Question

In which of the following functions Rolle's theorem is applicable?

• A. $f\left(x\right)=\left|x\right|,-2\le x\le 2$
• B. $f\left(x\right)=\tan x,0\le x\le \mathrm{\pi }$
• C. $f\left(x\right)=1+{\left(x-2\right)}^{\frac{2}{3}},1\le x\le 3$
• D. $f\left(x\right)=x{\left(x-2\right)}^2,0\le x\le 2$

Answer 1

The correct option is D

$f\left(x\right)=x{\left(x-2\right)}^2,0\le x\le 2$

$f\left(x\right)=x{\left(x-2\right)}^2,0\le x\le 2$$f\left(x\right)=x{\left(x-2\right)}^2,0\le x\le 2$$f\left(x\right)=x{\left(x-2\right)}^2,0\le x\le 2$

Explanation for the correct option:

Option (D):

Rolle's theorem states that if a function $f$ is continuous on the closed interval $\left[a,b\right]$ and differentiable on the open interval $\left(a,b\right)$ such that$f\left(a\right)=f\left(b\right)$, then $f\text{'}\left(x\right)=0$ for some $x$ with $a\le x\le b$

$f\left(x\right)=x{\left(x-2\right)}^2,0\le x\le 2$

$f\left(x\right)$ is continuous on $\left[0,2\right]$ and differentiable on $\left(0,2\right)$

$f\left(0\right)=f\left(2\right)=0$

Therefore Rolle's theorem is applicable

Explanation for the incorrect option:

Option (A):

$f\left(x\right)=\left|x\right|,-2\le x\le 2$ is not differentiable at $x=0$ hence Rolle's theorem is not applicable

Option(B):

$f\left(x\right)=\tan x,0\le x\le \mathrm{\pi }$ at$\frac{\mathrm{\pi }}{2}$,$f\left(x\right)$ is not defined, therefore not discontinuous

Option(C):

$f\left(x\right)=1+{\left(x-2\right)}^{\frac{2}{3}},1\le x\le 3$at $x=1$,$f$ real-valued, hence Rolle's theorem is not applicable

Hence, option (D) is the correct answer