## Sweta Kumari is teaching live on Unacademy Plus

Consider a relation scheme R (A, B, C, D, E, H) on which the following functional dependencies hold: {A->B, BC->D, E->C, D->AL what are the candidate keys of R? (A) AE, BE (B) AE, BE, DE (C) AEH, BEH, BCH (D) AEH, BEH, DEH GATE 2007 Let R-(A, B, C, D, E, F) be a relation scheme with the following dependencies: C->F, E->A, EC->D, A->B. Which of the following is a key for R? (a) CD (b) EC(c)AE (d) AC GATE 1999

Consider a relation scheme R (A, B, C, D, E, H) on which the following functional dependencies hold: {A->B, ->D, E->C, D- ). What are the candidate keys of R? (A) AE, BE (B) AE, BE, DE (C) AEH, BEH, BCH (D) AEH, BEH, DEH A set of attributes S is candidate key of relation R if the closure of S is all attributes of R and there is no subset of S whose closure is all attributes of R. Closure of AEH, i.e. AEH+(ABCDEH) closure of BEH, i.e. BEH-={ABCDEH) Closure of DEH, i.e. DEH+ ABCDEH) GATE 2007 Let R- (A, B, C, D, E, F) be a relation scheme with the following dependencies: C->F, E-3A, EC->D, A->B. Which of the following is a key for R? (a) cD GATE 1999 ECFA E->A) Closure set of EC covers all the attributes of the relation R. (b) EC(c) AE (d) AC

From the following instance of a relation scheme R Given the following relation instance. (A, B, C), we can conclude that X Y Z 1 4 2 1 5 3 1 6 3 3 2 2 Which of the following functional dependencies are satisfied by the instance? (a) XY ->Z and Z-> Y (b) YZ ->X and Y-> Z (c) YZ-> X and X ->Z (d) XZ-> Y and Y->IX GATE 2000 1 GATE 2002 0 3 (A) A functionally determines B and B functionally determines C (B) A functionally determines B and B does not functionally determine C (C) B does not functionally determine C (D) A does not functionally determine B and B does not functionally determine C

From the following instance of a relation scheme R Given the following relation instance. (A, B, C), we can conclude that X Y Z 1 4 2 1 5 3 1 6 3 3 2 2 Which of the following functional dependencies are satisfied by the instance? GATE 2000 1 GATE 2002 0 3 For the value 1 of B, it(a) XY->Zand Z -> Y gives two different values of C i.e. 1 & o. (c) YZ->Xand X->z (b) YZ->Xand Y->Z (A) A functionally determines B and B functionally determines C (B) A functionally determines B and B does not functionally determine C (C) B does not functionally determine C (D) A does not functionally determine B and B does not functionally determine C (d) XZ-> Y and Y->IX (a), its given Z->Y, it means that the value of Z uniquely determines the value of Y. But here the value 2 of Z, gives 2 different values of Yi.e. 4 and 2. (c), its given X->Z, But value 1 of X, gives 2 values of Z i.e. 2 and 3. (d), its given Y-- . Now take FD XZ->Y, here (13) cannot uniquely determine the value of Y. (1,3) gives two values for Y i.e. 5 and 6. Therefore this FD (XZ->Y) is not satisfied by the instance.