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

p + q = 0, r is any real number

For $-\frac{π}{2}<x≤0,\,f(x)=-p\sin x+qe^{-x}-rx^3$

So, $f'(0-)=\underset{x→0^-}{\lim}\frac{f(x)-f(0)}{x-0}=\underset{x→0^-}{\lim}\left[-\frac{p\sin x}{x}-q\left(\frac{e^{-x}-1}{-x}\right)-rx^2\right]$

$=-p-q$

For $0<x≤\frac{π}{2},\,f(x)=p\sin x+qe^{x}+rx^3$

$f'(0+)=\underset{x→0^+}{\lim}\frac{f(x)-f(0)}{x-0}=\underset{x→0^+}{\lim}\left[\frac{p\sin x}{x}+q\left(\frac{e^{x}-1}{x}\right)+rx^2\right]$

$=p+q$

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

$p + q = −p − q ⇒ p + q = 0$