See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

f(f(2)) – f(f(1)) = 0  (3a + b)(5a2 + 3ab + 2ac + b) = 0 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 17  3a + b = 0 ... (i) or 5a2 + 3ab + 2ac + b = 0 … (ii) f(f(3)) – f(f(1)) = 0  2(4a + b)(10a2 + 4ab + 2ac + b) = 0  4a + b = 0 ... (iii) or 10a2 + 4ab + 2ac + b = 0 … (iv) Case I : When b = – 3a put it in IV we get c = 3 a 2  (not possible) Case II : When b = – 4a put it in II we get c = 7a 2 2   a = 2, c = 7 + 2 & b = – 8 Case III : When 5a2 + 3ab + 2ac + b = 0 & 10a2 + 4ab + 2ac + b = 0  5a2 + ab = 0  b = – 5a (not possible) So a = 2, c = 7 + 2 & b = – 8 Now 20 8 20   (  0) 