bt0063 math

BT 0063

Mathematics for IT (with Solved Problems)

Contents

Unit 1

Set Theory 1

Unit 2

Mathematical Logic 28

Unit 3

Modern Algebra 53

Unit 4

Trigonometry 70

Unit 5

Limits and Continuity 110

Unit 6

Differentiation 145

Unit 7

Integrations 192

Unit 8

Differential Equations 231

Unit 9

Complex Numbers 250

Edition: Spring 2009

BKID – B0947 3rd

Jan. 2009

Unit 10

Matrices and Determinants 269

Unit 11

Infinite Series 306

Unit 12

Probability 326

Unit 13

Basics Statistics 366

Reference 415

Prof. V. B. Nanda Gopal Director & Dean Directorate of Distance Education Sikkim Manipal University of Health, Medical & Technological Sciences (SMU DDE)

Board of Studies Dr. U. B. Pavanaja (Chairman) Mr. Nirmal Kumar Nigam General Manager – Academics HOP- IT Manipal Universal Learning Pvt. Ltd. Sikkim Manipal University – DDE Bangalore. Manipal.

Prof. Bhushan Patwardhan Dr. A. Kumaran Chief Academics Research Manager (Multilingual) Manipal Education, Bangalore Microsoft Research Labs India, Bangalore.

Dr.Harishchandra Hebbar Mr. Ravindranath.P.S. Director Director (Quality) Manipal Centre for Info. Sciences, Manipal Yahoo India, Bangalore

Dr. N. V. Subba Reddy Dr. Ashok Kallarakkal HOD-CSE Vice President Manipal Institute of Technology, Manipal IBM India, Bangalore.

Dr. Ashok Hegde Mr. H. Hiriyannaiah Vice President Group Manager MindTree Consulting Ltd., Bangalore. EDS Mphasis, Bangalore.

Dr. Ramprasad Varadachar Mr. Ashok Kumar K Director, Computer Studies Additional Registrar Dayanand Sagar College of Engg. Bangalore Sikkim Manipal University – DDE, Manipal.

Mr. M. K. N. Prasad Controller of Examinations Sikkim Manipal University – DDE, Manipal.

Content Preparation Team Content Writing Content Editing Mr. Deepak Shetty Dr. Kuncham Syam Prasad Assistant Professor (Mathematics) Associate Professor-Dept. of Mathematics Sikkim Manipal University – DDE, Manipal Institute of Technology, Manipal.

Mathematical Content Editing Mr. Deepak Shetty Assistant Professor (Mathematics) Sikkim Manipal University – DDE, Manipal.

Edition: Spring 2009 This book is a distance education module comprising a collection of learning material for our students. All rights reserved. No part of this work may be reproduced in any form by any means without permission in writing from Sikkim Manipal University of Health, Medical and Technological Sciences, Gangtok, Sikkim. Printed and published on behalf of Sikkim Manipal University of Health, Medical and Technological Sciences, Gangtok, Sikkim by Mr.Rajkumar Mascreen, GM, Manipal Universal Learning Pvt. Ltd., Manipal – 576 104. Printed at Manipal Press Limited, Manipal.

BLOCK INTRODUCTION

In this book Mathematics for IT we study the different concepts of

mathematics with well illustrated examples.

Unit 1: This deals with the concept of sets. The different ideas in set

theory is discussed in this unit in a simple manner.

Unit 2: In this unit we learn about statement, truth values of a

statement, compound statements, algebra of statement and use

Venn diagrams in logic.

Unit 3: This unit deals with the theory of groups which is a branch of

algebra and is very important in the development of

mathematics.

Unit 4: This unit deals with the concept of Trigonometry, where we

study Radian measure, Area of sector of a circle, Trigonometric

functions, standard angles, allied angles, compound angles and

other related concepts.

Unit 5: In this unit we are recalling the concepts of numbers and

studying the Limits of a Function of a discrete variable. We will

also be studying the concept of continuity which is essential for

describing a process that goes on without any change.

Unit 6: In this unit we study the rate of change of a continuous function.

The rate of growth of any object with respect to time can be

studied using derivatives.

Unit 7: In this unit we study integration as the inverse operation of

differentiation. It generalizes the process of summation. Also the

area under the graph of a function is discussed here.

Unit 8: In this unit we study a branch of mathematics called as

Differential Equations. Here the technique of converting a given

problem into Differential Equations which is then solved and the

solution to the problem is found out is discussed.

Unit 9: In this unit the concept of complex numbers is discussed with

many illustrative examples. The idea of modulus of a complex

numbers, exponential form of a complex numbers and De

Moivere’s Theorem is explained in a simple manner.

Unit 10: In this we study a powerful tool in mathematics called as

Matrices and Determinants. The concept is mainly developed for

solving equations.

Unit 11: This unit deals with the concept of Infinite Series. Here we study

Convergence and Divergence of Series. The concept of Binomial

Series, Exponential Series, Logarithmic Series is discussed here

in a simple manner

Unit 12: This unit deals with the concept of probability. The idea of

Conditional Probability and Independence of Events is discussed

here. Baye’s theorem and its applications is also explained here.

Unit 13: In this unit of Basic Statistics we study Measure of Central

Tendency, Standard Deviation, Discrete Series and Continuous

Series. Also the idea of standard deviation and Variance is dealt

with examples wherever necessary.

Mathematics for IT Unit 1

Sikkim Manipal University Page No.: 1

Unit 1 Set Theory

Structure

1.1 Introduction

Objectives

1.2 Sets and Their Representations

1.3 The Empty Set

1.4 Finite and Infinite Sets

1.5 Equal and Equivalent Sets

1.6 Subsets

1.7 Power Set

1.8 Universal Set

1.9 Venn Diagrams

1.10 Complement of a Set

1.11 Operations on Sets

1.12 Applications of Sets

1.13 Cartesian Product of Sets

1.14 Summary

1.15 Terminal Questions

1.16 Answers

1.1 Introduction

The concept of set is basic in all branches of mathematics. It has proved to

be of particular importance in the foundations of relations and functions,

sequences, geometry, probability theory etc. The study of sets has many

applications in logic philosophy, etc.

The theory of sets was developed by German mathematician Georg Cantor

(1845 – 1918 A.D.). He first encountered sets while working on problems on

trigonometric series. In this unit, we discuss some basic definitions and

operations involving sets.



Objectives:

At the end of the unit you would be able to

understand the concepts of sets

perform the different operations on sets

write the Power set of a given set

1.2 Sets and their Representations

In every day life, we often speak of collection of objects of a particular kind

such as pack of cards, a herd of cattle, a crowd of people, cricket team, etc.

In mathematics also, we come across various collections, for example,

collection of natural numbers, points in plane, prime numbers. More

specially, we examine the collections:

i) Odd natural numbers less than 10, i.e., 1, 3, 5, 7, 9

ii) The rivers of India

iii) The vowels in the English alphabet, namely a, e, I, o, u

iv) Prime factors of 210, namely 2, 3, 5 and 7

v) The solutions of a equation x2 – 5x + 6 = 0 viz, 2 and 3

We note that each of the above collections is a well defined collection of

objects in the sense that we can definitely decide whether a given object

belongs to a given collection or not. For example, we can say that the river

Nile does not belong to collection of rivers of India. On the other hand, the

river Ganga does belong to this collection. However, the following

collections are not well defined:

i) The collection of bright students in Class XI of a school

ii) The collection of renowned mathematicians of the world

iii) The collection of beautiful girls of the world

iv) The collection of fat people



For example, in (ii) above, the criterion for determining a mathematician as

most renowned may vary from person to person. Thus, it is not a well

defined collection.

We shall say that a set is a well defined collection of objects. The following

points may be noted:

i) Objects, elements and members of a set are synonymous terms. These

are undefined

ii) Sets are usually denoted by capital letters A, B, C, X, Y, Z etc.

iii) The elements of a set are represented by small letters a, b, c, x, y, z etc.

If a is an element of a set A, we say that „a belongs to A‘. The Greek symbol

is used to denote the phrase „belongs to‟. Thus, we write A. If b is not

an element of a set A, we write b A and read „b does not belong to A‘.

Thus, in the set V of vowels in the English alphabet, a V but l V. In the

set P of prime factors of 30, 3 P but 15 P.

There are two methods of representing a set:

i) Roster or tabular form

ii) Set builder form.

i) In roster form, all the elements of a set are listed, the elements being

separated by commas and are enclosed within braces { }. For example,

the set of all even positive integers less than 7 is described in roster

form as {2, 4, 6}. Some more examples of representing a set in roster

form are given below:

a) The set of all natural numbers which divide 42 is {1, 2, 3, 6, 7, 14, 21, 42}.

Note that in roster form, the order in which the elements are listed is

immaterial. Thus, the above set can also be represented as

{l, 3, 7, 21, 2, 6, 14, 42}.

b) The set of all vowels in the English alphabets is {a, e, i, o, u}.



c) The set of odd natural numbers is represented by {1, 3, 5,. . .}. The

three dots tell us that the list is endless.

It may be noted that while writing the set in roster form an element is

not generally repeated, i.e., all the elements are taken as distinct.

For example, the set of letters forming the word “SCHOOL” is

{S, C, H, O, L}.

ii) In set builder form, all the elements of a set possess a single common

property which is not possessed by any element outside the set. For

example, in the set ―{a, e, i, o, u}‖ all the elements possess a common

property, each of them is a vowel in the English alphabet and no other

letter possesses this property. Denoting this set by V, we write

V = {x : x is a vowel in the English alphabet}.

It may be observed that we describe the set by using a symbol x for

elements of the set (any other symbol like the letters y, z etc. could also be

used in place of x). After the sign of „colon‟ write the characteristic property

possessed by the elements of the set and then enclose the description

within braces. The above description of the set V is read as „The set of all

x such that x is a vowel of the English alphabet‟. In this description the

braces stand for „the set of all‟, the colon stands for „such that‟.

For example, the following description of a set

A = {x : x is a natural number and 3 < x < 10)

is read as “the set of all x such that x is a natural number and 3 < x < 10‖.

Hence, the numbers 4, 5, 6, 7, 8 and 9 are the elements of set A.

If we denote the sets described above in (a), (b) and (c) in roster form by A,

B and C, respectively, then A, B and C can also be represented in set

builder form as follows

A = {x : x is a natural number which divides 42}

B = {y : y is a vowel in the English alphabet}

C = {z : z is an odd natural number}.



Example: Write the set of all vowels in the English alphabet which precede q.

Solution: The vowels which precede q are a, e, i, o. Thus A = {a, e, i, o} is

the set of all vowels in the English alphabet which precede q.

Example: Write the set of all positive integers whose cube is odd.

Solution: The cube of an even integer is also an even integer. So, the

members of the required set can not be even. Also, cube of an odd integer

is odd. So, the members of the required set are all positive odd integers.

Hence, in the set builder form we write this set as {x : x is an odd positive

integer} or equivalently as

{2k + 1 : k 0, k is an integer}

Example: Write the set of all real numbers which can not be written as the

quotient of two integers in the set builder form.

Solution: We observe that the required numbers can not be rational

numbers because a rational number is a number in the form q

p, where p, q

are integers and q 0. Thus, these must be real and irrational. Hence, in set

builder form we write this set as

{x : x is real and irrational}

Example: Write the set 7

6,

6

5,

4

3,

3

2,

2

1 in the set builder form.

Solution: Each member in the given set has the denominator one more

than the numerator. Also, the numerators begin from 1 and do not exceed 6.

Hence, in the set builder form the given set is

6n1andnumbernaturalaisn,1n

nx:x



Example: Match each of the sets on the left described in the roster form

with the same set on the right described in the set builder form:

i) { L, I, T, E) a) {x : x is a positive integer and is a divisor of 18}

ii) {0) b) {x : x is an integer and x2 – 9 = 0}

iii) {1, 2, 3, 6, 9, 18} c) {x : x is an integer and x + 1 = 1}

iv) {3, – 3} d) {x : x is a letter of the word LITTLE}

Solution: Since in (d), there are six letters in the word LITTLE and two

letters T and L are repeated, so (i) matches (d). Similarly (ii) matches (c) as

x + 1 = 1 implies x = 0. Also, 1, 2, 3, 6, 9, 18 are all divisors of 18. So,

(iii) matches (a). Finally, x2 – 9 = 0 implies. x = 3, –3. So, (iv) matches (b).

Example: Write the set {x : x is a positive integer and x2 < 40} in the roster

form.

Solution: The required numbers are 1, 2, 3, 4, 5, 6. So, the given set in the

roster form is {1, 2, 3, 4, 5, 6}.

1.3 The Empty Set

Consider the set

A = {x : x is a student of Class XI presently studying in a school}

We can go to the school and count the number of students presently

studying in Class XI in the school. Thus, the set A contains a finite number

of elements.

Consider the set {x : x is an integer, x2 + 1 = 0}. We know that there is no

integer whose square is –1. So, the above set has no elements.

We now define set B as follows:

B = {x : x is a student presently studying in both Classes X and XI}.

We observe that a student cannot study simultaneously in both Classes X

and XI. Hence, the set B contains no element at all.



Definition: A set which does not contain any element is called the empty

set or the null set or the void set.

According to this definition B is an empty set while A is not. The empty set is

denoted by the symbol ‗ ‘. We give below a few examples of empty sets.

i) Let P = {x: 1 < x < 2, x is a natural number }.

Then P is an empty set, because there is no natural number between

1 and 2.

ii) Let Q = {x : x2 - 2 = 0 and x is rational}.

Then, Q is the empty set, because the equation x2 - 2 = 0 is not satisfied

by any rational number x.

iii) Let R = {x : x is an even prime number greater than 2}

Then R is the empty set, because 2 is the only even prime number.

iv) Let S = {x : x2 = 4, and x is an odd integer}. Then, S is the empty set,

because equation x2 = 4 is not satisfied by any value of x which is an

odd integer.

1.4 Finite and Infinite Sets

Let A = {1, 2, 3, 4, 5), B = {a, b, c, d, e, f} and C = {men in the world}.

We observe that A contains 5 elements and B contains 6 elements. How

many elements does C contain ? As it is, we do not know the exact number

of elements in C, but it is some natural number which may be quite a big

number. By number of elements of a set A, we mean the number of distinct

elements of the set and we denote it by n(A). If n(A) is a natural number,

then A is a finite set, otherwise the set A is said to be an infinite set. For

example, consider the set, N, of natural numbers. We see that n(N), i.e., the

number of elements of N is not finite since there is no natural number which

equals n(N). We, thus, say that the set of natural number is an infinite set.



Definition: A set which is empty or consists of a definite number of

elements is called finite. Otherwise, the set is called infinite.

We shall denote several set of numbers by the following symbols:

N : the set of natural numbers

Z : the set of integers

Q : the set of rational numbers

R : the set of real numbers

Z+ : the set of positive Integers

Q+ : the set of positive rational numbers

R+ : the set of positive real numbers

We consider some examples:

i) Let M be the set of days of the week. Then M is finite.

ii) Q, the set of all rational numbers is infinite.

iii) Let S be the set of solution (s) of the equation x2 - 16 = 0. Then S is finite.

iv) Let G be the set of all points on a line. Then G is infinite.

When we represent a set in the roster form, we write all the elements of the

set within braces { }. It is not always possible to write all the elements of an

infinite set within braces { } because the number of elements of such a set is

not finite. However, we represent some of the infinite sets in the roster form

by writing a few elements which clearly indicate the structure of the set

followed (or preceded) by three dots.

For instance, {1, 2, 3, 4, ... } is the set of natural numbers, {1, 3, 5, 7, 9, .. . }

is the set of odd natural numbers and {..., – 3, –2, –1, 0, 1, 2, 3, ... } is the

set of integers. But the set of real numbers cannot be described in this form,

because the elements of this set do not follow any particular pattern.



1.5 Equal and Equivalent Sets

Given two sets A and B. If every element of A is also an element of B and if

every element of B is also an element of A, the sets A and B are said to be

equal. Clearly, the two sets have exactly the same elements.

Definition: Two sets A and B are said to be equal if they have exactly the

same elements and we write A = B. Otherwise, the sets are said to be

unequal and we write A B

We consider the following examples:

i) Let A = {1, 2, 3, 4, } and B = {3, 1, 4, 2).

ii) Then A = B.

iii) Let A be the set of prime numbers less than 6 and P the set of prime

factors of 30. Obviously, the set A and P are equal, since 2, 3 and 5

are the only prime factors of 30 and are less than 6.

Let us consider two sets L = {1, 2, 3, 4} and M = {1, 2, 3, 8}. Each of them

has four elements but they are not equal.

Definition: Two finite sets A and B are said to be equivalent if they have the

same number of elements. We write A B.

For example, let A = {a, b, c, d, e} and B = {1, 3, 5, 7, 9}. Then A and B are

equivalent sets.

Obviously, all equal sets are equivalent, but all equivalent sets are not

equal.

Example: Find the pairs of equal sets, if any, giving reasons:

A = {0}, B = {x : x > 15 and x < 5}, C = {x : x - 5 = 0}, D = {x:x2 = 25}

E = {x : x is a positive integral root of the equation x2 – 2x – 15 = 0}

Solution: Since 0 A and 0 does not belong to any of the sets B, C, D and

E. Therefore, A B, A C, A D, A E. B = but none of the other sets



are empty. Hence B C, B D and B E. C = {5} but –5 D, hence C D.

Since E = {5}), C = E. D = {–5, 5} and E = {5}. Therefore D E. Thus, the

only pair of equal sets is C and E.

1.6 Subsets

Consider the sets S and T, where S denotes the set of all students in your

school and T denotes the set of all students in your class. We note that

every element of T is also an element of S. We say that T is a subset of S.

Definition: If every element of a set A is also an element of a set B, then A

is called a subset of B or A is contained in B. We write it as A B.

If at least one element of A does not belong to B, then A is not a subset of

B. We write it as A B.

We may note that for A to be a subset of B, all that is needed is that every

element of A is in B. It is possible that every element of B may or may not be

in A. If it so happens that every element of B is also in A, then we shall also

have B A. In this case, A and B are the same sets so that we have A B

and B A which implies A = B.

It follows from the definition that every set A is a subset of itself, i.e., A A.

Since the empty set has no elements, we agree to say that is a subset of

every set. We now consider some examples

i) The set Q of rational numbers is a subset of the set R of real numbers

and we write Q R.

ii) If A is the set of all divisors of 56 and B the set of all prime divisors of

56, then B is a subset of A, and we write B A.

iii) Let A = {1, 3, 5} and B = {x : x is an odd natural number less than 6},

then A B and B A and hence A = B.



iv) Let A = {a, e, i, o, u}, B = {a, b, c, d}. Then A is not a subset of B. Also

B is not a subset of A. We write A B and B A.

v) Let us write down all the subsets of the set {1, 2}. We know is a

subset of every set. So is a subset of {1, 2}. We see that {1}, {2} and

{l, 2} are also subsets of {1,2}. Thus the set {1,2} has, in all, four

subsets, viz. , {1}, {2} and {1,2}.

Definition: Let A and B be two sets. If A B and A B, then A is called a

proper subset of B and B is called a superset of A. For example, A= {1, 2, 3}

is a proper subset of B = {1, 2, 3, 4}.

Definition: If a set A has only one element, we call it a singleton. Thus {a }

is a singleton.

1.7 Power Set

In example (v) of Section 1.6, we found all the subsets of the set {1, 2}, viz.,

, {1}, {2} and {1, 2}. The set of all these four subsets is called the power set

of {1, 2}.

Definition: The collection of all subsets of a set A is called the power set of

A. It is denoted by P(A). In P(A), every element is a set.

Example (v) of section 1.6, if A = {1, 2}, then P(A)={ , {1}, {2}, {1,2}. Also,

note that, n[P(A)] = 4 = 22.

In general, if A is a set with n(A) = m, then it can be shown that

n[P(A)] = 2m > m = n(A).

1.8 Universal Set

If in any particular context of sets, we find a set U which contains all the sets

under consideration as subsets of U, then set U is called the universal set.

We note that the universal set is not unique.



For example, for the set Z of all integers, the universal set can be the set Q

of rational numbers or, for that matter, the set R of real numbers.

For another example, in the context of human population studies, the

universal set consists of all the people in the world.

Example: Consider the following sets : , A = {1, 3), B = {1, 5, 9},

C = {1, 3, 5, 7, 9}, Insert the correct symbol or between each pair of sets

(i) — B, (ii) A — B (iii) A — C (iv) B — C.

Solution:

i) B as is a subset of every set.

ii) A B as 3 A and 3 B..

iii) A C as 1, 3 A also belongs to C.

iv) B C as each element of B also belongs to C.

Example: Let A = {1, 2, 3, 4}, B = {1, 2, 3} and C = {2, 4}. Find all sets X

such that

(i) X B and X C (ii) X A and X B.

Solution:

i) X B means that X is a subset of B, and the subsets of B are , {1}, {2},

{3}, {1,2}, {1,3}, {2,3} and {1,2,3} . X C means that X is a subset of C,

and the subsets of C are , {2}, {4} and {2, 4}. Thus, we note that X B

and X C means that X is a subset of both B and C. Hence, X = , {2}.

ii) X A, X B means that X is a sub set of A but X is not a subset of B.

So, X is one of these {4}, {1,2,4}, {2,3,4}, {l,3,4}, {1,4}, {2,4}, {3,4},

{1,2,3,4}.

Note: A set can easily have some elements which are themselves sets. For

example, {1, {2,3}, 4} is a set having {2,3} as one element which is a set and

also elements 1,4 which are not sets.



Example: Let A, B and C be three sets. If A B and B C, is it true that

A C? If not, give an example.

Solution: No. Let A = {1}, B = C = { { 1 }, 2}. Here A B as A = {1} and

B = C implies B C. But A C as 1 A and 1 C.

Note that an element of a set can never be a subset of it.

1.9 Venn Diagrams

Most of the relationships between sets can be represented by means of

diagrams. Figures representing sets in the form of enclosed region in the

plane are called Venn diagrams named after British logician John Venn

(1834—1883 A.D.). The universal set U is represented by the interior of a

rectangle.Other sets are represented by the interior of circles.

Fig. 1.1

Fig. 1.1 is a Venn diagram representing sets A and B such that A B.

Fig. 1.2



In Fig.1.2, U = {1, 2, 3, ..., 10} is the universal set of which A = {2,4,6,8,10}

and B = {4,6} are subsets. It is seen that B A. The reader will see an

extensive use of the Venn diagrams when we discuss the operations on

sets.

1.10 Complement of a Set

Let the universal set U be the set of all prime numbers. Let A be the subset

of U which consists of all those prime numbers that are not divisors of 42.

Thus A = {g x : x U and x is not a divisor of 42}. We see that 2 U but

2 A, because 2 is a divisor of 42. Similarly 3 U but 3 A, and 7 U but

7 A. Now 2, 3 and 7 are the only elements of U which do not belong to A.

The set of these three prime numbers, i.e., the set {2, 3, 7} is called the

complement of A with respect to U, and is denoted by A . So we have

A = {2, 3, 7}. Thus, we see that A ={x : x U and x A). This leads to the

following definition.

Definition: Let U be the universal set and A is a subset of U. Then the

complement of A with respect to (w.r,t.) U is the set of all elements of U

which are not the elements of A. Symbolically we write A to denote the

complement of A with respect to U. Thus A = {x:x U and x A}. It can be

represented by Venn diagram as

Fig. 1.3



The shaded portion in Fig. 1.3 represents A .

Example: Let U = {1,2,3,4,5,6,7,8,9,10} and A= {1,3,5,7,9}. Find A .

Solution: We note that 2, 4, 6, 8, 10 are the only elements of U which do

not belong to A. Hence A = {2, 4, 6, 8, 10}.

Example: Let U be the universal set of all the students of Class XI of a

co-educational school. Let A be the set of all girls in the Class Xl. Find A .

Solution: As A is the set of all girls, hence A is the set of all boys in the

class.

1.11 Operations on Sets

In earlier classes, you learnt how to perform the operations of addition,

subtraction, multiplication and division on numbers. You also studied certain

properties of these operations, namely, commutativity, associativity,

distributivity etc. We shall now define operations on sets and examine their

properties. Henceforth, we shall refer all our sets as subsets of some

universal set.

a) Union of Sets: Let A and B be any two sets. The union of A and B is

the set which consists of all the elements of A as well as the elements of B,

the common elements being taken only once. The symbol ‗ ‟ is used to

denote the union. Thus, we can define the union of two sets as follows.

Definition: The union of two sets A and B is the set C which consists of all

those elements which are either in A or in B (including those which are in

both).

Symbolically, we write A B = {x:x A or x B} and usually read as

‗A union B‟.



The union of two sets can be represented by a Venn diagram as shown in

Fig. 1.4.

Fig. 1.4

The shaded portion in Fig. 1.4 represents A B.

Example: Let A = {2, 4, 6, 8} and B = {6, 8, 10, 12}. Find A B.

Solution: We have A B = {2, 4, 6, 8, 10, 12}.

Note that the common elements 6 and 8 have been taken only once while

writing A B.

Example: Let A = {a, e, i, o, u} and B = {a, i, u}. Show that A B = A.

Solution: We have A B = {a, e, i, o, u} = A.

This example illustrates that the union of a set A and its subset B is the set

A itself, i.e., if B A , then A B = A.

Example: Let X = {Ram, Shyam, Akbar} be the set of students of Class XI

who are in the school Hockey team. Let Y = {Shyam, David, Ashok} be the

set of students from Class XI who are in the school Football team. Find

X Y and interpret the set.



Solution: We have X Y = {Ram, Shyam, Akbar, David, Ashok}. This is the

set of students from Class XI who are either in the Hockey team or in the

Football team.

Example: Find the union of each of the following pairs of sets:

i) A = {1, 2, 3, 4}; B = {2, 3, 5}

ii) A = {x : x Z+ and x2 > 7}; B = {1, 2, 3}

iii) A = {x : x Z+ }; B = {x : x Z and x < 0}

iv) A = {x : x N and 1 < x 4}; B = {x:x N and 4 < x < 9}

Solution:

i) A B = {1, 2, 3, 4, 5}

ii) A = {3, 4, 5,... }, B = {1, 2, 3}. So, A B = {1, 2, 3, 4, 5,... } = Z+

iii) A={1, 2, 3,. . .}, B ={x:x is a negative integer}. So A B={x:x Z, x 0}.

iv) A = {2, 3, 4}, B = {5, 6, 7, 8}. So, A B = {2, 3, 4, 5, 6, 7, 8}.

b) Intersection of Sets: The intersection of sets A and B is the set of all

elements which are common to both A and B. The symbol is used to

denote the intersection.

Thus, we have the following definition.

Definition: The intersection of two sets A and B is the set of all those

elements which belong to both A and B. Symbolically, we write A B =

{x:x A and x B} and read as „A intersection B‘.

The intersection of two sets can be represented by a Venn diagram as

shown in Fig. 1.5.



Fig. 1.5

The shaded portion represents A B.

If A B = , then A and B are said to be disjoint sets. For example, let

A = {2, 4, 6, 8} and B = {1, 3, 5, 7}. Then, A and B are disjoint sets, because

there is no element which is common to A and B. The disjoint sets can be

represented by Venn diagram as shown in Fig. 1.6.

Fig. 1.6



Example: Let A = {2, 4, 6, 8} and B = {6, 8, 10, 12}. Find A B.

Solution: We see that 6, 8 are the only elements which are common to both

the sets A and B. Hence A B = {6, 8}.

Example: Consider the sets X and Y of Example 17. Find X Y.

Solution: We see that the element “Shyam” is the only element common to

both the sets X and Y. Hence, X Y = { Shyam }.

SAQ 1: Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = {2, 3, 5, 7}. Find A B

and prove that A B = B.

SAQ 2: Let A = A = {x : x Z+} ; B = {x : x is a multiple of 3, x Z}:

C = {x:x is a negative integer}; D = {x:x is an odd integer}. Find (i) A B,

(ii) A C, (iii) A D, (iv) B C, (v) B D, (vi) C D.

c) Difference of Sets: The difference of sets A and B, in this order, is the

set of elements which belong to A but not to B. Symbolically, we write

A — B and read as „A difference B‘. Thus A — B = {x : x A and x B} and

is represented by Venn diagram in Fig.1.7. The shaded portion represents

A — B.

Fig. 1.7



SAQ 3: Let V = {a, e, i, o, u} and B = {a, i, k, u}. Find V – B and B – V.

SAQ 4: Let A = {1, 2, 3, 4, 5, 6} and B = {2, 4, 6, 8}. Find A – B and

B – A.

1.12 Applications of Sets

Let A and B be finite sets. If A B = , then

n(A B) = n(A) + n(B) (1)

The elements in A B are either in A or in B but not in both as

A B = . So (1) follows immediately.

In general, if A and B are finite sets, then

n(A B) = n(A) + n(B) – n(A B) (2)

Fig. 1.8

Note that the sets A – B, A B and B – A are disjoint and their union is

A B (Fig 1.8). Therefore

n(A B) = n(A – B) + n(A B) + n(B – A)

= n(A – B) + n(A B) + n(B – A) +n(A B) – n(A B)

= n(A) + n(B) – n(A B).

which verifies (2).

If A, B and C are finite sets, then

n(A B C) = n(A) + n(B) + n(C) –n(A B) –n(B C)

– n(A C) + n(A B C) (3)



In fact, we have

n(A B C) = n(A) + n(B C) – n(A (B C )) [by (2)]

= n(A) + n(B) + n(C) – n(B C) – n (A (B C)) [by (2)]

Since A (B C) = (A B) (A C), we get

n[A (B C)] = n(A B) + n (A C) – n[A B A C)]

= n(A B) + n (A C) – n[A B C)]

Therefore

n(A B C) = n(A) + n(B) + n(C) – n(B C) – n(A B)

– n(A C) + n(A B C).

This proves (3).

Example: If X and Y are two sets such that n(X Y) = 50, n(X) = 28 and

n(Y) = 32, find n(X Y).

Solution: By using the formula

n(X Y) = n(X) + n(Y) – n(X Y),

we find that

n(X Y) = n(X) +n(Y) – n(X Y)

= 28 + 32 –50 = 10..

Alternatively, suppose n(X Y) = k, then

Fig. 1.9



n(X – Y) = 28 – k, n(Y – X) = 32 – k. (by Venn diagram in Fig 1.9)

This gives 50 = n(X Y) = (28 – k) + k + (32 – k).

Hence, k = 10

Example: In a school there are 20 teachers who teach mathematics or

physics. Of these, 12 teach mathematics and 4 teach physics and

mathematics. How many teach physics?

Solution Let M denote the set of teachers who teach mathematics and

P denote the set of teachers who teach physics. We are given that

n(M P) = 20, n(M) = 12, n(M P) = 4. Therefore

n(P) = n(M P) – n(M) + n(M P) = 20 – 12 + 4 = 12.

Hence, 12 teachers teach physics.

SAQ 5: In a group of 50 people, 35 speak Hindi, 25 speak both English and

Hindi and all the people speak at least one of the two languages. How many

people speak only English and not Hindi ? How many people speak

English?

1.13 Cartesian Product of Sets

Let A, B be two sets. If a A, b B, then (a, b) denotes an ordered pair

whose first component is a and the second component is b. Two ordered

pairs (a, b) and (c, d) are said to be equal if and only if a = c and b = d.

In the ordered pair (a, b), the order in which the elements a and b appear in

the bracket is important. Thus (a, b) and (b, a) are two distinct ordered pairs

if a ≠ b. Also, an ordered pair (a, b) is not the same as the set {a, b}.

Definition: The set of all ordered pairs (a, b) of elements a A , b B is

called the Cartesian Product of sets A and B and is denoted by A x B. Thus

A B = {(a, b): a A, b B}.



Let A = {a1, a2}, B = {b1, b2, b3}. To write the elements of A x B, take a1 A

and write all elements of B with a1, i.e., (a1, b1), (a, b2), (a1, b3). Now take

a2 ε A and write all the elements of B with a2, i.e., (a2, b1), (a2, b2), (a2, b3).

Therefore, A x B will have six elements, namely, (a1, b1), (a1, b2), (a1, b3),

(a2, b1), (a2, b2), (a2, b3).

Remarks:

i) If A = or B = , then A B = .

ii) If A ≠ and B ≠ , then A B ≠ . Thus, A B ≠ if and only if A ≠

and B ≠ . Also, A B ≠ B A.

iii) If the set A has m elements and the set B has n elements, then A B

has mn elements.

iv) If A and B are non-empty sets and either A or B is an infinite set, so is

A x B.

v) If A = B, then A B is expressed as A2.

vi) We can also define, in a similar way, ordered triplets. If A, B and C are

three sets, then (a ,b, c), where a A, b B and c C, is called an

ordered triplet. The Cartesian Product of sets A, B and C is defined as

A B C = {(a, b, c): a A, b B, c C}. An ordered pair and ordered

triplet are also called 2-tuple and 3-tuple, respectively. In general, if

A1, A2,.. ., An are n sets, then (a1,a2,..., an) is called an n-tuple where

ai Ai, i = 1, 2 n and the set of all such n-tuples, is called the Cartesian

product of A1, A2, ……, An. It is denoted by A1 x A2 x. . .x An. Thus

A1 A2 ….. An = {(a1, a2, …. an): a1 A1, 1 i n}}.

Example: Find x and y if (x + 2, 4) = (5, 2x + y).

Solution: By definition of equal ordered pairs, we have

x + 2 = 5 (1)

2x + y = 4 (2)

Solving (1) and (2), we get x = 3, y = –2.



Example: Let A = {1, 2, 3} and B = {4, 5}. Find A x B and B x A and show

that A B ≠ B A.

Solution: We have

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

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

Note that (1, 4) A B and (1, 4) B A. Therefore, A B B A.

Example: Let A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}. Find

i) A (B C) ii) (A B) (A C)

iii) A (B C) iv) (A B) (A C)

Solution:

i) We have B C = {4}. Therefore, A x (B C) = {(1, 4), (2, 4), (3, 4)}.

ii) We note that

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

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

Therefore (A B) (A C) = {(1, 4), (2, 4), (3, 4)}.

iii) Clearly B C = {3, 4, 5, 6}. Thus

A (B C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3),

(3, 4), (3, 5), (3, 6)}

iv) In view of (ii), we see that

(A B) (A C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 3), (3, 4), (3, 5), (3, 6)}.

In view of the assertion in Example 3 above, we note that

A (B C) = (A B) (A C)

and A (B C) = (A B) (A C).

SAQ 6: Let A and B be two sets such that n(A) = 5 and n(B) = 2.

If (a1, 2), (a2, 3), (a3, 2), (a4, 3), (a5, 2) are in A B and a1, a2, a3, a4 and

a5 are distinct. Find A and B.



1.14 Summary

This unit tells us about sets and their representations. We study the

concepts of Empty sets, Finite and Infinite sets, Equal sets. All the

concepts discussed is well illustrated by standard examples. The

different operations on sets like complement of Set, Operation on

Sets and Applications of sets is discussed here.


1. Which of the following pairs of sets are equal ? Justify your answer.

i) A, the set of letters in “ALLOY” and B, the set of letters in “LOYAL”

ii) A = {n : n Z and n2 4} and B = {x:x R and x2 – 3x + 3x + 2 = 0} .

2. State which of the following sets are finite and which are infinite:

i) {x : x N and (x – 1) (x – 2) = 0}

ii) {x : x N and x2 = 4}

iii) {x : x N and 2x – 1 = 0 }

iv) {x : x N and x prime}

v) {x : x N and x odd}

3. If A and B are two non-empty sets such that A B = B A, show that

A = B

1.16 Answer

Self Assessment Questions

1. We have A B = {2, 3, 5, 7} = B.

We note that if B A , then A B = B.



2. A = {x:x is a positive integer}, B = {3n : n Z};

i) A B = {3, 6, 9, 12,...} = {3n:n Z+}.

ii) A C =

iii) A D = {1, 3, 5, 7,...}

iv) B C = {– 3, –6, –9,. . . } = {3n : n is a negative integer}

v) B D = {. . ., –15, –9, –3, 3, 9, 15,...}

vi) C D = {–1, –3, –5, –7,...}

3. We have V – B = {e, o}, since the only elements of V which do not

belong to B are e and o. Similarly B – V = {k}

4. We have A – B = {1, 3, 5}, as the only elements of A which do not

belong to B are 1, 3 and 5. Similarly, B – A = {8}.

We note that A – B B –A

5. Let H denote the set of people speaking Hindi and E the set of people

speaking English. We are given that n(H E) = 50, n(H) = 35,

n(H E) = 25. Now

n(H E) = n(H) + n(E – H).

So 50 = 35 + n(E – H), i.e. , n(E – H) = 15.

Thus, the number of people who speak only English but not Hindi is 15.

Also, n(H E) = n(H) + n(E) – n(H E) implies

50 = 35 + n(E) – 25,

which gives n(E) = 40.

Hence, the number of people who speak English is 40.

6. Since a1, a2, a3, a4, a5 A and n(A) = 5, A = {a1, a2, a3, a4, a5}. Also 2, 3

B and n(B) = 2. Therefore, B = {2, 3}.



Terminal Quesitons

1. i) A = {A, L, L, O, Y}, B = {L, O, Y, A, L}. Then A, B are equal sets as

repetition of elements in a set do not change a set. Thus A = {A, L,

O, Y} = B.

ii) A = {–2, –1, 0, 1, 2,}, B = (1, 2). Since 0 A and 0 B, A and B are

not equal sets.

2. i) Given set = {1, 2}. Hence, it is finite.

ii) Given set = {2}. Hence, it is finite.

iii) Given set = . Hence, it is finite.

iv) The given set is the set of all prime numbers and since the set of

prime numbers is infinite, hence the given set is infinite.

v) Since there are infinite number of odd numbers, hence the given set

is infinite

3. Let a A. Since B ≠ , there exists b B. Now, (a, b) A B = B A

implies a B. Therefore, every element in A is in B giving A B.

Similarly, B A. Hence A = B



Unit 2 Mathematical Logic

Structure

2.1 Introduction

Objectives

2.2 Statements

2.3 Basic Logical Connectives

2.4 Conjunction

2.5 Disjunction

2.6 Negation

2.7 Negation of Compound Statements

2.8 Truth Tables

2.9 Tautologies

2.10 Logical Equivalence

2.11 Applications

2.12 Summary


2.14 Answers

2.1 Introduction

Logic is the study of general patterns of reasoning, without reference to

particular meaning or contexts. If an object is either black or white, and if it is

not black, then logic leads us to the conclusion that it must be white.

Observe that logical reasoning from the given hypotheses cannot reveal

what „black‟ or „white‟ mean, or why an object can not be both.

Logic can find applications in many branches of sciences and social

sciences. Logic, infact is the theoretical basis for many areas of computer

science such as digital logic circuit design, automata theory and artificial

intelligence.



In this chapter, we shall learn about statements, truth values of a statement,

compound statements, basic logical connectives, truth tables, tautologies,

logical equivalence, duality, algebra of statements, use of Venn diagrams in

logic and finally, some simple applications of logic in switching circuits.

Objectives:


understand the ideas in Mathematical Logic

identify a proposition

apply the concept of Mathematical Logic in circuits

2.2 Statements

A statement is a sentence which is either true or false, but not both

simultaneously.

Note: A sentence which is both true and false simultaneously is not a

statement, rather, it is a paradox.

Example:

(a) Each of the following sentences:

i) New Delhi is in India.

ii) Two plus two is four.

iii) Roses are red.

iv) The sun is a star.

v) Every square is a rectangle.

is true and so each of them is a statement.

(b) Each of the following sentences:

i) The earth is a star.

ii) Two plus two is five.

iii) Every rectangle is a square.

iv) 8 is less than 6.

v) Every set is a finite set.

is false and so each of them is a statement.



Example:

a) Each of the sentences:

i) Open the door.

ii) Switch on the fan.

iii) Do your homework.

can not be assigned true or false and so none of them is a statement.

Infact, each of them is a command.

b) Each of the sentences:

i) Did you meet Rahman?

ii) Where are you going?

iii) Have you ever seen Taj Mahal?


Infact, each of them is a question.

c) Each of the sentences:

i) May you live long!

ii) May God bless you!


Infact, each of them is optative.

d) Each of the sentences:

i) Hurrah! We have won the match.

ii) Alas! I have failed.


In fact, each of them is exclamatory.

e) Each of the sentences:

i) Good morning to all.

ii) Wish you best of luck.


In fact, each them is a wish.



f) Each of the sentences:

i) Please do me a favour. T

ii) Give me a glass of water.


In fact, each them is a request.

g) Each of the following sentences:

i) x is a natural number

ii) He is a college student.

is an open sentence because the truth or falsity of (xv) depends on x

and that of xvi) depends on the reference for the pronoun he. We may

observe that for some values of x like x = 1, 2,….. etc, (xv) may be true

and for some other values like ...,3

1,

2

1x etc, (xv) is false. Similarly,

(xvi) may be true or false. However, at a particular point of time or

situation they are either true or false. Since, we are interested only in the

fact that it is true or false, sentences (xv) and (xvi) can be considered as

statements.

Note: The statements (xv) and (xvi) in Example 2 are also called open

statements.

It is useful to have some notation to represent statements. Let us represent

the statements by lower case letter like p, q, r, s, ….. Thus, a statement

„New Delhi is city may be represented or denoted by p and we write

p : New Delhi is a city.

similarly, we may denote a statement „2 + 3 = 6‟ by q and write

q : 2 + 3 = 6.

Truth value of a statement: The truth or falsity of a statement is called its

truth value. Every statement must be either true or false. No statement can

be both true and false at the same time. If a statement is true, we say that

its truth value is TRUE or T and if it is false we say that its truth value is

FALSE or F.



Example: The statements in Example 1(a) have the truth value T while the

statements in Example 1(b) have the truth value F.

Compound statements: A statement is said to be simple, if it cannot be

broken down into two or more sentences. The statements that we

considered in Example 1(a) and (b) are all simple statements.

New statements that can be formed by combining two or more simple

statements are called compound statements. Thus, a compound statement

is the one which is made up of two or more simple statements.

Example:

a) The statement “Roses are red and Violets are blue” is a compound

statement which is a combination of two simple statements “Roses are

red” and “Violets are blue”.

b) The statement “Gita is sick or Rehana is well” is a compound statement

made up of two simple statements “Gita is sick” and “Rehana is well”.

c) The statement “It is raining today and 2 + 2 = 4” is a compound

statement composed of two simple statements “It is raining today” and

“2 + 2 = 4”.

Simple statements, which on combining, form compound statements, are

called sub-statements or component statements of the compound

statements. The compound statements S consisting of sub-statements

p, q, r,... is denoted by S (p, q, r,...).

A fundamental property of a compound statements is that its truth value is

completely determined by the truth value of each of its sub-statements

together with the way in which they are connected to form the compound

statement.



2.3 Basic Logical Connectives

There are many ways of combining simple statements to form compound

statements. The words which combine simple statements to form compound

statements are called connectives. In the English language, we combine two

or more statements to form a new statement by using the connectives „and‟,

„or‟, etc. with a variety of meanings. Because the use of these connectives in

English language is not always precise and unambiguous, it is necessary to

define a set of connectives with definite meanings in the language of logic,

called object language. We now define connectives for object language

which corresponds to the connectives discussed above. Three basic

connectives (logical) are conjunction which corresponds to the English word

„and‟ ; disjunction which corresponds to the word „or‟ ; and negation which

corresponds to the word „not‟.

Throughout we use the symbol „ ‟ to denote conjunction ; „ ‟ to denote

disjunction and the symbol „~„ to denote negation.

Note:. Negation is called a connective although it does not combine two or

more statements. In fact, it only modifies a statement.

2.4 Conjunction

If two simple statements p and q are connected by the word „and‟, then the

resulting compound statement “p and q” is called a conjunction of p and q

and is written in symbolic form as “p q“.

Example: Form the conjunction of the following simple statements:

p : Dinesh is a boy.

q : Nagma is a girl.

Solution: The conjunction of the statement p and q is given by

p q : Dinesh is a boy and Nagma is a girl.



Example: Translate the following statement into symbolic form

“Jack and Jill went up the hill.”

Solution: The given statement can be rewritten as

“Jack went up the hill and Jill went up the hill”

Let p : Jack went up the hill and q : Jill went up the hill.

Then the given statement in symbolic form is p q.

Example: Write the truth value of each of the following four statements:

i) Delhi is in India and 2 + 3 = 6.

ii) Delhi is in India and 2 + 3 = 5.

iii) Delhi is in Nepal and 2 + 3 = 5.

iv) Delhi is in Nepal and 2 + 3 = 6.

Solution: In view of (D1) and (D2) above, we observe that statement (i) has

the truth value F as the truth value of the statement “2 + 3 = 6” is F. Also,

statement (ii) has the truth value T as both the statement “Delhi is in India”

and “2 + 3 = 5” has the truth value T. Similarly, the truth value of both the

statements (iii) and (iv) is F.

2.5 Disjunction

If two simple statements p and q are connected by the word „or‟, then the

resulting compound statement “p or q” is called disjunction of

p and q and is written in symbolic form as “p q”.

Example: Form the disjunction of the following simple statements:

p : The sun shines.

q : It rains.

Solution: The disjunction of the statements p and q is given by

p q : The sun shines or it rains.



Example: Write the truth value of each of the following statements:

i) India is in Asia or 2 + 2 = 4.

ii) India is in Asia or 2 + 2 =5.

iii) India is in Europe or 2 + 2 = 4.

iv) India is in Europe or 2 + 2 = 5.

Solution: In view of (D3) and (D4) above, we observe that only the last

statement has truth value F as both the sub-statements “India is in Europe”

and “2 + 2 = 5” have the truth value F. The remaining statements (i) to (iii)

have the truth value T as at least one of the sub-statements of these

statements has the truth value T.

2.6 Negation

An assertion that a statement fails or denial of a statement is called the

negation of the statement. The negation of a statement is generally formed

by introducing the word “not” at some proper place in the statement or by

prefixing the statement with “It is not the case that” or “It is false that”.

The negation of a statement p in symbolic form is written as “~ p”.

Example: Write the negation of the statement

p : New Delhi is a city.

Solution: The negation of p is given by

~ p : New Delhi is not a city

or ~ p : It is not the case that New Delhi is a city.

or ~ p : It is false that New Delhi is a city

Example: Write the negation of the following statements:

p : I went to my class yesterday.

q : 2 + 3 = 6

r : All natural numbers are integers.



Solution: Negation of the statement p is given by

~ p : I did not go to my class yesterday.

or

It is not the case that I went to my class yesterday.

or

It is false that I went to my class yesterday.

or

I was absent from my class yesterday.

The negation of the statement q is given by

~q : 2 + 3 ≠ 6

or

It is not the case that 2 + 3 = 6

or

It is false that 2 + 3 = 6

The negation of the statement r is given by

~ r : Not all natural numbers are integers.

or

There exists a natural number which is not an integer.

or

it is not the case that all natural numbers are integers.

or

It is false that all natural numbers are integers.

Regarding the truth value of the negation ~ p of a statement p. we have

(D5) : ~ p has truth value T whenever p has truth value F.

(D6) : ~ p has truth value F whenever p has truth value T.

Example: Write the truth value of the negation of each of the following

statements::

i) p : Every square is a rectangle.

ii) q : The earth is a star.

iii) r :2 + 3 < 4



Solution: In view of (D5) and (D6), we observe that the truth value of ~p is F

as the truth value of p is T. Similarly, the truth value of both ~q and ~r is T

as the truth value of both statements q and r is F

2.7 Negation of compound statements

I) Negation of conjunction: Recall that a conjunction p q consists of

two sub-statements p and q both of which exist simultaneously.

Therefore, the negation of the conjunction would mean the negation of

at least one of the two sub-statements. Thus, we have

(D7): The negation of a conjunction p q is the disjunction of the

negation of p and the negation of q. Equivalently, we write

~ ( p q) = ~ p v ~ q

Example: Write the negation of each of the following conjunctions:

a) Paris is in France and London is in England.

b) 2 + 3 = 5 and 8 < 10.

Solution:

(a) Write p : Paris is in France and q : London is in England.

Then, the conjunction in (a) is given by p q.

Now ~ p : Paris is not in France, and

~ q : London is not in England.

Therefore, using (D7), negation of p q is given by

~ p q = Paris is not in France or London is not in England.

(b) Write p : 2+3 = 5 and q :8 < 10.

Then the conjunction in (b) is given by p q.

Now ~ p : 2 + 3 ≠ 5 and 108:q~

Then, using (D7), negation of p q is given by

~ p q = 2 + 3 ≠ 5 or .108



(II) Negation of disjunction: Recall that a disjunction p q is consisting of

two sub-statements p and q which are such that either p or q or both

exist. Therefore, the negation of the disjunction would mean the

negation of both p and q simultaneously. Thus, in symbolic form, we

have

(D8): The negation of a disjunction p q is the conjunction of the

negation of p and the negation of q. Equivalently, we write

~ (p q) = ~ p ~ q

Example: Writ the negation of each of the following disjunction:

a) Ram is in class X or Rahim is in Class XII

b) 7 is greater than 4 or 6 is less than 7

Solution:

a) Let p : Ram is in class X and q : Rahim is in Class XII.

Then, the disjunction in (a) is given by p q.

Now ~ p : Ram is not in Class X.

~ q : Rahim is not in Class XII.


~ p q : Ram is not in Class X and Rahim is not in Class XII.

b) Write p : 7 is greater than 4, and q : 6 is less than 7.


~ p q : 7 is not greater than 4 and 6 is not less than 7.

(III) Negation of a negation: As already remarked the negation is not a

connective but a modifier. It only modifies a given statement and applies

only to a single simple statement. Therefore, in view of (D5) and (D6), for

a statement p, we have

(D9) : Negation of negation of a statement is the statement itself

Equivalently, we write

~ (~p) = p



Example: Verify (D9) for the statement

p : Roses are red.

Solution: The negation of p is given by

~ p : Roses are not red.

Therefore, the negation of negation of p is

~ (~ p) : It is not the case that Roses are not red.

or

It is false that Roses are not red.

or

Roses are red.

Many statements, particularly in mathematics, are of the type “If p then

q”. Such statements are called conditional statements and are denoted

by p → q read as „p implies q‟.

Another common statement is of the form “p if and only if q”. Such

statements are called bi-conditional statements and are denoted by

p ↔ q.

Regarding the truth values of p → q and p ↔ q , we have

a) the conditional p → q is false only if p is true and q is false.

Accordingly, if p is false then p → q is true regardless of the truth

value of q.

b) the bi-conditional p ↔ q is true whenever p and q have the same

truth values otherwise it is false.

One may verify that p → q = (~ p) q

2.8 Truth Tables

A truth table consists of rows and columns. The initial columns are filled with

the possible truth values of the sub-statements and the last column is filled with



the truth values of the compound statement S (the truth values of S depends on

the truth values of the sub-statements entered in the initial columns)

Example: Construct the truth table for ~p.

Solution: Note that one simple statement ~p is consisting of only one

simple statement p. Therefore, there must be 2‟ (= 2) rows in the truth table.

It is necessary to consider all possible truth values of p.

In view of (D5) above, recall that p has the truth value T if and only if ~p has

the truth value F. Therefore, the truth table for ~p is given by

Table 21 Truth table for ~ p

p ~ p

T F

F T

Example: Construct the truth table for p (~p)

Solution: Note that the compound statement p (~p) is consisting of only

one simple statement p. Therefore, there must be 2‟ (= 2) rows in the truth

table. It is necessary to consider all possible truth values of p.

Table 2.2

p ~ p p (~p)

T

F

Step 1: Enter all possible truth values of p. namely, T and F in the first

column of the truth table (Table 2.2).

Table 2.3

p ~ p p (~ p)

T F

F T



Step 2: Using (D5) and (D6), enter the truth values of ~ p in the second

column of the truth table (Table 2.3).

Table 2.4

p ~ p p (~ p)

T F F

F T F

Step 3: Finally, using (D2) enter the truth values of p (~ p) in the last

column of the truth table (Table 2.4)

Example: Construct the truth table for p q.

Solution: The compound statement p q is consisting of two simple

statements p and q. Therefore, there must be 22(= 4) rows in the truth table

of p q. Now enter all possible truth values of statements p and q namely

TT, TF, FT and FF in first two columns of Table 2.5.

Table 2.5

P q p q

T T

T F

F T

F F

Then, in view of (D1) and (D2) above, enter the truth values of the compound

statement p q in the truth table (Table 18.6) to complete the truth table.

Table 2.6: Truth table for p q

P q p q

T T T

T F F

F T F

F F F



Example: Construct the truth table for p q.

Table 2.7: Truth table for p q.

P q p q

T T T

T F T

F T T

F F F

Solution: In view of (D3) and (D4) above, recall that the compound

statement p q has the truth value F if and only if both p and q have the

truth value F; otherwise p q has truth value T. Thus, the truth table for p

q is as given in Table 2.7.

a) ~ [p (~q)]

b) (p q) (~ p)

c) ~[(~p) (~q)]

Solution:

a) Truth table for ~ [p (~ q)] is given by

Table 2.8: Truth table for ~ [p (~ q)]

p q ~p p (~q) ~[p (~q)]

T T F F T

T F T T F

F T F F T

F F T F T

b) Truth table for (p q) (~p) is given by

Table 2.9: Truth table for (p q) (~ p)

p q p q ~p (p q) (~ p)

T T T F F

T F F F F

F T F T F

F F F T F



c) Truth table for ~ [(~p) v (~q)] is given by

Table 2.10 : Truth table for ~ [(~p) (~q)]

p q ~p ~q (~ p) (~ q) ~ [( ~ p)] [(~q)]

T T F F F T

T F F T T F

F T T F T F

F F T T T F

2.9 Tautologies

A statement is said to be a tautology if it is true for all logical possibilities. In

other words, a statement is called tautology if its truth value is T and only T

in the last column of its truth table. Analogously, a statement is said to be a

contradiction if it is false for all logical possibilities. In other words, a

statement is called contradiction if its truth value is F and only F in the last

column of its truth table. A straight forward method to determine whether a

given statement is tautology (or contradiction) is to construct its truth table.

Example: The statement p (~p) is a tautology since it contains T in the

last column of its truth table (Table 2.11)

Table 2.11: Truth table for p (~p)

p ~p p (~p)

T F T

F T T

Example: The statement p (~p) is a contradiction since it contains F in the

last column of its truth table (Table 2.12)

Table 2.12: Truth table for p (~ p)

p ~p p (~p)

T F F

F T F



Remark: The negation of a tautology is a contradiction since it is always

false, and the negation of a contradiction is a tautology since it is always

true.

SAQ 1: Show that

a) ~ [p (~p)] is a contradiction.

b) ~ [p (~p)] is a tautology.

Example: Show that

a) (p q) (~ p) is a tautology.

b) (p q) (~ p) is a contradiction.

Solution:

a) The truth table for (p q) (~ p) is given by

Table 2.15: Truth table for (p q) (~ p)

P q p q ~p (p q) (~ p)

T T T F T

T F T F T

F T T T T

F F F T T

Since the truth table for (p q) (~ p) contains only T in the last column,

it follows that (p q) (~ p) is a tautology.

b) Recall Table 2.9 which is the truth table for (p q) (~ p) and observe

that it contains only F in the last column. Therefore, (p q) (~ p) is a

contradiction.



2.10 Logical Equivalence

Two statements S1 (p, q, r, ...) and S2 (p, q, r, ...) are said to be logically

equivalent, or simply equivalent if they have the same truth values for all

logical possibilities is denoted by

S1 (p, q, r,...) ≡ S2 (p, q, r,...).

In other words, S1 and S2 are logically equivalent if they have identical truth

tables (by identical truth tables we mean the entries in the last column of the

truth tables are same).

Example: Show that ~ p q is logically equivalent to (~p) (~ q).

Solution: The truth tables for both the statements are

Table 2.16: Truth table for ~ (p q) Table 2.17: Truth table for (~ p) (~q)

p q p q ~(p q) p q ~p ~q (~ p) (~q)

T T T F T T F F F

T F F T T F F T T

F T F T F T T F T

F F F T F F T T T

Now, observe that the entries (truth values) in the last column of both the

tables are same. Hence, the statement ~(p q) is equivalent to the

statement (~ p) (~q).

Remark: Consider the statements:

p : Mohan is a boy.

q : Sangita is a girl.

Now, we have

~(p q) ≡ (~ p) (~q).

Therefore, the statement

“It is not the case that Mohan is a boy and Sangita is a girl”

has the same meaning as the statement

“Mohan is not a boy or Sangita is not a girl”.



Example: Let

p : The South-West monsoon is very good this year and

q : Rivers are rising.

Give verbal translation of ~ [(~p) (~q)].

Solution: we have

~(p q) ≡ (~ p) (~q)

Therefore, the statement ~ [(~p) (~q)] is the same as the negation of the

statement ~(p q) which is the same as the conjunction p q. Thus, the

verbal translation for ~ [(~p) (~q)] is

“The South-West monsoon is very good this year and rivers are rising”

Example: Prove the following:

a) ~ [p (~ q)] (~p) q

b) ~ [(~ p) q] p (~q)

c) ~ (~p) p

Solution:

a) The truth tables for ~ [p (~q)] and (~p) q are given by

Table 2.18: Truth table for~ [p (~q)] Table 2.19: Truth table for (~p) q

p q ~q p (~q) ~ [p (~ q)] p q ~p (~p) ~ q

T T F T F T T F F

T F T T F T F F F

F T F F T F T T T

F F T T F F F T F

The last column of the two tables are the same.

b) It follows in view of the truth Table 2.20



Table 2.20: Truth table for p (~q) and ~ [(~ p) q]

p q ~p ~q (~ p) q p (~ q) ~ [( ~ p) q]

T T F F F T T

T F F T F T T

F T T F T F F

F F T T F T T

c) The assertion follows in view of Table 2.21

Table 2.21: Truth table for ~(~p)

p ~p ~(~p)

T F T

F T F

2.11 Applications

The logic that we have discussed so far is called two-value logic because

we have considered only those statements which are having truth values

True or False. A similar situation exists in various electrical and mechanical

devices. Claude Shannon, in late 1930‟s, was first to notice an analogy

between the operations of switching devices and the operations of logical

connectives. He used this analogy with great success to solve problems of

circuit design.

Observe that an electric switch which is used for turning „on‟ and „off‟ an

electric light is a two-state device. We shall now explain various electric

networks with the help of logical connectives. For this, first we discuss how

an electric switch works. Observe that, in Fig. 2.1, we have shown two

positions of a simple switch.



Fig. 2.1

In (a) when switch is closed (i.e. on), current can flow from one terminal to

the other. In (b), when the switch is open (i.e. off), current can not flow.

Let us now consider the example of an electric lamp controlled by switch.

Such a circuit is given in Fig. 2.2.

Fig. 2.2

Observe that when the switch s is open, no current flows in the circuit and

therefore, the lamp is „off‟. But when switch s is closed, the lamp is „on‟.

Thus the lamp is on if and only if the switch s is closed.

If we denote the statements as

p : The switch s is closed

l : The lamp l is „on‟

then, by using logic, the above circuit can be expressed as p ≡ l.

Next, consider an extension of the above circuit in which we have taken two

switches s1 and s2 in series as shown in Fig. 2.3.



Fig. 2.3

here, observe that the lamp is „on‟ if and only if both the switches s1 and s2

are closed.

If we denote the statements as:

p : the switch s1 is closed.

q : the switch s2 is closed.

l : the lamp l is ‟on‟.

then the above circuit can be expressed as p q 1.

Now, we consider a circuit in which two switches s1 and s2 are connected in

parallel (Fig. 2.4).

Fig. 2.4



SAQ 2: Express the following circuit in Fig. 2.5 in symbolic form of logic.

Fig. 2.5

2.12 Summary

In this unit we study the truth values of a statements. The different basic

logical connectives are discussed in detail with some standard examples.

Compound statements and the negation are clearly explained . The concept

of Tautology, Contradiction and Logical Equivalence is discussed in detail

with example wherever necessary. The applications of mathematical logic to

switching circuits is dealt with standard examples.


1. Define Tautology and Contradiction

2. Draw the truth tables of Conjunction, disjunction and Biconditional



2.14 Answers


1. a) The truth table of ~ [p (~p)] is given by

Table 2.13: Truth table for ~ [p (~p)]

P ~p p (~p) ~ [p (~p)]

T F T F

F T T F

Since it contains only F in the last column of its truth table, it follows

that

~ [p (~ p)] is a contradiction.

b) The truth table of ~ [p (~ p)] is given by

Table 2.14: Truth table for ~ [p (~ p)]

P ~p p (~ p) ~ [p (~ p)]

T F F T

F T F T

Since it contains only T in the last column of its truth table, it follows

that ~ [p (~ p)] is a tautology.

2. Observe that the lamp is „on‟ if and only if either s1 and s2 both are

closed or s1 and s2 both are open or only s1 is closed.

If we denote the statements as

p : The switch s1 is closed

q : The switch s2 is closed

l : The lamp l is „on‟

then



~p: The switch s1 is open.

or

The switch s1 is closed.

~ q: The switch s2 is open.

or

The switch s2 is closed.

Therefore, the circuit in Fig. 2.5 in symbolic form of logic may be

expressed as

p [(~ p) (~ q)] (p q) 1



Unit 3 Modern Algebra

Structure

3.1 Introduction

Objectives

3.2 Binary Operation

3.3 Addition Modulo n

3.4 Multiplication Modulo n

3.5 Semigroup

3.6 Properties of Groups

3.7 Subgroup

3.8 Summary


3.10 Answers

3.1 Introduction

The theory of groups which is a branch of Abstract Algebra is of paramount

importance in the development of mathematics.

The idea of group was first given by the French Mathematician Evariste

Galois in 1832 who died at the age of 21 years in a duel. The group theory

was later developed by an English Mathematician Arthur Cayley. He defined

the notion of an abstract group with a general structure which could be

applied to numerous particular cases. The theory of groups has applications

in Quantum Mechanics and other branches of mathematics.

Objectives:


apply the concepts of Algebraic Structure in practical problems

understand Binary Operations and its applications in group theory



3.2 Binary Operation

Let G be a non-empty set. Then G G = {(x, y): x, y G}. A function of

G G in to G is said to be a binary operation on the set G. The image of an

ordered pair (x, y) under f is denoted by x f y.

The symbols +, x, 0, *, …. Are very often used as the binary operations on a

set.

Thus * is a binary operation on the set G if for every a, b G implies a * b G.

Hence a binary operation * combines any two elements of G to give an

element of the same set G.

Examples:

1. If Z is the set of integers then usual addition (+) is the binary operation

on Z. For if M and n are two integers then m + n is again an integer i.e.

for every m, n Z, m + n Z.

In particular – 5, 3 Z, implies – 5 + 3 = –2 Z, etc.

Similarly the usual multiplication is the binary operation on the set Q of

rationals, for the product of two rational numbers is again a rational

number.

2. Let E be the set of even integers. i.e., E = {0, 2, 4, 6, ….} and O be

the set of odd integers i.e. O = { 1, 3, 5, ….}. Clearly the usual

addition is the binary operation on E whereas it is not a binary operation

on O. Because the sum of two even integers is even but the sum of two

odd integers is not an odd integer.

Also the usual subtraction is not a binary operation on the set N of

natural numbers.

Algebraic Structure

A non-empty set with one or more binary operations is called an algebraic

structure. If * is a binary operation on G then (G, *) is an algebraic structure.



For example the set of integers Z is an algebraic structure with usual

addition as the binary operation. Similarly (Q, .), (E, +) are algebraic

structures.

Group

A non-empty set G is said to be a group with respect to the binary operation

* if the following axioms are satisfied.

1. Closure law. For every a, b G, a * b G.

2. Associative law. For every a, b, c G

a * (b * c) = (a * b) * c

3. Existence of identity element. There exists an element e G such that

a * e = e * a = a for every a G.

Here e is called the identity element

4. Existence of inverse. For every a G there exists an element b G

such that

a * b = b * a = e. Here b is called the inverse of a and is denoted by

b = a–1. A group G with respect to the binary operation * is denoted by

(G, *). If in a group (G, *), a * b = b * a for every a, b, G then G is said

to be commutative or Abelian group named after Norwegian

mathematician Niels Henrik Abel (1802 – 1820).

Finite and Infinite Groups

A group G is said to be finite if the number of elements in the set G is finite,

otherwise it is said to be an infinite group. The number of elements in a finite

group is said to be the order of the group G and is denoted by O(G).

Example: Prove that the set Z of integers is an abelian group with respect

to the usual addition as the binary operation.

1. Closure law. We know that the sum of two integers is also an integer.

Hence for every m, n Z, m + n Z.



2. Associative law. It is well known that the addition of integers is

associative. Therefore (m + n) + p = m + (n + p) for every m, n, p Z.

3. Existence of identity element. There exists 0 Z such that

m + 0 = 0 + m = m for every m Z. Hence 0 is called the additive

identity.

4. Existence of inverse. For every m Z there exists – m Z such that

m + (–m) = (–m) + m = 0.

Here – m is called the additive inverse of m or simply the negative of m.

Therefore (Z, +) is a group.

5. Commutative law. We know that the addition of integers is commutative

i.e., m + n = n + m for every m, n Z. Hence (Z, +) is an abelian group.

Since there are an infinite elements in Z, (Z, +) is an infinite group.

Similarly we can prove that the set Q of rationals, the set R of reals and

the set C of complex numbers are abelian groups with respect to usual

addition.

Example: Prove that the set Q0 of all non-zero rationals forms an abelian

group with respect to usual multiplication as the binary operation.

Now Q0 = Q – {0}

Solution:

1. Closure law. Let a, b Q0 i.e. a and b are two non-zero rationals. Then

their product a b is also a non-zero rational. Hence a b Q0.

Since a, b are two arbitrary elements of Q0, we have for every

a, b, Q0, ab Q0.

2. Associative law. We know that the multiplication of rationals is

associative. i.e.,, a(b c) = (a b) c for every a, b, c Q0.

3. Existence of identity element. There exists 1 Q0 such that

a.1 = 1 . a = a for every a Q0. Here 1 is called the multiplicative identity

element.



4. Existence of inverse. Let a Q0. Then a is a non-zero rational.

Therefore a

1 exists and is also a rational 0.

Also 1a.a

1

a

1.a for every a Q0.

a

1 is the multiplicative inverse of a.

Therefore (Q0, .) is a group.

Further, it is well-known that the multiplication of rationals is

commutative i.e., ab = ba for every a, b Q0.

Hence (Q0, .) is an abelian group.

Similarly we can show that the set R0 of non-zero reals and the set C0 of

non-zero complex numbers are abelian groups w.r.t. usual multiplication.

1. The set N of natural numbers is not a group w.r.t. usual addition, for

there does not exist the identity element 0 in N and the additive inverse

of a natural number is not a natural number i.e., for example

2 N but – 2 N. Also N is not a group under multiplication because

5 N but .N5

1

2. The set of integers is not a group under multiplication for 2 Z but

.Z2

1

3. The set of rationals, reals and complex numbers (including 0) do not

form groups under multiplication for multiplicative inverse of 0 does not

exist.

SAQ 1: Prove that the fourth roots of unity form an abelian group with

respect to multiplication.



3.3 Addition Modulo n

Let n be a positive integer a and b be any two integers. Then “addition

modulo n of two integers a and b”, written a + n b, is defined as the least

non-negative remainder when a + b is divided by n. If r is the remainder

when a + b is divided by n, then

A + n b = r where 0 r < n.

In other words, if a + b r (mod n), 0 r < n. Then a + n b = r.

For example,

7 + 5 10 = 2 since 7 + 10 = 17 2 (mod 5)

15 + 7 11 = 5 since 15 + 11 = 26 5 (mod 7)

17 + 8 21 = 38 since 17 + 21 = 38 6 (mod 8)

12 + 5 8 = 0 since 12 + 8 = 20 0 (mod 5)

1 + 7 1 = 2 since 1 + 1 = 2 2 (mod 7)

Properties:

1. Commutative since a + b and b + a leave the same remainder when

divided by n, a + n b = b + n a.

For example 5 + 7 6 = 4 = 6 + 7 5

2. Associative since a + (b + c) and (a + b) + c leave the same remainder

when divided by n, a + n (b + n c) = (a + n b) + n c.

For example 4 +6 (3 + 6 5) = (4 + 6 3) + 6 5

Example: Prove that the set Z4 = {0, 1, 2, 3} is an abelian group w.r.t.

addition modulo 4.

Solution: Form the composition table w.r.t. addition modulo 4 as below:



+4 0 1 2 3

0 0 1 2 3

1 1 2 3 0

2 2 3 0 1

3 2 0 1 2

Since 1 + 3 = 4 0 (mod 4), 3 + 3 = 6 2 (mod 4) 2 + 3 = 5 1 (mod 4) etc.

1. Closure law. From the above composition table for all a, b G, a +4 b

also belongs to Z4.

2. Associative law. Since a + (b + c) and (a + b) + c leave the same

remainder when divided by 4, we have

(a + 4 (b +4 c) = (a +4 b) +4 c.

3. Existence of identity element. From the above table, we observe that

0 Z4 satisfies a + 4 0 = 0 +4 a = a for every a Z4.

0 is the identity element.

4. Existence of inverse. From the above table, the inverses of 0, 1, 2, 3 are

respectively 0, 3, 2, 1 because 0 +4 0 = 0, 1 + 4 3 = 0, 2 +4 2 = 0, and

3 + 41 = 0.

Hence (z4, +4) is a group

Further, since a + b and b + a leave the same remainder when divided

by 4, a + 4 b = b +4 a.

(Z4, +4) is an abelian group.

Similarly, we can show that the set of remainders of 5 viz.

Z5 = {0, 1, 2, 3, 4} from an abelian group under addition (mod 5).

In general the set of remainders of a positive integer m.

Zm = {0, 1, 2, …. (m –1) form an abelian group under addition

(mod m).



3.4 Multiplication modulo n

Let n be a positive integer an a, b any two integers. Then multiplication

modulo n of two integers a and b, written a n b, is defined as the least non-

negative remainder when ab is divided by n. If r is the remainder when ab is

divided by n. If r is the remainder when ab is divided by n then a n b = r,

where 0 r < n. In other words, if ab r (mod n), 0 r < n then a xn b = r.

For example,

7 5 3 = 1 since 7 . 3 = 21 1 (mod 5)

9 7 5 = 3 since 9 . 5 = 45 3 (mod 7)

12 8 7 = 4 since 12 . 7 = 84 4 (mod 8)

2 7 3 = 6 since 2 . 3 = 6 6 (mod 7)

14 46 = 0 since 14 . 6 = 84 0 (mod 4)

Properties

1. Commutative: Since ab and ba leave the same remainder when divided

by n,

a n b = b n a

For example 5 7 4 = 4 7 5

2. Associative: Since a(bc) and (ab)c leave the same remainder when

divided by n

a n (b n c) = (a n b) n c

For example 3 7 (4 7 5) = (3 7 4) 7 5

Example: Prove that the set 4,3,2,15Z is an abelian group under

multiplication modulo 5.

Solution: Form the composition table w.r.t. multiplication modulo 5 as

below:



x5 1 2 3 4

1 1 2 3 4

2 2 4 1 3

3 3 1 4 2

4 4 3 2 1

Since 2 . 3 = 6 1 (mod 5)

2 . 4 = 8 3 (mod 5)

4 . 4 = 16 1 (mod 5) etc.

1. Closure law. Since all the elements entered in the above table are the

elements of ,5Z closure law holds good i.e. for all a, b G, a 5 b also

belongs to .5Z

2. Associative law. Since a (bc) and (ab) c leave the same remainder when

divided by 5 we have for every a, b, c ,5Z

a 5 (b 5 c) = (a 5 b) 5 6.

3. Existence of identity element. From the above table, we observe that

1 5Z satisfies a 5 1 = 1 5 a = a for every a 5Z .

1 is the identity element.

4. Existence of inverse. Also the inverses of 1, 2, 3, 4 are respectively

1, 3, 2, 4 because 1 5 1 = 1, 2 5 3 = 1, 3 5 2 = 1, and 4 5 4 = 1.

Therefore ( 5Z x5) is an abelian group.

Similarly, we can show that the non-zero remainders of 7 viz.

7Z = {1, 2, 3, 4, 5, 6} form an abelian group under multiplication

(mod 7). In general, the non-zero remainders of a positive integer

p viz. pZ = {1, 2, 3, …… (p – 1)} form a group under multiplication

(mod p) if and only if p is a prime number.



Note: The set 6Z = {1, 2, 3, 4, 5} does not form a group under multiplication

(mod 6) for 2, 3 G, but 2 6 3 = 0 G. This is because 6 is not a prime

number.

3.5 Semigroup

A non-empty set G is said to be a semigroup w.r.t. the binary operation if the

following axioms are satisfied.

1. Closure: For every a, b, G, a * b G

2. Associative: For every a, b, c G, a * (b * c) = (a * b) * c.

Examples:

1. The set N of all natural numbers under addition is a semigroup because

for every a, b, c N

(i) a + b N, and (ii) a + (b + c) = (a + b) + c. The set, N is semigroup

under multiplication also.

2. The set Z of integers is a semigroup under multiplication because for

every a, b Z, a + b Z and for every a, b, c Z, a(bc) = (ab) c. Note

that every group is a semigroup but a semigroup need not be a group.

For example, the set N of all natural numbers is a semigroup under

multiplication (also under addition) but it is not a group. Similarly Z, the

set of integers is an example of a semigroup but not a group under

multiplication.

3.6 Properties of Groups

For the sake of convenience we shall replace the binary operation * by dot .

in the definition of the group. Thus the operation dot . may be the operation

of addition or multiplication or some other operation. In what follows by ab

we mean a . b or a * b. With this convention, we rewrite the definition of the

group.



Definition: A non-empty set G is said to be a group w.r.t. the binary

operation. if the following axioms are satisfied.

1. Closure property: For every a, b G, ab G

2. Associative property: For every a, b, c G, a (bc) = (ab) c.

3. Existence of identity element: There exists an element e G such that

ae = ea = a for every a G. Here e is called the identity element.

4. Existence of inverse: For every a G there exists an element b G such

that ab = ba = e. Here b is called the inverse of a i.e., b = a–1 Further,

5. If ab = ba for every a, b G then G is said to be an abelian group or a

commutative group.

Theorem: The identity element in a group is unique.

Proof: Let e and e be the two identity elements of a group G. Then by

definition, for every a G.

ae = ea = a

and aaeea

Substitute ea in (1) and a = e in (2). Then we obtain

eeeee

and eeeee

Hence eeee

The identity element in a group is unique.

Theorem: In a group G the inverse of an element is unique

Proof: Let b and c be the two inverses of an element a in G.

Then by definition ab = ba = e

ac = ca = e

Now consider, b = be

= b(ac)

= (ba) c

= ec

= c



Therefore inverse of every element in a group is unique

Theorem: If a is any element of a group G, then (a–1)–1 = a.

Proof: Since a–1 is the inverse of a, we have aa–1 = a–1a = e

This implies that a is an inverse of a–1, but inverse of every element is unique

aa11

Thus the inverse of the inverse of every element is the same element.

Theorem: If a and b are any two elements of a group G then

.abab111

Proof:

Consider, (ab) (b–1 a–1) = 11 abba

= 11 abba

= 1eaa

= aa–1

= e

Similarly we can prove that eabab 11

Hence eabababab 1111

Therefore 11 ab is the inverse of ab,

i.e., .abab 111

Corollary: If a, b, c belong to a group G then (abc)–1 = c–1 b–1 a–1 etc.

Note: If (ab)–1 = a–1 b–1 for all a, b G, the G is abelian.

For, (ab)–1 = a–1 b–1 implies 111

11aab

i.e. 1111 abab

= ba for all a, b G

Hence G is abelian.



Theorem: (Cancellation laws).

If a, b, c are any three elements of a group G, then

ab = ac implies b = c (left cancellation law)

ba = ca implies b = c (right cancellation law)

Proof: Since a is an element of a group G, there exists a–1 G there exists

a–1 G such that aa–1 = a–1 a = e, the identity element

Now ab = ac

acaaba 11

caabaa 11

eb = ec

b = c

Similarly ba = ca

11 acaaba

11 aacaab

be = ce

b = c

Theorem: If a and b are any two elements of a group G, then the equations

ax = b and ya = b have unique solutions in G.

Proof:

i) Since .Ga,Ga 1

Now Ga 1 and b G implies Gba 1

(closure axiom) and

.bebbaabaa 11

Hence x = a–1 b satisfies the equation ax = b and hence is a solution. If x1,

x2 are the two solutions of the equation, ax = b then ax1 = b and ax2 = b.

ax1 = ax2

x1 = x2

Hence the solution is unique.



ii) Also b G, a–1 G implies ba–1 G and bbeaababa 11

y = ba–1 satisfies the equation ya = b and hence is a solution. If y1, y2

are two solutions of the equation ya = b then y1a = b and y2a = b

y1a = y2a

y1 = y2

Therefore the solution is unique

SAQ 2: Prove that in a group G if a2 = a then a = e, the identity element.

Note: Any element a which satisfies a2 = a is called the idempotent element

in a group. Thus e is the only idempotent element in G.

Example: If in group G, (ab)2 = a2b2 for every a, b G prove that G is

abelian.

Solution:

Now 222

baab

(ab) (ab) = (a . a) (b . b)

a[b(ab)] = a[a(bb)] (Associative)

b (ab) = a (bb) (Left cancellation law)

(ba) b = (ab) b (Associative)

ba = ab (Right cancellation law)

Hence G is an abelian group.

Example: Show that if every element of a group G is its own inverse then G

is abelian.

Solution: Let a, b G then a–1 = a and b–1 = b

Clearly ab G (ab)–1 = ab by hypothesis

i.e. b–1 a–1 = ab

i.e. ba = ab since b–1 = b, a–1 = a

G is abelian.



3.7 Subgroup

A non-empty subset H of a group G is said to be a subgroup of G if under

the operation of G, H itself forms a group.

If e be the identity element of a group G, Then H = { e } and H = G are

always subgroups of G. These are called the trivial or improper subgroups.

If H is a subgroup of G and H {e} and H G then H is called a proper

subgroup.

Examples:

1. We know that the set Z of integers forms a group under addition.

Consider a subset E = {2x : x Z} = {0, 2, 4, …. } of Z. Then E also

forms a group under addition.

Therefore E is a subgroup of Z.

Similarly F = {3x : x Z} = {0, 3, 6, 9, ….. } is a subgroup of z.

2. Clearly the multiplicative group H = {1, –1} is a subgroup of the

multiplicative group G = {1 –1, i, –i}.

3. Let G = {1, 2, 3, 4, 5, 6} be a subset of G. Now it is clear from the

following composition table that H also forms a group under x7.

X7 1 2 4

1 1 2 4

2 2 4 1

4 4 1 2

Therefore H is a subgroup of G.

Theorem: A non-empty subset H of a group G is a subgroup of G if and

only if

i) for every a, b H implies ab H

ii) for every a H implies a–1 H



Note: Union of two subgroups need not be subgroups for, let H = {0, 2, 3, 4,

….} and K = {0, e, 6,….} be two subgroups of the group of integers Z, so that

H K = {0, 2, 3, 4, 6, …. }.

Now 2, 3 H K but 2 + 3 = 5 H K because 5 is neither a multiple of 2

nor a multiple of 3.

3.8 Summary

In this unit we studied clearly that the rectangular array of numbers is

denoted by matrix, also we know that determinant is a square matrix which

is associated with a real number. Then we studied that a set which satisfies

certain rules is called as a group. Here we studied sub group, semi group

etc. with well illustrated examples.


1. Prove that a non-empty subset H of a group G is a subgroup of G if and

only if for every a, b H implies ab–1 H.

2. Prove that the intersection of two subgroups of a group is again a

subgroup.

3.10 Answers


1. Roots of the equation x4 = 1 are called the fourth roots of unity and they

are 1, –1, i, – i. Let G = {1, – 1 i, – i }.

From the composition table w.r.t. usual multiplication as follows:

. 1 –1 i – i

1 1 –1 i – i

–1 –1 1 – i i

i i – i –1 1

– i – i I 1 –1



1. Closure Law. Since all the elements written in the above composition

table are the elements of G, we have for all a, b, G, ab G.

2. Associative Law. We know that the multiplication of complex

numbers is associative and G is a subset of the set of complex

numbers

Hence a(bc) = (ab) c for all a, b, c G.

3. existence of identity element. From the composition table it is clear

that there exists 1 G satisfying a . 1 = 1 . a = a for every a G.

Therefore 1 is the identity element.

4. Existence of inverse. From the composition table we observe that

the inverses of 1, – 1, i – i are 1, -1, -i, i.

Thus for every a G there exists a–1 G such that a a–1 = a–1 a = 1,

the identity element. Hence (G, .) is a group.

Further multiplication of complex numbers is commutative.

Therefore ab = ba for every a, b G.

Also we observe that the elements are symmetric about the principal

diagonal in the above composition table. Hence commutative law

holds good.

Therefore (G, .) is an abelian group.

Note that G is a finite group of order 4.

2. Now e.aa.aaa2 since a = ae a = e



Unit 4 Trigonometry

Structure

4.1 Introduction

Objectives

4.2 Radian or Circular Measure

4.3 Trigonometric Functions

4.4 Trigonometrical ratios of angle when is acute

4.5 Trigonometrical ratios of certain standard angles

4.6 Allied Angles

4.7 Compound Angles

4.8 Multiple and Sub-multiple angle

4.9 Summary


4.11 Answers

4.1 Introduction

This unit of Trigonometry gives us an idea of circular measure. The different

Trigonometric functions are studied here. Some of the standard angles and

their Trigonometric ratios are discussed in detail. The basic knowledge allied

angles and compound angles are explained in a simple manner.

Objectives:


understand the concepts of Trigonometrical functions

use allied and compound angles in calculations

4.2 Radian or Circular Measure

A radian is the angle subtended at the centre of a circle by an arc equal to

the radius of the circle. O is the centre of a circle. A and B are points on the



circle such that arc AB = radius OA. Then BOA is called one radian or one

circular measure. We write c1BOA

Radian is a constant angle and 180π c

Consider a circle whose centre is O and radius r. A and B are points on the

circle such that arc AB = OA = r. Join OA, OB and draw OC to OA.

c1BOA , 1BOA right angle and arc AB = r. We know that arc 4

1AC

(circumference of the circle) = 2/rr24

1 . In a circle the arcs are

proportionated to the angles subtended by them at the centre.

anglerightangleright 1

12.,e.i

1

1

2/r

r;

COA

BOA

ACarc

ABarc.,e.i

cc

1c = 2/ 1 right angle, which is constant



Radian is a constant angle

Further we have, 1C = 2 1 right angle

C = 2 90 = 180

Note:

i) C = 180 mans radians are equal to 180

Hereafter, this is written as = 180 .

For example 3602,454/,902/ and so on.

In each of these cases the unit ‘radians’ on the left side is understood.

ii) 7/22

180180l c

11

630

11

790

547157' (nearly)

Here is the real number which is the ratio of circumference of a circle to its

diameter. Its approximate value is 22/7.

1 radian = 547157 (approximately)

Clearly 1 radian is < 60

Examples:

1) Express 2.53 radians in degrees

radians = 180

9.144~53.222

753.218053.218053.2 radiansradians

2) Express 144 into radians

For 180 = radians

5

4

180

144144

C

radiansFor



It is better to remember the following:

1) 180/xx radians

2) x radians = /x180

Length of an arc of a circle

Consider a circle who centre is O and radius r. A and B are points on the

circle such that arc AB = r. ,1BOA C P is a point on the circle such that

arc PA = s and CAOP

rs1r

s

BOA

AOP

ABarc

PAarc

Hence the length of an arc of a circle is equal to the product of the radius of

the circle and the angle in radians subtended at the centre by the arc.

Note:

s = the arc length of the circle; r = the radius of the circle

= angle in radians subtended by s at the centre

Area of a sector of a circle

The portion of the circle bounded by two radii, say, OA, OB and the arc

AB is called the sector BOA . Consider a circle whose centre is O and

radius r. Let AOB be the sector of angle C

S

B

1C

OC

A

P



radiansπ2

BOA

CircletheofArea

AOBsectorofArea

θr2

1AOBsectorofAreaπ2

θ

rπ

AOBsectorofArea 2

c

c

2

Worked Examples

1. Express 792 in radians and 7 /12 in degrees

radians180/xx

c5/22radians180/)792(792

10512

1807

12

7

2. The angles of a triangle are in the ratio 2:3:5 find them (i) in radians

(ii) in degrees.

A : B : C = 2 : 3 : 5 A = 2K, B = 3K, C = 5K

i) A + B + C = 10 K = K = /10

210

5C,

10

3B;

510

2A

The angles are 2

,10

3,

5 in radians

ii) A + B + C = 180 = 10K = 180 K = 18

A = 36 B = 54 C = 90 . The angles are 36 , 54 , 90

3. An arc of a circle subtends 15 at the centre. If the radius is 4 cms, find

the length of the arc and area of the sector formed.

r

B

A

c

Y O



= 15 , r = 4 cms to find s

12180

1515

cc

s = r = 4( /12) = /3 = 22/21 cm

Area of the sector .cm.sq3/212/42

1r2

1 22

4. A spaceship moves in a circular orbit of radius 7200 km round the earth.

How far does it travel while sweeping an angle of 100 ?

9

5

180

100100,km7200r

cc

S = r = (7200) (5 /9) = (800) 5 = (4000 ) km.

The spaceship travels through a distance of (4000 ) km.

SAQ 1: A circular wheel is rotating at the rate of 25 revolutions per minute. If

the radius of the wheel is 50 cms, find the distance covered by a point on

the rim in one second (Take = 3.1416)

4.3 Trigonometric Functions

Consider a circle whose centre is the origin and radius is r. Let the circle cut

X-axis at A and A and Y-axis at B and .B P(x, y) is any point on the circle.

Join OP and draw PM to X-axis. OP = r, OM = x, MP = y, POX .

The six trigonometrical functions (ratios) of angle are defined as given below:

O

B

Y

x M A X

P(x y)

y



Sine of angle r

ysin cosecant of angle = cosec 0y

y

r

Cosine of angle r

xcos secant of angle 0x

x

r tangent of angle

0xx

ytan cotangent of angle 0y

y

xcot

Since 222 yxr we have 22 yxr . So we have

0xx

ytan,

yx

xcos,

yx

ysin

2222

0xx

yxsec,0y

y

xcot

22

0yy

yxeccos

22

Note

i) Reciprocal relations

1y

r:

r

yeccos,sin

sin and cosec are reciprocal to each other.

sin

1eccos,

eccos

1sin

Similarly we have,

tan

1cot,

cos

1tan,

cos

1sec,

sin

1cos

ii) sin

coscotand

cos

sintantan

x

y

rx

ry

cos

sin

iii) The above definitions of trigonometric functions hold good whatever may

be the position of the point P(x, y) on the circle. We shall discuss this in

detail later.



iv) Identities

a) 1cossin 22

b) 22 sectan1

c) 22 eccoscos1

a) From the figure 222 ryx

on dividing by r2,

1r

y

r

x.,e.i1

r

y

r

x22

2

2

2

2

But

1sincostherefore

sinr

yandcos

r

x

22

squaredcosasreadisitandasw ritten is 22coscos

Thus for all value of , cos2 + sin2 = 1

b) 222 ryx . If x ≠ 0, we can divide by x2.

2

2

2

2

x

r

x

y1

22

x

r

x

y1 But sec

x

r,tan

y

x

2222sectan1.,e.isectan1

c) .ryx 222 If y ≠ 0 we can divide y2.

22

2

2

2

2

y

r1

y

x.,e.i

y

r1

y

x

But eccosy

rcot

y

x

2222eccos1cot.,e.ieccos1cot

Thus

222222 eccoscot1,sectan1,1sincos



4.4 Trigonometrical ratios of angle when is acute

The revolving line, starting from OX rotates through an acute angle and

comes to the position OA. Draw AB to X-axis. In the triangle OAB,

BOA90OBA The side opposite to i.e., AB is called opposite side.

The side opposite to 90 i.e., OA is called the hypotenuse and OB is called

the adjacent side. The six trigonometrical ratio of are defined as

OA

ABsin

hypotenuse

sideopposite

AB

OBcot

sideopposite

sideadjacent

OA

OBcos

hypotenuse

sideadjacent

OB

OAsec

sideadjacent

hypotenuse

OB

ABtan

sideadjacent

sideopposite

AB

OAeccos

sideopposite

hypotenuse

These definitions hold good whenever is one of the acute angles of a right

angled triangle.

Y A

O B X

A

O B



The following identities should be memorized.

1) 1sincos 22 2)

22 sectan1

3) 22 eccoscos1 4) eccos/1sin,sin/1eccos

5) sec/1cos,cos/1sec 6) cot/1tan,tan/1cot

7) cos/sintan 8) sin/coscot

Worked Examples:

1. Show that 2222

cottan7seccoseccossin

sec.cos22

sec2

coseccos.sin22

eccos2

sinLHS

cos/1.cos2sin/1.sin22

sec2

eccos2

cos2

sinLHS

cos/1sec;sin/1eccos222

tan12

cot11

RHScottan7 22 1cossin 22

22 tan1sec

22 cot1eccos

2. Prove that cos

sin1

sin1

cos

1sectan

1sectan

1sectan

1sectanLHS write 1 in numerator as

22 tansec

22 sectan1

1sectan

tansecsectan 22

1sectan

tansectansecsectan

1sectan

tansec1sectan

cos

1sin

cos

1

cos

sinsectan …………………..(1)



Now sin1cos

sin1sin1

cos

sin1

sin1cos

sin1 2

22

22

sin1cos

1cossin

sin1

cos

sin1cos

cos 2

…………………………(2)

From (1) and (2) the result follows

3. Show that cosec Acos1/Acos1AcotA

LHS = cosec A – cot A

Asin

Acos

Asin

1

Asin

Acos1

Acos1Acos1Acos1AsinAcos1Acos1

Acos1 22

RHSAcos1

Acos1

Acos1

Acos1.

Acos1

Acos1

4. Show that Acos1Asin12AcosAsin12

Acos2AcosAsin2Asin2AcosAsin1LHS 22

Acos2Acos2Asin2Asin211

= 2 – 2 sin A – 2 sin A cos A + 2 cos A

= 2(1 – sin A – sin A cos A + cos A)

RHS = 2[1 – sin A + cos A – sin A cos A]

LHS = RHS

5. If ,cosbsinaandysinbcosax show that 2222 bayx

sinbcosax …….. (1)



cosbsinay …….. (2)

Square and add (1) and (2)

2222 cosbsinasinbcosayx

cosb.sina22cosbsinasinb.cosa2sinbcosa 2222222

222222 cossinbsincosa

2222 ba1b1a

2222 bayx

6. If 22

sinbtcosax then show that (x – a) (b – x) = (a – b)2 sin2 t cos2 t

x – a = a cos2 t + b sin2 t – a

= b sin2 t – a + a cos2 t

= b sin2 t – a (1 – cos2 t)

= b sin2 t – a sin2 t

= sin2 t. (b – a) (1)

b – x = b – (a cos2 t + b sin2 t) = b – a cos2 t – b sin2 t

= b(1 – sin2 t) – a cos2 t

= b cos2 t – a cos2 t

= cos2 t (b – a) (2)

Multiply (1) and (2),

(x – a) (b – x) = sin2 t (b – a) . cos2 t(b – a)

= (b – a)2 sin2 t cos2 t

= (a – b)2 sin2 t cos2 t

7. Express all the trigonometrical ratios of angle A in terms of sin A where

A is acute.

I Method

Take sin ypotenuseopp.side/h1

xA

Let PQR be the right angled triangle in which 90QRP

,AQRP Then PR = x, PQ = 1



222

x1QRPQQR

2

x1QR

From the figure

Asin11

x1Acos 2

2

(1)

Asin1

Asin

x1

xAtan

22 (2)

Asin

Asin1

x

x1Acos

22

(3)

Asec1

1

x1

1Asec

22 (4)

Asin

1

x

1ecAcos (5)

AsinAsinand (6)

Thus we have expressed all trigonometric al ratios in terms of A.

Y

P

Q R

A

1 x

2x1



II Method

We know that

Asin1AcosorAsin1Acos1AsinAcos 22222 (1)

Since Acos

AsinAtan we have tan

Asin1

AsinA

2 (2)

Asin

Asin1

Atan

1Acot

2

(3)

Asin1

1

Acos

1Asec

2 (4)

Asin

1ecAcos (5)

sin A = sin A (6)

8. Express all the trigonometrical ratios of angle A in terms of sec A

I Method

Take sec sideadjacent

hypotenuse

1

xA

Let PQR be the right angled triangle, in which

1QR,xPQ,ARQP,90QRP

Now PR2 = PQ2 – QR2 = x2 – 1

P

Q R

A

1

1x 2 x



1xPR 2

sec A = sec A (1)

Asec

1Asec

x

1xAsin

22

(2)

Asec

1

x

1Acos (3)

1

1Asec

1

1xAtan

22

(4)

1Asec

1

1x

1Acot

22 (5)

1Asec

Asec

1x

xecAcos

22 (6)

II Method

We know that 1 + tan2 A = sec2 A tan2 A = sec2 A – 1

1AsecAtan 2 (1)

1Asec

1

Atan

1Acot

2 (2)

Asec

1Acos (3)

Asec

1Asec

Asec

11AsecAcosAtanAsin

22

(4)

1Asec

Asec

Asin

1Aeccos

2 (5)

sec A = sec A (6)



SAQ 2: If 900and)5/4(cos find the value of

cotsin4

eccos2cos3

SAQ 3: If cos = (m2 – n2)/ (m2 + n2) find the values of cot and cosec in

terms of m and n.

4.5 Trigonometrical ratios of certain standard angles

To remember the trigonometrical ratios of the standard angles the following

table is useful.

0 30 45 60 90

Sin 0 1/2 2/1 2

3 1

Cos 1 2

3 2/1 1/2 0

Tan 0 3/1 1 3

Worked examples

1. Find the value of 03cos60cos90sin45sin2

30sec45tan260tan22

222

3

4

4/7

3/41

4/12/3

3/423

1.2/11.2/13

3/2123.E.G

322

222

2. Find the value of

3/2

sin24/2

tan4/36/2

eccos3/2

eccos3/2

sec3/4

2222

2/3214/3223/4

12

7

12

1894864

2

3

4

34

3

16

3. Find ‘x’ from the equation, 30tan60tan60sin45cos8

45sec30eccosx 2222



Given equation is, 22

22

22

3/13

2/32/18

22x

3/83/x8.,e.i3/134/32/18

x8

x = 1

4. Find x from (2x – 3) (cosec2 /3 – sin2 /4) = x tan2 ( /4) – sec2 ( /6) – 2

(2x – 3) cosec2 60 - sin2 45 ) = x tan2 45 - sec2 30 - 2

23/21x2/13/23x22222

(2x – 3) (4/3 – ½) = x – 4/3 – 2

(2x – 3) (5/6) = x – 10/3

5(2x – 3) = 6x – 20; 10x – 15 = 6x – 20; i.e., 4x = – 5.

x = – (5/4).

5. Show that

2

3/cot1

3/cot1

6/cos1

6/cos1

32

32

2/3

2/31

30cos1

30cos1LHS

222

13

13

3/11

3/11

60cot1

60cot1RHS

RHSLHS32

32

324

324

3213

3213

SAQ 4: Prove that sec 30 tan 60 + sin 45 cosec 45 + cos 30 cot 60 = 2

7

4.6 Allied Angles

To connect the trigonometrical ratios of with those of

Thus tantanandcoscos,sinsin



Similarly cotcotandsecsec,eccoseccos

sin (90 – ) = cos cosec (90 – ) = sec

cos (90 – ) = sin sec (90 – ) = cosec

tan (90 – ) = cot cot (90 – ) = tan

To connect the trigonometrical ratios of 90 + with those

sin (90 + ) = cos cosec (90 + ) = sec

cos (90 + ) = –sin sec (90 + ) = –cosec

tan (90 + ) = –cot cot (90 + ) = –tan

To express the trigonometrical ratios of 180 – in terms of those of

sin (180 – ) = sin cosec (180 – ) = cosec

cos (180 – ) = –cos sec (180 – ) = –sec

tan (180 – ) = –tan cot (180 – ) = –cot

To express the trigonometrical ratios of 180 + in terms of those of

Thus

sin (180 + ) = sin

cos (180 + ) = –cos

tan (180 + ) = tan

Taking reciprocals,

cosec (180 + ) = –cosec

sec (180 + ) = –sec

cot (180 + ) = cot

Coterminal angles

Two angles are said to be coterminal angles, if their terminal sides are one

and the same. and 360 + are coterminal angles. – and 360 – are

coterminal angles.

sin (360 – ) = sin (– ) = – sin cosec (360 – ) = cosec (– ) = cosec

cos (360 – ) = cos (– ) = cos sec (360 – ) = sec (– ) = –sec



tan (360 – ) = tan (– ) = – tan cot (360 – ) = cot (– ) = –cot

sin (360 + ) = sin cosec (360 + ) = cosec

cos (360 + ) = cos sec (360 + ) = sec

tan (360 + ) = tan cot (360 + ) = cot

Worked Examples

1. Find the value of (a) sin 120 (b) sec 300 (c) tan 240 (d) cos 1770

(e) cosec 1305 (f) cosec (-1110 )

a) sin 120 = 2

360sin60180sin

b) sec 300 = 260sec60360sec

c) tan 240 = 360tan6018tan

d) cos 1770 = 2/330cps309020cos

e) cosec 1305 = 225eccos2253603eccos

= 245eccos45180eccos

f) cosec (-1110 ) = 1110eccos

= 230eccos3003603eccos

2. Find the value of 270cos360sec270sin45tan3

240tan210cot240cot120tan 2222

tan 120 = 360tan60180tan

cot 240 = 3/160cot60180cot

cot 210 = 330cot30180cot

tan 240 = 360tan60180tan

sin 270 = 0270cos,1360sec,1



3

8

3

91

01113

333/13.E.G

2222

3. Find the value of 4/7cot3/5sin

3/4sin4/5cot4/3tan 2

145tan45180tan135tan4/3tan

145cot45180cot225cot4/5cot

2/360sin60180sin240sin3/4sin



2

3

2/3

4/3

12/3

2/311.E.G

2

4. Show that 223120cos135sin

480cos135sin

sin 135 = sin (180 - 45 ) = sin 45 = 2/1

cos 480 = cos (6 90 - 60 ) = - cos 60 = -1/2

cos 120 = -1/2

22322

22

2

1

2

1

2

1

2

1

LHS

5. Find x given that

3

5tan

4

7cot

3

4eccos

4

5cos

4

4sec

3

5sinx 222222


260sec60180sec240sec3/4sec



2145cos45180cos225cos4/5cos

3/260eccos60180eccos240eccos3/4seccos


360tan60360tan300tan3/5tan

The given equation is

22

22

2

2

313

2

2

12

2

3x

3

2x2x3.,e.i31

3

4

2

1x3

6. Simply: A90cotAsec

A90eccosA180secA180tan

AsecAtanAsec

AsecAsecAtan.E.G

7. Show that

2

cot2

3sin

2sin

2

3cotcos

2sin

2

3cotcos

2sinLHS

tancostancoscos 2

2

cot2

3sin

2sinRHS

tancostancoscos 2

RHSLHS



8. If 2

and5

3cos find the value of

eccos4tan3

sec3sin5

OP is the bounding line for .

r

x

5

3cos

4y.,e.i5r3x

So coordinates of P = (–3, 4)

4

5eccos,

3

4

3

4

x

ytan,

3

5sec,

5

4

4

ysin

4

54

3

43

3

53

5

45

.E.G

9

1

9

1

54

54

SAQ 5: If 360270and5

13sec find the value of

cos9sin4

cos3sin2

Note: As tan is given to be negative, the angle lies in II quadrant and IV

quadrant.

(–3, 4) P

Y

X M – 30

5 4



4.7 Compound Angles

The sum of difference of angles like A + B, A – B, A + B – C etc, are known

as compound angles.

The trigonometrical ratios of A + B, A – B can be expressed in term of those

of A and B. It should be noted that sin (A + B) sin A + sin B, cos (A + B)

cos A + cos B etc. This can be easily verified by the example

.30sin60sin3060sin

i) BsinAcosBcosAsinBAsin

ii) BsinAsinBcosAcosBAcos

iii) BtanAtan1

BtanAtan)BA(tan

iv) BsinAcosBcosAsinBAsin

v) BsinAsinBcosAcosBAcos

vi) BtanAtan1

BtanAtan)BA(tan

Worked Examples

1. Find the values of sin 75 , cos 75 and tan 75 .

We know that sin (A + B) = sin A cos B + cos A sin B.

Put A = 45 , B = 30

sin (45 + 30 ) = sin 45 cos 30 + cos 45 sin 30

i.e., 22

13

2

1

2

1

2

3

2

175sin

22

1375sin

We know that cos (A + B) = cos A cos B – sin A Sin B.

Put A = 45 , B = 30

cos (45 + 30 ) = cos 45 cos 30 - sin 45 sin 30



22

13

2

1

2

1

2

3

2

175cos.,e.i

Thus, we have 22

1375cos

BtanAtan1

BtanAtanBAtan Put A = 45 , B = 30

13

13

3/1.11

3/11

30tan45tan1

30tan45tan3045tan

3213

132

3275tan

2. Find the values of sin 15 , cos 15 , tan 15

30sin45cos30cos45sin3045sin15sin

22

13

2

1

2

1

2

3

2

1

22

1315sin

cos 15 = cos (45 – 30 ) = cos 45 cos 30 + sin 45 sin 30

22

13

2

1

2

1

2

3

2

1

22

1315cos

30tan45tan1

30tan45tan3045tan15tan

323/111

3/11



3. Show that (i) BcotAcot

1BcotAcotBAcot

(ii) AcotBcot

1BcotAcot)BAcot(

i) BsinAcosBcosAsin

BsinAsinBcosAcos

BAsin

BAcos)BAcot(

Divide both N & D by sin A sin B

AcotBcot

1BcotAcot

BsinAsin

BsinAcos

BsinAsin

BcosAsin

BsinAsin

BsinAsin

BsinAsin

BcosAcos

BAcot

BcotAcot

1BcotAcotBAcot

ii) BsinAcosBcosAsin

BsinAsinBcosAcos

BAsin

BAcosBAcot

Divide both N and D by sin A sin B

AcotBcot

1BcotAcot

BsinAsin

BsinAcos

BsinAsin

BcosAsin

1BsinAsin

BcosAcos

BAcot

AcotBcot

1BcotAcotBAcot

4. If 2

3BandA

2,

5

4Bcos,

13

5Asin find

i) sin (A + B) ii) cos (A + B) iii) sin (A – B) iv) cos (A – B).

Find also the quadrants in which the angles A + B and A – B lie.

Asin1Acos 2 (– ve sign is taken cosine is negative in II

quadrant)



13/12169/14413/512

13

12Acos

Bcos1Bsin 2 ( B is in the III quadrant)

5

325/161

Thus we have sin A = 5/13 sin B = -3/5

cos A= -12/13 cos B = -4/5

i) sin (A + B) = sin A cos B + cos A sin B

= 65

16

5

3

13

12

5

14

13

5

ii) cos (A + B) = cos A cos B – sin A sin B

= 65

63

5

3

13

5

5

4

13

12

iii) sin (A – B) = sin A cos B – cos A sin B

= 65

56

5

3

13

12

5

4

13

5

iv) cos (A – B) = cos A cos B + sin A sin B

= 65

33

5

3

13

5

5

4

13

12

Now, since sin (A + B) and cos (A + B) are both positive, A + B is in the

first quadrant. Since sin (A – B) is negative and cos (A – B) is positive,

A – B is in the fourth quadrant.

5. Prove that sin (A + B) sin (A – B) = sin2A – sin2B

LHS = sin (A + B) sin (A – B)

= (sin A cos B + cos A sin B) (sin A cos B – cos A sin B)

= sin2 A cos2 B – cos2 A sin2 B

= sin2 A (1 – sin2 B) – (1 – sin2 A) sin2 B

= sin2 A – sin2 A sin2 - sin2 B + sin2 A sin2 B

= sin2 A – sin2 B = RHS



6. If 4

BA show that (1 + tan A) 1 + tan B) = 2. Deduce that

.128

tan

4

tanBAtan BtanAtan1

BtanAtan.,e.i

tan A + tan B = 1 – tan A tan B …… (1)

LHS = (1 + tan A) (1 + tan B)

= 1 + tan A + tan B + tan A tan B

= 1 + (1 – tan A tan B) + tan A tan B using (1)

= 2

(1 + tan A) (1 + tan B) = 2 ……(2)

Put A = B in (2). (1 + tan A)2 = 2

i.e., 2Atan1 12Atan.,e.i

But 8

A,4

A2 128

tan

7. If ,7

1BAtan,

3

1Atan find tan B.

BtanAtan1

BtanAtanBAtan

Btan3/11

Btan3/1

7

1.,e.i

Btan3

Btan31

7

1 7 + 21 tan B = 3 – tan B tan B = -2/11

8. Show that cot 2 + tan = cosec 2

cosθsin2θ

sinθsin2θcosθcos2θ

cos

sin

2sin

2costan2cotLHS

RHS2eccos2sin

1

cos2sin

cos

cos.2sin

2cos



SAQ 6: If ,asinn1

acosasinntan

2 then show that tann1tan

4.8 Multiple and Sub-multiple angle

The angles 2A, 3A, 4A etc., are called multiple angles. And 2

A3,

3

A,

2

A etc.,

are called submultiple angles.

sin 2A = 2 sin A cos A

cos 2A = cos2 A – sin2 A = 1 – 2 sin2 A

= 2 cos A2 – 1

tan 2A = Atan1

Atan22

To prove that

i) Atan1

Atan1Acos

2

22

ii) Atan1

Atan2Asin

2

2

iii) cot A – tan A = 2 cot 2A

iv) cot A + tan A = 2 cosec 2A

v) AtanA2cos1

A2cos1 2

i) AsinAcos

AsinAcos

Acos

Asin1

Acos

Asin1

A2tan1

A2tan1RHS

22

22

2

2

2

2

LHSA2cos1

A2cos



ii) AsinAcos

AcosAsin2

Acos

Asin1

Acos

Asin2

Atan1

Atan2RHS

22

2

22

LHSA2sin1

A2sin

iii) AcosAsin

AsinAcos

Acos

Asin

Asin

AcosAtanAcotLHS

22

A2sin

A2cos2

AcosAsin2

Acos2

AcosAsin

A2cos 2

iv) AcosAsin

AsinAcos

Acos

Asin

Asin

AcosAtanAcotLHS

22

A2sin

2

AcosAsin2

2

AcosAsin

1

= 2 cosec 2A = RHS

v) RHSAtanAcos2

Asin2

1Acos21

Asin211

A2cos1

A2cos1LHS 2

2

2

2

2

4.8.1 Function of half angles

2

cos2

sin2sin

2

sin2

coscos 22

2

sin21cos 2

12

cos2cos 2



2tan1

2tan2

tan2

To prove that i) sin 3A = 3 sin A – 4 sin3A ii) cos 3A = 4 cos3 A – 3 cos A

iii) Atan31

AtanAtan3A3tan

2

3

i) sin 3A = sin(2A + A

= sin 2 A cos A + cos 2 A sinA

= (2 sin A cos A) cos A + (1 – 2 sin2 A) sin A

( sin 2A = 2 sin A cos A, cos 2A = 1 – 2 sin2 A)

= 2 sin A cos2 A + sin A – 2 sin3 A

= 2 sin A(1 – sin2 A) + sin A – 2 sin3 A

= 2 sin A – 2 sin3 a + sin A – 2 sin3 A

= 3 sin A – 4 sin3 A

sin 3A = 3 sin A – 4 sin3 A

ii) cos 3A = cos(2A + A)

= cos 2 A cos A – sin 2A sin A

= (2 cos2 A – 1) cos A – 2 sin A cos A sin A

( cos 2A = 2 cos2 A – 1, sin 2A = 2 sin A cos A)

= 2 cos3 A – cos A – 2 cos A sin2 A

= 2 cos3 A – cos A – 2 cos A (1 – cos2 A)

= 2 cos3 A – cos A – 2 cos A + 2 cos3 A

= 4 cos3 A – 3 cos A

cos3A = 4 cos3 A – 3 cos A

iii) tan (3A) = tan(2A + A)

= AtanA2tan1

AtanA2tan



=

AtanAtan1

Atan21

AtanAtan1

Atan2

2

2

=

Atan1

AtanAtan2Atan1

Atan1

)Atan1(AtanAtan2

2

2

2

2

= Atan2Atan1

AtanAtanAtan222

3

= Atan31

AtanAtan32

3

Atan31

AtanAtan3A3tan

2

3

4

1518sin

4

1536cos

4

1572cos

4

1554sin

Worked Examples

1. Show that A45tanA2sin1

A2cos



AcosAsin2AsinAcos

AsinAcos

A2sin1

A2cosLHS

22

22

2

AsinAcos

AsinAcosAsinAcos

Atan1

Atan1

AsinAcos

AsinAcos (by dividing both N & D by cos A)

RHSLHSAtan1

Atan1

Atan45tan1

Atan45tanA45tanRHS

2. Show that AtanA2sinA2cos1

A2sinA2cos1

Acosasin2A2sin&1Acos2

Asin21A2cos

AcosAsin21Acos22

AcosAsin2Asin211LHS

2

2

2

2

AcosAsin2Acos2

AcosAsin2Asin22

2

RHSAtanAcos

Asin

AsinAcosAcos2

)AcosA(sinAsin2

3. If Bsin

Bcos1Atan show that tan 2A = tan B.

Give that ,Bsin

Bcos1Atan

i.e.,

2

Bcos

2

Bsin2

2

B2sin2

2

Bcos

2

Bsin2

2

Bsin211

Atan

2

2

Btan

2

Bcos

2

Bsin

2

tantanB

A



Now, BB

B

A

AA tan

2tan1

2tan2

tan1

tan22tan

22

BtanA2tan

4. If 7

1tanand

3

1tan show that

42

4

3

9

83

2

9

11

3

12

tan1

tan22tan

1

28

2528

25

7

1

7

31

7

1

4

3

tan2tan1

tan2tan2tan

2

12tan 4

2

5. If t2

tan show that i) 2t1

t2sin ii)

2

2

t1

t1cos

i) sin = 2

cos2

sin2

=

2tan1

2tan2

2sin

2cos

2cos

2sin2

222

[by dividing both N & D by 2

cos2

= 2t1

t2



ii) cos = 2

sin2

cos 22

=

2sin

2cos

2sin

2cos

22

22

=

2tan1

2tan1

2

2

2

cos2byD&Nbothdividingby

= 2

2

t1

t1

Aliter:

adj

oppt

2tan

In the ABC where C = 90 , 2

B

AC = t, BC = 1 2t1AB

B C

A

2t1

0/2



22 t1

1cos,

t1

t

2sin

i) 2

cos2

sin2sin

222 t1

t2

t1

1

t1

t2

ii) 2

2

2

2

2

22

t1

t1

t1

t

t1

1

2sin

2coscos

SAQ 7: Show that 2eccos45tan1

45tan12

2

4.9 Summary

In this unit we study the concept of radian measure and its applications. The

basics of Trigonometric functions and trigonometric ratios of standard angles is

discussed here with examples. Allied angles and compound angles is a

explained in a simple manner with illustrative examples wherever necessary.


1. If ABC is any triangle show that 2

CBsin

2

Acos

2. Find the value of in between 0 and 360 and satisfying the equation

03

1tan

3. Show that the value of tan 3 cot can not lie between 3

1 and 3.

4. Show that sec (45 + A) sec [45 - A) = 2 sec 2A.

5. Show that cos24cos222



4.11 Answers


1. Angle through which the wheel turns in one minutes = 25 2 = 50

Angle described in one sec .cm50r,6

5

60

50

s = r = 5 /6 50 = 130.9 cm

Distance covered by a point in the rim in 1 sec is 130.9 cms.

2. Since cos = (4/5) and (adj/hyp), consider a right angled triangle ABC

in which C = 90 , B = , BC = 4, AB = 5; clearly AC = 3.

cos = 4/5, sin = 3/5

cosec = 5/3, cot = 4/3

Substituting these in the given expression

2

43

4

86

15/)4036(

15/)5036(

)3/8()5/12(

)3/10()5/12(

)3/4(2)5/3(4

)3/5(2)5/4(3.E.G

3. ABC is a triangle in which 90C and B

A

B C 4

5

3

A

B C m2 – n

2

m2 + n

2

2mn



Since cos 22

22

nm

nm

BC = m2 – n2 and AB = m2 + n2

AC2 = AB2 – BC2

= (m2 + n2)2 – (m2 – n2)2

= m4 + n4 + 2 m2n2 – (m4 + n4 – 2 m2n2)

= 4m2n2

mn2nm4AC 22

Now mn2

nm

opp

hypeccos,

mn2

nm

hyp

adjcot

2222

4. LHS = sec 30 tan 60 + sin 45 cosec 45 + cos 30 cot 60

3

1

2

32

2

13

3

2

RHS2

7

2

13

2

112

5. r

x

13

5cosimplies

5

13sec

x = 5, r = 13

Now 222222 xry,ryx

= 132 – 52 = 122

y = 12.

Here we have to take y = 12

So P (5, –12)

13

5cos;

13

12

r

ysin

P (5, –12)

Y

–2

M 5 O

13

X



31

3

93

9

4548

1524

13

59

13

124

13

53

13

122

.E.G

6. asinn1

acosasinntan

2

atannasec

atann22

(Divide Nr and Dr. by cos2 a)

atann11

atann

atannatan1

atanntan

222

LHS =

2

2

tann11

ntann1

tann11

tanntan

tantan1

tantantan

RHStann1tanntann11

tanntann1tan22

3

7. 45sin45cos

45sin45cos

45cos

45sin1

45cos

45sin1

LHS22

22

2

2

2

2

2eccos2sin

1

290cos

1

452cos

1

Terminal Questions

1. In any triangle ABC we have A + B + C = 180

2

CB90

2

AorCB180A

Thus, 2

CBsin

2

CB90cos

2

Acos



2. 3

1tan

We know that 3

130tan

Consider

3

130tan30180tan

3

1150tan.,e.i …(1)

3

130tan30360tan

3

1330tan …(2)

From (1) and (2) it follows that = 150 , = 330

3. sayxtan31

tan3

tan31tan

tantan3cottan

2

2

2

33

2

2

1x3

3x1x3

1x3

3xtan

Since tan2 is positive, either ,3xor3

1x so x cannot lie between

.3and3

1

4. A4590secA45secLHS

= sec (45 + A) cosec (45 + A)

= A45cosA45sin2

2

A45sinA45cos

1

= A2sec2A2cos

2

A290sin

2

A452sin

2



5. LHS = 12cos2222 2

= 2cos4222cos422 22

= 2cos122cos22

= cos2cos41cos212 22



Unit 5 Limits and Continuity

Structure

5.1 Introduction

Objectives

5.2 The Real Number System

5.3 The Concept of Limit

5.4 Concept of Continuity

5.5 Summary


5.7 Answers

5.1 Introduction

In this chapter you will be recalling the properties of number. You will be

studying the limits of a function of a discrete variable, represented as a

sequence and the limit of function of a real variable. Both these limits

describe the long term behaviour of functions. You will be studying

continuity which is essential for describing a process that goes on without

abrupt changes. You will see a good number of examples for understanding

the concepts clearly. As mathematics is mastered only by doing, examples

are given for practice.

You are familiar with numbers and using them in day – to – day life. Before

introducing the concept of limits let us refresh our memory regarding various

types of numbers.

Objectives:


understand the concept of limit.

apply the concept of continuity in problems.

find whether a given function is continuous or not.



5.2 The Real Number System

You are using numbers like i32,i1,i,i,,7

4,

4

3,3,2 etc.

The last two numbers – 1 + i and 2 – 3i are complex numbers. The rest of

them are real numbers.

The numbers 1, 2, 3, …… are called natural numbers.

N = {1, 2, 3, ……..}

The numbers …………………….. – 3, –2, –1, 0, 1, 2, 3, …… are integers.

,........2,1,0Z or ....................,3,2,1,0,1,2,3.....,

The set of quotients of two integers, the denominator not equal to 0 are

called rational numbers and the set of rational numbers is denoted by Q.

Usually these numbers are represented as points on a horizontal line called

the real axis. (Refer to Fig. 5.1)

Fig. 5.1 Representation of numbers

After representing the integers and rational numbers. So there are no gaps

in the real line and so it is called “continuum”.

We can also represent the relations “greater than” or “less than”

geometrically. If a < b, then a lies to the left of b in the real line (and b lies to

the right of a).

The modulus function

The modulus function simply represents the numerical value of a number. It

is defined as follows:

7

4

– 3 – 2 – 1 0 1 2 3

4

3 2



0xifx

0xifxx

For example, 00,22,22

Note: baba

SAQ: (Self Assessment Questions).

Choose the right answer

1. If a > b, then isba

A) Positive

B) Negative

C) Zero

2. Choose the right answer

abba is equal to

A) ba2

B) 0

C) 2 (a – b)

D) 2 (b – a)

An Important Logical Symbol

In Mathematics, we use symbols instead of sentences. For example, “3 is

greater than 2” is written as 3 > 2. Throughout the test we used the symbol

(read as “implies”)

If x > 2, then x > 4 is written as (x > 2) (2x > 4).

Generally „If P, then Q” is written as

P Q. (P is given and Q is the conclusion)

Note: P Q is different from Q P. Q P is called the converse of P Q.



The distance function

If a and b are two real numbers. Then the distance between a and b is

defined as | a – b|. Refer Figure 5.2. Why we choose |a – b| as the distance

between a and b should be clear from figure 5.2. When b > a, then the

distance is b – a; when a > b, it is a – b. Both a – b and b – a are equal to

|a – b|. So wherever a and b are on the real line, the distance is |a – b|.

Figure 5.2 Distance function

The distance function satisfies the following properties.

1. ba0ba

2. abba

3. bccaba

5.3 The Concept of Limit

The concept of limit is an important concept in Mathematics, which is used

to describe the long term behaviour of a phenomenon. You might have

heard about the half life of a radioactive substance. It is the time required for

a radioactive substance to lose half is radioactivity. This is used in carbon

dating. Carbon dating is a method of calculating the age of a very old object

a b

b – a

b a

a – b



by measuring the radioactive carbon it contains. Thus the long term state of

an old object is described by the concept of limit.

Function of a discrete variable and a continuous variable

The Concept of limit is associated with functions. A function from a set A to

a set B is a rule which assigns, to each element of A a unique (one and only

one) element of B. Examples of functions are the marks of students in a

class of 100 in a particular subject or the B.P. or sugar level in blood of a

particular student in the class.

There are two types of functions. The first is a function from the set N of

natural numbers (i.e., N = {1, 2, 3,….}. If the students in a class are

numbered as 1, 2, ……., 100.

Then the mark in a particular subject is a function from {1 2, ….., 100} to {0,

1, 2, …., 100) if the marks are given as percentages. This is called a

function of a discrete variable. (Don‟t be afraid of the term variable. It is a

simply a symbol which can be replaced by value you choose).

Definition A function of a discrete variable is a function from N or a subset

of N to the set R of all real numbers.

The second type of functions refer to functions from R to R. It is called a

function of a continuously. So it can be treated as a function of a continuous

real variable t, t denoting time.

Definition: A function of a continuous real variable or simply a function of a

real variable is a function from R to R. (or [a, b]). Here [a, b] denote the set

of all real numbers between the numbers a and b).

Functions of a discrete variable

We have defined a function of a discrete variable as a function from N

{1, 2, 3, …} or subset of N to the set R of real numbers. A convenient way to



representing this function is by listing the images of 1, 2, 3, etc. If f denotes

the function then the list.

F(1), F(2), F(3), ….. ………………… (*)

Represents the function f usually f(1), f(2) are written as a1, a2 etc.

The list given in (*) is called a sequence.

In a sequence the order of the elements appearing in it is important. A

common example of a sequence is a queue you see in a reservation

counter. Then a1 is the person standing in front of the counter getting his

reservation done. a2 is the person behind a1 etc. the order of persons in the

queue is important. You won‟t certainly be happy if the order of the persons

in the queue is changed.

The limit of a sequence

From the above discussion, two points should be clear to you.

1. A sequence is an arrangement of real numbers as the first element,

second element etc.

2. A sequence represents a function of a discrete variable.

We denote a sequence by (an) and an denotes the nth term.

Assume that you have a string of length 1 cm. Denote it by a1. Cut the string

into two halves and throw away one half. Denote the remaining half by a2.

Then 2

1a2 . Repeat the process indefinitely.

Then

3

4

2

32

1a,

2

1a etc. After (n + 1) repetitions, you are left with a

string of length .2

1a

n

1n Intuitively you feel that the string becomes

smaller and smaller and you are left with a string whose length is nearly

zero in the long run. At the same time you realize that you will have “some

bit” of positive length at any time. Also you can make the string as small as

you please provided you repeat the process sufficient number of times. In



this case we say that “an tends to 0 as n tends to infinity” “an tends to 0”

means 0an is as small as we please/ “n lends to infinity” means we

repeat the process sufficient number of times.

Now we are in a position to define the limit of a sequence (an)

Definition: Let (an) be a real sequence. Then (an) tends to a number a, if

given a positive number , (pronounced as epsilon), there exists a natural

number n0 such that

aan for all 0nn ……………. (1.1)

In this case we write .aaItoraa nan

n We also say (an) converges to a.

Note: aan is the numerical value of an – a. For example | 2 | = 2 and

| – 3 | = 3, n0 is a “stage”.

n > n0 means after a certain stage aan simply means that an comes

as close to a as we choose.

Example: Show that .n

1awhere0a nn

Solution: .n

10aSo.

n

10

n

10a nn

Let be a given positive number.

n

10an when

1n .

Let n0 be the smallest natural number .1

(For example, if ,7.1471

take n0 = 148). Then .n

1

0

If n > n0, then .n

1

n

1

0

Hence 0an when .nn 0



This proves (1.1) with 0 in place of a.

Hence .0)a( n

Example: Show that 0an where nn

2

1a .

Solution: As in Example, .2

10a

nn

For ,2

1n

we require 1

lognor1

2 2n

By choosing n0 to be the smallest natural number greater than 1

log2 , we

see that (1.1) is satisfied for n0.

Hence .02

1n

You can see that several similar sequences tend to 0. Some of them are

,.log

1........,

4

1,

3

1.......,

n

1,

n

1,

n

1nnn432

…… (1.2)

Algebra of limits of sequences

If (an) and (bn) are two sequences, then we can get a new sequence by

“adding them”. Define cn = an + bn. Then (cn) is a sequence and we can write

(cn) = (an + (bn). We can also subtract one sequence (bn) from another

sequence (an), multiply two sequences etc. We can also multiply a

sequence (an) by a constant k.

Let us answer the following questions.

1. What happens to the limit of sum of two sequences ?

2. What happens to the limit of difference, product of two sequences ?

We summarize the results as a theorem.

Theorem( ): It (an) and (bn) and two sequences converging to a and b

respectively, then

a) (an + bn) → a + b



b) (an – bn) → a – b

c) (kan) → ka

d) (anbn) → ab

e) b

a

b

a

n

n provided bn 0 for all n and b 0.

Proof: We prove only a) > 0 be a positive number.

As (an) → a, we can apply (5.5) by taking 2

in place of . Thus we get a

natural number n0 such that

2aan for all 0nn ………….. (1.3).

Similarly, using the convergence of (bn), we an get n1 such that

2bbn for all 1nn …………….. (1.4).

Let m = maximum of n0 and n1. Then (5.3) and (5.4) are simultaneously true

for n > m.

Thus,

2bb,

2aa nn for all mn ………………… (1.5)

From (1.5) we get

22bbaabbaababa nnnnnn

for n > m.

Thus (1.1) holds good for the sequence (an + bn) in place of (an) and a + b

(in place of a)

Hence (an + bn) → a + b

Note: The choice of m may puzzle you. When 2

aan for all n > 1000,



then certainly 2

aan for all n > 1001, 1002 etc. So 2

aan for

all n > m, m being greater than 1000.

Remark: The other subdivisions can be proved similarly. As you are more

interested in applications you need not get tied down by the technical details

of the proof. In mathematics, there is a “commandment”, Though Shall Not

Divide By Zero”. If bn = 0, then n

n

b

a is not defined. So is

b

a when b = 0. So

when you apply (2), see to it that the conditions are satisfied. (bn 0 and

b 0). With this theoretical foundation, you are in a position to find limits of

sequences.

Worked Examples

W.E.: Show that a constant sequence is convergent (A sequence

(A sequence (an) is a constant sequence if an = k for all n).

Solution an = k for all n. Consider kan .

0kkkan

As 0 < , for every positive number , for all n > 1, that is, n0 = 1 for all >

0. So a constant sequence converges to its constants value.

W.E.: Find the nth term of the sequence ...............,5

7,

4

6,

3

5,2,3

Solution: To discover a pattern in the terms of the sequence start from the

third term.

3

32

3

5a3

4

42

4

6a4

5

52

5

7a5

6

62

6

8a6

So a positive choice is ,n

n2an when an is written in this way



,31

12a1 ,

3

5

3

32a,

2

22a 32 etc.

To find the limit of the sequence (if it exists), with 1n

2

n

n

n

2an

Taking n

1bn and cn = 1, we get an = 2bn + cn

As 0n

1bn and ,11cn

1102an . Hence the given sequence converges to 1.

W.E.: Evaluate 2

2

n n4n32

nn2It

Solution As we know that 0n

1 and ,0

n

12

we try to write the nth

term of the given sequence in terms of .n

1and

n

12

2

2

nn4n32

nn2a

2

2

2

2

n

n4n32

n

nn2

4n

3

n

2

1n

1

n

2

2

2

As 0n

1,

n

12

and (1) → 1,

110021Itn

1It

n

1It21

n

1

n

2It

22n



Similarly,

4403024n

3

n

1It

2n

By theorem ( ),

4

1

n4n32

nn2It

2

2

n

Note: It is difficult to prove that a given sequence is not convergent. For

proving convergence we start with > 0 and find a stage no satisfying ( ).

To prove that a sequence is not convergent we have to prove that ( ) does

not hold good for every stage for a particular > 0 and this is certainly

difficult. However we can prove that certain sequences are not convergent

indirectly as the following example shows.

W.E.: Show that the sequence 1, –1, 1, –1, 1, –1 , …… is not

convergent.

Solution: We can write the sequence as (an) where

evenisnwhen1

oddisnwhen1an

Suppose (an) → a for some real number a. the number has to satisfy one

and only one of the following conditions: a < –1, –1 < a < 1, a > 1.

(See Fig. 5.3 representing these cases).

Fig. 5.3: Illustration of W.E.

In case 1aa,1 n if n is odd. So we cannot prove ( ) in this case

(for > 2).

a –5

1 a –5

1 1 –5

a

Case 5

Case 2

Case 3



In case 2aa,3 n if n is even and we cannot prove ( ) for > 2.

In case 2, if a is closer to 1, then 1aan for even n. If a is close to –1,

then 1aan for odd n. If a = 1, then 1aa,3 n for even n. So ( )

cannot hold good for > 2 or = 2. So the given sequence is not convergent.

Try to answer the following questions before you proceed to the next section.

S.A.Q. 3: Which of the following sets can be arranged as a sequence?

a) The passengers in a 3 ties coach

b) The people attending a meeting in a beach

c) The people living in Karnataka

d) The students of I M.Sc. Biotechnology in a college

The Limit of a Function of a Real Variable

You are now familiar with natural numbers and real numbers. The natural

numbers appear as “discrete” points along the real line and we are able to

fix some element say 1 as the first natural number, 2 as the second natural

number etc. So the natural numbers appear as the terms of a sequence. But

it is not possible to arrange the real numbers as a sequence. If a real

number a is the nth element and b is the (n+1)th element, where will you

place 2

ba. It appears between the nth element and the (n+1)th element. If

you take 2

ba as the (n+1)th element, where will you place ?

2

baa

2

1

So you feel initiatively that real numbers can not be arranged as a sequence.

When we consider numbers between a and b. We consider points lying

between the points representing the numbers a and b. The numbers lying

between two numbers a and b from an “interval”. So “interval” on the real line

is the basic concept. Usually we define a function of a real variable on an

interval. We define various types of “intervals” as follows (refer to Table 5.1)



Table 5.1 Intervals

Set notation Interval Graphical representation

}bxa|Rx{

(a, b)

}bxa|Rx{

[a, b]

}bxa|Rx{

[a, b)

}bxa|Rx{

(a, b]

}xa|Rx{ [a, )

}xa|Rx{ (a, )

}xx|Rx{ (– , a]

}ax|Rx{ (– , a)

R (– , )

Note: Here (a represents inclusion of all numbers > a. [a represents the

inclusion of all numbers > a].

is not a number. It simply represents the inclusion of all “large” –ve

numbers.

represents the inclusion of all “large” positive real numbers.

(a, b) and [a, b] are called open interval and closed interval.

So it is natural to represent R as the interval ,

a b

a b

a b

a b

a

a

a

a



S.A.Q.4

a) Find all natural numbers in the intervals [3, ), (3, ), (– , 3) and

(– , 3]

b) Find all integers lying in the intervals given in a

c) Find all numbers in [2, 2], (2, 2)

Example; Represent 23x/Rx as an interval

Solution: 23x represents two inequalities x – 3 < 2, – (x – 3) < 2.

When x – 3 < 2, x < 5

When – (x – 3) < 2, x – 3 > – 2, or x > 3 – 2 = 1

Hence the given set is {1, 5}

We can also arrive at this interval geometrically (see Fig 5.4)

Figure 5.4

S.A.Q.5: Represent the sets 23x/Rx ,

23x/Rx,23x/Rx as intervals.

Now we have enough background to define the limit of a function f of a real

variable x. In the case of functions of a discrete variable or sequence, we

defined )n(fItoraItn

nn

. This limit represented the long term behaviour

of f. In the case of a function of a real variable, we can discuss the

behaviour of f(x), when the variable x comes close to a real number a. In

other ways we will be defining ).x(fItax

We want to write the statement “x

2 5 3 4 5

2 2



comes close to a” rigorously. The geometric representation of real numbers

can be used for this purpose. When do you say that your house is near your

college? When the distance between your house and your college is small.

In the same way, we can say that “x is close to “when | x – a| is small. If

“smallness” is defined by a distance of say 0.1, then x is close to a if

|x – a| < 0.1 of course the measure of “smallness” is relative. For a person

living in Mangalore, Manipal is not near Mangalore. For a person living in

US, Mangalore and Manipal are near to each other. So “smallness” is

decided by the choice of a positive number (This was done in defining the

limit of a sequence also)

Before giving a rigorous definition of limit, let us consider two examples.

Consider F(x) = 1 + x. Let us try to see what happens when x is close to 1.

We evaluate f(x), when x = 0.9, 0.99, 0.999, 1.1, 1.01, 1.001

F(0.9) = 1.9 f(0.99) = 1.99 f(0.999) = 1.999

F(1.1) = 2.1 f(1.01) = 2.01 f(1.001) = 2.001

Note all these values are near the value 2.

Consider another function .1x,1x

1xxg

2

(Why don‟t we define g(1) ? If we put x = 1 in ,1x

1x2

then we get 0

0 which

is not defined).

As in the case of f(x), we compute some function values.

99.199.0g9.119.0

19.09.0g

2

g(0.999) = 1.999 g(1.1) = 2.1

g(1.01) = 2.01 g(1.001) = 2.001



Note: These values are close to 2. Hence we can say that x close to

1 both f(x) and g(x) are closed to 2 and we can take 2 as

xfIt1x

or xgIt1x

.

Now let us formulate a rigorous definition of xfItax

Definition: Let f be a function of a real variable. Then IxFItax

if given a

positive number There exists a positive number such that

Ixfax0 …………. (1.6)

Let us analyze the definition.

We have two choice of positive numbers ( and ) and two conditions

Ixfandax0 Given any positive number there exists a

positive number such that the condition P "ax0" Implies the

condition "Ixf"Q

The choice of depends on the given number . The function f need not be

defined at x = a.

The condition P says that x is close to a.

The condition Q says that f(x) is close to I

We can also express the definition geometrically.

Given > 0, there exists > 0 such that

I,IxFa,aa,ax

In figure 5.5, the point (a, I) is measured as 0, meaning that the functional

value of f at x = a is not known or defined.



Figure 5.5: Definition of limit of function

The images of points in a,a)a,a( under the function f is a subset

of the interval (I – , I + ) along the vertical axis.

W.E.: Evaluate 6x2It3x

Solution: Here f(x) = 2x – 6

Choose any > 0. We have to choose a such that (5.5) is satisfied. We

have to guess the value of I. When x comes close to 3.2x – 6 should come

close to 2(3) – 6 = 0. Take I = 0. Then

06x20)x(f

3x2

2

3x

So, choose ,2

then



0)x(f2

3x0

Hence, 06x2It3x

Note: f(3) = 2(3) – 6 = 0. In this case the function f is defined at x = 3 and

f(3) coincides with xfIt3x

.

W.E.: Evaluate 1x2It 2

0x

Solution: Let 0 . As in the previous problem, we can guess the value of I

it is 11)0(2 2

11x21xf 2

2x2

2

x

Hence for a given 0 , the corresponding is chosen as 2

and condition

(1.6) holds good. Hence 11x2It 2

0x

W.E.: Evaluate xIt0x

Solution: Proceed as in the previous example. In this problem 2 and

0xIt0x

General Remark While evaluating ,xfItax

if f(a) is defined or it is not of

the form ,0

0 and f(x) is defined by a single expression, it will turn out that

.afxfItax

If f(a) is not defined or f is given by two different expressions.



Then we have to guess the value of the limit and prove condition (1.6).

In some problems, f(x) may be given as quotient of two expressions but it

may reduce to an easier function on simplification. In such cases the

problem will reduce to an easier one.

W.E.: Evaluate 2x

2x3x2It

2

2x

Solution: When x = 2, f(x) is of the form .0

0 So we try to see when x – 2 is a

factor of

2x2 – 3x – 2, 2x2 – 3x – 2 = 2x2 – 4x + x – 2 = 2x (x – 2) + 5 (x – 2) = (2x + 5)

(x – 2).

When x 2, x does not assume the value 2. So,

2x

2x1x2

2x

2x3x2 2

= 2x + 1 on canceling x – 2, since x – 2 0.

So the given limit reduces to

51x2It2x

(as in W.E.)

Algebra of Limits of Functions

It is not necessary that we use the – definition for every problem. We

study important properties of limits of functions as a theorem. We can

evaluate limits using this theorem (noted as a proposition).

Proposition:

a) axItax

b) kkItax

, where k is a constant

c) 22

axaxIt



d) 33

axaxIt

e) nn

axaxIt

f) axItax

when a > 0

Theorem: Let k be a constant, f and g functions having limit at a and n a

positive integer. Then the following hold good.

a) kkItax

b) xfItkxkfItaxax

c) xgItxfItxgxfItaxaxax

d) xgItxfItxgxfItaxaxax

e) xgIt.xfItxg.xfItaxaxax

f) ,xIt

xfIt

xg

xfIt

ax

ax

axprovided 0xgIt

ax

g) n

ax

n

axxfItxfIt

h) xfItxfItaxax

provided xfItax

is positive

You need not prove these results. It is enough if you clearly understand the

theorem and proposition and apply them for evaluating limits.


SAQ 6: Evaluate the following limits

a) n2

1n2It

n b)

1n

nIt

n c) 100It

n



d) 2

2

n n2n35

7n4n3It e)

2

2

n nn1

nnIt f)

3

3

n n1

n1It

g) 2

2

n nn1

n2It h)

2

2

n n

n9n32It

SAQ 7: Evaluation the following limits

a) 2

1xxx1It b)

21x xx1

1It c) 2

0xx4x32It

d) 2x

4xIt

2

2x e)

1x

1xIt

3

1x f)

2

2

1x xx1

x3x2It

g) 2x

2x3xIt

2

2x h)

1x

1xIt

21x

S.A.Q.8 If ,mxgItandIxfItaxax

evaluate the following

a) xg3xf2Itax

b) xgxfxgxfItax

c) 22

axxgxfIt d)

2ax xg1

xfIt

e) xg4

xg2xfIt

ax if m is positive f) xf2xgIt

ax if I, m > 0.

5.4 Concept of Continuity

In mathematics and sciences, we use the word “continuous” to describe a

process that goes on without abrupt changes. For example, the growth of a

plant, the water level in a tank and the speed of a moving car in a four-base

highway are exhibiting continuous behaviour.

Before defining continuous functions, let us look at the graphs of three

functions.



Figure 5.6: Two discontinuous functions and a continuous function

The first graph has a break at x = a in the second graph also there is a

break at x = a. If you ignore the point corresponding to x = a, there is no

break for the break occurs at x = a the third function has no break. So it

should be intuitively clear to you that the first two functions are not

continuous while the third function is continuous.

Let us formulate a rigorous definition of continuity.

Definition: Let f be a function of a real variable defined in an open interval

containing a. Then f is continuous at a if afxfItax

Note: In order to define continuity at a, we need three conditions.

1) xfItax

exists

2) f(a) is defined

3) )a(fxfItax

Even if one of them fails, then the function f is not continuous at a.

Now look at Fig. 5.4. The first function say f has no limit at a, i.e., xfItax

does not exist. For the second function, xfItax

exists but .afxfItax

The third function is continuous.



Example: Define f as follows:

2xif3

2xif2x

4xxf

2

Is f continuous at 2?

Solution:

2x

2x2xIt

2x

4xIt

2x

2

2x

2xIt2x

= 4

Note: We can cancel (x – 2) since it is non zero.

So xfIt2x

exists.

But f(2) = 3 4 xfIt2x

Hence the function f is not continuous 2.

Example: Define f as follows.

2xif4

2xif2x

4x

xf

2

Is f continuous at 2?

Solution: From above example, we have .4xfIt2x

A s f(2) = 4, f is

continuous at 2.

Sometimes function may be defined by two different expressions. In such

cases the following method of proving continuity will be useful. For that we

need the concept of left limit and right limit.



Definition: Let a function f be defined in an interval (b, a) where b < a).

Then IxfItax

(called the left limit) if given > 0. There exists > 0

such that Ixf,ax ………… (1.7).

Definition; Let a function f be defined in an interval (a, b), where b > a.

Then IxfItax

called the right limit if given > 0, there exists > 0

such that Ixfa,ax ………….. (1.8)

Note: We can prove that (1.7) and (1.8) implies (1.6). Hence if the left and

right limits exist and are equal then

)x(fIt)x(fItxfItaxaxax

When a function is defined by two different expressions, we have to

evaluate the left and right limits.

IxfItax

if both the left and right limits exist and are equal.

Worked Examples:

W.E.: Test whether f is continuous at x = 3 where f is defined by

3xif2

3xif4x3xf

Solution: As f is defined by two expressions one for ]3,( and another

for (3, ), we evaluate the left and right limits.

54334x3ItxfIt3xxx

22ItxfIt3xxx

As the left and right limits are not equal, xfIt3x

does not exist. Hence f is

not continuous at 3.



When a function f is continuous at every point of an interval (a, b) we say

that the function is continuous on (a, b). In particular if a function is

continuous at every real number. Then we say that a function is continuous

on R.

Using propositions, we can prove that the functions, k (a constant), x, x2,

……..xn, where n > 2 are continuous on R.

The function x is continuous on (0, )

The function x

1 is continuous on ,00,

Using previous theorem, we can prove the following theorem.

Theorem:

a) If f and g are continuous at a, then

i) f(x) + g(x) ii) f(x) – g(x) iii) f(x) g(x) are continuous at a

b) If f and g are continuous at a and g(a) 0, then g

f is continuous at a

c) If f is continuous at a, f(x) > 0 for x in an open interval containing a,

xf is continuous at a.

Worked Examples

W.E.: Test the continuity of the function f at all real points where f is defined

by

3xfor1x2

3xforxxf

2

Solution: If a < 3, then f(x) is defined by the expression x2 in an open

interval containing a. So f is continuous for all a < 3.

So it remains to test continuity only at 3.

93xItxfIt 22

3x3x



71321x2ItxfIt3x3x

As xfIt,xfItxfIt3x3x3x

does not exist. So f is not continuous at 3.

Thus f is continuous at all real points except 3.

W.E.: Test the continuity of the function f where f is defined by

2xif7

2xif|2x|

2x

xf

Solution: When x < 2, |x – 2| is negative. So |x – 2| = – (x – 2)

12x

2xxf

When x > 2, |x – 2| is positive. So |x – 2| = x – 2

12x

2xxf

f(2) = 0. Thus

2xif1

2xif0

2xif1

xf

As in the previous worked example, f is continuous for all a < 2 and all a > 2.

11ItxfIt2x2x

11ItxfIt2x2x

As xfIt,xfItxfIt2x2x2x

does not exist. So f is continuous at all

points except 2.

Not all functions are continuous. In some cases xfItax

may not exist. You

may ask a question: how to establish that xfItax

does not exist. One



sample case is where the left and right limits exist but are not equal. We have

examples for this case. The worst case is when neither of the two limits exist.

For proving the non-existence of limits, we use the following theorem.

Theorem: A function f is continuous at a if and only if the following

conditions holds good.

afxfax nn

To prove the non existence of the limit it is enough to construct a sequence

(xn) converging to a such that f(xn) does not converge to f(a).

W.E. : Show that the function f defined by

0xif0

0xifx

1

xf is not continuous at 0.

Solution: Let .n

1xn Then 0xn (obvious) f(xn) = n and so f(xn) does

not converge to 0 since f(xn) indefinitely increases and so cannot approach

0.


S.A. Q. 9 Verify whether the following functions f is continuous at a

a) 1a1xif5

1xif3x2xf

b) 1a1xif9

1xif5x4xf

c) 1a1xif9

1xif5x4xf

d) 2a2xifx1

2a2xifx1xf 2

e) 3a3xif

3xif13x

9xxf

2



f) 3a

3xif6

3xif3x

9xxf

2

g) 2a

3xif7

3xif3x

9xxf

2

h) 3a

3xif6

3xif9x

3x

xf 2

S.A.Q. 10 Show that the function f defined by

1xif0

1xif1x

1xf is not continuous at 1.

S.A.Q. 11 Show that the following functions are continuous on R.

a) 1xif

1xif

2x2

5x6xxf

2

2

b) 1xif4

1xifxxx1xf

32

c) 1xifx4

1xifx41xf

2

3

d) 1xif1x

1xif2x3xxf

3

2

5.5 Summary

In this unit we studied the basics of real number system then the concept of

limit was discussed which was further extended to the concept of continuity.

All definitions and properties of the above mentioned concepts is given very

clearly with sufficient number of examples wherever necessary.




1. Find all natural numbers in the following intervals

a) [4, 9) b) (4, 9] c) (4, 9) d) [f, 9]

e) (4, ) f) (9, ) g) [9, ) h) (4, ) (9, )

2. Find nn

aIt when

a) nn

2

3a b) 1

3

2a

nn c) nnn

3

2

2

3a

d) !n

1an e)

2n

1an

3. Show nn

aIt does not exist when

a) an = n b) an = 2n c) an = n!

4. Evaluate nn

aIt when

a) 7n4

2n3an b)

n

1nan c)

n

n32an

d) 2

2

nn7n43

nn2a e)

2

2

nnn1

nn1a f)

2

2

nn

nn1a

g) n23

n2an h)

2

2

nnn23

n3n21a i)

3n24n

1n3na

2

2

n j) 2n

nna

3

2

n k) n1nan

5. Evaluate xfItan

when f(x) is equal to

a) 3x2 b)

43 xx c) 543 xxx2

d) 2x54x32 e) 2x54

x32 f)

4

9x2

6. Evaluate the following limits

a) 1x2It 2

1x b) x231x2It

1x



c) 21x x5

x74It d) 3x3It

3x

e) 3x2

1It

1x f)

2x3x

1x2xIt

2

2

3x

g) 1x

1xIt

2

3

1x h)

1x

2xxIt

2

2

1x

7. If IxfItan

and ,mxgItan

evaluate

a) xgxfItan

b) xgxf2xg2xfItan

c) xg3xf2Itan

(when I, m, > 0)

d) 2an xg

xgxfIt when m > 0

8. Show that the following functions are continuous at a

a) 22 xx1xf for all x in R, a = 0

b) 2x1

1xf for all x in R a = 0

c) 2x21

x32xf for all x in R a = 0

9. Show that the following functions are continuous at a

a) 4a

4xif8

4xif4x

16xxf

2

b) 1a1xifx2x34

1xifx4x32xf

2

2

c) 1a1xifx1

1xifx1xf

3

2



10. Show that the following functions are not continuous at a

a) 2a

2xif7

2xif2x

8x2xf

2

b) 1a

1xif1

1xifx1

x1xf

2

c) 1a1xifx4x3

1xifx4x31xf

2

2

d) 1a1xif1

1xif1xf

e) 0a

0xif1

0xif0

0xif1

xf

5.7 Answers


1. A

2. A

3. a) and d) can be arranged as a sequence according to chart and

attendance register

4. a) {3, 4, 5, …..} b) {4, 5, 6, ……}

c) {……. –3, –2, –1, 0, 1, 2,} d) {…. –3, –2, –1, 0, 1, 2, 3}

5. [1, 5], (– , 1) (5, ), (– , 1] [5, )

6. a) 1 b) 1 c) 100 d) 2

3

e) 1 f) –1 g) 2 h) 9



7. a) 3 b) 3

1 c) 2 d) 4

e) 3 (Hint: x3 – 1 = (x – 1) (x2 + x + 1)) f) 3

5 g) 1

h) 2

1

8. a) 2I + 3m b) I2 – m2 c) 22 mI d) 2m1

1

e) m4

m2I f) m2I

9. a), b), c), d) continuous e) discontinuous f), g) continuous,

d) not continuous

10. Take n

11xn ,0

n

11 but f(xn) = n f(xn) does not converge. So f

is not continuous at 1.

11. a) For a < 1, f(x) = x2 – 6x + 5. So f is continuous for a < 1. As f(x) = 2x2

– 2, f is continuous for a > 1.

.022xfIt,0561xfIt1x1x

Hence .0xfIt1x

Also f(d) = 1 – 6 + 5 = 0

b), c), d) Similar.

Terminal Questions

1. a) {4, 5, 6, 7, 8} b) {5, 6, 7, 8, 9} c) {5, 6, 7, 8}

d) {4, 5, 6, 7, 8, 9}

e) {4, 5, 6, …..} f) 10, 11, 12,…..} g) {9, 10, 11, …..}

h) {9, 10, 11, ….. }

2. a) 0 b) 1 c) 0 d) 0 e) 0

3. a) As n increases an increases. So kan cannot be made less than

a fixed number . So nn

aIt does not exists.



4. a) 3

4 b) 1 c) 3 d)

7

1

e) – 1 f) 1 g) 2

1 h) 3

i) Write an as 2

1

12

11It.

n

41

n

11

.

n

32

n

31

n

2

2

j) 01

0

01

00It.

n

21

n

1

n

1a

n

3

2

n

k) n1n

n1nn1nn1nan

n

1n1

1.

n

1

n

1n1n

1

n1n

1

5. 011

0aIt n

n

a) 2a3 b) a3 + a4 c) 2a3 – a4 + a5 d) (2 + 3a) (4 + 5a2)

e) 2a54

a32 f)

4

9a2

6. a) 2(–1)2 – 1 = 1 b) (2 + 1) (3 – 2) = 3 c) 6

11 d) 6

e) 5

1 f) 8

299

169 g) 1

h) .1x

2x

1x1x

2x1x

1x

2xx2

2

Hence answer is .2

3



7. a) Im b) (l + 2m) (2I – m) c) m3I2 d) 2m

mI

10. a) 8xfIt2x

2x2x2

2x

8x2

2x

2

f(2) = 7. So f is not

continuous at x=2.

b) .2x1Itx1

x1ItxfIt

1x

2

1x1x But f(1) = 1

c) Let limit at 1 = 8; right limit at 1 = 2

d) The left and right limits are 1 and –1 respectively.

e) The left and right limits are –1 and 1.



Unit 6 Differentiation

Structure

6.1 Introduction

Objectives

6.2 Differentiation of Powers of x

6.3 Differentiation of ex and log x

6.4 Differentiation of Trigonometric Functions

6.5 Rules for Finding Derivatives

6.6 Different types of Differentiation

6.7 Logarithmic Differentiation

6.8 Differentiation by Substitution

6.9 Differentiation of Implicit Functions

6.10 Differentiation from Parametric Equation

6.11 Differentiation from First Principles

6.12 Summary


6.14 Answers

6.1 Introductions

In the last chapter, we studied continuous functions. In this chapter we will

study the rate of change of a continuous function. Consider a car moving

from Manipal to Mangalore. Then the distance traveled by the car from

Manipal is a continuous function of time. (We assume that the car does not

stop midway.) If you need to know the rate at which the distance is changing

(increasing) then you need the concept of derivatives. Consider the height of

a growing plant. It is a continuous function of time. The rate of growth of the

plant can be studied using derivatives.



Objectives:


differentiate the given functions using suitable rules

find the derivative from first principles

6.2 Differentiation of Power of x

Let f be a function defined in an open interval containing c and continuous in

that interval.

Definition:

a) The derivative or differential coefficient of f at c is defined as the limit

h

cfhcfIt

0h

If it exists as a real number. It is denoted by .cf

b) If cf exists then we say that the function f is differentiable at c.

Note: The process of finding a derivative is called differentiation and the

subject dealing with differentiation is called differential calculus.

Example: Find whether 1f exists when f(x) = 4x – 3

Solution:

h

1fh1fIt1f

0h

h

343h14It

0h

44Ith

4hIt

0h0h

Hence 1f = 4

SAQ 1: What is 2f for the function f defined in above example.



REMARK We have defined cf at c. When we evaluate cf for various

values of c, we get a function xf . It is defined by

h

xfhxfItxf

0h …………………. (6.1)

Sometimes we defined xf in a different way. Take x + h as t. Then h = (x

+ h) – x = t – x. As x.tor0xt0,h

Hence xt

xftfItxf

xt ……………… (6.2)

We will be using (6.1) or (6.2) as per our convenience.

Now we find xf for some standard functions.

Example: If f(x) = k find xf .

Solution:

xt

xftfItxf

xt

xt

kkIt

xt

xt

0It

xt

0Itxt

= 0

Hence 0xf

Note: 1n23n2n1nnn x.........xtxttxtxt

This can be verified by expanding the R.H.S. (Right hand side). This is used

in next example.

Example: If f(x) = xn when n is a positive integer then 1nnxxf .



Solution:

xt

xftfItxf

xt

xt

xtIt

nn

xt

xt

x....xttxtIt

1n2n1n

xt

1n23n2n1n

xtx....xtttIt

1n23n2n1n x.....x.xx.tt

Since 2nk1forxt kk

1nnx

Hence 1nnxxf

Example: If xxf when x > 0 find xf .

Solution:

h

xfhxfItxf

0h

h

xhxIt

0h

By this time you will have noticed that finding a derivative always involves

taking the limit of a quotient where both numerator and denominator are

approaching zero. Our task is to simplify this quotient so that we cab cancel

a factor h from numerator and denominator, there by allowing us to evaluate

the limit by substitution. In the present example, this can be accomplished

by rationalizing the numerator.



xhx

xhx.

h

xhxItxf

0h

xhxh

xhxIt

0h

xhxh

hIt

0h

xhxh

1It

0h

x2

1

xx

1

Since xhx

Thus, ,f the derivative of f, is given by x2

1xf for x > 0.

SAQ 2: Find cf when

a) f(x) = 4, c = 2

b) f(x) = 4, c = 4

c) f(x) = x3 c = 1

d) f(x) = x c = 1

e) f(x) = x, c = 1

f) f(x) = x, c = 0

6.3 Differentiation of ex and log x

e denote a constant called exponential limit. It is defined as follows:

x

1

0x

x

xx1Itor

x

11Ite ………….. (6.3)



Its value is 2.7182818285…….

ex is called the exponential function. We have another function log x which is

the ‘inverse’ of ex. By inverse of a function we mean the following:

If we apply a function and then its inverse to a real value x or the inverse of

a function and the function to x, then we get back x. That is

xexelog xlogx

If you are familiar with logarithms you can easily understand that log x = loge

x and log e = 1.

We prove the following limit.

Example: Show that alogh

1aIt e

h

0h………….. (6.4)

Solution: If we put 0has,y1ah so does y.

Also alog

y1logelogy1logh

e

eae

alogy1log

yIt

h

1aIt e

e0y

h

0h

y1

e

e

0yy1log

alogIt

Since y1

e y1log

y1logy

1e

But as ey1,0y y1

alogh

1aIt e

h

0h



If we apply 6.4 with e in place of a we get

1elogh

1eIt e

h

0h

1h

1eIt

h

0h ……………… (6.5)

Example: If xexf show that xexf

Solution: h

eeItxf

xhx

0h

h

1eeIt

hx

0h

h

1eIte

h

0h

x

1.ex

xe

xexf

Example: If f(x) = log x x > 0, show that x

1xf

Solution:

h

xloghxlogItxf

0h

h

x

hxlog

It0h

b

alogblogalogSince

x

h1log

h

x.

x

1It

0h

hx

0h x

h1log

x

1It since log xm = m log x



hx

h

x x

h1logIt

x

1 since

h

xwhen h → 0

hx

h

x x

h1Itlog

x

1 ( log x is continuous)

elogx

1 Using (6.3)

x

1

x

1xf

SAQ 3: Find cf when

a) f(x) = ex, c = 0

b) f(x) = ex, c = 1

c) f(x) = ex c = log 2

d) f(x) = ex c = log e

e) f(x) = log x c = 1

f) f(x) = log x c = e

g) f(x) = logx c = log 2

h) f(x) = log x c = log log 2

6.4 Differentiation of Trigonometric Functions

Trigonometric functions from another class of functions which are frequently

used in applications. For answering questions regarding rotating wheels,

weather cycles and periodic phenomenon, we require trigonometric function.

For those unfamiliar with trigonometric functions, we introduce the basic

trigonometric functions sin x and cos x (called sine and cosine functions).

These are defined in fig. 6.1



Figure 6.1. sine and cosine functions

The solid curve is the graph of y = sin x; dotted curve that of y = cos x from

Fig. 6.1, it is clear that sine and cosine functions are continuous on R.

Four more functions are defined as follows:

xcos

1xsec

xsin

1ecxcos

xsin

xcosxcos

xcos

xsinxtan

………….. (6.6)

Two important trigonometric limits:

1x

xsinIt

0x 0

x

xcos1It

0x………. (6.7)

Example: If f(x) = sin x, then cosxxf

Solution:

h

xsinhxsinItxf

0h

h

xsinsinhcoshxsinIt

0h

h

sinhxcos

h

cosh1xsinIt

0h

x – Measured in radius



h

sinhItxcos

h

cosh1Itxsin

0h0h

But 1h

sinhIt

0h and 0

h

cosh1It

0h from 6.7

Thus 1.xcos0.xsinxf

= cos x

Example: If f(x) = cos x, show that .xsinxf

Solution: h

xcoshxcosItxf

0h

h

xcossinhxsincoshxcosIt

0h

h

sinhxsin

h

cosh1xcosIt

0h

1.xsin0.xcos

xsin

SAQ 4: Find the value of xf at c when

a) f(x) = sinx, 4

πc

b) f(x) = sin x, πc

c) f(x) = sin x, 2

3πc

d) f(x) = cos x, 2π,2

3ππ,,

2

π0,c

SAQ 5: If f(x) = sin x, find cf when 2

π,

4

π0,c

SAQ 6: If f(x) = cos x, find cf when .π2ππ,2

3πc



6.5 Rules for Finding Derivatives

So far you have seen the derivatives of elementary functions like xn, ex, log

x, sin x and cos x. But more functions can be formed by

Adding two function

Subtracting one function from another function

Multiplying two functions

Taking quotient of two function

Applying one function after another function (composition of two functions)

Taking the inverse of a function

In this section we discuss the rules for differentiating functions obtained by

applying the operations mentioned above.

Before deriving rules for differentiation, let us look at the relation between

continuous functions and functions having derivatives.

Theorem: If f has a derivative at c, then f is continuous at c.

Proof: We need to show that cfxfItcx

cx,cx.cx

cfxfcfxf

Therefore, cx.cx

cfxfcfItxfIt

cxcx

cxIt.cx

cfxfItcfIt

cxcxcx

.0cfcf

= f(c)

Note: The converse of above theorem is not true. Consider the function

xxf (Refer fig 6.2) So f(x) = –x if x < 0

= x if x > 0.



Figure 6.2.The graph of y = | x |

11Ith

0hIt

h

0fx0fIt

0h0h0h

11Ith

0hIt

h

0fh0fIt

0h0h0h

h

0fh0fIt

0h does not exist

Theorem: (Constant function rule) If f(x) = k, being a constant, then

0xf

Proved in above Example

Theorem: (Identify function rule) If f(x) = x then 1xf

Proved in above Example.

Theorem: (Constant multiple rule) If g(x) = k f(x), then xfkxg



Proof:

h

xkfhxkfIt

h

xghxgItxg

0h0h

h

xfhxfItk.

h

xfhxfk.It

0h0h

xf.k

Theorem: (Sum and difference rule)

xgxfxgf

xgxfxgf

Proof: h

g(x)f(x)h)g(xhxfItit(x)g)'f

0h

h

xghxg

h

xfhxfIt

0h

h

xghxgIt

h

xfhxfIt

0h0h

xgxf

h

g(x)f(x)h)g(xhxfItit(x)g)'f

0h

h

xghxg

h

xfhxfIt

0h

h

xghxgIt

h

xfhxfIt

0h0h

xgxf

Theorem: (Product rule)

xfxgxgxfxf.g



Proof:

Let F(x) = f(x) g(x) Then

h

xFhxFItxF

0h

h

xgxfhxghxfIt

0h

h

xgxfxghxfxghxfhxghxfIt

0h

h

xfhxf.xg

h

xghxg.hxfIt

0h

h

xfhxfIt.xg

h

xghxgIt.hxfIt

0h0h0h

xfxgxgxf

Theorem: (Quotient rule)

.0xgif,xg

xgxfxfxgx

g

f2

Proof:

Let Then.xg

xfxF

h

xg

xf

hxg

hxf

ItxF0h

hxgxg

1.

h

hxgxfhxfxg

0hIt

hxgxg

1.

h

xghxg.xf

h

xfhxfxg

0hIt

xgxg

1.xgxfxfxg (By above theorem)



Definition: If f and g are two functions of a real variable then

.xfgxfg

Note: In g of the function f appearing rightmost is applied first and then g is

applied to the functional value obtained after applying f.

For example, sin 2x is obtained by applying the multiple function 2x to x;

then sine function is applied to 2x.

Note: sin .xsin2x2 So the order of applying functions is important in g of

If y = f(x), then the mathematician Gottfried Wilhelm Van Leibniz used the

symbol .xffordx

dy

Issac Newton and Leibniz are the two mathematician who developed

calculus). He denoted h by ,x which is change in the value of x and the

difference ,ybyxfxxf the corresponding change in the value of y.

Then

x

yIt

dx

dy

0x …………. (6.8)

Theorem: (Chain rule) If f(u) and u = g(x), then dx

du.

du

dy

dx

dy

Proof:

)x(g)xx(gu

)u(f)uu(fy

As g is continuous, 0x implied 0u

x

ytI

dx

dy

0x

x

u.

u

yIt

0x

x

uIt.

u

yIt

0u0u



As 0x we have .0u So we can replace 0x by 0u

= dx

du.

dy

dy

Example: If y = tan x, evaluate dx

dy

Solution:

2xcos

xcosdx

dxsinxsin

dx

dxcos

xcos

xsin

dx

dxtan

dx

d

xcos

xsinxsinxcos.xcos2

xsecxcos

1

xcos

xsinxcos 2

22

22

Example: If y = sec x, evaluate dx

dy

Solution:

2xcos

xcosdx

d11

dx

dxcos

xcos

1

dx

dxsec

dx

d

xcos

xsin.

xcos

1

xcos.xcos

xsin

xcos

xsin10xcos2

xtanxsec

SAQ 7: If y = cot x, show that xeccosdx

dy 2

SAQ 8: If y = cos ecx, show that .xcotecxcosdx

dy

Example: If y = xm show that 1mmxdx

dy

Where m is a negative integer.



Solution: Write m = –n where n is a positive integer. Then nx

1y

nx

1

dx

d

dx

dy

n2

nn

x

xdx

d11

dx

dx

(by applying Quotient rule with u = 1 v = xn)

n2

1nn

x

nx0.x

1nnx

1mmx

Example: If n,m,xy nm

being integers, n > 0 find .dx

dy

Solution: As .xxy,n,m,xy m

n

nm

nnm

By chain rule, .mxdx

dyny 1m1n

So

1n

nm

1m

1n

1m

x

x

n

m

y

x

n

m

dx

dy

n

n1nm1m

n

1nm

1m

xn

m

x

x

n

m

1

n

m

n

mm1m

n

11m1m

xn

mx

n

mx

n

m



Derivatives of Inverse Function

If y = f(x) and x can be written as yx such that

.yyfandxxf Then is called the inverse of f.

We have already seen that log x and ex are inverse functions of each other.

By Chain rule .1xdx

dxf

dx

d So .1

dx

dy

dy

dx Hence

dy

dx

1

dx

dy …………………. (6.9)

Example: If sin–1 x denotes the inverse of the function

2

1

x1

1xsin

dx

d,xsiny

Solution: Let y = sin–1 x. By definition of inverse function, x = sin y. By (6.9),

22 x1

1

ysin1

1

ycos

1

dy

dx

1

dx

dy

Now we are ready to differentiate functions which appear in several

applications.

Table 6.1 gives the differentiate coefficient of standard functions. Using

table 6.1, product rule, quotient rule and chain rule we can differentiate

many functions.



f(x) xf

1. k 0

2. xn nxn–1

3. x

1

2x

1

4. x x2

1

5. ax 0a,,aloga ex

6. ex ex

7. log x 0x,x

1

8. sin x cos x

9. cos x –1 sin x

10. tan x sec2x

11. cot x –cosec2x

12. sec x sec x tan x

13. cosec x –cosec x cot x

14. xsin 1 1x1,x1

1

2

15. cos–1 x 1x1,x1

1

2

16. tan–1 x 2x1

1

17. cot–1 x 2x1

1

18. sec–1 x 1x,1xx

1

2

19. cosec–1 x 1x,1xx

1

2

Table 6.1: Table of derivatives



Worked Examples:

W.E.: If dx

dyfind

2

xx

9

xx97y

2

Solution:

2

xx

9

xx97

dx

d

dx

dy 2

23

xdx

d

2

1

9

x2190

1

23

x2

3

2

1

9

x29

21

x4

3

9

x29

4

x3

9

x29

W.E.: Find the derivative of ex tan x

Solution:

xtandx

dextan.e

dx

dxtane

dx

d xxx

xsecextan.e 2xx

xsecxtane 2x

W.E.: If xcosxsin

xcosxsiny find

dx

dy



Solution:

xcosxsin

xcosxsin

dx

d

dx

dy

xcosxsin

xcosxsindx

dxcosxsinxcosxsin

dx

dxcosxsin

2

xcosxsin

xsinxcosxcosxsinxsinxcosxcosxsin

2

22

xcosxsin

xcosxsin2xsinxcosxsinxcosxcosxsin

xcosxsin

cosxsin2xsinxcosxcosxsin2xsinxcos 2222

1xcosxsincesincosxsin

11 22

2

2

xcosxsin

2

W.E.: Find the derivative of x2tanx5

x3tane2

x2

x2tanx5

x3tane

dx

d2

x2

22

2x2x22

x2tanx5

x2tanx5dx

dx3tanex3tane

dx

dx2tanx5

x2tanx5

2.x2secx2.5x3tanex3tan2.e3.x3sec.ex2tanx52

2x2x22x22

22

2x222x2

x2tanx5

x2sec2x10x3tanex3tan2x3sec2x2tanx5e



22

22222x2

x2tanx5

x2secx3tan2x3tanx10x3tanx2tan2x3tanx10x3secx2tan3x3secx15e

22

2222x2

x2tanx5

x3tanx2tan2x3tanx10x10x3secx2tan23x3secx15e

22

222x2

x2tanx5

x3tanx2tan2x3tan1xx10x3secx2tanx3secx15eW.

W.E.: Find the differential coefficient of x.t.r.waxxlog 22

Solution:

22 axxlogdx

d

x2.ax2

11

axx

1

2222

22

22

22 ax

xax

axx

1

22 ax

1

W.E: Evaluate dx

dy when

xx1

xx1logy

2

2

Solution:

xx1logxx1logdx

d 22

(Since BlogAlogB

Alog )



xx1dx

d

xx1

1xx1

dx

d.

xx1

1 2

2

2

2

1x2.x12

1

xx1

11x2.

x12

1

xx1

1

2222

22

2

2

2

2 x1xx1

x1x

x1

x1x

xx1

1

22 x1

1

x1

1

2x1

2

W.E.: Find the derivative of sin (log x) + log sin x

Solution: xsinlogxlogsindx

d

xsindx

d

xsin

1xlog

dx

d.xlogcos

xcos.sin

1

x

1.xlogcos

xcotxlogcosx

1

W.E.: Find the differential coefficient of x.t.r.we32sin x21



Solution:

x21 e32sindx

d

x2

2x2

e32dx

d.

e321

1

2.e.30.

e321

1 x2

2x2

2x2

x2

e321

e6

W.E.9: Find dx

dy when xx1tany 21

Solution: xx1tandx

d

dx

dy 21

xx1dx

d.

xx11

1 2

22

1x2.x12

1.

x1x2xx11

1

2222

2

2

22 x1

x1x

x1x2x22

1

2

2

22 x1

x1x

xx1.x12

1



/

/

xx1x12

xx1.1

22

2

2x12

1

W.E.: Evaluate the derivative of 1x2

1sec

2

1

Solution: If y = sec–1 x, x = sec y So x

1ycos implying .

x

1cosy 1

1x2cosdx

d

1x2

1sec

dx

d 21

2

1

1x2dx

d.

1x21

1 2

22

0x4.x41x41

1

24

42 x4x4

x4

22 x1x4

x4

2x1x2

x4

2x1

2

SAQ 9: If x2x1x3y 22 find dx

dy



SAQ 10: Find the differential coefficient of 4

5

x

3x4xsin2

SAQ 11: Differentiate .x.t.r.wxsinxex

SAQ 12: If dx

dyfindxsec2xlog5siny 10

SAQ 13: Find the derivative of (3 sec x – 4 cos ecx) (2 sin x + 5 cos x)

SAQ 14: Differentiate x.t.r.wxcosxsin

xcosxsin

SAQ 15: If dx

dyfind

baxcos

ey

bx

SAQ 16: Find the differential coefficient of 3x

1x21x

SAQ 17: Find the derivative of 1x

1x2

2

SAQ 18: If dx

dyfindx2xsiny 21

SAQ 19: Differentiate x.t.r.wx1

x1cos 1

SAQ 20: If dx

dyfindetany x1

SAQ 21: Find the derivative of 2x1 ecot

SAQ 22: Find xlogtanyifdx

dy 1

SAQ 23: If xtancotxcottany 11



SAQ 24: Find cbxaxyifdx

dy 2

SAQ 25: Find the differential coefficient of x4cosx3sine x2

SAQ 26: Differentiate x.t.r.wesinlog x1

SAQ 27: Prove that

a) 1x

11xxlog

dx

d

2

2

b) 1x

11xxlog

dx

d

2

2

6.6 Different types of Differentiation

In the earlier sections we had formulas for differentiating a given function of

x i.e. y = f(x). But direct application of formulas may require quite a good

amount of computational effort. Also we may not be able to express y as a

function of x. We adapt the following methods to tackle these problems.

1. Logarithmic differentiation

2. Differentiation by substitution

3. Differentiation of implicit functions

4. Differentiation from parametric equations

6.7 Logarithmic Differentiation

When y is a product or quotient of functions or of the form xg

xf we can

take logarithms and get a sum of logarithm of functions or product of two

functions. So differentiation becomes easier.

W.E.: Differentiate x.t.r.wx xsin



Solution: Put xsinxy

AlogBAlogxlogxsinxlogylog Bxsin

xlogxsindx

dylog

dx

d

xsindx

dxlogxlog

dx

dxsin

dx

dy

y

1 (using product rule)

xcos.xlogx

1.xsin

dx

dy

y

1

So xcos.xlogxsinx

1y

dx

dy

Thus xcos.xlogxsinx

1xx

dx

d xsinxsin

W.E.: If dx

dyfind

7x4x

2x2xy

2

Solution:

7x4x

2x2xlogylog

2

7x44log2x2xlog 2

BlogAlogB

Alog

7xlog4xlog2xlog2xlog 2

7xlog4xlog2xlog2xlog 21

2

7xlog4xlog2

12xlog2xlogylog 2

Differentiating w.r.t. x,



7xlog4xlog2

12xlog2xlog

dx

dylog

dx

d 2

7xdx

d

7x

14x

dx

d.

4x2

12x

dx

d

2x

12x

dx

d.

2x

1

dx

dy

y

1 2

2

1.7x

1

4x2

11.

2x

1x2.

2x

1

dx

dy

y

12

1.7x

1

4x2

11.

2x

1

2x

x2y

dx

dy2

W.E.: If ,axy nx show that xlog1nydx

dy

Solution: nxaxlogylog

So log y = log a + logxnx = loga + nx log x

xlognxdx

dalog

dx

d

dx

dy

y

1

xlogxdx

dn0

dx

dy

y

1

xlog.1x

1.xn

dx

dy

y

1

.xlog1nydx

dy

6.8 Differentiation by Substitution

Sometimes a function becomes simpler if we introduce a new variable which

is a function of x. Mostly we will be using trigonometric functions for

substitution. It is better to remember the following formulas.

2cos1sin

7x

1

4x2

1

2x

1

2x

x2y

dx

dy2



22 tan1sec

2tan1

tan22tan

cossin2tan1

tan22sin

2

2222

2sin211cos2sincos

tan1

tan22cos

The following table gives the appropriate substitution

For ,xa 22 let x = sin or a cos

For ,xa 22 let x = a tan

For tanxlet,x1

x2or

x1

x222

For tanxlet,x1

x12

2

W.E.: If 2

1

x1

x2siny show that

2x1

2

dx

dy

Solution: Put x = tan . Then xtan 1

2sintan1

tan2

x1

x222

xtan222sinsinx1

x2siny 11

2

1

xtan2y 1

2

11

x1

1.2xtan

dx

d2xtan2

dx

d

dx

dy

2x1

2

dx

dy



W.E.: Differentiate x.t.r.wa

xsin

2

a

2

xaxy 1

222

Solution: Put x = a sin

a

sinasin

2

a

2

sinaasinay 1

222

a

sinasin

2

a

2

sinaasin

2

a 122222

.2

acos.sin

2

a 22

dx

d.

2

a

dx

d.2cos2.

4

a

dx

dy 22

dx

d

2

a

dx

d2cos

2

a 22

2cos1dx

d

2

a2

22

cos2.dx

d.

2

a

dx

dy since cos 2 = 2 cos2 – 1

dx

d.cosa

dx

dy 22

x = a sin

Differentiating w.r.t. , we get cosad

dx

22 xa

1

cosa

1

dx

d0

22

2

2

22

xa

1.

a

x1a

dx

dy

222

22

xa

1.

a

x1a



22

22

xa

1.xa

22 xadx

dy

6.9 Differentiation of Implicit Functions

Sometimes y may not be given as a function of x but y and x may be related

by some relations. In such cases we differentiate the given relation w.r.t. x

and then evaluate dx

dy from the resulting equation.

W.E.: Find ifdx

dy

a) 0cfy2gx2yx 22

b) 32 a2x4ay27

Solution:

a) Differentiating the given relation w.r.t.x

0dx

dcfy2gx2yx

dx

d 22

00dx

dy.f21.g2

dx

dyy2x2

0dx

dyfygx2

gxdx

dyfy

Hence fy

gx

dx

dy

b) Differentiating 32 a2x4ay27 w.r.t. x,

32 a2x4

dx

day27

dx

d



32 a2x

dx

d4y

dx

da27

a2xdx

da2x3.4

dx

dyy2.a27

2

1.a2x12dx

dyay54

2

ay45

a2x21

dx

dy

9

22

ay9

a2x2

dx

dy2

W.E.: If ,yxyxnmnm Prove that

x

y

dx

dy

Solution: Taking logarithms on both sides, we get

nmnm yxlogyxlog

yxlognmylogxlog nm

yxlognmylognxlogm

Differentiating this w.r.t. x,

dx

dy1

yx

1.nm

dx

dy

y

1.n

x

1.m

dx

dy

y

n

dx

dy

yx

nm

yx

nm

x

m

dx

dy

yyx

yxnnmy

yxx

nmxyxm

dx

dy

yxy

ynnxynmy

yxx

nxxmmyxm

dx

dy.

yxy

nxmy

yxx

nxmy.,e.i



On canceling yx

nxmy we get

x

y

dx

dy

6.10 Differentiation from Parametric Equation

Sometimes x and y may be functions of another variable t we may have to

find .dx

dy In this section we give a method for finding

dx

dy in such cases.

tf

tg

dx

dythen,tgyandtfIfx …………. (6.10)

W.E.: Find dx

dy when tsinay,tcosaZ 33

Solution:

tcosdt

d.tcos3.atcosa

dt

d

dt

dx 23

tsintcosa3tsin.tcosa3dt

dx 22

tcostsina3tsindt

dtsin3.atsina

dt

d

dt

dy 223

ttantcos

tsin

tsintcosa3

tcostsina3

dt

dxdt

dy

dx

dy2

2

W.E.: Find dx

dy when sinay,

2tanlogcosax

Solution: 2

1.

2sec

2tan

1sina

d

dx 2



2

1.

2cos

1.

2sin

2cos

sina2

2cos

2sin2

1sina

sin

1sina

sin

1sina

2

sin

1cosa

2

cosad

dy

tancos

sin

sin

cosa

cosa

d

dxd

dy

dx

dy2

W.E.: If dx

dyfindcos1ayandsinax

Solution:

cos1ad

dx

sinasin0ad

dy

2sin2

2cos

2sin2

cos1

sin

cos1a

sina

dx

dy

2



2

cot

2sin

2cos

SAQ 28: Differentiate xx w.r.t. x

SAQ 29: Differentiate xlog1 xtan

SAQ 30: Differentiate x.t.r.w1x3x

2xx1 2

SAQ 31: Differentiate the following functions w.r.t. x

a) 2

1

x1

x2tan

b) 2

21

x1

x1cos

c) xcos1

xcos1tan 1

SAQ 32: Find ifdx

dy

a) 64yxy8x 33

b) xy = tan xy

SAQ 33: If ,ex yxy show that 2

xlog1

xlog

dx

dy

SAQ 34: Find ifdx

dy

a) at2y,atx 2

b) t

cy,ctx



6.11 Differentiation from first Principles

If we evaluate dx

dy by using only the definition and not the known formulas,

then we call this differentiation from first principles. We illustrate this method

with a few examples.

W.E. : Differentiate tan x from first principles

Solution: Let y = tan x

Let x be a small increment in x and y , the corresponding increment in y.

Then xxtanyy

xtanxxtany

xcos

xsin

xxcos

xxsin

xcosxxcos

xsinxxcosxcosxxsin

xcosxxcos

xxxsin

xcosxxcos

xsin

xcosxxcos

1.

x

xsin.

x

y

By definition, x

yIt

dx

dy

0x

xcos.xxcos

1It

x

xsinIt

0x0x

1sin

Itxcos.xcos

1.1

0

xsecxcos

1 2

2



W.E.: Differentiate sec x from first principles

Solution: Let y = sec x

Let x be a small increment in x and y be the small increment in y.

Then xxsecyy

xsecxxsecy

xcos

1

xxcos

1

xcos.xxcos

xxcosxcos

xcosxxcos

2

xxxsin

2

xxxsin2

xcos.xxcos

2

xsin

2

xxsin2

x.xcosxxcos

2

xsin

2

xxsin2

x

y

By definition x

yIt

dx

dy

0x

2

xcosxxcos

2

xsin

2

xxsin

It0x

,0xAs

2

x2

x

Itxcosxxcos

2

xxsin

It

0also2

x02

x0x

xcos

xsin.

xcos

11

sinIt1

xcos.xcos

xsin

0

= sec x tan x



W.E.: Differentiate log (2x + 3) from this principles

Solution: Let y = log (2x + 3)

Let x be a small increment in x and y the corresponding increment in y.

Then 3xx2logyy

3x2log3x2x2logy

3x2

x23x2logdy

3x2

x21log

2

3x2.

3x2

x2

3x2

x21log

Itx

yIt

dx

dy

0x0x

3x2

x21log

3x2

x2

1It

3x2

2

0x

3x2

x2

1

03x2

x2 3x2

x21Itlog

3x2

2

eh1Itelog3x2

2h1

0x

3x2

2

W.E.: Differentiate 3x

2 from first principles

Solution: Let 3x

2y

Let be a small increment in x and y the corresponding increment in y.



Then 3xx

2yy

3x

2

3xx

2y

3x3xx

3xx3x2

3x3xx

x2

3x3xx

x2It

x

yIt

dx

dy

0x0x

20x 3x

1.2

3x3xx

1It2

2

3x

2

SAQ 35: Differentiate cot x and cosecx from first principles

SAQ 36: Differentiate x2 + 2x + 3 from first principles

SAQ 37: Differentiate x

1x from first principles

6.12 Summary

In this unit we studied how to differentiate different types of function. The

rules used to differentiate these functions are illustrated with standard

examples. Different types differentiation are illustrated step by step with

clear cut examples. Lastly to find the derivative of a function by using the

definition is discussed.




1. Differentiate the following functions w.r.t. x.

a) 5x4

a3 b)

x

1xx3 34 c) 257 x5x2x


a) 1x3x17x 32 b) xlogxsinex

c) xlogx8 d) 2x cos x = x2 sin x

3. Differentiate the following functions w.r.t.x.

a) 9x3x4

12

b) 1x2

1x3x2 2

c) xcos

xcosxsin d)

1x

xsinxcosx2

4. Find dx

dy if y is equal to

a) xxsin 2 b) 222

x32x3

c) 4x3

1x2

d) xsinxcos 22 e) 1x

2xcos

2

22

5. Differentiate the following w.r.t. x.

a) xetanlog b) 41 xseclog

c) 21 xtanxsin d) 31 x21x2sin

6. If dx

dyfindxsinxcosey 23ax

7. If dx

dyfind,xsinxy xxsin

8. If dx

dyfind,1x2xy

1x2

(Hint: 22 1x1x2x



9. If dx

dyfind,

x1x2

x32xy

3


a) 2

1

x1

xsin b)

2

21

x1

x1tan c)

x

1x1tan

21

(For a), c), x = tan ; for b) x2 = tan

Note:

.14

tan

tan4

tan1

tan4

tan

tan1

tan1

So

211 xtan44

tantany

11. Find dx

dy when

a) tan (x + y) + tan (x – y) = 1

b) x4 + y4 = 4a2 x3 y3

c) xy = yx

12. If sin y = x sin (a + y), Prove that asin

yasin

dx

dy 2

(Hint: Use sin A cos B – cos A sin B = sin (A – B))

13. Find dx

dy when

a) x = a sec , y = b tan

b) x = 2 sin t, y = cos 2t

c) x = a ( + sin ), y = a (1 – cos )

14. Differentiate the following function from first principles

a) sin 3x b) e5x + 3 c) x

1x



6.14 Answers


1. 42f

2. a), b), 0; c) 3, d) 11, e), f) 1

3. a) 1, b) e, c) 2, d) 3, d) 1, f) e

1 g) ,

2log

1 h)

2loglog

1

4. a) ,2

1 b) –1 c) 0 d) 0, 0,1,0,1,

2

1

5. 0,2

1,1

6. 1, 0, 0

7. xsin

xcosxcosxsinxsin

xsin

xcos

dx

dxcot

dx

d2

8. xcotecxcos1sin

xcos10xsin

xsin

1

dx

d2

9. 2x2x18x12 23

10. 5

4

x

12x20xcos2

11. xcosxxsinxxsinex

12. xtanxsec2x

log:Answer,

10log

xlogxlog 10

10

13. On simplifying xeccos20xsec6dx

dy.xcot207xtan6y 22

14. 2

22

xcosxsin

xcosxsinxsinxcos

dx

dy

15. baxcosbbaxsina.baxcos

e2

bx



16. 2

2

3x

2x12x2

17. 22 1x

x4

18. 22 x2x1

2x2

19. 2

2

x1

x11

x1

1x11x1

dx

dy

xx1

1 on simplifying

20. x2

x

e1

e

21. 2x2

2x

e1

xe2

22. 2

xlog1x

1

23. .xcottanxtan

1tanxtancot 111

Hence y = 2tan–1 (cot x) Answer: - 2

24. baxcbxaxn2n2

25. 2 sin 3x cos 4x = sin 7x – sin x

(by a Standard trigonometric formula) xsinx7sin2

ey

x2

xsin2x7sin2xcosx7cos72

e

dx

dy x2



26. x2x1

x

e1esin

e

27. a)1x

1

1xx1x

x1x

1x2

x2.11.

1xx

1

dx

dy

222

2

22

b) is similar

28. xx (1 + log x)

29. x

xtanlog

xtanx1

xlogxtan

1

12

xlog1

30. 3x

1

2x

x

1x2

1

1x3x

2xx1 2

31. a) 2x1

2

dx

dy,tanx

b) 2x1

2

dx

dy,tanx

c) Use 2

1

dx

dy.etc

2

xsin2xcos1 2

32. x

yis

y3x8

y8x32

2

33. Taking logarithms, ,yxxlogy So .xlog1

xy Use quotient

value.

34. a) t

1

b) t

1

35. xeccos 2

36. 2x + 2

37. 2x

11



Terminal Questions

1. a) 6x

a15 b)

2

23

x

1x3x12 c) 346 x10x10x7

2. a) 5x4 + 4x2 + 2x – 51 b) x

xsinxlogxcosxlogxsine x

c) x

xlog24 d) xsinx4xcosx2 2

3. a) 22 9x3x4

3x8 b)

2

2

1x2

5x4x4

c) sec2 x d) 22

3

1x

xsinx3xxcos2

4. a) (2x + 1 cos (x2 + x) b) 22 x9x49x32x32

c) 2

4x3

11x31x d) 222 xsinxsinx2xcosx2sin

e) 2x

2xsin

2x

2xcos

2x

x162

2

2

2

22

5. a) xx

x

ecosesin

e b)

412 xsec1xx

4

c) 2

11

x1

xsin2xtanxcosxtan d)

33

32

x21x21x22

x10x38

6. xcot2xtan3axsinxcose 23ax

7. xsinxlogxcotxxsinxlogxcosx

xsinx xxsin

8. 1x

1xlog

1x

1x21x2x

1x2



9. x2

1

x1

x

x322

3

x

3

x1x2

x32x3

10. a) 2x1

1 b)

4x1

x2 c)

2x12

1

11. a) yxsecyxsec

yxsecyxsec22

22

b) 322

322

xa3yy

xyx3x

12. xxlogyx

yylogxy

13. a) eccosa

b b) –2 sin t c)

2tan

14. a) 3 cos 3x b) 5e5x+3 c) 2x

11



Unit 7 Integrations

Structure

7.1 Introduction

Objectives

7.2 Integration of Standard Functions

7.3 Rules of Integration

7.4 More Formulas in Integration

7.5 Definite Integrals

7.6 Summary


7.8 Answers

7.1 Introduction

Most of the mathematical operations we come across occur in inverse pairs.

For example, addition and subtraction, multiplication and division, squaring

and taking square roots are such pairs. In this chapter we study integration

as the inverse operation of differentiation. Integrals also have independent

interpretation. It generalizes the process of summation. It can be used for

evaluating the area under the graph of a function.

Objectives:


integrate standard functions

apply the concept of definite integrals in the process of summation



7.2 Integration of standard function

Definition: If f(x) is a function of a real variable, then g(x) is the integral of

f(x) if .xfxgdx

d It is denoted by .dxxf

Remark: Integration is the operation of determining a function whose

derivative is the given function.

Note: Clearly

dxxfdx

dxf …… (7.1)

dxxfxf …… (7.2)

(7.1) and (7.2) can be stated as follows:

Differential coefficient of integral of f(x) = integral of differential coefficient of

f(x) = f(x)

Remark: If f(x) = x2 and g(x) = x2 + 2, then .x2xgxf That is, the

derivative of a function remains the same if a constant is added to it.

So while writing the integral of f(x), we need to add c, an arbitrary constant

(by arbitrary constant we mean any real value).



In table 7.1 we list the integrals of some of standard function.

ckxkdx

c2

xxdx

2

cx

dxx3

32

1nwhenc1n

xdxx

1nn

cxlogdxx

1

cedxe xx

cxcosxdxsin

cxsinxdxcos

cxtanxdxsec2

cxcotxdxeccos 2

cxsecxdxtanxsec

cxeccosxdxcotecxcos

cxtanx1

dx 1

2

cxsin

x1

dx 1

2

c1xxlog

1x

dx 2

2

cxsec1xx

dx 1

2

Table 7.1: Table of integrals



We get formulas in Table 7.1 by simply reading Table 2.1 backwards.

We frequently apply constant function rule (Rule 1) and sum and difference

rule (Rule 2) for integration which are similar to differentiation. However we

do not have “product rule and quotient rule” in integration. So problems in

integration are more difficult their problems in differentiation.

Rule 1: (Constant function rule)

dxxfcdxxcf where c is a constant

Rule 2: (Sum and difference rule)

dxxgdxxfdxxgxf

Example: Evaluate dxxf when f(x) equals

a) 4x b) 3

2

x

cbxax

c)

2

x

1x d)

x

1excos3xsin2 x

e) 1xx

5

x1

4

x1

3

222

Solution:

a) c3

xc

14

xdxx

3144

cx3

13

b) dxx

c

x

bx

x

axdx

x

cbxax333

2

3

2

dxx

1cdx

x

1bdx

x

1adxcxbxax

32

321 (by rules 1 and 2)



k13

cx

12

bxxlog

1312

kx2

c

x

bxlog

2

(Note: We use k for arbitrary constant since c appears in the function)

c) dxx

1.x2

x

1xdx

x

1x

2

22

dx2dxx

1dxx

2

2 (by rule 2)

dx2dxxdxx 22 (by rule 1)

cx21

x

3

x 13

cx2x

1

3

x3

d) dxx

1excos3xsin2 x

dxx

1dxexdxcos3xdxsin2

x (by rule 2)

dxx

1dxexdxcos3xdxsin2 x (by rule 1)

cxlogexsin3xcos2 x

cxlogexsin3xcos2x

e) dx1xx

5

x1

4

x1

3

222

dx

1xx

5dx

x1

4dx

x1

3

222 (by rule 2)



1xx

dx5

x1

dx4

x1

dx3

222

(by rule 1)

.cxsec5x1xlog4xsin3 121

S.A.Q.1: Integrate 2

221x

x1

1xsecxsinxxe

7.3 Rules of Integration

We have the following rules for integrating complicated function.

Rule 1: Constant function rule

Rule 2: Sum and difference rule

Rule 3: By substitution

Rule 4: Integration by parts

Sum and difference rules:

We have already given rules 1 and 2 and used it for integrating functions in

7.1. In trigonometry some products can be expressed as um or difference of

two simpler trigonometric functions. The following formulas are useful in

integration.

2 Sin A Cos B = Sin (A + B) + Sin (A – B)

2 Cos A Sin B = sin (A + B) – Sin (A – B)

2 Cos A Cos B = Cos (A + B) – Cos (A – B)

2 sin A Sin B = Cos (A – B) – Cos (A + B)

Worked Example: Evaluate dxx2cosx5sin



Solution:

dxx2cosx5sin22

1dxx2cosx5sin

dxx2x5sinx2x5sin2

1

dxx3sin2

1dxx7sin

2

1

c3

x3cos

2

1

7

x7cos

2

1

c6

x3cos

14

x7cos

Worked Example: Integrate Sin 10x Sin 2x w.r.t. x

Solution:

dxx2sinx10sin22

1dxx2x10sin

dxx2x10cosx2x10cos2

1

dxx12cosx8cos2

1

dxxdxx 12cos2

18cos

2

1

c12

x12sin.

2

1

8

x8sin.

2

1

c24

x12sin.

2

1

x

x8sindxx2sinx10sin

Worked Example: Evaluate xdxsin2

Solution:

dxx2cos2

1dx1

2

1dx

2

x2cos1xdxsin2

c4

x2sin

2

xc

2

x2sin.

2

1

2

x



Integration by substitution:

Now we state rule 3.

Rule 3 (Substitution rule) If x can be written as a function g(t) of t, then

dttgtgfdxxf

Proof: As .tgdt

dx,tgx

Hence dttgdx (Don’t think that this step is obtained by cross

multiplication. dx and dt are called differentials and dttgdx is the

relation connecting the differentials).

So dxxfdttgtgf .

Note If f(x) cannot be integrated using known formulas, we try to write f(x)dx

as g(t)dt. Part of f(x) is written as f(g(t)) and the remaining part together with

dx is written as .dttg

Example: Evaluate .1n,dxbaxn

Solution: Let t = ax + b. Then .adx

dt So dt = adx

dta

1dx

c1n

t

a

1dtt

a

1dt

a

1tdxbax

1nnnn

c1na

ba1n

Note: If cxgdxxf then

cbaxga

1dxbaxf …………………. (7.3)



So using table (7.1) we get a list of formulas for dxbaxsin etc.

For example:

cbaxsina

1dxbacos

cbaxtana

1dxbaxsec2

cea

1dxe baxbax

Example: Integrate the following w.r.t. x

a) n

x23 b) x23sin

c) x47sec2 d) 3

4x

e

Solution:

a) c1n.2

x23dxx23

1nn

c1n2

x231n

b) cx23cos2

1dxx23sin

c2

x23cos

c) cx47tan4

1dxx47sec2

c4

x47tan



d) C

3

1

edxe

3

4x

3

4x

ce3 3

4x

Working Example 4: Integrate the following w.r.t. x

a) axx b) bxa

x c) Sin3x d)

3

2

bxa

x e) Sin2 3x

Solution: Denote the required integrals by 1

a) 1dxaxx (say)

Put x + a = t2 so x = t2 – a, dx = 2t dt

dtatt2tdttatl 2422

c3

at

5

t2dttadtt2

3524

c3

axa

5

ax2

2

3

2

5

b) ldxbxa

x (say)

Put a + bx = t2 b

dtt2.1dx

b

atx

2

dtatb

2dt

b

t2.

t

1.

b

atl

2

2

2

dt1adttb

2 2

2



cat3

t

b

2 3

2

cbxaa3

bxa

b

22

12

3

2

c) dxxsinxsinxdxsin 23

dxxsinxcos1 2

Let t = cosx; dt = –sinx dx

dtt1l 2

dttdt1 2

c3

t_t

3

c3

xcosxcos

3

d) ldxbxa

x3

2

(say)

Put a + bx = t. So dtb

1dx

b

atx

dtt

at

b

1dt

b

1

t

b

1t

l3

2

33

2

dtt

at2at

b

13

22

3

dtta2dttadttb

1 2321

3



c1

at2

2

ta

2

tatlog

b

1 12222

3

cbxa

a2

)bxa(2

abxalog

b

12

2

3

e) dxx3.2cos12

1dxx3sin2

dxx6cosdx12

1

cx6sin6

1x

2

1

Your skill in integration lies in spotting a suitable variable t – g(x) so that

dxxgdt can be “located” in f(x) dx. We illustrate the method of

substitution by more examples, so that you can master this technique

thoroughly.

Hereafter I stand for the integral to be evaluated.

Example: Integrate the following functions w.r.t. x

a) n1n xsinx b) n

xlogx

1 c)

2

x1tan

x1

e

Solution:

a) If t = xn then dt = nxn – 1 dx and this appears in dxxsinx n1n . So let

t = xn, then dt = nxn-1 dx.

cn

ttcos

n

dttsindxxxsinI 1nn

Cn

xcos n



b) Idxxlogx

1n

(Say)

Put log x = t so dtdxx

1

dttdtt

1dx

x

1

xlog

1I n

nn

c1n

xlogc

1n

t1n1n

cxlog1n

11n

c) Idxx1

xe2

1tan

(Say)

Put tan–1 x = t So dtdxx1

12

dxx1

1xedx

x1

xeI

2

1tan

2

1tan

cxtanecedte 1tt

Working Example: Integrate the following w.r.t. x

a) 6

2

1 x

x b)

2xxe c) x

xsin

Solution:

a) Idxx1

x6

2

(Say)

Put x3 = t so 3x2 dx = dt; dt3

1dxx2



cttan3

1

t1

dt3

1

x1

dxxI 1

223

2

cxtan3

1 31

b) Idxxe2x (Say)

Put dt2

1xdx;dtxdx2;tx2

c1

e

2

1dte

2

1dt

2

1exdxeI

ttt2x

c2

e2

x

c) Idxx

xsin (Say)

Put dt2dxx

1;dtdx

x2

1;tx

tdtsin2dt2tsindxx

1xsinI

cxcos2ctcos1

Working Example: Evaluate

a) dtan b) dsec c) dxxsinx

xcos12

d) dxxax 332 e) dcot f) deccos



Solution:

a) dcos

sindtan

Let y = cos ; dy = – sin d

cylogy

dy

y

dydtan

cseclogccoslogccoslog1

b) dtansec

tansecsecdsec

Let y = sec + tan ; dy = (sec tan + sec2 ) d

dtansecsecdy

cylogy

dydsec

dtanseclog

c) Idxxsinx

xcos1 (Say)

Put y = x + sinx; dy = (1 + cos x) dx

c1

ydyy

y

dy

xsinx

dxxcos1I

12

22

xsinx

1

d) Idxxax 332 (Say)

Put ;tdt3

2dxx;tdt2dxx3;txa 22233

Also txa 2

133

dt3

2tdt

3

2tdxx.xaI 2t233



ct9

2c

3

t

3

2 33

cxa9

2 2

3

33

e) I = log sin + c (check it your self)

f) I = – log (cosec + cot ) + c (Check)

Note: Remember the integrals of tan , sec , cot and cosec .

Working Example: Integrate the following w.r.t. x

a) xx e1e1

1 b)

xsin5xcos4

xsin3xcos2

Solution:

a)

x

xxx

e

11e1

dx

e1e1

dx

I

e1

dxe

e

1ee1

dx2x

x

x

xx

Let dxedy;e1y xx

c1

ydyy

y

dyI

12

2

ce1

1x

b) 1dxxsin5xcos4

xsin3xcos2

xcos5xsin4xsin5xcos4dx

d

Let numerator = I (denominator) + dx

dm (denominator)



Put xcos5xsin4mxsin5xcos4Ixsin3xcos2

We have 41 + 5m = 2 and 51 – 4m = 3

20 + 25m = 10

20 – 16 m = 12

41

2m,2m41

41

92

41

1082

41

102

41

252m5241

441

2329

I

41

2mand

41

23I

xcos5xcos4

dxxsin5xcos4dx

d

41

2xcos5xcos4

41

23

I

dxxsin5xcos4

xsin5xcos4d

41

2dx

41

23

cxsin5xcos4log41

2

41

23

S.A.Q.2: Integrate the following functions w.r.t. x

a) 3x4cot3x4eccos b) 5

3x2

1 c)

2x3

1

d) x117eccos 2 e) xe 45

S.A.Q.3: Evaluate the following integrals

a) x43cos

dxx43tan b) dx

xcos1

xcos1

c) dxx2sin1 d) dx3 x



(Hint: b) etc2

xsin2xcos1 2 c) xcosxsin2x2sin

d) 3logxx3log ex3

S.A.Q. 4: Integrate the following functions w.r.t. x

a) x

log b) xx esine c) 11xx 43 d)

x

e x

2

tan

cos

S.A.Q.5: Integrate the following functions w.r.t. x

a) x

x

xe

xe

2cos

1 b)

4

3

1

24

x

xx c)

xxx log

1 d) xsecxtan3

(Hint: b) dxx1

x2dx

x1

x4Evaluate

44

3

c) xlog1t d) )xsect


a) xsin 1 b) xxe c) )1x(logx d) xtanx 1


a) 2x

xlog b)

2

x

x1

xe c)

2

22

x

axlog d) xlogx3

Integration by parts

By now you might have noticed that there are not many general rules for

integration as in the case of differentiation. Even simple functions like logx

cannot be integrated with the rules we have come across so far.

In this section we give a method of integration called “Integration by parts”.

Using this method we can integrate many functions.



Rule 4: If u and v are functions of x,

vduuvudv ….. (7.4)

This follows by integrating the product rule

dx

duv

dx

dvuuv

dv

d

Integrating both sides we get

dxdx

duvdx

dx

dvuuv

vduudv

(7.4) Follows from the above identity while applying integration by parts, we

“locate” dv in f(x) dx. That is we integrate that part of f(x) and term it as v.

The remaining part is taken as u.

Working Example: Evaluate dxxlog

Solution: Let u = log x dv = dx

Then x

dxduor

x

1

dx

du

V = x.

duvuvdxxlogdvu

x

dxxxlogx

cxxlogx

Working Example: Evaluate dxxa 22



Solution: Let 22 xau and dv = dx

v = x

duvuvdxxa

dvu

22

dxxa2

x2.xxax

22

22

dxxa

xxax

22

222

dxxa

axaxax

22

22222

22

22222

xa

dxadxxaxax

222222222 xaxlogadxxaxaxdxxa

2222222 xaxlogaxaxdxxa2

cxaxlog2

axa

2

xdxxa 22

22222


a) dxxsinx 12 b) dxxsinx 2

Solution

a) dxxsinxI 12

Let 3

xvdxxdvxsinu

321

dxx1

1.

3

xxsin

3

xvduuvudvI

2

31

3



xdxx1

x

3

1xsin

3

xI

2

21

3

Put 1 – x2 = t2 in the second term; x2 = 1 – t2; – 2x dx = –2tdt; xdxv= –tdt

dtt1t

tdtt1xdx

x1

x 22

2

2

c3

ttdttdt1

32

c3x13

x1c3t

3

t 22

12

2

Hence c2x3

x1

3

1xsin

3

xI 22

12

13

cx2x19

1xsin

3

x 2213

b) xdxsinxI 2

Let u = x dv = sin2 x dx

Then du = dx

2

x2sinx

2

1dxxcos1

2

1xdxsinv 2

4

x2sin

2

xv

dx4

x2sin

2

xx

4

x2sin

2

xI

c2

x2cos

4

x

4

x2sinx

2

x 22

C2

x2cosxx2sinxx2

4

1 22

c2

x2cosx2sinxx

4

1 2



7.4 More formulas in integration

Using the method of substitution or integration by parts, we can derive the

following formulas (see table 7.2)

1) c

a

xtan

a

1

ax

dx 1

22

2) c

ax

axlog

a2

1

ax

dx22

3) c

xa

xalog

a2

1

xa

dx22

4) cxaxlog

xa

dx 22

22

5) caxxlog

ax

dx 22

22

6) c

a

xsin

xa

dx 1

22

7) c

2

xax

a

xsin

2

adxxa

221

222

8) c

2

xaxaxxlog

2

adxxa

2222

222

9) c

2

xaxaxxlog

2

adxax

2222

222

Table 7.2 Additional formulas for integration

Now we are in a position to integrate functions having quadratic factors or

square root of quadratic factors in numerator or denominator of functions.

Integral of functions of the form xsindxcos

xsinbxcosa

Method

Step 1: Let numerator = A (Denominator) + dx

dB (Denominator)



Step 2: Find the values of A and B

Step 3: Split the function and integrate

We have already seen this in W.E. 7(b)

Completion of squares

All the subsequent methods use a technique called “Completing the

square”. It is simply writing a quadratic expression in the form, a2 + x2,

a2 – x2, x2 – a2.

Example: Consider x2 + 3x + 2. We can write

2x2

32x2x3x 22

222

2

32

2

3x

2

32x

4

98

2

3x

2

222

2

1

2

3x

4

1

2

3x

Example: Consider 2 – 7x – x2

x7x2xx72 22

222

2

7

2

7x

2

72x2

2

2

7x

4

492

2

2

7x

4

57

22

2

7x

2

57



Integration of functions of the form cbxax

mIx2

Method:

Step 1: Write Bcbxaxdx

dAmIx 2



Working Example: Evaluate dxx7x6

4x2

Solution:

Here 6x2x7x6dx

d 2

Let x + 4 = A (–2x + 6) + B = –2 Ax + 6A + B

1 = –2A; 4 = 6A + B

7342

164A64B;

2

1A

x6x7

dx7dx

x7x6

6x2

2

1dx

x7x6

4x222

93x7

dx7

x7x6

x7x6d

2

1

22

2

22

2

3x2

dx7x7x6log

2

1

c3x2

3x2log

22

7x7x6log

2

1 2


12

Method: Write cbxax2 in the form 222222 xaoraxorax

and integrate.




a) 2x4x4

dx2

b) 10x13x3

dx2

Solution:

a) 2x4x4

dx2

2

1xx

dx

4

1

2

22

2

1

2

1x

dx

4

1

c

2

12

1x

tan2

1.

4

1 1

c1x2tan8

1 1

b)

3

10

3

x13x

dx

3

1

10x13x3

dx

22

22

6

13

3

10

6

13x

dx

3

1

36

169

3

10

6

13x

dx

3

12

22

6

17

6

13x

dx

3

1



c

6

17

6

13x

6

17

6

13x

log

617.

2

1.

3

1

c5x

3

2x

log17

1

c5x3

2x3log

17

1


mIx

2

Method:


dAIm 2



Working Example: Find dx

x2x6

5x6

2

Solution:

Let Bx2x6dx

dA5x6 2

Bx41A5x6

BAAx45x6

BA5;A46

2

13

2

35A5B;

2

3A

dxx2x6

2

13x41

2

3

dxx2x6

5x6

22



22 x2x6

dx

2

13dx

x2x6

x41

2

3

2

2

12

2

x2

x32

dx

2

13

x2x6

x2x6d

2

3

2

2

12

4

1x

16

13

dx

22

13x2x62.

2

3

22

4

1x

4

7

dx

22

132x2x63

c

4

74

1x

sin22

13x2x63 12

c7

1x4sin

22

13x2x63 12

Integrals of functions of the form cbxax

1

2

Method: Write cbxax 2 in form 222222 axorxaorxa

and integrate

Working Example: Evaluate 2xx3

dx

2

Solution:

3

2

3

xx

dx

3

1

2xx3

dx

22



12.3

12.2

36

1

6

1x

dx

3

1

2

22

6

5

6

1x

dx

3

1

c6

5

6

1x

6

1xlog

3

122

c3

2

3

xx

6

1xlog

3

1 2

c2xx33

1

6

1xlog

3

12

12

Integration of functions of the form cbxaxmIx 2

Method:


dAmIx 2



Working Example: Find dx1xx2x3 2

Solution: Let B1xxdx

dA2x3 2

B1x2A2x3

3x – 2 = 2Ax + A + B

3 = 2A; –2 = A + B

2

7

2

32A2B;

2

3A



dx1xx2

71x2

2

3dx1xx2x3 22

dx1xx2

7dx1x21xx

2

3 22

dx3

4

2

1x

2

71xxd1xx

2

32

12

221

2

c1xx2

1xlog

2

1

4

31xx

2

1x

2

7

23

1xx

2

3 222

32

\/

c1xx2

1xlog

15

211xx1x2

2

71xx 222

32

Integration of functions of the form cbxax2

Method: Write cbxax 2 in the form

222222 axorxaorxa and integrate.

Working Example: Find dxx2x1 2

Solution: dxx2

x

2

12I

21

2

dx4

1x

16

1

2

12

2

12

dx4

1x

16

92

2

12



dx4

1x

4

32

2

122

c

4

34

1x

sin16

9.

2

1

4

1x

16

9

4

1x2 1

2

12

c3

1x4sin

32

9x

2

x

2

1

4

1x42 12

Integration using partial fractions

Consider the function .1x2x

1 Expanding the denominator as

,232x2 we can integrate the function using the method given in 7.3.3.

We can also integrate the same function using partial fractions.

2x

1

1x

1

1x2x

1x2x

1x2x

1

Now it is easy to integrate 2x

1

1x

1. We illustrate this method using an

example.

Working Example: Evaluate dx2x1x

6x7

Solution: Let us write the given function as 2x

B

1x

A

Then 2x

B

1x

A

2x1x

6x7

Multiplying both sides by (x – 1) (x – 2), we get

7x – 6 = A(x – 2) + B(x – 1) ……………… (*)

Put x = 1 in (*). We get 7(1) – 6 = A(1 – 2)



That is, 1 = – A of 1A

Put x = 2 in (*). We get 7(2) – 6 = B(2 – 1)

That is, 8 = B 8B

dx2x

B

1x

AI

dx2x

8dx

1x

1

= log (x – 1) + 8 log (x – 2) + c

7.5 Definite integrals

dxxf is called the indefinite integral in integral calculus. You will be

curious to know why it is called indefinite integral. Rieman defined a definite

integral first. It is the form .dxxf

b

a

If

b

a

agbgdxxfthen,dxxfxg

Note that the definite integral

b

a

dxxf is a real number whereas dxxf is

a function.

Working Example 18: Evaluate

1

0

2xx1

dx

Solution:

cx215

x215log

5

1

xx1

dx2

(Check)



Hence

c1215

1215log

5

1c

0215

0215log

5

1

xx1

dx1

02

35

15log

5

1

15

15log

5

1

Note: The arbitrary constant c gets cancelled while finding g(0) – g(a).

S.A.Q 8: Integrate the following functions w.r.t. x

a) 8x3x

3x42

b) 1xx

2x32

S.A.Q 9: Integrate the following functions w.r.t. x

a) 2xx8

1 b)

1x2x

3x4

2


a) 2xx8

1 b)

20x8x

1

2


a) 2xx23 b) 16x9 2


a) 2xx32

1 b)

2x1x

x2

21x

C

1x

B

2x

AasfunctiongiventheWriteb)(Hint

S.A.Q. 14: Evaluate the following definite integrals

a) dx3x2x

1x1

0

b)

1

0

44 dx3x23x5



7.6 Summary

In this unit we studied the standard forms of integration. The different rules

of integration is given and explained with the help of standard examples. All

the formula concerned with integration is followed by good examples. The

concept of definite integral and its application is explained clearly with good

examples.


1. Integration of the following w.r.t. x

a) xsinx1

12

b) xsecxcosx2

7x 2

c) x4

5x4x3 2

d)

2

x

1x

e) xcostamxxsec

2. Evaluate the following integrals

a) dxx5cos3 2 b) dxxsin1

c) dxx3cosx5cos d) dxx5cosx7sin

3. Integrate the following functions w.r.t. x

a) xcosxsin

122

b) xcos1

xsin2

c) xsin1 2 d) x42x e1e

(Hint: a) sin2x + cos2 x=1 b) sin2x=1 – cos2x c) sin2x=2 sin x cos x

d) Expand and integrate.)


a) 32 xcosx3 b) x

xlogsin

c) xlogx

1 d)

x2

x

xecos

x1e




a) 2

x

1

x

e6 b)

x2

x

e94

e

c) 0nxtanxsec

xsecn

d) xsin

xcos3


a) a

xcos 1 b) x sin2x

c) xtan 1 d) xlogx n

(Hint: a) a

xcosu 1 b) u = x c) u = tan-1 x d) u = log x)


a) x tan2 x b) ex(sin x + cos x) c) x2 sin x

(Hint: a) u = x b) use integration by parts to

xx exdsinxdxsine and proceed c) take u = x2, evaluate

dxxsinx

8. Integrate the following functions w.r.t.x

a) 3xx2

1x32

b) 35x2x

1x52


a) 10x13x3

12

b) 10x4x4

12


a) 2xx43

x2 b)

5x4x

7x6




a) 10x3x

1

2 b)

2xx1

1

12. Integrate the following function w.r.t. x

a) 6x4x2 b) x1x2


a) x311x2

3x2 b)

1x21x

1x2

2

(Hint: x2 – 1 = (x + 1) (x – 1). Write the given function as

1x2

C

1x

B

1x

A)


a)

04

dxx21

x3 b)

1

08

3

x1

x

7.8 Answers

Self Assessment Question

1. xtanxtanxcosx

2xloge 1

3

x

2. a) 4

3x4eccos b)

43x28

1

c) 3

2x3log d)

11

x117cot

e) 4

e x45



3. a) 4

x43sec b) x

2

xtan2

c) xcosxsin d) 3log

x3

4. a) 2

xlog2

b) xecos

c) 2

34 11x

6

1 d) xtane

5. a) xxetan b) 214 xtanx1log

c) xlog1log d) xsec3

sec x3

6. a) 211 x1xsinx;xsinu b) 1xe;xu x

c) 2

x

4

x1xlog1x

2

1;1xlogu

22

d) xxtan1x2

1 12

7. a) x

xlog1;xlogu b)

x1

e;

x1

dxdv,xeu

x

2

x Answ er

c) a

xtan

a

2

x

axlog;axlogu 1

2222 d)

16

x

4

xlogx 44

8. a) 23

3x2tan

23

188x3xlog2 12

b) 3

1x2tan

3

11xxlog

2

3 12

9. a) 3

5x2tan

3

2 1 b) 297x2

297x2log

29

1



10. a) 33

1x2sin

2

3xx8 12 b) 1x2x1xlog71x2x4 22

11. a) 33

1x2sin 1 b) 20x8x4xlog 2

12. a) 2

1xsin2xx231x

2

1 12 b) 16x9x3log1616x9x36

1 22

13. a) 1x

2xlog b)

1x

1.

3

1

2x

1xlog

9

2

14. a) The indefinite integral is 2 log (x + 3) – log (x + 2) + c

Answer: 2 log 4 – 3 log 3 + log 2

b) Answer: 2

6253

2

5502

10

30312

10

31 555

Terminal Questions:

1. a) xcosxtan 1 b) xtanxsinxlog2

7

2

x

c) xlog4

5xx

8

3 2 d) xlogx22

x 2

e) xsinxsec

2. a) x10sin20

3x

2

3 b)

2

x

4cos22

c) x2sin4

1x8sin

10

1 d)

2

x2cos

12

x12cos

3. a) xcotxtan b) xsinx

c) xsinxcos d) 4

e

3

e2

2

e x4x3x2

4. a) sin (x3) b) – cos (log x)

c) log (log x) d) tan (xex)



5. a) x

1

e6 b) 2

e3tan

6

1 x1

c) n

xtanxsecn

1 d)

5

xsin1xsin2

2

6. a) 221 xaa

xcosx b)

4

x2sin

2

x2cosx

c) 21

x1log2

1xtanx d)

2

1n1n

1n

x

1n

xlogx

7. a) 2

xxcoslogxtanx

2

b) xsinex

c) xcos2xsinx2xcosx2

8. a) 23

1x4tan

232

13xx2log

4

3 12

b) 5x

7xlog

2

135x2xlog

2

5 2

9. a) 15x3

2x3log

17

1 b)

2

1x2tan

6

1 1

10. a) 7

2xsin4xx432 12

b) 20x9x2

9xlog3420x9x6 22

11. a) 10x3x2

3xlog 2 b)

5

1x2sin 1

12. a) 6x4x2xlog6x4x2

2x 22

b) 3

1x2sin9xx21x22

8

1 12



13. a) x31log15

111x2log

5

2

b) 1x12log6

51xlog1xlog

3

1

14. a) The indefinite integral is cx2tan4

23 21 (Put t = x2 and

integrate).

Answer:

.8

230.

4

23

2.

4

230tan

4

23tan

4

23 11

b) cxsin4

1 41 is the indefinite integral (Put t = x4 and integrate).

Answer:

.8

02

.4

10sin

4

11sin

4

1 11



Unit 8 Differential Equations

Structure

8.1 Introduction

Objectives

8.2 First Order Differential Equations

8.3 Practical Approach to Differential Equations

8.4 First Order and First Degree Differential Equations

8.5 Homogeneous Equations

8.6 Linear Equations

8.7 Bernoulli’s Equation

8.8 Exact Differential Equations

8.9 Summary


8.11 Answers

8.1 Introduction

Differential equation is a branch of Mathematics which finds its application in

a variety of fields. First of all a given problem is converted to differential

equations, which is then solved and the solution to the problem is found out.

Objectives:


to find the solution of Differential Equations

apply differential equations in practical situations

8.2 First order Differential Equations

Definitions

A differential equation is an equation which involves differential coefficients

or differentials.



Thus,

i) ,0dyedxe yx ii) 0xndy

xd 2

2

2

iii) dx/dy

x

dx

dy.xy iii) c

dx

yd

dx

dy1

2

22

32

v) u2y

uy

x

u.x

are all examples of differential equations.

An ordinary differential equation is that in which all the differential

coefficients have reference to a single independent variable. Thus the

equation (i) to (iv) are all ordinary differential equations.

A partial differential equation is that in which there are two or more

independent variables and partial coefficients with respect to any of them.

Thus equation (v) is an example for the partial differential equation.

The order of a differential equation is the order of the highest derivative

appearing in it.

The degree of a differential equation is the degree of the highest derivative

occurring in it, after the equation has been expressed in a form free from

radicals and fractions as far as the derivatives are concerned.

Thus from the examples above,

i) is of the order and first degree;

ii) is the second order and first degree;

iii) can be written as ,1dx

dy.x

dx

dyy

2

and is clearly a first order but

second degree equation.



8.3 Practical Approach to Differential Equations

Differential equations arise from many problems in oscillations of

mechanical, electrical systems, bending of beams, conduction of heat,

velocity of chemical reactions etc., and as much play a very important role in

al modern scientific and engineering studies.

The approach of an engineering student to the study of differential equations

has got to be practical unlike that a student of mathematics, who is only

interested in solving the differential equations without knowing as to how the

differential equations are formed and how their solutions are physically

interpreted.

Thus for the applied mathematics, the study of the differential equation

consists of 3 phases:

i) Formation of the differential equation from the given problem

ii) Solution of this differential equation, evaluating the arbitrary constants

from the given conditions, and

iii) Physical interpretation of the solution

Formation of a differential equation

An ordinary differential equation is formed in an attempt to eliminate certain

arbitrary constants from a relation in the variables and constants. In applied

mathematics, every geometrical or physical problem when translated into

mathematical symbols gives rise to a differential equation.

Examples

From the differential equation of simple harmonic motion given by t,

ntcosAx

Solution: To eliminate the constant A and , differentiating it twice, we have,

ntsinnAdt

dx.



and xnntcosAndt

xd. 22

2

2

thus, ,0xndt

xd. 2

2

2

is the desired differential equation.

Example: Form the differential equation of all circles of radius a.

Solution: The general equation of a circle with centre at (h, k) is given by

(x – h)2 + (y – k)2 = a2 …….. (1)

Where h and k, the co-ordinates of the centre, and a are the constants.

Differentiate twice, we have,

0dx

dy.kyhx0

dx

dy.ky2hx2

and 0dx

dy

dx

ydky1

2

2

2

Then, 2

22

dx

yd

dx

dy1ky

And x – h = -(y – k) dx

dy

2

22

dx

yd

dx

dy1

dx

dy

Substituting these in (1), and simplifying, we get,

,a

dx

yd

dx

dy1

2

2

2

32

It states that the radius of curvature of a circle at any point is constant



Solution of a differential equation

A solution (or integral) of a differential equation is a relation between the

variables which satisfies the given differential equation.

For example, ntcosAx ------------ (1)

Is a solution of 0xndt

xd. 2

2

2

------------ (2)

The general or complete solution of a differential equation is that in which

the number of arbitrary constants is equal to the order of the differential

equation. Thus, (1) is a general solution of (2) as the number of arbitrary

constants (A, ) is the same as the order of (2).

A particular solution is that which can be obtained from the general

solution by giving particular values to the arbitrary constants

For example, ,4

ntcosAx

is the particular solution of the equation (2) as it can be derived from the

general solution by putting .4

8.4 First Order and First Degree Differential Equations

One can represent the general and particular solution of a differential

equation geometrically. But it is not possible to solve a family of parabola

y = x2 + c, in this approach. We shall, however, discuss some special

methods of solution which are applied to the following types of equations:

(i) Equations where variables are separable

(ii) Homogeneous equations

(iii) Linear equations

(iv) Exact equations



Variables separable

If in any equation if it is possible to collect all functions of x and dx on one

side and all the functions of y and dy on the other side, then the variables

are said to be separable. Thus the general form of such equation is f(y), dy

= (x) dx.

Integrating both the sides, we get,

cdxxdyyf as its solution.

Example:

Solve y22y2x3 exedx

dy

Solution: Given equation is 2x3y2 xeedx

dy

dxxedye 2x3y2

Integrating on both the sides,

cdxxedye 2x3y2

c3

x

3

e

2

e 3x3y2

13x3y2 cxe2e3

8.5 Homogeneous Equations

Homogeneous Equations are of the form y,x

y,xf

dx

dy

Where f(x, y) and (x, y) are homogeneous functions of the same degree in

x and y.

To solve a homogeneous equation,

i) Put y = vx, then ,dx

dv.xv

dx

dy

ii) Separate the variables v and x and then integrate.



Example:

Solve xydy2dxyx 22

Solution: The given equation is ,xy2

yx

dx

dy 22

which is homogeneous in x

and y.

Put y = vx then dx

dv.xv

dx

dy

Then the given problem becomes, v2

v1

dx

dv.xv

2

or v2

v31v

v2

v1

dx

dv.x

22

Separating the variables, we get, x

dxdv

v31

v22

Integrating both the sides

cx

dxdv

v31

v22

or cx

dxdv

v31

v6

3

12

or cxlogv31log3

1 2

or c3v31logxlog3 2

or c3v31xlog 23

or 1c3223 cex/y31x

Hence the required solution is 122 cy3xx

Equations reducible to homogeneous form

The equations of the form cybxa

cbyax

dx

dy ---------- (1)



can be reduced to the homogeneous form as follows:

Case 1: when b

b

a

a

Putting x = X + h, y = Y + k (h, k being constants)

So that, dx = dX = dY, becomes,

ckbhaYbXa

cbkahbYaX

dX

dY ----------- (2)

Choose h, k so that (2) may becomes homogeneous.

Put ah + bk + c = 0,

and 0ckbha

So that,

abba

1

acac

k

cbcb

h

or

abba

cbcbh ---------- (3)

abba

acack

Thus when 0,abba then (2), becomes

YbXa

bYaX

dX

dY --------- (4)

Which is a homogeneous in X, Y and can be solved by putting Y = vX.

Case II: When b

b

a

a

i.e. ,0abba the above method fails as h and k become infinite or in

determinant

Now, )say(m

1

b

b

a

a

bmb,ama and (1) becomes



cbyaxm

cbyax

dx

dy

Put ax + by = t, so that dx

dy

dx

dyba

or adx

dt

b

1

dx

dy

therefore we have

cmt

bccatbam

cmt

bcbta

dx

dtcmt

cta

dx

dt

b

1

so that the variables are separable. In this solution, putting t = ax + by, we

get the required solution of (1).

Examples

1. Solve 4xy

2xy

dx

dy

Solution: The given equation is 4xy

2xy

dx

dy [this is where

]b

b

a

a --------- (1)

Putting x = X + h, y = Y + k (h, k being constants)

So that dx = dX, dy = dY, (1) becomes

)4hk(XY

)2hk(XY

dX

dY --------- (2)

Put x + h – 2 = 0 and k – h – 4 = 0,

So that, h = –1, k = 3.

Therefore (2) becomes, XY

XY

dX

dY

Which is homogeneous in X and Y.

Put Y = vX, then dx

dvXv

dX

dY



Therefore (3) becomes, 1v

1v

dX

dvXv

Or

X

dXdv,

vv21

1v1v

vv21v

1v

1v

dX

dvX

2

2


cX

dXdv

vv21

v22

2

12

Or cXlogvv21log2/1 2

cXlogvv21log2

1 2

Or

c2XlogX

Y

X

Y21log 2

2

2

Or log (X2 + 2XY – Y2) = –2c

Or X2 + 2XY – y2 = e – 2c = c

Putting X = x – h = x + 1, and Y = y – k – 3, equation (4) becomes,

c3y3y1x21x22

Which is the required solution.

8.6 Linear Equations

A differential equation is said to be linear if the different variable and its

differential coefficients occur only in the first degree and are not multiplied

together.

Thus the standard form of a linear equation of the first order, commonly

known as Leibnitz’s linear equation, is



Qpydx

dy ---------- (1)

Where P, Q are any functions of x.

To solve the equation, multiply both the sides by Pdx

e so that we get,

PdxPdxPdx

QePeyedx

dy

PdxPdx

Qeyedx

d,.e.i


cdxQeyePdxPdx

As the required solution

Example

1. Solve 2x3 1xey

dx

dy1x

Solution: The given equation is, 2x3 1xey

dx

dy1x

Dividing both the sides by (x + y), given equation becomes,

1x3 1xe

1x

y

dx

dy ---------- (1)

Which is Leibnitz’s equation.

Here 1x

1P

Therefore, 1

1xlog1xlog1x

dxPdx

And 1x

1e

e

11xlogPdx

Thus the solution of (1) is,



cdx1x

11xeye x3Pdx

ce3

1cdxe

1x

y x3x3

1xce3

1y x3

8.7 Bernoulli’s Equation

The equation ,QyPydx

dy n -------- (1)

Where P, Q are functions of x, is reducible to the Leibnitz’s linear equation

and is usually called the Bernoulli’s equation.

To solve (1), divide both the sides of yn, so that

QPydx

dyy n1n -------- (2)

Put ,1 zy n so that

dx

dz

dx

dyyn1 n

QPzdx

dz

n1

1

(2) Becomes,

)n1(Qz)n1(Pdx

dz

Which is Leibnitz’s linear in z and can be solved easily.

Examples

1. Solve 63 yxydx

dyx

Solution: The given equation is, 63 yxydx

dyx -------- (1)



Dividing throughout by xy6,

25

6 xx

y

dx

dyy -------- (2)

Put y–5 = z, so that dx/dzdx/dyy5 6

Therefore (2) becomes, 2xx

z

dx

dz

5

1

Or 2x5zx

5

dx

dz --------- (3)

Which is Leibnitz’s linear in z, and the intermediate form I.F. is,

I.F. = 55xlogdx

x

5

Pdxxee

Therefore the solution of (3) is,

cdxF.Ix5F.Iz 2

cdxxx5zx 525

Or c2

x.5xy

255 [since z = y–5]

Dividing both sides by ,xy 55 we get,

1 = (2.5 + cx2) x3 y5

8.8 Exact Differential Equations

1. Definition: A differential equation of the form

,0dyy,xNdxy,xM

Is said to be exact if its left hand member is the exact differential of

some function u(x, y).

i.e., 0NdyMdxdu

Its solution therefore, is u(x, y) = c.



2. Theorem: The necessary and sufficient condition for the differential

equation Mdx + Ndy = 0 to be exact is

x

N

y

M

Condition is necessary:

The equation M(x, y) dx + N(x, y) dy = 0 will be exact, if Mdx + Ndy

du ----- (1)

Where u is some function of x and y.

But dyy

udx

x

udu ------- (2)

Equating coefficients of dx and dy in (1) and (2), we get

x

uM and

y

uN

xy

u

y

M 2

and yx

u

x

N 2

But, yx

u

xy

u 22

(Assumption)

x

N

y

M

Which is the necessary condition of exactness.

Condition is sufficient:

i.e., if ,x

N

y

M then Mdx + Ndy = 0 is exact.

Let ,uMdx where u is supposed constant while performing integration.

Then

x

uM.,e.i

x

uMdx

y ------- (3)

xy

u

y

M 2

since x

N

y

M (given)



Or y

u

xyx

u

x

N and

yx

u

xy

u 22

Integrating both the sides, w.r.t x (taking y as constant

yfy

uN , where f(y) is a function of y alone. -------- (4)

dyyfy

udx

x

uNdyMdx [by (3) and (4).]

dyyfdyy

udx

x

u -------- (5)

dyyfuddyyfdu

Which shows that Mdx + Ndy = 0 is exact.

Method of solution: By equation (5), equation Mdx + Ndy = 0 becomes

0dyyfud

Integrating cdyyfu

But yfandttanycons

cMdxu =terms of N not containing x.

The solution of Mdx + Ndy = 0 is

)yconst(

Mdx (terms of N not containing x) dy = c

Provided x

N

y

M

Examples

1. Solve 0xycosxxsin

yysinxcosy

dx

dy

Solution: Given equation can be written as,

(y cos x + sin y + y) dx + (sin x + x cos y + x) dy = 0



Here,

M = y cos x + sin y + y

N = sinx + x cos y + x

x

N1ycosxcos

y

M

Thus the equation is exact and its solution is

cdy)xcontainingnotNofterms(Mdx

)yconst(

i.e., cdy0dxysinxcosy

Or y sin x + (sin y + y)x = c


1. Solve ycosxy2sinxdx

dy 23

2. Solve dyxytandxy1 12

3. Solve 0dy5y6x4dx4x2y3

8.9 Summary

In this unit we study the first order differential equations. The practical

approach to differential equations is clearly explained with suitable

examples. Solving first order first degree differential equation by the variable

separable method is discussed here. Equations reducible to homogeneous

form is discussed here with example. The linear equations, Bernoulli’s

equation and exact differential equation is solved here in a simple manner

with proper examples.




1. Derive the necessary and sufficient condition for the differential equation

Mdx + Ndy to be exact

2. Briefly describe Bernoulli’s equation

8.11 Answers


1. Dividing throughout by cos2y,

3

2

2 xycos

ycosysinx2

dx

dyysec

32 xytanx2dx

dyysec -------- (1)

Put tan y = z, so that dx

dz

dx

dyysec2

Therefore equation (1) becomes,

,xxz2dx

dz 3

Which is Leibnitz’s linear equation in z.

2xxdx2Pdx

eee.F.I

Therefore the solution is,

ce1x2

1cdxxee

2x232x2x

Replacing z by tan y, we get

2x2 e.c1x

2

1ytan

Which is the required form



2. This equation contains y2 and tan–1 y and is, therefore not a linear in y,

but since only x occurs, it can be written as,

xytandy

dxy1 12

2

1

2 y1

ytan

y1

x

dy

dx

Which is leibnitz’s equation in x.

Therefore, we have intermediately form I.F. as,

yeee.F.I1tan

dy2y1

1

Pdx

Thus the solution is,

cdyF.Iy1

ytanF.Ix

2

1

Or cdy.ytane.y1

ytane.x 1

2

1y1tan

dty1/dy

ytantPut2

1

Therefore

cdt.e.tye.x t1tan

cdt.e.1e.t tt

cee.t tt

cytane.1ytan 11

Or ye.c1ytanx1tan1

3. The given equation is, 5y3x22

4y3x2

dx

dy -------- (1)

Putting 2x + 3y = t, so that dx

dt

dx

dy32



5t2

4t72

dx

dt

3

1

(1) becomes, 5t2

22t7

5t2

12t32

dx

dt

dxdt22t7

5t2

Integrating both the sides, cdxdt22t7

5t2

Or cxdt22t7

1.

7

9

7

2

Or cx22t7log49

9t

7

2

Putting t = 2x + 3y, we have

14(2x + 3y) – 9 log (14x + 21y + 22) = 49x + 49 c

Or c22y21x14log9y42x21

Which is the required solution

Terminal Questions

1. Refer to Section 16.8




Unit 9 Complex Numbers

Structure

9.1 Introduction

Objectives

9.2 Complex Numbers

9.3 Conjugate of a Complex Number

9.4 Modulus of a Complex Number

9.5 Geometrical Representation of Complex Number

9.6 Exponential Form of a Complex Number

9.7 De Moivere’s* Theorem

9.8 nth Roots of a Complex Number

9.9 Summary


9.11 Answers

9.1 Introduction

We recall that, if x and y are real numbers and 1i then x + iy is called

a complex number. The complex numbers were first introduced by Cardan

(1501 – 1576). Two hundred years later Euler (1707 – 1783) and John

Bernoulli recognized the complex numbers introduced by Cardan and

studied their properties in detail. In 1983, Sir William Rowan Hamilton (1805

– 1865) an Irish mathematician introduced the complex number as an

ordered pair of real numbers. In this chapter we begin the study of complex

numbers as ordered pairs.



Objectives:


understand the concept of complex numbers.

apply De Moivere’s Theorem in finding the roots of complex

numbers.

9.2 Complex Numbers

Let C denote the set of all ordered pairs of real numbers.

That is, .Ry,x;y,xC

On this set C define addition “+” and multiplication “.” by,

(x1, y1) + (x2 + y2) = (x1 + x2, y1 + y2) … (1)

(x1, y1) . (x2, y2) = (x1x2 – y1y2, x1y2 + x2y1) … (2)

Then the elements of C which satisfy the above rules of addition and

multiplication are called complex numbers. If z = (x, y) is a complex number

then x is called the real part and y is called t he imaginary part of the

complex number z and they are denoted by x = Re z and y = Im z. If (x1, y1)

and (x2, y2) are two complex numbers then (x1, y1) = (x2, y2) if and only if

x1 = x2 and y1 = y2.

(a) Properties of addition

1. Closure law: If z1 = (x1, y1), z2 = (x2, y2) then from (1) z1 + z2 = (x1, y2) +

(x2, y2)

= (x1 + x2, y1 + y2), which is also an ordered pair of real numbers. Hence

z1 + z2 C. Therefore for every z1, z2 C, z1 + z2 C.

2. Commutative law: z1 + z2 = z2 + z1 for every z1, z2 C

Consider z1 + z2 = (x1, y1) + (x2, y2) = (x1 + x2, y1 + y2)

= (x2 + x1, y2 + y1) = (x2, y2) + (x1, y1) = z2 + z1.

3. Associative law: z1 + (z2 + z3) = (z1 + z2) + z3 for every z1, z2, z3 C

Proof of this is similar to (2)



4. Existence of identity element: There exists an element (0, 0) C such

that,

(x, y) + (0, 0) = (x + 0, y + 0) = (x, y).

for every (x, y) C. Here (0, 0) is called the additive identity element

of C.

5. Existence of inverse: For every (x, y) C there exists (–x, –y) C

such that

(x, y) + (–x, –y) = (x – x, y – y) = (0, 0).

Hence (–x, –y) is the additive inverse of (x, y).

Thus we have shown that the set C is an abelian group w.r.t. the

addition of complex numbers defined by (1).

(b) Properties of multiplication

1. Closure law: If z1 = (x1, y1), z2 = (x2, y2) C then from (2)

z1z2 = (x1, y1) (x2, y2) = (x1x2 – y1y2, x1y2, x1y2 + x2y1), which is also an

ordered pair of real numbers. Hence z1z2 is also a complex number.

Thus for every z1, z2 C, z1z2 C.

2. Commutative law: z1z2 = z2z1 for every z1, z2 C.

Now z1z2 = (x1, y1) (x2, y2) = (x1x2 – y1y2, x1y2 + x2y1) ….. (i)

and z2z1 = (x2, y2) (x1, y1) = (x2x1 – y2y1, x2y1 + x1y2)

= (x1x2 – y1y2, x1y2 + x2y1) ….. (ii)

From (i) and (ii) z1z2 = z2z1.

3. Associative law: z1(z2z3) = (z1z2) z3, for every z1, z2, z3 C Proof is

similar to (2)

4. Existence of identity element: There exists (1, 0) C such that

(x, y) (1, 0) = (x . 1 – y . 0, x . 0 + 1 . y) = (x, y) for every (x, y) C.

Here (1, 0) is called the multiplicative identity element.

5. Existence of inverse: Let z = (x, y) (0, 0), be a complex number. Let

(u, v) be the inverse of (x, y).



Then (u, v) . (x, y) = (1, 0), the identity element.

i.e. (ux – vy, uy + vx) = 1, 0).

Hence ux – vy = 1, and uy + vx = 0.

Solving for u and v, we get, 2222 yx

yv,

yx

xu

Hence Cyx

y,

yx

x2222

is the multiplicative inverse of (x, y).

Thus we have shown that the set of non-zero complex numbers forms an

abelian group w.r.t. the multiplication defined by (2).

Also we can prove that the multiplication is distributive over addition.

(c) Distributive law: For all z1, z2, z3 C

i) z1 (z2 + z3) = z1z2 + z1z3 (left distributive law)

ii) (z2 + z3) z1 = z2z1 + z3z1 (right distributive law)

The complex numbers whose imaginary parts are equal to zero possess the

following properties.

(x1, 0) + (x2, 0) = (x1 + x2, 0).

and (x1, 0) . (x2, 0) = (x1 x2, 0).

Which are essentially the rules for addition and multiplication of real

numbers. We identify the complex number (x, 0) with the real number x.

Denote the complex number (0, 1) by i.

Now i2 = I . I = (0, 1) (0, 1) = (0 . 0 – 1 . 1, 0 . 1 + 1 . 0)

= (–1, 0) = –1.

Hence i2 = –1.

With this convention we shall show that the ordered pair (x, y) is equal to x + iy.

For, (x, y) = (x, 0) + (0, y)

= (x, 0) + (0, 1) (y, 0)

= x + iy



Since (x, 0) = x, (y, 0) = y and (0, 1) = i.

Because of the extreme manipulative convenience we shall continue to use

the notation x + iy for the complex number (x, y).

9.3 Conjugate of a Complex Number

Let z = x + iy be a complex number. Then the complex number x – iy is called

the complex conjugate or simply, the conjugate of z and is denoted by .z

Thus, if z = x + iy then .iyxz

For example, if z = 3+4i then .i43z

Clearly ,zz

,zRe2x2iyxiyxzz

and .zlmi2iy2iyxiyxzz

Also, iyx.iyxz.z

= x2 – i2y2

=x2 + y2, which is a real number.

Thus the product of two conjugate complex numbers is a real number.

Theorem: For all z1, z2 C

1. 2121 zzzz

i.e., the conjugate of a sum is equal to the sum of the conjugates.

2. 2121 z.zz.z

i.e., the conjugate of a product is equal to the product of the conjugates.

3. 0z,z

z

z

z2

2

1

2

1

i.e., the conjugate of a quotient is equal to the quotient of then

conjugates.



9.4 Modulus of a Complex Number

If z = x + iy is a complex number then 22 yx is called the modulus or

absolute value of z and is denoted by | z |.

Thus .yxz 22

Clearly | z | is a non-negative real number i.e., .0z

then .516943z22

We can easily verify the following:

1. 2

zz.z 2. zz 3. .zzRez

Theorem: For all z1, z2 C

1. 2221 z.zz.z

i.e., modulus of a product is equal to the product of their moduli.

2. 0z,z

z

z

z2

2

1

2

1

i.e., modulus of a quotient is equal to the quotient of the moduli.

3. 2121 zzzz

4. 2121 zzzz

9.5 Geometrical Representation of Complex Number

A complex number x + iy can be represented by a point P(x, y) in the

Cartesian plane with x as the abscissa and y as the ordinate. Thus every

point on the x-axis corresponds to a real number and every point on the y-

axis corresponds to a pure imaginary number (iy) and vice versa. Hence x-

axis is called the real axis and y-axis, the imaginary axis. And the plane

whose points are represented by complex numbers is called the complex



plane or Argand plane named after the French mathematician J.R. Argand

(1768 – 1822). Although the geometric representation of complex numbers

is usually attributed to J.R. Argand but it was Casper Wessel of Norway

(1745 – 1818) who first gave the geometric representation of complex

numbers.

Now draw PM perpendicular to the x-axis. Let POX and OP = r. Clearly

OM = x and MP = y.

Now )1...(

sinryr

y

OP

MPsin

cosrxr

x

OP

OMcos

Hence x + iy = r (cos + i sin )

Thus every complex number z = x + iy can be represented in the form

.sinicosr This form of a complex number is called the polar form or

the trigonometric form.

Squaring and adding the equations given in (1), we get

.rsin2rcosryx 222222

,yxr 22 which is the modulus of the complex number z = x + iy.

Thus z represents the distance of the point z from the origin.



The angle is called the argument or the amplitude of z and is denoted by

= arg z or = amp z.

Since sin (2n + ) = sin , cos (2n + ) = cos , when n is any integer, is

not unique. The value of satisfying – < < is called the principal value

of the argument.

Note:

1. If z1 = x1 + iy1 and z2 = x2 + iy2

Then z1 – z2 = (x1 – x2) + i(y1 – y2).

,yyxxzz2

112

2121 which is the distance between

the points z1 and z2.

2. cos + I sin is briefly denoted by cis

Theorem: 1. 2121 cisciscis

2. 212

1 ciscis

cis

Theorem:

1. amp z1z2 = amp z1 + z1 + amp z2

2. 211

1 zampzampz

zamp

Remark:

1. To find the amplitude of a complex number we use the following rule:

sin cos

+ + (say)

+ – –

– + –

– – –( – )



For example,

i) if 2

1cos,

2

3sin then

3

ii) If 2

1cos,

2

3sin then

3

2

3

iii) 2

1cos,

2

3sin then

3

iv) if 2

1cos,

2

3sin then .

3

2

3

2. The value of the amplitude must satisfy the equations r

xcos and

,r

ysin where .yxr 22 Some times we combine these

equations dividing one by another. In that case we get x

ytan or

.x

ytan 1 Because of the difference in principal values of

111 tanandcos,sin the value of the argument is not necessarily the

principal value of .x

ytan 1 For example,

4

3sini

4

3cos2i1

and so 4

3i1amp but .

41tan 1

9.6 Exponential Form of a Complex Number

If x is real, it can be proved, in the advanced mathematics that the functions

ex, sin x, cos x etc. can be expressed in the form of an infinite series.

i.e. ......!3

x

!2

x

!1

x1e

32x … (1)

.....!7

x

!5

x

!3

xxxsin

753

… (2)



....!6

x

!4

x

!2

x1xcos

642

… (3)

Assuming that (1) holds good for a complex number also, replacing x by ix

in (1) we get,

....!3

xi

!2

xi

!1

ix1e

3322ix

....!7

x

!5

x

!3

xxi......

!6

x

!4

x

!2

x1

753642

= cos x + i sin x

Thus xsinixcoseix

This is called the Euler’s formula.

We know that a complex number z = x + iy can be expressed in the polar

form as

sinicosriyxz

where .x

ytanandyxr 122

The complex number can also be written in the form

.,eriyxz i

This is called the exponential form of a complex number.

Example: Express the following complex numbers in the polar form and

hence find their modulus and amplitude.

1) i3 2) 1 – i 3) 3i1

Solution:

1) Let sinicosri3 .

On equating the real and imaginary parts, we get

3cosr and r sin = 1.



Squaring and adding, 222222 13sinrcosr

2ror4r 2

Hence 2

1sinand

2

3cos

Hence .6

6

sini6

cos2i3

Therefore modulus = 2 and amp 6

i3

2. Let 1 – I = r (cos + i sin )

r cos = 1, r sin = 1

2ror,2r.e.i11sinrcosr 22222 .

2

1sin,

2

1cos

Therefore 4

4

sini4

cos2i1

i.e. Modulus = .4

i1amp,2

3. Let .sinicosr3i1

Hence r cos 3sinr,1

2ror,4ror,31sinrcosr 22222

Hence ,2

1cos ,

2

3sin

3

2

3

3

2sini

3

2cos23i1

Modulus = 2, amp 3

23i1



Example: If a = cos + i sin , 20 prove that 2

cotia1

a1

Solution:

sinicos1

sinicos1.S.H.L

2cos

2sini2

2sin2

2cos

2sini2

2cos2

2

2

2cosi

2sin

2sini

2cos

2sin2

2cos2

Multiplying and dividing b y i,

2sini

2cos

2sini

2cos

2coti

.S.H.R2

coti


1. Find the smallest positive integer n such that .1i1

i1n

Example: If idc

ibaiyx prove that

22

22222

dc

bayx

Solution:

Now idc

ibaiyx . Taking the conjugate on both sides

We get, idc

ibaiyx



Multiplying, idcidc

ibaibaiyxiyx

22

22222

22

2222

dc

bayx

dc

bayx

9.7 De Moivere’s* Theorem

If n is any integer then

(cos + i sin )n = cos n + i sin n …….. (1)

And if n is a rational fraction say q

p then q

p

sinicos has q values and

one of its values is .q

psini

q

pcos

Proof: Case (i) Let n be a positive integer.

In this case we shall prove (1) by mathematical induction.

If n = 1 then (cos + i sin )1 = cos 1 . + I sin 1 . .

= cos + i sin .

Hence (1) is true for n = 1. Assume that (1) is true for n = m,

i.e., (cos + i sin )m = cos m + i sin m (Induction hypothesis)

Multiplying both sides of (2) by cos + i sin we get

(cos + i sin )m+1 = (cos m + i sin m ) cos + i sin )

= (cos m cos - sin m sin ) + i (sin m cos + cos m sin )

= cos (m + ) + i sin (m + )

= cos (m + 1) + i sin (m + 1)

Hence the theorem is true for n = m + 1.

Hence by mathematical induction the theorem is true for all positive

integers n.

Case (ii). Let n be a negative integer.



n = – m, where m is a positive integer.

Consider (cos + i sin )n = (cos + i sin )-m

m

sinicos

1

)i(casefrommsinimcos

1

msinimcosmsinimcos

msinimcos

1imsinimcos

msinimcos 2

22

= cos m – i sin m

= cos (–m) + i sin (–m)

= cos n + i sin n .

Case (iii). Let n be a rational fraction i.e., ,q

pn where p and q are integers

and q > 0.

Let q

psini

q

pcosz

qq

q

psini

q

pcosz

q

psini

q

pqcos

= cos p + i sin p

zq = (cos + i sin )p, from Cases (i) and (ii),

which is an algebraic equation of degree q. Hence from fundamental

theorem of algebra it has q roots. Therefore taking qth root on both sides, we

get QP

sinicosz



Hence q

p

sinicos has q values and one of them is

.q

psini

q

pcosz

This completes the proof of the theorem.

Note: Replacing by – in (1), we get

(cos – i sin )n = cos n – i sin n


2. Simplify

42

35

5sini5cos.4sini4cos

2sini2cos.3sini3cos

Example: Prove that ,223i113i1 1n3n3n3n3 where n is

any integer

Solution:

Expressing 3i1 in the polar form, we get

3

2sinisu

3

2cos23i1

Taking conjugate on both sides, we get

3

2sini

3

2cos23i1

L.H.S. =

n3n3

3

2sini

3

2cos2

3

2sini

3

2cos2

n2sinin2cos2n2sinin2cos2 n3n3

Using De Moivre’s theorem

1n3n3 2n2cos2.2 since cos 2n = 1

= R.H.S.



Example: Prove that

2

ncos.

2cos2sinicos1sinicos1 n1nnn

Where n is any integer.

Solution:

n2

n2

2cos

2sin2i

2cos2

2cos

2sin2i

2cos2.S.H.L

nn

2sini

2cos

2cos2

2sini

2cos

2cos2

2

nsini

2

ncos

2cos2

2

nsini

2

ncos

2cos2 nnnn

From De Moivre’s theorem.

.S.H.R2

ncos

2cos2

2

ncos2.

2cis2 n1nnn

9.8 nth Roots of a Complex Number

If zn = a, where a is a non-zero complex number and n, is a positive integer

then z is called the nth root of a. Since the given equation is of degree n,

there are n roots of the equation. Hence solving zn = a, we obtain n, nth roots

of a.

Example: Find the cube roots of i3 and represent them on the Argand

plane. Also find their continued product.

Let sinicosri3

.1sinrand3cosr

Squaring and adding r2 cos2 + r2 sin2 = 3+1

r2 = 4, r = 2



Hence;

2

1sin;

2

3cos

6 (Principal value)

6n2

6n2cis.2

6n2sini

6n2cos2i3

3

1

3

1

6n2cis2i3

6n2

3

1cis.2 3

1

18

1n12cis.2 3

1

Substituting n = 0, 1, 2 (or any three consecutive values of n), we obtain the

cube roots of i3

They are 18

25cis2,

18

13cis2,

18cis2 3

131

31

i.e., .250cis2,130cis2,10cis2 31

31

31

To represent these roots on the Argand plane consider a circle whose

centre is at the origin and whose radius is .2 31

Since modulus of each of

these roots is ,2 31

these roots lie on the circle.



In above figure the points A, B, C represent the cube roots of .i3

Since C,B,A,120AOCCOBBOA are the vertices of an equilateral

triangle

Continued product = 18

25cis2.

18

13cis2.

18cis2 3

1

3

1

3

1

18

25

18

13

18cis2

3

3

1

6

2cis26

13cis2

.i36

cis2

Note:

1. The cube roots of unity are 2,,1 where .3

2cis Also,

.01 2

2. The fourth roots of unity are .i,i,1,1

3. In general the nth roots of unity are 1n2 ..........,,,1 where

.n

2cis



9.9 Summary

In this we discuss about the concept of complex numbers in detail. The idea

of representing a complex number in Polar Form is explained in a simple

manner. The method of finding the roots of a Complex Numbers using De

Moivere’s Theorem is well illustrated.


1. State and prove De Movere’s Theorem

2. Find the Cube roots of the Complex Numbers 1+i and express it in the

Argand Diagram

9.11 Answers


1. Now i1i1

i1

i1

i12

.i2

i2

i1

ii212

2

Therefore nn

i2

i2

i1

i1

Now by inspection n = 4 is the smallest positive integer such that .1i n

2. Given expression 42

35

5cis.4cis

2cis.3cis

4524

3253

cis.cis

cis.cis

208

615

cis.cis

cis,cis

21208615

ciscis

= cis 21

= cos 21 + i sin 21



Unit 10 Matrices and Determinants

Structure

10.1 Introduction

Objectives

10.2 Definition of a Matrix

10.3 Operations on Matrices

10.4 Square Matrix and Its Inverse

10.5 Determinants

10.6 Properties of Determinants

10.7 The Inverse of a Matrix

10.8 Solution of Equations Using Matrices and Determinants

10.9 Solving equations using determinants

10.10 Summary


10.12 Answers

10.1 Introduction

The theory of matrices, introduced by French mathematician Cayley in

1957, is presently a powerful tool in the study of branches of Mathematics,

Physical sciences, biological sciences and business applications. The

concept was initially developed for solving equations.

Objectives:


solve determinant using their properties

find the solution of equations using matrices and determinant

10.2 Definition of a Matrix

Definition: A matrix A is a rectangular array of numbers arranged as m

horizontal lists, called rows, each list having n elements; the vertical lists are

called columns.



It is written as

mn1m

n22221

n11211

a.............a

.

.

.

.

aaa

a.........aa

A

Note: The element in the ith row and jth column is aij. So A is also written as

jia nj1

mi1

or simply .aij A is called an m n matrix. We also

write the (I, j)th entry as (A)ij. When m = n, A is called a square matrix (also

called an n – square matrix A)

Example

3864

5601

1432

A ,

605

413

201

D,

1

8

7

5

C,312B

are 3 4,1 3, 4 1 and 3 3 matrices respectively.

Definition: Two matrices ijij bBandaA are equal if (i) the number

of rows of A and B are the same (ii) the number of columns of A and B are

the same (iii) ijij ba for all i, j.

Example: Find the values of a, b, c, d if

1014

26

dcdc2

b2a2b2a2



Solution: Equating the corresponding entries of the matrix, we get

2a + 2b = 6 2a – 2b = 2

2c + d = 14 c – d = 10

By adding the first two equations we get 4a = 8. so a = 2.

2b = 6 – 2a = 6 – 4 = 2. So b = 1

Similarly,

3c = 24. So c = 8, d = 14 – 2c = 14 – 16 = –2

S.A.Q.1 How many entries are there in an m n matrix ?

10.3 Operations on Matrices

In this section we define sum of two matrices, difference of two matrices, the

scalar multiple of a matrix A by a scalar (real number) k.

An m n matrix is said to be of size (m, n)

Definition: If ijij bBandaA are two m n matrices then A + B is

defined as an m n matrix as follows:

nj1mi1ijij baBA

Note: For getting the sum of A and B, we add to an entry in A, the entry in B

in the same place. We can add only two matrices of the same size.

Definition: If ijij bBandaA are two m n matrices then A – B is

defined as an m n matrix as follows:

nj1mi1ijkakA

Example: If 765

432A and

321

987B find A + B, A – B, 2A + 3B



Solution: 321

987

765

432BA

1086

13119

372615

948372

444

555

372615

948372

321

987

765

432BA

321

9873

765

4322B3A2

332313

938373

726252

)4(2)3(2)2(2

963

272421

141210

864

914612310

278246214

231813

353025

We list some special matrices in the next example.

Example:

a)

000

000

is called the zero matrix.

We can write it as O. We can have zero matrix of any order.

b)

100

010

001

is called then n – square unit matrix.

(Note: The number of rows is equal to the number of columns in the unit

matrix. It is denoted by In or simply I when n is understood.)



c)

3000

0100

0040

0002

is called a diagonal matrix. A square matrix is diagonal

matrix if only the entries on the diagonal are nonzero and other entries

are 0 s.

d) A diagonal matrix having the same number along the diagonal is called

a scalar matrix. A scalar matrix is simply kIn for some scalar k and some

positive integer n.

Definition: If ijaA is an m n matrix then the transpose of A denoted by

AT is defined as mi1nj1ij

T aA

Note: The transpose of an m n matrix is an n m matrix.

Just as addition of real numbers is commutative, the addition of matrices is

also commutative. As far as addition is concerned matrices behave like

numbers. The following theorem lists properties of addition of matrices.

Theorem: If A, B, C and m n matrices and k and I are scalars.

a) A + (B + C) = (A + B) + C

b) A + 0 = 0 + A where 0 is the m n zero matrix

c) A + (–A) = (–A) + A = 0 (Here –A denotes (–1) A)

d) A + B = B + A

e) k(A + B) = kA + kB

f) (k + I) A = kA + IA

g) (kI) A = k (lA) = l(kA)

h) lA = A

i) (A + B)T = AT + BT

j) AATT

k) nT

n II



The theorem can be proved by using the definition of sum of two matrices

and scalar multiple of a matrix.

If A and B are two matrices AB is defined only when the number of columns

of A = number of rows of B. We defined AB in Definition 4.7.

Definition: If A is an m n matrix and B is an n p matrix then the product

of A and B, denoted by AB, is an m p matrix and is defined by

nkink22iik1iik ba......babaAB

for ,pk1,mi1

Note (AB)ik can be understood as follows.

in2i1i a........,a,a is the ith row of A,

nk

k2

ik

b

b

b

Is the kth column of B and both

these have n elements. For calculating (AB)ik, multiply the respective

elements of ith row of A and kth column of B and add them. The resulting

number is (AB)ik.

Example: Find AB when

101

102A and

10

64

21

B

Solution A is a 2 3 matrix and B is a 3 2 matrix. So AB is a 2 2 matrix.

11

52

116021014011

116022014012AB

S.A.Q. 2 Find BA for matrices A and B given in above example

We have seen that A + B = B + A when A and B are matrices of the same

size. But AB BA in general. It can happen that one of the products is



defined whereas the other product is not defined. Let us illustrate this with

an example.

Example: Find two matrices A and B such that

a) AB is defined but BA is not

b) BA is defined but AB is not

c) Both are defined but AB BA

d) Both are defined and AB = BA

Solution:

a) Assume 102

321A and

210

043

124

B

Then AB is defined as the number of columns of A = 3 = number of

rows of B.

Number of columns of B = 3 number of rows of A. Hence BA is not

defined.

b) If

0

3

4

A and 102

321B then BA is defined, as number of

columns of B = 3 = number of rows of A.

Number of columns of A = 1 number of rows of rows of B. Hence

AB is not defined.

c) Assume 32

102

321A and

2310

53

24

B

115022013042

135221033241AB

38

152



102

321

10

53

24

BA

113001202110

153305232513

123402242214

102

1467

1088

Hence AB BA

d) Consider

043

312

210

A and

100

010

001

B

Then

100

010

001

043

312

210

AB

100403001403000413

130102031102030112

120100021100020110

043

312

210

043

312

210

100

010

001

BA

013020411010312000

003120401110302100

003021401011302001



043

312

210

Thus AB = BA

Theorem 4.2 lists the properties of multiplication.

Theorem: If A is an m n matrix and B is an n p matrix and k is any

scalar, then

a) (AB)T = BTAT

b) AAlandAAl mn

c) k(AB) = (kA)B = A(kB)

d) OA = O, BO = O where the four zero matrices are k m, k n, p t

and n t matrices respectively (for some k and t).

The proof of theorem 10.2 follows from the definition of product of two

matrices.

We can define the product of 3 matrices A, B, C when

Cofrow sofNumberBofcolumnsofNumberAnd

Bofrow sofNumberAofcolumnsofNumber……………… (10.1)

The following theorem describes the properties of product of three matrices.

Theorem 10.3 Let A, B, C be 3 matrices. Then the following hold good

whenever the sums and products of matrices appearing below are defined.

a) (AB) C = A(BC) (Associative law)

b) A(B + C) = AB + AC (Left distributive law)

c) (B + C) A = BA + CA (Right distributive law)

Theorem 10.3 follows from the definition. The proof is technical in nature

and it is enough if you remember these properties.



S.A.Q. 3 If

26

24

42

A and

01

64

21

B find A + B, 2A – 3B, 3B – 2A, (A –

B)T and (B – A)T.

S.A.Q. 4 If 210

064A and

111

642B

Verify that (A + B)T = AT + BT

S.A.Q. 5 Find a matrix A such that

41

22

12

14A3

S.A.Q. 6 If 43

21YX and

01

23YX find X and Y.

S.A.Q. 7 A matrix A is said to be symmetric if A = AT. Show that A + AT is

symmetric for a 3 3 matrix A

S.A.Q. 8 If

240

121

312

A and

12

31

21

B

Show that

100

55

105

AB Does BA exist ?

S.A.Q.9 If

425

313

132

A

show that (A – I) (A + 2I) =

111111

333

101010



10.4 Square Matrix and its Inverse

We know that a square matrix is an n n matrix for some integer n. The set

of n n square matrices satisfy some additional properties.

In theorem 4.2 we saw that AIn = InA = A for any n n square matrix A. We

can multiply 2n n matrices and the product is an n matrix. In general we

can define AA, AAA etc.

We define powers of a square matrix as follows. We define A0 = In,

timesn

n32 a.....AAA,.....,AAAA,AAA ….. (10.2)

The set of all n n matrices satisfy the properties of indices (powers) of

numbers.

Note Am An = Am+n = An . Am ......... (10.3)

If a is a non zero real number then we know that .1aa

1

a

1a

A similar property holds good for some square matrices. In the case of

numbers a

1 is called the reciprocal of a. But in the case of matrices it is

called the inverse of a square matrix.

Definition: A square matrix A is invertible (or non singular) if there exists a

square matrix B such that AB = BA = In ………….. (10.4) B is called the

inverse of A and is denoted by A – 1.

Note: If A has an inverse then A is called as invertible matrix.

Example: Let 21

53A If

31

52B then

31

52

21

53AB

10

01

32511221

35531522



Hence 31

52 is the inverse of A.

Theorem: If B is the inverse of a square matrix A then A is the inverse of

the matrix B.

Proof: As B is the inverse of A, by definition (4.8), AB = BA = In ……….

(10.5)

So BA = AB = In ………………………….. (10.6)

From (10.6) we see that A satisfies the condition for the inverse of B. Hence

A is the inverse of B.

Note In the case of numbers a

1 is the reciprocal of a and a is the reciprocal

of a

1. Theorem 10.4 guarantees this property for square matrices.

We are going to see a method for finding the inverse of a matrix. However

you will have a formula for the inverse of a 2 2 matrix (Example 10.7)

Example: If ,dc

baA then

ac

bd

bcad

1A 1 ………….. (10.7)

For the present, you can verify that

I10

01

bcad

bcad

adbc0

0bcad

bcad

1

ac

bd

dc

ba

bcab

1AA 1

S.A.Q. 10 If ,01

10A show that

01

10A 1

S.A.Q. 11 Verify that

215

5311

12726

is the inverse of

194

283

121

10.5 Determinants

The determinant of an n – square matrix A is a unique number associated

with A and is denoted by det (A) or | A|. | A | is called a determinant of order n.



If

nn2n1n

n22221

n11211

aaa

aaa

aaa

A

then | A | is denoted by

nn2n1n

n22221

n11211

aaa

aaa

aaa

As the definition of | A | is complex for a general n – square matrix A, we

define determinant of orders 1, 2, 3 and then extend it for a general n –

square matrix A.

Evaluation of determinants

Definition: The determinants of orders 1, 2, 3 are defined as follows

a) 1111 aa

b) 211222112221

1211aaaa

aa

aa

(We can understand the determinant in the following way).

We (1) multiply the elements in the diagonal from left to right (ii) multiply the

elements in the diagonal from right to left (iii) subtract product got in (ii) from

the product got in (i)

c) 3231

222113

3331

232112

3332

232211

333231

232221

131211

aa

aaa

aa

aaa

aa

aaa

aaa

aaa

aaa

Note: We can calculate the value of a determinant of order 3 as follows:

1. Consider the first element a11 in the first row. Attach the sign + (plus)

2. Delete the row and column in which a11 appears; that is first row and first

column. We get 3332

2322

aa

aa

3. Multiply + a11 and the value of 3332

2322

aa

aa

4. Consider the second element a12 in first row. Attach the sign – (minus)



5. Delete the row and second column in which a12 appears; that is the first

row and second column. We get 3331

2321

aa

aa

6. Multiply – a12 and the value of 3331

2321

aa

aa

7. Consider the third element a13 in the first row. Attach the sign + (plus)

8. Delete the row and column in which a13 appears; that is the first row and

third column. We get 3231

2221

aa

aa

9. Multiply + a13 and the value of 3231

2221

aa

aa

Add the values got in steps 3, 6 and 9. This is the value of the given

determinant.

Note: We usually denote a determinant by the symbol (read as Delta).

Example: Evaluate the determinant

402

741

320

Solution:

02

413

42

712

40

740

= 0 – 2[1(4) – 2(7)] + 3[1(0) – 2(4)]

= 0 – 2 (4 – 14) + 3(0 – 8)

= –2(–10) – 24

= 20 – 24

= –4

Thus 4



Evaluation of a determinant in term of any row or column

Recall the definition 10.9(c). we obtained

3231

222113

3331

232112

3332

232211

aa

aaa

aa

aaa

aa

aaaA

The 2 – order determinants 3231

2221

3331

2321

3332

2322

aa

aa,

aa

aa,

aa

aa are called

the minors of a11, a12, a13. We can denote the minors by M11, M12, and M13.

If we attach the signs, these are called cofactors of a11, a12, a13. We denote

them by A11, A12, A13. Then (10.7) can be written as

| A | = a11A11, a12A12, a13A13.

We can also define minors of a21, a22, a23, a31, a32, a33 in a similar manner.

For example.

32a31

121123

a

aaM

(M23 is got by deleting the second row and third column of | A |)

The cofactors can be defined in a similar manner using the rule of signs

given by (10.8)

………………………….. (10.8)

Any cofactor is got by multiplying the minor and its sign given in (4.8). For

example cofactor for a32 is – M32.

So 2321

131132

aa

aaA

Thus we can expand in term of any row or column in a similar way. (10.9)

gives the expansion in term of various rows and columns.



333323231313

323222221212

313121211111

333332323131

232322222121

131312121111

AaAaAa

AaAaAa

AaAaAa

AaAaAa

AaAaAa

AaAaAa

…………………………………. (10.9)

You may wonder why so many expansions given in (10.9) are necessary. If

a row or column has many zeros then evaluating by the elements of that row

or column makes the evaluation simpler.

Example:: Evaluate

541

004

421

Solution: As the second row has two zeros, we expand by the elements of

the second row.

232322222121 AaAaAa

232221 A0A0A4

54

4214

(Note: For (2, 1) position, the sign is –. See (4.8)).

= – 4[2(5) – 4(4)]

= – 4(10 – 16)

= – 4(–6)

= 240

The evaluation of a determinant of order n is similar.

For example, if

44434241

34333231

24232221

14131211

aaaa

aaaa

aaaa

aaaa

A

Then | A | = a11A11 + a12A12 + a13A13 + a14A14



The signs of cofactors can be defined by

………………………… (10.10)

Example: Evaluate

0321

1004

4842

2741

Solution As the third row has two zeros we expand by the elements of third

row. The signs of cofactors of A determined by using (10.10)

| A | = a31A31 + a32A32 + a33A33 + a34A34

= 4A31 + 0A32 + 0A33 + 1A34

321

842

741

11

032

484

274

14

21

427

31

824

32

8411

32

842

02

447

03

484

= 4[4(0 – 12) – 7 (0 – 8) + 2(12 – 16)] – 1[1 (12 – 16) – 4(6 – 8) + 7(4 – 4)]

= 4[4(–12) –7 (–8) + 2 (–4)] –1[1 (–4) –4 (–2) + 7(0)]

= 4 (– 48 + 56 – 8) –1 (–4 + 8)

= 4(–56 + 56) –1 (4)

= 4(0) – 4

= – 4

Thus | A | = –4



S.A.Q. 12: Evaluate the following determinants

a)

210

213

121

b)

011

101

110

c)

612

347

465

S.A.Q. 13: Evaluate the following determinants

a)

2100

4111

2130

1210

b)

1111

0110

1010

1100

c)

6127

3474

4651

0001

10.6 Properties of Determinants

In this section we list some properties of determinants. These properties

enable us to evaluate a determinant in an easier way.

Property 1: If

333231

232221

131211

aaa

aaa

aaa

then

333231

232221

131211

aaa

aaa

mamama

m

Let A11, A12, a13 denote the cofactors of a11, a12, a13 in . These are also t he

cofactors of ma11, ma12, ma13, in m .

So

333231

232221

131211

aaa

aaa

mamama

= ma11 A11 + ma12A12 + ma13A13

= m (a11A11 + a12A12 + a13A13)

= m .

Note: This property holds good when any row or column of is multiplied by

m. This property essentially means that any common factor of a row or

column can be taken outside the determinant.



Remark: If A is a matrix then mA is got by multiplying each entry of A by m.

In the case of determinant m is got by multiplying the entries of a single

row or column by m.

Example:

531

142

532

6

531

142

656362

531

142

301812

Property 2: If |A| = det (a), then det (AT) = |A|

Proof: If dc

baB then

db

caBT

So | B | = ad – bc = |BT|

So the second order minors of A and AT have the same value. As the sign of

a cofactor is the same in both A and AT, the value det (AT) through

expanding along the first column is equal to det (A) through expanding along

the first row.

Hence dt (AT) = |A|

Property 3: If two rows or columns of a determinant are interchanged

then the value of the is unchanged but the sign is changed.

Proof: Let

333231

232221

131211

aaa

aaa

aaa

A If the second and third rows are

interchanged we get

232221

333231

131211

aaa

aaa

aaa

B

2221

323113

2321

333112

2322

333211

aa

aaa

aa

aaa

aa

aaaB



= a11 (a32a23 – a22a33) – a12 (a31a23 – a21a33) + a13 (a31a22 – a32a21)

223132211323313321122332332211 aaaaaaaaaaaaaaa

3231

222113

3331

232112

3332

232211

aa

aaa

aa

aaa

aa

aaa

|A|

Property 4: If a determinant has two identical rows then the value of the

determinant is 0.

Proof: If we interchange two identical rows then the value of the new

determinant is . (By property 3). But the new determinant is the same as

. So – = or = 0.

Example: Evaluate

9741

9789

7452

9741

Solution: The first and fourth rows of are identical. By property 4, = 0.

Property: The value of a determinant remains the same when multiple of

some rows are added to a particular row. The same is true for columns.

Note: For example,

ihg

fed

likfclhkeblgkda

ihg

fed

cba

Since we add k times the second row and l times the third row to the first

row.

Example: Evaluate

521

642

1074



Solution The first entry in the first row (R1) is 4. To make it 0, we subtract 4

times the third row (R3). Thus R1 of is replaced by R1 – 4R3 (This is

indicated on the right of the determinant). Similarly if we subtract 2R3 from

R2 we get 0, as the first element in R2. In the resulting determinant the first

column has two zeros and a one. This makes the evaluation (along the first

column) easier.

521

642

1074

3

32

31

R

R2R

R4R

521

526224122

5410247144

521

400

1010

40

1011A0A0 2111 (by expanding along the first column)

= 0 + 0 +1 (4 – 0)

= 4

Example: Evaluate

c1111

1b111

11a11

1111

Solution

c1111

1b111

11a11

1111



14

13

12

1

RR

RR

RR

R

c000

0b00

00a0

1111

413121 A0A0A0

c00

0b0

00a

1

00

b00

c0

000

c0

0ba

= a (bc – 0)

= abc.

Example: Show that

accbba

baaccb

baaccb

111

222222

Solution 222222 baaccb

baaccb

111

13

12

1

222222 CC

CC

C

baaccb

cabacb

111

13122222 A0A0caba

caba1

2222 bacacaba

= (a – b) (a – c) (a + c) – (a – c) (a – b) (a + b)

= (a – b) (a – c) [a + c – a – b]

= (a – b) (a – c) (c – b)

= (a – b) [– (c – a)] [–(b – c)]

= (a – b) (b – c) (c – a)



S.A.Q. 14 Evaluate 1

2

2

cc1

bb1

aa1

S.A.Q. 15 Prove that 0

0cb

c0a

ba0

S.A.Q. 16 Evaluate

14x4x3x

13x3x2x

12x2x1x

S.A.Q. 17 Evaluate the following determinants

a)

7654

6543

5432

4321

b)

0111

1011

1101

1110

10.7 The Inverse of a Matrix

In this section we give a method of finding the inverse of a matrix.

Definition: If

333231

232221

131211

aaa

aaa

aaa

A then the adjoint matrix of A (denoted by

Adj A) is given by Adj

T

333231

232221

131211

AAA

AAA

AAA

A

Example 10.17 Find the adjoint of matrix

543

432

321

A



Solution

1445354

431A11

21210153

421A12

19843

321A12

21210154

321A21

49553

311A22

264143

211A23

19843

321A31

264142

311A32

14332

21133A

T

333231

232221

131211

AAA

AAA

AAA

AAdj

T

121

242

121



121

242

121

We use Adj A for evaluating the inverse of a matrix.

A square matrix is invertible if and only if .0A When ,,0A the

inverse of a matrix A is given by

AAdjA

1A 1 ………………………. (10.11)

If |A| = 0, the matrix a is called singular; otherwise it is non singular. So a

matrix is invertible if and only if it is nonsingular.

Example: Find the inverse of dc

ba

Solution A11 = d, A12 = –c, A21 = –b, A22 = a.

So Adj ac

bd

ab

cdA

T

|A| = ad – bc. Hence

ac

bd

bcad

1A 1 ………. (10.12)

Example: Find the inverse of

112

111

111

A

Solution

;321112

111A;0

11

111A 1211

;211111

111A;3

12

111A 2113



;121112

111A;3

12

111A 2322

;011111

111A;211

11

111A 3231

21111

111A33

Adj

213

033

220

202

132

330

A

T

|A| = a11A11 + a12A12 + a13A13

= 1 (0) = 1(3) + 1(3)

= 6

13

6

1

2

1

02

1

2

13

1

3

10

213

033

220

6

1

|A|

AadjA 1

S.A.Q.18 Find the inverse of

113

111

132

A

S.A.Q. 19 Test whether A-1 exists when

321

642

101

A

S.A.Q. 20 Find the inverse of

431

341

331

A



10.8 Solution of Equations using Matrices and Determinants

Matrices are useful in representation of data. For example if we want to

classify the students of a class in terms of gender and their grades then we

can use a matrix for representing the information. Suppose we have three

grades A, B, C. Then the matrix 232221

131211

aaa

aaa represents.

The classified data:

a11 denotes the number of male students who got grade A.

a12 denotes the number of male students who got grade B

a13 denotes the number of male students who got grade C.

a21 denotes the number of female students who got grade A.

a22 denotes the number of female students who got grade B.

a23 denotes the number of female students who got grade C.

Solving linear equations using matrices

We can also use matrices for solving n equations in n variables. The ideas

is to represent n equations as a single matrix equation and then solve the

matrix equation. The next example illustrates this.

Example: Solve x + y = 3

2x + 3y = 8

Solution The given system of equations is equivalent to the single matrix

equation AX = B where

8

3B

y

xX

32

11A

Multiplying both side of AX = B by A–1. we get X = A–1 B. By (10.12)

12

13

12

13

1231

1A 1

2

1

8132

8133

8

3

12

13X

Hence x = 1, y = 2



Example: Solve the equations

x + y + z = 6

x + 2y + 3z = 14

–x + y – z = –2

Solution: The given systems of equations is equivalent to A X = B where

2

14

6

B

z

y

x

x

111

321

111

A

53211

321A11

231111

311A12

;32111

211A13

;211111

111A21

;01111

111A22

;211111

111A23

;12332

111A31

;213131

111A32

11221

111A33

|A|=a11A11 + a12A12 + a13A13

= 1(–5) + 1(–2) + 1(3)



= –5 – 2 + 3

= –7 + 3

= –4.

T

1

121

202

325

4

1

|A|

AAdjA

123

202

125

4

1A 1

2

14

6

123

202

125

4

1BAX 1

2114263

2214062

2114265

4

1

22818

4012

22830

12

8

4

4

1

3

2

1

Hence x = 1, y = 2, z = 3



10.9 Solving equations using determinants

We can also solve a system of n linear equations in n variables using

determinants. The method is provided by Cramer’s rule. Cramer’s rule for

three equations in three variables.

Consider the system of three linear equation in three variables x, y, z.

a11x + a12y + a13z = b1

a21x + a22y + a23z = b2

a31x + a32y + a33z = b3

Let be the co-efficient determinant i.e., the determinant of the coefficients

of the variables x, y, z such that 0

aaa

aaa

aaa

333231

232221

131211

(If = 0 the system has no unique solution).

By Cramer’s rule we have

33231

22221

11211

33331

23221

131111

33323

23222

13121

baa

baa

baa

zaba

aba

aba

yaab

aab

aab

x

or

1x 2y 3z

Note: As in the previous section the system of equations Ax = B. Then |A| =

. Now 1 is obtained by replacing the first column of by B. Similarly 2

and 3 are obtained by replacing the second and third columns of by B

respectively.

Example: Solve

2x + 3y + 4z = 20

x + y + 2z = 9

3x + 2y + z = 10



Solution: In this example,

1023

911

2032

1103

291

4202

1210

219

4320

123

211

432

321

= 2(1 – 4) – 3 (1 – 6) + 4(2 – 3) = 5

1 = 20(1 – 4) – 3(9 – 20) + 4 (18 – 10) = 5

2 = 2(9 – 20) – 20(1 – 6) + 4 (10 – 27) = 10

3 = 2(10 – 18) – 3 (10 – 27) + 20(2 – 3) = 15

Hence

.3z,2y,1x 321

S.A.Q. 21. Solve the following system of equations using matrices

a) 2x + 3y – z = 9 b) 2x – y + 3z = – 9

x + y + z = 9 x + y + z = 6

3x – y – z = –1 x – y + z = 2

S.A.Q. 22 Solve the following system of equations using Cramer’s rule

a) 5x – 6y + 4z = 15 b) x + y + z = 9

7x + 4y – 3z = 19 2x + 5y + 7z = 52

2x + y + 6z = 46 2x + y – z = 0

10.10 Summary

In this unit we discuss about the concept of matrices and determinants. The

different types matrices is defined, the concept of inverse of matrix is well

defined with good examples, Determinants, the different properties of

determinants and solving equations using matrices and determinants is

explained with the help of standard examples.



10.11 Terminal questions

1. Find the values of x, y, z and t satisfy the matrix relationship

201x25

5t241

1t35x45y2

2t4z3x

2. Find the values for x, y, z that satisfy the matrix relationship

3zy

2x4

z21

62

zy

x23

3. Find a matrix A satisfying

15

43A2

15

23

4. If 111

321B,

011

021A and

11

11

21

C

Show that AB = AC. (In the case of real numbers, ab = ac will imply that

b = c. But this is not so for matrices as this example shows)

5. If

012

130

201

A and

120

302

013

B evaluate AB – BA.

6. If 63

42A , show that AA 42

7. If 13

21A , show that A2 – 2A – 5I = 0

8. If 13

21A find AAT and ATA

9. If 34

21A Evaluate A2 and A3.

10. If 12

64A

31

52 find A.



11. If 23

12

47

35A find A

12. 13

42

35

23A

23

12If , find A.

13. Evaluate the following determinants

a)

25169

1694

941

b)

cfg

fbh

gha

c)

abc

cab

bca


a)

3232

3102

0121

3242

b)

0521

1130

2421

3112

c)

2032

1423

3212

2124

15. Show that accbba

1baab

1acca

1cbbc


a)

201041

10631

4321

1111

b)

351551

201041

10631

4321

17. Show that 0

bac1

acb1

cba1

18. Prove that 0

baabba

accaac

cbbccb

22

22

22



19. Solve the following system of equations using (i) matrices (ii) determinants

a) x + 2y – z = 3 b) 2x + 3y – z = 9

3x – y + 2z = 2 x + y + z = 9

2x – 2y + 3z = 2 3x – y – z = –1

c) a + b + z = 6 d) 2a + 3b + c = 8

a + 2b + 3c = 14 4a + b + c = 6

–a + b – z = –2 a + b + c = 3

10.12 Answers


1. mn entries

2.

101

1002

300

3. 242

701,

242

701

620

64

84

,

415

144

21

,

25

88

63

5. 11

12

6. 22

21Y,

21

02X

8. BA does not exist since the number of columns of B = 2 3 = the

number of rows of A.

12. a) – 9 b) 2 c) 419

13. a) – 9 b) 2 c) 419

(Expand using the first column, first column and first row respectively.)

14. By row operations R2 – R1 and R3 – R1, 22

22

acac

abcb.

So = (a – b) (b – c) (c – a) on simplification.



16. The row operations are R2 – R1 and R3 – R2. Answer is –2.

17. a) Answer 0; the row operations are R1, R, – R1, R3 – R2, R4. b) –3.

The row operations R1, R2, R3 – R2, R4, – R3 reduce to

.

110

011

111

1 Apply R1, R2 – R1, R3.

18.

1114

314

440

16

1

19. As |A| = 0, A–1 does not exist.

20.

101

011

337

21. a) x = 2, y = 3, z = 4 b) x = 1, y = 2, z = 3

22. a) x = 3, y = 4, z = 6 b) x = 1, y = 3, z = 5

Terminal Questions

1. x = –2, y = –5, z = –8, t = –7

2. x = 4, y = 1, z = 3

3. 10

33

5.

156

1412

980

8. 51

110,

101

15

9. 178

49A2 ,

6760

307A3



10. 80

232

12

64

31

52A

1

(Use (10.7))

11. 12

11

47

35

23

12A

1

(Use (10.7))

12. Let 13

42D,

35

23C,

23

12B

Then BAC = D. So A = B–1 DC–1, Using (10.7),

1834

1324A,

35

23c,

23

12B 11

13. a) –8 b) abc + 2fgh – af2 – bg2 – ch2 c) a3 + b3 + c3 – 3abc

14. a) – 15 (Row operations: R1, R2, R3 –R1, R4 – R1)

b) 102 (Row operations: R1 + 2R4, R2 + R4, R3, R4

c) 87 (column operations: C1 + 4C3, C2 + 2C3, C3, C4 + 2C3).

15. Apply row operations R1, R2 – R1, R3 – R1. Expand using last column.

16. a) and b) answer 1 (Apply row operations R1, R2 – R1, R3 – R1, R4 – R1)

17. Apply R1, R2 – R1, R3 – R1, we get .caac

baab This is equal to (b –

a) (a – c) – (a – c) (c – a) – (b – a) (a – c) – (–1) (–1) (b – a) (a – c) = 0

18. Multiplying R1 by a, R2 by b and R3 by c and then dividing by abc we

get

bacabccba

acbabcbca

cbaabccab

abc

1

22

22

22

bcca1ab

abbc1ca

acab1bc

abc

cba 222

(By taking out abc from columns 1 and 2)



13 CC

abbcca1ab

caabbc1ca

bcacab1bc

abc

11ab

11ca

11bc

cabcababc (by taking out ab + bc + ca from C3)

19. a) x = –1, y = 4, z = 4 b) x = 2, y = 3, z = 4

c) a = 1, b = 2, c = 3 c) a = 1, b = 2, c = 0



Unit 11 Infinite Series

Structure

11.1 Introduction

Objectives

11.2 Convergence and Divergence

11.3 Series of Positive Terms

11.4 Binomial Series

11.5 Exponential Series

11.6 Logarithmic Series


11.7 Summary


11.9 Answers

11.1 Introduction

Infinite series: If un is a real sequence, then an expression of the form

u1 + u2 + u3 + …. + …………. …………. + un + ………….

Which can be written also as 1

n

1n

n uororu is called an INFINITE

SERIES.

Example, 1) ......4

13

12

11

2) 1 + 2 + 3 + 4 + ………

Objectives:


understand the properties of infinite series

test the convergence or the divergence of an infinite series



Partial Sum

The expression u1 + u2 + u3 + … + …… ….. + un + ….. involves addition of

infinitely many term. To give meaning to this expression we define its

sequence of partial sums „Sn‟ by

Sn = u1 + u2 + u3 + … + …………… + un

We know that an infinite series is given by

1n

n321 ...........u.........uuu

n321n3213212,11 u.......uuuS,.......uuuS,uuSuS

are called partial sums.

S1 is called the 1st partial sum, S2 is called the 2nd partial sum, ……. Sn is the

nth partial sum.

The sequence (Sn) is called the sequence of partial sums, then we say

nu converges, diverges, Oscillates according as its sequence of partial

sums Sn, converges, diverges or oscillates.

Examples

1. The expression 1+(–1) + 1 + (–1) + ….. + (–1)n+1 + …… (i)

Or as it is usually written as 1 – 1 + 1 – 1 + 1 – 1 + ….. is a series. The

meaning of expression (i) is that from the terms 1, –1, +1, –1, ….. (–

1)n+1, ….. we form the partial sums,

S1 = 1, S2 = 1 – 1 = 0, S3 = 1 – 1 + 1 = 1, …….

S1 = 1 – 1 + ……. + (–1)n+1 = 2

1n11

……………… (ii)

2. The expression ........2

1.........

8

1

4

1

2

11

1n

is a series. This

series can also be written as, ...,.........2

1,.........

8

1,

4

1,

2

1,1

1n

the partial sums,



,.......2

12S,......

431S,

211S,1S

1n

n321

General properties of Series

Following are the fundamental rules or properties of a series:

1. The coverage or divergence of an infinite series remains unaffected by

the addition or removal of a finite number of the terms; for the sum of

these terms being the finite quantity addition or removal doesnot change

the nature of its sum.

2. If a series in which all the terms are positive is convergent, the series

remain convergent even when some or all of its terms are negative; for

the sum is clearly the greatest when all the terms are positive.

3. The convergence or divergence of an infinite series remains unaffected

by multiplying each term by a finite number.

11.2 Convergence and divergence

Convergence of the infinite series

A series is called convergent if the sequence of its partial sums has a finite

limit, this limit is termed as the sum of the convergent series.

An infinite series 1n

nu is said to be convergent if nlim Sn = l where l is a

unique real number.

Examples

1) Show that 1n

2n21n......

2

1......

2

1

2

11

2

1 is convergent

series.



a2

2

11

2

11

limSlimn

nn

nunique real number.

The given series is a convergent series.

Divergence of the infinite series

If a sequence of its partial sums has no finite limit, then the series is called

divergent. A divergent series has no sum.

An infinite series 1n

nu is said to be divergent.

If .Slim nn

Example

Show that .......................n.................4321u1n

n is a

divergent

Since 2

1nnlimSlimn

nn

and hence the given series is divergent.

Necessary condition for convergence of a series

The series,

u1 + u2 + u3 + ….+ ……….. + un + ….. (1)

can converge only when the term un (the general term of the series) tends

to zero.

i.e nlim un = 0

if the general term un does not tend to zero, then the series diverges.



Examples:

a) The series, 0.0 + 0.44 + 0.444 + 0.4444 + ….. diverges because the

general term un does not tend to zero.

b) The series 1 – 1 + 1 – 1 + ……….. diverges because the general term un

does not tend to zero (and has no limit at all).

The remainder of a series

Let us consider and infinite series

u1 + u2 + u3 + ….+ ……. +um + um+1 + um+2 + ….. ----------- (I)

If we discard first m terms of a series, we get the series,

um+1 + um+2 + ……….. ----------- (II)

which converges (or diverges) if the series (I) converges (or diverges) and

diverges if the series (I) diverges. Therefore, while finding the convergence

of a series we can distinguish between a few terms.

When the series (I) converges, the sum Rm = um+1 + um+2 + …… Of series

(II) is called the remainder or (remainder term) of the first series. (R1 = u2 +

u3 + … + ……… ….. is the first remainder. R2 = u3 + u4 + …. + ……. is the

second, etc.) the remainder Rm is the error committed by substituting the

partial sum Sn (or the sum S of the series (I)). The sum S of the series and

the remainder Rm are connected by

S = Sm + Rm.

As m the reminder term of the series approaches to zero. It is of

practical importance that this approach be “sufficiently rapid”, that is, that the

remainder Rm should become less than the permissible error, for m not too

great. Then we say that the series (I) converges rapidly, otherwise that

series is said to converge slowly.

For example consider the series ..........4

1

3

1

2

11 converges very

slowly. Summing the first 20 terms, we get the value of the sum of the series



only to within 0.5 10–1; to attain the accuracy up to 0.5 10–4; we have to

take at leat 19,999 terms.

Examples

1. ..............4.3

1

3.2

1

2.1

1

Solution:

1nn

1................

4.3

1

3.2

1

2.1

1Sn

Here 1n

1

n

1

1nn

1un

1n

1

n

1u...........,

4

1

3

1u,

3

1

2

1u,

2

1

1

1u n321

1n

1

n

1......

4

1

3

1

3

1

2

1

2

11Sn

1n

11 all the other terms cancel]

11n

11limSlim nnn which is a unique finite quantity.

The given series is convergent.

2. Show that the given series divergent. .....1111111n

n

Solution: Sn = –1+1 – 1+1 – 1+1 …….. to n terms

= 0 or –1 according as n is even or odd.

1or0Slim nn

The given series oscillates between 2 finite values 0 and –1.

3. Test for divergence of the following series: 12 + 22 + 32 + …… +

Solution: 6

1n21nnn.......321S 2222

n



6

1n21nnlimSlim nnn

given series diverges to + .

4. Test for divergence the following series: 1 + 2 + 3 + ….. + n + ……..

Solution: 2

1nnn.............321Sn

1nnlim2

1

2

1nnlimSlim nnnn

given series in divergent.

5. Show that the series 1 + r + r2 + r3 + ………. +

(i) converges if | r | < 1

(ii) diverges if r > 1 and

(iii) oscillates if r < 1

Solution: Let Sn = 1 + r + r2 + r3 + ………. + rn – 1

Case (i), when | r | < 1, ,0lim nn r since it is a G.P. series,

r1

r

r1

1

r1

r1S

nn

n

r1

1Slim nn

given series is convergent.

Case (ii), when r = 1,

Sn = 1+ 1 + 1 + 1 + ……. + 1 = n

And, nn Slim

given series is divergent.

Case (iii), when r = –1, the series becomes

Sn = 1 – 1 + 1 – 1 + …….

Which is an oscillatory series.

(ii) when r < –1, let r = –p, so that p > 1,



then nnn p1r

and p1

p11

r1

r1S

nnn

n

as nn Plim

orSlim nn , from this n is even or odd.

Hence the series oscillates.

11.3 Series of Positive terms

If the terms of a series of nu are positive, then its sequence of partial

sums

n321n u........uuuS is monotonically increasing for

0uSS 1nn1n for all the values of n.

un of positive terms converges or diverges to according as Sn is

bounded or unbounded. If Sn is bounded then 1n1n uSS for all n, gives

.0uSlim 1n

Theorem1: A positive term series either converges to a positive number or

diverges to , according as its sequence of partial sums is bounded or not.

Proof:

Let n321nn a..........aaaSa

1nn3211n aa.......aaaS

0aSS 1nn1n

nS is monotonically increasing.

According to nS we have the following 2 possibilities.

(i) nS is bounded, or



(ii) nS is unbounded above

(i) If nS is bounded, then nS is bounded above. Hence nS is a

monotonically increasing sequence which is bounded above.

nS is convergent.

aSlim nn unique real number.

na is convergent.

(ii) If nS is unbounded above

The nS is a monotonically increasing sequence which is unbounded

above

nS diverges to +

nn Slim

na diverges to +

Therefore na converges to diverges to

Theorem 2: Necessary constant for the convergence of a series of a

positive terms. If a an series is convergent then nlim an = 0. The

converse is not true.

Theorem 3: The nature of the series is not altered by the multiplication of all

the terms of the series by the same non-zero constant C.

Theorem 4: The nature of the series is not altered by addition of a finite

number of terms to the series or by removing a finite number of terms from

the beginning.

Theorem 5: If na and nb are 2 series which converge to l and m

respectively then the series nn ba converges to l m.



11.4 Binomial Series

According to the Binomial theorem, we have,

n32nx.......x

!3

2n1nnx

!2

1nnx

!1

n1x1

If the right hand side is extended to ,

to.......x!3

2n1nnx

!2

1nnx

!1

n1x1 32n

Become a finite series and this series is called as Binomial series.

The Binomial series is absolutely convergent if | x | < 1, and when the series

is convergent, the sum of the finite series is (1 + x)n.

Replacing x by – x,

to.......x!3

2n1nnx

!2

1nnx

!1

n1x1 32n

While finding the sum of the Binomial series we can use some of the

following cases.

1. When n = –1,

132 x1..........xxx1

2. When n = –1, and x is changed to –x,

1 + x + x2 + x3 + …….. = (1 + x)–1

3. When n = –2,

1 – 2x + 3x2 – 4x3 + ………….. = (1 + x)–2

4. When n = –2, and x is changed to –x,

1 + 2x + 3x2 + 4x3 + ………….. = (1 + x)–2

5. When q

pn where p and q are integers and q 0, we get

to.........q

x

!3

q2pqpp

q

x

!2

qpp

q

x

!1

p1x1

32

q

p



6. When q

pn , x is replaced by –x, we get,

to...........q

x

!3

q2pqpp

q

x

!2

qpp

q

x

!1

px1

32

q

p

7. When ,q

pn we get,

to.........q

x

!3

q2pqpp

q

x

!2

qpp

q

x

!1

p1x1

32

q

p

8. When ,q

pn x is replaced by – x, we get,

to.........q

x

!3

q2pqpp

q

x

!2

qpp

q

x

!1

p1x1

32

q

p

Examples

1. Solve the following: .........150.100.50

33.18.3

100.50

18.3

50

3

Solution: Let ..........150.100.50

33.18.3

100.50

18.3

50

3S

S can be written as,

........50

1

!3

33.18.3

50

1

!.2

18.3

50!.1

1.3S

32

q

p

x11S

10

3

50

15x

50

1

q

x,15q,3P

5

1

5

1

15

3

7

10

10

7

10

311S



17

10S

5

1

2. Solve: ....................3

1

3.2.1

7.5.3

3

1.

2.1

5.3

1

32

Solution: Comparing the given series with one of the general Binomial

series, we get

3

1

q

x,2q,3p

3

2x

But the power of 3

1 is not equal to the number factors. Hence we have

to multiply and divide by 3,

Let ...............3

1

3.2.1

7.5.3

3

1.

2.1

5.3

1

3S

2

....................3

1

!3

7.5.3

3

1.

!2

5.3

3

1

!1

33S

32

.................3

1

!3

7.5.3

3

1.

!2

5.3

3

1

!1

3113

32

.to................3

1

!3

7.5.3

3

1.

!2

5.3

3

1

!1

3133

32

q

p

x133

2

3

3

2133



2

3

3

133

2

3

333

25

33S

11.5 Exponential series

The exponential function ex expressed as an infinite series in the form

......!n

x...........

!3

x

!2

x

!1

x1e

n32x

This series is convergent for all values of x.

In finding the sum of the exponential series, the following are to be used.

I. 0n

32nx ...........

!3

x

!2

x

!1

x1

!n

xe

II. Putting x = 1, in form (I) we get,

0n

n

...........!3

1

!2

1

!1

11

!n

xe

III. By changing x to – x, in form (I) we get,

0n

32nx ...........

!3

x

!2

x

!1

x1

!n

xe

IV. Putting x = –1 in form (I), it gives,

0n

n1 ...........

!3

1

!2

1

!1

11

!n

1e

V. By adding (I) and (III), results in,

0n

n242xx

!n2

x...............

!4

x

!2

x1

2

ee



VI. Subtracting (III) from (I),

0n

1n253xx

!1n2

x...............

!5

x

!3

x

!1

x

2

ee

VII. Putting x =1 in (V),

0n

11

!n2

1.......

!4

1

!2

11

2

ee

VIII. Putting x = 1 in (VI),

0n

11

!1n2

1....................

!5

1

!3

1

!1

1

2

ee

Note: In exponential series it should be carefully observed whether the

summation is from 0 to or 1 to or 2 to etc.

Examples

Problem 1. to..........!7

6

!5

4

!3

2

The solution is:

1n

!1n2

n2S

1n

!1n2

11n2

1n

!1n2

1

!1n2

11n2

1n

!1n2

1

!1n2

1

1n 1n!1n2

1

!1n2

1



......!7

1

!5

1

!3

1.......

!6

1

!4

1

!2

1

12

ee1

2

ee 11

1e

or e

1s

Problem 2. .........!4

7.6.3

!3

5.4.2

!2

3.2.1

Solution: The given series can be written as,

1n

23

1n!1n

n2n4

!1n

1n2n2nS

Consider 1nn1ndn1nc1nban2n4 23

dndncncnbbna 32

bandcbcndn 23

.2b042b0dcb,2c,4d

2aor02a0ba

1nn1n4n1n21n22n2n4 23

1n1n

23

!1n

1nn1n4n1n21n22

!1n

n2n4

1n 1n1n1n

!2n

14

!1n

12

!n

12

!1n

12

e4e21e2!1

11e2

S = 6e + 2



11.6 Logarithmic Series

The series:

......n

x1......

4

x

3

x

2

x1

n1n

432

is called the logarithmic series and is denoted by 1n

n1n

n

x1

Theorem: If un is a positive term series and xu

ulognlim

1n

nn (finite

or infinite).

Then the series 1n

nu

(i) Converges if x > 1

(ii) Diverges if x < 1

(iii) May converge or diverge if x < 1

The logarithmic series convergent if –1 < x < 1. When it is convergent the

sum of the logarithmic series is given by,

1. x1log.................4

x

3

x

2

xx e

432

2. Replacing x by – x, in (1) we get,

x1log...........................4

x

3

x

2

xx e

432

x1log...........................4

x

3

x

2

xx e

432

3. Adding (1) and (2) we get,

x1logx1log..................5

x

3

xx2 ee

53

x1

x1log..............

5

x

3

xx2 e

53



4. Subtracting (2) from (1) we get,

x1logx1log...............6

x

4

x

2

x2 ee

642

Therefore, 2e

642

x1log.................5

x

4

x

2

x2 where x2 < 1

5. Putting x = 1 in (1) it gives,

2log...........4

1

3

1

2

11 e

Note: Usually the sum of the logarithmic series is found by resolving the nth

term into partial fractions.

Examples

1. Solve to...........7

x

5

x

3

x1

32

Solution:

Let to.......7

x

5

x

3

x1S

32

to......7

x

5

x

3

x1

642

to.......7

x

5

x

3

xx

x

1753

x1

x1log

2

1

x

1e

x1

x1log

x2

1S e

2. ........7.6.5

1

5.4.3

1

3.2.1

1



Solution:

Let .........7.6.5

1

5.4.3

1

3.2.1

1S

1n

1n2n21n2

1

Consider, 1n2

C

n2

B

1n2

A

1n2n21n2

1

therefore, 1 = A (2n) (2n + 1) + B(2n – 1) (2n + 1) + C(2n) (2n – 1)

Put n = 0, 1 = B(–1) B = –1

Put m = ½, 1 = 2A or A = 2

1 .

Put n = –1/2, then C = (–1) (–2) = 2 or 2C = 1.

Substituting these vales in 1n

,1n2n21n2

1 we get,

1n1n

1n2n2

1

n2

1

1n22

1

1n2n21n2

1

1n1n

1n2

1

n2

1

2

1

n2

1

1n2

1

2

1

Splitting n2

1 as

n2

1

2

1

n2

1

2

1

......5

1

4

1

3

1

2

1

2

1........

4

1

3

1

2

1

1

1

2

1S

2log12

12log

2

1

2

12log


1. Solve ......1296

412

72

12

6

21



2. Show that harmonic series of order ......3

1

2

1

1

1

n

1p

pppp

converges for p > 1 and diverges for 1p

11.7 Summary

In this unit initially we discussed about the partial sum and general

properties of series. Then we studied different rules for convergence or

divergence of series. Lastly in this unit we studied binomial series,

exponential series, logarithmic series with properly illustrated examples.


1. Write the general properties of a series

2. Explain the binomial series

11.9 Answers


1. Let ............1296

412

72

12

6

21S

S can be written as

..........6

1

!3

412

6

1

!2

12

6

1

!1

21S

32

Comparing with the expansion of ,1/ qp

x

6

1

q

x,3q,2p

2

1x

q/p

x1S



3

32

32

4

1

2

1

2

11

3 4

1S

2. But the above test, this series will converge or diverge according as

1px

dx is finite. If

p1

1mlim

x

dxlim

x

dx,1p

p1

n

m

1pn

1p

,p1

1 i.e. finite for p > 1

for p < 1

If

11p

xlogx

dx,1p

Therefore the series converges for p > 1 and p < 1.

Terminal Questions

1. Refer to Section 11.1.3




Unit 12 Probability

Structure

12.1 Introduction

Objectives

12.2 Concept of Probability

12.3 Sample Space and Events

12.4 Three Approaches to Probability

12.5 Kolmogorov’s Axiomatic Approach to Probability

12.6 Conditional Probability and Independence of Events

12.7 Baye’s theorem

12.8 Summary


12.10 Answers

12.1 Introductions

Even in day – to – day life uncertainty plays an important role. When we are

unable to forecast the future with certainty, we make statements like

“probably it will rain in the evening”, “Ram has a better chance of winning

the elections” etc. Although we are not sure of the happening of some event

we make statements like those mentioned above.

It is interesting to note that the seed of probability theory was thrown when a

French nobleman Anokine Gombould (1607 – 1684) sought an explanation

from the mathematician Blaise Pascal (1623 – 1662) regarding the frequent

occurrence of some combinations of number in the roll of dice. Pierre de

Fermat (1601 – 1655) and Blaise Pascal were working on this problem.

Another problem was posed to Blaise Pascal. If a game of change is

stopped in the middle, how should the two players divide the stake ? This

was another problem leading to the concept of probability, J. Bernoullis



(1654 – 1705) first treatise on probability was published in 1718. Other

mathematicians who were instrumental in the development of probability

were chebyshev (1821 – 1894), A. Markov (1856 – 1920), Liapounoff (who

enunciated central limit theorem), De Moivre, T. Bayes and P.S. Laplace.

Initially ideas of probability and statistics were used to explain natural

phenomena. Now it is an indispensable tool in many decisions regarding

business also.

Objectives:


understand the idea of Probability

apply Baye’s theorem in problems

12.2 Concept of Probability

Consider the following statements.

1. A particular medicine is effective except for one out of 1000 patients.

2. There is likely to be moderate to heavy rains in most part of Karnataka

3. Getting a head or a tail in the toss of a coin are equally likely

4. When a single die is rolled, any number from 1 to 6 is equally likely

5. Only 3 out of 2 million parts is likely to be defective

In all the above statements the outcome is not certain. But we are able to list

all possible outcomes. For example, we are not certain about the number

likely to be seen in a roll of a single die but we know that only one of the six

numbers 1, 2, 3, 4, 5, 6 will definitely be seen. Statement 4 is about the

likelihood of something to happen. From statement 1 we are not able to say

to whom the medicine is likely to be ineffective but we can say that it is

ineffective only for one patient when the medicine is administered to 1000

patients. Let us consider statement 3. Although we cannot say whether we



get a tail or head, we can say that we will be getting head half the number of

times when a coin is tossed several times.

Before proceeding to study the rigorous definition of probability let us

understand that Probability is a numerical measure of the likelihood of an

event to happen.

It is a number between 0 and 1, 0 and 1 representing the impossibility and

certainty.

Figure 12.1 Probability of an event

For example, non occurrence of rain is more likely in summer and

occurrence of rain is more likely in rainy season. Head or tails are equally

likely in the toss of a coin.

12.3 Sample Space and Events

For defining probability we need the definition of an experiment. (to be more

precise random experiment).

Definition: An experiment is a process that generates well – defined

outcomes.

When we perform an experiment we call it a trial.

0 0.2 0.5 0.8 1

Impossible event Non

occurrence more likely

Occurrence and nonoccurrence

equal more likely

Occurrence more likely

Certain event or

deterministic event



For each trial there is one and only one outcome among the several well –

defined outcomes.

Example: Find all the possible outcomes of the following experiments

a) Tossing a single coin

b) Tossing two coins

c) Tossing three coins

d) Roll of a single die

e) Roll of two dice

f) Play a one day cricket match

Solution: The possible outcomes are

a) H, T (H and T denotes head and tail respectively)

b) HH, HT, TH, TT

c) HHH, HHT, THH, HTH, HTT, TTH, THT, TTT

d) 1, 2, 3, 4, 5, 6

e) 11, 12, 13, 14, 15, 16

21, 22, 23, 25, 25, 26

31, 32, 33, 34, 35, 36

41, 42, 43, 44, 45, 46

51, 52, 53, 54, 55, 56

61, 62, 63, 64, 65, 66

f) win, defeat, tie

Definition: The sample space for an experiment is the set of all possible

outcomes.

Example: Find the sample space for the following experiments.

a) Number of heads in a toss of two coins

b) Sum on the roll of two dice

c) Sum on the roll of three dice

d) Play a cricket game



Solution: the sample space is

a) {2, 1, 0}

b) {2, 3, 4, 5, 6, 7, 8, 10, 11, 12}

c) {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18}

d) {win, defeat, tie}

Definition: An event in an experiment is a subset of the sample space.

Definition: A single outcome is called an elementary event.

Note: Any event consists of elementary events.

Example: Write down the following events as a subset of the respective

sample space

a) Getting at least one head in a toss of two coins

b) Getting a sum of 6 or more in a roll of two dice

c) Getting two defective parts when five parts are inspected

d) India winning at least one in 3 matches played against Australia.

Solution:

a) {HT, TH, HH} {HT, TH, HH, TT}

b) {6, 7, 8, 9, 10, 11, 12} {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

c) {2} {0, 1, 2, 3, 4, 5}

d) {1, 2, 3} {0, 1, 2, 3}

S.A.Q. 1 Write the following events as a subset of the respective sample

space.

a) Getting even number of heads in a toss of 3 coins

b) Winning in alternate matches when 5 matches are played

c) Getting a sum of 4 or 10 in a roll of two dice

d) Winning at least one match when three matches are played



12.4 Three approaches to Probability

As the concept of probability was developed for more than two centuries,

the statisticians defined it in several ways. In 1933 the Russian

mathematician A. N. Kolmogorov developed probability theory using the

axiomatic approach which was used to define various mathematical objects

in the last century. The axiomatic approach unified all the three approaches

of probability. We study three approaches to probability which were

developed before Kolmogorov’s unified approach in this section.

Kolmogorov’s axiomatic approach is developed in the next section.

Classical Probability

The classical probability is also called mathematical, or a priori probability.

We need a few preliminary definitions before defining the classical

probability of an event.

Definition: The outcomes of an experiment are equally likely if there is no

reason to expect one outcome in preference to other outcomes.

For example, Head or Tail are equally likely in tossing a single coin. Any

number from 1 to 6 is equally likely in the roll of a die.

Definition: Two or more events are mutually exclusive. If Head falls then

Tail cannot fall and vice versa. The appearance of 1, 2, 3, 4, 5, 6 are

mutually exclusive. For when outcome of the appearance of 1 occurs, the

remaining cannot occur.

Definition: A collection of events is collectively exhaustive if they, when

taken together constitute the entire sample space.

For example, the events 1, 2, 3, 4, 5, 6 are collectively exhaustive. So also

Head and Tail in the toss of a single coin.



Definition: If the outcomes of an experiment are, equally likely, collectively

exhaustive and mutually exclusive then the probability of an event E is

defined by

experimenttheofoutcomesofnumberTotal

EofhappeningthetofavourableoutcomesofNumberEofyProbabilit

Note: The classical probability is called a priori probability since we are able

to calculate the probability of an event in advance (that is, without repeating

the experiment). Of course this definition is applicable only when the

outcomes of an experiment are mutually exclusive, collectively exhaustive

and equally likely. In the case of the toss of a coin, if we can assume the

validity of these three conditions then we say that the coin is unbiased.

Similarly we defined an unbiased die.

Example: Find the classical probability of the following events.

a) Getting at least one Head in a toss of two coins

b) Getting a sum of 10 in a roll of two dice.

Solution: Let E denote the given event.

a) HT, TH, HH are the 3 outcomes favorable to the event E and the total

number of outcomes is 4. So 75.04

3)E(P

b) (4,6), (5,5) and (6,4) are the 3 outcomes favorable to the event and the

total number of outcomes is 36. Hence 12

1

36

3)E(P

Note A classical way of monitoring possibility is as follows: If odds in favour

of E are x : y or x to y, then yx

x)E(P

Statistical or empirical Probability

In this approach we repeat the experiment a large number of times and

define the probability of an event.



Definition: If n trials are performed and m trials are favourable to the

occurrence of an event E, then the probability of the event E is defined by

Probability of n

mItE

n

Note: We assume that such a limit exists.

Example: A bag contains 3 red balls, 4 green balls and 5 blue balls. Find

the probability of choosing 2 red balls, 1 green ball and one blue ball.

Solution: Denote the required event by E. As there are 12 balls in all, the

total number of choosing 4 balls is C (12, 4). We can choose 2 red balls

from 3 red balls in C (3, 2) ways. One green ball can be chosen in C(4, 1)

ways and one blue ball can be chosen in C(5, 1) ways. By multiplication

principle, the number of outcomes favorable to E is C(3, 2) , C(4, 1) C(5, 1).

So

4,12C

1,5C1,4C2,3CEP

4.3.2.19.10.11.12

5.4.3

33

4

Example: There are 25 cards having the numbers 1, 2, …….., 25, written in

them. If one card is chosen what is the probability that the number in the

card is divisible by 3 or 7.

Solution: The numbers divisible by 3 are 3, 6, 9, 12, 15, 18, 21, 24 and

those divisible by 7 are 7, 14 (and 21 which is already considered). So the

number of favourable outcomes is 10.

4.025

10EP



Bayesian or subjective probability

In the last two approaches either we make certain assumptions about the

outcomes or we assume that we can have a large number of trials. When

we have an experiment that can be performed only once or only a few

times, the earlier methods fail. In such cases we resort to subjective

approach.

Definition: The subjective probability of an event is the probability assigned

to an event by an individual based on the evidence available to him, if there

is any.

Many of the social and managerial decisions are concerned with specific

unique situations. In such cases the decision makes has to frame subjective

probability for these events.

When a new product is developed, the marketing manager makes prediction

based on subjective probability framed by him.

S.A.Q. 2 A bag contains 3 red, 6 yellow and 7 blue balls. What is the

probability that the two balls drawn are yellow and blue ?

S.A.Q. 3 A ball is drawn from a bag containing 10 black and 7 white balls.

What I the probability that it is white ?

S.A.Q. 4 One number is chosen from each of the two sets {1, 2, 3, 4, 5, 6, 7,

8, 9} and {2, 4, 6, 8, 10}. What is the probability that the sum of these two

numbers is 13 ?

12.5 Kolmogorov’s Axiomatic Approach to Probability

As we saw earlier, Kolmogorov proposed axiomatic theory of probability in

1933. In axiomatic approach to any theory, a minimal set of properties are

taken as axioms (by an axiom we mean an assumption). They are taken as



a basis and all other properties are deduced from the axioms. This is how

many of modern mathematical objects are defined.

Before proceeding further, recall the definition of a sample space and events

discussed in earlier sections.

Definition: Let S be a sample spacer and be a collection of events

defined in the sample space S.

Then the probability of an event A is defined by a function P: B → R (R is

the set of all real numbers) Satisfying the following axioms.

(A1) for each AP,A satisfies .1AP0

(A2) P(S) = 1

(A3) If A1, A2, ….., An, ……. Is a sequence of mutually exclusive (disjoint

events in then

1i

i

1i

i APAP

Note: In most of the applications we take only a finite number of disjoint

events A1, ………., An. In this case, (A3) reduces to

n

1i

n21in21 AP......APAPAPA..........AAP (12.1)

In particular if A and B are mutually exclusive then

BPAPBAP (12.2)

Using the axiomatic approach, we can deduce all properties of probability

which hold good for classical and statistical probabilities.

We derive a few important properties of probability using (A1) (A2) (A3) or

(12.1). In most of the proofs (12.1) or (12.2) is used.

Property 12.1 The probability of an impossible event is zero.



Proof: The impossible event is the empty set . We know that SS

and the union S is a disjoint union. Then

SPSP

PSP (by 12.2)

Canceling P(S) on both sides. We get .0P

Property 12.2: If A is the complement of the event A, that is ,ASA

then AP1AP

Proof: We know that AAs (see figure 12.2)

Figure 12.2 The complementary event

So APAPSP

As 1APAP,Aby1SP 2 or

AP1AP

Theorem: (Addition theorem) If A and B are any two events, then

)3.12.......(....................BAPBPAPBAP

Proof: We write BA as a disjoint union.

A A



Figure 12.3 BA as disjoint union

From Fig 12.3, we see that BABAA

and the union is disjoint.

By (12.2), BAPBAPAP ……… (12.4)

Similarly BAABB

And the union is disjoint. By (12.2), BAPABPBP ……. (12.5)

As ABBABABA and the union on RHS is disjoint.

ABPBAPBAPBAP

BAPBAPABPBAPBAP

BAPBPAP by (12.4) and (12.5)

Note: (12.4) and (12.5) are useful for doing problems.

Thus (12.3) is proved

Corollary (Extended Addition theorem)

CBAPACPCBPBAPCPBPAPCBAP

…. (12.6)

Note: Fig. 12.5 will help you to prove (12.6)

A B

A – B A B B – A



Property 12.3 If ,AB then APBP

Proof: From Fig. 12.4, we see that

Figure 12.4, Venn Diagram for property 12.3

BABA and the union is disjoint. Hence

BABPAP

BAPBP by (12.2)

As APBPorBPAP,Aby0BAP 1

Example: If A, B, C are any three events, write the following events using

C,B,A,C,B,A and set operations.

a) only A occurs

b) A and B occur but not C

c) All the three events occur

d) None of them occur

e) At least one of them occur

f) At least two of them occur

g) Exactly one of them occurs

h) Exactly two of them occur

A – B

A B



Solution: We represent A, B, C and their complements in Fig. 12.5

Figure 12.5: Three events

Now using the Venn diagram given in Fig 12.5, we can represent the events

as follows.

a) CBA b) CBA

c) CBA d) CBA

e) CBA f) ACCBBA

g) CBACBACBA

h) CBACBACBA

Example: If two dice are thrown what is the probability that the sum is a)

greater than 9 b) neither 3 or 9 c) less than 4

Solution:

a) The favourable outcomes are (4,6), (5,5), (6,4), (5,6), (6,5), (6,6). So

.6

1

36

6EP

C

B A

CBA

CBA

CBA

CBA

CBA

CBA



b) Let A and B denote the events that the sum is 3 and 9 respectively.

Then 36

2AP and .

36

4BP As A and B are mutually exclusive

.0BAP

So 6

1

36

6

36

4

36

2BAP .

P(Sum is neither 3 or 9).

BAP

BAP (By De Morgan’s law BABA

BAP 1

6

11

6

5

c) The favorable outcomes are 11, 12, 21. So probability is .12

1

36

3

Example: A bag contains two red balls, three blue balls and five green balls.

Three balls are drawn at random. Find the probability that

a) the three balls are of different colours

b) two balls are of the same colour

c) all the three are of the same colour

Solution Let E denote the given event..

a) We can choose one red ball in C (2, 1) ways, etc. So

3,10C

1,5C1,3C1,2CEP

4

13.2.1.

8.9.10

5.3.2



b) Let E1, E2, E3 denote the events that two balls among the three are red,

blue and green respectively.

15

13.2.1.

8.9.10

8.1

3,10C

1,8C2,2CEP 1

Similarly

40

73.2.1.

8.9.10

7.3

3,10C

1,7C2,3CEP 2

12

53.2.1.

8.9.10

5.

2.1

4.5

3,10C

1,5C2,5CEP 3

So P(E) = P(E1) + P(E2) + P(E3)

12

5

40

7

15

1

120

50218

120

79

c) As there are only two red balls, the chosen three balls are of the same

colour only if they are all blue or green.

Let E1, E2 denote the events that the three balls are blue and green

respectively.

120

13.2.1.

8.9.10

1

3,10C

3,3CEP 1

12

13.2.1.

8.9.10

3.2.1.

2.1

4.5

3,10C

3,5CEP 2

So P(E) = P(E1) + P(E2)

12

1

120

1

120

101

120

11



Example: Two dice are rolled. If A is the event that the number in the first die is

odd and B is the event that the number in the second die is at least 3, find

ABP,BAP,BAP,BAP .

Solution: 3

2BP,

2

1AP

The outcomes favourable to BA are 13, 14, 15, 16, 33, 34, 35, 36, 53,

54, 55, 56. So .3

1

36

12BAP

BAPBPAPBAP

3

1

3

2

2

1

6

5

BAPAPBAP (By (12.4))

3

1

2

1

6

1

BAPBPABP (By (12.5))

3

1

3

2

3

1

S.A.Q.5 A bag contains 6 white and 10 black balls. Three balls are drawn at

random. Find the probability that

a) all the three are black

b) none of them is black

c) two of them are black

d) two of them are white

e) all the three are of the same colour



.

S.A.Q.6 The following table gives a distribution of wages of 1,000 workers:

Wages (in Rs.)

120 – 140 140 – 160 160 – 180 180 – 200 200 – 220 220 – 240 240 – 260

No. of workers

9 118 478 200 142 35 18

An individual is selected at random from the above group. What is the

probability that his wages are (i) under Rs. 160 (ii) above Rs. 200, (iii)

between Rs. 169 to 200 ?

S.A.Q. 7 Let a sample space be S = {a1, a2, a3}. Which of the following

defines probability space on S?

i) ,4

1aP 1 ,

3

1aP 2 .

2

1aP 3

ii) ,3

2aP 1 ,

3

1aP 2 .

3

2aP 3

iii) ,0aP 1 ,3

1aP 2

3

2aP 3

S.A.Q. 8 Out of numbers 1 to 100, one is selected at random. What is the

probability that it is divisible by 4 or 5?

S.A.Q. 9 The chance of an accident in a factory in a year is 1 in 5 in Bomay,

2 in 20 in Poona, 10 in 120 in Nagpur. Find the chances that a accident may

happen in (i) at least one of them (ii) all of them.

S.A.Q. 10 A person is known to hit a target in 3 out of 5 shots, whereas

another person is known to hit in 2 out of 3 shots. Find the probability that

the target being hit in all when they both try.

S.A.Q. 11 A six faced dice is biased that is twice as likely to show an even

number as an odd number when it is thrown. What is the probability that the

sum of the two numbers is even ?



S.A.Q. 12 The odd in favour of one student passing a test are 3 : 6. The

odds against another student passing at are 3:5. What are odds that (i) both

pass, (ii) both fail ?

S.A.Q. 13 If two dice are thrown, what is the probability that the sum is (a)

greater than 8, and (b) neither 7 or 11 ?

S.A.Q. 14 Let A and B two events such that .8

5BPand

4

3AP

Show that i) 4

3BAP

ii) 8

5BAP

8

3

S.A.Q. 15 From a group of children, 5 boys and 3 girls, three children are

selected at random. Calculate the probabilities that the selected group

contain i) no girl, ii) only one girl, iii) one particular girl, (iv) at least one girl,

and v) more girls than boys.

S.A.Q. 16 According to the census Bureau, deaths in the United States

occur at a rate of 2, 425,000 per year. The National Centre for Health

statistics reported that the three leading causes of death during 1997 were

heart disease (725, 790), cancer (537, 390) and stroke (159, 877). Let H, c

and represent the events that a person dies of heart disease, cancer and

stroke, respectively.

a) Use the data to estimate P(H), P(C) and P(S).

b) Are the events H and C mutually exclusive. Find .CHP

c) What is the probability that a person dies from heart disease or cancer?

d) What is the probability that a person dies from cancer or a stroke?

e) Find the probability that someone dies from a cause other than one of

these three.



12.6 Conditional Probability and Independence of Events

Let us consider a class having both boys and girls. Suppose a girl is

selected. What is the probability that the selected girl gets a first class? The

required probability is the probability of B (getting a first class) given that

A(The selected student is a girl) has already happened. Such a probability is

called conditional probability.

Definition 12.12 Let A and B be two events in the same sample space and

P(B) > 0. Then the conditional probability P(A/B) is the probability for A to

happen given that B has already happened.

The following theorem gives us a method of finding the conditional probability.

Theorem 12.2 (Multiplicative law for probability and conditional probability)

For any two events A and B,

,B/APBPBAP provided P(B) > 0.

= P(A) P(B/A), provided P(A) > 0.

Proof Let the total number of outcomes be N. Let nA, nB, nAB denote the

number of outcomes favourable to the events A, B and BA respectively.

Then N

nBP,

N

nAP BA and .

N

nBAP AB Let us calculate P(A/B)

using the classical approach. As B has already happened, the total number

of outcomes is nB. The number of outcomes favourable to A given B is the

number of outcomes favourable to .BA Hence

B

AB

n

nB/AP

N

nN

n

B

AB

BP

BAP



So B/APBPBAP proving the first identity. The second identity

can be proved similarly.

Example 12.11 The following table shows the distribution of blood types in a

state in India

A B AB O

Rh+ 33% 11% 4% 42%

Rh- 3% 2% 2% 3%

Find the following probabilities

a) The probability that a person has type O blood.

b) The probability that a person is Rh-

c) The probability that a married couple are both Rh+

d) The probability that a married couple have type AB blood

e) The probability that a person has type B blood given that the person is

Rh-

f) The probability that a person has type B blood given that the person is

Rh+

Solution Let A, B, AB, O, Rh+ and Rh- denote the events that a person has

type A blood etc.

a) 36.033.033.0RhOPRhOPOP

b) RhPORhABPRhBPRhAPRhP )(

= 0.03 + 0.02 + 0.02 + 0.03 = 0.10

c) P (Married couple are both Rh+)

= P[(husband is Rh+) (wife is Rh+)]

= [P (husband is Rh+)][ P(wife is Rh+) ]assuming independent.

= [1 – P (Rh – )] [1 – P(Rh – )]

= (0.9) (0.9) = 0.81

d) P(a couple have AB)

= P(husband has AB) P(wife has AB)



= (0.04 + 0.02) (0.04 + 0.02)

= 0.0036

e)

3.010.0

03.0

RhP

RhOPRh/OP

f)

122.090.0

11.0

RhP

RhBPRh/BP

Let A be the event that it rains heavily in Sikkim and B be the event that you

will score a first class in Bio informatics. Obviously the events A and B have

no dependence among themselves. We formulate this in the following

definition.

Definition: Two events A and B are independent if the occurrence or non –

occurrence of one does not affect the occurrence of the other. This happen

when

APB/AP and BPA/BP ……………….. (12.7)

Note: We know that BP

BAPB/AP

P(A/B) = P(A), ( as P(A B) = P(A) P(B))

Also,

BPAP

BAPA/BP

when BP.APBAP

So P(A/B) = P(A) implies P(B/A) = P(B)

So we note the following

A and B are independent if P(A B) = P(A) P(B) ……………………… (12.8)

So (12.8) can be taken as the working definition of independence of two

events.

Example: A bag has 20 blue balls and 10 green balls. Two balls are taken

from the bag one after the other. Find the probability that both are blue if.

i) The first ball is not replaced before taking out the second ball

ii) The first ball is replaced before taking out the second



Solution Let A denote the event that the first ball is blue and B be the event

that the second ball is blue. So we have to find P(A B)

As the bag has 20 blue balls and the total number of balls is 30,

3

2

30

20

1,30C

1,20CAP

i) When the first ball is not replaced and the second ball is taken out there

are 19 balls. So

29

19

1,29

1,20

C

CAP

So P(A B) = P(B/A) P(A) (By theorem 12.2)

3

2.

29

19

87

38

ii) When the first ball is replaced before the second ball is taken out, there

are 20 blue balls and 30 balls in all before the second choice

So 30

20

1,30C

1,20CA/BP

So 9

4

30

20.

30

20APA/BPBAP

Note in the above example, A and B are not independent in (i) but

independent in (ii)

Example 12.13 If A and B are independent show that

a) BandA are independent

b) BandA are independent

Solution As A and B are independent

BPAPBAP



We know that BABAA and the union is disjoint. Hence

BAPBAPAP

BAPAPBAP

(BPAPAP A and B are independent)

= P(A) [1 – P(B)]

BPAP (by property 12.2)

Hence A and B are independent

b) lawsMorganDebyBABABAPBAP ')(

= 1 – P (A B)

= 1 – [P(A) + P(B) – P(A B)] (by Addition theorem)

= 1 – [P(A) + P(B) – P(A) P(B)] (since A and B are independent)

= 1 – P(A) – P(B) + P(A) P(B)

= 1 – P(A) – P(B) [1 – P(A)]

= [1 – P(A)] [1 – P(B)]

BPAP

Hence BandA are independent.

Example: One third of the students in a class are girls and the rest are

boys. The probability that a girl gets a first class is 0.4 and that of a boy is

0.3. If a student having first class is selected, find the probability that the

student is a girl.

Solution: Let A, B and C denote the event that a student is a boy, a girl and

a student having first class. We are given the following

,3

2AP ,

3

1BP

10

4B/CPand

10

3A/CP

So .5

1

3

2.

10

3/ APACPCAP Similarly

30

4BCP



BACPCP since 5BA

= P((C A) (C B)) by Demorgan’s law

= P(C A) + P(C B) by Addition theorem

30

13

30

4

10

3

We are required to find P(B/C)

13

4

30

1330

4

))/(

CP

CBPCBP

Example: The probability that a 60 – year old man to be alive for 5 years is

0.80 and the same probability for a 55 – year old woman is 0.85. Find the

probability that a couple of ages 60 and 50 respectively will be alive for the

next 5 years.

Solution: We assume that the age expectation of the couple are

independent. Let A, B denote the probability that the husband and wife will

be alive for the next 5 years.

P(both will be alive for next 5 years)

= P(A B)

= P(A) P(B)

= (0.80) (0.85)

= 0.68

We can extend the concept of independence to more than two events.

Definition: Three events A, B and C are mutually independent if the

occurrence or non occurrence of any one of the events does not affect the

occurrence of other events.



Note: When A, B, C are mutually independent then A and B are

independent etc. So the working definition of 3 mutually independent events

A, B, C can be given as follows.

A, B, C are mutually independent if

P(A B) = P(A) P(B), P(B C) = P(B) P(C),

P(C A) = P(C) P(A) and

P(A B C) = P(A) P(B) P(C) ….. (12.9)

Example: A difficult problem is given to the students of 1st, 2nd and 3rd rank

by a professor. The probability that these students solve the problem are

5

2,

2

1,

4

3 respectively. Find the probability that the problem is solved.

Solution Let A denote the event that A solves the problem etc. Let us find

the probability that the problem is not solved by any of them (Assume

independence of A, B, C and hence )CandB,A

Then CPBPAPCBAP

5

21

2

11

4

31

5

3

2

1

4

1

40

3

Probability that the problem is solved = CBAP1CBAP

CBAP1

laws'MorganDeby

CBACBA

40

31



40

37

S.A.Q. 17 If P(A) = 0.25, P(B/A) = 0.5 and P(A/B) = 0.25 find

BAPandBAP,BAP,BAP,BAP,BAP

S.A.Q. 18 An article consists of two parts A and B. The probabilities of

defect in A and B are 0.08 and 0.04. What is the probability that the

assembled part will not have any defect?

S.A.Q. 19 From a bag containing 3 red and 4 black balls two balls are drawn

in succession without replacement. Find the probability that both the balls

are (i) red (ii) black (iii) of the same colour.

12.7 Baye’s theorem

We have seen that subjective probability is used when some event may

happen only once or a few times. But after assuming subjective probability

we may get some new information. This information can be used to revise

the subjective probability. Baye’s theorem is used for revising probability on

the basis of new information.

Reverend Thomas Bayes (1702 – 1761), a Christian Priest, enunciated

Baye’s theorem which has significant applications in many areas of

business administration especially marketing.

Theorem (Baye’s theorem)

If E1, E2, ……, En are mutually exclusive events with P(Ei) > 0, i = 1, 2, ….. n

then for any arbitrary event A which is a subset of n

1i

iE

such that P(A) > 0,

we have



n

1i

ii

iii

E/APEP

E/APEPA/EP

Note: P(E1). ……, P(En) are called a priori or prior probabilities. A denotes

some new information. Then we revise the probabilities P(Ei) as P(Ei/A). The

revised probabilities are called posteriori probabilities.

This process is illustrated in Fig. 12.6

Figure 12.6 Prior and posterior probabilities

Example: A company has three plants A, B and C manufacturing the same

spare part in the ratio 30:45:25. The percentage of defective parts in the

plants are 3%, 2% and 5% respectively. A part is chosen at random and

found to be defective. What is the probability that it is manufactured by

plants A, B or C ?

Solution Let A, B, C denote the event that it is manufactured in plants A, B

and C respectively.

Let P(D) be the probability that a spare part is defective. Given

P(A) = 0.3 P(B) = 0.45, P(C) = 0.25

P(D/A) = 0.03 P(D/B) = 0.02 P(D/C) = 0.05

Probability that the chosen defective part is manufactured by plant

A = P(A/D)

C/DPCPB/DPBPA/DPAP

A/DPAP

05.025.002.045.003.03.0

03.03.0

Prior Probabilities

New information

Application of Baye’s Theorem

Posterior Probabilities



= 0.295

Similarly,

P(B/D) = 0.295

P(C/D) = 0.410

Example: A physician believes that the people in a particular region are

prone to diseases A and B. He estimates that P(A) = 0.6 and P(B) = 0.4.

The diseases have symptoms S1, S2 and S3. Given that the patient has the

diseases, the probabilities for him to have symptoms S1, S2 and S3 are given

in the following table.

Disease Symptoms

S1 S2 S3

A 0.15 0.10 0.15

B 0.80 0.15 0.03

If a patient has symptom S1, find the probability that he has disease A.

Solution We are given that

P(A) = 0.6 P(B) = 0.4

P(S1/A) = 0.15 P(S2/A) = 0.10 P(S3/A) = 0.15

P(S1/B) = 0.80 P(S2/B) = 0.15 P(S3/B) = 0.03

The required probability

= P(A/S1)

B/SPBPA/SPAP

A/SPAP

11

1

80.04.015.06.0

15.06.0

= 0.738



S.A.Q.20 The contents of vessels I, II and III are as follows:

1. white, 2 black and 1 red balls

2. white, 1 black and 1 red balls, and

4. white, 5 black and 3 red balls.

One vessel is chosen at random and two balls are drawn. They happen to

be white and red. What is the probability that they come from vessels I, II or

III ?

S.A.Q. 21 A company has three machines M1, M2, M3 which produces 20%,

30% and 50% of the products respectively. Their respective defective

percentages are 7, 3 and 5. From these products one is chosen and

inspected. It is defective. What is the probability that it has been made by

machine M2 ?

12.8 Summary

In this unit we discussed about the concept of probability. The different basic

term of probability is well defined with examples. Different types of

probability, Baye’s theorem and its application is discussed with clear cut

examples.


1. Five persons are selected from a group of 8 men, 6 women and 6

children. Find the probability that 3 of the 5 persons selected are

children.

2. A bag contains 5 white, 6 black and 3 green balls. Find the probability

that a ball drawn at random is white or green.



3. Find the probability of getting (i) only one head and (ii) at least 4 heads

in six tosses of an unbiased coin.

4. Find the chance of getting more than 15 in rolling three dice

5. A five-digit number is formed from the digits 2, 3, 5, 6, 8, no repetition

being allowed. Find the probability that the number is (i) odd (ii) even.

6. A sample space has 5 elementary events E1, E2, E3, E4, E5. If P(E3) =

0.4, P(E4) = 2, P(E5) and P(E1) = P(E2) = 0.15, determine P(E4), P(E5),

P({E4, E5}), P({E1, E2, E3})

7. A problem is probability is given to three students A, B and C. The

probabilities that they solve the problem are 4

1,

3

1,

2

1 respectively. Find

the probabilities that

i) A alone solves the problem

ii) Just two of them solve the problem

iii) The problem is solved

(Mention the assumptions you make in solving the problem).

8. A large company dumps its chemical waste in a local river. The

probability that either a fish or an animal dies on drinking the water is

.21

11 The probability that only a fish dies is

3

1and the probability that

only an animal dies in .7

2 What is the probability, (i) that both will dies ?

(ii) none of them will die ?

9. In a college, the percentage of students reading the new Indian Express,

Deccan Herald and both are 20%, 30% and 15%. Find the probability

that a student of the college

i) reads at least one newspaper

ii) reads none of them

iii) reads only Deccan Herald



10. A team of 6 students is to be selected from a class consisting of 7 boys

and 4 girls. Find the probability that the team consists of a) exactly two

girls b) at least 2 girls.

11. 40 candidates appeared for examination in Papers A and B. 16 students

passed in paper A, 14 passed in Paper B and 16 failed in both. If one

student is selected at random what is the probability that he

a) passed in both papers

b) failed only in A

c) failed in A or B

12. In a town, only 80% of the children born reach the age of 15 and only

85% of them reach the age of 30. 2.5% of persons aged 30 die in one

year. What is the probability that a person will reach the age of 31.

13. If A, B, C are mutually exclusive and collectively exhaustive and 2 P(C)

= 3P (A) = 6P(B) find P(A), P(B), P(C).

14. Two vessels contain 20 white, 12 red and 18 black balls; 6 white, 14 red

and 30 black balls respectively. One ball is taken out from each vessel.

Find the probability that a) both are red b) both are of the same colour

15. The probability that a candidate passes in Biostatistics is 0.6 and that

the probability that he passes in Genetics 50.5. What is the probability

that he passes in only one of the papers ? (Mention the assumption you

make).

16. The following table gives the frequency distribution of 50 professors

according to age and their salaries.

Age in years

Salary

10000 – 15000 15000 – 20000 20000 – 25000 25000 – 30000

20 – 30 16 6 – –

30 – 40 4 10 4 4



40 – 50 – 4 18 12

50 – 60 – – 12 10

If a professor is chosen at random find the probability that

a) he is in the age group 30 – 40 years and earns more than 20,000

b) he earns in the range 15000 – 20000 and less than 40 years old.

17. In a class of 20 boys and 40 girls, half the boys and half the girls have

two – wheelers. Find the probability that a randomly selected student is

a boy or has a two wheeler.

18. Three fair (unbiased) coins are tossed. If the first coin shows a head find

the probability of getting all heads.

19. If 2

1BAPand

4

1BP,

3

1AP find P B/A and P(A/B).

20. If ,6

1B/A(P,

2

1BP,

3

1AP find P(B/A) and P(B/A).

21. The records of 400 examinees are given below.

Score Educational Qualification

B.A. B.Sc B.Com Total

Below 50 90 30 60 180

Between 50 & 60 20 70 70 160

Above 60 10 30 20 60

Total 120 130 150 400

If an examinee is selected from this group find

i) The probability that he is a commerce graduate

ii) The probability that he is a science graduate given that his score is

above 60



iii) The probability that his score is below 50 given that he is a B.A

graduate.

22. There are 3 boxes containing respectively 1 white, 2 red, 3 black balls; 2

white, 3 red, 1 black ball; 3 white, 1 red and 2 black b alls. A box is

chosen at random and from it two balls are drawn at random. The two

balls are 1 red and 1 white. What is the probability that they come from

(i) the first box (ii) second box (iii) third box?

23. An item is manufactured by three machines M1, M2 and M3. Out of the

total manufactured during a specified production period, 50% are

manufactured on M1, 30% on M2 and 20% on M3.

It is also known that 2% of the item produced by M1 and M2 are

defective, white 3% of those manufactured by M3 are defective. All the

items are put into one bin. From the bin, one item is drawn at random

and is found to be defective. What is the probability that it was made on

M1, M2 or M3.

12.10 Answers


1. a) {TTT, HHT, HTH, THH}

b) {WLWLW, LWLWL}

c) {13, 22, 31, 46, 55, 64}

d) {WLL, LWL, LLW, WWL, WLW, LWW, WWW}

2. 20

7

2,16C

1,7C1,6C

3. 17

7

4. The sum 13 can be obtained from the pairs (3, 10), (5, 8), (7, 6), (9, 4).

The total number of pairs that can be chosen is 9(5) = 45.



5. Answer: 45

4

a) 14

3

3,16C

3,10C

b) 28

1

3,16C

3,6C

c)

56

27

3,16

1,62,10

C

CC

d) 56

15

e) 4

1

28

1

14

3

6. Total number of workers = 1000. (i) 0.127 (ii) 0.195 (iii) 0.596

7. i) As P(S) = P(a1) + P(a2) + P(a3) 1, answer is No.

ii) As No,3

1aP 3

iii) P(S) = 1; Yes

8. Let A and B denote divisibility by 4, 5. Outcomes favourable to A are 4,

8, …… , 100. So P(A) = 0.25. Similarly P(B) = 0.2 and P(A B) = 0.05.

Answer: 0.4

9. i) Prob (accident occurs in none of them)

.50

33

12

11,

10

9,

5

4CBAP

Answer: 50

17

ii) 600

1CBAP

10. Assume independence. (In case of independence, P(A B) = P(A)

P(B). See later sections).



Answer: 15

13

11. Let P (odd number) = a. Treat the die as a coin showing ‘even’ or ‘odd’.

Then 2a + a = 1 or .3

1a Prob (sum is even = Prob (both are even) +

Prob (both are odd) = .9

5

3

1.

3

1

3

2.

3

2

12. .3

1

63

3AP

For the second, odds for passing are 5 : 3. So

.8

5BP Assume independence i)

24

5

8

5.

3

1 ii)

4

1

8

3.

3

2 .

13. a) Favourable outcomes are: 36, 45, 54, 63, 46, 55, 64, 56, 65,

66.

Answer: .18

5

b) .18

1

6

111P7P11or7P Answer: .

9

7

14. As ,BAPAP (i) follows. .8

5BPBAP

18

5

4

3BAPBPAPBAP

(since P(A B) < 1 and so – P (A B) > –1). So (ii) follows.

15. (i) 28

5

3,8C

2,5C

(ii)

28

15

3,8C

2,51,3C

(iii) 8

3

3,8

2,7C

(iv) 28

23

28

51



(v) 7

2

56

15

56

1

16. a) 2425000

159877SP

2425000

537390CP

000,425,2

725790HP

b) Yes. They are also independence. Answer: P(H) P(C)

c) P(H) + P(C)

d) P(C) + P(S)

e) P(H C S) = P(H) + P(C) + P(S) – P(H) P(C) – P(C) P(S) – P(H)

P(S) + P(H) P(C) P(S). (We assume independence of H, C, S)

17. P(A B) = 0.125, P(B) = 0.5,

So P(A B) = 0.25 + 0.5 – 0.125 = 0.625.

875.0BAP1BAP.375.0BAP1BAP

.375.0BAPBPBPBAP

125.0BAPAPBAP

18. Let x, y be the events that A and B do not have any defect respectively.

P(x) = 0.92 and P(y) = 0.96. Assuming independence, the answer is

(0.92) (0.96) = 0.8832.

19. Let A and B denote the events that the ball is red in first and second

attempt (i) 7

1

6

2.

7

3A/BPAPBAP (ii)

7

2 (iii)

7

3

7

2

7

1

20. P(E1) = P(E2) = P(E3) = 1

118

33A/EP

11

2E/AP

3

1E/AP

5

1E/AP 1321

118

55/2 AEP So

.118

30

118

55

118

331A/EP 3

21. Answer:

1875.0

05.05.003.03.007.02.0

3.03.0



Terminal Questions

1.

3876

455

5,20C

2,14C3,6C

2. 7

4

3. One head can appear in 6 ways.

(i) 32

3 (ii)

32

11

64

6,6C5,6C4,6C

4. We should get a sum of 16, 17, 18. Answer: .108

5

5. An odd number ends in 3 or 5. Number of favourable outcomes is 2(4!).

Answer:

5

3;

5

2

!5

!42

6. Let P(E5) = x. Then 0.15 + 0.15 + 0.4 + 2x + x = 1. Hence x = 0.1

Answers: 0.2, 0.1, 0.3, 0.7

7. a) 4

1CBAP

b) 12

5CBAPCBAPCBAP

c) .4

1CBAP

4

3CBAP

8. Given that ,7

2APand

3

1FP,

21

11AFP use addition

theorem,

i) 21

2AFP ii)

21

19

21

21AFP

9. Given that P(A) = 0.2, P(B) = 0.3 and P(A B) = 0.15

(i) P(A B) = 0.35



(ii) 65.0BAP

(iii) 15.0BAP

10. a) 11

5

b)

66

53

6,11C

2,7C4,4c3,7C3,4c4,7C2,4C

11. Given that .40

24BAP,Soand

40

14BP,

40

16AP

a) 20

3 b)

5

1BAP c)

20

17BAP

12. P (Fifteen) = 0.8, P (thirty/ Fifteen) = 0.85 and P (thirty one/ thirty) =

0.975.

Answer: (0.8) (0.85) (0.975) = 0.663

13. AP2

3AP

2

1APCPBPAP1

So, ,3

1AP ,

6

1BP .

2

1CP

14. a) 625

42

50

14.

50

12

b) 625

207

50

30.

50

18

50

6.

50

20

50

14.

50

12

15. Assume independence of B and G. Answer:

5.0GBPGBP

16. Total number of professors = 100

a) 08.0100

44 b) 0.16

17. Given that 2

1WP,

61

60

10WBP,

3

1BP



Answer: 3

2WBP

18. Let A denote getting H is the first toss and B denote getting H in second

and third tosses.

Answer: 4

1

2

1

8

1AP/BAPA/BP

19. 12

1BAP

4

1A/BP and

3

1B/AP

20. 12

1BAP

4

1

3

1

12

1AP/BAPA/BP

12

5

12

1

2

1 ABPBPABP

8

5

3

2

12

5AP

12

5A/BP

21. (i) 8

3 (ii)

2

160Score/SP (iii)

4

3A/50ScoreP

22. 3

1EPEPEP 321

15

2E/AP 1

15

6E/AP 2

15

3E/AP 3

,11

2A/EP 1

11

6A/EP 2 Hence

11

3A/EP 3

23. P(E1) = 0.5 P(E2) = 0.3 P(E3) = 0.2 P(A/E1) = 0.02 etc

P(E1/A) = 0.454, P(E2/A) = 0.273, P(E3/A) = 0.273



Unit 13 Basic Statistics

Structure

13.1 Introduction

Objectives

13.2 Measures of Central Tendency

13.3 Standard Deviation

13.4 Discrete Series

13.5 Methods: Deviation taken from Assumed Mean

13.6 Continuous Series

13.7 Combined Standard Deviation

13.8 Coefficient of Variation

13.9 Variance

13.10 Summary


13.12 Answers

13.1 Introduction

In this unit we discuss some of the concepts of Basic Statistics. The single

value, which is representative of a set of values, may be used to give an

indication of the general size of the members in a set, the word ‘average’

often being used to indicate the single value. The Statistical term used for

‘average’ is the arithmetic mean or just the mean. Other measures of central

tendency may be used and these include the median and the modal values.

The standard deviation of a set of data gives an indication of the amount of

dispersion, or the scatter, of members of the set from the measure of central

tendency.



Objectives

At the end of the unit you would be able to understand

the concept of central tendency and its applications

how to calculate standard deviation and variation of a given set of data.

13.2 Measures of central tendency

Measures of central tendency

This chapter is devoted to different measures used to summarize the data.

Different measures discussed are Mean, Median and Mode. Along with

these three fundamental and trivial measures, two other measures,

Geometric Mean, Harmonic Mean are clearly introduced. Definition, method

of computation, Interpretation and uses form the structure of the explanation

for each measure.

Generally, in a frequency distribution, the values cluster around a central

value. This property of concentration of the values around a central value is

called Central Tendency. The central value around which there is

concentration is called measure of central tendency (measure of location,

average).

Generally, a simple comparison of frequency distribution is made by

comparing their measures of central tendency.

For a frequency distribution, five important measures of central tendency are

defined.

They are:

1. Arithmetic Mean (A.M.)

2. Median

3. Mode

4. Geometric Mean (G.M.)

5. Harmonic Mean (H.M.)



Depending upon the need and nature of study, proper measure is chosen.

However, a measure of central tendency is considered to be good if it has

some of the following qualities:

Desired qualities of an ideal measure of central tendency.

1. It should be easy to understand. Its computation procedure should be

simple

2. It should be rigidly defined

3. It should be based on all the values

4 It should not be affected too much by abnormal extreme values

5. It should be capable of further algebraic treatment so that it could be

used in further analysis of the data

6. It should be stable. That is, the measure should be such that sampling

variation in the value of the measure should be least.

Arithmetic Mean (Mean)

Arithmetic mean of a set of values is obtained by dividing the sum of the

values by the number of values in the set. Arithmetic mean of the values

isx,..........,x,x n21

n

x

n

x.......xxx n21

If the observations n21 x,..........,x,x have frequencies ,f...........,f,f n21

the arithmetic mean is

N

fx

f.........ff

xf.......xfxfx

n21

nn2211

(for discrete frequency distribution)

Where N = f is the total frequency.

Thus, for a raw data, the arithmetic mean is

n

xx



For a tabulated data (discrete or continuous), it is

N

fxx

Example 1: Heights of six students are 163, 173, 168, 156, 162 and 165

cms. Find the arithmetic mean.

Solution: The arithmetic mean of the heights is

6

165162156168173163

n

xx

.cms5.1646

987

Example 2: In a one-day cricket match, a bowler bowls 8 overs. He gives

away 3, 5, 12, 0, 4, 1, 3 and 7 runs in these overs. Find the mean run rate

per over.

Solution: The mean run rate is

n

xx

valuesofNumber

valuesofSum

8

731401253

375.48

35 runs per over.

Example 3: In an office there are 84 employees. Their salaries are as given

below:

Salary (Rs.)

2430 2590 2870 3390 4720 5160

Employees 4 28 31 16 3 2

i) Find the mean salary of the employees

ii) What is the total salary of the employees ?



Solution:

Salary (Rs.) (x)

Employees (f)

fx

2430

2590

2870

3390

4720

5160

Total

4

28

31

16

3

2

84

9720

72520

88970

54240

14160

10320

249930

(i) The mean salary of the employees is

36.2975.Rs84

249930

N

fxx

(ii) Total salary of the employees is

fx = Rs. 249930.

Example 4: A survey of 128 smokers revealed the following frequency

distribution of daily expenditure on smoking of these smokers. Find the

mean daily expenditure

Expenditure (Rs.) 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80

No. of smokers 23 44 35 12 9 3 2



Solution:

Expenditure (Rs.)

Frequency (f) Mid-value (x) fx

10 – 20

20 – 30

30 – 40

40 – 50

50 – 60

60 – 70

70 – 80

23

44

35

12

9

3

2

15

25

35

45

55

65

75

345

1100

1255

540

495

195

150

Total 128 - 4050

The mean is

64.31.Rs128

4050

N

fxx

The mean daily expenditure is Rs. 31.64.

Change of Origin and Scale

Let x1, x2, ………., xn be n values. Let ‘a’ be a constant. Then x1 – a, x2 – a,

…….., xn – a are the values of x1, x2,……….. xn with origin shifted to ‘a’. If ‘c’

is a positive constant,

c

ax

c

ax

c

ax n .,,........., 21

are the values x1, x2, ………., xn with origin shifted to a and scale changed

by c. Thus, c

axu

is the variable x with origin shifted to a and scale

changed by c.

Here c

axu

therefore, x = a + cu

And so, N

fucaucax



However, if c = 1, n

uauax

Deviations: Let x1, x2, x3, …….., xn be n values. Let ‘a’ be a constant. Then

ax,.....,ax,ax,ax n321 are the deviations of the values from

the constant a. The squares of these deviations, namely,

22

3

2

2

2

1 ,,.........,, axaxaxax n are the squared deviations of

the values.

Thus, xx,..........,xx,xx,xx n321 are the deviations from the

arithmetic mean.

2n2

32

22

1 xx,..........,xx,xx,xx are the squared deviations

from the arithmetic mean. The deviations may be positive, negative or zero.

But, the squared deviations will never be negative.

Properties of Arithmetic Mean:

Arithmetic mean has the following important properties:

1. Algebraic sum of the deviations of a set of values from their arithmetic

mean is zero

That is, 0xx

2. Sum of the squared deviations of a set of values is minimum when

deviations are taken around the arithmetic mean.

3. Let 1x be the arithmetic mean of a set of n1 values. And let 2x , be the

arithmetic mean of another set of n2 values. Then, the arithmetic

mean of the two sets of values put together is

21

2211

nn

xnxnx

(combined arithmetic mean)

Example 5: The mean of marks scored by 30 girls of a class is 44%. The

mean for 50 boys is 42%. Find the mean for the whole class.



Solution: Here %.42xand%44x,50n,30n 2121 The combined

arithmetic mean is

21

2211

nn

xnxnx

%75.4280

21001230

5030

42504430

Example 6: Average (mean) weight of a type of screws is 10.4 gms. A

packet of 100 such screws is mixed with another packet of 150 screws of

another type. In the mixture, average weight is found to be 10.9 gms. Find

the average weight of the second packet of screws.

Solution:

Here .gms9.10xand.gms4.10x,150n,100n 121

The mean weight of the second set of screws 2x can be calculated by

using the relation.

21

2211

nn

xnxnx

150100

x1504.101009.10 2

250

x15010409.10 2

Therefore, 168510402509.10x150 2

And so, .gms23.11150

1685x2

Thus, the mean weight of screws in the second packet is .gms23.11x2

Merits of arithmetic mean:

1. It is rigidly defined

2. The logic behind its computation can be easily understood. It can be

easily computed.



3. It can be easily adopted for further statistical analysis

4. It is based on all the values

5. It is more stable than any other average

6. It can be calculated even when some of values are equal to zero or

negative.

Demerits of arithmetic mean:

1. It is highly affected by abnormal extreme values

2. Since it is based on all the values, even if one of the values is missing, it

cannot be calculated.

3. Sometimes, the arithmetic mean may be a value which is not assumed

by the variable.

Median

Median of a set of values is the middle most value when they are arranged

in the ascending order of magnitude. (Such an arrangement is called an

array). It is a value that is greater than half of the values and lesser than the

remaining half. The median is denoted by M.

In the case of a raw data and also a discrete frequency distribution, the

mean is-

th

2

1nM

value in the arrayed series

In the case of a continuous frequency distribution, the median is –

f

cmN

IM2

Where I : lower limit of the median class

c : width of the median class

f : frequency of the mean class.



m : less-than cumulative frequency up to/ (cumulative

frequency corresponding to the class preceding the median

class)

N : Total frequency

Median class is the class which contains the median.

Example 7

The following data relates to the number of children of 25 couples. Find the

median

No. of children per couple: 2, 0, 5, 2, 5 , 1, 0, 0, 3, 4, 2, 1, 1, 2, 3, 0,

1, 2, 7, 2, 2, 1, 3, 4, 1.

Solution:

The arrayed series (ascending series) is:

0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5,

7

Here, n =25. Therefore, median is

th

2

1nM

value in the arrayed series

th

2

125

value = 13th value

= 2 children per couple

Merits of median:

1. The logic behind its computation is easily understood. It can be easily

computed.

2. Even when some of the extreme values are missing, it can be

computed.

3. It is not affected by abnormal extreme values

4. It can be used for the study of qualitative data also.



Demerits of median:

1. It is not based on all the values

2. It cannot be used in deep statistical analysis.

Mode

Mode is the value which has the highest frequency. It is the most

frequently occurring value. It is denoted by Z.

In the case of raw data, and also in the case of a discrete frequency

distribution, mode is the value with highest frequency.

In the case of a continuous frequency distribution, mode is

21

1

fff2

cffIZ

Where l : lower limit of the modal class

f : frequency of the modal class

c : width of the modal class

f1 : frequency of the class preceding the modal class

f2 : frequency of the class succeeding the modal class

Modal class is the class width containing the mode.

Generally, modal class will be the class with highest frequency. But

sometimes, it may be a class other than the class with highest

frequency. In such a situation, mode is obtained by using the formula –

21

2

ff

cflZ

Most of frequency distributions have only one value with highest frequency.

Such frequency distributions are unimodal – they have only one mode. On

the other hand, if in a frequency distribution, there is more than one value

with highest frequency, such a distribution is multimodal – it will have more

than one mode. If there are two modes, the distribution is bimodal.



However, for a distribution which has more than one mode, it is said to be ill

– defined.

Example 8:

The following are the number of children for 20 couples. Find the mode.

No. of children per couple: 2, 3, 6 , 3, 4, 0, 5, 2 , 2, 4, 3, 2, 1, 0, 4, 2, 2, 1, 1, 3

Solution:

The data should be tabulated first

No. of children No. of couples

Tally marks Frequency

0 II 2

1 III 3

2 IIII I 6

3 IIII 4

4 III 3

5 I 1

6 I 1

Total 20

Here, the value 2 has the highest frequency.

Therefore, mode is Z = 2 children/ couple.

Example 9:

For the following distribution, find the mode

Percentage marks 10 – 19 20 – 29 30 – 39 40 – 49 50 – 59 60 – 69 70 – 79

No. of students 8 19 29 36 25 13 4

Solution:



Here, the class intervals are of inclusive type. Firstly, they should be

converted into the exclusive type.

Modified class Frequency

9.5 – 19.5

19.5 – 29.5

29.5 – 39.5

49.539.5

49.5 – 59.5

59.5 – 69.5

69.5 – 79.5

8

19

29 (f1)

f36

25 (f2)

13

4

Total 134

Modal class

Since 36 is the highest frequency and it is far higher than the other

frequencies, the class interval 39.5 – 49.5 is the modal class. Thus l = 39.5,

f = 36, f1 = 29, f2 = 25 and c = 10.

The model is –

2529362

1029365.39

fff2

cfflZ

21

1

%4.4318

705.39

Merits and demerits of mode

The merits and demerits of mode are the same as merits and demerits of

median. In addition, one demerit of mode can be listed. It is –

For some frequency distribution, mode is ill – defined.

13.3 Standard Deviation



Definition: Standard Deviation is the root mean square deviation of the

values from their arithmetic mean.

Standard Deviation is the abbreviation and (read, sigma) is the symbol.

Mean square deviation of the value from their arithmetic mean is variance

and is denoted by 2. Standard Deviation is the positive square root of

variance. Karl Pearson introduced the concept of standard deviation in

1893. Standard Deviation is also called mean square deviation. It is a

mathematical deficiency of mean deviation to ignore negative sign. Standard

deviation possesses most of the desirable properties of a good measure of

dispersion. It is the most widely used absolute measure of dispersion.

The corresponding relative measure is coefficient of variation. It is very

popular and so extensively used.

100meanArithmatic

DeviationdardtanSiationvaroftCoefficien

Formulae

Method Individual Observation

Discrete Series Continuous Series

1. Actual

Mean

N

xx2

N

xxf2

N

xmf2

2. Direct

method

22

N

x

N

x

22

N

fx

N

fx

22

N

fm

N

fm

3. Assumed

mean

22

N

d

N

d

22

N

fd

N

fd

22

N

fd

N

fd

4. Step

Deviation

22

N

'd

N

'dC

22

N

'fd

N

'fdC

22

N

'fd

N

'fdC

Individual Observations



Method 1: Deviation taken from actual mean

Standard Deviation, N

X 2 where .XXx This method is simple

when XX values are integers

Steps:

1. Form a table with the given values, x in the first column

2. Find out the arithmetic mean,

N

xX

3. Find out the deviation of each values from the actual mean and call it x

i.e, find .XXx Enter those values in the next column

4. Find out the squares of the deviation of the values from the actual mean,

i.e, find x2. Enter those values in the next column.

5. Find out the mean of the squared deviation of the values from their

arithmetic mean i.e., find .

2

N

x.

6. Find out the square root of N

x 2

. It is the standard deviation.

Example 1: Consider 77, 73, 75, 70, 72, 76, 75, 72, 74, 76. Give standard

deviation for the numbers given above.



Solution:

X 74:X

XXx

x2

77 3 9

73 –1 1

75 1 1

70 –4 16

72 –2 4

76 2 4

75 1 1

72 –2 4

74 0 0

76 2 4

740x 0XX 44x2

Method 2: Direct Method

Without taking any deviations, the standard deviation can directly be calculated by the formula.

Standard Deviation

22

N

X

N

X

This method can be used for all kinds of data. This formula is used later for

correcting the mistakes in the calculations.

Steps:

1. Form a table with the given values, x, in the first column

2. Find the square of each X and write in the next column under the title X2.

3. Find the totals X and 2X and N, the number of values

Arithmetic mean: N

XX

10

740

= 74

Standard Deviation

N

x2

=10

44

= 4.4

= 2.10



4. Substitute in the above formula and simplify

Example: 2

10 students of B.Com class of a college have obtained the following marks

in statistics out of 100. Calculate the standard deviation

S. No : 1 2 3 4 5 6 7 8 9 10

Marks : 5 10 20 25 40 42 45 48 70 80

Solutions:

S. No Marks X X2

1 5 25

2 10 100

3 20 400

4 25 625

5 40 1600

6 42 1764

7 45 2025

8 48 2304

9 70 4900

10 80 6400

Total 385X 201432X

Method 3: Deviations taken from assumed mean

This is same as the one followed in the calculation of arithmetic mean. But

the formula is as follows:

Standard Deviation,

22

N

d

N

d

d = X – A is preferred when XX are fractions.

Standard Deviation

22

N

X

N

X

2

10

385

10

20143

25.3820143

25.148230.2014

05.532

07.23



Steps:

1. Form a table with the given values, X, in the first column.

2. Assume any value as ‘A’ if it is root specified in a problem. It is

preferable to assume a value in between the minimum value and the

maximum value of X as A.

3. Find out the observation of each value from the assumed mean A and

call it d. i.e., find d = X – A and write them in the next column

4. Write the squares of the deviations, d2, in the next column

5. Find d and d2 and identify N, the number of values substitute them in

the above formula and simplify.

Example 3:

For the data below, calculate standard deviation:

40, 50, 60, 70, 80, 90, 100

Solution:

X d = X – A A = 70

X

40 – 30 900

50 – 20 400

60 – 10 100

70 0 0

80 10 100

90 20 400

100 30 900

Total 0d 2800d 2

Method 4 Step Deviation Method:

Standard Deviation

22

N

d

N

d

2

7

0

7

2800

20400

= 20



This is same as the one followed in the calculation of arithmetic mean. But

the formula is as follows:

Standard Deviation,

22

N

d

N

dC

C

AXd

This method is preferred when XX are fractions and there is common

difference between X.

Steps:

1. Form a table with the given values X, in the first column

2. Choose A and C as mentioned under Arithmetic mean

3. Find out the step deviation corresponding to each X. i.e., find

C

AXd

and write those values in the next column.

4. Write the squares of ,d,e.i.d2 in the next column.

5. Find d’ and d’2 and identify N, the number of values. Substitute them

in the above formula and simplify.

Example 4: Given below are the marks obtained by 5 B.Sc. students

Roll No : 101 102 103 104 105

Marks : 10 30 20 25 15

Calculate Standard Deviation

Solution:

Roll No. Marks

X

C

AX'd

A = 20

C = 5

d'2

101 10 – 2 4

Standard Deviation 22 ''

N

d

N

dC

2

5

0

5

105

2025

25

= 5 1.4142

=7.07



102 30 2 4

103 20 0 0

104 25 1 1

105 15 –1 1

Total – d’ = 0 d’2 = 10

Note: A problem can be solved by any one method.

13.4 Discrete Series:

13.4.1 Method 1: Deviations taken from actual mean

Standard Deviation, N

fx 2

Where x = XX and N = f

Steps:

1. Find out the arithmetic mean, .X

2. Find out the deviation of each X from the actual mean and call it x. i.e.,

find x = XX

3. Find out x2 values

4. Multiply each x2 by the corresponding f and get fx2

5. Find fx2

6. Divide fx2 by N and find the square root of the quotient. i.e, findN

fx 2

Example 5: Calculate the standard deviation of the following series.

x 6 9 12 15 18

f 7 12 13 10 8



Solution:

X f fx 12X

XXX

x2 fx2

6 7 42 – 6 36 252

9 12 108 – 3 9 108

12 13 156 0 0 0

15 10 150 3 9 90

18 8 144 6 36 288

Total N = 50 fx = 600 – – fx2 = 738

Arithmetic mean = N

fxX

50

600

= 12.00

Standard Deviation N

fx 2

50

738

36.14

= 3.84

Method 2: Direct Method

Under this method, the formula becomes the following

Standard Deviation,

22

N

fx

N

fx

Steps:



1. From a table with the given values, x and the frequencies, f in the first

two columns

2. Multiply each x by the corresponding f to find fx. Write all such fx

values in the next column.

3. Multiply each fx by the corresponding x to find fx2 (It is not (fx)2. That

is, fx should not be squared) such fx2 value in the next column.

4. Find N(=f), fx and fx2.

5. Substitute in the above formula and simplify.

Example 6: Calculate the standard deviation

No. of goals scored in a match : (x) 0 1 2 3 4 5

No. of Matches : (f) 1 2 4 3 0 2

Solution:

X f fx fx2

0 1 0 0

1 2 2 2

2 4 8 16

3 3 9 27

4 0 0 0

5 2 10 50

Total N = 12 fx = 29 fx2 = 95

13.5 Method 3: Deviation taken from assumed measure

The formula is as follows:

Standard Deviation,

22

N

fd

N

fd

d = X – A, A – Assumed mean, N = f

Steps:

Standard Deviation

22

N

fx

N

fx

2

12

20

12

95

24167.29167.7

8404.59167.7

0763.2

= 1.44



1. Form a table with the given values X and the frequencies, f is the first

two columns

2. Choose the value for A, assumed mean, if it is not specified.

3. Subtract A from each X and form the next column with d = X – A values

4. Multiply each d by the corresponding f and enter all such products in the

next column under the title fd.

5. Multiply each fd by the corresponding d and enter all such products in

the next column under the title fd2 (these are not the squares of fd

values)

6. Find N ( = f), fd and fd2


Example 7:

Calculate standard deviation from the following data:

x : 6 9 12 15 18

f : 7 12 19 10 2

Solution:

Let X: 6, 9, 12, 15 and 18.

X f d = X – A

A = 12 fd fd2

6 7 – 6 – 42 252

9 12 – 3 – 36 108

12 19 0 0 0

15 10 3 30 90

18 2 6 12 72

Total N = 50 – fd = – 36 fd2 = 522



Standard Deviation

22

N

fd

N

fd

2

50

36

50

522

272.044.10

5184.04400.10

9216.9

= 3.15

13.5.1 Method 4: Step Deviation Method

This following formula is used.

Standard Deviation,

22

N

'fd

N

'fdC

fN;C

AX'd

Steps:

1. Form a table with the given values, x and the frequencies, f in the first

two columns

2. Choose the value for A.

3. Find out C

AX'd

corresponding to each X and enter them in the next

column

4. Multiply each d by the corresponding f to get df . Enter them in the

next column.

5. Multiply each df by the corresponding d and get .df 2 Enter them in

the next column.

6. Find 2dfanddf,fN




Example 8:

The weekly salaries of a group of employees are given in the following table.

Find the mean and standard deviation of the salaries.

Salaries (in Rs.) : 75 80 85 90 95 100

No. of persons : 3 7 18 12 6 4

Solution:

Salary (in Rs.)

x

No. of persons

f C

AXd

A = 85; C = 5

fd’ fd2

75 3 – 2 – 6 12

80 7 – 1 – 7 7

85 18 0 0 0

90 12 1 12 12

95 6 2 12 24

100 4 3 12 36

Total N = 50 – fd = 23 d2 = 91

Arithmetic mean

N

fdCAX

50

23585

= 85 + 2.3

= 87.30 (Rs.)

Standard Deviation

22

N

fd

N

fdC



2

50

23

50

915

246.082.15

2116.08200.15

6084.15

= 5 1.27

= 6.35 (Rs.)

13.6 Continuous Series

When X in the formulae for the calculation of standard deviation of discrete

series is replaced by m the corresponding formulae for continuous series

are obtained.

The calculations start with the mid values (m) of the class intervals and

class frequencies (f) when less than or more than frequencies are given,

class interval and class frequencies are to be found first.

Method 1: Deviations taken from actual mean.

The formula is as follows:

Standard deviation,

N

Xmf2

Steps:

1. Form a table with class intervals and class frequencies in the first two

column.

2. Find the mid values (m) and write them the next column

3. Find the products of f and m and write them in the next column

4. Find N

fmX

where X,fN may be found by other formula also.



5. Subtract X from each m. Enter the resulting Xm values in the next

column

6. Write 2Xm in the next column

7. Find 2Xmf and write them in the next column

8. Find 2Xmf

9. Divide 2Xmf by N and take the square root to get the standard

deviation.

Example 9: Find the standard deviation

Class Intervals : 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50

Frequency : 2 5 9 3 1

Solution:

Class Interval

Frequency f

Mid value m

fm Xm

23X 2Xm 2Xmf

0 – 10 2 5 10 – 18 324 648

10 – 20 5 15 75 – 8 64 320

20 – 30 9 25 225 2 4 36

30 – 40 3 35 105 12 144 432

40 – 50 1 45 45 22 484 484

N = 20 – fm = 460 – – 1920Xmf2

Arithmetic Mean 2320

460

N

fmX

Standard Deviation

N

Xmf2

20

1920



96

= 9.80

Method 2: Direct method

Under this method, the form of the formula is as follows:

Standard Deviation

22

N

fm

N

fm

Steps:

1. Form a table with class intervals and frequencies in the first two

columns.

2. Find the mid values (m) and write them in the next column.

3. Find the products of f and m and write those fm values in the next

column

4. Find the products of m and fm and write those fm2 values in the next

column.

5. Find N(=f), fm and fm2

6. Substitute in the formula and simplify.

Example 10:

The following data were obtained while observing the life span of a few neon

lights of a company calculate S.D.

Life span (years) : 4 – 6 6 – 8 8 – 10 10 – 12 12 – 15

No. of neon lights : 10 17 32 21 20

Solution:

Life span (Years)

No. of Neon lights (f)

Mid value (m)

fm fm2

4 – 6 10 5 50 250

6 – 8 17 7 119 833

8 – 10 32 9 288 2592

10 – 12 21 11 231 2541



12 – 14 20 13 260 3380

N = 100 - fm = 948 fm2 = 9596

Standard Deviation,

22

N

fm

N

fm

2

100

948

100

9596

248.996.95

8704.899600.95

0896.6

= 2.47

13.7 Combined Standard Deviation

When two or three groups merge, the mean and standard deviation of the

combined group are calculated as follows:

13.7.1 Case 1: Merger of two groups

Size Mean SD

Group I N1 1X 1

Group II N2 2X 2

That is

N1 – Number of items in the first group

N2 – Number of items in the second group

1X - Mean of items in the first group

2X - Mean of items in the second group

1 - Standard deviation of items in the first group



2 - Standard deviation of items in the second group.

The mean of the combined group

21

221112

NN

XNXNX

The standard deviation of the combined group

21

222

211

222

211

12NN

dNdNNN

12221211 XXdandXXd

Example 11:

The mean and standard deviation of 63 children on an average test are

respectively 27.6 and 7.1. To them are added a new group of 26 who have

less training and whose mean is 19.2 and standard deviation is 6.2. How will

the value of combined group differ from those of the original 63 children as

to mean and standard deviation?

Solution:

Given number of children Mean Mark S.D. of marks

N1 = 63 6.27X1 1.71

N2 = 26 2.19X2 2.62

Combined mean 21

221112

NN

XNXNX

2663

2.19266.2763

89

2.4998.1738

89

0.2238

= 25.15

Combined standard deviation:



21

222

211

222

211

12NN

dNdNNN

26.63

95.52645.2632.6261.7632222

89

4650.9201575.37844.99983.3175

89

8925.5473

5044.61

= 7.84

13.7.2 Case 2: Merger of three groups

Size Mean SD

Group I N1 1X 1

Group II N2 2X 2

Group III N3 3X 3

The mean of the combined groups

321

332211123

NNN

XNXNXN

The standard deviation of the combined groups

321

333

222

211

333

222

211

123NNN

dNdNdNNNN

Where 123331232212311 XXdXXdXXd

Uses

Standard deviation is the best absolute measure of dispersion. It is a part of

many statistical concepts such as skewness, kurtosis, correlation,

regression, estimation sampling, tests of significance and statistical quality



control. Not only in statistics but also in biology, education, psychology and

other disciplines standard deviation is of immense use.

Merits

1. Standard deviation is rigidly defined

2. It is calculated on the basis of the magnitudes of all the items

3. It could be manipulated further. The combined standard deviation can

be calculated

4. Mistakes in its calculation can be corrected. The entire calculation

need not be redone.

5. Coefficient of variation is based on standard deviation. It is the best

and most widely used relative measure of dispersion

6. It is free from sampling fluctuations. This property of sampling stability

has brought it in dispensable place in tests of significance

7. It reduces the complexity in the approach of normal distribution by

providing standard normal variable

8. It is the most important absolute measure of dispersion. It is used in all

the areas of statistics. It is widely used in other disciplines such as

psychology, education and biology as well.

9. Scientific calculators show the standard deviation of any series.

10. Different forms of the formula are available.

2.7.5 Demerits

1. Compared with other absolute measures of dispersion, it is difficult to

calculate

2. It is not simple to understand

3. It gives more weightage to the items away from the mean than those

near the mean as the deviations are squared.

2.8 Coefficient of variation

Coefficient of variation = 100MeanArithmetic

DeviationStandard



C.V. is the abbreviation

100.M.A

.D.S.V.C

100X

Karl Pearson gave this definition. Like all other relative measures of

dispersion, it is a pure number. All relative measures of dispersion are free

from units of measurement such as kg., metre, litre, etc. The variations in

two or more series (groups or sets of data) are compared on the basis of a

relative measure of dispersion.

For example, an Indian may have different income at various periods of

time. His income is quoted in dollars. The variations in their incomes can be

compared by using any relative measure of dispersion.

Coefficient of variation is the more widely used relative measure of

dispersion and the best measure of central tendency. It is a percentage.

While comparing two or more groups, the group which has less coefficient of

variation is less variable or more consistent or more stable or more uniform

or more homogeneous.

Example 12:

Calculate the coefficient of variation of the following:

40, 41, 45, 49, 50, 51, 55, 59, 60, 60



Solution:

X 51X

XX

2XX

Mean N

XX

00.5110

510

N

XX.D.S

2

4.5010

504

= 7.10

100X

.V.C

10000.51

10.7

= 13.92

40 – 11 121

41 –10 100

45 – 6 36

49 – 2 4

50 – 1 1

51 0 0

55 4 16

59 8 64

60 9 81

60 9 81

510x – 504XX

13.9 Variance

Definition

Variance is the mean square deviation of the values from their arithmetic

mean 2 (read, sigma square) is the symbol. Standard deviation is the

positive square root of variance and is denoted by . The term of variance

was introduced by R.A. Fisher in the year 1913. It is used much in sampling,

analysis of variance, etc. In analysis of variance, total variation is split into a

few components. Each component is due to one factor of variation. The

significance of the variation is then tested.



Formulae

These formulae can be compared with those under standard deviation

Method Individual Discrete Continuous

Observations Series Series

1. Actual mean

N

XX2

N

XXf2

N

Xmf2

2. Direct Method 22

N

x

N

X

22

N

fX

N

fX

22

N

fm

N

fm

3. Assumed mean 22

N

d

N

d

22

N

fd

N

fd

N

fd

N

fd 22

4. Step deviation

22

2

N

d

N

dC

22

2

N

df

N

dfC

22

2

N

df

N

dfC

Individual Observations

Example 13: Number of goals scored by a team in different matches.

Calculate variance.

Goals X 7.1:X

XX 2XX

2 0.3 0.09 Mean

N

XX

7.110

17

Variance,

N

XX2

2

10

10.16

= 1.61

0 – 1.7 2.89

1 – 0.7 0.49

3 1.3 1.69

0 – 1.7 2.89

4 2.3 5.29

3 1.3 1.69

1 – 0.7 0.49



1 – 0.7 0.49

2 0.3 0.09


1. Why is that standard deviation considered to be the most popular

measure of dispersion?

2. Calculate the standard deviation from the following data:

14, 22, 9, 15, 20, 17, 12, 11

3. The table below gives the marks obtained by 10 B.Com. students in

statistics examination. Calculate standard deviation.

Numbers: 1 2 3 4 5 6 7 8 9 10

Narks: 43 48 65 57 31 60 37 48 78 59

4. Calculate standard deviation from the following:

Marks : 10 20 30 40 50 60

No. of students : 8 12 20 10 7 3

5. Compute the standard deviation from the following data:

Class : 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70

Frequency : 8 12 11 14 9 7 4

13.10 Summary

In this unit we discussed the concept of standard deviation, the different

types of formulas are discussed with good examples. The concept of

variance is discussed next with examples.


1. Explain coefficient of variation (co-variance) and given formulae

2. Prices of a particular commodity in five years in two cities are below:

Price in City A Price in City B

20 10

22 20

19 18



23 12

16 15

From the above data find variance and covariance and the city which

had more stable prices?

3. Define Skewness and how many type of formulae?

4. Calculate (i) Karl – Pearson’s coefficient of Skewness and (ii) Bowley’s

coefficient of Skewness for the data given below:

Mid Value : 20 30 40 50 60 70 80

Frequency : 1 12 55 91 55 12 1

5. Calculate Kelly’s coefficient of Skewness

Class : 30 – 49 50 – 69 70 – 89 90 – 109 100 – 1290130 – 149 150 – 169

Frequency : 25 40 50 100 80 50 25

13.12 Answers


1. Karl Person introduced the concept of standard deviation in 1893. It is

the most important measure of dispersion and is widely used in many

statistical formulae. Standard deviation is also called Root-mean square

deviation or Mean Error or Mean Square Error

The reason is that it is the square-root of the means of the squared

deviation from the arithmetic mean. It provides accurate result. In this

method the drawback of ignoring algebraic sign (in mean deviation) is

overcome by taking the square of deviations, there by making all the

deviations as positive.

It is defined as positive square-root of the arithmetic mean of the

squares of the deviation of the given observation from their arithmetic

mean. The standard deviation is denoted by the Greek letter (Sigma).

Deviation taken from actual mean method.



2. Calculate of standard deviation from actual mean

Value (X) 15X

XX

2XX

14 – 1 1

22 7 49

9 – 6 36

15 0 0

20 5 25

17 2 4

12 – 3 9

11 – 4 16

10X 140XX

158

120X

N

XXor

N

X22

18.45.178

140



Alternatively

We can find out standard deviation by using variable directly, i.e. no

deviation is fount out.

Values = x X2

14 196

22 484

9 81

15 225

20 400

17 289

12 144

11 121

X = 120 X2 = 1940

22

N

X

N

X

2

8

120

8

1940

2255.242

5.17

= 4.18

Deviation taken from assumed mean method

The formula

22

N

d

N

d



3. Deviation from assumed mean

R. Numbers Marks x d = X – A d2

(1 – A = 50)

1 43 – 7 49

2 48 – 2 4

3 65 15 225

4 57 7 49

5 31 – 19 361

6 60 10 100

7 37 – 13 169

8 48 – 2 4

9 78 28 784

10 59 9 81

N = 10 d = 26 d2 = 1826

22

N

d

N

d

22

6.26.18210

26

10

1826

76.66.182

84.175

= 13.26



4. Marks (x) f fx 30.8x

xxd

d

2 fd

2

10 8 80 – 20.8 432.64 3461.12

20 12 240 – 10.8 116.64 1399.68

30 20 600 – 0.8 0.64 12.80

40 10 400 9.2 84.64 846.40

50 7 350 19.2 368.64 2580.48

60 3 180 29.2 852.64 2557.92

X = 210 N = 60 fx = 1850 fd2 = 10,858.40

Mean 8.3060

1850

N

fxX

Standard deviation N

fd 2

60

64.10858

= 13.45

Another method

Marks f d = x – 30 fd fd2

10 8 – 20 – 160 3200

20 12 – 10 – 120 1200

30 20 0 0 0

40 10 10 100 1000

50 7 20 140 2800

60 3 30 90 2700

X = 210 N = 60 fx = 1850 fd = 50 fd2 = 10,900

Standard deviation

22

N

fd

N

fd

fd = 50; fd2 = 10,900; N = 60



2

60

50

60

10900

69.067.181

45.1398.180

45.13

Method 3:

Marks f 10

30xd

df 2

df

10 8 – 2 – 16 32

20 12 – 1 –12 12

30 20 0 0 0

40 10 1 10 10

50 7 2 14 28

60 3 3 9 27

5df 109df2

CN

df

2

df2

;109df 2 ;5df N = 60 C = 10

1060

5

60

1092

100069.010817

1081.1

= 1.345 10

= 13.45



5.

Class (x) Mid value

10

35X

C

AXd

f fd fd2

0 – 10 5 – 3 8 – 24 72

10 – 20 15 – 2 12 – 24 48

20 – 30 25 – 1 17 – 17 17

30 – 40 35 0 14 0 0

40 – 50 45 1 9 9 9

50 – 60 55 2 7 14 28

60 – 70 65 3 4 12 36

N = 71 fd = – 30 fd2 = 210

CN

fdAX

A = 35 30fd N = 71 C = 10

1071

3035

= 35 – 4.225 = 30.775

Standard deviation CN

fd

N

fd22

1071

30

71

2102

104225.0957.22

107785.2

= 1.667 10 = 16.67



Terminal Questions

1. The standard deviation is an absolute measure of dispersion. Coefficient

being considered as the “percentage variation in mean, standard

deviation being considered as the total variation in the mean. That is it

shows the relationship between the standard deviation and the

arithmetic mean expressed in terms of percentage.

100mean

deviationstandardvarianceoftCoefficien

(or) Covariance = 100X

2. Calculation of coefficient of variation

Price deviation from dx2 y deviation from dy2

x dx20x dy15y

20 0 0 10 – 5 25

22 2 4 20 5 25

19 – 1 1 18 3 9

23 3 9 12 – 3 9

16 – 4 16 15 0 0

x = 100 dx = 0 dx2 = 30 dy = 0 dy2 = 68

City A City B

205

100

N

xx

155

75

N

yy

20x 15y

N

dx 2

x

N

dy 2

y

5

30

5

68

6 6.13



45.2x 69.3x

100x

.V.C x

100y

.V.Cy

10020

45.2 100

15

69.3

C.V. = 12.25 C.V. = 24.6

Variance 2 = 2.45 Variance 2 = 3.69

City A had more stable prices than in city B because the coefficient of

variations is lower in city A

3. “Skewness is the degree of asymmetry, or departure from symmetry, of

a distributuion”

1. Karl Pearson’s coefficient of Skewness

σS.D.

ZXor

deviationstandard

modemean

PKS

(or)

σ

ZX3

deviationstandard

modemean3

PKS

It can be used when mode is ill defined.

2. Bowley’s coefficient of Skewness

1

13

BKQ

M2QQS

3Q

3. Kelly’s coefficient of Skewness

10

1090

KKP

M2PPS

90P

19

19

DD

M2DD



4. Mid P frequency C

M.A.d df 2df class cf

value f A = 50 C = 10 interval

20 1 – 3 – 3 9 15 – 25 1

30 12 – 2 – 24 48 25 – 35 13

40 55 – 1 – 55 55 35 – 45 68

50 91 0 0 0 45 – 55 159

60 55 1 55 55 55 – 65 214

70 12 2 24 48 65 – 75 226

80 1 3 3 9 75 – 85 227

Total N = 227 – 0 224df 2 – –

i) A.M. 5005010227

050C

N

dfAX

93.9227

0

227

22410

N

fd

N

dfC.D.S

22

Greatest frequency = 91 model class interval: 45 – 55

L = 45 f1 = 91 f0 = 55 f2 = 55

5555912

559110

fff2

ffh

201

01

Mode

5072

360

110182

3610

Karl Pearson’s coefficient of skewness

093.9

5050ZXS

PK

ii) ;75.564

227

4

N Q1 class interval: 35 – 45

L = 35 h = 10 f = 55 C = 13

C

4

N

f

hLQ1



1375.5055

1035

95.4275.4355

1035

Q2 (or) median

5.1132

227

2

N Median class interval: 45 – 55

L = 45 f = 91 h = 10 c = 68

C

2

N

f

hLM

685.11391

1045

= 57. 05

Q3: 25.1704

N3 Q3 class interval: 55 – 65

L = 55 h = 10 f = 55 Q = 159

C

4

N3

f

hLQ3

15925.17055

1055

= 57.05

Bowley’s Skewness 13

13

BKQQ

M2QQS

954205.57

50295.4205.57

010.14

0



5.

Class frequency True class cumulative Intervals frequency

30 – 49 25 29.5 – 49.5 25

50 – 69 40 49.5 – 69.5 65

70 – 89 50 69.5 – 89.5 115

90 – 109 100 89.5 – 109.5 215

110 – 129 80 109.5 – 129.5 295

130 – 149 50 129.5 – 149.5 345

150 – 169 25 149.5 – 169.5 370

N = 370

.37100

37010

100

N10 37th cumulative frequency is included in the class

interval

49.5 – 69.5. It is P10 class interval

L10 = 49.5 h10 = 20 f10 = 40 C10 = 25

10

10

101010 C

100

N10

f

hLP

25.3740

205.49

= 49.5 + 6 = 55.5

1852

370

2

N 89.5 – 109.5 is the class interval

L = 89.5 h = 20 f = 100 C = 115

C

2

N

f

hLM

115185100

205.89

= 89.5 + 14 = 103.5



333100

37090

100

N90

129.5 – 149.5 is P90 class interval

90

90

909090 C

100

N90

f

hLL

29533350

205.129

7.1442.155.129

Kelly’s coefficient of Skewness

1090

1090

KKPP

M2PPS

5.557.144

5.10325.557.144

0762.02.89

8.6



References:

1. Algebra and Trigonometry by Richard Brown

2. Integral calculus by Shanthi Narayan

Publication – S. Chand & Co.

3. Differential calculus by Shanthi Narayan


4. Problems in Calculus of one variable by I. A. Maron

Publication – CBS Publishers

5. Trigonometry by S.L. Loney


6. Applied & Computational Complex Analysis by Peter Henrici

7. Mathematical Analysis by K.G. Binmore.

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