Oswaal NDA-NA (National Defence Academy / Naval Academy) 12 Solved Papers (2017-2023) Mathematics For 2024 Exam [3 ed.] 9359582204, 9789359582207

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For 2024 Exam

HIGHLY RECOMMENDED

NDA-NA

NDA / NA NATIONAL DEFENCE ACADEMY / NAVAL ACADEMY

YEAR-WISE

12

YEAR-WISE SOLVED PAPERS 2017 - 2023

SOLVED

MATHEMATICS As per Latest Exam Pattern Issued by UPSC

MATHEMATICS

The ONLY book you need to Crack NDA-NA

2017-2023

1

2

3

100% Updated

Extensive Practice

Concept Clarity

Valuable Exam Insights

Exam Analysis

with Fully Solved Apr. & Sep. 2023 Papers

with more than 1800+ Questions & 2 Sample Question Papers

with Detailed Explanations, Mind Maps & Mnemonics

with Tips to crack NDA/NA Exam in the first attempt

with Last 5 Years’ Chapter-wise Trend Analysis

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4

5

3rd EDITION, YEAR 2023-24

ISBN

“9789359582207”

NDA/NA

SYLLABUS COVERED

PUBLISHED BY

C OPYRIG HT

RESERVED BY THE PUBLISHERS

All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without written permission from the publishers. The author and publisher will gladly receive information enabling them to rectify any error or omission in subsequent editions.

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DI SC L A IMER This book is published by Oswaal Books and Learning Pvt Ltd (“Publisher”) and is intended solely for educational use, to enable students to practice for examinations/tests and reference. The contents of this book primarily comprise a collection of questions that have been sourced from previous examination papers. Any practice questions and/or notes included by the Publisher are formulated by placing reliance on previous question papers and are in keeping with the format/pattern/guidelines applicable to such papers. The Publisher expressly disclaims any liability for the use of, or references to, any terms or terminology in the book, which may not be considered appropriate or may be considered offensive, in light of societal changes. Further, the contents of this book, including references to any persons, corporations, brands, political parties, incidents, historical events and/or terminology within the book, if any, are not intended to be offensive, and/or to hurt, insult or defame any person (whether living or dead), entity, gender, caste, religion, race, etc. and any interpretation to this effect is unintended and purely incidental. While we try to keep our publications as updated and accurate as possible, human error may creep in. We expressly disclaim liability for errors and/or omissions in the content, if any, and further disclaim any liability for any loss or damages in connection with the use of the book and reference to its contents”.

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PREFACE 

“We fight to win and win with a knockout because there are no runners up in war.” — General JJ Singh

The National Defence Academy is an iconic institution and hallmark of global excellence in the sphere of military education. Over the years it has emerged as a unique military academy, attracting the best of youth from our nation and also from friendly foreign countries and transforming them into officers and gentlemen. National Defence Academy or NDA exam is conducted twice a year by Union Public Service Commission for admission to the Army, Navy, and Air Force wings of NDA and Indian Naval Academy Course (INAC). In 2023, 4.5 Lacs students applied for the NDA examination, the opportunity you get from the Indian Armed Forces is just limitless, which helps in enhancing your personality traits. For a youngster who is aspiring to get a job full of challenges and excitement, then there is no better job than the defence. This book aims to make aspirants exam-ready, boost their confidence and help them achieve better results in NDA. By making learning Simple, we are also making better careers and a better life for every student. Every day we are moving ahead pursuing our noble cause of spreading knowledge. This set of solved question papers is designed to enrich students with ample and exam-oriented practice so that they can clear NDA examinations with extraordinary results. Not one or two but 12 Previous Year Solved Question Paper (2017 to 2023) to focus on polishing every topic. Thorough studying of this book will boost my confidence and familiarise me with exam patterns. Some benefits of studying from Oswaal NDA 12 Previous year solved question papers: 1. 2. 3. 4. 5. 6.

100% updated with Fully Solved Paper of April & September 2023. Concept Clarity with detailed explanations of 2017 (I) to 2023 Papers. Extensive Practice with 1440+ Questions and Two Sample Question Papers. Crisp Revision with Mind Maps. Expert Tips helps you get expert knowledge master & crack NDA/NA in first attempt. Exam insights with 5 Year-wise (2023-2019) Trend Analysis, empowering students to be 100% exam ready.

Our Heartfelt Gratitude Finally, we would like to thank our authors, editors, and reviewers. Special thanks to our students who send us suggestions and constantly help improve our books. To stay true to our motto of ‘Learning Made Simple’, we constantly strive to present information in ways that are easy to understand as well as remember. Wish you all Happy Learning! All the Best!! TEAM OSWAAL

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Tips to Crack NDA in the First Attempt

CONTENTS g g g g g g g g

Tips to Crack NDA in the First Attempt Syllabus Scheme of Examination Height and Weight Standards NDA vs CDS: Know All the Similarities & Differences Trend Analysis from (2023-2019) NDA/NA 2023 - Solved Paper - I NDA/NA 2023 - Solved Paper - II

4 - 4 6 - 6 7 - 7 8 - 10 11 - 11 12 - 12 15 - 38 39 - 64

 Mind Maps

1 - 19

 Mnemonics

20 - 27

 NDA/NA 2022 - Solved Paper - I

28 - 69

 NDA/NA 2022 - Solved Paper - II

70 - 110

 NDA/NA 2021 - Solved Paper - I

111 - 157

 NDA/NA 2021 - Solved Paper - II

158 - 213

 NDA/NA 2020 - Solved Paper - I

214 - 226

 NDA/NA 2019 - Solved Paper - I

227 - 239

 NDA/NA 2019 - Solved Paper - II

240 - 251

 NDA/NA 2018 - Solved Paper - I

252 - 264

 NDA/NA 2018 - Solved Paper - II

265 - 277

 NDA/NA 2017 - Solved Paper - I

278 - 290

 NDA/NA 2017 - Solved Paper - II

291 - 304

 NDA/NA Sample Question Paper - I

305 - 316

 NDA/NA Sample Question Paper - II

317 - 328 qqq

Keep yourself updated! For the latest NDA updates throughout the Academic year 2023-24 Scan the QR code below

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Syllabus PAPER-I MATHEMATICS (Code No. 01) (Maximum Marks - 300) 1. ALGEBRA: Concept of set, operations on sets, Venn diagrams. De Morgan laws, Cartesian product, relation, equivalence relation. Representation of real numbers on a line. Complex numbers—basic properties, modulus, argument, cube roots of 19 unity. Binary system of numbers. Conversion of a number in decimal system to binary system and vice-versa. Arithmetic, Geometric and Harmonic progressions. Quadratic equations with real coefficients. Solution of linear inequations of two variables by graphs. Permutation and Combination. Binomial theorem and its applications. Logarithms and their applications. 2. MATRICES AND DETERMINANTS: Types of matrices, operations on matrices. Determinant of a matrix, basic properties of determinants. Adjoint and inverse of a square matrix, Applications-Solution of a system of linear equations in two or three unknowns by Cramer’s rule and by Matrix Method. 3. TRIGONOMETRY: Angles and their measures in degrees and in radians. Trigonometrical ratios. Trigonometric identities Sum and difference formulae. Multiple and Sub-multiple angles. Inverse trigonometric functions. Applications-Height and distance, properties of triangles. 4. ANALYTICAL GEOMETRY OF TWO AND THREE DIMENSIONS: Rectangular Cartesian Coordinate system. Distance formula. Equation of a line in various forms. Angle between two lines. Distance of a point from a line. Equation of a circle in standard and in general form. Standard forms of parabola, ellipse and hyperbola. Eccentricity and axis of a conic. Point in a three dimensional space, distance between two points. Direction Cosines and direction ratios. Equation two points. Direction Cosines and direction ratios. Equation of a plane and a line in various forms. Angle between two lines and angle between two planes. Equation of a sphere. 5. DIFFERENTIAL CALCULUS: Concept of a real valued function–domain, range and graph of a function. Composite functions, one to one, onto and inverse functions. Notion of limit, Standard limits—examples. Continuity of functions—examples, algebraic operations on continuous functions. Derivative of function at a point, geometrical and physical interpretation of a derivative—applications. Derivatives of sum, product and quotient of functions, derivative of a function with respect to another function, derivative of a composite function. Second order derivatives. Increasing and decreasing functions. Application of derivatives in problems of maxima and minima. 6.  INTEGRAL CALCULUS AND DIFFERENTIAL EQUATIONS: 20 Integration as inverse of differentiation, integration by substitution and by parts, standard integrals involving algebraic expressions, trigonometric, exponential and hyperbolic functions. Evaluation of definite integrals— determination of areas of plane regions bounded by curves—applications. Definition of order and degree of a differential equation, formation of a differential equation by examples. General and particular solution of a differential equations, solution of first order and first degree differential equations of various types—examples. Application in problems of growth and decay. 7. VECTOR ALGEBRA: Vectors in two and three dimensions, magnitude and direction of a vector. Unit and null vectors, addition of vectors, scalar multiplication of a vector, scalar product or dot product of two vectors. Vector product or cross product of two vectors. Applications—work done by a force and moment of a force and in geometrical problems. 8. STATISTICS AND PROBABILITY: Statistics: Classification of data, Frequency distribution, cumulative frequency distribution—examples. Graphical representation—Histogram, Pie Chart, frequency polygon— examples. Measures of Central tendency—Mean, median and mode. Variance and standard deviation—determination and comparison. Correlation and regression. Probability : Random experiment, outcomes and associated sample space, events, mutually exclusive and exhaustive events, impossible and certain events. Union and Intersection of events. Complementary, elementary and composite events. Definition of probability—classical and statistical—examples. Elementary theorems on probability—simple problems. Conditional probability, Bayes’ theorem—simple problems. Random variable as function on a sample space. Binomial distribution, examples of random experiments giving rise to Binominal distribution. qqq (6)

Scheme of Examination

1. The subjects of the written examination, the time allowed and the maximum marks allotted to each subject will be as follows:— Subject

Code

Duration

Maximum Marks

Mathematics

01

2½ Hours

300

General Ability Test

02

2½ Hours

600

Total

900

SSB Test/Interview:

900

2. THE PAPERS IN ALL THE SUBJECTS WILL CONSIST OF OBJECTIVE TYPE QUESTIONS ONLY. THE QUESTION PAPERS (TEST BOOKLETS) OF MATHEMATICS AND PART “B” OF GENERAL ABILITY TEST WILL BE SET BILINGUALLY IN HINDI AS WELL AS ENGLISH. 3. In the question papers, wherever necessary, questions involving the metric system of Weights and Measures only will be set. 4. Candidates must write the papers in their own hand. In no circumstances will they be allowed the help of a scribe to write answers for them. 5. The Commission have discretion to fix qualifying marks in any or all the subjects at the examination. 6.  The candidates are not permitted to use calculator or Mathematical or logarithmic table for answering objective type papers (Test Booklets). They should not therefore, bring the same inside the Examination Hall.

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Height and Weight Standards

For Female Candidates joining NDA (Army): Age (yrs)

Height (cm)

Minimum weight for all ages

Age: 17 to 20 yrs

Age: 20 + 01 day - 30 yrs

Age : 30 + 01 Day - 40 yrs

Age: Above yrs

Weight (kg)

Weight (kg)

Weight (kg)

Weight (kg)

Weight (kg)

140

35.3

43.1

45.1

47.0

49.0

141

35.8

43.7

45.7

47.7

49.7

142

36.3

44.4

46.4

48.4

50.4

143

36.8

45.0

47.0

49.1

51.1

144

37.3

45.6

47.7

49.8

51.8

145

37.8

46.3

48.4

50.5

52.6

146

38.4

46.9

49.0

51.2

53.3

147

38.9

47.5

49.7

51.9

54.0

148

39.4

48.2

50.4

52.6

54.8

149

40.0

48.8

51.1

53.3

55.5

150

40.5

49.5

51.8

54.0

56.3

151

41.0

50.2

52.4

54.7

57.0

152

41.6

50.8

53.1

55.4

57.8

153

42.1

51.5

53.8

56.2

58.5

154

42.7

52.2

54.5

56.9

59.3

155

43.2

52.9

55.3

57.7

60.1

156

43.8

53.5

56.0

58.4

60.8

157

44.4

54.2

56.7

59.2

61.6

158

44.9

54.9

57.4

59.9

62.4

159

45.5

55.6

58.1

60.7

63.2

160

46.1

56.3

58.9

61.4

64.0

161

46.7

57.0

59.6

62.2

64.8

162

47.2

57.7

60.4

63.0

65.6

163

47.8

58.5

61.1

63.8

66.4

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40

...CONTD. For Male Candidates joining NDA (Army): Height requirement varies as per the stream of entry. Weight should be proportionate to height as per the chart given below:Age (yrs)

Minimum weight Age: 17 to for all ages 20 yrs

Age: 20 + 01 day - 30 yrs

Age : 30 + 01 Day - 40 yrs

Age: Above 40 yrs

Height (cm) 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174

Weight (kg) 35.3 35.8 36.3 36.8 37.3 37.8 38.4 38.9 39.4 40.0 40.5 41.0 41.6 42.1 42.7 43.2 43.8 44.4 44.9 45.5 46.1 46.7 47.2 47.8 48.4 49.0 49.6 50.2 50.8 51.4 52.0 52.6 53.3 53.9 54.5

Weight (kg) 45.1 45.7 46.4 47.0 47.7 48.4 49.0 49.7 50.4 51.1 51.8 52.4 53.1 53.8 54.5 55.3 56.0 56.7 57.4 58.1 58.9 59.6 60.4 61.1 61.9 62.6 63.4 64.1 64.9 65.7 66.5 67.3 68.0 68.8 69.6

Weight (kg) 47.0 47.7 48.4 49.1 49.8 50.5 51.2 51.9 52.6 53.3 54.0 54.7 55.4 56.2 56.9 57.7 58.4 59.2 59.9 60.7 61.4 62.2 63.0 63.8 64.6 65.3 66.1 66.9 67.7 68.5 69.4 70.2 71.0 71.8 72.7

Weight (kg) 49.0 49.7 50.4 51.1 51.8 52.6 53.3 54.0 54.8 55.5 56.3 57.0 57.8 58.5 59.3 60.1 60.8 61.6 62.4 63.2 64.0 64.8 65.6 66.4 67.2 68.1 68.9 69.7 70.6 71.4 72.3 73.1 74.0 74.8 75.7

Weight (kg) 43.1 43.7 44.4 45.0 45.6 46.3 46.9 47.5 48.2 48.8 49.5 50.2 50.8 51.5 52.2 52.9 53.5 54.2 54.9 55.6 56.3 57.0 57.7 58.5 59.2 59.9 60.6 61.4 62.1 62.8 63.6 64.3 65.1 65.8 66.6

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...CONTD. Age (yrs)

Minimum weight Age: 17 to for all ages 20 yrs

Age: 20 + 01 day - 30 yrs

Age : 30 + 01 Day - 40 yrs

Age: Above 40 yrs

Height (cm) 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210

Weight (kg) 55.1 55.8 56.4 57.0 57.7 58.3 59.0 59.6 60.3 60.9 61.6 62.3 62.9 63.6 64.3 65.0 65.7 66.4 67.0 67.7 68.4 69.1 69.9 70.6 71.3 72.0 72.7 73.4 74.2 74.9 75.6 76.4 77.1 77.9 78.6 79.4

Weight (kg) 70.4 71.2 72.1 72.9 73.7 74.5 75.4 76.2 77.0 77.9 78.7 79.6 80.4 81.3 82.2 83.0 83.9 84.8 85.7 86.6 87.5 88.4 89.3 90.2 91.1 92.0 92.9 93.8 94.8 95.7 96.7 97.6 98.6 99.5 100.5 101.4

Weight (kg) 73.5 74.3 75.2 76.0 76.9 77.8 78.6 79.5 80.4 81.3 82.1 83.0 83.9 84.8 85.7 86.6 87.6 88.5 89.4 90.3 91.3 92.2 93.1 94.1 95.0 96.0 97.0 97.9 98.9 99.9 100.9 101.8 102.8 103.8 104.8 105.8

Weight (kg) 76.6 77.4 78.3 79.2 80.1 81.0 81.9 82.8 83.7 84.6 85.6 86.5 87.4 88.4 89.3 90.3 91.2 92.2 93.1 94.1 95.1 96.0 97.0 98.0 99.0 100.0 101.0 102.0 103.0 104.0 105.1 106.1 107.1 108.2 109.2 110.3

Weight (kg) 67.4 68.1 68.9 69.7 70.5 71.3 72.1 72.9 73.7 74.5 75.3 76.1 76.9 77.8 78.6 79.4 80.3 81.1 81.9 82.8 83.7 84.5 85.4 86.2 87.1 88.0 88.9 89.8 90.7 91.6 92.5 93.4 94.3 95.2 96.1 97.0

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NDA vs CDS: Know All the Similarities & Differences The National Defence Academy (NDA) and the Combined Defence Services (CDS) Exams are gateways to tri-services of the Indian Armed Forces. Though both the exams are conducted by the Union Public Service Commission, i.e. UPSC, there are many similarities and differences in the recruitment, training, salary, perks and promotion opportunities, etc. For those who are planning to join Indian Army, Navy or Air Force, it is essential to know the differences and similarities in NDA and CDS. The similarities are given below: Parameter

NDA

CDS

16.5-19.5 Years

19-25 Years

Men only

Men & Women

10+2

Degree

Scheme of Examination

Written + SSB

Written + SSB

Frequency of the Exam

Twice/Year

Twice/Year

Age Eligibility Educational Qualification

Duration of Training

4-4.5 Years 3 Yrs. at NDA and 1 Yr. at IMA (For Army cadets) 3 Yrs. at NDA and 1 Yr. at Naval Academy (For Naval cadets)/ 3 Yrs. at NDA and 1 & 1/2 Yrs. at AFA Hyderabad (For AF cadets)

18 months for IMA Cadets 37-40 months for Navy Officers 74 months for Air Force Officers

Training Centres

National Defence Academy, Khadakwasla, Pune Indian Military Academy, Dehradun Indian Naval Academy, Ezhimala Indian Air Force Academy, Hyderabad

Indian Military Academy (IMA), Dehradun for Army Cadets Indian Naval Academy, Ezhimala for Navy Cadets Indian Air Force Academy, Hyderabad for Air Force Officers Officers Training Academy (OTA), Chennai

Degrees awarded

Army Cadets - B.Sc./B.Sc. (Computer)/BA /B.Tech. degree Naval Cadets - B.Tech. degree Air Force Cadets - B.Tech. degree

Army Cadets in IMA - PG Diploma in ‘Military and Defence Management OTA Chennai – Post Graduate Diploma in Defence Management and Strategic Studies

Rank assigned after training Stipend during training

Lieutenant

Lieutenant

Rs. 21,000/- p.m. (fixed)

Rs. 21,000/- p.m. (fixed)

Promotional Avenues Rank

Min. Commissioned Service for Promotion NDA Officer

CDS Officer

Lieutenant

On Commission

On Commission

Captain

02 Years

02 Years

Major

06 years

06 years

Lieutenant Colonel

13 years

13 years

Colonel(Selection)

15 years

15 years

Colonel (Time Scale)

26 years

26 years

Brigadier

On Selection

23 years

Major General

On Selection

25 years

Lieutenant General

On Selection

28 years

General

On Selection

No restrictions

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qqq

Trend Analysis (2023-2019) Units No.

Chapter Name

Number of Question(s) in 2023

2022

2022

2021

2020

2019

I

I

II

I

II

I

I

II

1.

Algebra

23

30

29

25

33

20

27

30

2.

Matrices & Determinants

11

11

9

11

10

8

8

5

3.

Trigonometery

17

16

17

19

7

24

22

16

4.

Analytical Geometry of Two and Three Dimensions

15

14

11

15

15

15

16

10

5.

Differential Calculus

15

10

14

11

17

15

11

26

6.

Integral Calculus and Differential Equations

14

14

17

14

15

13

11

8

7.

Vector Algebra

5

5

5

6

5

5

5

5

8.

Statistics and Probability

20

20

18

19

18

20

20

18

9.

Mathematical Induction

–

–

–

–

–

–

–

–

10.

Speed, Distance & Time

–

–

–

–

–

–

–

–

11.

Applied Mathematics

–

–

–

–

–

–

–

2

120

120

120

120

120

120

120

120

Total

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BHUBANESHWAR M/s Pragnya, 8847888616, 9437943777, Padmalaya, 9437026922, Bidyashree, 9937017070, Books Godown, 7894281110 BARIPADA Trimurti Book World, 9437034735 KEONJHAR Students corner, 7008435418

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Mahavir Stationers, 9429474177 College Store, (ISC) NO CALL 02637-258642, 9825099121

AMBALA PATIALA

VADODARA

Umakant Book Sellers & Stationer, 9624920709

HARYANA

FEROZPUR LUDHIANA

ROHTAK

Manish Traders, 9812556687, Swami Kitab Ghar, 9355611088,

CHANDIGARH

REWARI

Sanjay book depot, 9255447231

Kashi Ram Kishan lal, 9289504004, 8920567245 Natraj Book Distributors, 7988917452

AJMER KOTA

BHUNA

Khurana Book Store, 9896572520

BHILWARA

JAMMU

JAIPUR

Sahitya Sangam, 9419190177

UDAIPUR

Nakoda Book Depot, (01482) 243653, 9214983594, Alankar Book Depot, 9414707462 Ravi Enterprises, 9829060694, Saraswati Book House, (0141) 2610823, 9829811155, Goyal Book Distt., 9460983939, 9414782130 Sunil Book Store, 9828682260

Crown Book Distributor & Publishers, (0651) 2213735, 9431173904, Pustak Mandir, 9431115138, Vidyarthi Pustak Bhandar, 9431310228

AGARTALA

Book Corner, 8794894165, 8984657146, Book Emporium, 9089230412

KARNATAKA

COIMBATORE

SURAT

BALLABGARH HISAR

BOKARO RANCHI DUMKA

PUNJAB

Shopping Point, 9824108663

JALANDHAR

Babu Ram Pradeep Kumar, 9813214692

JHARKHAND

Bokaro Student Friends, (0654) 2233094, 7360021503, Bharati Bhawan Agencies, 9431740797

JODHPUR

Renuka Book Distributor, (0836) 2244124, Vidyamandir Book Distributors, 9980773976 CHENNAI

BANGLORE

Krishna book house, 9739847334, Hema Book Stores, 9986767000,

BELLERI

Chatinya book centre, 9886064731

PUDUCHERRY

ERNAKULAM

Academic Book House, (0484) 2376613, H & C Store, 9864196344, Surya Book House, 9847124217, 9847238314 Book Centre, (0481) 2566992 Academic Book House, (0471) 2333349, 9447063349, Ponni Book Stall, 9037591721

TRICHY

KOTTAYAM TRIVANDRUM CALICUT

Sapna Book House Pvt. Ltd., 9980513242, Hema Book World, (Chamrajpet) (ISC) 080-40905110, 9945731121

Aman Book Stall, (0495) 2721282,

MADHYA PRADESH

CHHINDWARA

Pustak Bhawan, ( E & C ), 8982150100

GWALIOR

Agarwal Book Depot, 9425116210

Cheap Book Store, 9872223458, 9878258592, City Book Shop, 9417440753, Subhash Book Depot, 9876453625, Paramvir Enterprises, 9878626248 Sita Ram book Depot, 9463039199, 7696141911 Amit Book, 9815807871, Gupta Brothers, 9888200206, Bhatia Book Centre, 9815277131 Mohindra Book Depot, 9814920226

RAJASTHAN

Laxmi General Store, Ajmer, 0145- 2428942 9460652197 Vardhman Book Depot, 9571365020, 8003221190 Raj Traders, 9309232829

Second Hand Book Stall, 9460004745

TRIPURA

TAMIL NADU

HUBLI

KERALA

Bharat Book Depot, 7988455354 Goel Sons, 9463619978, Adarsh Enterprises, 9814347613

SALEM

THENI MADURAI VELLORE

HYDERABAD

Majestic Book House, (0422) 2384333, CBSC Book Shop, 9585979752

Arraba Book Traders, (044) 25387868, 9841459105, M.R. Book Store (044) 25364596, Kalaimagal Store, (044) 5544072, 9940619404, Vijaya Stores, 9381037417, Bookmark It-Books & Stat. Store, 7305151653, M.K. Store, 9840030099, Tiger Books Pvt. Ltd., 9710447000, New Mylai Stationers, 9841313062, Prince Book House, Chennai, 0444-2053926, 9952068491, S K Publishers & Distributors, 9789865544, Dharma Book Shop, 8667227171 Sri Lakshmi Book Seller, 7871555145 Pattu book centre, 9894816280

P.R.Sons Book Seller, 9443370597, Rasi Publication, 9894816280 Maya Book Centre, 9443929274 Selvi Book Shoppe, 9843057435, Jayam Book Centre, 9894658036 G.K book centre and collections, 9894517994

TELANGANA

Sri Balaji Book Depot, (040) 27613300, 9866355473, Shah Book House, 9849564564 Vishal Book Distributors, 9246333166, Himalaya Book World, 7032578527

( 13 )

0808

AHMEDABAD

UTTARAKHAND

GORAKHPUR

Central Book House, 9935454590, Friends & Co., 9450277154, Dinesh book depot, 9125818274, Friends & Co., 9450277154

DEHRADUN

Inder Book Agencies, 9634045280, Amar Book Depot , 8130491477, Goyal Book Store, 9897318047, New National Book House, 9897830283/9720590054

JHANSI

Bhanu Book Depot, 9415031340

MUSSORIE

Ram Saran Dass Chanda kiran, 0135-2632785, 9761344588

KANPUR

Radha News Agency, 8957247427, Raj Book Dist., 9235616506, H K Book Distributors, 9935146730, H K Book Distributors, 9506033137/9935146730

UTTAR PRADESH

LUCKNOW

AGRA

Sparsh Book Agency, 9412257817, Om Pustak Mandir, (0562) 2464014, 9319117771,

MEERUT

Ideal Book Depot, (0121) 4059252, 9837066307

ALLAHABAD

Mehrotra Book Agency, (0532) 2266865, 9415636890

NOIDA

Prozo (Global Edu4 Share Pvt. Ltd), 9318395520, Goyal Books Overseas Pvt.Ltd., 1204655555 9873387003

AZAMGARH

Sasta Sahitya Bhandar, 9450029674

PRAYAGRAJ

Kanhaiya Pustak Bhawan, 9415317109

ALIGARH

K.B.C.L. Agarwal, 9897124960, Shaligram Agencies, 9412317800, New Vimal Books, 9997398868, T.I.C Book centre, 9808039570

MAWANA

Subhash Book Depot, 9760262264

BULANDSHAHAR

Rastogi Book Depot, 9837053462/9368978202

BALRAMPUR

Universal Book Center, 8933826726

KOLKATA

BAREILLY

Siksha Prakashan, 9837829284

RENUKOOT

HARDOI

Mittal Pustak Kendra, 9838201466

Sanjay Publication, 8126699922 Arti book centre, 8630128856, Panchsheel Books, 9412257962, Bhagwati Book Store, (E & C), 9149081912

Vyapar Sadan, 7607102462, Om Book Depot, 7705871398, Azad Book Depot Pvt. Ltd.,

7317000250, Book Sadan, 9839487327, Rama Book Depot(Retail), 7355078254, Ashirwad Book Depot, 9235501197, Book.com, 7458922755, Universal Books,

9450302161, Sheetla Book Agency, 9235832418, Vidyarthi Kendra Publisher & Distributor Pvt Ltd, (Gold), 9554967415, Tripathi Book House, 9415425943

WEST BENGAL Oriental Publishers & Distributor (033) 40628367, Katha 'O' Kahini, (033) 22196313, 22419071, Saha Book House, (033), 22193671, 9333416484, United Book House, 9831344622, Bijay Pustak Bhandar, 8961260603, Shawan Books Distributors, 8336820363, Krishna Book House, 9123083874

Om Stationers, 7007326732

DEORIA

Kanodia Book Depot, 9415277835

COOCH BEHAR

S.B. Book Distributor, Cooch behar, 9002670771

VARANASI

Gupta Books, 8707225564, Bookman & Company, 9935194495/7668899901

KHARAGPUR

Subhani Book Store, 9046891334

MATHURA

Sapra Traders, 9410076716, Vijay Book House , 9897254292

SILIGURI

Agarwal Book House, 9832038727, Modern Book Agency, 8145578772

FARRUKHABAD

Anurag Book Agencies, 8844007575

DINAJPUR

Krishna Book House, 7031748945

NAJIBABAD

Gupta News Agency, 8868932500, Gupta News Agency, ( E & C ), 8868932500

MURSHIDABAD

New Book House, 8944876176

DHAMPUR

Ramkumar Mahaveer Prasad, 9411942550

Entrance & Competition Distributors PATNA

BIHAR

CUTTAK

A.K.Mishra Agencies, 9437025991

Metro Books Corner, 9431647013, Alka Book Agency, 9835655005, Vikas Book Depot, 9504780402

BHUBANESHWAR

M/s Pragnya, 9437943777

CHATTISGARH KORBA

Kitab Ghar, 9425226528, Shri Ramdev Traders, 9981761797

PUNJAB JALANDHAR

Cheap Book Store, 9872223458, 9878258592

DELHI

RAJASTHAN

DELHI

Singhania Book & Stationer, 9212028238, Radhey Book depot, 9818314141, The KOTA Book Shop, 9310262701, Mittal Books, 9899037390, Lov Dev & Sons, 9999353491

Vardhman Book Depot, 9571365020, Raj Traders, 9309232829

NEW DELHI

Anupam Sales, 9560504617, A ONE BOOKS, 8800497047

JAIPUR

HARYANA AMBALA

BOKARO

Goyal Book Distributors, 9414782130

UTTAR PRADESH

Bharat Book Depot, 7988455354

AGRA

BHAGWATI BOOK STORE, 9149081912, Sparsh Book Agency, 9412257817, Sanjay Publication, 8126699922

JHARKHAND

ALIGARH

New Vimal Books, 9997398868

Bokaro Student Friends Pvt. Ltd, 7360021503

ALLAHABAD

Mehrotra Book Agency, (532) 2266865, 9415636890

MADHYA PRADESH

GORAKHPUR

Central Book House, 9935454590

INDORE

Bhaiya Industries, 9109120101

KANPUR

Raj Book Dist, 9235616506

CHHINDWARA

Pustak Bhawan, 9827255997

LUCKNOW

Azad Book Depot PVT LTD, 7317000250, Rama Book Depot(Retail), 7355078254 Ashirwad Book Depot , 9235501197, Book Sadan, 8318643277, Book.com , 7458922755, Sheetla Book Agency, 9235832418

MAHARASHTRA

PRAYAGRAJ

Format Center, 9335115561, Garg Brothers Trading & Services Pvt. Ltd., 7388100499

NAGPUR

Laxmi Pustakalay and Stationers, (0712) 2727354

PUNE

Pragati Book Centre, 9850039311

MUMBAI

New Student Agencies LLP, 7045065799

ODISHA

Inder Book Agancies, 9634045280

WEST BENGAL KOLKATA

Bijay Pustak Bhandar Pvt. Ltd., 8961260603, Saha Book House, 9674827254 United Book House, 9831344622, Techno World, 9830168159

Trimurti Book World, 9437034735

0808

BARIPADA

UTTAR PRADESH DEHRADUN

( 14 )

NDA / NA

MATHEMATICS

I

National Defence Academy / Naval Academy

Time : 2 : 30 Hours

QUESTION PAPER

2023 Total Marks : 300

Instructions :

1. This Test Booklet contains 120 items (questions). Each item comprises four responses (answers). You will select the response which you want to mark on the Answer Sheet. In case you feel that there is more than one correct response, mark the response which you consider the best. In any case, choose ONLY ONE response for each item. 2. You have to mark all your responses ONLY on the separate Answer Sheet provided. See directions in the Answer Sheet. 3. All items carry equal marks. 4. Penalty for wrong answers : THERE WILL BE PENALTY FOR WRONG ANSWERS MARKED BY A CANDIDATE IN THE OBJECTIVE TYPE QUESTION PAPERS. (i) There are four alternatives for the answer to every question. For each question for which a wrong answer has been given by the candidate, one-third of the marks assigned to that question will be deducted as penalty. (ii) If a candidate gives more than one answer, it will be treated as a wrong answer even if one of the given answers happens to be correct and there will be same penalty as above to that question. (iii) If a question is left blank, i.e., no answer is given by the candidate, there will be no penalty for that question.

1. If ω is a non-real cube root of 1, then what is the value of (a)

3

1−ω ω + ω2

? (b)

2

4 (c) 1 (d) 3 2. What is the number of 6-digit numbers that can be formed only by using 0, 1, 2, 3, 4 and 5 (each once); and divisible by 6 ? (a) 96 (b) 120 (c) 192 (d) 312 3. What is the binary number equivalent to decimal number 1011 ? (a) 1011 (b) 111011 (c) 11111001 (d) 111110011 4. Let A be a matrix of order 3 × 3 and |A| = 4. If |2 adj(3A)| = 2α3β then what is the value of (α + β) ? (a) 12 (b) 13 (c) 17 (d) 24 5. If α and β are the distinct roots of equation x2 − x +1 = 0, then what is the value of (a) (c) 1

3

(b) (d)

2 1 3

α100 + β100 α100 − β100

?

6. Let A and B be symmetric matrices of same order, then which one of the following is correct regarding (AB − BA)? 1. Its diagonal entries are equal but nonzero 2. The sum of its non-diagonal entries is zero Select the correct answer using the code given below: (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 7. Consider the following statements in respect of square matrices A, B, C each of same order n : 1. AB = AC ⇒ B = C if A is non-singular 2.  If BX = CX for every column matrix X having n rows then B = C Which of the statements given above is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 8. The system of linear equations x + 2y + z = 4, 2x + 4y + 2z = 8 and 3x + 6y + 3z = 10 has (a) a unique solution (b) infinite many solutions (c) no solution (d) exactly three solutions 9. Let AX = B be a system of 3 linear equations with 3-unknowns. Let X1 and X2 be its two distinct solutions. If the combination aX1 + bX2 is a solution of AX = B; where a, b are real numbers, then which one of the following is correct ?

16

Oswaal NDA/NA Year-wise Solved Papers

(a) a = b (c) a + b = 0

(b) a + b = 1 (d) a − b = 1

10. What is the sum of the roots of the equation 0 x−a x−b 0 0 x−c = 0 ? x+b x+c 1 (a) a + b + c (c) a + b − c

(b) a − b + c (d) a − b − c

11. If 2 − i 3 where i = −1 is a root of the equation x2 + ax + b = 0, then what is the value of (a + b) ? (a) − 11 (b) − 3 (c) 0 (d) 3 12. If z =

1+i 3

where i = −1 , then what is the 1−i 3 argument of z ? 2π π (a) (b) 3 3 5π (c) 4π (d) 3 6 13. If a, b, c are in AP, then what is x +1 x + 2 x +3 x + 2 x + 3 x + 4 equal to ? x+a x+b x+3 (a) −1 (c) 1

(b) 0 (d) 2

14. logxa, ax and logbx are in GP, then what is x equal to ? (a) loga(logba) (b) logb(logab) (c) log a ( log b a ) 2 1

b

(d) log b ( log a b ) 2

1

15. If 2 c , 2 ac , 2 a are in GP, then which one of the following is correct? (a) a, b, c are in AP (b) a, b, c are in GP (c) a, b, c are in HP (d) ab, be, ca are in AP 5 16. The first and the second terms of an AP are 2 23 and respectively. If nth term is the largest 12 negative term, what is the value of n ? (a) 5 (b) 6 (c) 7 (d) n cannot be determined

17. For how many integral values of k, the equation x2 − 4x + k = 0, where k is an integer has real roots and both of them lie in the interval (0, 5) ? (a) 3 (b) 4 (c) 5 (d) 6 18. In an AP, the first term is x and the sum of the first n terms is zero. What is the sum of next m terms ? mx( m + n) mx( m + n) (a) (b) n −1 1−n (c) nx( m + n) m −1

(d) nx( m + n) 1−m

19. Consider the following statements : 1. (25)! + 1 is divisible by 26 2. (6)! + 1 is divisible by 7 Which of the above statements is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 20. If z is a complex number such that z − 1 is z +1 purely imaginary, then what is |z| equal to ? 2 1 (b) 3 2 (c) 1 (d) 2 (a)

21. How many real numbers satisfy the equation |x − 4| + |x − 7| = 15 ? (a) Only one (b) Only two (c) Only three (d) Infinitely many 22. A

mapping

f (x) =

f

:

A

→

B

defined

as

2x + 3 , x ∈ A . If f is to be onto, then 3x + 5

what are A and B equal to ? (a) A = R \ {− 5 } and B = R \ {− 2 } 3 3 5 (b) A = R and B = R \ {− } 3 3 (c) A = R \ {− } and B = R \ {0} 2 5 2 (d) A = R \ {− } and B = R \ { } 3 3 23. α and β are distinct real roots of the quadratic equation x2 + ax + b = 0. Which of the following statements is/are sufficient to find α ? 1. α + β = 0, α2+ β2 = 2 2. αβ2 = −1, a = 0

17

SOLVED PAPER – 2023-I

Select the correct answer using the code given below: (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 24. If the sixth term in the binomial expansion 8

 −8  of  x 3 + x 2 log10 x  is 5600, then what is the     value of x ? (a) 6 (b) 8 (c) 9 (d) 10 25. How many terms are there in the expansion of (3x − y)4(x + 3y)4 ? (a) 9 (b) 12 (c) 15 (d) 17 26. p, q, r and s are in AP such that p + s = 8 and qr =15. What is the difference between largest and smallest numbers ? (a) 6 (b) 5 (c) 4 (d) 3 27. Consider the following statements for a fixed natural number n : 1. C(n, r) is greatest if n = 2r 2. C(n, r) is greatest if n = 2r − 1 and n = 2r + 1 Which of the statements given above is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 28. m parallel lines cut n parallel lines giving rise to 60 parallelograms. What is the value of (m + n)? (a) 6 (b) 7 (c) 8 (d) 9 29. Let x be the number of permutations of the word ‘PERMUTATIONS’ and y be the number of permutations of the word ‘COMBINATIONS’. Which one of the following is correct ? (a) x = y (b) y = 2x (c) x = 4y (d) y = 4x 30. 5-digit numbers are formed using the digits 0, 1, 2, 4, 5 without repetition. What is the percentage of numbers which are greater than 50,000 ? (a) 20% (b) 25% (c) 100 % 3

(d) 110 % 3

Consider the following for the next two (02) items that follow : Let sinβ be the GM of sinα and cosα; tanγ be the AM of sinα and cosα. 31. What is cos2β equal to ? (a) (cosα − sinα)2 (b) (cosα + sinα)2 2 (d) (cos α − sin α ) 2 32. What is the value of sec2γ?

(c) (cosα − sinα)3

(a)

3 − sin 2α 5 + 2 sin 2α

(b)

5 + sin 2α 3 − sin 2α

(c)

3 − 2 sin 2α 4 + sin 2α

(d)

3 − sin 2α 4 + 3 sin 2α

Consider the following for the next two (02) items that follow : A flagstaff 20 m long standing on a pillar 10 m high subtends an angle tan−1(0.5) at a point P on the ground. Let q be the angle subtended by the pillar at this point P 33. If x is the distance of P from bottom of the pillar, then consider the following statements: 1.  x can take two values which are in the ratio 1:3 2.  x can be equal to height of the flagstaff Which of the statements given above is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 34. What is a possible value of tanθ ? (a)

3 4

(b)

2 3

1 (c) 1 (d) 4 3 Consider the following for the next two (02) items that follow : The perimeter of a triangle ABC is 6 times the AM of sine of angles of the triangle. Further BC = 3 and CA = 1. 35. What is the perimeter of the triangle ? (a) 3 +1 (b) 3 +2 (c)

3 +3 (d) 2 3 +1

36. Consider the following statements : 1. ABC is right angled triangle 2. The angles of the triangle are in AP Which of the statements given above is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2

18

Oswaal NDA/NA Year-wise Solved Papers

Consider the following for the next two (02) items that follow :

(a) 1 (c) 17

sin 2 A  sin A  1 π Let x  where 0 < A ≤ 2 sin A

Consider the following for the next two (02) items that follow :

37. What is the minimum value of x ? (a) 1 (b) 2 (c) 3 (d) 4

Consider the following for the next two (02) items that follow : 3 bc

39. What is the nature of the triangle ? (a) Equilateral (b) Isosceles (c) Right angled triangle (d) Scalene but not right angled 40. If c = 8, what is the area of the triangle ? (a) 4 3

In a triangle PQR, P is the largest angle and 1 . Further the in-circle of the triangle 3 touches the sides PQ, QR and RP at N, L and M respectively such that the lengths PN, QL and RM are n, n + 2, n + 4 respectively where n is an integer. cosP =

38. At what value of A does x attain the minimum value ? (a) π (b) π 6 4 π π (c) (d) 3 2

In the triangle ABC, a2+b2 + c2 = ac +

(b) 3 (d) 34

(b) 6 3

(c) 8 3 (d) 12 3 Consider the following for the next two (02) items that follow : Consider the function f(x) = |x − 2| + |3 − x| + |4 − x|, where x ∈ R. 41. At what value of x does the function attain minimum value ? (a) 2 (b) 3 (c) 4 (d) 0 42. What is the minimum value of the function ? (a) 2 (b) 3 (c) 4 (d) 0 Consider the following for the next two (02) items that follow : Consider the sum S = 0! + 1! + 2! +3! +4! + . . . . + 100! 43. If the sum S is divided by 8, what is the remainder ? (a) 0 (b) 1 (c) 2 (d) Cannot be determined 44. If the sum S is divided by 60, what is the remainder ?

45. What is the value of n ? (a) 4 (b) 6 (c) 8 (d) 10 46. What is the length of the smallest side ? (a) 12 (b) 14 (c) 16 (d) 18 Consider the following for the next two (02) items that follow : Given that sinx + cosx + tanx + cotx + secx + cosecx = 7 47. The given equation can be reduced to (a) sin22x − 44 sin2x + 36 = 0 (b) sin22x + 44 sin2x − 36 = 0 (c) sin22x − 22 sin2x +18 = 0 (d) sin22x + 22 sin2x − 18 = 0 48. If sin2x = a − b c , where a and b are natural numbers and c is prime number, then what is the value of a − b + 2c ? (a) 0 (b) 14 (c) 21 (d) 28 Consider the following for the next two (02) items that follow : A quadratic equation is given by (3 + 2 2 )x2 − (4 + 2 3 )x + (8 + 4 3 ) = 0 49. What is the HM of the roots of the equation ? (a) 2 (b) 4 (c) 2 2 (d) 2 3 50. What is the GM of the roots of the equation ? (a) (b) (c) (d)

 2

 6  3  2  1  6  3  2  1  6  3  2  1 2

6  3  2 1

19

SOLVED PAPER – 2023-I

Consider the following for the next two (02) items that follow : a b aα + b b c bα + c Let ∆(a, b, c, α) = aα + b bα + c 0 51. If ∆(a, b, c, α) = 0 for every α > 0, then which one of the following is correct ? (a) a, b, c are in AP (b) a, b, c are in GP (c) a, 2b, c are in AP (d) a, 2b, c are in GP 52. If ∆(7, 4, 2, α) = 0, then α is a root of which one of the following equations ? (a) 7x2 + 4x + 2 = 0 (b) 7x2 − 4x + 2 = 0 (c) 7x2 + 8x + 2 = 0 (d) 7x2 − 8x + 2 = 0 Consider the following for the next two (02) items that follow : Given that m(θ) = cot2θ+ n2tan2θ + 2n, where n is a fixed positive real number. 53. What is the least value of m(θ)? (a) n (b) 2n (c) 3n (d) 4n 54. Under what condition does m attain the least value ? (a) n = tan2θ (b) n = cot2θ 2 (c) n = sin θ (d) n = cos2θ Consider the following for the next two (02) items that follow : A quadrilateral is formed by the lines x = 0, y = 0, x + y= 1 and 6x + y = 3. 55. What is the equation of diagonal through origin ? (a) 3x + y = 0 (b) 2x + 3y = 0 (c) 3x − 2y = 0 (d) 3x + 2y = 0 56. What is the equation of other diagonal ? (a) x + 2y − 1 = 0 (b) x − 2y − 1 = 0 (c) 2x + y + 1 = 0 (d) 2x + y − 1 = 0 Consider the following for the next two (02) items that follow : P(x, y) is any point on the ellipse x2 + 4y2 = 1. Let E, F be the foci of the ellipse. 57. What is PE + PF equal to ? (a) 1 (b) 2 (c) 3 (d) 4 58. Consider the following points :  3  1.  , 0  2   

  2.  3 , 1   2 4  

 3 1 3.  , −   2 4  

Which of the above points lie on latus rectum of ellipse ? (a) 1 and 2 only (b) 2 and 3 only (c) 1 and 3 only (d) 1, 2 and 3 Consider the following for the next two (02) items that follow : The line y = x partitions the circle (x − a)2 + y2 = a2 in two segments. 59. What is the area of minor segment ? 2

2 (π − 1)a (a) (π − 2)a (b) 4 4 2 (π − 2)a (π − 1)a2 (c) (d) 2 2 60. What is the area of major segment ? 2 (a) (3π − 2)a 4

2 (b) (3π + 2)a 4 2 (3π + 2)a2 (c) (3π − 2)a (d) 2 2

Consider the following for the next two (02) items that follow : Let A(l, −1, 2) and B(2, 1, −1) be the end points of the diameter of the sphere x2 +y2 + z2 + 2ux + 2vy + 2wz − 1 = 0. 61. What is u + v + w equal to ? (a) − 2 (b) − 1 (c) 1 (d) 2 62. If P(x, y, z) is any point on the sphere, then what is PA2 + PB2 equal to ? (a) 15 (b) 14 (c) 13 (d) 6·5 Consider the following for the next two (02) items that follow : Consider two lines whose direction ratios are (2, − 1, 2) and (k, 3, 5). They are inclined at an angle π . 4 63. What is the value of k ? (a) 4 (b) 2 (c) 1 (d) − 1 64. What are the direction ratios of a line which is perpendicular to both the lines ? (a) (1, 2, 10) (b) (− 1, − 2, 10) (c) (11, 12, − 10) (d) (11, 2, − 10)

20

Oswaal NDA/NA Year-wise Solved Papers

Consider the following for the next two (02) items that follow : → Let → a = 3^ i + 3^ j +3^ k and c→=^ j −^ k . Let b be → → → such that → a · b =27 and → a × b = 9c → 65. What is b equal to ? (a) 3^ i + 4^ j +2^ k ^ ^ ^ (c) 5 i − 2 j +6 k

(b) 5^ i + 2^ j +2^ k ^ ^ ^ (d) 3 i + 3 j +4 k

→ 66. What is the angle between ( → a + b ) and→ c π π (a) (b) 2 3 π π (c) (d) 4 6 Consider the following for the next two (02) items that follow : Let a vector → a = 4^ i − 8^ j +^ k make angles α, β, γ with the positive directions of x, y, z axes respectively. 67. What is cosα equal to ? (a)

1 3

(b)

4 9

2 (c) 5 (d) 3 9 68. What is cos2β + cos2γ equal to ? (a) − 32 81

(b) −

16 81

16 32 (c) (d) 81 81 Consider the following for the next two (02) items that follow : The position vectors of two points A and B are ^− i ^and j ^+ j ^ k respectively. 69. Consider the following points : 1. (− 1, − 3, 1) 2. (− 1, 3, 2) 3. (− 2, 5, 3) Which of the above points lie on the line joining A and B ? (a) 1 and 2 only (b) 2 and 3 only (c) 1 and 3 only (d) 1, 2 and 3 → 70. What is the magnitude of AB ? (a) 2 (b) 3 (c) 6 (d) 3

Consider the following for the next three (03) items that follow : Let f(x) = Pex + Qe2x + Re3x, where P, Q, R are real numbers. Further f(0) = 6, f '(ln 3) = 282 and ln 2

∫0

f ( x )dx = 11

71. What is the value of Q ? (a) 1 (b) 2 (c) 3 (d) 4 72. What is the value of R ? (a) 1 (b) 2 (c) 3 (d) 4 73. What is f'(0) equal to ? (a) 18 (b) 16 (c) 15 (d) 14 Consider the following for the next two (02) items that follow : Suppose E is the differential equation representing family of curves y2 = 2cx + 2c c where c is a positive parameter. 74. What is the order of the differential equation ? (a) 1 (b) 2 (c) 3 (d) 4 75. What is the degree of the differential equation ? (a) 2 (b) 3 (c) 4 (d) Degree does not exist Consider the following for the next three (03) items that follow : cos x

x 2

Let f ( x ) = 2 sin x x tan x x

1 2x 1

76. What is f(0) equal to ? (a) − 1 (b) 0 (c) 1 (d) 2 77. What is lim

x →0

(a) − 1 (c) 1 78. What is lim

x →0

(a) − 1 (c) 1

f (x) equal to ? x (b) 0 (d) 2 f (x) x2

equal to ? (b) 0 (d) 2

21

SOLVED PAPER – 2023-I

Consider the following for the next two (02) items that follow : Let f(x) = sin[π2]x + cos[−π2]x where [.] is a greatest integer function π 79. What is f   equal to ? 2 (a) − 1 (b) 0 (c) 1 (d) 2

(a) − 1 2

(b) − 1

(c) 1

(d)

1 (a) − 2 (c) 1 2

1

(a)

2

x

π

x

π

2

1 + cos x

dx and I 2 = ∫0

1 + sin 2 x

dx

I1 + I 2 81. What is the value of I − I ? 1 2 (b) π π+1 (d) π−1

(a) 1 2

(c) π /2

82. What is the value of 8 I12 ? (b) π2 (d) π4

(a) π (c) π3

83. What is the value of I2 ? 2 (a) π (b) π

2 2

2

(c) 3π

(d) π

2 2

4 2

Consider the following for the next two (02) items that follow : b

x

a

x

Let l = ∫

dx , a < b

84. What is l equal to when a < 0 < b ? (a) a + b (b) a − b (c) b − a

(d)

(a + b) 2

85. What is l equal to when a < b < 0 ? (a) a + b (b) a − b (c) b − a

(d)

(b) −1 (d) 2

88. What is the derivative of f°f(x), where 1 < x < 2?

Consider the following for the next three (03) items that follow : 0

86. What is the derivative of f(x) at x = 0·5 ? (a) −2 (b) −1 (c) 1 (d) 2 87. What is the derivative of f(x) at x = 2 ?

π 80. What is f   equal to ? 4

Let I1 = ∫

Consider the following for the next three (03) items that follow : Let f(x) = |lnx|, x ≠ 1

(a + b) 2

1 lnx

(c) −

(b)

1 xlnx

1 1 (d) − lnx xlnx

Consider the following for the next two (02) items that follow :  x + 6, x ≤ 1 Let f ( x ) =  px + q , 1 < x < 2  5x , x ≥ 2  and f(x) is continuous 89. What is the value of p ? (a) 2 (b) 3 (c) 4 (d) 5 90. What is the value of q ? (a) 2 (b) 3 (c) 4 (d) 5 91. Consider the following statements : 1. f(x) = lnx is increasing in (0, ∞) 2. g(x) = ex+ex is decreasing in (0, ∞) Which of the statements given above is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 92. What is the derivative of sin2x with respect to cos2x ? (a) −1 (b) 1 (c) sin2x (d) cos2x 93. For what value of m with m < 0, is the area bounded by the lines y = x, y = mx and x = 2 equal to 3 ? (a) − 1 (b) −1 2 3 (c) − 2

(d) −2

22

Oswaal NDA/NA Year-wise Solved Papers

94. What is the derivative of cosec(x°) ? (a) −cosec(x°) cot(x°)

103. A biased coin with the probability of getting head equal to 1 is tossed five times. What is 4 the probability of getting tail in all the first four tosses followed by head ?

 cos ec( x) cot( x) (b)  180 (c)

 cos ec( x) cot( x) 180

(d) 

81 81 (a) (b) 1024 512

 cos ec( x ) cot( x ) 180

(a) y = x2/ 2 + c

(b) y = 2x + 4

27 81 (c) (d) 1024 256 104. A coin is biased so that heads comes up thrice as likely as tails. In four independent tosses of the coin, what is probability of getting exactly three heads ?

(c) y = x2 + 1

2 (d) y  ( x  x ) 2

27 81 (b) (a) 64 256

95. A

solution

of

the

differential

equation

2

dy  dy   dx   x dx  0 is  

96. If f(x) = x2 + 2 and g(x) = 2x − 3, then what is (fg)(1) equal to ? (a) 3 (b) 1 (c) −2 (d) −3 97. What is the range of the function f(x) = x + |x| if the domain is the set of real numbers ? (a) (0, ∞) (b) [0, ∞) (c) (−∞, ∞) (d) [1, ∞) 98. If f(x) = x(4x2 − 3), then what is f(sinθ) equal to ? (a) − sin3θ (b) − cos3θ (c) sin3θ (d) − sin4θ 99. What is lim

x 5

5x equal to ? x5

(a) − 1 (c) 1

(b) 0 (d) Limit does not exist

100. What is lim

x 1

(a) − 1 (c) 3

x9  1 x3  1

equal to ? (b) − 3 (d) Limit does not exist

101. The mean and variance of five obser-vations are 14 and 13.2 respectively. Three of the five observations are 11, 16 and 20. What are the other two observations ? (a) 8 and 15 (b) 9 and 14 (c) 10 and 13 (d) 11 and 12 102. Let A and B be two independent events such that P(A') = 0.7, P(B') = k, P(A ∪ B) = 0.8. What is the value of k ? 5 4 (a) (b) 7 7 2 1 (c) (d) 7 7

27 (d) 9 (c) 256 256 105. Let X and Y be two random variables such that X + Y = 100. If X follows Binomial distribution with parameters n = 100 and 4 p = , what is the variance of Y? 5 1 (a) 1 (b) 2 1 (c) 16 (d) 16 106. If two lines of regression are x + 4y + 1 = 0 and 4x + 9y + 7 = 0, then what is the value of x when y = −3? (a) −13 (b) −5 (c) 5 (d) 7 107. The central angles p, q, r and s (in degrees) of four sectors in a Pie Chart satisfy the relation 9p = 3q = 2r = 6s. What is the value of 4p – q ? (a) 12 (b) 24 (c) 30 (d) 36 108. The observations 4, 1, 4, 3, 6, 2, 1, 3, 4, 5, 1, 6 are outputs of 12 dices thrown simultaneously. If m and M are means of lowest 8 observations and highest 4 observations respectively, then what is (2m + M) equal to ? (a) 10 (b) 12 (c) 17 (d) 21 109. A bivariate data set contains only two points (–1, 1) and (3, 2). What will be the line of regression of y on x ? (a) x – 4y + 5 = 0 (b) 3x + 2y - 1 = 0 (c) x + 4y + 1 = 0 (d) 5x – 4y + l = 0

23

SOLVED PAPER – 2023-I

110. A die is thrown 10 times and obtained the following outputs: 1, 2, 1, 1, 2, 1, 4, 6, 5, 4 What will be the mode of data so obtained ? (a) 6 (b) 4 (c) 2 (d) 1 111. Consider the following frequency distribution: x

1

2

3

5

f

4

6

9

7

What is the value of median of the distribution? (a) 1 (b) 2 (c) 3 (d) 3-5 112. For data –1, 1, 4, 3, 8, 12, 17, 19, 9, 11; if M is the median of first 5 observations and N is the median of last five observations, then what is the value of 4M – N ? (a) 7 (b) 4 (c) 1 (d) 0 113. Let P, Q, R represent mean, median and mode. R then If for some distribution 5= P 4= Q , 2 P+Q what is equal to ? 2 P + 0.7 R (a) 1 12

(b) 1 7

(c) 2 9

1 (d) 4

114. If G is the geometric mean of numbers 1, 2, 22, 23,. . ., 2n-1, then what is the value of 1 + 2log2G? (a) 1 (b) 4 (c) n - 1 (d) n 115. If H is the harmonic mean of numbers 1, 2, 22, 23, . . . , 2n-1, then what is n/H equal to ? 1 1 (a) 2 − n −1 2  n 1 (b) 2 2 1 1 2− n (c) 2  n 1 (d) 2 2

116. Let P be the median, Q be the mean and R be the mode of observations x1, x2, x3, ... xn. Let S   i  1 ( 2 xi  a ) 2 n

S takes minimum value,

when a is equal to (a) P (c) 2Q

(b) Q 2 (d) R

117. One bag contains 3 white and 2 black balls, another bag contains 2 white and 3 black balls. Two balls are drawn from the first bag and put it into the second bag and then a ball is drawn from the second bag. What is the probability that it is white ? 6 (b) 33 (a) 7 70 1 3 (d) (c) 70 10 118. Three dice are thrown. What is the probability that each face shows only multiples of 3 ? 1 1 (b) (a) 18 9 1 1 (c) (d) 3 27 119. What is the probability that the month of December has 5 Sundays ?

(a) 1

(b)

1 4

2 3 (d) (c) 7 7 120. A natural number n is chosen from the first 50 natural numbers. What is the probability that n

50  50 ? n

47 23 (b) (a) 50 25 24 (d) 49 (c) 25 50

24

Oswaal NDA/NA Year-wise Solved Papers

ANSWER KEY Q No

Answer Key

1

a

Topic

Chapter

Cube root of unity

Complex Numbers

2

d

Number of ways

Permutations and Combinations

3

Bonus

Binary operation

Sets

4

b

Adjoint of a matrix

Matrices

5

d

Cube root of unity

Complex Numbers

6

b

Properties of matrices

Matrices

7

d

Properties of determinants

Determinants

8

b

System of equations

Determinants

9

b

Properties of determinants

Determinants

10

b

Expansion of determinant

Determinants

11

d

Roots of Equations

Complex Numbers

12

b

Argument

Complex Numbers

13

b

Expansion of determinant

Determinants

14

c

Geometric Progression

Sequence and Series

15

a

Geometric Progression

Sequence and Series

16

b

Arithmetic Progression

Sequence and Series

17

a

Nature of roots

Quadratic Equations

18

b

Suum of n terms

Sequence and Series

19

b

Factorial

Permutations and Combinations

20

c

Modulus

Complex Numbers

21

b

Roots of Equations

Equations

22

d

Onto Functions

Relations and Functions

23

a

Roots of Equations

Quadratic Equations

th

24

d

N term

Binomial Theorem

25

c

Binomial Expansion

Binomial Theorem

26

a

Arithmetic Progression

Sequence and Series

27

c

Combinations

Permutations and Combinations

28

d

Combinations

Permutations and Combinations

29

c

Number of permutations

Permutations and Combinations

30

b

Number of ways

Permutations and Combinations

31

a

Trigonometric Identities

Trigonometry

32

b

Trigonometric Identities

Trigonometry

33

a

Height and Distance

Trigonometry

34

c

Height and Distance

Trigonometry

35

c

Triangle

Trigonometry

36

c

Triangle

Trigonometry

37

c

Arithmetic and Geometric Progression

Trigonometry

38

d

Minimum Value

Trigonometry

39

c

Triangle property

Trigonometry

25

SOLVED PAPER – 2023-I

Topic

Chapter

Q No

Answer Key

40

c

Area of triangle

Trigonometry

41

b

Extreme values

Continuity and Differentiability

42

a

Extreme values

Continuity and Differentiability

43

c

Factorial

Permutations and Combinations

44

d

Factorial

Permutations and Combinations

45

c

Triangle

Trigonometry

46

a

Triangle

Trigonometry

47

a

Trignometric Relation

Trigonometry

48

d

Trignometric Relation

Trigonometry

49

b

Harmonic Mean

Sequence and Series

50

a

Geometric Mean

Sequence and Series

51

b

Expansion of determinant

Determinants

52

c

Properties of determinants

Determinants

53

d

Trigonometric expressions

Trigonometry

54

b

Trigonometric expressions

Trigonometry

55

c

Equation of a line

Straight lines

56

d

Equation of a line

Straight lines

57

b

Ellipse

Conic Section

58

d

Ellipse

Conic Section

59

a

Circle

Conic Section

60

b

Circle

Conic Section

61

a

Sphere

3D Geometry

62

b

Sphere

3D Geometry

63

a

Direction ratios

Three Diimensional Geomtery

64

d

Direction ratios

Three Diimensional Geomtery

65

b

Product of two vectors

Vector Algebra

66

a

Product of two vectors

Vector Algebra

67

b

Direction cosines

3D Geometry

68

a

Direction cosines

3D Geometry

69

b

Line

3D Geometry

70

c

Line

3D Geometry

71

b

Definite integral

Calculus

72

c

Definite integral

Calculus

73

d

Differentiation

Calculus

74

a

Order and degree

Differential equations

75

b

Order and degree

Differential equations

76

b

Evaluation of limits

Limits

77

b

Evaluation of limits

Limits

78

a

Evaluation of limits

Limits

79

b

Trigonometric functions

Trigonometry

80

d

Trigonometric Functions

Trigonometry

81

Bonus

Definite Integral

Calculus

26

Oswaal NDA/NA Year-wise Solved Papers

Topic

Chapter

Q No

Answer Key

82

d

Definite Integral

Calculus

83

b

Definite Integral

Calculus

84

a

Definite Integral

Calculus

85

c

Definite Integral

Calculus

86

a

Differentiation

Calculus

87

c

Differentiation

Calculus

88

d

Differentiation

Calculus

89

b

Continuity

Calculus

90

c

Continuity

Calculus

91

a

Increasing-decreasing functions

Calculus

92

a

Differentiation

Calculus

93

a

General Equation of a line

Straight Lines

94

b

Differentiation

Calculus

95

a

Variable separable

Differential Equations

96

d

Operations on functions

Functions

97

b

Range

Functions

98

a

Value of a function

Functions

99

d

Evaluation of limits

Limits

100

c

Evaluation of limits

Limits

101

c

Mean and variance

Statistics

102

c

Independent events

Probability

103

b

Independent events

Probability

104

b

Independent events

Probability

105

c

Binomial distribution

Probability

106

c

Regression

Statistics

107

d

Angles

Trigonometry

108

a

Mean

Statistics

109

a

Regression

Statistics

110

a

Mode

Statistics

111

c

Median

Statistics

112

d

Median

Statistics

113

d

Mean, median, mode

Statistics

114

d

Geometric mean

Sequence and Series

115

b

Harmonic mean

Sequence and Series

116

c

Derivative

Continuity and Differentiability

117

b

Total Probability

Probability

118

c

Probability

Probability

119

c

Probability

Probability

120

b

Probability

Probability

NDA / NA

MATHEMATICS

SOLVED PAPER

I

National Defence Academy / Naval Academy

2023

ANSWERS WITH EXPLANATION 1. Option (a) is correct. Explanations: We have,

4. Option (b) is correct. Explanations: |2 adj (3A)| = 23 |adj (3A)| (i) Now, |3A| = 33 |A|= 33.4 = 33.22 |adj (3A)|=|3A|3–1 = |3A|2 =|33.22|2 = 36.24 from (i), we have |2 adj (3A)| = 23.24.36 = 27.36 = 2α.3β ⇒    α = 7 and β = 6 ∴ α + β = 7 + 6 = 13

1− ω 1− ω = = −1 + ω 2 ω+ ω −1  −1 + 1 3  = −1 +   2   2

=

2  −3   3   =   +   2   2 

3

2. Option (d) is correct. Explanations: For number to be divisible by 6, the number should be divisible by 2 and 3 both. Now, number is divisible by 2 if units place digit is 0, 2, or 4: Also, sum of all digits = 0 + 1 + 2 +3 + 4 + 5 = 15 Case I: If units digit is 0; then no. of ways = 5! = 120 Case II: If units digit is either 2 or 4, then no. of ways = 2 × 4! × 4 = 192 So, total number of 6 digit number formed = 120 + 192 = 312 3. Option (Bonus) is correct. Explanations: To covert 1011 decimal number, we have, Divisible by 2

Quotient

Remainder

Binary Bit

1011 ÷ 2

505

1

1

505 ÷ 2

252

1

1

252 ÷ 2

126

0

0

126 ÷ 2

63

0

0

63 ÷ 2

31

1

1

31 ÷ 2

15

1

1

15 ÷ 2

7

1

1

7÷2

3

1

1

3÷2

1

1

1

1÷2

0

1

1

1011 = (1111110011)

5. Option (d) is correct. Explanations: We have, x2 – x + 1 x=

1 ± 3i ⇒ x = −ω or − ω 2 2

So, α = –w and b = –w2 α 1 ω + β1 ω ω1ω + ω 2 ω = α 1 ω − β1 ω ω1ω − ω 2 ω =

1 + ω 1ω 1+ ω = 1−ω 1 − ω 1ω

( (

( −ω ) + −ω 2 ±100 + β100 = ±100 − β100 ( −ω )100 − −ω 2 100

=

=

=

( ) + (1 − ω )

ω100 + 1 + ω100 ω

100

1+ ω = 1−ω

1 + 3i 3 + 3i

100

=

) )

100 100

1 + ω100 1 + ω 3 × 33 ω = 1 − ω100 1 − ω 3 × 33 ω

 −1 − 3i  1+   2    −1 + 3i  1+   2   =

1+ 3 9+3

=

1 3

6. Option (b) is correct. Explanations: When A and B be symmetric matrices then (AB – BA) is skew symmetric.

28

Oswaal NDA/NA Year-wise Solved Papers

7. Option (d) is correct. Explanations:

=

 3 5   K  7 3   K      =    7 3   2 K   3 5   2 K 

  Now, tan θ =    

13K  13K  13K  = 13K     



10. Option (b) is correct. 0 x−a x−b 0 0 x−c = 0 x+b x+c 1

⇒ 0 – (x – a)(0 – (x – c)(x + b)) + (x – b)(0 – 0) = 0 ⇒ (x – a) (x + b) (x – c) = 0 ⇒ x = a, x = – b or x = c Sum of roots = a – b + c 11. Option (d) is correct. 2

Explanations: 2 – i 3 is a root of x + ax + b. So, 2 + i 3 is also the root of x2 + ax + b. Sum of roots = 4 –a = 4 ⇒ a = –4 Product of roots = 4 + 3 = 7 ⇒ b = 7 So, a + b = –4 + 7 = 3

1+i 3 1−i 3

×

1+i 3 1+i 3

=

1−3+ 2 3i 1+ 3

(i) (a, b, c in AP)

x+1 1 2 x+1 1 2 = x+2 1 2 = 1 0 0 x+a b−a c−a x+a b−a c−a

9. Option (b) is correct. Explanations: We know that if X1 and X2 are solution of system of equations AX = B, B = 0 then aX1 + bX2 is also solution iff a + b = 1

z=

π 2π = 3 3

13. Option (b) is correct. Explanations: We have, 2b = a + c x+1 x+2 x+3 Let ∆ = x + 2 x + 3 x + 4 x+a x+b x+c

10 3

12. Option (b) is correct. Explanations: We have,

  =− 3   

)

(

So, the linear equations have infinity many solutions.

Explanations:

3 2 −1 2

⇒ θ = tan −1 − 3 = π −

 3 5  7 3  7 3  ≠  3 5      So, both statements are wrong. 8. Option (b) is correct. Explanations: We have, x + 2y + z = 4 2x + 4y + 2z = 8 ⇒ 2(x + 2y + z) = 8 ⇒ x + 2y + z = 4 and 3x + 6y + 3z = 10 ⇒ 3(x + 2y + z) = 10 ⇒ x + 2y + z =

3 −2 + 2 3i −1 i = + 4 2 2

(x + 1)(0 – 0)–1(c – a – 0) + 2(b – a – 0) = a – c + 2b – 2a = –a – c + a + c [Using (i)] =0 14. Option (c) is correct. Explanations: Since, logxa, ax, logbx are in G.P

( ) = (log a)(log x )

∴ ax

2

⇒ a2 x =

x

b

log a log x = log b a log x log b

Taking log both sides, we get 2x loga = log (logba) 1 loga(logba) 2 15. Option (a) is correct. x=

Explanations: 21/c, 2b/ac, 21/a are in G.P 22b/ac = 21/c.21/a = 22b/ac = 21/c+1/a 2b 1 1 = + = = 2b = a + c ac c a Hence, a, b, c are in A.P 16. Option (b) is correct. Explanations: We have, 5  −7  + ( n − 1)    12  2 30 37 ⇒ n −1 = ⇒n= 7 7 an = 0 =

So, largest negative term will be for integer n =6

29

SOLVED PAPER – 2023-I

17. Option (a) is correct. Explanations: We have, f(x) = x2 – 4x + x has real roots D > 0 = (4)2 – 4k, 1 > 0 = 16 – 4k > 0 k < 4 (i) Now, roots of above equation are lying in the internal (0, 5). f(0) > 0 = k > 0 (ii) and f(5) > 0 = 25 – 20 + k > 0 = k > –5 (iii) from (i), (ii), and (iii) we have, k = (0, 4) Possible integral values of l are 1, 2 and 3 i.e. 3 is number. 18. Option (b) is correct. Explanations: We have a = x, Sn = 0 n  2 a + ( n − 1) d  = 0 ⇒ 2 x + ( n − 1) d = 0 2  −2 x  ⇒d=  n − 1  m+n  2 x + ( m + n − 1) d  − 0 = 2  m+n = [ 2x + md − 2x ] 2  m + n   −2 x  = m  2   n − 1  ⇒

=

mx ( m + n )

=

If

x 2 + y 2 − 1 + 2iy 2

x + 1 + 2x + y

2

( i2 = –1)

z −1 is purely imaginary number, then z+1

 z − 1 =0 Re   z + 1  ⇒ x2 + y2 = 0 ⇒ x2 + y2 = 1 ⇒ |z|2 = 1 or |z| = 1 Thus the value of |z| = 1 21. Option (b) is correct. Explanations: We have, |x – 4|+|x – 7| = 15 There are two cases arise. Case I: When x < 4 –x + 4 – x + 7 = 15 ⇒ n = –2 Case II: When x > 7 So, only 2 Solution possible. 22. Option (d) is correct. Explanations: f(x) is onto 5 3 So, A = {x = R–(–5/3)} Let, y = 2x + 3/3x + 5 ⇒ 3xy + 5y = 2x + 3 5y  = x = 3 – –2 3y 3x + 5 = 0 ⇒ x = −

2 3 B = {y = R – (2 – 3)}

3y – 2 = 0 = y =

1−n

19. Option (b) is correct. Explanations: (1) as 5! = 120 and 5! + 1 = 121 has 1 at unit place. so, 25! + 1 also has 1 at units place. 25! +1 is not divisible by 26. (2) 6! = 720 6! + 1 = 721, which is divisible by 7. So, only (2) is true. 20. Option (c) is correct. Explanations: Let x = x + iy z − 1 x + iy − 1 = then z + 1 x + iy + 1 =

=

( x − 1) + iy ( x + 1) − iy × ( x + 1) + iy ( x + 1) − iy x 2 + x + ixy − x − 1 + iy + ixy + iy − i 2 y 2 ( x + 1)2 − i 2 y 2

23. Option (a) is correct. Explanations: We have, α + β = 0 (i) α2 + β2 = 2 (α + β)2 – 2ab = 2 2αβ = –2 [Using (i)] αβ = –1 Now, (α – β)2 = α2 + β2 – 2αβ = 2 – 2(–1) = 4 α – β = +2 (ii) Solving (i) and (ii), we get α = 1 and α + β = +1 So, only (1) is sufficient to find x. 24. Option (d) is correct. Explanations: We have, T5+1 = 5600 8C5(x–8/3)8–5 = (x2log10x)5 = 5600 56.x–8.x10 (log10x)5 = 5600

x2(log10x)5= (10)2.(log1010)5 So, x = 10

30 25. Option (c) is correct. Explanations: We have, (3x – y)4(x + 3y)4 = [(3x – y)(x + 3y)]4   = (3x2 + 9xy – xy – 3y2)4   = (3x2 + 8xy – 3y2)4 Here, r = 3 and n = 4 Required number of terms = n+r–1Cr–1    = 4+3–1C3–1    = 6C2 = 15 26. Option (a) is correct. Explanations: Let P = 1 – 3d, q = a – d, r = a + d Then, P+S=8 a – 3d + a + 3d = 8 ⇒ a = 4 Also, qr = 15   = a2 – d2 = 15 = d2 = 16 – 15 d=+1 If d = + 1 and a = 4, then Largest number = 7 and smallest number = 1 Required difference = 7 – 1 = 6 27. Option (c) is correct. Explanations: Both statements are true. 28. Option (d) is correct. Explanations: Selection of 2 parallel lines from m lines = mC2 Selection of 2 parallel lines from n lines = nC2 No. of parallelograms formed = mC2.nC2 = 60 = mC2.nC2 = 5C2 × 4C2 = mC2.nC2 ∴ m = 5 and n = 4 So, m + n = 5 + 4 = 9 29. Option (c) is correct. Explanations: No. of permutations of the word PERMUTATIONS = 12!/2! (T accurs twice) No. of permutations of the word COMBINATIONS =12!/2! 2! 2! (As 0, I, M occurs twice) y = 12!/2!. 1/4 = 4y = x 30. Option (b) is correct. Explanations: Total 5 digit numbers that can be formed using 0, 1, 2, 4 and 5 without repetition = 4 × 4! = 96 No. of 5 digit numbers greater that 50000 = 1 × 4! = 24 (Ten thousand should be filled by 5 only) Required percentage = 24/96 × 100 = 25%

Oswaal NDA/NA Year-wise Solved Papers

(31-32.) We have sin2β = sinα cosα(i) and 2 tanγ = sinα + cosα(ii) 31. Option (a) is correct. Explanations: Now, cos2β = 1 – 2sin2β = 1 – 2 sinα cosα = (sinα – cosα)2 32. Option (b) is correct. Explanations: cos 2 γ =

1 − tan 2 γ 1 + tan 2 γ

⇒ sec 2 γ =

1 + tan 2 γ = 1 − tan 2 γ

 sin α + cos α  1+    2

2

 sin α + cos α  1−   2

2

5 + 2 sin α cos α 3 − 2 sin α cos α 5 + sin 2α ⇒ sec 2 γ = 3 − sin 2α =

(33-34). From the given question, figure should be as follows: C 20 m

A 10 m α P

θ

B

Let AB be the pillar and a be the angle formed by flagstaff. 33. Option (a) is correct. Explanations: It is given that, AB 10 = AP x 30 tan (θ + α ) = x 30 tan θ + tan α ⇒ = 1 − tan θ tan α x 10 1 + 30 x 2 = ⇒ x  10   1  1−     x   2 tan θ =

31

SOLVED PAPER – 2023-I

20 + x 30 = 2 x − 10 x ⇒ 20 x + x 2 − 60 x + 300 = 0 ⇒

⇒ x − 40 x + 300 = 0 2

⇒ ( x − 30 ) ( x − 10 ) = 0 ⇒ x = 30 or x = 10 Ratio of two values of x = 1 : 3 And x ≠ 20 m So, only (1) is correct. 34. Option (c) is correct. Explanations: 10 10 or tanθ = Now, tanθ = 10 30   tanθ =

1 or 1 3

(35-36). Let A, B, C be the angle of ∆ABC Now, sin A sin B sin C = = =k a b c sin A = ak, sin B = bk and sin c = ck It is given that,  sin A + sin B + sin C  a+b+c = 6×   3 ⇒ 2k = 1 ⇒ k =

1 2

sin A 1 3 π = k ⇒ sin A = BC = ⇒A= a 2 2 3 1 1 sin B π = k ⇒ sin B = AC = ⇒ B = 2 2 6 b π π π   C = π− +  =  3 6 2 So,

35. Option (c) is correct. Explanations: Perimeter of triangle = 3 +1+2=3+ 3 36. Option (c) is correct. Explanations: C = π/3 = 1/2 (π/2 + π/6) C = 1/2 (A+B) A, C, B are in A.P Both (1) and (2) are true. 37. Option (c) is correct. Explanations: We have, 2

sin A + sin A + 1 sin A 1 = sin A + 1 + sin A x=

Now , sin A + ⇒ s in A + ⇒x≥3

1 ≥2 sin A

( AM > GM )

1 +1≥ 3 sin A

Minimum value of x = 3 38. Option (d) is correct. Explanations: Now, x = 3 sin2A + sin A + 1 = 3 sin A sin A – 2 sinA + 1 = 0 (sin A–1)2 = 0 sin A = 1 = A = π/2 39. Option (c) is correct. Explanations: We know that a2 = b2 + c2 – 2bc cos A b2 = c2 + a2 – 2ca cos B c2 = a2 + b2 – 2ab cos C Adding above equations, we get a2 + b2 + c2 = 2a2 + 2b2 + 2c2 = 2bc cosA = 2ca cosB – 2ab cosC a2 + b2 + c2 = 2ab cosC + 2bc cosA + 2ac cosB Now, it is given that, a2 + b2 + c2 = ac + 3 bc 2ab cosC + 2bc cosA + 2ac cosB = ac + On comparing, we get ABC is right angled triangle.

3 bc

40. Option (c) is correct. Explanations: Now, area of ABC = 1/2 × AC × BC

= 1/2 × 4 3 × 4

 3 AC  1 BC = BC = 4 and = = AC = 4 3   =  2 8 2 8 =8 3 (41-42). We have, f(x) = |x – 2|+|3 – x|+|4 – x| f (x ) = x − 2 + 3 − x + 4 − x

 − x + 2 + 3 − x + 4 − x , x ∈ ( −∞ , 2 )  x + 2 + 3 − x + 4 − x , x ∈[22 , 3) ⇒ f (x ) =   x + 2 + 3 − x + 4 − x , x ∈[3, 4 )  x + 2 + 3 − x + 4 − x,x ≥ 4  9 − 3x , x < 2  5 − x , x ∈[ 2 , 3) ⇒ f (x ) =   x − 1, x ∈[3, 4 )  3x − g , x ≥ 4

32

Oswaal NDA/NA Year-wise Solved Papers

47. Option (a) is correct. Explanations: We have, sin x cos x 1 1 + + + =7 sinx + cosx + cos x sin x cos x sin x

 −3 , x < 2  −1 , x ∈( 2 , 3 ) ⇒ f 1 (x ) =   1 , x ∈( 3, 4 )  3,x ≥ 4 41. Option (b) is correct. Explanations: Since sign changes from negative to positive a x = 3 f(x) is minimum at x = 3 42. Option (a) is correct. Explanations: Minimum value of f(x)= f(3) = |3 – 2|+|3 – 3|+|4 – 3| =1+0+1=2 43. Option (c) is correct. Explanations: Given, s = 0! + 1! + 2! +...+ 100! From 41 onwards every terms has 4 × 2, which is divisible by 8. Remaining sum = 0! + 1! + 2! + 3! = 1 + 1 + 2 + 6 = 10 Now, remainder when 10 is divisible by 8 is 2 so, required remainder = 2 44. Option (d) is correct. Explanations: Similarly from 5! onwards every terms has 10, which is divisible by 60 Remainder = 0! + 1! + 2! + 3! + 4! = 1 + 1 + 2 + 6 + 24 = 34 45. Option (c) is correct. Explanations: We have PN = PM (Tangents from an external point) PN = PM = n Similarly, QL = QN = n + 2 and, RM = RL = n + 4 So, sides of triangle are, PQ = 2n + 2, QR = 2n + 6, PR = 2n + 4 Now, cos P = 1/3

( PQ)

2

⇒

⇒

⇒ ⇒

+ ( PR ) − (QR ) 2

2

2 PQ ⋅ PR

=

1 3

( 2n + 2 )2 + ( 2n + 4 )2 − ( 2n + 6 )2 2 2 ⋅ ( 2n + 2 ) ( 2n + 4 )

(

4 ( n + 1) + ( n + 2 ) − ( n + 3 ) 2

2

4 ( n + 1) ( 2 n + 4 )

2

=

3

2n + 6n + 4 n = 8, or n = –2 46. Option (a) is correct. Explanations: Length of smallest side = 2n + 2 = 18

Squaring both sides, we get, 2

2   ⇒ (1 + sin 2 x )  1 +  = 7 −  sin x

∴α + β =

4+2 3 3+2 2

and αβ =

2αβ HM of α and β = α+β = =

(

2⋅ 8 + 4 3 4+2 3

)× 4−2

4−2 3

(

8+4 3 3+2 2

3

2 32 − 16 3 + 16 3 − 24

) = 16 = 4

16 − 12 4 50. Option (a) is correct. Explanations: GM of α and β = αβ =

8+4 3 3+2 2

= 2 =

1 3

2 sin 2 x

⇒ sin22x – 44 sin2x + 36 = 0 48. Option (d) is correct. Explanations: sin22x – 44 sin2x + 36 = 0 a = 22, b = 8 and c = 7 So, a – b + 2c = 22 – 8 + 14 = 28 49. Option (b) is correct. Explanations: Let a and b are the roots of the given equation

( (

=

(

2 4+2 3

)

3+2 2

) 2 2 + 1) 3 +1

2

 3 +1 2 − 1 = 2 ×  2 −1  2+1

)=1

2

1 sin x + cos x + =7 sin x.cos x sin x.cos x

2  2  ⇒ (sin x + cos x )  1 +  =7−  sin 2 x sin x 

  = 2  

1 3

n2 + 1 + 2n + n2 + 4 + 4 n − n2 − n − 6n

sinx + cosx +

(

)

6 − 3 + 2 −1

51. Option (b) is correct. Explanations: ∴

a b aα + b b c bα + c = 0 aα + b bα + c 0

33

SOLVED PAPER – 2023-I

0 0

⇒

b c

aα 2 + 2bα + c bα + c

aα + b bα + c = 0 0

= 0 – 0 + (aa2 + 2ba + c)(b2a + bc – aca – bc) =0 = b2a + aca = 0 or b2 – ac = 0 = b2 = ac So, a, b, c are in G.P 52. Option (c) is correct. Explanations: (7, 4, 2, a) = 0 7a2 + 8a + 2 = 0 So, a is root of the equation, 7x2 + 8x + 2 = 0 53. Option (d) is correct. Explanations: We have, m(0) = cotθ2 + n2tan2θ + 2n = (cotθ + n tanθ)2 m(θ) > 0 cot θ + n tan θ ≥ n 2 = (cotθ + n tanθ)2 > 4n ∴ Minimum value of m(θ) = 4n 54. Option (b) is correct. Explanations: (cotθ + n tanθ)2 – 4x = 0 ⇒ (cotθ – n tanθ)2 – 4x = 0 ⇒ cotθ = n tanθ ⇒ x = cot2θ 55. Option (c) is correct. Explanations: Equation of line of the quadrilateral is, x = 0, y = 0, x + y = 1 and 6x + y = 3 Point of intersection of these lines are Now,

1   2 3 A  , 0 , B ( 0 , 0 ) , C ( 0 , 1) , D  ,  2   5 5 So, equation of diagonal passes through B is ,

35 (x − 0) 25 2y = 3x ⇒ 3x – 2y = 0 56. Option (d) is correct. Explanations: Equation of diagonal AC is BD = y – 0 =

1−0  1 y−0 =  x −  0 −1 2 2 1  ⇒ y = −2  x −  ⇒ y = −2 x + 1 ⇒ 2 x + y − 1 = 0  2 57. Option (b) is correct. Explanations: The given ellipse is, 2

2

y x + =1 1  1 2   2

As we know, sum of distances of any point P from two is, PE + PE = 2a = 2 58. Option (d) is correct. Explanations: Equation of latus return of ellipse is x = 2 3 /2 So, points 1, 2 and 3 will be on it. 59. Option (a) is correct. Explanations: Given equation of circle is (x – a)2 + y2 = a2 Now, y = x intersect it 2 parts Point of intersection of line and circle is, (0, 0) and (a, a) a

So, required area =

∫

0

a

a 2 − ( x − a ) dx − ∫ xdx 2

0

a

2 x−a a2  x − a  x   2 ax − x 2 + sin −1  = −     a   2  2  2 0

=

a2 a2 a2 sin −1 ( 0 ) − sin −1 ( −1) − 2 2 2

=

a2 a2 π a2 ×0+ × − 2 2 2 2

=

a2 (π − 2) 4

60. Option (b) is correct. Explanations: Area of major segment = πr2 – Area of minor segment = πa 2 −

πa 2 a 2 − 4 2

3πa 2 a 2 a 2 = + ( 3π + 2 ) 4 2 4 61. Option (a) is correct. Explanations: End points of diameter are A(1, –1, 2) and B(2, 1, –1) =

3 Centre of sphere =  , 0 , 2 and radius =

1  2

2

3   2 1  − 1 + ( 0 + 1) +  − 2 2 2

14 1 9 +1+ = = 4 4 4 Equation of space is

=

⇒ x2 + y2 + z2 +

7 2

9 1 7 + − 3x − z = 4 4 2

⇒ x2 + y2 + z2 + 9/4 + 1/4 – 3x – z = 7/2 ⇒ x2 + y2 + z2 – 3x – z – 1 = 0

2

34

Oswaal NDA/NA Year-wise Solved Papers

So, from given equation of sphere we have 2a = –3, 2v = 0 and 2w = –1 ⇒ a = –3/2, v = 0, w = –1/2 So, u + v + w = 4/2 = –2 62. Option (b) is correct. Explanations: PA2 + PB2 = AB2 = (2 – 1)2 + (1 + 1)2 + (–1 – 2)2 = 1 + 4 + 9 = 14 63. Option (a) is correct. Explanations: dr1 (2, –1, 2) and (k, 3, 5) indicated at π/4 π 2 k − 3 + 10 ∴ cos = 4 4 + 1 + 4 k 2 + 9 + 25

⇒

1 2

=

2k + 7

3 k 2 + 34

⇒ 9 (k2 + 34) = 2(2k + 7)2 ⇒ 9k2 + 306 = 2(4k2 + 49 + 28k) ⇒ 9k2 + 306 – 8k2 – 98 – 56k = 0 ⇒ k2 – 56k + 209 = 0 ⇒ k2 – 52k – 4k + 208 = 0 ⇒ (k – 52)(k – 4) = 0 ⇒ k = 52 or k =4 64. Option (d) is correct. Explanations: Let the drs of line perpendicular to given lines be (a, b, c) Then, 2a – b + 2c = 0 and 4a + 3b + 5c = 0 a b c = = −5 − 6 8 − 10 6 − 4 a b c = = ⇒ −11 −2 10 So, (11, 2, –10) as the required drs. 65. Option (b) is correct.  Explanations: Let b = ai = bj = ck   Then, a . b = 27 ⇒

⇒ 3a + 3b + 3c = 27 ⇒ a + b + c = 9 Also,

66. Option (a) is correct. Explanations: Now, a + b = 8i + 5j + 5k    c = j – k (a+b).c = |a + b||c| cosq, where θ is the required angle

⇒ 0 = 8 2 + 5 2 + 5 2 1 + 1 cos θ π ⇒ cos θ = 0 ⇒ θ = 2 67. Option (b) is correct. Explanations: We have, a = 4i – 8j + k ∴ cosα =

4

=

4 2 + 8 2 + 12

4 9

68. Option (a) is correct. Explanations: Also, cosβ = 8/9 and cosγ = 1/9 Now, cos2β + cos2γ = 2cos2b – 1 + 2cos2γ – 1 −32  64 1  = 2 +  − 2 =  81 81  81 69. Option (b) is correct. Explanations: We have, A = (1, –1, 0) and B = (0, 1, 1) Equation of line AB is, x −1 y +1 z −0 = = 0 −1 1+1 1−0 1−x y+1 z = = 1 2 1 Now, only (2) and (3) satisfy this equation. 70. Option (c) is correct. Explanations: We have, A = ˆi + ˆj + 0.kˆ and B = oi + j + k AB = (0 – 1)i + (1+1)i + (1 – 0) = i + 2j + k AB = 12 + 2 2 + 1 = 6

(i)

i j k ⇒ 3 3 3 = 9 ( j − k) a b c ⇒ i(3c – 3b) – j(3c – 3a) + k(3b – 3a) = 9(j – k) ⇒ 3c – 3b = 0, 3c – 3A = 9, 3b – 3a = –9 ⇒ c = b, a – c = 3, a – b = 3 ⇒ c = a – 3, b = a – 3 From (i), we have a+a –3+a–3=9 ⇒ 3a = 15 ⇒ a = 5 b=5–3=2=c So, b = 5i + 2j + 2k

(71-73). We have, f(x) = Pex + Qe2x + Re3x It is given that, f(0) = 6 P + Q + R = 6 ln 2

∫

(i)

f ( x ) dx = 11

0 ln 2

 Qe 2 x Re 3 x  ⇒  Pe x + +  2 3   0

= 11

71. Option (b) is correct. 72. Option (c) is correct. 73. Option (d) is correct. f(0) = P + 2Q +3R = 1 + 4 + 9 = 14

35

SOLVED PAPER – 2023-I

(74-76). We have, y2 = 2c x + 2c = y4 + 4y2(y)2x2 – 4y3(y)x – 4y3.(y)3 = 0 74. Option (a) is correct. 76. Option (b) is correct. Explanations: We have,

87. Option (c) is correct. Explanations: F’(2) = 1/2 88. Option (d) is correct. Explanations:

x 1 cos x f (x) lim = lim 2 sin x x x 2 = 0 x→0 x→0 x x 1 tan x

x2

1 1 1 = 2 1 2 = −1 0 1 1

90. Option (c) is correct. Explanations: Also, p = 3, q = 4 91. Option (a) is correct. Explanations: Only (1) is true

79. Option (b) is correct. Explanations: We have, f(x) = sin[p2]x + cos[–p2]x = sin9x + cos(–10x) = sin9x + cos10x = 1 + (–1) = 0

92. Option (a) is correct. Explanations: du = 2 sin x cos x = sin 2 x dx dv and = 2 cos x( − sin x ) = − sin 2 x dx sin 2 x du du dv = = −1 ∴ = dv dx dx − sin 2 x ⇒

80. Option (d) is correct. Explanations: 9π 5π  π f   = sin + cos  4 4 2 1 π π  = sin  2 π +  + 0 = sin =  4 4 2 81. Option (Bonus) is correct.

So, 81 and 83 is bonus.

d  −1 1 −1 ⋅ =  fof ( x ) = dx ln x x x ln x

89. Option (b) is correct. Explanations: f(x) is continuous 7 = P + q(i) 10 = 2P + q(ii) Solving (i) and (ii) we get P=3

cos x 1 1 = lim 2 sin x x 1 2 x →0 tan x 1 1

Explanations: Since I1 = I 2 =

, x < 0.1  ln( −ln x )  −ln( −ln x ) , x ∈( 0.1, 1)   −ln( −ln x ) , x ∈(1, 2 ) ∴

78. Option (a) is correct. Explanations: lim

83. Option (b) is correct.

86. Option (a) is correct. Explanations: f’(0.5) = –1/0.5 = –2

77. Option (b) is correct. Explanations:

x →0

Explanations: 8I =

 π2  8 π4 8 = π4  = 8 2 2 

85. Option (c) is correct. Explanations: Now, a < b < 0 l = – (–b) + (–a) = b – a

1 2x 1

cos 0 0 1 ∴ f ( 0 ) = 2 sin 0 0 0 = 0 tan 0 0 1

f (x)

2

2

84. Option (a) is correct. Explanations: Now, a < 0 < b l = b – (–a) = a + b

75. Option (b) is correct.

cos x x f ( x ) = 2 sin x x 2 tan x x

82. Option (d) is correct.

π2 2 2

93. Option (a) is correct Let the equation of line segments of ABC are given A = (2, 2), B(2, 2 m) and C(0, 0) Since area of ABC = 3 = |1/2(0 + 2(2m – 0) + 2(0 – 2)|= 3 = 4m – 4 = ±6 m = 5/2 or m = –1/2 Qm 0, q > 0; then what is the value of p–2 + q–2? (a) 1 (b) 2 1 1 (c) (d) 2 2 2   3  28. What is 1 + sin 2 cos−1    equal to?  17   

8 25 (a) (b) 17 17 (c)

9 47 (d) 17 17

) cot ( π sin θ ) , 0 < θ < π ; then what 29. If tan ( π cos θ= 2 π 2 is the value of 8 sin  θ +  ? 4  (a) 16 (c) 1

(b) 2 1 (d) 2

1 π 1 30. If tan α = , sin β = ; 0 < α , β < , then what 7 2 10 is the value of cos( α + 2β) ? (a) − (c)

1 1 − (b) 2 2

1 2

1 (d) 2

Consider the following for the next two (02) items that follow: Consider the equation (1 – x)4 + (5 – x)4 = 82. 31. What is the number of real roots of the equation? (a) 0 (b) 2 (c) 4 (d) 8 32. What is the sum of all the roots of the equation? (a) 24 (b) 12 (c) 10 (d) 6 Consider the following for the next three (03) items that follow: Consider equation –I : z3 + 2z2 + 2z + 1 = 0 and equation – II : z1985 + z100 + 1 = 0. 33. What are the roots of equation–I? (a) 1, ω, ω2 (b) –1, ω, ω2 2 (c) 1, –ω, ω (d) –1, –ω, –ω2 34. Which one of the following is a root of equation-II? (a) –1 (b) –ω (c) –ω2 (d) ω 35. What is the number of common roots of equation-I and equation-II? (a) 0 (b) 1 (c) 2 (d) 3 Consider the following for the next two (02) items that follow: A quadratic equation is given by (a + b) x2 – (a + b + c) x + k = 0, where a, b, c are real. 36. If k =

c ,( c ≠ 0) , then the roots of the equation are: 2

(a) Real and equal (b) Real and unequal (c) Real iff a > c (d) Complex but not real 37. If k = c, then the roots of the equation are: b a+c (a) and a+b a+b

a+c b and − (b) a+b a+b (c) 1 and

c a+b

(d) –1 and −

c a+b

Consider the following for the next three (03) items that follow: Let (1 + x)n = 1 + T1x + T2x2 + T3x3 + … + Tnxn. 38. What is T1 + 2T2 + 3T3 + … + nTn equal to? (a) 0 (b) 1 (c) 2n (d) n2n–1 39. What is 1 – T1 + 2T2 – 3T3 + …+ (–1)nnTn equal to? (a) 0 (b) –2n–1 (c) n2n–1 (d) 1 40. What is T1 + T2 + T3 + ... + Tn equal to? (a) 2n (b) 2n – 1 (c) 2n – 1 (d) 2n + 1

42 Consider the following for the next two (02) items that follow: Let f(x) = x2 – 1 and gof(x) = x − x + 1 . 41. Which one of the following is a possible expression for g(x)? (a) x + 1 − 4 x + 1 (b) x +1 − 4 x +1 +1 (c) x + 1 + 4 x + 1 (d) x +1− x +1 +1 42. What is g(15) equal to? (a) 1 (b) 2 (c) 3 (d) 4 Consider the following for the next two (02) items that follow: 1 Let a function f be defined on R – [0] and 2 f ( x ) + f   x = x + 3. 43. What is f(0.5) equal to? 2 1 (a) (b) 3 2 (c) 1 (d) 2 44. If f is differentiable, then what is f ’(0.5) equal to?

1 2 (a) (b) 4 3 (c) 2 (d) 4 Consider the following for the next (02) items that follow: A function is defined by x +1 2 3 x+4 6 f(x) = 2 3 6 x+9 45. The function is decreasing on:

 28   28  (a)  − , 0  (b) 0, 3     3   50   56  (c) 0,  (d) 0, 3   3   46. The function attains local minimum value at: 28 (a) x = − (b) x = –1 3 28 (c) x = 0 (d) x = 3 Consider the following for the next (02) items that follow: Given that 4x2 + y2 = 9. 47. What is the maximum value of y? 3 (a) (b) 3 2 (c) 4 (d) 6 48. What is the maximum value of xy? 9 3 (a) (b) 4 2

4 2 (c) (d) 9 3

Oswaal NDA/NA Year-wise Solved Papers Consider the following for the next (02) items that follow: A function is defined by f(x) = π + sin 2 x . 49. What is the range of the function? (a) [0, 1] (b) [ π, π + 1] (c) [ π − 1, π + 1] (d) [ π − 1, π − 1] 50. What is the period of the function? (a) 2p (b) p π (c) (d)  The function is non2 periodic Consider the following for the next (02) items that follow: A parabola passes through (1, 2) and satisfies the dy 2 y differential equation = , x > 0, y > 0 . dx x 51. What is the directrix of the parabola? 1 1 y= (a) y = − (b) 8 8 (c) x = −

1 1 x= (d) 8 8

52. What is the length of latus rectum of the parabola? 1 (a) 1 (b) 2

1 1 (c) (d) 4 8 Consider the following for the next (02) items that follow: a x −1 + b x −1 Let f(x) = and g(x) = x – 1. 2 53. What is lim x →1 (a)

f (x) − 1 equal to? g( x )

ln( ab ) ln( ab ) (b) 4 2

(c) ln (ab) 54. What is lim f ( x ) x →1

(d) 2ln (ab) 1 g(x)

equal to?

(a) ab

(b) ab

(c) 2ab

(d)

ab 2

Consider the following for the next (02) items that follow: Let f(x) = 2 − x + 2 + x . 55. What is the domain of the function? (a) (–2, 2) (b) [–2, 2] (c) R –(–2, 2) (d) R – [–2, 2] 56. What is the greatest value of the function?

6 (a) 3 (b) (c) 8 (d) 4 Consider the following for the next (02) items that follow: Let f(x) = |x| and g(x) = [x] – 1, where [.] is the greatest integer function.

43

SOLVED PAPER – 2023 (II) Let h(x) =

f ( g ( x )) . g ( f ( x ))

57. What is lim+ h( x ) equal to? x →0 (a) –2 (b) –1 (c) 0 (d) 1 58. What is lim h( x ) equal to? x →0 − (a) –2 (b) –1 (c) 0 (d) 2 Consider the following for the next (02) items that follow:  x−3  x − 3 + a; x < 3  Let f(x) =  a − b ; 3 and x=  x−3  + b; x > 3  x − 3

f(x) be continuous at x = 3. 59. What is the value of a? (a) –1 (b) 1 (c) 2 (d) 3 60. What is the value of b? (a) –1 (b) 1 (c) 2 (d) 3 Consider the following for the next (02) items that follow: 2π

Let I =

sin 4 x + cos4 x dx 1 + 3x −2 π

∫

4 4 61. What is ∫ (sin x + cos x )dx equal to? 0

3π 3π (b) 4 8

3π 2 62. What is I equal to? (c)

(d) 3p

3π (a) 0 (b) 4 3π (c) (d) 3p 2 Consider the following for the next (02) items that follow: x β) be the roots of the equation x2 - 8x + q = 0. If a2 - β2 = 16, then what is the value of q? (a) -15 (b) -10 (c) 10 (d) 15 21. What is the maximum value of n such that 5n divides (30! + 35!), where n is a natural number? (a) 4 (b) 6 (c) 7 (d) 8 22. What is the value of 2(2 × 1) + 3 (3 × 2 × 1) + 4 (4 × 3 × 2 × 1)+ 5(5 × 4 × 3 × 2 × 1) + ...... ...... ...... + 9(9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) + 2? (a) 11! (b) 10! (c) 10 + 10! (d) 11 + 10! 23. If A = {{1, 2, 3}}, then how many elements are there in the power set of A? (a) 1 (b) 2 (c) 4 (d) 8 24. If a, b, c are in GP where a > 0, b > 0, c > 0, then which of the following are correct? (1) a2, b2, c2 are in GP 1 1 1 are in GP (2) , , a b c (3) a , b , c are in GP Select the correct answer using the code given below: (a) 1 and 2 only (b) 2 and 3 only (c) 1 and 3 only (d) 1, 2 and 3 a+b b+c are in HP, , b, 2 2 then which one of the following is correct? (a) a, b, c are in AP (b) a, b, c are in GP (c) a + b, b + c, c + a are in GP (d) a+ b, b + c, c + a are in AP 25. If

26. What is value of cot2 15° + tan2 15°? (a) 12 (b) 14 (c) 8 3 (d) 4 27. In a triangle ABC, sin A - cos B - cos C = 0. What is angle B equal to? π π (a) (b) 6 4

π (c) 3 28. If α + β =

(d)

π 2

π and 2tan a = 1, then what is tan 2b 4

equal to? 1 2 (a) (b) 3 3 3 3 (c) (d) 4 5 29. If tan(45° + θ) = 1 + sin 2θ, where π π − < θ < , then what is the value of cos 2θ? 4 4 1 (a) 0 (b) 2 (c) 1 (d) 2

30. Let sin 2θ = cos 3θ, where θ is acute angle. What is the value of 1 + 4sin θ? 5 −1 ) (given that sin 18° = 4 (a) 3

(b) 2

(c) 5

(d) 3 5 , then what can be the value of 31. If tan q = 12 sin q ? 5 5 (a) but cannot be 13 13 5 5 (b) but cannot be 13 13 5 5 (c) or 13 13 (b) None of the above 32. What is the value of 7π 5π + cos4 ? 8 8 3 (a) 2 3 (c) 8

cos4

3 4 3 (d) 16 (b)

  2π 2π 33. What is sin  + θ  − sin  − θ  equal to? 4 4     (a) sin 2q (b) cos 2q (c) 2sin q (d) 2cos q 34. A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. At a point on the plane the angles of elevation of the bottom and top of the flagstaff are q and 2q respectively. What is the height of the tower?

31

solved PAPER - 2022 (I)

(a) hcos q hcos 2q (c)

(b) hsin q (d) hsin 2q

35. The shadow of a tower becomes x metre longer, when the angle of elevation of sun changes from 60° to q. If the height of the tower is 3 x metre, then which one of the following is correct ? (a) 0 < q < 30° (b) 30° < q < 45° (c) 45° < q < 60° (d) 60° < q < 90° π −1  1  −1  x  36. If tan   + tan   = , where 0 < x < 6, 2 3   4   then what is x equal to? (a) 1 (b) 2 (c) 3 (d) 4 37. If 3sin–1 x + cos–1 x = π, then what is x equal to? (a) 0

(b)

1 2

1 1 (c) (d) 2 3 38. If tana + tanb = 1 - tana tanb, where tana tanb ≠ 1, then which of the following is one of the values of (a +b) ? π π (a) (b) 4 6 π π (c) (b) 3 2 39. If (1 + tan q)(l + tan 9q) = 2, then what is the value of tan(10q)? (a) 0 (b) 1 (c) 2 (d) Infinite 40. What is the value of sin 0° + sin 10° + sin 20° + sin 30° + ... + sin 360° ? (a) -1 (b) 0 (c) 1 (d) 2 41. Consider all the subsets of the set A = {1, 2, 3, 4}. How many of them are supersets of the set {4}? (a) 6 (b) 7 (c) 8 (d) 9 42. Consider the following statements in respect of two non-empty sets A and B: (1) x ∉ (A∪B) ⇒ x ∉ A or x ∉ B (2) x ∉ (A∩B) ⇒ x ∉ A and x ∉ B Which of the above statements is/are correct? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 43. Consider the following statements in respect of two non-empty sets A and B:

(1) A∪B = A∪B if A = B (2) ADB = ϕ if A = B Which of the above statements is/are correct? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 44. Consider the following statements in respect of the relation R in the set IN of natural numbers defined by xRy if x2- 5xy +4y2 = 0: (1) R is reflexive (2) R is symmetric (3) R is transitive Which of the above statements is/are correct? (a) 1 only (b) 2 only (c) 1 and 2 only (d) 1, 2 and 3 45. Consider the following statements in respect of any relation R on a set A: (1) R is reflexive, then R–1 is also reflexive (2) If R is symmetric, then R–1 is also symmetric (3) If R is transitive, then R–1 is also transitive Which of the above statements are correct? (a) 1 and 2 only (b) 2 and 3 only (c) 1 and 3 only (d) 1, 2 and 3 1 46. What is the principal argument of where 1+i i = −1 ? 3π − (a) 4 π (c) 4

π 4 3π (d) 4 (b) −

 −3 1  −  47. What is the modulus of  2   2 1 (a) 4 (c) 1

200

?

1 2 (d) 2200 (b)

48. Consider the following statements: n! (1) is divisible by 6, where n > 3 3! n! (2) + 3 is divisible by 7, where n > 3 3! Which of the above statements is/are correct? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 49. In how many ways can a team of 5 players be selected out of 9 players so as to exclude two particular players? (a) 14 (b) 21 (c) 35 (d) 42

32 Oswaal NDA/NA Year-wise Solved Papers present. If half of the substance decays in 100 years, then what is the decay constant (proportionality constant)?

2n

1 50. In the expansion of  x +  , what is the x  (n + 1)th term from the end (when arranged in descending powers of x)? (a) C(2n, n)x (b) C(2n, n - 1)x (c) C(2n, n) (d) C(2n, n- 1) 51. If the sum of the first 9 terms of an AP is equal to sum of first 11 terms, then what is the sum of the first 20 terms? (a) 20 (b) 10 (c) 2 (d) 0 52. If the, 5th term of an AP is

1 and its 10th term 10

1 then what is the sum of first 50 terms? 5 (a) 25 (b) 25·5 (c) 26 (d) 26·5 is

53. What is (1110011)2 ÷ (10111)2 equal to? (a) (101)2 (b) (1001)2 (c) (111)2 (d) (1011)2 3

3

54. If x + y = (100010111)2 and x + y = (11111)2, then what is (x - y)2 + xy equal to? (a) (1101)2 (b) (1001)2 (c) (1011)2 (d) (1111)2 55. Consider the inequations 5x - 4y + 12 < 0, x + y < 2, x < 0 and y > 0. Which one of the following points lies in the common region? (a) (0, 0) (b) (- 2, 4) (c) (- 1, 4) (d) (- 1, 2) 56. Consider the following statements in respect of the function y = [x], x∈(- 1, 1) where [.] is the greatest integer function: (1) Its derivative is 0 at x = 0·5 (2) It is continuous at x = 0 Which of the above statements is/are correct? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 57. What is the degree of the differential equation 4

2  d2 y 3  dy  1+   = 2  ?  dx   dx   

4 (a) 3 (c) 3

ln 2 ln 5 (a) (b) 100 100 ln 10 2 ln 2 (c) (d) 100 100 59. What is the domain of the function

f ( x ) = 1 − ( x − 1) ?

(a) (0, 1) (c) (0, 2)

(d) 4

58. A radioactive substance decays at a rate proportional to the amount of substance

(b) [- 1, 1] (d) [0, 2]

60. The area of the region bounded by the parabola y2 = 4kx, where k > 0 and its latus rectum is 24 square units. What is the value of k? (a) 1 (b) 2 (c) 3 (d) 4 π

61. What is

dx

∫04 ( sin x + cos x )2

1 - (a) 2

62. What is

equal to?

1 2 3 (d) 2 (b)

(c) 1

∫ (sin x )

−1 / 2

(cos x )−3 / 2 dx

equal to?

(a) tan x + c

(b) 2 tan x + c

cot x + c (c)

(d) 2 tan x + c

63. If I1 = ∫

x

e dx −x

x

and I 2 = ∫

e +e I1 + I2 equal to? x (a) + c 2 (c) ln(ex + e–x) + c 64. What is

−1

x

∫−2 | x |dx

dx e

2x

+1

, then what is

(b) x + c (d) ln(ex - e–x) + c equal to?

(a) - 2

(b) - 1

(c) 1

(d) 2

65. How many extreme values does sin4x + 2x, π where 0 < x < have? 2 (a) 1 (b) 2 (c) 4

(b) 2

2

(d) 8

66. What is the maximum value of the function 1 π f (x) = , where 0 < x < ? tan x + cot x 2

33

solved PAPER - 2022 (I)

1 (a) 4 (c) 1

1 2 (d) 2 (b)

1  1 1  67. If 4 f ( x ) − f   =  2 x +   2 x −  , then what x x x      is f(2) equal to? (a) 0 (b) 1 (c) 2 (d) 4 68. If f(x) = 4x + 3, then what is fofof (- 1) equal to? (a) -1 (b) 0 (c) 1 (d) 2 dy at (1, 1) equal to? dx (a) -1 (b) 0 (c) 1 (d) 4 dy at x = 1? 70. If y = (xx)x, then what is the value of dx 1 (a) (b) 1 2 (c) 2 (d) 4 69. If xyyx = 1, then what is

71. Let y = [x + 1], - 4 < x < - 3 where [.] is the greatest integer function. What is the derivative of y with respect to x at x = - 3·5? (a) - 4 (b) - 3·5 (c) - 3 (d) 0 dy = (ln5)y with y(0) = ln5, then what is y(1) dx equal to? (a) 0 (b) 5 (c) 2ln5 (d) 5ln5

72. If

76. In which one of the following intervals is the x3 7x2 − + 6 x + 5 decreasing? 3 2 (a) (- ∞, 1) only (b) (1, 6) (c) (6, ∞) only (d) (- ∞, 1) ∪ (6, ∞) function f ( x ) =

m + 2nx + 1 x vanishes at x = 2, then what is the value of m + 8n? (a) -2 (b) 0 (c) 2 (d) Cannot be determined due to insufficient data 77. If the derivative of the function f(x) =

78. What is the area included in the first quadrant between the curves y = x and y = x3? 1 1 (a) square unit (b) square unit 8 4 1 (c) square unit (d) 1 square unit 2 79. If xy = 4225 where x, y are natural numbers, then what is the minimum value of x + y? (a) 130 (b) 260 (c) 2113 (d) 4226 dy - 2y = 0 represent? dx (a) A family of straight lines (b) A family of circles (c) A family of parabolas (d) A family of ellipses 80. What does the equation x

73. Consider the following in respect of the function f(x) = 10x: (1) Its domain is (- ∞, ∞) (2) It is a continuous function (3) It is differentiable at x = 0

81. If the points with coordinates (- 5, 0), (5p2, 10p) and (5q2, 10q) are collinear, then what is the value of pq where p ≠ q? (a) - 2 (b) - 1 (c) 1 (d) 2

Which of the above statements are correct? (a) 1 and 2 only (b) 2 and 3 only (c) 1 and 3 only (d) 1, 2 and 3

82. What is the equation of the straight line which passes through the point (1, - 2) and cuts off equal intercepts from the axes? (a) x + y - 1 = 0 (b) x - y - 1 = 0 (c) x + y + 1 = 0 (d) x - y - 2 = 0

74. What is lim x3 (cosec x)2 equal to? x→0

1 2 (d) Limit does not exist

(a) 0

(b)

(c) 1 75. What is lim

x →1

(a) 0 (c) 6

x3 − 1 x −1

equal to? (b) 3 (d) Limit does not exist

83. What is the equation of the circle which touches both the axes in the first quadrant and the line y - 2 = 0? (a) x2 + y2 - 2x - 2 y- 1= 0 (b) x2 + y2 + 2x + 2y + 1 = 0 (c) x2 + y2 -2x - 2y + 1 = 0 (d) x2 + y2 - 4x - 4y + 4 = 0

34 Oswaal NDA/NA Year-wise Solved Papers 84. What is the equation of the para­bola with focus (- 3, 0) and directrix x - 3 = 0? (a) y2 = 3x (b) x2 = 12y (c) y2 = 12x (d) y2 = - 12x 85. What is the distance between the foci of the ellipse x2 + 2y2 = 1? (a) 1 (b) 2 (c) 2

(d) 2 2

86. Let a, b, c be the lengths of sides BC, CA, AB respectively of a triangle ABC. If p is the perimeter and q is the area of the triangle, then A what is p(p - 2a) tan   equal to? 2 (a) q (b) 2q (c) 3q (d) 4q 87. A straight line passes through the point of intersection of x + 2y + 2 = 0 and 2x- 3y- 3 = 0. It cuts equal inter­cepts in the fourth quadrant. What is the sum of the absolute values of the intercepts? (a) 2 (b) 3 (c) 4 (d) 6 88. Under which one of the following conditions are the lines ax + by + c = 0 and bx + ay + c = 0 parallel (a ≠ 0, b ≠ 0)? (a) a - b = 0 only (b) a + b = 0 only (c) a2 - b2 = 0 (d) ab + 1 = 0 89. What is the equation of the locus of the midpoint of the line segment obtained by cutting the line x + y = p, (where p is a real number) by the coordinate axes? (a) x - y = 0 (b) x + y = 0 (c) x - y = p (d) x + y = p 90. If the point (x, y) is equidistant from the points (2a, 0) and (0, 3a) where a > 0, then which one of the following is correct? (a) 2x - 3y = 0 (b) 3x - 2y = 0 (c) 4x- 6y + 5a = 0 (d) 4x - 6y - 5a = 0

Consider the following for the next three (03) items that follow: The plane 6x + ky + 3z-12 = 0 where k ≠ 0 meets the coordinate axes at A, B and C respectively. The equation of the sphere passing through the origin and A, B, C is x2 + y2 + z2 - 2x - 3y - 4z = 0. 91. What is the value of k? (a) 3 (b) 4 (c) 6 (d) 12

92. If p is the perpendicular distance from the centre of the sphere to the plane, then which one of the following is correct? (a) 0 < p < 0.5 (b) 0.5 < p < 1 (c) 1< p < 1.5 (d) p > 1.5 93. What is the equation of the line through the origin and the centre of the sphere? (a) x = y = z (b) 2x = 3y = 4z (c) 6x = 3y = 4z (d) 6x = 4y = 3z

Consider the following for the next two (02) items that follow: 2x 2 y z + + = 2 pass through the Let the plane 3 3 k point (2, 3, - 6). 94. What are the direction ratios of a normal to the plane? (a) (b) (c) (d) 95. If p, q and r are the intercepts made by the plane on the coordinate axes respectively, then what is (p + q + r) equal to? (a) 10 (b) 11 (c) 12 (d) 13 96. If 4i + j − 3k and pi + q j − 2 k are collinear vectors, then what are the possible values of p and q respectively? (a) 4, 1 (b) 1, 4 2 8 8 2 (d) , (c) , 3 3 3 3   

97. If a , b , c are the position vectors of the vertices A, B, C respectively of a triangle ABC and G is the centroid of the triangle, then what is AG equal to?       a+b+c 2a - b - c (a) (b) 3 3       b + c − 2a a - 2b - 2 c (c) (d) 3 3 98. Consider the following statements: (1) Dot product over vector addition is distributive (2) Cross product over vector addition is distributive (3) Cross product of vectors is associative Which of the above statements is/ are correct?: (a) 1 only (b) 2 only (c) 1 and 2 only (d) 1, 2 and 3

35

solved PAPER - 2022 (I)

   99. Let a , b , c be three non-zero vectors such that    a × b = c . Consider the following statements:    (1) a is unique if b and c are given    (2) c is unique if a and b are given Which of the above statements is/are correct? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2   100. Let a and b be two unit vectors such that     a - b Median> Mode (b) Mean > Mode > Median (c) Median > Mean > Mode (d) Mode > Median > Mean 103. The variance of five positive observations is 3.6. If four of the observations are 2, 2, 4, 5 then what is the remaining observation? (a) 4 (b) 5 (c) 7 (d) 9



The algebraic sum of the deviations of a set of values x1, x2, x3, ... xn measured from 100 is - 20 and the algebraic sum of the deviations of the same set of values measured from 92 is 140. 106. What is the mean of the values? (a) 91 (b) 96 (c) 98 (d) 99 107. What is the algebraic sum of the deviations of the same set of values measured from 99? (a) 0 (b) 10 (c) 20 (d) 40

108. If the algebraic sum of the deviations of the same set of values measured from y is 180, then what is the value of y? (a) 80 (b) 85 (c) 90 (d) 95 Consider the following data for the next three (03) items that follow: The marks obtained by 51 students in a class are in AP with its first term 4 and common difference 3. 109. What is the mean of the marks? (a) 67 (b) 71 (c) 75 (d) 79

110. What is the median of the marks? (a) 79·5 (b) 79 (c) 78·5 (d) 77 111. What is the sum of the deviations measured from the median? (a) - 1 (b) 0 (c) 1 (d) 2

104. What is the arithmetic mean of 50 terms of an AP with first term 4 and common difference 4? (a) 50 (b) 51 (c) 100 (d) 102 105. What is the coefficient of mean deviation of 21, 34, 23, 39, 26, 37, 40, 20, 33, 27 (taken from mean)? (a) 0·11 (b) 0·22 (c) 0·33 (d) 0·44

Consider the following for the next three (03) items that follow:

Consider the following data for the next three (03) items that follow: There are 90 applicants for a job. Some of them are graduates. Some of them have less than three years experience.

At least 3 years

Number of graduates

Number of non-graduates

18

9

36

27

experience Less than 3 years expenence

Let G be the event that the first applicant interviewed is a graduate and T be the event

36 Oswaal NDA/NA Year-wise Solved Papers that first applicant interviewed has at least 3 years experience.

(

)

112. What is P G ∩ T equal to? 1 (a) 5

(b)

2 5

3 (c) 5

(d)

4 5

(

)

113. What is P G| T equal to? 2 (a) 7

(b)

3 7

4 (c) 7

(d)

5 7

(

)

(b)

1 3

3 3 (c) (d) 5 4 Consider the following data for the next three (03) items at follow: The incidence of suffering from a disease among workers in an industry has a chance of 1 33 % . 3 115. What is the probability that exactly 3 out of 6 workers suffer from a disease? 80 (a) 729 10 (c) 243

10 81 160 (d) 729 (b)

665 (a) 729

(b)

64 729

4 (c) 243

(d)

1 729

117. What is the probability that at least one out of 6 workers suffer from a disease? 728 (a) 729

(b)

665 729

653 (c) 729

(d)

596 729



114. What is P T | G equal to? 1 (a) 4

116. What is the probability that no one out of 6 workers suffers from a disease?

Consider the following frequency distribution for the next three (03) items that follow: Class Frequency

0-20 20-40 40-60 60-80 80-100 17

p+q

32

p - 3q

19

The total frequency is 120. The mean is 50. 118. What is the value of p? (a) 25 (b) 26 (c) 27 (d) 28 119. What is the value of q? (a) 1 (c) 3

(b) 2 (d) 4

120. If the frequency of each class is doubled, then what would be the mean? (a) 25 (b) 50 (c) 75 (d) 100

37

solved PAPER - 2022 (I)

Answers Q. No.

Answer Key

Topic Name

Chapter Name

1

(c)

Properties of Determinants

Matrices and Determinants

2

(b)

Properties of Determinants

Matrices and Determinants

3

(d)

Properties of Determinants

Matrices and Determinants

4

(c)

Algebra of Matrices

Matrices and Determinants

5

(d)

Properties of Adjoint Matrices

Matrices and Determinants

6

(d)

Properties of Determinants

Matrices and Determinants

7

(b)

Properties of Adjoint Matrices

Matrices and Determinants

8

(b)

Algebra of Matrices

Matrices and Determinants

9

(d)

Properties of Determinants

Matrices and Determinants

10

(c)

Properties of Inverse

Matrices and Determinants

11

(b)

Number of Terms of Binomial Expansion

Algebra

12

(c)

Binomial Theorem

Algebra

13

(d)

Coefficients of Binomial Expansion Algebra for the Integral Index

14

(c)

Coefficients of Binomial Expansion Algebra for the Integral Index

15

(a)

Permutations and Combinations

Algebra

16

(b)

Permutations and Combinations

Algebra

17

(d)

Root and Coefficients

Algebra

18

(a)

Root and Coefficients

Algebra

19

(c)

Root and Coefficients

Algebra

20

(d)

Root and Coefficients

Algebra

21

(c)

Permutations and Combinations

Algebra

22

(b)

Permutations and Combinations

Algebra

23

(b)

Set Theory and Relations

Algebra

24

(d)

Geometric Progression

Algebra

25

(b)

Harmonic Progression

Algebra

26

(b)

Trigonometric Identities

Trigonometry

27

(d)

Trigonometric Equations

Trigonometry

28

(c)

Trigonometric Equations

Trigonometry

29

(c)

Trigonometric Equations

Trigonometry

30

(c)

Trigonometric Equations

Trigonometry

31

(c)

Trigonometric Equations

Trigonometry

38 Oswaal NDA/NA Year-wise Solved Papers Q. No.

Answer Key

Topic Name

Chapter Name

32

(b)

Trigonometric Identities

Trigonometry

33

(a)

Trigonometric Identities

Trigonometry

34

(c)

Heights and Distances

Trigonometry

35

(b)

Heights and Distances

Trigonometry

36

(a)

Inverse Trigonometric Identities

Trigonometry

37

(c)

Inverse Trigonometric Identities

Trigonometry

38

(b)

Trigonometric Identities

Trigonometry

39

(b)

Trigonometric Identities

Trigonometry

40

(b)

Trigonometric Identities

Trigonometry

41

(c)

Set Theory and Relations

Algebra

42

(d)

Set Theory and Relations

Algebra

43

(c)

Set Theory and Relations

Algebra

44

(a)

Set Theory and Relations

Algebra

45

(d)

Set Theory and Relations

Algebra

46

(b)

Arguments of Complex Number

Algebra

47

(c)

Modulus of Complex Number

Algebra

48

(d)

Permutations and Combinations

Algebra

49

(b)

Permutations and Combinations

Algebra

50

(c)

Binomial Theorem

Algebra

51

(d)

Arithmetic Progression

Algebra

52

(b)

Arithmetic Progression

Algebra

53

(a)

Binary System

Algebra

54

(b)

Binary System

Algebra

55

(d)

Inequalities

Algebra

56

(a)

Continuity and Differentiability

Differential Calculus

57

(d)

Differential Equations

Integral Calculus and Differential Equations

58

(a)

Differential Equations

Integral Calculus and Differential Equations

59

(d)

Function

Differential Calculus

60

(c)

Area under Curves

Integral Calculus & Differential Equations

61

(b)

Basics of Definite Integration

Integral Calculus & Differential Equations

62

(b)

Indefinite Integration

Integral Calculus & Differential Equations

39

solved PAPER - 2022 (I)

Q. No.

Answer Key

Topic Name

Chapter Name

63

(b)

Basics of Integration

Integral Calculus & Differential Equations

64

(b)

Properties of Definite Integration

Integral Calculus & Differential Equations

65

(b)

Maxima and Minima

Differential Calculus

66

(b)

Maximum and Minimum Value

Trigonometry

67

(d)

Basics of Function

Differential Calculus

68

(a)

Composite Function

Differential Calculus

69

(a)

Logarithmic Differentiation

Integral Calculus & Differential Equations

70

(b)

Logarithmic Differentiation

Integral Calculus & Differential Equations

71

(d)

Basics of Differentiation

Integral Calculus & Differential Equations

72

(d)

Differential Equation

Integral Calculus & Differential Equations

73

(d)

Differentiability

Differential Calculus

74

(a)

Limit of Trigonometric Functions

Differential Calculus

75

(c)

Rationalization Method

Differential Calculus

76

(b)

Increasing and Decreasing Functions

Differential Calculus

77

(d)

Basic Differentiation

Integral Calculus & Differential Equations

78

(b)

Area under Curves

Integral Calculus & Differential Equations

79

(a)

Maxima and Minima

Differential Calculus

80

(c)

Differential Equations

Integral Calculus & Differential Equations

81

(c)

Area of Triangle

Matrices and Determinants

82

(c)

Equation of Straight Line

Analytical Geometry of 2 & 3 Dimensions

83

(c)

Equations of Circle

Analytical Geometry of 2 & 3 Dimensions

84

(d)

Basic of Parabola

Analytical Geometry of 2 & 3 Dimensions

85

(b)

Basic of Ellipse

Analytical Geometry of 2 & 3 Dimensions

86

(d)

Half Angle Formula and Area of Triangle

Analytical Geometry of 2 & 3 Dimensions

40 Oswaal NDA/NA Year-wise Solved Papers Q. No.

Answer Key

Topic Name

Chapter Name

87

(a)

Family of Straight Line

Analytical Geometry of 2 & 3 Dimensions

88

(c)

Parallel and Perpendicular Condition

Analytical Geometry of 2 & 3 Dimensions

89

(a)

Locus of Point

Analytical Geometry of 2 & 3 Dimensions

90

(c)

Distance Formula

Analytical Geometry of 2 & 3 Dimensions

91

(b)

Sphere

Analytical Geometry of 2 & 3 Dimensions

92

(b)

Perpendicular Distance from Point to Plane

Analytical Geometry of 2 & 3 Dimensions

93

(d)

Equation of Straight Line

Analytical Geometry of 2 & 3 Dimensions

94

(a)

Direction Ratio of Plane

Analytical Geometry of 2 & 3 Dimensions

95

(b)

Direction Ratio of Plane

Analytical Geometry of 2 & 3 Dimensions

96

(c)

Collinear Vectors

Vector Algebra

97

(c)

Centroide of Triangle

Vector Algebra

98

(c)

Properties of Dot and Cross Product of Vectors

Vector Algebra

99

(b)

Definition of Cross Product of Two Vectors

Vector Algebra

100

(d)

Properties of Dot Product of Vectors

Vector Algebra

101

(c)

Basics of Probability

Probability & Statistics

102

(d)

Normal Distribution and Skewed

Probability & Statistics

103

(c)

Variance

Probability & Statistics

104

(d)

Central Tendency

Probability & Statistics

105

(b)

Mean Deviation

Probability & Statistics

106

(d)

Deviation

Probability & Statistics

107

(a)

Deviation

Probability & Statistics

108

(c)

Deviation

Probability & Statistics

109

(d)

Central Tendency

Probability & Statistics

110

(b)

Central Tendency

Probability & Statistics

111

(b)

Central Tendency

Probability & Statistics

112

(b)

Conditional Probability

Probability & Statistics

113

(c)

Conditional Probability

Probability & Statistics

41

solved PAPER - 2022 (I)

Q. No.

Answer Key

Topic Name

Chapter Name

114

(d)

Conditional Probability

Probability & Statistics

115

(d)

Binomial Distribution

Probability & Statistics

116

(b)

Binomial Distribution

Probability & Statistics

117

(b)

Binomial Distribution

Probability & Statistics

118

(c)

Central Tendency

Probability & Statistics

119

(a)

Central Tendency

Probability & Statistics

120

(b)

Central Tendency

Probability & Statistics

NDA / NA

MATHEMATICS

I

National Defence Academy / Naval Academy





2. Option (b) is correct. Explanation:

1 p q 1 p q ∆1 = 1 q r = 1 q r 1 r p 1 r p



−

1 1 1 = p q r q r p



1 1 1 = q r p = ∆2 r p q







Now, ∆1 + ∆2 = 2∆2



1 0 = 2 q r−q r p−r

a

b

Let ∆ = a

2

b

2

a

3

b3

c

1 c = abc a

1 b

1 c

a2

b2

c2

2

c3

(Taking common a, b, c from C1, C2 and C3 respectively) [Applying C1 → C1 - C2 and C2 → C2 - C3]

= abc



0 a−b

0 b−c

1 c

( a − b )( a + b ) ( b − c )( b + c ) = abc(a - b)(b - c)



0 1

0 1

c3

1 c

a + b b + c c3

 Applying C 2 → C 2 − C1    C3 → C3 − C1   and







1 1 1 = 2q r p r p q





 Applying C1 ↔ C 2    C 2 ↔ C3   and





2022

AnSWERS wITH eXPLANATION 1. Option (c) is correct. Explanation:



Solved Paper

0 p−q q−r







= 2[(r - q)(q - r) - (p - q)(p - r)] = -2[p2 + q2 + r2 - pq - qr - rp] = -[2p2 + 2q2 + 2r2 - 2pq - 2qr - 2rp] = -[(p - q)2 + (q - r)2 + (r - p)2] < 0 Hence, value of ∆1 + ∆2 is always negative. Hints: • Use |A-| = |A| • Make sum of completing square of 2p2 + 2q2 + 2r2 - 2pq - 2qr - 2rp and use property sum of square of number



[ Taking common (a - b), (b - c) from C1 and C2 respectively] Expand through R1 = abc(a - b) (b - c)(b + c - a - b) = abc(a - b) (b - c) (c - a) = 6 × 2 = 12 Hints: •

Use properties of determinant and take common facter abc, (a - b) (b - c) (c - a) from determinant.

3. Option (d) is correct. Explanation:

Let

a b c b c a =0 ∆= c a b

43

SOLVED PAPER - 2022 (I)

[Applying C1 → C1 + C2 + C3]







⇒

a+b+c b c a+b+c c a = 0 a+b+c a b



1 b c ⇒ (a + b + c) 1 c a = 0 1 a b



[Take a + b + c common from C1] [Applying R2 → R2 - R1 and R3 → R3 - R1]



⇒ (a + b + c) 0 c − b a − c = 0 0 a−b b−c [Expands through C1] = 0 2 2 ⇒ (a + b + c) (-a - b - c2 + ab + bc + ca) = 0 ⇒ -(a + b + c)(a2 + b2 + c2 - ab - bc - ca) = 0 ⇒ -(a3 + b3 + c3 - 3abc) = 0 ⇒ a + b + c = 0 or a2 + b2 + c2 - ab - bc - ca = 0 or a3 + b3 + c3 = 3abc Hence, all statements 1, 2, 3 are correct.

1



b

c

Hint: •

•

Use properties of determinant and take common factor (a + b + c) (a2 - b2 - c2 + ab + bc + ca). Use algebraic identing (a3 + b3 + c3 - 3abc) = (a + b + c) (a2 - b2 - c2 - ab - bc - ca).

4. Option (c) is correct. Explanation: Statement 1:



CA =

 m2 m m n] =  [ −m  −m 2



Statement 3:



m C(A + B) =   [ m − n n − m ]  −m 







 m 2 − mn mn − m 2    CA + CB = 2 2  −m + mn −mn + m 



∴ C(A + B) = CA + CB So, statement 3 is true. Hint: •

 −mn −m 2  m CB =   [ −n − m ] =   m 2   mn  −m  CA ≠ CB So, statement 1 is not true. Statement 2.

 m AC = [ m n ]   = m 2 − mn   −m  m BC = [ −n − m ]   =  −mn + m 2   −m  ∴ AC = BC So, statement 2 is true.

Use multiplication rule of matrices.

5. Option (d) is correct. Explanation:

 sin θ 2 cos θ sin θ − 2 cos θ  A dy A =  − cos θ 2 sin θ −2 sin θ − cos θ   0  0 2



 2 sin θ cos θ 0    A (adj A) =  −2 cos θ sin θ 0   −1 1 1 



1

sin θ − cos θ 0   2 sin θ 0  =  2 cos θ sin θ − 2 cos θ −2 sin θ − cos θ 2 



mn   −mn 

 m 2 − mn mn − m 2  =   2 2  −m + mn −mn + m 

sin θ − cos θ 0   2 sin θ 0  ×  2 cos θ sin θ − 2 cos θ −2 sin θ − cos θ 2 

é 2 sin 2 q + 2 cos2 q ù 0 0 ê ú ê ú 0 2 cos 2 q + 2 sin 2 q 0 = ê ú ê ú 2 2 ú ê 0 0 2 sin q + 2 cos qú ëê û





2 0 0    =  0 2 0  = 2I  0 0 2 

44 Oswaal NDA/NA Year-wise Solved Papers Shortcut:

We know that for a square matrix A of order n, A(adj A) = adj(A) A = |A|I



 2 sin θ cos θ 0    ∴ |A| =  −2 cos θ sin θ 0   −1 1 1 



= 2sin2 θ + 2cos2 θ [Expand through C3] =2 ∴ A(adj A) = |A| I = 2I

•



C13   C 23   C33 

C12 C 22 C32

Where Cij is cofactor of aij.

6. Option (d) is correct. Explanation: 2 cos 2θ



2 cos 2θ

2

6

2

∆ = 1 − 2 sin θ 2 cos θ − 1 3 k 2k 1

Let

• •

2

1 1 Let, A =   and B = 1 1



1 1  1 −1 0 0  AB =    =   (Null matrix) 1 1  −1 1  0 0 



But A and B are not null matrix. So, statement 1 is not correct. Let, AB = I ⇒ B = A-1 We, know that AA-1 = I = A-1 A ⇒ AB = BA So, statement 2 is correct. • •

2

Use cos 2θ = 2cos2 θ - 1 = 1 - 2sin2 θ Use property of determinant that if two rows or column of determinant are identical then its value is zero. If determinant of matrix is zero then matrix is called singular matrix.

7. Option (b) is correct. Explanation: We know that for a square matrix A of order n, A(adj A) = (adj A)A = |A|I given that B = adj A ∴ AB = BA

If all element of matrix is zero then it is called null matrix. If AB = I then B = A-1.

9. Option (d) is correct. Explanation:

1 0 0    Given that A = I3 = 0 1 0  0 0 1 



1 0 0    And B = A′ = 0 1 0  = I 0 0 1 



Now, C = A + B = 2I ∴ |C| = |2I| = 23 |I| = 8

Hints:

•

 1 −1  −1 1   



Hints:

[ cos 2θ = 2cos θ - 1 = 1 - 2sin θ] =0 [ R1 and R2 are identical] Hence, for all real value of k, the values determinant of given matrix is zero. i.e., for any k given matrix is singular. • •

Use A(adj A) = (adj A)A = |A|I If adj = constant for i = j then matrix A is called scalar matrix.

8. Option (b) is correct. Explanation:

2 cos 2θ 2 cos 2θ 6 cos 2θ 3 = cos 2θ k 2k 1



So, statement 1 is correct. AB = |A|I is scalar matrix but not null matrix So, statement 2 is correct and statement 3 is not correct. Hints:

Hints:  C11  Use adj A = C 21  C31



Shortcut:

Given, A = I3 and B = A′ = I′ = I ∴ C = 2I ⇒ |C| = 23 |I| = 8

45

SOLVED PAPER - 2022 (I)

Hints: • •

Use I′ = I Use |kA| = kn |A|, where order of square matrix A is n.

10. Option (c) is correct. Explanation: Given that |A| ≠ 0 and |B| ≠ 0  AB = A ⇒ A-1 AB =A-1 A  [Pre multiply by A-1 both sides] ⇒ IB = I ⇒ B = I …(i) -1 -1 and BA = B ⇒ B BA = B B  [Pre multiply by A-1 both sides] ⇒ IA = I ⇒ A = I …(ii) Statement 1 A2 = I2 = I = A (Statement 1 is correct.) Statement 2 AB2 = II2 = I and A2B = I2I = I [∵ In = I] ∴ AB2 = A2B (Statement 2 is correct.)

12. Option (c) is correct. Explanation: We know that coefficient of middle term of expansion (x + y)n has the highest value.



 10 + 1 + 1  Middle term of (x + y)10 is   2  

Hints: • •

• Premultiply by A BA = B • Use In = I.

-1

and B

in AB = A and

9





9

9

2  2 4    1 + x  1 − x  = 1 − 2  x       [ (a + b) (a - b) = a2 - b2] We know that number of terms in expansion (a + b)n is n + 1. 4   ∴ Number of terms in expansion  1 − 2  x   9 + 1 = 10

9

is

Hints: n n

n

Coefficient of middle term of expansion (x + y)n has the highest value. Coefficient of kth term = coefficient of (n - k + 2)th term

13. Option (d) is correct. Explanation: Given that,

11. Option (b) is correct. Explanation:

term =

6th term ∴ The coefficient of the 6th term has the highest value. So, statement 1 is correct. We know that in expansion (x + y)n coefficient of kth term = coefficient of (n - k + 2)th term ∴ In expansion (x + y)10 Coefficient of 3rd term = coefficient of (10 - 3 + 2)th = coefficient of 9th term. So, statement 2 is correct.

Hints: -1

th

• Use a b = (ab) • Use (a + b) (a - b) = a2 - b2 • Number of terms in expansion (a + b)n is n+1

3n

C2n =

3n

C 2 n−7







[ If n C x = n C y then x + y = n or x = y] ∴ 2n + 2n - 7 = 3n ⇒ 4n - 3n = 7 ⇒ n=7



∴

n

Cn− 5 = 7 C 2 =

7×6 2×1

= 21

Hints: •

Use If n C x = n C y then x + y = n or x = y

•

Use n Cr =

n! (n − r )!r !

46 Oswaal NDA/NA Year-wise Solved Papers 14. Option (c) is correct. Explanation:

We, know that n Cr = n Cr−1



∴



51

+





=









51

51

C 21 51

C 22 +

C 25 -

C 51−30 -

51

51

C 23 -

C 26 +

51

C 51− 29 +

+

51

=

51

51

51

C 51− 26 -

C30 -

51















Now, 51 C 51 -





+

51

C 26 -

51

C 24 51

C 27 -

51

C 28 +

C 29 C30



51

51

C 26 +

51

C 27 -

51

C 28

+

51

C 29 -

51

C30

C 29 + 51

51

C 26 + 51

C 51− 27

C 28 51

51

C 27 -

C 29 -

51

51

C 28

C30 = 0

[ n Cn = n C0 ]

Hints: n Use n Cr = Cr−1

•

Use n Cn = n C0 = 1

15. Option (a) is correct. Explanation: For odd numbers between 300 and 400. When digits are not repeated. ⋅

3



⋅

1 × 8 × 4 Here one choice 3 for hundred place; 4 choice {1, 5, 7, 9} for unit place and remaining 8 choice for ten’s place ∴ Total odd numbers between 300 and 400, when digits are not repeated = 1 × 8 × 4 = 32. Hints: • • •

1

Use for odd number unit place digits will be 1, 3, 5, 7, 9. 3 is fixed at hundred place. Use non repetition case.

2

3

4

5

Since, vowels not occupy the even positions. So, there are 3 odd places are available of vowels. ∴ Number of words = 3 C 2 × 2! × 3! = 36 Hints: • • •

C 27

C0 = 0

•



51

C 51− 28 -

+ 51

16. Option (b) is correct. Explanation: Given, Word: TIGER Vowels : I, E

First select number of places for vowels. Than arrange number of vowels Also arrange number of consonant

17. Option (d) is correct. Explanation: Given that, α and β are roots of the equation x2 + px + q = 0 ∴ α + β = -p and α⋅β = q

α 3 + β3 = ( α + β ) − 3αβ ( α + β ) 3



Now,





= − p 3 − 3q ( − p )





= 3pq − p 3



α 3 ⋅β3 = ( αβ ) = q 3 3 Since, α and β3 are roots of the equation x2 +mx + n = 0 ∴ α3 + β3 = -m ⇒ m = - (α3 + β3) = p3 - 3pq 3 3 α ⋅β = n ⇒ n = q3 So, m + n = p3 + q3 - 3pq 3

Hints: • Use, If α and β are roots of equation ax2 + bx = c = 0 then, sum of roots (α + β) = b c and product of roots (α⋅β) = a a • Use algebraic identity a3 + b3 = (a + b)3 - 3ab(a + b) −

47

SOLVED PAPER - 2022 (I)

18. Option (a) is correct. Explanation: Given that, α and β are roots of the equation x2 - ax - bx + ab - c = 0 i.e., x2 - (a + b)x + ab - c = 0 ∴ α+β=a+b

b c [ for ax2 + bx + c = 0; α + β = − , α⋅β = ] a a α⋅β = ab - c ⇒ ab = α⋅β + c Quadratic equation whose roots are a and b is x2 - (a + b)x + ab = 0 ∴ x2 - (α + β)x + α⋅β - c = 0 ⇒ x2 - αx - βx + αβ - c = 0 Hints: • Use, If α and β are roots of the equation −b c and α⋅β = a a • Use, quadratic equation whose roots are α and β is x2 - (α + β)x + α⋅β = 0 ax2 + bx + c = 0 then α + β =

19. Option (c) is correct. Explanation: Let, α be two equal roots of the equation x2 - ax - bx - cx + bc + ca = 0 i.e. x2 - (a + b + c)x + bc + ca = 0 ∴ α+α=a+b+c

a+b+c …(i) 2 α⋅α = bc + ca ⇒ α2 = c(a + b)…(ii) From (i) and (ii) 4c(a + b) = [(a + b) + c]2 ⇒ (a + b)2 + c2 - 2c(a + b) = 0 ⇒ [(a + b) - c]2 = 0 ⇒ a+b-c=0 ⇒

α=

Shortcut:

For equal roots (a + b + c)2- 4×1×(bc + ca) = 0 ⇒ [(a + b) - c]2 = 0 a+b-c=0

Hints: • Use, if quadratic equation ax2 + bx + c = 0 has equal roots then b2 - 4ac = 0 • Use (a + b)2 + c2 - 2c(a + b) = [(a + b) - c]2 20. Option (d) is correct. Explanation: Given that, α and β are roots of the equation x2 - 8x + q = 0 ∴ α + β = 8 and α⋅β = q Now, (α - β)2 = (α + β)2 - 4αβ = 64 - 4q α - β = 64 − 4q ( α > β) α - β2 = (α + β)(α - β) = 16



Now,







⇒ 64 − 4q = 2 ⇒ 64 - 4q = 4 ⇒ q = 15

2

8

(

)

64 − 4q = 16

Hints: •

Use, If α and β are roots of the equation ax2 + bx + c = 0 then α + β =

•

−b c and α⋅β = a a

Use (a - b)2 = (a + b)2 - 4ab and a2 - b2 = (a + b)(a - b)

21. Option (c) is correct. Explanation: 30! + 35! = 30! + 35⋅34⋅33⋅32⋅31⋅30! = 30!(1 + 35⋅34⋅33⋅32⋅31) ∴ (1 + 35⋅34⋅33⋅32⋅31) not divisible by 5 ∴ Only 30! Is divisible by 5n So, maximum value of n such that 5n divides 30!

 30   30   30  =  + 2+ 3  5  5  5  =6+1+0=7 Maximum value of n = 7

Hints: •

If p is prime number then the highest power of p in n! is given by

n  n   n    +  2  +  3  + ....  p   p   p  •

[x] is greatest integer less than equal to x e.g. [2.5] = [2.99] = 2

48 Oswaal NDA/NA Year-wise Solved Papers 22. Option (b) is correct. Explanation: 2(2×1) + 3(3×2×1) + 4⋅(4×3×2×1) + …… + 9(9 × 8 × 7….. × 1) + 2 = 2⋅2! + 3⋅3! + 4⋅4! + ……. + 9.9! + 2 = 2⋅2! + 2! + 3⋅3! + 4⋅4! + …. 5⋅5! = 3⋅2! + 3⋅3! + 4⋅4! + ….9⋅9! = 3! + 3⋅3! + 4⋅4! + …..9⋅9! = 4⋅3! + 4⋅4! + …..9⋅9! = 4! + 4⋅4! + …..9⋅9! Same way we solve, then we get = 9! + 9⋅9! = 10⋅9! = 10! Hints: • •

Use, n! = n(n - 1)! Use n! + n⋅n! = (n + 1)n! = (n + 1)!

23. Option (b) is correct. Explanation: Given that, A = {{1, 2, 3}} ∴ n(A) = 1 Number of subset = 21 = 2 So, number of element in power set of A = 2 Hints: •

Use, if n(A) = n then number of element in p(A) = 2n. including with φ

24. Option (d) is correct. Explanation: Given that, a, b, c are in G.P ∴ b2 = ac …(i) Squaring both sides, we get (b2)2 = a2c2 ⇒ a2, b2, c2 are also in G.P. So, statement 1 is correct From (i) 1

2

=

1 1 1 1 ⇒  = × b ac a c  







⇒



So, statement 2 is correct. from (i) b2 = ac ⇒ b = ac



⇒ b = a c ( a > 0, b > 0> c > 0)

b

2

1 1 1 , , are also in G.P. a b c

( )

2



⇒ a , b , c are also in G.P. So, statement 3 is correct. Hints: •

Use, if a, b, c are in G.P. then b2 = ac

25. Option (b) is correct. Explanation:

2 1 1 We know that if a, b, c are in H.P, then = + b a c a+b b+c Given that, , b, are in H.P. 2 2



2 ( a + 2b + c ) 2 2 2 ∴ = + = b a + b b + c ( a + b )( b + c )



⇒ ab + ac + b2 + bc = ab + 2b2 + ca ac = b2 ∴ a, b, c are in G.P. Hints: • •

2 1 1 + Use, if a, b, c are in H.P then = b a c Use, if a, b, c are in G.P then b2 = ac

26. Option (b) is correct. Explanation: cot2 15° + tan2 15° = cosec2 15° - 1 + sec2 15° - 1

=



1 2

sin 15°

+

1 2

cos 15°

-2

( tan2 θ = sec2 θ - 1, cot2 θ = cosec2 θ - 1



sec θ =

1 1 and cosec θ = ) cos θ sin θ

sin 2 15° + cos2 15°





−2 = 1 4 sin 2 15° ⋅ cos2 15° 4





=



[ sin2 θ + cos2 θ = 1 and sin 2θ = 2sin θ⋅cos θ]





(

=

)

4 2

sin 30°

−2

4 − 2 = 16 - 2 = 14 1 4

Hints: •

Use trigonometric identities convert given expression in form of sin θ and cos θ and further solve.

49

SOLVED PAPER - 2022 (I)

27. Option (d) is correct. Explanation: Given that, sin A - cos B - cos C = 0 sin A = cos B + cos C



tan A + tan B   ∵ tan ( A + B ) =   1 − tan A ⋅ tan B   ⇒ tan α + tan β = 1 - tan α⋅tan β



⇒



⇒ 1 + 2tan β = 2 - tan β

π ] 2



⇒ tan β =

1 A A π A  B−C  sin ⋅ cos = 2 cos  −  ⋅ cos   2 2 2 2 2   2 



Now,

B+C  B−C  ⋅ cos   2  2 



⇒ sin⋅A = 2cos



[In ∆ABC A + B + C =



⇒



[ sin 2θ = 2sin θ⋅cos θ] 1 A A A B−C sin ⋅ cos = 2 sin ⋅ cos 2 2 2 2 2



⇒



⇒



⇒



⇒

A+C B = 2 2



⇒

π B B − = 2 2 2



⇒ B =

cos



=

1 3 2 tan β 1 − tan 2 β 2 3

1 1− 9

=

2 9 3 × = 3 8 4

Hints:

A B C = − 2 2 2

•

Use tan (A + B) = tan A + tan B 1 − tan A ⋅ tan B

•

Use tan 2θ =

2 tan θ 1 − tan 2 θ

29. Option (c) is correct. Explanation: Given that, tan(45° + θ) = 1 + sin 2θ

π 2



Hints: • Use, cos x + cos y = 2cos

• Use in ∆ABC, A + B + C = 28. Option (c) is correct. Explanation: Given that, 2 tan α = 1

1 ⇒ tan α = 2



and



⇒



⇒

π 4

tan (α + β) = tan

tan α + tan β =1 1 − tan α ⋅ tan β

⇒

x+y x−y ⋅ cos 2 2

θ θ ⋅ cos 2 2

α+β=

tan 2β =



A B−C = cos 2 2

• Use sin θ = 2 sin

1 tan β + tan β = 1 2 2

π 4

π 2

2 tan θ tan 45° + tan θ -1= 1 − tan 45° ⋅ tan θ 1 + tan 2 θ tan A + tan B   ∵ tan ( A + B ) = 1 − tan A ⋅ tan B     sin 2θ = 2 tan θ  2   1 + tan θ







⇒



⇒



⇒ tan3 θ + tan2 θ = 0 ⇒ tan2 θ(tan θ + 1) = 0 ⇒ tan θ = 0 or tan θ = -1 (Not possible)



2 tan θ 1 + tan θ −1 = 1 − tan θ 1 + tan 2 θ

2 tan θ 2 tan θ = 1 − tan θ 1 + tan 2 θ 2 ⇒ tan θ(1 + tan θ) = tan θ(1 - tan θ)

 ⇒ Now,

π  ∵ − 4 < θ <  θ = 0° cos 2θ = cos 0° = 1

π 4 

50 Oswaal NDA/NA Year-wise Solved Papers Hints: • • •

Hints:

Use tan(A + B) = tan A + tan B 1 − tan A ⋅ tan B 2 tan θ Use sin 2θ = 1 + tan 2 θ Simplify and solve the trigonometric equation

30. Option (c) is correct. Explanation: Given that sin 2θ = cos 3θ ⇒ sin 2θ = sin(90° - 3θ) ⇒ 2θ = 90° - 3θ ⇒ θ = 18° Now, 1 + 4sin θ = 1 + 4sin 18°

= 1 + 4⋅





=



Use tan θ =

• •

Use h2 = p2 + b2 Value of tan θ in 2nd and 4th quadrant is negative.

32. Option (b) is correct. Explanation: 2

2



cos4

7π   5π  7π 5π  +  cos2 + cos4 =  cos2  8   8  8 8 





7π 5π  7π 5π  − cos2 =  cos2 + 2 cos2 ⋅ cos2 8 8  8 8 

2



5 −1 4  5 − 1 ∵ sin 18° =  4  

= 1+ 5 −1



p p and sin θ = h b

•

  7 π 5π   7 π 5π   =  − sin  + ⋅ sin  −  8  8    8  8 



+

+

5

Hints:

1 7π 5π  2 cos ⋅ cos  2  8 8 

3π π  ⋅ sin  =  − sin 2 4 



1  7 π 5π   7 π 5π   cos  + + cos  −    2 8  8   8  8

• •

Use cos θ = sin(90° - θ) Solve angle θ

•

Use sin 18° =

5 −1 4

31. Option (c) is correct. Explanation:

Given that,



tan θ = −



Let,







=

2

1 1 3 + = 2 4 4

Hints: • • •

p = 5k and b = 12k 2

2

2



H = p + b = 25k + 144 k = 13k Since value of tan θ is -ve that represent θ lies in 2nd and 4th quadrant.



∴

sin θ = ±



1 1 1  = + 0 +  2 2 2

5 p = 12 b

2

1 3π π 1   =  − ( −1 ) ⋅  + 2 cos 2 + cos 4  2   



p 5 = ± h 13

2

2

2



2

Use cos2 x - cos2 y = (-1)sin(x + y)⋅sin(x - y) Use 2cos x⋅cos y = cos(x + y) + cos(x - y) Use a2 + b2 = (a - b)2 + 2ab

33. Option (a) is correct. Explanation:

π  π  sin 2  + θ  − sin 2  − θ  4 4    







 π  π  sin  4 + θ  − sin  4 − θ       

 π  π  = sin  + θ  + sin  − θ   4 4     

2

2

51

SOLVED PAPER - 2022 (I)



π π π  π   4 + θ + 4 − θ  4 + θ − 4 + θ  ⋅ cos   = 2 sin  2 2 π π π  π   4 + θ + 4 − θ  4 + θ − 4 + θ     ⋅ 2 cos × sin 2 2







= 2 sin



π π ⋅ cos θ ⋅ 2 cos ⋅ sin θ 4 4

π π = 2 sin ⋅ cos ⋅ 2 sin θ ⋅ cos θ 4 4 = sin

π ⋅ sin 2θ = sin 2θ 2

Shortcut:

We, know that



sin2 x - sin2 y = sin(x + y)⋅sin(x - y)



π  π  ∴ sin 2  + θ  − sin 2  − θ  4  4 



π  π π π  = sin  + θ + − θ  ⋅ sin  + θ − + θ  4 4 4  4 



= sin



BC PC ⇒ PC = BC cot θ



In ∆APC







⇒



⇒



⇒



[ tan θ⋅cot θ = 1]



⇒



⇒ BC =



tan θ =

tan 2θ =

BC + h PC

tan 2θ⋅PC = BC + h

2 tan θ 1 − tan 2 θ 2 1 − tan 2 θ

⋅ cot θ BC = BC + h BC - BC = h

1 + tan 2 θ 1 − tan 2 θ

BC = h

1 − tan 2 θ



1 + tan 2 θ ⇒ BC = hcos 2θ





π ⋅ sin 2θ = sin 2θ 2

 1 − tan 2 θ  ∵ cos 2θ =  1 + tan 2 θ   Hints: •

Draw diagram condition

•

Use tan θ⋅cot θ = N

•

Use cos 2θ =

Hints: • Use a2 - b2 = (a + b)(a + b) A+B A−B ⋅cos 2 2

• Use sin A + sin B = 2sin

A+B A−B ⋅sin 2 2

and sin A - sin B = 2cos • Use sin 2θ = 2sin θ⋅cos θ

according

1 + tan 2 θ

35. Option (b) is correct. Explanation:

Let, AB be a tower of height

Explanation:



In ∆ABP





Let, BC is vertical tower and AB is flag staff of height h.

tan 60° =

3x PB

A

3 x metre.

A

h

3x

B

P



In ∆PBC

 2

to

1 − tan 2 θ

34. Option (c) is correct.

⋅h

Q

C



⇒

3=

x P

3x PB

B

given

52 Oswaal NDA/NA Year-wise Solved Papers

⇒ PB = x



In ∆ABQ



tan θ =







1


0 (a) ∴ (0, 0) not lies in the common region (b) -2 + 4 = 2, So, (-2, 4) not lies in the common region (c) -1 + 4 > 2, So, (-1, 4) not lies in the common region (d) -1 + 2 < 2, and 5(-1) -4(2) = -13 < 0

Explanation:



1310 = (1101)

So,

3

1

3

x + y = (100010111)2









= 1 × 28 + 1 × 2 4 + 1 × 2 2 + 1 × 2 1 + 1 × 2 0





= 256 + 16 + 4 + 2 + 1 = 279



⇒ x3 + y3 = 279















Now, (x - y)2 + xy = x2 + y2 - 2xy + xy





So, (-1, 2) lies in the common region Hints: •

Substitute all given option in all given in equations and check validity

56. Option (a) is correct.

x + y = (11111)2 = 1 × 24 + 1 × 23 + 1 × 22 + 1 × 2 1 + 1 × 2 0

Explanation: y = {x}, x ∈(-1, 1)

= 16 + 8 + 4 + 2 + 1 = 31

−1 x ∈ ( −1, 1) f(x) =  x ∈ [0, 1) 0

∴

= x2 + y2 - xy





=





=

( x + y ) ( x 2 − xy + y 2 ) x+y 279 =9 31

Division by 2 9 2 4 2 2 2

Quotient 4

2

1

1

x3 + y3 = x+y

Remainder 1

0

0

–1

0 0.5 1 –1



It is clear from figure that f(x) is differentiable at x = 0.5 and f′(x) = 0



But discontinuous at x = 0 Hints: •

Use [x] = -1, x ∈ (-1, 0)



and [x] = 0, x ∈ [0, 1)

•

If L.H.L = R.H.L = f(a) then f(x) is continuous at n = a

•

If L.H.D = R.H.D then f′(x) is exists.

57

SOLVED PAPER - 2022 (I)

57. Option (d) is correct.



∴ 1 - (x - 1)2 ≥ 0

Explanation:



⇒ 0 - x2 + 2x ≥ 0



⇒

x2 - 2x ≤ 0



⇒

x(x - 2) ≤ 0





+ – 0



∴ Domain = [0, 2]

4



2  d2 y 3  dy  1+  =  2   dx   dx   



3

  dy 2   d2 y  1 +    =  2   dx    dx    



⇒



∴ Degree = 4

4

• Degree of differential equation is the power of highest order derivative when differential equation in the form of polynomial.

Explanation:





⇒



⇒ ln p = -kt + c When, t = 0 c = lnp



Put,



•

To solve in equation use wavy curve method

Required area = 2.2 k

dp α−p dt





Use y =

dp ∫ p = −k ∫ dt

p 2

p ln   = -100 k + lnp 2 ⇒ lnp - ln2 = -100k + lnp

∴



3 2  k ⋅ x 2  = 6 3   0



⇒



⇒



⇒

k2 = 9



⇒

k = 3

1

ln 2 k = 100

3

k2 ⋅k2 = 6 ×

dp = rate of change of radio active dt substance

Use

•

Take negative sign for decays

•

Use variable separable to solve it

f(x) =

3 =9 2

[ k > 0]

Hints:

Explanation:

11 (k, 0)

0

•

Focus of parabola y2 = 4kx is (k, 0)

•

Equation of latus rectum is x = k

61. Option (b) is correct. Explanation: π 4 0

dx

π



∫ ( sin x + cos x )2

=





=

59. Option (d) is correct.

x dx = 24

11

Hints: •

∫

k

t = 100 and substance

⇒

k

0





f ( x ) is define when f(x) ≥ 0

•

60. Option (c) is correct.



dp ⇒ = -kp dt (where p is radio active substance initially)





2

Explanation:

58. Option (a) is correct. Given that

+

Hints:

Hints:



–

1 − ( x − 1)

2

1 4 dx ∫ 2 0 2  1 1  sin x + cos x   2  2  π

1 4 2 ∫0

dx π  cos  − x  4   2

58 Oswaal NDA/NA Year-wise Solved Papers π



1 4 π  sec 2  − x  dx ∫ 0 2 4 



=



1 π  4 = − tan  − x   2 4 0



= −



1 1 [0 − 1] = 2 2

Hints: •







 e2x 1  + 2x Now, I1 + I2 = ∫  2 x  dx  e +1 e +1





=





= ∫ 1 dx = x + c

=

∫

Use a   sin x  2  2 a +b  a2 + b 2    b + cos x   2 2   a +b

asin x + bcos x =



I=





=

=



=

∫ ( sin x )

∫ ∫



Let, tan x = t



⇒

−

1 2

( cos x )

−3 2

1

∫



2 tan x

1 ex

e2x

∫ e 2 x + 1 dx

e2x + 1

∫ e 2 x + 1 dx

64. Option (b) is correct.



1

ex +

dx =

• Simplify function in I1 and add I1 + I2.

Explanation:

ex

Hints:

62. Option (b) is correct.



∫ e x + e −x dx

I1 =

π



ex



sin x ⋅ cos x ⋅ cos x

Explanation: dx

dx

1 cos x sec 2 x tan x

sin x ⋅ cos x ⋅ cos x cos2 x

 x, x ≥ 0 −x , x < 0



 |x| = 



∴ I =

−1

∫

−2

dx

dx



=



x dx = x

−1

x

∫ −x dx

−2

−1

∫ ( −1) dx

−2





= ( −1 ) [ x ]−2





= (-1) [-1 + 2] = -1

−1

Hints: •

⋅ sec2 x dx = dt

 x, x ≥ 0 |x| =  −x , x < 0

65. Option (b) is correct. 2

sec x



⇒



I = 2 ∫ dt = 2t + c = 2 tan x + c

tan x

dx = 2dt

Hints: •

Make positive power

•

Convert given function in the form of tan x and sec x.

•

By substitution solve it

63. Option (b) is correct. Explanation:

Explanation: f(x) = sin 4x + 2x



Let















∴ 4x = π −





f′(x) = 4cos 4x + 2 = 0 cos 4x = −

x=

1 2

π   ∵ 0 < x < 2    ⇒ 0 < 4 x < 2 π 

π π or 4x = π + 3 3

π π or 6 3

59

SOLVED PAPER - 2022 (I)

Hints: • For extreme value f′(x) = 0



⇒

f(x) = x2





f(2) = (2)2 = 4 Hints:

66. Option (b) is correct. Explanation:

1 1 f(x) = = sin x cos x tan x + cot x + cos x sin x



sin x cos x





=





=



We know that 0 ≤ sin 2x ≤ 1



⇒

≤

2

2

sin x + cos x

= sin x⋅cos x

1 sin 2x 2



⇒ 0 ≤ f(x) ≤



1 ∴ Maximum value = 2

1 2

Use -1 ≤ sin θ ≤ 1

67. Option (d) is correct. Explanation:

Given that,



1  1 1  4f(x) - f   =  2 x +  2 x −  x x x      = 4x2 −

1

Find f(x)

1 and further solve it x

68. Option (a) is correct.

Explanation:



Given that, f(x) = 4x + 3



∴





= f(f(4(-1) + 3)) = f(f(-1))





= f(-4 + 3) = f(-1) = -4 + 3 = -1

fofof(-1) = f(f(f(-1)))

Use fofof(x) = f(f(f(x)))

69. Option (a) is correct.

Hints: •

•

•

π  ∵ 0 < x < 2   



Replace x by

Hints:

1 1 sin 2x ≤ 2 2



•

Explanation: xy yx = 1



Given that,



Taking log both side



ln(xy yx) = ln 1



ln xy + ln yx = 0







Differentiate w.r.t or x







Put, x = 1 and y = 1, we get







y ln x + x ln y = 0 dy y x dy ln x + + 1.ln y + =0 dx x y dx

dy dy ln 1 + 1 + ln 1 + =0 dx dx dy 0 + 1 + 0 + =0 dx dy = -1 dx









Replace x by



4 1 4 f   − f ( x ) = 2 − x 2 …(ii) x x



from 4(i) + (ii)

70. Option (b) is correct.

4 1 16 f ( x ) − 4 f   =16 x 2 − 2 x x 4 1 4 f   − f (x ) = 2 − x2 x x

Explanation:



x2

…(i)

Hints: • Taking log both sides

1 x

15 f = ( x ) 15x

• Use logrithumic properties

2



log (m ⋅ n) = log m + log n log mn = nlog m

( )

y = xx

x







Taking log both side

= xx

2

60 Oswaal NDA/NA Year-wise Solved Papers

ln y = x2 ln x



Differentiate w.r.t or x





1 dy 1 = 2 x ln x + x 2 ⋅ …(i) y dx x



When x = 1, y = 1



∴



1

73. Option (d) is correct. Explanation: It is clear from graph that domain of f(x) is (-∞, ∞).

dy = 2 ln 1 + 1 dx

Hints: • Use (xm)n = xmn

y’



It is continuous function unique tangent can be drawn at x = 0.

• Taking log both sides

Hints: • Draw the graph of 10x and check statement.

• Use properties of logritham

log mn = nlog m

71. Option (d) is correct. Explanation:



-4 < x < -3



⇒

-3 < x + 1 < -2



So,





y = [x + 1 ] = -3 dy =0 dx



Explanation:

lim x3 (cosec x)2

n →0

= lim x ⋅





n →0

∵ lim

θ→0

Hints:

• If -4 < x < -3 then -3 < x + 1 < -2

•

Explanation:

74. Option (a) is correct.

Hints:

72. Option (d) is correct. dy = (ln 5)y dx



1 ∫ y dy = ( ln 5 ) ∫ dx



x

0

x’

dy =0+1=1 dx



10x

y

sin θ =1=1 θ

Use ∵ lim

θ→0

x2 sin 2 x

= 0.1 = 0

sin θ θ = lim =1 θ→0 sin θ θ

75. Option (c) is correct. Explanation:

lim

x →1

x3 − 1 x −1

= lim

( x − 1) ( x 2 + x + 1) x −1

x →1

×

( x − 1) ( x 2 + x + 1) (





= lim





= (1 + 1 + 1)

x +1 x +1 x +1

)

x −1



ln y = (ln 5)x + c



put x = 0, y = ln5







∴ lny = (ln5)x + ln(ln5)

Hints:



Put x = 1

•

Rationalize the denominator



lny = ln5 + ln(ln5)

•

Factorize numerator by using formula



lny = ln[5⋅ln5]



a3 - b3 = (a - b) (a2 + ab + b2)





ln(ln 5) = c

y = 5ln 5 Hints: •

Solve differential equation by variable separable method

x →1

(

)

1 +1 = 6

76. Option (b) is correct. Explanation:



f(x) =

x3 7x2 − + 6x + 5 3 2

61

SOLVED PAPER - 2022 (I)



f′(x) = x2 - 7x + 6 = 0

2

x - 6x - x + 6 = 0







x(x - 6) -1(x - 6) = 0



(x - 6) (x - 1) = 0







x = 1 or x = 6 – + + – – 6 1

∴ f(x) is decreasing on (1, 6).



Required area





1 1 =  ∫ x dx − ∫ x 3 dx   0  0





 2 x =   2 





1 1 1 =  −  = square unit 4 2 4 Hints:

Hints: •

Equate f′(x) = 0 and find critical point

•

If f′(x) > 0 in interval (a, b) then increasing

•

If f′(x) < 0 in interval (a, b) then decreasing.

77. Option (d) is correct. Explanation:

Given that, f′(2) = 0











f(x) =

m + 2nx + 1 x

f′(x) = −

m x2

+ 2n

m f′(2) = − + 2n = 0 4 8n - m = 0 Cannot be determined value of m + 8n due to insufficient data. Hints: •

f(x) is vanish at x = 2 i.e., f′(2) = 0

78. Option (b) is correct.



•

Find inter section point of given curves

xy = 4225







⇒



Let, S = x + y = x +







⇒

x2 = 4225



⇒

x = 65







∴ x + y is minimum at x = 65



Minimum value = 65 +

y=

4225 x 4225 x

dS 4225 =1=0 dx x2

d 2S dt

2

=

( x, y ∈ N)

2 × 4225 x

3

=

2 × 4225

( 65 )3

>0

4225 = 130 65

Hints: 3

y = x …(ii) y



Draw the graph y = x and y = x3

Explanation:

y = x …(i)



•

79. Option (a) is correct.

Explanation:

1 1   x4   −     0  4  0 

•

Equate f′(x) = 0 and find critical point

•

If f″(x) > 0 at critical point then f(x) is minimum.

80. Option (c) is correct. Explanation:

–1 x´

0

1

x



x



⇒



⇒

y´



Solving equation (i) and (ii)



We get x = 0, -1, 1



dy - 2y = 0 dx dy 2y = dx x

1 dx x lny = ln x2 + ln c = ln cx2 1

∫ y dy

= 2∫

62 Oswaal NDA/NA Year-wise Solved Papers

y = cx2 Represent a family of parabolas



Since it is passes through (1, -2)



∴ 1 - 2 = a

Hints:



⇒

•



∴ Required equation is





Solve differential equation by variable separable method.

a = -1 x+y+1=0

81. Option (c) is correct.

Hints:

Explanation:

•



Given that, (-5, 0), (5p2, 10p) and (5q2, 10q) are collinear. −5



∴

0

1

intercept is

5p

10 p 1 = 0

Explanation:

5q

2

10 q 1



50 p 2

⇒

q

2



Since, circle which touches both the axes in the first quadrant and the line y - 2 = 0.

p 1 =0

y=2

2

q 1

1

Applying C1 → C1 + C3 0

x y + =1 a a

83. Option (c) is correct.

2

−1 0 1

Equation of straight line cuts off equal

1 1

0 1

0

x´

x

p2 + 1 p 1 = 0

⇒

q2 + 1 q 1

⇒

q(p2 + 1) - p(q2 + 1) = 0



⇒

p2q + q - pq2 - p = 0



⇒

p2q - pq2 + q - p = 0



⇒

pq(p - q) - (p - q) = 0



⇒ (p - q)(pq - 1) = 0



⇒

pq - 1 = 0



⇒

pq = 1

y1

1

x2

y2

1 =0

x3

y3

1

Then centre of circle is (1, 1) and radius is 1 unit.



∴ Equation of circle is



(x - 1)2 + (y - 1)2 = 1



⇒

x2 + y2 - 2x - 2y + 1 = 0

Hints:

( p ≠ q)

Hints: • (x1, y1), (x2, y2) and (x3, y3) is collinear if x1



•

When circle touches x-axis and y = 2 then diameter is 2 units

•

Equation of circle with centre (h, k) and radius r is (x - h)2 + (y - k)2 = r2

84. Option (d) is correct. Explanation:

Given that, focus (-3, 0) and directrix







∴

82. Option (c) is correct.



∴ Equation of parabola is

Explanation:





y2 = 4(-3)x



Since, straight line cuts off equal intercepts from the axes.



⇒

y2 = -12x



∴



⇒

x y + =1 a a x+y=a

x - 3 = 0 of parabola a = -3 and axis is x-axis

Hints: •

Focus and direction of parabola y2 = 4ax is (a, 0) and x - a = 0 respectively.

63

SOLVED PAPER - 2022 (I)

85. Option (b) is correct.

Explanation:

Explanation:



Equation of straight line passes through point of intersection of two given lines is







⇒ (1 + 2λ)x + (2 - 3λ)y + (2 - 3λ) = 0



Since, it cuts equal intercepts in the fourth quadrant





x-intercept =





y-intercept =



∴ x-intercept = 1



Sum of absolute values of the intercepts = |-1| + |1| = 2



Given, equation ellipse is







x2 + 2y2 = 1 y2 x2 =1 + 1  1 2    2

⇒



∴

a = 1 and b =



We know that 2

2

1 2



b =a -c



⇒

1 = 1 - c2 2



⇒

c2 =

1 2

•

1

⇒



Distance between the foci = |2c|





2

1 2

=

2

Explanation:



Given that, a, b, c are sides of triangle ABC, perimeter = a + b + c = p and ar ∆ABC = ∆ = q We know that A ∆ tan = 2 s (s − a)







(where s = semiperimeter)







⇒

A s(s - a) tan =∆ 2

⇒

pp A  =q − a  tan  22 2 

⇒

p(p - 2a)

A = 4q 2

Hints: •

Use formula tan

87. Option (a) is correct.

− ( 2 − 3λ )

( 2 − 3λ )

= -1

Family of straight lines passes through the point of intersection of two given lines is L1 + λL2 = 0

88. Option (c) is correct. Explanation:

Since, lines ax + by + c = 0 and bx + ay + c = 0 are parallel.



∴

a b = b a



⇒

a2 = b2



⇒





86. Option (d) is correct.

1 + 2λ

Hints:



= 2⋅

− ( 2 − 3λ )

2



c= ±

x + 2y + 2 + λ(2x - 3y - 3) = 0

A ∆ = 2 s (s − a)

a2 - b2 = 0 ( a ≠ 0 and b ≠ 0)

Hints: •

Two lines a1x +b1y + c1 = 0 and a2x + b2y + c2 = 0 are parallel if

a1 a2

89. Option (a) is correct. Explanation:

Given, lines is x + y = p P

A

(h, k) p

B

=

b1 b2

.

64 Oswaal NDA/NA Year-wise Solved Papers

⇒

x y + =1 p p



∴ Centre of x2 + y2 + z2 - 2x - 3y - 4z = 0



 3 is  1,  .  2



Let (h, k) he the coordinate of line segment AB.



p p ∴  ,  = (h, k) 2 2

Explanation:



⇒





∴ equation of locus is y = x

 12  Since, B  0, , 0 lies on sphere  k 



⇒



 12   12  ∴ 0 +   + 0 - 0 - 3   - 0 = 0  k   k 







⇒

h=k

91. Option (b) is correct.

2

x-y=0

Hints: • •

x y + =1 a b Mid-point of (x1, y1) and (x2, y2) is Use intercept form of line

 x1 + x 2 y1 + y 2 ,  2  2

  . 

 12   12   k   k − 3 = 0    k=4

92. Option (b) is correct. Explanation:

90. Option (c) is correct.



Explanation:

Let, P(x, y), A(2a, 0) and B(0, 3a)



According to question 2

 3  Distance of  1, , 2  to the plane  2 









2

3 6 ( 1 ) + 4   + 3 ( 2 ) − 12 2 = 36 + 16 + 9



AP = BP ⇒ AP = BP



(x - 2a)2 + (y - 0)2 = (x - 0)2 + (y - 3a)2



⇒ x2 - 4ax + 4a2 + y2 = x2 + y2 - 6ay + 9a2



⇒ 4ax - 6ay + 5a2 = 0

Hints:



⇒ 4x - 6y + 5a = 0

• A perpendicular distance from point (x1, y1, z1) to the plane ax + by + cz + d = 0

Hints: •

(x

2

− x1

) + (y

2

− y1



For questions 91 to 93.



Given equation of plane is



6x + ky + 3z - 12 = 0



⇒



∴ A(2, 0, 0), B(0,

).

6

=



= 0.74

65

ax1 + by1 + cz1 + d a2 + b 2 + c 2

2

x y z + + =1 2 12 4 k

12 , 0) and C(0, 0, 4) k

We know that Centre of sphere



Distance between (x1, y1) and (x2, y2) is 2

=

x 2 + y2 + z2 + 2gx + 2fy + 2hz + c = 0 is (-g, -f, -h).

93. Option (d) is correct. Explanation:

Equation of line passes through (0, 0, 0) and  3   1, 2 , 2   







y−0 x−0 z−0 = = 3 1 2 2 6x = 4y = 3z

Hints: •

Equation of line passes through two point is

x − x1 x 2 − x1

=

y − y1 y 2 − y1

=

z − z1 z2 − z1

65

SOLVED PAPER - 2022 (I)



[For Q. 94 to Q. 95]



Equation of plane is



⇒

97. Option (c) is correct. 2x 2 y z + + =2 k 3 3



x y z + + =2 k 3 3 2 2 is passes through (2, 3, -6) is







⇒







∴



3x + 2y + z = 6

4 3 × 2 −6 + + =q k 3 3 4 =2 k k=2

Explanation:    Given that a , b and i are the position vectors of the vertices. A, B, C respectively of triangle ABC ∴ Centroide (G) of triangle ABC    a+b +c = 3     a+b +c  Now, AG = - a 3    b + c − 2a = 3

94. Option (a) is correct.

98. Option (c) is correct. Explanation: Statement 1         a ⋅ b + c = a ⋅b + a ⋅c

Explanation:



So, but product over vector addition is distribute





∴ Statement 1 is correct.



Statement 2        a × b + c = a × b + a × c (By properties)

2y z + =2 3 3

x+

(

Direction ratios of plane is 3, 2, 1

(

)

)

95. Option (b) is correct.



Explanation:



So, cross product over vector addition is distributive.



∴ Statement 2 is correct.



Statement 3          a × b × c = (a ⋅ c )b − a ⋅ b c





Given that, vectors 4ˆi + ˆj − 3kˆ and piˆ + qjˆ − 2 kˆ are collinear



∴



∴ Statement 3 is correct.



4 1 3 ∴ = = p q 2

99. Option (b) is correct.



⇒

4 3 1 3 = and = p 2 p 2



⇒

p=



3x + 2y + z = 6







x y z + + =1 2 3 6 ∴ p = 2, q = 3, r = 6



Now, p + q + r = 2 + 3 + 6 = 11

96. Option (c) is correct. Explanation:

8 and q = 2 3 3

Hints: •

If a1ˆi + b1 yˆ + c1 kˆ and a2ˆi + b2 yˆ + c 2 kˆ are collinear then

a1 a2

=

b1 b2

=

c1 c2

.

(

)

( )    ( a × b ) × c = ( a ⋅ c ) b − ( b ⋅ c ) a       a × (b × c ) ≠ (a × b ) × c

Explanation:       c Given that, a ≠ 0, b ≠ 0, c ≠ 0 and a × b =     ⇒ a ⊥ c and b ⊥ c    If vector b and c are given and vector c   is perpendicular to vector b then c is also perpendicular to all vector which is coplanar   with vector b , then a is not unique. So, statement 1 is correct.   If vectors a and b are given then cross product two vector is unique vector.

So, statement 2 is correct.

66 Oswaal NDA/NA Year-wise Solved Papers 100. Option (d) is correct. Explanation:   Given that, a = b = 1

Made

  a −b < 2



and



⇒

 2 a −b < 4



⇒

( a − b ) ⋅ ( a − b ) < 4



⇒

2 2   a + b - 2 a b cos 2θ < 4



⇒

1 + 1 - 2⋅1⋅1⋅cos 2θ < 4



⇒

-cos 2θ < 1



⇒ 2sin2 θ - 1 < 1



⇒ sin2 θ < 1



⇒





Median Mean

Left (Negative0 skewed frequency distribution. Mean < Median < Mode 103. Option (c) is correct. Explanation: Let x be the 5th observation

-1 < sin θ < 1

Hints: •





( a − b ) ⋅ ( a − b ) =



2+2+4+5+x -  5  

2

49 + x 2 ( 13 + x ) − 5 25 ⇒ 3.6 × 25 = 5(49 + x2) - (13 + x)2 2

2 2   a + b − 2 a b cos θ

Two digits taken from 1, 2, 3, 4, 5 in 5 C2 ways ∴ n(S) = 5 C 2 Product of two digits whose last digit is zero i.e. (2, 5), (4, 5), (5, 2), (5, 4) n(E) = 4 4 × 2! 2 4 ∴ p(E) = 5 = = 5.4 5 C2



3.6 =

⇒ 90 = 245 + 5x2 - 169 - x2 - 26x ⇒ 2x2 - 13x - 7 = 0 1 ⇒ x = 7, x = − 2 ( observation is positive) ∴ x=7 Hints:

Hints: •

2 2 + 2 2 + 4 2 + 52 + x 2 5



2   Use a = a ⋅ b

101. Option (c) is correct. Explanation:

∴ Variance =

When r things is taken out from n things is n Cr .

102. Option (d) is correct. Explanation: If the distribution of data is skewed to the left, then the mean is less than the median and median is less than mode.

∑ xi N

•

Mean =

•

Variance =

∑ xi 2  ∑ xi  −  N  N 

2

104. Option (d) is correct. Explanation:

Given that, a = 4 and d = 4



∴ S50 =







Mean of 50 terms =

50 50 [8 + 49(4)] = × 204 2 2

= 50 × 102 50 × 102 = 102 50

67

SOLVED PAPER - 2022 (I)

⇒ x1 + x2 + x3 + ….. + xn = 92n + 140 …(ii) From, (i) and (ii) 100x - 20 = 92n + 140 8n = 160 n = 20 ∴ x1 + x2 + x3 ………… + xn = 2000 - 20 = 1980 106. Option (d) is correct. Explanation:

Hints: •

Use sum of n term of A.P

•

Sn =

•

Mean =

n (2a + (n - 1)d) 2 Sum of observation Number of observation

105. Option (b) is correct. Explanation:





21 + 34 + 23 + 39 + 26 + 37 + 40 + 20 + 33 + 27 Mean = 10



300 = 30 10 Mean deviation from mean



= 21 − 30 + 34 − 30 + 23 − 30



= 1980 - 20 × 99 = 0 108. Option (c) is correct. Explanation: Sum of the deviations from y

+ 20 − 30 + 33 − 30 + 27 − 30

9+4+7+9+4 + 7 + 10 + 10 + 3 + 3 = 10 66 = = 6.6 10









Coefficient of mean deviation





=

∑ xi N

•

Mean deviation from mean =

•

Coefficient

of

mean

∑ xi − x N

deviation

•

Sum of deviation of n observations from ‘a’ is





= ∑ xi − an



[For Q. 109 to Q. 111] Given that, a = 4 and d = 3



∴



∑ xi = S51 = =



51 [8 + 50 × 3] 2

51 × 158 = 51 × 79 2

109. Option (d) is correct. =

Mean deviation Mean

⇒ 20y = 1800 = 90 Hints:

Mean deviation 6.6 = = 0.22 Mean 30

Mean ( x ) =

= 1980 - 20y = 180





Hints: •



Explanation: Sum of the deviations from 99

+ 39 − 30 + 26 − 30 + 37 − 30 + 40 − 30



x1 + x 2 + x3 ..... + xn n

1980 = = 99 20 107. Option (a) is correct.

=



Mean =

[For Q. 106 to Q. 108] According to question (x1 - 100) + (x2 - 100) + …..+ (xn - 100) = -20 ⇒ (x1 + x2 + x3 + ….. + xn) - n × 100 = -20 ⇒ x1 + x2 + x3 + ….. + xn = 100n - 20 ….(i) Similarly

Explanation:



Mean marks =

∑ xi n

51 × 79 = 79 51 110. Option (b) is correct.

=



Explanation: Median of the marks = middle term



=

( 51 + 1)th 2

term = 26th term

68 Oswaal NDA/NA Year-wise Solved Papers ∴ Median = T26 = 4 + 25 × 3 = 79 111. Option (b) is correct. Explanation: Sum of deviation from median = ∑ xi - Median × n

115. Option (d) is correct. Explanation: Given that, number of trial n = 6 3

= 51 × 79 - 51 × 79 = 0

Hint:

116. Option (b) is correct.

•

Explanation:

Use sum of deviation of x observation from ‘a’ = ∑ xi - n × a

112. Option (b) is correct. Explanation: n(S) = 90







G → Graduate



T → At least 3 years experience



∴



∴

0



1 2 P(x = 0) = 6 C0     3 3





( ) 36 2 p (G ∩ Τ) = = 90 5

n G ∩ Τ = 36



)

P ( x ≥ 1) = 1 - P ( x = 0 )





( )

•

( )

)

[For Q. 118 to Q. 120]

( )



n Τ ∩ G = 27, n G = 9 + 27 = 36



∴

(

)

p Τ ∩G Τ 27 3 p   = = = 36 4 p G G

Hint:

( )

P ( E ∩ F) E P  = P ( F) F

[For Q. 115 to Q. 117]



1 1 P(suffering from a disease) = P = 33 % = 3 3 P(not suffering from a disease) = q = 1 -

C.I

fi

xi

fixi

0-20

17

10

170

20-40

p+q

30

30p + 30q

40-60

32

50

1600

60-80

p - 3q

70

70p - 210q

90

1710

80-100 19





Binomial distribution p → probability of success

114. Option (d) is correct. Explanation:

(

64 665 = 729 729

n → no. of trial

)

p G∩Τ 36 4 G p  = = = 63 7 Τ p Τ

∴

•

= 1−

Hints: P(r) = nCrpr qn-r, r= 0, 1, 2, 3

n G ∩ Τ = 36, n Τ = 36 + 27 = 63

(

64 729

=



Explanation:

(

6

117. Option (b) is correct. Explanation:

113. Option (c) is correct.



3

160 1 2 ∴ P(x = 3) = 6 C3 ⋅     = 729 3 3



Total

1 2 = 3 3



68 + 2p - 2q

3480 + 100p - 180q

According to question ⇒

∑ fi = 68 + 2p - 2q = 120 p - q = 26

…(i)

69

SOLVED PAPER - 2022 (I)



Mean =

∑ fi xi ∑ fi



50 =

3480 + 100 p − 180 q 120



⇒ 6000 = 3480 + 100p - 180q



⇒ 5p - 9q = 126 Solving equation (i) and (ii) we get

p = 27, and q = 1 118. Option (c) is correct. 119. Option (a) is correct.

120. Option (b) is correct. Explanation: If the frequency of each class is doubled then

…(ii)



Mean =





∑ 2 fi xi 2 ∑ fi xi ∑ fi xi = = ∑ 2 fi 2 ∑ fi ∑ fi

= 50 (Mean remains same) Hints: • •

∑ fi xi ∑ fi If the frequency of each class multiply by Mean =

same constant then mean remains same.

NDA / NA

MATHEMATICS

National Defence Academy / Naval Academy Time : 2:30 Hour

Ii

question Paper

2022 Total Marks : 300

Important Instructions : 1. This test Booklet contains 120 items (questions). Each item is printed in Mathematics. Each item comprises four responses (answer's). You will select the response which you want to mark on the Answer Sheet. In case you feel that there is more than one correct response, mark the response which you consider the best. In any case, choose ONLY ONE response for each item. 2. You have to mark all your responses ONLY on the separate Answer Sheet provided. 3. All items carry equal marks. 4. Before you proceed to mark in the Answer Sheet the response to various items in the Test Booklet, you have to fill in some particulars in the Answer Sheet as per instructions. 5. Penalty for wrong answers : THERE WILL BE PENALTY FOR WRONG ANSWERS MARKED BY A CANDIDATE IN THE OBJECTIVE TYPE QUESTION PAPERS. (i) There are four alternatives for the answer to every question. For each question for which a wrong answer has been given by the candidate, one third of the marks assigned to that question will be deducted as penalty. (ii) If a candidate gives more than one answer, it will be treated as a wrong answer even if one of the given answers happens to be correct and there will be same penalty as above to that question. (iii) If a question is left blank, i.e., no answer is given by the candidate, there will be no penalty for that question.

1. How many four-digit natural numbers are there such that all of the digits are odd? (a) 625 (b) 400 (c) 196 (d) 120 n

2. What is ∑ 2 r C ( n, r ) equal to? r=0

n

(a) 2 (c) 22n

(b) 3n (d) 32n

3. If different Permutations of the letters of the word ‘MATHEMATICS’ are listed as in a dictionary, how many words (with or without meaning) are there in the list before the first word that starts with C ? (a) 302400 (b) 403600 (c) 907200 (d) 1814400 4. Consider the following statements : 1. If f is the subset of Z ×Z defined by f= {(xy, x – y); x,y ∈Z}, then f is a function from Z to Z. 2. If f is the subset of N×N defined by f ={(xy, x + y); x,y∈N},then f is a function from N to N. Which of the statements given above is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2

5. Consider the determinant a11 a12 a13 ∆ = a21 a22 a23 a31 a32 a33 If a13= yz, a23=zx, a33=xy and the minors of a13, a23, a33 are respectively (z – y), (z – x), (y – x)then what is the value of ∆ ? (a) (z – y)(z – x)(y – x) (b) (x – y)(y – z)(x – z) (c) (x – y)(z – x)(y – z)(x + y + z) (d) (xy + yz + zx)(x + y + z) 0 0  1   6. If A =  0 cos θ sin θ  , Then which of the  0 sin θ − cos θ    following are correct ? 1. A + adjA is a null matrix 2. A-1 + adjA is a null matrix 3. A – A-1 is a null matrix Select the correct answer using the code given below: (a) 1 and 2 only (b) 2 and 3 only (c) 1 and 3 only (d) 1, 2 and 3

71

SOLVED PAPERS: 2022 (II) 7. If X is a matrix of order 3 × 3, Y is a matrix of order 2 × 3 and Z is a matrix of order 3 × 2, then which of the following are correct ? 1. (ZY)X is a square matrix having 9 entries. 2. Y(XZ)is a square matrix having 4 entries. 3. X(YZ)is not defined. Select the correct answer using the code given below: (a) 1 and 2 only (b) 2 and 3 only (c) 1 and 3 only (d) 1, 2 and 3 8. For how many quadratic equations, the sum of roots is equal to the product of roots ? (a) 0 (b) 1 (c) 2 (d) Infinitely many 9. Consider the following statements: 1. The set of all irrational numbers between 2 and 5 is an infinite set. 2. The set of all odd integers less than 100 is a finite set. Which of the statements given above is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 10. Consider the following statements : 1. 2+4 + 6 + ... + 2n = n2 + n 2. The expression n2+ n + 41 always gives a prime number for every natural number n Which of the above statements is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 11. Let p,q(p>q) be the roots of the quadratic equation x2+ bx + c = 0 where c > 0. If p2 + q2– 11pq = 0, then what is p –q equal to ? (b) 3c (a) 3 c (d) 9c (c) 9 c 12. What is the diameter of a circle inscribed in a regular polygon of 12 sides, each of length 1 cm? (b) 2 + 2 cm (a) 1 + 2 cm (d) 3 + 3 cm (c) 2 + 3 cm 13. Let A = {7, 8, 9, 10, 11, 12, 13, 14, 15, 16} and let f: A → N be defined by f(x) = the highest prime factor of x. How many elements are there in the range of f? (a) 4 (b) 5 (c) 6 (d) 7 14. Let R be a relation from N to N defined by R= {(x,y):x,y ∈N and x2=y3}.Which of the following are not correct ? 1. (x, x) ∈R for all x ∈N

2. (x,y)∈R ⇒ (y,x)∈R 3. (x,y) ∈ R and(y,z) ∈R⇒(x, z)∈ R Select the correct answer using the code given below: (a) 1 and 2 only (b) 2 and 3 only (c) 1 and 3 only (d) 1, 2 and 3 15. Consider the following : 1. A ∩ B = A ∩ C ⇒ B = C 2. A ∪ B = A ∪ C⇒B = C Which of the above is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 Consider the following for the next three (03) items that follow : 1 + i sin θ where i = −1 Let z = 1 − i sin θ 16. What is the modulus of z? (a) 1

(b)

(c) l + sin2θ

(d)

2 1 + sin 2 θ 1 − sin 2 θ

17. What is angle θ such that z is purely real? ( 2n + 1) π nπ (a) (b) 2 2 (c) nπ (d) 2nπ only where n is an integer 18. What is angle θ such that z is purely imaginary? ( 2n + 1) π nπ (a) (b) 2 2 (c) nπ (d) 2nπ where n is an integer Consider the following for the next three (03) items that follow : Let P be the sum of first n positive terms of an increasing arithmetic progression A. Let Q be the sum of first n positive terms of another increasing arithmetic progression B. Let P: Q = (5n + 4): (9n + 6) 19. What is the ratio of the first term of A to that of B? (a) 1/3 (b) 2/5 (c) 3/4 (c) 3/5 20. What is the ratio of their 10th terms ? (a) 11/29 (b) 22/49 (c) 33/59 (d) 44/69 21. If d is the common difference of A, and D is the common difference of B, then which one of the following is always correct ? (a) D > d (b) D< d (c) 7D >12d (d) None of the above

72

Oswaal NDA/NA Year-wise Solved Papers



Consider the following for the next three (03) items that follow : Consider the Binomial Expansion of (p + qx)9: 22. What is the value of q if the coefficients of x3 and x6are equal ? (a) p (b) 9p 1 (c) (d) p2 p

a21 a31 a11 30. What is the value of a23 a33 a13 ? a22 a32 a12

23. What is the ratio of the coefficients of middle terms in the expansion (when expanded in ascending powers of x)? (a) pq (b) p/q (c) 4p/5q (d) 1/(pq) 24. Under what condition the coefficients of x2and x4are equal ? (a) p:q = 7:2 (b) p2:q2 = 7:2 (c) p : q = 2 : 7 (d) p2: q2 = 2 : 7 Consider the following for the next three (03) items that follow : Consider the word ‘QUESTION’: 25. How many 4-letter words each of two vowels and two consonants with or without meaning, can be formed ? (a) 36 (b) 144 (c) 576 (d) 864 26. How many 8-letter words with or without meaning, can be formed such that consonants and vowels occupy alternate positions ? (a) 288 (b) 576 (c) 1152 (d) 2304 27. How many 8-letter words with or without meaning, can be formed so that all consonants are together ? (a) 5760 (b) 2880 (c) 1440 (d) 720 Consider the following for the next three (03) items that follow : Let ∆ be the determinant of a matrix A, where



 a11 a12 a13    A   a21 a22 a23  and C11, C12, C13 be the cofactors  a31 a32 a33 

of a11, a12, a13 respectively. 28. What is the value of a11C11 + a12C12 + a13C13? (a) 0 (b) 1 (c) ∆ (d) –∆ 29. What is the value of a21C11 + a22C12 + a23C13? (a) 0 (b) 1 (c) ∆ (d) –∆



(a) 0 (b) 1 (c) ∆ (d) –∆ Consider the following for the next three (03) items that follow: Let f(x) be a function satisfying f(x + y) = f(x)f(y) for all x,y ∈ N such that f (1) = 2 : n

31. If ∑ f ( x ) = 2044 , then what is the value of n ? x=2

(a) 8 (c) 10

(b) 9 (d) 11 5

32. What is

∑ f ( 2x − 1) equal to?

x =1

(a) 341 (c) 1023 33. What is

6

∑2

(b) 682 (d) 1364 x

f ( x ) equal to?

x =1

(a) 1365 (b) 2730 (c) 4024 (d) 5460 Consider the following for the next three (03) items that follow : A university awarded medals in basket ball, football and volleyball. Only x students (x y). The angles of elevation of the top of the tower from P and Q are 15° and 75° respectively. 43. At what height is the top of the tower above the ground level ? x-y x-y (a) (b) 2 3 4 3 (c)

x-y x-y (d) 4 2 44. If θ is the inclination of the tower to the horizontal, then what cotθ equal to? 3 (x − y) 3 (x − y) (a) 2 + x + y (b) 2 − x + y (c)

(c) 2 +

x −y



3 (x + y)

(d) 2 −

x −y

45. What is the length of the tower ? (a)



(c)

x−y 2 3 x−y

(b)

3 (d) 1 4

Consider the following for the next three (03) items that follow ABC is a triangular plot with AB= 16 m, BC = 10 m and CA = 10 m. A lamp post is situated at the middle point of the side AB. The lamp post subtends an angle 45° at the vertex B. 40. What is the height of the lamp post ? (a) 6 m (b) 7 m (c) 8 m (d) 9 m

3 (x + y)

2 3 x−y



(d)

4 3 x−y



4 3

 1 + 2 + 

3 ( x + y )   x − y 

 3 ( x + y )  1 + 2 −  x − y    1 + 2 + 

2

3 ( x + y )   x − y 

 3 ( x + y )  1 + 2 −  x − y  

2

2

2

46. What is the value of cosec  − 73π  ? 3   2 2 (b) (a) 3 3 (c) 2

(d) -2

74

Oswaal NDA/NA Year-wise Solved Papers

47. What is the value of  5π   7π   11π  π  cos  cos   ? + cos  + 2 cos      17   17   17   17  (a) 0 (b) 1 π   6π  cos   (c) 4 cos    17   17   11π  π  (d) 4 cos   cos  17   17    3π  48. What is the value of tan  ?  8  (a) 2 - 1 (b) 2 + 1 (c) 1 - 2

(d) −

(

2 +1

)

49. What is tan–1 cot(cosec–1 2) equal to ? π π (a) (b) 8 6 (c)

π 4

(d)

π 3

50. In a triangle ABC, a = 4, b = 3, c = 2. What is cos3C equal to ? (a) 7 (b) 11 128 128 7 (c) (d) 11 64 64 51. What is cos36° - cos72° equal to ? (a) 5 (b) - 5 2 2 1 1 (c) (d) 2 2 25 52. If sec x = and x lies in the fourth quadrant, 24 then what is the value of tan x + sin x ? (a) - 625 168 625 (c) 168

(b) - 343 600 343 (d) 600

53. What is the value of tan2 165° + cot2 165°? (a) 7 (b) 14 (c) 4 3 (d) 8 3 54. What is the value of 5π  5π    sin  2nπ + sin  2nπ − , where n ∈Z ?  6  6    3 1 (a) – (b) – 4 4 1 3 (c) (d) 4 4

55. If l + 2 (sin x + cos x)(sin x – cos x) = 0 where 0 1 1 2 2 1 1 , , , , (c) (d) 6 6 6 6 6 6 65. Consider the following statements : 1. The direction ratios of y-axis can be < 0, 4, 0> 2. The direction ratios of a line perpendicular to z-axis can be < 5, 6, 0 > Which of the statements given above is /are correct ? (a) 1 only (b) 2 only (d) Neither 1 nor 2 (c) Both 1 and 2     66. PQRS is a parallelogram. If PR = a and QS = b ,  then what  is PQ equal to?   (a) a + b (b) a - b   a+b a-b (c) (d) 2 2     67. Let a and bare two unit vectors such that a + 2b  and 5a - 4b are perpendicular. What is the angle between a and b ? π π (a) (b) 6 4 (c)

π 3

(d)

π 2

   68. Let a , b and c be unit vectors on the same  lying     plane. What is 3a + 2b × ( 5a − 4 c ) . b + 2 c equal to ? (a) -8 (b) -32 (c) 8 (d) 0 69. What are the values of x for which the angle 2 between the vectors 2 x ˆi + 3xjˆ + kˆ and 2 ˆi − 2 ˆj + x kˆ is obtuse ?

{(

}(

)

)

(b) x< 0 (a) 0 2 (d) 0 ≤ x ≤ 2 70. The position vectors of vertices A, B and C of triangle ABC are respectively ˆj + kˆ, 3ˆi + ˆj + 5kˆ and 3ˆj + 3kˆ. What is angle C equal to ? π π (a) (b) 6 4 (c)

π 3

(d)

π 2

71. Let z = [y] and y = [x] – x, where [.] is the greatest integer function. If x is not an integer but positive, then what is the value of z ? (a) –1 (b) 0 (c) 1 (d) 2 72. If f(x) = 4x + 1and g(x) = kx + 2such that fog(x) = gof(x), then what is the value of k ? (a) 7 (b) 5 (c) 4 (d) 3 73. What is the minimum value of the function f(x) = log10(x2 + 2x + 11) ? (a) 0 (b) 1 (c) 2 (d) 10

( ) (1 + lnx) dx equal to ?

x 74. What is ∫ x

2

(b) 1 x2x + c 2 2x (c) 2x + c (d) 1 xx+ c 2 x 75. What is  e {1 + lnx + xlnx}dx equal to? (a) xexlnx + c (a) x2exlnx + c (c) x + exlnx + c (d) xex + lnx + c (a) x2x + c

( cos x )

1.5

76. What is ∫

− ( sin x )

1.5

sin x. cos x

(a)

sin x − cos x + c

(b)

sin x + cos x + c

(c) 2 sin x + 2 cos x + c (d)

1 1 sin x + cos x + c 2 2

dxequal to?

76

Oswaal NDA/NA Year-wise Solved Papers

77. If y =

x x 2 − 16 2

− 8 ln x + x 2 − 16 , then what

dy equal to ? dx 2 (a) x x - 16

is

2 (b) x - x - 16

2

2 (d) 4 x - 16 (c) x - 16 78. If y = (xx)x, then which one of the following is correct ? dy (a) + xy (1 + 2lnx) = 0 dx dy – xy (1 + 2lnx) = 0 (b) dx dy (c) – 2xy (1 + lnx) = 0 dx dy + 2xy (1 + lnx) = 0 (d) dx 79. What is the maximum value of 3 (sin x - cos x) + 4(cos3 x – sin3 x) ? (a) 1 (b) 2 (c) 3 (d) 2

80. What is the area of the region (in the first quadrant) bounded by y = 1 − x 2 y = x and y = 0? (a)

π 4

(b) π 6 π (d) 12

(c) π 8 81. What is the area of the region bounded by x – |y| = 0 and x – 2 = 0 ? (a) 1 (b) 2 (c) 4 (d) 8 f (α ) + f ( β ) 82. If f(α) = sec 2 α − 1 , then what is 1 − f (α ) f ( β ) equal to? (a) f(α – β) (c) f(α) (β)

(b) f(α + β) (d) f(αβ)

2 83. If f(x) = ln(x + 1 + x ), then which one of the following is correct ? (a) f(x) + f(-x) = 0 (b) f(x) – f(–x) = 0 (c) 2f(x)= f(-x) (d) f(x) = 2f(-x)

(c) 2 (d) Limit does not exist 4 x − 2π 85. What is lim equal to ? π cos x x→ 2

(a) –4 (c) 2

(b) –2 (d) 4

x2 + x + x

86. If f(x) = , then what is lim f ( x ) equal x→0 x to ? (a) 0 (b) 1 (c) 2 f ( x ) does not exist (d) xlim →0 87. What is lim

h→0

sin 2 ( x + h ) − sin 2 x h

equal to?

(a) sin2x (b) cos2x (c) sin 2x (d) cos 2x 88. Let f(x) be a function such that f ’(x)= g(x) and f ”(x) = -f(x). Let h(x)= {f(x)}2 + {g(x)}2. Then consider the following statements: 1. h(3)= 0 2. h(l) = h(2) Which of the statements given above is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 2 dy   1 x − x + 89. If y = ln 2  2 at x = 0  , then what is dx  x + x + 1 equalto?

(a) -2 (b) 0 (c) 1 (d) 2 4 8   d 1+x +x 3 90. If   = ax + bx , then which one of dx  1 − x 2 + x 4  the following is correct ? (a) a = b (b) a = 2b (c) a + b = 0 (d) 2a = b 91. Under which one of the following conditions does the functionf(x) = (p sec x)2 + (q cosec x)2 attains minimum value ? q q (a) tan2 x = (b) cot2 x = p p (c) tan2 x = pq (d) cot2 x = pq 7

84. What is lim

x→0

(a)

1

2 2 1 (b) 2 2

x 1 − cos 4 x

equal to ?

92. Where does the function f ( x ) = ∑ ( x − j ) j =1 attains its minimum value ? (a) x = 3.5 (b) x = 4 (c) x = 4.5 (d) x = 5

2

93. Consider the following statements in respect of  x + 1, 0 < x ≤ 3 the function f ( x ) =  x=0 1,

77

SOLVED PAPERS: 2022 (II) 1. The function attains maximum value only at x = 3 2. The function attains local minimum only at x = 0 Which of the statements given above is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 1  1 94. What is ∫0 ln  − 1  dx equal to? x  (a) -1 (b) 0 (c) 1 (d) ln2 95. If

∫ ( sin π /2

4

0

)

x + cos4 x dx = k ,  than what is the

20 π

value of ∫ 0 (a) k (c) 20k 96. What is to?

∫

( sin

π /2

−π / 2

4

(e

)

x + cos4 x dx ? (b) l0k (d) 40k

cos x

)

sin x + e sin x cos x dx equal

e2 - 1 e2 + 1 (b) e e 1 - e2 (c) (d) 0 e 97. What is the area of the region enclosed in the first quadrant by x2+ y2 = π2, y = sin x and x=0? 3 3 (a) π − 1 (b) π − 2 4 4 3 2 π π (c) (d) − 1 −2 2 4 98. Consider the following statements : 1. The degree of the differential equation dy  dy  + cos   = 0 is 1. dx  dx  (a)

2.



The order of the differential equation 3  d2 y   dy   2  + cos   = 0 is 2. dx  dx   

Which of the statements given above is/are correct? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 99. What is the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis ? dy dy (a) x (b) x + 2 y = 0 − 2y = 0 dx dx dy dy (c) y (d) y + 2 x = 0 − 2x = 0 dx dx

100. What is the solution of the differential equation (dy – dx) + cos x(dy + dx) = 0? x (a) y = tan  – x +c 2 1 x (b) y = tan  – x +c 2 2 x (c) y = 2tan  – x +c 2 x (d) y = tan  – 2x +c 2 101. Let x be the mean of squares of first n natural numbers and y be the square of mean of first x 55 n natural numbers.If = , then what is the y 42 value of n? (a) 24 (b) 25 (c) 27 (d) 30 102. What is the probability of getting a composite number in the list of natural numbers from 1 to 50? (b) 17 (a) 7 10 25 18 33 (c) (d) 25 50 103. If n >7, then what is the probability that C(n, 7) is a multiple of 7? 1 (a) 0 (b) 7 1 (c) (d) 1 2 104. Two numbers x and y are chosen at random from a set of first 10 natural numbers. What is the probability that (x + y) is divisible by 4 ? (a) 1 (b) 2 5 9 8 (c) (d) 7 45 45 105. A number x is chosen at random from first n natural numbers. What is the probability that 1 the number chosen satisfies x + > 2 ? x 1 1 (a) (b) ( 2n ) n (c)

( n − 1)

(d) 1 n 106. Three fair dice are tossed once. What is the probability that they show different numbers that are in AP ? (a) 1 (b) 1 12 18 1 (c) 1 (d) 72 36

78

Oswaal NDA/NA Year-wise Solved Papers

107. If P(A) = 0.5, P(B) = 0.7 and P(A∩B) = 0.3, than what is the value of P(A’∩B’) + P(A’∩B) + P(A∩B’)? (a) 0.6 (b) 0.7 (c) 0.8 (d) 0.9 108. Five coins are tossed once. What is the probability of getting at most four tails? (a) 31 (b) 15 32 16 29 (c) (d) 7 32 8 109. Three fair dice are probability of getting equal to 15 ? (a) 19 216 (c) 17 216

thrown. What is the a total greater than or (b) 1 12 (d) 5 54

110. The probability that a person hits a target is 0.5. What is the probability of at least one hit in 4 shots ? (a) 1 (b) 1 8 16 15 (c) (d) 7 16 8 111. A box contains 2 white balls, 3 black balls and 4 red balls. What is the number of ways of drawing 3 balls from the box with at least one black ball ? (a) 84 (b) 72 (c) 64 (d) 48 112. During war one ship out of 5 was sunk on an average in making a certain voyage. What is the probability that exactly 3 out of 5 ships would arrive safely ? (b) 32 (a) 16 625 625 64 128 (c) (d) 625 625 113. A card is drawn from a pack of 52 cards. A gambler bets that it is either a spade or an ace. The odds against his winning are (a) 9 : 4 (b) 35 : 17 (c) 17 : 35 (d) 4 : 9 114. The coefficient of correlation between ages of husband and wife at the time of marriage for a given set of 100 couples was noted to be 0.7. Assume that all these couples survive to celebrate the silver jubilee of their marriage.

The coefficient of correlation at that point of time will be (a) 1 (b) 0.9 (c) 0.7 (d) 0.3 115. The completion of a construction job may be delayed due to strike. The probability of strike is 0.6. The probability that the construction job gets completed on time if there is no strike is 0.85 and the probability that the construction job gets completed on time if there is a strike is 0.35. What is the probability that the construction job will not be completed on time ? (a) 0.35 (b) 0.45 (c) 0.55 (d) 0.65 Consider the following for the next two (02) items that follow: The mean and standard deviation (SD) of marks obtained by 50 students of a class in 4 subjects are given below : Subject Mean Marks

Mathematics 40

Physics Chemistry Biology 28

38

36

SD 15 12 14 16 116. Which one of the following subjects shows highest variability of marks ? (a) Mathematics (b) Physics (c) Chemistry (d) Biology 117. What is the coefficient of variation of marks in Mathematics ? (a) 37.5% (b) 38.0% (c) 38.5% (d) 39.0% Consider the following for the next three (03) items that follow :

Consider the following grouped frequency distribution : Class

0-10 10-20 20-30 30-40 40-50 50-60

2 4 6 4 3 Frequency 1 118. What is the median of the distribution? (a) 34 (b) 34.5 (c) 35 (d) 35.5 119. What is mean deviation about the median ? (a) 11.4 (b) 11.1 (c) 10.8 (d) 10.5 120. What is the mean deviation about the mean ? (a) 10.15 (b) 10.65 (c) 11.15 (d) 11.65 nnn

79

SOLVED PAPERS: 2022 (II)

Answers Mathematics Q No

Answer Key

Topic Name

Chapter Name

1

(a)

Permutation

Permutation and combination

2

(b)

Binomial Theorem for Positive Integral Index

Binomial theorem and its applications

3

(c)

Permutation

Permutation and combination

4

(d)

Basics of function

Functions

5

(a)

Determinant of a Square Matrix

Matrices and determinants

6

(d)

Inverse of a Matrix

Matrices and determinants

7

(d)

Algebra of Matrices

Matrices and determinants

8

(d)

Relation between Roots and Coefficients

Quadratic equations

9

(a)

Basics of Sets

Set Theory

10

(a)

Series of Natural Numbers and other Miscellaneous Series

Sequences and Series

11

(a)

Relation between Roots and Coefficients

Quadratic Equations

12

(c)

Basics of circle

Circle

13

(c)

Basics of function

Functions

14

(d)

Algebra of Relations

Relations

15

(d)

Algebra of Sets

Set Theory

16

(a)

Modulus and Argument of Complex Numbers

Complex numbers

17

(c)

Basics of Complex Numbers

Complex numbers

18

(b)

Basics of Complex Numbers

Complex numbers

19

(d)

Arithmetic Progressions

Sequences and Series

20

(c)

Arithmetic Progressions

Sequences and Series

21

(a)

Arithmetic Progressions

Sequences and Series

22

(a)

Binomial Theorem for Positive Integral Index

Binomial theorem and its applications

23

(b)

Properties of Binomial Coefficients

Binomial theorem and its applications

24

(b)

Binomial Theorem for Positive Integral Index

Binomial theorem and its applications

25

(d)

Permutations

Permutation and combination

26

(c)

Permutations

Permutation and combination

80

Oswaal NDA/NA Year-wise Solved Papers

Q No

Answer Key

Topic Name

Chapter Name

27

(b)

Permutations

Permutation and combination

28

(c)

Determinant of a Square Matrix

Matrices and determinants

29

(a)

Properties of Determinants

Matrices and determinants

30

(d)

Properties of Determinants

Matrices and determinants

31

(c)

Geometric Progressions

Sequences and Series

32

(b)

Geometric Progressions

Sequences and Series

33

(d)

Geometric Progressions

Sequences and Series

34

(c)

Algebra of Sets

Set Theory

35

(d)

Algebra of Sets

Set Theory

36

(a)

Algebra of Sets

Set Theory

37

(a)

Symmetric and SkewSymmetric Matrices

Matrices and determinants

38

(d)

Symmetric and SkewSymmetric Matrices

Matrices and determinants

39

(a)

Determinant of a Square Matrix

Matrices and determinants

40

(c)

Basics of Trigonometry

Trigonometric Ratios, Functions and Identities

41

(b)

Inradius, Exradii and Circumradius

Properties of Triangle

42

(d)

Relations Between Sides and Angles of a Triangle

Properties of Triangle

43

(a)

Heights and Distances

Heights and Distances

44

(d)

Heights and Distances

Heights and Distances

45

(b)

Heights and Distances

Heights and Distances

46

(b)

Trigonometric Functions and Trigonometric Ratios, Properties Functions and Identities

47

(a)

Trigonometric Identities

Trigonometric Ratios, Functions and Identities

48

(b)

Trigonometric Identities

Trigonometric Ratios, Functions and Identities

49

(d)

Properties of Inverse Trigonometric Functions

Inverse Trigonometric Functions

50

(a)

Relations Between Sides and Angles of a Triangle

Properties of Triangle

51

(c)

Trigonometric Identities

Trigonometric Ratios, Functions and Identities

52

(b)

Basics of Trigonometry

Trigonometric Ratios, Functions and Identities

53

(b)

Trigonometric Functions and Trigonometric Ratios, Properties Functions and Identities

54

(a)

Trigonometric Functions and Trigonometric Ratios, Properties Functions and Identities

81

SOLVED PAPERS: 2022 (II)

Q No

Answer Key

Topic Name

Chapter Name

55

(d)

Trigonometric Equations

Trigonometric Equations

56

(c)

Straight Line and its Equations

Point and Straight Line

57

(b)

Straight Line and its Equations

Point and Straight Line

58

(a)

Straight Line and its Equations

Point and Straight Line

59

(c)

Interaction between Circle and a Line

Circle

60

(a)

Basics of Ellipse

Ellipse

61

(c)

Basics of Parabola

Parabola

62

(b)

Point in Cartesian Plane

Point and Straight Line

63

(b)

Sphere

Three Dimensional Geometry

64

(c)

Planes in 3D

Three Dimensional Geometry

65

(c)

Direction Cosines and Direction Ratios

Three Dimensional Geometry

66

(d)

Addition of Vectors

Vector Algebra

67

(c)

Scalar and Vector Products

Vector Algebra

68

(d)

Triple Products

Vector Algebra

69

(a)

Scalar and Vector Products

Vector Algebra

70

(d)

Scalar and Vector Products

Vector Algebra

71

(a)

Types of Functions

Functions

72

(a)

Composite Function

Functions

73

(b)

Maxima and Minima

Application of Derivatives

74

(b)

Integration by Substitution

Indefinite Integration

75

(a)

Integration by Parts

Indefinite Integration

76

(c)

Integration by Substitution

Indefinite Integration

77

(c)

Basics of Indefinite Integrals

Indefinite Integration

78

(b)

Logarithmic Differentiation

Differential Coefficient

79

(b)

Range of Trigonometric Expressions

Trigonometric Ratios, Functions and Identities

80

(c)

Area Bounded by Curves

Area under curves

81

(c)

Area Bounded by Curves

Area under curves

82

(b)

Basics of Functions

Functions

83

(a)

Even and Odd Functions

Functions

84

(d)

Basics of Limits

Limits

85

(a)

Methods of Evaluation of Limits

Limits

86

(d)

Basics of Limits

Limits

87

(c)

Methods of Evaluation of Limits

Limits

88

(c)

Rules of Differentiation

Differential Coefficient

82

Oswaal NDA/NA Year-wise Solved Papers

Q No

Answer Key

Topic Name

Chapter Name

89

(b)

Rules of Differentiation

Differential Coefficient

90

(d)

Basics of Differentiation

Differential Coefficient

91

(a)

Maxima and Minima

Application of Derivatives

92

(b)

Maxima and Minima

Application of Derivatives

93

(b)

Maxima and Minima

Application of Derivatives

94

(b)

Basics of Definite Integrals

Definite Integration

95

(d)

Properties of Definite Integrals

Definite Integration

96

(a)

Properties of Definite Integrals

Definite Integration

97

(b)

Area Bounded by Curves

Area under curves

98

(b)

Basics of Differential Equations

Differential Equations

99

(b)

Formation of Differential Equations

Differential Equations

100

(c)

Variable Separable Form

Differential Equations

101

(c)

Series of Natural Numbers and other Miscellaneous Series

Sequences and Series

102

(b)

Basics of Probability

Probability

103

Bonus

Basics of Probability

Probability

104

(b)

Basics of Probability

Probability

105

(c)

Basics of Probability

Probability

106

(b)

Basics of Probability

Probability

107

(b)

Addition and Multiplication Theorems of Probability

Probability

108

(a)

Addition and Multiplication Theorems of Probability

Probability

109

(d)

Basics of Probability

Probability

110

(c)

Algebra of Probabilities

Probability

111

(c)

Combinations

Permutation and Combination

112

(d)

Bernoulli Trials and Binomial Probability Distribution

113

(a)

Basics of Probability

Probability

114

(c)

Corelation

Statistics

115

(b)

Total Probability Theorem

Probability

116

(d)

Measures of Dispersion

Statistics

117

(a)

Measures of Dispersion

Statistics

118

(c)

Measures of Central Tendency

Statistics

119

(d)

Measures of Dispersion

Statistics

120

(b)

Measures of Dispersion

Statistics

NDA / NA

MATHEMATICS

II

National Defence Academy / Naval Academy



Hint: Recall Fundamental principle of counting. 2. Option (b) is correct. Solution : As we know by n

binomial

theorem,

(1 + x ) = ∑ x c ( n, r ) n

r

r =0

Put x = 2 in above equation, we get n

(1 + 2)n = ∑ 2 r c( n , r )



r =0

⇒



n

∑ 2 r c( n , r ) = 3 n

r =0

n



2022

AnSWERS wITH eXPLANATION

1. Option (a) is correct. Solution : We have to form 4-digit natural numbers using 1, 3, 5, 7, 9. So for each digit we have 5 possible numbers. ⇒ Total number of 4-digit numbers possible =5×5×5×5 = 625



Solved Paper

Hint: Use (1 + x )n = ∑ x r n Cr r =0

3. Option (c) is correct. Solution : Given alphabets are MATHEMATICS. In the dictionary the words are arranged alphabetically. The words in the dictionary before the first word that starts with C will start with A. So we will fix letter A at first place and arrange 2M, A, 2T, H, E, I, C, S. 10 ! ⇒ Number of required words = 2!2! = 907200 Hint: The number of permutations of n things taken all at a time when ‘p’ of them



are of same, ‘q’ of them are of same and rest n! are different is p!q!

4. Option (d) is correct. Solution : Statement 1: f = {(xy, x – y), x, y ∈ z} Let a, b ∈ z and x = a, y = b ⇒ xy = ab and x – y = a – b So image of ab is a – b(1) Now, as a, b ∈ z ⇒ –a, –b ∈ z Let x = –a and y = –b ⇒ xy = (–a)(–b) = ab and x – y = (–a) – (–b) = –a + b So image of ab is –a + b(2) From (1) and (2), we get ab has two images a – b and –a + b. And as we know a function is a relationship between inputs and outputs where each input is related to exactly one output. So f is not a function. Similarly, Statement 2: f = {(xy; x + y), x, y ∈ N} Let a, b ∈ N ⇒ 4a, 4b ∈ N and x = 4a, y = 4b. ⇒ xy = 16ab and x + y = 4a + 4b. So image of 16ab is 4a +4b(3) Also 8a, 2b ∈ N and x = 8a, y = 2b ⇒ xy = 16 ab and x + y = 8a + 2b So image of 16 ab is 8a + 2b(4) From (3) and (4), f is also not a function. So both of the statements are incorrect. Hint: Use definition of function which states that function is a relationship between input and output such that each input is related to exactly one output.

84

Oswaal NDA/NA Year-wise Solved Papers

5. Option (a) is correct. Solution : a11 a12 a13 Given:   a21 a22 a23 a31 a32 a33

a13 = yz, a23 = zx, a33 = xy M13 = z – y, M23 = z – x, M33 = y – x ⇒ Cofactor of a13 = C13 = (–1)4M13 = (z – y) C23 = (–1)5M23 = –(z – x) C33 = (–1)6M33 = (y – x) Now, expanding the determinant along column 3 we get ∆ = a13C13 + a23C23 + a33C33 = (yz)(z – y) + zx(x – z) + xy (y – x) = z2y – y2z + x2z – z2x + xy2 – x2y = z2(y – x) – z (y2 – x2) + xy(y – x) = (y – x)(z2 – z(y + x) + xy) = (y – x)(z2 – zy – xz + xy) = (y – x)(z (z – y) – x (z – y)) = (y – x)(z – y)(z – x) = (z – y)(z – x)(y – x) Hint: Use ∆ = a13C13 + a23C23 + a33C33 where C denotes the cofactor.

6. Option (d) is correct. Solution : 0 0  1 Given: A  0 cos  sin   0 sin   cos  





0 0 0  Now, A+adj A  0 0 0  = Null Matrix 0 0 0  1

1  ( adj A ) | A|



Now, A



|A| = 1(–cos2θ – sin2θ) – 0 + 0 = –1









0 0 0  A 1  adj ( A )  0 0 0   Null matrix 0 0 0  AA



⇒ A 1

0 0   1   ( 1)  0  cos   sin    0  sin  cos  

1

0 0 0   0 0 0   Null matrix 0 0 0 

7. Option (d) is correct. Solution : Order of X is 3 × 3 Order of Y is 2 × 3 Order of Z is 3 × 2 Now, (ZY) X = ([ ]3×2 [ ]2×3) [ ]3×3 = ([ ]3×3)[ ]3×3 = [ ]3×3 (ZY) X has 9 entries. Similarly, Y(XZ) = [ ]2×3 ([ ]3×3 [ ]3×2) = [ ]2×3 [ ]3×2 = [ ]2×2 Y(XZ) has 4 entries Now, X(YZ) = [ ]3×3 ([ ]2×3 [ ]3×2) = [ ]3×3 [ ]2×2 X(YZ) is not defined as number of columns of the first matrix is not equal to the number of rows of the second matrix. Hint: The multiplication of two matrices is possible if number of columns of the first matrix is equal to the number of rows of the second matrix.

−1 0 0    adj A =  0 − cos θ − sin θ  0 − sin θ cos θ    0 0   1   ⇒ adj A   0  cos   sin    0  sin  cos  

0 0  1   0 cos  sin   0 sin   cos  

8. Option (d) is correct Solution: Let the quadratic equation be ax2 + bx + c = 0 and roots be a and β. b c ⇒   and   a a

Now, a + b = ab b c  ⇒ a a ⇒b+c=0 \ There are infinite possible values for which b + c = 0 so infinetly many quadratic equations possible.

85

SOLVED PAPER - 2022 (II) Hint: For a quadratic equation ax2 + bx + c = -b c 0 sum of roots = and product of roots = . a a 9. Option (a) is correct. Solution: Statement 1: The set of all irrational numbers between 2 and 3 is an infinite set. As we know, there are infinite irrational numbers between two irrational numbers. So Statement 1 is true. Statement 2: The set of odd integers less then 100 is a finite set. Let the given set be X. X ∈ {..., –5, –3, –1, 1, 3, 5, ... 99} So it is an infite set. So statement 2 is false. Hint: The number of irrational numbers between two irrational numbers are infinite. 10. Option (a) is correct Solution: 2 + 4 + 6 + ... + 2n = x x = 2(1 + 2 + 3 + ... + n)

2[n( n + 1)] x= 2



⇒ x = n(n + 1) ⇒ x = n2 + n Let y = n2 + n + 41 ⇒ y = n(n + 1) + 41 for n = 40 y = 40(40 + 1) + 41 ⇒ y = 40(41) + 41 ⇒ y = 40(41) + 41 y is not a prime number. So only Statement 1 is correct. Hint: Sum of first n natural numbers is n( n + 1) 2

11. Option (a) is correct Solution: p, q are the roots of x2 + bx + c = 0 p + q = –b and pq = c Now, p2 + q2 – 11pq = 0 ⇒ p2 + q2 – 2pq – 9pq = 0 ⇒ (p – q)2 – 9pq = 0

⇒ (p – q)2 = 9pq ⇒ (p – q)2 = 9c ⇒p–q= 3 c



Hint: Use sum of roots of ax2 + bx + c = 0 is -b c and product of roots is . a a 12. Option (c) is correct. Solution: Given a regular polygon of 12 sides interior angle of a regular polygon of n sides =



( n  2 )  180 n

So, interior angle of given regular polygon =



(12  2 )  180 = 150° 12

The interior angle is bisected by the radius of the inscribed circle. OAP 



150  75 2

O

A



P

B

Let OP = r and AB = 1 1 ⇒ AP = 2



\ In ∆OAP,. tan 75° = ⇒ 2  3  2r



Diameter = (2 +

r 12

3 ) cm

Hint: Interior angle of a regular polygon of n (n  2)  180 sides is n

13. Option (c) is correct. Solution: A = {7, 8, 9, 10, 11, 12, 13, 14, 15, 16} f: A → N, f(x) = The highest prime factor of x f (7) = 7 f (8) = 2 { 8 = 2 × 2 × 2} f (9) = 3 { 9 = 3 × 3} f (10) = 5 { 10 = 5 × 2)

86

Oswaal NDA/NA Year-wise Solved Papers f (11) = 11 { 11 = 11 × 1} f (12) = 3 { 12 = 3 × 2 × 2} f (13) = 13 { 13 = 13 × 1} f (14) = 7 { 14 = 7 × 2} f (15) = 5 { 15 = 5 × 3} f (16) = 2 { 16 = 2 × 2 × 2 × 2} Range of f = {2, 3, 5, 7, 11, 13} Number of elements = 6

Hint: Find highest prime factor of each element of A. 14. Option (d) is correct. Solution: R = {(x, y) : x, y ∈ N and x2 = y3} For x = y, x2 ≠ x3 ⇒ (x, x) ∉ R Now, (x, y) ∈ R ⇒ x2 = y3 ⇒ x3 = (x2)(x) ⇒ x3 = y3x ⇒ x3 ≠ y2 ⇒ ( y, x) ∉ R Now, (x, y) ∈ R and (y, z) ∈ R ⇒ x2 = y3 and y2 = z3 ⇒ x2 = y(z3) ⇒ x2 ≠ z2 ⇒ (x ; z) ∉ R 2



Hints: (1) Intersection of two sets contain elements present in both sets. (2) Union of two sets contain all elements present in both sets. 16. Option (a) is correct. Solution: 1  i sin  z 1  i sin 

15. Option (d) is correct. Solution: Let A = {2, 4, 6, 8, 10} B = {2, 4, 12} C = {2, 4, 14} ⇒ A ∩ B = {2, 4} and A ∩ C = {2, 4} ⇒A∩B=A∩C But B ≠ C So statement 1 is incorrect. Let A = {2, 4, 6, 8, 10} B = {2, 4, 6} C = {2, 4, 8} ⇒ A ∪ B = A and A ∪ C = A

Now, | z | ⇒ | z |



1  i sin  1  i sin 

|1  i sin | |1  i sin |

⇒ |z|

1  sin 2 

1  sin 2  ⇒ |z| = 1 17. Option (c) is correct. Solution: 1  i sin  (1  i sin ) z  1  i sin  (1  i sin ) ⇒ z=

3

Hint: Let x = 2 ⇒ 2 ≠ 2 ⇒ (x, x) ∉ R Also, (8, 4) ∈ R as 82 = 43 but 42 ≠ 83 ⇒ (4, 8) ∉ R

⇒A∪B=A∪C But B ≠ C So statement 2 is also incorrect.



⇒ z

(1 + i sin θ)2 (1 − i 2 sin 2 θ) (1  sin 2 )  ( 2 sin )i 1  sin 2 

z is purely real if img (z) = 0 2 sin  0 ⇒ 1  sin 2  ⇒ sin θ = 0 ⇒ θ = nπ Hint: Simplify z and put img (z) = 0.

18. Option (b) is correct. Solution: 1  i sin  1  i sin  z  1  i sin  1  i sin 

⇒ z

(1  sin 2 )  2i sin  1  sin 2 

z is purely imaginary, if Re(z) = 0. ⇒ 1 – sin2θ = 0 ⇒ sinθ = ±1

87

SOLVED PAPER - 2022 (II)

⇒ q=

(2n + 1)π 2

Hint: Simplify z and put Re(z) = 0. 19. Option (d) is correct. Solution: Given: P = sum of first n positive terms of arithmetic progression A. Q = Sum of first n positive terms of another arithmetic progression B. Let first term of A be a and first term of Q be b. For n = 1, P = a and Q = b. P 5n  4   Q 9n  6 P a 9 For n = 1, = = Q b 15

⇒

a 3 = b 5

Hint: For n = 1, P = First term of A and Q = First term of B. 20. Option (c) is correct. Solution:

Let first term of arithmetic progression A be a and common difference be d1.



And first term of arithmetic progressoin B be b and common difference be d2. n [ 2 a  ( n  1)d1 ] 2



So, P 



And, Q  



n [ 2b  ( n  1)d2 ] 2

P 5n  4  Q 9n  6

⇒



Now, 10th term of A, (T10)A = a + 9d1.



10th term of B, (T10)B = b + 9d2. (T ) a  9d1 Now, 10 A  (T10 )B b  9d2





⇒

(T10 ) A 2 a  18d1  (T10 )B 2b  18d2

Put n = 19 in equation (i), we get

(T10 ) A 33 = (T10 )B 59

Hints: 1. nth term of AP is Tn = a + (n – 1)d, where a = first term and d = common difference. n 2. Sum of nth term of A.P. is Sn  [ 2 a  ( n  1)d ] 2 where a = first term and d = common difference. 21. Option (a) is correct. Solution: Let first term of A be a and first term of B be b. Given: d = Common difference of A D = Common difference of B. P 5n  4   Q 9n  6

a 9 3 For n = 1, = = ...(i) b 15 5 ⇒ad

2 a 12 + d 21 7

Hint: Put n = 1 and n = 2 in given relation and solved further.

2 a  ( n  1)d1 5n  4  ...(i) 2b  ( n  1)d2 9n  6







2 a  18d1 5(19 )  4 99 33    2b  18d2 9(19 )  6 177 59

22. Option (a) is correct. Solution: Given: The binomial expansion of (p + qx)9 general term Tr 1  9 Cr ( P )9  r ( qx )r  9 Cr ( P )9  r ( q )r x r



Now, coefficient of x3 = coefficient of x6



⇒ 9 C3 ( P )9  3 ( q )3  9 C6 ( P )9  6 ( q )6



⇒ 9 C3 p 6 q 3 = 9 C3 p 3 q 6

88

Oswaal NDA/NA Year-wise Solved Papers ⇒ p 3 = q3 ⇒p=q

Hint: Find the general term of given expansion and then find the coefficient of x3 and x6. 23. Option (b) is correct. Solution: Given expansion of (p + qx)9 The middle term of given expansion are 9 + 1 2 9+3 and 2 ⇒ 5th and 6th term.  General term Tr 1  9 Cr ( P )9  r ( qx )r

9

5 4 4

9

4 5 5

⇒ T5 = C4 ( P ) q x and T6 = C5 ( P ) q x 9 Coefficient of T5 C4 P 5 q 4 = ⇒ 9 Coefficient of T6 C5 P 4 q 5

= =

9 9

C4 p C4 q

p q

Hint: The middle terms in the expansion of n1 (a + bx)n for n is odd are    2  th n  3   terms.  2   

th

and

24. Option (b) is correct. Solution:

Tr 1  9 Cr ( p )9  r ( qx )r

Coefficient of x2 = coefficient of x4 9

C2 p7 q 2 = 9 C4 p 5 q 4

⇒ ⇒

p2 q

2

p2 q

2

= =

9 9

C4 C2

=

9 × 8 ×7 × 6 2 × 4 ×3× 2 9×8

7 2

Hint: The general term of expansion of (a + bx)n is T  n C ( a )n  r ( bx )r r 1 r 25. Option (d) is correct. Solution:



Given word is QUESTION Vowels are UEIO and consonants are QSTN. We have to form 4 letter word using 2 vowels and 2 consonants. So firstly we will choose 2 vowels and 2 consonants and then arrange them So required words = 4 C2 × 4 C2 × 4 !







43 43   43 2 2 2

= 864 Hint: Choose 2 vowels and 2 consonents and then arrange them.

26. Option (c) is correct. Solution: There are 4 vowels UEIO and 4 consonants QSTN in QUESTION For alternate positions of vowels and consonants there are 2 possibilities. VCVCVCVC or CVCVCVCV So we will arrange 4 vowels and 4 consonants seperately. So required words are 2 × 4! × 4! = 2 × 24 × 24 = 1152

Hint: For alternate positions of vowels and consonents 2 possibilities are VCVCVCVC or CVCVCVCV.

27. Option (b) is correct. Solution: There are 4 vowels UEIO and 4 consonants QSTN. Let us consider 4 consonants as one alphabet. We will arrange these 5 alphabets and then we will arrange 4 consonants. So required words = 5! × 4! = 120 × 24 = 2880 Hint: Consider 4 consonants as one alphabet and then arrange them. 28. Option (c) is correct. Solution:  a11 a12 a13  Given A   a21 a22 a23   a31 a32 a33 

89

SOLVED PAPER - 2022 (II)

c11, c12 and c13 are cofactors of a11, a12, a13. As we know, |A| = a11c11 + a12c12 + a13c13 ∆ = a11c11 + a12c12 + a13c13 int: Use formula to find the value of a H determinant.



a21 ⇒   ( 1) a23 a22



a21 ⇒ a23 a22

29. Option (a) is correct. Solution:

 a11 Given A   a21   a31

a12 a22 a32

a13  a23  a33 

c11 = (–1)2 (a22a33 – a23a32) c12 = (–1)3 (a21a33 – a31a23) c13 = (–1)4 (a21a32 – a22a31) ⇒ a21c11 + a22c12 + a23c13 = a21(a22a33 – a23a32) – a22 (a21a33 – a31a23) + a23 (a21a32 – a22a31) ⇒ a21c11 + a22c12 + a23c13 = 0 Hint: Use Cij = (–1)i+j det (Mij). where Cij is cofactor and Mij is minor.

30. Option (d) is correct. Solution: a11 a12 a13 Given   a21 a22 a23 a31 a32 a33





Transposing the elements we get, a11 a21 a31   a12 a22 a32 a13 a23 a33 Applying c1 ↔ c2, we get a21 a11 a31 ⇒   ( 1) a22 a12 a32 a23 a13 a33 Again applying c2 ↔ c3, we get a21 a31 a11 ⇒   a22 a32 a12 a23 a33 a13 Applying R2 ↔ R3, we get

a31 a33 a32

a31 a33 a32

a11 a13 a12

a11 a13   a12

Hint: Apply row and column transformations to get the desired result. 31. Option (c) is correct. Solution: f(x + y) = f(x) f(y) and f(1) = 2 ⇒ f(2) = f(1) f(1) = 22, f(3) = f(2) f(1) = 23, f(4) = 24 ⇒ f(n) = 2n.

Now,

n

 f ( x )  2044

x 2



⇒ 22 + 23 + .... + 2n = 2044.



2 2 ( 2 n1  1) ⇒  2044 2 1 



⇒ 2n–1 = 511 – 1 = 510 ⇒n–1=9 ⇒ n = 10

 a( r n  1)   Sum of GP  r  1  

Hint: Use sum of a GP with a as first term, r as common ratio and n number of terms is a( r n - 1) r -1 32. Option (b) is correct. Solution:

f(x + y) = f(x) f(y) and f(1) = 2



⇒ f(2) = f(1) f(1) = 22, f(3) = f(2) f(1) = 23, f(4) = 24



⇒ f(n) = 2n. ∴





5

 f ( 2x  1)  f (1)  f (3)  f ( 5)  f (7 )  f (9)

x 1

= 2 + 2 3 + 25 + 27 + 29 = =

2( 2 2 )5 - 1 22 - 1



2(1024 - 1) 3

a( r n  1)    Sum of GP  r  1  

90



Oswaal NDA/NA Year-wise Solved Papers

=

2 × 1023 3

= 682 Hint: Use sum of GP =

a( r n - 1) , where a = r -1

first term and r = common ratio. 33. Option (d) is correct. Solution: f(x + y) = f(x) f(y) and f(1) = 2 ⇒ f(2) = f(1) f(1) = 22, f(3) = f(2) f(1) = 23, f(4) = f(3) f(1) = 24



⇒ f(n) = 2n. 6

 2 x f ( x )  2 f (1)  2 2 f ( 2)  23 f (3)  2 4 f ( 4 )

x 1

 2 5 f ( 5)  2 6 f ( 6 )



= 2(2) + 22(22) + 23(23) + 24(24) + 25(25) + 26(26)



= 22 + 24 + 26 + 28 + 210 + 212 a( r n  1)    Sum of GP  r  1  

2 2 ( 212 - 1)



=



4 × 4095 = 3



= 5460

22 - 1



a( r n - 1) , where a = r -1 first term and r = common ratio.

Hint: Use sum of GP =

34. Option (c) is correct. Solution: Let B, F, V represents Basketball, football and volleyball and a, b, c, p, q, r, x represented in venn diagram. B p q

x c



Hint: Use venn diagram to solve it. 36. Option (a) is correct. Solution: Medals in exactly one of the sport = a + b + c. Putting p + q + r = 40 – 7x in equation (1) of question No. 34. we get. a + b + c = 14x – 40 + 7x = 21x – 40 Hint: Use venn diagram to solve it. 37. Option (a) is correct. Solution:  0 sin 2  cos2    2  A  cos  0 sin 2   and A = P + Q  2  2 0   sin  cos 

A=

1 (A + AT) + 1 (A – AT) 2 2



where

1 (A + AT) is a symmetric matrix. 2

b



⇒P=

r

1 (A + AT) 2  0 sin 2  cos2    0 cos2  sin 2         1  2 sin 2     sin 2  0 cos2    0  cos  2  2    2 0   0  cos2  sin 2    sin  cos  

 0 sin 2  cos2    0 cos2  sin 2          1 2 0 cos2    sin 2     sin 2  0  cos  a + b + c + p + q + r + x = 15x 2  2    2 0   0  cos2  sin 2  ⇒ a + b + c + p + q + r = 14x....(1)   sin  cos   & a + p + q + x = 5x V



35. Option (d) is correct. Solution: Using final answer of question No. 34 Medals in at least two sports = p+ q+r+x = 40 – 7x +x = 40 – 6x

F a



& a + p + q = 4x....(2) & b + p + r + x = 4x + 15 & b + p + r = 3x + 15 ....(3) & c + q + r + x = x + 25 & c + q + r = 25. ....(4) Adding above (2)+(3)+(4) equations, we get a + b + c + 2 (p + q + r ) = 7x + 40 (5) Eq. (5) – (1), we get p + q + r = 7x + 40 – 14x = 40 – 7x Hint: Use venn diagram to solve it.



∴



91

SOLVED PAPER - 2022 (II)  sin 2   cos2  cos2   sin 2   0   1 = cos2   sin 2  sin 2   cos2   0 2 2  2 2 2 0 sin   cos  cos   sin    1 1  0 2 2   1 1 =  0 2 2 1 1   0  2 2  1 1 Hint: Use A  ( A  AT )  ( A  AT ) where 2 2

1 ( A + AT ) is a symmetric matrix. 2 38. Option (d) is correct. Solution:  0 sin 2  cos2    2  A  cos  sin 2   and A = P + Q 0  2  2 0   sin  cos 

1 (A + AT) + 1 (A – AT) 2 2 1 where (A + AT) is a skew symmetric matrix. 2

A =

⇒Q=

1 (A – AT) 2

 0 sin 2  cos2    0 cos2  sin 2         1  2 0 cos2    sin 2     sin 2  0  cos  2  2    2 0   0  cos2  sin 2    sin  cos    sin   cos  cos   sin   0  1 2 2 sin 2   cos2   0 = cos   sin  2 2  2 2 2 0 sin   cos  cos   sin     cos 2 cos 2   0 1 0  cos 2  =  cos 2 2   cos 2 cos 2 0  1 1    0 2 2    1 1  cos 2 0   =  2 2  1 1   0   2 2  2

2

2

1 1 Hint: Use A  ( A  AT )  ( A  AT ) where 2 2

1 ( A + AT ) is a skew symmetric matrix. 2 39. Option (a) is correct. Solution:



 0 sin 2  cos2    2  A  cos  sin 2   0  2  2 0   sin  cos 



|A| = 0 – sin2θ(0 – sin4θ) + cos2θ(cos4θ – 0)



⇒ |A| = sin6θ + cos6θ



⇒ |A| = (sin2θ + cos2θ) (sin4θ + cos4θ – sin2θ cos2θ)



⇒ |A| = (1) [sin2θ + cos2θ)2 – 3 sin2θ cos2θ]



⇒ |A| = 1 – 3(sinθ cosθ)2



3 2 ⇒ |A| = 1  sin 2 4



For |A| to be minimum, sin22θ = 1



⇒ |A| = 1 

3 1  4 4

Hint: Use trigonometric identities to solve it.

and

algebraic

40. Option (c) is correct. Solution: ABC is a plot. Let OL be the lamp post. ∠OBL = 45° and ∠BOL = 90° A

2

L

b

O c



B

a

C

OL OL = In ∆ BOL, tan 45° = OB 8 ⇒ OL = 8. Hint: Use tan  

Perpendicular Base

41. Option (b) is correct. Solution: From the given question no. 40,

92

Oswaal NDA/NA Year-wise Solved Papers As we know, by sin rule. AB BC AC = = = 2R sin C sin A sin B





abc Where R is a circumradius and R  where 4   s( s  a )( s  b )( s  c ) and s  Here s 



16  10  10  18 2

AB  CB  CA 2

B

Leaning Tower

⇒   18  2  8  8 ⇒ ∆ = 48 ( AB)( BC )(CA ) 16  10  10 25 ⇒R=   4 4  48 3 ⇒





O



AB 25 50  2R  2   sin c 3 3

Hint: Use sin rule

a b c  abc     2  sin A sin B sin c  4 

where   s( s  a )( s  b )( s  c ) and s

abc 2

42. Option (d) is correct. Solution: As we know, by cosine rule b 2  c 2  a2 a2  c 2  b 2 , cos B  , 2bc 2 ac a2  b 2  c 2 cos C  2 ab cos A 

=

2

2

2

2

(10 )  (16 )  (10 ) (10 )  (16 )  (10 )  2(10 )(16 ) 2(10 )(16 ) 

(10 )2  (10 )2  (16 )2 2(10 )(10 )



= 100  256  100  100  256  100 320 320 100  100  256  200



4 4 7 =   5 5 25



=

20  20  7 33 = 25 25

2

A

y

75°

Q

x

15° x–y

P

Let AB be learning tower which leans towards north. And OB = height of top of the tower above ground level. OB In ∆OBQ, tan 75° = OA + y



⇒ OA + y = OB cot 75°



In ∆OBP, tan 15° =



⇒ OA + x = OB cot 15° Equation (ii) – Equation (i), we get x – y = OB (cot 15° – cot 75°)



⇒ OB 



⇒ OB 

From the given question no. 40, ⇒ cos A + cos B + cos C 2

b 2  c 2  a2 2bc

43. Option (a) is correct. Solution:

  18(18  16 )(18  10 )(18  10 )





Hint: Use cosine rule, cos A =

...(i)

OB OA + x ...(ii)

xy (2  3 )  (2  3 ) xy 2 3

Hint: Find OB using the concept of height and distance.

44. Option (d) is correct. Solution: From the given question No.43.

h

xy 2 3



h OA + x ⇒ OA + x = h cot 15°



Put value of h in equation (ii), we get



⇒ OA  x 



⇒ 2 3OA  ( 2  3 )x  y( 2  3 )  2 3x



In ∆OBP, tan 15° =

xy 2 3

 (2  3 )

...(ii)

93

SOLVED PAPER - 2022 (II)

⇒ 2 3OA  x( 2  3 )  y( 2  3 )



⇒ OA 



OA Now, cot   n

1 2 3

 x( 2 

1

⇒ cot   2 3

3 )  y( 2  3 )

 2(x  y)  1 2 3

⇒ cot   2  3

3(x  y)



...(iii)



(x  y) xy

Using equation No (iii) of question no. 43. 1 2( x  y )  3 ( x  y ) 2 3 ⇒ cot   1 (x  y) 2 3





OB AB



⇒ AB = OB cosec θ x−y ⇒ AB = cosec θ 2 3 As we know 1 + cot2θ = cosec2θ  (x  y)  ⇒ cosec2θ = 1   2  2  xy  

...(iii)

2

 (x  y)  1  2  3  xy  

⇒ cosecθ =



Put value of cosecθ in equation (iii), we get AB 

2 3

3 ( x  y )   1  2   x  y  

Hint: Use cosec (−θ) = −cosec θ and cosec (2π + θ) = cosec θ in Ist quadrant.



 5 7    7  5   17  17   17  17  ⇒ A  2 cos   cos   2  2     11       2 cos   cos  17     17 



   11     6  ⇒ A  2 cos   cos    2 cos   cos  17  17 17 17        



     6   11   ⇒ A  2 cos   cos    cos    17    17   17  



    51  ⇒ A  2 cos   2 cos   cos    17  2  34 



⇒ A = 0

2

46. Option (b) is correct. Solution:  73  Let A  cosec     3   73  ⇒ A  cosec   { cosec(−θ) = − cosec θ}  3 



∵ cos



Hint: Use



C D CD cos C  cos D  2 cos  cos     2   2 

2



xy

 3

CD C D  cos C + Cos D = 2 cos   cos  2  2    

⇒ In ∆OAB, sin  



 2  ⇒ A     3

(x  y) ⇒ cot   2  3 xy







OA h





⇒ A   cosec

{ cosec(2π + θ) = −cosec θ in Ist quadrant.}

47. Option (a) is correct. Solution: Let  5   7   11    A  cos    cos    2 cos  cos     17   17   17   17 

Now, cot  











  ⇒ A   cosec  24    3 

(x  y)

45. Option (b) is correct. Solution: From the given question no. 43.







 0 2

48. Option (b) is correct Solution:  3  Let A  tan    8 

  ⇒ A  tan    2 8



π ⇒ A = cot   8 ⇒ A



cos   / 8  sin   / 8 

  π  cot θ  ∵ tan  − θ=  2   

94



Oswaal NDA/NA Year-wise Solved Papers

⇒ A

1 / 2  1  cos  / 4  2 cos2 θ  1  cos 2θ  ∵   2 sin 2 θ   1  cos 2θ 







⇒ A ⇒ A



⇒

1 / 2  1  cos  / 4 

A

1  cos  / 4 1  cos  / 4 11 / 2 11 / 2

 

 2  1 

2 1

 2  1

49. 

Hint: Use 2 cos2θ = 1 + cos2θ and 2 sin2θ = 1 – cos2θ. Option (d) is correct. 5 +1 {cos 36°= Solution: 4 Let A = tan−1cot (cosec−12) & 5 −1 } ⇒ A = tan−1[cot (cot−1 3 )] sin 18° = 4 h b  ∵ cos ec  , cot    p p  ⇒ A = tan−1( 3 )]{ cot (cot−1a) = a} ⇒ A = π/3



⇒ cosC 

16  9  4 24



⇒ cosC =

7 8



7 7 ∵ cos 3C  4    3   8 8



⇒ cos 3C 

343  336 128



⇒ cos 3C =

7 128

3

2 1

⇒ A  2 1





Hint: Use cosec θ 

Hyp. Base and cot θ = perp. perp.

50. Option (a) is correct. Solution: As we know, cos 3C = 4 cos3C – 3cos C



Hint: Use cos 3θ = 4 cos3θ – 3 cosθ and cosine rule.

51.

Option (c) is correct. Solution: Let A = cos 36° − cos 72° ⇒ A = cos 36° – 18° { cos (90° − θ) = sin θ}



 5 1  5 1 ⇒ A   4   4     



⇒ A



A





Given: sec x =





a=4

Also by cosine rule, cos C  ⇒ cosC 

4

2

 3  2 2

2  4  3 

2

C

a2  b 2  c 2 2 ab

x 24

5 −1 4

C ? B

25 24

⇒ Perpendicular =

 25 2   24 2

=7

7 7 and sin x = ⇒ tan x = 25 24  x lies in the fourth quadrant where sin θ and tan θ are negative 7 7 ⇒ tan x   and sin x  24 25

B

25





b=3

5 +1 sin 18° = 4

Hint: use cos 36°=

52. Option (b) is correct Solution:

A

c=2

1 1 1   4 4 2



⇒ tan x  sin x 



7 7 343   24 25 600

Hint: tan θ and sin θ are negative when θ lies in fourth quadrant.

95

SOLVED PAPER - 2022 (II) 53.

Option (b) is correct. Solution: Let A = tan2165° + cot2165° ⇒ A = tan2(180° − 15°) + cot2(180° − 15°) ⇒ A = tan215° + cot215°



  2  3  2

2



⇒ A = 2 3



⇒ A = 434 3 434 3



⇒ A = 14



  π  ⇒ A  sin  π    6  



⇒ A   sin 2



1 ⇒ A 4

55.

Option (c) is correct. Solution: Given: Line passes through origin Line makes 75° with positive x axis

Hint: use tan 15  2  3 and cot 15  2  3

As we know, sin θ is −ve in fourth quadrant and + ve in first quadrant. 5   5   ⇒ A   sin    sin  6  6  



56.

Hints: (1) use cos2θ = cos2θ – sin2θ. (2) General solution of cos x = cos θ is x = 2nπ ± θ.

Y

54. Option (a) is correct. Solution: 5  5    Let A  sin  2n   sin  2n  6  6    



2

75°



Slope = m = tan 75°



⇒ m  2 3  line passes through origin so equation of the line is y = mx.



⇒ y  2 3 x





Option (d) is correct. Solution: 1 + 2 (sin x + cos x) (sin x − cos x) = 0 ⇒ 2(sin2x – cos­x) = −1 1 ⇒ cos 2 x = 2  3



⇒ cos 2 x  cos



π ⇒ 2 x 2 nπ   3



⇒ x  n   / 6



For n = 0, x 

  6



For n = 1, x 

5 7  , 6 6



For n = 2, x 

11 6



y  2  3 1  2  3 



Hint: use the sign convention of trigonometric ratios in four quadrants.



. Statement 1: At x = 1,



 6

X

1 2 3



1   ⇒ The line passes through  1,   2 3 



So statement 1 is correct



Statement 2: y  2  3 x



If x is +ve then y is also +ve always and if x is –ve then y is also –ve always so the line entirely lies is 1st and 3rd quadrant.







Hint: Slope of a line is tan θ where θ is the angle made by the line with positive x-axis.

57. Option (b) is correct. Solution: Y

(0, b) B • • (3, 4) P

{ x ∈ (0°, 360°)}

0•



A • (a, 0)

X

Let the line be AB with A ≡ (a, 0) and B ≡ (0, b) and p is mid point of AB. By mid point formula, 0a 3 a6 2

96

Oswaal NDA/NA Year-wise Solved Papers b0 b8 2



and 4 



By intercept form of line, equation of AB is x y  1 a b



⇒



⇒ 4x + 3y – 24 = 0

x y  1 6 8

Hint: use mid point formula and intercept form of equation of line.

58. Option (a) is correct. Solution:

Y

B 4 C

O

X

4

AB = BC = 8



OB = OA = 4



⇒ B ≡ (0, 4) & A ≡ (0, −4)



∠OCB = 30°



∴ In Δ OCB, tan 30 =



⇒ OC = 4 3



2 ⇒ 2 g c  4



⇒ g2 = 4 ⇒g=±2 and intercept on y axis = 6



⇒ 2 f2 c  6



⇒ f 2 = 9 ⇒f=±3 Now, centre of the circle is (−g, −f) ⇒ centre ≡ (± 2, ± 3)  Intercepts are made with positive axes so center lies in I quadrant. ⇒ centre ≡ (2, 3) Among all the options (2, 3) lies on 3x – 4y + 6 = 0 Hint: If the general equation of the circle is x2 + y2 + 2gx + 2fy +c = 0 the intercept on x axis = 2 g 2 - c and intercept on y axis = 2 f2 -c .

4 OC

60. Option (a) is correct. Solution:

Y



• (1,6)

04

1



Now, slope of AC 



The required line has slope through (8, 0)



⇒ The equation of line is y  0 



⇒



⇒ x  3y  8  0







∴ C  4 3,0

Option (c) is correct Solution: Let the equation of the circle be x2 + y2 + 2gx + 2fy + c = 0  circle passes through origin. ∴ c = 0. ⇒ The equation of circle becomes x2 + y2 + 2gx + 2fy = 0 Now, intercept on x axis = 4



A



59.

4 3 0



• (3,2)

3

•

1

X

and passes

3 1 3

x  8

3y  x  8

Hint: The equation of line having slope m and passing through (x1, y1) is (y – y1) = m (x – x1)



 center ≡ (0, 0) and major axis is on y axis.



⇒ Equation of ellipse is



 Ellipse passes through (3, 2) and (1, 6)



⇒

9 a

2



4 b

2

 1 and

1 a

2



x2 a2 36 b2



y2 b2

1

1

a  b

97

SOLVED PAPER - 2022 (II)



⇒



⇒



⇒

9 a

2



8 a

2

a

2

b

2

= =

4 b

2



1 a

2



36 b

2

32 b

2

1 4



Now, eccentricity e  1 



1 e  1 4 e=



a

2

Hint: Use eccentricity e  x2

equation of ellipse is

a2



y2 b2

1  a2 b2

, where

 1 ; b > a.

D • 30° •

 p 3p   So, A   , 2 2 



 Point A lies on parabola x 2 = 3 y



3 p    3 p 2 2



⇒



⇒p=6 Now length of latus rectum, q = 3 \ p = 2 3q



(0, 0)



60°

•A





∆ABC is equilateral. ∠ACD = 30° = ∠BCD Let coordinates of A be (x, y) CD In ∆ACD, sin 60° = AC 3 y ⇒ = 2 p 3 p 2



⇒ y=



In ACD, cos60° =



⇒

1 x = 2 p

AD AC

2

p2 3 = p 4 2

62. Option (b) is correct. Solution: Given points A(2, 4, 6), B(–2, –4, –2), C(4, 6, 4) and D(8, 14, 12)

x 2  3y

C





61. Option (c) is correct. Solution: Equation of parabola is x 2 = 3 y

B • 60°

⇒ x=

b2

3 2

p 2





Now, AB  ( 4 )2  ( 8 )2 ( 8 )2  144  12 BC  ( 6 )2  (10 )2  ( 6 )2  172 CD  ( 4 )2  ( 8 )2  ( 8 )2  144  12 DA  ( 6 )2  ( 10 )2  ( 6 )2  172 AC  ( 2 )2  ( 2 )2  ( 2 )2  2 3 BD  (10 )2  (18 )2  (14 )2  620  AB = CD and BC = DA and AC ≠ BD.  Given points are the vertices of parallelogram ABCD. So, statement 1 is incorrect. Now, midpoint of 24 46 64 AC   , ,   ( 3, 5, 5)  2 2 2  midpoint of  8  2 14  4 12  2  BD   , ,   ( 3, 5, 5)  2 2 2  So statement 2 is correct. Hints: 1. For rectangle AB = CD, BC = DA and AC = BD. 2. Midpoint of two points (x1, y1, z1) and (x2, y2,  x  x 2 y1  y 2 z1  z2  z2) is given by  1 , ,   2 2 2 

98

Oswaal NDA/NA Year-wise Solved Papers

63. Option (b) is correct. Solution: Equation of sphere is x2 + y2 + z2 – 4x – 6y – 8z – 16 = 0. As we know plane is the tangent of the sphere. So, z-axis cannot be tangent of sphere. So, statement 1 is incorrect. Now, coordinates of the centre of sphere = (2, 3, 4) Equation of plane is x + y + z – 9= 0. 2+3+4–9=0 \ Point (2, 3, 4) lies on the plane x + y + z – 9 = 0 So, statement 2 is correct. Hints: 1.  Plane is the tangent of the sphere. 2. Coordinates of the centre of the sphere x2 + y2 + z2 + 2gx + 2fy + 2kz + c are (–g, –f, –k). 64. Option (c) is correct. Given: Plane cuts intercepts 2, 2, 1 on the coordinate axes. x y z \ Equation of plane is    1 2 2 1 ⇒ x + y + 2z = 2 So, direction ratios of the normal to plane = {1, 1, 2} Direction cosines of the normal to plane

=



1 1 2 , , k k k



2

2

2

where k  1  1  2  6 So, direction cosines of normal to plane  1 1 2  , , =    6 6 6 Hints: 1.  If plane cuts intercepts a, b, c on the coordinate axes, then equation of plane in x y z   1 a b c 2. Direction ratios of the normal to plane Ax + By + Cz + d = 0 are A, B, C.

65. Option (c) is correct. Solution: As we As we know that cosines of y-axis = {cos90°, cos0°, cos90°} = {0, 1, 0} \ Direction ratios of y-axis = {0, k, 0}; k ∈ R



So, statement 1 is correct. Direction cosines of z-axis = {cos90°, cos90°, cos0°} = {0, 0, 1} \ Direction ratios of z-axis = {0, 0, l); l ∈ R Now, 5(0) + 6(0) + 0(l) = 0 \ line whose direction ratios are {5, 6, 0} is perpendicular to z-axis. So, statement 2 is also correct.



Hints: 1.  Direction cosines of y-axis = {0, 1, 0} 2.  Direction cosines of z-axis = {0, 0, 1} 3. If a1a2 + b1b2 + c1c2 = 0, then lines whose direction ratios are {a1, b1, c1} and {a2, b2, c2}, are perpendicular. 66. Option (d) is correct. Solution:     Given: PR = a and QS = b P

Q a b

S



    Let PQ = x and QR = y     Then SR = x and PS = y



R

By triangle law of vector addition.    PQ  QR  PR    x  y  a ...(i) Applying triangle law of vector addition in ∆QRS    QR  RS  QS    y  x  b ...(ii) Equation (i)–Equation (ii), we get    2x  a  b    ab ⇒ x 2    a − b ⇒ PQ = 2

Hint: Apply triangle law of addition in ∆PQR and ∆QRS and solve further.

99

SOLVED PAPER - 2022 (II) 69. Option (a) is correct. Solution:  Let A  2 x 2 i  3x j  k  B  i  2 j  x 2 k   Let angle between A and B be θ   A, B \ cos     | A || B |

67. Option (c) is correct. Solution:   a | |= b| 1 Given: |=     And vectors ( a + 2b ) and ( 5a - 4 b ) are perpendicular.     ⇒ ( a  2b )  ( 5a  4 b )  0 2     2 ⇒ 5 a  4 a  b  10b  a  8 b  0     ⇒ 5(1) − 4 a ⋅ b + 10 a ⋅ b − 8(1) = 0   ⇒ 6a  b  3



  1 ⇒ ab  2   1 ⇒ | a || b |cos   , where θ = Angle between 2   a and b





⇒ cos  



⇒ 

1 2







⇒ cos  



For obtuse angle, cosθ < 0 \ 3x2 – 6x < 0 ⇒ x(x – 2) < 0 ⇒0 1 

is continuous, then what is the value of (a + b) ? (a) 5 (b) 10 (c) 15 (d) 20 94. Consider the following statements in respect of the function f(x) = sin x : (1) f(x) increases in the interval (0, π).  5π  (2) f(x) decreases in the interval  , 3π  .  2  Which of the above statements is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 95. What is the domain of the function f(x) = 3x ? (a) (-∞, ∞) (b) (0, ∞) (c) [0, ∞) (d) (-∞, ∞) - {0}

97. What is the degree of the following differential equation ? x = 1+

d2 y

dx 2 (a) 1 (b) 2 (c) 3 (d) Degree is not defined 98. Which one of the following differential equations has the general solution y = aex + be-x ? d2 y (a) 2 + y = 0 dx

(b)

d2 y dx 2

−y =0

d2 y dy −y =0 (c) 2 + y = 1 (d) dx dx 99. What is the solution of the following differential equation ? (a) ex + ey = c (c) ex - ey = c 100. What is

∫e

 dy  ln   + y = x  dx  (b) ex+y = c (d) ex-y = c

( 2 ln x + ln x 2 ) dx equal to ?

x4 (a) + c 4

(b)

x3 +c 3

2x 5 x5 +c (d) (c) + c 5 5 101. Consider the following measures of central tendency for a set of N numbers  : (1) Arithmetic mean (2) Geometric mean Which of the above uses/use all the data ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 102. The numbers of Science, Arts and Commerce graduate working in a company are 30, 70 and 50 respectively. If these figures are represented by a pie chart, then what is the angle corresponding to Science graduates ? (a) 36° (b) 72° (c) 120° (d) 168°

118 Oswaal NDA/NA Year-wise Solved Papers 103. For a histogram based on a frequency distribution with unequal class intervals, the frequency of a class should be proportional to (a) the height of the rectangle (b) the area of the rectangle (c) the width of the rectangle (d) the perimeter of the rectangle 104. The coefficient of correlation is independent of (a) change of scale only (b) change of origin only (c) both change of scale and change of origin (d) neither change of scale nor change of origin 105. The following table gives the frequency distribution of number of peas per pea pod of 198 pods  : Number of peass

Frequency

1

4

2

33

3

76

4

50

5

26

6

8

7

1

What is the median of this distribution ? (a) 3 (b) 4 (c) 5 (d) 6 106. If M is the mean of n observations x1 - k, x2 - k, x3 - k, xn -... k, where k is any real number, then what is the mean of x1, x2, x3, ..., xn ? (a) M (b) M + k (c) M - k (d) kM 107. What is the sum of deviations of the variate values 73, 85, 92, 105, 120 from their means ? (a) - 2 (b) - 1 (c) 0 (d) 5 108. Let x be the HM and y be the GM of two positive numbers m and n, if 5x = 4y, then which one of the following is correct ? (a) 5m = 4n (b) 2m = n (c) 4m = 5n (d) m = 4n 109. If the mean of a frequency distribution is 100 and the coefficient of variation is 45%, then what is the value of the variance ? (a) 2025 (b) 450 (c) 45 (d) 4.5

110. Let two events A and B be such that P(A) = L and P(B) = M. Which one of the following is correct ? L + M −1 (a) P ( A|B) < M L + M −1 (b) P ( A|B) > M L + M −1 (c) P ( A|B) ≥ M L + M −1 M 111. For which of the following sets of numbers do the mean, median and mode have the same value ? (a) 12, 12, 12, 12, 24 (b) 6, 18, 18, 18, 30 (c) 6, 6, 12, 30, 36 (d) 6, 6, 6, 12, 30

P ( A|B) = (d)

112. The mean of 12 observation is 75. If two observations are discarded, then the mean of the remaining observations is 65. What is the mean of the discarded observations ? (a) 250 (b) 125 (c) 120 (d) Cannot be determined due to insufficient data 113. If k is one of the roots of the equation x (x + 1) + 1 = 0, then what is its other root ? (a) 1 (b) - k 2 (c) k (d) - k2 114. The geometric mean of a set of observation is computed as 10. The geometric mean obtained when each observation xi is replaced by 3xi4 is (a) 810 (b) 900 (c) 30000 (d) 81000 5 1 1 , P ( A ∩ B ) = and P ( A ) = , 6 3 2 then which of the following is/are correct ? (1) A and B are independent events. (2) A and B are mutually exclusive events. 115. If P ( A ∪ B ) =



Select the correct answer using the code given below. (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 116. The average of a set of 15 observations is recorded, but later it is found that for one observation, the digit in the tens place was wrongly recorded as 8 instead of 3. After correcting the observation, the average is

119

SOLVED PAPER - 2021 (I)

1 10 (b) increased by 3 3 10 (c) reduced by (d) reduced by 50 3 117. A coin is tossed twice. If E and F denote occurrence of head on first toss and second toss respectively, then what is P(E ∪ F) equal to ? 1 1 (a) (b) 4 2 3 1 (c) (d) 4 3 2 118. In a binomial distribution, the mean is and 3 5 variance is . What is the probability that 9 random variable X = 2 ? 5 25 (a) (b) 36 36 25 25 (c) (d) 54 216 (a) reduced by

119. If the mode of the scores 10, 12, 13, 15, 15, 13, 12, 10, x is 15, then what is the value of x ? (a) 10 (b) 12 (c) 13 (d) 15 120. If A and B are two events such that P ( A ) = and P ( B ) =

3 4

5 , then consider the following 8

statements  : 3 . 4 5 (2) The maximum value of P(A ∩ B) is . 8 Which of the above statements is/are correct/ (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 (1) The minimum value of P(A ∪ B) is

120 Oswaal NDA/NA Year-wise Solved Papers

Answers Q. No. Answer Key

Topic Name

Chapter Name

1

(a)

Algebra of Complex Numbers

Complex Numbers

2

(c)

Properties of Logarithms

Logarithm and its Applications

3

(d)

Properties of Determinants

Matrices and Determinants

4

(b)

Properties of Binomial Coefficients

Binomial Theorem

5

(b)

Properties of Determinants

Matrices and Determinants

6

(c)

Trigonometric Equations

Trigonometric Equations and Inequations

7

(a)

Properties of Determinants

Matrices and Determinants

8

(d)

Nature of Roots

Theory of Equation

9

(b)

Properties of Logarithms

Logarithm and its Applications

10

(b)

Algebra of Complex Numbers

Complex Numbers

11

(d)

Algebra of Matrices

Matrices and Determinants

12

(c)

Basics of Matrices

Matrices and Determinants

13

(c)

Determinant of a Square Matrix

Matrices and Determinants

14

(d)

Properties of Determinants

Matrices and Determinants

15

(c)

Trigonometric Identities

Trigonometric Ratios and Identities

16

(d)

Trigonometric Functions and Properties

Trigonometric Ratios and Identities

17

(c)

Trigonometric Functions and Properties

Trigonometric Ratios and Identities

18

(c)

Determinant of a Square Matrix

Matrices and Determinants

19

(c)

Trigonometric Identities

Trigonometric Ratios and Identities

20

(a)

Properties of Determinants

Matrices and Determinants

21

(b)

Inverse Trigonometric Equations and Inequalities

Inverse Trigonometric Functions

22

(b)

Trigonometric Functions and Properties

Trigonometric Ratios and Identities

23

(d)

Relations between Sides and Angles of a Triangle

Properties of Triangle

24

(b)

Trigonometric Identities

Trigonometric Ratios and Identities

25

(a)

Trigonometric Identities

Trigonometric Ratios and Identities

26

(b)

Measure of an Interior Angle of a Regular Polygon

Polygon

27

(d)

Trigonometric Identities

Trigonometric Ratios and Identities

28

(d)

Heights and Distance

Heights and Distance

29

(b)

Heights and Distance

Heights and Distance

30

(b)

Trigonometric Identities

Trigonometric Ratios and Identities

31

(a)

Identities of Inverse Trigonometric Functions

Inverse Trigonometric Functions

121

SOLVED PAPER - 2021 (I)

Q. No. Answer Key

Topic Name

Chapter Name

32

(a)

Trigonometric Identities

Trigonometric Ratios and Identities

33

(b)

Relations between Sides and Angles of a Triangle

Properties of Triangle

34

(c)

Half-Angle Formula and the Area of a Triangle

Properties of Triangle

35

(c)

Types of Sets

Set Theory and Relations

36

(d)

Subsets and Power Set

Set Theory and Relations

37

(b)

Basics of Relations

Set Theory and Relations

38

(c)

Types of Functions

Functions

39

(a)

Properties of Conjugate, Modulus and Complex Numbers Argument of Complex Numbers

40

(c)

Algebra of Complex Numbers

Complex Numbers

41

(b)

Basics of Complex Numbers

Complex Numbers

42

(a)

Quadratic Equation and Its Solutions

Theory of Equation

43

(c)

Quadratic Equation and Its Solutions

Theory of Equation

44

(c)

Combination

Permutation and Combination

45

(a)

General Term, Middle Term and Greatest Term in Binomial Expansion

Binomial Theorem

46

(c)

Properties of Binomial Coefficients

Binomial Theorem

47

(b)

Binomial Theorem for Positive Integral Index

Binomial Theorem

48

(b)

Arithmetic Progression

Sequence and Series

49

(c)

Geometric Progression

Sequence and Series

50

(d)

Fundamental Principles of Counting

Permutation and Combination

51

(b)

Basics of Functions

Functions

52

(c)

Arithmetic Progression

Sequence and Series

53

(b)

Geometric Progression

Sequence and Series

54

(a)

Properties of Determinants

Matrices and Determinants

55

(d)

Determinant of a Square Matrix

Matrices and Determinants

56

(d)

Basics of Circle

Circle

57

(a)

Basics of 2D Coordinate Geometry

Straight Line

58

(c)

Interaction between Two Straight Lines

Straight Line

59

(d)

Family of Straight Lines

Straight Line

60

(b)

Straight Line and a Point

Straight Line

61

(a)

Basics of 2D Coordinate Geometry

Straight Line

62

(b)

Interaction between Two Straight Lines

Straight Line

63

(a)

Interaction between Two Straight Lines

Straight Line

64

(d)

Basics of Hyperbola

Hyperbola

65

(d)

Basics of Ellipse

Ellipse

66

(b)

Interaction between Two Lines in 3D

Three Dimensional Geometry

122 Oswaal NDA/NA Year-wise Solved Papers Q. No. Answer Key 67

(b)

Topic Name Basics of 3D Geometry

Chapter Name Three Dimensional Geometry

68

(c)

Dot Product of Two Vectors

Vector Algebra

69

(c)

Equation of a Line in 3D

Three Dimensional Geometry

70

(c)

Plane and a Point

Three Dimensional Geometry

71

(c)

Dot Product of Two Vectors

Vector Algebra

72

(c)

Scalar and Vector Triple Products

Vector Algebra

73

(c)

Dot Product of Two Vectors

Vector Algebra

74

(c)

Scalar and Vector Triple Products

Vector Algebra

75

(d)

Cross Product of Two Vectors

Vector Algebra

76

(c)

Methods of Evaluation of Limits

Limits

77

(c)

Variable Separable Form

Differential Equation

78

(d)

Properties of Definite Integral

Definite Integration

79

(b)

Methods of Evaluation of Limits

Limits

80

(a)

Properties of Definite Integral

Definite Integration

81

(b)

Area under the Curve

Area under Curves

82

(c)

Tangent and Normal

Application of Derivatives

83

(a)

Maxima and Minima

Application of Derivatives

84

(d)

Differentiability of a Function

Continuity and Differentiability

85

(d)

Integration using Trigonometric Identities

Indefinite Integration

86

(b)

Integration using Trigonometric Identities

Indefinite Integration

87

(a)

A.M., G.M., H.M. and their Relations

Sequence and Series

88

(a)

Basics of Differentiation

Differential Coefficient

89

(b)

Methods of Differentiation

Differential Coefficient

90

(a)

Trigonometric Identities

Trigonometric Ratios and Identities

91

(a)

Methods of Differentiation

Differential Coefficient

92

(b)

Basics of Limits

Limits

93

(a)

Continuity of a Function

Continuity and Differentiability

94

(b)

Monotonicity

Application of Derivatives

95

(a)

Basics of Functions

Functions

96

(a)

Basics of Differential Equations

Differential Equation

97

(a)

Basics of Differential Equations

Differential Equation

98

(b)

Basics of Differential Equations

Differential Equation

99

(c)

Variable Separable Form

Differential Equation

100

(d)

Methods of Integration

Indefinite Integration

101

(c)

Measures of Central Tendency

Statistics

102

(b)

Basics of Statistics

Statistics

103

(b)

Basics of Statistics

Statistics

104

(c)

Correlation and Covariance

Statistics

105

(a)

Measures of Central Tendency

Statistics

106

(b)

Measures of Central Tendency

Statistics

123

SOLVED PAPER - 2021 (I)

Q. No. Answer Key 107

(c)

Topic Name

Chapter Name

Measures of Central Tendency

Statistics

108

(d)

Measures of Central Tendency

Statistics

109

(a)

Measures of Dispersion

Statistics

110

(c)

Conditional Probability

Probability

111

(b)

Measures of Central Tendency

Statistics

112

(b)

Measures of Central Tendency

Statistics

113

(c)

Roots of Unity

Complex Numbers

114

(c)

Measures of Central Tendency

Statistics

115

(a)

Algebra of Probability

Probability

116

(c)

Measures of Central Tendency

Statistics

117

(c)

Algebra of Probability

Probability

118

(d)

Bernoulli Trials and Binomial Distribution

Probability

119

(d)

Measures of Central Tendency

Statistics

120

(c)

Algebra of Probability

Probability

NDA / NA

MATHEMATICS

I

National Defence Academy / Naval Academy



 1−i    1+i 









n2

⇒



⇒

 1−i 1−i  ×    1+i 1−i 

⇒

 1 + (i )2 − 2i   2 2    1 − (i ) 

⇒

 1 − 1 − 2i   1 − ( −1)   

⇒

 −2i     2 

•

x=

=1

n2

=1



n2

3. Option (d) is correct. Explanation: a1

n2

=1 n2

∆ = a2 a3

Given:

=1 n =4 n=2 2

As we know logaa = 1 ∴ cos x = sin x π  sin  − x  = sin x 2 

b1 b2 b3

c1 c2 c3

Now,

1−i Rationalize the and solve further 1 +i 2 using i = -1

logcosxsin x = 1; 0 < x
1 47. Option (b) is correct. Explanation: Given: Expansion of (1 + x)2n Now, as we know the coefficient of 1st term = 2nC0 =



2n ! = =1 2n !

Similarly, last term coefficient = =

2n ! 2n !(2n − 2n)!



=

2n ! =1 2n !

2n

C2n

∴ Sum of coefficient of first and last term =1+1=2 Hint: • Use coefficient of (x+1)th term in (1 + a)n is given by nCx.

48. Option (b) is correct. Explanation: Let, first term of an A.P. = a = 2 Sum of first five terms = S5 Sum of next five terms = S10 - S5 Now,



Now, Sn =



difference. \ S10 = 5S5



⇒



⇒ 5[4 + 9d] =

S5 =

1 (S10 - S5) 4

⇒ 4S5 = S10 - S5

n [2a + (n - 1) d] where d = common 2

5 10 [2(2) + (10 - 1) d] = 5 × [2(2) + (5 - 1) d] 2 2 25 [4 + 4d] 2

⇒ 2(4 + 9d) = 5(4 + 4d) ⇒ 8 + 18d = 20 + 20d ⇒ 2d = - 12 ⇒ d=-6 10 \ S10 = [2(2) + (10 - 1)(-6)] 2 = 5[4 - 54] = -250



Hint: n • Use Sn = [2a + (n - 1)d] where 2 n = number of terms, a = first term and

•





⇒ S10 = 5S5

2n ! 0!(2n − 0)!







d = common difference. 1 Solve S5 = (S10 - S5) 4

49. Option (c) is correct. Explanation: Let us assume a G.P. a, b, c, d, e with common ratio be r. Statement 1: Let the non zero multiplier be s. New sequence will be as, bs, cs, ds, es.

Q

b c d e = = = =r a b c d



Q

bs cs ds es = = = =r as bs cs ds



∴ as, bs, cs, ds, es is also a G.P. Statement 2: Let each of the G.P. is divided by t.



⇒ New sequence will be



Q



⇒



∴

a b c d e , , , , . t t t t t

b c d = = = a b c bt ct d t = = at bt ct

e = r d et = =r d t

a b c d e , , , , is also a G.P. t t t t t

137

SOLVED PAPER - 2021 (I)



So, both the statements are true. Hint: • Assume a, b, c, d, e be a G.P. and use the concept that if common ratio of each consecutive terms are equal then the sequence is G.P.

50. Option (d) is correct. Explanation: Given: 1, 2, 3, 4, 5 are given digits. 5-digit numbers are formed without repetition of given digits. ⇒ All digits 1, 2, 3, 4, 5 will be present in the number. Now, prime numbers is a number that has only two factors one and itself. Also, divisibility rule of 3 says that a number is divisible by 3 when sum of the digits of the number is divisible by 3. Now, the sum of the digits of 5-digit number made by 1, 2, 3, 4, 5 will be 1 + 2 + 3 + 4 + 5 = 15 ∵ 15 is divisible by 3, so every 5-digit number formed will also be divisible by 3. ⇒ No prime numbers can be formed. Hint: • Use the divisibility rule of 3. 51. Option (b) is correct. Explanation: Given: f(x + 1) = x2 - 3x + 2 Let x+1=t⇒x=t-1 ⇒ f(t) = (t - 1)2 - 3(t - 1) + 2 ⇒ f(t) = t2 + 1 - 2t - 3t + 3 + 2 ⇒ f(t) = t2 - 5t + 6 or f(x) = x2 - 5x + 6 Shortcut: f(x + 1) = x2 - 3x + 2 ⇒ f(x) = (x - 1)2 - 3(x - 1) + 2 ⇒ f(x) = x2 - 5x + 6 Hint: • Assume x + 1 = t and replace x by t - 1 in the given functional equation. 52. Option (c) is correct. Explanation: Given: x2, x, -8 are in A.P. As we common difference of an A.P. is equal. ⇒ x2 - x = x - (-8)



⇒ x2 - x = x + 8 ⇒ x2 - 2x - 8 = 0 ⇒ (x - 4)(x + 2) = 0 ⇒ x = 4, -2 \ x ∈ {-2, 4} Hint: • Recall the definition of an A.P.

53. Option (b) is correct. Explanation: Given: Third term of a G.P. = 3 Let a be the first term and r be the common ratio. ⇒ T3 = ar3–1 ⇒ T3 = ar2=3...[i] So, the first five terms of G.P. = a, ar, 3, ar3, ar4 Product = (a)(ar)(3)(ar3)(ar4) = 3a4r8 = 3(ar2)4 = 3(3)4 = 243 [from (i)] Hint: • Use the nth term of a G.P. with a as first term and r as the common ratio is arn–1. 54. Option (a) is correct. Explanation: Given: aij = 2(i + j) After put the values of i and j from 1 to 3, matrix A will be.

4 6 8  A =  6 8 10   8 10 12  4



⇒

6

8

|A| = 6 8 10 8 10 12

2 3 4 3



⇒



Applying C2 → C2 - C1 & C3 → C3 - C2, we get



⇒

|A| = 2 3 4 5 4 5 6

2 1 1 |A| = 2 3 1 1 4 1 1 3

|A| = 0  [∵ Two columns are identical]

138 Oswaal NDA/NA Year-wise Solved Papers Hint: • Find matrix using aij = 2(i + j). • If two columns/rows are identical, then the value of determinant is zero. 55. Option (d) is correct. Explanation: Given: 4 elements 2, 4, 6, 8. Now, total number of determinants possible = 4! = 24 ways. 2 8 = (2 × 4) - (6 × 8) = - 40 6 4



Let one way be



Now, we will get - 40 in three other ways also.



So in 4 ways we will get determinant = - 40 and similarly if we interchange any row or column then determinant gets negative.



⇒

8 2 = (8) × (6) - (2)(4) = +40 4 6



So, 4 determinants will have +40 as its value. Similarly, for every determinant there exists another determinant with negative sign. ⇒ Sum of determinants = 0 Hint: • Use fundamental principle of counting to find total number of ways. • Recall the properties of determinants.

56. Option (d) is correct. Explanation: Given: 4x2 + 4y2 - 20x + 12y - 15 = 0 x2 + y2 - 5x + 3y -

15 =0 4



⇒



As we know radius of circle x2 + y2 + 2gx + 2f + y + c = 0 is r =

g2 + f 2 − c

3 15 5 ,f= ,c=2 4 2 25 9 15 + + \ radius r = 4 4 4

Here g =

r=



⇒

r = 3.5 units

Hint: • Radius of the circle x2 + y2 + 2gx + 2fy + c = 0 is

g2 + f 2 − c .

57. Option (a) is correct. Explanation: Given: Parallelogram has three consecutive vertices (-3, 4), (0, -4) & (5, 2). (x , y) D

(5, 2) C

O

⇒ As



⇒

4 8 2 6 4 6 , , 6 2 8 4 8 2

2 8 = - 40 6 4



49 7 = 4 2





A (–3, 4)

B (0, –4)



Let fourth vertex be (x, y) As we know diagonals of parallelogram bisect each other. ∴ Mid point of AC = mid point of BD



⇒



⇒



⇒



⇒



⇒ x = 2 and y = 10 ∴ Coordinates of fourth vertex is (2, 10).

 5−3 4+2  x+0 y−4 , ,    = 2   2 2   2 x y−4 (1, 3) =  ,  2 2  y−4 x = 1 and =3 2 2 x = 2 and y - 4 = 6

Hint: • Diagonals of parallelogram bisect each other. 58. Option (c) is correct. Explanation: Given lines: y + px = 1 and y - qx = 2 ⇒ y = - px + 1 and y = qx + 2 Compare line (i) with y = mx + c, we get Slope of line (i) is m1 = - p Compare line (ii) with y =mx + c, we get Slope of the line (ii) is m2 = q

…(i) …(ii)

139

SOLVED PAPER - 2021 (I)

As we know product of slopes of two perpendicular lines is -1. ⇒ - pq = -1 ⇒ pq - 1 = 0 Hint: • Use two lines are perpendicular if the product of their slopes is -1. 59. Option (d) is correct. Explanation: Given line Ax + 2By + C = 0 Since, A, B & C are in A.P. ⇒ 2B = A + C Put 2B = A + C in the equation (i), we get Ax + (A + C)y + C = 0 ⇒ A(x + y) + C(y + 1) = 0

⇒

x+y+

C (y + 1) = 0 A



(x1, y1) and (x2, y2) is given by m =







Hint: • If l, m, n are in A.P. then 2m = l + n. • L1 + λL2 = 0 represents family of lines passing through the intersection point of L1 = 0 and L2 = 0. 60. Option (b) is correct. Explanation:

P (–4, 2) Given point



 4−4 2−2 , Now, midpoint of PQ =   = (0, 0)  2 2  As we know equation of line passing through the point (x1, y1) having a slope ‘m′ is given by y - y1 = m(x - x1) ∴ Equation of mirror line is y - 0 = 2(x - 0) ⇒ y = 2x

61. Option (a) is correct. Explanation: Statement-1: As we know if three points P (x1, y1), Q (x2, y2) & R (x3, y3) are colinear, then x 1 y1 1

x2 y2 x3 y3

1 =0 1

Let P (p, p - 3), Q = (q + 3, q) & R = (6, 3)

p q+3 Now, 6  (4, –2) Q Image

line mirror The line mirror is perpendicular to the line joining two points P (-4, 2) and Q (4, -2) and it passes through the mid point of PQ.

−2 − 2 −4 1 = = − 4 − ( −4) 8 2

Hint: • Line mirror is perpendicular to the line joining two points P(-4, 2) and Q(4, -2) and it passes through the mid point of PQ. • If two lines are perpendicular, then product of their slopes is -1. • Equation of line passing through the point (x1, y1) and having a slope ‘m′ is given by y - y1 = m(x - x1)

…(ii)

As we know L1 + λL2 = 0 represents family of lines passing through the intersection point of L1 = 0 & L2 = 0. ∴ Equation (ii) passes through intersection point of x + y = 0 …(iii) and y + 1 = 0 …(iv) On solving equation (iii) & (iv), we get x = 1 and y = -1 ∴ Ax + 2By + C = 0 always passes through the point (1, -1).

So, slope of a line PQ, m1 =

y 2 − y1 x 2 − x1

As we know product of the slopes of perpendicular lines is -1. ∴ Slope of line mirror = 2

…(i)

As we know slope of a line joining two points

p−3 1 q 3

1 = p(q - 3) - (p - 3)(q + 3 1 6) + 1(3q + 9 - 6q)

= pq - 3p - (pq - 3p - 3q + 9) - 3q + 9 =0 ∴ Given points lies on a straight line. So, statement 1 is true. Statement-2: Given points (p, p - 3), (q + 3, q) and (6, 3).

140 Oswaal NDA/NA Year-wise Solved Papers



As we can see that, for any value of p and q it is not necessary that the points lies in the first quadrant only. So, statement 2 is false.

63. Option (a) is correct. Explanation: (4, 2) A

Hint: • Three points P (x1, y1), Q (x2, y2) & R (x3,

x1

y1

1

y3) are colinear if x 2 x3

y2

1 =0 1

y3

62. Option (b) is correct. Explanation: Given lines x - 2 = 0 …(i) x - y - 2 = 0 …(ii) Since, line (i) is parallel to y-axis. ∴ Slope of line (i) is m1 → ∞ Compare line (ii) with y = mx + c, we get

3

∴ Slope of line (ii) is m2 = 3 As we know angle between two lines having the slopes m1 and m2 is given by



m=

 m1 − m2  θ = tan–1    1 + m1m2 



θ = tan–1



⇒

θ = tan–1



⇒

θ = tan–1



(0, 0)







1 + m2 m1

0+ 3

{∵ m1 → ∞}

1

Hint: • Angle between two lines y = m1x + c1 and y

Parallel lines have equal slope.

Slope of AC =

2−0 1 = 4−0 2

⇒ Slope of BD = -2 As we know equation of line passing through the point (x1, y1) and having a slope ‘m′ is given by y - y1 = m(x - x1). ∴ Equation of line BD is y - 0 = -2(x - 0) ⇒ y + 2x = 0

on the hyperbola

⇒ θ = 30° or 150° ∴ Acute angle is 30°.

•

As we know diagonals of square bisect each other at right angle. \ AC ⊥ BD As we know product of slopes of perpendicular lines is -1. 1 \ Slope of BD = Slope of AC

64. Option (d) is correct. Explanation: As we know parametric coordinates of any point

3

= m2x + c2 is given by θ = tan −1

C

Hint: • Diagonals of square bisect each other at right angle. • If two lines are perpendicular, then product of their slopes is -1. • Equation of line passing through the point (x1, y1) having a slope ‘m′ is given by y - y1 = m(x - x1).

m2 m1

1−0

D



∴ Angle between line 1 and line 2 is

1−

B

m1 − m2 . 1 + m1m2

y2 x2 − = 1 is (b tan θ, a sec θ) . a2 b 2



Here, given point (3 tan θ, 2 sec θ) . \ a = 2, b = 3



As we know ecentricity of hyperbola is given by



e=

1+

b2 a2

y2 x2 − =1 a2 b 2

141

SOLVED PAPER - 2021 (I)

9 4



\ cos θ =

13 4



⇒ cos θ =

13 2



⇒ cos θ =



⇒ cos θ =



⇒



\

e=

1+



⇒

e=



⇒

e=

Hint: • Parametric coordinates of any point on the 2

hyperbola •

2

y x − 2 = 1 is (b tan θ, a sec θ) . 2 a b

Eccentricity of hyperbola is e =

1+

b2 . a2

65. Option (d) is correct. Explanation: As we know the eccentricity of a circle is 0 i.e. e = 0 for a circle. So, statement 1 is true. We also know that eccentricity of a parabola is 1 i.e. e = 1 for a parabola. So, statement 2 is true. We also know that eccentricity of an ellipse is always less than 1 i.e. e < 1 for an ellipse. So, statement 3 is also true. Hint: • The eccentricity ‘e′ of a conic section is defined to be the distance from any point on the conic section to its focus, divided by the perpendicular distance from that point to the nearest direction. If e = 0, the conic is circle If e = 1, the conic is parabola If e < 1, the conic is ellipse If e > 1, the conic is hyperbola, 66. Option (b) is correct. Explanation: Given: Direction ratios of two line {6, 3, 6} and {3, 3, 0}. As we know if two lines have direction ratios {a1, b1, c1} and {a2, b2, c2}, then angle between two lines is given by a1 a2 + b1b2 + c1 c 2 cos θ = 2 a1 + b1 2 + c1 2 ⋅ a2 2 + b2 2 + c 2 2 Here, a1 = 6, b1 = 3, c1 = 6, a2 = 3, b2 = 3, c2 = 0. Let angle between two lines be θ.

θ=

(6)(3) + (3)(3) + (6)(0) 36 + 9 + 36 ⋅ 9 + 9 + 0 18 + 9 (9)(3 2 ) 27 9(3 2 )

1 2 π 4

Hint: • If angle between two lines having direction ratios {a1, b1, c1} and {a2, b2, c2} be θ, the

a1 a2 + b1b2 + c1 c 2

cos θ =

2

a1 + b1 2 + c1 2 ⋅ a2 2 + b2 2 + c 2 2

67. Option (b) is correct. Explanation: Given line x - 1 = 2(y + 3) = 1 - z y + 3 z −1 x −1 = = 1 −2 2



⇒



⇒ Direction ratios of line = {2, 1, -2}



[∵ Direction ratios of line



is (a, b, c)] As we know if direction ratios of the line is {a, b, c}, then direction cosine of the line will be {l, m, n}

a  , =  2 2 2  a +b +c

b 2

2

a +b +c

2

,

c

  a +b +c  2

2

2

∴ Direction cosine of given line is {l, m, n} 2  , =   4 +1+ 4



x − x1 y − y1 z − z1 = = a b c

1 4 +1+ 4

,

−2

  4 +1+ 4 

2 1 −2  ⇒ {l, m, n} =  , ,  3 3 3  4

4

 2   1   −2  Now, l4 + m4 + n4 =   +   +   3 3  3 



=

16 1 16 + + 81 81 81



=

33 11 = 81 27

4

142 Oswaal NDA/NA Year-wise Solved Papers Hint: • If a, b & c are the direction ratios of the line then direction cosine of the line is {l, m, n}

a  , =  2 2 2  a +b +c

b 2

2

a +b +c

2

,

c

  a +b +c  2

2

2

68. Option (c) is correct. Explanation: Given points A (1, 7, -5) and B (-3, 4, -2) As we know projection of two lines joining the two points P (x1, y1, z1) and Q (x2, y2, z2) on the line having direction cosines l, m & n is given by P′Q′ = |l(x2 - x1) + m(y2 - y1) + n(z2 - z1)| Now, direction cosine of y-axis is {l, m, n} = {0, 1, 0} ∴ Projection of the line segment joining A & B on y-axis is A′B′ = |0(-3 - 1) + 1(4 - 7) + 0(-2 + 5)|

A′B′ = 3

Hint: • Projection of two line joining the two points P (x1, y1, z1) and Q (x2, y2, z2) on the line having direction cosines l, m & n is given by P′Q′ = |l(x2 - x1) + m(y2 - y1) + n(z2 - z1)| • Direction cosine of y-axis is 0, 1, 0. 69. Option (c) is correct. Explanation: Given: The line joining the points (k, 1, 3) and (1, -2, k + 1) also passes through the piont (15, 2, -4) As we know equation of line passing through the points (x1, y1, z1) and (x2, y2, z2) is given by

y − y1 z − z1 x − x1 = = y − y z x 2 − x1 2 1 2 − z1

⇒

−1 −7 15 − k = = 3 k−2 1−k



⇒

−1 −7 −1 15 − k = = or 3 k−2 3 1−k



⇒ 45 - 3k = -1 + k or -21 = -k + 2 ⇒ 4k = 46 or k = 23



23 or k = 23 2 So, there are two possible values of k is possible.

⇒

k=

Hint: • Equation of line passing through the points (x1, y1, z1) and (x2, y2, z2) is given by

x − x1 y − y1 z − z1 = = x 2 − x 1 y 2 − y 1 z 2 − z1 70. Option (c) is correct. Explanation: P (0, 0, 0)



x+ y +z=3

Q

Let P be the given point and Q be the foot of the perpendicular. Given: Equation of plane

x+y+z=3 ∴ Direction ratios of the normal of plane are {1, 1, 1}.

As we know equation of line passing through the point (x1, y1, z1) having direction ratios a, b, c is given by

z − z1 y − y1 x − x1 = = c b a y−0 x−0 z−0 = = 1 1 1



∴ Equation of line PQ is



⇒

x−k y −1 z−3 = = 1−k −2 − 1 ( k + 1) − 3



Let the coordinates of point Q be (λ, λ, λ)



Since, Q lies in the plane x + y + z = 3

x−k y −1 z −3 = = 1−k −3 k−2



\

λ+λ+λ=3



⇒

λ=1



∴ Coordinate of Q is (1, 1, 1).

∴ Equation of line passing through the points (k, 1, 3) and (1, -2, k + 1) is







⇒



Since, line passes through the piont (15, 2, -4)



\

2 − 1 −4 − 3 15 − k = = −3 k−2 1−k

x=y=z=λ

143

SOLVED PAPER - 2021 (I)

Hint:

•

72. Option (c) is correct. Explanation:      Given: c= a + b , a= b ≠ 0

P (0, 0, 0)



x+ y +z=3

Q

Use equation of line PQ is x = y = z. Assume the coordinates of point Q on line and find the value of parametric coefficient by satisfying the point Q in the plane.

71. Option (c) is correct. Explanation:  r aiˆ + bjˆ is equally inclined to Given: Vector = both x and y axes and magnitude of the vector is 2 units.  r =2 ⇒

a2 + b 2 = 2 a2 + b2 = 4 …(1)  ˆ ˆ r ai + bj is equally inclined to both ∵ Vector = x and y axes. aiˆ + bjˆ .ˆi Let θ be the angle between the and 2 2 a + b .(1) both x and y axis. aiˆ + bjˆ .ˆj aiˆ + bjˆ .ˆi ⇒ cos θ = = 2 2 a 2 + b .(1) a 2 + b .(1) a b ⇒ cos θ = = 2 2 ⇒ a=b Put a = b in the equation (1), we get a 2 + a2 = 4 ⇒ 2a2 = 4 ⇒ a=± 2 Q a=b \ b=± 2

⇒ ⇒

(

(



)

)

(

)

So, value of a & b are ± 2 & ± 2 respectively. Hint: •

Magnitude

 = r •

of

x2 + y2 .

vector

is

  If angle between two vectors a and b is θ then cos θ =

a ⋅b a b

  Statement-1: As we know if p and q are  

0. perpendicular, then p ⋅ q =        Now, c ⋅ ( a − b ) = ( a + b ) ⋅ ( a − b )         = a ⋅a − a ⋅b + b ⋅a − b ⋅b 2     2 = a − a ⋅b + a ⋅b − b 2 2 = a −b

  = 0 ∵ a = b     ⇒ c is perpendicular to ( a − b ). So, statement 1 is true. Statement-2:        Now, c ⋅ ( a × b ) = ( a + b ) ⋅ ( a × b )       = a ⋅ (a × b ) + b ⋅ (a ⋅ b )       =  a a b  + b a b  =0+0 =0    c ⇒ is perpendicular to ( a × b ) . So, statement 2 is also true. Hint:   • If p and q are perpendicular, then   p⋅q = 0. • •

      2 Use x ⋅ x =x and x ⋅ y = y ⋅ x .       Use a ⋅ (b × c ) = a b c 

73. Option (c) is correct. Explanation:     a +b = a −b = 4 Given:  2  2 a +b = a −b ⇒ 2           ⇒ ( a + b ) ⋅ ( a + b ) = ( a − b ) ⋅ ( a − b ) ∵ x = x ⋅ x                   ⇒ a ⋅ a + a ⋅ b + b ⋅ a + b ⋅ b = a ⋅ a − a ⋅ b − b ⋅ a + b ⋅ b 2     2 2     2 ⇒ a + a ⋅ b + a ⋅ b + b = a − a ⋅ b − a ⋅ b + b   ⇒ 4a ⋅ b = 0   ⇒ a ⋅b = 0   ⇒ a⊥b

144 Oswaal NDA/NA Year-wise Solved Papers Hint: • Squaring both the sides of given equation 2   x = x ⋅ x and and simplify using     x⋅y = y⋅x .   • If two vector a and b are perpendicular   then a ⋅ b = 0. 74. Option (c) is correct. Explanation:    Given: a , b and c are coplanar.    ⇒  a b c  = 0       Let A = (2 a × 3b ) ⋅ 4 c + (5b × 3c ) ⋅ 6 a       ⇒ A =  2 a 3b 4 c  +  5b 3c 6 a  {∵ ( p × q ) ⋅ r =[ p q r ]}    ⇒ A = (2 × 3 × 4) [ a b c ] + (5 × 3 × 6) [ b c a ]   ∵ [a= b c ] [b= c a ] 0  ⇒ A = 0 + 0 Hint: • •

    If a , b and c are coplanar, then [ a b c ] = 0.     [a b c ] . Use ( a × b ) ⋅ c =

75. Option (d) is correct. Explanation: Statement-1: As we know the cross product     of two vectors p and q is given by p × q =   ˆ. p q sin θn   Let a = b = 1     a × b = a b sin θ ⇒   a × b = sin q ⇒ As we know the range of sin θ is [-1, 1]. So, it is not necessarily true that the cross product of two unit vectors is always a unit vector. So, statement 1 is false. Statement-2:   Let, a and b are two unit vectors.   a =1& b =1 ⇒  As we know the dot product of two vector p      p ⋅ q p q cos θ. and q is given by=     \ a ⋅ b = a b cos θ

  a ⋅ b = cos q



⇒



Q -1 ≤ cos θ ≤ 1

So, it is not necessarily true that dot product of two vectors is always a unit vector. So, statement 2 is false. Statement-3:   Let a and b are two unit vectors.   a = b =1 ⇒ 2 2     a + b = a + b + 2a ⋅ b Now,     a + b = 1 + 1 + 2a ⋅ b ⇒     a + b = 2 1+ a ⋅b  ⇒ …(1)   2 2    a − b = a + b − 2a ⋅ b Now,     a − b = 1 + 1 − 2a ⋅ b ⇒     a − b = 2 1− a ⋅b  ⇒ …(2)   If angle between vectors a and b is 90°, then   a ⋅ b = 0.     a +b = a −b = 2 ⇒ So, the magnitude of sum of two unit vector is not always greater than the magnitude of their difference. So, statement 3 is also false. Hint: • •

•

  The cross product of two vectors p and q     ˆ. p × q p q sin θn is given by =   The dot product of two vectors p and q     p ⋅ q p q cos θ. is given by=   p + q=

2 2   p + q + 2p ⋅ q

  p − q=

2 2   p + q − 2p ⋅ q

76. Option (c) is correct. Explanation: ax − x a = -1 x a − aa



Given: lim



LHS limit has

x →a

0 indeterminate form. 0

As we know if

lim x →a

f (x) g(x)

indeterminate form, then lim x →a

has

0 0

or

f (x) f '( x ) = lim g ( x ) x →a g '( x )

∞ ∞

145

SOLVED PAPER - 2021 (I)



\

d x (a − x a ) dx lim = -1 x →a d ( x a − aa ) dx



⇒ lim x →a

a x log e a − ax a −1 = -1 ax a −1 − 0

d n  d x  x = log e a , ( x ) nx n−1  ∵ dx ( a ) a=  dx

⇒

 a x log e a  lim  − 1 = -1 x → a  ax a − 1  lim

⇒

x →a

78. Option (d) is correct. Explanation:

b

g ( x )dx =

∫

…(i)

g (b x )dx

\ I =

∫

a



⇒ I =

∫

a



Adding equation (i) & equation (ii), we get

Hint:



Integrating both sides of above equation, we get



⇒

= ∫ dt

⇒ ln(x + 1) = t + C ∵ At t = 0, x = 0 ∴ ln(1) = 0 + C ⇒ C = 0 \ ln(x + 1) = t Now, at x = 24 m t = ln(24 + 1) ⇒ t = ln25 ⇒ t = ln52 ⇒ t = 2 ln5

…(ii)

f (x) + f (a − x) dx f (x) + f (a − x)

∫



⇒ 2I = [ x ]0 ⇒ 2I = a - 0 = a



⇒ I =

a

0

1 ⋅ dx a

a 2

Hint:

dx Given: =x+1 dt

⇒

f (x) dx  f (a − x) + f (x)

⇒ 2I =

77. Option (c) is correct. Explanation:



0

f ( a − ( a − x )) dx f ( a − x ) + f ( a − ( a − x ))



f (x) f ′( x ) = lim form then lim . x→a g( x ) x → a g ′( x )

dx = dt x +1

0

∫



d x d n ( a ) = a x log e a . ( x ) = nx n−1 , dx dx

0

2I =

∞ 0 f (x) If lim has or indeterminate x →a g ( x ) ∞ 0

dx

∫

f (a − x) dx  f (x) + f (a − x)



⇒ logea = 0 ⇒ a=1

∫ x +1

a

0

a

a log e a = -1 + 1 = 0 ax a −1





∫

I=

Let

As we know

a log e a =0 a ⋅ a a −1

Use

dx = dt , integrating both the sides x +1 and solve further using the basic formulae of indefinite integration from definion.

Use



⇒

•

•

x



•

Hint:

∫

b

g (= x )dx

∫

b

g (b − x )dx .

•

Use

•

If f ( x ) = F( x ) , then

0

0

∫

b

a

f ( x )dx = F(b) - F(a).

79. Option (b) is correct. Explanation: Let

x3 + x2 x →−1 x 2 + 3 x + 2

L = lim

Given limit has

0 indeterminate form. 0

As we know if

lim x →a

f (x) g(x)

indeterminate form, then lim x→a



\ L = lim

x →−1

has

0 0

or

∞ ∞

f (x) f ′( x ) = lim x → a g( x ) g ′( x )

d 3 (x + x 2 ) dx

d 2 ( x + 3x + 2) dx

146 Oswaal NDA/NA Year-wise Solved Papers 3x 2 + 2 x  d n  ∵ ( x ) = nx n−1  x →−1 2 x + 3  dx 



⇒ L = lim



3( −1)2 + 2( −1) ⇒ L = 2( −1) + 3



⇒ L =

a



= dx \ ∫ [ f ( x ) + f ( −x )]



⇒

0

3−2 =1 −2 + 3

x3 + x2  Let L = lim 2 x →−1 x + 3 x + 2

0   0 form 

[Applying L′hospital rule]

Hint: •

If lim x →a

∞ f (x) 0 has or indeterminate ∞ 0 g(x)

form then lim x→a

•



−a

0

I=

⇒ I =

∫

∫

0

−a

g ( x )dx + ∫ g ( x )dx 0



c

I=

∫

0

⇒ I =

∫

a



a

0



⇒ I =

∫

a

0

⇒ I =

a

b

f ( x )dx + ∫ f ( x )dx ; a < c < b c

a

f ( x )dx = − ∫ f ( x )dx b

b

f ( x )dx = ∫ f (t)dt a

y=

16 − x 2

⇒ y2 = 16 - x2 ⇒ x2 + y2 = 16 ⇒ It represents circle of radius 4 units with centre at (0, 0). Y



(4, 0)

X

So, Area bounded by y =



a

0

1 [π(4)2 ]. 2

⇒ A =

1 π(16) = 8π sq. units 2

Hint: • Draw the graph of curve & identify the required region and use area of circle is πr2, where ‘r′ is the radius of circle.

a

g ( −t)dt + ∫ g ( x )dx

16 − x 2 , y ≥ 0 and the

[∵ Area of circle of radius ‘r′ is πr2]

0

82. Option (c) is correct. Explanation: Given: y = -x3 + 3x2 + 2x - 27

a

g ( −x )dx + ∫ g ( x )dx 0



As we know slope of the curve y = f(x) is



\

dy d = (-x3 + 3x2 + 2x - 27) dx dx



⇒

dy = -3x2 + 6x + 2 dx

a

∫ [ g(x) + g(−x)]dx 0

∫

b

a

c

a



− g ( −t)dt + ∫ g ( x )dx

∵ b h( x )dx = b h(t)dt  ∫a  ∫a 



b

∵ b h( x )dx = − a h( x )dx  ∫b  ∫a 



b

x-axis is A =

∵ h= ( x )dx ∫ h( x )dx + ∫ h( x )dx ; a < c < b   ∫a  a c Put x = - t in first integral, we get b

∫

∫

Given:

g ( x )dx a

−a

2.

f= ( x )dx

a

∫ [ f (x) + f (−x)]dx = ∫ g(x)dx

Let

b

a

(–4, 0)

a

a

∫



f (x) f ′( x ) = lim . x → a g( x ) g ′( x )

80. Option (a) is correct. Explanation: Given:

1.

3.

d n ( x ) = nx n−1 dx

Use

g(x) = f(x)

81. Option (b) is correct. Explanation:

3x 2 + 2 x x →−1 2 x + 3

⇒ L = lim

 ⇒ L = 1

0

Hint: • Use properties of definite integration

Shortcut:



a

∫ [ g(x) + g(−x)]dx

dy . dx

147

SOLVED PAPER - 2021 (I)

 d n n −1  ∵ dx ( x ) = nx 

Hint: • Area of ∆ABC =

As we know quadratic expression ax2 + bx + c;

∴

•

dy −6 has maximum value of x = =1 dx 2( −3)

Hint: • •

dy . dx Quadratic expression f(x) = ax2 + bx + c;

Slope of the curve y = f(x) is

a < 0 has maximum value at x = −

b . 2a

83. Option (a) is correct. Explanation:

maximum when x = y = z. 84. Option (d) is correct. Explanation: Given: f(x) = e|x| Let us generalize the f(x) using the definition of modulus function.  e x ; x ≥ 0 So, f(x) =  − x e ; x < 0 Differentiate the f(x) w.r.t. x, we get

 ex ; x≥0  f ′(x) =  − x d ( −x ); x < 0 e  dx



A c 60°

a+b+c , & sides of ∆ are a, b, c. 2 Use if x + y + z = constant, then xyz is

where s =

b a < 0 has maximum value at x = . 2a



 d  ∵ dx [ f ( g ( x )) = f ′( g ( x )) ⋅ g ′( x )

b



B

D a

C

Perimeter of ∆ABC = 24 ⇒ a + b + c = 24 By Heron′s formula,  Area of ∆ABC =

s( s − a)( s − b)( s − c )

a+b+c 2 For maximum area, s(s - a)(s - b)(s - c) should be maximum. Since, s is a constant (because the perimeter is fixed). So, for maximum area, (s - a) (s - b)(s - c) should be maximum. Now, (s - a) + (s - b) + (s - c) = 3s - (a + b + c) = 36 - 24 = 12 = constant. As we know by AM-GM inequality, a product xyz is maximum, if sum x + y + z is a constant, when x = y = z.

where s =

∴ For maximum area, s - a = s - b = s - c ⇒ a=b=c ⇒ a = b = c = 8 [∵ a + b + c = 24] ⇒ ∆ABC is equilateral triangle. ⇒ ∠A = ∠B = ∠C = 60° Now, length of altitude = c sin 60°

⇒ Length of altitude = 8∙

3 = 4 3 cm 2

s( s − a)( s − b)( s − c ) ;



x  e ; x ≥ 0 f ′(x) =  − x −e ; x < 0

⇒

Let us check differentiability of f(x) at x = 0. Now, L.H.D. = f ′(0–) = -e–0 = -1 R.H.D. = f ′(0+) = e0 = 1 Q L.H.D. ≠ R.H.D. \ f ′(0) does not exist. Hint: • • •

 x; x ≥ 0 Use |x| =  −x ; x < 0

Function f(x) is differentiable at x = a if L.H.D. = R.H.D. = Finite at x = a. d [ f ( g ( x ))] = f ′( g ( x )) ⋅ g ′( x ) dx

85. Option (d) is correct. Explanation: Let

I=

dx

∫ sec x + tan x dx 1 sin x + cos x cos x



⇒ I =

∫



⇒ I =

∫ 1 + sin x dx

cos x

148 Oswaal NDA/NA Year-wise Solved Papers t ⇒ cos xdx = dt Let 1 + sin x =

1

∫ t dt



\ I =



⇒ I = ln|t| + C ⇒ I = ln|1 + sinx| + C



⇒ I = ln



⇒



⇒ I = ln (sec x + tan x ) + ln cos x + C



⇒ I = ln sec x + tan x + ln



⇒ I = ln sec x + tan x − ln sec x + C

1 + sin x ⋅ cos x + C cos x

I = ln (sec x + tan x ) ⋅ cos x + C

Hint: • Simplify

1 +C sec x

As we know, A.M. ≥ G.M.



⇒



⇒



⇒ P ≤ 100 ∴ Maximum value of P is 100.

x+y ≥ 2

Hint: • Use A.M. ≥ G.M. 88. Option (a) is correct. Explanation: Let f(x) = sin(lnx) + cos(lnx) d [sin(lnx) + cos(lnx)] dx d d f ′(x) = {sin(lnx)} + {cos(lnx)} dx dx

⇒

1 & cosθ



⇒

sin θ and then solve further using cos θ substitution method.



⇒

Use ln(ab) = lna + lnb and lnab = blna.



the

given

integral

using

tan θ =

86. Option (b) is correct. Explanation: Let

I=

dx ∫ sec 2 (tan −1 x)



As we know 1 + tan2θ = sec2θ



\ I =

dx ∫ 1 + tan 2 (tan −1 x)



⇒ I =

dx ∫ 1 + [tan(tan −1 x)]2



⇒ I =

∫1+ x



⇒ I = tan–1x + C

dx

2

[∵ tan(tan–1x) = x]

Hint: • Simplify using 1 + tan2θ = sec2θ. • Use tan(tan–1x) = x. • Recall basic formulae of indefinite integration from definition. 87. Option (a) is correct. Explanation: Given: x + y = 20 P = xy

xy

20 ≥ (P)1/2 2 ⇒ 10 ≥ (P)1/2



trigonometric identities sec θ =

•



f ′(x) =

f ′(x) = cos(lnx)

{

∵

d d (lnx) - sin(lnx) (lnx) dx dx

}

d [ f ( g ( x ))] = f ′( g ( x )) ⋅ g ′( x ) dx



⇒

1 1 f ′(x) = cos(lnx)   - sin(lnx)∙   x x



⇒

f ′(x) =

cos(ln x ) − sin(ln x ) x

Put x = e in the above equation, we get cos(ln e ) − sin(ln e ) e cos1 − sin1 f ′(e) = [∵ lne = 1] e f ′(e) =



⇒ Hint: • •

d [ f ( g ( x ))] = f ′( g ( x )) ⋅ g ′( x ) dx d d (sin x ) = cos x , (cos x ) = − sin x Use dx dx d 1 & (ln x ) = dx x

Use

89. Option (b) is correct. Explanation: Given: x = etcos t & y = et sin t dx / dt dx Now, = dy / dt dy

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Now,

dx d t = [e cos t] dt dt

Hint: • Simplify the given expression using trigo-

d dx t d (cos t) + cos t e t = e dt dt dt

⇒

[∵ (uv)′ =



⇒

dx = −e t sin t + cos t ⋅ e t dt



⇒

dx t = e (cos t − sin t) dt

• u′ v + v′u]

d t dy Now, = [e sin t ] dt dt

⇒

d dy t d (sin t) + sin t ( e t ) = e dt dt dt



⇒

dy = e t cos t + sin t ⋅ e t dt



⇒

dy t = e (sin t + cos t) dt



\

dx e t (cos t − sin t) = t dy e (sin t + cos t)



⇒

dx cos t − sin t = dy sin t + cos t



cos0 − sin 0  dx  ⇒   = =1  dy t =0 sin 0 + cos0



Hint: • •

dx dx dt = Use dy dy dt Use product rule of differentiation (uv)′ = uv′ + vu′.

90. Option (a) is correct. Explanation: Let f(x) = sin 2x ∙ cos 2x

⇒

1 f(x) = × (2 sin 2x ∙ cos 2x) 2



⇒

f(x) =



As we know the range of sinx is [-1, 1]



\ [f(x)]max =

1 sin 4x[∵ sin 2α = 2 sin α ∙ cos α] 2 1 2

2α 2 sin α ⋅ cos α . nometric formula sin= Range of sin x is [-1, 1].

91. Option (a) is correct. Explanation: d( e x ) x d( e ) = dx d e d( x e ) (x ) dx

d( e x ) ex = e −1 e d( x ) ex

⇒

d n  d x  x = and ( x ) nx n−1  ∵ dx ( e ) e=  dx



⇒

d( e x ) ex = e −1 e d( x ) ex ⋅ x



⇒

xe x d( e x ) = e d( x ) ex e

Hint: •

d( e x ) d( e ) = dx Use d( x e ) d( x e ) dx

•

Use

x

d n d x ( x ) = nx n−1 (e ) = e x & dx dx

92. Option (b) is correct. Explanation: Given: lim x →−1



⇒

f (x) + 1 3 =x2 − 1 2

lim{ f ( x ) + 1}

x →−1

lim{x 2 − 1}

=-

x →−1

3 2

3 lim( x 2 − 1) 2 x →−1 3 lim f ( x ) + 1 = - {( −1)2 − 1} x →−1 2



⇒ lim{ f ( x ) + 1} = -



⇒



⇒



⇒

x →−1

{

}

lim f ( x ) + 1 = 0

x →−1

lim f ( x ) = -1

x →−1

Hint: • Simplify the given limit using algebra of limits and solve further.

150 Oswaal NDA/NA Year-wise Solved Papers 93. Option (a) is correct. Explanation:  a + bx , x < 1  Given: f(x) =  5, x = 1 b − ax , x > 1 



As we know if function is continuous at a point then limiting value of a function is equal to the functional value at that point. So, for f(x) to be continuous at x = 1



⇒



⇒ \

lim f ( x ) = lim f ( x ) = f(1) +

x →1−

x →1

lim( a + bx ) = lim(b − ax ) = 5

x →1−

x →1+

a+b=b-a=5 a+b=5

Hint: • If g(x) is a continuous at a point then limiting value of g(x) is equal to the functional value of g(x) at that point. 94. Option (b) is correct. Explanation: Given: f(x) = sin x ⇒ f ′(x) = cos x  

π π   and f ′(x) < 0, ∀x ∈  , π  2  2



Q f ′(x) > 0, ∀x ∈  0,



∴ f(x) is not strictly increasing function in the interval (0, π). So, statement-1 is false.



Q f ′(x) < 0 ∀x ∈ 



 5π  , 3π   2 

 5π  ∴ f(x) decreases in the interval x ∈  , 3π   2  So, statement 2 is true. Hint: • If function is increasing, then first derivative of function is positive. • If function is decreasing, then first derivative of function is negative.

95. Option (a) is correct. Explanation: Given function f(x) = 3x

Let y = 3x Take logarithm on both sides log y = log 3x ⇒ log y = x log 3

⇒



⇒

log y log 3 x = log3y

x=

Because

log a = log(b)(a) log b

From here we can clearly say that domain of f(x) = 3x is R i.e. (-∞, ∞) Hint: • Domain for exponential function, y = ax is R where a ∈ R+ & a ≠ 1. • Use properties of logarithm:

log a = logba log b



1.



2. log am = m log a

96. Option (a) is correct. Explanation: Order of differential equation is equal to the number of arbitrary (independent) constants in it. Given, solutions of differential equation is y2 + 2cy - cx + c2 = 0 Here number of arbitrary constant is 1 which is c. ∴ Order of differential equation is 1. Hint: • Order of differential equation is equal to the number of arbitrary constant in it. 97. Option (a) is correct. Explanation: • Degree of differential equation is defined when differential equation can be expressed in the form of a polynomial. • When differential equation is expressed in form of polynomial, then degree is the power of highest order derivative in the differential equation. Given differential equation,

x=

1+

d2y dx 2

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Squaring both the sides, we get d y dx 2 ∵ Power of highest order derivative is 1. ∴ Degree of differential equation is 1.



2

x2 = 1 +

Hint: Degree of differential equation is the power of highest order derivative when differential equation express in form of polynomial. 98. Option (b) is correct. Explanation: Given: y = aex + be–x Differentiating both the sides of the above equation w.r.t. x,

dy d d ( ae x ) + (be − x ) = dx dx dx

dy = aex - be–x dx Again differentiating above equation w.r.t. x,

⇒



d2y d d ( ae x ) − (be − x ) = dx 2 dx dx



⇒

d2y = aex + be–x dx 2



⇒

d2y =y dx 2



⇒

d2y -y=0 dx 2

Hint: • Use the method of eliminating the arbitrary constants by differentiation it two times (as two constants are present). 99. Option (c) is correct. Explanation:

 dy  Given: Differential equation ln   + y = x.  dx 



⇒



⇒



⇒



⇒

 dy  ln   = x - y  dx  dy = ex–y dx dy ex = y dx e y e dy = exdx



Integrating both the sides, we get

∫ e dy y



x = ∫ e dx

⇒ ey + C1 = ex + C2 ⇒ ex - ey = C[where C = C1 - C2] This is the required solution. Hint: • Use variable separable method to find the required solution.

100. Option (d) is correct. Explanation: Let

( 2 ln x + ln x ) dx I = ∫e 2



⇒ I =

∫e

(ln x 2 + ln x 2 )



⇒ I =

∫e

(ln( x 2 ⋅ x 2 )

dx [∵ logabm = mlogab]

dx [∵ logab + logac = logabc]





∫ e dx ⇒ I = ∫ x dx [∵ alogax = x]



⇒ I =



4

ln( x )

⇒ I =

4

x5 +C 5

Hint: • Use principle properties of logarithm, logabm = mlogab, logab + logac = logabc and alogax = x. 101. Option (c) is correct. Explanation: Arithmetic mean of n numbers is, Σai a + a +…+ an A.M. = = 1 2 n n

Geometric mean of n numbers is,



G.M. = n

n

∏ a= i =1



i

n a1 ⋅ a2 ⋅ ... ⋅ an

So, we can clearly see that both A.M. and G.M. uses all the data. Shortcut: Σai n



A.M. of n numbers =



G.M. of n numbers = n Πai



So, both uses all the data.

152 Oswaal NDA/NA Year-wise Solved Papers 2. The coefficient of correlation is independent of change of scale, which means some value is multiplied or divided to observations. 3. The coefficient of correlation is independent of change of origin of the variable X and Y, which means some value has been added or subtracted in observations. So, it is independent of both change of scale and change of origin.

Hint: • Recall the formula of A.M. of n numbers and G.M. of n numbers. 102. Option (b) is correct. Explanation: Given: Science Arts Commerce No. of graduates 30 70 50 Total number of graduates = 30 + 70 + 50 = 150 Now, when these are represents on pie chart, then 150 graduates are represented by 360°   1 graduate is represented by 360 = 12 150 5 12 30 graduates are represented by × 30° = 72° 5 Hint: • Recall the concept of representation of data on a pie chart. 103. Option (b) is correct. Explanation: In a histogram, it is the area of the bar that denotes the value. While constructing a histogram with non-uniform (unequal) class intervals (widths), we must ensure that the area of the rectangles are proportional to the class frequency. ⇒ The frequency of a class should be proportional to the area of the rectangle. Shortcut: From the definition of histogram, the frequency of a class is proportional to the area of rectangle and not its height. Hint: • Recall the properties of histogram. 104. Option (c) is correct. Explanation: In simple regression analysis, the coefficient of correlation, is a statistic which indicates an association between the independent variables and dependent variables. By the properties of coefficient of correlation. 1. The coefficient of correlation lies between -1 and 1.

Hint: • Recall the properties of coefficient of correlation. 105. Option (a) is correct. Explanation: Given: Σfi = 198

For odd

Σf, median =

n+1 2



For even

Σf, median =

n 2

198 = 99 2 99 is closest to frequency 76 which is for 3 number of peas.

Median =



∴ Median = 3. Hint: • Recall the definition of median of frequency distribution.

106. Option (b) is correct. Explanation: Given = M is mean of x1, -k, x2 - k, …, xn - k n



⇒ M =

∑ (x i =1

i

− k)

n ( x1 − k ) + ( x 2 − k ) + ... + ( x n − k ) ⇒ M = n x1 + x 2 + ... + x n nk − ⇒ M = n n



⇒ M = M′ - k



Now, required mean = M′ =



⇒ M′ = M + k

x1 + x 2 +…+ x n n

Hint: •

Use mean =

Sum of observations No. of observations

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107. Option (c) is correct. Explanation: Given: Values are 73, 85, 92, 105, 120 Mean = x =

73 + 85 + 92 + 105 + 120 5 475 5



⇒

x =



⇒

x = 95



Required value =

5

∑ (x i =1



i

⇒



⇒ 2t2 - 5t + 2 = 0 ⇒ 2t2 - 4t - t + 2 = 0 ⇒ (t - 2)(2t - 1) = 0



⇒

t = 2 or



⇒

m 1 = 2 or n 2



⇒



⇒

− x)

= (x1 - x ) + (x2 - x ) + (x3 - x ) + (x4 - x ) + (x5 - x ) = (73 - 95) + (85 - 95) + (92 - 95) + (105 - 95) + (120 - 95) = -22 - 10 - 3 + 10 + 25 =0

•

Use mean = x =

Σx i . n

108. Option (d) is correct. Explanation: Given: H.M. of m & n = x G.M. of m & n = y 5x = 4y

x 4 = y 5



⇒



∵ H.M. of m & n =



and G.M. of m & n =



2mn = x. m+n

4 2mn = 5 (m + n) mn



⇒

2 mn = 5 m+n



n  5  m + ⇒   = 2 mn   mn



m n 5 + = n m 2

⇒

Let

m =t n

xy .

109. Option (a) is correct. Explanation: Given: Mean = 100 Coefficient of variance = 45% σ Q CV = µ  [σ = standard deviation, µ = mean] σ = C.V. × µ 45 × 100 ⇒ σ= 100 ⇒ σ = 45 Variance = σ2 = (45)2 = 2025 Hint: • Use coefficient of variance

x 2mn = y (m + n) mn

⇒

m 1 = 4 or n 4 m = 4n or n = 4m

geometric mean of x and y =

mn = y



1 2

Hint: 2xy • Use harmonic mean of x & y = and x+y

Shortcut: Sum of deviation from mean is zero. Hint:

t+

1 5 = t 2







=

Standard deviation Mean

110. Option (c) is correct. Explanation: Given: P(A) = L, P(B) = M  A  P(A ∩ B) P  = P(B) B By addition theorem of probability  A  P(A) + P(B) − P(A ∪ B) P  = P(B) B By Axiomatic approach of probability,





P(A ∪ B) ∈ [0, 1]

⇒ P(A ∪ B) ≤ 1

154 Oswaal NDA/NA Year-wise Solved Papers

L + M −1 A P  ≥ M B

⇒ Hint: •

Use

conditional

probability,

A P  B

P(A ∩ B) = and addition theorem of P(B) •

probability, P(A ∪ B) = P(A) + P(B) - P(A ∩ B) Use the Axiomatic approach of probability, probability lies between 0 and 1.

111. Option (b) is correct. Explanation: As we know, the mean or average of a data set is found by adding all the numbers in the data set and then dividing by number of values in the set.





x=

Σx i 12



⇒ 75 =

Σx i 12



Σxi = 75 × 12 ⇒ Mean after observation a and b are discarded = 65. Σx i − ( a + b ) ⇒ 65 = 10 ⇒ Σxi - (a + b) = 650 ⇒ (a + b) = (75 × 12) - 650 ⇒ a + b = 250





Hint:

x =

Mode is the number that occurs most often in the data set. (1) 12, 12, 12, 12, 24

Mean =

12 + 12 + 12 + 12 + 24 72 = 5 5

Median = 12 Mode = 12 (2) 6, 18, 18, 18, 30 Mean =

6 + 18 + 18 + 18 + 30 90 = = 18 5 5

Median = 3rd value = 18 Mode = 18 Mean = Mode = Median Hint: • Mean is found by adding all the numbers of data set and then dividing by number of values in the set. • Median is the middle value when a data set is in ascending order. • Mode is the number that occurs most often in data set. 112. Option (b) is correct. Explanation: Given: Mean of 12 observations = 75 Let two discarded observations be a and b.

a+b 250 = 2 2 ⇒ Mean = 125

Mean of a & b =

Σx i n Median is the middle value when a data set is in ordered form (least to greatest)





•

Use mean = x =

Σx i where n is number n

of observations. 113. Option (c) is correct. Explanation: Given: k is one root of x(x + 1) + 1 = 0 ⇒ x2 + x + 1 = 0 By Sridharacharya rule, roots of ax2 + bx + c = 0 are

x=

−b ± b 2 − 4 ac 2a



⇒

x=

−1 ± 1 − 4 2



⇒

x=

−1 ± i 3 2

Let

k=

−1 + i 3 −1 − 3i and β = 2 2



k2 =

( −1 + i 3)2 4



⇒

k2 =

1 − 3 − 2 3i 4



⇒

k2 =

−2 − 2 3i 4



⇒

k2 =



⇒

−1 − 3i 2 2 k =β

155

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By addition theorem of probability, P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Shortcut: x(x + 1) + 1 = 0 ⇒ x2 + x + 1 = 0

⇒

x=

−1 ± 1 − 4 2

−1 ± i 3 2 2 ⇒ Roots are ω and ω Let k = ω2 and β = ω k2 = (ω2)2 = ω3ω = ω k2 = β =









⇒ P(B) =



Statement 1: A and B are independent events. As we know, if P(A ∩ B) = P(A) ∙ P(B) then A and B are independent events.



114. Option (c) is correct. Explanation: Let number of observation = n





−b ± b 2 − 4 ac . 2a

x=

⇒



G.M. = n x1 ⋅ x 2 …x n

⇒ 10 = n x1 ⋅ x 2 ⋅…x n 4

4

4

New observations are 3x1 , 3x2 , 3x3 , … 3xn

n 4 = n (3) ⋅ ( x1 ⋅ x 2 ⋅ x 3 ⋅…⋅ x n )



= 3 n x1 ⋅ x 2 ⋅ x 3 ⋅….x n



= 3(10)4 = 30000

(

)

Hint: • Use G.M. = n x1 ⋅ x 2 ⋅…⋅ x n 115. Option (a) is correct. Explanation: 5 Given: P(A ∪ B) = 6



1 3

1 2 By complement rule of probability,



P(A ∩ B) = P(

)=

P(A) = 1 - P( A )

1 1 ⇒ P(A) = 1 = 2 2

1 2 1 × = = P(A ∩ B) 2 3 3

∴ A and B are independent events. So, statement 1 is correct. Statement-2: A and B are mutually exclusive events. As we know, if P(A ∩ B) = 0 then A and B are mutually exclusive events. 1 ∵ P(A ∩ B) = ≠ 0 3 So, A and B are not mutually exclusive events. So, statement 2 is incorrect. Hint: • Use the definition of independent events and mutually exclusive events. • Use addition theorem of probability and complement rule of probability to find P(A) and P(B).

4

4 4 4 4 New G.M. = n 3x1 ⋅ 3x 2 ⋅ 3x 3 ⋅…⋅ 3x n



5 1 1 5−3+2 4 2 − + = = = 6 2 3 6 6 3

P(A)∙P(B) =



Hint: • Roots of ax2 + bx + c = 0 are

5 1 1 = + P(B) 6 2 3



4

116. Option (c) is correct. Explanation: Given: Set of 15 observations n = 15 As we know, Σx Average = x = i n Σx x= i ⇒ 15

Σxi = 15 x ⇒ Let xa was the observation recorded wrongly xa′ was corrected observation.

Let new average == x′

Σx i ′ Σx i ′ = n 15

Σx i − x a + x a ′ 15 As x a ′ and xa has difference of only tens digit, it must be 3 instead of 8.

=

156 Oswaal NDA/NA Year-wise Solved Papers

⇒

xa - xa′ = 8(10) - 3(10) = 50



\

x′ =



⇒

x′ = x −

50 15

x′ = x −

10 3



Σx i ( x a − x a ′ ) − 15 15

2 1 = 4 2 E ∩ F = {(H, H)}



1 4 By addition theorem of probability,

⇒



∴ New average is reduced by

P(E ∪ F) = P(E) + P(F) - P(E ∩ F)

10 . 3



Shortcut:

x =



New x = x′ =

Σx i ′ 15

old

and

new

⇒

x′ =



⇒

x′ = x −

50 15



⇒

x′ = x −

10 3



So, new average is reduced by

118. Option (d) is correct. Explanation: 2 5 , Variance = 3 9 Let n = number of events, p = probability of occurrence of events q = probability of non-occurrence of event

10 . 3

Hint: • •

Σx i n Difference between old and observation = 8(10) - 3(10) = 50.

P(E) =

2 1 = 4 2



Given: Mean =



Q Mean = np =

Variance = npq =

Average = x =

new

117. Option (c) is correct. Explanation: Given: E : Head on first toss F : Head on second toss Coin is tossed twice. Sample space = {(H, H), (T, H), (H, T), (T, T)} E = {(H, H), (H, T)}

3 4

Hint: • Use addition theorem of probability, P(A) + P(B) - P(A ∩ B) = P(A ∪ B)

Σx i ( x a − x a ′ ) − 15 15



3 1 1 1 + − = 4 2 2 4

P(E ∪ F) =

= 8(10) - 3(10) = 50



=

Shortcut: Sample space = {(H, H), (H, T), (T, H), (T, T)} E ∪ F = {(H, H), (H, T), (T, H)}

Σx i 15

Difference between observation = xa - xa′

P(F) =

P(E ∩ F) =



Let

F = {(H, H), (T, H)}



⇒

q=

2 3 5 9

5/9 5 = 2/3 6

Also, p = 1 - q = 1 and np = n=

1 5 = 6 6

2 3 2/3 =4 1/6



⇒



Q P(X = 2) = nC2(p)2(q)n–2 2



1 5 = 4C2     6 6



=

2

4×3 1 5× 5 × × 2 6×6 6×6

157

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=

25 216

Required probability =

25 216

Hint: • For binomial distribution, mean = np and variance = npq where n is number of trails, p is probability of success and q is probability of failure. • P(X = r) = nCr(p)r(q)n–r 119. Option (d) is correct. Explanation: Given: Scores are 10, 12, 13, 15, 15, 13, 12, 10, x. Mode is 15. As we know, the mode of n observations is the number that has the highest frequency. Here, frequency of score ‘12′ = 2 Frequency of score ‘13′ = 2 Frequency of score ‘15′ = 2 So, for mode to be 15, its frequency must be the highest so x = 15. ⇒ Frequency of score ‘15′ = 3 \ x = 15 Shortcut: For mode to be ‘15′, its frequency must be highest ⇒ x = 15.

Hint: • Recall that mode of n observations is the number that has the highest frequency. 120. Option (c) is correct. Explanation: Given:

P(A) =

5 3 and P(B) = 8 4

Statement 1: As we know min∙(P(A ∪ B)) = max∙(P(A), P(B)) Q P(A) > P(B) min∙(P(A ∪ B)) =

3 4



⇒



So, statement 1 is correct. Statement 2: As we know, max∙(P(A ∩ B)) = min∙(P(A), P(B)) Q P(B) < P(A)



⇒ max.(P(A ∩ B)) =



So, statement 2 is also correct.

5 8

Hint: • Use min(P(A ∪ B) = max∙(P(A), P(B)) and max∙(P(A ∩ B)) = min∙(P(A), P(B))

NDA / NA

MATHEMATICS

National Defence Academy / Naval Academy

question Paper

iI

2021

Time  : 2 :30 Hour

Total Marks  : 300

Important Instructions  : 1. This ‘test Booklet contains 120 items (questions). Each item is printed in English. Each item comprises four responses (answer,s). You will select the response which you want to mark on the Answer Sheet. In case you feel that there is more than one correct response, mark the response which you consider the best. In any case, choose ONLY ONE response for each item. 2. You have to mark all your responses ONLY on the separate Answer Sheet provided. 3. All items carry equal marks. 4. Before you proceed to mark in the Answer Sheet the response to various items in the Test Booklet, you have to fill in some particulars in the Answer Sheet as per instructions. 5. Penalty for wrong answers : THERE WILL BE PENALTY FOR WRONG ANSWERS MARKED BY A CANDIDATE IN THE OBJECTIVE TYPE QUESTION PAPERS. (i) There are four alternatives for the answer to every question. For each question for which a wrong answer has been given by the candidate, one·third of the marks assigned to that question will be deducted as penalty. (ii) If a candidate gives more than one answer, it will be treated as a wrong answer even if one of the given answers happens to be correct and there will be same penalty as above to that question. (iii) If a question is left blank, i.e., no answer is given by the candidate, there will be no penalty for that question.

1. If x2 + x + 1 = 0, then what is the value of 199

200

x +x (a) - 2 (c) 1

201

+x

 ? (b) 0 (d) 3

2. If x, y, z are in GP, then which of the following is/ are correct? (1) ln(3x), ln(3y), ln(3z) are in AP (2) xyz + ln(x), xyz + ln(y), xyz + ln(z) are in HP

Select the correct answer using the code given below : (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 3. If log10 2, log10 (2x - 1), log10 (2x + 3) are in AP, then what is x equal to ? (a) 0 (b) 1 (c) log2 5 (d) log5 2 4. Let S = {2, 3, 4, 5, 6, 7, 9}. How many different 3-digit numbers (with all digits different) from S can be made which are less than 500 ? (a) 30 (b) 49 (c) 90 (d) 147

5. If p = (1111 …up to n digits), then what is the value of 9p2 + p ? (a) 10n p (b) 2p· 10n (d) 10n p + 1 (c) 10n p - 1

6. The quadratic equation 3x2 - (k2 + 5k)x + 3k2 - 5k = 0 has real roots of equal magnitude and opposite sign. Which one of the following is correct ?

(a) 0 1, equal to? an − b n (a) -1 (b) 0 (c) 1 (d) Limit does not exist n →∞

x  , 0 0 and (3k - 5) < 0 5 k > 0 and k < 3 This satisfies 5 Hence, 0 < k < 3 Roots are equal & opposite in sign Hence, sum of roots = 0

k 2 + 5k =0 3 ⇒ k(k + 5) = 0 ⇒ k = 0, k = -5 Now, we have 3 answers for k 5 (i) 0 < k < 3 (ii) k = 0, -5



From (i) & (ii) we don’t have any common solution for k. Hence, no such value of k exists. Shortcut : ⇒

k 2 + 5k =0 3 k = 0, k = -5…(i)

α−α =

Product roots < 0 3k ≤ 5k < 0 ⇒ k (3K − 5) < 0 3 + − + −∞

0

5

3

 5 k ←  0,  from(i) and (ii) no any value of k.  3 7. Option (d) is correct. Explanation : Given: an = n(n!) a1 = 1(1!) = 1 a2 = 2(2!) = 4 a3 = 3(3!) = 18 a4 = 4(4!) = 96 a5 = 5(5!) = 600

174 Oswaal NDA/NA Year-wise Solved Papers

a6 = 6(6!) = 4320 a7 = 7(7!) = 35280 a8 = 8(8!) = 322560 a9 = 9(9!) = 3265920 a10 = 10(10!) = 36288000 Now, to find let I = a1 + a2 + a3 + ………. a10 = 1(1!) + 2(2!) + 3(3!) + …………. + 10(10!) [put all values] I = 1 + 4 + 18 + …………….. + 36288000 = 39916799 = 39916800 - 1 I = 11! - 1 [value of 11! = 39916800] Hence, a1 + a2 + a3 + ………… a10 = 11! - 1 Hints : (1) Find value of an and add. (2) Use value of 11!. Shortcut :

(1!+ 1.1!) + 2.2! + 3.3! + ……. + 10.10!)1!=(2!+2.2!)+3.3!+......+10.10!- 1 [n!+nn!=(n+1)!]

8. Option (b) is correct. Explanation : Quadratic equation is x2 + px + q = 0 …(i) Roots given are p, q …(ii) We know, Sum of roots = -

b a







⇒

p+q=-



⇒

p + q = -p …(iii)



Also, product of roots =



⇒

pq =



⇒

pq = q



⇒ p=1 Put in (iii), ⇒ 1 + q = -1 ⇒ q = -1 - 1

p 1

[from (i), (ii)]

c a

q 1

[from (i), (ii)]



⇒ q = -2 Only one value of q exist Hints : (1) Use product and sum formula. (2) Solve equations.

9. Option (d) is correct. Explanation : Consider,

3d + 5 g 4 a + 7 g 6 g 4b + 7 h 6h

A = 3e + 5h 3 f + 5i

4 c + 7i

6i



Taking, 6 from C3



3d + 5 g A = 6 3e + 5h 3 f + 5i



C1 → C1 - 5C3, C2 → C2 - 7C3



3d 4 a g A = 6 3e 4 b h 3 f 4c i



Taking 3 from C1, 4 from C2



d a g A = 6 × 3 × 4 e b h f c i



d A = 72 a g



[∵ transpose of matrix] R1 ↔ R 2



a A = 72 × -1 d g







A = -72 D



e b h

4a + 7 g 4b + 7 h 4 c + 7i

g h i

f c i

b e

c f

h

i

[Interchange of Rows]

a

b

c

[given d g

e h

f =D i

Hints : (1) Add, subtract of row and column. (2) Interchange of column.

175

solved PAPER - 2021 (ii)

10. Option (a) is correct. Explanation : 1 1 1 We know, if p, q, r are in A.P., then , , are in p q r H.P. 1 1 1 in H.P. Given, , , b+c c+a a+b

So, (b + c), (c + a), (a + b) are in A.P. ⇒ (b + c) + (a + b) = 2(c + a) ⇒ 2b + a + c = 2c + 2a ⇒ 2b = a + c Hence, a, b, c are in A.P. Statement 1 is correct. Now, (b + c)2, (c + a)2, (a + b)2 given to check for G.P. condition. ⇒ (b + c)2 × (a + b)2 = [(c + a)2]2 ⇒ (b + c)2 × (a + b)2 = [c + a]4 Which is not possible since, 2b = c + a from A.P. Hence, statement 2 is incorrect. Hint : (1) Use A.P., H.P. G.P. Properties.

11. Option (a) is correct. Explanation :

Given :



We know that,

1 a  A=   0 1 n







1 a  An =   0 1  1 na  ⇒ An =   0 1   1 100 a  ⇒ A100 =  1  0  1 50 a  ⇒ A50 =   0 1   1 25a  ⇒ A25 =   0 1  Using all values in given equation. Directly apply  1 100 a   1 50 a   1 25a  ⇒   - 0 1  - 2 0 1  0 1      



0 50 a   2 50 a  ⇒   -   0 0  0 2 



⇒ -2I Hint :  1 na  (1) Use An =   0 1 

12. Option (a) is correct. Explanation : Consider,

a −b a − b − c A = −a b −a + b − c − a −b − a − b + c



R1 → R1 + R2



A = −a b −a + b − c − a −b − a − b + c



Expand along R1 for determinant



A = 0 + 0 - 2c



A = -2c[ab + ab] Apply in given equation, ⇒ -4abc - kabc = 0 ⇒ -4abc = kabc ⇒ k = -4

0

−2c

0

−a b − a −b

Hints : (1) Apply R1 → R1 + R2 (2) Expand along R1. 13. Option (a) is correct. Explanation :

Consider, Given

8 n+7

∑ in

n =1



Now, we get by expansion, let I = i1 + i2 + i3 + ………….. + i8n+7 We know, if consider i1 + i2 + i3 + i4, we get: ⇒ i-1-i+1=0 [value of i1 = i, i2 = -1, i3 = -i, i4 = +1] Similarly, in expansion consecutive 4 terms get cancel out

176 Oswaal NDA/NA Year-wise Solved Papers

So, In (8n + 7) terms we are left with only 3 terms. {(8n + 7) divisible by 4 gives remainder 3} Hence, I = i1 + i2 + i3 = i + (-1) + (-i) = -1 Hints : (1) Consecutive 4 terms cancel out. (2) Remainder 3 terms.



Option (c) is correct. Explanation : Given : z = x + iy …(i) We know, conjugate z = x - iy …(ii) Now, consider





zz + z + 4 ( z + z ) - 48 = 0 We know, |z| = Modulus of complex number





|z| = x 2 + y 2 Put (i), (ii), (iii) in equation



⇒ (x + iy)(x - iy) +



+ 4(x + iy + x - iy) - 48 = 0 2 2 2 2 ⇒ x - (iy) + (x + y ) + 4(2x) - 48 = 0 {i2 = -1} ⇒ x2 + y2 + x2 + y2 + 8x - 48 = 0 ⇒ 2x2 + 2y2 + 8x - 48 = 0 ….(iv) Compare (iv) with general equation of circle x2 + y2 + 2gx + 2fy + c = 0 Hence, it is equation of circle.



2

x2 + y2

)

+ 2



14.

(

Now, consider J = (a + ib) (a - ib) J = a(a - ib) + ib(a - ib) J = a2 - abi + abi - i2b2 J = a2 - b2i2 J = a2 + b2 [i2 = -1] So, put (a + ib) (a - ib) = a2 + b2 in (i), we get I = [(a + ib) + (a - ib)]



15. Option (a) is correct. Explanation :

I = (2a + ib - ib) + 2 a 2 + b 2

2

a − ib

)

2

( a + ib )( a − ib ) 2



2a + 2 a 2 + b 2 =  a + ib + a − ib  Taking square root both sides, we get





2a + 2 a2 + b 2 =

2

a + ib + a − ib

(1) Add and Subtract ib. (2) a2 + b2 = (a + ib) (a - ib). (3) Use a2 + b2 + 2ab= (a + b)2. 16.

−1 ]



I = (a + ib + a - ib) + 2 a 2 + b 2



I = [(a + ib) + (a - ib)] + 2 a 2 + b 2 …(i)

Option (c) is correct. Explanation : Given: Quadratic equation ax2 + bx + c = 0 We know,



c a −b Sum of roots = a Since, sin θ, cos θ are roots given, we get



⇒ sin θ + cos θ =

−b …(i) a



⇒ sin θ cos θ =

c …(ii) a



Now, squaring both sides of eq.(i)



 −b  (sin θ + cos θ)2 =    a 



⇒ sin2 θ + cos2 θ + 2sin θ cos θ =



(1) Use conjugate, modulus (2) Compare with circle



) +(

Hints :

Hints :

Let I = 2a + 2 a 2 + b 2 [Add and Subtract ib, where i =

a + ib

I =  a + ib + a − ib  {using (a + b)2 = a2 + b2 + 2ab} Hence,

2



(

+ 2

…(iii)

=



( a + ib )( a − ib )



Product of roots =

b2 a2

2

177

solved PAPER - 2021 (ii)



⇒ sin2 θ + cos2 θ +







2c b2 = 2 a a

[using (ii)]

⇒ 1 + 2

⇒

b

2

2c = 2 a a

a - b + 2ac = 0

Hints : (1) Use product and Sum of roots. (2) Squaring of equation. Option (a) is correct. Explanation : Given: C(n, 4), C(n, 5), C(n, 6) in A.P. We know, formula for combination



C(n, r) = nCr =



⇒ C(n, 4) =



⇒ C(n, 5) =



⇒

6! ( n − 6 ) ! + 4 ! ( n − 4 ) ! 2 = 5! ( n − 5 ) ! 4 ! ( n − 4 ) ! × 6! ( n − 6 ) !

n! 4 !(n − 4 )!



⇒

n! 5! ( n − 5 ) !







⇒



= 6!(n - 6)! + 4(n - 4)! ⇒ 2[4!(n - 4)! × 6 × (n - 6) ] = 6!(n - 6)! + 4(n - 4)! ⇒ 4!(n - 4)! × 12 × (n - 6) = 6 × 5 × 4! × (n - 6) × (n - 5)(n - 4)! + 4(n - 4)! ⇒ 4!(n - 4)! × 6 × (n - 6) = 4!(n - 4)! [6 × 5(n - 6)(n - 5) + 1] ⇒ 12(n - 6) = 30(n - 6)(n - 5) + 1 ⇒ 12n - 72 = 30[n2 - 5n - 6n + 30] + 1 ⇒ 12n - 72 = 30[n2 - 11n + 30] + 1 ⇒ 12n - 72 = 30n2 - 330n + 900 + 1 ⇒ 12n - 72 = 30n2 - 330n + 901 ⇒ 30n2 - 342n + 973 = 0, we get ⇒ n=7

n! , we get r !(n − r )!

n! 6! ( n − 6 ) !

Since, given in A.P.



2 × C(n, 5) = C(n, 4) + C(n, 6)



2 nC5 = nC4 + nC6



⇒

2n ! n! n! = + n − 5 !5! n − 4 !4 ! n − ( ) ( ) ( 6 ) !6! 2

( n − 5 ) !5!

=

1

+

1

( n − 4 ) !4 ! ( n − 6 ) !6!

2 1 = ( n − 5 )( n − 6 ) !5! ( n − 4 )( n − 5 )( n − 6 ) !4 ! + 1 ( n − 6 ) !6!



12n − 48 − 30 =1 ( n − 4 )( n − 5 )

2 1 1 = + 4 !(n − 4 )! 6! ( n − 6 ) ! 5! ( n − 5 ) !



⇒

⇒

⇒

⇒ C(n, 6) =











⇒

⇒ 12n - 48 - 30 = n2 - 9n + 20 ⇒ n2 - 21n + 98 = 0 ⇒ (n - 7)(n - 14) = 0 ⇒ n = 7 or n = 14 Hence, n = 7 is correct answer. 2 × n! n! n! ⇒ = + 5! ( n − 5 ) ! 4 ! ( n − 4 ) ! 6! ( n − 6 ) !

2

17.

12 30 =1 − ( n − 5 ) ( n − 4 )( n − 5 )



Multiplying 6!(n - 6)! in all terms



⇒

6! 2 × 6! = +1 5! × ( n − 5 ) ( n − 4 )( n − 5 ) 4 !



⇒

12 30 +1 = ( n − 5 ) ( n − 4 )( n − 5 )

2 4 ! ( n − 4 ) ! × 6! ( n − 6 ) ! 5! ( n − 5 ) !

= 6!(n - 6)! + 4!(n - 4)! 2 x 4 ! ( n − 4 ) ! × 6 × 5! × ( n − 6 ) × ( n − 5 ) ! 5! ( n − 5 ) !

Hints : (1) Use combination formula. (2) Solve equation.

178 Oswaal NDA/NA Year-wise Solved Papers 18.

Option (a) is correct. Explanation : Consider, word LUCKNOW, It contain 2 vowels (O, U) and 5 constant (L, C, K, N, W) Now, we have to make 4 letter, it must Contain 2 vowels and only 2 consonant out of 5, we get ⇒ Ways to select consonant = 5C2 ⇒ Ways to select vowels = 2C2 Finally, 4 letter words = 2C2 × 5C2 × 4! = 1 × 10 × 24 = 240 Hints : (1) Select vowel, consonant. (2) 4 Letter word.

19.

Option (b) is correct. Explanation : Given: Twenty distinct points on circle. First, we have join any two points to make line. Using concept of combination, we get Number of lines(l) = 20C2







n  n!  Cr =  r ! ( n − r ) !  







[n! = n(n - 1)(n - 2)! ……] l = 190 Statement 1 is incorrect. Now, to make triangle, we have to connect 3 points. Number of triangles (T) = 20C3



l=

l=

20! 18! × 2!

20 × 19 2



Statement 2 is correct. Hint : (1) Use combination concept.

20. Option (d) is correct. Explanation : 21



 a2 b 2  Given: expansion of  2 + 2 + 2  = I b  a   Now, we know (x + y)2 = x2 + y2 + 2xy Using this, we get







….(i)

2



a2 b 2 a b b + a = 2 + 2 + 2 b a   Putting in (i),  a2 b 2   2 + 2 + 2 a  b 



21

21

 a b 2  =   +   , we get   b a   42



a b I =  +  b a We know, number of terms in expansion of (x + y)n is (n + 1) Hence, number of terms = 42 + 1 = 43

Hints : (1) Use (a + b)2 formula. (2) Number of terms is (n + 1). 21.

Option (b) is correct. Explanation : Consider, system of equations: 2k2x + 3y - 1 = 0 7x - 2y + 3 = 0 6kx + y + 1 = 0 Since, solution given is consistent

2k 2 7

3 −1 −2 3 = 0 1 1



20! T = 3! ( 20 − 3 ) !





20! T = 3! × 17 !



Expand along row 1



T =

20 × 19 × 18 × 17 ! 3 × 2 × 17 !



2k2



T =

20 × 19 × 18 3× 2



T = 1140



2k2[-2 -3] -3[7 - 18k] -1[7 + 12k] = 0 2k2(-5) - 21 + 54k - 7 - 12k = 0 -10k2 + 42k - 28 = 0



6k

−2 3 7 3 7 -3 -1 1 1 6k 1 6k

−2 =0 1

…(i) …(ii) ...(iii)

179

solved PAPER - 2021 (ii)



-2[5k2 - 21k + 14] = 0 ⇒ 5k2 - 21k + 14 = 0 For quadratic equation, consider a = 5, b = -21, c = 14 By quadratic formula





k=



21 ± 161 k= 10



−b ± b 2 − 4 ac , we get 2a





Since, logarithmic and constant are non-period function. We find period on basis of sin2 x. ...(i) Now, we know, period of sin x is 2π. We find period of sin2x, By trigonometric formula: 2sin2 x = 1 - cos 2x



⇒ sin2 x =







Hints : (1) Use consistent solution condition. (2) Solve equation. 22. Option (a) is correct. Explanation :









 −2 1  Consider, A =  3 1  −   2 2  We know,

1 A = |A| –1



[Adjoint of A] …(i)

Now, we know   a b  = , then if B      c d     Adjo int of B =  d −b    −c a       By this we get:  −1   2 −1  Adjoint of A =    − 3 −2   2   −2 1    Now, Since A =  3 − 1   2 2 

1 3  1  |A| =  −2 × − − × 1 = − 2 2 2  

1 2  A–1 =   3 4  23. Option (b) is correct. Explanation : Period of composite functions such that f[g(x)] is t if f(x) is non period function and g(x) is period of t.

Hence,



1 1 − cos 2x 2 2

[period of cos x is 2π and period of cos 2x is π] 2 ⇒ period of sin x is π ...(ii) Hence, Period of f(x) = log(2 + sin2 x) = π [By (i), (ii)] Hints : (1) Period of composite function. (2) Period of sin2 x is π.

24. Option (d) is correct. Explanation : Given: sin(A + B) = 1 π 2



sin(A + B) = sin



On comparing



A + B =



Now,



sin(A - B) =



sin(A - B) = sin



A - B =



On solving (i) and (ii), we get



A =



Now, to find



π 3 = 3 = 3 I = π 1 1 tan 6 3

π …(i) 2

2sin (A - B) = 1

I=

1 2 π 6

π …(ii) 6 π π ,B= 3 6 tan A tan B

tan

180 Oswaal NDA/NA Year-wise Solved Papers

⇒

tan A : tan B = 3 : 1

Hint :

sin 90° = 1.

25. Option (b) is correct. Explanation :



Absolute values of roots = |x|, we get Sum of absolute values: S = |+1| + |-1| + |+3| + |-3| S = + 1 + 1 + 3 + 3 S = 8 Hints : (1) Solve quadratic equation (2) Sum of absolute values









Number of triangles that can be formet from any 10 vertices =10C3 10 ⋅ 9 ⋅ 8 =120 3 ⋅ 2 ⋅1 Number of triangles that can be formed with only one side of triangles .....(1, 2, 9), (2, 3, 5) .... (2, 3, 10) .... y =(10 - 4)×10=60 Numbe of triangles that can be formed with two consecutive sides of polygon. e.g., {(1, 2, 3), (2, 3, 4) ...... (10, 1, 2)} =10 ∴Number of triangly that can be formed with no common side with any of the sides of the polygon =120 - 60 - 10=50

27.

Hints :

Hints : (1) Find all triangle, (2) Find number of triangle with one side common. (3) Find number o triangle with two side common. 26.

Option (c) is correct. Explanation : Given: Quadratic equation: x4 - 10x2 + 9 = 0 Put x2 = y, we get ⇒ y2 - 10y + 9 = 0 ⇒ y2 - 9y - y + 9 = 0 ⇒ y(y - 9) -1(y - 9) = 0 ⇒ (y - 1) (y - 9) = 0 ⇒ y = 1, 9 2 But y = x , we get ⇒ x2 = 1, x2 = 9 ⇒ x = +1, -1, +3, -3

Option (c) is correct. Explanation : In expansion of (1 + x)n: Coefficient of first term = nC0 = 1 Coefficient of second term = nC1 = n Coefficient of nth term = nCn–1 = n Coefficient of (n + 1)th term = nCn = 1 Given that coefficient’s are p, q, r, s: n C0 = p = 1 ⇒ n ⇒ C1 = q = n n ⇒ Cn–1 = r = n n ⇒ Cn = s = 1 Now, to find let I = ps + qr I = 1 × 1 + n × n = 1 + n2 (1) Coefficient of first, second, nth, (n + 1)th term. (2) Value of nCr =

n! r !(n − r )!

28. Option (c) is correct. Explanation : Given:

sin–1 x + sin–1 y + sin–1 z =



We write



sin–1 x + sin–1 y + sin–1 z =



On comparing, we get



sin–1 x =



3 π 2 π π π + + 2 2 2

π π π , sin–1 y = , sin–1 z = 2 2 2 ⇒ x = 1, y = 1, z = 1 1000

Now, to find I = x +y Using values from (i)

1001

+z

1002

…(i)

181

solved PAPER - 2021 (ii)



I = 1 + 1 + 1 = 3 I = 3 Hints : (1) sin 90° = 1 (2) Split

29.

3 π and get value 2

Option (a) is correct. Explanation : Given : sin x + sin y = cos x + cos y Use formulas, we get

30. Option (d) is correct. Explanation :

 0 2 Consider, A =    −2 0 



Now, we know



1 0  I =   0 1 



⇒

m 0  mI =   …(i)  0 m



2n   0 nA =  …(ii) − 2 n 0   Consider, given equation (mI + nA)2 = A {By (i), (ii)}



⇒

 m 0   0 2n    0 2   0 m  +  −2n 0   =  −2 0       



⇒

 m 2n   0 2  −2n m  =  −2 0     

⇒

x+y x+y sin  = cos     2   2 



⇒

 m 2n   m 2n   0 2   −2n m   −2n m  =  −2 0       

⇒

x y x y sin  +  = cos  +  2 2 2 2





⇒

x y sin  +  2 2 = 1 x y cos  +  2 2



x y [Divide cos  +  both sides] 2 2

 A+B  A−B  sin A + sin B= 2 sin  2  cos  2         A+B  A − B  = 2 cos  cos A + cos B  cos  2   2     







x+y x−y cos  ⇒ 2 sin     2   2 







On concelling out, we get





x+y x−y cos  = 2 cos     2   2 

x y tan  +  = 1 2 2



⇒



 sin θ  tan θ  ∵ cos= θ   Hints : (1) Use formulae of sin (A + B), cos (A + B) sin θ = tan q (2) Use formulae cos θ



⇒

2

2



 m2 − 4n2 2mn + 2mn   0 2  ⇒   =    −2mn − 2mn −4 n 2 + m 2   −2 0  On comparing, we get m2 - 4n2 = 0 …(iii) 2mn + 2mn = 2 …(iv) Now, from (iv), ⇒ 4mn = 2



⇒

mn =



⇒

m=



Put in (iii),



⇒



⇒



⇒

1 2 1 …(v) 2n

2

 1  2  2n  - 4n = 0  

1 4n

2

- 4n2 = 0

1 − 16n 4 4n2

=0

182 Oswaal NDA/NA Year-wise Solved Papers

⇒ 1 - 16n4 = 0



⇒ 1 = 16n4



⇒

1 = n4 16



⇒

1 4 2 =n  



⇒

n=



Put in (ii), we get







Hence, m + n = 1 +



  3 2  +  −1  4 3   I = cot tan    1 − 3 × 2   4 3   

4

m=1





  = cot tan −1  

 17    12      1   2  





  17   = cot tan −1     6  





  6  = cot cot −1     17   











1 3 = 2 2

(1) Make matrix and Satisfy equations. (2) Solve by comparing. 31. Option (a) is correct. Explanation : 3 3  Given: cot sin −1 + cot −1  5 2  Let



 9 + 8   12      1 − 1  2   

1 2

Hints :





  −1 = cot tan  



3 3  I = cot sin −1 + cot −1  5 2 



Now, consider sin −1

3 =x 5



⇒

3 = sin x 5



By pythogoras theorem, we get: P2 + B2 = H2

So,









tan x =

3 4

3 4 3 2 Cot −1 = tan −1 2 3 x = tan–1

1  −1 cot −1  ∵ tan x = x  =

6 17

Hints : (1) Comment function in tan θ. (2) Use formula tan (A + B). 32. Option (c) is correct. Explanation : Given: 4sin2 x = 3 3 4



⇒ sin2 x =



⇒ sin x =



[Taking square root] ⇒ sin x = sin 60° ⇒ x = 60°, 120° Now, to find I = tan 3x I = tan 3 × 60° I = tan 180° We can write I = tan (180° + 0°) [tan(180° + θ) = tan θ] I = tan 0° = 0 I = tan 3 × 120° = tan 360°

3 2

183

solved PAPER - 2021 (ii)

Hints : (1) Get x by solving. (2) Put in given equation.







I = -





33.

Option (c) is correct. Explanation : Given: for AP, first term = p third term = q fifth term = 3 Let for A.P. first term = a, difference = d Now, for nth term an = a + (n - 1)d ⇒ a1 = a = p Third term, we get: ⇒ a3 = a + (3 - 1)d = q a3 = a + 2d = q …(i) Fifth term, we get ⇒ a5 = a + (5 - 1)d = 3 ⇒ a5 = p + 4d = 3 …(ii) Now, to find Let I = pq I = pq I = (3 - 4d)(a + 2d) [from (i) & (ii)] = (3 - 4d)(3 - 4d + 2d) = (3 - 4d)(3 - 2d) = 3(3 - 2d) - 4d(3 - 2d) = 9 - 6d - 12d + 8d2 = 9 - 18d + 8d2 Now, we have to find I minimum I = 9 - 18d + 8d2 Differentiate with respect to d





dI = 0 - 18 + 8 × 2d dd







dI = -18 + 16d dd





Put

dI =0 dd



-18 +16d = 0 16d = 18 18 9 d= = 16 8



At d =

9 , critical point minimum, value of I 8

 9  9 I = 9 - 18  +  + 8  +   8  8

d2I d2d

2

9 , to check 8

= +16, which is Positive

Hints : (1) Use an formula for nth term. (2) Find pq, make in respect of d (3) Differentiate to find critical point 34.

Option (a) is correct. Explanation : Given: Equation is x3 - 8 = 0 Now, x3 - 8 = 0 ⇒ x3 = 8 ⇒ x3 = 23 ⇒ x=2 By property of complex number, that says for x3 - 1 = 0, roots are 1, ω, ω2 We get: x = 2, 2ω, 2ω2

′

′



Statement 1: Roots are non-collinear correct. We can see from figure, roots not on one line. So, true. Statement 2: Roots lie on circle inside this triangle but circle is of Radius 2. As, complex roots is 2. So, false. Hence, statement 1 correct only. Hints : (1) Use x3 - 1 = 0. (2) Find roots, Make figure.

35. Option (c) is correct. Explanation : Given: sec x ⋅cosec x = p …(i)

184 Oswaal NDA/NA Year-wise Solved Papers

We know,



sec x =

1 cos x



cosec x =

1 sin x



Put value in (i),



1 1 =p × cos x sin x

⇒

1 =p sin x cos x [Multiple and Divide by 2]



⇒







⇒



⇒



Now, we know value of sine function maximum is +1 and minimum is -1. ⇒ sin 2x, maximum is +1 and minimum is -1



Now,

2 =p 2 sin x cos x 2 =p sin 2x [using sin 2x = 2sin x cos x]

p=

2 sin 2x



⇒ 2cos2 θ + cos θ - 1 = 0 ⇒ 2cos2 θ + 2cos θ - cos θ - 1 = 0 [Middle term split] ⇒ 2cos θ(cos θ + 1) - 1(cos θ + 1) = 0 ⇒ (cos θ + 1)(2cos θ - 1) = 0



⇒ cos θ = -1 or cos θ =







But, given that 0 < θ
60

= 65

Hint :

g2 + f 2 − c

Hint :

(1) Matrix multiplication. Option (d) is correct. Explanation : Consider, General equation of circle x2 + y2 + 2gx + 2fy + c = 0 …(i) Satisfy all three given points in (i), we get: By (5, -8): 10g - 16f + c + 89 = 0 …(ii) By (-2, 9): -4g + 18f + c + 85 = 0 …(iii) By (2, 1): 4g + 2f + c + 5 = 0 …(iv)

…(v)

Hints :





51.

Now, subtract (iii) from (ii), we get 14g - 34f + 4 = 0 ⇒ 2(7g - 17f + 2) = 0 ⇒ 7g - 17f + 2 = 0 Now, subtract (iii) from (iv), we get ⇒ 8g - 16f - 80 = 0 ⇒ 8(g - 2f - 10) = 0 ⇒ g - 2f - 10 = 0 Solving, (v) and (vi), By substitution method, we get g = 58, f = 24 We know, centre of circle is given by (-g, -f) = (-58, -24)



Radius is

g2 + f 2 − c

53.

Option (c) is correct. Explanation : Given: Two coordinate are A(0, 0), B(2, 2) Let third coordinate is C(x, y) Since, equilateral triangle given: AB = BC = AC Now, Length of AB:



AB =

( 2 − 0 )2 + ( 2 − 0 )2



AB =

8 …(i)

190 Oswaal NDA/NA Year-wise Solved Papers

Length of BC:



BC =



Length of AC:

( x − 2 )2 + ( y − 2 )2 …(ii)



AC =

(x − 0)



AC = Now, put AB = AC

x 2 + y 2 …(iii)



8 = x2 + y2 [By (i), (iii)] Square both sides ⇒ 8 = x2 + y2 …(iv) Now, put BC = AC







Square both sides ⇒ (x - 2)2 + (y - 2)2 = x2 + y2 ⇒ x2 + 4 - 4x + y2 + 4 - 4y = x2 + y2 ⇒ 8 - 4x - 4y = 0 ⇒ -(4x + 4y) = -8 ⇒ x + y = 2 …(v) Now, put y = 2 - x in (iv), we get ⇒ 8 = x2 + (2 - x)2 ⇒ 8 = x2 + 4 + x2 - 4x ⇒ 8 = 2x2 - 4x + 4 ⇒ 2x2 - 4x - 4 = 0 ⇒ 2(x2 - 2x - 2) = 0 ⇒ x2 - 2x - 2 = 0 By quadratic formula





x=





x=





2± 4+8 x= 2×1





x=





x=





x = 1± 3

( x − 2 )2 + ( y − 2 )2

=

2

+ ( y − 0)

2



put x = 1 ± 3 in (v), we get







ence, (x, y) = ( 1 ± 3 , 1 ∓ 3 ) H Statement 1 is correct, since third coordinate is irrational. Now, we have three coordinates and one is irrational. So, length of sides will be irrational. Hence, area is also irrational. Statement 2 is correct.



y = 1∓ 3

Hints : (1) Find length of sides and put equal. (2) By quadratic formula, get coordinates.

x2 + y2

54. Option (d) is correct. Explanation :

(1 + 3 ) ,



Third coordinate is x=



from Q. 53 Difference of coordinates is :









= 1+ 3 −1+ 3





= 2 3

(

y=

(1 − 3 )

) (

x - y = 1+ 3 - 1− 3

)

Hint : Take coordinates we find in Q. 53.



Option (c) is correct. Explanation : Given: ABCD is a parallelogram A(1, 3) B(-1, 2) C(3, 5), D(x, y) Since, oposite sides are parallel lines so slope equal. Slope of AB = Slope of CD



⇒

2 ± 12 2



2±2 3 2

 y 2 − y1  = slope    x 2 − x1 



⇒



⇒ 3 - x = 10 - 2y ⇒ x - 2y = -7 …(i)

−b ± b 2 − 4 ac 2a − ( −2 ) ±

( −2 )2 − 4 × 1 × −2 2a

55.

5−y 3−2 = 1 − ( −1 ) 3−x

5−y 1 = 2 3−x

191

solved PAPER - 2021 (ii)



Slope of AD = Slope of BC



⇒



y−3 5−2 = 3+1 x −1 y−3 3 = 4 x −1

⇒

⇒ 3x - 3 = 4y - 12 ⇒ 3x - 4y = -9 …(ii) On solving (i), (ii), we get x = 5, y = 6 So, coordinate D(5, 6) Now, for equation of BD : Let B(x1, y1) = (-1, 2) D(x2, y2) = (5, 6) By point-slope equation : y − y1 y - y1 = 2 (x - x1) x 2 − x1



⇒

y-2=

6−2 (x + 1) 5+1



⇒

y-2=

(x + 1)



⇒ 3y - 6 = 2x + 2 ⇒ 2x - 3y + 8 = 0 Hence, equation of BD is 2x - 3y + 8 = 0 Hints :





=

1 [-3 -3(-4) + (-5) -6] 2





=

1 [-3 + 12 - 5 - 6] 2





=

1 1 [12 - 14] = × -2 = -1 2 2



Area of ∆ABC = |-1| = +1 Now, Area of ∆ACD A(x1, y1) = (1, 3) C(x2, y2) = (3, 5) D(x3, y3) = (5, 6) Again using (i)



Area =





=

1 [-1 - 3(-2) + 18 - 25] 2





=

1 [-1 + 6 - 7] 2





=

1 [-2] = -1 2



Area of ∆ACD = |-1| = +1 Hence, Area of parallelogram = 1 + 1 =2

(1) Slope of parallel lines are equal. (2) Find by slope-point form. 56. Option (c) is correct. Explanation : We know, coordinates are A(1, 3), B(-1, 2), C(3, 5), D(5, 6). Area of ∆ABC : A(x1, y1) = (1, 3) B(x2, y2) = (-1, 2) C(x3, y3) = (3, 5)

x1 1 Area = x 2 2 x3













=

=

y1

1

y2

1 …(i)

y3

1

1 [1(5 - 6) -3(3 - 5) + (3)(6) - (5)(5)] 2

Shortcut :

1 3 1 Area of ABC = −1 2 1 3 5 1



=|(2 - 5) - 3(-1 -3) + 1(-5 - 5)| =|- 3 + 12 -10|=|- 1|=1 Area of parallelogram =2×Arq ∆ABC =2×1=2

57. Option (d) is correct. Explanation :

1 [x1(y2 - y3) - y1(x2 - x3) 2 + x2y3 - x3y2] 1 [1(2 - 5) - 3(-1 -3) + (-1)(5) 2 - 3 × 2]



First we find coordinate B : AB is x - 2 = 0

192 Oswaal NDA/NA Year-wise Solved Papers

BC is y + 1 = 0 So, x = 2, y = -1, we get : Coordinate B(2, -1)



Now, slope of line AC =







Since, line AC and line BD are perpendicular So, slope of line AC × Slope of line BD = -1



⇒





⇒ Coordinate A is (2, 1).



(2) Coordinate of B :

Put y = -1 in (iii), we get:

−Coefficient of x Coefficient of y

⇒ x - 2 - 4 = 0 ⇒

x = 6,

−1 2

So,

B = (6, -1)

=

−1 × m = -1 2 m=2

⇒ Therefore, for line BD, one point B(2, -1) and slope is 2. By slope-point form of line : ⇒ y - y1 = m(x - x1) ⇒ y - (-1) = 2(x - 2) ⇒ y + 1 = 2x - 4 ⇒ 2x - y - 5 = 0 Hence, equation of altitude is 2x - y - 5 = 0 Hints : (1) Get coordinate B. (2) Use slope-line form.

58. Option (a) is correct. Explanation :

Now, Circumcentre of circle will be mid-points of line AC A(2, 1), B(6, -1)

Mid-point =



=





x = 2 …(i) y = -1 …(ii) x + 2y - 4 = 0 …(iii) Since, x = 2, y = -1 for AB and AC respectively, we get ∆ABC is right angle at B. Now, Intersection of AB and AC is A and Intersection of BC and AC is C. (1) Coordinate of A :

put x = 2 in (iii), we get 2y - 2 = 0

y=1

2 + 6 1 + ( −1 ) , 2 2 = (4, 0)

Hence, circumcentre is (4, 0). Hints : (1) Triangle ABC is a right angle triangle. (2) Midpoint of hypotenuse (AC) is circumcenter of triangle ABC. 59.



x1 + x 2 y1 + y 2 , 2 2





Option (b) is correct. Explanation : We have the end points of the latus rectum. Focus lies on the midpoint of the segment joining the end points of the latus rectum. So, the focus of the parabola is fixed. We have the length of the latus rectum. So, the distance between the focus and the directrix is fixed. Directrix is parallel to the latus rectum. So, the slope of directrix is fixed. So, we can have two lines which are parallel to latus rectum and are at a fixed distance from the focus. So, we have one focus and two possible directrices. So, a maximum of two parabolas can be drawn. Hints : (1) Make parabola on given points. (2) From both quadrant, two kind of parabola passes.

193

solved PAPER - 2021 (ii)

60. Option (c) is correct. Explanation : Statement 1: (–2, 4) A

Y z=7 X z=7 Z

C (–2, 0)

O



B (–2, –4)





Consider, parabola y2 = -4ax Point A and B on Latus Rectum. So, (+a, -2a) = (-2, 4) (+a, +2a) = (-2, -4) On comparing: a = -2 Also, we get focus (a, 0) = (-2, 0) Since, focus is at (-2, 0) From figure and focus point : End point of parabola O is (0, 0). Hence, parabola passes through origin. Statement 1 is correct. Statement 2:

We know, points of Latus Rectum: (+a, -2a) = (-2, 4) (a, +2a) = (-2, -4) On comparing, a = -2 So, focus is (a, 0) = (-2, 0) Hence, statement 2 is correct. Hints : (1) Make figure of both parabola. (2) Use formula of Latus Rectum.

61. Option (d) is correct. Explanation : Given, point z = 7

It is clear if point moving on Z-axis means it is parallel to XY-plane. Hint :

Make diagram for plane.

62. Option (c) is correct. Explanation : Statement 1: It is properties of line that in space it can have infinitely many direction ratios. Since, in space line is free to move in any direction and can revolve in any direction x, y, z. Hence, direction ratios are infinitely. It is correct. Statement 2: Incorrect, because sum of squares of direction cosines is always 1. Example,

(1) cos α =



(2) cos β =



(3) cos γ =



vx vx2 + v y2 + vz2 vy vx2

+ v y2 + vz2

vz vx2

+ v y2 + vz2

Square and add both sides. cos2 α + cos2 β + cos2 γ = 1 Hint : (1) Properties of direction cosines.

63.

Option (c) is correct. Explanation : Given : xy plane divides line segment. Consider, two points : (x1, y1, z1) = (-1, 3, 4) (x2, y2, z2) = (2, -5, 6) Now, xy plane divides by :



Ratio =

− z1 z2

=

−4 −2 = 6 3

194 Oswaal NDA/NA Year-wise Solved Papers

Since, division in negative. So, externally division. Hence, in ratio 2 : 3 externally. Hint : (1) Division by Z-coordinates.

XY-plane.

So,

only

64. Option (c) is correct. Explanation : Consider, xy axis.









cos α =



Now, similarly for AC and AG:



cos β =







cos β =



cos β =



cos β =



Now, for find let I = cos 2α + cos 2β, we get



I = 2cos2 α - 1 + 2cos2 β - 1

(By (i), (ii)) 6 12 = …(iv) 7 14

12 × 12 + 6 × 6 + 4 × 0 2

12 + 6 2 + 4 2 × 12 2 + 6 2 + 0 2 (By (ii) and (iii)) 144 + 36 144 + 36 + 24 × 144 + 36 180 196 × 180 180 …(v) 14

2



2  180  6 = 2  + 2 -2  14  7   [By (iv), (v)]





=2×

Hints :





=

(1) Make circle an xy-axis. (2) Use xz, yz axis and make sphere by each circle.

72 90 + -2 49 49





=

64 49

Circles touching coordinate axes in xy-axis are 4. Now, for sphere we have to consider 3D plane means x, y, z axis. There are 8 quadrants in 3D, so for a given radius there will be 8 possible spheres which will touch all 3 axis.

65.

Option (b) is correct. Explanation : Consider, angles given between lines. Direction cosines of lines by formula, we get: ⇒ AB = (12 - 0, 0 - 0, 0 - 0) ⇒ AB = (12, 0, 0) …(i) ⇒ AG = (12 - 0, 6 - 0, 4 - 0) ⇒ AG = (12, 6, 4) …(ii) ⇒ AC = (12 - 0, 6 - 0, 0 - 0) ⇒ AC = (12, 6, 0) …(iii) Now, for angle between AB and AG: Direction cosines of AB = (a1, b1, c1) = (12, 0, 0) Direction cosines of AG = (a2, b2, c2) = (12, 6, 4)



cos α =





cos α =

a1 a2 + b1 b2 + c1 c 2 a12

+ b12 + c12 × a22 + b22 + c 22

144 12 × 12 2 + 6 2 + 4 2



36 180 +2× -2 49 196

Hints : (1) Find direction cosines. (2) Put in formula of angle. 66. Option (c) is correct. Explanation :    Given: a , b , c are unit vectors    a = b = c =1 So,    Now, also given a × b perpendicular to c

…(i)



Statement 1: Take LHS     ˆ a × b = a b sin θ n







ˆ = 1 × 1 sin θ n  = sin θ c    [ c is perpendicular to a × b and of unit modulus]

195

solved PAPER - 2021 (ii)



   Hence, a × b = sin θ c Statement 2: Take LHS      a ⋅ b × c = [ ab c ]  = −[ a c b ]  = [c a b ]    = c ⋅ [a × b ] =0     Given, a ⊥ b × c

(

)

⇒



⇒













2 2 = 4 a −4 a





=0





⇒



⇒

(

) (

) (

) (



 Put b = ˆi in (ii), we get  2a + ˆi = ˆi − 2 ˆj ⇒  ⇒ a = −ˆj



So,







Angle between ˆi (x-axis) and ˆj (y-axis) is 90° i.e.,

 a = −ˆj   b = i

π . 2

68. Option (a) is correct. Explanation :    Given : a + b perpendicular to a

(

)

2  2 = - 4 a +  2 a  [By (i), (ii)]

69. Option (c) is correct. Explanation :    Given: a , b , c are coplanar. Since, they are co-planar so scalar triple product is zero, we get :   a b c  = 0      a ⋅ b × c = 0 …(i) Now, Statement 1: Consider,    I = a × b × c

(

)

(



(1) Multiply (i) by 2.   (2) Subtract and find a , b .

)

(1) Use given conditions. (2) Put in formula.



Hints :

(

Hints :

)

5bˆ = 5iˆ  b = ˆi

=0





Subtract (ii) from (iii)   2 a + 6bˆ − 2 a + bˆ = 6ˆi − 2 ˆj − ˆi − 2 ˆj



( a + b ) ⋅ a

    a⋅a + b ⋅a = 0   2   ∵ a ⋅ b =−b ⋅ a  a + a ⋅ b = 0      2 a ⋅ b = a  ⇒ …(i)   b = 2 a …(ii) Now, given    To find let I = 4a + b ⋅ b     I = 4a ⋅ b + b ⋅ b   2 = 4a ⋅ b + b



67. Option (d) is correct. Explanation : Consider, given vectors :   a + 3b = 3iˆ − ˆj …(i)   2a + b = ˆi − 2 ˆj …(ii) Multiply (i) by 2   2 a + 6b = 6ˆi − 2 ˆj …(iii)







(1) Use angle given between them.



So,



Hint :







)

  = ( a ⋅ c ) b − ( b ⋅ c )a         Now, ( a ⋅ c ) b − ( b ⋅ c )a ⋅ a × b       = ( a ⋅ c ) b a b  − b ⋅ c  a a b      = 0 - 0=0 …(ii)

{

}(

)

( )



 =  b  a b 0      So, a × b × c is co planar with a and b .

(

)

 a ab  (∵ =

)

196 Oswaal NDA/NA Year-wise Solved Papers

Statement 2 is correct. Statement 2: Consider,





     I =  a × b × c  ⋅  a × b    I = 0

d + (1 +x)(1 +x4)(1 + x8)(1 + x16) dx (1 + x2) d + (1 +x2)(1 +x4)(1 + x8)(1 + x16) dx (1 + x)















(

)

[From (ii)] Hint : (1) Use triple product rule.

70. Option (c) is correct. Explanation :  Given: A = 2 − 1 ˆi − ˆj  B = ˆi + 2 + 1 ˆj  AB = 1 − 2 − 1  ˆi +  2 + 1 − ( −1 )  ˆj     ˆ ˆ = 2 − 2  i +  2 + 2  j      2 2 AB = 2 − 2 + 2 + 2

(

)

(

(













=  AB =





 AB = 2 3

)

(

) (

)

)

[Magnitude of vector] 4+2−4 2 +4+2+4 2

 (1) Make AB . (2) Find magnitude.

71. Option (b) is correct. Explanation : Product rule of differentiation says :

( )

d n n −1   dx x = nx   

=1

(1) Use differentiation Rule of product. 72.

Option (a) is correct. Explanation : Given : y = cos x⋅cos 4x⋅cos 8x We find differentiation by product rule.



d d d d (a ⋅ b ⋅ c) = a ⋅b (c) + b⋅c (a) + a⋅c (b) dx dx dx dx



Similarly,



12

Hints :

=0+1

Hint :

)

(

dy ( x = 0) = 0 + 0 + 0 + 0 + (1 × 1 × 1 × 1 × 1) dx



…(i)

d d (cos x ⋅cos 4x⋅cos 8x) = cos x ⋅cos 4x (cos 8x) dx dx d d (cos x) + cos x cos 8x (cos 4x) + cos 8x⋅cos 4x dx dx ⇒ cos x cos 4x × -sin 8x × 8 + cos 8x cos 4x



× - sin x + cos x cos 8x × -sin 4x × 4 ⇒ -8 cos x cos 4x sin 8x - sin x cos 4x cos 8x - 4sin 4x cos x cos 8x We get : dy = -8 cos x cos 4x sin 8x - sin x cos 4x cos 8x dx - 4sin 4x cos x cos 8x ….(ii) 1 dy Now, to find I = y dx

d d d (a ⋅ b) = (a) (b) + (b) (a) dx dx dx Now,



y = (1 +x)(1 +x2)(1 + x4)(1 + x8)(1 + x16) Differentiate with respect to x dy d = (1 +x)(1 +x2)(1 + x4)(1 + x8) (1 + x16) dx dx



Use value from (i) and (ii),



I =



[-8cos x cos 4x sin 8x - sin x cos 4x cos 8x - 4sin 4x cos x cos 8x] −8 sin 8 x sin x 4 sin 4 x ⇒ − − cos 8 x cos x cos 4 x



d (1 +x)(1 +x2)(1 + x4)(1 + x16) + (1 + x8) dx



+ (1 +x)(1 +x2)(1 + x8)(1 + x16)

d (1 + x4) dx





1 cos x cos 4 x cos 8 x

⇒ -8tan 8x - tan x - 4tan 4x

197

solved PAPER - 2021 (ii)







Put



⇒ -8tan 2π - tan



⇒ 0 - tan



⇒





 sin θ  tan θ  ∵ cos= θ  

x=

π 4



 bn  an 1 + n   a  ⇒ lim n→∞  bn  an 1 − n   a 



  b n  an 1 +      a   lim ⇒ n→∞   b n  an 1 −      a  

π = -1 4

n

π   1 ∵ tan 4 =   Hints : (1) Find

Taking an common

π - 4tan π [∵ tan π = 0 and 4 tan 2π = 0]

π -0 4

-tan



dy using product rule. dx

(2) Put value of x =

π . 4





b 1+  a ⇒ lim n n→∞ b 1−  a Put limit, n → ∞, we get ∞





b 1+  a = ∞ b 1−  a





=





=1

Shortcut :

73.

log y = log cos x +log cos 4x +log cos 8x 1 dy tan x − 4 tan 4 x − 8 sin 8 x = y dx

Hint :

Use composition of function.

74. Option (c) is correct. Explanation :

Hints :

Put x=II =-1 -4.0 - 8.0=-1

Option (c) is correct. Explanation : Given : fof(x) = x4 f[f(x)] = x4 [composition of function] f[f(x)] = (x2)2 On comparing, we get f(x) = x2 Differentiate with respect x f′(x) = 2x Now, f′(x) at x = 1 f′(1) = 2 × 1 = 2

Given : lim

n→∞

n

a +b

n

an − b n

b 1+0   b < a , So < 1 a 1−0  

(1) Take common an. ∞

b (2) Use   = 0 a 75. Option (d) is correct. Explanation : Given :

x  1 + , 0 < x < 2 f(x) =  2 k  kx , 2 ≤ x < 4



lim f(x) exist mean at x = 2, we get :

x→ 2



⇒ Left hand limit = Right hand limit Now, Case I: Right hand limit for x=2+h f(x) taken = kx f(2 + h) = lim k(2 + h)



[put limit] = 2k …(i)

h→0

198 Oswaal NDA/NA Year-wise Solved Papers

Case II: Left hand limit for x = 2 - h







2−h f (2 - h) ⇒ lim 1 + h→0 2k



⇒ 1 +



⇒ 1 +



f(x) taken as 1 +

x 2k

[put limit]

2 2k



1 …(ii) k Using limit exist condition and (i), (ii), we get



⇒ 2k = 1 +



1 =1 k ⇒ 2k2 - 1 = k ⇒ 2k2 - k - 1 = 0 ⇒ 2k2 - 2k + k - 1 = 0 ⇒ 2k(k - 1) +1(k - 1) = 0 ⇒ (k - 1)(2k + 1) = 0



⇒



Option given (d), k = 1



1 k





 +x , x ≥ 0 g(x) =  −x , x < 0 (1) Left hand derivative

− (0 − h) − 0 lim h →0 −h (middle term split)

+h lim h →0 − h

= -1 −1 2

(1) Solve limits by LHL, RHL. (2) Satisfy on given condition.



Hence, Left hand limit = Right hand limit = g(0) So, g(x) continuous at x = 0 Hence, f(x) = g(x) - 1 ⇒ f(x) = |x| - 1 is continuous at x = 1 Statement 1 is correct. Statement 2: Again,

f (0 − h) − f (0) lim h →0 −h

Hints :

76.





⇒ 2k -

k = 1,

(3) g(x) at x = 0

g(0) = +0 = 0

Option (a) is correct. Explanation : Let g(x) = |x| Statement 1: +x , x ≥ 0 g(x) =  −x , x < 0



f (0 + h) − f (0) lim h →0 h

(0 + h) − 0 lim h →0 +h = +1

Hence, Left hand derivative ≠ Right hand derivative Hence, f(x) = g(x) - 1 also not differentiable at x=0 Statement 2 is incorrect. Hints : (1) Find continuity and differentiability at x = 0. (2) Use continuity and differentiability rule.

(1) Left hand limit (x < 0)

lim -(x - h) h→0 lim -x + h

(2) Right hand derivative

=0

77. Option (c) is correct. Explanation :





Consider,



Here, [x] is greatest integer function. Now, to find right hand limit :

h→0

(2) Right hand limit (x > 0) lim +(x + h)

h→0

lim +x + h h→0

=0

f(x) =

[x] x

199

solved PAPER - 2021 (ii)

lim

[1 + h ]

h →0

1+ h









[ In greatest Integer function, gives greatest possible integer less than or equal to the number. Example, [1 ⋅ 3] = 1

1 lim h →0 1 + h







Put limit, we get

Hints : (1) Greatest integer function give greatest possible integer less than or equal to the number. (2) Solve limits. 78. Option (b) is correct. Explanation :  1  f(x) = sin  2  x 

Consider,



Statement 1: Given f(0) = 0 To check continuity to x = 0 (1) Right hand limit (x > 0)

 1 lim sin  h →0  ( 0 + h )2 

   



[Put limit, h = 0]

1 = sin   0 = sin ∞

We know, sine function value lies between -1 and +1. Hence, -1 ≤ sin x ≤ + 1 So, -1 ≤ sin x ∞ ≤ 1 Right hand limit ≠ f(0) Hence, f(x) not continuous at x = 0. Statement 2: 2 To check continuity at x = π (1)

At x =



=

2 π

1 2

(2) Right hand limit

    1  lim sin  h →0  4 + h2 + 4h  π  π 



   1  = sin    4 + 0  π 



= sin



=



 1  lim sin  2  h →0 h 

 2  π f  = sin   4  π

    1   lim sin  2  h →0    2   π + h    

1 = 1 1+0









1 f(x) = sin    x2 

[Put limit, h = 0]

π 4

1 2

(3) Left hand limit

  1  lim sin  2 h →0   2 − h   π  

      

    1  lim sin  h →0  4 + h2 − 4h    π  π

[Put limit, h = 0]

π = sin   4 1 = 2

So,



 2  Left hand limit = Right hand limit = f    π

200 Oswaal NDA/NA Year-wise Solved Papers

Hence, f(x) is continuous at x =



Statement 2 is correct.

2 π

.



Hints : (1) Check continuity at x = 0. (2) Since value of sin x lies between -1 and +1. 79.

Option (b) is correct. Explanation : To find Range of f(x) = 1 - sin x We know, value of sine function lies between -1 and +1, we get : ⇒ -1 ≤ sin x ≤ + 1 [Multiply by -1] ⇒ -1 ≤ -sin x ≤ + 1 [Add 1] ⇒ 1 - 1 ≤ 1 - sin x ≤ 1 + 1 ⇒ 0 ≤ 1 - sin x ≤ 2 ⇒ 0 ≤ f(x) ≤ 2 Hence, f(x) = 1 - sin x lies between 0 and 2, we get : Range : [0, 2] Hint : (1) Consider Range of sin x.

80. Option (a) is correct. Explanation : y = cos–1 cos (x) ⇒ cos y = cos x

⇒

d d (cos y) = (cos x) dx dx



⇒

-sin y



⇒



dy = -sin x dx dy sin x = dx sin y

Now, at x = − π 4



  π  y = cos−1 cos  −    4   π  = cos  cos  4  [cos(- θ) = cos θ] −1

=



∴

 dy   dx  π  x = − 4

π 4

 π sin  −   4 = π sin   4 [sin (-θ) = -sin θ]



π 4 = π sin 4 = -1 π  Hence, slope of tangent at  x = −  = -1 4  − sin

Hints : (1) At x = −

π , y = -cos–1(cos x). 4

(2) Differentiation,

dy = -1. dx

81.

Option (d) is correct. Explanation : Given : f(x) = 1 + x2 + x4 We have to find w.r.t x2, we get :



I =



Let y = x2, we get



I =



I = y +



 n x n +1   ∫ x dx =  n + 1  



[put y = x2]

∫(

) ( )

+ x2 + x4 d x2

∫ (1 + y + y

2

) dy

y2 y3 + +c 2 3



(x ) + (x ) I = x2 +



I = x2 +

2

2

2

2

3

x4 x6 + +c 2 3

Hints : (1) Put y = x2 and Modify function. (2) Integrate according to function.

3

+c

201

solved PAPER - 2021 (ii)

82.

Option (b) is correct. Explanation : Given: f(x) = x2 + 1, interval = (1, 2) Differentiate with Repeat to x f′(x) = 2x put f′(x) = 0 for critical values ⇒ 2x = 0 ⇒ x=0 f″(x) = 2 < 0 ∴ f(1) is maximum at x = 0 ∴ minimum value = 2 In the interval (1, 2) Statement 1 incorrect. Statement 2 correct.

π



[Integration of cos x is sin x] π I = sin - sin 0 [put limits] 2 I = 1 Hints : (1) eln x = x property. (2) Definite integration.

85. Option (b) is correct. Explanation : Given :

∫



I=

Hints :



Consider,





I =

If f′(a) = 0 and f″(a) > 0 then f′(x) is minimum at x= a.



Now, given

4

I=

∫ f ′ ( x ) dx 1

4



I =  f ( x )  1







Putting limits I = f(4) - f(1) I = 0

[Integrate of differentials function gives function itself.]

[using (i)]

Hints : (1) Integration of differentiate function. (2) Using given values. 84. Option (c) is correct. Explanation :







Option (b) is correct. Explanation : Given: f(1) = f(4) …(i) Mean value of f(x) at x = 1, 4 is equal



I = [ sin x ]02



83.





Consider, I =

1 − sin 2x dx

[Putting values, sin2 x + cos2 x = 1 sin 2x = 2sin x cos x]







I=







I =



I = ∫ (cos x − sin x ) dx



∫

∫

sin 2 x + cos2 x − 2 sin x cos x dx [using (A + B)2 = A2 + B2 - 2AB]

2 ∫ ( cos x − sin x ) dx

 ∵ 0 < x < 

 I = sin x + cos x = 1 Comparing using (i) cos x + sin x + c = A sin x + B cos x + c We get: A=1 B = 1 So, ⇒ A + B = 1 + 1 = 2 ⇒ A + B = 2 ⇒ A + B - 2 = 0

π 4 

Hints : π 2

∫e

ln ( cos x )

0

π 2

∫ cos x dx 0

(1) Integrate using formula. (2) Compare with given equation to find value.

dx …(i)

We know, eln x = x …(ii) Using (ii) in (i), we get

I =

1 − sin 2x dx = A sin x + B cos x + C…(i)

86. Option (b) is correct. Explanation :

Equation of ellipse is Consider,

x

2

a

2

+

y

2

b2

x2

a2 =1

+

y2 b2

= 1

…(i)

202 Oswaal NDA/NA Year-wise Solved Papers

Differentiate with respect to x



⇒



⇒



Again differentiate with respect to x



⇒

2x a

2

+

2 y dy =0 b 2 dx

+

y dy = 0 b 2 dx

x a 1 a2

+

2

87. Option (b) is correct. Explanation :

1  d 2 y dy dy  + ×  =0 y b 2  dx 2 dx dx 



⇒

2 1  d 2 y  dy   1  + y    = − 2 b 2  dx 2  dx   a



⇒

 d 2 y  dy 2  b 2 − y 2 +    = 2 a  dx  dx  

⇒

d 2 y  dy  b2 −y 2 −   = 2 a dx  dx 



⇒

2 − y dy d 2 y  dy  −y 2 −   = x dx dx  dx 







2



⇒



⇒

y



(1) Do differentiation. (2) Make differential equation using value of constant.

[By (ii)]

y dy d 2 y  dy  +  = 2 x dx dx dx  

88. Option (b) is correct. Explanation :

2

y

Let circle of Radius = r Centre (r, r) Equation of circle of radius r : ⇒ (x - r)2 + (y - r)2 = r2 In above equation number of aribitrary constant =1 Hence, the order of differential equation = 1 and degree =1 Hints :

2



r (r, r) r

…(ii)





2



To find differential equation, Differentiate both side with respect to x. We get:



Now,



⇒

dy  1  = 0 − B − 2  dx  x 



⇒

dy B = 2 dx x

d2 y

dy  dy  ⇒ xy 2 + x   − y =0 dx dx dx  

Order of differential equation is the highest derivative of differential equation. Hence, order = 2 Hints : (1) Differentiation two times. (2) Remove constant using equation.

previous

y=A-

Equation of ellipse is x2 a



2

+

x2 b2

= 1

Since, it has two arbitrary constant So, order = 2

B x

y=A-

1 1   A & B = constant Differentiation of x is − 2  x  

Shortcut :

B x

d 2 y  dy  y dy =0 +  dx 2  dx  x dx

y1 =

B



⇒



Differentiate again with respect to x.



⇒

 −2  y2 = B ×  3  x 



⇒

y2 =

x2

…(i)

−2B x3

203

solved PAPER - 2021 (ii)



⇒



⇒



⇒ ⇒

get

−2 B × x x2

y2 =

−2 × y1 x xy2 = -2y1

y2 =

[using (i)]

xy2 + 2y1 = 0

Hints : (1) Differentiate two times. (2) Put value from y1 in y2. 89. Option (a) is correct. Explanation : Consider, π



I =



y - y1 = m(x - x1) ⇒ y - 1 = 1(x - 0) ⇒ y - 1 = x …(i) Given that line touches x-axis, so y = 0 in (i) ⇒ 0 - 1 = x ⇒ x = -1 Hence, it meets curve at (-1, 0).

(1) Find slope

x

∫ log tan 2 dx …(i) π

I =

Slope =

Hint :

0



dy (0, 1) = e° = 1 dx Now, by slope point form



π

x

∫ log tan  2 − 2  dx

dy and equation. dx

91. Option (a) is correct. Explanation : Consider,

∫ f ( a − x ) dx 



1 x Differentiate with respect to x



x I = ∫ log cot dx …(ii) 2 0







Add (i) and (ii)



I + I =

0





a =  ∫ f ( x ) dx  0



a



0



π

π

∫ log tan 0



x dx + 2

π

x

∫ log cot 2 dx 0

2I = 0 I = 0 Hint :

90.

a Use  ∫ f ( x ) dx =  0

a

0



Option (b) is correct. Explanation : Tangent to the curve at (0, 1) Consider, y = ex Differentiate with respect to x



dy = ex dx Slope of tangent at (0, 1)



We know,





∫ f ( a − x ) dx 



dy (0, 1) gives slope of tangent, we dx



f(x) = x +

f′(x) = 1 -

1 x2

( )



d n n −1   dx x = nx    Find critical point by f′(x) = 0, we get



⇒ 1 -



1



=0 x2 ⇒ x2 = 1 ⇒ x = ±1 Now, critical points are +1, -1



Consider,

f′(x) = 1 -



x2 Again differentiate with respect to x





f″(x) =



Check at x = +1, -1 Now, At x = +1, f″(x) = is (+)ve, x = -1, f″(x) is negative. By rule of second differention, we get At x = -1, f(x) is maximum and at x = +1, f(x) is minimum Hence, at x = -1, f(x) = -2 and at x = +1, f(x) = +2

1

2 x3

204 Oswaal NDA/NA Year-wise Solved Papers

So, maximum value is -2 and minimum value is +2. Statement 1: It is correct. Statement 2: It is incorrect.



dy = 0, we get dx 2(8 - x2) = 0





Hints :



(1) Find critical point. (2) By rule of second differentiation.

Put x = 2 2 in (i),



⇒

y=

16 − 8



⇒

y=

8



⇒

y= 2 2



Hence, we get x = 2 2 , y = 2 2



Now,

92. Option (c) is correct. Explanation : y A 2 y x

2 C





Put

x= 2 2

(

)

2 8 − x2 dA = dx 16 − x 2 Differentiate with repeat to x

(

16 − x 2 × ( −2 x ) − 8 − x 2

B x

1

×

)

× −x



Let Radius (r) = 2 units of circle and for Rectangle, Let x = length and y = breadth Now, In ∆ABC, By Pythagoras theorem ⇒ x2 + y2 = 42 ⇒ x2 + y2 = 16 ⇒ y2 = 16 - x2



⇒ y = 16 − x 2 …(i) Now, Area of rectangle = length × breadth A = xy





A = x 16 − x 2 Differentiate with respect to x



is negative. dx 2 Hence, by second differential rule if second derivative is negative, then function maximum. Hence, maximum area of rectangle:



  1 dA = x × × ( 0 − 2x ) dx  2 16 − x 2 



A = 2 2 × 2 2 A = 8 square units





2

d A







dx 2



−2 x 2 dA = + 16 − x 2 2 dx 2 16 − x 2













(

−x + 16 − x dA = dx 16 − x 2 2

dA 16 − 2 x = dx 16 − x 2

(

2

)

2 8−x dA = dx 16 − x 2

2

)

(

Put x = 2 2 in

16 − x 2

d2 A dx 2

)

2

, we get

d2 A

Hints :

+  16 − x 2 × 1  

= 2⋅

16 − x 2

(1) Make figure and area of rectangle. (2) Differentiate and get maximum area of rectangle. 93. Option (a) is correct. Explanation : Consider,

I =







I =

∫x

dx

(x

2

+1

)

[Multiply and Divide by x, we get]

∫ x2

x

(x

2

+1

)

dx

205

solved PAPER - 2021 (ii)







I =

[Multiply and Divide by 2, we get] 1 2x dx ∫ 2 2 x x2 + 1

(

….(i)

)

2



Put t=x Differentiate both sides dt = 2x dx Put values in (i), we get



I =







I =

1 dt 2 ∫ t ( t + 1)

[Add and Subtract t in Numerator, we get] 1 ( t + 1) − t dt 2 ∫ t ( t + 1)



1  ( t + 1) t  −  dt I = ∫  2  t ( t + 1 ) t ( t + 1 ) 



I =

1 1 1  − dt ∫  2  t t + 1 



I =

1 [log |t| - log |t + 1| + c 2



I = 1 log

2

t +c t +1



I =

⇒



Divide (ii) by (iii), we get



⇒



⇒

du = dv

ev

dv dx

ex



ev × ex du = dv ex



⇒



⇒

[using (iii)]

du = ev dv

du

( )

d e

x

x

= e e

[using (i)]

Hints : (1) Differentiate both separately. (2) Divide both result.

[put t = x2]

 x2  1 log  2 +c 2  x + 1 

Hints : (1) Multiply and Divide by x and 2. (2) Put t = x2. 94. Option (a) is correct. Explanation : x

u = e e , v = ex …(i)



Consider,



Now, u = e e = ev [using (i)] Differentiate with respect to x [Differentiation of ex is ex itself and use change rule]



⇒

du dv = ev × dx dx



⇒

du dv = ev …(ii) dx dx



Now,

x

v = ex

dv = ex …(iii) dx



95.

Option (b) is correct. Explanation : Consider, f(x) = x3 + x2 + kx Differentiate with respect to x f′(x) = 3x2 + 2x + k Now, given that f(x) has no local extremum, we get f′(x) ≠ 0 2 ⇒ 3x + 2x + k ≠ 0 ⇒ 3x2 + 2x + k > 0 or 3x2 + 2x + k < 0 {For quadratic equation if ax2 + bx + c ≠ 0 then, discriminant < 0} Now, By using above condition, we get ⇒ D -2x ⇒ k < 2x Using given interval (1, ∞)

207

solved PAPER - 2021 (ii)



first, put x = 1 ⇒ k loga y when a = . 2 Then the relation is (a) reflexive only (b) symmetric only (c) transitive only (d) both symmetric and transitive

Directions for the following three (03) items : Consider the following Venn diagram, where X, Y and Z are three sets. Let the number of elements in Z be denoted by n(Z) which is equal to 90. Y

X a

16 12

b

18

17

c

12. What is the value of the determinant i

2

4

i

6

i9

i 12

i

i

3

i

i 8 where i = −1  ? i 15

(a) 0 (c) 4i a  13. Let A =  h  g

(b) -2 (d) -4i h b f

g x  f  and B =  y  , then what is    z  c 

AB equal to ?  ax + hy + gz    (a) y     z  ax + hy + gz   hx + by + fz  (b)     z  ax + hy + gz   hx + by + fz  (c)    gx + fy + cz  (d) [ax + hy + gz hx + by + fz gx + fy + cz]

Z 16. If the number of elements in Y and Z are in the ratio 4 : 5 then what is the value of b ? (a) 18 (b) 19 (c) 21 (d) 23 17. What is the value of n(X) + n(Y) + n(Z) - n(X ∩ Y) - n(Y ∩ Z) - n(X ∩ Z) + n(X ∩ Y ∩ Z) ? (a) a + b + 43 (b) a + b + 63 (c) a + b + 96 (d) a + b + 106 18. If the number of elements belonging to neither X, nor Y, nor Z is equal to p, then what is the number of elements in the complement of X ? (a) p + b + 60 (b) p + b + 40 (c) p + a + 60 (d) p + a + 40

Directions for the following two (02) items : Read the following information and answer the two items that follow : tan 3A 1 Let = K, where tan A ≠ 0 and K ≠ . tan A 3

19. What is tan2 A equal to ? K+3 (a) 3K − 1

(b)

K -3 3K - 1

3K - 3 (c) K -3

(d)

K+3 3K + 1

216 Oswaal NDA/NA Year-wise Solved Papers 20. For real values of tan A, K cannot lie between

28. What is the value of cos 48° - cos 12° ?

1 (a) and 3 3

5 -1 (a) 4

1 and 2 2 1 1 (d) and 7 (c) and 5 7 5 Directions for the following two (02) items : Read the following information and answer the two items that follow : ABCD is a trapezium such that AB and CD are parallel and BC is perpendicular to them. Let ∠ADB = θ, ∠ABD = α, BC = p and CD = q. 21. Consider the following : (1) AD sin θ = AB sin α (2) BD sin θ = AB sin (θ + α) (b)

Which of the above is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 22. What is AB equal to ?

( p2 + q 2 ) sin θ (a) p cos θ + q sin θ ( p2 + q 2 ) sin θ (c)

q cos θ + p sin θ



(b)

(d)

( p2 − q 2 ) cos θ p cos θ + q sin θ

( p2 − q 2 ) cos θ q cos θ + p sin θ

cos 17° − sin 17° , then what is the cos 17° + sin 17° value of θ ? (a) 0° (b) 28° (c) 38° (d) 52° 23. If tan θ =

24. A and B are positive acute angles such that cos 2B = 3 sin2 A and 3 sin 2A = 2 sin 2B. What is the value of (A + 2B) ? π π (a) (b) 6 4 π π (c) (d) 3 2 25. What is sin 3x + cos 3x + 4 sin3 x - 3 sin x + 3 cos x - 4 cos3 x equal to ? (a) 0 (b) 1 (c) 2 sin 2x (d) 4 cos 4x 26. The value of ordinate of the graph of y = 2 + cos x lies in the interval (a) [0, 1] (b) [0, 3] (c) [- 1, 1] (d) [1, 3] 27. What is the value of 8 cos 10°. cos 20° . cos 40° ? (a) tan 10° (b) cot 10° (c) cosec 10° (d) sec 10°

(b)

1- 5 4

5 +1 1- 5 (c) (d) 2 8 29. Consider the following statements : (1) If ABC is a right-angled triangle, rightangled at A and if sin B =

1 , then cosec C = 3. 3

(2) If b cos B = c cos C and if the triangle ABC is not right-angled, then ABC must be isosceles. Which of the above statements is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2

(d) Neither 1 nor 2

30. Consider the following statements : (1) If in a triangle ABC, A = 2B and b = c, then it must be an obtuse-angled triangle. (2) There exists no triangle ABC with A = 40°, a = sin 40° cosec 15°. c Which of the above statements is/are correct ? (a) 1 only (b) 2 only B = 65° and

(c) Both 1 and 2

(d) Neither 1 nor 2

Directions for the following three (03) items :



Read the following information and answer the three items that follow : Let a sin2 x + b cos2 x = c; b sin2 y + a cos2 y = d and p tan x = q tan y. 31. What is tan2 x equal to ? c-b (a) a-c

(b)

a-c c-b

c-a (c) c-b

(d)

c-b c-a

d-a equal to ? b-d (a) sin2 y (b) cos2 y 2 (c) tan y (d) cot2 y 32. What is

33. What is

p2 q2

equal to ?

(b − c ) (b − d ) (a) (a − d)(a − c)

(b)

(a − d )(c − a) (b − c ) (d − b)

(d − a) (c − a) (c) (b − c ) (d − b)

(d)

(b − c ) (b − d ) (c − a)(a − d )

217

solved PAPER - 2020 (I)

40. If sin x + sin y = cos y - cos x, where 0 < y < x <

Directions for the following three (03) items : Read the following information and answer the three items that follow : Let tn = sinn q + cosn q.

34. What is

t3 - t5 equal to ? t5 - t7 t3 (b) t5

t5 (c) t7

(d)

t1 t7

35. What is t12 - t2 equal to ? (a) cos 2θ (b) sin 2θ (c) 2 cos θ (d) 2 sin θ 36. What is the value of t10 where θ = 45° ? (a) 1

1 (b) 4

1 1 (c) (d) 16 32 Directions for the following three (03) items : Read the following information and answer the three items that follow : Let α = β = 15°. 37. What is the value of sin α + cos β ? 1 (a) 2

(b)

1 2 2

3 3 (c) (d) 2 2 2 38. What is the value of sin 7α - cos 7β ? 1 (a) 2

(b)

1 2 2

3 3 (c) (d) 2 2 2 39. What is sin (α + 1°) + cos (β + 1°) equal to ? 3 cos 1° + sin 1° (a) 1 (b) 3 cos 1° − sin 1° 2 1 (c) ( 3 cos 1° + sin 1° ) 2 1 (d) ( 3 cos 1° + sin 1° ) 2

x−y then what is tan   equal to ?  2  (a) 0 (c) 1

t1 (a) t3

π , 2

1 2 (d) 2

(b)

41. If A is a matrix of order 3 × 5 and B is a matrix of order 5 × 3, then the order of AB and BA will respectively be (a) 3 × 3 and 3 × 3 (b) 3 × 5 and 5 × 3 (c) 3 × 3 and 5 × 5 (d) 5 × 3 and 3 × 5 42. If p2, q2 and r2 (where p, q, r > 0) are in GP, then which of the following is/are correct ? (1) p, q and r are in GP. (2) ln p, ln q and ln r are in AP.

Select the correct answer using the code given below : (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 43. If cot α and cot β are the roots of the equation x2 - 3x + 2 = 0, then what is cot (α + β) equal to ? 1 (a) 2 (c) 2

1 3 (d) 3 (b)

44. The roots α and β of a quadratic equation, satisfy the relations α + β = α2 + β2 and αβ= α2β2. What is the number of such quadratic equations ? (a) 0 (b) 2 (c) 3 (d) 4 45. What is the argument of the complex number 1−i 3

, where i = −1  ? 1+i 3 (a) 240° (b) 210° (c) 120° (d) 60° 46. What is the modulus of the complex number cos θ + i sin θ , where i = −1 ? cos θ − i sin θ 1 (a) 2

(b) 1

3 (c) 2

(d) 2

218 Oswaal NDA/NA Year-wise Solved Papers 47. Consider the proper subsets of {1, 2, 3, 4}. How many of these proper subsets are superset of the set {3} ? (a) 5 (b) 6 (c) 7 (d) 8 48. Let p, q and r be three distinct positive real p q r number, If D = q r p , then which one of r p q the following is correct ? (a) D < 0 (b) D ≤ 0 (c) D > 0 (d) D ≥ 0 49. What is the sum of last five coefficients in the expansion of (1+ x)9 when it is expanded in ascending powers of x ? (a) 256 (b) 512 (c) 1024 (d) 2048 50. Consider the following in respect of a nonsingular matrix of order 3 : (1) A (adj A) = (adj A) A (2) |adj A| = |A| Which of the above statements is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 51. The center of the circle (x - 2a) (x - 2b) + (y - 2c) (y - 2d) = 0 is (a) (2a, 2c) (b) (2b, 2d) (c) (a + b, c + d) (d) (a - b, c - d) 52. The point (1, -1) is one of the vertices of a square. If 3x + 2y = 5 is the equation of one diagonal of the square, then what is the equation of the other diagonal ? (a) 3x - 2y = 5 (b) 2x - 3y = 1 (c) 2x - 3y = 5 (d) 2x + 3y = -1 53. Let P(x, y) be any point on the ellipse 25x2 + 16y2 = 400. If Q(0, 3) and R(0, - 3) are two points, then what is (PQ + PR) equal to ? (a) 12 (b) 10 (c) 8 (d) 6 54. If the circumcenter of the triangle formed by the lines x + 2 = 0, y + 2 = 0 and kx + y + 2 = 0 is (- 1, - 1), then what is the value of k ? (a) - 1 (b) - 2 (c) 1 (d) 2 2

55. In the parabola, y = x, what is the length of the chord passing through the vertex and inclined to the x-axis at an angle θ ?

(a) sin θ . sec2 θ (b) cos θ . cosec2θ (c) cot θ . sec2 θ (d) 2 tan θ . cosec2 θ 56. Under which condition, are the points (a, b), (c, d) and (a - c, b - d) collinear ? (a) ab = cd (b) ac = bd (c) ad = bc (d) abc = d 57. Let ABC be a triangle. If D(2, 5) and E(5, 9) are the mid-points of the sides AB and AC respectively, then what is the length of the side BC ? (a) 8 (b) 10 (c) 12 (d) 14 58. If the foot of the perpendicular drawn from the point (0, k) to the line 3x - 4y - 5 = 0 is (3, 1), then what is the value of k ? (a) 3 (b) 4 (c) 5 (d) 6 59. What is the obtuse angle between the lines whose slopes are 2 − 3 and 2 + 3  ? (a) 105° (b) 120° (c) 135° (d) 150° 60. If 3x - 4y - 5 = 0 and 3x - 4y + 15 = 0 are the equations of a pair of opposite sides of a square, then what is the area of the squares ? (a) 4 square units (b) 9 square units (c) 16 square units (d) 25 square units 61. What is the length of the diameter of the sphere whose centre is at (1, -2, 3) and which touches the plane 6x - 3y + 2z - 4 = 0 ? (a) 1 unit (b) 2 units (c) 3 units (d) 4 units 62. What is the perpendicular distance from the point (2, 3, 4) to the line (a) 6 units (c) 3 units

x−0 y−0 z−0 = =  ? 1 0 0 (b) 5 units (d) 2 units

63. If a line has direction ratios < a + b, b + c, c + a >, then what is the sum of the squares of its direction cosines ? (a) (a + b + c)2 (b) 2(a + b + c) (c) 3 (d) 1 64. Into how many compartments coordinate planes divide the space ? (a) 2 (b) 4 (c) 8 (d) 16

do

the

219

solved PAPER - 2020 (I)

65. What is the equation of the plane which cuts an intercept 5 units on the z-axis and it parallel to xy-plane ? (a) x + y = 5 (b) z = 5 (c) z = 0 (d) x + y + z = 5 66. If a is a unit vector in the xy-plane making an angle 30° with the positive x-axis, then what is a equal to ? 3 i + j (a) 2

3 i - j (b) 2

i + 3 j (c) 2

(d)

i - 3 j 2

 67. Let A be a point in space such that OA = 12,  where O is the origin. If OA is inclined at angles 45° and 60° with x-axis and y-axis respectively,  then what is OA equal to ? 6i + 6 j ± 2 k (a) (b) 6i + 6 2 j ± 6 k (c) 6 2 i + 6 j ± 6 k

(d) 3 2 i + 3 j ± 6 k

68. Two adjacent sides of a parallelogram are 2i − 4 j + 5k and i - 2 j - 3k . What is the magnitude of dot product of vectors which represent its diagonals ? (a) 21 (b) 25 (c) 31 (d) 36  2  2  69. If a × b + a . b = 144 and a = 4, then what is  b equal to ? (a) 3 (b) 4 (c) 6 (d) 8   70. If the vectors a = 2i − 3 j + k , b = i + 2 j − 3k and  c = j + pk are coplanar, then what is the value of p ? (a) 1 (c) 5

(b) - 1 (d) - 5

x + x2 + x3 − 3 equal to ? 71. What is lim x →1 x −1 (a) 1 (b) 2 (c) 3 (d) 6 72. The radius of a circle is increasing at the rate of 0.7 cm/sec. What is the rate of increase of its circumference ? (a) 4.4 cm/sec (b) 8.4 cm/sec (c) 8.8 cm/sec (d) 15.4 cm/sec

x4 − 1 x3 − k 3 , where k ≠ 0, then = lim 2 x →1 x − 1 x →k x − k 2 what is the value of k ?

73. If lim

2 4 (a) (b) 3 3 8 (c) (d) 4 3 74. The order and degree of the differential equation 2

  dy 2  3 dy = 1 +    dx are respectively k dx   dx   (a) 1 and 1 (b) 2 and 3 (c) 2 and 4 (d) 1 and 4

∫

75. What is lim

sin x log ( 1 − x )

x →0

equal to ?

x2

(a) - 1

(b) Zero

1 e 76. If f(x) = 3x2 - 5x + p and f(0) and f(1) are opposite in sign, then which of the following is correct ? (a) - 2 < p < 0 (b) - 2 < p < 2 (c) 0 < p < 2 (d) 3 < p < 5 (d) -

(c) - e

77. If eqj = c + 4qj, where c is an arbitrary constant and ϕ is a function of θ, then what is ϕdθ equal to ? (a) θdϕ (b) - θdϕ (c) 4θdϕ (d) -4θdϕ 78. If p(x) = (4e)2x, then what is

∫ p ( x ) dx equal to ?

p (x) +c (a) 1 + 2 ln 2

(b)

p (x) +c 2 ( 1 + 2 ln 2 )

2p ( x ) +c (c) 1 + ln 4

(d)

p (x) +c 1 + ln 2

π/4

79. What is the value of

∫ ( tan

3

x + tan x ) dx  ?

0

1 (a) 4 (c) 1

1 2 (d) 2 (b)

80. Let y = 3x2 + 2. If x changes from 10 to 10.1, then what is the total change in y ? (a) 4.71 (b) 5.23 (c) 6.03 (d) 8.01

220 Oswaal NDA/NA Year-wise Solved Papers dx

sin x , where x ∈ R, is to be continuous x at x = 0, then the value of the function at x = 0 (a) should be 0 (b) should be 1 (c) should be 2 (d) cannot be determined

1  xn  (a) ln  n +c n  x +1

 xn + 1  (b) ln  n  + c  x 

 xn  ln  n (c) +c  x +1

(d)

82. The solution of the differential equation dy = (1 + y2) dx is (a) y = tan x + c (b) y = tan (x + c) -1 (d) tan-1 (y + c) = 2x (c) tan (y + c) = x

89. What is the minimum value of |x - 1|, where x ∈ R ? (a) 0 (b) 1 (c) 2 (d) - 1

81. If f ( x ) =

83. What is

∫ (e

log e x

+ sin x ) cos x equal to ? 2

sin x + x cos x + (a)

sin x +c 2

sin x − x cos x + (b)

sin 2 x +c 2

x sin x + cos x + (c)

sin 2 x +c 2

sin 2 x +c 2 84. What is the domain of the function f(x) = cos-1 (x - 2) ? (a) [- 1, 1] (b) [1, 3] (c) [0, 5] (d) [- 2, 1] x sin x − x cos x + (d)

85. What is the area of the region enclosed between the curve y2 = 2x and the straight line y = x ? 1 (a) (b) 1 2 2 (c) (d) 2 3 86. If f(x) = 2x - x2, then what is value of f(x + 2) + f(x - 2) when x = 0 ? (a) - 8 (b) - 4 (c) 8 (d) 4 87. If xmyn = am+n then what is

dy equal to ? dx

my (a) nx

(d) -

mx (c) ny

(d) -

my nx ny mx

88. What is

∫ x (x

n

+ 1)

equal to ?

1  xn + 1  ln  +c n  xn 

90. What is the value of k such that integration of 3x 2 + 8 − 4 k with respect to x, may be a rational x function ? (a) 0 (b) 1 (c) 2 (d) - 2 91. Consider the following statements for f(x) = e-|x| : (1) The function is continuous at x = 0. (2) The function is differentiable at x = 0. Which of the above statements is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 92. What is the maximum value of sin x . cos x ? (a) 2 (b) 1 1 (c) (d) 2 2 2 −x x 3 +3 −2 equal to ? 93. What is lim x →0 x (a) 0 (b) - 1 (c) 1 (d) Limit does not exist 94. What is the derivative of tan-1 x with respect to cot-1 x ? (a) -1 (b) 1 1 x (d) 2 (c) 2 x +1 x +1 95. The function u(x, y) = c which satisfies the differential equation x(dx - dy) + y(dy - dx) = 0, is (a) x2 + y2 = xy + c (b) x2 + y2 = 2xy + c 2 2 (d) x2 - y2 = 2xy + c (c) x - y = xy + c π  96. What is the minimum value of 3cos  A +   3 where A ∈ R ? (a) - 3 (b) - 1 (c) 0 (d) 3

221

solved PAPER - 2020 (I)

97. Consider the following statements : (1) The function f(x) = ln x increases in the interval (0, ∞). (2) The function f(x) = tan x increases in the  π π interval  − ,  .  2 2 Which of the above statements is/are correct ? (a) 1 only (b) 2 only (c) Both 1and 2 (d) Neither 1 nor 2 98. Which one of the following is correct in respect 1  ? x −1 (a) The domain is {x ∈  |x ≠ 1} and the range is the set of reals. (b) The domain is {x ∈  |x ≠ 1}, the range is {y ∈  |y ≠ 0} and the graph intersects y-axis at (0, -1). (c) The domain is the set of reals and the range is the singleton set {0}. (d) The domain is {x ∈  |x ≠ 1} and the range is the set of points on the y-axis. of the graph of y =

99. What is the solution of the differential equation  dy  ln   = x  ?  dx  (a) y = ex + c (c) y = ln x + c

(b) y = e-x + c (d) y = 2 ln x + c

100. Let l be the length and b be the breadth of a rectangle such that l + b = k. What is the maximum area of the rectangle ? (a) 2k2 (b) k2 k2 k2 (c) (d) 2 4 101. The numbers 4 and 9 have frequencies x and (x - 1) respectively. If their arithmetic mean is 6, then what is the value of x ? (a) 2 (b) 3 (c) 4 (d) 5 102. If three dice are rolled under the condition that no two dice show the same face, then what is the probability that one of the faces is having the number 6 ? 5 (a) 6

(b)

5 9

1 (c) 2

(d)

5 12

103. If P ( A ∪ B ) =

5 ( 1 , P A ∩ B ) = and 6 3

1 , then which one of the following 2 is not correct ? 2 P (B) = (a) 3 (b) P(A ∩ B) = P(A)P(B) (c) P(A ∪ B) > P(A)+P(B) (d) P(not A and not B) = P(not A) P(not B)

P(not A) =

104. The sum of deviations of n number of observations measured from 2.5 is 50. The sum of deviations of the same set of observations measured from 3.5 is - 50. What is the value of n ? (a) 50 (b) 60 (c) 80 (d) 100 105. A data set of n observations has mean 2M, while another data set of 2n observations has mean M. What is the mean of the combined data sets ? 3M (a) M (b) 2 2M 4M (d) (c) 3 3 Directions for the following three (03) items : Read the following information and answer the three items that follow : Marks

Number of students Physics

Mathematics

10 - 20

8

10

20 - 30

11

21

30 - 40

30

38

40 - 50

26

15

50 - 60

15

10

60 - 70

10

6

106. The difference between number of students under Physics and Mathematics is largest for the interval (a) 20 - 30 (b) 30 - 40 (c) 40 - 50 (d) 50 - 60 107. Consider the following statements : (1) Modal value of the marks in Physics lies in the interval 30 - 40. (2) Median of the marks in Physics is less than that of marks in Mathematics.

222 Oswaal NDA/NA Year-wise Solved Papers Which of the above statements is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 108. What is the mean of marks in Physics ? (a) 38.4 (b) 39.4 (c) 40.9 (d) 41.6 109. What is the standard deviation of the observations

- 6 , - 5 , - 4 , - 1, 1,

4,

5 , 6  ?

(a) 2

(b) 2

(c) 2 2

(d) 4

110. If

∑x

i

= 20 ,

∑x

2 i

= 200 and n = 10 for an

observed variable x, then what is the coefficient of variation ? (a) 80 (b) 100 (c) 150 (d) 200 111. What is the probability that February of a leap year selected at random, will have five Sundays ? 1 (a) 5

(b)

1 7

2 (c) (d) 1 7 112. The arithmetic mean of 100 observations is 40, Later, it was found that an observation ‘53’ was wrongly read as ‘83’. What is the correct arithmetic mean ? (a) 39.8 (b) 39.7 (c) 39.6 (d) 39.5 113. A husband and wife appear in an interview for two vacancies for the same post. The probability of the husband’s selection is

1 and that of the 7

1 wife’s selection is . If the events are indepen5 dent, then the probability of which one of the 11  ? 35 (a) At least one of them will be selected (b) Only one of them will be selected (c) None of them will be selected (d) Both of them will be selected following is

114. A dealer has a stock of 15 gold coins out of which 6 are counterfeits. A person randomly picks 4 out of the 15 gold coins. What is the probability that all the coins picked will be counterfeits ? 1 (a) 91

(b)

4 91

6 (c) 91

(d)

15 91

115. A committee of 3 is to be formed from a group of 2 boys and 2 girls. What is the probability that the committee consists of 2 boys and 1 girl ? 2 (a) 3

(b)

1 4

3 1 (c) (d) 4 2 116. In a lottery of 10 tickets numbered 1 to 10, two tickets are drawn simultaneously. What is the probability that both the tickets drawn have prime numbers ? 1 (a) 15

(b)

1 2

2 (c) 15

(d)

1 5

117. Let X and Y represent prices (in `) of a commodity in Kolkata and Mumbai respectively. = X 65 = , Y 67 , σX = 2.5, σY = It is given that 3.5 and r(X, Y) = 0.8. What is the equation of regression of Y on X ? (a) Y = 0.175X - 5 (b) Y = 1.12X - 5.8 (c) Y = 1.12X - 5 (d) Y = 0.17X + 5.8 118. Consider a random variable X which follows Binomial distribution with parameters n = 10 1 . Then Y = 10 - X follows Binomial 5 distribution with parameters n and p respectively given by and p =

1 5, (a) 5 10 , (c)

3 5

(b) 5,

2 5

(d) 10 ,

4 5

223

solved PAPER - 2020 (I)

119. If A and B are two events such that P(A) = 0.6, P(B) = 0.5 and P(A ∩ B) = 0.4, then consider the following statements : (1) P ( A ∪ B ) = 0.9. (2) P ( B | A ) = 0.6. Which of the above statements is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2

For elaborated Solutions

9 (a) 29

(b)

10 29

19 (c) 29

(d)

28 29

 

Finished Solving the Paper ? Time to evaluate yourself !

SCAN THE CODE

120. Three cooks X, Y and Z bake a special kind of cake, and with respective probabilities 0.02, 0.03 and 0.05, it fails to rise. In the restaurant where they work, X bakes 50%, Y bakes 30% and Z bakes 20% of cakes. What is the proportion of failures caused by X ?

SCAN

224 Oswaal NDA/NA Year-wise Solved Papers

Answers Q. No.

Answer Key

Topic Name

Chapter Name

1

(c)

Properties of Matrix

Matrix

2

(a)

Independent Term

Binomial Theorem

3

(b)

Sum of Coefficients

Binomial Theorem

4

(c)

Properties of Combination

Combination

5

(d)

Singular Matrix

Matrix

6

(c)

Binary Number

Binary Number

7

(a)

Properties of Logs

Logarithm

8

(c)

Properties of Logs

Logarithm

9

(b)

Properties of Matrices

Determinant

10

(d)

Inequations

Function

11

(c)

Types of Relation

Relation and Function

12

(d)

Properties of Iota

Determinant

13

(c)

Properties of Matrix

Matrix

14

(b)

Permutation

Permutation and Combination

15

(c)

Combination

Permutation and Combination

16

(c)

Properties of Sets

Sets

17

(d)

Properties of Sets

Sets

18

(a)

Properties of Sets

Sets

19

(b)

Properties of Trigonometry

Trigonometry

20

(a)

Properties of Trigonometry

Trigonometry

21

(c)

Properties of Triangle

Trigonometry

22

(a)

Properties of Triangle

Trigonometry

23

(b)

Properties of Trigonometry

Trigonometry

24

(d)

Properties of Trigonometry

Trigonometry

25

(a)

Identities

Trigonometry

26

(d)

Domain and Range

Trigonometry

27

(b)

Identities

Trigonometry

28

(b)

Values

Trigonometry

29

(b)

Properties of Triangle

Trigonometry

30

(d)

Properties of Triangle

Trigonometry

31

(a)

Identities

Trigonometry

32

(c)

Identities

Trigonometry

33

(b)

Identities

Trigonometry

34

(a)

Identities

Trigonometry

35

(b)

Identities

Trigonometry

36

(c)

Identities

Trigonometry

37

(d)

Values

Trigonometry

225

solved PAPER - 2020 (I)

Q. No.

Answer Key

Topic Name

Chapter Name

38

(d)

Values

Trigonometry

39

(c)

Values

Trigonometry

40

(c)

Formulas

Trigonometry

41

(c)

Order of Matrix

Matrix

42

(c)

A.P. and G.P.

Sequence and Series

43

(b)

Properties of Roots

Quadratic Equation

44

(d)

Properties of Roots

Quadratic Equation

45

(a)

Argument

Complex Number

46

(b)

Modulus

Complex Number

47

(c)

Super Set

Sets

48

(a)

Value

Determinant

49

(a)

Sum of Coefficients

Binomial Theorem

50

(a)

Properties of Matrix

Matrix

51

(c)

Circle

Circle

52

(c)

Straight Line

Straight Line

53

(b)

Properties of Ellipse

Ellipse

54

(c)

Equation of Line

Straight Line

55

(b)

Length of Chord

Parabola

56

(c)

Collinear Points

2D

57

(b)

Length of Side

2D

58

(c)

Slopes

Straight Line

59

(b)

Slopes

Straight Line

60

(c)

Area of Square

2D

61

(d)

Radius of Circle

3D

62

(b)

Distance of a Point

3D

63

(d)

Direction Cosine

3D

64

(c)

Octants

3D

65

(b)

Equation of Plane

3D

66

(a)

Position Vector

Vector

67

(c)

Position Vector

Vector

68

(c)

Product of Vector

Vector

69

(a)

Properties of Vector

Vector

70

(b)

Coplanar Vector

Vector

71

(d)

Limit

Limits

72

(a)

Rate of Change

Application of Derivative

73

(c)

Limit

Limits

74

(b)

Order and Degree

Differential Equation

75

(a)

Limit

Limits

76

(c)

Properties of Function

Function

77

(b)

Differential Equation

Differentiation

78

(b)

Indefinite Integration

Integration

226 Oswaal NDA/NA Year-wise Solved Papers Q. No.

Answer Key

Topic Name

Chapter Name

79

(b)

Definite Integration

Integration

80

(c)

Rate of Change

Application of Derivative

81

(b)

Continuity

Continuity and Differentiability

82

(b)

Solution

Differential Equation

83

(c)

Indefinite Integration

Application of Integration

84

(b)

Domain and Range

Function

85

(c)

Area under Curves

Integration

86

(a)

Value of Function

Function

87

(b)

Differential Coefficient

Differentiation

88

(a)

Indefinite Integration

Integration

89

(a)

Modulus Function

Modulus Function

90

(c)

Indefinite Integration

Integration

91

(a)

Continuity

Continuity and Differentiability

92

(c)

Maximum Value

Trigonometry

93

(a)

Limit

Limits

94

(a)

Differential Coefficient

Differentiation

95

(b)

Solution

Differential Equation

96

(a)

Minimum Value

Trigonometry

97

(c)

Increasing and Decreasing

Application of Derivative

98

(b)

Domain and Range

Function

99

(a)

Solution

Differential Equation

100

(d)

Maxima and Minima

Application of Derivative

101

(b)

Mean

Statistics

102

(c)

Probability

Probability

103

(c)

Probability

Probability

104

(d)

Standard Deviation

Statistics

105

(d)

Mean

Statistics

106

(c)

Class Interval

Statistics

107

(a)

Median

Statistics

108

(c)

Mean

Statistics

109

(b)

Standard Deviation

Statistics

110

(d)

Coefficient of Variation

Statistics

111

(b)

Probability

Probability

112

(b)

Mean

Statistics

113

(a)

Probability

Probability

114

(a)

Probability

Probability

115

(d)

Probability

Probability

116

(c)

Probability

Probability

117

(b)

Equation of Regression

Statistics

118

(d)

Probability

Probability

119

(d)

Probability

Probability

120

(b)

Probability

Probability

NDA / NA

MATHEMATICS

National Defence Academy / Naval Academy

question Paper

i

2019

Time : 2:30 Hour

Total Marks : 300

Important Instructions : 1. This test Booklet contains 120 items (questions). Each item is printed in English. Each item comprises four responses (answer's). You will select the response which you want to mark on the Answer Sheet. In case you feel that there is more than one correct response, mark the response which you consider the best. In any case, choose ONLY ONE response for each item. 2. You have to mark all your responses ONLY on the separate Answer Sheet provided. 3. All items carry equal marks. 4. Before you proceed to mark in the Answer Sheet the response to various items in the Test Booklet, you have to fill in some particulars in the Answer Sheet as per instructions. 5. Penalty for wrong answers : THERE WILL BE PENALTY FOR WRONG ANSWERS MARKED BY A CANDIDATE IN THE OBJECTIVE TYPE QUESTION PAPERS. (i) There are four alternatives for the answer to every question. For each question for which a wrong answer has been given by the candidate, one·third of the marks assigned to that question will be deducted as penalty. (ii) If a candidate gives more than one answer, it will be treated as a wrong answer even if one of the given answers happens to be correct and there will be same penalty as above to that question. (iii) If a question is left blank, i.e., no answer is given by the candidate, there will be no penalty for that question.

1. What is the nth term of the sequence 25, - 125, 625, - 3125, ... ? (a) ( -5)2n-l (b) (- 1)2n 5n+1 (c) (- l)2n-1 5n+1 (d) (- l)n-1 5n+1 2. Suppose X = {1, 2, 3, 4} and R is a relation on X. If R= {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}, then which one of the following is correct ? (a) R is reflexive and symmetric, but not transitive (b) R is symmetric and transitive, but not reflexive (c) R is reflexive and transitive, but not symmetric (d) R is neither reflexive nor transitive, but symmetric 3. A relation R is defined on the set N of natural numbers as xRy ⇒ x2 - 4xy + 3y2 = 0. Then which one of the following is correct ? (a) R is reflexive and symmetric, but not transitive (b) R is reflexive and transitive, but not symmetric (c) R is reflexive, symmetric and transitive (d) R is reflexive, but neither symmetric nor transitive

4. If A = {x ∈ Z : x3 - l = 0} and B = {x ∈ Z : x2 + x + 1 = 0}, where Z is set of complex numbers, then what is A ∩ B equal to ? (a) Null set  −1 + 3i −1 − 3i  (b) ,   2 2    −1 + 3i −1 − 3i  (c) ,   4 4    1 + 3i 1 − 3i  (d) ,   2   2 5. Consider the following statements for the two non-empty sets A and B :

(

) (

)

(1) ( A ∩ B) ∪ A ∩ B ∪ A ∩ B = A ∪ B

( (

(2) A∪ A∩B

)) = A ∪ B

which of the above statements is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 6. Let X be a non-empty set and let A, B, C be subsets of X. Consider the following statements : (1) A ⊂ C ⇒ (A ∩ B) ⊂ (C ∩ B),  (A ∪ B) ⊂ ( C ∪ B)

228 Oswaal NDA/NA Year-wise Solved Papers (2) (A ∩ B) ⊂ (C ∩ B) for all sets B ⇒ A ⊂ C (3) (A ∪ B) ⊂ ( C ∪ B) for all sets B ⇒ A ⊂ C which of the above statements is/are correct ? (a) 1 and 2 only (b) 2 and 3 only (c) 1 and 3 only (d) 1, 2 and 3 3 2 0    7. If B =  2 4 0  , then what is adjoint of B  1 1 0  equal to ?

14. The equation px2 + qx + r = 0 (where p, q, r, all are positive) has distinct real roots a and b. Which one of the following is correct ? (a) a > 0, b > 0 (b) a < 0, b < 0 (c) a > 0, b < 0 (d) a < 0, b > 0 15. If A= {l, {l, μ}}, then the power set of A is (a) {φ, {φ}, {l}, {l, m}} (b) {φ, {l},{{l, m}}, {l,{l, m}}} (c) {φ, {l},{l, m}, {l,{l, m}}} (d) {{l}, {l, m}, {l,{l, m}}}

8. What are the roots of the equation |x2 - x - 6| = x + 2 ? (a) - 2, 1, 4 (b) 0, 2, 4 (c) 0, 1, 4 (d) - 2, 2, 4

Consider the following for the next 02 (two) items that follow : In a school, all the students play at least one of three indoor games - chess, carrom and table tennis. 60 play chess, 50 play table tennis, 48 play carrom, 12 play chess and carrom, 15 play carrom and table tennis, 20 play table tennis and chess. 16. What can be the minimum number of students in the school ? (a) 123 (b) 111 (c) 95 (d) 63

0 1  9. If A =   , then the malrix A is a/an 1 0  (a) Singular matrix (b) Involutory matrix (c) Nilpotent matrix (d) Idempotent matrix

17. What can be the maximum number of students in the school ? (a) 111 (b) 123 (c) 125 (d) 135

 0 0 0  0 0 0 (a)    −2 −1 8 

0 0 −2  (b) 0 0 −1  0 0 8 

0 0 2  0 0 1  (c)   0 0 0 

(d) It does not exist

x -3i 1 10. If y 1 i = 6 + 11i, then what are the 0 2i -i values of x and y respectively ? (a) - 3, 4 (b) 3, 4 (c) 3, - 4 (d) – 3, – 4 11. The common roots of the equations z3 + 2z2 + 2z + 1 = 0 and z2017 + z2018 + 1 = 0 are (a) - 1, w (b) 1, w2 (c) - 1, w2 (d) w, w2 12. If C(20, n + 2) = C(20, n - 2), then what is n equal to ? (a) 8 (b) 10 (c) 12 (d) 16 13. There are 10 points in a plane. No three of these points are in a straight line. What is the total number of straight lines which can be formed by joining the points ? (a) 90 (b) 45 (c) 40 (d) 30



18. If A is an identity matrix of order 3, then its inverse (A-1) (a) is equal to null matrix (b) is equal to A (c) is equal to 3A (d) does not exist 19. A is a square matrix of order 3 such that its determinant is 4. What is the determinant of its transpose ? (a) 64 (b) 36 (c) 32 (d) 4 20. From 6 programmers and 4 typists, an office wants to recruit 5 people. What is the number of ways this can be done so as to recruit at least one typist ? (a) 209 (b) 210 (c) 246 (d) 242 21. What is the number of terms in the expansion of [(2x - 3y}2 (2x + 3y)2]2 ? (a) 4 (b) 5 (c) 8 (d) 16

229

solved PAPER - 2019 (I)

22. In the expansion of (1 + ax)n, the first three terms are respectively 1, 12x and 64x2. What is n equal to ? (a) 6 (b) 9 (c) 10 (d) 12

31. What is the value of sin 34° cos 236° − sin 56° sin 124°  ? cos 28° cos 88° + cos 178° sin 208° (a) - 2 (b) - 1 (c) 2 (d) 1

23. The numbers 1, 5 and 25 can be three terms (not necessarily consecutive) of (a) only one AP (b) more than one but finite numhers of APs (c) infinite number of APs (d) finite number of GPs

32. tan 54° can be expressed as

24. The sum of (p + q)th and (p - q)th terms of an AP is equal to (a) (2p)th term (b) (2q)th term (c) Twice the pth term (d) Twice the qth term 25. If A is a square matrix of order n > 1, then which one of the following is correct ? (a) det (- A ) = det A (b) det (- A ) = (- 1)n det A (c) det (- A ) = - det A (d) det (- A ) = n det A 26. What is the least value of 25 cosec2 x + 36 sec2 x ? (a) 1 (b) 11 (c) 120 (d) 121

Consider the following for the next 02 (two) items : Let A and B be (3 × 3) matrices with det A = 4 and det B = 3. 27. What is det (2AB) equal to ? (a) 96 (b) 72 (c) 48 (d) 36



A complex number is given by z =

1 + 2i 1 − (1 − i )

29. What is the modulus of z ? (a) 4 (b) 2 1 2 30. What is the principal argument of z ? π 4

(b)

π (c) 2

(d) p

sin 36° cos 9° + sin 9° (c) (d) cos 36° cos 9° − sin 9° Consider the following for the next 03 (three) items : If p = X cos q - Y sin q , q = X sin q + Y cos q and p2 + 4pq + q2 = AX2 + BY2, 0 ≤ θ ≤

π . 2

33. What is the value of q ? π π (b) (a) 2 3 π π (c) (d) 4 6 34. What is the value of A ? (a) 4 (b) 3 (c) 2 (d) 1 35. What is the value of B ? (a) - 1 (b) 0 (c) 1 (d) 2

Consider the following for the next 02 (two) items : It is given that cos (q - a) = a, cos (q - b) = b.

a 1 - b 2 - b 1 - a 2 (d) a 1 − b 2 + b 1 − a 2 (c)

Consider the following for the next 02 (two) items :

(a) 0

sin 9° − cos 9° sin 9° + cos 9°

ab + 1 − a 2 1 − b 2 (b) ab - 1 - a 2 1 - b 2 (a)



(d)

(b)

36. What is cos (a - b) equal to ?

28. What is det (3AB-1) equal to ? (a) 12 (b) 18 (c) 36 (d) 48

(c) 1

sin 9° + cos 9° (a) sin 9° − cos 9°

2

.

37. What is sin2 (a - b) + 2ab cos (a - b) equal to ? (a) a2 + b2 (b) a2 - b2 (c) b2 - a2 (d) - (a2 + b2) 38. If sin a + cos a = p, then what is cos2 (2a) equal to ? (a) p2 (b) p2 - 1 (c) p2(2 - p2) (d) p2 + l

π (a) 4

4 5 π + sec −1 −  ? 5 4 2 π (b) 2

(c) p

(d) 0

39. What is the value of sin −1

230 Oswaal NDA/NA Year-wise Solved Papers −1 40. If sin

2p 1 + p2

− cos−1

1 − q2 1 + q2

= tan −1

2x 1 + x2

, then

what is x equal to ? p+q (a) 1 + pq

(b)

p−q 1 + pq

pq (c) 1 + pq

(d)

p+q 1 − pq

1 1 and tan ϕ = , then what is the 3 2 value of (q + ϕ) ? π (a) 0 (b) 6 π π (c) (d) 4 2

41. If tan θ =

3 42. If cos A = , then what is the value of 4  3A  A sin   sin    ? 2  2  5 (a) 8

(b)

5 16

5 5 (c) (d) 24 32 43. What is the value of tan 75° + cot 75° ? (a) 2 (b) 4 2 3 (d) 4 3 (c) 44. What is the value of cos 46° cos 47° cos 48° cos 49° cos 50° ... cos 135° ? (a) - 1 (b) 0 (c) 1 (d) Greater than 1 45. If sin 2q = cos 3q, where 0 < θ
(b) < 1, 0, 0 > (c) < 0, 1, 0 > (d) < 0, 0, 1 >   66. If a = i − 2 j + 5k and b = 2i + j − 3k then what is     b − a ⋅ 3a + b equal to ?

(

)(

(a) 106 (c) 53

)

(b) - 106 (d) - 53

67. If the position vectors of points A and B are 3i − 2 j + k and 2i + 4 j − 3k respectively, then  what is the length of AB  ? 14 (a)

(b) 29

43 (c)

(d) 53

68. If in a right-angled triangle ABC, hypotenuse AC = p, then what is       AB ⋅ AC + BC ⋅ BA + CA ⋅ CB equal to ? (a) p2

(b) 2p2

p2 (c) 2

(d) p

69. The sine of the angle between vectors   a = 2i − 6 j − 3k and b = 4i + 3 j − k is 1 (a) 26

(b)

5 26

5 1 (c) (d) 26 26 70. What is the value of l for which the vectors 3i + 4 j − k and - 2i + λ j + 10 k are perpendicular ? (a) 1 (c) 3

(b) 2 (d) 4

71. What is the derivative of sec2 (tan-1 x) with respect to x ? (a) 2x (b) x2 + 1 (c) x + 1 (d) x2 72. If f(x) = 1og10 (1 + x), then what is 4f(4) + 5f(1) log10 2 equal to ? (a) 0 (b) 1 (c) 2 (d) 4

232 Oswaal NDA/NA Year-wise Solved Papers 73. A function f defined by f ( x ) = ln is (a) an even function (b) an odd function (c) Both even and odd function (d) Neither even nor odd function

(

x2 + 1 − x

)

74. The domain of the function f defined by f(x) = logx 10 is (a) x > 10 (b) x > 0 excluding x = 10 (c) x ≥ 10 (d) x > 0 exduding x = 1 75. lim

1 − cos3 4 x

x →0

(a) 0 (c) 24

x2

is equal to

d2 y (c) 2 − 4 y = 0 dx

(d)

79. If f(x) = sin (cos x), then f ′(x) is equal to (a) cos (cos x) (b) sin (- sin x) (c) (sin x) cos (cos x) (d) (- sin x) cos (cos x) 80. The domain of the function

(2 − x ) (x − 3) (b) [0, ∞) (d) (2, 3)

81. The solution of the differential equation dy = cos (y - x) + 1 is dx (a) ex[sec (y - x) - tan (y - x)] = c (b) ex[sec (y - x) + tan (y - x)] = c (c) exsec (y - x) tan (y - x) = c (d) ex = csec (y - x) tan (y - x)

∫

sin x − cos x dx is equal to

0

(a) 0 (c) 2 2

( (d) 2 ( (b) 2

dx 2 d2 y dx 2

+ 2y = 0 + 4y = 0

84. A given quantity of metal is to be cast into a half cylinder (i.e., with a rectangular base and semicircular ends). If the total surface area is to be minimum, then the ratio of the height of the half cylinder to the diameter of the semicircular ends is (a) p : (p + 2) (b) (p + 2) : p (c) 1 : 1 (d) None of the above

∫e

sin x

cos x dx is equal to

(a) e + l (c) e + 2

(b) e - 1 (d) e

86. If f ( x ) = to ?

x−2 , x ≠ - 2, then what is f -1(x) equal x+2

4 (x + 2) (a) x−2

x+2 (b) 4 ( x − 2 )

x+2 (c) x−2

(d)

87. What is

) 2 + 1) 2 −1

2 (1 + x ) 1−x

∫ ln ( x ) dx equal to ? 2

(a) 2x ln(x) - 2x + c

(b)

2 +c x 2 ln ( x )

− 2x + c x 88. The minimum distance from the point (4, 2) to y2 = 8x is equal to (c) 2x ln(x) + c

(d)

(a) 2

(b) 2 2

(c) 2

(d) 3 2

89. The differential equation of the system of circles touching the y-axis at the origin is x 2 + y 2 − 2 xy (a)

dy =0 dx

x 2 + y 2 + 2 xy (b)

dy =0 dx

x 2 − y 2 + 2 xy (c)

dy =0 dx

x 2 − y 2 − 2 xy (d)

dy =0 dx

π/2

82.

d2 y

0

78. The number of real roots for the equation x2 + 9 |x| + 20 = 0 is (a) Zero (b) One (c) Two (d) Three

(a) (0, ∞) (c) [2, 3]∞

(b)

85.

(b) 12 (d) 36

77. If f(x) = 31+x, then f(x) f(y) f(z) is equal to (a) f(x + y + z) (b) f(x + y + z + 1) (c) f(x + y + z + 2) (d) f(x + y + z + 3)

f (x) =

d2 y (a) 2 + y = 0 dx

π/2

76. For r > 0, f(r) is the ratio of perimeter to area of a circle of radius r. Then f(1) + f(2) is equal to (a) 1 (b) 2 (c) 3 (d) 4



83. If y = acos 2x + bsin 2x, then

233

solved PAPER - 2019 (I)

90. Consider the following in respect of the differential equation :

2

d2 y

 dy  + 2   + 9y = x 2 dx  dx 

(1) The degree of the difterential equation is 1. (2) The order of the differential equation is 2. Which of the above statements is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 91. What is the general solution of the differential dy x + = 0  ? equation dx y (a) x2 + y2 = c (c) x2 + y2 = cxy

(b) x2 - y2 = c (d) x + y = c

92. The value of k which makes sin x x ≠ 0 f (x) =  continuous at x = 0, is x=0  k (a) 2 (b) 1 (c) - 1 (d) 0



93. What is the minimum value of a2x + b2y where xy = c2 ? (a) abc (b) 2abc (c) 3abc (d) 4abc 94. What is

∫e

x ln ( a )

ax (a) + c ln ( a )

dx equal to ? (b)

ex +c ln ( a )

ex ae x (d) (c) +c +c ln ( ae ) ln ( a ) 95. What is the area of one of the loops between the curve y = c sin x and x-axis ? (a) c (b) 2c (c) 3c (d) 4c 96. If sin q + cos q = 2 cos q, then what is (cos q - sin q) equal to ? − 2 cos θ (a)

(b) − 2 sin θ

2 sin q (c)

(d) 2 sin q

97. In a circle of diameter 44 cm, the length of a chord is 22 cm. What is the length of minor arc of the chord ? 484 (a) cm 21

(b)

242 cm 21

121 (c) cm 21

44 (d) cm 7

1 1 and tan θ = , then in which 2 3 quadrant does q lie ? (a) First (b) Second (c) Third (d) Fourth

98. If sin θ = −

99. How many three-digit even numbers can be formed using the digits 1, 2, 3, 4 and 5 when repetition of digits is if allowed ? (a) 36 (b) 30 (c) 24 (d) 12 100. The angle of elevation of a tower of height h from a point A due South of it is x and from a point B due East of A is y. If AB = z, then which one of the following is correct ? (a) h2(cot2 y - cot2 x) = z2 (b) z2(cot2 y - cot2 x) = h2 (c) h2(tan2 y - tan2 x) = z2 (d) z2(tan2 y - tan2 x) = h2 101. From a deck of cards, cards are taken out with replacement. What is the probability that the fourteenth card taken out is an ace ? 1 (a) 51

(b)

4 51

1 1 (c) (d) 52 13 102. If A and B are two events such that P(A) = 0·5, P(B) = 0·6 and P(A ∩ B) = 0·4, then

(

)

what is P A ∪ B equal to ? (a) 0·9 (c) 0·5

(b) 0·7 (d) 0·3

103. A problem is given to three students A, B and C whose probabilities of solving the problem 1 3 1 , respectively. What is the and 2 4 4 probability that the problem will be solved if they all solve the problem independently ?

are

29 (a) 32

(b)

27 32

25 23 (c) (d) 32 32 104. A pair of fair dice is rolled. What is the probability that the second dice lands on a higher value than does the first ? 1 (a) 4

(b)

1 6

5 (c) 12

(d)

5 18

234 Oswaal NDA/NA Year-wise Solved Papers 105. A fair coin is tossed and an unbiased dice is rolled together. What is the probability of getting a 2 or 4 or 6 along with head ? 1 (a) 2

(b)

1 3

1 1 (c) (d) 4 6 106. If A, B, C are three events, then what is the probability that at least two of these occur together ? (a) P(A ∩ B) + P(B ∩ C) + P(C ∩ A) (b) P(A ∩ B) + P(B ∩ C) + P(C ∩ A)  - P(A ∩ B ∩ C) (c) P(A ∩ B) + P(B ∩ C) + P(C ∩ A)  - 2P(A ∩ B ∩ C) (d) P(A ∩ B) + P(B ∩ C) + P(C ∩ A)  - 3P(A ∩ B ∩ C) 107. If two variables X and Y are independent, then what is the correlation coefficient between them ? (a) 1 (b) -1 (c) 0 (d) None of the above 108. Two independent events A and B are such that 2 1 P(A ∪ B) = and P(A ∩ B) = . If P(B) < P(A), 3 6 then what is P(B) equal to ? 1 (a) 4

1 (b) 3

1 1 (c) (d) 2 6 109. The mean of 100 observations is 50 and the standard deviation is 10. If 5 is subtracted from each observation and then it is divided by 4, then what will be the new mean and the new standard deviation respectively ? (a) 45, 5 (b) 11·25, 1·25 (c) 11·25, 2·5 (d) 12·5, 2·5 110. If two fair dice are rolled then what is the conditional probability that the first dice lands on 6 given that the sum of numbers on the dice is 8 ? 1 (a) 3

(b)

1 4

1 (c) 5

1 (d) 6

111. Two symmetric dice flipped with each dice having two sides painted red, two painted black, one painted yellow and the other painted white. What is the probability that both land on the same colour ? 3 (a) 18

(b)

2 9

5 1 (c) (d) 18 3 112. There are n socks in a drawer, of which 3 socks are red. If 2 of the socks are chosen randomly and the probability that both selected socks are 1 red is , then what is the value of n ? 2 (a) 3 (b) 4 (c) 5 (d) 6 113. Two cards are chosen at random from a deck of 52 playing cards. What is the probability that both of them have the same value ? 1 (a) 17

(b)

3 17

5 (c) 17

(d)

7 17

114. In eight throws of a die, 5 or 6 is considered a success. The mean and standard deviation of total number of successes is respectively given by 8 16 , (a) 3 9 4 4 , (c) 3 3

8 (b) , 3 4 (d) , 3

4 3 16 9

115. A and B are two events such that A and B are mutually exclusive. If P(A) = 0·5 and P(B) = 0·6, then what is the value of P(A|B) ? 1 1 (a) (b) 5 6 2 1 (c) (d) 5 3 116. Consider the following statements : (1) The algebraic sum of deviations of a set of values from their arithmetic mean is always zero. (2) Arithmetic mean > Median > Mode for a symmetric distribution. Which of the above statements is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2

235

solved PAPER - 2019 (I)

117. Let the correlation coefficient between X and Y be 0·6. Random variables Z and W are defined Y . What is the correlation 3 coefficient between Z and W ? (a) 0·1 (b) 0·2 (c) 0·36 (d) 0·6 as Z = X + 5 and W =

118. If all the natural numbers between 1 and 20 are multiplied by 3, then what is the variance of the resulting series ? (a) 99·75 (b) 199·75 (c) 299·25 (d) 399·25

For elaborated Solutions

1 (a) 4 1 (b) 2 3 (c) 4 (d) It cannot be determined 120. If A and B are two events, then what is the probability of occurrence of either event A or event B ? (a) P(A) + P(B) (b) P(A ∪ B) (c) P(A ∩ B) (d) P(A) P(B)

 

Finished Solving the Paper ? Time to evaluate yourself !

SCAN THE CODE

119. What is the probability that an interior point in a circle is closer to the centre than to the circumference ?

SCAN

236 Oswaal NDA/NA Year-wise Solved Papers

Answers Q. No.

Answer Key

Topic Name

Chapter Name

1

(d)

Sequence & Series

Algebra

2

(a)

Set Theory and Relations

Algebra

3

(d)

Set Theory and Relations

Algebra

4

(b)

Set Theory and Relations

Algebra

5

(a)

Set Theory and Relations

Algebra

6

(d)

Set Theory and Relations

Algebra

7

(a)

Matrices and Determinants

Algebra

8

(d)

Theory of Equation

Algebra

9

(b)

Matrices and Determinants

Algebra

10

(a)

Matrices and Determinants

Algebra

11

(d)

Complex Number

Algebra

12

(b)

Binomial Theorem

Algebra

13

(b)

Permutation and Combination

Algebra

14

(b)

Theory of Equation

Algebra

15

(b)

Set Theory and Relations

Algebra

16

(b)

Set Theory and Relations

Algebra

17

(b)

Set Theory and Relations

Algebra

18

(b)

Matrices and Determinants

Algebra

19

(d)

Matrices and Determinants

Algebra

20

(c)

Permutations and Combinations

Algebra

21

(b)

Binomial Theorem

Algebra

22

(b)

Binomial Theorem

Algebra

23

(c)

Sequence and Series

Algebra

24

(c)

Sequence and Series

Algebra

25

(b)

Matrices and Determinants

Algebra

26

(d)

Sequence and Series

Algebra

27

(a)

Matrices and Determinants

Algebra

28

(c)

Matrices and Determinants

Algebra

29

(c)

Complex Numbers

Algebra

30

(a)

Complex Numbers

Algebra

31

(a)

Trigonometric Ratios and Identities

Trigonometry

237

solved PAPER - 2019 (I)

Q. No.

Answer Key

Topic Name

Chapter Name

32

(c)

Trigonometric Ratios and Identities

Trigonometry

33

(c)

Trigonometric Ratios and Identities

Trigonometry

34

(b)

Trigonometric Ratios and Identities

Trigonometry

35

(a)

Trigonometric Ratios and Identities

Trigonometry

36

(a)

Trigonometric Ratios and Identities

Trigonometry

37

(a)

Trigonometric Ratios and Identities

Trigonometry

38

(c)

Trigonometric Ratios and Identities

Trigonometry

39

(d)

Inverse Trigonometric Functions

Trigonometry

40

(b)

Inverse Trigonometric Functions

Trigonometry

41

(c)

Trigonometric Ratios and Identities

Trigonometry

42

(b)

Trigonometric Ratios and Identities

Trigonometry

43

(b)

Trigonometric Ratios and Identities

Trigonometry

44

(b)

Trigonometric Ratios and Identities

Trigonometry

45

(b)

Trigonometric Ratios and Identities

Trigonometry

46

(a)

Trigonometric Ratios and Identities

Trigonometry

47

(b)

Sequence and Series

Algebra

48

(a)

Trigonometric Ratios and Identities

Trigonometry

49

(b)

Trigonometric Ratios and Identities

Trigonometry

50

(c)

Properties of Triangle

Trigonometry

51

(d)

Straight Line

Coordinate Geometry

52

(b)

Circle

Coordinate Geometry

53

(a)

Ellipse

Coordinate Geometry

54

(d)

Hyperbola

Coordinate Geometry

55

(a)

Parabola

Coordinate Geometry

56

(d)

Straight Line

Coordinate Geometry

57

(c)

Straight Line

Coordinate Geometry

58

(b)

Straight Line

Coordinate Geometry

59

(a)

Straight Line

Coordinate Geometry

60

(a)

Circle

Coordinate Geometry

61

(b)

Three Dimensional Geometry

Vectors and 3D Geometry

62

(c)

Three Dimensional Geometry

Vectors and 3D Geometry

63

(a)

Three Dimensional Geometry

Vectors and 3D Geometry

64

(a)

Three Dimensional Geometry

Vectors and 3D Geometry

238 Oswaal NDA/NA Year-wise Solved Papers Q. No.

Answer Key

Topic Name

Chapter Name

65

(d)

Three Dimensional Geometry

Vectors and 3D Geometry

66

(b)

Vector Algebra

Vectors and 3D Geometry

67

(d)

Vector Algebra

Vectors and 3D Geometry

68

(a)

Vector Algebra

Vectors and 3D Geometry

69

(b)

Vector Algebra

Vectors and 3D Geometry

70

(d)

Vector Algebra

Vectors and 3D Geometry

71

(a)

Differential Coefficient

Calculus

72

(d)

Logarithm and its Applications

Algebra

73

(b)

Functions

Calculus

74

(d)

Logarithm and its Applications

Algebra

75

(c)

Limits

Calculus

76

(c)

Functions

Calculus

77

(c)

Functions

Calculus

78

(a)

Theory of Equation

Algebra

79

(d)

Differential Coefficient

Calculus

80

(c)

Functions

Calculus

81

(a)

Differential Equation

Calculus

82

(b)

Definite Integration

Calculus

83

(d)

Differential Equation

Calculus

84

(a)

Application of Derivatives

Calculus

85

(b)

Definite Integration

Calculus

86

(d)

Functions

Calculus

87

(a)

Indefinite Integration

Calculus

88

(b)

Application of Derivatives

Calculus

89

(c)

Differential Equation

Calculus

90

(c)

Differential Equation

Calculus

91

(a)

Differential Equation

Calculus

92

(d)

Continuity and Differentiability

Calculus

93

(b)

Application of Derivatives

Calculus

94

(a)

Indefinite Integration

Calculus

95

(b)

Area under Curves

Calculus

96

(c)

Trigonometric Ratios and Identities

Trigonometry

97

(a)

Circle

Coordinate Geometry

239

solved PAPER - 2019 (I)

Q. No.

Answer Key

Topic Name

Chapter Name

98

(c)

Trigonometric Ratios and Identities

Trigonometry

99

(c)

Permutation and Combination

Algebra

100

(a)

Height and Distance

Trigonometry

101

(d)

Probability

Statistics and Probability

102

(d)

Probability

Statistics and Probability

103

(a)

Probability

Statistics and Probability

104

(c)

Probability

Statistics and Probability

105

(c)

Probability

Statistics and Probability

106

(c)

Probability

Statistics and Probability

107

(c)

Statistics

Statistics and Probability

108

(b)

Probability

Statistics and Probability

109

(c)

Statistics

Statistics and Probability

110

(c)

Probability

Statistics and Probability

111

(c)

Probability

Statistics and Probability

112

(b)

Probability

Statistics and Probability

113

(a)

Probability

Statistics and Probability

114

(b)

Probability

Statistics and Probability

115

(b)

Probability

Statistics and Probability

116

(a)

Statistics

Statistics and Probability

117

(d)

Statistics

Statistics and Probability

118

(c)

Statistics

Statistics and Probability

119

(a)

Probability

Statistics and Probability

120

(b)

Probability

Statistics and Probability

NDA / NA

MATHEMATICS

National Defence Academy / Naval Academy

Ii

question Paper

2019

Time  : 2 :30 Hour

Total Marks  : 300

Important Instructions  : 1. This test Booklet contains 120 items (questions). Each item is printed in English. Each item comprises four responses (answer's). You will select the response which you want to mark on the Answer Sheet. In case you feel that there is more than one correct response, mark the response which you consider the best. In any case, choose ONLY ONE response for each item. 2. You have to mark all your responses ONLY on the separate Answer Sheet provided. 3. All items carry equal marks. 4. Before you proceed to mark in the Answer Sheet the response to various items in the Test Booklet, you have to fill in some particulars in the Answer Sheet as per instructions. 5. Penalty for wrong answers  : THERE WILL BE PENALTY FOR WRONG ANSWERS MARKED BY A CANDIDATE IN THE OBJECTIVE TYPE QUESTION PAPERS. (i) There are four alternatives for the answer to every question. For each question for which a wrong answer has been given by the candidate, one·third of the marks assigned to that question will be deducted as penalty. (ii) If a candidate gives more than one answer, it will be treated as a wrong answer even if one of the given answers happens to be correct and there will be same penalty as above to that question. (iii) If a question is left blank, i.e., no answer is given by the candidate, there will be no penalty for that question.

1. If both p and q belong to the set (1, 2, 3, 4), then how many equations of the form px2 + qx + 1 = 0 will have real roots ? (a) 12 (b) 10 (c) 7 (d) 6 2. What is the value of 1 - 2 + 3 - 4 + 5 - ...... + 101  ? (a) 51 (b) 55 (c) 110 (d) 111 3. If A, B, and C are subsets of a given set, then which one of the following relations is not correct ? (a) A ∪ (A ∩ B) = A ∪ B (b) A ∩ (A ∪ B) = A (c) (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C) (d) (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C) 4. If the sum of first n terms of a series is (n + 12), then what is its third term ? (a) 1 (b) 2 (c) 3 (d) 4 5. What is the value of k for which the sum of the squares of the roots of 2x2 - 2(k - 2) x - (k + 1) = 0 is minimum ? (a) -1 (b) 1

3 (c) (d) 2 2 6. If the roots of the equation a(b - c)x2 + b (c - a)x + c (a - b) = 0 are equal, then which one of the following is correct ? (a) a, b and c are in AP (b) a, b and c are in GP (c) a, b and c are in HP (d) a, b and c do not follow any regular pattern 7. |x2 - 3x + 2| > x2 - 3x + 2, then which one of the following is correct ? (a) x ≤ 1 or x ≥ 2 (b) 1≤x≤2 (c) 1 b} (b) A = {x|x < a} and B = {x|x > b} (c) A = {x|x < a} and B = {x|x < b} (d) A = {x|x > a} and B = {x|x < b} 29. If the constant term in the expansion of 10

k    x − 2  is 405, then what can be the values x   of k ? (a) ± 2 (b) ± 3 (c) ± 5 (d) ± 9

30. What is C(47, 4) + C(51, 3) + C(50, 3) + C(49, 3) + C(48, 3) + C(47, 3) equal to ? (a) C(47, 4) (b) C(52, 5) (c) C(52, 4) (d) C(47, 5) 31. Let a, b, c be in AP and k ≠ 0 be a real number, Which of the following are correct ? (1) ka, kb, kc are in AP (2) k - a, k - b, k - c are in AP a b c (3) , , are in AP k k k Select the correct answer using the code given below : (a) 1 and 2 only (b) 2 and 3 only (c) 1 and 3 only (d) 1, 2 and 3 32. How many two-digit numbers are divisible by 4 ? (a) 21 (b) 22 (c) 24 (d) 25

33. Let Sn be the sum of the first n terms of an AP. If S2n = 3n + 14n2, then what is the common difference ? (a) 5 (b) 6 (c) 7 (d) 9 34. If 3rd, 8th and 13th terms of a GP are p, q and r respectively, then which one of the following is correct ? (a) q2 = pr (b) r2 = pq (c) pqr = 1 (d) 2q = p + r 35. What is the solution of x ≤ 4, y ≥ 0 and x ≤ - 4, y ≤ 0 ? (a) x ≥ - 4, y ≤ 0 (b) x ≤ 4, y ≥ 0 (c) x ≤ - 4, y = 0 (d) x ≥ - 4, y = 0 36. If xlog7 x > 7 where x > 0, then which one of the following is correct ? (a) x ∈ (0, ∞)

1  (b) x ∈  , 7  7 

 1 1  x ∈  0 ,  ∪ ( 7 , ∞ ) (d) x ∈  , ∞  (c)  7 7  37. How many real roots does the equation x2 + 3|x| + 2 = 0 have ? (a) Zero (b) One (c) Two (d) Four 38. Consider the following statements in respect of the quadratic equation 4(x - p)(x - q) - r2 = 0, where p, q and r are real numbers : (1) The roots are real (2) The roots are equal if p = q and r = 0 Which of the above statements is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 39. Let S = {2, 4, 6, 8, .........20}, What is the maximum number of subsets does S have ? (a) 10 (b) 20 (c) 512 (d) 1024 40. A binary number is represented by (cdccddcccddd)2, where c > d, what is its decimal equivalent ? (a) 1848 (b) 2048 (c) 2842 (d) 2872

29 where 0 < θ < 90°, then what is 21 the value of 4sec θ + 4tan θ ? (a) 5 (b) 10 (c) 15 (d) 20 41. If cosec θ =

243

solved PAPER - 2019 (II)

42. Consider the following statements : (1) cos θ + sec θ can never be equal to 1.5. (2) tan θ + cot θ can never be less than 2. Which of the above statements is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 43. A ladder 9 m long reaches a point 9 m below the top of a vertical flagstaff. From the foot of the ladder, the elevation of the flagstaff is 60°. What is the height of the flagstaff ? (a) 9 m (b) 10.5 m (c) 13.5 m (d) 15 m 44. What is the length of the chord of a unit circle which subtends an angle θ at the centre ? θ sin   (a) 2

θ (b) cos   2

θ 2 sin   (c) 2

θ (d) 2 cos   2

  1  45. What is tan 2 tan −1    equal to ?  3   2 (a) 3

(b)

3 4

3 1 (c) (d) 8 9 46. What is the scalar projection of         a = i − 2 j + k on b = 4i − 4 j + 7 k 6 (a) 9

(b)

19 9

9 6 (c) (d) 19 19 47. If the magnitude of the sum of two non-zero vectors is equal to the magnitude of their difference, then which one of the following is correct ? (a) The vectors are parallel (b) The vectors are perpendicular (c) The vectors are anti-parallel (d) The vectors must be unit vectors 48. Consider the following equations for two   vectors a and b :     2 2 a+b . a−b = a − b (1)

(

)(

  a+b (2)

(

)

 

2

)( a − b ) = a

2 −b

2  2 2 2 a.b + a × b = a b (3) Which of the above statements are correct ? (a) 1, 2 and 3 (b) 1 and 2 only (c) 1 and 3 only (d) 2 and 3 only 49. Consider the following statements:   (1) The magnitude of a × b is same as the area   of a triangle with sides a and b          (2) If a × b = 0 where a ≠ 0 , b ≠ 0, then a = λb Which of the above statements is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2   50. If a and b are unit vectors and θ is the angle θ between them, then what is sin 2   equal to ? 2 2

a+b (a) 4

(b)

a-b

2

4

2

2

a+b a-b (c) (d) 2 2 51. The equation ax + by + c = 0 represents a straight line (a) for all real numbers a, b and c (b) only when a ≠ 0 (c) only when b ≠ 0 (d) only when at least one of a and b is non-zero 52. What is the angle between the lines xcos α + ysin α = a and xsin β - ycos β = a ? (a) β - α (b) p + β - α

( π − 2β + 2α ) ( π + 2β − 2α ) (d) (c) 2 2 53. What is the distance between the points P(m cos 2α, msin 2α) and Q(mcos 2β, msin 2β) ? (a) |2msin (α - β)| (b) |2mcos (α - β)| (c) |msin (2α - 2β)| (d) |msin (2α - 2β)| 54. An equilateral triangle has one vertex at

(

)

(- 1, - 1) and another vertex at − 3 , 3 . The third vertex may lie on

(

− 2, (a) (c) (1, 1)

)

2

(b)

(

2, − 2

(d) (1, - 1)

)

244 Oswaal NDA/NA Year-wise Solved Papers 55. If the angle between the lines joining the end points of minor axis of the ellipse with one of its foci is

x2 a

2

+

y

2

b

2

=1

π , then what is the 2

eccentricity of the ellipse ? 1 (a) 2

(b)

3 (c) 2

(d)

61. What is the minimum value of ?

a2 2

b2

+

cos x sin 2 x (a) (a + b)2 (c) a2 + b2

where a > 0 and b > 0 ? (b) (a - b)2 (d) |a2 + b2|

62. If the angles of a triangle ABC are in AP and 1 2 1 2 2

56. A point on a line has coordinates (p + 1, p - 3, 2p ) where p is any real number. What are the direction cosines of the line ? 1 1 1 (a) , , 2 2 2 1 1 1 (b) , , 2 2 2 1 1 1 (c) , , 2 2 2 (d) Cannot be determined due to insufficient data x −1 y −3 z + 2 = = has 57. A point on the line 1 2 7 coordinates (a) (3, 5, 4) (b) (2, 5, 5) (c) (-1, -1, 5) (d) (2, -1, 0) x−4 y−2 z−k = = lies on the plane 1 1 2 2x - 4y + z = 7, then what is the value of k ? (a) 2 (b) 3 (c) 5 (d) 7

58. If the line

59. A straight line passes through the point (1, 1, 1) makes an angle 60° with the positive direction of z-axis, and the cosine of the angles made by it with positive directions of the y-axis and the x-axis are in the ratio 3 : 1 . What is the acute angle between the two possible positions of the line ? (a) 90° (b) 60° (c) 45° (d) 30° 60. If the points (x, y, - 3), (2, 0, - 1) and (4, 2, 3) lie on a straight line, then what are the values of x and y respectively ? (a) 1, - 1 (b) - 1, 1 (c) 0, 2 (d) 3, 4

b : c =

3 : 2 , then what is the measure of

angle A ? (a) 30° (c) 60°

(b) 45° (d) 75°

63. If tan A - tan B = x and cot B - cot A = y, then what is the value of cot (A - B) ? 1 1 y x

1 1 (a) + x y

(b)

xy (c) x+y

(d) 1 +

1 xy

64. What is sin (α + β) - 2sin α cos β + sin (α - β) equal to ? (a) 0 (b) 2sin α (c) 2sin β (d) sin α + sin β 65. If 2tan A = 3tan B = 1, then what is tan (A - B) equal to ? 1 (a) 5

(b)

1 (c) 7

(d)

1 6

1 9 66. What is cos 80° + cos 40° - cos 20° equal to ? (a) 2 (b) 1 (c) 0 (d) - 19 67. If angle C of a triangle ABC is a right angle, then what is tan A + tan B equal to ? a2 - b 2 (a) ab

(b)

a2 bc

b2 (c) ca

(d)

c2 ab

A A 68. What is cot   − tan   equal to  ? 2 2 (a) tan A (b) cot A (c) 2tan A (d) 2cot A 69. What is cot A + cosec A equal to ? A (a) tan   2

A (b) cot   2

A (c) 2 tan   2

A (d) 2 cot   2

245

solved PAPER - 2019 (II)

70. What is tan 25° tan 15° + tan 15° tan 50° + tan 25° tan 50° equal to ? (a) 0 (b) 1 (c) 2 (d) 4 71. What is the area of the region bounded by |x| < 5, y = 0 and y = 8? (a) 40 square units (b) 80 square units (c) 120 square units (d) 160 square units 72. Consider the following statements in respect of 1 the function f(x) = sin   for x ≠ 0 and f(0) = 0 : x lim f(x) exists (1) x→0

(2) f(x) is continuous at x = 0 Which of the above statement is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 73. What is the value of lim

x →0

1 (a) 4

sin x°  ? tan 3x°

1 (b) 3

1 (c) (d) 1 2 74. What is the degree of the differential equation 2  4  d 3 y  dy  2 d y + − x  4  = 0 ?   dx 3  dx   dx 

(a) 1 (c) 3

(b) 2 (d) 4

75. Which one of the following is the second degree polynomial function f(x) where f(0) = 5, f(- 1) = 10 and f(1) = 6 ? (a) 5x2 - 2x + 5 (c) 3x2 - 2x + 5

(b) 3x2 - 2x - 5 (d) 3x2 - 10x + 5



Directions for the following three (03) items : Read the following information and answer the three items that follow : A curve y = memx where m > 0 intersects y-axis at a point P. 76. What is the slope of the curve at the point of intersection P ? (a) m (b) m2 (c) 2m (d) 2m2 77. How much angle does the tangent at P make with y-axis ? (a) tan-1 m2

(b) cot-1 (1 + m2)

  1 −1 1 + m4 sin −1  (c)  (d) sec  4   1+m  78. What is the equation of tangent to the curve at P ? (a) y = mx + m (b) y = - mx + 2m (c) y = m2x + 2m (d) y = m2x + m

Directions for the following two (02) items : Read the following information and answer the two items that follow : Let f(x) = x2, g(x) = tan x and h(x) = In x.

79. For x = (a) 0

π , what is the value of [ho(gof)] (x)  ? 2 (b) 1

π π (d) (c) 4 2 80. What is [fo(fof)] (2) equal to ? (a) 2 (b) 8 (c) 16 (d) 256 81. What is

∫ 2x

dx 2

− 2x + 1

equal to ?

tan −1 ( 2 x − 1 ) +c (a) 2 (b) 2 tan-1 (2x - 1) + c tan −1 ( 2 x + 1 ) +c (c) 2 (d) tan-1 (2x - 1) + c 82. What is

dx

∫ x (1 + In x )

n

equal to (n ≠ 1)  ?

1 +c (a) ( n − 1) (1 + ln x )n−1 1−n +c (b) (1 + ln x )1−n n+1 +c (c) (1 + ln x )n+1 1

− (d)

( n − 1) (1 + ln x )n−1

+c

83. Which one of the following is the differential equation that represents the family of curves

y=

1 2

2x − c

where c is an arbitrary constant ?

dy (a) = 4 xy 2 dx

(b)

dy 1 = dx y

246 Oswaal NDA/NA Year-wise Solved Papers dy dy = −4 xy 2 (c) = x 2 y (d) dx dx Directions for the following two (02) items : Read the following information and answer the two items that follow : Consider the equation xy = ex-y dy at x = 1 equal to ? dx (a) 0 (b) 1 (c) 2 (d) 4

84. What is

85. What is

d2 y dx 2

at x = 1 equal to ?

(a) 0 (c) 2

(b) 1 (d) 4



Directions for the following three (03) items  : Read the following information and answer the three items that follow : Consider the function f(x) = g(x) + h(x)  4x  x where g(x) = sin   and h(x) = cos    5  4 86. What is the period of the function g(x) ? (a) π (b) 2π (c) 4π (d) 8π 87. What is the period of the function h(x) ? 5π (a) π (b) 2 3π 5π (c) (d) 2 2 88. What is the period of the function f(x) ? (a) 10π (b) 20π

(c) 40π



Directions for the following two (02) items :



Read the following information and answer the two items that follow :



Consider the function



f(x) = 3x4 - 20x3 - 12x2 + 288x + 1

(d) 80π



I1 =

∫

π

0

xdx and I 2 = 1 + sin x

94. What is the value of I1 ? (a) 0 (c) π

∫

π

0

( π − x ) dx

1 − sin ( π + x )

π 2 (d) 2π

(b)

95. What is the value of I1 + I2 ? (a) 2π (b) π π (c) (d) 0 2 96. The differential equation which represents the family of curves given by tan y = c(1 - ex) is (a) extan ydx + (1 - ex)dy = 0 (b) extan ydx + (1 - ex)sec2 ydy = 0 (c) ex(1 - ex)dx + tan ydy = 0 (d) extan ydy + (1 - ex)dx = 0 97. What is the derivative of 2(sin x) with respect to sin x ? 2 2 (a) sin x 2(sin x) ln 4 (b) 2sin x 2(sin x) ln 4 2 2 (c) ln (sin x) 2(sin x) (d) 2sin xcos x 2(sin x) 98. For what value of k is the function

(d) (- 4, - 3)

90. In which one of the following intervals is the function decreasing ? (a) (- 2, 3) (b) (3, 4) (c) (4, 6)

Directions for the following three (03) items : Read the following information and answer the three items that follow : Let f(x) = x2 + 2x - 5 and g(x) = 5x + 30 91. What are the roots of the equation g[f(x)] = 0 ? (a) 1, - 1 (b) - 1, - 1 (c) 1, 1 (d) 0, 1 92. Consider the following statements : (1) f[g(x)] is a polynomial of degree 3. (2) g[g(x)] is a polynomial of degree 2. Which of the above statements is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 93. If h(x) = 5f(x) - xg(x), then what is the derivative of h(x) ? (a) - 40 (b) - 20 (c) - 10 (d) 0 Directions for the following two (02) items  : Read the following information and answer the two items that follow : Consider the integrals

2

89. In which one of the following intervals is the function increasing ? (a) (- 2, 3) (b) (3, 4) (c) (- 3, -2)



(d) (6, 9)



 2 x + 1 , x < 0 4   continuous  ? f ( x ) = k , x = 0  2  x + 1  , x > 0   2

247

solved PAPER - 2019 (II)

1 (a) 4

(b)

1 2

(c) 1

(d) 2

106. For the variable x and y, the two regression lines are 6x + y = 30 and 3x + 2y = 25. What are the value of x , y and r respectively ?

99. What is the area of the region enclosed between the curve y2 = 2x and the straight line y = x ?

20 35 (a) , , - 0.5 3 9

2 (a) square units 3

(b)

1 (c) square units 3

(d) 1 square unit

35 20 35 20 (c) , (d) , , - 0.5 , 0.5 9 3 9 3 107. The class marks in a frequency table are given to be 5, 10, 15, 20, 25, 30, 35, 40, 45, 50. The class limits of the first five classes are (a) 3-7, 7-13, 13-17, 17-23, 23-27 (b) 2.5-7.5, 7.5-12.5, 12.5-17.5, 17.5-22.5, 22.5-27.5 (c) 1.5-8.5, 8.5-11.5, 11.5-18.5, 18.5-21.5, 21.5-28.5 (d) 2-8, 8-12, 12-18, 18-22, 22-28

4 square units 3

x 3 5x 2 − + 6 x + 7 increases in the 3 2 interval T and decreases in the interval S, then which one of the following is correct ? (a) T = (- ∞, 2) ∪ (3, ∞) and S = (2, 3) (b) T = φ and S = (- ∞ , ∞) (c) T = (- ∞, ∞) and S = φ (d) T = (2, 3) and S = (- ∞ , 2) ∪ (3, ∞)

100. If f(x) =

101. A coin is biased so that heads comes up thrice as likely as tails. For three independent tosses of a coin, what is the probability of getting at most two tails ? (a) 0.16 (b) 0.48 (c) 0.58 (d) 0.98 102. A bag contains 20 books out of which 5 are defective. If 3 of the books are selected at random and removed from the bag in succession without replacement, then what is the probability that all three books are defective ? (a) 0.009 (b) 0.016 (c) 0.026 (d) 0.047

(b)

20 35 , , 0.5 3 9

108. The mean of 5 observations is 4.4 and variance is 8.24. If three of the five observations are 1, 2 and 6, then what are the other two observations ? (a) 9, 16 (b) 9, 4 (c) 81, 16 (d) 81, 4 109. If a coin is tossed till the first head appears, then what will be the sample space ? (a) {H} (b) {TH} (c) {T, HT, HHT, HHHT, ................} (d) {H, TH, TTH, TTTH, ................} 110. Consider the following discrete frequency distribution :

x 1 2 3 4 5 6 7 8 f 3 15 45 57 50 36 25 9

103. The median of the observations 22, 24, 33, 37, x + 1, x + 3, 46, 47, 57, 58 in ascending order is 42. What are the values of 5th and 6th observations respectively ? (a) 42, 45 (b) 41, 43 (c) 43, 46 (d) 40, 40

What is the value of median of the distribution ? (a) 4 (b) 5 (c) 6 (d) 7

104. Arithmetic mean of 10 observations is 60 and sum of squares of deviations from 50 is 5000. What is the standard deviation of the observations ? (a) 20 (b) 21 (c) 22.36 (d) 24.70

5 (a) 12

105. If p and q are the roots of the equation x2 - 30x + 221 = 0, what is the value of p3 + q3 ? (a) 7010 (b) 7110 (c) 7210 (d) 7240

111. Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on them is a prime number ? (b)

1 2

7 2 (c) (d) 12 3 112. If 5 of a Company’s 10 delivery trucks do not meet emission standards and 3 of them are chosen for inspection, then what is the probability that none of the trucks chosen will meet emission standards ?

248 Oswaal NDA/NA Year-wise Solved Papers 1 (a) 8

(b)

1 (c) 12

(d)

3 8

1 4 113. There are 3 coins in a box. One is a two-headed coin; another is a fair coin; and third is biased coin that comes up heads 75% of time. When one of the three coins is selected at random and flipped, it shows heads. What is the probability that is was the two-headed coin ? 2 (a) 9

(b)

1 3

4 5 (c) (d) 9 9 114. Consider the following statements : (1) If A and B are mutually exclusive events, then it is possible that P(A) = P(B) = 0.6. (2) If A and B are any two events such that

(

)

P(A|B) = 1, then P B | A = 1. Which of the above statements is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 115. If a fair die is rolled 4 times, then what is the probability that there are exactly 2 sixes ? 5 (a) 216

(b)

25 216

125 (c) 216

(d)

175 216

For elaborated Solutions

117. If the range of a set of observations on a variable X is known to be 25 and if Y = 40 + 3X, then what is the range of the set of corresponding observations on Y ? (a) 25 (b) 40 (c) 75 (d) 115 118. If V is the variance and M is the mean of first 15 natural numbers, then what is V + M2 equal to ? 124 (a) 3

(b)

SCAN

148 3

248 124 (c) (d) 3 9 119. A car travels first 60 km at a speed of 3v km/hr and travels next 60 km at 2v km/hr. What is the average speed of the car ? (a) 2.5v km/hr (b) 2.4v km/hr (c) 2.2v km/hr (d) 2.1v km/hr 120. The mean weight of 150 students in a certain class is 60 kg. The mean weight of boys is 70 kg and that of girls is 55 kg. What are the number of boys and girls respectively in the class ? (a) 75 and 75 (b) 50 and 100 (c) 70 and 80 (d) 100 and 50

 

Finished Solving the Paper ? Time to evaluate yourself !

SCAN THE CODE

116. Mean of 100 observations is 50 and standard deviation is 10. If 5 is added to each observation, then what will be the new mean and new standard deviation respectively ? (a) 50, 10 (b) 50, 15 (c) 55, 10 (d) 55, 15

249

solved PAPER - 2019 (II)

Answers Q. No.

Answer Key

Topic Name

Chapter Name

1

(c)

Quadratic Equation

Algebra

2

(a)

Sequence and Progression

Algebra

3

(a)

Set and Relation

Algebra

4

(a)

Sequence and Progression

Algebra

5

(c)

Quadratic Equation

Algebra

6

(c)

Sequence and Progression

Algebra

7

(c)

Function

Algebra

8

(b)

Sequence and Progression

Algebra

9

(b)

Set and Relation

Calculus

10

(d)

Coordinate Geometry

Geometry

11

(c)

Determinants

Algebra

12

(d)

Determinants

Algebra

13

(a)

Determinants

Algebra

14

(a)

Matrices

Algebra

15

(c)

Sequence and Progression

Algebra

16

(c)

Quadratic Equation

Algebra

17

(c)

Complex Number

Algebra

18

(d)

Complex Number

Algebra

19

(c)

Probability

Algebra

20

(c)

Set and Relation

Algebra

21

(c)

Complex Number

Algebra

22

(b)

Quadratic Equation

Algebra

23

(c)

Permutation and Combination

Algebra

24

(c)

Binomial Expansion

Algebra

25

(a)

Binomial Expansion

Algebra

26

(c)

Matrices

Algebra

27

(d)

Permutation and Combination

Algebra

28

(b)

Sets and Relation

Algebra

29

(b)

Binomial Expansion

Algebra

30

(c)

Permutation and Combination

Algebra

31

(d)

Sequence and Progression

Algebra

32

(b)

Sequence and Progression

Algebra

33

(c)

Sequence and Progression

Algebra

34

(a)

Sequence and Progression

Algebra

35

(c)

Function

Calculus

36

(c)

Function

Calculus

37

(a)

Function

Calculus

38

(c)

Quadratic Equation

Algebra

39

(d)

Set and Relation

Calculus

250 Oswaal NDA/NA Year-wise Solved Papers Q. No.

Answer Key

40

(d)

Topic Name Set and Relation

Chapter Name Calculus

41

(b)

Trigonometric Ratios and Identities

Trigonometry

42

(c)

Trigonometric Ratios and Identities

Trigonometry

43

(c)

Height and Distance

Trigonometry

44

(c)

Trigonometric Ratios and Identities

Trigonometry

45

(b)

Trigonometric Ratios and Identities

Trigonometry

46

(b)

Vector Algebra

Algebra

47

(b)

Vector Algebra

Algebra

48

(a)

Vector Algebra

Algebra

49

(b)

Vector Algebra

Algebra

50

(b)

Vector Algebra

Algebra

51

(d)

Point and Straight line

Coordinate Geometry

52

(c)

Trigonometric Ratio & Identities

Trigonometry

53

(a)

Point and Straight line

Coordinate Geometry

54

(c)

Point and Straight line

Coordinate Geometry

55

(b)

Ellipse

Coordinate Geometry

56

(a)

Point and Straight line

Coordinate Geometry

57

(b)

3D

Coordinate Geometry

58

(d)

3D

Coordinate Geometry

59

(b)

3D

Coordinate Geometry

60

(a)

3D

Coordinate Geometry

61

(a)

Trigonometric Ratios and Identities

Trigonometry

62

(d)

Trigonometric Ratios and Identities

Trigonometry

63

(a)

Trigonometric Ratios and Identities

Trigonometry

64

(a)

Trigonometric Ratios and Identities

Trigonometry

65

(c)

Trigonometric Ratios and Identities

Trigonometry

66

(c)

Trigonometric Ratios and Identities

Trigonometry

67

(d)

Trigonometric Ratios and Identities

Trigonometry

68

(d)

Trigonometric Ratios and Identities

Trigonometry

69

(b)

Trigonometric Ratios and Identities

Trigonometry

70

(b)

Trigonometric Ratios and Identities

Trigonometry

71

(b)

Application of Integral

Calculus

72

(d)

Limit

Calculus

73

(b)

Limit

Calculus

74

(a)

Differential Equation

Calculus

75

(c)

Quadratic Equation

Algebra

76

(b)

Application of Derivative

Calculus

77

(c)

Application of Derivative

Calculus

78

(d)

Application of Derivative

Calculus

79

(a)

Function

Calculus

80

(d)

Function

Calculus

251

solved PAPER - 2019 (II)

Q. No.

Answer Key

81

(d)

Topic Name Integration

Chapter Name Calculus

82

(d)

Integration

Calculus

83

(d)

Differential Equation

Calculus

84

(a)

Continuity and Differentiability

Calculus

85

(b)

Continuity and Differentiability

Calculus

86

(d)

Function

Calculus

87

(c)

Function

Calculus

88

(d)

Function

Calculus

89

(a)

Application of Derivative

Calculus

90

(b)

Application of Derivative

Calculus

91

(b)

Function

Calculus

92

(d)

Function

Calculus

93

(b)

Differentiation

Calculus

94

(c)

Integration

Calculus

95

(a)

Integration

Calculus

96

(b)

Differential Equation

Calculus

97

(a)

Continuity and Differentiability

Calculus

98

(a)

Continuity and Differentiability

Calculus

99

(a)

Application of Derivative

Calculus

100

(a)

Application of Derivative

Calculus

101

(d)

Probability

Algebra

102

(a)

Probability

Algebra

103

(b)

Statistics

Algebra

104

(a)

Statistics

Algebra

105

(b)

Nature Equation

Algebra

106

(d)

Statistics

Algebra

107

(b)

Statistics

Algebra

108

(b)

Statistics

Algebra

109

(d)

Probability

Algebra

110

(a)

Statistics

Algebra

111

(a)

Probability

Algebra

112

(c)

Probability

Algebra

113

(c)

Probability

Algebra

114

(b)

Probability

Algebra

115

(b)

Probability

Algebra

116

(c)

Statistics

Algebra

117

(c)

Statistics

Algebra

118

(c)

Statistics

Algebra

119

(b)

Applied Mathematics

Algebra

120

(b)

Applied Mathematics

Algebra

NDA / NA

MATHEMATICS

National Defence Academy / Naval Academy

i

Time : 2:30 Hour

question Paper

2018 Total Marks : 300

Important Instructions : 1. This test Booklet contains 120 items (questions). Each item is printed in English. Each item comprises four responses (answer's). You will select the response which you want to mark on the Answer Sheet. In case you feel that there is more than one correct response, mark the response which you consider the best. In any case, choose ONLY ONE response for each item. 2. You have to mark all your responses ONLY on the separate Answer Sheet provided. 3. All items carry equal marks. 4. Before you proceed to mark in the Answer Sheet the response to various items in the Test Booklet, you have to fill in some particulars in the Answer Sheet as per instructions. 5. Penalty for wrong answers : THERE WILL BE PENALTY FOR WRONG ANSWERS MARKED BY A CANDIDATE IN THE OBJECTIVE TYPE QUESTION PAPERS. (i) There are four alternatives for the answer to every question. For each question for which a wrong answer has been given by the candidate, one·third of the marks assigned to that question will be deducted as penalty. (ii) If a candidate gives more than one answer, it will be treated as a wrong answer even if one of the given answers happens to be correct and there will be same penalty as above to that question. (iii) If a question is left blank, i.e., no answer is given by the candidate, there will be no penalty for that question.

1. If n ∈ N, then 121n - 25n + 1900n - (-4)n is divisible by which one of the following ? (a) 1904 (b) 2000 (c) 2002 (d) 2006 2. If n = (2017)!, then

1 1 1 1 + + + .... + is log 2 n log 3 n log 4 n log 2017 n

equal to ? (a) 0 (b) 1 n (c) (d) n 2 3. In the expansion of (1 + x)43, if the coefficients of (2r + 1)th and (r + 2)th terms are equal, then what is the value of r (r ≠ 1) ? (a) 5 (b) 14 (c) 21 (d) 22 4. What is the principal argument of (- 1 - i ), where i = -1  ? π π (a) (b) − 4 4 3π 3π − (d) (c) 4 4 5. Let α and β be real numbers and z be a complex number. If z2 + αz + β = 0 has two distinct nonreal roots with Re(z) = 1, then it is necessary that

(a) β ∈ (- 1, 0) (c) β ∈ (1, ∞)

(b) |b| = 1 (d) β ∈ (0, 1)

6. Let A and B be subsets of X and C = (A ∩ B′) ∪ (A′ ∩ B) where A′ and B′ are complements of A and B respectively in X. What is C equal to ? (a) (A ∪ B′) - (A ∩ B′) (b) (A′ ∪ B) - (A′ ∩ B) (c) (A ∪ B) - (A ∩ B) (d) (A′ ∪ B′) - (A′ ∩ B′) 7. How many numbers between 100 and 1000 can be formed with the digits 5, 6, 7, 8, 9, if the repetition of digits is not allowed ? (a) 3 5 (b) 53 (c) 120 (d) 60 8. The number of non-zero integral solutions of the equation |1 - 2i|x = 5x is, where i = (a) Zero (No solution) (b) One (c) Two (d) Three

-1

9. If the ratio of AM to GM of two positive numbers a and b is 5 : 3, then a : b is equal to (a) 3 : 5 (b) 2 : 9 (c) 9 : 1 (d) 5 : 3 10. If the coefficients of am and an in the expansion of (1 + a)m + n are α and β, then which one of the following is correct ? (a) α = 2β (b) α = β (c) 2α = β (d) α = (m + n)β

253

solved PAPER - 2018 (I)

11. If x + log15 (1 + 3x) = xlog15 5 + log15 12, where x is an integer, then what is x equal to ? (a) - 3 (b) 2 (c) 1 (d) 3 12. How many four-digit numbers divisible by 10 can be formed using 1, 5, 0, 6, 7 without repetition of digits ? (a) 24 (b) 36 (c) 44 (d) 64

Consider the information given below and answer the two items (02) that follow : In a class, 54 students are good in Hindi only, 63 students are good in Mathematics only and 41 students are good in English only. There are 18 students who are good in both Hindi and Mathematics. 10 students are good in all three subjects. 13. What is the number of students who are good in either Hindi or Mathematics but not in English ? (a) 99 (b) 107 (c) 125 (d) 130 14. What is the number of students who are good in Hindi and Mathematics but not in English ? (a) 18 (b) 12 (c) 10 (d) 8 15. If α and β are different complex numbers with |α| = 1, then what is (a) |β| (c) 1

α −β 1 − αβ

equal to ?

(b) 2 (d) 0

16. The equation |1 - x| + x2 = 5 has (a) a rational root and an irrational root (b) two rational roots (c) two irrational roots (d) no real roots

1 (a) log 100 ! N

(b)

99 (c) log 100 ! N

(d)

1 log 99 ! N 99 log 99 ! N

20. The modulus-amplitude form of

3 + i , where

i = −1 is π π  2  cos + i sin  (a) 3 3 

π π  (b) 2  cos + i sin  6 6 

π π π π   4  cos + i sin  (d) 4  cos + i sin  (c) 3 3 6 6   21. What is the number of non-zero terms in the expansion of

( 1 + 2 3x ) + ( 1 − 2 3x )

simplification) ? (a) 4 (c) 6

11

11

(after

(b) 5 (d) 11

22. What is the greatest integer among the following by which the number 55 + 75 is divisible ? (a) 6 (b) 8 (c) 11 (d) 12 23. If x = 1 - y + y2 - y3 + .... up to infinite terms, where |y| < 1, then which one of the following is correct ? 1 1 x= (b) x = (a) 1+ y 1− y x= (c)

y 1+ y

(d) x =

y 1− y

24. What is the inverse of the matrix

 cos θ sin θ 0    A =  − sin θ cos θ 0  ?  0 0 1  

 cos θ − sin θ 0   cos θ 0 − sin θ      1 0  (a)  sin θ cos θ 0  (b)  0  0  sin θ 0 cos θ  0 1    

17. The binary number expression of the decimal number 31 is (a) 1111 (b) 10111 (c) 11011 (d) 11111

0 0  1  cos θ sin θ 0      0 cos θ − sin θ (c)   (d)  − sin θ cos θ 0   0 sin θ cos θ   0 0 1    

18. What is i1000 + i1001 + i1002 + i1003 equal to (where i = −1 ) ? (a) 0 (b) i (c) - i (d) 1

25. If A is a 2 × 3 matrix and AB is a 2 × 5 matrix, then B must be a (a) 3 × 5 matrix (b) 5 × 3 matrix (c) 3 × 2 matrix (d) 5 × 2 matrix

19. What is

1 1 1 1 + + + ..... + log 2 N log 3 N log 4 N log 100 N



equal to (N ≠ 1) ?

254 Oswaal NDA/NA Year-wise Solved Papers 1 2 2 26. If A =   and A - kA - I2 = O, where I2 is 2 3   the 2 × 2 identity matrix, then what is the value of k ? (a) 4 (b) -4 (c) 8 (d) -8 27. What is the number of triangles that can be formed by choosing the vertices from a set of 12 points in a plane, seven of which lie on the same straight line ? (a) 185 (b) 175 (c) 115 (d) 105 28. What is C(n, r) + 2C(n, r - 1) + C(n, r - 2) equal to ? (a) C(n + 1, r) (b) C(n - 1, r + 1) (c) C(n, r + 1) (d) C(n + 2, r) 29. Let [x] denote the greatest integer function. What is the number of solutions of the equation x2 - 4x + [x] = 0 in the interval [0, 2] ? (a) Zero (No solution) (b) One (c) Two (d) Three 30. A survey of 850 students in a University yields that 680 students like music and 215 like dance. What is the least number of students who like both music and dance ? (a) 40 (b) 45 (c) 50 (d) 55 31. What is the sum of all two-digit numbers which when divided by 3 leave 2 as the remainder ? (a) 1565 (b) 1585 (c) 1635 (d) 1655 32. If 0 < a < 1, the value of log10 a is negative. This is justified by (a) Negative power of 10 is less than 1 (b) Negative power of 10 between 0 and 1 (c) Negative power of 10 is positive (d) Negative power of 10 is negative 33. The third term of a GP is 3. What is the product of the first five terms ? (a) 216 (b) 226 (c) 243 (d) Cannot be determined due to insufficient data 3 , z are in AP; x, 3, z are in GP; then which 2 one of the following will be in HP ?

34. If x,

(a) x, 6, z (c) x, 2, z

(b) x, 4, z (d) x, 1, z

35. What is the value of the sum

11

∑ (i

n

)

+ i n +1 , where i = −1 ?

n=2

(a) i (c) -2i 36. If sin x =

(b) 2i (d) 1 + i 1 5

, sin y =

1 10

, where 0 < x
2 55. What is the equation of the circle which passes through the points (3, - 2) and (- 2, 0) and having its centre on the line 2x - y - 3 = 0 ? (a) x2 + y2 + 3x + 2 = 0 (b) x2 + y2 + 3x + 12y + 2 = 0 (c) x2 + y2 + 2x = 0 (d) x2 + y 2 = 5 20   2 56. What is the ratio in which the point C  − , −  7 7   divides the line joining the points A(- 2, - 2) and B(2, - 4) ? (a) 1 : 3 (b) 3:4 (c) 1 : 2 (d) 2:3 57. What is the equation of the ellipse having foci (± 2, 0) and the eccentricity

1  ? 4

x2 y2 x2 y2 (a) + = 1 (b) + =1 64 60 60 64 x2 y2 x2 y2 (c) + = 1 (d) + =1 20 24 24 20

58. What is the equation of the straight line parallel to 2x + 3y + 1 = 0 and passes through the point (- 1, 2) ? (a) 2x + 3y - 4 = 0 (b) 2x + 3y - 5 = 0 (c) x + y - 1 = 0 (d) 3x - 2y + 7 = 0 59. What is the acute angle between the pair of straight lines

2 x + 3 y = 1 and

3x + 2 y = 2  ?

 1   1  tan −1  tan −1  (a)  (b)  2 6  2  1  tan −1  (c) tan-l (3) (d)   3 60. If the centroid of a triangle formed by (7, x), (y, - 6) and (9, 10) is (6, 3), then the values of x and y are respectively (a) 5, 2 (b) 2, 5 (c) 1, 0 (d) 0, 0 61. A straight line with direction cosines (0, 1, 0) is (a) parallel to x-axis (b) parallel to y-axis (c) parallel to z-axis (d) equally inclined to all the axes 62. (0, 0, 0), (a, 0, 0), (0, b, 0) and (0, 0, c) are four distinct points. What are the coordinates of the point which is equidistant from the four points ? a+b+c a+b+c a+b+c , , (a)   3 3 3   (b) (a, b, c) a b c (c) 2, 2, 2   a b c (d) 3, 3, 3   63. The points P(3, 2, 4), Q(4, 5, 2), R(5, 8, 0) and S(2, - 1, 6) are (a) vertices of a rhombus which is not a square (b) non-coplanar (c) collinear (d) coplanar but not collinear 64. The line passing through the points (1, 2, - 1) and (3, - 1, 2) meets the yz-plane at which one of the following points ? 7 5   7 1 (a)  0 , − 2 , 2  (b)  0, 2 , 2      7 5 7 5   (c)  0 , − 2 , − 2  (d)  0, − 2 , 2     

283

solved PAPER - 2017 (I)

65. Under which one of the following conditions are the lines x = ay + b; z = cy + d and x = ey + f; z = gy + h perpendicular? (a) ae + cg - 1 = 0 (b) ae + bf - 1 = 0 (c) ae + cg + 1 = 0 (d) ag + ce + 1 = 0    66. If a = i − j + k , b = 2i + 3 j + 2 k and c = i + m j + nk  are three coplanar vectors and c = 6 , then which one of the following is correct ? (a) m = 2 and n = ± 1 (b) m = ± 2 and n = - 1 (c) m = 2 and n = - 1 (d) m = ± 2 and n = 1 67. Let ABCD be a parallelogram whose diagonals intersect at P and let O be the origin. What is     OA+OB+OC+OD equal to ?   (a) 2OP (b) 4OP   (c) 6OP (d) 8OP 68. ABCD is a quadrilateral whose diagonals are AC and BD. Which one of the following is correct?     (a) BA+CD = AC+DB     (b) BA+CD = BD+CA     (c) BA+CD = AC+BD     (d) BA+CD = BC+ AD       69. a × b = c and b × c = a then which one of the following is correct ?      (a) a , b , c are orthogonal in pairs and a = c  and b = 1    a , b , c are non-orthogonal to each other (b)    a , b , c are orthogonal in pairs but a ≠ c (c)    a , b , c are orthogonal in pairs but b ≠ 1 (d)   70. If a = 2i + 3 j + 4 k and b = 3i + 2 j + λ k are perpendicular, then what is the value of l ? (a) 2 (b) 3 (c) 4 (d) 5 71. What is lim

x →0

e x − (1 + x ) x2

equal to?

1 (a) (b) 1 2 (c) None of the above 3 (d) 73. What is

∫x

(

dx x7 + 1

)

equal to ?

1 1 x7 − 1 x7 + 1 (a) ln + c (b) ln +c 2 7 x7 + 1 x7 (c) ln

1 x7 − 1 x7 + c (d) ln +c 7x 7 x7 + 1

74. The function f : X → Y defined by f (x) = cos x, where x ∈ X, is one-one and onto if X and Y are respectively equal to (a) [0, p] and [- 1, 1]  π π (b)  − 2 , 2  and [- 1, 1]   (c) [0, p] and (- 1, 1) (d) [0, p] and [0, 1] 75. If f ( x ) =

f (a) x equal to ? , then what is x −1 f ( a + 1)

 ( a )  (b) (a) f(a2) f −  ( a + 1 )   

1 (c) f(- a) f   (d) a 76. What is

∫

( x e−1 + ex−1 ) dx equal to? xe + ex

x2 ln (x + e) + c (a) + c (b) 2 1 (c) ln (xe + ex) + c (d) ln x e + e x + c e 77. Let f : [- 6, 6] → be defined by f(x) = x2 - 3. Consider the following : (1) ( fofof )(- 1) = ( fofof )(1) (2) ( fofof )(- 1) - 4( fofof )(1) = ( fof )(0)

(

)

Which of the above is/are correct ? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2

1 (a) 0 (b) 78. Let f(x) = px + q and g(x) = mx + n. Then f (g(x)) 2 (c) 1 (d) 2 = g(f(x)) is equivalent to (a) f(p) = g(m) (b) f(q) = g(n) π/2 dθ 72. What is ∫ equal to? (c) f(n) = g(q) (d) f(m) = g(p) 0 1 + cos θ

284 Oswaal NDA/NA Year-wise Solved Papers The function f(x) satisfies f(x + p) = f(x) for all real x.

79. If F ( x ) = 9 − x 2 , then what is lim

F ( x ) − F (1)



Statement 2 : sin2 (x + p) = sin2 x for all real x. Which one of the following is correct in respect of the above statements ? (a) Both the statements are true and Statement 2 is the correct explanation of Statement 1 (b) Both the statements are true but Statement 2 is not the correct explanation of Statement 1 (c) Statement 1 is true but Statement 2 is false (d) Statement 1 is false but Statement 2 is true

x −1

x →1

equal to?

1 1 (a) (b) 8 4 2 1 1 (c) (d) 2 2 2 2 80. What is

d2x dy 2

d y (a) −   dx 2    2

equal to?

−1

 dy   dx   

−3

d y (b)    dx 2    2

 d 2 y   dy −3  d2 y  (c) −     (d)   dx 2   dx   dx 2      x , 81. Let f ( x ) :  0,

0 , and g ( x ) :  x,

−1

 dy   dx   

−2

−1

x is rational x is irrational x is rational x is irrational

If f :  →  and g :  → , then (f - g) is (a) one-one and into (b) neither one-one nor onto (c) many-one and onto (d) one-one and onto 82. What is the length of the longest interval in which the function f(x) = 3 sin x - 4 sin3 x is increasing ? π π (a) (b) 2 3 3π (c) (d) p 2 83. If xdy = y(dx + ydy); y(1) = 1 and y(x) > 0, then what is y(- 3) equal to ? (a) 3 (b) 2 (c) 1 (d) 0 84. What is the maximum value of the function f(x) = 4 sin2 x + 1 ? (a) 5 (b) 3 (c) 2 (d) 1 85. Let f(x) be an indefinite integral of sin2 x. Consider the following statements : Statement 1 :

86. What are the degree and order respectively of the differential equation 2

2

 dy   dx  y = x  +   dx   dy  (a) 1, 2 (b) 2, 1 (c) 1, 4 (d) 4, 1 87. What is the differential equation corresponding to y2 - 2ay + x2 = a2 by eliminating a ? (a) (x2 - 2y2)p2 - 4pxy - x2 = 0 (b) (x2 - 2y2)p2 + 4pxy - x2 = 0 (c) (x2 + 2y2)p2 - 4pxy - x2 = 0 (d) (x2 + 2y2)p2 - 4pxy + x2 = 0 dy . where p = dx 88. What is the general solution of the differential equation ydx - (x + 2y2)dy = 0 ? (a) x = y2 + cy (b) x = 2cy2 2 (c) x = 2y + cy (d) None of the above 89. Let f(x + y) = f(x)f(y) for all x and y. Then what is f ′(5) equal to [where f(x) is the derivative of f(x)] ? (a) f(5)f(0) (b) f(5) - f(0) (c) f(5) f(0) (d) f(5) + f(0) 90. If f(x) and g(x) are continuous functions satisfying f(x) = f(a - x) and g(x) + g(a - x) = 2, then what is

a

∫0 f ( x ) g ( x ) dx

equal to?

a

a

0

0

(a) g ( x ) dx (b) ∫ ∫ f ( x ) dx a

(c) 0 2 ∫ f ( x ) dx (d) 0

91. What is the solution of the differential equation



 dy  ln   − a = 0 ?  dx 

285

solved PAPER - 2017 (I)

(a) y = xea + c (b) x = yea + c (c) y = ln x + c (d) x = ln y + c

97. What is the maximum area of a triangle that can be inscribed in a circle of radius a ?

92. Let f(x) be defined as follows :

3a 2 a2 (a) (b) 4 2

2 x + 1, −3 < x < −2  f ( x ) = x − 1, −2 ≤ x < 0 x + 2 , 0≤x