$f (x) = \{\begin{array}{cc} x^{2} (\frac{e^{1 / x} - e^{- 1 / x}}{e^{1 / x} + e^{- 1 / x}}), & x \neq 0 \\ 0, & x = 0 \end{array}$ Then

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$f (x)$ is continuous but non-differentiable at $x = 0$

$f^{'} (0) = 2$

$f (x)$ is differntiable at $x = 0$

$f (x)$ is discontinuous at $x = 0$

answer is C.

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Detailed Solution

$A t x = 0 lim_{x \to 0^{-}} f (x) = lim_{h \to 0} h^{2} (\frac{e^{- 1 / h} - e^{1 / h}}{e^{- 1 / h} + e^{1 / h}}) = 0$

$lim_{x \to 0 +} f (x) = lim_{h \to 0} h^{2} (\frac{e^{1 / h} - e^{- 1 / h}}{e^{1 / h} + e^{- 1 / h}}) = 0$

And $f (0) = 0$ , $f (x)$ is continuous at $x = 0 .$

Also $L.H.D. = lim_{h \to 0} h \frac{e^{2 / h} - 1}{e^{2 / h} + 1} = 0$ and

$R.H.D. = lim_{h \to 0} \frac{1 - e^{- 2 / h}}{1 + e^{2 / h}} = 0$

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