If $f(x)$ is an even function, and $f(g(x))$ is an even function, then $g(x)$ MUST be?

Question

If $f(x)$ is an even function, and $f(g(x))$ is an even function, then $g(x)$ must be:

(a) odd

(b) even

(c) general

(d) even or odd

(e) even or general

(f) even or odd or general

This question is in one of my college textbooks and the answer to it was "even or odd", what I have a problem with is the use of the word MUST, because $g(x)$ doesn't have to be either odd or even for the function $f(g(x))$ to be even. An example of this would be as follows:

$$ f(x) = \cos(x) $$

$$ g(x) = x + \pi $$

Then:

$$ h(x) = f(g(x)) = \cos(x + \pi) = -\cos(x) $$

We can see that the function $h(x)$ is even, now is there something that I don't understand? or simply the choice of the word "must" was poor?

Answer 1

$g$ could be anything.

Take $f\equiv 0$.You have $f(x)=f(-x)$ for all $x$ and also $f(g(x))=f(g(-x))$ for any function $g$.

EDIT: As Dominique pointed out in his comment, you can take $f$ to be any constant.

Answer 2

I believe that $g(x)=x$ (odd) and $f(x)=|x|$ (for $f(x)=|x|$ is an even function) are perfect counter-examples, aren't they? :-)