Counterexample of option (c): Consider any arbitrary graph with a pendant vertex (say $\mathcal{A}$). Also, let $e_{max}$ of this graph will be the pendant edge (edge incident on $\mathcal{A}$). There is only one way to connect pendant vertex to the rest of the graph, i.e., through $e_{max}$. Hence MST of such a graph must contain $e_{max}$.