Correct : a,b,d

Since all edge weights are unique, the MST of G is unique. The following properties apply:
Option A — True (Cycle Property)
The cycle property states that the maximum weight edge in any cycle is never part of any MST. If the heaviest edge in a cycle were included in a spanning tree, removing it and adding any other edge from that cycle would produce a lighter spanning tree. Since edge weights are unique, this heavy edge is excluded from every MST. A is true.
Option B — True
Take any cycle C and its minimum weight edge e. Suppose e is not in some MST T. Then T must contain an alternative path between e''s two endpoints. That path together with e forms a new cycle. In this new cycle, e is still the minimum weight edge. By the cycle property, the heaviest edge in this new cycle — call it f — cannot be in any MST. But f is part of T, giving a contradiction. So the minimum weight edge of every cycle must be in every MST. B is true.
Option C — False
Consider a leaf vertex in a star-shaped graph that has only one incident edge. That single edge is simultaneously the largest and only edge incident on that vertex. Since removing it disconnects the graph, every spanning tree must include it. So the largest edge incident on a vertex can and must be in the MST. C is false.
Option D — True (Cut Property)
For any vertex v, the minimum weight edge incident on v is the lightest crossing edge for the cut ({v}, V−{v}). The cut property guarantees that the minimum weight crossing edge of any cut is in every MST. So the minimum weight edge incident on every vertex must be in every MST. D is true.
Correct answer: A, B, D ✓