Evasiveness of acyclicity of undirected graph
Consider any algorithm for acyclicity. We will use the following adversary:


  Whenever the algorithm asks for an edge, say that the edge exists if it connects two different connected components. (That is, if it doesn't close a cycle.)


You can show that the graph constructed by this dialog is always a forest, and moreover the constructed graph has the same connected components as the query graph (which contains all queried edges). In particular, when the last edge is asked, the graph is a tree (assuming $n \geq 3$; otherwise every graph is acyclic), and the graph is acyclic if and only if this edge does not belong to the graph.