Given the following two languages : L1 = {an bn | n \(\geq\) 1} ∪ {a} L2 = {w…

Correct statement: L1 is not deterministic and L2 is deterministic.

Reason for L1 (a^n b^n ∪ {a}):

• After reading the first symbol 'a', a machine must decide whether the input is the single string "a" or the beginning of a longer string of the form a^n b^n. A deterministic pushdown automaton (DPDA) cannot make that future-looking choice without nondeterminism.

• Any DPDA that, on the first 'a', commits to pushing expecting b's will fail to accept the single string "a"; any DPDA that commits to accepting immediately cannot handle strings that require matching b's. Thus recognizing this union requires nondeterministic choices, so the language is not deterministic.

Reason for L2 ( { w C w^R | w ∈ {a,b}* }):

• The marker C clearly separates the two halves. A DPDA can deterministically operate in two phases:

  1) Push phase: while reading the prefix w (before C), push each symbol onto the stack.

  2) Pop-and-match phase: upon reading C switch to a state that pops one stack symbol for each input symbol and checks it matches; accept if the stack is correctly emptied at the end.

• All transitions are determined by the current state, the input symbol (or the marker C), and the top of stack, so this is a deterministic procedure.

Conclusion: L1 requires nondeterminism and thus is not a deterministic context-free language; L2 is recognized by a deterministic PDA. Therefore the correct choice states that L1 is not deterministic and L2 is deterministic.