Question 27

1. Analyze Grammar 1 ($S_1 \to a S_1 b \mid \varepsilon$):

• Derivations:
  $$S_1 \implies \varepsilon$$
  $$S_1 \implies a S_1 b \implies a b$$
  $$S_1 \implies a S_1 b \implies a^2 S_1 b^2 \implies a^2 b^2$$
  In general: $S_1 \implies^* a^n b^n$ for all $n \ge 0$.
• Thus, $L_1 = \{a^n b^n \mid n \ge 0\}$.
• By the Pumping Lemma for regular languages, $L_1$ requires an unbounded stack to match the number of $a$'s and $b$'s. It cannot be accepted by any finite automaton. Hence, $L_1$ is not regular (statement $P$ is FALSE).

2. Analyze Grammar 2 ($S_2 \to a b S_2 \mid \varepsilon$):

• Derivations:
  $$S_2 \implies \varepsilon$$
  $$S_2 \implies ab S_2 \implies ab$$
  $$S_2 \implies ab S_2 \implies (ab)^2 S_2 \implies (ab)^2$$
  In general: $S_2 \implies^* (ab)^n$ for all $n \ge 0$.
• Thus, $L_2 = \{(ab)^n \mid n \ge 0\} = (ab)^*$.
• The regular expression $(ab)^*$ describes $L_2$, which is accepted by a simple 2-state DFA. Hence, $L_2$ is regular (statement $Q$ is TRUE).

Therefore, $P$ is false and $Q$ is true, corresponding to option (C).