This question tests your understanding of the equivalence between DFAs and NFAs. Consider NFA M = ({q_1, q_2), {0. 1), delta, q_1, {q_1}) for delta defined as: Draw both the state diagrams for M and for a DFA M' equivalent to M based on the construction of Theorem 1.39 in the text (recall the latter proves that DFAs and NFAs are equivalent)