These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.8

Question 1.

Verify Rolle’s theorem for the function

f(x) = x² + 2x – 8, x ∈ [- 4, 2]

Solution:

Now f(x) = x² + 2x – 8 is a polynomial

∴ It is continuous and derivable in its domain x ∈ R.

Hence it is continuous in the interval [- 4, 2] and derivable in the interval (- 4, 2)

f(-4) = (- 4)² + 2(- 4) – 8 = 16 – 8 – 8 = 0,

f(2) = 2² + 4 – 8 = 8 – 8 = 0

Conditions of Rolle’s theorem are satisfied.

f'(x) = 2x + 2

∴ f’ (c) = 2c + 2 = 0

or c = – 1, c = – 1 ∈ [- 4, 2]

Thus f’ (c) = 0 at c = – 1.

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 example?

(i) f(x) = [x] for x ∈ [5, 9]

(ii) f (x) = [x] for x ∈ [-2, 2]

(iii) f (x) = x² – 1 for x ∈ [1, 2]

Solution:

(i) In the interval [5, 9], f (x) = [x] is neither continuous nor derivable at x = 6, 7, 8 Hence Rolle’s theorem is not applicable

(ii) f (x) = [x] is not continuous and derivable at – 1, 0, 1. Hence Rolle’s theorem is not applicable.

(iii) f(x) = (x² – 1), f(1) = 1 – 1 = 0,

f(2) = 22 – 1 = 3

f(a) ≠ f(b)

Though it is continous and derivable in the interval [1,2].

Rolle’s theorem is not applicable.

In case of converse if f (c) = 0, c ∈ [a, b] then conditions of rolle’s theorem are not true.

(i) f (x) = [x] is the greatest integer less than or equal to x.

∴ f(x) = 0, But fis neither continuous nor differentiable in the interval [5, 9].

(ii) Here also, theough f (x) = 0, but f is neither continuous nor differentiable in the interval [- 2, 2].

(iii) f (x) = x² – 1, f'(x) = 2x. Here f'(x) is not zero in the [1, 2], So f (2) ≠ f’ (2).

Question 3.

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:

For Rolle’s theorem

If (i) f is continuous in [a, b]

(ii) f is derivable in [a, b]

(iii) f (a) = f (b)

then f’ (c) = 0, c ∈ (a, b)

∴ f is continuous and derivable

i.e., f'(c) ≠ 0. Hence \(\frac{f(5)-f(-5)}{10}\)

but f (c) ≠ 0 ⇒ f(a) ≠ f(b)

⇒ f(-5) ≠ f(5)

Question 4.

Verify Mean Value Theorem, if

f (x) = x² – 4x – 3 in the interval [a, b], where a = 1 and b = 4.

Solution:

f (x) = x² – 4x – 3. It being a polynomial it is continuous in the interval [1, 4] and derivable in (1,4), So all the condition of mean value theorem hold.

then f’ (x) = 2x – 4,

f'(c) = 2c – 4

f(4) = 16 – 16 – 3 = – 3,

f(1) = 1 – 4 – 3 = – 6

∴ f'(c) = 0 \(\frac{f(b)-f(a)}{b-a}\) = \(\frac{f(4)-f(1)}{4-1}\)

⇒ 2c – 4 = \(\frac{-3-6}{4-1}\)

⇒ 2c – 4 = 1 ⇒ c = \(\frac{5}{2}\) ∈ (1, 4)

∴ Mean Value Theorem is verified for f(x) on (1, 4)

Question 5.

Verify Mean Value Theorem, if f (x) = x^{3} – 5x^{2} – 3x in the interval [a, b], where a = 1 and b = 3. Find all c ∈ (1, 3) for which f’ (c) = 0.

Solution:

f (x) = x^{3} – 5x^{2} – 3x

f'(x) = 3x² – 10x – 3

Since f'(x’) exists, f(x) is continous on [1, 3]

f(x) is differentiable on (1, 3)

f'(c) = 3c² – 10c – 3

f(b) = f(3) = – 27

f(a) = f(1) = – 7

∴ f'(c) = 0 \(\frac{f(b)-f(a)}{b-a}\)

⇒ 3c² – 10c – 3 = \(\frac{-27-7}{3-1}\)

⇒ 3c² – 10c – 3 = – 10

⇒ 3c² – 10c + 7 = 0

⇒ (c – 1)(3c – 7) = 0 ⇒ c = 1 or c = \(\frac{7}{3}\)

\(\frac{7}{3}\) ∈ (1, 3)

∴ Mean Value Theorem is verified for f(x) on (1, 3)

Question 6.

Examine the applicability of Mean Value theroem for all three functions given in the above Question 2.

Solution:

(i) F (x)= [x] for x ∈ [5, 9], f (x) = [x] in the interval [5, 9] is neither continuous, nor differentiable.

(ii) f (x) = [x], for x ∈ [-2, 2],

Again f (x) = [x] in the interval [-2, 2] is neither continous, nor differentiable.

(iii) f(x) = x² – 1 for x ∈ [1,2], It is a polynomial.

Therefore it is continuous in the interval [1,2] and differentiable in the interval (1,2)

f (x) = 2x, f(1) = 1 – 1 = 0 ,

f(2) = 4 – 1 = 3, f'(c) = 2c

∴ f'(c) = 0 \(\frac{f(b)-f(a)}{b-a}\)

2c = \(\frac{3-0}{2-1}\) = \(\frac{3}{1}\)

∴ c = \(\frac{3}{2}\) which belong to (1, 2)