Official solution
\[A = {\sin ^2}x + {\cos ^4}x\]
\[ = {\sin ^2}x + {\cos ^2}x\left( {1 - {{\sin }^2}x} \right)\]
\[ = {\sin ^2}x + {\cos ^2}x - {1 \over 4}{\left( {2\sin x.\cos x} \right)^2}\]
\[ = 1 - {1 \over 4}{\sin ^2}\left( {2x} \right)\]
Now \[0 \le {\sin ^2}\left( {2x} \right) \le 1\]
\[ \Rightarrow 0 \ge - {1 \over 4}{\sin ^2}\left( {2x} \right) \ge - {1 \over 4}\]
\[ \Rightarrow 1 \ge 1 - {1 \over 4}{\sin ^2}\left( {2x} \right) \ge 1 - {1 \over 4}\]
\[ \Rightarrow 1 \ge A \ge {3 \over 4}\]