Question:

Rolle's theorem is applicable in the interval $[-2, 2]$ for the function

Updated On: Apr 26, 2024
  • $f \left(x\right)=x^{3}$
  • $f \left(x\right)=4X^{4}$
  • $f \left(x\right)=2X^{3}+3$
  • $f \left(x\right)=\pi\left|x\right|$
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The Correct Option is B

Solution and Explanation

If we take $f(x)=4 x^{4}$, then
(i) $f(x)$ is continuous in $(-2,2)$
(ii) $f(x)$ is differentiable in $(-2,2)$
(iii) $f(-2)=f(2)$
So, $f(x)=4 x^{4}$ satisfies all the conditions of Rolle's theorem therefore $\exists$ a point $c$ such that $f'(c)=0$
$\Rightarrow 16 c^{3}=0 \Rightarrow c=0 \in(-2,2)$
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Concepts Used:

Continuity

A function is said to be continuous at a point x = a,  if

limx→a

f(x) Exists, and

limx→a

f(x) = f(a)

It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.

If the function is undefined or does not exist, then we say that the function is discontinuous.

Conditions for continuity of a function: For any function to be continuous, it must meet the following conditions:

  • The function f(x) specified at x = a, is continuous only if f(a) belongs to real number.
  • The limit of the function as x approaches a, exists.
  • The limit of the function as x approaches a, must be equal to the function value at x = a.