`sqrt( sin^(-1) ( log_2 x ) ` exists for

Options

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

Answer

To determine the values of \( x \) for which the expression \( \sqrt{\sin^{-1}(\log_2 x)} \) exists, we need to analyze the conditions under which the expression is defined.

### Step-by-Step Solution:

1. **Identify the Expression**:
We have the expression \( \sqrt{\sin^{-1}(\log_2 x)} \). For the square root to exist, the argument of the square root must be non-negative:
\[ \sin^{-1}(\log_2 x) \geq 0 \]

2. **Condition for the Inverse Sine Function**:
The function \( \sin^{-1}(y) \) is defined for \( y \) in the range \([-1, 1]\). Therefore, we need:
\[ -1 \leq \log_2 x \leq 1 \]

3. **Solving the Inequalities**:
We will solve the inequalities separately.

- **First Inequality**:
\[ \log_2 x \geq -1 \]
Converting this from logarithmic to exponential form gives:
\[ x \geq 2^{-1} = \frac{1}{2} \]

- **Second Inequality**:
\[ \log_2 x \leq 1 \]
Again, converting this from logarithmic to exponential form gives:
\[ x \leq 2^1 = 2 \]

4. **Combining the Results**:
From the two inequalities, we combine the results:
\[ \frac{1}{2} \leq x \leq 2 \]
This means \( x \) must be in the closed interval \([1/2, 2]\).

5. **Conclusion**:
Therefore, the expression \( \sqrt{\sin^{-1}(\log_2 x)} \) exists for:
\[ x \in \left[\frac{1}{2}, 2\right] \]