Solution

The correct answer is option4

Only the language L2 is regular.

Key Points

\({L_1} = \{ {a^{{{̇ z}^z}}}|̇ Z\;is\;an\;integer\} \)

∴ L₁ = {a¹, a^4, a²⁷, a²⁵⁶, .....}

L2 = {azż | Ż ≥ 0}

∴ L₂ = {ε, a¹¹, a²², a³³, ......}

L3 = {ωω | ω ϵ {a, b}* }

L₃ = set of all strings starting and ending with the same word.

Explanation:

L₁ does not contain any pattern to form a loop. Hence it's not possible to construct a DFA for L1.

L₂ is a regular language since a DFA can be constructed using a loop that after every 11 symbols, the automata reaches the final state.

In L₂, once the word 'ω' is scanned, there is no way to compare the next word with the starting symbol of the first word.