Question

The number of values of $x\in [0,2]$ at which the real function $f\left(x\right)=|x-\frac{1}{2}|+|x+1|+\tan x$ is not finitely differentiable is

• A. $2$
• B. $3$
• C. $1$
• D. 0

Answer 1

The correct option is D 0

$f\left(x\right)=|x-\frac{1}{2}|+|x+1|+\tan x$

$=\{\begin{array}{c}\frac{1}{2}-x+x+1+\tan x,0\le x\le \frac{1}{2}\\ x-\frac{1}{2}+x+1+\tan x,\frac{1}{2}<x\le 2\end{array}$

$=\{\begin{array}{c}\frac{3}{2}+\tan x,0\le x\le \frac{1}{2}\\ 2x+\frac{1}{2}+\tan x,\frac{1}{2}<x\le 2\end{array}$

$f^{\prime }\left(x\right)=\{\begin{array}{c}{\sec }^{2}x,0\le x\le \frac{1}{2}\\ 2+{\sec }^{2}x,\frac{1}{2}<x\le 2\end{array}$

Clearly at $x=\frac{1}{2},f\left(x\right)$ is not differentiable in $[0,2]$.