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 differentiable at x=0.
2. The function f(x) is differentiable at `x=(pi)/(2)`.
Which of the above statements is /aer correct ?

Options

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

Explanation

To determine the differentiability of the function \( f(x) \) defined piecewise as follows:

\[ f(x) = \begin{cases} x + \pi & \text{for } x \in [-\pi, 0) \\ \pi \cos x & \text{for } x \in [0, \frac{\pi}{2}] \\ \left( x - \frac{\pi}{2} \right)^2 & \text{for } x \in \left( \frac{\pi}{2}, \pi \right] \end{cases} \]

we need to check the differentiability at \( x = 0 \) and \( x = \frac{\pi}{2} \).

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

**Left-hand derivative at \( x = 0 \):**

The left-hand derivative is given by:

\[ f'(0^-) = \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} \]

Since \( h \) approaches from the left, we use the first piece of the function:

\[ f(0) = \pi \cos(0) = \pi \]

Thus, we have:

\[ f'(0^-) = \lim_{h \to 0^-} \frac{(0 + h) + \pi - \pi}{h} = \lim_{h \to 0^-} \frac{h}{h} = 1 \]

**Right-hand derivative at \( x = 0 \):**

The right-hand derivative is given by:

\[ f'(0^+) = \lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h} \]

Using the second piece of the function:

\[ f'(0^+) = \lim_{h \to 0^+} \frac{\pi \cos(h) - \pi}{h} \]

Using the fact that \( \cos(h) \approx 1 - \frac{h^2}{2} \) as \( h \to 0 \):

\[ f'(0^+) = \lim_{h \to 0^+} \frac{\pi(1 - \frac{h^2}{2}) - \pi}{h} = \lim_{h \to 0^+} \frac{-\frac{\pi h^2}{2}}{h} = \lim_{h \to 0^+} -\frac{\pi h}{2} = 0 \]

**Conclusion for \( x = 0 \):**

Since \( f'(0^-) = 1 \) and \( f'(0^+) = 0 \), the left-hand and right-hand derivatives are not equal. Therefore, \( f(x) \) is **not differentiable at \( x = 0 \)**.

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

**Left-hand derivative at \( x = \frac{\pi}{2} \):**

Using the second piece of the function:

\[ f\left(\frac{\pi}{2}\right) = \pi \cos\left(\frac{\pi}{2}\right) = 0 \]

The left-hand derivative is given by:

\[ f'\left(\frac{\pi}{2}^-\right) = \lim_{h \to 0^-} \frac{f\left(\frac{\pi}{2} + h\right) - f\left(\frac{\pi}{2}\right)}{h} \]

Using the second piece:

\[ f'\left(\frac{\pi}{2}^-\right) = \lim_{h \to 0^-} \frac{\pi \cos\left(\frac{\pi}{2} + h\right) - 0}{h} = \lim_{h \to 0^-} \frac{\pi \cdot 0 - 0}{h} = 0 \]

**Right-hand derivative at \( x = \frac{\pi}{2} \):**

Using the third piece of the function:

\[ f'\left(\frac{\pi}{2}^+\right) = \lim_{h \to 0^+} \frac{f\left(\frac{\pi}{2} + h\right) - f\left(\frac{\pi}{2}\right)}{h} \]

\[ = \lim_{h \to 0^+} \frac{\left(\frac{\pi}{2} + h - \frac{\pi}{2}\right)^2 - 0}{h} = \lim_{h \to 0^+} \frac{h^2}{h} = \lim_{h \to 0^+} h = 0 \]

**Conclusion for \( x = \frac{\pi}{2} \):**

Since \( f'\left(\frac{\pi}{2}^-\right) = 0 \) and \( f'\left(\frac{\pi}{2}^+\right) = 0 \), the left-hand and right-hand derivatives are equal. Therefore, \( f(x) \) is **differentiable at \( x = \frac{\pi}{2} \)**.

### Final Conclusion:

1. The function \( f(x) \) is **not differentiable at \( x = 0 \)**.
2. The function \( f(x) \) **is differentiable at \( x = \frac{\pi}{2} \)**.

Thus, the correct answer is that only the second statement is true.