Question

If $f(x)=\begin{cases}\frac{x^{2} \log (\cos x)}{\log \left(1+x^{2}\right)}, & x \neq 0 \\ 0 & , x=0\end{cases}.$, then $f$ is

• A. discontinuous at zero
• B. continuous but not differentiable at zero
• C. differentiable at zero
• D. not continuous and not differentiable at zero

Answer 1

Answer: (C)

Given function,
$f(x)=\begin{cases}\frac{x^{2} \log (\cos x)}{\log \left(1+x^{2}\right)}, & x \neq 0 \\ 0, & x=0\end{cases}$
$\because$ LHL $($ at $x=0)=\displaystyle \lim _{h \rightarrow 0} \frac{\log (\cos (0-h))}{\frac{\log \left(1+h^{2}\right)}{h^{2}}}=\frac{\log (1)}{1}=0$
and $RHL$ ( at $x=0$ ) $=\displaystyle \lim _{h \rightarrow 0} \frac{\log (\cos h)}{\frac{\log \left(1+h^{2}\right)}{h^{2}}}=\frac{\log (1)}{1}=0$
and $f(0)=0$
$\therefore f(x)$ is a continuous at $x=0$.
Now, LHD $(\operatorname{at}(x=0)) $
$= \displaystyle \lim _{h \rightarrow 0} \frac{\frac{(0-h)^{2} \log (\cos (0-h))}{\log \left(1+h^{2}\right)}-0}{-h} $
$= \displaystyle \lim _{h \rightarrow 0} \frac{h^{2} \log (\cos h)}{\frac{\log \left(1+h^{2}\right)}{-h}}=\displaystyle \lim _{h \rightarrow 0} h \times \frac{\log (\cos h)}{\frac{\log \left(1+h^{2}\right)}{-h^{2}}} $
$= 0 \times \frac{\log (1)}{-1}=0 $
and $ \operatorname{RHD} (\text { at } x=0) $
$=\displaystyle \lim _{h \rightarrow 0} \frac{h^{2} \log (\cos h)}{\frac{\log \left(1+h^{2}\right)}{h}}$
$=\displaystyle \lim _{h \rightarrow 0} \frac{h \log (\cos h)}{\frac{\log \left(1+h^{2}\right)}{h^{2}}}=0 \times \frac{\log (1)}{1}=0 $
$\therefore f(x)$ is differentiable at $x=0$