(a) (1) by removing {(a,b)(e,f)(b,d)(a.e)(a,d)} graph is not disconnected , it will give 0 components

(2) by removing {(b,d)(c,f)(a,b)} graph will be disconnected into two components k1{bc} ,k2{aefd}

(b) if we want to disconnect then we should take any vertice which hav min degree and disconnect the edge , whch are connected to that min degree virtice , it will give one isolated vertice

therefore we can say that min-cut of graph = min degree of graph=2

(c) as i have done above question (b) lets min-cut have cardinality k , assume that min degree vertice have degree =k

in question it has been asked that atleast no . of edges so, lets assume that all the virtices have k(min degree) so , by hand shake theoram 2*edges = n *k => edges = (nk)/2