Question

Which of the following function fail to satisfy the condition of Rolle's theorem on interval $[-1,1]$.

• A. $f\left(x\right)=|x|+|\sin x|$
• B. $f\left(x\right)=\left\{x\right\}+\{-x\}$
• C. $f\left(x\right)=\left[\begin{array}{c}\frac{\tan (x^2-1)}{1-x^2}+\frac{{\log }_e|x|}{|x|-1},x\ne 1,-1,0\\ 0,x=0\end{array}\right]$
• D. $f\left(x\right)=|x|-|\sin x|$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $f\left(x\right)=|x|+|\sin x|$
B $f\left(x\right)=\left\{x\right\}+\{-x\}$
C $f\left(x\right)=\left[\begin{array}{c}\frac{\tan (x^2-1)}{1-x^2}+\frac{{\log }_e|x|}{|x|-1},x\ne 1,-1,0\\ 0,x=0\end{array}\right]$

$f\left(x\right)=|x|+|\sin x|$ is non-differentiable at $x=0$.
$f\left(x\right)=\left\{x\right\}+\{-x\}$ is discontinuous at $x=0$
$f\left(x\right)=\left[\begin{array}{c}\frac{\tan (x^2-1)}{1-x^2}+\frac{{\log }_e|x|}{|x|-1},x\ne 1,-1,0\\ 0,x=0\end{array}\right]$ is non differentiable at $x=1,-1,0$
$f\left(x\right)=|x|-|\sin x|$ is continuous & differentiable $\forall x\in [-1,1]$, so Rolle's theorem is applicable.