UGC NET CS 2017 Jan -paper-2

 Question 1
Consider a sequence F​ 00​ defined as : F​ 00​ (0) = 1, F​ 00​ (1) = 1
F​ 00​ (n) = ((10 ∗ F​ 00​ (n – 1) + 100)/ F​ 00​ (n – 2)) for n ≥ 2 Then what shall be the set of values of the sequence F​ 00​ ?
 A (1, 110, 1200) B (1, 110, 600, 1200) C (1, 2, 55, 110, 600, 1200) D (1, 55, 110, 600, 1200)
Question 1 Explanation:
Given data,
Sequence F​ 00​ defined as
F​ 00​ (0) = 1,
F​ 00​ (1) = 1,
F​ 00​ (n) = ((10 ∗ F​ 00​ (n – 1) + 100)/ F​ 00​ (n – 2)) for n ≥ 2
Let n=2
F​ 00​ (2) = (10 * F​ 00​ (1) + 100) / F​ 00​ (2 – 2)
= (10 * 1 + 100) / 1
= (10 + 100) / 1
= 110
Let n=3
F​ 00​ (3) = (10 * F​ 00​ (2) + 100) / F​ 00​ (3 – 2)
= (10 * 110 + 100) / 1
= (1100 + 100) / 1
= 1200
Similarly, n=4
F​ 00​ (4) = (10 * F​ 00​ (3) + 100) / F​ 00​ (4 – 2)
= (12100) / 110
= 110
F​ 00​ (5) = (10 * F​ 00​ (4) + 100) / F​ 00​ (5 – 2)
= (10*110 + 100) / 1200
= 1
The sequence will be (1, 110, 1200,110, 1).
 Question 2
Match the following :
 A a-i, b-ii, c-iii, d-iv B a-i, b-iii, c-iv, d-ii C a-ii, b-iii, c-iv, d-i D a-ii, b-ii, c-iii, d-iv
Question 2 Explanation:
Absurd→ Clearly impossible being contrary to some evident truth.
Ambiguous→ Capable of more than one interpretation or meaning.
Axiom→ An assertion that is accepted and used without a proof.
Conjecture→ An opinion preferably based on some experience or wisdom
 Question 3
The functions mapping R into R are defined as : f(x) = x​ 3​ – 4x, g(x) = 1/(x​ 2​ + 1) and h(x) = x​ 4​ . Then find the value of the following composite functions : hog(x) and hogof(x)
 A (x​ 2​ + 1)4 and [(x​ 3​ – 4x)​ 2​ + 1]​ 4 B (x​ 2​ + 1)4 and [(x​ 3​ – 4x)​ 2​ + 1]​ -4 C (x​ 2​ + 1)-4 and [(x​ 3​ – 4x)​ 2​ + 1]​ 4 D (x​ 2​ + 1)-4 and [(x​ 3​ – 4x)​ 2​ + 1]​ -4
Question 3 Explanation:
Step-1: Given data,
f(x) = x​ 3​ – 4x, g(x) = 1/(x​ 2​ + 1) and h(x) = x​ 4
hog(x)=h(1/(x​ 2​ + 1))
=h(1/(x​ 2​ )+1)​ 4
= 1/(x​ 2​ +1)​ 4
= (x​ 2​ +1)​ -4
hogof(x)= hog(x​ 3​ -4x)
= h(1/(x​ 3​ -4x)​ 2​ +1)
= h(1/(x​ 3​ -4x)​ 2​ +1)​ 4
= h((x​ 3​ -4x)​ 2​ +1)​ -4
So, option D id is correct answer.
 Question 4
How many multiples of 6 are there between the following pairs of numbers ?
0 and 100 and –6 and 34
 A 16 and 6 B 17 and 6 C 17 and 7 D 16 and 7
Question 4 Explanation:
Method-1:
0 and 100 → Counting sequentially:
0,6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96 Total=17
–6 and 34 → Counting sequentially: -6,0,6,12,18,24,30
Total=7
Method-2: 0 and 100 → Maximum number is 100. Divide ⌊100/6⌋ = 16+1 =17
Here, +1 because of 0.
–6 and 34 → Maximum number is 34. Divide ⌊34/6⌋ = 5+1+1 =7
Here, +1 because of 0 and +1 for -6
 Question 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 ?
 A deg(v) ≥n/2 for each vertex v. B |E(G)| ≥1/2(n – 1) (n – 2) + 2 C deg (v) + deg(w) ≥ n whenever v and w are not connected by an edge D All of the above
Question 5 Explanation:
With the help of dirac’s theorem, we can prove above three statements.
 Question 6
In propositional logic if (P → Q) ∧ (R → S) and (P ∨ R) are two premises such that
 A P ∨ R B P ∨ S C Q ∨ R D Q ∨ S E None of These
Question 6 Explanation:
Option-A: Let P be TRUE and R be false, then the conclusion PVR will be TRUE. Now if we make Q as false then premises (P→Q)∧(R→ S)will be false because P→ Q is false. Hence this option is not correct.
Option-B: Let P be TRUE and S be false then the conclusion PVS is TRUE. Now, if we make R as TRUE then the premises (P→ Q)∧(R→ S) will be false because (R→ S) will be false. Hence this option is not correct.
Option-C: Let Q be false, R be TRUE then conclusion QVR will be TRUE. Now if we make S as FALSE then Premises (P→ Q)∧(R→ S) will be FALSE because (R→ S) will be false. Hence, this option is not correct.
Option-D: Let Q be TRUE and S be FALSE then conclusion QVS will be TRUE. Now if we make R as TRUE then premises (P→ Q)∧(R→ S) will be FALSE because (R→ S) will be false.
Therefore None of the given options are correct.
Note: As per UGC NET key, given option D as correct answer.
 Question 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 ____.
 A Race condition B Saturation C Delay D High impedance
Question 7 Explanation:
→ 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 Saturation.
→ Emitter-coupled logic (ECL) is a high-speed integrated circuit bipolar transistor logic family. ECL uses an overdriven BJT differential amplifier with single-ended input and limited emitter current to avoid the saturated (fully on) region of operation and its slow turn-off behavior.
→ As the current is steered between two legs of an emitter-coupled pair, ECL is sometimes called current-steering logic (CSL), current-mode logic (CML) or current-switch emitter-follower (CSEF) logic.[
 Question 8
A binary 3 bit down counter uses J-K flip-flops, FF​ i​ with inputs J​ i​ , K​ i​ and outputs Q​ i​ , i=0,1,2 respectively. The minimized expression for the input from following, is
I. J​ 0​ = K​ 0​ = 0
II. J​ 0​ = K​ 0​ = 1
III. J​ 1​ = K​ 1​ = Q​ 0
IV. J​ 1​ = K​ 1​ =Q’0
V. J​ 2​ = K​ 2​ = Q​ 1​ Q​ 0
VI. J​ 2​ = K​ 2​ = Q’​ 1​ Q’​ 0
 A I,III,V B I,IV,VI C II,III,V D II,IV,VI
Question 8 Explanation:
In a JK flip-flop, Qn=Q(bar) iff J=K=1.
State sequence of down counter is as follows:
 Question 9
Convert the octal number 0.4051 into its equivalent decimal number.
 A 0.51001 B 0.2096 C 0.52 D 0.4192
Question 9 Explanation:
 Question 10
The hexadecimal equivalent of the octal number 2357 is :
 A 2EE B 2FF C 4EF D 4FE
Question 10 Explanation:
Step-1: Convert octal number into binary number
(2357)​ 8​ = (010 011 101 111)​ 2
Step-2: Divide 4 bits from LSB then will get hexadecimal number
0100 1110 1111
2 E F
(2EF)​ 16​ = (2357)​ 8
 Question 11
Which of the following cannot be passed to a function in C++ ?
 A Constant B Structure C Array D Header file
Question 11 Explanation:
→ Header files contain definitions of Functions and Variables, which is imported or used into any C++ program by using the pre-processor #include statement. Header file have an extension ".h" which contains C++ function declaration and macro definition.
→ Header file is a library file, we can not passed to a function in C++.
→ We can pass constant,Structure and Array
 Question 12
Which one of the following is correct for overloaded functions in C++ ?
 A Compiler sets up a separate function for every definition of function. B Compiler does not set up a separate function for every definition of function. C Overloaded functions cannot handle different types of objects. D Overloaded functions cannot have same number of arguments.
Question 12 Explanation:
→ Function overloading allows you to use the same name for different functions, to perform, either same or different functions in the same class.
→ Compiler sets up a separate function for every definition of function.
→ Two ways to use overloaded function is
1. By changing number of Arguments.
2. By having different types of argument.
There are 12 questions to complete.

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