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

Answer 1

The correct option is C

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

Explanation for the correct option:

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

Determining the applicability in $f\left(x\right)=x{(x-2)}^{2}in0\le x\le 2$

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

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

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

Thus it is a differentiable function and every differentiable function is continuous so it is continuous also.

Hence Rolle's theorem is applicable to it.

Therefore, the correct answer is option(C).

Explanation for in correct options:

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

Determining the applicability in $f\left(x\right)=\left|x\right|in-2\le x\le 2$

Checking differentiability at $0$

Left-hand derivative

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

Right-hand derivative

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

Thus Left hand derivative $\ne$ Right- hand derivative so it is not differentiable.

Hence Rolle's theorem is not applicable in this function.

Therefore, option(A) is incorrect.

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

Determining the applicability in $f\left(x\right)=\tan xin0\le x\le \mathrm{\pi }$

Since $\tan x$ is undefined at $\frac{\mathrm{\pi }}{2}$

So it is not differentiable in the interval $\left[0,\mathrm{\pi }\right]$

Hence Rolle's theorem is not applicable in this function.

Therefore, option(B) is incorrect.

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

Determining the applicability in $f\left(x\right)=1+{(x-2)}^{\frac{2}{3}}in1\le x\le 3$

Let $f\left(x\right)=0$

$$\begin{gathered} \Rightarrow 1+{(x-2)}^{\frac{2}{3}}=0 \\ \Rightarrow {(x-2)}^{\frac{2}{3}}=-1 \\ \Rightarrow {(x-2)}^{\frac{1}{3}}=\sqrt{-1}[\text{taking squre root both sides}] \\ \Rightarrow {(x-2)}^{\frac{1}{3}}=i\left[\because i=\sqrt{-1}\right] \\ \Rightarrow (x-2)=i^3[\text{cubing both sides}] \\ \Rightarrow (1-2)=i^3[\text{putting}x=1] \\ \Rightarrow -1=i^3 \end{gathered}$$

Thus we get an imaginary value at $x=1$

Hence Rolle's theorem is not applicable to it.

Therefore, option(D) is incorrect.

Hence, the correct answer is option (C).