Question

If $f\left(x\right)=|x|+|\sin x|$ for $x\in (-\frac{\pi }{2},\frac{\pi }{2})$, then its left hand derivation at $x=0$ is

• A. 0
• B. -1
• C. -2
• D. -3

Answer 1

The correct option is C -2

$\text{f}\left(x\right)=|x|+|\sin x|=\{\begin{array}{c}-x-\sin x,\text{if}-\frac{\pi }{2}<x\le 0\\ x+\sin x,\text{if}0<x<\frac{\pi }{2}\end{array}$

Hence left hand derivative at $x=0$ is

$\text{f'}(0-)=\frac{d}{dx}(-x-\sin x){\mid }_{x=0}=(-1-\cos x){\mid }_{x=0}=-2$