Question

The domain and range of \( f(x) = \sin^{-1}(x) \) are:

• A. \( [0, 2], \, \left\{ -\frac{\pi}{2}, \frac{\pi}{2} \right\} \)
• B. \( [-1, 2], \, \left\{ -\frac{\pi}{2}, 0, \frac{\pi}{2} \right\} \)
• C. \( [-1, 2], \, \left\{ -\frac{\pi}{2}, 0, \frac{\pi}{2} \right\} \)
• D. None of the above

Hint:

The domain of \( \sin^{-1}(x) \) is \( [-1, 1] \), and its range is \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \).

Answer 1

The Correct Option is B

Step 1: Recall the domain and range of \( \sin^{-1}(x) \).
The inverse sine function \( \sin^{-1}(x) \) is defined for values of \( x \) such that \( -1 \leq x \leq 1 \). Hence, the domain of \( f(x) = \sin^{-1}(x) \) is \( [-1, 1] \).

Step 2: Find the range of \( \sin^{-1}(x) \).
The range of the inverse sine function is: \[ \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \] This is because the range of the sine function is \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \).

Step 3: Conclusion.
Thus, the domain of \( f(x) = \sin^{-1}(x) \) is \( [-1, 1] \), and the range is \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \). Therefore, the correct answer is option (B).