Given the following two languages : \(L_1 = \{a^n b^n | n ≥ 0, n ≠ 100\}\)…

Answer: L1 is a context-free language, and L2 is not a context-free language.

• Reason for L1:

  The language {a^n b^n | n ≥ 0} is generated by the grammar S → a S b | ε, so it is context-free. L1 differs from this language only by removing the single string a^100 b^100. A finite set (in particular a single string) is regular, and context-free languages are closed under intersection with regular languages; equivalently, removing a regular set from a context-free language yields a context-free language. Therefore L1 is context-free.

• Reason for L2:

  Consider intersecting the language of all strings with equal counts of a, b, and c with the regular language a*b*c*. The intersection is {a^n b^n c^n | n ≥ 0}, which is a well-known non-context-free language. Since context-free languages are closed under intersection with regular languages, if the original language were context-free then the intersection would also be context-free, contradiction. Hence L2 is not context-free.

Conclusion: L1 is context-free; L2 is not context-free.