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Question:

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

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

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

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

Solution:

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

We have,

$f(x)=\sqrt{1-\sqrt{1-x^{2}}}$

Here, function will be defined for those values of $x$ for which

$1-x^{2} \geq 0$

$\Rightarrow 1 \geq x^{2}$

$\Rightarrow x^{2} \leq 1$

$\Rightarrow|x| \leq 1$

$\Rightarrow-1 \leq x \leq 1$

Therefore, function is continuous in $[-1,1]$

Now, we need to check the differentiability of $f(x)=\sqrt{1-\sqrt{1-x^{2}}}$ in the interval $(-1,1)$.

Now, we will check the differentiability at $x=0$

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

$=\lim _{x \rightarrow 0^{-}} \frac{\sqrt{1-\sqrt{1-x^{2}}-0}}{x}$

$=\lim _{x \rightarrow 0^{-}} \frac{\sqrt{1-\sqrt{1-x^{2}}}}{x}$

$=\lim _{h \rightarrow 0} \frac{\sqrt{1-\sqrt{1-(0-h)^{2}}}}{0-h}$

$=\lim _{h \rightarrow 0} \frac{\sqrt{1-\sqrt{1-h^{2}}}}{-h}=-\infty$

$(\mathrm{RHD}$ at $x=0)=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}$

$=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{1-\sqrt{1-x^{2}}}-0}{x}$

$=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{1-\sqrt{1-x^{2}}}}{x}$

$=\lim _{h \rightarrow 0} \frac{\sqrt{1-\sqrt{1-(0+h)^{2}}}}{0+h}$

$=\lim _{h \rightarrow 0} \frac{\sqrt{1-\sqrt{1-h^{2}}}}{h}=\infty$

So, the function is not differentiable at $x=0$.

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