JEE Main 2019 — Math solution

JEE Main2019Math

All the pairs (x, y) that satisfy the inequality $2^{s i n^{2} x - 2 s i n x + 5} . \frac{1}{4 ^{s i n^{2} y}} ⩽ 1$ also satisfy the equation :

A
sin x = |sin y|
B
sin x = 2 sin y
C
2 sin x = sin y
D
2 |sin x| = 3 sin y

Solution

$2^{s i n^{2} x - s i n x + 5} \leq 2^{2 s i n^{2}} \Rightarrow sin^{2} x - 2 sin x + 5 \leq 2 sin^{2} y \Rightarrow (sin x - 1)^{2} + 4 \leq 2 sin^{2} y$

it is true when sinx = 1

|siny| = 1

so sinx = |siny|

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