See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Sol.       2 2 sgn x k 1 x k 1 1        g g(x) 1   now let g(x) t  For More Material Join: @JEEAdvanced_2024 AITS-CRT-III (Paper-1)-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 13 g(x) 1 O 1 2 3 4 5 1 2 3 4 5 2 x 1 2 3 4 5 6 g(t) 1 t 0, , , , , ,      where 1 2 3 1 2; 2 3;3 4    and 4 5 6 , , 0  g(x) = 0 has 8 solutions g(x) = 1 has 8 solutions Hence total number of solutions is 16.