in Algorithms
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Kindly justify with an example

in Algorithms
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(c) is false if the given graph is connected and has (n-1) edges.
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Manu Thakur ,Sir, so every MST will have emax included in it... because every MST has n-1 edges ... am i right sir ?

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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 $e_{max}$.

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prateekdwv  SIr .. Thanks :)

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