i) R(A, B, C, D, E)
FD's ABC -> DEF, BC -> E, C ->B.
Now through the split rule we can write the functional dependencies given above is :-
ABC -> D, ABC -> E, ABC -> F, BC -> E, C -> B.
Now take the decomposition given above
R1(A, B, C, D, F)
FD's {ABC -> D, ABC -> F, ABC -> BC (Trivial functional dependency )}
R2(B, C, E)
FD's {BC -> E, C -> B}
Now let's check dependencies preserving:-
ABC -> D, ABC -> F, BC -> E, C -> B (these are present in the decomposition)
ABC -> BC and BC -> E (transitive dependency) hence ABC -> E
HENCE IT IS DEPENDENCY PRESERVING.
ii.)
R(A, B, C, D, E)
FD's ABC -> DEF, BC -> EF, EF ->B.
Now take the decomposition given above in the question.
R1(A, B, C, D, F)
FD's {ABC -> D, ABC -> F, ABC -> BC (Trivial functional dependency ), BC -> F}
R2(B, C, D, E)
now the functional dependency EF -> B, this is not satisfying because E and F are in different relations hence this is not dependency preserving decomposition.
HENCE ONLY i) IS SATISFYING DEPENDENCY PRESERVING DECOMPOSITION CONDITION.
ANS IS a) only (i)