Question

If $f(x)=\{\begin{array}{ll}\frac{x^2\log (\cos x)}{\log \left(1+x^2\right)},& x\ne 0\\ 0& ,x=0\end{array}.$, 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,
$f(x)=\{\begin{array}{ll}\frac{x^2\log (\cos x)}{\log \left(1+x^2\right)},& x\ne 0\\ 0,& x=0\end{array}$
$\because$ LHL $($ at $x=0)=\underset{h\to 0}{\lim }\frac{\log (\cos (0-h))}{\frac{\log \left(1+h^2\right)}{h^2}}=\frac{\log (1)}{1}=0$
and $RHL$ ( at $x=0$ ) $=\underset{h\to 0}{\lim }\frac{\log (\cos h)}{\frac{\log \left(1+h^2\right)}{h^2}}=\frac{\log (1)}{1}=0$
and $f(0)=0$
$\therefore f(x)$ is a continuous at $x=0$.
Now, LHD $(\mathrm{at}(x=0))$
$=\underset{h\to 0}{\lim }\frac{\frac{(0-h{)}^2\log (\cos (0-h))}{\log \left(1+h^2\right)}-0}{-h}$
$=\underset{h\to 0}{\lim }\frac{h^2\log (\cos h)}{\frac{\log \left(1+h^2\right)}{-h}}=\underset{h\to 0}{\lim }h\times \frac{\log (\cos h)}{\frac{\log \left(1+h^2\right)}{-h^2}}$
$=0\times \frac{\log (1)}{-1}=0$
and $\mathrm{RHD}(\text{ at }x=0)$
$=\underset{h\to 0}{\lim }\frac{h^2\log (\cos h)}{\frac{\log \left(1+h^2\right)}{h}}$
$=\underset{h\to 0}{\lim }\frac{h\log (\cos h)}{\frac{\log \left(1+h^2\right)}{h^2}}=0\times \frac{\log (1)}{1}=0$
$\therefore f(x)$ is differentiable at $x=0$