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

Option: 1

(-\infty, \infty)


Option: 2

(-\infty, 0) \cup(0, \infty)


Option: 3

(-1, \infty)


Option: 4

none of these


Answers (1)

best_answer

Clearly, f(x)is differentiable for all non-zerovalues of x. For \mathrm{ x \neq 0}  we have
\mathrm{ f^{\prime}(x)=\frac{x e^{-x^2}}{\sqrt{1-e^{-x^2}}} }
Now,

\mathrm{ (\text { LHD at } x=0) =\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{x-0} \\ }

\mathrm{ =\lim _{h \rightarrow 0} \frac{\sqrt{1-e^{-h^2}}}{-h}=\lim _{h \rightarrow 0}-\frac{\sqrt{1-e^{-h^2}}}{h} \\ }
\mathrm{ =-\lim _{h \rightarrow 0} \frac{\sqrt{\frac{h^2}{h^2}-1}}{h^2} \times \frac{1}{\sqrt{e^{h^2}}}=-1 }
\mathrm{ \text { and, } (\text { RHD at } x=0) =\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}=\lim _{h \rightarrow 0} \frac{\sqrt{1-e^{-h^2}}-0}{h} \\ }
\mathrm{ =\lim _{h \rightarrow 0} \sqrt{\frac{e^{h^2}-1}{h^2}} \times \frac{1}{\sqrt{e^{h^2}}}=1 }
So, f(x) is not differentiable at x=0. Hence, the' points of differentiability of f(x) are \mathrm{ (-\infty, 0) \cup(0, \infty) }.

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jitender.kumar

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