Consider the language
L = { $${a^n}|n \ge 0$$ } $$ \cup $$ { $${a^n}{b^n}|n \ge 0$$ }
and the following statements.

I. L is deterministic context-free.
II. L is context-free but not deterministic context-free.
III. L is not LL(k) for any k.

Which of the above statements is/are TRUE?

Consider the following types of languages: $${L_1}:$$ Regular, $${L_2}:$$ Context-free, $${L_3}:$$ Recursive, $${L_4}:$$ Recursively enumerable. Which of the following is/are TRUE?

$$\,\,\,\,\,\,\,{\rm I}.\,\,\,\,\,\,\,$$ $$\overline {{L_3}} \cup {L_4}$$ is recursively enumerable
$$\,\,\,\,\,{\rm I}{\rm I}.\,\,\,\,\,\,\,$$ $$\overline {{L_2}} \cup {L_3}$$ is recursive
$$\,\,\,{\rm I}{\rm I}{\rm I}.\,\,\,\,\,\,\,$$ $$L_1^ * \cap {L_2}$$ is context-free
$$\,\,\,{\rm I}V.\,\,\,\,\,\,\,$$ $${L_1} \cup \overline {{L_2}} $$ is context-free

For any two languages L₁ and L₂ such that L₁ is context-free and L₂ is recursively enumerable but not recursive, which of the following is/are necessarily true?

I. $${\overline L _1}$$ (complement of L₁) is recursive
II. $${\overline L _2}$$ (complement of L₂) is recursive
III. $${\overline L _1}$$ is context-free
IV. $${\overline L _1} \cup {L_2}$$ is recursively enumerable

Consider the following statements.

$$\,\,\,$$ $${\rm I}.\,\,\,\,\,\,\,\,\,$$ The complement of every Turing decidable language is Turing decidable
$$\,$$ $${\rm II}.\,\,\,\,\,\,\,\,\,$$ There exists some language which is in $$NP$$ but is not Turing decidable
$${\rm III}.\,\,\,\,\,\,\,\,\,$$ If $$L$$ is a language in $$NP,$$ $$L$$ is Turing decidable

Which of the above statements is/are true?