NDA/NA Pathfinder 2020-2021 PDF

Pathf nder NATIONAL DEFENCE ACADEMY & NAVAL ACADEMY ENTRANCE EXAMINATION  2020-21 EDITION Pathf nder NATIONAL DEFENCE ACADEMY & NAVAL ACADEMY ENTRANCE EXAMINATION More than Compiled & Edited by Arihant ‘Expert Team...

Pathf nder NATIONAL DEFENCE ACADEMY & NAVAL ACADEMY ENTRANCE EXAMINATION  2020-21 EDITION Pathf nder NATIONAL DEFENCE ACADEMY & NAVAL ACADEMY ENTRANCE EXAMINATION More than Compiled & Edited by Arihant ‘Expert Team’ ARIHANT PUBLICATIONS (INDIA) LIMITED Pathf nder ARIHANT PUBLICATIONS (INDIA) LIMITED All Rights Reserved © Publisher No part of this publication may be re-produced, stored in a retrieval system or by any means, electronic, mechanical, photocopying, recording, scanning, web or otherwise without the written permission of the publisher. Arihant has obtained all the information in this book from the sources believed to be reliable and true. However, Arihant or its editors or authors or illustrators don’t take any responsibility for the absolute accuracy of any information published and the damage or loss suffered thereupon. All disputes subject to Meerut (UP) jurisdiction only. Administrative & Production Offices Regd. Office ‘Ramchhaya’ 4577/15, Agarwal Road, Darya Ganj, New Delhi -110002 Tele: 011- 47630600, 43518550 Head Office Kalindi, TP Nagar, Meerut (UP) - 250002 Tel: 0121-7156203, 7156204 Sales & Support Offices Agra, Ahmedabad, Bengaluru, Bareilly, Chennai, Delhi, Guwahati, Hyderabad, Jaipur, Jhansi, Kolkata, Lucknow, Nagpur & Pune. ISBN 978-93-24196-19-4 PRICE `775.00 PO No : TXT-XX-XXXXXXX-X-XX Published by Arihant Publications (India) Ltd. For further information about the books published by Arihant, log on to www.arihantbooks.com or e-mail at [email protected] Follow us on CONTENTS NDA/NA Solved Paper 2020 I & II 1-30 NDA/NA Solved Paper 2019 II 1-29 NDA/NA Solved Paper 2019 I 1-27 NDA/NA Solved Paper 2018 II 1-32 NDA/NA Solved Paper 2018 I 33-62 NDA/NA Solved Paper 2017 II 1-14 MATHEMATICS 1. Set Theory 3-12 2. Relations and Functions 13-29 3. Complex Numbers 30-45 4. Binary Numbers 46-51 5. Sequences and Series 52-66 6. Quadratic Equations and Inequalities 67-82 7. Permutations and Combinations 83-92 8. Binomial Theorem 93-101 9. Logarithm 102-106 10. Matrices 107-114 11. Determinants 115-132 12. Measurement of Angles and Trigonometric Ratios 133-150 13. Inverse Trigonometric Functions 151-158 14. Height and Distance 159-167 15. Properties of Triangles 168-177 16. Coordinate System and Straight Lines 178-195 17. Circle 196-204 18. Conic Section 205-217 19. Three Dimensional Geometry 218-235 20. Limits, Continuity and Differentiability 236-252 21. Differentiation 253-268 22. Application of Derivatives 269-283 23. Indefinite Integrals 284-294 24. Definite Integrals 295-304 25. Areas Bounded by Regions 305-312 26. Differential Equations 313-326 27. Vector Algebra 327-344 28. Statistics 345-367 29. Probability 368-385 30. Miscellaneous 386-388 GENERAL ENGLISH 1. Spotting the Errors 391-433 2. Vocabulary 434-450 3. Antonyms 451-460 4. Synonyms 461-470 5. Sentence Improvement 471-480 6. Sentence Completion 481-491 7. Jumbled Sentences and Paragraphs 492-505 8. Comprehension 506-514 GENERAL SCIENCE PHYSICS 1. Measurement, Motion and Force 517-530 2. Work, Energy and Power 531-536 3. Centre of Mass and Rotational Motion 537-542 4. Gravitation 543-549 5. General Properties of Matter 550-557 6. Heat and Kinetic Theory of Gases 558-566 7. Thermodynamics 567-573 8. Optics 574-590 9. Oscillation and Waves 591-602 10. Electrostatics 603-610 11. Current Electricity 611-621 12. Magnetic Effects of Electric Current and Magnetism 622-633 13. Nucleus and Radioactivity 634-639 14. Modern Physics 640-646 CHEMISTRY 1. Physical and Chemical Changes 647-649 2. Elements, Mixtures and Compounds 650-655 3. Laws of Chemical Combination and Gas Laws 656-660 4. Concept of Atomic, Molecular and Equivalent Masses 661-667 5. Atomic Structure and Radioactivity 668-676 6. Periodic Classification of Elements 677-684 7. Chemical Bonding 685-689 8. Acids, Bases and Salts 690-696 9. Oxidation, Reduction and Electrochemistry 697-703 10. Non-Metals and Their Compounds 704-721 11. Some Important Chemical Compounds 722-734 BIOLOGY 1. Diversity in Living World 735-746 2. Cell and Cell Division 747-755 3. Constituents of Food (Biomolecules) 756-761 4. Structural Organisation of Plants and Animals 762-776 5. Plants Physiology and Reproduction 777-785 6. Human System -I 786-796 7. Human System-II 797-803 8. Health and Diseases 804-812 9. Economic Importance of Biology 813-819 10. Ecology, Biodiversity and Environment 820-828 GENERAL STUDIES 1. History 831-913 2. Geography 914-986 3. Indian Polity 987-1020 4. Indian Economy 1021-1046 5. General Knowledge 1047-1086 Pathf nder ABOUT THE EXAMINATION The National Defence Academy (NDA) is an iconic institution, a global brand of excellence in the sphere of military education. For recruitment to the Indian Army, Navy and Air force wings of Indian Army, there is prestigious National Defence Academy Entrance Examination. To join National Defence Academy, a candidate must appear in the entrance exam conducted by Union Public Service Commission (UPSC), twice a year. The UPSC is solely responsible for issuing guidelines for selection and the final conduct of the entrance examination. NATIONALITY A candidate must be either (i) Indian citizen, or (ii) A subject of Bhutan, or (iii) A subject of Nepal, or (iv) A Tibetan refugee who came over to India before 1st January, 1962 with the intention of permanently setting in India, or (v) A person of Indian origin who has migrated from Pakistan, Burma, Sri Lanka and East African countries of Kenya, Uganda, the United Republic of Tanzania, Zambia, Malawi, Zaire and Ethiopia and Vietnam with the intention of permanently setting in India. Provided that a candidate belonging to categories (ii), (iii), (iv) and (v) above shall be a person in whose favour a certificate of eligibility has been issued by the Government of India. Certificate of eligibility, will not , however, be necessary in the case of candidate who are Gorkha subjects of Nepal. AGE LIMIT, SEX AND MARITAL STATUS Only unmarried male candidates whose age is not less than 15 years and not exceeding 18 years on 1st January in accordance with the year of examination are eligible to apply. Candidates must not marry until they complete their full training. EDUCATIONAL QUALIFICATION For Indian Army, candidates must have done or appearing class 12th from a recognised board. For Air force and Navy and for 10+2 course at Naval Academy, candidates must have done or appearing class 12th with Physics and Mathematics from a recognised board. SCHEME OF EXAMINATION Subject No. of Ques. Duration Max. Marks Mathematics 120 2 —12 hours 300 General Ability Test 1 Part-A (English) 50 2 —2 hours 200 Part-B (General Studies) 100 400 Total 900 SSB TEST/INTERVIEW The papers in all the subjects will consists of objective type questions only. There will be 1/3rd negative marking for wrong answers. -0.83 for Maths -1.33 for English -1.33 for General Ability Test The SSB procedure consists of two stage- (a) Stage I comprises of Officer Intelligence Rating (OIR) test, Picture Perception and Description Test (PP and DT). (b) Stage II comprises of interview, Group Testing Officer tasks, Psychology tests and the Conference. SYLLABUS PAPER 1 Mathematics (Maximum Marks: 300) NUMBER ALGEBRA Concept of a 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 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 application. Logarithms and their applications. 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. 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. 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 of a plane and a line in various forms. Angle between two lines and angles between two planes. Equation of a sphere. 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 a 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 of another function, derivative of a composite function. Second order derivatives. Increasing and decreasing functions. Application of derivatives in problems of maxima and minima. INTEGRAL CALCULUS AND DIFFERENTIAL EQUATIONS 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. Definitions of order and degree of a differential equation, formation of a differential equation by examples. General and particular solution of a differential equation, solution of first order and first degree differential equations of various types – examples. Application in problems of growth and decay. VECTOR ALGEBRA Vectors in two and three dimensions, magnitude and direction of a vector. Unit and null vectors, addition of vectors, scalar multiplication of vector, scalar product or dot of two-vectors. Vector product and cross product of two vectors. Applications-work done by a force and moment of a force, and in geometrical problems. 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. PAPER 2 General Ability Test (Maximum Marks: 600) PART I English (Maximum Marks: 200) This portion of the question paper is designed to test the candidate’s understanding of English and work man like use of words. The syllabus covers various aspects like grammar and its usage, vocabulary, comprehension and cohesion in extended texts to test the candidate’s proficiency in English. PART II General Knowledge (Maximum Marks: 400) This portion broadly covers the subjects Physics, Chemistry, General Science, Social Studies, Geography and Current Events. The syllabus given below is designed to indicate the scope of these subjects included in the paper. The topics mentioned are not to be regarded as exhaustive and questions on topics of similar nature not specifically mentioned in the syllabus may also be asked. The candidate’s answers are expected to show their knowledge and intelligent understanding of the subject. SECTION ‘A’ Physics Physical properties and states of matter, Mass, Weight, Volume, Density and specific gravity, Principle of Archimedes, Pressure barometer. Motion of objects, Velocity and acceleration, Newton’s laws of motion, Force and momentum, Parallelogram of forces, Stability and equilibrium of bodies, Gravitation, Elementary ideas of Work, Power and Energy. Effects of heat, Measurement of temperature and heat, Change of state and latent heat, Modes of transference of heat. Sound waves and their properties, Simple musical instruments. Rectilinear propagation of light, Reflection and refraction, Spherical mirrors and lenses, Human eye. Natural and Artificial magnets, Properties of a magnet, Earth as a magnet. Static and current electricity, Conductors and non-conductors, Ohm’s law, Simple electrical circuits, Heating, Lighting and magnetic effects of current, Measurement of electrical power, Primary and secondary cells, Use of X-rays. General principles in the working of the following Simple pendulum, Simple pulleys, Siphon, Levers, Balloon, Pumps, Hydrometer, Pressure cooker, Thermos flask, Gramophone, Telegraphs, Telephone, Periscope, Telescope, Microscope, Mariner’s compass, Lightening conductors, Safety fuses. SECTION ‘B’ Chemistry Physical and chemical changes. Elements, mixtures and compounds, Symbols, Formulae and simple chemical equations, Law of chemical combination (excluding problems), Properties of air and water, Preparation and properties of hydrogen, oxygen, nitrogen and carbon dioxide, Oxidation and reduction, Acids, bases and salts. Carbon — Different forms, Fertilisers — Natural and artificial. Materials used in the preparation of substances like soap, glass, ink, paper, cement, paints, safety matches and gunpowder. Elementary ideas about the structure of atom, atomic, equivalent and molecular weights, valency. SECTION ‘C’ General Science Difference between the living and non-living. Basis of life-cells, Protoplasm’s and tissues. Growth and reproduction in plants and animals. Elementary knowledge of human body and its important organs. Common epidemics, their causes and prevention. Food-source of energy for man. Constituents of food, balanced diet. The solar system — Meteors and comets, eclipses. Achievements of eminent scientists. SECTION ‘D’ History, Freedom Movement, etc A broad survey of Indian History, with emphasis on culture and civilisation. Freedom movements in India. Elementary study of Indian Constitution. Elementary knowledge of Five Year Plans of India. Panchayati Raj, Cooperatives and community development, Bhoodan, Sarvodaya, National integration and welfare state, basic teachings of Mahatma Gandhi. Forces shaping the modern world. Renaissance, Exploration and discovery, War of American independence, French Revolution, Industrial Revolution and Russian Revolution. Impact of science and technology on society. Concept of one world, United Nations, Panchsheel, Democracy, Socialism and communism. Role of India in the present world. SECTION ‘E’ Geography The Earth, its shape and size. Latitudes and longitudes, Concept of time. International date line. Movements of Earth and their effects. Origin of Earth, rocks and their classification. Weathering — Mechanical and chemical, earthquakes and volcanoes. Ocean currents and tides. Atmosphere and its composition. Temperature and atmospheric pressure, Planetary winds, Cyclones and anticyclones. Humidity. Condensation and precipitation. Types of climate, Major natural regions of the world, Regional geography of India — Climate, natural vegetation, mineral and power resources. Location and distribution of agriculture including industrial activities. Important sea ports and main sea, land and air routes of India. Main items of imports and exports of India. SECTION ‘F’ Current Affairs Knowledge of important events that have happened in India in the recent years. Current important world events. Prominent personalities - both Indian and International including those connected with various cultural activities and sports. NDA/NA Solved Paper 2020 (I & II) 1 NDA /NA National Defence Academy/Naval Academy SOLVED PAPER 2020 (I & II) PAPER I : Mathematics 1−i i  1. If matrix A =  where ⇒ 20 − ( n + 2 ) = n − 2 i  Ê (a) The general term in the binomial  − i 1 − 10 ⇒ 20 = n + 2 + n − 2 expansion of  2 − x  2 is i = − 1, then which one of the x  ⇒ 2 n = 20 following is correct? 10 − r ⇒ n = 10 C r  2  2 (a) A is hermitian Tr +1 = 10 ( − x )r x  5. For how many values of k, is the (b) A is skew-hermitian (c) ( A )T + A is hermitian −20 + 2r + r  0 k 4 = C r (2 )10 − r x 10 ( −1)r matrix − k 0 − 5 singular? 2 (d) ( A )T + A is skew-hermitian   For independent of x, put Ê (c) We have r 5r − k k − 1 −20 + 2 r + = 0 ⇒ = 20 1 − i i  2 2 (a) Only one (b) Only two A=  − i 1 − i  ⇒ r=8 (c) Only four (d) Infinite 1 + i −i  ∴ T8 + 1 = 10C 8(2 )10 − 8 ( −1)8 Now, A =  Ê (d) The condition for singular matrix is  i 1 + i  10 × 9 0 k 4 = × 22 × 1 1 + i i  2 ×1 − k 0 −5 = 0 ( A )T =   −i 1 + i  = 180 −k k −1 Now, consider Expanding along R1, we get 3. If (1 + 2x − x 2 )6 = a 0 + a1x + a 2 x 2 1 + i i  0 − k( k − 5k ) + 4( − k 2 + 0) = 0 X = ( A) + A =  T  − i 1 + i  + K + a12 x 12 , then what is a 0 − a1 ⇒ 4k 2 − 4k 2 = 0 1 − i i  + a 2 − a 3 + a 4 − K + a12 equal to? +  ⇒ 0 = 0, ∀k ∈ R  −i 1 − i  (a) 32 (b) 64 Hence, for infinite values of k, given  2 2i  (c) 2048 (d) 4096 = matrix is singular.  Ê (b) We have,  −2 i 2  (1 + 2 x − x 2 )6 = a0 + a1 x + a2 x 2 6. The number (1101101 + 1011011)2  2 −2 i  X =  can be written in decimal system as 2 i 2  +....+ a12 x12 (a) (198)10 (b) (199)10  2 2i  Put x = − 1 both sides, we get (c) (200)10 (d) (201)10 ( X )T =   = X 5  −2 i 2  (1 − 2 − 12 )6 = a0 − a1 + a2 −....+ a12 Ê (c) Now, (1101101)2 = 1 × 2 + 0 × 2 6 Hence, X = ( A ) + A is a hermitian T ⇒ a0 − a1 + a2 −.....+ a12 = ( −2 ) = 64 6 + 1 × 2 4 + 1 × 2 3 + 0 × 2 2 + 1 × 21 matrix. + 1 × 20 4. If C (20, n + 2) = C (20, n − 2), then 2. The term independent of x in the what is n equal to? = 64 + 0 + 16 + 8 + 0 + 2 + 1 10 (a) 18 (b) 25 = ( 91)10  2  binomial expansion of  2 − x  and x  (c) 10 (d) 12 (1011011)2 = 1 × 2 6 + 1 × 2 5 + 0 × 2 4 is equal to Ê (c) Given, C(20, n + 2) = C (20, n − 2 ) ⇒ C (20, 20 − ( n + 2 )) = C (20, n − 2 ) + 1 × 2 3 + 1 × 2 2 + 0 × 21 + 1 × 2 0 (a) 180 (b) 120 (c) 90 (d) 72 [Q C ( n, r ) = C ( n, n − r )] = 64 + 32 + 0 + 8 + 4 + 0 + 1 = (109)10 2 NDA/NA Solved Paper 2020 (I & II) ∴ (1101101 + 1011011)2 Also given, AB = C i i2 i3 = (1101101)2 + (1011011)2 x + y y  2   3 Ê (d) Let ∆ = i 4 i 6 i8 ⇒  2x = = ( 91)10 + (109)10 = (200)10  x − y  −1 2    i 9 i 12 i 15 7. What is the value of 2 x + 2 y − y  3 i −1 − i ⇒  4x − x + y  = 2  1 1     = 1 −1 1 log 5 1024 − log 5 10 + log 5 3125? 10 5 2 x + y  3 i 1 −i ⇒  3 x + y = 2  (a) 0 (b) 1 (c) 2 (d) 3     [Q i 2 = − 1, i 3 = − i , i 4 = 1] 1 1 Ê (a) log 1024 − log 10 + log 5 3125 On equating the corresponding 10 5 5 5 = i ( i − 1) + 1 ( − i − i ) − i (1 + i ) elements, we get 1 1 2 x + y = 3 and 3 x + y = 2 [Expanding along R1] = log 5 2 10 − log 5( 5 × 2 ) + log 5 5 5 10 5 ⇒ x = − 1 and y = 5 = i 2 − i − 2 i − i − i 2 = − 4i = 10 5 log 5 2 − [log 5 5 + log 5 2 ] + log 5 5  −1 + 5 5  4 5 10 5 ∴ A=  =  −2 −6 a h g  x × − − − 13. Let A =  h b f  and B =  y , 2 1 1 5    [Q log mn = log m + log n] ∴The determinant of matrix A is    = log 5 2 − [1 + log 5 2 ] + 1 [Q log m m = 1] 4 5  g f c   z  =0 A = = −24 + 10 = −14 −2 −6 then what is AB equal to? 8. If x = logc (ab ), y = loga (bc ),  ax + hy + gz z = logb (ca ), then which of the. ≤ x ≤ 4.5, then which one of 10. If 15 the following is correct? (a)  y  following is correct?   (a) (2 x − 3) (2 x − 9) > 0  z  (a) xyz = 1  ax + hy + gz (b) (2 x − 3) (2 x − 9) < 0 (b) x + y + z = 1 (c) (2 x − 3) (2 x − 9) ≥ 0 (b)  hx + by + fz  (c) (1 + x )−1 + (1 + y)−1 + (1 + z)−1 = 1   (d) (2 x − 3) (2 x − 9) ≤ 0  z  (d) (1 + x )−2 + (1 + y)−2 (1 + z)−2 = 1. ≤ x ≤ 4.5 Ê (d) We have, 15  ax + hy + gz Ê (c) We have, x = logc ( ab ) (c)  hx + by + fz  y = log a( bc ) 3 9   ⇒ ≤ x≤ ⇒ 3≤ 2x≤ 9 z = log b (ca ) 2 2  gx + fy + cz  ⇒ (2 x − 3) (2 x − 9) ≤ 0 (d) [ax + hy + gz hx + by + fz Now, 1 + x = logc c + logc ( ab ) = logc ( abc ) 11. Let S = {1, 2, 3, K}. A relation R on gx + fy + cz] 1 + y = log a( abc ) S × S is defined by xRy if Ê (c) Now, and 1 + z = log b ( abc ) 1  a h g   x loga x > loga y when a =. Then Now, (1 + x )−1 + (1 + y)−1 + (1 + z)−1 2 AB =  h b f   y     = [logc ( abc )]−1 + [log a( abc )]−1 the relation is  g f c   z (a) reflexive only  ax + hy + gz + [log b ( abc )]−1 (b) symmetric only =  hx + by + fz  = 1 + 1 + 1 (c) transitive only   logc ( abc ) log a( abc ) log b ( abc ) (d) both symmetric and transitive  gx + fy + cz  = log c + log a + log b Ê (c) We have, S = {1, 2, 3,....} 14. What is the number of ways in log( abc ) log( abc ) log( abc ) and log a x > log a y which the letters of the word  log n  1 a = ∈ ( 0, 1) Q log m n = log m  Here, ‘ABLE’ can be arranged so that the   2 vowels occupy even places? log c + log a + log b log( abc ) ∴ log a x > log a y ⇒ x < y = = =1 (a) 2 (b) 4 (c) 6 (d) 8 log( abc ) log( abc ) Now, x R x ⇒ x < x which is not possible. So it is not reflexive relation. Ê (b) In a given word ‘ABLE’ Vowels are 9.  x +y y   2 Now, x R y ⇒ x < y { A, E}. Let A =   ,B=  1 2 3 4  2x x − y  − 1 But y |< x, so it is not symmetric relation. As, vowel occupy even places, so two 3 Now, x R y and y R z vowels occupy the places 2 and 4. and C =  . If AB = C , then what is ⇒ x < y and y < z ⇒ x < z ⇒ x R z 2 Therefore, the number of ways of Hence, it is transitive relation only. occupying the vowels in even places is the value of the determinant of the 2!. matrix A? 12. What is the value of the Now, we have two consonants and (a) − 10 (b) − 14  i i 2 i 3 these consonants occupy the odd determinant i 4 i 6 i 8 where (c) − 24 (d) − 34 places 1 and 3. Therefore, the number of x + y y    9 12 15  Ê (b) Given, A =  2 x x − y ways of occupying the consonants in   i i i  odd places is 2!. 2   3 B =   and C =   i = −1? ∴Total number of ways = 2 ! × 2 ! −  1 2  (a) 0 (b) − 2 (c) 4i (d) − 4i =2 ×2= 4 NDA/NA Solved Paper 2020 (I & II) 3 15. What is the maximum number of Ê 16. (c) Also given, Hence, for real values of tan A, K cannot lie between  , 3. n( Y ) 4 1 points of intersection of 5 = n( Z ) 5 3  non-overlapping circles? (a) 10 (b) 15 (c) 20 (d) 25 16 + 18 + 17 + b 4 ⇒ = Directions (Q. Nos. 21 and 22) Read 90 5 Ê (c) The maximum number of points of the following information and answer intersection of 5 non-overlapping circles ⇒ 51 + b = 72 the two items that follow. ⇒ b = 72 − 51 ABCD is a trapezium such that AB and CD = Selection of two circles × 2 = 21 are parallel and BC is perpendicular to [Q Two intersecting circles Ê 17.(d) Now, them. Let ∠ADB = θ, ∠ABD = α , BC = p cut at two points] n( X ) + n( Y ) + n( Z ) − n( X ∩ Y ) and CD = q. 5×4 − n( Y ∩ Z ) − n( X ∩ Z ) + n( X ∩ Y ∩ Z ) = 5C 2 × 2 = × 2 = 20 2 ×1 = n( X ∪ Y ∪ Z ) 21. Consider the following = a + 12 + 18 + 16 + b + 17 + c 1. AD sin θ = AB sin α Directions (Q. Nos. 16-18) = a + b + c + 63 2. BD sin θ = AB sin (θ + α ) Consider the following Venn diagram, Which of the above is/are correct? where X, Y and Z are three sets. Let the = a + b + 43 + 63 [Q c = 43] (a) 1 Only (b) 2 Only number of elements in Z be denoted by = a + b + 106 (c) Both 1 and 2 (d) Neither 1 nor 2 n(Z) which is equal to 90. Ê 18. (a) Complement of X Ê (c) We have, ∠ADB = θ, ∠ABD = α, = p + b + c + 17 X Y = p + b + 43 + 17 [Q c = 43] BC = p and CD = q = p + b + 60 D q C a 16 b Directions (Q. Nos. 19 and 20) Read θ 18 p -α 12 17 the following information and answer 90° the two items that follow. c tan 3A α Let = K , where tan A ≠ 0 A B Z tan A 1 1. In ∆ABD, use Sine rule, and K ≠. sinθ sinα 3 = 16. If the number of elements in Y and AB AD Z are in the ratio 4 : 5, then what is 19. What is tan 2 A equal to? ⇒ ADsinθ = ABsinα, which is correct. the value of b? K + 3 K −3 2. In ∆ABD, ∠A = π − (θ + α ) (a) (b) (a) 18 (b) 19 (c) 21 (d) 23 3K − 1 3K − 1 Use Sine rule in ∆ABD, 3K − 3 K + 3 sin A sinθ 17. What is the value of (c) (d) = K −3 3K + 1 BD AB n ( X ) + n (Y ) + n (Z ) − n ( X ∩ Y ) sin( π − (θ + α )] sinθ − n (Y ∩ Z ) − n ( X ∩ Z ) tan 3 A ⇒ = …(i) Ê (b) Given, =K BD AB + n ( X ∩ Y ∩ Z )? tan A ⇒ ABsin(θ + α ) = BDsinθ, (a) a + b + 43 (b) a + b + 63 3 tan A − tan3 A = =K which is correct. (c) a + b + 96 (d) a + b + 106 (1 − 3 tan2 A ) tan A Hence, both statements are 18. If the number of elements ⇒ 3 − tan2 A =K correct. belonging to neither X, nor Y, nor Z 1 − 3 tan2 A 22. What is AB equal to? is equal to p, then what is the ⇒ K − 3K tan2 A = 3 − tan2 A ( p2 + q 2 ) sin θ (a) number of elements in the ⇒ K − 3 = tan2 A( 3K − 1) p cos θ + q sin θ complement of X? ( p2 − q 2 ) cos θ K −3 (b) (a) p + b + 60 (b) p + b + 40 ⇒ tan2 A = p cos θ + q sin θ (c) p + a + 60 (d) p + a + 40 3K − 1 ( p2 + q 2 ) sin θ Solutions (16-18) (c) 20. For real values of tan A, K cannot q cos θ + p sin θ Given n( Z ) = 90 lie between ( p2 − q 2 ) cos θ ⇒ 12 + 18 + 17 + c = 90 1 1 (d) (a) and 3 (b) and 2 q cos θ + p sin θ ⇒ c = 90 − 47 = 43 3 2 X Y 1 (c) and 5 1 (d) and 7 Ê (a) In right angle, ∆BCD, 5 7 ∠B = 90° − α a 16 b Ê (a) For real values of tan A, K lies when BD = p2 + q 2 K −3 18 ≥ 0 and 3K − 1 ≠ 0 CD 12 17 3K − 1 and sin B = 1 BD ⇒ ( K − 3) ( 3K − 1) ≥ 0 and K ≠ q c 3 ⇒ sin( 90° − α ) = 1 p + q2 2 Z ⇒ K < and K ≥ 3 3 [Q ∠B = 90°−α ] 4 NDA/NA Solved Paper 2020 (I & II) q π = − 2 sin 30° sin18° ⇒ cosα = ⇒ 2B = − A p2 + q 2 2 1 5 −1 π = −2 × × BC ⇒ A + 2B = 2 4 and cos B = 2 BD 1− 5 = ⇒ cos( 90° − α ) = p 25. What is sin 3x + cos 3x + 4 sin 3 x 4 p +q 2 2 − 3 sin x + 3 cos x − 4 cos 3 x equal 29. Consider the following statements: p ⇒ sinα = to? 1. If ABC is a right-angled triangle, p2 + q 2 (a) 0 (b) 1 right-angled at A and if (c) 2 sin 2 x (d) 4 cos 4x 1 From eq. (i), sin B = , then cosec C = 3. Ê (a) sin 3 x + cos 3 x + ( 4sin x − 3sin x) 3 sin( π − (θ + α )) sinθ 3 = BD AB + ( 3 cos x − 4 cos 3 x ) 2. If b cos B = c cos C and if the BD sinθ triangle ABC is not right-angled, ⇒ AB = = sin 3 x + cos 3 x − sin 3 x − cos 3 x = 0 sin(θ + α ) then ABC must be isosceles. 26. The value of ordinate of the graph Which of the above statements p2 + q 2 sinθ = of y = 2 + cos x lies in the interval is/are correct? sinθ cos α + cos θ sinα (a) 1 Only (b) 2 Only (a) [0, 1 (b) [0, 3] (c) [− 1, 1] (d) [1, 3] (c) Both 1 and 2 (d) Neither 1 nor 2 [Q BD = p2 + q 2 ] Ê (d) We know that, 1 p2 + q 2 sinθ −1 ≤ cos x ≤ 1 Ê (b) 1. We have, sin B = = 3 q p ⇒ −1 + 2 ≤ 2 + cos x ≤ 1 + 2 C sinθ + cos θ p2 + q 2 p2 + q 2 ⇒ 1≤ y ≤ 3 ∴ y ∈ [1, 3] ( p + q )sinθ 2 2 = q sinθ + pcos θ 27. What is the value of cos 17 ° − sin 17 ° 8 cos 10° ⋅ cos 20° ⋅ cos 40° ? 23. If tan θ = , then cos 17 ° + sin 17 ° (a) tan 10° (b) cot 10° what is the value of θ? (c) cosec 10° (d) sec 10° B A (a) 0° (b) 28° (c) 38° (d) 52° Ê (b) 8cos 10° cos 20° cos 40° AC 1 Ê (b) We have, sin10° ⇒ = = 8 cos 10° cos 20° cos 40° × BC 3 cos 17 ° − sin17 ° sin10° tanθ = ⇒ AC = k and BC = 3k cos 17 ° + sin17 ° 4(2 sin10° cos 10° ) cos 20° cos 40° = 1 − tan17 ° sin10° Use pythagoras theorem in ∆ABC, = 4 sin20° cos 20° cos 40° 1 + tan17 ° = AB = ( BC )2 − ( AC )2 sin10° [Divide numerator and = ( 3k )2 − ( k )2 denominator by cos17 °] Q2 sin A cos A = sin2 A 2(2 sin20° cos 20° ) cos 40° = 9k 2 − k 2 ⇒ tanθ = tan( 45° − 17 ° ) = sin10°  tan 45° − tan17 °  = 8k 2 Q tan( 45° − 17 ° ) = 1 + tan 45° tan17 °  2 × sin 40° cos 40° =   sin10° = 2 2k ⇒ tanθ = tan28° sin 80° sin( 90° − 10° ) BC = = Now, cosec C = ⇒ θ = 28° sin10° sin10° AB cos 10° 3k 3 = = cot 10° = = , which is not correct. 24. A and B are positive acute angles sin10° 2 2k 2 2 such that cos 2B = 3 sin 2 A and 2. Suppose we consider ∆ABC is an 3 sin 2A = 2 sin 2B. What is the 28. What is the value of isosceles triangle. cos 48° − cos 12°? value of ( A + 2B )? A π π π π 5 −1 1− 5 (a) (b) (c) (d) (a) (b) 6 4 3 2 4 4 c b 5+1 1− 5 Ê (d) We have, cos 2 B = 3sin A 2 (c) (d) 2 8 and 3 sin2 A = 2 sin2 B 2 sin2 B 3 sin2 A Ê (b) cos 48° − cos 12 ° B C ∴ = 48° + 12 °   48° − 12 °  cos 2 B 3 sin2 A = − 2 sin  sin  ∴ ∠B = ∠C  2   2  sin2 B 2 × sin A cos A Also we have, b cos B = c cos C ⇒ 2 = cos 2 B sin2 A  C + D  Q cos C − cos D = − 2 sin 2   ⇒ b cos C = c cos C ⇒ tan2 B = cot A   [put B = C ]  C − D  ⇒ π tan2 B = tan − A  sin  ⇒ b = c , Which is correct. 2    2    NDA/NA Solved Paper 2020 (I & II) 5 30. Consider the following statements Ê (a) We have, (sin3 θ + cos 3 θ) 1. If in a triangle ABC, A = 2B and a sin x + b cos x = c 2 2 t3 − t5 − (sin5 θ + cos 5 θ) Now, = 2 t5 − t7 (sin θ + cos 5 θ) 5 b = c , then it must be an On dividing both sides by cos x, we get obtuse-angled triangle. a tan2 x + b(1) = c × sec 2 x − (sin7 θ + cos 7 θ) 2. There exists no triangle ABC ⇒ a tan x + b = c(1 + tan x ) 2 2 (sin3 θ − sin5 θ) + (cos 3 θ − cos 5 θ) with A = 40° , B = 65° and = [Q sec θ − tan θ = 1] 2 2 (sin5 θ − sin7 θ) + (cos 5 θ − cos 7 θ) a = sin 40° cosec 15°. ⇒ tan2 x( a − c ) = c − b sin3 θ(1 − sin2 θ) + cos 3 θ (1 − cos 2 θ) c = Which of the above statements c−b sin5 θ (1 − sin2 θ) + cos 5 θ (1 − cos 2 θ) ⇒ tan2 x = …(i) is/are correct? a−c sin3 θ cos 2 θ + cos 3 θ sin2 θ (a) 1 Only (b) 2 Only d −a = 32. What is equal to? sin5 θ cos 2 θ + cos 5 θ sin2 θ (c) Both 1 and 2 (d) Neither 1 nor 2 b−d 2 2 sin2 θ cos 2 θ (sinθ + cos θ) Ê (d) 1. We have, in ∆ABC, (a) sin y (b) cos y = A = 2 B and b = c (c) tan2 y (d) cot 2 y sin2 θ cos 2 θ (sin3 θ + cos 3 θ) Ê (c) We have, b sin y + a cos y = d 2 2 ⇒ Angles opposite to equal sides are sinθ + cos θ t1 = = equal. On dividing both sides by cos 2 y, we get sin3 θ + cos 3 θ t3 ∴ ∠C = ∠B b tan2 y + a(1) = d (sec 2 y) Also, A = 2 B = 2C 35. What is t 12 − t 2 equal to? ⇒ b tan y + a = d (1 + tan y) 2 2 (a) cos 2θ (b) sin 2θ In ∆ABC, ∠A + ∠B + ∠C = 180° [Q sec 2 θ − tan2 θ = 1] (c) 2 cos θ (d) 2 sin θ ⇒ 2C + C + C = 180° Ê (b) t 1 − t 2 = (sinθ + cos θ) 2 2 180° ⇒ tan2 y ( b − d ) = d − a ⇒ C = = 45° 4 d −a − (sin2 θ + cos 2 θ) ⇒ = tan2 y …(ii) ⇒ B = 45° and A = 90° b−d = sin2 θ + cos 2 θ + 2 sinθ cos θ Thus, it shows that ∆ABC is not an p2 33. What is equal to? − (sin2 θ + cos 2 θ) obtuse angle triangle. 2 q = 2 sinθ cos θ = sin2θ Hence, statement 1 is incorrect. (b − c ) (b − d ) ( a − d ) (c − a ) 2. We have A = 40°, B = 65° (a) (b) 36. What is the value of t 10 where (a − d ) (a − c ) ( b − c ) (d − b ) C θ = 45°? (d − a ) (c − a ) (b − c ) (b − d ) 1 1 1 (c) (d) (a) 1 (b) (c) (d) 75° ( b − c ) (d − b ) (c − a ) ( a − d ) 4 16 32 b a Ê (c) Now, t 10 = sin θ + cos θ 10 10 Ê (b) We have, p tan x = q tan y = (sin 45° )10 + (cos 45° )10 40° 65° c On squaring both sides, we get. [Put θ = 45°] A B p2 tan2 y 10 10 = =  1  +  In ∆ABC, 1  q 2 tan2 x    2  2 ∠A + ∠B + ∠C = 180° (d − a ) / ( b − d ) = = 2 5  = 4 = ⇒ 40° + 65° + ∠C = 180° 1 1 1 (c − b ) / ( a − c )  2  2 ⇒ ∠C = 75° 16 [Q from eq. (i) and (ii)] Use sine rule in ∆ABC, (d − a ) ( a − c ) a c = Directions (Q. Nos. 37-39) Read the = ( b − d ) (c − b ) following information and answer the sin 40° sin75° ( a − d ) (c − a ) three items that follow. a = sin 40° cos ec 75°, = (d − b ) ( b − c ) Let α = β = 15°. c Hence, Statement 2 is incorrect. 37. What is the value of sin α + cos β? Directions (Q. Nos. 34-36) Read the 1 1 3 3 following information and answer the (a) (b) (c) (d) Directions (Q. Nos. 31-33) Read the 2 2 2 2 2 2 following information and answer the three items that follow. Let t n = sin n θ + cosn θ Ê (d) sinα + cos β three items that follow. t3 − t5 =  cos β  × 2 Let a sin 2 x + b cos2 x = c, 1 1 sinα + 34. What is equal to?  2 2  b sin 2 y + a cos2 y = d t5 − t7 and p tan x = q tan y t1 t3 = 2 (sinα cos 45° + sin 45° cos β ) (a) (b) t3 t5 = 2 (sin15° cos 45° + sin 45° cos 15° ) 31. What is tan 2 x equal to? t t [Q α = β = 15° ] c−b a−c (c) 5 (d) 1 (a) (b) t7 t7 = 2 sin(15° + 45° ) a−c c−b 3 3 (c) c−a (d) c−b Ê (a) We have, = 2 sin 60° = 2 × = t n = sinn θ + cos n θ 2 2 c−b c−a 6 NDA/NA Solved Paper 2020 (I & II) cot α cot β − 1 38. What is the value of Ê (c) Given, sin x + sin y = cos y − cos x Now, cot(α + β ) = cot α + cot β sin 7α − cos 7β? ⇒ 2 sin x + y  x − y  cos   2 −1 1 1 1 3 3  2   2  = = (a) (b) (c) (d) 3 3 2 2 2 2 2 2 y + x  y − x = − 2 sin  sin  Ê (d) sin7α − cos 7β  2   2  44. The roots α and β of a quadratic x − y  x − y equation, satisfy the relations =  1 sin7 × 15° − 1 cos 7 × 15°  2 ⇒ cos   = sin   2 2   2   2  α + β = α 2 + β 2 and αβ = α 2β 2. x − y = 2 (sin105° cos 45° − cos 105° sin 45° ) ⇒ tan  =1 What is the number of such = 2 sin(105° − 45° )  2  quadratic equations? = 2 sin 60° (a) 0 (b) 2 (c) 3 (d) 4 41. If A is a matrix of order 3 × 5 and B Ê (d) Given, α + β = α + β 2 2 = 2 × 3 = 3 is a matrix of order 5 × 3, then the 2 2 order of AB and BA will and αβ = α 2β 2 39. What is sin (α + 1° ) + cos (β + 1° ) respectively be ⇒ αβ(1 − αβ ) = 0 (a) 3 × 3 and 3 × 3 ⇒ αβ = 0 and αβ = 1 equal to? (b) 3 × 5 and 5 × 3 (a) 3 cos 1° + sin 1° Consider α + β = α 2 + β 2 (c) 3 × 3 and 5 × 5 1 = (α + β )2 − 2αβ (b) 3 cos 1° − sin 1° (d) 5 × 3 and 3 × 5 2 1 Ê (c) Given, matrix A is of order 3 × 5 and When αβ = 0, (c) ( 3 cos 1° + sin 1° ) 2 matrix B is of order 5 × 3. α + β = (α + β )2 − 2 × 0 1 ∴Order of matrix AB = [ A ]3 × 5 [B]5 × 3 ⇒ (α + β ) = (α + β )2 (d) ( 3 cos 1° + sin 1° ) 2 = [ AB]3 × 3 ⇒ (α + β ) [1 − (α + β )] = 0 Ê (c) sin(α + 1° ) + cos(β + 1° ) and order of matrix BA = [B]5 × 3 [ A ]3 × 5 ⇒ (α + β ) = 0 = sin(α + 1° ) + cos(α + 1° ) [Q α = β ] = [BA ]5 × 5 and α + β =1 = sinα cos 1° + cos α sin1° 42. If p 2 , q 2 and r 2 (where p , q , r > 0) It implies that when αβ = 0, gives two + cos α cos 1° − sinα sin1° quadratic equations. are in GP, then which of the = cos 1° (sinα + cos α ) When αβ = 1 following is/are correct? + sin1° (cos α − sinα ) α + β = (α + β) 2 − 2αβ 3 1. p , q and r in GP. = cos1° × ⇒ (α + β ) = (α + β )2 − 2(1) 2 2. ln p , ln q and lnr are in AP. Select the correct answer using the ⇒ (α + β )2 − (α + β ) − 2 = 0 + sin1°  sinα  2 1 1 cos α − code given below ⇒ [(α + β ) − 2 ] [(α + β ) + 1] = 0  2 2  (a) 1 Only (b) 2 Only ⇒ α + β = 2, − 1  3 (c) Both 1 and 2 (d) Neither 1 nor 2 Q sinα + cos α =  Again it implies that, when αβ = 1, gives  2 2 2 2 Ê (c) 1. Given, p , q and r in GP. two quadratic equations. 3 ∴ q2 = p2r 2 ⇒ q 2 = pr Hence, there are total four such = cos 1° + 2 sin1° (sin 45° cosα …(i) 2 quadratic equations formed. ⇒ p, q and r in GP, hence Statement 1 − cos 45° sinα ) is correct. 45. What is the argument of the 3 = cos 1° + 2 sin1° sin( 45° − α ) ln p + ln r ln pr lnq 2 1−i 3 2 2. Now, = = complex number , where 2 2 2 3 1+i 3 = cos 1°+ 2 sin1° sin 30° [Q α = 15° ] [From eq. (i)] 2 lnq i = − 1? 3 2 =2 = lnq = cos 1° + sin1° 2 (a) 240° (b) 210° (c) 120° (d) 60° 2 2 ⇒ ln p, lnq and ln r are in AP, which is 1− i 3 1− i 3 3 1 correct statement. Ê (a) Let z = × = cos 1° + sin1° 1+ i 3 1− i 3 2 2 Hence, both statements are correct. 1 (1 − i 3 ) 2 = ( 3 cos 1° + sin1° ) = 2 43. If cot α and cot β are the roots of 12 − ( i 3 )2 the equation x 2 − 3x + 2 = 0, then 1 − 3 − 2 i 3 −2 − 2 i 3 40. If sin x + sin y = cos y − cos x , = = π what is cot (α + β ) equal to? 1+ 3 4 where 0 < y < x < , then what is 1 1 −1 − i 3 2 (a) (b) (c) 2 (d) 3 = 2 3  x − y 2 tan   equal to? Ê (b) Given, cot α and cot β are the roots of − 3 /2  2  Now, tanθ = y = x 2 − 3 x + 2 = 0. x −1 / 2 1 (a) 0 (b) 2 Now, sum of roots, cot α + cot β = 3 = 3 = 60° (c) 1 (d) 2 and product of roots, cot α cot β = 2 NDA/NA Solved Paper 2020 (I & II) 7 1 q r 3−1 2 Since, given complex number lies in IIIrd ∴ adj A = A = A , quadrant. D = ( p + q + r) 0 r−q p− r Which is incorrect statement. ∴ arg( z) = 180° + 60°= 240° 0 p−q q − r Hence, only Statement 1 is correct. 46. What is the modulus of the complex = ( p + q + r ) [1{( r − q ) (q − r ) 51. The centre of the circle ( x − 2a ) cos θ + i sin θ − ( p − r ) ( p − q )}] ( x − 2 b ) + ( y − 2c ) ( y − 2d ) = 0 is number , where cos θ − i sin θ = ( p + q + r ) [rq − r 2 − q 2 + qr (a) (2 a, 2c ) (b) (2 b, 2d ) (c) ( a + b, c + d ) (d) ( a − b, c − d ) i = − 1? − { p2 − pq − rp + rq }] 1 3 = ( p + q + r )[− p2 − q 2 Ê (c) Given equation ( x − 2 a) ( x − 2 b ) (a) (b) 1 (c) (d) 2 + ( y − 2c ) ( y − 2d ) = 0 is a circle, 2 2 − r 2 + pq + pr + rp] cos θ + i sinθ whose end points of a diameter are Ê (b) Let z =