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Differentiability — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiability1 Mark
MCQ
If $\text{f(x)}=\begin{cases}\frac{1}{1+\text{e}^{\frac{1}{\text{x}}}},&\text{x}\neq0\\0,&\text{x}=0\end{cases}$ 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 $x = 0$
  • ✓
    None of these.
✓

Answer

Correct option: D.
None of these.
$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{1}{1+\text{e}^{\frac{1}{\text{x}}}}=\lim\limits _{\text{x}\rightarrow0}\frac{1}{1+\text{e}^{\frac{-1}{\text{x}}}}=1\ \Big(\because\lim\limits_{\text{x}\rightarrow0}\text{e}^{\frac{-1}{\text{x}}}=0\Big)$
$\lim\limits_{\text{x}\rightarrow0^{-}}\text{f(x)}\neq\text{f}(0)$
Function is not continuous,
$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(-\text{h})-\text{f}(0)}{-\text{h}}$
$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}}=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{1}{1+\text{e}^{\frac{1}{\text{h}}}}-0}{-\text{h}}$
$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}}=\lim\limits_{\text{h}\rightarrow0}\frac{1}{-\Big(1+\text{e}^{\frac{1}{\text{h}}}\Big)\text{h}}=-\infty$
Similarly,
$\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}}=\infty$

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