Correct Option: D

LL is recursively enumerable means a TMTM accepts all strings in LL. LL is recursively enumerable means a TMTM accepts all strings in L¯L¯. So, we can always decide if a string is in LL or not, making LL recursive.
If a language L and its complement L are both recursively enumerable, then both languages are recursive. If L is recursive, then L is also recursive, and consequently both are recursively enumerable.
Proof : If L and L are both recursively enumerable, then there exist. Turing machines M and M̂ that serve as enumeration procedures for L and L, respectively.
The first will produce w₁, w₂, ..... in L, the second w₁̂ , w₂̂ , ....... in L. Suppose now we are given any . w ∈ &sum⁺ We first let M generate w₁ and compare it with w. If they are not the same, we let M̂ generate w₁̂ and compare again. If we need to continue, we next let M generate w₂, then M̂ generate w₂̂, and so on.
Any w ∈ ∑⁺ will be generated by either M or M̂ , so eventually we will get a match. If the matching string is produced by M, w belongs to L, otherwise it is in L̂ . The process is a membership algorithm for both L and L. The process is a membership algorithm for both L and L, so they are both recursive. For the converse, assume that L is recursive. Then there exists a membership algorithm for it. But this becomes a membership algorithm for L by simply complementing its conclusion. Therefore, L is recursive. Since any recursive language is recursively enumerable, the proof is completed.