[arlaness]

Let f(x) be an even function and g(x) be an odd function. Let h(x) =
f(x) + g(x). What can you say about h(x)?

(a) For all functions f(x) and g(x), h(x) is neither even nor odd.
(b) For all functions f(x) and g(x), h(x) is always even.
(c) There are functions f(x) and g(x) such that h(x) is both even and odd.
(d) There are non-zero functions f(x) and g(x) such that h(x) is even.
(e) All the statements from (a) to (d) are false.

can someone ans me this question? i'm confused as to which shld i choose as there are many contradictions.

[Yggdrasil]

For statements where it says "for all functions," you should test special cases to see whether it's true in those special cases.  For example, a good special case to test is f(x) = 0 and/or g(x) = 0.  Note that the zero function is both even and odd.  This can help you eliminate some choices.

[arlaness]

but it is said by mean individuals that 0 is considered an even number and ont an odd number.. which is why i'm confused.. i would like to choose (C) as my ans.. any one care to give comments?

[Yggdrasil]

Can you prove statement (c)?  (i.e. can you give an example of f(x) and g(x) such that h(x) is even or odd?)

[note: zero is an even integer.  An integer z is even if z has the form z = 2n, n an integer.  Zero is clearly of this form.  However, an interer z is odd if z has the form z = 2n + 1, n an integer.  Clearly no such n exists such that 0 = 2n +1, so zero is not an odd integer.  The zero function f(x) = 0, however, is both odd and even, since f(-x) = 0 = f(x) and -f(-x) = -0 = 0 = f(x).]

[arlaness]

as u said, since 0 is both even and odd, let's take f(x)=g(x)=0. frm h(x)=f(x)+g(x)=0 thus h(x) is both odd and even right?

[Yggdrasil]

Yup.  Just making sure you actually understood the answer and weren't guessing.

Good job.