Question

If $f\left(x\right)={\tan }^{-1}\left(\frac{1-\sqrt{x^2-1}}{\sqrt{x^2+2\sqrt{x^2-1}}}\right)+{\sin }^{-1}\left(\frac{\sqrt{x^2-1}}{|x|}\right)$, then the number of solution(s) of the equation $\tan \left(f\right(x\left)\right)=|x^2-2|$ is

Answer 1

$f\left(x\right)={\tan }^{-1}\left(\frac{1-\sqrt{x^2-1}}{\sqrt{x^2+2\sqrt{x^2-1}}}\right)+{\sin }^{-1}\left(\frac{\sqrt{x^2-1}}{|x|}\right)$
For the root to be defined, we get
$x^2-1\ge 0\Rightarrow x\in (-\infty ,-1]\cup [1,\infty )$
Putting $x=\sec \theta$, we get
$f\left(x\right)={\tan }^{-1}\left(\frac{1-\sqrt{{\sec }^2\theta -1}}{\sqrt{{\sec }^2\theta +2\sqrt{{\sec }^2\theta -1}}}\right)+{\sin }^{-1}\left(\frac{\sqrt{{\sec }^2\theta -1}}{|\sec \theta |}\right)\Rightarrow f\left(x\right)={\tan }^{-1}\left(\frac{1-|\tan \theta |}{\sqrt{{\sec }^2\theta +2|\tan \theta |}}\right)+{\sin }^{-1}\left(\frac{|\tan \theta |}{|\sec \theta |}\right)\Rightarrow f\left(x\right)={\tan }^{-1}\left(\frac{1-|\tan \theta |}{\sqrt{1+{\tan }^2\theta +2|\tan \theta |}}\right)+{\sin }^{-1}\left(|\sin \theta |\right)\Rightarrow f\left(x\right)={\tan }^{-1}\left(\frac{1-|\tan \theta |}{|1+|\tan \theta ||}\right)+{\sin }^{-1}\left(|\sin \theta |\right)$

When $\theta \in [0,\frac{\pi }{2}],$
$f\left(x\right)={\tan }^{-1}\left(\frac{1-\tan \theta }{1+\tan \theta }\right)+{\sin }^{-1}\left(\sin \theta \right)\Rightarrow f\left(x\right)={\tan }^{-1}\left[\tan (\frac{\pi }{4}-\theta )\right]+\theta \Rightarrow f\left(x\right)=\frac{\pi }{4}-\theta +\theta \Rightarrow f\left(x\right)=\frac{\pi }{4}$

When $\theta \in [\frac{\pi }{2},\pi ],$
$f\left(x\right)={\tan }^{-1}\left(\frac{1+\tan \theta }{1-\tan \theta }\right)+{\sin }^{-1}\left(\sin \theta \right)\Rightarrow f\left(x\right)={\tan }^{-1}\left[\tan (\frac{\pi }{4}+\theta )\right]+{\sin }^{-1}\left(\sin \theta \right)\Rightarrow f\left(x\right)=\frac{\pi }{4}+\theta -\pi +(\pi -\theta )\Rightarrow f\left(x\right)=\frac{\pi }{4}$

Now, $\tan \left(f\right(x\left)\right)=|x^2-2|$
$\Rightarrow |x^2-2|=\tan {45}^{\circ }\Rightarrow x^2-2=\pm 1\Rightarrow x^2=1,3\therefore x=\pm 1,\pm \sqrt{3}$

Hence, the number of solutions is $4.$