A function f(x) is defined as follows:
`f(x)={{:(x+pi," for ",x""in[-pi","0)),(picosx," for ",x""in[0","(pi)/(2)]),((x-(pi)/(2))^(2)," for ",x""in((pi)/(2)","pi]):}`
Consider the following statements :
1. The function f(x) is continuos at x=0.
2. The function f(x) is continuous at `x=(pi)/(2)`.
Which of the above statements is/are correct?

Options

A. 1 only
B. 2 only
C. Both 1 and 2
D. Neither 1 nor 2

Explanation

To determine the continuity of the function \( f(x) \) at the points \( x = 0 \) and \( x = \frac{\pi}{2} \), we will evaluate the left-hand limit, right-hand limit, and the function value at these points.

### Step 1: Check continuity at \( x = 0 \)

1. **Function Definition**:
- For \( x \in [-\pi, 0) \), \( f(x) = x + \pi \)
- For \( x \in [0, \frac{\pi}{2}] \), \( f(x) = \pi \cos x \)

2. **Value of the function at \( x = 0 \)**:
\[ f(0) = \pi \cos(0) = \pi \cdot 1 = \pi \]

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

4. **Right-hand limit as \( x \) approaches 0**:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (\pi \cos x) = \pi \cos(0) = \pi \cdot 1 = \pi \]

5. **Conclusion for \( x = 0 \)**:
Since \( f(0) = \pi \), \( \lim_{x \to 0^-} f(x) = \pi \), and \( \lim_{x \to 0^+} f(x) = \pi \), we have:
\[ \lim_{x \to 0} f(x) = f(0) = \pi \]
Therefore, \( f(x) \) is continuous at \( x = 0 \).

### Step 2: Check continuity at \( x = \frac{\pi}{2} \)

1. **Value of the function at \( x = \frac{\pi}{2} \)**:
\[ f\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2} - \frac{\pi}{2}\right)^2 = 0^2 = 0 \]

2. **Left-hand limit as \( x \) approaches \( \frac{\pi}{2} \)**:
\[ \lim_{x \to \left(\frac{\pi}{2}\right)^-} f(x) = \lim_{x \to \left(\frac{\pi}{2}\right)^-} (\pi \cos x) = \pi \cos\left(\frac{\pi}{2}\right) = \pi \cdot 0 = 0 \]

3. **Right-hand limit as \( x \) approaches \( \frac{\pi}{2} \)**:
\[ \lim_{x \to \left(\frac{\pi}{2}\right)^+} f(x) = \lim_{x \to \left(\frac{\pi}{2}\right)^+} \left(x - \frac{\pi}{2}\right)^2 = \left(\frac{\pi}{2} - \frac{\pi}{2}\right)^2 = 0^2 = 0 \]

4. **Conclusion for \( x = \frac{\pi}{2} \)**:
Since \( f\left(\frac{\pi}{2}\right) = 0 \), \( \lim_{x \to \left(\frac{\pi}{2}\right)^-} f(x) = 0 \), and \( \lim_{x \to \left(\frac{\pi}{2}\right)^+} f(x) = 0 \), we have:
\[ \lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right) = 0 \]
Therefore, \( f(x) \) is continuous at \( x = \frac{\pi}{2} \).

### Final Conclusion:
Both statements are correct:
1. The function \( f(x) \) is continuous at \( x = 0 \).
2. The function \( f(x) \) is continuous at \( x = \frac{\pi}{2} \).

Thus, the answer is that both statements are true.