Consider the following languages : \(L_1 = \{a^m b^n | m ≠ n\} \\ L_2 = \{a^m…

Answer: All three languages are context-free.

• L1 = {a^m b^n | m ≠ n}:

  Split L1 as the union of strings with m>n and strings with m<n.

  For m>n every string can be written as a^k a^n b^n = a^k (a^n b^n) with k≥1, which is the concatenation of a regular language a^+ and the CFL {a^n b^n}, hence a CFL. Similarly for m<n every string is (a^n b^n) b^+.

  Therefore L1 is a union of two CFLs and is context-free.

• L2 = {a^m b^n | m = 2n+1}:

  A simple context-free grammar that generates exactly this language is

  S -> a | aa S b

  This produces a (base case) and for each application of the recursive rule adds two a's and one b, yielding a^{2n+1} b^n, so L2 is context-free.

• L3 = {a^m b^n | m ≠ 2n}:

  Construct a pushdown automaton that on each a pushes two stack symbols and on each b pops one symbol. There are three possible end behaviours:

  1. If after processing all input the stack has leftover symbols, then m>2n, accept.

  2. If during processing some b causes an underflow (attempting to pop from an empty region), move to an accepting 'underflow' state and accept the remainder, which corresponds to m<2n.

  3. If stack is exactly balanced at the end, then m=2n and the PDA should reject; otherwise accept. This PDA construction shows L3 is context-free.

  Thus L3 is a CFL.

Conclusion: All three languages are context-free, so the correct choice is that each of L1, L2 and L3 is context-free.