Ncert Solution Class 12 Chapter Number 5

NCERT Solutions for Class 12 Science Math Chapter 5 Continuity And Differentiability are provided here with simple step-by-step explanations. These solutions for Continuity And Differentiability are extremely popular among Class 12 Science students for Math Continuity And Differentiability Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 12 Science Math Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation's NCERT Solutions. All NCERT Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate.

Page No 159:

Question 1:

Prove that the function is continuous at

Answer:

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{function}\mathrm{is}f\left(x\right)=5x-3 \\ \mathrm{At}x=0,f\left(0\right)=5\times 0-3=-3 \\ \underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\left(5x-3\right)=5\times 0-3=-3 \\ \therefore \underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right) \end{gathered}$$

Therefore, f is continuous at x = 0

Therefore, f is continuous at x = −3

Therefore, f is continuous at x = 5

Page No 159:

Question 2:

Examine the continuity of the function .

Answer:

Thus, f is continuous at x = 3

Page No 159:

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 ,

Therefore , f is continuous at x = 5

Case II I: 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.

Page No 159:

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.

Page No 159:

Question 5:

Is the function f defined by

c ontinuous 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.

T he left hand limit of f at x = 1 is,

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

Th erefore, 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

Page No 159:

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

T hen, the left hand limit of f at x = 2 is,

T he 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 .

Page No 159:

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,

T he 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 .

Page No 159:

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 .

Page No 159:

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,

Th erefore, the given function is a continuous function.

Hence, the given function has no point of discontinuity.

Page No 159:

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.

Page No 159:

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.

Page No 159:

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,

T he 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, f rom the above observation, it can be concluded that x = 1 is the only point of discontinuity of f .

Page No 159:

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 :

T he left hand limit of f at x = 1 is,

T he 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, f rom the above observation, it can be concluded that x = 1 is the only point of discontinuity of f .

Page No 160:

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 :

T he 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 s 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 :

T he 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 s 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

Page No 160:

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 :

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 :

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 s 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

Page No 160:

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 :

T he 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 :

T he left hand limit of f at x = 1 is,

T he 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.

Page No 160:

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,

Page No 160:

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

Page No 160:

Question 19:

Show that the function defined by is discontinuous at all integral point. Here denotes the greatest integer less than or equal to x .

Answer:

The given function is

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

Let n be an integer.

Then,

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

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

It is observed that the left and right hand limits of f at x = n do not coincide.

Therefore , f is not continuous at x = n

H ence, g is discontinuous at all integral points.

Page No 160:

Question 20:

Is the function defined by continuous at x = $\mathrm{\pi }$ ?

Answer:

The given function is

It is evident that f is defined at x =$\mathrm{\pi }$.

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

Page No 160:

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

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

I f 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

I f 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

Page No 160:

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 = np, n Î Z

Therefore, secant is continuous except at

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

Page No 160:

Question 23:

Find the point s 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 :

T he 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.

Page No 160:

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 :

$\Rightarrow -x^2\le x^2\sin \frac{1}{x}\le x^2$

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.

Page No 161:

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 :

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.

Page No 161:

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.

Page No 161:

Question 27:

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

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.

Page No 161:

Question 28:

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

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

Page No 161:

Question 29:

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 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

Th erefore, the required value of

Page No 161:

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

Th erefore, the values of a and b for which f is a continuous function are 2 and 1 respectively.

Page No 161:

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.

Page No 161:

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

I f x → c , then h → 0

h ( c ) = cos c

Theref ore, h ( x ) = cos x is a continuous function.

It is known that for real valued functions g and h,such that (g o h) is defined at c, if g is continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.

Therefore, is a continuous function.

Page No 161:

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

I f x → c , then k → 0

h ( c ) = sin c

Therefore , h is a continuous function.

It is known that for real valued functions g and h,such that (g o h) is defined at c, if g is continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.

Therefore, is a continuous function.

Page No 161:

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.

Th erefore, f has no point of discontinuity.

Page No 166:

Question 1:

Differentiate the functions with respect to x.

Answer:

$\mathrm{Let}f\left(x\right)=\sin \left(x^2+5\right),u\left(x\right)=x^2+5,\mathrm{and}v\left(t\right)=\sin t$
$\mathrm{Then},\left(vou\right)=v\left(u\left(x\right)\right)=v\left(x^2+5\right)=\tan \left(x^2+5\right)=f\left(x\right)$
Thus, f is a composite of two functions.

Alternate method

Page No 166:

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

Page No 166:

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

Page No 166:

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

Page No 166:

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

Page No 166:

Question 6:

Differentiate the functions with respect to x.

Answer:

The given function is.

Page No 166:

Question 7:

Differentiate the functions with respect to x.

Answer:

Page No 166:

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

Page No 166:

Question 9:

Prove that the function f given by

is notdifferentiable 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

Page No 166:

Question 10:

Prove that the greatest integer function defined byis 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

Page No 169:

Question 1:

Find :

Answer:

The given relationship is

Differentiating this relationship with respect to x , we obtain

Page No 169:

Question 2:

Find :

Answer:

The given relationship is

Differentiating this relationship with respect to x , we obtain

Page No 169:

Question 3:

Find :

Answer:

The given relationship is

Differentiating this relationship with respect to x , we obtain

U sing chain rule, we obtain and

From (1) and (2), we obtain

Page No 169:

Question 4:

Find :

Answer:

The given relationship is

Differentiating this relationship with respect to x , we obtain

Page No 169:

Question 5:

Find :

Answer:

The given relationship is

Differentiating this relationship with respect to x , we obtain

[Derivative of constant function is 0]

Page No 169:

Question 6:

Find :

Answer:

The given relationship is

Differentiating this relationship with respect to x , we obtain

Page No 169:

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

Page No 169:

Question 8:

Find :

Answer:

The given relationship is

Differentiating this relationship with respect to x , we obtain

Page No 169:

Question 9:

Find :

Answer:

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ y={\sin }^{-1}\left[\frac{2x}{1+x^2}\right] \\ \mathrm{put}x=\tan \theta \Rightarrow \theta ={\tan }^{-1}x \\ \mathrm{Now}, \\ y={\sin }^{-1}\left[\frac{2\tan \theta }{1+{\tan }^2\theta }\right] \\ \Rightarrow y={\sin }^{-1}\left(\sin 2\theta \right),\left(\mathrm{as}\sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }\right) \\ \Rightarrow y=2\theta ,\left(\mathrm{as}{\sin }^{-1}\left(\sin x\right)=x\right) \\ \Rightarrow y=2{\tan }^{-1}x \\ \Rightarrow \frac{dy}{dx}=2\times \frac{1}{1+x^2},\left\{\mathrm{because}\frac{d\left({\tan }^{-1}x\right)}{dx}=\frac{1}{1+x^2}\right\} \\ \Rightarrow \frac{dy}{dx}=\frac{2}{1+x^2} \end{gathered}$$

Page No 169:

Question 10:

Find :

Answer:

The given relationship is

It is known that,

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

Differentiating this relationship with respect to x , we obtain

Page No 169:

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

${\sec }^2\left(\frac{\mathrm{y}}{2}\right).\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{y}}{2}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}\right)$
$\Rightarrow {\sec }^2\frac{y}{2}\times \frac{1}{2}\frac{\mathrm{dy}}{\mathrm{dx}}=1$
$\Rightarrow \frac{dy}{dx}=\frac{2}{se{c}^2{\displaystyle \frac{y}{2}}}$
$\Rightarrow \frac{dy}{dx}=\frac{2}{1+{\tan }^2{\displaystyle \frac{y}{2}}}$
∴ $\frac{dy}{dx}=\frac{2}{1+x^2}$

Page No 169:

Question 12:

Find :

Answer:

The given relationship is

Differentiating this relationship with respect to x , we obtain

U sing chain rule, we obtain

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

Alternate method

⇒

Differentiating this relationship with respect to x , we obtain

Page No 169:

Question 13:

Find :

Answer:

The given relationship is

Differentiating this relationship with respect to x , we obtain

Page No 169:

Question 14:

Find :

Answer:

The given relationship is

Differentiating this relationship with respect to x , we obtain

Page No 169:

Question 15:

Find :

Answer:

The given relationship is

Differentiating this relationship with respect to x , we obtain

Page No 174:

Question 1:

Differentiate the following w.r.t. x:

Answer:

Let

By using the quotient rule, we obtain

Page No 174:

Question 2:

Differentiate the following w.r.t. x:

Answer:

Let

By using the chain rule, we obtain

Page No 174:

Question 3:

Differentiate the following w.r.t. x:

Answer:

Let

By using the chain rule, we obtain

Page No 174:

Question 4:

Differentiate the following w.r.t. x:

Answer:

Let

By using the chain rule, we obtain

Page No 174:

Question 5:

Differentiate the following w.r.t. x:

Answer:

Let

By using the chain rule, we obtain

Page No 174:

Question 6:

Differentiate the following w.r.t. x:

Answer:

Page No 174:

Question 7:

Differentiate the following w.r.t. x:

Answer:

Let

Then,

By differentiating this relationship with respect to x, we obtain

Page No 174:

Question 8:

Differentiate the following w.r.t. x:

Answer:

Let

By using the chain rule, we obtain

, x > 1

Page No 174:

Question 9:

Differentiate the following w.r.t. x:

Answer:

Let

By using the quotient rule, we obtain

Page No 174:

Question 10:

Differentiate the following w.r.t. x:

Answer:

Let

By using the chain rule, we obtain

Page No 178:

Question 1:

Differentiate the function with respect to x.

Answer:

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

Page No 178:

Question 2:

Differentiate the function with respect to x.

Answer:

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

Page No 178:

Question 3:

Differentiate the function with respect to x.

Answer:

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

Page No 178:

Question 4:

Differentiate the function with respect to x.

Answer:

u = x ^x

Taking logarithm on both the sides, we obtain

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

Differentiating both sides with respect to x, we obtain

Page No 178:

Question 5:

Differentiate the function with respect to x.

Answer:

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

Page No 178:

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

Page No 178:

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

Page No 178:

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

Page No 178:

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

Page No 178:

Question 10:

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

Page No 178:

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

Page No 178:

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

Page No 178:

Question 13:

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

Page No 178:

Question 14:

Find of function.

Answer:

The given function is

Taking logarithm on both the sides, we obtain

Differentiating both sides, we obtain

Page No 178:

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

Page No 178:

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

Page No 178:

Question 17:

Differentiate in three ways mentioned below

(i) By using product rule.

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

(iii By logarithmic differentiation.

Do they all give the same answer?

Answer:

(i)

(ii)

(iii)

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

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

Page No 179:

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

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

Differentiating both sides with respect to x, we obtain

Page No 181:

Question 1:

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

Answer:

The given equations are

Page No 181:

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

Page No 181:

Question 3:

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

x = sin t, y = cos 2t

Answer:

The given equations are x = sin t and y = cos 2t

Page No 181:

Question 4:

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

Answer:

The given equations are

Page No 181:

Question 5:

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

Answer:

The given equations are

Page No 181:

Question 6:

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

Answer:

The given equations are

Page No 181:

Question 7:

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

Answer:

The given equations are

Page No 181:

Question 8:

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

Answer:

The given equations are

Page No 181:

Question 9:

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

Answer:

The given equations are

Page No 181:

Question 10:

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

Answer:

The given equations are

Page No 181:

Question 11:

If

Answer:

The given equations are

Hence, proved.

Page No 183:

Question 1:

Find the second order derivatives of the function.

Answer:

Let

Then,

Page No 183:

Question 2:

Find the second order derivatives of the function.

Answer:

Let

Then,

Page No 183:

Question 3:

Find the second order derivatives of the function.

Answer:

Let

Then,

Page No 183:

Question 4:

Find the second order derivatives of the function.

Answer:

Let

Then,

Page No 183:

Question 5:

Find the second order derivatives of the function.

Answer:

Let

Then,

Page No 183:

Question 6:

Find the second order derivatives of the function.

Answer:

Let

Then,

Page No 183:

Question 7:

Find the second order derivatives of the function.

Answer:

Let

Then,

Page No 183:

Question 8:

Find the second order derivatives of the function.

Answer:

Let

Then,

Page No 183:

Question 9:

Find the second order derivatives of the function.

Answer:

Let

Then,

Page No 183:

Question 10:

Find the second order derivatives of the function.

Answer:

Let

Then,

Page No 183:

Question 11:

If, prove that

Answer:

It is given that,

Then,

Hence, proved.

Page No 184:

Question 12:

If findin terms of y alone.

Answer:

It is given that,

Then,

Page No 184:

Question 13:

If, show that

Answer:

It is given that,

Then,

Hence, proved.

Page No 184:

Question 14:

Ifshow that

Answer:

It is given that,

Then,

Hence, proved.

Page No 184:

Question 15:

If, show that

Answer:

It is given that,

Then,

Hence, proved.

Page No 184:

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.

Page No 184:

Question 17:

If, show that

Answer:

The given relationship is

Then,

Hence, proved.

Page No 186:

Question 1:

Verify Rolle's Theorem for the function

Answer:

The given function,, being a polynomial function, is continuous in [−4, 2] and is differentiable in (−4, 2).

∴ f (−4) = f (2) = 0

⇒ The value of f (x) at −4 and 2 coincides.

Rolle's Theorem states that there is a point c ∈ (−4, 2) such that

Hence, Rolle's Theorem is verified for the given function.

Page No 186:

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)

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at x = 5 and x = 9

⇒ f (x) is not continuous in [5, 9].

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

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

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

∴f is not differentiable in (5, 9).

It is observed that f does not satisfy all the conditions of the hypothesis of Rolle's Theorem.

Hence, Rolle's Theorem is not applicable for.

(ii)

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at x = −2 and x = 2

⇒ f (x) is not continuous in [−2, 2].

The differentiability of f in (−2, 2) is checked as follows.

Let n be an integer such that n ∈ (−2, 2).

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

∴f is not differentiable in (−2, 2).

It is observed that f does not satisfy all the conditions of the hypothesis of Rolle's Theorem.

Hence, Rolle's Theorem is not applicable for.

(iii)

It is evident that f, being a polynomial function, is continuous in [1, 2] and is differentiable in (1, 2).

∴f (1) ≠ f (2)

It is observed that f does not satisfy a condition of the hypothesis of Rolle's Theorem.

Hence, Rolle's Theorem is not applicable for.

Page No 186:

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.

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.

Page No 186:

Question 4:

Verify Mean Value Theorem, if in the interval, where and.

Answer:

The given function is

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

Mean Value Theorem states that there is a point c ∈ (1, 4) such that

Hence, Mean Value Theorem is verified for the given function.

Page No 186:

Question 5:

Verify Mean Value Theorem, if in the interval [a, b], where a = 1 and b = 3. Find all for which

Answer:

The given function f is

f, being a polynomial function, is continuous in [1, 3] and is differentiable in (1, 3) whose derivative is 3x ² − 10x − 3.

Mean Value Theorem states that there exist a point c ∈ (1, 3) such that

Hence, Mean Value Theorem is verified for the given function and is the only point for which

Page No 186:

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, if

(a) f is continuous on [a, b]

(b) f is differentiable on (a, b)

then, there exists some c ∈ (a, b) such that

Therefore, Mean Value Theorem is not applicable to those functions that do not satisfy any of the two conditions of the hypothesis.

(i)

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at x = 5 and x = 9

⇒ f (x) is not continuous in [5, 9].

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

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

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

∴f is not differentiable in (5, 9).

It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is not applicable for .

(ii)

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at x = −2 and x = 2

⇒ f (x) is not continuous in [−2, 2].

The differentiability of f in (−2, 2) is checked as follows.

Let n be an integer such that n ∈ (−2, 2).

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

∴f is not differentiable in (−2, 2).

It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is not applicable for .

(iii)

It is evident that f, being a polynomial function, is continuous in [1, 2] and is differentiable in (1, 2).

It is observed that f satisfies all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is applicable for.

It can be proved as follows.

Page No 191:

Question 1:

Answer:

Using chain rule, we obtain

Page No 191:

Question 2:

Answer:

Page No 191:

Question 3:

Answer:

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x , we obtain

Page No 191:

Question 4:

Answer:

Using chain rule, we obtain

Page No 191:

Question 5:

Answer:

Page No 191:

Question 6:

Answer:

Therefore , equation (1) becomes

Page No 191:

Question 7:

Answer:

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x , we obtain

Page No 191:

Question 8:

, for some constant a and b .

Answer:

B y using chain rule, we obtain

Page No 191:

Question 9:

Answer:

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x , we obtain

Page No 191:

Question 10:

, for some fixed and

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

Page No 191:

Question 11:

, for

Answer:

Differentiating both sides with respect to x, we obtain

Differentiating with respect to x , we obtain

Also,

Differentiating both sides with respect to x , we obtain

Substituting the expressions of in equation (1), we obtain

Page No 191:

Question 12:

Find , if

Answer:

Page No 191:

Question 13:

Find , if

Answer:

Page No 191:

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.}

Page No 191:

Question 15:

If , for some prove that

is a constant independent of a and b .

Answer:

It is given that,

Differentiating both sides with respect to x , we obtain

Hence, proved.

Page No 192:

Question 16:

If with prove that

Answer:

Then, equation (1) reduces to

$$\begin{gathered} \Rightarrow \sin \left(a+y-y\right)\frac{dy}{dx}={\cos }^2\left(a+y\right) \\ \Rightarrow \frac{dy}{dx}=\frac{{\cos }^2\left(a+y\right)}{\sin a} \end{gathered}$$

Hence, proved.

Page No 192:

Question 17:

If and, find

Answer:

Page No 192:

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,

Page No 192:

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.}

Page No 192:

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:

Differentiating both sides with respect to x , we obtain

Page No 192:

Question 21:

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

Answer:

$y=\left\{\begin{array}{l}\left|x\right|-\infty <x\le 1\\ 2-x1\le x\le \infty \end{array}\right.$

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.

Page No 192:

Question 22:

If , prove that

Answer:

Thus,

Page No 192:

Question 23:

If , show that

Answer:

It is given that,

View NCERT Solutions for all chapters of Class 12

Ncert Solution Class 12 Chapter Number 5

Source: https://www.meritnation.com/cbse-class-12-science/math/math-part-i-ncert-solutions/continuity-and-differentiability/ncert-solutions/46_1_1373_234_159_6748

Share :