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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main · Mathematics · Trigonometric Equations

All the pairs (x, y) that satisfy the inequality 2^ ^2 x-2 x+5 ,...

All the pairs (x,y)(x, y) that satisfy the inequality 2sin⁡2x−2sin⁡x+5,14sin⁡2y≤12^{\sqrt{\sin ^2 x-2 \sin x+5}}, \frac{1}{4\sin ^2 y} \leq 1 also satisfy the equation

Options

  1. A2∣sin⁡x∣=sin⁡y2|\sin x|=\sin y
  2. B2sin⁡x=sin⁡y2 \sin x=\sin y
  3. Csin⁡x=2sin⁡y\sin x=2 \sin y
  4. Dsin⁡x=∣sin⁡y∣\sin x=|\sin y|

Correct answer

D. sin⁡x=∣sin⁡y∣\sin x=|\sin y|

Step-by-step solution

2sin⁡2x−2sin⁡x+5⋅2−2sin⁡2y≤12^{\sqrt{\sin ^2 x-2 \sin x+5}} \cdot 2^{-2 \sin ^2 y} \leq 12sin⁡2x−2sin⁡x+5≤22sin⁡2ysin⁡2x−2sin⁡x+5≤2sin⁡2y(sin⁡x−1)2+4≤2sin⁡2y\begin{aligned} & 2^{\sqrt{\sin ^2 x-2 \sin x+5}} \leq 2^{2 \sin ^2 y} \\ & \sqrt{\sin ^2 x-2 \sin x+5} \leq 2 \sin ^2 y \\ & \sqrt{(\sin x-1)^2+4} \leq 2 \sin ^2 y\end{aligned}$\begin{aligned} & (\sin x-1)^2+4 \leq 4 \sin ^4 y \\ & -1 \leq \sin y \leq 1 \\ & 0 \leq \sin ^4 y \leq 1 \\ & \text { For } \sin y= \pm 1 \text { (maximum value) } \\ & (\sin x-1)^2+4=4\end{aligned}(sin⁡x−1)2=0sin⁡x=1sin⁡x=∣sin⁡y∣\begin{aligned} & (\sin x-1)^2=0 \\ & \sin x=1 \\ & \sin x=|\sin y|\end{aligned}$
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