Rolle's theorem is not applicable to the function $f(x)=|x|$ for $-2 \leq x \leq 2$ because
Answer & explanation
Correct answer: option 2
We have,
$f(x)=|x|= \begin{cases}x, & x \geq 0 \\ -x, & x<0\end{cases}$
∴ (LHD at x = 0) = $\lim\limits_{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}=\lim\limits_{x \rightarrow 0^{-}} \frac{-x-0}{x-0}=-1$
and, (RHD at x = 0) = $\lim\limits_{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}=\lim\limits_{x \rightarrow 0^{+}} \frac{x-0}{x-0}=1$
∴ (LHD at x = 0) ≠ (RHD at x = 0)
So, f(x) is not differentiable at x = 0.
Consequently, Rolle's is not applicable to the given function.