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Prove that a simple graph is connected if and only if it has a spanning tree.
Ans: First assume that a simple graph G has a spanning tree T. T consists of every node of G. By the definition of a tree, there is a path among any two nodes of T. As T is a subgraph of G, there is a path among each pair of nodes in G. Hence G is connected.
Here now let G is connected. If G is a tree then nothing to prove. If G is not a tree, it must consist of a simple circuit. Let G has n nodes. We can choose (n - 1) arcs from G in such type of a way that they not form a circuit. It results into a subgraph comprising all nodes and only (n - 1) arcs. So by definition this subgraph is a spanning tree.
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y=f(a^x) and f(sinx)=lnx find dy/dx Solution) dy/dx = (a^x)(lnx)f''(a^x), .........(1) but f(sinx) = lnx implies f(x) = ln(arcsinx) hence f''(x) = (1/arcsinx) (1/ ( ( 1-x
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