If i remove edge b/w B and E the graph is acyclic.
then apply DFS starts from vertex A.
It results two back edges D-C and E-B.
most cases it results only one back edge and in certain case like above it results two also.
Hence it results multiple back edges.
Once u refer Coreman text book for DFS traversal,then u can easily understand.
parent node time intervals are u/v
and child is x/y then backside lies in b/w parent.i.e)u<=x<=y<=v // condition for back edges.
here is a result given in cormen book.....
A directed graph G is acyclic iff depth first search of G yields no back edges
i think b is correct
X->YZ , Y->XZ , ...