2x - x2, Find when x decreases from 3 to 2.99

a) the small change in y when x change from 2

to 2.01 [Ans= - 0.02 ]

b) the small change in y when x change from 2

to 1.99 [Ans = 0.02]

131

5.2 Determining approximate values using differentiation

1. If y = f(x) then y  dy where x is small amount of change in x y is small change in y
x dx

Example 17

24 2 dy when y = 36. Hence

1 Given that y = , find the approximate value of 2.Given that y = 9x 3 , Find dx

x4

24 [ Ans 1.44] find the small change in x when y increases 36.0 to
(2.02) 4
36.3 [ Ans 3, 0.1]

Exercise 17

16 dy when x = 2. Find the 2.Given that y = 1 . Find the approximate value of
1. Given that y = , find
x4 dx 3x

approximate value of 16 [Ans : -2 , 1.04] 1 1
3 0.9
(1.98) 4 [ Ans 1 ]

30

132

SPM Questions

PAPER 1

YEAR 2003 4. Two variables, x and y are related by the
1. Given that y  14x(5  x) , calculate equation y  3x  2 . Given that y increases
x
(a) the value of x when y is maximum, at a constant rate of 4 units per second, find
(b) the maximum value of y. the rate of change of x when x  2.

[3 marks]

[3 marks]

2. Given that y  x2  5x , use differentiation to

find the small change in y when x increases YEAR 2005
5. Given that h(x)  1 , evaluate h”(1).
from 3 to 3.01 [3 marks]
(3x  5)2
[4 marks]

YEAR 2004 6. The volume of water, V cm3, in a container is
3. Differentiate 3x2 (2x  5)4 with respect to x.
given by V  1 h3  8h , when h cm is the
[3 marks] 3

height of the water in the container. Water is
poured into the container at the rate of 10 cm3
s1. Find the rate of change of the height of
water, in cm s1, at the instant when its height

is 2 cm.

[3 marks]

133

10 . The curve y = f(x) is such that dy
= 3 kx + 5,
where k is a constant.
YEAR 2006 dx

7. The point P lies on the curve y  (x  5)2 . It 9. Given that y  3x2  x  4,

is given that the gradient of the normal at P is (a) find the value of dy when x  1,
dx
 1 . Find the coordinates of P.
4 (b) express the approximate change in y,
[3 marks] in terms of p, when x changes from 1 to 1
+ p, where p is a small value.
[4 marks]

8. It is given that y  2 u 7 , when u  3x  5. Year 2008
3 12. Two variables, x and y , are related by the

Find dy in terms of x. [4 marks] equation y = 16 . Express, in terms oh h, the
dx x2

approximate change in y, when x changes from 4 to
4 + h, where h is a small value

[ 4 marks]

PAPER 2

YEAR 2007

1. The curve y  f (x) is such that dy  3kx  5, where k is a constant. The gradient of the curve at
dx

x  2 is 9 .

Find the value of k . [ 2 marks ]

134

2. The curve y  x2  32x  64 has a minimum point at x = p , where p is a constant.

Find the value of p . [ 3 marks ]

YEAR 2003
1. (a) Given that dy  2x  2 and y  6 when x  1, find y in terms of x.

dx

[3 marks]

(b) Hence, find the value of x if x2 d2y  (x 1) dy  y  8. [4 marks]
dx 2 dx

2. (a) Diagram 2 shows a conical container of diameter 0.6 m and height 0.5 m. Water is poured into the
container at a constant rate of 0.2 m3 s1.

0.6 m

0.5 m water
Diagram 2 135

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

water level is 0.4 m.

(Use   3.142; Volume of a cone = 1 r 2h ) [4 marks]
3

YEAR 2004
3. The gradient function of a curve which passes through A(1, 12) is 3x2  6x. Find

(a) the equation of the curve, [3 marks]

(b) the coordinates of the turning points of the curve and determine whether each of the turning points

is a maximum or a minimum. [5 marks]

4. Diagram 5 shows part of the curve y  3 which passes through A(1, 3).
(2x 1)2

y

 A(1, 3) [4 marks]

y 3
(2x 1) 2

Ox

Find the equation of the tangent to the curve at the point A.
136

YEAR 2007

1. A curve with the gradient function 2x  2 has a turning point at ( k , 8 ) .
x2

(a) Find the value of k . [ 3 marks ]

(b) Determine whether the turning point is a maximum or a minimum point .

[ 2 marks ]

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

ANSWERS (DIFFERENTIATION)

PAPER 1 8. 14(3x  5)6
1. (a) x  5
9. (a) 7 (b) 7p
2
(b) y  175 PAPER 2

2 1. (a) y  x2  2x  7
2. x  0.11 (b) x  3 or x  1
3. 6x(6x  5)(2x  5)3 5

4. 8 unit second1 2. (a) y  3x2  6x 10
5 (b) (2, 10)

5. 27 3. (a) p  3
8 (b) f (x)  x3  2x2  4

6. 0.8333 cm s1 4. y  12x 15
7. (7, 4)

137

CHAPTER 10 – SOLUTIONS OF TRAINGLE

Students will be able to:
1. Understand and use the concept of sine rule to solve problems.
1.1 Verify sine rule.
1.2 Use sine rule to find unknown sides or angles of a triangle.
1.3 Find the unknown sides and angles of a triangle involving ambiguous case.
1.4 Solve problems involving the sine rule

To solve the triangle means we must find three
interior angle and three sides of any triangle

A Sine Rule

Label the triangle with alphabets A,B,C and a,b,c . For any triangle ABC, whether it is an acute-angled or obtuse-
angle triangle, sine Rule state that :

abc
sin A SinB sin C

It is correct when the label of the triangle are correct.
So label on the triangle is vary importance to verify sine rule.

Homework Text Book Exercise 10.1.1 page 242 b)
Example 1 . Solve each of the following triangle

a)

138

Exercise 1 : Solve the triangles below

 a) Ans PQR =420 ,PQ=10.49 cm, 2. . Ans MKL =380,MK=17.66 3. Ans ABC=49.050,
 ACB =65.950, AB = 6.05 cm
PR=7.38 cm, ML=11.04 cm

R L 12 cm K C
1000
720 10 cm 5 cm 6 cm
660 Q 420 A 650 B

P M

Homework Text Book Exercise 10.1.2 page 243
1.3 Finding the unknown sides and angles of a triangle involving ambiguous case.

Ambiguous case (Ambiguous means having more than one value )
a) When two sides and an acute angle which is not an included angle are given with the acute angle
opposite the shorter of two given sides, then ambiguity arises.
b) Ambiguity means it is possible to form two triangle as shown below

A situation like this will not happen if
a) the angle given is an obtuse angle
b) the side that is opposite the given
angle is longer than the second side.

Example 2 : a) { Answer C =57.7o or 123.3o BC 12.8 cm or 23.5cm

a) ABC is a triangle with  B = 25o , AB = 20 cm and AC = 10 cm. Solve the triangle.

139

Exerice 2: For each of the following triangle , state if ambiguity arises. Hence, solve the triangles

a)  ABC :  B = 30o , c = 8cm and b = 6cm c)  ABC :  B= 64o 22' a = 13.3 cm and

b = 12 cm

 C=41o49' , 138o 11'  A=108o11' 11o49' a =11.4 cm , 2.46 cm  A = 87o 47', 92o13'  C = 27o51' 23o25' c = 6.22 , 5.29

b)  XYZ :  Y = 69o 10' , x = 8.2cm and y = 9.5 cm d)  PQR :  R = 28o , p = 8 cm and r = 11 cm

 X = 53o 47'  Z = 57o 3' z = 8.53  P = 19o58'  Q = 132o2' q = 17.4 cm
140
Homework Text Book Exercise 10.1.3 page 244

1.4 Solving problems involving the sine rule Example 4 L

Example 3

9.5 cm 7 cm

PS R 35o
J
PSR is a straight line as shown in the diagram above K

Given that SR = 6 cm , SQ = 7 cm , PQS = 23o and The diagram shows a triangle JKL,
a) Draw and label another triangle JLK ’ such that JK ‘ =
SRQ  650 , Calculate
7cm
a) SQR b) the length of PS
JL = 9.5 cm and LJK remains fixed at 35 o .

b) Calculate the obtuse angle of JKL
c) Find KLK 

Homework Text Book Exercise 10.1.4 page 245

Students will be able to:

2. Understand and use the concept of cosine rule to solve problems.
2.1 Verify cosine rule.
2.2 Use cosine rule to find unknown sides or angles of a triangle.
2.3 Solve problems involving cosine rule.
2.4 Solve problems involving sine and cosine rules.

2.1 Verify cosine rule

Cosine Rule

a 2 = b2 + c2 - 2bc CosA The cosine rule can be used to solve any triangle when
b2 = a2 + c2 - 2ac CosB a) two sides and the included angle are given,
c2 = a2 + b2 - 2ab CosC b) three sides are given

141

Example 5: Write the cosine rule for each of triangles below

a) C b) t c) b d) r
L p q
8 cm S a 
700 
12 cm e
c

BA

Homework Text Book Exercise 10.2.1 page 247

2.2 Using cosine rule to find unknown sides or angles of a triangle

Example 4

a) Given ABC such that  A = 50.5o , c = 4.5 cm and b) Find all the angle of triangle PQR. Given that p = 4 cm,

b = 6 . 5 cm . Find the value of a [ Ans a = 5.029 cm ] q = 6 cm, and r = 8 cm.

(Ans  R = 103o 29' ,  P = 28o 58'  Q =46o 33' )

Exercise 4 b) Given that ABC , a = 12.5 cm, b = 20.5 cm and
a) Given that ABC , b = 5 cm, c = 4 cm and  A = 66o .
 C = 124o 25' . Find the value of c and the remaining
Find the value of a [ Answer 4.97 ]
angle of the triangle.

[ Answer c = 29.43 cm,  A 20o 31',  C = 41o 25' ]

Homework Text Book Exercise 10.2.2 page 248

142

2.3 Solve problems involving cosine rule Example 6
2.4 Solve problems involving sine and cosine rules
From the diagram given , find i)  ACD
Example 5
ii) length of AB [ Answer 42.4 , 12.76 cm ]
From the diagram given , find  PQS and  RSQ

[ Answer 67.49o and 17.7o ]

Exercise 6 b) A

a) In the diagram below, ABC is a straight line. Find 15 cm
(a) length of BD (b) length of CD [Ans 3.538 cm 8.35 cm ]
B
D D C 12 cm
BCD is a straight line in the above diagram. Given that BC
4 cm 7
A5 00 = 12 cm, AC = 15 cm, CAD = 29o and ABC  680 .
a) Find BAC , b) the length of CD
00
[ answer 47.88o , 12.64 cm ]
B 6 cm C
L

Homework Text Book Exercise 10.2.4 page 250

143

Students will be able to:
3. Understand and use the formula for areas of triangles to solve problems.
3.1 Find the area of triangles using the formula 1 ab sin C or its equivalent.

2

3.2 Solve problems involving three-dimensional objects.

Area of Triangles 1

Area of ABC = ab sin C
2

1

= ac sin B

2

1

= bc sin A

2

Find the area of the following triangles b) Answer [ 57.22 cm2 ]
a) Answer 50.07 cm2

c) Answer [40.13 cm2 ] d) Answer [ 56.82 cm2 ]

Homework Text Book Exercise 10.3.1 page 254

3.2 Solve problems involving three-dimensional objects

a)Calculate BGE [ Answer 70.17 o ] b) Calculate i) Length of PS ii) PSQ

[Answer 16.76 cm , 83.41 o ]

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