Let `f: R to Q` be a continuous function such that f(3) = 10, then

Options

A. f(x) is always an even function
B. f (x) is always an odd function
C. nothing can be said about f (x)being even or odd
D. f (x) is an increasing function

Answer

To solve the problem, we need to analyze the function \( f: \mathbb{R} \to \mathbb{Q} \) given that it is continuous and satisfies the condition \( f(3) = 10 \). We want to determine if the function is always even, odd, or if we cannot say anything about it.

### Step-by-Step Solution:

1. **Understanding the Definitions**:
- A function \( f(x) \) is **even** if \( f(-x) = f(x) \) for all \( x \).
- A function \( f(x) \) is **odd** if \( f(-x) = -f(x) \) for all \( x \).

2. **Given Condition**:
- We know that \( f(3) = 10 \).

3. **Testing for Evenness**:
- To check if \( f \) is even, we need to see if \( f(-3) = f(3) \).
- Since \( f(3) = 10 \), for \( f \) to be even, we would need \( f(-3) = 10 \).

4. **Testing for Oddness**:
- To check if \( f \) is odd, we need to see if \( f(-3) = -f(3) \).
- Since \( f(3) = 10 \), for \( f \) to be odd, we would need \( f(-3) = -10 \).

5. **Conclusion on Evenness and Oddness**:
- We cannot determine \( f(-3) \) from the given information. The function could be either even, odd, or neither depending on its specific definition.
- Therefore, we cannot conclude that the function is always even or always odd.

6. **Considering Increasing or Decreasing**:
- The problem does not provide any information about the behavior of the function other than the value at one point.
- A continuous function can be increasing, decreasing, or constant. Thus, we cannot say anything about whether the function is increasing or decreasing.

7. **Final Conclusion**:
- Since we cannot determine if the function is even or odd, and we also cannot conclude whether it is increasing or decreasing, the answer is that we cannot say anything definitive about the function.

### Final Answer:
The correct option is that we cannot say anything about \( f(x) \) being even or odd.