If `f(x)={x^2 sin((pi x)/2), |x|<1`; `x|x|, |x|>=1` then `f(x)` is

Options

A. an even function
B. an odd function
C. a periodic function
D. None of these

Explanation

To determine whether the function \( f(x) \) is odd, even, or neither, we will analyze the function defined piecewise as follows:

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

### Step 1: Check for oddness in the first piece \( f(x) = x^2 \sin\left(\frac{\pi x}{2}\right) \) for \( |x| < 1 \)

To check if \( f(x) \) is odd, we need to evaluate \( f(-x) \):

\[ f(-x) = (-x)^2 \sin\left(\frac{\pi (-x)}{2}\right) = x^2 \sin\left(-\frac{\pi x}{2}\right) \]

Using the property of sine, \( \sin(-\theta) = -\sin(\theta) \):

\[ f(-x) = x^2 \left(-\sin\left(\frac{\pi x}{2}\right)\right) = -x^2 \sin\left(\frac{\pi x}{2}\right) = -f(x) \]

### Step 2: Conclusion for the first piece

Since \( f(-x) = -f(x) \) for \( |x| < 1 \), this part of the function is odd.

### Step 3: Check for oddness in the second piece \( f(x) = x |x| \) for \( |x| \geq 1 \)

Now, we evaluate \( f(-x) \) for this piece:

\[ f(-x) = (-x)(|-x|) = (-x)(x) = -x^2 \]

### Step 4: Conclusion for the second piece

For \( |x| \geq 1 \):

\[ f(-x) = -x^2 = -f(x) \]

This shows that this part of the function is also odd.

### Final Conclusion

Since both pieces of the function \( f(x) \) are odd, we conclude that the entire function \( f(x) \) is odd.

Thus, the final answer is:

**\( f(x) \) is an odd function.**

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