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 & x 2 + 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 = ${{{1 + \sqrt {17} } \over 2}}$