Correct answer : A
Explanation:-
Option A: e is the minimum weight edge among all the edges connecting X and Y, so there may be some other edge having weight less than e but belonging to X or Y. However, some edge connecting X and Y has to be included in the MST (otherwise the MST will become disconnected) and this edge will be e because it is the minimum among all the edges connecting X and Y.
Option B: There are two ways to view this option -
i) The path(s) between s and t containing e may not be the shortest path(s) between s and t
ii) The shortest path between s and t may not necessarily contain e.
This is because a path s->x->y->t connecting s and t, (where x->y is an edge connecting the partitions X and Y, s->x is a path from s to x and similarly y->t is a path from y to t) is not depending solely upon x->y. It is also depending upon s->x and y->t. Now even if we minimize x->y by choosing e, there is no guarantee that the other two paths are also minimized. So the minimum path will not necessarily contain e
Option C: e is not the only edge containing X and Y and hence a path from s to t can contain another edge.
Option D: Again, there can be paths from s to t which do not contain e but are more expensive.