JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

For the function f(x) = ^ -1 ( x+[x] 3 ) to be defined, the argument of the inverse sine function must lie in the interval [-1, 1]. -1 x+[x] 3 1 -3 x + [x] 3 Using the fractional part function \ x\ , we can write x = [x] + \ x\ , where 0 \ x\ -3 2[x] + \ x\ 3 We analyze the possible integer values for [x]: If [x] -2, then 2[x] + \ x\ -4 + \ x\ If [x] = -1, then 2[x] + \ x\ = -2 + \ x\ . Since 0 \ x\ If [x] = 0, then 2[x] + \ x\ = \ x\ . Since 0 \ x\ If [x] = 1, then 2[x] + \ x\ = 2 + \ x\ . Since 0 \ x\ If [x] 2, then 2[x] + \ x\ 4 + \ x\ 4 > 3, which does not satisfy the inequality. Taking the union of the valid intervals, the domain of f(x) is [-1, 0) [0, 1) [1, 2) = [-1, 2). Comparing this with the given domain [ , ), we get = -1 and = 2. Therefore, ^2 + ^2 = (-1)^2 + (2)^2 = 1 + 4 = 5. Answer: 5