## Rajasthan Board RBSE Class 12 Maths Chapter 7 Differentiation Ex 7.6

Question 1.

Verify the Rolle’s theorem, for the following functions :

Solution:

(i) Given function

∵ f(x) is polynomial in x.

∴ It is differentiable and continuous every-where.

Hence, f(x) satisfies Rolle’s theorem in interval [π/4, 5π/4]

From(1),

f'(x) = e^{x} (cos x + sin x) + (sin x – cos x).e^{x}

f'(x) = e^{x} (cos x + sin x + sin x – cos x)

Similarly,

e^{c} 2 sin c = 0

⇒ 2sinc = 0 ⇒ sin c = 0 ⇒ c = π

c = π ∈ (π/4, 5π/4),f(c) = 0

∴ Hence, Rolle’s theorem satisfied.

(ii) f(x) = (x – a)^{m} (x – b)^{n}, x ∈ [a, b], m, n ∈ N

Here, (x – a)^{m} and (x – b)^{n} both are polynomial. On its multiplication a polynomial of power (m + n) is obtained. A polynomial function is continuous everywhere always.

So, f(x) is continuous in [a, b]. Polynomial function is also differentiable.

∴ f'(x) = m(x- a)^{m-1} .(x – b)^{n} + n(x – d)^{m} (x – b)^{n-1}

= (x – a)^{m-1} (x – b)^{n-1} × [m(x – b) + n(x – a)]

= (x – a)^{m-1} (x – b)^{n-1} × [(m + n) x – mb – na]

Here,f'(x) exists.

∴ f(x) is differentiable in interval (a, b).

Again f(a) = (a – a)^{m} (a – b)^{n} = 0

f (b) = (b – a)m (b – b)^{m-n} = 0

∴ f(a) = f(b) = 0

Hence, Rolle’s theorem satisfies,then in interval (a, b) at least any one point is such that f'(c) = 0.

Function is not diffenentiable at x = 0. So, Rolle’s theorem does not satisfied.

(iv) Given function

f(x) = x^{2} + 2x – 8, x ∈ [- 4, 2]

Here, it is clear that function f(x) is continuous in interval [- 4, 2] and f’ (x) is finite and exist at every point of interval, (- 4, 2), hence given function is differentiable in interval (-4, 2).

∵ f (- 4) = 0 = f (2)

⇒ f (- 4) = f(2)

So, from above f'(x) satisfies all the three conditions of Rolle’s theorem in given interval.

(vi) Given function

f(x) = [x], x ∈ [- 2, 2]

∵ f(x) is not continuous in [- 2, 2] as greatest integer function is neither continuous nor differentiable.

Hence, Rolle’s theorem does not satisfied.

Question 2.

Prove Rolle’s theorem for following functions :

Solution:

(i) Given function

f(x) = x^{2} + 5x + 6, x ∈ [- 3, – 2]

∵ f(x) = x^{2} + 5x + 6 is a polynomial function.

Hence, it is continuous in [- 3, – 2],

Now f'(x) = 2x + 5, exists ∀ x ∈ (-3,-2)

∴ f (x) is differentiable in (- 3, – 2)

∵ f (- 3) = 0 = f (- 2)

⇒ f (- 3) = f (- 2)

All the condition of Rolle’s theorem is satisfied, then a point c ∈ (-3,-2) exists in such a way that f'(c) = 0.

All the conditions of Rolle’s theorem satisfied, then a point c in (0,π) exists in such a way that f'(c) = 0.

(iii) Given function

is finite and exist at every point in interval (0, 1).

Hence,f(x) is differentiable in (0, 1).

∵ f(0) = 0 = f(1)

⇒ f (0) = f (1)

So, f (x) satisfies all the conditions of Rolle’s theorem.

Hence, f'(c) = 0

(iv) Given function

f (x) = cos 2x, x ∈ [0, π]

⇒ f (x) = cos 2x is defined in [0,π].

∵ cosine is continuous in its domain.

So, it is continuous in [0, π].

Then, f'(x) = – 2 sin 2x exists,

where x ∈ (0, π)

∴ f (x) is differentiable in (0, π).

Now f(0) = cos 0 = 1

and f(π) = cos 2π = 1

∴ f(0) = f(π) = 1

Thus, all the conditions of Rolle’s theorem, are satisfied then at least one point c such that c ∈ (0, π) and f’ (c) = 0.

f'(c) = 0

∴ f'(c) = – 2 sin 2c = 0

⇒ sin 2c = 0 ⇒ 2c = π

c = π/2 is an element of (0,π)

∴ c = \(\frac { \pi }{ 2 } \) ∈ (0,π)

such that

f'(c) = 0

Thus for c = \(\frac { \pi }{ 2 } \), Rolle’s theorem satisfied.

Question 3.

Verify the Lagrange’s mean value theorem for the following functions :

Solution:

Clearly, f(x) is continuous in interval [0, 2] and f'(x) is finite and exists. So,f(x) is differentiable in (0, 2). Hence f(x) satisfies both conditions of Langrange’s mean value theorem.

∵ c is an imaginary number.

Hence, Langrange’s mean value theorem does not satisfied.

(iii) Given function

f(x) = x^{2} – 3x + 2, x ∈ [- 2, 3]

Clearly,f(x) is continuous in interval [-2, 3] and f'(x) is finite and exists in (-2, 3). So, f(x) is differentiable in (- 2, 3). Hence f(x) satisfies both conditions of Langrange’s mean value theorem.

Clearly,f(x) is continuous in [ 1,4] and f'(x) is finite and exist in interval (1,4). Hence,f(x) is differentiable in (1, 4). f(x) satisfies both conditions of Langrange’s mean value theorem.

Thus, Langrange’s mean value theorem satisfied.

## Leave a Reply