NTA Abhyas JEE Main2020MathematicsTrigonometric EquationsPractice
If x and y are the solutions of the equation 5 + 8 s i n 2 x = 2 y 2 - 8 y + 21 , then the least possible value of x 2 y 3 is
Options
- A2 π 2
- B4 π 2
- C9 π 2
- Dπ 2
Correct answer
A. 2 π 2
Step-by-step solution
∵ 5 + 8 s i n 2 x ∈ 5 , 13 Also, 2 y 2 - 8 y + 21 = 2 y - 2 2 + 13 ≥ 13 So, equality should hold true if, 5 + 8 s i n 2 x = 13 and 2 y 2 - 8 y + 21 = 13 ⇒ s i n x = ± 1 ,   y = 2 ⇒ s i n x = 2 n + 1 π 2 ,   y = 2 For least value, x = ± π 2 Hence, x 2 y 3 = π 2 4 · 8 = 2 π 2