The domain of definition of `f(x)=sqrt((log_(0.3)|x-2|)/(|x|))`, is

Options

A. `[1,2) cup (2,3]`
B. `[1,3]`
C. `R-(1,3]`
D. none of these

Answer

To find the domain of the function \( f(x) = \sqrt{\frac{\log_{0.3} |x-2|}{|x|}} \), we need to ensure that the expression inside the square root is positive. This leads us to the following conditions:

1. The logarithmic expression must be greater than zero:
\[ \log_{0.3} \left( \frac{|x-2|}{|x|} \right) > 0 \]

2. The denominator must not be zero:
\[ |x| \neq 0 \quad \Rightarrow \quad x \neq 0 \]

### Step 1: Solve the logarithmic inequality
The logarithm \( \log_{0.3}(y) > 0 \) implies that \( y < 1 \) because the base \( 0.3 < 1 \). Thus, we have:
\[ \frac{|x-2|}{|x|} < 1 \]
This can be rewritten as:
\[ |x-2| < |x| \]

### Step 2: Analyze the absolute value inequality
We will consider different cases based on the value of \( x \).

#### Case 1: \( x > 2 \)
In this case, \( |x-2| = x - 2 \) and \( |x| = x \):
\[ x - 2 < x \quad \text{(Always true)} \]

#### Case 2: \( 0 < x < 2 \)
Here, \( |x-2| = 2 - x \) and \( |x| = x \):
\[ 2 - x < x \quad \Rightarrow \quad 2 < 2x \quad \Rightarrow \quad x > 1 \]
Thus, for this case, \( 1 < x < 2 \).

#### Case 3: \( x < 0 \)
In this case, \( |x-2| = 2 - x \) and \( |x| = -x \):
\[ 2 - x < -x \quad \Rightarrow \quad 2 < 0 \quad \text{(Not possible)} \]
Thus, there are no valid solutions in this case.

### Step 3: Combine the results
From the analysis, we have:
- For \( x > 2 \): Valid for all \( x > 2 \).
- For \( 0 < x < 2 \): Valid for \( 1 < x < 2 \).

### Step 4: Exclude \( x = 0 \)
Since \( |x| \) cannot be zero, we exclude \( x = 0 \).

### Final Domain
Combining the valid intervals, the domain of \( f(x) \) is:
\[ (1, 2) \cup (2, \infty) \]

### Summary of the Domain
The domain of the function \( f(x) \) is:
\[ \boxed{(1, 2) \cup (2, \infty)} \]