Consider the language \(L = \{a^{n}\mid n \geq 0\} \cup \{a^{n}b^{n}\mid n…

Answer: Statements "L is deterministic context-free" and "L is not LL(k) for any k" are true; "L is context-free but not deterministic context-free" is false.

• Why L is deterministic context-free: A deterministic PDA works as follows. In a 'push' state push each input a. If EOF is reached while still in the push state, accept (this covers a^n). On the first b move to a 'pop' state and pop one a for each b; accept at EOF only if the stack has been popped exactly to the bottom marker (this covers a^n b^n). Any mismatch (extra b or leftover a in pop state) is rejected. All transitions are deterministic.

• Why L is not LL(k) for any k: For fixed k, consider the inputs a^{k+1} and a^{k+1}b^{k+1}. At the point after reading k a's the parser sees the same k-symbol lookahead in both cases, so a top-down parser with only k lookahead cannot decide whether to commit to the production that yields only a's or the production that yields a^n b^n. This ambiguity persists for every finite k, so L is not LL(k) for any k.