JEE MAIN - Mathematics (2011 - No. 21)
If $$A = {\sin ^2}x + {\cos ^4}x,$$ then for all real $$x$$:
$${{13} \over {16}} \le A \le 1$$
$$1 \le A \le 2$$
$${3 \over 4} \le A \le {{13} \over {16}}$$
$${{3} \over {4}} \le A \le 1$$
Explanation
$$A = {\sin ^2}x + {\cos ^4}x$$
$$ = {\sin ^2}x + {\cos ^2}x\left( {1 - {{\sin }^2}x} \right)$$
$$ = {\sin ^2}x + {\cos ^2}x - {1 \over 4}{\left( {2\sin x.\cos x} \right)^2}$$
$$ = 1 - {1 \over 4}{\sin ^2}\left( {2x} \right)$$
Now $$0 \le {\sin ^2}\left( {2x} \right) \le 1$$
$$ \Rightarrow 0 \ge - {1 \over 4}{\sin ^2}\left( {2x} \right) \ge - {1 \over 4}$$
$$ \Rightarrow 1 \ge 1 - {1 \over 4}{\sin ^2}\left( {2x} \right) \ge 1 - {1 \over 4}$$
$$ \Rightarrow 1 \ge A \ge {3 \over 4}$$
$$ = {\sin ^2}x + {\cos ^2}x\left( {1 - {{\sin }^2}x} \right)$$
$$ = {\sin ^2}x + {\cos ^2}x - {1 \over 4}{\left( {2\sin x.\cos x} \right)^2}$$
$$ = 1 - {1 \over 4}{\sin ^2}\left( {2x} \right)$$
Now $$0 \le {\sin ^2}\left( {2x} \right) \le 1$$
$$ \Rightarrow 0 \ge - {1 \over 4}{\sin ^2}\left( {2x} \right) \ge - {1 \over 4}$$
$$ \Rightarrow 1 \ge 1 - {1 \over 4}{\sin ^2}\left( {2x} \right) \ge 1 - {1 \over 4}$$
$$ \Rightarrow 1 \ge A \ge {3 \over 4}$$
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