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Edited By Ramraj Saini | Updated on Sep 14, 2023 07:52 PM IST | #CBSE Class 12th

**NCERT Solutions for Class 12 Maths Chapter 5 **are provided here. These NCERT solutions are created bu expert team at careers360 considering the latest syllabus of CBSE 2023-24. Questions based on the topics like continuity, differentiability, and relations between them are covered in the NCERT solutions for class 12 maths chapter 5. In NCERT Class 12 maths book, there are 48 solved examples to understand the concepts of continuity and differentiability class 12. If you are finding difficulties in solving them, you can take help from NCERT maths chapter 5 class 12 solutions.

This Story also Contains

- NCERT Continuity And Differentiability Class 12 Questions And Answers
- NCERT Continuity And Differentiability Class 12 Questions And Answers PDF Free Download
- Continuity And Differentiability Class 12 NCERT Solutions - Important Formulae
- NCERT Continuity And Differentiability Class 12 Questions And Answers (Intext Questions and Exercise)
- Topics of NCERT class 12 maths chapter 5 Continuity and Differentiability
- NCERT class 12 maths ch 5 question answer - Topics
- NCERT solutions for class 12 maths - Chapter wise
- NCERT solutions for class 12 subject wise
- NCERT Solutions class wise
- NCERT Books and NCERT Syllabus
- Tips to Use NCERT Solutions for Class 12 Maths Chapter 5

In NCERT class 11 Maths solutions, you have already learned the differentiation of certain functions like polynomial functions and trigonometric functions. In this chapter, you will get NCERT solutions for class 12 maths chapter 5 continuity and differentiability. If you are interested in the chapter 5 class 12 maths NCERT solutions then you can check NCERT solutions for class 12 other subjects.

- Class 12 Maths Chapter 5 continuity and differentiability Notes
- Ncert Exemplar Solutions For Class 12 Maths Chapter 5 continuity and differentiability

>> Continuity: A function f(x) is continuous at a point x = a if:

f(a) exists (finite, definite, and real).

lim(x → a) f(x) exists.

lim(x → a) f(x) = f(a).

>> Discontinuity: f(x) is discontinuous in an interval if it is discontinuous at any point in that interval.

Algebra of Continuous Functions:

Sum, difference, product, and quotient of continuous functions are continuous.

Differentiation:

The derivative of f(x) at x = a, denoted as f'(a), represents the slope of the tangent line to the graph.

Chain Rule:

If f = v o u, where t = u(x), and if both dt/dx and dv/dx exist, then: df/dx = dv/dt * dt/dx.

Derivatives of Some Standard Functions:

d/dx(x

^{n}) = nx^{n-1}d/dx(sin x) = cos x

d/dx(cos x) = -sin x

d/dx(tan x) = sec

^{2}xd/dx(cot x) = -csc

^{2}xd/dx(sec x) = sec x * tan x

d/dx(csc x) = -csc x * cot x

d/dx(a

^{x}) = a^{x}* ln(a)d/dx(e

^{x}) = e^{x}d/dx(ln x) = 1/x

**Mean Value Theorem:**

Mean Value Theorem states that if f(x) is continuous on [a, b] and differentiable on (a, b), then there exists some c in (a, b) such that: f'(c) = (f(b) - f(a)) / (b - a).

**Rolle's Theorem:**

Rolle's Theorem states that if f(x) is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there exists some c in (a, b) such that f'(c) = 0.

**Lagrange's Mean Value Theorem:**

Lagrange's Mean Value Theorem states that if f(x) is continuous on [a, b] and differentiable on (a, b), then there exists some c in (a, b) such that:

f'(c) = (f(b) - f(a)) / (b - a).

Free download C**ontinuity And Differentiability Class 12 NCERT Solutions **for CBSE Exam.

**NCERT Continuity And Differentiability Class 12 Solutions : Excercise: 5.1 **

** Question:1 ** . Prove that the function is continuous at and at

** Answer: **

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

** Answer: **

Given function is

at x = 3

Hence, function is continous at x = 3

** Question:3 ** Examine the following functions for continuity.

** Answer: **

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.

** Answer: **

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.

** Answer: **

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.

** Answer: **

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

** Question:4 ** . Prove that the function is continuous at x = n, where n is a positive integer

** Answer: **

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

** Question:5. ** Is the function f defined by

continuous at x = 0? At x = 1? At x = 2?

** Answer: **

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

** Question:6. ** Find all points of discontinuity of f, where f is defined by

** Answer: **

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

** Question:7. ** ** ** Find all points of discontinuity of f, where f is defined by

** Answer: **

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

** Question:8. ** Find all points of discontinuity of f, where f is defined by

** Answer: **

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 **

** Question:9. ** Find all points of discontinuity of f, where f is defined by

** Answer: **

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

** Question:10. ** ** ** Find all points of discontinuity of f, where f is defined by

** Answer: **

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

** Question:11. ** Find all points of discontinuity of f, where f is defined by

** Answer: **

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

** Question:12. ** Find all points of discontinuity of f, where f is defined by

** Answer: **

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

** Answer: **

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:14. ** Discuss the continuity of the function f, where f is defined by

** Answer: **

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

** Question:15 Discuss the continuity of the function f, where f is defined by **

** Answer: **

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

** Question:16. ** Discuss the continuity of the function f, where f is defined by

** Answer: **

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

** Question:17. ** Find the relationship between a and b so that the function f defined by

is continuous at x = 3.

** Answer: **

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

** Question:18. ** For what value of l is the function defined by

continuous at x = 0? What about continuity at x = 1?

** Answer: **

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

** Answer: **

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 = ?

** Answer: **

Given function is

Clearly, Given function is defined at x =

Hence, the function defined by continuous at x =

** Question:21. ** Discuss the continuity of the following functions:

a)

** Answer: **

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:

** Answer: **

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:

** Answer: **

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:22. ** Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

** Answer: **

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

** Answer: **

Given function is

Hence, the function is continuous

Therefore, no point of discontinuity

** Question:24. ** Determine if f defined by

is a continuous function?

** Answer: **

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

** Question:25 ** . Examine the continuity of f, where f is defined by

** Answer: **

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

** Question:26. ** Find the values of k so that the function f is continuous at the indicated point in Exercises

** Answer: **

Given function is

When

For the function to be continuous

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

** Question:27 ** . Find the values of k so that the function f is continuous at the indicated point in Exercises

** Answer: **

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

** Question:28 ** . Find the values of k so that the function f is continuous at the indicated point in Exercises

** Answer: **

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

** Question:29 ** Find the values of k so that the function f is continuous at the indicated point in Exercises

** Answer: **

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

** Question:30 ** Find the values of a and b such that the function defined by

is a continuous function.

** Answer: **

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

** Question:31. ** Show that the function defined by is a continuous function.

** Answer: **

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

** Question:32. ** Show that the function defined by is a continuous function.

** Answer: **

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.

** Answer: **

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

** Question:34. ** Find all the points of discontinuity of f defined by

** Answer: **

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

** NCERT class 12 maths chapter 5 question answer: Excercise: 5.2 **

** Question:1. ** Differentiate the functions with respect to x in

** Answer: **

Given function is

when we differentiate it w.r.t. x.

Lets take . then,

(By chain rule)

Now,

Therefore, the answer is

** Question:2. ** Differentiate the functions with respect to x in

** Answer: **

Given function is

Lets take then,

( By chain rule)

Now,

Therefore, the answer is

** Question:3. ** Differentiate the functions with respect to x in

** Answer: **

Given function is

when we differentiate it w.r.t. x.

Lets take . then,

(By chain rule)

Now,

Therefore, the answer is

** Question:4 ** . Differentiate the functions with respect to x in

** Answer: **

Given function is

when we differentiate it w.r.t. x.

Lets take . then,

take . then,

(By chain rule)

Now,

Therefore, the answer is

** Question:5. ** Differentiate the functions with respect to x in

** Answer: **

Given function is

We know that,

and

Lets take

Then,

(By chain rule)

-(i)

Similarly,

-(ii)

Now, put (i) and (ii) in

Therefore, the answer is

** Question:6. ** Differentiate the functions with respect to x in

** Answer: **

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

Therefore, the answer is

** Question:7. ** Differentiate the functions with respect to x in

** Answer: **

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 )

There, the answer is

** Question:8 ** Differentiate the functions with respect to x in

** Answer: **

Let us assume :

Differentiating y with respect to x, we get :

or

or

** Question:9 ** . Prove that the function f given by is not differentiable at x = 1.

** Answer: **

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

** Question:10. ** Prove that the greatest integer function defined by is not differentiable at

** Answer: **

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 class 12 maths chapter 5 question answer: Exercise: 5.3 **

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

** Answer: **

Given function is

We can rewrite it as

Now, differentiation w.r.t. x is

Therefore, the answer is

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

** Answer: **

Given function is

We can rewrite it as

Now, differentiation w.r.t. x is

Therefore, the answer is

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

** Answer: **

Given function is

We can rewrite it as

Now, differentiation w.r.t. x is

Therefore, the answer is

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

** Answer: **

Given function is

We can rewrite it as

Now, differentiation w.r.t. x is

Therefore, the answer is

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

** Answer: **

Given function is

We can rewrite it as

Now, differentiation w.r.t. x is

Therefore, the answer is

** Question:6 ** Find dy/dx in the following:

** Answer: **

Given function is

We can rewrite it as

Now, differentiation w.r.t. x is

Therefore, the answer is

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

** Answer: **

Given function is

Now, differentiation w.r.t. x is

Therefore, the answer is

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

** Answer: **

Given function is

We can rewrite it as

Now, differentiation w.r.t. x is

Therefore, the answer is

** Question:9 ** Find dy/dx in the following:

** Answer: **

Given function is

Lets consider

Then,

Now,

Our equation reduces to

Now, differentiation w.r.t. x is

Therefore, the answer is

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

** Answer: **

Given function is

Lets consider

Then,

Now,

Our equation reduces to

Now, differentiation w.r.t. x is

Therefore, the answer is

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

** Answer: **

Given function is

Let's consider

Then,

Now,

Our equation reduces to

Now, differentiation w.r.t. x is

Therefore, the answer is

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

** Answer: **

Given function is

We can rewrite it as

Let's consider

Then,

Now,

Our equation reduces to

Now, differentiation w.r.t. x is

Therefore, the answer is

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

** Answer: **

Given function is

We can rewrite it as

Let's consider

Then,

Now,

Our equation reduces to

Now, differentiation w.r.t. x is

Therefore, the answer is

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

** Answer: **

Given function is

Lets take

Then,

And

Now, our equation reduces to

Now, differentiation w.r.t. x

Therefore, the answer is

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

** Answer: **

Given function is

Let's take

Then,

And

Now, our equation reduces to

Now, differentiation w.r.t. x

Therefore, the answer is

** NCERT class 12 maths chapter 5 question answer: Exercise 5.4 **

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

** Answer: **

Given function is

We differentiate with the help of Quotient rule

Therefore, the answer is

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

** Answer: **

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:

** Answer: **

Given function is

Let

Then,

Now, differentiation w.r.t. x

-(i)

Put this value in our equation (i)

Therefore, the answer is

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

** Answer: **

Given function is

Let's take

Now, our function reduces to

Now,

-(i)

And

Put this value in our equation (i)

Therefore, the answer is

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

** Answer: **

Given function is

Let's take

Now, our function reduces to

Now,

-(i)

And

Put this value in our equation (i)

Therefore, the answer is

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

** Answer: **

Given function is

Now, differentiation w.r.t. x is

Therefore, answer is

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

** Answer: **

Given function is

Lets take

Now, our function reduces to

Now,

-(i)

And

Put this value in our equation (i)

Therefore, the answer is

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

** Answer: **

Given function is

Lets take

Now, our function reduces to

Now,

-(i)

And

Put this value in our equation (i)

Therefore, the answer is

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

** Answer: **

Given function is

We differentiate with the help of Quotient rule

Therefore, the answer is

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

** Answer: **

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)

Therefore, the answer is

**Class 12 Maths Chapter 5 NCERT solutions: Exercise: 5.5 **

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

** Answer: **

Given function is

Now, take log on both sides

Now, differentiation w.r.t. x

There, the answer is

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

** Answer: **

Given function is

Take log on both the sides

Now, differentiation w.r.t. x is

Therefore, the answer is

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

** Answer: **

Given function is

take log on both the sides

Now, differentiation w.r.t x is

Therefore, the answer is

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

** Answer: **

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,

Therefore, the answer is

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

** Answer: **

Given function is

Take log on both sides

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

Therefore, the answer is

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

** Answer: **

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,

Therefore, the answer is

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

** Answer: **

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,

Therefore, the answer is

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

** Answer: **

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,

Therefore, the answer is

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

** Answer: **

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,

Therefore, the answer is

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

** Answer: **

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,

Therefore, the answer is

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

** Answer: **

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,

Therefore, the answer is

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

** Answer: **

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,

Therefore, the answer is

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

** Answer: **

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,

Therefore, the answer is

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

** Answer: **

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,

Therefore, the answer is

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

** Answer: **

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,

Therefore, the answer is

** Question:16 ** Find the derivative of the function given by and hence find

** Answer: **

Given function is

Take log on both sides

NOW, differentiate w.r.t. x

Therefore,

Now, the vale of is

** Question:17 (1) ** Differentiate in three ways mentioned below:

(i) by using product rule

** Answer: **

Given function is

Now, we need to differentiate using the product rule

Therefore, the answer is

** Question:17 (2) ** Differentiate in three ways mentioned below:

(ii) by expanding the product to obtain a single polynomial.

** Answer: **

Given function is

Multiply both to obtain a single higher degree polynomial

Now, differentiate w.r.t. x

we get,

Therefore, the answer is

** Question:17 (3) ** Differentiate in three ways mentioned below:

(iii) by logarithmic differentiation.

Do they all give the same answer?

** Answer: **

Given function is

Now, take log on both the sides

Now, differentiate w.r.t. x

we get,

Therefore, the answer is

And yes they all give the same answer

** Question:18 ** 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: **

It is given that u, v and w are the functions of x

Let

Now, we differentiate using product rule w.r.t x

First, take

Now,

-(i)

Now, again by the product rule

Put this in equation (i)

we get,

Hence, by product rule we proved it

Now, by taking the log

Again take

Now, take log on both sides

Now, differentiate w.r.t. x

we get,

Hence, we proved it by taking the log

** Class 12 Maths Chapter 5 NCERT solutions: Exercise:5.6 **

** Answer: **

Given equations are

Now, differentiate both w.r.t t

We get,

Similarly,

Now,

Therefore, the answer is

** Answer: **

Given equations are

Now, differentiate both w.r.t

We get,

Similarly,

Now,

Therefore, answer is

** Answer: **

Given equations are

Now, differentiate both w.r.t t

We get,

Similarly,

Now,

Therefore, the answer is

** Answer: **

Given equations are

Now, differentiate both w.r.t t

We get,

Similarly,

Now,

Therefore, the answer is

** Answer: **

Given equations are

Now, differentiate both w.r.t

We get,

Similarly,

Now,

Therefore, answer is

** Answer: **

Given equations are

Now, differentiate both w.r.t

We get,

Similarly,

Now,

Therefore, the answer is

** Answer: **

Given equations are

Now, differentiate both w.r.t t

We get,

Similarly,

Now,

Therefore, the answer is

** Answer: **

Given equations are

Now, differentiate both w.r.t t

We get,

Similarly,

Now,

Therefore, the answer is

** Answer: **

Given equations are

Now, differentiate both w.r.t

We get,

Similarly,

Now,

Therefore, the answer is

** Answer: **

Given equations are

Now, differentiate both w.r.t

We get,

Similarly,

Now,

Therefore, the answer is

** Question:11 ** If , show that

** Answer: **

Given equations are

** differentiating with respect to x **

**Class 12 Maths Chapter 5 NCERT solutions:**** Exercise: 5.7 **

** Question:1 ** Find the second order derivatives of the functions given in Exercises 1 to 10.

** Answer: **

Given function is

Now, differentiation w.r.t. x

Now, second order derivative

Therefore, the second order derivative is

** Question:2 ** Find the second order derivatives of the functions given in Exercises 1 to 10.

** Answer: **

Given function is

Now, differentiation w.r.t. x

Now, the second-order derivative is

Therefore, second-order derivative is

** Question:3 ** Find the second order derivatives of the functions given in Exercises 1 to 10.

** Answer: **

Given function is

Now, differentiation w.r.t. x

Now, the second-order derivative is

Therefore, the second-order derivative is

** Question:4 ** Find the second order derivatives of the functions given in Exercises 1 to 10.

** Answer: **

Given function is

Now, differentiation w.r.t. x

Now, second order derivative is

Therefore, second order derivative is

** Question:5 ** Find the second order derivatives of the functions given in Exercises 1 to 10.

** Answer: **

Given function is

Now, differentiation w.r.t. x

Now, the second-order derivative is

Therefore, the second-order derivative is

** Question:6 ** Find the second order derivatives of the functions given in Exercises 1 to 10.

** Answer: **

Given function is

Now, differentiation w.r.t. x

Now, second order derivative is

Therefore, second order derivative is

** Question:7 ** Find the second order derivatives of the functions given in Exercises 1 to 10.

** Answer: **

Given function is

Now, differentiation w.r.t. x

Now, second order derivative is

Therefore, second order derivative is

** Question:8 ** Find the second order derivatives of the functions given in Exercises 1 to 10.

** Answer: **

Now, differentiation w.r.t. x

Now, second order derivative is

Therefore, second order derivative is

** Question:9 ** Find the second order derivatives of the functions given in Exercises 1 to 10.

** Answer: **

Now, differentiation w.r.t. x

Now, second order derivative is

Therefore, second order derivative is

** Question:10 ** Find the second order derivatives of the functions given in Exercises 1 to 10.

** Answer: **

Given function is

Now, differentiation w.r.t. x

Now, second order derivative is

Using Quotient rule

Therefore, second order derivative is

** Question:11 ** If prove that

** Answer: **

Given function is

Now, differentiation w.r.t. x

Now, the second-order derivative is

Now,

Hence proved

** Question:12 ** If Find in terms of y alone.

** Answer: **

Given function is

Now, differentiation w.r.t. x

Now, second order derivative is

-(i)

Now, we want in terms of y

Now, put the value of x in (i)

Therefore, answer is

** Question:13 ** If , show that

** Answer: **

Given function is

Now, differentiation w.r.t. x

-(i)

Now, second order derivative is

By using the Quotient rule

-(ii)

Now, from equation (i) and (ii) we will get

Now, we need to show

Put the value of from equation (i) and (ii)

Hence proved

** Question:14 ** If , show that

** Answer: **

Given function is

Now, differentiation w.r.t. x

-(i)

Now, second order derivative is

-(ii)

Now, we need to show

Put the value of from equation (i) and (ii)

Hence proved

** Question:15 ** If , show that ** Answer: **

Given function is

Now, differentiation w.r.t. x

-(i)

Now, second order derivative is

-(ii)

Now, we need to show

Put the value of from equation (ii)

Hence, L.H.S. = R.H.S.

Hence proved

** Question:16 ** If show that

** Answer: **

Given function is

We can rewrite it as

Now, differentiation w.r.t. x

-(i)

Now, second order derivative is

-(ii)

Now, we need to show

Put value of from equation (i) and (ii)

Hence, L.H.S. = R.H.S.

Hence proved

** Question:17 ** If show that

** Answer: **

Given function is

Now, differentiation w.r.t. x

-(i)

Now, the second-order derivative is

By using the quotient rule

-(ii)

Now, we need to show

Put the value from equation (i) and (ii)

Hence, L.H.S. = R.H.S.

Hence proved

**Class 12 Maths Chapter 5 NCERT solutions:**** Excercise: 5.8 **

** Question:1 ** Verify Rolle’s theorem for the function

** Answer: **

According to Rolle's theorem function must be

a ) continuous in given closed interval say [x,y]

b ) differentiable in given open interval say (x,y)

c ) f(x) = f(y)

Then their exist a such that

If all these conditions are satisfies then we can verify Rolle's theorem

Given function is

Now, being a polynomial function, is both continuous in [-4,2] and differentiable in (-4,2)

Now,

Similalrly,

Therefore, value of and value of f(x) at -4 and 2 are equal

Now,

According to roll's theorem their is point c , such that

Now,

And

Hence, Rolle's theorem is verified for the given function

** Answer: **

According to Rolle's theorem function must be

a ) continuous in given closed interval say [x,y]

b ) differentiable in given open interval say (x,y)

c ) f(x) = f(y)

Then their exist a such that

If all these conditions are satisfied then we can verify Rolle's theorem

Given function is

It is clear that Given function is not continuous for each and every point in [5,9]

Now, lets check differentiability of f(x)

L.H.L. at x = n ,

Now,

R.H.L. at x = n ,

We can clearly see that R.H.L. is not equal to L.H.L.

Therefore, the function is not differential in (5,9)

Hence, Rolle's theorem is not applicable for given function ,

** Answer: **

According to Rolle's theorem function must be

a ) continuous in given closed interval say [x,y]

b ) differentiable in given open interval say (x,y)

c ) f(x) = f(y)

Then their exist a such that

If all these conditions are satisfies then we can verify Rolle's theorem

Given function is

It is clear that Given function is not continuous for each and every point in [-2,2]

Now, lets check differentiability of f(x)

L.H.L. at x = n ,

Now,

R.H.L. at x = n ,

We can clearly see that R.H.L. is not equal to L.H.L.

Therefore, function is not differential in (-2,2)

Hence, Rolle's theorem is not applicable for given function ,

** Answer: **

According to Rolle's theorem function must be

a ) continuous in given closed interval say [x,y]

b ) differentiable in given open interval say (x,y)

c ) f(x) = f(y)

Then there exist a such that

If all these conditions are satisfied then we can verify Rolle's theorem

Given function is

Now, being a polynomial , function is continuous in [1,2] and differentiable in(1,2)

Now,

And

Therefore,

Therefore, All conditions are not satisfied

Hence, Rolle's theorem is not applicable for given function ,

** Question:3 ** If is a differentiable function and if does not vanish

anywhere, then prove that

** Answer: **

It is given that

is a differentiable function

Now, f is a differential function. So, f is also a continuous function

We obtain the following results

a ) f is continuous in [-5,5]

b ) f is differentiable in (-5,5)

Then, by Mean value theorem we can say that there exist a c in (-5,5) such that

Now, it is given that does not vanish anywhere

Therefore,

Hence proved

** Question:4 ** Verify Mean Value Theorem, if in the interval [a, b], where

a = 1 and b = 4.

** Answer: **

Condition for M.V.T.

If

a ) f is continuous in [a,b]

b ) f is differentiable in (a,b)

Then, there exist a c in (a,b) such that

It is given that

and interval is [1,4]

Now, f is a polynomial function , is continuous in[1,4] and differentiable in (1,4)

And

and

Then, by Mean value theorem we can say that their exist a c in (1,4) such that

Now,

And

Hence, mean value theorem is verified for the function

** Answer: **

Condition for M.V.T.

If

a ) f is continuous in [a,b]

b ) f is differentiable in (a,b)

Then, their exist a c in (a,b) such that

It is given that

and interval is [1,3]

Now, f being a polynomial function , is continuous in[1,3] and differentiable in (1,3)

And

and

Then, by Mean value theorem we can say that their exist a c in (1,4) such that

Now,

And

Hence, mean value theorem is varified for following function and is the only point where f '(c) = 0

** Answer: **

According to Mean value theorem function

must be

a ) continuous in given closed interval say [a,b]

b ) differentiable in given open interval say (a,b)

Then their exist a such that

If all these conditions are satisfies then we can verify mean value theorem

Given function is

It is clear that Given function is not continuous for each and every point in [5,9]

Now, lets check differentiability of f(x)

L.H.L. at x = n ,

Now,

R.H.L. at x = n ,

We can clearly see that R.H.L. is not equal to L.H.L.

Therefore, function is not differential in (5,9)

Hence, Mean value theorem is not applicable for given function ,

Similaly,

Given function is

It is clear that Given function is not continuous for each and every point in [-2,2]

Now, lets check differentiability of f(x)

L.H.L. at x = n ,

Now,

R.H.L. at x = n ,

We can clearly see that R.H.L. is not equal to L.H.L.

Therefore, function is not differential in (-2,2)

Hence, Mean value theorem is not applicable for given function ,

Similarly,

Given function is

Now, being a polynomial , function is continuous in [1,2] and differentiable in(1,2)

Now,

And

Now,

Now,

And

Therefore, mean value theorem is applicable for the function

**NCERT class 12 continuity and differentiability ncert solutions Miscellaneous Excercise **

** Question:1 ** Differentiate w.r.t. x the function in Exercises 1 to 11.

** Answer: **

Given function is

Now, differentiation w.r.t. x is

Therefore, differentiation w.r.t. x is

** Question:2 ** Differentiate w.r.t. x the function in Exercises 1 to 11.

** Answer: **

Given function is

Now, differentiation w.r.t. x is

Therefore, differentiation w.r.t. x is

** Question:3 ** Differentiate w.r.t. x the function in Exercises 1 to 11.

** Answer: **

Given function is

Take, log on both the sides

Now, differentiation w.r.t. x is

By using product rule

Therefore, differentiation w.r.t. x is

** Question:4 ** Differentiate w.r.t. x the function in Exercises 1 to 11.

** Answer: **

Given function is

Now, differentiation w.r.t. x is

Therefore, differentiation w.r.t. x is

** Question:5 ** Differentiate w.r.t. x the function in Exercises 1 to 11.

** Answer: **

Given function is

Now, differentiation w.r.t. x is

By using the Quotient rule

Therefore, differentiation w.r.t. x is

** Question:6 ** Differentiate w.r.t. x the function in Exercises 1 to 11.

** Answer: **

Given function is

Now, rationalize the [] part

Given function reduces to

Now, differentiation w.r.t. x is

Therefore, differentiation w.r.t. x is

** Question:7 ** Differentiate w.r.t. x the function in Exercises 1 to 11.

** Answer: **

Given function is

Take log on both sides

Now, differentiate w.r.t.

Therefore, differentiation w.r.t x is

** Question:8 ** , for some constant a and b.

** Answer: **

Given function is

Now, differentiation w.r.t x

Therefore, differentiation w.r.t x

** Question: 9 ** ** **

** Answer: **

Given function is

Take log on both the sides

Now, differentiate w.r.t. x

Therefore, differentiation w.r.t x is

** Question:10 ** , for some fixed a > 0 and x > 0

** Answer: **

Given function is

Lets take

Now, take log on both sides

Now, differentiate w.r.t x

-(i)

Similarly, take

take log on both the sides

Now, differentiate w.r.t x

-(ii)

Similarly, take

take log on both the sides

Now, differentiate w.r.t x

-(iii)

Similarly, take

take log on both the sides

Now, differentiate w.r.t x

-(iv)

Now,

Put values from equation (i) , (ii) ,(iii) and (iv)

Therefore, differentiation w.r.t. x is

** Question: 11 **

** Answer: **

Given function is

take

Now, take log on both the sides

Now, differentiate w.r.t x

-(i)

Similarly,

take

Now, take log on both the sides

Now, differentiate w.r.t x

-(ii)

Now

Put the value from equation (i) and (ii)

Therefore, differentiation w.r.t x is

** Question:12 ** Find dy/dx if

** Answer: **

Given equations are

Now, differentiate both y and x w.r.t t independently

And

Now

Therefore, differentiation w.r.t x is

** Question:13 ** Find dy/dx if

** Answer: **

Given function is

Now, differentiatiate w.r.t. x

Therefore, differentiatiate w.r.t. x is 0

** Question:14 ** If

** Answer: **

Given function is

Now, squaring both sides

Now, differentiate w.r.t. x is

Hence proved

** Question:15 ** If , for some c > 0, prove that is a constant independent of a and b.

** Answer: **

Given function is

- (i)

Now, differentiate w.r.t. x

-(ii)

Now, the second derivative

Now, put values from equation (i) and (ii)

Now,

Which is independent of a and b

Hence proved

** Question:16 ** If , with , prove that

** Answer: **

Given function is

Now, Differentiate w.r.t x

Hence proved

** Question:17 ** If and find

** Answer: **

Given functions are

and

Now, differentiate both the functions w.r.t. t independently

We get

Similarly,

Now,

Now, the second derivative

Therefore,

** Question:18 ** If , show that f ''(x) exists for all real x and find it.

** Answer: **

Given function is

Now, differentiate in both the cases

And

In both, the cases f ''(x) exist

Hence, we can say that f ''(x) exists for all real x

and values are

** Question:19 ** Using mathematical induction prove that for all positive integers n.

** Answer: **

Given equation is

We need to show that for all positive integers n

Now,

For ( n = 1)

Hence, true for n = 1

For (n = k)

Hence, true for n = k

For ( n = k+1)

Hence, (n = k+1) is true whenever (n = k) is true

Therefore, by the principle of mathematical induction we can say that is true for all positive integers n

** Question:20 ** Using the fact that and the differentiation,

obtain the sum formula for cosines.

** Answer: **

Given function is

Now, differentiate w.r.t. x

Hence, we get the formula by differentiation of sin(A + B)

** Answer: **

Consider f(x) = |x| + |x +1|

We know that modulus functions are continuous everywhere and sum of two continuous function is also a continuous function

Therefore, our function f(x) is continuous

Now,

If Lets differentiability of our function at x = 0 and x= -1

L.H.D. at x = 0

R.H.L. at x = 0

R.H.L. is not equal to L.H.L.

Hence.at x = 0 is the function is not differentiable

Now, Similarly

R.H.L. at x = -1

L.H.L. at x = -1

L.H.L. is not equal to R.H.L, so not differentiable at x=-1

Hence, exactly two points where it is not differentiable

** Question:22 ** If , prove that

** Answer: **

Given that

We can rewrite it as

Now, differentiate w.r.t x

we will get

Hence proved

** Question:23 ** If , show that

** Answer: **

Given function is

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

-(i)

Now, again differentiate w.r.t x

-(ii)

Now, we need to show that

Put the values from equation (i) and (ii)

Hence proved

**If you are looking for continuity and differentiability class 12 NCERT solutions of exercises then these are listed below.**

- Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.1
- Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.2
- Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.3
- Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.4
- Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.5
- Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.6
- Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.7
- Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.8
- Continuity And Differentiability Class 12 NCERT Solutions Miscellaneous Exercise

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

**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 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].

'Continuity and differentiability' is one of the very important and time-consuming chapters of the NCERT Class 12 maths syllabus. It contains 8 exercises with 121 questions and also 23 questions in the miscellaneous exercise. In this article, you will find all NCERT solutions for class 12 maths chapter 5 continuity and differentiability including miscellaneous exercises.

**Also read,**

NCERT exemplar solutions class 12 maths chapter 5

The main topics covered in chapter 5 maths class 12 are:

**Continuity**

A function is continuous at a given point if the left-hand limit, right-hand limit and value of function exist and are equal. In this class 12 NCERT topics elaborate concepts related to continuity, point of discontinuity, algebra of continuous function. Continuity and Differentiability class 12 solutions include a comprehensive module of quality questions.

**Differentiability**

This ch 5 maths class 12 discuss differentiability concepts of different functions including derivatives of composite functions, derivatives of implicit functions, derivatives of inverse trigonometric functions. To get command on these concepts you can refer to NCERT solutions for class 12 maths chapter 5.

**Exponential and Logarithmic Functions**

This ch 5 maths class 12 also includes concepts of exponential and logarithmic functions including natural log and their graphical representation. maths class 12 chapter 5 also contains fundamental properties of the logarithmic function. You can refer to class 12 NCERT solutions for questions about these concepts.

**Logarithmic Differentiation**

this class 12 ncert chapter discusses a special technique of differentiation known as logarithmic differentiation. to get command of these concepts you can go through the NCERT solution for class 12 maths chapter 5.

**Derivatives of Functions in Parametric Forms**

concepts to differentiate a function which is not implicit and explicit but given in the parametric form are explained in this chapter. Continuity and Differentiability class 12 solutions include problems to understand the concepts.

ch 5 maths class 12 also discuss in detail the concepts of second-order derivative, mean value theorem, Rolle's theorem. for questions on these concepts, you can browse NCERT solutions for class 12 chapter 5.

Topics enumerated in class 12 NCERT are very important and students are suggested to go through all the concepts discussed in the topics. Questions related to all the above topics are covered in the NCERT solutions for class 12 maths chapter 5

**Also read,**

- NCERT Exemplar Class 12 Chemistry Solutions
- NCERT Exemplar Class 12 Mathematics Solutions
- NCERT Exemplar Class 12 Biology Solutions
- NCERT Exemplar Class 12 Physics Solutions

Chapter 1 | NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions |

Chapter 2 | NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions |

Chapter 3 | NCERT solutions for class 12 maths chapter 3 Matrices |

Chapter 4 | NCERT solutions for class 12 maths chapter 4 Determinants |

Chapter 5 | NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability |

Chapter 6 | NCERT solutions for class 12 maths chapter 6 Application of Derivatives |

Chapter 7 | NCERT solutions for class 12 maths chapter 7 Integrals |

Chapter 8 | NCERT solutions for class 12 maths chapter 8 Application of Integrals |

Chapter 9 | NCERT solutions for class 12 maths chapter 9 Differential Equations |

Chapter 10 | NCERT solutions for class 12 maths chapter 10 Vector Algebra |

Chapter 11 | NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry |

Chapter 12 | NCERT solutions for class 12 maths chapter 12 Linear Programming |

Chapter 13 | NCERT solutions for class 12 maths chapter 13 Probability |

- NCERT solutions class 12 chemistry
- NCERT solutions for class 12 physics
- NCERT solutions for class 12 biology

NCERT solutions for class 12 maths chapter 5 continuity and differentiability are very helpful in the preparation of this chapter. But here are some tips to get command on this chapter.

- You should make sure that concepts related to 'limit' are clear to you as it forms the base for continuity.
- First, go for the theorem and solved examples of continuity given in the NCERT textbook then try to solve exercise questions. You may find some difficulties in solving them. Go through the NCERT solutions for class 12 maths chapter 5 continuity and differentiability, it will help you to understand the concepts in a much easy way.
- This chapter seems very easy but at the same time, the chances of silly mistakes are also high. So, it is advised to understand the theory and concepts properly before practicing questions of NCERT.
- Once you are good in continuity, then go for the differentiability. Practice more and more questions to get command on it.
- Differentiation is mostly formula-based, so practice NCERT questions, it won't take much effort to remember the formulas.

**Also check,**

- NCERT Exemplar Class 12 Solutions
- NCERT Exemplar Class 12th Maths Solutions
- NCERT Exemplar Class 12th Physics Solutions
- NCERT Exemplar Class 12th Chemistry Solutions
- NCERT Exemplar Class 12th Biology Solutions
- NCERT Exemplar Class 12 Maths Solutions Chapter 5

** Happy learning!!! **

1. What are the important topics in chapter Continuity and Differentiability?

Basic concepts of continuity and differentiability, derivatives of composite functions, derivatives of implicit functions, derivatives of inverse trigonometric functions, exponential and logarithmic functions, logarithmic differentiation, derivatives of functions in parametric form are the important topics in this chapter. Practice these class 12 maths ch 5 question answer to command the concepts.

2. What are the reasons to opt for ncert continuity and differentiability solution?

The maths chapter 5 class 12 NCERT solutions created by the experts at Careers360 offer numerous advantages to students preparing for their board exams. These solutions provide comprehensive explanations of each topic, which help students achieve high scores. Additionally, the solutions are based on the latest CBSE syllabus for the 2022-23 academic year. Furthermore, these solutions also assist students in preparing for other competitive exams such as JEE Main and JEE Advanced. For ease, Students can study continuity and differentiability pdf both online and offline

3. Which is the best book for CBSE class 12 maths ?

NCERT is the best book for CBSE class 12 maths. Most of the questions in CBSE board exam are directly asked from NCERT textbook. All you need to do is rigorous practice of all the problems given in the NCERT textbook.

4. How many exercises are included in ncert solutions class 12 maths chapter 5?

According to the given information, there are 8 exercises in NCERT Solutions for maths chapter 5 class 12 . The following is the number of questions in each exercise:

Exercise 5.1: 34 questions

Exercise 5.2: 10 questions

Exercise 5.3: 15 questions

Exercise 5.4: 10 questions

Exercise 5.5: 18 questions

Exercise 5.6: 11 questions

Exercise 5.7: 17 questions

Exercise 5.8: 6 questions

Additionally, there is a Miscellaneous Exercise with 23 questions.

5. What is the weightage of the chapter Continuity and Differentiability for CBSE board exam ?

Generally, Continuity and differntiability has 9% weightage in the 12th board final examination. if you want to obtain meritious marks or full marks then you should have good command on concepts that can be developed by practice therefore you should practice NCERT solutions and NCERT exercise solutions.

Application Date:20 November,2023 - 19 December,2023

Application Date:20 November,2023 - 19 December,2023

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9 minHave a question related to CBSE Class 12th ?

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

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Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

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Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available

3 Jobs Available

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available

.NET Developer Job Description: A .NET Developer is a professional responsible for producing code using .NET languages. He or she is a software developer who uses the .NET technologies platform to create various applications. Dot NET Developer job comes with the responsibility of creating, designing and developing applications using .NET languages such as VB and C#.

2 Jobs Available

2 Jobs Available

A DevOps Architect is responsible for defining a systematic solution that fits the best across technical, operational and and management standards. He or she generates an organised solution by examining a large system environment and selects appropriate application frameworks in order to deal with the system’s difficulties.

2 Jobs Available

Individuals who are interested in working as a Cloud Administration should have the necessary technical skills to handle various tasks related to computing. These include the design and implementation of cloud computing services, as well as the maintenance of their own. Aside from being able to program multiple programming languages, such as Ruby, Python, and Java, individuals also need a degree in computer science.

2 Jobs Available

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