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If    \mathrm{f(x)=\sqrt{1-\sqrt{1-x^2}}} , then f(x) is
 

Option: 1

continuous on [-1,1] and differentiable on (-1,1)


Option: 2

continuous on [-1,1] and differentiable on (-1,0) \cup(0,1)


Option: 3

continuous and differentiable on [-1,1]


Option: 4

none of these.


Answers (1)

We have, \mathrm{f(x)=\sqrt{1-\sqrt{1-x^2}}}. The domain of definition of \mathrm{f(x)~ is ~[-1,1} 
For \mathrm{x \neq 0, x \neq \pm 1}  we have
\mathrm{ f^{\prime}(x)=\frac{1}{\sqrt{1-\sqrt{1-x^2}}} \times \frac{x}{\sqrt{1-x^2}} }

Since f(x) is not defined on the right side of x=1 and on the left side of x=-1. Also, \mathrm{ f^{\prime}(x) \rightarrow \infty~ when ~x \rightarrow-1^{+}~or ~x \rightarrow 1^{-} } So, we check the differentiablility at x=0.

\mathrm{ \text { Now, (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)}{-h} }
\mathrm{ =\lim _{h \rightarrow 0} \frac{\sqrt{1-\sqrt{1-h^2}}-0}{-h}=-\lim _{h \rightarrow 0} \frac{\sqrt{1-\left(1-(1 / 2) h^2+(3 / 8) h^4+\right)_{\ldots}}}{h} \\ }
\mathrm{ =-\lim _{h \rightarrow 0} \sqrt{\frac{1}{2}-\frac{3}{8} h^2+\ldots . .}=-\frac{1}{\sqrt{2}} }
Similarly , (GHD at    \mathrm{ z=0,=\frac{1}{\sqrt{2} }
Hence, f(x) is not differentiable at z=0.

Posted by

Kshitij

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