R1 : AB+ = {ABCD} -------- CK , BC IS CK
AC+={ACD} ------------ NOT CK
FOR 2NF X->Y either subset of X is not CK or Y is prime attribute
AB→C ,C→A ,AC→D
since AB is CK not subset of CK ,C,AC is not subset of CK but C+={ACD}
here C can alone derive A which implies AC->D as PD
so R1 is not 2NF
FOR 3NF X->Y either X is superkey or Y is prime attribute
BUT AC->D here AC is not superkey and D is not prime attribute
Therefore R1 is not 3NF
R2: AB+ ={ABCD} ------- CK
BC+={BCD} ---------- NOT CK
AB->C ,BC->D here AB is CK A+={A},B+={B} AB+={ABCD} ,BC is not CK B+={B} ,C+={C},BC+={BCD}
No PD's in R2
Therefore R2 is in 2NF
BC->D here BC is not superkey and D is not prime attribute
R2 is not 3NF
So R1 is not 2NF ,R2 is 2NF
Therefore Option B
Let me know if anything wrong