Approved by the Government of Nepal, Ministry of Eduction, Curriculum Development
Centre (CDC), Sanothimi, Nepal as a reference material for school
Perfect
Optional
Mathematics
CLASS 9
N B Khatakho
Vidyarthi Pustak Bhandar
Publisher and Distributor
Bhotahity, Kathmandu
Perfect
Optional
Mathematics
CLASS 9
Publisher Vidyarthi Pustak Bhandar
Bhotahity / Kamalpokhari, Kathmandu
Copyright Tel : 01-4245834, 4227246, 4423333
Edition P.O. Box 12990, Email: [email protected]
ISBN Publisher
Computer
Printed First, 2012
Third Revised, 2020
978–99377–3–905–4
Dynamic Computer Services
Bhaktapur, 01-6614740
CONTENTS
1. Algebra .............................................................................................................. 1
1.1 Ordered Pairs .................................................................................................................. 2
1.2 Relations ......................................................................................................................... 5
1.3 Functions ......................................................................................................................... 13
1.4 Types of Functions .......................................................................................................... 20
1.5 Polynomials ..................................................................................................................... 25
1.6 Sequence and Series ........................................................................................................ 31
2. Limits ................................................................................................................ 42
2.1 Concept of Limit by Sequence of Numbers .................................................................... 43
2.2 Concept of Limit by Sequence of Diagrams ................................................................... 46
2.3 Concept of Limit by Sum of Infinite Series .................................................................... 51
2.4 Limit of a Function ......................................................................................................... 57
2.7 Techniques of Evaluating Limit ...................................................................................... 65
3. Matrices ............................................................................................................. 70
3.1 Representation and Types of Matrices ............................................................................ 71
3.2 Operation of Matrices ..................................................................................................... 77
4. Co-ordinate Geometry ..................................................................................... 91
4.1 Section Formula .............................................................................................................. 92
4.2 Locus of a Point .............................................................................................................. 104
4.3 Equation of Straight Lines .............................................................................................. 110
4.4 Slope Intercept Form of Equation of Straight Line ........................................................ 115
4.5 Double Intercept Form of Equation of Straight Line ..................................................... 119
4.6 Perpendicular Form of Equation of Straight Line .......................................................... 125
4.7 Reduction of the Linear Equation Ax + By + C = 0 ........................................................ 130
4.8 Special Forms of Lines .................................................................................................... 135
4.9 Length of Perpendicular ................................................................................................. 140
4.10 Area of Triangles and Quadrilaterals ............................................................................... 143
5 Trigonometry .................................................................................................... 149
5.1 Measurement of Angles .................................................................................................. 150
5.2 Length of Circular Arc ................................................................................................... 161
5.3 Trigonometric Ratios ...................................................................................................... 166
5.4 Conversion of Trigonometric Ratios ............................................................................... 176
5.5 Trigonometric Ratios of Any Angles .............................................................................. 182
5.6 Trigonometric Ratios of Standard Angles ....................................................................... 192
5.7 Compound Angles .......................................................................................................... 201
6. Vector ................................................................................................................. 212
6.1 Vectors and its Components ........................................................................................... 213
6.2 Magnitude and Direction of a Vector ............................................................................. 218
6.3 Types of Vectors .............................................................................................................. 223
6.4 Vector Operations ........................................................................................................... 228
7. TRANSFORMATION ..................................................................................... 239
7.1 Reflection ........................................................................................................................ 242
7.2 Rotation .......................................................................................................................... 253
7.3 Translation ...................................................................................................................... 263
7.4 Enlargement .................................................................................................................... 268
8. Statistics ............................................................................................................ 276
8.1 Partition Values ............................................................................................................... 277
8.2 Dispersion ....................................................................................................................... 288
Answer ............................................................................................................................ 298
Model Question .............................................................................................................. 315
PREFACE
Perfect Optional Mathematics is recently revised series of school-level textbooks based
on the current pedagogical principles on mathematics prepared for maintaining continuity
in learning mathematics considering the maturity, needs and interest of young learners
between 4 years and 15 or 16 years of age. This series provides ample opportunities for the
students to enhance their learning by encouraging/reinforcing creativity of the students.
For the purpose of making learning mathematics suitable as per the capacity of the students
of different age groups, appropriate exercises and activities have been included presenting
the concepts inductively through examples. Examples have been particularly given an
appropriate places for presenting mathematics as a subject matter which can be learnt and
applied in day-to-day life.The presentation and exercises essential for developing students’
creative thinking and problem solving skills have been given an important place.
The materials have been presented with special attention to meaningful learning of
mathematics based on different researches on teaching-training experiences, following
the curriculum developed by CDC. Therefore, meaningfulness and practicality of learning
have been taken into consideration rather than searching additional and complex subject
matters. Therefore, the current necessities of meaningful learning and pragmatic use of
mathematics for school-level children are the main features of this series which can be
integrated to their higher studies as well.
Perfect Optional Mathematics Class 9 prepared for the childere of grade 1, taking
their capacity and interest into consideration, is a revised link book of this series, which
embodies all the features discussed above. This book is also believed to make additional
contribution to enhance the learning of school-level mathematics meaningfully in the
present text.
Mr. Shanta Gautam, the Managing Director of Vidyarthi Pustak Bhandar deserves my
thanks for taking the responsibility of publishing this book. Likewise, I extend my thanks
go to Mr. Binod Kasula, Mr. Ramesh Konda for their contrubution in course of currently
revising and updating this book.
Any constructive suggestions for further improvement of this book will always be highly
acknowledged.
Author
1 ALGEBRA Algebra
Curriculum Contents Included Periods
17
a. Relation and Function
8
- Ordered pair 10
- Cartesian product
- Introduction of relation and types of relation, ways of representation of relation,
domain and range of relation
- Introduction of function and symbol, domain, co-domain, image, pre-image and
range of function
- Ways of representing function, testing function (vertical line test)
- Types of function - onto, into, one-to-one, many-to-one.
b. Polynomials
- Introduction (Algebraic term, Algebraic expression and polynomial)
- Degree of polynomial, standard polynomial and equal polynomial
- Operations of polynomials. (Addition, subtraction and multiplication)
c. Sequence and Series
- Introduction of sequence and general terms
- Introduction of series and sigma (S) notation
Objectives
On completion of this unit, students will be able to:
find the cartesian product of the given two sets.
define relation and function and state the method of representation of relation and function.
distinguish between relation and function.
state the types of relation and function and also represent it in the graph and by other methods.
state and find domain and range of relation.
state and find domain, co-domain, image, pre-image and range of the function.
define onto, into, one-to-one and many-to-one function.
introduce and distinguish polynomial function (linear, quadratic, cubic and biquadratic).
perform the operations of addition, subtraction and multiplication of polynomials.
define sequence and series.
find out the general terms of sequence and series according the given pattern and also write the expression
in notation.
use sigma notation (S) for the given series.
Specification Grid
Contents Knowledge Understanding Application Higher ability Total Total
Algebra 1 mark each 5 marks each Questions Marks
2 marks each 4 marks each
2 1 8 21
32
~1
1.1 Ordered Pairs
Introduction
A pair of objects in which the order plays an important role is called an ordered pair. We denote the ordered
pair made up of x and y by (x, y). We say x is the first element (1st component = antecendent) and y is the
second element (2nd component = consequent) of the ordered pair (x, y).
For example, (Kathmandu, Nepal) is an ordered pair in which the first component is the capital of the second
component. If the components of ordered pair are interchanged, it will change the sense of the pair also.
Hence, (Kathmandu, Nepal) and (Nepal, Kathmandu) are different ordered pairs. Similarly, the co-ordinates
( 2, – 3) is an order pair in which 2 is the x co-ordinate and – 3 is the y co-ordinate. If the components of the
ordered pair are interchanged, we get (– 3, 2) which is different from (2, – 3).
Equality of Ordered Pairs
Two ordered pairs (a, b) and (c, d) are said to be equal if a = c and b = d. For example, (2, 3) and 36, 3 are
equal because 2 = 6 and 3 = 3. But (2, 5) ≠ (5, 2) because 2 ≠ 5 and 5 ≠ 2.
3
Example 1:
If (x + 2, y) = (5, 4x – 8), find x and y.
We know that if (x + 2, y) = (5, 4x – 8) then
x + 2 = 5 and y = 4x – 8.
Now, x + 2 = 5 \x=3
and y = 4x – 8
or, y = 4 × 3 8 \ y = 4
Hence, x = 3 and y = 4.
Cartesian Product
If A and B are two non empty sets, the cartesian product of A and B is the set of ordered pairs (x, y) where
x∈A and y∈B. It is denoted by A×B and read as A cross B.
For example, if A = {1, 2, 3} and B = {a, b)
Tree diagram of A×B Ordered pairs
a (1, a)
1 (1, b)
b
a (2, a)
2 (2, b)
b
a (3, a)
3 (3, b)
b
2 ~ Perfect Optional Mathematics Class 9
A×B = {(1, a), (1, b), (2, a), (2, b), (3, a), (3, b)} Algebra
Cartesian product A×B can be constructed by the following table.
Set B
a b
(1, b)
Set A 1 (1, a) (2, b)
2 (2, a) (3, b)
3 (3, a)
Similarly, B×A can be constructed by the following tree diagram.
Tree diagram of B×A Ordered pairs
1 (a, 1)
a2 (a, 2)
3 (a, 3)
1 (b, 1)
b2 (b, 2)
3 (b, 3)
\ B×A = {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3)}.
Example 2:
If A = {2, 3, 4} and B = {5, 6, 7}, find the cartesian products A×B and B×A and represent them
graphically and with the help of a mapping diagram. Set B
Here, A = {2, 3, 4} and B = {5, 6, 7}
567
A×B = {(2, 5), (2, 6), (2, 7), (3, 5), (3, 6), (3, 7), Set A 2 (2, 5) (2, 6) (2, 7)
3 (3, 5) (3, 6) (3, 7)
(4, 5), (4, 6), (4, 7)}.
4 (4, 5) (4, 6) (4, 7)
Representing A×B in graph, Mapping diagram of A×B.
Y A B
10
9 25
8
7 36
6
5 47
4
3
2
1
X' O X 1 2 3 4 5 6 7 8 9 10
Y'
Ordered Pairs ~ 3
Similarly, Set A
23
B×A = {(5, 2), (5, 3), (5, 4), (6, 2), (6, 3), (6, 4), (5, 2) (5, 3) 4
(6, 2) (6, 3) (5, 4)
(7, 2), (7, 3), (7, 4)}. Set B 5 (7, 2) (7, 3) (6, 4)
6 (7, 4)
7
Representing B×A in graph. Mapping diagram of B×A.
Y
B A
10
9 52
8
7 63
6
5 74
4
3 EXERCISE : 1.1
2
1
X' O X 1 2 3 4 5 6 7 8 9 10
Y'
1. Identify which of the following ordered pairs are equal:
a. (a, b) and (b, a) b. (2, 3) and 2, 9
c. (2, 3) and (3, 2) 3
d. (23, 12) and (32, 12)
2. Find the values of x and y in each of following cases:
a. (x, y) = (2, x + 3) b. (x + 2, 8) = (4, y – 2)
c. (x + y, 4) = (8, x – y) d. (x – 3, 6) = (4, x + y)
3. a. If A = {a, b} and B = {p, q, r}, find A×B and B×A using a tree diagram.
b. If P = {– 2, 0, 3} and Q = {2, 4, 5}, find the cartesian product P×Q and Q×P and represent them
in a mapping diagram and graph.
c. If A×B = {(– 1, 3), (– 1, 5), (2, 3), (2, 5), find A and B.
d.
e. If A = {1, 2, 3} and B = {2, 3}, then show that A×B ≠ B×A.
If A = {2, 3}, B = {4, 5} and C = {3, 6}, find
4. a.
i. A × (B∪C) ii. B × (A∩C)
iii. (A∪B) × C iv. (A × B)∪(B × C)
The domain {a, b, c} and the co-domain {1, 2, 3} are given. Give the possible set of ordered pairs.
b. The domain {m, y} and the co-domain {l, o, v, e} are given. Give the possible set of ordered pairs.
5. a. If A = {x : x∈N, 7 < x ≤ 9}, then find A×A. Also show in arrow diagram.
b. If A = {x : x ≤ 4, x ∈ N}, then find A×A. Also show in arrow diagram.
6. a. If A = {x : x ≤ 3, x ∈ N} and B = {x : x ≤ 3, x ∈ W} then find A×B and B×A.
b. If A = {x : x ≤ 4, x ∈ N} and B = {x : 3 ≤ x < 5, x ∈ W} then find (A∩B)×A.
4 ~ Perfect Optional Mathematics Class 9
1.2 Relations Algebra
Introduction is capital of Nepal
Kathmandu China
The following statements express the relationship between
two elements. Beijing
a. Kathmandu is the capital of Nepal.
b. Beijing is the capital of China. Delhi India
c. Delhi is the capital of India.
In each of the above statements, two elements are connected by the phrase "is the capital of ". Here, "the
capital of" is a relation phrase. R B
Similarly, "is the square of " is a relation phrase. A is square of
a. 1 is square of 1. 1
1 2
4
b. 4 is square of 2. 93
c. 9 is square of 3. 5
If the first element and the second element in each of the above statements are the elements of set A and set
B then each of these statements defines a relation from the set A into the set B.
Let A and B be two non-empty sets then any subset of A×B is said to be a relation from the set A to the
set B. i.e. R = {(x, y); x∈A, y∈B}∈A×B.
Example 1:
Let A = {– 2, 4, 8} and B = {2, 4, 5), find the set of ordered pairs in the relation "is less than" from
A to B.
Here, A = {– 2, 4, 8} and B = {2, 4, 5}, then
A × B = {(– 2, 2), (– 2, 4), (– 2, 5), (4, 2), (4, 4), (4, 5), (8, 2), (8, 4), (8, 5)} and relation R = {(– 2, 2),
(– 2, 4), (– 2, 5), (4, 5)}.
Example 2:
Let A = {2, 3, 5} and B = {1, 2, – 5), find the cartesian porduct A×B and the relation defined by
R = {(x, y): x + y ≥ 5} from A to B.
Here, A = {2, 3, 5} and B = {1, 2, – 5}, then the cartesian product A×B is given by
A × B = {(2, 1), (2, 2), (2, – 5), (3, 1), (3, 2), (3, – 5), (5, 1), (5, 2), (5, – 5)}
The relation R = {(x, y): x + y ≥ 5} has the following ordered pairs
= {(3, 2), (5, 1), (5, 2)}
Ways of Representing a Relation
There are several ways of representing a relation. Some of them are as follows:
a. Set of ordered pairs (Roster form) b. Mapping diagram
Relations ~ 5
c. Table d. Graph
e. Description (Set builder form)
Study the following examples for various methods of representing a relation.
Example 3:
If A = {2, 5, 6} and B = {2, 8}, express the relation 'R' "is greater than" from the set A to the set B by
a. Set of ordered pairs b. Mapping diagram
c. Table d. Graph
e. Description
Here A = {2, 5, 6} and B = {2, 8}, then A×B = {(2, 2), (2, 8), (5, 2), (5, 8), (6, 2), (6, 8)} and the
relation R i.e. "is greater than" is represented by
a. Set of ordered pairs : R = {(5, 2), (6, 2)}
b. Mapping diagram B c. Table 5 6 d. Graph
A 2 2 Y
2
2 A 5
8 B
5 4
6 3
2
1
X' O 1 2 3 4 5 6 7 8 X
Y'
e. Description: R = {(x, y); x∈A, y∈B and x > y}
Domain and Range of a Relation
Let R be a relation from A to B. Then the set of all the first elements of the set of ordered pairs of R is called
the Domain of the Relation and the set of all the second elements of the set of ordered pairs is called the
Range of the relation.
Example 4:
If R = {(2, 4), (4, 5), (4, 6), (5, 6), (5, 7), (5, 8)} is a relation, find domain and range of R.
Here the given relation is
R = {(2, 4), (4, 5), (4, 6), (5, 6), (5, 7), (5, 8)}.
Domain of R = Set of the first elements of all the ordered pairs
= {2, 4, 5}
Range of R = Set of the second elements of all the ordered pairs
= {4, 5, 6, 7, 8}.
Types of Relation
The relation from set A to set B may be of the following types :
1. onto relation 2. into relation
3. one-to-one relation 4. one-to-many relation
5. many-to-one relation 6. inverse relation
6 ~ Perfect Optional Mathematics Class 9
1 Onto Relation
A relation R:A→B is said to be an onto relation if range of R = co-domain of R. Algebra
For example : let P = {1, 2, 3} and Q = {a, b, c} then a relation R1 is defined by R1= {(1, a), (2, b), (3, c)}.
Here, 1 is related to a, P R1 Q
2 is related to b and
3 is related with c. 1a
Now, Range of R1 = {a, b, c} and Co-domain of R1 = {a, b, c} 2 b
\ Co-domain of R1 = Range of R1. 3c
Hence, R1 is an onto relation.
2 Into Relation
A relation R:A→B is said to be an into relation if range of R is proper subset of co-domain of R.
For example: let M = {a, b, c, d} and N = {p, q, r} then a relation R2 is defined by {(a, p), (b, q), (c, p), (d, p)}.
Now, a is associated with p, M R2 N
b is associated with q, ap
bq
c is associated with p and c
dr
d is associated with p.
So, range of R2 = {p, q} and co-domain of R2 = {p, q, r}.
Now, range of R2 ⊂ co-domain of R2.
Hence, R2 is an into relation.
Example 5:
If A = {2, 3, 4, 5) and B = {3, 4, 5, 10, 11} are two sets, identify each of the following relations as
either onto or into relation.
a. R1= {(2, 3), (2, 10), (3, 5), (3, 11), (4, 4)}
b. R2 = {(2, 3), (4, 3), (4, 5), (5, 11)}
c. R3 = {(2, 3), (3, 4), (4, 10), (5, 11), (2, 5)}
d. R4 = {(2, 3), (2, 4), (5, 5)}
a. Here, the given relation R1 = {(2, 3), (2, 10), (3, 5), (3, 11), (4, 4)}.
A R1 B
Representing this relation in a mapping diagram, we have the
figure in the right side. The range of R1 = {3, 4, 5, 10, 11} and co- 2 3
domain of R1 = {3, 4, 5, 10, 11}. 3 4
As range of R1 = co-domain of R1, the relation R1 is an onto 4 5
relation. 5 10
11
b. Here, the given relation R2 = {(2, 3), (4, 3), (4, 5), (5, 11)}. A R2 B
Representing this relation in a mapping diagram, we have the
figure in the right side. The range of R2 = {3, 5, 11} and co-domain 2 3
of R2 = {3, 4, 5, 10, 11}. 3 4
As range of R2 ⊂ co-domain of R2, the relation R2 is an into 4 5
relation. 5 10
11
Relations ~ 7
c. Here, the given relation R3 = {(2, 3), (3, 4), (4, 10), (5, 11), (2, 5)}. A R3 B
Representing this relation in a mapping diagram, we have the
figure in the right side. The range of R3 = {3, 4, 5, 10, 11} and co- 2 3
domain of R3 = {3, 4, 5, 10, 11}. 3 4
As range of R3 = co-domain of R3, the relation R3 is an onto 4 5
relation. 5 10
11
d. Here, the given relation R4 = {(2, 3), (2, 4), (5, 5)}. A R4 B
Representing this relation in a mapping diagram, we have the
figure in the right side. The range of R4 = {3, 4, 5} and co-domain of 2 3
R4 = {3, 4, 5, 10, 11}. 3 4
As range of R4 ⊂ co-domain of R4, the relation R4 is an into 4 5
5 10
relation. 11
3. One-to-one Relation
A relation from set A to set B is said to be a one-to-one relation if different elements of set A are associated
or related to different elements of set B.
If A = {a, b, c} and B = {x, y, z} then a relation R1 is denoted by A R1 B
R1 = {(a, x), (b, y), (c, z)}.
Here, (a, x) ⇒ a is related to x, a∈A and x∈B. a x
(b, y) ⇒ b is related to y, b∈A and y∈B. b y
(c, z) ⇒ c is related to z, c∈A and z∈B. c z
\ R1 is called one-to-one relation, because different elements of set A are associated with unique elements
of set B.
There are two types of one-to-one relation
1. one-to-one onto relation 2. one-to-one into relation
Here, the relation R:A → B be a relation from set A to set B and is said to be a one-to-one onto relation, if
range of R = co-domain of R i.e. R(A) = B. i.e. if every element of set B is the image of at least one element
of set A.
For example, let A = {1, 2, 3} and B = {3, 4, 5}, a relation R1:A→B A R1 B
is defined by R1 = {(1, 3), (2, 4), (3, 5)}.
1 3
Here, Range of R1 = {3, 4, 5} and Co-domain of R1 = {3, 4, 5} 2 4
3 5
\ Co-domain of R1 = Range of R1.
Hence, R1 is a one-to-one onto relation.
Similarly, the relation R:P→Q be a relation from set P to set Q and is said to be a one-to-one into relation,
if the range of R is the proper subset of co-domain of R i.e. there is at least an element of set Q which is not
the image of any element of set P.
8 ~ Perfect Optional Mathematics Class 9
Again, let P = {4, 5, 6} and Q = {7, 8, 9, 10} and a relation R2:P→Q P R2 Q
is defined by R2 = {(4, 7), (5, 8), (6, 9)}.
Algebra
Here, Co-domain of R2 = {7, 8, 9, 10} and Range of R2 = {7, 8, 9}. 4 7
So, Range of R2 ⊂ Co-domain of R2. 5 8
∴ R2 is a one-to-one into relation. 6 9
10
4. One-to-many Relation
A relation R:A→B is said to be a one-to-many relation if at least an A R1 B
element of set A is associated with more than one element of set B.
For example, let A = {1, 2, 3} and B = {2, 3, 4, 5} then relation R1 is 1 2
defined by R1 = {(1, 2), (1, 3), (2, 4), (3, 5)} is known as one-to-many 2 3
relation. 3 4
5
Here, element 1 of set A is related to two elements 2 and 3 of set B. So, R1 is called a one-to-many relation.
There are two types of one-to-many relation.
1. one- to-many onto relation
2. one-to-many into relation
A relation R2:A→B is said to be one-to-many onto relation in which A R2 B
the element 4 of set A is associated with the elements 2 and 3 of set B
and range of R2 = co-domain of R2. Hence, R2 is a one-to-many onto 4 2
relation. 5 3
4
Similarly, a relation R3:P→Q is said to be one-to-many into relation in P R3 Q
which the element 5 of set P is associated with the elements 7 and 8 of
set Q and range of R3 is proper subset of co-domain of R3. Hence, R3 is 5 7
a one-to-many into relation. 6 8
9
10
5. Many-to-one Relation
A relation R:A→B said to be many-to-one relation if more than one element of set A is associated with a
unique element of set B.
For example, let A = {1, 2, 3} and B = {4, 5} then relation R1:A→B is A R1 B
defined by R1= {(1, 4), (2, 4), (3, 5)}. Here, the elements 1 and 2 of set
A are associated with the element 4 of set B. So, R1 is called a many-to- 1 4
one relation. There are two types of many-to-one relation: 2 5
3
1. many-to-one onto relation
2. many-to-one into relation
Relations ~ 9
A relation R2:A→B is said to be a many-to-one onto relation in which A R2 B
the elements a, b and d of set A are associated with the element q of set
B and range of set R2 = co-domain of R2. Hence, R2 is a many-to-one a q
onto relation. b r
c
d
Similarly, a relation R3:M→N is said to be a many-to-one into relation M R3 N
in which the elements p, q and s of set M are associated with the element
a of set N and range of R3 is proper subset of co-domain of R3. Hence, p a
R3 is a many-to-one into relation. q c
r d
s
Example 6:
If A = {a, b, c, d) and B = {p, q, r, s} are two sets, identify the types of the following relations.
a. G1 = {(a, p), (b, q), (c, r), (d, s)} b. G2 = {(a, q), (b, r), (c, s)}
c. G3 = {(a, p), (b, q), (b, r), (b, s)} d. G4 = {(a, p), (c, p), (b, s)}
a. Here, the given relation G1 = {(a, p), (b, q), (c, r), (d, s)}.
A G1 B
Representing this relation in a mapping diagram, we have the figure a p
b q
as shown. The elements of set A and set B correspond one-to-one. c r
The range of G1 = {p, q, r, s} and co-domain of G1 = {p, q, r, s}. s
As range of G1 = co-domain of G1, the relation G1 is a one-to-one
d
onto relation.
b. Here, the given relation is G2 = {(a, q), (b, r), (c, s)}. A G2 B
Representing this relation in a mapping diagram, we have the figure
as shown. The elements of set A and set B correspond one-to-one. a p
The range of G2 = {q, r, s} and co-domain of G2 = {p, q, r, s}. b q
As range of G2 ⊂ co-domain of G2, the relation G2 is a one-to-one c r
d s
into relation.
c. Here, the given relation G3 = {(a, p), (b, q), (b, r), (b, s)}. A G3 B
Representing this relation in a mapping diagram, we have the figure
as shown. The elements of set A and set B correspond one-to-many. a p
The range of G3 = {p, q, r, s} and co-domain of G3 = {p, q, r, s}. As b q
range of G3 = co-domain of G3, the relation G3 is a one-to-many c r
onto relation. d s
d. Here, the given relation is G4 = {(a, p), (c, p), (b, s)}. A G4 B
Representing this relation in a mapping diagram, we have the figure
as shown. The elements of set A and set B correspond many-to-one. a p
The range of G4 = {p, s} and co-domain of G4 = {p, q, r, s}. b q
As range of G4 ⊂ co-domain of G4, the relation G4 is a many-to- c r
d s
one into relation.
10 ~ Perfect Optional Mathematics Class 9
6. Inverse Relation Algebra
If R be a relation from the set A to set B, the inverse of a relation R is the relation obtained by interchanging
the components of each of the ordered pairs of the relation R. It is denoted by R– 1 and read as “R inverse.”
Symbolically, if R = {(x, y): x∈A, y∈B}, then
R– 1 = {(y, x): y∈B, x∈A}
Example 7:
Find the inverse of the following relations.
a. f= 31, 3 , 2, 1 , 4, 1 b. g = {(a, b), (p, q), (r, s), (u, v)}.
2 4
a. The given relation is f = 13, 3 , 2, 1 , 4, 1
2 4
The inverse relation is obtained by interchanging the components of all ordered pairs.
f – 1 = 3, 1 , 21, 2 , 41, 4 .
3
b. The given relation is g = {(a, b), (p, q), (r, s), (u, v)}
Hence its inverse relation is g– 1 = {(b, a), (q, p), (s, r), (v, u)}.
EXERCISE : 1.2
1. a. If A = {2, 3} and B = {2, 7, 9}, find the set of ordered pairs of the relation from A to B.
b. i. is less than ii. is not equal to
If A = {2, 4, 5} and B = {1, 2, 4}. Find the relation from A to B determined by the relation
c. Rn = {(x, y): x∈A and y∈B}.
i. x = y ii. x > y
d. iii. x is a factor of y iv. x is exactly divisible by y
If A = {1, 2} and B = {2, 4, 5}, find the relation from A to B in the form of set of ordered pairs
e. determined by R = {(x, y): x∈A, y∈B and y = 2x}. Also express the above relation by
i. Mapping diagram ii. Table
2. a.
iii. Graph iv. Description
If A = {1, 4, 8} and B = {1, 2, – 2, 5}, find the following set of ordered pairs
i. R = {(x, y): x = y2} ii. R = {(x, y): x > y}
Also express the above relation by a mapping diagram.
If A = {2, 5, 6}, find the relation in A × A determined by
i. x > y ii. x ≠ y
Find the domain, range and inverse relation in each of the above relations.
Find the domain and range of the following relations.
i. R1 = {(1, 5), (2, 5), (3, 8), (5, 10)}
ii. R2 = {(3, 5), (2, 3), (4, 1)}
Relations ~ 11
b. Graph the relation defined by R = {(x, y): y = 2x + 1}. If the replacement set for x or domain is
{0, 2, 3, 4, 5}, give the range.
c. Find the range of the following relations and represent them in a mapping diagram.
i. R = {(x, y): y = 3x} and x∈{0, 2, 3, 4}
ii. f = {(x, y): y = x2 + 2} and x∈{– 1, 0, 1, 2, 3}.
iii. g = {(x, 3x – 2): x∈{0, 1, 2, 3, 4}}.
3. If P = {1, 2, 3, 4, 5} and Q = {3, 5, 7, 8, 9, 10} are two sets, identify each of the following relations as
either onto or into relation.
a. R1 = {(1, 3), (2, 5), (3, 7), (4, 8), (5, 9), (5, 10)}
b. R2 = {(1, 5), (2, 10), (3, 5), (4, 8), (5, 7)}
c. R3 = {(2, 3), (2, 5), (3, 7), (3, 8), (4, 9), (5, 10)}
d. R4 = {(1, 5), (1, 8), (2, 5), (2,7), (3, 9), (5, 3)}
4. If A = {p, q, r, s} and B = {d, e, f, g} are two sets, identify the types of the following relations.
a. H1 = {(p, d), (q, f), (r, g), (s, e)} b. H2 = {(p, e), (q, g), (r, f)}
c. H3 = {(p, d), (p, e), (q, f), (r, g)} d. G4 = {(p, d), (q, d), (r, f), (s, g)}
e. H5 = {(p, f), (q, e), (q, g)} f. H6 = {(p, e), (q, d), (r, d), (s, d)}
5. a. If A = {2, 4, 5} and B = {1, 3, 4, 6} and R = {(x, y): x∈A, y∈B and y = x – 1}, write all the
elements of R. Also find R– 1.
b. Given A = {2, 3, 4, 5} and a relation R is defined by R = {(x, y)∈A×A : x > y} then
i. List the elements of R ii. Domain and range of R
iii. List the elements of R– 1 iv. Domain and range of R– 1.
c. Determine the domain and range of R defined by R = {(x, x2): x is an odd number less than 8}.
Also find R– 1 and mapping diagram of R and R– 1.
6. a. The adjoining arrow diagram represents a relation. A B
i. Express the relation in roster form.
ii. Express it in set builder form. – 1 – 2
iii. Find its inverse relation. 1 2
iv. Represent the inverse relation graphically. 2 4
3 6
4 8
b. Define the relation represented in each of the following arrow diagrams.
i. A B ii. A B
1 12 2
8 2 4
27 34 5
64 45
12 ~ Perfect Optional Mathematics Class 9
1.3 Functions Algebra
Introduction
Let us consider the sets A = {1, 2, 3, 5} and B = {a, b, c, d}. Then the cartesian product A×B is
A × B = {(1, a), (1, b), (1, c), (1, d), (2, a), (2, b), (2, c), (2, d), (3, a), (3, b), (3, c), (3, d), (5, a), (5, b), (5, c),
(5, d)}
Now, let us define the relations R1, R2, R3 and R4 as follows:
R1 = {(1, a), (2, b), (3, c), (5, d)}
R2 = {(1, a), (2, b), (3, b), (5, c)}
R3 = {(1, a), (2, b), (3, c)}
R4 = {(1, a), (2, b), (2, c), (3, d), (5, d)}
Representing these relations in mapping diagrams
A R1 B A R2 B
1a 1a
2b 2b
3c 3c
5d 5d
A R3 B A R4 B
1a 1a
2b 2b
3c 3c
5d 5d
In above relations R1 and R2, every element of the set A corresponds with a unique element of set B. In
relation R3, element 5 of the set A does not associate with any element of the set B. In relation R4, element 2
of the set A is associated with two elements b and c of the set B. Here, the relations R1 and R2 are functions
and R3 and R4 are not functions.
Definition
Let A and B are two non-empty sets. Then a relation from the set A to the set B is said to be a
function if every element of A corresponds with a unique element of B. The function from set A to
set B is symbolically written as f : A→B.
By definition, for every x∈A, there corresponds a unique element y = f(x)∈B.
The element y = f(x)∈B which has the correspondence with x, an element of A is said to be the image of
x under f and x is said to be the pre-image of y = f(x). The function is generally denoted by the letters f, g,
h, F, G, φ etc.
Example 1:
If A = {0, 1, 2, 3} and B = {w, x, y, z}, draw mapping diagrams for each of the following relations
and state whether each of the relations is a function or not.
Functions ~ 13
a. f = {(0, x), (1, y), (2, z), (3, w)}
b. g = {(1, w), (2, x), (3, y), (0, w)}
c. h = {(2, w), (0, y), (1, z), (3, x), (0, z)} A f B
a Mapping diagram of f = {(0, x), (1, y), (2, z), (3, w)} 0 w
1 x
2 y
3 z
As every element of A associates with a unique element of B, f is a function.
b. Mapping diagram of g = {(1, w), (2, x), (3, y), (0, w)} A g B
1w
2x
3y
0z
As every element of set A associates with a unique element of set B, g is a function.
c. Mapping diagram of h = {(2, w), (0, y), (1, z), (3, x), (0, z)} A h B
0w
1x
2y
3z
As the element 0 of the first set A associates with two elements y and z of the second set B, h is not a
function.
Example 2:
If A = {1, 2, 3} and B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, test whether the following relations from A to B
are functions or not.
a. y = f(x) = x2 b. y = f(x) = 2x + 5 c. y = f(x) = 3x
a. We have y = f(x) = x2, x∈{1, 2, 3}. A f(x) = x2 B
So, f(1) = 12 = 1 1 13
2 5 10
f(2) = 22 = 4 2 4 67
98
f(3) = 32 = 9 3
Representing the relation in mapping diagram,
From the above mapping diagram, each element of set A associates with a unique element of set B, hence
y = f(x) = x2 is a function.
b. We have, y = f(x) = 2x + 5, x∈{1, 2, 3} A f(x) = 2x + 5 B
So, f(1) = 2 × 1 + 5 = 7
f(2) = 2 × 2 + 5 = 9 1 72
f(3) = 3 × 2 + 5 = 11
2 9 37
10 4 9
3 15
Representing the relation in a mapping diagram.
From the above mapping diagram, the element 3 of the set A has no image in B, hence y = f(x) = 2x + 5 is
not a function.
14 ~ Perfect Optional Mathematics Class 9
c. We have, y = f(x) = 3x, x∈{1, 2, 3} A f(x) = 3x B
So, f(1) = 3 × 1 = 3
f(2) = 3 × 2 = 6 1 12 Algebra
f(3) = 3 × 3 = 9
2 3 4 10
6 58
3 97
Image and Pre-image of a Function
If f: A→B i.e. f is a function from set A to set B, and an element a of set A is maped with an element b of
set B then b is said to be Image of a and a is said to be the Pre-image of b. In fig.(ii) given below, the image
of 2 is 5 and pre-image of 7 is 3. Hence, any element of set A must have only one image in set B, but any
image in set B may have more than one pre-images in set A.
AB Af B
f(x) 2 5
Pre-image of b a f(a) = b Image of a 3 7
4 9
10
fig.(i) fig.(ii)
Domain, Co-domain and Range of a Function A B
If f: A→B i.e. f is a function from set A to set B, then the set x f(x)
A is said to be the Domain of f and the set B is said to be Range
the Co-domain of f. The set of elements of set B which are
the images of the elements of A, is known as the Range of f. y = f(x)
Domain Co-domain
Example 3:
Find the domain, co-domain and range of the following functions and images and pre-images of
the elements.
a. A f B b. f = {(a, 3), (b, 7), (q, 8) (r, 3)}
a 1
b 2
3
c 5
7
d
a. Domain, A = {a, b, c, d}, Co-domain, B = (1, 2, 3, 5, 7} and
Range of the function = {1, 2, 3}
Image of a = 1, Image of b and d = 2 and Image of c = 3.
Pre-image of 1 = a, Pre-image of 2 = b and d and
Pre-image of 3 = c.
b. Domain = {a, b, q, r}, Co-domain = {3, 7, 8} and
Range = {3, 7, 8}
Image of a and r = 3, Image of b = 7 and Image of q = 8.
Pre-image of 3 = a and r, Pre-image of 7 = b and
Pre-image of 8 = q.
Functions ~ 15
Example 4:
What element in the domain has the image 7 under the function f(x) = 2x – 3 ?
Here, given function is f(x) = 5
2x – 3
5
The given image is 7,
f(x) = 2x – 3 = 7
5
or, 2x – 3 = 35
or, 2x = 38
or, x = 19
Hence, the required element is 19. This is pre-image of 7.
Example 5:
What will be the pre-image of 4 in the function f(x) = x2?
Here, given function is f(x) = x2.
The given image is 4, i.e. f(x) = 4
or, x2 = 4
or, x2 – 4 = 0
or, (x + 2)(x – 2) = 16
Either x + 2 = 0 \ x = – 2
or, x – 2 = 0 \ x = 2
Hence, the pre-image of 4 is 2 or – 2.
Ways of Representing a Function
There are several ways of representing a function. The following are five different ways of representing
a function.
a. Set of ordered pairs b. Mapping diagram (Arrow diagram)
c. Table d. Graph (Lattice diagram)
e. Equation
Study the following examples for different ways of representing a function.
Example 6:
Let A = {1, 2, 3} and B = {1, 3, 4, 8, 9} and f:A→B such that f(1) = 1, f(2) = 4, f(3) = 9. Represent
the function f by the following ways.
a. Set of ordered pairs b. Mapping diagram
c. Table d. Graph
e. Equation
Here we have, A = {1, 2, 3}, B = {1, 3, 4, 8, 9} and f(1) = 1, f(2) = 4, f(3) = 9 then
a. An ordered pair of x and y is (x, y) so f = {(1, 1), (2, 4), (3, 9)}.
16 ~ Perfect Optional Mathematics Class 9
b. Mapping diagram: c. Representing f in a table form:
A1 2 3
A f B Algebra
11 B1 4 9
3
24
3 8
9
d. The points with ordered pairs (1, 1), (2, 4) and e. When x = 1, y = f(1) = 1 = 12
x = 2, y = f(2) = 4 = 22
(3, 9) are plotted. Y x = 3, y = f(x) = 9 = 32
\ The function f can be expressed as y = f(x) = x2.
The graph is shown as 12
10
8
6
4
2
X' O 2 4 6 8 10 X
Y'
Vertical Line Test
A function can be represented by graph. But all graphs may not be a function. So to examine whether the
given relation is a function or not, we use a vertical line test to its graph. A relation is a function if and
only if no vertical line can intersect its graph at more than one point. If any vertical line cuts the graph of a
relation at two or more points, there are two or more ordered pairs having the same x-coordinate. Hence, the
relation is not a function because in a function no two ordered pairs have the same x co-ordinate.
Example 6:
Using a vertical line test, examine whether the following graphs are functions or not.
a. Y b. Y
X' O X
X' O X
Y'
Y'
a. A vertical line drawn cuts the graph at two distinct points A Y
and B, hence it is not a function. A
X' O X
Y' B
Y
b. A vertical line drawn cuts the graph at only one point A and none
of the vertical lines cuts at the two or more points of the graph, so A
it is a function.
OX
X' Y'
Functions ~ 17
EXERCISE : 1.3
1. State whether each of the following mapping diagrams defines a function from set A to set B.
a. A B b. A B
1 aa 1
2 bb 2
3 cc 3
4 dd 4
c. A B d. A B
4 a6 2
6 5
8 b7 8
10 c4 9
– 2
2. Draw mapping diagrams for each of the following relations and tell whether each relation is a function
or not.
a. f = {(0, 3), (2, 7), (8, 10), (3, 6)} b. g = {(2, 8), (3, 7), (4, 8), (5, 10)}
c. h = {(4, 2), (6, 7), (4, 5), (3, 8)} d. h = {(x, y), (a, b), (c, d), (e, f), (u, v)}
3. If A = {2, 4, 6} and B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, test whether the following relations from set A to
set B are functions or not.
a. y = f(x) = x + 4 b. y = f(x) = 2x – 3
2
4x – 3
c. y = f(x) = x2 – 10 d. y = f(x) = 5
4. If A = {x : – 1 ≤ x ≤ 2, x∈Z} and B = {x : – 5 ≤ x ≤ 5, x∈Z} test whether each of the following relations
from set A to set B are functions or not.
a. y = f(x) = x – 4 b. y = f(x) = 2x + 4
c. y = f(x) = 2x2 – 4 d. y = f(x) = x2 – 5
5. Find the domain, co-domain and range of the following functions.
a. A B b. B f A
2 a4 p
4
8 b5 q
9 – 2
c 6 r
d 8
c. f = {(a, b), (b, q), (c, d), (d, r), (e, q)} d. h = {(1, a), (2, b), (3, c), (5, d), (7, a)}
6. Find the domain, co-domain, range of the following functions and images and pre-images of the
elements.
a. A B b. A B
a 42 a
b 54 b
c 68 c
d 77
c. f = {(a, 2), (b, 4), (c, 5), (d, 7)} d. g = {(2, a), (3, b), (4, c), (– 2, b), (– 1, c)}
18 ~ Perfect Optional Mathematics Class 9
7. a. If the function g is defined as y = g(x) = 2x2 – 3, x∈{0, – 2, 3, 5}, find the range of g.
b. If f(x) = x2 – 2x, find the values of f(3) and f(– 3). Algebra
c. What is the image of 5 under the function f(x) = (2x2 + 45)2?
x3 –
d. If a function f(x) = 3x – 4, find the value of f(3) – f(1).
e. What element in the domain has the image 1 under the function f(x) = 2xx+–13.
3x – 5
f. If the range of an element under the function f(x) = x+2 is 14, find the element.
8. Using vertical line test, state which of the following graph represents function.
a. Y b. Y
O
X' O X
X'
X
c. Y' Y'
Y d. Y
X' O X X' O X
Y' Y'
9. Study the following functions represented by mapping diagrams and state the types of functions.
a. A f B b. B g A
1a 4p
2b 5
3c – 2 q
4d
6 r
8
c. A h B d. A h B
a1 a1
b2 2
c3 c 3
d4
d4
10. a. Let A = {2, 3, 4} and B = {5, 7, 8, 9, 10} and f: A→B is such that f(2) = 5, f(3) = 7 and f(4) = 9.
Represent the function f by the following ways:
i. Set of ordered pairs ii. Mapping diagram
iii. Table iv. Graph
v. Equation
b. Let P = {1, 2, 3, 5} and Q = {4, 5, 6, 7, 8} and f: P→Q is such that f(x) = x + 3. Represent the
function f by:
i. Mapping diagram ii. Set of ordered pairs
iii. Graph
Functions ~ 19
1.4 Types of Functions
Types of Functions
There are mainly four types of functions. b. Into function
a. Onto function d. Many-to-one function.
c. One-to-one function
Onto Function
Let f be a function from set A to set B, then function f is said to be a onto function if every element of set B
is the image of at least one element of set A. i.e. Range = Co-domain.
Af B Af B
a 1 1a
2
b 2 3 c
c 3 4d
Into Function
Let g be a function from set A to set B, then function g is said to be an into function if there exists at least
one element of set B which is not associated with any element of set A. i.e. Range ⊂ Co-domain.
Ag B AgB
1a 1a
2b 2b
3c 3c
5d
d
One-to-one Function
Any function f from set A to set B is said to be a one-to-one, if different elements of set A have different
images in set B, i.e. no two or more elements of set A have the same image in set B.
Af B Af B
a 1 a3
b 2 b5
c 3 c8
d 8
9
Many-to-one Function
Any function h from set A to set B is said to be a many-to-one function, if two or more elements of set A
have the same image in set B.
20 ~ Perfect Optional Mathematics Class 9
Ah B AhB
1a 1a Algebra
2b
2 b 3c
3 4d
Note: From above definitions and mapping diagrams it is clear that the Onto function may be one-to-one or many-to-
one and Into function may be one-to-one or many-to-one.
Example 1:
Study the following functions represented by mapping diagrams.
i. A f B ii. A f B
a 2 4 8
9
b3 5 10
11
c4 6 3
iii. A f B iv. A f B
a1 49
b2 8
c3 63
d4 9
a. Examine which of the above diagrams represent onto and which represent into functions.
b. Examine which of the above diagrams represents one-to-one and which represent many-to-
one functions.
c. Examine which of the above diagram represents many-one onto and which represent many-
one into.
d. Examine which of the above diagram represents one-one onto and which represent one-one
into.
a.
From the given diagrams (iii) and (iv) are onto functions because all the elements of the second
set are busy and the diagrams (i) and (ii) represent into functions because Range ⊂ Co-domain
in both cases.
b. As every element of set A is associated with distinct element of set B in (ii) and (iii), they are one-to-
one functions and as two or more elements of A have same image in (i) and (iv), they are many-to-one
functions.
c. From the reasons given in (a) and (b) above we can conclude that figure (iv) is many-to-one onto and
fig (i) is many-to-one into.
d. From the reasons given in (a) and (b) above, we can conclude that figure (iii) is one-to-one onto and
figure (ii) is one-to-one into.
Types of Functions ~ 21
Example 2:
If f is a function such that f(x – 3) = f(x) + f(– 3) for all real number x, show that f(0) = 0 and
f(3) = – f(– 3). Again, when x = 3
We have the function,
f(x – 3) = f(x) + f(– 3) f(3 – 3) = f(3) + f(– 3)
When x = 0, f(0 – 3) = f(0) + f(– 3) or, f(0) = f(3) + f(– 3)
or, f(– 3) = f(0) + f(– 3) or, 0 = f(3) + f(– 3)
or, f(– 3) – f(– 3) = f(0) \ f(3) = – f(– 3).
\ f(0) = 0
Example 3:
If the images of 1 and 5 under the function f(x) = 3x + b are – a and 5a respectively, find the values
of a and b. And image of 5 is 5a ............ (2)
i.e. f(5) = 5a
Here, image of 1 under f(x) = 3x + b is – a or, 3 × 5 + b = 5a
or, 15 + b = 5a
i.e. f(1) = – a
or, 5a – b = 15
or, 3 × 1 + b = – a
or, 3 + b = – a
or, a + b = – 3 .............. (1)
Solving equations (1) and (2), we have
a + b = – 3
5a – b = 15
6a = 12
\ a =2
From equation (1), 2 + b = – 3 i.e. b = – 5
\ a = 2, b = – 5
Example 4:
A function is defined on the set of integers as follows :
x + 5 for – 2 ≤ x < 2
f(x) = 3x – 4 for 2 ≤ x ≤ 6
a. Find the value of f(0), f(2), f(1) + f(3)
b. Find the domain of the function.
Given function is f(x) = x + 5 for – 2 ≤ x < 2
3x – 4 for 2 ≤ x ≤ 6
a. Here, x = 0, lies in the interval – 2 ≤ x < 2,
f(x) = x + 5
or, f(0) = 0 + 5 = 5
And x = 2 lies in hte interval 2 ≤ x ≤ 6
22 ~ Perfect Optional Mathematics Class 9
f(x) = 3x – 4 Algebra
or, f(2) = 3 × 2 – 4 = 6 – 4 = 2
Similarly, f(1) + f(3) = (1 + 5) + (3 × 3 – 4)
= 6 + 9 – 4 = 11
b. As the function is defined on the set of integers, its domain is
A = {x: – 2 ≤ x ≤ 6, x∈N}
= {– 2, – 1, 0, 1, 2, 3, 4, 5, 6}
Example 5:
If f(x) = 1 cos x x, find the value of f πc and f πc .
+ sin 4 2
Here, f(x) = cos x x
1+ sin
cos πc 1 1
4
f πc = πc = 2 = 2 = 1
4 4 + 2+1
1+ sin 1 + 1 2 1
2
2
πc cos πc 0
2 2 +
and f = πc = = 0
1+ sin 2 1 1
EXERCISE : 1.4
1. a. If f : R→R is defined by y = f(x) = 2x + 3, find f(0), f(– 2), f 1 ,f – 7 , f(l + h) and f(1 + h) – f(1)
x 3 h
b. If the function f is defined as y = f(x) = 2x2 – 5, x∈R, find the value of
i. f(2) – f(3) ii. f(2) × f(4)
c. iii. f(0) – 21 f(2) iv. 4f(4) – 2f(3)
Let f : R→R be the function given by f(x) = x2 + x – 1, then find the value of
i. f(4) – f(3) ii. f(2) × f(3)
iii. Is f(4) – f(3) = f(1) ? iv. Is f(2) × f(3) = f(6) ?
2. a. If f (x) = x + 4 and g(x) = x2 – 2, x∈R, find
i. f(2) + g(3) ii. f(1) × g(0)
iii. 2f(1) + 3f 1 iv. gf((62))
2
b. If f(x) = x3 + 2 and g(x) = x2 – x + 3, x∈R, find
i. f(– 2) – g(2) ii. f(2) × g(– 3)
iii. 4f 1 + 3g (– 3) iv. f(5)
2 g(3)
Types of Functions ~ 23
3. a. If f is a function such that f(x + 2) = f(2) + f(x) for all real number x, show that f(0) = 0 and
f(– 2) = – f(2).
b. If f is a function such that f(x + 3) = f(x) + 3, prove that f(0) = f(3) – 3 and f(3) – f(– 3) = 6.
c. If f is a function such that f(x + 3) = f(x) × f(3) for all real number x, show that f(0) = 1 and f(– 3) = 1 .
f(3)
4. a. If 8 is the pre-image of a and 4 is the image of b under the function 2x + 3, find the values of a
and b.
b. If the image of a and 3 under the function f(x) = x + 3b are 5 and 6 respectively, find the values
of a and b
c. If the images of 2 and 5 under the function f(x) = 2x + b are a + 2 and 2a + 3 respectively, find
the values of a and b.
d. If the function f(x) = ax – b where a and b are real numbers, f(2) = 4 and f(– 1) = – 5, find a and b.
5. A function is defined on the set of integers as follows :
f(x) = x2 – 4 for x ≤ 3
5x – 1 for x > 3
a. Find images of – 3, 1 and 5. b. Find the pre-image of 19.
c. Find the values of f(– 2), f(4) – f(2) and f(3) + f(0).
6. A function is defined on the set of integers as follows
f(x) = 2x – 3 for – 3 ≤ x < 1
4x + 1 for 1 ≤ x ≤ 3
a. Find the domain of the function. b. Find the range of the function.
c. Find the values of f(– 1), f(2) and f(0) + f(3).
7. If f(q) = 1 cos θ θ, find the values of f pc and f pc .
– sin 2 4
8. If f(y) = cos y – sin yy, find the values of f pc and f pc .
cos y + sin 2 4
9. From the given graph of the function, find the function values : f(– 1), f(0), f(1) and f(3).
a. Y b. Y c. Y
7 X
6 7 7
6 6
5 5 5
4 4
4 3 3
3 2 2
1 1
2
X' O -3 -2 1 2 3 4 5 6 7 X' O -3 -2 1 2 3 4 5 6 7
1 X
-2 -2
X' O X -3 -2 1 2 3 4 5 6 7
-2
Y' Y' Y'
24 ~ Perfect Optional Mathematics Class 9
1.5 Polynomials Algebra
Introduction
A monomial is a constant, a variable, product of constant and variables or product of variables. In a
monomial, only whole numbers can be used as an exponent of a variable. So 2, – 8, 5, x, – 4x, x , xy, 3x2y2
2
etc. are monomials and 7x–1, x– 2, 2x1/3y, xy, x + 3 etc. are not monomials.
In a monomial 7x3y, the constant factor 7 is called the Numerical Coefficient of x3y, while x3 and y are the
Literal Coefficients of 7y and 7x3 respectively. The sum of exponents of variables in a monomial is called
the Degree of the Monomial. Thus 4 is the degree of the above monomial. Hence, the degree of a constant
monomial is zero.
For example,
the degree of 2x3y2z is 6.
the degree of 3x is 1.
the degree of 2y2z is 3.
A polynomial is monomial or the sum of monomials. So 3, 2 + x, x2 + y + 3, 2x2y + xy + y + 3 are the
examples of polynomials. Each monomial separated by plus or minus sign in a polynomial is called a term.
A polynomial with two terms is called a binomial. A polynomial of three terms is called a trinomial. 2x2 + y,
x + 3 are binomials and x + y – 7, x3 + x2y + x are trinomials. A polynomial on one variable is generally
denoted by p(x), f(x), q(x) etc. A polynomial on two variables x and y is generally denoted by p(x, y), f(x, y)
etc. The Degree of a Polynomial is the degree of the term of highest degree.
Hence, 4x4 + 3x3 + 7x is a polynomial of degree 4.
2x2y + 4x2y2 + 3y3 is a polynomial of degree 4.
x + y is a polynomial of degree 1.
If all the coefficients used in a polynomial p(x) belong to the set A, then we say p(x) is a polynomial over
a set A.
For example, p(x) = 2x2 – 7x – 3 is a polynomial over the integers.
f(x) = x2 – 12 x + 7 is a polynomial over the rationals.
Example 1:
Which of the following are polynomials?
x 1 3, 3 x + 2, x2 + 4, x y, x+7
+
In the first, fourth and fifth algebraic expressions, the exponents of the variables are not whole numbers.
Hence, they are not polynomials. The second and the third algebraic expressions are the polynomials
of degree 1 and 2 respectively.
Polynomials ~ 25
Example 2:
Find the degree of the following polynomials.
a. p(xy) = 7x2y + 3x3y2 – 8xy b. f(x) = 3x2 – 9x5 + 4x6
a. Given polynomial: p(xy) = 7x2y + 3x3y2 – 8xy
The term with highest degree is 3x3y2 and its degree = 3 + 2 = 5.
\ The degree of p(xy) is 5.
b. Given polynomial: f(x) = 3x2 – 9x5 + 4x6
The term with highest degree is 4x6 and its degree = 6.
\ The degree of polynomial f(x) = 6.
Standard form of a Polynomial
A polynomial in a variable x is called a standard form, if it is in simplified form and arranged in
ascending or descending order. For example,
a. f(x) = x5 + 4x4 – x3 – 3x2 + 5x + 3 (In descending order)
b. f(x) = 3 + 5x – 3x2 – x3 + 4x4 + x5 (In ascending order)
Equal Polynomials
Two polynomials p(x) and f(x) are said to be equal if their degrees and number of terms are same and
their corresponding like terms are equal.
For example, f(x) = 2x3 + 3x2 – 42 x + 5 and g(x) = 2x3 + 3x2 – 2x + 15 are equal.
3
Example 3:
Write the polynomial f(x) = 2x4 + 3(x2)3 + 6x2 + 23 x3 + 20 – 7x – x6 in standard form.
Given polynomial is
f(x) = 2x4 + 3(x2)3 + 6x2 + 23 x3 + 20 – 7x – x6
= 2x4 + 3x6 – x6 + 6x2 + 32 x3 + 20 – 7x
= 2x6 + 2x4 + 23 x3 + 6x2 – 7x + 20 which is the standard form.
Example 4:
If two polynomials f(x) = 3x3 – 4x2 + 6 and g(x) = 3x3 – 4x2 + m are equal, find the value of m.
The given two polynomials are
f(x) = 3x3 – 4x2 + 6 and
g(x) = 3x3 – 4x2 + m
As these two polynomials are equal, their corresponding terms are equal, hence m = 6.
26 ~ Perfect Optional Mathematics Class 9
Sum and Difference of Polynomials Algebra
The sum of two polynomials is the polynomial obtained by adding or subtracting their like terms according
to their signs. For examples, if f(x) = 4x3 – 7x + 3 and g(x) = 2x3 + x2 + 2x – 4 then the sum of f(x) and g(x)
is denoted by f(x) + g(x) and is given by
f(x) + g(x)
= (4x3 – 7x + 3) + (2x3 + x2 + 2x – 4)
= 4x3 + 2x3 + x2 – 7x + 2x + 3 – 4
= 6x3 + x2 – 5x – 1
Properties of Sum of Polynomials
The following properties are satisfied by the sum of the polynomials.
a. Closure Property
If f(x) and g(x) be two polynomials over a set A, then the sum
f(x) + g(x) is also a polynomial over the set A.
For example, f(x) = 2x3 – 7x + 3, g(x) = 2x2 – 4x – 4 then
f(x) + g(x) = 2x3 – 7x + 3 + 2x2 – 4x – 4
= 2x3 + 2x2 – 11x – 1
b. Commutative Property
If f(x) and g(x) are two polynomials then f(x) + g(x) = g(x) + f(x).
For example, f(x) = 4x3 – 7x + 3 and g(x) = 2x3 – 7x2 + x we have
f(x) + g(x) = 4x3 – 7x + 3 + 2x3 – 7x2 + x
= 6x3 – 7x2 – 6x + 3
and g(x) + f(x) = 2x3 – 7x2 + x + 4x3 – 7x + 3
= 6x3 – 7x2 – 6x + 3
c. Associative Property
If f(x), g(x) and h(x) be three polynomials,
then [f(x) + g(x)] + h(x) = f(x) + [g(x) + h(x)]
For example, f(x) = 2x2 – 7x + 3
g(x) = x2 + 4
h(x) = – 3x – 7
Now, [f(x) + g(x)] + h(x) and f(x) + [g(x) + h(x)]
= [2x2 – 7x + 3 + x2 + 4] – 3x – 7 = 2x2 – 7x + 3 + [x2 + 4 – 3x – 7]
= 3x2 – 7x + 7 – 3x – 7 = 2x2 – 7x + 3 + [x2 – 3x – 3]
= 3x2 – 10x = 2x2 – 7x + 3 + x2 – 3x – 3
= 3x2 – 10x.
Hence, [f(x) + g(x)] + h(x) = f(x) + [g(x) + h(x)]
Polynomials ~ 27
d. Existence of Additive Identity
If P(x) be a polynomial and 0 be a zero polynomial then
p(x) + 0 = p(x) = 0 + p(x)
Here, 0 is called additive identity.
For example, if p(x) = x3 + 7x2 – 3x + 4 and 0 = 0.x3 + 0.x2 + 0.x + 0, then
p(x) + 0 = x3 + 7x2 – 3x + 4 + 0.x3 + 0.x2 + 0.x + 0
= x3 + 7x2 – 3x + 4
= p(x)
Similarly, 0 + p(x) = p(x).
e. Existence of Additive Inverse
For any polynomial p(x) there exists a polynomial – p(x) such that
p(x) + {– p(x)} = 0 = – p(x) + p(x).
Here, – p(x) is called the additive inverse of p(x) and vice-versa.
For example, if f(x) = 2ax2 – 7x + b then – f(x) = – 2ax2 + 7x – b.
Now, f(x) + {– f(x)} = 2ax2 – 7x + b + {– 2ax2 + 7x – b}
= 2ax2 – 7x + b – 2ax2 + 7x – b
= 0.
Similarly, – f(x) + f(x) = 0.
Difference of two Polynomials
If f(x) and g(x) be two polynomials, then the difference of f(x) and g(x) denoted by f(x) – g(x) is defined as
f(x) – g(x) = f(x) + [– g(x)].
For example, if f(x) = 2mx2 – 3nx + 6 and g(x) = mx2 + 2nx – 4 then
f(x) – g(x) = (2mx2 – 3nx + 6) – (mx2 + 2nx – 4)
= 2mx2 – 3nx + 6 – mx2 – 2nx + 4
= mx2 – 5nx + 10
Example 5:
Find the sum of f(x) = 3x5 – 7x3 + 5x – 2 and g(x) = x5 + 5x4 – 3x2 – 4.
Here, f(x) = 3x5 – 7x3 + 5x – 2 and g(x) = x5 + 5x4 – 3x2 – 4.
Hence, f(x) + g(x) = 3x5 – 7x3 + 5x – 2 + x5 + 5x4 – 3x2 – 4
= 4x5 + 5x4 – 7x3 – 3x2 + 5x – 6.
Alternatively,
f(x) = 3x5 – 7x3 + 5x – 2
= 3x5 + 0.x4 – 7x3 + 0.x2 + 5x – 2
g(x) = x5 + 5x4 – 3x2 – 4
= x5 + 5x4 + 0.x3 – 3x2 + 0.x – 4
28 ~ Perfect Optional Mathematics Class 9
Adding two polynomials Algebra
f(x) = 3x5 + 0.x4 – 7x3 + 0.x2 + 5x – 2
g(x) = x5 + 5x4 + 0.x3 – 3x2 + 0.x – 4
f(x) + g(x) = 4x5 + 5x4 – 7x3 – 3x2 + 5x – 6
Example 6:
Find the additive inverse of the polynomial p(x) = x5 – 7x3 + 3x2 + 6.
Here, p(x) = x5 – 7x3 + 3x2 + 6
Let h(x) be the additive inverse of p(x), then p(x) + h(x) = 0.
or, h(x) = 0 – p(x)
or, h(x) = 0 – {x5 – 7x3 + 3x2 + 6}
= – x5 + 7x3 – 3x2 – 6.
Multiplication of Polynomials
For the multiplication of two polynomials, we multiply each term of one polynomial by every term of
the other and simplify the result.
Example 7:
Multiply the polynomial f(x) = x2 + 7x + 3 by polynomial g(x) = 2x – 3.
Here, f(x).g(x) = (x2 + 7x + 3) (2x – 3)
= x2(2x – 3) + 7x(2x – 3) + 3(2x – 3)
= 2x3 – 3x2 + 14x2 – 21x + 6x – 9
= 2x3 + 11x2 – 15x – 9
EXERCISE : 1.5
1. Find the degree and co-efficient of each of the following monomials:
a. 5x2 b. 3x2y c. x2y3
2. Find the numerical co-efficients and the literal co-efficients of x in the following monomials:
a. 4xy2 b. 3 xy3 c. 4xyz
3. Which of the following expressions are polynomials ?
a. x4 + 3x3 – 7x + 4 b. x2 + x – 7 2
c. x3 + 3x3 + 5x – 7
d. 4x2 + 3 x – 4 e. 4x3 + 3x2 – x1 + 4 f. 2x2 + 32 x – 7
4. Write the degree of the following polynomials:
a. 3x + 4x2 – 7 b. x4 + x2y3 – 7xy
c. x20 (x30 + y10) d. 4 + 3x – x3 + 5x4
5. Write each of the following polynomials in standard form:
a. x5 + 3x3 – 8 b. 3 + 2x + x4 – x3
c. 2y4 – 3y5 – 2y + y3 + 7 d. x2 – x(1 + x) + 3
e. 2x4(1 – x) + 6x5 + 7x(x + 3)
Polynomials ~ 29
6. Examine whether the following polynomials are equal:
a. f(x) = x3 + 3x2 – 7 and g(x) = 3x2 + x3 – 7
b. f(x) = 12 x3 + 7x + 4 and g(x) = 0.5x3 + 7x –4 23
c. f(x) = x4 – 41 x2 + 23 and g(x) = – 0.25x2 – x4 –
d. f(x) = ax2 + cx + d and g(x) = c + ax2 + bx
e. f(x) = 2 x2 + 7x – 4 and f(x) = 2x2 + 7 x – 4
7. a. If the polynomials f(x) = 2x3 – 7x2 + mx + 3 and g(x) = ax3 – 7x2 + 4x + 3 are equal, find the
values of m and a.
b. If the polynomials p(x) = x3 + (m + 2)x2 + x + 3 and q(x) = x3 + 4x2 + (n – 1)x + 3 are equal, find
the values of m and n.
8. Find f(x) + g(x) and f(x) – g(x) in each of the following cases:
a. f(x) = 3x2 + 7x – 3, g(x) = 4x2 – x + 2
b. f(x) = x3 – 7x + 2, g(x) = 2x3 + 4x2 – 7
c. f(x) = 2x4 – 3x2 + 3x, g(x) = x3 + 2x – x2 + 2
d. f(x) = 3x5 – 3x2 + 4x3 – 3, g(x) = 2x5 – 4x2– x + 4
9. Find the additive inverse of the following polynomials:
a. x3 – 4x2 – 3x b. x4 – 7x2 + 3x – 8
c. 4x2 – 3x3 – 7x – 3 d. 4x4 – 3 x3 + 2x + 5
10. Find the product of f(x) and g(x) in each of the following:
a. f(x) = 2x2 + 3, g(x) = x – 1 b. f(x) = 2x – 1, g(x) = 3x2 – x + 4
c. f(x) = x3 – 2x + 1, g(x) = x – 2 d. f(x) = 2 x2 – 2x + 3, g(x) = x – 1
11. a. Divide p(x) = x3 + 3x2 – 5x + 2 by g(x) = x – 2 and find the quotient and remainder.
b. Divide p(x) x3 + y3 by q(x) = x + y and find the quotient and remainder.
12. a. If the difference of p(x) = 4a x4 + 21 x3 – 52 x2 + 7b x – 1 and q(x) = 51 x4 + 112 x3 – 154 x – c is
b.
r(x) = 15 x4 + 172 x3 – 52 x2 + 21 x + 3, then find the values of a, b and c.
If the sum of p(x) = ax2 – 3x + b and q(x) = 3x3 – 12x2 + cx – 13 is r(x) = 3x3 – 8x2 + 3x – 5, then
find the value of a, b and c.
13. a. If p(x) = 21 x3 – 43 x2 + 25 x – 16, q(x) = 14 x3 + 21 x2 – 32 x + 2 and r(x) = 41 x3 – 54 x2 + 1165 x – 65, then find
3
p(x) – r(x) – q(x).
b. If p(x) = 2x3 – x + 7, q(x) = x2 + 7x – 16 and r(x) = – x2 – x, then find p(x) + q(x) – r(x).
14. a. 3 9 2 9
b.
What should be added to q(x) = 32 x3 – 91 x + 5 to get p(x) = 21 x2 + 79 x – 6?
What should be subtracted from (x + y)3 to get (x – y)3?
30 ~ Perfect Optional Mathematics Class 9
1.6 Sequence and Series Algebra
Introduction
Let us consider the following set of numbers.
a. 13, 16, 19, 22, 25, ... ... ... ... ...
b. 23, 19, 15, 11, 7, ... ... ... ... ...
c. 2, 4, 8, 16, 32, ... ... ... ... ...
d. 81, 27, 9, 3, 1, ... ... ... ... ...
e. 2, 3, 5, 8, 12, ... ... ... ... ...
f. 2, 6, 12, 20, 30, ... ... ... ... ...
Now we observe the above set of numbers carefully. In every set of numbers in the above case, the numbers
are arranged according to certain rule. In the set of numbers in (a), each number (except the first) is greater
than its previous number by 3. So the further numbers in the arrangement are 28, 31 and so on.
Similarly, in the remaining set of numbers, they are arranged with certain rule. The numbers in (b) are less
than its previous number by 4. Each number in (c) are double than its previous number. Each number in (d)
are one-third of its previous number. The second, third, fourth, fifth numbers in (e) are greater than their
previous numbers by 1, 2, 3 and 4 respectively. So, the further numbers in the arrangement are 17, 23 and so
on. Hence, every set of numbers in above case form sequence. For sequence, 2, 6, 18, 54, ... ... ... ... ..., each
of the numbers are known as a term of the sequence. So 2, 6, 18, 54 are the first, second, third and fourth
terms respectively. The first, second, third ... ... ... ... ... term of a sequence will be denoted by t1, t2, t3 ... ...
... ... ... tn. The nth term tn of the given sequence is known as general term.
Definition
A set of numbers each of which is formed according to certain rule is known as a sequence. Each
number of a sequence is known as a term.
The general term of above sequences (a) 13, 16, 19, 22, 25, ... ... ... and (f) 2, 6, 12, 20, 30, ... ... ... can be
generated as follows:
For sequence (a), 13, 16, 19, 22, 25 ... ... ...
t1 = 13 = 13 + 0 = 13 + 3 × 0 = 13 + 3 × (1 – 1)
t2 = 16 = 13 + 3 = 13 + 3 × 1 = 13 + 3 × (2 – 1)
t3 = 19 = 13 + 6 = 13 + 3 × 2 = 13 + 3 × (3 – 1)
t4 = 22 = 13 + 9 = 13 + 3 × 3 = 13 + 3 × (4 – 1)
t5 = 25 = 13 + 12 = 13 + 3 × 4 = 13 + 3 × (5 – 1)
... ... .... .... ...
tn = 13 + 3 × (n – 1) = 3n + 10
The general term can also be generated as,
t1 = 13
t2 = 16 = 13 + 3 = t1 + 3 = t2 – 1 + 3
t3 = 19 = 16 + 3 = t2 + 3 = t3 – 1 + 3
Sequence and Series ~ 31
t4 = 22 = 19 + 3 = t3 + 3 = t4 – 1 + 3
t5 = 25 = 22 + 3 = t4 + 3 = t5 – 1 + 3
... ... .... .... ...
tn = tn – 1 + 3
For sequence, (f) 2, 6, 12, 20, 30, ... ... ...
t1 = 2 = 1 × 2 = 1 × (1 + 1)
t2 = 6 = 2 × 3 = 2 × (2 + 1)
t3 = 12 = 3 × 4 = 3 × (3 + 1)
t4 = 20 = 4 × 5 = 4 × (4 + 1)
t5 = 30 = 5 × 6 = 5 × (5 + 1)
... ... .... .... ...
tn = n × (n + 1) or tn = n2 + n
Thus knowing the pattern of the arrangement of numbers in given sequence its nth term or general term
can be obtained. Also if we know the nth term or general term of any sequence, we can obtain its particular
terms by putting the values of n.
A sequence may have finite or infinite number of terms. A sequence having finite number of terms
is known as a Finite Sequence and a sequence having infinite numbers of terms is known as an Infinite
Sequence. The last term of a finite sequence is fixed and that of an infinite sequence is not fixed. The
following are the examples of finite and infinite sequences.
2, 7, 12, 17, … … , 32 and
4, 2, 1, 12, … … , 1 are finite sequences.
32
1, 3, 9, 27, 81, 243, ... ... ... ... ... and
13, 14, 51, 61, 17, 18, 91, ... ... ... ... ... are infinite sequences.
Series
Consider a sequence, 2, 4, 8, 16, 32, 64, 128, we can form their sum 2 + 4 + 8 + 16 + 32 + 64 + 128.
The expression 2 + 4 + 8 + 16 + 32 + 64 + 128 is said to be a series associated with the sequence 2, 4, 8, 16,
32, 64, 128. In general, the expression of the form a1 + a2 + a3 + ... ... + an is called a series associated with
the sequence a1, a2, a3, ... ... an. The sum of n terms of any series is denoted by Sn.
Singma Notation
If a1, a2, a3, ... ... ..., an is a finite sequence of n terms, the expression
a1+ a2 + a3 + ... ... ... + an is the sum of n terms of the sequence a1, a2, a3, ... ..., an. This sum can be written by
using summation sign or sigma sign in the following way.
n
a1 + a2 + a3 + ................ + an = ∑ ar
r =1
Here the symbol S (Greek letter) known as "sigma." We read the notation as "summation ar where r goes
from 1 to n" which means the sum of ar’s when r goes from 1 to n. Hence by substituting r = 1, 2, 3, ... ... ...
, n in ar, we get a1 + a2 + a3 + ... ... ... + an.
For infinite sequence b1, b2, b3, ... ... bn ... ..., the infinite series is written as
∞
b1 + b2 + b3 ... ... + bn + ... ... = ∑ br
r =1
32 ~ Perfect Optional Mathematics Class 9
Example 1:
Given a sequence 9, 5, 1, – 3, – 7, ... ... ... ... ... Algebra
a. Find the next three terms.
b. Find the 16th term.
c. Find the nth term.
d. Find the series associated with the given sequence.
The given sequence is
9 5 1 – 3 – 7
... ... ...
– 4 – 4 – 4 – 4
The pattern of the given sequence shows that each term is less than the previous term by 4.
a. t1 = 9 = 9 – 4 × 0 Now, the next three terms can be calculated as
t2 = 5 = 9 – 4 × 1 t6 = 9 – 4 × 5 = 9 – 20 = – 11
t3 = 1 = 9 – 4 × 2 t7 = 9 – 4 × 6 = 9 – 24 = – 15
t4 = – 3 = 9 – 4 × 3 t8 = 9 – 4 × 7 = 9 – 28 = – 19
t5 = – 7 = 9 – 4 × 4
\ The next three terms of the sequence are – 11, – 15 and – 19 .
b. 16th term = t16 = 9 – 4 × 15 = 9 – 60 = – 51
c. nth term = tn = 9 – 4 × (n – 1) = 13 – 4n
d. The series associated with the given sequence is
9 + 5 + 1 + (– 3) + (– 7) + (– 11) + ... ... ...
= 9 + 5 + 1 – 3 – 7 – 11 ... ... ...
Example 2:
If the general term of a sequence is given by tn = 2n + 3,
n
a. find the sum of the first five terms.
b. find the 12th and 16th terms.
a. Here, general term tn = 2n + 3.
n
Substituting n = 1, 2, 3, 4 and 5, we get
t1 = 2 × 1 + 3 = 5 t2 = 2 × 2 + 3 = 7
1 2 2
2 × 3 + 3 2 × 4 + 3 11
t3 = 3 = 3 t4 = 4 = 4
t5 = 2 × 5 + 3 = 13
5 5
The sum of the first five terms is
S5 = 5 + 7 + 3 + 11 + 13
2 4 5
Sequence and Series ~ 33
= 100 + 70 + 60 + 55 + 52
20
337
= 20
b. The 12th term of the sequence is t12 = 2 × 12 + 3 = 27
12 12
and the 16th term of the sequence is t16 = 2 × 16 + 3 = 1352 .
16
Example 3:
If t1, t2, t3, ... ... ..., tn be the n terms of a sequence defined as t2 = 7, tn– 1 = tn + 2, find the first three
terms.
Here, tn – 1 = tn + 2
for n = 2, t1 = t2 + 2
or, t1 = 7 + 2 = 9 (as t2 = 7)
for n = 3, t2 = t3 + 2
or, 7 = t3 + 2
\ t3 = 5
Hence, the first three terms are 9, 7, 5.
Example 4:
Find the sum of the following series. b. r∑ =412kk+–11
5
a. ∑ (2n – 3)
r = 1
a. 5
The given series is ∑ (2n – 3)
r =1
5
Here, ∑ (2n – 3) = (2 × 1 – 3) + (2 × 2 – 3) + (2 × 3 – 3) + (2 × 4 – 3) + (2 × 5 – 3)
r =1
= – 1 + 1 + 3 + 5 + 7 = 15
b. The given series is 4 2k – 1
k+1
∑
r =1
Hence, 4 2k – 1 = 2 × 1 – 1 + 2 × 2 – 1 + 2 × 3 – 1 + 2 × 4 – 1
k+1 1 + 1 3 3 + 1 4 + 1
∑
r =1
= 1 + 3 + 5 + 7 = 1 + 1 + 5 + 7
2 3 4 5 2 4 5
= 10 + 20 + 25 + 28 = 83
20 20
Example 5:
In the following pattern of numbers,
a. add one more figure in the same pattern.
34 ~ Perfect Optional Mathematics Class 9
b. find the formula for the nth term of the sequence. Algebra
a. Drawing one more figure in the same pattern.
b. For nth term of the sequence,
1st term (t1) = 1 = 1
2nd term (t2)
3rd term (t3) = 5 = 1 + 4 = 1 + (2 – 1) 4
4th term (t4)
... .... .... .... = 9 = 1 + 2 × 4 = + (3 – 1) 4
nth term (tn) = 13 = 1 + 3 × 4 = 1 + (4 – 1) 4
tn = 4n – 3
.... .....
Example 6:
= 1 + (n – 1) 4 = 1 + 4n – 4 = 4n – 3
Find the general term and write the following series with the sigma notation.
a. 15 + 12 + 9 + 6 + ... ... ... 15 terms.
b.
4+ 9 + 16 + 25 + ... ... ... 12 terms.
a. 8 27 24
The given series is 15 + 12 + 9 + 6 + ... ... ... 15 terms
t1 = 15 = 15 – 0 = 15 – 3 × 0 = 15 – 3 × (1 – 1)
t2 = 12 = 15 – 3 = 15 – 3 × 1 = 15 – 3 × (2 – 1)
t3 = 9 = 15 – 6 = 15 – 3 × 2 = 15 – 3 × (3 – 1)
t4 = 6 = 15 – 9 = 15 – 3 × 3 = 15 – 3 × (4 – 1)
... ... ... ... ... ...
tn = 15 – 3 × (n – 1) = 15 – 3n + 3 = 18 – 3n
15
Hence, 15 + 12 + 9 + 6 + ... ... ... 15 terms = ∑ (18 – 3n)
n =1
b. The given series is 4 + 9 + 16 + 25 + ... ... ... 12 terms.
8 27 64
t1 = 4 = 22 = (1 + 1)2
13 13
Sequence and Series ~ 35
t2 = 9 = 32 = (2 + 1)2
8 23 23
t3 = 16 = 42 = (3 + 1)2
27 33 33
t4 = 25 = 52 = (4 + 1)2
24 43 43
... ... ... ... ... ...
tn = (n + 1)2
n3
12
Hence, 4 + 9 + 16 + 25 + .... ... ... 12 terms = (n + 1)2
8 27 64 ∑ n
n =1
Relation between Sn and tn
Let t1 + t2 + t3 + t4 + ... ... + tn – 1 + tn be a linear series, then the sum of the first n terms is given by
Sn = t1 + t2 + t3 + t4 + ... ... + tn – 1 + tn ... ... (1)
Again, the sum of the first (n – 1) terms of the series is given by
Sn– 1 = t1 + t2 + t3 + t4 + ... ... + tn – 1 ... ... (2)
Subtracting equation (2) from equation (1), we have,
Sn – Sn– 1 = (t1 + t2 + t3 + t4 + ... ... + tn– 1 + tn) – (t1 + t2 + t3 + t4 + ... ... + tn – 1)
= t1 + t2 + t3 + t4 ... ... + tn – 1 + tn – t1 – t2 – t3 – t4 – ... ... – tn – 1
Sn – Sn – 1 = tn
Hence, the nth term (tn) of a linear series is equal to the difference of the sum of the first n terms (i.e. Sn) and
the sum of the first (n – 1) terms (i.e. Sn– 1).
Example 7:
The sum of the n terms of a given series is Sn = n2 – 3n
a. Find the values of S8 and S9. b. Find the 9th term.
c. Find the general term.
a. Here, we have Sn = n2 – 3n b. Hence the 9th term, t9 = S9 – S8
S9 = 92 – 3 × 9 = 81 – 27 = 54 \ t9 = 54 – 40 = 14
S8 = 82 – 3 × 8 = 64 – 24 = 40
c. The general term.
tn = Sn – Sn – 1
= (n2 – 3n) – {(n – 1)2 – 3(n – 1)}
= n2 – 3n – (n – 1)2 + 3(n – 1)
= n2 – 3n – n2 + 2n – 1 + 3n – 3
= 2n – 4 = 2(n – 2)
Hence, the general term (tn) = 2(n – 2)
36 ~ Perfect Optional Mathematics Class 9
General Term of a Linear Sequence Algebra
The sequence in which consecutive pair of terms have a common difference is known as linear sequence.
General term of linear sequence is given by tn = a + d(n – 1) where a = first term, d = common difference,
n = number of terms.
For example : let a sequence 5, 8, 11, 14, 17, 20, ... ...
Here, t1 = 5 = 5 + 3 × 0 = 5 + 3 × (1 – 1)
t2 = 8 = 5 + 3 × 1 = 5 + 3 × (2 – 1)
t3 = 11 = 5 + 3 × 2 = 5 + 3 × (3 – 1)
t4 = 14 = 5 + 3 × 3 = 5 + 3 × (4 – 1)
t5 = 17 = 5 + 3 × 4 = 5 + 3 × (5 – 1)
t6 = 20 = 5 + 3 × 5 = 5 + 3 × (6 – 1)
... ... ... ... ... ... ... ... ... ... ... ... ... ...
tn = 5 + 3 × (n – 1).
General Term of a Quadratic Sequence
The sequence in which the first difference is not constant but its the second difference is constant is called
a quadratic sequence. General term of quadratic sequence is given by
tn = an2 + bn + c, where a, b and c are constant and n = number of terms.
The general term of a quadratic sequence can be determined by two methods.
Method I
Suppose 3, 4, 7, 12, 19, 28, 39, ... ... ... ... , tn be a sequence.
3 4 7 12 19 28 39
First difference → 1 3 5 7 9 11
Second difference → 2 2 2 2 2
Here, first difference is not constant but second difference is constant.
So, given sequence is a quadratic sequence. Let the general term of the given sequence is
tn = an2 + bn + c. ... ... ... (1)
Now, when n = 1, from tn = an2 + bn + c,
t1 = a × 12 + b × 1 + c
or, 3 = a + b + c (\ t1 = 3 = 1st term)
or, c = 3 – a – b
When n = 2, from tn = an2 + bn + c
t2 = a × 22 + b × 2 + c
or, 4 = 4a + 2b + c (\ t2 = 4 = 2nd term)
Putting the value of c, we get
4 = 4a + 2b + 3 – a – b
or, 4 = 3a + b + 3
or, 4 – 3 – 3a = b
or, 1 – 3a = b ... ... ... (2)
Sequence and Series ~ 37
Again, when n = 3, from tn = an2 + bn + c
t3 = a × 32 + b × 3 + c
or, 7 = 9a + 3b + c (\ t3 = 7 = 3rd term)
Putting the value of c, we get,
7 = 9a + 3b + 3 – a – b
or, 7 = 8a + 2b + 3
or, 7 – 3 = 8a + 2b
or, 4 = 8a + 2b
or, 2 = 4a + b
Putting the value of b, we get
2 = 4a + 1 – 3a = a + 1
\ a = 1
Putting the value of a, in equation (2).
b = 1 – 3a = 1 – 3 × 1 = 1 – 3
\ b = – 2
Again, putting the value of a and b in equation (1), we get
c=3–a–b=3–1+2
\ c = 4
\ Required tn = an2 + bn + c
= 1 × n2 + (– 2)n + 4
\ tn = n2 – 2n + 4
Method II
We can find the gereral term of quadratic sequence as follows:
tn = an2 + bn + c, where a, b and c are constant and n = number of terms.
n → 1 2 3 4 5 6
36a + 6b + c
a + b + c 4a + 2b + c 9a + 3b + c 16a + 4b + c 25a + 5b + c
First sequence → 3a + b 5a + b 7a + b 9a + b 11a + b
Second sequence → 2a 2a 2a 2a
• First term of the third sequence = 2a = 2 times coefficient of n2
• First term of the second sequence = 3a + b = 3 times co-efficient of n2 + co-efficient of n
• First term of the first sequence = a + b + c = sum of co-efficients of n2, n and constant
For example,
For the above sequence 3, 4, 7, 12, 19, 28, 39, ... ... ... ... 19 28 39
First sequence → 3 4 7 12
Second sequence → 1 3 5 7 9 11
Third sequence → 2 2 2 2 2
38 ~ Perfect Optional Mathematics Class 9
Suppose the general term of the given sequence is tn = an2 + bn + c. Algebra
Hence, 2a = first term of the third sequence = 2, therefore, a = 1.
3a + b = the first term of the second sequence = 1, therefore, 3 + b = 1 and hence, b = – 2.
a + b + c = the first term of the first sequence = 3, therefore, 1 – 2 + c = 3 and hence, c = 4.
The required general term of the given sequence is tn = an2 + bn + c = n2 – 2n + 4.
Again, consider a sequence 3, 10, 25, 48, 79, 118, 165, ... ... ... ...
First sequence → 3 10 25 48 79 118 165
Second sequence → 7 15 23 31 39 47
Third sequence → 8 8 8 8 8
Suppose the general term of the given sequence is tn = an2 + bn + c.
Hence, 2a = first term of the third sequence = 8, therefore, a = 4.
3a + b = the first term of the second sequence = 7, therefore, 12 + b = 7 and hence, b = – 5.
a + b + c = the first term of the first sequence = 3, therefore, 4 – 5 + c = 3 and hence, c = 4.
The required general term of the given sequence is tn = an2 + bn + c = 4n2 – 5n + 4.
EXERCISE : 1.6
1. Find the 10th term and nth term of the following sequence :
a. – 5, – 1, 3, 7, ........... b. 35, 27, 19, 11, ................
2. From the given general terms of the following sequences, find the first four terms and the twelfth term.
a. tn = 3n + 2 b. tn = 2n2 – 3 c. tn = n(n + 1)
2
d. tn = (– 1)n(n + 1) e. tr = r(r + 1) f. tr = r2 + r + 1
n2 r+ 2
3. Find the first five terms and their sum in each of the following cases if the general terms are given
below.
a. an = 2n + 5 b. an = 2n + 11 c. an = (– 1)nn
2n – n+1
d. an = (– 1)n + 1.n2 + 1 e. an = 2n2 – 3n + 1 f. an = (n + 1)2 – 3n
4. a. The nth term of a sequence is (– 1)n – 1. 15n . Find its third and eighth terms.
n2 – 4
b. The nth term of the sequence is (– 1)n – 1. 75n , find its 5th and 6th terms.
n2 – n – 5
c. Find the nth term and tenth term of the sequence 13 – 2, 23 – 4, 33 – 6, 43 – 8, ...
5. If t1, t2, t3, ... ... ..., tn be the n terms of a sequence, find the first five terms in each of the following.
a. t1 = 3, tn + 1 = 2tn + 3 b. t1 = 21, tn + 1 = 2tn + 32 c. t1 = – 2, tn – 1 = 2tn – 3
d. t1 = 3, t2 = 8, tn= tn– 2 + 10 e. t3 = 8, tn – 1 = 3tn + 1 f. t6 = 13, tn = 2tn– 1 + 5
Sequence and Series ~ 39
6. a. Find the values of u3 and u4 if un = un – 1 and u1 = 1, u2 = 2 (n > 2).
un – 2 where un + 1 =1– u1n.
b. u1, u2, u3, ... ... are the terms of a sequence If u1 = 3, find the values of u2
and u3.
7. Find the sum of the following series:
6 5 10
a. ∑ (4n – 5) b. ∑ (n2 – 3) c. ∑ (k – 1)2 + 4
n =1 n =1 k =1
5 k2 + 41 e. r∑ =81 (– 1)r(2r2 + 3) 6
k +
d. ∑ f. ∑ xi – xi – 1
k =1 i =1
8. Study the following patterns, add one more pattern in the sequence and find the formula for the nth
term of the sequence.
a.
b.
c.
d.
e.
40 ~ Perfect Optional Mathematics Class 9
f. Algebra
h.
9. Find the general term and write the following series with sigma notation.
a. 20 + 15 + 10 + 5 + ... ... ... 10 terms
b. 32 + 48 + 72 + 108 + ... ... ... 13 terms
c. – 3 + 4 – 5 + 86, … … … 5 terms
5 6 7 15 terms
d. 12 + 14 13 + 16 32 + 19 + ... ... ...
10. The sum of the n terms of a given series is Sn = n2 + n.
a. Find the values of S7 and S8. b. Find the 8th term.
c. Find the general term.
11. The sum of the n terms of a given series is Sn = n6 (n + 1) (2n + 1).
a. Find the values of S9 and S10. b. Find the 10th term.
c. Find the general term.
12. The sum of the n terms of a given series is Sn = n(n + 1) 2
2
a. Find the values of S8, S9 and S10.
b. Find the 9th and 10th terms
c. Find the general term.
13. Find the general term of the following linear sequence.
a. 3, 8, 13, 18, 23, ... ... b. 13, 23, 33, 43, 53, ... ...
c. 2, 9, 16, 23, 30, ... ... d. 18, 16, 14, 12, 10, ... ...
14. Find the general term of the following quadratic sequence.
a. 10, 11, 13, 16, 20, ... ... b. 2, 5, 12, 23, 38, ... ...
c. 5, 15, 27, 41, 57, ... ... d. 8, 18, 26, 32, 36, ... ...
Sequence and Series ~ 41
2 LIMITS
Curriculum Contents Included Periods
10
Concept of limit
- From the sequence of numbers y = f(x)
- From the sequence of figures
- On the basis of the sum of the infinite series
- Value of a function
- Concept of the limit of a function
- Symbolic representation of the value of limit
- Meaning and definition of x→a
Objectives
On completion of this unit, students will be able to:
find general term of infinite series.
find the last figure of a sequence of figures.
find the value of a function at the given point.
distinguish the value and limit of a function.
state the meaning of x→a.
calculate the limiting value of the given function at the given point.
Materials
YY
y = f(x) = x2 – 4 f(x)→L
x – 2 L
0
f(2) = 0 f(x)→L
X' O X
X' O x→a– x→a+ X
Y' Y'
Knowledge Specification Grid
1 mark each
Contents Understanding Application Higher ability Total Total
Limits 1 5 marks each Questions Marks
2 marks each 4 marks each
– 2 5
–1
42 ~ Perfect Optional Mathematics Class 9
2.1 Concept of Limit by Sequence of Numbers
Limit of a Sequence Limit
Consider the sequence whose terms are 1, 12, 13, 41, … … . The general term of the sequence is 1n. As we take
more and more terms, each term is getting smaller in size. Here, we can make the terms as small as we like,
provided we go far enough along the sequence. Thus, although no term in the sequence is 0, the terms can
be made as close as we like to 0 by going far enough.
We say that the limit of the sequence n1: n = 1, 2, 3, ... ... is 0 and we write it as xl→im∞ n1 = 0.
It is important to emphasize that we are not putting n equal to ∞ in the sequence, since infinity is not
a number. It should be thought of as a convenient idea. The statement above says that the terms in the
sequence get as close to 0 as we please (and continue to be close to 0), by allowing n to be large enough.
1
n
(1,1)
1
0.50 (2,0.50)
0.33 0.25 (3,0.33) (4,0.25) (5,0.20) (6,0.16) (7,0.14)
0.20 0.16
0.14 n
0 1 2 3 4 5 6 7 8 9
Graph of the sequence 1
n
There are four different types of limits.
1. Limit → ∞ (infinity) 2. Limit → 0 (zero)
3. Limit → c (constant) 4. Limit → No limit
a. Limit tends to ∞ (infinity) i.e. lim → ∞
Let sequence { 1, 7, 13, 19, 25, ... ... ...}
When the value of term goes on increasing and positive, it can be made greater than any positive quantity,
however large we can say that the term approaches to infinty (∞) or the term tends to ∞ or term near by ∞.
Study the following examples.
SN Sequence Limit of sequence approaches
1. 1, 3, 5, 7, 9, ... ... +∞
2. 2, 4, 6, 8, 10, ... ... +∞
3. 200, 190, 180, 170, ... ... –∞
4. 10, 9, 8, 7, ... ... –∞
5. tn= 2n2 +∞
Concept of Limit by Sequence of Numbers ~ 43
b. Limit tends to 0 (zero) i.e. lim → 0
Let us consider a sequence of numbers 21, 41, 81, 1 , … … .
16
When the value of term goes on decreasing and can be made numerically less than any positive
quantity, however small, we say the term approaches zero. In other words, the term tends to zero.
Study the following examples.
SN Sequence Limit of sequence approaches
0
1. 1, 12, 14, 81, 116, ... ... 0
0
2. 0.1, 0.01, 0.001, 0.0001, ... ... 0
0
3. 110, 1100, 10100, 10100, ... ...
4. tn = 1
2n
5. tn= n 1 1
+
c. Limit tends to c (constant) i.e. lim → c
Let us consider a squence 12, 23, 34, 54, 65, … … .
In the above squence, the sequence of the numerators is 1, 2, 3, 4, 5, ... ... and the sequence of the
denominators is 2, 3, 4, 5, 6, ... ... . In every term of the given sequence, the numerator is less than the
denominator by 1 and the numerator and denominator both are increasing. Each term of the sequence
becomes closer to 1 than its previous term. So, we can say that the term of the sequence approaches 1.
In other words, the term tends to 1.
Study the following examples.
SN Sequence Limit of sequence approaches
1. 53, 75, 79, 191, 1113, … … . 1
2. 3.1, 3.01, 3.001, 3.0001, ... ... 3
3. 5.5, 5.05, 5.005, 5.0005, ... ... 5
4. 2.9, 2.99, 2.999, 2.9999, ... ... 3
5. 4.29, 4.299, 4.2999, 4.29999, ... ... 4.3
d. Limit tends to no limit
Let us consider a squence 1, – 1, 1, – 1, 1, – 1, 1, – 1, ... ... ...
Here, the term of squence continously goes to + 1 and – 1. The number can not converge in a single
number so it has no limit.
44 ~ Perfect Optional Mathematics Class 9
Study the following examples.
SN Sequence Limit of sequence approaches
1. 2, – 2, 2, – 2, 2, – 2, ... .. ... no limit
2. 21, – 12, 12, – 21, 12, – 21, ... ... no limit
EXERCISE : 2.1 Limit
1. If 1, 2, 3, 4, 5, 6, 7 ... ... is a sequence,
a. find the next two terms of the sequence.
b. what is the limit of the sequence?
2. If 3, 6, 9, 12, 15, ... ... is a sequence,
a. find the next three terms of the sequence.
b. what is the limit of the sequence?
3. Find the next two terms of the following sequences and also find the limit of its term.
a. 20, 19, 18, 17, ... ... b. 100, 90, 80, 70, ... ...
4. Find the next three terms of the following sequences and also find the limit of its term.
a. 230, 2030, 20300, 203000, ... ... b. 41, 18, 116, 312, ... ...
5. Find the next three terms of the following sequences and also find the limit of its term.
a. 2.1, 2.01, 2.001, 2.0001, 2.00001, ... ..
b. 5.29, 5.299, 5.2999, 5.29999, ... ...
6. Find the next two terms of the following sequences and also find the limit of its term.
a. 10, – 10, 10, – 10, 10, ... ...
b. 4, – 4, 4, – 4, 4, – 4, ... ...
7. a. If the general term of a sequence is tn = 5n, where n = {0, 1, 2, 3, 4, ... ...}, then find the first five
terms. Also find the limit of the term of the sequence.
b. If the general term of a sequence is tn = 2n2, where n = {1, 2, 3, 4, ... ...}, then find the first five
terms. Also find the limit of the term of the sequence.
1
8. a. If the general term of a sequence is tn = n + 2, where n = {0, 1, 2, 3, 4, ... ...}, then find the first
five terms. Also find the limit of the term of the sequence.
b. If the general term of a sequence is tn = 21n, where n = {1, 2, 3, 4, ... ...}, then find the first five
terms. Also find the limit of the term of the sequence.
9. a. If the general term of a sequence is tn = 2 + 31n, where n = {1, 2, 3, 4, ... ...}, then find the first five
terms. Also find the limit of the term of the sequence.
1
b. If the general term of a sequence is tn = 1 – n + 2, where n = {1, 2, 3, 4, ... ...}, then find the first
five terms. Also find the limit of the term of the sequence.
Concept of Limit by Sequence of Numbers ~ 45
2.2 Concept of Limit by Sequence of Diagrams
Introduction
In this unit, we discuss some activities to get the idea of limit. Consider the following circles with the same
radius and hence each of these circles has the same area i.e. A.
A AA A A
Now, let us draw an inscribed square with the area a in the first circle as shown. The number of sides of
the square is 4. Now again, what happens when we increase the number of sides of the polygon. Suppose
a regular polygon is drawn inside the second circle as shown. What is the differece between the area of the
circle and that of the polygon in the following two cases?
A AA A A
a4 a5
n=4 n=5
The difference between the area of a circle and the area of a pentagon is less than the difference between the
area of a circle and the area of a square. As we increase the number of sides of the polygon, the difference
(gap) between the polygon and the circle decreases.
A AA A A
a4 a5 a6 a8 a10
n=4 n=5 n=6 n=8 n = 10
In the similar way, let us increase the number of sides of the inscibed regular polygon as 6, 8, 10 and so on.
What effect do we have in the difference between the areas of the circle and the polygon. As the number of
sides of the inscribed regular polygon increases, its area increases and the difference between the area of
the cirlce and the area of the polygon decreases. If the number of sides of the polygon approaches to infinity
(i.e. n→∞), what happens ? Infinity means too huge number that we cannot imagine. Clearly, the polygon
almost covers the whole circle, i.e. the area of the polygon approaches the area of the circle. Symbolically,
we write it as an→A. So, when n→∞, then an→A. In such cases, to calculate the area of a polygon, we
introduce the new topic in mathematics that is called the limit. We can write the information of this example as
no. of lim Area of polygon = Area of circle or lim an = A
sides→∞ n→∞
46 ~ Perfect Optional Mathematics Class 9
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