+1 vote
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retagged | 137 views
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I am getting B
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F is the key of that relation ===> F is a single attribute ===> Given table itself in the 2NF

coming to the options, in Option A, by decomposing we loss AB->CDEF

remaining all options are loss-less and dependency preserving.

@MiNiPanda, i hope you should consider AB as Key

in that case option D is right, but not option B

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Yes Shaik I considered AB as a CK and that is why found B->D as a PFD. Since for this FD the relation is violating 2nf i made a separate relation R(BD)

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here AB,BC,AF,CF,DF,EF all are candidate keys....all the attributes are prime attributes.....how are u differentiating the options?
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himgta F is the CK so you can't take AF,CF,DF,EF as CKs..they are superkeys..

AB,BC,F are CKs. I couldn't find more than this..

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How you get D is correct?
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what is the problem you are facing with D?

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What is wrong with option B if we take AB as key?

if we take AB as key, Option B is in 2NF ? no it can't be in 2NF due to B->E is lead to PFD moreover it is loss-less decomposition but not dependency preserving ( D->E is not preserving )

Question is not standard... we can skip this question.... nothing is useful on discussing this question.
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@Minipanda,

–1 vote
I think option A is correct, since AB, BC, F are the candidate keys possible, B-->D and D-->E are in 1NF and 2NF respectivley. if we decompose correctly we get the relations which are in option A
by (37 points)