If `f(x)={sin^(-1)(sinx),xgt0`
`(pi)/(2),x=0,then
cos^(-1)(cosx),xlt0`

Options

A. x=0is apoint of maxima
B. x=0 is a point of minima
C. x=0 is a point of intersectionn
D. none of these

Explanation

To solve the problem, we need to analyze the function defined piecewise:

1. **Understanding the function**:
- The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} \sin^{-1}(\sin x) & \text{if } x > 0 \\ \frac{\pi}{2} & \text{if } x = 0 \\ \cos^{-1}(\cos x) & \text{if } x < 0 \end{cases} \]

2. **Evaluating \( f(x) \) for \( x > 0 \)**:
- For \( x > 0 \), \( f(x) = \sin^{-1}(\sin x) \).
- The range of \( \sin^{-1}(y) \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
- Therefore, for \( x \) in the interval \( (0, \frac{\pi}{2}) \), \( f(x) = x \).
- For \( x \) in the interval \( (\frac{\pi}{2}, \frac{3\pi}{2}) \), \( f(x) = \pi - x \).

3. **Evaluating \( f(x) \) for \( x = 0 \)**:
- For \( x = 0 \), we have \( f(0) = \frac{\pi}{2} \).

4. **Evaluating \( f(x) \) for \( x < 0 \)**:
- For \( x < 0 \), \( f(x) = \cos^{-1}(\cos x) \).
- The range of \( \cos^{-1}(y) \) is \( [0, \pi] \).
- For \( x \) in the interval \( (-\pi, 0) \), \( f(x) = -x \).
- For \( x \) in the interval \( (-2\pi, -\pi) \), \( f(x) = 2\pi + x \).

5. **Summary of the function**:
- Thus, we can summarize the function \( f(x) \) as:
\[ f(x) = \begin{cases} x & \text{if } 0 < x < \frac{\pi}{2} \\ \pi - x & \text{if } \frac{\pi}{2} < x < \frac{3\pi}{2} \\ \frac{\pi}{2} & \text{if } x = 0 \\ -x & \text{if } -\pi < x < 0 \\ 2\pi + x & \text{if } -2\pi < x < -\pi \end{cases} \]

6. **Conclusion**:
- The function is defined piecewise and behaves differently depending on the interval in which \( x \) lies.