Consider the following statements. I. The complement of every Turing decidable language is Turing decidable II. There exists some language which is in NP but is not Turing decidable III. If $L$ is a language in NP, $L$ is Turing decidable Which of the above statements is/are true?

Let us evaluate each statement:

• Statement I: *The complement of every Turing decidable language is Turing decidable.* Turing decidable (recursive) languages are closed under complementation. If a Turing machine $M$ decides $L$, a machine $M^{\prime}$ that swaps accept and reject states decides the complement $\overline{L}$. Thus, Statement I is TRUE.

• Statement II: *There exists some language which is in NP but is not Turing decidable.* By definition, the complexity class $\text{NP}$ is the class of languages decidable by a non-deterministic Turing machine in polynomial time ($O(n^k)$). Any language in $\text{NP}$ can also be decided by a deterministic Turing machine in exponential time ($O(2^{n^k})$), meaning $\text{NP} \subseteq \text{EXPTIME} \subset \text{Decidable}$. Thus, all languages in $\text{NP}$ are decidable. No undecidable language exists in $\text{NP}$. Thus, Statement II is FALSE.

• Statement III: *If $L$ is a language in NP, $L$ is Turing decidable.* Since $\text{NP} \subseteq \text{EXPTIME} \subset \text{Decidable}$, every language in $\text{NP}$ is indeed Turing decidable. Thus, Statement III is TRUE.

Hence, statements I and III are true, which corresponds to option (D).