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

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$f' (0) = 2$

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

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

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

answer is C.

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Detailed Solution

At $x = 0,$
$L . H . L = \lim_{x \to 0^{-}} f (x) = \lim_{h \to 0} f (0 - h) = \lim_{h \to 0} h^{2} (\frac{e^{- 1 / h} - e^{1 / h}}{e^{- 2 / h} + e^{1 / h}}) = \lim_{h \to 0} h^{2} (\frac{e^{- 2 / h} - 1}{e^{- 2 / h} + 1}) = 0 (\frac{0 - 1}{0 + 1}) = 0 R . H . L . = \lim_{x \to 0^{+}} f (x) = \lim_{h \to 0} f (0 + h) = \lim_{h \to 0} h^{2} (\frac{e^{1 / h} - e^{- 1 / h}}{e^{1 / h} + e^{- 1 / h}}) = \lim_{h \to 0} h^{2} (\frac{1 - e^{- 2 / h}}{1 + e^{- 2 / h}})$
$= 0 (\frac{1 - 0}{1 + 0}) = 0$ and $f (0) = 0$
$∴ L . H . L = R . H . L = f (0)$
Hence, $f (x)$ is continuous at $x = 0.$
Also, $L . H . D = \lim_{h \to 0} \frac{f (0 - h) - f (0)}{- h}$
$= \lim_{h \to 0} \frac{h^{2} \frac{e^{- 1 / h} - e^{1 / h}}{e^{- 1 / h} + e^{1 / h}} - 0}{- h}$
$= - \lim_{h \to 0} h \frac{e^{- 2 / h} - 1}{e^{- 2 / h} + 1} = 0$

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