Question

If $f(x)=\begin{cases}\frac{x \log \cos x}{\log \left(1+x^{2}\right)}, & x \neq0 \\ 0, & x=0\end{cases}$, then $f(x)$ is

• A. continuous as well as differentiable at x = 0
• B. continuous but not differentiable at x = 0
• C. differentiable but not continuous at x = 0
• D. neither continuous nor differentiable at x = 0

Answer 1

Answer: (A)

We have,
$Lf'(0)=\lim\limits _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}$
$=\lim\limits _{h \rightarrow 0} \frac{-h \log \cosh }{-h \log \left(1+h^{2}\right)}$
$=\lim\limits _{h \rightarrow 0} \frac{\log \cosh }{\log \left(1+h^{2}\right)}\left(\frac{0}{0}\right.$ form $)$
$=\lim\limits _{h \rightarrow 0} \frac{-\tan h}{2 h /\left(1+h^{2}\right)}=-1 / 2$
$Rf '(0)=\lim \limits_{ h \rightarrow 0} \frac{ f (0+ h )- f (0)}{ h }$
$=\lim \limits_{ h \rightarrow 0} \frac{ h \log \cosh }{ h \log \left(1+ h ^{2}\right)}$
$=\lim \limits_{h \rightarrow 0} \frac{\log \cos h}{\log \left(1+h^{2}\right)}\left(\frac{0}{0}\right.$ form $)$
$=\lim\limits _{h \rightarrow 0} \frac{-\tan h}{2 h /\left(1+h^{2}\right)}=\frac{-1}{2}$
Since $\text{Lf}'(0)=\text{Rf};(0)$, therefore $f ( x )$ is differentiable at $x =0$
Since differentiability $\Rightarrow $ continuity, therefore $f(x)$ is continuous at $x=0$.