Let 𝐿1 and 𝐿2 be two languages over a finite alphabet, such that 𝐿1∩𝐿2 and…

Given that L₂ is a regular language and the intersection L₁ ∩ L₂ is also regular.

Since every regular language is a context-free language, the intersection L₁ ∩ L₂ must be context-free. Therefore, the statement asserting that L₁ ∩ L₂ is context-free (Option C) is always true.

To verify why other options are not always true, consider the counter-example where L₂ = ∅ (the empty set). The empty set is regular. In this case, L₁ ∩ L₂ = ∅, which satisfies the condition regardless of what L₁ is. This means L₁ does not have to be regular or context-free, disproving options claiming constraints on L₁. Similarly, the union L₁ ∪ L₂ = L₁ in this case, so it is not necessarily regular.