`f (x) = {{:( tan^(-1) x , "," , |x| lt (pi)/(2)) , ((pi)/(2) - |x| , "," , |x| ge (pi)/(2)):}` then

Options

A. f(x) has no point of local maxima
B. f(x) has only one point of local maxima
C. f(x) has exactly two points of local maxima
D. f(x) has exactly two points of local minima

Answer

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

1. **Define the function**:
\[ f(x) = \begin{cases} \tan^{-1}(x) & \text{if } |x| < \frac{\pi}{2} \\ \frac{\pi}{2} - |x| & \text{if } |x| \geq \frac{\pi}{2} \end{cases} \]

2. **Identify the intervals**:
- For \( |x| < \frac{\pi}{2} \), the function is \( f(x) = \tan^{-1}(x) \).
- For \( |x| \geq \frac{\pi}{2} \), the function is \( f(x) = \frac{\pi}{2} - |x| \).

3. **Find the derivative**:
- For \( |x| < \frac{\pi}{2} \):
\[ f'(x) = \frac{1}{1+x^2} \]
- For \( |x| \geq \frac{\pi}{2} \):
\[ f'(x) = -1 \quad \text{(for } x \geq \frac{\pi}{2}\text{)} \quad \text{and} \quad f'(x) = 1 \quad \text{(for } x \leq -\frac{\pi}{2}\text{)} \]

4. **Analyze critical points**:
- For \( |x| < \frac{\pi}{2} \), since \( f'(x) > 0 \) for all \( x \) in this interval, there are no local maxima or minima.
- For \( |x| \geq \frac{\pi}{2} \), the function is linear and decreasing, thus no local maxima or minima exist in this region either.

5. **Check endpoints**:
- At \( x = \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = \frac{\pi}{2} - \frac{\pi}{2} = 0 \]
- At \( x = -\frac{\pi}{2} \):
\[ f\left(-\frac{\pi}{2}\right) = \frac{\pi}{2} - \frac{\pi}{2} = 0 \]

6. **Conclusion**:
- The function \( f(x) \) has only one point of local maxima at \( x = 0 \) where \( f(0) = 0 \) since it is the highest point in the interval \( |x| < \frac{\pi}{2} \).

Thus, the answer is that \( f(x) \) has only one point of local maxima.