JEE Main202430 Jan 2024Morning ShiftMathematicsDeterminantsActual
If f x = 2 cos 4 x 2 sin 4 x 3 + sin 2 2 x 3 + 2 cos 4 x 2 sin 4 x sin 2 2 x 2 cos 4 x 3 + 2 sin 4 x sin 2 2 x then 1 5 f ' ( 0 ) is equal to ________.
Options
- A0
- B1
- C2
- D6
Correct answer
A. 0
Step-by-step solution
Given: f x = 2 cos 4 x 2 sin 4 x 3 + sin 2 2 x 3 + 2 cos 4 x 2 sin 4 x sin 2 2 x 2 cos 4 x 3 + 2 sin 4 x sin 2 2 x Applying R 2 → R 2 - R 1 , R 3 → R 3 - R 1 ⇒ f x = 2 cos 4 x 2 sin 4 x 3 + sin 2 2 x 3 0 - 3 0 3 - 3 ⇒ f x = 9 2 cos 4 x 2 sin 4 x 3 + sin 2 2 x 1 0 - 1 0 1 - 1 ⇒ f x = 9 2 cos 4 x + 2 sin 4 x + 3 + sin 2 2 x Now, differentiating above function we get, ⇒ f ' x = 9 8 cos 3 x - sin x + 8 sin 3 x cos x + 2 sin 2 x cos 2 x 2 ⇒ f ' 0 = 9 0 + 0 + 0 ⇒ f ' 0 = 0