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

Options

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

Explanation

To find the domain of the function \( f(x) = \frac{\sin^{-1}(3 - x)}{\ln(|x| - 2)} \), we need to consider the conditions under which both the numerator and the denominator are defined.

### Step 1: Determine the conditions for the numerator
The numerator is \( \sin^{-1}(3 - x) \). The inverse sine function is defined for inputs in the range \([-1, 1]\). Therefore, we need:

\[ -1 \leq 3 - x \leq 1 \]

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:
\[ 3 - x \leq 1 \implies 3 - 1 \leq x \implies x \geq 2 \]

So, from the numerator, we have:
\[ 2 \leq x \leq 4 \]

### Step 2: Determine the conditions for the denominator
The denominator is \( \ln(|x| - 2) \). The logarithm function is defined for positive arguments, so we need:

\[ |x| - 2 > 0 \implies |x| > 2 \]

This gives us two cases to consider:

1. \( x > 2 \)
2. \( x < -2 \)

### Step 3: Combine the conditions
Now we combine the conditions from both the numerator and the denominator.

From the numerator, we have:
\[ 2 \leq x \leq 4 \]

From the denominator, we have:
1. \( x > 2 \) (which is satisfied by the range from the numerator)
2. \( x < -2 \) (which does not overlap with the range from the numerator)

Thus, the only valid condition is:
\[ 2 < x \leq 4 \]

### Conclusion
The domain of the function \( f(x) \) is:
\[ (2, 4] \]