21.
Which of the following is NOT a superkey in a
relational schema with attributes V, W, X, Y, Z and primary key V Y ?
(A) V X Y Z (B)
V W X Z
(C) V W X Y (D)
V W X Y Z
Answer: B
Explanation:
Option (B)
does not include Y which is a part of Primary Key.
A Super Key
is simply a non-minimal Candidate Key, that is to say one with additional
columns not strictly required to ensure uniqueness of the row.
A Primary
Key is a minimal Candidate Key, which is to say all constituent columns are
strictly required in order to ensure uniqueness.
22.
Which one of the following is NOT a part of
the ACID properties of database transactions?
(A) Atomicity (B)
Consistency
(C) Isolation (D)
Deadlock-freedom
Answer: D
Explanation:
ACID
(Atomicity, Consistency, Isolation, Durability) is a set of properties of
database transactions. A transaction is a very small unit of a program and it
may contain several low-level tasks. A transaction in a database system must
maintain Atomicity, Consistency, Isolation, and Durability in order to ensure
accuracy, completeness, and data integrity.
So option D
is the answer.
23.
A database of research articles in a journal
uses the following schema.
(VOLUME, NUMBER, STARTPGE, ENDPAGE, TITLE, YEAR,
PRICE)
The primary key is (VOLUME, NUMBER, STARTPAGE,
ENDPAGE) and the following functional dependencies exist in the schema.
(VOLUME, NUMBER, STARTPAGE, ENDPAGE) → TITLE
(VOLUME, NUMBER) → YEAR
(VOLUME, NUMBER, STARTPAGE, ENDPAGE) → PRICE
The database is redesigned to use the
following schemas.
(VOLUME, NUMBER, STARTPAGE, ENDPAGE, TITLE,
PRICE)
(VOLUME, NUMBER, YEAR)
Which is the weakest normal form that the new
database satisfies, but the old one does not?
(A) 1NF (B)
2NF
(C) 3NF (D)
BCNF
Answer: B
Explanation:
There are
partial dependencies in the given FD set. So, it is in 1 NF but not in 2NF.
The new
design is removing all the partial dependencies. So it is in 2NF.
So the
weakest normal form that the new database satisfies, but the old one does not
is 2NF.
24.
Which one of the following protocols is NOT
used to resolve one form of address to another one?
(A) DNS (B)
ARP
(C) DHCP (D)
RARP
Answer: C
Explanation:
The mapping
of host names to IP addresses is handled through a service called Domain Name
Service (DNS).
The Address
Resolution Protocol (ARP) allows a host to find the MAC address of a node with
an IP address on the same physical network, when given the node's IP address.
RARP
(Reverse ARP) allows a machine to use its physical address to determine its
logical address on the Internet.
Dynamic Host
Configuration Protocol (DHCP) is a protocol for assigning dynamic IP addresses
to devices on a network.
So DHCP is the
answer. All others are used to convert one address to other.
25.
Which of the following is/are example(s) of
stateful application layer protocols?
(i)
HTTP
(ii) FTP
(iii) TCP
(iv) POP3
A (i) and (ii) only
B (ii) and (iii) only
C (ii) and (iv) only
D (iv) only
Answer: C
Explanation:
A stateless protocol
is a communications protocol that treats each request as an independent
transaction. A stateless protocol does not require the server to retain session
information or status about each communications partner for the duration of
multiple requests.
A protocol
that requires keeping of the internal state on the server is known as a
stateful protocol.
HTTP is
stateless.
FTP is
stateful.
TCP is
stateful, but not an application layer protocol.
POP3 is
stateful.
So answer is
Option C.
26. The
coefficient of x12 in (x3 + x4 + x5
+ x6 + ...)3 is............
Answer: 10
27. Consider
the recurrence relation a1 = 8, an = 6n2 + 2n
+ an-1. Let a99 = k x 104. The value of K is .................
Answer: 198
Explanation:
an=6n2+2n+an−1
=6n2+2n+6(n−1)2+2(n−1)+an−2
=6n2+2n+6(n−1)2+2(n−1)+6(n−2)2+2(n−2)+......+a1
=6n2+2n+6(n−1)2+2(n−1)+6(n−2)2+2(n−2)+......+6.12+2.1
=6(n2+(n−1)2+...+22+12)+2(n+(n−1)+...+2+1)
=6×n(n+1)(2n+1)/6+2×n(n+1)/2
=n(n+1)(2n+1+1)
an=2n(n+1)2
for n=99, a99=2×99×(99+1)2=198×104
So, K=198
28. A
function f : N+ → N+, defined on the set of positive integers
N+, satisfies the following properties:
f (n) = f
(n/2) if n is even
f (n) = f
(n+5) if n is odd
Let R = {i |∃j : f(j) = i } be the set of distinct values
that f takes. The maximum possible size of R is .............
Answer: 2
Explanation:
Let us
assume: f(1) = x. Then,
f(2) =
f(2/2) = f(1) = x
f(3) =
f(3+5) = f(8) = f(8/2) = f(4/2) = f(2) = f(2/2) = f(1) = x
Similarly,
f(4) = x
f(5) =
f(5+5) = f(10) = f(10/2) = f(5) = y. [Let us assume: f(5) = y ]
f(10) = f(10+5)
= f(15) = f(15+5) = f(20) = f(20/2) = f(10) = f(10/2) = f(5) = y
So it will
have two values. All multiples of 5 will have value y and others will have
value x. It will have 2 different values.
29.
Consider the following experiment.
Step 1. Flip a fair coin twice.
Step 2. If the outcomes are (TAILS, HEADS)
then output Y and stop.
Step 3. If the outcomes are either (HEADS,
HEADS) or (HEADS, TAILS), then output N and stop.
Step 4. If the outcomes are (TAILS, TAILS),
then go to Step 1.
The probability that the output of the
experiment is Y is (up to two decimal places) ..............
Answer: 0.33
Explanation:
The output
will be Y if you get TH on your first trial with chance 1/4.
So P(Y1)=1/4
or
You got TT
on the first trial with chance 1/4 and then repeat and get TH on the second. So
P(Y2)=(1/4)(1/4) or
You got TT
on 1 and 2, and then got TH on the third.
So
P(Y3)=(1/4)(1/4)(1/4)
etc.
Since the
events are mutually exclusive, you can just add up the probabilities.
Thus P(Y)=P(Y1)+P(Y2)+P(Y3)+⋯
=1/4+(1/4)2+(1/4)3+⋯
It goes on
as infinite GP.
Sum of
infinite GP = a/(1-r)
here a= 1/4
and r = 1/4
=1/4/(1−1/4)=(1/4)/(3/4)=1/3≈0.33
30. Consider
the two cascaded 2-to-1 multiplexers as shown in the figure.
The minimal
sum of products form of the output X is
(A) P’Q’+PQR
(B) P’Q+QR
(C) PQ+ P’Q’R
(D) Q’R’+PQR
Answer: D
Explanation:
For 2:1 MUX,
output Y=S’I0+SI1
So, output
of MUX-1, f1=P’0+PR=PR
Output of
MUX-2, f2=Q’R’+Qf1=Q’R’+Q(PR)
= Q’R’+PQR
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