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
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)
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..
what is the problem you are facing with D?