Official solution
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 = \[{{{1 + \sqrt {17} } \over 2}}\]