Correct : a,b,d

The correct answers are A, B, and D.
Option A - 3NF dependency-preserving decomposition is always possible: TRUE
The 3NF synthesis algorithm guarantees that any relation can always be decomposed into 3NF in a way that is both lossless-join and dependency-preserving.
Option B - 1NF dependency-preserving decomposition is always possible: TRUE
1NF simply requires atomic values in each column. The original relation itself is already a valid 1NF decomposition (no decomposition needed), so it trivially preserves all dependencies.
Option C - BCNF dependency-preserving decomposition is always possible: FALSE
BCNF decomposition is not always dependency-preserving. A classic counterexample is the relation with attributes (A, B, C) and dependencies A→B and BC→A - no BCNF decomposition of this can preserve all functional dependencies.
Option D — BCNF dependency-preserving decomposition is not always possible: TRUE
This directly follows from the counterexample above. BCNF is a stronger condition than 3NF, and this extra strictness comes at the cost of losing dependency preservation in some cases.