Rolle's Theorem MCQ Quiz in தமிழ் - Objective Question with Answer for Rolle's Theorem - இலவச PDF ஐப் பதிவிறக்கவும்

f(1) = f(2)

⇒ 1 - a + b + 1 = 8 - 4a + 2b + 1

3a - b = 7      ---(i)

f'(x) = 3x² - 2ax + b

$\Rightarrow f^{\prime }\left(\frac{4}{3}\right)=0$

$\Rightarrow 3\times \frac{16}{9}-\frac{8}{3}a+b=0$

⇒ -8a + 3b = -16      ---(ii)

a = 5, b = 8

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) ${\int }_a^b{f}^{\prime }(x)dx$ = 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 ${\int }_1^2{\mathrm{f}}^{\prime }(\mathrm{x})\mathrm{d}\mathrm{x}$

⇒ ${\int }_1^2{\mathrm{f}}^{\prime }(\mathrm{x})\mathrm{d}\mathrm{x}$ = $[\mathrm{f}(\mathrm{x}){]}_1^2$ (Integration is the 'inverse' of differentiation)

⇒ ${\int }_1^2{\mathrm{f}}^{\prime }(\mathrm{x})\mathrm{d}\mathrm{x}$ = f(2) - f(1)

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

Therefore, f(2) = f(1)

⇒ ${\int }_1^2{\mathrm{f}}^{\prime }(\mathrm{x})\mathrm{d}\mathrm{x}$ = 0

∴  ${\int }_1^2{\mathrm{f}}^{\prime }(\mathrm{x})\mathrm{d}\mathrm{x}$ 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) = $\cos {\displaystyle \frac{\mathrm{x}}{2}}$ on [π, 3π].

f(π) = $\cos {\displaystyle \frac{\pi }{2}}$ = 0 and f(3π) = $\cos {\displaystyle \frac{3\pi }{2}}$ = 0.

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

f'(x) = ${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}}(\cos {\displaystyle \frac{\mathrm{x}}{2}})=-{\displaystyle \frac{1}{2}}(\sin {\displaystyle \frac{\mathrm{x}}{2}})$

⇒ f'(c) = $-{\displaystyle \frac{1}{2}}(\sin {\displaystyle \frac{\mathrm{c}}{2}})$ = 0

⇒ $\sin {\displaystyle \frac{\mathrm{c}}{2}}$ = 0

⇒ ${\displaystyle \frac{\mathrm{c}}{2}}$ = 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.