`f(x)=sqrt(sin^(- 1)(log_2x))` Find the domain

Options

A. `x in (1,2)`
B. `x in [1,2]`
C. `x in [2,oo)`
D. `x in (0,oo)`

Answer

To find the domain of the function \( f(x) = \sqrt{\sin^{-1}(\log_2 x)} \), we need to ensure that the expression inside the square root is non-negative and that the logarithm is defined. Let's break this down step by step.

### Step 1: Ensure the argument of the square root is non-negative
The expression inside the square root, \( \sin^{-1}(\log_2 x) \), must be greater than or equal to 0 for \( f(x) \) to be real. Therefore, we have:
\[ \sin^{-1}(\log_2 x) \geq 0 \]

### Step 2: Determine when \( \sin^{-1}(y) \geq 0 \)
The inverse sine function, \( \sin^{-1}(y) \), is non-negative when:
\[ y \geq 0 \]
Thus, we require:
\[ \log_2 x \geq 0 \]

### Step 3: Solve the inequality \( \log_2 x \geq 0 \)
Using the properties of logarithms, we can rewrite this inequality:
\[ x \geq 2^0 \]
This simplifies to:
\[ x \geq 1 \]

### Step 4: Ensure the argument of the inverse sine is within its valid range
The function \( \sin^{-1}(y) \) is defined for \( y \) in the interval \([-1, 1]\). Therefore, we also need:
\[ -1 \leq \log_2 x \leq 1 \]

### Step 5: Solve the inequalities for \( \log_2 x \)
1. For the lower bound:
\[ \log_2 x \geq -1 \implies x \geq 2^{-1} \implies x \geq \frac{1}{2} \]

2. For the upper bound:
\[ \log_2 x \leq 1 \implies x \leq 2^1 \implies x \leq 2 \]

### Step 6: Combine the inequalities
From the conditions we derived:
1. \( x \geq 1 \)
2. \( x \geq \frac{1}{2} \)
3. \( x \leq 2 \)

The most restrictive conditions are:
\[ 1 \leq x \leq 2 \]

### Conclusion
Thus, the domain of the function \( f(x) = \sqrt{\sin^{-1}(\log_2 x)} \) is:
\[ \boxed{[1, 2]} \]