JEE Main 2026 — 6 April, Morning Shift — Question 21

Given $f(x) = \sin^{-1}\left( \frac{x+[x]}{3} \right)$. Domain requires $-1 \leq \frac{x+[x]}{3} \leq 1$. Multiply by 3: $-3 \leq x+[x] \leq 3$. Let $n = [x]$. Then $n \leq x < n+1$. Inequality: $-3 \leq x+n \leq 3$. For $n = -1$: $-3 \leq x-1 \leq 3 \Rightarrow -2 \leq x \leq 4$.

Intersection with $[-1,0)$ gives $[-1,0)$. For $n = 0$: $-3 \leq x \leq 3$. Intersection with $[0,1)$ gives $[0,1)$. For $n = 1$: $-3 \leq x+1 \leq 3 \Rightarrow -4 \leq x \leq 2$.

Intersection with $[1,2)$ gives $[1,2)$. For $n = 2$: $-3 \leq x+2 \leq 3 \Rightarrow -5 \leq x \leq 1$.

Intersection with $[2,3)$ is empty. Thus domain $= [-1,2)$, so $\alpha = -1, \beta = 2$. Hence $\alpha^2 + \beta^2 = 1 + 4 = 5$.