Solve this

(b) $a=0, b=0 ; c \in R$

We have,

$f(x)=a|\sin x|+b e^{|x|}+c|x|^{3}$

$= \begin{cases}a \sin x+b e^{x}+c x^{3} & 0

Here, $f(x)$ is differentiable at $x=0$

Therefore, $(\mathrm{LHD}$ at $x=0)=(\mathrm{RHD}$ at $x=0)$

$\Rightarrow \lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}$

$\Rightarrow \lim _{x \rightarrow 0^{-}} \frac{-a \sin x+b e^{-x}-c x^{3}-b}{x}=\lim _{x \rightarrow 0^{+}} \frac{a \sin x+b e^{x}+c x^{3}-b}{x}$

$\Rightarrow \lim _{h \rightarrow 0} \frac{-a \sin (0-h)+b e^{-(0-h)}-c(0-h)^{3}-b}{0-h}=\lim _{h \rightarrow 0} \frac{a \sin (0+h)+b e^{(0+h)}+c(0+h)^{3}-b}{0+h}$

$\Rightarrow \lim _{h \rightarrow 0} \frac{a \sin h+b e^{h}+c h^{3}-b}{-h}=\lim _{h \rightarrow 0} \frac{a \sin h+b e^{h}+c h^{3}-b}{h}$

$\Rightarrow \lim _{h \rightarrow 0} \frac{a \cos h+b e^{h}+3 c h^{2}}{-1}=\lim _{h \rightarrow 0} \frac{a \cos h+b e^{h}+3 c h^{2}}{1}$           (By L'Hospital rule)

$\Rightarrow-(a+b)=a+b$

$\Rightarrow-2(a+b)=0$

$\Rightarrow a+b=0$

This is true for all value of $c$

$\therefore c \in \mathrm{R}$

In the given options, option (b) satisfies $a+b=0$ and $c \in \mathrm{R}$