Let $L_1$ and $L_2$ be two languages over a finite alphabet, such that $L_1 \cap L_2$ and $L_2$ are regular languages. Which of the following statements is/are always true?

We are given that $L_2$ and $L_1 \cap L_2$ are regular languages over a finite alphabet $\Sigma$.

1. Option (C) is always true:

   • Regular languages are closed under complementation. Since $L_2$ is regular, its complement $\overline{L_2} = \Sigma^* \setminus L_2$ is also a regular language.

   • Every regular language is a context-free language (regular languages are a proper subset of context-free languages).

   • Therefore, $\overline{L_2}$ is always context-free.

2. Counterexample for Options (A), (B), and (D):

   • Let $L_2 = \emptyset$, which is a regular language.

   • For any arbitrary language $L_1$, the intersection $L_1 \cap L_2 = L_1 \cap \emptyset = \emptyset$, which is also regular, satisfying the given conditions.

   • If we choose $L_1 = \{a^n b^n c^n \mid n \ge 0\}$ (or any non-context-free / undecidable language):

     • $L_1$ is not regular (disproving A).

     • $L_1 \cup L_2 = L_1 \cup \emptyset = L_1$, which is not regular (disproving B).

     • $L_1$ is not context-free (disproving D).

Hence, only statement (C) is always true.