NTA CUET (UG) 10 Mock Test Sample Question Papers Applied Maths/Mathematics (2024) 9789357288088, 9357288082

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For 2024 Exam

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MATHEMATICS/

APPLIED MATHEMATICS Section II (Domain Specific Subject)

Strictly ar per the Latest Examination Pattern issued by NTA

The ONLY book you need to Ace CUET (UG)

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With 450+ Explanations & Smart Answer Keys

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3rd EDITION

ISBN SYLLABUS COVERED

YEAR 2023-24 “9789357288088”

CUET (UG) CERTIFICATE OF COMMON UNIVERSITY ENTRANCE TEST

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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 National Testing Agency (NTA) has been established in November 2017 under the Societies Registration Act (1860) by the Ministry of Education as a premier, specialist, autonomous, and self-sustained testing organization to conduct entrance examinations for admission/fellowship in higher educational institutions. The Common University Entrance Test (CUET (UG) - 2022) is being introduced for admission into all UG Programmes in all Central Universities for the academic session 2023-24 under the Ministry of Education, (MoE). The Common University Entrance Test (CUET) will provide a common platform and equal opportunities to candidates across the country, especially those from rural and other remote areas, and help establish a better connection with the Universities. A single examination will enable the Candidates to cover a wide outreach and be part of the admissions process to various Central Universities. CUET – UG Computer Based Test (CBT) for the Central Universities is to be conducted by the National Testing Agency (NTA). The curriculum for CUET is based on the National Council of Educational Research and Training (NCERT) syllabus for class 12 only. CUET scores are mandatorily required while admitting students to undergraduate courses in 44 central universities. A merit list will be prepared by participating Universities/organizations. Universities may conduct their individual counselling on the basis of the scorecard of CUET (UG)-2023 provided by NTA.

A few benefits of studying from Oswaal Sample Question Papers • • • • •

Crisp Revision with On-Tips Notes & Updated Mind Maps Valuable Exam Insights with Latest Solved Papers 2023 100% Exam Readiness with 10 Solved Sample Question Papers Extensive Practice with 650+ NCERT-based MCQs Concept Clarity with 450+ Explanations & Smart Answer Keys

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. We promise to always strive towards ‘Making Learning Simple’ for all of you. Wish you all Happy Learning!

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-Team Oswaal Books

Contents l Oswaal Books Expert Tips to Crack CUET (UG) in the first Attempt l Latest Syllabus

l CUET Solved Paper 2023

l CUET Solved Paper 2022 l CUET Solved Paper 2021 (23rd SEP. 2021–Slot-2 UIQP02)

5 - 5 6 - 8

11 - 32

33 - 50 51 - 56

Sample Question Papers

1 - 6 7 - 13 14 - 19 20 - 25 26 - 32 33 - 39 40 - 46 47 - 52 53 - 58 59 - 64

l Sample Question Paper - 1 l Sample Question Paper - 2 l Sample Question Paper - 3 l Sample Question Paper - 4 l Sample Question Paper - 5 l Sample Question Paper - 6 l Sample Question Paper - 7 l Sample Question Paper - 8 l Sample Question Paper - 9 l Sample Question Paper - 10

Solutions 65 - 78 79 - 90 91 - 103 104 - 115 116 - 128 129 - 140 141 - 153 154 - 164 165 - 177 178 - 191

l Sample Question Paper - 1 l Sample Question Paper - 2 l Sample Question Paper - 3 l Sample Question Paper - 4 l Sample Question Paper - 5 l Sample Question Paper - 6 l Sample Question Paper - 7 l Sample Question Paper - 8 l Sample Question Paper - 9 l Sample Question Paper - 10 

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Books Expert Tips to Crack Oswaal BooksOswaal Expert Tips to Crack CUET (UG) in the First A empt CUET (UG) in the First Attempt

Excited about your UG but unsure if you will get admission to your preferred university? In a major announcement by the chairman of the University Grants Commission, the Naonal Tesng Agency will be conducng the Common Universies Entrance Test (CUET (UG) 2022) for undergraduate programs in Central Universies for the upcoming academic session. However, the UGC Chairperson also stated that CUET (UG) will not just be limited to admissions to Central Universies. Many prominent private universies have indicated that they would also like to adopt a common entrance exam for undergraduate admissions and take admissions on the basis of CUET (UG) scores. This makes CUET (UG) a very important examinaon in itself and hence it becomes mandatory to be aware of the ps & tricks that could help you ace the exam on the first a† empt.

The first step is to understand The pa† ern of the examinaon. CUET includes three secons, secon 1 includes queson based on languages, secon 2 includes 27 domain-based subjects and secon 3 includes General Test. The syllabus of the upcoming Common University Entrance Test, CUET 2022, will be completely based on the syllabus of class 12 th . No queson will be asked from class 11th syllabus.

While preparing for the exam, it is i m p o r t a n t t o i d e n f y t h e important topics and pracce important quesons from those t o p i c s . P r a c c e i m p o r t a n t q u e s o n s t h r o u g h O s w a a l Q u e so n B a n k a n d S a m p l e Q u e s o n P a p e r s , L i s n g topics also helps in idenfying the weak areas that need special effort and me. The aspirants can start preparing to focus on the areas that they consider to be tough, followed by the ones that are their strengths.

Make a habit of preparing notes f ro m t h e b e g i n n i n g o f t h e preparaon. It will not only help in making the study systemac but also make the revision of the syllabus easy even when you might have limited me to revise.

Collecng and preparing from the appropriate study material cannot be ignored as irrelevant. The books chosen by the aspirants to study from should be on the lines of the current syllabus and the ones that could help you with swi• revision before the examinaon.

Make sure to revise as much as possible. The revision will help the aspirants in keeping the concepts fresh in their minds unl the day of the final examinaons. They may refer to a few good pracce quesons and concise revision notes to achieve their desired results.

Devote a sufficient amount of me to all the secons of the examinaons. This requires a wellmade plan and an honest adherence to the said plan. Priorize the most important topics or the topics that the aspirants are not familiar with to be able to master them in me.

With this said, an important queson that is gaining ground amongst students who will be appearing for this exam is if they should take coaching to get themselves ready for the exams. The answer is a simple no, the exam will simply not require any coaching as it is completely based on the Class 12th syllabus which will be quite fresh in students' minds as they will be just out of school. All they need is a good revision and pracce of quesons from Oswaal Queson Bank and Sample Queson Papers for CUET (UG) preparaons.

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Latest Syllabus MATHEMATICS/APPLIED MATHEMATICS - 319 Note: There will be one Question Paper which will contain Two Sections i.e. Section A and Section B [B1 and B2]. Section A will have 15 questions covering both i.e. Mathematics/ Applied Mathematics which will be compulsory for all candidates Section B1 will have 35 questions from Mathematics out of which 25 questions need to be attempted. Section B2 will have 35 questions purely from Applied Mathematics out of which 25 question will be attempted. SECTION A 1. Algebra (i) Matrices and types of Matrices (ii) Equality of Matrices, transpose of a Matrix, Symmetric and Skew Symmetric Matrix (iii) Algebra of Matrices (iv) Determinants (v) Inverse of a Matrix (vi) Solving of simultaneous equations using Matrix Method 2. Calculus (i) Higher order derivatives (ii) Tangents and Normals (iii) Increasing and Decreasing Functions (iv) Maxima and Minima 3. Integration and its Applications (i) Indefinite integrals of simple functions (ii) Evaluation of indefinite integrals (iii) Definite Integrals (iv) Application of Integration as area under the curve 4. Differential Equations (i) Order and degree of differential equations (ii) Formulating and solving of differential equations with variable separable 5. Probability Distributions (i) Random variables and its probability distribution (ii) Expected value of a random variable (iii) Variance and Standard Deviation of a random variable (iv) Binomial Distribution 6. Linear Programming (i) Mathematical formulation of Linear Programming Problem (ii) Graphical method of solution for problems in two variables (iii) Feasible and infeasible regions (iv) Optimal feasible solution Section B1: Mathematics UNIT I: RELATIONS AND FUNCTIONS 1. Relations and Functions Types of relations: Reflexive, symmetric, transitive and equivalence relations. One to one and onto functions, composite functions, inverse of a function. Binary operations. 2. Inverse Trigonometric Functions Definition, range, domain, principal value branches. Graphs of inverse trigonometric functions. Elementary properties of inverse trigonometric functions.

UNIT II: ALGEBRA 1. Matrices Concept, notation, order, equality, types of matrices, zero matrix, transpose of a matrix, symmetric and skew symmetric matrices. Addition, multiplication and scalar multiplication of matrices, simple properties of addition, multiplication and scalar multiplication. Non-commutativity of multiplication of matrices and existence of non-zero matrices whose product is the zero matrix (restrict to square matrices of order 2). Concept of elementary row and column operations. Invertible matrices and proof of the uniqueness of inverse, if it exists; (Here all matrices will have real entries). 2. Determinants Determinant of a square matrix (up to 3 × 3 matrices), properties of determinants, minors, cofactors and applications of determinants in finding the area of a triangle. Adjoint and inverse of a square matrix. Consistency, inconsistency and number of solutions of system of linear equations by examples, solving system of linear equations in two or three variables (having unique solution) using inverse of a matrix. UNIT III: CALCULUS 1. Continuity and Differentiability Continuity and differentiability, derivative of composite functions, chain rule, derivatives of inverse trigonometric functions, derivative of implicit function. Concepts of exponential, logarithmic functions. Derivatives of log x and ex. Logarithmic differentiation. Derivative of functions expressed in parametric forms. Second-order derivatives. Rolle’s and Lagrange’s Mean Value Theorems (without proof) and their geometric interpretations. 2. Applications of Derivatives Applications of derivatives: Rate of change, increasing/ decreasing functions, tangents and normals, approximation, maxima and minima (first derivative test motivated geometrically and second derivative test given as a provable tool). Simple problems (that illustrate basic principles and understanding of the subject as well as real-life situations). Tangent and Normal. 3. Integrals Integration as inverse process of differentiation. Integration of a variety of functions by substitution, by partial fractions and by parts, only simple integrals of the type – dx



dx

dx

dx

dx

∫ x 2 ± a2 , ∫ x 2 ± a2 , ∫ a2 − x 2 , ∫ ax 2 + bx + c , ∫ ax 2 + bx + c , ( px + q )

( px + q )

∫ ax 2 + bx + c dx , ∫ ax 2 + bx + c dx , ∫

a 2 ± x 2 dx

and ∫ x 2 − a 2 dx , ∫ ax 2 + bx + c dx and

∫

ax 2 + bx + c dx

to be evaluated. Definite integrals as a limit of a sum. Fundamental Theorem of Calculus (without proof). Basic properties of definite integrals and evaluation of definite integrals.

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Contd... 4. Applications of the Integrals Applications in finding the area under simple curves, especially lines, arcs of circles/parabolas/el- lipses (in standard form only), area between the two above said curves (the region should be cleraly identifiable). 5. Differential Equations Definition, order and degree, general and particular solutions of a differential equation. Formation of differential equation whose general solution is given. Solution of differential equations by method of separation of variables, homogeneous differential equations of first order and first degree. Solutions of linear differential equation of the type –



dy + Py = Q, where P and Q are functions of x or dx constant dx dx + Px = Q, where P and Q are functions of y or dy

constant UNIT IV: VECTORS AND THREE-DIMENSIONAL GEOMETRY 1. Vectors Vectors and scalars, magnitude and direction of a vector. Direction cosines/ratios of vectors. Types of vectors (equal, unit, zero, parallel and collinear vectors), position vector of a point, negative of a vector, components of a vector, addition of vectors, multiplication of a vector by a scalar, position vector of a point dividing a line segment in a given ratio. Scalar (dot) product of vectors, projection of a vector on a line. Vector (cross) product of vectors, scalar triple product. 2. Three-dimensional Geometry Direction cosines/ratios of a line joining two points. Cartesian and vector equation of a line, coplanar and skew lines, shortest distance between two lines. Cartesian and vector equation of a plane. Angle between (i) two lines, (ii) two planes, (iii) a line and a plane. Distance of a point from a plane. Unit V: Linear Programming Introduction, related terminology such as constraints, objective function, optimization, different types of linear programming (L.P.) problems, mathematical formulation of L.P. problems, graphical method of solution for problems in two variables, feasible and infeasible regions, feasible and infeasible solutions, optimal feasible solutions (up to three non-trivial constrains). Unit VI: Probability Multiplications theorem on probability. Conditional probability, independent events, total probability, Baye’s theorem. Random variable and its probability distribution, mean and variance of haphazard variable. Repeated independent (Bernoulli) trials and Binomial distribution. Section B2: Applied Mathematics Unit I: Numbers, Quantification and Numerical Applications A. Modulo Arithmetic l Define modulus of an integer l Apply arithmetic operations using modular arithmetic rules

B. Congruence Modulo l Define congruence modulo l Apply the definition in various problems C. Allegation and Mixture l Understand the rule of allegation to produce a mixture at a given price l Determine the mean price of a mixture l Apply rule of allegation D. Numerical Problems l Solve real life problems mathematically E. Boats and Streams l Distinguish between upstream and downstream l Express the problem in the form of an equation F. Pipes and Cisterns l Determine the time taken by two or more pipes to fill or G. Races and Games l Compare the performance of two players w.r.t. time, l distance taken/ distance covered/ Work done from the given data H. Partnership l Differentiate between active partner and sleeping partner l Determine the gain or loss to be divided among the partners in the ratio of their investment with due l consideration of the time volume/ surface area for solid formed using two or more shapes I. Numerical Inequalities l Describe the basic concepts of numerical inequalities l Understand and write numerical inequalities UNIT II: ALGEBRA A. Matrices and types of matrices l Define matrix l Identify different kinds of matrices B. Equality of matrices, Transpose of a matrix, Symmetric and Skew symmetric matrix l Determine equality of two matrices l Write transpose of given matrix l Define symmetric and skew symmetric matrix UNIT III: CALCULUS A. Higher Order Derivatives l Determine second and higher order derivatives l Understand differentiation of parametric functions and implicit functions Identify dependent and independent variables B. Marginal Cost and Marginal Revenue using derivatives l Define marginal cost and marginal revenue l Find marginal cost and marginal revenue C. Maxima and Minima l Determine critical points of the function l Find the point(s) of local maxima and local minima and corresponding local maximum and local minimum values l Find the absolute maximum and absolute minimum value of a function

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Contd... UNIT IV: PROBABILITY DISTRIBUTIONS A. Probability Distribution l Understand the concept of Random Variables and its Probability Distributions l Find probability distribution of discrete random variable B. Mathematical Expectation l Apply arithmetic mean of frequency distribution to find the expected value of a random variable C. Variance l Calculate the Variance and S.D. of a random variable UNIT V: INDEX NUMBERS AND TIME BASED DATA A. Index Numbers l Define Index numbers as a special type of average B. Construction of Index numbers l Construct different type of index numbers C. Test of Adequacy of Index Numbers l Apply time reversal test UNIT VI: UNIT V: INDEX NUMBERS AND TIME BASED DATA A. Population and Sample l Define Population and Sample l Differentiate between population and sample l Define a representative sample from a population B. Parameter and Statistics and Statistical Interferences l Define Parameter with reference to Population l Define Statistics with reference to Sample l Explain the relation between Parameter and Statistic l Explain the limitation of Statistic to generalize the estimation for population l Interpret the concept of Statistical Significance and Statistical Inferences l State Central Limit Theorem l Explain the relation between Population Sampling Distribution-Sample UNIT VII: INDEX NUMBERS AND TIME-BASED DATA A. Time Series l Identify time series as chronological data



B. Components of Time Series l Distinguish between different components of time series C. Time Series analysis for univariate data l Solve practical problems based on statistical data and Interpret UNIT VIII: FINANCIAL MATHEMATICS A. Perpetuity, Sinking Funds l Explain the concept of perpetuity and sinking fund l Calculate perpetuity l Differentiate between sinking fund and saving account B. Valuation of Bonds l Define the concept of valuation of bond and related terms l Calculate value of bond using present value approach C. Calculation of EMI l Explain the concept of EMI l Calculate EMI using various methods D. Linear method of Depreciation l Define the concept of linear method of Depreciation l Interpret cost, residual value and useful life of an asset from the given information l Calculate depreciation UNIT IX: LINEAR PROGRAMMING A. Introduction and related terminology l Familiarize with terms related to Linear Programming Problem B. Mathematical formulation of Linear Program ming Problem l Formulate Linear Programming Problem C. Different types of Linear Programming Problems l Identify and formulate different types of LPP D. Graphical Method of Solution for problems in two Variables l Draw the Graph for a system of linear inequalities involving two variables and to find its solution graphically E. Feasible and Infeasible Regions l Identify feasible, infeasible and bounded regions F. Feasible and infeasible solutions, optimal feasible solution l Understand feasible and infeasible solutions l Find optimal feasible solution

Don't Stop Reading !

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You never know what might be asked in the exam. To download Chapter-wise Mind Maps scan the code below

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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

Sahitya Sangam, 9419190177 BOKARO

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

(9)

Contd... 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

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

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

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

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

Trimurti Book World, 9437034735

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

2705

BARIPADA

UTTAR PRADESH DEHRADUN

( 10 ) 0808

NAGPUR

CUET (UG) Exam Paper 2023 National Testing Agency Held on 26th May 2023

MATHEMATICS/APPLIED MATHEMATICS Solved

[This includes Questions pertaining to Domain Specific Subject only] Time Allowed: 45 Mins. 

Maximum Marks: 200

General Instructions : (i)

Section A will have 15 questions covering both i.e., Mathematics/Applied Mathematics which will be compulsory for all candidates. (ii) Section B1 will have 35 questions from Applied Mathematics out of which 25 questions need to be attempted. Section B2 will have 35 questions purely from Mathematics out of which 25 questions will be attempted. (iii) Correct answer or the most appropriate answer : Five marks (+ 5) (iv) Any incorrect option marked will be given minus one mark (– 1). (v) Unanswered/Marked for Review will be given no mark (0). (vi) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options. (vii) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question. (viii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (ix) Calculator / any electronic gadgets are not permitted.

Section - A



Mathematics/Applied Mathematics

1. The area enclosed by the ellipse (1) 15p (3) 18p Ans. Option (2) is correct

x2 9

2

+

y2

6

2

= 1 is:

(2) 54p 3 (4) π 2

 0 1 3  3. The matrix A=  1 0 0  is a  3 0 0  (1) (2) (3) (4)

Diagonal matrix Symmetric matrix Skew-symmetric matrix Scalar matrix

2 2 Explanation: Equation of Ellipse x  y  1 a2 b2 Ans. Option (3) is correct We know that Area of ellipse = pab Explanation: A matrix can be skew symmetric only x2 y2 On comparing given ellipse equation 2  2  1 if it is square and if the transpose of a matrix is equal 9 6 to the negative of itself, then the matrix is said to with above ellipse equation a=9 and b=6, then be skew symmetric matrix. This means that for a Area enclosed by the ellipse = pab matrix to be skew symmetric, A’=−A = p×9×6 Also, The diagonal elements of a skew symmetric = 54p matrix are equal to zero.

2. If

Here in question,

Ans. Option (3) is correct

 0 1 3    A =  1 0 0   3 0 0 

A is a square matrix of order 3, B=kA and |B|=x|A| then, (1) x=2k (2) x=k2 3 (3) x=k (4) x=3k

Explanation: From the properties of the determinants, we know that |kA|=kn |A|, where n is the order of the determinant. Here, n = 3, therefore, on comparing |B|=x|A| with k3|A| the answer x is k3

 0 1 3    A =  1 0 0   3 0 0 

OSWAAL CUET (UG) Sample Question Papers, Mathematics/App. Math.

12

 0 1 3    A' =  1 0 0   3 0 0 

A'

 0 1 3    = −  1 0 0   3 0 0 



x2−2x−x+2=0



x(x−2)−1(x−2)=0

(x−2)(x−1)=0

x=2, 1 15

2

4  2  dy  d y  1     2  is: dx    dx 

function ≤x is equal to: 1 1 (1) (2) (3) 1 4 2

3 4

Explanation:

Ans. Option (3) is correct Explanation: We know that degree of differential equation is the power of the highest derivative in a differential equation. Therefore, Degree of given differential equation is 2.

5. The differential equation dy + x = 0, represents dx y the family of curves: x (1) x2−y2=C (2) =C y (4) x2+y2=C

Ans. Option (4) is correct Explanation:

dy x =0 + dx y



dy x  dx y



−∫ydy = ∫xdx



y2 C x2   2 2 2 C x2 y2   2 2 2 (C1=C)

2 3x  0 , then the values of x are: x 1

(1) 1 and 3 (2) 1 and 2 (3) 2 and 3 (4) 3 and 0 Ans. Option (2) is correct Explanation:

1

15

0

0

1

1

15

0

1

 [x]dx =  [x]dx   [x]dx

=  [0]dx   [1]dx 0  [ x]115 =

= 0 + (1·5−1) = 0·5 = 1/2

8. If matrix A is of order 2 × 3 and B of order 3 × 2, then (1) (2) (3) (4)

AB, BA both are defined and are equal AB is defined but BA is not defined AB is not defined but BA is defined AB, BA are defined but are not equal

Explanation: We know that if A, B are, respectively m×n, k×l matrices, then both AB and BA are defined if and only if n = k and l=m and If AB and BA are both defined, it is not necessary that AB = BA.

on integrating both sides

6. If

15

Ans. Option (4) is correct

ydy = −xdx −ydy = xdx

C=x2+y2 x2+y2=C

(4) 0

Ans. Option (2) is correct 2. 4.

(3) xy=C

[x] denotes the greatest integer

0

4. The degree of the differential equation

1 2

(2)(1)−(x)(3−x)=0 2−3x+x2=0

7.  [x]dx where

A' = −A

1. 3.



2 3x 0 x 1

9. In

a box containing 100 bulbs. 10 are defective. Then the probability, that out of a sample of 5 bulbs none is defective, is: 5

1 (2)   2 Ans. Option (3) is correct 1(1) 10−1

5  9  (3)  9  (4)    10   10 

Explanation: The repeated selections of defective bulbs from a box are Bernoulli trials. Let X denote the number of defective bulbs out of a sample of 5 bulbs. Probability of getting a defective bulb, p = 10 = 1 100 10 1 9 \ q=1−p=1− = 10 10 Clearly, X has a binomial distribution with n = 5 and p = 1 10

13 13

Solved Paper-2023

C. Maximum value of f(x) = 6−x2 is 6 when x=0. Such that option C from list I belongs to option I of list II. D. There is no maximum value of f(x) = x3+1 Such that option D from list I belongs to option III of list II.

\ p (X=r)=n Cr qn−rpr  9  p( X  r )  5Cr    10 

5 r

 1     10 

r

p (none of the bulb is defective) = p(X=0)  9  p( X  0 )  5C0    10 

5 0

 1     10 

12. The mean number of heads in two tosses of a coin

0

is:

1 2 Ans. Option (3) is correct (1) 2

5



5  9   9   (1)   (1)    10  10   

2 10. If y = 1 , then d y2 at x=2 is: x +1 dx 3 2 2 (1) (2) (3) 2 9 27

(4)

3 8

So,

1 y = x +1

d.w.r to x again d.w.r to x

d2y

Xi

0

1

2

Pi

1/4

2/4

1/4

1 3

( x + 1) dx 2 1 d 2 y  = 2×  ( 2 + 1)3 dx 2 x  2 2 27

1 , then for x > 1, f(x) is: 1−x decreasing constant increasing neither decreasing nor increasing

13. If f(x) = (1) (2) (3) (4)

Ans. Option (1) is correct

11. Match List I with List II

Explanation:

LIST I

LIST II

A Maximum value of f(x)=−|x +1|+3 2+5

3 2

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



= −1×−2 ×

=

(4)

Mean = ∑XiPi = 0 ×

dy 1 = dx ( x  1)2



(3) 1

Explanation: Total possible outcomes = {TT, TH, HT, HH} No head P(0) = 1 (TT) One Head P(1) = 2 (HT, TH Two Head P(2) = 1 (HH)

Ans. Option (3) is correct Explanation:

(2)

I 6

B Minimum value of f(x)=(2x−1)

II 5

C Maximum value of f(x) = 6−x2

III no maximum value

D Maximum value of f(x) =x3+1

IV 3

Choose the correct answer from the options given below: (1) A-IV, B-II, C-I, D-III (2) A-III, B-IV, C-I, D-II (3) A-I, B-II, C-III, D-IV (4) A-II, B-III, C-IV, D-I

Ans. Option (1) is correct Explanation: A. Maximum value of f(x) = −|x+1|+3 is 3 when x=−1 Such that A is belongs to IV. B. Minimum value of f(x) = (2x−1)2 +5 is 5 when 1 x= . 2 Such that option B from list I belongs to option II of list II.

f(x) =

1 1−x

for x>1 ® −x < −1 1−x < 1−1 f(x) is decreasing 1−x < 0

14. In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed. and X= 1 if he is in favour. Then. E (X) is : 1 7 1 7 (1) (2) 2 (3) (4) 10 3 11

Ans. Option (1) is correct Explanation: Case 1: If selected member opposed 30 p(X=0) = 100 Case 2: If selected member favours 70 p(X=1) = 100 Required potability distribution is

X

0

1

p(X)

30 100

70 100

OSWAAL CUET (UG) Sample Question Papers, Mathematics/App. Math.

14

(1) (2) (3) (4)

Then

70 30 E(x) = 0× + 1× 100 100 =

70 100

x =0, y=20, Max Z= 60 x=10, y=15, Max Z=65 x=10, y=20. Max Z=70 x=15, y=10, Max Z=60

Ans. Option (2) is correct Explanation: At (x, y)= (10, 15)

7 = 10

Z = 2x+3y = 2(10) + 3(15) = 20+45 = 65 Here, objective function is Max. at x=10 and y=15.

15. The solution of a LPP with basic feasible solutions (0, 0), (10, 0), (0, 20). (10, 15) and objective function Max Z=2x+3y is:

Note: (1920) is not given solution of feasible solution.

Section - B1



Mathematics

Explanation: Here, X represents the number of sixes obtained when two dice are thrown simultaneously. Therefore, X can take the value of 0, 1 or 2 \ p(x=0) = p (not getting six on any of the dice) 5 5 = × 6 6

16. The linear constraints, for which the shaded area in the figure is the feasible region of an LPP. are: y 80

40

25 36 p(X=1) = p (six on first dice and number after than six on second dice) + p (number other than six on first dice and six on second dice)

30

=

1 5 5 1    6 6 6 6

=

5 5 + 36 36

=

10 36

70

=

60 50

(30, 20)

20 10

x

x' 10

20

30 40 50

60

p(X=2) = P (six on both the dice) 1 1 = × 6 6 1 = 36

y'

(1) (3)

x + y ≥ 50 2x + y ≤ 80 x, y ≥ 0 x + y ≤ 50 2x + y ≤ 80 x, y ≥ 0

(2) (4)

x + y ≤ 50 2x + y ≥ 80 x, y ≥ 0 x + y ≥ 50 2x + y ≥ 80 x, y ≥ 0

Ans. Option (3) is correct Explanation: Shaded region is bounded between Four Corner points (0,0), (40,0), (30, 20) and (0,50) Hence, x + y ≤ 50 2x + y ≤ 80 x; y ≥ 0 are required equations.

17. Two dice are thrown simultaneously. If X denotes the number of sixes, then the variance of X is: 5 7 1 2 (1) (2) (3) (4) 18 18 3 3 Ans. Option (2) is correct

X

0

1

2

p(X)

25 36

10 36

1 36

Mean (expected value) m = Xi p(Xi)

=0 × =

25 1 10 +1 × +2 × 36 36 36

2 12 10 1 + = = 36 36 36 3

Variance (s2) = ∑p(Xi) (Xi)2 − (Mean)2

∑p(Xi) (Xi)2 − ∑(Xi p(Xi))2

1  1  2 25 2 10 = ( 0 )  36  1  36  4  36    3     

2

Solved Paper-2023 =

Now Angle between the line and plane

10 1 1 + − 36 9 9

( 3)( 2 )  ( 2 )(10 )  ( 6 )( 11) sin q =

5 10 = = 18 36

2sin x + 2x sin2 x + x2 sin (sin x + x) + sin x + x sin2 x + 2sin x + x

19. Which

of the following statements is NOT CORRECT. (1) A row matrix has only one row. (2) A diagonal matrix has all diagonal elements equal to zero. (3) A symmetric matrix is a square matrix satisfying certain conditions. (4) A skew-symmetric matrix has all diagonal elements equal to zero.

Explanation: Because, we know that a square matrix in which every element except the diagonal elements are zero is called diagonal Matrix.

20. The

angle between the line

−8 21

1  8  q = sin    21 

1  8  (1) cos    21   21  (3) cos1    82 

1  8  (2) sin  21    21   (4) sin 1   82  

Ans. Option (2) is correct Explanation: We know that the Angle q between the line x  x1 y  y1 z  z1   and b1 c1 a1

plane a2x+b2y+c2z + D = 0 is, sin q =

21. Match List I with List II LIST I sin x dx A.  1  cos x 1

 1  tan x dx

B.



C.



a1a2 + b1b2 + c1c2 2 a 1 +b 21 + c 21 a 22 +b 22 + c 22

x  ( 2 ) y  3 z  ( 5)   3 2 6 plane 2x+10y −11z−5=0 when, a1=3, b1=2, c1=6 and a2=2, b2=10, c2=−11,

etan

1

LIST II I. etan

−1

x+

C

II. log (log x +1) + C

x

dx III. −log |1+cos x| + C 1 x 1 dx IV. x − 1 log|cos x −sin x + C x  x log x 2 2 2

Choose the correct answer from the options given below: (1) A-II, B-III, C-IV, D-I (2) A-III, B-IV, C-I, D-II (3) A-I, B-II, C-III, D-IV (4) A-IV, B-I, C-III, D-II

x2 y3 z5   3 2 6

and the plane 2x + 10y − 11z = 5 is:

49 225

sin q =

D.

Ans. Option (2) is correct

6  20  66 40 7  15

Explanation: If f: R→R, f(x) = sin x + x f(f(x)) = f (sin x + x) = sin (sin x + x) + sin x + x

( 3)  ( 2 )2  ( 6 )2 ( 2 )2  (10 )2  ( 11)2

sin q =

Ans. Option (3) is correct



2

sin q =

18. If f : R→R is defined by f(x) = sin x + x. then f(f (x)) is: (1) (2) (3) (4)

15 15

Ans. Option (2) is correct Explanation: A.

sin x

 1  cos x dx

1+cos x = t −sin xdx = dt 1   dt = t = − log t + C = − log |1+cos x| + C

let

   Here, A Match with III 1 B.  dx 1  tan x 1  sin x dx = 1 cos x cos x =  cos x  sin x dx 1 2 cos x = dx 2  cos x  sin x 1 cos x  cos x  sin x  sin x dx = 2 cos x  sin x

OSWAAL CUET (UG) Sample Question Papers, Mathematics/App. Math.

16

x = 0, x = 1 and x = −1 Therefore, given function is not defined at Three Values of x Hence, f(x) is discontinuous at Exactly three points.

1 (cos x  sin x )  (cos x  sin x ) dx = 2 (cos x  sin x ) 1  cos x  sin x cos x  sin x   =   dx 2   cos x  sin x cos x  sin x  1 cos x  sin x 1 = 1 dx   dx 2 2 cos x  sin x let cos x−sin x = t (−sin x−cos x)dx = dt −(cos x+sin x)dx = dt (cos x+sin x)dx = −dt 1 1 1 = 1 dx     dt 2 2 t 1 1 x  log | t |  C = 2 2

23. The

area of the region bounded by the lines x=2y +3, x=0, y=l and y=−1 is: (1) 4 sq. units (2) 6 sq. units 3 (3) 8 sq. units (4) sq. units 2 Ans. Option (2) is correct 1

Explanation: Required area, A=  ( 2 y  3)dy 1

1

 2y2   3y   =  2  1 1

1 x = − log | cos x−sin x|+C 2 2    Here, B Match with IV C.



etan

let tan-1x = t

1

x

1  x2

 y2  3y  =   1

= [1+3−1+3]



= 6 sq units.

dx

Y y=1

1 dx = dt 1 + x2

X'

X

x=

2y

+

−1 = etan x + C

    Here, C Match with I 1 D.  x  x log x dx

1 dx = dt x 1 = ∫ t dt = log |t|+ C

x −1 function f(x)= , x≠1, f(1)=1, is x( x 2 − 1) discontinuous at: (1) Exactly one point (2) Exactly two points (3) Exactly three points (4) No point

22. The

Ans. Option (3) is correct 2

x( x − 1)

LIST II I.

2

The value of i  j  k  j  k  i

II.

4

The value of a for which the vectors 2i  3 j  4 k and ai  6 j  8 k are

III.

0

The value of l for which the vectors 2i + j + k and 2i  4 j   k are

IV.

6

A.

The area of parallelogram determined by vectors 2i and 3 j

B. C.

D.

= log |1+log x|+ C

x −1

LIST I

 





collinear.

   Here, D Match with II

Explanation: f(x)=

Y'

24. Match List I with List II

1 =  x(1  log x) dx let 1+log x= t 1   0   dx = dt x 



y = –1

3

= ∫et dt = et + C

Thus given function

f(x) is discontinuous when f(x) is not defined when denominator is equal to zero. \ x (x2−1) = 0 x (x−1) (x+1) = 0

perpendicular

Choose the correct answer from the options given below: (1) A-I, B-II, C-III, D-IV (2) A-II, B-I, C-III, D-IV (3) A-III, B-IV, C-II, D-I (4) A-IV, B-I, C-II, D-III

Ans. Option (4) is correct Explanation:

A. Area of parallelogram = |2 i × 3 j | = |6 k | = 6

17 17

Solved Paper-2023

Here, A Match with IV

B.

k  k  i  i i  j   k   j  k   i =

Here, B Match with I

= 1+1 = 2

C. Vectors are collinear when

2 3 4   a 6 8 a = 4 Here, C Match with II D. Vectors are perpendicular when (2)(2) + (1)(−4)+(1)(l) = 0 4−4 + l = 0 l = 0 Here, D Match with III  cos  sin   2 25. If the matrix A=   , then A is equal   sin  cos   to:  cos 2 sin 2  (1)     sin 2 cos 2 

dV = 3a2 = DV = 3a2Da, da Now it is given that Da = 0.04a \ DV = 3a2Da DV = 3a2(0.04a) DV = 0.12a3 Approximate volume of a cube = V + DV = a3 + .12a3 = 1.12a3 m3 Þ

sin 2   cos 2 

 cos   sin  cos   sin   (4)   sin   sin  cos   sin   Ans. Option (1) is correct Explanation:

28. Match List I with List II

 cos  sin   A =   sin  cos   

LIST I

 cos  sin    cos  sin   A2 =       sin  cos     sin  cos   

sin  cos   sin  cos      sin  cos   sin  cos    sin 2   cos2 

 =

cos2   sin 2 

cos2   sin 2  2 sin  cos     =  2 sin  cos  cos2   sin 2    cos 2 sin 2  =     sin 2 cos 2 

(2) 12

(3) 16

LIST II I.



 2

II.



 6

A.

sin−1 x+cos−1 x, x ∈[−1,1]

B.

tan−1

C.

13   cos−1  cos  6  

III.

π 2

D.

 1 sin−1     2

IV.

π 6



3 −cos−1  3



Choose the correct answer from the options given

26. The maximum slope of the curve y = −x3 + 3x2 + 9x − 27 is: (1) 0

on increasing the side by 4% is: (1) 1.04a3 m3 (2) 1.004a3 m3 3 3 (3) 1.12a m (4) 1.12a2 m3

Explanation: Volume (V) of a cube with side a is given by, V=a3

2

 cos 2 (3)    sin 2

27. The approximate volume of a cube of side a metres

Ans. Option (3) is correct

 cos  sin    (2)    sin 2  cos2   2

and equate it to zero Þ y''=−6x+6 Now, y''=0 Þ −6x+6=0 Þ x=1 Now, Þ y'''=−6 Þ y'''(1)=−6 Þ y'''(1)0, is: dx y = 2x3 + cx½ y = x3 + cx½ y = 2x3 + cx−½ y = x3 + cx−½

1(−2l+5)+3(−4l+3)+2(10−3)=0 −2l+5−12l+9+14=0 −2l+5−12l+23=0 −14l=2B l=2

Ans. Option (3) is correct

50. A coin is tossed 7 times. The probability of getting

dy Explanation: 2x +y=14x3 dx dy 1  3 2x   y  = 14x dx 2 x   dy 1 + y = 7x2 dx 2 x On comparing with 1 dy + Py = as, P= and 2x dx

at least 4 heads is:

.... (i)

Q = 7x2 I.F = e∫Px 1 ∫ dx = e 2x = e1 / 2 log x = e log x = x

On Multiply

x on both sides in eq. (i)

dy x  y  7x2 x dx 2 x 1 dy + ×y=7x5/2 x dx 2 x x

dy ( x y)= 7x5/2 dx On Integrating both sides

y=

2x3

x +c

+

(2)

3 4

(3)

1 4

(4)

1 2

Ans. Option (4) is correct Explanation: Use the binomial distribution directly. Let us assume that the number of heads is represented by x (where a result of heads is regarded at success and in this case x≥4. Assuming that the coin is unbiased you have a probability of failure ‘q’ is 1/2 (where q is considered as failure). The number of trials is represented by the letter n and for this question n = 7, you have a probability of success ‘p’ (where p is considered as success is 1/2. Now just use the probability function for a binomial distribution p(X=x)=nCxPx qn−x Using the information in the problem we get, p(x≥4)=p(x=4)+p(x=5)+p(x=6)+p(x=7)



1 4 1 3 1 1 ) ( ) +7c ( )5( )2 5 2 2 2 2

+7c ( 6

1 6 1 1 1 ) ( ) +7c ( )7 7 2 2 2

1 7 2 ) [7c4+ 7c5+ 7c6+ 7c7] 1 =( )7[35+21+7+1] 2 1 [64] 128 =(

c x

x−½

  the vectors a  i  3 j  2 k , b  2i  j  k and  c  3i  5 j  2 k be coplanar. Then l is equal to:

49. Let

(1) −1 (3) −2

5 8

4

x7 / 2 x y=7 7 + c 2 x7 / 2

(1)

7c (

x y=7∫x5/2dy

y= 2

and

[abc]=0

48. The solution of the differentiable equation 2x (1) (2) (3) (4)

23 23

(2) 1 (4) 2

=

1 2

OSWAAL CUET (UG) Sample Question Papers, Mathematics/App. Math.

24

Section - B2



Applied Mathematics

51. Consider the following data: Commodity

Price year 2010

Price year 2016

Quantity Quantity year 2010 year 2016

A

1

2

10

13

B

5

10

12

16

C

6

10

15

18

The Laspeyre’s price index number for year 2016 with year 2010 as base year is: (1) 160 (2) 200 (3) 150

Given, Shopkeeper selling syrup at the rate of `250 per litre, \ Total sealing price = 250(10+x) According to the question. 3000 = 250(10+x) 3000 = 2500 +250x 500 = 250x x = 2 litre

54. Three persons A, B and C enter into a partnership to run a business. They invested their capitals in

(4) 170

the ratio

Ans. Option (1) is correct

share by 40%. If the total profit at the end of a year is `50,550, then A’s share in the profit is: (1) `8,000 (2) `10,000 (3) `20,000 (4) `12,000

Commodity

Explanation: P0 P1 Q0 Q1

A

1

2

B

5

C

6

10 13

P0Q1

P1Q0

13

26

20

10 12 16

80

160

120

10 15 18

108

180

150

∑P1Q1=366

∑P1Q0=290

∑P0Q1=201

P01=

P1Q1

∑ P1Q0 ×100 ∑ P0Q0

= 290 ×100 160

52. The present value of a perpetuity of `1200 payable at the beginning of each year, if money is worth 5% per annum is: (1) `25,500 (2) `24,000 (3) `24,200 (4) `25,200

Ans. Option (4) is correct Explanation: PV = `25,200 1200 R P = R+ = 1200 + 5 I = 1200 + 24000 = `25200

53. Pure

honey costs `300 per litre. A shopkeeper adds water to 10 litres of pure honey and sells the resulting syrup at `250 per litre. The quantity of water added by the shopkeeper is : (1) 2 litres (2) 5 litres (3) 3 litres (4) 1.5 litres

Ans. Option (1) is correct Explanation: Given, Quantity of pure honey = 10 litre. Per litre Price of pure honey = `300 Total selling price of 10 litre honey = 10 × 300 = `3000 After Adding x ltr of water into 10 ltr of honey, Now new quantity of syrup (mixture of water and honey) = (10+x) ltr

4 5 6 : : . After 5 months B increases his 3 2 5

Ans. Option (4) is correct Explanation: One month ratio of investment of 4 5 6 three persons A, B and C = : : 3 2 5

= 40 : 75 : 36

A's Investment for a year = 40x × 12

= 480x

B's Investment for a year = 75x × 5 +

140 × 75x × 7 100



= 375x+735x



= 1110x

C's Investment for a year = 36x × 12

= 432x

Yearly ratio of capital investment of three persons A, B and C = 480x : 1110x : 432x = 80 : 185 : 72 Given, Total profit at the end of a year = `50550 A's profit share

=

80 × 50550 80 + 185 + 72

80 × 50550 = 337 = `12000

55. Two positive numbers x and y whose sum is 25 and the product x3y2 is maximum are: (1) x=10, y=15 (2) x=15, y=10 (3) x =12, y=13 (4) x=16, y=9

Ans. Option (2) is correct Explanation: Given, x+y = 25 y = 25 − x  p = x3y2 p = x3(25−x)2 [from eq. (i)] d.w.r.t. x

... (i)

25 25

Solved Paper-2023

dp dx dp dx dp dx dp dx

= x3 2(25−x)(−1) + (25−x)2 (3x)2



x=3 in eq. (i)

a b − = −1 3 9 3a − b = −9 On solving eq (ii) and eq (iii) a = −4 and b = −3 (a, b) = (−4, −3)

= x2 (25−x)[−2x+75−3x] = x2 (25−x)(−5x+75) = 5x2 (25−x)(x−15)

... (ii)

dp On putting = 0 dx −5x2 (25−x)(x−15) = 0 x = 0, x = 25, x = 15 again differentiating eq. (ii) w.r.t. x d2p

on putting

= −5x2(25−x)+−5x2(x−15)(−1)

dx 2 + (25−x) (x−15)(−10x) d 2 p   = 0 dx 2  x  0



d 2 p  = −12250  dx 2  x 15



d 2 p   = +31250 dn 2  x  25

... (iii)

57. The

maximum value of Z=3x+y subject to the constraints x+y≤30, 2x+y≤40, x, y≥0 is: (1) 50 (2) 30 (3) 25 (4) 60

Ans. Option (4) is correct Explanation: when, X x+y = 30 Y

0

30

10

30

0

20

X

0

20

10

Y

40

0

20

2x+y = 40

For x+y≤30, 2x+y≤40 and x, y≥0 Y

from above d 2 p  = −12250 < 0  dx 2  x 15 Hence at x = 15 and y = 10 the product x3y2 is maximum.

40 20

b 56. If the function f(x)=alog x+ x +x has extreme values at x=1 and x=3, then (a, b) is: 3  1 (1)   ,   2 2  (3) (−2, −1)

0 (0,0)

(2) (4, 3) (4) (−4, −3)

Explanation: b +x x

a b f'(x) = − 2 +1 x x As we know, An extremum is calculated from the derivative of the function about a point where the derivative is equal to zero. Þ f'(x)=0 a b − +1 = 0 ... (i) x x2 Now it is given that the extremum is at x=1 and x=3 on putting x=1 in eq. (i) a−b = −1 ... (ii)

(10,20)

10

Ans. Option (4) is correct

f(x)=alog x+

(0,40)

(0,30) 30

(30,0)

X

10 20 30 40 50 (20,0) x+y = 30 2x+y = 40

(x, y)

z=3x+y

(0, 0)

0

(20, 0)

60

(10, 20)

50

(0, 30)

30

Here, at x=20 and y=0 Maximum value of z=3x+y is 60.

58. If 57 = x(mod 5). Then the least positive value of x is: (1) 57

(2) 5

(3) 4

(4) 2

Ans. Option (4) is correct Explanation: We know that when 57 is divided by 5, we get 2 as reminder,  Hence x=2

OSWAAL CUET (UG) Sample Question Papers, Mathematics/App. Math.

26

1

59. If x=log t and y = (1)

2 t2



(2)

t

4 t

2

2

, then

d2y dx 2

(3) −



77 x 12 Now, time taken by motor in Downstream Distance = Speed of motor boat in Downstream



is:

1 t

(4) −

4 t2

Ans. Option (2) is correct

=

Explanation: Given, x=log t dx 1 =  d.w.r.t. x dt t and d.w.r.t. t

... (i)

y= 1 t2

dy 2   dt t 3 on dividing eq (i) by eq (ii) we get,

d.w.r.t. x

d y dx 2

= =

4 3

t 4

 2  ×

D 14.4 x + x

13.4 x 77 × 15.4 x 12 134 1 × = 2 12 =



= 5 hours 35 minutes.

61. A

simple random sample consists of five observations 2, 4, 6, 7, 6. The point estimate of population standard deviation is: (1) 4 (2) 2.5 (3) 5 (4) 2

Ans. Option (4) is correct

dy 2  dx t 2 2



... (ii)

dy 2 t   dx t 3 1



D = 13.4 ×

Explanation: 2 t3



dt dx

t [from eq (i)] 1

t2

60. The speed of a motor boat in still water is 14.4 times the speed of the current of water. If the motor boat covers a certain distance upstream in 6 hours 25 minutes, then the time taken by the motor boat to come back is: (1) 5 hours 35 minutes (2) 5 hours 25 minutes (3) 5 hours 10 minutes (4) 5 hours 55 minutes

Ans. Option (1) is correct Explanation: Let the speed of current of water = x km/hr speed of motor boat in still water = 14.4x km/hr Speed in upstream = (14.4x−x) km/hr Speed of motor boat in down stream = (14.4x×x) km/hr time taken by boat in upstream = 6 hr 25 min 25 = 6+ 60 77 = hr 12 Speed of motor boat in upstream = Distance time Distance 14.4 x−x = 77 12 12D 13.4 x = 77

2+4+6+7+6 Mean = 5 25 = 5 x- = 5 Standard deviation =

2  ( xi  x ) f

2 2 2 2 2 = ( 2  5)  ( 4  5)  ( 6  5)  (7  5)  ( 6  5) 5

9+1+1+ 4 +1 = 5 16 = 1.78 = 5 62. The price relatives and weights of a set of commodities are given as: Commodity Price Relative Weight

A

B

C

150

130

180

x

3x

y

If the sum of weights is 30 and the index for the set is 144, then the values of x and y are: (1) x=6, y=8 (2) x=8, y=4 (3) x=6, y=6 (4) x=5, y=10

Ans. Option (3) is correct Explanation: Commodity

Weight (w)

Price relative (p)

wp

A

x

150

150x

B

3x

130

390x

C

y

180

180y

Sum Sum of weights 4x + y = 30

4x+y

540x+180y

= 30 ... (i)

Solved Paper-2023 Index Number

= 144

540 x + 180 y = 144 30 9x + 3y = 72 3x + y = 24 ... (ii) On solving eq. (i) and eq. (ii) to get x and y, we get x = 6 and y = 6

63. Given the data for the sales of a product in a state is as follows:

2005

2006

2007

2008

2009

Sales(In lakh `)

150

130

160

170

200

The equation of the straight-line trend by method of least squares is: (1) 14 + 162x (3) 128 + 14x Ans. Option (4) is correct

(2) 126 + 15x (4) 162 + 14x

Sales (y) X = x − 2007

X2

XY

150

−2

4

−300

2006

130

−1

1

−130

2007

160

0

0

0

2008

170

1

1

170

2009

200

2

4

400

∑y = 810

∑x=0

2

∑x = 10 ∑xy = 140

 y 810  = 162 n 5  xy 140 = 2  = 14 10 x Y = a+bX = 162+14x a =

3y 3   7 9 3  2x  3     If  y  1 2 x  z 1   2 4 1 then the  3 z  1 2 5   25 2 5 

values of x, y and z are: (1) x=2, y=3, z=8 (2) x=2, y=−3, z=8 (3) x=−3, y=2, z=6 (4) x=−2, y=3, z=8 Ans. Option (2) is correct Explanation: On, comparing, 2x+3=7 2x=4 x=2 y+1=−2 y=−3 and 3z+1=25 3z = 24 z=8 x=2, y = −3, z=8

X  0 Test statistic t = /  19.5  20 0.5   2 = 2/8 2 / 64 phone calls per minute coming into the help line desk of a bank is 5. The probability that during one particular minute there will be only one phone call is: (1) 0.5e−5 (2) 5e−5 (3) e−5 (4) 25e−5

Ans. Correct Option is (2)

2005

64.

The population standard deviation is 2. The value of the test statistic is: (1) −2.5 (2) −2 (3) 2 (4) −1.5 Ans. Option (2) is correct

66. Between 3 p.m. and 5 p.m. the average number of

Explanation: Calculating for the straight line trend by least square method

n=5

65. Consider the following hypothesis test: H0 : m≥20 H1 : m 8 cm (3) Shortest-side ≤ 8 cm (4) Shortest-side ≥ 8 cm

Ans. Option (4) is correct Explanation: Let the Shortest side = x cm Longest side = 4x cm Third side = (4x−3) cm Given, Perimeter ≥ 69 x+ 4x + 4x−3≥69 9x≥72 x≥8 Shortest side ≥ 8

71. An

asset costing `2,00,000 is expected to have a useful life of 10 years and a final scrap valve of `40,000. The book value of the machine at the end of sixth year is: (1) `1,36,000 (2) `1,04,000 (3) `1,20,000 (4) `88,000

Ans. Option (2) is correct

29 29

Solved Paper-2023 Explanation: Annual Depreciation =

Cost Price - Scrap Value Estimated life of machine



 5   7  L.H.S. = 3    4    24   24  15 28 + 24 24

200000 − 40000 = 10

=



= 43 24

= 16000

Book value = C  ost Price − no. of years after which book value is to be computed × Annual Depreciation

= 200000 − 6 × 16000



= 200000 − 96000



= 1,04,000

72. The point on the straight line 3x + 4y = 8, which is closest to the origin is:

 13 17  (1)  ,   24 24   5 7  (3)  ,   24 24 

 24 32  (2)  ,   25 25   5 (4)  1,   4

Ans. Option (2) is correct





 13   17  L.H.S. = 3    4    24   24 

=

39 68 + 24 24

107 = 24 = R.H.S.  13 17   ,  Does not lie on line 3x+4y = 8  24 24   24 32  Now, at  ,   25 25  L.H.S. = 3  24   4  32       25   25  = 72 + 128 25 25 200 25 = 8

≠

8

 5 7   ,  Does not lie on line 3x+4y=8  24 24   5 Now at,  1,   4

L.H.S.

=

5 3(1) + 4   4 3+5  

= = 8 = R.H.S.  5  1,  Does lie on line 3x+4y = 8  4

 24 32  Now, Distance between  25 , 25  and (0,0)  

Explanation: Given, 3x + 4y = 8  13 17  at  ,   24 24 





2

d1 =  24  0    32  0    25   20 576 + 1024 625

=

= 1600 625

=

40 25

=

8 5



=

1.6 unit

 5 Now, Distance between  1,  and (0, 0)  4

d2 = (1  0 )2   5  0    4 

=

= R.H.S.  24 32   ,  Does lie on line 3x+4y = 8  25 25   5 7  Now, at  ,   24 24 

= 1 + 25 16 41 = 16 =

6.403124 4

2

2

30

OSWAAL CUET (UG) Sample Question Papers, Mathematics/App. Math. = 1.600781 24 32  Hence,  ,  is the point on the straight line  25 25  which is closest to the origin.

73. Match List I with List II LIST I A. A special characteristic of a population is known as a

LIST II I.

statistic

C. The uncertainty of a sampling III. Estimation process is expressed by D. The process by which one IV. Parameter makes the inferences about a population based on the information obtained from a sample is known as

Choose the correct answer from the options given below: (1) A-II, B-III, C-IV, D-I (2) A-I, B-IV, C-II, D-III (3) A-IV, B-I, C-II, D-III (4) A-IV, B-I, C-III, D-II

5  x Î  ,  3 



Here, A Match with III

3x + 5 ≥1 2 3x+5≥2 ⇒ 3x≥−3 x≥−1 ⇒ x Î (−1, ∞) Here, B Match with I C. 2x+5 0 So, f(x) is minium at x = 4 Hence, value of f(x) at x = 4 is f(4) = (4)2 – 8(4) + 17 = 16 – 32 + 17 = 33 – 32 = 1

Explanation: Let,

(1) enx (3) ny

Ans. Option (4) is correct.

Ans. Option (4) is correct. é x y zù ê ú A = ê 2 u v ú êë -1 6 w úû

25. If y = enx, then nth derivative of y is:



28. If m is mean of random variable X, with probability distribution x P(X = x)

0 4 9

1 4 9

2 1 9

39 39

Solved Paper-2022 Ans. Option (4) is correct.

then value of 9m + 4 is: (1) 4 (2) 9 (3) 10 (4) 17

Explanation: Given, P(X = x) = (0.6)x(0.4)1–x, x = 0, 1

Ans. Option (3) is correct. Explanation: x P(X = x) 4 0 9

0

1

4 9

2

1 9

2 9

Since, \

x.P(X = x)

4 9

Sx.P(X = x) =

Total

Here, p = 0.6, q = 0.4, n = 1 Variance = npq = 1 × 0.6 × 0.4 = 0.24

LIST I

6 9



A.

Quantity index

I.

Measures relative price change over a period of time.

B.

Time series

II.

Measures change in quantity of consumption of goods over a specific period of time.

C.

Price index

III.

Measures average value of goods for specific time period.

D.

Value index

IV.

29. In a game, a child will win `5 if he gets all heads or all tails when three coins are tossed simultaneously and he will lose `3 for all other cases. The expected amount to lose in the game is (1) `0 (2) `0.8 (3) `1 (4) `2

Ans. Option (3) is correct. Explanation: Let X be the amount received by the person. Then X can take values 5 and –3 such that P(X = 5) : Probability of getting all heads or all tails when three coins are tossed 2 1 P(X = 5) = = 8 4 P(X = –3) : Probability of getting one or two heads 6 3 P(X = –3) = = \ 8 4 Therefore, expected amount to win, on the average, per game = X = S|pixi 1 3 = 5 ´ + ( -3) ´ 4 4 5 4



LIST II

Mean(m) = Sx.P(X = x) 6 m = 9

6 Now, 9m + 4 = 9. + 4 = 6 + 4 = 10 9

31. Match List I with List II

Statistical observation taken at different points of time for specific period of time. Choose the correct answer from the options given below: (1) A-III, B-I, C-II, D-IV (2) A-II, B-III, C-I, D-IV (3) A-III, B-IV, C-I, D-II (4) A-II, B-IV, C-I, D-III Ans. Option (4) is correct.

32. Given that, Sp0q0 = 700, Sp0q1 = 1450, Sp1q0 = 855

and Sp1q1 = 1300. Where subscripts 0 and 1 are used for base year and current year respectively. The Laspeyer's price index number is: (1) 118.46 (2) 119.35 (3) 120.23 (4) 122.14 Ans. Option (4) is correct. Explanation: Laspeyre's price index Spi q0 ´ 100 = Sp 0 q 0

9 4

855 ´ 100 700

=

=

= –1 Thus, the person will, on average lose `1 per toss of the coin.

= 122.14

30. The Probability mass functions of Random variable

X is: P(X = x) = (0.6)x(0.4)1–x, x = 0, 1 The variance of X is: (1) 0.60 (2) 0.124 (3) 0.244 (4) 0.240



33. If y = a + b(x – 2005) fits the time series date x(year) :

2003

2004

2005

y (yield in tons) : 6 13 17 Then the value of a + b is: (1) 16 (2) 20.3 (3) 43 (4) 80.3

2006

2007

20

14

OSWAAL CUET (UG) Sample Question Papers, Mathematics/App. Math.

40

Ans. Option (2) is correct.



Explanation: x(year) 2003 2004 2005 2006 2007

y(yield X = xi – in tons) 2005 6 13 17 20 24

–2 –1 0 1 2

X2

Xy

4 1 0 1 4

–12 –13 0 20 48

a =



b =

Explanation: Option B, EMI =

\

SX

2

=

23 = 2.3 10

a + b = 18 + 2.3 = 20.3



34. Which of the following statements are correct?

(A) If discount rate > coupon rate, then present value of a bond > face value (B) An annuity in which the periodic payment begins on a fixed date and continues forever is called perpetuity (C) The issuer of bond pays interest at fixed interval at fixed rate of interest to investor is called coupon payment (D) A sinking fund is a fixed payment made by a borrower to a lender at a specific date every month to clear off the loan (E) The issues of bond repays the principle i.e., face value of the bond to the investor at a later date termed as maturity date Choose the correct answer from the options given below: (1) (A), (C), (E) only (2) (A), (B), (D) only (3) (B), (C), (E) only (4) (A), (B), (C) only

=

Choose the correct answer from the options given below:

40 , 000 ´ 0.005 1 - (1 + 0.005)-60

= 40 , 000 ´

0.005 1 - (1.005)-60

= 40,000 × 0.0194 é ù 0.005 = 0.0194 ú êGiven, -60 1 - (1.005) ë û





a down payment of `5000 and balance in EMI for 5 years. If Bank charges 6% per annum compounded monthly then monthly EMI is:

Explanation: Given, Cost of AC = `45,000 Down payment = `5000 \ Balance = `40,000 So, P = `40,000, 6 i = = 0.005 , 12 ´ 100 n = 5 × 12 = 60 P ´i EMI = 1 - (1 + i ) - n

(A) EMI in flat rate method, Principle + Interest EMI = Number of Payment (B) EMI in reducing balance method, i EMI = P ´ 1 + (1 + i )n



36. Mr. Dev wishes to purchase an AC for `45,000 with

Ans. Option (1) is correct.

35. Which of the following statements is true?

where P = Principle, i = interest rate, n = no. of payments (C) In sinking fund, a fixed amount at regular intervals is deposited. (D) Approximate Yield to Maturity Face Value + Present Value Coupen Payment + nt Number of Paymen = Face Value + Present Value 2

1 - (1 + i ) - n

æ ö 0.005 = 0.0194÷ ç use -60 1 - (1.005) è ø (1) `776 (2) `700 (3) `737 (4) `673

Ans. Option (3) is correct.

P ´i

where, P = Principle, i = interest rate n = no. of payments c + {( F - P.V .) / N } Option D, Approx. YTM = ( F + P.V .) 2 Where F is face value c is coupon rate P.V. is present value N is no. of payments

Sy 80 = = 18 (n = 5) n 5 SXy

(2) (B) and (C) only (4) (C) and (D) only

Ans. Option (3) is correct.

Sy = 80 SX = 0 SX2 = 10 SXy = 23 Now,

(1) (A) and (B) only (3) (A) and (C) only

= `776



37. The cost of a machine is `20,000 and its estimated useful life is 10 years. The scrap value of the machine, when its value depreciates at 10% p.a, is: use (0.9)10 = 0.35 (1) `9672 (2) `7000 (3) `6982 (4) `3500

Ans. Option (2) is correct.

Solved Paper-2022 Explanation: Amount on depreciation r ö æ = Original amount ç 1 è 100 ÷ø 10 ö æ = 20000 × ç 1 è 100 ÷ø

10

= 20000 × (0.9)10 = 20000 × 0.35 = `70000

n



38. One of the following is true of relation between

sample mean (x) and population mean (m). (1) |x – m| increases when increases the size of sample (2) x = m, for all sample sizes (3) |x – m| do not change with size of sample (4) |x – m| decreases when increases the size of sample





40. Corner points of the feasible region for an LPP, are

(0, 2), (3, 0), (6, 0) and (6, 8). If z = 2x + 3y is the objective function of LPP then max.(z) – min.(z) is equal to: (1) 30 (2) 24 (3) 21 (4) 9

x = `50,000 (part of money given to Shyam) Therefore, part of money given to Shushil = `1,50,000 Mean interest rate 50 , 000 ´ 6 + 1, 50 , 000 ´ 10 = 2 , 00 , 000



(0, 2)

Z = 6 ® Minimum

(3, 0)

Z=6

(6, 0)

Z = 12

(6, 8)

Z = 36 ® Maximum

\ max(³) – min(³) = 36 – 6 = 30

Read the below passage and solve the questions from Q.No. 41-Q.No.45. Passage Sitaram, a money lender lent a part of `200000 to Shyam at simple interest 6% p.a. and the remaining to Sushil at 10% p.a. at simple interest. Sitaram earned an annual interest income of `18000. Based on the given information answer the following question:

3, 00 , 000 + 1, 5, 00 , 000 = 9% 2 , 00 , 000

42. In what ratio did Sitaram lent the money at 6% p.a. and 10% p.a. respectively? (1) 1 : 3 (2) 3 : 1 (3) 2 : 3 (4) 3 : 5

Explanation: Money lent to Shyam = `50,000 Money lent to Shushil = `1,50,000 [from Q. 41] Ratio = 50,000 : 1,50,000 = 1:3

Explanation: Value of Z = 2x + 5y

=

Ans. Option (1) is correct.

Ans. Option (1) is correct. Corner point

(2) 8% p.a. (4) 16% p.a.

where A2 is income earned on Sushil's money Given, A1 + A2 = 18,000 \ 0.06x + (2,00,000 – x) × 0.1 = 18,000 0.06x – 0.1x + 20000 = 18000 0.04x = 2000

39. Below are the stages for Drawing statistical

Ans. Option (3) is correct.

(1) 6% p.a. (3) 9% p.a.

Explanation: Let part of money given as a loan to shyam be `x. 6 x´ = A1 where A1 is income earned on 100 Shyam's money Similarly 10 ( 2 , 00 , 000 - x ) ´ = A2 100

[Given (0.9)10 = 0.35]

inferences. (A) Sample (B) Population (C) Making Inference (D) Data tabulation (E) Data Analysis Choose the correct answer from the options given below: (1) (B), (D), (A), (E), (C) (2) (A), (B), (D), (C), (E) (3) (B), (A), (D), (E), (C) (4) (D), (B), (A), (C), (E)

41. What is the mean rate of interest?

Ans. Option (3) is correct.

Ans. Option (2) is correct.

41 41



43. How much money did Shyam borrow? (1) `150000 (3) `50000

(2) `75000 (4) `12000

Ans. Option (3) is correct. Explanation: See Q. 41

44. What amount of money is lent at 10% p.a. simple interest? (1) `20,000 (3) `75,000

(2) `50,000 (4) `1,50,000

Ans. Option (4) is correct. Explanation: Money is lent at 10% p.a. Money lent to Shushil = 1,50,000 [from Q.41]

OSWAAL CUET (UG) Sample Question Papers, Mathematics/App. Math.

42

45. What is the ratio of the interest paid by Shyam and Sushil respectively (1) 1 : 3 (3) 3 : 5



Ans. Option (3) is correct. Explanation: Total revenue of cable network provider after the increment is given by R(x) = (500 – x)(300 + x)

(2) 1 : 5 (4) 2 : 3

Ans. Option (2) is correct. Explanation: Amount of interest paid by 6 Shyam = 50 , 000 ´ = `3000 100 Amount of interest paid by Shushil 10 = `15,000 = 1, 50 , 000 ´ 100



maximum revenue is (1) 100 (3) 300 Ans. Option (4) is correct.



Explanation: If `x is monthly increase in subscription amount, then x subscribers will discontinue the service. So, no. of subscriber = 500 – x

\ R'(x) = –2x + 200 Put R'(x) = 0, we get –2x + 200 = 0 x = 100 Now, R"(x) = –2 < 0 maximum So, R(x) is maximum when x = 100. Hence, no. of subscribes which given the maximum revenue = 500 – 100 = 400

49. What is increase in changes per subscribes that yields maximum revenue? (1) 100 (2) 200 (3) 300 (4) 400

Ans. Option (1) is correct. Explanation: Increase in change per subscriber = x = 100

50. The maximum revenue generated is (1) `200000 (3) `160000

(2) `180000 (4) `150000

Ans. Option (3) is correct.

47. Total revenue 'R' is given by (in Rs.) (1) R = 300x + 300(500 – x) (2) R = (300 + x) (500 + x) (3) R = (300 + x) (500 – x) (4) R = 300x + 500(x + 1)

Explanation: R(100) = (500 – 100)(300 + 100) = 400 × 400 = `1,60,000

Section - B2



(2) 200 (4) 400

Explanation: Since, R(x) = (500 – x)(300 – x) = 1,50,000 + 500x – 300x – x2 = –x2 + 200x + 1,50,000

So, ratio of interest paid = 3000 : 15,000 = 1:5 Read the below passage and solve the questions from Q.No. 46–Q.No.50. Passage Item are based on the information below: A cable network provider in a small town has 500 subscribes and he used to collect `300 per month from each subscribes. He proposes to increase the monthly charges and it is believed from the past experiences that for every increase of `1, one subscribes will discontinue the service. Based on the above in formation, answer the following question: 46. If `x is the monthly increase in subscription amount, then the number of subscribes are (1) x (2) 500 – x (3) x – 500 (4) 500 ns. Option (2) is correct. A

48. The number of subscribes which gives the

51. Match List I with List II A. B.

C. D.

LIST I R = {(x, y): x and y are I. student of the same school} II. R = {(L1, L2) : L1 ^ L2, L1, L2 ÎL, where L is a set of all lines} III. A function f : R ® R defined f(x) = 2 – 3x is IV. A function f [0, 1] ® R defined by f(x) = 1 + x2 is f : [0, 1] ® R

Mathematics LIST II Symmetric

Choose the correct answer from the options given below: (1) A-I, B-IV, C-II, D-III (2) A-IV, B-I, C-III, D-II (3) A-I, B-IV, C-III, D-II (4) A-IV, B-I, C-II, D-III Ans. Option (2) is correct.

one-one

bijective Equivalence

Explanation: A. R = {(x, y): x and y are students of the same school} is an equivalence relation. B. R{(L1, L2): L1 ^ L2, L1, L2 ^ L where L is a set of all lines} only follows symmetric relation. C. A function f : R ® R such that f(x) = 2 – 3x is one-one onto relation. Hence, bijective D. A function f : [0, 1] ® R such that f(x) = 1 + x2 is one-one.

Solved Paper-2022

é

æ 3 ö ïü ù ÷ ý ú , is: è 2 ø þï úû

ì

ï 52. The value of tan -1 ê 2 sin í2 cos-1 ç êë

(1)

îï

p 3

(3) -

p 3

(2)

2p 3

(4)

p 6

écos a - sin a ù A' = ê cos a úû ë sin a é cos a sin a ù écos a - sin a ù AA' = ê úê cos a úû ë - sin a cos a û ë sin a

é1 0 ù = ê ú =I ë0 1 û

-1 é

æ pöö ù -1 æ ê 2 sin ç 2 cos çè cos ÷ø ÷ ú è 6 øû ë

é2

1

0ù ú 2 ú then |adj(A)| is equal to êë 0 4 -1úû



55. If A = êê 3 1

éæ pö ù = tan -1 êç 2 ´ sin ÷ ú è 3øû ë



(1) 11

(2) 12



(3) 225

(4) –225

é æ 3öù = tan -1 ê 2 ´ ç ÷ú è 2 ø ûú êë

Ans. Option (3) is correct.

= tan

-1 é

æ æ pöö ù ê 2 sin ç 2 çè ÷ø ÷ ú è 6 øû ë

Explanation: Given

-1 3 = tan



pö æ = tan -1 ç tan ÷ è 3ø

53. Match List I with List II LIST I

LIST II –1x

is

I.

é -p p ù ê 2 , 2 ú - {0} ë û

II.

æ -p p ö çè 2 , 2 ÷ø

A.

The range of sin

B.

The range of tan–1x is

C.

The range of cosec–1x III. is

D.

The range of sec–1x is

IV.

ìp ü [0 , p] - í ý î2þ

é -p p ù ê 2 , 2ú ë û

Choose the correct answer from the options given below: (1) A-IV, B-III, C-II, D-I (2) A-IV, B-I, C-III, D-II (3) A-I, B-IV, C-II, D-III (4) A-IV, B-II, C-I, D-III Ans. Option (4) is correct. é cos a sin a ù 54. If A = ê ú , then: ë - sin a cos a û (1) A'A = I (3) A'A = 2I

Ans. Option (1) is correct.

(2) A'A = 0 (4) A'A = –I

é2 1 0 ù ê ú A = ê 3 1 2 ú êë 0 4 -1úû

|A| = 2(–1 – 8) – 1(–3) = –18 + 3 = –15 ¹ 0 Since, |adj A| = |A|n – 1 where n = order of matrix Therefore, |adj A| = |–15|3 – 1 = (–15)2 = 225

p = 3



\

é cos2 a + sin 2 a - cos a sin a + cos a sin a ù ê ú = êë - sin a cos a + cos a sin a úû sin 2 a + cos2 a

é æ æ 3 öö ù Explanation: tan -1 ê 2 sin ç 2 cos -1 ç ÷÷ ú è 2 ø ø úû êë è



é cos a sin a ù A = ê ú ë - sin a cos a û



Ans. Option (1) is correct.

= tan

Explanation:

43 43



56. Identify the correct option (s)

(A) A modulus function is continuous at every point in its domain. (B) A modulus function may or may not be continuous at every point in its domain. (C) Every rational function is continuous in its domain. (D) If a function f is differentiable at a point then it is also continuous at that point. (E) If a function f is continuous at a point then it is also differentiable at that point. Choose the correct answer from the options given below: (1) (A) and (C) only (2) (B) and (E) only

(3) (A), (C) and (D) only (4) (C) and (E) only

Ans. Option (3) is correct. Explanation: E is incorrect as Differentiability Þ continuity, But converse is not true. B is incorrect as modules function is continuous at every point in its domain.

OSWAAL CUET (UG) Sample Question Papers, Mathematics/App. Math.

44



ì 1 - cos 2 x , x¹0 ï , then the value of k x 2 ï = 0 k , x î

will make function f continuous at x = 0 is: (1) 1 (2) –1 (3) 0 (4) No value

Ans. Option (1) is correct. Explanation: Since, f(x) is continuous at x = 0 1 - cos 2 x Þ k = lim x ®0 x 2



1 - (1 - 2 sin 2 x )

= lim





x 2

x ®0



= lim



= lim



= 1

x ®0

sin x x

2 ex 2 (3) 2xex tan ex

dx

2 2 (2) ex tan ex 2 2 (4) xex tan ex

B.

e log ò x

x

dx

dx = dt dx = 2tdt 2tdt I1 = ò 2 t +t

I2 =

ò

2 x +C 2

(

)

x -1 e

e log x

x

dx

=

x -

1 - +1 2

1 +1 2

+c

1

= 2 x + c dx (C) Let I3 = ò 2 4x - 9 =

1 dx 4 ò x2 - 9 4 1 4ò

1 = 4 x

1

= ò x 2 dx

LIST II

II. dx

dx

x = t

=

LIST I x

2 x

x

x2 +c = 1 2

60. Match List I with List II 1

1

1

òx+

-

(4) does not exist

I.

1 2x - 3 log +C 12 2x + 3

x = ò dx x

dy =0 dx

òx+

Explanation: (A) I1 =

(B) Let

Explanation: –1 –1 y = elogsin x + elogcos x, 0 < x < 1 –1 –1 y = sin x + cos x, 0 < x < 1 pù é -1 -1 p y= ê∵sin x + cos x = 2 ú ë û 2

A.

)

x +1 +C

1 dt t+1 = 2log(t + 1) + c \ I1 = 2 log| x + 1| + c

Ans. Option (1) is correct.



IV.

dx

(

= 2 ò

[Using chain rule of differentiation] 2 2 = 2xex .tanex –1 logsin x logcos–1x 59. If y = e +e , 0 < x < 1, then dy dy p (1) (2) =0 = dx dx 2

\

x

2 log

Ans. Option (3) is correct.

\

Explanation: Given, 2 y = log(sec ex ) 2 2 2 dy 1 (sec e x tan e x ).( e x ).2 x = x2 dx sec( e )

dy p (3) = dx 3

III.

Choose the correct answer from the options given below: (1) A-II, B-IV, C-I, D-III (2) A-III, B-II, C-IV, D-I (3) A-III, B-I, C-IV, D-II (4) A-I, B-II, C-III, D-IV

Þ

2 58. If y = log(sec ex ), then dy = 2 (1) x2ex tan 2

òe

Þ

x 2

x ®0

D.

dx

ò 4x2 - 9

put

2 sin x

Ans. Option (3) is correct.



C.

57. If f(x) = í

+C

dx æ 3ö x2 - ç ÷ è 2ø

2

3 x1 2 +c log 3 æ 3ö x+ 2. ç ÷ è 2ø 2

45 45

Solved Paper-2022 é ê Using ë

dx

1

x-a

ò x 2 - a2 = 2a log x + a

ù + cú û

1 2x - 3 log +c = 12 2x + 3 x

òe

(D) Let

I4 =

Put

x = t

dx

(1) sec x – tan x + x – C (2) sec x – tan2 x + C (3) sec x + tan x + x + C (4) sec x – tan x + C Explanation: Let

I =

=

ò tan x(sec x - tan x )dx ò tan x sec xdx - ò tan

2

xdx

t = 2 ò te dt

{ò (sec - 1)dx} = sec x - {ò (sec xdx - ò dx}

t t = 2 éëte - ò e dt ùû

= secx – [tanx – x + c] = secx – tanx +x – c

\

I4 =

= 2[tet – et] + c =

= sec x -

t

ò e ( 2t )dt

[Integration by parts] 2et(t

= 2 e

– 1) + c

x

( x - 1) + c

61. The order of the differential equation whose

general solution is y = ex(acosx + bsinx), where a and b are arbitrary constants is: (1) 1 (2) 3 (3) 2 (4) 6



a , then value of cos 2a+ cos 2b+ cos 2g is equal to:

(1) 3 (3) 2

(2) 0 (4) –1

Explanation: Since, cosa, cosb and cosg are  direction cosines of vector a , then cos2a + cos2b + cos2g = 1 Now, cos2a + cos2b + cos2g = (2cos2a – 1) + (2cos2b – 1) + (2cos2g – 1) = 2(cos2a + cos2b + cos2g) – 3 =2×1–3 =2–3 = –1

dx 2

64. If cosa, cosb, cosg are the direction cosines of vector

Ans. Option (4) is correct.

Explanation: General solution is y = ex(acosx + bsinx) ...(i) y' = ex(acosx + bsinx) + ex(–asinx + bcosx) or, y' = y – aexsinx + bexcosx or, y' = y + ex(–asinx + bcosx) ...(ii) Differentiating again w.r.t. x d2y

( ) ( ) ( )

65. The value of i. j ´ k + j. i ´ k + k. i ´ j is (1) 0 (3) 1

=

dy + ex(–asinx + bcosx) + ex(–acosx – bsinx) dx



=

dy + ex(–asinx + bcosx) – y ...(iii) dx

Ans. Option (4) is correct.

dx 2

d2y

-

dy dy = - 2y dx dx

dy + 2y = 0 2 dx dx Thus, order of differential equation is 2. d é 2a 62. f (sin 2 x )dx úù = dx êë ò0 û Þ

= i.i + j.j + k .k =1+1+1 =3

-2

(1) 2a (3) f(cos 2a)

Ans. Option (4) is correct.

(2) f(sin 2a) (4) 0

(2) –1 (4) 3

Explanation: We have, i.( j ´ k ) + j.(i ´ k ) + k .(i ´ j )

Subtracting eq. (ii) from eq. (iii), we get d2y

2 2

Ans. Option (3) is correct.



63. ò tan x(sec x - tan x )dx =

Ans. Option (1) is correct.

dx = dt 2 x dx = 2tdt

Þ





1

Þ









66. The corner points of the feasible region for an L.P.P

are (2, 0), (7, 0), (4, 5) and(0, 3) and z = 2x + 3y is the objective function. The difference of the maximum and minimum value of z is: (1) 19 (2) 4 (3) 23 (4) 14

Ans. Option (1) is correct.

OSWAAL CUET (UG) Sample Question Papers, Mathematics/App. Math.

46

Explanation:



Corner Points

Value of z = 2x + 3y

(7, 0)

z = 14

(4, 5)

z = 23 ® max



(0, 3)

z=9

Ans. Option (4) is correct.

are i + k and 2i + j + k is

(2) 2



(3) 4

(4) 3

Equation first and last term, 2r + 1 - 4 = 4r + 3 5

Explanation: Area of parallelogram whose     adjacent sides are a and b is given by | a ´ b |   Here, a = i + j and b = 2i + j + k

= i(1 - 0 ) - j(1 - 0 ) + k (1 - 2 ) = i - j - k   Thus, | a ´ b | = 12 + ( -1)2 + ( -1)2 = 3

68. If x(i + j + k ) is a unit vector then value of x is: (1) ± 3

(2) ±



(3) ±3

(4) ±

1 3

Ans. Option (4) is correct.

Þ

| x(i + j + k )| = 1

| x ||i + j + k | = 1

Þ | x | 12 + 12 + 12 = 1 Þ

|x| 3 = 1

Þ |x| = Þ

1 3



70. The distance between the point (3, 4, 5) and the

x-3 y-4 z-5 meets = = 1 2 2 the plane x + y + z = 17, is

point where the line



(1) 3

(2) 2



(3) 1

(4) 0

Ans. Option (1) is correct. Explanation: Let the point of intersection of line x-3 y-4 z-5 = = 1 2 2 and the plane x + y + z = 17 be (x0, y0, z0) As (x0, y0, z0) is the point of intersection of the line and plane, So it must satisfy both of the equations of line and plane x0 - 3 y -4 z -5 i.e., = 0 = 0 = k (let) 1 2 2

Explanation: Given

Þ 2r – 3 = 20 r + 15r Þ 18r = –18 Þ r = –1 Required point of intersection, from eq. (i) x = –1, y = –1 and z = –1

i j k   a ´ b = 1 1 0 2 1 1



(2) (1, –1, –1) (4) (–1, –1, –1)

Substitute value of x, y and z in line 2, we get 2r + 1 - 4 2r + 2 - 1 = = 4r + 3 5 2

Ans. Option (4) is correct.

So,

(1) (1, 1, 1) (3) (–1, 1, –1)

Explanation: Let x -1 y-2 z-3 = = =r Line 1 2 3 4 Þ x = 2r + 1, y = 3r + 2, z = 4r + 3 ...(i) x-4 y -1 z-0 Line 2 = = 5 2 1

67. The area of the parallelogram whose adjacent sides (1) 3

3

z-3 x -4 y -1 and = = z , is: 4 5 2

z = 4 ® min





2

(2, 0)

Difference = zmax – zmin = 23 – 14 = 19

69. The point of intersection the lines x - 1 = y - 2 =

1

x = ±

3 1 3

x0 - 3 y -4 z -5 = k, 0 = k, 0 =k 1 2 2 or, x0 = k + 3, y0 = 2k + 4, z0 = 2k + 5 Put these values in equation off plane, we get (k + 3) + (2k + 4) + (2k + 5) = 17 5k + 12 = 17 5k = 5 k = 1 Thus, required point of intersection is x0 = 4, y0 = 6, z0 = 7

Solved Paper-2022 Now, distance between (3, 4, 5) and (4, 6, 7) is d= =

2

2

(3 - 4 ) + ( 4 - 6) + ( 5 - 7 )

Solving eqs. (ii) and (iii), we get a b c = = 4 + 10 15 - 1 2 + 12

2

1+ 4 + 4

= 3 units



71. If events A and B are independent, then identify (A) A and B must be mutually exclusive



(B) The sum of their probabilities must be equal to 1



(C) P(A).P(B) = P(A Ç B)



(D) A' and B' are also independent

Choose the correct answer from the options given below:

(1) (A) and (B) only

(2) (B) and (C) only



(3) (C) and (D) only

(4) (A) and (D) only

Ans. Option (3) is correct. Explanation: (A) If two events are independent, they cannot be mutually exclusive. (B) To test if probability of independent events is 1 or not. Let A be the event of getting head 1 P(A) = \ 2





1 ¹1 3



is: (1) x – y – z = 0 (3) x + y + z = 0

x +1 y -3 z+2 = = , 2 1 -3

= 1 -

(2) 4x + y + z = 0 (4) 3x + 2y + 2z = 0

=

Ans. Option (3) is correct. Explanation: Any plane passing through (0, 7, –7) is a(x – 0) + b(y – 7) + c(z + 7) = 0 ...(i) If plane (i) contains thee given line, then it must passes through the point (–1, 3, –2) and must be parallel to the given line. If (i) passes through (–1, 3, –2), then a(–1 – 0) + b(3 – 7) + c(–2 + 7) = 0 a + 4b – 5c = 0 ...(ii) If (i) is parallel to thee given line, then (–3)a + 2.b + 1.c = 0 –3a + 2b + c = 0 ...(iii)

8 45 2 (4) 9 (2)

é 47 12 ù = 1 - ê - ú ë 45 45 û

72. The equation of plane passing through the point (0, 7, –7) and containing the line

4 15 1 (3) 3 (1)

é 27 + 20 12 ù - ú = 1 - ê 45 û ë 45

Thus, statement (B) is also incorrect.

3 4 and P(B) = , then P(A' Ç B') is equal to 5 9

Explanation: P(A' Ç B') = P(A È B)' = 1 – P(A È B) = 1 – [P(A) + P(B) – P(A Ç B)] = 1 – [P(A) + P(B) – P(A).P(B)] [ A and B are independent P(A Ç B) = P(A).P(B)] é3 4 3 4 ù = 1 - ê + - ´ ú ë5 9 5 9û

Here, A and B are independent events 1 1 4 Therefore, P(A) + P(B) = + = 2 6 12 =

73. If A and B are two independent events with P(A) =

Ans. Option (4) is correct.

Let B be the event of getting 5 on a die 1 \ P(B) = 6



a b c = = 14 14 14 a b c = = =k 1 1 1

\ a = k, b = k and c = k Substituting these values in eq. (i), we get k(x – 0) + k(y – 7) + k(z + 7) = 0 kx + by + kz – 7k + 7k = 0 kx + by + kz = 0 k(x + y + z) = 0 or (x + y + z) = 0

the correct statements

47 47

35 45

10 45

2 = 9



74. A line L: x - 2 = y - 3 = z - 1 is perpendicular to 1 2 -1 a plane (P), which passing through the point (4, 3, 9). If the mirror image of point 'S' on the line (L) in the given plane (P) is (2, 3, 1), then co-ordinates of point S, is: (1) (1, 0, 3) (2) (0, –1, 3) (3) (–2, –3, –1) (4) (4, 7, –1)

Ans. Option (2) is correct.

48

OSWAAL CUET (UG) Sample Question Papers, Mathematics/App. Math. Explanation: The direction ratios of given line are (i , 2 j , - k )

(D) P(X = 4) =

ATQ, Given line is perpendicular to the plane. Þ The direction ratios of plane are also (i, 2 j , - k ) \ The equation of plane passing through (4, 3, 9) is a(x – 4) + b(y – 3) + c(z – 9) = 0 1.(x – 4) + 2(y – 3) – 1(z – 9) = 0 x + 2y – z – 4 – 6 + 9 = 0 x + 2y – z = 1

75. A biased dice is thrown once. If X denotes the

number appearing on it and have probability distribution: x

1

P(X = x)

2

k

k 2

3

4

5

6

2k

8k2

1 – 5k

k 2

(E) P(X = 1) + P(X = 5) =

(B) P(X £ 2)

(A) P(X = 3)



Read the below passage and solve the questions from Q.No.76- Q.No.80. Passage In a school, a auditorium was used for its cultural activities. The shape of the floor of the auditorium is rectangular with dimensions x and y(x > y). has fixed perimeter p. 76. If x and y represent the length and breadth of the rectangular region, then: (1) p = x + y (2) p2 = x2 + y2 (3) p = 2(x + y) (4) p = x + 2y Ans. Option (3) is correct. Explanation: Perimeter of rectangular region = 2(x + y) \ p = 2(x + y) ...(i)

77. The area (A) of the floor, as a function of x can be

(B) P(X ³ 5) (D) P(X = 4) (E) P(X = 1) + P(X = 5) Choose the correct answer from the options given below: (1) C > D > B > A > E (2) E > C > D > A > B (3) E > C > A > B > D (4) C > E > A > B > D



Ans. Option (3) is correct.

Ans. Option (3) is correct.

(1) A( x ) = px +



(3) A( x ) =

1 (Rejecting k = 0) 8

x

1

2

3

4

5

6

P(X = x)

1 8

1 16

1 4

1 8

3 8

1 16

1 4

(B) P(X £ 2) = P(X = 1) + P(X = 2) 1 1 2+1 3 = = = + 8 16 16 16 (C) P(X ³ 5) = P(X = 5) + P(X = 6) 3 1 6+1 7 = = = + 8 16 16 16

(2) A( x ) =

px - 2 x 2 2

px + x 2 2

x2 (4) A( x ) = + px 2 2

[from eq. (i) of Q. 76]

Probability Distribution is:

(A) P(X = 3) =

x 2

Explanation: Area off region (floor) is given by A = xy ...(ii) Since, p = 2(x + y) p - 2x ...(iii) y = Þ 2

Þ 2k + k + 4k + 16k2 + 2 – 10k + k = 2 16k2 – 2k = 0 Þ Þ 2k(8k – 1) = 0 1 Þ k = 0 or k = 8 k =

expressed as:



Explanation: S[P(X = x)] = 1 k k Þ k + + 2 k + 8 k 2 + (1 - 5 k ) + = 1 2 2

Þ

1 3 4 1 + = = 8 8 8 2

Thus, E > C > A > B > D

where k > 0. Then consider the following statements:

1 8



\

æ p - 2x ö A(x) = x ç è 2 ÷ø

Þ

A(x) =

px - 2 x 2 2

...(iv)

78. The value of x, for which area of floor of auditorium



is maximum is: p (1) 4

(2)

p 2



(3) p

(4)

p 3

Ans. Option (1) is correct. Explanation: From eq. (iv) of Q. 77, we have

A =

px - 2 x 2 2

Solved Paper-2022 \ Put

dA dx dA dx



=

49 49

the ball is thrown also consists of the points A(1, 0, 2), B(3, –1, 1) and C(1, 2, 1) on it. The highest point of the ball takes, is D(2, 3, 1) as shown in the figure, Using this information answer the questions.

1 (p - 4x) 2

=0

1 (p - 4x) = 0 2 p x = Þ 4 dA 1 Also, = ( 0 - 4 ) = –2 < 0 dx 2 2 p Thus, area is maximum at x = 4 Þ



79. The value of y, for which area of floor of auditorium



is maximum is: p (1) 2

(2)



p (3) 4

p (4) 16

p 3



points A, B and C is: (1) 3x – 2y + 4z = – 11 (2) 3x + 2y + 4z = 11 (3) 3x – 2y – 4z = 11 (4) –3x + 2y + 4z = –11

Ans. Option (2) is correct.

Ans. Option (3) is correct. Explanation: From Q. 78, we have x = Substituting x =

81. The equation of the plane passing through the

Explanation: The equation off plane passing through three non-collinear points is given by

p 4

x - x1 x 2 - x1 x3 - x1

p in eq. (iii) of Q. 77 4

We get p p-2´ p - 2x 4 y = = 2 2



=



p 2 = p 2 4

p-

Þ

80. Maximum area of floor is: 2

(1)

p 4

(2)

p 16



(3)

p2 28

(4)

p2 64

Þ 3(x – 1) + 2y + 4(z – 2) = 11

Explanation: From Q. 78 and Q. 79, we have p p x = and y = 4 4 from eq. (ii) of Q. 77, we get A = xy =

82. The maximum heights of the ball from the ground is:

(1)

Ans. Option (2) is correct.



x -1 y z -2 2 -1 -1 = 0 0 2 -1

Þ (x – 1)(1 + 2) – y(–2 + 0) + (z – 2)(4 + 0) = 0 2





z - z1 z2 - z1 = 0 z3 - z1

x -1 y -0 z -2 Þ 3 - 1 -1 - 0 1 - 2 = 0 1-1 2 -0 1-2

p y = 4

\

y - y1 y 2 - y1 y3 - y1

p2 p p ´ = 16 4 4

Read the below passage and solve the questions from Q.No. 81-Q.No.85. Passage A ball is thrown upwards from the plane surface of the ground. Suppose the plane surface from which

(3)

5 29 6 29

units

(2)

units

(4)

7 29 8



29

units units

Ans. Option (1) is correct. Explanation: Height of the ball = Perpendicular distance from point (2, 3, 1) to the plane 3x + 2y + 4z = 11 So,

6 + 6 + 4 - 11 2

2

3 +2 +4

2

=

5 29

=

5 29

units

OSWAAL CUET (UG) Sample Question Papers, Mathematics/App. Math.

50

83. The equation of the perpendicular line drawn from



(2)



(3)



-5 29

Thus, coordinates of foot of perpendicular are éæ æ 5 ö ö æ æ 5 ö öù ö æ æ 5ö êç 3 èç - ø÷ + 2÷ , ç 2 èç - ø÷ + 3÷ , ç 4 èç - ø÷ + 1÷ ú øû ø è ø è 29 29 29 ëè

x -2 y -3 z -1 = = -3 2 -4

éæ -15 ö æ -10 ö æ -20 ö ù + 2÷ , ç + 3÷ , ç + 1÷ ú = êç ø è 29 ø è 29 øû ëè 29

x -2 y -3 z -1 = = 3 2 4 x +1 y +3 z -5 (4) = = 2 -2 -1

æ 43 77 9 ö = ç , , ÷ è 29 29 29 ø

Ans. Option (3) is correct. Explanation: D.R.s of perpendicular are [Since, perpendicular is parallel to the normal to the plane] Since, perpendicular is passing through the point (2, 3, 1), therefore its equation is x -2 y -3 z -1 = = 3 2 4

l =



the maximum height of the ball to the ground, is: x -1 y +3 z -5 (1) = = 2 1 -2

84. The co-ordinates of the foot of the perpendicular drawn from the maximum height of the ball to the ground are



æ 43 -77 -9 ö , (1) ç , è 29 29 29 ÷ø

æ 9 -1 -10 ö , (2) ç , è 7 ÷ø 7 7



æ 43 77 9 ö , (3) ç , è 29 29 29 ÷ø

7 19 ö æ 13 (4) ç - , - , - ÷ è 29 29 29 ø



85. The area of DABC is: 29 sq. units

(2)

1 29 sq. units 4

1 29 sq. units 16

(4)

1 29 sq. units 2



(1)



(3)

Ans. Option (4) is correct. Explanation: Clearly area of DABC 1   = | AB ´ AC | 2 =

i j k 1 = 2 -1 -1 2 0 2 -1

Ans. Option (3) is correct. Explanation: Let the coordinate of foot of perpendicular are (3l + 2, 2l + 3, 4l + 1) Since, this point lie on the plane 3x + 2y + 4z = 11, therefore we get 3(3l + 2) + 2(2l + 3) + 4(4l + 1) = 11 9l + 6 + 4l + 6 + 16l + 4 = 11 29l + 16 = 11 29l = 11 – 16

1 |( 2i - j - k ) ´ ( 0i + 2 j - k )| 2

=

1  |i(1 + 2 ) - j( -2 + 0 ) + k ( 4 + 0 )| 2

=

1  | 3i + 2 j + 4 k | 2

=

1 2 9 + 22 + 42 2

=

1 29 sq. units 2



CUET Question Paper 2021 National Testing Agency

UIQP02 23rd SEP 2021—SHIFT 2

MATHEMATICS/APPLIED MATHEMATICS Solved

(This includes Questions pertaining to Domain Specific Subject only)

Max. Marks : 80

Time allowed : 30 Min.

General Instructions:

(i) This paper consists of 25 MCQs, attempt any 20 out of 25. (ii) Correct answer or the most appropriate answer: Four marks (+4) . (iii) Any incorrect option marked will be given minus One mark (–1) . (iv) Unanswered/Marked for Review will be given No mark (0) . (v) If more than one option is found to be correct then Four marks (+4) will be awarded to only those who have marked any of the correct options . (vi) If all options are found to be correct then Four marks (+4) will be awarded to all those who have attempted the question . (vii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given Four marks (+4). (viii) Calculator / any electronic gadgets are not permitted . 1. Let P be any non-empty set containing p elements. Then, what is the number of relations on P? 2 (1) 2p (2) 2p 2 p (3) p (4) p Ans. Option (2) is correct. Explanation: Any subset of PXP will be a relation, as we know that with p elements 2p subsets are possible, so these are number of relations on a set with p elements will be 2p×p = 2p2 2. The domain of the function defined by f(x) = logx10 is (1) x < 10 excluding x = –10 (2) x > 10 (3) x ³ 10 (4) x > 10 excluding x = 1 Ans. Option (4) is correct. 3. Let det M denotes the determinant of the matrix M. Let A and B be 3 × 3 matrices with det A = 3 and det B = 4. Then the det (2AB) is (1) 24 (2) 42 (3) 96 (4) 48 Ans. Option (3) is correct. Explanation: A and B are square matrices of order 3 \ AB is a square materix of order 3 We know that if A is a square materix or order n then |kA| = kn|A| Hence |2AB| = 23|A||B| = 23 (3) (4) = 96 4. If 19th term of a non-zero arithmetic progression (AP) is zero, then its (49th term) : (29th term) is (1) 2 : 1 (2) 4 : 1 (3) 1 : 3 (4) 3 : 1

Ans. Option (4) is correct. Explanation: Let the first term and common difference of A.P. be a and a respectively Given a19 = 0 \ a + 18d = 0 a = –18d [an = a + (n – 1)d] a49 a + 48d −18d + 48d 30d = = = =3:1 a29 −18d + 28d 10d a + 28d 5. The value of k for which the system of equations x + ky + 3z = 0, 4x + 3y + kz = 0, 2x + y + 2z = 0 has non-trivial solution is 9 (1) k = 0 or (2) k = 10 2 (3) k < 9 (4) k > 0 Ans. Option (1) is correct. Explanation: x + ky + 3z = 0 ù 4x + 3y + kz = 0 ú has non-trival 2x + y + 2z = 0 û 1 k 3 4 3 k =0 2 1 2 3 k 4 k 4 3 1 -k +3 =0 1 2 2 2 2 1 6 – k – 8k + 2k – 6 = 0 2k2 – 9k = 0 k (2k – 9) = 0 \

k = 0 or k =

9 2

OSWAAL CUET (UG) Sample Question Papers, Mathematics/App. Math.

52

6. A card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is either Ace or a King? 4 1 (1) (2) 13 13 2 (3) (4) None of these 13 Ans. Option (3) is correct. Explanation: No. of kings in a pack = 4 No of ace in a pack = 4 \ favourable events n(E) = 4 + 4 = 8 Total events n(S) = 52 Probability of (Ace or King) Favourable events = Total events =

8 2 = 52 13

7. Let E1 and E2 be two independent events. Let P(E) denotes the probability of the occurrence of the event E. Further, let E1 and E2 denote the complements of E1 and E2, respectively. If 2 1 and P(E1 Ç E2') = , then P(E1) is P(E1' Ç E2) = 15 6 2 13 (1) (2) 15 15 2 1 (3) (4) 13 5 Ans. Option (4) is correct. Explanation: P (E1 Ç E2) =

2 (Given) 15



Let P(E1) = x 1 1   2 x− − x  x −  = 30 30 15   x 1 2 = x- x2 + 30 30 15 31x 5 2 −x + − = 0 30 30 or 30x2 – 31x + 5 = 0 30x2 – 25x – 6x + 5 = 0 (6x – 5) (5x – 1) = 0 5 if 6x – 5 = 0 then x = 6 if 5x – 1 = 0 then x = 8. If a =

1 5

2 sin θ 1 + sin θ − cos θ is , then 1 + cos θ + sin θ 1 + sin θ

1 a (3) a Ans. Option (3) is correct. (1)

(2) 1 – a (4) 1 + a

Explanation: a = =

=

2 sin θ 1 + sin θ − cos θ × 1 + cos θ + sin θ 1 + sin θ − cos θ 2 sin θ (1 + sin θ − cos θ )

(1 + sin θ )2 − cos2 θ 2 sin θ (1 + sin θ − cos θ ) 1 + sin 2 θ + 2 sin θ − cos2 θ 2 sin θ (1 + sin θ − cos θ )

P(E1) . P(E2) =

2 (Independent events) 15

=



2 15

1 + sin θ − cos θ = 1 + sin θ =a

[1 – P(E1)]. P(E2) =

2 sin θ (1 + sin θ )

P(E2) – P(E1) . P(E2) =

2 ...(i) 15

P(E1 Ç E1) =

1 6

P(E1) . P(E2) =

1 6

P(E1) (1 – P(E2) =

1 6

Explanation: tan–1 x + tan–1 y + tan–1 z =

P(E1) – P(E1) P(E2) =

1 ...(ii) 6

tan–1 x + tan–1 y =

P(E1) – P(E2) =

1 2 5−4 1 − = = 6 15 30 30

1 P(E2) = P(E1) – ...(iii) 30 Now, from equations (i) and (iii), 1 1   2 P( E1 ) − − P( E1 )  P ( E1 ) −  = 30 30  15 

9. If tan–1x + tan–1y + tan–1z =

p , then 2

(1) x + y+ z – xyz = 0 (2) xy + yz + zx – 1 = 0 (3) x + y + z + xyz = 0 (4) xy + yz + zx + 1 = 0 Ans. Option (2) is correct.

p – tan–1 z 2



 x+y  tan −1   = cot–1 z  1 − xy 



 x+y  1 tan −1   = tan −1   1 xy −   z

x+y 1 = 1 − xy z xz + yz = 1 – xy xy + yz + zx – 1 =0

p 2

Solved Paper-2021 10. The equation ax + by + c = 0 represents a straight line (1) for all real numbers a, b and c (2) only when b ¹ 0 (3) only when a ¹ 0 (4) only when at least one of a and b is non-zero Ans. Option (4) is correct. Explanation: Any equation of the form ax + by + c = 0 will represent a straight line on the xy-plane at least one of a and b is non-zero. 11. The equation of the circle passing through the foci x2 y2 + = 1. of the ellipse 16 9 (1) x2 + y2 – 6y – 5 = 0 (2) x2 + y2 – 6y + 7 = 0 2 2 (3) x + y – 6y – 7 = 0 (4) x2 + y2 – 6y + 5 = 0 Ans. Option (3) is correct. Explanation: Eqn. of ellipse

x2 y2 + =1 16 9

\ a = 4 and b = 3 b2 = a2(1 – e2) 9 = 16(1 – e2) 9 e2 = 1 − 16 =

7 16

\ e =

7 4



Focus (ae, 0) Û

(

-1 + 0 - 1 + 0 - 1 ..........0 = -50 1-0

R.H.L.= lim+ x ®0

[ x] + [ x 2 ] + [ x 3 ] + .......... + [ x 100 ] 1+ x

0 + 0 + 0 + 0.......... + 0 =0 1+0 L.H.L. ≠ R.H.S. So, Limit does not exist =

13. If f(x) = ax2 + 6x + 5 attains its maximum value at x = 1. then the value of a is (1) 0 (2) 5 (3) 3 (4) –3 Ans. Option (4) is correct. Explanation: For maximum/minimum value f'(x) = 0 2ax + 6 + 0 = 0 at x = 1 Þ 2a + b = 0 Þ a = –3 14. The area of the region bounded by the line y = 4 and the curve y = x2 is 32 (1) square units (2) 0 square unit 3 (3) 1 square unit (4) 32 square units Ans. Option (1) is correct. 4

7,0

)

Explanation: Required area = 2 ò0 xdy y y=x2 y=4 X

Circle's centre = (0, 3)

\ Radius of circle =

( 7 )2 + 32 = 4 units = 2 ò

Equation of circle (x – h)2 + (y – k)2 = r2 (x – 0)2 + (y – 3)2 = (4)2 x2 + y2 – 6y + 9 = 16 x2 + y2 – 6y – 7 = 0

3

2 3

3

2 3

= 2( 4 ) 2 ´

å [x r ]

Then lim r =1 x ® 0 1+ | x |

Explanation:

ydy

= 2 ( y ) 2 ´

100

(1) does not exist (3) is 1 Ans. Option (1) is correct.

4

0

12. Let [xr] denotes the greatest integer of xr and |x| denotes the modulus of x.

(2) is –1 (4) is 100

100

å [x r ]

lim r =1 1+ | x |

x ®0

2 3 100 L.H.L. = lim [ x ] + [ x ] + [ x ] + .......... + [ x ] 1-x x ®0-





=

53 53

2 3 100 = lim [ x ] + [ x ] + [ x ] + .......... + [ x ] 1-x x ®0-

=

4 ´8 3

=

32 units2 3

15. The equation of the tangent to the curve given by p x = asin3t, y = bcos3t at a point where t = is 2 (1) y = 1 (3) x = 0 Ans. Option (2) is correct.

(2) y = 0 (4) x = 1 p

Explanation: at t = 2

x = asin3

p =a 2

y = bcos3

p =0 2

OSWAAL CUET (UG) Sample Question Papers, Mathematics/App. Math.

54

x = asin3t

y = bcos3t

dx = 3asin2tcost dt

dy = -3b cos2 t sin t dt

\

dy / dt dy = dx / dt dx



b dy = − cot t a dx

æ dy ö b π çè dx ÷ø p = − cot a 2 t= 2 =0 Equ. of tangent

y – 0 = 0 (x – a) \

y = 0

16. The solution of the differential equation dx + Px = Q, where P and Q are constants or dy

a b c = = −2 − 3 −1 − 4 6 −1



a b c = = 5 −5 −5



a b c = = 1 −1 −1

    d = k i − j − k

(

\

æ dy ö ( x - x1 ) y – y1 = ç ÷ è dx ø x y 1 1



a + 2b – c = 0

2a – b + 3c = 0

−3b cos2 t sin t = 3a sin 2 t cos t

Explanation: Let d = ai + b j + ck     d ^ a and d ^ b

(

)(

)

k i − j − k . 2i + j + 6 k = 10

k(2 – 1 – 6) = 10 k =



10 = –2 −5

 d = – 2i + 2 j + 2 k  d = 22 + 22 + 22



functions of y, is given by

)

 δ.γ = 10

= 2 3 units

(1) xe ò

Pdx

= ò Qe ò

Pdx

dx + c

(2) ye ò

Pdy

= ò Qe ò

Pdy

dy + c

18. Let a , b and c be three unit vectors such that

(3) ye ò

Pdx

= ò Qe ò

Pdx

dx + c

3   a × ( b × c ) = ( b + c ). If b is not parallel to c , then 2

(4) xe ò

Pdy

= ò Qe ò

Pdy

dy + c

the angle between a and c is

Ans. Option (4) is correct. Explanation: Solutions of xe ò

pdy

∫ = ∫ Qe

pdy

dx + Px = Q is given by dy

(3) 3

(4)

 



(4)

3

 

( a.c ) b − ( a.b ) c =

)

( )



  3 a c cosθ =+ 2

cosq =+

Ans. Option (2) is correct. q =

3   b´c 2

(

3 3 b+ c 2 2

on comparing we get  3 a.c = + 2

(2) 2 3 1

p 6

Ans. Option (4) is correct.

(

  and b are both perpendicular to a vector d and   d.g = 10 , then the magnitude of d is

3 2

(2) 2p

   Explanation: a ´ b ´ c =

dy + c

    17. Let a = i + 2i - k , b = 2i - j + 3k , g = 2i + j + 6 k . If α

(1)

(1) 0 5p (3) 6

3 2 p 6

)

Solved Paper-2021 19. Which one of the following statements is correct for a moving body? (1) If its velocity changes, its speed must change and it must have some acceleration (2) If its speeds changes, its velocity must change and it must have some acceleration (3) If its speed changes but direction of motion does not change, its velocity will remain constant (4) None of the above Ans. Option (1) is correct. Explanation: When velocity changes then its magnitude (speed) and direction may change. Hence, the speed must change and must have acceleration. 20. a ball is thrown upward at a speed of 28 metre per second. What is the speed of ball one second before reaching maximum height? (Given that g = 10 metre per second2) (1) 10 metre per second (2) 1 metre per second (3) 2 metre per second (4) 18 metre per second Ans. Option (1) is correct. Explanation:

v = u + at

where

v = final velocity



t = time taken



u = initial velocity



a = acceleration

\

V = 4 – gt



0 = 28 – 10t

\

t = 2.8

(upwards)

V = u – gt

= 28 – 10 × 1.8 = 28 – 18 = 10 m/s 21. A bullet of mass m and velocity a is fired into a large block of wood of mass M. Then final velocity of the system is m M (1) a (2) a m+M m+M m+M a m

(4)

22. If the horizontal and vertical components of a force are negative, then that force is acting in between (1) North and East (2) North and West (3) South and West (4) South and East Ans. Option (3) is correct. Explanation: Force is acting between south and west. N

W

E

S

\ Velocity at 2.8 – 1 = 1.8 s

(3)

Momentum before collision = m × a + m × o = ma Momentum after collision = (M+m)v (where v is velocity of the system) By law of conservation of momentum ma = (M + m)v m a v = \ M+m

force

Mass height reach at in 2.8 s

55 55

m+M a M

Ans. Option (1) is correct. Explanation: Since there is no extra force other than the action and reaction force, so the linear momentum should be conserved.

23. Suppose we have block of 4 kilogram kept on a horizontal surface and we are applying a horizontal force of 10 newton. Let the coefficient of friction is 0.2. Find the force of friction. Assume that g = 10. (1) 4 newton (2) 8 newton (3) 1 newton (4) None of above Ans. Option (2) is correct. Explanation: From the figure If the block is in motion, applying the force 10 N then Force of friction (Fr) = mN = m mg = 0.2 × 4 × 10 = 8.0 N Here µ is coefficient of friction 24. The general solution of the differential equation dy x + = 0 is dx y (1) x2 + y2 = cxy (3) x2 – y2 = c Ans. Option (2) is correct.

(2) x2 + y2 = c (4) x + y = c

OSWAAL CUET (UG) Sample Question Papers, Mathematics/App. Math.

56

Explanation:

dy x + =0 dx y x dy = − y dx



ò



ydy = – xdx ydy = - xdx

ò

y2 −x 2 = + c1 2 2

x2 + y2 = 2c1 x2 + y2 = c (c = 2c1)  cos x sin x   25. Let A =  2  sin x cos2 x    2

2

 sin 2 x cos2 x  and B =  .  cos2 x sin 2 x   

Then the determinant of the matrix A + B is (1) 1 (2) 10 (3) 0 (4) 2 Ans. Option (3) is correct. Explanation: cos2 x sin 2 x  sin 2 x cos2 x  +  A+B=  2 sin x cos2 x  cos2 x sin 2 x  écos2 x + sin 2 x = ê 2 êësin x + cos2 x

sin 2 x + cos2 x ù ú cos2 x + sin 2 x úû

1 1 1 1 =1–1=0 =  =  1 1 1 1 

SAMPLE

1

Question Paper Maximum Marks : 200

Time : 45 Min.

General Instructions :

(i) Section A will have 15 questions covering both i.e., Mathematics/Applied Mathematics which will be compulsory for all candidates. (ii) Section B1 will have 35 questions from Mathematics out of which 25 questions need to be attempted. Section B2 will have 35 questions purely from Applied Mathematics out of which 25 questions will be attempted. (iii) Correct answer or the most appropriate answer : Five marks (+ 5) (iv) Any incorrect option marked will be given minus one mark (– 1). (v) Unanswered/Marked for Review will be given no mark (0). (vi) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options. (vii) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question. (viii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (ix) Calculator / any electronic gadgets are not permitted.

Section - A





Mathematics/Applied Mathematics

1. If

π  mx + 1 if x ≤ 2 f (x) =  , sin x + n, if x > p  2

p x= then 2 (1) m = 1, n = 0



(3) n =

mp 2

is continuous at



3. Assertion (A):

np +1 2 p (4) m= n= 2

2 3 2 2. If x x x + 3 = 0 , then the value of x is 4 9 1



(1) 3 (2) 0



(3) – 1

(4) 1

Directions : In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as:

(A) Both A and R are true and R is the correct explanation of A



(B) Both A and R are true but R is NOT the correct explanation of A



(C) A is true but R is false



(D) A is false and R is True

∫ 0

Reason (R):

p/2

∫ 0

(2) m =

π/2



4. Assertion (A):

Reason (R):

cos x π dx = sin x + cos x 4

sin x p dx = sin x + cos x 4 x2  d  dt  2x = dx  ∫0 t 2 + 4  x 4 + 4   dx

1

∫ x 2 + a2 = a tan

x  +c a

−1 

5. Let T be the set of all triangles in the Euclidean



plane, and let a relation R on T be defined as aRb if a is congruent to b " a, b Î T. Then R is (1) reflexive but not transitive (2) transitive but not symmetric (3) equivalence relation (4) None of these



6. The area of the region bounded by the y-axis, y = cos x and y = sin x, 0 ≤ x ≤ p/2 is 2 sq. units

(2) ( 2 + 1) sq. units

(3) ( 2 - 1) sq. units

(4) (2 2 - 1) sq. units

(1)

7. A ladder, 5 metre long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of 10 cm/sec, then the rate at which the angle between the floor

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

2

and the ladder is decreasing when lower end of ladder is 2 metre from the wall is : 1 1 radian/sec radian/sec (1) (2) 20 10



(3) 20 radian/sec



(4) 10 radian/sec

8. If A and B are two events such that P(A) ≠ 0 and



P(B|A) = 1, then (1) A ⊂ B (2) B⊂A (3) B = j (4) A = j 3p   9. The value of sin–1  cos  is 5   p (1) 10



(2)

p 10



(4)

11. The equation of normal to the curve 3x 2 - y 2 = 8 which is parallel to the line x + 3 y = 8 is



(1) 3x – y = 8

(2) 3x + y + 8 = 0



(3) x + 3 y ± 8 = 0

(4) x + 3 y = 0



12. Consider the non-empty set consisting of children











 (2) 7  5

16. The value of sin -1  cos  33   is  5  3 (1) 5  (3) 10

(4)

 10

(3)

2 x

2



2 (4) - 2 x

18. If the curve ay + x 2 = 7 and x3 = y, cut orthogonally

at (1, 1), then the value of a is : (1) 1 (2) 0 (3) –6 (4) 6 200 50    50 40  19. Let A =  and B =    , then |AB|  10 2  2 3 is equal to (1) 460 (2) 2000 (3) 3000 (4) – 7000

by the system of linear constraints are (0, 0), (0, 40), (20, 40), (60, 20), (60, 0). The objective function is Z = 4x + 3y. Compare the quantity in Column A and Column B



Column B

Maximum of Z 325 (1) The quantity in column A is greater. (2) The quantity in column B is greater. (3) The two quantities are equal. (4) The relationship cannot be determined on the basis of the information supplied. /2

14. The value of   / 2 ( x 3  x cos x  tan 5 x  1) dx is (1) 0 (3) p

(2) 2 (4) 1

15. Distance of the point (α, β, γ) from y-axis is (1) b

(2) |b|

(3) |b|+ |g|

(4)

2   2

Mathematics

 x2  d2y equals 17. If y = loge  2  , then dx 2 e  1 1 (1) (2) - 2 x x

13. The corner points of the feasible region determined

Section - B1





Column A

3p 5

3 5 10. If P(A|B) > P(A), then which of the following is correct : (1) P(B|A) < P(B) (2) P(A∩B) < P(A).P(B) (3) P(B|A) > P(B) (4) P(B|A) = P(B) (3) -





in a family and a relation R defined as aRb if a is brother of b. Then R is (1) symmetric but not transitive (2) transitive but not symmetric (3) neither symmetric nor transitive (4) both symmetric and transitive



20. If A and B are any two events such that P(A) + P(B) – P(A and B) = P(A), then



(1) P(B|A) = 1

(2) P(A|B) = 1



(3) P(B|A) = 0

(4) P(B|A) = 0



21. The sine of the angle between the straight line

x - 2 y - 3 z - 4  and the plane 2x – 2y + z = 5 is = = 3 4 5

(1)

10 6 5

(3)

2 3 5

(2)

4 5 2

(4)

2 10

22. The area of the region bounded by the curve x2 = 4y and the straight-line x = 4y – 2 is



(1)

3 sq. units 8

(2)

5 sq. units 8



(3)

7 sq. units 8

(4)

9 sq. units 8

Sample Question Papers

a -b b +c a



c -a a +b c



23. The value of determinant b - a c + a b is 3

3

3

(1) a + b + c



(2) 3bc

(3) a3 + b 3 + c 3 - 3abc (4) None of these 24. In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is

 1 (2)    2

(1) 10–1 5

 9 (3)    10 



(4)

5

9 10

25. The degree of the differential equation 2

2  d2y   dy   dy    dx   x sin  dx  is  2  dx  (1) 1 (2) 2 (3) 3 (4) not defined



26. The set of points where the function f given by f ( x ) = | 2 x - 1|sin x is differentiable is

1 2

(1) R



(3) (0,  )



(2) R –  

(4) none of these

27. The equation of tangent to the curve y(1 + x2) = 2 – x, where it crosses x-axis is : (1) x + 5 y = 2 (2) x – 5 y = 2

(3) 5x – y = 2 (4) 5 y + x = 2 28. If a relation R on the set {1, 2, 3} be defined by R = {(1, 2)}, then R is (1) reflexive (2) transitive (3) symmetric (4) None of these  29. The distance of the plane r.  2 ˆi + 3 ˆj − 6 kˆ = 1 7 7 7  from the origin is (1) 1 1 (3) 7





3+ 5 2 -3 + 5 (3) 2 cos 2 x (1)





(4) None of these

   30. The value of tan  1 cos1  5   is  3    2





(2) 7

31.

 (sin x  cos x )2 dx

(2)

3- 5 2

(4)

-3 - 5 2

is equal to

-1 +C sin x + cos x (2) log |sin x + cos x| + C (3) log |sin x – cos x| + C 1 (4) (sin x + cos x )2 (1)





3

32. The maximum number of equivalence relations on the set A = {1, 2, 3} are (1) 1 (2) 2 (3) 3 (4) 5

33. The mean of the numbers obtained on throwing a

die having written 1 on three faces, 2 on two faces and 5 on one face is (1) 1 (2) 2 8 (3) 5 (4) 3 34. If y = e–x (Acos x + Bsin x), then y is a solution of

(1)



(3)

d2y dx

2

d2y

d2y

dy + 2y = 0 dx

+2

dy =0 dx

+2

d2y dy + 2 y = 0 (4) 2 + 2 y = 0 dx dx

(2)

2 dx

-2

dx 2 35. The probability of obtaining an even prime number on each die, when a pair of dice is rolled is 1 (1) 0 (2) 3 1 1 (4) 12 36    36. Let a , b , and c be three unit vectors, out of which vectors b and c are non-parallel. If a and b are the angles which vector a makes with c  vectors b and      b respectively and a × (b × c ) = , then |  –  | is 2 equal to : (1) 30° (2) 90° (3) 60° (4) 45°



(3)

37. The distance of the point having position vector – i + 2 j + 6 k from the straight line passing through the point (2, 3, –4) and parallel to the vector   6 i + 3 j – 4 k is : (2) 4 3



(1) 7



(3) 2 13



1 x

38. If y  log 

(4) 6 2 2

1 x 

(1)

4x3

1-x 4

, then

dy is equal to dx (2)

-4 x

1 - x4

1

3 (4) -4 x 1 - x4   39. Let a = 3i + 2 j + 2 k and b = i + 2 j – 2 k be two vectors. If a vector perpendicular to both the    vectors a + b and a – b has the magnitude 12 then one such vector is : (1) 4(2i – 2j – k ) (2) 4(2i – 2j + k )



(3)



(3) 4(2i + 2j + k )





4 -x 4

(4) 4(2i + 2j – k )

40. The sum of the distinct real values of m, for which

the vectors, mi + j + k , i + m j + k , i + j + m k are coplanar, is : (1) –1 (2) 0 (3) 1 (4) 2

4

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

41. The feasible solution for a LPP is shown in given figure. Let Z = 3x – 4y be the objective function Minimum of Z occurs at

(6, 8)

(6, 5)

(5, 0) (0, 0) (1) (0, 0) (2) (0, 8) (3) (5, 0) (4) (4, 10)   42. Let  = (l – 2)a + b and  = (4l – 2)a + 3b be two given vectors where a and b are non collinear. The   value of l for which vectors α and β are collinear,

is  : (1) –4 (3) 4



(1) f is everywhere differentiable (2) f is everywhere continuous but not differentiable at x = np, n Î Z. (3) f is everywhere continuous but not differentiable  at x = (2n + 1) , n Î Z. 2 (4) none of these



(2) –3 (4) 3

43. Let f ( x ) = |sin x |, then



44. The integrating factor of differential equation dy  y sin x  1 is dx (1) cos x (3) sec x

cos x

(3) k = 1



(2) tan x (4) sin x





49. Which of the following matrix equations represent the information given above?

5 (1)  5 5 (2)  5 

4   x   40   8   y   80  4   x   40   8   y   80 

 5 4   x   40  (3)      5 8   y   80   5 4   x   40  (4)      5 8   y   80  50. The number of children who were given some money by Seema, is (1) 30 (2) 40 (3) 23 (4) 32

Section - B2

Applied Mathematics

16. If a ≡ b (mod n), then (1) n|a and n|b (3) n|(a – b)



(2) n|b only (4) None of these

17. Evaluate: (9 + 23) mod 12 = .......

48. The equations in terms of x and y are (1) 5x – 4y = 40 (2) 5x – 4y = 40 5x – 8y = –80 5x – 8y = 80 (3) 5x – 4y = 40 (4) 5x + 4y = 40 5x + 8y = –80 5x – 8y = –80



(2) 0 < k < 1 1 1 (4) k = or 3 3





(0, k) is 9 sq. units. Then, the value of k will be (1) 9 (2) 3 (3) –9 (4) 6





45. If the direction cosines of a line are k, k, k, then (1) k > 0



46. The area of a triangle with vertices (–3, 0), (3, 0) and

47. Refer to Q. 41 maximum of Z occurs at (1) (5, 0) (2) (6, 5) (3) (6, 8) (4) (4, 10) Read the following text and answer the following questions on the basis of the same: On her birthday, Seema decided to donate some money to children of an orphanage home. If there were 8 children less, everyone would have got `10 more. However, if there were 16 children more, everyone would have got `10 less. Let the number of children be x and the amount distributed by Seema for one child be y (in `).

(4, 10)

(0, 8)



(1) 2 (3) 32

(2) 8 (4) 12



18. If B > A, then which expression will have the highest value, given that A and B are positive integers. (1) A – B (2) A × B (3) A + B (4) Can't say

19. Tea worth ` 126 per kg and ` 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture

Sample Question Papers per kg will be: (1) ` 169.50 (2) ` 170 (3) ` 175.50 (4) ` 180 20. A = [ aif ]m × n is a square matrix, if (1) m < n (3) m = n

(2) m > n (4) None of these 2

21. I f A is a square matrix such that A = A, then (I + A)³ – 7A is equal to (1) A (2) I – A (3) I (4) 3A 22. Which of the given values of x and y make the following pair of matrices equal 5  0 y − 2  3x + 7  y + 1 2 − 3x  , 8 4     −1 , y=7 (1) x = (2) Not possible to find 3 (3) y = 7 , x =

−2 3

(4)



x=

−1 −2 , y= 3 3

23. Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. The restriction on n, k and p so that PY + WY will be defined are: (1) k = 3, p = n (2) k is arbitrary, p = 2 (3) p is arbitrary, k = 3 (4) k = 2, p = 3 24. If A and B are symmetric matrices of same order, AB – BA is a: (1) Skew-symmetric matrix (2) Symmetric matrix (3) Zero matrix (4) Identity matrix

25. The function f ( x ) = 2 x 3 − 3x 2 − 12 x + 4 , has (1) two points of local maximum (2) two points of local minimum (3) one maxima and one minima (4) no maxima or minima

26. The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 3x2 + 26x + 15. The marginal revenue, when x = 15 is : (1) ` 100 (2) ` 115 (3) ` 123 (4) None 27. The maximum profit that a company can make, if the profit function is given by p ( x ) = 41 − 72 x − 18 x 2 is : (1) 111 (3) 113

28. If y = x3 log x, then (1) 6x (3)

x 6

(2) 112 (4) 114 d4y is : dx 4 (2)

6 x

(4) log 6

29. If x is real, the minimum value of x2 – 8x + 17 is (1) –1 (2) 0 (3) 1 (4) 2

5

30. A candidate claims 70% of the people in her constituency would vote for her. If 1,20,000 valid votes are polled, then the number of votes she expects from her constituency is (1) 100000 (2) 84000 (3) 56000 (4) 36000 d2y 31. Given that x = at2 and y = 2at then is dx 2 –1 (1) –1 (2) 2 2at 3 2at 1 (4) –2a t2 t 32. The expectation of a random variable X (continuous or discrete) is given by.......... (1) ∑X f(x), ∫ X f(X) (2) ∑X2 f(X), ∫ X2 f(X) (3) ∑f(X), ∫ f(X) (4) ∑X f(X2), ∫ X f(X2)

(3)



33. Mean of a constant ‘a’ is ............... (1) 0 (2) a (3) a/2 (4) 1 34. Find the expectation of a random variable X. X

0

1

2

3

f(X)

1 6

2 6

2 6

1 6

(1) 0.5 (3) 2.5

(2) 1.5 (4) 3.5

35. Skewness of Normal distribution is ........ . (1) Negative (2) Positive (3) 0 (4) Undefined 36. Which of the following values is used as a summary measure for a sample, such as a sample mean ? (1) Population Parameter (2) Sample Parameter (3) Sample Statistic (4) Population mean 37. A simple random sample consist of four observation 1, 3, 5, 7. What is the point estimate of population standard deviation ? (1) 2.3 (2) 2.52 (3) 0.36 (4) 0.4 38. A price index which is based on the prices of the items in the composite, weighted by their relative index is called: (1) price relatives (2) Consumer price index (3) Weighted aggregative price index (4) Simple aggregative index 39. Which of the following is an example of line series problem ? (i) Estimating numbers of hotel rooms booking in next 6 months. (ii) Estimating the total sales in next 3 years of an insurance company. (iii) Estimating the number of calls for the next one week. (1) Only (iii) (2) (i) and (ii) (3) (i), (ii) and (iii) (4) (ii) and (iii)

6

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

40. In Paasche’s price index number weight is considered as (1) Quantity in base year (2) Quantity in current year (3) Prices in base year (4) Prices in current year 41. Moving average method is used for measurement of trend when: (1) Trend is linear (2) Trend is non-linear (3) Trend is curvilinear (4) None of these 42. The present value of a sequence of payment of ` 1000 made at the end of every 6 months and continuing forever, if money is worth 8% per annum compounded semi-annually is (1) 1000 (2) 2500 (3) 25,000 (4) 15,000

43. Assume that Shyam holds a perpetual bond that

Read the following text and answer the following questions on the basis of the same : Rohan has completed his MBA and now he wants to start a new business. So, he approaches to many banks. One bank is agreed to give loan to Rohan. So, Rohan has borrowed ` 5 lakhs from a bank on the interest rate of 12 per cent for 10 years. 46. EMI stands for: (1) Equated Monthly Installments (2) Emerging Monthly Installments (3) Easy Monthly Installments (4) None of the above 47. To calculate monthly installment, we use the following formula : (1) Installment Amount =

(1 + i )n (1 + i )n

× (P × i)

(1 + i )n

× (P × i)

generates an annual payment of ` 500 each year. He believes that the borrower is creditworthy and that an 8% interest rate will be suitable for this bond. The present value of this perpetuity is (1) ` 6520 (2) ` 6250 (3) ` 5620 (4) ` 2650

(2) Installment Amount =

44. Feasible region in the set of points which satisfy (1) The objective functions (2) Some the given constraints (3) All of the given constraints (4) None of these

48. Calculate monthly installment using (1.01)120 = 3.300 (1) ` 7100

(2) ` 7174

(3) ` 7147

(4) ` 7200

45. Z = 20x1 + 20x2, subject to x1 ≥ 0, x2 ≥ 0, x1 + 2x2 ≥ 8, 3x1 + 2x2 ≥ 15, 5x1 + 2x2 ≥ 20. The minimum value of Z occurs at

(1) ` 8,60,88



(1) (8, 0)



7 9 (3)  ,  2 4

 5 15  (2)  ,  2 4  (4) (0, 10)

(3) Installment Amount =

(1 + i )n − 1 (1 + i )n (1 + i )n −1

× (P × i)

(4) None of these



49. Find the amount of total payment made by Rohan.

(3) ` 8,60,000

(2) ` 8,80,880 (4) ` 8,60,880

50. Find the amount of interest paid by Rohan. (1) ` 3,60,88

(2) ` 3,60,880

(3) ` 3,60,00

(4) ` 3,600,88





Sample Question Papers

SAMPLE

Question Paper Maximum Marks : 200

7

2 Time : 45 Min.

General Instructions :

(i) Section A will have 15 questions covering both i.e., Mathematics/Applied Mathematics which will be compulsory for all candidates. (ii) Section B1 will have 35 questions from Mathematics out of which 25 questions need to be attempted. Section B2 will have 35 questions purely from Applied Mathematics out of which 25 questions will be attempted. (iii) Correct answer or the most appropriate answer : Five marks (+ 5) (iv) Any incorrect option marked will be given minus one mark (– 1). (v) Unanswered/Marked for Review will be given no mark (0). (vi) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options. (vii) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question. (viii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (ix) Calculator / any electronic gadgets are not permitted.

Section - A



Mathematics/Applied Mathematics 3





2

1. If A = [2 –3 4], B =  2  , X = [1 2 3] and Y =  3  ,  2  then AB + XY equals (1) [28] (2) [24] (3) 28 (4) 24

2. Let us define a relation R in R as aRb if a ³ b. Then R is (1) an equivalence relation (2) reflexive, transitive but not symmetric (3) symmetric, transitive but not reflexive (4) neither transitive nor reflexive but symmetric.

3. If P(A) = 0.4, P(B) = 0.8 and P(B|A) = 0.6, then P(A∪B) is equal to (1) 0.24 (3) 0.48

(2) 0.3 (4) 0.96 b



4. If f(a + b – x) = f(x), then  xf ( x ) dx is equal to a

b



 4 

(1)

ab f ( b  x ) dx 2  a

(3)

b

ba f ( x ) dx 2 a

(2)

ab b f ( b  x ) dx 2 a

ab b f ( x ) dx (4) 2 a

b 2 - ab b - c



bc - ac

c - a ab - a 2



to (1) abc( b - c )( c - a )( a - b )



(2) ( b - c )( c - a )( a - b )



(3) ( a + b + c )( b - c )( c - a )( a - b )



(4) None of these



bc - ac

5. The determinant ab - a 2 a - b b 2 - ab is equal

6. The points at which the tangents to the curve y = x 3 – 12 x + 18 are parallel to x-axis are : (1) (2, –2), (–2, –34) (2) (2, 34), (–2, 0) (3) (0, 34), (–2, 0) (4) (2, 2), (–2, 34)

7. The function f ( x ) =

4 - x2

4x - x3 (1) discontinuous at only one point (2) discontinuous at exactly two points (3) discontinuous at exactly three points (4) none of these

8. Let A = {1, 2, 3} and consider the relation R = (1,

1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}. Then R is (1) reflexive but not symmetric

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

8

(2) reflexive but not transitive (3) symmetric and transitive (4) neither symmetric, nor transitive



9. A box has 100 pens of which 10 are defective. What



is the probability that out of a sample of 5 pens drawn one by one with replacement at most one is defective? 5



Ê 9ˆ (1) Á ˜ Ë 10 ¯



(3)

1Ê 9 ˆ (2) Á ˜ 2 Ë 10 ¯ 5



1Ê 9 ˆ Á ˜ s 2 Ë 10 ¯

4

5

1Ê 9 ˆ Ê 9ˆ (4) Á ˜ + Á ˜ Ë 10 ¯ 2 Ë 10 ¯

4

1 10. The value of expression 2 sec−1 2 + sin −1   is 2





π (1) 6

5π (2) 6

7π (3) 6

(4) 1

11. The area of the quadrilateral ABCD, where A(0, 4, 1), B(2, 3, –1), C(4, 5, 0) and D(2, 6, 2), is equal to



(1) 9 sq. units

(2) 18 sq. units



(3) 27 sq. units

(4) 81 sq. units



(1) ( y ¢ )2 + x = y 2



2 2 (3) y ¢¢¢ + ( y ¢¢ ) + y = 0 (4) y ¢ = y



(2) y ¢y ¢¢ + y = sin x

13. The function f(x) = cot x is discontinuous on the set



(1) {x = np ; n ΠZ}



(2) {x = 2np ; n Œ Z} p Ï ¸ (3) Ìx = ( 2n + 1) ; n Œ Z ˝ Ó 2 ˛



np Ï ¸ (4) Ìx = ; n Œ Z˝ 2 Ó ˛



14. The value of



(1) 1



(3) –1



1

Ú0 tan

2x - 1 ˆ dx is ËÁ 1 + x - x 2 ¯˜

-1 Ê

(2) 0 p (4) 4

15. The tangent to the curve y = e2x at the point (0, 1) meets x-axis at :



(1) (0, 1)



(3) (2, 0)

Ê 1 ˆ (2) Á - , 0˜ Ë 2 ¯ (4) (0, 2)

Mathematics

16. Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer: (1) R is reflexive and symmetric but not transitive (2) R is reflexive and transitive but not symmetric (3) R is symmetric and transitive but not reflexive (4) R is an equivalence relation

17. The degree of differential equation 3

d 2 y Ê dy ˆ + Á ˜ + 6 y 5 = 0 is dx 2 Ë dx ¯ (1) 1 (2) 2 (3) 3 (4) 5

differential equation?



Section - B1



12. Which of the following is a second-order

differential equation



(1)



(3)



1 1 and , respectively. If 3 4 the probability of their making a common error 1 and they obtain the same answer, then the is, 20 probability of their answer to be correct is

a problem correctly are

1 (2) 40 (4)

19. y = ae mx + be - mx satisfies which of the following

10 13



dy + my = 0 dx d2y dx 2

- m2 y = 0

(2) (4)

dy - my = 0 dx d2y dx 2

+ m2 y = 0

2 λ −3 5 . Then A–1 exist if 1 1 3

20. If A = 0 2

18. A and B are two students. Their chances of solving

1 (1) 12 13 (3) 120



(1) l = 2 (3) l ≠ –2

(2) l ≠ 2 (4) None of these

21. If y = sin x + y , then dy is equal to dx

cos x (1) 2y - 1 sin x (3) 1 - 2y

(2)

cos x 1 - 2y

(4)

sin x 2y - 1

Directions : In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as (1) Both A and R are true and R is the correct explanation of A

Sample Question Papers



(2) Both A and R are true but R is NOT the correct explanation of A (3) A is true but R is false (4) A is false and R is True



22. Assertion (A) : The area enclosed by the circle



x2 + y2 = a2 is pa2.



9

24. In a college, 30% students fail in physics, 25% fail in mathematics and 10% fail in both. One student is chosen at random. The probability that she fails in physics if she has failed in mathematics is

1 (1) 10

(2)

2 5

9 1 (4) 20 3 25. The locus represented by xy + yz = 0 is

(3)



(1) A pair of perpendicular lines



(2) A pair of parallel lines



(3) A pair of parallel planes



(4) A pair of perpendicular planes



26. Two dice are thrown. If it is known that the sum of numbers on the dice was less than 6, the probability of getting a sum 3, is



(1) 1 18 1 (3) 5

Reason (R): The area enclosed by the circle



a

 4  xdy 0 a

 4

0

a 2  y 2 dy a



y 2 a2 y  4 a  y2  sin 1  2 2 a  0   a   a2  4    0  sin 1 1  0  2 2     a2  2 2  a 2



5 18

(4)

2 5

27. Let a ∈ R and the three vectors a = ai + j + 3 k ,   b = 2i + j – a k and c = ai – 2 j + 3 k . Then the set    S (a : a , b and c are coplanar)



(1) Contains exactly two numbers only one of which is positive



(2) Is empty



(3) Contains exactly two positive numbers



(4) Is singleton

4

23. Assertion (A): The area of the region bounded by 3 the curve y = x2 and the line y = 4 is . 32 Reason (R):

(2)



28. Let A = {1, 2, 3}. Then number of relations



containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is (1) 1 (2) 2



(3) 3



(4) 4 3

29. The function f ( x )  4 sin x  6 sin 2 x  12 sin x  100 is strictly

Since the given curve represented by the equation y = x2 is a parabola symmetrical about y-axis only, therefore, from figure, the required area of the region AOBA is given by



 3  (1) increasing in   ,  2 



 (2) decreasing in  ,   2 



    (3) decreasing in  ,  2 2 



  (4) decreasing in  0 ,   2

4

A  2  xdy 0 4

 2

0

y dy

4 2  2   y 3 / 2  0 3

4 8 3 32  3 





30. The value of

∫ sin

2

x.dx is



(1)

sin 2 x x – 4 2

(2)

x sin 2 x + +C 4 2



(3)

sin 2 x x + 4 2

x sin 2 x − +C (4) 2 4

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

10

31. Which of the given values of x and y make the following pair of matrices equal 55    0 0 yy22   3 3xx77  yy11 2233xx , , 8 8 44       



-1 (1) x = , y =7 3



(3) y = 7 , x =



-2 3



-1 -2 (4) x = ,y= 3 3



32. Distance between the two planes: 2x + 3y + 4z = 4



(1) 2 units (3) 8 units

(1) 2



(3)



2 x

2

(4)

29

unit

(2)

2 1 - x2

(4) 1 - x 2



4 (3) 3

(2)

 6

3π (4) 4

35. Let a, b and c be three unit vectors such that

a + b + c = 0. If l = a.b + b.c + c. a and d = a × b + b × c + c × a, then the ordered pair, (l, d) is equal to :



 −3  (1)  , 3c × b   2 

3  (2)  , 3a × c  2 



 −3  (3)  , 3a × b   2 

3  (4)  , 3b × c  2 



37. If

1 4 × 13 51

x 2 6 2 = then x is equal to 18 x 18 6

(1) 6

(2) ± 6

(3) –6

(4) 0

is equal to

sin( x - b ) +C sin( x - a )

(2) cosec( b - a )log

sin( x - a ) +C sin( x - b )



(3) cosec ( b - a )log

sin( x - b ) +C sin( x - a )



sin( x - a ) +C (4) sin( b - a )log sin( x - b )



41. Let x0 be the point of Local maxima of f(x) = a.(b × c),   a  xi  2 j  3k , b  2i  x j  k

where

and

 c = 7i  2 j  xk . Then the value of a.b + b.c + c.a at

x = x0 is : (1) –22 (3) –30

(2) –4 (4) 14 d2y 42. If x = t2 and y = t3 then is dx 2





(4)

x

∫ sin(x − a)sin( x − b )



52 playing cards with replacement. The probability, that both cards are queens, is

1 1 × (3) 13 17

40.

(2)

(1) sin( b - a )log



1 1 (2) 13  13

 2 π (4) 3

5 6 π (3) 4 (1)



36. Two cards are drawn from a well shuffled deck of

1 1 (1)  13 13





-1

 1   1  1 is cos 1    2 sin 1    4 tan 1   2  2  3   3

39. The principal value of cot -1 ( - 3 ) is



34. The principal value of

(1)

        vector and b × c = b × a , a . c = 0, then b . c is : 1 1 (1) (2) 2 3 1 1 (4) − (3) − 3 2



(2) 4 units

33. The derivative of cos-1 ( 2 x 2 - 1) w.r.t. cos–1 x is



  38. Let a = i – 2 j + k , b = i – j + k and c is nonzero



(2) Not possible to find

and 4x + 6y + 8z = 12 is





3 2 3 (3) 2t

3 4t 3 (4) 4 43. The planes : 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are (1) Perpendicular (2) Parallel (3) Intersect y-axis 5  (4) Passes through  0 , 0 ,   4 (1)

(2)

 cos   sin   , then A + A’ = I, then the cos  

44. If A =  sin   value of a is:

(1)

π 6

(2)

π 3

3π (3) p (4) 2

Sample Question Papers

45. The

smallest value of the x 3 - 18 x 2 + 96 x in [0, 9] is (1) 126 (2) 0 (3) 135 (4) 160



polynomial

46. If A and B are invertible matrices, then which of the following is not correct?



(1) adj A = A . A -1

(2) det( A -1 ) = [det( A )]



(3) (AB) = B -1 A -1

(4) (A + B) = B -1 + A -1

-1

-1

kg per bag Brand P

Brand Q

Nitrogen

3

3.5

Phosphoric acid

1

2

Potash

3

1.5

Chlorine

1.5

2

47. What is the total number of bags used by fruit



grower to minimise the amount of nitrogen ? (1) 160 (2) 190 (3) 140 (4) 130



48. If the grower wants to minimise the amount of



nitrogen added to the garden, how many bags of brand P should be used? (1) 40 (2) 50 (3) 100 (4) 60



49. If the grower wants to minimise the amount of

-1

Read the following text and answer the following questions on the basis of the same: A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine.

11



nitrogen added to the garden, how many bags of brand Q should be used? (1) 40 (2) 50 (3) 100 (4) 60



50. The Objective function to minimise the amount of



nitrogen added to garden? (1) Maximise Z = 3x + 4y (2) Minimise Z = 3x + 3.5y (3) Maximise Z = 4x + 3.5y (4) Minimise Z = 3x + 4y

Section - B2



Applied Mathematics

16. For any integers a and b, and positive integer n, then

a ≡ a(mod n) = ...... (1) 0

(2) n



(3) a

(4) 1

17. Assume the current time is 2 : 00 p.m.; what time (in a.m. or p.m.) will after 65 hours? (1) 7 a.m. (2) 7 p.m. (3) 3 a.m.

(4) 3 p.m.

21. If the matrix A is both symmetric and skew symmetric, then (1) A is a diagonal matrix (2) A is a zero matrix (3) A is a square matrix (4) None of these

0 1 2  then A is equal to 1 0

22. If A = 

18. If 0 < x < 1, which of the following is greatest. (1) x (2) x2

0 1 (1) 1 0  

(2)

1 0  1 0   

1 (3) x

0 1 (3) 0 1  

(4)

1 0 0 1  

(4)

1 x

2

19. Tap P alone fills a cistern in 2 hours; while tap Q alone fills the same cistern in 3 hours. A new tap R is attached to the bottom of the cistern which can empty the completely filled cistern in 6 hours. Sunny started all three taps together at 9 a.m. When will the tank be full? (1) 10.30 a.m. (2) 11.15 a.m. (3) 12 p.m.

(4) 9.45 a.m.

20. The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is: (1) 27 (2) 18 (3) 81

(4) 512

23. If x ≡ −4 (mod 3), then a solution for x is: (1) –2 (2) 12 (3) 19 (4) 35 24. The interval on which the function f (x) = 2x3 + 9x2 + 12x – 1 is decreasing is. (1) [–1, ∞] (2) [–2, –1] (3) [– ∞, –2] (4) [–1, 1] 25. Assume X, Y, Z W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. If n = p, then the order of the matrix 7X – 5Z is: (1) p × 2 (2) 2 × n (3) n × 3 (4) p × n

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

12

α

β 

26. If A =   is such that A² = I, then  γ −α 

(1) 1 + α² + βγ = 0 (3) 1 – α² – βγ = 0

(2) 1 – α² + βγ = 0 (4) 1 + α² – βγ = 0

0 0 4  27. The matrix P =  0 4 0  is a  4 0 0 

(1) square matrix (3) unit matrix

(2) diagonal matrix (4) None of these

28. A person can row in still water at the rate of 8 km/h. If it takes him thrice as long to row upstream as to row downstream then the speed of the stream is: (1) 2 km/h (2) 3 km/h (3) 4 km/h (4) 6 km/h 29. The variable cost of producing x units is V(x) = x2 + 2x. If the company incurs a fixed cost of ` 10,000, then the level of output where the average cost is minimum is (1) 10 units (2) 50 units (3) 100 units (4) 200 units 30. A sales promotion company sells tickets for ` 100 each to win a prize of ` 5 lakhs. If a person buys one of the 10,000 tickets sold, then his expected gain in rupees is (1) –50 (2) 0 (3) 50 (4) 100 31. An insurance company has found that 50% of its claims are for damages resulting from accidents. The probability that a random sample of 10 claims will contain fewer than 2 for accidents is

1 1024 11 (3) 1024 (1)

(2)

5 512

(4)

15 1024

32. During a pandemic, 10% of the patients who have the disease get complications. If 100 patients of a locality get infected by the disease, then the standard deviation of the number of patient getting complications is: (1) 10 (2) 9 (3) 6 (4) 3 33. An electrical supplier distributor has found the daily demand for fluorescent light bulbs is normally distributed with a mean of 432 and standard deviation of 86. Find the probability that the demand on a particular day exceeds 518 bulbs. (1) 0.1587 (2) 0.3413 (3) 0.7587 (4) 0.8413 34. The value of mortgage loans made by a certain bank is normally distributed with mean of ` 36 lacs and a standard deviation of ` 12 lacs. The probability that



a randomly selected mortgage loan is less than 54 lacs is (1) 85.26% (2) 93.32% (3) 97.42% (4) 98.04%

35. The prices of group of commodities is given in the following table: Commodities

A

B

C

D

p0 [Price (`) in 2019]

40

28

120

112

p1 [Price (`) in 2020]

50

35

135

120

The price index for 2020 taking 2019 as base year using simple aggregative method is: (1) 88.23% (2) 113.33% (3) 120.5% (4) 36% 36. For data regarding some commodities, the price indexes using Laspeyres and Paasches method are 118.4 and 117.5 respectively. The Fishers price index for the data is (1) 115.95 (2) 117.95 (3) 120.84 (4) 121.45 37. The price and quantities of certain commodities are shown in the following table: A

B

p0

1

1

q0

10

5

p1

2

x

q1 5 2 If ratio of Laspeyres (L) and Paasches (P) index number i.e., L : P = 28 : 27, then the value of x is (1) 2 (2) 3 (3) 4 (4) 5 38. To find the Index number by weighted average of price relatives, we use the formula p  ∑  p1  ( p0 q0 ) 0 (1) × 100 ∑ ( p0 q 0 ) p1 ( p0 q0 ) ∑ × 100 (2) ∑ ( p0 q 0 ) ∑ p0 ( p0 q0 ) × 100 (3) ∑ ( p0 q 0 )  p1  ∑  p  ( p1q2 ) 0 (4) × 100 ∑ ( p1q1 ) 39. To find the Index number by weighted average of price relatives, we use the formula (1) Laspeyres index only. (2) Paasches index only (3) Both Laspeyres and Paasches index numbers (4) Fishers ideal index

Sample Question Papers 40.

Which of the following is a branch of statistics ? (1) Descriptive Statistics (2) Inferential Statistics (3) Industrial Statistics (4) Both A and B

41. An Observed set of the population that has been selected for analysis is called. (1) a sample (2) a process (3) a forecast (4) a parameter. 42. The effective rate which is equivalent to a stated rate of 6% compounded semi-annually (1) 0.0609 (2) 0.9061 (3) 0.0062 (4) 0.9601 43. A machine costing ` 50,000 has a useful life of 4 years. The estimated scarp value is ` 10,000, then the annual depreciation is (1) ` 20,000 (2) ` 10,000 (3) ` 5,000 (4) ` 2,500 44. The value of a machine purchased two years ago, depreciates at the annual rate of 10%. If its present value is ` 97,200, then its value after 3 years is (1) ` 70,859 approx (2) ` 71,895 approx (3) ` 80,859 approx (4) ` 88,509 approx 45. The pdf of Poisson Distribution is given by ........... .

(1)

e − mm x x! x!

(2)

e −mx !

mx

e m mx x! Read the following text and answer the following questions on the basis of the same: A manufacturer produces two Models of bikesModel X and Model Y. Model X takes a 6 manhours to make per unit, while Model Y takes 10 man-hours per unit. There is a total of 450 man-hours available per week. Handling and (3)

mx e − m

(4)



13

Marketing costs are ` 2,000 and ` 1,000 per unit for Models X and Y respectively. The total funds available for these purposes are ` 80,000 per week. Profits per unit for Models X and Y are ` 1,000 and ` 500, respectively. The feasible region of L.P.P. is shown in the following graph.

E

46. The equation of line AB is (1) 3x + 5y = 225 (2) 2x + y = 80 (3) 5x + 3y = 225 (4) x + 2y = 80 47. The equation of line CD is (1) 3x + 5y = 225 (2) 2x + y = 80 (3) 5x + 3y = 225 (4) x + 2y = 80 48. The coordinates of point E are (1) (25, 30) (2) (30, 25) (3) (25, 25) (4) (30, 30) 49. The constraints for L.P.P. are: (1) 3x + 5y ≤ 225, 2x + y ≤ 80, x ≥ 0, y ≥ 0 (2) 3x + 5y ≥ 225, 2x + y ≤ 80, x ≥ 0, y ≥ 0 (3) 3x + 5y ≤ 225, 2x + y ≥ 80, x ≥ 0, y ≥ 0 (4) 3x + 5y ≤ 225, 2x + y ≤ 80, x ≤ 0, y ≤ 0 50. How many bikes of model X and model Y should the manufacturer produce so as to yield a maximum profit? (1) (0, 0) (2) (25, 30) (3) (25, 45) (4) (40, 25) 

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

14

SAMPLE

3

Question Paper Maximum Marks : 200

Time : 45 Min.

General Instructions :

(i) Section A will have 15 questions covering both i.e., Mathematics/Applied Mathematics which will be compulsory for all candidates. (ii) Section B1 will have 35 questions from Mathematics out of which 25 questions need to be attempted. Section B2 will have 35 questions purely from Applied Mathematics out of which 25 questions will be attempted. (iii) Correct answer or the most appropriate answer : Five marks (+ 5) (iv) Any incorrect option marked will be given minus one mark (– 1). (v) Unanswered/Marked for Review will be given no mark (0). (vi) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options. (vii) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question. (viii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (ix) Calculator / any electronic gadgets are not permitted.

Section - A



Mathematics/Applied Mathematics



1. The function f ( x ) = 2 x 3 - 3x 2 - 12 x + 4 , has



2. Area of the region in the first quadrant enclosed by



3. Which of the following is the principal value

(1) two points of local maximum (2) two points of local minimum (3) one maxima and one minima (4) no maxima or minima

the x-axis, the line y = x and the circle x 2 + y 2 = 32 is (1) 16p sq. units (2) 4p sq. units (3) 32p sq. units (4) 24p sq. units branch of cos–1x?    (1)   ,   2 2





4.

x9

 ( 4 x 2  1)6 dx

(1) e x



y x (3) e = e + C

=C 2

2

(2) e - y + e x = C 2

(4) e x + y = C 8. The feasible region for an LPP is shown in the given Figure. Let F = 3x – 4y be thet objective function. Maximum value of F is

is equal to

1  1 (1)  4  2  x 5 x

5

C

1 (1 + 4 )−5 + C 10 x



(2)

1 1  4  2  5 x

(12, 6)

5

C 5

1  1   4  C  10  x 2 5. Eight coins are tossed together. The probability of getting exactly 3 heads is (3)

2 -y



 (4) (0,  )    2

(3) [0, p]



(2) [– p, p]

1 7 (1) (2) 256 32 5 3 (3) (4) 32 32 6. If the set A contains 5 elements and the set B contains 6 elements, then the number of one-one and onto mappings from A to B is (1) 720 (2) 120 (3) 0 (4) None of these 2 dy 7. The general solution of = 2 xe x - y is dx

(0, 4)

(4)



(6, 0)

Sample Question Papers (1) 0 (3) 12



(2) 8 (4) –18



9. A box contains 3 orange balls, 3 green balls and 2



blue balls. Three balls are drawn at random from the box without replacement. The probability of drawing 2 green balls and one blue ball is 3 2 (1) (2) 28 21 1 167 (4) 28 168 10. Which of the following functions from Z into Z are bijections? (1) f(x) = x3 (2) f(x) = x + 2 (3) f(x) = 2x + 1 (4) f(x) = x2 + 1 (3)





11. Let A = {1, 2, 3, ...n} and B = {a, b}. Then the number of surjections from A into B is (1) nP2 (2) 2n – 2 n (3) 2 – 1 (4) None of these



12. If the volume of parallelopiped formed by the

vectors i + l j + k , j + l k and li + k is minimum, then l is equal to :



13. If A is an invertible matrix of order 2, then det (A–1) is equal to

1 det ( A )

(3) 1

(4) 0



14. The plane passing through the points (1, 2, 1),



(2, 1, 2) and parallel to the line, 2x = 3y, z = 1 also passes through the point : (1) (0, –6, 2) (2) (0, 6, –2) (3) (–2, 0, 1) (4) (2, 0, –1)



15. If A and B are symmetric matrices of same order, then

AB – BA is a: (1) Skew-symmetric matrix (2) Symmetric matrix (3) Zero matrix (4) Identity matrix

Mathematics

16. The solution of







18. If

(1) 0 (3) 12



(2) –16 (4) does not exist

x 3dx 1  x2

 a(1  x 2 )3 / 2  b 1  x 2  C , then



1 (1) a = , b = 1 3



(3) a =

-1 , b = -1 3

(2) a =

-1 , b =1 3

1 (4) a = , b = -1 3

19. Which of the following is the principal value branch of

   (1)  ,    2 2

 (2) [0, ]     2

   (3)   ,   2 2

  the non-zero vector a × c is :

(1) 10i – 5j (3) –14i + 5j

(2) –14i – 5j (4) –10i + 5j

21. The foot of the perpendicular drawn form the point (4, 2, 3) to the line joining the points (1, –2, 3) and (1, 1, 0) lies on the plane : (1) x – y – 2z = 1 (2) x – 2y + z = 1 (3) 2x + y – z = 1 (4) x + 2y – z = 1

22. The maximum value of sin x. cos x is



(1)



(3)



cosec–1x?

   (4)   ,   {0}  2 2    20. Let a = i + 2 j + 4 k , b = i + l j + 4 k and  c = 2i + 4 j + (l2 – 1) k be coplanar vectors. Then



(4) − 3



17. Refer to Q. 8, minimum value of F is



3

(2)





(3) −

3

(1) det (A)





1

1

(2)



dy  y  e  x , y( 0 )  0 is dx (1) y = e–x (x – 1) (2) y = xex –x (3) y = xe + 1 (4) y = xe–x



3

Section - B1





(1)

15

23. If

1 4 2

(2)

1 2

(4) 2 2

2 x 5 6 -2 = , then the value of x is 8 x 7 3

(1) 3 (3) ±6

(2) ±3 (4) 6

 24. Let a = 2i + l1 j + 3 k , b = 4i + (3 – l2) j + 6 k and

 c = 3i + 6 j + (l3 – 1) k be three vectors such that    b = 2 a and a is perpendicular to c . Then a

possible value of (l1, l2, l3) is :

(1) (1, 3, 1)

 −1  (2)  , 4 , 0  2  



1 (3)  , 4 , − 2  4  

(4) (1, 5, 1)



25. Area of the region bounded by the curve y = cos x

between x = 0 and x = p is (1) 2 sq. units (2) 4 sq. units (3) 3 sq. units (4) 1 sq. unit

16

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

26. Maximum slope of the curve y = -x 3 + 3x 2 + 9 x - 27 is : (1) 0 (3) 16



(2) 12 (4) 32



27. A die is thrown and a card is selected at random

from a deck of 52 playing cards. The probability of getting an even number on the die and a spade card is 1 1 (1) (2) 4 2 1 3 (3) (4) 4 8 28. If 3tan–1x + cot–1x = π, then x equals (1) 0 (2) 1 1 (3) –1 (4) 2



29. The distance of the point (1, –2, 3) from the plane

x – y + z = 5 measured parallel to the line x y z   is : 2 3 6

(1) 1 (2) 7 7 7 (3) (4) 1 5 30. Let A be a non-singular square matrix of order 3 × 3. Then |adj A| is equal to (1) |A|

(2) |A|2



(3) |A|3

(4) 3|A|





(0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). Let F = 4x + 6y be the objective function. The minimum value of F occurs at (1) (0, 2) only (2) (3, 0) only (3) the mid-point of the line segment joining the points (0, 2) and (3, 0) only (4)  any point on the line segment joining the points (0, 2) and (3, 0) 3 d f ( x ) = 4 x 3 - 4 such that f(2) = 0. Then f(x) is 32. If dx x 1 129 1 129 3 4 (1) x + 3 (2) x + 4 + 8 8 x x 1 129 1 129 (3) x 4 + 3 + (4) x 3 + 4 8 x 8 x

1 x

33. Let f : R ® R be defined by f(x) = , ∀x Î R. Then f is (1) one-one (3) bijective

(2) onto (4) f is not defined

34. The general solution of differential equation ( e x + 1) ydy = ( y + 1)e x dx is (1) (y + 1) = k (ex+ 1) (2) y + 1 = ex + 1 + k

(3)

2 3

36. The maximum value of

(4)

4 7

x

 1  is :  x 



(1) e

(2) ee



(3) e1/e

 1 (4)     e





1/e

37. Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. The restriction on n, k and p so that PY + WY will be defined are : (1) k = 3, p = n (2) k is arbitrary, p = 2 (3) p is arbitrary, k = 3 (4) k = 2, p = 3

 a 0 0   A  38. If  0 a 0  , then det (adj A) equals  0 0 a 

31. Corner points of the feasible region for an LPP are

35. Assume that in a family, each child is equally likely

to be a boy or a girl. A family with three children is chosen at random. The probability that the eldest child is a girl given that the family has at least one girl is 1 1 (1) (2) 3 2





(3) y = log {k (y +1) (ex + 1)} x +1 (4) y = log +k y +1

(1) a27 (2) a9 2

(3) a6

(4) a   39. Let a = 3i + j and b = 2i – j + 3 k . If b = b1 – b2, where b1 is parallel to a and b2 is perpendicular to a, then b1 × b2 is equal to?    ( 3i − 9 j + 5k ) (1) ( −3i + 9 j + 5k ) (2) 2 2 (3) 3i + 9j + 5k (4) 3i – 9j + 5k



40. If for some a ∈ R, the lines L1 : x  1  y  2  z  1

2 1 1 x 2 y 1 z 1 and L2 : are coplanar, then   5 1 

the line L2 passes through the point : (1) (2, –10, –2) (2) (10, –2, –2) (3) (10, 2, 2) (4) (–2, 10, 2)   41. Let a = i – j , b = i + j + k and c be a vector such       that a × c + b = 0 and a . c = 4, then | c |2 is



equal to : 19 (1) 2



(3) 8

(2) 9 (4)

17 2

Sample Question Papers 5 8  0  42. The matrix  5 0 12  is a   8 12 0  (1) diagonal matrix (2) symmetric matrix (3) skew symmetric matrix (4) scalar matrix Directions : In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as (1) Both A and R are true and R is the correct explanation of A (2) Both A and R are true but R is NOT the correct explanation of A (3) A is true but R is false (4) A is false but R is True



43. Assertion (A) : If A and B are two mutually 1 5 exclusive events with P( A ) = and P(B) = . 3 6 Then P( A / B ) is equal to 1 . 4 Reason (R) : If A and B are two events such that P(A) = 0.2, P(B) = 0.6 and P(A|B) = 0.2 then the value of P( A | B ) is 0.2.

44. Assertion (A) : The probability of an impossible

event is 1. Reason (R) : If A is a perfect subset of B and P(A) < P(B), then P(B – A) is equal to P(B) – P(A). 2/3 dx 45.  equals 0 4  9x 2 π  (1) (2) 6 12 (3)



 24

(4)

 4

17

Read the following text and answer the following questions on the basis of the same: Ms. Remka of city school is teaching chain rule to her students with the help of a flow-chart The chain rule says that if h and g are functions and f(x) = g(h(x)), then

Let f(x) = sin x and g(x) = x3

46. fog(x) = _______.



47. gof(x) = _______.



(1) sin x3 (3) sin 3x

(2) sin3 x (4) 3sin x

(1) sin x3 (2) sin3 x

(3) sin 3x (4) 3 sin x d (sin 3 x ) = _______. 48. dx (1) cos3 x (2) 3sin xcos x (3) 3sin2 x cos x (4) – cos3 x d 49. sin x3 _______. dx (1) cos (x3) (2) –cos (x3)

(3) 3x2 sin (x3) (4) 3x2 cos (x3)  d 50. (sin 2 x ) at x  is _______. dx 2



(1) 0 (3) 2

(2) 1 (4) –2

Section - B2



Applied Mathematics

16. If a ≡ b(mod n), then

(1) 46 minutes

(2) 12 minutes (4) 11 minutes

(1) a ≡ a(mod n)

(2) b ≡ a(mod n)

(3) 23 minutes

(3) b ≡ b(mod n)

(4) None of these

17. (186 × 93) mod 7 = a ...........

20. For Modular arithmetic, what value is the most important ?

(1) 1

(2) 2



(1) The modulo

(2) The given number

(3) 5

(4) 7



(3) The remainder

(4) The divisor

18. If p > q and r < 0, then which of the following is true? (1) pr < qr

(2) p – r < q – r

21. For a race a distance of 224 metres can be covered by P in 28 second and Q in 32 seconds. By what distance does P defeat Q eventually ?

(3) p + r < q + r

(4) None of these



(1) 26 m

(2) 32 m



(3) 24 m

(4) 28 m

19. Guddi’s swimming speed in still water to the speed of river is 7 : 1. She swims 4.2 km up the river in just 14 min. How much time will Guddi take to swim 18.4 km down the river?

22. A began a business with ` 85,000. He was joined afterwards by B with ` 42,500. If the profits at the

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

18

end of the year are divided in the ratio of 3 : 1, then B joined for ....... months only. (1) 2 (2) 3 (3) 5 (4) 8



23. Kishore's speed with the current is 15 km/hr and the speed of current is 2.5 km/hr. Kishore's speed against the current is: (1) 8.5 km/hr (2) 10 km/hr (3) 11.5 km/hr (4) None of these

3 7



24. Let A = 0 2 

−1 5  and B =

7 9 0   3 −5 6  , then  

the sum of (2, 2) entries of A and B is : (1) 3 (2) – 3 (3) 2 (4) – 5

0 0 3   25. The matrix A = 0 3 0  is a 0 0 3  (1) scalar matrix (3) square matrix





 −8 x   and A = A', then  y 5

26. If A = 

(1) x = 5, y = – 8 (3) x = y

28. If y = x + x + x + ...∞ then (1)



(3)





(2) x = – 8, y = 5 (4) None of these

If A is a symmetric matrix and n ∈ N. then An is (1) symmetric matrix (2) skew-symmetric matrix (3) a diagonal matrix (4) None of these

27.

(2) diagonal matrix (4) None of these

1 ( 2 y − 1) 1 (2 y)

29. If y =

(2)

dy = dx

1 ( 2 y + 1)

(4) (2y – 1)

f '( x ) f (x) , then and z = φ '( x ) φ( x )

f ''( x ) φ "( x ) 2( y − z ) − + {φ ' ( x )}2 = f ( x ) φ ( x ) f ( x )φ ( x )



d2 y



(1)



(3) y

2

dx d2 y dx 2

(2)

(4)

1 d2 y y dx 2 d2 y dy 2

30. The minimum value of f(x) = x3 – 3x in [0, 2] is: (1) 2 (2) – 2 (3) 0 (4) None of these

31. The derivative of m = 13x4 + 5x3 – 12y3 + 24x2 + y2 – 2x – 156 w.r.t. x is

(1)

dm = 52x3 + 15x2 + 48x – 2 dx



(2)

dm = 52x3 + 15x2 – 36y2 + 48x + 2x – 2 dx



(3)

dm = 52x3 + 15x2 – 36y2 + 48x + 2y – 3 dx



(4)

dm = – 36x3 + 2y dx

32. A weighted aggregate price index in which the weight for each variable is considered its currentperiod quantity is: (1) Aggregative index (2) Consumer Price index (3) Laspeyres Index (4) Paasche’s index 33. An index constructed to measure changes in quantities over a period of time is: (1) Quantity index (2) Time series index (3) Quality index (4) Value index 34. For calculating the weighted index number, which of the following uses quantities consumed in the base period as weights: (1) Fisher’s method (2) Paasche’s method (3) Laspeyres method (4) Aggregative method 35. What is the index number of the base period? (1) 200 (2) 300 (3) 10 (4) 100 36. Which of the following is not an example of a time series model? (1) Naive approach (2) Exponential smoothing (3) Moving average (4) None of these 37. Which of the following can't be a component for a time series plot? (1) Seasonality (2) Trend (3) Noise (4) None of these 38. Increase in the number of patients in the hospital due to heat stroke is: (1) Secular trend (2) Irregular variation (3) Seasonal variation (4) Cyclical variations 39. The graph of time series is called: (1) Histogram (2) Straight line (3) Historigram (4) Ogive 40. The sum of money is needed now, so as to get ` 6000 at the beginning of every month forever, then the money is worth 6% per annum compounded monthly is (1) ` 1000000 (2) ` 1206000 (3) ` 60,000 (4) ` 1600000

Sample Question Papers 41. The present value of perpetuity of ` 600 payable at end of every 6 month be ` 10,000. Find rate of interest. (1) 10% (2) 12% (3) 6%

(4) 12.5%

42. Assume that the year-end revenues of a business over a three period, are mentioned in the following table: Year End

31-12-2018

31-12-2021

9,000

13,000

Year End Revenue

19

Read the following text and answer the following questions on the basis of the same: For a random sample of 10 pigs fed on diet A, the increases in weight in pounds in a certain period were 10, 6, 16, 17, 13, 12, 8, 14, 15, 9 lbs for another random sample of 12 pigs fed on diet B, the increases in the same period were 7, 13, 22, 15, 12, 14, 18, 8, 21, 23, 10, 17 lbs

Calculate the CAGR of revenues over, three-years period spanning the "end" of 2018 to the "end" of 1

 13  3 2021. Given that   = 1.13  9  (1) 13%

(2) 14%

(3) 15%

(4) None of these

43. Z = 7x + y, subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0. The maximum value of Z occurs at (1) (3, 0)

1 5 (2)  ,  2 2

(3) (7, 0)

(4) (0, 5)

44. For the constraint of a linear optimizing function z = x1 + x2, given by x1 + x2 ≤ 1, 3x1 + x2 ≥ 3 and x1, x2 ≥ 0 (1) There are two feasible regions

(2) There are infinite feasible regions

(3) There is no feasible region (4) None of the above 45. Objective function of an L.P.P. is (1) a constant (2) a function to be optimized (3) a relation between the variables (4) None of these

46. Mean of increases in weight on diet A is (1) 12 pounds (2) 15 pounds (3) 20 pounds (4) 27 pounds 47. Sum of square of deviation for diet B is: (1) 120 (2) 314 (3) 150 (4) 214 48. The number of degrees of freedom is: (1) 10 (2) 12 (3) 15 (4) 20 49. The standard error of this is: (1) 4.65 (2) 4.55 (3) 46.5 (4) 4.10 50. Using the fact the 5% value of t for 20 degrees of freedom is 2.09, whether diets A and B differ significantly as regards the effect on increases in weight. (1) Significant (2) not significant (3) can’t say (4) None of these qq

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

SAMPLE

4

Question Paper Maximum Marks : 200

Time : 45 Min.

General Instructions :

(i) Section A will have 15 questions covering both i.e., Mathematics/Applied Mathematics which will be compulsory for all candidates. (ii) Section B1 will have 35 questions from Mathematics out of which 25 questions need to be attempted. Section B2 will have 35 questions purely from Applied Mathematics out of which 25 questions will be attempted. (iii) Correct answer or the most appropriate answer : Five marks (+ 5) (iv) Any incorrect option marked will be given minus one mark (– 1). (v) Unanswered/Marked for Review will be given no mark (0). (vi) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options. (vii) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question. (viii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (ix) Calculator / any electronic gadgets are not permitted.

Section - A



Mathematics/Applied Mathematics

1. Let f : R ® R be defined as f(x) = x4. Choose the correct answer.



(1) f is one-one onto



(2) f is many-one onto



(3) f is one-one but not onto



(4) f is neither one-one nor onto



2. The value of ∫ 2 log  4 + 3 sin x  dx is

p

0



(1) 2



(3) 0



 4 + 3 cos x 

3 4 (4) –2 (2)

x2 3. If f x 2 x and g x  1 then which of the 2 following can be a discontinuous function? (1) f(x) + g(x) (2) f(x) – g(x) (3) f(x).g(x)

(4)

g (x) f (x )

ab bc a

4. The value of determinant b  a c  a b is ca ab c 3

3

3



(1) a  b  c



(3) a3  b 3  c 3  3abc (4) None of these

(2) 3bc



5. Two events E and F are independent. If P(E) = 0.3,

P(E ∪ F) = 0.5, then P(E|F) – P(F|E) equals 3 2 (1) (2) 35 7 1 1 (3) (4) 7 70 6. The order of the differential equation d2y dy 2 x 2 2 - 3 + y = 0 is dx dx (1) 2 (2) 1 (3) 0 (4) not defined

7. The rate of change of area of circle with respect to



its radius r at r = 3 cm is : (1) 6p (2) 8p



(3) 12p



(4) 3p

  8. The lines r  (i  j ) + l ( 2i + k ) and r  ( 2i  j ) + m (i  j  k )



(1) Do not intersect for any values of l and m (2) Intersect when l =1 and m = 2



(3) Intersect when l = 2 and m =



(4) Intersect for all values of l and m

1 2

Sample Question Papers 9. If A and B are two independent events with



4 3 P(A) = and P(B) = , then P( A Ç B) equals 5 5 4 8 (1) (2) 15 45 1 2 (4) 3 9 10. The slope of the normal to the curve x2 + 3y + y2 = 5 at (1, 1) : −2 5 (1) (2) 5 2 −5 2 (3) (4) 5 2 (3)





11. A plane passing through the point (3, 1, 1) contains



two lines whose direction ratios are 1, –2, 2 and 2, 3, –1 respectively. If this plane also passes through the point (a, –3, 5), then a is equal to : (1) –5 (2) 10 (3) 5 (4) –10



12. Let f : R ® R be defined as f(x) = 3x. Choose the correct answer.



16. If the volume of a parallelopiped, whose coterminous edges are given by the vectors   a = i + j + n k , b = 2i + 4 j – n k and

 c = i + n j + 3 k , (n ≥ 0) , is 158 cu. units, then :   (1) a ⋅ c = 17 (3) n = 9

  (2) b ⋅ c = 10 (4) n =7

17. The area of a triangle with vertices (–3, 0), (3, 0) and



18. The plane which bisects the line joining the points



(4, –2, 3) and (2, 4, –1) at right angles also passes through the point : (1) (0, –1, 1) (2) (4, 0, 1) (3) (4, 0, –1) (4) (0, 1, –1)



19. The area of the region bounded by the ellipse x2 y2 + = 1 is 25 16

(1) 20p sq. units (3) 16p2 sq. units





x →a

x n − an is equal to x−a

(1) nan–1 (3) 0

(2) an–1 (4) 1

 x 2 − 2x − 3  14. f(x) =  x + 1  k 

x ≠ −1 x =−1

Find k when function is continuous at x = – 1 (1) k = 4 (2) k = – 2 (3) k = – 3 (4) k = – 4 1



15. The value of integral (1) 6 (3) 3



 x3 )3 dx is x4 (2) 0 (4) 4

1 (x 1 3

Mathematics

(0, k) is 9 sq. units. Then, the value of k will be (1) 9 (2) 3 (3) –9 (4) 6





13. lim

Section - B1



(1) f is one-one onto (2) f is many-one onto (3) f is one-one but not onto (4) f is neither one-one nor onto



21

Directions : In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is NOT the correct explanation of A (C) A is true but R is false (D) A is false but R is True

21. Assertion (A) : For the constraints of linear



22. Assertion (A): Z = 20x1 + 20x2, subject to x1 ³ 0,

optimizing function Z = x1 + x2 given by x1 + x2 £ 1, 3x1 + x2 ≥ 1, there is no feasible region. Reason (R): Z = 7x + y, subject to 5x + y £ 5, x + y ≥ 3, x ≥ 0, y ≥ 0. Out of the corner points of  1 5 feasible region (3, 0),  ,  , (7, 0) and (0,5), the  2 2 maximum value of Z occurs at (7, 0). x2 ³ 2, x1 + 2x2 ³ 8, 3x1 + 2x2 ³ 15, 5x1 + 2x2 ³ 20. Out of the corner points of feasible region (8, 0),

 5 15   7 9  and (0,10), the minimum value of  2 , 2  ,  2 , 4   

2

(2) 20p sq. units (4) 25p sq. units

20. Let T be the set of all triangles in the Euclidean

plane, and let a relation R on T be defined as aRb if a is congruent to b ∀ a, b ∈ T. Then R is (1) reflexive but not transitive (2) transitive but not symmetric (3) equivalence (4) None of these



7 Z occurs at  , 2 Reason (R) :

9 . 4 

Corner Points

Z = 20x1 + 20x2

(8, 0)

160

 5 15   2 , 4 

125

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

22

 7 9  2 , 4 

115 minimum

(0, 10)

200

23. If A and B are such events that P(A) > 0 and P(A) ≠ 1, then P (A|B) equals 1 - P( A| B) (2) (1) 1 – P (B | A) 1  P( A  B ) (3) (4) P(A )|P(B) P( B  )



24. Matrices A and B will be inverse of each other only if (1) AB = BA (3) AB = 0, BA = I

(2) AB = BA = 0 (4) AB = BA = I

25. The numbers of arbitrary constants in the particular solution of a differential equation of third order are  : (1) 3 (2) 2 (3) 1 (4) 0

 cos  sin   , then A + A’ = I, then the  sin cos  value of α is : p p (1) (2) 3 6 3p (3) p (4) 2 27. The set of points where the function f given by f  x  |2 x  1|sin x is differentiable is 1  (1) R (2) R –   2 



26. If A = 

(3) ( 0, ∞ )

(3)



(4) both symmetric and transitive.



32. The general solution of the differential equation

dy = e x + y is dx

-y x (1) e + e = C

(2) e x + e y = C



y -x (3) e + e = C

(4) e -x + e - y = C



33. The area of the region bounded by parabola y2 = x and the straight line 2y = x is

x − x tan −1 x + C −1

x − ( x + 1)tan x +C   29. A vector a = ai + 2 j – b k , (a, b ∈ R) lies in the   plane of the vectors, b  = i  + j and  c = i – j + 4 k . If a bisects the angle between b and c, then :   (1) a ⋅ i + 1 = 0 (2) a ⋅ k + 2 = 0  (3) a ⋅ i + 3 = 0 (4) a ⋅ k + 4 = 0 30. Let P(A) = 7 , P( B) = 9 and P( A ∩ B) = 4 . Then 13 13 13 P(A|B) is equal to (4)

6 (1) 13

4 13 5 (4) 9 (2)

4 (3) 9 31. Consider the non-empty set consisting of children in a family and a relation R defined as aRb if a is brother of b. Then R is (1) symmetric but not transitive.

4 sq. units 3

(1)



(2) 1 sq. units

1 2 (3) (4) sq. units 3 sq. units 3 34. If (a, b, c) is the image of the point (1, 2, –3) in the z x 1 y 3 line   , then a + b + c is : 2 1 2 (1) 2 (2) 3 (3) –1



(4) 1

35. The



number of distinct real roots of sin x cos x cos x p p cos x sin x cos x  0 in the interval   x  4 4 cos x cos x sin x is (1) 0 (2) 2 (3) 1 (4) 3



36. Let the volume of a parallelopiped whose



 coterminous edges are given by u = i + j + l k ,   v = i + j + 3 k and w = 2i + j + k be 1 cu. unit.   If q be the angle between the edges u and w , then cos q can be :

(1) ( x + 1)tan −1 x − x + C







28.  tan 1 x dx is equal to (2) x tan −1 x − x + C



(2) transitive but not symmetric. (3) neither symmetric nor transitive.

(4) none of these







(1)



(3)



5 7

(2)

5

(4)

(6 3 )

7 (6 6 ) 7 (6 3 )

37. A and B are events such that P(A) = 0.4, P(B) = 0.3 and P(A∪B) = 0.5, Then P(B∩A) equals

1 (1) 2 (2) 3 2 3 1 (3) 10 (4) 5



∫

38.



5x − 4 x 2

(1) sin −1 (3)



dx

=

8x − 5 +C 5

(2)



( 8 x − 5) 1 log ( 8 x + 5) 20

8x − 5 1 −1 8 x − 5 +C + C (4) 2 sin sin −1 5 2 5

Sample Question Papers

2

39. If cos  sin 1 + cos1x  = 0 , then x is equal to   5



2 5 (4) 1

1 5 (3) 0

(1)



(2)

40. Refer to Q.41, (Maximum value of Z + Minimum value of Z) is equal to (1) 13 (3) –13



(2) 1 (4) –17



41. If f x  = x 2 sin 1  , where x ≠ 0, then the value



of the function f at x = 0, so that the function is continuous at x = 0, is (1) 0 (2) –1 (3) 1 (4) none of these

x



42. The value of sin[2tan–1(0.75)] is equal to



43. The shortest distance between the lines

(1) 0.75 (3) 0.96



23

46. If P(A) = 2 ,P(B) = 3 and P(A Ç B)= 1 , then

5 10 P(A¢|B¢).P(B’|A’) is equal to

5

5 5 (1) (2) 7 6 25 (4) 1 (3) 42 Read the following text and answer the following questions on the basis of the same: The shape of a toy is given as f(x) = 6(2x4 – x2). To make the toy beautiful 2 sticks which are perpendicular to each other were placed at a point (2, 3), above the toy.

(2) 1.5 (4) sin1.5

x  1 y  2 z and x + y + z + 1 = 0,   0 1 1 2x – y + z + 3 = 0 is : 1 (1) 1 (2) 2

(3)



1



(4) 1 2

3





44. If A, B and C are angles of a triangle, then the 1 cos C cos B 1 cos A is equal to determinant cos C 1 cos B cos A (1) 0 (3) 1



(2) –1 (4) None of these

45. The magnitude of the projection of the vector

2i + 3 j + k on the vector perpendicular to the plane containing the vectors i + j + k and i + 2 j + 3  , is :



(1) 3 6

(2)



(3)

(4)

6

16. If a = b(mod m), then (1) a(mod m) = b(mod m) (2) a(mod a) = b(mod m) (3) a ≡ a(mod m) (4) None of these

2 3 3 2  

critical point? (1) ± 1/4 (3) ± 1

(2) ± 12 (4) None of these

48. Find the slope of the normal based on the position



of the stick. (1) 360 (2) –360 1 -1 (3) (4) 360 360 49. What will be the equation of the tangent at the critical point if it passes through (2, 3)? (1) x + 360y = 1082 (2) y = 360x – 717 (3) x = 717y + 360 (4) None of these



50. At which of the following intervals will f(x) be



 1   1  1   1     ,  (2)  , 0   ,  (1)    ,   2  2   2  2 



1 1 (3)  0 ,    ,   2  2 



k



47. Which value from the following may be abscissa of

increasing?

1   1   (4)    ,    0 ,   2   2

Section - B2

Applied Mathematics

17. The last two digits of the product 2345 × 6789 are : (1) 0, 5 (2) 5, 0 (3) 9, 5 (4) 5, 9

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

24

18. If a and b are positive integers and

(1) b > a (3) b = a

a−b 8 = , the 2.5 6.25

(2) b < a (4) b ≥ a



(1) –1

(2) 0

(3) 7

(4) 9

21. Two athletes Vijay and Samuel finish 100 meters race in 12 secs and 16 secs respectively. By how many meters does Vijay defeat Samuel? (1) 10.2 meters (2) 15 meters (3) 25 meters (4) 33.3 meters



rd of the capital, while A contributes as much 3 as B and C together contribute. The ratio of their capitals is: (1) 1 : 2 : 3 (3) 3 : 1 : 1

then what is the type of R as a matrix ?





(1) Row matrix

(2) Column matrix

(3) Square Matrix

(4) None of these

25. If matrix A = [1, 2, 3], then AA' = ........ (1) [12]

(2) [13]

(3) [14]

(4) [15]

3 − 2 , then (A + I) (A – I) is equal to 1 

26. If A + I =  4

− 5 − 4   − 9

(1)  8 

5 4  9 

(3)  8 





 −5 4  9 

(2)  −8 

− 5 − 4  − 9 

(4)  −8 

λ 1 2 27. If A =   , then for what value of λ, A = 0 ? − 1 − λ   (1) 0 (2) ±1 (3) – 1 (4) 1

dy is dx

(2) f '(x) =

10 x 2 + 470 x ( x + 47 )

(4) None of the above

(2) 3x2 ln x + x2 (4) 3x

30. The derivative of the following function: f(x) = 1963 (1) + ∞ (2) 1963 (3) – ∞ (4) 0

31. The least value of a so that the function f(x) = x2 +



32. Consider a binomial random variable X. If X1, X2,

ax + 1 is strictly increasing on [1, 2] is (1) 2 (2) – 2 (3) 1 (4) – 1

.... Xn are independent and identically distributed samples from the distribution of X with sum Y = n

∑ Xi

then the distribution of Y as n → ∞ can be

i =1

approximated as (1) Exponential (3) Binomial

23. Two pipes A and B can fill a cistern in 8 hours and

24. A is a matrix of the type 3 × 5 and R is a row of A,

29. Given y = x3 ln x, (1) 3x2 ln x (3) x2

(2) 3 : 2 : 1 (4) 2 : 1 : 1

12 hours respectively. The pipes when opened simultaneously takes 12 minutes more to fill the cistern due to leakage. Once the cistern is full, it will get emptied due to leakage in (1) 5 hours (2) 20 hours (3) 60 hours (4) 120 hours



22. A, B and C enter into a partnership. B contributes 1

5x 2 x + 47

5x 2 − 470 x (1) f '(x) = ( x + 47 )2 (3) f '(x) = 10x

20. The value of 5 8 11, where  is multiplication modulo is

28. The derivative of the following function:

f(x) =

19. In a kilometre race, A, B and C are three participants.

A can give B start of 50 m and C a start of 69 m. The start which B can allow C, is: (1) 17 m (2) 18 m (3) 19 m (4) 20 m





(2) Bernoulli (4) Normal

33. The area under the standard normal curve which lies to the right of z = – 0.68 is: (1) 1 – F(0.68) (2) F(0.68) – 1 (3) F(0.68) (4) F(– 0.68)



34. A specific characteristics of a sample is known as

................... . (1) population (3) statistic

(2) parameter (4) variance

35. A statement made about a population parameter for

testing purpose is called ............... . (1) statistic (2) parameter (3) hypothesis (4) level of significance

36. The optimal value of the objective function is

attained at the points (1) on X-axis (2) on Y-axis (3) which are corner points of the feasible region (4) None of these

37. The intermediate solutions of constraints must be

checked by substituting them back into (1) Objective function (2) Constraint equations (3) Not required (4) None of these

Sample Question Papers

38. Which of the following is a vertex of the positive

45. CAGR stands for : (1) Compound Aggregate Groth Rate (2) Compound Annual Growth Rate (3) Computed Annual Growth Rate (4) Computed Aggregate Growth Rate

39. The following commodities have the given price

Read the following text and answer the following questions on the basis of the same: Mr. Rama Roy a owner of a factory lives in Karnataka. In his factory, he manufactures razor 1 blades. There is a small chance that for any 500 blade to be defective. The blades are in the packets of 10.

region bounded by the inequalities, 2x + 3y = 6 and 5x + 3y = 15 (1) (0, 2) (2) (0, 0) (3) (3, 0) (4) All of these

25

indices relative to a base of 100. The weights are also given: Relative Index

Weight

Butter

Commodity

181

4

Bread

116

12

Tea

110

3

Bacon

152

7

The new index for this set of commodities is: (1) 132 (2) 133 (3) 134 (4) 135



40. Index number was first constructed in: (1) 1750

(2) 1760

(3) 1764

(4) 1770

41. Statement 1: An index number is a statistical

measure, designed to measure changes in a variable, or a group of related variables with respect to time, geographical location or other characteristics such as income, profession, etc. Statement 2: Index number is a single ratio (or a percentage) with measures the combined change of several variables between two different times, places or situations. (1) Only 1 (2) Only 2 (3) Both 1 and 2 (4) None of these

42. The most commonly use mathematical method for

measuring the trend is: (1) Moving average method (2) Semi average method (3) Method of least squares (4) None of these

43. Prosperity, Recession, and depression in a business

is an example of (1) Irregular Trend (3) Cyclical Trend

(2) Secular Trend (4) Seasonal Trend

44. If rate the return on an investment is positive, then

it indicates : (1) Profit (3) No profit no loss

(2) Loss (4) None



46. The probability that packet contain no defective blade

is : (1) e–0.02 (3) 0.02 e–0.02

(2) 0.02 (4) 0.002 e–0.02

47. The probability that packet contain one defective blade

is : (1) e–0.02 (3) 0.02e–0.002

(2) 0.02e–0.02 (4) 0.002e–0.02



48. Find the probability that packet contain two detective



49. Find the approximate number of packets containing



50. Find the approximate number of packets containing

blade, when it is given that e–0.02 = 0.9802. (1) 0.0002 (2) 0.002 (3) 0.2 (4) None of these

no defective blade, when there are 10000 packets in a consignment. (Use e–0.02 = 0.9802) (1) 9802 (2) 9800 (3) 2 (4) 196 one defective blade, when there are 20,000 packets in the consignment. (Use e–0.02 = 0.9802) (1) 196 (2) 392 (3) 932 (4) 619

qqq

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

SAMPLE

5

Question Paper Maximum Marks : 200

Time : 45 Min.

General Instructions : (i) Section A will have 15 questions covering both i.e., Mathematics/Applied Mathematics which will be compulsory for all candidates. (ii) Section B1 will have 35 questions from Mathematics out of which 25 questions need to be attempted. Section B2 will have 35 questions purely from Applied Mathematics out of which 25 questions will be attempted. (iii) Correct answer or the most appropriate answer : Five marks (+ 5) (iv) Any incorrect option marked will be given minus one mark (– 1). (v) Unanswered/Marked for Review will be given no mark (0). (vi) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options. (vii) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question. (viii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (ix) Calculator / any electronic gadgets are not permitted.

Section - A





cos t

t

sin t

t

1

1. If f(t)= 2 sin t t 2t , then lim f (2t ) is equal to tÆ0



(1) 0 (3) 2



2. If P( A ) =



Mathematics/Applied Mathematics

equal to 1 (1) 10

t

t

(2) –1 (4) 3 4 7 and P(A « B)= , then (P(B|A) is 5 10 (2)

1 8

7 17 (4) 8 20 3. The maximum number of equivalence relations on the set A = {1, 2, 3} are (1) 1 (2) 2 (3) 3 (4) 5 Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as. (A) Both A and R are true and R is the correct explanation of A

(3)

(B) Both A and R are true but R is NOT the correct explanation of A (C) A is true but R is false (D) A is false and R is True Ï kx Ô ,if x < 0 4. Consider the function f(x) = Ì x Ô 3 ,if x ≥ 0 Ó which is continuous at x = 0. Assertion (A): The value of k is – 3.

Ï - x ,if x < 0 Reason (R): x = Ì Ó x , if x ≥ 0 5. Assertion (A): |sin x| is continuous at x = 0. Reason (R): |sin x| is differentiable at x = 0.  3 10  6. If A =   , then write A–1. 2 7 



 7 − 10  (1)   −2 3  

− 2  7 (2)  − 10 3  



 3 − 10  (3)   −2 7  

 − 7 10  (4)   2 − 3 

Sample Question Papers 7. Corner points of the feasible region determined by



the system of linear constraints are (0, 3), (1, 1) and (3, 0). Let Z = px + qy, where p, q > 0. Condition on p and q so that the minimum of Z occurs at (3, 0) and (1, 1) is (1) p = 2q (2) p = q/2 (3) p = 3q (4) p = q



2  2a  −1 1 − a −1  2 x  cos +   = tan    2 2  1+ a   1 − x2  , 1+ a 

8. If sin −1 





where a,x Î ]0, 1[ , then the value of x is a (1) 0 (2) 2 (3) a



(4)

2a 1  a2

9. The probability that a student is not a swimmer is



1/5. Then the probability that out of five students, four are swimmers is 3

4 1 5 C3     (1) 5 5

1  4 5 (3) (4) None of these C1   5  5 10. The area of the region bounded by the circle x 2 + y 2 = 1 is (1) 2p sq. units (2) p sq. units (3) 3p sq. units (4) 4p sq. units  ( x + 3)2 − 36 , x≠3  11. If the function f(x) =  value x−3  , x=3 k 

(2) 12 (4) – 6

 −1 −1  x    − cos ( x p ) tan    1 p  and B = then A – B is  p −1  x  −1  sin   −tan ( px ) p   equal to

(1) I

(2) O



(3) 2I

(4)



1 I 2

13. Suppose P and Q are two different matrices of



order 3 × n and n × p then order of matrix P × Q is  ? (1) 3 × p (2) p × 3 (3) n × n (4) 3 × 3



14. If A = {1, 2, 3}, B = {x, y} then the number of



functions from A to B is : (1) 3 (2) 6 (3) 8 (4) 12

15. The mean of the numbers obtained on throwing a

die having written 1 on three faces, 2 on two faces and 5 on one face is (1) 1 (2) 2 (3) 5 (4) 8/3

Section - B1



Mathematics

16. The variables x and y in a linear programming



problem are called (1) decision variables (3) optimal variables



17. The maximum value of   is (θ is real number)

(2) linear variables (4) None of the above

1 1 1 1 1  sin  1 1  cos  1 1

3 (2) 2 2 3 (4) 4



1 (1) 2



(3)



, x≤2  5  If the function f(x) = ax + b , 2 < x < 10 is  21 , x ≥ 10 

18.



of k is : (1) 6 (3) – 12

 −1 −1  x    sin ( x ) tan    1  p  12. If A =   p −1  x  −1 sin   cot ( px ) p  

4

 4 1 (2)    5 5

4





27

2

continuous function, then the value of a + b is : (1) 1 (2) 5 (3) 4 (4) 3



19. The area of the region bounded by the curve

y = x + 1 and the lines x = 2 and x = 3 is 9 7 sq. units (1) (2) sq. units 2 2 11 13 (3) sq. units (4) sq. units 2 2

20. The order and degree of the differential equation

d 2 y  dy    dx 2  dx 

1/4

(1) 2 and 4 (3) 2 and 3

 x 1 / 5  0 respectively, are (2) 2 and 2 (4) 3 and 3

21. Let us define a relation R in R as aRb if a ≥ b. Then R is (1) an equivalence relation. (2) reflexive, transitive but not symmetric. (3) symmetric, transitive but not reflexive. (4) neither transitive nor reflexive but symmetric.

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

28

 22. The projection vector of a and b ,is

Ê a ◊ b ˆ  (1) Á  ˜ b ÁË b ˜¯ .  a b (3)  a

  a .b (2)  b  Ê a ◊ b ˆ (4) Á  2 ˜ b Á a ˜ Ë ¯ 23. Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is (1) 37/221 (2) 5/13 (3) 1/13 (4) 2/13



24. A plane P meets the coordinate axes at A, B and C

respectively. The centroid of ∆ABC is given to be (1, 1, 2). Then the equation of the line through this centroid and perpendicular to the plane P is : x − 1 y − 1 z − 2 (2) x − 1 = y − 1 = z − 2 = = (1) 2 1 2 2 1 1 (3)

x −1 y −1 z − 2 x −1 y −1 z − 2 = = = = (4) 1 2 2 1 1 2

25. The equation of normal to the curve 3x 2 – y 2 = 8 which is parallel to the line x + 3 y = 8 is



(1) 3x – y = 8

(2) 3x + y + 8 = 0



(3) x + 3 y ± 8 = 0

(4) x + 3 y = 0

-5 8 ˘ ˙ 0 12 ˙ is a ÍÎ -8 -12 0 ˙˚ (1) diagonal matrix (2) symmetric matrix (3) skew symmetric matrix (4) scalar matrix   27. If a , b and c are three vectors such that        a + b + c = 0 and a = 2 , b = 3 and c = 5, then È0



26. The matrix ÍÍ 5

 





 

 

the value of a . b + b . c + c . a is (1) 0 (2) 1 (3) –19 (4) 38 4 . A coin is tossed. 5 A reports that a head appears. The probability that actually there was head is 4 1 (1) (2) 5 2

28. Probability that A speaks truth is

(3)

1 5

(4)





(2, 2), (3, 3), (1, 2), (2, 3), (1,3)}. Then R is (1) reflexive but not symmetric (2) reflexive but not transitive (3) symmetric and transitive (4) neither symmetric, nor transitive

(0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). let F = 4x + 6y be the objective function. The minimum value of F occurs at (1) Only (0, 2) (2) Only (3, 0) (3) the mid-point of the line segment joining the points (0, 2) and (3, 0) (4) any point on the line segment joining the point (0, 2) and (3, 0)

0 x−a x−b 0 x+a x − c , then 31. If f ( x ) = x+b x+c 0 (1) f (a) = 0 (2) f (b) = 0 (3) f (0) = 0 (4) f (1) = 0

32. The solution of differential equation xdy − ydx = 0 represents (1) a rectangular hyperbola (2) parabola whose vertex is at origin (3) straight line passing through origin (4) a circle whose centre is at origin

33. If A and B are two events such that A⊂ B and P(B)

≠ 0, then which of the following is correct? (1) P(A|B) = P(B)/P(A) (2) P(A|B) < P(A) (3) P(A|B) ≥ P(A) (4) P(A|B) ≥ P(A) È -1 Ê 7 ˆ ˘ 34. The value of cot Í cos Á ˜ ˙ is Ë 25 ¯ ˚ Î 25 25 (1) (2) 24 7

(3)

24 25

(4)

7 24

35. The shortest distance between the lines

x +3 y+7 z-6 x-3 y-8 z-3 = = = and is : = = -3 2 4 -1 1 3 (1) 7 30 (2) 3 30 2

(3) 3

36. The

(

equation

y 1+x

2 5

29. Let A = {1, 2, 3} and consider the relation R = {1, 1),

30. Corner points of the feasible region for an LPP are



2

of

(4) 2 30 tangent to

the

) = 2 – x , where it crosses x-axis is :

(1) x + 5 y = 2

curve

(2) x – 5 y = 2

(3) 5x – y = 2 (4) 5x + y = 2 37. The corner points of the feasible region determined by the system of linear constraints are (0, 10), (5, 5), (15, 15), (0, 20). Let Z = px + qy, where p, q > 0. Then, the condition on p and q so that the maximum of Z occurs at both the points (15, 15) and (0, 20), is (1) p = q (2) p = 2q (3) q = 2p (4) q = 3p 2 ˆ Ê1 38. The value of the expression tan Á cos-1 ˜ is Ë2 5¯



(1) 2 + 5



(3)

5 +2 2

(2) 5 - 2 (4) 5+ 2

Sample Question Papers

 39. If a = 4 and −3 ≤ λ ≤ 2 , then the range of λ a is (1) [0, 8] (3) [0, 12]

3.

(2) [–12, 8] (4) [8, 12]

40. The points at which the tangents to the curve

4.

3



 y x – 12 x  18 are parallel to x-axis are : (1) (2, –2), (–2, –34) (2) (2, 34), (–2, 0) (3) (0, 34), (–2, 0) (4) (2, 2), (–2, 34)

41. The length of the perpendicular from the point x3 y2 z (2, –1, 4) on the straight line   , is : 10 1 7 (1) greater than 2 but less than 3 (2) less than 2 (3) greater than 4 (4) greater than 3 but less than 4



2  3 –1 5 . Then A exist if 1 1 3 (1) λ = 2 (2) λ ≠ 2 (3) λ ≠ –2 (4) None of these

42. If A  0 2



5.

43. Family y = Ax + A3 of curves is represented by the differential equation of degree  : (1) 1 (2) 2 (3) 3 (4) 4



continuous at x =



, x=p/2

p , then k equal to : 2

(1) 6 (3) 5



3

(2) – 6 (4) – 5

45. If matrix A = [aij]2 × 2, where aij= 1 if i ≠ j, and 0 if A2

i = j then is equal to (1) I (2) A (3) 0 (4) None of these Read the following text and answer the following questions on the basis of the same: Let’s say that we want to evaluate ∫[P(x)/Q(x)] dx, where P(x)/Q(x) is a proper rational fraction. In such cases, it is possible to write the integrand as a sum of simpler rational functions by using partial fraction decomposition. Post this, integration can be carried out easily. The following image indicates some simple partial fractions which can be associated with various rational functions: S. No. 1.

2.

Form of the rational function

px + q ,a≠b ( x − a )( x − b ) px + q

( x - a )2

Form of the partial fraction A B + x -a x -b A B + x - a ( x - a )2

A B C + + 2 x -b x - a ( x - a)

px 2 + qx + r 2

( x - a) ( x - b)

A Bx + C + , x - a x 2 + bx + c

px 2 + qx + r 2

( x - a )( x + bx + c )



In the above table, A, B and C are real numbers to be determined suitably. dx 46.  ( x  1)( x  2 ) =

x +1 +C x +2

x -1 (2) log +C +2 x

x +1 (3) log +C x -2

x +2 (4) log +C x +1

(1) log

( x + 1)2 +C (x + 2)



(1) log



( x )2 +C (3) log ( x + 1)



x

47. Integration of ∫ ( x + 1)( x + 2 )

 , x≠p/2  is 44. If the functions f(x) =  p − 2 x  

A B C + + x -a x -b x -c

px 2 + qx + r ( x - a )( x - b( x - c )

where, x2 +bx +c cannot be factorised further

k cos x



29

48.

1

 x 2  9 dx (1)



49.

2

(4) log ( x - 2 ) + C ( x + 1)

=

1 x -3 +C log 6 x +3

1 x +3 (3) +C log 6 x -3

( x + 2 )2 +C (2) log ( x + 1)

1

1 x -2 (2)  log +C 6 x +3 1 x -3 +C (4)  log 3 x +3

 e x  1 dx 



(1) log

ex - 1 +C 2

ex - 1 +C (2) log 2e x



(3) log

ex - 1 +C 2x

(4) log



50.

dx

 x( x 2  1)  1 (1) log x + log x 2 + 1 2 1 2 (2) log x - log x + 1 4 1 (3) log x - log x 2 + 1 2 1 (4) log x - log x 2 - 1 3

+C +C +C +C

ex - 1 ex

+C

30

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

Section - B2

16. If a ≡ b(mod m), then (1) (a + c) ≡ (b + c) (mod m) (2) (a – c) ≡ (b – c) (mod m) (3) (a ± c) ≡ (b ± c) (mod m) (4) All of the above 17. [(4 – 9) × 7] mod 4 is : (1) 0 (3) 1

(2) -1 (4) None

18. In what ratio must a grocer mix two varieties of

pulses costing ` 15 and ` 20 per kg respectively so as to get a mixture worth ` 16.50 kg?

(1) 3 : 7 (3) 7 : 3

Applied Mathematics

(2) 5 : 7 (4) 7 : 5

19. A and B invest in a business in the ratio 3 : 2. If 5% of

the total profit goes to charity and A's share is ` 855, the total profit is: (1) 500 (2) 1000 (3) 1500 (4) 2000

20. If the present time is 8.40 PM, then the time after 876

1 hours will be: 2

(1) 8.40 AM (3) 6.10 PM

(2) 9.10 AM (4) 10.40 PM

21. Let m ∈ Z+ consider the relation Rm defined as a Rm b iff a ≡ b (mod m), then Rm is (1) reflexive but not symmetric (2) symmetric but not transitive (3) reflexive, symmetric but not transitive (4) an equivalence relation 22. Three friends X, Y and Z agrees to invest for time periods in the ratio 2 : 3 : 4. If their profit sharing ratio is 6 : 7 : 8 then the ratio of their investments is (1) 4 : 5 : 6 (2) 9 : 7 : 6 (3) 8 : 7 : 6 (4) 12 : 21 : 32 23. A retailer buys 250 kg of rice, a part of which he sells at 10% profit and the remaining at 5% loss. If the net profit made by the retailer in the whole transaction is 7%, then the quantity of rice sold at 10% profit is

(1) 200 kg (3) 100 kg

(2) 150 kg (4) 50 kg

 a b −5    24. If matrix A =  c d 0  is skew symmetric, then 5 0 0   

value of 2a + b + c – 3d is: (1) 1 (2) –1 (3) 0 (4) 2

25. If aij =

1 (3i – 2j) and A = [aij]2 × 2 is: 2



 1  2 (1)  − 1   2



2 (3)  1   2

a

 2  1 

1 (2)  2   2

2   1 −  2 

x

 1 − (4)  2   1 

1 −  2  1  1 2  2 

T 26. If A =   and if xy = 1, then det (AA ) is equal to : y a  

(1) (a – 1)2 (3) a2 – 1



(2) (a2 + 1)2 (4) (a2 – 1)2

27. If a matrix has 7 elements, then the possible orders are: (1) 1 × 7 and 7 × 1 (2) 1 × 1 and 7 × 7 (3) 2 × 1 and 7 × 2 (4) None of these 28. A TV manufacturer tests a random sample of 6 picture tubes to determine any defect. Past experience suggests the probability of defective picture tube is 0.05. The probability that there is at least one defective picture tube in the sample is 6

19 (2) 1 −    20 



19 (1)    20 



6  19 6 3  19 5   1  (3) 1 −   +    (4)    20   20  10  20  

6

29. Given that

∑ p1q1 = 506,



∑ p0 q0 = 406, ∑ p1q0 = 456 and

∑ p0 q1 = 451, where subscript 0 and 1 are used for base year and current year respectively. The Paasche’s index number is (1) 112.19 (2) 112.31 (3) 117.31 (4) 108.52

30. Price index by Marshall Edgeworth method takes (1) q0 as weights (2) q1 as weights q0 + q1 (3) as weights (4) q0 q1 as weights 2 31. A time series consists of (1) Short term variations (2) Long term variations (3) Irregular variations (4) All of the above

Sample Question Papers 32. A set of observations recorded at an equal interval of time is called.

(1) Array data

(2) Data



(3) Geometric Series

(4) Time series data

33. In the measurement of the secular trend, the moving averages: (1) Smooth out the time series (2) Given the trend in a straight line (3) Measure the seasonal Variations (4) None of these 34. The total area under the normal distributed curve above the base line i.e.,

∞

∫−∞ f ( x )dx is



(1) 0

(2) 0.5



(3) 0.75

(4) 1

The probability that a student studies at least 3 hours on a particular day is



1 7

(2)

2 7

(3)

3 7

(4)

1 2

36. An automatic machine produces 20000 pins per day. On rare occasion it produces a perfect pin whose

1 . Assuming Poisson distribution, 10000

chance is

the mean and variance of the number of perfect pins are respectively

(1)



2,

2

(3) 2, 4

(2) 2, 2 (4) 4, 2

37. For a Poisson distribution with mean l, ∑ ∞k = 0

e −λλ k k!

is equal to

(1) –1

(2) 0

1 2

(4) 1

(3)

40. If y2 + xy + x2 – 2x = 0 then 2

d2y dx 2

=?



d 2 y  dy  dy (1) ( 2 y + x ) + +2 +2 = 0 2  dx  dx dx  



d2y dy  dy  (2) ( 2 y + x ) + 2  + +2 = 0 2 dx dx dx  



d2y dy  dy  (3) ( 2 y + x ) + 2  + 2 + 2 = 0 2 dx dx dx  



d2y dy  dy  (4) x + 2  + 2 + 2 = 0 2 dx  dx  dx

2

2

if x=0  k  kx if x = 1 or 2  P(X = x) =  k ( 5 − x ) if x = 3 or 4  0 otherwise

(1)

39. The demand function of a toy is, x = 75 − 3p and its total cost function is TC = 100 + 3x . For maximum profit the value of x is (1) 33 (2) 31 (3) 29 (4) 24

2

35. Let X denotes the number of hours a student devotes to self-study during a randomly selected school day. The probability that X takes the value x, where k is some unknown constant is



31

38. The total cost function is given by C(x) = x2 + 30x + 1500 .The marginal cost when 10 units are produced is: (1) ` 20 (2) ` 30 (3) ` 50 (4) ` 70

41. If y = a log |x| + bx2 + x has extreme values at x = – 1 and at x = 2, respectively, then values of a and b are: 1 1 (1) 2 , (2) −2 , 2 2 42.

(3) 2 , −



1 2

(4) −2 , −



1 2

If we reject the null hypothesis, we might be making. (1) Type-I error (2) Type-II error (3) A correct decision (4) A wrong decision.

43. A sample of 50 pens is taken at random out of 50 we found 15 pens are of Cello, 17 are of Parker and 18 are of Reynolds. What is the point estimate of population proportion of Parker. (1) 0.3 (2) 0.34 (3) 0.36 (4) 0.4 44. In maximization problem, optimal occuring at corner point yields the (1) Heighest value of Z. (2) Lowest value of Z. (3) Mid value of Z. (4) Mean value of Z.

solution

45. In LPP, the objective funtion is always (1) Linear (2) Quadratic (3) Cubic (4) Biquadratic Read the following text and answer the following questions on the basis of the same: Seema a married women has four siblings. Her father prepared his will. Seema received a good amount of money against her father’s will. Seema decided to invest this money in order to gain profit.

32

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH. She has purchased bonds with par value of ` 1000 for three years having nominal interest rate of 10 %.

46. Which of the following is correct: (1) Present value of Bond = Present value of Interest Payments + Present value of Maturity Payments (2) Present value of Bond = Present value of Interest Payments – Present value of Maturity Payments (3) Present value of Bond = Present value of Interest Payments × Present value of Maturity Payments (4) None of these 47. G iven C is a periodic coupon payment, P is the par value and r is the interest rate then which of the following is the correct mathematical form for bond valuation : C P N + (1) Bond Value = ∑ t =1 −t (1 + r ) (1 + r ) N

C

P

C

P

C

P

(2) Bond Value =

∑ t =1 (1 + r )t + (1 + r )N

(3) Bond Value =

∑ t =1 (1 + r )t + (1 + r )− N

(4) Bond Value =

∑ t =1 (1 + r )t − (1 + r )N

N

N

N

48. Find amount of periodic coupon payment, C. (1) ` 1000 (2) ` 10,000 (3) ` 100 (4) ` 114 49. A bond is said to be issued at premium when (1) Coupon rate > Required returns (2) Coupon rate = Required returns (3) Coupon rate (4) None of these 50. Find the bond value. (1) ` 1000 (3) ` 900

(2) ` 907 (4) ` 999

qqq

SAMPLE

Question Paper Maximum Marks : 200

6 Time : 45 Min.

General Instructions :

(i) Section A will have 15 questions covering both i.e., Mathematics/Applied Mathematics which will be compulsory for all candidates. (ii) Section B1 will have 35 questions from Mathematics out of which 25 questions need to be attempted. Section B2 will have 35 questions purely from Applied Mathematics out of which 25 questions will be attempted. (iii) Correct answer or the most appropriate answer : Five marks (+ 5) (iv) Any incorrect option marked will be given minus one mark (– 1). (v) Unanswered/Marked for Review will be given no mark (0). (vi) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options. (vii) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question. (viii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (ix) Calculator / any electronic gadgets are not permitted.

Section - A



Mathematics/Applied Mathematics π



3 5 1. The value of  2π ( x  x cos x  tan x  1) dx is



(1) 0



(3) π







(2) 2 (4) 1

3x + 2 3 , 2. Let f : R −   → R be defined by f ( x ) =

Then

2

5x − 3

5

−1 (1) f (x ) = f (x )

−1 (2) f (x ) = − f (x )

(4) f −1 (x ) = 1 f (x ) 19 3. Five dice are thrown simultaneously. If the occurrence of an odd number in a single dice is considered a success, find the probability of maximum three successes. (3)  fof  x =  x

(1)

13 12

(2)

13 15

(3)

11 16

(4)

13 16

4. The feasible region for an LPP is shown in the given figure.

Then, the minimum value of Z = 11x + 7y is (1) 21 (2) 47 (3) 20 (4) 31 π 5. Solve sin −1 ( 1 − x ) − 2sin −1 x = , then x is equal to 2 1 (1) 0, (2) 1, 1 2 2

(3) 0

(4)

1 2

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

34

(1) Both A and R are true and R is the correct explanation of A



(2) Both A and R are true but R is NOT the correct explanation of A



(3) A is true but R is false



(4) A is false and R is True 2 0 0 



6. Assertion (A): If A =  0 3 0  , then  0 0 4 

1 2   –1 A = 0  0 



 0  0 .   1 4 

0 1 3 0



(1) (0, 0)

(2) (0, 8)



(3) (5, 0)

(4) (4, 10)



4 12. Let f : R − −  → R be function defined as

 3 4x f (x) = . The inverse of f is map g : Range 3x + 4  4 f → R − −  given by  3

Reason (R): The inverse of an invertible diagonal matrix is a diagonal matrix.

7. Given A =  2

 4



Reason (R): A -1 =

1 ( adj A ) A

8. The integrating factor of differential equation (1 - x 2 )

(3)

dy - xy = 1 is dx (2) 2

1-x

x 1 + x2

(4)

1 log (1 - x 2 ) 2



(1) 0 (2) 1 3



(3) 1 12



(3) g ( y ) =



13.

4y 3  4y

cos 2 x

∫ (sin x + cos x )

2

4y (2) g ( y ) = 4 3y −



(4) g ( y ) =



dx is equal to

−1 +C sin x + cos x



(1)



(2) log |sin x + cos x| + C



(3) log |sin x – cos x| + C



(4)



3y 4 − 3y

1 (sin x + cos x )2

14. If A and B are invertible matrices, then which of the following is not correct?

(4) 1 36

10. Find the value of tan −1 3 − sec−1 (−2) is equal to π 3



(1) π

(2) −



π (3) 3

2π (4) 3



3y (1) g ( y ) = 3 − 4y

9. The probability of obtaining an even prime number on each die, when a pair of dice is rolled is





Assertion (A): 2A–1 = 9I – A

(1) – x



3  . 7 



(1) adj. A = A . A −1



(2) det A −1 = det ( A ) 



(3) ( AB )



(4) ( A + B ) = B −1 + A −1



11. The feasible solution for a LPP is shown in following figure. Let Z = 3x – 4y be the objective

( )

function. Minimum of Z occurs at

= B −1 A −1 −1

15. If 2x+ 3y = sin y, then dy is equal to : dx



(1)

−1

−1



2 cos y

(2)

2 cos y + 3

(4)

2 3 − cos y



2 (3) cos y − 3



Sample Question Papers

Section - B1



Mathematics

16. Area lying in the first quadrant and bounded by 2

2

circle x + y = 4 and the lines x = 0 and x = 2 is p 2 p (4) 4

(1) p



(2)

p 3

(3)

 

17. Solve that : tan −1  x  − tan −1 x − y is equal to y

(1)

 2



 3





(3)

(3) ecos x

19. The

curves x 3 – 3xy 2  2  0 3x y – y – 2  0 intersect at an angle of



(1) (3)

and

(2)

p 2

(4)

Z = 2x + y subject to constraints x + y ≤ 2, x ≥ 0, y ≥ 0 is (1) 4 (2) 3 (3) 1 (4) 0  24. The angle between two vectors a and b , with  magnitudes 3 and 4, respectively, and a .b = 2 3 is : p p (1) (2) 3 6

p 5p (4) 2 2 25. If A is a square matrix such that A2 = I, then (A − I)3 + (A + I)3 – 7A is equal to (1) A (2) I – A (3) I + A (4) 3A

3

p 4

p 3 p 6

20. The position vector of the point which divides the 

(2) 2x – z = 2 (4) x – y – z = 0

23. The maximum value of Z for the problem maximise



(4) esec x

two

2





x+y

(2)

(1) x + 3y + z = 4 (3) x – 3y – 2z = –2



 3 (4) 4 4 18. The integrating factor of differential equation dy + y tan x - sec x = 0 is dx (1) cos x (2) sec x

(3)

b



26. If f(a + b – x) = f(x), then  × f ( x )dx is equal to a

b



(1)

a+b ∫ f (b − x )dx 2 a b

(3) b − a f(x)dx ∫

(2)

ab b f ( b + x )dx 2 a

(4)

a+b b f ( x )dx 2 a

  join of points 2 a - 3b and a + b in the ratio 3 :1 is      7 a - 8b 3a - 2b (1) (2) 4 2   3a (3) (4) 5a 4 4 21. On using elementary row operation R1 → R1 − 3R2 in the following matrix equation



4 2 1 2 2 0  3 3  =  0 3   1 1  we have       5 7   1 7   2 0  = (1) 3 3   0 3   1 1    5 7   1 2   1 3  = (2) 3 3   0 3   1 1  

(1) xyz (2) x–1y–1z–1 (3) –x – y – z (4) –1

 5 7  1 2   2 = (3) 3 3  1 7   1  2 1 2  2 4 (4)  5 7  =  3 3   1    

35

0 1  0 1 



a

27. If x, y and z are all different from zero and 1x 1 1 then 1 1 y 1  0, 1 1 1z

value

of

x –1  y –1  z –1 is







22. The equation of a plane containing the line of intersection of the planes 2x – y – 4 = 0 and y + 2z – 4 = 0 and passing through the point (1, 1, 0) is :

2



p

  28. If f(x) = |cos x – sin x|, then f '   is equal to 3 (1)

3 −1 2

(2)

1− 3 2

(3)

3 +1 2

 3 +1 (4) −  2   

29. The plane through the intersection of the planes x + y + z = 1 and 2x + 3y – z + 4 = 0 and parallel to y-axis also passes through the point : (1) (–3, 0, –1) (2) (–3, 1, 1) (3) (3, 3, –1) (4) (3, 2, 1)

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

36

30. On

using elementary column operations C2 → C2 − 2C1 in the following matrix equation

 1 3   1 1  3 1   2 4  =  0 1   2 4  we have     

 1 5   1 1  3 5  (1)  =    0 4   2 2   2 0 



 1 5   1 1  3 5  (2)  =    0 4   0 1   0 2 



 1 5   1 3   3 1  (3)  2 0  =  0 1   2 4      

 1 5   1 1  3 5  (4)  =    2 0  0 1   2 0  31. If x = 2 at y = at2, where

d2y dx

2

at x =

1 is : 2

1 (1) 2a (3) 2a



(1) (0, 1)



(3) (2, 0)







(1) 1 (3)





3

(2) Is * commutative but not associative?



(3) Is * associative but not commutative?



a 2

y  e 2x at the point (0, 1)  1



(2)   2 , 0 (4) (0, 2)

2 5

(4)

(2)

1



40. The identity element for the binary operation * defined on Q ~ {0} as

of intersection of the planes x + y + z =1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0 is : (1) r × (i – k) + 2 = 0 (2) r . (i – k) – 2 = 0 (3) r . (i – k) + 2 = 0 (4) r × (i – k) – 2 = 0 p dy 36. If x = a sec2 q and y = a tan3 q, then at θ = dx 3

(1) Is * both associative and commutative?

(4) Is * neither commutative nor associative?     39. For any vector a, the value of ( a × i )2 + ( a × j )2 + ( a × k )2    ( a × i )2 + ( a × j )2 + ( a × k )2 is equal to 2 2 (1) a (2) 3a a is constant, then 2 2 (3) 4 a (4) 2 a

35. The vector equation of the plane through the line

is :





2i − 4 j + λ k are parallel is 2 3 (1) (2) 3 2 5 2

1 2

(4)

(4)

= a3 + b3. Choose the correct answer.

34. The value of λ for which the vectors 3i − 6 j + k and

(3)

(2) 0

(3) –1

(2) 1

33. The tangent to the curve



(1) 1

π 4 38. Consider a binary operation * on N defined as a * b

32. If the lines x = ay + b, z = cy + d and x = a¢ z + b¢,

meets x-axis at :

1





(4)

2x  1

 dx is 37. The value of 0 tan –1   1+ x  x 2 



y = c¢ z + d¢ are perpendicular, then : (1) cc¢ + a + a¢ = 0 (2) aa¢ + c + c¢ = 0 (3) ab¢ + bc¢ + 1 = 0 (4) bb¢ + cc¢ + 1 = 0





3 2





a*b=

ab ,∀a , b ∈ Q ~ {0} is 0 2

(1) 1

(2) 0

(3) 2

(4) None of these

41. The vectors λˆi + ˆj + 2 kˆ, ˆi + λˆj − kˆ and 2iˆ − ˆj + λkˆ are coplanar, if

(1) l = –2

(2) l = 0



(3) l = 1

(4) l = –1



42. The area of the region bounded by the curve x = 2y + 3 and the y lines y = 1 and y = –1 is,

3 (1) 4 sq. units (2) sq. units 2 (3) 6 sq. units (4) 8 sq. units

43. The solution of x (1) y =

ex k + x x

dy + y = e x is dx (2) y = xex + Cx

ey k + y y 44. The slope of tangent to the curve x = t2 +

(3) y = xex + k

(4) x =

3t – 8, y = 2t2 – 2t – 5 at the point (2, –1) is : 22 6 (2) 7 7 –6 (3) (4) – 6 7 dy 45. If y = log2 (log2 x), then is equal to : dx



(1)



(1)

log 2 e log e x

(2)

log 2 e x log e 2

(3)

log 2 e x log e x

(4)

log e x x log 2 e







Sample Question Papers Attempt any four sub-parts from each question.



37

47. What is the probability that B, C will hit and A will lose?

Each sub-part carries 1 mark. Read the following text and answer the following questions on the basis of the same: A coach is training 3 players. He observes that the player A can hit a target 4 times in 5 shots, player B can hit 3 times in 4 shots and the player C can hit 2 times in 3 shots.



(1)

1 10

(2)

3 10

7 4 (4) 10 10 48. What is the probability that ‘any two of A, B and C will hit?



(3)

(1)

1 30

(2)

11 30

(3)

17 30

(4)

13 30

49. What is the probability that ‘none of them will hit the target’?





46. Let the target is hit by A, B: the target is hit by B and, C: the target is hit by A and C. Then, the probability that A, B and, C all will hit, is (1)



(3)





4 5 2 5

3 5 1 (4) 5



(2)



16. If a ≡ b(mod n) and b ≡ c(mod n), then

(3) a ≡ c(mod n)

(4) None of these

17. When 783 × 657 × 594 × 432 × 346 × 251 is divided

by 5, then remainder is : (1) 3 (2) 5 (3) 0 (4) 1







this it takes 8 hours to fill the cistern. Had there not been a leak, it would take one hour less to fill the cistern. How much time does it take for the leak to completely empty the cistern? 1 (1) 48 hours (2) 55 hours 3





(3)

(1)

59 60

(2)

2 5

(3)

3 5

(4)

1 60

(4) 15 hours

19. If x < 5, then

(1) – x < – 5 (2) – x ≤ – 5 (3) – x > – 5 (4) – x ≤ – 5 20. 6 mod 9 + 11 mod 9 congruent to: (1) 8 mod 9 (2) 6 mod 9 (3) 7 mod 9 (4) 0 mod 9

21. If ` 85 is divided among 100 students such that

each boy got a rupee and each girl got ` 0.5, then number of girls are:

(1) 70 (3) 40

18. There is a leak at the bottom of a cistern. Due to

(3) 56 hours

1 60

Applied Mathematics

(2) b ≡ c(mod a)



(2)

Section - B2

(1) a ≡ a(mod n)

1 30

2 1 (4) 15 15 50. What is the probability that at least one of A, B or C will hit the target?



(1)



(2) 30 (4) 50

22. In what ratio must be kind of rice at ` 1.15 and `

1.24 per kg be mixed so that by selling at ` 1.50 per kg, 15% may be gained ?

(1) 2 : 5 (3) 1 : 5

(2) 5 : 2 (4) 2 : 3

23. For two distinct positive numbers x and y



(1) x + y > 2 xy



(3) xy >



x+y 2



(2)

x+y > xy 2

(4)

2xy > xy x+y

24. A river passing near a town floods it on an average

twice every 10 years. Assuming Poisson distribution find the probability that the town faces flooding at least once in 10 years.



(1) 0.0198

(2) 0.1353



(3) 0.5657

(4) 0.8647

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

38

25. The height of certain species of plant is normally distributed with mean of and standard deviation of 4 cm. the probability that the height of chosen at random lies between 10 30 cm



(1) 0.0062

(2) 0.5341



(3) 0.9876

(4) 0.9938



20 cm what is a plant cm and

26. Which one of the following is not the correct property of normal distribution ?

(3) Number of columns of A is equal to the number of rows of B

(4) A = B.

32. If A is a square matrix, then which of the following is not symmetric ?

(1) A + AT

(1) Continuous distribution



(2) Equality of central values (Mean, Mode and



(3) Standard deviation is the sole parameter of the



(4) Uni-modal distribution

Median) distribution

(3) ATA (4) A – AT

33. If A and B are symmetric matrices of order n, where



(1) A + B is skew-symmetric



(2) A + B is symmetric

(3) A + B is a diagonal matrix (4) A + B is a zero matrix



27. A die is thrown twice, the probability that 5 will



34. A chocolate bar sells for ` 20. If a shopkeeper sells 30 bars per day then, the by selling 30 bars, total revenue and average revenue respectively are:

come up at least once is:

11 36



(1)



(3)



(2)

35 36

7 12

(4) None of these

28. A random variable x has Poisson distribution with mean 2. Then P(x > 1.5) equals .....

2



(1)



(3) 1 −



e 3 2

e

(4)



(1) ` 600, ` 20 (2) ` 200, ` 60



(3) ` 200, ` 20 (4) ` 600, ` 60



respect x?

(3) 7x–2/3 – 3x–3 + 5

(4) 7x–1/3 – 3x–3 + 5

3 e

2

29. What is the other name for Bernoulli trials ?

Find −





(3) Nucleo experiment





(4) Three-way experiment

30. Which one of the following is not true about the

1 0 0  matrix  0 0 0  ?   0 0 5  (1) a scalar matrix

(2) a diagonal matrix

(3) an upper triangular matrix (4) a lower triangular matrix

(1) A and B are two matrices not necessarily of same order

1 2x

u ''( x ) . u '( x ) (2) −

(3) 2x

1 2x

(4) – 2x

37. Find the second derivative of the following function: f(x) = 5x2 (x + 47)



(1) f "(x) = 30x – 470

(2) f "(x) = 30x + 470



(3) f "(x) = 15x2 + 235

(4) f "(x) = 15x2 + 470x



38. A statement made about a population for testing purpose is called :



(1) Statistics



(3) Level of significance (4) Test-statistics



31. If A and B are two matrices such that A + B and AB are both defined, then

(1)



(2) Dichotomous experiment



36. Suppose that you have the following utility function: u( x ) = x





+ 5x with

(2) 7/3x–2/3 + 12x–5 + 5

(1) Two-way experiment



3 x4

(1) 7x–1/3 – 3x–5 + 5





35. What is the derivative of 7 3 x −



(2) 0

2

(2) AAT

(A ≠ B), then





(2) A and B are square matrices of same order

(2) Hypothesis

39. A statement whose validity is tested on the basis of a sample is called :



(1) Null hypothesis



(3) Simple hypothesis (4) Composite hypothesis

(2) Statistical hypothesis

Sample Question Papers

40. The constraint of an LP model restricts :



(1) Value of the objective function



(2) Value of the decision variable



(3) Use of the available resources



(4) All of the above



39

by equal payment at the end of each month for 5 years. Given that (1.005)60 = 1.3489, (1.005)21 = 1.1104.

41. The objective function for a LP model is 3x1 + 2x2; if x1 = 20 and x2 = 30, what is the value of the objective function ?



(1) 0

(2) 50



(3) 60

(4) 120



42. Index number is a special type of:

(1) Average

(2) Dispersion

(3) Correlation

(4) None of the above

43. Index number is always expressed in

46. The size of each monthly payment. is

(1) Percentage

(2) Ratio



(3) Proportion

(4) None of the above

44. In a straight line equation Y = a + bX; a is the .....

(1) ` 4833 (3) ` 4383

(1) X-intercept

(2) Slope



(3) Y-intercept

(4) None of these

47. The principal outstanding at beginning of 40th month

45. An orderly set of data arranged in accordance with their time of occurrence is called.

(1) ` 60069 (3) ` 96000

(1) Arithmetic Series (2) Geometric Series



(3) Harmonic Series (4) Time Series

(1) ` 480 (3) ` 840

Read the following text and answer the following questions on the basis of the same:

Ajay is a service man. He lives in a joint family. There are 6 members in his family. He is planning to purchase a car so he is searching for a bank for a loan. He take a loan of ` 250000 at the interest rate of 6% p.a. compounded monthly is to be amortize

(2) ` 8432 (4) ` 4338



48. The interest paid in 40th payment is (2) ` 408 (4) ` 482

49. Principal amount paid in 40th payment is

(1) ` 4325 (3) ` 4352

(2) ` 96097 (4) ` 98670

(2) ` 4359 (4) ` 4532

50. Total interest paid is

(1) ` 39961 (3) ` 31699

(2) ` 36991 (4) ` 39691 qqq

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

SAMPLE

7

Question Paper Maximum Marks : 200

Time : 45 Min.

General Instructions :

(i) Section A will have 15 questions covering both i.e., Mathematics/Applied Mathematics which will be compulsory for all candidates. (ii) Section B1 will have 35 questions from Mathematics out of which 25 questions need to be attempted. Section B2 will have 35 questions purely from Applied Mathematics out of which 25 questions will be attempted. (iii) Correct answer or the most appropriate answer : Five marks (+ 5) (iv) Any incorrect option marked will be given minus one mark (– 1). (v) Unanswered/Marked for Review will be given no mark (0). (vi) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options. (vii) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question. (viii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (ix) Calculator / any electronic gadgets are not permitted.

Section - A



Mathematics/Applied Mathematics 1

1. If f : R → R be given by f ( x ) = (3 − x 3 ) 3 ,,then fof(x) is



(1) x



(3) x

1/3





(2) x3

 sin 3x

, x≠0

value

of

the

determinant

x x + y x + 2y x + 2y x x + y is x + y x + 2y x

(4) (3 - x 3 )

2. If the functions f(x) =  x

5. The



(1) 9x 2 ( x + y )

(2) 9y 2 ( x + y )

at x = 0, then the value of k is :



(3) 3y 2 (x + y )

(4) 7x 2 ( x + y )



(1) 3

(2) 6



(3) 9

(4) 12





 k / 2





continuous

, x=0

dy 2 xy 1 + = is dx 1 + x 2 (1 + x 2 )2

3. Two events A and B will be independent, if

(1) A and B are mutually exclusive



(2) P(A′ B′) = [1 – P(A)] [1 – P(B)] (3) P(A) = P(B)



4.

∫

cos x · dx (1 + sin x ) ( 2 + sin x )

(3) y log (1 + x 2 ) = C + tan -1 x

( 2 + sin x ) (1 + sin x ) + C (2) log +C (1) log ( 1 + sin x ) ( 2 + sin x ) 2



(3)

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

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

(4) P(A) + P(B) = 1

6. The solution of differential equation

 2 + sin x  1  2 + sin x  log   + C (4) log   +C x 2 1 + sin    1 + sin x 

(4) y(1 + x 2 ) = C + sin -1 x

7. If x is real, the minimum value of x 2 – 8 x + 17 is



(1) –1

(2) 0



(3) 1

(4) 2

Sample Question Papers 8. If P(A) = 1/2, P(B) = 0, then P(A|B) is (1) 0 (2) 1/2 (3) not defined (4) 1







(1) 1

(2) 2



(3) 3

(4) 4



9. Let f : R → R be defined as f(x) = x4. Choose the

correct answer. (1) f is one-one onto. (2) f is many-one onto. (3) f is one-one but not onto. (4) f is neither one-one nor onto.  10. If a and b ,are two collinear vectors, then which of the following are incorrect :   (1) b = la , for some scalar λ   (2) a = ±b   (3) the respective components of a and b ,are not proportional   (4) both the vectors a and b  ,have same direction, but different magnitudes

11. If A and B are events such that P(A|B) = P(B|A), then (1) A ⊂ B but A ≠ B (2) A = B (3) A ∩ B = φ (4) P(A) = P(B)

12.

(

)

(3) 0



equals



(2) log



Ê x - 1ˆ (3) log Á +C Ë x - 2 ˜¯



(4) log ( x - 1)( x - 2) + C

( x - 2)2 +C x -1 2





sin 15. If y = e

−1

x

and z = e

cos−1 x

, find the value of

dy dz

π

(1)

π 2

(2) e 2

(3)

π 6

π (4) e 6

Section - B1

Mathematics

(1) Both A and R are true and R is the correct explanation of A



(2) Both A and R are true but R is NOT the correct explanation of A



(3) A is true but R is false



(4) A is false and R is True



Reason (R): Generally AB ¹ BA

18. The general solution of



(1) ysec x = tan x + C



(2) ytan x = sec x + C



(3) tan x = ytan x + C



(4) xsec x = tan y + C

dy + y tan x = sec x is dx



(1) – 32

(2) – 30



(3) 20

(4) 30





(1) 1

(2) 2



(3) 3

(4) 4

1 1 with position vectors −ˆi + ˆj + 4 kˆ, ˆi + ˆj + 4 kˆ, 2 2 1 1 ˆi − ˆj + 4 kˆ and −ˆi − ˆj + 4 kˆ, respectively is 2 2 1 (1) (2) 1 2 (3) 2 (4) 4

22. The

smallest

value

of

the

polynomial

2



x – 18 x + 96 x in [0, 9] is (1) 126 (2) 0



(3) 135

19. Let A = {1, 2, 3}. Then number of equivalence



21. Area of a rectangle having vertices A, B, C and D

3



relations containing (1, 2) is

x + y ≤ 7, 2x – 3y + 6 ≥ 0, x ≥ 0, y ≥ 0 is



A2 = A, then (I + A)2 – 3A = I

17. Assertion (A): (A + B)2 ¹ A2 + 2AB + B2.

20. The minimum value of the objective function Z = 13x – 15y, subject to constraints

16. Assertion (A): If A is a square matrix such that

Reason (R): AI = IA = A



x dx

Ú ( x - 1)( x - 2)

2 (1) log ( x - 1) + C x-2

(4) 2 3





14.





Directions : In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.





π (2) 2





13. Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is

tan -1 3 - cot -1 - 3 is equal to

(1) p



41

(4) 160

 23. If θ is the angle between any two vectors a and b ,    then a.b = a ´ b when θ is equal to



(1) 0



(3)

p 2

(2)

p 4

(4) p

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

42

24. If P(A) = 0.4, P(B) = 0.8 and P(B|A) = 0.6, then

P(A∪B) is equal to (1) 0.24 (2) 0.3 (3) 0.48 (4) 0.96   25. ALet a and b be two-unit  vectors and q is the angle between them. Then a + b is a unit vector if (1) =



 4

(2) =

2  (4) = 3 2 26. The maximum value of sin x.cos x is 1 1 (1) (2) 4 2 (3)



the line x + y = 2

(3) 2p – 1 (4) 2(p + 2)



(2)

2 (3) x

(4) 1 – x2





28. The feasible region (shaded) for a LPP is shown in following figure.





1 2

2

a −x

2



2

1

(2)

1

(3)



2 1 − x2

dy







−1

(1) 2

a−x

−1 32. y = tan a + x Find dx

(1) −

27. The derivative of cos–1 (2x2 – 1) w.r.t. cos–1 x is :





(4) 2 2

2

31. Smaller area enclosed by the circle x 2 + y 2 = 4 and

(1) 2(p – 2) (2) p–2

 3

(3) =





a2 + x 2



2

2

a − x2

1

(4)

2

a2 + x 2

1 0 0  33. The matrix  0 2 0  is a  0 0 4  (1) identity matrix (2) symmetric matrix (3) skew-symmetric matrix (4) None of these

34. The solution of equation (2y −1)dx − (2x +3)dy = 0 is



(1)

2x - 1 =k 2y + 3

(2)

(3)

2x + 3 =k 2y - 1

(4)

2y + 1 =k 2x - 3

2x - 1 =k 2y - 1

35. A box has 100 pens of which 10 are defective. What

is the probability that out of a sample of 5 pens drawn one by one with replacement at most one is defective? 5

4  9 1 9  (1)  10  (2)   2  10 

Then, the maximum value of Z = 3x – 4y is : (1) 156 (2) 196 (3) 152 (4) 180 1 1 , then 29. If cos x > sin x 1

(1)



(3) 1  x
6}. Choose the correct answer. (1) (2, 4) ∈ R (2) (3, 8) ∈ R (3) (6, 8) ∈ R (4) (8, 7) ∈ R

5. The general solution of differential equation ( e x  1) y dy  ( y 1)e x dx is (1) (y + 1) = k (ex+ 1)

6. Three persons, A, B and C, fire at a target in turn,



(1) both continuous and differentiable on R–{–1} (2) continuous on R–{–1} and differentiable on R–{–1,1} (3) continuous on R–{1} and differentiable on R–{–1,1} (4) both continuous and differentiable on R–{1} Directions : In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.

54

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.



(A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is NOT the correct explanation of A (C) A is true but R is false (D) A is false and R is True



9. Let W be the set of words in the English dictionary.







A relation R is defined on W as R = {(x, y) Î W × W such that x and y have at least one letter in common}. Assertion (A): R is reflexive. Reason (R): R is symmetric. 2p 2p 10. Assertion (A): sin 1  sin    3  3

 p p  Reason (R): sin 1 (sin )  , if    ,   2 2 11. The feasible solution for a LPP is shown in following figure. Let Z = 3x – 4y be the objective function. Minimum of Z occurs at





12. Which of the following functions are decreasing on

(1) cos x (3) cos 3x

are drawn at random without replacement the probability of getting exactly one red ball is

45 196 15 (3) 56





(2) cos 2x (4) tan x

13. A bag contains 5 red and 3 blue balls. If 3 balls

(1)



14.

135 392 15 (4) 29

(2)

dx



is equal to 9x  4 x 2 1 1  9x  8  1  8 x  9  C  C (2) sin  (1) sin 1   9  2  8  9 (3)

1 1  9x  8   9x  8  sin 1   C (4) sin 1  C  9   9  3 2

15. The vertices B and C of a DABC lie on the line x  2 y  1 z such that BC = 5 units.   , 3 0 4 Then the area (in sq. units) of this triangle, given that the point A (1, –1, 2) is :

(1) 2 34

(2) 34

(3) 6

(4) 5 17

Section - B1





(2) (0, 8) (4) (4, 10)

 p  0, 2 





(1) (0, 0) (3) (5, 0)

16. Let f : [–1, 3] Æ R be defined as

Mathematics

 x  x  ,  1  x  1  f (x)   x  x , 1  x  2  x  2, 2  x  3 

where [t] denotes the greatest integer less than or equal to t. Then, ƒ is discontinuous at : (1) four or more points (2) only one point (3) only two points (4) only three points 3p  1  17. The value of cos  cos  is  2 p 3p (1) (2) 2 2 5p 7p (3) (4) 2 2 18. The vector in the direction of the vector ˆi  2 ˆj  2 kˆ that has magnitude 9 is ˆi  2 ˆj  2 kˆ (1) ˆi  2 ˆj  2 kˆ (2) 3 ˆ ˆ ˆ ˆ (3) (4) 3(i − 2 j + 2 k ) 9(i − 2 ˆj + 2 kˆ)



19. Area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x 2 + y 2 = 32 is



(1) 16 p sq. units

(2) 4 p sq. units

(3) 32 p sq. units (4) 24 p sq. units 7 9 20. Let P(A)  , P(B) = and P(AÇB)  4 . Then 13 13 13 P(A|B) is equal to 6 4 (1) (2) 13 13 4 5 (3) (4) 9 9



21. Let f : R → R be defined as f(x) = 3x. Choose the correct answer.



(1) f is one- one onto



(2) f is many-one onto



(3) f is one-one but not onto

(4) f is neither one-one nor onto

Sample Question Papers

22. The feasible region (shaded) for a LPP is shown in following figure

Then, the maximum value of Z = x + 2y is, (1) 1 (2) 5 (3) 9 (4) 8

23. If the events A and B are independent, then P(A∩B)

is equal to (1) P(A) + P(B) (2) P(A) – P(B) (3) P(A).P(B) (4) P(A)|P(B)

24. The plane passing through the point (4, –1, 2) and parallel to the Lines

x 2 y 2 z 1   3 1 2

x2 y3 z4 and also passes through the   1 2 3 point  : (1) (1, 1, –1) (3) (–1, –1, –1)



(2) (1, 1, 1) (4) (–1, –1, 1)

25. On which of the following intervals is the function f given by f(x) = x100 + sin x – 1 decreasing?



(1) (0, 1)

p  (2)  , p 2 



 p (3)  0,   2

(4) None of these



26. Solve sin tan 1 x  , x P(A ) B)) > P(A ) P(PA(A|B   P (A ) P(P A( B)B)  P(A )   ) B) > P(A ).P(B) P(PA(B  B P ( A B)) > P(A ).P(B) P (A   )B)  P(B) P(P A(A   P (B )  PP((BA|)A ) > P(B)  P(B |A ) > P(B)

67

 3   1  = sin  cos     5  

1 



Solutions

⇒

y = ±2

So, the points at which normal is parallel to the given line are (±2, ±2). Hence, the equation of normal at (±2, ±2) is 1 y  ( 2 )  1 [ x  ( 2 )] ( ) y   2   1 3[ x  ( 2 )] y  ( 2 )   3 [ x  ( 2 )] 3[ y  ( 2 )]  [3x  ( 2 )] 3[ y  ( 2 )]  [ x  ( 2 )] ⇒ 3[xy3(y2)]8  0[ x  ( 2 )] x  3y  8  0 x  3y  8  0 ∴ ⇒



Also, the given equation of the line is x + 3y = 8 3 y 8  x ⇒ x 8 y   ⇒ 3 3

1 Thus, slope of the line is − which should be 3 equal to slope of the equation of normal to the curve. On differentiating equation (i) with respect to x, we get

y2 = 4

⇒

11. Option (3) is correct. Explanation : We have, the equation of the curve is 3x 2 − y 2 = 8  ….(i)

– y2 = 8

dy 6x  2 y 0 dy 6x  2 y 0 dx dx dy 6 x 3x    Slope of the curve dy 6 x 3x      Slope dx of 2 ythe ycurve dx 2 y y Now, slope of normal to the curve Now, slope of normal to the curve 1 1    dy   dy   dx   dx  1 1    3x   3x   y   y  y y   3x 3x 1  y y 1            3x  3  3x  3   3 y  3x   3 y  3x  yx  yx

12. Option (2) is correct. Explanation: aRb Þ a is brother of b. This does not mean b is also a brother of a as b can be a sister of a. Hence, R is not symmetric. aRb Þ a is brother of b and bRc Þ b is a brother of c. So, a is brother of c. Hence, R is transitive.

13. Option (2) is correct. Explanation : Corner points

Corresponding value of Z = 4x + 3y

(0, 0)

0

(0, 40)

120

(20, 40)

200

(60, 20)

300 ← Maximum

(60, 0)

240

Hence, maximum value of Z = 300 < 325 So, the quantity in column B is greater.

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

68

14. Option (3) is correct.

a

∫−a f ( x ) dx = 0

Explanation : Let, I=Ú =Ú

p/2 -p / 2 p/2 -p / 2

( x 3 + x cos x + tan 5 x + 1) dx x dx + Ú 3

p/2 -p / 2

x cos x + Ú

p/2 -p / 2

tan xdx +

∫−a

= 2 [ x ]0

π/2

p/2

Ú - p / 2 1 ◊ dx



π/2

0

5

It is known that if f(x) is an even function, then a

I = 0 + 0 + 0 + 2∫

∴

=

1 ⋅ dx

2π =π 2

15. Option (4) is correct. Explanation :

a

f ( x ) dx = 2 ∫ f ( x ) dx

The foot of perpendicular from point P( α , β, γ )

0

on y-axis is Q(0, β, 0).

and if f(x) is an odd function, then

∴ Required distance,

PQ =

16. Option (4) is correct.

Þ

Explanation: Let, È 3p ˆ ˘ Ê 33p ˆ ˘ Ê -1 È = sin -1 Í cos Á ˙ = sin Í cos ÁË 6 p + ˜¯ ˙ ˜ Ë ¯ 5 ˚ 5 ˚ Î Î



Ê 3p ˆ ˘ = sin Í cos Á ˜ ˙ Ë 5 ¯˚ Î [ cos( 2np + q) = cos q]

dy and + 2x = 0 dx 2x dy and fi = dx a -2 Ê dy ˆ = = m1 fi Á ˜ Ë dx ¯ (1,1) a

pˆ Ê = sin -1 Á - sin ˜ Ë 10 ¯

[ sin -1 ( - x ) = - sin -1 x]

\

m1m2 = - 1

fi

Ê -2 ˆ ÁË a ˜¯ .3 = -1

\

17. Option (4) is correct. Given,

logee2

Þ

y = 2loge x –

Þ

y = 2loge x – 2

Þ

dy 2 = dx x

dy = 3x 2 dx

a= 6

19. Option (4) is correct. Explanation:

È 200 50 ˘ A = Í ˙ Î 10 2 ˚



È 50 40 ˘ B = Í ˙ Î2 3˚

Explanation: Ê x2 ˆ y = log e Á 2 ˜ Ëe ¯

dy dx

Since, the curve cuts orthogonally at (1, 1).

pˆ Ê = - sin -1 Á sin ˜ Ë 10 ¯

È Ê -p p ˆ ˘ -1 , ˙ Ísin (sin x ) = x , x Œ ËÁ 2 2 ¯˜ ˚ Î

3x 2 =

Ê dy ˆ and Á ˜ = 3.1 = 3 = m2 Ë dx ¯ (1,1)

˘ È Êp ˆ Ícos Á + x ˜ = - sin x ˙ Ë2 ¯ ÍÎ ˙˚

p =10

18. Option (4) is correct.

a.

È Ê p p ˆ˘ = sin -1 Í cos Á + ˜ ˙ Ë 2 10 ¯ ˚ Î



-2 d2y = 2 dx 2 x

3 Explanation : Given that, ay + x 2 = 7 and x = y On differentiating both equations with respect to x, we get

-1 È



α2 + γ 2

Section - B1



( α − 0)2 + (β − β)2 + ( γ − 0)2 =



È 200 50 ˘ È 50 40 ˘ AB = Í ˙Í ˙ Î 10 2 ˚ Î 2 3 ˚

È10000 + 100 8000 + 150 ˘ = Í 500 + 4 400 + 6 ˙˚ Î

Solutions È10100 8150 ˘ AB = Í 406 ˙˚ Î 504 |AB| = (10100)(406) – (504)(8150) = 4100600 – 4107600 = –7000

Points of intersection are (–1,

shown in the following figure :

20. Option (2) is correct.

A = = A = =

21. Option (4) is correct.

Equation of plane is 2 x - 2 y + z = 5.  Normal to the plane is n = 2ˆi - 2 ˆj + kˆ.

= = = = = =

The angle between line and plane is ‘θ’. Then,  b.n sin q =   b n = = = = sin q =

2

x 33 ˘ 2 11 ÈÈ xx 22 + 22 xx - x ˘˙ ÍÍ + 44 Î 22 33 ˙˚ -1 Î ˚ -1 11 ÈÈ 11 ˘˘ -2- 33 ˙˙ Í 88 ˚˚ 2 44 ÍÎÎ 11 ÈÈ 11 ˘˘ -2- 33 ˙˙ Í 88 ˚˚ 44 ÎÍÎ 2 99 sq. units units sq. 88

[ C1 Æ C1 + C2 and C2 Æ C2 + C3 ] a+c 1 a = (a + b + c) b + c 1 b c+b 1 c

50 9 3

[Taking ( a + b + c ) common from C2 ] [ R2 Æ R2 - R3 and R1 Æ R1 - R3 ]

15 2 1

a-b 0 a-c = (a + b + c) 0 0 b-c c+b 1 c

5 2 2 10

[Expanding along R2 ] = ( a + b + c )[( b - c ) ( a - b )] = ( a + b + c )( b - c )(( a - b )

22. Option (4) is correct. Explanation:

1 For x=−1, y = and for x = 2, y = 1 4

x 22 ˘˘

Explanation: We have a-b b+c a a+c b+c+a a b-a c+a b = b+c c+a+b b c-a a+b c c+b a+b+c c

32 + 4 2 + 52 4 + 4 + 1 6-8+5

x2 - x - 2 = 0 ( x - 2 )( x + 1) = 0 x = -1, 2

ÈÈ x + 2

23. Option (4) is correct.

(3ˆi + 4 ˆj + 5kˆ) ◊ (2ˆi - 2 ˆj + kˆ)

x2 = x + 2

22

dx ÚÚ ÍÍ x +4 2 -- x4 ˙˙˚ dx 4˚ -1 Î Î 4 -1

Explanation : We have, the equation of line as x-2 y-3 z-4 = = 3 4 5  This line is parallel to the vector, b = 3ˆi + 4 ˆj + 5kˆ



1 ) and (2, 1). 4

Graphs of parabola x 2 = 4 y and x = 4y − 2 are

Explanation : P(A ) + P(B) - P(A and B)=P(A ) P(A ) + P(B) - P(A and B)=P(A ) fi P(A )+P(B) - P(A « B) = P(A ) fi P(A )+P(B) - P(A « B) = P(A ) fi P (B ) - P (A « B ) = 0 fi P (B ) - P (A « B ) = 0 P (A « B ) = P (B ) fi P (A « B ) = P (B ) fi P (A « B ) P(A|B)= P(A « B) \ \ P(A|B)= P(B) P (B ) P (B ) = P (B ) = P (B ) P (B ) =1 =1

69



24. Option (3) is correct. Explanation :

The

repeated

selections

of

defective bulbs from a box are Bernoulli trials. Let X denotes the number of defective bulbs out of a sample of 5 bulbs. Probability of getting a defective bulb,

70

On differentiating with respect to x , we get

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH. 2 dy  y.( 0  2 x )  (1  x ).

 0 1 dx dy  2 xy  (1  x 2 )  1 dx dy 1  2 xy ...(ii)   dx 1  x2 Since, the given curve passes throu ugh x -axis, i.e., y0

10 100 1  10 q  1p

p



 1

1 10



9  10 Clearly, X has a binomial distribution with n = 5 and p  



1 . 10

1  2  0 1  dy   1   dy 2 0 dx  ( 2 , 0 )  11 22 5  dx  5 1  22 ( 2 , 0 )  Slope of the curve  Slope of 1 the curve  Slope of tangent to the curve=  1  Slope of tangent to the curve=  5  Equation of tangent to the curve 5passsing Equation  through (2, of 0) ta is ngent to the curve passsing through (2, 0) is 1 y  0   1 (x  2) y  0   5 (x  2) 25 x  y x 2 5  y 5 5 y  5x  2 5  5y  x  2   

P (X  x )  n C x q n  x p x 5x

 1  10 

x

P (none of the bulbs is defective) = P(X=0)  9  5 C0 .    10   9  1.    10   9    10 

5

5

5

25. Option (4) is correct.



 dy  expand sin   we get an infinite series in the  dx  dy increasing powers of . Therefore its degree is dx not defined.



 2 3 6  1 The distance of the plane r   ˆi  ˆj  kˆ  7 7 7  from the origin is 1.  [Since r  nˆ  d is the form of above equation, where d represents the distance of plane from the origin, i.e., d = 1]

Explanation: Given that, f ( x )  | 2 x  1|sin x

1 Thus, the given function is differentiable R –   . 2



27. Option (4) is correct. Explanation : Given that the equation of curve is y(1 + x2) = 2 – x ...(i) On differentiating with respect to x , we get dy  y.( 0  2 x )  (1  x 2 ).  0  1 dx dy  2 xy  (1  x 2 )  1 dx dy 1  2 xy ...(ii)   dx 1  x2 Since, the given curve passes throu ugh x -axis,

29. Option (1) is correct. Explanation:

26. Option (2) is correct.

The function sin x is differentiable. The function |2 x  1| is differentiable, except 2x  1  0 1  x 2

28. Option (2) is correct. Explanation: R on the set {1, 2, 3} is defined by R = {(1, 2)} That shows neither reflexive nor symmetric but transitive.

Explanation: The degree of above differential equation is not defined because when we



x2

So the curve passes through the point (2, 0).

 9  5 Cx    10 



0(1  x 2 )  2  x [By using Eq. (i)]



30. Option (2) is correct. Explanation:

1  5 1 x = tan  cos    3    2

5 Let cos-1 = q 3



cos q =

5 3 1 q 2

Þ

x = tan

Þ

q 2 x = q cos 2 sin

Solutions



q sin = 2

Þ

q cos = 2

12

5 3



2



1+



5 3

3 6 1 = 2 2 P (X = 2 ) = 6 1 = 3 P(X = 1) = 3 61 P (X = 5 ) = 1 = 6 2 2 P (X = 2 ) = 6 1 Therefore, the probability distribution is as = 3 follows. 1 X P(X = 5)1= 2 5 6 P(X) 1/2 1/3 1/6 Mean = E ( X ) P(X = 1) =

3- 5

=

3+ 5



3- 5

=

3+ 5



×

3- 5 3- 5

3- 5

=

( 3 )2 - ( 5 )2



3- 5 = 9- 5

=

3- 5 2

31. Option (2) is correct. Explanation: Let

I 

cos 2 x

=  pi xi

cos x  sin x 2

cos2 x  sin 2 x dx I   (cos x  sin x )2 (cos x  sin x )(cos x  sin x ) dx   (cos x  sin x )2 cos x  sin x   dx cos x  sin x Let cos x  sin x  t  (cos x  sin x ) dx  dt dt I    t  log t  C  log cos x  sin x  C

33. Option (2) is correct. Explanation : Let X be the random variable representing a number on the die. The total number of observations is 6. Therefore,

5 13

x =

R4 = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)} R5 = {(1, 2, 3) Û A × A = A2} \  Maximum number of equivalence relations on the set A = {1, 2, 3} = 5

5 3

1+

71

32. Option (4) is correct. Explanation: Given that, A = {1, 2, 3} Now, number of equivalence relations are as follows: R1 = {(1, 1), (2, 2), (3, 3)} R2 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)} R3 = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)}

1 1 1 = ×1  × 2 + × 5 2 3 6 1 2 5 = + + 2 3 6 3+ 4 + 5 = 6 12 = 6 =2

34. Option (3) is correct. Explanation: Given that, y = e–x (Acos x + Bsin x) On differentiating both sides w.r.t., x we get dy   e  x ( A cos x  B sin x ) dx  e  x (  A sin x  B cos x ) dy   y  e  x (  A sin x  B cos x ) dx Again, differentiating both sides w.r.t. x, we get

72

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.  dy d2y  dx dx 2  e  x (  A cos x  B sin x )



37. Option (1) is correct. Explanation:  r = ( 2i + 3 j − 4 k ) + λ ( 6i + 3 j − 4 k )

 e  x (  A sin x  B cos x )

  

C( i  2 j  6 k )

dy d2y  dy     y   y 2 dx dx  dx  d2y dy dy   y y dx dx dx 2 d2y dy  2  2y dx dx 2



B

d2y dy 2  2y  0 dx dx 2

A

AB : Projection of AC on AB  AC = ( −i + 2 j + 6 k ) − ( 2i + 3 j − 4 k )

35. Option (4) is correct. Explanation : When two dices are rolled, the number of outcomes is 36. The only even prime number is 2.

= −3i − j + 10 k   AB ⋅ AC  AB =  AB

Let E be the event of getting an even prime number on each die. ∴    E = {(2, 2)} 1 Þ P (E )  36



  ab  0

&

1   (a  c )  2

BC = AC 2 − AB 2 = 7

38. Option (2) is correct. Explanation: Given that,  1  x2  y  log    1  x2  





1 2



 p cos   cos    3

 

β = 60°

  a & b are unit vectors α −β

= 90° − 60° = 30°









dy d  d log 1  x 2    log 1  x 2    dx   dx dx  2 x 2x   1  x2 1  x2   2  2 x    1  x 2 1  x 2 

α = 90°

⇒



Differentiate with respect to x, we have

π cos α = cos   2

ac cos β =



y  log 1  x 2  log 1  x 2 .

ab cos α = 0

⇒

61

= 61  AC = 110

Explanation:

 b    a  (b  c )  2  b       ( a  c )b  ( a  b )c  2

−18 − 3 − 40

=

36. Option (1) is correct.

Compare

k j  4  3 2i 





4x 1  x4

39. Option (1) is correct. Explanation:

  a + b = 4i + 4 j

Solutions   a − b = 2i + 4 k     a vector ^ to both a + b & a − b     λ{( a + b ) × ( a − b )} i j k  c = 4 4 0 2 0 4  



(given)

| λ (16i − 16 j − 8 k )| = 12

⇒

λ 16 2 + 16 2 + 8 2 = 12 λ =

1 2

⇒

1  c = (16i − 16 j − 8 k ) 2

= 4( 2i − 2 j − k )



40. Option (1) is correct.

1 3

λ = −4 1 = 3µ 1 µ = 3

43. Option (2) is correct. Explanation: Given that, f (x ) =|sin x |

Explanation: For coplanar vectors

µ 1 1

1 µ 1

The functions |x| and sin x are continuous function for all real value of x. Thus, the function f ( x ) = |sin x |, is continuous function everywhere. Now, |x| is non-differentiable function at x = 0. f ( x ) = |sin x |, is non-differentiable Since

1 1 = 0 µ

µ ( µ 2 − 1) − 1( µ − 1) + 1(1 − µ ) = 0

⇒

(λ − 2) = ( 4λ − 2)

⇒  b→

From eqn. (i),







   β = ( 4 λ − 2 )a + 3 b ,    α = ( λ − 2 )a + b   α = µβ

Because α & β are collinear     ( λ − 2 )a + b = µ[( 4 λ − 2 )α + 3b ]  a→ λ − 2 = ( 4 λ − 2 )µ

= (16i − 16 j − 8 k )λ ...(i) ...(i)  | c | = 12 (given)

             

42. Option (1) is correct. Explanation:

is

function at sin x = 0

µ 3 − 3µ + 2 = 0

Thus, f is everywhere continuous but not differentiable at x = np, n Î Z.

2

µ ( µ − 1) + µ ( µ − 1) − 2( µ − 1) = 0 (By remainder theorem)



44. Option (3) is correct.

2

( µ + µ − 2 )( µ − 1) = 0

Explanation: Given that, dy cos x dy  y sin x  1 cos x dx  y sin x  1 dx dy  dy  y tan x  sec x dx  y tan x  sec x  dx

µ = 1, 1, − 2 Sum of distinct value of m is −2 + 1 = −1

41. Option (2) is correct.

Here, P = tan x and Q = sec x

Explanation: Corner points

73

Corresponding value of Z = 3x – 4y

(0, 0)

0

(5, 0)

15 ← Maximum

(6, 5)

–2

(6, 8)

–14

(4, 10)

–28

(0, 8) –32 ← Minimum Hence, the minimum of Z occurs at (0, 8) and its minimum value is (–32).

Pdx IF  e  tan xdx  e



 e log sec x IF  sec x

45. Option (4) is correct. Explanation : Since, direction cosines of a line are k, k and k. ∴ l = k, m = k and n = k

74

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

k2  k2  k2  1 1 k2  3

  





l 2  m2  n2 = 1

We know that,

k  

Explanation: Since,  5 −4   x   40   5 −8   y  =  −80        5 −4  x  Let kA =   , x = y 5 8 −    

1 3

46. Option (2) is correct. Explanation: We know that, area of a triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is given by ∆=

∴

⇒ ∴



50. Option (4) is correct.

∆=

x1 1 x2 2 x3

y1 y2 y3

∴

X = A −1B |A| = 5( −8 ) − ( −4 ) × 5 = −40 + 20 = −20

1 1 1

T

−3 0 1 1 3 0 1 2 0 k 1

18 =3 6

47. Option (1) is correct. Explanation: Maximum of Z occurs at (5, 0).



48. Option (1) is correct. Explanation: According to question, ( x  8 )( y  10 )  xy ( x  8 )( y  10 )  xy  xy  10 x  8 y  80  xy  xy  10 x  8 y  80  xy  5x  4 y  40  5x  4 y  40 and ( x  16 )( y  100 )  xy and ( x  16 )( y  100 )  xy  xy  10 x  16 y  160  xy  xy  10 x  16 y  160  xy  5x  8 y  80  5x  8 y  80



1 2   5  40  5  ( 80 )      1  40  1  ( 80 )  4  4 16  16     10  20 

...(i) ...(i) ...(ii) ...(ii)

49. Option (3) is correct. Explanation: Given equations are; 5x  4 y  40 5x  8 y  80  5 4   x   40         5 8   y   80 

...(iii)

 8 5  adj( A )    5 T 4  8 5  adj( A )   8 4   5    4  5 5   8 4   Adj( A ) A 1   5 5  |A| Adj( A ) A 1  1  8 4   |A|  20  5 5  1  8 4    2 1   20  5 5    25 51   1 1      45 45  1 1 X  A 1B   4 4   2 1  X  A 1B   40    25 51    1 1  80   5 5   40    4 4     1 1   80   4 4 

[Expanding along R1 ] 1 9 = [ −3( − k ) − 0 + 1(3k )] 2 18 = 3k + 3k = 6 k k=

 40  B =    −80  AX = B



 32      30   32  x   y    30      x = 32 y = 30

Hence, 32 children were given some money by Seema.

Solutions

Section - B2



75

16. Option (3) is correct.

(APPLIED MATHEMATICS)

21. Option (3) is correct.

Explanation: If a and b are integers and n > 0, then we write a ≡ b(mod n) to mean n|(a – b).

17. Option (2) is correct. Explanation: (9 + 23) mod 12 or, 32 mod 12 On dividing 32 by 12, we get remainder = 8 Hence, (9 + 23) mod 12 = 8

18. Option (4) is correct.

Explanation:

(I + A)

3

3 2             = I + A + 3 A + 3 A − 7 A 2             = I + A ⋅ A + 3 A + 3 A − 7 A

                   

  = I + A ⋅ A − A   = I + A 2 − A

 A 2 = A 

=I+A−A             = I 3 ∴(I + A) − 7A = I

 

Explanation: As,   B > A ⇒ A < B ⇒ A – B < 0 A + B > 0 and AB > 0 If A = 1, B = 4 then, AB < A + B If A = 2, B = 4 then, AB > A + B Thus, we can't say which one of A + B and AB has higher value.

− 7 A = I 3 + A 3 + 3I 2 A + 3 A 2 I − 7 A



22. Option (2) is correct. Explanation: It is given that 5  0 y − 2  3x + 7  y + 1 2 − 3x  = 8 4     Equating the corresponding elements, we get:

19. Option (3) is correct.

3x + 7 = 0

Explanation: Since first and second varieties are mixed in equal proportions.  126 + 135  So, their average price = `   2  

⇒

= ` 130.50

⇒

So, the mixture is formed by mixing two varieties, one at ` 130.50 per kg and the other at say ` x per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find x. By the rule of alligation, we have:

7 3 5= y−2 y=7

⇒

x=−

y +1 = 8 y=7 2 − 3x = 4

2 3 We find that on comparing the corresponding elements of the two matrices, we get two different values of x, which is not possible. Hence, it is not possible to find the values of x and y for which the given matrices are equal. ⇒

x=−

23. Option (1) is correct.

\

x − 153 = 1 22.50

⇒

x – 153 = 22.50

⇒

x = ` 175.50

20. Option (3) is correct. Explanation: It is known that a given matrix is said to be a square matrix if the number of rows is equal to the number of columns. Therefore, A = [ aif ]m×n is a square matrix, if m = n.

Explanation: Matrices P and Y are of the orders p × k and 3 × k, respectively. Therefore, matrix PY will be defined if k = 3. Consequently, PY will be of the order p × k. Matrices W and Y are of the orders n × 3 and 3 × k respectively. Since the number of columns in W is equal to the number of rows in Y, matrix WY is well-defined and is of the order n × k. Matrices PY and WY can be added only when their orders are the same. However, PY is of the order p × k and WY is of the order n × k. Therefore, we must have p = n. Thus, k = 3 and p = n are the restrictions on n, k, and p so that PY + WY will be defined.

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

76

24. Option (1) is correct. Explanation: Given that, A and B are symmetric matrices. ⇒ A = A' and B = B' Now, (AB – BA)' = (AB)' – (BA)' ...(i) ⇒ (AB – BA)' = B'A' – A'B' [By reversal law] ⇒ (AB – BA)' = BA – AB [From Eq. (i)] ⇒ (AB – BA)' = – (AB – BA)

25. Option (3) is correct. Explanation: We have, f (x) = 2x3 – 3x2 – 12x + 4 f ′(x) = 6x2 – 6x – 12 Now, f ′(x) = 0 ⇒ 6(x2 – x – 2) = 0 ⇒ 6(x + 1) (x – 2) = 0 ⇒ x = – 1 and x = +2 On number line for f ′(x), we get + – + –1 2 ence, x=−1 is point of local maxima and H x = 2 is point of local minima. So, f(x) has one maxima and one minima.

26. Option (2) is correct. Explanation: There R(x) = 3x2 + 26x + 15 dR = 6x + 26 dx



 dR  = 6(15) + 26  dx   at x = 15

⇒

= ` 116

27. Option (3) is correct. Explanation: p(x) = 41 – 72x – 18x2 \ p'(x) = – 72 – 36x 72 ⇒ x = − = −2 36 p''(x) = – 36 p "( −2) = −36 < 0 Also, By second derivative test, x = –2 is the point of local maxima of p. \ Maximum profit = p(–2) = 41 – 72 (–2) – 18 (–2)2 = 41 + 144 – 72 =113 Hence, the maximum profit that the company can make is 113 units.

28. Option (2) is correct. Explanation: If \



y = x3 log x d dy = (x3 log x) dx dx

dy = x2 (1 + 3 log x) dx

d2y

d d = x 2 (1 + 3 log x ) + (1 + 3 log x ) x 2 2 dx dx dx d2y = x (5 + 6 log x) 2 dx d3y d d = x ( 5 + 6 log x ) + ( 5 + 6 log x ) x 3 dx dx dx d3y

= 11 + 6 log x 3 dx d4y 6 = 4 dx x

29. Option (3) is correct. Explanation: Let, f ( x ) = x 2 − 8x + 17 On differentiating with respect to x, we get f´(x) = 2x – 8 So, f´(x) = 0 ⇒ 2x – 8 = 0 ⇒ 2x = 8 \ x = 4 Now, Again on differentiating w.r.t. x, we get f ’’ ( x ) = 2 > 0, ∀x So, x = 4 is the point of local minima. Minimum value of f(x) at x = 4 f(4) = 4 × 4 − 8 × 4 + 17 = 1

30. Option (2) is correct. Explanation: Expected number of votes = np 70 × 120000 = 100 = 84000

31. Option (1) is correct. Explanation:



dy dy / dt = dx dx / dt

2a = 2 at 1 = t d2y

2

dx



= −

= –

1 t

2

×

dt dx

1 2 at 3

32. Option (1) is correct. Explanation: The expectation of a random variable X is given by the summation (integral) of X times the function in its interval. If it is a continuous random variable, then summation is used and if it is discrete random variable, then integral is used.

Solutions 33. Option (2) is correct. Explanation: Let f(x) be the pdf of the random variable X. Now, E(q) = ∫af(x) = a ∫ f(x) = a.1 = a

34. Option (2) is correct.

2 4 3 + + 6 6 6 9 3 = = = 1.5 6 2

=

Explanation: Since the normal curve is symmetric about its mean, its skewness is zero.

37. Option (2) is correct. Explanation: The point estimation of population standard deviation is sample deviation. ( xi − x ) n −1 2

2

2

(1 − 4 ) + ( 3 − 4 ) + ( 5 − 4 ) + ( 7 − 4 ) 4 −1

2

Σx 16    x = n = 4 = 4    9+1+1+ 9 = 3

20 = 2.52 3

39. Option (3) is correct.

40. Option (2) is correct. Explanation: When trend is linear, then only we use moving average method for measurement.

42. Option (3) is correct. Explanation: The given annuity is a perpetuity. Cash flow Interest rate

case flow = ` 1000 interest rate =

(8, 0)

160

 5 15  2, 4   

125

7 9 2, 4  

115 (minimum)

(0, 10)

200

46. Option (1) is correct. 47. Option (2) is correct. Explanation: Formula to calculate monthly installment is: (1 + i )n (1 + i )n − 1

× (P × i)

Explanation:

Given,

41. Option (1) is correct.



Z = 20x1 + 20x2

48. Option (3) is correct.

Explanation: All the component are associated with time.

present value of perpetuity =

Corner Points

Installment Amount =

38. Option (3) is correct.

Here,

44. Option (3) is correct. 45. Option (3) is correct. Explanation:

36. Option (3) is correct.

2

500 500 = 8 0.08 100

= ` 6250

35. Option (3) is correct.

=

1000 0.04

present value =

Explanation: PV of perpetuity Annual Payment/Cash flow = Interest rate/yield

= 0 +

=

So,

= ` 25,000

 1  2  2  1 = 0   + 1  + 2   + 3    6  6  6  6

S =

4 = 0.04 100

43. Option (2) is correct.

Explanation: E(X)



=

77

8/2 100

  annual rate       12  i =  100      

  12     12    =   100      1 = 100 = 0.01 n = 10 × 12 = 120 P = ` 5,00,000

78

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH. Installment Amount = Installment Amount =

(1 + i )n (1 + i )n − 1

49. Option (4) is correct. × (P × i)

(1 + 0.01)120 (1 + 0.01)120 − 1



× (5,00,000 × 0.01) 3.300 = ×5,000 3.300 − 1 16 , 500 = 2.300

= ` 7173.91 ~ ` 7174 So, EMI that Rohan has to pay is ` 7147

Explanation: Total payment made by Rohan to the bank in 10 years = (EMI × Total tenure in months) = ` (7174 × 120) = ` 8,60,880

50. Option (1) is correct. Explanation: The total interest amount payable will be (Total payment – loan amount) = ` 3,60,88.

qqq

Solutions

2

SOLUTIONS OF Question Paper Section - A



1. Option (1) is correct.



Explanation : Given,





È2˘ Y = Í 3 ˙ Í ˙ Í4 ˙ Î ˚

È2˘ È3˘ Í ˙ Í ˙ AB + XY = ÈÎ2 -3 4 ˘˚ Í 2 ˙ + ÈÎ1 2 3˘˚ Í 3 ˙ ÍÎ 4 ˙˚ ÍÎ 2 ˙˚ = [6 – 6 + 8] + [2 + 6 + 12] = [8] + [20] = [28]

Let

I = I =

b

Úa

b

Úa

xf ( x )dx ( a + b - x ) f ( a + b - x ) dx



b

Úa ( a + b - x ) f ( x ) dx

fi

I =

fi

I = ( a + b )Ú f ( x ) dx - I

b

[Ussing Eq . ( i )]

a b

fi

I + I = ( a + b )Ú f ( x ) dx

fi

2 I = ( a + b )Ú f ( x ) dx

fi

a b a

Ê a + bˆ b I = Á f ( x ) dx Ë 2 ˜¯ Úa

5. Option (4) is correct. b 2 − ab b − c

bc − ac

b( b − a ) b − c

c( b − a )

2 bExplanation: − ab b − ab c −We bca 2−have aca − bb( b b−2 a−) abb −= ca( bc(−b a−)a ) a − b b( b − a ) 2 − bc bbc2 − − ab − bc bc((bb − bab − ac = ba(( bb − a) − ab a 2 ab − − aa−)) a 2ba − bc − ac c − a ab c( b − a−)a ) c − a a( b − a ) 2 2 2 = a( b − a ) − − − − ( − ab a a b b ab a b b b bc − ac c − a ab − a c( b − a ) c − a a( b − aa )) bc − ac c − a ab − a 2 c( b − a ) c − a a( b − ab) b − c c b b−c c 2 = −(bcb −bca ) a a − b b = ( b − a )2 ba ab − 2 = − ba ba c c − a a ( b − a ) ac ac − taking ( b − a ) commo c − a( b[On a− a ) common [On ctaking

2. Option (2) is correct.

taking ( bC− aeach] )3common from C1 and C3 each] [On taking (b – a) common[On from C each] 1 and from CCand 1





3

fromb − C1c and b −Cc3 each] cb − c 2 b−c (abb −−− bca )2bca − b = ( b − a ) a − b= = ( b − a )2 ac − − ba ac − − ba bac − a

3. Option (4) is correct. Explanation : Here, Here, = P( A ) 0.4, = P( B) 0.8 and= P( A | B) 0.6 = P( A ) 0.4, = P( B) 0.8 = P)( A | B) 0.6 P(Band ∩A  P(B|A ) = P(B ∩ A )  P(B|A ) = P(A ) ⇒ P( B ∩ A ) = P( P B(|AA)).P( A ) ⇒ P ( B ∩ A ) = P ( B | A ).P( A ) = 0.6×0.4 = 0.24 = 0.6×0.4 = 0.24  P(AÈB) = P(A ) + P(B) − P(A ∩ B)  P(AÈB) == 0.4 P(A+) + ) − P(A ∩ B) 0.8P(−B0.24 == 1.2 0.4 − + 0.24 0.8 −= 0.24 0.96 = 1.2 − 0.24 =0.96

...(i)

È b f ( x ) dx = b f ( a + b - x ) dx ˘ Úa ÍÎ Úa ˙˚

Explanation: Given that, aRb if a ³ b Þ aRa Þ a ³ a which is true Let aRb, a ³ b, then b ³ a which is not true as R is not symmetric. But aRb and bRc Þ a ³ b and b ³ c Þ a ³ c Hence, R is transitive.

4. Option (4) is correct. Explanation:

A = [2 -3 4 ],

È3˘ B = Í 2 ˙ , Í ˙ Í2˙ Î ˚ X = [1 2 3],





79

c−a a [ C1 → C1c−−C[a 3 ] C1 → C1 − C3 ] [ C → C − C ] 1 3 = 01 = 0 = 0 =0

[Since, two columns C1 and C2 are identical, so the value of determinant is zero.]

6. Option (4) is correct. Explanation : The equation of the curve is given by y = x 3 - 12 x + 18 On differentiating with respect to x, we get

b−c c a−b b c−a a

r1 r  0, 1

 Also,

80

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH. P( X  r )  n C r p r q n  r P( X  r )  P( r  0 )  P( r  1) 0

dy  3x 2 - 12 dx



So, the slope of line parallel to the x-axis,

5

1  9  9     5. .    10  10  10 

dy 0 dx 3x 2  12  0 12 x2  3  4 x  2 x  2,

   For

5

1 9   9       10  2  10 

= 2 sec −1 sec = 2×

⇒

4 − x2 , 4x − x 3



4π + π 6

=

5π 6

)

x 4 − x2 = 0

⇒ x ( 2 + x )( 2 − x ) = 0

= 2ˆi − ˆj − 2kˆ BC = ( 4 − 2)ˆi + (5 − 3)ˆj + ( 0 + 1)kˆ

x = 0, − 2, 2

⇒

hus, the given function is discontinuous at T exactly three points.

= 2ˆi + 2ˆj + kˆ CD = ( 2 − 4)ˆi + ( 6 − 5)ˆj + ( 2 − 0)kˆ

8. Option (1) is correct.

= −2ˆi + ˆj + 2kˆ DA = ( 0 − 2)ˆi + ( 4 − 6)ˆj + (1 − 2)kˆ

Explanation: Given that A = {1, 2, 3} and R = (1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}.  (1, 1), (2, 2), (3, 3) Î R Hence, R is reflexive. (1, 2) Î R but (2, 1) Ï R Hence, R is not symmetric. (1, 2) Î R and (2, 3) Î R Þ (1, 3) Î R Hence, R is transitive.

= −2ˆi − 2ˆj − kˆ Thus quadrilateral formed is parallelogram. ∴ Area of quadrilateral ABCD

= AB × BC = AB × BC ˆ ˆ ˆ ˆi i ˆj j kˆ k = 2 −1 −2 = 2 −1 −2 2 2 1 2 2 1 = 3ˆi − 6ˆj + 6kˆ     = 3ˆi − 6ˆj + 6kˆ = 9 + 36 + 36 = 9 + 36 + 36 = 9 sq. units = 9 sq. units

9. Option (4) is correct. Explanation : Here, 10 1 9 n = 5, p = = and q = 100 10 10  Also,

r1 r  0, 1

P( X  r )  n C r p r q n  r P( X  r )  P( r  0 )  P( r  1) 0

5

1

 1  9  1  9  5 C0      5C1      10   10   10   10  5

11. Option (1) is correct. Explanation : We have, A(0, 4, 1), B(2, 3, –1), C(4, 5, 0) and D(2, 6, 2) AB = ( 2 − 0)ˆi + (3 − 4)ˆj + ( −1 − 1)kˆ

4x − x 3 = 0

(

π π + sin −1 sin 3 6

π π + 3 6

=

Explanation: Given that,

⇒

4

 sec −1 (sec x ) = x and sin −1 (sin x ) = x   

7. Option (3) is correct.

then it is discontinuous if

4

10. Option (2) is correct.

So, the points are (2, 2) and (−2, 34).

                  f ( x ) =

4

1 2 sec −1 2 + sin −1   2

y  ( 2 )3  12  ( 2 )  18  34

1

Explanation: We have,

y  2 3  12  2  18  2 x  2 ,

and for

5

 1  9  1  9  5 C0      5C1      10   10   10   10 

1  9  9     5. .    10  10  10 

4

4

12. Option (2) is correct. Explanation: The second-order equation is y'y"+y =sin x.

differential

2I 

f ( x ) = cot x =

 

cos x sin x



It is discontinuous at sin x  0  x  n , n  Z



Explanation :



1  2x  1  1 tan 1  2 x  1  dx 1  0 tan    x  x 22  dx 0  11  xx 



1  x  (1  x )  1 tan 1  x  (1  x )  dx 1 0 tan   x(1  x )  dx  11  0 x (1  x )  1 1 [tan 1 x  tan 1 (1  x )] dx 1 1 0 [tan x  tan (1  x )] dx 0 1 1 [tan 1 (1  x )  tan n 11 (1  1  1 n (1  1  0 [tan (1  x )  tan 0 1  1 1 [tan (1  x )  tan 1 ( x )] dx 1 1 0 [tan (1  x )  tan ( x )] dx 0 1 1 [tan 1 (1  x )  tan 1 ( x )] dx 1 1 0 [tan (1  x )  tan ( x )] dx 0

 

I  I 



 

I  I 

 

I  I 

 

I  I 

 

 

I  I 

 

2I 

...(i) ...(i) x )] dx x )] dx

2x

Since, tangent to the curve y = e at the point (0, 1) meets x-axis, i.e., .  0  2x  1 1  x   2  1  So, the required point is   , 0  .  2 

...(ii) ...(ii)

1

0 (tan 1 x  tan 1 (1  x )



2I  0 I  0

Section - B1

16. Option (2) is correct. Explanation: Let R be the relation in the set {1, 2, 3, 4} is given by: R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)} (1) (1, 1), (2, 2), (3, 3), (4, 4) Î R Therefore, R is reflexive. (2) (1, 2) Î R but (2,1) Ï R. Therefore, R is not symmetric. (3) If (1, 3) Î R and (3, 2) Ï R then (1, 2) Î R. Therefore, R is transitive.



 dy   2.e 2.0  dx  ( 0 , 1)

 Slope of tangent to the curve.  Equation of tangent is y  1  2( x  0 )  y  2x  1

 tan 1 (1  x )  tan 1 x ) dx  

dy  e 2 x .2 dx  2.e 2 x

 2

Adding equations (i) and (ii), we obtain



15. Option (2) is correct.

14. Option (2) is correct. I  I 

2I  0 I  0

Explanation : The equation of the curve is given by y = e2x Since, it passes through the point (0, 1).

Thus, the given function is discontinuous at {x  n : n  Z}.

Let Let

81

 tan 1 (1  x )  tan 1 x ) dx

13. Option (1) is correct. Explanation: Given that,



1

1 (1  x ) 0 (tan 1 x  tanSolutions

17. Option (1) is correct. Explanation: 3



18. Option (4) is correct. Explanation : Let E1 = Event that both A and B solve the problem 

Let E2 = Event that both A and B got incorrect solution of the problem 

d 2 y  dy      6y5  0 dx 2  dx  We know that, the degree of a differential equation is exponent of highest order derivative. \ Degree = 1

1 1  3 4 1  12

P( E1 ) 

2 3  3 4 1  2

P( E2 ) 

Let E = Event that they got same answer Here,

82

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH. P( E / E1 ) P( E / E1 ) P( E / E2 ) P( E / E2 ) P( E1 / E) P( E1 / E)

         

     

1, 11, 1 20 20( E1  E) P P( E P1( E) E) P( E) P( E ).P( E / E ) 1 1 P(/EE1 ).)P( EP(/EE1))P( E / E ) P( E1 ).P( E 1 2 1 P( E1 ).1P( E / E1 )  P( E2 )P( E / E1 ) 1 1 12 1 1 12 1 1 1  1 1 1 12  1  2  20 121 2 20 1 12 1012 3 3 10120 120 120  13 12120 12  13 10 10 13 13

19. Option (3) is correct.

  

So, A−1 exists if and only if l≠

Again, differentiating both sides w.r.t. x, we get d 22 y d y2 dx 2 dx d 22 y  d y2  dx dx 2 d 22 y  d y2  dx 2 dx d 22 y  d y2  m 22 y  dx 2  m y dx

 m 22 ae mx  bm 2 e  mx  m ae mx  bm 2 e  mx  m 22 ( ae mn  be  mn )  m ( ae mn  be  mn )  m 22 y  m y  0  0

20. Option (4) is correct. Explanation: Given that, 2 λ −3 A= 0 2 5 1 1 3 Expanding along R1, A  2 6  5    5  3  2   2  5  6 We know that A–1 exists, if A is non-singular matrix, i.e., |A|≠0

21. Option (1) is correct. y = sin x + y y 2 = sin x + y

⇒

.

Differentiate with respect to x, we have

dy dy = cos x + dx dx dy ⇒ ( 2 y − 1) = cos x dx dy cos x ⇒ = dx 2 y − 1 ⇒



2y

22. Option (1) is correct. Explanation: Assertion (A) and Reason (R) both are correct, Reason (R) is the correct explanation of Assertion (A).

y = ae mx + be -mx dy = mae mx - bme -mx dx

−8 . 5

Explanation: Given that,

Explanation: Given that,

On differentiating both sides w.r.t. x, we get

2  5  6  0 5  8 8 λ 5



23. Option (4) is correct. Explanation: Assertion (A) is wrong. Reason (R) is the correct solution of Assertion (A).



24. Option (2) is correct. Explanation : Here, 30 P( Ph ) = 100 3 = , 10 25 P( M ) = 100 1 = 4 10 100 1  10

and P( M  Ph ) 



P( Ph  M )  Ph  P    M P( M ) 1  10 1 4 2  5

Solutions

25. Option (4) is correct.

Now, if we add any two pairs (3, 2) and (2, 3) (or both) to relation R, then relation R will become transitive. Hence, the total number of desired relations is one.

Explanation : We have,  

xy  yz  0 x( y  z )  0 x  0 and y  z  0



Above are equations of planes. Normal to the plane x = 0 is ˆi . And normal to the plane y + z = 0 is ˆj + kˆ. Now, ˆi ⋅ ( ˆj + kˆ) = 0

29. Option (2) is correct. Explanation : Given that, f ( x ) = 4 sin 3 x - 6 sin 2 x + 12 sin x + 100 On differentiating with respect to x, we get

So, planes are perpendicular.

f ( x )  12 sin 2 x.cos x  12 sin x.cos x  12 cos x

26. Option (3) is correct.



 f ( x )  12 cos x[sin 2 x  1(1  sin x )]  1  sin x  0 and sin 2 x  0  sin 2 x  1  sin x  0

E2 = Event that the sum of numbers on the dice is 3. ∴   E1 = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2,3), (3, 1), (3, 2), (4,1)} ⇒ n( E1 ) = 10

    Hence, f ( x )  0 , when cos x  0 , i.e., x   , .  2 2 

    So, f ( x ) is increasing when x   ,  2 2 

and E2={(1, 2), (2, 1)} ⇒ n( E2 ) = 2

  3  and f ( x )  0 , when cos x  0 , i.e., x   ,  .  2 2

2  Required probability  10 1  5

  3  Hence, f ( x ) is decreasing when x   ,   2 2

     3  Since  ,     ,  2   2 2 

27. Option (2) is correct. Explanation: For coplanar

α 2 α ⇒

1 3 1 −α = 0 −2 3

  Hence, f (x) is decreasing in  ,  2

Let

2

−3α − 18 = 0

∫ sin

2

x.dx

1 = ∫ (1 − cos 2 x )dx 2 x sin 2 x +c = − 2 4

28. Option (1) is correct. Explanation: The given set is A = {1, 2, 3}. The smallest relation containing (1, 2) and (1, 3), which is reflexive and symmetric, but not transitive is given by: R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (3, 1)} This is because relation R is reflexive as (1, 1), (2, 2), (3, 3) Î R. Relation R is symmetric since (1, 2), (2, 1) Î R and (1, 3), (3, 1) Î R. But relation R is not transitive as (3, 1), (1, 2) Î R, but (3, 2) Ï R.

I =

1 − cos 2 x dx = ∫ 2

⇒ No real value of α ⇒ Empty set

30. Option (4) is correct. Explanation :

α ( 3 − 2α ) − 1( 6 + α 2 ) + 3( −4 − α ) = 0

⇒

 12[sin 2 x.cos x  sin x. cos x  cos x]  12 cos x[sin 2 x  sin x  1]

Explanation : Let, E1 = Event that the sum of numbers on the dice was less than 6 and



83



31. Option (2) is correct. Explanation : It is given that 3 x + 7 5  0 y − 2   =  4   y + 1 2 − 3 x  8 Equating the corresponding elements, we get

84

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH. 3x  7  0   

And,

7 3 5  y2 y7 y 1  8 y7 2  3x  4 x

Thus,

32. Option (4) is correct. Explanation : Distance between two parallel planes, Ax + By + Cz = d1 and Ax+ By + Cz = d2 is d1 − d2 A + B2 + C 2

2



2x + 3y +4z = 4 Comparing with Ax + By + Cz = d1 A = 2, B = 3, C = 4, d1 = 4 And now, 4x + 6y + 8z = 12 2(2x + 3y + 4z) = 12 Dividing by 2 2x + 3y + 4z = 6 Comparing with Ax + By + Cz = d2 A = 2, B = 3, C = 4, d2 = 6 So, Distance between the two planes = = =

2

   a+b +c = 0    Squaring ( a + b + c )2 = 0    a 2 + b 2 + c 2 + 2( a.b + b .c + c .a ) = 0    ⇒ 1 + 1 + 1 + 2 ( a.b + b .c + c .a ) = 0 ⇒

4 -6

2 + 32 + 4 2 -2

4 + 9 + 16 2 29

)

(

)

2

du 4x =− 4 dx 1 − 4x + 4x 2 − 1 4x du ⇒ =− dx −4x 4 + 4x 2 2 du ⇒ =− dx 1 − x2

3 2

      = a×b + a×b −0 −0 + a×b   = 3( a × b )   a ×a = 0     b ×b = 0       a × b = −b × a 

u = cos−1 2x 2 − 1

du 4x =− dx 1 − 2x 2 − 1

a⋅b + b⋅c + c⋅a = −

   = (| a | 1= ,| b | 1,| c | = 1, ( given)        d = a×b + b ×c + c ×a   b = − a − c     c = −a − b            = a × b + b × ( −a − b ) + ( −a − b ) × a           = a×b −b ×a −b ×b − a×a −b ×a

Explanation: Let

⇒

35. Option (3) is correct. Explanation:

33. Option (1) is correct.

(

34. Option (3) is correct.  1   1  1 cos 1  1   2 sin 1  1   4 tan 1  1  cos 11  12   2 sin 11  12   4 tan 11  13  cos  2   2 sin  2   4 tan  3   2   2     3     cos 1  cos    2 sin 1  sin    4 tan 1  tan   1   1  1     3   2 sin  sin 6   4 tan  tan  6   cos  cos   cos1  cos 3   2 sin 1  sin 6   4 tan 1  tan 6      3   6 6   2  4  3 2  6 4  6   3 2 6 4 6 2  2  4   23  2  6 4  6  2   26  4   8 6  8 6  86  6 4  46  43  3 3

We find that on comparing the corresponding elements of the two matrices, we get two different values of x, which is not possible. Hence, it is not possible to find the values of x and y for which the given matrices are equal.

du =2 dv

Explanation :

2 x 3



v = cos−1 x dv 1 =− dx 1 − x2

⇒



36. Option (1) is correct. Explanation : Required probability =

4 4 1 1 × = × 52 52 13 13



dx I dx sin( )sin( x  b ) x  a I sin( )sin( xsin( x  a  b )b  a )Solutions 85 1  dx  1 sin( ) b  a  sin(b  a )  sin( x  a )sin( x  b ) dx sin(b1  a ) sin( x xa )sin( sin( a  x xb )b )    1 sin( x a  x  b ) dx  sin(b  a )  sin( x  a )sin( x  b ) dx sin(b1  a ) sin( sin{(xxa )sin( a )  (xx  bb))}   a )  ( x  b )} dx 1 sin{( x   sin(b  a )  sin( x  a )sin( x  b ) dx sin(xxaa)cos( )sin(xx  bb))  sin(b  a ) sin( sin( x a )cos( b)  x  b)  cos( x x a)sin( 1 dx   sin(cxos(ax)sin(  a )sin( 1 x  bx)  b ) dx  sin(b  a )  sin(b1  a ) sin( x  a )sin( x  b )  [cot( x  b )  cot( x  a )]dx  1  sin(b  a )  [cot( x  b )  cot( x  a )]dx sin(b1  a )  [log sin( x  b )  log sin( x  a ) ]  C 1  sin(b  a ) [log sin( x  b )  log sin( x  a ) ]  C sin(b  a ) sin( x  b )  cosec( b  a ) log sin( x  b )  C  cosec( b  a ) log sin( x  a )  C sin( x  a )

37. Option (2) is correct. Explanation: Given that, xx 22 66 22   18 18 18 xx 18 66      



xx 22 2 xx 2

 36  36  36  36

    xx  

36  36 36  36 00   66

38. Option (3) is correct. Explanation:

  b×c = b ×a Cross both the sides by a    a × (b × c ) = a × (b × a )           ( a.c )b − ( a.b )c = ( a.a )b − ( a.b )a     0 − ( a b ) c = a 2 b − ( 4 )a (∵ a ⋅ b = 4 )



41. Option (1) is correct. Explanation :

   f ( x ) = a ⋅ (b × c )

a b = 1 ⋅ 1 + ( −2 )( −1) + 1 ⋅ 1

x −2 3 = −2 x −1 7 −2 x

[(( a . c ) = 0 , given )] = 4

  Substitute the values of a & b vectors which are given  −4 c = 6(i − j + k ) − 4(i − 2 j + k )

= x( x 2 − 2 ) + 2( −2 x + 7 ) + 3( 4 − 7 x ) = x 3 − 2 x − 4 x + 14 + 12 − 21x

1     ( −i − j − k ) c = 2

= x 3 − 27 x + 26

 1 b .c = − 2

a( x − 1) + b( y − 2 ) + c( z − 1) = 0 For maxima & minima, a( 2 − 1) + b(1 − 2 ) + c( 2 − 1) = 0

39. Option (1) is correct.

a−b+c = 0

Explanation : cot 11 ((   33 )) cot cot 1 (  cot3 ) cot cot  cot   cot cot 

   3  3  3    cot cot    cot 66  6   cot      cot      6 6 cot     55  6   cot   cot   cot  cot 5  cot   cot 66 55  6( 0 ,  )       5   (0, ) 6 6     (0, ) 5 6  11 (  3 ) is 5   Principal value of cot  Principal value of cot (  3 ) is 5 1  Principal value of cot (  3 ) is 66 6 Let Let Let      



= 2x

           

40. Option (3) is correct. Explanation : Let, dx I sin( x  a )sin( x  b ) 1 sin(b  a )  dx sin(b  a )  sin( x  a )sin( x  b ) 1 sin( x  a  x  b )  dx sin(b  a )  sin( x  a )sin( x  b ) 1 sin{( x  a )  ( x  b )}  dx sin(b  a )  sin( x  a )sin( x  b )

...(i)

= 3y, z 1

x y z −1 = = 3 2 0 Max for, ⇒

3a + 2b + 0 c = 0

...(ii)

a−b+c = 0 = ( −3i − 2 j + 3k )( −2i − 3 j − k ) + ( −2i − 3 j − k ) (7i − 2 j − 3k ) + (7i − 2 j − 3k )( −3i − 2 j + 3k )

a = −2λ , b = +3λ , c = 5λ = 25 − 47 = −22

42. Option (2) is correct. Explanation: Given that, x = t2 and y = t3 dy dx Then, = 3t 2 = 2t and dt dt Thus,

dy 3t 2 3t = = dx 2t 2

86

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH. ⇒

=



d 2 y d  dy  d  3t  =  =   dx 2 dx  dx  dx  2 



3  dt  3  1  3  =  = 2  dx  2  2t  4t

43. Option (2) is correct. Explanation : Angle between two planes A1x + B1y + C1z = d1 and A2x + B2y + C2z = d2 is given by cos q =



+ B12 + C12 A22 + B22 + C22

Given that plane, 2x – 1y + 4z = 5 Comparing with A1x + B1y +C1z = d1 A1 = 2, B1 = –1, C1 = 4, d1 = 5 5x – 2.5y + 10z = 16 Multiplying by 2 on both sides, 10x – 5y + 20z = 12 Comparing with A2x + B2y + C2z = d2 A2 = 10, B2 = –5, C2 = 20, d2 = 12

So, cos q = = = = =

cos   cos



 

We shall now calculate the value of f(x) at these points and at the end points of the interval [0, 9], i.e., at x = 4 and x = 8 and at x = 0 and at x = 9. f ( 4 ) = 4 3 − 18 × 4 2 + 96 × 4 = 64 − 288 + 384 = 160

( 2 × 10) + ( −1 × −5) + ( 4 × 20)



20 + 5 + 80 105 21 525

and f ( 0 ) = 0 3 − 18 × 0 2 + 96 × 0 =0 Thus, we conclude that absolute minimum value of f(x) in [0, 9] is 0 occurring at x = 0.

105 21 × 25 × 21 105

46. Option (4) is correct. Explanation: Since, A and B are invertible matrices, so, we can say that

So, cos θ = 1 ∴ θ = 0° Since angle between the planes is 0°. Therefore, the planes are parallel.

Also, 

44. Option (2) is correct. Explanation: cos α − sin α Given that, A =    sin α cos α 

Equating corresponding entries, we have

f ( 8 ) = 8 3 − 18 × 8 2 + 96 × 8 = 128 f ( 9 ) = 9 3 − 18 × 9 2 + 96 × 9 = 729 − 1458 + 864 = 135

4 + 1 + 16 100 + 25 + 400

Also A + A’ = I cos α − sin α  cos α sin α  1 0 ⇒   = +  sin α cos α  − sin α cos α 0 1 2 cos α 0  1 0  ⇒   = 2 cos α 0 1  0

 3

 3

45. Option (2) is correct.

22 + ( −1)2 + 42 102 + ( −5)2 + 202

21 × 5 21 105 = 21 × 5 =1





Explanation : Given that, the smallest value of polynomial is  f(x) = x 3 − 18x 2 + 96x On differentiating with respect to x, we get f '( x ) = 3x 2 − 36 x + 96    So, f´(x) = 0 ⇒ 3x2 – 36x + 96 = 0 2 ⇒ 3(x – 12x + 32) = 0 ⇒ (x – 8)(x – 4) = 0 ⇒ x = 8, 4 ∈ [0, 9]

A1 A2 + B1B2 + C1C2 A12

2 cos   1 1 cos   2

( AB)1  B 1 A 1 1 A 1  ( adj A ) A

...(i)

adj A  A 1 .| A |

...(ii)

1

Also,

det ( A )



det ( A )1

1

 [det ( A )] 1  [det ( A )]

 det ( A ).det ( A )1  1 ...(iii) From equation (iii),, we conclude that it is true. 1 Again , ( A  B)1  adj ( A  B) |( A  B)| 

( A  B)1  B 1  A 1

...(iv)

47. Option (3) is correct. Explanation: Let the fruit grower use x bags of brand P and y bags of brand Q.

Solutions The problem can be formulated as follows: Minimise Z = 3x + 3.5y ...(i)

87

The corner points are A(240, 50), B(20, 140), and C(40, 100)

Subject to the constraings, x + 2y ³ 240 ...(ii) x + 0.5y ³ 90 ...(iii) 1.5x + 2y £ 310 ...(iv) x, y ³ 0 ...(v) The feasible region determined by the system of constraints is as follows:

Corner points

Z = 3x + 3.5y

A(140, 50)

595

B(20, 140)

550

C(40, 100)

470

¬ Minimum

The minimum value of Z is 470 at (40, 100). Thus, 40 bags of brand P and 100 bags of brand Q should be added to the garden to minimise the amount of nitrogen. The minimum amount of nitrogen added to the garden is 470 kg.

16. Option (1) is correct.

Explanation : a ≡ a(mod n) means n|(a – a) = n|0 = 0.

17. Option (1) is correct. Explanation: We know that time repeats after every 24 hours So, we find 65 (mod 24) Q 65 = 24 × 2 + 17 So, 65 ≡ 17 (mod 24) \65 hours is equivalent to 17 hours. Now, 2 : 00 p.m. + 17 hours = 7 : 00 a.m.

18. Option (4) is correct. between 0 and 1, x2 < x < 1

.

1

1 > > 1, also, 1 > x > 0 (given) 2 x x Therefore, Thus,

1 x2

1 x2

>

48. Option (1) is correct.



49. Option (3) is correct.



50. Option (2) is correct.

Section - B2



Explanation: 0 < x < 1



1 >1>x >0 x

is greatest

19. Option (1) is correct. Explanation: Cistern filled or work done by 1 Tap P in 1 hour = 2

Cistern filled or work done by Tap Q in 1 hour 1 = 3 Tank emptied or work done by Tap R in 1 hour 1 = 6 Tank filed or work done by all three pipes in 1 1 1 1 2 hour = + − = 2 3 6 3 So Tank gets completely filled in =

3 hours 2

= 1.5 hours = 1 hr 30 min So time will be 9.00 a.m. + 1 hour 30 min = 10.30 a.m.

20. Option (4) is correct. Explanation: The given matrix of the order 3 × 3 has 9 elements and each of these elements can be either 0 or 1. Now, each of the 9 elements can be filled in two possible ways. Therefore, by the multiplication principle, the required number of possible matrices is 29 = 512.

21. Option (2) is correct. Explanation: If A is both symmetric and skewsymmetric matrices, then we should have,

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

88

Now,

A ' = A and A ' = − A ⇒ A = −A ⇒ A+A=0 ⇒ 2A = 0 ⇒ A=0 Therefore, A is a zero matrix.

α 2 + βγ 0  1 0   =  βγ + α 2  0 1   0 On comparing the corresponding elements, we have: α 2 + βγ = 1

22. Option (4) is correct.

⇒ α 2 + βγ − 1 = 0

Explanation:

⇒ 1 − α 2 − βγ = 0

2

A = A⋅A 0 = 1 1 = 0

1  0 1  ⋅  0  1 0  0 1 

27. Option (1) is correct. Explanation: We know that, in a square matrix number of rows is equal to the number of 0 0 4 columns. So, the matrix P =  0 4 0  is a    4 0 0 

23. Option (4) is correct. Explanation: Since 3|(x + 4) is true for x = 35

24. Option (4) is correct. Explanation : Given that, f(x) = 2x3 + 9x2 + 12x – 1 f '(x) = 6x2 + 18x + 12 = 6(x2 + 3x + 2) = 6(x + 2)(x + 1) So, f '(x) ≤ 0, for decreasing on drawings number line as below

A2 = I

square matrix.

28. Option (3) is correct. Explanation: Let x be the speed of the stream \ 8 + x = 3(8 − x) ⇒ 4x =16 ⇒ x = 4 km/h

29. Option (3) is correct. Explanation:

TC = VC + FC

= x2 + 2x + 10000 AC = x + 2 +

We see that f(x) is decreasing in [– 2, – 1].

25. Option (2) is correct. Explanation: Matrix X is of the order 2 × n. Therefore, matrix 7X is also of the same order. Matrix Z is of the order 2 × p, i.e., 2 × n [Since n = p] Therefore, matrix 5Z is also of the same order. Now, both the matrices 7X and 5Z are of the order 2 × n. Thus, matrix 7X – 5Z is well-defined and is of the order 2 × n.

10000 d( AC ) = 1 − dx x2



= 0 ⇒

x = 100

30. Option (1) is correct. Explanation: Prize (xi)

pi

xipi

500000

1 10000

50

0

9999 10000

0

26. Option (3) is correct. Explanation: α β  A=   γ −α  ∴ A2 = A ⋅ A



α β  α β  = ⋅   γ −α   γ −α   α 2 + βγ αβ − αβ  = 2  αγ − αγ βγ + α  α 2 + βγ 0  =  0 βγ + α2  

10000 x

So,

∑ xi pi = 50

Net expected gain = 50 – 100 = – 50 So gain is – 50.

31. Option (3) is correct. Explanation: P(r < 2) = P(0 or 1) 10

10

1  2

= C 0 

10

10 1   + C1   2

Solutions =

1 + 10 1024

=

11 1024

38. Option (1) is correct. Explanation:

p 

∑  p 1  ( p0 q 0 ) 

0 ( ∑ p0 q 0 )

32. Option (4) is correct. Explanation:

n = 100, p =



q =



s =

Explanation: Time reversal Test is satisfied by Fishers ideal index.

1

40. Option (4) is correct.

10

9

Explanation: Study of statistics can be categorized in two types : (i) Descriptive Statistics (ii) Inferential Statistics.

10 npq

100 ×

=

41. Option (1) is correct. 1

9

× 10 10

= 3

33. Option (1) is correct.

Explanation: A selection of a group of individuals from a population in such a way that it represents the population is called as sample and the number of individuals in the sample is called the sample size.

42. Option (1) is correct.

Explanation: P(x > 518) = 1 – p(x < 518)

= 1 – P(z < 1) = 1 – 0.8413 = 0.1587

34. Option (2) is correct.

35. Option (2) is correct.

n = 2 half years So, effective rate = (1 + i)n – 1 6   =  1 + −1 2 × 100   2

6   =  1 + −1 200   = (1.03)2 – 1 = 1.0609 – 1 = 0.0609

43. Option (2) is correct.

Explanation:

∑ p1 × 100 = p ∑ 0

Explanation: Given, r = 6% p.a 6 ⇒ i = semi annually 100 × 2

2

Explanation: P(x < 54) = P(z < 1.5) = 0.9332 = 93.32 %

340 300

Explanation: Annual depreciation =

Original cost − Scrap value Useful life

=

50 , 000 − 10 , 000 4

= 113.34%

36. Option (2) is correct. Explanation: F P01 =



L P ( P01 × P01 )

118.4 × 117.5

= = 117.95

37. Option (3) is correct. Explanation: Since, ∴ ⇒ ⇒

∑ p1q0

×

× 100

39. Option (4) is correct.





89

∑ p0 q0 ∑ p1q1

=

28 27

9x + 36 = 40 + 8x x = 4

40 , 000 = ` 10,000 4

44. Option (1) is correct. Explanation: Given, P = ` 97,200, i = 10% p.a. ⇒

L : P = 28 : 27,

∑ p0 q1



=

i =

10 = 0.1 100

So, Value after 3 years = 97,200 × (1 – 0.1)3 = 97,200 × 0.729 = ` 70,858.80 = ` 70,859 (approx.)

90

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

45. Option (1) is correct. 46. Option (1) is correct. Explanation: From the given graph OA = 75 and OB = 45. x y + The equation of line AB is = 1 75 45 i.e.,

3x + 5y = 225

47. Option (2) is correct. Explanation: From the given graph OC = 40 and OD = 80. x y + The equation of line CD is = 1 40 80 i.e., 2x + y = 80 Explanation: On solving the equations of lines AB and CD, we get the coordinates of point E i.e., (25, 30).

49. Option (1) is correct. Explanation: Let the manufacturer produces x number of model X and y number of model Y bikes. Model X takes 6 man-hours to make per unit and model Y takes 10 man-hours to make per unit. There is total of 450 man-hours available per week.



6x + 10y ≤ 450

⇒  

3x + 5y ≤ 225



50. Option (2) is correct. Explanation: The objective function for given L.P.P. is

48. Option (1) is correct.

∴

For models X and Y, handling and marketing costs are ` 2,000 and ` 1,000, respectively, total funds available for these purposes are ` 80,000 per week. ∴ 2,000x + 1,000y ≤ 80,000 ⇒       2x + y ≤ 80  ...(ii)  Also, x ≥ 0, y ≥ 0 Alternate Solution: As (0, 0) lies in the region 3x + 5y ≤ 225 and also (0, 0) lies in the region 2x + y ≤ 80, therefore the constraints for the L.P.P. are : 3x + 5y ≤ 225, 2x + y ≤ 80, x ≥ 0, y ≥ 0

...(i)



Z = 1000x + 500y

From the shaded feasible region, it is clear that coordinates of corner points are (0, 0), (40, 0), (25, 30) and (0, 45). Corner points

Value of Z

(0, 0)

0

(40, 0)

40,000

(25, 30)

40,000

(0, 45)

22,500

So, the manufacturer should produce 25 bikes of model X and 30 bikes of model Y to get a maximum profit of ` 40,000. qq

3

SOLUTIONS OF Question Paper Section - A

1. Option (3) is correct.

From the above figure, area of the shaded region,

Explanation : We have, 3

2

f ( x ) = 2x − 3x − 12x + 4 2

f '( x ) = 6x − 6x − 12 Now,

+

2

Hence, x=−1 is point of local maxima and x = 2 is point of local minima.

2. Option (2) is correct. Explanation: We have y = 0, y = x and the circle x 2 + y 2 = 32 in the first quadrant. Solving y = x with the circle x 2 + x 2 = 32 x 2 = 16 x = 4

4

( 4 2 )2 − x 2 dx 4 2

= 8 + [ 8 π − 8 − 4 π]

So, f(x) has one maxima and one minima.

0

∫

( )

 6( x 2  x  2 )  0  6( x  1)( x  2 )  0  x  1 and x  2

–1

4 2

2  4  4 2  x2   x x  2 2 sin −1 =   +  (4 2 ) − x + 2 4 2   2 0  2  4 4   0 + 16 sin −1 1 − ( 4 2 )2 − 16 2   16   2  = +   2   −16 sin −1 4   4 2 π  16 π =8+ − 2 16 − 16  2 4 

f '( x )  0

On number line for f’(x), we get + –

4

A = ∫ xdx +

= 4 π sq. units



3. Option (3) is correct. Explanation: As we know that the principal value of cos–1x is [0,π]. Y

(In the first quadrant)

5 2 2

When x = 4, y = 4 for the point of intersection of the circle with the x-axis. Put y = 0

3 2   2 1

x 2 + 0 = 32 x = ±4 2

X'

So, the circle intersects the x-axis at ( ± 4 2 , 0) .

–1 O

X

 2 –

3 2 –2 5 2





Y'

y = cos–1x



4. Option (4) is correct. Explanation : Let I  

x9

 ( 4 x 2  1)6 dx 

x9 1 x 4  2   x  dx 12 

6

dx

92

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. dy x9 x 2 − yMATH. I  



 ( 4 x 2  1)6 dx  

x9 1  x 12  4  2   x 

dx

2

= 2 xe x ⋅ e − y

dx 6

6



1 1 C 10  t 5 



1  1 4 2 10  x 

∫e ∫e



C

dy = ∫ e t dt

⇒

e y = et + C

⇒

e y = ex + C

2

8. Option (3) is correct.

Corner points

3

 1  1  Required probability = 8C3      2  2



8 3

8!  1  5 ! 3 !  2  87 6 1   3  2 28 7  32

6. Option (3) is correct.



Given that, m = 5 and n = 6 \m¹n Number of one-one and onto mapping = 0

2

2 dy = 2 xe x dx x2

(12, 6)

12 ← Maximum

9. Option (1) is correct.

10. Option (2) is correct. Explanation: For bijection on Z, f(x) must be oneone and onto. Function f(x) = x2 + 1 is many-one as f(1) = f(–1) Range of f(x) = x3 is not Z for x Î Z. Also f(x) = 2x + 1 takes only values of type = 2k + 1 for x Î k Î Z But f(x) = x + 2 takes all integral values for x Î Z Hence f(x) = x + 2 is bijection of Z.

elements, respectively, then the number of oneone and onto mapping from A to B is n! if m = n 0, if m ¹ n

= 2 xe x ⋅ e − y

0

= P(G) ×P(G) ×P( B) + P( B) ×P(G) ×P(G) +P(G) ×P( B) ×P(G) 3 2 2 2 3 2 3 2 2 = × × + × × + × × 8 7 6 8 7 6 8 7 6 3 = 28

Explanation: We know that, if A and B are two non-empty finite sets containing m and n

Explanation : Given that, 2 dy = 2 xe x − y dx

(0, 0)

Explanation : Probability of drawing 2 green balls and one blue ball



7. Option (3) is correct.

Corresponding value of F = 3x – 4y

(0, 4) –16 ← Minimum Hence, the maximum value of F is 12.

8

y

y

5

1 1 Here, n = 8, r = 3, p = and q = 2 2

ey

2

dy = 2 ∫ xe x dx

Explanation: The feasible region as shown in the figure, has objective function F = 3x – 4y.

5. Option (2) is correct.

⇒

y

Put x2    = t in RHS integral, we get 2 xdx = dt

P( X = r ) = n Cr ( p )r q n -r



2

On integrating both sides, we get

Explanation : We know that, probability distribution



2 dy = 2 xe x dx

e y dy = 2 xe x dx

⇒

1 Put 4  2  t x 2 dx  dt  x3 1 1 dx   dt  2 x3 1 dt I    6  2 t 1  t 6 1     C 2  6  1  

ey

⇒

dx 1  x3  4  2   x 

= 2 xe



11. Option (2) is correct. Explanation: Total number of functions from A to B = 2n Number of into functions = 2 Number of surjections from A to B = 2n – 2



d b A| A | | | | A 1 |  a c | A| | A| 



12. Option (2) is correct.



= 1(1 − 0 ) − λ ( 0 − λ 2 ) + 1( 0 − λ )



for minimum V dV = 3λ 2 − 1 dλ

for



a−b+c = 0 and this plane parallel to line = 2x

1 3

From eqn. (i) & (ii),

13. Option (2) is correct.

+b a c = = = λ (let) 0−2 3−0 2+3 a = −2λ , b = +3λ , c = 5λ

a b  As matrix A is of order 2, let A =   c d

Substituting in eqn. of plane −2λ ( x − 1) + 3λ ( y − 2 ) + 5λ ( z − 1) = 0 2x − 3 y − 5z + 9 = 0

 d b  Then, |A| = ad – bc and adj A =    c a 

A 1



d b 1 2 c a | A|

1 .| A | | A |2 1  | A| 16. Option (4) is correct. 1 det ( A 1 )   det ( A ) Explanation: Given that,



by option, (3) satisfies it.

 

dy + y = e -x dx

a−b+c = 0

3a + 2b + 0 c = 0

Explanation: Given that A is an invertible 1 matrix, A–1 exists and A–1 = adj. A. | A|

b   d | A | | A | 1   adj. A    c a  | A| | A | | A |   b d | A| | A| | A 1 |  a c | A| | A|

= 3y, z 1

direction numbers of ^ plane are a, b, c & line direction number's 3, 2, 0 ⇒ 3a + 2b + 0 c = 0 ...(ii)

, it is minimum

Now

...(i)

x y z −1 = = 3 2 0

= 6λ

λ =

14. Option (3) is correct.

Also passing through (2, 1, 2) a( 2 − 1) + b(1 − 2 ) + c( 2 − 1) = 0

for Maximum or Minimum dV = 0 dλ 1 ⇒ λ = ± 3 dλ 2

1 .| A | | A |2 1  | A| 1 det ( A 1 )  det ( A )

Explanation: Plane passing through (1, 2, 1) a( x − 1) + b( y − 2 ) + c( z − 1) = 0

= 1 + λ3 − λ

d V

93



Explanation : Volume of parallelopiped 1 λ 1 V = 0 1 λ λ 0 1

2

b dSolutions 1 2 c a | A|

15. Option (1) is correct. Explanation : Given that, A and B are symmetric matrices. Þ A = A’ and B = B’ Now, (AB – BA)’ = (AB)’ – (BA)’ ...(i) Þ                  (AB – BA)’ = B’A’ – A’B’ [By reversal law] Þ                  (AB – BA)’ = BA – AB [From Eq. (i)] Þ                  (AB – BA)’ = –(AB – BA) Þ (AB – BA) is a skew-symmetric matrix.

Section - B1 which is a linear differential equation. Here, P = 1 and Q = e–x dx IF = e ∫

= ex

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

94



The general solution is

Explanation:

   a , b , c are coplanar

 e  e dx  C  dx  C x

y  ex  

ye x 



ye x  x  C

20. Option (4) is correct.

x

1 2 1 λ

...(i)

eqn. (i) becomes y ⋅ e = x ⇒ y = xe x



4 λ −1

R 2 → R 2 − R1 , R 3 → R 3 − 2 R1

−x

1 2 0 λ −2

17. Option (2) is correct. Explanation: Minimum value of F is −16 at (0, 4).



0

 

  1  x2 1  x2 2 xdx 2 xdx I I

Put Put    

         

 

a  a 

x3 x 3 dx 1  x 2 dx 1  x2 a(1  x 2 )3 / 2  b a(1  x3 2 )3 / 2  b x  1x3 x 2 dx  1  x 2 dx x2  x  x12 xx 2 dx  1  x 2 dx t2 t2 2tdt 2tdt 2 t(t  1)  t(t 2t 1) dt 3 t dt t t33  t  C tC 13 2 3/2 13 (1  x 2 )3 / 2  (1  x )  31 13 and b  1 and b  1 3

 

λ −9

0

λ = 2, ± 3 i j k   a×c = 1 2 4 2 4 3 = −10i + 5 j

1  x2  C 1  x2  C

(Using l = ±3, we get a is parallel to c.)

(4, 2, 3)

Line AB, 1  x2  C 1  x2  C

Y

1

–2 –1 O

 2

Y'

y = cosec–1x

λ =

2

X

2 3

Foot of perpendicular = Q(1, 0, 1)

Which satisfy option (3). 22. Option (2) is correct.



–



Line AB and line PQ are perpendicular 3 × 0 + ( 4 − 3λ ) × 3 + 3λ ( −3) = 0

∴

2 3 2

X'

x −1 y + 2 z −3 = λ (let) = = 0 3 −3

Any point on the line (1, 3l – 2, –3l + 3) Let this point is foot of perpendicular Q (1, 3l – 2, –3l + 3) Direction ratio’s of PQ {4 – 1, 2 – 3l + 2, 3 + 3l – 3} = {3, 4 − 3λ , 3λ }

Explanation: As we know that the principal π π value of cosec–1x is  − ,  − {0} .  2 2

 2

21. Option (3) is correct. Explanation:

19. Option (4) is correct.



= 0

2

Explanation: Let,

  I  I 

4 0

⇒

18. Option (4) is correct.

I  I 

= 0

2

2

When x = 0 and y = 0 then, 0 = 0 + C Þ C = 0

4 4

Explanation : Let us assume that, f ( x ) = sin x.cos x Now, we know that 1 sin x.cos x = 1 sin 2 x sin x.cos x = 2 sin 2 x 2 1 ∴ f '( x ) = 1 .cos 2 x.2 = cos 2 x ∴ f '( x ) = 2 .cos 2 x.2 = cos 2 x Now, f '( x ) = 02 Now, f '( x ) = 0 ⇒ cos 2 x = 0 ⇒ cos 2 x = 0

Solutions ⇒

cos 2 x = cos



π 2

From the figure, area of the shaded region, A 

 cos xdx 0

 2 sin x 

 2 sq. units

26. Option (2) is correct. Explanation : Given that, 

and

2 x 2 − 40 = 18 + 14 ⇒ 2 x 22 − 40 = 18 + 14 22 xx 22 − 40 ===18 +++14 18 14 x2 = 32+ 40 2 x −−240 40 18 14 2 x 222 = 32 + 40 ⇒ + 22 xx 22 = 32 40 72 == 32 2xx = 32 ++ 40 40 72 x 222 = 72 2 72 xx 22 = ⇒ 2 = 72 22 xx2 == 36 2 x 22 = 36 xxx2 = = ±36 6 36 xx == ± 36 6 xx = ± 6 x == ±±66

   Now,

y   x 3  3x 2  9 x  27 dy  3x 2  6 x  9 dx  Slope of the curve d2y dx 2 d2y dx 2 6( x  1) x

 6 x  6  6( x  1)  0  0  10

d3y  6  0 dx 3

So, the maximum slope of given curve is at x = 1.

∴

24. Option (2) is correct.

∴   b = 2a

3 − λ2 = 2λ1 2λ1 + λ2 = 3

...(i)

  a^c



27. Option (3) is correct. Explanation : Let, E1 = Event for getting an even number on die and E2 = Event that a spade card is selected 

  a⋅c = 0

P( E1 )  

( 2i + λ1 j + 3k )[3i + 6 j + ( λ3 −1)k ] = 0

and

6 + 6λ1 + 3( λ3 − 1) = 0 2λ1 + λ3 = −1

 dy  = −3 × 12 + 6 × 1 + 9   dx ( x = 1) = 12

4i + ( 3 − λ2 ) j + 6 k = 2( 2i + λ1 j + 3k )

⇒



 2

cos x dx  /2 0

23. Option (3) is correct.

Explanation:

 0

Explanation: Given that 2x 5 6 −2 ∴ = , 8 x 7 3



25. Option (1) is correct. Explanation : We have y = cos x , x = 0, x = π

    xx  44 dd Also, ff ″″(( xx ))  .cos 22 xx  .sin 22 xx  dx .cos  22.sin Also, dx    2 sin 2.     ff ″″(( xx ))at at xx     2 sin 2. 4 44 4     22 sin sin 2 2  22   00    point of maxima.   xx   4 is is point of maxima. 4       11 .sin.2.  ff  4   .sin.2. 4 2  4 2 4 1 1   22

95



...(ii)

From equation (i) and (ii) by trial method, it satisfied the option (2).

P( E2 ) 

Then ,

P( E1  E2 )  

3 6 1 2 13 52 1 4 P( E1 )  P( E2 ) 1 1 1   2 4 8

96

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

28. Option (2) is correct.



Explanation: We know that,

Explanation: Given that,           3tan–1x + cot–1x = π Now, we have,           3tan–1x + cot–1x = π –1 ⇒   2tan x + (tan–1x + cot–1x) = π Using this property, we get

π  −1 −1  tan x + cot x = 2    π −1 2tan x = π − 2 π 2tan −1 x = 2 π −1 tan x = 4

A|| 00 0  || A 00  | A | 0 (( adj adj A A))A A= =|| A A||II = =  00 || A A|| 00  ( adj A ) A =| A | I =  0 | A | 0   0 A|| 0 |A  00 | 00 ||A |A A|| 00 0 ||A | 0 00 |(adj adj A A))A A| = 0 | A A|| 00 |( ⇒ |( adj A ) A ||== 00 ||A | 0 0 0 |A A|| 00 00 ||A | 11 000 000 1 |adj adj A A|| || A A| = | A A||333 00 11 00 = =|A A||333 ((II )) ⇒ ||adj A || A ||==||A | 0 1 0 = ||A | (I ) 00 000 111 0 |adj adj A A| = |A A||222 ||adj A ||==||A |

π ⇒ 2tan −1 x +   = π 2 ⇒                ⇒          ⇒            



∴

Corresponding value of F = 4x+6y (0, 2) 12 ← Minimum (3, 0) 12 ← Minimum (6, 0) 24 (6, 8) 72 ← Maximum (0, 5) 30 Hence, minimum value of F occurs at any points on the line segment joining the points (0, 2) and (3, 0). Corner points

29. Option (4) is correct. Explanation: Distance of point (1, –2, 3) from x – y + z = 5 measured parallel to x y z = = 2 3 −6 line passing through (1, –2, 3) and parallel to the given line x −1 y+2 z−3 = = =λ 2 3 −6 Any point on the line which is also on the plane

31. Option (4) is correct. Explanation  :

π x = tan   4 x=1

       ⇒ ⇒     

30. Option (2) is correct.



32. Option (1) is correct. Explanation : It is given that, d 3 f (x) = 4x3 - 4 dx x \ Anti-derivative of 4x 3 −

satisfies plane

x−y+z = 5

2 λ + 1 − 3λ + 2 − 6 λ + 3 = 5

f (x) f (x) f (x) f (x)

 

f ( x )  4  x 4   3  x 3   C f ( x )  4  4   3  3   C  3   4 1 f ( x )  x 44  13  C f (x)  x  x3  C x

−7 λ = −1

λ =

1 7

distance of (1, –2, 3) and a point on the plane ( 2λ + 1, 3λ − 2 , −6λ + 3) is

( 2λ + 1 − 1)2 + ( 3λ − 2 + 2 )2 + ( −6λ + 3 − λ )2 =

4 λ 2 + 9λ 2 + 36λ 2

1 = 7× =1 7

Also, 

 

= 7λ 1  λ = 7   

3

 44 xx 33  x34 dx dx  x4 4  x 33dx  3   x 44  dx 4  x dx  3   x  dx  x 3   x4 

 

x = 2λ + 1, y = 3λ − 2 , z = −6λ + 3

 

   

3 = f (x) x4

f (2)  0 1 f ( 2 )  ( 2 )44  3  C ( 2 )3 0 16  1  C  0 8

1  C    16   8  129 C  8

1 129 f ( x )  x 44  3  8 x3

Solutions

33. Option (4) is correct. Explanation: We have, f(x) =

E2 = Event that the eldest child is a girl, then E2 = {(G, B, B), (G, G, B), (G, B, G), (G, G, G)}

1 , "xÎR x

For x = 0, f(x) is not defined. Hence, f(x) is a not defined function.



E1  E2  {(G, B, B),(G, G, B),(G, B, G), (G, G, G)} P( E1  E2 )  P( E2 | E1 )  P( E1 ) 4  8 7 8 4  7

34. Option (3) is correct. Explanation: Given differential equation

             

( e x  1) ydy ( e x  1) ydy dy dy dx dx dx dx dy dy dx dx dy dy dx

       

dx  dy  dy dx dx  dy  dy dx dx  dy  dy  y   1 y y  dy     1  y  dy 

( y  1)e x dx ( yx  1)e x dx e (1  y ) x (ee x(11)yy) ( e xx  1) y ( e  1) y (eexx(11)yy) e x (1x  y ) e y y x  e x (e1 y y )  e x (1y y ) e xy(1  y ) ye x (1  y )  1 y y  (1 yy )e x 1  y (1  y )e x y  1 1 1 y y  1  e1x  1  y  x e x  y  e  1 x  x 1 1 y y  e e    1  yx  e x   e   xe x  dx  e  1  e x  1  dx  

On integrating both sides, we get On integrating both sides,xwe get y e dy   e x x dx  1 y ydy 1  e dx  1 y  1 xe x e 1 y 1   e x x dx   1 1 y y 1dy 1  e dx dy    1 y  1 xe x e 1   1dy   1 dy   e x x dx   1dy   1  y dy   1  e x dx 1 y 1e  y  log |(1  y )  log|(1+ee x )  log k  y  log |(1  y )  log|(1+ee x )  log k x  y  log(1  y )  log(1  e )   y  log(1  y )  log(1  e x )  log( k ) log( k ) x  y  log k (1  y )(1  e ) x  y  log k (1  y )(1  e )

 



97

 

35. Option (4) is correct. Explanation : We have, S = {B, B, B), (G, G, G), (B, G, G), (G, B, G), (G, G, B), (G, B, B), (B, G, B), (B, B, G)} E1 = Event that a family has at least one girl, then E1 = {(G, B, B), (B, G, B), (B, B, G), (G, G, B), (B, G, G), (G, B, G), (G, G, G)}



36. Option (3) is correct. Explanation: Let  

 Now,   

 1 y     x

x

1 x 1 dy 1  1 1  x. .   2   log .1 . 1  x  x y dx x 1  1  log x log y  x.log

dy dx dy dx 1 log x 1 x

1    1   log  1 .      x x

x

 0  1  log e  e

x 

1 e

 1 Hence, the maximum value of f    ( e )1 / e .  e

37. Option (1) is correct. Explanation : Matrices P and Y are of the orders p × k and 3 × k, respectively. Therefore, matrix PY will be defined if k = 3. Consequently, PY will be of the order p × k. Matrices W and Y are of the orders n × 3 and 3 × k respectively. Since the number of columns in W is equal to the number of rows in Y, matrix WY is well-defined and is of the order n × k. Matrices PY and WY can be added only when their orders are the same. However, PY is of the order p × k and WY is of the order n × k. Therefore, we must have p = n. Thus, k = 3 and p = n are the restrictions on n, k, and p so that PY + WY will be defined.

98

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

38. Option (3) is correct.

Point on L1 (–1, 2, 1) & on L2 (–2, –1, –1) direction ratio’s of L1 L2{–1, –3, –2} Lines are coplanar

Explanation:  a 0 0 A =  0 a 0     0 0 a 



−1 −3 2 −1 α 5 −α

Det(A) = a(a × a – 0 × 0) – 0 + 0 = a3 Det(adj A) = (a3)2 = a6

−1( −1 − 5 + α ) + 3( 2 − α ) − 2(10 − 2α + α ) = 0 6 − α + 6 − 3α − 20 + 2α = 0

39. Option (1) is correct. Explanation:

−2α = 8

α = −4     3i  j ,   2i  j  3k

Point on L2

 β1 is parallel to a 



41. Option (1) is correct. Explanation:     a×c +b = 0     Let c = c1 i + c 2 j + c3 k    ⇒ b = c×a i i + j + k = c 1

= i( 3λ − 2 ) + j( λ + 1) − 3k 2  0

 2 is  to  given

1

[i( 3  2 )  j(   1)  3k ]( 3i  j )  0 10   5  0 1   2

i + j + k = c3 i − c3 j + k ( − c1 − c 2 )

⇒

c3 = −1 & − c1 − c 2 = 1   a.c = 4

(i − j )( c1 i + c 2 j + c3 k ) = 4   ⇒ c1 − c 2 = 4

 1  3  2  i  j  3k 2 2

Solving eqns. (i) & (ii),

i 3 2 1 2

=

j 1 2 3 2

c1 =

k

⇒

0 3

−3  9  5  i+ j+ k 2 2 2

x +1 y − 2 z −1 = L1 : = 2 −1 1 L2 :

x+2 y +1 z+1 = = α 5 −α 1

...(i)

...(ii)

3 5 & c2 = − 2 2

3 5  c = i − j − k 2 2  c =

9 25 + +1 4 4

=

38 19 = 4 2

19 2 c = 2

40. Option (1) is correct. Explanation:

j k c 2 c3 −1 0

⇒

 1   1  ( 3i  j ) 2

  1   2 

&

x+2 y +1 z+1 = =λ = −4 9 +1

(–4l –2, 9l – 1, l – 1) Check by option 1 satisfies

  1        1   2       2     2      ( 3i  j )  ( 2i  j  3k )



−2 1 = 0 1



42. Option (3) is correct. Explanation: We know that, in a square matrix, if bij = 0 when i ≠ j then it is said to be a diagonal matrix. Here, b12, b13…. ≠ 0 so the given matrix is not a diagonal matrix.         

Solutions 5 0  0 B5  8 12 0 5  B '   5 0  8 12

Now,

Put 3x = t Þ 3dx = dt dx 1 dt  2   2 3 ( 2 )  ( t )2 ( 2 )  ( 3 x )2

8  12  0  8   12  0 



1  3x  tan 1    2 6  F( x ) By second fundamental theorem of calculus, we obtain 2/3 dx  2 0 4  9x 2  F  3   F(0)

So, the given matrix is a skew-symmetric matrix, since we know that in a square matrix B, if B’ = −B, then it is called skew-symmetric matrix.

1 3 2 1  tan 1     tan 1 ( 0 ) 6 2 3 6 1 1  tan (1) 6  1   tan 1  tan  6 4    24

43. Option (2) is correct. Explanation: Assertion (A) is correct. P( A Ç B ) P( B ) P( A ) P ( A |B ) = P( B ) P ( A|B) =

[since, given A and B are two mutually exclusive events] 5   1  6  .  A P     B 1   1  3  1  6 2 3 1   4







But Reason (R) is correct. From Basic Theorem of Probability, P(B – A) = P(B) – P(A), this is true only if the condition given in the question is true.



∫ 4 + 9x 2

49. Option (4) is correct. Explanation: d d (sin x 3 ) = cos x 3 (x3 ) dx dx = 3x 2 cos x 3



50. Option (4) is correct. Explanation: d d (sin 2 x ) = cos 2 x ( 2 x ) dx dx = 2 cos 2 x d π (sin 2 x ) = 2 cos 2 × = 2 cos π π 2 dx x=

45. Option (3) is correct. Explanation : dx

48. Option (3) is correct. Explanation: d d (sin 3 x ) = 3 sin 2 x (sin x ) dx dx = 3 sin 2 x cos x

44. Option (4) is correct. Explanation : Assertion (A) is wrong. If the probability of an event is 0, then it is called as an impossible event.

47. Option (2) is correct. Explanation: gof(x) = g(f(x)) = g(sin x) = sin3 x

For independent events,



46. Option (1) is correct. Explanation: fog(x) = f(g(x)) = f(x3) = sin (x3)

Reason (R) is also correct.

P ( A | B ) = P( A ) = 0.2.

t 1 1 tan 1  3  2 2



5 8  0   0 12    5  8 12 0   B



4

=

dx

∫ ( 2 )2 + ( 3 x )2

99

= 2 ( −1) = −2

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

100

Section - B2

16. Option (2) is correct.

20. Option (3) is correct.

Explanation: If a ≡ b(mod n) then n|(a – b) or, n|–(b – a) or, n|(b – a) or, b ≡ a(mod n)

17. Option (1) is correct.

19. Option (1) is correct.

\

X : Y = 7 : 1 X = 7Y Distance Speed = Time

Upstream speed =

4.2 km 14 min

= 0.3 km/min

Upstream speed = X – Y

= 7Y – Y = 0.3 km/min \

Y = 0.05 km/min

Downstream speed = X + Y = (7Y + Y) = 8Y = 8 × 0.05 = 0.4 km/min

Downstream time =

=



Speed of P =

224 = 8 m/sec 28

and

Speed of Q =

224 = 7 m/s 32

Difference in time taken = 4 sec Therefore, distance covered by P in that time = 32 m

Explanation: Since given p > q and r < 0 then pr < qr (inequality sign is reversed when multiplied by a negative integer)





= 8 m/s × 4 s

18. Option (1) is correct.

\

21. Option (2) is correct. Explanation:

Explanation: We know that, a . b(mod n) ≡ a(mod n) . b(mod n) So, (186 × 93)mod 7 ≡ 186(mod 7) . 93(mod 7) ≡ 4(mod 7) . 2(mod 7) ≡ 4.2 (mod 7) ≡ 8(mod 7) ≡ 1(mod 7)

Explanation:

Explanation: Modulo arithmetic is the arithmetic of remainders. Mode arithmetic considers remainder when an integer is divided by another integer.

Downstream time =

Distance Speed 18.4 km 0.4 km/min 184 4

= 46 minutes

22. Option (4) is correct. Explanation: Suppose B joined for x months . ( 85000 × 12 ) Then, =3 ( 42500 × x ) or,

x =

( 85000 × 12 ) =8 ( 42500 × 3)

So, B joined for 8 months.

23. Option (2) is correct. Explanation: Kishore's speed in still water = (15 – 2.5) km/hr = 12.5 km/hr ishore's speed against the current K = (12.5 – 2.5) km/hr = 10 km/hr

24. Option (2) is correct. Explanation: Here, (2, 2) entry of A is 2 and (2, 2) entry of B is –5. Therefore, sum of (2, 2) entries of A and B = 2 + (–5) = –3

25. Option (3) is correct. 26. Option (3) is correct. Explanation: Since given A = A' ⇒

 −8 x   −8 y   y 5 =  x 5    

⇒ By the equality of matrices, we get x = y.

Solutions 27. Option (1) is correct. 28. Option (1) is correct. Explanation:

y =

x + x + x + ...∞

y =

x+y

f ′′( x ) φ′′( x ) 1 d2 y − = 2 f ( x ) φ( x ) y dx +

We can write it as ⇒

+

dy dy 2y = 1+ dx dx dy 1 = dx ( 2 y − 1)

⇒

Explanation: Given, f(x) = x3 – 3x , x ∈ [0, 2] f ′(x) = 3x2 – 3 \ Now, Put f ′(x) = 0 ⇒ 3x2 – 3 = 0 ⇒ x2 = 1 ⇒ x = ± 1 But x ∈ [0, 2], so x = 1 is the only required point. f(1) = 1 – 3 = – 2, f(0) = 0, f(2) = 8 – 6 = 2 Therefore, minimum value of f(x) = – 2.

Explanation: y = f ’(x) = =

f′(x) =

f (x) f ′( x ) and z = φ( x ) φ′( x )

dy dx φ( x ) × f ′( x ) − f ( x )φ′( x ) φ2 ( x )

φ( x ) f ′( x ) − f ( x )φ′( x ) φ2 ( x ) d2 y

31. Option (1) is correct.

f′′(x) = dx 2

Explanation: Given m = 13x4 + 5x3 – 12y3 + 24x2 + y2 – 2x – 156

d [φ( x ) f ′( x ) − f ( x )φ′( x )] dx −[φ( x ) f ′( x ) − f ( x )φ′( x )].2φ( x )φ′( x )

φ2 ( x ) =

φ4 ( x )



d2 y

dx 2



d [φ( x ) f ′( x ) − f ( x )φ′( x )] dx = φ2 ( x ) −

2



d y = dx

2φ( x )φ′( x )[φ( x ) f ′( x ) − f ( x )φ′( x )] φ4 ( x )

φ′( x ) f ′( x ) + φ( x ) f ′′( x ) − f ′( x )φ′( x ) − f ( x )φ′′( x )] φ4 ( x )



−

2φ′( x )[φ( x ) f ′( x ) − f ( x )φ′( x )] 3

φ (x)



d2 y dx 2

=

2 ( y −z ) {φ ′( x )}2 f ( x )φ( x )

30. Option (2) is correct.

29. Option (2) is correct.



2φ′( x )  ( y − z )φ( x )φ′( x )    f (x) φ2 ( x )  

f ′( x ) φ′′( x ) 1 d2 y − = 2 f ( x ) φ( x ) y dx

Squaring we get ⇒ y2 = x + y On differentiating, we get ⇒

101

φ( x ) f ′′( x ) − f ( x )φ′′( x )] φ2 ( x ) −

2φ′( x )[φ( x ) f ′( x ) − f ( x )φ′( x )] φ3 ( x )

\

dm = 52x3 + 15x2 + 48x – 2 dx

32. Option (4) is correct.

33. Option (1) is correct.

34. Option (1) is correct. 35. Option (4) is correct. 36. Option (1) is correct. 37. Option (4) is correct. 38. Option (3) is correct. Explanation: In seasonal variation, tendency movements are due to the nature which repeat themselves periodically in every season.

39. Option (3) is correct.

102

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

40. Option (2) is correct.

44. Option (3) is correct.

Explanation: Given R = ` 6000, r =

6 % = 0.5 per month 12

So,

0.5 = 0.005 100

i =

P = R +

So,

Explanation: Clearly from graph there is no feasible region.

R i

= 6000 +

6000 0.005

= 6000 + 1200000 = ` 1206000

45. Option (2) is correct.

41. Option (2) is correct. Explanation: Let rate of interest be r% per annum, r then i = 200 Given,

R = ` 600 and P = ` 10,000 R P = i

Using ⇒

i =

R 600 = P 10 , 000

⇒

10 + 6 + 16 + 17 + 13 + 12 + 8 + 14 + 15 + 9 = 10 =

120 = 12 pounds 10

Explanation: Sum of squares of deviations of diet B = Σ( y − y )2 Here

= 12%

42. Option (1) is correct.

=

Explanation: The CAGR of the revenues over the three years period spanning the "end" of 2018 to "end" of 2021 is 1

1

 Final value  n  13000  3  Initial Value  − 1 =  9000  − 1     = 1.13 – 1 = 0.13 = 13%

43. Option (3) is correct. Explanation: Corner Points (3, 0)

Explanation: Mean of increases in weight on Σx diet A = x = n1

47. Option (2) is correct.

600 r = 10 , 000 200 120 , 000 r = 10 , 000

⇒

46. Option (1) is correct.

Z =7x + y 21

1 5 2, 2  

6

(7, 0) (0, 5)

49 (maximum) 5

y=

Σy n2

7 + 13 + 22 + 15 + 12 + 14 + 18 + 21 + 23 + 10 + 17 12

180 12 = 15 is pounds. Σ( y − y )2 = 64 + 4 + 49 + 0 + 9 + 1 + 9 + 49 + 36 + 64 + 25 + 4 = 314 =

48. Option (4) is correct. Explanation: The number of degrees of freedom (n) = n1 + n2 – 2  = 10 + 12 – 2  = 20

49. Option (1) is correct. Explanation: Q Standard error,

S =

Σ( x − x )2 + Σ( y − y )2 n1 + n2 − 2

Solutions 120 + 314 20 [ S (x – x– )2 = 120 and S (y – y–)2 = 314 ]

=

= 21.7 = 4.65

50. Option (2) is correct. Explanation: Here t =

x  y n1n2 n1 + n2 S

=

103

12 ~ 15 12 × 10 4.65 12 + 10

= 1.5 for 20 degrees of freedom we are given that the value of t at 5% level of significance is 2.09 the calculated value t is less than this value. Hence the difference between the sample means is not significant.



OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

SOLUTIONS OF Question Paper Section - A



1. Option (4) is correct. Explanation: We know that f : R ® R is defined as f(x) = x4. Let x, y Î R such that f(x) = f(y) Þ x4 = y4 Þ x = ± y \ f(x) = f(y) does not imply that x = y. For example, f(1) = f(–1) = 1 \ f is not one-one. Consider an element 2 in co-domain R. It is clear that there does not exist any x in domain R such that f(x) = 2.

continuous functions, then by using the algebra of continuous functions , the functions f(x) + g(x), f(x) – g(x), f(x).g(x) are also continuous functions but

2. Option (3) is correct. Explanation: p

 4 + 3 sin x  I = ∫ 2 log  …(i)  dx       0  4 + 3 cos x 

Let

 p  p  4 + 3 sin  2 − x      dx I = ∫ 2 log  0 p    4 + 3 cos  2 − x      a f ( x ) dx = a f ( a − x ) dx  ∫0  ∫0   p



 4 + 3 cos x  = ∫ 2 log   dx …(ii) ...(ii) 0  4 + 3 sin x 



Adding equations (i) and (ii), we obtain p   4 + 3 sin x   4 + 3 cos x   2I = ∫ 2 log   + log  4 + 3 sin x   dx 0 x 4 3 cos +      p + x + x 4 3 sin 4 3 cos   ⇒ 2I = ∫ 2 log  ×  dx 0  4 + 3 cos x 4 + 3 sin x  p

⇒ 2I = ∫ 2 log 1dx 0

p

⇒ 2I = ∫ 2 0dx 0

⇒ I = 0

3. Option (4) is correct. Explanation: 2 Since f ( x ) = 2x and g ( x ) = x + 1 are 2

g (x) f (x)

is discontinuous function

at x = 0.

\ f is not onto. Hence, function f is neither one-one nor onto.

4

4. Option (4) is correct. Explanation: We have a−b b+c a a+c b+c+a a b−a c+a b = b+c c+a+b b c−a a+b c c+b a+b+c c [ C1 → C1 + C2 and C2 → C2 + C3 ] a+c 1 a = (a + b + c) b + c 1 b c+b 1 c [Taking ( a + b + c ) common from C2 ] a−b 0 a−c 0 b−c = (a + b + c) 0 c+b 1 c [ R2 → R2 − R3 and R1 → R1 − R3 ] = ( a + b + c ) – [( b − c ) ( a − b )] [Expanding along R2 ] = ( a + b + c )( b − c )( a − b )

5. Option (3) is correct. Explanation : We have, P ( E ) = 0. 3 and P ( E ∪ F ) = 0. 5 Also, E and F are independent. Now, P( E ∪ F ) = P( E ) + P( F ) − P( E ∩ F ) P( E ∪0F.5) = ⇒ = 0P.(3E+) +P(PF()F−) − 0.P 3P( E( F∩) F ) ⇒ 0.5 = 00..35 +− P ( F ) − 0 . 3P( F ) 0.3 ⇒ P( F ) = 0.5 − 0.3 ⇒ P( F ) = 0..7 2 0..7 = 2 7 = 7 ( E ) − P( F ) ∴ P( E / F ) − P( F / E ) = P ∴ P( E / F ) − P( F / E(as ) =EPand ( E) −F Pare ( F )independent)) (as E 3and 2F are independent)) = − 3 2 = 10 − 7 10 1 7 = 1 70 = 70

0.5 − 0.3 0..7 2 = 7 ∴ P( E / F ) − P( F / E ) = P( E ) − P( F ) (as E and F are independent)) 3 2 = − 10 7 1 = 70 ⇒



P( F ) =

Solutions

6. Option (1) is correct.

Þ

7. Option (1) is correct.

Slope of the normal −1 +5 = dy 2 dx

Explanation: A = pr2 dA = 2pr dr

11. Option (3) is correct. Explanation: Plane passing through (3, 1, 1) a( x − 3) + b( y − 10 ) + c( z − 1) = 0 Containing line whose direction ratio's are 1, –2, 2 and 2, 3, –1 So direction ratio’s of plane & lines are ^ a n 2b  2 c  0 ...(i)

dA dr r = 3 = 2p × 3 = 6p

8. Option (1) is correct.

2 a + 3b n c = 0

Explanation: If lines intersect then

...(ii)

Solving eqn. (i) & (ii), a b c = = =λ 2−6 4 +1 3+4

  r = r

(i − j ) + l( 2i + k ) = ( 2i − j ) + m(i + j − k ) Compare coefficient of i → −1 + 2l = 2 + m j → −1 = −1 + m k → l = −m

a = −4 λ , b = 5λ , c = 7 λ Hence eq. of plane −4 λ ( x − 3) + 5λ ( y − 1) + 7 λ ( z − 1) = 0

...(i)

4x − 5y − 7 z = 0

...(ii)

Also passes through (a, –3, 5) 4α + 15 − 35 = 0

...(iii)

From eqn. (ii) & eqn. (iii), we get ⇒ l = 0=m

α = 5

but this value does not satisfy eqn. (i) So no value of l and m satisfy option (1).

− 2x dy = 3 + 2y dx  dy  −2  dx  =  1,1 5

d2y dy −3 + y = 0 dx 2 dx The highest order derivative present in the d2y given differential equation is . Therefore, its dx 2 order is two. 2x2



10. Option (2) is correct. Explanation: x2 + 3y + y2 = 5 dy Þ 2x + (3 + 2y) =0 dx

Explanation  :



105

Explanation: f : R ® R is defined as f(x) = 3x. Let x, y Î R such that f(x) = f(y) Þ 3x = 3y Þ x = y \ f is one-one. Also, for any real number y in co-domain R, there y y y exists in R such that f   = 3   = y. 3 3 3 \ f is onto. Hence, function f is one-one and onto.

9. Option (4) is correct. Explanation : Since A and B are independent events, A’ and B’ are also independent. Therefore, P( A′ ∩ B′) = P( A′) ⋅ ( B′) = (1 − P( A )(1 − P( B))  3  4  = 1 −  1 −   5  9  2 5 = ⋅ 5 9 2 = 9

12. Option (1) is correct.



13. Option (1) is correct. Explanation: x n − an This is a condition lim = nan–1 x →a x − a

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

106

14. Option (4) is correct. Explanation:

=∫

1 sin −1    3

lim f ( x ) = f(– 1) x →−1

Þ Þ Þ



π

x 2 − 2x − 3 lim =k x →−1 x+1 ( x + 1) ( x − 3) lim =k x →−1 x+1 lim ( x − 3) = k x →−1

Let I = ∫1

(

)

1 sin −1    3

sin

=∫

 1  3 

1 sin −1    3

1 sin −1    3



=∫

1 sin −1    3

sin 4 θ



cos θ dθ

2

( sin θ ) 3 ( cos θ ) 3 cos θ dθ sin 4 θ

(t ) 3 dt

3 8  = −  ( t ) 83  0  83  = −  (t )3  2 2 38 8  02 2 = − ( t ) 83  0 83   = − ( t ) 3  2 2 83  2 2 8  = −  − 2 2 83   83  = −  − 2 2 3  8  8   3 =  8 83   83  =  8 3  83  4   = (8 ) 34  83   = ( 8 ) 3  83  = [16]  83 = 16 = 38 ×[ 2 ] = 36 × 2 =6

( ) ( ) ( ) ( )

2

( sin θ ) 3 ( cos θ ) 3 cos θ dθ sin 2 θ sin 2 θ





1

( sin θ ) 3 (1 − sin 2 θ ) 3

1



cos θ dθ

sin 4 θ

1

π

= ∫2



1 3

1

π 2

π 2

sin θ  sin θ 3

1

2

0

1 3

dx x4 Also, let x = sin θ ⇒ dx = cos θ dθ 1 −1  1  When x = , θ = sin   and when x = 1, 3 3 p θ= 2 I

5

0

∴ I = −∫ 2

3

 2

( cot θ ) 3 cosec2θ dθ

1 When θ = sin −1   , t = 2 2 and 3 t= θ

Explanation: x − x3

5

( sin θ ) 3

Let, cotθ = t ⇒ −cosec 2θd θ = dt

15. Option (1) is correct.

1

( cos θ ) 3 cosec2θ dθ 5

= ∫2



k=−4

5

π 2

16. Option (2) is correct.

Section - B2

Explanation: Volume of parallelopiped 1 1 n 158 = 2 4 −n 1 n 3 158 = 1(12 + n 2 ) − 1( 6 + n) + n( 2n − 4 )

158 = 12 + n 2 − 6 − n + 2n 2 − 4 n 158 = 3n 2 − 5n + 6 3n 2 − 5n − 152 = 0

( 3n + 19 )( n − 8 ) = 0 n = but ⇒ So,

−19 or 8 3

n ≥ 0 n  a  b  c   b ⋅c

= 8 = i + j + 8 k = 2i + 4 j − 8 k = i + 8 j + 3k = 2 + 32 − 24 = 10

when θ =

π , 2

y2 42

Explanation: We know that, area of a triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is given by x1 y1 1 1 ∆ = x2 y2 1 2 x3 y3 1

5

4 2 5 − x 2 dx 5 0

Explanation: P (4, –2, 3)

Plane passes through midpoint of PQ R (3, 1, 1) direction ratio's of PQ {4 − 2 , − 2 − 4 , 3 + 1} = {2 , −6 , 4} ⇒ Direction ratios of plane {2, –6, 4} Equation of plane whose direction ratio’s are (2, –6, 4) passes (3, 1, 1) 2( x − 3) − 6( y − 3) + 4( z − 1) = 0 2x − 6 y + 4 z − 4 = 0

5

=

 16  x 2 52 2 −1 x  5 − x − sin  2 5  2 5  0

=

 16  52 n −1 1 − 0 − 0  0 + sin 5  2 

16 25 π . . 5 2 2 = 20π sq. units =



18. Option (3) is correct.

20. Option (3) is correct.

Explanation: Consider that aRb, if a is congruent ∀ü= ∈ Then, aRa ⇒ a ≅ a, Which is true for all a ∈ T So, R is reflexive, Let aRb ⇒ a ≅ b ⇒b≅a⇒b≅a ⇒ ü So, R is symmetric.  Let aRb and bRc ⇒ b ≅ b and b ≅ c ⇒ a ≅ c ⇒ aRc So, R is transitive.  Hence, R is equivalence relation. 21. Option (1) is correct.

19. Option (1) is correct. x2 y2 + = 1 , which is 52 4 2 ellipse with its axes as coordinate axes. Explanation: We have

Reason (R) is also correct. Corner Points y2 42

= 1−

x2

52  x2  y 2 = 16 1 −   25    4 2 2 y = 5 −x 5

to b,

…(i)

…(ii)

…(iii)

Explanation: Assertion (A) is correct. Clearly from the graph below that there is no feasible region.

Point (4, 0, –1) by option (3) satisfies this plane.

107

A = 4∫

[Expanding along R1 ] 1 9 = [ −3( − k ) − 0 + 1(3k )] 2 ⇒ 18 = 3k + 3k = 6k



Solutions

From the figure, area of the shaded region,

−3 0 1 1 ∴ ∆= 3 0 1 2 0 k 1

18 =3 6

x2

52  x2  y 2 = 16 1 −   25    4 2 2 y = 5 −x 5

17. Option (2) is correct.

∴ k=

= 1−

Z = 7x + y

(3, 0)

21

1 5 2, 2  

6

(7, 0)

49 maximum

(0, 5)

5

108

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

22. Option (1) is correct. Explanation: Assertion (A) and Reason (R) both are correct and Reason (R) is the correct explanation of Assertion (A).



23. Option (3) is correct.

hus, the given function is differentiable R – T  1 .   2

Explanation: Let,

Explanation : We have, P( A ) > 0 and P( B) ≠ 1 P( A′ ∩ B′) P( A′ / B′) = P( B′) 1 − P( A ∪ B ) = P( B′)

 I

 x tan 1

24. Option (4) is correct.



26. Option (2) is correct.



27. Option (2) is correct.

1

x dx

1 1 2  dx 2  (1  x ) x 1 2 x  dx 2 x (1  x )

Put x = t2 ⇒ dx = 2t dt ∴

= x tan −1

t dt t(1 + t 2 ) t2 x −∫ dt 1 + t2 1   x − ∫ 1 − dt + t 2  1 

I = x tan −1 x − ∫ = x tan −1

25. Option (4) is correct. Explanation : In the particular solution of a differential equation, there are no arbitrary constants.

 1  tan

 tan 1 x  x 

Explanation: We know that if A is a square matrix of order m, and if there exists another square matrix B of the same order m, such that AB = BA = I, then B is said to be the inverse of A. In this case, it is clear that A is the inverse of B. Thus, matrices A and B will be the inverse of each other only if AB = BA = I.

28. Option (1) is correct.

= x tan −1 x − x + tan −1 t + C = x tan −1 x − x + tan −1 x + C



= ( x + 1)tan −1 x − x + C

29. Option (2) is correct.

Explanation:    Explanation: a is bisector of b and c Given that,  a = λ ( b + c )  cos   sin   A =  sin  cos        i + j i − j + 4 k  αi + 2 j − βk = λ  +  Also A + A’ = I 18   2  cos   sin    cos  sin    1 0       λ  4i + 2 j + 4 k   sin  cos     sin  cos    0 1  =   3 2   0  1 0  2 cos    0  0 1 2 cos  Compare     4λ 2λ 4λ Equating corresponding entries, we have =α, = 2, = −β 3 2 3 2 3 3 ⇒ 2 cos α = 1  ⇒ α = 4 & β = −4 1 ⇒ cos α =  2 & a = 4i + 2 j + 4 k p ⇒ cos α = cos but from this, no option match. 3 So consider p ∴ α=  3 a = λ ( b − c )  i + j (i − j + 4 k )  = λ −   2 18  

Explanation: Given that, f ( x ) =| 2x − 1|sin x The function sin x is differentiable.

αi + 2 j − βk =

The function | 2x − 1| is differentiable, except 2x − 1 = 0 1 ⇒ x= 2

⇒

⇒

 α =

λ  2i + 4 j − 4 k    3 2  

2λ

,2=

4λ

3 2 3 2  α = 1, β = 2

, −β =

−4 λ 3 2

Solutions

109

 a = i + 2 j − 2 k

So,

which will satisfy option (2).

30. Option (4) is correct. Explanation : Here,

and and and

P( A ) P P(( A A )) P( B ) P P(( B B)) P( A ∩ B ) P ∩B P(( A A∩ B)) P( A′| B) P(( A A′′|| B B)) P

= = = = = = = = = = = = = = = = = = = = =



7 , 13 77 , 13 9 , 13 13 99 13 4 13 13 44 13 P( A′ ∩ B) 13 ∩) B P(( A A B)) P(′′ B ∩ P P ( B ) − P( B ) P P( B ) ( A ∩ B ) −PP P ( B P( B(( A A ) ∩ P( B)) − ∩B B)) P ( B ) 9 4 − P( B ) 99 13 44 13 − − 9 13 13 13 13 99 13 5 13 13 955 99

A =

4



∫  0

4



34. Option (1) is correct. Explanation:

32. Option (1) is correct. Explanation:

x +1 y−3 z = =λ = 2 −2 −1

dy = ex+ y dx = ex ⋅ e y dy = e x dx ey e − y dy = e x dx

⇒ ⇒

Any point on the line R ( 2λ − 1, − 2λ + 3, − λ ) direction ratio’s of PR {2λ − 1 − 1, − 2λ + 3 − 2 , − λ + 3}

Integrating both sides, we get:

= {2λ − 2 , − 2λ + 1, − λ + 3}

−y x ∫ e dy = ∫ e dx

⇒ ⇒

Line PR & given line are perpendicular to each other

−e − y = e x + k

⇒



x x −  dx 2

2 1 x2  =  x3 / 2 − .  2 2   3 0 2 3 2 16 = .(4) − −0 3 4 166 = −4 3 4 = sq. unit 3

31. Option (2) is correct. Explanation: aRb ⇒ a is brother of b. This does not mean b is also a brother of a as b can be a sister of a. Hence, R is not symmetric. aRb ⇒ a is brother of b and bRc ⇒ b is a brother of c. So, a is brother of c. Hence, R is transitive.



From the figure, area of the shaded region,

e x + e − y = −k x

e +e

−y

=C

( where,C = − k )

33. Option (1) is correct. Explanation : When y 2 = x and 2y = x Solving we get y 2 = 2 y ⇒ y = 0, 2 and when y = 2, x = 4 So, points of intersection are (0, 0) and (4, 2). Graphs of parabola y2 = x and 2y = x are as shown in the following figure :

∴( 2λ − 2 )2 + ( −2λ + 1) ( −2 ) + ( −λ + 3)( −1) = 0

λ = 1

⇒ Point R (1, 1, –1)

R is midpoint of P & Q ∴

⇒

1+ a 2+b −3 + c = 1, = 1, = −1 2 2 2

a+b+c = 2

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

110

35. Option (3) is correct.

or When λ = 4 ,

Explanation: We have,

(i + j + 4 k )( 2i + j + k ) = ( 1 + 1 + 16 4 + 1 + 1 ) cos θ

sin x cos x cos x cos x sin x cos x = 0 cos x cos x sin x pplying C1 → C1+C2+C3 A

cosxx++sin sinxx 22cos 2 cos x + cos 2 cos x + cos xx cosxx++cos cosxx 22cos

cosxx cos sinxx sin cosxx cos

cos θ = =

cosxx cos cos cos xx==00 sinxx sin

cos θ =

Ontaking taking( 2( 2cos cosxx++sin sinxx) )common commonfrom from On theCC,1 ,we weget get the 1 cosxx cos cosxx 11 cos Þ( 2( 2cos cosxx++sin sinxx) )11 sin sinxx cos cosxx Þ 1 cos x sin 1 cos x sin xx) )



[RR2® ®RR2--RR1and andRR3® ®RR3--RR]1 ] [ 2 2 1 3 3 1 cosxx cosxx 11 cos cos Æ ( 2 cos ü sin ) 0 sin co s x Þ ( 2 cos x + sin x ) 0 sin x - cos x 00 ==00 (sinxx--cos cosxx) ) 00 00 (sin =0 Expanding along C1,

( 2 cos x + sin x )[1.(sin x − cos x )2 ] = 0

⇒ ( 2 cos x + sin x )(sin x − cos x )2 = 0 2 cos x = − sin x Either 1 ⇒ cos x = − sin x 2 ⇒ tan x = −2 But here for, we get − 1 ≤ tan x ≤ 1. So, no solution is possible and for (sin x − cos x )2 = 0, sin x = cos x p tan x = 1 = tan 4 ...(i)

⇒



p 4 Thus, one distinct real root exists. 36. Option (4) is correct. ∴

x=

Explanation: Volume of Parallelopiped   [u v w] = 1cu.unit 1 1 λ   1 1 3  = ±1    2 1 1  1(1 − 3) − 1(1 − 6 ) + λ (1 − 2 ) = ± 1

λ = 2 or λ = 4     u.w = |u|| w |cos θ

⇒ When

λ = 2,

(i + j + 2 k )( 2i + j + k ) = ( 1 + 1 + 4 4 + 1 + 1 ) cos θ

cos θ =

5 6



7 6 18 7 3 2 23 7 6 3

37. Option (4) is correct. Explanation : We have, P( A ) = 0.4 , P( B) = 0.3 and P( A ∪ B) = 0.5 Now , P( A ∪ B ) = P( A ) + P( B ) − P( A ∩ B ) ⇒ P( A ∩ B) = 0.4 + 0.3 − 0.5 = 0.2 ′ ∴ P( B ∩ A ) = P( A ) − P( A ∩ B ) P( A ) = 0.4 , = 0.4 − 0.2 P( B) = 0.3 = 0.2 and P( A ∪ B) = 0.5 1 = P( A ) + P( B ) − P( A ∩ B ) Now , P( A ∪ B ) = 5 P( A ∩ B) = 0.4 + 0.3 − 0.5 ⇒ = 0.2 P( B′ ∩ A ) = P( A ) − P( A ∩ B) ∴ = 0.4 − 0.2 = 0.2 1 = 5

38. Option (3) is correct. Explanation: I =

5x − 4 x 2

=

1 2∫

=

1 2

∫

=

1 2

∫







dx

∫

dx 5x − x2 4 dx 25  2 5x 25  − x − + 64  4 64  dx 2

5   8  − x −   

= 1 sin −1 2

x− 5 8

8x − 5 1 +C ⇒ sin −1 2 5

5 8

+C

5 8 

2

Solutions

39. Option (2) is correct.

  48   = sin sin −1     50      24   = sin sin −1     25   

Explanation: Given that, 2   cos  sin −1 + cos−1 x  = 0 5   2 + cos−1 x = cos−1 0 ⇒             sin 5     sin −1

⇒       sin −1 ⇒  

2 π + cos−1 x = cos−1 cos 5 2



              

⇒

40. Option (4) is correct.

Shortest distance of (1, –1, 0) from plane x + y + z+ 1 = 0 is 1 + ( −1) + 0 + 1 1 = 2 2 2 3 1 +1 +1

41. Option (1) is correct.

Applying C1→aC1+bC2+cC3, − a + b cos C + c cos B cos C cos B a cos C − b + c cos A −1 cos A −1 a cos B + b cos A − c cos A

Explanation: Given that,

1 f ( x ) = x 2 sin x      Thus, 1  f ( 0 ) = lim  x 2 sin  x →0 x   an oscillating value  ⇒ f (0) = 0 ×    between − 1 and 1 

42. Option (3) is correct.

lso, by projection rule in a triangle, we know A that a = b cos C + c cos B, b = c cos A + a cos C and c = a cos B + b cos A Using above equation in column first, we get − a + a cos C cos B 0 cos C cos B b−b −1 cos A = 0 −1 cos A = 0 c − c cos A −1 0 cos A −1

Explanation: We have,

[ Since, determinant having all elements of any column or row gives value of determinant as zero]

⇒ f (0) = 0



44. Option (1) is correct. Explanation: −1 cos C cos B −1 cos A We have, cos C cos B cos A −1

Explanation: Maximum value of Z + Minimum value of Z = 15 − 32 = −17

λ = 0 x + y + z+1 = 0

Hence plane

2 5

2 5

43. Option (3) is correct.

It is parallel to given line 0(1 + 2λ ) − 1(1 − λ ) + 1(1 + λ ) = 0

π  −1 2 −1 sin 5 + cos x = 2   

 x =



[ sin sin–1 θ = θ]

Explanation: Plane through line of intersection is ( x + y + z + 1) + λ ( 2 x − y + z + 3) = 0

p 2 − sin −1 2 5

−1 −1 ⇒               cos x = cos





2 π + cos−1 x = 5 2

−1          cos x =

24 25 = 0   .96 =

−1

⇒  

111

3  sin[ 2 tan −1( 0.75)] = sin  2 tan −1   4 3    −1 2 ⋅ 4  = sin  sin  9  1 +  16             3    −1 2  = sin sin 25    16          



45. Option (4) is correct. Explanation: Plane ^ to i + j + k and i + 2 j + 3k i j k

1 1 1 = i( 3 − 2 ) − j( 3 − 1) 1 2 3 + k ( 2 − 1) = i − 2 j + k

112

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

Projection of 2i + 3j + k on i – 2j + k is (i − 2 j + k )( 2i + 3 j + k ) 1+ 4 +1

Explanation: Slope of the normal based on the position of the stick −1 = f ′( x )

3 2

=

46. Option (3) is correct. Explanation : We have, 2 P( A ) = 2 , P( A ) = 5 , 53 P( B ) = 3 P( B) = 10 10 1 P( A ∩ B ) = 1 and P( A ∩ B ) = 5 and 5 P( A′ ∩ B′) P( A′ ∩ B′) P( A′ | B′) ⋅ P( B′ | A′) = P( A′ ∩ B′) ⋅ P( A′ ∩ B′) P( A′ | B′) ⋅ P( B′ | A′) = P( B′) ⋅ P( A′)

f ′( x ) = 6 [8 x 3 − 2 x ] f ′( 2 ) = 6 [8 × 8 − 2 × 2] = 6 [64 − 4] = 360 ∴

P( B′) P ( A′ ) 2 ( P(( A ∪ B )′)) = ( P(( A ∪ B )′))2 = P( A′)P( B′) P( A′)P( B′) 2 (1 − P( A ∪ B )) = (1 − P( A ∪ B ))2 = (1 − P( A ))(1 − P( B )) (1 − P( A ))(1 − P( B )) =

(1 − P( A ) + P( B ) − P( A ∩ B ))

 − 1 1   2 =

2

2





−1 360

49. Option (2) is correct. 360 360 dy dy ( x − x′) dx ( x − x′) dx ( x − 2 ) 360 360 ( x − 2 ) 360 x − 7220 360 x − 7220 360 x − 717 360 x − 717

50. Option (2) is correct. Explanation: For increasing f ′( x ) f ′( x ) 3 6 (8x 3 − 2x ) 6 (8x − 2x ) i.e., x( 4 x 22 − 1) i.e., x( 4 x − 1) 4 x 22 − 1 ⇒ ⇒ 4x − 1 and x and x 4 x 22 4x ⇒ x 22 x ⇒

2

3

=

Slope =

Explanation: We have dy  = dy  dx ( 2 , 3 ) = dx ( 2 , 3 ) ∴ ( y − y′) = ∴ ( y − y′) = ( y − 3) = ( y − 3) = y−3 = y−3 = y = y =

(1 − P( A ))(1 − P( B ))

  2 3 1  1− + −    5 10 5   = 3   1  1 − 1 −   2   10 

48. Option (4) is correct.

7 ⋅ 5 10 25

42

47. Option (2) is correct. Explanation: f ( x ) = 6( 2 x 4 − x 2 )

> > > > > > > > > > > > > >

f ′( x ) = 6 [ 8 x 3 − 2 x ]

⇒ ⇒

f ″( x ) = 6 [ 24 x 2 − 2] f ″( 5) = 6 [ 24 × 25 − 2] = 6 [600 − 2] = 3588

x > x >  −1   1  i.e., x ∈  −1 , 0  ∪  1 , ∞  i.e., x ∈  2 , 0  ∪  2 , ∞   2  2  and and

x > x >

0 0 0 0 0 0 0 0 0 0 1 1 1 1 4 41 1 2 21 −1 −2 2

Section - B2



16. Option (1) is correct.



17. Option (1) is correct. Explanation: Since 2345 = 45(mod 100) and 6789 ≡ 89(mod 100) So, 2345 × 6789 ≡ 45 × 89 (mod 100) ≡ 4005 (mod 100) ≡ 05 (mod 100) Hence, the last two digits of the product 2345 × 6789 are 0 and 5.



18. Option (2) is correct. Explanation: Given, a−b 8 = 2 .5 6 . 25 6.25 × 8 a – b = 2.5 ⇒ a – b = 20 ⇒ a = 20 + b Hence, a > b or b < a

Solutions

19. Option (4) is correct. Explanation: A : B : C = 1000 : (1000 – 50) : (1000 – 69) = 1000 : 950 : 931 In a 950 m race, B can give C a start of (950 – 931) m = 19 m In a 1000 m race. B can give C a start of  19   950 × 1000  = 20 m  



=

Time taken by leakage to empty the tank = 120 h

21. Option (3) is correct. Explanation: Since Vijay is faster by 4 seconds. ∴

He beats Samuel by =

100 ×4 16

= 25 meters

25. Option (3) is correct. Explanation:

Explanation: 5811 = (5 ×11)mod 8 = 55 mod 8 = 7

24. Option (1) is correct. Explanation: Since A is a matrix of type 3 × 5, therefore, each row of A contains 5 elements. Hence, R is a row matrix of the type 1 × 5.



20. Option (3) is correct.



AA′ = [1 2 3] [1 2 3]′



1   = [1 2 3]  2   3 

= [1 × 1 + 2 × 2 + 3 × 3] = [14]

26. Option (1) is correct. Explanation:

22. Option (2) is correct.

3 − 2  4 1 

Explanation: Let total capital be = x &

A + I = 



let C’s contribution = y, B’s contribution =

x , 3



x

A’s Contribution = + y. 3

1 0 A+  = 0 1

⇒ x = 6y

3 − 2 4 1     3 − 2 1 0 −   4 1  0 1

⇒

A = 

⇒

A = 

Now (A + B + C)’s contribution = x Hence their contributions are

2 − 2  4 0 

 2 − 2 1 0  –    4 0  0 1

2y + y : 2y : y

A – I = 

i.e., in the ratio 3 : 2 : 1.

1 120

23. Option (4) is correct.

1 − 2

=   4 − 1

Explanation: Portion of cistern filled by both pipes in 1 hour

3 − 2 1 − 2   4 1  4 − 1

=

1 1 + 8 12

∴ (A + I)(A – I) = 

=

5 . 24

= 

 3 − 8 − 6 + 2  4 + 4 − 8 − 1

Time taken by both pipes to fill the cistern

− 5 − 4

= 4 h 48 minutes

=    8 − 9

Time taken to fill tank due to leakage = 5 h Work done by leakage in 1 h

5 1 = − 24 5



27. Option (2) is correct. Explanation:



1 λ − 1 − λ  

A = 

113

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

114

A2 = 0

1 λ ⇒  − 1 − λ  

⇒ 2 + a ≤ 2x + a ≤ 4 + a ⇒ 2 + a ≤ f ′(x) ≤ 4 + a For f(x) to be strictly increasing on [1, 2], 2 + a≥0 ⇒a≥–2 Hence, the least value of a = –2.

1 λ  −1 − λ   

λ 2 − 1 0   =   0 −1 + λ 2 



=0 ⇒ λ2 – 1 = 0 ⇒ λ2 = 1 λ = ± 1

32. 33. 34. 35.

28. Option (4) is correct. f ′(x) =

=

=

=

Option (3) is correct. Option (3) is correct. Option (3) is correct. Explanation: Statistical hypothesis is some assumption or statement, which may or may not be true, about a population.

Explanation: Apply the quotient rule:

Option (3) is correct.

g ′( x )h( x ) − g ( x )h′( x )



h( x )2

36. Option (3) is correct. Explanation: The optimal value of the objective function is attained at the corner points of feasible region.

10 x( x + 47 ) − 5x 2 (1) ( x + 47 )2

37. Option (2) is correct. 38. Option (4) is correct.

10 x 2 + 470 x − 5x 2 ( x + 47 )2

Explanation: Here (0, 2), (0, 0) and (3, 0) all are vertices of feasible region.

5x 2 + 470 x ( x + 47 )2

29. Option (2) is correct. Explanation: Using the product rule,

d dv du (uv ) = u + v dx dx dx 

...(1)

u = x3 v = ln x



39. Option (4) is correct.

du = 3x2 dx dv 1 = dx x

Explanation: New index 4

∑i =1 (Relative index)i =

Substituting into equation (1), =

dy 1 = x 3 × + ln x × 3x 2 dx x

= x2 + 3x2 ln x

=

30. Option (4) is correct. =

Explanation: The derivative of a constant function is always zero.

=

31. Option (2) is correct. Explanation: Given, f(x) = x2 + ax + 1, x ∈ [1, 2] Differentiating w.r.t. x, we get f ′(x) = 2x + a Now, x ∈ [1, 2] ⇒ 1 ≤ x ≤ 2 ⇒ 2 ≤ 2x ≤ 4

× (Index no. Weight)i 4

∑i =1 (weight)i

181 × 4 + 116 × 12 + 110 × 3 + 152 × 7 4 + 12 + 3 + 7 724 + 1392 + 330 + 1064 26 3510 26 1755 13

= 135

40. 41. 42. 43.

Option (3) is correct. Option (3) is correct. Option (3) is correct. Option (3) is correct.

Solutions Explanation: There are 4 phases through which trade cycles are passed. They are prosperity, recession, depression, and recovery. In economic terms, these 4 stages are called economic fluctuations.



Explanation: Positive rate of return represents profit where as negative rate of return shows loss.

45. Option (2) is correct. 46. Option (1) is correct.



47. Option (2) is correct. Explanation: P(one defective blade) = P(X = 1) e −0.02 ( 0.02 )1 = 1! = 0.02e–0.02

=

e −0.02 ( 0.02 )2 2!

=

0.9802 × 0.0004 2

= 0.000196 = 0.0002

49. Option (1) is correct. Explanation: P (no defective blade) = e–0.02 = 0.9802 So, the approximate number of packets containing no defective blade = 10,000 × 0.9802 = 9802

Explanation: Here given N = 1000, n = 10, 1 p= = 0.002 500 ⇒ m = np = 10 × 0.002 ⇒ m = 0.02 P(No defective blade) = P(X = 0) = e–0.02 (0.002)0 = e–0.02

48. Option (1) is correct. Explanation: P(two defective blade) = P(X = 2)

44. Option (1) is correct.



115



50. Option (2) is correct. Explanation:  P(one defective blade) = 0.02 e–0.02 = 0.02 × 0.9802 = 0.0196 So, the approximate number of packets containing 1 defective blade = 20,000 × 0.0196

= 392. qqq

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

116

5

SOLUTIONS OF Question Paper Section - A



1. Option (1) is correct.

R1 = {(1,1) , ( 2, 2 ) , ( 3, 3)}

R2 = {(1,1) , ( 2, 2 ) , ( 3, 3) , (1, 2 ) , ( 2,1)}

Explanation: We have,

R3 = {(1,1) , ( 2, 2 ) , ( 3, 3) , (1, 3) , ( 3,1)}

cos t t 1 f (t ) = 2 sin t t 2t , sin t t t



R4 = {(1,1) , ( 2, 2 ) , ( 3, 3) , ( 2, 3) , ( 3, 2 )}

{

= cos t (t 2 − 2t 2 ) − 2 sin t (t 2 − t ) + sin t ( 2t 2 − t ) = −t 2 cos t − (t 2 − t ) 2 sin t + ( 2t 2 − t ) sin t



t →0

f (t ) ( −t 2 cos t ) t sin t = lim + lim 2 2 t →0 t →0 t t2 t sin t = − lim cos t + lim t →0 t →0 t = −1 + 1

x →0

sin t   = 1 and cos 0 = 1  lim t →0 t  

=0



− x ,if x < 0 . x =  x , if x ≥ 0 This is the definition for modulus function and hence true. Hence R is true. Since f is continuous at x = 0, lim− f ( x ) = lim+ f ( x ) = f ( 0 )



∴ lim

4. Option (1) is correct. Explanation:

= – t 2 cos t − t 2 .2 sin t + t.2 sin t + 2t 2 sin t – t sin t = −t 2 cos t + t sin t



Here

kx kx = lim = −k | x | x →0 − − x − k = 3 or k = −3. = lim− x →0

∴

Explanation : P( A ) = P( A ∩ B ) = ∴

x →0

f ( 0 ) = 3, LHL = lim− f ( x ) x →0

2. Option (3) is correct.



}

R5 = (1, 2, 3) ⇔ A × A = A 2 \ Maximum number of equivalence relation on the set A = {1, 2, 3} = 5

Expanding along C1,

P( B | A ) =

=

=

4 , 5 7 10 P( A ∩ B ) P( A ) 7 10 4 5 7 8

3. Option (4) is correct. Explanation: Given that, A = {1,2,3} Now, number of equivalence relations are as follows:

Hence, A is true. R is the correct explanation of A.

5. Option (3) is correct. Explanation: Since sin x and |x| are continuous functions in R, |sin x| is continuous at x = 0. Hence A is true. − sin x , if x < 0 sin x =  if x ≥ 0 sin x , f ( 0 ) = |sin 0 | = 0 − sin x − 0 LHD = f ′( 0 − ) = lim x →0 x = −1 sin x − 0 RHD = f ′( 0 + ) = lim x →0 x =1

Solutions Using this property, we have

At x = 0, LHD ¹ RHD. So f(x) is not differentiable at x = 0. Hence, R is false.

2   2x   −1 −1  2x  −1  1 − x  tan −1  = = =  2 tan x sin   1 + x 2  cos 2  2     1 − x   1+ x  

6. Option (1) is correct.



Explanation:

A–1 =



3 10





adj A A

|A| = = 21 – 20 = 1 2 7  7 − 10  adj A =  −2 3    7 − 10  adj A A–1 = =  3  | A| − 2

⇒

Explanation :

(0, 3) (1, 1) (3, 0)

Corresponding value of Z = px + qy; p, q > 0 3q p+q 3p

So, condition of p and q, so that the minimum of Z occurs at (3, 0) and (1, 1) is p + q = 3p ⇒ 2p = q q ∴ p = 2

8. Option (4) is correct.

2 × tan −1

2a 2x = tan −1 1 − a2 1 − x2

 2a  2 1 − a 2   2x  ⇒ tan −1  = tan −1  2 2  1− x   2a  1−  2  1− a   2a  ⇒ x  1 − a2  =  

9. Option (1) is correct. Explanation: The repeated selection of students who are swimmers are Bernoulli trial. Let X denotes the number of students, out of 5 students, who are swimmers. Probability of students who are not swimmers, 1 q= 5 1 4 ∴ p =1 − q =1 − = 5 5 Clearly, X has a binomial distribution with n = 4 5 and p = . 5

7. Option (2) is correct.

Corner points



⇒

 3 10  A=  2 7  



117

1 Cx   5 (4 students are swimmers) P P(X= x= )

x Cx q n − x p=

n

1 4 C4   .   5  5

4 .  5

x

4

5

=P(X= 4= )



5− x

5

10. Option (2) is correct. 1, which is a Explanation : We have, x 2 + y 2 = circle having centre at (0, 0) and radius ‘1’ unit. ⇒

y2 = 1 − x2 y = 1 − x2

Explanation: We have,  1 − a2   2a   2x  + cos−1  = sin −1  tan −1  2  2  2  + + 1 a 1 a   1− x    Let we assume that a = tanθ ⇒ θ = tan-1a 2    2 tan θ   2x  −1 1 − tan θ sin −1  + cos = tan −1      1 + tan 2 θ   1 + tan 2 θ   1 − x2   

 2x  ⇒ sin −1 sin 2θ + cos−1 cos 2 θ = tan −1    1 − x2   2x  ⇒ 2θ + 2θ = tan −1    1 − x2   2x  tan −1  ⇒ 4 tan −1 a =   1 − x2   2x  tan −1  ⇒ (2 × 2)tan −1 a =   1 − x2 



From the figure, area of the shaded region, 1

A = 4 ∫ 12 − x 2 dx 0

x 12 x = 4  12 − x 2 − sin −1  2 1   2   12 π = 4 0 + × − 0 − 0  2 2   = π sq. units

1

0

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

118

11. Option (2) is correct. Explanation: Given :  ( x + 3)2 − 36 , x≠3  f(x) =  x−3  k, x = 3  and f(3) = k But it is a continuous function, so lim f ( x ) = k x →3 2



2



[ Using a2 – b2 = (a+b) (a – b)] ⇒ lim ( x + 9 ) = k x →3 ⇒ k = 3 + 9 = 12



12. Option (4) is correct.

 −1 −1  x   sin ( xπ ) tan    1  π  A=   π  −1  x  −1  sin   cot ( πx ) π    



and





 −1 −1  x    − cos ( xπ ) tan    1 π  B=   π x   −1 −1  sin   − tan ( πx ) π    

1 −1 −1  sin xπ + cos xπ π A−B=   1  −1 x x − sin −1    sin π π  π

(



1  π     π 2 =     0  =



 0  π  2

1 x x  tan −1 − tan −1   π  π π   1 cot −1 πx + tan −1 x  π 

(

P(X)

1 1 0    2 0 1 

1

2

5

1/2

1/3

1/6

Mean = E( X )

)

= ∑ pi xi



1 1 1 = ×1 + ×2 + ×5 2 3 6 1 2 5 = + + 2 3 6 3+ 4 + 5 12 = = =2 6 6



Section - B1

16. Option (1) is correct. Explanation: The variables used in a Linear Programming Problem are called decision variables.



15. Option (2) is correct.

X

π   −1 −1  sin x + cos x = 2 and    tan −1 x + cot −1 x = π    2



)

14. Option (3) is correct.

Explanation: Let X be the random variable representing a number on the die. The total number of observations is 6. Therefore, 3 1 P(X = 1) = = 6 2 2 1 P (X = 2 ) = = 6 3 1 P (X = 5 ) = 6 Therefore, the probability distribution is as follows.

Explanation: We have,



13. Option (1) is correct.

Explanation: Given A = {1, 2, 3} \ n(A) = 3 and B = {x, y} \ n(B) = 2 \ The number of function from A to B is = 23 = 8

⇒ lim ( x + 3 + 6 ) − ( x + 3 − 6 ) = k x →3 x−3





1 I 2

Explanation: order of P = 3 × n order of Q = n × p since the number of columns of P is equal to number of Rows of Q order of PQ is 3 × P

\ lim ( x + 3) − ( 6 ) = k x →3 x−3



=

1 1 1 1 1+sinθ 1 1+cosθ 1 1 [Applying C1→C2–C3 and C2→C1–C3] ∆=

17. Option (1) is correct. Explanation: Given that,

=

0 0 1 0 sin θ 1 cos θ 0 1

= − sin θ.cos θ 1 = − .2 sin θ.cos θ 2

=

0 0 1 0 sin θ 1 cos θ 0 1

Solutions

= − sin θ.cos θ 1 = − .2 sin θ.cos θ 2 1 = − sin 2θ 2

Given that, ⇒

Explanation: Given : x≤2 5, f(x) = ax + b , 2 < x < 10  21, x ≥ 10 

d2y

1/4

 dy  + 2  dx  dx  

= −x1 / 5 = −x1 / 5

 dy   dx   

1/2

 d2y  =  x1 / 5 + 2   dx   Again, on squaring both sides, we have

2

dy  1 / 5 d 2 y  = x + 2  dx  dx 

4

 dy   dx   

Here, lim+ f ( x ) = lim+ ( ax + b ) x →2 x →2 =2a + b and lim− f ( x ) = 5(given) x →2 But f(x) is continuous, then lim f ( x ) = lim− f ( x ) 2+ x →2 \ 2a + b = 5 Similarly, lim − f ( x ) = lim + f ( x )



1/4

 dy  + 2  dx  dx  

 d2y  = −  x1 / 5 + 2   dx   On squaring both sides, we get ⇒

Order = 2, degree = 4

21. Option (2) is correct. Explanation: Given that, aRb if a ≥ b ⇒ aRa ⇒ a ≥ a which is true Let aRb, a ≥ b, then b ≥ a which is not true as R is not symmetric. But aRb and bRc ⇒ a ≥ b and b ≥ c ⇒ a≥c

...(i)

x →10

i.e., 10a + b = 21 From eq. (i) and (ii), we get a =2 and b = 1 \ a + b = 2 + 1 = 3.

d2y

1/4

18. Option (4) is correct.

x →10

20. Option (1) is correct. Explanation :

1 So, maximum value of ∆ is when sin 2θ = −1 2

119

...(ii)

19. Option (1) is correct.

Hence, R is transitive.

Explanation:



22. Option (1) is correct.

  Explanation: Projection vector of a on b is given by,    b ·b =a⋅  b   = a ⋅   



From the figure, area of the shaded region, 3

A = ∫ ( x + 1) dx 2

3

 x2  =  + x 2   2 9 4   =  +3 − − 2  2 2  7 = sq. units 2

 b  b

  ⋅b   

23. Option (4) is correct. Explanation: Let X denotes the number of aces obtained. Therefore, X can take any of the values of 0, 1, or 2. I n a deck of 52 cards, 4 cards are aces. Therefore, there are 48 non-ace cards. ∴ P (X = 0) = P (0 ace and 2 non-ace cards) C0 ×48 C2 1128 = 52 C2 1326 P (X = 1) = P (1 ace and 1 non-ace cards) =       



4

C1 ×48 C1 192 = 52 C 1326          2 =

4

P (X = 2) = P (2 ace and 0 non-ace cards)

OSWAAL CUET(UG) Sample Question Papers,

120

=

4

C2 ×48 C0 6 = 52 C2 1326

     Thus, the probability distribution is a follows : X P(X)

1

2

3

1128/1326

192/1326

6/1326

dy 66dx x 3x dy = 6xx == 33xx == Slope Slope of the the curve dy = dx = 22yy = yy = Slope of of the curve curve dx dx 2 y y Now, slope of normal to the curve Now, Now, slope slope of of normal normal to to the the curve curve 1 1 y 1 1 = − = − = −− yy 1  = −  1MATH. == −−  dy = MATHEMATICS/APP.  dy  = −  33xx  = − 333xxx dy  dx dx   3yx   dx   yy  y 1   ∴ −− yy  == −−11 ∴∴   ∴ −  333xxx  = − 333  − 3y = −3x ⇒ −− 33yy == −−33xx ⇒⇒ ⇒ ⇒ yyy === xxx ⇒⇒ ⇒ n substituting the value of the given equation O of the curve, we get 3x2 – x2 = 8 ⇒ 2x2 = 8 ⇒ x2 = 4 ⇒ x = ±2 For x=2 3(2)2 – y2 = 8 ⇒ y2 = 4 ⇒ y = ±2 and for x = –2, 3(2)2 – y2 = 8 ⇒ y2 = 4 ⇒ y = ±2 So, the points at which normal is parallel to the given line are (±2, ±2). Hence, the equation of normal at (±2, ±2) is

Then, E(X ) = ∑ pi xi

1128 192 6 + 1× + 2× 1326 1326 1326 204 2 = = 1326 13 = 0×





24. Option (2) is correct. Explanation: Plane passes through A, B, C Let A (a, 0, 0), B (0, b, 0), Centroid (1, 1, 2) a+0+0 0+b+0 1+ 1= , , 3 3 ⇒

a=3

So plane

⇒ b=3

C (0, 0, c) 2=

0+0+c 3

⇒ c=6

x y z + + = 1 3 3 6

⇒

2x + 2 y + z = 6

⇒

line through centroid (1, 1, 2) on ^ to plane 2x + 2 y + z = 6 So direction ratio's of line {2, 2, 1} x −1 y −1 z − 2 line = = 2 2 1

∴

1 y − ( ±2) =1− [ x − ( ±2)] y − ( ±2) = − 1[ x3 − ( ±2)] y − ( ±2) = −3 [ x − ( ±2)] 3[ y − ( ±2)] = 3 −[ x − ( ±2)] 3[ y − ( ±2)] = −[ x − ( ±2)] 3[xy+−3(y±2±)]8== −0[ x − ( ±2)] x + 3y ± 8 = 0 x + 3y ± 8 = 0

26. Option (3) is correct. Explanation: We know that, in a square matrix, if bij, when i ≠ j then it is said to be a diagonal matrix. Here, b12, b13…. ≠ 0 so the given matrix is not a diagonal matrix.  0 −5 8  Now, B =  5 0 12   −8 −12 0 

25. Option (3) is correct. Explanation: We have, the equation of the curve is 3x 2 − y 2 = 8  ….(i) Also, the given equation of the line is x + 3y = 8 . ⇒ 3y = 8 − x

 0 5 −8  B ' =  −5 0 −12   −8 12 0 

x 8 ⇒ y=− + 3 3

1 which should be 3 equal to slope of the equation of normal to the

Thus, slope of the line is − curve.

On differentiating equation (i) with respect to x, we get dy 6x − 2 y =0 dx dy 6x 3x = = = Slope of the curve ⇒ dx 2 y y Now, slope of normal to the curve

1 1 y =− =− =− dy 3x    3x   dx      y   1  y  ∴ −  = − 3  3x  ⇒ − 3y = −3x ⇒ y=x



                   

 0 −5 8   = −  5 0 12   −8 −12 0  = −B

o, the given matrix is a skew-symmetric S matrix, since we know that in a square matrix B, if B’ = −B, then it is called skew-symmetric matrix.

27. Option (3) is correct. Explanation : 2 2 2     Here, a + b + c = 0 and a = 4, b = 9, c = 25        ∴ (a + b + c ) ⋅ (a + b + c ) = 0             ⇒ a2 + a ⋅ b + a ⋅ c + b ⋅ a + b 2 + b ⋅ c + c ⋅ a +     c ⋅b + c2 = 0          ⇒ a 2 + b 2 + c 2 + 2( a ⋅ b + b ⋅ c + c ⋅ a ) = 0     [∵ ∵ a ⋅ b = b ⋅ a]

       ∴ (a + b + c ) ⋅ (a + b + c ) = 0             ⇒ a2 + a ⋅ b + a ⋅ c + b ⋅ a + b 2 + b ⋅ c + c ⋅ a +     c ⋅b + c2 = 0          ⇒ a 2 + b 2 + c 2 + 2( a ⋅ b + b ⋅ c + c ⋅ a ) = 0     [∵ ∵ a ⋅ b = b ⋅ a]       4 + 9 + 25 + 2( a ⋅ b + b ⋅ c + c ⋅ a ) = 0 ⇒ −38       ⇒ a ⋅b + b ⋅c + c ⋅a = 2 = −19

28. Option (1) is correct.

Solutions

30. Option (4) is correct. Explanation: Putting (0,2) is 4x+ 6y = 4 × 0 + 6 × 2 = 12 Putting (3, 0) is F = 4 × 3 + 6 × 0 = 12 Putting (6, 0) is F = 4 × 6 + 6 × 0 = 24 Putting (6, 8) is F = 4 × 6 + 6 × 8 = 72 Putting (0, 5) is F = 4 × 0 + 6 × 5 = 30 Hence minimum value of F occurs at (0, 2) and (3, 0)



31. Option (3) is correct. Explanation: Given that,

Explanation: Let E1 and E2 be the events such that, E1 : A speaks truth E2 : A speaks false



Let X be the event that a head appears.



P( E1 ) =

4 1 = 5 5 If a coin is tossed, then it may result in either head (H) or tail (T).



0 0 0 0 a−b a−b ⇒ f ( a) = a + a 0 0 a − c = 2a a−c 0 0 a+b a+c a+b a+c

P( E2 ) = 1 − P( E1 ) = 1 −

= [( a − b ){2 a.( a + c )}] ≠ 0

he probability of getting a head is 1/2, whether T A speaks truth or not.

0 0 0 b−a b−b b−a ∴ f (b ) = b + a 0 0 b−c = b+a b−c 0 2b b + c 0 b+b b+c

1 2 The probability that there is actually a head is given by,





∴ P( X | E1 ) = P( X | E2 ) =

P( E1 ).P( X |E1 ) P( E1 ).P( X |E1 ) + P( E2 ).P( X |E2 ) 4 1 1 4 . . 5 2 = = 2 5 4 1 1 1 14 1 . + . + 5 2 5 2 2  5 5  4 4 =5= 1 5 29. Option (1) is correct.

= (b − a)[ 2b(b − c )] = 2b(b − a )(b − c ) ≠ 0 0 − a −b ∴ f ( 0) = a 0 − c b c 0

P( E1 | X ) =

Explanation: Given that A = {1, 2, 3} and R = {1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1,3)}.  (1, 1), ( 2, 2), (3, 3) ∈ R Hence, R is reflexive. ( 1, 2) ∈ R but ( 2, 1) ∉ R Hence, R is not symmetric. (1, 2) ∈ R and ( 2, 3) ∈ R ⇒ (1, 3) ∈ R Hence, R is transitive.



0 x−a x−b f (x) = x + a 0 x−c x+b x+c 0

0 x−a x−b 0 f (x) = x + a x−c 0 x+b x+c

4 5

Therefore,

121





= a (bc) – b(ac)   = abc – abc   = 0

32. Option (3) is correct. Explanation: Given that, ⇒ ⇒

xdy − ydx = 0 xdy = ydx dy dx = y x

On integrating both sides, we get ⇒ ⇒

log y = log x + log C log y = log Cx y = Cx

which is a straight line passing through the origin.

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

122

33. Option (3) is correct.



Explanation: If A ⊂ B, then A ∩ B=A ⇒ P (A ∩ B ) = P (A ) Also, P(A )1

LHD at x = 0,

2

2

x≤0 0 < x ≤1

2 x x ≤ 0  0 < x ≤1 f ′( x ) = 1 2 x x > 1 

e=x

⇒

165

= – ve

1/e Maximum value y = e

x →1

Not differentiable at x = 1 OR By graph, there are two sharp point at x = 0, 1 so not differentiable.

f(x) = max{x, x2} y

x=e x'

0 y'

1

x

166

OSWAAL CUET(UG) Sample Question Papers, MATHEMATICS/APP. MATH.

3. Option (4) is correct.



Explanation: We have, S = {B, B, B), (G, G, G), (B, G, G), (G, B, G), (G, G, B), (G, B, B), (B, G, B), (B, B, G)} E1 = Event that a family has at least one girl, then E1 = {(G, B, B), (B, G, B), (B, B, G), (G, G, B), (B, G, G), (G, B, G), (G, G, G)} E2 = Event that the eldest child is a girl, then E2 = {(G, B, B), (G, G, B), (G, B, G), (G, G, G)} ( G, B, B ) , ( G, G, B ) , ( G, B, G ) ,  ∴ E1 ∩ E2 =   ( G, G, G )  4 P( E1 ∩ E2 ) 8 4 P( E2 | E1 ) = ∴ = = 7 7 P( E1 ) 8





Explanation: We have, P(A) = 0.4, P(B) = 0.3, and P(C) = 0.2

∴ Probability of two hits = P( A ) ⋅ P( B) ⋅ P(C′) + P( A ) ⋅ P( B′) ⋅ P(C ) + P( A′) ⋅ P( B) ⋅ P(C ) = 0.4 × 0.3 × 0.8 + 0.4 × 0.7 × 0.2 + 0.6 × 0.3 × 0.2 = 0.096 + 0.056 + 0.036 = 0.188

7. Option (2) is correct. Explanation:

∫x

2

dx dx = + 2x + 2 ∫ x 2 + 2x + 1 + 1

(

=∫

4. Option (3) is correct. Explanation: R = {(a, b) : a = b − 2, b > 6} Now, Since b > 6, (2, 4) ∉ R Also, as 3 ≠ 8 − 2, (3, 8) ∉ R ∴ And, as 8 ≠ 7 – 2 (8, 7) ∉ R ∴ Now, consider (6, 8), We have 8 > 6 and also, 6 = 8 − 2. ∴ (6, 8) ∈ R



6. Option (2) is correct.



8. Option (3) is correct. Explanation:

⇒

dy e x (1 + y ) dx (e x + 1) y = x ⇒ = dx (e + 1) y dy e x (1 + y ) x

dx e y y = + dy e x (1 + y ) e x (1 + y ) dx y y = + dy 1 + y (1 + y )e x

⇒ ⇒ ⇒

dx y  1 = 1 +  dy 1 + y  e x 

⇒

dx y  ex + 1  =   dy 1 + y  e x 

 ex   y   dx   dy =  x 1+ y   e +1 On integrating both sides, we get y ex = dy ∫ 1 + y ∫ 1 + e x dx ⇒

⇒

1+ y −1 ex = dy ∫ 1+ y ∫ 1 + e x dx

⇒ ∫1dy − ∫

1 ex dy = ∫ dx 1+ y 1 + ex

⇒ y − log | (1 + y ) = log|(1+e x ) + log k ⇒ ⇒

y = log(1 + y ) + log(1 + e x ) + log(k ) y = log {k (1 + y )(1 + e x )}

1 dx ( x + 1)2 + (1)2

= tan −1 ( x + 1) + C

p −1  4 + tan x −1 ≤ x ≤ 1  f (x) =  x < −1 or  1 (| x | −1)  2 x >1 1 x < −1  2 ( − x − 1)  π =  + tan −1 x −1 ≤ x ≤ 1 4 1 x >1  2 ( x − 1) 

5. Option (3) is correct. Explanation: Given differential equation (e x + 1) ydy = ( y + 1)e x dx

)

LHL at x = −1 lim

1

x →−1 2

( − x − 1) = 0

RHL at x = −1 π

lim

x →−1 4

+ tan −1 x = 0 π + tan −1 ( −1) = 0 4 Continuous at x = –1

f ( −1) = LHL at x = 1 lim

π

x→1 4

+ tan −1 1 =

π π π + = 4 4 2

RHL at x = 1 1 lim ( x − 1) = 0

x →1 2

Not continuous at x = 1  1 x < −1 − 2   1 f ′( x ) =  −1 ≤ x ≤ 1 2 1 + x 1 x >1  2

Solutions

∴f1(x)=cos x is strictly decreasing in interval

LHD at x = −1 −1  −1  lim   = x→−1  2  2 RHD at x = −1 lim

x→−1

1 1 1+  2

2

≠

−1 2

Þ (y, x) Î R \ R is symmetric

10. Option (4) is correct.

⇒

π  −2 sin 2x < 0 on  0,  f2 '( x ) = 2  is strictly decreasing at interval ∴ f2 ( x ) = cos 2x

 π  0, 2   

(3) Let f3(x)=cos3x ∴ f3 '( x ) = −3 sin 3x Now, f3 '( x ) = 0 ⇒ sin 3x = 0 ⇒ 3x = π π  x ∈  0,  2  π x= 3

⇒

π π divides the interval x = into 3 3  π π π two disjoint intervals, i.e.,  0,  and  ,   3 3 2 . The point x =

Explanation:  −π π  The principal value branch of sin–1x is  ,   2 2 Let x = sin q Þ q = sin–1x sin–1(sin q) = sin–1 x = q  −π π  sin–1(sin q) = q, if q Î  ,  .  2 2 Hence R is true.

2 π  −π π  2π  2π  ∉ , sin −1  sin  ≠ , since 3  2 2  3  3  Hence A is false.

11. Option (2) is correct. Explanation: Value of Z

(0, 0) 0 (5, 0) 15 (6, 5) – 12 (6, 8) – 14 (4, 10) – 28 (0, 8) – 32 (Mini.) Hence, the maximum of z occurs at (0, 8).

0