Question

Domain of the function \[ f(x) = \sqrt{\sin^{-1}(2x) + \frac{\pi}{6}} \text{ for real values of } x \]

• A. \([- \frac{1}{4}, \frac{1}{2}]\)
• B. \([- \frac{1}{2}, \frac{1}{2}]\)
• C. \([- \frac{1}{4}, \frac{1}{4}]\)
• D. \([- \frac{1}{2}, \frac{1}{9}]\)

Hint:

For functions involving inverse trigonometric functions, always ensure the argument of the inverse function is within its valid domain and that the expression inside the square root is non-negative for real-valued outputs.

Answer 1

The Correct Option is A

Step 1: Understand the structure of the function.
We are given the function \( f(x) = \sqrt{\sin^{-1}(2x) + \frac{\pi}{6}} \). For the function to have real values, the expression inside the square root must be non-negative, i.e.
\[ \sin^{-1}(2x) + \frac{\pi}{6} \geq 0 \]

Step 2: Conditions for the inverse sine function.
The domain of \( \sin^{-1}(y) \) is restricted to \( y \in [-1, 1] \), meaning that for the expression \( \sin^{-1}(2x) \) to be valid, we must have \( -1 \leq 2x \leq 1 \). Solving for \( x \), we get:
\[ - \frac{1}{2} \leq x \leq \frac{1}{2} \]
Thus, the domain of \( 2x \) is restricted to this interval.

Step 3: Solve the inequality for the square root condition.
Now, we solve the condition \( \sin^{-1}(2x) + \frac{\pi}{6} \geq 0 \). Subtract \( \frac{\pi}{6} \) from both sides to get:
\[ \sin^{-1}(2x) \geq - \frac{\pi}{6} \]
Since \( \sin^{-1}(2x) \) is defined for \( 2x \in [-1, 1] \), we examine the possible range of values of \( \sin^{-1}(2x) \), which lies between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \).

Step 4: Analyze the limits of the domain.
For the inequality \( \sin^{-1}(2x) \geq - \frac{\pi}{6} \), the smallest value of \( \sin^{-1}(2x) \) is \( -\frac{\pi}{2} \) when \( 2x = -1 \). We require the condition \( \sin^{-1}(2x) \geq -\frac{\pi}{6} \), so solving for \( x \), we get:
\[ 2x \geq \sin\left(- \frac{\pi}{6}\right) = -\frac{1}{2} \]
Thus, we have: \[ x \geq - \frac{1}{4} \]

Step 5: Combine the conditions.
From Step 2, we already have the condition \( -\frac{1}{2} \leq x \leq \frac{1}{2} \). Combining this with the condition from Step 4, we get the domain of \( x \) as:
\[ - \frac{1}{4} \leq x \leq \frac{1}{2} \]
Thus, the domain of the function is \( [- \frac{1}{4}, \frac{1}{2}] \).

Step 6: Conclusion.
Therefore, the correct answer is option (A). Final Answer: \[ \boxed{[- \frac{1}{4}, \frac{1}{2}]} \]