- Option : B
- Explanation :
- An OODBMS avoids the “impedance mismatch” problem.
**Correct.** - An OODBMS doesn't avoids the “phantom” problem.
- An OODBMS doesn't provides higher performance concurrency control than most relational databases. Since it is distributed.
- An OODBMS provides faster access to individual data objects once they have been read from disk.
**Correct**
SO, option (B) is correct.

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- Option : D
- Explanation : RA
_{1}: T_{1}⨝_{T1.P = T2.A}T_{2}where ⨝is natural join symbol. It will result 3 tuples:

RAP = A R C 10 5 6 10 5 5 25 6 3 _{2}: T_{1}⟕_{T1.P = T2.A}T_{2}where ⟕ is left outer join symbol. It will result in 4 tuples.

RAP = A R C 10 5 6 10 5 5 15 8 Null 25 6 3 _{3}: T_{1}⨝_{T1.P = T2.A}and T_{1}.R = T_{2}.C^{T2}. It will result in 1 tuple.

So, option (D) is correct.P = A R = C 10 5

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- Option :
- Explanation :

Decomposition of R into R_{1}(C, A) and R_{2}(A, B) is lossless. Because C → A, A → B. So, C → AB can be derived and there is no loss. Decomposition of R into R_{1}(A, B) and R_{2}(B, C) is lossy. Because A → B, C → B are derived but we can't derive C → AB, So it is lossy. So, option (C) is correct.

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