JEE Advanced Mathematics Question (2020) — Solution

Step-by-step solution

lim x → π 2 8 2   sin 2 x · cos x cos x 2 - cos 7 x 2 + cos 5 x 2 - 2 · 2 cos 2 x + cos 3 x 2 = lim x → π 2 8 2   sin 2 x · cos x cos x 2 - cos 3 x 2 + cos 5 x 2 - cos 7 x 2 - 2 2 cos 2 x = lim x → π 2 16 2 sin x cos x · cos x 2 sin x sin x 2 + 2 sin 3 x sin x 2 - 2 2 cos 2 x = lim x → π 2 16 2 sin x cos x · cos x 2 sin x 2 sin x + sin 3 x - 2 2 cos 2 x = lim x → π 2 16 2 sin x   cos 2 x 2 sin x 2 2 sin 2 x · cos x - 2 2 cos 2 x = lim x → π 2 16 2 sin x 2 sin x 2 . 4 sin x - 2 2 = 16 2 4 2 - 2 2 = 16 2 2 2 = 8 .