Mathematics - Functions Question with Solution | TestHub

$f\left(x\right)={\cos }^{-1}\left(\frac{2{\sin }^{-1}\left(\frac{1}{4x^2-1}\right)}{\pi }\right)$

Case I: $-1\le \frac{1}{4x^2-1}\le 1$

$\Rightarrow 4x^2-1\ge 1$ or $4x^2-1\le -1$ and $4x^2-1\ne 0$

$\therefore x^2\ge \frac{1}{2}$ or $x^2\le 0\Rightarrow x=0$ and $x\ne \pm \frac{1}{2}$

$\therefore x\in \left(-\infty ,-\frac{1}{\sqrt{2}}\right]\cup \left[\frac{1}{\sqrt{2}},\infty \right)\cup \left\{0\right\}$

Case II: $-1\le \frac{2{\sin }^{-1}\left(\frac{1}{4x^2-1}\right)}{\pi }\le 1$

$\Rightarrow -\frac{\mathrm{\pi }}{2}\le {\sin }^{-1}\left(\frac{1}{4x^2-1}\right)\le \frac{\mathrm{\pi }}{2}$ which is always true.

Hence, $x\in \left(-\infty ,-\frac{1}{\sqrt{2}}\right]\cup \left[\frac{1}{\sqrt{2}},\infty \right)\cup \left\{0\right\}$