Consider the following for the next two items that 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 statement:
1. The function f(x) is continuous at x = 0.
2. The function f(x) is continuous at `x=(pi)/(2)`
Which of the above statements is / are correct?

Options

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

Answer

To determine the continuity of the function \( f(x) \) at the specified points \( x = 0 \) and \( x = \frac{\pi}{2} \), we will analyze the left-hand limit and right-hand limit at these points and check if they are equal to the function value at those points.

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

1. **Identify the left-hand limit as \( x \) approaches 0**:
- For \( x \) in the interval \([- \pi, 0)\), the function is defined as \( f(x) = x + \pi \).
- Therefore, the left-hand limit is:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x + \pi) = 0 + \pi = \pi. \]

2. **Identify the right-hand limit as \( x \) approaches 0**:
- For \( x \) in the interval \([0, \frac{\pi}{2}]\), the function is defined as \( f(x) = \pi \cos x \).
- Therefore, the right-hand limit is:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (\pi \cos x) = \pi \cos(0) = \pi \cdot 1 = \pi. \]

3. **Check if the left-hand limit equals the right-hand limit**:
- Since both limits are equal:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = \pi. \]
- Therefore, \( f(0) = \pi \) (since \( f(0) = \pi \cos(0) = \pi \)).
- Thus, \( f(x) \) is continuous at \( x = 0 \).

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

1. **Identify the left-hand limit as \( x \) approaches \( \frac{\pi}{2} \)**:
- For \( x \) in the interval \([0, \frac{\pi}{2})\), the function is defined as \( f(x) = \pi \cos x \).
- Therefore, the left-hand limit is:
\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}^-} (\pi \cos x) = \pi \cos\left(\frac{\pi}{2}\right) = \pi \cdot 0 = 0. \]

2. **Identify the right-hand limit as \( x \) approaches \( \frac{\pi}{2} \)**:
- For \( x \) in the interval \((\frac{\pi}{2}, \pi]\), the function is defined as \( f(x) = \left(x - \frac{\pi}{2}\right)^2 \).
- Therefore, the right-hand limit is:
\[ \lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{x \to \frac{\pi}{2}^+} \left(x - \frac{\pi}{2}\right)^2 = \left(\frac{\pi}{2} - \frac{\pi}{2}\right)^2 = 0. \]

3. **Check if the left-hand limit equals the right-hand limit**:
- Since both limits are equal:
\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}^+} f(x) = 0. \]
- Therefore, \( f\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2} - \frac{\pi}{2}\right)^2 = 0 \).
- Thus, \( f(x) \) is continuous at \( x = \frac{\pi}{2} \).

### Conclusion

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