## UGC NET CS 2005 Dec-Paper-2

Question 1 |

T is a graph with n vertices. T is connected and has exactly n-1 edges, then :

T is a tree | |

T contains no cycles | |

Every pairs of vertices in T is connected by exactly one path | |

All of these |

Question 1 Explanation:

This is little bit tricky question.

Step-1:

n= number of vertices

n-1 = number of edges

Example: n=5 vertices and n-1=4 edges

Step-2: The above graph T won’t have cycle then we are calling as tree. Here, every pairs of vertices in T is connected by exactly one path.

Note: The above properties is nothing but minimum spanning tree properties.

Step-1:

n= number of vertices

n-1 = number of edges

Example: n=5 vertices and n-1=4 edges

Step-2: The above graph T won’t have cycle then we are calling as tree. Here, every pairs of vertices in T is connected by exactly one path.

Note: The above properties is nothing but minimum spanning tree properties.

Question 2 |

If the proposition ¬ P → Q is true, then the truth value of the proportion ¬ PV (P → Q) is:

True | |

Multi - Valued | |

Flase | |

Can not determined |

Question 2 Explanation:

We can also write (¬p → q) into (p ∨ q)

Here, we can minimize the boolean form into

¬p ∨ (p → q)

= ¬p ∨ (¬p ∨ q)

= ¬p ∨ q

In question, they are not discussed about the truth values of p, it implies that ¬p ∨ q also be True or False. So, we cannot be determined.

Here, we can minimize the boolean form into

¬p ∨ (p → q)

= ¬p ∨ (¬p ∨ q)

= ¬p ∨ q

In question, they are not discussed about the truth values of p, it implies that ¬p ∨ q also be True or False. So, we cannot be determined.

Question 3 |

Let A and B be two arbitrary events, then :

P(A∩B) = P(A) P(B) | |

P(P∪B) = P(A) + P(B) | |

P(A∪B) ≤ P(A) + P(B) | |

P(A / B) = P(A∩B) + P(B) |

Question 3 Explanation:

Option-A is happens when A and B are independent.

Option-B is happens when A and B are mutually exclusive.

Option-C is not happens.

Option-D is P(A∪B) ≤ P(A) + P(B) is true because P(A∪B) = P(A) + P(B) - P(A∩B).

Option-B is happens when A and B are mutually exclusive.

Option-C is not happens.

Option-D is P(A∪B) ≤ P(A) + P(B) is true because P(A∪B) = P(A) + P(B) - P(A∩B).

Question 4 |

Which sentence can be generated by

S→d/bA

A→d/ccA

S→d/bA

A→d/ccA

bccddd | |

aabccd | |

ababccd | |

abbbd | |

None of the above |

Question 4 Explanation:

Here, there is no terminal symbol ‘a’, so the answer never be option B,C,D.

Option-A is also wrong because the string generated by grammar is bccccd.

Option-A is also wrong because the string generated by grammar is bccccd.

Question 5 |

Regular expression a+b denotes the set :

{a} | |

{ε, a, b} | |

{a, b} | |

None of these |

Question 5 Explanation:

The regular expression a+b denotes the set {a, b}.

Question 6 |

Which of the following is divisible by 4 ?

100101100 | |

1110001110001 | |

11110011 | |

10101010101010 |

Question 6 Explanation:

Option-A: (100101100)

300 is divisible by 4

Option-B: (1110001110001)

7281 is not divisible by 4

Option-C: (11110011)

243 is not divisible by 4

Option-D: (10101010101010)

10,922 is not divisible by 4.

_{2} = (300)_{10}300 is divisible by 4

Option-B: (1110001110001)

_{2} = (7281)_{ 10}7281 is not divisible by 4

Option-C: (11110011)

_{2} = (243)_{ 10}243 is not divisible by 4

Option-D: (10101010101010)

_{2} = (10,922)_{ 10}10,922 is not divisible by 4.

Question 7 |

A half-adder is also known as :

AND Circuit | |

NAND Circuit | |

NOR Circuit | |

EX-OR Circuit |

Question 7 Explanation:

→ The half adder adds two single binary digits A and B. It has two outputs, sum (S) and carry (C). The carry signal represents an overflow into the next digit of a multi-digit addition. The value
of the sum is 2C + S. The simplest half-adder design incorporates an XOR gate for S and an AND gate for C. The Boolean logic for the sum (in this case S) will be A′B + AB′ whereas for the carry (C) will be AB.

Question 8 |

Consider the following sequence of instructions :

a=a⊕b, b=a⊕b, a=b⊕a

This Sequence

a=a⊕b, b=a⊕b, a=b⊕a

This Sequence

retains the value of the a and b | |

complements the value of a and b | |

swap a and b | |

negates values of a and b |

Question 8 Explanation:

⇒ a=a⊕b

⇒ b=a ⊕b

= a⊕b ⊕b (Substitute (1))

= 0 ⊕ b

= b

⇒ a = b⊕a

= b⊕a⊕b (Substitute (1))

= a⊕b⊕b

= a⊕0

= a

⇒ b=a ⊕b

= a⊕b ⊕b (Substitute (1))

= 0 ⊕ b

= b

⇒ a = b⊕a

= b⊕a⊕b (Substitute (1))

= a⊕b⊕b

= a⊕0

= a

Question 9 |

Consider the following circuit :

to make it a Tautology the ? should be :

to make it a Tautology the ? should be :

NAND gate | |

AND gate | |

OR gate | |

EX-OR gate |

Question 9 Explanation:

Method 1:

(X+Y) + (XY)’

= X+Y+X’+Y’

= (X+X’)+(Y+Y’)

= 1

Method 2:

(X+Y) + (XY)’

= X+Y+X’+Y’

= (X+X’)+(Y+Y’)

= 1

Method 2:

Question 10 |

When an inventor is placed between both inputs of an S-R flip flop, the resulting flip flop is :

JK flip-flop | |

D-flip-flop | |

T flip-flop | |

None of these |

Question 10 Explanation:

Given question is ambiguous. It is not mentioned how the inverter is connected. Placing NOT gate in different positions gives different solutions.

Question 11 |

What is the output of the following C-program

main()

{

printf(''%d%d%d'', sizeof(3.14f), sizeof(3.14), sizeof(3.141));

}

main()

{

printf(''%d%d%d'', sizeof(3.14f), sizeof(3.14), sizeof(3.141));

}

4 4 4 | |

4 8 10 | |

8 4 8 | |

8 8 8 | |

None of the above |

Question 11 Explanation:

The sizeof operator will print number of bytes of a data type.

sizeof(3.14f) → It will consider as float data type. The float data type size is 4 bytes.

sizeof(3.14) → It will consider as double data type. The double data type size is 8 bytes.

sizeof(3.141) → It will consider as double data type. The double data type size is 8 bytes.

Output= 4 8 8

Note: The exact size of each of these 3 types depends on the C compiler implementation (or) platform

sizeof(3.14f) → It will consider as float data type. The float data type size is 4 bytes.

sizeof(3.14) → It will consider as double data type. The double data type size is 8 bytes.

sizeof(3.141) → It will consider as double data type. The double data type size is 8 bytes.

Output= 4 8 8

Note: The exact size of each of these 3 types depends on the C compiler implementation (or) platform

Question 12 |

The bitwise OR of 35 with 7 in C will be :

35 | |

7 | |

42 | |

39 |

Question 12 Explanation:

Step-1: First we have to convert into binary numbers

(35)

(7)

Step-2: Perform OR operation on both binary numbers.

0010 0011

0000 0111

---------------

0010 0111

----------------

Step-3: Convert result into Decimal number

(0010 0111)

(35)

_{10} = ( 0010 0011 ) _{2}(7)

_{10} = (0000 0111)_{ 2}Step-2: Perform OR operation on both binary numbers.

0010 0011

0000 0111

---------------

0010 0111

----------------

Step-3: Convert result into Decimal number

(0010 0111)

_{2} = (39)_{ 10}