JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let I = _ -2 ^ 2 (| x| + [x x] ) dx = I_1 + I_2, where I_1 = _ -2 ^ 2 | x|\, dx and I_2 = _ -2 ^ 2 [x x]\, dx. Evaluating I_1: Since | x| is even, I_1 = 2 _ 0 ^ 2 | x|\, dx. For x [0, 2], 2 I_1 = 2 _ 0 ^ 2 x\, dx = 2 [- x ]_0^2 = 2(1 - 2) = 2 - 2 2 Evaluating I_2: Let g(x) = x x. Then g(-x) = (-x) (-x) = x x = g(x), so g(x) is even, and hence [x x] is also even. I_2 = 2 _ 0 ^ 2 [x x]\, dx Behaviour of y = x x on [0, 2]: At x = 0, y = 0. At x = 2, y = 2 2 1.818. y' = x + x x > 0 on [0, 2], so y is strictly increasing on this interval. Therefore, x x takes each value in [0, 1.818] exactly once, and crosses 1 at some unique (0, 2) with = 1. For x [0, ): 0 x x For x [ , 2]: 1 x x I_2 = 2 ( _ 0 ^ 0\, dx + _ ^ 2 1\, dx ) = 2(2 - ) = 4 - 2 Combining: I = I_1 + I_2 = (2 - 2 2) + (4 - 2 ) = 6 - 2 2 - 2 Given I = 2(3 - 2) + = 6 - 2 2 + , so: 6 - 2 2 - 2 = 6 - 2 2 + = -2 Now: ( 2 ) = -2 ( -2 2 ) = -2 (- ) = 2 Since = 1: ( 2 ) = 2 1 = 2 Hence, the correct option is (2)\ 2.