First find the intersection:
$R_1\cap R_2=\{A,B\}$.
Therefore, the decomposition becomes lossless if
$AB\to R_1$ or $AB\to R_2$
Under the original FD set:
From $B\to EF$, starting with $AB$ we obtain $A,B,E,F$.
But we cannot obtain $C$ or $D$.
So the original decomposition is indeed lossy.
Now consider A:
Add $D\to ABCDEF$.
Starting from $AB$, we still cannot obtain $D$.
Therefore, this new FD cannot be activated.
So A does not help.
Now consider B:
Add $E\to C$.
From $AB : B\to E$.
Then $E\to C$.
Then $C\to D$.
Therefore, $(AB)^+=\{A,B,C,D,E,F\}$.
So $AB$ determines both component schemas.
Hence the decomposition becomes lossless.
Thus B is correct.
Now consider C:
Add $B\to D$.
From $AB : B\to D$ and $B\to E,F$.
Therefore, $AB\to ABDE$.
But $ABDE=R_1$.
Hence, $(R_1\cap R_2)\to R_1$.
That alone is sufficient for losslessness.
Thus C is also correct.
