If f(x)= 1- 1-x^2 , then f(x) is : - Vidyadip

MCQ

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

A
Continuous on $[-1, 1]$ and differentiable on $(-1, 1)$
✓
Continuous on $[-1, 1]$ and differentiable on $(-1,0)\cup(0,1)$
C
Continuous and differentiable on $[-1, 1]$
D
None of these.

✓

Answer

Correct option: B.

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

We have,
$\text{f(x)}=\sqrt{1-\sqrt{1-\text{x}^2}}$
Here, function will be defined for those values of $x$ for which
$1-\text{x}^2\geq0$
$\Rightarrow1\geq\text{x}^2$
$\Rightarrow\text{x}^2\leq1$
$\Rightarrow|\text{x}\leq1|$
$\Rightarrow-1\leq\text{x}\leq1$
Therefore, function is continuous in $-1, 1$
Now, we need to check the differentiability of $\text{f(x)}=\sqrt{1-\sqrt{1-\text{x}^2}}$ in the interval $-1, 1$
Now,
We will check the differentiable at $x = 0$
$\text{LHL}$ at $x = 0$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\sqrt{1-\sqrt{1-\text{x}^2}}-0}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\sqrt{1-\sqrt{1-\text{x}^2}}}{\text{x}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{1-\sqrt{1-(0-\text{h})^2}}}{0-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{1-\sqrt{1-\text{h}^2}}}{-\text{h}}$
$=-\infty$
$\text{RHL}$ at $x = 0$
$=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\sqrt{1-\sqrt{1-\text{x}^2}}-0}{\text{x}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{1-\sqrt{1-(0+\text{h})^2}}}{0+\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{1-\sqrt{1-\text{h}^2}}}{\text{h}}$
$=\infty$
So, the function is not differentiable at $x = 0.$

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