If f(x)=p|sin x|+qe^|x|+r|x|^3 and if f(x) is differentiable at x = 0…

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 must have

$p + q = −p − q ⇒ p + q = 0$