JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The given region is defined by the inequalities y - |x|, y |x x|, and y 0. Since replacing x with -x leaves the inequalities unchanged, the region is symmetric with respect to the y-axis. We can find the area of the region in the first quadrant (x 0) and multiply it by 2. For x 0, the inequalities become: y - x y x x y 0 The condition y 0 and y - x restricts x to the interval [0, ]. In this interval, x 0, so |x x| = x x. To find the intersection of the curves y = - x and y = x x, we equate them: x x = - x x(1 + x) = By inspection, x = 2 is a solution since 2 (1 + 1) = . For x [0, 2 ], x x - x. For x [ 2 , ], x x - x. The area in the first quadrant, A_1, is given by: A_1 = _ 0 ^ /2 x x \, dx + _ /2 ^ ( - x) \, dx Evaluating the first integral using integration by parts: _ 0 ^ /2 x x \, dx = [-x x]_ 0 ^ /2 - _ 0 ^ /2 (- x) \, dx = 0 + [ x]_ 0 ^ /2 = 1 Evaluating the second integral: _ /2 ^ ( - x) \, dx = [ - ( - x)^2 2 ]_ /2 ^ = 0 - ( - ^2 8 ) = ^2 8 Thus, the area in the first quadrant is: A_1 = 1 + ^2 8 The total area of the region is twice the area in the first quadrant: Total Area = 2 A_1 = 2 ( 1 + ^2 8 ) = 2 + ^2 4 Answer: 2 + ^2 4