If f:A→B, A={1,2,3,4}, B={0,1}, and f(x)=0 for even x while f(x)=1 for odd x, what is f⁻¹({0})?

Correct answer: {2,4}

The inverse image or preimage f⁻¹({0}) contains the elements of the domain A whose function value belongs to the set {0}. We inspect each domain element: 1 is odd, so f(1)=1; 2 is even, so f(2)=0; 3 is odd, so f(3)=1; and 4 is even, so f(4)=0. Therefore the inputs that map to 0 are 2 and 4, giving f⁻¹({0})={2,4}. Option A is correct. Option B is the preimage of {1}, option C is not a subset of the given domain, and option D incorrectly includes inputs whose images are 1. A preimage consists of domain elements, not necessarily elements of the codomain.