JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

For the domain of f(x) = ^ -1 ( 4x+2[x] 3 ), the argument must satisfy: -1 4x+2[x] 3 1 -3 4x + 2[x] 3 Let x = I + f, where I = [x] Z and f = \ x\ [0, 1). Substituting x = I + f into the inequality: -3 4(I + f) + 2I 3 -3 6I + 4f 3 Since 0 f If I 1, 6I + 4f 6(1) + 0 = 6 > 3, which gives no solution. If I -2, 6I + 4f Thus, the only possible values for I are 0 and -1. Case 1: I = 0 -3 6(0) + 4f 3 -3 4f 3 - 3 4 f 3 4 Since f [0, 1), we get 0 f 3 4 . Therefore, x = I + f [0, 3 4 ]. Case 2: I = -1 -3 6(-1) + 4f 3 3 4f 9 3 4 f 9 4 Since f [0, 1), we get 3 4 f Therefore, x = I + f [-1 + 3 4 , -1 + 1 ) = [- 1 4 , 0 ). Taking the union of the intervals from both cases, the domain is x [- 1 4 , 3 4 ]. Comparing this with [ , ], we get = - 1 4 and = 3 4 . Finally, evaluating the required expression: 12( + ) = 12 (- 1 4 + 3 4 ) = 12 ( 2 4 ) = 12 ( 1 2 ) = 6 Answer: 6