taking up option A first :
We have, R(A, B, C, D) and the Functional Dependency set = {A→B, B→CD}.
Now we will try to decompose it such that the decomposition is a Lossless Join, Dependency Preserving and new relations thus formed are in BCNF.
We decomposed it to R1(A, B) and R2(B, C, D). This decomposition satisfies all three properties we mentioned prior.
taking up option B :
we have, R(A, B, C, D) and the Functional Dependency set = {A→B, B→C, C→D}.
we decomposed it as R1(A, B), R2(B, C) and R3(C, D). This decomposition too satisfies all properties as decomposition in option A.
taking up option D :
we have, R(A, B, C, D) and the Functional Dependency set = {A→BCD}.
This set of FDs is equivalent to set = {A→B, A→C, A→D} on applying decomposition rule which is derived from Armstrong's Axioms.
we decomposed it as R1(A, B), R2(A, C) and R3(A, D). This decomposition also satisfies all properties as required.
taking up option C :
we have, R(A, B, C, D) and the Functional Dependency set = {AB→C, C→AD}.
we decompose it as R1(A, B, C) and R2(C, D). This preserves all dependencies and the join is lossless too, but the relation R1 is not in BCNF. In R1 we keep ABC together otherwise preserving {AB→C} will fail, but doing so also causes {C→A} to appear in R1. {C→A} violates the condition for R1 to be in BCNF as C is not a superkey. Condition that all relations formed after decomposition should be in BCNF is not satisfied here.
We need to identify the INCORRECT, Hence mark option C.