Official solution
Given that,
\[{\sin ^{ - 1}}x = 2{\sin ^{ - 1}}a\]
We know,
\[ - {\pi \over 2} \le {\sin ^{ - 1}}x \le {\pi \over 2}\]
\[\therefore\] \[\,\,\,\] \[ - {\pi \over 2} \le 2{\sin ^{ - 1}}a \le {\pi \over 2}\]
\[ \Rightarrow - {\pi \over 4} \le {\sin ^{ - 1}}a \le {\pi \over 4}\]
\[ \Rightarrow \sin \left( { - {\pi \over 4}} \right) \le a \le \sin \left( {{\pi \over 4}} \right)\]
\[ \Rightarrow - {1 \over {\sqrt 2 }} \le a \le {1 \over {\sqrt 2 }}\]
\[\therefore\] \[\,\,\,\,\,\,\,\,\left| a \right| \le {1 \over {\sqrt 2 }}\]