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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Trigonometric Equations

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

If x and y are the solutions of the equation 5 + 8sin^ 2 x = 2y^ 2...

If xx and yy are the solutions of the equation 5+8sin2x=2y2−8y+215 + 8sin^{2}x = 2y^{2} - 8y + 21, then the least possible value of x2y3x^{2}y^{3} is

Options

  1. A2π22{\pi}^{2}
  2. B4π24{\pi}^{2}
  3. C9π29{\pi}^{2}
  4. Dπ2{\pi}^{2}

Correct answer

A. 2π22{\pi}^{2}

Step-by-step solution

∵5+8sin2x∈[5,13]∵ 5 + 8si{n}^{2}x \in \left[5, 13\right]
Also, 2y2−8y+21=2(y−2)2+13≥132{y}^{2} - 8y + 21 = 2{\left(y - 2\right)}^{2} + 13 \ge 13
So, equality should hold true if,
5+8sin2x=135 + 8si{n}^{2}x = 13 and 2y2−8y+21=132{y}^{2} - 8y + 21 = 13
⇒sin⁡x=±1,y=2⇒ \sin x = \pm 1, y = 2
⇒sin⁡x=(2n+1)π2,y=2⇒ \sin x = \left(2n + 1\right)\dfrac{\pi}{2}, y = 2
For least value, x=±π2x = \pm \dfrac{\pi}{2}
Hence, x2y3=π24⋅8=2π2{x}^{2}{y}^{3} = \dfrac{{\pi}^{2}}{4} \cdot 8 = 2{\pi}^{2}
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