Language $L_1$ is defined by the grammar: $S_1 \to aS_1b \mid \varepsilon$ Language $L_2$ is defined by the grammar: $S_2 \to abS_2 \mid \varepsilon$ Consider the following statements: $P$: $L_1$ is regular $Q$: $L_2$ is regular Which one of the following is $\textbf{TRUE}$?

1. Language $L_1$: The production $S_1 \to aS_1b \mid \varepsilon$ generates the language:

   $$L_1 = \{a^n b^n \mid n \ge 0\}$$

   $L_1$ is a standard non-regular context-free language (it requires a stack to match the number of $a$'s and $b$'s). Therefore, statement $P$ is false.

2. Language $L_2$: The production $S_2 \to abS_2 \mid \varepsilon$ is a right-linear grammar that generates:

   $$L_2 = \{(ab)^n \mid n \ge 0\} = (ab)^*$$

   Since $L_2$ can be expressed by a regular expression $(ab)^*$, it is regular. Therefore, statement $Q$ is true.

Hence, $P$ is false and $Q$ is true, making (C) the correct option.