The question is not precise here. Since, minimal DFA has two states exactly one must be final. Because if
- No state is final, no strings can be accepted and for this only one state is required in the minimal DFA
- If both the states are final, then they can be merged to a single state and hence it won’t be a minimal DFA
Now, with one final state and two states
- if we make a transition on first $a$ to final state and stay there for any remaining number of $a’s$, the language we get is $L=\{a^n \mid n > 0\}$ which is $a^* – \{\epsilon\}$
- Like in above we do the transition must if the initial state is made final, then the only string accepted is $\epsilon$
The above two cases are ignored in the given options.
The remaining possibility is for each input symbol $a,$ the DFA transitions between the first and second states. Then,
- if initial state is final we get $L = \{a^n \mid n \text{ is even} \}$
- if initial state is not final we get $L = \{a^n \mid n \text{ is odd} \}$
So, none of the options is correct though D is the best option to pick.