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(B) The domain of $\sin^{-1}(u)$ is $u \in [-1, 1]$.Thus, $-1 \le \frac{x+[x]}{3} \le 1$, which implies $-3 \le x+[x] \le 3$.Case $1$: If $x \in [-2,

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$5$

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(B) The domain of $\sin^{-1}(u)$ is $u \in [-1, 1]$.Thus, $-1 \le \frac{x+[x]}{3} \le 1$, which implies $-3 \le x+[x] \le 3$.Case $1$: If $x \in [-2, -1)$, then $[x] = -2$. So, $-3 \le x - 2 \le 3 \implies -1 \le x \le 5$. Since $x \in [-2, -1)$, the intersection is $[-1, -1)$ (empty).Case $2$: If $x \in [-1, 0)$, then $[x] = -1$. So, $-3 \le x - 1 \le 3 \implies -2 \le x \le 4$. The intersection with $[-1, 0)$ is $[-1, 0)$.Case $3$: If $x \in [0, 1)$, then $[x] = 0$. So, $-3 \le x \le 3$. The intersection with $[0, 1)$ is $[0, 1)$.Case $4$: If $x \in [1, 2)$, then $[x] = 1$. So, $-3 \le x + 1 \le 3 \implies -4 \le x \le 2$. The intersection with $[1, 2)$ is $[1, 2)$.Case $5$: If $x \in [2, 3)$, then $[x] = 2$. So, $-3 \le x + 2 \le 3 \implies -5 \le x \le 1$. The intersection with $[2, 3)$ is empty.Combining these, the domain is $[-1, 2)$.Thus, $\alpha = -1$ and $\beta = 2$.Therefore, $\alpha^2 + \beta^2 = (-1)^2 + (2)^2 = 1 + 4 = 5$.

Let $[\cdot]$ denote the greatest integer function. If the domain of the function $f(x) = \sin^{-1} \left( \frac{x+[x]}{3} \right)$ is $[\alpha, \beta)$, then $\alpha^2 + \beta^2$ is equal to:

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