JEE Mains · Maths · STD 11 - Trigonometrical equations
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 \(sin\, x = 2\, sin\, y\)
Answer & Solution
Correct Answer
(B) \(\sin \,x = \left| {\sin \,y} \right|\)
Step-by-step Solution
Detailed explanation
\(2^{\sqrt{\sin ^{2} x-2 \sin x+5}}-4^{-\sin ^{2} y} \leq 1\) \( \Rightarrow \,{2^{\sqrt {{{(\sin \,x - 1)}^2} + 4} }}\, \leqslant {4^{{{\sin }^2}y}}\)…
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