Domain of Inverse Sine Function with Greatest Integer Function

To find the domain of the function $f(x) = \sin^{-1}\left(\frac{x + [x]}{3}\right)$ we use the property that the domain of $\sin^{-1}(y)$ is $-1 \le y \le 1$.

Therefore, the function $f(x)$ is defined if and only if: $-1 \le \frac{x + [x]}{3} \le 1$

Multiplying the entire inequality by $3$: $-3 \le x + [x] \le 3$

Let $g(x) = x + [x]$. Since $g(x)$ is a strictly increasing function, we analyze its behavior across different intervals of $x$:

1. For $x < -1$: If $x \in [-2, -1)$, then $[x] = -2$. $g(x) = x + (-2) = x - 2$ Since $-2 \le x < -1$, we have $-4 \le g(x) < -3$. This violates the condition $g(x) \ge -3$. Thus, no values of $x < -1$ belong to the domain.

2. For $x \in [-1, 0)$: Here, $[x] = -1$, so $g(x) = x - 1$. Since $-1 \le x < 0$, we have: $-2 \le g(x) < -1$ This satisfies $-3 \le g(x) \le 3$. Hence, $x \in [-1, 0)$ is part of the domain.

3. For $x \in [0, 1)$: Here, $[x] = 0$, so $g(x) = x$. Since $0 \le x < 1$, we have: $0 \le g(x) < 1$ This satisfies $-3 \le g(x) \le 3$. Hence, $x \in [0, 1)$ is part of the domain.

4. For $x \in [1, 2)$: Here, $[x] = 1$, so $g(x) = x + 1$. Since $1 \le x < 2$, we have: $2 \le g(x) < 3$ This satisfies $-3 \le g(x) \le 3$. Hence, $x \in [1, 2)$ is part of the domain.

5. For $x \ge 2$: At $x = 2$, $[x] = 2$, so $g(2) = 2 + 2 = 4 > 3$. For all $x \ge 2$, $g(x) \ge 4 > 3$, which violates $g(x) \le 3$.

Combining the valid intervals, the domain of $f(x)$ is $x \in [-1, 2)$.

Comparing this interval with the endpoints of the domain $[\alpha, \beta]$, we get: $\alpha = -1 \quad \text{and} \quad \beta = 2$

Now, calculating $\alpha^2 + \beta^2$: $\alpha^2 + \beta^2 = (-1)^2 + (2)^2 = 1 + 4 = 5$

Thus, the correct option is B.