Consider the following languages :

L₁ = {a^m bⁿ | m ≠ n}

L₂ = {am bn | m = 2n + 1}

L₃ = {am bn | m ≠ 2n}

Which one of the following statement is correct ? 

The correct answer is option 4.

Key Points

L1 = {am bn | m ≠ n}

• It means m could be greater than (m>n) or less than 'n' (m<n)but m will not be equal to 'n'(m=n).
• Since for accepting L1, we need storage so then a number of 'a' can be stored in it and when b's come as input a's and b's can be compared.
• Since finite automata do not have a storage element hence language is not regular. 

L2 = {am bn | m = 2n + 1}

L2 rewrite as {a2n+1bn }.

Here skip the first 'a'. If the input string reaches an empty symbol then it accepts the language(string: a) or if the input symbol is 'a' stores all a's in the stack. After every 'b' pop the two a's until it reaches the empty symbol then it accepts the language.

Hence the language L₂  is CFL.

L3 = {am bn | m ≠ 2n}

It means m could be greater than (m>2n) or less than 'n' (m<2n) but m will not be equal to 'n'(m=n). Store all a's in the stack and for two b's pop one 'a'. If if there any a's in the stack (m>2n) or the top of the stack is Z₀ and the input symbol is not reached to the empty symbol (m<2n) then it is accepting the language. 

Hence the language L3  is CFL.

Hence the correct answer is L1, L2 and L3 are context-free languages.