GATE 2016 CS Set 1-5

33) A database of research articles in a journal uses the following schema.

( VOLUME , NUMBER , START PAGE , ENDPAGE , TITLE , YEAR , PRICE )

The primary key is (VOLUME , NUMBER , START PAGE , 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?

1NF
2NF
3NF
BCNF

Answer: B

Explanation:

The Old relation schema design is in 1NF because it has functional dependency: Volume, Number → Year as a partial dependency so the original database design is in 1NF but not in 2NF.

Now, the new relation schema design removes all partial dependencies, so it is in 2NF as well as in 3NF. Therefore, 2NF is the weakest normal form that the new database satisfies, but the old database does not.

Hence option(B) is the right answer.

34) Which one of the following protocols is NOT used to resolve one form of address to another one?

DNS
ARP
DHCP
RARP

Answer: C

Explanation:

A) DNS - It converts domains or hostname to IP address

B) ARP - It converts IP to MAC address

C) RARP - It converts MAC to IP address

D) DHCP - It assigns IP dynamically. So, it doesn't resolve any addresses.

Therefore, option (C) is the correct Answer.

35) Which of the following is/are example(s) of stateful application layer protocols?

(i) HTTP
(ii) FTP
(iii) TCP
(iv) POP3

(i) and (ii) only
(ii) and (iii) only
(ii) and (iv) only
(iv) only

Answer: C

Explanation:

A Stateless Protocol doesn't maintain information about a transaction after a transaction is processed. There is no tight dependency between client and server.

Example: HTTP

A Stateful Protocol kept information about a transaction even after a transaction has been processed. There is a tight dependency between client and server.

Example: FTP, POP3

Therefore option (C) is the right answer.

36) The coefficient of x¹² in (x³ + x⁴ + x⁵ + x⁶ + • • • )³ is __________.

Answer: C

Explanation:

The coefficient of x¹² is:
(x⁴)³ having coefficient ³C₀ = 1
(x³)²(x⁶) having coefficient ³C₁ = 3
(x³)(x⁴)(x⁵) having coefficient ³C₂ * ²C₁ = 6

The coefficient of x¹² is: 1+ 3 + 6 = 10
Therefore option (C) will be the correct answer.

37) Consider the recurrence relation a₁ = 8, a_n = 6n² + 2n + a_n-1. Let a₉₉ = K × 10⁴ . The value of K is __________.

Answer: B

Explanation:

a_n = 6n² + 2n + a_n-1
= 6n² + 2n + 6(n-1)²+ 2(n-1) + a_n-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)
= 2n(n + 1)²

So, for n = 99,
a₉₉ = 2 × 99 × (99 + 1) 2 = 198 × 10⁴
Hence K = 198

Therefore option (B) is the correct answer.

38) 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: A

Explanation:

Assume: f(1) = a. Then,
f(2) = f(2/2) = f(1) = a
f(3) = f(3+5) = f(8) = f(8/2) = f(4/2) = f(2/1) = f(1) = a
Again, f(4) = f(4/2) = f(2) = a
f(5) = f(5+5) = f(10/2) = f(5) = b.
....
f(10) = f(10 + 5) = f(15) = b

Hence the given function will have two values, i.e. a and b. All multiples of 5 will have value b and others will have value a. So the above function has two values only.

Therefore option (A) is the right answer.

39) 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) __________.

0.33
0.43
0.25
0.27

Answer: A

Explanation:

According to the question,
We get TH on our first trial with chance 1/4. So P(Y₁) = 1/4 or
We got TT on the first trial with chance 1/4 then repeat and get TH on the second trial. So P(Y₂) = (1/4)(1/4) or
We got TT on a first and second trial, and then got TH on the third trial. So P(Y₃) = (1/4)(1/4)(1/4) etc. Since the events are mutually exclusive, Thus
P(Y) = P(Y₁)+P(Y₂)+P(Y₃)+............
= 1/4+(1/4)₂+ (1/4)₃+.....

= 1/4 Gate 2016 CS set 1

= 1/4 / 1-1/4 (GP =a / (1-r))
= 1/4 / 3/4
= 1/3
= 0.33

Therefore option (A) is the correct answer.

40) Consider the two cascaded 2-to-1 multiplexers as shown in the figure.

Gate 2016 CS set 1

The minimal sum of products form of the output X is

P Q + PQR
PQ + QR
PQ + P QR
Q R + PQR

Answer: D

Explanation:

Solving first multiplexer (f1), we get
= 0P' + RP = RP
Then for Solving second multiplexer (f2), we get
= Q'R'+Qf1 = Q'R' + PQR
Therefore option (D) is the correct answer.