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Trigonometrical Equations
hard
All the pairs $(x, y)$ that satisfy the inequality ${2^{\sqrt {{{\sin }^2}{\kern 1pt} x - 2\sin {\kern 1pt} x + 5} }}.\frac{1}{{{4^{{{\sin }^2}\,y}}}} \leq 1$ also Satisfy the equation
A
$2\left| {\sin \,x} \right| = 3\sin \,y$
B
$\sin \,x = \left| {\sin \,y} \right|$
C
$2\,sin\, x = sin\, y$
D
$sin\, x = 2\, sin\, y$
(JEE MAIN-2019)
Solution
$2^{\sqrt{\sin ^{2} x-2 \sin x+5}}-4^{-\sin ^{2} y} \leq 1$
$ \Rightarrow \,{2^{\sqrt {{{(\sin \,x – 1)}^2} + 4} }}\, \leqslant {4^{{{\sin }^2}y}}$
$\sqrt {\frac{{\underbrace {{{(\sin \,x – 1)}^2}}_{ \geqslant 0} + 4}}{{ \geqslant 2}}} \leqslant \underbrace {2\,{{\sin }^2}\,y}_{ \leqslant 2}$
this is possible only if $\sin x=1$ and $|\sin y|=1$
Standard 11
Mathematics
