Correct : a

A vertex cover S of size k must cover every edge in the graph — for each edge at least one of its endpoints must be in S.
Option A — True: degree of v ≥ k+1 forces v ∈ S
If v has k+1 or more neighbours and v is not in S, then every single edge from v to each of its k+1 neighbours must be covered by the neighbour''s side. That means all k+1 neighbours must be in S. But S has only k vertices total — it cannot hold k+1 vertices just for v''s neighbours. This contradiction means v must be in S. Option A always guarantees v ∈ S.
Option B — False: being on a path of length k+1 does not force v ∈ S
A path of length k+1 has k+1 edges and k+2 vertices. Its minimum vertex cover picks every alternate vertex, needing only ⌊(k+1)/2⌋ + 1 vertices. The vertex v could be one of the endpoints or an unselected alternate vertex and still not be needed in the cover. B does not always force v ∈ S.
Option C — False: being on a cycle of length k+1 does not force v ∈ S
A cycle of length k+1 can be covered by picking ⌊(k+1)/2⌋ vertices and skipping others. Vertex v can be one of the skipped vertices. C does not always force v ∈ S.
Option D — False: being in a clique of size k does not force v ∈ S
A clique of size k needs only k−1 vertices in its cover — all vertices except one. Vertex v could be the one excluded from the cover. D does not always force v ∈ S.
Correct answer: A ✓