See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Given g(f(x)) = x  g(x) is inverse of f(x) g(f(x)) = x      g f x f x 1         1 g f x f x   ... (i)     1 1 g f 0 f 0 3       1 g 1 3   Now h(g(g(x)) = x  h(g(g(f(x)))) = f(x)  h(g(x)) = f(x) ... (ii)  h(g(1)) = f(1) = 5 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 16 Now from equation (ii) h(g(x)) = f(x)  h(g(f(x))) = f(f(x))  h(x) = f(f(x)) ... (iii)  h(x) = f (f(x)) . f(x)  h(0) = f(f(0)) f(0) = f(1) . 3 = 18 and g(h(g(x))) = g(f(x)) = x  g(h(g(7)) = 7.