Approved by the Government of Nepal, Ministry of Education, Science and Technology,
Curriculum Development Centre, Sanothimi, Bhaktapur as an Additional Learning Material
vEMexdcAaenlTtaiHn OEMptAioTnICaSl
9Book
Author
Piyush Raj Gosain
Hukum Pd. Dahal Editors P. L. Shah
Tara Bdr. Magar
vedanta
Vedanta Publication (P) Ltd.
jb] fGt klAns;] g kf| = ln=
Vanasthali, Kathmandu, Nepal
+977-01-4982404, 01-4962082
[email protected]
www.vedantapublication.com.np
EvMexdcAaenlTtaiHn OEMptAioTnICaSl
9Book
Author
Piyush Raj Gosain
All rights reserved. No part of this publication may
be reproduced, copied or transmitted in any way,
without the prior written permission of the publisher.
¤ Vedanta Publication (P) Ltd.
First Edition: B.S. 2077 (2020 A. D.)
Second Edition: B.S. 2078 (2021 A. D.)
Layout and Design
Pradeep Kandel
Printed in Nepal
Published by:
Vedanta Publication (P) Ltd.
j]bfGt klAns];g k|f= ln=
Vanasthali, Kathmandu, Nepal
+977-01-4982404, 01-4962082
[email protected]
www.vedantapublication.com.np
Preface
The series of Vedanta Excel in Opt. Mathematics for class 9 and 10 is completely based on the
contemporary pedagogical teaching learning activities and methodologies. It is an innovative
and unique series in the sense that the contents of each textbooks of the series are written and
designed to ful ill the need of integrated teaching learning approaches.
Vedanta Excel in Opt. Mathematics has incorporated applied constructivism, the latest trend of
learner centered teaching padagogy. Every lesson of the series is written and designed in such
a manner that makes the classes automatically constructive and the learner actively participate
in the learning process to construct knowledge themselves, rather than just receiving ready
made information from their instructor. Even teachers will be able to get enough opportunities
to play the role of facilitators and guides shifting themselves from the traditional methods
imposing instractions. The idea of the presentation of every mathematical item is directly or
indirectly re lected from the writer's long expenena, more than two decades, of teaching optional
mathematics.
Each unit of Vedanta Excel in Opt. Mathematics series is provided with many more worked out
examples, arranged in the hierarchy of the learning objectives and they are re lective to the
corresponding exercises.
Vedanta Excel in Opt. Mathematics class 9 covers the latest syllabus of CDC, the government of
Nepal, on the subject. My honest efforts have been to provide all the essential matters and practice
materials to the users. I believe that the book serves as a staircase for the students of class 10.
The book contains practice exercises in the form of simple to complex including the varieties of
problems. I have tried to establish relationship between the examples and the problems set for
practice to the maximum extent.
In the book, every chapter starts with review concepts of the same topic that the students have
studied in previous classes. Discussion questions in each topic are given to warm up the students
for the topic. Questions in each exercise are catagorized into three groups - Very Short Questions,
Short Questions and Long Questions.
The project works are also given at the end of exercise as required. In my experience, the students
require more practices on Trigonometry and Vector Geometry, the examples and the exercise
questions are given to ful ill it in the corresponding topics. The latest syllabus of the subject
speci ication grid and a model question issued by CDC are given at the end of the book.
My hearty thanks goes to Mr. Hukum Pd. Dahal, Tara Bahadur Magar and P.L. Shah, the series
editors, for their invaluable efforts in giving proper shape to the series. I am also thankful to my
college Mr. Gyanendra Shrestha who helped me a lot during the preparation of the book.
I am also thankful to my respected parents and my family members for their valuable support to
bring the book out in this form. I would also like to express my hearty gratitude to all my friends,
colleagues and beloved students who always encouraged me to express my knowledge, skill and
experience in the form of books. I am highly obliged to all my known and unknown teachers who
have laid the foundation of knowledge upon me to be such a person.
Last but not the least, I am hearty thankful to Mr. Pradeep Kandel, the computer and designing
senior of icer of the publication for his skill in designing the series in such an attractive form.
Efforts have been made to clear the subject matter included in the book. I do hope that teachers
and students will best utilize the series.
Valuable suggestions and comments for its further improvement from the concerned will be
highly appreciated. Piyush Raj Gosain
CONTENT
Unit 1 Relations and Functions 5
Unit 2 Polynomials 42
Unit 3 Sequence and Series 52
Unit 4 Limit 67
Unit 5 Matrix 85
Unit 6 Coordinate Geometry 114
Unit 7 Equations of Straight Lines 129
Unit 8 Trigonometry: Measurement of Angles 173
Unit 9 Trigonometric Ratios 188
Unit 10 Trigonometric Ratios of Some Standard Angles 206
Unit 11 Trigonometric Ratios of Any Angle 214
Unit 12 Vector 241
Unit 13 Transformation 269
Unit 14 Statistics : Partition Values 305
Unit 15 Statistics : Measure of Dispersion 318
Syllabus
Specification Grid
Model Questions
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
Relations and Functions 1
1.0 Review
Discuss the followings in your class:
(a) Name any three words that represent relationship between your family members.
(b) State a relation between a set of real numbers and a set of natural numbers.
(c) Plot points (1, 1), (2, 4), (3, 9) in a graph paper. Is there any relation between
x-coordinateand y-coordinate in there points?
(d) From the given equation, y = 2x + 1, fill in the table that satisfies the equation.
x0 0 ............... ............... ............... ...............
y 1 ............... ............... ............... ............... ...............
1.1 Ordered Pair
Let us take two sets A = {a, b} and B ={b, a}, what is difference between them ? Can we
write A = B ?
Here, A and B contain the same elements. They are equal sets. So, we can write A =B. The
order of elements in a set is not necessary.
A set consisting of any two elements is called a pair. Some examples of pairs are given below:
(a) A pair of shoes, a pair of socks etc. Y
(b) {Kathmandu, Nepal} 7
6
5
(c) {2, 4} etc. 4 (1, 4)
In a pair the order of occurrence of the elements is not important. 3
It means the pair of two elements is unordered. 2 (4, 1)
X' 1
Again, let us take two pairs (1, 4) and (4, 1), Do they represent
O 123456 X
same point? Plot them in a graph. Y'
Here, (1, 4) and (4, 1) have different meanings. (1, 4) ≠ (4, 1), (1, 4) and (4, 1) are ordered pairs.
Definition : A pair having one element as the first and the other as the second element is
called an ordered pair. An ordered pair having 'a' as the first element and 'b' as the second
element is denoted by (a, b). The ordered pairs (a, b) and (b, a) are different.
In an ordered pair (a, b), the first element 'a' is called antecedent and the second element 'b'
is called consequent.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 5
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
In sense of coordinates, in ordered pair (a, b), 'a' is x-coordinate or abscissa and b is
y-coordinate or ordinate.
Note :
1. {1, 2} is a set of two elements 1 and 2.
2. (1, 2) is an ordered pair with the first element 1 and the second element 2.
Equality of ordered pairs
Let us take ordered pairs (4, 5) and 2=50,250255an.dT5h=ey255a.re equal ordered pairs as their
corresponding elements are equal, i.e.,
4
Definition : Two ordered pairs are said to be equal when their corresponding components
are equal, i.e., their antecedents and consequents are equal.
If (a, b) = (x, y), then a = x and b = y.
Worked Out Examples
Example 1. Find the values of x and y from the following conditions :
Solutions :
(a) (4x, 9) = 16, 3 (b) (3x + 4, 4y + 7) = (16, 23)
y (d) (x, y) = (2x + 8, 3y + 6)
(c) (x + y, x – y) = (5, 3)
(a) Here, (4x, 9) = 16, 3 , equating the corresponding elements of ordered
y
pairs, we get,
4x = 16
or, x = 4
and 9 = 3
y
or, 9y = 3
or, y = 3 = 1
9 3
1
Hence, x = 4 and y = 3
(b) Here, (3x + 4, 4y + 7) = (16, 23)
Equating the corresponding elements of the equal ordered pairs,
we get,
3x + 4 = 16
or, 3x = 16 – 4
or, 3x = 12
? x = 4
and 4y + 7 = 23
6 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
or, 4y = 23 – 7
or, 4y = 16
? y=4
Hence, x = 4 and y = 4.
(c) (x + y, x – y) = (5, 3)
Equating the corresponding elements of equal ordered pair, we get,
x + y = 5 ............... (i)
x – y = 3 .................. (ii)
Adding equations (i) and (ii), we get
2x = 8 or, x = 4
Substituting the value of x in equation (i) we get,
y=5–x
or, y = 5 – 4 ? y=1
? x = 4 and y = 1
(d) Here, (x, y) = (2x + 8, 3y + 6)
Equating the corresponding elements of equal ordered pairs,
we get,
x = 2x + 8 or, x – 2x = 8
or, –x = 8
? x = –8
and y = 3y + 6 or, y – 3y = 6
or, –2y = 6
? y = –3
? x = –8 and y = –3
Example 2. If (p, 27) and (1, q) are the elements of the set {(x, y) : y = 3x + 4}, find the
Solution : values of p and q.
Here, the ordered pairs (p, 27) and (1, q) are
the elements of the set {(x, y) : y = 3x + 4}.
So, (p, 27) {(x, y) : y = 3x + 4}
We put x = p and y = 27, then we get
27 = 3p + 4
or, 27 – 4 = 3p
or, 3p = 23 ? p = 23
3
and also (1, q) {(x, y) : y = 3x + 4}
we put, x = 1, y = q, q = 3.1 + 4 = 7
Hence, p = 23 and q = 7
3
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 7
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
Exercise 1.1
Very Short Questions :
1. (a) Define a pair.
(b) Define an ordered pair.
(c) Write a difference between {1, 2} and (1, 2)
(d) Give an example of each of a pair and an ordered pair.
(e) Define equality of two ordered pairs.
2. Which are equal ordered pairs in each of the following:
(a) (4, 5) and (4, 5) (b) 21 , 5 and 7, 25
3 5
(c) (4 + 9, 21 ÷ 7) and (13, 3) (d) (1.5, 4.5) = 3 , 45
2 10
3. Find the values of x and y in the following conditions :
(a) (x, 4) = (7, 2y) (b) (x + 4, y + 2) = (7, 4)
(c) (x – 3, y + 4) = (3, 6) (d) (2x – 5, 4) = (19, 2y – 6)
(e) (4x + 8, 5x – 7) = (16, 3) (f) x + 1, y + 1 = 5 , 9
Short Questions : 3 4 3 5
4. Find the values of x and y in the following conditions :
(a) (2x + y, 2) = (1, x –y) (b) (x + y, x – y) = (7, 3)
(c) (2x + y , 2x – y) = (16, 1) (d) (2x+y, 3x–y) = (16, 9)
5. Find the missing elements in the following ordered pais for the relation f(x) = x2 + 3:
(a) (2, ...) , (b) (..., 19) (c) (–4, ...) (d) (..., 39)
Long Questions :
6. (a) If the ordered pairs (p, –1) and (5, q) belong to {(x, y): y = 2x – 3}, find the values
of p and q.
(b) If the ordered pairs (13, p) and (q, 2) belong to {(x, y): x – 3y = 10}, find the values
of p and q.
7. (a) If x {3, 4, 5, 6} and y (1, 2), find the ordered pairs such that x + y < 8.
(b) If x {1, 2, 3, 4, 5} and y (6, 7), find the ordered pairs such that x + y > 8.
Project work
8. Write any ten ordered pairs in which the first element is country the second element is
its capital.
1. (b) {1, 2} a set, (1, 2) ordered pair 2. All equal ordered pairs
3. (a) x = 7, y = 2 (b) x = 3, y = 2 (c) x = 6, y = 2 (d) x = 12, y = 5
(e) x = 2, y = 2 31 4. (a) x = 1, y = –1 (b) x = 5, y = 2
(f) x = 2, y = 20
(c) x = 2, y = 2 (d) x = 3, y = 1 5. (a) 7 (b) ± 4 (c) 19 (d) ± 6
6. (a) p = 1, q = 7 (b) p = 1, q = 16 7. (a) (4, 1), (4, 2), (5, 1 (5, 2), (6, 1)
(b) (2, 7), (3, 6), (3, 7) (4, 6), (4, 7), (5, 6), (5, 7)
8 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
1.2. Cartesian Product
Let A = {a, b} and B ={x, y, z} be two non–empty sets. Let us find all possible ordered pairs
from the set A to the set B as given below:
Set A Set B Ordered pairs
x (a, x)
ay (a, y)
z (a, z)
x (b, x)
by (b, y)
z (b, z)
All the above ordered pairs can be represented by a set {(a, x), (a, y), (a, z), (b, x), (b, y),(b, z)}
This set is represented by A × B which is known as the cartesian product of A and B.
? A × B = {(a, x), (a, y), (a, z), (b, x), (b, y), (b, z)}
Definition : Let A and B be two non–empty sets. Then the cartesian product A × B (read as
A cross B) is defined as the set of all possible ordered pairs (x, y) such that x A and y B.
Symbolically,
we write, A × B = {(x, y) : x A and y B}
Note :
1. If A × B = I , then A = I or B = I
2. If A × B = B × A, then A = B
3. If the cardinalities of A and B are respectively n(A) and n(B),
then n(A × B) = n(A) × n(B).
4. In general, A × B ≠ B × A.
Ways of Representation of Cartesian Product
The following are the ways of representation of cartesian product of two non-empty sets A
and B. Let A = {2, 4} and B = {3, 4, 5}
1. Set of ordered pairs :
A × B = {(2, 3), (2, 4), (2, 5), (4, 3), (4, 4), (4, 5)}
2. Set builder method
A × B = {(x, y) : x A or y B}
3. Tree diagram method:
Set A Set B Set A Set B (4, 3)
3
3 (2, 3)
24 (2, 4) 4 4 (4, 4)
5 (2, 5) 5 (4, 5)
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 9
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
4. Tabulation Method:
Sets B
×3 4 5
A 2 (2, 3) (2, 4) (2, 5)
4 (4, 3) (4, 4) (4, 5)
5. Arrow Diagram Method:
f
AB
23
4
45
6. Lattice or Graphic Method:
Y
7
6
5
4
3
2
1
X' O 1 2 3 4 5 6 X
Y'
Worked out Examples
Example 1. If A = {1, 2, 3} and B = {4, 5}, find:
Solution :
(a) A × B (b) B × A
Example 2.
Solution : (c) n(A × B)
(d) n (A) × n (B) Also show that : n(A × B) = n(A)× n (B)
Here, A = {1, 2, 3}, B = {4, 5}
(a) A × B = {1, 2, 3} × {4, 5} = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}
(b) B × A = {4, 5} × {1, 2, 3} = {(4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3)}
(c) n(A × B) = 6 (d) n(A) × n(B) = 3 × 2 = 6
Now, n(A × B) = n(A) × n(B)
or, 6 = 3 × 2
? 6 = 6 Proved.
Let A = {4, 5} and B = {5, 4}, show that A × B = B × A.
Here, A = {4, 5}, B = {5, 4}
Now, A × B = {4, 5} × (5, 4) = {(4, 5), (4, 4), (5, 5), (5, 4)}
B × A= {5, 4} × {4, 5} = {(5, 4), (5, 5), (4, 4), (4, 5)}
= {(4, 5), (4, 4), (5, 5), (5, 4)}
? A × B = B × A.
Note : A × B = B × A is true only when A = B.
10 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
Example 3. If A = {x : x ≤ 4, x N} and B = {x : x2 – 1 = 0}, then find A × B and B × A.
Solution : Here, A = {x : x ≤ 4, x N} = {1, 2, 3, 4}
B = {x : x2 – 1 = 0} = {x : x2 = 1 } = {x : x = ± 1} = {–1, 1 }
Now, A × B= {1, 2, 3, 4} × {– 1, 1}
={(1, –1), (1, 1), (2, –1), (2, 1), (3, –1), (3, 1), (4, –1) (4, 1)}
Also, B × A= {–1, 1} × {1,2, 3, 4}
= {(–1, 1), (–1, 2), (–1, 3), (–1, 4), (1, 1), (1, 2), (1,3), (1, 4)}
Example 4. If A× B = {(3, 5), (3, 6), (3, 7), (4, 5), (4, 6), (4, 7)}, then find A and B.
Solution : Here, A × B = {(3, 5), (3, 6), (3, 7), (4, 5), (4, 6), (4, 7)} by definition of
cartesian product A× B,
we have,
A = the set of the first elements of cartesian product. = {3, 4}
B = The set of the second elements of the cartesian product.
= {5, 6, 7}
Example 5. If A = {1, 2}, B = {3, 4}, C = {4, 5}.
Then verify that A × (B C) = (A × B) (A× C)
Solution : Here, A = {1, 2}, B = {3, 4}, C = {4, 5}
Now, A × B = {1, 2} × {3, 4} = {(1, 3), (1, 4), (2, 3), (2, 4)}
A × C = {1, 2} × {4, 5} = {(1, 4), (1, 5), (2, 4), (2, 5)}
B C = {3, 4, 5}
LHS = A × (B C) = {1, 2} × {3, 4, 5}
= {(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)}
? RHS = (A × B) (A × C) = {(1, 3), (1, 4), (2, 3), (2, 4), (1, 5), (2, 5)}
? LHS = RHS, Verified.
Exercise 1.2
Very Short Questions :
1. (a) Define cartesian product of sets A and B.
(b) If the cardinality of set A is 3 and that of B is 4, what is the cardinality of A×B?
(c) Under which condition A × B is not defined.
(d) Under which condition A× B = B × A.
2. (a) If A = {1, 2, 3}, B = {7}, find A × B.
(b) If A = {1, 2}, B = {6, 7}, find A × B.
(c) If P = {1, 2, 3}, Q = {10}, find n(P × Q).
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 11
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
Short Questions :
3. (a) If A = {4, 5} and B = {8, 10}, find A × B and B × A.
(b) If A = {1, 4} and B = {4, 1}, then verify that A × B = B × A.
(c) If A = {5, 10} and B = {10, 5} verify, that A × B = B × A.
(d) If A = {2, 3} and B = {4, 5}, verify that A × B ≠ B × A.
(e) If P = {a, b, c} and Q = {x, y, z}, find P × Q and n(P × Q).
Long Questions :
4. (a) Let A × B = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6)} then, find A, B, n(A), n(B).
(b) If P × Q = {(a, x), (a, y), (a, z), (b, x), (b, y), (b, z)} find P and Q. Also find n(P),
n(Q), n(p × Q). Also show that n(P × Q) = n (P) × n (Q).
5. (a) Let A = {a, b}, B = {b, c, d} and C = {d, e}. Then, find:
(i) A × (B C) (ii) (A × B) (A × C)
(iii) (A – B) × (B – C) (iv) (A B) × (B C).
(b) If A = {4, 5}, B = {4, 5} and C = {5, 6} verify that A × (B C) = (A × B) (A × C)
6. If A = {x : x ≤ 4, x N}, B= {x : x ≤ 3, x N}, and C = {x : 3 ≤ x ≤ 5, x N}, then
verify the following
(a) A × (B C) = (A × B) (A × C) (b) A× (B C) = (A × B) (A × C)
(c) (A – B) × C = (A × C) – (B × C)
7. Let A = {1, 2, 3, 4} and B = {3, 4, 5}, then find A × B and represent them in the
following ways:
(a) Set of ordered pairs (b) Set Builder form
(c) Tree diagram method (d) Arrow diagram method
(e) Tabular method
1. (b) 12 (d) B = A 2.(a) {(1, 7), (2, 7), (3, 7)}
(b) {(1, 6), (1, 7), (2, 6), (2, 7) (c) 3
3. (a) A × B = {(4, 8), (4, 10), (5, 8), (5, 10)}, B × A = {(8, 4), (8, 5), (10, 4), (10, 5)}
(e) {(a, x), (a, y), (a, z), (b, x), (b, y), (b, z), (c, x), (c, y), (c, z)}, n(P × Q ) = 9
4. (a) A = {1, 2, 3}, B = {5, 6), n(A) = 3, n(B) = 2
(b) P = {a, b}, Q = (x, y, z), n(P) = 2, n(Q) = 3, n(P × Q) = 6
5. (a) (i) (a, d), (b, d)}
(ii) {(a, b), (a, c), (a, d), (b, b), (b, c), (b, d), (a, e), (b, e)}
(iii) {(a, b), (a, c)} (iv) {(b, d)}
7. (a) {(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5), (4, 3), (4, 4), (4, 5)}
(b) {(x, y) : x A and y B} (c), (d), (e) show to the teacher.
12 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
1.3. Relation
Introduction :
Let us study the following tables.
(a) x 1 2 3456
y 1 4 9 16 25 36
(b) A Nepal Bhutan Pakistan Shrilanka China
B Kathmandu Thimpu Islamabad Colombo Beijing
Is there any relation between the elements of the table ?
Obviously, there is relation between the elements of the table.
The table (a) shows that y is square of x. i.e. y = x2
The table (b) shows that the countries and their capital.
Let A = {1, 2, 3} and B = {1, 4, 9}
Then, A × B = {1, 2, 3} × {1, 4, 9}
= {(1, 1), (1, 4), (1, 9), (2, 1), (2, 4), (2, 9), (3, 1), (3, 4), (3, 9)}
Let R1 = {(x, y) : x + y ≤ 5}
= {(1, 1), (1, 4), (2, 1), (3, 1)}
R2 = {(x, y) : y = x2}
= {(1, 1), (2, 4), (3, 9)}
R3 = {(x, y) : y = x} = {(1, 1)}
Here, R1, R2 and R3 are subsets of the cartesian product A × B. These are called relation from
set A to set B.
Definition : A relation is defined as a subset of cartesian product of two non–empty sets. The
relation from set A to set B is denoted by R : A o B, where R ≤ (A ×B).
Also, we write R = {(x, y) : x A and y B} and xRy.
xRy is read as x is related to y and (x, y) R.
Ways of Representation of Relations:
A relation R from set A to set B can be represented in the following ways:
(a) Set of ordered pairs form. (b) Set builder form or rule method
(c) Tabulation method (d) Arrow diagram method.
(e) Graphic or Lattice method.
Let A = {1, 2, 3} and B = {1, 4, 9}. Let us define a relation R : A o B such that y = x2. The
relation from set A to set B can be represented in the following ways:
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 13
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
S.N Ways of Representation Representation
1. Set of ordered pair form R = {(1, 1), (2, 4), (3, 9)}
2. Set builder form R = {(x, y) : y = x2 , x A and y B}
3. Tabulation Method
4. Arrow diagram method R = {(x, y) : y = x2, x A and y B}
5. Graphic or Lattice method x 123
y 149
R B
A
1
1 4
2 9
3
Y
9 (3, 9)
8
7
6
5
4 (2, 4)
3
2
1 (1, 1)
X' O 1 2 3 4 5 6 X
Y'
Worked out Examples
Example 1. Let A {1, 2, 3} and B = {4, 5, 6}, find A × B and then find the following
Solution : relations from set A to set B. (a) x is less than y (b) x is equal to y (c) x is
half of y.
Given, A = {1, 2, 3}, B = {4, 5, 6}
Then A × B= {1, 2, 3} × {4, 5, 6}
= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(a) Let R1 = {(x, y) : x < y}
= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(b) Let R2 = {(x, y) : x = y} = I, there is no relation of this type.
"x is equal to y" relation is not defined from A to B.
(c) Let R3 = (x, y) : x = 1 y = {(2, 4), (3, 6)}
2
14 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
Example 2. Let A = {1, 2, 3} and B = {2, 4, 6} and a relation P is defined from set A to B
Solution : defined by y = 2x. Then represent the relation R in the following ways:
(a) Set of ordered pair (b) Set builder form
(c) Tabulation method (d) Arrow diagram method
(e) Graphic method
Here, A = {1, 2, 3}, B = {2, 4, 6}
A × B= {1, 2, 3} × {2, 4, 6}
= {(1, 2), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 2), (3, 4), (3, 6)}
Let R = {(x, y) : y = 2x}
Then, the relation R is presented in the following ways :
(a) Set of ordered pair. R = {(1, 2), (2, 4), (3, 6)}
(b) Set builder form R = {(x, y) : y = 2x}
(c) Tabulation Method
x1 2 3
y1 4 6
(d) Arrow diagram Method
R
AB
11
24
36
(e) Graphic method
Y
7
6
5
4
3
2
1
Y' O 1 2 3 4 5 6 X
Y'
Exercise 1.3 15
Very Short Questions :
1. (a) Define relation in terms of cartesian product.
(b) What are the ways of representation of relations?
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
2. If A = {a, b, c} and B = {x, y, z}. Which of the following, relations are relations from
set A to set B?
(a) R1 = {(a, x), (b, y), (c, z)}
(b) R2 = {(a, x), (b, y), (c, z), (a, y), (b, z)}
(c) R3 = {(a, x), (b, z), (x, c)}
Short Questions :
3. (a) If P = {1, 2} and Q = {4, 6}, find P × Q and relation from P to Q
such that x + y ≥ 4.
(b) If M = {1, 2, 3} and N = {2, 4, 5}, find a relation from M to N
such that x + y ≤ 10.
Long Questions :
4. If P × Q = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4)}, then find the
following relations:
(a) is greater than, i.e., {(x, y) : x > y)}
(b) is less than, i.e., {(x, y) : x < y)}
(c) is equal to, i.e., {(x, y) : x = y)}
(d) is square of, i.e., {(x, y) : x = y2)}
5. If P = {1, 2, 3} and Q = {2, 4, 6}, then from P × Q, find the following relations.
(a) is greater than
(b) is less than
(c) is double of
(d) is half of
6. Let P = {4, 8} and Q = {1, 2, 3, 4}. Find P × Q, and from it find the following
relations: (a) x + y = 8
(b) x + y ≥ 8
(c) x + y ≤ 8
(d) y= 1 x
2
7. Let R = {(1, 2), (2, 4), (3, 6), (4, 8)} be a relation from set A to set B. Then represent the
following relations in the following ways:
(a) Set of ordered pair (b) Mapping diagram
(c) Graphically (d) Tabulation
(e) Set builder form
Project Work
8. Write the name of your family members. Show the relation of each member with the
other members.
16 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
2. R1 and R2 3.(a) {(1, 4), (1, 6), (2, 4), (2, 6)}
(b) {(1, 2), (1, 4), (1, 5), (2, 2), (2, 4), (2, 5), (3, 2), (3, 4), (3, 5)}
4. (a) {(3, 2)} (b) {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}
(c) {(2, 2), (3, 3)} (d) not defined 5.(a) {(3, 2)}
(b) {(1, 2), (1, 4), (1, 6), (2, 4), (2, 6), (3, 4), (3, 6)} (c) not defined (d) {(2, 4), (3, 6)}
6. (a) {(4, 4)} (b) {(4, 4), (8, 1), (8, 2), (8, 3), (8, 4)}
(c) {(4, 1), (4, 2), (4, 3), (4, 4)} (d) {(4, 2), (8, 4)}
7. (a) {(1, 2), (2, 4), (3, 6), (4, 8)}
(b) A B (d) x 1 2 3 4
1 2 y2468
2 4
3 6
4 8
1.4. Domain and Range of a Relation
Let R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)} be a relation from set A to set B
where A = {1, 2, 3, 4, 5} and B = {2, 3, 4, 5, 6}.
Then the set of all the first element of each ordered pair is called domain of relation R.
? Domain of R = {1, 2, 3, 4, 5}
The set of all the second element of each ordered pair is called range of relation R.
? Range of R = {2, 3, 4, 5, 6}
Definition : If R is a relation defined from set A to set B then the set of all the first elements
of each of ordered pair of R is called domain and the set of all the second elements of each
of ordered pair of R is called range.
Types of Relations
There are various types of relations.
(a) Identity relation
Let A be non-empty set. Then, the relation R = {(a, a):a A} is known as the identity
relation on A and is denoted by IA.
Let A = {1, 2, 3, 4},
Then IA = {(1, 1), (2, 2), (3, 3), (4, 4)} is an identity relation.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 17
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
(2) Universal Relation
Let A be a non-empty set, then A × A is a relation known as the universal relation
on A.
Let A = {a, b, c}
Then A × A = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)} is a universal
relation on A.
(3) Reflexive Relation
A relation R on a set A is said to be reflexive if every element of A is related to itself.
Thus R is reflexive iff aRa i.e. (a, a) R for all a A.
Let A = {1, 3, 4}, then R = {(1, 1), (3, 3), (4, 4)} is a reflexive relation.
But R' = {(1, 1), (3, 3), (4, 1), (1, 3)} is not a reflexive relation because 4 A and
(4, 4) R'.
Again considering a set A = {a, b, c} if any one of the set (a, a), (b, b), (c, c) is absent in
the relation, the relation cannot be reflexive.
Hence, universal relation on a non-empty set is a reflexive. The identity relation on a
non-empty set is always a reflexive but reflexive relation is not necessarily an identity
relation.
(4) Transitive Relation
A relation R on a non–empty set is said to be transitive iff (a, b) R and
(b, c) R (a, c) R for all a, b, c, A.
Thus for a transitive relation on R on A, aRb and bRc aRc for all a, b, c A
Let A = {1, 2, 3}
Then, R = {(1, 2), (2, 3), (1, 3)} is a transitive relation.
(5) Symmetric Relation
A relation R on the set A is said to be symmetric if (a, b) R (b, a) R for all
a, b A. Thus for symmetric relation R on set A,
aRb bRa for all a, b A.
Let A = {a, b, c, d}
Then, R = {(a, b), (b, a), (c, c)} is a symmetric relation because (a, b) R, (b, a) R,
(c, c) R.
But R' = {(b, b), (b, a), (c, c), (d, a)} is not a symmetric because (a, d) R' and
(d, a) R (a, d) R
(6) Equivalence Relation
A relation R on a set A is called an equivalence relation iff
(i) It is reflexive, i.e., for every a A, (a, a) R
(ii) It is symmetric. For all a, b, c A, (a, b) R, (b, c) R (a, c) R
(iii) It is transitive i.e. a, b A, (a, b) R (b, a) R
18 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
Equivalence relations are generally denoted by the symbol ~.
Example: Similarity of triangle in the plane geometry is an equivalence relation.
Inverse Relation:
Let R = {(1, 1), (2, 4), (3, 9)} be a relation defined from set A to set B.
R
AB
11
24
39
Then interchanging the elements of each ordered pair of R1 we get a relation from set B to
set. This relation is called inverse relation of R and it is denoted by R–1.
? R–1 = {(1, 1) (4, 2), (9, 3)}
Definition : Let R be a relation defined from set A to set B, then the relation obtained by
interchanging the elements of each ordered pair of R is called inverse of relation R. It is
denoted by R–1.
Symbolically,
If R = {(x, y): x A and y B}, then
R–1 = {(y, x) : x A and y B}.
Worked Out Examples
Example 1. R = {(1, 3), (2, 4), (3, 5), (4, 6)} be a relation. Then find the domain and range
Solution : of R. Also find inverse of relation R.
Here, R = {(1, 3), (2, 4), (3, 5), (4, 6)}
Example 2. Domain of R = {1, 2, 3, 4}
Solution : Range of R = {3, 4, 5, 6}
Inverse of relation R is R–1 = {(3, 1), (4, 2), (5, 3), (6, 4)}
Let R = {(x, y): y = 2x, 1 ≤ x ≤ 10, x N}, Find the domain and range of R.
Also find R–1.
Here, R = {(x, y) : y = 2x, 1 ≤ x ≤ 4, x N}
Domain of R = {1, 2, 3, 4}
Put x = 1, 2, 3, 4 successively in y = 2x,
We get, y = 2, 4, 6, 8
? Range of R = {2, 4, 6, 8}
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 19
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
Example 3. To find inverse of R, we have,
Solution : R = {(1, 2), (2, 4), (3, 6), (4, 8)}
? R–1 = {(2, 1), (4, 2), (6, 3), (8, 4)}
Three relations R1, R2 and R3 are defined on set A = {–1, 0, 1} as follows :
(a) R1 = {(–1, –1), (0, 0), (1, 1)}
(b) R2 = {(–1, 0), (0, 1), (1, –1), (0, –1), (1, 0), (–1, 1)}
(c) R3 = {(0, 0), (0, 1), (1, 1), (0, 1), (1, 0)}
Find which of the relations on A is reflexive, symmetric, or transitive.
(a) Here, R1 = {(–1, –1), (0, 0), (1, 1)}
Since each of element of the set is related to itself, R1 is a reflexive.
(b) Here, R2 = {(–1, 0), (0, 1), (1, –1), (0, –1), (1, 0), (–1, 1)}
Since, (–1, 0) R2 and (0, –1) R2
(1. 0) R2 and (0, 1) R2
(1, – 1) R2 and (–1, 1) R2.
Hence, R2 is a symmetric relation.
Again, (–1, 0), (0, 1), (–1, 1) R2 . R2 in also a transitive relation.
(c) Here, R3 = {(0, 0), (0, 1), (1, 1), (1, 0)}
Since (0, 1) R3 and (1, 0) R3.
? R3 is symmetric relation.
Exercise 1..4
Very Short Questions :
1. (a) Define domain and range of a relation.
(b) Define the following relations:
(i) Reflexive Relation (ii) Symmetric Relation
(iii) Transitive Relation (iv) Equivalent relation
Short Questions :
2. Find the domain and range of the following relations:
(a) R1 = {(a, x), (b, y), (c, z)} (b) R2 = {(2, 4), (4, 8), (8, 16)}
(c) R3 = {(1, 2), (2, 4), (3, 6), (4, 8)}
3. From the adjoint figure, write the relation R in set of ordered pairs. Then find the
domain and range of R. Also find the inverse of R.
20 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
R
26
4 12
5
6 10
4. (a) If R = {(x, y) : y = x , x = 2, 4, 6}, find the range of R.
2
(b) If R = {(x, y) : y = x + 3, and x = 1, 2, 3}, find the range of R.
5. (a) A relation R : A o B, defined by:
R = {(x, y) : y = x – 1, x N, x ≤ 5}.
Find the domain and Range of R.
(b) A relation R : A o B is defined by:
R= {(x, y) : y = 2x + 5, 0 ≤ x ≤ 5, x W}.
Find the domain and range of R.
6. (a) If A = {1, 2, 3} and a relation. R is defined on A as R = {(1, 1), (2, 2), (3, 3),
(1, 3), (2, 3), (3, 1), (3, 2), (1, 2), (2, 1)}. Check whether R is reflexive, symmetric
or transitive relation.
(b) Let R = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}, Show that R is
reflexive, symmetric, and transitive relation and it is in equivalence relation.
Long Questions :
7. Let a relation R:AoB be defined by R = {(x, y) : x + y = 8}, where x A = {1, 2, 3, 4}.
(a) Write the relation in set of ordered pair form.
(b) Find the range of R.
(c) Find the inverse relation of R.
(d) Show the relation R on mapping diagram.
8. Three relations are defined on A = {1, 2, 3}:
(a) R1 = {(1, 2), (2, 3), (1, 3)}
(b) R2 = {(1, 1), (2, 2), (3, 3)}
(c) R3 = {(1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)}
Write the types of each of above relations with reasons.
9. Let A = {1, 2, 3}, find reflexive, symmetric, transitive, and equivalence relation.
10. Let R = {(1, 1), (2, 1), (1, 3), (2, 1), (2, 2). (2, 3), (3, 1), (3, 2), (3, 3)}. Show that R is
equivalence relation.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 21
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
2. (a) D = {a, b, c}, R = {x, y, z} (b) D = {2, 4, 8}, R = {4, 8, 16}
(c) D = {1, 2, 3, 4}, R = {2, 4, 6, 8}
3. R = {(2, 6), (4, 6), (6, 12), (5, 10)}, D = {2, 4, 6, 5}, R = {6, 12, 10}
R–1 = {(6, 2), (6, 4), (10, 5), (12, 6)} 4.(a) R = {1, 2, 3} (b) {4, 5, 6}
5. (a) D = {1, 2, 3, 4, 5}, R = {0, 1, 2, 3, 4}
(b) D = {0, 1, 2, 3, 4, 5}, R = {5, 7, 9, 11, 13, 15}
6. (a) reflexive, symmetric and transitive 7.(a) {(1, 7), (2, 6), (3, 5), (4, 4)}
(b) R = {4, 5, 6, 7} (c) R–1 = {(7, 1), (6, 2), (5, 3), (4, 4)}
8. (a) transitive (b) reflexive (c) symmetric
9. R1 = {(1, 1), (2, 2), (3, 3)} - reflexive
R2 = {(1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)) - symmetric
R3 = {(1, 2), (2, 3), (1, 3)} - transitive
R4 = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)} - equivalence
1.5. Functions
A. An elephant goes to a river everyday and it drinks 20 litre water. Let us tabulate the
number of days and quantity of water that the elephant drinks.
Days 1 2 3 ........ 10 15 20
Quantity of water (liter) 20 40 60 ........ 200 ........ .........
If x be the number of days and y be the quantity of water drunk by elephant, then the
relation between x and y can be written as y = 20x.
Y
Water (liter) 220
200
180 X
160
140
120
100
80
60
40
20
X' O 1 2 3 4 5 6 7 8 9 10 11
Days
Y'
Is there any relation between the number of days and the quantity of water drunk by
elephant ?
22 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
B. Let us observe the following relations defined from the set A to set B.
R1 R2
AB AB
1 12 4
2 43
3 94 8
Fig. (i) B Fig. (ii) B
R3 2 R4 4
A 4 A 5
6
1 1
2 2
3 3
Fig. (iii) Fig. (iv)
R5 B
A
14
2
35
Fig. (v)
Above all stated arrow/mapping diagrams are relations. But all these relations may not be
functions. Let us discuss each of above mapping diagrams :
Fig (i) Each element of the first set A is associated with a unique element of the second
Fig (ii) set B. Hence, R1 represents a function.
Fig (iii)
Each element of the first set A is associated with unique element of the second set
B. Hence, R2 represents a function from set A to set B.
One element 3 of the first set A is not associated with any element of the second
set B. Hence, R3 does not represent a function from set A to set B. All the elements
must be associated with unique element of the second set.
Fig (iv) An element '1' of set A is associated with two elements of 4 and 5 of the second
set B. Hence, R4 does not represent a function from set A to set B. If one element of
the first set is associated with more than one element of the second set the relation
will not be a function.
Fig (v) All the elements of the first set A are mapped with unique element of the second
set B. Hence, R5 represents a function. There may be some elements in the second
set which is not associated with any element of the first set.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 23
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
Definition : Let A and B be two non–empty sets. Then, a relation from set A to set B is said
to be a function if every element of the first set A associates with unique elements of the
second set B. Symbolically, we write f : A o B to represent a function f from set A to set B.
For every x A, there corresponds a unique element y = f(x) B. In this function, y = f(x)
B is known as image of x under f and x is known as pre–image of y. Functions are denoted
by f, g, h, k etc.
Alternative definition : A function is a special type of relation in which every element of
the first set has one and only one image in the second set. In a function, no two or more
ordered pairs have the same first element.
Note :
In a function f : A o B
1. Each element of set A must be associated with unique element of set B.
2. Two or more elements of set A may be associated with same element of B.
3. There may be some elements of the set B which are not associated with any element
of the set B.
Domain and Range of a Function :
Let f be a function defined from set A to set B, i.e., f : A o B. Then the set A is called domain
and set B is the co–domain of f. The set of elements of B which are the images of the elements
of set A is known as the range of f.
Example : Let A = {1, 2, 3, 4}, B = {1, 4, 9, 16} Then, f : A o B be is defined by
f = {(1, 1), (2, 4), (3, 9), (4, 16)}
Find the domain co-domain and range of f.
Solution : Here, f = {(1, 1), (2, 4), (3, 9), (4, 16)}
Domain of f = {1, 2, 3, 4}
Codomain of f = {1, 4, 9, 16}
Range of f = {1, 4, 9, 16}
Worked Out Examples
Example 1. Which of the following relations are functions? Give reasons.
Solution :
(a) f = {(1, 2), (2, 4), (3, 6)}
(b) g = {(1, 2), (1, 3), (1, 4), (3, 2), (4, 1)}
(a) Here, f = {(1, 2), (2, 4), (3, 6)}. The relation f is function because every
element of the first component has a unique image.
(b) Here, g = {(1, 2), (1, 3), (1, 4), (3, 2), (4, 1)}. The relation g does not
represent a function because the first component of the three order
pairs have same image.
24 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
Example 2. From the following mapping diagrams, state with reasons which represent a
function.
(a) f (b) g
A BP Q
4
1 12
2 43 6
3 94
(c) h (d) k
M NI
J
a x1 9
b y2
c 16
3
Solution : (a) The relation f is function because every element of the first set A is
Example 3. associated with unique elements of the second set B.
Solution:
(b) The relation g is a function because every element of the first set P is
associated with unique element of the second set Q.
(c) The relation h does not represent a function because one of the element
c of M has no image in the second set N.
(d) The relation k represents a function because each element of the first
set I is associated with unique element of the second of the set J.
Let A = {1, 2, 3, 4} and B = {1, 4, 9, 16} and a function f : A o B is defined
f = {(1, 1), (2, 4), (3, 9), (4, 16)}.
Represent the function in the following methods:
(a) Equation method
(b) Set builder form method
(c) Arrow diagram method
(d) Table method
(e) Set of order pair
(f) Graphical method
Here, f = {(1, 1), (2, 4), (3, 9), (4, 16)}.
Let us represent the above function in the following ways:
(a) Equation method :
y = x2
(b) Set builder form method :
f = {(x, y) : y = x2}
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 25
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
(c) Arrow diagram method : 34
A fB 9 16
11
24 Y X
39 18
4 16 16
14
(d) Table method : 12
x12 10
y14 8
6
(e) Set of ordered pair : 4
Given function is in this form, 2
f = {(1, 1), (2, 4), (3, 9), (4, 16)}
O 123456
(f) Graphical method :
X'
Y'
Exercise 1.5(A)
Very Short Questions :
1. (a) Define a function.
(b) Define domain and range of a function.
(c) If f is a function defined from set A to set B, write it in set relation.
2. (a) Write the range of function f = {(1, 1), (4, 16), (3, 9)}.
(b) Write the range of function g = {(2, 4), (3, 9), (4, 16)}
3. (a) In the given mapping diagram f B
f A
1
A B1 8
27
1 12 64
2 43
fig. (i) fig. (ii)
(i) What is the image of 2 ?
(ii) What is the pre-image of 4?
26 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
(b) In the mapping diagram (ii), find domain, co-domain, and range of the function.
Short Questions :
4. Which of the following relations are functions ? State with reasons.
(a) f = {(a, x), (b, y), (c, z)}
(b) g = {(4, 1), (4, 2), (4, 4), (2, 3)}
(c) h = {(1, 5), (2, 6), (3, 7)}
(d) k = {(1, 2), (2, 3), (4, 5), (2, 8)}
5. State whether the following mapping diagrams represent the functions or not. Give
reasons as well.
(a) f (b) y
A B
C D
14 x
2 8a y
3 12 z
(c) h (d) k
E f FX Y
0 2 2 1
1 8 4 2
4 10 6 3
5 4
(e) Q
P x
a y
b
c
6. Let f : A o B be a function with A = {1, 2, 3, 4} and B = 1 , 1, 3 , 2 .
2 2
Find the function defined is double of. Then represent the function in the following
ways :
(a) Arrow Diagram (b) Set - builder form
(c) Formula (d) Set of ordered pairs (e) Graph
Project Work
7. List any five examples of functions used in our daily life with reasons.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 27
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
1. (c) f : A o B 2.(a) Range = {1, 9, 16} (b) Range = {4, 9, 16}
3. (a) (i) 4 (ii) 2 (b) Domain = {1, 2, 3}, Range = {1, 8, 27}, Co-dom = {1, 8, 27}
4. (a) Function (b) Not function (c) Function (d) Function
(e) Not function 5.(a) Function (b) Not function (c) Not function
(d) Function (e) Function
6. (a) A f (b) f = {(x, y) : x A and y B}
B
1 (c) f= (x, y) : y = 1 x
2 1 2
3 2
4 1 (d) f = 1, 1 , (2, 1), 3, 3 , (4, 2)
3 2 2
2
2 (f) Show to your teacher.
Vertical Line Test
All relations are not functions. The relations can be tested whether it is a function. The
following are some ways of testing them:
(a) For a relation to be a function, all the elements of the domain must have image in the
Co-domain.
(b) If a relation is expressed in set of ordered pair form, examine whether the first element
of all the ordered pair are different or not. If all the first elements of the ordered pair are
different, the relation is a function.
(c) Each element of the first set must have only one image.
There is a special way of testing of a function if it is represented in diagrams. This
method is called vertical line test.
For this test, a vertical line is drawn in the graph at any point. If the vertical line cuts the
graph at only one point, it is a function and if it cuts at more than one point, it is not a
function.
Worked Out Examples
Example 1. State which of the following graphs represent functions by using vertical line test.
(a) Y (b) Y
X' O X
X
X' O
Y' Y'
28 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
Solution : (a) Since the vertical line (dotted line) cuts at only one point, the curve
(parabola) represents a function.
(b) Since the vertical line cuts the curve (circle) at two points, the curve
does not represent any function.
Exercise 1.5(B)
By using the vertical line test, find which of the following represent function.
(a) Y (b) Y
X' O X X' O X
(c) Y' Y'
Y
(d) Y
X' O X X' O X
X
Y' Y' X
(e) Y (f) Y 29
X' O X X' O
Y'
Y' (h) Y
(g) Y O
X' O X X' Y'
Y'
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 (j) Relations and Functions
(i) Y
Y
X' O X X' O X
X
Y' Y'
(k) Y (l) Y
X' OX X' O
(m) Y' Y'
Y
(n) Y
X' O X
X' O X
Y' Y'
1. (a) Function (b) Function (c) Not function (d) Not function
(e) Not function (f) Not function (g) Function (h) Function
(i) Function (j) Not function (k) Function (l) Function
(m) Function (n) Function
Types of Functions A f
There are different types of functions: B
(a) Onto Function : 1 2
2 4
Let f : A o B be a function as shown in 3 6
the arrow diagram. In this function,
Co-domain = {2, 4, 6}
30 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
Range = {2, 4, 6} C g D
a x
Here, co-domain is equal to range of the function. b fig. (ii) y
This type of function is called onto function. c
Definition : A function f : A o B is said to be an onto
function if the range is equal to co-domain of the function.
It means that every elements of the co-domain is mapped
with the elements of the domain.
Example : In the diagram, g is an onto function as co-domain = Range = {x, y}
(b) Into Function : A f B
1 1
In mapping diagrams, some elements of the second 2 fig. (i) 8
set are not mapped with the elements of domain. In 3 27
figure (i) 64 of co-domain B does not have its pre– g 64
image in set A. This type of function is called into C
function. Again, in figure (ii), y and z do not have a fig. (ii) D
their pre - images in the first set C. This is also into b x
function. c y
.z
Definition : A function f : A o B is said to be an into
function if the range is proper subset of co-domain. It
means that some elements of the second set do not have
their pre-images in the first set.
In above diagram (i),
Co–domain = {1, 8, 27, 64}
Range = {1, 8, 27}
i.e., Range is proper subset of co-domain. Similarly, fig (ii) is the range of g is proper
subset of co–domain.
(c) One to one function : B g
f x CD
y
A z 12
24
a 39
b 16
c
(Fig-i) (Fig-ii)
In above mapping diagrams, there is one to one correspondence between the elements
of the domain and co-domain. Such type of functions are called one to one function. As
shown in the diagram, f and g both are one to one function.
Definition : A function f : A o B is said to be one to one function if every element of the
domain has distinct image in the co-domain. In above diagrams, f is one to one onto function
and g is one to one into function. Why ?
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 31
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
(d) Many to one Function : B g
f CD
A
1 2 a
2 6 bx
3 9 c
dy
(Fig-i)
(Fig-ii)
In above diagrams, two or more elements of the domain are mapped with unique element of
the co-domain. Such type of functions are called many to one function.
In above mapping diagrams,
(i) The elements 1 and 2 of domain have the same image 2. Hence, f is many to one function.
(ii) The elements a, b, and c have same image x; hence, g is also many to one function.
In above diagrams, f is many to one into function and g is many to one onto function.
Give reason.
Definition : A function f : A o B is said to be many to one function if at least two elements
of the domain have a unique image in co-domain.
Algebraic Functions :
(a) Constant Function :
A function f : A o B is said to be a constant function if every element of the domain A
have the same image.
Y
f y=5
A
B
1
2 1 X' O X
3
Y'
Here, the line y = 5 ie. f(x) = 5 represents a constant function as f(1) = 5, f(3) = 5, i.e.
for all values of x, the image y is constant. It is also a linear function represented by
y=c or, f(x) = c
(b) Identity function :
A function f : A o B is said to be an identity function if every element of f maps into
itself. The identity function has same domain and range.
Y
f y=x
1 1 X' O X
2 2
3 3 fig. (ii) Y'
fig. (i)
32 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
In fig (i), f = {(1, 1), (2, 2), (3, 3)}, every element is associated with itself. Hence, f is an
identify function.
In fig (i), the line y = x passes through the points as (1, 1,), (2, 2), (–1, –1) and so on. It
also shows that every element is associated with itself.
(c) Linear Function :
A algebraic function of the first degree of form y = mx + c, which represents a straight
is called linear function. Examples : y = 2x + 6, y = 4x etc.
f Y
CD y = 2x + 6
–1 4
06 10 (2, 10)
18
2 10 (1, 8)
(-1, 4)5 (0, 6)
X' (-3, 0) O 5 X
(Fig-i) Y' (Fig-ii)
In mapping diagram (i), f = {(–1, 4) (0, 6), (1, 8), (2, 8)} represents a straight line as it is
plotted in graph which (Fig ii)
Worked Out Examples
Example 1. Classify the following functions shown in the diagrams with reasons:
(a) Y (b) x
(0, 4) a
X' b
(3, 0) c
OX d
4x + 3y = 12
Y'
Solution : Fig (a) : It is one to one onto function as every element of the domain has
distinct image in codomain and range and codomain are equal.
Fig (b) : It represent a many to one function as a, b, c, d of the domain has
same image x.
Exercise 1.5(C) 33
Very Short Questions
1. Define the following functions:
(a) Many to one function.
(b) One to one function
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
(c) Onto function
(d) Into function
(e) Linear Function
(f) Constant function
(g) Bijective Function
2. Stating reasons, classify the following functions as shown in diagrams:
(a) fB (b) f D
A C x
1 3 a y
5 b z
2 6 c w
3
(c) f B (d)
A x 1
a y 2 1
b z 3
c
d
(e) Y (f) Y y = 4
X' (4, 0) X X' O X
O
(0, -4) Y'
Y'
Y
(g) Y (h)
(0, 3)
X' y=x X' (2, 0)
(-3, -3)
(3, 3) O
O (1, 1) X X
Y' Y'
3. Draw graphs of the following functions and also state the type of the function with
reasons:
(a) f = {(1, 2), (2, 3), (3, 4)}
(b) g = {(4, 10), (5, 10), (6, 10), (7, 10)}
(c) h = {(1, 1), (2, 2), (3, 3), (4, 4)}
(d) k = {(1, 1), (2, 4), (3, 9), (4, 16)}
34 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
2. (a) One to one onto (b) One to one into
(c) Many to one onto (d) Many to one onto
(e) Linear function (f) Constant function
(g) Linear function (h) Linear function
(b) Constant function
3. (a) Linear function (d) Quadratic function
(c) Identity function
Functional Value
Let us consider a linear function defined by,
y = f(x) = 2x + 3
i.e. f(x) = 2x + 3
Putting the values of x = 1, 2, 3, 4 successively, we get,
f(1) = 2.1 + 3 = 5 f(2) = 2.2 + 3 = 7
f(3) = 2.3 + 3 = 9 f(4) = 2.4 + 3 = 11
The values 5, 7, 9, 11 are called functional values of f(x) at x = 1, x = 2, x = 3, x = 4
respectively. The set values of x is called range of the function. We can also write above
values as,
For x = 1, y = 5 For x = 2, y = 7
For x = 3, y = 9 For x = 4, y = 11
The above function can be written as,
f = {(1, f(1)), (2, f(2)), (3, f(3)), (4, f(4))} Y
or, f = {(1, 5) (2, 7), (3, 9), (4, 11)} 11 y = 2x + 3
10
The relation between x and y is called functional X
relationship. The above ordered pairs can be plotted 9
in a graph. 8
7
Real valued Functions : 6
5
A function f : A o B is called real values function if each X' 4
3
2
1
O 123456
element 'a' of set A associates with a unique real number Y'
f(a) of set B. Here, A and B subsets of the set of real
numbers R. Real valued functions are simply called real function. Here, we study only real
valued function.
Example : f : R o R defined by f(a) = x2 + 3x + 2, x R is a real function. For every value
of x, f(x) gives a real value. For x = 4 R, f (4) = 42 + 3.4 + 2 = 30 R.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 35
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
Worked out Examples
Example 1. If f(0) = 1, f(1) = 2, f(2) = 3, f(3) = 4. Write function f in set of ordered pairs
Solution : and hence the domain and range of the function.
Example 2. Here, f(0) = 1, f(1) = 2, f(2) = 3, f(3) = 4. in set of ordered pair, f can be
Solution : written as, f = {(0, f(0)), (1, f(1)), (2, f(2)), (3, f(3))}
Example 3. = {(0, 1), (1, 2), (2, 3), (3, 4)}
Solution :
Domain of f = {0, 1, 2, 3}
Example 4.
Solution : Range of f = {1, 2, 3, 4}
Let f : A o B be a function with A = {–2, –1, 0, 1, 2} and f(x) = x + 1. Find
the domain and range of the function f.
Here, f(x) = x + 1
Now, x A = {–2, –1, 0, 1, 2}
f(–2) = –2 + 1 = – 1
f(–1) = –1 + 1 = 0
f (0) = 0 + 1 = 1
f (1) = 1 + 1 = 2
f (2) = 2 + 1 = 3
Domain A is itself, i.e., domain of f = {1, 2, 3, 4}
Range = {–1, 0, 1, 2, 3}
If f(x) = 4x + 1, find the pre-image of 9.
Here, f (x) = 4x + 1
9 is image of x. i.e. x is pre – image of y or f(x)
f(x) = 4x + 1
or, 9 = 4x + 1
or, 9 – 1 = 4x
? x=2 Hence, pre-image of 9 is 2.
If f(x + 3) = 4x – 3, find the value of f(2).
Here, f(x + 3) = 4x – 3,
or, f(x + 3) = 4(x + 3) – 15
Replacing x + 3 by x, we get,
f (x) = 4x – 15
Put x = 2, we get,
f(2) = 4 . 2 – 15 = 8 – 15 = – 7
36 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
Alternative Method:
Here, f(x + 3) = 4x – 3.
Replacing x by x – 3, we get,
f(x – 3 + 3) = 4(x – 3) – 3
or, f(x) = 4x – 15
Put x = 2, f(2) = 4.2 – 15 = – 7
Example 5. If f(x + 3) = f(x) + f(3), x R then prove that : (i) f(0), (ii) f(–3) = –f(3).
Solution : Here, f(x + 3) = f(x) + f(3)
(i) Put x = 0, we get,
Example 6.
Solution : f(0 + 3) = f(0) + f(3)
or f(3) = f(0) + f(3)
? f(0) = 0 ....... (i)
(ii) Again put x = –3, we get,
f(–3 + 3) = f(–3) + f(3)
or, f(0) = f(–3) + f(3) Proved.
by using (i), we get,
0 = f(–3) + f(3)
? f (–3) = – f(3)
If f(x) = px + q, f(–3) = –4 and f(3) = 2, find the values of p and q.
Here, f(x) = px + q
Now, f(–3) = – 3p + q
or, – 4 = –3p + q ........... (i)
f(3) = 2
or, 2 = 3p + q .............. (ii)
adding (i) and (ii), we get, q = –1
put the value of q in equation (i), we get,
–4 = –3p – 1
or, p = 1
Required values of p and q are respectively 1 and –1.
Example 7. If f(x) = 3x2 – 2x + 5, and g(x) = 2x2 + 3x – 1 and f(x) = g(x), find the value
Solution : of x.
Here, f(x) = 3x2 – 2x + 5
g(x) = 2x2 + 3x – 1
But, given that f(x) = g(x)
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 37
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
3x2 – 2x + 5 = 2x2 + 3x – 1 i.e. x = 2
or, x2 – 5x + 6 = 0
or, x2 – 3x – 2x + 6 = 0
or, x(x – 3) – 2 (x – 3) = 0
or, (x – 3) (x – 2) = 0
Either : (x – 3) = 0 i.e. x = 3
or, x – 2 = 0
? x = 2, 3
Example 8. For what value of element of domain has the image 1 under the function
Solution : f(x) = 3x2 – 8x + 5x R ?
Example 9. Here, f(x) = 3x2 – 8x + 5
Solution :
For Image f(x) = 1, we write
i.e. f(x) = 3x2 – 8x + 5
or, 1 = 3x2 – 8x + 5
or, 3x2 – 8x + 4 = 0
or, 3x2 – 6x – 2x + 4 = 0
or, 3x (x–2) – 2(x –2) = 0
or, (x – 2) (3x – 2) = 0
Either x – 2 = 0 i.e. x=2
or, 3x – 2 = 0 i.e. x = 2
3
Let f : R o R be defined by:
3 + 2x, for – 2 ≤ x < 0
f(x) 3 – 2x 3
–3 – 2x for 0 ≤ x < 2
3
for x ≥ 2
3
Find f – 3 , f(0) , f(2)
2
2
For – 3 ≤ x < 0, we take,
f(x) = 3 + 2x
when, x = – 2 ,
3
2 2 4 5
f – 3 = 3 + 2 – 3 = 3 – 3 = 3
For 0 ≤ x < 2 , we take,
3
f(x) = 3 – 2x
38 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
when, x = 0 f(0) = 3 – 2.0 = 3
For x ≥ 2 , we take f(x) = –3 – 2x
3
when x = 2, f (2) = – 3 – 2.2 = –7
Exercise 1.5(D)
Very Short Questions
1. (a) If f(x) = 4x + 3, then find f(2) and f(–2)
(b) If g(x) = 2x2 + 4, then find g(–2) and f(2).
(c) If h(x) = 4x3 + 5, then find h(–2) and h(4).
2. (a) If f(x) = 3x + 2 , find (1) and f(2).
2x + 4
x–2
(b) If f(x) = x+3 , find f(4) and f(–4).
(d) If f(x) = x2 – 3x + 1 , find f(1) and f(2).
x + 1
Short Questions
3. (a) If f(2) = 4, f(3) = 9, f(4), = 16, then find the domain and range of the function.
(b) If g(1) = 4, g(2) = 5, g(3) = 6, g(4) = 7, then find the domain and range of the
function.
4. (a) If g(–1) = 1, g(–2) = 4, g(2) = 4, g(1) = 1, write the function g in set of ordered pair
and find the domain and range of the function.
(b) If f 1 = 1, f 1 = 2 , f 2 = 4 , write the function f in set of ordered pair and find
2 3 3 3 3
the domain and range of the function.
5. (a) Let f: A o B be a function with A = {1, 2, 3, 4} and f(x) = x2 + 1, then find the
range of the function.
(b) Let f : A o B be a function with A = {–2, –1, 0, 1, 2} and f(x) = x2 + 1. Find the
range of the function.
6. (a) If f(x) = 3x + 4, find the image of 4.
(b) If f(x) = 7x + 5, find the image of –1.
(c) If f(x) = 4x + 1, find the pre-image of 13.
(d) If g(x) = 2x + 5, find the pre-image of 15.
7. (a) Let f(x) = x + 4 be a function defined in the interval –5 ≤ x ≤ 5, find f(–2), f(2),
f(3), f(7).
(b) If g(x) = x2 + 5 defined – 6 ≤ x ≤ 7, find g(2), g(7), g(8).
8. (a) A function f : N o N is defined by f(x) = x + 3, is f one to one and onto or both ?
What is the pre-image of 100.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 39
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
(b) If f : N o N is denoted by f(x) = x2 + 1. What types of the function is f ? Write with
examples.
9. (a) If f = (x, y) : y = x and x {2, 4, 6}, find the range of f.
2
(b) If f(x) = {(x, y) : y= x + 3}, and x {1, 2, 3}, find the range of f.
10. If the domain of function f is D, find the range of the function in the following :
(a) f(x) = 2x + 4, D = {1, 2, 3, 4}
(b) f(x) = 3x + 2, D = {–1, 0, 1, 2 }
(c) f(x) = 1 – 4x, D = {–2, –1, 0, 1, 2}
Long Questions
11. (a) If f(x + 5) = f(x) + f(5), x R, prove that :
(i) f(0) (ii) f(–5) = – f(5)
(b) If f(x + 7) = f(x) + f(7), x R, prove that :
(i) f(0) = 0 (ii) f(7) = –f(7)
(c) If f(x + 3) = f(x) + f(9), x R, prove that :
(i) f(6) = 0 (ii) f(3) = – f(9)
(d) If f(x + 8) = f(x) + f(10), x R, prove that :
(i) f(2) = 0 (ii) f(–6) = –f(10)
(e) If f(x + 4) = f(x) + f(4), then prove that :
(i) f(0) = 0 (ii) f(–4) = f(4)
12. (a) If f(x + 2) = 4x + 3, find f(x) and f(2).
(b) If f(3x + 2) = 6x + 7, find f(x) and f(3).
(c) If f(x + 2) = 5x – 10, find f(x) and f(5).
(d) If f(x + 1) = 4x + 8, find f(x) and f(5).
(e) If f(2x – 3) = 4x + 5, find f(x) and f(–2).
13. (a) If f(x) = ax + b, f(5) = 26 and f(2) = 14, find the values of a and b.
(b) If f(x) = mx + c, f(4) = 11 and f(5) = 13, find the value of m and c.
14. If f(x) = 4x2 – 2x + 5 and g(x) = 7x2 – x + 3 and f(x) = g(x), find the values of x.
15. If f(x) = 9x, then prove that : f(m + n + p) = f(m). f(n). f(p).
16. For what value of domain has its 2 under the function f(x) = x2 + 6x + 7
17. A function g : R o R is defined by
x–1 for 0 ≤ x < 4
g(x) = 2x + 1 for 4 ≤ x < 6
3x + 10 for 6 ≤ x < 8
Find the values of f(2), f(5) and (7). Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
40
vedanta Excel in Opt. Mathematics - Book 9 Relations and Functions
18. IF f(x) = 2x2 + 5, find
(a) f(a + h)
(b) f(a + h) – f(a) , h > 0.
h
1. (a) 11, –5 (b) 12, 12 (c) –27, 261
2. (a) 5 , 1 (b) 2 , 6 (c) – 1 , – 1
6 7 2 3
3. (a) Domain = {2, 3, 4}, Range = {4, 9, 16}
(b) Domain = {1, 2, 3, 4}, Range = {4, 5, 6, 7}
4. (a) {(–1, 1), (–2, 4), (2,4), (1, 1)}, Domain = {–1, –2, 2, 1}, Range = {1, 4}
(b) 1 , 1 , 1 , 2 , 2 , 4 , Domain = 1 , 1 , 2 , Range = 1, 2 , 4
2 3 3 3 3 2 3 3 3 3
5. (a) {2, 5, 10, 17} (b) {1, 2, 5} 6.(a) 16 (b) –2
(c) 3 (d) 5 7.(a) 2, 6, 7 f(7) is not defined.
(b) 9, 54, g(8) is not defined. 8.(a) one-one into function, 97
(b) one-one into function 9.(a) {1, 2, 3} (b) {4, 5, 6}
10. (a) {6, 8, 10, 12} (b) {–1, 2, 5, 8} (c) {–7, –3, 1, 5, 9}
12. (a) f(x) = 4x – 5, 3 (b) f(x)= 2x + 3, 9 (c) f(x) = 5x – 20, 5
(d) f(x) = 4x + 4, 24 (e) f(x) = 2x + 11, 7
13. (a) a = 4, b = 6 (b) m = 2,c = 3 14. 2 , –1 16. –1, –5
17. 1, 11, 31 3
18. 2a2 + 4ah + 2h2 + 5, 2(2a + h)
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 41
vedanta Excel in Opt. Mathematics - Book 9 Polynomials
Polynomials 2
2.0 Review
Discuss the answer of the following questions:
(a) Define rational and irrational numbers with examples.
(b) If f(x) = x2 – 25, find f(4), f(5) and f(6).
(c) Write the index of x in x13.
(d) What is the index of x in 5 x20 ?
(e) Write the degree of algebraic expression p(x) = x2y + xy + y2.
(f) Write the indices of each term of given expression : 3 x + x + 1 + x4 + x2
x
Among these indices write rational and irrational.
2.1 Introduction to polynomials
Algebraic expressions of functions are classified into following types:
(a) On the Basis Coefficient :
Function Coefficients Nature of coefficients
Integers
f(x) = 4x2 + 3x – 5 4, 3, –5 Rational
f(x) 1 x2 – 4 x + 8 1 , – 4 , 8
4 5 4 5
f(x) = x2 + 4x + 7 1, 4, 7 Positive integers
f(x) = 3x3 – 4x2 + 7x + 8 3, – 4, 7, 8 Real numbers
(b) On the Basis of Terms
Functions Number of terms Name
f(x) = 7x 1 Monomial
f(x) = 5x + 3 2 Binomial
f(x) = x2 + 4x + 3 3 Trinomial
42 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Polynomials
(c) On the Basis of Indices Degree Name of Function
0 Constant
Function 1 Linear
f(x) = 7
f(x) =4x + 7 2 Quadratic
3 Cubic
f(x) = 2x2 + 3x + 1 6 polynomial
f(x) =4x3 + 7x2 + 3x + 6
f(x) = x6 + 2x4 + 3x2 + 8
Polynomials have always positive integer indices. If index is zero, it is called constant
polynomial. In above table, indices are either zero or positive
Definition : An algebraic expression having non–negative integer as the power of variables
in each term is called a polynomial. Polynomials may have one or more variables.
Example :
(a) 4, 2x, 2x + 3, 4x2 + 3x + 8, etc. are examples of polynomials.
(b) 4x2 + x + 8 is not a polynomial as index of x in second term is 1 which is not positive
integer. 2
(c) 7x3 + 8x2 + 7x + 9 is a polynomial as index of x in each term is positive integer.
(d) 16x2 + 3x + 5 +in8xteigsenr.ot a polynomial as the index of x in the fourth term is – 1 which
is not a positive
Standard From of Polynomial
Let us consider the following polynomials:
p(x) = x4 + 3x3 + 4x2 + 7x + 8 (descending order of power of x)
q(x) = 8 + x + 3x2 + 4x3 + 7x4 (Ascending order of power of x).
The polynomials in such forms in which the terms are arranged in ascending or descending
powers of variables are called standard forms.
The polynomial in the variable x can be written in standard form as :
p(x) = anxn + an–1xn–1 + an–2xn–2 + ........ + a0, an z 0
or, p(x) = a° + a1x + a2x2 + ............ + anxn, an ≠ 0. Then,
(i) If an ≠ 0, then n is called the degree of polynomial p(x).
(ii) anxn, an–1xn–1, ...... a° are called the terms of the polynomials p(x)
(iii) an, an–2, ...... , are called the coefficient of the polynomials.
(iv) anxn is called the leading term and an is called leading coefficient of the polynomial.
(v) If all the coefficients an , an–1 , ..... are zero, then P(x) is called zero polynomial.
(vi) The degree of polynomial is zero if and only if it is a non–zero constant polynomial.
Example P(x) = 7.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 43
vedanta Excel in Opt. Mathematics - Book 9 Polynomials
Degree of a polynomial
The highest power (or exponent or index) of the variables of the term of the polynomial is called
the degree of polynomial. If the polynomial involves two or more variables, the highest sum of
the powers of the term involved in the polynomial is called the degree of the polynomial.
Examples :
1. p(x) = x4 + x3 + 6x2 + 4x + 8 is a polynomial of degree 4.
2. p(x, y) = x3y2 + x2y + xy2 + y3 is the polynomial of degree 5(as degree of x is 3 and
degree of y is 2 in the first term 3 + 2 = 5)
Equal Polynomials
Two polynomials are said to be equal if and only if (iff),
(i) The degree of both of the polynomials are the same.
(ii) The coefficients of the corresponding terms are the same.
Examples : f(x) = x2 + 4x + 7 and 2 x2 + 8x + 7 are equal polynomial as the corresponding
2 2
coefficients are equal and we write f(x) = g(x).
If f(x) = anxn + an–1xn–1 + ....... + a° and g(x) = bmxm + bmxm–1 + ...... + b° are equal polynomials,
then we write an = bm, an–1 = bm–1 , ...... a° = b°.
Coefficients : A number or alphabet which is the multiple of a variable in an algebraic/
polynomial term is known as coefficient. We use two types of coefficients.
Literal Coefficient : The alphabet used as the coefficient is called literal coefficient. For
example : in 4xy, y is the literal coefficient of 4x and x is the literal coefficient of 4y.
Numerical Coefficient : The number used for the coefficient is called numerical coefficient
For example, In 7xy, 7 is the numerical coefficient of xy.
In 8xy, 8 is the numerical coefficient of xy
y is the literal coefficient of 8x.
x is the literal coefficient of 8y,
Worked Out Examples
Example 1. Which of the following algebraic expressions are polynomials ? Give your reasons.
Solution :
(a) 5x2 – 7x + 8 (b) 5x2 + 8x + 16 (c) 4x2 + 8x + 1
x
14
(d) 4x3 + x2 + x + x + 8 (e) 9
(a) 5x2 – 7x + 8 is a polynomial because exponents of x in each term are
whole numbers.
(b) 5x2 + 8x + 16 is a polynomial because the exponents of x in all the
terms are whole numbers.
(c) 4x2 + 8x + x–1 is not a polynomial because the exponent of x in the
third term is –1 which is not whole number, i.e., negative integer.
44 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Polynomials
(d) 4x3 + x2 + x + x + 8, is not a polynomial because x has exponent 1
which is not a whole number. 2
(e) 14 can be written as 14 x°, which is called a constant polynomial.
9 9
Example 2. Find the numerical coefficient of the following polynomials:
Solution :
Example 3. (a) 14x2 (b) 7 x2y2 (c) xyz
Solution : 8
Example 4. (a) Numerical coefficient of 14x2 = 14
Solution :
(b) Numerical coefficient of 7 x2y2 = 7
Example 5. 8 8
Solution : (c) Numerical coefficient of xyz = 1.
Find the literal coefficient of x2 in the following polynomials:
(a) 17x2y2 (b) 25x2yz (c) 12x2y2z2
(a) Literal coefficient of x2 in 17x2y2 = y2
(b) Literal coefficient of x2 in 25x2yz = yz
(c) Literal coefficient of x2 in 12x2y2z2 = y2z2
Find the degree of polynomials of the following polynomials.
(a) 4x3 + 3x2 + 5x + 2 (b) 17x4 – 18x2 + 9x2 + 5
(c) x2 + x2y2 + y3 (d) x3y2 + y2z + z2
(a) Since the highest power of x is 3, the degree of polynomial is 3.
(b) Since the highest power of x is 4, the degree of the polynomial is 4.
(c) Here, the highest power of sum of x and y in the second term is
(2 + 2) = 4, the degree of polynomial is 4.
(d) Here, the highest power of sum of x and y in the first term is
(3 + 2) = 5, the degree of polynomial is 5.
Write the given polynomials in descending power of x(simplify if necessary).
Also state the leading term.
(a) x3 + 2x(x – x2) + 5x2 + 5x + 15 (b) 20x4 (1 –x2) + 6x4 + 5(x – 1) + 8
(a) Here, x3 + 2x(x – x3) + 5x2 + 5x + 15
= x3 + 2x2 – 2x4 + 5x2 + 5x + 15
= –2x4 + x3 + 7x2 + 5x + 15
which is the polynomial in descending power of x. Leading term is –2x4.
(b) Here, 20x4 (1 – x2) + 6x4 + 5(x – 1) + 8
= 20x4 – 20x6 + 6x4 + 5x – 5 + 8
= –20x6 + 26x4 + 5x + 3
which is the polynomial in descending power of x. Leading term of given
polynomial is – 20 x6.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 45
vedanta Excel in Opt. Mathematics - Book 9 Polynomials
Example 6. Let p(x) = x4 + 3x3 + 7x2 + 6x + 8 and q(x) = ax4 + bx3 + cx2 + dx + e are
Solution: equal polynomial. Find the values of a, b, c, d, e.
Here, p(x) = q(x)
i.e. x4 + 3x3 + 7x2 + 6x + 8 = ax4 + bx3 + cx2 + dx + e
Equating the coefficients of corresponding term, we get,
a = 1, b = 3, c = 7, d = 6, e = 8
Exercise 2.1
Very Short Questions
1. (a) Define a polynomial.
(b) Define the degree of a polynomial.
(c) Give an example of each of the following type of polynomial.
(i) Constant polynomial (ii) Linear polynomial
(iii) Quadratic polynomial (iv) cubic polynomial
2. Which of the following algebraic expressions are polynomials ? Give your reasons.
(a) 4x2 + 3x + 6 (b) x + 4 x (c) 4 + 3
(d) x3 + 3x2 + 6x + 8 (e) x2 + 6 x + 8 x
(f) 3x2y + 6xy + 5xy2
(g) 4x2 + x + 1 (h) 1 + x2 + 4 (i) 7 x4 + 3x2 + 3 x
x x 3
4 1 37
(j) x3 + 5 (k) 4 (x + 1) + 9
3. Find the numerical coefficient of x in the following polynomials.
(a) –0.9x (b) 6 3x (c) 3 x (d) 4xy2 (e) 5x + 9
4 13
4. Find the literal coefficients of
(a) x in 7xy (b) y in 3x2y (c) z in x3y3z3
2
5. Find the coefficients of y in the following:
(a) 6xy (b) – 3x2y (c) 2xyz (d) xy
8
6. Find the degree of the following monomials:
(a) 4x3 (b) 6x2 (c) 7x2y (d) xyz (e) x2y2z
2
7. Find the degree of the following polynomials:
(a) 2x2 + 8x – 7 (b) x3 + 6x2y2 + xy2
(c) 6x + y5 (d) 5xyz + 3y2 + x2
8. Find the degree of polynomial in x-
(a) 5x3 – 3x2y + xy2 (b) 2y2 + 6y + 15 (c) x6 + y6 + 36 + 2
46 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Polynomials
Short Questions
9. Which of the following polynomials are equal:
(a) f(x) = 3x3 – 4x2 + 6, g(x) = 6 – 4x2 + 3x3
(b) f(x) = 3 x3 – 5x2 + 0.25, g(x) = 0.75x3 + 1 – 5x2
4 4
4x3 8x2 8x 1
(c) f(x0 = 2x3 + 4x2 + 8x + 0.25 , g(x) = 2 + 2 + 3 + 4
10. Write the following polynomials in standard form (in descending order):
(a) 2x3 + 7x2 + 8x + x4 (b) 3x3 + 7x2 + 4x4 + 5
(c) 4x3 + 2x – 3 + 4x2
11. Write the following polynomials in standard form (in ascending order):
(a) 7x2 + 8x3 + 9x + 75 (b) 3x2 + 4x6 + 9x3 + 3x + 7
(c) 0.75x3 + 25x2 + 6x + 9x5
12. (a) If 4x3 + 6x2 + 8x + 9 = ax3 + bx2 + cx + d, find the values of a, b, c, and d. Also
find the product and the sum of them.
(b) If p(x) = ax5 + bx4 + 7x3 + 6x2 + 9 and q(x) = 5x5 + 6x4 + 7x3 + 6x2 + 9 and
p(x) = q(x), find the values of a and b.
2. (a) Polynomial (b) Not a polynomial (c) Not a polynomial
(d) Polynomial (e) Not a polynomial (f) Polynomial
(g) Not a polynomial (h) Not a polynomial (i) Not a polynomial
(j) Polynomial (k) Polynomial 3. (a) –0.9
(b) 6 3 (c) 3 (d) 4 (e) 5
4. (a) y 4 (c) x3y3 13
5. (a) 6x (b) x2 (c) 2xz
6. (a) 3 (c) 3
(b) –3x2 (d) x
(b) 2 8
(d) 3 (e) 5
7. (a) 2 (b) 4 (c) 5 (d) 3
8. (a) 3 (b) 0 (c) 6 9. (a) equal
(b) equal (c) equal 10. (a) x4 + 2x3 + 7x2 + 8x
(b) 4x4 + 3x3 + 7x2 + 5 (c) 4x3 + 4x2 + 2x – 3
11. (a) 75 + 9x + 7x2 + 8x3 (b) 7 + 3x + 3x2 + 9x3 + 4x6
(c) 6x + 25x2 + 0.75x3 + 9x5
12. (a) a = 4, b = 6, c = 8, d = 9, product = 1728, sum= 27 (b) a = 5 , b = 6
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 47
vedanta Excel in Opt. Mathematics - Book 9 Polynomials
2.2 Operations on polynomials
Addition, subtraction, multiplication, and division are four algebraic operations that can be
done on polynomials. In this section, we study only addition, subtraction, and multiplication.
Addition of Polynomials
Let p(x) = x3 + 7x2 + 3x + 9
q(x) = 2x3 – 8x2 + 8x + 11
Then we can add them as follows:
p(x) + q(x) = (x3 + 7x2 + 3x + 9) + 2x3 – 8x2 + 8x +11
= (1 + 2) x3 + (7 – 8)x2 + (3 + 8) x + 9 + 11
= 3x3 – x2 + 11x + 20
Properties of Addition of Polynomials
Let p(x), q(x), and r(x) be three polynomials over real numbers (R). Then the following are
the properties of addition of polynomials:
S.N. Properties Mathematical Forms
(a) Closure Property p(x) + q(x) is again a polynomial
(b) Commutative Property p(x) + q(x) = q(x) + p(x)
(c) Associative Property {p(x) + q(x)} + r(x) = p(x) + {q(x) + r(x)}
(d) Additive Identity p(x) + 0 = p(x)
(e) Additive Inverse p(x) + {– p(x)} = 0, 0 is zero polynomial
q(x) + {–q(x)} = 0, 0 is zero polynomial
r(x) + {– (r (x)} = 0
Difference of Two Polynomials.
Let p(x) = 3x2 + 7x + 8,
and q(x) = 4x2 + 8x + 11
Then the difference of p(x) and q(x) is p(x) – q(x).
p(x) – q(x) = (3x2 + 7x + 8) – (4x2 + 8x + 11)
= (3 – 4)x2 + (7 – 8) x + (8 – 11)
= –x2 – x – 3
Difference of two polynomial is again a polynomial.
Multiplication of Polynomials.
Let p(x) = x2 + 3x + 7 and q(x) = 2x2 + 4x + 3.
Then, the product of p(x) and q(x) is p(x) q(x).
48 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel in Opt. Mathematics - Book 9 Polynomials
Now, p(x). q(x) = (x2 + 3x + 7) (2x2 + 4x + 3)
= x2 (2x2 + 4x + 3) + 3x (2x2 + 4x + 3) + 7(2x2 + 4x + 3)
= 2x4 + 4x3 + 3x2 + 6x3 + 12x2 + 9x + 14x2 + 28x + 21
= 2x4 + 10x3 + 29x2 + 37x + 21
Let p(x), q(x) and r(x) be three polynomials over real numbers (R). None of them is zero
polynomial.
S.N. Properties Mathematical Forms
(a) Closure Property P(x) . q(x) is again is also a polynomial
(b) Commutative Property p(x) . q(x) = q(x). p(x)
(c) Associative Property {p(x) . q(x)} r(x) = p(x) . {q(x) . r(x)}
(d) Multiplicative Identity p(x) .1 = 1. p(x) = p(x)
Steps of Multiplication of two Polynomials:
(a) Multiply each term of one polynomial by term of other polynomial.
(b) Then, add or subtract like terms.
(c) Rearrange the terms in ascending or descending order of powers of variables (in
standard forms of polynomials).
Worked Out Examples
Example 1. Find the sum of the given polynomials: p(x) = x3 – 5x2 + 7x + 2 and
Solution: q(x) = 4x3 + 8x2 + 6x + 5
Here, p(x) = x3 – 5x2 + 7x + 2
Example 2. and q(x) = 4x3 + 8x2 + 6x + 5
Solution : sum of p(x) and q(x) is given by,
p(x) + q(x) = (x3 – 5x2 + 7x + 2)+ (4x3 + 8x2 + 6x + 5)
= (1 + 4)x3 + (–5 + 8)x2 + (7 + 6)x + (2 + 7)
= 5x3 + 3x2 + 13x + 9
Let p(x) = 2x2 – 3x + 8 and q(x) = 6x3 – 4x2 + 7x – 10 subtract q(x) from p(x).
Here, p(x) = 2x2 – 3x + 8
q(x) = 6x3 – 4x2 + 7x – 10
Now, p(x) – q(x) = (2x2 – 3x + 8) – (6x3 – 4x2 + 7x – 10)
= 2x2 – 3x + 8 – 6x3 + 4x2 – 7x + 10
= – 6x3 + 6x2 – 10x + 18
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 49
vedanta Excel in Opt. Mathematics - Book 9 Polynomials
Example 3. Find the product of f(x) = x2 + x + 1 and g(x) = x – 1.
Solution : Here, f(x).g(x) = (x2 + x + 1) (x – 1)
= x(x2 + x + 1) – 1 (x2 + x + 1)
= x3 + x2 + x – x2 – x – 1 = x3 – 1
Example 4. If f(x) = 5x + 2, g(x) = 2x2 + 6x + 1 and h(x) = x3 + x2, then verify
{f(x) + g(x)} + h(x) = f(x) + {g(x) + h(x)}
Solution : Here, f(x) = 5x + 2, g(x) = 2x2 + 6x + 1, h(x) = x3 + x2
LHS = {f(x) + g(x)} + h(x)
= {(5x + 2) + (2x2 + 6x + 1)} + (x3 + x2)
= (5x + 6x + 2x2 + 2 + 1) + x3 + x2
= 2x2 + 11x + 3 + x3 + x2
= x3 + 3x2 + 11x + 3
RHS = f(x) + {g(x) + h(x)}
= (5x + 2) + {(2x2 + 6x + 1) + (x3 + x2)}
= (5x + 2) + (2x2 + 6x + 1 + x3 + x2)
= 5x +2 + x3 + 3x2 + 6x + 1
= x3 + 3x2 + 11x + 3
? LHS = RHS, Proved.
Exercise 2.2
Very Short Questions
1. Find the sum of p(x) and q(x) in the following:
(a) p(x) = x2 + 3, q(x) = 2x2 + 7
(b) p(x) = 3x3 – 8x2 + 9x + 8, q(x) = 4x3 + 4x2 – 5x – 8
(c) p(x) = – 4x2 + 8 and q(x) = 4x2 – 8
2. Subtract the following polynomials:
(a) 4x2 – 3x + 2 from 3x4 + 2x2 + 6x – 8
(b) 4x2 + 5x + 7 from 3x2 – 2x + 2
3. Find the product of the polynomials p(x) and q(x) in the following:
(a) p(x) = x + y, q(x) = x – y (b) p(x) = x2 + x + 1, q(x) = x – 1
(c) p(x) = x + y, q(x) = x2 – xy + y2
Short Questions
4. (a) If p(x) = x + 3, q(x) = x + 6, verify that p(x) + q(x) = q(x) + p(x)
(b) If p(x) = x + 8, q(x) = x + 12, verify that p(x) + q(x) = q(x) + p(x)
50 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.
Vedanta Opt. Math 9 Final (2078)
Discover the best professional documents and content resources in AnyFlip Document Base.
Search