Determine Domain of Inverse Sine Function

The function given is $f(x) = \sin^{-1}\left(\frac{x + [x]}{3}\right)$. The domain of $\sin^{-1}(y)$ is $[-1, 1]$.

Therefore, the argument must satisfy:

$ -1 \le \frac{x + [x]}{3} \le 1 $

Solve the Inequality

Multiply by 3:

$ -3 \le x + [x] \le 3 $

We analyze this in two parts:

1. Inequality 1: $x + [x] \le 3$
   • For $0 \le x < 1$, $[x]=0$. So, $x \le 3$. This holds true for $x \in [0, 1)$.
   • For $1 \le x < 2$, $[x]=1$. So, $x + 1 \le 3 \implies x \le 2$. This holds true for $x \in [1, 2)$.
   • For $x \ge 2$, $x+[x] > 3$. No solutions here.
   • Combining these, the solution for $x \ge 0$ is $[0, 2)$.
2. Inequality 2: $x + [x] \ge -3$
   • For $-1 \le x < 0$, $[x]=-1$. So, $x - 1 \ge -3 \implies x \ge -2$. This holds true for $x \in [-1, 0)$.
   • For $-2 \le x < -1$, $[x]=-2$. So, $x - 2 \ge -3 \implies x \ge -1$. This contradicts $x < -1$, so no solutions here.
   • For $x < -2$, $x+[x] < -3$. No solutions here.
   • Combining these, the solution for $x < 0$ is $[-1, 0)$.

The overall domain is the union of solutions from both cases: $[-1, 0) \cup [0, 2) = [-1, 2)$.

Find Alpha and Beta

The domain is given as $[\alpha, \beta)$. Comparing with $[-1, 2)$, we have:

• $\alpha = -1$
• $\beta = 2$

Calculate Alpha Squared Plus Beta Squared

Calculate $\alpha^2 + \beta^2$:

$ \alpha^2 + \beta^2 = (-1)^2 + (2)^2 $

$ \alpha^2 + \beta^2 = 1 + 4 $

$ \alpha^2 + \beta^2 = 5 $