The natural domain of, `sqrt(sin^(-1) (2x) + (pi)/(6) )` for all `x in R `, is

Options

A. `[-1/4 , 1/2]`
B. `[-1/4 ,1/4]`
C. `[-1/2 ,1/2]`
D. `[-1/2 ,1/4]`

Answer

To find the natural domain of the function \( f(x) = \sqrt{\sin^{-1}(2x) + \frac{\pi}{6}} \), we need to ensure that the expression inside the square root is non-negative. This means we need to satisfy the following condition:

\[ \sin^{-1}(2x) + \frac{\pi}{6} \geq 0 \]

### Step 1: Isolate the inverse sine function
We start by isolating the inverse sine function:

\[ \sin^{-1}(2x) \geq -\frac{\pi}{6} \]

### Step 2: Determine the range of the inverse sine function
The function \( \sin^{-1}(y) \) is defined for \( y \) in the interval \([-1, 1]\). Therefore, we need to ensure that \( 2x \) falls within this range:

\[ -1 \leq 2x \leq 1 \]

### Step 3: Solve for \( x \)
Now, we solve the inequalities for \( x \):

1. From \( 2x \geq -1 \):
\[ x \geq -\frac{1}{2} \]

2. From \( 2x \leq 1 \):
\[ x \leq \frac{1}{2} \]

### Step 4: Combine the results
Combining these two results, we find:

\[ -\frac{1}{2} \leq x \leq \frac{1}{2} \]

### Step 5: Verify the condition for the square root
Next, we need to check if the condition \( \sin^{-1}(2x) + \frac{\pi}{6} \geq 0 \) holds within this interval.

The minimum value of \( \sin^{-1}(2x) \) occurs at \( x = -\frac{1}{2} \):
\[ \sin^{-1}(2(-\frac{1}{2})) = \sin^{-1}(-1) = -\frac{\pi}{2} \]
Thus,
\[ -\frac{\pi}{2} + \frac{\pi}{6} = -\frac{3\pi}{6} + \frac{\pi}{6} = -\frac{2\pi}{6} = -\frac{\pi}{3} < 0 \]

The maximum value occurs at \( x = \frac{1}{2} \):
\[ \sin^{-1}(2(\frac{1}{2})) = \sin^{-1}(1) = \frac{\pi}{2} \]
Thus,
\[ \frac{\pi}{2} + \frac{\pi}{6} = \frac{3\pi}{6} + \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} > 0 \]

### Conclusion
The function \( f(x) \) is defined for \( x \) in the interval \([-0.5, 0.5]\) where the expression under the square root is non-negative.

Thus, the natural domain of the function is:

\[ \boxed{[-\frac{1}{2}, \frac{1}{2}]} \]