JEE Main 2019 (Online) 10th April Morning ShiftQuestion 1
All the pairs $(x,y)$ that satisfy the inequality $2^{\sqrt{\sin^{2}x-2\sin x+5}}\cdot\frac{1}{4^{\sin^{2}y}}\le1$ also satisfy the equation :
Trigonometric Ratio & Identities
Solution
Concept. The exponent on the left is at least $2$ and the one on the right is at most $2$; both must be exactly $2$. Formulas used. (1) $\sin^2x-2\sin x+5=(\sin x-1)^2+4\ge4$Step 1 - Rewrite. The inequality is $2^{\sqrt{\sin^2x-2\sin x+5}}\le2^{2\sin^2y}$, i.e. $$\sqrt{(\sin x-1)^2+4}\le2\sin^2y.$$ Step 2 - Squeeze. By (1) the left side is at least $2$, and the right side is at most $2$. So both equal $2$: $\sin x=1$ and $\sin^2y=1$. Step 3 - Check the options. $|\sin x|=|\sin y|=1$. (A), (B), (D) fail (for example $2\sin x=\sin y$ would need $2=\pm1$). Answer: (C)