Rolle's Theorem MCQ Quiz in मल्याळम - Objective Question with Answer for Rolle's Theorem - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

(i) Rolle's Theorem:

Let f(x) be defined in [a, b] such that 

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

(ii) f(x) is differentiable in (a, b)

(iii) f(a) = f(b)

then there exists at least one point c ∈ (a, b) such that f'(c) = 0 

(ii) \(\int^b_af'(x) dx\) = f(b) - f(a) 

Here, f(a) is the lower limit value of the integral and f(b) is the upper limit value of the integral.

Calculation:

According to the question we have to find the value of \(\rm \int^2_1f'(x) dx\)

⇒ \(\rm \int^2_1f'(x) dx\) = \(\rm[ f(x)] _{1}^{2}\) (Integration is the 'inverse' of differentiation)

⇒ \(\rm \int^2_1f'(x) dx\) = f(2) - f(1)

Since , f(x) satisfies the requirements of Rolle’s theorem [1, 2].

Therefore, f(2) = f(1)

⇒ \(\rm \int^2_1f'(x) dx\) = 0

∴  \(\rm \int^2_1f'(x) dx\) equal to 0.

Concept:

Rolle's theorem states that if a function f(x) is continuous in the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b), then f′(c) = 0 for some c ∈ [a, b].

Calculation:

The given function is f(x) = \(\rm \cos \dfrac{x}{2}\) on [π, 3π].

f(π) = \(\rm \cos \dfrac{\pi}{2}\) = 0 and f(3π) = \(\rm \cos \dfrac{3\pi}{2}\) = 0.

Since, f(π) = f(3π), there must exist a c ∈ [π, 3π] such that f'(c) = 0.

f'(x) = \(\rm \dfrac{d}{dx}\left (\cos \dfrac{x}{2} \right )=-\dfrac{1}{2}\left (\sin \dfrac {x}{2} \right )\)

⇒ f'(c) = \(\rm -\dfrac{1}{2}\left (\sin \dfrac {c}{2} \right )\) = 0

⇒ \(\rm \sin \dfrac {c}{2}\) = 0

⇒ \(\rm \dfrac {c}{2}\) = nπ

⇒ c = 2nπ, where n is an integer.

We want c ∈ [π, 3π], therefore c = 2π.

Given:

Y = f(x) = x(x - 3)² = x³ – 6x² + 9x ⇒ Polynomial

F(x) is

i) Continuous on [0, 3]

And ii) differentiable on (0, 3)

iii) f(0) = f(3) = 0

Thus, all three conditions of Rolle’s Theorem are satisfied.

By Rolle’s Theorem, C ϵ (0, 3) such that f’(c) = 0

f(C) = C (C - 3)²

f(C) = C [C² – 6c + 9]

f(C) = C³ – 6c² + 9C

Differentiating,

f’(C) = 3C² – 12C + 9 = 0

∴ C = 1, 3

But, C = 3 ∉ (0, 3)

∴ C = 1

If x = C = 1, then y (1) = f (1) = 1 (1 - 3)² = 4

∴ At a point (x, y) = (1, 4) on given curve, the tangent is parallel to x-axis.