See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

If we could show that  f "' x 0  then all choices follow since    f 0 f ' 0 f " 0 0    Indeed   2 2 8 f ' x 1 x sec x, f ' 0 0 3     2 16 f " x x 2sec x tan x 3    f " 0 0   4 2 2 16 f "' x 2sec x 4sec x tan x 3      2 2 2 16 2sec x sec x 2tan x 3    0 x 6     For More Material Join: @JEEAdvanced_2024 AITS-CRT-II (Paper-2)-PCM(Sol.)-JEE(Advanced)/2023 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 11   2 2 2 max 2 sec x sec x 2tan x      2 2 4 2 16 2 3 3 3 3                f "' x 0  Whence all choices follow.