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

Options

A. `x =0` is a point of maxima
B. `f (x)` is continous `AA x in R`
C. glolab maximum vlaue of `f (x) AA x in R` is `pi`
D. global minimum vlaue of `f (x) AA x in R` is 0

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} \sin^{-1}(\sin x) & \text{if } x > 0 \\ \frac{\pi}{2} & \text{if } x = 0 \\ \cos^{-1}(\cos x) & \text{if } x < 0 \end{cases} \]

### Step 1: Analyze \( f(x) \) for \( x > 0 \)

For \( x > 0 \):
\[ f(x) = \sin^{-1}(\sin x) \]
Since \( \sin^{-1}(\sin x) = x \) for \( x \) in the interval \( (0, \frac{\pi}{2}) \), we can conclude that:
\[ f(x) = x \quad \text{for } 0 < x < \frac{\pi}{2} \]

### Step 2: Analyze \( f(x) \) at \( x = 0 \)

At \( x = 0 \):
\[ f(0) = \frac{\pi}{2} \]

### Step 3: Analyze \( f(x) \) for \( x < 0 \)

For \( x < 0 \):
\[ f(x) = \cos^{-1}(\cos x) \]
Since \( \cos^{-1}(\cos x) = -x \) for \( x \) in the interval \( (-\frac{\pi}{2}, 0) \), we can conclude that:
\[ f(x) = -x \quad \text{for } -\frac{\pi}{2} < x < 0 \]

### Step 4: Check continuity at \( x = 0 \)

To check if \( f(x) \) is continuous at \( x = 0 \), we need to evaluate the left-hand limit (LHL) and right-hand limit (RHL):

- **Right-hand limit** as \( x \to 0^+ \):
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0 \]

- **Left-hand limit** as \( x \to 0^- \):
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} -x = 0 \]

Since both limits equal \( 0 \) and \( f(0) = \frac{\pi}{2} \), we see that:
\[ \text{LHL} = 0 \neq f(0) = \frac{\pi}{2} \quad \Rightarrow \quad f(x) \text{ is not continuous at } x = 0. \]

### Step 5: Determine the nature of the point at \( x = 0 \)

Next, we need to check the derivatives to find if it is a maximum or minimum:

- **Derivative for \( x > 0 \)**:
\[ f'(x) = 1 \quad \text{for } x > 0 \]

- **Derivative for \( x < 0 \)**:
\[ f'(x) = -1 \quad \text{for } x < 0 \]

At \( x = 0 \):
- \( f'(0^+) = 1 \) (positive)
- \( f'(0^-) = -1 \) (negative)

Since \( f'(0^+) > 0 \) and \( f'(0^-) < 0 \), this indicates that \( x = 0 \) is a point of local maximum.

### Conclusion

Thus, we conclude:
- \( f(x) \) is not continuous at \( x = 0 \).
- \( x = 0 \) is a point of local maximum.