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Tuesday, 24 January 2017

CBSE UGC NET Computer Science Paper II January 2017 - Part 1

1.       Consider a sequence F00 defined as:
Then what shall be the set of values of the sequence F00?
(1) (1, 110, 1200)
(2) (1, 110, 600, 1200)
(3) (1, 2, 55, 110, 600, 1200)
(4) (1, 55, 110, 600, 1200)
Answer: 1
Explanation:
We have given, F00(0) = 1, F00(1) = 1
F00(2) = (10*F00(1) + 100)/F00(0) = 110
F00(3) = (10*F00(2) + 100)/F00(1) = 1200
F00(4) = (10*F00(3) + 100)/F00(2) = 110

Since the values repeats after the first three values, the set of values of F00 will be (1,110,1200).
2.       Match the following:
List-I                  List-II
a. Absurd          i. Clearly impossible being
contrary to some evident truth.
b. Ambiguous   ii. Capable of more than one
interpretation or meaning.
c. Axiom            iii. An assertion that is accepted
and used without a proof.
d. Conjecture    iv. An opinion Preferably based
on some experience or wisdom.
Codes:
     a   b   c    d
(1) i    ii   iii   iv
(2) i    iii  iv   ii
(3) ii   iii  iv   i
(4) ii   i    iii   iv
Answer: 1
3.       The functions mapping R into R are defined as:
f(x) = x3-4x, g(x)=1/(x2+1) and h(x)=x4
Then find the value of the following composite functions:
hog(x) and hogof(x)
(1) (x2+1)4 and [(x3-4x)2+1]4
(2) (x2+1)4 and [(x3-4x)2+1]- 4
(3) (x2+1)- 4 and [(x3-4x)2+1]4
(4) (x2+1)‑ 4 and [(x3-4x)2+1]- 4
Answer: 4
Explanation:
hog(x) = h(1/(x2+1))
= [(1/(x2+1))]4 = (x2+1)- 4
hogof(x) = hog(x3-4x)
= hog(x3-4x)

= [(x3-4x)2+1]- 4 [since hog(x) = (x2+1)- 4]
4.       How many multiples of 6 are there between the following pairs of numbers?
0 and 100          and     -6 and 34
(1) 16 and 6
(2) 17 and 6
(3) 17 and 7
(4) 16 and 7
Answer: 3
Explanation:
Number of multiples of 6 between 1 and 100 = 100/6 = 16
Since the range starts from zero, we need to take zero too. [zero is a multiple of every integer (except zero itself)].
So, answer = 16+1 = 17
Number of multiples of 6 between 1 and 34 = 34/6 = 5
Since the range is -6 to 34, we need to take -6 and zero.

So, answer = 5+2 = 7
5.       Consider a Hamiltonian Graph G with no loops or parallel edges and with |V(G)|=n≥3. Then which of the following is true?
(1) deg(v) ≥ n/2 for each vertex v.
(2) |E(G)| ≥ 1/2(n-1)(n-2)+2
(3) deg(v)+deg(w) ≥ n whenever v and w are not connected by an edge.
(4) All of the above
Answer: 4

Explanation:
Dirac's theorem: A simple graph with n vertices (n ≥ 3) is Hamiltonian if every vertex
has degree n/2 or greater.
Ore's theorem: deg(v) + deg(w) ≥ n for every pair of distinct non-adjacent vertices v
and w of G (*), then G is Hamiltonian.

6.       In propositional logic if (P→Q)˄(R→S) and (P˅R) are two premises such that
Y is the premise:
(1) P˅R
(2) P˅S
(3) Q˅R
(4) Q˅S
Answer: 4
7.       ECL is the fastest of all logic families. High Speed in ECL is possible because transistors are used in difference amplifier configuration, in which they are never driven into ...............
(1) Race condition
(2) Saturation
(3) Delay
(4) High impedance
Answer: 2
Explanation:

Emitter-coupled logic (ECL) is the fastest of all logic families and therefore is used in 
applications where very high speed is essential. High speeds have become possible in 
ECL because the transistors are used in difference amplifier configuration, in which 
they are never driven into saturation and thereby the storage time is eliminated.
8.       A binary 3-bit down counter uses J-K flip-flops, FFi with inputs Ji, Ki and outputs Qi, i=0,1,2 respectively. The minimized expression for the input from following, is
I. J0=K0=0
II. J0=K0=1
III. J1=K1=Q0
IV. J1=K1=Q'0
V. J2=K2=Q1Q0
Vl. J2=K2=Q'1Q'0
(1) I, Ill, V
(2) I, IV, VI
(3) Il, III, V
(4) Il, IV, Vl
Answer: 4
9.       Convert the octal number 0.4051 into its equivalent decimal number.
(1) 0.5100098
(2) 0.2096
(3) 0.52
(4) 0.4192
Answer: 1
Explanation:
(0.4051)8 = 4x8-1+0x8-2+5x8-3+1x8-4

= 0.5100098
10.    The hexadecimal equivalent of the octal number 2357 is:
(1) 2EE
(2) 2FF
(3) 4EF
(4) 4FE
Answer: 3
Explanation:
(2357)8 can be converted into binary just digit by digit.
= 010 011 101 111
Now we can regroup the bits into groups of 4 and convert to hexadecimal.
= 0100 1110 1111
= 4EF

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2 comments:

  1. So fast in publishing the answer within the couple of days after written the
    exam. Thanks a lot very much.

    ReplyDelete
  2. Question 4 answer is 1) 16 and 6

    ReplyDelete