Correct Answer - Option 1 : in 1NF but not in 2NF
The correct answer is option 1
Explanation:
Since attribute D is not a part of any FD, it must be a part of the candidate key.
AD+ = {ABCDEFGH}
BD+ = {ABCDEFGH}
ED+ = {ABCDEFGH}
FD+ = {ABCDEFGH}
CD+, GD+, and HD+ do not all the attributes. So, they can not be candidate keys.
Candidate keys are AD, BD, ED, and FD.
A → BC, B → CFH, and F → EG are partial dependencies. So, the relation is not in 2NF.
Hence the relation is in 1NF but not in 2NF.