If f(x)=x⋅a1/x−a−1/xa1/x+a−1/x,x≠0(a>0) and f(0)=0 then

limx→0+ f(x)=limx→0+ xa1/x−a−1/xa1/x+a−1/x=limx→0+ x1−a−2/x1+a−2/xalso, limx→0− f(x)=limx→0− xa2/x−1a2/x−1=0So f is continuous at x = 0f′(0+)=limx→0+ ha1/h−a−1/hha1/h+a−1/h=limx→0+ 1−a−2/h1+a−2/h=1Similarly f′(0−)=−1 Hence, f is not differentiable at x = 0

If f(x)=x⋅a1/x−a−1/xa1/x+a−1/x,x≠0(a>0) and f(0)=0 then

limx→0+ f(x)=limx→0+ xa1/x−a−1/xa1/x+a−1/x=limx→0+ x1−a−2/x1+a−2/xalso, limx→0− f(x)=limx→0− xa2/x−1a2/x−1=0So f is continuous at x = 0f′(0+)=limx→0+ ha1/h−a−1/hha1/h+a−1/h=limx→0+ 1−a−2/h1+a−2/h=1Similarly f′(0−)=−1 Hence, f is not differentiable at x = 0

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If $f (x) = x \cdot \frac{(a^{1 / x} - a^{- 1 / x})}{a^{1 / x} + a^{- 1 / x}}, x \neq 0 (a > 0)$ and $f (0) = 0$ then

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$f$ is differentiable at $x = 0$

None of these

$f$ is not continuous at $x = 0$

$f$ is not differentiable at $x = 0$

answer is B.

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Detailed Solution

$\begin{array}{l} lim_{x \to 0 +} f (x) = lim_{x \to 0 +} \frac{x (a^{1 / x} - a^{- 1 / x})}{a^{1 / x} + a^{- 1 / x}} \\ = lim_{x \to 0 +} \frac{x (1 - a^{- 2 / x})}{1 + a^{- 2 / x}} \end{array}$

also, $lim_{x \to 0 -} f (x) = lim_{x \to 0 -} \frac{x (a^{2 / x} - 1)}{a^{2 / x} - 1} = 0$

So $f$ is continuous at $x = 0$

$\begin{array}{l} f^{'} (0 +) = lim_{x \to 0 +} \frac{h (a^{1 / h} - a^{- 1 / h})}{h (a^{1 / h} + a^{- 1 / h})} \\ = lim_{x \to 0 +} \frac{1 - a^{- 2 / h}}{1 + a^{- 2 / h}} = 1 \end{array}$