Suppose [tex]f(x)[/tex] is a function such that if [tex]p \ \textless \ q[/tex], then [tex]f(p) \ \textless \ f(q)[/tex]. Which statement best describes [tex]f(x)[/tex]?

A. [tex]f(x)[/tex] can be odd or even.
B. [tex]f(x)[/tex] can be odd but cannot be even.
C. [tex]f(x)[/tex] can be even but cannot be odd.
D. [tex]f(x)[/tex] cannot be odd or even.



Answer :

Let's analyze the given information which states that for any two numbers \(p\) and \(q\) such that \(p < q\), \(f(p) < f(q)\). This tells us that \(f(x)\) is a strictly increasing function.

Now, let's consider what it means for a function to be even or odd:

1. Even function: A function \(f(x)\) is even if \(f(-x) = f(x)\) for all \(x\).

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

A function that is strictly increasing means that as \(x\) increases, \(f(x)\) always increases and never decreases.

### Considering whether \(f(x)\) can be an even function:

For \(f(x)\) to be even, the function must be symmetric about the y-axis such that \(f(-x) = f(x)\). However, this symmetry would imply an interval \([-a, a]\) where \(a > 0\):

- If \(p = -a\) and \(q = a\), then \(p < q\) as \(-a < a\).
- Since \(f(x)\) is strictly increasing, it must hold that \(f(-a) < f(a)\).

This would contradict our requirement for even functions since \(f(-a) \neq f(a)\). Thus, a strictly increasing function \(f(x)\) cannot be even because it cannot satisfy \(f(-x) = f(x)\) for all \(x\).

### Considering whether \(f(x)\) can be an odd function:

For \(f(x)\) to be odd, the function must be symmetric about the origin such that \(f(-x) = -f(x)\). Let's consider positive and negative values:

- If \(p = -a\) and \(q = 0\), with \(a > 0\), then:
- \(f(-a) < f(0)\).

- Similarly, if \(p = 0\) and \(q = a\), then:
- \(f(0) < f(a)\).

Since \(f(x)\) is strictly increasing, these relationships hold naturally. Suppose we include \(f(-a) = -f(a)\):
- If \(p < q\), then for \( f(p) < f(q)\), if \(p\) is negative and \(q\) is positive, \(f(p)\) will be a negative value when mirrored over the origin for odd functions, implying \(f(-p) = -f(p)\).

Thus, a strictly increasing function \(f(x)\) can indeed be odd because it can satisfy \(f(-x) = -f(x)\) for all \(x\).

### Conclusion:

Given the strict increasing nature of \(f(x)\) eliminates the possibility of \(f(x)\) being even while still allowing the possibility of \(f(x)\) being odd, the best statement that describes \(f(x)\) is:

- \(f(x)\) can be odd but cannot be even.

So the correct answer is:

[tex]\[ \boxed{2} \][/tex]