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 given problem, we need to analyze the function defined piecewise as follows:

1. **Function Definition**:
- For \( x > 0 \): \( f(x) = \sin^{-1}(\sin x) \)
- For \( x = 0 \): \( f(0) = \frac{\pi}{2} \)
- For \( x < 0 \): \( f(x) = \cos^{-1}(\cos x) \)

2. **Understanding the Function**:
- The function \( \sin^{-1}(\sin x) \) for \( x > 0 \) simplifies to \( x \) when \( 0 < x < \frac{\pi}{2} \) because the range of \( \sin^{-1} \) is \([- \frac{\pi}{2}, \frac{\pi}{2}]\).
- For \( x = 0 \), the function value is given as \( \frac{\pi}{2} \).
- The function \( \cos^{-1}(\cos x) \) for \( x < 0 \) simplifies to \( -x \) when \( -\frac{\pi}{2} < x < 0 \) because the range of \( \cos^{-1} \) is \([0, \pi]\).

3. **Finding Limits**:
- We need to evaluate the limits of \( f(x) \) as \( x \) approaches 0 from the left and right.
- **Limit from the right** (\( x \to 0^+ \)):
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \sin^{-1}(\sin x) = 0 \]
- **Limit from the left** (\( x \to 0^- \)):
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \cos^{-1}(\cos x) = \cos^{-1}(1) = 0 \]

4. **Value at \( x = 0 \)**:
- From the problem statement, we have \( f(0) = \frac{\pi}{2} \).

5. **Comparing Values**:
- We find:
- \( \lim_{x \to 0^+} f(x) = 0 \)
- \( \lim_{x \to 0^-} f(x) = 0 \)
- \( f(0) = \frac{\pi}{2} \)

6. **Conclusion**:
- Since \( f(0) = \frac{\pi}{2} \) is greater than both the left-hand limit and right-hand limit, we conclude that \( x = 0 \) is a local maximum point.

### Final Result:
The maximum point of the function occurs at \( x = 0 \) with \( f(0) = \frac{\pi}{2} \).

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