Mathematics - Functions Question with Solution | TestHub

$f\left(x\right)={\sin }^{-1}\left(\frac{\left|x\right|+5}{x^2+1}\right)$

$\Rightarrow \begin{array}{l}-1\le \frac{\left|x\right|+5}{x^2+1}\le 1\end{array}, \because x\in \left[-1,1\right] \mathrm{for} {\sin }^{-1}x$ to exist.

Case $1$

$-x^2-1\le \left|x\right|+5$

$\Rightarrow x^2+1+\left|x\right|+5\ge 0$, true for all $x\in R$.

Case $2$

$\left|x\right|+5\le x^2+1$

$\Rightarrow x^2-\left|x\right|-4\ge 0$

Let $\left|x\right|=t\Rightarrow t^2-t-4\ge 0$

$\Rightarrow \left(\left|x\right|+\frac{\sqrt{17}-1}{2}\right)\left(\left|x\right|-\frac{\sqrt{17}+1}{2}\right)\ge 0$

$\Rightarrow x\in \left(-\infty ,-\frac{\sqrt{17}+1}{2}\right]\cup \left[\frac{\sqrt{17}+1}{2},\infty \right)$

$\therefore a=\frac{1+\sqrt{17}}{2}$