## Introduction

UGC NET Exam is one of the popular exams in India for people interested in research. Previous Year Questions are an excellent option to learn about the exam pattern. By solving the PYQs, you will get a basic idea about your preparation. You can evaluate your weak areas and work on them to perform better in the examination. In this article, we have given the questions of UGC NET 2012 December Paper-III. We have also explained every problem adequately to help you learn better.

We have discussed the first 25 questions of the paper in this article and for the other questions you can visit this link __December 2012 Paper-III Part-2__ and __December 2012 Paper-III Part-3__.

## Questions 1 to 25

**1. Eco system is a Frame work for**

** (A) Building a Computer System**

** (B) Building Internet Market**

** (C) Building Offline Market**

** (D) Building Market**

**Answer: B**

Explanation:

The ecosystem approach is a theoretical paradigm for dealing with ecosystem problems and its goal is to build an internet market.

**2. The efficiency (E) and speed up (sp) for Multiprocessor with p processors satisfies :**

** (A) E ≤ p and sp ≤ p**

** (B) E ≤ 1 and sp ≤ p**

** (C) E ≤ p and sp ≤ 1**

** (D) E ≤ 1 and sp ≤ 1**

**Answer: B**

Explanation:

Efficiency E can never exceed 1 or 100%, and even 1 is just theoretical. Multiprocessor speed up (sp) can never have more processors (p). As a result, E1 will always be less than 1 and sp should be less than or equal to p.

**3. Match the following :**

** List – I List – II**

**a. Critical region 1. Hoares Monitor**

**b. Wait/signal 2. Mutual exclusion**

**c. Working set 3. Principal of locality**

**d. Deadlock 4. Circular wait**

** Codes :**

** a b c d**

** (A) 2 1 3 4 **

** (B) 1 2 4 3**

** (C) 2 3 1 4**

** (D) 1 3 2 4**

**Answer: B**

Explanation:

Hoares Monitor has proposed a variant of critical regions.

On the semaphore, two standard operations, wait and signal, are defined that implements mutual exclusion.

A working set is detected during the Circular wait process. In this one process is waiting for a resource held by a second process, which is waiting for a resource held by a third process, and so on.

The principle of locality describes a processor's tendency to repeatedly visit the same set of memory locations over a short period of time and when each process holds a resource and waits for another process to hold that resource, a deadlock occurs.

**4. The technique of temporarily delaying outgoing acknowledgments so that they can be hooked onto the next an outgoing data frame is known as**

** (A) Bit stuffing**

** (B) Piggy backing**

** (C) Pipelining**

** (D) Broadcasting**

**Answer: B**

Explanation:

Piggybacking is a technique that momentarily delays acknowledgment in order to connect to the next outgoing data frame. Because some bandwidth is allotted for acknowledgment in non-piggybacking systems, it saves a lot of channel bandwidth

**5. ______ is process of extracting previously non-known valid and actionable information from large data to make crucial business and strategic decisions.**

** (A) Data Management**

** (B) Data base**

** (C) Data Mining**

** (D) Meta Data**

**Answer: C**

Explanation:

Data mining is a computer technology that uses artificial intelligence, machine learning, statistics, and database systems to find patterns in large amounts of data.

**6. The aspect ratio of an image is defined as**

** (A) The ratio of width to its height measured in unit length.**

** (B) The ratio of height to width**

**measured in number of pixels.**

** (C) The ratio of depth to width**

**measured in unit length.**

** (D) The ratio of width to depth**

**measured in number of pixels**

**Answer: A**

Explanation:

The proportionate connection between an image's width and height is described by its aspect ratio. It's frequently written as 16:9, which is two integers separated by a colon.

**7. Which of the following features will characterize an OS as multiprogrammed OS ?**

** (a) More than one program may be**

**loaded into main memory at the**

**same time.**

** (b) If a program waits for certain**

**event another program is**

**immediately scheduled.**

** (c) If the execution of a program**

**terminates, another program is**

**immediately scheduled.**

** (A) (a) only**

** (B) (a) and (b) only**

** (C) (a) and (c) only**

** (D) (a), (b) and (c) only**

**Answer: D**

Explanation:

At the same time, many programs were loaded into the main memory. If a program is delayed until a specific event occurs, another program is instantly arranged and if a program's execution is interrupted, another program is instantly scheduled. These are some features of multi-programmed OS.

**8. Using RSA algorithm, what is the**

**value of cipher text C, if the plain text**

**M = 5 and p = 3, q = 11 & d = 7 ?**

** (A) 33**

** (B) 5**

** (C) 25**

** (D) 26**

**Answer: D**

Explanation:

Given

p=3,

q=11

As we know,

n=(p X q) = (3 X 11)=33

m=(p-1)X(q-1) = (2 X 10)=20

Find a little odd integer e that is close to m in size.

When e=3, GCD(3,20)=1. We set e=3 because e should be small and prime. d is set to 7.

Public key = (e,n). (e and n have known values.)

We use the public key to encrypt a message using the function E(s) = se (mod n)

Here s stands for the message and e and n stand for the public key integer pair. In the above question, the plain text M = 5. The plain text needs to be encrypted using the above formula.

=53(mod 33)

= 125 (mod 33)

= 26

As a result, E(s) = 26 is the encrypted message.

**9. You are given an OR problem and a XOR problem to solve. Then, which one of the following statements is true ?**

** (A) Both OR and XOR problems can be solved using single layer**

**perception.**

** (B) OR problem can be solved using single layer perception and XOR**

**problem can be solved using self organizing maps.**

** (C) OR problem can be solved using radial basis function and XOR**

**problem can be solved using single layer perception.**

** (D) OR problem can be solved using single layer perception and XOR**

**problem can be solved using radial basis function.**

**Answer: D**

Explanation:

In Single Perceptron / Multi-layer Perceptron(MLP), we only have linear separability because they are composed of input and output layers(some hidden layers in MLP)⁃ For example, AND, OR functions are linearly separable & XOR function is not linearly separable.

**10. Match the following :**

**List – I List – II**

**a. Application Layer 1. TCP**

**b. Transport layer 2. HDLC**

**c. Network layer 3. HTTP**

**d. Data link layer 4. BGP**

** Codes :**

** a b c d**

** (A) 2 1 4 3**

** (B) 3 4 1 2**

** (C) 3 1 4 2**

** (D) 2 4 1 3**

**Answer: C**

Explanation:

HTTP is an application-layer protocol for transmitting hypermedia documents like HTML.

TCP stands for Transport Control Protocol. It establishes a secure virtual circuit between applications using the transport-layer protocol.

TCP/IP is the basis of BGP in networking. It controls the Network Layer via the OSI Transport Layer.

HDLC (High-level Data Link Control) is a group of data link layer communication protocols.

Must Read __HDLC Protocol__

**11. The time complexities of some standard graph algorithms are given. Match each algorithm with its time complexity? (n and m are no. of nodes and edges respectively)**

**a. Bellman Ford algorithm 1. O (m log n)**

**b. Kruskals algorithm 2. O (n3)**

**c. Floyd Warshall algorithm 3. O(mn)**

**d. Topological sorting 4. O(n + m)**

** Codes :**

** a b c d**

** (A) 3 1 2 4**

** (B) 2 4 3 1**

** (C) 3 4 1 2**

** (D) 2 1 3 4**

**Answer: A**

Explanation:

Bellman-Ford algorithm runs in time O(mn) since the initialization takes O(m) for each of m-1 passes and the for loop in the algorithm takes O(n) time. Hence its time complexity is O(mn).

Kruskal's algorithm's time complexity is O(m log n)

Floyd-Warshall algorithm has a time complexity O(n3).

The time complexity of topological sort using Kahn's algorithm is O(n+m).

**12. Let V1 = 2I – J + K and V2 = I + J – K, then the angle between V1 & V2 and a vector perpendicular to both V1 & V2 shall be :**

** (A) 90° and (–2I + J – 3K)**

** (B) 60° and (2I + J + 3K)**

** (C) 90° and (2I + J – 3K)**

** (D) 90° and (–2I – J + 3K)**

**Answer: D**

Explanation:

As we know-

cosθ = (u⇠· v⃗) / (||u⃗|| ||v ⃗ ||).

The dot product of vectors is (u⃗ · v⃗) . = [2 -1 1][1 1 -1]

calculated as (2-1-1) as 0 .

Length of U is = (||u⃗||)=((2)^{2}+-1^{2}+1^{2})^{½ }

Length of V is = (||v ⃗ ||).=(1^{2}+1^{2}+(-1)^{2})^{1/2}

On calculating we get Cosθ=0

hence θ is 90° (as Cos 90° = 0)

Now Vector perpendicular to both shall be-

*u*×*v* =i(u2v3-u3v2)-j(u1v3-v1u3)+(u1v2-u2v1)K.

=i(1-1)-j(-2-1)+(2+1)k

On calculating we get the vector as (0i+3j+3k)

Let’s calculate the dot product of-

2i-j+k and (0i+3j+3k)

= (2i-j+k).(0i+3j+3k) = 0

So these two vectors are perpendicular as their dot product is zero.

And the dot product of (0i,3j,3k) with (I,j,-k) is also zero.

Hence we can say that (0i,3j,3k ) is perpendicular to both vectors.

**13. Consider a fuzzy set A defined on the interval X = [0, 10] of integers by the membership Junction**

** μ🇦(x) = x/x + 2**

** Then the cut corresponding to α= 0.5 will be**

** (A) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}**

** (B) {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}**

** (C) {2, 3, 4, 5, 6, 7, 8, 9, 10}**

** (D) { } **

**Answer: C**

Explanation:

Substitute the values in the membership function starting from 0. On substituting you will find that at 2 the membership function will result in 0.5. Hence the answer will lie between 2 to 10.

On substituting value from 0 to 10 in membership function =x/x+2

we get

0/2= 0, 1/3= 0.33, 2/4=0.5 , 3/5=0.6 .........10/12 =0.83

here α =0.5 so the first two elements will not be included in the result as their degree of belongingness < 0.5

**14. Let T(n) be the function defined by T(n) = 1 and T(n) = 2T (n/2) + n, which of the following is TRUE ?**

** (A) T(n) = O(√ n)**

** (B) T(n) = O(log2n)**

** (C) T(n) = O(n)**

** (D) T(n) = O(n²)**

**Answer: C**

Explanation:

n(logьa) = n which is = n^(1-.5) = O(sqrt n)

then by applying case 1 of master method we get T(n) = Θ(n)

**15. In classful addressing, an IP address 123.23.156.4 belongs to ______ class**

**format.**

** (A) A**

** (B) B**

** (C) C**

** (D) D**

**Answer: A**

Explanation:

There is a 7-bit network address in class A.

We know that

=2ⁿ-2

=2^7 - 2 = 126

Hence all the addresses from 1.x.x.x to 126.x.x.x belong to class A

**16. The Mandelbrot set used for the construction of beautiful images is based on the following transformation :**

** Xₙ+1 = xₙ²+ z**

** Here,**

** (A) Both x & z are real numbers.**

** (B) Both x & z are complex numbers.**

** (C) x is real & z is complex.**

** (D) x is complex & z is real.**

**Answer: B**

Explanation:

All (complex) c-values for which the associated orbit of 0 under x2 + c does not escape to infinity constitute the Mandelbrot set.

**17. Which of the following permutations can be obtained in the output using a**

**stack of size 3 elements assuming that input, sequence is 1, 2, 3, 4, 5 ?**

** (A) 3, 2, 1, 5, 4**

** (B) 5, 4, 3, 2, 1**

** (C) 3, 4, 5, 2, 1**

** (D) 3, 4, 5, 1, 2**

**Answer: 2**

On inserting the elements into the stack as per the given order 1,2,3,4,5 if the input contains any order which doesn't follow this is invalid...

For D

insert 1,2,3 into the stack now top symbol is similar to the input symbol just pop it...

After popping 3 top of the stack is 2 not matched with the remaining first input symbol...

So insert 4 into the stack. Now matched pop it...

Similarly, insert 5 and pop it...

Now the I input symbol is 1 but the top of the stack is 2...so this sequence is not followed

For B

1,2,3,4,5 we need to insert but our stack size is 3 so not possible...

For A

Push 1,2,3 now top matches so pop 3,2,1...

Push 4,5 top matches with input so pop 5,4

For C

push 1,2,3, top match with input 3 so pop it...

Now push 4, top matches with input so pop 4

Now push 5, top matches with input so pop 5,2,1

**18. In a Linear Programming Problem, suppose there are 3 basic variables and 2 non-basic variables, then the possible number of basic solutions are**

** (A) 6**

** (B) 8**

** (C) 10**

** (D) 12**

**Answer: C**

Explanation:

The basic variables are known as the m variables that can take any value other than zero.

The decision variables that remain after the first step of the simplex method in linear programming are known as non-basic variables.

Factorial(n)/(factorial(m)*factorial(n-m)) is the total number of possible solutions, where

M is the fundamental variable with non-zero values.

N=3+2

M=3

(5*4*3*2*1)/((3*2)(2))=10.

**19. Identify the following activation function :**

** Φ(V) = Z + 1 / (1 + exp (– x * V + Y)) ,**

** Z, X, Y are parameters**

** (A) Step function**

** (B) Ramp function**

** (C) Sigmoid function**

** (D) Gaussian function**

** Answer: C**

Explanation:

The sigmoid function also called the logistic function or logistic curve is a common S-shaped curve. The logistic function with parameters (z = 1, y = 0, x = 1) provides the standard logistic function which when substituted to the given equation yields a sigmoid curve equation that is

f(x) = 1 /(1+ e-🗶) and on solving the above equation Φ(V) = Z + 1 / (1 + exp (– x * V + Y)) using standard parameters we get the same sigmoid curve equation. Hence the answer is Sigmoid Function.

**20. The no. of ways to distribute n distinguishable objects into k distinguishable boxes, so that ni objects are placed into box i, i = 1, 2, …. K equals which of the following?**

** (A) n! / ( n1! + n2! + ..... + nk!)**

** (B) (n1! + n2! + ..... + nk!) / ( n1! n2! n3! ..... nk!)**

** (C) n! / (n1! n2! n3! ..... nk!)**

** (D) n1! n2! ..... nk! / (n1! – n2! – n3! ..... – nk!** )

**Answer: C**

Explanation:

An r-permutation is an ordered arrangement of r items from a set of different objects. From a set of n distinct objects the number of r-permutations is P(n, r) = n(n − 1)(n − 2)· · ·(n − r + 1) = n! (n − r)! The number of permutations of n objects where there are n1 indistinguishable type 1 objects, n2 indistinguishable type 2 objects, . . . , nk indistinguishable type k objects, where n1 + n2 + · · · + nk = n is n!/ n1! n2! ... nk!

**21. How many solutions do the following equation have x1 + x2 + x3 = 11 where x1 ≥1 , x2 ≥ 2, x3 ≥ 3**

** (A) C(7, 11)**

** (B) C(11, 3)**

** (C) C(14, 11)**

** (D) C(7, 5) **

**Answer: D**

Explanation:

Given that-

x1+x2+x3=11

Here let's say we need to put 11 similar balls into x1,x2, and x3 types of boxes. So the number of boxes is n = 3

As x_{1}≥1, x_{2}≥2, x_{3}≥3 now allocate 1,2,3 from 11 directly . so remaining ball = 11-(1+2+3) = 5 =k

(n+k-1)C_{k } = (3+5-1)C5 =C(7,5).

**22. Which provides an interface to the TCP/IP suit protocols in Windows95 and Windows NT ?**

** (A) FTP Active-X Control**

** (B) TCP/IP Active-X Control**

** (C) Calinsock Active-X Control**

** (D) HTML Active-X Control**

**Answer: C**

Explanation:

ActiveX is a deprecated Microsoft software framework that extends it's earlier... However, most ActiveX controls only function on Windows. Calinsock Active-X Control provides an interface to the TCP/IP suit protocols in Windows95 and Windows NT

**23. What are the final values of Q1 and Q0**

**after 4 clock cycles, if initial values are**

**00 in the sequential circuit shown**

**below :**

** (A) 11**

** (B) 10**

** (C) 01**

** (D) 00**

**Answer: D**

Explanation:

The main function of the T flip flop is to toggle the state of input T when it is 1, and it does not toggle the state if input T is 0. In the circuit shown above, the clock symbolizes that it is positive edge triggering (low to high). Now, we need to find out Q0Q1 after 4 clock cycles.

Initially, Q1 and Q2 are in zero state

When the first positive edge clock has triggered the state of Q0 and Q1 is changed to 1 because input T is 1 and the previous state was 0

When the next positive edge clock has triggered the state of Q0 is toggled to 0 because input T is 1 and its previous state was 1. Whereas Q1 remains at 1 as the clock value is 0.

When the third positive edge clock has triggered the state of Q0 is toggled to 1 because input T is 1 and its previous state was 0. Whereas Q1 is changed to 0 because input T is 1 and the previous state was 1.

When the last positive edge clock has triggered the state of Q0 is changed to 0 because input T is 1 and the previous state was 1 and State Q1 remains the same because clock value is 0.

**24. If dual has an unbounded solution, then it's corresponding primal has**

** (A) no feasible solution**

** (B) unbounded solution**

** (C) feasible solution**

** (D) none of these**

**Answer: A**

Explanation:

If the primal has the unbounded solution, then the dual is infeasible; If the dual has an unbounded solution, then the primal is infeasible.

**25. The number of distinct bracelets of five beads made up of red, blue, and green**

**beads (two bracelets are indistinguishable if the rotation of one yield another) is,**

** (A) 243**

** (B) 81**

** (C) 51**

** (D) 47**

**Answer: C**

Explanation:

The ways to put k things into n ordered slots = k!/(k−n)!

here k =3 and n ordered slots can be understood as:

When all the beads of red color RRRRR it is 1 ordered slot

RRBBR here are 2 colors are used so it is 2 ordered slot.

Let R =x, G=y, B=z

possible combinations from these variables can be:

xxxxx: 3!/(3−1)! = 3

xxxxy: 3!/(3−2)! = 6

xxxyy: 3!/(3−2)! = 6

xxxyz: 3!/(3−3) = 6

xxyyz: 3!/(3−3)! = 6

xxyxy: 3!/(3−2)! = 6

xxyxz: 3!/(3−3)! = 6

xxyzy: 3!/(3−3)! = 6

xyzyz: 3!/(3−2)! = 6

The number of distinct bracelets=** **3 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 = 51