Question

In which of the following functions, Rolle’s theorem is applicable

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

Answer 1

The correct option is D

$f\left(x\right)=x{(x–2)}^2\text{in}0\le x\le 2$

Explanation for the correct option:

By Rolle’s Theorem, for a function $f:[a,b]\to R$ if

(a) $f$ is continuous on $[a,b]$

(b) $f$ is differentiable on $(a,b)$ $f\left(a\right)=f\left(b\right)$ then, there exists some $c\in (a,b)$such that$f\text{'}\left(c\right)=0$

Therefore, Rolle’s Theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.

Option (D): $f\left(x\right)=x{(x–2)}^2\text{in}0\le x\le 2$

Determining applicability of Rolle’s theorem for the given equation

$f\left(x\right)=x{(x–2)}^{2}in0\le x\le 2$

Differentiating$f\left(x\right)$ w.r.t $x$ we get

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=x\frac{d}{dx}{(x-2)}^2+{(x-2)}^2\frac{d}{dx}\left(x\right) \\ \Rightarrow f\text{'}\left(x\right)=2x(x-2)+{(x-2)}^2 \\ \Rightarrow f\text{'}\left(0\right)=4\left(\text{exist}\right)\mathbf{[}\mathbf{\text{put}}\mathbf{}\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{}\mathbf{]} \end{gathered}$$

Similarly $\forall x\in \left(0,2\right)$, $f\text{'}\left(x\right)$ exist.

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

And polynomial functions are always differentiable and continuous

Hence Rolle's theorem is applicable to it.

Therefore, option (D) is the correct answer.

Explanation for incorrect Options:

Option (A): $f\left(x\right)=\left|x\right|\text{in}–2\le x\le 2$

Determining applicability of Rolle’s theorem in

$f\left(x\right)=\left|x\right|in–2\le x\le 2$

Checking the differentiability at $0\in \left(-2,2\right)$

Left hand derivative

$$\begin{gathered} \because \underset{h\to 0}{\lim }f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x-h)-f\left(x\right)}{h} \\ \underset{h\to 0^{-}}{\lim }f\text{'}\left(0\right)=\underset{h\to 0^{-}}{\lim }\frac{f(0-h)-f\left(0\right)}{h} \\ =\underset{h\to 0^{-}}{\lim }\frac{\mid 0-h\mid -\mid 0\mid }{h}\mathbf{}\left[\because f\left(x\right)=\left|x\right|\right] \\ =\underset{h\to 0^{-}}{\lim }\frac{\left|-h\right|}{h} \\ =\underset{h\to 0^{-}}{\lim }\frac{-h}{h}\left[\because \left|x\right|=-xifx<0\right] \\ =-1 \end{gathered}$$

Right hand derivative

$$\begin{gathered} \underset{h\to 0^{+}}{\lim }f\text{'}\left(0\right)=\underset{h\to 0^{+}}{\lim }\frac{f(0-h)-f\left(0\right)}{h} \\ =\underset{h\to 0^{+}}{\lim }\frac{\mid 0-h\mid -\mid 0\mid }{h}\left[\because f\left(x\right)=\left|x\right|\right] \\ =\underset{h\to 0^{+}}{\lim }\frac{\left|-h\right|}{h} \\ =\underset{h\to 0^{+}}{\lim }\frac{h}{h}\left[\because \left|x\right|=xifx>0\right] \\ =1 \end{gathered}$$

So, Left hand derivative $\ne$Right-hand derivative

Thus it is not differentiable at $0$.

Hence Rolle's' theorem is not applicable.

Therefore, option (A) is the incorrect answer.

Option (B): $f\left(x\right)=\tan x\text{in}0\le x\le \pi$

Determining applicability of Rolle’s theorem in

$f\left(x\right)=\tan xin0\le x\le \pi$

Since $\tan x$ is undefined at $\frac{\mathrm{\pi }}{2}\in \left[0,\mathrm{\pi }\right]$

So, it is not continuous

Thus Rolle's theorem is not applicable on it.

Therefore, option (B) is the incorrect answer.

Option (C): $f\left(x\right)=1+{(x–2)}^{\frac{2}{3}}\text{in}1\le x\le 3$

Let us find the derivative of the function so,

$f\text{'}\left(x\right)=\frac{2}{3{(x-2)}^{\left({\displaystyle \frac{1}{3}}\right)}}$

And at $x=2$ the derivative don't exist because the value will be infinite

Hence Rolle's theorem is not applicable on it

Therefore, option (C) is the incorrect answer.

Hence, option (D) is the correct answer.