Let `f(x)={{:( 0,x=0),(x^(2) sin pi//2x,|x| lt 1),(x|x|, |x| ge 1):}`. Then , f(x) is

Options

A. an even function
B. an odd function
C. neither an even function nor an add function
D. f'(x) is an even function

Answer

To determine the nature of the function \( f(x) \) given by:

\[ f(x) = \begin{cases} 0 & \text{if } x = 0 \\ x^2 \sin\left(\frac{\pi}{2} x\right) & \text{if } |x| < 1 \\ x |x| & \text{if } |x| \geq 1 \end{cases} \]

we will analyze the function step by step.

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

For \( x = 0 \):
\[ f(0) = 0 \]

### Step 2: Check the behavior for \( |x| < 1 \)

For \( |x| < 1 \):
\[ f(x) = x^2 \sin\left(\frac{\pi}{2} x\right) \]

Now, we will check \( f(-x) \):
\[ f(-x) = (-x)^2 \sin\left(\frac{\pi}{2} (-x)\right) = x^2 \sin\left(-\frac{\pi}{2} x\right) = -x^2 \sin\left(\frac{\pi}{2} x\right) \]

### Step 3: Check if \( f(x) \) is odd

To check if \( f(x) \) is odd, we need to see if \( f(-x) = -f(x) \):
\[ f(-x) = -x^2 \sin\left(\frac{\pi}{2} x\right) \quad \text{and} \quad -f(x) = -\left(x^2 \sin\left(\frac{\pi}{2} x\right)\right) = -x^2 \sin\left(\frac{\pi}{2} x\right) \]

Thus, for \( |x| < 1 \):
\[ f(-x) = -f(x) \]

### Step 4: Check the behavior for \( |x| \geq 1 \)

For \( |x| \geq 1 \):
\[ f(x) = x |x| = x^2 \quad \text{(since } |x| = x \text{ for } x \geq 1 \text{ and } |x| = -x \text{ for } x \leq -1\text{)} \]

Now, we will check \( f(-x) \):
\[ f(-x) = -x |x| = -x^2 \quad \text{(for } x \geq 1, -x \leq -1\text{)} \]

### Step 5: Check if \( f(x) \) is odd for \( |x| \geq 1 \)

For \( |x| \geq 1 \):
\[ f(-x) = -x^2 \quad \text{and} \quad -f(x) = -x^2 \]

Thus, for \( |x| \geq 1 \):
\[ f(-x) = -f(x) \]

### Conclusion

Since \( f(-x) = -f(x) \) holds true for all \( x \), we conclude that \( f(x) \) is an odd function.

### Step 6: Check the derivative

To check if \( f'(x) \) is even, we differentiate \( f(x) \) in both cases:

1. For \( |x| < 1 \):
\[ f'(x) = 2x \sin\left(\frac{\pi}{2} x\right) + x^2 \frac{\pi}{2} \cos\left(\frac{\pi}{2} x\right) \]

2. For \( |x| \geq 1 \):
\[ f'(x) = 2x \]

Now checking \( f'(-x) \):
- For \( |x| < 1 \):
\[ f'(-x) = 2(-x) \sin\left(-\frac{\pi}{2} x\right) + (-x)^2 \frac{\pi}{2} \cos\left(-\frac{\pi}{2} x\right) = -2x \sin\left(\frac{\pi}{2} x\right) + x^2 \frac{\pi}{2} \cos\left(\frac{\pi}{2} x\right) \]

- For \( |x| \geq 1 \):
\[ f'(-x) = 2(-x) = -2x \]

Thus, we find that \( f'(x) \) is even.

### Final Answer

- \( f(x) \) is an odd function.
- \( f'(x) \) is an even function.