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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.
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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.
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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
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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}.
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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}.
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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
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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.
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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.
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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.
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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
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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.
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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.
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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.
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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
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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
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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‟.
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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.
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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.
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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
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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
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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)
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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
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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}.
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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.
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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.
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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.
1.15 Terminal Questions
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.
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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}.
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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
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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
2.13 Terminal Questions
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.
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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:
At the end of the unit you would be able to
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.
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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?
can not be assigned true or false and so none of them is a statement.
Infact, each of them is a question.
c) Each of the sentences:
i) May you live long!
ii) May God bless you!
can not be assigned true or false and so none of them is a statement.
Infact, each of them is optative.
d) Each of the sentences:
i) Hurrah! We have won the match.
ii) Alas! I have failed.
can not be assigned true or false and so none of them is a statement.
In fact, each of them is exclamatory.
e) Each of the sentences:
i) Good morning to all.
ii) Wish you best of luck.
can not be assigned true or false and so none of them is a statement.
In fact, each them is a wish.
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f) Each of the sentences:
i) Please do me a favour. T
ii) Give me a glass of water.
can not be assigned true or false and so none of them is a statement.
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.
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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.
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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.
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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.
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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.
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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
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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
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(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.
Then, using (D8), negation of p q is given by
~ 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.
Then, using (D8), negation of p q is given by
~ 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
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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
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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
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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
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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
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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
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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.
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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”.
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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
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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.
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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.
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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
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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.
2.13 Terminal Questions
1. Define Tautology and Contradiction
2. Draw the truth tables of Conjunction, disjunction and Biconditional
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2.14 Answers
Self Assessment Questions
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
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~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
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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
3.9 Terminal Questions
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:
At the end of the unit you would be able to
apply the concepts of Algebraic Structure in practical problems
understand Binary Operations and its applications in group theory
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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.
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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.
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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.
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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.
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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:
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+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).
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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:
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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.
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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.
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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
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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.
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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.
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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.
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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
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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.
3.9 Terminal Questions
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
Self Assessment Questions
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
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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
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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
4.10 Terminal Questions
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:
At the end of the unit you would be able to
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
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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
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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
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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
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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
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= 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
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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.
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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
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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
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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)
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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)
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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
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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
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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
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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)
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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
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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
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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
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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
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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
2/360sin60360sin300sin3/5sin
145cot45360cot315cot4/7cot
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
2/360sin60360sin300sin3/5sin
260sec60180sec240sec3/4sec
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2145cos45180cos225cos4/5cos
3/260eccos60180eccos240eccos3/4seccos
145cot45360cot315cot4/7cot
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
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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
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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
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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
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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)
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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
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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
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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
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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
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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
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=
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
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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
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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
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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
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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.
4.10 Terminal Questions
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
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4.11 Answers
Self Assessment Questions
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
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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
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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
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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
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5. LHS = 12cos2222 2
= 2cos4222cos422 22
= 2cos122cos22
= cos2cos41cos212 22
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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
5.6 Terminal Questions
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:
At the end of the unit you would be able to
understand the concept of limit.
apply the concept of continuity in problems.
find whether a given function is continuous or not.
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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
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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.
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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
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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
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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
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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
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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
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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,
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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
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,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
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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
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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)
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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
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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
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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
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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.
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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
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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.
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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
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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.
Self Assessment Questions
SAQ 6: Evaluate the following limits
a) n2
1n2It
n b)
1n
nIt
n c) 100It
n
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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.
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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.
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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.
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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.
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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
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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
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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.
Self Assessment Questions
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
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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.
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5.6 Terminal Questions
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
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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
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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
Self Assessment Questions
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
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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.
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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
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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.
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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
6.13 Terminal Questions
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.
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Objectives:
At the end of the unit you would be able to
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.
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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 .
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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.
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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)
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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
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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
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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
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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
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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
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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.
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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
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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
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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)
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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
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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.
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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
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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.
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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
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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
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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
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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 )
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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
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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
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/
/
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
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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
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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
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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,
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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.
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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.
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6.13 Terminal Questions
1. Differentiate the following functions w.r.t. x.
a) 5x4
a3 b)
x
1xx3 34 c) 257 x5x2x
2. Differentiate the following functions w.r.t. x.
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
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9. If dx
dyfind,
x1x2
x32xy
3
10. Differentiate the following functions w.r.t. x.
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
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6.14 Answers
Self Assessment Questions
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
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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
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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
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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
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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
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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
7.7 Terminal Questions
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:
At the end of the unit you would be able to
integrate standard functions
apply the concept of definite integrals in the process of summation
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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).
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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
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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)
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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)
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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
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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
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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)
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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
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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
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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
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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
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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
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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
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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
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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)
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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
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(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
S.A.Q. 6: Integrate the following functions w.r.t. x
a) xsin 1 b) xxe c) )1x(logx d) xtanx 1
S.A.Q. 7: Integrate the following functions w.r.t. x
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.
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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
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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
Working Example: Evaluate
a) dxxsinx 12 b) dxxsinx 2
Solution
a) dxxsinxI 12
Let 3
xvdxxdvxsinu
321
dxx1
1.
3
xxsin
3
xvduuvudvI
2
31
3
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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
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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)
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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
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Integration of functions of the form cbxax
mIx2
Method:
Step 1: Write Bcbxaxdx
dAmIx 2
Step 2: Find the values of A and B
Step 3: Split the function and integrate
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
Integration of functions of the form cbxax
12
Method: Write cbxax2 in the form 222222 xaoraxorax
and integrate.
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Working Example: Evaluate
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
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c
6
17
6
13x
6
17
6
13x
log
617.
2
1.
3
1
c5x
3
2x
log17
1
c5x3
2x3log
17
1
Integration of functions of the form cbxax
mIx
2
Method:
Step 1: Write Bcbxaxdx
dAIm 2
Step 2: Find the values of A and B
Step 3: Split the function and integrate
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
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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
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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:
Step 1: Write Bcbxaxdx
dAmIx 2
Step 2: Find the values of A and B
Step 3: Split the function and integrate
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
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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
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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)
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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)
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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
S.A.Q. 11: Integrate the following functions w.r.t. x
a) 2xx8
1 b)
20x8x
1
2
S.A.Q. 12: Integrate the following functions w.r.t. x
a) 2xx23 b) 16x9 2
S.A.Q. 13: Integrate the following functions w.r.t. x
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
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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.
7.7 Terminal Questions
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.)
4. Integrate the following functions w.r.t. x
a) 32 xcosx3 b) x
xlogsin
c) xlogx
1 d)
x2
x
xecos
x1e
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5. Integrate the following functions w.r.t. x
a) 2
x
1
x
e6 b)
x2
x
e94
e
c) 0nxtanxsec
xsecn
d) xsin
xcos3
6. Integrate the following functions w.r.t. x
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)
7. Integrate the following functions w.r.t. 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
9. Integrate the following functions w.r.t. x
a) 10x13x3
12
b) 10x4x4
12
10. Integrate the following functions w.r.t. x
a) 2xx43
x2 b)
5x4x
7x6
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11. Integrate the following functions w.r.t. x
a) 10x3x
1
2 b)
2xx1
1
12. Integrate the following function w.r.t. x
a) 6x4x2 b) x1x2
13. Evaluate the following integrals
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)
14. Evaluate the following integrals
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
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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
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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)
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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
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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
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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
8.10 Terminal Questions
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:
At the end of the unit you would be able to
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.
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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.
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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.
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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
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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
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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.
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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)
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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
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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
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Therefore (3) becomes, 1v
1v
dX
dvXv
Or
X
dXdv,
vv21
1v1v
vv21v
1v
1v
dX
dvX
2
2
Integrating both the sides
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
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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
Integrating both the sides
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,
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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)
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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.
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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)
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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
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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
Self Assessment Questions
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.
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8.10 Terminal Questions
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
Self Assessment Questions
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
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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
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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
2. Refer to Section 16.7
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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
9.10 Terminal Questions
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.
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Objectives:
At the end of the unit you would be able to
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)
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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).
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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
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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.
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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
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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.
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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)
+ – –
– + –
– – –( – )
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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)
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....!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.
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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
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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
Self Assessment Questions
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
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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.
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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
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
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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
Self Assessment Questions
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.
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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
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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.
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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
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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.
9.10 Terminal Questions
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
Self Assessment Questions
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
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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
10.11 Terminal Questions
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:
At the end of the unit you would be able to
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.
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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
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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
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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.)
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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
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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
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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
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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
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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.
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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
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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
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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.
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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)
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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
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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.
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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
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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
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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.
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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
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= 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
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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
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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)
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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
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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
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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
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;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
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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
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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)
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= –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
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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
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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.
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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.
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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
14. Evaluate the following determinants
a)
3232
3102
0121
3242
b)
0521
1130
2421
3112
c)
2032
1423
3212
2124
15. Show that accbba
1baab
1acca
1cbbc
16. Evaluate the following determinants
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
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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
Self Assessment Questions
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.
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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
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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)
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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
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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
Self Assessment Questions
11.7 Summary
11.8 Terminal Questions
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:
At the end of the unit you would be able to
understand the properties of infinite series
test the convergence or the divergence of an infinite series
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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,
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,.......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.
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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.
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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
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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
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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,
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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
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(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.
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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
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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
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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
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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
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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
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......!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
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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
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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
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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
Self Assessment Questions
1. Solve ......1296
412
72
12
6
21
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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.
11.8 Terminal Questions
1. Write the general properties of a series
2. Explain the binomial series
11.9 Answers
Self Assessment Questions
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
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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
2. Refer to Section 11.4
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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
12.9 Terminal Questions
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
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(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:
At the end of the unit you would be able to
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
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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
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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
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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
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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.
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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.
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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.
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 ?
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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.
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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
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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)
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= (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
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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
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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
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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.
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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
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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
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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
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= 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
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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.
12.9 Terminal Questions
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.
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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
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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
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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
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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
Self Assessment Questions
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.
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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).
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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
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(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
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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
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(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
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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
Mathematics for IT Unit 13
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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
13.11 Terminal Questions
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.
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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.)
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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
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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 ?
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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
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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
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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.
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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.
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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.
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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.
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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.
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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:
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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
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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
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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.
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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
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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
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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
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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
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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
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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:
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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
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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
7. Substitute in the above formula and simplify.
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
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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
7. Substitute in the above formula and simplify.
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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
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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.
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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
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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
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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
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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:
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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
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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
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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
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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.
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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
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1 – 0.7 0.49
2 0.3 0.09
Self Assessment Questions
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.
13.11 Terminal Questions
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
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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
Self Assessment Questions
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.
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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References:
1. Algebra and Trigonometry by Richard Brown
2. Integral calculus by Shanthi Narayan
Publication – S. Chand & Co.
3. Differential calculus by Shanthi Narayan
Publication – S. Chand & Co.
4. Problems in Calculus of one variable by I. A. Maron
Publication – CBS Publishers
5. Trigonometry by S.L. Loney
Publication – S. Chand & Co.
6. Applied & Computational Complex Analysis by Peter Henrici
7. Mathematical Analysis by K.G. Binmore.
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