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4 Answers

4 votes
4 votes

here A is key....

So A->BC is in BCNF...

C->D ...is not in BCNF as C is not a super key ...its also not in 3NF as D is not a prime atribute...but its in 2nf as C is not a part of Key..

so overall relation is in 2NF

answer is option A

2 votes
2 votes
(A)2NF

R(A B C D)     HERE A is the CK

so for A->BC  A is the SUPER KEY so  it is in BCNF

BUT C->D    (NP->NP) so TRANSITIVE DEPENDENCY EXIST SO IT IS NOT IN 3NF

SO it is in 2NF
1 votes
1 votes
Given, functional dependencies are F={A→ BC, C→ D}
Finding candidate key of given relation is relation is
A+={A,B,C} because we don’t have an A element in right hand side.
A→ BC dependency A is the candidate key and also part of FD. So, it is lossless.
C→ D is following Transitive dependency. So, it never become 3NF.
According to given choices, A is the correct choice.
1 votes
1 votes
ANS : A  2NF

R(A B C D)     HERE A is the Candidate key

so for A→BC  A is the SUPER KEY so  it is in BCNF

BUT C→D    (NP->NP) so TRANSITIVE DEPENDENCY EXIST SO IT IS NOT IN 3NF

SO it is in 2NF
Answer:

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