If [x] denotes the greatest integer `le x`, then domain of
`f(x) = 1/sqrt([x]^(2) -7[x] + 12)` is

Options

A. `(-infty, 3) cup [5, infty)`
B. `(-infty,3] cup (5, infty)`
C. `R - {3,4}`
D. `(-infty, 3) cup [4, infty)`

Answer

To find the domain of the function \( f(x) = \frac{1}{\sqrt{[\![x]\!]^2 - 7[\![x]\!] + 12}} \), where \([\![x]\!]\) denotes the greatest integer less than or equal to \(x\), we need to ensure that the expression inside the square root is positive, and the denominator is not equal to zero.

### Step 1: Set up the inequality
We need to solve the inequality:
\[ [\![x]\!]^2 - 7[\![x]\!] + 12 > 0 \]

### Step 2: Factor the quadratic
To factor the quadratic, we look for two numbers that multiply to \(12\) and add to \(-7\). The numbers \(-3\) and \(-4\) work, so we can factor it as:
\[ ([\![x]\!] - 3)([\![x]\!] - 4) > 0 \]

### Step 3: Analyze the intervals
Next, we find the critical points from the factors:
- The critical points are \(3\) and \(4\).
- We will test the intervals determined by these points: \( (-\infty, 3) \), \( (3, 4) \), and \( (4, \infty) \).

### Step 4: Test the intervals
1. **For \(x < 3\)** (e.g., \(x = 2\)):
\[ [\![2]\!] = 2 \implies (2 - 3)(2 - 4) = (-1)(-2) = 2 > 0 \]
This interval is valid.

2. **For \(3 < x < 4\)** (e.g., \(x = 3.5\)):
\[ [\![3.5]\!] = 3 \implies (3 - 3)(3 - 4) = (0)(-1) = 0 \]
This interval is not valid since it equals zero.

3. **For \(x > 4\)** (e.g., \(x = 5\)):
\[ [\![5]\!] = 5 \implies (5 - 3)(5 - 4) = (2)(1) = 2 > 0 \]
This interval is valid.

### Step 5: Combine the intervals
From our analysis, the valid intervals for \([\![x]\!]\) are:
- \( (-\infty, 3) \)
- \( (4, \infty) \)

### Step 6: Write the domain in terms of \(x\)
Since \([\![x]\!]\) can take integer values, we need to express the domain of \(f(x)\) in terms of \(x\):
- For \(x < 3\), \([\![x]\!]\) can be \(2, 1, 0, -1, \ldots\) (all integers less than 3).
- For \(x > 4\), \([\![x]\!]\) can be \(5, 6, 7, \ldots\) (all integers greater than 4).

Thus, the domain of \(f(x)\) is:
\[ (-\infty, 3) \cup (4, \infty) \]

### Final Answer
The domain of \(f(x)\) is \( (-\infty, 3) \cup (4, \infty) \).