JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The given region is defined by the inequalities: x 0 0 y 6 - x y 0 and x 6 - y y^2 4x - 3 x y^2 + 3 4 From these inequalities, for a given y 0, the value of x ranges from 0 to (6 - y, y^2 + 3 4 ). To find the point where the two bounding curves intersect, we equate them: 6 - y = y^2 + 3 4 24 - 4y = y^2 + 3 y^2 + 4y - 21 = 0 (y + 7)(y - 3) = 0 Since y 0, the intersection occurs at y = 3. For 0 y 3, the right boundary is the parabola x = y^2 + 3 4 . For 3 y 6, the right boundary is the line x = 6 - y. The total area A can be calculated by integrating with respect to y: A = _ 0 ^ 3 y^2 + 3 4 dy + _ 3 ^ 6 (6 - y) dy Evaluating the first integral: _ 0 ^ 3 y^2 + 3 4 dy = 1 4 [ y^3 3 + 3y ]_ 0 ^ 3 = 1 4 (9 + 9) = 18 4 = 9 2 Evaluating the second integral: _ 3 ^ 6 (6 - y) dy = [ 6y - y^2 2 ]_ 3 ^ 6 = (36 - 18) - (18 - 9 2 ) = 18 - 13.5 = 9 2 Total Area = 9 2 + 9 2 = 9 Answer: 9