The domain of the function `f(x)=1/(sqrt([x]^2-2[x]-8))` is, where [*] denotes greatest integer function

Options

A. `(-2, 5]`
B. `(-2, 5)`
C. `(-oo, -2), uu(5, oo)`
D. none of these

Answer

To find the domain of the function \( f(x) = \frac{1}{\sqrt{[x]^2 - 2[x] - 8}} \), where \([x]\) denotes the greatest integer function, we need to ensure that the expression under the square root is positive. This is because the square root must be defined and cannot be zero since it is in the denominator.

### Step-by-Step Solution:

1. **Identify the Condition for the Square Root**:
We need to solve the inequality:
\[ [x]^2 - 2[x] - 8 > 0 \]

2. **Factor the Quadratic Expression**:
We can factor the quadratic:
\[ [x]^2 - 2[x] - 8 = ([x] - 4)([x] + 2) \]
Therefore, we need:
\[ ([x] - 4)([x] + 2) > 0 \]

3. **Find the Critical Points**:
The critical points occur when the expression equals zero:
\[ [x] - 4 = 0 \quad \Rightarrow \quad [x] = 4 \]
\[ [x] + 2 = 0 \quad \Rightarrow \quad [x] = -2 \]

4. **Test Intervals**:
We will test the intervals determined by the critical points \(-2\) and \(4\):
- **Interval 1**: \( (-\infty, -2) \)
- **Interval 2**: \( (-2, 4) \)
- **Interval 3**: \( (4, \infty) \)

5. **Evaluate Each Interval**:
- For \( [x] < -2 \) (e.g., \( [x] = -3 \)):
\[ (-3 - 4)(-3 + 2) = (-7)(-1) = 7 > 0 \quad \text{(Valid)} \]
- For \( -2 < [x] < 4 \) (e.g., \( [x] = 0 \)):
\[ (0 - 4)(0 + 2) = (-4)(2) = -8 < 0 \quad \text{(Not Valid)} \]
- For \( [x] > 4 \) (e.g., \( [x] = 5 \)):
\[ (5 - 4)(5 + 2) = (1)(7) = 7 > 0 \quad \text{(Valid)} \]

6. **Combine Valid Intervals**:
The valid intervals for \([x]\) are:
\[ [x] < -2 \quad \text{or} \quad [x] > 4 \]

7. **Translate Back to \(x\)**:
- If \([x] < -2\), then \(x < -2\).
- If \([x] > 4\), then \(x \geq 5\).

8. **Final Domain**:
Therefore, the domain of the function \( f(x) \) is:
\[ (-\infty, -2) \cup [5, \infty) \]