JEE Main202127 Jul 2021Morning ShiftMathematicsDeterminantsActual
Let f x = sin 2 x - 2 + cos 2 x cos 2 x 2 + sin 2 x cos 2 x cos 2 x sin 2 x cos 2 x 1 + cos 2 x , x ∈ 0 , π . Then the maximum value of f x is equal to
Correct answer
0
Step-by-step solution
Given, f x = sin 2 x - 2 + cos 2 x cos 2 x 2 + sin 2 x cos 2 x cos 2 x sin 2 x cos 2 x 1 + cos 2 x = - 2 - 2 0 2 0 - 1 sin 2 x cos 2 x 1 + cos 2 x R 1 → R 1 - R 2 &   R 2 → R 2 - R 3 = - 2 cos 2 x + 2 2 + 2 cos 2 x + sin 2 x = 4 + 4 cos 2 x - 2 cos 2 x - sin 2 x ⇒ f x = 4 + 2 cos 2 x We know, - 1 ≤ cos 2 x ≤ 1 So, f x max = 4 + 2 = 6