If $f\left(x\right)=p|\sin x|+q{e}^{\left|x\right|}+r|x{|}^3$ and  if$f\left(x\right)$ is s differentiable at $x=0$,then 

For $-\pi /2<x\le 0,f\left(x\right)=-p\sin x+q{e}^{-x}$$-r{x}^3$

so  $f^{\mathrm{\prime }}(0-)=\underset{x\to 0-}{lim} \frac{f\left(x\right)-f\left(0\right)}{x-0}$

$\begin{array}{r}=\underset{x\to 0}{lim} \left[-\frac{p\sin x}{x}-q\left(\frac{e^{-x}-1}{-x}\right)-r{x}^2\right]\\ =-p-q.\end{array}$

For $0<x<\pi /2,f\left(x\right)=p\sin x+q{e}^x+r{x}^3$

$\begin{array}{l}{f}^{\mathrm{\prime }}(0+)=\underset{x\to 0+}{lim} \frac{f\left(x\right)-f\left(0\right)}{x-0}\\ =\underset{x\to 0+}{lim} \left[\frac{p\sin x}{x}+q\left(\frac{e^x-1}{x}\right)-r{x}^2\right]\\ =p+q.\end{array}$

For $f$ to be differentiable at $x = 0,$ we must have 

$p+q=-p-q\Rightarrow p+q=0$