## UGC NET CS 2017 Jan -paper-2

Question 1 |

Consider a sequence F 00 defined as : F

F

_{ 00} (0) = 1, F_{00} (1) = 1F

_{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} ?(1, 110, 1200) | |

(1, 110, 600, 1200) | |

(1, 2, 55, 110, 600, 1200) | |

(1, 55, 110, 600, 1200) |

Question 1 Explanation:

Given data,

Sequence F

F

F

F

Let n=2

F

= (10 * 1 + 100) / 1

= (10 + 100) / 1

= 110

Let n=3

F

= (10 * 110 + 100) / 1

= (1100 + 100) / 1

= 1200

Similarly, n=4

F

= (12100) / 110

= 110

F

= (10*110 + 100) / 1200

= 1

The sequence will be (1, 110, 1200,110, 1).

Sequence F

_{00} defined asF

_{00} (0) = 1,F

_{00} (1) = 1,F

_{00} (n) = ((10 ∗ F_{00} (n – 1) + 100)/ F_{00} (n – 2)) for n ≥ 2Let 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) + 1_{00}) / 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-i, b-ii, c-iii, d-iv | |

a-i, b-iii, c-iv, d-ii | |

a-ii, b-iii, c-iv, d-i | |

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

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)(x ^{2} + 1)4 and [(x^{ 3} – 4x)^{ 2} + 1]^{ 4} | |

(x ^{2} + 1)4 and [(x^{ 3} – 4x)^{ 2} + 1]^{ -4} | |

(x ^{2} + 1)-4 and [(x^{ 3} – 4x)^{ 2} + 1]^{ 4} | |

(x ^{2} + 1)-4 and [(x^{ 3} – 4x)^{ 2} + 1]^{ -4} |

Question 3 Explanation:

Step-1: Given data,

f(x) = x

hog(x)=h(1/(x

=h(1/(x

= 1/(x

= (x

hogof(x)= hog(x

= h(1/(x

= h(1/(x

= h((x

So, option D id is correct answer.

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

0 and 100 and –6 and 34

16 and 6 | |

17 and 6 | |

17 and 7 | |

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 ?

deg(v) ≥n/2 for each vertex v. | |

|E(G)| ≥1/2(n – 1) (n – 2) + 2 | |

deg (v) + deg(w) ≥ n whenever v and w are not connected by an edge | |

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

P ∨ R | |

P ∨ S | |

Q ∨ R | |

Q ∨ S | |

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.

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 ____.

Race condition | |

Saturation | |

Delay | |

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.[

→ 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

II. J

III. J

IV. J

V. J

VI. J

I. J

_{0} = K_{0} = 0II. J

_{0} = K_{0} = 1III. 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}I,III,V | |

I,IV,VI | |

II,III,V | |

II,IV,VI |

Question 8 Explanation:

In a JK flip-flop, Q

State sequence of down counter is as follows:

_{n}=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.

0.5100098 | |

0.2096 | |

0.52 | |

0.4192 |

Question 9 Explanation:

Question 10 |

The hexadecimal equivalent of the octal number 2357 is :

2EE | |

2FF | |

4EF | |

4FE |

Question 10 Explanation:

Step-1: Convert octal number into binary number

(2357)

Step-2: Divide 4 bits from LSB then will get hexadecimal number

0100 1110 1111

2 E F

(2EF)

(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++ ?

Constant | |

Structure | |

Array | |

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

→ 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++ ?

Compiler sets up a separate function for every definition of function. | |

Compiler does not set up a separate function for every definition of function. | |

Overloaded functions cannot handle different types of objects. | |

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.

→ 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.