A function f(x) given by
`f(x)={{:(x^(2)sin""(pix)/(2), |x| lt1),(x|x|,|x| ge1):}is`

Options

A. Even function
B. Odd function
C. Periodic function
D. Neither even nor odd

Answer

To determine whether the function \( f(x) \) is odd or even, we need to analyze the function defined piecewise:

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

### Step 1: Check if the function is odd
A function \( f(x) \) is odd if \( f(-x) = -f(x) \) for all \( x \).

#### Case 1: \( |x| < 1 \)
For \( |x| < 1 \), we have:
\[ f(x) = x^2 \sin\left(\frac{\pi x}{2}\right) \]
Now, we calculate \( f(-x) \):
\[ f(-x) = (-x)^2 \sin\left(\frac{\pi (-x)}{2}\right) = x^2 \sin\left(-\frac{\pi x}{2}\right) \]
Using the property of sine (i.e., \( \sin(-\theta) = -\sin(\theta) \)):
\[ f(-x) = x^2 \left(-\sin\left(\frac{\pi x}{2}\right)\right) = -x^2 \sin\left(\frac{\pi x}{2}\right) = -f(x) \]
Thus, for \( |x| < 1 \), \( f(-x) = -f(x) \).

#### Case 2: \( |x| \geq 1 \)
For \( |x| \geq 1 \), we have:
\[ f(x) = x |x| \]
Now, we calculate \( f(-x) \):
\[ f(-x) = -x |-x| = -x |x| = -f(x) \]
Thus, for \( |x| \geq 1 \), \( f(-x) = -f(x) \).

### Conclusion
Since in both cases \( f(-x) = -f(x) \), we conclude that the function \( f(x) \) is an odd function.

### Final Answer
The given function \( f(x) \) is an odd function.

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