The domain of the function `f(x)=(sin^(-1)(3-x))/(log_(e)(|-x|-2))`, is

Options

A. [2,4]
B. `(2,3) cup (3,4]`
C. `[2,3)`
D. `(-oo,-3) cup [2,oo)`

Answer

To find the domain of the function \( f(x) = \frac{\sin^{-1}(3-x)}{\log_{e}(|-x|-2)} \), we need to consider the restrictions imposed by both the inverse sine function and the logarithm.

### Step 1: Analyze the inverse sine function
The expression inside the inverse sine function, \( 3 - x \), must satisfy the condition:
\[ -1 \leq 3 - x \leq 1 \]
This can be split into two inequalities:
1. \( 3 - x \geq -1 \)
2. \( 3 - x \leq 1 \)

**Solving the first inequality:**
\[ 3 - x \geq -1 \implies 3 + 1 \geq x \implies x \leq 4 \]

**Solving the second inequality:**
\[ 3 - x \leq 1 \implies 3 - 1 \leq x \implies x \geq 2 \]

Thus, from the inverse sine function, we have:
\[ 2 \leq x \leq 4 \]

### Step 2: Analyze the logarithmic function
The logarithm \( \log_{e}(|-x| - 2) \) requires:
\[ |-x| - 2 > 0 \implies |-x| > 2 \implies |x| > 2 \]
This gives us two cases:
1. \( x > 2 \)
2. \( x < -2 \)

### Step 3: Combine the conditions
From the inverse sine function, we have:
\[ 2 \leq x \leq 4 \]
From the logarithmic function, we have:
- For \( x > 2 \): This is valid for \( x \) in the interval \( (2, 4] \).
- For \( x < -2 \): This does not intersect with the interval \( [2, 4] \).

### Step 4: Exclude points where the logarithm is undefined
The logarithm is also undefined when its argument equals 1:
\[ |-x| - 2 \neq 1 \implies |-x| \neq 3 \implies |x| \neq 3 \]
This means \( x \neq 3 \) and \( x \neq -3 \). However, since we are only considering \( x \) in the interval \( [2, 4] \), we only need to exclude \( x = 3 \).

### Final Domain
Combining all the conditions, we find that the domain of \( f(x) \) is:
\[ [2, 4] \setminus \{3\} = [2, 3) \cup (3, 4] \]

### Conclusion
Thus, the domain of the function \( f(x) \) is:
\[ \boxed{[2, 3) \cup (3, 4]} \]