1)Relation R in BCNF with atleast one simple candidate key is also in 4NF
if multivalue attribute exist in the table then every candidate key will be compound key
so here there is one simple candidate key no multivalue attribute present so surely its in 4NF also.
2)Relation R in BCNF with all candidate key simple is also in 4NF
same reason as above.
3)Relation R in 3NF with only one compound candidate key is also in BCNF
if relation R is in 3NF and atmost one compound CK then it is also in BCNF
if more than one compound Ckey will present then prime to prime will be possible which will violate the BCNF property but with one compound CK there will be no such possiblity
e.g. R(ABCDE) {AB->C C->D D->E E>A}
CK{AB EB DB CB} Here C->D D->E E>A violates the BCNF
4)Relation R in 3NF with atleast one simple candidate key is also in 4NF
There is no MVD in the relation
But there can be possiblity of two compound candidate key then it will not be in BCNF
R(ABCDE) {AB->C C->BD D->E E>A}
CK{AB C EB DB } Here D->E E>A violates the BCNF so 4NF will also not satisfied.
option 4 is false.