Question

If \[f\left( x \right) = \sqrt{1 - \sqrt{1 - x^2}},\text{ then } f \left( x \right)\text {  is }\] 

पर्याय

• continuous on [−1, 1] and differentiable on (−1, 1)

• continuous on [−1, 1] and differentiable on
  \[\left( - 1, 0 \right) \cup \left( 0, 1 \right)\]

• continuous and differentiable on [−1, 1]

• none of these

Answer 1

\[\left( b \right) \text{continuous on }\left[ - 1, 1 \right] \text { and differentiable on} \left( - 1, 0 \right) \cup \left( 0, 1 \right)\]

We have, 
\[f\left( x \right) = \sqrt{1 - \sqrt{1 - x^2}}\]
\[\text{Here, function will be defined for those values of x for which}\]
\[1 - x^2 \geq 0\]
\[ \Rightarrow 1 \geq x^2 \]
\[ \Rightarrow x^2 \leq 1\]
\[ \Rightarrow \left| x \right| \leq 1\]
\[ \Rightarrow - 1 \leq x \leq 1\]
\[\text{Therefore, function is continuous in} \left[ - 1, 1 \right]\]

\[\text{Now, we need to check the differentiability of }f\left( x \right) = \sqrt{1 - \sqrt{1 - x^2}} in the interval \left( - 1, 1 \right) . \]
\[\text{Now, we will check the differentiability at} x = 0\]
\[\left( \text { LHD at x } = 0 \right) = \lim_{x \to 0^-} \frac{f\left( x \right) - f\left( 0 \right)}{x - 0}\]
\[ = \lim_{x \to 0^-} \frac{\sqrt{1 - \sqrt{1 - x^2}} - 0}{x}\]
\[ = \lim_{x \to 0^-} \frac{\sqrt{1 - \sqrt{1 - x^2}}}{x}\]
\[ = \lim_{h \to 0} \frac{\sqrt{1 - \sqrt{1 - \left( 0 - h \right)^2}}}{0 - h}\]
\[ = \lim_{h \to 0} \frac{\sqrt{1 - \sqrt{1 - h^2}}}{- h} = - \infty \]
\[\left(\text {  RHD at x }= 0 \right) = \lim_{x \to 0^+} \frac{f\left( x \right) - f\left( 0 \right)}{x - 0}\]
\[ = \lim_{x \to 0^+} \frac{\sqrt{1 - \sqrt{1 - x^2}} - 0}{x}\]
\[ = \lim_{x \to 0^+} \frac{\sqrt{1 - \sqrt{1 - x^2}}}{x}\]
\[ = \lim_{h \to 0} \frac{\sqrt{1 - \sqrt{1 - \left( 0 + h \right)^2}}}{0 + h}\]
\[ = \lim_{h \to 0} \frac{\sqrt{1 - \sqrt{1 - h^2}}}{h} = \infty \]

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