Optional Math 9

According to new curriculum in compliance with
Curriculum Development Centre (CDC) .
Approved by CDC.

PRIME Optional
Mathematics

Pragya Books &
Distributors Pvt. Ltd.

Author Editors

Dirgha Raj Mishra LN Upadhyaya
Rajkumar Mathema

DN Chaudhary
Narayan Shrestha

Khem Timsina
J.N. Aryal

Kadambaba Pradhan
Dinesh Silwal

Pragya Books & Distributors Pvt. Ltd.

Lalitpur, Nepal
Tel : 5200575
email : [email protected]

© Author

Author Dirgha Raj Mishra

Editors LN Upadhyaya
Rajkumar Mathema
DN Chaudhary
Narayan Shrestha
Khem Timsina
J.N. Aryal
Kadambaba Pradhan
Dinesh Silwal

First Edition 2076 B.S. (2019 A.D.)
Revised Edition 2077 B.S. (2020 A.D.)

Price

ISBN 978-9937-9170-5-6

Typist Sachin Maharjan
Sujan Thapa

Layout and Design Desktop Team

Printed in Nepal

Preface

Prime Optional Mathematics series is a distinctly outstanding mathematics
series designed according to new curriculum in compliance with Curriculum
Development Centre (CDC) to meet international standard in the school level
additional mathematics. The innovative, lucid and logical arrangement of the
contents make each book in their series coherent. The representation of ideas in
each volume makes the series not only unique but also a pioneer in the evaluation
of activity based mathematics teaching.

The subject is set in an easy and child-friendly pattern so that students will
discover learning mathematics is a fun thing to do even for the harder problems.
A lot of research, experimentation and careful graduation have gone into the
making of the series to ensure that the selection and presentation is systematic,
innovative, and both horizontally and vertically integrated for the students of
different levels.

Prime Optional Mathematics series is based on child-centered teaching
and learning methodologies, so that the teachers can find teaching this series
equally enjoyable.

I am optimistic that, this series shall bridge the existing inconsistencies
between the cognitive capacity of children and the subject matter.

I owe an immense dept of gratitude to the publishers (Pragya Books team)
for their creative, thoughtful and inspirational support in bringing about the
series. Similarly, I would like to acknowledge the tremendous support of editors
team, teachers, educationists and well-wishers for their contribution, assistance
and encouragement in making this series a success. I would like to express my
special thanks to Sachin Maharjan (Wonjala Desktop) for his sincere support
of designing part of the book and also Mr. Gopal Krishna Bhattarai to their
memorable support to prepare this series.

I hope this series will be another milestone in the advancement of teaching
and learning Mathematics in Nepal. We solicit feedback and suggestions from
teachers, students and guardians alike so that I can refine and improvise the series
in the future editions.

– Author

Contents Page

S.N. Units 1
2
1. Algebra 10
1.1 Ordered Pair 17
1.2 Relation 30
1.3 Function 36
1.4 Polynomials 44
1.5 Operation on Polynomial 59
1.6 Sequence and Series 79
2. Limit and Continuity 113
3. Matrices 169
4. Co-ordinate Geometry 241
5. Trigonometry 269
6. Vector Geometry 309
7. Transformation 341
8. Statistics
Model questions

Unit 1 Algebra

1. Algebra

1.1 Ordered Pair
1.2 Relation
1.3 Function
1.4 Polynomial
1.5 Operation on Polynomial
1.6 Sequence and Series

Specification Grid Table

K(1) U(2) A(4) HA(5) TQ TM Periods

No. of Questions 2 3 2 1 8 21 33
Weight 2 6 8 5

K = Knowledge, U = Understanding, A = Application, HA = Higher ability,
TQ = Total Question, TM = Total Marks

Objectives : At the end of the lesson
• Students understand the ordered pairs & Cartesian product and its representation.
• Students are able to find relation.
• Students are able to identify domain, co-domian and range.
• Students are able to find the problems involving functions and their types.
• Students are able to solve the equal ordered pairs & equation involving functions.
• Students are able to find the problems involving sequence and series.
• Students are able to solve the related problems of sequence and series.
• Students are able to find the problems involving polynomial & its types & operations.

Materials Required: 1
• Function machine
• Arrow diagram chart
• Graph board
• Chart board
• Chard paper
• Model of sequence of numbers

PRIME Opt. Maths Book - IX

1.1 Ordered Pair

Ordered pair :

Let us consider the different pattern of writing two elements 1 and 2 in pair

respectively 1 2 1, 2 (1 2) (1, 2)* {1, 2} [1, 2] 1
2.

Among the above pairs, the pair indicated by * is called an ordered pair where the
elements are kept inside ( ) and separated by comma.

Here (1, 2) is called the order pair of 1 and 2 respectively where there is the important
role of order of the elements 1 and 2.

Let us consider any two sets A = {1, 2, 3} and B = {2, 3, 4, 5, 6} then the two elements
are taken first from set A and second from set B as (1, 3), (2, 4), (3, 5) where in each
pairs second is more than the first by 2 which is the order of such pairs. They are
called ordered pairs.

The pair of elements taken in definite order enclosed in a
parenthesis ( ) and separated by comma (,) is called ordered
pair.

Examples : (x, y), (2, 3), (0, 0) etc.

Here, In the order pair (x, y)
x - component is called antecedent.
y - component is called consequence.

Note :

i. Some time notation (1, 2) is also used to represent an order pair.

ii. If order (position) of 1 and 2 interchange, new ordered pair is formed i.e.

(2, 1) is not same as (1, 2).



It is applicable in different activities other than numerical values too.

Some of the examples can be taken as,

• Ordered pair of capitals with respect to countries are :
(Kathmandu, Nepal), (Tokyo, Japan), (Thimpu, Bhutan)

• Ordered pair of holy books with respect to religion are :
(Geeta, Hindu), (Kuran, Muslim), (Bibal, Christian)

• Ordered pair of districts with respect to headquarter are :
(Jhapa, Chandragadi), (Morang, Biratnagar), (Rautahat, Gaur)

2 PRIME Opt. Maths Book - IX

Equal ordered pairs :

Any two ordered pairs having same antecedents and equal
consequences are called equal ordered pairs.

In the ordered pairs (3, 4) and (4 – 1, 7 – 3), the antecedent are both 3 and consequence
are both 4. Hence they are called equal ordered pairs.

If ordered pairs (a, b) and (x, y) are equal, then a = x and b = y.

Cartesian product :

Let us consider A = {a, b} and B = {p, q, r} where set of all possible ordered pairs from
the set A to the set B can be taken as {(a, p), (a, q), (a, r), (b, p), (b, q), (b, r)} which is
called the Cartesian product A × B (reads A cross B).

The set of all possible ordered pairs (x, y) taken from non - empty
set A to non - empty set B is called the Cartesian product A × B
where x ∈ A and y ∈ B.

i.e. A × B = { (x, y) : x ∈ A, y ∈ B }

In the above example, the Cartesian products A × B and B × A are;

A × B = {(a, p), (a, q), (a, r), (b, p), (b, q), (b, r)}

B × A = { (p, a), (p, b), (q, a), (q, b), (r, a), (r, b)}

Here, A × B ≠ B × A
But, Taking the cardinality of sets,

n(A × B) = n(A) × n(B) = 2 × 3 =6

n(B ×A) = n(B) × n(A) = 3 × 2 =6

\ n(A × B) = n(B × A)

Representation of Cartesian product :

The Cartesian product A × B of the non - empty set A to the set B can be expressed
in different ways which are discussing below. For the representation x - components
and y - components of the set of ordered pairs have to be taken from the sets A and
B respectively. But for the Cartesian product B × A, x - component and y - component
should be taken from the set B to A respectively.

Let us consider A = {1, 2, 3} and B = {6, 7, 8}.
i) Set builder form :
A × B = {(x, y) : x ∈ A and y ∈ B }
B × A = {(x, y) : x ∈ B and y ∈ A}

PRIME Opt. Maths Book - IX 3

ii) Ordered pairs form (listing form)
A × B = {(1, 6), (1, 7), (1, 8), (2, 6), (2, 7), (2, 8), (3, 6), (3, 7), (3, 8)}

B × A = {(6, 1), (6, 2), (6, 3), (7, 1), (7, 2), (7, 3), (8, 1), (8, 2), (8, 3)}

iii) Tabular form : B×A
A×B

AB 6 7 8 A 1 2 3
1 (1, 6) (1, 7) B
(1, 8) (6, 1) (6, 2) (6, 3)
2 (2, 6) (2, 7) (2, 8) 6 (7, 1) (7, 2) (7, 3)
(3, 8) (8, 1) (8, 2) (8, 3)
3 (3, 6) (3, 7) 7

8

iv) Tree diagram form : A×B B A B×A
AB (1, 6) 6 1 – (6, 1)
(1, 7) 7 2 – (6, 2)
6– (1, 8) 8 3 – (6, 3)
1 7– (2, 6) 1 – (7, 1)
v) (2, 7) A 2 – (7, 2)
8– (2, 8) 3 – (7, 3)
6– (3, 6) 1 – (8, 1)
2 7– (3, 7) 2 – (8, 2)
8– (3, 8) 3 – (8, 3)
6–
3 7– B×A A
8–
1
Arrow diagram (Balloon diagram) 2
A A×B B 3

1 66
2 77
3 88



4 PRIME Opt. Maths Book - IX

vi) Graphical form
Y

10

9 Y
8

77

66

55

44

33

22

11

O 1 2 3 4 5 X O 1 2 3 4 5 6 7 8 9 10 X

In the above examples, Cartesian products A × B ≠ B × A

But For cardinality of sets,

n(A × B) = n(A) × n(B) = 3 × 3 =9

n(B × A) = n(B) × n(A) = 3 × 3 =9

\ n(A × B) = n(B × A)

Worked out Examples

1. Write down the any five ordered pairs of headquarters with respect to provinces of
Nepal. Also show in arrow diagram.

Solution :
Here, x - component represents the temporary headquarters.
y - components represents the province numbers.­
Ordered of them are : (Biratnagar, Province No. 1), (Janakpur, Province No. 2),

(Hetauda, Province No. 3), (Pokhara, Province No. 4), (Butwal, Province No. 5),
(Surkhet, Province No. 6), (Dhangadi, Province No. 7)

Arrow diagram : Province No. 1
Province No. 2
Biratnagar Bagmati
Janakpur
Hetauda Gandaki
Pokhara Province No. 5
Butwal Karnali
Sudur Paschim
Surkhet
5
Godawari


PRIME Opt. Maths Book - IX

2. If (2x – y, x + 3) and (4, 6) are the equal ordered pairs, find the value of ‘x’ and ‘y’.

Solution :

Here, Equal ordered pairs are : (2x – y, x + 3) = (4, 6)

By equating the antecedent and consequences, respectively.

x + 3 = 6 and 2x – y = 4

or, x = 6 – 3 and 2x = y + 4

or, x = 3 and y = 2x – 4

or, y = 2 × 3 – 4

\ x = 3 \ y=2

y = 2

3. If (3 x + y, 16) = (27, 2 2x + y), find the value of ‘x’ and ‘y’.

Solution :
The equal ordered pairs are :

(3x + y, 16) = (27, 22x+ y)

or, (3x+y, 24) = (33, 22x + y)

By equating the antecedent and consequences, we have

3x+y = 33 and 24 = 22x + y

x + y = 3 ..........................(i) and 2x + y = 4 .....................(ii)

From equation (i)
x + y = 3
or, x = 3 – y ...............................(iii)

From equation (ii)
2x + y = 4
or, 2(3 – y) + y = 4
or, 6 – 2y + y = 4
or, 6 –4 = y
\ y = 2
Substituting the value of ‘y’ in equation (iii), we get
x = 3 – y = 3 –2 = 1
\ x = 1
y = 2

4. If A = {1, 2, 3}, B = {4, 5, 6}, find A × B and B × A. Also prove that n (A × B) = n(B × A)
Solution :
A = {1, 2, 3},
B = {4, 5, 6}
A × B = {(x, y) : x ∈ A and y ∈ B}
= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
B × A = {(x, y) : x ∈ B and y ∈ A}
= {(4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3)}

6 PRIME Opt. Maths Book - IX

Here,
n(A × B) = 3 × 3 = 9
n (B × A) = 3 × 3 = 9
∴ n (A × B) = n (B × A)

5. If A × B = {(a, x), (a, y), (b, x), (b, y)}, find the sets A and B. Also find A × A and B × B.
Solution:
A × B = {(a, x), (a, y), (b, x), (b, y)}
Set A = {set of antecedents} = {a, b}
Set B = {set of consequences} = {x, y}

Again, = {a, b} × {a, b}
A × A = {(a, a), (a, b), (b, a), (b, b)}
= {x, y} × {x, y}
B × B = {(x, x), (x, y), (y, x), (y, y)}


6. If A = {a, b, c} and A × B = {(...., x), (...., y), (a, ....), (...., x), (b, ....), (...., z), (c, ....), (c, ....),
(c, ....)}, find the set B, Complete A × B. Also show A × B in arrow diagram.

Solution : A = {a, b, c}
A × B = {(...., x), (...., y), (a, ....), (...., x), (b, ....), (...., z), (c, ....), (c, ....), (c, ....)}
By comparing A and A × B, we get B = {x, y, z}
Also, A × B = (a, x), (a, y), (a, z), (b, x), (b, y), (b, z), (c, x), (c, y), (c, z)}

Arrow diagram: B
A A×B x
a

by

c z


PRIME Opt. Maths Book - IX 7

Exercise 1.1

1. i) What is ordered pair? Write down one example.

ii) What do you mean by antecedent and consequence?

iii) What is Cartesian product ? Write down with a suitable example.
iv) Write down any five ordered pairs of temples with respect to location. Also

show in arrow diagram.

v) From the given sets write down the ordered pairs of capitals with respect to

country.

Rangoon Turkey

Madrid Ethiopia

Adis Ababa Spain

Bagdad Myanmar

Instanbul Iraq

2. i) Write down any four ordered pairs of capital with respect to counties of SAARC.

ii) If x = a and y = b in equal ordered pairs, write down the ordered pairs.
do12wannAd o21rdineraecdaprtaeirssia. n
iii) If a b product A × B, find the sets A and B. Also
write ×B
in
iv) Complete the following ordered pairs by filling the gap
(Bhanubhakta, .................), (................., Aanshukabi), (Lekhnath Poudel, ................)
v) Complete the following ordered pairs by taking an order ‘more than by 2’.

(2, ...), (5, ...), (..., 9) (..., 12), (15, ...)

3. Find the value of ‘x’ and ‘y’ from the given equal ordered pairs.

i) (x + 2y, 5) = (10, 2y – 1) ii) (3x –2, 4 –y) = (2 – x, 2x + 1)

iii) (23x+y, 9) = (32, 35x – 3) iv) (x + y, 6) = (6, 2x – y)

v) (2x, y + 3) = (y + 3, 3x – 4)

4. If A = {a, b}, B = {1, 2, 3}, find
i) A × B and show in arrow diagram.

ii) B × A and show in arrow diagram.
iii) Prove that A × B ≠ B × A
iv) Prove that n(A × B) = n(B × A)

5. i) If A = {x : x ∈ (1, 2)}, B = {y : y ∈ 3 < N ≤ 6}, find A × B. Also show in tabular form.
ii) If A × B = {(....., 3), (5, .....), (....., 4), (....., 4), (5, .....), (2, .....)} and A = {2, 5}, find B.

Also find A × B and B × A and prove that A × B ≠ B × A.
iii) If A = {1, 2}, B = {2, 3}, C = {3, 4, 5}, find A × (B ∪ C). Also show in tabular form.
iv) If A = {4, 5}, B = {6, 7, 8}, find A × (A ∪ B). Also show in tree diagram.
v) If P(Q ∪ R) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6)} find the sets

P and Q ∪ R.

8 PRIME Opt. Maths Book - IX

6. Prime more creative questions :
i) Find the value of ‘x’ and ‘y’ from the equal ordered pairs (2x+y, 9) and (256, 3x –y)
ii) If ordered pair (3, a) belongs to the ordered pair (x, y) such that y = 5x –7, find

the value of ‘a’.
iii) If ordered pair (m, 8) belongs to the ordered pair (x, y) such that y = 3x + 2, find

the value of m.

iv) If (2, p) and (q, – 1) are the two ordered pairs of the members of (x, y) such that
2x + y = 7, find the value of p and q.

v) If A = {1, 2, 3}, B = {4, 5} and C = {6, 7, 8}, prove that the Cartesian products.

A × (B ∪ C) = ( A × B) ∪ (A × C)

1. Show to your teacher. Answer iii) x = 1, y = 2
iii) m = 2
2. Show to your teacher. ii) x = 1, y = 1
v) x = 4, y = 5
3. i) x = 4, y = 3
iv) x = 4, y = 2 ii) a = 8

4. Show to your teacher.

5. Show to your teacher.

6. i) x = 5, y = 3
iv) p = 3, q = 4

PRIME Opt. Maths Book - IX 9

1.2 Relation

If A = {1, 2, 3}, B = {4, 5, 6} and A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4),
(3, 5), (3, 6)}
Here,
Taking the sub set of ordered pairs from A × B as R = {(1, 6), (2, 5), (3, 4)}

Where sum of antecedent and consequence elements is always 7.
i.e. x + y = 7
It is called the relation from the set A to the set B which is the sub - set of Cartesian
product A × B.

The set of related pairs (x, y) taken from the non - empty
sets A to B which is the sub - set of Cartesian product A × B is
called the relation R : A → B where x and y associated under
the rule given by R.

Let us consider R = {(1, 6), (2, 5), (3, 4)},
6 is called the image of 1 under R and we write R(1) = 6
5 is called the image of 2 under R and we write R(2) = 5
4 is called the image of 3 under R and we write R(3) = 4

Similarly, 1 is called the pre-image of 6.
2 is called the pre-image of 5 and
3 is called the pre-image of 4.

Domain, Co - domain and Range

In the above example, the relation (x, y) is the x + y = 7 which is R = {(1, 6), (2, 5),
(3, 4)}
Here, A × B is called the Cartesian product. Set A is called the domain.
Set B is called the co - domain. Set {4, 5, 6} taken from set B is called the range.

If R be the relation (x, y) from the set A to the set B.
i.e. R = {(x, y) : x ∈ A and y ∈ B}

• The set of all the elements ‘x’ of the relation R = {(x, y) : x ∈ A and y
∈ B } are taken from the main set A which is called domain.

• The set of all the elements ‘y’ of the relation R = {(x, y) : x ∈ A and y
∈ B } are taken from the another set B which is called range.

• The set of all the elements of the set B from which set of elements ‘y’
are taken is called the co - domain.

• The range is the sub set of co - domain.

10 PRIME Opt. Maths Book - IX

Inverse relation :

Let us take a relation R = {(1, 3), (2, 4), (3, 5), (4, 6)} taken from the set A to the set B.
The set of ordered pairs of the relation ‘R’ can be taken reversely as {(3, 1), (4, 2), (5, 3),
(6, 4)} which is the relation taken from the set B to the set A is called the inverse relation
(R–1) of the relation R.

If R is a relation from the set A to the set B, the new relation
taken from the set B to the set A is called the inverse relation
R–1.

i.e. R–1 = {(x, y): x ∈ B and y ∈ A for R}

Representation of relation:
1. Ordered pair form :

R = {Set of ordered pairs (x, y)}
2. Description form :
R = Set of ordered pairs (x, y) where x is from the set A and y is from the set B
3. Set builder form :
R = {(x, y) : x ∈ A and y ∈ B and relation of x and y }
4. Arrow diagram :

R



5. Tabular form :
R= x • • • • • •

y••••••
6. Graphical form :

Y

O X

PRIME Opt. Maths Book - IX 11

Worked out Examples

1. If A = {a, b, c, d} and B = {p, q, r, s }, find the relation R : A → B where R shows the
relation of the elements according to their position respectively. Also show in arrow
diagram.

Solution :
A = {a, b, c, d}
B = { p, q, r, s}
Then,
R = {(x, y) : x ∈ A and y ∈ B }
= {(a, p), (b, q), (c, r), (d, s)}

Arrow diagram :
R

a p
b q
c r
d s


2. If A = {1, 2, 3 4}, find the relation R is the square root of in set builder form and in
ordered pair. Also show in tabular form.

Solution :

A = {1, 2, 3, 4}
R = {(x, y) : x ∈ A and y = x2}
= {(1, 1), (2, 4), (3, 9), (4, 16)}


Table :

R= x 1 2 3 4

y 1 4 9 16

3. If A = {2, 3, 4, 5}, find a relation R : A × A and x + y ≤ 6. Also show in mapping diagram.
Solution :

A = {2, 3, 4, 5}
Then,
A × A = {(2, 2), (2, 3), (2, 4), (2, 5), (3, 2), (3, 3), (3, 4), (3, 5), (4, 2), (4, 3),

(4, 4), (4, 5), (5, 2), (5, 3), (5, 4), (5, 5)}.

The relation,
R = {(x, y) : x ∈ A, y ∈ A and x + y ≤ 6)}
= {(2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (4, 2)}.

12 PRIME Opt. Maths Book - IX

Mapping diagram :
R

A A

2 2
3 3
4 4
5 5

4. If R = {(x, y) : x + y = 7, x, y∈ N}. Also show in graph. 5 67
Solution : 2 1 0∉N
N = {1, 2, 3, 4, 5, .....}
R = {(x, y) : x + y = 7, x, y ∈ N}
Then,

x1234

y 6 5 4 3
\ R = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

Graph :
Y

R

O X

5. If a relation R = {(x, y) : y > x, x ∈ A and y ∈ B }, A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, find the
inverse relation. Also show in arrow diagram.

Solution :

A = {1, 2, 3, 4}

B = {3, 4, 5, 6}
Then, A × B = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4),
(3, 5), (3, 6), (4, 3), (4, 4), (4, 5), (4, 6)}

From the ordered pairs of A × B, we get the relation y > x,
R = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5),
(3, 6), (4, 5), (4, 6)}

Inverse Relation : 13
R-1 = {(3, 1), (4, 1), (5, 1), (6, 1), (3, 2), (4, 2), (5, 2), (6, 2), (4, 3), (5, 3),
(6, 3), (5, 4), (6, 4)}

PRIME Opt. Maths Book - IX

Arrow diagram of R-1
B R–1
A

3 1
4 2
5 3
6 4

Exercise 1.2

1. i) What is relation?
ii) Define the term domain and co-domain.
iii) What is the difference between range and co-domain?
iv) What do you mean by inverse relation?
v) If R = {(x, y) : x ∈ A, y ∈ B}, write down its inverse.

2. Which of the following relations are the relation in A × B where A = {1, 2, 3, 4}, B = {4,

5, 6, 7} 5), 6), (6, 7)} RR42
i) RRR513 = {(3, 4), (4, 4), (5, 5), (7, 6)} ii)
iii) = {(4, 3), (5, 6)} (6, iv) = {(1, 4), (2, 5), (3, 6), (4, 7)}
v) = {(3, 5), (4, = {((3, 5), (4, 6), (2, 7)}

3. If A = {a, b, c, d}, find A × A and find which of the followings are the relation.
i) RRR315 = {(a, a), (b, b), (c, c), (d, d)} ii) RR42
iii) = {(a, 1), (b, 2), c, 3), (d, 4)} iv) = {(a, b), (b, c), (c, d), (d, a)}
v) = {(d, a), (c, b), (b, a), (d, c)} = {(1, a), (2, b), (3, c), 4, d)}

4. If A = {1, 2, 3, 4}, B = {5, 6, 7, 8} find the relation R from the set A to B and represent

it in the followings.

i) Arrow diagram ii) Graphical representation
iii) Ordered pair form iv) Set builder form
v) Tabular form

5. If the sets A and B represents the numbers of articles and cost respectively. Find
the relation R and it’s inverse relation by calculating A × B. Also show R-1 in the

followings.

i) In the set of ordered pairs ii) In tabular form

iii) In description form iv) In Mapping diagram

v) In graphical form
Where A = {12, 15, 25, 30} and B = {Rs.200, Rs.400, Rs.450, Rs.550} are the articles

and their costs respectively.

14 PRIME Opt. Maths Book - IX

6. Find domain, range relation in ordered pairs and inverse relation from the followings.
i) x 1 2 3 4 5 6 7

y 3 5 7 9 11 13 15

ii) A R B iii) Y

6
a 35

b 44

c 53
62
d 71

O 123456 X

iv) R = {(x, y) : x, y ∈ N, x + y = 6} v) R = {(x, y) : y = x2, 1 ≤ x < 6}

7. If A = {1, 2, 3, 4}, find the relation R in ordered pair form from A × A under the
following conditions.

i) The relation is ‘is equal to’ Also show in arrow diagram
ii) The relation is ‘greater than’ Also show in graph
iii) The relation is ‘less than’ Also show in set builder form
iv) The relation is ‘x + y < 5’ Also show in tabular form.
v) The relation is ‘x + y ≥ 5’ Also show in mapping diagram.

8. If A = {3, 4, 5, 6}, B = {7, 8, 9, 10, 11}, find ‘R’ in ordered pair form from

A× B under the following conditions.
i) RRR135 = {(x, y) : x + y= 14} ii) RR24
iii) = {(x, y) : x + y≥ 15} iv) = {(x, y) : x + y ≤ 13} x by 5}
v) = {(x, y) : y is more than
= {(x, y) : y < 2x}

9. Prime creative questions :
i) If A = {2, 3, 4, 5}, Relation R = {(x, y) : y = 3x}, find R and show in arrow diagram.
ii) If a relation R = {(x, y) : x + y = 10, x, y ∈ N}, find R and show in graph.
iii) If a relation R = {(x, y) : x2 + y2 = 25, x, y ∈ I}. Find the relation R and represent it

in mapping.
iv) If the linear relation R = {(x, y) : y = ax + b} has 5 and 8 as the range for the

domain elements 1 and 2 respectively. Find the value of ‘a’ and ‘b’.

x –1 2 3 is from the relation R = {(x, y) : y = ax + b}, find the
v) If y 17 9

value of ‘a’ and ‘b’.

10. Project Work:
Write down two sets of name of vegetables and their cost per kilogram respectively and

find their Cartesian product. Also write down the relation of vegetables in arrow diagram.

PRIME Opt. Maths Book - IX 15

Answer

1. Show to your teacher.

2. Show to your teacher.

3. Show to your teacher.

4. Show to your teacher.

5. Show to your teacher.
6. i) Domain = {1, 2, 3, 4, 5}
Range = {3, 5, 7, 9, 11, 13, 15}
R = {(1, 3), (2, 5), (3, 7), (4, 9), (5, 11), (5, 2), (7, 15)}
R-1 = {(3, 1), (5, 2), (7, 3), (9, 4), (11, 5), (13, 6), (15, 7)}
ii) Domain = {a, b, c, d}
Range = {3, 4, 5, 6, 7}
R = {(a, 3), (b, 3), (b, 4), (c, 5), (c, 6), (d, 7)}
R-1 = {(3, a), (3, b), (4, b), (5, c), (6, c), (7, d)}
iii) Domain = {1, 2, 3, 4, 5, 6}
Range = {4, 5, 6}
R = {(1, 5), (2, 6), (3, 4), (4, 4), (5, 6)}
R-1 = {(5, 1), (6, 2), (4, 3), (4, 4), (6, 5)}
iv) Domain = {1, 2, 3, 4, 5}
Range = {1, 2, 4, 5, 6}
R = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
R-1 = {(5, 1), (4, 2), (3, 3), (2, 4), (1, 5)}
v) Domain = {1, 2, 3, 4, 5}
Range = {1, 4, 9, 16, 25}
R = {(1, 1), (2, 4), (3, 9), (4, 16), (5, 25)}
R-1 = {(1, 1), (4, 2), (9, 3), (16, 4), (25, 5)}
7. i) R = {(1, 1), (2, 2), (3, 3), (4, 4)}
ii) R = {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}
iii) R = {(2, 1), (3, 1), (4, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)}
iv) R = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)}
v) R = {(1, 4), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 5)}
8. i) RRRRRR25314======{{{{{{((((((244333, ,,,,,617871))))1), ,,,,()((((3,5345(, ,4,,,98179,)))01), ,,,()0(((4,)535(,,,,5,1(981,52)10),),,)90,((,())53(5,,,6,((,1966,1)10,,5,81)9)(,)))}6(}},,4,(d76,i7)a,,)1g(,r06(a)4,m,8,(8)6,),(,16(1,49),}9),)(,6(,51, 07)),,
ii) (5, 8), (6, 7)}
iii) (6, 11)}
iv)
v)
9. i)
ii) R = {(1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)}, diagram
iii) R = {(0, ±5), (±3, ±4), (±4, ±3), (±5, 0)}

iv) a = 3, b = 2

v) a = 2, b = 3

16 PRIME Opt. Maths Book - IX

1.3 Function

Let us consider a non - empty sets A = {Sita, Pranav, Pranisha} and B = {(School, NGO,
Municipality}. If all the members of the set A has got only one job in the organization
mentioned in set B. The relation in such condition from the set A to the set B is called the
function where all of them may got job in same organization as well as in different but
there should not be more than one organization to a function from the set A to the set B.

• The set of ordered pairs (x, y) taken from the non - empty
set A to B where every element of first set is associated
with one and only one elements of second set is called the
function f : A → B.

• A function f is at special type of relation from the set A to the
set B where each element of domain uniquely related with
the element of co - domain.
It is denoted by f : A → B and defined as y = f(x)

Some informations about function:

• f is called the function from the non - empty set A to B which is written as f : A → B.
• Function f : A → B can be written in the form of ordered pairs (x, y) and the function
‘f’ can be written as y = f(x).
• Function y = f(x) can be written in the form of algebra as well.
• All the elements of set A should be associated with the elements of B to be a function.
• Any element of A can associate with only one element of B to be a function.


Image, pre - image, domain, co - domain and range of a function

In a function f : A → B from the set A to the set B defined by y = f(x), x ∈ A, y ∈ B.
• The set of elements of A from which elements x are taken is called domain.
• The set of elements of B from which elements y are taken is called co - domain.
• The elements y in the function y = f(x) is called image of x.
• The element x in the function y = f(x) is called pre - image of y.
• The set of elements ‘y’ taken from the set B satisfying y = f(x) is called range.

PRIME Opt. Maths Book - IX 17

Examples of functions in arrow diagram are as follows.
Af B

ap
bq
cr
ds
t
Here,
• f = {(a, p), (b, p), (c, q), (d, r)} is a function from the set A to B.
• Set A = {a, b, c, d} is domain
• Set B = {p, q, r, s, t} is co - domain
• Set of range = {p, q, r}
• Element p is the image of ‘a’ and ‘b’
• Elements ‘a’ and ‘b’ are the pre - images of the element ‘p’.

Representation of function :
1. Set of ordered pairs.

f = {(1, 2), (2, 3), (3, 4), (4, 5)}

2. Algebraic equation
In the above set of ordered pairs of a function ‘f’.

f = {(1, 2), (2, 3), (3, 4), (4, 5)}
It can be represented in equation where the relation is ‘more than by 1’.

i.e. y = f(x) = x + 1

3. Tabular form :
In the example; f = {(1, 2), (2, 3), (3, 4), (4, 5)}

x12345

y23456

4. Arrow diagram (Mapping)
The function : f = {(1, 2), (2, 3), (3, 4), (4, 5)} can be represented in mapping as,

f

12

23

34

45


18 PRIME Opt. Maths Book - IX

5. Graphical representation :
The function : f = {(1, 2), (2, 3), (3, 4), (4, 5)} can be represented in graph as,

Y

5
4
3
2
1
O 123456 X

Types of function :
1. One to one function (Injective function)

The function in which for different elements of domain
have different images in co - domain is called a one to one
function. In the other words : if each image has only one pre
- image in domain, the function is called one to one function.
Mathematically : x1, x2 ∈ A and x1 ≠ x2 ∃ y1, y2 ∈ B in f : A
→ B such that y1 = f(x1) ≠ y2 = f(x2) then f is called one-one
function.

Types of one-one function.
i) If all the elements of co - domain have pre - images, the function is said to be one

to one and onto function (adjective function).

f

ax

by

cz

ii) If at least one element of co - domain does not have pre - image in domain, it is

said to be one to one and into function.
f

ap
bq
cr

s

PRIME Opt. Maths Book - IX 19

2. Many to one function :

A function in which more than one elements of domain are
associated with one and only one element of co - domain, the
function is called many to one function. In the other words, if at
least one element of co domain has more than one pre- images,
then the function is called the many to one function.

Types of many to one function.
i) If all the elements of co - domain have pre - images, the function is called many

to one and onto function.
f

ap
b
cq
d
ii) If at least one element of co - domain doesn’t have pre - images, the function is
said to be many to one and into function.

f

a p
b q
c r
d s
e


3. Algebraic function f(x) :

The function y = f(x) in the form of algebraic equation is
called algebraic function.

Example : y = f(x) = ax + b

Types of algebraic function.
i) Constant function.
The algebraic function in the form of f(x) = c (constant) is called constant

function.

ii) Linear function
The algebraic function in the form of linear equation f(x) = ax + b, a ≠ 0 is called

linear function.

20 PRIME Opt. Maths Book - IX

iii) Quadratic function :
The algebraic function in the form of quadratic equation f(x) = ax2 + bx + c, a ≠ 0

is called quadratic function.
iv) Cubic function
The algebraic function in the form of cubic equation f(x) = ax3 + bx2 + cx + d, a ≠

0 is called cubic function.
v) Biquadratic function
The algebraic function in the form of biquadratic form f(x) = ax4 + bx3 + cx2 + dx

+ e, a ≠ 0 is called biquadratic function.
vi) Identity function
The algebraic function in the form of algebraic equation y = f(x) = x, is called

identity function.
vii) Trigonometric function
The function in the form of trigonometric equation like f(x) = Sinx, f(x) = Tanx

etc is called trigonometric function.
viii) Exponential function :
The algebraic function in the form of exponential equation like f(x) = ex or f(x) =

ax, a > 0 is called exponential function.

Vertical line test for a function :

After plotting the ordered pairs of a relation in a graph paper, different types of curve,
and straight line can be obtained. If a straight line vertically drawn (parallel to y - axis)
cuts the graph of relation at a single point only, it is a function otherwise it represents
relation only.

• If a vertical line drawn (parallel to y - axis cuts) the graph
of relation at a single point only, it is a function.

• If it cuts the graph at two or more than two points, it will
not a function. It represents relation only.

Examples : YA ii) AY
i)
y=x2 f
X’ Vertical line X

OX X’ O

Vertical line B Y’
Y’ B (It is a function)

(It is a function)

PRIME Opt. Maths Book - IX 21

iii) AY iv) YA

X’ O X X’ O X

Vertical line f f Vertical line
B Y’ Y’ B

(It is not a function) (It is not a function)

v) AY

f

X’ O X

Vertical line

B Y’
(It is a function)

Worked out Examples

1) If f(x) = 2x2 – 3, domain = {–2, –1, 0, 1, 2} find range. Show the function in arrow

diagram.

Solution :
f(x) = 2x2 – 3

domain = { –2, –1, 0, 1, 2}
Then,
f(–2) = 2(–2)2 – 3 = 5
f(–1) = 2(–1)2 – 3 = –1
f(0) = 2(0)2 – 3 = – 3
f(1) = 2(1)2 – 3 = –1

f(2) = 2(2)2 – 3 = 5
\ Range = {–3, –1, 5}

Arrow diagram f

–2 –3
–1 –1
0 5
1
2 PRIME Opt. Maths Book - IX

22

2. If image of a function f(x) = 2x – 3 is 1, find the pre-image of it.
Solution :

f(x) = 2x – 3

image = 1
Then,
f(x) = 1

or, 2x – 3 = 1

or, 2x = 4

\ x = 2
\ Pre- image of ‘1’ is ‘2’.

3. If f(x) = 2Cos x, range = {2, 3, 1}, find domain. Also show in mapping.
Solution :

f(x) = 2Cos x

range = { 2, 3 , 1}

Now,
Taking, f(x) = 2
or, 2Cos x= 2

or, Cos x = 1

or, Cos x = Cos 0°

\ x = 0°
Again, Taking,
2Cos x = 3

or, Cos x = 3
2

or, Cos x = Cos 30°

\ x = 30°

Again, Taking,

2Cos x = 1

or, Cos x = 1
2

or, Cos x = Cos 60°

\ x = 60°
\ Domain = {0°, 30°, 60°}

Arrow diagram :
f

0° 1
30° 3
60°
2

PRIME Opt. Maths Book - IX 23

4. If f(2x + 3) = 3x – 1, find f (x + h) – f (x)
Solution : h

Let 2x + 3 = a

x= a–3
2

\ f(2x + 3) = 3x – 1

or, f(a) = 3a a –3 k – 1
2

= 3a – 9 – 2
2

= 3a – 11
2

Since ‘a’ is dummy suffix we exchange ‘a’ by ‘x’
3a – 11
\ f(x) = 2

Now, f(x + h) = 3 (x + h) – 11
2

= 3x + 3h – 11
2

f (x + h) – f (x) 3x + 3h – 11 – 3x – 11
h 2 2
and =
h

3x + 3h – 11 – 3x + 11

= 2
h

= 3h Alternative Method
2h

= 3 f(2x + 3) = 3 (2x + 3) – 1 – 9
2 2 2

= 3 (2x + 3) – (1 + 9 )
2 2

= 3 (2x + 3) – 11
2 2

= 3 x– 11
2 2

= 3x – 11
2

f(x + h) = 3 (x – h) – 11
2

f (x + h) – f (x) 3x + 3h – 11 – 3x – 11
h 2 2
=
h

3x + 3h – 11 – 3x + 11

= 2
h

3h
= 2h

= 3
2

24 PRIME Opt. Maths Book - IX

5. If f(x) = x2 – 5, g(x) = 3x + 5 and f(x) = g(x), find the value of ‘x’.

Solution :

f(x) = x2 – 5

g(x) = 3x + 5
Then,

f(x) = g(x)

or, x2 – 5 = 3x + 5

or, x2 – 3x – 10 = 0

or, x2 – (5 –2) x – 10 = 0

or, x2 – 5x + 2x – 10 = 0

or, x(x – 5) + 2(x – 5) = 0

or, (x – 5) (x + 2) = 0

Either Or

x – 5 = 0 x+2=0

\ x = 5 x=–2

\ x = 5 or – 2

6. If f(x + k) = f(x) + f(k), prove that f(–k) = – f(k). Also find f(2k)
Solution :

f(x + k) = f(x) + f(k)
Taking x = 0,
f(0 + k) = f(0) + f(k)

or, f(k) = f(0) + f(k)

\ f(0) = 0

Again, Taking x = –k
f(– k + k) = f(– k) + f(k)
or, f(0) = f(–k) + f(k)
or, 0 = f(–k) + f(k)
or, –f(k) = f(– k)
\ f(–k) = –f(k)

Again, Taking x = k,
f(x + k) = f(x) + f(k)
or, f(k + k) = f(k) + f(k)
or, f(2k) = 2f(k)
\ f(2k) = 2f(k)

PRIME Opt. Maths Book - IX 25

Exercise 1.3

1. i) What do you mean by function?
ii) Define domain and rage in a function y = f(x).
iii) If f = {(1, 2), (3, 4), (5, 6), (7, 8)}, find its inverse function.
iv) Define image and pre-image in a function y = f(x).
v) How can you say that the relation is a function by using vertical line test?

2. i) If f = {(3, 2), (4, 3), (5, 2), (6, 1)}, find domain and range. Also show in mapping
diagram.

ii) If a function ‘f’ is defined as the mapping given below, find domain and range of
the function.
f

2p

4q

6r

8 s

iii) If a function ‘f’ is defined as the table given below. Represent it in graph. Also
find domain and range of the function.

x345678
y 5 6 7 8 9 10

iv) Which of the given diagram represents a function ? Why?

a) Y b) Y

R

R

X’ O X X’ OX

Y’ Y’

v) Which of the given curves are the functions? Write down with reason.

a) Y b) Y
R

R

X’ O X X’ O X

26 Y’ PRIMYE’ Opt. Maths Book - IX

3. Which of the followings relations are the functions ? Mention the reason also ?
R R
i) ii)

1 5 a3
2 6 b4
3 7 c5
4 86

iii) R iv) R

a pa
b q xb
c r yc
d
zd

v) R

p f
aq
br
cs

4. Which types of functions are given below ?
i) f ii)

2a a 4
3b b 5
4c 6

iii) f iv) f

pa a3
q b 4
r b c 5

6

7

PRIME Opt. Maths Book - IX 27

v) f

1

p2
q3

4

5

5. i) Find the image of an element 3 in a function f(x) = 4x – 3.
ii) If domain = {–2, 1, 3} in a function f(x) = 2x + 5, find the range.
iii) Find the pre - image of an element 5 of a function f(x) = 4x – 3.

iv) Find the domain element of a function f(x) = 2x + 7 whose range is 3.

v) What is the range of a function f(x) = x2 – 3 whose domain is {–2}?


6. i) If domain = {–2, –1, 0, 1, 2} of a function, f(x) = 2x2 – 3, find range. Also show the

function in arrow diagram.
ii) If f(x) = 3x + 2 is a function and domain = {0, 1, 2, 3}, find the range. Also show

the function in arrow diagram.
iii) If range = {5, 7, 11} of a function f(x) = 2x + 7, find the domain. Also show the

function in graph.
iv) If range of a function g(x) = 5x – 3 is {–3, 2, 7}, find the domain. Also show the

function in graph. Which type of function is it? Why.
v) If domain of a function f(x) = x2 + 2 is {–2, –1, 1, 2} find the range. Also show the

function in arrow diagram. Which types of function is it? Why?

7. i) If f(x) = 2x + 5, find the value of f(2), f(–3), f(a), f(a + 2) and f(x + 2).
ii) If g(x) = 3x – 1, find g(x + h), g(2x + 3) and g(4x – 3).

iii) If f(x) = 3x + 2, find f (x + h) –f (x)
h
iv) If f(x + 3) = f(x) + f(3), prove that f(0) = 0 and f(–3) = –f(3)

v) If f(x + a) = f(x) + f(a), prove that f(–a) = –f(a).

8. i) If f(x + 2) = x – 3, find f(x) and f(2x + 3)
ii) If f(x – 2) = 2x + 1, find f(x) and f(x + h)

iii) If f(x + 3) = 2x – 5, find f(x) and f (x + h) – f (x)
h
iv) If f(2x + 1) = 3x + 2, find f(x) and f(x + 2)
v) If f(3x – 2) = 2x + 5, find f(x) and f(2x – 5).

9. PRIME more creative questions :
i) If f(x) = x2 + 2x – 7, g(x) = 8, find the value of ‘x’ where f(x) = g(x)
ii) If f(x) = x2 – 5x, g(x) = x – 2 and f(x) + g(x) = 10, find the value of ‘x’.
iii) If f(x) = 2Sinx + 1, domain = {0°, 30°, 60°, 90°}, find range. Also show in arrow

diagram.

28 PRIME Opt. Maths Book - IX

iv) If f(x) = 2Cos x – 1, domain = {0°, 30°, 60°, 90°}, find range. Also show in Mapping
diagram.

v) If f(x) = 1 + 2Sin x, range = {1, 2, 3}, find the domain. Also show in arrow diagram.

10. Project work
Collect the types of functions with examples in a chart paper and present it into your

classroom as the instructor.

Answer

1. Show to your teacher.
2. Show to your teacher.
3. Show to your teacher.
4. Show to your teacher.

5. i) 9 ii) Range = {1, 7, 11} iii) 2
iv) – 2 v) 1

6. i) Range = {–3, –1, 5}; arrow diagram
ii) Range = {2, 5, 8, 11}; arrow diagram
iii) Domain = {–1, 0, 2}; graph
iv) Domain = {0, 1, 2}; graph; one to one onto
v) Range = {3, 6}; arrow diagram; many to one onto

7. i) 9 –1, 2a + 5, 2a + 9, 2x + 9
ii) 3x + 3h – 1, 6x + 8, 12x – 10
iii) 3

8. i) x – 5, 2x – 2 ii) 2x + 5, 2x + 2h + 1 4x + 9
iii) 2x – 11, 2 3x + 1 3x + 7 + 3
iv) 2 , 2 v) 2x 3 19 ,

9. i) 3, – 5 ii) –2, 6
iii) Range = {1, 2, 3 + 1, 3}; arrow diagram
iv) Range = {–1, 0, 1, 3 – 1}; mapping diagram
v) Domain = {0°, 30°, 90°}; arrow diagram

PRIME Opt. Maths Book - IX 29

1.4 Polynomials

Enjoy the recalls:

In previous classes we have already discussed about the variables, constant, index, base

etc which are used in algebra, where
The composition of coefficient, variable and index of the variable with multiplication

form is called algebraic term.

Examples : 3x2, 2xy, 3 x3, 2x2 etc.
2 y2

Let us consider an example of algebraic term 2x3y in x.

Here,
2 is called numeral coefficient.
x is called base.

3 is called power (index exponent)
y is called literal coefficient of x.

The combination of the algebraic terms with + or – sign is called algebraic expression.
Examples : 2x2 – 3x + 5, 2x + 3, 2x etc.

In expressions the index of the variable may be positive as well as negative. But the
expressions having index (exponent) non-negative integer of the variables is called
polynomial.

The algebraic expression having non–negative integer
as the power of the variable and having coefficient a
real number is called polynomial. It is denoted by p(x),

f(x), g(x) etc.

Standard form of polynomial

The Polynomials are written according to their degree in ascending order of the variables.
The polynomial of degree ‘n’ can be generalized as below which is the standard form of
the polynomial in x.

p(x) = a0xn + a1xn-1 + a2xn–2 + ...................................... an–1x + anx0

• Where, n is non–negative integer and
• an0,isa1c,aal2le, .d....t.h...e...d...e..g...r..e..e. aonfatrheetphoelyrenaolmniuaml pb(exr)s coefficient & a0 ≠ 0.

• aS0oxmn, ea1oxfn-t1h.e.. are the terms of the polynomial. are
examples of standard form of polynomials

x5 – 2x4 – 3x2 – 2x + 7

x4 – 2x3 + 3x2 – 5x + 2

3x3 + 2x2 – 5x + 3

30 PRIME Opt. Maths Book - IX

Degree of polynomial

The highest power (exponent) of the variables used in a polynomial is taken as degree of
the polynomial. In the above examples the three polynomials are of degree fifth, fourth,
and third respectively.

Types of Polynomials :
i) According to degree

• The polynomial having degree ‘1’ is called linear polynomial.

p(x) = ax + b (First degree polynomial)
• The polynomial having degree ‘2’ is called quadratic polynomial.

p(x) = ax2 + bx + c. (Second degree polynomial)
• The polynomial having degree ‘3’ is called cubic polynomial.

p(x) = ax3 + bx2 + cx + d. (Third degree polynomial)
• The polynomial having degree ‘4’ is called biquadratic polynomial.

p(x) = ax4 + bx3 + cx2 + dx + e. (Fourth degree polynomial)

ii) According to number of terms.

• Monomial → Having only one term in the expression.

f(x) = 2x

• Binomial → Having two terms in the expression.

f(x) = 3x2 + 2x

• Trinomial → Having three terms in the expression.

f(x) = 3x2 + 2x + 1

• Multinomial → Having more terms in the expression.

f(x) = 3x3 + 2x2 + 5x + 2

iii) Polynomials according to coefficients:
• Polynomials over integer : x4 – 3x2 + 2x – 5
3 5 7
• Polynomial over rational numbers : x5 – 2 x4 + 3 x3 + 2 x2 + 5

• Polynomial over real number : 5x3 – 3x2 + 5x + 7

Other information on polynomials
• The polynomial may have more than one variable also.
• P(xy) = ax2y + bxy + cxy2
• p(xyz) = ax2yz + bxy2z + cxyz2.

PRIME Opt. Maths Book - IX 31

Equal polynomials:

The polynomials having degree and coefficient of the corresponding terms are same are

called equal polynomials.
• No of terms should be equal.

• Degree of the variable should be equal.

• Coefficient of the variables should be equal.

• Type of variable should be equal.

p(x) = 2x3 + 5x2 – 3x + 2

q (x) = 4 x3 + 5x2 – 6x + 8
2 2 4

= 2x3 + 5x2 – 3x + 2

Here, p(x) = q(x)

Worked out Examples

1. Which of the followings expressions are the polynomials? Why?

i) x4 – 2x3 + 3x2 – 5x + 2
3
ii) 3x3 – 4x2 + 2x – x +5

iii) 7 x3 + x2 – 3 x + 7

Solution :

In the given expressions,

i) x3 – 2x3 + 3x2 – 5x + 2 is a polynomial because all the variable of the terms have

positive integer (whole number) as the index.

ii) 3x3 – 4x2 + 2x – 3 + 5 is not the polynomial because index of the variable ‘x’ of
x
a term in negative.

iii) 7x3 + x2 – 3x + 7 is a polynomial because index of the variable used in all the

terms a re-equal.

2. Write down the degree of the polynomial x4 – 3x3 + 2x2 + 5x – 3. Also write down its

types.

Solution :
The given polynomial is,
x4 – 3x3 + 2x2 + 5x – 3
The maximum index of the variable used in it is ‘4’.
Hence it’s degree is ‘4’.
Also the polynomial having degree 4 is called biquadratic polynomial.

32 PRIME Opt. Maths Book - IX

3. Write down the polynomials in standard form. Also write dwon the conclusion which
you find in the polynomials.

x3 + 2x4 – 5x + 7 + 3x2 and 3x2 + 7 – 5x + 2x4 + x3.

Solution :
The standard form of, x3 + 2x4 – 5x + 7 + 3x2 is 2x4 + x3 + 3x2 – 5x + 7
The standard form of, x3 + 2x4 – 5x + 7 + 3x2 is 2x4 + x3 + 3x2 – 5x + 7
Conclusion : The number of terms and coefficient of variable ‘x’ used in both of them

are same.

Hence, they are equal polynomials.

4. If the polynomials (m + 1)x3 – 3x2 + nx + 3 and 5x3 – 3x2 + 2x + 3 are equal polynomials,
find the value of ‘m’ and ‘n’.

Solution :
The equal polynomials are:

(m + 1)x3 – 3x2 + nx + 3 and 5x3 – 3x2 + 2x + 3
Comparing the corresponding coefficients,

i.e. m + 1 = 5 and n=2

\ m = 4 and \ n=2

\ m = 4

n=2

5. If the polynomials p(x) = x3 + 2x2 – 3x + 1 and q(x) = x3 + x2 – 2x + 7 are equal at x = a,
find the value of a.

Solution :
p(x) = x3 + 3x2 – 3x + 1
q(x) = x3 + x2 – 2x + 7
Taking,
p(x) = q(x) for x = a,
or, a3 + 2a2 – 3a + 1 = a3 + a2 – 2a + 7
or, a2 – a – 6 = 0
or, a2 – (3 – 2)a – 6 = 0
or, a2 – 3a + 2a – 6 = 0

or, a(a – 3) + 2(a – 3) = 0

or, (a – 3) (a + 2) = 0

Either, OR
or a – 3 = 0 a+2=0

\ a = 3 a=–2

\ a = 3, – 2

PRIME Opt. Maths Book - IX 33

Exercise 1.4

1. Answer the following questions.
i) What is polynomial? Write down its types according to degree of polynomial.
ii) What is algebraic expression? Write down the types of polynomial according to

number of terms of the polynomials.
iii) What do you mean by equal polynomials? Explain with an example.
iv) Write down numeral coefficient, literal coefficient, base and power of the

polynomial 3x3y2.
v) Write down the degree of the polynomial 3x4 – 2x3 + 5x2 – 2x + 7. Also write

down its type according to degree of the polynomial.

2. Which of the algebraic expressions are the polynomials from the followings? Write

down with reasons.

i) 4x3 + 5x2 – 3x + 2 ii) 3 x3 + 4x + 2x

iii) x12 – 3 + 4 +x2 iv) x4 ( 1 + 2 + 3 )+ 10
x x3 x2 x

v) 37 x3 + 3x2 – 5 x–7+ 5 vi) 2 x3 + 3x2 – 5x +3
2 x

3. Write down the numeral and literal coefficients of the polynomials.

i) 3xy of y ii) 2x2y of x
iii) 38 xyz of xy 2xy + 3
iv) 2 of y

3x2 yz + 2
v) 2

4. Write down the following polynomials in standard form and write down the degree

of the polynomials.

i) 3x – 5 – 3x2 + x4 – 2x3 ii) 7 – 2x2 – 3x + 5x3

iii) 3 – 5x2 – x3 + 2x4 – 3x + x5 iv) 3x – 2 + 5x2 – x5 + 2x3 – 3x4

v) 3x4 + 2x5 – 7 – 3x – 2x2 + x3

5. Write down the types of the polynomials according to degree and number of terms.

i) 3x – 2x2 – 5 + 2x4 – x3 ii) 5 + 3x2 – 2x + 3x3

iii) –3 + 2x + x2 iv) 7 – 3x

v) 2

6. Prove that the following polynomials are equal.
i) 3x3 – 4x2 + 3x – 2 and 9 x3 – 2 + 3x – (2x)2

ii) p(x) = 25 x2 – 4x + 5 and q(x) = 5 – 22x + 5x2
iii) f(x) = 4x4 – 9x2 – 10x + 7 + x3 and g(x) = x3 – (3x)2 + (2x2)2 – 10x + 7
iv) 3x3 + x2 – 5 + 2x and 2x – 5 + x4 + 3 27 x3
v) 12x3 + mx2 – 2mx + 7 and (2m + 2)x3 + 5x2 – (3n + 1)x + 7 where m = 5 and n = 3.

34 PRIME Opt. Maths Book - IX

PRIME more creative questions.
7. If the following polynomials are equal, find the value of ‘a’ and ‘b’.
i) p(x) = (2a + 1)x3 – 7x2 + 3x + 2 and q(x) = 7x3 + (2b – 3)x2 + 3x + 2
ii) f(x) = 10x4 – 3x3 + 5x2 – 12x + 5 and g(x) = 10x4 – (3a – 3)x3 + 5x2 + 3bx + 5.
iii) p(x) = 5x2a – 1 + (3b + 2)x2 – 7x + 3.
q(x) = 5x3 + 8x2 – 7x + 3.
iv) f(x) = 2ax4 – 3x3 + bx2 – 2
g(x) = 4x4 – 3x3 – 2
v) p(x) = (3a – 2)x4 – 2x3b – 5 – 3x2 + 2x + 5
q(x) = 7x4 – 3x2 + 2x + 5
vi) If p(x) = q(x) at x = a where p(x) = x3 + 4x2 + 3x – 2 and q(x) = x3 + 3x2 + x + 13,

find the value of a.

1. Show to your teacher. Answer iii) a = 2, b = 2
2. Show to your teacher. vi) –5, 3
3. Show to your teacher. ii) a = 2, b = –4
4. Show to your teacher. v) a = 3, b = 2
5. Show to your teacher.
6. Show to your teacher.
7. i) a = 3, b = –2
iv) a = 2, b = 0

PRIME Opt. Maths Book - IX 35

1.5 Operation on Polynomial

Let us consider some of the examples which are already discussed in previous classes.
x × x = x1 + 1 = x2
x + x = (1 + 1)x = 2x
x3 ÷ x2 = x3 – 2 = x
3x3 × 2x2 = 6x5
2x2 + x3 = 2x2 + x3

1. Addition and subtraction :

Like terms of polynomials can be added and subtracted.

Example : –3x3, 2x3 and 3 x3 can be added.
2

Example : p(x) = x3 + 3x2 – 2x + 5

q(x) = 2x3 – 5x2 + x + 7
Then,

p(x) + q(x) = (x3 + 3x2 – 2x + 5) + (2x3 – 5x2 + x + 7)

= x3 + 3x2 – 2x + 5 + 2x3 – 5x2 + x + 7

= (x3 + 2x3) + (3x2 – 5x2) + (–2x + x) + (5 + 7)

= 3x3 – 2x2 – x + 12

p(x) – q(x) = (x3 + 3x2 – 2x + 5) – (2x3 – 5x2 + x + 7)
= x3 + 3x2 – 2x + 5 – 2x3 + 5x2 – x – 7
= –x3 + 8x2 – 3x – 2

2. Multiplication:

The coefficients of the variables should be multiplied
and index of the variables should be added during

multiplication of the polynomials.

Example : p(x) = 2x3 – 3x2 + 2x – 5

q(x) = x2 – 3x + 2
Then,

p(x) × q(x) = (2x3 – 3x2 + 2x – 5) (x2 – 3x + 2)

= x2(2x3 – 3x2 + 2x – 5) – 3x(2x3 – 3x2 + 2x – 5) + 2(2x3 – 3x2 + 2x – 5)

= 2x5 – 3x4 + 2x3 – 5x2 – 6x4 + 9x3 – 6x2 + 15x + 4x3 – 6x2 + 4 x – 10

= 2x5 – 9x4 + 15x3 – 17x2 + 19x – 10

36 PRIME Opt. Maths Book - IX

3. Division :

The coefficients of the variables should be divided and
index of the variables should be subtracted during

division of the polynomials.

Example : p(x) = 4x4 – 6x3 + 8x2 – 10x

q(x) = 4x
Then,

p(x) ÷ q(x) = (4x4 – 6x3 + 8x2 – 10x) ÷ (4x)

= 4x4 – 6x3 + 8x2 – 10x
4x 4x 4x 4x

= x3 – 3 x3–1 + 2x2–1 – 5 x1–1
2 2

= x3 – 3 x2 + 2x – 5
2 2

Let us taking another example:
p(x) = x3 – 3x2 + 4x – 2
q(x) = x + 3
then,
p(x) ÷ q(x)

x + 3 ) x2 – 3x2 + 4x – 2 (x2 – 6x + 22
x2 + 3x2

––
– 6x2 + 4x – 2
– 6x2 – 18x
++
22x – 2
22x + 66
––

– 68

\ Quotient = (x2 – 6x + 22)
Remainder = – 68

PRIME Opt. Maths Book - IX 37

4. Synthetic division:
The algebric expressions can be divided by taking the coefficients of the variables

according to the alternative way of division other than the direct division. It is
discussing here in this topic.

This is the way of division of polynomials by using
the coefficient of the variables after arranging the
polynomial in standard form and by taking the
opposite sign of the constant term of the divisor.

i.e. For (ax3 + bx2 + cx + d) ÷ (x – k)

k abc d

ka k2a + kb k3a + k2b + kc

a ka + b k2a + kb + c k3a + k2b + kc + d

x2 x1 x0

Here,
Divisor = x – k [Taking ‘k’ for division]
Dividend = ax3 + bx2 + cx + d [Taking constant coefficients a, b, c and d for division]
Quotient = ax2 + (ka + b)x + (k2a + bk + c)
Remainder = k3a + k2b + kc + d

Example :
p(x) = x3 – 3x2 + 2x – 3
q(x) = x + 2
p(x) ÷ q(x)

Taking x = – 2 –3 2 –3
–2 1 –2 10 –24
–5 12 –27
1

x2 x1 x0

Here,
Dividend = p(x) = x3 – 3x2 + 2x – 3
Divisor = q(x) = x + 2
Quotient = Q(x) = x2 – 5x + 12
Remainder = R = – 27

38 PRIME Opt. Maths Book - IX

Worked out Examples

1. If f(x) = x3 + 2x2 + 3x – 2 and g(x) 3x3 + 5x2 – 7x – 1, find f(x) + g(x).
Solution :
f(x) = x3 + 2x2 + 3x – 2
g(x) = 3x3 + 5x2 – 7x – 1
Then,
f(x) + g(x) = (x3 + 2x2 + 3x – 2) + (3x3 + 5x2 – 7x – 1)
= x3 + 2x2 + 3x – 2 + 3x3 + 5x2 – 7x – 1
= 4x3 + 7x2 – 4x – 3

2. What must be subtracted from the polynomial 5x3 – 3x2 + 2x + 5 to get the polynomial
2x3 – x2 + 3x – 2.

Solution :
Let, the subtracted polynomial be ‘K’.
Then, by the question,
(5x3 – 3x2 + 2x + 5) – K = 2x3 – x2 + 3x – 2
or, (5x3 – 3x2 + 2x + 5) – (2x3 – x2 + 3x – 2) = K
or, 5x3 – 3x2 + 2x + 5 – 2x3 + x2 – 3x + 2 = K
∴ K = 3x3 – 2x2 – x + 7
∴ The subtracted polynomial is, 3x3 – 2x2 – x + 7

Required polynomial
= (subtracted from the polynomial) – (to get the polynomial)

3. Find the product of (x2 – 3x + 2) and sum of the polynomials (x3 + 2x2 – 3x + 2) and
2x3 – 3x2 + x + 2.

Solution :
Sum of x3 + 2x2 – 3x + 2 and 2x3 – 3x2 + x + 2 is,
= x3 + 2x2 – 3x + 2 + 2x3 – 3x2 + x + 2
= 3x3 – x2 – 2x + 4

Then, Product of (x2 – 3x + 2) and 3x3 – x2 – 2x + 4 is
= x2(3x3 – x2 – 2x + 4) – 3x(3x3 – x2 – 2x + 4) + 2(3x3 – x2 – 2x + 4)
= 3x5 – x4 – 2x3 + 4x2 – 9x4 + 3x3 + 6x2 – 12x + 6x3 – 2x2 – 4x + 8
= 3x5 – 10x4 + 7x3 + 8x2 – 16x + 8

PRIME Opt. Maths Book - IX 39

4. What should be added with 5x3 – 3x2 + 2x + 5 to get 2x3 – x2 + 3x – 2?

Solution :
The required polynomial can be obtained as,
Required polynomial = (to get the polynomial) – (added with the polynomial)
= (2x3 – x2 + 3x – 2) – (5x3 – 3x2 + 2x + 5)

= 2x3 – x2 + 3x – 2 – 5x3 + 3x2 – 2x – 5

= –3x3 + 2x2 + x – 7

5. Divide x3 – y3 + z3 + 3xyz by x – y + z
Solution :

x – y + z x2 – y3 + z3 + 3xyz x2 + xy + y2 – xz + yz + z2
x2 – x2y + x2z

–+ –
x2 y – x2z – y3 + z3 + 3xyz
x2 y – xy2 + xyz
–+–
xy2 – x2z – y3 + z3 + 2xyz
xy2 – y3 + y2z
–+–
– x2 z + z3 – y2z + 2xyz
– x2 z + xyz – xz2
+ –+
xyz + xz2 – y2 z + z3
xyz – y2 z + yz2
–+ –
xz2 – yz2 + z3
xz2 – yz2 + z3
–+ –

×
\ Quotient = Q(x) = x2 + y2 + z2 + xy + yz – zx
Remainder = R = 0

6. Divide : 4x3 + 2x2 – 5x – 3 by 2x – 1 using synthetic division method.

Solution :

p(x) = x3 + 2x2 – 5x – 3

Divisor = 2x – 1 = 2(x – 1 ) 1 4 2 –5 –3
2 2 22
1
Taking x = 2 – 3
2
Applying synthetic division method.

2 4 4 –3 – 9
2

Quotient = Q(x) = 2x2 + 2x – 3 Common 2 2 – 3 – 9
2 2 2
9
Remainder = R = – 2

40 PRIME Opt. Maths Book - IX

Exercise 1.5

1. Add the following polynomials:

i) 2x3 + 3x2 – 5x + 7 and x3 – 7x2 – 3x + 2

ii) x4 – 7x3 – 2x + 2 and 3x3 + 4x4 – 7 + 2x2 – 3x

iii) 32 x3 + 7 x2 – 5x + 2 and 3 + x – 4 x2 + 1 x3
3 3 2

iv) 34 x2 – 1 x3 + 3 x4 – 7 + 2x and 4 x4 – 3 x3 – 7 x2 + 3x + 2
2 7 7 2 3

v) 1 x3 + 3 x2 – 5x + 2 = and 2 2 – 3x + 2 x2 + 5 x3
35 53

2. Subtract the followings.

i) 3x3 + 5x – 2x2 + 7 and 2x3 – x – 4x2 – 2

ii) x4 – 7x3 – 3 + 2x2 – 5x and – 7 – 3x2 + 2x – 9x3 + x4

iii) 53 x3 – 3 x2 + 2x + x4 – 1 and 2 x3 – 7 x2 – 2x4 + 2 – 3x
2 3 3 2 3

iv) 57 x4 – 2 + 3 x3 – 7x + 1 x2 and 5 x2 + 5 – 3 x4 + 11 x3 – 3x.
7 2 2 2 7 5 2

v) 1 + 3 x2 – 5x + 1 x3 and 4 x3 + 9 x2 – 4 + 2x
52 3 325

3. Find the following polynomials.
i) If p(x) = 3x3 + 2x2 – 5x + 2 and q(x) = 2x3 – 3x2 – 2x + 3, find p(x) + q(x).
ii) If p(x) + q(x) = 5x3 – 3x2 + 2x – 5 and p(x) = 3x3 – x2 + 3x – 2, find q(x).
iii) What must be subtracted from the polynomial 4x4 – 3x3 + 2x2 – 5x + 1 to get

3x4 – x3 – 5x2 + 2x + 3 ?

iv) What must be added to the polynomial x3 – 3x2 – 2x + 3 to get x4 + 3x3 – x2 + 3x – 2?

v) What must be subtracted from the sum of x4 + 2x3 – 3x2 + 2x – 5 and 2x4 + x3 – x2 – 3x + 2

to get x4 – x3 + 2x2 – x – 2?

4. i) If f(x) = 3x3 + 2x2 – x + 2 and g(x) = x4 – x3 + x2 + x – 5, find f(x) + g(x). Also write
down the types of polynomial of the result according to degree and number of
terms.

ii) A polynomial x3 + x2 – 3x – 1 is subtracted from p( ) results x4 – x3 + 2x2 – x + 2,
find the polynomial p(x).

iii) If p(x) = (x2 + 2x – 3) and q( ) = (2x2 – x + 2). Find the value of p(x) × q(x). Also
write down the type of polynomial according to degree.

iv) Multiply the polynomial x2 + 2x – 3 and 3x3 – 2x2 + 3x – 5.
v) If f(x) = (x2 + 2), g(x) = 2x2 – x + 3, what must be subtracted from the product of

f(x) and g(x) to get x4 – x3 + 2x2 – x – 2?

5. Find p(x). q(x) from the followings.
i) P(x) = 2x3 – 3x2 + x – 2 and q(x) = x2 – 2x + 1.
ii) P(x) = x3 – 5x2 + 2x – 4 and q(x) = 2x2 – 3x + 2.
iii) P(x) = f(x) + g(x), q(x) = g(x), f(x) = x2 + 3x + 2 and g(x) = 2x3 – 3x2 + 2x – 5

PRIME Opt. Maths Book - IX 41

iv) P(x) = 2f(x) – g(x), q(x) = f(x), f(x) = x2 + 2x + 3 and g(x) = 2x3 – 3x2 + 5x – 3
v) 2p(x) = 4x3 – 6x2 + 4x – 2 & 3q(x) = 3x2 – 6x + 3
6. Divide the following polynomials:
i) (x3 – 3x2 – 2x + 2) by (x – 2)
ii) x4 + 3x3 – 4x2 + 2x – 1 by x + 3
iii) p(x) = 2x3 + 5x2 – 28x – 15, q(x) = 2x + 1; p(x) ÷ q(x)
iv) (x3 + 3x2y + 3xy2 + y3) by (x + y)
v) p(x) = x3 – y3 by q(x) = x – y; p(x) ÷ q(x)
7. Divide the followings using synthetic division method.
i) x3 – 2x2 – 3x – 2 by x – 1
ii) 2x3 + 3x2 – 5x + 1 by x + 2
iii) x4 + 5x3 – 2x2 – 3x – 1 by x – 2
iv) 4x3 – 3x + 2 by 2x – 1
v) 2x3 + 5x2 – 3x – 7 by 2x + 3
8. Prime more creative question.
i) Divide x3 + y3 + z3 – 3xyz by x + y + z.
ii) Divide x4 + x3 – 2x2 – 16x – 5 by x2 – 3x + 2
iii) Multiply the sum of x3 + 2x2 + 3x – 2 and x2 – x + 3 by x2 – 3x + 2
iv) Divide p(x) = x4 – 2x2 – 7 by P(x) = x + 3, by using synthetic division method.
v) Divide p(x) by 2x – 1 where p(x) = 2x3 – 3x2 + 5x + 3
9. Project work
Prepare the types of polynomials and operation on polynomials in a chart paper and

present it in your classroom.

42 PRIME Opt. Maths Book - IX

1. i) 3x3 – 4x2 – 8x + 9 Answer
iii) 2x3 + x2 – 4x + 5
v) 2 3x3 + 5x2 – 8x + 3 2 ii) 5x4 – 4x3 + 2x2 – 5x – 5
iv) x4 – 2x3 – x2 + 5x – 5

2. i) x3 + 2x2 + 6x + 9 ii) 2x3 + 5x2 – 7x + 4
iii) 3x4 + x3 + 2x2 + 5x – 1 iv) 2x4 – 4x3 – 2x2 – 4x – 1
v) – 3x3 – 3 2x2 – 7x + 5

3. i) 5x3 – x2 – 7x + 5 ii) 2x3 – 2x2 – x – 3
iii) x4 – 2x3 + 7x2 – 7x – 2 iv) x4 + 2x3 + 2x2 + 5x – 5
v) 2x4 + 4x3 – 6x2 – 1

4. i) x4 + 2x3 + 3x2 – 3; Biquadratic, multinomial
ii) x4 – 2x3 + x2 + 2x + 3
iii) 2x4 + 3x3 – 6x2 + 7x – 6 ; Biquadratic.
iv) 3x5 + 4x4 – 10x3 + 7x2 – 19x + 15
v) x4 + 5x2 – x + 8

5. i) 2x5 – 7x4 + 9x3 – 7x2 + 5x – 2
ii) 2x5 – 13x4 + 21x3 – 24x2 + 16x – 8
iii) 4x6 – 10x5 + 20x4 – 35x3 + 29x2 – 31x + 15.
iv) –2x5 + x4 + 3x3 + 22x2 + 15x + 27.
v) 2x5 – 7x4 + 10x3 – 8x2 + 4x – 1

6. i) quotient = x2 – x – 4, remainder = – 6
ii) quotient = x3 – 4x + 14, remainder = –43
iii) quotient = x2 + 2x – 15, remainder = 0
iv) quotient = x2 + 2xy + y2, remainder = 0
v) quotient = x2 + xy + y2, remainder = 0

7. i) quotient = x2 – x – 4, remainder = –6
ii) quotient = 2x2 – x – 3, remainder = 7
iii) quotient = x3 + 7x2 + 12x + 21 remainder = 41
iv) quotient = 2x2 + x – 1, remainder = 1
v) quotient = x2 + x – 3, remainder = 2

8. i) quotient = x2 + y2 + z2 – xy – yz – zx, remainder = 0
ii) quotient = x2 + 4x + 8, remainder = –21
iii) x5 – 5x3 + x2 + x + 2
iv) quotient = x3 – 3x2 + 7x – 21, remainder = 56
v) quotient = x2 – x + 2, remainder = 5

PRIME Opt. Maths Book - IX 43

1.6 Sequence and Series
Sequence

The arrangement of the numberals in the specific order can be written in different ways
like,
1, 5, 9, 13, ..., ..., ..., ..., ..., ..., ..., ..., ..., ...
1, 4, 9, 16, 25, ..., ..., ..., ..., ..., ..., ..., ..., ..., ...
2, 6, 18, 54, ..., ..., ..., ..., ..., ..., ..., ..., ..., ... etc.

Such type of arrangement of the numbers is called sequence of numbers.

The array of the numberal elements in a specific order under
a certain rule is called sequence.

Examples :
• 2, 5, 8, 11, 14, ..., ..., ..., ..., ..., ..., ..., ..., ..., ...
Here, Ascending order with the rule of each terms increased by 3.

• 3, 6, 12, 24, 48, ..., ..., ..., ..., ..., ..., ..., ..., ..., ...
Here, Ascending order with the rule of multiples the each terms by 2.

• 1, 2, 4, 7, 11, 16, ..., ..., ..., ..., ..., ..., ..., ..., ..., ...
Here, Ascending order with the rule of more than each terms by 1, 2, 3, 4, successively
..., ..., ..., ..., ..., ..., ..., ..., ..., ...

• 12, 22, 3,2, 42, ..., ..., ..., ..., ..., ..., ..., ..., ..., ...
Here, Ascending order with the rule of square of natural numbers from 1.

• 100, 90, 80, 70, 60, ..., ..., ..., ..., ..., ..., ..., ..., ..., ...
Here, Descending order with the rule of decreased the each term by 10.

Finite and infinite sequence:

The sequence having finite number of terms which can be
counted is called finite sequence.

Example : PRIME Opt. Maths Book - IX
2, 5, 9, 14, 20, 27
Here are 6 terms in the sequence,
Hence, it is finite sequence.

44

The sequence having infinite number of terms (which cannot
be counted) is called infinite sequence.

Example :
3, 5, 8, 12, 17, ..., ..., ..., ..., ..., ..., ..., ..., ..., ...
Here, last term is not defined and number of terms can not be counted. Hence, it is

called infinite sequence.

Series:

The arrangement of the numbers (terms) in definite order which are joint by either (+)
or (–) sign is called series.

The sum of the terms of a sequence is called a series.
The every terms of the sequence can be written by using sum or
difference (+ve or –ve sign) which is called series.

Above examples of sequence can be expressed into series by using +ve or –ve sign in
each terms as,

2 + 5+ 8 + 11 + ... + ... + ... + ... + ... + ... + ... + ...
3 + 6 + 12 + 24 + 48 + ... + ... + ... + ... + ... + ... + ... + ...
12 + 22 + 32 + 42 + 52 + ... + ... + ... + ... + ... + ... + ... + ...

Finite sequence can be expressed into finite series only as,
1 + 5 + 9 + 13 + 17
Infinite sequence can be expressed into infinite series only as,
1 + 2 + 4 + 7 + 11 + 16 + ... + ... + ... + ... + ... + ... + ... + ...

Progression

The sequence and series of the numbers written in definite
order either in ascending or descending order is called

progression.

The progression having rule of addition with a constant number is called arithmetic
progression (AP).
Examples :
2 + 7 + 12 + 17 + 22 + ... + ... + ... + ... + ... + ... + ... + ...
1, 5, 9, 13, 17, 21, ..., ..., ..., ..., ..., ..., ..., ..., ..., ...
80, 74, 68, 62, 56, 50, ..., ..., ..., ..., ..., ..., ..., ..., ..., ...

PRIME Opt. Maths Book - IX 45

The progression having rule of multiplication with a constant number is called geometric
progression (GP).
Examples :
2, 6, 18, 54, ..., ..., ..., ..., ..., ..., ..., ..., ..., ...
3, 6, 12, 24, 48, ..., ..., ..., ..., ..., ..., ..., ..., ..., ...
128, 64, 32, 16, 8, ..., ..., ..., ..., ..., ..., ..., ..., ..., ...

nth term of the sequence and series:
Let us discuss different examples to find the nth term of the sequence.
2, 4, 6, 8, ..., ..., ..., ..., ..., ..., ..., ..., ..., ... 2n

1, 3, 5, 7, .........................., 2n – 1
12 + 22 + 32 + ... + ... + ... + ... + ... + ... + ... + ... + n2
0, 2, 6, 12, ..., ..., ..., ..., ..., ..., ..., ..., ..., ... n2 – n.

Here, taking above examples,

Terms 1st example 2nd example 3rd example 4th example
2×1–1 12 12 – 1
1st 2 × 1 2×2–1 22 22 – 2
2×3–1 33 33 – 3
2nd 2 × 2 2×4–1 44 44 – 4
.............. ..............
3rd 2 × 3 2n – 1 .............. n2 – n
n2
4th 2 × 4

So on ..............

nth term 2n

From the table the last term (nth term) can be generalized by comparing the number of

terms which is,

nth term (tn) = 2n (1st example) tn = 2n – 1 (2nd example)
tn = n2 (3rd example) tn = n2 – n (4th example)

• The sequence whose terms are written in order
according to the rule of linear equation (y = ax + b) is

called linear sequence.
• The sequence whose terms are written in order

according to the rule of quadratic equation (y = ax2 + bx

+ c) is called quadratic sequence.

46 PRIME Opt. Maths Book - IX