Answer -(b) Turing-recognizable but may not be decidable
Turing-recognizable languages (recursive enumerable languages) are closed under the UNION operation. so, (L1 U L2) will be Turing-recognizable language.
but, we think for (L1 U L2) will decidable : then according to definition of decidable(recursive)languages holding both the following conditions---
- (L1 U L2 ) will be Turing-recognizable.
- (L1 U L2)' will be Turing-recognizable .
here, (L1 U L2 ) will be Turing-recognizable, but (L1 U L2)' will not Turing-recognizable. because,
(L1 U L2)' = L1' ⋂ L2'
L1' and L2' will not be Turing recognizable because Turing-recognizable languages (recursive enumerable languages) are not closed under the COMPLEMENT operation. so, (L1 U L2)' will not be Turing-recognizable .
so, L1 U L2 may not be decidable .