11 Math Ncert
11 Math Ncert
11 Math Ncert
Foreword
The National Curriculum Framework (NCF), 2005, recommends that childrens life
at school must be linked to their life outside the school. This principle marks a
departure from the legacy of bookish learning which continues to shape our system
and causes a gap between the school, home and community. The syllabi and textbooks
developed on the basis of NCF signify an attempt to implement this basic idea. They
also attempt to discourage rote learning and the maintenance of sharp boundaries
between different subject areas. We hope these measures will take us significantly
further in the direction of a child-centred system of education outlined in the National
Policy on Education (1986).
The success of this effort depends on the steps that school principals and
teachers will take to encourage children to reflect on their own learning and to
pursue imaginative activities and questions. We must recognise that given space,
time and freedom, children generate new knowledge by engaging with the information
passed on to them by adults. Treating the prescribed textbook as the sole basis of
examination is one of the key reasons why other resources and sites of learning are
ignored. Inculcating creativity and initiative is possible if we perceive and treat
children as participants in learning, not as receivers of a fixed body of knowledge.
These aims imply considerable change in school routines and mode of
functioning. Flexibility in the daily time-table is as necessary as rigour in implementing
the annual calendar so that the required number of teaching days are actually devoted
to teaching. The methods used for teaching and evaluation will also determine how
effective this textbook proves for making childrens life at school a happy experience,
rather than a source of stress or boredom. Syllabus designers have tried to address
the problem of curricular burden by restructuring and reorienting knowledge at
different stages with greater consideration for child psychology and the time available
for teaching. The textbook attempts to enhance this endeavour by giving higher
priority and space to opportunities for contemplation and wondering, discussion in
small groups, and activities requiring hands-on experience.
The National Council of Educational Research and Training (NCERT) appreciates
the hard work done by the Textbook Development Committee responsible for this
book. We wish to thank the Chairperson of the advisory group in Science and
Mathematics, Professor J.V. Narlikar and the Chief Advisor for this book
Professor P.K. Jain for guiding the work of this committee. Several teachers
contributed to the development of this textbook; we are grateful to their principals
for making this possible. We are indebted to the institutions and organisations which
have generously permitted us to draw upon their resources, material and personnel.
We are especially grateful to the members of the National Monitoring Committee,
appointed by the Department of Secondary and Higher Education, Ministry of Human
Resource Development under the Chairpersonship of Professor Mrinal Miri and
Professor G.P. Deshpande, for their valuable time and contribution. As an organisation
committed to the systemic reform and continuous improvement in the quality of its
products, NCERT welcomes comments and suggestions which will enable us to
undertake further revision and refinement.
New Delhi
20 December 2005
Director
National Council of Educational
Research and Training
Acknowledgements
The Council gratefully acknowledges the valuable contributions of the following
participants of the Textbook Review Workshop: P. Bhaskar Kumar, P.G.T., Jawahar
Navodaya Vidyalaya, Ananthpur, (A.P.); Vinayak Bujade, Lecturer, Vidarbha Buniyadi
Junior College, Sakkardara Chowk Nagpur, Maharashtra; Vandita Kalra, Lecturer,
Sarvodaya Kanya Vidyalaya Vikashpuri District Centre, New Delhi; P.L. Sachdeva
Deptt. of Mathematics, Indian Institute of Science, Bangalore, Karnataka; P.K.Tiwari
Assistant Commissioner (Retd.), Kendriya Vidyalaya Sangathan; Jagdish Saran,
Department of Statistics, University of Delhi; Quddus Khan, Lecturer, Shibli National
P.G. College Azamgarh (U.P.); Sumat Kumar Jain, Lecturer, K.L. Jain Inter College
Sasni Hathras (U.P.); R.P. Gihare, Lecturer (BRC), Janpad Shiksha Kendra Chicholi
Distt. Betul (M.P.); Sangeeta Arora, P.G.T., A.P.J. School Saket, New Delhi; P.N.
Malhotra, ADE (Sc.), Directorate of Education, Delhi; D.R. Sharma, P.G.T., J.N.V.
Mungespur, Delhi; Saroj, P.G.T. Government Girls Sr. Secondary School, No. 1,
Roop Nagar, Delhi, Manoj Kumar Thakur, P.G.T., D.A.V. Public School, Rajender
Nagar, Sahibabad, Ghaziabad (U.P.) and R.P. Maurya, Reader, DESM, NCERT,
New Delhi.
Acknowledgements are due to Professor M. Chandra, Head, Department of
Education in Science and Mathematics for her support.
The Council acknowledges the efforts of the Computer Incharge, Deepak Kapoor;
Rakesh Kumar, Kamlesh Rao and Sajjad Haider Ansari, D.T.P. Operators; Kushal Pal
Singh Yadav, Copy Editor and Proof Readers, Mukhtar Hussain and Kanwar Singh.
The contribution of APCOffice, administration of DESM and Publication
Department is also duly acknowledged.
Contents
1.
Foreword
iii
Sets
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
1.11
1.12
1
1
1
5
6
7
9
12
12
13
14
18
21
Introduction
Sets and their Representations
The Empty Set
Finite and Infinite Sets
Equal Sets
Subsets
Power Set
Universal Set
Venn Diagrams
Operations on Sets
Complement of a Set
Practical Problems on Union and Intersection of Two Sets
2.
30
30
30
34
36
3.
Trigonometric Functions
3.1 Introduction
3.2 Angles
3.3 Trigonometric Functions
3.4 Trigonometric Functions of Sum and Difference of Two Angles
3.5 Trigonometric Equations
49
49
49
55
63
74
4.
86
86
87
88
5.
97
97
97
98
102
104
108
6.
Linear Inequalities
6.1 Introduction
6.2 Inequalities
6.3 Algebraic Solutions of Linear Inequalities in One Variable
and their Graphical Representation
6.4 Graphical Solution of Linear Inequalities in Two Variables
6.5 Solution of System of Linear Inequalities in Two Variables
116
116
116
7.
134
134
134
138
148
8.
Binomial Theorem
8.1 Introduction
8.2 Binomial Theorem for Positive Integral Indices
8.3 General and Middle Terms
160
160
160
167
9.
177
177
177
179
181
186
191
194
118
123
127
203
203
204
212
220
225
236
236
236
239
242
247
255
268
268
281
281
281
284
298
303
321
321
321
324
329
335
339
15. Statistics
15.1 Introduction
15.2 Measures of Dispersion
15.3 Range
15.4 Mean Deviation
15.5 Variance and Standard Deviation
15.6 Analysis of Frequency Distributions
347
347
349
349
349
361
372
ix
269
269
271
273
16. Probability
16.1 Introduction
16.2 Random Experiments
16.3 Event
16.4 Axiomatic Approach to Probability
383
383
384
387
394
412
412
412
414
416
419
421
421
421
425
Answers
433
Supplementary Material
466
xi
ISBN 81-7450-486-9
First Edition
February 2006 Phalguna 1927
Reprinted
October 2006 Kartika 1928
November 2007 Kartika 1929
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Chapter
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SETS
In these days of conflict between ancient and modern studies; there
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1.1 Introduction
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Georg Cantor
(1845-1918)
In everyday life, we often speak of collections of objects of a particular kind, such as,
a pack of cards, a crowd of people, a cricket team, etc. In mathematics also, we come
across collections, for example, of natural numbers, points, prime numbers, etc. More
specially, we examine the following 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) Various kinds of triangles
(v) Prime factors of 210, namely, 2,3,5 and 7
(vi) The solution of the equation: x2 5x + 6 = 0, viz, 2 and 3.
We note that each of the above example is a well-defined collection of objects in
MATHEMATICS
the sense that we can definitely decide whether a given particular object belongs to a
given collection or not. For example, we can say that the river Nile does not belong to
the collection of rivers of India. On the other hand, the river Ganga does belong to this
colleciton.
:
:
:
:
:
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We give below a few more examples of sets used particularly in mathematics, viz.
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In roster form, all the elements of a set are listed, the elements are 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}.
SETS
$Note In roster form, the order in which the elements are listed is immaterial.
Thus, the above set can also be represented as {1, 3, 7, 21, 2, 6, 14, 42}.
(b)
(c)
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Note It may be noted that while writing the set in roster form an element is not
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generally repeated, i.e., all the elements are taken as distinct. For example, the set
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, namely, each of them
is a vowel in the English alphabet, and no other letter possess this property. Denoting
this set by V, we write
V = {x : x is a vowel in English alphabet}
It may be observed that we describe the element of the set by using a symbol x
(any other symbol like the letters y, z, etc. could be used) which is followed by a colon
: . After the sign of colon, we write the characteristic property possessed by the
elements of the set and then enclose the whole 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 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 x lies between 3 and 10. Hence, the numbers 4, 5, 6, 7,
8 and 9 are the elements of the set A.
If we denote the sets described in (a), (b) and (c) above in roster form by A, B,
C, respectively, then A, B, 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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(ii)
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MATHEMATICS
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}.
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1 2 3 4 5 6
Example 4 Write the set { , , , , , } in the set-builder form.
2 3 4 5 6 7
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Example 3 Write the set A = {1, 4, 9, 16, 25, . . . }in set-builder form.
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Solution We see that each member in the given set has the numerator one less than
the denominator. Also, the numerator begin from 1 and do not exceed 6. Hence, in the
set-builder form the given set is
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Example 5 Match each of the set on the left described in the roster form with the
same set on the right described in the set-builder form :
(i) {P, R, I, N, C, A, L} (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 PRINCIPAL}
Solution Since in (d), there are 9 letters in the word PRINCIPAL and two letters P and I
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 and so (iii) matches (a). Finally, x2 9 = 0
implies x = 3, 3 and so (iv) matches (b).
EXERCISE 1.1
1.
SETS
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4.
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2.
5.
(ii) B = {x : x is an integer,
6.
1
9
<x< }
2
2
(iii) C = {x : x is an integer, x2 4}
(iv) D = {x : x is a letter in the word LOYAL}
(v) E = {x : x is a month of a year not having 31 days}
(vi) F = {x : x is a consonant in the English alphabet which precedes k }.
Match each of the set on the left in the roster form with the same set on the right
described in set-builder form:
(i) {1, 2, 3, 6}
(a) {x : x is a prime number and a divisor of 6}
(ii) {2, 3}
(b) {x : x is an odd natural number less than 10}
(iii) {M,A,T,H,E,I,C,S} (c) {x : x is natural number and divisor of 6}
(iv) {1, 3, 5, 7, 9}
(d) {x : x is a letter of the word MATHEMATICS}.
MATHEMATICS
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Definition 1 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 an empty set. The
empty set is denoted by the symbol or { }.
We give below a few examples of empty sets.
(i) Let A = {x : 1 < x < 2, x is a natural number}. Then A is the empty set,
because there is no natural number between 1 and 2.
(ii) B = {x : x2 2 = 0 and x is rational number}. Then B is the empty set because
the equation x2 2 = 0 is not satisfied by any rational value of x.
(iii) C = {x : x is an even prime number greater than 2}.Then C is the empty set,
because 2 is the only even prime number.
(iv) D = { x : x2 = 4, x is odd }. Then D is the empty set, because the equation
x2 = 4 is not satisfied by any odd value of x.
Let
A = {1, 2, 3, 4, 5},
B = {a, b, c, d, e, g}
and
C = { men living presently in different parts of 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 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 S,
we mean the number of distinct elements of the set and we denote it by n (S). If n (S)
is a natural number, then S is non-empty finite set.
Consider the set of natural numbers. We see that the number of elements of this
set is not finite since there are infinite number of natural numbers. We say that the set
of natural numbers is an infinite set. The sets A, B and C given above are finite sets
and n(A) = 5, n(B) = 6 and n(C) = some finite number.
Definition 2 A set which is empty or consists of a definite number of elements is
called finite otherwise, the set is called infinite.
Consider some examples :
(i) Let W be the set of the days of the week. Then W is finite.
(ii) Let S be the set of solutions of the equation x2 16 = 0. Then S is finite.
(iii) Let G be the set of 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 possible to write all the elements of an infinite set within
braces { } because the numbers of elements of such a set is not finite. So, we represent
SETS
some infinite set in the roster form by writing a few elements which clearly indicate the
structure of the set followed ( or preceded ) by three dots.
For example, {1, 2, 3 . . .} is the set of natural numbers, {1, 3, 5, 7, . . .} is the set
of odd natural numbers, {. . .,3, 2, 1, 0,1, 2 ,3, . . .} is the set of integers. All these
sets are infinite.
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Note All infinite sets cannot be described in the roster form. For example, 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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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, then the sets A and B are said to be equal.
Clearly, the two sets have exactly the same elements.
Definition 3 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}. Then A = B.
(ii) Let A be the set of prime numbers less than 6 and P the set of prime factors
of 30. Then A and P are equal, since 2, 3 and 5 are the only prime factors of
30 and also these are less than 6.
Note A set does not change if one or more elements of the set are repeated.
$
For example, the sets A = {1, 2, 3} and B = {2, 2, 1, 3, 3} are equal, since each
MATHEMATICS
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Solution Since 0 A and 0 does not belong to any of the sets B, C, D and E, it
follows that, A B, A C, A D, A E.
Since B = but none of the other sets are empty. Therefore B C, B D
and B E. Also C = {5} but 5 D, hence C D.
Since E = {5}, C = E. Further, D = {5, 5} and E = {5}, we find that, D E.
Thus, the only pair of equal sets is C and E.
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Example 8 Which of the following pairs of sets are equal? Justify your answer.
(i) X, 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 + 2 = 0}.
Solution (i) We have, X = {A, L, L, O, Y}, B = {L, O, Y, A, L}. Then X and B are
equal sets as repetition of elements in a set do not change a set. Thus,
X = {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.
EXERCISE 1.2
1.
2.
3.
SETS
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1.6 Subsets
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4.
Consider the sets : X = set of all students in your school, Y = set of all students in your
class.
We note that every element of Y is also an element of X; we say that Y is a subset
of X. The fact that Y is subset of X is expressed in symbols as Y X. The symbol
stands for is a subset of or is contained in.
10
MATHEMATICS
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(i) The set Q of rational numbers is a subset of the set R of real numbes, and
we write Q R.
(ii) If A is the set of all divisors of 56 and B the set of all prime divisors of 56,
then B is a subset of A and we write B A.
(iii) Let A = {1, 3, 5} and B = {x : x is an odd natural number less than 6}. Then
A B and B A and hence A = B.
(iv) Let A = { a, e, i, o, u} and B = { a, b, c, d}. Then A is not a subset of B,
also B is not a subset of A.
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 superset of A. For example,
A = {1, 2, 3} is a proper subset of B = {1, 2, 3, 4}.
If a set A has only one element, we call it a singleton set. Thus,{ a } is a
singleton set.
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Solution (i)
(ii)
(iii)
(iv)
(ii) A . . . B
(iii) A . . . C
(iv) B . . . C
Solution No. Let A = {1}, B = {{1}, 2} and C = {{1}, 2, 3}. Here A B as A = {1}
and B C. But A C as 1 A and 1 C.
Note that an element of a set can never be a subset of itself.
1.6.1 Subsets of set of real numbers
As noted in Section 1.6, there are many important subsets of R. We give below the
names of some of these subsets.
The set of natural numbers N = {1, 2, 3, 4, 5, . . .}
The set of integers
Z = {. . ., 3, 2, 1, 0, 1, 2, 3, . . .}
The set of rational numbers Q = { x : x =
p
, p, q Z and q 0}
q
SETS
11
p
which is read Q is the set of all numbers x such that x equals the quotient q , where
p and q are integers and q is not zero. Members of Q include 5 (which can be
1
7
11
5
, 3
(which can be expressed as ) and .
2
2
3
7
The set of irrational numbers, denoted by T, is composed of all other real numbers.
Thus T = {x : x R and x Q}, i.e., all real numbers that are not rational.
5
1
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expressed as ) ,
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1.6.2 Intervals as subsets of R Let a, b R and a < b. Then the set of real numbers
{ y : a < y < b} is called an open interval and is denoted by (a, b). All the points
between a and b belong to the open interval (a, b) but a, b themselves do not belong to
this interval.
The interval which contains the end points also is called closed interval and is
denoted by [ a, b ]. Thus
[ a, b ] = {x : a x b}
We can also have intervals closed at one end and open at the other, i.e.,
[ a, b ) = {x : a x < b} is an open interval from a to b, including a but excluding b.
( a, b ] = { x : a < x b } is an open interval from a to b including b but excluding a.
These notations provide an alternative way of designating the subsets of set of
real numbers. For example , if A = (3, 5) and B = [7, 9], then A B. The set [ 0, )
defines the set of non-negative real numbers, while set ( , 0 ) defines the set of
negative real numbers. The set ( , ) describes the set of real numbers in relation
to a line extending from to .
On real number line, various types of intervals described above as subsets of R,
are shown in the Fig 1.1.
Fig 1.1
12
MATHEMATICS
The number (b a) is called the length of any of the intervals (a, b), [a, b],
[a, b) or (a, b].
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Consider the set {1, 2}. Let us write down all the subsets of the set {1, 2}. We
know that is a subset of every set . So, is a subset of {1, 2}. We see that {1}
and { 2 }are also subsets of {1, 2}. Also, we know that every set is a subset of
itself. So, { 1, 2 } is a subset of {1, 2}. Thus, the set { 1, 2 } has, in all, four
subsets, viz. , { 1 }, { 2 } and { 1, 2 }. The set of all these subsets is called the
power set of { 1, 2 }.
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Definition 5 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.
Thus, as in above, 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.
Usually, in a particular context, we have to deal with the elements and subsets of a
basic set which is relevant to that particular context. For example, while studying the
system of numbers, we are interested in the set of natural numbers and its subsets such
as the set of all prime numbers, the set of all even numbers, and so forth. This basic set
is called the Universal Set. The universal set is usually denoted by U, and all its
subsets by the letters A, B, C, etc.
For example, for the set of all integers, the universal set can be the set of rational
numbers or, for that matter, the set R of real numbers. For another example, in human
population studies, the universal set consists of all the people in the world.
EXERCISE 1.3
1.
SETS
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5.
6.
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2.
7.
8.
9.
13
Fig 1.2
14
MATHEMATICS
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In earlier classes, we have learnt how to perform the operations of addition, subtraction,
multiplication and division on numbers. Each one of these operations was performed
on a pair of numbers to get another number. For example, when we perform the
operation of addition on the pair of numbers 5 and 13, we get the number 18. Again,
performing the operation of multiplication on the pair of numbers 5 and 13, we get 65.
Similarly, there are some operations which when performed on two sets give rise to
another set. We will now define certain operations on sets and examine their properties.
Henceforth, we will refer all our sets as subsets of some universal set.
1.10.1 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 and all the elements of B, the common elements
being taken only once. The symbol is used to denote the union. Symbolically, we
write A B and usually read as A union B.
Example 12 Let A = { 2, 4, 6, 8} and B = { 6, 8, 10, 12}. Find A B.
Solution We have, A B = { a, e, i, o, u } = A.
This example illustrates that union of sets A and its subset B is the set A
itself, i.e., if B A, then A B = A.
Example 14 Let X = {Ram, Geeta, Akbar} be the set of students of Class XI, who are
in school hockey team. Let Y = {Geeta, David, Ashok} be the set of students from
Class XI who are in the school football team. Find X Y and interpret the set.
Solution We have, X Y = {Ram, Geeta, Akbar, David, Ashok}. This is the set of
students from Class XI who are in the hockey team or the football team or both.
SETS
15
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(iv) A A = A
(Idempotent law)
(v) U A = U
(Law of U)
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(iii) A = A
1.10.2 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.
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}.
Example 15 Consider the sets A and B of Example 12. Find A B.
Solution We see that 6, 8 are the only elements which are common to both A and B.
Hence A B = { 6, 8 }.
Example 16 Consider the sets X and Y of Example 14. Find X Y.
Solution We see that element Geeta is the only element common to both. Hence,
X Y = {Geeta}.
Example 17 Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = { 2, 3, 5, 7 }. Find A B and
hence show that A B = B.
Solution We have A B = { 2, 3, 5, 7 } = B. We
note that B A and that A B = B.
Definition 7 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}
The shaded portion in Fig 1.5 indicates the
Fig 1.5
intersection of A and B.
16
MATHEMATICS
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(i)
(iii)
(ii)
(iv)
(v)
SETS
17
1.10.3 Difference of sets The difference of the 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 minus B.
Example 18 Let A = { 1, 2, 3, 4, 5, 6}, B = { 2, 4, 6, 8 }. Find A B and B A.
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Fig 1.8
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Fig 1.9
EXERCISE 1.4
8.
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7.
(i) A B
(ii) A C
(iii) B C
(iv) B D
(v) A B C
(vi) A B D
(vii) B C D
Find the intersection of each pair of sets of question 1 above.
If A = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13}, C = {11, 13, 15}and D = {15, 17}; find
(i) A B
(ii) B C
(iii) A C D
(iv) A C
(v) B D
(vi) A (B C)
(vii) A D
(viii) A (B D)
(ix) ( A B ) ( B C )
(x) ( A D) ( B C)
If A = {x : x is a natural number }, B = {x : x is an even natural number}
C = {x : x is an odd natural number}andD = {x : x is a prime number }, find
(i) A B
(ii) A C
(iii) A D
(iv) B C
(v) B D
(vi) C D
Which of the following pairs of sets are disjoint
(i) {1, 2, 3, 4} and {x : x is a natural number and 4 x 6 }
(ii) { a, e, i, o, u } and { c, d, e, f }
(iii) {x : x is an even integer } and {x : x is an odd integer}
If A = {3, 6, 9, 12, 15, 18, 21}, B = { 4, 8, 12, 16, 20 },
C = { 2, 4, 6, 8, 10, 12, 14, 16 }, D = {5, 10, 15, 20 }; find
(i) A B
(ii) A C
(iii) A D
(iv) B A
(v) C A
(vi) D A
(vii) B C
(viii) B D
(ix) C B
(x) D B
(xi) C D
(xii) D C
If X= { a, b, c, d } and Y = { f, b, d, g}, find
(i) X Y
(ii) Y X
(iii) X Y
If R is the set of real numbers and Q is the set of rational numbers, then what is
R Q?
State whether each of the following statement is true or false. Justify your answer.
(i) { 2, 3, 4, 5 } and { 3, 6} are disjoint sets.
(ii) { a, e, i, o, u } and { a, b, c, d }are disjoint sets.
(iii) { 2, 6, 10, 14 } and { 3, 7, 11, 15} are disjoint sets.
(iv) { 2, 6, 10 } and { 3, 7, 11} are disjoint sets.
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5.
6.
MATHEMATICS
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10.
11.
12.
Let U be the universal set which consists of all prime numbers and A be the subset of
U which consists of all those prime numbers that are not divisors of 42. Thus,
A = {x : x U and x is not a divisor of 42 }. We see that 2 U but 2 A, because
2 is 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
SETS
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Definition 8 Let U be the universal set and A a subset of U. Then the complement of
A 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 }. Obviously A = U A
We note that the complement of a set A can be looked upon, alternatively, as the
difference between a universal set U and the set A.
Example 20 Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {1, 3, 5, 7, 9}. Find A.
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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 }.
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Solution Since A is the set of all girls, A is clearly the set of all boys in the class.
Note If A is a subset of the universal set U, then its complement A
$
subset of U.
is also a
set U, we have
( A ) = A
( A B ) = { 1, 6 } = A B
It can be shown that the above result is true in general. If A and B are any two
subsets of the universal set U, then
( A B ) = A B. Similarly, ( A B ) = A B . These two results are stated
in words as follows :
MATHEMATICS
2. De Morgans law:
Fig 1.10
(ii) A A =
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(i) (A B) = A B (ii) (A B ) = A B
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EXERCISE 1.5
1.
2.
3.
4.
5.
6.
SETS
7.
21
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Fig 1.11
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Note that the sets A B, A B and B A are disjoint and their union is A B
(Fig 1.11). 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)
(iii) If A, B and C are finite sets, then
n ( A B C ) = n ( A ) + n ( B ) + n ( C ) n ( A B ) n ( B C)
n(A C)+n(A B C)
... (3)
In fact, we have
n ( A B C ) = n (A) + n ( B C ) n [ A ( B C ) ]
[ by (2) ]
= n (A) + n ( B ) + n ( C ) n ( B C ) n [ A ( B C ) ]
[ by (2) ]
Since A ( B C ) = ( A B ) ( A C ), we get
n [ A ( B C ) ] = n ( A B ) + n ( A C ) n [ ( A B ) (A C)]
= n ( A B ) + n ( A C ) n (A B C)
Therefore
n ( A B C ) = n (A) + n ( B ) + n ( C ) n ( A B ) n ( B C)
n(A C)+n(A B C)
This proves (3).
Example 23 If X and Y are two sets such that X Y has 50 elements, X has
28 elements and Y has 32 elements, how many elements does X Y have ?
22
MATHEMATICS
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Solution Let M denote the set of teachers who teach mathematics and P denote the
set of teachers who teach physics. In the statement of the problem, the word or gives
us a clue of union and the word and gives us a clue of intersection. We, therefore,
have
n ( M P ) = 20 , n ( M ) = 12 and n ( M P ) = 4
We wish to determine n ( P ).
Using the result
n ( M P ) = n ( M ) + n ( P ) n ( M P ),
we obtain
20 = 12 + n ( P ) 4
Thus
n ( P ) = 12
Hence 12 teachers teach physics.
Example 25 In a class of 35 students, 24 like to play cricket and 16 like to play
football. Also, each student likes to play at least one of the two games. How many
students like to play both cricket and football ?
Solution Let X be the set of students who like to play cricket and Y be the set of
students who like to play football. Then X Y is the set of students who like to play
at least one game, and X Y is the set of students who like to play both games.
Given
n ( X) = 24, n ( Y ) = 16, n ( X Y ) = 35, n (X Y) = ?
Using the formula n ( X Y ) = n ( X ) + n ( Y ) n ( X Y ), we get
35 = 24 + 16 n (X Y)
SETS
Thus,
i.e.,
23
n (X Y) = 5
5 students like to play both games.
Example 26 In a survey of 400 students in a school, 100 were listed as taking apple
juice, 150 as taking orange juice and 75 were listed as taking both apple as well as
orange juice. Find how many students were taking neither apple juice nor orange juice.
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Solution Let U denote the set of surveyed students and A denote the set of students
taking apple juice and B denote the set of students taking orange juice. Then
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Example 27 There are 200 individuals with a skin disorder, 120 had been exposed to
the chemical C1, 50 to chemical C2, and 30 to both the chemicals C1 and C2. Find the
number of individuals exposed to
(i)
Solution Let U denote the universal set consisting of individuals suffering from the
skin disorder, A denote the set of individuals exposed to the chemical C1 and B denote
the set of individuals exposed to the chemical C2.
Here
Fig 1.13
24
MATHEMATICS
Thus, the number of individuals exposed to chemical C2 and not to chemical C1 is 20.
(iii) The number of individuals exposed either to chemical C1 or to chemical C2, i.e.,
n(AB) =n(A)+n(B)n (AB)
= 120 + 50 30 = 140.
3.
4.
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EXERCISE 1.6
6.
7.
8.
Miscellaneous Examples
Example 28 Show that the set of letters needed to spell CATARACT and the
set of letters needed to spell TRACT are equal.
Solution Let X be the set of letters in CATARACT. Then
X = { C, A, T, R }
Let Y be the set of letters in TRACT. Then
Y = { T, R, A, C, T } = { T, R, A, C }
Since every element in X is in Y and every element in Y is in X. It follows that X = Y.
Example 29 List all the subsets of the set { 1, 0, 1 }.
SETS
25
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Solution Let U be the set of consumers questioned, S be the set of consumers who
liked the product A and T be the set of consumers who like the product B. Given that
n ( U ) = 1000, n ( S ) = 720, n ( T ) = 450
So
n(ST)=n(S)+n(T)n(ST)
= 720 + 450 n (S T) = 1170 n ( S T )
Therefore, n ( S T ) is maximum when n ( S T ) is least. But S T U implies
n ( S T ) n ( U ) = 1000. So, maximum values of n ( S T ) is 1000. Thus, the least
value of n ( S T ) is 170. Hence, the least number of consumers who liked both products
is 170.
Example 33 Out of 500 car owners investigated, 400 owned car A and 200 owned
car B, 50 owned both A and B cars. Is this data correct?
Solution Let U be the set of car owners investigated, M be the set of persons who
owned car A and S be the set of persons who owned car B.
Given that
n ( U ) = 500, n (M ) = 400, n ( S ) = 200 and n ( S M ) = 50.
Then n ( S M ) = n ( S ) + n ( M ) n ( S M ) = 200 + 400 50 = 550
But S M U implies n ( S M ) n ( U ).
This is a contradiction. So, the given data is incorrect.
Example 34 A college awarded 38 medals in football, 15 in basketball and 20 in
cricket. If these medals went to a total of 58 men and only three men got medals in all
the three sports, how many received medals in exactly two of the three sports ?
26
MATHEMATICS
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1.
2.
3.
4.
5.
6.
7.
Decide, among the following sets, which sets are subsets of one and another:
A = { x : x R and x satisfy x2 8x + 12 = 0 },
B = { 2, 4, 6 }, C = { 2, 4, 6, 8, . . . }, D = { 6 }.
In each of the following, determine whether the statement is true or false. If it is
true, prove it. If it is false, give an example.
(i) If x A and A B , then x B
(ii) If A B and B C , then A C
(iii) If A B and B C , then A C
(iv) If A B and B C , then A C
(v) If x A and A B , then x B
(vi) If A B and x B , then x A
Let A, B, and C be the sets such that A B = A C and A B = A C. Show
that B = C.
Show that the following four conditions are equivalent :
(i) A B(ii) A B = (iii) A B = B (iv) A B = A
Show that if A B, then C B C A.
Assume that P ( A ) = P ( B ). Show that A = B
Is it true that for any sets A and B, P ( A ) P ( B ) = P ( A B )? Justify your
answer.
SETS
12.
13.
14.
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11.
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16.
27
Summary
This chapter deals with some basic definitions and operations involving sets. These
are summarised below:
A set is a well-defined collection of objects.
A set which does not contain any element is called empty set.
A set which consists of a definite number of elements is called finite set,
otherwise, the set is called infinite set.
Two sets A and B are said to be equal if they have exactly the same elements.
A set A is said to be subset of a set B, if every element of A is also an element
of B. Intervals are subsets of R.
A power set of a set A is collection of all subsets of A. It is denoted by P(A).
MATHEMATICS
The union of two sets A and B is the set of all those elements which are either
Historical Note
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in A or in B.
The intersection of two sets A and B is the set of all elements which are
common. The difference of two sets A and B in this order is the set of elements
which belong to A but not to B.
The complement of a subset A of universal set U is the set of all elements of U
which are not the elements of A.
For any two sets A and B, (A B) = A B and ( A B ) = A B
If A and B are finite sets such that A B = , then
n (A B) = n (A) + n (B).
If A B , then
n (A B) = n (A) + n (B) n (A B)
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The modern theory of sets is considered to have been originated largely by the
German mathematician Georg Cantor (1845-1918). His papers on set theory
appeared sometimes during 1874 to 1897. His study of set theory came when he
was studying trigonometric series of the form a1 sin x + a2 sin 2x + a3 sin 3x + ...
He published in a paper in 1874 that the set of real numbers could not be put into
one-to-one correspondence wih the integers. From 1879 onwards, he publishd
several papers showing various properties of abstract sets.
Cantors work was well received by another famous mathematician Richard
Dedekind (1831-1916). But Kronecker (1810-1893) castigated him for regarding
infinite set the same way as finite sets. Another German mathematician Gottlob
Frege, at the turn of the century, presented the set theory as principles of logic.
Till then the entire set theory was based on the assumption of the existence of the
set of all sets. It was the famous Englih Philosopher Bertand Russell (18721970 ) who showed in 1902 that the assumption of existence of a set of all sets
leads to a contradiction. This led to the famous Russells Paradox. Paul R.Halmos
writes about it in his book Nave Set Theory that nothing contains everything.
The Russells Paradox was not the only one which arose in set theory.
Many paradoxes were produced later by several mathematicians and logicians.
SETS
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29
Chapter
2.1 Introduction
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Fig 2.1
31
brackets and grouped together in a particular order, i.e., (p,q), p P and q Q . This
leads to the following definition:
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Definition 1 Given two non-empty sets P and Q. The cartesian product P Q is the
set of all ordered pairs of elements from P and Q, i.e.,
P Q = { (p,q) : p P, q Q }
If either P or Q is the null set, then P Q will also be empty set, i.e., P Q =
From the illustration given above we note that
A B = {(red,b), (red,c), (red,s), (blue,b), (blue,c), (blue,s)}.
Again, consider the two sets:
A = {DL, MP, KA}, where DL, MP, KA represent Delhi,
03
Madhya Pradesh and Karnataka, respectively and B = {01,02,
03}representing codes for the licence plates of vehicles issued 02
by DL, MP and KA .
01
If the three states, Delhi, Madhya Pradesh and Karnataka
were making codes for the licence plates of vehicles, with the
DL MP KA
restriction that the code begins with an element from set A,
Fig 2.2
which are the pairs available from these sets and how many such
pairs will there be (Fig 2.2)?
The available pairs are:(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03),
(KA,01), (KA,02), (KA,03) and the product of set A and set B is given by
A B = {(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03), (KA,01), (KA,02),
(KA,03)}.
It can easily be seen that there will be 9 such pairs in the Cartesian product, since
there are 3 elements in each of the sets A and B. This gives us 9 possible codes. Also
note that the order in which these elements are paired is crucial. For example, the code
(DL, 01) will not be the same as the code (01, DL).
As a final illustration, consider the two sets A= {a1, a2} and
B = {b1, b2, b3, b4} (Fig 2.3).
A B = {( a1, b1), (a1, b2), (a1, b3), (a1, b4), (a2, b1), (a2, b2),
Fig 2.3
Remarks
(i) Two ordered pairs are equal, if and only if the corresponding first elements
are equal and the second elements are also equal.
32
MATHEMATICS
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x = 2 and y = 3.
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Solving we get
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Solution Since the ordered pairs are equal, the corresponding elements are equal.
Therefore
x + 1 = 3 and y 2 = 1.
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(iii) Since,
(iv) Using the sets A B and A C from part (ii) above, we obtain
(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)}.
33
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Solution
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Example 6 If A B ={(p, q),(p, r), (m, q), (m, r)}, find A and B.
EXERCISE 2.1
2 5 1
x
1. If + 1, y = , , find the values of x and y.
3 3 3
3
2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of
elements in (AB).
3. If G = {7, 8} and H = {5, 4, 2}, find G H and H G.
4. State whether each of the following statements are true or false. If the statement
is false, rewrite the given statement correctly.
(i) If P = {m, n} and Q = { n, m}, then P Q = {(m, n),(n, m)}.
(ii) If A and B are non-empty sets, then A B is a non-empty set of ordered
5.
6.
7.
8.
9.
34
MATHEMATICS
10. The Cartesian product A A has 9 elements among which are found (1, 0) and
(0,1). Find the set A and the remaining elements of A A.
2.3 Relations
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Consider the two sets P = {a, b, c} and Q = {Ali, Bhanu, Binoy, Chandra, Divya}.
The cartesian product of
P and Q has 15 ordered pairs which
can be listed as P Q = {(a, Ali),
(a,Bhanu), (a, Binoy), ..., (c, Divya)}.
We can now obtain a subset of
P Q by introducing a relation R
between the first element x and the
Fig 2.4
second element y of each ordered pair
(x, y) as
R= { (x,y): x is the first letter of the name y, x P, y Q}.
Then R = {(a, Ali), (b, Bhanu), (b, Binoy), (c, Chandra)}
A visual representation of this relation R (called an arrow diagram) is shown
in Fig 2.4.
Fig 2.5
35
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x is the square of y.
(i) In set-builder form, R = {(x, y): x
is the square of y, x P, y Q}
(ii) In roster form, R = {(9, 3),
Fig 2.6
(9, 3), (4, 2), (4, 2), (25, 5), (25, 5)}
The domain of this relation is {4, 9, 25}.
The range of this relation is { 2, 2, 3, 3, 5, 5}.
Note that the element 1 is not related to any element in set P.
The set Q is the codomain of this relation.
Note The total number of relations that can be defined from a set A to a set B
is the number of possible subsets of A B. If n(A ) = p and n(B) = q, then
n (A B) = pq and the total number of relations is 2pq.
Example 9 Let A = {1, 2} and B = {3, 4}. Find the number of relations from A to B.
Solution We have,
A B = {(1, 3), (1, 4), (2, 3), (2, 4)}.
Since n (AB ) = 4, the number of subsets of AB is 24. Therefore, the number of
relations from A into B will be 24.
Remark A relation R from A to A is also stated as a relation on A.
EXERCISE 2.2
1.
36
2.
MATHEMATICS
4.
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Fig 2.7
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3.
6.
7.
8.
9.
2.4 Functions
In this Section, we study a special type of relation called function. It is one of the most
important concepts in mathematics. We can, visualise a function as a rule, which produces
new elements out of some given elements. There are many terms such as map or
mapping used to denote a function.
Definition 5 A relation f from a set A to a set B is said to be a function if every
element of set A has one and only one image in set B.
In other words, a function f is a relation from a non-empty set A to a non-empty
set B such that the domain of f is A and no two distinct ordered pairs in f have the
same first element.
If f is a function from A to B and (a, b) f, then f (a) = b, where b is called the
image of a under f and a is called the preimage of b under f.
37
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Example 10 Let N be the set of natural numbers and the relation R be defined on
N such that R = {(x, y) : y = 2x, x, y N}.
What is the domain, codomain and range of R? Is this relation a function?
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Solution The domain of R is the set of natural numbers N. The codomain is also N.
The range is the set of even natural numbers.
Since every natural number n has one and only one image, this relation is a
function.
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Example 11 Examine each of the following relations given below and state in each
case, giving reasons whether it is a function or not?
(i) R = {(2,1),(3,1), (4,2)}, (ii) R = {(2,2),(2,4),(3,3), (4,4)}
(iii) R = {(1,2),(2,3),(3,4), (4,5), (5,6), (6,7)}
Solution (i)
(ii)
(iii)
Definition 6 A function which has either R or one of its subsets as its range is called
a real valued function. Further, if its domain is also either R or a subset of R, it is
called a real function.
Example 12 Let N be the set of natural numbers. Define a real valued function
f (1) = 3
f (2) = 5
f (3) = 7
38
MATHEMATICS
no N
C
tt E
o R
be T
re
pu
bl
is
he
Fig 2.8
(ii)
Fig 2.9
39
The graph is a line parallel to x-axis. For example, if f(x)=3 for each xR, then its
graph will be a line as shown in the Fig 2.9.
Polynomial function A function f : R R is said to be polynomial function if
for each x in R, y = f (x) = a0 + a1x + a2x2 + ...+ an xn, where n is a non-negative
integer and a0, a1, a2,...,anR.
he
(iii)
is
no N
C
tt E
o R
be T
re
pu
y = f(x) = x2
bl
y = f (x) = x
16
16
Fig 2.10
40
MATHEMATICS
Example 14 Draw the graph of the function f :R R defined by f (x) = x3, xR.
no N
C
tt E
o R
be T
re
pu
bl
is
he
Solution We have
f(0) = 0, f(1) = 1, f(1) = 1, f(2) = 8, f(2) = 8, f(3) = 27; f(3) = 27, etc.
Therefore, f = {(x,x3): xR}.
The graph of f is given in Fig 2.11.
Fig 2.11
f (x)
, where f(x) and g(x) are
g (x)
1
,
x
x R {0}. Complete the Table given below using this definition. What is the domain
and range of this function?
y =
1
x
1.5
...
...
...
...
...
1.5
...
...
...
y=
1
x
1.5
1 0.5
0.5 0.67 1 2
0.25
0.5
1.5
0.67
0.5
41
no N
C
tt E
o R
be T
re
pu
bl
is
he
The domain is all real numbers except 0 and its range is also all real numbers
except 0. The graph of f is given in Fig 2.12.
FigFig
2.122.12
Fig 2.13
1,if x > 0
f (x) = 0 ,if x = 0
1,if x < 0
is called the signum function. The domain of the signum function is R and the range is
42
MATHEMATICS
no N
C
tt E
o R
be T
re
pu
is
Fig 2.14
bl
x
f(x) = x , x 0and 0for x = 0
he
the set {1, 0, 1}. The graph of the signum function is given by the Fig 2.14.
Fig 2.15
2.4.2 Algebra of real functions In this Section, we shall learn how to add two real
functions, subtract a real function from another, multiply a real function by a scalar
(here by a scalar we mean a real number), multiply two real functions and divide one
real function by another.
(i) Addition of two real functions Let f : X R and g : X R be any two real
functions, where X R. Then, we define (f + g): X R by
43
he
is
(iv) Multiplication of two real functions The product (or multiplication) of two real
functions f:X R and g:X R is a function fg:X R defined by
(fg) (x) = f(x) g(x), for all x X.
This is also called pointwise multiplication.
bl
(v) Quotient of two real functions Let f and g be two real functions defined from
no N
C
tt E
o R
be T
re
pu
f
XR where X R. The quotient of f by g denoted by g is a function defined by ,
f
f ( x)
, provided g(x) 0, x X
( x) =
g ( x)
g
2
Solution We have,
2
2
(f + g) (x) = x + 2x + 1, (f g) (x) = x 2x 1,
f
x2
1
(
x
)
=
,x
(fg) (x) = x (2x + 1) = 2x + x ,
2x + 1
g
2
2
f
negative real numbers. Find (f + g) (x), (f g) (x), (fg) (x) and (x).
g
Solution We have
(f + g) (x) =
(fg) x =
x + x, (f g) (x) =
x( x ) =
3
x2
x x,
1
f
x
(
x
)
=
= x 2,x 0
and g
x
44
MATHEMATICS
EXERCISE 2.3
(ii) f(x) =
(i) f(x) = x
4.
9 x2 .
is
3.
he
2.
bl
1.
9C
+ 32.
5
t(10) (iv) The value of C, when t(C) = 212.
no N
C
tt E
o R
be T
re
pu
(ii) t(28)
(iii)
Miscellaneous Examples
Fig 2.16
45
he
is
Solution
bl
Example 20 Let f = {(1,1), (2,3), (0, 1), (1, 3)} be a linear function from Z into Z.
Find f(x).
no N
C
tt E
o R
be T
re
pu
Solution Since f is a linear function, f (x) = mx + c. Also, since (1, 1), (0, 1) R,
f (1) = m + c = 1 and f (0) = c = 1. This gives m = 2 and f(x) = 2x 1.
Example 21 Find the domain of the function f ( x) =
x2 + 3x + 5
x2 5x + 4
Solution Since x 5x + 4 = (x 4) (x 1), the function f is defined for all real numbers
except at x = 4 and x = 1. Hence the domain of f is R {1, 4}.
Example 22 The function f is defined by
1 x, x < 0
1 , x=0
f (x) =
x +1, x > 0
= 1 (1) = 2; etc,
Fig 2.17
46
MATHEMATICS
x2 + 2 x + 1
.
x 2 8 x + 12
is
f (1.1) f (1)
.
(1.1 1)
bl
2. If f (x) = x , find
he
x 2 , 0 x 2
The relation g is defined by g ( x) =
3x , 2 x 10
Show that f is a function and g is not a function.
no N
C
tt E
o R
be T
re
pu
4. Find the domain and the range of the real function f defined by f (x) =
( x 1) .
5. Find the domain and the range of the real function f defined by f (x) = x 1 .
x2
=
R
f
x
,
:
x
of f.
7. Let f, g : RR be defined, respectively by f(x) = x + 1, g(x) = 2x 3. Find
f
f + g, f g and g .
47
he
Summary
is
A B = {(a, b): a A, b B}
In particular R R = {(x, y): x, y R}
no N
C
tt E
o R
be T
re
pu
bl
The range of the relation R is the set of all second elements of the ordered
pairs in a relation R.
MATHEMATICS
he
(kf) (x)
Historical Note
bl
f
f ( x)
(x) = g ( x) , x X, g(x) 0
g
is
(f.g) (x)
no N
C
tt E
o R
be T
re
pu
48
Chapter
TRIGONOMETRIC FUNCTIONS
A mathematician knows how to solve a problem,
he can not solve it. MILNE
3.1 Introduction
The word trigonometry is derived from the Greek words
trigon and metron and it means measuring the sides of
a triangle. The subject was originally developed to solve
geometric problems involving triangles. It was studied by
sea captains for navigation, surveyor to map out the new
lands, by engineers and others. Currently, trigonometry is
used in many areas such as the science of seismology,
designing electric circuits, describing the state of an atom,
predicting the heights of tides in the ocean, analysing a
musical tone and in many other areas.
In earlier classes, we have studied the trigonometric
Arya Bhatt
ratios of acute angles as the ratio of the sides of a right
(476-550)
angled triangle. We have also studied the trigonometric identities and application of
trigonometric ratios in solving the problems related to heights and distances. In this
Chapter, we will generalise the concept of trigonometric ratios to trigonometric functions
and study their properties.
3.2 Angles
Angle is a measure of rotation of a given ray about its initial point. The original ray is
Vertex
Fig 3.1
50
MATHEMATICS
called the initial side and the final position of the ray after rotation is called the
terminal side of the angle. The point of rotation is called the vertex. If the direction of
rotation is anticlockwise, the angle is said to be positive and if the direction of rotation
is clockwise, then the angle is negative (Fig 3.1).
The measure of an angle is the amount of
rotation performed to get the terminal side from
the initial side. There are several units for
Fig 3.2
measuring angles. The definition of an angle
suggests a unit, viz. one complete revolution from the position of the initial side as
indicated in Fig 3.2.
This is often convenient for large angles. For example, we can say that a rapidly
spinning wheel is making an angle of say 15 revolution per second. We shall describe
two other units of measurement of an angle which are most commonly used, viz.
degree measure and radian measure.
th
1
3.2.1 Degree measure If a rotation from the initial side to terminal side is
of
360
a revolution, the angle is said to have a measure of one degree, written as 1. A degree is
divided into 60 minutes, and a minute is divided into 60 seconds . One sixtieth of a degree is
called a minute, written as 1, and one sixtieth of a minute is called a second, written as 1.
Thus,
1 = 60,
1 = 60
Some of the angles whose measures are 360,180, 270, 420, 30, 420 are
shown in Fig 3.3.
Fig 3.3
TRIGONOMETRIC FUNCTIONS
51
3.2.2 Radian measure There is another unit for measurement of an angle, called
the radian measure. Angle subtended at the centre by an arc of length 1 unit in a
unit circle (circle of radius 1 unit) is said to have a measure of 1 radian. In the Fig
3.4(i) to (iv), OA is the initial side and OB is the terminal side. The figures show the
angles whose measures are 1 radian, 1 radian, 1
(i)
1
1
radian and 1 radian.
2
2
(ii)
(iii)
(iv)
Fig 3.4 (i) to (iv)
l
or l = r .
r
52
MATHEMATICS
P
2
1 A 0
1
2
Fig 3.5
3.2.4 Relation between degree and radian Since a circle subtends at the centre
an angle whose radian measure is 2 and its degree measure is 360, it follows that
2 radian = 360
or
radian = 180
22
, we have
7
1 radian =
Also
1 =
180
= 57 16 approximately.
The relation between degree measures and radian measure of some common angles
are given in the following table:
Degree
30
45
60
90
180
270
360
Radian
3
2
TRIGONOMETRIC FUNCTIONS
53
Notational Convention
Since angles are measured either in degrees or in radians, we adopt the convention
that whenever we write angle , we mean the angle whose degree measure is and
whenever we write angle , we mean the angle whose radian measure is .
Note that when an angle is expressed in radians, the word radian is frequently
Radian measure =
Degree measure
180
Degree measure =
180
Radian measure
40 20 = 40
121
1
121
degree =
radian =
radian.
3
180
540
3
40 20 =
121
radian.
540
6 radians =
1080 7
180
degree
6 degree =
22
= 343
7
degree
11
= 343 + 38 +
Hence
= 343 +
2
minute
11
7 60
minute [as 1 = 60]
11
[as 1 = 60]
= 343 + 38 + 10.9
= 34338 11 approximately.
6 radians = 343 38 11 approximately.
Example 3 Find the radius of the circle in which a central angle of 60 intercepts an
arc of length 37.4 cm (use
22
).
7
54
MATHEMATICS
by r =
r=
60
radian =
180
3
l
, we have
37.43 37.437
=
= 35.7 cm
22
Example 4 The minute hand of a watch is 1.5 cm long. How far does its tip move in
40 minutes? (Use = 3.14).
Solution In 60 minutes, the minute hand of a watch completes one revolution. Therefore,
in 40 minutes, the minute hand turns through
or
2
2
of a revolution. Therefore, = 360
3
3
4
radian. Hence, the required distance travelled is given by
3
l = r = 1.5
4
cm = 2 cm = 2 3.14 cm = 6.28 cm.
3
Example 5 If the arcs of the same lengths in two circles subtend angles 65and 110
at the centre, find the ratio of their radii.
Solution Let r1 and r2 be the radii of the two circles. Given that
1 = 65 =
13
65 =
radian
36
180
2 = 110 =
and
22
110 =
radian
36
180
Let l be the length of each of the arc. Then l = r11 = r22, which gives
r1 22
13
22
r1 =
r2 , i.e., r =
36
36
13
2
Hence
r1 : r2 = 22 : 13.
EXERCISE 3.1
1.
TRIGONOMETRIC FUNCTIONS
2.
4.
11
16
22
).
7
5
7
(iv)
3
6
A wheel makes 360 revolutions in one minute. Through how many radians does
it turn in one second?
Find the degree measure of the angle subtended at the centre of a circle of
(i)
3.
55
(ii) 4
(iii)
22
).
7
In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of
minor arc of the chord.
If in two circles, arcs of the same length subtend angles 60 and 75 at the
centre, find the ratio of their radii.
Find the angle in radian through which a pendulum swings if its length is 75 cm
and th e tip describes an arc of length
(i) 10 cm
(ii) 15 cm
(iii) 21 cm
radius 100 cm by an arc of length 22 cm (Use
5.
6.
7.
,
2
Fig 3.6
56
MATHEMATICS
cos 0 = 1
sin 0 = 0,
=0
2
cos = 1
cos
=1
2
sin = 0
sin
3
3
=0
sin
= 1
2
2
cos 2 = 1
sin 2 = 0
Now, if we take one complete revolution from the point P, we again come back to
same point P. Thus, we also observe that if x increases (or decreases) by any integral
multiple of 2, the values of sine and cosine functions do not change. Thus,
sin (2n + x) = sin x , n Z , cos (2n + x) = cos x , n Z
Further, sin x = 0, if x = 0, , 2 , 3, ..., i.e., when x is an integral multiple of
cos
and cos x = 0, if x =
multiple of
3
5
,
,
, ... i.e., cos x vanishes when x is an odd
2
2
2
. Thus
2
sin x = 0 implies x = n,
, where n is any integer
1
, x n, where n is any integer.
sin x
1
cosec x =
sec x
tan x
cot x
sin x
TRIGONOMETRIC FUNCTIONS
57
(why?)
1 + cot2 x = cosec2 x
(why?)
3
2
sin
1
2
1
2
3
2
cos
3
2
1
2
1
2
tan
1
3
not
defined
not
defined
Fig 3.7
58
MATHEMATICS
3
3
) a and b are both negative and in the fourth quadrant (
< x < 2) a is
2
2
positive and b is negative. Therefore, sin x is positive for 0 < x < , and negative for
( < x <
3
, negative for < x <
and also
2
2
2
3
< x < 2. Likewise, we can find the signs of other trigonometric
2
II
III
IV
sin x
cos x
tan x
cosec x
sec x
cot x
3.3.2 Domain and range of trigonometric functions From the definition of sine
and cosine functions, we observe that they are defined for all real numbers. Further,
we observe that for each real number x,
1 sin x 1 and 1 cos x 1
Thus, domain of y = sin x and y = cos x is the set of all real numbers and range
is the interval [1, 1], i.e., 1 y 1.
TRIGONOMETRIC FUNCTIONS
59
1
Since cosec x = sin x , the domain of y = cosec x is the set { x : x R and
x n , n Z} and range is the set {y : y R, y 1 or y 1}. Similarly, the domain
, sin x
2
3
, sin x decreases from 0 to 1and finally, in
2
3
to 2.
the fourth quadrant, sin x increases from 1 to 0 as x increases from
2
Similarly, we can discuss the behaviour of other trigonometric functions. In fact, we
have the following table:
third quadrant, as x increases from to
I quadrant
II quadrant
sin
increases from 0 to 1
decreases from 1 to 0
cos
tan
cot
sec
increases from 1 to increases from to1 decreases from 1to decreases from to 1
III quadrant
IV quadrant
increases from to 0
Remark In the above table, the statement tan x increases from 0 to (infinity) for
0<x<
60
MATHEMATICS
3
, 2) and assumes arbitrarily large negative values as
2
x approaches to 2. The symbols and simply specify certain types of behaviour
cosec x decreases for x (
of functions and variables.
We have already seen that values of sin x and cos x repeats after an interval of
2. Hence, values of cosec x and sec x will also repeat after an interval of 2. We
Fig 3.8
Fig 3.9
Fig 3.10
Fig 3.11
TRIGONOMETRIC FUNCTIONS
Fig 3.12
61
Fig 3.13
shall see in the next section that tan ( + x) = tan x. Hence, values of tan x will repeat
after an interval of . Since cot x is reciprocal of tan x, its values will also repeat after
an interval of . Using this knowledge and behaviour of trigonometic functions, we can
sketch the graph of these functions. The graph of these functions are given above:
Example 6 If cos x = 3 , x lies in the third quadrant, find the values of other five
5
trigonometric functions.
Now
3
5
, we have sec x =
5
3
sin2 x + cos2 x = 1, i.e., sin2 x = 1 cos2 x
or
sin2 x = 1
16
9
=
25
25
4
5
Since x lies in third quadrant, sin x is negative. Therefore
Hence
sin x =
sin x =
4
5
5
4
62
MATHEMATICS
Further, we have
tan x =
sin x 4
cos x 3
=
and cot x =
= .
cos x 3
sin x 4
Example 7 If cot x =
5
, x lies in second quadrant, find the values of other five
12
trigonometric functions.
Solution
Since cot x =
12
5
, we have tan x =
5
12
Now
sec2 x = 1 + tan2 x = 1 +
Hence
sec x =
144 169
=
25
25
13
5
13
,
5
cos x
5
13
Further, we have
sin x = tan x cos x = (
and
cosec x =
12
5
12
) (
)=
5
13
13
1
13
=
.
sin x 12
31
.
3
31
3
= sin (10 + ) = sin
=
.
3
3
3
2
TRIGONOMETRIC FUNCTIONS
63
EXERCISE 3.2
Find the values of other five trigonometric functions in Exercises 1 to 5.
1. cos x =
1
, x lies in third quadrant.
2
2. sin x =
3
, x lies in second quadrant.
5
3. cot x =
3
, x lies in third quadrant.
4
13
, x lies in fourth quadrant.
5
5
5. tan x =
, x lies in second quadrant.
12
4. sec x =
7.
cosec ( 1410)
19
3
9.
sin (
8. tan
10. cot (
11
)
3
15
)
4
64
3.
MATHEMATICS
Consider the unit circle with centre at the origin. Let x be the angle P4OP1and y be
the angle P1OP2. Then (x + y) is the angle P4OP2. Also let ( y) be the angle P4OP3.
Therefore, P 1 , P 2 , P 3 and P 4 will have the coordinates P 1 (cos x, sin x),
P2 [cos (x + y), sin (x + y)], P3 [cos ( y), sin ( y)] and P4 (1, 0) (Fig 3.14).
Fig 3.14
Consider the triangles P1OP3 and P2OP4. They are congruent (Why?). Therefore,
P1P3 and P2P4 are equal. By using distance formula, we get
P 1P 32
Also,
P 2P 42
(Why?)
TRIGONOMETRIC FUNCTIONS
65
cos (
x ) = sin x
2
If we replace x by
cos (
x ) = cos
cos x + sin
sin x = sin x.
2
2
2
x ) = cos x
2
Using the Identity 5, we have
6.
sin (
x ) = cos x = cos x.
2
2 2
7. sin (x + y) = sin x cos y + cos x sin y
We know that
sin (
cos ( + x ) = sin x
2
x) = cos x
cos (
sin ( + x ) = cos x
2
x) = sin x
sin (
66
MATHEMATICS
+ x) = cos x
cos (
x) = cos x
cos (2
+ x) = sin x
sin (
x) = sin x
sin (2
Similar results for tan x, cot x, sec x and cosec x can be obtained from the results of sin
x and cos x.
10. If none of the angles x, y and (x + y) is an odd multiple of
tan (x + y) =
, then
2
tan x + tan y
1 tan x tan y
tan ( x y) =
tan x tan y
1 + tan x tan y
tan x tan ( y )
tan x tan y
=
1 tan x tan ( y ) 1 tan x tan y
cot x cot y 1
cot ( x + y) = cot y +cot x
TRIGONOMETRIC FUNCTIONS
67
Since, none of the x, y and (x + y) is multiple of , we find that sin x sin y and
sin (x + y) are non-zero. Now,
cot ( x + y) =
cot x cot y 1
cot y cot x
cot x cot y + 1
if none of angles x, y and xy is a multiple of
cot y cot x
We know that
cos (x + y) = cos x cos y sin x sin y
Replacing y by x, we get
cos 2x = cos2x sin2 x
= cos2 x (1 cos2 x) = 2 cos2x 1
Again,
cos 2x = cos2 x sin2 x
= 1 sin2 x sin2 x = 1 2 sin2 x.
We have
cos 2 x sin 2 x
cos 2x = cos x sin x =
cos 2 x sin 2 x
2
1 tan 2 x
, x n , where n is an integer
2
2
1+ tan x
2tan x
x n , where n is an integer
2
1 + tan x
2
We have
sin (x + y) = sin x cos y + cos x sin y
Replacing y by x, we get sin 2x = 2 sin x cos x.
Again
sin 2x =
2sin x cos x
cos 2 x sin 2 x
68
MATHEMATICS
2tan x
1tan 2 x
2tan x
if 2 x n , where n is an integer
2
1 tan x
2
We know that
tan x tan y
tan (x + y) = 1 tan x tan y
Replacing y by x , we get tan 2 x
2 tan x
1 tan 2 x
if 3x n , where n is an integer
2
1 3 tan x
2
We have tan 3x =tan (2x + x)
19. tan 3 x
2tan x
tan x
1 tan 2 x
tan 2 x tan x
=
1 tan 2 x tan x 1 2tan x . tan x
1 tan 2 x
TRIGONOMETRIC FUNCTIONS
20.
1 tan 2 x 2tan 2 x
1 3tan 2 x
x+ y
x y
cos
2
2
x+ y
x y
sin
2
2
x+ y
x y
cos
2
2
x+ y
x y
sin
2
2
We know that
cos (x + y) = cos x cos y sin x sin y
and
cos (x y) = cos x cos y + sin x sin y
Adding and subtracting (1) and (2), we get
cos (x + y) + cos(x y) = 2 cos x cos y
and
cos (x + y) cos (x y) = 2 sin x sin y
Further sin (x + y) = sin x cos y + cos x sin y
and
sin (x y) = sin x cos y cos x sin y
Adding and subtracting (5) and (6), we get
sin (x + y) + sin (x y) = 2 sin x cos y
sin (x + y) sin (x y) = 2cos x sin y
Let x + y = and x y = . Therefore
x
and y
2
Substituting the values of x and y in (3), (4), (7) and (8), we get
cos + cos = 2 cos
cos
cos cos = 2 sin
sin
2
sin + sin = 2 sin
cos
2
2
... (1)
... (2)
...
...
...
...
(3)
(4)
(5)
(6)
... (7)
... (8)
69
70
MATHEMATICS
sin sin = 2 cos
sin
2
2
Since and can take any real values, we can replace by x and by y.
Thus, we get
cos x + cos y = 2 cos
sin x + sin y = 2 sin
Remark
21. (i)
(ii)
(iii)
(iv)
x y
x y
x y
x y
cos
sin
; cos x cos y = 2 sin
,
2
2
2
2
x y
x y
x y
x y
cos
sin
; sin x sin y = 2 cos
.
2
2
2
2
5
L.H.S. = 3sin sec 4sin cot
6
3
6
4
=3
2 4 sin 1 = 3 4 sin
6
2
6
1
= 1 = R.H.S.
2
Example 11 Find the value of sin 15.
=34
Solution We have
sin 15 = sin (45 30)
= sin 45 cos 30 cos 45 sin 30
=
1
3 1 1
3 1
.
2 2
2 2 2 2
13
.
12
TRIGONOMETRIC FUNCTIONS
Solution We have
tan
13
= tan
12
tan
= tan
12
12
4 6
1
tan tan
4
6
=
=
1
1 tan tan
4
6
1
3 3 1 2 3
1
3 1
3
tan 3x
tan 2 x tan x
1 tan 2 x tan x
or
tan 3x tan 3x tan 2x tan x = tan 2x + tan x
or
tan 3x tan 2x tan x = tan 3x tan 2x tan x
or
tan 3x tan 2x tan x = tan 3x tan 2x tan x.
Example 15 Prove that
cos x cos x 2 cos x
4
4
Solution Using the Identity 20(i), we have
71
72
MATHEMATICS
L.H.S.
cos x cos x
4
4 x 4 x
4 x ( 4 x)
2cos
cos
2
2
= 2 cos
cos x = 2
cos x =
2
4
2 cos x = R.H.S.
cos 7 x cos 5 x
cot x
sin 7 x sin 5 x
L.H.S.
Solution We have
L.H.S.
cos5 x cos x
cos5 x cos x
2sin 3 x cos 2 x 2sin 3 x
sin 3 x (cos 2 x 1)
2sin 3x sin 2x
sin 3x sin 2x
1 cos 2 x
2sin 2 x
= tan x = R.H.S.
sin 2 x
2sin x cos x
TRIGONOMETRIC FUNCTIONS
73
EXERCISE 3.3
Prove that:
1. sin2
1
+ cos2 tan2
3
6
4
2
2. 2sin2
7
3
cos 2
+ cosec2
6
6
3 2
2 3
cosec 3tan 2 6
2cos 2 2sec2 10
4. 2sin
6
6
6
4
4
3
5. Find the value of:
(i) sin 75
(ii) tan 15
2
3. cot
4
4
7.
tan x
2
4
1 tan x
1 tan x
tan x
4
8.
cos ( x) cos ( x)
cot 2 x
sin ( x) cos x
2
10. sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
11.
cos 9 x cos 5x
sin 17 x sin 3x
sin x sin y
cos x cos y
tan
sin x sin 3x
2
sin x cos x
sin 2 x
cos10 x
x y
2
2 sin x
17.
19.
21.
sin 5x sin 3x
cos 5x cos 3x
sin x sin 3x
cos x cos 3x
tan 4 x
tan 2 x
cot 3x
74
MATHEMATICS
tan 4 x
4tan x (1 tan 2 x)
1 6 tan 2 x tan 4 x
24.
3
2
3
sin sin
and sin
.
3
3
3
2
3
2
2
and
.
3
3
1
3
6
6
6
3
and
tan 2 tan
6
6
3
Thus
tan
5
11
1
tan
.
6
6
3
5
11
and
.
6
6
We will now find the general solutions of trigonometric equations. We have already
TRIGONOMETRIC FUNCTIONS
75
seen that:
sin x =0 gives x = n, where n Z
, where n Z.
2
We shall now prove the following results:
cos x =0 gives x = (2n + 1)
which gives
cos
x y
2
x y
= 0 or sin
x y
2
x y
2
sin
x y
2
=0
=0
x y
= n, where n Z
2
2
2
i.e.
x = (2n + 1) y or x = 2n + y, where nZ
Hence
x = (2n + 1) + (1)2n + 1 y or x = 2n +(1)2n y, where n Z.
Combining these two results, we get
x = n + (1)n y, where n Z.
Therefore
= (2n + 1)
or
sin
x y
2
x y
=0
or
2 sin
sin
x y
2
x y
2
sin
x y
2
=0
x y
i.e.
= n, where n Z
2
x = 2n y or x = 2n + y, where n Z
Hence
x = 2n y, where n Z
Therefore
= n or
=0
, then
2
76
MATHEMATICS
Proof
or
which gives
sin (x y) = 0
(Why?)
Therefore
x y = n, i.e., x = n + y, where n Z
sin x = sin
3
2
sin sin
3
3
3
4
, which gives
3
x n ( 1) n
Note
= sin
4
, where n Z.
3
4
3
is one such value of x for which sin x
. One may take any
3
2
x 2n
1
2
cos
2
3
, where n Z.
3
TRIGONOMETRIC FUNCTIONS
or
tan2 x tan x
6
Therefore
2 x n x
5
, where nZ
6
5
, where nZ.
6
Example 23 Solve sin 2x sin 4x + sin 6x = 0.
x n
or
1
2
or cos 2 x
Therefore
sin 4x = 0
i.e.
Hence
4 x n or 2 x 2n
i.e.
, where nZ
3
n
or x n , where nZ.
4
6
2 1 sin 2 x 3 sin x 0
or
2 sin 2 x 3 sin x 2 0
or
(2sinx 1) (sinx 2) 0
Hence
sin x =
But
1
or sin x = 2
2
sin x = 2 is not possible (Why?)
Therefore
sin x =
1
7
= sin
.
2
6
77
78
MATHEMATICS
x n ( 1)n
7
, where n Z.
6
EXERCISE 3.4
Find the principal and general solutions of the following equations:
1.
tan x
2. sec x = 2
3. cot x 3
4. cosec x = 2
Find the general solution for each of the following equations:
5. cos 4 x = cos 2 x
6. cos 3x + cos x cos 2x = 0
7. sin 2x + cos x = 0
8. sec2 2x = 1 tan 2x
9. sin x + sin 3x + sin 5x = 0
Miscellaneous Examples
Example 25 If sin x =
, cos y =
5
find the value of sin (x + y).
12
13
cos2 x = 1 sin2 x = 1
9
25
... (1)
16
25
4
cos x = .
5
Since x lies in second quadrant, cos x is negative.
Therefore
Hence
cos x =
Now
sin2y = 1 cos2y = 1
i.e.
sin y =
5
13
144
169
25
169
5
13
. Substituting
TRIGONOMETRIC FUNCTIONS
sin( x y )
Example 26
Solution
3 12 4 5
5 13 5 13
36 20
56
.
65 65
65
Prove that
x
9x
5x
cos 2 x cos cos 3x cos
sin 5 x sin
.
2
2
2
We have
L.H.S. =
x
1
9x
2
2
2
1
x
x
9x
9x
2
2
2
1
5x
3x
15x
3x 1
5x
15x
cos cos
cos
cos = cos cos
2
2
2
2
2 2
2
2
5 x 15 x 5 x 15 x
2 2 2 2
1
2sin
sin
= 2
2
2
5x
5x
= sin 5x sin sin 5x sin
= R.H.S.
2
2
. Then 2 x .
4
8
2 tan x
Now
tan 2 x
or
2tan
8
tan
4 1 tan 2
8
Let y = tan
2y
. Then 1 =
1 y2
8
1 tan 2 x
.
8
79
80
MATHEMATICS
or
y2 + 2y 1 = 0
Therefore
Since
y=
2 2 2
2
1 2
2 1.
8
tan
x
x
x
3
3
Example 28 If tan x = , < x <
, find the value of sin , cos and tan .
4
2
2
2
2
Solution Since x
x 3
.
2 2 4
Also
Therefore, sin
3
, cos x is negative.
2
x
2
x
2
is negative.
Now
sec2 x = 1 + tan2 x = 1
Therefore
cos2 x =
Now
2 sin 2
Therefore
sin2
or
sin
x
2
x
2
16
9
16
or cos x =
25
4
5
25
16
(Why?)
x
4 9
= 1 cos x = 1 .
5 5
2
=
9
10
(Why?)
10
Again
2cos2
Therefore
cos2
x
2
x
= 1+ cos x = 1
=
1
10
4
5
1
5
TRIGONOMETRIC FUNCTIONS
or
cos
Hence
tan
Example 29
Solution
x
2
sin
=
cos
x
2
x
10
(Why?)
10
= 3.
10 1
3
3
Prove that cos2 x + cos2 x cos 2 x .
3
3
We have
L.H.S. = 1 cos 2 x
2
2
2
1 cos 2 x
1 cos 2 x
3
3 .
2
2
1
2
2
3 cos 2 x cos 2 x
cos 2 x
2
3
3
1
2
3 cos 2 x 2cos 2 x cos
2
3
2
3
2
3
1
3
3 cos 2 x cos 2 x = R.H.S.
2
2
9
3
5
cos
cos
cos 0
13
13
13
13
1.
2 cos
2.
81
82
MATHEMATICS
x y
2
x y
2
5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x
6.
x
2
, cos
x
2
and tan
x
2
x
2
cos
3x
4
8. tan x = , x in quadrant II
3
10. sin x =
1
9. cos x = , x in quadrant III
3
1
, x in quadrant II
4
Summary
Radian measure =
Degree measure
180
180
Degree measure =
cos2 x + sin2 x = 1
1 + tan2 x = sec2 x
1 + cot2 x = cosec2 x
cos (2n + x) = cos x
sin (2n + x) = sin x
sin ( x) = sin x
cos ( x) = cos x
Radian measure
radians, then
TRIGONOMETRIC FUNCTIONS
cos ( 2 x ) = sin x
sin ( 2 x ) = cos x
sin (x + y) = sin x cos y + cos x sin y
sin (x y) = sin x cos y cos x sin y
+ x = sin x
2
cos ( x) = cos x
sin + x = cos x
2
sin ( x) = sin x
cos ( + x) = cos x
sin ( + x) = sin x
cos (2 x) = cos x
sin (2 x) = sin x
cos
y) is an odd multiple of
, then
2
tan x tan y
1 tan x tan y
tan x tan y
y) is a multiple of , then
cot x cot y 1
cot (x + y) = cot y cot x
cot (x y) =
cot x cot y 1
cot y cot x
1 tan 2 x
1 tan 2 x
83
84
MATHEMATICS
2 tan x
1 tan 2 x
2tanx
tan 2x = 1 tan 2 x
sin 3x = 3sin x 4sin3 x
cos 3x = 4cos3 x 3cos x
3tan x tan 3 x
tan 3x = 1 3tan 2 x
x y
x y
cos
2
2
x y
x y
sin
2
2
x y
x y
cos
2
2
x y
x y
sin
2
2
(i) 2cos x cos y = cos ( x + y) + cos ( x y)
sin x
= 0 gives x = n, where n Z.
cos x
= 0 gives x = (2n + 1)
, where n Z.
2
TRIGONOMETRIC FUNCTIONS
Historical Note
The study of trigonometry was first started in India. The ancient Indian
Mathematicians, Aryabhatta (476), Brahmagupta (598), Bhaskara I (600) and
Bhaskara II (1114) got important results. All this knowledge first went from
India to middle-east and from there to Europe. The Greeks had also started the
study of trigonometry but their approach was so clumsy that when the Indian
approach became known, it was immediately adopted throughout the world.
In India, the predecessor of the modern trigonometric functions, known as
the sine of an angle, and the introduction of the sine function represents the main
contribution of the siddhantas (Sanskrit astronomical works) to the history of
mathematics.
Bhaskara I (about 600) gave formulae to find the values of sine functions
for angles more than 90. A sixteenth century Malayalam work Yuktibhasa
(period) contains a proof for the expansion of sin (A + B). Exact expression for
sines or cosines of 18, 36, 54, 72, etc., are given by
Bhaskara II.
The symbols sin1 x, cos1 x, etc., for arc sin x, arc cos x, etc., were
suggested by the astronomer Sir John F.W. Hersehel (1813) The names of Thales
(about 600 B.C.) is invariably associated with height and distance problems. He
is credited with the determination of the height of a great pyramid in Egypt by
measuring shadows of the pyramid and an auxiliary staff (or gnomon) of known
height, and comparing the ratios:
H h
= tan (suns altitude)
S s
Thales is also said to have calculated the distance of a ship at sea through
the proportionality of sides of similar triangles. Problems on height and distance
using the similarity property are also found in ancient Indian works.
85
Chapter
he
PRINCIPLE OF
MATHEMATICAL INDUCTION
is
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4.1 Introduction
bl
One key basis for mathematical thinking is deductive reasoning. An informal, and example of deductive reasoning,
borrowed from the study of logic, is an argument expressed
in three statements:
(a) Socrates is a man.
(b) All men are mortal, therefore,
(c) Socrates is mortal.
If statements (a) and (b) are true, then the truth of (c) is
established. To make this simple mathematical example,
we could write:
(i) Eight is divisible by two.
G . Peano
(ii) Any number divisible by two is an even number,
(1858-1932)
therefore,
(iii) Eight is an even number.
Thus, deduction in a nutshell is given a statement to be proven, often called a
conjecture or a theorem in mathematics, valid deductive steps are derived and a
proof may or may not be established, i.e., deduction is the application of a general
case to a particular case.
In contrast to deduction, inductive reasoning depends on working with each case,
and developing a conjecture by observing incidences till we have observed each and
every case. It is frequently used in mathematics and is a key aspect of scientific
reasoning, where collecting and analysing data is the norm. Thus, in simple language,
we can say the word induction means the generalisation from particular cases or facts.
87
In algebra or in other discipline of mathematics, there are certain results or statements that are formulated in terms of n, where n is a positive integer. To prove such
statements the well-suited principle that is usedbased on the specific technique, is
known as the principle of mathematical induction.
4.2 Motivation
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Fig 4.1
When the first tile is pushed in the indicated direction, all the tiles will fall. To be
absolutely sure that all the tiles will fall, it is sufficient to know that
(a) The first tile falls, and
(b) In the event that any tile falls its successor necessarily falls.
This is the underlying principle of mathematical induction.
We know, the set of natural numbers N is a special ordered subset of the real
numbers. In fact, N is the smallest subset of R with the following property:
A set S is said to be an inductive set if 1 S and x + 1 S whenever x S. Since
N is the smallest subset of R which is an inductive set, it follows that any subset of R
that is an inductive set must contain N.
Illustration
Suppose we wish to find the formula for the sum of positive integers 1, 2, 3,...,n, that is,
a formula which will give the value of 1 + 2 + 3 when n = 3, the value 1 + 2 + 3 + 4,
when n = 4 and so on and suppose that in some manner we are led to believe that the
formula 1 + 2 + 3+...+ n =
n ( n + 1)
is the correct one.
2
How can this formula actually be proved? We can, of course, verify the statement
for as many positive integral values of n as we like, but this process will not prove the
formula for all values of n. What is needed is some kind of chain reaction which will
88
MATHEMATICS
have the effect that once the formula is proved for a particular positive integer the
formula will automatically follow for the next positive integer and the next indefinitely.
Such a reaction may be considered as produced by the method of mathematical induction.
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Suppose there is a given statement P(n) involving the natural number n such that
(i) The statement is true for n = 1, i.e., P(1) is true, and
(ii) If the statement is true for n = k (where k is some positive integer), then
the statement is also true for n = k + 1, i.e., truth of P(k) implies the
truth of P (k + 1).
Then, P(n) is true for all natural numbers n.
Property (i) is simply a statement of fact. There may be situations when a
statement is true for all n 4. In this case, step 1 will start from n = 4 and we shall
verify the result for n = 4, i.e., P(4).
Property (ii) is a conditional property. It does not assert that the given statement
is true for n = k, but only that if it is true for n = k, then it is also true for n = k +1. So,
to prove that the property holds , only prove that conditional proposition:
If the statement is true for n = k, then it is also true for n = k + 1.
This is sometimes referred to as the inductive step. The assumption that the given
statement is true for n = k in this inductive step is called the inductive hypothesis.
For example, frequently in mathematics, a formula will be discovered that appears
to fit a pattern like
1 = 12 =1
4 = 22 = 1 + 3
9 = 32 = 1 + 3 + 5
16 = 42 = 1 + 3 + 5 + 7, etc.
It is worth to be noted that the sum of the first two odd natural numbers is the
square of second natural number, sum of the first three odd natural numbers is the
square of third natural number and so on.Thus, from this pattern it appears that
1 + 3 + 5 + 7 + ... + (2n 1) = n2 , i.e,
the sum of the first n odd natural numbers is the square of n.
Let us write
P(n): 1 + 3 + 5 + 7 + ... + (2n 1) = n2.
We wish to prove that P(n) is true for all n.
The first step in a proof that uses mathematical induction is to prove that
P (1) is true. This step is called the basic step. Obviously
1 = 12, i.e., P(1) is true.
The next step is called the inductive step. Here, we suppose that P (k) is true for some
89
n( n + 1) (2n + 1)
.
6
n(n + 1)(2n + 1)
6
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P(n) : 12 + 22 + 32 + 42 ++ n2 =
bl
12 + 22 + 32 + 42 ++ n2 =
is
he
positive integer k and we need to prove that P (k + 1) is true. Since P (k) is true, we
have
1 + 3 + 5 + 7 + ... + (2k 1) = k2
... (1)
Consider
1 + 3 + 5 + 7 + ... + (2k 1) + {2(k +1) 1}
... (2)
[Using (1)]
= k2 + (2k + 1) = (k + 1)2
1(1 + 1) (2 1 + 1) 1 2 3
=1 which is true.
=
6
6
Assume that P(k) is true for some positive integer k, i.e.,
For n = 1,
P(1): 1 =
k (k + 1) (2k + 1)
6
We shall now prove that P(k + 1) is also true. Now, we have
(12 +22 +32 +42 ++k2 ) + (k + 1) 2
12 + 22 + 32 + 42 ++ k2 =
k (k + 1)(2k + 1)
+ (k + 1)2
6
k ( k + 1) (2k + 1) + 6( k + 1) 2
6
... (1)
[Using (1)]
( k + 1) (2k 2 + 7 k + 6)
=
6
(k + 1)(k + 1 + 1){2(k + 1) + 1}
6
Thus P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true
for all natural numbers n.
=
90
MATHEMATICS
he
2 k + 1 > 2k = k + k > k + 1
is
i.e.,
... (1)
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1
1
1
1
n
+
+
+ ... +
=
1.2 2.3 3.4
n(n + 1) n + 1 .
bl
1
1
1
1
n
P(n): 1.2 + 2.3 + 3.4 + ... + n(n + 1) = n + 1
1 1 1
= =
, which is true. Thus, P(n) is true for n = 1.
1.2 2 1 +1
1
1
1
1
k
+
+
+ ... +
=
1.2 2.3 3.4
k (k + 1) k + 1
... (1)
1
1
1
1
1
+
+
+ ... +
+
1.2 2.3 3.4
k (k + 1) (k + 1) (k + 2)
1
1
1
1
1
+
+ ... +
+
= +
k (k + 1) ( k + 1) ( k + 2)
1.2 2.3 3.4
k
1
= k + 1 + (k + 1) ( k + 2)
[Using (1)]
91
(k 2 + 2k + 1)
k +1
k +1
k (k + 2) + 1
( k + 1)2
=
=
=
=
=
(k + 1) (k + 2)
(k + 1) (k + 2)
( k + 1) ( k + 2 ) k + 2 ( k + 1) + 1
Thus P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical
induction, P(n) is true for all natural numbers.
bl
is
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... (1)
... (2)
... (3)
92
MATHEMATICS
he
bl
is
We note that P(n) is true for n = 1, since 2.7 + 3.5 5 = 24, which is divisible by 24.
no N
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... (1)
We have
= 7 24q 6.5k + 30
= 7 24q 6 (5k 5)
... (2)
The expression on the R.H.S. of (1) is divisible by 24. Thus P(k + 1) is true whenever
P(k) is true.
Hence, by principle of mathematical induction, P(n) is true for all n N.
93
12 + 22 + ... + n2 >
2
We note that P(n) is true for n = 1 since 1 >
he
n3
, nN
3
13
3
is
k3
3
...(1)
bl
i.e.
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= 12 + 22 + ... + k 2 + ( k + 1) >
2
k3
2
+ ( k + 1)
3
1 3
[k + 3k2 + 6k + 3]
3
1
1
[(k + 1)3 + 3k + 2] >
(k + 1)3
3
3
[by (1)]
Therefore, P(k + 1) is also true whenever P(k) is true. Hence, by mathematical induction
P(n) is true for all n N.
Example 8 Prove the rule of exponents (ab)n = anbn
by using principle of mathematical induction for every natural number.
... (1)
94
MATHEMATICS
[by (1)]
he
Prove the following by using the principle of mathematical induction for all n N:
(3n 1)
.
2
is
1. 1 + 3 + 32 + ... + 3n 1 =
bl
n( n + 1)
2. 1 + 2 + 3 + +n =
.
2
3
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1
1
1
2n
3. 1 + (1 + 2) + (1 + 2 + 3) + ... + (1 + 2 + 3 + ...n) = (n + 1) .
4. 1.2.3 + 2.3.4 ++ n(n+1) (n+2) =
5. 1.3 + 2.32 + 3.33 ++ n.3n =
n(n + 1) (n + 2) (n + 3)
.
4
(2n 1)3n +1 + 3
.
4
n(n + 1)(n + 2)
6. 1.2 + 2.3 + 3.4 ++ n.(n+1) =
.
3
n(4n 2 + 6n 1)
.
3
8. 1.2 + 2.22 + 3.22 + ...+n.2n = (n1) 2n + 1 + 2.
9.
1 1 1
1
1
+ + + ... + n = 1 n .
2 4 8
2
2
10.
1
1
1
1
n
+
+
+ ... +
=
2.5 5.8 8.11
(3n 1) (3n + 2) (6n + 4) .
11.
1
1
1
1
n(n + 3)
+
+
+ ... +
=
1.2.3 2.3.4 3.4.5
n(n + 1)(n + 2) 4(n + 1)(n + 2) .
a ( r n 1)
.
r 1
7 (2n + 1)
2
1 + ... 1 +
= ( n + 1) .
n2
9
he
3 5
13. 1 + 1 +
1 4
15. 12 + 32 + 52 + + (2n1)2 =
n(2n 1)(2n + 1)
.
3
bl
1
1
1
1
n
16. 1.4 + 4.7 + 7.10 + ... + (3n 2)(3n + 1) = (3n + 1) .
is
1 1 1 1
14. 1 + 1 + 1 + ...1 + = (n + 1) .
1 2 3 n
1
1
1
1
n
+
+
+ ... +
=
3.5 5.7 7.9
(2n + 1)(2n + 3) 3(2n + 3) .
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17.
95
18. 1 + 2 + 3 ++ n <
1
(2n + 1)2.
8
19. n (n + 1) (n + 5) is a multiple of 3.
20. 102n 1 + 1 is divisible by 11.
21. x2n y2n is divisible by x + y.
22. 32n+2 8n 9 is divisible by 8.
23. 41n 14n is a multiple of 27.
24. (2n + 7) < (n + 3)2.
Summary
MATHEMATICS
for the case n = 1 is examined. Then assuming the truth of P(k) for some
positive integer k, the truth of P (k+1) is established.
he
Historical Note
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96
Chapter
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5.1 Introduction
bl
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W. R. Hamilton
(1805-1865)
1
complex number. For example, 2 + i3, ( 1) + i 3 , 4 + i are complex numbers.
11
For the complex number z = a + ib, a is called the real part, denoted by Re z and
b is called the imaginary part denoted by Im z of the complex number z. For example,
if z = 2 + i5, then Re z = 2 and Im z = 5.
Two complex numbers z1 = a + ib and z2 = c + id are equal if a = c and b = d.
98
MATHEMATICS
Example 1 If 4x + i(3x y) = 3 + i ( 6), where x and y are real numbers, then find
the values of x and y.
Solution We have
3
33
and y = .
4
4
is
... (1)
he
4x + i (3x y) = 3 + i (6)
Equating the real and the imaginary parts of (1), we get
4x = 3, 3x y = 6,
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(i) The closure law The sum of two complex numbers is a complex
number, i.e., z1 + z2 is a complex number for all complex numbers
z1 and z2.
(ii) The commutative law For any two complex numbers z 1 and z 2 ,
z1 + z2 = z2 + z1
(iii) The associative law For any three complex numbers z 1 , z 2 , z 3 ,
(z1 + z2) + z3 = z1 + (z2 + z3).
(iv) The existence of additive identity There exists the complex number
0 + i 0 (denoted as 0), called the additive identity or the zero complex
number, such that, for every complex number z, z + 0 = z.
(v) The existence of additive inverse To every complex number
z = a + ib, we have the complex number a + i( b) (denoted as z),
called the additive inverse or negative of z. We observe that z + (z) = 0
(the additive identity).
5.3.2 Difference of two complex numbers Given any two complex numbers z1 and
z2, the difference z1 z2 is defined as follows:
z1 z2 = z1 + ( z2).
For example,
(6 + 3i) (2 i) = (6 + 3i) + ( 2 + i ) = 4 + 4i
and
(2 i) (6 + 3i) = (2 i) + ( 6 3i) = 4 4i
99
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1
z. = 1 (the multiplicative identity).
z
(vi) The distributive law For any three complex numbers z1, z2, z3,
(a) z1 (z2 + z3) = z1 z2 + z1 z3
(b) (z1 + z2) z3 = z1 z3 + z2 z3
5.3.4 Division of two complex numbers Given any two complex numbers z1 and z2,
z1
z1 = 6 + 3i and z2 = 2 i
Then
z1
1
= (6 + 3i )
= 6 + 3i )
z2
2i (
( 1)
2
+i
2
2
2
22 + ( 1)
2
+
1
(
)
100
MATHEMATICS
2+i 1
1
= ( 6 + 3i )
= 12 3 + i ( 6 + 6 ) = ( 9 + 12i )
5
5 5
5.3.5 Power of i we know that
i = ( 1) i = i ,
2
( )
i6 = i2
= ( 1) = 1
= ( 1) = 1 , etc.
1 i i
= i,
i 1 = =
i i 1
Also, we have
he
( )
i5 = i 2
i4 = i2
i 2 =
1
1
=
= 1,
2
1
i
is
( )
i 3 = i 2i = ( 1) i = i ,
1 1 i i
1 1
= = = i, i 4 = 4 = = 1
3
i i 1
1
i
i
4k
4k + 1
4k + 2
4k + 3
In general, for any integer k, i = 1, i
= i, i
= 1, i
=i
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i 3 =
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( 3 i) = ( 3)
2
i2 = 3 ( 1) = 3
( 3 i) = ( 3)
2
i2 = 3
3 i and 3 i .
3 is meant to represent
a =
3 i only, i.e.,
a
1 =
3 =
3i .
a i,
result also holds true when either a > 0, b < 0 or a < 0, b > 0. What if a < 0, b < 0?
Let us examine.
Note that
( 1) ( 1)
i 2 = 1 1 =
(by assuming
a b =
101
a b = ab = 0.
he
Therefore,
is
Proof We have,
(Distributive law)
z22
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= z12 + z1 z2 + z1 z2 + z22
(Distributive law)
bl
z12
= z12 + 2 z1 z2 + z22
( z1 z2 )2 = z12 2 z1 z2 + z22
(ii)
(iii)
(iv)
z12 z22 = ( z1 + z2 ) ( z1 z2 )
In fact, many other identities which are true for all real numbers, can be proved
to be true for all complex numbers.
Example 2 Express the following in the form of a + bi:
1
(i) ( 5i ) i
8
Solution
(i)
(ii)
( i ) ( 2i )
1
i
8
( 5i )
1
5 2 5
5 5
i =
i =
1) = = + i 0
(
8
8
8 8
8
3
1
1 2
1
i5 =
i
(ii) ( i ) ( 2i ) i = 2
888
256
8
( )
i=
1
i.
256
102
MATHEMATICS
)(
) (2
) = ( 3 + 2 i) (2 3 i)
2 i = ( 6 + 2 ) + 3 (1 + 2 2 ) i
3 i
= 6 + 3i + 2 6i
is
Solution We have, 3 + 2
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For example,
2 5i = 22 + ( 5) 2 = 29 ,
3 + i = 3 i , 2 5 i = 2 + 5 i , 3i 5 = 3i 5
Observe that the multiplicative inverse of the non-zero complex number z is
given by
and
z1 =
a
b
a ib
1
+i 2
= 2
=
2
2 =
a +b
a +b
a 2 + b2
a + ib
z z= z
or
(i)
(iii)
z1 z2 = z1 z2
z1 z2 = z1 z2
(ii)
(iv)
z
z1
= 1 provided z 0
2
z2
z2
z1 z1
z1 z2 = z1 z2 (v) z = z provided z2 0.
2
2
103
Let z = 2 3i
Then
z = 2 + 3i and
= 22 + ( 3)2 = 13
z1 =
2 + 3i 2 3
= + i
13
13 13
2 + 3i
2 + 3i 2 3
=
= + i
2
13
13 13
2 (3i )
2
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1
2 + 3i
=
2 3i (2 3i) (2 + 3i)
bl
z1 =
is
he
5 + 2i
1 2i
(ii) i35
5 + 2i 5 + 2i 1 + 2i
=
1 2i 1 2i 1 + 2i
35
(ii) i =
35
5 + 5 2i + 2i 2
1
( 2i )
3 + 6 2i 3(1 + 2 2i )
=
= 1 + 2 2i .
1+ 2
3
(i )
2 17
i
1 i
= 2 =i
i
i i
EXERCISE 5.1
( 5i )
3
i
5
2. i 9 + i 19
3. i 39
MATHEMATICS
5. (1 i) ( 1 + i6)
5
1 2
+ i 4 +i
2
5 5
8. (1 i)
1 7
1 4
7. + i + 4 + i + i
3 3
3 3
1
9. + 3i
3
10. 2 i
3
he
104
13. i
5 + 3i
14. Express the following expression in the form of a + ib :
12.
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(3 + i 5 ) (3 i 5 )
( 3 + 2 i) ( 3 i 2 )
bl
11. 4 3i
is
Find the multiplicative inverse of each of the complex numbers given in the
Exercises 11 to 13.
105
x 2 + y 2 is the distance between the point P(x, y) and the origin O (0, 0)
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(Fig 5.2). The points on the x-axis corresponds to the complex numbers of the form
a + i 0 and the points on the y-axis corresponds to the complex numbers of the form
Fig 5.2
0 + i b. The x-axis and y-axis in the Argand plane are called, respectively, the real axis
and the imaginary axis.
The representation of a complex number z = x + iy and its conjugate
z = x iy in the Argand plane are, respectively, the points P (x, y) and Q (x, y).
Geometrically, the point (x, y) is the mirror image of the point (x, y) on the real
axis (Fig 5.3).
Fig 5.3
106
MATHEMATICS
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modulus of z and is called the argument (or amplitude) of z which is denoted by arg z.
For any complex number z 0, there corresponds only one value of in
0 < 2. However, any other interval of length 2, for example < , can be
such an interval.We shall take the value of such that < , called principal
argument of z and is denoted by arg z, unless specified otherwise. (Figs. 5.5 and 5.6)
107
i.e.,
r = 4 = 2 (conventionally, r >0)
Therefore,
cos =
he
r 2 cos 2 + sin 2 = 4
Fig 5.7
bl
is
1
3
, sin =
, which gives =
2
3
2
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16
1+ i 3
16 1 i 3
1+ i 3 1 i 3
(
) = 16 (1 i 3 ) 4 1 i 3 = 4 + i4
=
(
)
1+ 3
1 (i 3 )
16 1 i 3
3 (Fig 5.8).
Let
4 = r cos , 4 3 = r sin
By squaring and adding, we get
2
2
2
16 + 48 = r cos + sin
which gives
Hence
r = 64, i.e., r = 8
2
cos = , sin =
2
=
3 3
2
2
+ i sin
Thus, the required polar form is 8 cos
3
3
Fig 5.8
108
MATHEMATICS
EXERCISE 5.2
he
Find the modulus and the arguments of each of the complex numbers in
Exercises 1 to 2.
1. z = 1 i 3
2. z = 3 + i
Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:
3. 1 i
4. 1 + i
5. 1 i
6. 3
7. 3 + i
8. i
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We are already familiar with the quadratic equations and have solved them in the set
of real numbers in the cases where discriminant is non-negative, i.e., 0,
Let us consider the following quadratic equation:
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b b 2 4ac b 4ac b 2 i
=
2a
2a
Note At this point of time, some would be interested to know as to how many
roots does an equation have? In this regard, the following theorem known as the
Fundamental theorem of Algebra is stated below (without proof).
A polynomial equation has at least one root.
Example 9 Solve x2 + 2 = 0
Solution We have, x2 + 2 = 0
or
x2 = 2 i.e., x = 2 = 2 i
Example 10 Solve x2 + x + 1= 0
Solution Here,
b2 4ac = 12 4 1 1 = 1 4 = 3
1 3 1 3i
=
2 1
2
109
5x2 + x + 5 = 0
he
12 4 5 5 = 1 20 = 19
Therefore, the solutions are
3. x2 + 3x + 9 = 0
6. x2 x + 2 = 0
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EXERCISE 5.3
is
1 19 1 19i
=
.
2 5
2 5
7.
9.
2x2 + x + 2 = 0
x2 + x +
=0
3x 2 2 x + 3 3 = 0
8.
10.
x2 +
+1 = 0
Miscellaneous Examples
(3 2i)(2 + 3i)
Example 12 Find the conjugate of (1 + 2i )(2 i) .
Solution We have ,
(3 2i)(2 + 3i)
(1 + 2i )(2 i)
6 + 9i 4i + 6 12 + 5i 4 3i
=
2 i + 4i + 2
4 + 3i 4 3i
63 16
48 36i + 20i +15 63 16i
i
=
=
25 25
16 + 9
25
(3 2i )(2 + 3i ) 63 16
Therefore, conjugate of (1 + 2i )(2 i ) is 25 + 25 i .
110
MATHEMATICS
1+ i
,
1 i
(ii)
1
1+ i
1+ i 1+ i 1+ i 1 1 + 2i
=
=i= 0 + i
=
1 i 1 i 1+ i
1+1
he
(i)
1 = r sin
1+ i
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Therefore, =
is
cos = 0, sin = 1
(ii) We have
1
1 i
1 i 1 i
=
=
=
1 + i (1 + i ) (1 i) 1 + 1 2 2
Let
1
1
= r cos ,
= r sin
2
2
Therefore
1
1
1
; cos =
, sin =
2
2
2
1
1
is
, argument is
.
2
1+ i
4
a + ib
Example 14 If x + iy = a ib , prove that x2 + y2 = 1.
Solution We have,
(a + ib)(a + ib)
a 2 b 2 + 2abi
a 2 b2
2ab
+ 2 2i
x + iy = (a ib)(a + ib) =
=
2
2
2
2
a +b
a +b
a +b
So that, x iy =
111
a2 b2
2ab
2 2i
2
2
a +b a +b
Therefore,
(a 2 b 2 )2
4a 2b 2
(a 2 + b 2 ) 2
+
= 2 2 2 =1
x + y = (x + iy) (x iy) =
(a 2 + b 2 ) 2 (a 2 + b 2 )2
(a + b )
2
he
is
3 + 2i sin
is purely real.
1 2i sin
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3 + 2i sin
(3 + 2i sin)(1 + 2i sin)
=
(1 2i sin)(1 + 2i sin)
1 2i sin
bl
Solution We have,
Thus
= n, n Z.
Solution We have, z =
Now, put
i 1
i 1
1
3
+
i
2 2
2(i 1) 1 3i 2 i + 3 1 + 3i
=
=
1+ 3
1 + 3i 1 3i
3 1
= r cos ,
2
3 +1
= r sin
2
3 1
+
2
3 +1
i
2
112
MATHEMATICS
cos =
( 3)
+ 1
= 2 4 = 2
4
4
2
3 1
3 +1
, sin =
2 2
2 2
5
(Why?)
+ =
4 6 12
Hence, the polar form is
is
Therefore, =
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5
5
2 cos + i sin
12
12
18 1 25
1. Evaluate: i + .
i
2 3 4i
1
3. Reduce
4. If x iy =
1 + 7i
(i)
(2 i)
(ii)
1 + 3i
1 2i
20
=0
3
8. 27 x 2 10 x + 1 = 0
he
2
2
3 1 3 +1
2
r =
+
=
2 2
2
7.
x2 2x +
3
=0
2
113
9. 21x 2 28 x + 10 = 0
z1 + z2 +1
10. If z1 = 2 i, z2 = 1 + i, find z z +1 .
1
2
bl
1
.
z1 z1
(ii) Im
is
z1 z2
(i) Re z ,
1
he
( x 2 + 1) 2
( x + i )2
11. If a + ib =
, prove that a2 + b2 = 2 x 2 + 1 2 .
2x2 + 1
1+ 2i
.
1 3i
14. Find the real numbers x and y if (x iy) (3 + 5i) is the conjugate of 6 24i.
no N
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1+ i 1 i
15. Find the modulus of 1 i 1 + i .
u v
+ = 4( x 2 y 2 ) .
x y
1+ i
20. If
= 1 , then find the least positive integral value of m.
1 i
= 2x .
MATHEMATICS
Summary
A number of the form a + ib, where a and b are real numbers, is called a
complex number, a is called the real part and b is called the imaginary part
of the complex number.
he
a
b
1
+i 2
or z 1, called the
2
2 , denoted by
a +b
a +b
z
2
bl
complex number
is
a2
b
+i 2
= 1 + i0 =1
multiplicative inverse of z such that (a + ib) 2
2
a + b2
a +b
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114
z , is given by
z = a ib.
b 4ac b 2 i
2a
115
Historical Note
x + y = 10, xy = 40.
is
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The fact that square root of a negative number does not exist in the real number
system was recognised by the Greeks. But the credit goes to the Indian
mathematician Mahavira (850) who first stated this difficulty clearly. He mentions
in his work Ganitasara Sangraha as in the nature of things a negative (quantity)
is not a square (quantity), it has, therefore, no square root. Bhaskara, another
Indian mathematician, also writes in his work Bijaganita, written in 1150. There
is no square root of a negative quantity, for it is not a square. Cardan (1545)
considered the problem of solving
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Chapter
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LINEAR INEQUALITIES
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In earlier classes, we have studied equations in one variable and two variables and also
solved some statement problems by translating them in the form of equations. Now a
natural question arises: Is it always possible to translate a statement problem in the
form of an equation? For example, the height of all the students in your class is less
than 160 cm. Your classroom can occupy atmost 60 tables or chairs or both. Here we
get certain statements involving a sign < (less than), > (greater than), (less than
or equal) and (greater than or equal) which are known as inequalities.
In this Chapter, we will study linear inequalities in one and two variables. The
study of inequalities is very useful in solving problems in the field of science, mathematics,
statistics, optimisation problems, economics, psychology, etc.
6.2 Inequalities
(i) Ravi goes to market with Rs 200 to buy rice, which is available in packets of 1kg.
The price of one packet of rice is Rs 30. If x denotes the number of packets of rice,
which he buys, then the total amount spent by him is Rs 30x. Since, he has to buy rice
in packets only, he may not be able to spend the entire amount of Rs 200. (Why?)
Hence
30x < 200
... (1)
Clearly the statement (i) is not an equation as it does not involve the sign of equality.
(ii) Reshma has Rs 120 and wants to buy some registers and pens. The cost of one
register is Rs 40 and that of a pen is Rs 20. In this case, if x denotes the number of
registers and y, the number of pens which Reshma buys, then the total amount spent by
her is Rs (40x + 20y) and we have
40x + 20y 120
... (2)
LINEAR INEQUALITIES
117
Since in this case the total amount spent may be upto Rs 120. Note that the statement
(2) consists of two statements
and
... (3)
... (4)
he
Statement (3) is not an equation, i.e., it is an inequality while statement (4) is an equation.
Definition 1 Two real numbers or two algebraic expressions related by the symbol
is
bl
3 < 5 < 7 (read as 5 is greater than 3 and less than 7), 3 < x < 5 (read as x is greater
than or equal to 3 and less than 5) and 2 < y < 4 are the examples of double inequalities.
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... (5)
ax + b > 0
... (6)
ax + b 0
... (7)
ax + b 0
... (8)
ax + by < c
... (9)
ax + by > c
... (10)
ax + by c
... (11)
ax + by c
... (12)
ax + bx + c 0
... (13)
ax2 + bx + c > 0
... (14)
Inequalities (5), (6), (9), (10) and (14) are strict inequalities while inequalities (7), (8),
(11), (12), and (13) are slack inequalities. Inequalities from (5) to (8) are linear
inequalities in one variable x when a 0, while inequalities from (9) to (12) are linear
inequalities in two variables x and y when a 0, b 0.
Inequalities (13) and (14) are not linear (in fact, these are quadratic inequalities
in one variable x when a 0).
In this Chapter, we shall confine ourselves to the study of linear inequalities in one
and two variables only.
118
MATHEMATICS
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Let us consider the inequality (1) of Section 6.2, viz, 30x < 200
Note that here x denotes the number of packets of rice.
Obviously, x cannot be a negative integer or a fraction. Left hand side (L.H.S.) of this
inequality is 30x and right hand side (RHS) is 200. Therefore, we have
For x = 0, L.H.S. = 30 (0) = 0 < 200 (R.H.S.), which is true.
For x = 1, L.H.S. = 30 (1) = 30 < 200 (R.H.S.), which is true.
For x = 2, L.H.S. = 30 (2) = 60 < 200, which is true.
For x = 3, L.H.S. = 30 (3) = 90 < 200, which is true.
For x = 4, L.H.S. = 30 (4) = 120 < 200, which is true.
For x = 5, L.H.S. = 30 (5) = 150 < 200, which is true.
For x = 6, L.H.S. = 30 (6) = 180 < 200, which is true.
For x = 7, L.H.S. = 30 (7) = 210 < 200, which is false.
In the above situation, we find that the values of x, which makes the above
inequality a true statement, are 0,1,2,3,4,5,6. These values of x, which make above
inequality a true statement, are called solutions of inequality and the set {0,1,2,3,4,5,6}
is called its solution set.
Thus, any solution of an inequality in one variable is a value of the variable
which makes it a true statement.
We have found the solutions of the above inequality by trial and error method
which is not very efficient. Obviously, this method is time consuming and sometimes
not feasible. We must have some better or systematic techniques for solving inequalities.
Before that we should go through some more properties of numerical inequalities and
follow them as rules while solving the inequalities.
You will recall that while solving linear equations, we followed the following rules:
Rule 1 Equal numbers may be added to (or subtracted from) both sides of an equation.
Rule 2 Both sides of an equation may be multiplied (or divided) by the same non-zero
number.
In the case of solving inequalities, we again follow the same rules except with a
difference that in Rule 2, the sign of inequality is reversed (i.e., < becomes >,
becomes and so on) whenever we multiply (or divide) both sides of an inequality by
a negative number. It is evident from the facts that
3 > 2 while 3 < 2,
8 < 7 while ( 8) ( 2) > ( 7) ( 2) , i.e., 16 > 14.
LINEAR INEQUALITIES
119
(ii) x is an integer.
is
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Rule 2 Both sides of an inequality can be multiplied (or divided) by the same positive
number. But when both sides are multiplied or divided by a negative number, then the
sign of inequality is reversed.
Now, let us consider some examples.
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30 x 200
<
(Rule 2), i.e., x < 20 / 3.
30
30
(i) When x is a natural number, in this case the following values of x make the
statement true.
1, 2, 3, 4, 5, 6.
The solution set of the inequality is {1,2,3,4,5,6}.
(ii) When x is an integer, the solutions of the given inequality are
..., 3, 2, 1, 0, 1, 2, 3, 4, 5, 6
The solution set of the inequality is {...,3, 2,1, 0, 1, 2, 3, 4, 5, 6}
or
120
MATHEMATICS
Example 4 Solve
he
Solution We have,
4x + 3 < 6x + 7
or
4x 6x < 6x + 4 6x
or
2x < 4
or x > 2
i.e., all the real numbers which are greater than 2, are the solutions of the given
inequality. Hence, the solution set is (2, ).
5 2x x
5.
3
6
is
Solution We have
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5 2x x
5
3
6
or
2 (5 2x) x 30.
or
10 4x x 30
or
5x 40, i.e., x 8
Thus, all real numbers x which are greater than or equal to 8 are the solutions of the
given inequality, i.e., x [8, ).
Example 5 Solve 7x + 3 < 5x + 9. Show the graph of the solutions on number line.
Fig 6.1
Example 6 Solve
3x 4 x + 1
Solution We have
3x 4 x + 1
1
2
4
or
or
3x 4 x 3
2
4
2 (3x 4) (x 3)
LINEAR INEQUALITIES
or
6x 8 x 3
or
5x 5 or x 1
121
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Fig 6.2
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Example 7 The marks obtained by a student of Class XI in first and second terminal
examination are 62 and 48, respectively. Find the minimum marks he should get in the
annual examination to have an average of at least 60 marks.
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Solution Let x be the marks obtained by student in the annual examination. Then
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62 + 48 + x
60
3
or
110 + x 180
or
x 70
Thus, the student must obtain a minimum of 70 marks to get an average of at least
60 marks.
Example 8 Find all pairs of consecutive odd natural numbers, both of which are larger
than 10, such that their sum is less than 40.
Solution Let x be the smaller of the two consecutive odd natural number, so that the
other one is x +2. Then, we should have
x > 10
and x + ( x + 2) < 40
Solving (2), we get
2x + 2 < 40
i.e., x < 19
From (1) and (3), we get
10 < x < 19
... (1)
... (2)
... (3)
Since x is an odd number, x can take the values 11, 13, 15, and 17. So, the required
possible pairs will be
(11, 13), (13, 15), (15, 17), (17, 19)
122
MATHEMATICS
EXERCISE 6.1
(ii) x is an integer.
(ii) x is an integer.
(ii) x is a real number.
(ii) x is a real number.
x x
x + + < 11
2 3
10.
x x
> +1
3 2
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1 3x 1
+ 4 ( x 6)
2 5
3
14. 37 (3x + 5) > 9x 8 (x 3)
3( x 2) 5(2 x)
5
3
13. 2 (2x + 3) 10 < 6 (x 2)
12.
x (5 x 2) (7 x 3)
<
4
3
5
16.
11.
15.
(2 x 1) (3 x 2) (2 x )
3
4
5
Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each
case on number line
17. 3x 2 < 2x + 1
x (5 x 2) (7 x 3)
2
3
5
Ravi obtained 70 and 75 marks in first two unit test. Find the minimum marks he
should get in the third test to have an average of at least 60 marks.
To receive Grade A in a course, one must obtain an average of 90 marks or
more in five examinations (each of 100 marks). If Sunitas marks in first four
examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain
in fifth examination to get grade A in the course.
Find all pairs of consecutive odd positive integers both of which are smaller than
10 such that their sum is more than 11.
Find all pairs of consecutive even positive integers, both of which are larger than
5 such that their sum is less than 23.
19. 3 (1 x) < 2 (x + 4)
21.
22.
23.
24.
18. 5x 3 > 3x 5
20.
LINEAR INEQUALITIES
123
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25. The longest side of a triangle is 3 times the shortest side and the third side is 2 cm
shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find
the minimum length of the shortest side.
26. A man wants to cut three lengths from a single piece of board of length 91cm.
The second length is to be 3cm longer than the shortest and the third length is to
be twice as long as the shortest. What are the possible lengths of the shortest
board if the third piece is to be at least 5cm longer than the second?
[Hint: If x is the length of the shortest board, then x , (x + 3) and 2x are the
lengths of the second and third piece, respectively. Thus, x + (x + 3) + 2x 91 and
2x (x + 3) + 5].
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In earlier section, we have seen that a graph of an inequality in one variable is a visual
representation and is a convenient way to represent the solutions of the inequality.
Now, we will discuss graph of a linear inequality in two variables.
We know that a line divides the Cartesian plane into two parts. Each part is
known as a half plane. A vertical line will divide the plane in left and right half planes
and a non-vertical line will divide the plane into lower and upper half planes
(Figs. 6.3 and 6.4).
Fig 6.3
Fig 6.4
A point in the Cartesian plane will either lie on a line or will lie in either of the half
planes I or II. We shall now examine the relationship, if any, of the points in the plane
and the inequalities ax + by < c or ax + by > c.
Let us consider the line
ax + by = c, a 0, b 0
... (1)
124
MATHEMATICS
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>
(as b > 0)
This means that the point (, ) lies in the half plane II.
Thus, any point in the half plane II satisfies ax + by > c, and conversely any point
satisfying the inequality ax + by > c lies in half plane II.
In case b < 0, we can similarly prove that any point satisfying ax + by > c lies in
the half plane I, and conversely.
Hence, we deduce that all points satisfying ax + by > c lies in one of the half
planes II or I according as b > 0 or b < 0, and conversely.
Thus, graph of the inequality ax + by > c will be one of the half plane (called
solution region) and represented by shading in the corresponding half plane.
Note 1 The region containing all the solutions of an inequality is called the
$
solution region.
2. In order to identify the half plane represented by an inequality, it is just sufficient
to take any point (a, b) (not online) and check whether it satisfies the inequality or
not. If it satisfies, then the inequality represents the half plane and shade the region
LINEAR INEQUALITIES
125
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which contains the point, otherwise, the inequality represents that half plane which
does not contain the point within it. For convenience, the point (0, 0) is preferred.
3. If an inequality is of the type ax + by c or ax + by c, then the points on the
line ax + by = c are also included in the solution region. So draw a dark line in the
solution region.
4. If an inequality is of the form ax + by > c or ax + by < c, then the points on the
line ax + by = c are not to be included in the solution region. So draw a broken or
dotted line in the solution region.
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MATHEMATICS
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126
Fig 6.7
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Fig 6.9
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EXERCISE 6.2
Fig 6.10
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127
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LINEAR INEQUALITIES
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128
MATHEMATICS
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Fig 6.12
Fig 6.13
LINEAR INEQUALITIES
129
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EXERCISE 6.3
2. 3x + 2y 12, x 1, y 2
3. 2x + y 6, 3x + 4y < 12
4. x + y 4, 2x y > 0
5. 2x y >1, x 2y < 1
6. x + y 6, x + y 4
7. 2x + y 8, x + 2y 10
8. x + y 9, y > x, x 0
9. 5x + 4y 20, x 1, y 2
12. x 2y 3, 3x + 4y 12, x 0 , y 1
13. 4x + 3y 60, y 2x, x 3, x, y 0
130
MATHEMATICS
Miscellaneous Examples
Example 16 Solve 8 5x 3 < 7.
or
10 5 3x 16
or
5x
5 3x
8
2
or 15 3x 11
11
3
11
x 5
3
Example 18 Solve the system of inequalities:
3x 7 < 5 + x
11 5 x 1
and represent the solutions on the number line.
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Solution We have 5
5 3x
8.
2
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Example 17 Solve 5
he
... (1)
... (2)
Fig 6.15
Thus, solution of the system are real numbers x lying between 2 and 6 including 2, i.e.,
2x<6
LINEAR INEQUALITIES
131
5
(F 32), where C and F represent temperature in degree
9
he
or
C=
5
(F 32), we get
9
30 <
5
(F 32) < 35,
9
9
9
(30) < (F 32) <
(35)
5
5
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Putting
is
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or
54 < (F 32) < 63
or
86 < F < 95.
Thus, the required range of temperature is between 86 F and 95 F.
Example 20 A manufacturer has 600 litres of a 12% solution of acid. How many litres
of a 30% acid solution must be added to it so that acid content in the resulting mixture
will be more than 15% but less than 18%?
or
30 x
12
15
+
(600) >
(x + 600)
100
100
100
and
30 x
12
18
+
(600) <
(x + 600)
100
100
100
or
and
or
or
i.e.
132
MATHEMATICS
Thus, the number of litres of the 30% solution of acid will have to be more than
120 litres but less than 300 litres.
3 4
5. 12 < 4
7x
18
2
4. 15 <
3x
2
5
6. 7
3( x 2 )
0
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3.
2. 6 3 (2x 4) < 12
( 3x +11 )
11 .
2
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7.
8.
9.
10.
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9
C + 32 ?
5
it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have
640 litres of the 8% solution, how many litres of the 2% solution will have to be added?
13. How many litres of water will have to be added to 1125 litres of the 45% solution
of acid so that the resulting mixture will contain more than 25% but less than 30% acid
content?
MA
100,
CA
LINEAR INEQUALITIES
133
Summary
Two real numbers or two algebraic expressions related by the symbols <, >,
or form an inequality.
Equal numbers may be added to (or subtracted from ) both sides of an inequality.
Both sides of an inequality can be multiplied (or divided ) by the same positive
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number. But when both sides are multiplied (or divided) by a negative number,
then the inequality is reversed.
The values of x, which make an inequality a true statement, are called solutions
of the inequality.
To represent x < a (or x > a) on a number line, put a circle on the number a and
dark line to the left (or right) of the number a.
To
represent x a (or x a) on a number line, put a dark circle on the number
a and dark the line to the left (or right) of the number x.
If an inequality is having or symbol, then the points on the line are also
included in the solutions of the inequality and the graph of the inequality lies left
(below) or right (above) of the graph of the equality represented by dark line
that satisfies an arbitrary point in that part.
If an inequality is having < or > symbol, then the points on the line are not
included in the solutions of the inequality and the graph of the inequality lies to
the left (below) or right (above) of the graph of the corresponding equality
represented by dotted line that satisfies an arbitrary point in that part.
The solution region of a system of inequalities is the region which satisfies all
the given inequalities in the system simultaneously.
Chapter
135
Let us name the three pants as P1, P2 , P 3 and the two shirts as S1, S2. Then,
these six possibilities can be illustrated in the Fig. 7.1.
Let us consider another problem
of the same type.
Sabnam has 2 school bags, 3 tiffin boxes
and 2 water bottles. In how many ways
can she carry these items (choosing one
each).
A school bag can be chosen in 2
different ways. After a school bag is
chosen, a tiffin box can be chosen in 3
different ways. Hence, there are
2 3 = 6 pairs of school bag and a tiffin
box. For each of these pairs a water
bottle can be chosen in 2 different ways.
Fig 7.1
Hence, there are 6 2 = 12 different ways in which, Sabnam can carry these items to
school. If we name the 2 school bags as B1, B2, the three tiffin boxes as T1, T2, T3 and
the two water bottles as W1, W2 , these possibilities can be illustrated in the Fig. 7.2.
Fig 7.2
136
MATHEMATICS
In fact, the problems of the above types are solved by applying the following
principle known as the fundamental principle of counting, or, simply, the multiplication
principle, which states that
If an event can occur in m different ways, following which another event
can occur in n different ways, then the total number of occurrence of the events
in the given order is mn.
The above principle can be generalised for any finite number of events. For
example, for 3 events, the principle is as follows:
If an event can occur in m different ways, following which another event can
occur in n different ways, following which a third event can occur in p different ways,
then the total number of occurrence to the events in the given order is m n p.
In the first problem, the required number of ways of wearing a pant and a shirt
was the number of different ways of the occurence of the following events in succession:
(i)
(ii)
In the second problem, the required number of ways was the number of different
ways of the occurence of the following events in succession:
(i)
(ii)
(iii)
Here, in both the cases, the events in each problem could occur in various possible
orders. But, we have to choose any one of the possible orders and count the number of
different ways of the occurence of the events in this chosen order.
Example 1 Find the number of 4 letter words, with or without meaning, which can be
formed out of the letters of the word ROSE, where the repetition of the letters is not
allowed.
Solution There are as many words as there are ways of filling in 4 vacant places
by the 4 letters, keeping in mind that the repetition is not allowed. The
first place can be filled in 4 different ways by anyone of the 4 letters R,O,S,E. Following
which, the second place can be filled in by anyone of the remaining 3 letters in 3
different ways, following which the third place can be filled in 2 different ways; following
which, the fourth place can be filled in 1 way. Thus, the number of ways in which the
4 places can be filled, by the multiplication principle, is 4 3 2 1 = 24. Hence, the
required number of words is 24.
137
Note If the repetition of the letters was allowed, how many words can be formed?
One can easily understand that each of the 4 vacant places can be filled in succession
in 4 different ways. Hence, the required number of words = 4 4 4 4 = 256.
Example 2 Given 4 flags of different colours, how many different signals can be
generated, if a signal requires the use of 2 flags one below the other?
Solution There will be as many signals as there are ways of filling in 2 vacant places
in succession by the 4 flags of different colours. The upper vacant place can
be filled in 4 different ways by anyone of the 4 flags; following which, the lower vacant
place can be filled in 3 different ways by anyone of the remaining 3 different flags.
Hence, by the multiplication principle, the required number of signals = 4 3 = 12.
Example 3 How many 2 digit even numbers can be formed from the digits
1, 2, 3, 4, 5 if the digits can be repeated?
Solution There will be as many ways as there are ways of filling 2 vacant places
in succession by the five given digits. Here, in this case, we start filling in units
place, because the options for this place are 2 and 4 only and this can be done in 2
ways; following which the tens place can be filled by any of the 5 digits in 5 different
ways as the digits can be repeated. Therefore, by the multiplication principle, the required
number of two digits even numbers is 2 5, i.e., 10.
Example 4 Find the number of different signals that can be generated by arranging at
least 2 flags in order (one below the other) on a vertical staff, if five different flags are
available.
Solution A signal can consist of either 2 flags, 3 flags, 4 flags or 5 flags. Now, let us
count the possible number of signals consisting of 2 flags, 3 flags, 4 flags and 5 flags
separately and then add the respective numbers.
There will be as many 2 flag signals as there are ways of filling in 2 vacant places
in succession by the 5 flags available. By Multiplication rule, the number of
ways is 5 4 = 20.
Similarly, there will be as many 3 flag signals as there are ways of filling in 3
vacant places
138
MATHEMATICS
EXERCISE 7.1
1.
How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5
assuming that
(i)
(ii)
2.
3.
4.
5.
6.
How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the
digits can be repeated?
How many 4-letter code can be formed using the first 10 letters of the English
alphabet, if no letter can be repeated?
How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if
each number starts with 67 and no digit appears more than once?
A coin is tossed 3 times and the outcomes are recorded. How many possible
outcomes are there?
Given 5 flags of different colours, how many different signals can be generated if
each signal requires the use of 2 flags, one below the other?
7.3 Permutations
In Example 1 of the previous Section, we are actually counting the different possible
arrangements of the letters such as ROSE, REOS, ..., etc. Here, in this list, each
arrangement is different from other. In other words, the order of writing the letters is
important. Each arrangement is called a permutation of 4 different letters taken all
at a time. Now, if we have to determine the number of 3-letter words, with or without
meaning, which can be formed out of the letters of the word NUMBER, where the
repetition of the letters is not allowed, we need to count the arrangements NUM,
NMU, MUN, NUB, ..., etc. Here, we are counting the permutations of 6 different
letters taken 3 at a time. The required number of words = 6 5 4 = 120 (by using
multiplication principle).
If the repetition of the letters was allowed, the required number of words would
be 6 6 6 = 216.
139
places
by
r vacant places
the n objects. The first place can be filled in n ways; following which, the second place
can be filled in (n 1) ways, following which the third place can be filled in (n 2)
ways,..., the rth place can be filled in (n (r 1)) ways. Therefore, the number of
ways of filling in r vacant places in succession is n(n 1) (n 2) . . . (n (r 1)) or
n ( n 1) (n 2) ... (n r + 1)
This expression for nPr is cumbersome and we need a notation which will help to
reduce the size of this expression. The symbol n! (read as factorial n or n factorial )
comes to our rescue. In the following text we will learn what actually n! means.
7.3.2 Factorial notation The notation n! represents the product of first n natural
numbers, i.e., the product 1 2 3 . . . (n 1) n is denoted as n!. We read this
symbol as n factorial. Thus, 1 2 3 4 . . . (n 1) n = n !
1=1!
12=2!
1 2 3 = 3 !
1 2 3 4 = 4 ! and so on.
We define 0 ! = 1
We can write 5 ! = 5 4 ! = 5 4 3 ! = 5 4 3 2 !
= 5 4 3 2 1!
Clearly, for a natural number n
n ! = n (n 1) !
= n (n 1) (n 2) !
[provided (n 2)]
= n (n 1) (n 2) (n 3) !
[provided (n 3)]
and so on.
140
MATHEMATICS
(ii) 7 !
(iii) 7 ! 5!
(i) 5 ! = 1 2 3 4 5 = 120
(ii) 7 ! = 1 2 3 4 5 6 7 = 5040
(iii) 7 ! 5! = 5040 120 = 4920.
7!
Example 6 Compute (i) 5!
Solution
(i) We have
and
(ii)
12!
(ii) 10! (2!)
7 6 5!
7!
=
= 7 6 = 42
5!
5!
12 11 10!
12!
=
= 6 11 = 66.
10! 2!
10! 2
n!
Example 7 Evaluate r ! n r ! , when n = 5, r = 2.
Solution
5!
We have to evaluate 2! 5 2 ! (since n = 5, r = 2)
We have
5!
5!
4 5
10 .
=
2 ! 5 2 ! 2! 3!
2
1 1
x
Example 8 If 8! 9! 10! , find x.
1
1
x
Solution We have 8! 9 8! 10 9 8!
Therefore
So
1
x
1
or
9 10 9
x = 100.
10
x
9 10 9
EXERCISE 7.2
1.
Evaluate
(i) 8 !
(ii) 4 ! 3 !
8!
3. Compute 6! 2!
2.
Is 3 ! + 4 ! = 7 ! ?
5.
n!
Evaluate n r ! , when
(i) n = 6, r = 2
141
1
1
x
4. If 6! 7! 8! , find x
(ii) n = 9, r = 5.
Pr =
n!
n r ! , 0 r n
Let us now go back to the stage where we had determined the following formula:
n
Pr = n (n 1) (n 2) . . . (n r + 1)
Thus
Pr
n n 1 n 2 ... n r 1 n r n r 1 ...3 2 1
n r n r 1 ...3 2 1
Pr
n!
n r ! ,
n!
n r ! , where 0 < r n
This is a much more convenient expression for nPr than the previous one.
n!
n!
0!
Counting permutations is merely counting the number of ways in which some or
all objects at a time are rearranged. Arranging no object at all is the same as leaving
behind all the objects and we know that there is only one way of doing so. Thus, we
can have
In particular, when r = n, n Pn
P0 = 1 =
n!
n!
n ! ( n 0)!
Pr
n!
,0 r n .
n r !
... (1)
142
MATHEMATICS
6!
6
NUMBER = P3 3! = 4 5 6 = 120. Here, in this case also, the repetition is not
allowed. If the repetition is allowed,the required number of words would be 63 = 216.
The number of ways in which a Chairman and a Vice-Chairman can be chosen
from amongst a group of 12 persons assuming that one person can not hold more than
one position, clearly
12
P2
12!
11 12 = 132.
10!
7.3.4 Permutations when all the objects are not distinct objects Suppose we have
to find the number of ways of rearranging the letters of the word ROOT. In this case,
the letters of the word are not all different. There are 2 Os, which are of the same kind.
Let us treat, temporarily, the 2 Os as different, say, O1 and O2. The number of
permutations of 4-different letters, in this case, taken all at a time
is 4!. Consider one of these permutations say, RO1O 2T. Corresponding to this
permutation,we have 2 ! permutations RO1O2T and RO2O1T which will be exactly the
same permutation if O1 and O 2 are not treated as different, i.e., if O1 and O2 are the
same O at both places.
Therefore, the required number of permutations =
Permutations when O1, O2 are
different.
4!
3 4 12 .
2!
RO1O 2 T
RO 2 O1 T
ROOT
T O1 O2 R
T O2 O1R
TOOR
R O1T O2
R O2 T O1
ROTO
T O1 R O2
T O 2 R O1
TORO
R T O1 O2
R T O2 O1
RTOO
T R O1 O 2
T R O 2 O1
TROO
O1 O2 R T
O2 O1 T R
OORT
O1 R O 2 T
O2 R O1 T
OROT
O1 T O 2 R
O2 T O1 R
OTOR
O1 R T O 2
O2 R T O1
ORTO
O1 T R O 2
O2 T R O1
OTRO
O1 O2 T R
O2 O1 T R
OOTR
143
Let us now find the number of ways of rearranging the letters of the word
INSTITUTE. In this case there are 9 letters, in which I appears 2 times and T appears
3 times.
Temporarily, let us treat these letters different and name them as I1, I2, T1 , T2 , T3.
The number of permutations of 9 different letters, in this case, taken all at a time is 9 !.
Consider one such permutation, say, I1 NT1 SI2 T2 U E T3 . Here if I1, I2 are not same
144
MATHEMATICS
and T1, T2 , T3 are not same, then I1, I 2 can be arranged in 2! ways and T1, T2, T3 can
be arranged in 3! ways. Therefore, 2! 3! permutations will be just the same permutation
corresponding to this chosen permutation I1NT1SI2T2UET3. Hence, total number of
different permutations will be
9!
2! 3!
n!
same kind and rest are all different = p! .
In fact, we have a more general theorem.
Theorem 4 The number of permutations of n objects, where p1 objects are of one
kind, p2 are of second kind, ..., pk are of kth kind and the rest, if any, are of different
kind is
n!
.
p1! p2! ... p k!
Example 9 Find the number of permutations of the letters of the word ALLAHABAD.
Solution Here, there are 9 objects (letters) of which there are 4As, 2 Ls and rest are
all different.
Therefore, the required number of arrangements =
9!
5 6 7 8 9
= 7560
4! 2!
2
Example 10 How many 4-digit numbers can be formed by using the digits 1 to 9 if
repetition of digits is not allowed?
Solution Here order matters for example 1234 and 1324 are two different numbers.
Therefore, there will be as many 4 digit numbers as there are permutations of 9 different
digits taken 4 at a time.
Therefore, the required 4 digit numbers = 9 P4 =
9!
9!
= = 9 8 7 6 = 3024.
9 4 ! 5!
Example 11 How many numbers lying between 100 and 1000 can be formed with the
digits 0, 1, 2, 3, 4, 5, if the repetition of the digits is not allowed?
Solution Every number between 100 and 1000 is a 3-digit number. We, first, have to
145
count the permutations of 6 digits taken 3 at a time. This number would be 6P3 . But,
these permutations will include those also where 0 is at the 100s place. For example,
092, 042, . . ., etc are such numbers which are actually 2-digit numbers and hence the
number of such numbers has to be subtracted from 6 P3 to get the required number. To
get the number of such numbers, we fix 0 at the 100s place and rearrange the remaining
5 digits taking 2 at a time. This number is 5P2 . So
= 6 P3 5 P2
6! 5!
3! 3!
= 4 5 6 4 5 = 100
Example 12 Find the value of n such that
n
(i)
P5 42 P3 , n 4
n
(ii)
P4 5
= , n> 4
P4 3
n 1
P5 42 n P3
n (n 1) (n 2) (n 3) (n 4) = 42 n(n 1) (n 2)
Since
n>4
so n(n 1) (n 2) 0
(n 3 (n 4) = 42
n2 7n 30 = 0
or
or
or
n2 10n + 3n 30
(n 10) (n + 3) = 0
n 10 = 0 or n + 3 = 0
or
n = 10
or n = 3
As n cannot be negative, so n = 10.
n
P4 5
P4 3
n 1
Therefore
3n (n 1) (n 2) (n 3) = 5(n 1) (n 2) (n 3) (n 4)
or
or
3n = 5 (n 4)
n = 10.
[as (n 1) (n 2) (n 3) 0, n > 4]
146
MATHEMATICS
4!
5!
6
4 r !
5 r 1!
or
5!
6 5!
4
r
!
5
r
1
5 r 5 r 1!
or
or
or
or
or
Hence
(6 r) (5 r) = 6
r2 11r + 24 = 0
r2 8r 3r + 24 = 0
(r 8) (r 3) = 0
r = 8 or r = 3.
r = 8, 3.
Example 14 Find the number of different 8-letter arrangements that can be made
from the letters of the word DAUGHTER so that
(i) all vowels occur together
(ii) all vowels do not occur together.
Solution (i) There are 8 different letters in the word DAUGHTER, in which there
are 3 vowels, namely, A, U and E. Since the vowels have to occur together, we can for
the time being, assume them as a single object (AUE). This single object together with
5 remaining letters (objects) will be counted as 6 objects. Then we count permutations
of these 6 objects taken all at a time. This number would be 6 P6 = 6!. Corresponding to
each of these permutations, we shall have 3! permutations of the three vowels A, U, E
taken all at a time . Hence, by the multiplication principle the required number of
permutations = 6 ! 3 ! = 4320.
(ii) If we have to count those permutations in which all vowels are never
together, we first have to find all possible arrangments of 8 letters taken all at a time,
which can be done in 8! ways. Then, we have to subtract from this number, the number
of permutations in which the vowels are always together.
Therefore, the required number
8 ! 6 ! 3 ! = 6 ! (78 6)
= 2 6 ! (28 3)
= 50 6 ! = 50 720 = 36000
Example 15 In how many ways can 4 red, 3 yellow and 2 green discs be arranged in
a row if the discs of the same colour are indistinguishable ?
Solution Total number of discs are 4 + 3 + 2 = 9. Out of 9 discs, 4 are of the first kind
147
(red), 3 are of the second kind (yellow) and 2 are of the third kind (green).
Therefore, the number of arrangements
9!
= 1260 .
4! 3! 2!
Let us fix P at the extreme left position, we, then, count the arrangements of the
remaining 11 letters. Therefore, the required number of words starting with P are
(ii)
12!
1663200
3! 4! 2!
11!
138600 .
3! 2! 4!
There are 5 vowels in the given word, which are 4 Es and 1 I. Since, they have
to always occur together, we treat them as a single object EEEEI for the time
being. This single object together with 7 remaining objects will account for 8
objects. These 8 objects, in which there are 3Ns and 2 Ds, can be rearranged in
8!
3! 2! ways. Corresponding to each of these arrangements, the 5 vowels E, E, E,
5!
ways. Therefore, by multiplication principle the
4!
required number of arrangements
=
(iii)
8! 5!
16800
3! 2! 4!
148
(iv)
MATHEMATICS
10!
= 12600
3! 2! 4!
EXERCISE 7.3
1. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is
repeated?
2. How many 4-digit numbers are there with no digit repeated?
3. How many 3-digit even numbers can be made using the digits
1, 2, 3, 4, 6, 7, if no digit is repeated?
4. Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4,
5 if no digit is repeated. How many of these will be even?
5. From a committee of 8 persons, in how many ways can we choose a chairman
and a vice chairman assuming one person can not hold more than one position?
6. Find n if n 1 P3 : nP4 = 1 : 9.
7. Find r if (i) 5 Pr 2 6 Pr1
(ii) 5 Pr 6 Pr 1 .
8. How many words, with or without meaning, can be formed using all the letters of
the word EQUATION, using each letter exactly once?
9. How many words, with or without meaning can be made from the letters of the
word MONDAY, assuming that no letter is repeated, if.
(i) 4 letters are used at a time,
(ii) all letters are used at a time,
(iii) all letters are used but first letter is a vowel?
10. In how many of the distinct permutations of the letters in MISSISSIPPI do the
four Is not come together?
11. In how many ways can the letters of the word PERMUTATIONS be arranged if the
(i) words start with P and end with S,
(ii) vowels are all together,
(iii) there are always 4 letters between P and S?
7.4 Combinations
Let us now assume that there is a group of 3 lawn tennis players X, Y, Z. A team
consisting of 2 players is to be formed. In how many ways can we do so? Is the team
of X and Y different from the team of Y and X ? Here, order is not important.
In fact, there are only 3 possible ways in which the team could be constructed.
149
Fig. 7.3
P2 = 4C2 2! or
4!
4 C2
4 2 ! 2!
150
MATHEMATICS
P3 5 C 3 3! or
5!
5C 3
5 3! 3!
Pr n Cr r! , 0 r n .
n!
n
Remarks 1. From above n r ! C r r ! , i.e.,
n
In particular, if r n , C n
Cr
n!
r ! n r ! .
n!
1.
n! 0!
2.
3.
As
n!
1 n C 0 , the formula
0! n 0 !
Cr
n!
is applicable for r = 0 also.
r ! n r !
Hence
n
4.
Cn r
Cr
n!
r ! n r ! , 0 r n.
n!
= n r ! r ! = n Cr ,
n r ! n n r !
n!
151
C a = nCb a = b or a = n b, i.e., n = a + b
Theorem 6 n C r n C r 1
Proof We have
n 1
Cr
C r n C r 1
n!
n!
r! n r ! r 1 ! n r 1 !
n!
n!
+
r r 1 ! n r ! r 1 ! n r 1 n r !
n!
1
1
r 1! n r ! r n r 1
n!
n r 1 r
n 1! n 1 C
=
r
r
1
!
n
r
!
r
n
r
1
r! n 1 r !
Example 17 If n C9 n C 8 , find
C17 .
Solution We have n C9 n C 8
i.e.,
n!
n!
9! n 9 ! n 8 ! 8!
or
1
1
9 n8
Therefore
or n 8 = 9
or
n = 17
C17 17 C17 1 .
5!
4 5
10 .
3! 2!
2
Now, 1 man can be selected from 2 men in 2C 1 ways and 2 women can be
selected from 3 women in 3C 2 ways. Therefore, the required number of committees
152
MATHEMATICS
2!
3!
2
3
= C1 C2 1! 1! 2! 1! 6 .
Example 19 What is the number of ways of choosing 4 cards from a pack of 52
playing cards? In how many of these
(i)
(ii)
(iii)
(iv)
(v)
Solution There will be as many ways of choosing 4 cards from 52 cards as there are
combinations of 52 different things, taken 4 at a time. Therefore
The required number of ways =
52
C4
52!
49 50 51 52
4! 48!
2 3 4
= 270725
(i) There are four suits: diamond, club, spade, heart and there are 13 cards of each
suit. Therefore, there are 13C 4 ways of choosing 4 diamonds. Similarly, there are
13
C 4 ways of choosing 4 clubs, 13 C4 ways of choosing 4 spades and 13C4 ways of
choosing 4 hearts. Therefore
The required number of ways = 13C 4 + 13C4 + 13 C4 + 13 C4.
13!
= 4 4! 9! 2860
(ii) There are13 cards in each suit.
Therefore, there are 13C 1 ways of choosing 1 card from 13 cards of diamond,
13
C 1 ways of choosing 1 card from 13 cards of hearts, 13C 1 ways of choosing 1
card from 13 cards of clubs, 13C1 ways of choosing 1 card from 13 cards of
spades. Hence, by multiplication principle, the required number of ways
= 13C 1 13 C1 13C 1 13 C1 = 134
(iii) There are 12 face cards and 4 are to be selected out of these 12 cards. This can be
done in 12 C4 ways. Therefore, the required number of ways =
12!
495 .
4! 8!
153
(iv) There are 26 red cards and 26 black cards. Therefore, the required number of
ways = 26C2 26C2
2
26!
2
=
325 = 105625
2! 24!
(v)
C 4 ways.
26
26!
= 29900.
4! 22!
EXERCISE 7.4
1.
2.
3.
4.
5.
6.
7.
8.
9.
Miscellaneous Examples
Example 20 How many words, with or without meaning, each of 3 vowels and 2
consonants can be formed from the letters of the word INVOLUTE ?
Solution In the word INVOLUTE, there are 4 vowels, namely, I,O,E,Uand 4
consonants, namely, N, V, L and T.
154
MATHEMATICS
Now, each of these 24 combinations has 5 letters which can be arranged among
themselves in 5 ! ways. Therefore, the required number of different words is
24 5 ! = 2880.
Example 21 A group consists of 4 girls and 7 boys. In how many ways can a team of
5 members be selected if the team has (i) no girl ? (ii) at least one boy and one girl ?
(iii) at least 3 girls ?
Solution (i) Since, the team will not include any girl, therefore, only boys are to be
selected. 5 boys out of 7 boys can be selected in 7C5 ways. Therefore, the required
number of ways
(ii)
7!
6 7
7
= C5 5! 2! 2 21
Since, at least one boy and one girl are to be there in every team. Therefore, the
team can consist of
(a) 1 boy and 4 girls
155
Example 22 Find the number of words with or without meaning which can be made
using all the letters of the word AGAIN. If these words are written as in a dictionary,
what will be the 50th word?
Solution There are 5 letters in the wordAGAIN, in which A appears 2 times. Therefore,
the required number of words =
5!
60 .
2!
To get the number of words starting with A, we fix the letter A at the extreme left
position, we then rearrange the remaining 4 letters taken all at a time. There will be as
many arrangements of these 4 letters taken 4 at a time as there are permutations of 4
different things taken 4 at a time. Hence, the number of words starting with
4!
= 12 as after placing G
2!
at the extreme left position, we are left with the letters A, A, I and N. Similarly, there
are 12 words starting with the next letter I. Total number of words so far obtained
= 24 + 12 + 12 =48.
A = 4! = 24. Then, starting with G, the number of words
6! 4 5 6
= 60, as when 1 is
3! 2!
2
fixed at the extreme left position, the remaining digits to be rearranged will be 0, 2, 2, 2,
4, 4, in which there are 3, 2s and 2, 4s.
Total numbers begining with 2
=
6!
3 4 5 6
= 180
2! 2!
2
6!
4 5 6 = 120
3!
156
MATHEMATICS
numbers also, which have 0 at the extreme left position. The number of such
arrangements
6!
(by fixing 0 at the extreme left position) = 60.
3! 2!
Note If one or more than one digits given in the list is repeated, it will be
understood that in any number, the digits can be used as many times as is given in
the list, e.g., in the above example 1 and 0 can be used only once whereas 2 and 4
can be used 3 times and 2 times, respectively.
Example 24 In how many ways can 5 girls and 3 boys be seated in a row so that no
two boys are together?
Solution Let us first seat the 5 girls. This can be done in 5! ways. For each such
arrangement, the three boys can be seated only at the cross marked places.
G G G G G .
There are 6 cross marked places and the three boys can be seated in 6P 3 ways.
Hence, by multiplication principle, the total number of ways
6!
3!
= 4 5 2 3 4 5 6 = 14400.
= 5! 6P3 = 5!
5.
6.
7.
8.
9.
10.
11.
157
listed as in a dictionary, how many words are there in this list before the first
word starting with E ?
How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9
which are divisible by 10 and no digit is repeated ?
The English alphabet has 5 vowels and 21 consonants. How many words with
two different vowels and 2 different consonants can be formed from the
alphabet ?
In an examination, a question paper consists of 12 questions divided into two
parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student
is required to attempt 8 questions in all, selecting at least 3 from each part. In
how many ways can a student select the questions ?
Determine the number of 5-card combinations out of a deck of 52 cards if each
selection of 5 cards has exactly one king.
It is required to seat 5 men and 4 women in a row so that the women occupy the
even places. How many such arrangements are possible ?
From a class of 25 students, 10 are to be chosen for an excursion party. There
are 3 students who decide that either all of them will join or none of them will
join. In how many ways can the excursion party be chosen ?
In how many ways can the letters of the word ASSASSINATION be arranged
so that all the Ss are together ?
Summary
n!
,
(n r )!
where 0 r n.
n! = 1 2 3 ...n
n! = n (n 1) !
The number of permutations of n different things, taken r at a time, where
repeatition is allowed, is nr .
The number of permutations of n objects taken all at a time, where p1 objects
158
MATHEMATICS
are of first kind, p2 objects are of the second kind, ..., p k objects are of the kth
n!
kind and rest, if any, are all different is p ! p !... p ! .
1
2
k
n!
, 0 r n.
r !( n r )!
Historical Note
The concepts of permutations and combinations can be traced back to the advent
of Jainism in India and perhaps even earlier. The credit, however, goes to the
Jains who treated its subject matter as a self-contained topic in mathematics,
under the name Vikalpa.
Among the Jains, Mahavira, (around 850) is perhaps the worlds first
mathematician credited with providing the general formulae for permutations and
combinations.
In the 6th century B.C., Sushruta, in his medicinal work, Sushruta Samhita,
asserts that 63 combinations can be made out of 6 different tastes, taken one at a
time, two at a time, etc. Pingala, a Sanskrit scholar around third century B.C.,
gives the method of determining the number of combinations of a given number
of letters, taken one at a time, two at a time, etc. in his work Chhanda Sutra.
Bhaskaracharya (born 1114) treated the subject matter of permutations and
combinations under the name Anka Pasha in his famous work Lilavati. In addition
to the general formulae for n C r and nP r already provided by Mahavira,
Bhaskaracharya gives several important theorems and results concerning the
subject.
Outside India, the subject matter of permutations and combinations had its
humble beginnings in China in the famous book IKing (Book of changes). It is
difficult to give the approximate time of this work, since in 213 B.C., the emperor
had ordered all books and manuscripts in the country to be burnt which fortunately
was not completely carried out. Greeks and later Latin writers also did some
scattered work on the theory of permutations and combinations.
Some Arabic and Hebrew writers used the concepts of permutations and
combinations in studying astronomy. Rabbi ben Ezra, for instance, determined
the number of combinations of known planets taken two at a time, three at a time
and so on. This was around 1140. It appears that Rabbi ben Ezra did not know
the formula for nC r. However, he was aware that nCr = nCn r for specific values
n and r. In 1321, Levi Ben Gerson, another Hebrew writer came up with the
formulae for nPr , nPn and the general formula for nC r.
The first book which gives a complete treatment of the subject matter of
permutations and combinations is Ars Conjectandi written by a Swiss, Jacob
Bernoulli (1654 1705), posthumously published in 1713. This book contains
essentially the theory of permutations and combinations as is known today.
159
Chapter
BINOMIAL THEOREM
Mathematics is a most exact science and its conclusions are capable of
absolute proofs. C.P. STEINMETZ
8.1 Introduction
In earlier classes, we have learnt how to find the squares
and cubes of binomials like a + b and a b. Using them, we
could evaluate the numerical values of numbers like
(98)2 = (100 2)2, (999)3 = (1000 1) 3, etc. However, for
higher powers like (98)5, (101)6, etc., the calculations become
difficult by using repeated multiplication. This difficulty was
overcome by a theorem known as binomial theorem. It gives
an easier way to expand (a + b)n, where n is an integer or a
rational number. In this Chapter, we study binomial theorem
for positive integral indices only.
Blaise Pascal
(1623-1662)
BINOMIAL THEOREM
161
Fig 8.1
Do we observe any pattern in this table that will help us to write the next row? Yes we
do. It can be seen that the addition of 1s in the row for index 1 gives rise to 2 in the row
for index 2. The addition of 1, 2 and 2, 1 in the row for index 2, gives rise to 3 and 3 in
the row for index 3 and so on. Also, 1 is present at the beginning and at the end of each
row. This can be continued till any index of our interest.
We can extend the pattern given in Fig 8.2 by writing a few more rows.
Fig 8.2
Pascals Triangle
The structure given in Fig 8.2 looks like a triangle with 1 at the top vertex and running
down the two slanting sides. This array of numbers is known as Pascals triangle,
after the name of French mathematician Blaise Pascal. It is also known as Meru
Prastara by Pingla.
Expansions for the higher powers of a binomial are also possible by using Pascals
triangle. Let us expand (2x + 3y)5 by using Pascals triangle. The row for index 5 is
1
5
10
10
5
1
Using this row and our observations (i), (ii) and (iii), we get
(2x + 3y) 5 = (2x) 5 + 5(2x) 4 (3y) + 10(2x) 3 (3y) 2 +10 (2x)2 (3y) 3 + 5(2x)(3y)4 +(3y) 5
= 32x5 + 240x4y + 720x3 y2 + 1080x2y3 + 810xy4 + 243y5.
162
MATHEMATICS
Now, if we want to find the expansion of (2x + 3y) 12, we are first required to get
the row for index 12. This can be done by writing all the rows of the Pascals triangle
till index 12. This is a slightly lengthy process. The process, as you observe, will become
more difficult, if we need the expansions involving still larger powers.
We thus try to find a rule that will help us to find the expansion of the binomial for
any power without writing all the rows of the Pascals triangle, that come before the
row of the desired index.
For this, we make use of the concept of combinations studied earlier to rewrite
the numbers in the Pascals triangle. We know that
Cr
n!
, 0 r n and
r!(n r )!
Fig 8.3
Pascals triangle
Observing this pattern, we can now write the row of the Pascals triangle for any index
without writing the earlier rows. For example, for the index 7 the row would be
C 0 7 C1 7C2 7C 3 7 C4 7C5 7C 6 7 C7.
Thus, using this row and the observations (i), (ii) and (iii), we have
(a + b) 7 = 7C 0 a7 + 7C1 a6b + 7C 2a5b 2 + 7C3a 4b 3 + 7C4a 3b4 + 7C5 a2b 5 + 7C 6ab6 + 7C7b 7
An expansion of a binomial to any positive integral index say n can now be visualised
using these observations. We are now in a position to write the expansion of a binomial
to any positive integral index.
BINOMIAL THEOREM
163
... (1)
k+1
C0 a k + 1 +
k+1
C1 akb +
k+ 1
C 2 ak 1 b2 + ...+
k+1
Ck+1 bk + 1
Now, (a + b) k + 1 = (a + b) (a + b) k
= (a + b) (kC 0 ak + kC 1ak 1 b + kC 2 ak 2 b2 +...+ kCk 1 abk 1 + kCk b k)
[from (1)]
= kC0 ak + 1 + kC1 akb + kC2ak 1b2 +...+ kCk 1 a2bk 1 + kCk abk + kC0 akb
+ kC1ak 1b2 + kC2ak 2b3+...+ kCk-1abk + kCkbk + 1
[by actual multiplication]
= kC0a k + 1 + (kC1+ kC 0) akb + (kC2 + kC1)a k 1b 2 + ...
+ (kCk+ kCk1) abk + kC kb k + 1
= k + 1C0a
(by using
k+1
k+ 1
+ k + 1C1a kb +
C 0=1, kC r + kCr1 =
k+ 1
k+1
Cr
and
Ck = 1= k + 1C k + 1)
Thus, it has been proved that P (k + 1) is true whenever P(k) is true. Therefore, by
principle of mathematical induction, P(n) is true for every positive integer n.
We illustrate this theorem by expanding (x + 2)6:
(x + 2)6 = 6C 0x6 + 6C 1x5.2 + 6 C2x42 2 + 6C 3x3.23 + 6 C4x2.2 4 + 6C5 x.2 5 + 6C 6.26.
= x6 + 12x5 + 60x4 + 160x3 + 240x2 + 192x + 64
Thus (x + 2)6 = x6 + 12x5 + 60x4 + 160x3 + 240x2 + 192x + 64.
164
MATHEMATICS
Observations
n
1.
The notation
n
n 0
k0
C k a n k b k stands for
(a b ) n n C k a n k b k .
k 0
2.
3.
4.
5.
(x2y) 5 =
Taking a = 1, b = x, we obtain
(1 + x) n = nC 0(1)n + nC1(1)n 1x + nC2 (1)n 2 x2 + ... + n Cnxn
= nC 0 + nC1x + nC 2x2 + nC3 x3 + ... + nC nxn
Thus
BINOMIAL THEOREM
165
Taking a = 1, b = x, we obtain
(1 x)n
2 3
Example 1 Expand x , x 0
x
3
2 3
3
3
3
x = 4C0 (x2)4 + 4C 1(x2)3 + 4C 2(x2 )2 + 4C 3(x2 ) + 4C4
x
x
x
x
x
2
= x8 + 4.x6 .
3
9
27 81
+ 6.x4 . 2 + 4.x2. 3 + 4
x
x
x
x
= x8 + 12x5 + 54x2 +
108 81
.
x x4
166
MATHEMATICS
(1.01)1000000 = (1 + 0.01)1000000
=
1000000
C0 +
1000000
Hence
= 5, we get
(1 + 5) n = n C0 + nC1 5 + nC2 52 + ... + nCn 5n
i.e.
i.e.
or
or
6n 5n = 25k+1
EXERCISE 8.1
Expand each of the expressions in Exercises 1 to 5.
5
1. (12x)
2 x
2.
x 2
3. (2x 3)6
BINOMIAL THEOREM
4.
x 1
3 x
5. x
x
167
2 )4 .
8.3
3
r 0
r n
C r 4n .
1.
2.
n 1 1
, i.e.,
n is even so n + 1 is odd. Therefore, the middle term is
2
th
th
n
1 term.
2
th
5th term.
(ii) If n is odd, then n +1 is even, so there will be two middle terms in the
168
MATHEMATICS
n 1
n 1
1 term. So in the expansion
term and
expansion, namely,
2
2
th
th
7 1
7 1
1 , i.e., 5 th term.
(2x y) 7, the middle terms are
, i.e., 4 th and
2
2
th
th
2 n 1 1
,
In the expansion of x , where x 0, the middle term is
x
2
2n
3.
th
1
It is given by C nx = 2nC n (constant).
x
2n
Therefore,
T18 = 50C17 2 33 a 17
Given that
T17 = T18
So
50
Therefore
50
i.e.,
a=
50
C 16 . 2
50
C 17 .2 33
C16 2
50
C 17
34
a 17
16
a
50!
17! . 33!
2 =1
16!34!
50!
1.3.5...(2n 1)
2n xn, where n is a positive integer.
n!
BINOMIAL THEOREM
169
th
2n
Solution As 2n is even, the middle term of the expansion (1 + x) is 1 ,
2
(2n )! n
x
n! n !
2n (2 n 1) (2n 2) ...4.3 .2 .1 n
x
n ! n!
1.2.3.4...(2n 2) (2n 1) (2 n) n
x
n! n!
[1.3.5...(2n 1)] n ! n n
2 .x
n! n!
1.3.5...(2n 1) n n
2 x
n!
9! 3 9 .8.7 3
.2 =
. 2 = 672.
3!6!
3.2
Example 8 The second, third and fourth terms in the binomial expansion (x + a) n are
240, 720 and 1080, respectively. Find x, a and n.
Solution Given that second term T2 = 240
170
MATHEMATICS
We have
T2 = nC 1xn 1 . a
So
Similarly
and
C 1xn1 . a = 240
... (1)
C 2x
... (2)
C 3xn3 a3 = 1080
... (3)
n2
a = 720
or
(n 1)! a
. 6
(n 2)! x
C2 xn 2 a 2 720
i.e.,
n
C1 xn 1 a 240
a
6
x ( n 1)
... (4)
2
x
(n 2)
... (5)
.
n 1
2 (n 2)
Hence, from (1), 5x4a = 240, and from (4),
Thus, n = 5
a 3
x 2
and
Cr 2
1
, i.e., n 8r + 9 = 0
C r 1
7
Cr 1
7
, i.e., n 7r + 1 = 0
Cr
42
... (1)
... (2)
BINOMIAL THEOREM
171
EXERCISE 8.2
Find the coefficient of
1. x5 in (x + 3) 8
2. a 5b7 in (a 2b)12 .
3. (x2 y) 6
5.
6.
1
Find the 13 term in the expansion of 9 x
, x 0.
3 x
18
th
x3
3
9.
10
8. 9 y
3
10. The coefficients of the (r 1)th, rth and (r + 1)th terms in the expansion of (x + 1)n
are in the ratio 1 : 3 : 5. Find n and r.
11. Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient
of xn in the expansion of (1 + x)2 n 1.
12. Find a positive value of m for which the coefficient of x2 in the expansion
(1 + x) m is 6.
Miscellaneous Examples
6
3 2
1
Example 10 Find the term independent of x in the expansion of x
.
3x
2
6r
Solution We have Tr + 1
3 2
= Cr x
2
6
3
= Cr
2
6
6 r
1
3x
x2
6 r
1 1
r
x 3
172
MATHEMATICS
r
= ( 1)
Cr
(3)6 2 r 12 3 r
x
(2) 6 r
(3) 6 8 5
.
Hence 5th term is independent of x and is given by ( 1)4 6 C4
(2) 6 4 12
Example 11 If the coefficients of ar 1, a r and ar + 1 in the expansion of (1 + a) n are in
arithmetic progression, prove that n2 n(4r + 1) + 4r2 2 = 0.
Solution The (r + 1)th term in the expansion is nCr ar . Thus it can be seen that ar occurs
in the (r + 1)th term, and its coefficient is n Cr . Hence the coefficients of ar 1 , ar and
ar + 1 are n Cr 1, nCr and n Cr + 1, respectively. Since these coefficients are in arithmetic
progression, so we have, nC r 1+ nC r + 1 = 2. nCr . This gives
n!
n!
n!
2
(r 1)!( n r 1)! ( r 1)!( n r 1)!
r !(n r )!
i.e.
1
1
or
1
( r 1)! ( n r 1)!
2
i.e.
1
r (r 1)!(n r ) (n r 1)!
1
1
(
n
r
)
(
n
r
1)
(
r
1)
(
r
)
1
( r 1)! ( n r 1)![ r ( n r )]
1
1
2
,
(n r 1) ( n r ) r (r 1) r ( n r )
or
r( r 1) (n r ) ( n r 1)
2
(n r ) ( n r 1) r ( r 1) r (n r )
or
r(r + 1) + (n r) (n r + 1) = 2 (r + 1) (n r + 1)
or
r2 + r + n 2 nr + n nr + r2 r = 2(nr r 2 + r + n r + 1)
BINOMIAL THEOREM
173
n2 4nr n + 4r2 2 = 0
n2 n (4r + 1) + 4r2 2 = 0
or
i.e.,
Example 12 Show that the coefficient of the middle term in the expansion of (1 + x)2n is
equal to the sum of the coefficients of two middle terms in the expansion of (1 + x) 2n 1.
Solution As 2n is even so the expansion (1 + x)2 n has only one middle term which is
2n
th
and
2
2
2n 1
2n 1
these terms are
C n 1 and
C n, respectively.
Now
th
2n 1
th
Cn 1 +
Cn = 2n Cn
2n 1
[As nCr 1+ n Cr =
C r]. as required.
n+1
174
MATHEMATICS
3
1
Example 15 Find the term independent of x in the expansion of x 3 , x > 0.
2 x
Solution We have Tr + 1 =
18
18
Cr
Cr
1
3
2 x
18 r
x 3
18 r
.
r
2 .x
r
3
18
1
Cr r . x
2
18 2 r
3
Since we have to find a term independent of x, i.e., term not having x, so take
We get r = 9. The required term is 18 C9
18 2 r
0.
3
1
.
29
Example 16 The sum of the coefficients of the first three terms in the expansion of
m
containing x3.
3
9 m (m 1)
559
2
BINOMIAL THEOREM
175
Now
3
12
12
r
2 = 12C ( 3)r . x12 3r
Tr + 1 = C r x
r
x
3 2
3 2
2
2
6. Find the value of a a 1
a
4
a2 1 .
7. Find an approximation of (0.99)5 using the first three terms of its expansion.
8. Find n, if the ratio of the fifth term from the beginning to the fifth term from the
n
4
1
end in the expansion of 2 4 is
3
6 :1 .
176
MATHEMATICS
x 2
2
2
10. Find the expansion of (3x 2ax + 3a )3 using binomial theorem.
Summary
integral n is given by Binomial
The expansion of a binomial forn anyn positive
n
n
n 1
n
n 2 2
Theorem, which is (a + b) = C 0a + C 1a
b + C2a
b + ...+
nC
n
1
n
n
a.b
+ C nb .
n 1
The
coefficients
of the expansions are arranged in an array. This array is
n
In
the
expansion
(a
+
b)
,
if
n
is
even,
then
the
middle
term
is
the
1
2
n
n 1
n1
term.If n is odd, then the middle terms are
and 1 terms.
2
2
th
th
Historical Note
The ancient Indian mathematicians knew about the coefficients in the
expansions of (x + y)n , 0 n 7. The arrangement of these coefficients was in
the form of a diagram called Meru-Prastara, provided by Pingla in his book
Chhanda shastra (200B.C.). This triangular arrangement is also found in the
work of Chinese mathematician Chu-shi-kie in 1303. The term binomial coefficients
was first introduced by the German mathematician, Michael Stipel (1486-1567) in
approximately 1544. Bombelli (1572) also gave the coefficients in the expansion of
(a + b) n, for n = 1,2 ...,7 and Oughtred (1631) gave them for n = 1, 2,..., 10. The
arithmetic triangle, popularly known as Pascals triangle and similar to the MeruPrastara of Pingla was constructed by the French mathematician Blaise Pascal
(1623-1662) in 1665.
The present form of the binomial theorem for integral values of n appeared in
Trate du triange arithmetic, written by Pascal and published posthumously in
1665.
Chapter
he
9.1 Introduction
is
no N
C
tt E
o R
be T
re
pu
bl
9.2 Sequences
178
MATHEMATICS
no N
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tt E
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be T
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pu
bl
is
he
The number of persons ancestors for the first, second, third, , tenth generations are
2, 4, 8, 16, 32, , 1024. These numbers form what we call a sequence.
Consider the successive quotients that we obtain in the division of 10 by 3 at
different steps of division. In this process we get 3,3.3,3.33,3.333, ... and so on. These
quotients also form a sequence. The various numbers occurring in a sequence are
called its terms. We denote the terms of a sequence by a1, a2, a3, , an, , etc., the
subscripts denote the position of the term. The nth term is the number at the nth position
of the sequence and is denoted by an. The nth term is also called the general term of the
sequence.
Thus, the terms of the sequence of persons ancestors mentioned above are:
a1 = 2, a2 = 4, a3 = 8, , a10 = 1024.
Similarly, in the example of successive quotients
a1 = 3, a2 = 3.3, a3 = 3.33, , a6 = 3.33333, etc.
A sequence containing finite number of terms is called a finite sequence. For
example, sequence of ancestors is a finite sequence since it contains 10 terms (a fixed
number).
A sequence is called infinite, if it is not a finite sequence. For example, the
sequence of successive quotients mentioned above is an infinite sequence, infinite in
the sense that it never ends.
Often, it is possible to express the rule, which yields the various terms of a sequence
in terms of algebraic formula. Consider for instance, the sequence of even natural
numbers 2, 4, 6,
Here
a1 = 2 = 2 1
a2 = 4 = 2 2
a3 = 6 = 2 3
a4 = 8 = 2 4
....
....
....
....
....
....
....
....
....
....
....
....
In fact, we see that the nth term of this sequence can be written as an = 2n,
where n is a natural number. Similarly, in the sequence of odd natural numbers 1,3,5, ,
the nth term is given by the formula, an = 2n 1, where n is a natural number.
In some cases, an arrangement of numbers such as 1, 1, 2, 3, 5, 8,.. has no visible
pattern, but the sequence is generated by the recurrence relation given by
a1 = a2 = 1
a3 = a1 + a2
an = an 2 + an 1, n > 2
179
9.3 Series
is
he
In the sequence of primes 2,3,5,7,, we find that there is no formula for the nth
prime. Such sequence can only be described by verbal description.
In every sequence, we should not expect that its terms will necessarily be given
by a specific formula. However, we expect a theoretical scheme or a rule for generating
the terms a1, a2, a3,,an, in succession.
In view of the above, a sequence can be regarded as a function whose domain
is the set of natural numbers or some subset of it of the type {1, 2, 3...k}. Sometimes,
we use the functional notation a(n) for an.
bl
is called the series associated with the given sequence .The series is finite or infinite
according as the given sequence is finite or infinite. Series are often represented in
no N
C
tt E
o R
be T
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pu
compact form, called sigma notation, using the Greek letter (sigma) as means of
indicating the summation involved. Thus, the series a1 + a2 + a3 + ... + an is abbreviated
n
as
ak .
k =1
Remark When the series is used, it refers to the indicated sum not to the sum itself.
For example, 1 + 3 + 5 + 7 is a finite series with four terms. When we use the phrase
sum of a series, we will mean the number that results from adding the terms, the
sum of the series is 16.
We now consider some examples.
Example 1 Write the first three terms in each of the following sequences defined by
the following:
(i) an = 2n + 5,
(ii) an =
n3
.
4
n3
1 3
1
1
= , a2 = , a3 = 0
. Thus, a1 =
4
4
2
4
180
MATHEMATICS
he
no N
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be T
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Solution We have
bl
a1 = 1, an = an 1 + 2 for n 2.
is
a1 = 1, a2 = a1 + 2 = 1 + 2 = 3, a3 = a2 + 2 = 3 + 2 = 5,
a4 = a3 + 2 = 5 + 2 = 7, a5 = a4 + 2 = 7 + 2 = 9.
Hence, the first five terms of the sequence are 1,3,5,7 and 9. The corresponding series
is 1 + 3 + 5 + 7 + 9 +...
EXERCISE 9.1
Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth
terms are:
1.
an = n (n + 2)
4. an =
2n 3
6
n
n +1
3. an = 2n
an = (1)n1 5n+1
6. a n = n
2. an =
5.
n2 + 5
.
4
Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth
terms are:
7.
an = 4n 3; a17, a24
9. an = (1)n 1n3; a9
n2
8. an = 2n ; a7
n( n 2)
; a20 .
10. an =
n+3
181
Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the
corresponding series:
11. a1 = 3, an = 3an 1 + 2 for all n > 1
12. a1 = 1, an =
an 1
,n2
n
he
is
an +1
an , for n = 1, 2, 3, 4, 5
bl
Find
no N
C
tt E
o R
be T
re
pu
= a + (n 1) d
MATHEMATICS
Sn =
We can also write, Sn =
n
[ 2a + (n 1)d ]
2
n
[a + l ]
2
he
182
... (1)
... (2)
bl
Solution We have am = a + (m 1) d = n,
and
an = a + (n 1) d = m
Solving (1) and (2), we get
(m n) d = n m, or d = 1,
and
a= n+m1
Therefore
a p = a + (p 1)d
= n + m 1 + ( p 1) (1) = n + m p
Hence, the pth term is n + m p.
is
Example 4 In an A.P. if mth term is n and the nth term is m, where m n, find the pth
term.
no N
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be T
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... (3)
... (4)
1
n( n 1)Q , where P and Q
2
Sn = a1 + a2 + a3 +...+ an1 + an = nP +
1
n (n 1) Q
2
Therefore
S1 = a1 = P, S2 = a1 + a2 = 2P + Q
So that
a2 = S2 S1 = P + Q
Hence, the common difference is given by d = a2 a1 = (P + Q) P = Q.
Example 6 The sum of n terms of two arithmetic progressions are in the ratio
(3n + 8) : (7n + 15). Find the ratio of their 12th terms.
Solution Let a1, a2 and d1, d2 be the first terms and common difference of the first
and second arithmetic progression, respectively. According to the given condition, we
have
or
n
[ 2a1 + ( n 1 )d1 ] 3n + 8
2
=
n
7 n + 15
2
+
1
a
(
n
)d
[ 2
2]
2
or
2a1 + (n 1)d1 3n + 8
=
2a2 + (n 1)d 2 7 n + 15
Now
is
a1 + 11d1
12 th term of first A.P.
7
= th
=
a2 + 11d 2 12 term of second A.P. 16
no N
C
tt E
o R
be T
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pu
Therefore
... (1)
bl
2a1 + 22d1 3 23 + 8
=
2a2 + 22d 2 7 23 + 15
183
he
Example 7 The income of a person is Rs. 3,00,000, in the first year and he receives an
increase of Rs.10,000 to his income per year for the next 19 years. Find the total
amount, he received in 20 years.
Solution Here, we have an A.P. with a = 3,00,000, d = 10,000, and n = 20.
Using the sum formula, we get,
20
[600000 + 19 10000] = 10 (790000) = 79,00,000.
2
Hence, the person received Rs. 79,00,000 as the total amount at the end of 20 years.
S20 =
9.4.1 Arithmetic mean Given two numbers a and b. We can insert a number A
between them so that a, A, b is an A.P. Such a number A is called the arithmetic mean
(A.M.) of the numbers a and b. Note that, in this case, we have
A a = b A,
i.e., A =
a+b
2
We may also interpret the A.M. between two numbers a and b as their
average
a+b
. For example, the A.M. of two numbers 4 and 16 is 10. We have, thus
2
constructed an A.P. 4, 10, 16 by inserting a number 10 between 4 and 16. The natural
184
MATHEMATICS
he
question now arises : Can we insert two or more numbers between given two numbers
so that the resulting sequence comes out to be an A.P. ? Observe that two numbers 8
and 12 can be inserted between 4 and 16 so that the resulting sequence 4, 8, 12, 16
becomes an A.P.
More generally, given any two numbers a and b, we can insert as many numbers
as we like between them such that the resulting sequence is an A.P.
Let A1, A2, A3, , An be n numbers between a and b such that a, A1, A2, A3, ,
An, b is an A.P.
Here, b is the (n + 2) th term, i.e., b = a + [(n + 2) 1]d = a + (n + 1) d.
ba
n +1
no N
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be T
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pu
A1 = a + d = a +
is
ba
.
n +1
bl
d=
This gives
A2 = a + 2d = a +
2(b a )
n +1
A3 = a + 3d = a +
3(b a )
n +1
.....
.....
.....
.....
.....
.....
An = a + nd = a +
.....
.....
n (b a )
n +1 .
Example 8 Insert 6 numbers between 3 and 24 such that the resulting sequence is
an A.P.
Solution Let A1, A2, A3, A4, A5 and A6 be six numbers between 3 and 24 such that
3, A1, A2, A3, A4, A5, A6, 24 are in A.P. Here, a = 3, b = 24, n = 8.
Therefore, 24 = 3 + (8 1) d, so that d = 3.
A2 = a + 2d = 3 + 2 3 = 9;
Thus
A1 = a + d = 3 + 3 = 6;
A3 = a + 3d = 3 + 3 3 = 12; A4 = a + 4d = 3 + 4 3 = 15;
A5 = a + 5d = 3 + 5 3 = 18; A6 = a + 6d = 3 + 6 3 = 21.
Hence, six numbers between 3 and 24 are 6, 9, 12, 15, 18 and 21.
185
EXERCISE 9.2
1
(pq +1), where p q.
2
If the sum of a certain number of terms of the A.P. 25, 22, 19, is 116. Find the
last term.
Find the sum to n terms of the A.P., whose kth term is 5k + 1.
If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants,
find the common difference.
The sums of n terms of two arithmetic progressions are in the ratio
5n + 4 : 9n + 6. Find the ratio of their 18th terms.
If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then
find the sum of the first (p + q) terms.
Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.
terms is
no N
C
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be T
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pu
6.
1
1
and qth term is , prove that the sum of first pq
p
q
bl
is
he
7.
8.
9.
10.
11.
Prove that
a
b
c
(q r ) + (r p) + ( p q) = 0
p
q
r
12. The ratio of the sums of m and n terms of an A.P. is m2 : n2. Show that the ratio
of mth and nth term is (2m 1) : (2n 1).
13. If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value
of m.
14. Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
15. If
a n + bn
is the A.M. between a and b, then find the value of n.
a n 1 + b n 1
16. Between 1 and 31, m numbers have been inserted in such a way that the resulting
sequence is an A. P. and the ratio of 7th and (m 1)th numbers is 5 : 9. Find the
value of m.
186
MATHEMATICS
17. A man starts repaying a loan as first instalment of Rs. 100. If he increases the
instalment by Rs 5 every month, what amount he will pay in the 30th instalment?
18. The difference between any two consecutive interior angles of a polygon is 5.
If the smallest angle is 120 , find the number of the sides of the polygon.
Let us consider the following sequences:
(i) 2,4,8,16,..., (ii)
1 1 1 1
, , ,
(iii) .01,.0001,.000001,...
9 27 81 243 ...
he
In (i), we have
bl
is
In each of these sequences, how their terms progress? We note that each term, except
the first progresses in a definite order.
and so on.
no N
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be T
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1 a
1 a
1 a
1
and so on.
In (ii), we observe, a1 = , 2 = , 3 = , 4 =
9 a1 3 a2 3 a3 3
Similarly, state how do the terms in (iii) progress? It is observed that in each case,
a3
a
a
a1 =ratio
2, 2to=the
2, term
=immediately
2, 4 = 2 preceding
every term except the first term bears a constant
a1
a2
a3
1
it. In (i), this constant ratio is 2; in (ii), it is and in (iii), the constant ratio is 0.01.
3
Such sequences are called geometric sequence or geometric progression abbreviated
as G.P.
A sequence a1, a2, a3, , an, is called geometric progression, if each term is
non-zero and
ak + 1
ak
= r (constant), for k 1.
1
and 0.01, respectively.
3
As in case of arithmetic progression, the problem of finding the nth term or sum of n
terms of a geometric progression containing a large number of terms would be difficult
without the use of the formulae which we shall develop in the next Section. We shall
use the following notations with these formulae:
a = the first term, r = the common ratio, l = the last term,
geometric progression (i), (ii) and (iii) above are 2,
187
is
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9.5.1 General term of a G .P. Let us consider a G.P. with first non-zero term a and
common ratio r. Write a few terms of it. The second term is obtained by multiplying
a by r, thus a2 = ar. Similarly, third term is obtained by multiplying a2 by r. Thus,
a3 = a2r = ar2, and so on.
We write below these and few more terms.
1st term = a1 = a = ar11, 2nd term = a2 = ar = ar21, 3rd term = a3 = ar2 = ar31
4th term = a4 = ar3 = ar41, 5th term = a5 = ar4 = ar51
Do you see a pattern? What will be 16th term?
a16 = ar161 = ar15
Therefore, the pattern suggests that the nth term of a G.P. is given by
bl
an = ar n1 .
no N
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be T
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Thus, a, G.P. can be written as a, ar, ar2, ar3, arn 1; a, ar, ar2,...,arn 1... ;according
as G.P. is finite or infinite, respectively.
The series a + ar + ar2 + ... + arn1 or a + ar + ar2 + ... + arn1 +...are called
finite or infinite geometric series, respectively.
9.5.2. Sum to n terms of a G .P. Let the first term of a G.P. be a and the common
ratio be r. Let us denote by Sn the sum to first n terms of G.P. Then
Sn = a + ar + ar2 +...+ arn1
... (1)
Case 1
If r = 1, we have Sn = a + a + a + ... + a (n terms) = na
If r 1, multiplying (1) by r, we have
rSn = ar + ar2 + ar3 + ... + arn
Subtracting (2) from (1), we get (1 r) Sn = a arn = a(1 rn)
Case 2
a (1 r n )
1 r
... (2)
a ( r n 1)
r 1
Example 9 Find the 10th and nth terms of the G.P. 5, 25,125, .
Solution Here a = 5 and r = 5. Thus, a10 = 5(5)101 = 5(5)9 = 510
and an = arn1 = 5(5)n1 = 5n .
This gives
Sn =
or
Sn =
188
MATHEMATICS
Example11 In a G.P., the 3rd term is 24 and the 6th term is 192.Find the 10th term.
Solution Here, a3 = ar 2 = 24
... (1)
he
and
... (2)
Dividing (2) by (1), we get r = 2. Substituting r = 2 in (1), we get a = 6.
Hence
a10 = 6 (2)9 = 3072.
Example12 Find the sum of first n terms and the sum of first 5 terms of the geometric
2
. Therefore
3
bl
is
series
no N
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tt E
o R
be T
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pu
2 n
1
n
a (1 r n ) 3 3 1 2
=
Sn =
=
2
3
1 r
1
3
a =2 ar45 = 192
1 6+ + + ...
3 9
2 5
211
211
3
In particular, S5 = 3 1 =
=
.
243
81
3
3 3
Example 13 How many terms of the G.P. 3, , ,... are needed to give the
2 4
3069
sum 512 ?
Solution Let n be the number of terms needed. Given that a = 3, r =
a (1 r n )
1 r
Since
Sn =
Therefore
3069
=
512
1
)
2n = 6 1 1
n
1
2
1
2
3(1
1
3069
and Sn =
2
512
189
1
3069
= 1 n
2
3072
or
or
he
1
3069
3
1
=
=
n = 1
2
3072 3072 1024
2n = 1024 = 210, which gives n = 10.
or
13
and their product is 1.
12
a
, a, ar be the first three terms of the G.P. Then
r
no N
C
tt E
o R
be T
re
pu
a
13
+ ar + a =
r
12
bl
Solution Let
is
a
(a ) (ar ) = 1
r
From (2), we get a3 = 1, i.e., a = 1 (considering only real roots)
and
... (1)
... (2)
1
13
1 r =
or 12r2 + 25r + 12 = 0.
r
12
3
4
or .
4
3
4
3
3
3
4
4
and , 1,
for r =
,
Thus, the three terms of G.P. are : , 1, for r =
3
4
4
4
3
3
Example15 Find the sum of the sequence 7, 77, 777, 7777, ... to n terms.
This is a quadratic in r, solving, we get r =
Solution This is not a G.P., however, we can relate it to a G.P. by writing the terms as
Sn = 7 + 77 + 777 + 7777 + ... to n terms
=
7
[9 + 99 + 999 + 9999 + ...to n term]
9
7
[(10 1) + (102 1) + (103 1) + (104 1) + ...n terms]
9
190
MATHEMATICS
7
[(10 + 102 + 103 + ...n terms) (1+1+1+...n terms)]
9
7 10 (10n 1)
7 10(10n 1)
n
n .
=
9 10 1
9
9
he
is
a (r n 1)
Sn =
r 1
bl
We have
S10 = 2(210 1) = 2046
Hence, the number of ancestors preceding the person is 2046.
no N
C
tt E
o R
be T
re
pu
9.5.3 Geometric Mean (G .M.) The geometric mean of two positive numbers a
b = ar
n +1
b n + 1 .
r =
a
or
Hence
2
b n +1
G1 = ar = a ,
b n +1
2
G 2 = ar = a ,
a
a
b n +1
G 3 = ar = a ,
a
3
b n +1
G n = ar n = a
a
Example17 Insert three numbers between 1 and 256 so that the resulting sequence
is a G.P.
Solution Let G1, G2,G3 be three numbers between 1 and 256 such that
1, G1,G2,G3 ,256 is a G.P.
191
he
Therefore
256 = r4 giving r = 4 (Taking real roots only)
For r = 4, we have G1 = ar = 4, G2 = ar2 = 16, G3 = ar3 = 64
Similarly, for r = 4, numbers are 4,16 and 64.
Hence, we can insert 4, 16, 64 between 1 and 256 so that the resulting sequences are
in G.P.
a+b
and G = ab
2
Thus, we have
a +b
a + b 2 ab
ab =
2
2
no N
C
tt E
o R
be T
re
pu
AG=
bl
A=
is
Let A and G be A.M. and G.M. of two given positive real numbers a and b, respectively.
Then
a b
0
2
From (1), we obtain the relationship A G.
=
... (1)
Example 18 If A.M. and G.M. of two positive numbers a and b are 10 and 8,
respectively, find the numbers.
a+b
=10
2
A.M. =
and
G.M. = ab = 8
... (1)
... (2)
a + b = 20
... (3)
ab = 64
... (4)
2
Putting the value of a and b from (3), (4) in the identity (a b) = (a + b)2 4ab,
we get
(a b)2 = 400 256 = 144
or
a b = 12
... (5)
Solving (3) and (5), we obtain
a = 4, b = 16 or a = 16, b = 4
Thus, the numbers a and b are 4, 16 or 16, 4 respectively.
192
MATHEMATICS
EXERCISE 9.3
5 5 5
, , , ...
2 4 8
Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.
The 5 th, 8 th and 11 th terms of a G.P. are p, q and s, respectively. Show
that q2 = ps.
The 4th term of a G.P. is square of its second term, and the first term is 3.
Determine its 7th term.
Which term of the following sequences:
5.
(a)
2 ,2 2 ,4 ,... is 128 ?
(c)
1 1 1
1
, , ,... is
?
3 9 27
19683
(b)
is
4.
bl
2.
3.
he
no N
C
tt E
o R
be T
re
pu
2
7
, x, are in G.P.?
7
2
Find the sum to indicated number of terms in each of the geometric progressions in
Exercises 7 to 10:
7. 0.15, 0.015, 0.0015, ... 20 terms.
6. For what values of x, the numbers
8.
7 , 21 , 3 7 , ... n terms.
9. 1, a, a2, a3, ... n terms (if a 1).
10. x3, x5, x7, ... n terms (if x 1).
11
11. Evaluate
(2 + 3k ) .
k =1
39
12. The sum of first three terms of a G.P. is
and their product is 1. Find the
10
common ratio and the terms.
13. How many terms of G.P. 3, 32, 33, are needed to give the sum 120?
14. The sum of first three terms of a G.P. is 16 and the sum of the next three terms is
128. Determine the first term, the common ratio and the sum to n terms of the G.P.
15. Given a G.P. with a = 729 and 7th term 64, determine S7.
16. Find a G.P. for which sum of the first two terms is 4 and the fifth term is
4 times the third term.
17. If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x,
y, z are in G.P.
193
18. Find the sum to n terms of the sequence, 8, 88, 888, 8888 .
19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8,
1
.
2
Show that the products of the corresponding terms of the sequences a, ar, ar2,
arn 1 and A, AR, AR2, ARn 1 form a G.P, and find the common ratio.
Find four numbers forming a geometric progression in which the third term is
greater than the first term by 9, and the second term is greater than the 4th by 18.
If the pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that
aq r br pcP q = 1.
If the first and the nth term of a G.P. are a and b, respectively, and if P is the
product of n terms, prove that P2 = (ab)n.
Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from
22.
23.
24.
is
21.
bl
20.
he
1
.
rn
25. If a, b, c and d are in G.P. show that
(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2 .
26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
no N
C
tt E
o R
be T
re
pu
a n+1 + b n+1
may be the geometric mean between
a n + bn
a and b.
28. The sum of two numbers is 6 times their geometric mean, show that numbers
)(
29. If A and G be A.M. and G.M., respectively between two positive numbers,
prove that the numbers are A ( A + G )( A G ) .
30. The number of bacteria in a certain culture doubles every hour. If there were 30
bacteria present in the culture originally, how many bacteria will be present at the
end of 2nd hour, 4th hour and nth hour ?
31. What will Rs 500 amounts to in 10 years after its deposit in a bank which pays
annual interest rate of 10% compounded annually?
32. If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then
obtain the quadratic equation.
194
MATHEMATICS
n (n + 1)
2
he
We shall now find the sum of first n terms of some special series, namely;
(i) 1 + 2 + 3 + + n (sum of first n natural numbers)
(ii) 12 + 22 + 32 + + n2(sum of squares of the first n natural numbers)
(iii) 13 + 23 + 33 + + n3(sum of cubes of the first n natural numbers).
Let us take them one by one.
(See Section 9.4)
no N
C
tt E
o R
be T
re
pu
bl
is
k =1
k =1
n3 = 3 k 2 3 k + n
n
k =1 + 2 + 3 + ... + n =
k =1
Hence Sn =
k 2 = 3 n3 +
k =1
n (n + 1)
2
3n (n + 1)
n = 1 (2n3 + 3n 2 + n)
2
6
n (n +1)(2n +1)
6
(iii) Here Sn = 13 + 23 + ...+n3
We consider the identity, (k + 1)4 k4 = 4k3 + 6k2 + 4k + 1
Putting k = 1, 2, 3 n, we get
=
bl
is
he
195
no N
C
tt E
o R
be T
re
pu
= 4 k3 + 6 k2 + 4 k + n
k =1
k =1
k =1
... (1)
k =
k =1
n (n + 1)
2
and
k2 =
k =1
n (n + 1) (2n + 1)
6
4 k 3 = n 4 + 4 n 3 + 6n 2 + 4n
k =1
or
6n (n + 1) (2n + 1) 4n (n + 1)
n
6
2
Hence,
or
On subtraction, we get
Sn = 5 + 11 + 19 + 29 + ... + an1 + an
Sn =
5 + 11 + 19 + ... + an2 + an1 + an
196
MATHEMATICS
0 = 5 + [6 + 8 + 10 + 12 + ...(n 1) terms] an
or
an = 5 +
(n 1)[12 + (n 2) 2]
2
k =1
k =1
k =1
2
2
Sn = ak = ( k + 3k + 1) = k + 3 k + n
n(n + 2)(n + 4)
n(n +1) (2n +1) 3n(n + 1)
+
+n =
.
3
6
2
is
Hence
he
= 5 + (n 1) (n + 4) = n2 + 3n + 1
no N
C
tt E
o R
be T
re
pu
bl
Example 20 Find the sum to n terms of the series whose nth term is n (n+3).
Sn =
k =1
k =1
k =1
ak = k 2 + 3 k
EXERCISE 9.4
2. 1 2 3 + 2 3 4 + 3 4 5 + ...
3. 3 12 + 5 22 + 7 32 + ...
1
1
1
4. 1 2 + 2 3 + 3 4 + ...
5. 52 + 62 + 72 + ... + 202
6. 3 8 + 6 11 + 9 14 + ...
Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by
8. n (n+1) (n+4).
10. (2n 1) 2
9. n2 + 2n
197
Miscellaneous Examples
Example21 If p th, q th, r th and s th terms of an A.P. are in G.P, then show that
(p q), (q r), (r s) are also in G.P.
Solution Here
... (1)
... (2)
he
ap = a + (p 1) d
aq = a + (q 1) d
ar = a + (r 1) d
... (3)
as = a + (s 1) d
Given that ap, aq, ar and as are in G.P.,
ap
ar aq ar q r
=
=
aq a p aq p q (why ?)
ar as ar as r s
= =
=
aq ar aq ar q r
no N
C
tt E
o R
be T
re
pu
Similarly
is
... (5)
bl
aq
So
... (4)
(why ?)
... (6)
Let a x = b y = c z = k Then
a = kx , b = ky and c = kz.
Since a, b, c are in G.P., therefore,
b2 = ac
Using (1) in (2), we get
k2y = kx + z, which gives 2y = x + z.
Hence, x, y and z are in A.P.
Solution
... (1)
... (2)
... (1)
198
MATHEMATICS
But L.H.S.
= (a2p2 2abp + b2) + (b2p2 2bcp + c2) + (c2p2 2cdp + d2),
which gives (ap b)2 + (bp c)2 + (cp d)2 0
... (2)
ap b = 0, bp c = 0, cp d = 0
b c d
= = =p
a b c
is
or
he
Since the sum of squares of real numbers is non negative, therefore, from (1) and (2),
we have,
(ap b)2 + (bp c)2 + (cp d)2 = 0
bl
Example 24 If p,q,r are in G.P. and the equations, px2 + 2qx + r = 0 and
d e f
, , are in A.P.
p q r
no N
C
tt E
o R
be T
re
pu
2q 4q 2 4rp
2p
q
q
but
is also root of
p
p
q
q
d
+ 2e
+ f = 0,
p
p
or
2e d f
d 2e fp
= +
+ = 0, or
q p r
p q pr
Hence
d e f
, , are in A.P.
p q r
... (1)
199
f(1) = 3 and
bl
is
he
1. Show that the sum of (m + n)th and (m n)th terms of an A.P. is equal to twice
the mth term.
2. If the sum of three numbers in A.P., is 24 and their product is 440, find the
numbers.
3. Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that
S3 = 3(S2 S1)
4. Find the sum of all numbers between 200 and 400 which are divisible by 7.
5. Find the sum of integers from 1 to 100 that are divisible by 2 or 5.
6. Find the sum of all two digit numbers which when divided by 4, yields 1 as
remainder.
7. If f is a function satisfying f (x +y) = f(x) f(y) for all x, y N such that
x =1
no N
C
tt E
o R
be T
re
pu
8. The sum of some terms of G.P. is 315 whose first term and the common ratio are
5 and 2, respectively. Find the last term and the number of terms.
9. The first term of a G.P. is 1. The sum of the third term and fifth term is 90.
Find the common ratio of G.P.
10. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers
in that order, we obtain an arithmetic progression. Find the numbers.
11. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times
the sum of terms occupying odd places, then find its common ratio.
12. The sum of the first four terms of an A.P. is 56. The sum of the last four terms is
112. If its first term is 11, then find the number of terms.
13. If
a + bx b + cx c + dx
( x 0) , then show that a, b, c and d are in G.P.
=
=
a bx b cx c dx
14. Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P.
Prove that P2Rn = Sn.
15. The pth, qth and rth terms of an A.P. are a, b, c, respectively. Show that
(q r )a + (r p )b + (p q )c = 0
1 1 1 1 1 1
16. If a + , b + , c + are in A.P., prove that a, b, c are in A.P.
b c c a a b
17. If a, b, c, d are in G.P, prove that (an + bn), (bn + cn), (cn + dn) are in G.P.
18. If a and b are the roots of x2 3x + p = 0 and c, d are roots of x2 12x + q = 0,
where a, b, c, d form a G.P. Prove that (q + p) : (q p) = 17:15.
200
MATHEMATICS
19. The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n. Show
)(
that a : b = m + m 2 n 2 : m m 2 n 2
).
bl
is
he
1 1 1
20. If a, b, c are in A.P.; b, c, d are in G.P. and , , are in A.P. prove that a, c, e
c d e
are in G.P.
21. Find the sum of the following series up to n terms:
(i) 5 + 55 +555 +
(ii) .6 +. 66 +. 666+
22. Find the 20th term of the series 2 4 + 4 6 + 6 8 + ... + n terms.
23. Find the sum of the first n terms of the series: 3+ 7 +13 +21 +31 +
24. If S1, S2, S3 are the sum of first n natural numbers, their squares and their
25. Find the sum of the following series up to n terms:
no N
C
tt E
o R
be T
re
pu
13 13 + 23 13 + 23 + 33
+
+
+ ...
1
1+ 3
1+ 3 + 5
1 22 + 2 32 + ... + n (n + 1) 2 3n + 5
=
12 2 + 22 3 + ... + n 2 (n + 1) 3n + 1 .
A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to
pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid
amount. How much will the tractor cost him?
Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to
pay the balance in annual instalment of Rs 1000 plus 10% interest on the unpaid
amount. How much will the scooter cost him?
A person writes a letter to four of his friends. He asks each one of them to copy
the letter and mail to four different persons with instruction that they move the
chain similarly. Assuming that the chain is not broken and that it costs 50 paise to
mail one letter. Find the amount spent on the postage when 8th set of letter is
mailed.
A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually.
Find the amount in 15th year since he deposited the amount and also calculate the
total amount after 20 years.
A manufacturer reckons that the value of a machine, which costs him Rs. 15625,
will depreciate each year by 20%. Find the estimated value at the end of 5 years.
150 workers were engaged to finish a job in a certain number of days. 4 workers
dropped out on second day, 4 more workers dropped out on third day and so on.
28.
29.
30.
31.
32.
201
It took 8 more days to finish the work. Find the number of days in which the work
was completed.
Summary
he
is
Let a1, a2, a3, ... be the sequence, then the sum expressed as a1 + a2 + a3 + ...
no N
C
tt E
o R
be T
re
pu
bl
is called series. A series is called finite series if it has got finite number of
terms.
An arithmetic progression (A.P.) is a sequence in which terms increase or
decrease regularly by the same constant. This constant is called common
difference of the A.P. Usually, we denote the first term of A.P. by a, the
common difference by d and the last term by l. The general term or the nth
term of the A.P. is given by an = a + (n 1) d.
The sum Sn of the first n terms of an A.P. is given by
n
n
Sn = 2a + ( n 1) d = ( a + l ) .
2
2
a+b
i.e., the
2
sequence a, A, b is in A.P.
common ratio by r. The general or the nth term of G.P. is given by an= arn 1.
The sum Sn of the first n terms of G.P. is given by
202
MATHEMATICS
) or a (1 r ) , if r 1
a rn 1
Sn =
r 1
1r
he
The geometric mean (G.M.) of any two positive numbers a and b is given by
Historical Note
is
no N
C
tt E
o R
be T
re
pu
bl
Evidence is found that Babylonians, some 4000 years ago, knew of arithmetic and
geometric sequences. According to Boethius (510), arithmetic and geometric
sequences were known to early Greek writers. Among the Indian mathematician,
Aryabhatta (476) was the first to give the formula for the sum of squares and cubes
of natural numbers in his famous work Aryabhatiyam, written around
499. He also gave the formula for finding the sum to n terms of an arithmetic
sequence starting with p th term. Noted Indian mathematicians Brahmgupta
(598), Mahavira (850) and Bhaskara (1114-1185) also considered the sum of squares
and cubes. Another specific type of sequence having important applications in
mathematics, called Fibonacci sequence, was discovered by Italian mathematician
Leonardo Fibonacci (1170-1250). Seventeenth century witnessed the classification
of series into specific forms. In 1671 James Gregory used the term infinite series in
connection with infinite sequence. It was only through the rigorous development of
algebraic and set theoretic tools that the concepts related to sequence and series
could be formulated suitably.
Chapter
10
STRAIGHT LINES
G eometry, as a logical system, is a means and even the most powerful
means to make children feel the strength of the human spirit that is
of their own spirit. H. FREUDENTHAL
10.1 Introduction
We are familiar with two-dimensional coordinate geometry
from earlier classes. Mainly, it is a combination of algebra
and geometry. A systematic study of geometry by the use
of algebra was first carried out by celebrated French
philosopher and mathematician Ren Descartes, in his book
La Gomtry, published in 1637. This book introduced the
notion of the equation of a curve and related analytical
methods into the study of geometry. The resulting
combination of analysis and geometry is referred now as
analytical geometry. In the earlier classes, we initiated
Ren Descartes
the study of coordinate geometry, where we studied about
(1596 -1650)
coordinate axes, coordinate plane, plotting of points in a
plane, distance between two points, section formulae, etc. All these concepts are the
basics of coordinate geometry.
Let us have a brief recall of coordinate geometry done in earlier classes. To
recapitulate, the location of the points (6, 4) and
(3, 0) in the XY-plane is shown in Fig 10.1.
We may note that the point (6, 4) is at 6 units
distance from the y-axis measured along the positive
x-axis and at 4 units distance from the x-axis
measured along the negative y-axis. Similarly, the
point (3, 0) is at 3 units distance from the y-axis
measured along the positive x-axis and has zero
distance from the x-axis.
Fig 10.1
We also studied there following important
formulae:
204
MATHEMATICS
PQ
x1 y2 y1
2
3 6 2 0 4 2
9 16 5 units.
II. The coordinates of a point dividing the line segment joining the points (x1, y1)
m x 2 n x1 m y 2 n y1
.
,
m n
mn
For example, the coordinates of the point which divides the line segment joining
A (1, 3) and B (3, 9) internally, in the ratio 1: 3 are given by x
and y
1.( 3) 3.1
0
1 3
1.9 3. 3
0.
1 3
x1 x2 y1 y 2
,
.
2
2
IV. Area of the triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) is
1
x1 y 2 y 3 x 2 y 3 y 1 x 3 y1 y 2 .
2
For example, the area of the triangle, whose vertices are (4, 4), (3, 2) and ( 3, 16) is
54
1
4( 2 16) 3(16 4) (3)(4 2)
27.
2
2
Remark If the area of the triangle ABC is zero, then three points A, B and C lie on
a line, i.e., they are collinear.
In the this Chapter, we shall continue the study of coordinate geometry to study
properties of the simplest geometric figure straight line. Despite its simplicity, the
line is a vital concept of geometry and enters into our daily experiences in numerous
interesting and useful ways. Main focus is on representing the line algebraically, for
which slope is most essential.
STRAIGHT LINES
205
... (1)
MQ y2 y1
.
MP x2 x1
... (2)
206
MATHEMATICS
y2 y1
.
x2 x1
m = tan
= tan ( 180 MPQ) = tan MPQ
=
y2 y1
MQ
y y
.
2 1 =
x2 x1
MP
x1 x2
Consequently, we see that in both the cases the slope m of the line through the points
(x1, y1) and (x2, y2) is given by m
y2 y1
.
x2 x1
tan = tan .
Fig 10. 4
STRAIGHT LINES
207
Hence, two non vertical lines l1 and l2 are parallel if and only if their slopes
are equal.
If the lines l1 and l2 are perpendicular (Fig 10.5), then = + 90.
Therefore,tan = tan ( + 90)
= cot =
1
m2 = m
1
i.e.,
or
1
tan
m1 m2 = 1
i.e.,
1
or, m1 m2 = 1.
m1
(b)
(c)
(d)
4 ( 2) 6
3
.
1 3
4
2
(b) The slope of the line through the points (3, 2) and (7, 2) is
2 ( 2) 0
0.
73
4
(c) The slope of the line through the points (3, 2) and (3, 4) is
208
MATHEMATICS
4 ( 2) 6
, which is not defined.
33
0
(d) Here inclination of the line = 60. Therefore, slope of the line is
m
m = tan 60 =
3.
10.2.3 Angle between two lines When we think about more than one line in a plane,
then we find that these lines are either intersecting or parallel. Here we will discuss the
angle between two lines in terms of their slopes.
Let L1 and L2 be two non-vertical lines with slopes m1 and m2, respectively. If 1
and 2 are the inclinations of lines L1 and L2, respectively. Then
tan 2 tan 1
m m1
2
1 tan 1 tan 2 1 m1m2
(as 1 + m1m2 0)
m2 m1
, as 1 + m1m2 0
1 m1m2
Fig 10. 6
STRAIGHT LINES
Case I If
209
m 2 m1
is positive, then tan will be positive and tan will be negative,
1 m1 m 2
m 2 m1
is negative, then tan will be negative and tan will be positive,
1 m1 m 2
m 2 m1
, as 1 m1m 2 0
1 m1m 2
... (1)
1
and slope of one of the lines is , find
4
2
tan
m2 m1
1 m1m 2
... (1)
, m2 = m and = .
4
2
tan
4
1
1 m
2
or 1
1
2
1
1 m
2
1
1
m
2 1 or
2 1.
1
1
1
m
1
m
2
2
m
which gives
1
2
1
Therefore m 3 or m .
3
210
MATHEMATICS
1
3 or . Fig 10.7 explains the
3
reason of two answers.
Fig 10. 7
Example 3 Line through the points (2, 6) and (4, 8) is perpendicular to the line
through the points (8, 12) and (x, 24). Find the value of x.
Solution Slope of the line through the points ( 2, 6) and (4, 8) is
m1
86
2 1
4 2 6 3
Slope of the line through the points (8, 12) and (x, 24) is
24 12 12
x8
x 8
Since two lines are perpendicular,
m1 m2 = 1, which gives
m2
1 12
1 or x = 4 .
3 x 8
10.2.4 Collinearity of three points We
know that slopes of two parallel lines are
equal. If two lines having the same slope
pass through a common point, then two
lines will coincide. Hence, if A, B and C
are three points in the XY-plane, then they
will lie on a line, i.e., three points are
collinear (Fig 10.8) if and only if slope of
AB = slope of BC.
Fig 10. 8
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211
Example 4 Three points P (h, k), Q (x1, y1) and R (x2, y2) lie on a line. Show that
(h x1) (y2 y1) = (k y1) (x2 x1).
Solution Since points P, Q and R are collinear, we have
Slope of PQ = Slope of QR, i.e.,
y1 k
y y
2 1
x1 h
x2 x1
or
k y1 y 2 y1
h x1 x2 x1 ,
or
8 2 D 8
or
30 T 3
or
D = 2(T + 1),
which is the required relation.
Fig 10.9
6 (T 3) 3 (D 8)
EXERCISE 10.1
1. Draw a quadrilateral in the Cartesian plane, whose vertices are ( 4, 5), (0, 7),
(5, 5) and ( 4, 2). Also, find its area.
2. The base of an equilateral triangle with side 2a lies along the y-axis such that the
mid-point of the base is at the origin. Find vertices of the triangle.
3. Find the distance between P (x1, y1) and Q (x2, y2) when : (i) PQ is parallel to the
y-axis, (ii) PQ is parallel to the x-axis.
4. Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
5. Find the slope of a line, which passes through the origin, and the mid-point of the
line segment joining the points P (0, 4) and B (8, 0).
212
MATHEMATICS
6. Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and
(1, 1) are the vertices of a right angled triangle.
7. Find the slope of the line, which makes an angle of 30 with the positive direction
of y-axis measured anticlockwise.
8. Find the value of x for which the points (x, 1), (2,1) and (4, 5) are collinear.
9. Without using distance formula, show that points ( 2, 1), (4, 0), (3, 3) and (3, 2)
are the vertices of a parallelogram.
10. Find the angle between the x-axis and the line joining the points (3,1) and (4,2).
11. The slope of a line is double of the slope of another line. If tangent of the angle
between them is
1
, find the slopes of the lines.
3
12. A line passes through (x1, y1) and (h, k). If slope of the line is m, show that
k y1 = m (h x1).
13. If three points (h, 0), (a, b) and (0, k) lie on a line, show that
a b
1.
h k
14. Consider the following population and year graph (Fig 10.10), find the slope of the
line AB and using it, find what will be the population in the year 2010?
Fig 10.10
10.3
We know that every line in a plane contains infinitely many points on it. This relationship
between line and points leads us to find the solution of the following problem:
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213
How can we say that a given point lies on the given line? Its answer may be that
for a given line we should have a definite condition on the points lying on the line.
Suppose P (x, y) is an arbitrary point in the XY-plane and L is the given line. For the
equation of L, we wish to construct a statement or condition for the point P that is
true, when P is on L, otherwise false. Of course the statement is merely an algebraic
equation involving the variables x and y. Now, we will discuss the equation of a line
under different conditions.
10.3.1 Horizontal and vertical lines If a horizontal line L is at a distance a from the
x-axis then ordinate of every point lying on the line is either a or a [Fig 10.11 (a)].
Therefore, equation of the line L is either y = a or y = a. Choice of sign will depend
upon the position of the line according as the line is above or below the y-axis. Similarly,
the equation of a vertical line at a distance b from the y-axis is either x = b or
x = b [Fig 10.11(b)].
Fig 10.11
Fig 10.12
214
MATHEMATICS
y y0
, i.e., y y 0 m x x 0
x x0
...(1)
Fig 10.13
Thus, the point (x, y) lies on the line with slope m through the fixed point (x0, y0),
if and only if, its coordinates satisfy the equation
y y0 = m (x x0)
Example 7 Find the equation of the line through ( 2, 3) with slope 4.
Solution Here m = 4 and given point (x0 , y0) is ( 2, 3).
By slope-intercept form formula
(1) above, equation of the given
line is
y 3 = 4 (x + 2) or
4x + y + 5 = 0, which is the
required equation.
10.3.3 Two-point form Let the
line L passes through two given
points P1 (x1, y1) and P2 (x2, y2).
Let P (x, y) be a general point
on L (Fig 10.14).
The three points P1, P2 and P are
collinear, therefore, we have
slope of P1P = slope of P1P2
i.e.,
y y1 y 2 y1
,
x x1 x 2 x1
Fig 10.14
or y y 1
y 2 y1
( x x1 ).
x 2 x1
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215
Thus, equation of the line passing through the points (x1, y1) and (x2, y2) is given by
y y1
y 2 y1
( x x1)
x 2 x1
... (2)
Example 8 Write the equation of the line through the points (1, 1) and (3, 5).
Solution Here x1 = 1, y1 = 1, x2 = 3 and y2 = 5. Using two-point form (2) above
for the equation of the line, we have
y 1
or
5 1
x 1
3 1
10.3.4 Slope-intercept form Sometimes a line is known to us with its slope and an
intercept on one of the axes. We will now find equations of such lines.
Case I Suppose a line L with slope m cuts the y-axis at a distance c from the origin
(Fig10.15). The distance c is called the yintercept of the line L. Obviously,
coordinates of the point where the line meet
the y-axis are (0, c). Thus, L has slope m
and passes through a fixed point (0, c).
Therefore, by point-slope form, the equation
of L is
y c m ( x 0 ) or y mx c
Thus, the point (x, y) on the line with slope
Fig 10.15
m and y-intercept c lies on the line if and
only if
y mx c
...(3)
Note that the value of c will be positive or negative according as the intercept is made
on the positive or negative side of the y-axis, respectively.
Case II Suppose line L with slope m makes x-intercept d. Then equation of L is
y m( x d )
... (4)
Students may derive this equation themselves by the same method as in Case I.
Example 9 Write the equation of the lines for which tan =
inclination of the line and (i) y-intercept is
1
, where is the
2
3
(ii) x-intercept is 4.
2
216
MATHEMATICS
1
3
and y - intercept c = .
2
2
1
3
x or 2 y x 3 0 ,
2
2
y
which is the required equation.
1
and d = 4.
2
1
( x 4) or 2 y x 4 0 ,
2
y0
i.e.,
b0
(x a) or ay bx ab ,
0a
x y
1.
a b
x y
1
a b
Fig 10.16
... (5)
Example 10 Find the equation of the line, which makes intercepts 3 and 2 on the
x- and y-axes respectively.
Solution Here a = 3 and b = 2. By intercept form (5) above, equation of the line is
x y
1
3 2
or
2x 3 y 6 0 .
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217
10.3.6 Normal form Suppose a non-vertical line is known to us with following data:
(i) Length of the perpendicular (normal) from origin to the line.
(ii) Angle which normal makes with the positive direction of x-axis.
Let L be the line, whose perpendicular distance from origin O be OA = p and the
angle between the positive x-axis and OA be XOA = . The possible positions of line
L in the Cartesian plane are shown in the Fig 10.17. Now, our purpose is to find slope
of L and a point on it. Draw perpendicular AM on the x-axis in each case.
Fig 10.17
In each case, we have OM = p cos and MA = p sin , so that the coordinates of the
point A are (p cos , p sin ).
Further, line L is perpendicular to OA. Therefore
The slope of the line L =
Thus, the line L has slope
1
1
cos
.
slope of OA
tan
sin
cos
and point A p cos , p sin on it. Therefore, by
sin
218
MATHEMATICS
y p sin
cos
x p cos or
sin
or
x cos + y sin = p.
Hence, the equation of the line having normal distance p from the origin and angle
which the normal makes with the positive direction of x-axis is given by
x cos + y sin = p
... (6)
Example 11 Find the equation of the line whose perpendicular distance from the
origin is 4 units and the angle which the normal makes with positive direction of x-axis
is 15.
Solution Here, we are given p = 4 and
= 150 (Fig10.18).
Now
cos 15 =
and
sin 15 =
3 1
2 2
3 1
(Why?)
2 2
3 1
3 1
x
y 4 or
2 2
2 2
This is the required equation.
x cos 150 y sin 150 4 or
Fig 10.18
3 1 x
3 1 y 8 2.
K 273
373 273
100
F 32 or K 273 F 32
212 32
180
5
F 32 273
9
which is the required relation.
or
... (1)
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219
5
F 32 273
9
or F 32
273 9
491.4
5
or F= 459.4 .
... (3)
5
2297
and c =
.
9
9
5
2297
F
9
9
which is the required relation. When K = 0, (4) gives F = 459.4.
K
... (4)
m and c. For finding these two constants, we need two conditions satisfied by the
equation of line. In all the examples above, we are given two conditions to determine
the equation of the line.
EXERCISE 10.2
In Exercises 1 to 8, find the equation of the line which satisfy the given conditions:
1. Write the equations for the x-and y-axes.
2. Passing through the point ( 4, 3) with slope
1
.
2
220
MATHEMATICS
8. Perpendicular distance from the origin is 5 units and the angle made by the
perpendicular with the positive x-axis is 300.
9. The vertices of PQR are P (2, 1), Q (2, 3) and R (4, 5). Find equation of the
median through the vertex R.
10. Find the equation of the line passing through (3, 5) and perpendicular to the line
through the points (2, 5) and (3, 6).
11. A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides
it in the ratio 1: n. Find the equation of the line.
12. Find the equation of a line that cuts off equal intercepts on the coordinate axes
and passes through the point (2, 3).
13. Find equation of the line passing through the point (2, 2) and cutting off intercepts
on the axes whose sum is 9.
14.
15.
16.
17.
18.
2
with the
3
positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis
at a distance of 2 units below the origin.
The perpendicular from the origin to a line meets it at the point (2, 9), find the
equation of the line.
The length L (in centimetre) of a copper rod is a linear function of its Celsius
temperature C. In an experiment, if L = 124.942 when C = 20 and L= 125.134
when C = 110, express L in terms of C.
The owner of a milk store finds that, he can sell 980 litres of milk each week at
Rs 14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear
relationship between selling price and demand, how many litres could he sell
weekly at Rs 17/litre?
P (a, b) is the mid-point of a line segment between axes. Show that equation
Find equation of the line through the point (0, 2) making an angle
of the line is
x y
2.
a b
19. Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find
equation of the line.
20. By using the concept of equation of a line, prove that the three points (3, 0),
( 2, 2) and (8, 2) are collinear.
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221
Therefore, any equation of the form Ax + By + C = 0, where A and B are not zero
simultaneously is called general linear equation or general equation of a line.
10.4.1 Different forms of Ax + By + C = 0 The general equation of a line can be
reduced into various forms of the equation of a line, by the following procedures:
(a) Slope-intercept form If B 0, then Ax + By + C = 0 can be written as
A
C
x or y mx c
B
B
... (1)
A
C
and c .
B
B
We know that Equation (1) is the slope-intercept form of the equation of a line
where
whose slope is
A
C
, and y-intercept is .
B
B
If B = 0, then x =
x-intercept is
C
,which is a vertical line whose slope is undefined and
A
C
.
A
1 or
1
C
C
a b
A
B
where
a=
... (2)
C
C
and b = .
A
B
We know that equation (2) is intercept form of the equation of a line whose
x-intercept is
C
C
and y-intercept is .
A
B
A
B
C
p
cos sin
222
MATHEMATICS
cos
which gives
Ap
Bp
and sin
.
C
C
2
sin cos
2
Now
2
p
or
C
2
2
A B
cos
Therefore
Ap
C
or p
A
A B
2
Bp
C
C
2
A B
2
and sin
B
2
A B
2
A
A B
2
, sin
B
A B
2
and p
C
2
A B
2
3
5
x
4
2
... (1)
3
.
4
5
.
2
or
x
y
1
10 5
3 2
... (2)
x y
10
1 , we have x-intercept as a =
and
a b
3
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223
of p and .
Solution Given equation is
3x y 8 0
Dividing (1) by
... (1)
12 2 , we get
3
1
x y 4 or cos 30 x sin 30 y 4
... (2)
2
2
Comparing (2) with x cos + y sin = p, we get p = 4 and = 30.
3y x 6 0 .
y 3 x 5 0 or y 3 x 5
and
... (1)
3 y x 6 0 or y 3 x 2 3
... (2)
1
.
3
tan
m 2 m1
1 m1m 2
... (3)
1 3
1
2 3
3
which gives = 30. Hence, angle between two lines is either 30 or 180 30 = 150.
Example 16 Show that two lines a1 x b1 y c1 0 and a 2 x b 2 y c 2 0 ,
where b1, b2 0 are:
224
MATHEMATICS
(i) Parallel if
a1
b1
a2
b2
and
y a1 x c1
b1
b1
... (1)
y a2 x c2
b2
b2
... (2)
b1
a 2 , respectively. Now
b2
a1 a 2
a1
a
.
2 or
b1 b2
b1
b2
a1 a2
. 1 or a a + b b = 0
1 2
1 2
b1 b2
Example 17 Find the equation of a line perpendicular to the line x 2 y 3 0 and
passing through the point (1, 2).
Solution Given line x 2 y 3 0 can be written as
1
3
x
2
2
m2
...(1)
1
. Therefore, slope of the line perpendicular to line (1) is
2
1
2
m1
Equation of the line with slope 2 and passing through the point (1, 2) is
y ( 2) 2(x 1) or y = 2 x ,
which is the required equation.
STRAIGHT LINES
225
Fig10.19
the line meets the x-and y-axes at the points Q and R, respectively. Then, coordinates
of the points are
Q , 0 and R
A
is given by
1
area (PQR) PM.QR , which gives PM =
2
1
C C C
x1 0 y1 0 y 1 0
2
B A B
or
2 area (PQR)
QR
2 area (PQR)
2
1 C
C C2
x1 y 1
2 B
A AB
C
. A x1 B y 1 C , and
AB
C
QR 0
C
0
B
C
AB
2
2
A B
... (1)
226
MATHEMATICS
PM
or
A x1 B y 1 C
2
2
A B
A x1 B y1 C
.
2
2
A B
Thus, the perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x1, y1)
is given by
d
A x1 B y1 C
2
2
A B
c1
A m , 0 as shown in Fig10.20.
Fig10.20
Distance between two lines is equal to the length of the perpendicular from point
A to line (2). Therefore, distance between the lines (1) and (2) is
c1
c2
m
1 m2
or d =
c1 c2
1 m2
c1 c2
1 m2
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227
C1 C 2
A 2 B2
Ax1 By1 C
A B
2
3.3 4 5 26
32 4
3
.
5
75
32 4
2
.
5
EXERCISE 10.3
1. Reduce the following equations into slope - intercept form and find their slopes
and the y - intercepts.
(i) x + 7y = 0,
(ii) 6x + 3y 5 = 0,
(iii) y = 0.
2. Reduce the following equations into intercept form and find their intercepts on
the axes.
(i) 3x + 2y 12 = 0, (ii) 4x 3y = 6,
(iii) 3y + 2 = 0.
3. Reduce the following equations into normal form. Find their perpendicular distances
from the origin and angle between perpendicular and the positive x-axis.
(iii) x y = 4.
(i) x 3y + 8 = 0, (ii) y 2 = 0,
4. Find the distance of the point (1, 1) from the line 12(x + 6) = 5(y 2).
5. Find the points on the x-axis, whose distances from the line
x y
1 are 4 units.
3 4
228
MATHEMATICS
7. Find equation of the line parallel to the line 3 x 4 y 2 0 and passing through
the point (2, 3).
8. Find equation of the line perpendicular to the line x 7y + 5 = 0 and having
x intercept 3.
9. Find angles between the lines
3 x y 1and x 3 y 1.
10. The line through the points (h, 3) and (4, 1) intersects the line 7 x 9 y 19 0.
at right angle. Find the value of h.
11. Prove that the line through the point (x1, y1) and parallel to the line Ax + By + C = 0 is
A (x x1) + B (y y1) = 0.
12. Two lines passing through the point (2, 3) intersects each other at an angle of 60o.
If slope of one line is 2, find equation of the other line.
13. Find the equation of the right bisector of the line segment joining the points (3, 4)
and (1, 2).
14. Find the coordinates of the foot of perpendicular from the point (1, 3) to the
line 3x 4y 16 = 0.
15. The perpendicular from the origin to the line y = mx + c meets it at the point
(1, 2). Find the values of m and c.
16. If p and q are the lengths of perpendiculars from the origin to the
lines x cos y sin k cos 2 and x sec + y cosec = k, respectively, prove
that p2 + 4q2 = k2.
17. In the triangle ABC with vertices A (2, 3), B (4, 1) and C (1, 2), find the equation
and length of altitude from the vertex A.
18. If p is the length of perpendicular from the origin to the line whose intercepts on
the axes are a and b, then show that
1
1 1
2 2.
2
p
a b
Miscellaneous Examples
Example 20 If the lines 2 x y 3 0 , 5 x ky 3 0 and 3 x y 2 0 are
concurrent, find the value of k.
Solution Three lines are said to be concurrent, if they pass through a common point,
i.e., point of intersection of any two lines lies on the third line. Here given lines are
2x + y 3 = 0
... (1)
5x + ky 3 = 0
... (2)
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3x y 2 = 0
229
... (3)
x
y
1
or x 1, y 1 .
23 9 4 23
Therefore, the point of intersection of two lines is (1, 1). Since above three lines are
concurrent, the point (1, 1) will satisfy equation (2) so that
5.1 + k .1 3 = 0 or k = 2.
Example 21 Find the distance of the line 4x y = 0 from the point P (4, 1) measured
along the line making an angle of 135 with the positive x-axis.
Solution Given line is 4x y = 0
In order to find the distance of the
line (1) from the point P (4, 1) along
another line, we have to find the point
of intersection of both the lines. For
this purpose, we will first find the
equation of the second line
(Fig 10.21). Slope of second line is
tan 135 = 1. Equation of the line
with slope 1 through the point
P (4, 1) is
... (1)
(1, 4)
Fig 10.21
y 1 = 1 (x 4) or x + y 5 = 0
... (2)
Solving (1) and (2), we get x = 1 and y = 4 so that point of intersection of the two lines
is Q (1, 4). Now, distance of line (1) from the point P (4, 1) along the line (2)
= the distance between the points P (4, 1) and Q (1, 4).
=
1 4 2 4 12 3
2 units .
Example 22 Assuming that straight lines work as the plane mirror for a point, find
the image of the point (1, 2) in the line x 3 y 4 0 .
Solution Let Q (h, k) is the image of the point P (1, 2) in the line
x 3y + 4 = 0
... (1)
230
MATHEMATICS
Fig10.22
Therefore, the line (1) is the perpendicular bisector of line segment PQ (Fig 10.22).
Hence Slope of line PQ =
so that k 2 1
h 1 1
3
1
,
Slope of line x 3 y 4 0
... (2)
or 3h k 5
h 1 k 2
,
will satisfy the equation (1) so that
2
2
h 1 k 2
3
4 0 or h 3k 3
2
2
Solving (2) and (3), we get h =
6
7
and k = .
5
5
6 7
Hence, the image of the point (1, 2) in the line (1) is , .
5 5
Example 23 Show that the area of the triangle formed by the lines
y m1 x c1 , y m 2 x c 2
c c
and x = 0 is
1
2 m1 m 2
... (3)
STRAIGHT LINES
231
c2 c1
m1 m2
and y
m1c2 m2c1
m1 m2
Fig 10.23
c2 c1
m1c2 m2 c1
c 2 c1
m c m2 c1
c c
1
m c m2 c1
c2 2 1 c2 c1 0 c1 1 2
0 1 2
2
m1 m2 2 m1 m 2
m1 m2
m1 m2
232
MATHEMATICS
4 3 2
.
4
We are given that the mid point of the segment of the required line between (1, 1)
and (2, 2) is (1, 5). Therefore
or
1 = 51 + 4 and 2
1 2
1 and 1 2 5 ,
2
2
or
1 2 2 and
5 1 4
2
4 32
4
5,
or 1 2 2 and 20 1 3 2 20
... (3)
1 =
26
222
20
26
4
and 2 =
and hence, 1 5.
.
23
23
23
23
222
5
5
y 5 1
( x 1) or y 5 23
(x 1)
26
1 1
1
23
or
107x 3y 92 = 0,
which is the equation of required line.
Example 25 Show that the path of a moving point such that its distances from two
lines 3x 2y = 5 and 3x + 2y = 5 are equal is a straight line.
Solution Given lines are
3x 2y = 5
(1)
and
(2)
3x + 2y = 5
Let (h, k) is any point, whose distances from the lines (1) and (2) are equal. Therefore
3h 2k 5
94
3h 2k 5
94
or 3h 2k 5 3h 2k 5 ,
STRAIGHT LINES
233
5
. Thus, the point (h, k) satisfies the
3
5
, which represent straight lines. Hence, path of the point
3
x y
1 is
3 4
4 units.
5. Find perpendicular distance from the origin to the line joining the points (cos, sin )
and (cos , sin ).
6. Find the equation of the line parallel to y-axis and drawn through the point of
intersection of the lines x 7y + 5 = 0 and 3x + y = 0.
7. Find the equation of a line drawn perpendicular to the line
8.
9.
10.
11.
12.
x y
1 through the
4 6
234
MATHEMATICS
13. Show that the equation of the line passing through the origin and making an angle
with the line y mx c is
y
m tan
.
x
1 m tan
14. In what ratio, the line joining (1, 1) and (5, 7) is divided by the line x + y = 4?
15. Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line
2x y = 0.
16. Find the direction in which a straight line must be drawn through the point (1, 2)
so that its point of intersection with the line x + y = 4 may be at a distance of
3 units from this point.
17. The hypotenuse of a right angled triangle has its ends at the points (1, 3) and
( 4, 1). Find an equation of the legs (perpendicular sides) of the triangle.
18. Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the
line to be a plane mirror.
19. If the lines y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find
the value of m.
20. If sum of the perpendicular distances of a variable point P (x, y) from the lines
x + y 5 = 0 and 3x 2y +7 = 0 is always 10. Show that P must move on a line.
21. Find equation of the line which is equidistant from parallel lines 9x + 6y 7 = 0
and 3x + 2y + 6 = 0.
22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the
reflected ray passes through the point (5, 3). Find the coordinates of A.
23. Prove that the product of the lengths of the perpendiculars drawn from the
x
y
cos sin 1is b 2 .
a
b
24. A person standing at the junction (crossing) of two straight paths represented by
the equations 2x 3y + 4 = 0 and 3x + 4y 5 = 0 wants to reach the path whose
equation is 6x 7y + 8 = 0 in the least time. Find equation of the path that he
should follow.
points
Summary
Slope (m) of a non-vertical line passing through the points (x1, y1) and (x2, y2)
y 2 y1 y1 y 2
, x 1 x 2.
x 2 x1 x1 x 2
If a line makes an angle with the positive direction of x-axis, then the slope
of the line is given by m = tan , 90.
Slope of horizontal line is zero and slope of vertical line is undefined.
is given by m
STRAIGHT LINES
Two lines are parallel if and only if their slopes are equal.
Two lines are perpendicular if and only if product of their slopes is 1.
Three points A, B and C are collinear, if and only if slope of AB = slope of BC.
Equation of the horizontal line having distance a from the x-axis is either
y = a or y = a.
Equation of the vertical line having distance b from the y-axis is either
x = b or x = b.
The point (x, y) lies on the line with slope m and through the fixed point (xo, yo),
if and only if its coordinates satisfy the equation y y o = m (x xo).
Equation of the line passing through the points (x1, y1) and (x2, y2) is given by
y y1
y 2 y1
( x x1).
x 2 x1
The point (x, y) on the line with slope m and y-intercept c lies on the line if and
only if y mx c .
If a line with slope m makes x-intercept d. Then equation of the line is
y = m (x d).
Equation of a line making intercepts a and b on the x-and y-axis,
respectively, is
x y
1.
a b
The equation of the line having normal distance from origin p and angle between
Ax1 B y1 C
.
2
2
A B
Distance between the parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0,
is given by d
C1 C 2
2
2
A B
235
Chapter
11
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CONIC SECTIONS
Let the relation of knowledge to real life be very visible to your pupils
11.1 Introduction
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Fig 11. 1
237
Fig 11. 2
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CONIC SECTIONS
Fig 11. 3
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MATHEMATICS
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Fig 11. 5
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Fig 11. 4
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Fig 11. 6
Fig 11. 7
CONIC SECTIONS
239
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In the following sections, we shall obtain the equations of each of these conic
sections in standard form by defining them based on geometric properties.
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Fig 11. 8
11.3 Circle
Fig 11. 9
Fig 11. 10
Definition 1 A circle is the set of all points in a plane that are equidistant from a fixed
point in the plane.
The fixed point is called the centre of the circle and the distance from the centre
to a point on the circle is called the radius of the circle (Fig 11.11).
MATHEMATICS
Fig 11. 12
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Fig 11. 11
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240
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The equation of the circle is simplest if the centre of the circle is at the origin.
However, we derive below the equation of the circle with a given centre and radius
(Fig 11.12).
Given C (h, k) be the centre and r the radius of circle. Let P(x, y) be any point on
the circle (Fig11.12). Then, by the definition, | CP | = r . By the distance formula,
we have
(x h) 2 + (y k ) 2 = r
i.e.
(x h)2 + (y k)2 = r2
This is the required equation of the circle with centre at (h,k) and radius r .
Example 1 Find an equation of the circle with centre at (0,0) and radius r.
Solution Here h = k = 0. Therefore, the equation of the circle is x2 + y2 = r2.
Example 2 Find the equation of the circle with centre (3, 2) and radius 4.
Solution Here h = 3, k = 2 and r = 4. Therefore, the equation of the required circle is
(x + 3)2 + (y 2)2 = 16
Example 3 Find the centre and the radius of the circle x2 + y2 + 8x + 10y 8 = 0
CONIC SECTIONS
241
Example 4 Find the equation of the circle which passes through the points (2, 2), and
(3,4) and whose centre lies on the line x + y = 2.
EXERCISE 11.1
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In each of the following Exercises 1 to 5, find the equation of the circle with
1. centre (0,2) and radius 2
1
1 1
) and radius
12
2 4
3. centre ( ,
a2 b2 .
In each of the following Exercises 6 to 9, find the centre and radius of the circles.
6. (x + 5)2 + (y 3)2 = 36
7. x2 + y2 4x 8y 45 = 0
8. x2 + y2 8x + 10y 12 = 0
9. 2x2 + 2y2 x = 0
10. Find the equation of the circle passing through the points (4,1) and (6,5) and
whose centre is on the line 4x + y = 16.
11. Find the equation of the circle passing through the points (2,3) and (1,1) and
whose centre is on the line x 3y 11 = 0.
12. Find the equation of the circle with radius 5 whose centre lies on x-axis and
passes through the point (2,3).
13. Find the equation of the circle passing through (0,0) and making intercepts a and
b on the coordinate axes.
14. Find the equation of a circle with centre (2,2) and passes through the point (4,5).
15. Does the point (2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?
242
MATHEMATICS
11.4 Parabola
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Fig 11. 13
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CONIC SECTIONS
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We will derive the equation for the parabola shown above in Fig 11.15 (a) with
focus at (a, 0) a > 0; and directricx x = a as below:
Let F be the focus and l the directrix. Let
FM be perpendicular to the directrix and bisect
FM at the point O. Produce MO to X. By the
definition of parabola, the mid-point O is on the
parabola and is called the vertex of the parabola.
Take O as origin, OX the x-axis and OY
perpendicular to it as the y-axis. Let the distance
from the directrix to the focus be 2a. Then, the
coordinates of the focus are (a, 0), and the
equation of the directrix is x + a = 0 as in Fig11.16.
Fig 11.16
Let P(x, y) be any point on the parabola such that
PF = PB,
... (1)
where PB is perpendicular to l. The coordinates of B are ( a, y). By the distance
formula, we have
PF =
(x a) 2 + y 2 and PB =
Since PF = PB, we have
(x + a) 2
(x a) 2 + y 2 = (x + a ) 2
i.e. (x a)2 + y2 = (x + a)2
or x2 2ax + a2 + y2 = x2 + 2ax + a2
or y2 = 4ax ( a > 0).
244
MATHEMATICS
... (2)
(x a) 2 + y 2
(x + a) 2 = PB
(x a ) 2 + 4ax
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PF
... (3)
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Thus, from (2) and (3) we have proved that the equation to the parabola with
vertex at the origin, focus at (a,0) and directrix x = a is y2 = 4ax.
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Discussion In equation (2), since a > 0, x can assume any positive value or zero but
no negative value and the curve extends indefinitely far into the first and the fourth
quadrants. The axis of the parabola is the positive x-axis.
Similarly, we can derive the equations of the parabolas in:
Fig 11.15 (b) as y2 = 4ax,
Fig 11.15 (c) as x2 = 4ay,
Fig 11.15 (d) as x2 = 4ay,
From the standard equations of the parabolas, Fig11.15, we have the following
observations:
1.
2.
3.
Parabola is symmetric with respect to the axis of the parabola.If the equation
has a y2 term, then the axis of symmetry is along the x-axis and if the
equation has an x2 term, then the axis of symmetry is along the y-axis.
When the axis of symmetry is along the x-axis the parabola opens to the
(a) right if the coefficient of x is positive,
(b) left if the coefficient of x is negative.
When the axis of symmetry is along the y-axis the parabola opens
(c) upwards if the coefficient of y is positive.
(d) downwards if the coefficient of y is negative.
CONIC SECTIONS
245
AC = FM = 2a
Hence
AF = 2a.
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Fig 11.17
Fig 11.18
Fig 11.19
Thus, the focus of the parabola is (2, 0) and the equation of the directrix of the parabola
is x = 2 (Fig 11.19).
Length of the latus rectum is 4a = 4 2 = 8.
246
MATHEMATICS
Example 6 Find the equation of the parabola with focus (2,0) and directrix x = 2.
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Solution Since the focus (2,0) lies on the x-axis, the x-axis itself is the axis of the
parabola. Hence the equation of the parabola is of the form either
y2 = 4ax or y2 = 4ax. Since the directrix is x = 2 and the focus is (2,0), the parabola
is to be of the form y 2 = 4ax with a = 2. Hence the required equation is
y2 = 4(2)x = 8x
Example 7 Find the equation of the parabola with vertex at (0, 0) and focus at (0, 2).
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Solution Since the vertex is at (0,0) and the focus is at (0,2) which lies on y-axis, the
y-axis is the axis of the parabola. Therefore, equation of the parabola is of the form
x2 = 4ay. thus, we have
x2 = 4(2)y, i.e., x2 = 8y.
Example 8 Find the equation of the parabola which is symmetric about the y-axis, and
passes through the point (2,3).
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Solution Since the parabola is symmetric about y-axis and has its vertex at the origin,
the equation is of the form x2 = 4ay or x2 = 4ay, where the sign depends on whether
the parabola opens upwards or downwards. But the parabola passes through (2,3)
which lies in the fourth quadrant, it must open downwards. Thus the equation is of
the form x2 = 4ay.
Since the parabola passes through ( 2,3), we have
22 = 4a (3), i.e., a =
1
3
1
x2 = 4 y, i.e., 3x2 = 4y.
3
EXERCISE 11.2
In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the
parabola, the equation of the directrix and the length of the latus rectum.
1. y2 = 12x
2. x2 = 6y
3.
y2 = 8x
4. x2 = 16y
5. y2 = 10x
6.
x2 = 9y
In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the
given conditions:
CONIC SECTIONS
247
11. Vertex (0,0) passing through (2,3) and axis is along x-axis.
12. Vertex (0,0), passing through (5,2) and symmetric with respect to y-axis.
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11. 5 Ellipse
Fig 11.20
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The mid point of the line segment joining the foci is called the centre of the
ellipse. The line segment through the foci of the ellipse is called the major axis and the
line segment through the centre and perpendicular to the major axis is called the minor
axis. The end points of the major axis are called the vertices of the ellipse(Fig 11.21).
Fig 11.21
Fig 11.22
We denote the length of the major axis by 2a, the length of the minor axis by 2b
and the distance between the foci by 2c. Thus, the length of the semi major axis is a
and semi-minor axis is b (Fig11.22).
248
MATHEMATICS
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Fig 11.23
b2 + c2
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b2 + c 2 = 2 b2 + c2
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2 b 2 + c 2 = 2a, i.e.,
or
a 2 = b2 + c2 , i.e.,
a = b2 + c2
c=
a 2 b2 .
11.5.3 Eccentricity
Fig 11.24
Fig 11.25
Definition 5 The eccentricity of an ellipse is the ratio of the distances from the centre
of the ellipse to one of the foci and to one of the vertices of the ellipse (eccentricity is
denoted by e) i.e., e =
c
.
a
CONIC SECTIONS
249
Then since the focus is at a distance of c from the centre, in terms of the eccentricity
the focus is at a distance of ae from the centre.
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(a)
Fig 11.26
on the x-axis or y-axis. The two such possible orientations are shown in Fig 11.26.
We will derive the equation for the ellipse shown above in Fig 11.26 (a) with foci
on the x-axis.
Let F1 and F2 be the foci and O be the midpoint of the line segment F1F2. Let O be the origin
and the line from O through F2 be the positive
x-axis and that through F1as the negative x-axis.
Let, the line through O perpendicular to the
x-axis be the y-axis. Let the coordinates of F1 be
( c, 0) and F2 be (c, 0) (Fig 11.27).
Let P(x, y) be any point on the ellipse such
that the sum of the distances from P to the two
x2 y2
+
=1
foci be 2a so given
a 2 b2
PF1 + PF2 = 2a.
... (1)
Fig 11.27
Using the distance formula, we have
( x + c) 2 + y 2 +
i.e.,
( x c) 2 + y 2 = 2a
( x + c) 2 + y 2 = 2a
( x c) 2 + y 2
250
MATHEMATICS
( x c) 2 + y 2 + ( x c) 2 + y 2
(x + c)2 + y2 = 4a2 4a
which on simplification gives
c
x
a
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( x c) 2 + y 2 = a
Squaring again and simplifying, we get
x2
y2
+
=1
a2
b2
(Since c2 = a2 b2)
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x2
y2
+
= 1.
a2
b2
bl
i.e.,
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x2
y2
+ 2
=1
a2
a c2
... (2)
Conversely, let P (x, y) satisfy the equation (2) with 0 < c < a. Then
x2
y =b
a 2
Therefore, PF1
( x + c) 2 + y 2
a2 x2
( x + c) 2 + b 2
2
a
a2 x2
( x + c) 2 + (a 2 c 2 )
(since b2 = a2 c2)
2
a
cx
c
a+ =a+ x
a
a
Similarly
PF2 = a x
a
CONIC SECTIONS
Hence
PF1 + PF2 = a +
c
c
x + a x = 2a
a
a
251
... (3)
x2
y2
+
= 1, satisfies the geometric condition and so
a2
b2
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x2 y 2
+
= 1.
a 2 b2
Discussion From the equation of the ellipse obtained above, it follows that for every
point P (x, y) on the ellipse, we have
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x2
y2
=
1
1, i.e., x2 a2, so a x a.
a2
b2
Therefore, the ellipse lies between the lines x = a and x = a and touches these lines.
Similarly, the ellipse lies between the lines y = b and y = b and touches these
lines.
Similarly, we can derive the equation of the ellipse in Fig 11.26 (b) as
x2 y 2
+
=1 .
b2 a 2
From the standard equations of the ellipses (Fig11.26), we have the following
observations:
1. Ellipse is symmetric with respect to both the coordinate axes since if (x, y) is a
point on the ellipse, then ( x, y), (x, y) and ( x, y) are also points on the ellipse.
2. The foci always lie on the major axis. The major axis can be determined by
finding the intercepts on the axes of symmetry. That is, major axis is along the x-axis
if the coefficient of x2 has the larger denominator and it is along the y-axis if the
coefficient of y2 has the larger denominator.
252
MATHEMATICS
Fig 11. 28
x2 y2
+
= 1 , we have
a 2 b2
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x2 y2
+
=1
a 2 b2
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of the ellipse
he
(ae)2 l 2
+ 2 =1
a2
b
l2 = b2 (1 e2)
But
e2 =
c2 a 2 b2
b2
=
=
1
a2
a2
a2
Therefore
l2 =
b4
b2
l
=
,
i.e.,
a2
a
Since the ellipse is symmetric with respect to y-axis (of course, it is symmetric w.r.t.
both the coordinate axes), AF2 = F2B and so length of the latus rectum is
2b 2
.
a
Example 9 Find the coordinates of the foci, the vertices, the length of major axis, the
minor axis, the eccentricity and the latus rectum of the ellipse
x2 y 2
+
=1
25 9
x2
y2
Solution Since denominator of
is larger than the denominator of
, the major
25
9
CONIC SECTIONS
253
x2 y2
+
= 1 , we get
a 2 b2
a = 5 and b = 3. Also
c = a 2 b 2 = 25 9 = 4
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Therefore, the coordinates of the foci are ( 4,0) and (4,0), vertices are ( 5, 0) and
(5, 0). Length of the major axis is 10 units length of the minor axis 2b is 6 units and the
2
2b
18
4
=
and latus rectum is
.
5
a
5
Example 10 Find the coordinates of the foci, the vertices, the lengths of major and
minor axes and the eccentricity of the ellipse 9x2 + 4y2 = 36.
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eccentricity is
Solution The given equation of the ellipse can be written in standard form as
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x2 y 2
+
=1
4
9
x2
y2
is larger than the denominator of
, the major axis is
9
4
along the y-axis. Comparing the given equation with the standard equation
x2 y 2
+
= 1 , we have b = 2 and a = 3.
b2 a2
Also
c=
and
e=
a 2 b2
94= 5
c
5
=
3
a
Hence the foci are (0, 5 ) and (0, 5 ), vertices are (0,3) and (0, 3), length of the
major axis is 6 units, the length of the minor axis is 4 units and the eccentricity of the
5
.
3
Example 11 Find the equation of the ellipse whose vertices are ( 13, 0) and foci are
( 5, 0).
Solution Since the vertices are on x-axis, the equation will be of the form
ellipse is
x2 y2
+
= 1 , where a is the semi-major axis.
a2 b2
254
MATHEMATICS
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Example 12 Find the equation of the ellipse, whose length of the major axis is 20 and
foci are (0, 5).
Solution Since the foci are on y-axis, the major axis is along the y-axis. So, equation
is
x2 y 2
+
=1.
b2 a 2
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20
= 10
2
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a = semi-major axis =
c 2 = a2 b2 gives
52 = 102 b2 i.e., b2 = 75
Therefore, the equation of the ellipse is
x2
y2
+
=1
75 100
Example 13 Find the equation of the ellipse, with major axis along the x-axis and
passing through the points (4, 3) and ( 1,4).
Solution The standard form of the ellipse is
and
x2
y2
= 1. Since the points (4, 3)
+
a2
b2
16
9
+ 2 =1
2
a
b
... (1)
1
16
+ 2 =1
2
a
b
.(2)
2
Solving equations (1) and (2), we find that a =
247
247
2
and b =
.
7
15
CONIC SECTIONS
255
x2
y2
+
= 1 , i.e., 7x2 + 15y2 = 247.
247 247
7 15
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EXERCISE 11.3
x2
y2
+
=1
25 100
3.
5.
x2 y2
+
=1
49 36
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4.
2.
x2 y 2
+
=1
4 25
6.
x2 y 2
+
=1
16 9
x2
y2
+
=1
100 400
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1.
x2 y 2
+
=1
36 16
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In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length
of major axis, the minor axis, the eccentricity and the length of the latus rectum of the
ellipse.
8. 16x2 + y2 = 16
9. 4x2 + 9y2 = 36
In each of the following Exercises 10 to 20, find the equation for the ellipse that satisfies
the given conditions:
10.
11.
12.
13.
11.6 Hyperbola
Definition 7 A hyperbola is the set of all points in a plane, the difference of whose
distances from two fixed points in the plane is a constant.
MATHEMATICS
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Fig 11.29
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The term difference that is used in the definition means the distance to the
farther point minus the distance to the closer point. The two fixed points are called the
foci of the hyperbola. The mid-point of the line segment joining the foci is called the
centre of the hyperbola. The line through the foci is called the transverse axis and
the line through the centre and perpendicular to the transverse axis is called the conjugate
axis. The points at which the hyperbola
intersects the transverse axis are called the
vertices of the hyperbola (Fig 11.29).
We denote the distance between the
two foci by 2c, the distance between two
vertices (the length of the transverse axis)
by 2a and we define the quantity b as
b =
c2 a2
Fig 11.30
CONIC SECTIONS
257
11.6.1 Eccentricity
c
is called the eccentricity of the
a
hyperbola. Since c a, the eccentricity is never less than one. In terms of the
eccentricity, the foci are at a distance of ae from the centre.
Definition 8 Just like an ellipse, the ratio e =
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(b)
(a)
Fig 11.31
We will derive the equation for the hyperbola shown in Fig 11.31(a) with foci on
the x-axis.
Let F1 and F2 be the foci and O be the mid-point of the line segment F1F2. Let O
be the origin and the line through O
through F2 be the positive x-axis and
that through F 1 as the negative
x-axis. The line through O
perpendicular to the x-axis be the
y-axis. Let the coordinates of F1 be
( c,0) and F2 be (c,0) (Fig 11.32).
Let P(x, y) be any point on the
hyperbola such that the difference
of the distances from P to the farther
point minus the closer point be 2a.
So given, PF1 PF2 = 2a
Fig 11.32
258
MATHEMATICS
i.e.,
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(x + c)2 + y2 = 4a2 + 4a
and on simplifying, we get
cx
a=
a
is
(x c) 2 + y 2
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x2 y 2
=1
a2 b2
bl
x2
y2
=1
a2 c2 a2
i.e.,
(Since c2 a2 = b2)
x2 y 2
= 1 1.
a 2 b2
Conversely, let P(x, y) satisfy the above equation with 0 < a < c. Then
Therefore,
= b
PF1 = +
= +
Similarly,
PF2 = a
x2 a2
2
a
(x + c) 2 + y 2
x2 a2
c
(x + c) 2 + b 2
x
= a+
2
a
a
a
x
c
c
c
x becomes negative. Thus, PF2 =
x a.
a
a
c
x > a. Therefore,
a
CONIC SECTIONS
259
c
cx
x
+ a = 2a
a
a
Also, note that if P is to the left of the line x = a, then
Therefore
PF1 PF2 = a +
he
c
PF1 = a + x , PF2 = a x .
a
a
x2 y 2
= 1 , lies on the
a 2 b2
x2 y 2
=1.
a 2 b2
bl
along x-axis is
is
hyperbola.
Thus, we proved that the equation of hyperbola with origin (0,0) and transverse axis
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Discussion From the equation of the hyperbola we have obtained, it follows that, we
x2
y2
=
+
1
have for every point (x, y) on the hyperbola, 2
1.
a
b2
i.e,
x
1, i.e., x a or x a. Therefore, no portion of the curve lies between the
a
y2 x2
=1
a2 b2
Hyperbola is symmetric with respect to both the axes, since if (x, y) is a point on
the hyperbola, then ( x, y), (x, y) and ( x, y) are also points on the hyperbola.
260
MATHEMATICS
2.
The foci are always on the transverse axis. It is the positive term whose
x2 y 2
=1
9 16
y 2 x2
=1
25 16
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As in ellipse, it is easy to show that the length of the latus rectum in hyperbola is
(i)
x2 y 2
= 1 , (ii) y2 16x2 = 16
9 16
x2 y 2
x2 y 2
=1
a2 b2
Here, a = 3, b = 4 and c =
a 2 + b 2 = 9 + 16 = 5
Therefore, the coordinates of the foci are ( 5, 0) and that of vertices are ( 3, 0).Also,
c 5
2b 2 32
=
=
=
The eccentricity e =
. The latus rectum
a 3
a
3
y 2 x2
=1
16 1
y 2 x2
= 1 , we find that
a 2 b2
CONIC SECTIONS
261
(0,
he
2b2 1
17
c
= .
. The latus rectum =
The eccentricity e = =
a
4
a
2
Example 15 Find the equation of the hyperbola with foci (0, 3) and vertices
11
).
2
y 2 x2
=1
a 2 b2
11
), a =
2
11
2
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Solution Since the foci is on y-axis, the equation of the hyperbola is of the form
25
.
4
x2
y2
Example 16 Find the equation of the hyperbola where foci are (0, 12) and the length
of the latus rectum is 36.
Solution Since foci are (0, 12), it follows that c = 12.
2b 2
= 36 or b2 = 18a
Length of the latus rectum =
a
c2 = a2 + b2; gives
144 = a2 + 18a
a2 + 18a 144 = 0,
i.e.,
a = 24, 6.
So
Since a cannot be negative, we take a = 6 and so b2 = 108.
Therefore
x2
y2
= 1 , i.e., 3y2 x2 = 108
36 108
262
MATHEMATICS
EXERCISE 11.4
In each of the Exercises 1 to 6, find the coordinates of the foci and the vertices, the
eccentricity and the length of the latus rectum of the hyperbolas.
x2 y 2
y2 x2
=1
=1
2.
3. 9y2 4x2 = 36
16 9
9 27
4. 16x2 9y2 = 576
5. 5y2 9x2 = 36
6. 49y2 16x2 = 784.
In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given
conditions.
7. Vertices ( 2, 0), foci ( 3, 0)
8. Vertices (0, 5), foci (0, 8)
9. Vertices (0, 3), foci (0, 5)
10. Foci ( 5, 0), the transverse axis is of length 8.
11. Foci (0, 13), the conjugate axis is of length 24.
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1.
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4
.
3
Miscellaneous Examples
Hence
AB = 2y = 2 30 = 60 cm.
Fig 11.33
Example 18 A beam is supported at its ends by
supports which are 12 metres apart. Since the load is concentrated at its centre, there
CONIC SECTIONS
263
Fig 11.34
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Solution Let the vertex be at the lowest point and the axis vertical. Let the coordinate
axis be chosen as shown in Fig 11.34.
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The equation of the parabola takes the form x2 = 4ay. Since it passes through
3
3
36 100
= 300 m
6,
, we have (6)2 = 4a
, i.e., a =
12
100
100
i.e.
x 2 = 4 300
x=
24
1
2
m. Coordinates of B are (x,
).
100
100
2
= 24
100
= 2 6 metres
Fig 11.35
MATHEMATICS
From
PBQ, cos =
x
9
From
PRA, sin =
y
6
Since
cos2 + sin2 = 1
2
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264
x2 y2
+ =1
81 36
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or
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x y
+ =1
9 6
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1.
2.
3.
The cable of a uniformly loaded suspension bridge hangs in the form of a parabola.
The roadway which is horizontal and 100 m long is supported by vertical wires
attached to the cable, the longest wire being 30 m and the shortest being 6 m.
Find the length of a supporting wire attached to the roadway 18 m from the
middle.
4.
5.
A rod of length 12 cm moves with its ends always touching the coordinate axes.
Determine the equation of the locus of a point P on the rod, which is 3 cm from
the end in contact with the x-axis.
Find the area of the triangle formed by the lines joining the vertex of the parabola
x2 = 12y to the ends of its latus rectum.
6.
7.
A man running a racecourse notes that the sum of the distances from the two flag
posts from him is always 10 m and the distance between the flag posts is 8 m.
Find the equation of the posts traced by the man.
8.
CONIC SECTIONS
265
Summary
In this Chapter the following concepts and generalisations are studied.
in the plane.
he
A circle is the set of all points in a plane that are equidistant from a fixed point
is
A parabola is the set of all points in a plane that are equidistant from a fixed
line and a fixed point in the plane.
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The equation of the parabola with focus at (a, 0) a > 0 and directrix x = a is
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y2 = 4ax.
Latus rectum of a parabola is a line segment perpendicular to the axis of the
parabola, through the focus and whose end points lie on the parabola.
Length of the latus rectum of the parabola y2 = 4ax is 4a.
An ellipse is the set of all points in a plane, the sum of whose distances from
two fixed points in the plane is a constant.
x2 y 2
+ =1.
a 2 b2
x2 y2
2b 2
+
=
1
is
.
a 2 b2
a
The eccentricity of an ellipse is the ratio between the distances from the centre
of the ellipse to one of the foci and to one of the vertices of the ellipse.
A hyperbola is the set of all points in a plane, the difference of whose distances
from two fixed points in the plane is a constant.
x2 y 2
The equation of a hyperbola with foci on the x-axis is : a 2 b 2 = 1
266
MATHEMATICS
x2 y 2
2b 2
=
1
is
:
.
a 2 b2
a
he
The eccentricity of a hyperbola is the ratio of the distances from the centre of
bl
Historical Note
is
the hyperbola to one of the foci and to one of the vertices of the hyperbola.
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CONIC SECTIONS
267
is
(x )
( a b )
bl
y =
he
made formal use of the two axes and gave the equation of a circle as
( y a)2 + (b x)2 = r
He gave the best exposition of the analytical geometry of his time. Monge
(1781) gave the modern point-slope form of equation of a line as
y y = a (x x)
and the condition of perpendicularity of two lines as aa + 1 = 0.
S.F. Lacroix (17651843) was a prolific textbook writer, but his contributions
to analytical geometry are found scattered. He gave the two-point form of
equation of a line as
1 + a2
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a a
His formula for finding angle between two lines was tan = 1 + aa . It is, of
course, surprising that one has to wait for more than 150 years after the invention
of analytical geometry before finding such essential basic formula. In 1818, C.
Lame, a civil engineer, gave mE + mE = 0 as the curve passing through the
points of intersection of two loci E = 0 and E = 0.
Many important discoveries, both in Mathematics and Science, have been
linked to the conic sections. The Greeks particularly Archimedes (287212 B.C.)
and Apollonius (200 B.C.) studied conic sections for their own beauty. These
curves are important tools for present day exploration of outer space and also for
research into behaviour of atomic particles.
Chapter
12
is
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INTRODUCTION TO THREE
DIMENSIONAL GEOMETRY
12.1 Introduction
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269
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270
MATHEMATICS
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accordingly. Thus, to each point P in the space there corresponds an ordered triplet
(x, y, z) of real numbers.
Conversely, given any triplet (x, y, z), we would first fix the point L on the x-axis
corresponding to x, then locate the point M in the XY-plane such that (x, y) are the
coordinates of the point M in the XY-plane. Note that LM is perpendicular to the
x-axis or is parallel to the y-axis. Having reached the point M, we draw a perpendicular
MP to the XY-plane and locate on it the point P corresponding to z. The point P so
obtained has then the coordinates (x, y, z). Thus, there is a one to one correspondence
between the points in space and ordered triplet (x, y, z) of real numbers.
Alternatively, through the point P in the
space, we draw three planes parallel to the
coordinate planes, meeting the x-axis, y-axis
and z-axis in the points A, B and C, respectively
(Fig 12.3). Let OA = x, OB = y and OC = z.
Then, the point P will have the coordinates x, y
and z and we write P (x, y, z). Conversely, given
x, y and z, we locate the three points A, B and
C on the three coordinate axes. Through the
points A, B and C we draw planes parallel to
Fig 12.3
the YZ-plane, ZX-plane and XY-plane,
respectively. The point of interesection of these three planes, namely, ADPF, BDPE
and CEPF is obviously the point P, corresponding to the ordered triplet (x, y, z). We
observe that if P (x, y, z) is any point in the space, then x, y and z are perpendicular
distances from YZ, ZX and XY planes, respectively.
Note The coordinates of the origin O are (0,0,0). The coordinates of any point
on the x-axis will be as (x,0,0) and the coordinates of any point in the YZ-plane will
be as (0, y, z).
Remark The sign of the coordinates of a point determine the octant in which the
point lies. The following table shows the signs of the coordinates in eight octants.
Table 12.1
s
ant
Oct n a t e s
rdi
Coo
II
III
IV
VI
VII
VIII
271
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Solution From the Table 12.1, the point (3,1, 2) lies in second octant and the point
(3, 1, 2) lies in octant VI.
EXERCISE 12.1
2.
A point is in the XZ-plane. What can you say about its y-coordinate?
3.
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(1, 2, 3), (4, 2, 3), (4, 2, 5), (4, 2, 5), ( 4, 2, 5), ( 4, 2, 5),
(3, 1, 6) (2, 4, 7).
4.
(i) The x-axis and y-axis taken together determine a plane known as_______.
(ii) The coordinates of points in the XY-plane are of the form _______.
Fig 12.4
... (1)
Also, triangle ANQ is right angle triangle with ANQ a right angle.
272
MATHEMATICS
Therefore
From
... (2)
PA = y2 y1, AN = x2 x1 and NQ = z2 z1
2
Therefore
PQ =
( x2 x1 ) 2 +( y 2 y1 ) 2 +( z 2 z1 ) 2
he
Hence
This gives us the distance between two points (x1, y1, z1) and (x2, y2, z2).
2
2
2
x2 + y 2 + z 2 ,
is
which gives the distance between the origin O and any point Q (x2, y2, z2).
bl
Example 3 Find the distance between the points P(1, 3, 4) and Q ( 4, 1, 2).
Solution The distance PQ between the points P (1,3, 4) and Q ( 4, 1, 2) is
(4 1) 2 + (1 + 3) 2 + (2 4) 2
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PQ =
25 + 16 + 4
45 = 3 5 units
Example 4 Show that the points P (2, 3, 5), Q (1, 2, 3) and R (7, 0, 1) are collinear.
Solution We know that points are said to be collinear if they lie on a line.
Now,
and
PQ =
(1 + 2) 2 + (2 3) 2 + (3 5) 2 = 9 + 1 + 4 = 14
QR =
(7 1) 2 + (0 2) 2 + (1 3) 2 = 36 + 4 + 16 = 56 = 2 14
PR =
(7 + 2) 2 + (0 3) 2 + (1 5) 2 = 81 + 9 + 36 = 126 = 3 14
Example 5 Are the points A (3, 6, 9), B (10, 20, 30) and C (25, 41, 5), the vertices
of a right angled triangle?
Solution By the distance formula, we have
AB2 = (10 3)2 + (20 6)2 + (30 9)2
= 49 + 196 + 441 = 686
CA 2
273
We find that
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Example 6 Find the equation of set of points P such that PA2 + PB2 = 2k2, where
A and B are the points (3, 4, 5) and (1, 3, 7), respectively.
Solution Let the coordinates of point P be (x, y, z).
Here
PA2 = (x 3)2 + (y 4)2 + ( z 5)2
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EXERCISE 12.2
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2. Show that the points (2, 3, 5), (1, 2, 3) and (7, 0, 1) are collinear.
3. Verify the following:
(i) (0, 7, 10), (1, 6, 6) and (4, 9, 6) are the vertices of an isosceles triangle.
(ii) (0, 7, 10), (1, 6, 6) and ( 4, 9, 6) are the vertices of a right angled triangle.
(iii) (1, 2, 1), (1, 2, 5), (4, 7, 8) and (2, 3, 4) are the vertices of a parallelogram.
4. Find the equation of the set of points which are equidistant from the points
(1, 2, 3) and (3, 2, 1).
5. Find the equation of the set of points P, the sum of whose distances from
A (4, 0, 0) and B ( 4, 0, 0) is equal to 10.
In two dimensional geometry, we have learnt how to find the coordinates of a point
dividing a line segment in a given ratio internally. Now, we extend this to three dimensional
geometry as follows:
Let the two given points be P(x1, y1, z1) and Q (x2, y2, z2). Let the point R (x, y, z)
divide PQ in the given ratio m : n internally. Draw PL, QM and RN perpendicular to
MATHEMATICS
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mz 2 +nz1
m+n
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z=
Fig 12.5
bl
m PR SP
SL PL
NR PL z z1
=
=
=
=
=
n QR QT QM TM QM NR z 2 z
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274
This implies
y =
my2 + ny1
mx2 + nx1
and x =
m+n
m+n
Hence, the coordinates of the point R which divides the line segment joining two points
P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m : n are
,
,
m+n
m + n
m+n
If the point R divides PQ externally in the ratio m : n, then its coordinates are
obtained by replacing n by n so that coordinates of point R will be
mn
mn
mn
Case 1 Coordinates of the mid-point: In case R is the mid-point of PQ, then
m : n = 1 : 1 so that x =
x1 + x2
y + y2
z + z2
,y = 1
and z = 1
.
2
2
2
These are the coordinates of the mid point of the segment joining P (x1, y1, z1)
and Q (x2, y2, z2).
275
Case 2 The coordinates of the point R which divides PQ in the ratio k : 1 are obtained
by taking k =
m
which are as given below:
n
he
k x 2 + x1 ky2 + y1 kz2 + z1
,
,
1+k
1+k
1+k
Generally, this result is used in solving problems involving a general point on the line
passing through two given points.
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Example 7 Find the coordinates of the point which divides the line segment joining
the points (1, 2, 3) and (3, 4, 5) in the ratio 2 : 3 (i) internally, and (ii) externally.
2(3) +3(1) 9
2(4) + 3(2) 2
= , y=
=
2+3
5
2+3
5
, z=
2(5) + 3(3) 1
=
2+3
5
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x=
bl
Solution (i) Let P (x, y, z) be the point which divides line segment joining A(1, 2, 3)
and B (3, 4, 5) internally in the ratio 2 : 3. Therefore
9 2 1
, ,
5 5 5
(ii) Let P (x, y, z) be the point which divides segment joining A (1, 2, 3) and
B (3, 4, 5) externally in the ratio 2 : 3. Then
Thus, the required point is
x =
2(3) + (3)(1)
2(4) + (3)(2)
2(5) + (3)(3)
= 3, y =
= 14 , z =
= 19
2 + (3)
2 + (3)
2 + (3)
Example 8 Using section formula, prove that the three points ( 4, 6, 10), (2, 4, 6)
and (14, 0, 2) are collinear.
Solution Let A ( 4, 6, 10), B (2, 4, 6) and C(14, 0, 2) be the given points. Let the
point P divides AB in the ratio k : 1. Then coordinates of the point P are
2k 4 4k 6 6k
,
,
k 1
k 1
k
10
1
Let us examine whether for some value of k, the point P coincides with point C.
On putting
2k 4
=14 , we get k = 3
k +1
2
When
and
MATHEMATICS
4k + 6
3
=
k = , then
k +1
2
3
4( ) + 6
2
=0
3
+1
2
3
6( ) + 10
6k + 10
2
=
= 2
3
k +1
+1
2
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276
is
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Example 9 Find the coordinates of the centroid of the triangle whose vertices are
(x1, y1, z1), (x2, y2, z2) and (x3, y3, z3).
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Solution Let ABC be the triangle. Let the coordinates of the vertices A, B,C be
(x1, y1, z1), (x2, y2, z2) and (x3, y3, z3), respectively. Let D be the mid-point of BC.
Hence coordinates of D are
x2 + x3 y2 + y3 z 2 + z3
,
,
2
2
2
Let G be the centroid of the triangle. Therefore, it divides the median AD in the ratio 2 : 1.
Hence, the coordinates of G are
x2 + x3
y2 + y3
z2 + z3
2 2 + x1 2 2 + y1 2 2 + z1
,
,
2 +1
2 +1
2 +1
or
x1 + x2 + x3 y1 + y2 + y3 z1 + z2 + z3
,
,
3
3
3
Example 10 Find the ratio in which the line segment joining the points (4, 8, 10) and
(6, 10, 8) is divided by the YZ-plane.
Solution Let YZ-plane divides the line segment joining A (4, 8, 10) and B (6, 10, 8)
at P (x, y, z) in the ratio k : 1. Then the coordinates of P are
4 + 6k 8 + 10k 10 8k
,
,
k +1 k +1 k +1
4 + 6k
=0
k +1
2
3
Therefore, YZ-plane divides AB externally in the ratio 2 : 3.
k =
he
or
277
EXERCISE 12.3
Find the coordinates of the point which divides the line segment joining the points
( 2, 3, 5) and (1, 4, 6) in the ratio (i) 2 : 3 internally, (ii) 2 : 3 externally.
2.
Given that P (3, 2, 4), Q (5, 4, 6) and R (9, 8, 10) are collinear. Find the ratio
in which Q divides PR.
3.
Find the ratio in which the YZ-plane divides the line segment formed by joining
the points (2, 4, 7) and (3, 5, 8).
4.
Using section formula, show that the points A (2, 3, 4), B (1, 2, 1) and
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1.
5.
1
C 0, ,2 are collinear.
3
Find the coordinates of the points which trisect the line segment joining the points
P (4, 2, 6) and Q (10, 16, 6).
Miscellaneous Examples
Example 11 Show that the points A (1, 2, 3), B (1, 2, 1), C (2, 3, 2) and
D (4, 7, 6) are the vertices of a parallelogram ABCD, but it is not a rectangle.
Solution To show ABCD is a parallelogram we need to show opposite side are equal
Note that.
4 + 16 + 16 = 6
AB =
(1 1) 2 +(2 2) 2 +(1 3) 2 =
BC =
(2 + 1) 2 +(3 + 2) 2 +(2 + 1) 2 =
CD =
( 4 2) 2 +(7 3) 2 +(6 2) 2 =
4 + 16 + 16 = 6
DA =
(1 4) 2 +(2 7) 2 +(3 6) 2 =
9 + 25 + 9 = 43
9 + 25 + 9 =
43
Now, it is required to prove that ABCD is not a rectangle. For this, we show that
diagonals AC and BD are unequal. We have
278
MATHEMATICS
AC
(2 1) 2 +(3 2) 2 +(2 3) 2 = 1 + 1 + 1= 3
BD
he
Note We can also show that ABCD is a parallelogram, using the property that
$
diagonals AC and BD bisect each other.
( x + 2) 2 + ( y 1) 2 + ( z 4) 2
( x 3) 2 + ( y 4) 2 + ( z + 5) 2 = ( x + 2) 2 + ( y 1) 2 + ( z 4) 2
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or
( x 3) 2 + ( y 4) 2 + ( z + 5) 2 =
bl
Now
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Example 12 Find the equation of the set of the points P such that its distances from
the points A (3, 4, 5) and B ( 2, 1, 4) are equal.
or
10 x + 6y 18z 29 = 0.
Example 13 The centroid of a triangle ABC is at the point (1, 1, 1). If the coordinates
of A and B are (3, 5, 7) and (1, 7, 6), respectively, find the coordinates of the
point C.
Solution Let the coordinates of C be (x, y, z) and the coordinates of the centroid G be
(1, 1, 1). Then
x + 31
y 5+ 7
z+76
= 1, i.e., x = 1;
= 1, i.e., y = 1;
= 1, i.e., z = 2.
3
3
3
279
5. A point R with x-coordinate 4 lies on the line segment joining the points
P(2, 3, 4) and Q (8, 0, 10). Find the coordinates of the point R.
[Hint Suppose R divides PQ in the ratio k : 1. The coordinates of the point R are given
he
8k + 2 3 10k + 4
by
,
,
].
k +1 k +1 k +1
6.If A and B be the points (3, 4, 5) and (1, 3, 7), respectively, find the equation of the
set of points P such that PA2 + PB2 = k2, where k is a constant.
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Summary
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system are three mutually perpendicular lines. The axes are called the x, y
and z-axes.
The three planes determined by the pair of axes are the coordinate planes,
called XY, YZ and ZX-planes.
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The three coordinate planes divide the space into eight parts known as octants.
The coordinates of a point P in three dimensional geometry is always written
in the form of triplet like (x, y, z). Here x, y and z are the distances from the
YZ, ZX and XY-planes.
(i)
Distance between two points P(x1, y1, z1) and Q (x2, y2, z2) is given by
PQ = ( x2 x1 )2 + ( y2 y1 )2 + ( z2 z1 )2
The coordinates of the point R which divides the line segment joining two
points P (x1 y1 z1) and Q (x2, y2, z2) internally and externally in the ratio m : n
are given by
and
m+n
m+n
mn
mn ,
m+n
mn
respectively.
The coordinates of the mid-point of the line segment joining two points
x1 + x2 y1 + y2 z1 + z2
,
,
P(x1, y1, z1) and Q(x2, y2, z2) are
.
2
2
2
280
MATHEMATICS
The coordinates of the centroid of the triangle, whose vertices are (x1, y1, z1)
he
x1 + x2 + x3 y1 + y2 + y3 z1 + z2 + x3
,
,
(x2, y2, z2) and (x3, y3, z3), are
.
3
3
3
Historical Note
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bl
is
Chapter
13
he
bl
13.1 Introduction
is
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MATHEMATICS
Table 13.1
t
0
1
1.5
1.8
1.9
1.95
2
2.05
2.1
2.2
2.5
3
4
(19.6 0 ) m = 9.8 m / s
.
( 2 0) s
bl
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is
(19.6 4.9 ) m
= 14.7 m/s
( 2 1) s
s
0
4.9
11.025
15.876
17.689
18.63225
19.6
20.59225
21.609
23.716
30.625
44.1
78.4
he
282
Table 13.2
t1
1.5
1.8
1.9
1.95
1.99
9.8
14.7
17.15
18.62
19.11
19.355
19.551
From Table 13.2, we observe that the average velocity is gradually increasing.
As we make the time intervals ending at t = 2 smaller, we see that we get a better idea
of the velocity at t = 2. Hoping that nothing really dramatic happens between 1.99
seconds and 2 seconds, we conclude that the average velocity at t = 2 seconds is just
above 19.551m/s.
This conclusion is somewhat strengthened by the following set of computation.
Compute the average velocities for various time intervals starting at t = 2 seconds. As
before the average velocity v between t = 2 seconds and t = t2 seconds is
=
283
Table 13.3
29.4
2.5
2.2
2.1
2.05
2.01
is
t2
he
The following Table 13.3 gives the average velocity v in metres per second
between t = 2 seconds and t2 seconds.
no N
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bl
Here again we note that if we take smaller time intervals starting at t = 2, we get
better idea of the velocity at t = 2.
In the first set of computations, what we have done is to find average velocities
in increasing time intervals ending at t = 2 and then hope that nothing dramatic happens
just before t = 2. In the second set of computations, we have found the average velocities
decreasing in time intervals ending at t = 2 and then hope that nothing dramatic happens
just after t = 2. Purely on the physical grounds, both these sequences of average
velocities must approach a common limit. We can safely conclude that the velocity of
the body at t = 2 is between 19.551m/s and 19.649 m/s. Technically, we say that the
instantaneous velocity at t = 2 is between 19.551 m/s and 19.649 m/s. As is
well-known, velocity is the rate of change of displacement. Hence what we have
accomplished is the following. From the given data of distance covered at various time
instants we have estimated the rate of
change of the distance at a given instant
of time. We say that the derivative of
the distance function s = 4.9t2 at t = 2
is between 19.551 and 19.649.
An alternate way of viewing this
limiting process is shown in Fig 13.1.
This is a plot of distance s of the body
from the top of the cliff versus the time
t elapsed. In the limit as the sequence
of time intervals h1, h2, ..., approaches
zero, the sequence of average velocities
approaches the same limit as does the
Fig 13.1
sequence of ratios
284
MATHEMATICS
C1B1 C2 B2 C3 B3
,
,
, ...
AC1 AC2 AC3
he
where C1B1 = s1 s0 is the distance travelled by the body in the time interval h1 = AC1,
etc. From the Fig 13.1 it is safe to conclude that this latter sequence approaches the
slope of the tangent to the curve at point A. In other words, the instantaneous velocity
v(t) of a body at time t = 2 is equal to the slope of the tangent of the curve s = 4.9t2 at
t = 2.
13.3 Limits
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lim f ( x ) = 0
bl
is
The above discussion clearly points towards the fact that we need to understand limiting
process in greater clarity. We study a few illustrative examples to gain some familiarity
with the concept of limits.
Consider the function f(x) = x2. Observe that as x takes values very close to 0,
the value of f(x) also moves towards 0 (See Fig 2.10 Chapter 2). We say
x 0
(to be read as limit of f (x) as x tends to zero equals zero). The limit of f (x) as x tends
to zero is to be thought of as the value f (x) should assume at x = 0.
In general as x a, f (x) l, then l is called limit of the function f (x) which is
f ( x) = l .
symbolically written as lim
xa
Consider the following function g(x) = |x|, x 0. Observe that g(0) is not defined.
Computing the value of g(x) for values of x very
near to 0, we see that the value of g(x) moves
towards 0. So, lim
g(x) = 0. This is intuitively
x0
clear from the graph of y = |x| for x 0.
(See Fig 2.13, Chapter 2).
Consider the following function.
x2 4
, x2.
x2
Compute the value of h(x) for values of
x very near to 2 (but not at 2). Convince yourself
that all these values are near to 4. This is
somewhat strengthened by considering the graph
of the function y = h(x) given here (Fig 13.2).
h( x) =
Fig 13.2
285
he
In all these illustrations the value which the function should assume at a given
point x = a did not really depend on how is x tending to a. Note that there are essentially
two ways x could approach a number a either from left or from right, i.e., all the
values of x near a could be less than a or could be greater than a. This naturally leads
to two limits the right hand limit and the left hand limit. Right hand limit of a
function f(x) is that value of f(x) which is dictated by the values of f(x) when x tends
to a from the right. Similarly, the left hand limit. To illustrate this, consider the function
bl
is
1, x 0
f ( x) =
2, x > 0
lim f ( x) =1 .
x0
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x 0+
Fig 13.3
In this case the right and left hand limits are different, and hence we say that the
limit of f (x) as x tends to zero does not exist (even though the function is defined at 0).
Summary
We say xlim
f(x) is the expected value of f at x = a given the values of f near
a
f near x to the right of a. This value is called the right hand limit of f(x) at a.
If the right and left hand limits coincide, we call that common value as the limit
of f(x) at x = a and denote it by lim
f(x).
x a
Illustration 1 Consider the function f(x) = x + 10. We want to find the limit of this
function at x = 5. Let us compute the value of the function f(x) for x very near to 5.
Some of the points near and to the left of 5 are 4.9, 4.95, 4.99, 4.995. . ., etc. Values
of the function at these points are tabulated below. Similarly, the real number 5.001,
286
MATHEMATICS
5.01, 5.1 are also points near and to the right of 5. Values of the function at these points
are also given in the Table 13.4.
Table 13.4
4.9
4.95
4.99
4.995
5.001
5.01
5.1
f(x)
14.9
14.95
14.99
14.995
15.001
15.01
15.1
he
bl
lim f ( x ) = 15 .
x 5
is
From the Table 13.4, we deduce that value of f(x) at x = 5 should be greater than
14.995 and less than 15.001 assuming nothing dramatic happens between x = 4.995
and 5.001. It is reasonable to assume that the value of the f(x) at x = 5 as dictated by
the numbers to the left of 5 is 15, i.e.,
Similarly, when x approaches 5 from the right, f(x) should be taking value 15, i.e.,
no N
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lim f ( x ) = 15 .
x 5+
Hence, it is likely that the left hand limit of f(x) and the right hand limit of f(x) are
both equal to 15. Thus,
lim f ( x ) = lim+ f ( x ) = lim f ( x ) = 15 .
x 5
x 5
x 5
Illustration 2 Consider the function f(x) = x3. Let us try to find the limit of this
function at x = 1. Proceeding as in the previous case, we tabulate the value of f(x) at
x near 1. This is given in the Table 13.5.
Table 13.5
0.9
0.99
f(x)
0.729
0.970299
0.999
0.997002999
1.001
1.01
1.1
1.003003001
1.030301
1.331
From this table, we deduce that value of f(x) at x = 1 should be greater than
0.997002999 and less than 1.003003001 assuming nothing dramatic happens between
287
x = 0.999 and 1.001. It is reasonable to assume that the value of the f(x) at x = 1 as
dictated by the numbers to the left of 1 is 1, i.e.,
lim f ( x ) = 1 .
x 1
Similarly, when x approaches 1 from the right, f(x) should be taking value 1, i.e.,
he
lim f ( x ) = 1 .
x 1+
x 1
x 1
x 1
is
Hence, it is likely that the left hand limit of f(x) and the right hand limit of f(x) are
both equal to 1. Thus,
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1.9
1.95
1.99
1.999
2.001
2.01
2.1
f(x)
5.7
5.85
5.97
5.997
6.003
6.03
6.3
x 2
x2
x 2
fact.
Fig 13.4
288
MATHEMATICS
value (3, in this case) everywhere, i.e., its value at points close to 2 is 3. Hence
lim f ( x ) = lim+ f ( x ) = lim f ( x ) = 3
x 2
x2
x2
Graph of f(x) = 3 is anyway the line parallel to x-axis passing through (0, 3) and
is shown in Fig 2.9, Chapter 2. From this also it is clear that the required limit is 3. In
he
0.99
0.999
f(x)
1.71
1.9701
1.997001
1.01
1.1
1.2
2.31
2.64
bl
0.9
2.0301
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f ( x ) . We
Illustration 5 Consider the function f(x) = x2 + x. We want to find lim
x 1
x 1
x 1
x 1
Fig 13.5
x 1
x 1
Then
Also
x 1
x 1
x 1
x 1
x 1
x 1
x 1
289
Illustration 6 Consider the function f(x) = sin x. We are interested in lim sin x ,
x
he
is
Further, this is supported by the graph of f(x) = sin x which is given in the Fig 3.8
x
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Table 13.8
bl
0.1
2
0.01
2
+ 0.01
2
+ 0.1
2
f(x)
0.9950
0.9999
0.9999
0.9950
Illustration 7 Consider the function f(x) = x + cos x. We want to find the lim f (x).
x0
0.1
0.01
0.001
0.001
0.01
0.1
f(x)
0.9850
0.98995
0.9989995
1.0009995
1.00995
1.0950
x 0
x 0
x 0
290
MATHEMATICS
1
for x > 0 . We want to know lim f (x).
x0
x2
Here, observe that the domain of the function is given to be all positive real
numbers. Hence, when we tabulate the values of f(x), it does not make sense to talk of
x approaching 0 from the left. Below we tabulate the values of the function for positive
x close to 0 (in this table n denotes any positive integer).
From the Table 13.10 given below, we see that as x tends to 0, f(x) becomes
larger and larger. What we mean here is that the value of f(x) may be made larger than
any given number.
Table 13.10
0.1
f(x)
100
10n
10000
102n
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Mathematically, we say
0.01
bl
is
he
lim f ( x ) = +
x 0
We also remark that we will not come across such limits in this course.
f ( x ) , where
Illustration 9 We want to find lim
x 0
x 2, x < 0
f ( x) = 0 , x = 0
x + 2, x > 0
As usual we make a table of x near 0 with f(x). Observe that for negative values of x
we need to evaluate x 2 and for positive values, we need to evaluate x + 2.
Table 13.11
0.1
0.01
0.001
0.001
0.01
0.1
f(x)
2.1
2.01
2.001
2.001
2.01
2.1
From the first three entries of the Table 13.11, we deduce that the value of the
function is decreasing to 2 and hence.
lim f ( x ) = 2
x 0
291
From the last three entires of the table we deduce that the value of the function
is increasing from 2 and hence
lim f ( x ) = 2
x 0+
x 1
x =1
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x + 2
f ( x) =
0
is
where
Fig 13.6
bl
f ( x) ,
Illustration 10 As a final illustration, we find lim
x 1
he
Table 13.12
0.9
0.99
0.999
1.001
1.01
1.1
f(x)
2.9
2.99
2.999
3.001
3.01
3.1
As usual we tabulate the values of f(x) for x near 1. From the values of f(x) for
x less than 1, it seems that the function should take value 3 at x = 1., i.e.,
lim f ( x ) = 3 .
x 1
Similarly, the value of f(x) should be 3 as dictated by values of f(x) at x greater than 1. i.e.
lim f ( x ) = 3 .
x 1+
x 1
x 1
x 1
Fig 13.7
292
MATHEMATICS
note that in general, at a given point the value of the function and its limit may be
different (even when both are defined).
he
13.3.1 Algebra of limits In the above illustrations, we have observed that the limiting
process respects addition, subtraction, multiplication and division as long as the limits
and functions under consideration are well defined. This is not a coincidence. In fact,
below we formalise these as a theorem without proof.
Theorem 1 Let f and g be two functions such that both lim f (x) and lim g(x) exist.
x a
x a
is
Then
(i) Limit of sum of two functions is sum of the limits of the functions, i.e.,
lim [f(x) + g (x)] = lim f(x) + lim g(x).
x a
x a
bl
x a
(ii) Limit of difference of two functions is difference of the limits of the functions, i.e.,
lim [f(x) g(x)] = lim f(x) lim g(x).
x a
x a
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x a
(iii) Limit of product of two functions is product of the limits of the functions, i.e.,
lim [f(x) . g(x)] = lim f(x). lim g(x).
x a
x a
x a
(iv) Limit of quotient of two functions is quotient of the limits of the functions (whenever
the denominator is non zero), i.e.,
lim
x a
$ Note
f ( x)
g ( x)
lim f ( x )
x a
lim g ( x )
xa
lim ( . f ) ( x ) = .lim f ( x ) .
x a
x a
In the next two subsections, we illustrate how to exploit this theorem to evaluate
limits of special types of functions.
293
x a
x a
x a
lim x n = a n
x a
he
lim f ( x )
x a
a0 + a1 x + a2 x 2 + ... + an x n
= lim
x a
is
bl
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(Make sure that you understand the justification for each step in the above!)
g ( x)
A function f is said to be a rational function, if f(x) = h x , where g(x) and h(x)
( )
lim f ( x ) = lim
x a
x a
g ( x)
h( x)
lim g ( x )
xa
lim h ( x )
xa
g (a)
h(a)
However, if h(a) = 0, there are two scenarios (i) when g(a) 0 and (ii) when
g(a) = 0. In the former case we say that the limit does not exist. In the latter case we
can write g(x) = (x a)k g1 (x), where k is the maximum of powers of (x a) in g(x)
Similarly, h(x) = (x a)l h1 (x) as h (a) = 0. Now, if k > l, we have
lim f ( x ) =
xa
lim g ( x )
xa
lim h ( x )
xa
lim ( x a ) g1 ( x )
k
xa
lim ( x a ) h1 ( x )
l
xa
294
MATHEMATICS
lim ( x a )
xa
( k l )
g1 ( x )
lim h1 ( x )
x a
0.g1 ( a )
h1 ( a )
=0
(iii)
x ( x + 1)
(ii) lim
x 3
he
x3 x 2 + 1
Example 1 Find the limits: (i) lim
x 1
lim 1 + x + x 2 + ... + x10 .
x 1
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x ( x + 1) = 3 ( 3 + 1) = 3 ( 4 ) = 12
(ii) lim
x 3
bl
(i) lim
[x3 x2 + 1] = 13 12 + 1 = 1
x1
is
Solution The required limits are all limits of some polynomial functions. Hence the
limits are the values of the function at the prescribed points. We have
(iii)
x 1
= 1 1 + 1... + 1 = 1 .
(i) lim
x 1 x + 100
x2 4
lim
(iii) x 2 3
2
x 4x + 4x
x3 4 x 2 + 4 x
(ii) lim
x 2
x2 4
x3 2 x 2
lim
(iv) x 2 2
x 5x + 6
1
x2
3
(v) lim
.
2
2
x 1 x x
x 3x + 2 x
Solution All the functions under consideration are rational functions. Hence, we first
0
, we try to
0
rewrite the function cancelling the factors which are causing the limit to be of
the form
0
.
0
295
x2 + 1
12 + 1
2
=
=
x 1 x + 100
1 + 100 101
0
.
0
he
(ii)
x ( x 2)
x3 4 x 2 + 4 x
lim
Hence
= lim
2
x 2 ( x + 2 )( x 2 )
x 2
x 4
2
is
as x 2
( x + 2)
2 ( 2 2)
2+2
0
=0.
4
no N
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x ( x 2)
bl
= lim
x 2
Hence lim
x 2
0
.
0
( x + 2 )( x 2 )
x2 4
lim
2
=
x2
x ( x 2)
x3 4 x 2 + 4 x
( x + 2) = 2 + 2 = 4
= lim
x 2 x ( x 2 )
2 ( 2 2) 0
Hence
0
.
0
x2 ( x 2)
x3 2 x 2
lim 2
= lim
x 2 ( x 2 )( x 3)
x 2 x 5 x + 6
( 2) 4
x2
=
=
= 4.
( x 3) 2 3 1
2
= lim
x 2
296
MATHEMATICS
x2
1
= x x 1 x x 1 x 2
) ( )(
)
(
bl
x2 4x + 3
= x x 1 x 2
( )(
)
is
x2 4x + 4 1
= x x 1 x 2
)(
)
(
0
.
0
no N
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Evaluating the function at 1, we get it of the form
Hence
he
1
1
x2
x2
=
2
x 2 x x3 3x 2 + 2 x x ( x 1) x x 3 x + 2
x2 2
x2 4x + 3
1
lim
lim 2
3
=
x 1 x x
x 3 x 2 + 2 x x 1 x ( x 1)( x 2 )
( x 3)( x 1)
= lim
x 1 x ( x 1)( x 2 )
x3
1 3
= lim
=
x 1 x ( x 2 )
1(1 2 ) = 2.
We remark that we could cancel the term (x 1) in the above evaluation because
x 1.
Evaluation of an important limit which will be used in the sequel is given as a
theorem below.
Theorem 2 For any positive integer n,
xn an
= na n 1 .
xa x a
lim
Remark The expression in the above theorem for the limit is true even if n is any
rational number and a is positive.
297
Thus, lim
he
Example 3 Evaluate:
x15 1
x 1 x10 1
Solution (i) We have
(i) lim
1+ x 1
x
bl
(ii) lim
is
n 1
= na
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x 0
lim
x 1
15
10
x15 1 lim x 1 x 1
=
x 1
x 1
x10 1
x 1
x15 1
x10 1
lim
lim
= x 1
x 1 x 1 x 1
3
2
Then
lim
x 0
y 1
1+ x 1
= lim
y 1 y 1
x
1
y 2 12
= lim
y 1
y 1
1
1
1 2 1
(1)
(by the remark above) =
2
2
298
MATHEMATICS
he
f (x) g( x) for all x in the domain of definition, For some a, if both lim
f(x) and
x a
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bl
is
lim g(x) exist, then lim f(x) lim g(x). This is illustrated in Fig 13.8.
x a
x a
x a
Fig 13.8
Fig 13.9
cos x <
sin x
<1
x
(*)
299
Proof We know that sin ( x) = sin x and cos( x) = cos x. Hence, it is sufficient
.
2
In the Fig 13.10, O is the centre of the unit circle such that
1
x
1
OA.CD < ..(OA)2 < OA.AB .
2
2
2
i.e., CD < x . OA < AB.
From OCD,
Fig 13.10
is
he
bl
i.e.,
CD
AB
(since OC = OA) and hence CD = OA sin x. Also tan x =
and
OA
OA
hence
AB = OA. tan x. Thus
OA sin x < OA. x < OA. tan x.
Since length OA is positive, we have
sin x < x < tan x.
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sin x =
x
1
<
. Taking reciprocals throughout, we have
sin x cos x
sin x
<1
x
which complete the proof.
cos x <
sin x
=1.
x 0 x
(i) lim
1 cos x
=0.
x 0
x
(ii) lim
sin x
is sandwiched between the
x
function cos x and the constant function which takes value 1.
300
MATHEMATICS
Then
is
x
x
2sin 2
sin
2 = lim 2 .sin x
2
x 0
x
x
2
bl
1 cos x
lim
= lim
x 0
x 0
x
he
x
To prove (ii), we recall the trigonometric identity 1 cos x = 2 sin2 .
2
no N
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x
sin
2
x
= lim .lim sin = 1.0 = 0
x 0
x 0
x
2
2
Example 4 Evaluate:
Solution (i)
lim
x 0
x
.
2
(i) lim
sin 4 x
x 0 sin 2 x
(ii) lim
x 0
tan x
x
sin 4 x 2 x
sin 4 x
= lim
.
.2
x
0
sin 2 x
4 x sin 2 x
sin 4 x sin 2 x
= 2.lim
x 0
4x 2x
sin 4 x
sin 2 x
lim
= 2. 4lim
x 0
2
x
0
4x
2x
x
0 . This
2
301
sin x
sin x
1
tan x
. lim
= lim
= lim
= 1.1 = 1
x
0
x
0
x
0
cos x
x cos x
x
x
A general rule that needs to be kept in mind while evaluating limits is the following.
he
f ( x)
exists and we want to evaluate this. First we check
Say, given that the limit lim
xa g ( x )
g ( x)
q ( x ) , where q(x) 0.
f ( x)
p (a)
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lim
p ( x)
bl
f ( x)
is
the value of f (a) and g(a). If both are 0, then we see if we can get the factor which
is causing the terms to vanish, i.e., see if we can write f(x) = f1 (x) f2(x) so that
f1 (a) = 0 and f2 (a) 0. Similarly, we write g(x) = g1 (x) g2(x), where g1(a) = 0 and
g2(a) 0. Cancel out the common factors from f(x) and g(x) (if possible) and write
Then
x a
g ( x)
q(a) .
EXERCISE 13.1
4x + 3
x 4 x 2
4. lim
7. lim
x 2
3x 2 x 10
x2 4
22
x
2. lim
x
7
r 2
3. lim
r 1
( x + 1)5 1
5. lim
x10 + x5 + 1
x 1
6. lim
x 0
8. lim
x 4 81
2 x2 5x 3
9. lim
ax + b
cx + 1
x 1
x 3
x 0
10. lim
z 1
z 3 1
1
z 6 1
1 1
+
12. lim x 2
x 2 x + 2
ax 2 + bx + c
,a+b+c 0
x 1 cx 2 + bx + a
11. lim
13. lim
x 0
sin ax
bx
14. lim
x 0
sin ax
, a, b 0
sin bx
MATHEMATICS
sin ( x )
15. lim
x ( x )
16. lim
x 0
cos 2 x 1
x 0 cos x 1
cos x
x
17. lim
18. lim
ax + x cos x
b sin x
x sec x
19. lim
x 0
20. lim
sin ax + bx
a, b, a + b 0 ,
ax + sin bx
(cosec x cot x)
21. lim
x 0
x 0
lim
tan 2 x
x
2
bl
22.
x
2
is
x 0
he
302
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2 x + 3, x 0
f ( x ) and lim f ( x ) , where f ( x ) =
23. Find lim
x 0
x 1
3 ( x + 1) , x > 0
x 2 1, x 1
=
f
x
lim
f
x
(
)
2
24. Find x 1 ( ) , where
x 1, x > 1
| x |
, x0
f ( x ) , where f ( x ) = x
25. Evaluate lim
x 0
0, x = 0
x
, x0
f ( x ) , where f ( x ) = | x |
26. Find lim
x 0
0, x = 0
f ( x ) , where f ( x ) = | x | 5
27. Find lim
x 5
a + bx ,
28. Suppose f ( x ) = 4,
b ax ,
x <1
x =1
x >1
303
29. Let a1, a2, . . ., an be fixed real numbers and define a function
f ( x ) = ( x a1 ) ( x a2 ) ... ( x an ) .
What is xlim
f (x) ? For some a a1, a2, ..., an, compute lim
f (x).
xa
a
he
x2 1
= , evaluate lim f ( x ) .
x 1
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x 1
f ( x) 2
bl
is
x + 1, x < 0
x=0 .
30. If f ( x ) = 0,
x 1, x > 0
32.
mx 2 + n ,
x<0
f ( x)
If f ( x ) = nx + m , 0 x 1 . For what integers m and n does both lim
x 0
3
x >1
nx + m ,
f ( x ) exist?
and lim
x 1
13.5 Derivatives
We have seen in the Section 13.2, that by knowing the position of a body at various
time intervals it is possible to find the rate at which the position of the body is changing.
It is of very general interest to know a certain parameter at various instants of time and
try to finding the rate at which it is changing. There are several real life situations
where such a process needs to be carried out. For instance, people maintaining a
reservoir need to know when will a reservoir overflow knowing the depth of the water
at several instances of time, Rocket Scientists need to compute the precise velocity
with which the satellite needs to be shot out from the rocket knowing the height of the
rocket at various times. Financial institutions need to predict the changes in the value of
a particular stock knowing its present value. In these, and many such cases it is desirable
to know how a particular parameter is changing with respect to some other parameter.
The heart of the matter is derivative of a function at a given point in its domain
of definition.
304
MATHEMATICS
f (a + h) f (a)
h 0
h
provided this limit exists. Derivative of f (x) at a is denoted by f(a).
f (2) = lim
f ( 2 + h ) f ( 2)
h 0
3( 2 + h ) 3( 2)
h
6 + 3h 6
3h
= lim = lim 3 = 3 .
h
0
h
h h 0
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h 0
h 0
bl
= lim
= lim
is
Solution We have
he
Example 6 Find the derivative of the function f(x) = 2x2 + 3x 5 at x = 1. Also prove
that f (0) + 3f ( 1) = 0.
Solution We first find the derivatives of f(x) at x = 1 and at x = 0. We have
f ' ( 1) = lim
h 0
f ( 1 + h ) f ( 1)
h
2 ( 1 + h )2 + 3 ( 1 + h ) 5 2 ( 1)2 + 3 ( 1) 5
= lim
h 0
h
= lim
h 0
and
f ' ( 0 ) = lim
h 0
2h 2 h
= lim ( 2h 1) = 2 ( 0 ) 1 = 1
h 0
h
f (0 + h ) f (0)
h
2 ( 0 + h )2 + 3 ( 0 + h ) 5 2 ( 0 )2 + 3 ( 0 ) 5
= lim
h 0
h
305
2h 2 + 3h
= lim ( 2h + 3) = 2 ( 0 ) + 3 = 3
h 0
h 0
h
= lim
Clearly
f ' ( 0 ) + 3 f ' ( 1) = 0
he
Remark At this stage note that evaluating derivative at a point involves effective use
of various rules, limits are subjected to. The following illustrates this.
Example 7 Find the derivative of sin x at x = 0.
f ( 0 + h ) f ( 0)
f (0) = lim
h 0
h
sin ( 0 + h ) sin ( 0 )
h 0
sin h
=1
h 0 h
= lim
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bl
= lim
is
Solution Since the derivative measures the change in function, intuitively it is clear
that the derivative of the constant function must be zero at every point. This is indeed,
supported by the following computation.
f ' ( 0 ) = lim
f (0 + h) f (0)
h 0
Similarly
f ' ( 3) = lim
h0
f ( 3 + h ) f ( 3)
h
= lim
33
0
= lim = 0 .
h 0 h
h
= lim
33
=0.
h
h 0
h 0
Fig 13.11
306
MATHEMATICS
f (a + h) f (a)
h 0
f (a + h) f (a)
he
h
From the triangle PQR, it is clear that the ratio whose limit we are taking is
precisely equal to tan(QPR) which is the slope of the chord PQ. In the limiting process,
as h tends to 0, the point Q tends to P and we have
QR
h0
Q P PR
h
This is equivalent to the fact that the chord PQ tends to the tangent at P of the
curve y = f(x). Thus the limit turns out to be equal to the slope of the tangent. Hence
= lim
is
lim
f ( a ) = tan .
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For a given function f we can find the derivative at every point. If the derivative
exists at every point, it defines a new function called the derivative of f . Formally, we
define derivative of a function as follows.
Definition 2 Suppose f is a real valued function, the function defined by
lim
f ( x + h) f ( x)
h 0
h
wherever the limit exists is defined to be the derivative of f at x and is denoted by
f(x). This definition of derivative is also called the first principle of derivative.
Thus
f ' ( x ) = lim
f ( x + h) f ( x)
h 0
h
Clearly the domain of definition of f (x) is wherever the above limit exists. There
are different notations for derivative of a function. Sometimes f (x) is denoted by
d
dy
f ( x ) ) or if y = f(x), it is denoted by
(
. This is referred to as derivative of f(x)
dx
dx
or y with respect to x. It is also denoted by D (f (x) ). Further, derivative of f at x = a
is also denoted by
d
f ( x)
dx
or
df
dx
df
or even .
dx x = a
h 0
f ( x + h) f ( x)
h
= lim
307
10 ( x + h ) 10 ( x )
h 0
10h
(10 ) = 10 .
= lim
h 0
h 0 h
he
= lim
h 0
( x + h )2 ( x )2
( h + 2x ) = 2x
= lim
h 0
bl
= lim
h 0
is
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Example 11 Find the derivative of the constant function f (x) = a for a fixed real
number a.
Solution We have, f (x) = lim
f ( x + h) f ( x)
h 0
= lim
h 0
aa
0
= lim = 0 as h 0
h 0 h
h
1
x
f ( x + h) f ( x)
h 0
1
1
= lim ( x + h) x
h 0
h
1 x ( x + h)
= lim
h 0 h
x ( x + h )
1
1 h
= lim
= lim
h
0
h 0 h x ( x + h )
x ( x + h)
1
x2
308
MATHEMATICS
d
d
d
f ( x) + g ( x) = f ( x) + g( x) .
dx
dx
dx
bl
(ii)
he
is
(i)
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d
d
d
f ( x ) g ( x ) =
f ( x) g ( x) .
dx
dx
dx
(iii)
d
d
d
f ( x ) . g ( x ) =
f ( x) . g ( x) + f ( x) . g ( x)
dx
dx
dx
(iv)
=
dx g ( x)
( g ( x) )2
The proofs of these follow essentially from the analogous theorem for limits. We
will not prove these here. As in the case of limits this theorem tells us how to compute
derivatives of special types of functions. The last two statements in the theorem may
be restated in the following fashion which aids in recalling them easily:
Let u = f ( x ) and v = g (x). Then
( uv )
= uv + uv
309
u u v uv
=
v2
v
h 0
x+hx
h 0
h
he
= lim
is
=1 .
= lim1
h0
df ( x)
d
=
dx
dx
( x + ... + x )
bl
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d
d
x +...+
x (ten terms)
dx
dx
= 1 + ... + 1 (ten terms) = 10.
We note that this limit may be evaluated using product rule too. Write
f(x) = 10x = uv, where u is the constant function taking value 10 everywhere and
v(x) = x. Here, f(x) = 10x = uv we know that the derivative of u equals 0. Also
derivative of v(x) = x equals 1. Thus by the product rule we have
=
df
d
d
d
( x.x ) = ( x ) .x + x. ( x )
=
dx
dx
dx
dx
= 1.x + x.1 = 2 x .
More generally, we have the following theorem.
h 0
f ( x + h) f ( x)
h
= lim
h0
( x + h )n x n
h
310
MATHEMATICS
n1
+... + h
C0 x n +
( C )x
n
n 1
h + ... +
C n h n and
). Thus
n1
df ( x)
( x + h ) xn
= lim
dx
h 0
h
h 0
nx n 1 + ... + h n 1 = nx n 1 .
= lim
h 0
is
= lim
h nx n 1 + .... + h n 1
he
( )
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d n
d
x.x n 1
x =
dx
dx
bl
Alternatively, we may also prove this by induction on n and the product rule as
follows. The result is true for n = 1, which has been proved earlier. We have
d
d
( x ) . x n1 + x. x n1 (by product rule)
dx
dx
= 1.x
n 1
= x n 1 + ( n 1) x n 1 = nx n 1 .
Remark The above theorem is true for all powers of x, i.e., n can be any real number
(but we will not prove it here).
13.5.2 Derivative of polynomials and trigonometric functions We start with the
ai s are all real numbers and an 0. Then, the derivative function is given by
df ( x)
= nan x n 1 + ( n 1) an 1 x x 2 + ... + 2a2 x + a1 .
dx
Proof of this theorem is just putting together part (i) of Theorem 5 and Theorem 6.
Solution A direct application of the above theorem tells that the derivative of the
311
( 50 )( 51)
2
= 1275.
he
x +1
x
Solution Clearly this function is defined everywhere except at x = 0. We use the
quotient rule with u = x + 1 and v = x. Hence u = 1 and v = 1. Therefore
is
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bl
df ( x) d x + 1 d u u v uv 1( x ) ( x + 1)1
1
=
=
= 2
= =
2
2
dx
dx x dx v
v
x
x
f ( x + h) f ( x)
sin ( x + h ) sin ( x )
df ( x)
= lim
= lim
h 0
h 0
dx
h
h
2x + h h
2cos
sin
2 2 (using formula for sin A sin B)
= lim
h 0
h
h
sin
h
0
h
2
f ( x + h) f ( x)
tan ( x + h ) tan ( x )
df ( x)
= lim
= lim
h 0
h 0
dx
h
h
1 sin ( x + h ) sin x
lim
= h0 h cos x + h cos x
(
)
312
MATHEMATICS
= lim
h 0
h cos ( x + h ) cos x
1
= sec 2 x .
cos 2 x
bl
= 1.
sin h
1
.lim
h h0 cos ( x + h ) cos x
is
= lim
h 0
he
sin ( x + h x )
= lim
(using formula for sin (A + B))
h 0 h cos ( x + h ) cos x
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df ( x) d
= ( sin x sin x )
dx
dx
EXERCISE 13.2
1.
2.
3.
4.
x 3 27
1
x2
5. For the function
(iii)
f ( x) =
(ii)
( x 1)( x 2 )
(iv)
x +1
x 1
x100 x99
x2
+
+. . .+
+ x +1 .
100 99
2
313
( ax
+b
(iii)
xn an
for some constant a.
xa
3
4
(ii)
x 3 ( 5 + 3 x )
(v)
x 4 3 4 x 5
) ( x 1)
+ 3x 1
5
9
(iv) x 3 6 x
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(iii)
(5x
bl
(i) 2 x
(vi)
xa
xb
he
( x a) ( x b)
is
(i)
2
x2
x + 1 3x 1
(ii) sec x
Miscellaneous Examples
Example 19 Find the derivative of f from the first principle, where f is given by
(i) f (x) =
2x + 3
x2
(ii)
f (x) = x +
1
x
f ( x ) = lim
h 0
f ( x + h) f ( x)
h
2( x + h) + 3 2x + 3
x2
= lim x + h 2
h 0
h
MATHEMATICS
= lim
h 0
( 2 x + 2h + 3)( x 2 ) ( 2 x + 3)( x + h 2 )
h ( x 2 )( x + h 2 )
= lim
h 0
( 2 x + 3)( x 2 ) + 2h ( x 2 ) ( 2 x + 3)( x 2 ) h ( 2 x + 3)
h ( x 2 )( x + h 2 )
7
7
=
= lim
h 0 ( x 2 ) ( x + h 2 )
( x 2 )2
is
1
1
x+h+
x+
f ( x + h) f ( x)
x+h
x
= lim
= lim
h 0
h 0
h
h
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f ( x)
bl
(ii)
he
314
= lim
h 0
1
1
1
h+
h
x + h x
1
x x h
1
1
lim
h
lim
h
1
+
=
= h 0 h
x ( x + h ) h0 h
x ( x + h )
1
1
1
=1 2
= lim
h 0
x
x ( x + h )
h 0
= lim
h 0
f ( x + h) f ( x)
h
sin x cos h + cos x sin h + cos x cos h sin x sin h sin x cos x
h
= lim
h0
h 0
( cos h 1)
sin h
( cos h 1)
sin x
( cos x sin x ) + lim
+ lim cos x
h
0
h 0
h
h
h
= cos x sin x
h0
= lim
( x + h ) sin ( x + h ) x sin x
h 0
h 0
= lim
= lim
is
f ( x + h) f ( x)
h 0
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x sin x ( cos h 1)
bl
f ' ( x ) = lim
= lim
he
= lim
(ii)
315
h 0
+ lim h0 x cos x
sin h
+ lim ( sin x cos h + sin h cos x )
h 0
h
= x cos x + sin x
Example 21 Compute derivative of
Solution (i) Recall the trigonometric formula sin 2x = 2 sin x cos x. Thus
df ( x)
d
d
( 2sin x cos x ) = 2 ( sin x cos x )
=
dx
dx
dx
= 2 ( sin x ) cos x + sin x ( cos x )
= 2 cos 2 x sin 2 x
wherever it is defined.
cos x
. We use the quotient rule on this function
sin x
dg d
d cos x
= (cot x) =
dx dx
dx sin x
MATHEMATICS
he
316
sin 2 x + cos 2 x
= cosec 2 x
2
sin x
bl
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d
d 1
dg
(cot x) =
=
dx
dx tan x
dx
is
1
. Here, we use the fact
tan x
that the derivative of tan x is sec2 x which we saw in Example 17 and also that the
derivative of the constant function is 0.
Alternatively, this may be computed by noting that cot x =
sec 2 x
= cosec 2 x
tan 2 x
x5 cos x
sin x
(ii)
x + cos x
tan x
x5 cos x
. We use the quotient rule on this function wherever
sin x
it is defined.
h( x) =
317
h( x) =
x + cos x
wherever it is defined.
tan x
bl
is
he
x 5 cos x + 5 x 4 sin x + 1
=
(sin x) 2
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)
8
Find the derivative of the following functions (it is to be understood that a, b, c, d,
p, q, r and s are fixed non-zero constants and m and n are integers):
(i) x
(ii) ( x) 1
(iii) sin (x + 1)
(iv) cos (x
2. (x + a)
3. (px + q) + s
x
4. ( ax + b )( cx + d )
ax + b
5.
cx + d
1
x
6.
1
1
x
7.
8.
ax + b
px + qx + r
2
1+
9.
px 2 + qx + r
ax + b
10.
1
ax + bx + c
2
a
b
2 + cos x
4
x
x
11. 4 x 2
12. (ax + b) n
14. sin (x + a)
16.
cos x
1 + sin x
318
MATHEMATICS
17.
sin x + cos x
sin x cos x
18.
sec x 1
sec x + 1
19. sin n x
20.
a + b sin x
c + d cos x
21.
sin( x + a)
cos x
23.
(x
24.
( ax
+ sin x
25.
( x + cos x ) ( x tan x )
4 x + 5sin x
26.
3x + 7cos x
28.
x
1 + tan x
29.
) ( p + q cos x )
x 2 cos
4
27.
sin x
( x + sec x ) ( x tan x )
30.
x
sin n x
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Summary
he
is
+ 1 cos x
bl
The expected value of the function as dictated by the points to the left of a
point defines the left hand limit of the function at that point. Similarly the right
hand limit.
Limit of a function at a point is the common value of the left and right hand
limits, if they coincide.
xa
x a
xa
xa
f ( x)
f ( x ) lim
xa
lim
=
xa g ( x)
g ( x)
lim
x a
lim
lim
x 0
319
sin x
=1
x
1 cos x
=0
x 0
x
The derivative of a function f at a is defined by
f ( a + h) f ( a )
h
Derivative of a function f at any point x is defined by
he
lim
f (a ) = lim
(u v ) = u v
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(uv ) = u v + uv
bl
df ( x)
f ( x + h) f ( x )
= lim
h
0
dx
h
For functions u and v the following holds:
f ( x) =
is
h 0
u u v uv
provided all are defined.
=
v2
v
d
(sin x) = cos x
dx
d
(cos x) = sin x
dx
Historical Note
In the history of mathematics two names are prominent to share the credit for
inventing calculus, Issac Newton (1642 1727) and G.W. Leibnitz (1646 1717).
Both of them independently invented calculus around the seventeenth century.
After the advent of calculus many mathematicians contributed for further
development of calculus. The rigorous concept is mainly attributed to the great
320
MATHEMATICS
y f ( x + i ) f ( x)
, and called the limit for
for = 0. He wrote x =
i
he
sin
bl
is
Before 1900, it was thought that calculus is quite difficult to teach. So calculus
became beyond the reach of youngsters. But just in 1900, John Perry and others
in England started propagating the view that essential ideas and methods of calculus
were simple and could be taught even in schools. F.L. Griffin, pioneered the
teaching of calculus to first year students. This was regarded as one of the most
daring act in those days.
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Today not only the mathematics but many other subjects such as Physics,
Chemistry, Economics and Biological Sciences are enjoying the fruits of calculus.
Chapter
14
he
MATHEMATICAL REASONING
is
There are few things which we know which are not capable of
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mathematical reasoning and when these can not, it is a sign that our
knowledge of them is very small and confused and where a mathematical
reasoning can be had, it is as great a folly to make use of another,
as to grope for a thing in the dark when you have a candle stick
standing by you. ARTHENBOT
14.1 Introduction
George Boole
(1815 - 1864)
14.2 Statements
322
MATHEMATICS
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is
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When we read these sentences, we immediately decide that the first sentence is
false and the second is correct. There is no confusion regarding these. In mathematics
such sentences are called statements.
On the other hand, consider the sentence:
Women are more intelligent than men.
Some people may think it is true while others may disagree. Regarding this sentence
we cannot say whether it is always true or false . That means this sentence is ambiguous.
Such a sentence is not acceptable as a statement in mathematics.
A sentence is called a mathematically acceptable statement if it is either
true or false but not both. Whenever we mention a statement here, it is a
mathematically acceptable statement.
While studying mathematics, we come across many such sentences. Some examples
are:
Two plus two equals four.
The sum of two positive numbers is positive.
All prime numbers are odd numbers.
Of these sentences, the first two are true and the third one is false. There is no
ambiguity regarding these sentences. Therefore, they are statements.
Can you think of an example of a sentence which is vague or ambiguous? Consider
the sentence:
The sum of x and y is greater than 0
Here, we are not in a position to determine whether it is true or false, unless we
know what x and y are. For example, it is false where x = 1, y = 3 and true when
x = 1 and y = 0. Therefore, this sentence is not a statement. But the sentence:
For any natural numbers x and y, the sum of x and y is greater than 0
is a statement.
Now, consider the following sentences :
How beautiful!
Open the door.
MATHEMATICAL REASONING
323
is not a statement. The sentence is correct (true) on a Thursday but not on other
days. The same argument holds for sentences with pronouns unless a particular
person is referred to and for variable places such as here, there etc., For
example, the sentences
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Would you call this a statement? Note that the period mentioned in the sentence
above is a variable time that is any of 12 months. But we know that the sentence is
always false (irrespective of the month) since the maximum number of days in a month
can never exceed 31. Therefore, this sentence is a statement. So, what makes a sentence
a statement is the fact that the sentence is either true or false but not both.
While dealing with statements, we usually denote them by small letters p, q, r,...
For example, we denote the statement Fire is always hot by p. This is also written
as
p: Fire is always hot.
Example 1 Check whether the following sentences are statements. Give reasons for
your answer.
(i) 8 is less than 6.
(iii) The sun is a star.
(v) There is no rain without clouds.
Solution (i) This sentence is false because 8 is greater than 6. Hence it is a statement.
(ii) This sentence is also false since there are sets which are not finite. Hence it is
a statement.
(iii) It is a scientifically established fact that sun is a star and, therefore, this sentence
is always true. Hence it is a statement.
(iv) This sentence is subjective in the sense that for those who like mathematics, it
may be fun but for others it may not be. This means that this sentence is not always
true. Hence it is not a statement.
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MATHEMATICS
he
The above examples show that whenever we say that a sentence is a statement
we should always say why it is so. This why of it is more important than the answer.
EXERCISE 14.1
1.
Which of the following sentences are statements? Give reasons for your answer.
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2.
Give three examples of sentences which are not statements. Give reasons for the
answers.
We now look into method for producing new statements from those that we already
have. An English mathematician, George Boole discussed these methods in his book
The laws of Thought in 1854. Here, we shall discuss two techniques.
As a first step in our study of statements, we look at an important technique that
we may use in order to deepen our understanding of mathematical statements. This
technique is to ask not only what it means to say that a given statement is true but also
what it would mean to say that the given statement is not true.
the statement.
Let us consider the statement:
MATHEMATICAL REASONING
325
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Note While forming the negation of a statement, phrases like, It is not the
$
case or It is false that are also used.
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7 is rational.
Solution (i) This statement says that in a rectangle, both the diagonals have the same
length. This means that if you take any rectangle, then both the diagonals have the
same length. The negation of this statement is
It is false that both the diagonals in a rectangle have the same length
This means the statement
There is atleast one rectangle whose both diagonals do not
have the same length.
(ii) The negation of the statement in (ii) may also be written as
It is not the case that 7 is rational.
7 is not rational.
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MATHEMATICS
Example 3 Write the negation of the following statements and check whether the
resulting statements are true,
(i) Australia is a continent.
(ii) There does not exist a quadrilateral which has all its sides equal.
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Solution (i)
MATHEMATICAL REASONING
327
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something wrong with the wiring. That means the given statement is actually made up
of two smaller statements:
q: There is something wrong with the bulb.
r: There is something wrong with the wiring.
connected by or
Now, suppose two statements are given as below:
p: 7 is an odd number.
q: 7 is a prime number.
These two statements can be combined with and
r: 7 is both odd and prime number.
This is a compound statement.
This leads us to the following definition:
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(iii) All rational numbers are real and all real numbers are complex.
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MATHEMATICS
p: 0 is a positive number.
q: 0 is a negative number.
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Example 5 Find the component statements of the following and check whether they
are true or not.
(i) A square is a quadrilateral and its four sides equal.
(ii) All prime numbers are either even or odd.
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(v)
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(iii) A person who has taken Mathematics or Computer Science can go for
MCA.
p: A square is a quadrilateral.
We know that both these statements are true. Here the connecting word is and.
(ii)
Both these statements are false and the connecting word is or.
(iii) The component statements are
Both these statements are true. Here the connecting word is or.
(iv) The component statements are
The first statement is true but the second is false. Here the connecting word is and.
(v)
MATHEMATICAL REASONING
p:
q:
329
2 is a rational number.
2 is an irrational number.
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The first statement is false and second is true. Here the connecting word is or.
(vi) The component statements are
p: 24 is a multiple of 2.
q: 24 is a multiple of 4.
r: 24 is a multiple of 8.
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EXERCISE 14.2
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All the three statements are true. Here the connecting words are and.
Thus, we observe that compound statements are actually made-up of two or more
statements connected by the words like and, or, etc. These words have special
meaning in mathematics. We shall discuss this mattter in the following section.
(iii)
(iv)
(iii)
Some of the connecting words which are found in compound statements like And,
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MATHEMATICS
Or, etc. are often used in Mathematical Statements. These are called connectives.
When we use these compound statements, it is necessary to understand the role of
these words. We discuss this below.
14.4.1 The word And Let us look at a compound statement with And.
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r: 42 is divisible by 6.
s: 42 is divisible by 7.
Here, we know that the first is false while the other two are true.
We have the following rules regarding the connective And
1.
2.
statements is false (this includes the case that some of its component
statements are false or all of its component statements are false).
(iii) All living things have two legs and two eyes.
MATHEMATICAL REASONING
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Both these statements are true, therefore, the compound statement is true.
(ii) The component statements are
p: 0 is less than every positive integer.
q: 0 is less than every negative integer.
The second statement is false. Therefore, the compound statement is false.
(iii) The two component statements are
p: All living things have two legs.
q: All living things have two eyes.
Both these statements are false. Therefore, the compound statement is false.
Now, consider the following statement.
p: A mixture of alcohol and water can be separated by chemical methods.
This sentence cannot be considered as a compound statement with And. Here the
word And refers to two things alcohol and water.
This leads us to an important note.
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Note Do not think that a statement with And is always a compound statement
$
as shown in the above example. Therefore, the word And is not used as a connective.
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MATHEMATICS
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Solution (i) Here Or is inclusive since a person can have both a passport and a
voter registration card to enter a country.
(ii) Here also Or is inclusive since school is closed on holiday as well as on
Sunday.
(iii) Here Or is exclusive because it is not possible for two lines to intersect
and parallel together.
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(iv) Here also Or is exclusive because a student cannot take both French and
Sanskrit.
2.
r: 125 is a multiple of 8.
Both q and r are false. Therefore, the compound statement p is false.
MATHEMATICAL REASONING
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Example 8 Identify the type of Or used in the following statements and check
whether the statements are true or false:
(i)
2 is a rational number or an irrational number.
(ii) To enter into a public library children need an identity card from the school
or a letter from the school authorities.
p: 2 is a rational number.
q: 2 is an irrational number.
Here, we know that the first statement is false and the second is true and Or is
exclusive. Therefore, the compound statement is true.
(ii) The component statements are
p: To get into a public library children need an identity card.
q: To get into a public library children need a letter from the school authorities.
Children can enter the library if they have either of the two, an identity card or the
letter, as well as when they have both. Therefore, it is inclusive Or the compound
statement is also true when children have both the card and the letter.
(iii) Here Or is exclusive. When we look at the component statements, we get that
the statement is true.
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MATHEMATICS
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14.4.3 Quantifiers Quantifiers are phrases like, There exists and For all.
Another phrase which appears in mathematical statements is there exists. For example,
consider the statement. p: There exists a rectangle whose all sides are equal. This
means that there is atleast one rectangle whose all sides are equal.
A word closely connected with there exists is for every (or for all). Consider
a statement.
p is an irrational number.
This means that if S denotes the set of all prime numbers, then for all the members p of
p is an irrational number.
is
the set S,
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EXERCISE 14.3
1. For each of the following compound statements first identify the connecting words
and then break it into component statements.
(i) All rational numbers are real and all real numbers are not complex.
(ii) Square of an integer is positive or negative.
(iii) The sand heats up quickly in the Sun and does not cool down fast at night.
(iv) x = 2 and x = 3 are the roots of the equation 3x2 x 10 = 0.
MATHEMATICAL REASONING
335
2. Identify the quantifier in the following statements and write the negation of the
statements.
(i) There exists a number which is equal to its square.
(ii)
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To apply for a driving licence, you should have a ration card or a passport.
(iii)
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(ii)
14.5 Implications
In this Section, we shall discuss the implications of if-then, only if and if and only if .
The statements with if-then are very common in mathematics. For example,
consider the statement.
r: If you are born in some country, then you are a citizen of that country.
When we look at this statement, we observe that it corresponds to two statements p
and q given by
p : you are born in some country.
q : you are citizen of that country.
Then the sentence if p then q says that in the event if p is true, then q must be true.
One of the most important facts about the sentence if p then q is that it does
not say any thing (or places no demand) on q when p is false. For example, if you are
not born in the country, then you cannot say anything about q. To put it in other words
not happening of p has no effect on happening of q.
Another point to be noted for the statement if p then q is that the statement
does not imply that p happens.
There are several ways of understanding if p then q statements. We shall
illustrate these ways in the context of the following statement.
r: If a number is a multiple of 9, then it is a multiple of 3.
Let p and q denote the statements
p : a number is a multiple of 9.
q: a number is a multiple of 3.
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MATHEMATICS
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other statements which can be formed from a given statement with if-then.
For example, let us consider the following if-then statement.
If the physical environment changes, then the biological environment changes.
Then the contrapositive of this statement is
If the biological environment does not change, then the physical environment
does not change.
Note that both these statements convey the same meaning.
To understand this, let us consider more examples.
(ii) If you are born in India, then you are a citizen of India.
(ii) If you are not a citizen of India, then you were not born in India.
MATHEMATICAL REASONING
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Example 11 For each of the following compound statements, first identify the
corresponding component statements. Then check whether the statements are
true or not.
(i) If a triangle ABC is equilateral, then it is isosceles.
(ii) If a and b are integers, then ab is a rational number.
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MATHEMATICS
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Example 12 Given below are two pairs of statements. Combine these two statements
using if and only if .
(i) p: If a rectangle is a square, then all its four sides are equal.
q: If all the four sides of a rectangle are equal, then the rectangle is a
square.
(ii) p: If the sum of digits of a number is divisible by 3, then the number is
divisible by 3.
q: If a number is divisible by 3, then the sum of its digits is divisible by 3.
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Solution (i) A rectangle is a square if and only if all its four sides are equal.
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(ii) A number is divisible by 3 if and only if the sum of its digits is divisible by 3.
EXERCISE 14.4
1. Rewrite the following statement with if-then in five different ways conveying
the same meaning.
If a natural number is odd, then its square is also odd.
(ii)
If the two lines are parallel, then they do not intersect in the same plane.
(iii)
(iv)
(v)
(ii)
(iii)
(iv)
MATHEMATICAL REASONING
339
4. Given statements in (a) and (b). Identify the statements given below as
contrapositive or converse of each other.
If you live in Delhi, then you have winter clothes.
(i)
If you do not have winter clothes, then you do not live in Delhi.
(ii)
(b)
(i)
(ii)
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In this Section, we will discuss when a statement is true. To answer this question, one
must answer all the following questions.
What does the statement mean? What would it mean to say that this statement is
true and when this statement is not true?
The answer to these questions depend upon which of the special words and
phrases and, or, and which of the implications if and only, if-then, and which
of the quantifiers for every, there exists, appear in the given statement.
Here, we shall discuss some techniques to find when a statement is valid.
We shall list some general rules for checking whether a statement is true or not.
Rule 1 If p and q are mathematical statements, then in order to show that the
statement p and q is true, the following steps are followed.
Step-1 Show that the statement p is true.
Step-2 Show that the statement q is true.
If p and q are mathematical statements , then in order to show that the statement
p or q is true, one must consider the following.
Case 1 By assuming that p is false, show that q must be true.
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MATHEMATICS
In order to prove the statement if p then q we need to show that any one of the
following case is true.
Case 1 By assuming that p is true, prove that q must be true.(Direct method)
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q : xy is odd
To check the validity of the given statement, we apply Case 1 of Rule 3. That is
assume that if p is true, then q is true.
p is true means that x and y are odd integers. Then
x = 2m + 1, for some integer m. y = 2n + 1, for some integer n. Thus
xy = (2m + 1) (2n + 1)
= 2(2mn + m + n) + 1
This shows that xy is odd. Therefore, the given statement is true.
Suppose we want to check this by using Case 2 of Rule 3, then we will proceed
as follows.
We assume that q is not true. This implies that we need to consider the negation
of the statement q. This gives the statement
q : Product xy is even.
This is possible only if either x or y is even. This shows that p is not true. Thus we
have shown that
q p
$Note
Example 14 Check whether the following statement is true or false by proving its
contrapositive. If x, y such that xy is odd, then both x and y are odd.
Solution Let us name the statements as below
MATHEMATICAL REASONING
341
p : xy is odd.
q : both x and y are odd.
We have to check whether the statement p q is true or not, that is, by checking
its contrapositive statement i.e., q p
he
Now q : It is false that both x and y are odd. This implies that x (or y) is even.
Then x = 2n for some integer n.
is
Therefore, xy = 2ny for some integer n. This shows that xy is even. That is p is true.
Thus, we have shown that q p and hence the given statement is true.
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Now what happens when we combine an implication and its converse? Next, we
shall discuss this.
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7 is irrational
Solution In this method, we assume that the given statement is false. That is
we assume that
such that
a
7 = , where a and b have no common factors. Squaring the equation,
b
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MATHEMATICS
a2
a2 = 7b2 7 divides a. Therefore, there exists an integer c such
b2
that a = 7c. Then a2 = 49c2 and a2 = 7b2
Hence, 7b2 = 49c2 b2 = 7c2 7 divides b. But we have already shown that
7 divides a. This implies that 7 is a common factor of both of a and b which contradicts
our earlier assumption that a and b have no common factors. This shows that the
he
we get 7 =
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Example 16 By giving a counter example, show that the following statement is false.
If n is an odd integer, then n is prime.
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Solution The given statement is in the form if p then q we have to show that this is
false. For this purpose we need to show that if p then q. To show this we look for an
odd integer n which is not a prime number. 9 is one such number. So n = 9 is a counter
example. Thus, we conclude that the given statement is false.
In the above, we have discussed some techniques for checking whether a statement
is true or not.
Note In mathematics, counter examples are used to disprove the statement.
$
However, generating examples in favour of a statement do not provide validity of
the statement.
EXERCISE 14.5
MATHEMATICAL REASONING
Which of the following statements are true and which are false? In each case
give a valid reason for saying so.
(i) p: Each radius of a circle is a chord of the circle.
(ii) q: The centre of a circle bisects each chord of the circle.
(iii) r: Circle is a particular case of an ellipse.
(iv) s: If x and y are integers such that x > y, then x < y.
(v)
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t : 11 is a rational number.
Miscellaneous Examples
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Solution Or used in the given statement is inclusive because it is possible that it rains
and you are in the river.
The component statements of the given statement are
p : you are wet when it rains.
q : You are wet when you are in a river.
Here both the component statements are true and therefore, the compound statement
is true.
Example 18 Write the negation of the following statements:
(i) p: For every real number x, x2 > x.
(ii) q: There exists a rational number x such that x2 = 2.
(iii) r: All birds have wings.
(iv) s: All students study mathematics at the elementary level.
Solution (i) The negation of p is It is false that p is which means that the condition
x2 > x does not hold for all real numbers. This can be expressed as
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MATHEMATICS
(iv) The negation of the given statement is s: There exists a student who does not
study mathematics at the elementary level.
Example 19 Using the words necessary and sufficient rewrite the statement The
integer n is odd if and only if n2 is odd. Also check whether the statement is true.
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Solution The necessary and sufficient condition that the integer n be odd is n2 must be
odd. Let p and q denote the statements
p : the integer n is odd.
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q : n2 is odd.
To check the validity of p if and only if q, we have to check whether if p then q
and if q then p is true.
Case 1 If p, then q
If p, then q is the statement:
If the integer n is odd, then n2 is odd. We have to check whether this statement is
true. Let us assume that n is odd. Then n = 2k + 1 when k is an integer. Thus
n2 = (2k + 1)2
= 4k2 + 4k + 1
Therefore, n2 is one more than an even number and hence is odd.
Case 2 If q, then p
If q, then p is the statement
If n is an integer and n2 is odd, then n is odd.
We have to check whether this statement is true. We check this by contrapositive
method. The contrapositive of the given statement is:
If n is an even integer, then n2 is an even integer
n is even implies that n = 2k for some k. Then n2 = 4k2. Therefore, n2 is even.
Example 20 For the given statements identify the necessary and sufficient conditions.
t: If you drive over 80 km per hour, then you will get a fine.
Solution Let p and q denote the statements:
p : you drive over 80 km per hour.
q : you will get a fine.
The implication if p, then q indicates that p is sufficient for q. That is driving over
80 km per hour is sufficient to get a fine.
Here the sufficient condition is driving over 80 km per hour:
Similarly, if p, then q also indicates that q is necessary for p. That is
MATHEMATICAL REASONING
345
When you drive over 80 km per hour, you will necessarily get a fine.
Here the necessary condition is getting a fine.
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1.
6.
7.
Check the validity of the statements given below by the method given against it.
(i) p: The sum of an irrational number and a rational number is irrational (by
contradiction method).
(ii) q: If n is a real number with n > 3, then n2 > 9 (by contradiction method).
Write the following statement in five different ways, conveying the same meaning.
p: If a triangle is equiangular, then it is an obtuse angled triangle.
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MATHEMATICS
Summary
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Historical Note
The first treatise on logic was written by Aristotle (384 B.C.-322 B.C.). It
was a collection of rules for deductive reasoning which would serve as a basis
for the study of every branch of knowledge. Later, in the seventeenth century,
German mathematician G. W. Leibnitz (1646 1716) conceived the idea of using
symbols in logic to mechanise the process of deductive reasoning. His idea was
realised in the nineteenth century by the English mathematician George Boole
(18151864) and Augustus De Morgan (18061871) , who founded the modern
subject of symbolic logic.
Chapter
15
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STATISTICS
15.1 Introduction
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MATHEMATICS
x=
1
n
xi
i =1
Also, the median is obtained by first arranging the data in ascending or descending
order and applying the following rule.
th
he
n +1
If the number of observations is odd, then the median is
observation.
th
and
is
n
If the number of observations is even, then median is the mean of
2
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+ 1 observations.
2
We find that the mean and median of the runs scored by both the batsmen A and
B are same i.e., 53. Can we say that the performance of two players is same? Clearly
No, because the variability in the scores of batsman A is from 0 (minimum) to 117
(maximum). Whereas, the range of the runs scored by batsman B is from 46 to 60.
Let us now plot the above scores as dots on a number line. We find the following
diagrams:
For batsman A
For batsman B
Fig 15.1
Fig 15.2
We can see that the dots corresponding to batsman B are close to each other and
are clustering around the measure of central tendency (mean and median), while those
corresponding to batsman A are scattered or more spread out.
Thus, the measures of central tendency are not sufficient to give complete
information about a given data. Variability is another factor which is required to be
studied under statistics. Like measures of central tendency we want to have a
single number to describe variability. This single number is called a measure of
dispersion. In this Chapter, we shall learn some of the important measures of dispersion
and their methods of calculation for ungrouped and grouped data.
STATISTICS
349
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The dispersion or scatter in a data is measured on the basis of the observations and the
types of the measure of central tendency, used there. There are following measures of
dispersion:
(i) Range, (ii) Quartile deviation, (iii) Mean deviation, (iv) Standard deviation.
In this Chapter, we shall study all of these measures of dispersion except the
quartile deviation.
15.3 Range
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Recall that, in the example of runs scored by two batsmen A and B, we had some idea
of variability in the scores on the basis of minimum and maximum runs in each series.
To obtain a single number for this, we find the difference of maximum and minimum
values of each series. This difference is called the Range of the data.
In case of batsman A, Range = 117 0 = 117 and for batsman B, Range = 60 46 = 14.
Clearly, Range of A > Range of B. Therefore, the scores are scattered or dispersed in
case of A while for B these are close to each other.
Thus, Range of a series = Maximum value Minimum value.
The range of data gives us a rough idea of variability or scatter but does not tell
about the dispersion of the data from a measure of central tendency. For this purpose,
we need some other measure of variability. Clearly, such measure must depend upon
the difference (or deviation) of the values from the central tendency.
The important measures of dispersion, which depend upon the deviations of the
observations from a central tendency are mean deviation and standard deviation. Let
us discuss them in detail.
Recall that the deviation of an observation x from a fixed value a is the difference
x a. In order to find the dispersion of values of x from a central value a , we find the
deviations about a. An absolute measure of dispersion is the mean of these deviations.
To find the mean, we must obtain the sum of the deviations. But, we know that a
measure of central tendency lies between the maximum and the minimum values of
the set of observations. Therefore, some of the deviations will be negative and some
positive. Thus, the sum of deviations may vanish. Moreover, the sum of the deviations
from mean ( x ) is zero.
Also
Sum of deviations
0
Mean of deviations = Number of observations = n = 0
Thus, finding the mean of deviations about mean is not of any use for us, as far
as the measure of dispersion is concerned.
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MATHEMATICS
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M.D.(a) =
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Remark Mean deviation may be obtained from any measure of central tendency.
However, mean deviation from mean and median are commonly used in statistical
studies.
Let us now learn how to calculate mean deviation about mean and mean deviation
about median for various types of data
15.4.1 Mean deviation for ungrouped data Let n observations be x1, x2, x3, ...., xn.
The following steps are involved in the calculation of mean deviation about mean or
median:
Step 1 Calculate the measure of central tendency about which we are to find the mean
deviation. Let it be a.
Step 2 Find the deviation of each xi from a, i.e., x1 a, x2 a, x3 a,. . . , xn a
Step 3 Find the absolute values of the deviations, i.e., drop the minus sign (), if it is
there, i.e., x1 a , x2 a , x3 a , ...., xn a
Step 4 Find the mean of the absolute values of the deviations. This mean is the mean
deviation about a, i.e.,
n
M.D.(a ) =
Thus
and
xi a
i =1
M.D. ( x ) =
M.D. (M) =
1
n
1
n
xi x
i =1
, where x = Mean
xi M , where M = Median
i =1
STATISTICS
351
Note In this Chapter, we shall use the symbol M to denote median unless stated
$
otherwise.Let us now illustrate the steps of the above method in following examples.
6 + 7 + 10 + 12 + 13 + 4 + 8 + 12
72
=
= 9
8
8
is
x=
he
Example 1 Find the mean deviation about the mean for the following data:
6, 7, 10, 12, 13, 4, 8, 12
bl
Step 2 The deviations of the respective observations from the mean x , i.e., xi x are
6 9, 7 9, 10 9, 12 9, 13 9, 4 9, 8 9, 12 9,
or
3, 2, 1, 3, 4, 5, 1, 3
no N
C
tt E
o R
be T
re
pu
M.D. ( x ) =
xi x
i =1
3 + 2 + 1 + 3 + 4 + 5 + 1 + 3 22
=
= 2.75
8
8
Note Instead of carrying out the steps every time, we can carry on calculation,
$
step-wise without referring to steps.
Example 2 Find the mean deviation about the mean for the following data :
12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5
Solution We have to first find the mean ( x ) of the given data
x =
1 20
200
xi =
= 10
20 i =1
20
The respective absolute values of the deviations from mean, i.e., xi x are
2, 7, 8, 7, 6, 1, 7, 9, 10, 5, 2, 7, 8, 7, 6, 1, 7, 9, 10, 5
352
MATHEMATICS
20
Therefore
xi x
and
M.D. ( x ) =
= 124
i =1
he
124
= 6.2
20
Example 3 Find the mean deviation about the median for the following data:
3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21.
Now
th
or 6th observation = 9
bl
11 + 1
Median =
is
Solution Here the number of observations is 11 which is odd. Arranging the data into
ascending order, we have 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21
The absolute values of the respective deviations from the median, i.e., xi M are
no N
C
tt E
o R
be T
re
pu
6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12
11
xi M = 58
Therefore
and
i =1
M.D. ( M ) =
1 11
1
xi M = 58 = 5.27
11 i =1
11
15.4.2 Mean deviation for grouped data We know that data can be grouped into
two ways :
(a) Discrete frequency distribution,
(b) Continuous frequency distribution.
Let us discuss the method of finding mean deviation for both types of the data.
(a) Discrete frequency distribution Let the given data consist of n distinct values
x1, x2, ..., xn occurring with frequencies f1, f2 , ..., fn respectively. This data can be
represented in the tabular form as given below, and is called discrete frequency
distribution:
x : x1 x2 x3 ... xn
f : f1
f2
f3 ... fn
STATISTICS
353
x=
xi fi
i =1
n
fi
1
N
xi fi ,
i =1
he
i =1
x f
i =1
frequencies fi and N =
f
i =1
is
where
bl
Then, we find the deviations of observations xi from the mean x and take their
absolute values, i.e., xi x for all i =1, 2,..., n.
no N
C
tt E
o R
be T
re
pu
After this, find the mean of the absolute values of the deviations, which is the
required mean deviation about the mean. Thus
n
M.D. ( x ) =
f
i =1
xi x
n
f
i =1
1 n
=
f i xi x
N i =1
(ii) Mean deviation about median To find mean deviation about median, we find the
median of the given discrete frequency distribution. For this the observations are arranged
in ascending order. After this the cumulative frequencies are obtained. Then, we identify
, where
2
N is the sum of frequencies. This value of the observation lies in the middle of the data,
therefore, it is the required median. After finding median, we obtain the mean of the
absolute values of the deviations from median.Thus,
M.D.(M) =
1
N
f
i =1
xi M
Example 4 Find mean deviation about the mean for the following data :
xi
2
5
6
8
10
12
fi
2
8
10 7
8
5
Solution Let us make a Table 15.1 of the given data and append other columns after
calculations.
354
MATHEMATICS
Table 15.1
fi
f ix i
xi x
f i xi x
5.5
11
40
2.5
20
10
60
1.5
15
56
0.5
3.5
10
80
2.5
20
12
60
4.5
40
300
6
i =1
92
f i xi = 300 ,
i =1
xi x = 92
no N
C
tt E
o R
be T
re
pu
i =1
22.5
bl
N = f i = 40 ,
is
he
xi
1
N
Therefore
x=
and
M. D. ( x ) =
i =1
fi xi =
1
N
1
300 = 7.5
40
6
fi
i =1
1
92 = 2.3
40
xi x =
Example 5 Find the mean deviation about the median for the following data:
xi
12
13
15
21
22
fi
Solution The given observations are already in ascending order. Adding a row
corresponding to cumulative frequencies to the given data, we get (Table 15.2).
Table 15.2
xi
12
13
15
21
22
fi
c.f.
12
14
18
23
27
30
STATISTICS
355
Median is the mean of the 15th and 16th observations. Both of these observations
lie in the cumulative frequency 18, for which the corresponding observation is 13.
Therefore, Median M =
13 + 13
= 13
Table 15.3.
Table 15.3
f i xi M
8
30
28
20
10
32
27
We have
Therefore
M. D. (M) =
i =1
i =1
1
N
fi
i =1
xi M = 149
no N
C
tt E
o R
be T
re
pu
= 30 and
is
fi
10
bl
xi M
he
Now, absolute values of the deviations from median, i.e., xi M are shown in
xi M
1
149 = 4.97.
30
40-50
17
50-60
6
(i) Mean deviation about mean While calculating the mean of a continuous frequency
distribution, we had made the assumption that the frequency in each class is centred at
its mid-point. Here also, we write the mid-point of each given class and proceed further
as for a discrete frequency distribution to find the mean deviation.
Let us take the following example.
356
MATHEMATICS
Example 6 Find the mean deviation about the mean for the following data.
Marks obtained
Number of students
14
Table 15.4
Number of
students
Mid-points
fi
xi
f ix i
xi x
30
fi xi x
is
Marks
obtained
he
Solution We make the following Table 15.4 from the given data :
15
30
60
20-30
25
75
20
60
30-40
35
280
10
80
no N
C
tt E
o R
be T
re
pu
bl
10-20
40-50
14
45
630
50-60
55
440
10
80
60-70
65
195
20
60
70-80
75
150
30
60
40
1800
Here
Therefore
x=
and
M.D. ( x ) =
i =1
fi = 40, fi xi = 1800,
N =
i =1
fi
i =1
400
xi x = 400
1 7
1800
fi xi =
= 45
N i =1
40
1 7
1
fi xi x = 400 = 10
N i =1
40
Shortcut method for calculating mean deviation about mean We can avoid the
tedious calculations of computing x by following step-deviation method. Recall that in
this method, we take an assumed mean which is in the middle or just close to it in the
data. Then deviations of the observations (or mid-points of classes) are taken from the
STATISTICS
357
he
assumed mean. This is nothing but the shifting of origin from zero to the assumed mean
on the number line, as shown in Fig 15.3
is
Fig 15.3
no N
C
tt E
o R
be T
re
pu
bl
If there is a common factor of all the deviations, we divide them by this common
factor to further simplify the deviations. These are known as step-deviations. The
process of taking step-deviations is the change of scale on the number line as shown in
Fig 15.4
Fig 15.4
The deviations and step-deviations reduce the size of the observations, so that the
computations viz. multiplication, etc., become simpler. Let, the new variable be denoted
by d i =
xi a
, where a is the assumed mean and h is the common factor. Then, the
h
fi di
i
x = a + =1
h
N
Let us take the data of Example 6 and find the mean deviation by using stepdeviation method.
358
MATHEMATICS
Take the assumed mean a = 45 and h = 10, and form the following Table 15.5.
Table 15.5
Number of Mid-points di =
students
xi 45
10
fi di
xi x
xi
10-20
15
30
20-30
25
20
30-40
35
10
40-50
14
45
50-60
55
60-70
65
70-80
75
Therefore
80
10
80
20
60
30
60
bl
400
7
fi di
x = a + i =1
h
N
= 45 +
and
60
no N
C
tt E
o R
be T
re
pu
40
60
is
fi
f i xi x
he
Marks
obtained
M .D. (x ) =
1
N
0
10 = 45
40
7
fi
i =1
xi x =
400
= 10
40
Note The step deviation method is applied to compute x . Rest of the procedure
$
is same.
(ii) Mean deviation about median The process of finding the mean deviation about
median for a continuous frequency distribution is similar as we did for mean deviation
about the mean. The only difference lies in the replacement of the mean by median
while taking deviations.
Let us recall the process of finding median for a continuous frequency distribution.
The data is first arranged in ascending order. Then, the median of continuous
frequency distribution is obtained by first identifying the class in which median lies
(median class) and then applying the formula
STATISTICS
359
N
C
2
Median = l +
h
f
where median class is the class interval whose cumulative frequency is just greater
N
, N is the sum of frequencies, l, f, h and C are, respectively the lower
2
limit , the frequency, the width of the median class and C the cumulative frequency of
the class just preceding the median class. After finding the median, the absolute values
he
than or equal to
M.D. (M) =
bl
n
fi xi M
N i =1
The process is illustrated in the following example:
Then
is
of the deviations of mid-point xi of each class from the median i.e., xi M are obtained.
no N
C
tt E
o R
be T
re
pu
Example 7 Calculate the mean deviation about median for the following data :
Class
0-10
Frequency
15
16
Solution Form the following Table 15.6 from the given data :
Table 15.6
xi Med.
f i xi Med.
23
138
13
15
13
91
15
28
25
45
30-40
16
44
35
112
40-50
48
45
17
68
50-60
50
55
27
54
Class
Frequency
Mid-points
fi
Cumulative
frequency
(c.f.)
0-10
10-20
20-30
50
xi
508
360
MATHEMATICS
th
N
or 25th item is 20-30. Therefore, 2030 is the median
2
he
N
C
h
Median = l + 2
f
25 13
10 = 20 + 8 = 28
15
Thus, Mean deviation about median is given by
Median = 20 +
1
N
fi
i =1
xi M =
1
508 = 10.16
50
no N
C
tt E
o R
be T
re
pu
M.D. (M) =
bl
Therefore,
is
EXERCISE 15.1
Find the mean deviation about the mean for the data in Exercises 1 and 2.
1. 4, 7, 8, 9, 10, 12, 13, 17
2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Find the mean deviation about the median for the data in Exercises 3 and 4.
3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Find the mean deviation about the mean for the data in Exercises 5 and 6.
5. x i
5
10
15
20
25
fi
6. x i
fi
10
30
50
70
90
24
28
16
Find the mean deviation about the median for the data in Exercises 7 and 8.
7. x i
10
12
15
fi
15
21
27
30
35
8. x i
fi
STATISTICS
361
no N
C
tt E
o R
be T
re
pu
bl
is
he
Find the mean deviation about the mean for the data in Exercises 9 and 10.
9. Income 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800
per day
Number
4
8
9
10
7
5
4
3
of persons
10. Height
95-105
105-115 115-125 125-135 135-145 145-155
in cms
Number of
9
13
26
30
12
10
boys
11. Find the mean deviation about median for the following data :
Marks
0-10
10-20
20-30
30-40
40-50
50-60
Number of
6
8
14
16
4
2
Girls
12. Calculate the mean deviation about median age for the age distribution of 100
persons given below:
Age
16-20 21-25 26-30 31-35 36-40 41-45 46-50 51-55
Number
5
6
12
14
26
12
16
9
[Hint Convert the given data into continuous frequency distribution by subtracting 0.5
from the lower limit and adding 0.5 to the upper limit of each class interval]
15.4.3 Limitations of mean deviation In a series, where the degree of variability is
very high, the median is not a representative central tendency. Thus, the mean deviation
about median calculated for such series can not be fully relied.
The sum of the deviations from the mean (minus signs ignored) is more than the
sum of the deviations from median. Therefore, the mean deviation about the mean is
not very scientific.Thus, in many cases, mean deviation may give unsatisfactory results.
Also mean deviation is calculated on the basis of absolute values of the deviations and
therefore, cannot be subjected to further algebraic treatment. This implies that we
must have some other measure of dispersion. Standard deviation is such a measure of
dispersion.
Recall that while calculating mean deviation about mean or median, the absolute values
of the deviations were taken. The absolute values were taken to give meaning to the
mean deviation, otherwise the deviations may cancel among themselves.
Another way to overcome this difficulty which arose due to the signs of deviations,
is to take squares of all the deviations. Obviously all these squares of deviations are
362
MATHEMATICS
non-negative. Let x1, x2, x3, ..., xn be n observations and x be their mean. Then
(x1 x)2
(x2
x)2
.......
(xn x)2
(xi x)2 .
i 1
If this sum is zero, then each ( xi x ) has to be zero. This implies that there is no
n
If
(x x)
i =1
he
is small , this indicates that the observations x1, x2, x3,...,xn are
(x x)
i =1
bl
is
close to the mean x and therefore, there is a lower degree of dispersion. On the
contrary, if this sum is large, there is a higher degree of dispersion of the observations
i
is a reasonable indicator
no N
C
tt E
o R
be T
re
pu
(x x)
i =1
(y
i =1
= 2
15 (15 + 1) (30 + 1)
= 5 16 31 = 2480
6
n (n + 1) (2n + 1)
. Here n = 15)
6
STATISTICS
If
(x x)
i =1
363
he
will tend to say that the set A of six observations has a lesser dispersion about the mean
than the set B of 31 observations, even though the observations in set A are more
scattered from the mean (the range of deviations being from 25 to 25) than in the set
B (where the range of deviations is from 15 to 15).
This is also clear from the following diagrams.
Fig 15.5
no N
C
tt E
o R
be T
re
pu
bl
is
Fig 15.6
Thus, we can say that the sum of squares of deviations from the mean is not a proper
measure of dispersion. To overcome this difficulty we take the mean of the squares of
1 n
the deviations, i.e., we take
( xi x ) 2 . In case of the set A, we have
n i =1
1
1
1750 = 291.67 and in case of the set B, it is
2480 = 80.
6
31
This indicates that the scatter or dispersion is more in set A than the scatter or dispersion
in set B, which confirms with the geometrical representation of the two sets.
Mean =
1
( xi x ) 2 as a quantity which leads to a proper measure
of dispersion. This number, i.e., mean of the squares of the deviations from mean is
called the variance and is denoted by 2 (read as sigma square). Therefore, the
variance of n observations x1, x2,..., xn is given by
364
MATHEMATICS
2=
1 n
( xi x ) 2
n i =1
he
15.5.1 Standard Deviation In the calculation of variance, we find that the units of
individual observations xi and the unit of their mean x are different from that of variance,
since variance involves the sum of squares of (xi x ). For this reason, the proper
measure of dispersion about the mean of a set of observations is expressed as positive
square-root of the variance and is called standard deviation. Therefore, the standard
deviation, usually denoted by , is given by
is
1 n
( xi x ) 2
n i =1
... (1)
bl
Let us take the following example to illustrate the calculation of variance and
hence, standard deviation of ungrouped data.
no N
C
tt E
o R
be T
re
pu
Solution From the given data we can form the following Table 15.7. The mean is
calculated by step-deviation method taking 14 as assumed mean. The number of
observations is n = 10
Table 15.7
xi
di =
xi 14
2
(x i x )
81
49
10
25
12
14
16
18
20
25
22
49
24
5
5
81
330
STATISTICS
365
d
Mean x = assumed mean +
1
Variance ( ) =
n
2
and
h =
14 +
5
2 = 15
10
10
( xi x )2 = 101 330 = 33
he
Therefore
i =1
i =1
f1,
f2,
f3 ,. . . , fn
bl
f:
is
... (2)
no N
C
tt E
o R
be T
re
pu
where N = fi .
i =1
Example 9 Find the variance and standard deviation for the following data:
xi
11
17
20
24
32
fi
xi
fi
fi xi
xi x
( xi x ) 2
4
8
11
17
20
24
32
3
5
9
5
4
3
1
30
12
40
99
85
80
72
32
420
10
6
3
3
6
10
18
100
36
9
9
36
100
324
f i ( xi x ) 2
300
180
81
45
144
300
324
1374
366
MATHEMATICS
N = 30,
i =1
i =1
fi xi = 420, fi ( xi x )
= 1374
1
420 = 14
30
1
variance ( ) =
N
2
fi (xi
he
x )2
i =1
is
Hence
i =1
x=
1
1374 = 45.8
30
bl
Therefore
fi xi
and
no N
C
tt E
o R
be T
re
pu
1
N
fi ( xi
i =1
x)2 ,
i =1
f i ( xi x ) 2 =
1
= N
fi xi 2 +
1
= N
fi xi 2 +
i =1
n
i =1
1
N
fi ( xi 2 +
i =1
x 2 fi
i =1
2x
i =1
x 2 2 x xi )
fi xi
n
n
x 2 f i 2 x xi fi
i =1
i =1
STATISTICS
i =1
2
f i xi + x 2 2 x 2 =
n
fi xi
1 n
2
2 =
fi xi i =1N
N i 1
1
N
1
N
fi xi 2
i =1
he
or
n
n
= 1 N fi xi 2 f i xi
N 2 i =1
i =1
N fi xi f i xi
i =1
i =1
n
is
1 n
1 n
2
Here
or
x
f
x
xi fi = Nx
=
f
x
x
x
x
+
N
2
.
N
i i
i i
N i =1
N i =1
i =1
bl
367
... (3)
no N
C
tt E
o R
be T
re
pu
Example 10 Calculate the mean, variance and standard deviation for the following
distribution :
Class
Frequency
12
15
Solution From the given data, we construct the following Table 15.9.
Table 15.9
Class
Frequency Mid-point
(f i )
(x i)
f ix i
(xi x ) 2
f i(x i x ) 2
30-40
35
105
729
2187
40-50
45
315
289
2023
50-60
12
55
660
49
588
60-70
15
65
975
135
70-80
75
600
169
1352
80-90
85
255
529
1587
90-100
95
190
1089
2178
50
3100
10050
Mean x =
1
N
fx
i =1
1
2
Variance =
N
( )
=
and
3100
= 62
50
i i
fi (xi x )2
i =1
he
Thus
MATHEMATICS
1
10050 = 201
50
is
368
13
18
23
fi
10
15
10
bl
xi
no N
C
tt E
o R
be T
re
pu
xi
fi
f ix i
x i2
f ix i2
21
63
10
80
64
640
13
15
195
169
2535
18
10
180
324
3240
23
138
529
3174
48
614
9652
1
N
N fi xi ( f i xi )
1
48
48 9652 (614)2
1
463296 376996
48
STATISTICS
369
1
293.77 = 6.12
48
Therefore, Standard deviation ( ) = 6.12
=
1
times (h being the
h
is
he
15.5.4. Shortcut method to find variance and standard deviation Sometimes the
values of xi in a discrete distribution or the mid points xi of different classes in a
continuous distribution are large and so the calculation of mean and variance becomes
tedious and time consuming. By using step-deviation method, it is possible to simplify
the procedure.
yi =
xi A
or xi = A + hyi
... (1)
no N
C
tt E
o R
be T
re
pu
i.e.
bl
fx
i
x =
We know that
i =1
... (2)
f ( A + hy )
i
i =1
n
n
n
n
1
fi A + h fi yi = N A fi + h fi yi
N i =1
i =1
i =1
i =1
= A.N +h
N
Thus
Now
fi yi
i =1
because
fi = N
i =1
x=A+h y
... (3)
2
2
Variance of the variable x, x = fi ( xi x )
N i =1
1
=
N
fi (A + hyi A h y )2
i =1
MATHEMATICS
1
N
fi h2 (yi y )2
h2
N
fi (yi y )2 = h
i =1
n
i =1
he
370
2
2
x2 = h y
i.e.
x = h y
From (3) and (4), we have
n
N fi yi fi yi
i =1
i =1
... (5)
bl
h
x =
N
... (4)
is
or
Let us solve Example 11 by the short-cut method and using formula (5)
no N
C
tt E
o R
be T
re
pu
Examples 12 Calculate mean, Variance and Standard Deviation for the following
distribution.
Classes
Frequency
12
15
Class
Frequency
fi
Mid-point yi=
xi 65
10
y i2
fi y i
f i y i2
xi
30-40
35
27
40-50
45
14
28
50-60
12
55
12
12
60-70
15
65
70-80
75
80-90
85
12
9 0-100
95
18
15
105
N=50
STATISTICS
50
2
N f i yi ( fi yi
(10 )2
50 105 (15) 2
2
(50)
1
[5250 225] = 201
25
is
h2
N2
50
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EXERCISE 15.2
bl
2 =
Variance
fi yi h = 65 15 10 = 62
he
x= A+
Therefore
371
Find the mean and variance for each of the data in Exercies 1 to 5.
1. 6, 7, 10, 12, 13, 4, 8, 12
2. First n natural numbers
3. First 10 multiples of 3
4.
5.
xi
10
14
18
24
28
30
fi
12
xi
92
93
97
98
102
104
109
fi
60
61
62
63
64
65
66
67
68
fi
12
29
25
12
10
Find the mean and variance for the following frequency distributions in Exercises
7 and 8.
7.
Classes
Frequencies
10
372
8.
MATHEMATICS
Classes
Frequencies
0-10
10-20
20-30
30-40
40-50
15
16
9. Find the mean, variance and standard deviation using short-cut method
15
he
No. of
children
is
Height
in cms
10. The diameters of circles (in mm) drawn in a design are given below:
No. of circles
33-36
37-40
15
17
41-44
21
45-48
49-52
bl
Diameters
22
25
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[ Hint First make the data continuous by making the classes as 32.5-36.5, 36.5-40.5,
40.5-44.5, 44.5 - 48.5, 48.5 - 52.5 and then proceed.]
In earlier sections, we have studied about some types of measures of dispersion. The
mean deviation and the standard deviation have the same units in which the data are
given. Whenever we want to compare the variability of two series with same mean,
which are measured in different units, we do not merely calculate the measures of
dispersion but we require such measures which are independent of the units. The
measure of variability which is independent of units is called coefficient of variation
(denoted as C.V.)
The coefficient of variation is defined as
C.V. =
100 , x 0 ,
where and x are the standard deviation and mean of the data.
For comparing the variability or dispersion of two series, we calculate the coefficient
of variance for each series. The series having greater C.V. is said to be more variable
than the other. The series having lesser C.V. is said to be more consistent than the
other.
STATISTICS
373
15.6.1 Comparison of two frequency distributions with same mean Let x1 and 1
be the mean and standard deviation of the first distribution, and x2 and 2 be the
mean and standard deviation of the second distribution.
1
C.V. (1st distribution) = x 100
1
and
2
C.V. (2nd distribution) = x 100
2
he
Then
is
Given x1 = x2 = x (say)
C.V. (1st distribution) =
1
x
100
... (1)
bl
Therefore
100
... (2)
x
It is clear from (1) and (2) that the two C.Vs. can be compared on the basis of values
C.V. (2nd distribution) =
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and
of 1 and 2 only.
Thus, we say that for two series with equal means, the series with greater standard
deviation (or variance) is called more variable or dispersed than the other. Also, the
series with lesser value of standard deviation (or variance) is said to be more consistent
than the other.
Let us now take following examples:
Example 13 Two plants A and B of a factory show following results about the number
of workers and the wages paid to them.
A
B
No. of workers
5000
6000
Rs 2500
Rs 2500
Variance of distribution
of wages
81
100
374
MATHEMATICS
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Solution Given
is
Example 14 Coefficient of variation of two distributions are 60 and 70, and their
standard deviations are 21 and 16, respectively. What are their arithmetic means.
x1 100
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bl
Let x1 and x2 be the means of 1st and 2nd distribution, respectively. Then
21
21
100 or x1 = 100 = 35
x1
60
Therefore
60 =
and
i.e.
70 =
x2 100
16
16
100 or x2 = 100 = 22.85
70
x2
Example 15 The following values are calculated in respect of heights and weights of
the students of a section of Class XI :
Height
Weight
Mean
Variance
162.6 cm
52.36 kg
23.1361 kg2
127.69 cm2
Can we say that the weights show greater variation than the heights?
Solution To compare the variability, we have to calculate their coefficients of variation.
Given
Therefore
Also
127.69cm = 11.3 cm
STATISTICS
375
Therefore
Standard deviation of weight = 23.1361 kg = 4.81 kg
Now, the coefficient of variations (C.V.) are given by
and
(C.V.) in weights =
Mean
11.3
100
he
Standard Deviation
100 = 6.95
162.6
4.81
52.36
100 = 9.18
is
(C.V.) in heights =
EXERCISE 15.3
From the data given below state which group is more variable, A or B?
no N
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1.
bl
Marks
2.
70-80
Group A
17
32
33
40
10
Group B
10
20
30
25
43
15
From the prices of shares X and Y below, find out which is more stable in value:
X
3.
35
54
52
53
56
58
52
50
51
49
Y 108
107
105
105
106
107
104
103
104
101
Firm A
Firm B
586
648
Rs 5253
Rs 5253
100
121
of wages
376
MATHEMATICS
4.
No. of matches
50
xi = 212 ,
i =1
50
xi2 = 902.8 ,
i =1
50
yi = 261 ,
i =1
yi2 = 1457.6
i =1
bl
50
is
5.
he
For the team B, mean number of goals scored per match was 2 with a standard
deviation 1.25 goals. Find which team may be considered more consistent?
The sum and sum of squares corresponding to length x (in cm) and weight y
(in gm) of 50 plant products are given below:
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Miscellaneous Examples
Solution Let the observations be x1, x2, ..., x20 and x be their mean. Given that
variance = 5 and n = 20. We know that
( )
Variance
1
=
n
20
(xi x ) , i.e., 5 =
2
i =1
1 20
(xi x )2
20 i =1
20
or
(xi x )2 = 100
i =1
... (1)
If each observation is multiplied by 2, and the new resulting observations are yi , then
yi = 2xi i.e., xi =
1
yi
2
1 20
1 20
1 20
y
=
x
.
2
2
i 20 i = 20 xi
n i =1
i =1
i =1
Therefore
y=
i.e.
y=2x
or
x=
1
y
2
STATISTICS
20
1
1
2 yi 2 y = 100 , i.e.,
i =1
Thus the variance of new observations =
20
(y
i =1
377
y ) 2 = 400
1
400 = 20 = 22 5
20
he
Note The reader may note that if each observation is multiplied by a constant
k, the variance of the resulting observations becomes k2 times the original variance.
Now
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or
Therefore
1+ 2 + 6 + x + y
5
22 = 9 + x + y
x + y = 13
Mean x = 4.4 =
bl
is
Example17 The mean of 5 observations is 4.4 and their variance is 8.24. If three of
the observations are 1, 2 and 6, find the other two observations.
Also
... (1)
1 5
2
variance = 8.24 = ( xi x )
n i =1
1
( 3.4 )2 + ( 2.4 )2 + (1.6 )2 + x 2 + y 2 2 4.4 (x + y ) + 2 ( 4.4 )2
5
or 41.20 = 11.56 + 5.76 + 2.56 + x2 + y2 8.8 13 + 38.72
... (2)
Therefore
x2 + y2 = 97
But from (1), we have
... (3)
x2 + y2 + 2xy = 169
From (2) and (3), we have
2xy = 72
... (4)
Subtracting (4) from (2), we get
x2 + y2 2xy = 97 72 i.e. (x y)2 = 25
or
xy= 5
... (5)
So, from (1) and (5), we get
x = 9, y = 4 when x y = 5
or
x = 4, y = 9 when x y = 5
Thus, the remaining observations are 4 and 9.
Example 18 If each of the observation x1, x2, ...,xn is increased by a, where a is a
negative or positive number, show that the variance remains unchanged.
i.e. 8.24 =
378
MATHEMATICS
Solution Let x be the mean of x1, x2, ...,xn . Then the variance is given by
1
n
(xi x )
i =1
yi = n (xi + a)
i =1
i =1
1 n
1
x
a =
+
= n
n
i =1
i =1
x +
i =1
na
=x+a
n
... (2)
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i.e.
y = x+a
Thus, the variance of the new observations
is
1
n
bl
y=
1
1 n
2
2 = n (yi y ) = n
i =1
2
1
=
n
(x + a
i =1
... (1)
he
12 =
x a) 2
(xi x )2 = 12
i =1
Thus, the variance of the new observations is same as that of the original observations.
Note We may note that adding (or subtracting) a positive number to (or from)
each observation of a group does not affect the variance.
Example 19 The mean and standard deviation of 100 observations were calculated as
40 and 5.1, respectively by a student who took by mistake 50 instead of 40 for one
observation. What are the correct mean and standard deviation?
Solution Given that number of observations (n) = 100
We know that
i.e.
x=
1 n
xi
n i =1
40 =
1 100
xi or
100 i =1
100
xi = 4000
i =1
STATISTICS
i.e.
Thus
379
correct sum
3990
=
= 39.9
100
100
1 n 2
1 n
xi 2 xi
n i =1
n i =1
Standard deviation =
Also
1
n
5.1 =
x (x )
2
i =1
Incorrect xi (40) 2
2
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i.e.
is
Correct mean =
bl
Hence
or
100
1
26.01 =
100
i =1
Incorrect
1600
i =1
Therefore
i =1
n
Correct xi = Incorrect
Now
i =1
x
i =1
(50)2 + (40)2
Therefore
=
=
Correct
(Correct mean)
161701
100
(39.9)
1617.01 1592.01
25
=5
he
= 4000 50 + 40 = 3990
380
MATHEMATICS
4.
5.
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6.
he
3.
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2.
The mean and variance of eight observations are 9 and 9.25, respectively. If six
of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.
The mean and variance of 7 observations are 8 and 16, respectively. If five of the
observations are 2, 4, 10, 12, 14. Find the remaining two observations.
The mean and standard deviation of six observations are 8 and 4, respectively. If
each observation is multiplied by 3, find the new mean and new standard deviation
of the resulting observations.
Given that x is the mean and 2 is the variance of n observations x1, x2, ...,xn.
Prove that the mean and variance of the observations ax1, ax2, ax3, ...., axn are
a x and a2 2, respectively, (a 0).
The mean and standard deviation of 20 observations are found to be 10 and 2,
respectively. On rechecking, it was found that an observation 8 was incorrect.
Calculate the correct mean and standard deviation in each of the following cases:
(i) If wrong item is omitted.
(ii) If it is replaced by 12.
The mean and standard deviation of marks obtained by 50 students of a class in
three subjects, Mathematics, Physics and Chemistry are given below:
bl
1.
Subject
7.
Mathematics
Physics
Chemistry
Mean
42
32
40.9
Standard
deviation
12
15
20
which of the three subjects shows the highest variability in marks and which
shows the lowest?
The mean and standard deviation of a group of 100 observations were found to
be 20 and 3, respectively. Later on it was found that three observations were
incorrect, which were recorded as 21, 21 and 18. Find the mean and standard
deviation if the incorrect observations are omitted.
Summary
xi x
n
M.D. (M) =
xi M
n
STATISTICS
381
fi
xi x
fi
xi M
, where N = f i
N
N
Variance and standard deviation for ungrouped data
M.D. ( x ) =
M.D. (M) =
1
1
2
(xi x ) 2 ,
=
( xi x )
n
n
Variance and standard deviation of a discrete frequency distribution
he
2 =
1
1
2
2
fi ( xi x ) ,
=
fi ( xi x )
N
N
Variance and standard deviation of a continuous frequency distribution
bl
is
2 =
2
1
1
2
N fi xi2 ( fi xi )
fi ( xi x ) ,
=
N
N
Shortcut method to find variance and standard deviation.
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2 =
2 =
2
h2
N f i yi2 ( fi yi ) ,
2
where yi =
h
N
N f i yi2 ( fi yi ) ,
2
xi A
h
100, x 0.
x
For series with equal means, the series with lesser standard deviation is more consistent
or less scattered.
Historical Note
Statistics is derived from the Latin word status which means a political
state. This suggests that statistics is as old as human civilisation. In the year 3050
B.C., perhaps the first census was held in Egypt. In India also, about 2000 years
ago, we had an efficient system of collecting administrative statistics, particularly,
during the regime of Chandra Gupta Maurya (324-300 B.C.). The system of
collecting data related to births and deaths is mentioned in Kautilyas Arthshastra
(around 300 B.C.) A detailed account of administrative surveys conducted during
Akbars regime is given in Ain-I-Akbari written by Abul Fazl.
382
MATHEMATICS
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Chapter
16
he
PROBABILITY
Where a mathematical reasoning can be had, it is as great a folly to
16.1 Introduction
bl
is
make use of any other, as to grope for a thing in the dark, when
you have a candle in your hand. JOHN ARBUTHNOT
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3
1
i.e., . Here the
6
2
384
MATHEMATICS
laid down some axioms to interpret probability, in his book Foundation of Probability
published in 1933. In this Chapter, we will study about this approach called axiomatic
approach of probability. To understand this approach we must know about few basic
terms viz. random experiment, sample space, events, etc. Let us learn about these all,
in what follows next.
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In our day to day life, we perform many activities which have a fixed result no matter
any number of times they are repeated. For example given any triangle, without knowing
the three angles, we can definitely say that the sum of measure of angles is 180.
We also perform many experimental activities, where the result may not be same,
when they are repeated under identical conditions. For example, when a coin is tossed
it may turn up a head or a tail, but we are not sure which one of these results will
actually be obtained. Such experiments are called random experiments.
An experiment is called random experiment if it satisfies the following two
conditions:
(i) It has more than one possible outcome.
Example 1 Two coins (a one rupee coin and a two rupee coin) are tossed once. Find
a sample space.
Solution Clearly the coins are distinguishable in the sense that we can speak of the
first coin and the second coin. Since either coin can turn up Head (H) or Tail(T), the
possible outcomes may be
PROBABILITY
385
he
Note The outcomes of this experiment are ordered pairs of H and T. For the
$
sake of simplicity the commas are omitted from the ordered pairs.
is
Example 2 Find the sample space associated with the experiment of rolling a pair of
dice (one is blue and the other red) once. Also, find the number of elements of this
sample space.
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Solution Suppose 1 appears on blue die and 2 on the red die. We denote this outcome
by an ordered pair (1,2). Similarly, if 3 appears on blue die and 5 on red, the outcome
is denoted by the ordered pair (3,5).
In general each outcome can be denoted by the ordered pair (x, y), where x is
the number appeared on the blue die and y is the number appeared on the red die.
Therefore, this sample space is given by
S = {(x, y): x is the number on the blue die and y is the number on the red die}.
The number of elements of this sample space is 6 6 = 36 and the sample space is
given below:
{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
Example 3 In each of the following experiments specify appropriate sample space
(i)
(ii)
A boy has a 1 rupee coin, a 2 rupee coin and a 5 rupee coin in his pocket. He
takes out two coins out of his pocket, one after the other.
A person is noting down the number of accidents along a busy highway
during a year.
Solution (i) Let Q denote a 1 rupee coin, H denotes a 2 rupee coin and R denotes a 5
rupee coin. The first coin he takes out of his pocket may be any one of the three coins
Q, H or R. Corresponding to Q, the second draw may be H or R. So the result of two
draws may be QH or QR. Similarly, corresponding to H, the second draw may be
Q or R.
Therefore, the outcomes may be HQ or HR. Lastly, corresponding to R, the second
draw may be H or Q.
So, the outcomes may be RH or RQ.
386
MATHEMATICS
Thus, the sample space is S={QH, QR, HQ, HR, RH, RQ}
(ii) The number of accidents along a busy highway during the year of observation
can be either 0 (for no accident ) or 1 or 2, or some other positive integer.
Thus, a sample space associated with this experiment is S= {0,1,2,...}
he
Example 4 A coin is tossed. If it shows head, we draw a ball from a bag consisting of
3 blue and 4 white balls; if it shows tail we throw a die. Describe the sample space of
this experiment.
is
Solution Let us denote blue balls by B1, B2, B3 and the white balls by W1, W2, W3, W4.
Then a sample space of the experiment is
S = { HB1, HB2, HB3, HW1, HW2, HW3, HW4, T1, T2, T3, T4, T5, T6}.
bl
Here HBi means head on the coin and ball Bi is drawn, HWi means head on the coin
and ball Wi is drawn. Similarly, Ti means tail on the coin and the number i on the die.
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Example 5 Consider the experiment in which a coin is tossed repeatedly until a head
comes up. Describe the sample space.
Solution In the experiment head may come up on the first toss, or the 2nd toss, or the
3rd toss and so on till head is obtained. Hence, the desired sample space is
S= {H, TH, TTH, TTTH, TTTTH,...}
EXERCISE 16.1
In each of the following Exercises 1 to 7, describe the sample space for the indicated
experiment.
1. A coin is tossed three times.
2. A die is thrown two times.
3. A coin is tossed four times.
4. A coin is tossed and a die is thrown.
5. A coin is tossed and then a die is rolled only in case a head is shown on the coin.
6. 2 boys and 2 girls are in Room X, and 1 boy and 3 girls in Room Y. Specify the
sample space for the experiment in which a room is selected and then a person.
7. One die of red colour, one of white colour and one of blue colour are placed in a
bag. One die is selected at random and rolled, its colour and the number on its
uppermost face is noted. Describe the sample space.
8. An experiment consists of recording boygirl composition of families with 2
children.
(i) What is the sample space if we are interested in knowing whether it is a boy
or girl in the order of their births?
PROBABILITY
387
11.
12.
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13.
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10.
bl
9.
What is the sample space if we are interested in the number of girls in the
family?
A box contains 1 red and 3 identical white balls. Two balls are drawn at random
in succession without replacement. Write the sample space for this experiment.
An experiment consists of tossing a coin and then throwing it second time if a
head occurs. If a tail occurs on the first toss, then a die is rolled once. Find the
sample space.
Suppose 3 bulbs are selected at random from a lot. Each bulb is tested and
classified as defective (D) or non defective(N). Write the sample space of this
experiment.
A coin is tossed. If the out come is a head, a die is thrown. If the die shows up
an even number, the die is thrown again. What is the sample space for the
experiment?
The numbers 1, 2, 3 and 4 are written separatly on four slips of paper. The slips
are put in a box and mixed thoroughly. A person draws two slips from the box,
one after the other, without replacement. Describe the sample space for the
experiment.
An experiment consists of rolling a die and then tossing a coin once if the number
on the die is even. If the number on the die is odd, the coin is tossed twice. Write
the sample space for this experiment.
A coin is tossed. If it shows a tail, we draw a ball from a box which contains 2 red
and 3 black balls. If it shows head, we throw a die. Find the sample space for this
experiment.
A die is thrown repeatedly untill a six comes up. What is the sample space for
this experiment?
he
(ii)
14.
15.
16.
16.3 Event
We have studied about random experiment and sample space associated with an
experiment. The sample space serves as an universal set for all questions concerned
with the experiment.
Consider the experiment of tossing a coin two times. An associated sample space
is S = {HH, HT, TH, TT}.
Now suppose that we are interested in those outcomes which correspond to the
occurrence of exactly one head. We find that HT and TH are the only elements of S
corresponding to the occurrence of this happening (event). These two elements form
the set E = { HT, TH}
We know that the set E is a subset of the sample space S . Similarly, we find the
following correspondence between events and subsets of S.
MATHEMATICS
Description of events
Number of tails is exactly 2
Number of tails is atleast one
Number of heads is atmost one
Second toss is not head
Number of tails is atmost two
Number of tails is more than two
Corresponding subset of S
A = {TT}
B = {HT, TH, TT}
C = {HT, TH, TT}
D = { HT, TT}
S = {HH, HT, TH, TT}
he
388
is
The above discussion suggests that a subset of sample space is associated with
an event and an event is associated with a subset of sample space. In the light of this
we define an event as follows.
Definition Any subset E of a sample space S is called an event.
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16.3.2 Types of events Events can be classified into various types on the basis of the
elements they have.
1. Impossible and Sure Events The empty set and the sample space S describe
events. In fact is called an impossible event and S, i.e., the whole sample space is
To understand these let us consider the experiment of rolling a die. The associated
sample space is
S = {1, 2, 3, 4, 5, 6}
Let E be the event the number appears on the die is a multiple of 7. Can you
write the subset associated with the event E?
Clearly no outcome satisfies the condition given in the event, i.e., no element of
the sample space ensures the occurrence of the event E. Thus, we say that the empty
set only correspond to the event E. In other words we can say that it is impossible to
have a multiple of 7 on the upper face of the die. Thus, the event E = is an impossible
event.
Now let us take up another event F the number turns up is odd or even. Clearly
PROBABILITY
389
F = {1, 2, 3, 4, 5, 6,} = S, i.e., all outcomes of the experiment ensure the occurrence of
the event F. Thus, the event F = S is a sure event.
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2. Simple Event If an event E has only one sample point of a sample space, it is
called a simple (or elementary) event.
In a sample space containing n distinct elements, there are exactly n simple
events.
For example in the experiment of tossing two coins, a sample space is
S={HH, HT, TH, TT}
There are four simple events corresponding to this sample space. These are
E1= {HH}, E2={HT}, E3= { TH} and E4={TT}.
3. Compound Event If an event has more than one sample point, it is called a
Compound event.
For example, in the experiment of tossing a coin thrice the events
E: Exactly one head appeared
F: Atleast one head appeared
G: Atmost one head appeared etc.
are all compound events. The subsets of S associated with these events are
E={HTT,THT,TTH}
F={HTT,THT, TTH, HHT, HTH, THH, HHH}
G= {TTT, THT, HTT, TTH}
Each of the above subsets contain more than one sample point, hence they are all
compound events.
16.3.3 Algebra of events In the Chapter on Sets, we have studied about different
ways of combining two or more sets, viz, union, intersection, difference, complement
of a set etc. Like-wise we can combine two or more events by using the analogous set
notations.
Let A, B, C be events associated with an experiment whose sample space is S.
390
MATHEMATICS
or
Event A or B = A B
= { : A or B}
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Therefore
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3. The Event A and B We know that intersection of two sets A B is the set of
those elements which are common to both A and B. i.e., which belong to both
A and B.
If A and B are two events, then the set A B denotes the event A and B.
Thus,
A B = { : A and B}
For example, in the experiment of throwing a die twice Let A be the event
score on the first throw is six and B is the event sum of two scores is atleast 11 then
A = {(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}, and B = {(5,6), (6,5), (6,6)}
so A B = {(6,5), (6,6)}
Note that the set A B = {(6,5), (6,6)} may represent the event the score on the first
throw is six and the sum of the scores is atleast 11.
4. The Event A but not B We know that AB is the set of all those elements
which are in A but not in B. Therefore, the set AB may denote the event A but not
B.We know that
A B = A B
Example 6 Consider the experiment of rolling a die. Let A be the event getting a
prime number, B be the event getting an odd number. Write the sets representing
the events (i) Aor B (ii) A and B (iii) A but not B (iv) not A.
Solution Here
Obviously
(i) A or B = A B = {1, 2, 3, 5}
(ii) A and B = A B = {3,5}
PROBABILITY
391
16.3.4 Mutually exclusive events In the experiment of rolling a die, a sample space is
S = {1, 2, 3, 4, 5, 6}. Consider events, A an odd number appears and B an even
number appears
Clearly the event A excludes the event B and vice versa. In other words, there is
no outcome which ensures the occurrence of events A and B simultaneously. Here
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Now 3 A as well as 3 B
Therefore, A and B are not mutually exclusive events.
E1 E 2 E 3 ... E n = E i = S
i =1
then E1, E2, ...., En are called exhaustive events.In other words, events E1, E2, ..., En
are said to be exhaustive if atleast one of them necessarily occurs whenever the
experiment is performed.
Further, if Ei Ej = for i j i.e., events Ei and Ej are pairwise disjoint and
n
E i = S , then events E1, E2, ..., En are called mutually exclusive and exhaustive
i =1
events.
392
MATHEMATICS
he
Example 7 Two dice are thrown and the sum of the numbers which come up on the
dice is noted. Let us consider the following events associated with this experiment
A: the sum is even.
B: the sum is a multiple of 3.
C: the sum is less than 4.
D: the sum is greater than 11.
Which pairs of these events are mutually exclusive?
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Solution There are 36 elements in the sample space S = {(x, y): x, y = 1, 2, 3, 4, 5, 6}.
Then
A = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4),
(4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)}
B = {(1, 2), (2, 1), (1, 5), (5, 1), (3, 3), (2, 4), (4, 2), (3, 6), (6, 3), (4, 5), (5, 4),
(6, 6)}
C = {(1, 1), (2, 1), (1, 2)} and D = {(6, 6)}
We find that
A B = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6)}
Therefore, A and B are not mutually exclusive events.
Similarly A C , A D , B C and B D .
Thus, the pairs, (A, C), (A, D), (B, C), (B, D) are not mutually exclusive events.
Also C D = and so C and D are mutually exclusive events.
PROBABILITY
393
EXERCISE 16.2
1. A die is rolled. Let E be the event die shows 4 and F be the event die shows
even number. Are E and F mutually exclusive?
2. A die is thrown. Describe the following events:
(ii)
(iv)
(vi)
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Also find A B, A B, B C, E F, D E, A C, D E, E F, F
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3. An experiment involves rolling a pair of dice and recording the numbers that
come up. Describe the following events:
A: the sum is greater than 8, B: 2 occurs on either die
C: the sum is at least 7 and a multiple of 3.
Which pairs of these events are mutually exclusive?
4. Three coins are tossed once. Let A denote the event three heads show, B
denote the event two heads and one tail show, C denote the event three tails
show and D denote the event a head shows on the first coin. Which events are
(i) mutually exclusive?
(ii) simple?
(iii) Compound?
5. Three coins are tossed. Describe
(i) Two events which are mutually exclusive.
(ii) Three events which are mutually exclusive and exhaustive.
(iii) Two events, which are not mutually exclusive.
(iv) Two events which are mutually exclusive but not exhaustive.
(v) Three events which are mutually exclusive but not exhaustive.
6. Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die.
B: getting an odd number on the first die.
C: getting the sum of the numbers on the dice 5.
Describe the events
(i) A
(ii) not B
(iii) A or B
(iv) A and B
(v) A but not C
(vi) B or C
(vii) B and C
(viii) A B C
7. Refer to question 6 above, state true or false: (give reason for your answer)
(i) A and B are mutually exclusive
(ii) A and B are mutually exclusive and exhaustive
(iii) A = B
394
MATHEMATICS
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It follows from (iii) that P() = 0. To prove this, we take F = and note that E and
are disjoint events. Therefore, from axiom (iii), we get
P (E ) = P (E) + P () or
P(E) = P(E) + P () i.e. P () = 0.
1
2
to each
P(H) =
and P(T) =
(1)
Clearly this assignment satisfies both the conditions i.e., each number is neither
less than zero nor greater than 1 and
PROBABILITY
P(H) + P(T) =
395
1
1
+
=1
2
2
1
1
, and probability of T =
2
2
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1
3
and P(T) =
... (2)
4
4
Does this assignment satisfy the conditions of axiomatic approach?
If we take
P(H) =
is
3
.
4
4
We find that both the assignments (1) and (2) are valid for probability of
H and T.
In fact, we can assign the numbers p and (1 p) to both the outcomes such that
0 p 1 and P(H) + P(T) = p + (1 p) = 1
This assignment, too, satisfies both conditions of the axiomatic approach of
probability. Hence, we can say that there are many ways (rather infinite) to assign
probabilities to outcomes of an experiment. We now consider some examples.
Example 9 Let a sample space be S = {1, 2,..., 6}.Which of the following
assignments of probabilities to each outcome are valid?
Outcomes 1
2
3
4
5
6
and probability of T =
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(a)
1
6
1
6
1
6
1
6
1
6
1
6
(b)
(c)
1
8
2
3
1
3
1
3
(d)
1
12
1
12
1
6
1
6
1
6
3
2
(e)
0.1
0.2
0.3
0.4
0.5
0.6
1
4
1
3
Solution (a) Condition (i): Each of the number p(i) is positive and less than one.
Condition (ii): Sum of probabilities
=
1 1 1 1 1 1
+ + + + + =1
6 6 6 6 6 6
396
MATHEMATICS
(d)
3
> 1, the assignment is not valid
2
Since, sum of probabilities = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 = 2.1, the assignment
is not valid.
Since p(6) =
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(e)
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16.4.1 Probability of an event Let S be a sample space associated with the experiment
examining three consecutive pens produced by a machine and classified as Good
(non-defective) and bad (defective). We may get 0, 1, 2 or 3 defective pens as result
of this examination.
A sample space associated with this experiment is
S = {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG},
where B stands for a defective or bad pen and G for a non defective or good pen.
Let the probabilities assigned to the outcomes be as follows
Sample point:
BBB BBG BGB GBB BGG GBG GGB GGG
Probability:
1
8
1
8
1
8
1
8
1
8
1
8
1
8
1
8
Let event A: there is exactly one defective pen and event B: there are atleast two
defective pens.
Hence
Now
and
1 1 1 3
+ + =
8 8 8 8
P(B) = P(i ), i B
1 1 1 1 4 1
+ + + = =
8 8 8 8 8 2
Let us consider another experiment of tossing a coin twice
= P(BBG) + P(BGB) + P(GBB) + P(BBB) =
PROBABILITY
P(HH) =
397
1
1
2
9
, P(HT) = , P(TH) = , P(TT) =
4
7
7
28
1 9 4
+
=
4 28 7
For the event F: exactly two heads, we have F = {HH}
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Clearly this assignment satisfies the conditions of axiomatic approach. Now, let
us find the probability of the event E: Both the tosses yield the same result.
Here
E = {HH, TT}
Now
P(E) = P(wi), for all wi E
P(F) = P(HH) =
1
4
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and
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= P(HH) + P(TT) =
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Let all the outcomes are equally likely to occur, i.e., the chance of occurrence of each
simple event must be same.
i.e.
P(i) = p, for all i S where 0 p 1
n
Since
or
np = 1 i.e., p =
1
n
Let S be a sample space and E be an event, such that n(S) = n and n(E) = m. If
each out come is equally likely, then it follows that
P(E) =
m
n
16.4.3 Probability of the event A or B Let us now find the probability of event
A or B, i.e., P (A B)
Let A = {HHT, HTH, THH} and B = {HTH, THH, HHH} be two events associated
with tossing of a coin thrice
Clearly
A B = {HHT, HTH, THH, HHH}
Now
398
MATHEMATICS
Also
3
8
and
3
8
Therefore
P(A) + P(B) =
is
It is clear that
3 3 6
+ =
8 8 8
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1 1 1 1 4 1
P ( A B) = + + + = =
8 8 8 8 8 2
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The points HTH and THH are common to both A and B . In the computation of
P(A) + P(B) the probabilities of points HTH and THH, i.e., the elements of A B are
included twice. Thus to get the probability P(A B) we have to subtract the probabilities
of the sample points in A B from P(A) + P(B)
P(A B) = P(A) + P(B) P(i ) , i A B
i.e.
= P( A) + P(B) P(A B)
Since
we have
A B = (AB) (A B) (BA) ,
... (1)
[ P(i )i (A B)]
= P(A B) + [ P(i ) i A B]
= P(A B) + P(A B) .
[using (1)]
PROBABILITY
399
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Fig 16.1
If A and B are disjoint sets, i.e., they are mutually exclusive events, then A B =
Therefore
P(A B) = P () = 0
1
10
400
Now
MATHEMATICS
1 1 1 1
4 2
+ + + = =
10 10 10 10 10 5
Also event not A = A = {1, 3, 5, 7, 9, 10}
Now
P(A) = P(1) + P(3) + P(5) + P(7) + P(9) + P(10)
=
he
6 3
=
10 5
3
2
= 1 = 1 P(A)
5
5
Also, we know that A and A are mutually exclusive and exhaustive events i.e.,
A A = and A A = S
or
P(A A) = P(S)
Now
P(A) + P(A) = 1,
by using axioms (ii) and (iii).
or
P( A ) = P(not A) = 1 P(A)
We now consider some examples and exercises having equally likely outcomes
unless stated otherwise.
P(A) =
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Thus,
Example 10 One card is drawn from a well shuffled deck of 52 cards. If each outcome
is equally likely, calculate the probability that the card will be
(i) a diamond
(ii) not an ace
(iii) a black card (i.e., a club or, a spade)
(iv) not a diamond
(v) not a black card.
Solution When a card is drawn from a well shuffled deck of 52 cards, the number of
possible outcomes is 52.
(i) Let A be the event 'the card drawn is a diamond'
Clearly the number of elements in set A is 13.
Therefore, P(A) =
13 1
=
52 4
1
4
4
1 12
=1 =
52
13 13
PROBABILITY
401
26 1
=
52 2
1
.
2
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i.e. P(C) =
(iv) We assumed in (i) above that A is the event card drawn is a diamond,
so the event card drawn is not a diamond may be denoted as A' or not A
(v)
1 1
=
2 2
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1 3
=
4 4
The event card drawn is not a black card may be denoted as C or not C.
1
2
Example 11 A bag contains 9 discs of which 4 are red, 3 are blue and 2 are yellow.
The discs are similar in shape and size. A disc is drawn at random from the bag.
Calculate the probability that it will be (i) red, (ii) yellow, (iii) blue, (iv) not blue,
(v) either red or blue.
Solution There are 9 discs in all so the total number of possible outcomes is 9.
Let the events A, B, C be defined as
A: the disc drawn is red
4
9
(ii) The number of yellow discs = 2, i.e., n (B) = 2
Hence
P(A) =
2
9
(iii) The number of blue discs = 3, i.e., n(C) = 3
Therefore, P(B) =
402
MATHEMATICS
3 1
=
9 3
(iv) Clearly the event not blue is not C. We know that P(not C) = 1 P(C)
Therefore,
P(C) =
1 2
=
3 3
(v) The event either red or blue may be described by the set A or C
Since, A and C are mutually exclusive events, we have
P(A or C) = P (A C) = P(A) + P(C) =
4 1 7
+ =
9 3 9
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P(not C) = 1
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Therefore
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Example 12 Two students Anil and Ashima appeared in an examination. The probability
that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination
is 0.10. The probability that both will qualify the examination is 0.02. Find the
probability that
(a) Both Anil and Ashima will not qualify the examination.
(b)
(c)
Solution Let E and F denote the events that Anil and Ashima will qualify the examination,
respectively. Given that
P(E) = 0.05, P(F) = 0.10 and P(E F) = 0.02.
Then
(a)
The event both Anil and Ashima will not qualify the examination may be
expressed as E F.
Since, E is not E, i.e., Anil will not qualify the examination and F is not F, i.e.,
Ashima will not qualify the examination.
Also
Now
or
PROBABILITY
403
(c) The event only one of them will qualify the examination is same as the event
either (Anil will qualify, and Ashima will not qualify) or (Anil will not qualify and Ashima
will qualify) i.e., E F or E F, where E F and E F are mutually exclusive.
Therefore, P(only one of them will qualify) = P(E F or E F)
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Example 13 A committee of two persons is selected from two men and two women.
What is the probability that the committee will have (a) no man? (b) one man? (c) two
men?
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Solution The total number of persons = 2 + 2 = 4. Out of these four person, two can
be selected in 4 C 2 ways.
No men in the committee of two means there will be two women in the committee.
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(a)
Therefore
(b)
P ( no man ) =
2
4
C2 1 2 1 1
=
=
43
6
C2
One man in the committee means that there is one woman. One man out of 2
can be selected in 2 C1 ways and one woman out of 2 can be selected in 2 C1 ways.
Together they can be selected in 2 C1 2 C1 ways.
Therefore
(c)
P ( One man ) =
C1 2 C1 2 2 2
=
=
4
23 3
C2
Hence
P ( Two men ) =
2
4
C2
1
1
= 4
=
C2
C2 6
EXERCISE 16.3
1. Which of the following can not be valid assignment of probabilities for outcomes
of sample Space S = {1 , 2 , 3 , 4 , 5 , 6 , 7 }
MATHEMATICS
Assignment
(a)
(b)
(c)
(d)
0.1
0.01
0.05
0.03
0.01
0.2
0.6
1
7
0.1
0.1
1
7
0.2
0.2
1
7
0.3
0.3
1
7
0.4
0.4
1
7
0.6
0.1
1
7
0.7
0.3
1
7
0.5
0.2
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404
1
2
3
4
5
6
15
14
14
14
14
14
14
14
2. A coin is tossed twice, what is the probability that atleast one tail occurs?
3. A die is thrown, find the probability of following events:
(i) A prime number will appear,
(ii) A number greater than or equal to 3 will appear,
(iii) A number less than or equal to one will appear,
(iv) A number more than 6 will appear,
(v) A number less than 6 will appear.
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(e)
2
is the probability of an event, what is the probability of the event not A.
11
10. A letter is chosen at random from the word ASSASSINATION. Find the
probability that letter is (i) a vowel (ii) a consonant
9. If
PROBABILITY
405
1
5
1
15
...
(ii)
0.35
...
0.25
0.6
(iii)
0.5
0.35
...
0.7
3
1
and P(B) = . Find P(A or B), if A and B are mutually exclusive
5
5
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1
3
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(i)
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11. In a lottery, a person choses six different natural numbers at random from 1 to 20,
and if these six numbers match with the six numbers already fixed by the lottery
committee, he wins the prize. What is the probability of winning the prize in the
game? [Hint order of the numbers is not important.]
12. Check whether the following probabilities P(A) and P(B) are consistently defined
(i) P(A) = 0.5, P(B) = 0.7, P(A B) = 0.6
(ii) P(A) = 0.5, P(B) = 0.4, P(A B) = 0.8
13. Fill in the blanks in following table:
P(A B)
P(A)
P(B)
P(A B)
events.
19.
20.
1
1
1
, P(F) =
and P(E and F) = , find
4
2
8
406
MATHEMATICS
21. In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for
both NCC and NSS. If one of these students is selected at random, find the
probability that
The student opted for NCC or NSS.
The student has opted neither NCC nor NSS.
The student has opted NSS but not NCC.
he
(i)
(ii)
(iii)
Miscellaneous Examples
A before B?
A first and B last?
A just before B?
(ii)
(iv)
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(i)
(iii)
(v)
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Example 14 On her vacations Veena visits four cities (A, B, C and D) in a random
order. What is the probability that she visits
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Solution The number of arrangements (orders) in which Veena can visit four cities A,
B, C, or D is 4! i.e., 24.Therefore, n (S) = 24.
Since the number of elements in the sample space of the experiment is 24 all of these
outcomes are considered to be equally likely. A sample space for the
experiment is
S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB
BACD, BADC, BDAC, BDCA, BCAD, BCDA
CABD, CADB, CBDA, CBAD, CDAB, CDBA
(i)
Thus
(ii)
P(E) =
n(E)
n ( S)
12 1
=
24 2
n ( F) 4 1
Therefore, P ( F ) = n S = 24 = 6
( )
Students are advised to find the probability in case of (iii), (iv) and (v).
PROBABILITY
407
Example 15 Find the probability that when a hand of 7 cards is drawn from a well
shuffled deck of 52 cards, it contains (i) all Kings (ii) 3 Kings (iii) atleast 3 Kings.
C7
48
Hence
C 4 48 C3
1
=
52
7735
C7
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52
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Therefore
P (3 Kings) =
C3 48 C 4
9
=
52
1547
C7
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9
1
46
+
=
1547 7735 7735
P ( B C) + P ( A B C)
... (1)
Now
P (E ) = P ( B C)
= P ( B) + P ( C ) P ( B C )
... (2)
P ( A E ) = P ( A B ) + P ( A C ) P ( A B) ( A C )
408
MATHEMATICS
= P ( A B ) + P ( A C ) P [ A B C]
... (3)
(b)
What is the probability that A, B and C finish first, second and third,
respectively.
What is the probability that A, B and C are first three to finish (in any order)
(Assume that all finishing orders are equally likely)
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(a)
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P ( A B) P ( A C ) + P ( A B C )
5!
= 5 4 3 = 60 sample points, each with
( 5 3)!
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Solution If we consider the sample space consisting of all finishing orders in the first
1
.
60
A, B and C finish first, second and third, respectively. There is only one finishing
order for this, i.e., ABC.
a probability of
(a)
(b)
1
.
60
A, B and C are the first three finishers. There will be 3! arrangements for A, B
and C. Therefore, the sample points corresponding to this event will be 3! in
number.
So
3! 6
1
=
=
60 60 10
1. A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles
are drawn from the box, what is the probability that
(i) all will be blue? (ii) atleast one will be green?
2. 4 cards are drawn from a well shuffled deck of 52 cards. What is the probability
of obtaining 3 diamonds and one spade?
PROBABILITY
409
3. A die has two faces each with number 1, three faces each with number 2 and
one face with number 3. If die is rolled once, determine
(i) P(2)
(ii) P(1 or 3)
(iii) P(not 3)
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4. In a certain lottery 10,000 tickets are sold and ten equal prizes are awarded.
What is the probability of not getting a prize if you buy (a) one ticket (b) two
tickets (c) 10 tickets.
5. Out of 100 students, two sections of 40 and 60 are formed. If you and your friend
are among the 100 students, what is the probability that
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6. Three letters are dictated to three persons and an envelope is addressed to each
of them, the letters are inserted into the envelopes at random so that each envelope
contains exactly one letter. Find the probability that at least one letter is in its
proper envelope.
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7. A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A B) = 0.35.
Find (i) P(A B) (ii) P(A B) (iii) P(A B) (iv) P(B A)
Name
Harish
Sex
M
Age in years
30
2.
Rohan
33
3.
Sheetal
46
4.
Alis
28
5.
Salim
41
9. If 4-digit numbers greater than 5,000 are randomly formed from the digits
0, 1, 3, 5, and 7, what is the probability of forming a number divisible by 5 when,
(i) the digits are repeated? (ii) the repetition of digits is not allowed?
10. The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e.,
from 0 to 9. The lock opens with a sequence of four digits with no repeats. What
is the probability of a person getting the right sequence to open the suitcase?
410
MATHEMATICS
Summary
In this Chapter, we studied about the axiomatic approach of probability. The main
features of this Chapter are as follows:
no N
C
tt E
o R
be T
re
pu
bl
is
he
(iii) P(A) =
(ii)
P ( ) for all S = 1
i
of the outcome i.
n(A)
, where n(A) = number of elements in
n(S)
PROBABILITY
411
Historical Note
no N
C
tt E
o R
be T
re
pu
bl
is
he
INFINITE SERIES
is
A.1.1 Introduction
he
Appendix
no N
C
tt E
o R
be T
re
pu
a1 + a2 + a3 + . . . + an + . . . =
bl
As discussed in the Chapter 9 on Sequences and Series, a sequence a1, a2, ..., an, ...
having infinite number of terms is called infinite sequence and its indicated sum, i.e.,
a1 + a2 + a3 + ... + an + ... is called an infinte series associated with infinite sequence.
This series can also be expressed in abbreviated form using the sigma notation, i.e.,
k
k =1
In this Chapter, we shall study about some special types of series which may be
required in different problem situations.
In Chapter 8, we discussed the Binomial Theorem in which the index was a positive
integer. In this Section, we state a more general form of the theorem in which the
index is not necessarily a whole number. It gives us a particular type of infinite series,
called Binomial Series. We illustrate few applications, by examples.
We know the formula
(1 + x)n = n C0 + n C1 x + . . . + n C n xn
(1 + x )m = 1 + mx +
m ( m 1)
1.2
x2 +
m ( m 1)( m 2 )
1 . 2. 3
x 3 + ...
INFINITE SERIES
413
Remark 1. Note carefully the condition | x | < 1, i.e., 1< x < 1 is necessary when m
is negative integer or a fraction. For example, if we take x = 2 and m = 2, we
obtain
or
( 2 )( 3)
1.2
( 2 )2 + ...
he
(1 2 )2 = 1 + ( 2 )( 2 ) +
1= 1 + 4 + 12 + . . .
This is not possible
( a + b)
b
b
m
= a 1 + = a 1 +
a
a
bl
Consider
is
2. Note that there are infinite number of terms in the expansion of (1+ x)m, when m
is a negative integer or a fraction
b
a + ...
no N
C
tt E
o R
be T
re
pu
b m ( m 1)
m
= a 1 + m a + 1.2
m
m 1
= a + ma b +
m ( m 1)
1.2
a m 2b 2 + ...
b
< 1 or equivalently when | b | < | a |.
a
m ( m 1)( m 2 ) ... ( m r + 1) a m r b r
1.2.3...r
We give below certain particular cases of Binomial Theorem, when we assume
2. (1 x) 1 = 1 + x + x2 + x3 + . . .
3. (1 + x) 2 = 1 2 x + 3x2 4x3 + . . .
4. (1 x) 2 = 1 +2x + 3x2 + 4x3 + . . .
Example 1 Expand 1
2
1
2
, when | x | < 2.
414
MATHEMATICS
Solution We have
1
2
1
1 3
x
x 2
2
2 2 + ...
= 1+
+
1 2
1 .2
2
= 1+
x 3x 2
+
+ ...
4 32
he
x
1
2
is
bl
ak +1
= r (constant) for k = 1, 2, 3, ..., n1. Particularly, if we take a1 = a, then the
ak
no N
C
tt E
o R
be T
re
pu
resulting sequence a, ar, ar2, ..., arn1 is taken as the standard form of G.P., where a is
first term and r, the common ratio of G.P.
Earlier, we have discussed the formula to find the sum of finite series
a + ar + ar2 + ... + arn 1 which is given by
Sn =
a 1 rn
1 r
).
In this section, we state the formula to find the sum of infinite geometric series
a + ar + ar2 + ... + arn 1 + ... and illustrate the same by examples.
Let us consider the G.P. 1,
Here a = 1, r =
2 4
,
,...
3 9
2
. We have
3
n
2
1
n
3 = 3 1 2
Sn =
2
3
1
3
... (1)
2
Let us study the behaviour of as n becomes larger and larger.
3
INFINITE SERIES
n
2
3
415
10
20
0.6667
0.1316872428
0.01734152992
0.00030072866
he
2
We observe that as n becomes larger and larger, becomes closer and closer to
3
n
is
2
zero. Mathematically, we say that as n becomes sufficiently large, becomes
3
n
bl
2
sufficiently small. In other words, as n , 0 . Consequently, we find that
3
no N
C
tt E
o R
be T
re
pu
a 1 rn
Sn =
1 r
a
ar n
=
1 r 1 r
ar n
0 . Therefore,
In this case, r n 0 as n since | r | <1 and then
1 r
a
as n .
1 r
Symbolically, sum to infinity of infinite geometric series is denoted by S. Thus,
Sn
S=
we have
a
1 r
For example
(i)
1+
(ii) 1
1 1
1
1
+
+ + ... =
=2
1
2 2 2 23
1
2
1 1
1
+ 2 3 + ... =
2 2
2
1
1
2
=
=
1 3
1
1 1+
2
2
416
MATHEMATICS
5 5 5
, ,
,....
4 16 64
5
1
and r = . Also | r | <1 .
4
4
he
Solution Here a =
is
5
5
Hence, the sum to infinity is 4 = 4 = 1 .
1
5
1+
4
4
no N
C
tt E
o R
be T
re
pu
bl
Leonhard Euler (1707 1783), the great Swiss mathematician introduced the number
e in his calculus text in 1748. The number e is useful in calculus as in the study of the
circle.
Consider the following infinite series of numbers
1 1 1 1
+ + + + ...
... (1)
1! 2! 3! 4!
The sum of the series given in (1) is denoted by the number e
Let us estimate the value of the number e.
Since every term of the series (1) is positive, it is clear that its sum is also positive.
Consider the two sums
1+
and
1 1 1
1
+ + + ... + + ...
3! 4! 5!
n!
... (2)
1
1
1
1
+ 3 + 4 + .... + n1 + ...
2
2
2
2
2
... (3)
Observe that
1 1
1 1
1 1
= and 2 = , which gives < 2
4
3! 2
2
3! 6
1 1
1 1
1
1
<
=
and 3 = , which gives
8
4! 23
2
4! 24
1
1
1 1
1
1
=
and 4 = , which gives < 4 .
16
5! 2
2
5! 120
INFINITE SERIES
417
1 1
1
1
1
1 1 1
Adding
1 1
1 + + on both sides of (4), we get,
1! 2!
bl
1
1 1 1 1 1
1 + + + + + + ... + + ...
n!
1! 2! 3! 4! 5!
... (4)
is
Therefore
he
1
1
< n 1 , when n > 2
n! 2
We observe that each term in (2) is less than the corresponding term in (3),
no N
C
tt E
o R
be T
re
pu
1 1 1
1
1
1
1! 2! 2
... (5)
1 1
1
1
1
= 1 + 1 + + 2 + 3 + 4 + ... + n1 + ...
2
2
2
2 2
= 1+
1
1
2
=1+ 2 = 3
Left hand side of (5) represents the series (1). Therefore e < 3 and also e > 2 and
hence 2 < e < 3.
x x 2 x3
xn
+
+
+ ... +
+ ...
n!
1! 2! 3!
x x 2 x3
+
+
+ ...
1! 2! 3!
MATHEMATICS
2 x +3
= 1+
( 2 x + 3) ( 2 x + 3) 2
1!
( 2 x + 3)n
n!
2!
+ ...
(3+ 2 x) n
. This can be expanded by the
n!
Binomial Theorem as
C2 3n 2 22
. Therefore, the coefficient of x2 in the whole
n!
series is
n
n2
no N
C
tt E
o R
be T
re
pu
n ( n 1) 3
C 2 3n 2 2 2
2
=
n!
n!
n=2
bl
is
1 n n
2
n
3 + C1 3n 1 ( 2 x ) + n C2 3n 2 ( 2 x ) + ... + ( 2 x ) .
n!
n=2
he
418
= 2
3n 2
[using n! = n (n 1) (n 2)!]
n = 2 ( n 2 )!
3 32 33
2
= 1 + + + + ...
1! 2! 3!
= 2e3 .
Thus 2e is the coefficient of x2 in the expansion of e2x+3.
Alternatively e2x+3 = e3 . e2x
3
2 x (2 x) 2 (2 x)3
e3 1 +
+
+
+ ...
= 1!
2!
3!
2
3 2
e
.
= 2e 3
Thus, the coefficient of x2 in the expansion of e2x+3 is
2!
Example 4 Find the value of e2, rounded off to one decimal place.
x x 2 x3
xn
+
+
+ ... +
+ ...
n!
1! 2! 3!
INFINITE SERIES
419
Putting x = 2, we get
2 2 2 23 2 4 25 2 6
+
+
+
+
+
+ ...
1! 2! 3! 4! 5! 6!
= 1+ 2 + 2 +
4 2 4
4
+ + +
+ ...
3 3 15 45
he
e2 = 1 +
is
bl
2 2 2 23 2 4 2 5 2 2 2 2 3
e2 < 1 + +
+ + + 1 + + 2 + 3 + ...
6
1! 2! 3! 4! 5! 6 6
no N
C
tt E
o R
be T
re
pu
4 1 1
4 1
2
= 7 + = 7.4 .
= 7 + 1 + + + ... = 7 +
1
15 3 3
15
5
1
Thus, e2 lies between 7.355 and 7.4. Therefore, the value of e2, rounded off to one
decimal place, is 7.4.
Another very important series is logarithmic series which is also in the form of infinite
series. We state the following result without proof and illustrate its application with an
example.
Theorem If | x | < 1, then
x 2 x3
+
...
2
3
The series on the right hand side of the above is called the logarithmic series.
log e (1 + x ) = x
1 1 1
log e 2 = 1 + + ...
2 3 4
420
MATHEMATICS
log e 1 + px + qx 2 = ( + ) x
2 + 2 2 3 + 3 3
x +
x ...
2
3
he
2 x 2 3 x3
2 x 2 3 x 3
x
...
x
+
...
Solution Right hand side =
2
3
2
3
(
(1 + px + qx )
2
= log e 1 + ( + ) x + x
2
bl
= log e
is
= log e (1 + x ) + log (1 + x )
Here, we have used the facts + = p and = q . We know this from the
no N
C
tt E
o R
be T
re
pu
given roots of the quadratic equation. We have also assumed that both | x | < 1 and
| x | < 1.
Appendix
MATHEMATICAL MODELLING
A.2.1 Introduction
Much of our progress in the last few centuries has made it necessary to apply
mathematical methods to real-life problems arising from different fields be it Science,
Finance, Management etc. The use of Mathematics in solving real-world problems
has become widespread especially due to the increasing computational power of digital
computers and computing methods, both of which have facilitated the handling of
lengthy and complicated problems. The process of translation of a real-life problem
into a mathematical form can give a better representation and solution of certain
problems. The process of translation is called Mathematical Modelling.
Here we shall familiaries you with the steps involved in this process through
examples. We shall first talk about what a mathematical model is, then we discuss the
steps involved in the process of modelling.
A.2.2 Preliminaries
Mathematical modelling is an essential tool for understanding the world. In olden days
the Chinese, Egyptians, Indians, Babylonians and Greeks indulged in understanding
and predicting the natural phenomena through their knowledge of mathematics. The
architects, artisans and craftsmen based many of their works of art on geometric
prinicples.
Suppose a surveyor wants to measure the height of a tower. It is physically very
difficult to measure the height using the measuring tape. So, the other option is to find
out the factors that are useful to find the height. From his knowledge of trigonometry,
he knows that if he has an angle of elevation and the distance of the foot of the tower
to the point where he is standing, then he can calculate the height of the tower.
So, his job is now simplified to find the angle of elevation to the top of the tower
and the distance from the foot of the tower to the point where he is standing. Both of
which are easily measurable. Thus, if he measures the angle of elevation as 40 and
the distance as 450m, then the problem can be solved as given in Example 1.
422
MATHEMATICS
Example 1 The angle of elevation of the top of a tower from a point O on the ground,
which is 450 m away from the foot of the tower, is 40. Find the height of the tower.
Solution We shall solve this in different steps.
Step 1 We first try to understand the real problem. In the problem a tower is given and
its height is to be measured. Let h denote the height. It is given that the horizontal
distance of the foot of the tower from a particular point O on the ground is 450 m. Let
d denotes this distance. Then d = 450m. We also know that the angle of elevation,
denoted by , is 40.
The real problem is to find the height h of the tower using the known distance d
and the angle of elevation .
Step 2 The three quantities mentioned in the problem are height,
distance and angle of elevation.
So we look for a relation connecting these three quantities.
This is obtained by expressing it geometrically in the following
way (Fig 1).
AB denotes the tower. OA gives the horizontal distance
from the point O to foot of the tower. AOB is the angle of
elevation. Then we have
tan =
h
or h = d tan
d
... (1)
Fig 1
Step 3 We use Equation (1) to solve h. We have = 40. and d = 450m. Then we get
h = tan 40 450 = 450 0.839 = 377.6m
Step 4 Thus we got that the height of the tower approximately 378m.
Let us now look at the different steps used in solving the problem. In step 1, we
have studied the real problem and found that the problem involves three parameters
height, distance and angle of elevation. That means in this step we have studied the
real-life problem and identified the parameters.
In the Step 2, we used some geometry and found that the problem can be
represented geometrically as given in Fig 1. Then we used the trigonometric ratio for
the tangent function and found the relation as
h = d tan
So, in this step we formulated the problem mathematically. That means we found
an equation representing the real problem.
MATHEMATICAL MODELLING
423
In Step 3, we solved the mathematical problem and got that h = 377.6m. That is
we found
Solution of the problem.
In the last step, we interpreted the solution of the problem and stated that the
height of the tower is approximately 378m. We call this as
Interpreting the mathematical solution to the real situation
In fact these are the steps mathematicians and others use to study various reallife situations. We shall consider the question, why is it necessary to use mathematics
to solve different situations.
Here are some of the examples where mathematics is used effectively to study
various situations.
1. Proper flow of blood is essential to transmit oxygen and other nutrients to various
parts of the body in humanbeings as well as in all other animals. Any constriction
in the blood vessel or any change in the characteristics of blood vessels can
change the flow and cause damages ranging from minor discomfort to sudden
death. The problem is to find the relationship between blood flow and physiological
characteristics of blood vessel.
2. In cricket a third umpire takes decision of a LBW by looking at the trajectory of
a ball, simulated, assuming that the batsman is not there. Mathematical equations
are arrived at, based on the known paths of balls before it hits the batsmans leg.
This simulated model is used to take decision of LBW.
3. Meteorology department makes weather predictions based on mathematical
models. Some of the parameters which affect change in weather conditions are
temperature, air pressure, humidity, wind speed, etc. The instruments are used to
measure these parameters which include thermometers to measure temperature,
barometers to measure airpressure, hygrometers to measure humidity,
anemometers to measure wind speed. Once data are received from many stations
around the country and feed into computers for further analysis and interpretation.
4. Department of Agriculture wants to estimate the yield of rice in India from the
standing crops. Scientists identify areas of rice cultivation and find the average
yield per acre by cutting and weighing crops from some representative fields.
Based on some statistical techniques decisions are made on the average yield of
rice.
How do mathematicians help in solving such problems? They sit with experts in
the area, for example, a physiologist in the first problem and work out a
mathematical equivalent of the problem. This equivalent consists of one or more
equations or inequalities etc. which are called the mathematical models. Then
424
MATHEMATICS
solve the model and interpret the solution in terms of the original problem. Before
we explain the process, we shall discuss what a mathematical model is.
A mathematical model is a representation which comprehends a situation.
An interesting geometric model is illustrated in the following example.
Example 2 (Bridge Problem) Konigsberg is a town on the Pregel River, which in the
18th century was a German
town, but now is Russian. Within
the town are two river islands
that are connected to the banks
with seven bridges as shown
in (Fig 2).
People tried to walk around
the town in a way that only
crossed each bridge once, but it
proved to be difficult problem.
Leonhard Euler, a Swiss
Fig 2
mathematician in the service of
the Russian empire Catherine the Great, heard about the problem. In 1736 Euler proved
that the walk was not possible to do. He proved this by inventing a kind of diagram
called a network, that is made up of vertices
(dots where lines meet) and arcs (lines) (Fig3).
He used four dots (vertices) for the two
river banks and the two islands. These have
been marked A, B and C, D. The seven lines
(arcs) are the seven bridges. You can see that
3 bridges (arcs) join to riverbank, A, and 3 join
to riverbank B. 5 bridges (arcs) join to island
C, and 3 join to island D. This means that all
the vertices have an odd number of arcs, so
Fig 3
they are called odd vertices (An even vertex
would have to have an even number of arcs joining to it).
Remember that the problem was to travel around town crossing each bridge only
once. On Eulers network this meant tracing over each arc only once, visiting all the
vertices. Euler proved it could not be done because he worked out that, to have an odd
vertex you would have to begin or end the trip at that vertex. (Think about it). Since
there can only be one beginning and one end, there can only be two odd vertices if you
are to trace over each arc only once. Since the bridge problem has 4 odd vertices, it
just not possible to do!
MATHEMATICAL MODELLING
425
426
MATHEMATICS
parameters involved in the problem. For example, in the case of pendulum, the factors
are period of oscillation (T), the mass of the bob (m), effective length (l ) of the pendulum
which is the distance between the point of suspension to the centre of mass of the bob.
Here, we consider the length of string as effective length of the pendulum and acceleration
due to gravity (g), which is assumed to be constant at a place.
So, we have identified four parameters for studying the problem. Now, our purpose
is to find T. For this we need to understand what are the parameters that affect the
period which can be done by performing a simple experiment.
We take two metal balls of two different masses and conduct experiment with
each of them attached to two strings of equal lengths. We measure the period of
oscillation. We make the observation that there is no appreciable change of the period
with mass. Now, we perform the same experiment on equal mass of balls but take
strings of different lengths and observe that there is clear dependence of the period on
the length of the pendulum.
This indicates that the mass m is not an essential parameter for finding period
whereas the length l is an essential parameter.
This process of searching the essential parameters is necessary before we go
to the next step.
2. Mathematical description This involves finding an equation, inequality or a
geometric figure using the parameters already identified.
In the case of simple pendulum, experiments were conducted in which the values
of period T were measured for different values of l. These values were plotted on a
graph which resulted in a curve that resembled a parabola. It implies that the relation
between T and l could be expressed
T2 = kl
It was found that k =
... (1)
4 2
. This gives the equation
g
T = 2
l
g
... (2)
Finding the solution The mathematical formulation rarely gives the answer directly.
Usually we have to do some operation which involves solving an equation, calculation
or applying a theorem etc. In the case of simple pendulums the solution involves applying
the formula given in Equation (2).
MATHEMATICAL MODELLING
427
The period of oscillation calculated for two different pendulums having different
lengths is given in Table 1
Table 1
225 cm
275cm
3.04 sec
3.36 sec
The table shows that for l = 225 cm, T = 3.04 sec and for l = 275 cm, T = 3.36 sec.
Interpretation/Validation
A mathematical model is an attempt to study, the essential characteristic of a real life
problem. Many times model equations are obtained by assuming the situation in an
idealised context. The model will be useful only if it explains all the facts that we would
like it to explain. Otherwise, we will reject it, or else, improve it, then test it again. In
other words, we measure the effectiveness of the model by comparing the results
obtained from the mathematical model, with the known facts about the real
problem. This process is called validation of the model. In the case of simple
pendulum, we conduct some experiments on the pendulum and find out period of
oscillation. The results of the experiment are given in Table 2.
Table 2
Length (cms)
Time (secs)
385
275
225
3.371
3.056
230
275
225
3.352
3.042
Now, we compare the measured values in Table 2 with the calculated values given in
Table 1.
The difference in the observed values and calculated values gives the error. For
example, for l = 275 cm, and mass m = 385 gm,
error = 3.371 3.36 = 0.011
which is small and the model is accepted.
Once we accept the model, we have to interpret the model. The process of
describing the solution in the context of the real situation is called interpretation
of the model. In this case, we can interpret the solution in the following way:
(a) The period is directly proportional to the square root of the length of the
pendulum.
428
MATHEMATICS
(b) It is inversely proportional to the square root of the acceleration due to gravity.
Our validation and interpretation of this model shows that the mathematical model
is in good agreement with the practical (or observed) values. But we found that there
is some error in the calculated result and measured result. This is because we have
neglected the mass of the string and resistance of the medium. So, in such situation we
look for a better model and this process continues.
This leads us to an important observation. The real world is far too complex to
understand and describe completely. We just pick one or two main factors to be
completely accurate that may influence the situation. Then try to obtain a simplified
model which gives some information about the situation. We study the simple situation
with this model expecting that we can obtain a better model of the situation.
Now, we summarise the main process involved in the modelling as
(a) Formulation
(b) Solution
(c) Interpretation/Validation
The next example shows how modelling can be done using the techniques of finding
graphical solution of inequality.
Example 3 A farm house uses atleast 800 kg of special food daily. The special food is
a mixture of corn and soyabean with the following compositions
Table 3
Material
Cost per Kg
Corn
.09
.02
Rs 10
Soyabean
.60
.06
Rs 20
The dietary requirements of the special food stipulate atleast 30% protein and at most
5% fibre. Determine the daily minimum cost of the food mix.
Solution Step 1 Here the objective is to minimise the total daily cost of the food which
is made up of corn and soyabean. So the variables (factors) that are to be considered
are
x = the amount of corn
y = the amount of soyabean
z = the cost
Step 2 The last column in Table 3 indicates that z, x, y are related by the equation
z = 10x + 20y
... (1)
The problem is to minimise z with the following constraints:
MATHEMATICAL MODELLING
429
(a)
The farm used atleast 800 kg food consisting of corn and soyabean
i.e., x + y 800
... (2)
(b) The food should have atleast 30% protein dietary requirement in the proportion
as given in the first column of Table 3. This gives
0.09x + 0.6y 0.3 (x + y)
... (3)
(c) Similarly the food should have atmost 5% fibre in the proportion given in
2nd column of Table 3. This gives
0.02x + 0.06 y 0.05 (x + y)
... (4)
We simplify the constraints given in (2), (3) and (4) by grouping all the coefficients
of x, y.
Then the problem can be restated in the following mathematical form.
Statement Minimise z subject to
x + y 800
0.21x .30y 0
0.03x .01y 0
This gives the formulation of the model.
Step 3 This can be solved graphically. The shaded region in Fig 5 gives the possible
solution of the equations. From the graph it is clear that the minimum value is got at the
Fig 5
430
MATHEMATICS
B (t )
P (t ) is called the birth rate for the time interval t to t + 1.
2.
D (t)
P (t) is called the death rate for the time interval t to t + 1.
Assumptions
1. The birth rate is the same for all intervals. Likewise, the death rate is the same
for all intervals. This means that there is a constant b, called the birth rate, and a
constant d, called the death rate so that, for all t 0,
b=
B (t )
P (t )
and
d=
D (t )
P (t )
... (1)
2. There is no migration into or out of the population; i.e., the only source of population
MATHEMATICAL MODELLING
431
... (4)
for t = 0, 1, 2, ... The constant 1 + b d is often abbreviated by r and called the growth
rate or, in more high-flown language, the Malthusian parameter, in honor of Robert
Malthus who first brought this model to popular attention. In terms of r, Equation (4)
becomes
P(t) = P(0)r t
,
t = 0, 1, 2, ...
... (5)
P(t) is an example of an exponential function. Any function of the form cr t, where c
and r are constants, is an exponential function.
Equation (5) gives the mathematical formulation of the problem.
Step 2 Solution
Suppose the current population is 250,000,000 and the rates are b = 0.02 and d = 0.01.
What will the population be in 10 years? Using the formula, we calculate P(10).
P(10) = (1.01)10 (250,000,000)
= (1.104622125) (250,000,000)
= 276,155,531.25
Step 3 Interpretation and Validation
Naturally, this result is absurd, since one cant have 0.25 of a person.
So, we do some approximation and conclude that the population is 276,155,531
(approximately). Here, we are not getting the exact answer because of the assumptions
that we have made in our mathematical model.
The above examples show how modelling is done in variety of situations using
different mathematical techniques.
432
MATHEMATICS
START
<
ASSUMPTIONS/AXIOMS
FORMULATION
SOLUTION
INTERPRETATION
<
VALIDATION
NO
<
YES
SATISFIED
<
STOP
ANSWERS
EXERCISE 1.1
no N
C
tt E
o R
be T
re
pu
bl
is
he
EXERCISE 1.2
1.
2.
3.
4.
5.
(i),
(i)
(i)
(i)
(i)
(iii), (iv)
Finite
(ii)
Infinite (ii)
Yes
(ii)
No
(ii)
(v) Finite
(v) Infinite
EXERCISE 1.3
1.
2.
3.
4.
5.
6.
(i)
(vii)
(i)
(i),
(i)
(iii)
1
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vi) True
(iv)
( 4, 6]
(ii) ( 12, 10)
(iii) [ 0, 7 )
(iv)
[ 3, 4 ]
7. (i) { x : x R, 3 < x < 0 }
(ii) { x : x R, 6 x 12 }
(iii) { x : x R, 6 < x 12 }
(iv) { x R : 23 x < 5 } 9. (iii)
434
MATHEMATICS
EXERCISE 1.4
1.
(i) X Y = {1, 2, 3, 5 }
(iii)
(ii) A B = { a, b, c, e, i, o, u }
A B = {x : x = 1, 2, 4, 5 or a multiple of 3 }
3.
he
2. Yes, A B = { a, b, c }
is
(vii) { 3, 4, 5, 6, 7, 8, 9, 10 }
(i) X Y = { 1, 3 } (ii) A B = { a }
6.
(i) { 7, 9, 11 }
(ii) { 11, 13 }
(v)
(vi) { 7, 9, 11 }
(iii) { 3 }
(iv)
(iii)
(iv) { 11 }
(vii)
no N
C
tt E
o R
be T
re
pu
(viii) { 7, 9, 11 }
7.
9.
(i) B
(v) { 2 }
(ix) {7, 9, 11 }
(vii) { 20 }
(x) { 5, 10, 15 }
(x) { 7, 9, 11, 15 }
(ii) C
(iii) D
(iv)
(vi) { x : x is an odd prime number }
8.
(v)
bl
5.
(iii)
(viii) { 4, 8, 12, 16 }
10. (i) { a, c }
(ii) {f, g }
11. Set of irrational numbers 12. (i) F
(ii) F
(iii) { b , d }
(iii) T (iv) T
EXERCISE 1.5
(i)
(iv)
2. (i)
(iv)
3. (i)
(ii)
(iii)
(iv)
1.
{ 5, 6, 7, 8, 9}
(ii) {1, 3, 5, 7, 9 }
(iii) {7, 8, 9 }
{ 5, 7, 9 }
(v) { 1, 2, 3, 4 }
(vi) { 1, 3, 4, 5, 6, 7, 9 }
{ d, e, f, g, h}
(ii) { a, b, c, h }
(iii) { b, d , f, h }
{ b, c, d, e )
{ x : x is an odd natural number }
{ x : x is an even natural number }
{ x : x N and x is not a multiple of 3 }
{ x : x is a positive composite number and x = 1 ]
ANSWERS
(iv)
EXERCISE 1.6
2. 5
3. 50
4. 42
5. 30
6. 19
7. 25, 35
8. 60
bl
1. 2
9
}
2
he
(xi) { x : x N and x
is
(v)
(vi)
(vii)
(viii)
435
no N
C
tt E
o R
be T
re
pu
1. A B, A C, B C, D A, D B, D C
2. (i) False
(ii) False
(iii) True
(iv) False
(v) False
(vi) True
7. False
12. We may take A = { 1, 2 }, B = { 1, 3 }, C = { 2 , 3 }
13. 325
14. 125
15. (i) 52, (ii) 30
16. 11
EXERCISE 2.1
436
MATHEMATICS
EXERCISE 2.2
he
is
bl
3. R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}
5.
(i) R = {(1, 1), (1,2), (1, 3), (1, 4), (1, 6), (2 4), (2, 6), (2, 2), (4, 4), (6, 6),
(3, 3), (3, 6)}
(ii) Domain of R = {1, 2, 3, 4, 6}
(iii) Range of R = {1, 2, 3, 4, 6}
no N
C
tt E
o R
be T
re
pu
4.
9. Domain of R = Z
Range of R = Z
EXERCISE 2.3
1.
2.
3.
4.
(i) t (0) = 32
5.
(i) Range = ( , 2)
412
(iii)
5
(ii) Range = [2, )
(ii) t (28) =
t (10) = 14
(iii) Range = R
(iv) 100
ANSWERS
437
he
2.
4.
5.
6.
7.
11. No
12.
EXERCISE 3.1
5
36
(i) 39 22 30
19
72
(ii) 229 5 29
(ii)
3. 12
4. 12 36
4
3
(iii) 300
(iii)
26
9
(iv) 210
(iv)
no N
C
tt E
o R
be T
re
pu
(i)
bl
10.
is
f
x +1
3
, x
x=
2x 3
2
g
1.
2.
7.
(i)
2
15
(ii)
1
5
5.
20
3
(iii)
6. 5 : 4
7
25
EXERCISE 3.2
1. sin x =
3
2
1
, cosec x =
, sec x = 2 , tan x = 3 , cot x =
2
3
3
5
4
5
3
4
2. cosec x = , cos x = , sec x = , tan x = , cot x =
3
5
4
4
3
4
5
3
5
4
3. sin x = , cosec x = , cos x = , sec x = , tan x =
5
4
5
3
3
4. sin x =
12
13
5
12
5
, cosec x = , cos x = , tan x = , cot x =
13
12
13
5
12
MATHEMATICS
5. sin x =
6.
5
13
12
13
12
, cosec x = , cos x = , sec x = , cot x =
13
5
13
12
5
1
2
7. 2
8.
9.
3
2
10. 1
he
438
EXERCISE 3.3
(i)
3 +1
2 2
(ii) 2
is
5.
, , n + , n Z
3 3
3
2.
, , 2n , n Z
3 3
3
no N
C
tt E
o R
be T
re
pu
1.
bl
EXERCISE 3.4
3.
5 11
5
,
, n +
,nZ
6 6
6
5.
x=
7.
x = n + ( 1)n
8.
x=
n
or x = n, n Z
3
4.
7 11
7
,
, n + (1)n
,nZ
6 6
6
6.
x = (2n + 1) , or 2n , n Z
4
3
9.
x=
or (2n + 1) , n Z
6
2
n
n 3
, or
+
,nZ
2
2
8
, or n ,n Z
3
3
8.
5 2 5
,
,2
5
5
9.
6
3
,
, 2
3
3
10.
8 + 2 15 8 2 15
,
, 4 + 15
4
4
ANSWERS
439
EXERCISE 5.1
2. 0
5. 2 7 i
6.
13. i
14.
7.
17
5
+i
3
3
8.
22 107
i
3
27
11.
4
3
+i
25 25
12.
7 2
i
2
EXERCISE 5.2
2
3
5
3
i
14
14
2 . 2,
5
6
2 cos
+ i sin
4
4
no N
C
tt E
o R
be T
re
pu
1 . 2,
is
242
26i 10.
27
19 21i
5 10
bl
9.
4. 14 + 28 i
3. i
he
1. 3
4.
3
3
2 cos + i sin
4
4
5.
3.
3
3
2 cos
+ i sin
4
4
6. 3 (cos + i sin )
EXERCISE 5.3
1. 3 i
5.
9.
3 11 i
2
2.
1 7 i
4
3.
3 3 3 i
2
4.
6.
1 7 i
2
7.
1 7 i
2 2
8.
2 2 1 i
2
10.
1 7 i
2 2
1 7i
2
2 34 i
2 3
440
MATHEMATICS
(i)
6.
2 4
i
3 3
2
15. 2
2
i
2
7 . 1
12. (i)
3
3
2 cos + i sin
4
4
8.
2
, (ii) 0
5
17. 1
5
2
i
27 27
1
3
2 4
13.
9.
he
3
3
5.
10.
307 + 599 i
442
2
14
i
3
21
is
3.
14. x = 3, y = 3
bl
1. 2 2 i
18. 0
20. 4
no N
C
tt E
o R
be T
re
pu
EXERCISE 6.1
1.
(i) {1, 2, 3, 4}
2.
(i) No Solution
(ii) {... 4, 3}
3.
(i) {... 2, 1, 0, 1}
(ii) (, 2)
4.
(ii) (2, )
5. (4, )
6. ( , 3)
9. ( , 6)
10. (, 6)
11. (, 2]
12. ( , 120]
14. (, 2]
15. (4, )
16. (, 2]
13. (4, )
17. x < 3,
19.
7. (, 3]
18.
x 1,
20. x
x > 1,
8. (, 4]
2
,
7
25. 9 cm
ANSWERS
441
EXERCISE 6.2
2.
3.
4.
no N
C
tt E
o R
be T
re
pu
bl
is
he
1.
5.
6.
442
MATHEMATICS
8.
9.
10.
no N
C
tt E
o R
be T
re
pu
bl
is
he
7.
EXERCISE 6.3
1.
2.
he
3.
4.
5.
6.
7.
8.
is
bl
no N
C
tt E
o R
be T
re
pu
ANSWERS
443
444
MATHEMATICS
10.
11.
12.
no N
C
tt E
o R
be T
re
pu
bl
is
he
9.
13.
14.
ANSWERS
445
is
he
15.
2. (0, 1]
4. ( 23, 2]
80 10
,
5.
3
3
bl
3. [ 4, 2]
11
1, 3
no N
C
tt E
o R
be T
re
pu
6.
7. (5, 5)
8. (1, 7)
9. (5, )
10. [ 7, 11]
12. More than 320 litres but less than 1280 litres.
13. More than 562.5 litres but less than 900 litres.
14. 9.6 MA 16.8
EXERCISE 7.1
2. 108
5. 8
6. 20
3. 5040
4. 336
446
MATHEMATICS
EXERCISE 7.2
1. (i) 40320, (ii) 18
5. (i) 30, (ii) 15120
2. 30, No
3. 28
4. 64
EXERCISE 7.4
2. (i) 5, (ii) 6
6. 778320
3. 210
7. 3960
4. 40
8. 200
no N
C
tt E
o R
be T
re
pu
1. 45
5. 2000
9. 35
4. 120, 48
8. 40320
is
504
2. 4536
3. 60
56
6. 9
7. (i) 3, (ii) 4
(i) 360, (ii) 720, (iii) 240
10. 33810
(i) 1814400, (ii) 2419200, (iii) 25401600
bl
1.
5.
9.
11.
he
EXERCISE 7.3
1. 3600
4. 907200
8. 4C 1 48 C 4
2. 1440
5. 120
9. 2880
EXERCISE 8.1
32 40 20
5 3 x5
x
x
5
x
8
32
x5 x3
6
5
4
3. 64 x 576 x + 2160 x 4320 x3 + 4860 x2 2916 x + 729
2.
4.
x5 5 x3 10
10
5
1
+
+
+ 3+ 5
x+
243 81 27
9 x 3x
x
5.
x6
6 x 4 15 x 2
6. 884736
9. 9509900499
20
15
x2
6
1
4
x
x6
7. 11040808032
10. (1.1)
10000
> 1000
8. 104060401
ANSWERS
447
EXERCISE 8.2
4.
2.
105 9 35 12
x ; x
8
48
12. m = 4
7.
101376
3.
5. 1760 x9y3
6. 18564
8. 61236 x5y5
10. n = 7; r = 3
9
7
is
2. a =
5. 396 6
7. 0.9510
171
no N
C
tt E
o R
be T
re
pu
bl
3.
he
1. 1512
16 8 32 16
x 2 x3 x 4
+ 2 3 + 4 4x +
+ +
5
x x
2
2 16
x
x
10. 27x6 54ax5 + 117a2x4 116a3x3 + 117a4x2 54a5x + 27a6
9.
EXERCISE 9.1
1 2 3 4 5
, , , ,
2 3 4 5 6
1. 3, 8, 15, 24, 35
2.
1 1 1 5
7
4. , , , and
6 6 2 6
6
6.
3 9 21
75
, , , 21 and
2 2 2
2
9. 729
10.
12. 1,
7. 65, 93
3. 2, 4, 8, 16 and 32
8.
49
128
360
23
1 1 1 1
1 1 1 1
, , ,
; 1+ + + +
+ ...
2 6 24 120
2 6 24 120
448
MATHEMATICS
13. 2, 2, 1, 0, 1;
3 5
8
14. 1, 2, , and
2 3
5
2 + 2 + 1 + 0 + (1) + ...
EXERCISE 9.2
2. 98450
4. 5 or 20
6. 4
179
321
14. 11, 14, 17, 20 and 23
17. Rs 245
18. 9
8. 2q
10. 0
9.
15. 1
is
n
( 5n + 7 )
2
13. 27
16. 14
7.
5 5
,
220 2n
4. 2187
no N
C
tt E
o R
be T
re
pu
2. 3072
bl
EXERCISE 9.3
1.
8.
7
2
11.
22 +
13. 4
6. 1
3 + 1 3 2 1
9.
3 11
3 1
2
14.
he
1. 1002001
1 ( a )n
1+ a
7.
1
20
1 ( 0.1)
x3 1 x 2 n
10.
1 x
5 2
2 5 5 2
12. r = or ; Terms are ,1, or ,1,
2 5
5 2 2 5
16 16 n
;2;
2 1
7
7
15. 2059
4 8 16
80
8
, ,
,...or 4, 8,16, 32 , 64, ..
10n 1 n
18.
3 3 3
81
9
19. 496
20. rR
21. 3, 6, 12, 24 26. 9 and 27
16.
1
30. 120, 480, 30 (2n)
2
32. x2 16x + 25 = 0
27.
n=
n
( n + 1) ( n + 2 )
3
2.
EXERCISE 9.4
1.
n ( n + 1) ( n + 2 ) ( n + 3)
4
ANSWERS
6. 3n (n + 1) (n + 3)
8.
9.
n ( n + 1)
12
( 3n
n
n +1
4.
n ( n + 1)
7.
+ 23n + 34
5. 2840
2
( n + 2)
12
he
n
( n + 1) 3n2 + 5n + 1
6
n
( n + 1) ( 2n + 1) + 2 2n 1
6
n
( 2n + 1) ( 2n 1)
3
10.
is
3.
449
4. 8729
8. 160; 6
12. 11
5. 3050
9. 3
6. 1210
10. 8, 16, 32
no N
C
tt E
o R
be T
re
pu
2. 5, 8, 11
7. 4
11. 4
bl
23.
50 n
5n
2n 2
10 1
1 10 n
, (ii)
81
9
3 27
21. (i)
n 2
n + 3n + 5
3
27. Rs 16680
31. Rs 5120
25.
28. Rs 39100
32. 25 days
22. 1680
n
2n 2 + 9n + 13
24
29. Rs 43690
EXERCISE 10.1
1.
121
square unit.
2
2. (0, a), (0, a) and 3a,0 or (0, a), (0, a), and
3. (i) y2 y1 , (ii) x2 x1
7.
11. 1 and 2, or
8. x = 1
15
4. , 0
2
3a,0
5.
1
2
10. 135
1
1
and 1, or 1 and 2, or and 1
2
2
14.
1
, 104.5 Crores
2
450
MATHEMATICS
EXERCISE 10.2
4.
) (
6.
x 3y + 2 3 = 0
3 +1 x
2. x 2y + 10 = 0
3. y = mx
5. 2x + y + 6 = 0
3 1 y = 4
3 1
9. 3x 4y + 8 = 0
3 x + y = 10
11. (1 + n)x + 3(1 + n)y = n +11
13. x + 2y 6 = 0, 2x + y 6 = 0
10. 5x y + 20 = 0
8.
is
12. x + y = 5
3x + y 2 = 0 and 3x + y + 2 = 0
L=
16.
.192
( C 20 ) + 124.942
90
15. 2x 9y + 85 = 0
bl
14.
5x + 3y + 2 = 0
7.
he
1. y = 0 and x = 0
no N
C
tt E
o R
be T
re
pu
EXERCISE 10.3
1.
(i)
1
1
5
5
y = x + 0, , 0; (ii) y = 2 x + , 2, ; (iii) y = 0x + 0, 0, 0
7
7
3
3
2.
(i)
x y
+ = 1, 4 ,6;
4 6
(iii)
x
y
3
(ii) 3 + 2 =1, 2 , 2;
2
2
2
y = , intercept with y-axis = and no intercept with x-axis.
3
3
(i) x cos 120 + y sin 120 = 4, 4, 120 (ii) x cos 90 + y sin 90 = 2, 2, 90;
3.
65
1 p+r
units, (ii)
units.
17
2 l
5. ( 2, 0) and (8, 0)
6. (i)
7. 3x 4y + 18 = 0
8. y + 7x = 21
9. 30 and 150
22
9
10.
12.
4. 5 units
) (
3 + 2 x + 2 3 1 y = 8 3 + 1 or
) (
3 2 x + 1 + 2 3 y = 1 + 8 3
ANSWERS
68 49
14. ,
25 25
13. 2x + y = 5
1
5
,c =
2
2
he
17. y x = 1,
15. m =
2.
3. 2 x 3 y = 6, 3 x + 2 y = 6
8 32
4. 0, , 0 ,
3 3
x=
5
22
7. 2x 3y + 18 = 0
no N
C
tt E
o R
be T
re
pu
6.
bl
sin ( )
2 sin
2
is
5.
8. k2 square units
9. 5
11. 3x y = 7, x + 3y = 9
14. 1 : 2
15.
23 5
units
18
1 5 2
7
19.
13
22. ,0
5
EXERCISE 11.1
1. x2 + y2 4y = 0
2. x2 + y2 + 4x 6y 3 = 0
3. 36x2 + 36y2 36x 18y + 11 = 0
4. x2 + y2 2x 2y = 0
5. x2 + y2 + 2ax + 2by + 2b2 = 0
7. c(2, 4), r =
451
65
6. c(5, 3), r = 6
8. c(4, 5), r =
10. x2 + y2 6x 8y + 15 = 0
53 9. c (
1
1
, 0) ; r =
4
4
11. x2 + y2 7x + 5y 14 = 0
452
MATHEMATICS
13. x2 + y2 ax by = 0
14. x2 + y2 4x 4y = 5
15. Inside the circle; since the distance of the point to the centre of the circle is less
than the radius of the circle.
EXERCISE 11.2
2. F (0,
he
3
3
), axis - y - axis, directrix y = , length of the Latus rectum = 6
2
2
is
9
9
) , axis - y - axis, directrix y = , length of the Latus rectum = 9
4
4
no N
C
tt E
o R
be T
re
pu
6. F (0,
5
, length of the Latus rectum = 10
2
bl
5
2
7. y2 = 24x
10. y2 = 8x
8. x2 = 12y
11. 2y2 = 9x
9. y2 = 12x
12. 2x2 = 25y
EXERCISE 11.3
16
3
21
;
5
8
5
20
,
6
9
2
7
;
4
ANSWERS
453
3
;
2
Latus rectum = 5
Latus rectum =
13
;
7
he
72
7
3
;
2
is
bl
Latus rectum = 10
no N
C
tt E
o R
be T
re
pu
4
3
15
;
4
1
2
2 2
;
3
8
3
10.
x2 y 2
+
=1
25 9
11.
x2
y2
+
=1
144 169
12.
x2 y 2
+
=1
36 20
13.
x2 y 2
+
=1
9
4
14.
x2 y 2
+
=1
1
5
15.
x2
y2
+
=1
169 144
16.
x2 y 2
+
=1
64 100
17.
x2 y 2
+
=1
16 7
18.
x2 y 2
+
=1
25 9
5
;
3
454
19.
MATHEMATICS
x2 y 2
+
=1
10 40
20. x + 4y = 52 or
x2 y 2
+
=1
52 13
9
5
; Latus rectum =
4
2
2. Foci (0 6), Vertices (0, 3); e = 2; Latus rectum = 18
1. Foci ( 5, 0), Vertices ( 4, 0); e =
13
; Latus rectum = 9
2
6
2 14
14
4 5
), Vertices (0,
); e =
; Latus rectum =
5
5
3
3
no N
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5. Foci (0,
5
64
; Latus rectum =
3
3
bl
is
he
EXERCISE 11.4
49
65
; Latus rectum =
2
4
7.
x2 y 2
=1
4
5
8.
y 2 x2
=1
25 39
9.
y 2 x2
=1
9 16
10.
x2 y2
=1
16 9
11.
y2 x2
=1
25 144
12.
x2 y2
=1
25 20
13.
x2 y2
=1
4 12
14.
x2 9 y 2
=1
49 343
15.
y 2 x2
=1
5 5
x2 y 2
+
=1
81 9
8. 8 3a
6. 18 sq units
4. 1.56m (approx.)
7.
x2 y 2
+
=1
25 9
ANSWERS
455
EXERCISE 12.1
1. y and z - coordinates are zero
3. I, IV, VIII, V, VI, II, III, VII
4. (i) XY - plane
(ii) (x, y, 0)
2. y - coordinate is zero
(iii) Eight
43 (iii) 2 26 (iv) 2 5
5. 9x2 + 25y2 + 25z2 225 = 0
is
1. (i) 2 5 (ii)
4. x 2z = 0
he
EXERCISE 12.2
EXERCISE 12.3
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bl
4 1 27
1. (i) , , (ii) ( 8,17 ,3 )
2. 1 : 2
5 5 5
3. 2 : 3
5. (6, 4, 2), (8, 10, 2)
1. (1, 2, 8)
3. a = 2, b =
2. 7, 34 , 7
16
, c=2
3
6.
x2 + y 2 + z 2 2x 7 y + 2z =
k 2 109
2
EXERCISE 13.1
1. 6
5.
9. b
13.
a
b
1
2
22
2.
7
3.
6. 5
7.
10. 2
14.
a
b
11
4
11. 1
15.
4.
19
2
8.
108
7
12.
16.
1
4
MATHEMATICS
a +1
b
0
22. 2
Limit does not exist at x = 1
Limit does not exist at x = 0
0
28. a = 0, b = 4
17. 4
21.
24.
25.
27.
29.
19. 0
18.
20. 1
23. 3, 6
26. Limit does not exist at x = 0
he
456
x a1
30. lim
f (x) exists for all a 0.
x a
is
31. 2
of m and n.
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EXERCISE 13.2
bl
1. 20
2. 99
3. 1
4. (i) 3x2
(ii) 2x 3
(iii)
2
x3
(iv)
( x 1)2
8.
a b
(iii)
( x b )2
nx n anx n 1 x n + a n
( x a )2
9 . (i) 2
(v)
11.
2
(ii) 4ax ax + b
(i)
(iii)
(v)
(vii)
(iii)
2
x(3x 2)
12 36
+
2
(vi)
( x + 1) (3x 1)2
x5
x10
cos 2x
5sec x tan x 4sin x
3cosec2 x 5 cosec x cot x
2sec2 x 7sec x tan x
3
( 5 + 2x )
x4
10. sin x
(iv) 15x4 +
24
x5
ANSWERS
457
ps
ad bc
5.
8.
apx 2 2bpx + ar bq
( px
( x 1)
6.
+ qx + r
apx 2
9.
2ax b
, x 0,1
7.
2bpx bq ar
ax b
12. na ( ax + b )
n 1
ax 2
10.
4a
x5
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11.
cx d
2. 1
13.
( ax + b )n1 ( cx + d )m1 mc ( ax + b ) + na ( cx + d )
20.
22.
sin x cos x
bc cos x ad sin x bd
c d cos x
sec x 1
cos a
cos 2 x
21.
23.
24.
q sin x ax 2
25.
tan 2 x x cos x
26.
2sec x tan x
18.
sin x
p q cos x 2a x cos x
x tan x 1 sin x
bx c
is
qr
x2
2b
sin x
x3
bl
3.
1
x2
he
1. (i) 1 (ii)
14.
cos (x+a)
16.
1
1 sin x
19.
n1
n sin x cos x
MATHEMATICS
x cos
27.
29.
30.
2 sin x x cos x
4
sin 2 x
x sec x 1 sec 2 x
1 tanx x sec 2 x
28.
1 tanx
he
458
sin x n x cos x
sin n 1 x
is
EXERCISE 14.1
no N
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be T
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pu
bl
(i) This sentence is always false because the maximum number of days in a
month is 31. Therefore, it is a statement.
(ii) This is not a statement because for some people mathematics can be easy
and for some others it can be difficult.
(iii) This sentence is always true because the sum is 12 and it is greater than 10.
Therefore, it is a statement.
(iv) This sentence is sometimes true and sometimes not true. For example the
square of 2 is even number and the square of 3 is an odd number. Therefore,
it is not a statement.
(v) This sentence is sometimes true and sometimes false. For example, squares
and rhombus have equal length whereas rectangles and trapezium have
unequal length. Therefore, it is not a statement.
(vi) It is an order and therefore, is not a statement.
(vii) This sentence is false as the product is (8). Therefore, it is a statement.
(viii) This sentence is always true and therefore, it is a statement.
(ix) It is not clear from the context which day is referred and therefore, it is not
a statement.
(x) This is a true statement because all real numbers can be written in the form
a + i 0.
2. The three examples can be:
(i) Everyone in this room is bold. This is not a statement because from the
context it is not clear which room is referred here and the term bold is not
precisely defined.
(ii) She is an engineering student. This is also not a statement because who
she is.
(iii) cos2 is always greater than 1/2. Unless, we know what is, we cannot
say whether the sentence is true or not.
1.
ANSWERS
459
EXERCISES 14.2
(i) Chennai is not the capital of Tamil Nadu.
(ii)
(i)
(ii)
(iii)
he
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3.
2 is a complex number.
All triangles are equilateral triangles.
The number 2 is not greater than 7.
Every natural number is not an integer.
The negation of the first statement is the number x is a rational number.
which is the same as the second statement This is because when a number
is not irrational, it is a rational. Therefore, the given pairs are negations of
each other.
The negation of the first statement is x is an irrational number which is
the same as the second statement. Therefore, the pairs are negations of
each other.
Number 3 is prime; number 3 is odd (True).
All integers are positive; all integers are negative (False).
100 is divisible by 3,100 is divisible by 11 and 100 is divisible by 5 (False).
is
(ii)
(iii)
(iv)
(v)
2. (i)
bl
1.
EXERCISE 14.3
460
MATHEMATICS
No. The negation of the statement in (i) is There exists real number x and
y for which x + y y + x, instead of the statement given in (ii).
4. (i) Exclusive
(ii) Inclusive
(iii) Exclusive
3.
(v)
(i)
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2.
is
(i)
(ii)
(iii)
(iv)
bl
1.
he
EXERCISE 14.4
(ii)
(iii)
(iv)
(v)
3.
(i)
(ii)
ANSWERS
he
(iii)
(iv)
4. a (i)
(ii)
b (i)
(ii)
EXERCISE 14.5
(i) False. By definition of the chord, it should intersect the circle in two points.
(ii) False. This can be shown by giving a counter example. A chord which is not
a dimaeter gives the counter example.
(iii) True. In the equation of an ellipse if we put a = b, then it is a circle
(Direct Method)
(iv) True, by the rule of inequality
bl
is
5.
461
no N
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(i)
(ii)
(iii)
(iv)
2. (i)
1.
462
MATHEMATICS
EXERCISE 15.1
bl
is
he
3.
2. 8.4
3. 2.33
4. 7
5. 6.32
6. 16
7. 3.23
8. 5.1
11. 10.34
12. 7.35
no N
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1. 3
9. 157.92
10. 11.28
EXERCISE 15.2
1. 9, 9.25
n + 1 n2 1
,
2.
2
12
5. 100, 29.09
6. 64, 1.69
9. 93, 105.52, 10.27
3. 16.5, 74.25
7. 107, 2276
10. 5.55, 43.5
4. 19, 43.4
8. 27, 132
EXERCISE 15.3
1. B
4. A
2. Y
5. Weight
3. (i) B, (ii) B
1. 4, 8
2. 6, 8
3. 24, 12
5. (i) 10.1, 1.99 (ii) 10.2, 1.98
6. Highest Chemistry and lowest Mathematics
7. 20, 3.036
ANSWERS
463
EXERCISE 16.1
he
is
no N
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4.
5.
6.
7.
8.
9.
10.
11.
12.
bl
1.
2.
or
3.
13.
14.
15.
16.
EXERCISE 16.2
1. No.
2. (i) {1, 2, 3, 4, 5, 6}
(ii) (iii) {3, 6}
(iv) {1, 2, 3} (v) {6}
(vi) {3, 4, 5, 6}, AB = {1, 2, 3, 4, 5, 6}, AB = , BC = {3, 6}, EF = {6},
DE = ,
A C = {1, 2,4,5}, D E = {1,2,3}, E F = , F = {1, 2}
3. A = {(3,6), (4,5), (5, 4), (6,3), (4,6), (5,5), (6,4), (5,6), (6,5), (6,6)}
B = {(1,2), (2,2), (3, 2), (4,2), (5,2), (6,2), (2,1), (2,3), (2,4), (2,5), (2,6)}
C ={(3,6), (6,3), (5, 4), (4,5), (6,6)}
A and B, B and C are mutually exclusive.
4.
5.
464
MATHEMATICS
(iii) Getting at most two tails, and getting exactly two tails
(iv) Getting exactly one head and getting exactly two heads
(v) Getting exactly one tail, getting exactly two tails, and getting exactly
three tails
$Note
no N
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6. A = {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
B = {(1, 1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}
C = {(1, 1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}
(i) A = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)} = B
(ii) B = {(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} = A
(iii) AB = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5),
(3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (2,1), (2,2), (2,3), (2,5),
(2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4),
(6,5), (6,6)} = S
(iv) A B =
(v) A C = {(2,4), (2,5), (2,6), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3),
(6,4), (6,5), (6,6)}
(vi) B C = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (3,1), (3,2),
(3,3), (3,4), (3,5), (3,6), (4,1), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}
(vii) B C = {(1,1), (1,2), (1,3), (1,4), (3,1), (3,2)}
(viii) A B C = {(2,4), (2,5), (2,6), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2),
(6,3), (6,4), (6,5), (6,6)}
7. (i) True (ii) True (iii) True (iv) False (v) False (vi) False
EXERCISE 16.3
1. (a) Yes
(b) Yes
(c) No
(d) No
3. (i)
1
2
1
5
(ii) (iii) (iv) 0 (v)
2
3
6
6
5. (i)
1
1
(ii)
12
12
6.
3
5
(e) No
2.
3
4
4. (a) 52 (b)
1
1
1
(ii)
(c) (i)
52
13
2
ANSWERS
465
7. Rs 4.00 gain, Rs 1.50 gain, Re 1.00 loss, Rs 3.50 loss, Rs 6.00 loss.
P (Losing Rs 3.50) =
9.
1
1
, P (Losing Rs 6.00) = .
4
16
1
3
1
7
1
1
3
1
7
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
8
8
2
8
8
8
8
8
8
9
11
10. (i)
6
13
(ii)
7
13
11.
1
38760
is
8. (i)
1
1
3
, P(Winning Rs 1.50) = , P (Losing Re. 1.00) =
16
4
8
he
P ( Winning Rs 4.00) =
7
15
14.
4
5
no N
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13. (i)
bl
12. (i) No, because P(AB) must be less than or equal to P(A) and P(B), (ii) Yes
15. (i)
5
8
3
8
(ii)
18. 0.6
21. (i)
19
30
11
30
(ii)
(iii)
16. No
19. 0.55
20. 0.65
2
15
20
30
C3 . 13 C1
52
C4
C5
60
C5
3. (i)
1
1
5
(ii)
(iii)
2
2
6
4. (a)
5. (a)
17
33
6.
(ii) 1
(b)
16
33
C5
60
C5
13
1. (i)
2.
999
1000
33
83
(ii)
3
8
10.
C2
10000
C2
2
3
9990
(b)
1
5040
8.
4
5
(c)
9990
C10
C10
10000