The function f(x) given by
`f(x)={{:(x^(4)tan""(pix)/(2),|x| lt1),(x|x|,|x| ge1):} is `

Options

A. an odd function
B. an even function
C. a periodic function with period `(2pi)/(sqrt(5))`
D. none of these

Explanation

To determine whether the function \( f(x) \) is odd, even, or periodic, we will analyze the function given by:

\[ f(x) = \begin{cases} x^4 \tan\left(\frac{\pi x}{2}\right) & \text{if } |x| < 1 \\ x |x| & \text{if } |x| \geq 1 \end{cases} \]

### Step 1: Analyze the function for \( |x| < 1 \)

For \( |x| < 1 \):
\[ f(x) = x^4 \tan\left(\frac{\pi x}{2}\right) \]

Now, we will find \( f(-x) \):
\[ f(-x) = (-x)^4 \tan\left(\frac{\pi (-x)}{2}\right) = x^4 \tan\left(-\frac{\pi x}{2}\right) \]

Using the property of the tangent function, \( \tan(-\theta) = -\tan(\theta) \):
\[ f(-x) = x^4 \left(-\tan\left(\frac{\pi x}{2}\right)\right) = -x^4 \tan\left(\frac{\pi x}{2}\right) = -f(x) \]

### Step 2: Analyze the function for \( |x| \geq 1 \)

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

Now, we will find \( f(-x) \):
- If \( x \geq 1 \):
\[ f(-x) = -(-x)^2 = -x^2 = -f(x) \]

- If \( x \leq -1 \):
\[ f(-x) = -(-x)^2 = -x^2 = -f(x) \]

### Step 3: Conclusion

From both cases, we can conclude that:
\[ f(-x) = -f(x) \quad \text{for all } x \]

This means that the function \( f(x) \) is an **odd function**.

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

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