See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Let log10 x = t  1 – 8t2 = t – 2t2  6t2 + t – 1 = 0  1 1 t or t 2 3   log10 x = 1/2 1/2 1 1 x 10 2 10       and 1/3 10 1 log x x 10 3     Then 3 2 3 4 1 110 and 1    . 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 14 I =    2 2x 1 sin x dx 1 cos x      Using king property and add      2 2x 1 sinx 2 x 1 sin x 2I dx 1 cos x          2 x sinx 2I 4 dx 1 cos x       2 x sin x I 2 1 cos x      dx 2 0 x sinx 4 dx 1 cos x     Using king property then add. 2 0 sinx 2I 4 dx 1 cos x      Put cosx = t  sinx dx = – dt 1 2 1 dt I 2 1 t     = 1 1 1 2 tan t |    2 2 4 4            .