Here is a NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability. This solutions covers all questions of Maths Chapter 2 Inverse Trigonometric Functions Class 12 as per CBSE Board guidelines from the latest NCERT book for class 12 maths. Following topics and sub-topics in Class 12 Maths Chapter 2 Inverse Trigonometric Functions are covered.
5.1 Introduction
5.2 Continuity
5.2.1 Algebra of continuous functions
5.3. Differentiability
5.3.1 Derivatives of composite functions
5.3.2 Derivatives of implicit functions
5.3.3 Derivatives of inverse trigonometric functions
5.4 Exponential and Logarithmic Functions
5.5. Logarithmic Differentiation
5.6 Derivatives of Functions in Parametric Forms
5.7 Second Order Derivative
5.8 Mean Value Theorem.

Continuity and Differentiability NCERT Solutions – Class 12 Maths

NCERT Solutions for Class 12 Maths Chapter 5

Exercise 5.1 : Solutions of Questions on Page Number : 159
Q1 :Prove that the functionis continuous at
Answer :

Therefore, f is continuous at x= 0

Therefore, f is continuous at x= – 3

Therefore, f is continuous at x= 5


Q2 :Examine the continuity of the function.
Answer :

Thus, f is continuous at x= 3


Q3 :Examine the following functions for continuity.
(a)
(b)

(c)
(d)

Answer :
(a) The given function is
It is evident that fis defined at every real number kand its value at k is k – 5.
It is also observed that,

Hence, f is continuous at every real number and therefore, it is a continuous function.
(b) The given function is
For any real number k≠ 5,
we obtain
Hence, f is continuous at every point in the domain of fand therefore, it is a continuous function.
(c) The given function is
For any real number c≠ – 5, we obtain
Hence, f is continuous at every point in the domain of fand therefore, it is a continuous function.
(d) The given function is
This function fis defined at all points of the real line.
Let cbe a point on a real line. Then, c< 5 or c= 5 or c > 5
Case I: c< 5
Then, f (c) = 5 – c
Therefore, fis continuous at all real numbers less than 5.
Case II :
c= 5
Then,


Therefore, f is continuous at x= 5
Case III: c> 5


Q4 :Prove that the function is continuous at x= n, where n is a positive integer.
Answer :
The given function is f(x) = xn
It is evident that fis defined at all positive integers, n, and its value at n is nn.
Therefore, f is continuous at n, where n is a positive integer.


Q5 :Is the function fdefined by
continuous at x= 0? At x= 1? At x= 2?
Answer :
The given function f is
At x= 0,
It is evident thatf is defined at 0 and its value at 0 is 0.

Therefore, f is continuous at x= 0
At x = 1,
f is defined at 1 and its value at 1 is 1.
The left hand limit of f at x = 1 is,
The right hand limit of f at x = 1 is,
Therefore,f is not continuous at x= 1
At x = 2,
f is defined at 2 and its value at 2 is 5.


Q6 :Find all points of discontinuity of f, where f is defined by
Answer :
The given function fis
It is evident that the given function fis defined at all the points of the real line.
Let c be a point on the real line. Then, three cases arise.
(i) c< 2 (ii) c> 2
(iii) c= 2
Case (i) c< 2
Therefore, f is continuous at all points x, such that x< 2
Case (ii) c> 2

Therefore, f is continuous at all points x, such that x> 2
Case (iii) c= 2

Then, the left hand limit of f atx = 2 is,
The right hand limit of fat x = 2 is,

It is observed that the left and right hand limit of fat x = 2 do not coincide.
Therefore, f is not continuous at x= 2
Hence, x = 2 is the only point of discontinuity of f.


Q7 :Find all points of discontinuity of f, where f is defined by

Answer :
The given function f is
The given function f is defined at all the points of the real line.
Let c be a point on the real line.
Case I:

Therefore, f is continuous at all points x, such that x< – 3
Case II:


Therefore, f is continuous at x= – 3
Case III:


Therefore, f is continuous in ( – 3, 3).
Case IV: If c= 3, then the left hand limit of f atx = 3 is,
The right hand limit of f atx = 3 is,
It is observed that the left and right hand limit of fat x = 3 do not coincide. Therefore, f is not continuous at x= 3
Case V:
Therefore, f is continuous at all points x, such that x> 3
Hence, x = 3 is the only point of discontinuity of f.


Q8 :Find all points of discontinuity of f, where f is defined by

Answer :
The given function f is
It is known that,
Therefore, the given function can be rewritten as

The given function fis defined at all the points of the real line.
Let c be a point on the real line.
Case I:

Therefore, f is continuous at all points x< 0
Case II:
If c= 0, then the left hand limit of f at x = 0 is,

The right hand limit of f at x = 0 is,

It is observed that the left and right hand limit of fat x = 0 do not coincide. Therefore, f is not continuous at x= 0
Case III:

Therefore, f is continuous at all points x, such that x> 0
Hence, x = 0 is the only point of discontinuity of f.


Q9 :Find all points of discontinuity of f, where f is defined by

Answer :
The given function fis

It is known that,
Therefore, the given function can be rewritten as

Let c be any real number. Then,

Also,
Therefore, the given function is a continuous function.
Hence, the given function has no point of discontinuity.


Q10 :Find all points of discontinuity of f, where f is defined by
Answer :
The given function fis

The given function fis defined at all the points of the real line.
Let c be a point on the real line.
Case I:

Therefore, f is continuous at all points x, such that x< 1
Case II:

The left hand limit of f atx = 1 is,

The right hand limit of f atx = 1 is,

Therefore, f is continuous at x= 1
Case III:


Therefore, f is continuous at all points x, such that x> 1
Hence,the given function f has no point of discontinuity.


Q11 : Find all points of discontinuity of f, where f is defined by

Answer :
The given function fis
The given function fis defined at all the points of the real line.
Let c be a point on the real line.
Case I:


Therefore, f is continuous at all points x, such that x< 2
Case II:


Therefore, f is continuous at x= 2
Case III:
Therefore, f is continuous at all points x, such that x> 2
Thus, the given function fis continuous at every point on the real line.
Hence, f has no point of discontinuity.


Q12 : Find all points of discontinuity of f, where f is defined by

Answer :
The given function fis
The given function fis defined at all the points of the real line.
Let c be a point on the real line.
Case I:

Therefore, f is continuous at all points x, such that x< 1
Case II:
If c= 1, then the left hand limit of fat x = 1 is,

The right hand limit of fat x = 1 is,

It is observed that the left and right hand limit of fat x = 1 do not coincide. Therefore, f is not continuous at x= 1
Case III:


Therefore, f is continuous at all points x, such that x> 1
Thus, from the above observation, it can be concluded that x= 1 is the only point of discontinuity of f.


Q13 : Is the function defined by

a continuous function?
Answer :
The given function is

The given function fis defined at all the points of the real line.
Let c be a point on the real line.
Case I:


Therefore, f is continuous at all points x, such that x< 1
Case II:

The left hand limit of f at x = 1 is,

The right hand limit of fat x = 1 is,

It is observed that the left and right hand limit of fat x = 1 do not coincide. Therefore, f is not continuous at x= 1
Case III:


Therefore, f is continuous at all points x, such that x> 1
Thus, from the above observation, it can be concluded that x= 1 is the only point of discontinuity of f.


Q14 : Discuss the continuity of the function f, where f is defined by
Answer :
The given function is
The given function is defined at all points of the interval [0, 10].
Let c be a point in the interval [0, 10].
Case I:

Therefore, f is continuous in the interval [0, 1).
Case II:

The left hand limit of f at x = 1 is,

The right hand limit of fat x = 1 is,

It is observed that the left and right hand limits of f at x= 1 do not coincide.
Therefore, f is not continuous at x= 1
Case III:

Therefore, f is continuous at all points of the interval (1, 3).
Case IV:

The left hand limit of f at x = 3 is,

The right hand limit of fat x = 3 is,

It is observed that the left and right hand limits of f at x= 3 do not coincide.
Therefore, f is not continuous at x= 3
Case V:

Therefore, f is continuous at all points of the interval (3, 10].
Hence, f is not continuous at x = 1 and x = 3


Q15 :Discuss the continuity of the function f, where f is defined by
Answer :
The given function is
The given function is defined at all points of the real line.
Let c be a point on the real line.
Case I:

Therefore, f is continuous at all points x, such that x < 0
Case II:

The left hand limit of f at x= 0 is,
The right hand limit of fat x = 0 is,

Therefore, f is continuous at x= 0
Case III:

Therefore, f is continuous at all points of the interval (0, 1).
Case IV:

The left hand limit of f at x = 1 is,
The right hand limit of fat x = 1 is,
It is observed that the left and right hand limits of f at x= 1 do not coincide. Therefore, f is not continuous at x= 1
Case V:

Therefore, f is continuous at all points x, such that x> 1
Hence, f is not continuous only at x = 1


Q16 :Discuss the continuity of the function f, where f is defined by
Answer :
The given function fis
The given function is defined at all points of the real line.
Let c be a point on the real line.
Case I:

Therefore, f is continuous at all points x, such that x < – 1
Case II:

The left hand limit of f at x = – 1 is,

The right hand limit of fat x = – 1 is,


Therefore, f is continuous at x= – 1
Case III:

Therefore, f is continuous at all points of the interval ( – 1, 1).
Case IV:

The left hand limit of f at x = 1 is,

The right hand limit of fat x = 1 is,


Therefore, f is continuous at x= 2
Case V:


Therefore, f is continuous at all points x, such that x> 1
Thus, from the above observations, it can be concluded that f is continuous at all points of the real line.


Q17 :Find the relationship between a and b so that the function f defined by
is continuous at x = 3.
Answer :
The given function fis
If f is continuous at x= 3, then

Therefore, from (1), we obtain

Therefore, the required relationship is given by,


Q18 :For what value of is the function defined by
continuous at x = 0? What about continuity at x= 1?
Answer :
The given function f is
If f is continuous at x= 0, then

Therefore, there is no value of λ for which fis continuous at x= 0
At x = 1,
f(1) = 4x + 1 = 4 × 1 + 1 = 5

Therefore, for any values of λ, f is continuous at x= 1


Q19 :Show that the function defined byis discontinuous at all integral point. Here denotes the greatest integer less than or equal to x.
Answer :
The given function is
It is evident that g is defined at all integral points.
Let n be an integer.
Then,
The left hand limit of f at x = n is,

The right hand limit of f at x = n is,

It is observed that the left and right hand limits of f at x = n do not coincide.
Therefore, f is not continuous at x= n
Hence, g is discontinuous at all integral points.


Q20 :Is the function defined by continuous at x = π?
Answer :
The given function is
It is evident that f is defined at x = π.


Therefore, the given function f is continuous at x = π


Q21 :Discuss the continuity of the following functions.
(a) f(x) = sin x + cos x
(b) f(x) = sin x – cos x
(c) f(x) = sin x x cos x
Answer :
It is known that if g and h are two continuous functions, thenare also continuous.
It has to proved first that g(x) = sin x and h(x) = cos x are continuous functions.
Let g (x) = sin x
It is evident that g(x) = sin x is defined for every real number.
Let c be a real number. Put x= c + h
If x → c, then h → 0

Therefore, g is a continuous function.
Let h (x) = cos x
It is evident that h(x) = cos x is defined for every real number.
Let c be a real number. Put x= c + h
If x → c, then h → 0
h (c) = cos c

Therefore, h is a continuous function.
Therefore, it can be concluded that
(a) f(x) = g(x) + h(x) = sin x + cos x is a continuous function
(b) f(x) = g(x) – h (x) = sin x – cos x is a continuous function
(c) f(x) = g(x) × h (x) = sin x ×cos x is a continuous function


Q22 :Discuss the continuity of the cosine, cosecant, secant and cotangent functions,
Answer :
It is known that if g and h are two continuous functions, then

It has to be proved first that g(x) = sin x and h(x) = cos x  are continuous functions.
Let g (x) = sin x
It is evident that g(x) = sin x is defined for every real number.
Let c be a real number. Put x= c+ h
If xc, then h
0

Therefore, g is a continuous function.
Let h(x) = cos x
It is evident that h(x) = cos xis defined for every real number.
Let c be a real number. Put x= c+ h
If x ® c, then h ® 0
h (c) = cos c

Therefore, h(x) = cos x is continuous function.
It can be concluded that,

Therefore, cosecant is continuous except at x = np, n î Z

Therefore, secant is continuous except at


Therefore, cotangent is continuous except at x = np, n î Z


Q23 :Find the points of discontinuity of f, where
Answer :
The given function f is
It is evident that fis defined at all points of the real line.
Let c be a real number.
Case I:

Therefore, f is continuous at all points x, such that x < 0
Case II:

Therefore, f is continuous
at all points x, such that x> 0
Case III:

The left hand limit of fat x = 0 is,

The right hand limit of fat x = 0 is,

Therefore, f is continuous at x= 0
From the above observations, it can be concluded that f is continuous at all points of the real line.
Thus, f has no point of discontinuity.


Q24 : Determine if f defined by
is a continuous function?
Answer :
The given function f is
It is evident that f is defined at all points of the real line.
Let c be a real number.
Case I:

Therefore, f is continuous at all points x ≠0
Case II:



Therefore, f is continuous at x= 0
From the above observations, it can be concluded that f is continuous at every point of the real line.
Thus, f is a continuous function.


Q25 :Examine the continuity of f, where f is defined by
Answer :
The given function f is
It is evident that f is defined at all points of the real line.
Let c be a real number.
Case I:

Therefore, f is continuous at all points x, such that x ≠0
Case II:


Therefore, f is continuous at x= 0
From the above observations, it can be concluded thatf is continuous at every point of the real line.
Thus, f is a continuous function.


Q26 :Find the values of k so that the function f is continuous at the indicated point.
Answer :
The given function f is
The given function f is continuous at, if f is defined at
and if the value of the f at
equals the limit of f at
.
It is evident that f is defined atand


Therefore, the required value of k is 6.


Q27 :Find the values of k so that the function fis continuous at the indicated point.
Answer :
The given function is
The given function fis continuous at x= 2, if fis defined at x= 2 and if the value of fat x = 2 equals the limit of fat x = 2
It is evident that f is defined at x= 2 and

Therefore, the required value of.


Q28 :Find the values of k so that the function fis continuous at the indicated point.
Answer :
The given function is
The given function fis continuous at x= p, if fis defined at x= p and if the value of fat x= pequals the limit of fat x= p
It is evident that f is defined at x= p and

Therefore, the required value of


Q29 :Find the values of k so that the function fis continuous at the indicated point.
Answer :
The given function f is
The given function fis continuous at x= 5, if fis defined at x= 5 and if the value of fat x = 5 equals the limit of fat x = 5
It is evident that f is defined at x= 5 and


Therefore, the required value of


Q30 :Find the values of aand b such that the function defined by
is a continuous function.
Answer :
The given function f is
It is evident that the given function fis defined at all points of the real line.
If fis a continuous function, then fis continuous at all real numbers.
In particular, fis continuous at x = 2 and x = 10
Since f is continuous at x = 2, we obtain

Since f is continuous at x = 10, we obtain

On subtracting equation (1) from equation (2), we obtain
8a= 16
⇒ a= 2
By putting a= 2 in equation (1), we obtain
2 ×2 + b = 5
⇒ 4 + b = 5
⇒ b= 1
Therefore, the values of aand b for which f is a continuous function are 2 and 1 respectively.


Q31 :Show that the function defined by f (x) = cos (x2) is a continuous function.
Answer :
The given function is f (x) = cos (x2)
This function fis defined for every real number and fcan be written as the composition of two functions as,
f= g o h, where g(x) = cos x and h(x) = x2
It has to be first proved that g (x) = cos xand h (x) = xare continuous functions.
It is evident that gis defined for every real number.
Let c be a real number.
Then, g (c) = cos c
Therefore, g (x) = cos x is continuous function.
h(x) = x2
Clearly, h is defined for every real number.
Let k be a real number, then h (k) = k2
Therefore, h is a continuous function.
It is known that for real valued functions g and h,such that (g o h) is defined at c, if g is continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.
Therefore, is a continuous function.


Q32 :Show that the function defined by is a continuous function.
Answer :
The given function is
This function fis defined for every real number and fcan be written as the composition of two functions as,
f= g o h, where
It has to be first proved that are continuous functions.
Clearly, g is defined for all real numbers.
Let c be a real number.
Case I:
Therefore, g is continuous at all points x, such that x < 0
Case II:
Therefore, g is continuous at all points x, such that x> 0
Case III:
Therefore, g is continuous at x= 0
From the above three observations, it can be concluded that gis continuous at all points.
h (x) = cos x
It is evident that h(x) = cos x is defined for every real number.
Let c be a real number. Put x= c + h
If x → c, then h →0
h (c) = cos c
Therefore, h (x) = cos x is a continuous function.
It is known that for real valued functions g and h,such that (g o h) is defined at c, if g is continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.
Therefore, is a continuous function.


Q33 :Examine that is a continuous function.
Answer :
This function fis defined for every real number and  fcan be written as the composition of two functions as,
f= g o h, where
It has to be proved first that are continuous functions.
Clearly, g is defined for all real numbers.
Let c be a real number.
Case I:
Therefore, g is continuous at all points x, such that x < 0
Case II:
Therefore, g is continuous at all points x, such that x> 0
Case III:
Therefore, g is continuous at x= 0
From the above three observations, it can be concluded that gis continuous at all points.
h (x) = sin x
It is evident that h(x) = sinx is defined for every real number.
Let c be a real number. Put x= c + k
If x→ c, then k →0
h (c) = sin c
Therefore, h is a continuous function.
It is known that for real valued functions g and h,such that (g o h) is defined at c, if g is continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.
Therefore, is a continuous function.


Q34 :Find all the points of discontinuity of f defined by.
Answer :
The given function is
The two functions, gand h, are defined as
Then, f = g – h
The continuity of g and h is examined first.

Clearly, g is defined for all real numbers.
Let c be a real number.
Case I:
Therefore, g is continuous at all points x, such that x < 0
Case II:
Therefore, g is continuous at all points x, such that x> 0
Case III:

Therefore, g is continuous at x= 0
From the above three observations, it can be concluded that gis continuous at all points.
Clearly, h is defined for every real number.
Let c be a real number.
Case I:
Therefore, h is continuous at all points x, such that x < – 1
Case II:
Therefore, h is
continuous at all points x, such that x> – 1
Case III:
Therefore, h is continuous at x= – 1
From the above three observations, it can be concluded that his continuous at all points of the real line.


Exercise 5.2 : Solutions of Questions on Page Number : 166


Q1 :Differentiate the functions with respect to x.


Q2 :Differentiate the functions with respect to x.
Answer :

Thus, f is a composite function of two functions.
Put t = u (x) = sin x
By chain rule,
Alternate method


Q3 :Differentiate the functions with respect to x.
Answer :
Thus, f is a composite function of two functions, u and v.
Put t = u (x) = ax + b

Hence, by chain rule, we obtain

Alternate method


Q4 :Differentiate the functions with respect to x.
Answer :
Thus, f is a composite function of three functions, u, v, and w.

Hence, by chain rule, we obtain

Alternate method


Q5 :Differentiate the functions with respect to x.
Answer :
The given function
is, where g (x) = sin (ax + b) and
h (x) = cos (cx + d)

∴ g is a composite function of two functions, u and v.

Therefore, by chain rule, we obtain
∴h is a composite function of two functions, p and q.
Put y = p (x) = cx + d
Therefor, by chain rule, we obtain


Q6 :Differentiate the functions with respect to x.
Answer :
The given function is.


Q7 :Differentiate the functions with respect to x.
Answer :


Q8 :Differentiate the functions with respect to x.
Answer :
Clearly, f is a composite function of two functions, u and v, such that
By using chain rule, we obtain
Alternate method


Q9 :Prove that the function f given by
is not differentiable at x = 1.
Answer :
The given function is

It is known that a function f is differentiable at a point x = c in its domain if both

are finite and equal.
To check the differentiability of the given function at x = 1,
consider the left hand limit of f at x = 1

Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1


Q10 :Prove that the greatest integer function defined by is not
differentiable at x = 1 and x = 2.
Answer :
The given function f is

It is known that a function f is differentiable at a point x = c in its domain if both

are finite and equal.
To check the differentiability of the given function at x = 1, consider the left hand limit of f at x = 1


Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at
x = 1
To check the differentiability of the given function at x = 2, consider the left hand limit
of f at x = 2

Since the left and right hand limits of f at x = 2 are not equal, f is not differentiable at x = 2


Exercise 5.3 : Solutions of Questions on Page Number : 169


Q1 :Find :
Answer :
The given relationship is
Differentiating this relationship with respect to x, we obtain


Q2 :Find :
Answer :
The given relationship isDifferentiating this relationship with respect to x, we obtain


Q3 :Find :
Answer :
The given relationship is
Differentiating this relationship with respect to x, we obtain

Using chain rule, we obtain
and

From (1) and (2), we obtain


Q4 :Find :
Answer :
The given relationship is

Differentiating this relationship with respect to x, we obtain


Q5 :Find :
Answer :
The given relationship is

Differentiating this relationship with respect to x, we obtain
[Derivative of constant function is 0]


Q6 :Find :
Answer :
The given relationship is Differentiating this relationship with respect to x, we obtain


Q7 :Find :
Answer :
The given relationship is
Differentiating this relationship with respect to x, we obtain

Using chain rule, we obtain

From (1), (2), and (3), we obtain


Q8 :Find :
Answer :
The given relationship is
Differentiating this relationship with respect to x, we obtain


Q9 :Find :
Answer :
The given relationship is
Differentiating this relationship with respect to x, we obtain

The function,, is of the form of .
Therefore, by quotient rule, we obtain
Also,
From (1), (2), and (3), we obtain


Q10 :Find :
Answer :
The given relationship is
It is known that,
Comparing equations (1) and (2), we obtain

Differentiating this relationship with respect to x, we obtain


Q11 :Find :
Answer :
The given relationship is,

On comparing L.H.S. and R.H.S. of the above relationship, we obtain

Differentiating this relationship with respect to x, we obtain.


Q12 :Find :
Answer :
The given relationship is



Differentiating this relationship with respect to x, we obtain

Using chain rule, we obtain

From (1), (2), and (3), we obtain

Alternate method



Differentiating this relationship with respect to x, we obtain


Q13 :Find :
Answer :
The given relationship is

Differentiating this relationship with respect to x, we obtain


Q14 :
Find :
Answer :
The given relationship is

Differentiating this relationship with respect to x, we obtain


Q15 :Find :
Answer :
The given relationship is

Differentiating this relationship with respect to x, we obtain


Exercise 5.4 : Solutions of Questions on Page Number : 174


Q1 :Differentiate the following w.r.t. x:
Answer :
Let
By using the quotient rule, we obtain


Q2 :Differentiate the following w.r.t. x:
Answer :
Let

By using the chain rule, we obtain


Q3 :Differentiate the following w.r.t. x:
Answer :
Let
By using the chain rule, we obtain


Q4 :Differentiate the following w.r.t. x:
Answer :
Let
By using the chain rule, we obtain


Q5 :Differentiate the following w.r.t. x:
Answer :
Let
By using the chain rule, we obtain


Q6 :Differentiate the following w.r.t. x:
Answer :


Q7 :Differentiate the following w.r.t. x:

Answer :
Let
Then,
By differentiating this relationship with respect to x, we obtain


Q8 :Differentiate the following w.r.t. x:
Answer :
Let
By using the chain rule, we obtain

, x > 1


Q9 :Differentiate the following w.r.t. x:
Answer :
Let
By using the quotient rule, we obtain


Q10 :Differentiate the following w.r.t. x:
Answer :
Let
By using the chain rule, we obtain


Exercise 5.5 : Solutions of Questions on Page Number : 178


Q1 :Differentiate the function with respect to x.
Answer :
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect to x, we obtain


Q2 :Differentiate the function with respect to x.

Answer :

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain


Q3 :Differentiate the function with respect to x.
Answer :
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect to x, we obtain


Q4 :Differentiate the function with respect to x.
Answer :
u = xx
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect to x, we obtain

v = 2sin x
Taking logarithm on both the sides with respect to x, we obtain
Differentiating both sides with respect to x, we obtain


Q5 :Differentiate the function with respect to x.
Answer :
Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain


Q6 :Differentiate the function with respect to x.
Answer :
Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain


Q7 :Differentiate the function with respect to x.
Answer :
u = (log x)x

Differentiating both sides with respect to x, we obtain

Differentiating both sides with respect to x, we obtain

Therefore, from (1), (2), and (3), we obtain


Q8 :Differentiate the function with respect to x.
Answer :

Differentiating both sides with respect to x, we obtain

Therefore, from (1), (2), and (3), we obtain


Q9 :Differentiate the function with respect to x.
Answer :

Differentiating both sides with respect to x, we obtain

Differentiating both sides with respect to x, we obtain

From (1), (2), and (3), we obtain


Q10 :Differentiate the function with respect to x.

Answer :

Differentiating both sides with respect to x, we obtain

Differentiating both sides with respect to x, we obtain

From (1), (2), and (3), we obtain


Q11 :Differentiate the function with respect to x.
Answer :

Differentiating both sides with respect to x, we obtain

Differentiating both sides with respect to x, we obtain

From (1), (2), and (3), we obtain


Q12 :Find of function.
Answer :
The given function is
Let xy = u and yx = v
Then, the function becomes u + v = 1

Differentiating both sides with respect to x, we obtain

Differentiating both sides with respect to x, we obtain

From (1), (2), and (3), we obtain


Q13 :Find of function.
Answer :
The given function is
Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain


Q14 :Find of function.

Answer :
The given function is
Taking logarithm on both the sides, we obtain

Differentiating both sides, we obtain


Q15 :Find of function.
Answer :
The given function is
Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain


Q16 :Find the derivative of the function given by
and hence find
.
Answer :
The given relationship is

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain


Q17 :Differentiate in three ways mentioned below
(i) By using product rule.
(ii) By expanding the product to obtain a single polynomial.
(iii By logarithmic differentiation.
Do they all give the same answer?
Answer :

(i)
(ii)
(iii)
Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

From the above three observations, it can be concluded that all the results of are same.


Q18 :If u, v and w are functions of x, then show that
in two ways-first by repeated application of product rule, second by logarithmic differentiation.
Answer :
Let
By applying product rule, we obtain

By taking logarithm on both sides of the equation, we obtain

Differentiating both sides with respect to x, we obtain


Exercise 5.6 : Solutions of Questions on Page Number : 181


Q1 :If x and y are connected parametrically by the equation, without eliminating the parameter, find.

Answer :
The given equations are


Q2 :If x and y are connected parametrically by the equation, without eliminating the parameter, find.
x = a cos θ, y = b cos θ
Answer :
The given equations arex = a cos θ and y = b cos θ


Q3 :If x and y are connected parametrically by the equation, without eliminating the parameter, find.
x = sin t, y = cos 2t
Answer :
The given equations are x = sin t and y = cos 2t


Q4 :If x and y are connected parametrically by the equation, without eliminating the parameter, find.
Answer :
The given equations are


Q5 :If x and y are connected parametrically by the equation, without eliminating the parameter, find.
Answer :
The given equations are


Q6 :If x and y are connected parametrically by the equation, without eliminating the parameter, find.
Answer :
The given equations are


Q7 :If x and y are connected parametrically by the equation, without eliminating the parameter, find.
Answer :
The given equations are



Q8 :If x and y are connected parametrically by the equation, without eliminating the parameter, find.
Answer :
The given equations are


Q9 :If x and y are connected parametrically by the equation, without eliminating the parameter, find.
Answer :
The given equations are


Q10 :If x and y are connected parametrically by the equation, without eliminating the parameter, find.
Answer :
The given equations are


Q11 :If
Answer :
The given equations are



Hence, proved.


Exercise 5.7 : Solutions of Questions on Page Number : 183


Q1 :Find the second order derivatives of the function.
Answer :
Let
Then,


Q2 :Find the second order derivatives of the function.
Answer :
Let
Then,


Q3 :Find the second order derivatives of the function.
Answer :
Let
Then,


Q4 :Find the second order derivatives of the function.
Answer :
Let
Then,


Q5 :Find the second order derivatives of the function.
Answer :
Let
Then,


Q6 :Find the second order derivatives of the function.
Answer :
Let
Then,


Q7 :Find the second order derivatives of the function.
Answer :
Let
Then,


Q8 :Find the second order derivatives of the function.
Answer :
Let
Then,


Q9 :Find the second order derivatives of the function.
Answer :
Let
Then,


Q10 :Find the second order derivatives of the function.
Answer :
Let
Then,


Q11 :If, prove that
Answer :
It is given that,
Then,
Hence, proved.


Q12 :If findin terms of y alone.
Answer :
It is given that,
Then,


Q13 :If, show that
Answer :
It is given that,
Then,
Hence, proved.


Q14 :Ifshow that
Answer :
It is given that,
Then,
Hence, proved.


Q15 :If, show that
Answer :
It is given that,
Then,
Hence, proved.


Q16 :If, show that
Answer :
The given relationship is
Taking logarithm on both the sides, we obtain
Differentiating this relationship with respect to x, we obtain
Hence, proved.


Q17 :If, show that
Answer :
The given relationship is
Then,
Hence, proved.


Exercise 5.8 : Solutions of Questions on Page Number : 186


Q1 :Verify Rolle’s Theorem for the function
Answer :
The given function,, being a polynomial function, is continuous in [ – 4, 2] and is differentiable in ( – 4, 2).
∴ f ( – 4) = f (2) = 0
⇒ The value of f (x) at – 4 and 2 coincides.
Rolle’s Theorem states that there is a point c ∈ ( – 4, 2) such that
Hence, Rolle’s Theorem is verified for the given function.


Q2 :Examine if Rolle’s Theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle’s Theorem from these examples?
(i)
(ii)
(iii)
Answer :
By Rolle’s Theorem, for a function, if
(a) f is continuous on [a, b]
(b) f is differentiable on (a, b)
(c) f (a) = f (b)
then, there exists some c ∈ (a, b) such that
Therefore, Rolle’s Theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.
(i)
It is evident that the given function f (x) is not continuous at every integral point.
In particular, f(x) is not continuous at x = 5 and x = 9
⇒ f (x) is not continuous in [5, 9].
The differentiability of f in (5, 9) is checked as follows.
Let n be an integer such that n ∈ (5, 9).
Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n
∴f is not differentiable in (5, 9).
It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s Theorem.
Hence, Rolle’s Theorem is not applicable for.
(ii)It is evident that the given function f (x) is not continuous at every integral point.
In particular, f(x) is not continuous at x = – 2 and x = 2
⇒ f (x) is not continuous in [ – 2, 2].
The differentiability of f in ( – 2, 2) is checked as follows.
Let n be an integer such that n ∈ ( – 2, 2).
Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n∴f is not differentiable in ( – 2, 2).
It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s Theorem.
Hence, Rolle’s Theorem is not applicable for.
(iii)
It is evident that f, being a polynomial function, is continuous in [1, 2] and is differentiable in (1, 2).
∴f (1) ≠ f (2)
It is observed that f does not satisfy a condition of the hypothesis of Rolle’s Theorem.
Hence, Rolle’s Theorem is not applicable for.


Q3 :If is a differentiable function and if does not vanish anywhere, then prove that.
Answer :
It is given that is a differentiable function.
Since every differentiable function is a continuous function, we obtain
(a) f is continuous on [ – 5, 5].
(b) f is differentiable on ( – 5, 5).
Therefore, by the Mean Value Theorem, there exists c ∈ ( – 5, 5) such that
It is also given that does not vanish anywhere.
Hence, proved.


Q4 :Verify Mean Value Theorem, if in the interval, where and.
Answer :
The given function is
f, being a polynomial function, is continuous in [1, 4] and is differentiable in (1, 4) whose derivative is 2x – 4.
Mean Value Theorem states that there is a point c ∈ (1, 4) such that
Hence, Mean Value Theorem is verified for the given function.


Q5 :Verify Mean Value Theorem, if in the interval [a, b], where a = 1 and b = 3. Find all for which
Answer :
The given function f is
f, being a polynomial function, is continuous in [1, 3] and is differentiable in (1, 3) whose derivative is 3×2 – 10x – 3.
Mean Value Theorem states that there exist a point c ∈ (1, 3) such that
Hence, Mean Value Theorem is verified for the given function andis the only point for which


Q6 :Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.
Answer :
Mean Value Theorem states that for a function, if
(a) f is continuous on [a, b]
(b) f is differentiable on (a, b)
then, there exists some c ∈ (a, b) such that
Therefore, Mean Value Theorem is not applicable to those functions that do not satisfy any of the two conditions of the hypothesis.
(i)
It is evident that the given function f (x) is not continuous at every integral point.
In particular, f(x) is not continuous at x = 5 and x = 9
⇒ f (x) is not continuous in [5, 9].
The differentiability of f in (5, 9) is checked as follows.
Let n be an integer such that n ∈ (5, 9).
Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n
∴f is not differentiable in (5, 9).
It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value Theorem.
Hence, Mean Value Theorem is not applicable for .
(ii)
It is evident that the given function f (x) is not continuous at every integral point.
In particular, f(x) is not continuous at x = – 2 and x = 2
⇒ f (x) is not continuous in [ – 2, 2].
The differentiability of f in ( – 2, 2) is checked as follows.
Let n be an integer such that n ∈ ( – 2, 2).
Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n
∴f is not differentiable in ( – 2, 2).
It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value Theorem.
Hence, Mean Value Theorem is not applicable for .
(iii)
It is evident that f, being a polynomial function, is continuous in [1, 2] and is differentiable in (1, 2).
It is observed that f satisfies all the conditions of the hypothesis of Mean Value Theorem.
Hence, Mean Value Theorem is applicable for.
It can be proved as follows.


Exercise Miscellaneous : Solutions of Questions on Page Number : 191


Q1 :
Answer :
Using chain rule, we obtain


Q2 :
Answer :


Q3 :
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect to x, we obtain


Q4 :
Answer :
Using chain rule, we obtain


Q5 :
Answer :


Q6 :
Answer :
Therefore, equation (1) becomes


Q7 :
Answer :
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect to x, we obtain


Q8 :, for some constant a and b.
Answer :
By using chain rule, we obtain


Q9 :
Answere :
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect to x, we obtain


Q10 :
, for some fixed and
Answer :
Differentiating both sides with respect to x, we obtain
Differentiating both sides with respect to x, we obtain
s= aa
Since a is constant, aais also a constant.

From (1), (2), (3), (4), and (5), we obtain


Q11 :
, for
Answer :
Differentiating both sides with respect to x,we obtain
Differentiating with respect to x, we obtain
Also,
Differentiating both sides with respect to x, we obtain
Substituting the expressions of in equation (1), we obtain


Q12 :Find, if
Answer :


Q13 :Find, if
Answer :


Q14 :If, for, – 1 < x<1, prove that
Answer :
It is given that,
Differentiating both sides with respect to x, we obtain
Hence, proved.


Q15 :If, for some prove that
is a constant independent of a and b.
Answer :
It is given that,
Differentiating both sides with respect to x, we obtain
Hence, proved.


Q16 :If with prove that
Answer :
Hence, proved.


Q17 :If and, find
Answer :


Q18 :If, show that exists for all real x, and find it.
Answer :
It is known that,
Therefore, when x ≥0,
In this case, and hence,
When x < 0,
In this case, and hence,
Thus, for, exists for all real xand is given by,


Q19 :Using mathematical induction prove that for all positive integers n.
Answer :
For n= 1,
∴P(n) is true for n= 1
Let P(k) is true for some positive integer k.
That is,
It has to be proved that P(k+ 1) is also true.
Thus, P(k + 1) is true whenever P (k) is true.
Therefore, by the principle of mathematical induction, the statement P(n) is true for every positive integer n.
Hence, proved.


Q20 :Using the fact that sin (A + B) = sin Acos B + cos A sin B and the differentiation, obtain the sum formula for cosines.
Answer :
Differentiating both sides with respect to x, we obtain


Q21 : Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer ?
Answer :


Q22 :If, prove that
Answer :
Thus,


Q23 :If, show that
Answer :
It is given that,