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3 Problems of GATE 2014 (in Hindi)
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Sweta Kumari is teaching live on Unacademy Plus

Sweta Kumari
YouTuber (channel : GATE NoteBook) | Verified Educator | 2+ year Online Teaching Experience

U
Unacademy user
bhaiya what is difference between endomycium and perimysium
see Endomysium is connective tissue that wraps each individual muscle fibre. Perimysium is connective tissue that wraps bundles of muscle fibres - the "bundles" being known as fasicles. Epimysium is connective tissue that wraps the whole muscle. I hope you get it ! if still having problems than ask me where .
Ankita Singh
a year ago
thanks
  1. For (Student Name, Student Age) to be a key for this instance, the value Xshould NOT be equal to The maximum number of superkeys for the relation schema R(E,F,G,H) with E as the key is (A)5 (B)6 (C) 7 (D) 8 GATE 2014 Student Student Student Student Name Emai Shankar Shankar@ 1287 Swati swati@ee 19 GATE 2014 CPI ID Age 9.4 math 9.5 Shankar shankar@ 9.4 19 cse swati@m ech ganesh@ ivil 9876 Swati 18 9.3 8765 Ganesh 19 8.7


  2. For (Student Name, Student Age) to be a key for this instance, the value Xshould NOT be equal to The maximum number of superkeys for the relation schema R(E,F,G,H) with E as the key is (A)5 (B)6 (C) 7 (D) 8 Maximum no. of possible superkeys for a table with n attributes 2(n-1) Here, n 4. So, the possible superkeys 241 8 The possible superkeys are:E, EH, EG, EF, EGH, EFH, EFG, EFGH GATE 2014 Student Student Student Student Name Emai Shankar Shankar@ 1287 Swati swati@ee 19 GATE 2014 CPI ID Age 9.4 math 9.5 Shankar shankar@ 9.4 19 cse swati@m ech ganesh@ ivil 9876 Swati 18 9.3 8765 Ganesh 19 8.7 There is already an entry with same name and age as 19. So the age of this entry must be something other than 19


  3. Given the following two statements: S1: Every table with two single-valued attributes is in 1NF, 2NF, 3NF and BCNF. S2: AB->C, D->E, E->C is a minimal cover for the set of functional dependencies AB->C, D->E, AB->E, E->C Which one of the following is CORRECT? (A) S1 is TRUE and S2 is FALSE. (B) Both S1 and S2 are TRUE. (C) S1 is FALSE and S2 is TRUE. (D) Both S1 and S2 are FALSE. GATE 2014


  4. Given the following two statements: S1: Every table with two single-valued attributes is in 1NF, 2NF, 3NF and BCNF. S2: AB->C, D->E, E->C is a minimal cover for the set of functional dependencies AB->C, D->E, AB->E, E->C. Which one of the following is CORRECT? (A) S1 is TRUE and S2 is FALSE. (B) Both S1 and S2 are TRUE. (C) S1 is FALSE and S2 is TRUE. (D) Both S1 and S2 are FALSE. S2: AB->C, D->E, E->C is a minimal cover for the set of functional dependencies AB->C, D->E, AB->E, E->C. As we know Minimal Cover is the process of eliminating redundant Functional Dependencies and Extraneous attributes in Functional Dependency Set. So each dependency of F (AB->C, D->E, AB->E, E->C) should be implied in minimal cover. As we can see AB->Eis not covered in minimal cover since AB]t ABC in the given cover (AB->C, D->E, E->C) Hence, S2 is false. GATE 2014 S1: Every table with two single-valued attributes is in 1NF, 2NF, 3NF and BCNF. A relational schema R is in BCNF iff in Every non-trivial Functional Dependency X->Y, X is Super Key. If we can prove the relation is in BCNF then by default it would be in 1NF, 2NF, 3NF also. Let R(AB) be a two attribute relation, then 1. IfB->A) exists then BCNF since (B) AB R 2. If (A->B,B->Aexists then BCNFsince A and B both are Super Key now. 3. If {No non trivial Functional Dependency) then default BCNF. 4. Hence it's proved that a Relation with two single -valued attributes is in BCNF hence its also in 1NF, 2NF, 3NF.