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 ensure that both the numerator and denominator are well-defined.

### Step 1: Analyze the numerator \( \sin^{-1}(3-x) \)

The function \( \sin^{-1}(y) \) is defined for \( y \) in the interval \([-1, 1]\). Therefore, we need to set up the inequality:

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

#### Solving the inequalities:

1. **For the left side**:
\[ 3 - x \geq -1 \implies 3 + 1 \geq x \implies x \leq 4 \]

2. **For the right side**:
\[ 3 - x \leq 1 \implies 3 - 1 \leq x \implies x \geq 2 \]

Combining these results, we have:

\[ 2 \leq x \leq 4 \]

### Step 2: Analyze the denominator \( \ln(|x|-2) \)

The logarithmic function \( \ln(y) \) is defined for \( y > 0 \). Thus, we need:

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

This gives us two cases:

1. **Case 1**: \( x > 2 \)
2. **Case 2**: \( x < -2 \)

However, since we already have \( 2 \leq x \leq 4 \) from the numerator, we only consider the first case \( x > 2 \).

### Step 3: Combine the conditions

From the two steps, we have:

- From the numerator: \( 2 \leq x \leq 4 \)
- From the denominator: \( x > 2 \)

Combining these, we find:

\[ 2 < x \leq 4 \]

### Step 4: Exclude points where the denominator is zero

We also need to ensure that the denominator does not equal zero.

Setting the denominator to zero:

\[ \ln(|x|-2) = 0 \implies |x| - 2 = 1 \implies |x| = 3 \]

This gives us two points:

1. \( x = 3 \)
2. \( x = -3 \)

Since we are only considering \( x \geq 2 \), we exclude \( x = 3 \).

### Final Domain

Thus, the domain of \( f(x) \) is:

\[ (2, 3) \cup (3, 4] \]

### Summary of the Domain

The domain of the function \( f(x) = \frac{\sin^{-1}(3-x)}{\ln(|x|-2)} \) is \( (2, 3) \cup (3, 4] \).

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