Domain of definition of the function
`f(x)=sqrt(sin^(-1)(2x)+pi/6)` for real valued of x, is

Options

A. `[-1/4,1/2]`
B. `[-1/2,1/2]`
C. `(-1/2,1/9)`
D. None of these

Answer

To find the 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 and that the argument of the sine inverse function is within its valid range.

### Step 1: Determine the range of the sine inverse function
The function \( \sin^{-1}(y) \) is defined for \( y \) in the interval \([-1, 1]\). Therefore, for \( \sin^{-1}(2x) \) to be defined, we need:
\[ -1 \leq 2x \leq 1 \]
Dividing the entire inequality by 2 gives:
\[ -\frac{1}{2} \leq x \leq \frac{1}{2} \]

### Step 2: Ensure the expression inside the square root is non-negative
Next, we need to ensure that the expression inside the square root is non-negative:
\[ \sin^{-1}(2x) + \frac{\pi}{6} \geq 0 \]
This can be rearranged to:
\[ \sin^{-1}(2x) \geq -\frac{\pi}{6} \]

### Step 3: Solve the inequality involving sine
Taking the sine of both sides (noting that sine is an increasing function in the range of \([- \frac{\pi}{2}, \frac{\pi}{2}]\)):
\[ 2x \geq \sin\left(-\frac{\pi}{6}\right) \]
Since \( \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2} \), we have:
\[ 2x \geq -\frac{1}{2} \]
Dividing by 2 gives:
\[ x \geq -\frac{1}{4} \]

### Step 4: Combine the constraints
Now we have two inequalities to consider:
1. \( -\frac{1}{2} \leq x \leq \frac{1}{2} \)
2. \( x \geq -\frac{1}{4} \)

To find the domain, we take the intersection of these two intervals:
- The first interval is \( [-\frac{1}{2}, \frac{1}{2}] \).
- The second interval is \( [-\frac{1}{4}, \infty) \).

The intersection of these two intervals is:
\[ [-\frac{1}{4}, \frac{1}{2}] \]

### Final Answer
Thus, the domain of the function \( f(x) \) is:
\[ \boxed{[-\frac{1}{4}, \frac{1}{2}]} \]