Answer

Correct Option: (i) $\{a^n b^n \mid n \ge 0\}$

Explanation

• (ii) $\{a,b\}^*$ → Regular, regex $(a+b)^*$, DFA with 1 state.
• (iii) Finite languages → Regular: any finite set is regular (union of finitely many strings).
• (iv) Languages accepted by DFA → Definition of regular.
• (i) $\{a^n b^n\}$ → Classic non-regular. Requires counting unbounded $n$ and matching $a$'s with $b$'s. Pumping Lemma: let $p$ be pumping length, $w = a^p b^p$, pumping $y$ (all $a$'s) breaks equality $|w|_a = |w|_b$.

It is Context-Free (PDA can push $a$'s, pop for $b$'s) but not regular.

Final Answer

NOT regular is $\{a^n b^n\}$ → Option (i).