If `f(x)={(x",", "if x is rational "),(-x",","if x is irrational"):}`, then

Options

A. f(x) is an odd function
B. f(x) is continuous at `x=(1)/(2)`
C. f(x) is continuous at x = 0
D. f(x) is periodic function

Explanation

To determine the properties of the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} x & \text{if } x \text{ is rational} \\ -x & \text{if } x \text{ is irrational} \end{cases} \]

we will analyze whether the function is odd, continuous, and periodic.

### Step 1: Check if \( f(x) \) is an odd function

A function \( f(x) \) is odd if \( f(-x) = -f(x) \) for all \( x \).

1. **Case 1: \( x \) is rational**
- Then \( f(x) = x \).
- Now, \( -x \) is also rational, so \( f(-x) = -x \).
- Check:
\[ f(-x) = -x = -f(x) \]
- This condition holds.

2. **Case 2: \( x \) is irrational**
- Then \( f(x) = -x \).
- Now, \( -x \) is also irrational, so \( f(-x) = x \).
- Check:
\[ f(-x) = x = -(-x) = -f(x) \]
- This condition also holds.

Since both cases satisfy the condition for odd functions, we conclude that \( f(x) \) is an odd function.

### Step 2: Check for continuity

A function is continuous at a point \( c \) if:
\[ \lim_{x \to c} f(x) = f(c) \]

We will check continuity at \( x = 0 \) and \( x = \frac{1}{2} \).

1. **At \( x = 0 \)**:
- \( f(0) = 0 \) (since 0 is rational).
- Calculate the limit:
- For rational \( x \):
\[ \lim_{x \to 0, x \text{ rational}} f(x) = \lim_{x \to 0} x = 0 \]
- For irrational \( x \):
\[ \lim_{x \to 0, x \text{ irrational}} f(x) = \lim_{x \to 0} -x = 0 \]
- Since both limits equal \( f(0) \):
\[ \lim_{x \to 0} f(x) = 0 = f(0) \]
- Thus, \( f(x) \) is continuous at \( x = 0 \).

2. **At \( x = \frac{1}{2} \)**:
- \( f\left(\frac{1}{2}\right) = \frac{1}{2} \) (since \( \frac{1}{2} \) is rational).
- Calculate the limit:
- For rational \( x \):
\[ \lim_{x \to \frac{1}{2}, x \text{ rational}} f(x) = \lim_{x \to \frac{1}{2}} x = \frac{1}{2} \]
- For irrational \( x \):
\[ \lim_{x \to \frac{1}{2}, x \text{ irrational}} f(x) = \lim_{x \to \frac{1}{2}} -x = -\frac{1}{2} \]
- Since the limits are not equal:
\[ \lim_{x \to \frac{1}{2}} f(x) \neq f\left(\frac{1}{2}\right) \]
- Thus, \( f(x) \) is not continuous at \( x = \frac{1}{2} \).

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

A function \( f(x) \) is periodic if there exists a non-zero \( p \) such that \( f(x + p) = f(x) \) for all \( x \).

- Given \( f(x) \) takes different values based on whether \( x \) is rational or irrational, we can see that:
- For \( x = 1 \) (rational), \( f(1) = 1 \).
- For \( x = \sqrt{2} \) (irrational), \( f(\sqrt{2}) = -\sqrt{2} \).
- The values do not repeat in a consistent manner across intervals.

Thus, \( f(x) \) is not periodic.

### Conclusion

- \( f(x) \) is an odd function.
- \( f(x) \) is continuous at \( x = 0 \) but not at \( x = \frac{1}{2} \).
- \( f(x) \) is not periodic.