Database Management System Questions and Answers

Answer: C

Correct answer: mn and 0

A join is a subset of the Cartesian product, so at most every pair of rows matches, giving \( m \times n \) tuples; and at fewest no pair matches, giving 0. The extremes are the product and the empty relation.

Answer: D

Correct answer: None of these

In a B+ tree of order m each node holds at least \( \lceil m/2 \rceil \) pointers, so the height is bounded by

\[ h \le \log_{\lceil m/2 \rceil}(n) \]

The base is \( \lceil m/2 \rceil \), not m - the minimum fill, since the worst case is the sparsest tree. None of the three expressions offered has that form.

Answer: A

Correct answer: With finer degree of granularity of locking a high degree of concurrency is possible

Locking a smaller unit - a row rather than a table - leaves more of the data free for other transactions, so concurrency rises; the cost is more locks to manage. The other statements are false: plain locking does not by itself prevent non-serializable schedules, field-level locking is possible, and an exclusive lock is refused while a shared lock is held.

Answer: D

Correct answer: The definition of a view should not have GROUP BY clause in it

A view definition may perfectly well contain GROUP BY - that is one of the commonest uses of a view, to present a summary. What such a view cannot generally do is accept updates, which is the true statement the first option makes.

Answer: D

Correct answer: None of these .

All three schemes exist to guarantee serializability: two-phase locking and the tree protocol enforce it by the order in which locks are taken, and timestamp ordering by comparing transaction timestamps. So none of them is the odd one out.

Answer: D

Correct answer: All of the above

A data model serves all three purposes: it records how each user sees the data, it describes the data in its own right independently of any storage decision, and it shows how the same data is used across applications.

Answer: B

Correct answer: disjoint

If an entity may belong to at most one of the lower-level sets, the specialisation is disjoint; if it may belong to several, it is overlapping. Partial and total answer a different question - whether every higher-level entity must belong to some lower-level set at all.

Answer: C

Correct answer: natural join

A natural join matches on the columns the two relations share and then keeps only one copy of each such column. A theta join keeps both copies, since its condition need not be equality at all.

Answer: D

Correct answer: EXISTS

EXISTS takes a subquery and yields true if it returns at least one row, so NOT EXISTS tests emptiness. DISTINCT and UNIQUE concern duplicates and NULL tests a single value.

Answer: C

Correct answer: both (A) & (B)

In the SQL standard UNIQUE by itself permits nulls, but in the sense these questions use it - as the guarantee that the column identifies rows - it is taken to require distinct values and no nulls, which is what makes it equivalent in effect to a key.