The domain of `f(x)=cos^(-1)((2-|x|)/4)+[l log(3-x)]^1` is `[-2,6]` (b) `[-6,2)uu(2,3)` `[-6,2]` (d) `[-2,2]uu(2,3)`

Options

A. `[-2,6]`
B. `[-6,2) cup (2,3)`
C. `[-6,2]`
D. `[-2,2] cup (2,3)`

Explanation

To find the domain of the function \( f(x) = \cos^{-1}\left(\frac{2 - |x|}{4}\right) + \log(3 - x) \), we need to consider the restrictions imposed by both the inverse cosine function and the logarithmic function.

### Step 1: Analyze the Inverse Cosine Function
The function \( \cos^{-1}(y) \) is defined for \( y \) in the interval \([-1, 1]\). Therefore, we need to ensure that:
\[ -1 \leq \frac{2 - |x|}{4} \leq 1 \]

#### Step 1.1: Solve the Inequalities
1. **Upper Bound:**
\[ \frac{2 - |x|}{4} \leq 1 \]
Multiply both sides by 4:
\[ 2 - |x| \leq 4 \]
Rearranging gives:
\[ -|x| \leq 2 \quad \Rightarrow \quad |x| \geq -2 \quad \text{(always true)} \]

2. **Lower Bound:**
\[ \frac{2 - |x|}{4} \geq -1 \]
Multiply both sides by 4:
\[ 2 - |x| \geq -4 \]
Rearranging gives:
\[ -|x| \geq -6 \quad \Rightarrow \quad |x| \leq 6 \]

This means:
\[ -6 \leq x \leq 6 \]

### Step 2: Analyze the Logarithmic Function
The logarithmic function \( \log(3 - x) \) is defined for \( 3 - x > 0 \), which simplifies to:
\[ x < 3 \]

### Step 3: Combine the Conditions
Now we combine the results from the two analyses:
1. From the inverse cosine function: \( -6 \leq x \leq 6 \)
2. From the logarithmic function: \( x < 3 \)

The combined condition is:
\[ -6 \leq x < 3 \]

### Step 4: Exclude Points Where the Logarithm is Undefined
We also need to ensure that \( 3 - x \neq 1 \) (to avoid the logarithm being undefined):
\[ 3 - x \neq 1 \quad \Rightarrow \quad x \neq 2 \]

### Final Domain
Thus, the domain of \( f(x) \) is:
\[ [-6, 2) \cup (2, 3) \]

### Conclusion
The correct option for the domain of \( f(x) \) is:
**(b) \([-6, 2) \cup (2, 3)\)**

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