**NCERT Solutions For Class 12 Maths Chapter 5** gives detailed solutions for every question. It was prepared by the best teachers in India. This is the advantage of getting a better score on the exam. Continuity and Differentiability class 12 NCERT pdf will help students get a high score in the exam. This article gives step-wise solutions for better understanding. **NCERT Solutions For Class 12 Maths Chapter 5 pdf** will help the students as well as students who are preparing for JEE Mains, JEE, and other competitive exams. **Class 12 Maths** solutions were given in detailed explanation for easy understanding. **NCERT Solutions For Class 12 Maths** gives you the best results. By preparing for chapter 5 you can easily attempt the questions in the exam.

## NCERT Solutions For Class 12 Maths Chapter 5 – Continuity and Differentiability

Section Name |
Topic Name |

5.1 |
Introduction |

5.2 |
Continuity |

5.3 |
Algebra of Continuous Functions |

5.4 |
Exponential and Logarithmic Functions |

5.5 |
Logarithmic Differentiation |

5.6 |
Derivatives of Implicit Functions in Parametric Forms |

5.7 |
Derivatives of Inverse Trigonometric Functions |

5.8 |
Mean Value Theorem |

### Class 12 Maths NCERT Solutions Chapter 5 Continuity and Differentiability Ex 5.1 Introduction

**Question 1:
**Prove that the function is continuous at x=0, at x=-3, and at x=5

**Answer:**

Therefore,

*f*is continuous at

*x*= 0

Therefore, *f *is continuous at *x* = −3

Therefore, *f* is continuous at *x* = 5

**Question 2:
**Examine the continuity of the function

**Answer:**

Thus, *f* is continuous at *x* = 3

**Question 3:
**

Examine the following functions for continuity.

(a) (b) (c) (d)

**Answer:
**(a) The given function is

It is evident that

*f*is defined at every real number

*k*and 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 *f* and 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 *f* and therefore, it is a continuous function.

(d) The given function is

This function *f* is defined at all points of the real line.

Let *c* be 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, *f* is continuous at all real numbers less than 5.

Case II : *c* = 5

Then f(c)= f(5)=(5-5)=0

Therefore, *f *is continuous at *x* = 5

Case III: *c* > 5

Therefore, *f* is continuous at all real numbers greater than 5.

Hence,* f *is continuous at every real number and therefore, it is a continuous function.

**Question 4:
**Prove that the function is continuous at

*x*=

*n*, where n is a positive integer

**Answer:**

The given function is *f* (*x)* = *x*^{n}

It is evident that *f* is defined at all positive integers, *n*, and its value at *n* is *n*^{n}.

Therefore, *f *is continuous at *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:
**The given function

*f*is

At *x* = 0,

It is evident that* f *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.

Therefore, *f* is continuous at *x *= 2

**Question 6:
**

Find all points of discontinuity of *f*, where *f* is defined by

**Answer:
**The given function

*f*is

It is evident that the given function *f* is 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 *at* x *= 2 is,

The right hand limit of *f* at *x* = 2 is,

It is observed that the left and right hand limit of *f* at *x* = 2 do not coincide.

Therefore, *f* is not continuous at *x* = 2

Hence, *x* = 2 is the only point of discontinuity of *f*.

**Question 7:
**

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 *at* x *= 3 is,

The right hand limit of *f *at* x *= 3 is,

It is observed that the left and right hand limit 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 *x*, such that *x* > 3

Hence, *x* = 3 is the only point of discontinuity of *f*.

**Question 8:
**

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 *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* < 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 *f* at *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*.

**Question 9:
**

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

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.

**Question 10:
**

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* < 1

Case II:

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

The right hand limit of *f *at* x *= 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.

**Question 11:
**

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* < 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 *f* is continuous at every point on the real line.

Hence, *f *has no point of discontinuity.

**Question 12:
**

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* < 1

Case II:

If *c* = 1, then the left hand limit of *f* at *x* = 1 is,

The right hand limit of *f* at *x *= 1 is,

It is observed that the left and right hand limit 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 *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*.

**Question 13:
**Is the function defined by

a continuous function?

**Answer:**

The given function 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* < 1

Case II:

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

The right hand limit of *f* at *x *= 1 is,

It is observed that the left and right-hand limit 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 *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*

**Question 14:
**

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:

If c=1 then f(3)=3

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

The right hand limit of *f* at *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:

If c=3, then f(c)=5

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

The right hand limit of *f* at *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

**Question 15:
**

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:

If c=0, then f(c)=f(0)=0

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

The right hand limit of *f* at *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:

If c=1, then f(c)=f(1)=0

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

The right hand limit of *f* at *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

**Question 16:
**

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

**Answer:
**The given function

*f*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 *< −1

Case II:

If c=-1, then f(c)=f(-1)=-2

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

The right hand limit of *f* at *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:

If c=1, then f(c)=f(1)=2×1=2

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

The right hand limit of *f* at *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.

**Question 17:
**Find the relationship between

*a*and

*b*so that the function

*f*defined by

is continuous at

*x*= 3.

**Answer:**

The given function

*f*is

If

*f*is continuous at

*x*= 3, then

Therefore, from (1), we obtain

Therefore, the required relationship is given by,

**Question 18:
**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 *f* is continuous at *x* = 0

At *x* = 1,

*f* (1) = 4*x* + 1 = 4 × 1 + 1 = 5

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

**Question 19:
**Show that the function defined by g(x)=x-[x] is discontinuous at all integral point. Here [x] denotes the greatest integer less than or equal to

*x*.

**Answer:**

The given function is g(x)=x-[x]

It is evident that *g* is defined at all integral points.

Let *n* be an integer.

Then,

g(n)=n-[n]=n-n=0

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.

**Question 20:
**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

**Question 21:
**

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* × cos x

**Answer:
**It is known that if

*g*and

*h*are two continuous functions, then g+ h, g- h, and g.h

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

**Question 22:
**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 *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 continuous function.

It can be concluded that,

Therefore, cosecant is continuous except at *x *= *n*p, *n *Î **Z
**Therefore, secant is continuous except at

Therefore, cotangent is continuous except at

*x*=

*n*p,

*n*Î

**Z**

**Question 23:
**

Find the points of discontinuity of *f*, where

**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 all points *x*, such that* x* > 0

Case III:

If c=0, then f(c)=f(0)=0+1=1

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

The right hand limit of *f* at *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.

**Question 24:
**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:

⇒-x2≤x2sin1x≤x2

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.

**Question 25:
**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:

If c=0, then f(0)=-1

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.

**Question 26:
**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 at and

Therefore, the required value of *k* is 6.

**Question 27:
**Find the values of

*k*so that the function

*f*is continuous at the indicated point. at x=2

**Answer:**

The given function is

The given function

*f*is continuous at

*x*= 2, if

*f*is defined at

*x*= 2 and if the value of

*f*at

*x*= 2 equals the limit of

*f*at

*x*= 2

It is evident that

*f*is defined at

*x*= 2 and

Therefore, the required value of

**Question 28:
**Find the values of

*k*so that the function

*f*is continuous at the indicated point. at x=n

**Answer:**

The given function is

The given function

*f*is continuous at

*x*= p, if

*f*is defined at

*x*= p and if the value of

*f*at

*x*= p equals the limit of

*f*at

*x*= p

It is evident that

*f*is defined at

*x*= p and

Therefore, the required value of

**Question 29:**

Find the values of *k *so that the function *f* is continuous at the indicated point.

at x=5

**Answer:
**The given function

*f*is

The given function

*f*is continuous at

*x*= 5, if

*f*is defined at

*x*= 5 and if the value of

*f*at

*x*= 5 equals the limit of

*f*at

*x*= 5

It is evident that

*f*is defined at

*x*= 5 and

Therefore, the required value of

**Question 30:
**Find the values of

*a*and

*b*such that the function defined by

is a continuous function.

**Answer:**

The given function

*f*is

It is evident that the given function *f* is defined at all points of the real line.

If *f* is a continuous function, then *f* is continuous at all real numbers.

In particular, *f* is 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

8*a* = 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 *a* and *b* for which* f* is a continuous function are 2 and 1 respectively.

**Question 31:
**Show that the function defined by

*f*(

*x*) = cos (

*x*

^{2}) is a continuous function.

**Answer:**

The given function is *f *(*x*) = cos (*x*^{2})

This function *f* is defined for every real number and *f* can be written as the composition of two functions as,

*f* = *g o h*, where *g* (*x*) = cos *x* and *h* (*x*) = *x*^{2
}

It has to be first proved that *g *(*x*) = cos *x* and *h* (*x*) = *x*^{2} are continuous functions.

It is evident that *g* is 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*) = *x*^{2}

Clearly, *h* is defined for every real number.

Let *k* be a real number, then* h *(*k*) = *k*^{2
}

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.

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

**Answer:**

The given function is

This function

*f*is defined for every real number and

*f*can 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 *g* is 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,

continuous function.

**Question 33:
**Examine that is a continuous function.

**Answer:**

This function

*f*is defined for every real number and

*f*can 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 *g* is continuous at all points.

*h *(*x*) = sin *x*

It is evident that *h* (*x*) = sin* x* 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,

continuous function.

**Question 34:
**Find all the points of discontinuity of

*f*defined by

**Answer:**

The given function is

The two functions, *g* and *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 *g* is 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* h* is continuous at all points of the real line.

*g* and *h* are continuous functions. Therefore, *f *= *g* − *h *is also a continuous function.

Therefore, *f *has no point of discontinuity.

### NCERT Solutions For Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2 Continuity

**Question 1:
**

Differentiate the functions with respect to *x*.

**Answer:
**

Let f(x)=sinx2+5, ux=x2+5, and v(t)=sint

Then, vou=vux=vx2+5=tanx2+5=f(x)

Thus, *f* is a composite of two functions.

Alternate method

**Question 2:
**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

**Question 3:
**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

**Question 4:
**

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

**Question 5:
**

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
*Therefore, by chain rule, we obtain

**Question 6:
**

Differentiate the functions with respect to *x*.

**Answer:
**The given function is

**Question 7:
**

Differentiate the functions with respect to *x*.

**Answer:
**

**Question 8:
**

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

**Question 9:
**

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

**Question 10:
**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

### NCERT Solutions For Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 Algebra of Continuous Functions

**Question 1:
**

Find

**Answer:
**The given relationship is

Differentiating this relationship with respect to *x*, we obtain

**Question 2:
**

Find

**Answer:
**The given relationship is

Differentiating this relationship with respect to *x*, we obtain

**Question 3:
**

Find

**Answer:
**The given relationship is

Differentiating this relationship with respect to *x*, we obtain

Using chain rule, we obtain** **

From (1) and (2), we obtain

**Question 4:
**

Find

**Answer:
**The given relationship is

Differentiating this relationship with respect to

*x*, we obtain

**Question 5:
**

Find

**Answer:
**The given relationship is

Differentiating this relationship with respect to *x*, we obtain

[Derivative of constant function is 0]

**Question 6:
**

Find

**Answer:
**The given relationship is

Differentiating this relationship with respect to *x*, we obtain

**Question 7:
**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

**Question 8:
**

Find

**Answer:
**The given relationship is

Differentiating this relationship with respect to *x*, we obtain

**Question 9:
**Find

**Answer:**

We have,y = sin-12×1 + x2put x = tan θ ⇒ θ = tan-1xNow, y = sin-12 tan θ1 + tan2θ⇒y = sin-1sin 2θ, as sin 2θ=2 tan θ1 + tan2θ⇒y = 2θ, as sin-1sin x=x⇒y = 2 tan-1x⇒dydx = 2 × 11 + x2, because dtan-1xdx=11 + x2⇒dydx = 21 + x2

**Question 10
**

Find

**Answer:
**The given relationship is

…..(1)

It is known that,

Comparing equations (1) and (2), we obtain

Differentiating this relationship with respect to *x*, we obtain

**Question 11:
**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

sec2y2.ddxy2=ddxx

⇒sec2y2×12dydx=1

⇒dydx=2sec2y2

⇒dydx=21+tan2y2

∴

dydx=21+x2

**Question 12:
**

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

**Question 13:
**

Find

**Answer:
**The given relationship is

Differentiating this relationship with respect to

*x*, we obtain

**Question 14:
**Find

**Answer:**

The given relationship is

Differentiating this relationship with respect to *x*, we obtain

**Question 15:
**Find

**Answer:**

The given relationship is

Differentiating this relationship with respect to *x*, we obtain

### NCERT Solutions For Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4 Exponential and Logarithmic Functions

**Question 1:
**

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

**Answer:
**Let y=

By using the quotient rule, we obtain

**Question 2:
**

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

**Answer:
**Let y=

By using the chain rule, we obtain

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

*x*:

**Answer:**

Let y=

By using the chain rule, we obtain

**Question 4:
**

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

**Answer:
**Let y=

By using the chain rule, we obtain

**Question 5:
**

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

**Answer:
**Let y=

By using the chain rule, we obtain

**Question 6:
**

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

**Answer:
**

**Question 7: **

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

**Answer:
**Let y=

then,

By differentiating this relationship with respect to *x*, we obtain

**Question 8:
**

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

**Answer:
**Let y= log (log x)

By using the chain rule, we obtain

x>1

**Question 9:
**

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

**Answer:
**Let y=

By using the quotient rule, we obtain

**Question 10:
**

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

**Answer:
**Let y=

By using the chain rule, we obtain

### NCERT Solutions For Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Logarithmic Differentiation

**Question 1:
**

Differentiate the function with respect to *x*.

**Answer:
**Let y=

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to *x*, we obtain

**Question 2:
**

Differentiate the function with respect to *x*.

**Answer:
**Let y=

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to *x*, we obtain

**Question 3:
**

Differentiate the function with respect to *x*.

**Answer:
**Let y=

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to *x*, we obtain

**Question 4:
**

Differentiate the function with respect to *x*.

**Answer:
**

*u *= *x*^{x}

Taking logarithm on both the sides, we obtain

log u= x log x

Differentiating both sides with respect to *x*, we obtain

*v* = 2^{sin }^{x}

Taking logarithm on both the sides with respect to *x*, we obtain

log v=sin x. log 2

Differentiating both sides with respect to *x*, we obtain

**Question 5:
**

Differentiate the function with respect to *x*.

Answer:

Let y=

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to *x*, we obtain

**Question 6:
**

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

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

**Question 7:
**

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

**Question 8:
**

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

**Question 9:
**

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

**Question 10:
**

Differentiate the function with respect to *x*.

**Answer:
**Let y=

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

**Question 11:
**

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

**Question 12:
**Find of function.

**Answer:**

The given function is

Let

*x*

^{y}=

*u*and

*y*

^{x}=

*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

**Question 13:
**Find of function

**Answer:**

The given function is

Taking logarithm on both the sides, we obtain

x log y= y log x

Differentiating both sides with respect to *x*, we obtain

**Question 14:
**Find of function.

**Answer:**

The given function is

Taking logarithm on both the sides, we obtain

y log cos x= x log cos y

Differentiating both sides, we obtain

**Question 15:
**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

**Question 16:
**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

**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:
**Let

By applying product rule, we obtain

log y= log u+ log v+ log w

Differentiating both sides with respect to

*x*, we obtain

### NCERT Solutions For Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.6 Derivatives of Implicit Functions in Parametric Forms

**Question 1:
**If

*x*and

*y*are connected parametrically by the equation, without eliminating the parameter, find

**Answer:**

The given equations are

**Question 2:
**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 are

*x*=

*a*cos

*θ*and

*y*=

*b*cos

*θ*

**Question 3:
**If

*x*and

*y*are connected parametrically by the equation, without eliminating the parameter, find

*x*= sin

*t*,

*y*= cos 2

*t*

**Answer:**

The given equations are

*x*= sin

*t*and

*y*= cos 2

*t*

**Question 4:
**If

*x*and

*y*are connected parametrically by the equation, without eliminating the parameter, find

**Answer:**

The given equations are

**Question 5:
**If

*x*and

*y*are connected parametrically by the equation, without eliminating the parameter, find

**Answer:**

The given equations are

**Question 6:
**If

*x*and

*y*are connected parametrically by the equation, without eliminating the parameter, find

**Answer:**

The given equations are

**Question 7:
**If

*x*and

*y*are connected parametrically by the equation, without eliminating the parameter, find

**Answer:**

The given equations are

**Question 8:
**If

*x*and

*y*are connected parametrically by the equation, without eliminating the parameter, find

**Answer:**

The given equations are

**Question 9:
**If

*x*and

*y*are connected parametrically by the equation, without eliminating the parameter, find

**Answer:**

The given equations are

**Question 10:
**If

*x*and

*y*are connected parametrically by the equation, without eliminating the parameter, find

**Answer:**

The given equations are

**Question 11:
**If

**Answer:**

The given equations are

Hence, proved

### NCERT Solutions For Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7 Derivatives of Inverse Trigonometric Functions

**Question 1:
**Find the second order derivatives of the function.

**Answer:**

Let y=

Then,

**Question 2:
**Find the second order derivatives of the function.

**Answer:**

Let y=

Then,

**Question 3:
**Find the second order derivatives of the function.

x. cos x

**Answer:**

Let y= x. cos x

Then,

**Question 4:**

Find the second order derivatives of the function. log x

**Answer:
**Let y= log x

Then,

**Question 5:
**Find the second order derivatives of the function.

**Answer:**

Let y=

then,

**Question 6:
**Find the second order derivatives of the function.

**Answer:**

Let y=

then,

**Question 7:
**Find the second order derivatives of the function.

**Answer:**

Let y=

then,

**Question 8:
**Find the second order derivatives of the function.

**Answer:**

Let y=

then,

**Question 9:
**Find the second order derivatives of the function.

log(log x)

**Answer:**

Let y= log(log x)

then,

**Question 10:
**Find the second order derivatives of the function.

sin(log x)

**Answer:**

Let y= sin(log x)

then,

**Question 11:
**If y= 5 cos x-3 sin x, prove that

**Answer:**

It is given that, y= 5 cos x-3 sin x

then,

Hence, proved.

**Question 12:
**If find, in terms of

*y*alone.

**Answer:**

It is given that,

then,

**Question 13:
**If y= 3cos(log x)+4sin(log x), show that

**Answer:**

It is given that, y= 3cos(log x)+4sin(log x)

then,

Hence, proved.

**Question** **14:**

If show that,

**Answer:
**It is given that,

then,

**Question 15:
**If show that,

**Answer:**

It is given that,

then,

Hence, proved.

**Question 16:
**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.

**Question 17:
**If show that,

**Answer:**

The given relationship is

then,

### NCERT Solutions For Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8 Mean Value theorem

**Question 1:**

Verify Rolle’s Theorem for the function ,x∈[4-2]

**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.

**Question 2:
**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) f(x)=[x] for x∈[5,9]

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].

also f(5)=[5]=5 and f(9)=[9]
∴f(5)≠f(9)

The differentiability of *f* in (5, 9) is checked as follows.

Let *n *be an integer such that *n* ∈ (5, 9).

the left hand limit of f at x=n is

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

f(x)=[x]for x∈[5,9]
(ii) f(x)=[x] for x∈[-2,2]

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

f(x)=[x] for x∈[-2,2]
(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

**Question 3:
**If f:[-5,5]→R is a differentiable function and if does not vanish anywhere, then prove that

**Answer:**

It is given that f:[-5,5]→R

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.

**Question 4:
**Verify Mean Value Theorem, if in the interval [a,b], where a=1 and b=4

**Answer:**

The given function is

*f*, being a polynomial function, is continuous in [1, 4] and is differentiable in (1, 4) whose derivative is 2

*x*− 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.

**Question 5:
**Verify Mean Value Theorem, if in the interval [

*a*,

*b*], where

*a*= 1 and

*b*= 3. Find all c∈(1,3)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

*x*

^{2}− 10

*x*− 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

**Question 6:
**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 f:[a, b]→R, 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) f(x)=[x] for x∈[5,9]

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

f(x)=[x] for x∈[5,9]
(ii) f(x)=[x] for x∈[-2,2]

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

f(x)=[x]for x∈[-2,2]
(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.

### NCERT Solutions For Class 12 Maths Chapter 5 Continuity and Differentiability Miscellaneous Solutions

**Question 1:
**

**Answer:**

Let y=

Using chain rule, we obtain

**Question 2:
Answer:
**

**
Question 3:
Answer:
**Let y=

Taking logarithm on both the sides, we obtain

log y=3 cos 2x log 5x

Differentiating both sides with respect to *x*, we obtain

**Question 4:
Answer:
**Let y=

Using chain rule, we obtain

**Question 5:
Answer:
**

Question 6:

Answer:

Therefore, equation (1) becomes

**Question 7:
Answer:
**

Taking logarithm on both the sides, we obtain

log y=log x.log(log x)

Differentiating both sides with respect to *x*, we obtain

**Question 8:
**for some constant

*a*and

*b*

**Answer:**

Let y=

By using chain rule, we obtain

**Question 9:
Answer:
**

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to *x*, we obtain

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

**Answer:**

Differentiating both sides with respect to *x*, we obtain

Differentiating both sides with respect to *x*, we obtain

*s* = *a*^{a}

Since *a* is constant, *a*^{a} is also a constant.

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

**Question 11:
**for x>3

**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

**Question 12:
**Find if

**Answer:**

**Question 13:
**Find if

**Answer:**

**Question 14:
**If for, −1 <

*x*<1, prove that

**Answer:**

_{It is given that,
}

_{Differentiating both sides with respect to }_{x}_{, we obtain
}

_{Hence, proved.}

**Question 15:
**If for some c>0. prove that

*a*and A

**Answer:**

It is given that,

Differentiating both sides with respect to

*x*, we obtain

Hence, proved.

**Question 16:
**If cos y = x cos(a + y), with cos a≠±1, prove that

**Answer:**

⇒sina+y-ydydx=cos2a+y⇒dydx=cos2a+ysinaHence, proved.

**Question 17:
**If and find

**Answer:**

**
Question 18:
**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

*x*and is given by,

**Question 19:
**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.}

_{
}**Question 20:
**Using the fact that sin (

*A*+

*B*) = sin

*A*cos

*B*+ cos

*A*sin

*B*and the differentiation, obtain the sum formula for cosines.

**Answer:**

sin(A+B)=sin A cos B+ cos A sin B

Differentiating both sides with respect to *x*, we obtain

**Question 21:
**Does there exist a function which is continuos everywhere but not differentiable at exactly two points? Justify your answer ?

**Answer:**

y=x -∞<x≤1 2-x 1≤x≤∞

It can be seen from the above graph that, the given function is continuos everywhere but not differentiable at exactly two points which are 0 and 1.

**Question 22:
**If prove that

**Answer:**

thus,

**Question 23:
**If -1≤x≤1 show that

**Answer:**

It is given that,

Hence proved.