# Solve this

Question:

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

Solution:

(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 $x=0)$

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

$\Rightarrow \lim _{x \rightarrow 0^{-}} \frac{a-b x+c x^{4}-a}{x}=\lim _{x \rightarrow 0^{+}} \frac{a+b x+c x^{4}-a}{x}$

$\Rightarrow \lim _{h \rightarrow 0} \frac{a-b(0-h)+c(0-h)^{4}-a}{0-h}=\lim _{h \rightarrow 0} \frac{a+b(0+h)+c(0+h)^{4}-a}{0+h}$

$\Rightarrow \lim _{h \rightarrow 0} \frac{a+b h+c h^{4}-a}{-h}=\lim _{h \rightarrow 0} \frac{a+b h+c h^{4}-a}{h}$

$\Rightarrow \lim _{h \rightarrow 0} \frac{b h+c h^{4}}{-h}=\lim _{h \rightarrow 0} \frac{b h+c h^{4}}{h}$

$\Rightarrow \lim _{h \rightarrow 0}\left(-b-c h^{3}\right)=\lim _{h \rightarrow 0}\left(b+c h^{3}\right)$

$\Rightarrow-b=b$

$\Rightarrow 2 b=0$

$\Rightarrow b=0$

Leave a comment

Click here to get exam-ready with eSaral