If $$L_1$$​ is a regular language and $$L_2$$​ is a context free language, which of the following statement is correct?

Correct option is B

Given:
· $$L_1$$​ is a Regular Language (RL)
· $$L_2$$​ is a Context-Free Language (CFL)
We need to determine which statement is always true.
A fundamental closure property of context-free languages is:
​$$\boxed{\text{CFL} \cap \text{Regular Language} = \text{CFL}}$$​​
Since $$L_1$$​ is regular and $$L_2$$​ is context-free,
​$$L_1\cap L_2$$​ is guaranteed to be a context-free language. This can be established by constructing a product of:
· a finite automaton recognizing $$L_1,$$​ and
· a pushdown automaton (PDA) recognizing $$L_2.$$​​
The resulting PDA keeps track of both the finite-state information of $$L_1$$​ and the stack information of $$L_2.$$​ Hence, it recognizes $$L_1\cap L_2.$$​​
Why the Other Options Are Incorrect?
(a) Any subset of $$L_1\cup L_2$$​ is context-free — Incorrect
Although the union $$L_1\cup L_2$$​ is context-free because CFLs are closed under union, not every subset of a CFL is necessarily context-free. For example, a CFL can contain a non-context-free language as a subset.
(c) $$L_1\cap L_2$$​ is a regular language — Incorrect
The intersection of a regular language with a CFL is guaranteed to be context-free, but it need not be regular. For example, let:
$$L_1=\{0,1\}^*$$​ which is regular, and $$L_2=\{0^n1^n\mid n\geq0\}$$​ which is context-free but not regular. Then:
$$L_1\cap L_2=L_2=\{0^n1^n\mid n\geq0\}$$​ which is not regular.
(d) $$L_1\cap L_2$$​ is accepted by a deterministic PDA — Incorrect
A context-free language is generally accepted by a non-deterministic PDA (NPDA). Not every CFL is deterministic context-free. Therefore, we cannot guarantee that $$L_1\cap L_2$$​ is accepted by a DPDA.
Information Booster
1. Regular languages are closed under: union, intersection, complement, difference, concatenation, Kleene star, etc.
2. Context-free languages are closed under: union, concatenation, Kleene star, and intersection with a regular language.
3. CFLs are not closed under intersection with another CFL.
4. CFLs are not closed under complement.
5. Every regular language is a CFL, but every CFL is not necessarily regular.
Additional Knowledge
​$$\boxed{RL\cap CFL=CFL}$$​​
but generally:
​$$CFL_1\cap CFL_2\notin CFL$$​​
and:
​$$CFL\not\subseteq DCFL$$​​
Thus, the guaranteed statement is:
​$$\boxed{\text{(b) }L_1\cap L_2\text{ is a context-free language}}$$​​