Let `f(x) = (1)/(sqrt(|x-1|-[x]))` where[.] denotes the greatest integer funciton them the domain of `f(x)` is

Options

A. ` (-1, 1 )`
B. `(-oo, 1)`
C. `(-oo, - 1 )`
D. none of these

Explanation

To find the domain of the function \( f(x) = \frac{1}{\sqrt{|x-1| - [x]}} \), where \([x]\) denotes the greatest integer function, we need to ensure that the expression inside the square root is positive. The following steps outline the solution:

### Step 1: Identify the condition for the square root
The expression inside the square root must be greater than zero:
\[ |x - 1| - [x] > 0 \]
This implies:
\[ |x - 1| > [x] \]

### Step 2: Analyze the absolute value
The absolute value \( |x - 1| \) can be expressed in two cases:
1. When \( x \geq 1 \): \( |x - 1| = x - 1 \)
2. When \( x < 1 \): \( |x - 1| = 1 - x \)

### Step 3: Case 1: \( x \geq 1 \)
In this case, we have:
\[ x - 1 > [x] \]
Since \([x] = n\) where \(n\) is the greatest integer less than or equal to \(x\), we can write:
\[ x - 1 > n \]
This can be rearranged to:
\[ x > n + 1 \]
Since \(n \leq x < n + 1\), we can deduce:
\[ n \leq x < n + 1 \implies n + 1 > n \implies x > n + 1 \]
This condition is satisfied for any \(x \geq 1\).

### Step 4: Case 2: \( x < 1 \)
In this case, we have:
\[ 1 - x > [x] \]
Here, \([x] = n\) where \(n\) is the greatest integer less than \(x\). Since \(x < 1\), \(n\) can only be \(0\) or negative integers. Thus:
\[ 1 - x > n \]
For \(n = 0\):
\[ 1 - x > 0 \implies x < 1 \]
This condition is always satisfied for \(x < 1\).

For \(n < 0\):
\[ 1 - x > n \implies x < 1 - n \]
However, since \(n < 0\), this condition will not restrict \(x\) further.

### Step 5: Combine the results
From both cases, we see that:
- For \(x \geq 1\), the condition is satisfied.
- For \(x < 1\), the condition is also satisfied.

However, at \(x = 1\):
\[ |1 - 1| - [1] = 0 - 1 = -1 \quad (\text{not valid}) \]
Thus, \(x = 1\) is not included in the domain.

### Conclusion
The domain of \(f(x)\) is:
\[ (-\infty, 1) \]