See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

       x 2 2 2 0 f x f t f t dt 2025          Differentiate w.r.t x 2f(x)f(x) = f(x)2 + f(x)2  (f(x) – f(x)2 = 0  f(x) = cex f(0) = c =  2025  x f x 2025 e  (f(x) is strictly increasing function)   x f x e 2025  x = 0  x = ln y 0 y = ex (0, 1) Area = 1 x 0 e e dx 1    and area = e 1 lny dy  apply Kings property area =   e 1 ln 1 e y dy    and 2 3 2 3 x dx 1 sinx 1 sin x        = 3 3 I  2 2 x I dx 1 sinx 1 sin x        = 2 2 2 2 0 x x dx 1 sinx 1 sin x 1 sinx 1 sin x          3 2 0 x dx 3     hence 3 3 I 1  