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)
       Engineering-Mathematics       Combinatorics
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
       Engineering-Mathematics       Calculus
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
       Engineering-Mathematics       Combinatorics
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
       Engineering-Mathematics       Graph-Theory
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
       Engineering-Mathematics       Propositional-Logic
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
       Data-Communication       ideal-op-amp
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
       Digital-Logic-Design       Sequential-Circuits
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
0.5100098
B
0.2096
C
0.52
D
0.4192
       Digital-Logic-Design       Number-Systems
Question 9 Explanation: 
Question 10
The hexadecimal equivalent of the octal number 2357 is :
A
2EE
B
2FF
C
4EF
D
4FE
       Digital-Logic-Design       Number-Systems
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
       Programming-in-c++       Functions
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.
       Programming-in-c++       Properties
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.
Question 13
Which of the following storage classes have global visibility in C/C++ ?
A
Auto
B
Extern
C
Static
D
Register
       Programming-in-c++       Storage-Classes
Question 13 Explanation: 
→ Register and auto variables have limited scope (the block in which they are declared) and limited lifetimes (as for automatic variables). However, in some applications it may be useful to have data which is accessible from within any block and/or which remains in existence for the entire execution of the program. Such variables are called global variables. The scope of external variables is global.
Question 14
Which of the following operators cannot be overloaded in C/C++ ?
A
Bitwise right shift assignment
B
Address of
C
Indirection
D
Structure reference
       Programming-in-c++       Properties
Question 14 Explanation: 
→ We can’t overload structure reference in C and C++.
→ Bitwise right shift assignment(>>), Address of(&) and Indirection(*) can be overloaded in C and C++.
→ The . (dot) operator and the → (arrow) operator are used to reference individual members of classes, structures, and unions.
Question 15
If X is a binary number which is power of 2, then the value of X & (X – 1) is :
A
11....11
B
00.....00
C
100.....0
D
000......1
       Digital-Logic-Design       Number-Systems
Question 15 Explanation: 
Given data,
→ X is binary number which is power of 2. It means, we have to take powers of 2 numbers only.
Ex: 1,2,4,8,16,32,..,
Let X=4
X=4 equivalent binary number is 100
X-1=3 equivalent binary number is 011
100
011
-----
000 (AND operation)
-----
Ex-2:
X=8 and X-1=7
8 binary value is
1000
7 binary number is 0111
--------
0000(AND operation)
--------
So, Option B is correct answer.
Question 16
An attribute A of data type varchar (20) has value ‘Ram’ and the attribute B of data type char (20) has value ‘Sita’ in oracle. The attribute A has _______ memory spaces and B has _______ memory spaces.
A
20,20
B
3,20
C
3,4
D
20,4
       Database-Management-System       SQL
Question 16 Explanation: 
VARCHAR is variable length and CHAR is fixed length.
Given, varchar(20) and has value ‘Ram’ it means 3 spaces.
char(20) and has value ‘Sita’ but char is fixed spaces. It will take 20 spaces.
Question 17
Integrity constraints ensure that changes made to the database by authorized users do not result into loss of data consistency. Which of the following statement(s) is (are) true w.r.t. the examples of integrity constraints ?
(A) An instructor Id. No. cannot be null, provided Instructor Id No. being primary key.
(B) No two citizens have same Adhar-Id.
(C) Budget of a company must be zero.
A
(A), (B) and (C) are true.
B
(A) false, (B) and (C) are true.
C
(A) and (B) are true; (C) false.
D
(A), (B) and (C) are false
       Database-Management-System       Constraints
Question 17 Explanation: 
Primary key, every table have at least one attribute as “primary key”. No primary key should ever have null values.
Statement-1: TRUE: Here, instructor ID is primary key. So, they mentioned it is not null.
Statement-2: TRUE: No two citizens have same Adhar-Id. It is clear example of primary key.
Statement-3: FALSE: Budget of a company must be zero. It is violated constraint of “not null”. Because entity integrity constraints should not accept not null but referential integrity constraints will accept null value.
Question 18
Let M and N be two entities in an E-R diagram with simple single value attributes.
R​ 1​ and R​ 2​ are two relationship between M and N, where as
R​ 1​ is one-to-many and R​ 2​ is many-to-many.
The minimum number of tables required to represent M, N, R​ 1​ and R​ 2​ in the relational model are _______.
A
4
B
6
C
7
D
3
       Database-Management-System       ER-Model
Question 18 Explanation: 
R​ 1​ and R​ 2​ two relationships between M and N
R​ 1​ is one to many.
R​ 2​ is many to many.
→ M and N have separate table because they need to store multiple values.
→ R​ 2​ also have separate table by considering Primary keys M and N as foreign keys.
→ R​ 1​ is converted to many side table i.e., N as Primary key and M as Foreign key.
So, totally we need 3 tables to store the value.
Question 19
Consider a schema R(MNPQ) and functional dependencies M → N, P → Q. Then the decomposition of R into R​ 1​ (MN) and R​ 2​ (PQ) is________.
A
Dependency preserving but not lossless join
B
Dependency preserving and lossless join
C
Lossless join but not dependency preserving
D
Neither dependency preserving nor lossless join.
       Database-Management-System       Functional-Dependency
Question 19 Explanation: 
Definition of lossless decomposition:​ Let R be the relational schema decomposed into R​ 1
and R​ 2​ . Given decomposition is lossless only if
1. R​ 1​ U R​ 2​ =R
2. R​ 1​ U R​ 2​ =φ
3. R​ 1 ∩ R​ 2​ → R​ 1 ​ (or) R​ 1 ∩ R​ 2​ → R​ 2
→ In this schema, there is no common key attribute between R​ 1​ and R​ 2​ . So, this relation is lossy relation.
Definition of Functional dependency preserving:
→ Let R be the relational schema with functional dependency set F is decomposed into R​ 1​ , R​ 2​ ,..,R​ n ​ with functional dependency sets F​ 1​ ,F​ 2​ ,...,F​ n​ respectively. In general F​ 1​ ,F​ 2​ ,...,F​ n ​ can be ⊆ F.
So, dependencies are preserved in the given decomposition.
Question 20
​ The order of a leaf node in a B​ +​ tree is the maximum number of children it can have. Suppose that block size is 1 kilobytes, the child pointer takes 7 bytes long and search field value takes 14 bytes long. The order of the leaf node is ________.
A
16
B
63
C
64
D
68
E
None of the above
       Database-Management-System       B-and-B+-Trees
Question 20 Explanation: 
Excluded for evaluation. No record pointer is given.
Question 21
Which of the following is true for computation time in insertion, deletion and finding maximum and minimum element in a sorted array ?
A
Insertion – 0(1), Deletion – 0(1), Maximum – 0(1), Minimum – 0(l)
B
Insertion – 0(1), Deletion – 0(1), Maximum – 0(n), Minimum – 0(n)
C
Insertion – 0(n), Deletion – 0(n), Maximum – 0(1), Minimum – 0(1)
D
Insertion – 0(n), Deletion – 0(n), Maximum – 0(n), Minimum – 0(n)
       Data-Structures       Arrays
Question 21 Explanation: 
→ Insertion will take O(n). Let, 1,2,3,4,5. Suppose we want to insert an array. It will traverse end of the array. So, it takes worst case O(n) and best case O(1). But we always consider worst case time complexity.
→ Deletion will take O(n). Let, 1,2,3,4,5. Suppose we want to delete last element in the array, It will traverse end of the array. So, it takes worst case O(n) and best case O(1). But we always consider worst case time complexity.
→ Maximum will take O(1) because we know location of maximum element. So it requires only constant amount.
→ Minimum will take O(1) because we know location of maximum element. So it requires only constant amount.
Question 22
The seven elements A, B, C, D, E, F and G are pushed onto a stack in reverse order, i.e., starting from G. The stack is popped five times and each element is inserted into a queue.Two elements are deleted from the queue and pushed back onto the stack. Now, one element is popped from the stack. The popped item is ________.
A
A
B
B
C
F
D
G
       Data-Structures       Queues-and-Stacks
Question 22 Explanation: 
Step-1: Insert all elements into stack in reverse order then the stack is look like

Step-2: Stack is pooped five elements then stack is look like

Step-3: Popped five elements are inserted into a queue then queue is look like

Step-4: Two elements are deleted from the queue

Step-5: deleted queue elements are pushed back onto the stack

Top of the stack is B.
Question 23
Which of the following is a valid heap ?
A
A
B
B
C
C
D
D
       Data-Structures       Heap-Tree
Question 23 Explanation: 
Option A is violated max heap property because 8 is greater than his root.
Option B is satisfied max heap property.
Option C is violated max heap property because 7 is greater than his root.
Option D is violated max heap property because 9,10,8,7 elements are greater than his root.
Question 24
If h is chosen from a universal collection of hash functions and is used to hash n keys into a table of size m, where n ≤ m, the expected number of collisions involving a particular key x is less than _______.
A
1
B
1/n
C
1/m
D
n/m
       Data-Structures       Hashing
Question 24 Explanation: 
If h is chosen from a universal collection of hash functions and is used to hash n keys into a table of size m, where n ≤ m, the expected number of collisions involving a particular key x is less than 1. It is direct theorem from universal hashing.
Question 25
Which of the following statements is false ?
(A) Optimal binary search tree construction can be performed efficiently using dynamic programming.
(B) Breadth-first search cannot be used to find connected components of a graph.
(C) Given the prefix and postfix walks of a binary tree, the tree cannot be re-constructed uniquely.
(D) Depth-first-search can be used to find the connected components of a graph.
A
A
B
B
C
C
D
D
       Data-Structures       Graphs
Question 25 Explanation: 
→ For connected components we are using Depth first search but not breadth first search.
→ A connected component (or just component) of an undirected graph is a subgraph in which any two vertices are connected to each other by paths, and which is connected to no additional vertices in the supergraph.
1. Kosaraju’s algorithm for strongly connected components.
2. Tarjan’s Algorithm to find Strongly Connected Components Time complexity for both algorithms are O(V+E).
Question 26
Match the following Layers and Protocols for a user browsing with SSL :

A
a-(iv), b-(i), c-(ii), d-(iii)
B
a-(iii), b-(ii), c-(i), d-(iv)
C
a-(ii), b-(iii), c-(iv), d-(i)
D
a-(iii), b-(i), c-(iv), d-(ii)
       Computer-Organization       Hardware Devices
Question 26 Explanation: 
HTTP→ Application layer
TCP→ Transport layer
IP→ Network layer
PPP→ Data link layer
Question 27
The maximum size of the data that the application layer can pass on to the TCP layer below is __________.
A
2​ 16​ bytes
B
2​ 16​ bytes + TCP header length
C
2​ 16​ bytes - TCP header length
D
2​ 15​ byte.
E
None of the above
       Computer-Networks       TCP/IP-Layers
Question 27 Explanation: 
Application Layer - Any size
Transport Layer - 65515 byte
Network layer - 65535 byte
Data link layer - 1500 byte
Note: Actually it works on any size, but given options are wrong. Excluded for evaluation.
Question 28
A packet whose destination is outside the local TCP/IP network segment is sent to _____.
A
File server
B
DNS server
C
DHCP server
D
Default gateway
       Computer-Networks       TCP/IP
Question 28 Explanation: 
→ A gateway is a network node that serves as an access point to another network, often involving not only a change of addressing, but also a different networking technology.
→ More narrowly defined, a router merely forwards packets between networks with different network prefixes. The networking software stack of each computer contains a routing table that specifies which interface is used for transmission and which router on the network is responsible for forwarding to a specific set of addresses.
→ If none of these forwarding rules is appropriate for a given destination address, the default gateway is chosen as the router of last resort.
→ The default gateway is specified by the configuration setting often called default route.
Question 29
Distance vector routing algorithm is a dynamic routing algorithm. The routing tables in distance vector routing algorithm are updated _____.
A
automatically
B
by server
C
by exchanging information with neighbour nodes
D
with backup database
       Computer-Networks       Routing
Question 29 Explanation: 
→ Distance vector routing algorithm is a dynamic routing algorithm. The routing tables in distance vector routing algorithm are updated by exchanging information with neighbour nodes.
→ Distance-vector protocols update the routing tables of routers and determine the route on which a packet will be sent by the next hop which is the exit interface of the router and the IP address of the interface of the receiving router.
→ Distance is a measure of the cost to reach a certain node. The least cost route between any two nodes is the route with minimum distance.
Question 30
In link state routing algorithm after construction of link state packets, new routes are computed using:
A
DES algorithm
B
Dijkstra’s algorithm
C
RSA algorithm
D
Packets
       Computer-Networks       Routing
Question 30 Explanation: 
→ Link State Routing(LSR) algorithm after construction of link state packets, new routes are computed using Dijkstra’s algorithm.
→ Each node independently runs an algorithm over the map to determine the shortest path from itself to every other node in the network; generally some variant of Dijkstra's algorithm is used.
→ This is based around a link cost across each path which includes available bandwidth among other things.
Question 31
Which of the following strings would match the regular expression : p+ [3 – 5]∗ [xyz]?
I. p443y
II. p6y
III. 3xyz
IV. p35z
V. p353535x
VI. ppp5
A
I, III and VI only
B
IV, V and VI only
C
II, IV and V only
D
I, IV and V only
       Theory-of-Computation       Regular-Expression
Question 31 Explanation: 
This problem looks very difficult but solving procedure is very easy.
Given regular expression is p+[3-5]*[xyz]
It means we have to start with p.
[3-5]* means (3+4+5)* It can generate these numbers many number of times.
[xyz] → ends with either x or y or z or any combination but not violate order of occurrences
I. p443y → matched because it satisfied above regular expression.
Il. p6y → not matched because 6 not present in regular expression.
III. 3xyz → not matched because it not starts with p.
IV. p35z → matched because it satisfied above regular expression.
V. p353535x → matched because it satisfied above regular expression.
Vl. ppp5 → not matched because it is not end with x or y or z or any combination.
Question 32
Consider the following assembly language instructions :
mov al, 15
mov ah, 15
xor al, al
mov cl, 3
shr ax, cl
add al, 90H
adc ah, 0
What is the value in ax register after execution of above instructions ?
A
0270H
B
0170H
C
01E0H
D
0370H
       Computer-Organization       Machine-Instructions
Question 32 Explanation: 
Step-1: mov al, 15 → move 15 to lower part of 'ax' register
mov ah, 15 → move 15 to higher part of 'ax' register
XOR al,al → Convert 15 into binary number and perform XOR operation with ‘ax’
register. The final binary value is 0000111100000000
move cl,3 → move 3 to lower part of 'cx' register.
shr ax, cl → will shift right and rotate content of 'ax' register
The 'ax' register content looks like 0000000111100000
add al, 90H → add hexadecimal 90 to al
0000000111100000
0000000010010000
---------------------------
0000001001110000
-
---------------------------
adc ah,0 → addition with carry which does not affect 'ax' register.
Step-2: Finally the content of ‘ax’ register will be 0270H.
Question 33
Consider the following statements related to compiler construction :
I. Lexical Analysis is specified by context-free grammars and implemented by pushdown automata.
II. Syntax Analysis is specified by regular expressions and implemented by finite-state machine.
Which of the above statement(s) is/are correct ?
A
Only I
B
Only II
C
Both I and II
D
Neither I nor II
       Compiler-Design       Compilers
Question 33 Explanation: 
FALSE: Lexical Analysis is specified by regular grammars and implemented by finite automata.
FALSE: Syntax Analysis is specified by context-free grammars and implemented by pushdown automata.
Question 34
The contents of Register (BL) and Register (AL) of 8085 microprocessor are 49H and 3AH respectively. The contents of AL, the status of carry flag (CF) and sign flag (SF) after executing ‘SUB AL, BL’ assembly language instruction, are
A
AL = 0FH; CF = 1; SF = 1
B
AL = F0H; CF = 0; SF = 0
C
AL = F1H; CF = 1; SF = 1
D
AL = 1FH; CF = 1; SF = 1
       Computer-Organization       Microprocessor
Question 34 Explanation: 
Initially BL=0100 1001 and AL=0011 1010
Step-1: We have to perform subtraction of AL and BL
AL= 0011 1010 ( Here, MSB and Second MSB bit taken borrow 1)
BL= 0111 1001
------------------
1111 0001
------------------
Step-2: Final result have to be stored in AL. AL = F1H; CF = 1; SF = 1
So, option C is the correct answer.
Question 35
Which of the following statement(s) regarding a linker software is/are true ?
I A function of a linker is to combine several object modules into a single load module.
II A function of a linker is to replace absolute references in an object module by symbolic references to locations in other modules.
A
Only I
B
Only II
C
Both I and II
D
Neither I nor II
       Operating-Systems       Linker-and-Loader
Question 35 Explanation: 
TRUE: A function of a linker is to combine several object modules into a single load module. → A linker combines one or more object files and possible some library code into either some executable, some library or a list of error messages.
FALSE: A function of a linker is to replace absolute references in an object module by symbolic references to locations in other modules.
Question 36
There are three processes P​ 1​ , P​ 2​ and P​ 3​ sharing a semaphore for synchronizing a variable. Initial value of semaphore is one. Assume that negative value of semaphore tells us how many processes are waiting in queue. Processes access the semaphore in following order :
(a) P​ 2​ needs to access
(b) P​ 1​ needs to access
(c) P​ 3​ needs to access
(d) P​ 2​ exits critical section
(e) P​ 1​ exits critical section
The final value of semaphore will be :
A
0
B
1
C
-1
D
-2
       Operating-Systems       Process-Synchronization
Question 36 Explanation: 
There are 3 processes P​ 1​ , P​ 2​ and P​ 3​ . Given, Initial value of semaphore S is 1.
Step-1: P​ 2​ needs to access, we are allocating critical section to P​ 2​ . S value become 0.
Step-2: P​ 1​ needs to access but processor not available. So, S value become -1. Step-3: P​ 3​ needs to access but processor not available. So, S value become -2. Step-4: P​ 2​ exits critical section. So, we are allocating processor to P​ 1​ then S value is -1. Step-5: P​ 1​ exits critical section. So, we are allocating processor to P​ 3​ then S value is 0.
Question 37
In a paging system, it takes 30 ns to search translation Lookaside Buffer (TLB) and 90 ns to access the main memory. If the TLB hit ratio is 70%, the effective memory access time is :
A
48ns
B
147ns
C
120ns
D
84ns
       Operating-Systems       Memory-Management
Question 37 Explanation: 
Effective memory access(EMA)=Hit ratio*(TLB access time + Main memory access time) +(1–hit ratio) * (TLB access time + 2 * main memory time)
EAM=0.7*(30+90)+0.3(30+(2*90))
=0.7*120 + 0.3(30+(180))
=0.7*120 + 0.3*210
= 84 + 63
= 147
Question 38
Match the following w.r.t. Input/Output management :
A
a-iii, b-iv, c-i, d-ii
B
a-ii, b-i, c-iv, d-iii
C
a-iv, b-i, c-ii, d-iii
D
a-i, b-iii, c-iv, d-ii
       Computer-Organization       Hardware Devices
Question 38 Explanation: 
Device controller→ Performs data transfer
The Device Controller works like an interface between a device and a device driver. I/O units (Keyboard, mouse, printer, etc.) typically consist of a mechanical component and an electronic component where electronic component is called the device controller.
Device driver → Processing of I/O request
Device drivers are modules that can be plugged into an OS to handle a particular device or category of similar devices.
Interrupt handler→ Extracts information from the controller register and store it in data buffer
Interrupt handlers have a multitude of functions, which vary based on what triggered the interrupt and the speed at which the interrupt handler completes its task. For example, pressing a key on a computer keyboard.
Kernel I/O subsystem→ I/O scheduling
Kernel provide many services related to I/O. Several services like scheduling, buffering, caching, spooling, device reservation and error handling. These are provided by the kernel’s I/O subsystem and build on the hardware and device driver infrastructure.
Question 39
Which of the following scheduling algorithms may cause starvation ?
a. First-come-first-served
b. Round Robin
c. Priority
d. Shortest process next
e. Shortest remaining time first
A
a, c and e
B
c, d and e
C
b, d and e
D
b, c and d
       Operating-Systems       Process-Scheduling
Question 39 Explanation: 
→ Starvation is the name given to the indefinite postponement of a process because it requires some resource before it can run, but the resource, though available for allocation, is never allocated to this process.
1. Priority: A process ready to run for CPU can wait indefinitely because of low priority
2. Shortest process next : longest process possibility to get starvation
3. Shortest remaining time first: longest process possibility to get starvation
Solution: Aging
→ FCFS and Round Robin scheduling never happen indefinite blocking.
Question 40
Distributed operating systems consist of:
A
Loosely coupled O.S. software on a loosely coupled hardware.
B
Loosely coupled O.S. software on a tightly coupled hardware.
C
Tightly coupled O.S. software on a loosely coupled hardware.
D
Tightly coupled O.S. software on a tightly coupled hardware.
       Operating-Systems       Distributed-Operating-System
Question 40 Explanation: 
→ Distributed System is a collection of independent computers which can cooperate, but which appear to users of the system as a uniprocessor computer.
1. Tightly coupled: Short delay in communication between computers, high data rate (e.g., Parallel computers working on related computations)
Eg: Tightly coupled operating system is software
2. Loosely coupled: Large delay in communications, Low data rate (Distributed Systems working on unrelated computations)
Eg: Loosely coupled operating system is hardware.
Question 41
Software Engineering is an engineering discipline that is concerned with:
A
how computer systems work
B
theories and methods that underlie computers and software systems.
C
all aspects of software production
D
all aspects of computer-based systems development, including hardware, software and process engineering.
       Software-Engineering       Software-Engineering
Question 41 Explanation: 
→ Software Engineering is an engineering discipline that is concerned with all aspects of software production → It follows some basic activities are software specification, software development, software validation and software evolution.
Question 42
Which of the following is not one of three software product aspects addressed by McCall’s software quality factors ?
A
Ability to undergo change
B
Adaptability to new environments
C
Operational characteristics
D
Production costs and scheduling
       Software-Engineering       Software-design
Question 42 Explanation: 
McCall’s software quality factors
1. Product operation factors − Correctness, Reliability, Efficiency, Integrity, Usability.
2. Product revision factors − Maintainability, Flexibility, Testability.
3. Product transition factors − Portability, Reusability, Interoperability.
Question 43
Which of the following statement(s) is/are true with respect to software architecture ?
S1 : Coupling is a measure of how well the things grouped together in a module belong together logically.
S2 : Cohesion is a measure of the degree of interaction between software modules.
S3 : If coupling is low and cohesion is high then it is easier to change one module without affecting others.
A
Only S1 and S2
B
Only S3
C
All of S1, S2 and S3
D
Only S1
       Software-Engineering       Coupling-and-Cohesion
Question 43 Explanation: 
→ Cohesion is the indication of the relationship within a module.
→ Coupling is the indication of the relationships between modules.
→ If coupling is low and cohesion is high then it is easier to change one module without affecting others.
→ High cohesion within modules and low coupling between modules are often regarded as related to high quality in software architecture as well as object oriented programming languages.
Question 44
The prototyping model of software development is:
A
a reasonable approach when requirements are well-defined
B
a useful approach when a customer cannot define requirements clearly.
C
the best approach to use for projects with large development teams.
D
a risky model that rarely produces a meaningful product.
       Software-Engineering       Software-design
Question 44 Explanation: 
→ The prototyping model of software development is a useful approach when a customer cannot define requirements clearly.
Advantage:
1. We can develop the software where requirements are unclear
2. Customer satisfaction
Disadvantage:
1. Who pay cost of prototype
2. Required the design expertise
Question 45
A software design pattern used to enhance the functionality of an object at run-time is:
A
Adapter
B
Decorator
C
Delegation
D
Proxy
       Software-Engineering       Software-design
Question 45 Explanation: 
→ Software design pattern used to enhance the functionality of an object at run-time is Decorator.
→ Decorator will attach additional responsibilities to an object dynamically keeping the same interface.
→ Decorators provide a flexible alternative to subclassing for extending functionality.
Question 46
Match the following:
A
a-i, b-ii, c-iii, d-iv
B
a-i, b-iii, c-ii, d-iv
C
a-iii, b-ii, c-iv, d-i
D
a-ii, b-iii, c-i, d-iv
Question 46 Explanation: 
Affiliate Marketing: Vendors ask partners to place logos on partner’s site. If customers click, come to vendors and buy.
Viral Marketing: Spread your brand on the net by word-of-mouth. Receivers will send your information to their friends.
Group Purchasing: Aggregating the demands of small buyers to get a large volume. Then negotiate a price.
Bartering Online: Exchanging surplus products and services with the process administered completely online by an intermediary. Company receives “points” for its contribution.
Question 47
________ refers loosely to the process of semi-automatically analyzing large databases to find useful patterns.
A
Datamining
B
Data warehousing
C
DBMS
D
Data mirroring
       Database-Management-System       Relational-databases
Question 47 Explanation: 
→ Data mining refers loosely to the process of semi-automatically analyzing large databases to find useful patterns.
→ Data mining is the process of discovering patterns in large data sets involving methods at the intersection of machine learning, statistics, and database systems.
→ Data mining is the analysis step of the "knowledge discovery in databases" process, or KDD
Question 48
Which of the following is/are true w.r.t. applications of mobile computing ?
(A) Travelling of salesman
(B) Location awareness services
A
(A) true; (B) false.
B
Both (A) and (B) are true.
C
Both (A) and (B) are false.
D
(A) false; (B) true.
Question 48 Explanation: 
Mobile computing main applications:
1.Portability
2. Connectivity
3.Social Interactivity
4. Individuality
Travelling of salesman applicable to mobility of application.
Question 49
In 3G network, W-CDMA is also known as UMTS. The minimum spectrum allocation required for W-CDMA is _______.
A
2 MHz
B
20 KHz
C
5 KHz
D
5 MHz
       Computer-Networks       3-G-Network
Question 49 Explanation: 
→ In 3G network, W-CDMA is also known as UMTS. The minimum spectrum allocation required for W-CDMA is 5 MHz.
→ IMT-DS uses a version of CDMA called wideband CDMA or W-CDMA. W-CDMA uses a 5.MHz bandwidth. It was in europe, and it is compatible with the CDMA used in IS-95.
→ IMT-TC uses a combination of W-CDMA and TDMA. The standard tries to reach the IMT-2000 goals by adding TDMA multiplexing to W-CDMA.
Question 50
Which of the following statements is/are true w.r.t. Enterprise Resource Planning (ERP) ?
(A) ERP automates and integrates majority of business processes.
(B) ERP provides access to information in a Real Time Environment.
(C) ERP is inexpensive to implement
A
(A), (B) and (C) are false.
B
(A) and (B) false; (C) true.
C
(A) and (B) true; (C) false.
D
(A) true ; (B) and (C) are false.
Question 50 Explanation: 
Enterprise resource planning (ERP) is the integrated management of core business processes, often in real-time and mediated by software and technology.
→ The most fundamental advantage of ERP is that the integration of a myriad of business processes saves time and expense.
→ Management can make decisions faster and with fewer errors. Data becomes visible across the organization.
TRUE: ERP automates and integrates majority of business processes.
TRUE: ERP provides access to information in a Real Time Environment.
FALSE: ERP is expensive to implement.
There are 50 questions to complete.
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