The domain of the function `f(x) = sqrt((4-x^(2))/([x]+2))` where [x] denotes the greatest integer less than or equal to x,is

Options

A. `[-1, 2]`
B. `(-oo, -2 )`
C. `(-oo, -2) uu [-1,2]`
D. none of these

Explanation

To find the domain of the function \( f(x) = \sqrt{\frac{4 - x^2}{[x] + 2}} \), where \([x]\) denotes the greatest integer less than or equal to \(x\), we need to ensure that the expression inside the square root is non-negative and the denominator is not zero.

### Step 1: Ensure the expression inside the square root is non-negative
We require:
\[ \frac{4 - x^2}{[x] + 2} \geq 0 \]

This inequality holds when both the numerator and denominator are either both positive or both negative.

### Step 2: Analyze the numerator
The numerator \(4 - x^2\) is non-negative when:
\[ 4 - x^2 \geq 0 \implies x^2 \leq 4 \implies -2 \leq x \leq 2 \]

### Step 3: Analyze the denominator
The denominator \([x] + 2\) must be positive:
\[ [x] + 2 > 0 \implies [x] > -2 \]
Since \([x]\) is the greatest integer less than or equal to \(x\), this means:
- If \(x < -2\), then \([x] \leq -3\) which gives \([x] + 2 \leq -1\) (not valid).
- If \(-2 \leq x < -1\), then \([x] = -2\) which gives \([x] + 2 = 0\) (not valid).
- If \(-1 \leq x < 0\), then \([x] = -1\) which gives \([x] + 2 = 1\) (valid).
- If \(0 \leq x < 1\), then \([x] = 0\) which gives \([x] + 2 = 2\) (valid).
- If \(1 \leq x < 2\), then \([x] = 1\) which gives \([x] + 2 = 3\) (valid).
- If \(x = 2\), then \([x] = 2\) which gives \([x] + 2 = 4\) (valid).

### Step 4: Combine the results
From the analysis:
- The numerator \(4 - x^2 \geq 0\) gives us the interval \([-2, 2]\).
- The denominator \([x] + 2 > 0\) is valid for intervals \([-1, 2]\).

### Step 5: Final domain
The intersection of the intervals gives us the domain:
\[ [-1, 2] \]

Thus, the domain of the function \( f(x) \) is:
\[ [-1, 2] \]