JEE MAIN - Mathematics (2019 - 10th April Morning Slot - No. 7)
All the pairs (x, y) that satisfy the inequality
$${2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }}.{1 \over {{4^{{{\sin }^2}y}}}} \le 1$$
also satisfy the equation
$${2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }}.{1 \over {{4^{{{\sin }^2}y}}}} \le 1$$
also satisfy the equation
sin x = |sin y|
sin x = 2sin y
2 sin x = sin y
2 |sin x | = 3 sin y
Explanation
$${2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }} \le {2^{2{{\sin }^2}y}}$$
$$ \Rightarrow $$ $$\sqrt {{{\sin }^2}x - 2\sin x + 5} \le 2{\sin ^2}y$$
$$ \Rightarrow \sqrt {{{\left( {\sin x - 1} \right)}^2} + 4} \le 2{\sin ^2}y$$
it is true when sinx = 1, |siny| = 1
so sinx = |siny|
$$ \Rightarrow $$ $$\sqrt {{{\sin }^2}x - 2\sin x + 5} \le 2{\sin ^2}y$$
$$ \Rightarrow \sqrt {{{\left( {\sin x - 1} \right)}^2} + 4} \le 2{\sin ^2}y$$
it is true when sinx = 1, |siny| = 1
so sinx = |siny|
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