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

To find the domain of the function $f(x) = \sin^{-1} \left( \frac{x + [x]}{3} \right)$, we must first consider the fundamental properties of the inverse sine function. For any expression $\sin^{-1}(u)$ to be defined, the argument $u$ must satisfy the condition $-1 \le u \le 1$.

Substituting the given expression into this condition, we obtain the inequality:
$$ -1 \le \frac{x + [x]}{3} \le 1 $$
Multiplying the entire inequality by 3, we get:
$$ -3 \le x + [x] \le 3 $$
Let's define $g(x) = x + [x]$. The greatest integer function $[x]$ is a step function that stays constant between integers and jumps at every integer. Therefore, $g(x)$ is a monotonically non-decreasing function.

Step 1: Determining the lower bound $\alpha$.
We need to find the smallest $x$ such that $x + [x] \ge -3$.
Let's test the interval $[-2, -1)$. For any $x$ in this range, $[x] = -2$.
The inequality becomes $x - 2 \ge -3 \Rightarrow x \ge -1$.
Since we were looking in the interval $x<-1$, only the boundary point $x = -1$ satisfies this. For any $x<-1$, the sum $x + [x]$ will be less than $-3$. For example, at $x = -1.1$, $[x] = -2$ and $x + [x] = -3.1<-3$. Thus, the lower bound is $\alpha = -1$.

Step 2: Determining the upper bound $\beta$.
We need to find the range of $x$ such that $x + [x] \le 3$.
Let's test the interval $[1, 2)$. For any $x$ in this range, $[x] = 1$.
The inequality becomes $x + 1 \le 3 \Rightarrow x \le 2$.
Since the entire interval $[1, 2)$ is less than or equal to 2, every $x$ in $[1, 2)$ satisfies the condition. However, let's check $x = 2$. At $x = 2$, $[x] = 2$, so $x + [x] = 2 + 2 = 4$. Since $4>3$, $x = 2$ is excluded from the domain. Therefore, the domain ends just before $x = 2$. Thus, $\beta = 2$.

Step 3: Final Calculation.
The domain is the interval $[ -1, 2 )$. Comparing this with $[\alpha, \beta)$, we have $\alpha = -1$ and $\beta = 2$.
The required value is $\alpha^2 + \beta^2$:
$$ \alpha^2 + \beta^2 = (-1)^2 + 2^2 = 1 + 4 = 5 $$
Hence, the correct answer is 5, which corresponds to option 2.