AM015 20212022

Example 3

If m  x   e2 x  3, n  x   ln(3  x)

Find m  n and n  m.
H en c e, sh o w th a t m  n ( 2 )  n  m (1 1).

LL

Example 4 Example 5
Find f(x) if given g(x) = x – 1 and
gf(x) = 3 + 2x Find g(x) if given f(x) = x + 1 and
gf(x) = x2 + 2x + 2
Solution

LL

SOLUTION Conclusion

f  g  g  f.

LL

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FUNCTIONS AND GRAPHS Learning Outcomes:
5.3 Inverse Functions
At the end of the lesson, students are able
L to:

(a) use algebraic approach or horizontal
line test to determine whether a
function is one to one.

(b) determine the inverse of a function
(if exist)

Learning Outcomes: NOTES

(c) determine domain and range of an An inverse function of f exists
inverse function. only if the function f is one to one

(d) sketch the graph of a function and its and onto function.
inverse on the same axes
Algebraic Approach

Algebraic Approach Algebraic Approach

EXAMPLE 1:- EXAMPLE 2:-

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Graphical Approach Graphical Approach

(horizontal line test) (horizontal line test)

If a straight line that is parallel to the EXAMPLE 1:-
x-axis is drown and it at most cuts
the graph at only 1 point, then the y
function is one to one.
x

Graphical Approach

(horizontal line test)

EXAMPLE 2:- Using the Property

y

x

EXAMPLE 1

Algebraic Approach

is one to one function,
therefore the inverse function is denoted
by f -1 exists with

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EXAMPLE
EXAMPLE

RELATION BETWEEN GRAPH f(x) AND GRAPH f-1(x) EXAMPLE 2

The graph of f -1 is the reflection of the graph f on the line y = x Given

y (a) Find f-1
(b) Sketch the graph of f and f -1 on the
EXAMPLE : f(x) = 2x + 1
same axes
x (c) Determine the domain and range for

functions f and f-1 .

Line y = x

Algebraic Approach Using the Property

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The graph of f(x) and f-1(x)

y

x

EXAMPLE 3 Using the Property

Given

(a) Find f-1
(b) Sketch the graph of f and f -1 on the

same axes
(c) Determine the domain and range for

functions f and f-1 .

Algebraic Approach The graph of f(x) and f-1(x)
y

x

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EXAMPLE 4
Given
(a) Find f-1
(b) Sketch the graph of f and f -1 on the

same axes
(c) Determine the domain and range for

functions f and f-1 .

(b) y

(a)

x

EXAMPLE 4
Given f(x) = ex .
(a) Compute f -1.
(b) Sketch the graphs of f and f-1

on the same axes.
Thus, state the domain and
range for f and f-1.

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b) y

x

Relationship between an exponential EXAMPLE 5
and a logarithmic functions
Given f(x) = e-x + 2 .
Both graph are the reflection of each (a) Find f -1
other about the line y = x
(b) State the domain and range
Thus one is the inverse of the other for f -1
functions
(c) Sketch f and f-1 on the same
diagram

Solution Solution

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EXAMPLE 6 Solution

Given f(x) = ln(x+3).

(a) Find f -1
(b) State the domain and range for f -1
(c) Sketch f and f-1 on the same diagram

Solution Solution

FUNCTIONS AND GRAPHS Objectives
5.4 Polynomials
(a)Perform algebraic operations of
L polynomials.

(b) Apply the remainder and factor theorems

(c) Determine the roots of the equations and
the zeroes of a polynomial.

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Polynomials Examples of polynomial functions:
A polynomial P(x) of degree n is defined as degree 4
degree 3
where are real numbers degree 2

with n is a positive integer and . 44

43

Examples of nonpolynomial functions: The Algebraic Operations on Polynomials

45 Addition and Subtraction
The addition and subtraction of the
polynomial can be performed by collecting
the like terms.

Example 1 :
Given

Determine: (a) P(x) + Q(x) (b) P(x) – Q(x)

46

Solution Multiplication

Example 2 :

Given (b) P(x)Q(x)
Determine: (a) 3P(x)

Solution:

48

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Solution NOTE :

If P(x) is a polynomial of degree m and
Q(x) is a polynomial of degree n,
then the product of

P(x)Q(x) is a polynomial of degree (m + n)
From previous

50

Long Division : The division of polynomials
In the division of integer, can be expressed in the form of

=3+ or
• the quotient is 3,
• the remainder is 5, P(x) = divisor (quotient) + remainder

: 52

• the divisor is 9.

51

Example 3 : Solution
By using long division, find

dividend divisor

53

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Example 4 : Solution
Find

55

Example 5 : Solution

Divide by .

.

57

NOTE : CONCLUSION
A polynomial P(x) of degree n is defined as
From the previous example, we have:

linear constant

OPERATION ON POLYNOMIALS

quadratic linear (a) Addition (b) Subtraction

(c) Multiplication (d) Division

So, we can conclude that The division of polynomials can be expressed
in the form of

the degree of the remainder is one degree

less than the degree of the divisor. 59 60

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The Remainder Theorem NOTE :

When a polynomial P(x) is divided by x-a, then, the 1. If P(x) is divided by x + a = x - (-a),
remainder is P(a). then R =

Proof
Let P(x) be a polynomial of degree n (n ≥ 2).

P(x) = Q(x) (x – a) + R

divisor constant 2. If P(x) is divided by ax – b = ,
then R =
When x = a,

P(a) = Q(a) (a – a) + R

(a – a) = 0

then the remainder, R = P(a)

Example 6: Example 7 :

Find the remainders when When is divided by
is divided by :
the remainder is 2. Determine k.
Solution:
Solution:

Example 8 :

The polynomial P(x) gives a remainder

of 1 when divided by a

remainder of 3 when divided by .

Determine the remainder when P(x) is

divided by .

Solution:

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The Factor Theorem Example 10 :
If P(a) = 0, then (x – a) is a factor of P(x)

When is

Example 9 : is a factor of divided by , the
Show that
remainder is 15 and
Solution:

is one of the factor . Determine

the values of a and b.

Solution Solution

ZERO OF A POLYNOMIAL ROOT OF A POLYNOMIAL
If P(a) = 0, then “a” is a zero of If P(a) = 0, then x = a is a root of a
polynomial equation P(x) = 0.
polynomial P(x).
EXAMPLE EXAMPLE
If
If then
then

Thus, are the zeroes of P(x). are the roots of P (x) = 0.

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Example 11 : Solution
Given

(a) show that is a factor of P(x),

(b) factorise P(x) completely ,

(c) find all the zeroes of P(x),

(d) find the roots of P(x) = 0.

Solution Solution

Example 12 : Solution

-1 is one of the zeroes of the polynomial
P(x) = x4 + mx3 + nx2 –x + 6 and P(x) is
divisible by x - 1. Find the values of m and
n. Hence, find all the roots of P(x) = 0.

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Solution Solution

Example 13 : Solution
Find all the roots of x3 + 3x2 + x – 1 = 0.

Solution:

Solution CONCLUSION
The Remainder Theorem

• When a polynomial P(x) is divided by a
x - a , the remainder is P(a).

The Factor Theorem
If P(a) = 0, then (x – a) is a factor of P(x)

14

TUTORIAL

TOPIC 5.0 FUNCTIONS AND GRAPHS

5.1 Functions

1. Sketch the graph of each of the following functions and determine its domain
and range.

(a) f : x  x2  2
(b) f : x  x
(c) f : x  3  x2
(d) f : x  x  2
(e) f : x  x3  3
(f) f : x  7  x
(g) f : x  7  x
(h) f : x  x  2  3
(i) f : x  x3  2
Ans: Check your answer using https://www.meta-calculator.com/

2. Sketch the graph of each of the following piecewise function and determine its

domain and range.

(i) f  x   x, x  0
 x, x  9


x2, x  0

(ii) g  x  3,0  x  2

2x, x  2

Ans: Check your answer using https://www.meta-calculator.com/

TUTORIAL

3. Sketch the graph of each of the following functions. From the graph,
determine the domain and the range.

(a) h x  ex  2
(b) f  x  3  ex
(c) h x  log3x 1
(d) f  x  2  log x  2
(e) g  x  lnx
(f) k  x  ln x  3

Ans: Check your answer using https://www.meta-calculator.com/

5.2 Composite functions

(a) Given f  x  1 2x , g  x  x2  3 , h x  x  2 and k  x  5 . Find

x

(i)  f g  x Ans: 2x2  5

(ii)  g f  x Ans: 4x2  4x  4

(iii) h g  x Ans: x2 1

(iv)  f k  x Ans: 1  10
x

(v) h2  x Ans: x  2  2

(vi)  g k 10 Ans: 13
(vii) the value of if h f c  5 4

Ans: 13

(b) Given f  x  x  2 and g  x  5 . Express the following functions as

x
composite functions of and .

(i) p x  5  2 Ans:  f g  x

x

(ii) q(x)  5 Ans:  g f  x
x2

(iii) r  x  x  4 Ans:  f f  x

(iv) s x  x TUTORIAL

Ans:  g g  x

(c) Given f  x  3  2x . Find g  x if Ans: g  x  x2 1
(i)  f g  x  4  x2
(ii)  g f  x  1 2x , x  2 2

2 x Ans: gx  2x  2  1

,x
x 1

(d) If h x  x  2 and h k  x  x2  3x  2 , find k  x .
Ans: k  x  x2  3x

(e) If u  x  2 and vu x  3  5, find v x .

xx

Ans: v x  3 x  5

2

5.3 Inverse Functions

1. Show whether the following functions are one-to-one. For functions that are
one-to-one, find their inverse and state the domain and range.

(a) f  x  1 5x, x 
(b) g  x  2x2  x  3, x 
(c) h x  5  x, x  5
(d) p x  x2  4, x  0
(e) m x  x2  4, x  2
(f) n x  x2  2x  2, x  1
(g) k  x  3x 1 , x  1

3
Ans: Refer answer to lecturer

TUTORIAL

2. Given three functions f  x  2 x  8, x  R , g  x  e2x4, x  &

3

h x  ln(8  3x), x  8 .

3
(a) Show that f(x), g(x) and h(x) are one to one function.
(b) Find the inverse of each function if exist.
Ans: Refer answer to lecturer

3. The function f is defined by f  x  3x2  2, x  0 .

(a) State the range of and sketch its graph
(b) Explain why the inverse of exist and sketch its graph on the same

axes.
Ans: Refer to lecturer

4. Function f is defined by f  x  ln 2x  5 .

 (a) Show that f 1  x  1 ex  5 and determine the domain and range of
2
f 1 .
(b) Sketch the graphs of f and f 1 on the same diagram. Indicate the

relationship between f and f 1 clearly.
Ans: Refer to lecturer

5. The functions and are defined as follows

f  x  ln x 1, x  1
g  x  x2  2x, x  1

(a) State the range of and find the inverse function of .

Ans: f 1  x  ex 1,,

(b) Determine whether the function is one-to one and state the range of

g . Ans: 1,

(c) Find the composite function of f g . TUTORIAL

 (d) Find the values of x if f 1 g  x  e3 1. Ans: 2ln  x 1

Ans: x  3 or 1

6. Given a function

f  x  3 1 3x, x  m.
Find the most value of m. Show that f 1  x exist. Hence or otherwise, find
f 1  x .

Ans: Refer to lecturer

7. The function f and g are defined as

f x  2 1 x

g  x  x2  2x, x  1

(a) If a function h is defined as h(x)   f (x) for x , sketch h(x). State the
 g ( x)

domain and range. Ans: D:  ,1   3 ,  
 2 

R: ,

(b) Show that h(x) is not one to one function.

 (c) Find f 1  x and g1  x . Hence, evaluate f 1 4 and g1 f 1 2

Ans: 3 , 5
2

TUTORIAL

5.4 Polynomials

1. If P  x  5x  2 , Q x  3x2  5x 1 and R x  x3  4x2  x  3 , find the

following

(a) P 3 Ans: 17

(b) Q 2 Ans: 3

(c) R 4 Ans: 135

2. Simplify each of the following. Ans: 8x2  6x  3

(a) 5x2 10x  3  3x2  4x  6

   (b) 7 3x3  6x2  4x  4 4x3  3x2  9 Ans: 37x3  30x2  28x  36

(c) 2x2  x 12   x2  3x 1 Ans: x2  4x 13

   (d) 5 x3  4x2  3x 8  3 x3  5x2 10x  6 Ans: 2x3  35x2 15x  58

3. Expand and simplify each of the following.

(a) 4x3  97  3x Ans: 12x4  28x3  27x  63

(b)  x2  6x  42x  9 Ans: 2x3  3x2  46x  36

(c) 6x  5 x  22 Ans: 6x3 19x2  4x  20

TUTORIAL

4. Find the quotient and remainder of the following functions by using long

division. Write the answers in the form P x  Q x.D x  R  x .

15x2 16x  9 Ans: P x  5x  23x  2  5
(a)

3x  2

3x3 13x2  20x 18 Ans: P x   x2  5x 103x  2  38
(b)

3x  2

8x3  2x2  29x  3 Ans: P x  4x 12x2  x  6  4x  3
(c) 2x2  x  6

x3  27 Ans: P x   x2  3x  9 x  3
(d)

x3

5. Given that P x  x3  5x  3 . Find the remainder when P  x is divided by

(a)  x  2 Ans: 21

(b) 2x 1 Ans: 3
8

6. Find the value of a if P x  2x3  x2  a has a remainder 9 when divided

by  x  2 . Ans: a  3

7. The polynomial 2x3  ax2  3x  b has a remainder 13 when divided by

 x 1 , and a remainder 5 when divided by 2x 1 . Find the values of a

and b . Ans: a  15, b  7

8. When x3  px2  qx  5 is divided by  x 1 x  2 , the remainder is 2x  7 .

Find the values of p and q . Ans: p  2 , q  1

TUTORIAL

9. When the polynomial ax3  bx2  x  3 is divided by  x 1 and  x  2 , the

remainders are 5 and 17 respectively. Calculate the values of a and b .
Ans: a  1, b  2

10. When 2x3  x2  px  q is divided by x2  x  6 , the remainder is 10x  5 .

Find the values of p and q . Ans: p  3 , q  11

11. Prove that 2x 1 is a factor of P x  2x3  3x2  8x  3. Hence, factorize

P  x completely. Ans: 2x 1 x 1 x  3

12. Find the value of p if  x  2 is a factor of P x  6x3 17x2  px 12 .
Hence, find all the factors of P  x .

Ans: p  4 , The factors are  x  2 , 3x  2 and 2x  3

13. Find the values of a and b if  x  3 and  x 1 are factors of
P x  4x3  9x2  ax  b .State the third factor of P  x . Ans: 4x 1

14. Show that 4 is a zero of the polynomial P x  3x3 8x2 15x  4 . By
factorizing P  x completely, find the other two zeroes.

Ans: The zeroes are  1 , 1 and 4
3

15. Find all the zeroes of P x  x3  4x2  x  6 .

Ans: The zeroes are 1, 2 and 3

TUTORIAL

16. Show that x 1 is a factor of P x  3x3  7x2  2x 8 . Factorize P  x
completely and determine the roots of the equation P  x  0 .

Ans: The roots are x  2, x   4 and x  1
3

17. Given that 2 is a zero of the polynomial P x  2x3  5x2  x  6 . Factorize
P  x completely and hence, solve the equation P  x  0.

Ans: x  2, x   3 and x  1
2

6.1 Arithmetic Sequence and OBJECTIVES
Series
(a) Apply n-th term and sum of the
1 first n-th term formula in various
situation

(b) Solve problem involving arithmetic
sequence and series in business and
economics

SEQUENCES AND SERIES SEQUENCES AND SERIES

An arithmetic sequence is a sequence of An arithmetic series is the sum an arithmetic
numbers such that the difference sequence
between the consecutive terms is
constant

2, 4, 6, 8, 10 Finite sequence 2 + 4 + 6 +8 + 10 Finite series
1, 3, 5, 7, ... Infinite sequence 1 + 3 + 5 + 7 + ... Infinite series

3 4

SEQUENCES AND SERIES Given a, a + d, a + 2d, a + 3d, …
T1 = a + 0d = a + (1 – 1)d
2, 4, 6, 8, 10 Finite sequence T2 = a + 1d = a + (2 – 1)d
2 + 4 + 6 +8 + 10 Finite series T3 = a + 2d = a + (3 – 1)d
1, 3, 5, 7, ... Infinite sequence
1 + 3 + 5 + 7 + ... Infinite series .
.
5 .

Tn = a + (n – 1)d

The nth term of an arithmetic sequence is
given by

Tn = a + ( n – 1 )d

1

• To find the common difference of an Example 1
arithmetic sequence is
Given the first term, a and common
d  Tn  Tn1 difference, d of an arithmetic sequence.
Find the n-th terms in the sequence

(a) a=-1.4, d=0.6

(b) a=44, d=-2

7

Solution Example 2

The 11th term of an arithmetic sequence
is 52 and the 19th term is 92. Find the
1000th term.

Solution

9

Example 4
4n  5

The nth term of a sequence is 12 . Identify
that the sequence is an arithmetic sequence .

Solution

Tn1  Tn 1

2

Example 5 SUM OF A FINITE ARITHMETIC SERIES

Find the number of terms between 100 and Sn  n 2a  (n  1)d 
500 that are divisible by 7.
2
Solution

Sn  n a  l ,

2

l is the last term

Example 6 Example 7

20 In an arithmetic sequence, the 10th term is 3
and the sum for the first six terms is 76.5.
Find  2n  4 Find
n 1 (a)the first term and the common difference,
(b) the number of terms to be taken so that
Solution:
its sum is zero.

Solution:

17 18

3

Relationship Between Tn and Sn Example 7

Tn = Sn - Sn - 1 The sum of the first n terms of an arithmetic
series is given by Sn = 2n + n2. Find
(a) an expression for the nth term and the

common difference.
(b) the sum of the 11th term to the 20th term.

Example 8
Adriana needs RM 8500. She decides to save

RM 100 on the first month and increase her
saving by RM 20 for subsequent month.
Find how long she need to achieve her
goals

4

Solution: CONCLUSION

The nth term of an arithmetic sequence is

given by Tn = a + ( n – 1 )d

Sn  n 2a  (n  1)d 

2

Sn  n a  l
2

25

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6.2 Geometric sequence and OBJECTIVES
Series
(a) Apply the n-th term and sum of the n-th
term formula in various situation

(b)Solve problem involving geometric
sequences and series in business and
economics

A Geometric Sequence is a sequence of A Geometric Series is a sum of a geometric
numbers in which each term can be obtained sequence
from the previous term by multiplying by a
constant number called the common ratio. 2+6+18+54+⋯
0+5+2.5+1.25,⋯
2,6,18,54,⋯

0,5,2.5,1.25,⋯

Let a be the first term and r be the common ratio.

a , a r , a r2 , a r3 , . . . The nth term is
T1  a  ar11
ar  ar21 Tn  ar n1
T2  ar2  ar31
T3  ar3  ar41
T4 

. The common ratio,
.
. r  T2  T3 ... Tn

Tn  arn1 T1 T2 Tn1

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Example 1

Given sequence 3, -6, 12, -24, … . Find

(a)the nth term (b) the 9th term

If the kth term is 192, find the value of k.

Solution:

Example 2 Solution:  3 n1
 2
3 n1 Tn  5 
2  
The nth term of a sequence is Tn  5 


(a) Identify that the sequence is a

geometric sequence .

(b) Find the value of n where

Tn  1215
32

5 3  n 1 Example 2
 2 
Tn   10
(b)
Find 3n1

n1

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Solution: The Sum of Geometric Series

Sn  a(1 rn ) or Sn  a(rn 1)
1 r r 1

Example 3

The first term of a geometric sequence is 27 and
the common ratio is 4 . Find

3

(a) the sum of the first 5 terms,
(b) the least number of terms the sequence

can have if the sum exceeds 550.
Solution:

a(rn 1)
Sn  r 1

Example 4
Each year the price of a car depreciates
by 10% of the value at the beginning of
the year. If the original price of the car
was RM 50 000, find its price after 10 years

3

Solution: 7/7/2022
Solution:
Example 5
The first term of a geometric sequence is 6.
The last term is 1458 and the sum of all term
is 2184. Find the common ratio of this
geometric sequence

Solution:

Relationship Between Tn and Sn Example 6

Tn = Sn - Sn - 1 Sn  Tn  1

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Solution: Solution:

Solution: Solution:

CONCLUSION

a, ar, ar2 ,..., arn1

nth term, Tn  ar n1
The Sum of Geometric Series

Sn  a(1 r n ) S  a, where r 1
1 r 1r

5

TUTORIAL

TOPIC 6.0 SEQUENCES & SERIES

6.1 Arithmetic sequence and Series

1 Write down the term indicated in square brackets in each of the following
arithmetic sequences.

(i) 6,11,16,21,...;12thterm Ans: 61

(ii) 5 p, 3 p.7 p,2 p,...;nthterm Ans: p  n  4 
4 
4 24

2 Find the number of terms in each of the following arithmetic sequences

(i) 3.2,4.3,5.4,...,31.8 Ans: 27terms

(ii) 11 ,15 ,2 1 ,...,19 5 Ans: 29terms
66 2 6

3 Find the sum, as far as the term indicated in square brackets, of each of the
following arithmetic sequence.

(i) 1 11  21,...;24thterm Ans: 2784

(ii) 31  2 2  2,...;99thterm Ans: 2904

33

4 The 8th term and 12th term of an arithmetic sequence are 48 and 62
respectively. Find

(i) the first term and common difference Ans: a   47 ,d   7
22

(ii) the 20th term Ans: 90

(iii) the th term Ans: Tn   7 n  20
2

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TUTORIAL

5 The sum of the sixth term and tenth terms of an arithmetic sequence is 24 .
The eleventh term exceeds the eighth term by 8 . Calculate the common
difference and the first term.
Ans: a   20 ,d  8
33

6 Find the sum of an arithmetic series of 36 terms where the first term and the
last term are 7 and 214 respectively.

Ans: 3978

7 The fourth term of an arithmetic series is 9 and the sum of the first four terms

is 21 . Find the th term, Tn of this arithmetic sequence. Hence, find the
biggest value of such that Tn  80 .

Ans: Tn  5 n 1,n  32
2

8 Isahak starts with a monthly salary of RM1250 for the first year and receives
annual increment of RM80. How much is the salary for the n-th year of
service? How much will he receive monthly for his tenth year of service.

Ans: Tn  80n 1170, RM1970

9 In a contest, all ten finalists were given cash prizes. The first winner was given
RM800, the second RM740, the third RM680 and so on. Calculate the total
amount of money awarded to all the finalists.

Ans: RM5300

10 The purchase value of an office computer is RM12500. Its annual depreciation
is RM1875. Find the value of the computer after 6 years.

Ans: RM1250

11 Mei Ling saves her money to buy a new car. The first month she saves RM
600. Every month her saving increases by RM 100. Find
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TUTORIAL

(i) how much money she will save in the 11th month,

(ii) the total saving after 11 month

Ans: (i) RM1600 (ii) RM12100

6.2 Geometric Sequence and Series

1 Write down the term indicated in square brackets in each of the following
geometric sequence.

(i) 3,9,27,...;6thterm Ans: T6  729

(ii) 1 , 1 , 1 ,...nthterm Ans: Tn  1  2 n1 or Tn  3 2 n
4  3  8  3 
469

2 Find the number of terms in each of the following geometric sequences

(i) 0.12,0.48,1.92,...,491.52 Ans: 7

(ii) 2 1 ,3 3 , 5 1 ,..., 25 161 Ans: 7
4 8 16 256

3 Find the sum, as for as the term indicated in square brackets, of each of the
following geometric sequences.

(i) 6 12,24,...;10thterm Ans: 2046

(ii) 1.11.211.331, ...;5thterm Ans: 6.71561

4 Given the geometric sequence 5,15,45,135,.... Find

(i) the eighth term and the tenth term Ans: 10935,98415

(ii) the sum of the first eighth terms Ans: 16400

3

TUTORIAL

6 Find the number of terms in the series 2  6 18...39366 . Calculate the

sum of all the terms. Ans: 59048

7 Find the minimum number of terms that must be taken from the sequence
3,12,48,192,... so that the sum is more than 300.

Ans: n  5

8 A young graduate is offered a job with a salary of RM24000 for the first year
and a 5% raise each year after that. If that 5% raise continues every year, find
the amount of money earned in 30 years.

Ans: RM1594532.34

9 Ah Chen saves RM50 in January, RM 100 I February, RM 200 in March an so
forth by doubling his savings every month for 1.5 years. Find

(i) how much he saves in the month of October in the first year

Ans: RM25600

(ii) how much he saved after 1.5 years

Ans: RM13107150

10 If the sum of the first n terms of a geometric sequence is Sn  6n 1, find the
first term and the second term. Hence, write down the n th term of the
geometric sequence

 Ans: T1  5,T2  30,Tn  5 6n1

11 The sum of the first n terms of a sequence is 1   3 n . Write an expression for
 4 

the nth term of the sequence. Show that the sequence is geometric sequence.

Ans: 1  3 n
3  4 

12 The price of a new car is RM 60000. Its price decreases by 10% every year.
Find
4

TUTORIAL
(i) the price of the car after 5 years,
(ii) the percentage decreases in the price of the car in 5 years’ time.

Ans: (i) RM35429.40 (ii) RM738.73

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TOPIC 7 LEARNING OUTCOMES

MATHEMATICS OF FINANCE At the end of the lesson, students should be able to :
7.1 SIMPLE INTEREST (a) determine simple interest and the total amount.
(b) determine exact and ordinary interest.
(c) solve problem involving period of time, simple

interest rate and principal.

INTRODUCTION INTRODUCTION

Interest can be defined in two ways. Definition 2 : Interest is charge incurred when a loan or credit is
Definition 1 : Interest is the money earned when the money is obtained.

invested. Example : You borrowed RM1000 from the same bank for a year and
Example : You deposited RM1000 in a bank for a year and you find you paid back RM1080 at the end of that year.
The additional amount of RM80 is the charged or interest
that at the end of year 1, you have RM1050 in your account. you need to pay when you borrow RM1000.
The additional amount of RM50 is the interest you earned
when you invested RM1000.

SIMPLE INTEREST The simple interest, I for a principal, P for n years at an annual rate r is
given by the formula:
Simple interest is the interest that is calculated based on the
principal for the entire period it is borrowed or invested. I  Prt

The amount of the simple interest due at the end of the term P = Principal
depends on the principal, the length of the term and the rate of t = time or term in years
interest. r = simple interest rate per year
I = simple interest

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The simple amount is the sum of the original principal and the interest earned. EXACT SIMPLE INTEREST
The simple amount formula is given as It is calculated based on 365 days per year or 366 days for the
leap year.
P = Principal S PI
t = time or term in years S  P  Prt ORDINARY SIMPLE INTEREST
r = simple interest rate per year It is calculated based on 360 days per year. It is considered
S = amount (future value) S  P1 rt there are only 30 days for a month. It will increase the amount
I = simple interest of interest due to the lender.

EXACT AND ORDINARY SIMPLE INTEREST Example 1:
If the term in years given in number of days : Find the interest and the future amount for RM1000 deposited
in a bank if the simple interest is 8% per annum for two years.
Exact time : t  number of days or t  number of days

365 366

Approximate time : t  number of days

360

Solution : Solution :

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Example 2: Solution :
If Maria borrowed a sum of RM46,500 for a period of 21
months at 20% per annum, how much she must repay the bank?
How much interest must be paid?

Solution : Example 3:

Given RM2000 at 8% simple interest for 175 days.
Compute
(a) exact interest
(b) ordinary interest

Solution : (b) ordinary interest Example 4:

(a) exact interest How long will it take for a sum of money to double at a simple interest rate of 5% per year?

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Solution : Solution :

Example 5: Solution :
Mariam took loan from a bank that offers 8% simple interest
per annum. After 3 years, she had to pay the amount of
interest RM5400. How much the principal amount borrow by
her?

Solution : TOPIC 7

MATHEMATICS OF FINANCE
7.2 SIMPLE DISCOUNT

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LEARNING OUTCOMES SIMPLE DISCOUNT

At the end of the lesson, students should be able to : Simple discount is the amount that the bank deducts from a loan in
(a) determine the simple discount and proceed. advance.
(b) determine the equivalent simple interest rate to
The amount the borrower receives after the deduction is called the
simple discount rate. proceeds.
(c) solve problems involving finding proceed, period of
So in a bank discount, the amount borrowed (before deducting the
time, amount and simple discount rate. charges) is the future value.

Simple discount is also known as bank discount.

SIMPLE DISCOUNT EQUIVALENT SIMPLE INTEREST

D  Sdt Proceed  S  D An interest r% and a discount rate d% are said to be equivalent if the
two rates given the same present value for an amount due in the future.
Proceed  S1 dt

S = amount (money that need to be returned) The equivalent simple interest rate to simple discount rate
D = simple discount
d = simple discount rate per year
t = time or term in years

If the discount period is in number of days ; r d r  Discount
1 dt Proceed × Time
t =

Example 1: Solution :

Halim borrowed RM25000 for 3 months and he was charged a
bank discount rate of 6.5% per year. What is the bank’s
discount and Abdul Halim proceed?

5

Solution : 7/7/2022
Solution :
Example 2:
Amin needs RM15000 now. How much should he borrow from
a bank which charges 5% discount rate for 3 years. Determine
the simple interest rate that is equivalent to the discount rate.

Solution :

TOPIC 7 LEARNING OUTCOMES

MATHEMATICS OF FINANCE At the end of the lesson, students should be able to:
(a) determine the compound interest and compound amount.
7.3 COMPOUND INTEREST (b) solve problems involving compound amount, period of time,

interest rate, present value and effective rate.
(c) solve problems involving more than one investment and/or

more than one nominal rate with different compounding
rate.

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COMPOUND INTEREST COMPOUND AMOUNT

A common procedure for computing interest is by compounding Definition : The compound or future value or accumulated value is the final sum
interest. Under this procedure, the interest is reinvested. The interest at the end of the period.
earned each period is added to the principal for purposes of computing The future value (compound amount) for the nth period may be expressed as
interest for the next period. The amount of interest computed using this
procedure is called the compound interest. S  P1 in i  j , n  tm
m
Compound interest is the interest earned when the invested money or
principal is reinvested. P = original principal

t = time or term in years

j = nominal rate

m = frequency (number of compounding periods per year)

S = compound amount (future value)

COMPOUND INTEREST ADJUSTED PERIOD

The compound interest, I is the difference between the future value and The adjusted period, m is the number of times the interest is calculated
the original principal. per annum.

I SP Daily, m = 365 times per annum
Weekly, m = 52 times per annum
S = compound amount (future value) Monthly, m = 12 times per annum
P = original principal Quarterly, m = 4 times per annum
Semi annually, m = 2 times per annum

EFFECTIVE INTEREST RATE Example 1:
Suppose that RM1000 is invested in a bank, which earns
Two compound interest rates are said to be equivalent if both yield the interest at a rate of 8% per year compounded annually. What
same future value at the end of one year for any investment RM P. will the account balance be after 10 years and the amount of
interest earned?
re  (1 i)m 1

(is the yearly interest rate that is equivalent to the interest rate
compounded more frequent in a year)

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Solution : Solution :

Example 2: Solution :
How long will it take to increase RM1800 to RM3002 at the
rate of 9% per year compounded every three months?

Solution :

Example 3:
Find the nominal interest rate if RM1000 invested becomes
RM3248.03 compounded quarterly in 8 years?

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Solution : Solution :

Example 4: Solution :
Johan borrowed some amount from a bank at a rate of 3%
compounded annually. If he finished paying his loan by
paying RM8023.53 at the end of 4 years, what is the amount
of loan that he had taken? Hence, calculate the total interest
that he has to pay.

Solution : Example 5:

Find the effective rate for:
(a) 16 % compounded quarterly,
(b) 15% compounded monthly.

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Solution : Solution :

Example 6: Solution : RM1000 RM1000 S
Ahmad deposits RM1000 now in a bank at an interest rate of 0 2 10
8% compounded quarterly. He will invest another RM1000
after two years. What will be the future value of his account
ten years from now?

Solution :

TOPIC 7

MATHEMATICS OF FINANCE
7.4 ANNUITY

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LEARNING OUTCOMES ANNUITY

At the end of the lesson, students should be able to: An annuity is a series of periodic payments or deposits, usually made
(a) Determine in equal amount at equal intervals of time. The payments are
computed by the compound interest method such as annually,
(i) the amount of annuity or future value semiannually, quarterly or monthly.
(ii) the present value
(iii) the regular or periodic payment; and Examples of annuity are life insurance premium, pension, periodic
(iv) the total interest. saving and buying of house on installment plan.
(b) Solve problems involving mixed compound interest and annuity.

FUTURE VALUE OF THE ANNUITY FUTURE ANNUITY

Payments R R R ….. R RR Therefore, the formula for the future annuity is:
n – 2 n – 1 n amount accumulated
Period 0 1 2 3
R(1 + i)0
Sn   (1 i)n 1 
R(1 + i)1 R i 

R(1 + i)2 
:
: Sn = future value
R = the regular or periodic payment
R(1 + i)n-3
i = the interest rate per compounding period
R(1 + i)n-2 n = number of compounding periods

R(1 + i)n-1

Sn

FUTURE ANNUITY Example 1:

Interest, I  S n  R tm A teenager plans to deposit RM50 in a savings account at the end of
each quarter for the next 6 years. Interest is earned at a rate of 8%
Sn = future value compounded quarterly. What should her account balance be 6 years
R = the regular or periodic payment from now? How much interest will she earn?
t = time or term in years
m = number of compounding periods per year 11