# UGC NET Computer Science Solved Paper II DECEMBER 2004

1. AVA = A is called:

(A) Identity law             (B) De Morgan’s law

(C) Idempotent law      (D) Complement law

Explanation

1. Commutative Laws:

For any two finite sets A and B;

(i) A U B = B U A

(ii) A ∩ B = B ∩ A

2. Associative Laws:

For any three finite sets A, B and C;

(i) (A U B) U C = A U (B U C)

(ii) (A ∩ B) ∩ C = A ∩ (B ∩ C)

Thus, union and intersection are associative.

3. Idempotent Laws:

For any finite set A;

(i) A U A = A

(ii) A ∩ A = A

4. Distributive Laws:

For any three finite sets A, B and C;

(i) A U (B ∩ C) = (A U B) ∩ (A U C)

(ii) A ∩ (B U C) = (A ∩ B) U (A ∩ C)

Thus, union and intersection are distributive over intersection and union respectively.

5. De Morgan’s Laws:

For any two finite sets A and B;

(i) A – (B U C) = (A – B) ∩ (A – C)

(ii) A – (B ∩ C) = (A – B) U (A – C)

De Morgan’s Laws can also we written as:

(i) (A U B)’ = A’ ∩ B’

(ii) (A ∩ B)’ = A’ U B’

More laws of algebra of sets:

6. For any two finite sets A and B;

(i) A – B = A ∩ B’

(ii) B – A = B ∩ A’

(iii) A – B = A ⇔ A ∩ B = ∅

1. If f(x) =x+1 and g(x)=x+3 then fofofof is:

(A) g       (B) g+1

(C) g4     (D) None of the above

Explanation:

If f(x)=x+1 and g(x)=x+3, then fog(x)=x+4 and gof(x)=x+4

fof(x)=x+2, fofof(x)=x+3 and fofofof(x)=x+4.=x+3+1=g+1

1. The context-free languages are closed for:

(i) Intersection                          (ii) Union

(iii) Complementation             (iv) Kleene Star

then

(A) (i) and (iv)               (B) (i) and (iii)

(C) (ii) and (iv)              (D) (ii) and (iii)

Explanation:

Context-free languages are closed under the following operations. That is, if L and P are context-free languages, the following languages are context-free as well:

• the union {\displaystyle L\cup P} of L and P
• the reversal of L
• the concatenation {\displaystyle L\cdot P} of L and P
• the Kleene star {\displaystyle L^{*}} of L
• the image {\displaystyle \varphi (L)} of L under a homomorphism {\displaystyle \varphi }
• the image {\displaystyle \varphi ^{-1}(L)} of L under an inverse homomorphism {\displaystyle \varphi ^{-1}}
• the cyclic shift of L (the language {\displaystyle \{vu:uv\in L\}})

Context-free languages are not closed under complement, intersection, or difference. However, if L is a context-free language and D is a regular language then both their intersection {\displaystyle L\cap D} and their difference {\displaystyle L\setminus D} are context-free languages.

ref: https://en.wikipedia.org/wiki/Context-free_language

1. The following lists are the degrees of all the vertices of a graph:

(i) 1, 2, 3, 4, 5                (ii) 3, 4, 5, 6, 7

(iii) 1, 4, 5, 8, 6              (iv) 3, 4, 5, 6

then, which of the above sequences are graphic?

(A) (i) and (ii)

(B) (iii) and (iv)

(C) (iii) and (ii)

(D) (ii) and (iv)

Explanation:

Rest can’t be graphs as the number of vertices with odd degree in a graph should be even.

In list (i), number of vertices with odd degree is 3.

In list (ii), number of vertices with odd degree is 3.

In list (iii), number of vertices with odd degree is 2.

In list (iv), number of vertices with odd degree is 2.

1. If Im denotes the set of integers modulo m, then the following are fields with respect to the operations of addition modulo m and multiplication modulo m:

(i) Z23      (ii) Z29

(iii) Z31    (iv) Z33

Then

(A) (i) only

(B) (i) and (ii) only

(C) (i), (ii) and (iii) only

(D) (i), (ii), (iii) and (iv)

Explanation:

Basically, a field is a thing where you can add, subtract, multiply and divide. It is a bit tricky to see that the first three examples (Z23, Z29, Z31) are indeed fields. In fact, ZpZp happens to be a field always when pp is prime, and this result follows from Fermat’s little theorem.

But let us look at the fourth example. Assume you can divide the elements by 11, then you have

a contradiction. (The latter equality holds because 33=033=0 modulo 3333.) A similar argument shows you that ZqZq cannot be a field if qq is any composite number.

1. An example of a binary number which is equal to its 2’s complement is:

(A) 1100                        (B) 1001

(C) 1000                        (D) 1111

Explanation:

Option a : 1100 and 1s compliment 0011 +1 = 0100

and apply to all only c is the option which gives answer.

1. When a tri-state logic device is in the third state, then:

(A) it draws low current

(B) it does not draw any current

(C) it draws very high current

(D) it presents a low impedance

Explanation:

Tristate means that the pin is used: to output a high state (1) or output a low state (2) AND it can be directed to enter a high-z/high impedance state (3). The last state, high impedance, is designed to prevent other active devices from driving this line simultaneously–never connect two active outputs together. Things that are not tristated would be a dedicated input or output lines. Any bi-directional lines (both input and output) are always tristated: example data bus.

1. An example of a connective which is not associative is:

(A) AND             (B) OR

(C) EX-OR        (D) NAND

check in detail: https://en.wikipedia.org/wiki/Logical_connective

1. Essential hazards may occur in:

(A) Combinational logic circuits

(B) Synchronous sequential logic circuits

(C) Asynchronous sequential logic circuits working in the fundamental mode

(D) Asynchronous sequential logic circuits working in the pulse mode

1. The characteristic equation of a T flip-flop is:

(A) Qn+1=TQ’n + T’ Qn

(B) Qn+1=T+Qn

(C) Qn+1=TQn

(D) Qn+1= T’Q’n

The symbols used have the usual meaning.

FLIP-FLOP NAME FLIP-FLOP SYMBOL CHARACTERISTIC TABLE CHARACTERISTIC EQUATION EXCITATION TABLE
SR
 S R Q(next) 0 0 Q 0 1 0 1 0 1 1 1 ?
Q(next) = S + R’QSR = 0
 Q Q(next) S R 0 0 0 X 0 1 1 0 1 0 0 1 1 1 X 0
JK
 J K Q(next) 0 0 Q 0 1 0 1 0 1 1 1 Q’
Q(next) = JQ’ + K’Q
 Q Q(next) J K 0 0 0 X 0 1 1 X 1 0 X 1 1 1 X 0
D
 D Q(next) 0 0 1 1
Q(next) = D
 Q Q(next) D 0 0 0 0 1 1 1 0 0 1 1 1
T
 T Q(next) 0 Q 1 Q’
Q(next) = TQ’ + T’Q
 Q Q(next) T 0 0 0 0 1 1 1 0 1 1 1 0

1. Suppose x and y are two Integer Variables having values 0x5AB6 and 0x61CD respectively. The result (in hex) of applying bitwise operator AND to x and y will be:

(A) 0x5089        (B) 0x4084

Explanation:
0x5AB6=0101 1010 1011 0110
0x61CD=0110 0001 1100 1101
—————————————
AND     =0100 0000 1000 0100
= 0x4084

1. Consider the following statements,

int i=4, j=3, k=0;

k=++i – –j + i++ – –j +j++;

What will be the values of i, j and k after the statement.

(A) 7, 2, 8           (B) 5, 2, 10

(C) 6, 2, 8          (D) 4, 2, 8

1. What is the value of the arithmetic expression (Written in C)

2*3/4-3/4* 2

(A) 0       (B) 1

(C) 1.5    (D) None of the above

Explanation:

6/4 – 0 = 1

1. A function object:

(A) is an instance of a class for which operator () is a member function.

(B) is an instance of a class for which operator → is a member function.

(C) is a pointer to any function

(D) is a member function of a class

1. Polymorphism means:

(A) A template function

(B) Runtime type identification within a class hierarchy

(D) Virtual inheritance

1. The E-R model is expressed in terms of:

(i) Entities

(ii) The relationship among entities

(iii) The attributes of the entities

Then

(A) (i) and (iii)

(B) (i), (ii) and (iii)

(C) (ii) and (iii)

(D) None of the above

1. Specialization is a …………… process.

(A) Top – down             (B) Bottom -Up

(C) Both (A) and (B)    (D) None of the above

1. The completeness constraint has rules:

(A) Supertype, Subtype

(B) Total specialization, Partial specialization

(C) Specialization, Generalization

(D) All of the above

1. The entity type on which the …………….. type depends is called the identifying owner.

(A) Strong entity           (B) Relationship

(C) Weak entity            (D) E – R

1. Match the following:

(i) 5 NF              (a) Transitive dependencies eliminated

(ii) 2 NF                         (b) Multivalued attribute removed

(iii) 3 NF            (c) Contains no partial functional dependencies

(iv) 4 NF            (d) Contains no join dependency

(A) i-a, ii-c, iii-b, iv-d

(B) i-d, ii-c, iii-a, iv-b

(C) i-d, ii-c, iii-b, iv-a

(D) i-a, ii-b, iii-c, iv-d

1. What item is at the root after the following sequence of insertions into an empty splay tree:

1, 11, 3, 10, 8, 4, 6, 5, 7, 9, 2 ?

(A) 1       (B) 2

(C) 4       (D) 8

Explanation:

1. Suppose we are implementing quadratic probing with a Hash function, Hash (y)=X mode 100. If an element with key 4594 is inserted and the first three locations attempted are already occupied, then the next cell that will be tried is:

(A) 2       (B) 3

(C) 9       (D) 97

Explanation:

because Hash(y)= 4594 mod 100=94.
But first 3 location are attempted are already occupy i.e 94,95,96 So 97 is answer.

1. Weighted graph:

(A) Is a bi-directional graph

(B) Is directed graph

(C) Is graph in which number associated with arc

(D) Eliminates table method

1. What operation is supported in constant time by the doubly linked list, but not by the singly linked list?

(C) First                         (D) Retrieve

1. How much extra space is used by heap sort?

(A) O(1)              (B) O(Log n)

(C) O(n)             (D) O(n2)

Explanation:

Answer is O(1) and Bubble sort is an “in place” algorithm. Other than a temporary “switch” variable, no extra space is required.

1. Error control is needed at the transport layer because of potential error occurring …………..

(A) from transmission line noise

(B) in router

(C) from out of sequence delivery

(D) from packet losses

1. Making sure that all the data packets of a message are delivered to the destination is ……………. control.

(A) Error             (B) Loss

(C) Sequence (D) Duplication

1. Which transport class should be used with a perfect network layer?

(A) TP0 and TP2          (B) TP1 and TP3

(C) TP0, TP1, TP3       (D) TP0, TP1, TP2, TP3, TP4

1. Which transport class should be used with residual-error network layer?

(A) TP0, TP2                 (B) TP1, TP3

(C) TP1, TP3, TP4       (D) TP0, TP1, TP2, TP3, TP4

1. Virtual circuit is associated with a ………………… service.

(A) Connectionless     (B) Error-free

(C) Segmentation        (D) Connection-oriented

1. Which activity is not included in the first pass of two pass assemblers?

(A) Build the symbol table

(B) Construct the intermediate code

(C) Separate mnemonic opcode and operand fields

(D) None of the above

1. Which of the following is not collision resolution technique?

(C) Both (A) and (B)    (D) Indexing

Explanation:

click below for detail in techniques

collision resolution techniques

1. Code optimization is responsibility of:

(A) Application programmer

(B) System programmer

(C) Operating system

(D) All of the above

1. Which activity is included in the first pass of two pass assemblers?

(A) Build the symbol table

(B) Construct the intermediate code

(C) Separate mnemonic opcode and operand fields

(D) None of these

In the first pass of a two-pass assembler, it will “prepare” for the second pass, ie. it’ll build the symbol table which is later used in the second pass to generate code.

1. In two pass assembler the symbol table is used to store:

(A) Label and value    (B) Only value

(C) Mnemonic              (D) Memory Location

1. Semaphores are used to:

(A) Synchronise critical resources to prevent deadlock

(B) Synchronise critical resources to prevent contention

(C) Do I/o

(D) Facilitate memory management

1. In which of the following storage replacement strategies, is a program placed in the largest available hole in the memory?

(A) Best fit         (B) First fit

(C) Worst fit       (D) Buddy

1. Remote computing system involves the use of timesharing systems and:

(A) Real time processing        (B) Batch processing

(C) Multiprocessing                 (D) All of the above

1. Non modifiable procedures are called

(A) Serially useable procedures

(B) Concurrent procedures

(C) Reentrant procedures

(D) Topdown procedures

1. Match the following

(a) Disk scheduling                 (1) Round robin

(b) Batch processing               (2) Scan

(c) Time sharing                       (3) LIFO

(d) Interrupt processing          (4) FIFO

(A) a-3, b-4, c-2, d-1

(B) a-4, b-3, c-2, d-1

(C) a-2, b-4, c-1, d-3

(D) a-3, b-4, c-1, d-2

1. The main objective of designing various modules of a software system is:

(A) To decrease the cohesion and to increase the coupling

(B) To increase the cohesion and to decrease the coupling

(C) To increase the coupling only

(D) To increase the cohesion only

1. Three essential components of a software project plan are:

(A) Team structure, Quality assurance plans, Cost estimation

(B) Cost estimation, Time estimation, Quality assurance plan

(C) Cost estimation, Time estimation, Personnel estimation

(D) Cost estimation, Personnel estimation, Team structure

1. Reliability of software is dependent on:

(A) Number of errors present in software

(B) Documentation

(C) Testing suties

(D) Development Processes

1. In transform analysis, input portion is called:

(A) Afferent branch                 (B) Efferent branch

(C) Central Transform             (D) None of the above

1. The Function Point (FP) metric is:

(A) Calculated from user requirements

(B) Calculated from Lines of code

(C) Calculated from software’s complexity assessment

(D) None of the above

1. Data Mining can be used as …………….. Tool.

(A) Software     (B) Hardware

(C) Research    (D) Process

1. The processing speeds of pipeline segments are usually:

(A) Equal           (B) Unequal

(C) Greater        (D) None of these

1. The cost of a parallel processing is primarily determined by:

(A) Time complexity

(B) Switching complexity

(C) Circuit complexity

(D) None of the above

1. A data warehouse is always ………………..

(A) Subject oriented    (B) Object oriented

(C) Program oriented (D) Compiler oriented

1. The term ‘hacker’ was originally associated with:

(A) A computer program

(B) Virus

(C) Computer professionals who solved complex computer problems.

(D) All of the above