NCERT Solutions for Class 12 Maths Chapter 5 – Continuity and Differentiability Ex 5.8

NCERT Solutions for Class 12 Maths Chapter 5 – Continuity and Differentiability Ex 5.8

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.

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.

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.

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.

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 3x² − 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

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.