The domain of the function f ( x ) = ^ - 1 ( |x| + 5 x ^ 2 + 1 ) is...

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

$⇒$$\begin{array}{ll}- 1 \le \dfrac{\left|x\right| + 5}{{x}^{2} + 1} \le 1\end{array}$$, ∵ x \in \left[- 1, 1\right]for\sin^{- 1}x$ to exist.

Case $1$

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

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

Case $2$

\left|x\right| + 5 \le {x}^{2} + 1$</p><p>$⇒ {x}^{2} - \left|x\right| - 4 \ge 0$</p><p>Let \left|x\right| = t ⇒ {t}^{2} - t - 4 \ge 0$

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

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

$∴ a = \dfrac{1 + \sqrt{17}}{2}$