A) continuous on \[[-1,1]\]
B) differentiable on \[(-1,0)\cup (0,1)\]
C) Both (a) and (b)
D) None of the above
Correct Answer: A
Solution :
\[f(x)=\sqrt{1-\sqrt{1-{{x}^{2}}}}\]clearly, \[f(x)\] is continuous on\[[-1,1].\] The domain of \[f(x)\]is \[[-1,1].\] For\[x\ne 0,x\ne \pm 1,\] we have \[f(x)=\frac{1}{2\sqrt{1-\sqrt{1-{{x}^{2}}}}}.\frac{x}{\sqrt{1-{{x}^{2}}}}\] Since, \[f(x)\]is not defined on the right side of \[x=1\]and on the left side of \[x=-1.\]Also, \[f(x)\to \infty \]when\[x\to 1\] or \[x\to -1.\]So, we check the differentiability at \[x=0.\] Now, \[\angle f(0)=\underset{x\to \bar{0}}{\mathop{\lim }}\,\frac{f(x)-f(0)}{x-0}\] \[\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0-h)-f(0)}{-h}=-\frac{1}{\sqrt{2}}\] Similarly, \[Rf(0)=-\frac{1}{\sqrt{2}}\] Hence,\[f(x)\] is not differentiate at \[x=0.\]
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