To solve this problem, we need to understand the properties of odd and even functions.

Odd Function Property:

A function $f(x)$ is said to be odd if for all $x$ in the domain of $f$: $f(-x) = -f(x)$

Now, let's analyze each option:

1. Option A: $f(f(x))$
   Since $f(x)$ is odd, we know that $f(-x) = -f(x)$. However, this property does not give us information about whether $f(f(x))$ will be even or odd because it depends on the nature of $f(f(x))$ itself. If $f(x)$ is odd, $f(f(x))$ is not necessarily even.

2. Option B: $f(x - 1)$
   Shifting an odd function by any constant does not guarantee that the resulting function will be even. So, $f(x - 1)$ is not necessarily even.

3. Option C: None of these
   This option will be true if none of the other options result in an even function.

4. Option D: $f(x + 1)$
   Similarly to option B, shifting the function by a constant does not make an odd function even. Thus, $f(x + 1)$ is not necessarily even.

5. Option E: $-f(x)$
   The function $-f(x)$ is the negative of an odd function. However, negating an odd function still results in an odd function (since $-(-f(x)) = f(x)$). So, $-f(x)$ is still an odd function and not necessarily even.

Conclusion:

The correct answer is C. None of these because none of the given options must necessarily be even.

Would you like more details or have any questions?

Related Questions:

1. What are the properties of even functions?
2. How do you determine if a function is even, odd, or neither?
3. Can a function be both even and odd?
4. What happens when you add or multiply two odd functions?
5. How does the graph of an odd function differ from an even function?

Tip:

Remember that shifting, negating, or composing functions doesn't necessarily change their parity (odd or even nature).