See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

  x 0 d d 1 g x f t dt dx dx x   TG ~ @bohring_bot TG ~ @bohring_bot | @HeyitsyashXD AITS-PT-I (Paper-1)-PCM(Sol.)-JEE(Advanced)/2025 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 13    x 2 0 d 1 1 g x f x f t dt dx x x           ….. (1) and as g(2) = 0, we must have   2 0 1 f 2 f t dt 0 4 2        2 0 f t dt 2f 2 10    Now,    x 2 2 2 0 d d 1 1 g x f t dt f x dx x dx x             =     x 3 2 2 0 2 1 1 1 f t dt f x f x f x x x x x          2 2 2 0 d 1 f 2 f 2 3 g 2 f t dt 0 4 2 2 2 dx        Hence, g(x) has a local maximum at x = 2