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The set of points where the function $f(x)=\sqrt{1-e^{-x^2}}$ is differentiable is

$\begin{array}{1 1}(a)\;(-\infty,\infty)&(b)\;(-\infty,0)\cup (0,\infty)\\(c)\;(-1,\infty)&(d)\;\text{None of these}\end{array}$

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$f(x)$ is clearly differentiable for each $x\neq 0$ because
$f'(x)=\large\frac{xe^{-x^2}}{\sqrt{1-e^{-x^2}}}$ exists for $x\neq 0$
Now $f'(0^-)=\lim\limits_{x\to 0^-}\large\frac{f(x)-f(0)}{x-0}$
$\Rightarrow \lim\limits_{\large h\to 0^+}\large\frac{f(0-h)-f(0)}{0-h-0}$
$\Rightarrow -\lim\limits_{\large h\to 0}\large\frac{\sqrt{1-e^{-h^2}}}{h}$
$\Rightarrow -\lim\limits_{\large h\to 0}\large\frac{\sqrt{e^{h^2-1}}}{h^2}\times \frac{1}{\sqrt{e^{h^2}}}$
$\Rightarrow -1$
$f'(0^+)=\lim\limits_{x\to 0^+}\large\frac{f(x)-f(0)}{x-0}$
$\qquad\;\;\;=\lim\limits_{h\to 0^+}\large\frac{f(0+h)-f(0)}{h}$
$\Rightarrow 1$
$\therefore f$ is not differentiable at $x=0$
Hence (d) is the correct answer.
answered Jan 6, 2014 by sreemathi.v
 

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