In the following relation R(ABCDEF) which of the following FD

Step 1: Split FDs with multiple attributes on the right-hand side.

• A → B, A → C, A → D, A → E, A → F (from A → BCDEF)

• BC → A, BC → D, BC → E, BC → F (from BC → ADEF)

• B → F

• D → E

Step 2: Minimize left-hand sides (check for extraneous attributes).

For BC → ... check if B or C is extraneous. Compute closures:

• C+ (using the FDs) = C, so C alone does not imply A; B+ = B,F, so B alone does not imply A. Therefore neither B nor C is extraneous; BC remains as the left-hand side.

Step 3: Remove redundant FDs (check each FD against the set excluding it).

• A → B: If A → B is removed, A+ (using the other FDs) does not include B, so A → B is essential.

• A → C: If removed, A+ does not include C, so A → C is essential.

• BC → A: If removed, BC+ does not include A using the remaining FDs, so BC → A is essential.

• BC → D: keep BC → D so that A → D can be derived (A gives B and C which yield BC, and BC → D then yields D).

• B → F: essential because B+ without it does not include F.

• D → E: essential because D+ without it does not include E.

Notes on other FDs:

• A → D, A → E, and A → F are derivable from the chosen minimal set (for example A→B and A→C give BC, BC→D gives D, and D→E gives E; A→B and B→F give F). Thus they are not required explicitly in the minimal cover.

• BC → E and BC → F are also derivable through BC → D and D → E or through BC → A and then A → F (via A→B and B→F), so they are not required explicitly.

Final minimal cover (one valid minimal cover equivalent to the original set):

• A → B

• A → C

• BC → A

• BC → D

• B → F

• D → E

Therefore, A→B must be present in the minimal cover (the other choices suggested in the options are either derivable or not implied by the original FDs).