In lossless decomposition, the intersection of two decomposed relations will be the superkey of at least one decomposed relation.
R1∩R2 = B . It is not a super-key for any decomposed table .Hence lossy decomposition.
In Dependency preserving decomposition , All original dependencies must be implied in sub-relations
R1(A, B, D) preserves A -> B and A->D. But A-> C is not preserved by R1 or R2. hence not dependency preserving
Answer (C)
