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Saturday, 15 July 2017

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

CBSE UGC NET Computer Science Paper II January 2017 Solved Paper


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

Absurd- बेतुका
Ambiguous-अनेकाथी
Axiom-स्वयंसिद्ध
 Conjecture-अनुमान से निर्णय करना


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