Question

If $f\left(x\right)={\sin }^{-1}\sqrt{\frac{\sqrt{1+x^2}-1}{2\sqrt{1+x^2}}},$ then which of the following is (are) CORRECT?

• A. $f^{\prime }(-1)=\frac{-1}{4}$
• B. Range of $f\left(x\right)$ is $[0,\frac{\pi }{2}]$
• C. $f^{\prime }\left(x\right)$ is an odd function
• D. $\underset{x\to 0}{lim}\frac{f\left(x\right)}{x}=\frac{1}{2}$

Answer 1

Answer: (A), (C)

$f^{\prime }\left(x\right)$ is an odd function

Put $x=\tan \theta$ where $\theta \in (-\frac{\pi }{2},\frac{\pi }{2})$
$f\left(x\right)={\sin }^{-1}\left(\sqrt{\frac{\sec \theta -1}{2\sec \theta }}\right)$
$\Rightarrow f\left(x\right)={\sin }^{-1}\left(\sqrt{\frac{1-\cos \theta }{2}}\right)$
$\Rightarrow f\left(x\right)={\sin }^{-1}\left|\sin \frac{\theta }{2}\right|$

$f\left(x\right)=\{\begin{array}{cc}-{\sin }^{-1}\left(\sin \frac{\theta }{2}\right);& \frac{-\pi }{2}<\theta <0\\ {\sin }^{-1}\left(\sin \frac{\theta }{2}\right);& 0\le \theta <\frac{\pi }{2}\end{array}$

$\therefore f\left(x\right)=\{\begin{array}{cc}-\frac{1}{2}{\tan }^{-1}x;& -\infty <x<0\\ \frac{1}{2}{\tan }^{-1}x;& 0\le x<\infty \end{array}$

$f^{\prime }\left(x\right)=\{\begin{array}{cc}\frac{-1}{2(1+x^2)};& -\infty <x<0\\ \frac{1}{2(1+x^2)};& 0\le x<\infty \end{array}$

$\therefore f^{\prime }(-1)=\frac{-1}{4}$
Range of $f\left(x\right)$ is $[0,\frac{\pi }{4}]$
Also $f^{\prime }\left(x\right)$ is an odd function.

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x}\ne \frac{1}{2}$
as $f\left(0^{+}\right)=\frac{1}{2}$ and $f\left(0^{-}\right)=\frac{-1}{2}$