Answer

Correct Option: (ii) Multiple accepting states

Explanation

DFA definition: $M = (Q,\Sigma,\delta,q_0,F)$

• $q_0$: Exactly one start state → (i) multiple start states false. (NFA can have multiple, DFA has one).
• $F \subseteq Q$: Set of final/accepting states. Can be 0, 1, or multiple → (ii) true. Example even number of 0's DFA has 1 accept, but language $(a+b)^*ab$ needs 1, language “ends with 00 or 11” needs 2.
• Transitions: $\delta: Q \times \Sigma \rightarrow Q$ is total deterministic, no $\epsilon$-moves. $\epsilon$-transitions allowed only in $\epsilon$-NFA → (iii) false.

Final Answer

DFA can have multiple accepting states → Option (ii).