Vedanta Opt. Excel in Mathematics Book 8 Final (2078)

vEMexdcAaenlTtaiHn AEdMdAitTiIoCnSal

8Book

Author
Piyush Raj Gosain

Hukum Pd. Dahal Editors P. L. Shah
Tara Bdr. Magar

vedanta

Vedanta Publication (P) Ltd.
jb] fGt klAns];g k|f= ln=

Vanasthali, Kathmandu, Nepal
+977-01-4982404, 01-4962082
[email protected]
www.vedantapublication.com.np

EvMexdcAaenlTtaiHn AEMddAiTtiIoCnSal

8Book

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

Vedanta Excel in Additional Mathematics for class 8 is completely based on the contemporary
pedagogical teaching learning activities and methodologies. It is an innovative and unique in
the sense that the contents of the book are written and designed to ful ill the need of integrated
teaching learning approaches.
Vedanta Excel in Additional Mathematics has incorporated applied constructivism the latest trend
of learner centered teaching pedagogy. 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 experience, more than two decades, of teaching optional
mathematics.
Each unit of Vedanta Excel in Additional 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 Additional Mathematics for class 8 covers all the required contents for the
students of class 8 so that it laid foundation for optional mathematics in class 9 and 10. It also
helps the student for their compulsory mathematics. My honest efforts have been to provide all
the essential matters and practice materials to the students. It is believed that the book serves as
a staircase for the students of class 8. 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.
The book aims to give an elementary knowledge of Relation and Functions, Polynomials, Real
number System and Surds, Trigonometry, Coordinates Geometry, Matrix, Vectors, Transformations,
Statistics. Special emphasis has been given to all the chapters as all of them are entirely new to the
students. Questions in each exercise are catagorized into two groups - Short Questions and Long
Questions.
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
colleague 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 Relation and Function .....................................................................7-26

1.1 Ordered Pair ................................................................................................. 7
1.2 Equality of ordered pairs ............................................................................. 7
1.3 Cartesian Product ...................................................................................... 10
1.4 Ways of Representation of Cartesian Product .......................................... 10
1.5 Relation - Introduction .............................................................................. 14
1.6 Ways of Representation of a Relation ....................................................... 15
1.7 Domain and Range of a Relation .............................................................. 16
1.8 Type of Relations ....................................................................................... 16
1.9 Inverse Relation ......................................................................................... 17
1.10 Function - Introduction ............................................................................. 20
1.11 Domain, Co-domain, Range, Image and Pre-image of a Function ........... 21
1.12 Testing of a Function ................................................................................. 21
1.13 Ways of Representation of a Function ...................................................... 22
1.14 Functional Values ...................................................................................... 22

Unit 2 Polynomials ...................................................................................... 27-38

2.1 Review ....................................................................................................... 27
2.2 Polynomials ............................................................................................... 27
2.3 Degree of Polynomials ............................................................................... 27
2.4 Types of Polynomials ................................................................................ 28
2.5 Standard form of Polynomials .................................................................. 28
2.6 Equal Polynomials ..................................................................................... 28
2.7 Operation of Polynomials ......................................................................... 28
2.8 Synthetic Division ..................................................................................... 30

Unit 3 Real Number System and Surds .............................................. 39-49

3.1 Introduction ............................................................................................... 39
3.2 Real Numbers System ............................................................................... 39
3.3 Surds .......................................................................................................... 41
3.4 Different types of Surds ............................................................................ 42
3.5 Operation of Surds .................................................................................... 43
3.6 Rationalization of Surds ........................................................................... 44

Unit 4 Measurement of Angles ............................................................... 50-63

4.1 Angle .......................................................................................................... 50
4.2 Measurement of an Angle ......................................................................... 50

4.3 Measurement of Angles of Polygons ......................................................... 58

Unit 5 Trigonometry ................................................................................... 64-98

5.1 Trigonometry ............................................................................................. 64
5.2 Reference Angle ........................................................................................ 64
5.3 Pythagoras Theorem .................................................................................. 65
5.4 Converse of Pythagoras Theorem ............................................................. 67
5.5 Trigonometric Ratios.................................................................................. 70
5.6 Operations on Trigonometric Ratios ......................................................... 78
5.7 Trigonometric Identities ............................................................................ 83
5.8 Equation and Identity ............................................................................... 86
5.9 Trigonometric Identity .............................................................................. 87
5.10 Methods of Proving a Trigonometric Identity .......................................... 87
5.11 Conversion of trigonometric ratios ........................................................... 93

Unit 6 Trigonometric Ratios of Some Standard Angles ............ 99-110

6.1 Trigonometric Ratio of 45° ........................................................................ 99
6.2 Trigonometric Ratio of 30° and 60° ......................................................... 100
6.3 Trigonometric Ratios of 0° and 90° ......................................................... 101
6.4 Trigonometric Ratios of Complementary Angles ................................... 107

Unit 7 Solution of Right-angled Triangle and Height and Distance .... 111

7.1 Solution of Right-angled Triangle ........................................................... 111
7.2 Height and Distance ................................................................................ 117

Unit 8 Coordinates Geometry ............................................................126-152

8.1 Distance formula ..................................................................................... 126
8.2 Section formula ....................................................................................... 135
8.3 Slope ........................................................................................................ 142
8.4 Locus ........................................................................................................ 146

Unit 9 Equations of Straight Lines ...................................................153-167

9.1 Equations of straight lines parallel to coordinate Axes ......................... 153

9.2 Equation of straight line in slope intercept from (y=mx + c) ............... 156
x y
9.3 Equation of a Straight Line in double intercepts form : a + b = 1 ........ 160

9.4 Equation of a straight line in Normal or Perpendicular

form (x cos D+ y sin D = p) .................................................................... 164

Unit 10 Matrix .............................................................................................168-184

10.1 Introduction ............................................................................................. 168
10.2 Definition of Matrix ................................................................................ 168

10.3 Rows and Columns of a Matrix .............................................................. 169
10.4 Order of a Matrix ..................................................................................... 169
10.5 Types of Matrices .................................................................................... 171
10.6 Equal Matrices ......................................................................................... 172
10.7 Operation of Matrices - Addition and Subtraction ................................ 175
10.8 Multiplication of Matrix by a Scalar ...................................................... 176
10.9 Transpose of a Matrix .............................................................................. 177
10.10 Properties of Matrix Addition ................................................................. 179

Unit 11 Vectors ...........................................................................................185-193

11.1 Introduction ............................................................................................. 185
11.2 Directed Line Segment ............................................................................ 185
11.3 Components of a Vector .......................................................................... 186
11.4 Magnitude of a Vector ............................................................................. 187
11.5 Types of Vector ........................................................................................ 188
11.6 Operations of Vectors .............................................................................. 189

Unit 12 Transformations ........................................................................194-210

12.1 Introduction ............................................................................................. 194
12.2 Reflection ................................................................................................. 195
12.3 Use of Coordinates in Reflection ............................................................ 195
12.4 Rotation - Introduction ........................................................................... 200
12.5 Use of Coordinates in Rotation ............................................................... 201
12.6 Translation - Introduction ....................................................................... 206
12.7 Translation Using Translation Vector ..................................................... 207

Unit 13 Statistics ........................................................................................211-230

13.1 Introduction ............................................................................................. 211
13.2 Collection of Data .................................................................................... 211
13.3 Frequency Table ....................................................................................... 211
13.4 Grouped and Continuous Data ............................................................... 212
13.5 Cumulative Frequency Table .................................................................. 213
13.6 Histogram ................................................................................................ 215
13.7 Measures of Central Tendency ................................................................ 216
13.8 Arithmetic Mean ..................................................................................... 216
13.9 Median ..................................................................................................... 221
13.10 Quartiles .................................................................................................. 223
13.11 Mode ........................................................................................................ 226
13.12 Range ....................................................................................................... 228

Model Questions ................................................................................................. 231

Relation and Function

1Relation and Function

1.1 Ordered Pair

A pair of two elements with a fixed order is known as an ordered pair. An ordered
pair of two elements x and y is denoted by (x, y), where x is called the first component
or antecedent and y is called the second component or consequent.

In coordinate geometry, (x, y) represents a point in the cartesian plane, where x is
called x-coordinate or abscissa and y is called y-coordinate or ordinate.

Ordered pairs play an important role in our daily life.

Example : (7, 14)

Here, the first component = 7

the second component = 14

We can represent (7, 14) in the cartesian plane on graph paper.

Note :

(i) {a, b} and {b, a} are same in set, we write {a, b} = {b, a}

(ii) (a, b) and (b, a) are different in meaning, we write (a, b) z b, a)
(iii) (a, b) = (b, a) if and only if a = b.

1.2 Equality of ordered pairs

Two ordered pairs (x, y) and (m, n) are said to be equal ordered pairs if x = m and
y = n.

For example : (4, 5 ) and 20 , 15 are equal ordered pairs.
5 3

It is to be noted that two ordered pairs (a, b) and (b, a) are different as their

corresponding elements are different.

vedanta Excel in Additional Mathematics - Book 8 7

Relation and Function

Worked Out Examples

Example 1. Which of the following ordered pairs are equal?
Solution:
(a) (4, 5) and (5, 4)

(b) (1, 2) and 10 , 10
10 5

(a) Here, (4, 5) and (5, 4) ordered pairs are not equal as 4 ≠ 5 and

5 ≠ 4.

(b) Here, (1, 2) and 10 , 10 = (1, 2) are equal ordered pairs as
10 5

their corresponding elements are equal.

Example 2. If (x + 4, y – 4) = (2, 5), find the values of x and y.
Solution:
Here, (x + 4, y – 4) = (2, 5)

Since, the two equal ordered pairs are equal, their corresponding
elements are equal.

? x+4=2 and y – 4 = 5

or, x = 2 – 4 and y = 5 + 4

? x=–2 and y = 9

Example 3. From set A = {4, 5}, find all the possible ordered pairs.
Solution: Here, A = {4, 5},

the ordered pairs can be obtained by the following ways.

Components 4 5
4 (4, 4) (4, 5)
5 (5, 4) (5, 5)

Hence, the possible ordered pairs are(4, 4), (4, 5), (5, 4) and (5, 5)

Exercise 1.1

Short Questions :
1. (a) Define an ordered pair.

(b) In ordered pair (m, n), what are m and n called?
(c) Under which condition two ordered pairs (a, b) and (p, q) are equal?
(d) Why are (1, 2) and (2, 1) not equal ordered pairs?

8 vedanta Excel in Additional Mathematics - Book 8

Relation and Function

2. Which of the following ordered pairs are equal?

(a) (2, 3) and (3, 2)

(b) (7, 6) and (2 + 5, 3 + 3)

(c) (1, 6) and (6, 1)

(d) (-2, -3) and (– 10 , – 6 )
52

3. Find the values of x and y if,

(a) (x, y) = (6, 7) (b) (2x, 6y) = (4, 12)

(c) (x + 2, y – 2) = (5, 3) (d) (2x + 3, 2y – 4) = (9, 4)

(e) (x + y, x – y) = (4, 2) (f) (x + y , x – y) = (5, 3)

Long Questions :

4. Find the values of x and y from the following equal ordered pairs.

(a) (x + y, 2x – y) = (13, 2)

(b) (5x – 3y, 3x – y) = (16, 12)

(c) (3x – 2y, 7) = (11, x + y)

5. Find all the possible ordered pairs from the following sets.

(a) A = {a, b} (b) P = {p, q}

(c) Q = {6, 7} (d) R = {1, 2, 3}

(e) S = {7, 8, 9}

1. Show to your teacher.

2. Show to your teacher.

3. (a) x = 6, y = 7 (b) x = 2, y = 2 (c) x = 3, y = 5

(d) x= 3, y = 4 (e) x = 3 and y = 1

4. (a) x = 5, y = 8 (b) x = 5, y = 3 (c) x = 5, y = 2

5. (a) (a, a), (a, b), (b, a), (b, b) (b) (p, p), (p, q), (q, p), (q, q)

(c) (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)

(d) (7, 7), (7, 8), (7, 9), (8, 7), (8, 8), (8, 9), (9, 7), (9, 8), (9, 9)

vedanta Excel in Additional Mathematics - Book 8 9

Relation and Function

1.3 Cartesian Product
Cartesian Product of two sets :

Let A and B be two non-empty sets. Then, the cartesian product of sets A and B is
denoted by A × B (read as A cross B ) and is defined as the set of all ordered pairs
(x, y) whose first element x is from set A and second element y is from set B.

Symbolically we write, A × B = {(a, b) : a  A and b  B}

Similarly, B × A = {(b, a) : b  B and a  A}

For example Let A = {1, 2} and B = {3, 4}, find A × B and B × A.

Solution: Here, A = {1, 2} and B = {3, 4}

A×B = {1, 2} × {3, 4}

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

Also, B × A = {3, 4} × {1, 2}

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

Properties if Cartesian Product

For the non-empty sets, we have:

(a) A × B = B × A if A = B

(b) If A = {a, b} and B = { } = I, then

A×B=I i.e. A × B is not defined.

(c) n(A × B) = n(A) × n(B) = n(B) × n(A) = n(B × A)

1.4 Ways of Representation of Cartesian Product

A Cartesian product A × B can be represented by using any one of the following
ways:
(a) Tabulation Method
(b) Listing Method / Set of ordered pair
(c) Set Builder Method
(d) Arrow or Mapping Diagram
(e) Graphical Method
For example : Let A = {1, 2} and B = {3, 4}, then we can show A × B in the
following ways:

10 vedanta Excel in Additional Mathematics - Book 8

Relation and Function

(a) Tabulation Method :

Set A 1 Set B 4
2 3 (1, 4)
(1, 3) (2, 4)
(2, 3)

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

(b) Listing Method / Set of Ordered Pair.
A × B = {1, 2} × {3, 4}

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

(c) Set Builder Method
A × B = {(x, y) : x  A and y  B}

(d) Arrow or Mapping Diagram
AB

13
24

(e) Graphical Method
Here, A × B = {(1 3), (1, 4), (2, 3), (2, 4)}

Y

X' O X

Y'

Worked out Examples

Example 1. Let A = {4, 5} and B = {1, 2}, find
Solution:
(i) A × B (ii) B × A

Also show that (iii) A × B z B × A (iv) n(A × B) = n(A) × n(B)

Here, A = {4, 5) and B = {1, 2}. Then

(i) A × B = {4, 5} × {1, 2} = {(4, 1), (4, 2), (5, 1), (5, 2)}

(ii) B × A = {1, 2} × {4, 5} = {(1, 4), (1, 5), (2, 4), (2, 5)}

vedanta Excel in Additional Mathematics - Book 8 11

Relation and Function

(iii) From result of A × B and B × A, we observe that

(4, 1) z (1, 4) i.e. 4 z 1

(4, 2) z (2, 4) i.e. 2 z 4

Hence, A × B z B × A

(iv) n(A) = 2, n(B) = 2, n(A × B) = 4

So, n(A) × n(B) = n(A × B)

i.e. 2 × 2 = 4 (True)

Hence, n(A × B) = n(A) × n(B) Proved.

Example 2. If A × B = {(a, x), (a, y), (b, x), (b, y)}, then find A and B.

Solution: Here, A × B = {(a, x), (a, y), (b, x), (b, y)}

A is the set of all the first element of the ordered pair and B is
the set of all the second elements.

Hence, A = {a, b} and B = {x, y}

Exercise 1.2

Short Questions :

1. Find A × B if,

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

(b) A = {3, 4} and B = {2, 5}

(c) A = {1, 2, 3} and B = {4, 5}

(d) A = {1, 2, 5} and B = {3, 4, 7}

2. Find B × A if,

(a) A = {u, v} and B = {w, x}

(b) A = {0, 1, 2} and B = {3, 5}

(c) A = {4, 5, 6} and B = {1, 2, 3}

(d) A = {2, 4, 6} and B = {1, 3, 5}

3. If A = {a, b, c} and B = {d, c} and C = {m, n}, then find

(a) A × B (b) B × C (c) A × C (d) C × A

12 vedanta Excel in Additional Mathematics - Book 8

Relation and Function

Long Questions :
4. If A = {2, 5} and B = {3, 4}, then find A × B and represent it in the following ways:

(a) Tabulation method
(b) Listing method
(c) Arrow diagram or mapping diagram
(d) Graphic method
5. Find the given arrow diagram, list the elements of A × B.

(a) A B (b) A B
4 2
5 81
72 3

3 4

6. (a) If P × Q = {(4, 5), (4, 6), (4, 7), (5, 5), (5, 6), (5, 7)}, find P and Q. Also show
that n(P × Q) = n(P) × n(Q).

(b) If R × S = {1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4)}, find R and S. Also verify
that n(R × S) = n(R) × n(S)

1. (a) {(a, x), (a, y), (b, x), (b, y)} (b) {(3, 3), (3, 5), (4, 2), (4, 5)}

(c) {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}

(d) {(1, 3), (1, 4), (1, 7), (2, 3), (2, 4), (2, 7), (5, 3), (5, 4), (5, 7)}

2. (a) {(w, w), (w, v), (x, x), (x, v)} (b) {(3, 0), (3, 1), (3, 2), (5, 0), (5, 1), (5, 2)}

(c) {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

(d) {(1, 2), (1, 4), (1, 6), (3, 2), (3, 4), (3, 6), (5, 2), (5, 4), (5, 6)}

3. (a) {(a, d), (a, c), (b, d), (b, c), (c, d), (c, c)} (b) {(d, m), (d, n), (c, m), (c, n)}

(c) {(a, m), (a, n), (b, m), (b, n), (c, m), (c, n)}

(d) {(m, a), (m, b), (m, c), (n, a), (n, b), (n, c)}

4. (a) Show to your teacher

5. (a) {(4, 8), (4, 7), (5, 8), (5, 7)}

(b) {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4)}

6. (a) P = {4, 5}, Q = {5, 6, 7} (b) R = {1, 2}, S = {2, 3, 4}

vedanta Excel in Additional Mathematics - Book 8 13

Relation and Function

1.5 Relation - Introduction

There may be various types of relations in our daily life. Let us consider some of
them as given below:
(i) Kathmandu is the capital of Nepal.
(ii) Dasharath was father of Ram.
(iii) 20 is greater than 10.
In above relations, (i) shows the relation of "the capital and country", (ii) shows the
relation of "father and son" and (iii) shows "greater than" relation.
Hence the word relation shows an association of two objects, or people, or numbers
and so on. Here, we focus our attention on mathematical relation.
Let A and B be any two non-empty sets. Then, a relation R from A to B is a non-
empty subset of the cartesian product A × B. It is defined by:

R = {(x, y) : x  A and y  B} Ž A × B
For example : Let A = {1, 2} and B = {3, 4}, then

A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
Then, let us define the following relations from A × B.
(i) R1 = {(x, y) : x + y = 5}

= {(1, 4), (2, 3)}
(ii) R2 = {(x, y) : x < y}

= {1, 3), (1, 4), (2, 3), (2, 4)}
(iii) R3 = {(x, y) : x > y} = I
In all of above relations, R1, R2 and R3, all of them are subsets of A × B.
Hence, a relation is a subset of cartesan product A × B.

Worked out Examples

Example 1. If A = {1, 2, 3} and B = {1, 3, 5}, then find the set of ordered pairs in
Solution: the relation "is greater than" from A to B.

Here, A = {1, 2, 3} and B = {1, 3, 5}

Then,

A × B = {1, 2, 3} × {1, 3, 5}

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

14 vedanta Excel in Additional Mathematics - Book 8

Relation and Function

Let R1 = the set of the ordered pair whose first component is greater
than the second components

i.e. R1 = {(x, y) : x > y}
= {(2, 1), (3, 1)}

1.6 Ways of Representation of Relation

There are various ways in which a relation can be expressed. They are as following:

(a) Roster form (b) Set builder form

(c) Arrow or mapping diagram (d) Tabulation method

(e) Graphical method

Example 2. If A = {1, 4, 9, 16}, B = {1, 2, 3, 4} and a relation R : A o B is defined
Solution: by "is square of", then represent the relation in the following ways:

(a) Roster method (b) Arrow diagram method

(c) Tabulation (d) Graphical method

Here, A = {1, 4, 9, 16} and B = {1, 2, 3, 4}, then,

A × B = {1, 4, 9, 16} × {1, 2, 3, 4}

= {(1, 1), (1, 2), (1, 3), (1, 4), (4, 1), (4, 2), (4, 3), (4, 4),

(9, 1), (9,2), (9,3), (9,4), (16, 1), (16, 2), (16, 3), (16, 4)}

Now, R : A o B i.e. R is a relation from A to B.

Let us represent above relations in the following ways:

(a) Roster method

R = {(x, y) : x is square of y}

= {(x, y) : x = y2}

= {(1, 1), (4, 2), (9, 3), (16, 4)}

(b) Arrow diagram method
R

11
42
93
16 4

R{(x, y) : x = y2}

vedanta Excel in Additional Mathematics - Book 8 15

Relation and Function

(c) Tabulation method

x14 9 16
y12 3 4

(d) Graphical method
Here, R = {(1, 1), (4, 2), (3, 9), (4, 16)}
Let us show each point of R in a graph :

Y

(9, 3) (16, 4)
X
(4, 2)
(1, 1)

X' O

Y'

1.7 Domain and Range of Relation

Let R : A o B be a relation from A to B.

Suppose R = {(1, 1), (2, 4), (3, 9), (4, 16)}

In an arrow diagram, we can show R as in A R B
diagram given below :

Then, the set all the first component of R is 1 1
called domain. 2 4

? Domain of R = {1, 2, 3, 4} 3 9

Its elements are called pre-images. 4 16

The set of all the second components of R is
called range.

? Range of R = {1, 4, 9, 16}

Its elements are also called images.

1.8 Type of Relations

Let A and B be two non-empty sets. Then a relation R from A to B may be any one
of the following types:

(a) One to one relation:
A relation R from set A to set B is called one to one of different element set A
are related to different elements of set B.

Example : Let R : A o B be defined by

16 vedanta Excel in Additional Mathematics - Book 8

Relation and Function

R = {(1, 5), (2, 6), (3, 7) A R B
1 5
Here, R is a one to one relation because different
element of set A has different image in set B. 2 6

(b) Many to one relation : 37

A relation R from set A set B is called many to one

relation if two or more elements A are associated with unique element of set B.

R = {(a, x), (b, x), (c, x)}

R B
A x
a
b
c

Hence, R is called many to one relation as three elements of A are associated
with unique element x of A.

(c) One to many relation :

A relation R : A o B is called one to many relation if at least one element of set
A is associated with two or more elements of set B.

Here, R = {(a, x), (a, y), (b, y), (b, z)} A R B
a x
"a of A is associated with two elements x and y of B y
and b is associated with y and z of B". Hence, R is b z
called one to many relation.

1.9 Inverse Relation

Let R = {(a, x), (b, y), (c, z)}. Then, interchange the element of each of the ordered
pair, we get a new relation R–1 = {(x, a), (y, b), (z, c)}.

Thus, new relation R–1 is called inverse relation of R.

If R = {(x, y) : x  A and y  B}

then, R–1 = {(y, x) : y  B and x  A}

R B R–1 B
A A a
x b
ax y c
z
by

cz

vedanta Excel in Additional Mathematics - Book 8 17

Relation and Function

Example 3. If R = {(1, 2), (2, 3), (3, 4), (4, 5)}, then find the inverse relation of R.
Solution: Here, R = {(1, 2), (2, 3), (3, 4), (4, 5)}

R–1 = {(2, 1), (3, 2), (4, 3), (5, 4)}

Exercise 1.3

Short Questions :

1. (a) Define a relation.

(b) State any five ways of representation of relation.

(c) Define one to one relation with an example.

(d) Define many to one relation with an example.

(e) Define one to many relation with an example.

2. Write the domain and range of the following relation,

(a) R1 = {(1, 1), (2, 4), (3, 9}
(b) R2 = {(a, x), (b, y), (z, c)}
(c) R3 = {(1, x), (1, y), (1, z), (2, x), (2, y), (2, z)}
(d) R4 = {(1, 1), (2, 3), (3, 27)}
3. Find the inverse relation of each of the following relations:

(a) R1 = {(1, 1), (2, 4), (3, 9)}
(b) R2 = {(x, a), (y, b), (z, c)}
(c) R3 = {(1, 8), (2, 9), (3, 10)}
(d) R4 = {(2, 1), (3, 2), (4, 3)}
4. Find the domain, range and inverse of relation R from the given figure

(i.e. R : A o B)

(a) R (b) R

10 20 1 1
20 40 2 8
30 60 3 27
4 64

18 vedanta Excel in Additional Mathematics - Book 8

Relation and Function

Long Questions :

5. Let A = {1, 2, 3} and B = {2, 3, 4}, find the relation from A to B determined by
the relation R = {(x, y) : x  A and y  B}.

(a) x = y (b) x < y

(c) x > y (d) y = x2

6. If P = {1, 4, 9} and Q = {1, 2, 3} and a relation R : P o Q is defined by "is square
of". Represent the relation in the following ways:

(a) Roster method (b) Set builder form

(c) Mapping diagram method (d) Graphical method

(e) Tabulation method

7. If A = {1, 4, 9} and B = {1, 16, 81} and a relation R : A o B is defined by
"square of". Represent R in the followings ways:

(a) Set of ordered pairs (b) Set builder form

(c) Arrow diagram (d) Tabulation method

2. (a) D = {1, 2, 3}, R = {1, 4, 9} (b) D = {a, b, c}, R = {x, y, z}

(c) D = {1, 2}, R = {x, y z} (d) D = {1, 2, 3}, R = {1, 3, 27}

3. (a) R1–1 = {(1, 1), (4, 2), (9, 3)} (b) R2–1 = {(a, x), (b, y), (c, z)}

(c) R3–1 = {(8, 1), (9, 2), (10, 3)} (d) R4–1 = {(1, 2), (2, 3), (3, 4)}

4. (a) D = {10, 20, 30}, R = {20, 40, 60}, R–1 = {(20, 10), (40, 20), (60, 30)}

(b) D = {1, 2, 3, 4}, R = {1, 8, 27, 64}, R–1 = {(1,1), (8,2), (27,3), (64,4)}

5. (a) {(2, 2), (3, 3)} (b) {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}

(c) {(3, 2)} (d) {(2, 4)} 6. (a) {(1, 1), (4, 2), (9, 3)}

(b) {(x, y) : x = y2} (c), (d) Show to your teacher

(e) x 1 4 9
y12 3

7. (a) {(1, 1), (4, 16), (9, 81)} (b) {(x, y) : y = x2, x  A and y  B}
(c) Show to your teacher

(d) x 149
y 1 16 81

vedanta Excel in Additional Mathematics - Book 8 19

Relation and Function

1.10 Function - Introduction

Let A and B be two non-empty sets. A relation from A to B that is called a function
if each element of A maps a unique element of set B.

For examples:

(a) f B (b) g B
A a A x
x b 1 y
y c 2 z
z 3

Here, f is a function because each Here, g is a function because each
element of A maps to unique element of A maps to unique
element of B. element of B.

(c) h B (d) k B
A a A x
1 a
2 b b y
3 c

Here, h is not a function because Here, k is not a function because one
3 of A does not map any element element b of A maps two element x
of B. and y. One to many relation is not
a function.

A function from set A to set B is written as f : A o B such that x  A and y  B,
where y = f(x). We say that f is a function of x.

The element y  B is called the image of x under f. The element x  A is called the
pre-image of y under f.

Definition: Let A and B be two non-empty sets. Then, a relation from set A to B
is said to be a function if every element of the first set A associates with unique
element of the second set B.

A function is a special type of relation in which every element 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.

20 vedanta Excel in Additional Mathematics - Book 8

Relation and Function

1.11 Domain, Co-domain, Range, Image and Pre-image
of a Function

Let f : A o B be a function from A to B. Then A is called domain and B is called co-

domain of function f. A f
In the mapping diagram, f is a function from A to B. B

Domain of f = {1, 2, 3} 1 1
2 4
Co-domain of f = {1, 4, 9, 16} 3 9
16
Range of f = {1, 4, 9}, Range is subset of
co-domain.

Here, 1 is image of 1 and 1 is pr e-image of 1.

4 is image of 2 and 2 is pre-image of 4.

9 is image of 3 and 3 is pre-image of 9.

1.12 Testing of a Function

All relations are not functions. The relations can be tested whether it is a function
or not. 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 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. Then,
the method is called vertical line test.

For this test, a vertical line is drawn through any point on it. If the vertical line cuts
the graph at only one point, it is a function. If it cuts at more than one point, it is
not a function.

YY

X' X
O
X' O X

Y' (i) Y' (ii)

vedanta Excel in Additional Mathematics - Book 8 21

Relation and Function Y
Y

X' X X' X
O O

Y' (iii) Y' (iv)

In above figure, the vertical line test shows that the diagrams (i) and (ii) represent

functions because the vertical line cuts the diagrams only at a point. The diagrams

in (iii) and (iv) do not represent functions because the vertical line cuts the diagrams

at two points.

(iv) Some elements of co-domain may not have their pre-image on the first set A.

1.13 Ways of Representation of Function

A function can be represented by various methods as in a relation. It is represented
in the following ways:

(a) Set of ordered pairs (b) Tabulation method

(c) Arrow diagram or mapping diagram method

(d) Formula method or Equation method (e) Graphical method

1.14 Functional Values

Let us consider a linear function defined by
y = f(x) = 4x + 3

i.e. f(x) = 4x + 3
Put the value of x = 1, 2, 3, 4, 5, ......, etc. successively, we get

f(1) = 4 . 1 + 3 = 7
f(2) = 4 . 2 + 3 = 11
f(3) = 4 . 3 + 3 = 15
f(4) = 4 . 4 + 3 = 19
f(5) = 4 . 5 + 3 = 23
The values 7, 11, 15, 19, and 23 are called functional values of f(x) at x = 1, 2, 3, 4,
and 5 respectively. The set of values of x are called domain and the set of values of
y are called range of the functions.

22 vedanta Excel in Additional Mathematics - Book 8

Relation and Function

Worked out Examples

Example 1. Which of the following relation are functions?
Solution:
(a) f = {(1, 2), (2, 3), (3, 4)}
Example 2.
(b) g = {(4, 5), (6, 7), (8, 9)}

(c) h = {(1, 2), (1, 4), (1, 6)}

(d) k = {(a, 1), (a, 2), (a, 3), (b, 4)}

(a) Here, f = {(1, 2), (2, 3), (3, 4)}
Each pre-image element has distinct image.
Hence, f is a function.

(b) Here, g = {(4, 5), (6, 7), (8, 9)}
No first element is repeated in the ordered pairs.
Hence, g represents a function.

(c) Here, h = {(1, 2), (1, 4), (1, 6)}
In ordered pairs, 1 is repeated three times. It means an element
1 has three images.
Hence, h is not a function.

(d) Here, k = {(a, 1), (a, 2), (a, 3), (b, 4)}
The element a is repeated three times in the ordered pairs.
Hence, k is not a function.

Which of the following relations are functions? Give your reasons.

(a) f (b) g

1 41 2
2 62 4
3 83 6
8
4

Solution: (a) Here, f represents a function because every element of domain
has unique image in co-domain.

(b) Here, g is not a function because the element 4 has no image in
the co-domains.

Example 3. Which of the following diagrams represent function? Draw vertical
lines to test the functions.

vedanta Excel in Additional Mathematics - Book 8 23

Relation and Function (b) Y (c) Y

(a) Y

X' X X' X X' OX
O O

Y' Y' Y'

Solution: (a) A vertical line drawn cuts the diagram at a point.
Example 4. Hence, the diagram represents a function.

(b) A vertical line cuts the diagram at two points.
Hence, the diagram does not represent a function.

(c) A vertical line drawn cuts the diagram at a point.
Hence, the diagram represents a function.

If a function f(x) = 3x + 5 and domain = {1, 2, 3}, find the range.

Also show the function in a mapping diagram.

Solution: Here, f(x) = 3x + 5

Domain of f = {1, 2, 3} = set of values of x.

Then, put x = 1, 2, 3 successively, we get:

f(1) = 3 . 1 + 5 = 3 + 5 = 8

f(2) = 3 . 2 + 5 = 6 + 5 = 11

f(3) = 3 . 3 + 5 = 9 + 5 = 14 f
? f = {8, 11, 14}

Showing the function in a 1 8
mapping diagram. 2 11
14
3

Example 5. If f is a function such that f(x + 3) = f(x) + f(3) for all real number x,
Solution: then show that (i) f(0) = 0 (ii) f(–3) = –f(3)

Here, f(x + 3) = f(x) + 5

(i) Putting x = 0, we get,

f(0 + 3) = f(0) + f(3)

or, f(3) = f(0) + f(3)

or, f(0) = f(3) – f(3)

? f(0) = 0

(ii) Again putting x = –3, we get,

f(–3 + 3) = f(–3) + f(3)

or, f(0) = f(–3) + f(3)

by using f(0) = 0, we get

f(–3) = –f(3) Proved

24 vedanta Excel in Additional Mathematics - Book 8

Relation and Function

Exercise 1.4

Very Short Questions :

1. (a) Define function.

(b) State any four ways of representing function.

2. Which of the following relations are functions? Give your reasons.

(a) f1 = {(1, 2), (2, 3), (3, 4)} (b) f2 = {(1, 3), (2, 3), (3, 4)}
(c) g1 = {(2, 4), (3, 9), (4, 16)} (d) g2 = {(2, 3), (2, 4), (2, 5)}
(e) g3 = {(1, 2), (1, 3), (1, 4)}

3. Which of the following relations are functions? Give your reason.

(a) R1 (b) R2

14 ax
25 by
35 c

(c) R3 (d) R4

ax 1 4
by 2 5
3 6
z 7

(e) R5 (f) R6

1 9 a x
2 8 b
3 11 c
4

Long Questions :

4. Draw the mapping diagrams for the following relations and state whether the
relations are functions or not.

(a) R1 = {(1, 4), (2, 5), (3, 6), (4, 7)} (b) R2 = {(1, 2), (2, 4), (3, 6), (4, 8)}
(c) R3 = {(1, 2), (2, 3), (1, 4), (3, 3)} (d) R4 = {(a, x), (b, x), (c, x)}
(e) R5 = {(1, 2), (1, 4), (3, 6)}

vedanta Excel in Additional Mathematics - Book 8 25

Relation and Function

5. Find the elements of the range of each of the following relations and also draw

mapping diagrams for them.

(a) y = f(x) = 2x + 3, Domain = {1, 2, 3, 4}
x + 4
(b) y = f(x) = 2 , Domain = {2, 3, 4, 5, 6}

(c) y = f(x) = x2 + 3, x  {2, 3, 4, 5}

(d) y = f(x) = 7x + 3, x  {–2, –1, 0, 1, 2}

6. If the function f is given by f(x) = 2x + 3, find the values of the following.

(a) f(3 + h) – f(3) (b) f(a + h) – f(a)
h h

7. If the function g is given by g(x) = x2,

(a) prove that g(–4) = g(4) (b) find the value of g(4) + g(–4)

(c) find the value of g(2) . g(3)

(d) find the value of g(a + h) – g(a)
h

8. (a) If f(x + 5) = f(x + 5) for all real number x, then, prove that

(i) f(0) = 0 (ii) f(–5) = –f(5)

(b) If f(x – 3) = f(x) + f(–3) for all real number x, then prove that

(i) f(0) (ii) f(3) = –f(–3)

(c) If f(x) = ax , then show that

f(m + n + p) = f(m) . f(x) . f(p)

Project Work

9. Let y = f(x) = mx + c be a linear function. Then, put m = 1 and c = 2. Then
the equation of the function is y = x + 2. Complete the table and plot graph of
the function.

x – 2 – 1 0 1 2 3 ...... ......

y ...... ...... ...... ...... ...... ...... 6 7

2. (a) Function (b) Function (c) Function

(d) Not a function (e) Not a function

3. (a) Function (b) Function (c) Not a function

(d) Function (e) Not a function (f) Function

4. (a) Function (b) Function (c) Not a function

(d) Function (e) Not a function 5.(a) {5, 7, 9, 11}

(b) 3, 7 , 4, 9 , 5 (c) {7, 12, 19, 28} (d) {–11, –4, 3, 10, 17}
2 2

6. (a) 2 (b) 7 7.(b) 32 (c) 36 (d) 2a + h

26 vedanta Excel in Additional Mathematics - Book 8

Polynomials

2Polynomials

2.1 Review

Let us review definition of the following basic terms:

Variable : A symbol which may takes, any value during a problem is known
as a variable. Variables are generally denoted by x, y, z, etc.

Constant : A symbol which takes the same value throughout
the mathematical operation is known as a constant.
Examples: 1, 2, 3, a, b, c, etc.

Term : A mathematical quantity which contains either a constant,
variable, or a product or ratio of both is called a term. Example:
Expression : 2, x2, 4x3, xy, 4xz2y, etc.
A group of terms connected either by + or – sign is called an
expression. eg 4x2 + 3, 5x + 6y – 7, x2 + 5x + 6, etc.

2.2 Polynomials

An algebraic expression in which the exponent of variable in each term is a non-
negative integer is called a polynomial.

A polynomial with variable x is denoted by f(x), g(x), h(x), etc.

A polynomial with variable y is denoted by f(y), g(y), h(y), etc.

Examples: 4x2 + 6x – 5. It is a polynomial as powers of x are positive integer.
1
x2 + x + 1. It is not a polynomial as its second term has power of x as 2 .

2.3 Degree of Polynomials

The highest degree of the given polynomial is known as its degree.

Examples: In x + 2, the degree of x is 1. It is a polynomial of degree 1.

In 3x2 + 5x + 6, the highest degree of x is 2. It is a polynomial of degree 2.

In x3 – 6x2 + 11x + 6, the highest degree x is 3. It is a polynomial of
degree 3.

vedanta Excel in Additional Mathematics - Book 8 27

Polynomials

2.4 Types of Polynomials

Polynomials can be categorized into following types, on the basis of number of terms
present in the polynomials.
(a) Monomial : A polynomial with only one term is called a monomial.

Examples : 2x, 5y, etc.
(b) Binomial : A polynomial with two terms is called a binomial.

Examples : 3x2 + 5x, 5x + 3, x3 + 9, etc.
(c) Trinomial : A polynomial with three terms is known as a trinomial.

Examples : 2x2 + 7x + 8, x3 + 10x + 15, y4 + 2y2 + 15, etc.
(d) Multinomial : A polynomial with more than three terms is called multinomial.

Examples : 3x4 + 4x3 + 4x2 + 7x + 8, y3 + 4y2 + 15y + 4, etc.

2.5 Standard form of polynomials

If the terms of a given polynomials are arranged either in ascending or in
descending power of the variable involved, then, the polynomial is said to be in
standard form.

Examples: (a) x2 + 5x + 6

(b) x3 – 3x2y + 3xy2 – y3

2.6 Equal Polynomials

Two polynomials in the same variable are said to be equal if all their corresponding
terms, degree and the number of terms are the same.

Example : 2x2 + 6x + 4 is equal to 4x2 + 12x + 8
2 2 2

2.7 Operation of Polynomials

While polynomials are operated, it is exactly the same as operating the algebraic
expression and all the four basic operations are possible.

(i) Addition : Two polynomials can be added to get a new polynomial by adding
the coefficients of like terms.

Examples: f(x) = x2 + 4x + 6

g(x) = 2x2 + 7x + 5

28 vedanta Excel in Additional Mathematics - Book 8

Polynomials

f(x) + g(x) = (x2 + 4x + 6) + (2x2 + 7x + 5)

= x2 + 2x2 + 4x + 7x + 6 + 5

= 3x2 + 11x + 11

(ii) Subtraction : Two polynomials can be subtracted to get a new polynomial by
subtracting the coefficients of the like terms.

Examples: f(x) = 3x2 + 5x + 6

g(x) = 2x2+ 3x + 5

g(x) – f(x) = (2x2 + 3x + 5) – (3x2 + 5x + 6)

= 2x2 – 3x2 + 3x – 5x + 5 – 6 = – x2 – 2x – 1

(iii) Multiplication : Two polynomials are multiplied to get a new polynomial by
following the law of indices, i.e. multiplicative law.

Examples: f(x) = 2x + 3

g(x) = 3x – 2

f(x) . g(x) = (2x + 3) (3x – 2)

= 2x(3x – 2) + 3(3x – 2)

= 6x2 – 4x + 9x – 6 = 6x2 + 5x – 6

(iv) Division : A polynomial with lower degree divides another polynomial with a
degree higher than it to get quotient and remainder.

Example : f(x) = x + 3

g(x) = x2+ 5x + 6

Here g(x) can not divide f(x) as g(x) has higher degree. We divide g(x) by f(x).

So, x + 3 x2 + 5x + 6 x + 2

Hence, –x2 +– 3x
2x + 6

– 2x +– 6
0

Quotient = x + 2

Remainder = 0

Hence, it can be concluded that Dividend = Divisor × Quotient + Remainder. This
mathematical statement is called division algorithm. It can be written as, f(x) = (x + a)
Q (x) + R where (x + a) is the divisor, Q(x) is the quotient and R is the remainder.

Note : Generally, the division of two polynomials is not a polynomial.

Let P(x) = x3 + x2 + 7x + 5,

vedanta Excel in Additional Mathematics - Book 8 29

Polynomials

q(x) = x

Then, p(x) ÷ q(x) = x3 + x2 + x + 5 = x2 + x + 7 + 5x–1
x

Here, the quotient is not a polynomial.

2.8 Synthetic Division

Usually, division is performed in traditional method. But it can also be performed
by only writing the coefficient of powers of variable in descending order. It is then
divided by comparing the divisor with x – a and a becomes the divisor. Then the
quotient is determined with a degree less than the dividend and the last number
becomes the remainder.

Example : Divide g(x) = x2 + 5x + 10 by f(x) = x – 2
Solution:
Comparing the divisor f(x) = x – 2 with x – a, we get a = 2.

Now, writing the coefficients of the dividend in descending powers of
x, we get

Coefficient of x2 Coefficient of x Constant
21 5 10

2 14

1 7 24 = R

Hence, Quotient, Q(x) = x + 7

Remainder, R = 24

Process of synthetic division

1. Check whether the dividend has higher power or not. If not, division is not
possible.

2. Compare the divisor with (x – a) and get the value of 'a' which acts as a new
divisor.

3. Write down the dividend in descending powers of x (variable) till the constant.
Mention 0 (zero) wherever any term is missing in order.

4. Draw a line to separate the divisor and the dividend.

5. Bring down the first or leading coefficient of the dividend as it is.

6. Multiply it with 'a', place it right below the next coefficient, and add it with it
to get a new value. Repeat the process until you reach the constant value.

7. The value of the sum of the constant column becomes the remainder and the
remaining values represents the quotient with a degree lower than the dividend.

30 vedanta Excel in Additional Mathematics - Book 8

Polynomials

Let's observe another example

Divide : x3 + 9x – 20 by x – 1
Solution: Comparing x – 1 the divisor with x – a we get, a = 1.

Writing the coefficients of x is descending order, we get,

Coefficient of x3 Coefficient of x2 Coefficient of x Constant
– 20
11 09
10
11
– 10 = R
1 1 10

? Quotient, Q(x) = x2 + x + 10
Remainder, R = – 10

Worked out Examples

Example 1. Find the sum of the following polynomials:
Solution:
(a) f(x) = x2 + 3x + 6 and g(x) = x2 – 7x + 1

(b) f(x) = –2 x3 + 3 x2 – 9 x – 1 and g(x) = 5 x3 – 8 x2 + 7 x + 5
3 5 2 2 3 5 2

(a) Here, f(x) = x2 + 3x + 6 and g(x) = x2 – 7x + 1

f(x) + g (x) = (x2 + 3x + 6) + (x2 – 7x + 1)

= x2 + x2 + 3x – 7x + 6 + 1

= 2x2 – 4x + 7

(b) Here, f(x) = –2 x3 + 3 x2 – 9 x – 5
3 5 2 2
5 8 5
and g(x) = 3 x3 – 3 x2 + 2 x + 5

? f(x) + g (x) = – 2 x3 + 3 x2 – 9 x – 5 + 5 x3 – 8 x2 + 5 x + 8
3 5 2 2 3 5 2

= –2 x3 + 5 x3 + 3 x2 – 8 x2 – 9 x + 5 x – 2 + 8
3 3 5 5 2 2 3

= –2 + 5 x3 + 3 – 8 x2 + –9 + 5 x+ –5 + 8
3 3 5 5 2 2 2

= 3 x3 + –5 x2 + –4 x+ 11
3 5 2 2

= x3 – x2 – 2x + 11
2

vedanta Excel in Additional Mathematics - Book 8 31

Polynomials

Example 2. Subtract as indicated.
Solution: (a) f(x) = x2 – 7x + 10 from g(x) = x2 – x – 6
(b) g(x) = 3x3 + 8x2 – 2x – 9 from f(x) = 2x3 + 11x2 + 5
(a) g(x) = x2 – x – 6 and f(x) = x2 – 7x + 10
So, g(x) – f(x) = (x2 – x – 6) – (x2 – 7x + 10)

= x2 – x – 6 – x2 + 7x – 10
= x2 – x2 – x + 7x – 6 – 10
= 6x – 16
(b) f(x) = 2x3 + 11x2 + 5 and g(x) = 3x3 + 8x2 – 2x – 9
So, f(x) – g(x) = (2x3 + 11x2 + 5) – (3x3 + 8x2 – 2x – 9)
= 2x3 + 11x2 + 5 – 3x3 – 8x2 + 2x + 9
= 2x3 – 3x3 + 11x2 – 8x2 + 2x + 5 + 9
= – x3 + 3x2 +2x + 14

Example 3. What should be added to the polynomial x3 + 6x2 + 6 to give the
Solution: polynomial 5x3 + 2x2 – 20x + 12 ?

Let A be added to the polynomial x3 + 6x2 + 6 to give the polynomial
5x3 + 2x2 – 20x + 12

So, A + x3 + 6x2 + 6 = 5x3 + 2x2 – 20x + 12

or, A = 5x3 – x3 + 2x2 – 6x2 – 20x + 12 – 6

= 4x3 – 4x2 – 20x + 6

Hence, 4x3 – 4x2 – 20x + 6 must be added to x3 + 6x2 + 6 to give

5x3 + 2x2 – 20x + 12.

Example 4. What should be subtracted from 2x2 + x – 6 to give the result x2 + 8x + 22?
Solution: Let 'A' be subtracted from 2x2 + x – 6 to get x2 + 8x + 22
So, 2x2 + x – 6 – A = x2 + 8x + 22
or, 2x2 + x – 6 – x2 – 8x – 22 = A
or, A = x2 – 7x – 28
Hence, x2 – 7x – 28 must be subtracted from 2x2 + x – 6 to get x2 + 8x + 22.

Example 5. Find the product of the following polynomials:
(a) f (x) = x + 5 and g (x) = x – 6
(b) f (x) = x – 4 and g(x) = x2 + x – 6

32 vedanta Excel in Additional Mathematics - Book 8

Polynomials

Solution: (a) f (x) . g (x) = (x + 5) (x – 6)
= x (x – 6) + 5(x – 6)
= x2 – 6x + 5x – 30
= x2 – x – 30

(b) f (x) . g (x) = (x – 4) (x2 + x – 6)
= x (x2 + x – 6) – 4(x2 + x – 6)
= x3 + x2 – 6x – 4x2 – 4x + 24
= x3 – 3x2 – 10x + 24

Example 6. Divide the polynomials f (x) by g (x). Find out the quotient Q(x) and
Solution: the remainder R(x).

(a) f (x) = x2 + 4x + 4 and g (x) = x + 2

(b) f (x) = x3 + 5x2 + x – 20 and g (x) = x + 2

(a) Here,
x+2

x + 2 x2 + 4x + 4
x2 + 2x

––
2x + 4
2x + 4

––
0

? Quotient; Q (x) = x + 2.

Remainder, R = 0

(b) Here,
x2 + 3x – 5

x + 2 x3 + 5x2 + x – 20
x3 + 2x2

––
3x2 + x
3x2 + 6x

––
– 5x – 20
– 5x – 10
++
– 10

? Quotient; Q(x) = x2 + 3x – 5

Remainder, R = – 10

vedanta Excel in Additional Mathematics - Book 8 33

Polynomials

Example 7. Using synthetic division method, divide the following polynomials
as indicated:

(a) f(x) = x2 + 5x + 10 by g (x) = x + 4

(b) f (x) = x3 + 5x2 + x – 8 by g(x) = x – 2

(c) f(x) = 3x3 – 4x2 + 11x + 10 by g (x) = 3x + 2

Solution: (a) Here, Comparing g(x) with x – a, we get,

a=–4

Now, writing the coefficients of dividend in descending powers of x,

Coefficient of x2 Coefficient of x Constant
–4 1 5 10

–4 –4

1 1 6=R

Hence, Quotient; Q (x) = x + 1, remainder; R = 6

(b) Here, Comparing g(x) with x – a, we get,

a=2

Writing the coefficient of dividend in descending powers of x, we get,

Coefficient of x3 Coefficient of x2 Coefficient of x Constant

21 5 1 –8

2 14 30

1 7 15 22 = R

Hence, Quotient; Q(x) = x2 + 7x + 15

Remainder; R = 22

(c) Comparing g (x) with x – a we get, a = – 2
3

At this level, if the comparison is found to be hard, simply

equate divisor to be zero.

Here, 3x + 2 = 0

or, x = – 2
3

Also, it can be written as

3x + 2 = 3 x + 2 =3 x– – 2
3 3

34 vedanta Excel in Additional Mathematics - Book 8

Polynomials

Writing the coefficients of dividend in descending powers of x, we get,

– 2 Coefficient of x3 Coefficient of x2 Coefficient of x Constant
3 3 – 4 11 10

– 2 4 – 10

3 – 6 15 0=R

Hence, Quotient, Q(x) = 3x2 – 6x + 15 = x2 – 2x + 5
3

Remainder, R = 0

Exercise 2

Very Short Questions :
1. Answer the following questions:

(a) Define variable with examples.
(b) Define constant with examples.
(c) Define polynomial. Give two examples of it.
(d) Define monomial, binomial, and trinomials with examples.
2. Which of the following are polynomials? Tick the polynomial and cross the one
which is not a polynomial.

(a) 2x + 4 (b) 10
(d) x2 + x + 8
(c) x2 – 9x + 8 (f) 3x2 + 7x + 5

(e) 2x2 + 1 +1
x2

(g) 7y3 – 4y + y (h) x3 – 3x2y + 10xy2 – y3
(i) 8x2 – 3x32 + 1
(j) x2y – x + 22
y

(k) 3x2 + x x + 4 (l) x2 + 4 + 18
x

3. In the following polynomials given below, find the degree and the number of
variables.

S.N. Polynomial Degree No of variable
(a) 2x2 + 5x + 12
(b) 4x2y + 5xy + 6
(c) 4x + 9
(d) 4x2 + 2x + 10
(e) 2x3 – 18x2 + 2x – 2

vedanta Excel in Additional Mathematics - Book 8 35

Polynomials

(f) 4x2y2 – 8xy2 + 3
(g) 4x2y3+6xy2 + 8xy + 4

4. Name the polynomials by the number of terms present in it. Also calculate its
degree.

S.N. Polynomials Name Degree
(a) 8
(b) 7x
(c) 3x + 4
(d) 3x2 – 9
(e) 2x2 – x – 7
(f) 5x3 – 2x – 5
(g) 8x4 – 3x – 1

5. Given below are some polynomials. Write down the polynomials in standard form.

S.N. Polynomials Standard form
(a)
(b) – x + 4x2 – 12
(c)
(d) 5 + 13x
(e)
(f) 2 + 5x2 – 9x + 7x3

(g) 10 + 4x3 – 8x + 7x2+ 10x4

1 + y2 + y4 – y – y3

x – x2 + x3
3 7 5

14x + x3 + 2x6 – 8x2

Short Questions :

6. Find the sum of the following:

(a) f(x) = 2x + 3 and g(x) = 9x + 7

(b) f(x) = x2 + 5x + 6 and g(x) = 2x2 – 3x + 1

(c) f(x) = x3 – 5x2 + x + 2 and g(x) = x3 – 3x2 + 2x + 1

(d) f(x) = x3 + x2 + 3x + 4 and g(x) = x2 + x – 7

(e) f(x) = 4x4 – 3x2 + 2 and g(x) = x4 – 2x2 + 6

(f) f(x) = x4 – 3x3 + 2x + 6 and g(x) = x4 – 3x + 2

(g) f(x) = 1 x3 + 6 x2 – 7x and g(x) = 13 x3 – 11 x2 + 5x – 16
3 5 3 5
x3 x2 x3 x2 x
(h) f(x) = 3 + 2 +x and g(x) = 2 + 3 + 4

7. Find the difference of the following, i.e. subtract f(x) from g(x):

36 vedanta Excel in Additional Mathematics - Book 8

Polynomials

(a) f(x) = 9x – 6 and g(x) = 15x + 12.

(b) f(x) = x2 – x – 6 and g(x) = 2x2 – 9x + 7

(c) f(x) = x3 – 3x2 +3x – 1 and g(x) = x3 + 3x2 + 3x + 1

(d) f(x) = 6x2 + x – 2 and g(x) = 8x3 + 3x2 + 7

(e) f(x) = 3x4 – 3x + 2 and g(x) = x4 – 2x2 + 3

(f) f(x) = x4 – 5x + 3 and g(x) = x4 – 3x3 + 2x + 6

(g) f(x) = x3 + x2 + x and g(x) = x3 + x2 + x
3 2 4 2 3 4
9 5 8 7
(h) f(x) = 2 x3 + 3 x2 – 2 x– 1 and g(x) = 3 x3 – 5 x2 + 2 x+5
3 5 2
8. Find the product of the following:

(a) f(x) = x + 3 and g(x) = x – 3

(b) f(x) = x – 4 and g(x) = x – 6

(c) f(x) = x – 8 and g(x) = x2 + 5x + 3

(d) f(x) = x2– x – 4 and g(x) = x + 3

(e) f(x) = x2 + 3x + 3 and g(x) = x2 + 3x + 8

(f) f(x) = x2 + 3x + 4 and g(x) = x2 – x + 2

(g) f(x) = x2 – 2x + 3 and g(x) = x3 – 3x2 + 3x – 3

(h) f(x) = x2 + 4x + 1 and g(x) = x3 – x2 + 5

Long Questions :

9. (a) What should be added to 2x2 + 3x + 4 to get 3x2 + 2x – 5?

(b) What should be added to 3x3 – 7x – x2 + 8 to get 4x3 + 2x2 – 2x – 5?

(c) What should be added to y4 + y3 + 2y2 – 10 to get y6 – 2y4 + y3 + 2y2 – 4?

10. (a) What should be subtracted from 2x2 + x – 10 to get 2x2 – 7x + 22 ?

(b) What should be subtracted from 2x3 + 11x2 + 22 to get x3 – 19x2 + 2x – 2?

(c) What should be subtracted from y7 – 3y6 + 3y2 + 6 to get y7 – 4y6 + 5y2 + y + 5?

11. Divide the polynomial f(x) by g(x) in the following :

(i) by division method (ii) by synthetic division method

(a) f(x) = x2 – 5x + 8 and g(x) = x – 4

(b) f(x) = x3 – 3x2 + 3x – 5 and g(x) = x – 2

(c) f(x) = x2 + 7x + 9 and g(x) = x + 3

(d) f(x) = 4x3 + 9x2 + 3x – 7 and g(x) = x + 1

vedanta Excel in Additional Mathematics - Book 8 37

Polynomials

(e) f(x) = 4x4 – 8x2 + 3x – 12 and g(x) = x + 4
(f) f(x) = x3 + 13x + 20 and g(x) = x + 4
(g) f(x) = x3 – 19x – 40 and g(x) = x – 3
(h) f(x) = 3x3 – 13x2 + 22 and g(x) = x + 2
(i) f(x) = 4x3 – 5x + 6 and g(x) = 2x – 4
(j) f(x) = 3x3 – 4x2 + 11x + 5 and g(x) = 3x + 3

2. (a) polynomial (b) polynomial (c) polynomial (d) polynomial

(e) Not polynomial (f) polynomial (g) Not polynomial (h) polynomial

(i) Not polynomial (j) Not polynomial (k) Not polynomial

(l) Not polynomial 3. and 4. Show to your teacher

5. (a) 4x2 – x – 12 (b) 13x + 5 (c) 7x3 + 5x2 – 9x + 2

(d) 10x4 + 4x3 + 7x2 – 8x + 10 (e) y4 – y3 + y2 – y + 1

(f) x3 – x2 + x (g) 2x6 + x3 – 8x2 + 14x
5 7 3
6. (a) 11x + 10 (b) 3x2 + 2x + 7 (c) 2x3 – 8x2 + 3x + 3

(d) x3 + 2x2 + 4x – 3 (e) 5x4 – 5x2 + 8 (f) 2x4 – 3x3 – x + 9

(g) 14 x3 – x2 – 2x – 16 (h) 5 x3 + 5 x2 + 5 x
3 6 6 4
7. (a) 6(x + 3) (b) x2 – 8x + 13 (c) 2(3x2 + 1) (d) 8x3 – 3x2 – x + 9

(e) –2x4 – 2x2 + 3x + 1 (f) – 3x3 + 7x + 3 (g) x62(x–1) (h) x3 – 11 x2 + 8x + 11
5 2
8. (a) x2 – 9 (b) x2 – 10x + 24

(c) x3 – 3x2 – 37x – 24 (d) x3 + 2x2 – 7x – 12

(e) x4 + 6x3 + 20x2 + 33x + 24 (f) x4 + 2x3 + 3x2 + 2x + 8

(g) x5 – 5x4 + 12x3 – 18x2 + 15x – 9 (h) x5 + 3x4 – 3x3 + 4x2 + 20x + 5

9. (a) x2 – x – 9 (b) x3 + 3x2 + 5x – 13 (c) y6 – 3y4 + 6

10. (a) 8x – 32 (b) x3 + 30x2 – 2x + 24 (c) y6 – 2y2 – y + 1

11. (a) Q(x) = x – 1, R = 4 (b) Q(x) = x2 – x + 1, R = –3

(c) Q(x) = x + 4, R = – 3 (d) Q(x) = 4x2 + 5x – 2, R = – 5

(e) Q(x) = 4x2 – 16x2 + 56x – 221, R = 872 (f) Q(x) = x2 – 4x + 29, R = –96

(g) Q(x) = x2 + 3x – 10, R = – 70 (h) Q(x) = 3x2 – 19x + 38, R = – 54
(i) Q(x) = 2x2 + 4x + 121, R = 28
(j) Q(x) = x2 – 7x + 6, R = –13
3

38 vedanta Excel in Additional Mathematics - Book 8

Real Number System and 3
Surds

3.1 Introduction

In mathematics, a real number is a value of a continuous quantity that can represent

a distance along a line. The adjective real in this context was introduced by Rene

Descartes, who distinguised between real and imaginary roots of polynomials. The

real numbers include all the rational numbers such as the integer 5 and the fraction
4
o5f , and all the irrational numbers, such as 3 (1.732050808 .....), the square root
3. Real numbers can be used to measure quantities such as time, mass, energy,

velocity, and many more.

Real numbers can be thought of as points on an infinitely long line called the
real number line or simply the number line, where the points corresponding to
integers are equally spaced. Any real number can be determined by a possibly
infinite decimal representation such as 2.832....., where the points corresponding
to integer and equally spaced. Any real can be determined by a possibly decimal
representation such as that of 8.632 where each consecutive digit is measured in
units one tenth the size of the previous one.

3

–5 –4 –3 –2 –1 0 1 2 3 4 5

Real numbers can be thought of as points infinitely long number line

These descriptions of the real number are not sufficiently rigorous by the modern
standards of pure mathematics.

3.2 Real Numbers System

The set of all rational and irrational number taken together form a new system of
numbers known as real number system. One important property of the real number
is that they are ordered. Between any two real numbers a and b, only one relation
a = b or, a > b or, a < b holds.

vedanta Excel in Additional Mathematics - Book 8 39

Real Number System and Surds

The diagram for the real number system is given below :
Real Numbers (R)

Rational Numbers (Q) Irrational Numbers (Ir)

Integers (Z) Fractions Positive Negative

Positive Integers Zero Negative Integers

Let us define different types of real numbers.

Natural Number (N)
The counting numbers are called natural numbers. The set of natural numbers is
denoted by N.

N = {1, 2, 3, .......}

Whole Numbers
Counting numbers including zero are called whole numbers. The set of whole
numbers is denoted by W.

W = {0, 1, 2, 3, ........}

Integers
The natural numbers with their additive inverse (or negatives) including 0 are called
integers. It is denoted by Z, Z = {...., –2, – 1, 0, 1, 2, .....}

Natural Numbers

–5 –4 –3 –2 –1 0 1 2 3 4 5

Integers Whole Numbers

From the above number line, we have N  W  I
Every natural number is a whole number and every whole number is an integer.

Rational Numbers

Any number which can be expressed as the ratio of two integers (denominator being
not zero) is known as rational number. The set of rational number is denoted by Q.

40 vedanta Excel in Additional Mathematics - Book 8

Real Number System and Surds

Q= x : x = p , p, q  z, q z 0
q
1 1 7
Examples : 4 , 3 , 7, 5, 2 , etc.

Terminating and repeated decimal numbers are rational number.

Irrational Number

The numbers which are not rational are called irrational numbers. 5, 3, 2 etc.
are examples of irrational numbers. They are denoted by Q or Ir.

Q= x : x z p , p, q  z, q z 0
q

Example : S (pi) is a famous irrational number.

S = 3.142857143....... (and more)

We cannot write down a simple fraction that equals pi.

The popular approximation of pi is 22 = 3.142857143............
7

3.3 Surds

Surds are numbers left in root form to express its exact value. It has an infinite
number of non-recurring decimals. Therefore, surds are irrational numbers when
we cannot simplify a number to remove a square root (or cube root etc.), it is a surd.

Example : 2 (square root of 2) cannot be simplified further; so, it is a surd.

Example : 4 (square root of 4) can be simplified (to 2); so, it is not a surd.

Let us take some more examples:

Number Simplified As a decimal Remarks
2 2 1.41421356..... Surd
1.73205080..... Surd
3 3
2 2 Not a surd
4 0.5 Not a surd
1
1 2 2.2239800..... Surd
4 1.24573094..... Surd
3 11 3 11 1.709972947..... Surd
53 53 Not a surd
35 35 5
3 125 5

The surds have a decimal which goes on forever without repeating and they are
irrational numbers.

vedanta Excel in Additional Mathematics - Book 8 41

Real Number System and Surds

From above table, we have,

2 = 1.4421356..... (Non-terminating and non-repeated decimal number)

3 = 1.73205080 (Non-terminating and non-repeated decimal number)

Non-terminating and non-repreated decimal numbers are called irrational numbers.
Surds are also irrational numbers.

Definition: An irrational root of a rational number is called a surd. In other words,
the root of a rational number whose value cannot be obtained exactly is called a
surd. If n p be a surd, the sign ' ' is called radical sign, p is called radicand and n is
the order (or degree) of the surd.

Order of Surds
The surd n p can also be written as p1/n. Here, n is called order (or degree) of the surd.
The order of the surd is also called order (or degree) of the surds. The order of the
surd is the number which shows the root.

n p is the surd of order n.

Examples : (i) 3 2 = 21/3 is a surd of order 3.

(ii) 5 21 = 211/5 is a surd of order 5.

3.4 Different types of Surds

(i) Like and Unlike Surds:
The surds with the same order and same radicand are called like surds.

Examples : (i) 5 , 2 5 , – 5, etc. are like surds.
(ii) 3 5 , 53 5, – 23 5, etc. are like surds.

The surds which have different radicand are unlike surds. The degree of unlike
surds may be same or different.

(ii) Pure and Mixed Surds:
A surd having only one irrational factor is known as a pure surd.
Example : (i) 3 and 3 5 are pure surds

(ii) 3 7 and 3 9 are pure surds

A surd having a product of a rational and irrational numbers is called mixed surd.
Example : (i) 3 2 and 53 5 are mixed surds

(ii) 43 7 and 10 7 are mixed surds

(iii) Simple and Compound Surds:
A surd having only one term is called a simple surd.

Example : 7, 4 5, 7 2, etc. are simple surds.

42 vedanta Excel in Additional Mathematics - Book 8

Real Number System and Surds

A surd containing two or more terms is called a compound surd.

Example : 3 + 5, 7 – 2 5 are compound surds.

(iv) Conjugate Surds:
Two binomial surds which are different only in sign (+ or –) are called conjugate
surds.

Example : (i) 7 + 5 and 7 – 5 are conjugate surds.

(ii) 2 3 – 2 2 and 2 3 + 2 2 are conjugate surds.

Laws of Surd

Let a and b be two positive real numbers and m, n and p be integers. Then,

(i) n an = a e.g. 3 33 = 3

(ii) n a . n b = n ab e.g. 5 × 3 = 5 × 3 = 15

(iii) n a = n a e.g. 3 4 = 3 4 = 3 2
n b b 3 2 2

(iv) m n a = mn a = n m a e.g. 4 3 4 = 4×3 4 = 12 4

(v) n ap = n m ap m = mn apm e.g. 3 2 = 3×4 24 = 12 16 = 24×=112 = 1

23

3.5 Operation of Surds

(i) Addition and Subtraction of Surds

Two or more like surds can be added or subtracted. To add or subtract surds,
follow the given ways :

(a) Express all the like surds into simplest form.

(b) Add or subtract the coefficients of like terms keeping the irrational factor
same.

Examples : (i) 7 + 3 7 =(1 + 3) 7 = 4 7

(ii) 4 3 – 2 3 = (4 – 2) 3 = 2 3

(iii) 18 + 50 = 9 × 2 + 25 × 2

= 3 2 + 5 2 = (3 + 5) 2 = 8 2

(iv) 32 + 8 – 18 = 16 × 2 + 4 × 2 – 9 × 2

=4 2+2 2–3 2

= (4 + 2 – 3) 2 = 3 2

(ii) Multiplication and Division of Surds:
Two or more surds of same order can be multiplied.

Example: (i) 3 × 2 = 3×2= 6

vedanta Excel in Additional Mathematics - Book 8 43

Real Number System and Surds

(ii) 3 5 × 3 25 = 3 5 × 25 = 3 53 = 5

(iii) 3 7 × 4 7 = 12 7 × 7

= 12 × 7 = 84

Note: n a . n b = n ab

A surd can be divided by another surd of the same order.

Example: (i) 4 7 ÷ 2 7 = 4 7 = 2
(ii) 33 72 ÷ 3 9 2 7

= 33 72 = 3 3 72
9 9
3

= 33 8 = 3 × 2 = 6

3.6 Rationalization of Surds

When the product of two surds is a rational number, each of them is said to be the
rationalizing factor of the other.

Example: 3 + 2 is rationalizing factor of 3 – 2.

Then, 3 + 2 × ( 3 – 2 ) = 3 – 2 = 1.

The process from which a surd is changed to a rational number by multiplying
it with a suitable factor is called the rationalization of the surd.

The following are some basic steps for rationalization.

(i) Multiply the given surd by a simplest rationalizing factor.

Example : 4 2 × 2 = 4 × 2 = 8

Here 2 is a simplest rationalizing factor.

(ii) Multiply a binomial surd by its conjugate.

Example : 3 + 2, its conjugate is 3 – 2

Now, ( 3 + 2) × ( 3 – 2 ) = ( 3 )2 – ( 2 )2 = 3 – 2 = 1

Here, 3 – 2 is a rationalizing factor of 3 + 2.

(iii) When a surd is in the form of quotient, multiply both the numerator and denominator
by conjugate of the denominator to make denominator a rational number.

Example : 2 + 5
5– 2
(multiplying numerator and denominator by 5 + 2)

2 + 25× 5+2 = ( 5 + 2)2 = ( 5 )2 + 2 . 5 . 2 + 22
5– 5+2 ( 5 )2 – 22 5–4

= 5 + 45 + 4 = 9 + 4 5
1

44 vedanta Excel in Additional Mathematics - Book 8

Real Number System and Surds

Worked out Examples

Example 1. Add : 4 7 + 8 7
Solution: Here, 4 7 + 8 7
= (4 + 8) 7 = 12 7

Example 2. Subtract : 8 5 – 5 5
Solution: Here, 8 5 – 5 5
= (8 – 5) 5 = 3 5

Example 3. Multiply :
Solution:
(a) 5 3 × 4 2 (b) 3 3 × 2

(a) Here, 5 5 × 4 2

= 20 3 × 2 = 20 6
(b) Here, 3 3 × 2

LCM of 3 and 2 = 6

3 3 × 2 = 3×2 32 × 2×3 23 = 6 9 × 8 = 6 72

Example 4. Divide : 3 625 ÷ 3 5
Solution: Here, 3 625 ÷ 3 5
Example 5.
Solution: = 3 625 = 3 625 = 3 125 = 3 53 = 5
35 5
Example 6.
Solution: Compare surds 3 4 and 4 5

Here, 3 4 and 4 5

LCM of 3 and 4 = 3 × 4 = 12
Now, 3 4 = 3×4 44 = 12 256

4 5 = 4×3 53 = 12 125

Comparing the radicands, we get 256 > 125

? 12 256 > 12 125 i.e. 3 4 > 4 5

Arrange the surds in ascending order 2, 3 4 and 4 6
Given surds are 2, 3 4 and 4 6

LCM of 2, 3 and 4 = 12
Now, 2 = 2 2 = 2×6 26 = 12 64

vedanta Excel in Additional Mathematics - Book 8 45

Real Number System and Surds

3 4 = 3×4 44 = 12 256
4 6 = 4×3 63 = 12 216

Arranging in ascending order,
2, 4 6, 3 4

Example 7. Simplify : (b) 4 81 – 4 2401 + 4 625
Solution: (a) 8 – 128 + 32

(a) Here, 8 – 128 + 32

= 4 × 2 – 64 × 2 + 16 × 2

= 2 2–8 2+4 2

= (2 + 4) 2 – 8 2

= 6 2–8 2

= –2 2

(b) Here, 4 81 – 4 2401 + 4 54
= 4 34 – 4 74 + 4 54

= 3–7+5=8–7=1

Example 8. Find the value of (3 5 – 4 3)2
Solution: Here, (3 5 – 4 3)2
= (3 5)2 – 2 . 3 5 . 4 3 + (4 3 )2
= 45 – 24 5 × 3 + 48
= 93 – 24 15
= 3(31 – 8 15 )

Example 9. Rationalize denominator : 3 + 6
Solution: 3 – 6
3+ 6
Here, 3– 6

= 3+ 6 × 3 + 6
3– 6 3 + 6

= (3 + 6 )2
32 – ( 6 )2
9+2.3. 6+6
= 9–6

= 15 + 6 6
3

46 vedanta Excel in Additional Mathematics - Book 8

Real Number System and Surds

= 3(5 + 2 6)
3

= 5+2 6

Example 10. Simplify : 6 – 12 + 2 108
3 12
6 12
Solution: Here, 3 – 12 + 2 108

= 6 × 3 – 12 × 3 + 2 36 × 3
3 3 23 3

= 63 – 12 3 + 12 3
3 6

= 2 3 – 2 3 + 12 3

= 12 3

Example 11. Simplify : 2 6 3– 43 2 + 32
+ 6+ 6+ 3

Solution: Here, 6 3– 43 2 + 32
2+ 6+ 6+ 3

= 6 3 × 2 – 3 – 43 2 × 6– 2 + 32 3 × 6– 3
2+ 2 – 3 6+ 6– 2 6+ 6– 3

= 12 – 6 3 – 4 18 – 4 6+3 12 – 3 6
4–3 6–2 6–3

= 12 – 6 3 – 4( 9×2– 6 ) + 3( 4×3– 6)
4 3

= 12 – 6 3 – 3 2 + 6 + 2 3 – 6

= 12 – 4 3 – 3 2

Exercise

Short Questions (b) 3 7 + 5 7
1. Add : (d) 23 4 + 73 32 + 3 108

(a) 3 5 + 2 5 (b) 4 8 – 50
(c) 12 + 27 (d) 4 xy5 – 4 x5y
2. Subtract :
(a) 5 7 – 2 7 (b) 2 3 + 5 3 – 4 3
(c) 27 – 3 48
3. Simplify :
(a) 10 5 – 7 5 + 2 5

vedanta Excel in Additional Mathematics - Book 8 47

Real Number System and Surds

(c) 5 2 + 3 2 – 5 2 (d) 2 11 + 4 11 – 10 11

4. Multiply :

(a) 5 × 3 (b) 4 × 3

(c) 2 × 3 × 12 (d) 20 × 125 × 45

5. Divide :

(a) 3 12 ÷ 3 (b) 3 27 ÷ 3 3 (c) 44 32 ÷ 4 2

6. Find the value of :

(a) ( 3 + 2 )2 (b) (2 3 + 3 2 )2

(c) (3 5 – 4 3 )2 (d) (4 5 – 5 3 )2

7. Compare the given surds : (b) 4 5 and 3 4
(a) 3 25 and 7 (d) 4 3 and 3 4
(c) 3 18 and 6 144

8. Find the rationalizing factor of :

(a) 7 (b) 2 5 (c) 125

(d) 3 49 (e) 3 16

9. Rationalize the denominator :

(a) 1 (b) 1 (c) 32
5 (e) 15 39

(d) 2 343 1 (f) 1
77 3–1 5– 2

(g) 1
5– 3

Long Questions :

10. Simplify :

(a) 5 125 – 6 45 + 2 5 (b) 20 + 4 80 – 2 180

(c) 4 90 – 6 160 + 245 – 5 125 (d) 3 40 + 33 625 – 3 320

11. Arrange the following surds in ascending order :

(a) 3 5, 2 and 7 (b) 3 7, 4 5 and 3

(c ) 5, 3 11 and 26 3

12. Rationalize the denominators and simplify :

(a) 7+ 3 (b) 2+ 5 (c) 5+ 2
7– 3 2– 5 5– 2

48 vedanta Excel in Additional Mathematics - Book 8

Real Number System and Surds

(d) 5– 3 (e) 7– 5
5+ 3 7+ 5

13. Simplify :

(a) 2 6– 5 (b) 5+2 3 (c) m+n– m–n
3 5–2 6 7+4 3 m+n+ m–n

14. Simplify :

(a) 3+ 2 + 3– 2 (b) 5+ 3 + 5– 3
3– 2 3+ 2 5– 3 5+ 3

(c) 7– 5 + 7+ 5
7+ 5 7– 5

15. Simplify :

(a) 32 3 + 43 2 – 42 (b) 6 3+ 32 3 – 43
6+ 6+ 6– 2 2+ 6+ 6– 2

1. (a) 5 5 (b) 10 7 (c) 5 3 (d) 193 4
(c) –9 3 (d) (y – x)4 xy
2. (a) 3 7 (b) 3 2

3. (a) 5 5 (b) 3 5 (c) 3 2 (d) –4 11

4. (a) 15 (b) 2 3 (c) 6 2 (d) 150 5
(b) 3 9
5. (a) 6 (c) 8

6. (a) 5 + 2 6 (b) 6(5 + 2 6) (c) 3(31 – 8 15) (d) 5(31 – 8 15)
(c) 3 18 > 4 144 (d) 4 3 < 3 4
7. (a) 3 25 > 7 (b) 4 5 < 3 4 (d) 3 7 (e) 3 4
(c) 5
8. (a) 7 (b) 5

9. (a) 5 (b) 15 (c) 36 (d) 2
5 15 3
1 1 1
(e) 2 ( 3 + 1) (f) 3 ( 5 + 2) (g) 2 ( 5+ 3)

10. (a) 9 5 (b) 6 5 (c) –6(2 10 + 3 5) (d) 133 5

11. (a) 2 < 3 5 < 7 (b) 4 5 < 3 < 3 7 (c) 3 11 < 5 < 26 3

12. (a) 26 +7 3 (b) – 9 – 4 5 (c) 7+2 10 (d) 4 – 15 (e) 6 – 35
23 3
9 +4 30 m– m2 – n2
13. (a) 21 (b) 11 – 6 3 (c) n
(b) 8
14 (a) 10 (b) –2 16 (c) 12

15. (a) 3 2 – 2 6 – 2

vedanta Excel in Additional Mathematics - Book 8 49

Measurement of Angles 4

Measurement of Angles

4.1 Angle

An angle is said to be formed when two line segments meet C
at a common point. The common point is called vertex.
B
In figure, AC and AB are two line segments meeting each A Q
other at A. Here, A is called the vertex of angle BAC. Positive angle

Let OP be a revolving line which may revolve in both O P
direction, i.e. clockwise and anti-clockwise.
Negative angle
If OP revolves in anti-clockwise direction and traces R
an angle ‘POQ such that ‘POQ is a positive angle.

Again, if OP moves in clockwise direction and traces
an angle ‘POR such that ‘POR is a negative angle.

4.2 Measurement of an Angle

We use degree system of
measurement for measuring an
angle. For this, we use a protractor
to measure the angle where a half
circle protractor is divided into 180
small parts. 1 part called 1 degree.

The following are the system of
measurement of angles.

(a) Sexagesimal / Degree / English system

(b) Centesimal / Grade / French system

(c) Radian / Circular system.

(a) Sexagesimal system

This system has been the latest system of measurement of angles. It is said to be

50 vedanta Excel in Additional Mathematics - Book 8