Let L₁ and L₂ be two languages over a finite alphabet, such that L₁∩L₂ and L₂ are regular. Which of the following statements is/are always true?

The correct option is: L₂ is context-free.

Let us analyze the given information:
We are given two languages, L₁ and L₂, over a finite alphabet, such that:
1. L₁ ∩ L₂ is regular.
2. L₂ is regular.

According to the Chomsky hierarchy of formal languages, there is a strict containment relationship between different classes of languages. Specifically, the class of regular languages is a subset of the class of context-free languages. This relationship can be expressed as:
$$\text{Regular Languages}\subset \text{Context-Free Languages}$$
Since the problem statement explicitly establishes that L₂ is a regular language, it mathematically guarantees that L₂ is also a context-free language. Therefore, the statement "L₂ is context-free" must always be true.

To show why the other statements are not always true, we can consider a counterexample:
Let L₂ = ∅ (the empty set), which is a regular language.
Let L₁ be any non-context-free language (for example, the copy language $\{ww\mid w\in {\{a,b\}}^{*}\}$).
In this case, the intersection of the two languages is:
$$L_1\cap L_2=L_1\cap \varnothing =\varnothing$$
Since the empty set is regular, the condition that L₁ ∩ L₂ and L₂ are regular is satisfied. However, under these conditions:
- L₁ is not regular.
- L₁ is not context-free.
- L₁ ∪ L₂ = L₁ ∪ ∅ = L₁, which is not regular.
This demonstrates that the only statement that is guaranteed to always hold true is that L₂ is context-free.