# NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability: In CBSE class 11 you have already learnt the differentiation of certain functions like polynomial functions and trigonometric functions. In this article, you will get NCERT solutions for class 12 maths chapter 5 continuity and differentiability. Questions based on the topics like continuity, differentiability, and relations between them are covered in the solutions of NCERT for class 12 maths chapter 5 continuity and differentiability. In the NCERT textbook, there are 48 solved examples to understand the concepts. First, try to solve examples and understand the concept, then it will be easy to solve NCERT exercise questions. If you are finding difficulties in solving them, you can take help from CBSE NCERT solutions for class 12 maths chapter 5 continuity and differentiability article. Here, you will also learn the differentiation of inverse trigonometric functions, exponential functions, and logarithmic functions. Check all NCERT solutions at a single place which will help you to learn CBSE science and maths.

This chapter "continuity and differentiability" is a continuation of the differentiation of functions that you have already learnt in NCERT class XI. This chapter alone has 9% weightage in the 12th board final examination and the next chapters of calculus(44 % weightage in the final exam) also depend on the concepts of this chapter. This chapter is very important for the exam point of view as well as the application point of view. To get command on this chapter you should solve the miscellaneous exercise also. In this solutions of NCERT for class 12 maths chapter 5 continuity and differentiability article, you will get solutions for miscellaneous exercise too. This chapter contains 8 exercises with 121 questions. You will find all these CBSE NCERT solutions for class 12 maths chapter 5 continuity and differentiability explained in a detailed manner.

The mathematical definition of Continuity and Differentiability -

Let f be a real function and c be a point in the domain of f. Then f is continuous at c if. A function f is differentiable at a point c in its domain if it is continuous at point c. A function is said to be differentiable in an interval [a, b] if it is differentiable at every point of [a, b]. NCERT class 12 maths chapter 5 continuity and differentiability is one of the very important and time-consuming chapters. It contains 8 exercises with 121 questions and also 23 questions in the miscellaneous exercise. In this article, you will find all solutions of NCERT for class 12 maths chapter 5 continuity and differentiability including miscellaneous exercise.

## Topics of NCERT class 12 maths chapter 5 Continuity and Differentiability

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 Form

## Solutions of NCERT for class 12 maths chapter 5 Continuity and Differentiability Excercise: 5.1

Given function is

Hence, function is continous at x = 0

Hence, function is continous at x = -3

Hence, function is continuous at x = 5

Question:2.Examine the continuity of the function

Given function is

at x = 3

Hence, function is continous at x = 3

Given function is

Our function is defined for every real number say k
and value at x = k ,
and also,

Hence, the function  is continuous  at every real number

Question:3 b) Examine the following functions for continuity.

Given function is

For every real number  k ,
We get,

Hence, function   continuous for every real value of x,

Question:3 c) Examine the following functions for continuity.

Given function is

For every real number  k ,
We gwt,

Hence, function   continuous for every real value of x ,

Question:3 d) Examine the following functions for continuity.

Given function is

for x > 5  , f(x) =  x - 5
for x < 5 , f(x) = 5 - x
SO, different cases are their
case(i)   x > 5
for every real number k > 5   , f(x) = x - 5 is defined

Hence, function  f(x) = x - 5 is continous for x > 5

case (ii)      x < 5
for every real number k < 5   , f(x) = 5 - x is defined

Hence, function  f(x) = 5 - x is continous for x < 5

case(iii)    x = 5
for x  = 5   , f(x) = x - 5 is defined

Hence, function  f(x) = x - 5 is continous for x = 5

Hence, the function  is continuous for each and every real number

GIven function is

the function  is defined for all positive integer, n

Hence,  the function is continuous at x = n, where n is a positive integer

Given function is

function is defined at x = 0 and its value is 0

Hence , given function is continous at x = 0

given function  is defined for x = 1
Now, for x = 1 Right-hand limit and left-hand limit are not equal

R.H.L  L.H.L.
Therefore, given function is not continous at x =1
Given function is defined for x = 2 and its value at x = 2 is 5

Hence, given function is continous at x = 2

Given function is

given function is defined for every real number k
There are different cases for the given function
case(i)   k > 2

Hence, given function is continuous for each value of k > 2

case(ii)   k < 2

Hence, given function is continuous for each value of k < 2

case(iii)  x = 2

Right hand limit at x= 2  Left hand limit at x = 2
Therefore, x = 2 is the point of discontinuity

Given function is

GIven function is defined for every real number k
Different cases are their
case (i)  k < -3

Hence, given function is continuous for every value of k < -3

case(ii)    k = -3

Hence, given function is continous for x = -3

case(iii)   -3 < k < 3

Hence, for every value of k in -3 < k < 3 given function is continous

case(iv)  k = 3

Hence. x = 3 is the point of discontinuity

case(v)  k > 3

Hence, given function is continuous for each and every value of k > 3

Given function is

if x > 0 ,
if x < 0 ,
given function is defined for every real number k
Now,
case(i) k < 0

Hence, given function is continuous for every value of k < 0
case(ii)  k > 0

Hence, given function is continuous for every value of k > 0
case(iii)  x = 0

Hence, 0 is the only point of discontinuity

Given function is

if x < 0 ,
Now, for any value of x, the value of our function is -1
Therefore, the given function is continuous for each and every value of x
Hence, no point of discontinuity

Given function is

given function is defined for every real number k
There are different cases for the given function
case(i)   k > 1

Hence, given function is continuous for each value of k > 1

case(ii)   k < 1

Hence, given function is continuous for each value of k < 1

case(iii)  x = 1

Hence, at x = 2 given function is continuous
Therefore, no point of discontinuity

Given function is

given function is defined for every real number k
There are different cases for the given function
case(i)   k > 2

Hence, given function is continuous for each value of k > 2

case(ii)   k < 2

Hence, given function is continuous for each value of k < 2

case(iii)  x = 2

Hence, given function is continuous at x = 2
There, no point of discontinuity

Given function is

given function is defined for every real number k
There are different cases for the given function
case(i)   k > 1

Hence, given function is continuous for each value of k > 1

case(ii)   k < 1

Hence, given function is continuous for each value of k < 1

case(iii)  x = 1

Hence, x = 1  is the point of discontinuity

Question:13. Is the function defined by

a continuous function?

Given function is

given function is defined for every real number k
There are different cases for the given function
case(i)   k > 1

Hence, given function is continuous for each value of k > 1

case(ii)   k < 1

Hence, given function is continuous for each value of k < 1

case(iii)  x = 1

Hence, x = 1 is the point of discontinuity

Given function is

GIven function is defined for every real number k
Different cases are their
case (i)  k < 1

Hence, given function is continous for every value of k < 1

case(ii)    k = 1

Hence, given function is discontinous at x = 1
Therefore, x = 1 is he point od discontinuity

case(iii)   1 < k < 3

Hence, for every value of k in 1 < k < 3 given function is continous

case(iv)  k = 3

Hence. x = 3 is the point of discontinuity

case(v)  k > 3

Hence, given function is continous for each and every value of k > 3
case(vi)  when k < 3

Hence, for every value of k in k < 3 given function is continous

Given function is

Given function is satisfies for the all real values of x
case (i)  k < 0

Hence, function is continuous for all values of x < 0

case (ii)  x = 0

L.H.L at x= 0

R.H.L. at x = 0

L.H.L. = R.H.L. = f(0)
Hence, function is continuous at x = 0

case (iii)  k > 0

Hence , function is continuous for all values of x > 0

case (iv) k < 1

Hence , function is continuous for all values of x < 1

case (v)  k > 1

Hence , function is continuous for all values of x > 1

case (vi)  x = 1

Hence, function is not continuous at x = 1

Given function is

GIven function is defined for every real number k
Different cases are their
case (i)  k < -1

Hence, given function is continuous for every value of k < -1

case(ii)    k = -1

Hence, given function is continous at x = -1

case(iii)  k > -1

Hence, given function is continous for all values of  x > -1

case(vi)   -1 < k < 1

Hence, for every value of k in -1 < k < 1 given function is continous

case(v)  k = 1

Hence.at  x =1 function is continous

case(vi)  k > 1

Hence, given function is continous for each and every value of k > 1
case(vii)  when k < 1

Hence, for every value of k in k < 1 given function is continuous

Therefore, continuous at all points

Given function is

For the function to be continuous at x = 3 , R.H.L. must be equal to L.H.L.

For the function to be continuous

Given function is

For the function to be continuous at x = 0 , R.H.L. must be equal to L.H.L.

For the function to be continuous

Hence, for no value of function is continuous at x = 0

For x  = 1

Hence, given function is continuous at x =1

Given function is

Given is defined for all real numbers k

Hence, by this, we can say that  the function defined by is discontinuous at all integral points

Question:20. Is the function defined by continuous at x = ?

Given function is

Clearly, Given function is defined at x =

Hence, the function defined by continuous at x =

Given function is

Given function is defined for all real number
We, know that if two function g(x) and h(x) are continuous then g(x)+h(x) , g(x)-h(x) , g(x).h(x) allare continuous
Lets take g(x) = sin x   and    h(x) = cos x
Let suppose  x = c + h
if

Hence, function  is a continuous function
Now,
h(x) = cos x
Let suppose  x = c + h
if

Hence, function  is a continuous function
We proved independently that sin x and cos x is continous function
So, we can say that
f(x) = g(x) + h(x) = sin x + cos x is also a continuous function

Question:21. b) Discuss the continuity of the following functions:

Given function is

Given function is defined for all real number
We, know that if two function g(x) and h(x) are continuous then g(x)+h(x) , g(x)-h(x) , g(x).h(x) allare continuous
Lets take g(x) = sin x   and    h(x) = cos x
Let suppose  x = c + h
if

Hence, function  is a continuous function
Now,
h(x) = cos x
Let suppose  x = c + h
if

Hence, function  is a continuous function
We proved independently that sin x and cos x is continous function
So, we can say that
f(x) = g(x) - h(x) = sin x - cos x is also a continuous function

Question:21 c) Discuss the continuity of the following functions:

Given function is

Given function is defined for all real number
We, know that if two function g(x) and h(x) are continuous then g(x)+h(x) , g(x)-h(x) , g(x).h(x) allare continuous
Lets take g(x) = sin x   and    h(x) = cos x
Let suppose  x = c + h
if

Hence, function  is a continuous function
Now,
h(x) = cos x
Let suppose  x = c + h
if

Hence, function  is a continuous function
We proved independently that sin x and cos x is continous function
So, we can say that
f(x) = g(x).h(x) = sin x .cos x is also a continuous function

We, know that if two function g(x) and h(x) are continuous then

Lets take g(x) = sin x   and    h(x) = cos x
Let suppose  x = c + h
if

Hence, function  is a continuous function
Now,
h(x) = cos x
Let suppose  x = c + h
if

Hence, the function  is a continuous function
We proved independently that sin x and cos x is a continous function
So, we can say that
cosec x =   is also continuous except at
sec x  =   is also continuous except at
cot x =   is also continuous except at

Question:23. Find all points of discontinuity of f, where

Given function is

Hence, the function is continuous
Therefore, no point of discontinuity

Given function is

Given function is defined for all real numbers k
when x = 0

Hence, function is continuous at x = 0
when

Hence, the given function is continuous for all points

Given function is

Given function is defined for all real number
We, know that if two function g(x) and h(x) are continuous then g(x)+h(x) , g(x)-h(x) , g(x).h(x) allare continuous
Lets take g(x) = sin x   and    h(x) = cos x
Let suppose  x = c + h
if

Hence, function  is a continuous function
Now,
h(x) = cos x
Let suppose  x = c + h
if

Hence, function  is a continuous function
We proved independently that sin x and cos x is continous function
So, we can say that
f(x) = g(x) - h(x) = sin x - cos x is also a continuous function

When x = 0

Hence, function is also continuous at x = 0

Given function is

When

For the function to be continuous

Therefore, the values of k so that the function f is continuous is 6

Given function is

When x = 2
For the function to be continuous
f(2) = R.H.L. = LH.L.

Hence,  the values of k so that the function f is continuous at x= 2 is

Given function is

When x =
For the function to be continuous
f() = R.H.L. = LH.L.

Hence,  the values of k so that the function f is continuous at x=  is

Given function is

When x = 5
For the function to be continuous
f(5) = R.H.L. = LH.L.

Hence,  the values of k so that the function f is continuous at x= 5 is

Given continuous function is

The function is continuous so

By solving  equation (i) and (ii)
a = 2 and b = 1
Hence, values of a and b such that the function defined by          is a continuous function is 2 and 1 respectively

Given function is

given function is defined for all real values of x
Let x = k + h
if

Hence, the function   is a continuous function

Given function is

given function is defined for all values of x
f = g o h ,  g(x) = |x| and h(x) = cos x
Now,

g(x) is defined for all real numbers k
case(i)  k < 0

Hence, g(x) is continuous when k < 0

case (ii) k > 0

Hence, g(x) is continuous when k > 0

case (iii) k = 0

Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x
Now,
h(x) = cos x
Let suppose  x = c + h
if

Hence, function  is a continuous function
g(x) is continuous , h(x) is continuous
Therefore, f(x) = g o h is also continuous

Question:33. Examine that sin | x| is a continuous function.

Given function is
f(x) = sin |x|
f(x) = h o g  , h(x) = sin x and g(x) = |x|
Now,

g(x) is defined for all real numbers k
case(i)  k < 0

Hence, g(x) is continuous when k < 0

case (ii) k > 0

Hence, g(x) is continuous when k > 0

case (iii) k = 0

Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x
Now,
h(x) = sin x
Let suppose  x = c + h
if

Hence, function  is a continuous function
g(x) is continuous , h(x) is continuous
Therefore, f(x) = h o g is also continuous

Given function is

Let g(x) = |x|  and h(x)  = |x+1|
Now,

g(x) is defined for all real numbers k
case(i)  k < 0

Hence, g(x) is continuous when k < 0

case (ii) k > 0

Hence, g(x) is continuous when k > 0

case (iii) k = 0

Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x

Now,

g(x) is defined for all real numbers k
case(i)  k < -1

Hence, h(x) is continuous when k < -1

case (ii) k > -1

Hence, h(x) is continuous when k > -1

case (iii) k = -1

Hence, h(x) is continuous when k = -1
Therefore, h(x) = |x+1| is continuous for all real values of x
g(x) is continuous and h(x) is continuous
Therefore, f(x) = g(x) - h(x) = |x| - |x+1| is also continuous

## CBSE NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability Excercise: 5.2

Given function is

when we differentiate it w.r.t. x.
Lets take  . then,

(By  chain rule)

Now,

Given function is

Lets take   then,

( By chain rule)

Now,

Given function is

when we differentiate it w.r.t. x.
Lets take  . then,

(By  chain rule)

Now,

Given function is

when we differentiate it w.r.t. x.
Lets take  .  then,

take . then,

(By  chain rule)

Now,

Given function is

We know that,

and
Lets take
Then,

(By chain rule)

-(i)
Similarly,

-(ii)
Now, put (i) and (ii) in

Given function is

Differentitation w.r.t. x is

Lets take
Our functions become,
and
Now,
( By chain rule)

-(i)
Similarly,

-(ii)
Put (i) and (ii) in

Give function is

Let's take

Now,  take

Differentiation w.r.t. x
-(By chain rule)

So,
( Multiply and divide by  and multiply and divide  by  )

Let us assume   :

Differentiating y with respect to x, we get  :

or

or

Given function is

We know that any function is differentiable when both
and         are finite and equal
Required condition for function to be differential at x  = 1 is

Now, Left-hand limit of a function at x = 1 is

Right-hand limit of a function at x = 1 is

Now, it is clear that
R.H.L. at x= 1    L.H.L. at x= 1
Therefore, function  is not differentiable at x = 1

x = 1 and x =  2.

Given function is

We know that any function is differentiable when both
and         are finite and equal
Required condition for function to be differential at x  = 1  is

Now, Left-hand limit of the function at x = 1 is

Right-hand limit of the function at x = 1 is

Now, it is clear that
R.H.L. at x= 1    L.H.L. at x= 1  and L.H.L. is not finite as well
Therefore, function  is not differentiable at x = 1
Similary, for x = 2
Required condition for function to be differential at x  = 2  is

Now, Left-hand limit of the function at x = 2 is

Right-hand limit of the function at x = 1 is

Now, it is clear that
R.H.L. at x= 2    L.H.L. at x= 2  and L.H.L. is not finite as well
Therefore, function  is not differentiable at x = 2

## NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability Exercise: 5.3

Question:1. Find dy/dx  in the following:

Given function is

We can rewrite it as

Now, differentiation w.r.t. x is

Question:2. Find dy/dx  in the following:

Given function is

We can rewrite it as

Now, differentiation w.r.t. x is

Question:3. Find dy/dx  in the following:

Given function is

We can rewrite it as

Now, differentiation w.r.t. x is

Question:4. Find dy/dx  in the following:

Given function is

We can rewrite it as

Now, differentiation w.r.t. x is

Question:5. Find dy/dx  in the following:

Given function is

We can rewrite it as

Now, differentiation w.r.t. x is

Question:6 Find dy/dx  in the following:

Given function is

We can rewrite it as

Now, differentiation w.r.t. x is

Question:7. Find dy/dx  in the following:

Given function is

Now, differentiation w.r.t. x is

Question:8. Find dy/dx  in the following:

Given function is

We can rewrite it as

Now, differentiation w.r.t. x is

Question:9 Find dy/dx  in the following:

Given function is

Lets consider
Then,

Now,

Our equation reduces to

Now, differentiation w.r.t. x is

Question:10. Find dy/dx  in the following:

Given function is

Lets consider
Then,

Now,

Our equation reduces to

Now, differentiation w.r.t. x is

Question:11. Find dy/dx  in the following:

Given function is

Let's consider
Then,

Now,

Our equation reduces to

Now, differentiation w.r.t. x is

Question:12. Find dy/dx  in the following:

Given function is

We can rewrite it as

Let's consider
Then,

Now,

Our equation reduces to

Now, differentiation w.r.t. x is

Question:13. Find dy/dx  in the following:

Given function is

We can rewrite it as

Let's consider
Then,

Now,

Our equation reduces to

Now, differentiation w.r.t. x is

Question:14. Find dy/dx  in the following:

Given function is

Lets take
Then,

And

Now, our equation reduces to

Now, differentiation w.r.t. x

Question:15. Find dy/dx  in the following:

Given function is

Let's take
Then,

And

Now, our equation reduces to

Now, differentiation w.r.t. x

## Solutions of NCERT for class 12 maths chapter 5 Continuity and Differentiability Exercise 5.4

Question:1. Differentiate the following w.r.t. x:

Given function is

We differentiate with the help of Quotient rule

Question:2. Differentiate the following w.r.t. x:

Given function is

Let
Then,

Now, differentiation w.r.t. x
-(i)

Put this value in our equation (i)

Question:3. Differentiate the following w.r.t. x:

Given function is

Let
Then,

Now, differentiation w.r.t. x
-(i)

Put this value in our equation (i)

Question:4. Differentiate the following w.r.t. x:

Given function is

Let's take
Now, our function reduces to

Now,
-(i)
And

Put this value in our equation (i)

Question:5. Differentiate the following w.r.t. x:

Given function is

Let's take
Now, our function reduces to

Now,
-(i)
And

Put this value in our equation (i)

Question:6Differentiate the following w.r.t. x:

Given function is

Now, differentiation w.r.t. x is

Question:7. Differentiate the following w.r.t. x:

Given function is

Lets take
Now, our function reduces to

Now,
-(i)
And

Put this value in our equation (i)

Question:8  Differentiate the following w.r.t. x:

Given function is

Lets take
Now, our function reduces to

Now,
-(i)
And

Put this value in our equation (i)

Question:9. Differentiate the following w.r.t. x:

Given function is

We differentiate with the help of Quotient rule

Question:10. Differentiate the following w.r.t. x:

Given function is

Lets take
Then , our function reduces to

Now, differentiation w.r.t. x is
-(i)
And

Put this value in our equation (i)

## CBSE NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability Exercise: 5.5

Question:1 Differentiate the functions w.r.t. x.

Given function is

Now, take log on both sides

Now, differentiation w.r.t. x

### Question:2. Differentiate the functions w.r.t. x.

Given function is

Take log on both the sides

Now, differentiation w.r.t. x is

### Question:3  Differentiate the functions w.r.t. x.

Given function is

take log on both the sides

Now, differentiation w.r.t x is

## Question:4  Differentiate the functions w.r.t. x.

Given function is

Let's take
take log on both the sides

Now, differentiation w.r.t x is

Similarly, take
Now, take log on both sides and differentiate w.r.t. x

Now,

## Question:5 Differentiate the functions w.r.t. x.

Given function is

Take log on both sides

Now, differentiate w.r.t. x  we get,

### Question:6 Differentiate the functions w.r.t. x.

Given function is

Let's take
Now, take log on both sides

Now, differentiate w.r.t. x
we get,

Similarly, take
Now, take log on both sides

Now, differentiate w.r.t. x
We get,

Now,

## Question:7 Differentiate the functions w.r.t. x.

Given function is

Let's take
Now, take log on both the sides

Now, differentiate w.r.t. x
we get,

Similarly, take
Now, take log on both sides

Now, differentiate w.r.t. x
We get,

Now,

## Question:8 Differentiate the functions w.r.t. x.

Given function is

Lets take
Now, take log on both the sides

Now, differentiate w.r.t. x
we get,

Similarly, take
Now, differentiate w.r.t. x
We get,

Now,

### Question:9 Differentiate the functions w.r.t. x

Given function is

Now, take
Now, take log on both sides

Now, differentiate it w.r.t. x
we get,

Similarly, take
Now, take log on both the sides

Now, differentiate it w.r.t. x
we get,

Now,

## Question:10  Differentiate the functions w.r.t. x.

Given function is

Take
Take log on both the sides

Now, differentiate w.r.t. x
we get,

Similarly,
take
Now. differentiate it w.r.t. x
we get,

Now,

### Question:11 Differentiate the functions w.r.t. x.

Given function is

Let's take
Now, take log on both sides

Now, differentiate w.r.t. x
we get,

Similarly, take
Now, take log on both the sides

Now, differentiate w.r.t. x
we get,

Now,

## Question:12  Find dy/dx of the functions given in Exercises 12 to 15

.

Given function is

Now, take
take log on both sides

Now, differentiate w.r.t  x
we get,

Similarly, take
Now, take log on both sides

Now, differentiate w.r.t. x
we get,

Now,

## Question:13 Find dy/dx of the functions given in Exercises 12 to 15.

Given function is

Now, take
take log on both sides

Now, differentiate w.r.t  x
we get,

Similarly, take
Now, take log on both sides

Now, differentiate w.r.t. x
we get,

Now,

### Question:14 Find dy/dx of the functions given in Exercises 12 to 15.

Given function is

Now, take log on both the sides

Now, differentiate w.r.t  x

By taking similar terms on the same side
We get,

### Question:15 Find dy/dx of the functions given in Exercises 12 to 15.

Given function is

Now, take  take log on both the sides

Now, differentiate w.r.t  x

By taking similar terms on same side
We get,