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2 Answers

Best answer
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First, we need to find the candidate keys.

From the given functional dependencies, only $AB$ is the candidate key.

We have a partial dependency here,

$A \rightarrow C$

So, it is not in $2NF$.

Hence, option (A) 1NF but not in 2NF

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C is not fully functionally dependent on candidate key AB because A-->C. so it is not in 2NF.So answer is A
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Bikram asked Aug 26, 2017
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Decompose the following table into BCNF:$R(ABCD)$$A \rightarrow C$$C \rightarrow A$$AB \rightarrow D$The result is:(AC) (ABD)(AC) (CBD)(AB)(BAD)Both (A) & (B)