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Contents
PART I
Foreword
Preface
v
vii
1
...
1 Introduction
1
...
3 Types of Functions
1
...
5 Binary Operations
1
1
2
7
12
19
2
...
1 Introduction
2
...
3 Properties of Inverse Trigonometric Functions
33
33
33
42
3
...
1 Introduction
3
...
3 Types of Matrices
3
...
5 Transpose of a Matrix
3
...
7 Elementary Operation (Transformation) of a Matrix
3
...
Determinants
4
...
2 Determinant
4
...
4 Area of a Triangle
4
...
6 Adjoint and Inverse of a Matrix
4
...
Continuity and Differentiability
5
...
2 Continuity
5
...
4 Exponential and Logarithmic Functions
5
...
6 Derivatives of Functions in Parametric Forms
5
...
8 Mean Value Theorem
147
147
147
161
170
174
179
181
184
6
...
1 Introduction
6
...
3 Increasing and Decreasing Functions
6
...
5 Approximations
6
...
1
...
1
...
2
...
2
...
2
...
It may
be very hard to define mathematical beauty but that is just as true of
beauty of any kind, we may not know quite what we mean by a
beautiful poem, but that does not prevent us from recognising
one when we read it
...
H
...
1 Introduction
Recall that the notion of relations and functions, domain,
co-domain and range have been introduced in Class XI
along with different types of specific real valued functions
and their graphs
...
Let A be
the set of students of Class XII of a school and B be the
set of students of Class XI of the same school
...
However, abstracting from
this, we define mathematically a relation R from A to B as an arbitrary subset
of A × B
...
In general, (a, b) R, we do not bother whether there is a recognisable
connection or link between a and b
...
In this chapter, we will study different types of relations and functions, composition
of functions, invertible functions and binary operations
...
2 Types of Relations
In this section, we would like to study different types of relations
...
Thus, the empty set ✄ and A × A are two
extreme relations
...
This is the empty set, as no pair (a, b) satisfies the condition
a – b = 10
...
These two extreme examples lead us to the
following definitions
...
e
...
Definition 2 A relation R in a set A is called universal relation, if each element of A
is related to every element of A, i
...
, R = A × A
...
Example 1 Let A be the set of all students of a boys school
...
Solution Since the school is boys school, no student of the school can be sister of any
student of the school
...
It is also
obvious that the difference between heights of any two students of the school has to be
less than 3 meters
...
Remark In Class XI, we have seen two ways of representing a relation, namely
roaster method and set builder method
...
We may also use this notation, as and when convenient
...
One of the most important relation, which plays a significant role in Mathematics,
is an equivalence relation
...
Definition 3 A relation R in a set A is called
(i) reflexive, if (a, a) ✂ R, for every a ✂ A,
(ii) symmetric, if (a1, a2) ✂ R implies that (a2, a1) ✂ R, for all a1, a2 ✂ A
...
RELATIONS AND FUNCTIONS
3
Definition 4 A relation R in a set A is said to be an equivalence relation if R is
reflexive, symmetric and transitive
...
Show that R is an equivalence relation
...
Further,
(T1, T2) ✂ R ✞ T1 is congruent to T2 ✞ T2 is congruent to T1 ✞ (T2, T1) ✂ R
...
Moreover, (T1, T2), (T2, T3) ✂ R ✞ T1 is congruent to T2 and T2 is
congruent to T3 ✞ T1 is congruent to T3 ✞ (T1, T3) ✂ R
...
Example 3 Let L be the set of all lines in a plane and R be the relation in L defined as
R = {(L1, L2) : L1 is perpendicular to L2}
...
Solution R is not reflexive, as a line L1 can not be perpendicular to itself, i
...
, (L1, L1)
✟ R
...
R is not transitive
...
1
L2 is perpendicular to L3, then L1 can never be perpendicular to
L3
...
e
...
Example 4 Show that the relation R in the set {1, 2, 3} given by R = {(1, 1), (2, 2),
(3, 3), (1, 2), (2, 3)} is reflexive but neither symmetric nor transitive
...
Also, R is not symmetric,
as (1, 2) ✂ R but (2, 1) ✟ R
...
Example 5 Show that the relation R in the set Z of integers given by
R = {(a, b) : 2 divides a – b}
is an equivalence relation
...
Further, if (a, b) ✂ R, then
2 divides a – b
...
Hence, (b, a) ✂ R, which shows that R is
symmetric
...
Now, a – c = (a – b) + (b – c) is even (Why?)
...
This
shows that R is transitive
...
4
MATHEMATICS
In Example 5, note that all even integers are related to zero, as (0, ± 2), (0, ± 4)
etc
...
, do not lie in R
...
Therefore, the set E of all even integers and the set O of all odd integers are subsets of
Z satisfying following conditions:
(i) All elements of E are related to each other and all elements of O are related to
each other
...
(iii) E and O are disjoint and Z = E ✠ O
...
Similarly, O is the equivalence class containing 1 and is denoted by [1]
...
Infact, what we have seen above is true
for an arbitrary equivalence relation R in a set X
...
(ii) no element of Ai is related to any element of Aj , i ✡ j
...
The subsets Ai are called equivalence classes
...
For example, consider a subdivision of the set Z given
by three mutually disjoint subsets A1, A2 and A3 whose union is Z with
A1 = {x ✂ Z : x is a multiple of 3} = {
...
}
A2 = {x ✂ Z : x – 1 is a multiple of 3} = {
...
}
A3 = {x ✂ Z : x – 2 is a multiple of 3} = {
...
}
Define a relation R in Z given by R = {(a, b) : 3 divides a – b}
...
Also, A1 coincides with the set of all integers in Z which are related to zero, A2
coincides with the set of all integers which are related to 1 and A3 coincides with the
set of all integers in Z which are related to 2
...
In fact, A1 = [3r], A2 = [3r + 1] and A3 = [3r + 2], for all r ✂ Z
...
Show that R is an equivalence
relation
...
RELATIONS AND FUNCTIONS
5
Solution Given any element a in A, both a and a must be either odd or even, so
that (a, a) ✂ R
...
Similarly, (a, b) ✂ R and (b, c) ✂ R ✞ all elements a, b, c, must be
either even or odd simultaneously ✞ (a, c) ✂ R
...
Further, all the elements of {1, 3, 5, 7} are related to each other, as all the elements
of this subset are odd
...
Also, no element of the subset {1, 3, 5, 7} can be
related to any element of {2, 4, 6}, as elements of {1, 3, 5, 7} are odd, while elements
of {2, 4, 6} are even
...
1
1
...
, 13, 14} defined as
R = {(x, y) : 3x – y = 0}
(ii) Relation R in the set N of natural numbers defined as
R = {(x, y) : y = x + 5 and x < 4}
(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as
R = {(x, y) : y is divisible by x}
(iv) Relation R in the set Z of all integers defined as
R = {(x, y) : x – y is an integer}
(v) Relation R in the set A of human beings in a town at a particular time given by
(a) R = {(x, y) : x and y work at the same place}
(b) R = {(x, y) : x and y live in the same locality}
(c) R = {(x, y) : x is exactly 7 cm taller than y}
(d) R = {(x, y) : x is wife of y}
(e) R = {(x, y) : x is father of y}
2
...
3
...
4
...
5
...
6
MATHEMATICS
6
...
7
...
8
...
Show that all the
elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are
related to each other
...
9
...
Find the set of all elements related to 1 in each case
...
Give an example of a relation
...
(ii) Transitive but neither reflexive nor symmetric
...
(iv) Reflexive and transitive but not symmetric
...
11
...
Further, show that the set of
all points related to a point P ✡ (0, 0) is the circle passing through P with origin as
centre
...
Show that the relation R defined in the set A of all triangles as R = {(T1, T2) : T1
is similar to T2}, is equivalence relation
...
Which
triangles among T1, T2 and T3 are related?
13
...
What is the
set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?
14
...
Show that R is an equivalence relation
...
RELATIONS AND FUNCTIONS
7
15
...
Choose the correct answer
...
(B) R is reflexive and transitive but not symmetric
...
(D) R is an equivalence relation
...
Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b > 6}
...
(A) (2, 4) ✂ R
(B) (3, 8) ✂ R
(C) (6, 8) ✂ R
(D) (8, 7) ✂ R
1
...
along with their graphs have been given in Class XI
...
As the concept of function is of paramount importance in mathematics and
among other disciplines as well, we would like to extend our study about function from
where we finished earlier
...
Consider the functions f1, f2, f3 and f4 given by the following diagrams
...
2, we observe that the images of distinct elements of X1 under the function
f1 are distinct, but the image of two distinct elements 1 and 2 of X1 under f2 is same,
namely b
...
The above observations lead to the following definitions:
Definition 5 A function f : X ✌ Y is defined to be one-one (or injective), if the images
of distinct elements of X under f are distinct, i
...
, for every x1, x2 ✂ X, f (x1) = f (x2)
implies x1 = x2
...
The function f1 and f4 in Fig 1
...
2 (ii) and (iii) are many-one
...
e
...
The function f3 and f4 in Fig 1
...
2 (i) is
not onto as elements e, f in X2 are not the image of any element in X1 under f1
...
2 (i) to (iv)
Remark f : X ✌ Y is onto if and only if Range of f = Y
...
The function f4 in Fig 1
...
Example 7 Let A be the set of all 50 students of Class X in a school
...
Show that f is one-one
but not onto
...
Therefore,
f must be one-one
...
This implies that 51 in N is not roll number of any student of
the class, so that 51 can not be image of any element of X under f
...
Example 8 Show that the function f : N ✌ N, given by f (x) = 2x, is one-one but not
onto
...
Further,
f is not onto, as for 1 ✂ N, there does not exist any x in N such that f (x) = 2x = 1
...
Solution f is one-one, as f (x1) = f (x2) ✞ 2x1 = 2x2 ✞ x1 = x2
...
( ) = y
...
2
2
2
Fig 1
...
Solution f is not one-one, as f (1) = f (2) = 1
...
Also for 1 ✂ N, we
have f (1) = 1
...
Solution Since f (– 1) = 1 = f (1), f is not oneone
...
Therefore f is not onto
...
Fig 1
...
Note that if x1 is odd and x2 is even, then we will have
x1 + 1 = x2 – 1, i
...
, x2 – x1 = 2 which is impossible
...
Therefore,
both x1 and x2 must be either odd or even
...
Then
f (x1) = f (x2) x1 + 1 = x2 + 1 x1 = x2
...
Thus, f is one-one
...
Thus, f is onto
...
Solution Suppose f is not one-one
...
Also, the image of 3 under f can be
only one element
...
Hence, f must be one-one
...
Solution Since f is one-one, three elements of {1, 2, 3} must be taken to 3 different
elements of the co-domain {1, 2, 3} under f
...
Remark The results mentioned in Examples 13 and 14 are also true for an arbitrary
X is necessarily onto and an onto map
finite set X, i
...
, a one-one function f : X
f : X X is necessarily one-one, for every finite set X
...
In fact, this is a characteristic
difference between a finite and an infinite set
...
2
✍ ✌ R✍ defined by f (x) =
1
is one-one and onto,
x
where R is the set of all non-zero real numbers
...
Check the injectivity and surjectivity of the following functions:
(i) f : N
N given by f (x) = x2
Z given by f (x) = x2
(ii) f : Z
(iii) f : R
R given by f (x) = x2
N given by f (x) = x3
(iv) f : N
(v) f : Z
Z given by f (x) = x3
3
...
1
...
Show that the Modulus Function f : R ✌ R, given by f (x) = | x |, is neither oneone nor onto, where | x | is x, if x is positive or 0 and | x | is – x, if x is negative
...
Show that the Signum Function f : R ✌ R, given by
✁1, if x 0
✂
f ( x) ✄ ☎0, if x ✄ 0
✂ –1, if x ✆ 0
✝
is neither one-one nor onto
...
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function
from A to B
...
7
...
Justify your answer
...
Let A and B be sets
...
✟n ✞1
✠✠ 2 , if n is odd
9
...
✠ n , if n is even
✠☛ 2
State whether the function f is bijective
...
10
...
Consider the function f : A ✌ B defined by
✎ x✍2✏
✒
...
✓ x✍3✔
f (x) = ✑
11
...
Choose the correct answer
...
12
...
Choose the correct answer
...
12
MATHEMATICS
1
...
Consider the set A of all students, who appeared in Class X of a Board
Examination in 2006
...
In order to have confidentiality, the Board arranges to deface the
roll numbers of students in the answer scripts and assigns a fake code number to each
roll number
...
This gives rise to two functions f : A ✌ B and g : B ✌ C given by f (a) = the
roll number assigned to the student a and g (b) = the code number assigned to the roll
number b
...
Thus, by the
combination of these two functions, each student is eventually attached a code number
...
Then the composition of
f and g, denoted by gof, is defined as the function gof : A ✌ C given by
gof (x) = g(f (x)),
x
✂ A
...
5
Example 15 Let f : {2, 3, 4, 5} ✌ {3, 4, 5, 9} and g : {3, 4, 5, 9} ✌ {7, 11, 15} be
functions defined as f (2) = 3, f (3) = 4, f (4) = f (5) = 5 and g (3) = g (4) = 7 and
g (5) = g (9) = 11
...
Solution We have gof (2) = g (f (2)) = g (3) = 7, gof (3) = g (f (3)) = g (4) = 7,
gof (4) = g (f (4)) = g (5) = 11 and gof (5) = g (5) = 11
...
Show that gof ✡ fog
...
Similarly,
fog (x) = f (g (x)) = f (3x2) = cos (3x2)
...
Hence,
gof ✡ fog
...
Solution We have
✖ (3x ✕ 4) ✗
7✘
✙✕4
21x ✣ 28 ✣ 20 x ✤ 28 41x
✖ 3x ✕ 4 ✗
✛ (5 x ✚ 7) ✜
✥
✥x
gof ( x) ✢ g ✘
✢
=
✙
15 x ✣ 20 ✤ 15 x ✣ 21 41
x
(3
4)
✕
x
5
7
✚
✛
✜
✖
✗
5✘
✙✚3
✛ (5 x ✚ 7) ✜
✖ (7 x ✕ 4) ✗
3✘
✙ ✕ 4 21x ✟ 12 ✟ 20 x ✡ 12 41x
✖ 7x ✕ 4 ✗
✛ (5 x ✚ 3) ✜
✠
✠x
(
)
fog
x
f
✢
✢
Similarly,
=
✘
✙
35 x ✟ 20 ✡ 35 x ✟ 21 41
✛ 5x ✚ 3 ✜
✖ (7 x ✕ 4) ✗
5✘
✙✚7
✛ (5 x ✚ 3) ✜
Thus, gof (x) = x, ✓ x ✔ B and fog (x) = x, ✓ x ✔ A, which implies that gof = IB
and fog = IA
...
Solution Suppose gof (x1) = gof (x2)
g (f (x1)) = g(f (x 2))
✧
f (x1) = f (x2), as g is one-one
✧
x1 = x2, as f is one-one
✧
Hence, gof is one-one
...
Solution Given an arbitrary element z ✔ C, there exists a pre-image y of z under g
such that g (y) = z, since g is onto
...
Therefore, gof (x) = g (f (x)) = g (y) = z, showing that gof
is onto
...
Are f and g both necessarily one-one
...
Then, gof (x) = x x, which shows that gof is one-one
...
Example 21 Are f and g both necessarily onto, if gof is onto?
Solution Consider f : {1, 2, 3, 4} ✌ {1, 2, 3, 4} and g : {1, 2, 3, 4} ✌ {1, 2, 3} defined
as f (1) = 1, f (2) = 2, f (3) = f (4) = 3, g (1) = 1, g (2) = 2 and g (3) = g (4) = 3
...
Remark It can be verified in general that gof is one-one implies that f is one-one
...
Now, we would like to have close look at the functions f and g described in the
beginning of this section in reference to a Board Examination
...
After the answer scripts are
examined, examiner enters the mark against each code number in a mark book and
submits to the office of the Board
...
Further, the process reverse to f assigns a roll
number to the student having that roll number
...
We observe that while composing f and g, to get gof, first f
and then g was applied, while in the reverse process of the composite gof, first the
reverse process of g is applied and then the reverse process of f
...
Show that there exists a function g : {a, b, c} ✌ {1, 2, 3}
such that gof = IX and fog = IY, where, X = {1, 2, 3} and Y = {a, b, c}
...
It is
easy to verify that the composite gof = IX is the identity function on X and the composite
fog = IY is the identity function on Y
...
Not only this, even the converse
is also true , i
...
, if f : X ✌ Y is a function such that there exists a function g : Y ✌ X
such that gof = IX and fog = IY, then f must be one-one and onto
...
The function g is called the inverse of f and
is denoted by f –1
...
This fact significantly helps for proving a
function f to be invertible by showing that f is one-one and onto, specially when the
actual inverse of f is not to be determined
...
Show that f is invertible
...
Solution Consider an arbitrary element y of Y
...
This shows that x ✁
g ( y) ✁
( y 3)
...
Now, gof (x) = g (f (x)) = g (4x + 3) =
4
4
✆ ( y ☎ 3) ✝ ✞ 4 ( y ☎ 3) ✟ 3 = y – 3 + 3 = y
...
Example 24 Let Y = {n2 : n ✂ N } ✍ N
...
Show that
f is invertible
...
Solution An arbitrary element y in Y is of the form n2, for some n ✂ N
...
This gives a function g : Y ✌ N , defined by g (y) =
y
...
Hence, f is invertible with f –1 = g
...
Show that
f : N ✌ S, where, S is the range of f, is invertible
...
Solution Let y be an arbitrary element of range f
...
This gives x ✑
✔✒ y ✖ 6 ✓ ✖ 3✕
2
, as y ✗ 6
...
2
gof (x) = g (f (x)) = g (4x2 + 12x + 15) = g ((2x + 3)2 + 6)
=
(2 x ☛ 3) 2 ☛ 6 ☞ 6 ✆ ☞ 3✞
✟✝
✓
and
✖
✖
✚
fog (y) = f
✑✎
y ✕ 6 ✏ ✕ 3✒ ✔
2
✠
✍
2
✓
✗ ✘✖
✗
✖
✛
✚
2 x ☛ 3 ☞ 3✡
✍ x
2
✙
2
Hence,
✣✜
y ✦ 6 ✢ ✦ 3 ✧ 3✤✥
✧
2
3 ✗✗
✙
6
✛
2
=
✔
2 ✑✎ y ✕ 6 ✏ ✕ 3✒
6 ★ ✜ y ✦ 6✢
2
✧
6 = y – 6 + 6 = y
...
This implies that f is invertible with f –1 = g
...
Show that ho(gof ) = (hog) of
...
((hog) o f ) (x) = (hog) ( f (x)) = (hog) (2x) = h ( g (2x))
= h(3(2x) + 4) = h(6x + 4) = sin (6x + 4),
✩
x ✬ N
...
This result is true in general situation as well
...
Proof We have
and
Hence,
ho(gof ) (x) = h(gof (x)) = h(g (f (x))),
✩
x in X
(hog) of (x) = hog (f (x)) = h(g (f (x))),
✩
x in X
...
Example 27 Consider f : {1, 2, 3} ✌ {a, b, c} and g : {a, b, c} ✌ {apple, ball, cat}
defined as f (1) = a, f (2) = b, f (3) = c, g(a) = apple, g(b) = ball and g(c) = cat
...
Find out f –1, g–1 and (gof)–1 and show that
(gof) –1 = f –1o g–1
...
Let
f –1: {a, b, c}
(1, 2, 3} and g–1 : {apple, ball, cat}
{a, b, c} be defined as
–1
–1
f {a} = 1, f {b} = 2, f –1{c} = 3, g –1{apple} = a, g –1{ball} = b and g –1{cat} = c
...
Now, gof : {1, 2, 3}
gof (1) = apple, gof (2) = ball, gof (3) = cat
...
It is easy to see that (g o f)–1 o (g o f) = I{1, 2, 3} and
(gof) o (gof)–1 = ID
...
Now, f –1og–1 (apple)= f –1(g–1(apple)) = f –1(a) = 1 = (gof)–1 (apple)
f –1og–1 (ball) = f –1(g–1(ball)) = f –1(b) = 2 = (gof)–1 (ball) and
f –1og–1 (cat) = f –1(g–1(cat)) = f –1(c) = 3 = (gof)–1 (cat)
...
The above result is true in general situation also
...
Then gof is also
invertible with (gof)–1 = f –1og–1
...
Now,
(f –1og –1) o (gof) = ((f –1og–1) og) of, by Theorem 1
= (f –1o(g–1og)) of, by Theorem 1
= (f –1 o IY) of, by definition of g–1
= IX
...
✌
✌
✌
✌
✌
✌
Example 28 Let S = {1, 2, 3}
...
Find f –1, if it exists
...
(b) Since f (2) = f (3) = 1, f is not one-one, so that f is not invertible
...
18
MATHEMATICS
EXERCISE 1
...
Let f : {1, 3, 4} ✌ {1, 2, 5} and g : {1, 2, 5} ✌ {1, 3} be given by
f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}
...
2
...
Show that
(f + g) o h = foh + goh
(f
...
(goh)
3
...
3
(4 x 3)
2
2
, x ✂ , show that fof (x) = x, for all x ✄
...
If f (x) =
5
...
Show that f : [–1, 1] ✌ R, given by f (x) =
x
( x ☎ 2)
is one-one
...
(Hint: For y ✆ Range f, y = f (x) =
x
x✝2
, for some x in [–1, 1], i
...
, x =
2y
)
(1 ✞ y )
7
...
Show that f is invertible
...
8
...
Show that f is invertible with the
inverse f –1 of f given by f –1(y) =
real numbers
...
Consider f : R+ ✌ [– 5, ✎) given by f (x) = 9x2 + 6x – 5
...
3
✠
10
...
Show that f has unique inverse
...
Then for all y ✡ Y,
fog1(y) = 1Y(y) = fog2(y)
...
11
...
Find
f –1 and show that (f –1)–1 = f
...
Let f : X ✌ Y be an invertible function
...
e
...
13
...
Let f : R –
1
3 3
(3 ☛ x ) ,
☞
✑✏
✓
4✍
✒
3✔
✌
g ( y) ✢
(C)
g ( y) ✣
3y
3✘ 4y
4y
3 ✏ 4y
(D) (3 – x3)
...
The inverse of
3x ✕ 4
4✗
✚ given by
3✜
(B)
g ( y) ✢
(D)
g ( y) ✣
4y
4 ✘ 3y
3y
4 ✏ 3y
1
...
The main feature of these
operations is that given any two numbers a and b, we associate another number a + b
a
, b ✤ 0
...
When we need to add three numbers, we first add two numbers
and the result is then added to the third number
...
If we want to
have a general definition which can cover all these four operations, then the set of
numbers is to be replaced by an arbitrary set X and then general binary operation is
nothing but association of any pair of elements a, b from X to another element of X
...
We denote
✍ (a, b) by a ✍ b
...
Further, show that division is a binary
operation on the set R of nonzero real numbers
...
But ÷: R × R ✌ R, given by (a, b) ✌
operation, as for b = 0,
a
, is not a function and hence not a binary
b
a
is not defined
...
Example 30 Show that subtraction and division are not binary operations on N
...
Similarly, ÷ : N × N ✌ N, given by (a, b) ✌ a ÷ b
3
✟ N
...
✍
:R× R
✌
R given by (a, b)
✌
Solution Since ✍ carries each pair (a, b) to a unique element a + 4b2 in R, ✍ is a binary
operation on R
...
Show that ✠ : P × P ✌ P
given by (A, B) ✌ A ✠ B and ☛ : P × P ✌ P given by (A, B) ✌ A ☛ B are binary
operations on the set P
...
Similarly, the intersection operation ☛ carries
each pair (A, B) in P × P to a unique element A ☛ B in P, ☛ is a binary operation on P
...
Solution Since ✂ carries each pair (a, b) in R × R to a unique element namely
maximum of a and b lying in R, ✂ is a binary operation
...
Remark ✂ (4, 7) = 7, ✂ (4, – 7) = 4, ✁ (4, 7) = 4 and ✁ (4, – 7) = – 7
...
For example
consider A = {1, 2, 3}
...
1)
...
Table 1
...
This can be
generalised for general operation ✍ : A × A ✌ A
...
, an}
...
Conversely, given any operation table having n rows and n columns with each entry
being an element of A = {a1, a2,
...
One may note that 3 and 4 can be added in any order and the result is same, i
...
,
3 + 4 = 4 + 3, but subtraction of 3 and 4 in different order give different results, i
...
,
3 – 4 ✡ 4 – 3
...
Thus, addition and
multiplication of 3 and 4 are meaningful, but subtraction and division of 3 and 4 are
meaningless
...
22
MATHEMATICS
This leads to the following definition:
Definition 11 A binary operation ✍ on the set X is called commutative, if a ✍ b = b ✍ a,
for every a, b ✂ X
...
Solution Since a + b = b + a and a × b = b × a, ✁ a, b ✂ R, ‘+’ and ‘×’ are
commutative binary operation
...
Similarly, 3 ÷ 4 ✡ 4 ÷ 3 shows that ‘÷’ is not commutative
...
Solution Since 3 ✍ 4 = 3 + 8 = 11 and 4 ✍ 3 = 4 + 6 = 10, showing that the operation ✍
is not commutative
...
The expression a ✍ b ✍ c may be interpreted as
(a ✍ b) ✍ c or a ✍ (b ✍ c) and these two expressions need not be same
...
Therefore, association of three numbers 8, 5 and 3 through
the binary operation ‘subtraction’ is meaningless, unless bracket is used
...
Thus, association of 3 or even more than 3 numbers through addition is
meaningful without using bracket
...
Example 36 Show that addition and multiplication are associative binary operation on
R
...
Division is not associative on R
...
However, subtraction and division are not
associative, as (8 – 5) – 3 ✡ 8 – (5 – 3) and (8 ÷ 5) ÷ 3 ✡ 8 ÷ (5 ÷ 3)
...
Solution The operation ✍ is not associative, since
(8 ✍ 5) ✍ 3 = (8 + 10) ✍ 3 = (8 + 10) + 6 = 24,
while
8 ✍ (5 ✍ 3) = 8 ✍ (5 + 6) = 8 ✍ 11 = 8 + 22 = 30
...
✍ an which is not
ambiguous
...
✍ an is ambiguous
unless brackets are used
...
RELATIONS AND FUNCTIONS
23
For the binary operation ‘+’ on R, the interesting feature of the number zero is that
a + 0 = a = 0 + a, i
...
, any number remains unaltered by adding zero
...
This leads
to the following definition:
Definition 13 Given a binary operation ✍ : A × A ✌ A, an element e ✂ A, if it exists,
a ✂ A
...
But there is no identity element for the operations
✌ R and ÷ : R✁ × R✁ ✌ R✁
...
Further, there is no element
e in R with a – e = e – a, a
...
Hence, ‘–’ and ‘÷’ do not have identity element
...
In fact the addition operation on N does not have
any identity
...
Similarly, for the multiplication operation on R, given any a ✡ 0 in R, we can choose
in R such that a ×
1
a
1
1
= 1(identity for ‘×’) = × a
...
Example 39 Show that – a is the inverse of a for the addition operation ‘+’ on R and
1
is the inverse of a ✡ 0 for the multiplication operation ‘×’ on R
...
1
1
1
Similarly, for a ✡ 0, a × = 1 = × a implies that is the inverse of a for multiplication
...
a
Solution Since – a ✟ N, – a can not be inverse of a for addition operation on N,
although – a satisfies a + (– a) = 0 = (– a) + a
...
Similarly, for a ✡ 1 in N,
Examples 34, 36, 38 and 39 show that addition on R is a commutative and associative
binary operation with 0 as the identity element and – a as the inverse of a in R a
...
4
1
...
In the event that ✍ is not a binary operation, give justification for this
...
or associative
...
Consider the binary operation ✄ on the set {1, 2, 3, 4, 5} defined by
a ☎ b = min {a, b}
...
RELATIONS AND FUNCTIONS
25
4
...
2)
...
(Hint: use the following table)
Table 1
...
Let ✍ be the binary operation on the set {1, 2, 3, 4, 5} defined by
a ✍ b = H
...
F
...
Is the operation ✍ same as the operation ✍ defined
in Exercise 4 above? Justify your answer
...
Let ✍ be the binary operation on N given by a ✍ b = L
...
M
...
Find
(ii) Is ✍ commutative?
(i) 5 ✍ 7, 20 ✍ 16
(iii) Is ✍ associative?
(iv) Find the identity of ✍ in N
(v) Which elements of N are invertible for the operation ✍?
7
...
C
...
of a and b a binary
operation? Justify your answer
...
Let ✍ be the binary operation on N defined by a ✍ b = H
...
F
...
Is ✍ commutative? Is ✍ associative? Does there exist identity for this binary
operation on N?
9
...
10
...
11
...
Find the identity element for ✍ on
A, if any
...
State whether the following statements are true or false
...
(i) For an arbitrary binary operation ✍ on a set N, a ✍ a = a a ✂ N
...
Consider a binary operation ✍ on N defined as a ✍ b = a3 + b3
...
(A) Is ✍ both associative and commutative?
(B) Is ✍ commutative but not associative?
(C) Is ✍ associative but not commutative?
(D) Is ✍ neither commutative nor associative?
Miscellaneous Examples
Example 41 If R1 and R2 are equivalence relations in a set A, show that R1 ☛ R2 is
also an equivalence relation
...
This implies that (a, a) ✂ R1 ☛ R2, a, showing R1 ☛ R2 is reflexive
...
Similarly, (a, b) ✂ R1 ☛ R2 and
(b, c) ✂ R1 ☛ R2 ✞ (a, c) ✂ R1 and (a, c) ✂ R2 ✞ (a, c) ✂ R1 ☛ R2
...
Thus, R1 ☛ R2 is an equivalence relation
...
Show that R is an equivalence relation
...
This shows that R is
reflexive
...
This
shows that R is symmetric
...
Thus, R
u
u
is transitive
...
ub = va ✞ xv
Example 43 Let X = {1, 2, 3, 4, 5, 6, 7, 8, 9}
...
Show that
R1 = R2
...
Therefore,
(x, y) ✂ R1 ✞ x – y is a multiple of 3 ✞ {x, y} ✝ {1, 4, 7} or {x, y} ✝ {2, 5, 8}
or {x, y} ✝ {3, 6, 9} ✞ (x, y) ✂ R2
...
Similarly, {x, y} ✂ R2 ✞ {x, y}
✝ {1, 4, 7} or {x, y} ✝ {2, 5, 8} or {x, y} ✝ {3, 6, 9} ✞ x – y is divisible by
3 ✞ {x, y} ✂ R1
...
Hence, R1 = R2
...
Define a relation R in X given by
R = {(a, b): f(a) = f(b)}
...
Solution For every a ✂ X, (a, a) ✂ R, since f (a) = f (a), showing that R is reflexive
...
Therefore, R is
symmetric
...
Hence, R is an equivalence
relation
...
(a) a
✍b=1
a, b
✂N
(b) a ✍ b =
(a ✁ b )
2
Solution
(a) Clearly, by definition a ✍ b = b ✍ a = 1,
(a ✍ b) ✍ c = (1 ✍ c) =1 and a ✍ (b ✍ c) = a ✍ (1) = 1,
R is both associative and commutative
...
Also
a, b, c ✂ N
...
Further,
2
2
(a
a✆b
✍ b) ✍ c = ✟✡☞ 2 ✠☛✎ ✍ c
...
=
2
But
a, b
a
4
b✆ c
✍ (b ✍ c) = a ✘ ✡☞✟ 2 ☛✎✠
b✙c
a✙
2 ✚ 2a ✙ b ✙ c ✛ a ✙ b ✙ 2c
=
Hence, ✍ is not associative
...
28
MATHEMATICS
Example 46 Find the number of all one-one functions from set A = {1, 2, 3} to itself
...
Therefore, total number of one-one maps from {1, 2, 3} to itself is
same as total number of permutations on three symbols 1, 2, 3 which is 3! = 6
...
Then show that the number of relations containing (1, 2)
and (2, 3) which are reflexive and transitive but not symmetric is four
...
Now, if we add
the pair (2, 1) to R1 to get R2, then the relation R2 will be reflexive, transitive but not
symmetric
...
However, we can not add any two pairs out of (2, 1),
(3, 2) and (3, 1) to R1 at a time, as by doing so, we will be forced to add the remaining
third pair in order to maintain transitivity and in the process, the relation will become
symmetric also which is not required
...
Example 48 Show that the number of equivalence relation in the set {1, 2, 3} containing
(1, 2) and (2, 1) is two
...
Now we are left with only 4 pairs namely (2, 3), (3, 2),
(1, 3) and (3, 1)
...
Thus, the only
equivalence relation bigger than R1 is the universal relation
...
Example 49 Show that the number of binary operations on {1, 2} having 1 as identity
and having 2 as the inverse of 2 is exactly one
...
e
...
Since 1 is the identity for the
desired binary operation ✍, ✍ ✁✂ ✁✄ ☎ ✁✂ ✍ (1, 2) = 2, ✍ (2, 1) = 2 and the only choice
left is for the pair (2, 2)
...
e
...
Thus,
the number of desired binary operation is only one
...
✆x ✝
N
...
But IN + IN is not onto, as we can find an element 3
in the co-domain N such that there does not exist any x in the domain N with
(IN + IN) (x) = 2x = 3
...
Show that f and g are one-one, but f + g is not
☎✝ 2 ✆✞
one-one
...
But (f + g) (0) = sin 0 + cos 0 = 1 and
✎
✎
✟ ✠
(f + g) ☛ ☞ = sin ✏ cos ✑ 1
...
2
2
✌ 2✍
Miscellaneous Exercise on Chapter 1
1
...
Find the function g : R ✒ R such
that g o f = f o g = 1R
...
Let f : W ✒ W be defined as f (n) = n – 1, if n is odd and f (n) = n + 1, if n is
even
...
Find the inverse of f
...
3
...
4
...
5
...
6
...
(Hint : Consider f (x) = x and g (x) = | x |)
...
Give examples of two functions f : N ✒ N and g : N ✒ N such that g o f is onto
but f is not onto
...
Given a non empty set X, consider P(X) which is the set of all subsets of X
...
10
...
12
...
14
...
Is R an equivalence relation
on P(X)? Justify your answer
...
Show that X is the identity element for this operation and X is the only invertible
element in P(X) with respect to the operation ✍
...
, n} to itself
...
Find F–1 of the following functions F from S
to T, if it exists
...
Show that ✍ is commutative but not
associative, o is associative but not commutative
...
[If it is so, we say that the operation ✍ distributes
over the operation o]
...
Given a non-empty set X, let ✍ : P(X) × P(X) ✌ P(X) be defined as
A * B = (A – B) ✠ (B – A), A, B ✂ P(X)
...
(Hint : (A – ✄) ✠ (✄ – A) = A and (A – A) ✠ (A – A) = A ✍ A = ✄)
...
15
...
Are f and g equal?
2
Justify your answer
...
16
...
Then number of relations containing (1, 2) and (1, 3) which are
reflexive and symmetric but not transitive is
(A) 1
(B) 2
(C) 3
(D) 4
17
...
Then number of equivalence relations containing (1, 2) is
(B) 2
(C) 3
(D) 4
(A) 1
by f (x) = x2 – x, x
✂ A and
g ( x) ✑ 2 x ✒
RELATIONS AND FUNCTIONS
31
18
...
Then, does fog and gof coincide in (0, 1]?
19
...
The main features
of this chapter are as follows:
✟ Empty relation is the relation R in X given by R = ✠ ✡ X × X
...
✟ Reflexive relation R in X is a relation with (a, a) ☛ R ☞ a ☛ X
...
✟ Transitive relation R in X is a relation satisfying (a, b) ☛ R and (b, c) ☛ R
implies that (a, c) ☛ R
...
✟ Equivalence class [a] containing a ☛ X for an equivalence relation R in X is
the subset of X containing all elements b related to a
...
A function f : X ✌ Y is onto (or surjective) if given any y ☛ Y, ✏ x ☛ X such
that f (x) = y
...
The composition of functions f : A ✌ B and g : B ✌ C is the function
gof : A ✌ C given by gof (x) = g(f (x)) ☞ x ☛ A
...
A function f : X ✌ Y is invertible if and only if f is one-one and onto
...
This is the characteristic property of a
finite set
...
An element e X is the identity element for binary operation : X × X X,
if a e = a = e a a X
...
The element b is called inverse of a and is denoted by a–1
...
An operation on X is associative if (a b) c = a (b c) a, b, c in X
...
Descartes (1596-1650), who used the word ‘function’ in his manuscript
“Geometrie” in 1637 to mean some positive integral power xn of a variable x
while studying geometrical curves like hyperbola, parabola and ellipse
...
Later G
...
Leibnitz (1646-1716) in his manuscript “Methodus tangentium inversa, seu de
functionibus” written in 1673 used the word ‘function’ to mean a quantity varying
from point to point on a curve such as the coordinates of a point on the curve, the
slope of the curve, the tangent and the normal to the curve at a point
...
He was the first to use the phrase ‘function
of x’
...
But the general adoption of symbols like f, F, ,
...
Later on, Joeph Louis Lagrange
(1736-1813) published his manuscripts “Theorie des functions analytiques” in
1793, where he discussed about analytic function and used the notion f (x), F(x),
(x) etc
...
Subsequently, Lejeunne Dirichlet
(1805-1859) gave the definition of function which was being used till the set
theoretic definition of function presently used, was given after set theory was
developed by Georg Cantor (1845-1918)
...
✄
✄
—☎ —
✄✑
Chapter
2
INVERSE TRIGONOMETRIC
FUNCTIONS
Mathematics, in general, is fundamentally the science of
self-evident things
...
1 Introduction
In Chapter 1, we have studied that the inverse of a function
f, denoted by f –1, exists if f is one-one and onto
...
In Class XI, we
studied that trigonometric functions are not one-one and
onto over their natural domains and ranges and hence their
inverses do not exist
...
Besides, some elementary properties will also be discussed
...
D
...
The concepts of inverse trigonometric functions is also used in science and engineering
...
2 Basic Concepts
In Class XI, we have studied trigonometric functions, which are defined as follows:
sine function, i
...
, sine : R [– 1, 1]
cosine function, i
...
, cos : R [– 1, 1]
✂
✂
✁ , n ✥ Z} ✂ R
2
cotangent function, i
...
, cot : R – { x : x = n☎, n ✥ Z} ✂ R
✁
secant function, i
...
, sec : R – { x : x = (2n + 1) , n ✥ Z} ✂ R – (– 1, 1)
2
cosecant function, i
...
, cosec : R – { x : x = n☎, n ✥ Z} ✂ R – (– 1, 1)
tangent function, i
...
, tan : R – { x : x = (2n + 1)
34
MATHEMATICS
We have also learnt in Chapter 1 that if f : X✂Y such that f (x) = y is one-one and
onto, then we can define a unique function g : Y✂X such that g (y) = x, where x ✄ X
and y = f (x), y ✄ Y
...
The function g is called the inverse of f and is denoted by f –1
...
Thus, g –1 = (f –1)–1 = f
...
If we restrict its domain to ☎✝
and onto with range [– 1, 1]
...
, is one-one and its range is [–1, 1]
...
We denote the
inverse of sine function by sin–1 (arc sine function)
...
Corresponding to each such interval, we get a branch of the
✓ ✙✒ ✒ ✔
function sin–1
...
When we refer
to the function sin–1, we take it as the function whose domain is [–1, 1] and range is
✓ ✙✒ ✒ ✔
✓ ✙✒ ✒ ✔
–1
✕✗ 2 , 2 ✖✘
...
In other words, if y = sin–1 x, then
2
2
sin y = x
...
Thus, the graph of sin–1 function can be obtained from the graph of original
function by interchanging x and y axes, i
...
, if (a, b) is a point on the graph of
sine function, then (b, a) becomes the corresponding point on the graph of inverse
INVERSE TRIGONOMETRIC FUNCTIONS
35
of sine function
...
The graphs of y = sin x and
y = sin–1 x are as given in Fig 2
...
The dark portion of the graph of
y = sin–1 x represent the principal value branch
...
e
...
This can be visualised by looking the graphs of y = sin x and
y = sin–1 x as given in the same axes (Fig 2
...
Fig 2
...
1 (ii)
Fig 2
...
If we restrict the domain of cosine function
to [0, ☎], then it becomes one-one and onto with range [–1, 1]
...
, is bijective with range as
[–1, 1]
...
We denote the inverse of the cosine function by cos–1 (arc cosine function)
...
Corresponding to each such interval, we get a branch of the
function cos–1
...
We write
cos–1 : [–1, 1] ✂ [0, ☎]
...
The
graphs of y = cos x and y = cos–1 x are given in Fig 2
...
Fig 2
...
2 (ii)
Let us now discuss cosec–1x and sec–1x as follows:
1
, the domain of the cosec function is the set {x : x ✄ R and
Since, cosec x =
sin x
x ✝ n☎, n ✄ Z} and the range is the set {y : y ✄ R, y ✞ 1 or y ✆ –1} i
...
, the set
R – (–1, 1)
...
If we restrict the domain of cosec function to
✁
✟
✡☞ ✠ 2 , 2 ☛✌ – {0}, then it is one to one and onto with its range as the set R – (– 1, 1)
...
, is bijective and its range is the set of all real numbers R – (–1, 1)
...
The
be any of the intervals ☎ , ✆ {0} , ☎
☞✍ 2 2 ✌✎
2
2✞
2
2
✝
✞
✝
✂ ✁ ✁✄
function corresponding to the range ☎ , ✆ {0} is called the principal value branch
✝ 2 2✞
of cosec–1
...
3 (i), (ii)
...
3 (ii)
Fig 2
...
It means that sec (secant function) assumes
Also, since sec x =
all real values except –1 < y < 1 and is not defined for odd multiples of
restrict the domain of secant function to [0, ✓] – {
✑
2
✑
2
...
Actually, secant function restricted to any of the
✁
3✁
✄✂✆
}, [0, ✂] – ✝ ✞ , [☎, 2☎] – {
} etc
...
Thus sec can be defined as a function whose domain is R– (–1, 1) and
3✁
✁
✁
range could be any of the intervals [– ☎, 0] – {
}, [0, ☎] – { }, [☎, 2☎] – {
} etc
...
intervals [–☎, 0] – {
The branch with range [0, ☎] – {
✁
2
} is called the principal value branch of the
function sec–1
...
4 (i), (ii)
...
4 (ii)
Fig 2
...
It means that tan function
is not defined for odd multiples of
✁
2
...
Actually, tangent function
✝
✞
✡ ✟3✠ , ✟✠ ☛ ✡ ✟✠ , ✠ ☛ ✡ ✠ , 3 ✠ ☛
restricted to any of the intervals ☞
✌, ☞
✌, ☞
✌ etc
...
Thus tan–1 can be defined as a function whose domain is R and
✂ 3✁ ✁ ✄ ✂ ✁ ✁ ✄ ✂ ✁ 3✁ ✄
, ✆, ☎
, ✆ , ☎ , ✆ and so on
...
The branch with range ☎
is called the principal value branch of the function tan–1
...
5 (i), (ii)
...
5 (i)
Fig 2
...
It means that cotangent function is not
defined for integral multiples of ✓
...
In fact, cotangent function restricted
to any of the intervals (–✓, 0), (0, ✓), (✓, 2✓) etc
...
Thus
cot –1 can be defined as a function whose domain is the R and range as any of the
40
MATHEMATICS
intervals (–☎, 0), (0, ☎), (☎, 2☎) etc
...
The function with range (0, ☎) is called the principal value branch of
the function cot –1
...
6 (i), (ii)
...
6 (i)
Fig 2
...
✁
✄
✝✟ ✆ 2 , 2 ✞✠
[0, ☎]
sin–1
:
[–1, 1]
✂
cos –1
:
[–1, 1]
✂
cosec–1
:
R – (–1,1)
✂
sec –1
:
R – (–1, 1) ✂
tan–1
:
R
✂
✓ ✌✡ , ✡ ✔
✕ 2 2✖
✗
✘
cot–1
:
R
✂
(0, ☎)
☛ ✡ ✡☞
✍✏ ✌ 2 , 2 ✎✑ – {0}
✒
[0, ☎] – { }
2
INVERSE TRIGONOMETRIC FUNCTIONS
41
Note
1
1
...
In fact (sin x)–1 =
and
sin x
similarly for other trigonometric functions
...
Whenever no branch of an inverse trigonometric functions is mentioned, we
mean the principal value branch of that function
...
The value of an inverse trigonometric functions which lies in the range of
principal branch is called the principal value of that inverse trigonometric
functions
...
✟☛✌ ✡ ✞ ✞ ✠☞✍
✁✄✆ ✂☎✝ ✖
✏✒✔ ✎ ✑✓✕
✁✄✆ ✗ ✂☎✝
✁✄✆ ✗ ✂☎✝
✙ ✘ ✙✘ ✒✔✏ ✎ ✓✕✑ ✛✣✥ ✚✢ ✚ ✜✤✦ ✛✣✥ ✚ ✜✤✦
✧
✁✄✆ ✗ ✂☎✝ ✖
✏✒✔ ✎ ✑✓✕ ★
Solution Let sin–1
1
1
= y
...
2
2
We know that the range of the principal value branch of sin–1 is
,
2 2
1
1
...
Then,
3
2
1
cot y
cot
= cot
= cot
3
3
3
3
We know that the range of principal value branch of cot –1 is (0,
1
2
1
2
cot
=
...
1
☛✌✟ ✡ ☞✍✠
✯
✩✫ ✪✬
✭ ✮
☛✌✟ ✡ ☞✍✠
Find the principal values of the following:
1
...
tan–1 (
1
2
3)
2
...
cos–1
3
2
3
...
tan–1 (–1)
) and
42
MATHEMATICS
7
...
cot–1 ( 3)
1 ✁
✄
2✆
✂✝
☎
9
...
cosec–1 ( ✞ 2 )
Find the values of the following:
11
...
cos–1 ☛
+ 2 sin–1
✎
✒
✔
1✏
✓
2✕
13
...
tan–1
(A)
✘
✗
(B)
✙
✘
3
(C)
✘
3
(D)
2✘
3
2
...
It may be mentioned here that these results are valid within the principal
value branches of the corresponding inverse trigonometric functions and wherever
they are defined
...
In fact, they will be valid only for some values of x for which
inverse trigonometric functions are defined
...
Let us recall that if y = sin–1x, then x = sin y and if x = sin y, then y = sin–1x
...
We now prove
some properties of inverse trigonometric functions
...
(i) sin–1 = cosec–1 x, x ✰ 1 or x ✖ – 1
x
(ii) cos–1
1
= sec–1x, x
x
✰
1 or x
✖
–1
INVERSE TRIGONOMETRIC FUNCTIONS
1
= cot–1 x, x > 0
x
To prove the first result, we put cosec–1 x = y, i
...
, x = cosec y
1
Therefore
= sin y
x
1
Hence
sin–1 = y
x
1
= cosec–1 x
or
sin–1
x
Similarly, we can prove the other parts
...
(i) sin–1 (–x) = – sin–1 x, x ✄ [– 1, 1]
(ii) tan–1 (–x) = – tan–1 x, x ✄ R
(iii) cosec–1 (–x) = – cosec–1 x, | x | ✞ 1
Let sin–1 (–x) = y, i
...
, –x = sin y so that x = – sin y, i
...
, x = sin (–y)
...
3
...
e
...
(iii) tan–1
4
...
Then x = sin y = cos ✝ ✆ y ✟
✠2 ✡
Therefore
cos–1 x =
☛
2
☞y =
☛
2
☞ sin –1 x
43
44
MATHEMATICS
Hence
sin–1 x + cos–1 x =
2
Similarly, we can prove the other parts
...
(i) tan–1x + tan–1 y = tan–1
(ii) tan–1x – tan–1 y = tan–1
(iii) 2tan–1x = tan–1
2x
1 – x2
x+y
, xy < 1
1 – xy
x–y
, xy > – 1
1 + xy
, |x| < 1
Let tan–1 x = ✟ and tan–1 y = ✠
...
6
...
Now
sin–1
2 tan y
2x
–1
2 = sin
1 ✂ tan 2 y
1☞ x
= sin–1 (sin 2y) = 2y = 2tan–1 x
INVERSE TRIGONOMETRIC FUNCTIONS
Also cos–1
1✂ tan 2 y
1 x2
–1
=
cos
= cos–1 (cos 2y) = 2y = 2tan–1 x
1 ✄ tan 2 y
1 ✁ x2
(iii) Can be worked out similarly
...
Example 3 Show that
1
1
✟ x✟
(i) sin–1 ☎ 2 x 1 ✝ x 2 ✆ = 2 sin–1 x, ✞
2
2
1
✟ x ✟1
(ii) sin–1 ☎ 2 x 1 ✝ x 2 ✆ = 2 cos–1 x,
2
Solution
(i) Let x = sin ✠
...
We have
sin–1 ✡ 2 x 1 ☞ x 2 ☛ = sin–1 ✌ 2sin ✎ 1 ✏ sin 2 ✎ ✍
= sin–1 (2sin✠ cos✠ ) = sin–1 (sin2✠ ) = 2✠
= 2 sin–1 x
(ii) Take x = cos ✠, then proceeding as above, we get, sin–1 ✡ 2 x 1 ☞ x 2 ☛ = 2 cos–1 x
1
2
3
✑ tan –1 ✒ tan –1
2
11
4
Solution By property 5 (i), we have
Example 4 Show that tan–1
1 2
✕
1
15
✙1 3 = R
...
S
...
H
...
= tan
= tan
✖ tan
✖ tan ✔ 1
1 2
4
2
11
20
1✗ ✘
2 11
–1
✛ cos x ✜
✧
✧
, ★ ✩ x ✩ in the simplest form
...
x2 ❖ 1 =
sec 2 P ❑ 1 ◗ tan P
INVERSE TRIGONOMETRIC FUNCTIONS
–1
Therefore, cot
1
x2 1
= cot–1 (cot ✟) = ✟ = sec–1 x, which is the simplest form
...
Then ✟ = tan–1 x
...
H
...
= tan ☎
= tan–1 (tan3✟) = 3✟ = 3tan–1 x = tan–1 x + 2 tan–1 x
2x
= L
...
S
...
2
Prove the following:
✔ 1 1✕
1
...
3cos–1 x = cos–1 (4x3 – 3x), x ✖ ✗ , 1✘
✙2 ✚
3
...
2
7
1
✜ tan ✛1 ✢ tan ✛1
11
24
2
1
1
31
2 tan ✣1 ✤ tan ✣1 ✥ tan ✣1
2
7
17
Write the following functions in the simplest form:
1 ✧ x2 ★ 1
,x✩0
x
6
...
tan ✵ 1 ✹
5
...
tan ✫1 ✯✯
47
1
x
2
1
, |x| > 1
✶ cos x ✸ sin x ✷
✺, x < ✴
✼ cos x ✻ sin x ✽
48
9
...
MATHEMATICS
x
1
tan
a
2
✁
x2
, |x| < a
✄ 3a 2 x ✆ x 3 ☎
3
2 ✞,
✟ a ✆ 3ax ✠
tan ✂1 ✝
a > 0;
✡a
3
☛
x☛
a
3
Find the values of each of the following:
11
...
☞
✍
✖
✔
tan –1 ✒ 2 cos ✏ 2 sin –1
tan
1 ✎✌
✑
2 ✕ ✗✓
12
...
If sin ★ sin
✪
1
–1 ✥
✦ cos x ✩ ✧1 , then find the value of x
5
✫
x ✬1
–1 x ✭ 1 ✮
✭ tan
✯ , then find the value of x
x✬2
x✭2 4
Find the values of each of the expressions in Exercises 16 to 18
...
If tan
✱
–1
16
...
✤
tan –1 ★ tan
✪
3✷ ✥
4 ✩✫
✤
–1 3
–1 3 ✥
✦ cot
18
...
cos ✼ cos
✾
7✮
6
(A)
✺✹
20
...
7✹ ✻
✽ is equal to
6 ✿
❀ sin
(B)
✸1
5✮
6
1 ✻
(❀ ) ✽ is equal to
2 ✿
1
(B)
3
(C)
(C)
✮
3
1
4
(D)
✮
6
(D) 1
tan ✂1 3 ✆ cot ✂1 (✆ 3) is equal to
(A)
❁
(B)
✬
✮
2
(C) 0
(D) 2 3
INVERSE TRIGONOMETRIC FUNCTIONS
Miscellaneous Examples
1
Example 9 Find the value of sin (sin
3✁
)
5
1
Solution We know that sin ✂1 (sin x) ✄ x
...
e
...
e
...
e
...
6
Since x = – 1 does not satisfy the equation, as the L
...
S
...
6
negative, x ✓
Miscellaneous Exercise on Chapter 2
Find the value of the following:
✕
✙
–1
1
...
✕
✙
tan –1 ✗ tan
7✔ ✖
✘
6 ✚
Prove that
3
...
cos
5
13
65
–1
4
...
cos
63
5
3
✓ sin –1 ✛ cos –1
16
13
5
7
...
1
1
1
1
tan –1 ✛ tan ✜1 ✛ tan ✜1 ✛ tan ✜1 ✓
5
7
3
8 4
12
3
56
✛ sin –1 ✓ sin –1
13
5
65
52
MATHEMATICS
Prove that
9
...
cot –1 ✎
✎
✡ 1 ☞ x ✌ 1 ✌ x ☛ ✛ 1 –1
1
✢ x ✢ 1 [Hint: Put x = cos 2✣]
✏✏ ✍ ✌ cos x , ✜
4
2
2
✑ 1☞ x ☞ 1✌ x ✒
11
...
9✓ 9
1 9
2 2
✜ sin ✤1 ✥ sin ✤1
8 4
3 4
3
Solve the following equations:
13
...
tan –1
1 ✦ x 1 –1
✧ tan x,( x ★ 0)
1✩ x 2
15
...
sin–1 (1 – x) – 2 sin–1 x =
(A) 0,
17
...
In fact (sin x)–1 =
2
1
and
sin x
similarly for other trigonometric functions
...
For suitable values of domain, we have
y = sin–1 x
x = sin y
x = sin y
y = sin–1 x
–1
–1
sin (sin x) = x
sin (sin x) = x
✙
✙
sin–1
1
= cosec–1 x
x
cos–1 (–x) =
✠ – cos
cos–1
1
= sec–1x
x
cot–1 (–x) =
✠ – cot
tan–1
1
= cot–1 x
x
sec–1 (–x) =
✠ – sec
–1
x
–1
x
–1
x
53
54
MATHEMATICS
sin–1 (–x) = – sin–1 x
tan–1 (–x) = – tan–1 x
tan–1 x + cot–1 x =
cosec–1 (–x) = – cosec–1 x
✁
2
sin–1 x + cos–1 x =
✁
2
cosec–1 x + sec–1 x =
✂y
✄
x✆ y
1 ✝ xy
x
tan–1x + tan–1y = tan–1
tan–1x – tan–1y = tan–1
2tan–1 x = sin–1
2tan–1x = tan–1
1 xy
✁
2
2x
☎
1 x2
✟
✠
2x
1 x2
–1
2 = cos
1 x
1 x2
✞
Historical Note
The study of trigonometry was first started in India
...
D
...
D
...
D
...
D
...
All
this knowledge went from India to Arabia and then from there to Europe
...
In India, the predecessor of the modern trigonometric functions, known as
the sine of an angle, and the introduction of the sine function represents one of
the main contribution of the siddhantas (Sanskrit astronomical works) to
mathematics
...
D
...
A sixteenth century Malayalam work Yuktibhasa
contains a proof for the expansion of sin (A + B)
...
, were given by Bhaskara II
...
, for arc sin x, arc cos x, etc
...
W
...
C
...
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
INVERSE TRIGONOMETRIC FUNCTIONS
55
height, and comparing the ratios:
H
S
h
= tan (sun’s altitude)
s
Thales is also said to have calculated the distance of a ship at sea through
the proportionality of sides of similar triangles
...
—✁ —
Chapter
3
MATRICES
The essence of Mathematics lies in its freedom
...
1 Introduction
The knowledge of matrices is necessary in various branches of mathematics
...
This mathematical tool simplifies
our work to a great extent when compared with other straight forward methods
...
Matrices are not only used as a
representation of the coefficients in system of linear equations, but utility of matrices
far exceeds that use
...
Also, many physical operations such as magnification, rotation and
reflection through a plane can be represented mathematically by matrices
...
This mathematical tool is not only used in certain branches
of sciences, but also in genetics, economics, sociology, modern psychology and industrial
management
...
3
...
We may
express it as [15] with the understanding that the number inside [ ] is the number of
notebooks that Radha has
...
We may express it as [15 6] with the understanding that first number
inside [ ] is the number of notebooks while the other one is the number of pens possessed
by Radha
...
or
Radha
Notebooks
15
Pens
6
which can be expressed as:
Fauzia
10
2
Simran
13
5
In the first arrangement the entries in the first column represent the number of
note books possessed by Radha, Fauzia and Simran, respectively and the entries in the
second column represent the number of pens possessed by Radha, Fauzia and Simran,
58
MATHEMATICS
respectively
...
The
entries in the second row represent the number of pens possessed by Radha, Fauzia
and Simran, respectively
...
Formally, we define matrix as:
Definition 1 A matrix is an ordered rectangular array of numbers or functions
...
We denote matrices by capital letters
...
5
☛
☛ 3
✍
3
–1
5
1✟
2☞
✏1 ✒ x
☞
3 ✑
x3
2 ☞, C✓✔
✕
x
x✗
cos
tan
sin x ✒ 2
✖
5 ☞
☞
7 ✎
✡
In the above examples, the horizontal lines of elements are said to constitute, rows
of the matrix and the vertical lines of elements are said to constitute, columns of the
matrix
...
3
...
1 Order of a matrix
A matrix having m rows and n columns is called a matrix of order m × n or simply m × n
matrix (read as an m by n matrix)
...
We observe that A has
3 × 2 = 6 elements, B and C have 9 and 6 elements, respectively
...
, ain, while the jth column
consists of the elements a1j, a2j, a3j,
...
We can also call
it as the (i, j)th element of A
...
MATRICES
59
Note In this chapter
1
...
2
...
✁✄✆ ✂☎✝
We can also represent any point (x, y) in a plane by a matrix (column or row) as
x
y
(or [x, y])
...
1
Observe that in this way we can also express the vertices of a closed rectilinear
figure in the form of a matrix
...
✘✚
✜ ✚✚✚
✣✢
✙✛
✛✛
✤✛ ✗
Now, quadrilateral ABCD in the matrix form, can be represented as
X
✒ ✏✓✕
✎ ✑✔✖
A B C D
1 3 1 1
0 2
3
2
✍
2 4
A 1
or
Y
0
B 3 2
C 1 3
D 1 2
4 2
Thus, matrices can be used as representation of vertices of geometrical figures in
a plane
...
Example 1 Consider the following information regarding the number of men and women
workers in three factories I, II and III
Men workers
Women workers
I
30
25
II
25
31
III
27
26
Represent the above information in the form of a 3 × 2 matrix
...
Example 2 If a matrix has 8 elements, what are the possible orders it can have?
Solution We know that if a matrix is of order m × n, it has mn elements
...
Thus, all possible ordered pairs are (1, 8), (8, 1), (4, 2), (2, 4)
Hence, possible orders are 1 × 8, 8 ×1, 4 × 2, 2 × 4
Example 3 Construct a 3 × 2 matrix whose elements are given by aij
Solution In general a 3 × 2 matrix is given by
✏
✠ a11
A ✌ ☛ a21
☛
☛✍ a31
a12 ✡
a22 ☞
...
2
Now
aij
Therefore
a11 ✞
1
|1 ✟ 3 ✒ 1| ✞ 1
2
a12
✞
1
5
|1 ✟ 3 ✒ 2 | ✞
2
2
a21 ✞
1
1
| 2 ✟ 3 ✒ 1| ✞
2
2
a22
✞
1
| 2 ✟ 3✒ 2 | ✞ 2
2
a31 ✏
1
| 3 ✑ 3 ✓ 1| ✏ 0
2
a32
✏
1
3
| 3 ✑ 3✓ 2 | ✏
2
2
Hence the required matrix is given by A
✔
1
✖
✖1
✘✖
2
✖
✖0
✙
5✕
2✗
✗
2✗
...
2
MATRICES
61
3
...
(i) Column matrix
A matrix is said to be a column matrix if it has only one column
...
In general, A = [aij] m × 1 is a column matrix of order m × 1
...
For example, B ☛ ✌✠ ☞
✎
1
2
✡
5 2 3✍
✏1✟ 4
is a row matrix
...
(iii) Square matrix
A matrix in which the number of rows are equal to the number of columns, is
said to be a square matrix
...
For example
✒3
✔
3
A✖✔
✔2
✔4
✗
✑1
0✓
✕
3 2
3
1 ✕ is a square matrix of order 3
...
Note If A = [aij] is a square matrix of order n, then elements (entries) a11, a22,
...
Thus, if
Then the elements of the diagonal of A are 1, 4, 6
...
62
MATHEMATICS
(iv) Diagonal matrix
A square matrix B = [bij] m × m is said to be a diagonal matrix if all its non
diagonal elements are zero, that is a matrix B = [bij] m × m is said to be a diagonal
matrix if bij = 0, when i ☎ j
...
1 0 0☛
✁ 1 0✂
2 0✌ , are diagonal matrices
For example, A = [4], B ✄ ✆
, C ✍ ☞☞ 0
✝
✌
✞ 0 2✟
☞✎ 0 0 3✌✏
of order 1, 2, 3, respectively
...
For example
A = [3],
✁ 1 0✂,
B✄ ✆
1✝✟
✞0
✑ 3
✓
C✕✓ 0
✓
✖0
0
3
0
0✒
✔
0✔
✔
3✗
are scalar matrices of order 1, 2 and 3, respectively
...
In other words, the square matrix A = [aij] n × n is an
✘1 if i ✄ j
identity matrix, if aij ✄ ✙
...
When order is clear from the
context, we simply write it as I
...
Observe that a scalar matrix is an identity matrix when k = 1
...
MATRICES
63
(vii) Zero matrix
A matrix is said to be zero matrix or null matrix if all its elements are zero
...
We denote
0 ✄✆ ✂☎ 0 0 0 ✄✆
✂
☎0
zero matrix by O
...
3
...
1 Equality of matrices
Definition 2 Two matrices A = [aij] and B = [bij] are said to be equal if
(i) they are of the same order
(ii) each element of A is equal to the corresponding element of B, that is aij = bij for
all i and j
...
Symbolically, if two matrices A and B are equal, we write A = B
...
5
✠
a✡ ☛ ✠ 2
✡
c ✡✌ ☞✠ 3
Example 4 If
✏x ✍ 3
✒
✎6
✒
✒
✕b ✎ 3
0
✟
✡
6✡ ,
2 ✌✡
then x = – 1
...
Solution As the given matrices are equal, therefore, their corresponding elements
must be equal
...
1
2
1
...
2
...
If a matrix has 18 elements, what are the possible orders it can have? What, if it
has 5 elements?
4
...
Construct a 3 × 4 matrix, whose elements are given by:
(i) aij
☛
1
| ☞3i ✌ j |
2
(ii) aij
✆
2i ✂ j
6
...
Find the value of a, b, c and d from the equation:
✥ a✣b
★
✪ 2a ✣ b
2 a ✤ c ✦ ✥✣ 1 5 ✦
✧
3c ✤ d ✩✫ ★✪ 0 13✩✫
2✎
8 ✔✒
(iii)
✗x ✖ y ✖
✙
x✖ z
✙
✙
✜ y✖z
✗9 ✘
✚ ✙ ✚
✛ 5
✚ ✙ ✚
✚
✢ ✙
✜7 ✚
✢
z✘
MATRICES
65
8
...
Which of the given values of x and y make the following pair of matrices equal
5 ✂ ✁ 0 y ✆ 2✂
✁ 3x 7
✄ y 1 2 ✆ 3x☎ , ✄ 8
4 ☎✞
✝
✞ ✝
(A) x ✠
✟1
3
(C) y = 7,
, y✠7
x☛
✡2
(B) Not possible to find
(D) x ☛
✡1
, y☛
✡2
3
3
3
10
...
4 Operations on Matrices
In this section, we shall introduce certain operations on matrices, namely, addition of
matrices, multiplication of a matrix by a scalar, difference and multiplication of matrices
...
4
...
Each factory produces sport
shoes for boys and girls in three different price categories labelled 1, 2 and 3
...
Then the total production
In category 1 : for boys (80 + 90), for girls (60 + 50)
In category 2 : for boys (75 + 70), for girls (65 + 55)
In category 3 : for boys (90 + 75), for girls (85 + 75)
✌80 ☞ 90
This can be represented in the matrix form as ✎ 75 ☞ 70
✎
✑✎ 90 ☞ 75
60 ☞ 50✍
65 ☞ 55✏
...
We observe that the sum of
two matrices is a matrix obtained by adding the corresponding elements of the given
matrices
...
Thus, if A ✂ ✄
a11 a12
✆ a21
a22
a13
✁
☎
a23 ✝
is a 2 × 3 matrix and B ✂ ✄
b11 b12
✆b21
2×3 matrix
...
In general, if A = [aij] and B = [bij] are two matrices of the same order, say m × n
...
✟
Example 6 Given A ☛ ☞
✍
3 1
2 3
✡ 1✠
✌
0✎
and
✏2
B✔ ✒
✒ ✕2
✖✒
5 1✑
✓
3
1 ✓ , find A + B
2 ✗✓
Since A, B are of the same order 2 × 3
...
We emphasise that if A and B are not of the same order, then A + B is not
✚2
3✛
✦1 2 3✧
then A + B is not defined
...
For example if A ✜ ✢
2
...
3
...
2 Multiplication of a matrix by a scalar
Now suppose that Fatima has doubled the production at a factory A in all categories
(refer to 3
...
1)
...
We observe that
✡
170✡☞
the new matrix is obtained by multiplying each element of the previous matrix by 2
...
In other words, kA = k [aij] m × n = [k (aij)] m × n, that is, (i, j)th element of kA is kaij
for all possible values of i and j
...
5✍
✎
✏
A = ✎ 5 7 ✑3 ✏ , then
✎✒ 2 0 5 ✏✓
3 4
...
5✍ ✌ 9
✎
✏ ✎
✏
3A = 3 ✎ 5 7 ✑3 ✏ ✔ ✎3 5 21 ✑9 ✏
✎2 0 5✏ ✎ 6
0 15 ✏✓
✒
✓ ✒
Negative of a matrix The negative of a matrix is denoted by – A
...
68
MATHEMATICS
For example, let
A=
3
✂
✆ ☎5
1✁
, then – A is given by
x ✝✄
✟
– A = (– 1) A ✡ (✞1) ☛
3
✌ ✞5
✟ ✞3
☞✡☛
x✍ ✌ 5
1✠
✞1 ✠
☞
✞ x✍
Difference of matrices If A = [aij], B = [bij] are two matrices of the same order,
say m × n, then difference A – B is defined as a matrix D = [dij], where dij = aij – bij,
for all value of i and j
...
✟1 2 3✠
✟ 3 ✞1 3 ✠
Example 7 If A ✡ ☛
☞ and B ✡ ☛
☞ , then find 2A – B
...
4
...
Now
A + B = [aij] + [bij] = [aij + bij]
= [bij + aij] (addition of numbers is commutative)
= ([bij] + [aij]) = B + A
(ii) Associative Law For any three matrices A = [aij], B = [bij], C = [cij] of the
same order, say m × n, (A + B) + C = A + (B + C)
...
In other words, O is the
additive identity for matrix addition
...
So
– A is the additive inverse of A or negative of A
...
4
...
Solution We have 2A + 3X = 5B
or
2A + 3X – 2A = 5B – 2A
or
or
or
2A – 2A + 3X = 5B – 2A
O + 3X = 5B – 2A
3X = 5B – 2A
(Matrix addition is commutative)
(– 2A is the additive inverse of 2A)
(O is the additive identity)
or
X=
or
✟ ☛ 10 ✡ 10☞
✟ ☛ 2 ✡2 ☞
☛8 0 ☞ ✠
1✌ ✎
✍ 1 ✌ ✎ 20 10 ✏ ✖
✏
✎
✏
X ✑ ✌ 5 4 2 ✡ 2 4 ✡2 ✍ = ✌ ✎
✏
✏
✎
✏
3✌
3✌ ✎
✔ ✎✒ ✡25 5 ✏✓
✒✎ 3 6 ✓✏ ✕✍
✔ ✒✎ ✡5 1 ✓✏
1
(5B – 2A)
3
☛ ✡16 0 ☞ ✠
✎ ✡8 4 ✏ ✍
✎
✏✍
✎✒ ✡6 ✡12✏✓ ✍✕
70
MATHEMATICS
✠
☛ ✟2
✂ 10 16 10 ✁ 0✄
✂ 6 10✄ ☛
1☎
✆ 1☎
✆ ☛
= ☎ 20 8 10 ✁ 4 ✆ = ☎ 12 14 ✆ = ☛ 4
3
3
☎✝ 25 6 5 12 ✆✞
☎✝ 31 7 ✆✞ ☛
☛ ✟31
✌☛ 3
✟10 ✡
3 ☞
☞
14 ☞
3 ☞
☞
✟7 ☞
3 ✍☞
✎ 5 2✏
✎3 6 ✏
✔ and X ✗ Y ✒ ✓
✔
...
✕0 9✖ ✕ 0 ✗1✖
Solution We have ✘ X ✑ Y ✙ ✑ ✘ X ✗ Y ✙ ✒ ✓
✎ 8 8✏
✎8 8✏
(X + X) + (Y – Y) = ✓
✚ 2X ✒ ✓
✔
✔
✕ 0 8✖
✕0 8✖
or
or
X=
1 ✎ 8 8✏ ✎ 4 4✏
✒
2 ✓✕ 0 8✔✖ ✓✕ 0 4✔✖
Also
✛5 2✜ ✛ 3 6✜
(X + Y) – (X – Y) = ✣
✤✢✣
✤
✥ 0 9 ✦ ✥ 0 ✢1✦
or
✛ 5 ✢ 3 2 ✢ 6✜
✛ 2 ✢4 ✜
✚ 2Y ★ ✣
(X – X) + (Y + Y) = ✣
✤
✤
9 ✧ 1✦
✥ 0
✥ 0 10 ✦
or
Y=
1
2
✛ 2 ✢ 4✜ ✛1 ✢2✜
✣0 10✤ ★ ✣0 5 ✤
✥
✦ ✥
✦
Example 10 Find the values of x and y from the following equation:
5 ✏ ✎3 ✗ 4✏
✎x
✎7 6✏
✑✓
= ✓
✔
✔
✔
✕ 7 y ✗ 3✖ ✕1 2✖ ✕15 14✖
2✓
Solution We have
5 ✜ ✛ 3 ✢4 ✜
✛x
✎7 6✏
✎ 2 x 10 ✏ ✎ 3 ✗ 4✏ ✎ 7 6 ✏
✚ ✓
✤ ✧ ✣1 2 ✤ = ✓
✔
✔✑✓
✔✒✓
✔
7
y
3
✢
✥
✦ ✥
✦ ✕15 14✖
✕14 2 y ✗ 6✖ ✕1 2 ✖ ✕15 14✖
2✣
MATRICES
or
✂ 2 x 3 10 ✁ 4 ✄
6 ✄ ✂7 6✄
✂2x 3
✂7 6✄
✠
✟ ☎
☎ 14 1 2 y ✁ 6 2 ✆ = ☎
✆
2 y ✁ 4✆✞ ☎✝15 14✆✞
✝
✞
✝ 15
✝15 14✞
or
or
2x + 3 = 7
2x = 7 – 3
or
x=
i
...
71
4
2
x =2
and
and
and
and
2y – 4 = 14
2y = 18
(Why?)
18
2
y = 9
...
The sale (in Rupees) of these
varieties of rice by both the farmers in the month of September and October are given
by the following matrices A and B
...
(ii) Find the decrease in sales from September to October
...
Solution
(i) Combined sales in September and October for each farmer in each variety is
given by
72
MATHEMATICS
(ii) Change in sales from September to October is given by
(iii) 2% of B =
2
B = 0
...
02
=
Thus, in October Ramkishan receives Rs 100, Rs 200 and Rs 120 as profit in the
sale of each variety of rice, respectively, and Grucharan Singh receives profit of Rs
400, Rs 200 and Rs 200 in the sale of each variety of rice, respectively
...
4
...
Meera wants to buy 2 pens and 5 story
books, while Nadeem needs 8 pens and 10 story books
...
How much money does each need to spend? Clearly, Meera needs Rs (5 × 2 + 50 × 5)
that is Rs 260, while Nadeem needs (8 × 5 + 50 × 10) Rs, that is Rs 540
...
Now, the money required by Meera and Nadeem to make purchases will be
respectively Rs (4 × 2 + 40 × 5) = Rs 208 and Rs (8 × 4 + 10 × 40) = Rs 432
MATRICES
73
Again, the above information can be represented as follows:
Requirements Prices per piece (in Rupees) Money needed (in Rupees)
2
✂
☎8
5✁
10 ✄✆
4 ✝ 2 ✞ 40 ✝ 5 ✁
208✁
✂
✄✟✂
✄
☎ 8 ✝ 4 ✞ 10 ✝ 4 0✆ ☎ 432 ✆
4✁
✂ ✄
☎ 40 ✆
Now, the information in both the cases can be combined and expressed in terms of
matrices as follows:
Requirements Prices per piece (in Rupees) Money needed (in Rupees)
2
✂
☎8
5✁
10 ✄✆
5
✂
☎ 50
4✁
40✄✆
5 ✝ 2 ✞ 5 ✝ 50
✂
☎ 8 ✝ 5 ✞ 10 ✝ 5 0
=
✠ 260
☛
✌ 540
4 ✝ 2 ✞ 40 ✝ 5 ✁
8 ✝ 4 ✞ 10 ✝ 4 0✄✆
208 ✡
432 ☞✍
The above is an example of multiplication of matrices
...
Furthermore for getting the elements of the product matrix,
we take rows of A and columns of B, multiply them element-wise and take the sum
...
Let A = [aij] be an m × n matrix and B = [bjk] be an
n × p matrix
...
To get the (i, k)th element cik of the matrix C, we take the ith row of A and kth column
of B, multiply them elementwise and take the sum of all these products
...
ain] and the kth column of
B is
✎ b1k ✏
✑
b ✒
✑ 2k ✒
✑
✒
✑
✒
✑b ✒
✓ nk ✔
n
, then cik = ai1 b1k + ai2 b2k + ai3 b3k +
...
The matrix C = [cik]m × p is the product of A and B
...
✂1
☎
4 ✞☎ ✄
✄✝ 5 ✂ 4✞☎
2✁
This is a 2 × 2 matrix in which each
entry is the sum of the products across some row of C with the corresponding entries
down some column of D
...
Solution The matrix A has 2 columns which is equal to the number of rows of B
...
Now
✑ 6(2) ✏ 9(7)
6(6) ✏ 9(9) 6(0) ✏ 9(8)✒
✕
✖ 2(2) ✏ 3(7) 2(6) ✏ 3(9) 2(0) ✏ 3(8) ✗
AB ✓ ✔
=
✠12 ✘ 63
☞
✍ 4 ✘ 21
36 ✘ 81 0 ✘ 72 ✡
12 ✘ 27
0 ✘ 24✌✎
=
✑ 75
✔
✖ 25
117 72✒
39 24 ✕✗
MATRICES
75
Remark If AB is defined, then BA need not be defined
...
If A, B are, respectively m × n, k × l matrices, then both AB and BA are defined
if and only if n = k and l = m
...
Non-commutativity of multiplication of matrices
Now, we shall see by an example that even if AB and BA are both defined, it is not
necessary that AB = BA
...
Show that
2 5 ☎✞
✄✝ 2 1☎✞
✂2
1
Example 13 If A ✆ ✄
✝✂ 4
AB
✟
BA
...
Hence AB and BA are both
defined and are matrices of order 2 × 2 and 3 × 3, respectively
...
But
one may think that perhaps AB and BA could be the same if they were of the same
order
...
✛1
Example 14 If A ✢ ✣
✦0
and
✓0
BA ✕ ✖
✘1
0✜
✤
✥ 1✧
✛0
and B ✢ ✣
✑ 1✔
...
Thus matrix multiplication is not commutative
...
76
☎
MATHEMATICS
Note This does not mean that AB BA for every pair of matrices A, B for
which AB and BA, are defined
...
Zero matrix as the product of two non zero matrices
We know that, for real numbers a, b if ab = 0, then either a = 0 or b = 0
...
☞ ✡✌✎ ✠ ☛✍✏ ☞ ✡✌✎ ☛✍✏
✄ ✁✆✞ ✑ ✂✁✝✆✟✞ ✂✝✟ ✄ ✁✆✞ ✂✝✟
3 5
...
Thus, if the product of two matrices is a zero matrix, it is not necessary that one of
the matrices is a zero matrix
...
4
...
1
...
We have
(AB) C = A (BC), whenever both sides of the equality are defined
...
The distributive law For three matrices A, B and C
...
3
...
✓✗ ✕ ✒ ✖✔ ✗ ✕✓ ✔✖ ✗ ✓✕ ✒ ✔✖
✘✕✕ ✒ ✖✖✙ ✕✕✘✒ ✖✖✙ ✘ ✒ ✙
Now, we shall verify these properties by examples
...
1 2 3
2 0
4
2 1
, find
MATRICES
✂1
Solution We have AB ✝ ☎ 2
☎
☎✞ 3
1
0
1
1✄ ✂ 1 3 ✄ ✂ 1 ✁ 0 ✁ 1 3 ✁ 2 4 ✄ ✂ 2 1 ✄
3 ✆✆ ☎☎ 0 2 ✆✆ ✝ ☎☎ 2 ✁ 0 3 6 ✁ 0 ✁ 12 ✆✆ ✝ ☎☎ 1 18✆✆
2 ✆✟ ☎✞ 1 4 ✆✟ ☎✞3 ✁ 0 2 9 2 ✁ 8 ✆✟ ☎✞ 1 15✆✟
4 ✠ 0 6 ✡ 2 ✡ 8 ✠1 ☞
☛2 1☞
☛ 2✠ 2
☛1 2 3 ✡ 4☞ ✌
✍
✌
✍
(AB) (C) ✎ ✡1 18 ✌
✌
✍ ✏ 2 0 ✡2 1✍✑ ✎ ✌ ✡1 ✠ 36 ✡2 ✠ 0 ✡3 ✡ 36 4 ✠ 18✍
✌✏ 1 15✍✑
✌✏ 1 ✠ 30 2 ✠ 0 3 ✡ 30 ✡ 4 ✠ 15✍✑
✂4
☎
= ☎35
☎✞31
Now
7 ✄
39 22 ✆✆
27 11✆✟
4
4
2
2
☛ 1 3☞
☛ 1 ✠ 6 2 ✠ 0 3 ✡ 6 ✡4 ✠ 3☞
☛1 2 3 ✡4 ☞ ✌
✍
BC = ✌ 0 2✍ ✌
✌
✍ ✏2 0 ✡2 1✑✍ ✎ ✌ 0 ✠ 4 0 ✠ 0 0 ✡ 4 0 ✠ 2✍
✌✏ ✡1 ✠ 8 ✡2 ✠ 0 ✡3 ✡ 8 4 ✠ 4✍✑
✏✌ ✡1 4✍✑
✂7 2
☎
= ☎4 0
☎✞7 2
Therefore
1 ✄
4 2 ✆✆
11 8 ✆✟
3
☛1 1 ✡1 ☞ ☛ 7 2 ✡3 ✡1 ☞
A(BC) = ✌✌ 2 0 3 ✍✍ ✌✌ 4 0 ✡4 2 ✍✍
✌✏ 3 ✡1 2 ✍✑ ✌✏ 7 ✡2 ✡11 8 ✍✑
✂ 7✁4 7 2✁0✁ 2
☎
= ☎14 ✁ 0 ✁ 21 4 ✁ 0 6
☎✞ 21 4 ✁ 14 6 ✁ 0 4
3 4 ✁ 11
6 ✁ 0 33
9 ✁ 4 22
1✁ 2 8 ✄
2 ✁ 0 ✁ 24 ✆✆
3 2 ✁ 16 ✆✟
4 4 ✡7 ☞
☛4
✌35 ✡2 ✡39 22✍
= ✌
✍
...
Also, verify that (A + B)C = AC + BC
Solution Now,
So
7
✟ 0
✡
A + B ☞ ✡✌5 0
✡ 8 ✌ 6
✍
7
✟ 0
✡
0
✌5
✡
✡✍ 8 ✌ 6
(A + B) C =
0
Further
and
So
✟0
✡
1
✡
✡✍ 1
BC =
✟9 ✠
✡ ☛
12 ✏
✡ ☛
✡
✍30✎☛
AC + BC =
Clearly,
✟ 2
✡
✌2
✡
✡✍ 3
2 ☛☛
0 ☛✎
7✁
8 ✄✄
0✄✞
0 ✆ 12 ✑ 21
✄ ✂
☎ ✆12 ✑ 0 ✑ 24
✄ ✂
✄✞ ✂✝ 14 ✑ 16 ✑ 0
2✁
✂
✆2
✂
✂✝ 3
✟ 2
✡
✌2
✡
✡✍ 3
✠ ✟ 0 ✌ 14 ✏ 24 ✠ ✟10 ✠
☛ ✡
☛ ✡ ☛
☞ ✌10 ✏ 0 ✏ 30 ☞ 20
☛ ✡
☛ ✡ ☛
☛✎ ✡✍ 16 ✏ 12 ✏ 0 ☛✎ ✡✍ 28☛✎
9✁
✁
✄ ✂ ✄
☎ 12
✄ ✂ ✄
✄✞ ✂✝30✄✞
✠ ✟ 0 ✌ 2 ✏ 3✠ ✟ 1 ✠
☛ ✡
☛ ✡ ☛
☞ 2✏ 0✏ 6 ☞ 8
☛ ✡
☛ ✡ ☛
☛✎ ✡✍ 2 ✌ 4 ✏ 0☛✎ ✡✍✌ 2 ☛✎
✟ 1 ✠ ✟10 ✠
✡ ☛ ✡ ☛
8 ☞ 20
✡ ☛ ✡ ☛
✡
✍✌2 ✎☛ ✍✡ 28✎☛
(A + B) C = AC + BC
1
Example 18 If A
1 1✠
2
0 ✎☛
10☛☛
0 ☛✎
6
0
10☛☛
8✠
✂
✆6
0
✂
✂✝ 7 ✆ 8
AC =
8✠
2
✂
☎ 3 ✆2
✂
✂
2
✝4
Solution We have A 2
3✁
1✄✄ , then show that A3 – 23A – 40 I = O
1✞✄
2
✟1
✡
☞ A
...
The
cost per contact (in paise) is given in matrix A as
Cost per contact
A=
✒
✔
✔
✔✖
40
100
50
✓
✕
✕
✕✗
Telephone
Housecall
Letter
The number of contacts of each type made in two cities X and Y is given by
Telephone Housecall Letter
✙1000
B✛ ✜
✣3000
500 5000 ✚ ✘ X
...
80
MATHEMATICS
Solution We have
✂ 40,000 50,000 250, 000 ✄ ✁ X
BA = ☎
✆
✝120,000 + 100,000 +500,000 ✞ ✁ Y
✂ 340, 000 ✄ ✁ X
= ☎
✆
✝ 720,000 ✞ ✁ Y
So the total amount spent by the group in the two cities is 340,000 paise and
720,000 paise, i
...
, Rs 3400 and Rs 7200, respectively
...
2
✂2 4 ✄
✂1
✆ , B ✠ ☎✟2
3
2
✝
✞
✝
3✄
1
...
Compute the following:
✂a
(i) ☎
✝✟b
✓✒1
✕
(iii) ✕ 8
✕✘ 2
b✄
a ✞✆
4
5
8
✡a2 ☞ b2
✂a b✄
(ii)
✌ 2 2
☎
✆
✝b a ✞
✏✌ a ☞ c
✒6✔ ✓12 7 6✔
✚cos 2 x
16 ✖✖ ✗ ✕✕ 8 0 5✖✖ (iv) ✢
2
✤✢ sin x
5 ✖✙ ✕✘ 3 2 4✖✙
(iii) 3A – C
b2 ☞ c 2 ☛
✡ 2ab 2bc ☛
✍☞✌
✍
a ☞ b ✑✍ ✏✎2ac ✎2ab ✑
2
2
sin 2 x ✛ ✚ sin 2 x
cos 2 x ✛
✜
✣
✢
✣
cos 2 x ✥✣ ✤✢ cos 2 x sin 2 x ✥✣
3
...
✂a
(i) ☎
✝✟b
b ✄ ✂ a ✟b ✄
a ✞✆ ✝☎ b a ✞✆
✦ 1✧
★ 2✩
(ii) ★ ✩ [2 3 4]
✪★ 3✫✩
✦ 2 3 4 ✧ ✦1 ✬ 3
(iv) ★ 3 4 5✩ ★ 0 2
★
✩ ★
✪★ 4 5 6✫✩ ★✪ 3 0
✓ 2 ✒3✔
✓ 3 ✒1 3 ✔ ✕
(vi) ✕
1 0✖✖
✖
✕
✘✒1 0 2✙ ✕ 3 1✖
✘
✙
5✧
4✩✩
5✫✩
✂1 ✟2 ✄ ✂1 2 3✄
(iii) ☎
✆☎
✆
✝ 2 3 ✞ ✝ 2 3 1✞
✦ 2 1✧
✦ 1 0 1✧
★
(v) ★ 3 2✩✩ ★
✬1 2 1✫✩
★✪✬1 1✫✩ ✪
MATRICES
✁1
3✂
2
✄
4
...
Also, verify that A + (B – C) = (A + B) – C
...
If
✡3
✡
✡7
✌✡ 3
5✠
✟2
✡5
3☛
☛
✡
4☛
1
and B ☞ ✡
✡5
3☛
☛
✡
2☛
✡7
3 ✍☛
✌✡ 5
1
2
3
2
3
5
2
5
6
5
✠
1☛
☛
4☛
, then compute 3A – 5B
...
Simplify cos✎ ✓
✔
sin ✎✔✖
✕ ✏ sin ✎ cos ✎✖
✕ cos ✎
7
...
Find X, if Y = ✛
and 2X + Y = ✛
✜
✜
✢ 1 4✣
✢ ✤3 2 ✣
✙ 1 3 ✚ ✙ y 0 ✚ ✙ 5 6✚
9
...
Solve the equation for x, y, z and t, if 2 ✛
✑ 2✒
11
...
✕ 1 ✖ ✕5 ✖
y✓
x ✘ y✒
✑x y✒ ✑ x 6 ✒ ✑ 4
12
...
✗✓
✘✓
✔
✔
3 ✔✖
✕ z w✖ ✕ ✏1 2 w✖ ✕ z ✘ w
82
MATHEMATICS
✁ cos x
13
...
1☎✞
14
...
Find A2 – 5A + 6I, if A ✘ ✓✓ 2
0
1
✓✖1 ✏1
0✒ ✑ 1 2 3✒
1✔✔ ✓✓ 0 1 0✔✔
4✗✔ ✖✓ 1 1 0✔✗
1✒
3✔✔
0✔✗
✁1 0 2✂
16
...
If A ✙ ☞
✚✜
✛ 0
✢ tan ✤
✣
2 and I is the identity matrix of order 2, show that
18
...
A trust fund has Rs 30,000 that must be invested in two different types of bonds
...
Using matrix multiplication, determine how to divide Rs 30,000 among
the two types of bonds
...
The bookshop of a particular school has 10 dozen chemistry books, 8 dozen
physics books, 10 dozen economics books
...
Find the total amount the bookshop will receive
from selling all the books using matrix algebra
...
Choose the correct answer in Exercises 21 and 22
...
The restriction on n, k and p so that PY + WY will be defined are:
(A) k = 3, p = n
(B) k is arbitrary, p = 2
(C) p is arbitrary, k = 3
(D) k = 2, p = 3
22
...
5
...
Definition 3 If A = [aij] be an m × n matrix, then the matrix obtained by interchanging
the rows and columns of A is called the transpose of A
...
In other words, if A = [aij]m × n, then A✝ = [aji]n × m
...
5
...
These
may be verified by taking suitable examples
...
84
MATHEMATICS
Solution
(i) We have
A=
✂3
☎
✠4
3 2✄
✆
2 0✡
3 4✄
✂3
✆
3 2✄
3 2✆ ✞ A✝✁✝ ✟ ☎
✆ ✟A
0✡
✠4 2
✆
2 0✡
✂
☎
✞ A✝ ✟ ☎
☎
✠
Thus (A☛)☛ = A
(ii) We have
A=
Therefore
Now
So
Thus
(iii) We have
☞3
✍
✏4
3 2✌
2
✎,
0✑
B=
☞2
✍
✏1
(A + B)☛ =
✂
☎
☎
☎
✠
A☛ =
✗
✙
✙
✙
✣
A☛ + B☛ =
✂
☎
☎
☎
✠
✒1
☞5
2✌
✎✓A✔B✕✍
2 4✑
✏5
Thus
4
5
5✄
✆
3 ✖ 1 4✆
4
4✆✡
3 4✘
✗ 2 1✘
✚
3 2✚ , B✛ ✜ ✙✙✢1 2✚✚ ,
2 0✚✤
✙
✣
2 4✚✤
5
5✄
✆
3 ✖ 1 4✆
4
4✆✡
(A + B)☛ = A☛ + B☛
✦2
kB = k ✩
✫1
Then
3 ✒ 1 4✌
(kB)☛ =
✭ 2k
✯
✲k
✯
✯
✴ 2k
(kB)☛ = kB ☛
✥1
2
2✧
✦ 2k
✪★✩
4 ✬ ✫k
k
✮
2k ✰ ✳
✰
4k ✰✵
k
✭ 2
✯
✲1
✯
✯
✴ 2
✥k
2k ✧
2k
4 k ✪✬
1✮
2✰ ✳ kB✱
✰
4✰✵
✎
4✑
MATRICES
85
✄ ✂2 ☎
Example 21 If A ✞ ✆✆ 4 ✝✝ , B ✞ 1 3
✆
✟ 5✝
✠
✂6✁ ,
verify that (AB)✡ = B✡A✡
...
6 Symmetric and Skew Symmetric Matrices
Definition 4 A square matrix A = [aij] is said to be symmetric if A✡ = A, that is,
[aij] = [aji] for all possible values of i and j
...
5
3
✥1
3✜
✣
✥ 1 ✣ is a symmetric matrix as A✡ = A
1 ✣✧
Definition 5 A square matrix A = [aij] is said to be skew symmetric matrix if
A✡ = – A, that is aji = – aij for all possible values of i and j
...
Therefore 2aii = 0 or aii = 0 for all i’s
...
86
MATHEMATICS
For example, the matrix
0
e
B ☎ ✂ ✆e
✂
✂✝✆ f
0
✆g
f✁
g ✄ is a skew symmetric matrix as B✟= –B
✄
0 ✄✞
Now, we are going to prove some results of symmetric and skew-symmetric
matrices
...
Proof Let B = A + A✟, then
B✟ = (A + A✟)✟
= A✟ + (A✟ )✟ (as (A + B)✟ = A✟ + B✟ )
= A✟ + A (as (A✟)✟ = A)
= A + A✟ (as A + B = B + A)
= B
Therefore
B = A + A✟ is a symmetric matrix
Now let
C = A – A✟
C✟ = (A – A✟ )✟ = A✟ – (A✟)✟
= A✟ – A
(Why?)
(Why?)
= – (A – A✟) = – C
Therefore
C = A – A✟ is a skew symmetric matrix
...
Proof Let A be a square matrix, then we can write
1
1
A ✡ (A ☛ A✠) ☛ (A ☞ A✠)
2
2
From the Theorem 1, we know that (A + A✟ ) is a symmetric matrix and (A – A✟) is
a skew symmetric matrix
...
Thus, any square
2
matrix can be expressed as the sum of a symmetric and a skew symmetric matrix
...
4✂
4☎☎ as the sum of a symmetric and a
3☎✞
Solution Here
✡ 2 ✠1 1 ☛
B✟ = ☞✠ 2 3 ✠2✌
☞
✌
☞✍ ✠4 4 ✠3✌✎
Let
Now
Thus
Also, let
Then
✡ 4 ✠3
1
1☞
P = (B + B✏) ✑ ✠ 3 6
2
2☞
✍☞ ✠3 2
✡ 2 ✠3 ✠3 ☛
☞
2
2✌
☞
✌
✠3
P✟ = ☞
3 1 ✌= P
☞2
✌
☞ ✠3
✌
☞
1 ✠3 ✌
✌✎
✍☞ 2
P=
87
✡ 2 ✠3 ✠3 ☛
☞
2
2✌
✌
✠3☛ ☞
✠3
☞
✌
✌
3 1 ,
2 = ☞
✌
✌
2
✌
✠6 ✎✌ ☞ ✠3
☞
1 ✠3 ✌
☞✍ 2
✌✎
1
(B + B✒) is a symmetric matrix
...
2
✂
☎
2
☎
3
P+Q✝
☎ 2
☎
✁3
☎
☎
✟ 2
☎✁
Now
3
2
✁
3
1
3✄
2✆
✁
✂
☎
✆
☎
✆
☎
1
2
✁
0
1
1 ✞
✆
☎2
✆
☎
5
☎
✁3 ✆
✆
☎2
✠
✟
5✄
2✆
✁
✆
2
☎
✝ ✁
☎
✆
0
3
3
0✆
✆
✁
✂
✆
☎
✟
✁
1
1
2
3
✁2
✁
4✄
4✆ ✝ B
✆
✁ 3✆
✠
✆
✠
Thus, B is represented as the sum of a symmetric and a skew symmetric matrix
...
3
1
...
If A ✝
☎
☎
1
(ii) ✔
2
✖
✒
1 2 3✄
3
...
For the matrices A and B, verify that (AB)✤ = B✤A✤, where
✒✑
4
...
If (i) A ✄ ☎
☛ sin ✡
✑ ✏ cos ✡
(ii) If A ✌ ✍
7
...
✓1 ✕
0 1 ✗✗
✖✙ 1 ✓1 0 ✗✚
✔ 0
(ii) Show that the matrix A ✘ ✖✓1
✖
☛1
✑6
8
...
, verify that
(i) (A + A✠) is a symmetric matrix
(ii) (A – A✠) is a skew symmetric matrix
✔ 0
1
1
✛
✜
✛
✜
✢
✢
A
A
A
A
✣
✤
9
...
Express the following matrices as the sum of a symmetric and a skew symmetric
matrix:
(i)
(iii)
☛ 3 5☞
✍ 1 ✏1✎
✑
✒
(ii)
✦ 6 ✥2 2 ✧
★✥2 3 ✥1 ✩
★
✩
★✪ 2 ✥1 3 ✩✫
✔ 3 3 ✓1✕
✖✓2 ✓2 1✗
✖
✗
✖✙✓4 ✓5 2✗✚
(iv)
✁ 1
☎
✞ ✝1
5✂
2✟✆
90
MATHEMATICS
Choose the correct answer in the Exercises 11 and 12
...
If A, B are symmetric matrices of same order, then AB – BA is a
(A) Skew symmetric matrix
(B) Symmetric matrix
(C) Zero matrix
(D) Identity matrix
✂ cos
✞sin
12
...
7 Elementary Operation (Transformation) of a Matrix
There are six operations (transformations) on a matrix, three of which are due to rows
and three due to columns, which are known as elementary operations or
transformations
...
Symbolically the interchange
of ith and jth rows is denoted by Ri ✌ Rj and interchange of ith and jth column is
denoted by Ci ✌ Cj
...
6 7✑✕
(ii) The multiplication of the elements of any row or column by a non zero
number
...
The corresponding column operation is denoted by Ci ✗ kCi
✙
1
For example, applying C3 ✘ C3 , to B ✛ ✜
7
✤ ✣1
1
2
3
✦1
★
✢ , we get ★
★ ✪1
1✥
★✫
1✚
1✧
7✩
✩
1✩
3
7 ✩✬
2
(iii) The addition to the elements of any row or column, the corresponding
elements of any other row or column multiplied by any non zero number
...
MATRICES
91
The corresponding column operation is denoted by Ci ✡ Ci + kCj
...
3
...
In that case A is said to be invertible
...
✎3 ✠
2 ☞✍
✎6 ✏ 6 ✠ ✟ 1
☞✑☛
✎3 ✏ 4 ✍ ✌ 0
0✠
✑I
1☞✍
0✠
✑ I
...
e
...
A rectangular matrix does not possess inverse matrix, since for products BA
and AB to be defined and to be equal, it is necessary that matrices A and B
should be square matrices of the same order
...
If B is the inverse of A, then A is also the inverse of B
...
Proof Let A = [aij] be a square matrix of order m
...
We shall show that B = C
...
(1)
AC = CA = I
...
92
MATHEMATICS
Proof From the definition of inverse of a matrix, we have
(AB) (AB)–1 = 1
or
A–1 (AB) (AB)–1 = A –1 I
(Pre multiplying both sides by A–1)
or
(A–1A) B (AB)–1 = A –1
(Since A–1 I = A–1)
or
IB (AB)–1 = A –1
or
B (AB)–1 = A –1
or
B–1 B (AB)–1 = B–1 A–1
or
I (AB)–1 = B–1 A–1
(AB)–1 = B–1 A–1
Hence
3
...
1 Inverse of a matrix by elementary operations
Let X, A and B be matrices of, the same order such that X = AB
...
Similarly, in order to apply a sequence of elementary column operations on the
matrix equation X = AB, we will apply, these operations simultaneously on X and on the
second matrix B of the product AB on RHS
...
The matrix B will be the
inverse of A
...
Remark In case, after applying one or more elementary row (column) operations on
A = IA (A = AI), if we obtain all zeros in one or more rows of the matrix A on L
...
S
...
Example 23 By using elementary operations, find the inverse of the matrix
1
✆2
A=✂
2✁
✄
☎1✝
...
or
1
✂
✆2
2 ✁ 1 0✁
1
✄✞✂
✄ A, then ✂
☎1✝ ✆ 0 1✝
✆0
2✁
1 0✁
✄✞✂
✄ A (applying R2 ✡ R2 – 2R1)
☎ 5 ✝ ✆ ☎ 2 1✝
MATRICES
or
or
1 2✁
✂
☎0
1 ✄✆
=
✝1
✟
2
✟
☛5
✍1
✏5
✄ ✥ ✏
1✆
✏2
✏
✓5
1 0✁
✂
☎0
93
0✞
1
✠
R)
✡1 A (applying R2 ✌ –
✠
5 2
5☞
2✎
5✑
✑ A (applying R1 ✌ R1 – 2R2)
✒1 ✑
5 ✑✔
✍1
✏5
✏
✏2
✏
✓5
2✎
5✑
✑
Thus
A–1 =
✒1 ✑
5 ✑✔
Alternatively, in order to use elementary column operations, we write A = AI, i
...
,
1
✂
☎2
2✁
✄
✕ 1✆
= A✂
1 0✁
☎0
1 ✄✆
Applying C2 ✌ C2 – 2C1, we get
1
0✁
✂
☎2
Now applying C2 ✌
✖
✄
✕ 5✆
☎0
✕2✁
1 ✄✆
✍
1
✏
✏
✏
0
✏
✓
2✎
5✑
✑
✒1 ✑
5 ✑✔
= A✂
1
1
C2 , we have
5
✗1
✙
✛2
0✘
= A
1 ✚✜
Finally, applying C1 ✌ C1 – 2C2, we obtain
1 0✁
✂
☎0
Hence
1 ✄✆
= A
A–1 =
✍1
✏
5
✏
✏2
✏
✓5
✍1
✏5
✏
✏2
✏
✓5
2✎
5✑
✑
✒1 ✑
5 ✑✔
2✎
5✑
✑
✒1 ✑
5 ✔✑
94
MATHEMATICS
Example 24 Obtain the inverse of the following matrix using elementary operations
0 1 2✁
A ☎ ✂1
✂
✂
✆3
2 3✄
...
e
...
e
...
e
...
1✦★
0✣
P
...
Therefore, P–1 does not exist
...
4
Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17
...
✑
✒
✓ 2 3✔
✎ 2 1✏
2
...
✑
✒
✓ 2 7✔
✕ 2 3✖
4
...
✑
✒
✓ 7 4✔
✎ 2 5✏
6
...
✗
✘
✙ 5 2✚
✕ 4 5✖
8
...
✗
✘
✙2 7✚
✎ 3 ✍1✏
10
...
✑
✒
✓ 1 ✍ 2✔
✎ 6 ✍ 3✏
12
...
✗
✘
✙ ✛1 2 ✚
✕ 2 1✖
14
...
✙ 4 2✚
✢ 2 ✜3 3 ✣
✤
✥
15
...
✫★3
✫✭ 2
3 ★2 ✪
0 ★5 ✬✬
5
0 ✬✮
✩ 2 0 ★1 ✪
✫
✬
17
...
Matrices A and B will be inverse of each other only if
(A) AB = BA
(B) AB = BA = 0
(C) AB = 0, BA = I
(D) AB = BA = I
98
MATHEMATICS
Miscellaneous Examples
✁ cos
Example 26 If A ✄ ☎
✞ ✝ sin
sin ✂
✁ cos n
, then prove that A n ✄ ☎
✆
cos ✟
✞✝ sin n
sin n ✂
, n ✠ N
...
We have
✁ cos
P(n) : If A ✄ ☎
✞ ✝ sin
sin ✂
✁ cos n
✱ then A n ✄ ☎
✆
cos ✟
✞ ✝ sin n
sin ✂
✁ cos
✁ cos
1
, so A ✄ ☎
P(1) : A ✄ ☎
✆
cos ✟
✞ ✝ sin
✞ ✝ sin
Therefore,
the result is true for n = 1
...
So
✁ cos
P(k) : A ✄ ☎
✞ ✝ sin
sin n ✂
,n✠N
cos n ✟✆
sin ✂
cos ✆✟
✁ cos k
sin ✂
k
, then A ✄ ☎
✆
cos ✟
✞ ✝ sin k
sin k ✂
cos k ✆✟
Now, we prove that the result holds for n = k +1
Now
☛ cos ✡ sin ✡ ☞ ☛ cos k ✡ sin k ✡ ☞
k
Ak + 1 = A ✌ A ✍ ✎
✏✎
✏
✒ ✑ sin ✡ cos ✡ ✓ ✒ ✑ sin k ✡ cos k ✡ ✓
✁ cos cos k – sin sin k
= ☎
✞ ✝ sin cos k ✔ cos sin k
cos sin k ✔ sin cos k ✂
✝ sin sin k ✔ cos cos k ✆✟
☛ cos (✡ ✕ k✡ ) sin (✡ ✕ k ✡) ☞ ☛ cos( k ✕ 1)✡ sin (k ✕ 1)✡ ☞
= ✎
✏✍✎
✏
✒ ✑ sin (✡ ✕ k ✡) cos (✡ ✕ k✡) ✓ ✒ ✑ sin (k ✕ 1)✡ cos (k ✕ 1)✡ ✓
Therefore, the result is true for n = k + 1
...
cos n ✡ ✏✓
Example 27 If A and B are symmetric matrices of the same order, then show that AB
is symmetric if and only if A and B commute, that is AB = BA
...
Let
AB be symmetric, then (AB)✖ = AB
MATRICES
99
But
(AB)✝ = B✝ A✝= BA (Why?)
Therefore
BA = AB
Conversely, if AB = BA, then we shall show that AB is symmetric
...
✁2
Example 28 Let A ✄ ☎
✞3
1✂
✁ 5 2✂
✁ 2 5✂
,B✄ ☎
✆ , C ✄ ☎ 3 8✆
...
Solution Since A, B, C are all square matrices of order 2, and CD – AB is well
defined, D must be a square matrix of order 2
...
Then CD – AB = 0 gives
✌c d✍
or
✠ 2 5✡ ✠ a b ✡ ✠ 2 ✎1✡ ✠ 5 2✡
☛ 3 8☞ ☛ c d ☞ ✎ ☛ 3 4☞ ☛ 7 4☞ = O
✌
✍✌
✍ ✌
✍✌
✍
or
✠ 2a ✏ 5c 2b ✏ 5d ✡ ✠ 3 0 ✡ ✠ 0 0✡
☛
☞✎☛
☞ =☛
☞
✌ 3a ✏ 8c 3b ✏ 8d ✍ ✌ 43 22✍ ✌ 0 0✍
or
2b ✑ 5d ✂ ✁ 0 0 ✂
✁ 2a ✑ 5c 3
☎ 3a ✑ 8c 43 3b ✑ 8d 22✆ = ☎ 0 0✆
✞
✟ ✞
✟
By equality of matrices, we get
2a + 5c – 3 = 0
...
(2)
2b + 5d = 0
...
(4)
Solving (1) and (2), we get a = –191, c = 77
...
Therefore
✁ a b ✂ ✁ 191
D= ☎
✆✄☎
✞ c d ✟ ✞ 77
110 ✂
44 ✆✟
100
MATHEMATICS
Miscellaneous Exercise on Chapter 3
0 1✁
n
n
n–1
bA, where I is the identity
☎ , show that (aI + bA) = a I + na
0
0
✆
✝
1
...
2
...
If A ✂ ✄
1 1✠
1 1☛☛ , prove that
1 1☛✍
✘4 ✁
☎,
✘ 1✝
✏3n ✎1
✒
A n ✔ ✒3n✎1
✒ n ✎1
✖3
then prove that A n
3n✎1 3n✎1 ✑
✓
3n✎1 3n✎1 ✓ , n ✕ N
...
4
...
5
...
6
...
7
...
If A ✂ ✄
9
...
2 ✝☎
✩5
✥1
✧
✩ 1✤ 0
✧
✧
✫2
0 2✦ ✥ x✦
2 1 ★ ✧ 4★ ✪ O
★✧ ★
0 3★✬ ✧✫ 1★✬
☛
☛
z ✍☛
✛z
satisfy the equation
MATRICES
101
10
...
Annual sales are indicated below:
Market
Products
10,000
2,000
18,000
I
II
6,000
20,000
8,000
(a) If unit sale prices of x, y and z are Rs 2
...
50 and Rs 1
...
(b) If the unit costs of the above three commodities are Rs 2
...
00 and 50
paise respectively
...
✁1 2 3✂ ✁ 7
11
...
If A and B are square matrices of the same order such that AB = BA, then prove
by induction that ABn = BnA
...
Choose the correct answer in the following questions:
✁✠
✝☛
13
...
If the matrix A is both symmetric and skew symmetric, then
(B) A is a zero matrix
(A) A is a diagonal matrix
(C) A is a square matrix
(D) None of these
15
...
A matrix having m rows and n columns is called a matrix of order m × n
...
[aij]1 × n is a row matrix
...
A = [aij]m × m is a diagonal matrix if aij = 0, when i ✏ j
...
☎ j, a
ij
= k, (k is some
☎ j
...
A = [aij] = [bij] = B if (i) A and B are of same order, (ii) aij = bij for all
possible values of i and j
...
k(A + B) = kA + kB, where A and B are of same order, k is constant
...
✂ ✄a
✁
n
If A = [aij]m × n and B = [bjk]n × p , then AB = C = [cik]m × p, where cik
ij
b jk
j 1
(i) A(BC) = (AB)C, (ii) A(B + C) = AB + AC, (iii) (A + B)C = AC + BC
✝
If A = [aij]m × n, then A or AT = [aji]n × m
✝✝
✝
✝
✝ ✝ ✝
✝ ✝✝
(i) (A ) = A, (ii) (kA) = kA , (iii) (A + B) = A + B , (iv) (AB) = B A
✝
A is a symmetric matrix if A = A
...
Any square matrix can be represented as the sum of a symmetric and a
skew symmetric matrix
...
Inverse of a square matrix, if it exists, is unique
...
— C
...
STEINMETZ
4
...
We have also learnt that a system
of algebraic equations can be expressed in the form of
matrices
...
Now, this
1
2
system of equations has a unique solution or not, is
determined by the number a1 b2 – a2 b1
...
The number a1 b2 – a2 b1
P
...
Laplace
(1749-1827)
which determines uniqueness of solution is associated with the matrix A
✄ ✝✁☎ aa
1
2
b1
b2
✂✆
✞
and is called the determinant of A or det A
...
In this chapter, we shall study determinants up to order three only with real entries
...
4
...
104
MATHEMATICS
This may be thought of as a function which associates each square matrix with a
unique number (real or complex)
...
It is also denoted by | A | or det A or ✆
...
(ii) Only square matrices have determinants
...
2
...
2
...
–1 2
2 4
= 2 (2) – 4(–1) = 4 + 4 = 8
...
2
...
This is known as expansion of a determinant along
a row (or a column)
...
Consider the determinant of square matrix A = [aij]3 × 3
i
...
,
a 11
a12
a13
| A | = a21
a31
a22
a23
a32
a33
Expansion along first Row (R1)
Step 1 Multiply first element a11 of R1 by (–1)(1 + 1) [(–1)sum of suffixes in a11] and with the
second order determinant obtained by deleting the elements of first row (R1) and first
column (C1) of | A | as a11 lies in R1 and C1,
i
...
,
(–1)1 + 1 a11
a22
a23
a32
a33
Step 2 Multiply 2nd element a12 of R1 by (–1)1 + 2 [(–1)sum of suffixes in a12] and the second
order determinant obtained by deleting elements of first row (R1) and 2nd column (C2)
of | A | as a12 lies in R1 and C2,
i
...
,
(–1)1 + 2 a12
a21 a23
a31
a33
Step 3 Multiply third element a13 of R1 by (–1)1 + 3 [(–1)sum of suffixes in a ] and the second
order determinant obtained by deleting elements of first row (R1) and third column (C3)
of | A | as a13 lies in R1 and C3,
13
i
...
,
a21
(–1)1 + 3 a13 a
31
a22
a32
Step 4 Now the expansion of determinant of A, that is, | A | written as sum of all three
terms obtained in steps 1, 2 and 3 above is given by
a22
det A = |A| = (–1)1 + 1 a11 a
32
1
+ (–1)
or
3
a13
a23
a33
✁
(–1)1
2
a12
a21 a23
a31 a33
a21 a22
a31
a32
|A| = a11 (a22 a33 – a32 a23) – a12 (a21 a33 – a31 a23)
+ a13 (a21 a32 – a31 a22)
106
MATHEMATICS
= a11 a22 a33 – a11 a32 a23 – a12 a21 a33 + a12 a31 a23 + a13 a21 a32
– a13 a31 a22
...
Expansion along second row (R2)
a 11
| A | = a 21
a 31
✁
1
a21
✂ ✁
(–1) 2
3
a12
a13
a32
a33
a 32
a 33
(–1)2
a11
a23
a 13
a 23
✂ ✁
Expanding along R2, we get
2
| A | = (–1)
a 12
a 22
2
a22
a11
a13
a31 a33
a12
a31 a32
= – a21 (a12 a33 – a32 a13) + a22 (a11 a33 – a31 a13)
– a23 (a11 a32 – a31 a12)
| A | = – a21 a12 a33 + a21 a32 a13 + a22 a11 a33 – a22 a31 a13 – a23 a11 a32
+ a23 a31 a12
= a11 a22 a33 – a11 a23 a32 – a12 a21 a33 + a12 a23 a31 + a13 a21 a32
– a13 a31 a22
...
(3)
Clearly, values of | A | in (1), (2) and (3) are equal
...
Hence, expanding a determinant along any row or column gives same value
...
(ii) While expanding, instead of multiplying by (–1)i + j, we can multiply by +1 or –1
according as (i + j) is even or odd
...
Then, it is easy to verify that A = 2B
...
(iii) Let A =
✂
☎
Observe that, | A | = 4 (– 2) = 22 | B | or | A | = 2n | B |, where n = 2 is the order of
square matrices A and B
...
4 1 0
Solution Note that in the third column, two entries are zero
...
108
MATHEMATICS
Solution Expanding along R1, we get
✆= 0
0
– sin
sin
0
– sin ✁
– sin ✁ sin
cos ✁
0
– cos ✁
= 0 – sin ✝ (0 – sin ✞ cos ✝) – cos ✝ (sin ✝ sin ✞ – 0)
= sin ✝ sin ✞ cos ✝ – cos ✝ sin ✝ sin ✞ = 0
Example 5 Find values of x for which
Solution We have
3 x
3 2
✂
...
e
...
e
...
1
Evaluate the determinants in Exercises 1 and 2
...
2
–5
4
–1
2
...
If
✠ 1 2✡
A= ☛
☞ , then show that | 2A | = 4 | A |
✌ 4 2✍
(ii)
x2 – x ✟ 1 x – 1
x✟1
x ✟1
✎1 0 1✏
✑
✒
4
...
Evaluate the determinants
(i)
3 –1 –2
0 0 –1
3 –5
0
3 –4
(ii)
5
1
1 –2
2
3
1
– sin ✁
cos ✁
0
– sin
DETERMINANTS
(iii)
0 1 2
–1 0 –3
–2 3
2
0
(iv)
0
109
–1 –2
2 –1
3 –5
0
1 1 –2 ✁
6
...
Find values of x, if
(i)
2 4
5 1
x 2
18 x
(A) 6
8
...
3 Properties of Determinants
In the previous section, we have learnt how to expand the determinants
...
These properties are true for
determinants of any order
...
Property 1 The value of the determinant remains unchanged if its rows and columns
are interchanged
...
✠
Note If Ri = ith row and Ci = ith column, then for interchange of row and
columns, we will symbolically write Ci Ri
Let us verify the above property by example
...
Property 2 If any two rows (or columns) of a determinant are interchanged, then sign
of determinant changes
...
Note We can denote the interchange of rows by Ri
columns by Ci
Cj
...
–7
5
4
= – 28 (See Example 6)
5 –7
Interchanging rows R2 and R3 i
...
, R2
Expanding the determinant
✆
1
✆
1
✆
✠
R3, we have
2 –3 5
= 1 5 –7
6 0 4
along first row, we have
1
= 2
5
–7
0
4
– (–3)
1
–7
6
4
✁
5
1 5
6 0
= 2 (20 – 0) + 3 (4 + 42) + 5 (0 – 30)
= 40 + 138 – 150 = 28
112
MATHEMATICS
✆1 = – ✆
Clearly
Hence, Property 2 is verified
...
Proof If we interchange the identical rows (or columns) of the determinant ✆, then ✆
does not change
...
Example 8 Evaluate ✆
3 2 3
= 2 2 3
3 2 3
Solution Expanding along first row, we get
✆ = 3 (6 – 6) – 2 (6 – 9) + 3 (4 – 6)
= 0 – 2 (–3) + 3 (–2) = 6 – 6 = 0
Here R1 and R3 are identical
...
a1
b1
c1
Verification Let ✆ = a2
a3
b2
c2
b3
c3
and ✆1 be the determinant obtained by multiplying the elements of the first row by k
...
(ii) If corresponding elements of any two rows (or columns) of a determinant are
proportional (in the same ratio), then its value is zero
...
For example,
a1 ✁ ✂1
b1
a2 ✁ ✂ 2
b2
c1
c2
a1 ☎ ✄1
b1
Verification L
...
S
...
H
...
Similarly, we may verify Property 5 for other rows or columns
...
e
...
Verification
a1
Let
a2
a3
✆ = b1 b2 b3 and ✆1 =
c1
c2
c3
a1 ✝ k c1
a2 ✝ k c2
a3 ✝ k c3
b1
b2
b3
c1
c2
c3
,
where ✆1 is obtained by the operation R1 ☎ R1 + kR3
...
Symbolically, we write this operation as R1 ☎ R1 + k R3
...
(ii) If more than one operation like Ri ✄ Ri + kRj is done in one step, care should be
taken to see that a row that is affected in one operation should not be used in
another operation
...
a
a✁b
a✁b✁c
Example 11 Prove that 2a 3a ✁ 2b 4a ✁ 3b ✁ 2c ✂ a 3
...
Example 13 Evaluate
1 a bc
✆ = 1 b ca
1 c ab
Solution Applying R2 ✄ R2 – R1 and R3 ✄ R3 – R1, we get
1
a
bc
0
b
a
c
(
a
☎
☎ b)
✆=
0 c ☎ a b (a ☎ c)
Taking factors (b – a) and (c – a) common from R2 and R3, respectively, we get
1 a
bc
✆ = (b ✝ a ) (c ✝ a) 0 1 – c
0 1
–b
= (b – a) (c – a) [(– b + c)] (Expanding along first column)
= (a – b) (b – c) (c – a)
b
Example 14 Prove that
c
b
c
Solution Let ✆ =
a
c
a
b
a
c
a
b✂c
a
a
b
c✂a
b
c
c
a✂b
✁ 4 abc
b
DETERMINANTS
117
R1 ✄ R1 – R2 – R3 to ✆, we get
Applying
0
–2c
✆= b c
c
–2b
a
c
b
a
b
Expanding along R1, we obtain
✆= 0
c✁a
b
b
b
– (–2 c )
c
a ✁b
c a✁b
✂ (–2b)
b c✂a
c
c
= 2 c (a b + b2 – bc) – 2 b (b c – c2 – ac)
= 2 a b c + 2 cb2 – 2 bc2 – 2 b2c + 2 bc2 + 2 abc
= 4 abc
x
Example 15 If x, y, z are different and ☎ ✝ y
z
x2
y
2
z2
1 x3
1 y 3 ✝ 0 , then
1 z3
show that 1 + xyz = 0
Solution We have
x
x2
1 ✞ x3
2
3
✆ = y y 1✞ y
z z 2 1 ✞ z3
x
= y
z
x2 1
x
x2
x3
y2 1 ✞ y
y2
z2 1
z2
y 3 (Using Property 5)
z3
1 x
2
= (✟1) 1 y
1 z
1 x
= 1 y
1 z
z
x2
1 x
x2
y 2 ✞ xyz 1 y
y2
z
2
x2
y 2 (1✞ xyz )
z2
1 z
z
2
(Using C3 ✠ C2 and then C1 ✠ C2)
118
MATHEMATICS
1
x2
x
= 1 ✂ xyz ✁ 0
y2
y✄x
0 z✄x z
2
✄
✄
x2
x
(Using R2 ☎ R2–R1 and R3
☎
R3– R1)
2
Taking out common factor (y – x) from R2 and (z – x) from R3, we get
1 x
✆ = (1+xyz ) (y –x ) (z –x) 0 1
0 1
x2
y✝x
z✝x
= (1 + xyz) (y – x) (z – x) (z – y) (on expanding along C1)
Since ✆ = 0 and x, y, z are all different, i
...
, x – y ✞ 0, y – z ✞ 0, z – x ✞ 0, we get
1 + xyz = 0
Example 16 Show that
1✟ a
1
1 1✟ b
1
1
1
1
1✟ c
✠
☛ abc ☞ 1 ✟
✍
1 1 1✡
✟ ✟
☛ abc ✟ bc ✟ ca ✟ ab
a b c ✌✎
Solution Taking out factors a,b,c common from R1, R2 and R3, we get
1
1
1
✏1
a
a
a
1
1
1
✏1
L
...
S
...
H
...
a b c✕
Note Alternately try by applying C1 ✞ C1 – C2 and C3 ✞ C3 – C2, then apply
C1 – a C3
...
2
Using the property of determinants and without expanding in Exercises 1 to 7, prove
that:
1
...
3
...
5 9 86
5
...
MATHEMATICS
0 a
a 0
b
c ✁0
b
0
c
7
...
(i) 1 b b ✄ ☎ a ✂ b ✆☎ b ✂ c ✆☎ c ✂ a ✆
c2
1 c
(ii)
1
1
1
a
b
c ✟ ✝ a ✠ b ✞✝ b ✠ c ✞✝ c ✠ a ✞✝ a ✡ b ✡ c ✞
3
3
a
9
...
(i) 2 x x + 4 2x ✁ ☛ 5 x ✌ 4☞ ☛ 4 x ☞
2x
2x x + 4
(ii)
y+k
y
y
y
y+k
y
y
y
y+k
✏ k 2 ✍3y ✑ k ✎
a b c
2a
2a
3
2b
b c a
2b ✁ ✒ a ✌ b ✌ c ✓
11
...
13
...
1 ✄ x3 ✁
b
ab
ac
2
bc
✝1
cb
121
2
✄2b
2
2a
✂ ☎1 ✝
a2
2 3
✝b ✆
1 ✄ a2 ✄ b2
✂1 ✝
a2
✝b
2
✝c
2
c2 ✝ 1
Choose the correct answer in Exercises 15 and 16
...
Let A be a square matrix of order 3 × 3, then | kA | is equal to
(A) k| A |
(B) k 2 | A |
(C) k 3 | A |
(D) 3k | A |
16
...
(B) Determinant is a number associated to a matrix
...
(D) None of these
4
...
Now this expression can be written in the form of a determinant as
(x1, y1), (x2, y2) and (x3, y3), is given by the expression
x1
1
x
✞=
2
2
x3
y1 1
y2 1
...
122
MATHEMATICS
(ii) If area is given, use both positive and negative values of the determinant for
calculation
...
Example 17 Find the area of the triangle whose vertices are (3, 8), (– 4, 2) and (5, 1)
...
Solution Let P (x, y) be any point on AB
...
So
0 0 1
1
1 3 1 =0
2
x y 1
1
☛ y – 3 x ☞ = 0 or y = 3x,
2
which is the equation of required line AB
...
units, we have
This gives
1 3 1
1
0 0 1 =±3
2
k 0 1
This gives,
✌ 3k
2
✡ ✍3 ,
i
...
, k =
✎
2
...
3
1
...
Show that points
A (a, b + c), B (b, c + a), C (c, a + b) are collinear
...
Find values of k if area of triangle is 4 sq
...
(i) Find equation of line joining (1, 2) and (3, 6) using determinants
...
5
...
Then k is
(B) –2
(C) –12, –2
(D) 12, –2
(A) 12
4
...
Definition 1 Minor of an element aij of a determinant is the determinant obtained by
deleting its ith row and jth column in which element aij lies
...
Remark Minor of an element of a determinant of order n(n
order n – 1
...
7 8
Definition 2 Cofactor of an element aij , denoted by Aij is defined by
Aij = (–1)i + j Mij , where Mij is minor of aij
...
So M11 = Minor of a11= 3
M12 = Minor of the element a12 = 4
M21 = Minor of the element a21 = –2
1
–2
4
3
124
MATHEMATICS
M22 = Minor of the element a22 = 1
Now, cofactor of aij is Aij
...
Hence = sum of the product of elements of any row (or column) with their
corresponding cofactors
...
For example,
DETERMINANTS
✆ = a11 A21 + a12 A22 + a13 A23
= a11 (–1)1+1
a12
a13
a32
a33
a11
= a11
a31
a13
a13 = 0 (since R and R are identical)
1
2
a33
a12
a12
a32
+ a12 (–1)1+2
a11
a13
a31 a33
+ a13 (–1)1+3
a11
a12
a31 a32
Similarly, we can try for other rows and columns
...
4
Write Minors and Cofactors of the elements of following determinants:
1
...
(i) 0 1 0
0 0 1
(ii)
a
c
b d
1 0 4
3 5 –1
0 1
2
5 3 8
3
...
1 2 3
1 x
4
...
If ✆ = a21
a31
a12
a22
a32
yz
zx
...
6 Adjoint and Inverse of a Matrix
In the previous chapter, we have studied inverse of a matrix
...
To find inverse of a matrix A, i
...
, A–1 we shall first define adjoint of a matrix
...
6
...
Adjoint of the matrix A
is denoted by adj A
...
e
...
Theorem 1 If A be any given square matrix of order n, then
A(adj A) = (adj A) A =
where I is the identity matrix of order n
❆ ✒,
128
MATHEMATICS
Verification
a11
Let
A=
✂
a
✂ 21
✂
☎ a31
a12
a22
a32
a13 ✁
a23 ✄ ,
✄
a33 ✆✄
A11
then adj A =
✂
A
✂ 12
✂☎ A13
A 21
A 31 ✁
A 22
A32 ✄
A 23
A 33 ✄✆
✄
Since sum of product of elements of a row (or a column) with corresponding
cofactors is equal to | A | and otherwise zero, we have
A (adj A) =
✝A
✟
0
✟
✟ 0
✡
0
A
0
0✞
✠
0✠ = A
A ✠☛
1 0 0✁
✂
0
✂
✂☎ 0
1 0✄ = A I
✄
0 1 ✄✆
Similarly, we can show (adj A) A = A I
Hence A (adj A) = (adj A) A = A I
Definition 4 A square matrix A is said to be singular if A = 0
...
Definition 5 A square matrix A is said to be non-singular if A
Let
A=
☞1
✍
✏3
2✌
...
Hence A is a nonsingular matrix
We state the following theorems without proof
...
Theorem 3 The determinant of the product of matrices is equal to product of their
respective determinants, that is, AB = A B , where A and B are square matrices of
the same order
Remark We know that (adj A) A = A I =
✓
✕
✕
✕
✗
A
0
0
A
0
0
✔
✖
✖
A ✖✘
0
0
DETERMINANTS
129
Writing determinants of matrices on both sides, we have
A
(adj A) A = 0
0
0
A
0
0
0
A
1 0 0
i
...
3
|(adj A)| |A| = A 0 1 0
0 0 1
(Why?)
i
...
|(adj A)| |A| = | A |3 (1)
i
...
|(adj A)| = | A | 2
In general, if A is a square matrix of order n, then | adj (A) | = | A |n – 1
...
Proof Let A be invertible matrix of order n and I be the identity matrix of order n
...
So AB = I
or
A B =1
(since I 1, AB
A B)
A ✂ 0
...
Conversely, let A be nonsingular
...
Also find A–1
...
I
✄✆0 0 1☎✝
✟ 7 ✞3 ✞3✠ ✟ 7 ✞3 ✞3✠
1✡
1
☛
☛ ✡
✏ adj A = ✡ ✞1 1 0 ☛ = ✡ ✞1 1 0 ☛
1
A
✡☞ ✞1 0 1☛✌ ☞✡ ✞1 0 1✌☛
✒2 3 ✓
✒ 1 ✑ 2✓
and B ✔ ✕
Example 25 If A = ✕
✖
✖ , then verify that (AB)–1 = B–1A–1
...
Therefore, A–1 and B–1 both exist and are given by
A–1 = ✑
1 ✒ ✑ 4 ✑ 3✓ ✮1 ✒3 2✓
,B ✔ ✕
✖
11 ✗✕ ✑ 1 2 ✘✖
✗ 1 1✘
DETERMINANTS
B 1A 1 ☎ ✁
Therefore
1 ✂ 3 2✄
11 ✆✞ 1 1 ✝✟
✂✁
4
1
3✄
2 ✝✟
✁
✆
✁
✞
☎ ✁
1 ✂ ✁14
11 ✆✞ ✁5
5✄
✝
✁1
✟
✁
☎
131
1 ✂14 5✄
11 ✆✞ 5 1✝✟
Hence (AB)–1 = B–1 A–1
2 3✡
satisfies the equation A2 – 4A + I = O,
1 2☞✍
✌
where I is 2 × 2 identity matrix and O is 2 × 2 zero matrix
...
✠
Example 26 Show that the matrix A =
Solution We have A 2 ✎ A
...
5
Find adjoint of each of the matrices in Exercises 1 and 2
...
✆
3 4✝✟
✞
✂
✕
2
...
✆
✁4
✞
✂
3✄
✝
✁6
✟
✗
4
...
✁2
5
...
✠ 3 3 0 ✡
✠☛ 5 2 ✌1✡☞
✁ 1 5✂
6
...
✠ 4 ✌1 0✡
✠☛ ✌7 2 1✡☞
0
0 ✎
✍1
✏ 0 cos ✒
sin ✒ ✑
11
...
Let A = ✘
✙ and B =
✚ 2 5✛
✞ 1 2 3✟
✠
✡
7
...
✠ 0 2 ✌3✡
✠☛ 3 ✌2 4 ✡☞
✖ 6 8✗
✘ 7 9✙
...
✚
✛
✖ 3 1✗
13
...
Hence find A–1
...
For the matrix A = ✘
✙ , find the numbers a and b such that A2 + aA + bI = O
...
For the matrix A = ✠✠ 1 2 ✌3✡✡
☛✠ 2 ✌1 3 ☞✡
Show that A3– 6A2 + 5A + 11 I = O
...
✞ 2 ✌1 1 ✟
✠
✡
16
...
Let A be a nonsingular square matrix of order 3 × 3
...
If A is an invertible matrix of order 2, then det (A ) is equal to
1
(A) det (A)
(B) det (A)
(C) 1
(D) 0
DETERMINANTS
133
4
...
Consistent system A system of equations is said to be consistent if its solution (one
or more) exists
...
Note In this chapter, we restrict ourselves to the system of linear equations
having unique solutions only
...
7
...
Consider the system of equations
Let
✄✄✁
✝✄
✁✄
✝✄✄
a 1 x + b 1 y + c 1 z = d1
a2 x + b2 y + c 2 z = d 2
a3 x + b3 y + c 3 z = d 3
☎☎✂ ✆ ✄✁✄ ☎✂☎
☎✞ ✝✄ ☎✞
✂☎ ✁✄ ✂☎ ✁✄ ✂☎
☎☎✞ ✝✄✄ ☎☎✞ ✝✄✄ ☎☎✞
a1
b1
c1
x
A = a2
a3
b2
c2 , X
y and B
b3
c3
z
✆ ✝✄✄✄✁ ☎✂☎☎✞
d1
d2
d3
Then, the system of equations can be written as, AX = B, i
...
,
a1
b1
c1
a2
b2
c2
a3
b3
c3
x
d1
y = d2
d3
z
Case I If A is a nonsingular matrix, then its inverse exists
...
This method of solving system of equations is known as
Matrix Method
...
In this case, we calculate (adj A) B
...
If (adj A) B = O, then system may be either consistent or inconsistent according
as the system have either infinitely many solutions or no solution
...
A–1 =
Note that
✟
X = A–1B = –
Therefore
x✁
☎ ✆
✝ y✞
i
...
Hence
=
✠
✟5✁
2 ✞✆
1 2
11 ✝☎ ✟3
1 2
11 ☎✝✟ 3
✟ 5✁ 1✁
2 ✆✞ ☎✝7✆✞
1 ✡ ✠33☛ ✡ 3 ☛
☞
11 ✎✌ 11 ✏✍ ✎✌ ✠1✏✍
x = 3, y = – 1
Example 28 Solve the following system of equations by matrix method
...
Now
A11 = –1,
A12 = – 8,
A21 = –5,
A22 = – 6,
A31 = –1,
A32 = 9,
Therefore
–1
A =
✁ 1
1 ✄
8
17 ✄
✆✄ 10
5
6
1
135
A13 = –10
A23 = 1
A33 = 7
1✂
9☎
☎
7 ✝☎
✟ ✞1 ✞ 5 ✞1✠ ✟ 8 ✠
1 ✡
☛✡ ☛
X = A B = ✞ ✡ ✞8 ✞ 6 9 ☛ ✡ 1 ☛
17
☞✡ ✞10 1 7 ☛✌ ✡☞ 4☛✌
–1
So
✁ x✂
✄ y☎
✄ ☎ =
✄✆ z ☎✝
i
...
Hence
✁ 17✂ ✁1✂
1 ✄
34☎ ✍ ✄2☎
✄
☎ ✄ ☎
17
✄✆ 51☎✝ ✄✆3☎✝
x = 1, y = 2 and z = 3
...
If we multiply third number by 3 and add
second number to it, we get 11
...
Represent it algebraically and find the numbers using matrix method
...
Then, according to given conditions, we have
x+y+z=6
y + 3z = 11
x + z = 2y or x – 2y + z = 0
This system can be written as A X = B, where
✎ 1 1 1✏
✑
✒
A = ✑ 0 1 3✒ , X =
✓✑ 1 –2 1✒✔
✎ x✏
✑ y ✒ and B =
✑ ✒
✓✑ z ✔✒
✎6✏
✑11✒
✑ ✒
✓✑ 0 ✔✒
Here A ✗ 1 ✕1 ✘ 6✖ – (0 – 3) ✘ ✕ 0 – 1✖ ✗ 9 ✙ 0
...
6
Examine the consistency of the system of equations in Exercises 1 to 6
...
x + 2y = 2
2
...
x + 3y = 5
2x + 3y = 3
x+y=4
2x + 6y = 8
4
...
3x–y – 2z = 2
6
...
7
...
2x – y = –2
9
...
5x + 2y = 3
11
...
x – y + z = 4
3
3x + 2y = 5
x – 2y – z =
2x + y – 3z = 0
2
3y – 5z = 9
x+y+z=2
13
...
x – y + 2z = 7
x – 2y + z = – 4
3x + 4y – 5z = – 5
3x – y – 2z = 3
2x – y + 3z = 12
DETERMINANTS
2 –3
15
...
Using A–1 solve the system of equations
1
–2 ✄✆
2x – 3y + 5z = 11
3x + 2y – 4z = – 5
x + y – 2z = – 3
16
...
The cost of 2 kg onion,
4 kg wheat and 6 kg rice is Rs 90
...
Find cost of each item per kg by matrix method
...
c a b
Solution Applying C1 ✞ C1 + C2 + C3 to the given determinant, we get
a✟b✟c b c
1 b c
✝ = a ✟ b ✟ c c a = (a + b + c) 1 c a
1 a b
a✟b✟c a b
1
b
c
= (a + b + c) 0 c – b a – c (Applying R2✞ R2–R1,and R3✞ R3 –R1)
0 a–b b–c
= (a + b + c) [(c – b) (b – c) – (a – c) (a – b)] (Expanding along C1)
= (a + b + c)(– a2 – b2 – c2 + ab + bc + ca)
=
–1
(a + b + c) (2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca)
2
=
–1
(a + b + c) [(a – b)2 + (b – c)2 + (c – a)2]
2
which is negative (since a + b + c > 0 and (a – b)2 + (b – c)2 + (c – a)2 > 0)
138
MATHEMATICS
Example 31 If a, b, c, are in A
...
1
...
Without expanding the determinant, prove that b b
c c2
1 a2
bc
ca ✂ 1 b 2
ab
1 c2
cos ☎ cos ✝ cos ☎ sin ✝ – sin ☎
– sin ✝
cos ✝
0
...
Evaluate
sin ☎ cos ✝ sin ☎ sin ✝ cos ☎
4
...
x✟a
5
...
Prove that a ☛ ab
2
ab
b ☛ bc
✠ 0, a✡ 0
ac☛ c 2
ac
c
= 4a2b2c2
2
✎ 3 –1 1 ✏
✎ 1 2 –2✏
✑
✒
✑
✒
7
...
c3
141
142
MATHEMATICS
1
8
...
Verify that
5✆✄
3
1
–1
(i) [adj A] = adj (A–1)
x
y
x✝ y
x✝ y
x
x✝ y
x
y
9
...
Evaluate
1
x
(ii) (A–1)–1 = A
y
y
x+ y
Using properties of determinants in Exercises 11 to 15, prove that:
✞
11
...
✞
2
✟✠✡
✟
2
✡✠✞
✡
✡
2
✞ ✠✟
x
x 2 1 ✍ px 3
y
y 2 1 ✍ py 3 = (1 + pxyz) (x – y) (y – z) (z – x), where p is any scalar
...
3b
– c✠ a
– c+ b
– b ✠ c = 3(a + b + c) (ab + bc + ca)
3c
1 1✝ p
2 3✝ 2 p
1✝ p✝ q
4✝ 3 p✝ 2q = 1
3 6 ✝ 3 p 10 ✝ 6 p ✝ 3 q
16
...
= (☛ – ✌) (✌ – ☞) (☞ – ☛) (☞ +
✗
10
z
✘
4
15
...
17
...
P, then the determinant
x ✂ 2 x ✂ 3 x ✂ 2a
x ✂ 3 x ✂ 4 x ✂ 2b is
x ✂ 4 x ✂ 5 x ✂ 2c
(A) 0
(B) 1
(C) x
(D) 2x
✄ x 0 0☎
18
...
Then
19
...
✒
✒
|A | = |A|, where A = transpose of A
...
If any two rows or any two columns are identical or proportional, then value
of determinant is zero
...
Multiplying a determinant by k means multiply elements of only one row
(or one column) by k
...
A k 3 A
If elements of a row or a column in a determinant can be expressed as sum
of two or more elements, then the given determinant can be expressed as
sum of two or more determinants
...
DETERMINANTS
145
Area of a triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is given by
✁✂
y1 1
x1
1
x2
2
x3
y2 1
y3 1
Minor of an element aij of the determinant of matrix A is the determinant
obtained by deleting ith row and jth column and denoted by Mij
...
For example,
A = a11 A11 + a12 A12 + a13 A13
...
For example, a11 A21 + a12
A22 + a13 A23 = 0
✄a
A ✞ ✆a
✆✆
✟a
11
If
21
a12
a22
a32
☎
✝✝✝
✠
a13
a23 , then adj A
a33
✄A
✞ ✆✆A
✟✆A
11
12
A 21
A 22
☎
✝✝
✝✠
A31
A32 , where A ij is
A 23 A33
13
cofactor of aij
A (adj A) = (adj A) A = | A | I, where A is square matrix of order n
...
If AB = BA = I, where B is square matrix, then B is called inverse of A
...
A square matrix A has inverse if and only if A is non-singular
...
A system of equation is consistent or inconsistent according as its solution
exists or not
...
✂
✂
Historical Note
The Chinese method of representing the coefficients of the unknowns of
several linear equations by using rods on a calculating board naturally led to the
discovery of simple method of elimination
...
The Chinese, therefore, early developed the
idea of subtracting columns and rows as in simplification of a determinant
‘Mikami, China, pp 30, 93
...
But he used this device only in eliminating a
quantity from two equations and not directly in the solution of a set of simultaneous
linear equations
...
Hayashi, “The Fakudoi and Determinants in Japanese
Mathematics,” in the proc
...
Soc
...
Vendermonde was the first to recognise determinants as independent functions
...
Laplace (1772), gave general method of
expanding a determinant in terms of its complementary minors
...
In 1801, Gauss used determinants in his
theory of numbers
...
Also on the same day, Cauchy (1812) presented one on the same subject
...
He gave the proof of multiplication
theorem more satisfactory than Binet’s
...
Chapter
5
CONTINUITY AND
DIFFERENTIABILITY
The whole of science is nothing more than a refinement
of everyday thinking
...
1 Introduction
This chapter is essentially a continuation of our study of
differentiation of functions in Class XI
...
In this chapter, we introduce the
very important concepts of continuity, differentiability and
relations between them
...
Further, we introduce a
new class of functions called exponential and logarithmic
functions
...
We illustrate certain geometrically obvious
conditions through differential calculus
...
Sir Issac Newton
(1642-1727)
5
...
Consider
the function
f ( x)
✄ ✂☎✝1,2,ifif xx✁✆00
This function is of course defined at every
point of the real line
...
1
...
At the points near and to the
left of 0, i
...
, at points like – 0
...
01, – 0
...
At the points near
and to the right of 0, i
...
, at points like 0
...
01,
Fig 5
...
001, the value of the function is 2
...
In
particular the left and right hand limits do not coincide
...
Note that when we try to draw
the graph, we cannot draw it in one stroke, i
...
, without lifting pen from the plane of the
paper, we can not draw the graph of this function
...
This is one instance of function being not continuous at x = 0
...
Left and the right hand limits at x = 0
are both equal to 1
...
Again, we note that we
cannot draw the graph of the function without
lifting the pen
...
Naively, we may say that a function is
continuous at a fixed point if we can draw the
graph of the function around that point without
lifting the pen from the plane of the paper
...
2
Mathematically, it may be phrased precisely as follows:
Definition 1 Suppose f is a real function on a subset of the real numbers and let c be
a point in the domain of f
...
Recall that
if the right hand and left hand limits at x = c coincide, then we say that the common
value is the limit of the function at x = c
...
If f is not continuous at c, we say f is discontinuous at c and c is called a point
of discontinuity of f
...
Solution First note that the function is defined at the given point x = 1 and its value is 5
...
Clearly
lim f ( x ) ✁ lim (2 x ✂ 3) ✁ 2(1) ✂ 3 ✁ 5
1
x
Thus
1
x
lim f ( x ) ✁ 5 ✁ f (1)
x
1
Hence, f is continuous at x = 1
...
Solution First note that the function is defined at the given point x = 0 and its value is 0
...
Clearly
lim f ( x) ☎ lim x 2 ☎ 02
✄0
lim f ( x) ✁ 0 ✁ f (0)
x 0
x
Thus
✄0
x
☎0
Hence, f is continuous at x = 0
...
Solution By definition
f (x) =
✞ ✆ x, if x ✝ 0
✟ x, if x ✠ 0
✡
Clearly the function is defined at 0 and f (0) = 0
...
Hence, f is continuous at x = 0
...
✒✎ x3 ✏ 3,
✓
✒✔1,
if x ✑ 0
if x ☎ 0
150
MATHEMATICS
Solution The function is defined at x = 0 and its value at x = 0 is 1
...
Hence,
lim f ( x) = lim ( x 3 ✄ 3) ☎ 03 ✄ 3 ☎ 3
x
0
x
✁0
Since the limit of f at x = 0 does not coincide with f (0), the function is not continuous
at x = 0
...
Example 5 Check the points where the constant function f (x) = k is continuous
...
Let c be any real number
...
Example 6 Prove that the identity function on real numbers given by f (x) = x is
continuous at every real number
...
Also,
lim f ( x) = lim x ✞ c
x
✝c
x
✝c
Thus, lim f (x) = c = f (c) and hence the function is continuous at every real number
...
Definition 2 A real function f is said to be continuous if it is continuous at every point
in the domain of f
...
Suppose f is a function defined on a
closed interval [a, b], then for f to be continuous, it needs to be continuous at every
point in [a, b] including the end points a and b
...
As a consequence
x
✠a
x
✠b
of this definition, if f is defined only at one point, it is continuous there, i
...
, if the
domain of f is a singleton, f is a continuous function
...
Let c be a real number such that c < 0
...
Also
lim f ( x) = lim (✞ x) ✟ – c
x
✝c
x
✝c
(Why?)
Since lim f ( x) ✡ f (c ) , f is continuous at all negative real numbers
...
Then f (c) = c
...
Hence, f
x
✠c
is continuous at all points
...
Solution Clearly f is defined at every real number c and its value at c is c3 + c2 – 1
...
This means
x
✝c
f is a continuous function
...
x
Solution Fix any non zero real number c, we have
lim f ( x ) ✑ lim
x
✏c
x
1
✏c x
✑
1
c
1
Also, since for c ✎ 0, f (c) ✑ , we have lim f ( x) ✡ f (c ) and hence, f is continuous
x✠ c
c
at every point in the domain of f
...
152
MATHEMATICS
We take this opportunity to explain the concept of infinity
...
To carry out this analysis we follow the usual trick of
x
finding the value of the function at real numbers close to 0
...
We tabulate this in the following (Table 5
...
Table 5
...
3
1 3
...
0
...
1 = 10–1
0
...
001 = 10–3 10–n
1000 = 103
10n
We observe that as x gets closer to 0 from the right, the value of f (x) shoots up
higher
...
In symbols, we write
lim f ( x) ✂ ✄ ☎
x✁ 0
(to be read as: the right hand limit of f (x) at 0 is plus infinity)
...
Similarly, the left hand limit of f at 0 may be found
...
Table 5
...
3
– 1 – 3
...
– 0
...
2, we deduce that the
value of f (x) may be made smaller than any
given number by choosing a negative real
number very close to 0
...
Again, we wish to emphasise
that – ✆ is NOT a real number and hence the
left hand limit of f at 0 does not exist (as a real
number)
...
3 is a geometric representation
of the above mentioned facts
...
3
CONTINUITY AND DIFFERENTIABILITY
153
Example 10 Discuss the continuity of the function f defined by
✂ x 2, if x ✁ 1
✝ x ☎ 2, if x ✆ 1
f (x) = ✄
Solution The function f is defined at all points of the real line
...
Therefore, lim f ( x ) ✟ lim f ( x ✠ 2) ✟ c ✠ 2
x
✞c
x
✞c
Thus, f is continuous at all real numbers less than 1
...
Therefore,
lim f ( x) ✟ lim (x – 2) = c – 2 = f (c)
x
✞c
x
✞c
Thus, f is continuous at all points x > 1
...
4
do not coincide, f is not continuous at x = 1
...
The graph of the function is given in Fig 5
...
Example 11 Find all the points of discontinuity of the function f defined by
✍ x ☞ 2, if x ✌ 1
✎
f (x) = ✑ 0, if x ✏ 1
✎ x ✒ 2, if x ✓ 1
✔
Solution As in the previous example we find that f
is continuous at all real numbers x ✕ 1
...
Hence
x = 1 is the only point of discontinuity of f
...
5
...
5
154
MATHEMATICS
Example 12 Discuss the continuity of the function defined by
✂ x 2, if x ✁ 0
✝☎ x 2, if x ✆ 0
f (x) = ✄
Solution Observe that the function is defined at all real numbers except at 0
...
Case 2 If c ✟ D2, then lim f ( x ) ☞ lim (– x + 2)
x
☛c
x
☛c
= – c + 2 = f (c) and hence f is continuous in D2
...
Graph of this
function is given in the Fig 5
...
Note that to graph
Fig 5
...
Example 13 Discuss the continuity of the function f given by
✍✎ x, if x ✌ 0
2
✒✎ x , if x ✑ 0
f (x) = ✏
Solution Clearly the function is defined at
every real number
...
7
...
Let
D1 = {x ✟ R : x < 0}, D2 = {0} and
D3 = {x ✟ R : x > 0}
Fig 5
...
Case 2 At any point in D3, we have f (x) = x and it is easy to see that it is continuous
there (see Example 6)
...
The value of the function at 0 is f (0) = 0
...
This means that f is
x☎0
continuous at every point in its domain and hence, f is a continuous function
...
Solution Recall that a function p is a polynomial function if it is defined by
p(x) = a0 + a1 x +
...
Clearly this
function is defined for every real number
...
Since c is any real number, p is continuous at
every real number and hence p is a continuous function
...
Solution First observe that f is defined for all real numbers
...
8
...
Below we explore, if this is true
...
8
156
MATHEMATICS
Case 1 Let c be a real number which is not equal to any integer
...
e
...
Also f (c) = [c] and hence the function is continuous at all real
x
c
x
c
numbers not equal to integers
...
Then we can find a sufficiently small real number
r > 0 such that [c – r] = c – 1 whereas [c + r] = c
...
5
...
1 Algebra of continuous functions
In the previous class, after having understood the concept of limits, we learnt some
algebra of limits
...
Since continuity of a function at a point is entirely dictated by the limit of the function at
that point, it is reasonable to expect results analogous to the case of limits
...
Then
(1) f + g is continuous at x = c
...
(3) f
...
(4)
✆f✝
✞g✟
✠ ✡
is continuous at x = c, (provided g (c) ☛ 0)
...
Clearly it is defined at
x = c
...
x
c
(as f and g are continuous)
(by definition of f + g)
Proofs for the remaining parts are similar and left as an exercise to the reader
...
e
...
g) defined by (✝
...
g(x) is also
continuous
...
(ii) As a special case of (4) above, if f is the constant function f (x) = ✝ , then the
function
g
defined by
✁
g
( x) ✂
✁
g ( x)
is also continuous wherever g (x) ✄ 0
...
g
The above theorem can be exploited to generate many continuous functions
...
The following examples
illustrate this:
particular, the continuity of g implies continuity of
Example 16 Prove that every rational function is continuous
...
The domain of f is all real numbers except
points at which q is zero
...
f ( x) ☎
Example 17 Discuss the continuity of sine function
...
Now, observe that f (x) = sin x is defined for every real number
...
Put x = c + h
...
Therefore
lim f ( x ) = lim sin x
x✡ c
x✡ c
sin(c ☛ h)
= lim
h✞ 0
[sin c cos h ☛ cos c sin h]
= lim
h✞ 0
[sin c cos h] ☛ lim [cos c sin h]
= lim
h✞ 0
h✞ 0
= sin c + 0 = sin c = f (c)
Thus lim f (x) = f (c) and hence f is a continuous function
...
Example 18 Prove that the function defined by f (x) = tan x is a continuous function
...
This is defined for all real numbers such
cos x
that cos x ✂ 0, i
...
, x ✂ (2n +1)
...
Thus tan x being a quotient of two continuous functions is
continuous wherever it is defined
...
Recall that if f and g are two real functions, then
(f o g) (x) = f (g (x))
is defined whenever the range of g is a subset of domain of f
...
Theorem 2 Suppose f and g are real valued functions such that (f o g) is defined at c
...
The following examples illustrate this theorem
...
Solution Observe that the function is defined for every real number
...
Since both g and h are continuous functions, by Theorem 2,
it can be deduced that f is a continuous function
...
Solution Define g by g (x) = 1 – x + | x | and h by h (x) = | x | for all real x
...
Hence g being a sum
of a polynomial function and the modulus function is continuous
...
CONTINUITY AND DIFFERENTIABILITY
159
EXERCISE 5
...
Prove that the function f (x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5
...
Examine the continuity of the function f (x) = 2x2 – 1 at x = 3
...
Examine the following functions for continuity
...
Prove that the function f (x) = xn is continuous at x = n, where n is a positive
integer
...
Is the function f defined by
(c) f (x) =
☎ x, if x ✄ 1
f ( x) ✆ ✝
✞5, if x > 1
continuous at x = 0? At x = 1? At x = 2?
Find all points of discontinuity of f, where f is defined by
☎2 x ✟ 3, if x ✄ 2
✞2 x ✠ 3, if x > 2
6
...
✔| x |
✖ , if x ✕ 0
f ( x) ✗ ✘ x
✖✙ 0, if x ✗ 0
✩✪ x ✧ 1, if x ★ 1
2
✪✮ x ✧ 1, if x ✭ 1
10
...
f ( x) ✳ ✴
7
...
✚ x
✜ , if x ✛ 0
f ( x) ✢ ✣| x |
✜
✦✤1, if x ✥ 0
11
...
Is the function defined by
✺ x ✸ 5, if x ✹ 1
✿ x ✽ 5, if x ✾ 1
f ( x) ✻ ✼
a continuous function?
✯✲ x 3 ✰ 3, if x ✱ 2
2
✷✲ x ✵ 1, if x ✶ 2
f ( x) ✳ ✴
160
MATHEMATICS
Discuss the continuity of the function f, where f is defined by
14
...
✁3, if 0 x 1
✂
f ( x ) ✄ ✆ 4, if 1 ☎ x ☎ 3
15
...
Find the relationship between a and b so that the function f defined by
✒ ax ✏ 1, if x ✑ 3
✖bx ✏ 3, if x ✕ 3
f ( x) ✓ ✔
is continuous at x = 3
...
For what value of ✗ is the function defined by
✜✘✙ ( x 2 ✚ 2 x), if x ✛ 0
if x ✥ 0
✜✦ 4 x ✤ 1,
f ( x) ✢ ✣
19
...
21
...
23
...
Here [x] denotes the greatest integer less than or equal to x
...
cos x
Discuss the continuity of the cosine, cosecant, secant and cotangent functions
...
Determine if f defined by
★ 2 1
✪ x sin , if x ✰ 0
f ( x) ✫ ✬
x
✪✯ 0,
if x ✫ 0
is a continuous function?
CONTINUITY AND DIFFERENTIABILITY
161
25
...
26
...
f ( x) ☛ ☞
28
...
Find the values of a and b such that the function defined by
29
...
32
...
34
...
Show that the function defined by f (x) = cos (x2) is a continuous function
...
Examine that sin | x | is a continuous function
...
5
...
Differentiability
Recall the following facts from previous class
...
The derivative of f at c is
defined by
f (c ✬ h ) ✭ f (c)
h✫ 0
h
lim
162
MATHEMATICS
provided this limit exists
...
The
dx
function defined by
f ✄( x) ☎ lim
h
0
f ( x ✁ h) ✂ f ( x )
h
wherever the limit exists is defined to be the derivative of f
...
The process of finding
dx
dx
derivative of a function is called differentiation
...
The following rules were established as a part of algebra of derivatives:
(1) (u ± v)✠ = u✠ ± v✠
(2) (uv)✠ = u✠v + uv✠ (Leibnitz or product rule)
denoted by f ✠ (x) or
✞ u ✟✆ ✡ u✆v ✝ uv✆ , wherever v ✎ 0 (Quotient rule)
...
3
f (x)
xn
sin x
cos x
tan x
f ✠ (x)
nx n – 1
cos x
– sin x
sec2 x
Whenever we defined derivative, we had put a caution provided the limit exists
...
h
In other words, we say that a function f is differentiable at a point c in its domain if both
its answer
...
A function is said
h
0
h
h
to be differentiable in an interval [a, b] if it is differentiable at every point of [a, b]
...
Similarly, a function is said to be differentiable in an interval
(a, b) if it is differentiable at every point of (a, b)
...
Proof Since f is differentiable at c, we have
lim
x
c
But for x ☎ c, we have
f ( x) ✁ f ( c)
✂ f ✄(c )
x✁c
f (x) – f (c) =
Therefore
or
f ( x ) ✁ f (c)
...
( x ✟ c) ✡
☞✍
✞c ✌ x ✟ c
lim [ f ( x ) ✝ f (c)] = lim ☛
x
✆c
x
f ( x) ✏ f (c) ✒
...
0 = 0
lim [ f ( x)] ✝ lim [ f (c )] = lim ✑
✓
x✆ c
x✆ c
x
lim f ( x ) = f (c)
or
x
✘c
Hence f is continuous at x = c
...
We remark that the converse of the above statement is not true
...
Consider the left
hand limit
f (0 ✙ h) ✁ f (0) ✁ h
✂ ✂ ✁1
h
h
lim–
0
h
The right hand limit
lim✚
h
0
f (0 ✙ h) ✁ f (0) h
✂ ✂1
h
h
f (0 ✙ h) ✁ f (0)
h 0
h
does not exist and hence f is not differentiable at 0
...
Since the above left and right hand limits at 0 are not equal, lim
5
...
1 Derivatives of composite functions
To study derivative of composite functions, we start with an illustrative example
...
d
d
3
f ( x) =
✄ (2 x ✂ 1) ✁☎
dx
dx
d
(8 x 3 ✂ 12 x 2 ✂ 6 x ✂ 1)
dx
= 24x2 + 24x + 6
= 6 (2x + 1)2
f (x) = (h o g) (x)
=
Now, observe that
where g(x) = 2x + 1 and h(x) = x3
...
Then f(x) = h(t) = t3
...
2 = 3t2
...
We may formalise this observation in the following
theorem called the chain rule
...
e
...
Suppose t = u (x) and if both
dt
dv
and
exist, we have
dx
dt
df dv dt
✝
✆
dx dt dx
We skip the proof of this theorem
...
Suppose
f is a real valued function which is a composite of three functions u, v and w ; i
...
,
f = (w o u) o v
...
Reader is invited to formulate chain
rule for composite of more functions
...
Solution Observe that the given function is a composite of two functions
...
Observe that
dv
dt
cos t and
dt
dx
165
2 x exist
...
Thus
df
= cos t ✂ 2 x ✄ 2 x cos x 2
dx
Alternatively, We can also directly proceed as follows:
y = sin (x2) ✡
= cos x2
dy
dx
d
(sin x2)
dx
d 2
(x ) = 2x cos x2
dx
Example 22 Find the derivative of tan (2x + 3)
...
Then
(v o u) (x) = v(u(x)) = v(2x + 3) = tan (2x + 3) = f (x)
Thus f is a composite of two functions
...
Then
dv
dt
sec 2 t and
dt
☎ 2 exist
...
Solution The function f (x) = sin (cos (x2)) is a composition f (x) = (w o v o u) (x) of the
three functions u, v and w, where u(x) = x2, v(t) = cos t and w(s) = sin s
...
Hence by a generalisation of chain rule, we have
t = u(x) = x2 and s = v (t) = cos t
...
(– sin t)
...
cos (cos x2)
dx ds dt dx
2x
166
MATHEMATICS
Alternatively, we can proceed as follows:
y = sin (cos x2)
Therefore
dy
dx
d
d
sin (cos x2) = cos (cos x2)
(cos x2)
dx
dx
= cos (cos x2) (– sin x2)
d
(x2)
dx
= – sin x2 cos (cos x2) (2x)
= – 2x sin x2 cos (cos x2)
EXERCISE 5
...
2
...
sin (ax + b)
1
...
sec (tan ( x ))
sin ( ax ✁ b)
5
...
cos x3
...
cos ☎ x ✆
2 cot ✂ x 2 ✄
9
...
f (x) = | x – 1 |, x ✝ R
is not differentiable at x = 1
...
Prove that the greatest integer function defined by
f (x) = [x], 0 < x < 3
is not differentiable at x = 1 and x = 2
...
3
...
But it is not necessary that functions are always expressed in this form
...
In
the second case, it does not seem that there is an easy way to solve for y
...
When a
relationship between x and y is expressed in a way that it is easy to solve for y and
write y = f (x), we say that y is given as an explicit function of x
...
In this subsection, we learn to differentiate implicit
functions
...
dx
Solution One way is to solve for y and rewrite the above as
y=x–✟
dy
=1
dx
Alternatively, directly differentiating the relationship w
...
t
...
r
...
, x
...
dx
Solution We differentiate the relationship directly with respect to x, i
...
,
dy d
d
☎ (sin y) = (cos x)
dx
dx dx
which implies using chain rule
dy
dy
☎ cos y ✆
= – sin x
dx
dx
This gives
where
dy
sin x
= ✝
dx
1 ✞ cos y
y ✠ (2n + 1) ✟
168
MATHEMATICS
5
...
3 Derivatives of inverse trigonometric functions
We remark that inverse trigonometric functions are continuous functions, but we will
not prove this
...
Example 26 Find the derivative of f given by f (x) = sin–1 x assuming it exists
...
Then, x = sin y
...
r
...
x, we get
1 = cos y
dy
1
=
dx
cos y
which implies that
dy
dx
✁
1
cos (sin 1 x )
Observe that this is defined only for cos y ✂ 0, i
...
, sin–1 x ✂
☎
✄ ✄
, , i
...
, x ✂ – 1, 1,
2 2
i
...
, x ✆ (– 1, 1)
...
Recall that for x ✆ (– 1, 1), sin (sin–1 x) = x and hence
cos2 y = 1 – (sin y)2 = 1 – (sin (sin–1 x))2 = 1 – x2
Also, since y ✆
✞ ✝ ✝✟
✡✠ , ☛,
☞ 2 2✌
cos y is positive and hence cos y =
1✍ x2
Thus, for x ✆ (– 1, 1),
dy
dx
✎
1
1
✎
cos y
1 ✏ x2
Example 27 Find the derivative of f given by f (x) = tan–1 x assuming it exists
...
Then, x = tan y
...
r
...
x, we get
1 = sec2 y
dy
dx
which implies that
dy
1
✒
dx sec2 y
✒
1
1
✒
2
1 ✓ tan y 1 ✓ (tan (tan ✑1 x)) 2
✒
1
1 ✓ x2
CONTINUITY AND DIFFERENTIABILITY
169
Finding of the derivatives of other inverse trigonometric functions is left as exercise
...
4):
Table 5
...
3
dy
in the following:
dx
1
...
xy + y2 = tan x + y
2
...
x2 + xy + y2 = 100
3
...
x3 + x2y + xy2 + y3 = 81
7
...
sin2 x + cos2 y = 1
9
...
y = tan–1 ✓
✚ 1 ✜ x2 ✛
✥, 0 ✣ x ✣1
✧ 1✦ x2 ★
11
...
y ✢ sin ✙1 ✤
13
...
y ✷ sin ✶1 ✴ 2 x 1 ✗ x 2 ✵ , ✗
15
...
4 Exponential and Logarithmic Functions
Till now we have learnt some aspects of different classes of functions like polynomial
functions, rational functions and trigonometric functions
...
It needs to be
emphasized that many statements made
in this section are motivational and precise
proofs of these are well beyond the scope
of this text
...
9 gives a sketch of
y = f1(x) = x, y = f2(x) = x2, y = f3(x) = x3
and y = f4(x) = x4
...
Steeper the curve, faster is the rate of
growth
...
9
increment in the value of x (> 1), the
increment in the value of y = fn (x) increases as n increases for n = 1, 2, 3, 4
...
Essentially, this means that the graph of y = fn (x) leans more towards the y-axis as n
increases
...
If x increases from 1 to
2, f10 increases from 1 to 210 whereas f15 increases from 1 to 215
...
Upshot of the above discussion is that the growth of polynomial functions is dependent
on the degree of the polynomial function – higher the degree, greater is the growth
...
The answer is in affirmative and an example of such a function is
y = f (x) = 10x
...
For example, we can prove that 10x grows faster than f100 (x) = x100
...
Clearly f (x) is much greater than f100 (x)
...
But we will not attempt to give a proof of this here
...
CONTINUITY AND DIFFERENTIABILITY
171
Definition 3 The exponential function with positive base b > 1 is the function
y = f (x) = bx
The graph of y = 10 is given in the Fig 5
...
It is advised that the reader plots this graph for particular values of b like 2, 3 and 4
...
(2) Range of the exponential function is the set of all positive real numbers
...
(4) Exponential function is ever increasing; i
...
, as we move from left to right, the
graph rises above
...
In
other words, in the second quadrant, the graph approaches x-axis (but never
meets it)
...
In
the Appendix A
...
4 of Class XI, it was observed that the sum of the series
x
1 1
...
Using this e as the base we obtain an
extremely important exponential function y = ex
...
It would be interesting to know if the inverse of the exponential function exists and
has nice interpretation
...
1
Definition 4 Let b > 1 be a real number
...
Logarithm of a to base b is denoted by logb a
...
Let us
work with a few explicit examples to get a feel for this
...
In terms of
logarithms, we may rewrite this as log2 8 = 3
...
Also, 625 = 54 = 252 is equivalent to saying log5 625 = 4 or
log25 625 = 2
...
This function, called the
logarithmic function, is defined by
logb : R+ ✞ R
x ✞ logb x = y if by = x
172
MATHEMATICS
As before if the base b = 10, we say it
is common logarithms and if b = e, then
we say it is natural logarithms
...
In this
chapter, log x denotes the logarithm
function to base e, i
...
, ln x will be written
as simply log x
...
10 gives the plots
of logarithm function to base 2, e and 10
...
10
(1) We cannot make a meaningful definition of logarithm of non-positive numbers
and hence the domain of log function is R+
...
(3) The point (1, 0) is always on the graph of the log function
...
e
...
(5) For x very near to zero, the value
of log x can be made lesser than
any given real number
...
(6) Fig 5
...
It is of interest to observe
that the two curves are the mirror
images of each other reflected in the line y = x
...
11
Two properties of ‘log’ functions are proved below:
(1) There is a standard change of base rule to obtain loga p in terms of logb p
...
This means a = p, b✁ = p and b✂ = a
...
But then
log b p
logb a
(2) Another interesting property of the log function is its effect on products
...
Then b✄ = pq
...
But then b✄ = pq = b☎b✌ = b☎ + ✌
which implies ☛ = ☞ + , i
...
,
logb pq = logb p + logb q
A particularly interesting and important consequence of this is when p = q
...
In fact this is true for any real number n, but we will
not attempt to prove this
...
So the above equation is not true for non-positive real numbers
...
If
y > 0, we may take logarithm which gives us log y = log (elog x) = log x
...
Thus
y = x
...
One of the striking properties of the natural exponential function in differential
calculus is that it doesn’t change during the process of differentiation
...
Theorem 5
(1) The derivative of ex w
...
t
...
e
...
r
...
, x is
d x
(e ) = ex
...
e
...
dx
x
x
174
MATHEMATICS
Example 29 Differentiate the following w
...
t
...
Using chain rule, we have
dy
x d
= e ✁
(– x) = – e– x
dx
dx
(ii) Let y = sin (log x)
...
Using chain rule, we have
✄1
dy
=
dx
1 ✄ (e x )2
☎
x
d x
✄e
(e ) ✆
dx
1 ✄ e2 x
(iv) Let y = ecos x
...
4
Differentiate the following w
...
t
...
esin
3
...
sin (tan–1 e–x)
5
...
e
7
...
log (log x), x > 1
cos x
9
...
e
x
, x☛0
✡e
x2
✡
...
cos (log x + ex), x > 0
5
...
Logarithmic Differentiation
In this section, we will learn to differentiate certain special class of functions given in
the form
y = f (x) = [u(x)]v (x)
By taking logarithm (to base e) the above may be rewritten as
log y = v(x) log [u(x)]
CONTINUITY AND DIFFERENTIABILITY
175
Using chain rule we may differentiate this to get
1 dy
y dx
✁
dy
dx
✝
v( x)
1
...
log [u(x)]
u( x)
which implies that
v( x)
✆
✞ u✟( x ) ✡ v✟ ( x ) ✞ log ✂ u ( x ) ✄
☞
✌ u ( x)
✍
☎
y☛
The main point to be noted in this method is that f (x) and u(x) must always be
positive as otherwise their logarithms are not defined
...
r
...
x
...
r
...
x, we get
log y =
1☎ 1
2x
6x ✡ 4 ✆
1 dy
✡
✕
✔
= ☛
2
2
☞
2
(
3)
x
✕
4
3
x
x
✡
✡ 4x ✡ 5 ✍
✌
y dx
or
2x
6x ✖ 4 ✘
y✗ 1
dy
✙
✖
= ✚
2
2
✛
✙
2
(
3)
x
4
3
x
x
✖
✖ 4x ✖ 5✢
dx
✜
=
1
2
( x ✣ 3) ( x 2 ✤ 4) ✥ 1
2x
6x ✤ 4 ✦
✤
✣
2
2
2
✧
3 x ✤ 4 x ✤ 5 ✩ ( x ✣ 3) x ✤ 4 3 x ✤ 4 x ✤ 5 ★✪
Example 31 Differentiate ax w
...
t
...
Solution Let y = ax
...
r
...
x, we have
1 dy
y dx = log a
176
MATHEMATICS
or
dy
= y log a
dx
Thus
d x
(a ) = ax log a
dx
Alternatively
d x log a
d x
(a ) =
(e
)
dx
dx
e x log a
d
( x log a)
dx
= ex log a
...
Example 32 Differentiate xsin x, x > 0 w
...
t
...
Solution Let y = xsin x
...
sin x (log x ) ✁ log x (sin x )
y dx =
dx
dx
or
1 dy
1
(sin x) ✂ log x cos x
y dx =
x
or
dy
✄ sin x cos x log x ☎
✆
= y✝
✞✠
✟ x
dx
✄ sin x
☎
✆ cos x log x ✞
✟ x
✠
= xsin x✡1 ☛ sin x ☞ x sin x ☛ cos x log x
sin x
= x ✝
Example 33 Find
dy
, if yx + xy + xx = ab
...
Putting u = yx, v = xy and w = xx, we get u + v + w = ab
Therefore
du dv dw
✂ ✂
dx dx dx
0
Now, u = yx
...
r
...
x, we have
...
(2)
Also v = xy
Taking logarithm on both sides, we have
log v = y log x
Differentiating both sides w
...
t
...
(3)
Again
w = xx
Taking logarithm on both sides, we have
log w = x log x
...
r
...
x, we have
1 dw
d
d
✒
= x (log x) ✓ log x ✒ ( x)
w dx
dx
dx
= x✒
i
...
1
✓ log x ✒ 1
x
dw
= w (1 + log x)
dx
= xx (1 + log x)
...
xy–1 – yx log y
dx
(x
...
log x)
or
✂ log x
dy
=
dx
Therefore
✟[y
x
log y ✠ y
...
y x ✞1 ✠ x y log x
EXERCISE 5
...
r
...
x
...
cos x
...
cos 3x
2
...
(log x)cos x
4
...
(x + 3)
...
(x + 5)
6
...
(log x)x + xlog x
8
...
xsin x + (sin x)cos x
10
...
(x cos x)x + ( x sin x) x
Find
12
...
16
...
dy
of the functions given in Exercises 12 to 15
...
yx = xy
xy + yx = 1
(cos x)y = (cos y)x
15
...
Differentiate (x2 – 5x + 8) (x3 + 7x + 9) in three ways mentioned below:
(i) by using product rule
(ii) by expanding the product to obtain a single polynomial
...
Do they all give the same answer?
CONTINUITY AND DIFFERENTIABILITY
179
18
...
v
...
w + u
...
w+u
...
5
...
In such a situation, we say that the relation between
them is expressed via a third variable
...
More
precisely, a relation expressed between two variables x and y in the form
x = f (t), y = g (t) is said to be parametric form with t as a parameter
...
dy
dy dx
=
dx dt
dt
dy
dy
= dt
dx
dx
dt
or
g ✟(t )
dy
=
f ✟(t )
dx
Thus
Example 34 Find
✁
☎
✝
whenever
✠
☞
✍
as
dy
dt
☛
dx
dt
✄
✂
0✆
✞
g ✟(t ) and
dy
, if x = a cos ✒, y = a sin ✒
...
dx
Solution Given that x = at2, y = 2at
Example 35 Find
So
dx
= 2at and
dt
Therefore
dy
dy
dt
=
dx
dx
dt
dy
= 2a
dt
2a
2at
1
t
Example 36 Find
dy
, if x = a (✍ + sin ✍ ), y = a (1 – cos ✍)
...
2
2
2
Example 37 Find dy , if x 3 ✆ y 3 ✝ a 3
...
Then
2
2
2
2
3
3
x 3 ✞ y 3 = (a cos ✟) 3 ✠ ( a sin ✟) 3
2
2
2
2
= a 3 (cos ✟ ✠ (sin ✟) ✡ a 3
2
2
2
Hence, x = a cos3 ✍, y = a sin3 ✍ is parametric equation of x 3 ✞ y 3 ☛ a 3
Now
dx
dy
= – 3a cos2 ✍ sin ✍ and
= 3a sin2 ✍ cos ✍
d✁
d✁
CONTINUITY AND DIFFERENTIABILITY
dy
2
dy
d ✁ 3a sin cos ✁ ✂ tan ✁ ✂
=
dx ✂ 3a cos 2 sin
dx
d
Therefore
3
181
y
x
✄Note Had we proceeded in implicit way, it would have been quite tedious
...
6
If x and y are connected parametrically by the equations given in Exercises 1 to 10,
without eliminating the parameter, Find
1
...
x = sin t, y = cos 2t
dy
...
x = a cos ✍, y = b cos ✍
4
...
x = cos ✍ – cos 2✍, y = sin ✍ – sin 2✍
6
...
x =
sin 3 t
cos 2t
, y☎
cos3 t
cos 2t
t✝
✆
x ✞ a ✠ cos t ✟ log tan ✡ y = a sin t 9
...
x = a (cos ✍ + ✍ sin ✍), y = a (sin ✍ – ✍ cos ✍ )
8
...
If x ✎ a sin t , y ✎ a cos t , show that
✎✏
dx
x
5
...
Then
dy
= f ✑ (x)
...
r
...
x
...
r
...
x and
dx ✖ dx ✗
is denoted by
d2y
...
It is also
dx 2
182
MATHEMATICS
denoted by D2 y or y✎ or y2 if y = f (x)
...
d2y
, if y = x3 + tan x
...
Then
Example 38 Find
dy
= 3x2 + sec2 x
dx
Therefore
d
d2y
3x 2 ✂ sec 2 x ✁
=
dx
dx 2
= 6x + 2 sec x
...
dx 2
Solution We have
dy
= A cos x – B sin x
dx
and
Hence
d
d2y
(A cos x – B sin x)
2 =
dx
dx
= – A sin x – B cos x = – y
d2y
+y=0
dx 2
Example 40 If y = 3e2x + 2e3x, prove that
d2y
dy
✆5
✝ 6y ✞ 0
...
Then
dy
= 6e2x + 6e3x = 6 (e2x + e3x)
dx
Therefore
Hence
d2y
= 12e2x + 18e3x = 6 (2e2x + 3e3x)
dx 2
d2y
dy
✆5
+ 6y = 6 (2e2x + 3e3x)
2
dx
dx
– 30 (e2x + e3x) + 6 (3e2x + 2e3x) = 0
CONTINUITY AND DIFFERENTIABILITY
Example 41 If y = sin–1 x, show that (1 – x2)
d2y
dx 2
x
dy
✁0
...
Then
dy
=
dx
1
(1 ✂ x 2 )
(1 ✄ x 2 )
or
dy
☎1
dx
d ✆
dy ✝
2
✠ (1 ✞ x )
...
e
...
2 y1 y2 ✚ y12 (0 ✑ 2 x) ✓ 0
(1 – x2) y2 – xy1 = 0
EXERCISE 5
...
1
...
log x
2
...
x log x
3
...
cos x
6
...
e6x cos 3x
8
...
log (log x)
10
...
If y = 5 cos x – 3 sin x, prove that
dx 2
183
184
MATHEMATICS
12
...
dx 2
13
...
If y = Aemx + Benx, show that
d2y
dx2
(m ✁ n )
dy
✁ mny ✂ 0
dx
d2y
✄ 49 y
15
...
If e (x + 1) = 1, show that
dx 2 ✠ dx ✡
2
y
17
...
8 Mean Value Theorem
In this section, we will state two fundamental results in Calculus without proof
...
Theorem 6 (Rolle’s Theorem) Let f : [a, b] ☛ R be continuous on [a, b] and
differentiable on (a, b), such that f(a) = f(b), where a and b are some real numbers
...
In Fig 5
...
13, graphs of a few typical differentiable functions satisfying the
hypothesis of Rolle’s theorem are given
...
12
Fig 5
...
In each of the graphs, the slope becomes zero at least at one point
...
CONTINUITY AND DIFFERENTIABILITY
185
Theorem 7 (Mean Value Theorem) Let f : [a, b] ✞ R be a continuous function on
[a, b] and differentiable on (a, b)
...
Let us now understand a geometric interpretation of the MVT
...
14
...
The MVT states that
there is a point c in (a, b) such that the slope of the tangent at (c, f(c)) is same as the
slope of the secant between (a, f (a)) and (b, f (b))
...
tangent to the curve y = f (x) at (c, f (c))
...
14 it is clear that
Fig 5
...
Solution The function y = x2 + 2 is continuous in [– 2, 2] and differentiable in (– 2, 2)
...
Rolle’s
theorem states that there is a point c ✆ (– 2, 2), where f✄ (c) = 0
...
Thus at c = 0, we have f✄ (c) = 0 and c = 0 ✆ (– 2, 2)
...
Solution The function f (x) = x2 is continuous in [2, 4] and differentiable in (2, 4) as its
derivative f ✄ (x) = 2x is defined in (2, 4)
...
Hence
f (b) f (a ) 16 4
✁
✁6
b a
4 2
MVT states that there is a point c ✆ (2, 4) such that f ✠ (c) = 6
...
Thus at c = 3 ✆ (2, 4), we have f ✠ (c) = 6
...
8
1
...
2
...
Can
you say some thing about the converse of Rolle’s theorem from these example?
(ii) f (x) = [x] for x ✆ [– 2, 2]
(i) f (x) = [x] for x ✆ [5, 9]
2
(iii) f (x) = x – 1 for x ✆ [1, 2]
3
...
4
...
5
...
Find all c ✆ (1, 3) for which f ✂ (c) = 0
...
Examine the applicability of Mean Value Theorem for all three functions given in
the above exercise 2
...
r
...
x, the following function:
(i)
3x ☎ 2 ☎
1
2
(ii) esec x ✝ 3cos –1 x
2x ☎ 4
2
(iii) log7 (log x)
Solution
(i) Let y =
3x ☎ 2 ☎
1
2 x2 ☎ 4
1
✟ 12
2
= (3x ✡ 2) 2 ✡ (2 x ✡ 4)
Note that this function is defined at all real numbers x ☛
2
...
3
sec
(ii) Let y ✑ e
2
x
✒ 3cos
✏1
x
This is defined at every real number in [ ✕1,1] ✕ ✓0✔
...
log (log x)
(by change of base formula)
...
Therefore
(iii) Let y = log7 (log x) =
1 d
dy
(log (log x))
=
log 7 dx
dx
=
d
1
1
(log x )
✬
log 7 log x dx
=
1
x log 7 log x
188
MATHEMATICS
Example 45 Differentiate the following w
...
t
...
(i) cos – 1 (sin x)
(ii) tan
1✁
2 x ✟1
x
✎ 1✍ 4
✡
sin x ✂
☎
cos x ✞
(iii) sin ✠1 ☞
✄
✝1✆
☛
✌
✏
Solution
(i) Let f (x) = cos – 1 (sin x)
...
We may rewrite this function as
f (x) = cos –1 (sin x)
✒
✕✔
✣
2
✑1
= cos ✚ cos ✘✜
✖✓
✗ x✙ ✛
✢✤
✥
✦x
2
Thus
f ✧ (x) = – 1
...
Observe that this function is defined for all real
✝ 1 ✆ cos x ✞
numbers, where cos x ★ – 1; i
...
, at all odd multiplies of ✩
...
Thus f ✧(x) =
x ✮✬
✰
2 ✴✲
✲
✲
✲
✶
x ✖✓ x
✙ ✷
2 ✢ ✛✤ 2
x✹
✻ in both numerator and denominator as it
2✽
1
...
To find the domain of this function we need to find all
x ✙
✜1✿ 4 ✢
✕
(iii) Let f (x) = sin – 1 ✘
x such that ❁1 ❂
2 x ❀1
1 ❃ 4x
❂ 1
...
e
...
We
1
+ 2x which is true for all x
...
By putting 2x = tan ✍, this function may be
rewritten as
may rewrite this as 2
✏
2x ✄ 1 ✝
x✟
✡1 ✠ 4 ☛
✆
☎1
f (x) = sin ✞
2x ✓ 2
✑
✎1
= sin ✔
✒
✕
x 2
✔1 ✖ ☞ 2 ✌ ✘
✕
✗
2 tan ✚ ✜
2 ✣
✥ 1 ✤ tan ✚ ✦
✛
✙1
= sin ✢
= sin –1 [sin 2✍]
= 2✍ = 2 tan – 1 (2x)
Thus
f ✧ (x) = 2 ✪
=
1
1✫ ★2
x ✩2
✪
d
(2 x )
dx
2
x
✬ (2 )log 2
1 ✭ 4x
2 x ✮ 1 log 2
=
1 ✤ 4x
Example 46 Find f ✧(x) if f (x) = (sin x)sin x for all 0 < x < ✯
...
Taking
logarithms, we have
log y = log (sin x)sin x = sin x log (sin x)
Then
1 dy
d
=
(sin x log (sin x))
y dx
dx
= cos x log (sin x) + sin x
...
Clearly
y✄a
t
t
d t ☞1
dy
=
✡a t ☛ = a
dt
dt
= a
Similarly
✁
and x ✄ ✆ t ☎
1
t
d ✁ 1✂
✆ t ☎ ✝ ✌ log a
dt ✞ t ✟
1
t✍ ✎
t 1✑
dx
1✘
✗
= a ✤t ✛ ✥
dt
t✧
✦
✫
= a ✴t ✯
✶
dx
dt
✠
1✏
✓ log a
t2 ✕
✒
✔
a ✖1
1✬
t ✵✷
✜
a ✪1
d ✙ 1✚
✢t ✛ ✣
dt ★ t ✩
1✮
✭
✰ ✲1 ✱
✳
t2 ✹
✸
0 only if t ✠ ± 1
...
r
...
e cos x
...
We want to find du
dv
du
dv
= 2 sin x cos x and
= e cos x (– sin x) = – (sin x) e cos x
dx
dx
P
du / dx
...
r
...
x the function in Exercises 1 to 11
...
sin3 x + cos6 x
1
...
sin–1(x x ), 0 ✏ x ✏ 1
3
...
x
2 ,–2
6
...
(log x)log x, x > 1
8
...
3✌
✌
9
...
xx + xa + ax + aa, for some fixed a > 0 and x > 0
11
...
Find
, if y = sin–1 x + sin–1 1 ✖ x 2 , – 1 ✏ x ✏ 1
dx
12
...
If x 1 ✗ y ✗ y 1 ✗ x ✘ 0 , for , – 1 < x < 1, prove that
dy
1
✛✜
dx
✙1 ✢ x ✚ 2
15
...
191
192
MATHEMATICS
16
...
✁
dx
sin a
17
...
dx 2
18
...
19
...
20
...
21
...
f ( x ) g ( x ) h( x )
22
...
If y = ea cos✠ x , – 1 ✏ x ✏ 1, show that ✡1 ☞ x 2 ☛ d y ☞ x dy ☞ a 2 y ✌ 0
...
A function
is continuous if it is continuous on the whole of its domain
...
i
...
, if f and g are continuous functions, then
(f ± g) (x) = f (x) ± g (x) is continuous
...
g) (x) = f (x)
...
f ( x)
✑f✒
✔ g ✕ ( x) ✓ g ( x) (wherever g (x) ✂ 0) is continuous
...
CONTINUITY AND DIFFERENTIABILITY
193
Chain rule is rule to differentiate composites of functions
...
Here both f (x) and u (x) need to be positive for
this technique to make sense
...
Mean Value Theorem: If f : [a, b]
R is continuous on [a, b] and
differentiable on (a, b)
...
” — WHITEHEAD
6
...
In this chapter, we will study applications of the derivative in various disciplines, e
...
, in
engineering, science, social science, and many other fields
...
We will also use derivative to find intervals
on which a function is increasing or decreasing
...
6
...
In a similar fashion, whenever one quantity y varies with another
Recall that by the derivative
quantity x, satisfying some rule y
change of y with respect to x and
✞
✁ f (x) , then dydx (or f ✂(x)) represents the rate of
dy ☎
dx ✆✝ ✄ (or f ✂(x )) represents the rate of change
x x0
0
of y with respect to x at x x0
...
e
...
Let us consider some examples
...
Solution The area A of a circle with radius r is given by A = ✄r2
...
dr
dA
✂ 10☎
...
Example 2 The volume of a cube is increasing at a rate of 9 cubic centimetres per
second
...
Then, V = x3 and S = 6x2, where x is a function of time t
...
(1)
✞ 3 ✟ 36
☞✡
✌ x2 ✍ x
= 12x ✠ ☛
Hence, when
x = 10 cm,
dS
dt
3
...
At the instant, when the radius of the circular wave is 10 cm, how
fast is the enclosed area increasing?
Solution The area A of a circle with radius r is given by A = ✄r2
...
Therefore, when r = 10 cm,
dy
☎Note dx is positive if y increases as x increases and is negative if y decreases
as x increases
...
When x =10cm and y = 6cm, find
the rates of change of (a) the perimeter and (b) the area of the rectangle
...
y
Therefore
dA
dx
dy
✂ y ✎ x✂
=
dt
dt
dt
= – 3(6) + 10(2) (as x = 10 cm and y = 6 cm)
= 2 cm2/min
APPLICATION OF DERIVATIVES
197
Example 5 The total cost C(x) in Rupees, associated with the production of x units of
an item is given by
C (x) = 0
...
02 x2 + 30x + 5000
Find the marginal cost when 3 units are produced, where by marginal cost we
mean the instantaneous rate of change of total cost at any level of output
...
005(3x 2 ) ✁ 0
...
015(3 ) ✄ 0
...
135 – 0
...
015
Hence, the required marginal cost is Rs 30
...
Example 6 The total revenue in Rupees received from the sale of x units of a product
is given by R(x) = 3x2 + 36x + 5
...
Solution Since marginal revenue is the rate of change of total revenue with respect to
the number of units sold, we have
dR
6 x ✂ 36
dx
When
x = 5, MR = 6(5) + 36 = 66
Hence, the required marginal revenue is Rs 66
...
1
1
...
The volume of a cube is increasing at the rate of 8 cm3/s
...
The radius of a circle is increasing uniformly at the rate of 3 cm/s
...
4
...
How fast is the
volume of the cube increasing when the edge is 10 cm long?
5
...
At the instant when the radius of the circular wave is 8 cm, how fast is
the enclosed area increasing?
198
MATHEMATICS
6
...
7 cm/s
...
The length x of a rectangle is decreasing at the rate of 5 cm/minute and the
width y is increasing at the rate of 4 cm/minute
...
8
...
Find the rate at which the radius of
the balloon increases when the radius is 15 cm
...
A balloon, which always remains spherical has a variable radius
...
10
...
The bottom of the ladder is pulled
along the ground, away from the wall, at the rate of 2cm/s
...
A particle moves along the curve 6y = x3 +2
...
1
cm/s
...
The radius of an air bubble is increasing at the rate of
13
...
2
Find the rate of change of its volume with respect to x
...
Sand is pouring from a pipe at the rate of 12 cm3/s
...
How fast is the height of the sand cone increasing when the
height is 4 cm?
15
...
007x3 – 0
...
Find the marginal cost when 17 units are produced
...
The total revenue in Rupees received from the sale of x units of a product is
given by
R (x) = 13x2 + 26x + 15
...
Choose the correct answer in the Exercises 17 and 18
...
The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
(A) 10✄
(C) 8 ✄
(D) 11✄
(B) 12 ✄
APPLICATION OF DERIVATIVES
199
18
...
The marginal revenue, when x = 15 is
(A) 116
(B) 96
(C) 90
(D) 126
6
...
Consider the function f given by f (x) = x2, x ☎ R
...
1
...
1
First consider the graph (Fig 6
...
Observe that as we
move from left to right along the graph, the height of the graph continuously increases
...
Now consider the graph to the left of the origin and observe here that as we move
from left to right along the graph, the height of the graph continuously decreases
...
We shall now give the following analytical definitions for a function which is
increasing or decreasing on an interval
...
Then f is said to be
(i) increasing on I if x1 < x2 in I ✆ f (x1) ✝ f (x2) for all x1, x2 ☎ I
...
200
MATHEMATICS
(iii) decreasing on I if x1 < x2 in I ✆ f (x1) ✞ f (x2) for all x1, x2 ☎ I
...
For graphical representation of such functions see Fig 6
...
Fig 6
...
Definition 2 Let x0 be a point in the domain of definition of a real valued function f
...
Let us clarify this definition for the case of increasing function
...
Example 7 Show that the function given by f (x) = 7x – 3 is strictly increasing on R
...
Then
x1 < x2 ✆ 7x1 < 7x2
✆ 7x1 – 3 < 7x2 – 3 ✆ f (x1) < f (x2)
APPLICATION OF DERIVATIVES
201
Thus, by Definition 1, it follows that f is strictly increasing on R
...
The proof of this test requires the Mean Value Theorem studied in Chapter 5
...
Then
(a) f is increasing in [a,b] if f ✂(x) > 0 for each x ☎ (a, b)
(b) f is decreasing in [a,b] if f ✂(x) < 0 for each x ☎ (a, b)
(c) f is a constant function in [a,b] if f ✂(x) = 0 for each x ☎ (a, b)
Proof (a) Let x1, x2 ☎ [a, b] be such that x1 < x2
...
e
...
e
...
The proofs of part (b) and (c) are similar
...
Remarks
(i) f is strictly increasing in (a, b) if f ✂(x) > 0 for each x ☎ (a, b)
(ii) f is strictly decreasing in (a, b) if f ✂(x) < 0 for each x ☎ (a, b)
(iii) A function will be increasing (decreasing) in R if it is so in every interval of R
...
Solution Note that
f ✂(x) = 3x2 – 6x + 4
= 3(x2 – 2x + 1) + 1
= 3(x – 1)2 + 1 > 0, in every interval of R
Therefore, the function f is strictly increasing on R
...
✄
✂
✂
Solution Note that f (x) = – sin x
(a) Since for each x (0, ), sin x > 0, we have f (x) < 0 and so f is strictly
decreasing in (0, )
...
(c) Clearly by (a) and (b) above, f is neither increasing nor decreasing in (0, 2 )
...
However, since the function is continuous at
the end points 0 and , by Theorem 1, f is increasing in [ , 2 ] and decreasing in [0, ]
...
Now the point x = 2 divides the real line into two
disjoint intervals namely, (– , 2) and (2, ) (Fig 6
...
In the interval (– , 2),
f (x) = 2x – 4 < 0
...
Also, in the interval (2, ) , f ( x ) 0
and so the function f is strictly increasing in this
interval
...
3
✟
Note Note that the given function is continuous at 2 which is the point joining
the two intervals
...
Example 11 Find the intervals in which the function f given by f (x) = 4x3 – 6x2 – 72x + 30
is (a) strictly increasing (b) strictly decreasing
...
The
points x = – 2 and x = 3 divides the real line into
three disjoint intervals, namely, (– ✟, – 2), (– 2, 3)
Fig 6
...
In the intervals (– ✟, – 2) and (3, ✟), f ✂(x) is positive while in the interval (– 2, 3),
f ✂(x) is negative
...
However, f is neither increasing nor decreasing in R
...
Solution We have
or
f (x) = sin 3x
f ✂ (x) = 3cos 3x
Therefore, f ✂(x) = 0 gives cos 3x = 0 which in turn gives 3x ☛
✡ 3✡
✁
✄
(as x ☎ ✆ 0, ✝
2 2
✞ 2✠
,
☞
☞
☞
✁ 3 ✄
)
...
The point x ✌ divides the interval
✝
6
2
6
✞ 2✠
✁
✕
✘ ✍ ✍✏
into two disjoint intervals ✆ 0, ✖ and ✙ , ✒
...
5
✎ ✍✏
✑✓0, 2 ✒✔
✁
✕
✡
✡
Now, f ✜ ( x) ✛ 0 for all x ☎ ✆ 0, ✖ as 0 ✢ x ✣ ✤ 0 ✢ 3 x ✣ and f ✜ ( x) ✥ 0 for
6
✞
✗
6
2
✦
✦
✧
★
3✡
✡
✡ ✡
...
☞ 6 2✌
...
✕ 6✖
✕ 6 2✖
Example 13 Find the intervals in which the function f given by
f (x) = sin x + cos x, 0 ✗ x ✗ 2✘
is strictly increasing or strictly decreasing
...
4
4
4
✆ ✝
✖
✪ 4
Note that
✍ 5✍
Fig 6
...
2
1
...
2
...
3
...
Find the intervals in which the function f given by f (x) = 2x2 – 3x is
(a) strictly increasing
(b) strictly decreasing
5
...
Find the intervals in which the following functions are strictly increasing or
decreasing:
(b) 10 – 6x – 2x2
(a) x2 + 2x – 5
(c) –2x3 – 9x2 – 12x + 1
(d) 6 – 9x – x2
(e) (x + 1)3 (x – 3)3
7
...
8
...
9
...
206
MATHEMATICS
10
...
11
...
✁
✂
12
...
On which of the following intervals is the function f given by f (x) = x100 + sin x –1
strictly decreasing ?
✠ ✞✡
✠✞ ✡
(B) ☛ , ✞☞
(C) ☛ 0, ☞
(D) None of these
✌ 2✍
✌2 ✍
14
...
15
...
Prove that the function f given by
(A) (0,1)
f ( x) ✎ x ✏
1
is strictly increasing on I
...
Prove that the function f given by f (x) = log sin x is strictly increasing on ☛ 0, ☞
✌ 2✍
✁
✂
and strictly decreasing on ✄ , ☎
...
Prove that the function f given by f (x) = log cos x is strictly decreasing on
✠ 0, ✞ ✡
✠✞ ✡
☛ 2 ☞ and strictly increasing on ☛ 2 , ✞ ☞
...
Prove that the function given by f (x) = x3 – 3x2 + 3x – 100 is increasing in R
...
The interval in which y = x2 e–x is increasing is
(A) (– ✟, ✟)
(B) (– 2, 0)
(C) (2, ✟)
(D) (0, 2)
6
...
Recall that the equation of a straight line passing through a given point (x0, y0)
having finite slope m is given by
y – y0 = m (x – x0)
APPLICATION OF DERIVATIVES
207
Note that the slope of the tangent to the curve y = f (x)
at the point (x0, y0) is given by
dy
( ✂ f ✁( x0 ))
...
Therefore, the equation of the
Fig 6
...
e
...
Particular cases
(i) If slope of the tangent line is zero, then tan ✌ = 0 and so ✌ = 0 which means the
tangent line is parallel to the x-axis
...
✏
, then tan ✌ ✒ ✓, which means the tangent line is perpendicular to the
2
x-axis, i
...
, parallel to the y-axis
...
(ii) If
✎✑
Example 14 Find the slope of the tangent to the curve y = x3 – x at x = 2
...
dx ✖✗ x✔ 2
208
MATHEMATICS
Example 15 Find the point at which the tangent to the curve y
4 x ✁ 3 ✁ 1 has its
2
...
3
2
2
=
4x ✆ 3
3
4x – 3 = 9
x=3
So
or
or
4 x ✁ 3 ✁ 1
...
Now y
Therefore, the required point is (3, 2)
...
x✡3
Solution Slope of the tangent to the given curve at any point (x,y) is given by
dy
2
=
dx
( x ☛ 3) 2
But the slope is given to be 2
...
Thus, there are two tangents to the
given curve with slope 2 and passing through the points (2, 2) and (4, – 2)
...
Solution Differentiating
x2 y2
✂ ✄ 1 with respect to x, we get
4 25
x 2 y dy
☎
=0
2 25 dx
✆25 x
dy
= 4 y
dx
(i) Now, the tangent is parallel to the x-axis if the slope of the tangent is zero which
or
gives
✝25 x
✄ 0
...
Then
x2 y2
✂ ✄ 1 for x = 0 gives
4 25
4 y
y2 = 25, i
...
, y = ± 5
...
(ii) The tangent line is parallel to y-axis if the slope of the normal is 0 which gives
x2 y2
4y
✂ ✄ 1 for y = 0 gives x = ± 2
...
e
...
Therefore,
4 25
25 x
points at which the tangents are parallel to the y-axis are (2, 0) and (–2, 0)
...
Solution Note that on x-axis, y = 0
...
Thus, the curve cuts the x-axis at (7, 0)
...
Hence, the equation of the
20
tangent at (7, 0) is
y 0✁
1
( x 7)
20
20 y ✂ x ✄ 7 ☎ 0
or
2
2
Example 19 Find the equations of the tangent and normal to the curve x 3 ✆ y 3 ✝ 2
at (1, 1)
...
dx ✑✒ (1, 1)
So the equation of the tangent at (1, 1) is
y – 1 = – 1 (x – 1)
or
y+x–2=0
Also, the slope of the normal at (1, 1) is given by
✓1
slope of the tangent at (1,1)
=1
Therefore, the equation of the normal at (1, 1) is
y – 1 = 1 (x – 1)
or
y–x=0
Example 20 Find the equation of tangent to the curve given by
y = b cos3 t
x = a sin3 t ,
at a point where t ✁
✔
2
...
(1)
APPLICATION OF DERIVATIVES
dy
dx
or
211
dy
✁3b cos 2 t sin t ✁b cos t
dt
✂
=
dx
a sin t
3a sin 2 t cos t
dt
Therefore, slope of the tangent at t ☎
✄
2
is
✡
☛b cos
dy ✞
2 ☞0
dx ✟✠ t ✝ ✆ =
✡
a sin
2
2
Also, when t ☎
curve at t ☎
✄
2
✄
2
, x = a and y = 0
...
e
...
e
...
EXERCISE 6
...
Find the slope of the tangent to the curve y = 3x4 – 4x at x = 4
...
x✌2
3
...
4
...
2
...
Find the slope of the normal to the curve x ✎ a cos3 ✏, y ✎ a sin 3 ✏ at ✑ ☎
...
Find the slope of the normal to the curve x ✎ 1 ✒ a sin ✏, y ✎ b cos 2 ✏ at ✑ ☎
...
Find points at which the tangent to the curve y = x – 3x – 9x + 7 is parallel to
the x-axis
...
Find a point on the curve y = (x – 2)2 at which the tangent is parallel to the chord
joining the points (2, 0) and (4, 4)
...
Find the point on the curve y = x3 – 11x + 5 at which the tangent is y = x – 11
...
Find the equation of all lines having slope – 1 that are tangents to the curve
1
, x ☛ 1
...
Find the equation of all lines having slope 2 which are tangents to the curve
y
1
, x ☛ 3
...
Find the equations of all lines having slope 0 which are tangent to the curve
y
y✂
1
...
14
...
Find points on the curve
(v) x = cos t, y = sin t at t
✞
4
15
...
16
...
17
...
18
...
19
...
20
...
APPLICATION OF DERIVATIVES
213
21
...
22
...
23
...
24
...
25
...
Choose the correct answer in Exercises 26 and 27
...
The slope of the normal to the curve y = 2x2 + 3 sin x at x = 0 is
1
1
(C) –3
(D) ✞
3
3
2
27
...
5 Approximations
In this section, we will use differentials to approximate values of certain quantities
...
Let ✌ x denote a small
increment in x
...
We
define the following
(i) The differential of x, denoted by dx, is
defined by dx = ✌x
...
✑✎
✒ dx ✓
dy ✍ ✏
Fig 6
...
214
✌
✌
MATHEMATICS
✍✌✌ ✌
In case dx = x is relatively small when compared with x, dy is a good approximation
of y and we denote it by dy
y
...
8
...
8, we may note that the
differential of the dependent variable is not equal to the increment of the variable
where as the differential of independent variable is equal to the increment of the
variable
...
6
...
Let x = 36 and let x = 0
...
Then
y=
x
x
36
...
6 6
or
36
...
6) =
1
(0
...
05
2 36
36
...
05 = 6
...
1
✌
✏
✌ ✑✒ ✓ ✔ ✏ ✔
✌ ✌
✖
✡☞✝ ☛✎✞ ✟ ✕
✗ ✘ ✗ ✘✗
Example 22 Use differential to approximate (25) 3
...
Let x = 27 and let x = – 2
...
074
1
Thus, the approximate value of (25) 3 is given by
3 + (– 0
...
926
x3 )
3
(as y
✁
x)
APPLICATION OF DERIVATIVES
215
Example 23 Find the approximate value of f (3
...
Solution Let x = 3 and ✌x = 0
...
Then
f (3
...
Therefore
f (x + ✌x) = f (x) + ✌y
(as dx = ✌x)
✍ f (x) + f ✂(x) ✌x
2
or
f (3
...
02) (as x = 3, ✌x = 0
...
02)
= 45 + 0
...
46
Hence, approximate value of f (3
...
46
...
Solution Note that
V = x3
or
dV =
☎
✝
dV ✁
2
✆ ✄x = (3x ) ✌x
dx ✞
= (3x2) (0
...
06x3 m3
Thus, the approximate change in volume is 0
...
(as 2% of x is 0
...
03 cm,
then find the approximate error in calculating its volume
...
Then r = 9 cm and ✌r = 0
...
Now, the volume V of the sphere is given by
V=
or
4 3
✟r
3
dV
= 4✠r 2
dr
dV ✁
2
✆ ✄r ✡ (4☛r )✄r
dr ✞
= 4✠(9)2 (0
...
72✠ cm3
Thus, the approximate error in calculating the volume is 9
...
Therefore
dV =
☎
✝
216
MATHEMATICS
EXERCISE 6
...
Using differentials, find the approximate value of each of the following up to 3
places of decimal
...
3
(iii)
49
...
6
1
1
1
(iv) (0
...
999)10
(vi) (15) 4
1
1
1
(vii) (26) 3
(viii) (255) 4
(ix) (82) 4
1
1
1
(x) (401) 2
(xi) (0
...
57) 3
1
3
1
(xiii) (81
...
968) 2
(xv) (32
...
Find the approximate value of f (2
...
3
...
001), where f (x) = x3 – 7x2 + 15
...
Find the approximate change in the volume V of a cube of side x metres caused
by increasing the side by 1%
...
Find the approximate change in the surface area of a cube of side x metres
caused by decreasing the side by 1%
...
If the radius of a sphere is measured as 7 m with an error of 0
...
7
...
03 m, then find the
approximate error in calculating its surface area
...
If f(x) = 3x2 + 15x + 5, then the approximate value of f (3
...
66
(B) 57
...
66
(D) 77
...
The approximate change in the volume of a cube of side x metres caused by
increasing the side by 3% is
(A) 0
...
6 x3 m3 (C) 0
...
9 x3 m3
6
...
In fact, we will find the ‘turning points’ of the
graph of a function and thus find points at which the graph reaches its highest (or
APPLICATION OF DERIVATIVES
217
lowest) locally
...
Further, we will also find the absolute maximum and absolute minimum
of a function that are necessary for the solution of many applied problems
...
(i) The profit from a grove of orange trees is given by P(x) = ax + bx2, where a,b
are constants and x is the number of orange trees per acre
...
What is the maximum height the ball will
reach?
(iii) An Apache helicopter of enemy is flying along the path given by the curve
f (x) = x2 + 7
...
What is the nearest distance?
In each of the above problem, there is something common, i
...
, we wish to find out
the maximum or minimum values of the given functions
...
Definition 3 Let f be a function defined on an interval I
...
given by h( x )
60 ✁ x ✂
The number f (c) is called the maximum value of f in I and the point c is called a
point of maximum value of f in I
...
The number f (c), in this case, is called the minimum value of f in I and the point
c, in this case, is called a point of minimum value of f in I
...
The number f (c), in this case, is called an extreme value of f in I and the point c
is called an extreme point
...
9(a), (b) and (c), we have exhibited that graphs of certain particular
functions help us to find maximum value and minimum value at a point
...
218
MATHEMATICS
Fig 6
...
✞
☎
Solution From the graph of the given function (Fig 6
...
Also
f (x) 0, for all x R
...
Further, it may be observed
from the graph of the function that f has no maximum
value and hence no point of maximum value of f in R
...
10
Note If we restrict the domain of f to [– 2, 1] only,
then f will have maximum value(– 2)2 = 4 at x = – 2
...
✞
☎
Solution From the graph of the given function
(Fig 6
...
Therefore, the function f has a minimum value 0
and the point of minimum value of f is x = 0
...
Fig 6
...
APPLICATION OF DERIVATIVES
219
(ii) One may note that the function f in Example 27 is not differentiable at
x = 0
...
Solution The given function is an increasing (strictly) function in the given interval
(0, 1)
...
12) of the function f , it
seems that, it should have the minimum value at a
point closest to 0 on its right and the maximum value
at a point closest to 1 on its left
...
It is not possible to locate
such points
...
Also, if x1 is
x1 ✂ 1
✄ x1 for all x1 ✆ (0,1)
...
12
2
Therefore, the given function has neither the maximum value nor the minimum
value in the interval (0,1)
...
e
...
Infact, we have the following
results (The proof of these results are beyond the scope of the present text)
Every monotonic function assumes its maximum/minimum value at the end
points of the domain of definition of the function
...
✝Note By a monotonic function f in an interval I, we mean that f is either
increasing in I or decreasing in I
...
Let us now examine the graph of a function as shown in Fig 6
...
Observe that at
points A, B, C and D on the graph, the function changes its nature from decreasing to
increasing or vice-versa
...
Further, observe that at turning points, the graph has either a little hill or a little
valley
...
13
valleys
...
For this reason, the points A and C
may be regarded as points of local minimum value (or relative minimum value) and
points B and D may be regarded as points of local maximum value (or relative maximum
value) for the function
...
We now formally give the following definition
Definition 4 Let f be a real valued function and let c be an interior point in the domain
of f
...
(b) c is called a point of local minima if there is an h > 0 such that
f (c) ✝ f (x), for all x in (c – h, c + h)
The value f (c) is called the local minimum value of f
...
14(a)
...
e
...
e
...
This suggests that f ✂(c) must be zero
...
14
APPLICATION OF DERIVATIVES
✂
✂
221
✂
Similarly, if c is a point of local minima of f , then the graph of f around c will be as
shown in Fig 6
...
Here f is decreasing (i
...
, f (x) < 0) in the interval (c – h, c) and
increasing (i
...
, f (x) > 0) in the interval (c, c + h)
...
The above discussion lead us to the following theorem (without proof)
...
Suppose c I be any
point
...
Remark The converse of above theorem need
not be true, that is, a point at which the derivative
vanishes need not be a point of local maxima or
local minima
...
But 0 is neither a point of
local maxima nor a point of local minima (Fig 6
...
✂
✂
✂
Note A point c in the domain of a function
f at which either f (c) = 0 or f is not differentiable
is called a critical point of f
...
✂
Fig 6
...
✂✂
✂✂
✂
Theorem 3 (First Derivative Test) Let f be a function defined on an open interval I
...
Then
(i) If f (x) changes sign from positive to negative as x increases through c, i
...
, if
f (x) > 0 at every point sufficiently close to and to the left of c, and f (x) < 0 at
every point sufficiently close to and to the right of c, then c is a point of local
maxima
...
e
...
(iii) If f (x) does not change sign as x increases through c, then c is neither a point of
local maxima nor a point of local minima
...
15)
...
Similarly, if c is a point of local minima of f , then f(c) is a local minimum value of f
...
15 and 6
...
Fig 6
...
Solution We have
✂✂
f (x) = x3 – 3x + 3
or
f (x) = 3x2 – 3 = 3 (x – 1) (x + 1)
or
f (x) = 0 at x = 1 and x = – 1
Thus, x = ± 1 are the only critical points which could possibly be the points of local
maxima and/or local minima of f
...
Note that for values close to 1 and to the right of 1, f (x) > 0 and for values close
to 1 and to the left of 1, f (x) < 0
...
In the case of x = –1, note that
f (x) > 0, for values close to and to the left of –1 and f (x) < 0, for values close to and
to the right of – 1
...
✂
✂
✂
to the right (say 1
...
)
Close to –1
✁
Sign of f (x) = 3(x – 1) (x + 1)
Values of x
Close to 1
✂
✄✄
to the left (say 0
...
)
to the right (say 0
...
)
to the left (say 1
...
)
>0
☎✆
<0
0
0
APPLICATION OF DERIVATIVES
223
Example 30 Find all the points of local maxima and local minima of the function f
given by
f (x) = 2x3 – 6x2 + 6x +5
...
We shall now examine this point for local
maxima and/or local minima of f
...
Therefore, by first
derivative test, the point x = 1 is neither a point of local maxima nor a point of local
minima
...
Remark One may note that since f ✂(x), in Example 30, never changes its sign on R,
graph of f has no turning points and hence no point of local maxima or local minima
...
This test is often easier to apply than the first derivative test
...
Let f be twice differentiable at c
...
(ii) x = c is a point of local minima if f ✁ (c)
0 and f ✎(c) > 0
In this case, f (c) is local minimum value of f
...
In this case, we go back to the first derivative test and find whether c is a point of
local maxima, local minima or a point of inflexion
...
Example 31 Find local minimum value of the function f
given by f (x) = 3 + | x |, x ☎ R
...
So, second derivative test fails
...
Note that 0 is a critical point of f
...
Also
Fig 6
...
Therefore, by first derivative test,
x = 0 is a point of local minima of f and local minimum value of f is f (0) = 3
...
= 36x2 + 24x – 24 = 12 (3x2 + 2x – 1)
✆ f ☎☎(0)
✝
✟ f ☎☎(1)
✝ f ☎☎(✁2)
✠
or
✁12 ✄ 0
48 ✞ 0
84 ✞ 0
Therefore, by second derivative test, x = 0 is a point of local maxima and local
maximum value of f at x = 0 is f (0) = 12 while x = 1 and x = – 2 are the points of local
minima and local minimum values of f at x = – 1 and – 2 are f (1) = 7 and f (–2) = –20,
respectively
...
Solution We have
f (x) = 2x3 – 6x2 + 6x +5
or
✡✏ f ☛( x) ☞ 6 x 2 ✌ 12 x ✍ 6 ☞ 6( x ✌ 1) 2
✑
✏✒ f ☛☛( x) ☞ 12( x ✌ 1)
Now f ✂(x) = 0 gives x =1
...
Therefore, the second derivative test
fails in this case
...
We have already seen (Example 30) that, using first derivative test, x =1 is neither
a point of local maxima nor a point of local minima and so it is a point of inflexion
...
Solution Let one of the numbers be x
...
Let S(x)
denote the sum of the squares of these numbers
...
Also S✡✡ ✌ ✍ ☛ 4 ☞ 0
...
Hence the sum of squares of numbers is
2
minimum when the numbers are
15
15 15
and 15 ✑ ✞
...
2
2
Example 35 Find the shortest distance of the point (0, c) from the parabola y = x2,
where 0 ✒ c ✒ 5
...
Let D be the required distance
between (h, k) and (0, c)
...
(1)
Since (h, k) lies on the parabola y = x2, we have k = h2
...
e
...
Also when
2
2c ✑ 1
2c ✑ 1
, then D✢( k ) ✣ 0
...
2
2
Hence, the required shortest distance is given by
k✜
226
MATHEMATICS
2
2c 1 ✁ 2c 1 ✂
4c 1
✁ 2c 1 ✂
c✝ ✄
☎✆
✝✄
2
2
2
2
✞
✟
✞
✟
D✆
✠Note The reader may note that in Example 35, we have used first derivative
test instead of the second derivative test as the former is easy and short
...
If AP = 16 m, BQ = 22 m
and AB = 20 m, then find the distance of a point R on
AB from the point A such that RP2 + RQ2 is minimum
...
Then RB = (20 – x) m (as AB = 20 m)
...
18,
we have
RP2 = AR2 + AP2
Fig 6
...
Therefore
S✡(x) = 4x – 40
...
Also S✎(x) = 4 > 0, for all x and so S✎(10) > 0
...
Thus, the
distance of R from A on AB is AR = x =10 m
...
Solution The required trapezium is as given in Fig 6
...
Draw perpendiculars DP and
Fig 6
...
Let AP = x cm
...
Therefore, QB = x cm
...
Let A be the area of the trapezium
...
e
...
Since x represents distance, it can not be negative
...
Now
100 ✓ x 2 (✓4 x ✓ 10) ✓ (✓ 2 x 2 ✓ 10 x ✔ 100)
A✒(x) =
=
100 ✓ x
2 x 3 ✕ 300 x ✕ 1000
3
(✓ 2 x )
2 100 ✓ x 2
2
(on simplification)
(100 ✕ x 2 ) 2
or
A✒(5) =
2(5)3 ✕ 300(5) ✕ 1000
(100 ✕
3
(5) 2 ) 2
✖
✕2250
75 75
✖
✕30
75
✗0
Thus, area of trapezium is maximum at x = 5 and the area is given by
A (5) = (5 ✘ 10) 100 ✙ (5) 2 ✚ 15 75 ✚ 75 3 cm 2
Example 38 Prove that the radius of the right circular cylinder of greatest curved
surface area which can be inscribed in a given cone is half of that of the cone
...
Let a cylinder
with radius OE = x inscribed in the given cone (Fig 6
...
The height QE of the cylinder
is given by
228
MATHEMATICS
QE
EC
=
OA
OC
or
(since ✌QEC ~ ✌AOC)
QE
r x
=
h
r
x)
r
Let S be the curved surface area of the given
cylinder
...
20
2✄h
☎ ✆
✟✟S ( x) ✝ r ( r ✞ 2 x )
✠
✟S✆✆( x) ✝ ✞ 4✄h
✟✡
r
r
r
✍r✑
...
So x ☞ is a
2
2
✖ 2✗
point of maxima of S
...
Now S☛(x) = 0 gives x ☞
6
...
1 Maximum and Minimum Values of a Function in a Closed Interval
Let us consider a function f given by
f (x) = x + 2, x ✘ (0, 1)
Observe that the function is continuous on (0, 1) and neither has a maximum value
nor has a minimum value
...
However, if we extend the domain of f to the closed interval [0, 1], then f still may
not have a local maximum (minimum) values but it certainly does have maximum value
3 = f (1) and minimum value 2 = f (0)
...
Similarly, the minimum value 2 of f at x = 0 is called the absolute minimum
value (global minimum or least value) of f on [0, 1]
...
21 of a continuous function defined on a closed
interval [a, d]
...
21
minimum value is f (b)
...
Also from the graph, it is evident that f has absolute maximum value f (a) and
absolute minimum value f (d)
...
We will now state two results (without proof) regarding absolute maximum and
absolute minimum values of a function on a closed interval I
...
Then f has the
absolute maximum value and f attains it at least once in I
...
Theorem 6 Let f be a differentiable function on a closed interval I and let c be any
interior point of I
...
(ii) f ✂ (c) = 0 if f attains its absolute minimum value at c
...
Working Rule
Step 1: Find all critical points of f in the interval, i
...
, find points x where either
f ✁ ( x ) 0 or f is not differentiable
...
Step 3: At all these points (listed in Step 1 and 2), calculate the values of f
...
This maximum value will be the absolute maximum (greatest) value of
f and the minimum value will be the absolute minimum (least) value of f
...
Solution We have
f (x) = 2x3 – 15x2 + 36x + 1
or
f ✂(x) = 6x2 – 30x + 36 = 6 (x – 3) (x – 2)
Note that f ✂(x) = 0 gives x = 2 and x = 3
...
e
...
So
f (1) = 2 (13) – 15 (12) + 36 (1) + 1 = 24
f (2) = 2 (23) – 15 (22) + 36 (2) + 1 = 29
f (3) = 2 (33) – 15 (32) + 36 (3) + 1 = 28
f (5) = 2 (53) – 15 (52) + 36 (5) + 1 = 56
Thus, we conclude that absolute maximum value of f on [1, 5] is 56, occurring at
x =5, and absolute minimum value of f on [1, 5] is 24 which occurs at x = 1
...
Further note that f ✂(x) is not defined at x = 0
...
Now evaluating the value of f at critical points
8
1
and at end points of the interval x = –1 and x = 1, we have
8
4
1
f (–1) = 12(✁1) 3 ✁ 6( ✁1) 3
f (0) = 12 (0) – 6 (0) = 0
18
APPLICATION OF DERIVATIVES
4
1✁
3
f ✂ ✄ = 12 ✞✡ 1 ✟☛
8
☎ ✆
☞ 8✌
231
1
✝9
✞ 1 ✟3
✝ 6✡ ☛ ✠
4
☞ 8✌
4
1
f (1) = 12(1) 3 ✍ 6(1) 3 ✎ 6
Hence, we conclude that absolute maximum value of f is 18 that occurs at x = – 1
and absolute minimum value of f is
✏9
4
that occurs at x ✑
1
...
A soldier, placed at (3, 7), wants to shoot down the helicopter when it is
nearest to him
...
Solution For each value of x, the helicopter’s position is at point (x, x 2 + 7)
...
e
...
f (x) = (x – 3)2 + x4
f ✔(x) = 2(x – 3) + 4x3 = 2 (x – 1) (2x2 + 2x + 3)
Thus, f ✔(x) = 0 gives x = 1 or 2x2 + 2x + 3 = 0 for which there are no real roots
...
e
...
The value of f at this point is given by
f (1) = (1 – 3)2 + (1)4 = 5
...
Note that
5 is either a maximum value or a minimum value
...
Hence,
5 is the minimum value of
distance between the soldier and the helicopter
...
5
1
...
Find the maximum and minimum values, if any, of the following functions
given by
(i) f (x) = | x + 2 | – 1
(ii) g (x) = – | x + 1| + 3
(iii) h (x) = sin (2x) + 5
(iv) f (x) = | sin 4x + 3|
(v) h (x) = x + 1, x ☎ (– 1, 1)
3
...
Find
also the local maximum and the local minimum values, as the case may be:
(ii) g (x) = x3 – 3x
(i) f (x) = x2
(iii) h (x) = sin x + cos x, 0 ✁ x ✁
2
(iv) f (x) = sin x – cos x, 0 ✂ x ✂ 2✄
(v) f (x) = x3 – 6x2 + 9x + 15
(vii) g ( x) ✟
(vi) g ( x ) ✆
x 2
✝ , x✞ 0
2 x
1
(viii) f ( x) ✟ x 1 ✡ x , x ☛ 0
x ✠2
4
...
Find the absolute maximum value and the absolute minimum value of the following
functions in the given intervals:
(i) f (x) = x3, x ☎ [– 2, 2]
(ii) f (x) = sin x + cos x , x ☎ [0, ☞]
2
(iii) f (x) = 4 x ✎
1 2
✌ 9✍
x , x ✏ ✑ ✎2, ✒ (iv) f ( x ) ✟ ( x ✡ 1) 2 ✠ 3, x ✕[ ✡3,1]
2
✓ 2✔
6
...
Find both the maximum value and the minimum value of
3x4 – 8x3 + 12x2 – 48x + 25 on the interval [0, 3]
...
At what points in the interval [0, 2☞], does the function sin 2x attain its maximum
value?
9
...
Find the maximum value of 2x3 – 24x + 107 in the interval [1, 3]
...
APPLICATION OF DERIVATIVES
233
11
...
Find the value of a
...
Find the maximum and minimum values of x + sin 2x on [0, 2✄]
...
Find two numbers whose sum is 24 and whose product is as large as possible
...
Find two positive numbers x and y such that x + y = 60 and xy3 is maximum
...
Find two positive numbers x and y such that their sum is 35 and the product x2 y5
is a maximum
...
Find two positive numbers whose sum is 16 and the sum of whose cubes is
minimum
...
A square piece of tin of side 18 cm is to be made into a box without top, by
cutting a square from each corner and folding up the flaps to form the box
...
18
...
What should be
the side of the square to be cut off so that the volume of the box is maximum ?
19
...
20
...
21
...
A wire of length 28 m is to be cut into two pieces
...
What should be the length of the
two pieces so that the combined area of the square and the circle is minimum?
23
...
radius R is
27
24
...
25
...
26
...
234
MATHEMATICS
Choose the correct answer in the Exercises 27 and 29
...
The point on the curve x2 = 2y which is nearest to the point (0, 5) is
(A) (2 2,4)
(B) (2 2,0)
(C) (0, 0)
28
...
The maximum value of [ x( x ✂ 1) ✄ 1]3 , 0 ☎ x ☎ 1 is
1
✆ 1✝ 3
(A) ✞ ✟
✠ 3✡
(B)
1
2
(C) 1
(D) 0
Miscellaneous Examples
Example 42 A car starts from a point P at time t = 0 seconds and stops at point Q
...
Solution Let v be the velocity of the car at t seconds
...
Now v = 0 at P as well as at Q and at P, t = 0
...
Thus, the car will
reach the point Q after 4 seconds
...
Its semi-vertical angle is tan–1 (0
...
Water is poured
into it at a constant rate of 5 cubic metre per hour
...
Solution Let r, h and ✑ be as in Fig 6
...
Then tan
✄r☎
✝
...
5)
or
r
= 0
...
h
tan ✂1 ✆
So
or
✁
(given)
h
2
Let V be the volume of the cone
...
e
...
22
(by Chain Rule)
dh
dt
✛ 5 m3/h and h = 4 m
...
88
Example 44 A man of height 2 metres walks at a uniform speed of 5 km/h away from
a lamp post which is 6 metres high
...
236
MATHEMATICS
Solution In Fig 6
...
Then,
MS is the shadow of the man
...
✌MSN ~ ✌ ASB
Note that
or
MS
MN
=
AS
AB
or
Thus
So
AS = 3s (as MN = 2 and AB = 6 (given))
AM = 3s – s = 2s
...
23
5 km/h
...
2
Example 45 Find the equation of the normal to the curve x2 = 4y which passes through
the point (1, 2)
...
Now, slope of the tangent at (h, k) is given by
dy ✁
h
dx ✂✄ (h , k ) = 2
Hence, slope of the normal at (h, k) =
☎2
h
Therefore, the equation of normal at (h, k) is
✆2
( x ✆ h)
h
Since it passes through the point (1, 2), we have
y–k=
2☎k
☎2
h
(1 ☎ h) or k
2
2 ✝ (1 ☎ h)
h
...
(2)
APPLICATION OF DERIVATIVES
237
Since (h, k) lies on the curve x2 = 4y, we have
...
Substituting the values of h and k in (1),
we get the required equation of normal as
y 1✁
2
( x 2) or x + y = 3
2
Example 46 Find the equation of tangents to the curve
y = cos (x + y), – 2✄ ✝ x ✝ 2✄
that are parallel to the line x + 2y = 0
...
Thus, tangents to the
2
2
✕✔ ✖
✠ ☞3✟ ,0 ✡
and ✗ ,0✘
...
f (x) =
Solution We have
Therefore
f (x) =
3 4 4 3
36
x ✚ x ✚ 3x 2 ✛
x ✛ 11
10
5
5
f ✜(x) =
3
4
36
(4 x 3 ) ✘ (3x 2 ) ✘ 3(2 x) ✙
10
5
5
=
6
( x ✘ 1)( x ✙ 2)( x ✘ 3)
5
Now f ✜(x) = 0 gives x = 1, x = – 2, or x = 3
...
24)
...
24
Consider the interval (– ✢ , – 2), i
...
, when – ✢ < x < – 2
...
(In particular, observe that for x = –3, f ✜(x) = (x – 1) (x + 2) (x – 3) = (– 4) (– 1)
(– 6) < 0)
Therefore,
f ✜(x) < 0 when – ✢ < x < – 2
...
Consider the interval (– 2, 1), i
...
, when – 2 < x < 1
...
Thus,
f is strictly increasing in (– 2, 1)
...
e
...
In this case, we have
x – 1 > 0, x + 2 > 0 and x – 3 < 0
...
Thus,
f is strictly decreasing in (1, 3)
...
e
...
In this case, we have x – 1 > 0,
x + 2 > 0 and x – 3 > 0
...
Thus, f is strictly increasing in the interval (3, ✟)
...
✝ 4✞
Solution We have
f (x) = tan–1(sin x + cos x), x > 0
Therefore
f ✂(x) =
=
1
(cos x ✠ sin x )
1 ✡ (sin x ✡ cos x) 2
cos x ☛ sin x
2 ☞ sin 2 x
(on simplification)
✍ ✌✎
Note that 2 + sin 2x > 0 for all x in ✏✒ 0, ✑✓
...
e
...
4✓
Example 49 A circular disc of radius 3 cm is being heated
...
05 cm/s
...
2 cm
...
Then
A = ✄r2
dA
dr
= 2 r
dt
dt
or
(by Chain Rule)
dr
✁t ✂ 0
...
dt
Therefore, the approximate rate of increase in area is given by
Now approximate rate of increase of radius = dr =
dA =
☎ dr ✆
dA
(✁t ) = 2✝r ✟ ✞t ✠
dt
✡ dt ☛
= 2✄ (3
...
05) = 0
...
2 cm)
Example 50 An open topped box is to be constructed by removing equal squares from
each corner of a 3 metre by 8 metre rectangular sheet of aluminium and folding up the
sides
...
Solution Let x metre be the length of a side of the removed squares
...
25)
...
25
V (x) = x (3 – 2x) (8 – 2x)
= 4x3 – 22x2 + 24x
☞✑ V✌( x ) ✍ 12 x 2 ✎ 44 x ✏ 24 ✍ 4( x ✎ 3)(3x ✎ 2)
✒
✑✓ V✌✌( x) ✍ 24 x ✎ 44
Therefore
Now
V✔(x) = 0 gives x ✂ 3,
Thus, we have x ✂
2
...
Now V✖✖ ✟ ✠ ✗ 24 ✟ ✠ ✘ 44 ✗ ✘ 28 ✙ 0
...
e
...
The
100 ✝
✏x
✑
cost price of x items is Rs ✓ ✒ 500✔
...
Solution Let S (x) be the selling price of x items and let C (x) be the cost price of x
items
...
e
...
Also P✪ ( x) ✫
x2 x
✥ ✥ 500
100 5
✩1
✩1
...
Hence, the manufacturer can earn maximum
profit, if he sells 240 items
...
Using differentials, find the approximate value of each of the following:
1
1
(a) 17 ✁ 4
✂
✄
☎ 81✆
(b)
✟
✝ 33✞ 5
log x
has maximum at x = e
...
The two equal sides of an isosceles triangle with fixed base b are decreasing at
the rate of 3 cm per second
...
Find the equation of the normal to curve x2 = 4y which passes through the point
(1, 2)
...
Show that the function given by f ( x) ✠
5
...
6
...
f ( x) ✠
7
...
x2 y 2
8
...
9
...
If building of tank costs
Rs 70 per sq metres for the base and Rs 45 per square metre for sides
...
The sum of the perimeter of a circle and square is k, where k is some constant
...
APPLICATION OF DERIVATIVES
243
11
...
The total perimeter of the window is 10 m
...
12
...
2
Show that the maximum length of the hypotenuse is (a 3
2
3
b3 )2
...
Find the points at which the function f given by f (x) = (x – 2)4 (x + 1)3 has
(i) local maxima
(ii) local minima
(iii) point of inflexion
14
...
Show that the altitude of the right circular cone of maximum volume that can be
4r
...
Let f be a function defined on [a, b] such that f ✂(x) > 0, for all x ☎ (a, b)
...
17
...
Also find the maximum volume
...
Show that height of the cylinder of greatest volume which can be inscribed in a
right circular cone of height h and semi vertical angle ✑ is one-third that of the
a sphere of radius R is
4
✁h3 tan 2 ✆
...
19
...
Then the depth of the wheat is increasing at the rate of
(B) 0
...
5 m3/h
(C) 1
...
The slope of the tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point
(2,– 1) is
cone and the greatest volume of cylinder is
(A)
22
7
(B)
6
7
(C)
7
6
(D)
✝6
7
244
MATHEMATICS
21
...
The normal at the point (1,1) on the curve 2y + x2 = 3 is
(A) x + y = 0
(B) x – y = 0
(C) x + y +1 = 0
(D) x – y = 0
2
23
...
The points on the curve 9y = x , where the normal to the curve makes equal
intercepts with the axes are
(A)
✄ 4, ✂
✆
8✁
☎
3✝
(B)
✄ 4,
✆
✞8✁
☎
3✝
(C)
✟
☛ 4, ✡
✌
3✠
☞
8✍
(D)
✟
☛✡
✌
4,
3✠
☞
8✍
Summary
✎
If a quantity y varies with another quantity x, satisfying some rule y ✏ f ( x ) ,
then
dy
(or f ✑( x) ) represents the rate of change of y with respect to x and
dx
dy ✓
dx ✔✕ x✒ x0 (or f ✖(x0 ) ) represents the rate of change of y with respect to x at
x ✗ x0
...
e
...
A function f is said to be
(a) increasing on an interval (a, b) if
x1 < x2 in (a, b) ✚ f (x1) ✛ f (x2) for all x1, x2 ✜ (a, b)
...
x1 < x2 in (a, b)
Alternatively, if f (x) 0 for each x in (a, b)
The equation of the tangent at (x0, y0) to the curve y = f (x) is given by
✂ ✝
y
✆
☎
✞
✄ y ✟ dydx ✁✠✡
(x
0
( x0 , y0 )
✄x )
0
dy
does not exist at the point ( x0 , y0 ) , then the tangent at this point is
dx
parallel to the y-axis and its equation is x = x0
...
✍✎ ☛
x x0
Equation of the normal to the curve y = f (x) at a point ( x0 , y0 ) is given by
y
✏ y ✑ dy ✒✏1
dx ✓✔
(x
0
✏x )
0
( x0 , y0 )
If
dy
at the point ( x0 , y0 ) is zero, then equation of the normal is x = x0
...
Let y = f (x), x be a small increment in x and y be the increment in y
corresponding to the increment in x, i
...
, y = f (x + x) – f (x)
...
is a good approximation of ✕y when dx ✥ ✦ x is relatively small and we denote
it by dy ✧ ✕y
...
246
MATHEMATICS
First Derivative Test Let f be a function defined on an open interval I
...
Then
(i) If f (x) changes sign from positive to negative as x increases through c,
i
...
, if f (x) > 0 at every point sufficiently close to and to the left of c,
and f (x) < 0 at every point sufficiently close to and to the right of c,
then c is a point of local maxima
...
e
...
(iii) If f (x) does not change sign as x increases through c, then c is neither
a point of local maxima nor a point of local minima
...
Second Derivative Test Let f be a function defined on an interval I and
c I
...
Then
(i) x = c is a point of local maxima if f (c) = 0 and f (c) < 0
The values f (c) is local maximum value of f
...
(iii) The test fails if f (c) = 0 and f (c) = 0
...
Working rule for finding absolute maxima and/or absolute minima
Step 1: Find all critical points of f in the interval, i
...
, find points x where
either f (x) = 0 or f is not differentiable
...
Step 3: At all these points (listed in Step 1 and 2), calculate the values of f
...
This maximum value will be the absolute maximum
value of f and the minimum value will be the absolute minimum value of f
...
1
1
...
5
...
13
...
Neither reflexive nor symmetric nor transitive
...
Reflexive and transitive but not symmetric
...
(a) Reflexive, symmetric and transitive
...
(c) Neither reflexive nor symmetric nor transitive
...
(e) Neither reflexive nor symmetric nor transitive
...
Neither reflexive nor symmetric nor transitive
...
T1 is related to T3
...
The set of all lines y = 2x + c, c ✂ R
B
16
...
2
1
...
(i) Injective but not surjective
(ii) Neither injective nor surjective
(iii) Neither injective nor surjective (iv) Injective but not surjective
(v) Injective but not surjective
(ii) Neither one-one nor onto
...
(i) One-one and onto
9
...
Yes
11
...
A
EXERCISE 1
...
gof = {(1, 3), (3,1), (4,3)}
3
...
Inverse of f is f itself
ANSWERS
5
...
(iii) Yes, since h is one-one-onto
...
f –1 is given by f –1 (y) =
11
...
f –1 is given by f –1 (y) =
1 y
4
is given by f –1 (a) = 1, f –1 (b) = 2 and f –1 (c) = 3
...
(C)
14
...
4
1
...
(i) ☎ is neither commutative nor associative
(ii) ☎ is commutative but not associative
(iii) ☎ is both commutative and associative
(iv) ☎ is commutative but not associative
(v) ☎ is neither commutative nor associative
(vi) ☎ is neither commutative nor associative
3
...
(i) (2 * 3) * 4 = 1 and 2 * (3 * 4) = 1
(ii) Yes
(iii) 1
(iii) Yes
(iv) 1
5
...
(i) 5 * 7 = 35, 20 * 16 = 80
(ii) Yes
(v) 1
7
...
☎ is both commutative and associative; ☎ does not have any identity in N
9
...
11
...
12
...
B
270
MATHEMATICS
Miscellaneous Exercise on Chapter 1
1
...
11
...
19
...
The inverse of f is f itself
x4 – 6x3 + 10x2 – 3x
8
...
A
17
...
n!
12
...
No
EXERCISE 2
...
✂
6
2
...
3✄
4
13
...
5
...
✂
3
...
4
☎✄
11
...
B
✂
✂
4
...
6
3✄
4
12
...
2
1 ✆1
tan x
2
✆1 x
9
...
1 ✞ xy
5
...
4
21
...
✂
2
– sec–1 x
✆1
10
...
11
...
✂
1
5
15
...
17
6
19
...
0
4
14
...
✂
3
20
...
✂
6
15
...
✂
6
16
...
x✠
17
...
x✡
1
3
ANSWERS
EXERCISE 3
...
1 × 24, 2 × 12, 3 × 8, 4 × 6, 6 × 4, 8 × 3, 12 × 2, 24 × 1; 1 × 13, 13 × 1
3
...
(i) 3 × 4
9✁
2✄
2
4
...
(i)
✎
2 ✠
✎
✑5 4
7
5 ✠✠
3
2
2 ☛✠
6
...
a = 1, b = 2, c = 3, d = 4
8
...
B
10
...
2
1
...
(i)
✚ 2a
✜
✣ 0
(iii)
✌11
✎
16
✎
✎
✑5
7✔
7✖✘
✓8
✕
✗6
2b ✛
2a✢✤
(ii) A ✙ B =
7✔
(iv)
2✖✘
(ii)
11 0 ✍
5 21✏
10 9
✏
✏
✒
(iv)
✓1
✕
✗5
✓ ✙6
AB =✕
✗
1
✥ ( a ✧ b )2
★
2
★ (a ✪ c )
✫
✓1
✕
✗1
1✔
1✖✘
1✔
✖
✙3✘
26✔
19 ✖✘
(v) BA =
(b ✧ c) 2 ✦
✩
(a ✪ b) 2 ✬✩
✚11
✜
✣11
10✛
2 ✢✤
271
272
MATHEMATICS
3
...
5
...
(i)
8
...
✍5
X✛✏
✒1
0✎
✍ 2 0✎
,
✑ Y✛✏
✑
4✓
✒ 1 1✓
0 0 ✡☞
✍ ✌1
✒ ✌2
✌1✎
✑
✌1✓
✘1
4
8 ✡✡
6
(iii)
✍✌ 3 ✌ 4
✏
✒ 8 13
(vi)
✍14
✏
✒4
9 12✡☞
2 3✕
4 5✗
✗
0✚✗
2
✔ ✘1
✖
4
✖
✙✖ 1
✘2
1✎
9 ✑✓
✌6 ✎
5 ✑✓
0✕
✘1
3✗
2
0✚✗
✗
(ii) X
9
...
x = 3, y = 6, z = 9, t = 6
12
...
x = 3, y = – 4
15
...
k = 1
19
...
Rs 20160
21
...
B
EXERCISE 3
...
(i)
✩
5
✬
✮
1
2
✪
✫1
✭
✯
(ii)
✍1
✏
✒ ✌1
2✎
3✑✓
(iii)
✰ ✲1
✳
✳5
✳6
✵
3
5
6
2✱
✴
3✴
✲1✴
✶
ANSWERS
✞0 0 0✟ ✞ 0 a b✟
9
...
✄
☎
✆ 1 6✝
✍3
10
...
A
0
0
0
0✙
0✛
✛
0 ✛✤
3✧
2✩
✩
3✩
(iv)
✩
✩
0✩
✭✩
✁ 1 2✂ ✁ 0 3 ✂
☎✯✄
☎
✆ 2 2✝ ✆ 3 0 ✝
A✮✄
12
...
4
✰ 3
✲5
1
...
✒
✓
✕ ✔1 2 ✖
✍ 7 ✔3✎
3
...
✒
✓
✕ 5 ✔ 2✖
✍ 4 ✔ 1✎
5
...
✒
✓
✕ ✔1 2 ✖
✍ 2 ✔ 1✎
7
...
✒
✓
✕ ✔3 4 ✖
✍ 7 ✔10✎
9
...
✲
✲2
✲✵
1✱
2✳
✳
3✳
2 ✳✶
273
✰ ✴1
11
...
Inverse does not exist
...
✂
☎1
15
...
Inverse does not exist
...
✝2
✞
1
✠
✠
✠ ✝2
✠ 5
✠
✝3
✠
✠
☛ 5
5
4
25
1
25
✝3 ✟
5✡
✡
11 ✡
25 ✡
✡
9✡
25 ✡☞
17
...
D
Miscellaneous Exercise on Chapter 3
6
...
x = – 1
9
...
(a) Total revenue in the market - I = Rs 46000
Total revenue in the market - II = Rs 53000
(b) Rs 15000, Rs 17000
11
...
C
14
...
C
EXERCISE 4
...
(i) 1, (ii) x3 – x2 + 2
1
...
(i) – 12, (ii) 46, (iii) 0, (iv) 5
6
...
(i)
8
...
2
15
...
C
ANSWERS
275
EXERCISE 4
...
(i) 0, 8, (ii) 0, 8 4
...
(i)
5
...
4
1
...
(i) M11= 1, M12= 0, M13 = 0, M21 = 0, M22 = 1, M23 = 0, M31 = 0, M32 = 0, M33 = 1,
A11= 1, A12= 0, A13= 0, A21= 0, A22= 1, A23= 0, A31= 0, A32= 0, A33= 1
(ii)
M11= 11, M12= 6, M13= 3, M21= –4, M22= 2, M23= 1, M31= –20, M32= –13, M33= 5
A11=11, A12= – 6, A13= 3, A21= 4, A22= 2, A23= –1, A31= –20, A32= 13, A33= 5
3
...
(x – y) (y – z) (z – x)
5
...
5
✁4
1
...
13 ✑✓ 3 ✍1✔✒
✟ 3 1 ✞11✠
✡
☛
2
...
10 ✘
✘✚ 0
0
2 ✙✛
3✗
✟ ✞2 0 1 ✠
✖ ✕1 5
✕1 ✘
✙
✕ 4 23 12 ✙ 10
...
3 ✘
✡☞ 6 1 ✞2☛✌
✘✚ 1 ✕11 ✕ 6✙✛
13
...
a = – 4, b = 1
5
...
✢ ✜3 0 0 ✣
✜1 ✤
3 ✜1 0✥
✥
3 ✤
✤✦ ✜9 ✜2 3✥✧
0
0 ✠
✟1
✡0 cos ★ sin ★ ☛
11
...
A ✪ ✡ 9 ✞1 ✞ 4 ☛
11
☞✡ 5 ✞3 ✞1 ☛✌
276
16
...
B
18
...
6
1
...
Consistent
3
...
Consistent
5
...
Consistent
7
...
x✟
✞5
11
, y✟
12
11
9
...
x = 1, y ✟ , z ✟
2
2
10
...
x = 2, y = –1, z = 1
13
...
x = 2, y = 1, z = 3
✡ 0 1 ✠2 ☛
☞
✌
15
...
cost of onions per kg ✏ Rs 5
cost of wheat per kg ✏ Rs 8
cost of rice per kg ✏ Rs 8
Miscellaneous Exercise on Chapter 4
3
...
– 2(x3 + y3)
17
...
x✟
✞a
3
✡ 9 ✠3 5 ☛
☞
1 0✌
7
...
xy
16
...
A
19
...
1
2
...
5
...
8
...
12
...
15
...
Discontinuous at x = 3
Discontinuous at x = 0
9
...
No point of discontinuity
f is continuous at x = 1
13
...
For no value of ✝, f is continuous at x = 0 but f is continuous at x = 1 for any
value of ✝
...
f is continuous at x = ✞
21
...
Cosine function is continuous for all x ✂ R; cosecant is continuous except for
16
...
a b✁
x = n✞, n ✂ Z; secant is continuous except for x = (2n ✁ 1)
cotangent function is continuous except for x = n✞, n ✂ Z
23
...
24
...
k = 6
27
...
f is continuous for all x ✂ R
28
...
a = 2, b = 1
5
34
...
29
...
2
2
1
...
2
...
a cos (ax + b)
sec (tan x)
...
sec 2 x
2 x
5
...
10x4 sinx5 cosx5 cosx3 – 3x2 sinx3 sin2 x5
278
MATHEMATICS
2 2x
7
...
2
sin x
2 x
EXERCISE 5
...
cosx ✂ 2
3
2
...
sec 2 x ✝ y
x ✞ 2y ✝1
5
...
y sin xy
sin 2 y ☎ x sin xy 8
...
2
1 ✟ x2
10
...
✠2
1 ✟ x2
14
...
12
...
☎
a
2by ✆ sin y
6
...
✡
1✡ x2
EXERCISE 5
...
e x (sin x ☛ cosx)
, x ☞ n✌✍ n ✎ Z 2
...
3 x2 e x
e
9
...
e x ✙ 2 x e ✙ 3 x 2 e x ✙ 4 x 3e x ✙ 5 x 4 e x
x
4 xe
1 x2
4
...
– ex tan ex, e x ✖ (2n ✗ 1)
7
...
3
4
1
,x>1
x log x
( x sin x ✚ log x ✆ cos x) ,
✜1
✢
x ✛ 0 10
...
5
1
...
1
( x ✂ 1) ( x ✂ 2)
1
1
1
1
1 ✁
✄
✂
✂
✂
☎
2 ( x ✂ 3)( x ✂ 4)( x ✂ 5) ✝ x ✂ 1 x ✂ 2 x ✂ 3 x ✂ 4 x ✂ 5 ✞✆
✠
cos x
✡ sin x log (log x) ☞
✌ x log x
✍
4
...
✟
(log x )cos x ☛
5
...
(log x)x-1 [1 + log x
...
logx
1
1
8
...
✔
✘ x✓
✚
9
...
x x cosx [cos x
...
(x cos x)x [1 – x tan x + log (x cos x)] + (x sin x)
1
x
☎
✝
x cot x ✄ 1 ✂ log ( x sin x ) ✁
✆
x2
✞
12
...
y ✮ y ✰ x log y ✯
✱
✲
x ✳ x ✰ y log x ✴
14
...
y ( x ✶1)
x ( y ✵ 1)
16
...
5x4 – 20x3 + 45x2 – 52x + 11
EXERCISE 5
...
2t2
2
...
– 4 sin t
4
...
cos ✁ 2cos 2
2sin 2 ✁ sin
9
...
✄ cot
✂
2
7
...
tan t
10
...
7
1
...
✆
1
x2
2
...
– x cos x – 2 sin x
5
...
2ex (5 cos 5x – 12 sin 5x)
8
...
9 e6x (3 cos 3x – 4 sin 3x)
2x
(1 ✞ x 2 ) 2
sin (log x) ✡ cos (log x)
(1 ✟ log x)
10
...
– cot y cosec2 y
9
...
27 (3x – 9x + 5)8 (2x – 3)
3
...
3sinx cosx (sinx – 2 cos4 x)
3cos 2 x
☞
✁ 6sin 2 x log 5 x ✍
x
✏
4
...
✗ ✕
✘
3
2
✕ 4 ✗ x 2 x ✘ 7 (2 x ✘ 7) 2
✙
6
...
(log x )
✌✎ x ✟
✍✏ , x ✛ 1
x
8
...
(sinx – cosx)sin x – cos x (cosx + sinx) (1 + log (sinx – cos x)), sinx > cosx
10
...
2
✢ x2 ✆ 3
✣
✣
2
2 ✢ x
x x ✜3 ✤
✡ 2 x log x✥ ✡ ( x ✆ 3) x ✤
✡ 2 x log( x ✆ 3)✥
✦ x
✧
✦x✆3
✧
✔
✖
✖
✖
✚
ANSWERS
12
...
0
17
...
1
1
...
8
cm2/s
3
(b) 8✞ cm2/s
3
...
80✞ cm2/s
6
...
4✞ cm/s
(b) 2 cm2/min
7
...
1
✂
9
...
11
...
16
...
900 cm3/s
10
...
2✞ cm3/s
1
cm/s
48 ☞
17
...
Rs 20
...
D
EXERCISE 6
...
(a) ✝ , ✌ ✟
✠4 ✡
3✆
☎
(b) ✝ ✄ ✌, ✟
✠
4✡
5
...
(a) Strictly decreasing for x < – 1 and strictly increasing for x > – 1
3
3
and strictly increasing for x ✑ ✏
2
2
(c) Strictly increasing for – 2 < x < – 1 and strictly decreasing for x < – 2 and
x>–1
(b) Strictly decreasing for x ✎ ✏
(d) Strictly increasing for x ✑ ✏
9
9
and strictly decreasing for x ✎ ✏
2
2
282
MATHEMATICS
(e) Strictly increasing in (1, 3) and (3, ✡), strictly decreasing in (– ✡ , –1)
and (– 1, 1)
...
0 < x < 1 and x > 2
12
...
D
14
...
D
EXERCISE 6
...
764
2
...
11
4
...
(3, – 20) and (–1, 12)
2b
8
...
(2, – 9)
10
...
No tangent to the curve which has slope 2
...
1
12
...
(i)
(ii)
(iii)
(iv)
6
...
(i) (0, ± 4) (ii) (± 3, 0)
2
Tangent: 10x + y = 5;
Normal: x – 10y + 50 = 0
Tangent: y = 2x + 1;
Normal: x + 2y – 7 = 0
Tangent: y = 3x – 2;
Normal: x + 3y – 4 = 0
Tangent: y = 0;
Normal: x = 0
(v) Tangent: x + y ✂ 2 = 0; Normal x = y
15
...
19
...
22
...
(0, 0), (1, 2), (–1, –2)
(1, ± 2)
20
...
48x – 24y = 23
26
...
27
...
4
1
...
03
(iv) 0
...
035
(v) 0
...
8
(vi) 1
...
9629
(viii) 3
...
025
(xiii) 3
...
009
(xi) 0
...
948
(xiv) 7
...
00187
2
...
21
3
...
995
4
...
03 x3 m3
5
...
12 x2 m2
6
...
92 ✞ m3
7
...
16 ✞ m3
8
...
C
EXERCISE 6
...
(i) Minimum Value = 3
(ii) Minimum Value = – 2
(iii) Maximum Value = 10
(iv) Neither minimum nor maximum value
2
...
(i) local minimum at x = 0,
local minimum value = 0
(ii) local minimum at x = 1,
local minimum value = – 2
local maximum at x = – 1, local maximum value = 2
(iii) local maximum at x ✁
(iv) local maximum at x ✁
local minimum at x ☎
4
✂
4
,
local maximum value =
2
, local maximum value =
2
7✄
, local minimum value = – 2
4
(v) local maximum at x = 1,
local maximum value = 19
local minimum at x = 3,
local minimum value = 15
(vi) local minimum at x = 2,
local minimum value = 2
283
284
MATHEMATICS
(vii) local maximum at x = 0,
(viii) local maximum at x
2
,
3
local maximum value =
1
2
local maximum value =
2 3
9
5
...
Maximum profit = 49 unit
...
Minima at x = 2, minimum value = – 39, Maxima at x = 0, maximum value = 25
...
At x
✁
4
and
5✁
4
9
...
Maximum at x = 3, maximum value 89; maximum at x = – 2, maximum value = 139
11
...
Maximum at x = 2✞, maximum value = 2✞; Minimum at x = 0, minimum value = 0
13
...
45, 15
17
...
x = 5 cm
1
50 3
21
...
15
...
A
✁✡4
✁✡4
28
...
(a) 0
...
b 3 cm2/s
16
...
497
4
...
C
ANSWERS
6
...
(i) x < –1 and x > 1
8
...
length =
(ii)
2
✁x✁
3
2
(ii) – 1 < x < 1
9
...
(i) local maxima at x = 2 (ii) local minima at x ☎
2
7
(iii) point of inflection at x = –1
14
...
4✆ R 3
3 3
22
...
A
20
...
A
24
...
A
285
Appendix
1
PROOFS IN MATHEMATICS
Proofs are to Mathematics what calligraphy is to poetry
...
— VLADIMIR ARNOLD
A
...
1 Introduction
In Classes IX, X and XI, we have learnt about the concepts of a statement, compound
statement, negation, converse and contrapositive of a statement; axioms, conjectures,
theorems and deductive reasoning
...
A
...
2 What is a Proof?
Proof of a mathematical statement consists of sequence of statements, each statement
being justified with a definition or an axiom or a proposition that is previously established
by the method of deduction using only the allowed logical rules
...
Many a times, we prove a proposition directly from what is given in
the proposition
...
This leads to, two ways of proving a proposition
directly or indirectly and the proofs obtained are called direct proof and indirect proof
and further each has three different ways of proving which is discussed below
...
(i) Straight forward approach It is a chain of arguments which leads directly from
what is given or assumed, with the help of axioms, definitions or already proved
theorems, to what is to be proved using rules of logic
...
Solution x2 – 5x + 6 = 0 (given)
248
MATHEMATICS
✂
✂
✂
(x – 3) (x – 2) = 0 (replacing an expression by an equal/equivalent expression)
x – 3 = 0 or x – 2 = 0 (from the established theorem ab = 0
either a = 0 or
b = 0, for a, b in R)
x – 3 + 3 = 0 + 3 or x – 2 + 2 = 0 + 2 (adding equal quantities on either side of the
equation does not alter the nature of the
equation)
x + 0 = 3 or x + 0 = 2 (using the identity property of integers under addition)
x = 3 or x = 2 (using the identity property of integers under addition)
Hence, x2 – 5x + 6 = 0 implies x = 3 or x = 2
...
From the statement p, we deduced the statement r : “(x – 3) (x – 2) = 0” by
replacing the expression x2 – 5x + 6 in the statement p by another expression (x – 3)
(x – 2) which is equal to x2 – 5x + 6
...
e
...
The second one is by valid form of argumentation (rules of logic)
Next this statement r becomes premises or given and deduce the statement s
“ x – 3 = 0 or x – 2 = 0” and the reasons are given in the brackets
...
q
The symbolic equivalent of the argument is to prove by deduction that p
is true
...
This implies that “p q” is true
...
Solution Note that a function f is one-one if
f (x1) = f (x2) x1 = x2 (definition of one-one function)
Now, given that
f (x1) = f (x2), i
...
, 2x1+ 5 = 2x2 + 5
2x1+ 5 – 5 = 2x2 + 5 – 5 (adding the same quantity on both sides)
✂
✂
PROOFS IN MATHEMATICS
✂
✂
249
2x1+ 0 = 2x2 + 0
2x1 = 2x2 (using additive identity of real number)
2
2
x1 = x2 (dividing by the same non zero quantity)
2
2
✂
✂
x1 = x 2
Hence, the given function is one-one
...
The whole basis of proof of this method depends on the following axiom:
For a given subset S of N, if
(i) the natural number 1 ☎ S and
(ii) the natural number k + 1 ☎ S whenever k ☎ S, then S = N
...
We now consider some examples
...
Assume that P(k) is true, i
...
,
✁ cos k
P(k) : Ak = ✝
✠ ✟ sin k
sin k ✄
cos k ✞✡
250
MATHEMATICS
We want to prove that P(k + 1) is true whenever P(k) is true, i
...
,
✂ cos ( k 1) ✁ sin ( k
P(k + 1) : Ak+1 = ☎
✞ ✝ sin(k 1) ✁ cos (k
Now
Since P(k) is true, we have
1) ✁ ✄
1 ) ✁ ✟✆
Ak+1 = Ak
...
Hence, P(n) is true for all n ✒ 1 (by the principle of mathematical induction)
...
If the conditionals
r ✓ q;
s ✓ q;
and
t✓q
are proved, then (r ✔ s ✔ t) ✓ q, is proved and so p ✓ q is proved
...
It is
practically convenient only when the number of possible cases are few
...
From A draw AD a perpendicular to BC (BC produced if
necessary)
...
s : ABC is an obtuse angled triangle with ✝ C is obtuse
...
Hence, we prove the theorem by three cases
...
A1
...
From the right angled triangle ADB,
BD
= cos B
AB
i
...
BD = AB cos B
= c cos B
From the right angled triangle ADC,
CD
= cos C
AC
CD = AC cos C
= b cos C
Now
a = BD + CD
= c cos B + b cos C
Case (ii) When ✝ C is obtuse (Fig A1
...
From the right angled triangle ADB,
Fig A1
...
e
...
(1)
BD
= cos B
AB
i
...
BD = AB cos B
= c cos B
From the right angled triangle ADC,
CD
= cos ✝ ACD
AC
i
...
=
=
CD =
=
cos (180° – C)
– cos C
– AC cos C
– b cos C
Fig A1
...
e
...
3)
...
(2)
BC
= cos B
AB
i
...
BC = AB cos B
a = c cos B,
Fig A1
...
Thus, we may write
a = 0 + c cos B
= b cos C + c cos B
...
We assert that for any triangle ABC,
a = b cos C + c cos B
By case (i), r ✂ q is proved
...
By case (iii), t ✂ q is proved
...
e
...
Hence, from the proof by cases, (r
Indirect Proof Instead of proving the given proposition directly, we establish the proof
of the proposition through proving a proposition which is equivalent to the given
proposition
...
By rules of logic, we arrive at a
conclusion contradicting the assumption and hence it is inferred that the assumption
is wrong and hence the given statement is true
...
Example 5 Show that the set of all prime numbers is infinite
...
We take the negation of the statement
“the set of all prime numbers is infinite”, i
...
, we assume the set of all prime numbers
to be finite
...
, Pk (say)
...
, Pk
...
(1)
N is not in the list as N is larger than any of the numbers in the list
...
PROOFS IN MATHEMATICS
253
If N is a prime, then by (1), there exists a prime number which is not listed
...
But none of the
numbers in the list can divide N, because they all leave the remainder 1
...
Thus, in both the cases whether N is a prime or a composite, we ended up with
contradiction to the fact that we have listed all the prime numbers
...
Thus, the set of all prime numbers is infinite
...
✂
(ii) Proof by using contrapositive statement of the given statement
✂
Instead of proving the conditional p
~q
~ p
...
q, we prove its equivalent, i
...
,
The contrapositive of a conditional can be formed by interchanging the conclusion
and the hypothesis and negating both
...
x1 = x2
...
This is of the form
p q, where, p is 2x1+ 5 = 2x2 + 5 and q : x1 = x2
...
We can also prove the same by using contrapositive of the statement
...
e
...
✂✂
✂
✞
Now
Since “~ q
x1
2x1
✂
2x1+ 5
f (x1)
✞✞✂
✞✞
✞
✂
~ p”, is equivalent to “p
x2
2x 2
2x2 + 5
f (x 2 )
...
Example 7 Show that “if a matrix A is invertible, then A is non singular”
...
e
...
254
MATHEMATICS
If A is not a non singular matrix, then it means the matrix A is singular, i
...
,
|A| = 0
A–1 =
Then
adj A
does not exist as | A | = 0
|A|
Hence, A is not invertible
...
i
...
, ~ q ✂ ~ p
...
(iii) Proof by a counter example
In the history of Mathematics, there are occasions when all attempts to find a
valid proof of a statement fail and the uncertainty of the truth value of the statement
remains unresolved
...
The example to disprove the statement is called a counter example
...
Hence, this is
also a method of proof
...
This was once thought to be true on the basis that
n
22 ✁ 1 = 22 + 1 = 5 is a prime
...
2
22 ✁ 1 = 28 + 1 = 257 is a prime
...
But, eventually it was
22
shown that
5
1 = 232 + 1 = 4294967297
which is not a prime since 4294967297 = 641 × 6700417 (a product of two numbers)
...
n
Just this one example 22 ✁ 1 is sufficient to disprove the generalisation
...
5
Thus, we have proved that the generalisation “For each n, 22
(n ☎ N)” is not true in general
...
Proof We consider some functions given by
(i) f (x) = x2
(ii) g(x) = ex
(iii) h (x) = sin x
These functions are continuous for all values of x
...
This
makes us to believe that the generalisation “Every continuous function is differentiable”
may be true
...
This means that the
statement “Every continuous function is differentiable” is false, in general
...
Hence, “ (x) = | x |”
is called a counter example to disprove “Every continuous function is differentiable”
...
2
...
e
...
Roughly speaking
mathematical modelling is an activity in which we make models to describe the behaviour
of various phenomenal activities of our interest in many ways using words, drawings or
sketches, computer programs, mathematical formulae etc
...
Therefore, it is important to study mathematical modelling as a separate
topic
...
A
...
2 Why Mathematical Modelling?
Students are aware of the solution of word problems in arithmetic, algebra, trigonometry
and linear programming etc
...
Situational problems need physical insight
that is introduction of physical laws and some symbols to compare the mathematical
results obtained with practical values
...
Let us consider the
following problems:
(i) To find the width of a river (particularly, when it is difficult to cross the river)
...
(iii) To find the height of a tower (particularly, when it is not possible to reach the top
of the tower)
...
MATHEMATICAL MODELLING
257
(v) Why heart patients are not allowed to use lift? (without knowing the physiology
of a human being)
...
(vii) Estimate the yield of pulses in India from the standing crops (a person is not
allowed to cut all of it)
...
(ix) Estimate the population of India in the year 2020 (a person is not allowed to wait
till then)
...
In fact, you might have studied the methods
for solving some of them in the present textbook itself
...
A
...
3 Principles of Mathematical Modelling
Mathematical modelling is a principled activity and so it has some principles behind it
...
Some of the basic principles of
mathematical modelling are listed below in terms of instructions:
(i) Identify the need for the model
...
(iii) Identify the available relevent data
...
(vi) Identify
(a) the equations that will be used
...
(c) the solution which will follow
...
(b) utility of the model
...
258
MATHEMATICS
The above principles of mathematical modelling lead to the following: steps for
mathematical modelling
...
Step 2: Convert the physical situation into a mathematical model by introducing
parameters / variables and using various known physical laws and symbols
...
Step 4: Interpret the result in terms of the original problem and compare the result
with observations or experiments
...
Otherwise modify
the hypotheses / assumptions according to the physical situation and go to
Step 2
...
2
...
Solution Step 1 Given physical situation is “to find the height of a given tower”
...
2
...
Let PQ be an observer measuring the
height of the tower with his eye at P
...
Let ✂
be the angle of elevation from the eye of the observer to the top of the tower
...
2
...
(1)
Step 3 Note that the values of the parameters h, l and ✂ (using sextant) are known to
the observer and so (1) gives the solution of the problem
...
e
...
So from
☎PQB, we have
tan ✆ ✝
PQ h
✝ or l = h cot ✄
QB l
Step 5 is not required in this situation as exact values of the parameters h, l, ✂ and ✄
are known
...
Let the firm has purchase orders from
two clients F1 and F2
...
Solution Step 1 The physical situation is well identified in the problem
...
Then, A is of the form
P1 P2 P3
F1 ✞• • • ✟
A✠ ✡
F2 ☞• • • ☛✌
Let B be the matrix that represents the amount of raw materials R1, R2 and R3,
required to manufacture each unit of the products P1, P2 and P3
...
Example 3 Interpret the model in Example 2, in case
✞3 4 0 ✟
✞10 15 6 ✟
✠
✡
A=✠
✡ , B ☛ ✠7 9 3 ✡
10
20
0
☞
✌
✠☞5 12 7 ✡✌
and the available raw materials are 330 units of R1, 455 units of R2 and 140 units of R3
...
Since the amount of raw material required to manufacture
each unit of the three products is fixed, we can either ask for an increase in the
available raw material or we may ask the clients to reduce their orders
...
e
...
e
...
Thus, if the revised purchase orders of the clients are given by A1, then the firm
can easily supply the purchase orders of the two clients
...
Query Can we make a mathematical model with a given B and with fixed quantities of
the available raw material that can help the firm owner to ask the clients to modify their
orders in such a way that the firm makes the full use of its available raw material?
The answer to this query is given in the following example:
Example 4 Suppose P1, P2, P3 and R1, R2, R3 are as in Example 2
...
Step 2 Suppose the firm produces x units of P1, y units of P2 and z units of P3
...
Thus, the firm can produce 20 units of P1, 35
units of P2 and 5 units of P3 to make full use of its available raw material
...
Example 5 A manufacturer of medicines is preparing a production plan of medicines
M1 and M2
...
Further, it takes 3 hours to prepare enough material to fill 1000
bottles of M1, it takes 1 hour to prepare enough material to fill 1000 bottles of M2 and
there are 66 hours available for this operation
...
How should the manufacturer schedule his/her production in
order to maximise profit?
Solution Step 1 To find the number of bottles of M1 and M2 in order to maximise the
profit under the given hypotheses
...
Since profit is Rs 8 per bottle for M1 and Rs 7 per bottle
for M2, therefore the objective function (which is to be maximised) is given by
Z ✞ Z (x, y) = 8x + 7y
The objective function is to be maximised subject to the constraints (Refer Chapter
12 on Linear Programming)
x ✟ 20000
✠
✡
y ✟ 40000
✡
✡
x ☛ y ✟ 45000 ☞
3 x ☛ y ✟ 66000✡
✡
x ✌ 0, y ✌ 0
✡
✍
...
2
...
The co-ordinates of vertices O, P, Q, R, S and T are (0, 0), (20000, 0),
(20000, 6000), (10500, 34500), (5000, 40000) and (0, 40000), respectively
...
2
...
Hence, the manufacturer should produce 10500 bottles
of M1 medicine and 34500 bottles of M2 medicine in order to get maximum profit of
Rs 325500
...
Prepare a mathematical model to examine the profitability
...
264
MATHEMATICS
Step 2 Formulation: We are given that the costs are of two types: fixed and variable
...
g
...
g
...
Initially, we assume that the variable costs are directly proportional to the number of
units produced — this should simplify our model
...
For convenience,
we assume that all units produced are sold immediately
...
Then
C = a + bx
...
(2)
The profit P is then the difference between income and costs
...
(3)
We now have a mathematical model of the relationships (1) to (3) between
the variables x, C, I, P, a, b, s
...
Step 3 From (3), we can observe that for the break even point (i
...
, make neither profit
a
units
...
e
...
e
...
e
...
Further, if the break even point proves to be unrealistic, then another
model could be tried or the assumptions regarding cash flow may be modified
...
Thus, in order to gain profit, this should be positive and to get large gains, we need to
produce large quantity of the product and at the same time try to reduce the variable
cost
...
Brine containing 200 g of salt per litre flows into the tank at the rate of 25 litres per
minute and the mixture flows out at the same rate
...
What would be the amount of salt in the tank at
any time t?
Solution Step 1 The situation is easily identifiable
...
Further assume that y is a differentiable function
...
e
...
Now the inflow of brine brings salt into the tank at the rate of 5 kg per minute
(as 25 × 200 g = 5 kg) and the outflow of brine takes salt out of the tank at the rate of
y
✄ y ☎ y
✆
kg per minute (as at time t, the salt in the tank is
kg)
...
(1)
266
MATHEMATICS
This gives a mathematical model for the given problem
...
The solution of (1) is
given by
t
y e 40
t
200 e 40 ✁ C or y (t) = 200 + C e
✂ 40t
...
Note that when t = 0, y = 250
...
(3)
50
y ✝ 200
...
✆
t
Step 4 Since e 40 is always positive, from (3), we conclude that y > 200 at all times
Thus, the minimum amount of salt content in the tank is 200 kg
...
e
...
e
...
Limitations of Mathematical Modelling
Till today many mathematical models have been developed and applied successfully
to understand and get an insight into thousands of situations
...
are almost synonymous with mathematical modelling
...
The
reason behind this is that either the situation are found to be very complex or the
mathematical models formed are mathematically intractable
...
Due
to these fast and advanced computers, it has been possible to prepare more realistic
models which can obtain better agreements with observations
...
Infact, we can prepare reasonably accurate models to fit any data by choosing
five or six parameters / variables
...
Mathematical modelling of large or complex situations has its own special problems
...
Mathematical modellers from all disciplines —
mathematics, computer science, physics, engineering, social sciences, etc
...
—
—
CONSTITUTION OF INDIA
Part III (Articles 12 – 35)
(Subject to certain conditions, some exceptions
and reasonable restrictions)
guarantees these
Fundamental Rights
Right to Equality
⑨ before law and equal protection of laws;
⑨ irrespective of religion, race, caste, sex or place of birth;
⑨ of opportunity in public employment;
⑨ by abolition of untouchability and titles
...
Right against Exploitation
⑨ for prohibition of traffic in human beings and forced labour;
⑨ for prohibition of employment of children in hazardous jobs
...
Cultural and Educational Rights
⑨ for protection of interests of minorities to conserve their language, script and
culture;
⑨ for minorities to establish and administer educational institutions of their choice
...