NCERT Solutions for Class 12 Science Math Chapter 5 - Continuity And Differentiability

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

Answer:

Thus, f is continuous at x = 3

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 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.

Answer:

The given function is f (x) = xⁿ

It is evident that f is defined at all positive integers, n, and its value at n is nⁿ.

Therefore, f is continuous at n, where n is a positive integer.

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

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.

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.

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.

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.

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.

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.

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.

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.

Answer:

The given function is

The given function is defined at all points of the interval [0, 10].

Let c be a point in the interval [0, 10].

Case I:

Therefore, f is continuous in the interval [0, 1).

Case II:

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

The right hand limit of 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:

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

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 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

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:

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:

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.

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,

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) = 4x + 1 = 4 × 1 + 1 = 5

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

Answer:

The given function is

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

Let n be an integer.

Then,

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

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

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

Therefore, f is not continuous at x = n

Hence, g is discontinuous at all integral points.

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

Answer:

It is known that if g and h are two continuous functions, then

are also continuous.

It has to proved first that g (x) = sin x and h (x) = cos x are continuous functions.

Let g (x) = sin x

It is evident that g (x) = sin x is defined for every real number.

Let c be a real number. Put x = c + h

If x → c, then h → 0

Therefore, g is a continuous function.

Let h (x) = cos x

It is evident that h (x) = cos x is defined for every real number.

Let c be a real number. Put x = c + h

If x → c, then h → 0

h (c) = cos c

Therefore, h is a continuous function.

Therefore, it can be concluded that

(a) f (x) = g (x) + h (x) = sin x + cos x is a continuous function

(b) f (x) = g (x) − h (x) = sin x − cos x is a continuous function

(c) f (x) = g (x) × h (x) = sin x × cos x is a continuous function

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

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:

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.

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.

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.

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.

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.

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

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

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

8a = 16

⇒ a = 2

By putting a = 2 in equation (1), we obtain

2 × 2 + b = 5

⇒ 4 + b = 5

⇒ b = 1

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

Answer:

The given function is f (x) = cos (x²)

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) = x²

It has to be first proved that g (x) = cos x and h (x) = x² 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) = x²

Clearly, h is defined for every real number.

Let k be a real number, then h (k) = k²

Therefore, h is a continuous function.

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

Therefore, is a continuous function.

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, 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, is a continuous function.

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.

Answer:

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

Alternate method

Answer:

Thus, f is a composite function of two functions.

Put t = u (x) = sin x

By chain rule,

Alternate method

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

Answer:

Thus, f is a composite function of three functions, u, v, and w.

Hence, by chain rule, we obtain

Alternate method

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

Answer:

The given function is.

Answer:

Answer:

Clearly, f is a composite function of two functions, u and v, such that

By using chain rule, we obtain

Alternate method

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

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

Answer:

The given relationship is

Differentiating this relationship with respect to x, we obtain

Answer:

The given relationship is

Differentiating this relationship with respect to x, we obtain

Answer:

The given relationship is

Differentiating this relationship with respect to x, we obtain

Using chain rule, we obtain and

From (1) and (2), we obtain

Answer:

The given relationship is

Differentiating this relationship with respect to x, we obtain

Answer:

The given relationship is

Differentiating this relationship with respect to x, we obtain

[Derivative of constant function is 0]

Answer:

The given relationship is

Differentiating this relationship with respect to x, we obtain

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

Answer:

The given relationship is

Differentiating this relationship with respect to x, we obtain

Answer:

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

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

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

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

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

Answer:

The given relationship is

Differentiating this relationship with respect to x, we obtain

Answer:

The given relationship is

Differentiating this relationship with respect to x, we obtain

Answer:

The given relationship is

Differentiating this relationship with respect to x, we obtain

Answer:

Let

By using the quotient rule, we obtain

Answer:

Let

By using the chain rule, we obtain

Answer:

Let

By using the chain rule, we obtain

Answer:

Let

By using the chain rule, we obtain

Answer:

Let

By using the chain rule, we obtain

Answer:

Answer:

Let

Then,

By differentiating this relationship with respect to x, we obtain

Answer:

Let

By using the chain rule, we obtain

, x > 1

Answer:

Let

By using the quotient rule, we obtain

Answer:

Let

By using the chain rule, we obtain

Answer:

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

Answer:

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

Answer:

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

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

Answer:

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

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

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

Answer:

Differentiating both sides with respect to x, we obtain

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

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

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

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

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

Answer:

The given function is

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

Answer:

The given function is

Taking logarithm on both the sides, we obtain

Differentiating both sides, we obtain

Answer:

The given function is

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

Answer:

The given relationship is

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

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.

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

Answer:

The given equations are

Answer:

The given equations are x = a cos θ and y = b cos θ

Answer:

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

Answer:

The given equations are

Answer:

The given equations are

Answer:

The given equations are

Answer:

The given equations are

Answer:

The given equations are

Answer:

The given equations are

Answer:

The given equations are

Answer:

The given equations are

Hence, proved.

Answer:

Let

Then,

Answer:

Let

Then,

Answer:

Let

Then,

Answer:

Let

Then,

Answer:

Let

Then,

Answer:

Let

Then,

Answer:

Let

Then,

Answer:

Let

Then,

Answer:

Let

Then,

Answer:

Let

Then,

Answer:

It is given that,

Then,

Hence, proved.

Answer:

It is given that,

Then,

Answer:

It is given that,

Then,

Hence, proved.

Answer:

It is given that,

Then,

Hence, proved.

Answer:

It is given that,

Then,

Hence, proved.

Answer:

The given relationship is

Taking logarithm on both the sides, we obtain

Differentiating this relationship with respect to x, we obtain

Hence, proved.

Answer:

The given relationship is

Then,

Hence, proved.

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.

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.

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.

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.

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 3x² − 10x − 3.

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

– 10

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

Now, we have 
$$\begin{gathered} f\text{'}\left(c\right)=0 \\ \Rightarrow 3c^2-10c-3=0 \\ \Rightarrow c=\frac{-\left(-10\right)\pm \sqrt{{\left(-10\right)}^2-4\times 3\times \left(-3\right)}}{2\times 3} \\ \Rightarrow c=\frac{10\pm \sqrt{100+36}}{6} \\ \Rightarrow c=\frac{10\pm \sqrt{136}}{6} \\ \Rightarrow c=\frac{10\pm 2\sqrt{34}}{6} \\ \Rightarrow c=\frac{5\pm \sqrt{34}}{3} \\ \Rightarrow c=\frac{5+\sqrt{34}}{3}, \frac{5-\sqrt{34}}{3} \\ \Rightarrow c=3.61, -0.27 \end{gathered}$$

Since these values of c does not belong in the interval (1, 3), there is no value of c for which 

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.

Answer:

Using chain rule, we obtain

Answer:

Answer:

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

Answer:

Using chain rule, we obtain

Answer:

Answer:

Therefore, equation (1) becomes

Answer:

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

Answer:

By using chain rule, we obtain

Answer:

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

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

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

Answer:

Answer:

Answer:

_{It is given that,}

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

_{Hence, proved.}

Answer:

It is given that,

Differentiating both sides with respect to x, we obtain

Hence, proved.

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.

Answer:

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,

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.}

Answer:

Differentiating both sides with respect to x, we obtain

Answer:

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


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.

Answer:

Thus,

Answer:

It is given that,