The domain of definition of the real function `f(x)=sqrt(log_(12)x^(2))` of the real variable x, is

Options

A. `x gt 0`
B. `|x| gt 1`
C. `|x| gt 4`
D. `x gt 4`

Answer

To find the domain of the function \( f(x) = \sqrt{\log_{12}(x^2)} \), we need to consider the constraints imposed by both the square root and the logarithm.

### Step 1: Identify the constraints
1. **Square Root Constraint**: The expression inside the square root must be non-negative:
\[ \log_{12}(x^2) \geq 0 \]

2. **Logarithm Constraint**: The argument of the logarithm must be positive:
\[ x^2 > 0 \]

### Step 2: Analyze the logarithm constraint
The condition \( x^2 > 0 \) implies that \( x \) cannot be zero. Since \( x^2 \) is always positive for any \( x \neq 0 \), this constraint is satisfied for all \( x \) except \( x = 0 \).

### Step 3: Analyze the square root constraint
Now, we need to solve the inequality:
\[ \log_{12}(x^2) \geq 0 \]

Using the property of logarithms, we can rewrite this as:
\[ x^2 \geq 12^0 \]
Since \( 12^0 = 1 \), we have:
\[ x^2 \geq 1 \]

### Step 4: Solve the inequality
The inequality \( x^2 \geq 1 \) can be solved as follows:
\[ x \geq 1 \quad \text{or} \quad x \leq -1 \]

### Step 5: Combine the constraints
From the analysis, we have:
- \( x \) must be greater than or equal to 1, or less than or equal to -1.
- \( x \) cannot be zero.

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

### Final Answer
The domain of the function \( f(x) = \sqrt{\log_{12}(x^2)} \) is:
\[ (-\infty, -1] \cup [1, \infty) \]

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