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

### Continuity and Differentiability Exercise 5.8 Solutions

**1. Verify Rolle’s theorem for the function f (x) = x ^{2} + 2x – 8, x ∈ [– 4, 2].**

**Solution**

The given function, f(x) = x^{2} + 2x – 8, being a polynomial function, is continuous in [-4, 2] and is differentiable in (-4, 2).

f(-4) = (-4)^{2} + 2(-4) – 8 = 16 - 8 - 8 = 0

f(2) = 2^{2} + 2× 2 – 8 = 4 + 4 - 8 = 0

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

f(x) = x^{2} + 2x – 8

⇒ f '(x) = 2x + 2

∴ f '(c) = 0

⇒ 2c + 2 = 0

⇒ c = -1, where c = -1 ∈ (-4, -2)

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

**2. Examine if Rolle's Theorem is applicable to any of the following functions. Can you say some thing about the converse or Rolle's Theorem from these examples?(i) f(x) = x for x ∈ [5, 9] (ii) f(x) = x for x ∈ [-2, 2] (iii) f(x) = x**

^{2}- 1 for x ∈ [1, 2]

**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) f(x) = x for x ∈ [5, 9]

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).

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

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 f(x) = [x] for x ∈ [5, 9].

(ii) f(x) = [x] for x ∈ [-2, 2]

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).

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

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 f(x) = [x] for x ∈ [-2, 2] .

(iii) f(x) = x^{2} - 1 for x ∈ [1, 2]

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

f(1) = (1)^{2} - 1 = 0

f(2) = (2)^{2} - 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 f(x) = x^{2} - 1 for x ∈ [1, 2].

**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**

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

⇒ 10f '(c) = f(5) - f(-5)

It is also given that f'(x) does not vanish anywhere.

∴ f '(c) ≠ 0

⇒ 10 f ' (c) ≠ 0

⇒ f(5) - f(-5) ≠ 0

⇒ f(5) ≠ f(-5)

Hence, proved.

**4. Verify Mean Value Theorem, if f(x) = x ^{2} - 4x - 3 in the interval [a, b], where a = 1 and b = 4 .**

**Solution**

The given function is f(x) = x^{2} - 4x - 3

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

f(1) = 1^{2} - 4×1 - 3 = - 6, f(4) = 4^{2} - 4 ×4 - 3 = -3

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

f ' (c) = 1

⇒ 2c - 4 = 1

⇒ c = 5/2, where c = 5/2 ∈ (1, 4)

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

**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**

The given function f is f(x) = x^{3} – 5x^{2} – 3x

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

f(1) = 1^{3} - 5 × 1^{2} - 3 × 1 = -7, f(3) = 3^{3} - 5 × 3^{2} - 3 × 3 = -27

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

f '(c) = -10

⇒ 3c^{2} - 10c - 3 = 1-0

⇒3c^{2} - 10c + 7 = 0

⇒ 3c^{2} - 3c - 7c + 7 = 0

⇒ 3c(c - 1) - 7(c - 1) = 0

⇒ (c - 1)(3c- 7) = 0

⇒ c = 1, 7/3 [where c = 7/3 ∈ (1, 3) ]

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

**6. 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 f '(c) = [f(b) - f(a)]/(b -a)

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

(i) f(x) = [x] for x ∈ [5, 9]

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) in checked as follows.

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

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

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 f(x) = [x] for x ∊ [5, 9].

(ii) f(x) = [x] for x ∊ [-2, 2]

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).

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

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 f(x) = [x] for x ∊ [-2, 2] .

(iii) f(x) = x^{2} - 1 for x ∊ [1, 2]

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 f(x) = x^{2} - 1 for x ∊ [1, 2]

It can be proved as follows.

f(1) = 1^{2} - 1 = 0, f(2) = 2^{2} - 1 = 3

f '(x) = 2x

∴ f '(c) = 3

⇒ 2c = 3

⇒ c = 3/2 = 1.5, where 1.5 ∊ [1, 2].