The domain of the fuction `f(x)=2sin^(-1){log_(2)((1)/(2)x^(2))}` is

Options

A. `[-2, -1]uu(1,2]`
B. `(-2, -1]uu[1,2]`
C. `[-2, -1]uu[1,2]`
D. `(-2,-1)uu(1,2)`

Answer

To find the domain of the function \( f(x) = 2 \sin^{-1} \left( \log_2 \left( \frac{1}{2} x^2 \right) \right) \), we need to ensure that both the logarithmic function and the inverse sine function are defined.

### Step 1: Determine the condition for the logarithmic function
The logarithmic function \( \log_2 \left( \frac{1}{2} x^2 \right) \) is defined when its argument is greater than zero:
\[ \frac{1}{2} x^2 > 0 \]
Since \( \frac{1}{2} \) is positive, we only need to consider \( x^2 > 0 \). This is true for all \( x \) except \( x = 0 \). Thus, \( x \) must be any real number except zero:
\[ x \in \mathbb{R} \setminus \{0\} \]

### Step 2: Determine the condition for the inverse sine function
The function \( \sin^{-1}(y) \) is defined for \( y \) in the interval \([-1, 1]\). Therefore, we need:
\[ -1 \leq \log_2 \left( \frac{1}{2} x^2 \right) \leq 1 \]

### Step 3: Solve the inequalities
We will break this into two inequalities.

#### Inequality 1: \( \log_2 \left( \frac{1}{2} x^2 \right) \geq -1 \)
Taking the anti-logarithm:
\[ \frac{1}{2} x^2 \geq 2^{-1} \implies \frac{1}{2} x^2 \geq \frac{1}{2} \]
Multiplying both sides by 2:
\[ x^2 \geq 1 \]
This implies:
\[ x \leq -1 \quad \text{or} \quad x \geq 1 \]

#### Inequality 2: \( \log_2 \left( \frac{1}{2} x^2 \right) \leq 1 \)
Taking the anti-logarithm:
\[ \frac{1}{2} x^2 \leq 2^1 \implies \frac{1}{2} x^2 \leq 2 \]
Multiplying both sides by 2:
\[ x^2 \leq 4 \]
This implies:
\[ -2 \leq x \leq 2 \]

### Step 4: Combine the results
Now we combine the results from both inequalities:
1. From \( x^2 \geq 1 \): \( x \leq -1 \) or \( x \geq 1 \)
2. From \( x^2 \leq 4 \): \( -2 \leq x \leq 2 \)

The intersection of these two conditions gives us:
- For \( x \leq -1 \): the valid range is \( -2 \leq x \leq -1 \)
- For \( x \geq 1 \): the valid range is \( 1 \leq x \leq 2 \)

Thus, the domain of the function \( f(x) \) is:
\[ x \in [-2, -1] \cup [1, 2] \]

### Final Answer
The domain of the function \( f(x) \) is:
\[ [-2, -1] \cup [1, 2] \]