# CLASS 12 MATHS CHAPTER 5-CONTINUITY & DIFFERENTIABILITY

### Solve The Following Questions.

Question1. Prove that the function f(x) = 5x – 3 is continuous at x = 0, at  x = – 3 and x = 5

Solution :

Question2. Examine the continuity of the function f(x) = 2x2 – 1 at x = 3

Solution :

Thus, f is continuous at x = 3

Question3. Examine the following functions for continuity.

(a)

(c)

Solution :

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.

Question4. Prove that the function f(x) = xis continuous at x = n, where n is a positive integer.

Solution :
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.

Question5. Is the function f defined by

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

Solution :
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

Question6. Find all points of discontinuity of f, where f is defined by

Solution :

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.

Question7. Find all points of discontinuity of f, where f is defined by

Solution :
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.

Question8. Find all points of discontinuity of f, where f is defined by

Solution :

Question9. Find all points of discontinuity of f, where f is defined by

Solution :

Question10. Find all points of discontinuity of f, where f is defined by

Solution :

Therefore, f is continuous at all points x, such that x > 1

Hence, the given function has no point of discontinuity.

Question11. Find all points of discontinuity of f, where f is defined by

Solution :

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.

Question12. Find all points of discontinuity of f, where f is defined by

Solution :
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.

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.

Question13. Is the function defined by

a continuous function?

Solution :
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

Thus, from the above observation, it can be concluded that x = 1 is the only point of discontinuity of f.

Question14. Discuss the continuity of the function f, where f is defined by

f =

Solution :
The given function is f =

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 at all points of the interval (3, 10].

Hence, f is not continuous at x = 1 and x = 3

Question15. Discuss the continuity of the function f, where f is defined by

Solution :
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:

Question16. Discuss the continuity of the function f, where f is defined by

Solution :
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

Thus, from the above observations, it can be concluded that f is continuous at all points of the real line.

Question17. Find the relationship between a and b so that the function f defined by

is continuous at x = 3.

Solution :
The given function f is

If f is continuous at x = 3, then

18. For what value of λ is the function defined by

continuous at x = 0?

What about continuity at x = 1?

Solution :
The given function f is

If f is continuous at x = 0, then

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

Question19. Show that the function defined by is discontinuous at all integral point. Here [denotes the greatest integer less than or equal to x.

Solution :
The given function is

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

Let n be an integer.

Then,

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.

Question20. Is the function defined by continuous at x = π ?

Solution :
The given function is

It is evident that f is defined at x = π

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

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

Solution :
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

Question22. Discuss the continuity of the cosine, cosecant, secant and cotangent functions,

Solution :
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,

Question23. Find the points of discontinuity of f, where

Solution :
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 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.

Question24. Determine if f defined by

is a continuous function?

Solution :
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 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.

Question25. Examine the continuity of f, where f is defined by

Solution :
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 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.

Question26. Find the values of k so that the function f is continuous at the indicated point.

Solution :
The given function f is

The given function f is continuous at x = π/2 , if f is defined at x = π/2 and if the value of the f at x = π/2 equals the limit of f at x = π/2 .

It is evident that f is defined at x = π/2 and f( π/2) = 3

Therefore, the required value of k is 6.

Question27. Find the values of k so that the function f is continuous at the indicated point.

Solution :
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 f(2) = k(2)2 = 4k

Therefore, the required value of k is 3/4.

Question28. Find the values of k so that the function f is continuous at the indicated point.

Solution :
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 f(π) = kπ + 1

Therefore, the required value of k is -2/π

Question29. Find the values of k so that the function f is continuous at the indicated point.

Solution :
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 f(5) = kx + 1 = 5k + 1

Therefore, the required value of k is 9/5

Question30. Find the values of a and b such that the function defined by

is a continuous function.

Solution :
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

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

Question31. Show that the function defined by f (x) = cos (x2) is a continuous function.

Solution :
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, h is a continuous function.

Question32. Show that the function defined by f(x) = |cos x| is a continuous function.

Solution :
The given function is f(x) = |cos x|

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) = |x| and h(x) = cos x

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

Question33. Examine that sin|x| is a continuous function.

Solution :
Let, f(x) = sin|x|

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) = |x| and h (x) = sin x

It has to be proved first that g (x) = |x| and h (x) = sin x 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.

Question34. Find all the points of discontinuity of f defined by f(x) = |x| – |x + 1|.

Solution :
The given function is f(x) = |x| – |x + 1|

The two functions, g and h, are defined as

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.

### Solve The Following Questions.

Differentiate the functions with respect to x in Exercise 1 to 8.

Question1. sin (x2 + 5)

Solution :

Question2. cos(sin x)

Solution :

Question3. sin(ax + b)

Solution :

Question4. sec(tan (√x))

Solution :

Question5.

Solution :

h is a composite function of two functions, p and q.

Question6.

Solution :

Question7.

Solution :

Question8. cos(√x)

Solution :

Question9. Prove that the function f  given by  is not differentiable at x = 1

Solution :

Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1

Question10. Prove that the greatest integer function defined by f(x) = [x],0 < x < 3 is not differentiable at x = 1 and x = 2

Solution :

The given function f is f(x) = [x],0 < x < 3

It is known that a function f is differentiable at a point x = c in its domain if both

Since the left and right hand limits of f at x = 2 are not equal, f is not differentiable at x = 2

### Solve The Following Questions.

Find dy/dx in the following Exercise 1 to 15.

Question1. 2x + 3y = sin y

Solution :

The given relationship is2x + 3y = sin y

Differentiating this relationship with respect to x, we obtain

Question2. ax + by2 = cos y

Solution :

The given relationship is ax + by2 = cos y

Differentiating this relationship with respect to x, we obtain

Question3. xy + y2 = tanx + y

Solution :
The given relationship is xy + y2 = tanx + y

Differentiating this relationship with respect to x, we obtain

Question4. x2 + xy + y2 = 100

Solution :

The given relationship is x2 + xy + y2 = 100

Differentiating this relationship with respect to x, we obtain

Question 5. 2x + 3y = sin y

Solution :

The given relationship is 2x + 3y = sin y

Differentiating this relationship with respect to x, we obtain

Question6.

Solution :
The given relationship is

Differentiating this relationship with respect to x, we obtain

Question7. sin2 y + cos xy = π

Solution :
The given relationship is sin2 y + cos xy = π

Differentiating this relationship with respect to x, we obtain

Question8. sin2 x + cos2 y = 1

Solution :
The given relationship is  sin2 x + cos2 y = 1
Differentiating this relationship with respect to x, we obtain

Question9.

Solution :

We have,

Question10.

Solution :

Question11.

Solution :

Question12.

Solution :

Question13.

Solution :

Question14.

Solution :

Question15.

Solution :

### Solve The Following Questions.

Question1. Differentiate the following w.r.t. x:

Solution :

Question2. Differentiate the following w.r.t. x:

Solution :
Let y =

By using the chain rule, we obtain

Question3. Differentiate the following w.r.t. x:

Solution :
Let y =

By using the chain rule, we obtain

Question4. Differentiate the following w.r.t. x: sin (tan–1 e-x)

Solution :
Let, y = sin (tan–1 e-x)

By using the chain rule, we obtain

Question5. Differentiate the following w.r.t. x: log(cos ex)

Solution :
Let y = log(cos ex)

By using the chain rule, we obtain

Question6.  Differentiate the following w.r.t. x:

Solution :

Question7. Differentiate the following w.r.t. x:

Solution :
Let y =

Question 8. Differentiate the following w.r.t. x: log(log x), x > 1

Solution :
Let y = log (log x),x > 1

By using the chain rule, we obtain

Question9. Differentiate the following w.r.t. x:

Solution :
Let y =

By using the quotient rule, we obtain

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

Solution :
Let y =

By using the chain rule, we obtain

### Solve The Following Questions.

Question1. Differentiate the following w.r.t. x:

Solution :

Question2. Differentiate the following w.r.t. x:

Solution :
Let y =

By using the chain rule, we obtain

Question3. Differentiate the following w.r.t. x:

Solution :
Let y =

By using the chain rule, we obtain

Question4. Differentiate the following w.r.t. x: sin (tan–1 e-x)

Solution :
Let, y = sin (tan–1 e-x)

By using the chain rule, we obtain

Question5. Differentiate the following w.r.t. x: log(cos ex)

Solution :
Let y = log(cos ex)

By using the chain rule, we obtain

Question6.  Differentiate the following w.r.t. x:

Solution :

Question7. Differentiate the following w.r.t. x:

Solution :
Let y =

Question 8. Differentiate the following w.r.t. x: log(log x), x > 1

Solution :
Let y = log (log x),x > 1

By using the chain rule, we obtain

Question9. Differentiate the following w.r.t. x:

Solution :
Let y =

By using the quotient rule, we obtain

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

Solution :
Let y =

By using the chain rule, we obtain

### Solve The Following Questions.

Question1. Differentiate the function with respect to x.

cos x.cos 2x.cos3x

Solution :
Let y = cos x.cos 2x.cos3x

Taking logarithm on both the sides, we obtain

Question2. Differentiate the function with respect to x.

Solution :
Let y =

Taking logarithm on both the sides, we obtain

Question3. Differentiate the function with respect to x.

Solution :
Let, y =

Taking logarithm on both the sides, we obtain

log y = cos x .log(log x)

Differentiating both sides with respect to x, we obtain

Question4. Differentiate the function with respect to x.

Solution :

Question5. Differentiate the function with respect to x.

Solution :

Question6. Differentiate the function with respect to x.

Solution :

Differentiating both sides with respect to x, we obtain

Differentiating both sides with respect to x, we obtain

Question7. Differentiate the function with respect to x.

Solution :

Differentiating both sides with respect to x, we obtain

Differentiating both sides with respect to x, we obtain

Question8. Differentiate the function with respect to x.

Solution :

Differentiating both sides with respect to x, we obtain

Question9. Differentiate the function with respect to x.

Solution :
Let, y =

Differentiating both sides with respect to x, we obtain

Differentiating both sides with respect to x, we obtain

Question10. Differentiate the function with respect to x.

Solution :
Let, y =

Differentiating both sides with respect to x, we obtain

Differentiating both sides with respect to x, we obtain

Question11. Differentiate the function with respect to x.

Solution :
Let, y =

Differentiating both sides with respect to x, we obtain

Differentiating both sides with respect to x, we obtain

Question12. Find dy/dx of function.

xy+ yx = 1

Solution :
The given function is x+ yx = 1

Let xy = u and y= v

Then, the function becomes u + v = 1

∴du/dx + dv/dx = 1

Differentiating both sides with respect to x, we obtain

Differentiating both sides with respect to x, we obtain

Question13. Find dy/dx of function.

yx = xy

Solution :
The given function is yx = xy

Taking logarithm on both the sides, we obtain

x log y = y log x

Differentiating both sides with respect to x, we obtain

Question14. Find dy/dx of function.

(cos x)y = (cos y)x

Solution :
The given function is (cos x)y = (cos y)x

Taking logarithm on both the sides, we obtain

y log cos x = x log cos y

Differentiating both sides, we obtain

Question15. Find dy/dx of function.

xy = e(x-y)

Solution :
The given function is xy = e(x-y)

Taking logarithm on both the sides, we obtain

log(xy) = log(e(x-y))

Differentiating both sides with respect to x, we obtain

Question16. Find the derivative of the function given by  and hence find f'(1)

Solution :
The given relationship is

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

Question17. 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?

Solution :
Let, y =

(i)

(ii)

(iii) y =

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 dy/dx are same.

Question18. 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.

Solution :
Let y = u.v.w = u(v.w)

By applying product rule, we obtain

By taking logarithm on both sides of the equation y = u.v.w, we obtain

log y = log u + log v + log w

Differentiating both sides with respect to x, we obtain

### Solve The Following Questions.

If x  and y are connected parametrically by the equations given in Exercise 1 to 5, without eliminating the parameter, find dy/dx

Question1.

Solution :

Question2. x = a cosθ , y = b cosθ

Solution :
Given: x = a cosθ , y = b cos

Question3. x = sin t, y = cos 2t

Solution :
Given: x = sin t, y = cos 2t

Question4. x = 4t, y = 4/t

Solution :
Given:x = 4t, y = 4/t

Question5. x = cosθ – cos2θ, y = sinθ – sin2θ

Solution :
Given: x = cosθ – cos2θ, y = sinθ – sin2θ

If x and y are connected parametrically by the equations given in Exercises 6 to 10, without eliminating the parameter, find dy/dx

Question6. x = a (θ – sinθ), y = a(1+ cosθ)

Solution :
Given: x = a (θ – sinθ), y = a(1+ cosθ)

Question7.

Solution :
Given:

Question8.

Solution :
Given:

Question9. x = a sec θ, y = b tan θ

Solution :
Given: x = a sec θ and y = b tan θ

Question10. x = a (cos θ + θ sin θ), y = b (sin θ – θ  cosθ)

Solution :
Given: x = a (cos θ + θ sin θ) and y = b (sin θ – θ  cosθ)

Question11. If

Solution :
Given:

### Solve The Following Questions.

Find the second order derivatives of the functions given in Exercises 1 to 5.

Question1. x2 + 3x + 2

Solution :
Let y = x2 + 3x + 2

Question2. x20

Solution :
Let  x20

Question3. x.cos x

Solution :
Let x. cos x

Question4. log x

Solution :
Let log x

Question5. xlog x

Solution :
Let xlog x

Find the second order derivatives of the functions given in Exercises 6 to 10.

Question6. ex sin 5x

Solution :
Let ex sin 5x

Question7. e6x cos x

Solution :
Let e6x cos x

Question8. tan-1 x

Solution :
Let tan-1 x

Question9. log (log x)

Solution :

Let log (log x)

Question10. sin(log x)

Solution :
Let sin (log x)

Question11. If y = 5 cos x –  3 sin x prove that

Solution :
Let y = 5 cos x –  3 sin x

Question12. If y = cos-1 x Find  in terms of y alone.

Solution :
Given: y = cos-1 x

Question13. If y = 3 cos (log x) + 4 sin (log x), show that x2y2 + xy+ y = 0

Solution :
Given: y = 3 cos (log x) + 4 sin (log x)

Hence proved.

Question14. If y = Aemx + Benx show that

Solution :
Given: y = Aemx + Benx

Hence proved.

Question15. If  500e7x  + 600e-7x show that

Solution :
Given: 500e7x  + 600e-7x

Hence proved.

Question16. If  e(x + 1) = 1, show that

Solution :
Given: e(x + 1) = 1

Taking log on both sides,

Differentiating this relationship with respect to x, we obtain

Hence proved.

Question17. If y = (tan-1 x)2 show that (x2 + 1)2y2 + 2(x2 + 1)y= 2

Solution :
Given:  y = (tan-1 x)2

Hence proved.

### Solve The Following Questions.

Question1.Verify Rolle’s Theorem for the function f(x) = x2 + 2x – 8, x ∈ [– 4, 2].

Solution :
The given function,f(x) = x2 + 2x – 8 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 f'(c) = 0

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

Question2. 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)

Solution :
By Rolle’s Theorem, for a function f[a,b] →R, 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 f'(c) = 0

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].

Also, f(5) = [5] = 5 and f(9) = [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).

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].

Also, f(-2) = [-2] = -2 and f(2) = [2] = 2

∴ f(-2) ≠ f(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) = 12 – 1 = 0

f(2) = 22 – 1 = 3

∴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 .

Question3. If f:[-5,5] →R is a differentiable function and if f'(x) does not vanish anywhere, then prove that f(-5) ≠ f(5).

Solution :
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 f'(x) does not vanish anywhere.

Hence, proved.

Question4. Verify Mean Value Theorem, if f(x) = x2 – 4x – 3 in the interval [a,b], where a = 1 and b = 4.

Solution :
The given function is f(x) = x2 – 4x – 3  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 f'(c) = 1

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

Question5. Verify Mean Value Theorem, if f(x) = x3 – 5x2 – 3x in the interval [a, b], where a = 1 and b = 3. Find all c ∈ (1, 3) for which f'(c) = 0

Solution :
The given function f is f(x) = x3 – 5x2 – 3x f, being a polynomial function, is continuous in [1, 3] and is differentiable in (1, 3) whose derivative is 3×2 − 10x − 3.

Mean Value Theorem states that there exist a point c ∈ (1, 3) such that f'(c) = – 10

Hence, Mean Value Theorem is verified for the given function and c = 7/3 ∈ (1,3) is the only point for which f'(c) = 0

Question6. Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.

Solution :
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)

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.

### Solve The Following Questions.

Question1.

Solution :
Let

Using chain rule, we obtain

Question2.

Solution :

Let

Question3.

Solution :
Let,

Taking logarithm on both the sides, we obtain

log y = 3 cos 2x log(5x)

Differentiating both sides with respect to x, we obtain

Question4.

Solution :
Let,

Using chain rule, we obtain

Question5.

Solution :
Let y =

Question6.

Solution :
Let ,y =  …….(1)

Question7.

Solution :
Let, y =

Taking logarithm on both the sides, we obtain

log y = log x. log (log x)

Differentiating both sides with respect to x, we obtain

Question8. Differentiate w.r.t. x the function cos (a cos x + b sin x), for some constant a and b.

Solution :

Let, y = cos (a cos x + b sin x)

By using chain rule, we obtain

Question9.

Solution :
Let, y =

Taking logarithm on both the sides, we obtain

Question10. , for some fixed  a> 0 and x > 0

Solution :
Let y =

Since a is constant, aa is also a constant.

∴ ds/dx = 0                                 …..(5)

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

Question11. , for x > 3

Solution :

Question12. Find dy/dx , if

Solution :

Question13. Find

Solution :

Question14. If

Solution :
It is given that,

Differentiating both sides with respect to x, we obtain

Hence, proved.

Question15.

Solution :
It is given that,

Differentiating both sides with respect to x, we obtain

= – c

which is constant and independent of a and b

Hence, proved.

Question16. If cos y = x cos (a + y), with cos a ≠ ± 1, prove that  prove that

Solution :
It is given cos y = x cos (a + y)

Hence, proved.

Question17. If x = a (cos t + t sin t) and y = a (sin t – t cos t), find

Solution :
It is given that, x = a(cost + tsin t) and y = a (sin t – t cost)

Question18. If f (x) = |x|3 show that f”(x) exists for all real x, and find it.

Solution :
It is known that,

Therefore, when x ≥ 0, f(x) = |x|3 = x3

In this case, f'(x) = 3x and hence, f”(x) = 6x

When x < 0, f(x) = |x|3 = (-x3) = -x3

In this case, f'(x) = -3x2and hence, f”(x) = -6x

Thus, for f(x) = |x|3, f”(x) exists for all real x and is given by,

Question19. Using mathematical induction prove that for all positive integers n.

Solution :

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.

Question20. Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation, obtain the sum formula for cosines.

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

Differentiating both sides with respect to x, we obtain

Question21. Does there exist a function which is continuos everywhere but not differentiable at exactly two points? Justify your answer ?

Solution :
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.

Question22. If, prove that

Solution :

Question23. If, show that

Solution :
It is given that,

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