AM015 LECTURE & TUTORIAL NOTES

AM015
LECTURE&TUTORIALS

Ambilan 2021

KERJASAMA PENSYARAH
UNIT MATEMATIK

KOLEJ MATRIKULASI PERLIS

THERE IS
SOMETHING

everGyOOdDaINy

BY MDM SHAKIRAH
SANUSI

CONTENT:

CH1: NUMBER SYSTEM &
EQUATIONS
CH2: INEQUALITIES &
ABSOLUTE VALUES
CH3: SEQUENCES
CH4: MATRICES & SYSTEMS
OF
LINEAR EQUATIONS
CH5: FUNCTIONS & GRAPHS
CH6: POLYNOMIALS
CH7: LIMITS
CH8: DIFFERENTIATION
CH9: APPLICATIONS OF
DIFFERENTIATION

KNOW IS A MUST

ASSESSMENT:60%

1)UPS 20%:
UPS 1:CH1&CH2
UPS 2:CH3&CH4
UPS 3:CH5&CH6

2)PB 40%:

INDIVIDU 15%:CH3&CH4

KUMPULAN 20%:
CH6&CH7

DISKUSI 5%: CH7

PSPM 1:100M (40%)

-2HOURS 1 PAPER-

PART A(3Q): 25%
CH7: LIMITS
CH8: DIFFERENTIATION
CH9: APPLICATIONS OF
DIFFERENTIATION
PART B(7Q): 75%
CH1: NUMBER SYSTEM &
EQUATIONS
CH2: INEQUALITIES &
ABSOLUTE VALUES
CH5: FUNCTIONS & GRAPHS
CH8: DIFFERENTIATION
CH9: APPLICATIONS OF
DIFFERENTIATION

Ambilan2021:Maybe there is changes in terms
numbeR of Questions for Online PSPM

AM015/ 1. Number System and Equations

CHAPTER 1: NUMBER SYSTEM AND EQUATIONS

LECTURE 1 OF 3
At the end of lesson, student should be able to:

a) Define natural numbers (N), whole numbers (W), integers (Z), prime numbers,

rational numbers (Q) and irrational numbers ( Q ).

b) Represent the relationship of number sets in a real number system

diagrammatically showing N ⊂ W ⊂ Z ⊂ Q and Q ∪ Q =  .

c) Represent open, closed and semi-open intervals and their representations on
the number line.

d) Find 'union' and 'intersection' of two or more intervals with the aid of number
line.

e) Express the rules of indices
f) Solve equations involving indices.

1.1 REAL NUMBERS

The Real Number (  )

 The real number consists of rational numbers and irrational numbers.

Natural Numbers (N)
 is the set of counting number.

 including prime and non prime number N  1, 2,3,...

The Whole Numbers(W)

 are the set of natural numbers together with the number 0. W  0,1, 2,3,...

Prime Number
 are the natural number that greater than 1 and can only be divided by himself

and 1. Prime number  2,3,5,7,11...

Integers (Z)
 The whole numbers together with the negative of counting numbers form the
set of integers and denoted by Z.

Z  ...  3, 2, 1, 0,1, 2,3,...

Even Number

 Are the integers in the set ..., 6, 4, 2, 0, 2, 4, 6,... which can be

represented in the general form {2k : k Z}

Odd Number

 Are the integers in the set ..., 5, 3, 1, 1, 3, 5,... which can be

represented in the general form {2k 1: k Z}

A Rational Number ( Q )

 is any number that can be represented as a ratio (quotient or fraction) of two

integers and can be written as Q a ; a, b  Z, b  0 .
 
 b

 Rational number can also be expressed as terminating or repeating decimals.

Page 1 of 17

AM015/ 1. Number System and Equations
 For example 5, 3 , 0.25 and 0.333… .

2

 Irrational Number Q

 is the set of numbers whose cannot be written as a fraction.
 their decimal representations are neither terminating nor repeating.

 for example 3, 5 , 1.41421356…and  .

Example 1
Rewrite the following numbers into decimal numbers.

a) 0.13

b) 1.236
Example 2
Rewrite the following numbers into bar form.

a) 0.454545...

b) 2.5373737...

Page 2 of 17

AM015/ 1. Number System and Equations

Relationship of Number Sets

The set of real numbers, denoted by the symbol  , consists of all rational numbers

and irrational numbers. Relationships among the subsets of real numbers are
illustrated in the diagram below.

Real number

Rational Number Irrational Numbers

Integers Non- Integers

Negatives of natural Whole Numbers
numbers

Zero Natural numbers

prime numbers Non-prime numbers

Page 3 of 17

AM015/ 1. Number System and Equations

The relationship of number sets in a real number system can also be represented by
the following Venn Diagram


QQ

W 

From the diagram, we can see that:

1. N W  Z  Q  

2. Q Q  

Example 3 7 , 1 , 2, 0, 3, 4, 5.1212…}, identify the set of
Given S = {-9, 3

a) whole numbers

b) integers

c) rational numbers

d) irrational numbers

e) prime number

Page 4 of 17

AM015/ 1. Number System and Equations

The Number Line
The set of numbers that corresponds to all point on number lines is called the set of
real number. The real numbers on the number line are ordered in increasing
magnitude from the left to the right
For example for –3.5, 2 and  can be shown on a real number line as

3

4 3 2 1 01 2 34
3.5
2 
3
Example
Symbol Description 3=3
a=b a equal to b 4 < 4
a<b a less than b 5>0
a>b a greater than b

Note: The symbols ‘<’ or ‘>’ are called inequality sign

All sets of real numbers between a and b, where a  b can be written in the form of
intervals as shown in the following table.

Type of Interval Interval notation Solution set Representation on the number line
Closed ab
a,b x : a  x  b

Open a,b x : a  x  b a b
Half-open a b
Half-open a,b x : a  x  b a b
a,b x : a  x  b
Infinite interval a b
 ,b x : x  b a b
 ,b
a,  x : x  b
x : x  a
a,  x : x  a
x : x
(, )

Note:
The symbol  is not a numerical. When we write [a, ), we are simply referring to the
interval starting at a and continuing indefinitely to the right

Page 5 of 17

AM015/ 1. Number System and Equations
Intersection and Union for Two Intervals
Since intervals are sets of real numbers, we can combine two or more intervals by
using the set of operation of union and intersection.

a) the union of A and B, denoted by A  B , is the set of all elements which

belong to A or to B.

b) the intersection of two sets A and B, denoted by A  B , is the set of

elements which belong to both A and B.
Example 4
Solve the following using the number line

a) 0, 5   4, 7
b) 5, 1 2, 4

c) , 7 7, 

d) , 4 4, 

Example 5

Given A  (9,) , set B  x : 7  x 13, x  Z. Solve the following using the number

line

a) A  B

b) A  B

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AM015/ 1. Number System and Equations

1.2 INDICES

Index indicates the number of times the base is used as a factor.

an

Where
The number a is called the base, n is called the index,
an is read as ‘a’ to the power ‘n’.

Rules of Indices

 1. am  an  a mn 2. am  an  am  amn
an

 3. am n  amn 4.  abm  ambm

5.  a m  am ,b  0 6. a0  1, a  0
 b  bm

1 m m
am
a n  n am  na
7. am   8.

9. am  an  m  n

Example 6 b) 0.04 3
Without using calculator, evaluate: 2

3

a) 9 2

Example 7 b) (a 3b) 2
Simplify: a 8b 2

35  36
a) 34

Example 8

By using the rule of indices, evaluate ( 3)3  1 1 .

27 4 3 4

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AM015/ 1. Number System and Equations

Example 9 b) 27x1  9x1

Find the value of x for the following equation:

a) 5x  125

Example 10 b) 4x  6(2x )  16
Solve the following equation:

 a) 52x1  6 5x 1

Page 8 of 17

AM015/ 1. Number System and Equations

LECTURE 2 OF 3

At the end of lesson, student should be able to:

a) State the laws of logarithms such as:

i. log MN  loga M  loga N
a

ii. loga M  log M  loga N and
N a

iii. N  N loga M

loga M

b) Change the base of logarithm using loga M  logb M .
log a
b

c) Solve equations involving logarithms.

1.2 LOGARITHMS

Logarithm is a number y,(y  0) for any given base a a  0 and is written as

loga y  x where y  a x and x .

Common Logarithm

Common logarithm is a number y (y > 0) for base 10 and is written as log10 y  x
where y  10x and x .

Note : log10 y  log y  lg y

Natural Logarithm of x
The natural logarithms is a number y for base e and is written as loge y  x where
y  ex and x .

Note : loge y  ln y

Example 1
Write in the logarithmic form

a) 23  8

Example 2
Write in the indices form
a) log5 25  2

Page 9 of 17

AM015/ 1. Number System and Equations

The Laws of Logarithms

RULE 1

loga MN  loga M  loga N Product Rule

RULE 2

loga M  loga M  loga N Quotient rule
N

RULE 3

loga M N  N loga M Power Rule

The following table lists the general properties for logarithmic form.

Logarithms with base a Common logarithms Natural logarithms
ln1  0
1) log a 1  0 log1  0
log10 1 ln e 1
2) log a a  1

3) log a ax  x log10x  x lnex  x

4) aloga (x)  x 10log(x)  x eln(x)  x

5) log a  1    log a (N ) log  1    log(N ) ln  1    ln(N )
 N   N   N 

6) log a m  log a n  m  n logm  logn m  n ln m  ln n  m  n

Example 3

Express in terms of logx,logy and logz

a) logxyz

Example 4
Express the following in a form log a and log b.

a) log ab2

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AM015/ 1. Number System and Equations

Example 5
Write the following as single logarithms:
a) log2 8  log2 6  log2 12

Change of Base

loga M  logb M log a M  log M M  1
logb a log M a log M a

Example 6

Find the following expression to four decimal places.

a) log3 5 b) log5 10

Example 7 b) log2 x  log2  x  7  3

Solve the following equation:

 a) 22x3  25 2x  3  0

Page 11 of 17

c) log3 x  4logx 3 3  0 AM015/ 1. Number System and Equations

d) 2ln 4x  2  ln 9

Page 12 of 17

AM015/ 1. Number System and Equations

LECTURE 3 OF 3
At the end of lesson, student should be able to :

a) Explain the meaning of a surd and its conjugate.
b) Perform algebraic operations on surds.
c) Solve equations involving surds.

1.2 SURD
A number expressed in terms of root sign is radical or a surd. Surds cannot be
evaluated exactly.
For example,
“ 4  2 and 3 27  3” = radical.
“ 7 or 3 71 ” = irrational numbers (surds)

Rules of Surds

BIL PROPERTY EXAMPLE
1 a a a
2. a  b  ab a, b  0 5 5 5

2  3  2(3)  6

3. a  a ,a,  0, b  0 3 3
22
bb
2 2  5 2  2  5 2  7 2
4. a b  c b  a  c b

5 a b c b  a c b 6 2  3 2  6  3 2  3 2

 6  2
2 2  3 22 6 3

a  b  a  2 ab  b

7  a  b a  b  5  2 5  2

 2  2  2  2

a b 5 2

 a b 52

3

Remark: a b  a  b b)  3  4 2  3  2 2 

Example 1
Simplify:
a) 5 3  27

Page 13 of 17

AM015/ 1. Number System and Equations

Rationalizing The Denominator is a rational

  The conjugate of a  b is a  b where a  b a  b  a  b

number.

To rationalize a denominator
Multiply the numerator and the denominator of the same expression that will result in
the denominator to become a rational number.

If Denominator Multiply by To Obtain Denominator Free
contains the factor the conjugate
3 from surds
3
1 3  2  3 3 3
1 3
3

 12  2

3  1 3  2

3 2 3 2  32  2
5 3 5 3
2 927

   5 2  3 2  5  3  2

Example 2
Rationalize and simplify:

1 b) 17  5
17  5
a)

7 2

c) 5  3  1 3 d) 1 2  1 2
5  3 1 3 1 2 1 2

Page 14 of 17

AM015/ 1. Number System and Equations

Surd Equation
Example 3
Solve each of the following equation:

a) 3x 1 1  x

b) x  x  2  2  0

c) t  7  t  2  2t  3

Page 15 of 17

AM015/ 1. Number System and Equations

EXERCISE:

1. Represent the followings on the real number line and state its type.

a) [-1, 4] b) {x : 2  x  5}

c) [2, ) d) {x : x  0, x  }

e) (10, ) f) (,4)

2. Given X   5,10 and Y  0,. Find

a) X  Y Ans: 0,10
b) X  Y Ans: 5,

3. Simplify: 6y4
Ans: x 2
18x 2 y 5
a) 3x 4 y a4x

b)  a2b3 3  x  2b1  Ans:
 x1 y2  a2 y3 
   b10 y9

1 1 Ans: x 1
x
x2  x 2
Ans: 1
c) 1 x10 y9

x2 Ans: pq
q p
d) (3x 2 y 3 )3
27x 4

p 1  q 1
e) p 2  q 2

4.Find the value of x for the following equation:

2 Ans: x  8
Ans: x  1 or x  2
a) x3  4
2
b) 49x2  1
73x2

5.Write the following as single logarithms:

a) 1 log 25  2log 3  2log 6 Ans: log20
2

b) 2ln  x  7  ln x Ans: ln  x  72 
 
 x 

d) log a 3xy  5log a y  2 Ans: log a  3xa2 
 y4 
 

Page 16 of 17

6. Solve the following equation: AM015/ 1. Number System and Equations

 a) 4x1 5 2x 1 0 Ans: x  2or x  0
Ans: x  1.7095
b) 2x1  3x Ans: x  ln4
c) e2x 3ex  4  0 Ans: x  3
Ans: x  1.4307
d) x5e3lnx  4x  21 Ans: x  9
e) 5x  10 Ans: x  1

f) lnx  2  ln3x 16 2

g) log8 x3  log x 8  2 , x  8 Ans: 3 5
Ans:8 7
7. Simplify: Ans: 5 42
a) 45 Ans: 4 21  3
b) 6 7  2 7
c) 3 6  5 7

d) 3 4 7  3

8. Rationalize and simplify: Ans: 6 2
a) 12

2

b) 3 Ans: x  3
23 2

c) 2  5  5 Ans: 1  5 5
3 5 44

9. Solve each of the following equation: Ans: x  13
a) 2x 1  5  0

Page 17 of 17

AM015/1 .Number System And Equations

TUTORIAL CHAPTER 1 : NUMBER SYSTEM AND EQUATIONS

TUTORIAL 1 OF 3

1. Determine whether each statement is true or false

True/false

a) Every counting number is an integer
b) Zero is a counting number
c) Negative six is greater than negative three
d) Some of the integers is natural numbers

e) 3,5  x : 3  x  5

2. List the number describe and graph them on the number line

(a) The counting number smaller than 6 (b) The integer between -3 and 3

Given S  3,0, 7, 1 , e, 4,8
3
3. identify the set of

(a) Natural numbers (b) whole numbers (c) integers

(d) Rational numbers (e) irrational numbers

4. Write each of the following inequalities in interval notation or set notation and show them
on the real number line.

(a) x : 2  x  6 (b) x : 3  x  7 (c) x : x  3

(d) x : x  1 (e) (4, 4) (f) (,5]

(g) (2, 0)  (3, 6) (h) [6, 2)  [3, 7) (i)  ,33,

Page 1 of 6

AM015/1 .Number System And Equations

TUTORIAL 2 OF 3

1. Evaluate

1 2  100   3 11 1. 1
3 2
(a) 27 (b)  27  3 (c) 9 2.8 2 9 3 .27 2
8 9
(d) 1 (e) 1  2

22 3 6.3 3

2. Simplify the following expressions:

(a) 3n2 9n  27n 4n  2 n 1 n

(b) 83 164

n 1 (d) 5n1 10n  202n 23n

(c) 9 2  3n3  32 5

3. Solve the following equations:

(a) 3x  243 (b) 2x1  1 (c) 4x2 1281x
64 (f) (4x ) x  4  8x

(d) 2  1 (e) 7x2  4962x  0

4x 3 9

4. Solve the equations: (b) 32x1  26(3x )  9  0 (c) 4x  6(2x ) 16  0

(a) 2(22x )  5(2x )  2  0

5. Without using calculator, evaluate

(a) log 4 64 (b) log 1 4 (c) log 125 25 (d) ln e2
log 9
2
(g)
(e) e 2ln3 log 125
log 3
(f)

log 5

6. Simplify

(a) log2  log6  log4 (b) 2log3  log2 (c) 1 log 25  2log 3  2log 6
2

(d) ln y3  1 ln( x3 y6 )  5ln y (e) 2 ln x  4ln 1   3ln( xy)
3 y

7. Solve the equations:

(a) 22x  5 (b) 33x1  7 (c) (5x )(5x1 )  10 (d) e2lnx  9

(e) eln(1x)  2x (f) 3e2x  75 (g) 5x  40 (h) e2x  ex  2  0

e2

Page 2 of 6

8. Solve these equations: AM015/1 .Number System And Equations

(a) 2log(x  2)  log(2x  5) (b) 2ln x  ln3 ln(6  x)
(c) log 2 (2x  4)  2  log 2 (x2  6) (d) log( x2  6)  log( x 1)  1
(e) ln x  2  ln(1 x) (f) log 3 x  2log x 3  1
(h) log 2 2x  log 4 (x  3)
(g) log 2 x  log x 2  2
(i) log 2 (2x1  32 2)  x

TUTORIAL 3 OF 3

1.Express in terms of the simplest possible surds:

(a) 8 (b) 75 (c) 180 (d) 125

2.Simplify: (b) ( 2 1)( 2 1) (c) ( 3  2)( 3 1) (d) (2 5  3)(3 5  2)

(a) 2(3 2)

3.Rationalise the denominators and simplify in the form a  b c :

3 1 11
(a) 2  3 (b) 3  2 5 (c) 2  1 2 1

4.Solve the equations:

(a) 3x 1  x 1 (b) 4x  9 1  2 x (c) 4x 13  x 1  12  x

ANSWERS :TUTORIAL CHAPTER 1

TUTORIAL 1 OF 3

1 (a) True (b) false (c) false (d) true (e) false

2 (a) N = {1, 2, 3, 4, 5} graph (b) Z = {-2, -1, 0, 1, 2} graph

3 (d) 3, 0, 1 , 4, 8 e) 7,e
3
(a) 4,8 (b)0, 4,8 (c)3, 0, 4,8

4 (a) 2,6 ( b) 3,7 (c) ,3

(d) 1, (e) x : 4  x  4 (f) x : x  5

(g) x : 2  x  03  x  6 (h) x : 3  x  2 (i) { }

graph on real number line - your own answers

Page 3 of 6

AM015/1 .Number System And Equations

TUTORIAL 2 OF 3

11 9 27
(c) 1000
(a) 3 (b) 4 (d) 6 (e) 1

2 (a) 9 (b) 2n (c) 27 (d) 5
2

3 (c) 1 (d) -216 , 216 (e) -6, 2 ( f)  1 ,2
(a) -5 ( b) -7 3 2

4 (a) x  1,1 (b) x  2 (c) x  3

5 (a) 3 (b) 2 (c) 2 (d) 2 (e) 3e2 (f) 3 (g) 2
3

6 (d) ln x (e) ln y 
x
(a) log3 (b) log18 (c) log20

7 (a) 1.1610 (b) 0.9237 (c) 1.2153

(d) 3 (e) 1 (f) 1.6094 (g) 1.4756 (h) 0
3

8 (a) 3 (b) 3 (c) 5 e2 (f) 9 or 1 (i) 11
2 (e) (1  e2 ) 3 2
(d) 8, 2 (g)2 (h) 1

TUTORIAL 3 OF 3

1 a) 2 2 b) 5 3 c) 6 5 d) 5 5

2 a) 3 2  2 b) 1 c) 5  3 3 d) 24  5 5

3 b) 2 5  3 c) 2 2
11 11
a) 6  3 3

4 b) 25 c) 3
a) 5 4

Page 4 of 6

AM015/1 .Number System And Equations
EXTRA EXERCISE CHAPTER 1 : NUMBER SYSTEM AND EQUATIONS

1) (a) Simplify

 32n3 18 3 2n1

 5 3n 2

(b) Without using the calculator, evaluate

1

log10 8  log10 272  log10 5

3 log10 6  log10 5
2

(c) Solve

x  5  4x 13

2) Solve the equation 2log x 3  log 3 x  3  0

3) Given p  q 3  5 where p and q are integers. Find the value of p and q without
2 3

using calculator

4) (a) Solve the equation 9x  4  5(3x )

(b) Solve the equation 4x 16  6(2x )

5) (a) Solve the equation log 2 2x  log x 2  log 4 26

(b) Given 3  2 2  a  b 2 ,find a and b where a,b 

1

643 p6q 2r10

6) Simplify
2
8 3 p 2q 10r 2

7) Find the values of x which satisfies the equation log 2 (5  x)  log 2 (x  2)  3  log 2 (1 x)
8) Solve the equation 2(32x1)  3  7(3x )

11

9) (a) Evaluate   11  2  11  2
2 2
3  3 without using calculator.

Page 5 of 6

AM015/1 .Number System And Equations

(b) Show that

1  1 1
log p pq log q pq

10) Find the values of x that satisfies the equation (32x1)  32  28(3x )

11) Solve the equation log 2 x  log 4 8x  6

ANSWERS:EXTRA EXERCISE CHAPTER 1

1) (a)5 (b)1 (c)4 6) (pqr)8

2) x=9,x=3 7) x =3

3) p 175,q  100 8) x  1

4) (a) 0,1.262(3dp) 9) (a) 1 (b) shown
2
(b) 3
10) x=-1, x=2
5) (a) 2

(b) a  1,b  1

11) x=8

Page 6 of 6

AM015/ 2. Inequalities and Absolute Values 2021/2022

CHAPTER 2 : INEQUALITIES AND ABSOLUTE VALUES

LECTURE 1 OF 2

2.1 INEQUALITIES

At the end of lectures, you should be able to
(a) relate the properties of inequalities
(b) find linear inequalities
(c) find quadratic inequalities

A. Relate the properties of inequalities

Inequalities involved the use of symbols: , , , 

Suppose a , b , and c are any numbers, we can state the properties of inequalities as in the

following table:

If a  b , then Example

1 a  c  b  c,c  0 43

4 2  3 2

2 ac  bc,c  0 65
43

42  32

3 ac  bc,c  0 86
43

42  32

8  6

a b, c0 43
cc 43
2 2
* If both sides of an inequality are multiplied or divide by negative 2  1.5
number, the inequality sign must be reversed.
43
4 1  1 , where a,b  0 11
ab 43

B. Find linear inequalities

A linear inequalities in one variable, x , is defined as any relationship of the form:
ax b  0,ax b  0,ax b  0,ax b  0

where a and b are real numbers and a  0 .

The solution to the inequality must be expressed in solution set or interval form.

Page 1 of 8

AM015/ 2. Inequalities and Absolute Values 2021/2022

REMEMBER
These things will change (reverse) direction of the inequality:

 Multiplying or dividing both sides by a negative number
 Swapping left and right hand sides

Example
Solve the following inequalities:

(a) 3t +5  t  7

(b) 1 x  5  1 x  4
23

(c) 7  3x  2  8

Page 2 of 8

AM015/ 2. Inequalities and Absolute Values 2021/2022

EXERCISE (c) 2  x  4  9 (d) 1 3x  5x 3  x  9
Solve the following inequalities: 2

(a) 4x 5  2x+9 (b) 11 y  y  4
4

C. Find quadratic inequalities

A quadratic inequality is an inequality of the form ax2 bx  c  0 where a , b and c are
real number with a  0. The inequality symbols , ,  and  may also be used.

Quadratic inequalities can be solved by using

i. graphical approach; ii. table of signs; iii. positive number line

I. Graphical approach:

The graph of quadratic expression y  ax2  bx  c is sketched and points where the graph
cuts the x -axis, say p and q are noted.

Example
Solve the following inequalities by using graphical method.

(a) x2  6x 5  0 (b) 6 7x 3x2  0

Page 3 of 8

AM015/ 2. Inequalities and Absolute Values 2021/2022

II. Table of signs
Example
Solve the following quadratic inequalities by using table of signs

(a) (x 1)(x  2)  0

(b) x2 3x  4  0

III. Positive number line
Example
Solve the following quadratic inequalities by using positive number line

(a) x2  2x 15  0

(b) 2x2 10  x

EXERCISE (d) 6 4x  2x2 3x  3 2x
(a) (2x 1)(x 3)  4x (b) 3x(x 5)  2(2x 3)

MISCELLANEOUS EXERCISE

(a) x2  49 (b) 3x2 6x 5  0 (c) 0  x2 9  8x

Page 4 of 8

AM015/ 2. Inequalities and Absolute Values 2021/2022

LECTURE 2 OF 2

2.2 ABSOLUTE VALUES

At the end of lectures, you should be able to
(a) define the absolute value of a; present it on the number line and state the

properties of absolute values :

i. |a|≥0 ii. |-a|=|a| iii. |a+b|=|b+a| iv. |a-b|=|b-a| v. |ab|=|a||b| vi. a  a where |b| ≠ 0
bb

(b) solve absolute equations of the forms :

i.|ax+b|=c ii.|ax+b|=cx+d iii.|ax+b|=|cx+d| iv.|ax²+bx+c|=d

(c) solve absolute inequalities of the forms:

i. |ax+b|<cx+d ii. |ax+b|>cx+d

A. Define the absolute value of a and present it on the number line

The absolute value of a , a represents the distance of point a on the real number line from

the origin.

a  a a a

a 0 a

The absolute value of real number a , written as a can be defined as

a   a, if a0 or a  a2
a, if a0

Therefore, absolute value of every real number is non-negative, x  0 for every x  .

Example
Define the following absolute values

(a) x  5

(b) 5  x

Page 5 of 8

AM015/ 2. Inequalities and Absolute Values 2021/2022

Properties of absolute values

Properties Examples

1 a 0 5 50
0 0

2 a  a 5 5
5  5

3 ab  ba 53  8 8
35  8 8

4 ab  ba 53  2 2
35  2  2

5 ab  a b 53  15 15
5  3  53 15

5 5
a  a ,b 0 3 3
6 bb 5 5

3 3

B. Solve absolute equations of the form iii. |ax+b|=|cx+d|
i. |ax+b|=c iv. |ax2+bx+c|=d
ii. |ax+b|=cx+d

x  a  x  a or x  a

Example
Solve absolute equations of the form below

(a) 53 x 5  3

(b) 1 x  2x  5

Page 6 of 8

AM015/ 2. Inequalities and Absolute Values 2021/2022

(c) 2x  6  x 1

(d) x2  2x  4  4

EXERCISE (b) x2  2x 1 (c) 3x 5x  2  4

(a) x  4  8  x

Property of Absolute Value Inequalities:

Inequalities of Algebraic Graphical Interpretation
Absolute Value Interpretation -a a

,x  a a  0 a  x  a

x  a and x  a

x  a  x  a

x  a, a  0 x  a or x  a -a a
x  a,a  0 x  a  x  a -a a

a  x  a
x  a and x  a
x  a  x  a

x  a,a  0 x  a or x  a
x  a  x  a
-a a

Page 7 of 8

AM015/ 2. Inequalities and Absolute Values 2021/2022

C. Solve absolute inequalities of the forms

i. |ax+b|<cx+d ii. |ax+b|>cx+d

Example
Solve the following inequalities:

(a) 2x 5  9

(b) 2x 1  3x  2

(c) 5x  2  2x 1

EXERCISE (b) 16  2 6  x (c) 5x  6  4 1 BORANG MAKLUMBALAS
PENSYARAH & PELAJAR TERHADAP
(a) 2x  4  0
NOTA KULIAH BAB 2 AM015

https://goo.gl/forms/5Rr5vS
Q99qtFoue92

Page 8 of 8

AM015/3. SEQUENCES

CHAPTER 3: SEQUENCES

LECTURE 1 OF 2

At the end of lesson, student should be able to :
a) Express the nth term of a sequence.
b) Find the nth term and sum of the first n terms.
c) Solve problems involving arithmetic sequences in business and economics

DEFINITION OF A SEQUENCE
A sequence is a set of numbers arranged in a particular order.
Example,
(i) − 5, − 3, −1, 1, 3, 5,...
(ii) -1, 1, -3, 3, -5, 5
(iii) 3, 3, 3, 3, 3,…

Each number in the sequence is called a term.

Example : 3, 5, 7, 9

First term, T1 = 3
Second term, T2 = 5
T3 = 7
Third term,
Fourth term, T4 = 9

n Term Test Rule
13 2n+1 = 2×1 + 1 = 3
25 2n+1 = 2×2 + 1 = 5
37 2n+1 = 2×3 + 1 = 7

 Tn = 2n +1
The notation Tn represents the n th term, or general term of a sequence.

1 of 10

AM015/3. SEQUENCES

Example 1

Express the nth term of the following sequences.

a) 4, 7, 10, 13, 16… b) 1 , 3 , 5 , 7 , 9
2 4 6 8 10

Example 2

Given Tn = 1  , find the first three terms and the 12th term.
 
 3n + 1n=1

ARITHMETIC SEQUENCE

Definition of an Arithmetic Sequence
An arithmetic sequence or an arithmetic progression is a sequence of the form

a, a + d, a + 2d, a + 3d,... , a + (n −1)d , where a is the first term and d is the common

difference of the sequence.

a, a + d, a + 2d, a + 3d,...

First term: T1 = a

Second term: T2 = a + d = a + (2 −1)d

Third term: T3 = a + 2d = a + (3 − 1)d

The nth term of an arithmetic sequence is given by Tn = a + (n −1)d

The common difference, d = T2 −T1 OR
= T3 − T2 OR
= Tn −Tn−1

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AM015/3. SEQUENCES

Example 3
Given the arithmetic sequence 7 , 13 , 19 , … , 307

Find (i) the number of terms

(ii) the 20th term and

(iii) the nth term of the arithmetic sequence

Example 4
The 11th term of an arithmetic sequence is 52 and the 19th term is 92. Find
(a) the first term and common difference
(b) 100th term

Example 5
The nth term of an arithmetic sequence is 40 + 7n.
(a) Find the common difference.
(b) Which term of the sequence is 215 ?

3 of 10

AM015/3. SEQUENCES

The Sum of the First n Terms of Arithmetic Sequence (Arithmetic
Series)

The terms of an arithmetic sequence can be added to form an arithmetic series. Let

Sn = a + (a+d) + (a+2d) +...+ a+ (n-1)d ……..(1)

Writing the sum in reverse order

Sn = a+(n-1)d +...+ (a+2d)+ (a+d) + a ……(2)

thus, (1) + (2)
2Sn =[ 2a + ( n-1 )d + … + 2a + ( n-1 )d ]
There are n identical terms on the right side of this equation

2Sn = n 2a+(n-1)d

 Sn = n 2a + (n −1) d 
2

Since the last term, Tn = l = a + (n −1)d

Sn = n a + a + ( n-1) d 
2

 Sn = n a +l

2

The relationship between Tn and Sn is, Tn = Sn − Sn−1

Example: T5 = S5 − S4

Example 6
Find the sum of all integers between 100 and 200 that are multiples of 6.

4 of 10

AM015/3. SEQUENCES
Example 7
The tenth term of an arithmetic sequence is -25 and the sum of the first ten terms is
65. Find the first term , the common difference and the sum of the first twenty terms.

Example 8
The sum of the first 15 terms of Arithmetic Sequence is 255 and the sum of the next
15 terms is 705. Find the first term, the common difference and the 50th term.

APPLICATION OF ARITHMETIC SEQUENCES
Example 9
Iman settles her debt of RM4980 by paying RM50 at the end of the first month. And
for the following months, she pays RM8 more than the previous month. Find
(a) how long will Iman take to settle her debt,
(b) the amount of the last payment she makes.

5 of 10

AM015/3. SEQUENCES

EXERCISE

1. An arithmetic series has 48 terms where the first term and the last term are
-16 and 98 respectively. Find the sum of this series.
( Answer: 1968 )

2. The sum of the first n terms of an arithmetic sequence is Sn = 7n2 − 8n .
Find
a) the first term
b) the common difference
c) the 100 th term
( Answer: -1, 14, 1385 )

3. The sum of the first 8 terms of an arithmetic sequence is 60 and the sum of
the next 6 terms is 108. Find the 25th term of this arithmetic sequence.

(Answer: 153 )
4

4. Hisyam has to repay his debt of RM10 360 by monthly instalments. If the first
instalment is RM100 and he agree to increase the amount by RM20 each
month after the first instalment, find the number of months he takes to settle
his debt and the final instalment.
( Answer: 28, 640 )

LECTURE 2 OF 2
At the end of lesson, student should be able to :

a) Find the nth term and sum of the first n terms
b) Solve problems involving geometric sequences in business and economics

GEOMETRIC SEQUENCE

Definition of a Geometric Sequence

A sequence in which any term can be obtained from the previous term by multiplying
by a constant is called a geometric sequence or geometric progression.

The geometric sequence are given by a, ar, ar 2 , ar3 ,..., ar n−1,...
where a is the first term and r is the common ratio.

The first term, T1 = a
Second term, T2 = ar
Third term, T3 = ar2
Fourth term, T4 = ar3

So the n th term, Tn = arn−1

The n th term is called general term of a geometric sequence.
The common ratio, r is given by

r = T2 = T3 = ... = Tn . 6 of 10
T1 T2 Tn−1

AM015/3. SEQUENCES
Example 1
Find the sixth and nth term for the geometric sequence 4, -8, 16, -32, …

Example 2
The third term of a geometric sequence is 15 and the sixth term is 120. Find the first
term, the common ratio and tenth term.

Example 3
In a geometric progression, the second term exceeds the first term by 20 and the
fourth term exceeds the second term by 15. Find the possible values of the first term.

7 of 10

AM015/3. SEQUENCES

The Sum of the First n Terms of a Geometric Sequence
Consider the geometric sequence,
a, ar , ar 2, ar 3,...
Let the sum to n terms of a GS as Sn,
Sn = a + ar + ar 2 + ar 3 + ... + ar n−2 + ar n−1 − − − (1)

Multiply the equation by the common ratio r ,
r Sn = ar + ar 2 + ar 3 + ... + ar n−1+ ar n − − − − − − − (2)

(1) – (2), Sn − r Sn = a − ar n
or Sn (1 − r ) = a(1 − r n )

If r  1, Sn = a(1 − rn) and If r  1, Sn = a(r n − 1)
(1 − r) r −1

The relationship between Tn and Sn is, Tn = Sn − Sn−1

Example 4

Find the sum of a geometric series 1+ 1 + 1 + ... + 1
24 256

Example 5
The sum of the first 3 terms of a geometric progression is 7 and the sum of the next

4
three terms is 7 . Find the common ratio of the progression.

32

8 of 10

AM015/3. SEQUENCES
Example 6
In a geometric sequence, the first term is 7, the last term is 448, and the sum of all
the terms is 889, find the common ratio and the number of terms.

APPLICATION OF GEOMETRIC SEQUENCES
Example 7
An engineering company offers a position to Aiman with the starting pay RM18000
for one year and an increment of 10% yearly.
(a) What is Aiman’s salary in the 11th year?
(b) Find the total salary of Aiman after 11 years of service in the company.
(c) Find the number of years that Aiman had worked if the total salary is

RM571905.

9 of 10

AM015/3. SEQUENCES
EXERCISE
1. The first term of a geometric series is 27 and its common ratio is 4 . Find the

3
least number of terms the sequence can have if its sum exceeds 550.
( Answer: 8 )
2. The sum of the first three terms of a geometric series is 77 and the sum of
the first six terms is 693. Find the common ratio and the first term.
( Answer: 2, 11)

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AM015/3 SEQUENCES

CHAPTER 3 : SEQUENCES

TUTORIAL 1 OF 3
ARITHMETIC SEQUENCES
1. Which term of the arithmetic sequence 1, 4, 7, ... is 88 ?

2. Find the sum of each of the arithmetic series 1 + 5 + 9 + . . . + 401.

3. Find the sum of the odd numbers from 1 to 99 that are divisible by 3.

4. Find the largest and smallest of these numbers, if the sum of 25 consecutive odd
numbers is 1275.

5. If the sequence 4, x , y , 29 forms an arithmetic sequence, find the values of x
2

and y .

6. The 8th term and the 14th term of an arithmetic series are 31 and 55 respectively.
(a) Determine the first term and the common difference.
(b) Find the 20th term of the arithmetis series.

(c) Find the sum of the first 20 terms of the arithmetis series.

7. The difference between the tenth and the fifth term of an arithmetic series is 20,
and the sum of the first eight terms of series is 136. Find the first term and the
common difference. Hence, determine the value of n such that the n-th term is at
least 120.

8. The sum of the n terms of a series is 3 nn  7 .

2

(a) Write down an expression for the sum of the first n 1 terms.

(b) Show that the series is an arithmetic series. Hence, state the first terms and
the common difference.

9. In an arithmetic sequence, the 9th term is twice the 3rd term and the 15th term is 80.
Find the common difference and the sum of the terms from the 9th to the 15th
inclusive.

ANSWERS

1 30 2 101, 20301
3 867
4 27 and 75
5 x  15 , y 11
2 6 (a) a  3, d  4

7 a  3, d  4, n  31 (b) 79 (c) 820

9 d  40 , 466 2
93

8 (a) Sn1  3 n 1 n  6 , (b) shown, a 12, d  3
2

Page 1 of 5

AM015/3 SEQUENCES

TUTORIAL 2 OF 3

GEOMETRIC SEQUENCES

1. Find the common ratio, given that it is negative, of a Geometric Progression whose

first term is 8 and whose fifth term is 1 .
2

2. How many terms of the sequence 1, 2, 4, 8,… are required to give a sum of 16383.

3. Find the sum of the first n terms of the Geometric Progression 1, 5 , 25,... and find
4 16

the least value of n for which the sum exceeds 20.

4. In a Geometric Progression, the sum from the fifth term to the eighths term is twice
the sum of the first four terms. Find the common ratio of the progression.

5. The sum of the first four terms of a geometric series with common ratio  1 is 30.
2

Determine the tenth term.

6. The third term of a geometric sequence exceeds the second term by 6 and the
second term exceeds the first term by 9. Find the sum of the first four terms.

7. The third and the sixth term of a geometric series are 1 and 1 respectively.
3 81

Determine the values of the first term and the common ratio. Hence, find the sum
of the first seven terms of the series.

8. The sum of the first three terms of a geometric series is 77 and the sum of the first
six terms is 693. Find the common ratio and the first term.

9. Given a geometric series with common ratio more than one (r > 1). The ratio of
the sum of its first four terms to the sum of its first two terms is 10:1. Find the
common ratio. Hence, if the third term of the series is 54, find the first term.

1 a)  1 ANSWERS
2 2 14
41
3 4 5 n  r  24
 4  1 ,
Sn  n=9 6 a  27, S4  65
 8 r  2, a  11

5 3 Page 2 of 5
32

7 r  1 , a  3, S7  1093
3 243

9 r  3, a  6

AM015/3 SEQUENCES

EXERCISE

1. Determine the smallest integer n such that 1 4   4 2  ...   4 n  4.9
     5 
 5   5   

2. The first, second and third terms of a geometric series are , and 2 respectively.
The first, second and third term of an arithmetic series are , 2 and respectively.

Determine the values of and with < 0.

3. The ℎ term of a certain sequence is 2 − + 3.
(i) Find the sum of the first three terms.

(ii) Which term is 243?

4. The sum of the first n terms of a certain series is given by Sn  pn2  qn . The sum

of the first four terms and the sum of the first twelve terms are 68 and 588
respectively.
(i) Calculate the values of p and q .

(ii) Hence, find the nth term of the series and the first term.
(iii) Determine the type of sequence of this series.

5. The first term and the sum of all terms for a geometric sequence is 4 and -59048,
respectively. Given that the comman ratio of the sequence is -3, find the number of
terms of this geometric sequence.

6. (i) The first three terms of an arithmetic progression are 2x , x  4 and 2x  7
respectively. Find the value of x .

(ii) The first three terms of another sequence are also 2x , x  4 and 2x  7

respectively.

(a) Verify that when x  8 the terms form a geometric progression.
(b) Find the other possible value of x that also gives a geometric progression.

7. An arithmetic progression has first term log2 27 and common difference log2 x .

 (i) Show that the fourth term can be written as log2 27x3 .

(ii) Given that the fourth term is 6, find the exact value of x .

ANSWERS

1 17 2 = 1, = −1
3 i) 3 = 17 (ii) = 16 2
5 10
4 (i) p  4, q  1, (ii)Tn  8n  3,T1  5
74 (iii) d  8, Arithmetic sequences

(i) Shown (ii) 6 (i) 15 (ii) (a) verified, (b)  2
23
3

Page 3 of 5

AM015/3 SEQUENCES

TUTORIAL 3 OF 3

APPLICATIONS OF ARITHMETIC AND GEOMETRIC SEQUENCES

1. A man’s initial annual salary was RM600.00 and increased by RM45.00 a year.
How much does he expect to earn after 15 years?

2. Ahmad’s annual salary in 2013 is RM 30 000.00 per annum. His annual salary
increment is RM 1320.00. Find, in 2033 what is his
(a) the annual salary
(b) the total salary received.

3. Aiman takes an interest free loan to buy computer. He repays the loan in monthly
installments. For the first month, he repays RM 110.00. For the second month,
he repays RM 160.00. He repays RM 50.00 more every month until the loan is
completely settled. if his final monthly installment is RM 860.00, find
(a) the number of month to settled the loan
(b) the amount of the loan.

4. The price of a new car is RM42 000.00. If the value of the car depreciates at a rate
of 10% each year, find the value of the car after 12 years.

5. At the beginning of this year, Nurin deposited RM 10 000.00, in a bank that gives
interest rate at 6% per annum. Find the amount of the money should she receives
after 7 years.

6. Azmi is offered a post by an engineering company with the starting salary of

RM20000 yearly with the annual increment of 10 %.

(a) Find the expression to represent the second and third year’s income.
(b) Show that the sum of his income for the first n years is

RM 200 000 [(1.1)n – 1].
(c) How much is his salary at the 11th year? Find the sum of the income within 11

years of service in the company.
(d) Hence, how long is his service if the sum of his income is RM 635 450.00.

ANSWERS

1 RM1275 2 (a)RM56400, (b) RM907200
3 (a)16, 4 RM11862.04
(b) RM7760
5 RM15036.30
6 (a) T2  20000(1.1) , T3  20,000(1.1)2

(b) Sn  2000001.1n 1 , shown
(c) T11  RM51874.85,

S11  RM370623.34
(d) n  15

Page 4 of 5

AM015/3 SEQUENCES

EXERCISE

1. Julie bought an electric piano through an instalment plant with a down payment of
RM1000. She paid RM 100 for the first month with an increment of RM 25 every
month until the payment for the plan is settled. If the final monthly installment was
RM 675, find
(i) the number of month she took to settle the loan
(ii) the total amount she paid to the piano.

2. A 26 years old lady gets a job at a company. Her starting salary is RM1,400 and
the annual increment is RM90.
(i) What will her monthly salary be when she is 45 years old?
(ii) What will her age be when she gets a monthly salary of RM2,480?

3. Tower A has fifty floors. A cleaning company estimates that the charges of cleaning
one floor increases by 10% for each floor above the previous floor. If the charges
for cleaning the first floor is RM100.
(i) list in the form of sequence, the first five charges for the first five floors.
(ii) determine the type of sequence that can be used to estimate the cleaning
charges for any floor.
(iii) determine the charges to clean the top floor.
(iv) find the expression of total charges to clean the first n floors. Hence,
calculate the total charges to clean the first thirty floors.

4. An employee pays his debt monthly by paying RM 20 in the first month. For the
following month, he pays an additional RM 4 of the previous month. How many
months will it take for him to pay a debt of RM 3572? Hence, calculate the amount
he has to pay in the last month.

5. There are 20 rows of seats in a concert hall with 25 seats in the first row, 27 seats
in the second row, 29 seats in the third row and so on.

(i) Find the number of seats in the last row.
(ii) Determine the total number of seats in the hall.
(iii) If the price per ticket is RM500 for the first three rows and RM200 for the rest, how

much will be the total sales for a one-night concert if all seats are sold?

ANSWERS

1 (i) 24 (ii) RM10300 2 (i) 20 = 3110 (b) 38 years old

3 (i)T1  100,T2  110,T3  121,T4  133.1,T5  146.41, (ii) r 1.1, Geometric progression,

(iii) T50  RM10671.90 ,  (iv) Sn 1000 1.1n 1 , S30  RM16449.40

4 n  38,T38  RM168 5 (i) 63, (ii) 880, (iii) RM200300.00

Page 5 of 5