Let $f:\{0,1\}^n\to\{0,1\}$ be a boolean function computed by logical circuit comprising just binary AND and binary OR gates (assume that the circuit doesn't have any feedback). Let $PARITY:\{0,1\}^n\to\{0,1\}$ be the boolean function that outputs $1$ iff the total number of $1$'s in the input bits is odd. Similarly let, $MAJORITY:\{0,1\}^n\to\{0,1\}$ be the boolean function that outputs $1$ iff the total number of $1$'s in the input bits is at least as large as the total number of $0$'s in the input bits. which of the following is NOT possible?
$1.\ f(0,0,...,0)=f(1,1,...,1)=0\\ 2.\ f(0,0,...,0)=f(1,1,...,1)=1\\ 3.\ f\ \mathrm{is\ a\ MAJORITY\ function}\\ 4.\ f\ \mathrm{is\ a\ PARITY\ function}\\ 5.\ f\ \mathrm{outputs\ 1\ at\ exactly\ one\ assignment\ of\ the\ input\ bits}$
The fact that's bothering me most is that if $f$ is implemented using only AND and OR gates then $f(0,0,...,0)$ and $f(1,1,...,1)$ should ALWAYS be $0$ and $1$ respectively. In that case how can options $1$ and $2$ hold ?
Generally the constant $1$ and constant $0$ functions are considered to be available here; this would be equivalent to ignoring the input (i.e., no gates at all) and running a current (1) or no current (0) to the output. I would expect that the context of the problem or course would make this clear, but sometimes question writers are sloppy.
One can show that the functions under consideration are exactly the boolean functions which do not change from $1$ to $0$ if you change one of their inputs from $0$ to $1$; these are called monotone boolean functions. Parity does not have this property, but constant 0, constant 1, Majority, and the function which is $1$ only for the all-ones input do.