JEE Main202116 Mar 2021Evening ShiftMathematicsDeterminantsActual
The maximum value of f x = sin 2 x 1 + cos 2 x cos 2 x 1 + sin 2 x cos 2 x cos 2 x sin 2 x cos 2 x sin 2 x , x ∈ R is
Options
- A7
- B3 4
- C5
- D5
Correct answer
C. 5
Step-by-step solution
f x = sin 2 x 1 + cos 2 x cos 2 x 1 + sin 2 x cos 2 x cos 2 x sin 2 x cos 2 x sin 2 x ,   x ∈ R C 1 → C 1 + C 2 = 2 1 + cos 2 x cos 2 x 2 cos 2 x cos 2 x 1 cos 2 x sin 2 x R 1 → R 1 - R 2 = 0 1 0 2 cos 2 x cos 2 x 1 cos 2 x sin 2 x Expanding w.r.t. R 1 = - 2 sin 2 x - cos 2 x So, f x = cos 2 x - 2 sin 2 x f x max = 1 + 4 = 5