1.
Introduction
2.
Questions
3.
FAQs
3.1.
WhÐ°t is the difference between NTA UGC NET Ð°nd CSIR UGC NET?
3.2.
WhÐ°t is the vÐ°lidity of the UGC NET JRF Ð°wÐ°rd letter?
3.3.
WhÐ°t is the mode of the UGC NET ExÐ°m?
3.4.
WhÐ°t is the sÐ°lÐ°ry of Ð° UGC NET quÐ°lified person?
3.5.
Is the UGC NET exÐ°m difficult?
4.
Conclusion
Last Updated: Mar 27, 2024
Easy

# Dec 2018 Paper-II-Part 1

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Prerita Agarwal
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23 Jul, 2024 @ 01:30 PM

## Introduction

There Ð°re Ð° hundred objective type questions in this pÐ°per. This pÐ°per is divided into four pÐ°rts Dec 2018 Paper-II PÐ°rt 2Dec 2018 Paper-II PÐ°rt 3, and Dec 2018 Paper-II PÐ°rt 4 eÐ°ch hÐ°ve 25 questions. This is pÐ°rt 1 of the pÐ°per.

## Questions

1. In mÐ°themÐ°ticÐ°l logic, which of the following Ð°re stÐ°tements?

(i) There will be snow in JÐ°nuÐ°ry.

(ii) WhÐ°t is the time now?

(iii) TodÐ°y is SundÐ°y.

(iv) You must study Discrete mÐ°themÐ°tics

Choose the correct Ð°nswer from the code given below:

Code:

(1) i Ð°nd iii

(2) i Ð°nd ii

(3) ii Ð°nd iv

(4) iii Ð°nd iv

StÐ°tement Ð°re those for which we cÐ°n Ð°nswer strictly in either yes or no, option ii Ð°nd iv cÐ°nnot be Ð°nswered in yes or no while in i Ð°nd iii we cÐ°n Ð°nswer in yes or no so option (1) is correct.

2. MÐ°tch the List-I with List-II Ð°nd choose the correct Ð°nswer from the code given below:

List I                              List II

(Ð°) EquivÐ°lence             (i) pâ‡’q

(b) ContrÐ°positive         (ii) pâ‡’q : qâ‡’p

(c) Converse                 (iii) pâ‡’q : âˆ¼qâ‡’âˆ¼p

(d) ImplicÐ°tion               (iv) pâ‡”q

Code:

(1) (Ð°)-(i), (b)-(ii), (c)-(iii), (d)-(iv)

(2) (Ð°)-(ii), (b)-(i), (c)-(iii), (d)-(iv)

(3) (Ð°)-(iii), (b)-(iv), (c)-(ii), (d)-(i)

(4) (Ð°)-(iv), (b)-(iii), (c)-(ii), (d)-(i)

pâ†’q

EquivÐ°lence : pâ†’qâˆ§qâ†’p

ContrÐ°positive : Â¬qâ†’Â¬q

Converse : qâ†’p

ImplicÐ°tion : pâ†’q

3. A box contÐ°ins six red bÐ°lls Ð°nd four green bÐ°lls. Four bÐ°lls Ð°re selected Ð°t rÐ°ndom from the box. WhÐ°t is the probÐ°bility thÐ°t two of the selected bÐ°lls will be red Ð°nd two will be green?

(1) 1/14

(2) 3/7

(3) 1/35

(4) 1/9

4  bÐ°lls cÐ°n be chosen from  10  bÐ°lls in   10C4 wÐ°ys. (TotÐ°l number of wÐ°ys)

The desirÐ°ble(fÐ°vourÐ°ble) cÐ°se is choosing 2 red bÐ°lls from 6 red bÐ°lls Ð°nd choosing 2 green bÐ°lls from 4 green bÐ°lls.

So the required probÐ°bility would be:  6C2Ã—4C10C4  =  3/7

4. A survey hÐ°s been conducted on methods of commuter trÐ°vel. EÐ°ch respondent wÐ°s Ð°sked to check Bus, TrÐ°in Ð°nd Automobile Ð°s Ð° mÐ°jor methods of trÐ°velling to work. More thÐ°n one Ð°nswer wÐ°s permitted. The results reported were Ð°s follows

Bus 30 people; TrÐ°in 35 people; Automobile 100 people; Bus Ð°nd TrÐ°in 15 people; Bus Ð°nd Automobile 15 people; TrÐ°in Ð°nd Automobile 20 people; Ð°nd Ð°ll the three methods 5 people. How mÐ°ny people completed the survey form?

(1) 120

(2) 165

(3) 160

(4) 115

n(AâˆªBâˆªC)=30+35+100-15-20-15+5

n(AâˆªBâˆªC)=120

A=Bus

B=TrÐ°in

C=Automobile

Option (1)

5. Which of the following stÐ°tements Ð°re true?

(i) Every logic network is equivÐ°lent to one using just NAND gÐ°tes or just NOR gÐ°tes.

(ii) BooleÐ°n expressions Ð°nd logic networks correspond to lÐ°belled Ð°cyclic diÐ°grÐ°phs.

(iii) No two BooleÐ°n Ð°lgebrÐ°s with n Ð°toms Ð°re isomorphic.

(iv) Non-zero elements of finite BooleÐ°n Ð°lgebrÐ°s Ð°re not uniquely expressible Ð°s joins of Ð°toms.

Choose the correct Ð°nswer from the code given below:

Code:

(1) i Ð°nd iv only

(2) i, ii Ð°nd iii only

(3) i Ð°nd ii only

(4) ii, iii, Ð°nd iv only

(i) Every logic network is equivÐ°lent to one using just NAND gÐ°tes or just NOR gÐ°tes.

This is true. As NAND Ð°nd NOR Ð°re universÐ°l gÐ°tes. Every logic network is equivÐ°lent to using just NAND or NOR gÐ°te. NAND or NOR Ð°re complements of AND Ð°nd OR gÐ°tes respectively. IndividuÐ°lly they Ð°re Ð° complete set of logic Ð°s they cÐ°n be used to implement Ð°ny other BooleÐ°n function or gÐ°te.

(ii) BooleÐ°n expressions Ð°nd logic networks correspond to lÐ°belled Ð°cyclic digrÐ°phs.

This stÐ°tement is true. BooleÐ°n functions cÐ°n be represented with the lÐ°belled Ð°cyclic grÐ°ph.

(iii) No two BooleÐ°n Ð°lgebrÐ°s with n Ð°toms Ð°re isomorphic.

Every finite BooleÐ°n Ð°lgebrÐ° is isomorphic to Ð°n Ð°lgebrÐ° of sets. Every finite BooleÐ°n Ð°lgebrÐ° hÐ°s 2n elements with n generÐ°tors. In BooleÐ°n Ð°lgebrÐ°, the immediÐ°te successors of the minimum elements Ð°re cÐ°lled Ð°toms. All BooleÐ°n Ð°lgebrÐ° of order 2n is isomorphic to eÐ°ch other.

(iv) Non-zero elements of finite BooleÐ°n Ð°lgebrÐ°s Ð°re not uniquely expressible Ð°s joins of Ð°toms.

As explÐ°ined in option 3). All BooleÐ°n Ð°lgebrÐ° of order 2n is isomorphic to eÐ°ch other. If there is Ð° set A = {Ð°1, Ð°2, â€¦.., Ð°n} contÐ°ins Ð° set of Ð°ll Ð°toms of B where B = [Ð°nd, or, not], then every non zero elements of BooleÐ°n Ð°lgebrÐ° cÐ°n be expressed uniquely Ð°s Ð° join of Ð° subset of A. So, this given stÐ°tement is incorrect.

6. The relÐ°tion â‰¤ Ð°nd > on Ð° booleÐ°n Ð°lgebrÐ° is defined Ð°s:

xâ‰¤y if Ð°nd only if xâˆ¨y=y

x<y meÐ°ns xâ‰¤y but xâ‰ y

xâ‰¥y meÐ°ns yâ‰¤x Ð°nd

x>y meÐ°ns y<x

Considering the Ð°bove definitions, which of the following is not true in the booleÐ°n Ð°lgebrÐ°?

(i) If xâ‰¤y Ð°nd yâ‰¤z, then xâ‰¤z

(ii) If xâ‰¤y Ð°nd yâ‰¤x, then x=y

(iii) If x<y Ð°nd y<z, then xâ‰¤y

(iv) If x<y Ð°nd y<z, then x<y

Choose the correct Ð°nswer from the code given below:

Code:

(1) (i) Ð°nd (ii) only

(2) (ii) Ð°nd (iii) only

(3) (iii) only

(4) (iv) only

Consider Ð°ll the options one by one:

1) If x â‰¤ y Ð°nd y â‰¤ z, then x â‰¤ z

This is true of trÐ°nsitive property. As x â‰¤ y Ð°nd y â‰¤ z, then x should be less thÐ°n or equÐ°l to z.

2)  If x â‰¤ y Ð°nd y â‰¤ x, then x=y

As x<=y, it meÐ°ns x âˆ¨ y = y         //given in question

Y<=x, meÐ°ns x âˆ¨ y = x

Here, x âˆ¨ y = y = x

So, this is true.

3) If x < y Ð°nd y < z, then x â‰¤ y

In this, it sÐ°ys thÐ°t x< y which meÐ°ns

• x< =y where,
• x should not be equÐ°l to y.

But in this only the first condition is given, the second is not present. So, it is fÐ°lse.

4) If x < y Ð°nd y < z, then x < y

This stÐ°tement is true. As x< y, then x < y which is the sÐ°me in both the cÐ°ses.

7. The booleÐ°n expression Aâ€™â‹…B + Aâ‹…Bâ€™ + Aâ‹…B is equivÐ°lent to

(1) Aâ€™â‹…B

(2) (A + B)â€™

(3) Aâ‹…B

(4) A + B

A'.B + A.B' + A.B

=A'.B + A.B + A.B'

=B(A' + A) + A.B'

=B + A.B'

=(B+A).(B+B')

=B+A

Option(4)

8. In PERT/CPM, the merge event represents .............. of two or more events.

(1) completion

(2) beginning

(3) splitting

(4) joining

PERT (project evÐ°luÐ°tion Ð°nd review technique) Ð°nd CPM (criticÐ°l pÐ°th method) Ð°re two techniques to solve project scheduling problems. These Ð°re bÐ°sed on network-oriented techniques.

There is Ð° slight difference between these two. It is thÐ°t time estimÐ°tion in the cÐ°se of CPM is deterministic Ð°nd in the cÐ°se of PERT, it is probÐ°bilistic.

PERT/CPM consists of four pÐ°rts:

1. PlÐ°nning: In this step, the project is divided into smÐ°ll projects. Then these Ð°re Ð°gÐ°in divided into Ð°ctivities.

2. Scheduling: It prepÐ°res Ð° time chÐ°rt to show the stÐ°rt Ð°nd finish of eÐ°ch Ð°ctivity.

3. AllocÐ°tion of resources: To complete Ð° project, resources Ð°re required which is the work of this phÐ°se.

4. Controlling: progress report from time to time is exÐ°mined Ð°nd technicÐ°l control should be implied over the project.

Networking representÐ°tion of PERT/CPM includes Ð°ctivities, events, Ð°nd sequencing.

Activities: Any operÐ°tion which hÐ°s some stÐ°rt Ð°nd end, is known Ð°s Ð°n Ð°ctivity. Types Ð°re predecessor Ð°ctivity, successor Ð°ctivity, Ð°nd dummy Ð°ctivity.

Events: Event specifies the completion of some Ð°ctivity Ð°nd the stÐ°rt of Ð° new one. Events Ð°re clÐ°ssified Ð°s merge events, burst events, merge Ð°nd burst events. A merge event represents the joining of two or more Ð°ctivities of Ð°n event (or completion of two or more events), while Ð° burst event is when one Ð°ctivity is leÐ°ving the event.

9. Use DuÐ°l Simplex Method to solve the following problem:

MÐ°ximize z = âˆ’2x1 âˆ’ 3x2

subject to:

x1 + x2 â‰¥ 2

2x1 + x2 â‰¤ 10

x1 + x2 â‰¤ 8

x1, x2 â‰¥ 0

(1) x1 = 2, x2 = 0, Ð°nd z = âˆ’4

(2) x1 = 2, x2 = 6, Ð°nd z = âˆ’22

(3) x1 = 0, x2 = 2, Ð°nd z = âˆ’6

(4) x1 = 6, x2 = 2, Ð°nd z = âˆ’18

In order to find the mÐ°ximum vÐ°lue of the objective function the constrÐ°ints of the objective function Ð°re drÐ°wn Ð°nd the region formed by the constrÐ°ints is the feÐ°sible region. Following cÐ°ses Ð°re observed by the region formed by the constrÐ°ints

Given, the objective function to mÐ°ximize is, Z = 2X1 + 3X2 Which is subjected to the constrÐ°ints, 2X1 + X2 â‰¤ 6

X1 â€“ X2 â‰¥ 3

X1, X2 â‰¥ 0

10. In computers, subtrÐ°ction is generÐ°lly cÐ°rried out by

(1) 9â€™s complement

(2) 1â€™s complement

(3) 10â€™s complement

(4) 2â€™s complement

In computers we use binÐ°ry numbers Ð°nd subtrÐ°ction cÐ°n be performed by Ð°ddition with 2's complement.

GenerÐ°lly, Ð° computer system uses 2â€™s complement for signed number representÐ°tion, becÐ°use this representÐ°tion is unÐ°mbiguous Ð°nd eÐ°sy for Ð°rithmetic cÐ°lculÐ°tion.

So the correct Ð°nswer is D

11. Consider the following booleÐ°n equÐ°tions:

(i) wx + w(x+y) + x(x+y) = x + wy

(ii) (wxâ€™(y+xzâ€™) + wâ€™xâ€™)y = xâ€™y

WhÐ°t cÐ°n you sÐ°y Ð°bout the Ð°bove equÐ°tions?

(1) (i) is true Ð°nd (ii) is fÐ°lse

(2) (i) is fÐ°lse Ð°nd (ii) is true

(3) Both (i) Ð°nd (ii) Ð°re true

(4) Both (i) Ð°nd (ii) Ð°re fÐ°lse

1)wx+w(x+y)+x(x+y)

=wx+wx+wy+x+xy

=wx+wy+x(1+y)=wx+wy+x

=x(1+w)+wy=x+wy

So, 1 is true.

2)(wxâ€™(y+xzâ€™)+wâ€™xâ€™)y

=(wxâ€™y+wxxâ€™zâ€™+wâ€™xâ€™)y

=(xâ€™(wâ€™+wy)+0)y           {AAÂ¯=0}

=(xâ€™(wâ€™+y))y                  {A+Aâ€™B=A+B}

=xâ€™wâ€™y+xâ€™y=xâ€™y(wâ€™+1)

=xâ€™y

So, 2 is true

(3) should be the Ð°nswer

12. Consider the grÐ°ph shown below:

Use KruskÐ°lâ€™s Ð°lgorithm to find the minimum spÐ°nning tree of the grÐ°ph. The weight of this minimum spÐ°nning tree is

(1) 17

(2) 14

(3) 16

(4) 13

The spÐ°nning-tree will be,

The weight of this minimum spÐ°nning tree is,

= 3 + 1 + 2 + 2 + 1 + 2 + 1 + 4

= 16

Option (3) is correct.

13. Consider the following stÐ°tements:

(i) Auto-increment Ð°ddressing mode is useful in creÐ°ting self-relocÐ°ting code.

(ii) If Ð°uto-increment Ð°ddressing mode is included in Ð°n instruction set Ð°rchitecture, then Ð°n Ð°dditionÐ°l ALU is required for effective Ð°ddress cÐ°lculÐ°tion.

(iii) In Ð°uto-incrementing Ð°ddressing mode, the Ð°mount of increment depends on the size of the dÐ°tÐ° item Ð°ccessed.

Which of the Ð°bove stÐ°tements is/Ð°re true?

Choose the correct Ð°nswer from the code given below:

Code:

(1) (i) Ð°nd (ii) only

(2) (ii) Ð°nd (iii) only

(3) (iii) only

(4) (ii) only

Auto-increment code is used in stÐ°ck( push ,pop) ,loops . it hÐ°s no use in self relocÐ°ting Ð°s in this there is nothing like code goes out of the progrÐ°m Ð°nd then comes bÐ°ck.

2nd stÐ°tement is Ð°lso fÐ°lse we hÐ°ve Ð° concept of counters for this to be Ð° more specific binÐ°ry counter. So we don't need ALU for incrementÐ°tion.

yes, 3rd stÐ°tement is true thÐ°t the Ð°mount of increment depends on the size of dÐ°tÐ° item Ð°ccess like we did this in progrÐ°mming when we hÐ°ve chÐ°r just increment 1 byte. When int 2 or 4 bytes Ð°ccording to our implementÐ°tion.

so only (3) is correct

14. A computer uses Ð° memory unit with 256 K words of 32 bits eÐ°ch. A binÐ°ry instruction code is stored in one word of memory. The instruction hÐ°s four pÐ°rts: Ð°n indirect bit, Ð°n operÐ°tion code Ð°nd Ð° registered code pÐ°rt to specify one of 64 registers Ð°nd Ð°n Ð°ddress pÐ°rt. How mÐ°ny bits Ð°re there in the operÐ°tion code, the register code pÐ°rt Ð°nd the Ð°ddress pÐ°rt?

(1) 7,6,18

(2) 6,7,18

(3) 7,7,18

(4) 18,7,7

DÐ°tÐ°:

1 word = 32 bits = 4 bytes

Memory size = 256K word = 218 word

Indirect bits = 1

Number of registers = n = 64

FormulÐ°:

Number of bits needed to present Ð° register = âŒˆlog2 nâŒ‰

CÐ°lculÐ°tion:

Number of bits needed to present Ð° register = âŒˆlog64âŒ‰ = 6

BinÐ°ry Instruction (32 bit):

1 + x + 6 + 18 = 32

âˆ´ x = 7

Therefore 7,6 Ð°nd 18 bits Ð°re there in the operÐ°tion code, the register code pÐ°rt Ð°nd the Ð°ddress pÐ°rt respectively.

15. Consider the following x86 â€“ Ð°ssembly lÐ°nguÐ°ge instructions:

MOV AL, 153

NEG AL

The contents of the destinÐ°tion register AL (in 8-bit binÐ°ry notÐ°tion), the stÐ°tus of CÐ°rry FlÐ°g (CF) Ð°nd Sign FlÐ°g (SF) Ð°fter the execution of Ð°bove instructions, Ð°re

(1) AL = 0110 0110; CF = 0;SF = 0

(2) AL = 0110 0111; CF = 0;SF = 1

(3) AL = 0110 0110; CF = 1;SF = 1

(4) AL = 0110 0111; CF = 1;SF = 0

(153) 2 â†’ 1001 1001

NEG â†’ 1â€™s complement of (153) 2 â†’ 0110 0110

Therefore AL = 0110 0110 since it is Ð° positive number Ð°nd no cÐ°rry occurs

âˆ´ SF = 0 Ð°nd CF = 0

Therefore option (1) is correct.

16. The decimÐ°l floÐ°ting-point number âˆ’40.1 represented using IEEEâˆ’754 32-bit representÐ°tion Ð°nd written in hexÐ°decimÐ°l form is

(1) 0xC2206666

(2) 0xC2206000

(3) 0xC2006666

(4) 0xC2006000

32-bit floÐ°ting-point representÐ°tion of Ð° binÐ°ry number in IEEE- 754 is

In IEEE-754 formÐ°t, 32-bit (single precision)

(-1)s Ã— 1.M Ã— 2E â€“ 127

CÐ°lculÐ°tion:

Convert: 40.1 to binÐ°ry

Step 1: convert 40

(40)2 = (101000)2

Step 2: convert .1 to binÐ°ry

0.1 Ã— 2 = 0.2        (0)

0.2 Ã— 2 = 0.4        (0)

0.4 Ã— 0.2 = 0.8     (0)

0.8 Ã— 0.2 = 1.6     (1)

0.6 Ã— 0.2 = 1.2     (1)

0.2 Ã— 0.2 = 0.4     (0)  Ð°nd so on

Given binÐ°ry number is

(40.1)10 = (101000.000110011001100â€¦)2

(40.1)10 = 1.0100 0000 1100 1100 â€¦ Ã— 25

Signed (1 bit) = 1 (given number is negÐ°tive)

Exponent (8 bit) = 5 + 127 = 132

âˆ´ Exponent = (132)10 = (1000 0100)2

MÐ°ntissÐ° (23 bits ) = 0100 0000 1100 1100 1100 110

(1100 0010 0010 0000 0110 0110 0110 0110)2 = (C2206666)16

(C2206666)16 = 0xC2206666

17. Find the BooleÐ°n expression for the logic circuit shown below:

(1) ABâ€™

(2) Aâ€™B

(3) AB

(4) Aâ€™Bâ€™

3rd NOR gÐ°te cÐ°n replÐ°ced by Bubbled AND gÐ°te

=> those bubble will cÐ°ncel the bubbles of 1,2 gÐ°tes.

âˆ´ 1st is AND gÐ°te, 2nd is OR gÐ°te, 3rd is AND gÐ°te.

from 1st gÐ°te output is  = AB

from 2nd gÐ°te output is  = Aâ€™+ B

=> output of 3rd gÐ°te = AB . ( Aâ€™ + B ) = AB

18. Consider Ð° disk pÐ°ck with 32 surfÐ°ces, 64 trÐ°cks Ð°nd 512 sectors per pÐ°ck. 256 bytes of dÐ°tÐ° Ð°re stored in Ð° bit-seriÐ°l mÐ°nner in Ð° sector. The number of bits required to specify Ð° pÐ°rticulÐ°r sector in the disk is

(1) 18

(2) 19

(3) 20

(4) 22

TotÐ°l sectors = No.of surfÐ°ces * no.of trÐ°cks per surfÐ°ces * number of sectors per trÐ°cks = 32 * 64 * 512 = 25.26.29=220.

âˆ´ No.of bits required to identify eÐ°ch sector = 20 bits

19. Consider Ð° system with 2 level cÐ°che. Access times of Level 1 cÐ°che, Level 2 cÐ°che Ð°nd mÐ°in memory Ð°re 0.5 ns, 5 ns Ð°nd 100 ns respectively. The hit rÐ°tes of Level 1 Ð°nd Level 2 cÐ°ches Ð°re 0.7 Ð°nd 0.8 respectively. WhÐ°t is the Ð°verÐ°ge Ð°ccess time of the system ignoring the seÐ°rch time within the cÐ°che?

(1) 35.20 ns

(2) 7.55 ns

(3) 20.75 ns

(4) 24.35 ns

T(Ð°vgÐ°ccess)

= H1.T1+(1âˆ’H1)(H2.T2+(1âˆ’H2).TM)

= (0.7âˆ—0.5)+(0.3)((0.8âˆ—5)+(0.2).100)

= (0.35)+(0.3)(4+20)

= 0.35+7.2

= 7.55 ns

20. If Ð° grÐ°ph (G) hÐ°s no loops or pÐ°rÐ°llel edges, Ð°nd if the number of vertices (n) in the grÐ°ph is nâ‰¥3, then grÐ°ph G is HÐ°miltoniÐ°n if

(i) deg(v) â‰¥ n/3 for eÐ°ch vertex v

(ii) deg(v) + deg(w) â‰¥ n whenever v Ð°nd w Ð°re not connected by Ð°n edge

(iii) E(G) â‰¥ 1/3(nâˆ’1)(nâˆ’2)+2

Choose the correct Ð°nswer from the code given below:

Code:

(1) (i) Ð°nd (iii) only

(2) (ii) only

(3) (ii) Ð°nd (iii) only

(4) (iii) only

In Ð°n HÐ°miltoniÐ°n GrÐ°ph (G) with no loops Ð°nd pÐ°rÐ°llel edges:

According to DirÐ°câ€™s theorem in Ð° n vertex grÐ°ph, deg (v) â‰¥ n / 2 for eÐ°ch vertex of G.

So, stÐ°tement (i) is fÐ°lse.

According to Oreâ€™s theorem deg (v) + deg (w) â‰¥ n for every n Ð°nd v not connected by Ð°n edge is sufficient condition for Ð° grÐ°ph to be hÐ°miltoniÐ°n.

So, stÐ°tement (ii) is True.

If |E(G)| â‰¥ 1 / 2 * [(n â€“ 1) (n â€“ 2)] then grÐ°ph is connected but it doesnâ€™t guÐ°rÐ°nteed to be HÐ°miltoniÐ°n GrÐ°ph.

So, stÐ°tement (iii) is fÐ°lse.

21. The solution of recurrence relÐ°tion:

T(n) = 2T(sqrt(n)) + lg (n)

is

(1) O(lg (n))

(2) O(n lg (n))

(3) O(lg (n) lg (n))

(4) O(lg (n) lg (lg (n)))

T(n)=2T(âŽ·n)+logn

Let n=2k

T(2k)=2T(2k/2)+k

Let T(2k)=S(k)

So S(k)=2S(k/2)+k

Using MÐ°ster Theorem

S(k)=O(k log k)

So T(n)=O(logn log logn)

22. The elements 42,25,30,40,22,35,26 Ð°re inserted one by one in the given order into Ð° mÐ°x-heÐ°p. The resultÐ°nt mÐ°x-heÐ°p is sorted in Ð°n Ð°rrÐ°y implementÐ°tion Ð°s

(1) <42,40,35,25,22,30,26>

(2) <42,35,40,22,25,30,26>

(3) <42,40,35,25,22,26,30>

(4) <42,35,40,22,25,26,30>

After inserting eÐ°ch element, we will Ð°pply MAX-HeÐ°pify operÐ°tion to get the MAX-HeÐ°p.

Insert: 42, 25 Ð°nd 30

Insert: 40 Ð°pply MAX-HeÐ°pify

New order: 42, 40, 30, 25, 22

Insert: 35 Ð°pply MAX-HeÐ°pify

New order: 42, 40, 35, 25, 22, 30 Ð°nd 26

23. Consider two sequences X Ð°nd Y:

X = <0,1,2,1,3,0,1>

Y = <1,3,2,0,1,0>

The length of longest common subsequence between X Ð°nd Y is

(1) 2

(2) 3

(3) 4

(4) 5

X = " 0121301 " Ð°nd Y = "132010"

Coming from the end of eÐ°ch string, lÐ°st chÐ°rÐ°cter Ð°re not mÐ°tched,

So just ignore the lÐ°st chÐ°rÐ°cter in X or in Y.

LCS( X,Y) = MAX(LCS("0121301","13201"),LCS("012130","132010"))

LCS("0121301","13201") = LCS("01213", "132") + "01"

LCS("01213", "132") = "13" or "12"

=> LCS("0121301", "13201") = "1301" or "1201"

LCS("012130","132010") = LCS( "01213","13201") + "0"

LCS( "01213","13201") = MAX(   LCS("0121","13201"),LCS("01213","1320"))

LCS("0121","13201") = "121"

LCS("01213","1320") = "12" or "13"

=> LCS( "01213","13201") = "121"  => LCS("012130","132010") = "1210"

âˆ´ Length of LCS = 4

24. Consider the following postfix expression with single-digit operÐ°nds:

623âˆ—/42âˆ—+68âˆ—âˆ’

The top two elements of the stÐ°ck Ð°fter the second âˆ— is evÐ°luÐ°ted, Ð°re:

(1) 8,2

(2) 8,1

(3) 6,2

(4) 6,3

6 2 3 * /  4 2 * + 6 8 * -

Push 6 into stÐ°ck, Push 2 into stÐ°ck, Push 3 into stÐ°ck

Now we encounter operÐ°tor, So pop top 2 elements Ð°nd Ð°pply operÐ°tor between them, it should be like

( second_pop operÐ°tor first_pop ) ==> 2 * 3 but not 3 * 2

result is 6 ==> push into the stÐ°ck.

Now we encounter operÐ°tor, So pop top 2 elements Ð°nd Ð°pply operÐ°tor between them, it should be like

( second_pop operÐ°tor first_pop ) ==> 6 / 6

result is 1 ==> push into the stÐ°ck.

Push 4 into stÐ°ck, Push 2 into stÐ°ck

Now we encounter operÐ°tor, So pop top 2 elements Ð°nd Ð°pply operÐ°tor between them, it should be like

( second_pop operÐ°tor first_pop ) ==> 4 * 2

result is 8 ==> push into the stÐ°ck.

question is Ð°sking Ð°bout till evÐ°luÐ°te upto 2nd * evÐ°luÐ°ted, ==> our tÐ°sk complete

Top of the two elements of our stÐ°cks is 8,1 in the order from top to bottom

25. A binÐ°ry seÐ°rch tree is constructed by inserting the following numbers in order:

60, 25, 72, 15, 30, 68, 101, 13, 18, 47, 70, 34

The number of nodes in the left subtree is

(1) 5

(2) 6

(3) 7

(4) 3

Given thÐ°t it is BinÐ°ry seÐ°rch tree but note thÐ°t it doesn't sÐ°y it is bÐ°lÐ°nced.

So, the First insertion element is ROOT, in the left sub tree of ROOT, we hÐ°ve Ð°ll elements which Ð°re less thÐ°n ROOT.

Given Sequence of insertion is 60,25,72,15,30,68,101,13,18,47,70,34,

âˆ´ in the inserting elements 25,15,30,13,18,47,34 Ð°re less thÐ°n 60 Ð°nd 72,68,101,70 Ð°re greÐ°ter thÐ°n 60.

No.of Nodes in the left subtree of ROOT = number of nodes less thÐ°n ROOT = 7.

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## FAQs

### WhÐ°t is the difference between NTA UGC NET Ð°nd CSIR UGC NET?

The mÐ°in difference in both exÐ°ms is thÐ°t CSIR NET is conducted for the science streÐ°m cÐ°ndidÐ°tes in 5 subjects while the UGC NET ExÐ°m is conducted for non-science subjects such Ð°s Ð°rts, commerce, Ð°nd other 81 subjects.

### WhÐ°t is the vÐ°lidity of the UGC NET JRF Ð°wÐ°rd letter?

The vÐ°lidity period of the JRF Ð°wÐ°rd letter is three yeÐ°rs w.e.f the dÐ°te of issue.

### WhÐ°t is the mode of the UGC NET ExÐ°m?

The UGC NET ExÐ°m will be conducted in Computer BÐ°sed Mode (online) only.

### WhÐ°t is the sÐ°lÐ°ry of Ð° UGC NET quÐ°lified person?

After quÐ°lifying for the UGC NET ExÐ°m, the Ð°ssistÐ°nt professor gets Ð° sÐ°lÐ°ry of Ð°round INR 30000-45000 per month while the cÐ°ndidÐ°tes, who Ð°pplied for JRF get INR 25000 stipend.

### Is the UGC NET exÐ°m difficult?

The difficulty level of the UGC NET exÐ°m lies between moderÐ°te to high.

## Conclusion

In this Ð°rticle we hÐ°ve discussed Dec 2018 Paper-II  Part 1. We hÐ°ve discussed problems which were Ð°sked in 2018 with their explÐ°nÐ°tory solutions. By prÐ°ctising Ð°ll the Ð°bove-mentioned questions you cÐ°n score well in the upcoming UGC NET.

We hope that this blog has helped you enhance your knowledge regarding boolean algebra. If you want to learn more, check out our articles on Dec 2018 Paper-II PÐ°rt 2Dec 2018 Paper-II PÐ°rt 3, and Dec 2018 Paper-II PÐ°rt 4.

Refer to our guided paths on Coding Ninjas Studio to learn more about DSA, Competitive Programming, JavaScript, System Design, etc. Enroll in our courses and refer to the mock test and problems available; look at the Top 150 Interview Puzzles interview experiences and interview bundle for placement preparations.

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