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JEE Main 201910 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
10 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main 2019 · 10 April · Morning Shift · Mathematics · Trigonometric Equations

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

All the pairs (x,y)( x, y ), that satisfy the inequality 2sin⁡2x−2sin⁡x+5⋅14sin⁡2y≤1{2}^{\sqrt{\sin^{2}x - 2\sin x + 5}} \cdot \dfrac{1}{{4}^{\sin^{2}y}} \le 1 also satisfy the equation:

Options

  1. A2sin⁡x=sin⁡y2\sin x = \sin y
  2. Bsin⁡x=2sin⁡y\sin x = 2\sin y
  3. C∣sin⁡x∣=∣sin⁡y∣\left|\sin x\right| = \left|\sin y\right|
  4. D2∣sin⁡x∣=3sin⁡y2\left|\sin x\right| = 3\sin y

Correct answer

C. ∣sin⁡x∣=∣sin⁡y∣\left|\sin x\right| = \left|\sin y\right|

Step-by-step solution

Given inequality is
2sin⁡2x−2sinx+5≤4sin⁡2y{2}^{\sqrt{\sin^{2} x - 2sin x + 5}} \le {4}^{\sin^{2} y}
⇒sin⁡2x−2sinx+1+4≤2sin⁡2y⇒ \sqrt{\sin^{2} x - 2sin x + 1 + 4} \le 2\sin^{2} y
⇒(sinx−1)2+4≤2sin⁡2y⇒ \sqrt{{\left(sin x - 1\right)}^{2} + 4} \le 2\sin^{2} y
∵2sin⁡2y∈[0,2] and (sinx−1)2+4∈[2,22]∵ 2\sin^{2} y \in \left[0, 2\right] \text{ and } \sqrt{{\left(sin x - 1\right)}^{2} + 4} \in \left[2, 2\sqrt{2}\right]
∵∵ Only equality holds
⇒(sinx−1)2+4=2sin⁡2y=2⇒ \sqrt{{\left(sin x - 1\right)}^{2} + 4} = 2\sin^{2} y = 2
⇒(sinx−1)2+4=4sin⁡4y=4⇒ {\left(sin x - 1\right)}^{2} + 4 = 4\sin^{4} y = 4
⇒sin⁡2y=1⇒∣sin⁡y∣=1 and (sinx−1)2=0⇒ \sin^{2} y = 1 ⇒ \left|\sin y\right| = 1 \text{ and } {\left(sin x - 1\right)}^{2} = 0 ⇒sinx=1⇒ sin x = 1
∴∣sin⁡y∣=sinx∴ \left|\sin y\right| = sin x
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