Q62 / 125 · BITSAT 2016 · Mathematics

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

  • Correct answer
    A
    continuous as well as differentiable at $\mathrm{x}=0$
  • B
    continuous but not differentiable at $x=0$
  • C
    differentiable but not continuous at $\mathrm{x}=0$
  • D
    neither continuous nor differentiable at $\mathrm{x}=0$

Solution

We have, $$ \begin{aligned} \mathrm{Lf}^{\prime}(0) &=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0-\mathrm{h})-\mathrm{f}(0)}{-\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{-\mathrm{h} \log \cosh }{-\mathrm{h} \log \left(1+\mathrm{h}^{2}\right)} \\ =& \lim _{\mathrm{h} \rightarrow 0} \frac{\log \cosh }{\log \left(1+\mathrm{h}^{2}\right)}\left(\frac{0}{0} \text { form }\right) \end{aligned} $$ $\begin{aligned} &=\lim _{\mathrm{h} \rightarrow 0} \frac{-\tan \mathrm{h}}{2 \mathrm{~h} /\left(1+\mathrm{h}^{2}\right)}=-1 / 2 \\ \mathrm{Rf}^{\prime}(0) &=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0+\mathrm{h})-\mathrm{f}(0)}{\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{h} \log \cosh }{\mathrm{h} \log \left(1+\mathrm{h}^{2}\right)} \\ &=\lim _{\mathrm{h} \rightarrow 0} \frac{\log \cos \mathrm{h}}{\log \left(1+\mathrm{h}^{2}\right)}\left(\frac{0}{0} \text { form }\right) \\ &=\lim _{\mathrm{h} \rightarrow 0} \frac{-\tan \mathrm{h}}{2 \mathrm{~h} /\left(1+\mathrm{h}^{2}\right)}=\frac{-1}{2} \end{aligned}$ Since $\mathrm{Lf}^{\prime}(0)=\operatorname{Rf}^{\prime}(0),$ therefore $\mathrm{f}(\mathrm{x})$ is dif- ferentiable at $x=0$ Since differentiability $\Rightarrow$ continuity, there-fore $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$.

Hence, the correct answer is option: 'A'.

Relevant Questions