Question

The Rolle's theorem is applicable in the interval, $-1\le x\le 1$ for the function

• A. $f\left(x\right)=x$
• B. $f\left(x\right)=x^2$
• C. $f\left(x\right)=2x^3+3$
• D. $f\left(x\right)=\left|x\right|$

Answer 1

The correct option is B

$f\left(x\right)=x^2$

Explanation for the correct option:

Step 1.Condition of Rolle's theorem

According to Rolle's theorem we know that that if a function

1. $f\left(x\right)$ is continuous on the closed interval $\left[a,b\right]$
2. Differentiable on the open interval $\left(a,b\right)$ such that $f\left(a\right)=f\left(b\right)$
3. $f\text{'}\left(x\right)=0$ for some $x$ with $a\le x\le b.$

Step 2. Explanation for correct option

For option $\left(\mathrm{B}\right)$

$f\left(x\right)=x^2$ is

1. Continuous in the given interval
2. Differentiable in the given interval and $f\left(-1\right)=f\left(1\right)$
3. $f\text{'}\left(0\right)=0,0\in \left[-1,1\right]$

Therefore $f\left(x\right)=x^2$ satisfy Rolle's theorem.

So option $\left(\mathrm{B}\right)$ is correct

Step 3. Explanation for incorrect option

For option $\left(\mathrm{A}\right)$

$$\begin{gathered} f\left(x\right)=x \\ \Rightarrow f\text{'}\left(x\right)=1 \end{gathered}$$

$\therefore f\text{'}\left(x\right)\ne 0$

So Option $\left(\mathrm{A}\right)$ is discarded

For option $\left(\mathrm{C}\right)$

$f\left(x\right)=2x^3+3$

$\therefore f\left(-1\right)\ne f\left(1\right)$

So Option $\left(\mathrm{C}\right)$ is discarded

For option $\left(\mathrm{D}\right)$

$f\left(x\right)=\left|x\right|$ not deferentiable at $x=0$

So Option $\left(\mathrm{D}\right)$ is discarded

Hence, the correct option is option(B)