A periodic function with period $2\pi$ is defined by $f(x)=1$ in the interval $ a\lt x \lt b$ and $f(x)=0$ elsewhere.
Can the function be even or odd? If not why not and if so, for what values of $a$ and $b$?
I know that $f(-x)=f(x)$ is even and $f(-x)=-f(x)$ is odd. However, I can't see a way to apply this rule here to find a solution. I don't know any other way of figuring out whether a function is even or odd, all help is greatly appreciated!
On
It surely cannot be odd for any $a,b \in \mathbb{R}$ because $f(x) \geq 0$ for all $x \in \mathbb{R}$, and $f(x) = 1$ for some $x$, but never $f(x) = -1$.
By periodicity we can assume that $a,b \in [-\pi; \pi)$. Now $f$ is even for all values $a,b$ such that $a=-b$. That is because for all $c \in (-b,b)$ we get that f(c) = 1 = -f(c).
I suppose that your definition of the function is for the interval $[0,2\pi)$ and $0<a<b<2\pi$, so that the function has period $2\pi$. Now note that the function cannot be odd since it is always $\ge 0$. It can be even if $2\pi -b =a$.
If the definition is for an interval $[\alpha,\alpha+2\pi)$ you have to translate of $\alpha$ and you find: $b=2(\alpha+\pi)-a$ and the function is even if $\alpha=k\pi$.