Question

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

(a) continuous on [−1, 1] and differentiable on (−1, 1)
(b) continuous on [−1, 1] and differentiable on $\left(-1,0\right)\cup \left(0,1\right)$
(c) continuous and differentiable on [−1, 1]
(d) none of these

Answer 1

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

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ f\left(x\right)=\sqrt{1-\sqrt{1-x^2}} \\ \mathrm{Here},\mathrm{function}\mathrm{will}\mathrm{be}\mathrm{defined}\mathrm{for}\mathrm{those}\mathrm{values}\mathrm{of}x\mathrm{for}\mathrm{which} \\ 1-x^2\ge 0 \\ \Rightarrow 1\ge x^2 \\ \Rightarrow x^2\le 1 \\ \Rightarrow \left|x\right|\le 1 \\ \Rightarrow -1\le x\le 1 \\ \mathrm{Therefore},\mathrm{function}\mathrm{is}\mathrm{continuous}\mathrm{in}\left[-1,1\right] \end{gathered}$$
$$\begin{gathered} \mathrm{Now},\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{check}\mathrm{the}\mathrm{differentiability}\mathrm{of}f\left(x\right)=\sqrt{1-\sqrt{1-x^2}}\mathrm{in}\mathrm{the}\mathrm{interval}\left(-1,1\right). \\ \mathrm{Now},\mathrm{we}\mathrm{will}\mathrm{check}\mathrm{the}\mathrm{differentiability}\mathrm{at}x=0 \\ \left(\mathrm{LHD}\mathrm{at}\mathit{}x=0\right)=\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0} \\ =\underset{x\to 0^{-}}{\lim }\frac{\sqrt{1-\sqrt{1-x^2}}-0}{x} \\ =\underset{x\to 0^{-}}{\lim }\frac{\sqrt{1-\sqrt{1-x^2}}}{x} \\ =\underset{h\to 0}{\lim }\frac{\sqrt{1-\sqrt{1-{\left(0-h\right)}^2}}}{0-h} \\ =\underset{h\to 0}{\lim }\frac{\sqrt{1-\sqrt{1-h^2}}}{-h}=-\infty \\ \left(\mathrm{RHD}at\mathit{}x=0\right)=\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0} \\ =\underset{x\to 0^{+}}{\lim }\frac{\sqrt{1-\sqrt{1-x^2}}-0}{x} \\ =\underset{x\to 0^{+}}{\lim }\frac{\sqrt{1-\sqrt{1-x^2}}}{x} \\ =\underset{h\to 0}{\lim }\frac{\sqrt{1-\sqrt{1-{\left(0+h\right)}^2}}}{0+h} \\ =\underset{h\to 0}{\lim }\frac{\sqrt{1-\sqrt{1-h^2}}}{h}=\infty \\ \mathrm{So},\mathrm{the}\mathrm{function}\mathrm{is}\mathrm{not}\mathrm{differentiable}\mathrm{at}x=0. \end{gathered}$$