RBSE Solutions for Class 12 Maths Chapter 7 Differentiation Ex 7.6

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) = x² + 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) = x² + 5x + 6, x ∈ [- 3, – 2]
∵ f(x) = x² + 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) = x² – 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.

RBSE Solutions for Class 12 Maths