Let [·] denote the greatest integer function. If the domain of the...

For the function $f(x) = \sin^{-1}$\left(\dfrac{x+[x]}{3}\right)$to be defined, the argument of the inverse sine function must lie in the interval$[-1, 1]$.<br><br>$-1 \le \dfrac{x+[x]}{3} \le 1$<br><br>$-3 \le x + [x] \le 3$<br><br>Using the fractional part function$\{x\}$, we can write$x = [x] + \{x\}$, where$0 \le \{x\} < 1$. Substituting this into the inequality gives:<br><br>$-3 \le 2[x] + \{x\} \le 3$<br><br>We analyze the possible integer values for$[x]$:<br><br>If$[x] \le -2$, then$2[x] + \{x\} \le -4 + \{x\} < -3$, which does not satisfy the inequality.<br><br>If$[x] = -1$, then$2[x] + \{x\} = -2 + \{x\}$. Since$0 \le \{x\} < 1$, we have$-2 \le -2 + \{x\} < -1$, which satisfies the inequality. This gives$x \in [-1, 0)$.<br><br>If$[x] = 0$, then$2[x] + \{x\} = \{x\}$. Since$0 \le \{x\} < 1$, this satisfies the inequality. This gives$x \in [0, 1)$.<br><br>If$[x] = 1$, then$2[x] + \{x\} = 2 + \{x\}$. Since$0 \le \{x\} < 1$, we have$2 \le 2 + \{x\} < 3$, which satisfies the inequality. This gives$x \in [1, 2)$.<br><br>If$[x] \ge 2$, then$2[x] + \{x\} \ge 4 + \{x\} \ge 4 > 3$, which does not satisfy the inequality.<br><br>Taking the union of the valid intervals, the domain of$f(x)$is$[-1, 0) \cup [0, 1) \cup [1, 2) = [-1, 2)$.<br><br>Comparing this with the given domain$[\alpha, \beta)$, we get$\alpha = -1$and$\beta = 2$.<br><br>Therefore,$\alpha^2 + \beta^2 = (-1)^2 + (2)^2 = 1 + 4 = 5$.<br><br>Answer:$5