Free NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability solved by Expert Teachers as per NCERT (CBSE) Book guidelines and brought to you by CBSE Learning. These Continuity and Differentiability Exercise Questions with Solutions for Class 12 Maths covers all questions of Chapter Continuity and Differentiability Class 12 and help you to revise complete Syllabus and Score More marks as per CBSE Board guidelines from the latest NCERT book for class 12 maths. You can read and download NCERT Book Solution to get a better understanding of all topics and concepts
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
Exercise 5.1 : Solutions of Questions on Page Number : 159
Q1 :Prove that the function
is 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 by
is 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, then
are 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 x
c, 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) = x2 are 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 is
Differentiating 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
find
in terms of y alone.
Answer :
It is given that,
Then,
Q13 :If
, show that
Answer :
It is given that,
Then,
Hence, proved.
Q14 :If
show 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 
and
is 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,
