Question

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

• A. \( 2|\sin x| = \sin y \)
• B. \( 2\sin x = \sin y \)
• C. \( \sin x = 2\sin y \)
• D. \( \sin x = |\sin y| \)

Hint:

Inequalities involving absolute values or powers often reduce to simple equality forms under specific bounds. Try squaring both sides and substituting identities.

Answer 1

The Correct Option is D

Start with simplifying the inequality: \[ 2\sqrt{\sin^2x - 2\sin x + 5} \cdot \frac{1}{4^{\sin^2 y}} \leq 1 \] Let \( \sin x = s \), \( \sin y = t \). Then: \[ 2\sqrt{s^2 - 2s + 5} \cdot \frac{1}{4^{t^2}} \leq 1 \Rightarrow \sqrt{(s - 1)^2 + 4} \leq \frac{1}{2^{2t^2 - 1}} \] Now, test equality. Try \( s = 1 \Rightarrow \sqrt{4} = 2 \), so: \[ 2 \cdot 2 \cdot \frac{1}{4^{t^2}} = \frac{4}{4^{t^2}} \leq 1 \Rightarrow 4^{1 - t^2} \leq 1 \Rightarrow 1 - t^2 \leq 0 \Rightarrow t^2 \geq 1 \Rightarrow t = \pm 1 \Rightarrow \sin y = \pm 1 \] So \( \sin x = |\sin y| \). This holds generally over the domain under the constraint.