Let $f(x)=a+b|x|+c|x|^{4}$, where $a, b$, and $c$ are real constants. Then, $f(x)$ is differentiable at $x=0$, if
(a) $a=0$
(b) $b=0$
(c) $c=0$
(d) none of these
(b) $b=0$
We have,
$f(x)=a+b|x|+c|x|^{4}$
$f(x)= \begin{cases}a+b x+c x^{4} & x \geq 0 \\ a-b x+c x^{4} & x<0\end{cases}$
Here, $f(x)$ is differentiable at $x=0$
$\therefore(\mathrm{LHD}$ at $x=0)=(\mathrm{RHD}$ at $x=0)$
$\Rightarrow \lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}$
$\Rightarrow \lim _{x \rightarrow 0^{-}} \frac{a-b x+c x^{4}-a}{x}=\lim _{x \rightarrow 0^{+}} \frac{a+b x+c x^{4}-a}{x}$
$\Rightarrow \lim _{h \rightarrow 0} \frac{a-b(0-h)+c(0-h)^{4}-a}{0-h}=\lim _{h \rightarrow 0} \frac{a+b(0+h)+c(0+h)^{4}-a}{0+h}$
$\Rightarrow \lim _{h \rightarrow 0} \frac{a+b h+c h^{4}-a}{-h}=\lim _{h \rightarrow 0} \frac{a+b h+c h^{4}-a}{h}$
$\Rightarrow \lim _{h \rightarrow 0} \frac{b h+c h^{4}}{-h}=\lim _{h \rightarrow 0} \frac{b h+c h^{4}}{h}$
$\Rightarrow \lim _{h \rightarrow 0}\left(-b-c h^{3}\right)=\lim _{h \rightarrow 0}\left(b+c h^{3}\right)$
$\Rightarrow-b=b$
$\Rightarrow 2 b=0$
$\Rightarrow b=0$