QJEEM19Mad25b: JEE Main 2019 Mathematics - Class 11 JEEM19Mad25b Solution: All the pairs (x, y) that satisfy the inequality also satisfy…

Question

Verified Solution • 2025–26 Session moderate

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

Accepted Answer

{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|

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