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 R_{1}(A, B) and R_{2}(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 R_{1}(A, B), R_{2}(B, C) and R_{3}(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 R_{1}(A, B), R_{2}(A, C) and R_{3}(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 R_{1}(A, B, C) and R_{2}(C, D). This preserves all dependencies and the join is lossless too, but the relation R_{1 }is not in BCNF. In R_{1} we keep ABC together otherwise preserving {AB→C} will fail, but doing so also causes {C→A} to appear in R_{1}. {C→A} violates the condition for R_{1} 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**.