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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

JEE · Math · previous-year question

  1. A.sin x = |sin y|correct
  2. B.sin x = 2sin y
  3. C.2 sin x = sin y
  4. D.2 |sin x | = 3 sin y

Answer

A. sin x = |sin y|

Explanation

$${2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }} \le {2^{2{{\sin }^2}y}}$$ $$ \Rightarrow $$ $$\sqrt {{{\sin }^2}x - 2\sin x + 5} \le 2{\sin ^2}y$$ $$ \Rightarrow \sqrt {{{\left( {\sin x - 1} \right)}^2} + 4} \le 2{\sin ^2}y$$ it is true when sinx = 1, |siny| = 1 so sinx = |siny|

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