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

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

Since |x| + 5 & x² + 1 is always positive

So $${{\left| x \right| + 5} \over {{x^2} + 1}} \ge 0$$

That means this inequality $$ - 1 \le {{\left| x \right| + 5} \over {{x^2} + 1}}$$ always right. So we can ignore it.

So for domain :

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

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

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

$$ \Rightarrow $$ |x| $$ \ge $$ $${{{1 + \sqrt {17} } \over 2}}$$ or |x| $$ \le $$ $${{{1 - \sqrt {17} } \over 2}}$$

As $${{{1 - \sqrt {17} } \over 2}}$$ is < 0 and |x| always $$ \ge $$ 0. So
|x| $$ \le $$ $${{{1 - \sqrt {17} } \over 2}}$$ not possible.

$$ \therefore $$ |x| $$ \ge $$ $${{{1 + \sqrt {17} } \over 2}}$$

x $$ \in $$ $$\left( { - \infty , - {{1 + \sqrt {17} } \over 2}} \right) \cup \left( {{{1 + \sqrt {17} } \over 2},\infty } \right)$$

So, a = $${{{1 + \sqrt {17} } \over 2}}$$