The domain of definition of `f(x) = sqrt(sec^(-1){(1-|x|)/(2)})` is

Options

A. `(-oo, -3)`
B. `[3, oo)`
C. `phi`
D. `(-oo, -3] uu [3, oo)`

Explanation

To find the domain of the function \( f(x) = \sqrt{\sec^{-1}\left(\frac{1 - |x|}{2}\right)} \), we need to ensure that the expression under the square root is defined and non-negative.

### Step 1: Ensure the expression under the square root is non-negative
For \( f(x) \) to be defined, the argument of the square root must be non-negative:
\[ \sec^{-1}\left(\frac{1 - |x|}{2}\right) \geq 0 \]
The range of \( \sec^{-1}(y) \) is \( [0, \pi/2) \cup (\pi/2, \pi] \), which means \( \sec^{-1}(y) \) is defined for \( y \leq -1 \) or \( y \geq 1 \).

### Step 2: Determine the conditions for \( \frac{1 - |x|}{2} \)
We need to find when \( \frac{1 - |x|}{2} \) is either \( \leq -1 \) or \( \geq 1 \).

1. **Condition 1**: \( \frac{1 - |x|}{2} \geq 1 \)
\[ 1 - |x| \geq 2 \]
\[ -|x| \geq 1 \quad \Rightarrow \quad |x| \leq -1 \]
This condition is not possible since the absolute value cannot be negative.

2. **Condition 2**: \( \frac{1 - |x|}{2} \leq -1 \)
\[ 1 - |x| \leq -2 \]
\[ -|x| \leq -3 \quad \Rightarrow \quad |x| \geq 3 \]

### Step 3: Solve the inequality \( |x| \geq 3 \)
The inequality \( |x| \geq 3 \) means:
\[ x \leq -3 \quad \text{or} \quad x \geq 3 \]

### Step 4: Write the domain in interval notation
Thus, the domain of the function \( f(x) \) is:
\[ (-\infty, -3] \cup [3, \infty) \]

### Final Answer
The domain of the function \( f(x) = \sqrt{\sec^{-1}\left(\frac{1 - |x|}{2}\right)} \) is:
\[ \boxed{(-\infty, -3] \cup [3, \infty)} \]