Solve this

Question:

If $f(x)=a|\sin x|+b e^{|x|}+c|x|^{3}$ and if $f(x)$ is differentiable at $x=0$, then

(a) $a=b=c=0$

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

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

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

Solution:

(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}\$