Answer
If $f(x)$ and $g(x)$ are both even, $f(x)+g(x)$ is even.
If $f(x)$ and $g(x)$ are both odd, $f(x)+g(x)$ is odd.
If $f(x)$ is even and $g(x)$ is odd, $f(x)+g(x)$ is neither even nor odd.
Work Step by Step
If $f(x)$ and $g(x)$ are both even,
$f(x)=f(-x)$ (i)
$g(x)=g(-x)$ (ii)
Let $S(x)=f(x)+g(x)$
$=f(-x)+g(-x)$ (from i) and ii))
$=S(-x)$
$\therefore f(x)+g(x)$ is even.
If $f(x)$ and $g(x)$ are both odd,
$-f(x)=f(-x), f(x)=-f(-x)$ (i)
$-g(x)=g(-x), g(x)=-g(-x)$ (ii)
Let $S(x)=f(x)+g(x)$
$=-f(-x)-g(-x)$ (from i) and ii))
$=-(f(-x)+g(-x))$
$=-S(x)$
$\because S(-x)=-S(x),$
$\therefore f(x)+g(x)$ is odd.
If $f(x)$ is even and $g(x)$ is odd,
$f(x)=f(-x)$ (i)
$-g(x)=g(-x), g(x)=-g(-x)$ (ii)
Let $S(x)=f(x)+g(x)$
$=f(-x)-g(-x)$ (from i) and ii))
$S(-x)=f(-x)+g(-x)$
$\ne \pm S(x)$
$\because S(-x)\ne \pm S(x)$
$\therefore f(x)+g(x)$ is neither even nor odd.
