Question

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:

• A. 2
• B. 5
• C. 10
• D. 13

Hint:

Identify the values of $x$ that satisfy $-3 \le x + [x] \le 3$. Recall that $[x]$ is the greatest integer less than or equal to $x$, and use cases for integer values of $[x]$.

Answer 1

The Correct Option is B

The domain of the function $f(x) = \sin^{-1} \left( \frac{x + [x]}{3} \right)$ is determined by the constraint that the input to the inverse sine function must be within the range of the sine function itself, which is $[-1, 1]$.

Thus, we solve for $x$ in the following inequality:
$$ -3 \le x + [x] \le 3 $$
We can use the relation $x = [x] + \{x\}$, where $\{x\}$ is the fractional part of $x$ and $0 \le \{x\}<1$. Substituting this into the inequality, we get:
$$ -3 \le [x] + \{x\} + [x] \le 3 $$
$$ -3 \le 2[x] + \{x\} \le 3 $$

Now, let's examine possible integer values for $[x]$ to find the range of valid $x$ values:
1. If $[x] = -2$: The inequality becomes $-3 \le 2(-2) + \{x\} \Rightarrow -3 \le -4 + \{x\} \Rightarrow \{x\} \ge 1$. This is impossible since the fractional part must be less than 1.
2. If $[x] = -1$: The inequality becomes $-3 \le 2(-1) + \{x\} \le 3 \Rightarrow -3 \le -2 + \{x\} \le 3 \Rightarrow -1 \le \{x\} \le 5$. Since $0 \le \{x\}<1$, this is satisfied for all $x \in [-1, 0)$.
3. If $[x] = 0$: The inequality becomes $-3 \le 0 + \{x\} \le 3 \Rightarrow -3 \le \{x\} \le 3$. Since $0 \le \{x\}<1$, this is satisfied for all $x \in [0, 1)$.
4. If $[x] = 1$: The inequality becomes $-3 \le 2(1) + \{x\} \le 3 \Rightarrow -5 \le \{x\} \le 1$. Since $0 \le \{x\}<1$, this is satisfied for all $x \in [1, 2)$.
5. If $[x] = 2$: The inequality becomes $-3 \le 2(2) + \{x\} \le 3 \Rightarrow -7 \le \{x\} \le -1$. This is impossible since the fractional part cannot be negative.

Combining the valid intervals from $[x] = -1, 0, 1$, we get the domain $x \in [-1, 0) \cup [0, 1) \cup [1, 2)$, which is $x \in [-1, 2)$.
Given the domain is $[\alpha, \beta)$, we find $\alpha = -1$ and $\beta = 2$.
Calculating the required sum of squares:
$$ \alpha^2 + \beta^2 = (-1)^2 + 2^2 = 1 + 4 = 5 $$
The result is 5.