Q131 / 225 · JEE Main 2003 · Mathematics

If $f(x)= \begin{cases}x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}, & x \neq 0 \text { then } f(x) \text { is } \\ 0 & , x=0\end{cases}$

  • A
    discontinuous every where
  • B
    continuous as well as differentiable for all $x$
  • Correct answer
    C
    continuous for all $x$ but not differentiable at $x=0$
  • D
    neither differentiable nor continuous at $x=0$

Solution

$f(0)=0 ; f(x)=x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}$ R.H.L. $\lim _{h \rightarrow 0}(0+h) \mathrm{e}^{-2 / h}=\lim _{h \rightarrow 0} \frac{h}{\mathrm{e}^{2 / h}}=0$ L.H.L $\operatorname{Lim}_{h \rightarrow 0}(0-h) e^{-\left(\frac{1}{h}-\frac{1}{h}\right)}=0$ Therefore, $f(x)$ is continuous R.H.D. $\lim _{h \rightarrow 0} \frac{(0+h) e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}-h e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}}{h}=0$ L.H.D. $\operatorname{Lim}_{h \rightarrow 0} \frac{(0-h) \mathrm{e}^{-\left(\frac{1}{h}-\frac{1}{h}\right)}-h e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}}{-h}=1$ Therefore, L.H.D. $\neq$ R.H.D. $\mathrm{f}(\mathrm{x})$ is not differentiable at $\mathrm{x}=0$.

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

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