All the pairs $(\mathrm{x}, \mathrm{y})$ that satisfy the inequality $2^{\sqrt{\sin ^{2} x-2 \sin x+5}} \cdot \frac{1}{4^{\sin ^{2} y}} \leq 1$ also satisfy the equation:

(A) $2|\sin \mathrm{x}|=3 \sin \mathrm{y}$

(D) $\sin \mathrm{x}=2 \sin \mathrm{y}$

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Trigonometric Ratios and Identities

All the pairs

(x, y)

that satisfy the inequality

2^{s i n^{2} x - 2 s i n x + 5} \cdot \frac{1}{4 ^{s i n^{2} y}} \leq 1

also
satisfy the equation:

Answer: (B)

Write the inequality as

2^{s i n^{2} x - 2 s i n x + 5} \leq 2^{2 s i n^{2} y}

, so

sin^{2} x - 2 sin x + 5 \leq 2 sin^{2} y

. The left side is

(sin x - 1)^{2} + 4 \geq 2

, while the right side

2 sin^{2} y \leq 2

. The inequality can hold only if both equal

2

: that needs

sin x = 1

and

∣ sin y ∣ = 1

, which together give

sin x = ∣ sin y ∣

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option