Engineering Mathematics Exam Prep. Problems and Solutions 9781683929109

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Engineering Mathematics Exam Prep

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license, disclaimer of liability, and limited warranty By purchasing or using this book (the “Work”), you agree that this license grants permission to use the contents contained herein but does not give you the right of ownership to any of the textual content in the book or ownership to any of the information or products contained in it. This license does not permit uploading of the Work onto the Internet or on a network (of any kind) without the written consent of the Publisher. Duplication or dissemination of any text, code, simulations, images, etc. contained herein is limited to and subject to licensing terms for the respective products, and permission must be obtained from the Publisher or the owner of the content, etc., in order to reproduce or network any portion of the textual material (in any media) that is contained in the Work. Mercury Learning and Information (“MLI” or “the Publisher”) and anyone involved in the creation, writing, or production of the companion disc, accompanying algorithms, code, or computer programs (“the software”), and any accompanying Web site or software of the Work, cannot and do not warrant the performance or results that might be obtained by using the contents of the Work. The author, developers, and the Publisher have used their best efforts to ensure the accuracy and functionality of the textual material and/or programs contained in this package; we, however, make no warranty of any kind, express or implied, regarding the performance of these contents or programs. The Work is sold “as is” without warranty (except for defective materials used in manufacturing the book or due to faulty workmanship). The author, developers, and the publisher of any accompanying content, and anyone involved in the composition, production, and manufacturing of this work will not be liable for damages of any kind arising out of the use of (or the inability to use) the algorithms, source code, computer programs, or textual material contained in this publication. This includes, but is not limited to, loss of revenue or profit, or other incidental, physical, or consequential damages arising out of the use of this Work. The sole remedy in the event of a claim of any kind is expressly limited to replacement of the book, and only at the discretion of the Publisher. The use of “implied warranty” and certain “exclusions” vary from state to state and might not apply to the purchaser of this product.

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Engineering Mathematics Exam Prep Problems and Solutions

A. Saha, PhD D. Dutta, PhD S. Kar, PhD P. Majumder, PhD A. Paul, PhD

Mercury Learning and Information Boston, Massachusetts

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Reprint and Revision Copyright ©2023 by Mercury Learning and Information, An Imprint of DeGruyter, Inc. All rights reserved. Original Copyright © 2021 Khanna Publishers. This publication, portions of it, or any accompanying software may not be reproduced in any way, stored in a retrieval system of any type, or transmitted by any means, media, electronic display or mechanical display, including, but not limited to, photocopy, recording, Internet postings, or scanning, without prior permission in writing from the publisher. Publisher: David Pallai Mercury Learning and Information 121 High Street, 3rd Floor Boston, MA 02210 [email protected] www.merclearning.com 800-232-0223 A.Saha et al. Engineering Mathematics Exam Prep: Problems and Solutions. ISBN: 9781683929109 The publisher recognizes and respects all marks used by companies, manufacturers, and developers as a means to distinguish their products. All brand names and product names mentioned in this book are trademarks or service marks of their respective companies. Any omission or misuse (of any kind) of service marks or trademarks, etc. is not an attempt to infringe on the property of others. Library of Congress Control Number: 2023939598 212223 321  Printed on acid-free paper in the United States of America. Our titles are available for adoption, license, or bulk purchase by institutions, corporations, etc. For additional information, please contact the Customer Service Dept. at 800-232-0223(toll free). All of our titles are available in digital format at academiccourseware.com and other digital vendors. The sole obligation of Mercury Learning and Information to the purchaser is to replace the book, based on defective materials or faulty workmanship, but not based on the operation or functionality of the product.

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Contents CHAPTER 1: LINEAR ALGEBRA 1 1.1  MATRICES AND THEIR TYPES 1 1.1.1  Definition of a Matrix 1 1.1.2  Types of Matrices 1 1.2  ALGEBRA OF MATRICES 2 1.2.1 Negative, Sum, and Differences of Matrices  2 1.2.2  Multiplication of a Matrix by a Scalar 3 1.2.3  Transpose of a Matrix3 1.2.4 Multiplication of Matrices (Product of Matrices)3 1.3 DETERMINANT OF A SQUARE MATRIX 4 1.3.1  Definition of Determinant4 1.3.2  Properties of a Determinant4 1.3.3  Minors and Cofactors6 1.4 ADJOINT AND INVERSE OF A MATRIX 6 1.4.1  Adjoint of a Matrix6 1.4.2  Inverse of a Matrix7 1.5 VARIOUS TYPES OF REAL SQUARE MATRICES 7 1.5.1 Symmetric Matrix7 1.5.2 Skew-Symmetric Matrix7 1.5.3 Orthogonal Matrix8 1.5.4 Idempotent Matrix8 1.5.5 Involutary Matrix8 1.5.6 Nilpotent Matrix8 1.6 COMPLEX MATRICES AND THEIR TYPES 8 1.6.1 Complex Conjugate of a Matrix  8 1.6.2  Transposed Conjugate of a Matrix8 1.6.3  Unitary Matrix 9 1.6.4 Hermitian Matrix9 1.6.5 Skew-Hermitian Matrix9

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vi • Contents 1.7  RANK OF A MATRIX 9 1.7.1  Elementary Transformations 9 1.7.2 Equivalent Matrices10 1.7.3  Rank of a Matrix10 1.7.4  Determination of the Rank of a Matrix10 1.8 SYSTEM OF LINEAR EQUATIONS AND THEIR SOLUTIONS 11 1.8.1 Introduction 11 1.8.2 Methods for Solving Non-Homogeneous System of Linear Equations 11 1.8.2.1  Cramer’s Rule 11 1.8.2.2  Matrix Method 13 1.8.2.3  Rank Method 13 1.8.3 Homogeneous System of Linear Equations 15 1.9 EIGENVALUES AND EIGENVECTORS 15 1.9.1 Characteristic Roots (Eigenvalues) of a Matrix  15 1.9.2.  Trace of a Matrix16 1.9.3. Eigenvectors or Characteristic Vectors16 1.10 Vectors 17 1.10.1 Introduction 17 1.10.2 Linear Dependence and Linear Independence17 1.10.3  Inner Product and Norm of Vectors 18 1.10.4  Orthogonal and Orthonormal Vectors19 1.10.5  Basis and Dimension19 Fully Solved MCQs (Level-I) 19 Answer Key 22 Explanation22 Fully Solved MCQs (Level-II) 28 Answer Key 34 Explanation34 Previous Years Solved Papers (2000-2018) 52 Answer Key 71 Answer Key 71 Explanation71 Questions for Practice 102 Answer Key 108 Hints108 CHAPTER 2: CALCULUS113 2.1  FUNCTIONS AND LIMITS 113 2.1.1  Definition of a Function 113 2.1.2 Some Special Functions113 2.1.3  Introduction to Limits114

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Contents • vii

2.1.4  Definition of Limit114 2.1.5  Fundamental Theorems on Limits 114 2.1.6  Fundamental Formulas on Limits 115 2.1.7  The Sandwich Theorem115 2.1.8  Infinite Limits 115 2.1.9  Limits at Infinity116 2.1.10  Infinite Limits at Infinity116 2.2 CONTINUITY AND DIFFERENTIABILITY  116 2.2.1 Continuity 116 2.2.2 Discontinuity 117 2.2.3 Derivative 117 2.2.4  Computation of Derivatives 118 2.3  INDETERMINATE FORMS 118 2.3.1 Introduction 118 2.3.2  The L’Hospital Rule 118 2.4  MEAN VALUE THEOREMS 119 2.4.1  Rolle’s Theorem 119 2.4.2  Lagrange’s Mean Value Theorem 119 2.4.3  Cauchy’s Mean Value Theorem 119 2.5 INCREASING AND DECREASING FUNCTIONS 119 2.6 MAXIMA AND MINIMA OF FUNCTIONS OF A SINGLE VARIABLE 120 2.6.1  First Derivative Test 120 2.6.2  Second Derivative Test 120 2.6.3  Higher Order Derivative Test 120 2.7 INFINITE SERIES AND EXPANSION OF FUNCTIONS  121 2.7.1 Infinite Series121 2.7.2  Test for Convergence of Infinite Series  121 2.7.3 Taylor’s Theorem With Lagrange’s Form of Remainder 123 2.7.4  The Taylor Series 123 2.7.5  Maclaurin’s Series 123 2.8 INDEFINITE AND DEFINITE INTEGRALS 123 2.8.1  Indefinite Integral 123 2.8.2 Fundamental Formulas of Indefinite Integral 123 2.8.3 Advanced Formulas of Indefinite Integrals124 2.8.4  Definite Integral 124 2.8.5  Properties of Definite Integral 125 2.8.6 Definite Integral as a Limit of Sum125 2.8.7 Differentiation Under the Sign of Integration125 2.9 IMPROPER INTEGRALS, BETA, AND GAMMA FUNCTIONS 126 2.9.1  Improper Integral 126 2.9.2  Evaluation of Improper Integrals 126

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viii • Contents 2.9.3 Beta Function127 2.9.4 Gamma Function127 2.10 FUNCTIONS OF SEVERAL VARIABLES AND PARTIAL DERIVATIVES 128 2.10.1  Functions of Two Variables 128 2.10.2  Limit of Functions of Two Variables 128 2.10.3 Continuity of Functions of Two Variables 128 2.10.4  Partial Derivatives  129 2.10.5  Homogeneous Function 129 2.10.6  Euler’s Theorem 129 2.10.7 Total Differential and Total Derivative 129 2.10.8 Jacobian 130 2.11 MAXIMA AND MINIMA OF FUNCTIONS OF TWO VARIABLES 131 2.11.1 Introduction 131 2.11.2 Working Rule to Find the Maximum and Minimum Values of f ( x, y ) 131 2.11.3 Lagrange’s Method for Undetermined Multipliers 131 2.12 CHANGE OF ORDER OF INTEGRATION 132 2.13 DOUBLE AND TRIPLE INTEGRALS 133 2.13.1  Double Integrals 133 2.13.2  Triple Integrals 133 2.14  ARC LENGTH OF A CURVE 134 2.15 VOLUMES OF SOLIDS OF REVOLUTION 134 2.15.1  Working Formulas 134 2.16 SURFACE AREAS OF SOLIDS OF REVOLUTION 135 Fully Solved MCQs (Level-I) 135 Answer Key 141 Explanation141 Fully Solved MCQs (Level-II) 157 Answer Key 164 Explanation164 Previous Years Solved Papers (2000-2018) 191 Answer Key 208 Explanation209 Questions for Practice 241 Answer Key 246 Explanation246 CHAPTER 3: VECTORS249 3.1  BASIC CONCEPTS 249 3.1.1  Scalars and Vectors 249 3.1.2  Position Vector 249 3.1.3  Equal Vectors 249 3.1.4  Negative of a Vector 249

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Contents • ix

3.1.5  Unit Vectors 249 3.1.6  Sum and Difference of Two Vectors 249 3.1.7  Triangle Law of Addition 250 3.1.8  Product of a Vector with a Scalar 250 3.1.9  Collinear Vectors 250 3.1.10  Coplanar Vectors 250 3.1.11  Section Formula 250 3.2  PRODUCT OF VECTORS 250 3.2.1  Scalar Product (Dot Product) 250 3.2.2  Vector Product (Cross Product)250 3.2.3 Scalar Triple Product 251 3.2.4 Vector Triple Product 251 3.3 VECTOR DIFFERENTIATION AND INTEGRATION 251 3.3.1  Derivative of a Vector Function 251 3.3.2 General Rules for Vector Differentiation251 3.3.3  Velocity and Acceleration252 3.3.4  Vector Integration 252 3.4 GRADIENT, DIVERGENCE AND CURL 252 3.4.1  Del Operator 252 3.4.2  Gradient of a Scalar Point Function 252 3.4.3  Divergence of a Vector Point Function 252 3.4.4  Curl of a Vector Point Function 252 3.4.5  Vector Identities 253 3.4.6  Directional Derivative 253 3.5 LINE, SURFACE, AND VOLUME INTEGRALS 253 3.5.1  Line Integral 253 3.5.2  Surface Integral 254 3.5.3 Volume Integral255 3.6 GREEN’S, STOKES’, AND GAUSS DIVERGENCE THEOREM 255 3.6.1 Greens Theorem (in a Plane)255 3.6.2  Stokes’ Theorem 256 3.6.3  Gauss Divergence Theorem 256 Fully Solved MCQs 257 Answer Key 260 Explanation260 Previous Years Solved Papers (2000-2018) 269 Answer Key 276 Explanation277 Questions for Practice 290 Answer Key 292

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x • Contents CHAPTER 4: ORDINARY DIFFERENTIAL EQUATIONS 293 4.1  BASIC CONCEPTS 293 4.1.1  Definition of a Differential Equation 293 4.1.2  Classification of Differential Equations 293 4.1.3  Order of a Differential Equation 293 4.1.4  Degree of a Differential Equation 294 4.1.5  Formation of a Differential Equation  294 4.1.6  Solution of a Differential Equation294 4.2 LINEARLY DEPENDENT AND LINEARLY INDEPENDENT SOLUTIONS 295 4.2.1  Wronskian  295 4.2.2 Linearly Dependent Solutions295 4.2.3 Linearly Independent Solutions295 4.3  DIFFERENTIAL EQUATIONS OF 1ST ORDER AND 1ST DEGREE 296 4.3.1  General Form 296 4.3.2  Solution by Separation of Variables296 4.3.3  Homogeneous Differential Equation  296 4.3.4 Exact Differential Equations297 4.3.5 Linear Differential Equations298 4.4 LINEAR DIFFERENTIAL EQUATIONS OF 2ND ORDER299 4.4.1  General Form 299 4.4.2 Complementary Function (C.F) 299 4.4.3  Particular Integral (P.I) 300 4.4.4  Complete (General) Solution  301 4.4.5  Homogeneous Linear Differential Equations of Order Two 301 Fully Solved MCQs 302 Answer Key 304 Explanations304 Fully Solved MCQs 310 Answer Key 311 Explanations311 Previous Years Questions (2000-18) 318 Answer Key 327 Explanations327 Questions for Practice 344 Answer Key 346 Hints347 CHAPTER 5: PARTIAL DIFFERENTIAL EQUATIONS 5.1  BASIC CONCEPTS 5.1.1 Introduction 5.1.2  Order and Degree 

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349 349 349 349

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Contents • xi

5.1.3 Linear and No-Linear Partial Differential Equations 349 5.1.4 Formation of Partial Differential Equations350 5.2 CLASSIFICATION OF 2ND ORDER PARTIAL DIFFERENTIAL EQUATION  350 5.3 HEAT, WAVE, AND LAPLACE EQUATIONS 350 5.3.1  Solution by Separation of Variables 350 5.3.2 One-Dimensional Heat (Diffusion) Equation and Its Solution 350 5.3.3 One-Dimensional Wave Equation and Its Solution 351 5.3.4  The Laplace Equation and Its Solution351 Fully Solved MCQs 351 Answer Key 352 Explanation352 Fully Solved MCQs 353 Answer Key 354 Explanation354 Previous Years Solved Papers (2000-2018) 357 Answer Key 358 Explanation358 Questions for Practice 359 Answer Key 360 CHAPTER 6: LAPLACE TRANSFORMS 6.1 BASICS OF LAPLACE TRANSFORMS 6.1.1  Definition of the Laplace Transform 6.1.2  Linear Property of the Laplace Transform 6.1.3 Fundamental Formulas of the Laplace Transform 6.1.4  First Shifting Theorem 6.1.5 Some Advanced Formulas of the Laplace Transform 6.1.6  Change of Scale Property 6.2 LAPLACE TRANSFORM ON DERIVATIVES 6.3 LAPLACE TRANSFORM ON INTEGRALS 6.4 LAPLACE TRANSFORM ON PERIODIC FUNCTIONS 6.5 EVALUATION OF INTEGRALS USING LAPLACE TRANSFORMS 6.6 INITIAL AND FINAL VALUE THEOREMS 6.6.1  Initial Value Theorem 6.6.2  Final Value Theorem 6.7 FUNDAMENTALS OF INVERSE LAPLACE TRANSFORM 6.7.1 Definition of Inverse Laplace Transform 6.7.2 Useful Formulas on Inverse Laplace Transforms 6.8 IMPORTANT THEOREMS ON INVERSE LAPLACE TRANSFORMS

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xii • Contents 6.9 UNIT STEP FUNCTION AND UNIT IMPULSE FUNCTION 364 6.9.1 Unit Step Function 364 6.9.2 Second Shifting Theorem 365 6.9.3 Unit Impulse Function 365 6.10 SOLVING ORDINARY DIFFERENTIAL EQUATIONS 365 Fully Solved MCQs (Level-I) 365 Answer Key 368 Explanation368 Fully Solved MCQs (Level-II) 374 Answers Key 377 Explanation377 Previous Years Questions (2000-2018) 385 Answers Key 389 Explanations390 Questions for Practice 395 Answers Key 398 Explanation398 CHAPTER 7: NUMERICAL ANALYSIS 401 7.1  ERRORS AND APPROXIMATIONS 401 7.1.1  Rounding Off 401 7.1.2  Errors and their Computation401 7.2 CALCULUS OF FINITE DIFFERENCES 401 7.2.1  Forward Difference Operator 401 7.2.2  Backward Difference Operator 402 7.2.3  Shift Operator 402 7.3 INTERPOLATION 402 7.3.1 Newton’s Forward Difference Interpolation Formula402 7.3.2 Newton’s Backward Difference Interpolation Formula 402 7.3.3  Lagrange’s Interpolation Formula 403 7.3.4  Error in Interpolation 403 7.4  NUMERICAL DIFFERENTIATION 403 7.4.1 Differentiation Formula Based on Newton’s Forward Difference Formula 403 7.4.2 Differentiation Formula Based on Newton’s Backward Difference Formula 403 7.5  NUMERICAL INTEGRATION 403 7.5.1 Trapezoidal Rule403 7.5.2  Simpson’s 1/3rd Rule403 7.5.3 Weddle’s Rule404 7.5.4  Simpson’s 3/8th’s Rule404 7.6 SYSTEM OF LINEAR ALGEBRAIC EQUATIONS 404 7.6.1 Gauss Elimination Method404 7.6.2  LU Decomposition Method404 7.6.3  Gauss–Seidel Iteration Method 405

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Contents • xiii

7.7 SOLUTION OF ALGEBRAIC AND TRANSCENDAL EQUATIONS 405 7.7.1  Method of Bisection 405 7.7.2  Regula Falsi Method 405 7.7.3 Newton–Raphson Method405 7.8 NUMERICAL SOLUTION OF ORDINARY DIFFERENTIAL EQUATIONS 405 7.8.1  Euler’s Method 405 7.8.2  Modified Euler’s Method 405 7.8.3  Runge–Kutta Method 406 I.  Second-Order Runge–Kutta Method 406 II.  Fourth-Order Runge–Kutta Method 406 7.8.4  Predictor-Corrector Method 406 Fully Solved MCQs 406 Answer Key 408 Explanation408 Previous Years Solved Papers (2000-2018) 412 Answer Key 419 Explanation419 Questions for Practice 430 Answer Key 432 CHAPTER 8: COMPLEX ANALYSIS 433 8.1  BASICS OF COMPLEX ANALYSIS 433 8.1.1  Complex Number 433 8.1.2 Modulus and Amplitude of a Complex Number 433 8.1.3 Conjugate of a Complex Number 435 8.1.4 Properties of Modulus, Argument, and Conjugate 435 8.1.5 Sum, Difference, and Product of Two Complex Numbers 435 8.1.6 Cube Roots of Unity  435 8.1.7  De Moivre’s Theorem 435 8.1.8  Hyperbolic Functions 436 8.1.9  Logarithm of a Complex Number 436 8.2 CALCULUS OF COMPLEX VALUED FUNCTIONS 436 8.2.1  Function of a Complex Variable 436 8.2.2  Limit of a Complex Valued Function 436 8.2.3 Continuity of a Complex Valued Function 437 8.2.4 Derivative of a Complex Valued Function 437 8.2.5 Analytic Function437 8.2.6  Cauchy Riemann Equations 437 8.2.7 Conjugate Function  437 8.2.8  Harmonic Function 437 8.2.9 Construction of an Analytic Function (by Milne Thomson’s method) 438 8.2.10 Construction of Harmonic Conjugate 438

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xiv • Contents 8.3  COMPLEX INTEGRATION 438 8.3.1 Curves 438 8.3.2  Complex Line Integral439 8.3.3 Cauchy-Goursat Theorem439 8.3.4  Cauchy’s Integral Formula  439 8.3.5 Cauchy’s Integral Formula on Higher Order Derivatives 440 8.4 TAYLOR AND LAURENT SERIES 440 8.4.1  The Taylor Series440 8.4.2  The Laurent Series440 8.5 SINGULARITIES 440 8.5.1 Singular Point440 8.5.2  Types of Singularities441 8.5.2.1  Isolated singularity 441 8.5.2.2  Removable singularity 441 8.5.2.3  Essential singularity 441 8.5.3  Zeros and Poles  441 8.6  RESIDUES  442 8.6.1  Residue at a Simple Pole442 8.6.2  Residue at a Pole of Order “n” 442 8.6.3  Residue at Infinity443 8.6.4 Cauchy’s Residue Theorem443 Fully Solved MCQ’s 443 Answer Key 446 Explanation446 Fully Solved MCQ’s (Level-II) 452 Answer Key 455 Explanation456 Previous Years Solved Papers (2000-2018) 466 Answer Key474 Explanation474 Questions for Practice 490 Answer Key493 Explanation493 CHAPTER 9: PROBABILITY AND STATISTICS 495 9.1  BASICS OF PROBABILITY 495 9.1.1 Experiment 495 9.1.2  Random Experiment 495 9.1.3 Sample Space (Event Space)495 9.1.4 Event 496

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Contents • xv

9.1.5  Equally Likely Events 496 9.1.6  Mutually Exclusive Events 496 9.1.7  Mutually Exhaustive Events 496 9.1.8  Classical Definition Of Probability 496 9.1.9 Independent Events497 9.2 CONDITIONAL PROBABILITY AND BAYES’ THEOREM 497 9.2.1  Conditional Probability 497 9.2.2  Theorem on Total Probability498 9.2.3 Bayes’ Theorem498 9.3 RANDOM VARIABLE AND PROBABILITY DISTRIBUTION 498 9.3.1 Random Variable498 9.3.2  Types of Random Variable 498 9.3.3 Probability Mass Function (P.M.F) 498 9.3.4  Probability Distribution Function 499 9.3.5  Expectation or Mean500 9.3.6  Variance and Standard Deviation 500 9.4 SPECIAL TYPES OF PROBABILITY DISTRIBUTIONS 501 9.4.1 Binomial Distribution501 9.4.2 Poisson Distribution501 9.4.3  Normal Distribution 501 9.4.4  Geometric Distribution 502 9.4.5  Uniform (Rectangular) Distribution502 9.4.6 Gamma Distribution502 9.4.7 Exponential Distribution502 9.5  INTRODUCTION OF STATISTICS 502 9.5.1 Statistics 502 9.5.2 Scopes and limitations of Statistics 502 9.5.3 Frequency Distribution503 9.5.4  Mean (Arithmetic Mean) 503 9.5.5 Median 503 9.5.6 Mode 504 9.5.7 Standard Deviation (S.D) 504 9.5.8 Correlation 504 9.5.9 Regression 504 Fully Solved MCQs 505 Answers Key 506 Explanations506 Fully Solved MCQs 510 Answers Key 512 Explanations512 Previous Years Solved Papers (2000-2018) 517

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xvi • Contents Answer Key 533 Explanation533 Questions For Practice 558 Answer Key 562 Explanations562 CHAPTER 10: FOURIER SERIES 565 10.1  BASICS OF THE FOURIER SERIES 565 10.1.1  Definition of the Fourier Series 565 10.1.2 Dirichlet’s Condition  565 10.2 FOURIER SERIES OF EVEN AND ODD FUNCTIONS 566 10.2.1  Fourier Series of Even Function 566 10.2.2  Fourier Series for Odd Function 566 10.3  HALF RANGE FOURIER SERIES 566 10.3.1  Half Range Sine Series 566 10.3.2  Fourier Cosine Series 566 Fully Solved MCQs 566 Answer Key 567 Explanation567 Previous Years Solved Papers (2000-2018) 569 Answer Key 570 Explanation570 Questions for Practice 571 Answer Key 572 Explanation572 CHAPTER 11: GRAPH THEORY 573 11.1 GRAPHS 573 11.1.1  Definition of a Graph573 11.1.2 Incidence 573 11.1.3 Loops 574 11.1.4  Parallel Edges 574 11.1.5  Degree of a Vertex574 11.1.6  Directed Graph 575 11.1.7 In Degree and Out Degree575 11.1.8 Minimum Degree and Maximum Degree575 11.2  DIFFERENT TYPES OF GRAPHS 576 11.2.1 Mixed Graph576 11.2.2 Multi Graph576 11.2.3 Simple Graph576 11.2.4 Trivial Graph576 11.2.5 Null Graph576 11.2.6 K-regular Graph576

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Contents • xvii

11.2.7 Complete Graph577 11.2.8 Bipartite Graph577 11.2.9 Complete Bipartite Graph577 11.3  WALK AND PATH 577 11.3.1 Walk 577 11.3.2  Path and Circuit 577 11.4  MATRIX REPRESENTATION OF GRAPHS 578 11.4.1 Adjacent Matrix578 11.4.2  Incidence Matrix 579 11.4.3  Path Matrix 579 11.5  PLANAR GRAPHS AND EULER’S FORMULA 579 11.5.1  Planar Graph 579 11.5.2  Euler’s Formula 579 11.6 SUB GRAPHS AND ISOMORPHIC GRAPHS 580 11.6.1 Subgraphs 580 11.6.2  Isomorphic Graph 580 11.7  CONNECTEDNESS581 11.7.1  Connected Graph 581 11.7.2  Strongly Connected Graph 581 11.7.3 Weakly Connected Graph581 11.7.4 Component 581 11.7.5 Eulerian Graph581 11.7.6 Hamiltonian Graph582 11.8 VERTEX AND EDGE CONNECTIVITY 582 11.8.1  Cut Vertex 582 11.8.2 Cut Edge (Bridge)582 11.8.3 Cut Set583 11.8.4 Edge Connectivity583 11.8.5  Vertex Connectivity 583 11.9 GRAPH COLORING, MATCHING, AND COVERING 583 11.9.1 Vertex Coloring583 11.9.2  Chromatic Number  584 11.9.3 Matching 584 11.9.4 Covering 584 11.10 TREE 584 11.10.1 Definition 584 11.10.2 Spanning Tree585 Construction of Spanning Trees 585 (I) BFS (Breath First Search) Algorithm585 (II) DFS (Depth First Search) Algorithm586

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xviii • Contents 11.10.3  Minimal Spanning Tree586 (I) Prim’s Algorithm 586 (II) Kruskal’s Algorithm 587 11.10.4 Binary Tree588 11.10.5 Rooted Tree588 11.10.6  Traversal of a Tree 588 Fully Solved MCQs 589 Answer Key 592 Explanation592 Previous Years Solved Paper (2000-2018) 595 Answer Key 599 Explanation599 Questions for Practice 603 Answer Key 606 Hints606 APPENDIX A: GATE 2019 SOLVED PAPERS

607

APPENDIX B: GATE 2020 SOLVED PAPERS

623

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CHAPTER

1

Linear Algebra

1.1  MATRICES AND THEIR TYPES 1.1.1  Definition of a Matrix A set of mn numbers (real or complex) arranged in “m” rows and “n” columns is called a rectangular matrix or simply a matrix of order “m by n” (­denoted by m × n). Thus, æ a11 ç ç a21 A = ç ... ç ç ... ça è m1

a12 a22 ... ... am2

a13 .... a1n ö ÷ a23 ... a2n ÷ ... ... ... ÷ …(i) ÷ ... ... ... ÷ am3 ... amn ÷ø

is a matrix of order m×n. The numbers a11, a12,...amn are called elements of the matrix, and aij is the element of the matrix lying in the ith row and jth column. The matrix given in (i) can also be represented by

Remember: Two matrices A = [aij]m×n and B = [bij]p×q are said to be equal if m = p, n = q (i.e., they have same number of rows and columns) and the elements ­lying in the corresponding places of the two matrices are the same. Thus, for

æx

yö

æ 3 5ö

A= ç ÷, B = ç ÷; è0 2ø èa bø

A = B  ⇔  x = 3, y = 5, a = 0, b = 2. 1.1.2  Types of Matrices (a) Row matrix A matrix with only one row is called a row matrix. Example: [2 5 1] is a row matrix (order of this matrix is 1 × 3). (b) Column matrix

A = [aij]m×n or by A = (aij)m×n

A matrix with only one column is called a column matrix.

Examples:

Example: æ0

3 4

æ4

5ö ÷ 6 ÷ is a matrix of order 3 × 2. 1 ÷ø

A = ç è1

A = çç 3

ç2 è

5ö ÷ is a matrix of order 2 × 3, 1ø

The rows of a matrix are denoted by R1 (first row), R2 (second row) and so on. The columns of a m ­ atrix are denoted by C1 (first column), C2 (second column) and so on.

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é1 ù ê2 ú ê ú is a column matrix ê4 ú ê ú ë5 û

(order of this matrix is 4 × 1) (c) Square matrix A matrix that has equal number of rows and columns is called a square matrix. The elements aij of the square matrix A = [aij]n×n for which i = j are called diagonal elements and the

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2 • Engineering Mathematics Exam Prep line formed by the diagonal elements is called the principal diagonal of the matrix. If matrix A has “n” rows and “n” columns, then we say that A is a square matrix of order n. Then A can be represented by æ a11 ç ç a21 ç .... ç ç .... ça è n1

a12

a13

.... .... an2

.... .... an3

a22

a23

....

a1n ö ÷ .... a2n ÷ .... .... ÷ ÷ .... .... ÷ .... ann ÷ø

(g) Null matrix If all the elements of a matrix are zero, then it is called a null matrix or zero matrix. The null matrix of order m×n is denoted by Om×n. Example:

é0 0 ù é0 0 ù ê ú O2×2 = ê ú , O3´2 = ê0 0 ú , 0 0 ë û êë0 0 úû

etc. are null matrices.

For the sake of simplicity sometimes, we denote the null matrix by “O.”

where a11, a22, ..., ann are diagonal elements.

(h) Upper triangular matrix

Example:

A square matrix A = [aij]n×n is called an upper triangular matrix if aij = 0 " i > j.

1, 2, 4 are diagonal elements of the matrix æ1 ç ç 3 ç -1 è

0 5ö ÷ 2 1÷ 9 4 ÷ø

Example:

(d) Diagonal matrix A square matrix whose non-diagonal elements are all zero is called a diagonal matrix. Example:

æ4 ç ç0 ç0 è

0 7 0

Thus, for any upper triangular matrix, all the elements lying below the principal diagonal are zero.

0ö ÷ 0 ÷ is a diagonal matrix of order 3. - 1 ÷ø

The above diagonal matrix can also be written as diag (4, 7, - 1) or diag [-1, 7, 4]. (e) Scalar matrix



æ3 ç ç0 ç0 è

0 3 0

0ö ÷ 0 ÷ is a scalar matrix of order 3. 3 ÷ø

(f) Identity matrix A diagonal matrix whose diagonal elements are all “1,” is called an identity matrix. The identity matrix of order n is denoted by In or In×n or by I. Example:

æ1 è0

I1 = [1]1×1, I2 = ç

EMEP.CH01_3PP.indd 2

0ö ÷ , 1 ø2´2

æ1 ç I3 = ç 0 ç0 è

0 1 0

0ö ÷ 0 ÷ ,  etc. 1 ÷ø3´3

6 öæ 0 2 3 ö ÷ç ÷ 1 ÷ç 0 - 1 1 ÷ , ÷ç 0 2 øè 0 0 ø÷

etc. are all upper triangular matrices. (i) Lower triangular matrix: A square matrix A = [aij]n×n is called an lower triangular matrix if aij = 0 " i < j. Thus, for any lower triangular matrix, all the elements lying above the principal diagonal are zero. Example: æ4 0ö ç ÷, 2 7 ø çç è1

æ3 ç è1

If all the diagonal elements of a diagonal matrix are equal, then the matrix is called a scalar matrix. Example:

æ3 5 -1ö ç ÷, 0 -1 2 ø çç 0 è0

æ1 ç è0

0

6 0

0ö æ4 ÷ ç 0 ÷ ,ç 5 0 ÷ø çè1

0

0 0

0ö ÷ 0 ÷, 0 ÷ø

etc. are all lower triangular matrices.

1.2  ALGEBRA OF MATRICES 1.2.1 Negative, Sum, and Differences of Matrices (i) The negative of a matrix A = [aij]m×n is the matrix [-aij]m×n and is denoted by ‘-A’. Example: æ1

If A = ç è6

2 3

æ -1 5ö ÷ , then - A = ç -6 4ø è

-2 -3

-5ö ÷ - 4ø

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Linear Algebra • 3

(ii) The sum of two matrices A = [aij]m×n and B = [bij]m×n is denoted by A + B and defined by A + B = [aij + bij]m×n.

Properties:

Thus, the sum of two matrices A and B of same order is the matrix obtained by adding the corresponding elements of the two matrices A and B.

(i) k1 (A ± B) = k1A ± k1B

Example: æ4

5

6ö æ7 ÷ and B = ç 0ø è2 æ11 è4

Then A + B = ç

5 0

0 1

(iv) -k1A = k1(-A)

1ö ÷ 9ø

1.2.3  Transpose of a Matrix

7ö ÷ 9ø

(iii) The difference of two matrices A = [aij]m×n and B = [bij]m×n is denoted by A - B and is defined to be the sum A + ( - B).

æ4 è2

For A = ç

5ö æ -1 ÷ , B=ç 1ø è 0

The transpose of a matrix A is the matrix obtained from A by changing its rows into columns and columns into rows. The transpose of a matrix A is denoted by A’ or At or AT. Thus, if A is matrix of order m × n, then AT is of order n × m. Example:

æ1 If A= ç è4

Thus, A - B = A + (-B) = [aij - bij]m×n.



(ii) (k1 ± k2)A = k1A ± k2A (iii) k1(k2A) = (k1k2)A

If A = ç è2 -1

Example:

If A = [aij]m×n and B = [bij]m×n be two matrices and k1, k2 be scalars, then

0ö ÷; 3ø

2 5

3ö T ÷ , then A = 6ø

æ1 ç ç2 ç3 è

4ö ÷ 5÷ 6 ÷ø

Properties of transpose of a matrix

æ -5 æ5 5 ö A-B =ç ÷ and B - A = ç è2 - 2ø è -2

- 5ö ÷ 2ø

Remember: A - B ≠ B - A (in general)

If A = [aij]m×n and B = [bij]m×n are two matrices of same order, then (i) (AT)T = A (ii) (A ± B)T = AT ± BT

Properties of addition of matrices

(iii) (lA)T = lAT, where l is any number (­constant).

If A = [aij]m×n, B = [bij]m×n, C = [cij]m×n, are three matrices of same order, then

(iv) (lA ± mB)T = lAT ± mBT, where l, and m are scalars.

(i) A + B = B + A (Commutative property)

(v) (AB)T = BT AT provided AB is defined.

(ii)  A + (B + C) = (A + B) + C (Associative property)

Remember:

(iii)  A + O = O + A, where O is the null matrix of order m×n. (iv) A + ( - A) = ( - A) + A = O where “- A” is called the additive inverse of A w.r.t. the matrix addition. 1.2.2  Multiplication of a Matrix by a Scalar Let A = [aij]m×n be any matrix and k be any scalar quantity. Then the product kA is the matrix of ­order m×n defined by kA = [kaij]m×n. Example: æ4

If = 2, A = ç è6

EMEP.CH01_3PP.indd 3

æ8 5ö ÷ , then kA = ç12 7ø è

10 ö ÷. 14 ø

If A is a square matrix, then

1 1 1 ( A + AT ) + ( A - AT ), where ( A + AT ) 2 2 2 1 T is symmetric and ( A - A ) is skew-symmetric. 2



A=

Thus, every square matrix can be expressed as a sum of a symmetric and skew-symmetric matrix. 1.2.4 Multiplication of Matrices (Product of Matrices)

If the number of columns of a matrix A is equal to the number of rows of another matrix B, then we say that the product AB is defined. If A = [aij]m×n, B = [bjk]n×p, then AB = [cik]m×p

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4 • Engineering Mathematics Exam Prep n

where cik = å aijbjk , where the suffixes i, j, k ­ranges j =1

from 1 to m, 1 to n, 1 to p, respectively. In other words, if A be a matrix of order m×n and B be a matrix of order n × p, then the product AB is a matrix of order m×p and the element lying in the ith row and kth column in AB is the sum of the products of the elements of the ith row of A and the corresponding elements of the kth column of B. Sometimes we denote the matrix product by A.B or A × B also. Example: æ a1 For A = ç è b1



a2 b2

æ c1 c2 ö a3 ö ç ÷ ÷ , B = ç d1 d2 ÷ b3 ø2´3 çf ÷ è 1 f2 ø3´2

æ a1c1 + a2d1 + a3 f1

AB = ç è b1c1 + b2d1 + b3 f1

a1c2 + a2d2 + a3 f2 ö

÷ b1c2 + b2d2 + b3 f2 ø

Remember: (a) In the product AB, A is called pre-factor and B is called the post factor. In general, AB ≠ BA. (b)  A(BC) = (AB)C, provided the products AB and BC are defined (Associative property) (c)  A(B + C) = AB + AC, provided the products AB, AC are defined. (d)  A2 = AA, A3 = AAA and so on (provided A is a square matrix)

Here |A| is called the determinant of order “n.” One can use the symbols D, D1, D2, etc. to denote a determinant. We can find the value of determinant in the following manner: (i) If A =

a11 a21

a12 , then |A| = a11a22 - a12a21 a22

a11

a12

a13

a31

a32

a33

(ii) If A = a21 a22 a23 , then |A| = a11 (a22a33 - a32a23) - a12(a21a33 - a31a23)   + a13(a21a32 - a31a22) The above process of finding the value of a determinant is called expansion of a determinant. 1.3.2  Properties of a Determinant (i) A determinant remains unchanged by changing its rows into columns and columns into rows. Thus, the determinants have the same value.

a1

b1

a2

b2

and

a1

a2

b1

b2

(ii) The interchange of two rows (or columns) of determinant changes the sign of the determinant without changing its numerical value. Thus, if D1 =

a1

a2

b1

b2

and D2 =

a2 a1

b2

b1

(e) (A ± B) ≠ A ± 2AB + B (in general but equality holds if AB = BA)

(D2 is obtained by interchanging the first and second rows), then

(f) (A + B)(A - B) ≠ A2 - B2 (in general but equality holds if AB = BA)



2

2

2

D2 = a2b1 - a1b2 = - (a1b2 - a2b1) = -D1.

(g)  IA = AI = A, where A is a square matrix of order n and I is the identity matrix of order n.

The operation “interchange of R1 and R2” is denoted by R1 ↔ R2. Similar notation can be used for interchange of two columns.

1.3 DETERMINANT OF A SQUARE MATRIX

(iii) If two rows (or columns) of a determinant are identical, then the value of the determinant is zero.

1.3.1  Definition of Determinant Let A = [aij]n×n be a square matrix of order n. Then the determinant of A is denoted by |A| or det (A) and is defined by a11 a21 |A| = a31 ...... an1

EMEP.CH01_3PP.indd 4

a12 a22 a32 ...... an2

a13 a23 a33 ...... an3

...... ...... ...... ...... ......

a1n a2n a3n ...... ann

1 0 3 For example, in D = 5 -1 2 , 1 0 3

 R1(first row) and R3(third row) are identical, and so D = 0. (iv) If AB is defined, then |AB| = |A||B|. (v) If A be a square matrix of order “n,” then |kA| = kn|A|.

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Linear Algebra • 5

Example:

é1 Let k = 4 and A = ê ë0

3ù . So A is a square matrix 5úû

Example:

1 0 -1 Let, D = 2 3 4 . 0 1 5



Then, D = (15 - 4) - 0 - 2 = 9.

of order “2” and |A| = 5. Then |4A| =

4 ´ 1 4 ´ 3 4 12 = 4 ´ 0 4 ´ 5 0 20

= 80 = 42 × 5

= 4order of A × |A|.

1 0 -1 1 + 3 ´ ( -1) 0 -1 2 3 4 = 2 + 3´4 3 4 Again, D = 0 1 5 0 + 3´5 1 5

(vi) |An| = |A|n, n being a positive integer.

(by C1 ® C1 + 3C3 )

(vii) If every element of any row (or column) of a determinant is multiplied by a constant “k,” then the determinant is multiplied by the same constant “k.” Example: 1 If D = 5



(viii) The value of a determinant corresponding to a lower (or upper) triangular matrix is obtained by taking the product of all the diagonal elements. Example:

1 The value of the determinant 0 0

the product of 1, 5, and 6, i.e, 30.

2 5 0

3 - 4 is 6

(ix) The determinant value of the null and identity matrices are, respectively, 0 and 1. (x) If all elements of any row (or column) are zero, then the determinant value becomes zero. Example:

1 The determinant 0 0

2 5 0

3 - 4 has value “0.” 0

(xi) The value of a determinant remains unaltered by adding (or subtracting) “k” times the elements of any row (or column) to (or from) the corresponding elements of any other row (column), where “k” is any constant. In this case, we use the following notations: Ri → Ri + kRj, Ri → Ri - kRj, Ci → Ci + kCj, Ci → Ci - kCj.

EMEP.CH01_3PP.indd 5

Also, 1

2 , then 7

1 2 2 4 = (by multiplying R1 by 2) 2D = 2 5 7 5 7





-2 0 -1 = 14 3 4 = -2(15 - 4) - 0 - (14 - 45) = 9. 15 1 5 0

-1

D= 2 3

1

0

-1

4 = 2 - 2 ´1 3 - 2 ´ 0

0 1

5

0

4 - 2 ´ ( -1)

1

5

(by R2 ® R2 - 2R1 )

1 0 -1   = 0 3 6 = 15 - 6 = 9. 0 1 5

Remember: If the given determinant is of order 4 (or more), then use row (or column) operations to get only one nonzero element in the first row (or first column). Then the value of the determinant = non-zero element obtained in first row (first column) × its cofactor. Example:



1 0 If D = -1 1



D=

1 0

2 3 -1 2 1 2 . Then 5 1 0 0 -1 3 2 2

3 1

-1 5 1 1 0 -1

-1 2 0 3

=

1 0

2 2

3 1

-1 2

0 7 4 -1 0 -2 -4 4

[by R3 → R3 + R1, R4 → R4 - R1] = 1 × cofactor of “1” [since “1” is the only non-zero element in first column] 2 1 2 = 1 × (-1)1+1 7

4 -1 -2 -4 4



= 2(16 - 4) - 1(28 - 2) + 2(-28 + 8) = -42.

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6 • Engineering Mathematics Exam Prep 1.3.3  Minors and Cofactors The minor of an element in a determinant D is obtained by eliminations from D the row and the column containing the element. Thus, if aij be in ith row and jth column, then the minor of aij is denoted by Mij and the cofactor of aij denoted by Cij, is defined by Cij = (-1)i+j Mij Example:

1 2 3 Consider a determinant D = 4 7 5 6 9 8 Then 7 5 (i) the minor of “1” is (which is obtained 9 8

by omitting first row and first column, where the element 1 lies), (ii) the minor of “5” is 1

6

2 (which is obtained 9

by omitting second row and third column, where the element 5 lies), etc. (iii) the cofactor of “1” is (-1)1+1

7 9

5 7 , i.e., 8 9

(iv) the cofactor of “5” is (-1)2+3

1 6

1 2 , i.e., 6 9

[Cij]Tn×n is called the adjoint or adjugate of the matrix A and is denoted by adj(A). Thus adj(A) = [Cij]Tn×n. Example:

é1

3ù 6 úú . Then êë7 9 úû 5 (i) cofactor of “1” is (-1)1+1 8

Let A = êê4

2 5 8

6 , i.e., -3 9

(since “1” lies in first row and first column) (ii) cofactor of “2” is (-1)1+2

4 7

6 , i.e., + 6 9

(since “2” lies in first row and second column) (iii) cofactor of “3” is (-1)1+3

4 7

5 , i.e., -7 8

(since “3” lies in first row and third column) (iv) cofactor of “4” is (-1)2+1

2 8

3 , i.e., + 6 9

(since “4” lies in second row and first column)

5 8

(v) cofactor of “5” is (-1)2+2 2 9

1 7

3 , i.e., - 12 9

(since “5” lies in second row and second column) 1 7

2 , i.e., + 6 8

Remember:

(vi) cofactor of “6” is (-1)2+3

The sum of products of all elements of any row (or column) with their respective cofactors give the value of the determinant.

(since “6” lies in second row and third column)

a1



Thus, if

D = b1 c1



a2

b2

c2

a3

b3

c3

then considering first row, we can write

D = (a1 × cofactor of a1) + (a2 × cofactor of a2) + (a3 × cofactor of a3) Again, considering second column, we can write D = (a2 × cofactor of a2) + (b2 × cofactor of b2) + (c2 × cofactor of c2)

1.4 ADJOINT AND INVERSE OF A MATRIX 1.4.1  Adjoint of a Matrix If A = [aij]n×n be a square matrix of order “n” and Cij be the cofactor of aij in |A|. Then the matrix

EMEP.CH01_3PP.indd 6

(vii) cofactor of “7” is (-1)3+1

2 5

3 , i.e., - 3 6

(since “7” lies in third row and first column) (viii) cofactor of “8” is (-1)3+2

1 4

3 , i.e., + 6 6

(since “8” lies in third row and second column) (ix) cofactor of “9” is (-1)3+3

1 4

2 , i.e., - 3 5

(since “9” lies in third row and third column) Hence, adj(A) é cofactor of 1 cofactor of 2 = êêcofactor of 4 cofactor of 5 êë cofactor of 7 cofactor of 8 T - 7ö æ -3 6 æ -3 6 ç ÷ = 6 - 12 6 = ç 6 - 12 ç ÷ ç ç -3 ç -7 6 - 3 ÷ø 6 è è

T

cofactor of 3 ù ú cofactor of 6 ú cofactor of 9 úû - 3ö ÷ 6÷ - 3 ÷ø

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Linear Algebra • 7

Properties :

(d) (AT)–1 = (A–1)T

If A be a square matrix of order “n,” then

-1 (e) For A = ç ÷, A = ç | A |è -c c d è ø vided inverse exist.

æa

(i) A × (adj A) = (adj A) × A = |A|I I being the identity matrix of order “n.” (ii) |adj A| = |A| , n–1

2

(iii) |adj (adj A) = |adj (A)|n–1 = | A |( n-1) (iv) adj(adj A) = |A|

n–2

×A

(vi) adj (AT) = (adj A)T.

1   A = adj (A) provided |A| ≠ 0 | A| –1

æ1

ç7 è

2 5 8

3ö ÷ 6 ÷ , then 9 ÷ø

So A–1 does not exist.

æ1 Let us consider another matrix, B = ç 4 ç ç0 è



Then det (B) = 1(5 - 6) - 0 + 3(4 - 0) = 11.

0 3ö ÷ 5 6÷ 1 1 ÷ø

Therefore B–1 exist. ç \ adj(B) = ç 3 ç -15 è



T

-4

4ö

æ -1

3

1

-1 ÷ = ç -4 ç4 5 ÷ø è

1

6

÷

ç

-15 ö

-1

Properties: If A be a square matrix of order “n,” then (a) A × A = A × A = I –1

(b) (AB)–1 = B–1A–1 (c) (A–1)–1 = A

EMEP.CH01_3PP.indd 7

÷

6 ÷. 5 ÷ø

æ -1 3 -15 ö 1 1ç ÷ -1 Hence, B = adj(B) = ç -4 1 6 ÷. 11 ç B 5 ÷ø è 4 -1

–1

(f) If |A| ≠ 0, then A is called non-singular (inver­ tible) matrix; otherwise A is called singular.

1.5.1 Symmetric Matrix A square matrix A is called symmetric if At = A.



4ö æ1 3 ç ÷ ç 3 0 - 7 ÷ is symmetric. ç4 - 7 4 ÷ø è

Properties: (i) If A, B are symmetric matrices of same order, then kA, aA + bB, AB + BA, An (n ∈ N), are symmetric (where “k,” “a,” and “b” are real numbers). (ii) If A, B are symmetric matrices of same order, then AB is symmetric if and only if AB = BA.

det (A) = 1(45 - 48) - 2(36 - 42) + 3(32 - 35) = 0

æ -1

- bö ÷ proaø

Example:

1.4.2  Inverse of a Matrix Suppose A be a square matrix of order “n” and there exist an another square matrix B of the same order such that AB = BA = I, where I is the identity matrix of order “n,” then B is called the inverse of A, denoted by A–1 and is defined by

Let  A = çç 4

1 æ d

1.5 VARIOUS TYPES OF REAL SQUARE MATRICES

(v) adj (kA) = (adj A) × kn–1, where “k” is a real number

Example:

bö

(iii) For any square matrix A, the matrix A + AT is always symmetric. 1.5.2 Skew-Symmetric Matrix A square matrix A is called skew-symmetric if AT = -A. Example:

æ0 - 3 4ö ç ÷ 0 - 7 ÷ is skew-symmetric. ç3 ç -4 7 0 ÷ø è

Properties: (i) If A, B are skew-symmetric matrices of same order, then kA, aA ± bB, AB - BA, An (n ∈ N) are all skew-symmetric. (ii) If A and B are skew-symmetric matrices of same order, then AB is skew-symmetric if and only if AB + BA = 0. (iii) Diagonal elements of skew symmetric matrix are all zero. (iv) For any square matrix A, the matrix A - AT is always skew-symmetric.

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8 • Engineering Mathematics Exam Prep 1.5.3 Orthogonal Matrix A square matrix A is said to be orthogonal if and only if AAT = ATA = I Example:

æ cos a

A= ç è - sin a

sin a ö ÷ is orthogonal. cos a ø

Properties: (i) If A is orthogonal, then |A| = ±1 and hence A is non-singular. (ii) If A, B are orthogonal matrices of same order, then AT, AB, BA, and A–1 are all orthogonal matrices. (iii) If A is orthogonal, then AT = A–1. 1.5.4 Idempotent Matrix A square matrix “A” is called idempotent if A2 = A. Example:

æ 2 -2 The matrix çç -1 3 ç 1 -2 è

-4 ö ÷ 4 ÷ is idempotent. - 3 ÷ø

Properties: (a) If A, B are idempotent matrices of same order, then

æ0 0ö ÷ is a nilpotent matrix of index “2” è1 0 ø

(ii) ç

(since A2 = O).

1.6 COMPLEX MATRICES AND THEIR TYPES 1.6.1 Complex Conjugate of a Matrix If A = [aij]m×n be a complex matrix of order m × n, then the complex conjugate of A = [aij]m×n, is denoted by A and is defined by A = [ aij ]m´n . Thus, the complex conjugate matrix of a given matrix is obtained by replacing all the elements in the given matrix by their respective complex conjugates. Example: æ5

-8+i i ö ÷ 5 0 ÷ , then ç1 - i 2 - i ÷ø è

If A = çç 3 æ5

ç A = ç3

ç1 + i è

-8-i 5 2

-i ö ÷ 0÷ . i ÷ø

(i) AB is idempotent if and only if AB = BA

Properties:

(ii)  A + B is idempotent if and only if AB = BA = O

If A and B are complex matrices, then

(iii) I - A is idempotent

(ii) A + B = A + B

(b) If AB = A, BA = B, then A and B are both idempotent matrices.

(iii) kA = k ´ A, where “k” is any number (real or complex)

1.5.5 Involutary Matrix A square matrix “A” is called involutary if A2 = 1.

(iv) AB = A ´ B, provided AB is defined.

Example: æ -5 The matrix çç 3

(vi) A = A if and only if A is real.

ç1 è

-8 0 ö ÷ 5 0 ÷ is involutary. 2 - 0 ÷ø

1.5.6 Nilpotent Matrix A square matrix “A” is said to be nilpotent of index “k” if Ak = O for some positive number k “k” should be the least positive integer). Example:

é 1 -1 1 ù (i)  The matrix A = ê -3 3 -3 ú is nilpotent of ê ú êë -4 4 -4 úû

index “2” (since A2 = O)

EMEP.CH01_3PP.indd 8

(i) A = A

(v) A n = ( A )n

(vii) A = - A if and only if A purely imaginary 1.6.2  Transposed Conjugate of a Matrix The transposed conjugate of a complex matrix A is denoted by Aq and is defined by A q = ( A )T = AT . Example: -8+i æ5 ç 5 If A = ç 3 ç1 - i 2 è

i ö ÷ 0 ÷ , then - i ÷ø

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Linear Algebra • 9



æ 5 A = çç -8 - i ç -i è

3

i + 1ö ÷ 2 ÷. i ÷ø

5

q

0

Properties: If A and B are complex matrices of same order, then (i) (A )   = A. q q

(ii) (A + B) = A + B . q

q

q

(iii) (kA) = k ( A ) , where “k” is any number (real or complex). q

q

(iv) (AB)q = BqAq. (v) (An)q = (Aq)n. 1.6.3  Unitary Matrix A square complex matrix A is said to be unitary matrix if AAq = AqA = I. Example:

1 æ1 i ö A= ç ÷ is unitary. 2 è -i -1 ø

Properties: (i) If A is unitary then AT, A–1 are both unitary matrix. (ii) If A, B are unitary matrices of same order, then AB and BA are also unitary matrices. 1.6.4 Hermitian Matrix A square matrix “A” is called Hermitian if Aq = A. Example:

æ 1 A = çç 2 + i ç3 + i è

2-i 2 1 - 2i

3-i ö ÷ 1 + 2i ÷ is Hermitian. 4 ÷ø

Properties: (i) If A, B are Hermitian matrices of same order, then  kA, aA + bB, AAq, Aq A, AB + BA, A , An, A + Aq are also Hermitian matrices. (ii) If A, B are Hermitian matrices of same order, then AB is Hermitian if and only if AB = BA. 1.6.5 Skew-Hermitian Matrix A square matrix “A” is called skew-Hermitian if Aq = -A

EMEP.CH01_3PP.indd 9

Example:

æ 3i ç ç -3 + 4i ç -4 - 5i è

3 + 4i - 4i - 5 + 6i

4 - 5i ö ÷ 5 + 6i ÷ is skew-Hermitian. 0 ÷ø

Properties: (i)  A = [aij]n×n is skew-Hermitian if and only if aij = -a ji

(ii) In a skew-Hermitian matrix, all the elements lying in the principal diagonal are purely imaginary or zero. (iii) If A, B are skew-Hermitian matrices of same order, then kA, aA + bB, AB - BA, A - Aq, A are also skew-Hermitian matrices. (iv) If A is Hermitian, then iA is skew-Hermitian. (v) If A is skew-Hermitian, then iA is Hermitian. (vi) For any square matrix A, 1 1 A = ( A + A q ) + ( A - A q ). Thus, every 2

2

square matrix can be expressed as a sum of a Hermitian matrix 1 ( A + A q ) and a skew2

Hermitian matrix 1 ( A - A q ). 2

1.7  RANK OF A MATRIX 1.7.1  Elementary Transformations The following transformations are known as elementary transformations: (i) Interchange of two rows (or columns). Suppose we interchange first row and second row in a matrix, then this operation will be denoted by R1 ↔ R2. Again, if we interchange third column and fourth column in a matrix, then this operation will be denoted by C3 ↔ C4. (ii) The multiplication of the elements of a row (or column) by a non-zero number. Suppose we multiply each element of second row by “k,” then the operation is denoted by R2 → kR2. In a similar way, if we multiply each element of third column by “k,” then the operation is denoted by C3 → kC3. (iii)  The addition to the elements of a row (or column), the corresponding elements of a row (or column) multiplied by any number.

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10 • Engineering Mathematics Exam Prep Suppose we first multiply the second row by “k” and add it with the first row, then the corresponding operation is denoted by R1 → R1 + kR2. Similarly, if we first multiply the second column by “k” and add it with the first column, then the corresponding operation is denoted by C1 → C1 + kC2. 1.7.2 Equivalent Matrices Two matrices A and B are said to be equivalent if one can be obtained from the other by a sequence of elementary operations. The symbol ~ is used for equivalence. Thus, if A and B are equivalent matrices, then we write A~B. The relation “~” forms an equivalence relation. Example:

æ 3 ç If A = ç -3 ç -5 è

æ 3 ç A = ç -3 ç -5 è æ3 ç ç 0 ç -5 è

4 5ö ÷ - 4 6 ÷ , then 6 0 ÷ø

æ3 è1

Let us consider a square sub-matrix B = ç

4ö ÷. 1ø

Then det(B) = -1 ≠ 0. Thus, there exist a square sub-matrix B of order “2” whose determinant value is non-zero. Hence, rank (A) = 2. Properties: (i) If A is a matrix A of order m×n, then rank (A)≤ min {m, n}. (ii) rank (O) = 0 (iii) rank (In) = n, where In is the identity matrix of order “n.” (iv) rank (A) = n if A is non-singular square matrix of order “n” and vice versa. (v) if A is singular, then rank (A) < n and vice versa, provided A is a square matrix of order “n.”

4 5ö ÷ -4 6 ÷ 6 0 ÷ø 4 5ö ÷ 0 11 ÷ (by R2 ® R2 + R1 ) 6 0 ÷ø

æ 3 4 11 ö ç ÷  ç 0 0 11 ÷ (by C3 ® C3 + 2C1 ) ç -5 6 -10 ÷ è ø æ 3 4 11 ö ç ÷ = B (say), where B = ç 0 0 11 ÷ ç -5 6 -10 ÷ è ø

(vi) If A and B are matrices of same order, then r(A + B) ≤ r(A) + r(B). (vii) r(AB) ≤ min {rank (A), rank (B)}. (viii) rank (A) = rank (AT) and rank (A) = rank (AAT) (ix) rank (A) = rank (Aq) and rank (A) = rank (AAq) (x) Two equivalent matrices have same rank. (xi) Rank of a matrix remains the same when we multiply any non-singular matrix with it. (xii) The rank of a skew-symmetric matrix can’t be “1.”

Then A and B are equivalent.

(xiii) Let A be a square matrix of order “n.” Then

1.7.3  Rank of a Matrix A positive real number “r” is said to be the rank of a given matrix A of order m×n if



(i) there exist a square sub-matrix of order “r” whose determinant is non-zero (ii) the determinant value of every square submatrix of order higher than “r” is zero. The symbols r(A), r(A), rank(A) are used to denote the rank of a matrix A. Example: Consider a matrix A =

æ3 ç ç1 ç6 è

4 1

8

= 0. So rank of A cannot be “3.”

EMEP.CH01_3PP.indd 10

5 ö ÷ 2 ÷ . Then det (A) 10 ÷ø

ì n, if rank( A ) = n ï rank(adjA ) = í1, if rank( A ) = n - 1. ï î 0, if rank( A ) < n - 1

1.7.4  Determination of the Rank of a Matrix There are many forms namely “normal form,” “echelon form,” and “row reduced echelon form,” which are very useful in determining the rank of a matrix. But here we will use a secondary method (quickest method) to find the rank of a matrix discussed below: “Let A be a matrix of order m×n. Then apply elementary row transformations to get maximum number of zero rows. In this case, the number of non-zero rows appearing in the final equivalent

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Linear Algebra • 11

­matrix will be equal to the rank of the given matrix.” Example:

æ1 0 3 ö ç ÷ A = ç3 5 1 ÷ ç2 0 6÷ è ø æ1 0 3 ö ç ÷  ç 3 5 1 ÷ ( by R2 ® R2 - 2R1 ) ç0 0 0÷ è ø

On the other hand, no pair of values of x and y simultaneously satisfy both the equations in (iii). So it is inconsistent. Remember: The non-homogeneous system of linear equations given in (i) can also be rewritten as AX = B, where æ a11 ç a A = ç 21 ç .... çç è am1

which has only two non-zero rows. Therefore, rank(A) = 2.

A linear equation in “n” unknowns x1, x2, x3,...xn is an equation of the form

a1x1 + a2x2 + a3x3 + ... + anxn = b.

If b = 0, then the equation is called a homogeneous equation; otherwise it is called a non-homogeneous equation.



ü ï a21x1 + a22 x2 + a23 x3 + ....... + a2n xn = b2 ï ý ........................................................... ï am1x1 + am2 x2 + am3 x3 + ....... + amn xn = bm þï ..............................(i )

If there exist at least one set of values of x1, x2, x3, ..., xn satisfying all the equations in (i), then the system of equations given by (i), is said to be consistent; otherwise inconsistent. In other words, the system of equations given by (i) is consistent if it has a solution; otherwise, it is inconsistent.



x-y=2 ü ý …(ii) 2x + y = 4 þ



x+y=2 ü ý …(iii) 2x + 2 y = 3 þ

The system of equations given by (ii) has a solution x = 2, y = 0 and so it is consistent.

EMEP.CH01_3PP.indd 11

é a11 ê ê a21 éë A : B ùû = ê a31 ê ê ..... êa ë m1

a12 a22 a32 ..... am2

..... a1n b1 ù ú ..... a2n b2 ú ..... a3n b3 ú . ú ..... ..... .....ú ..... amn bm úû

1.8.2 Methods for Solving NonHomogeneous System of Linear Equations 1.8.2.1  Cramer’s Rule Let us consider a system of equations consisting 3 non-homogeneous linear equations in 3 unknowns given by:

a11x + a12 y + a13 z = b1 ü ï a21x + a22 y + a23 z = b2 ý …(i) a31x + a32 y + a33 z = b3 þï Let

Let us consider the following system of equations:

........... a1n ö ÷ ........... a2n ÷ , ........... .... ÷ ÷ ........... amn ÷ø

Here A is called the coefficient matrix. The augmented matrix, denoted by [A: B], is defined by

A system of “m” non-homogeneous linear equations in “n” unknowns is given by a11x1 + a12 x2 + a13 x3 + ....... + a1n xn = b1

a13 a23 .... am3

é x1 ù é b1 ù ê ú ê ú ê x2 ú ê b2 ú X = ê ... ú , B = ê ... ú ê ú ê ú ê ... ú ê ... ú êx ú êb ú ë nû ë mû

1.8 SYSTEM OF LINEAR EQUATIONS AND THEIR SOLUTIONS 1.8.1 Introduction

a12 a22 .... am2

a11 D = a21 a31



a12 a22 a32

a11 b1 D2 = a21 b2 a31 b3

a13 b1 a23 , D1 = b2 a33 b3 a13 a11 a23 , D3 = a21 a33 a31

a12 a22 a32

a13 a23 , a33

a12 b1 a22 b2 . a32 b3

Then,

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12 • Engineering Mathematics Exam Prep Case-I:

1 1 1 D = 1 2 3 = 0, 2 2 2

If D ≠ 0, then (i) has an unique solution and so consistent. In this case, the solution is given by x=

D D1 D , y= 2, z= 3 . D D D

6 1 1 D1 = 10 2 3 = 0, 12 2 2

Example: Let us consider the system of equations:

1 6 1 D2 = 1 10 3 = 0, 2 12 2

2x - z = 1, 2x + 4y - z = 1, x - 8y - 3z = -2. 2 0 -1 Then D = 2 4 -1 = -20 ¹ 0. 1 -8 -3

So the given system of equations has a unique solution. 1 0 -1 D1 = 1 4 -1 = -20, -2 -8 -3 2 1 -1 D2 = 2 1 -1 = 0, 1 -2 -3



2 0 1 D3 = 2 4 1 = -20. 1 -8 -2

1 1 6 D3 = 1 2 10 = 0. 2 2 12

Since D = D1 = D2 = D3 = 0, hence, the system has an infinite number of solutions and so is consistent. Remark: If we consider a system of equations consisting two non-homogeneous linear equations in two unknowns given by



Then

Hence, by Cramer’s rule, the solution is x=

D D1 D = 1, y = 2 = 0, z = 3 = 1. D D D

D=

Case-II: If D = 0 and at least one of D1, D2, D3 be zero, then (i) has no solution and so inconsistent. Thus, the following sub-cases arise: (i) If D = 0, D1 ≠ 0, then (i) has no solution

a11x + a12 y = b1 üï ý a21x + a22 y = b2 ïþ .....................(i )



D1 =

Case-III:

b1 b2

a12 , a22 a12 a b , D2 = 11 1 a22 a21 b2

Then Q

(ii) If D = 0, D2 ≠ 0, then (i) has no solution (iii) If D = 0, D3 ≠ 0, then (i) has no solution

a11 a21

If D ≠ 0, then (i) has an unique solution and so consistent. In this case, the solution is D D given by x = 1 , y = 2 . D

Q

D

If D = 0 and at least one of D1, D2 be zero, then (1) has no solution and so inconsistent. Thus, the following sub-cases arise:

If D = D1 = D2 = D3 = 0, then (i) has an infinite number of solutions and so consistent.

(i) If D = 0, D1 ≠ 0, then (i) has no solution

Example:

(ii) If D = 0, D2 ≠ 0, then (i) has no solution

Let us consider the system of equations:

Example:



x + y + z = 6, x + 2y + 3z = 10, 2x + 2y + 2z = 12.

Then

EMEP.CH01_3PP.indd 12

Let us consider the system of equations: 2x + y = 1, 4x + 2y = 5

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Linear Algebra • 13

Then

D=

2 1 2 1 = 0, but D2 = = 6 ¹ 0. 4 2 4 5

so A–1 exist and the solution is given by: X = A–1B. Now, X = A -1 B 1 adj( A ) ´ B = A

So the system has no solution. Q

If D = D1 = D2 = 0, then (i) has an infinite number of solutions and so consistent.

1.8.2.2  Matrix Method Let us consider a system of equations consisting 3 non- homogeneous linear equations in 3 unknowns given by



a11x + a12 y + a13 z = b1 ü ï a21x + a22 y + a23 z = b2 ý a31x + a32 y + a33 z = b3 ïþ



é -44 ù é 11 ù ê ú ê ú é x ù ê -32 ú ê 8 ú -42 ú ê 21 ú ê ú Þ ê yú = ê = ê -32 ú ê 16 ú êë z úû ê ú ê ú ê 2 ú ê -1 ú êë -32 úû êë 16 úû

The above system of equations can be rewritten as: AX = B, where



æ a11 a12 a13 ö ç ÷ A = ç a21 a22 a23 ÷ ça ÷ è 31 a32 a33 ø éx ù éb1 ù ê ú ê ú X = ê y ú , B = êb2 ú êë z úû êëb3 úû a11

a12

a13

a31

a32

a33

If | A |= a21

a22

a23 ¹ 0,

æ -26 -6 2 ö é 0 ù 1 ç ÷ ê ú -11 -5 7 ÷ ´ ê 7 ú ( -32) çç ÷ è -17 1 5 ø êë -1úû é0 - 42 - 2ù -1 ê ú 0 - 35 - 7 ú = 32 ê êë 0 + 7 - 5 úû

=



\x =

11 21 -1 ,y= ,z= . 8 16 16

1.8.2.3  Rank Method Let us consider a system of equations consisting “m” non- homogeneous linear equations in “n” unknowns given by a11x1 + a12 x2 + a13 x3 + ....... + a1n xn = b1

then we say that A–1 exist and the solution is given by: X = A–1B. Example:

ü ï a21x1 + a22 x2 + a23 x3 + ....... + a2n xn = b2 ï ý ........................................................... ï am1x1 + am2 x2 + am3 x3 + ....... + amn xn = bm þï .....................(2)

Let us consider the system of equations:



x - y + z = 0,  2x + 3y - 5z = 7,  3x - 4y - 2z = -1.

Then the above system of equations can be re-written as AX = B (discussed earlier).

The above system of equations can be re-written as AX = B, where Here





EMEP.CH01_3PP.indd 13

æ1 ç A = ç2 ç3 è

-1 1ö æxö æ0 ö ÷ ç ÷ ç ÷ 3 - 5 ÷ , X = ç y ÷ , B = ç7 ÷. çz ÷ ç -1 ÷ - 4 - 2 ÷ø è ø è ø

1 -1 1 Here A = 2 3 -5 = -32 ¹ 0. 3 -4 -2

Case-I: If rank (A) = rank ([A : B]) = n, then the system given by (2) has an unique solution and so the system is consistent. Example: Let us consider the system of equations: 2x - z = 1, 2x + 4y-z = 1, x - 8y - 3z = -2.

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14 • Engineering Mathematics Exam Prep [ A : B] æ 2 0 -1 1 ö ç ÷ = ç 2 4 -1 1 ÷ ç 1 -8 -3 -2 ÷ è ø

æ1 ç  ç2 ç2 è æ1 ç  ç0 ç0 è





-8 -3 -2 ö ÷ 4 -1 1 ÷ (by R1 « R3 ) 0 -1 1 ÷ø -8 -3 -2 ö ÷ 20 5 5 ÷ 16 5 5 ÷ø (by R2 ® R2 - 2R1 , R3 ® R3 - 2R1 ) æ1 ç  ç0 ç0 è (by

-8 -3 -2 ö ÷ 20 5 5 ÷ -4 0 0 ÷ø R3 ® R3 - R2 )



which has 3 non-zero rows.



So rank ([A:B]) = 3.



é1 [A : B] = ê ë1 é1 ê ë0



which has two non-zero rows.



So rank ([A:B]) = 2.

Thus, rank(A) = rank ([A:B]) = 2 < 3 (= number of unknowns). Hence, the system has an infinite number of solutions and so the system is consistent. Case-III: If rank (A) ≠ rank ([A:B]), then the system given by (2) has no solution and so the system is inconsistent. Example: Let us consider the system of equations: 2x + y - z = 12, x - y - 2z = -3, 3y + 3z = 10. æ2 ç [ A : B] = ç1 ç0 è æ0 ç  ç1 ç0 è

1 -1 12 ö ÷ -1 -2 -3 ÷ 3 3 10 ÷ø 3 3 18 ö ÷ -1 -2 -3 ÷ 0 3 10 ÷ø (by R1 ® R1 - 2R2 )

Ignoring the last column of the final equivalent matrix of [A:B], we see that there are three nonzero rows. So rank(A) = 3. Thus, rank(A) = rank ([A:B]) = 3 (=number of unknowns). Hence, the system is consistent and has the unique solution. Case-II:

æ0 ç  ç1 ç0 è (by

If rank (A) = rank ([A:B]) < n, then the system given by (2) has an infinite number of solutions and so the system is consistent. Example: Let us consider the system of equations:

x + y + z = 6, x + 2y + 3z = 1 é1 1 1 ù é1 1 1 6 ù Here A = ê ú , [A : B] = ê ú. ë1 2 3û ë1 2 3 1 û é1 1 1 ù Then A = ê ú ë1 2 3û



é1 1 1 ù ê ú (by R2 ® R2 - R1 ) ë0 1 2 û



which have 2 non-zero rows.



Hence, rank(A) = 2.

EMEP.CH01_3PP.indd 14

1 1 6ù ú 2 3 1û 1 1 6ù ú (by R2 ® R2 - R1 ) 1 2 -5û



æ0 ç  ç1 ç0 è (by

0 0 8ö ÷ -1 -2 -3 ÷ 3 3 10 ÷ø R1 ® R1 - R3 ) 3 3 10 ö ÷ -1 -2 -3 ÷ 0 0 8 ÷ø R1 « R3 )



which has 3 non-zero rows.



So rank ([A : B]) = 3.

Ignoring the last column of the final equivalent matrix of [A: B], we see that there are two non-zero rows. So rank (A) = 2. Thus, rank (A) ≠ rank ([A:B]). Hence, the system is inconsistent.

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Linear Algebra • 15

1.8.3 Homogeneous System of Linear Equations A homogeneous system of m linear equations in n unknowns is given by a11x1 + a12 x2 + a13 x3 + ....... + a1n xn = 0



ü ï a21x1 + a22 x2 + a23 x3 + ....... + a2n xn = 0 ï ý …(i) ........................................................... ï am1x1 + am2 x2 + am3 x3 + ....... + amn xn = 0 ïþ ...................(1.8.3.1)

The above homogeneous system can also be rewritten as AX = O, where



éa11 ê a A = ê 21 ê.... ê êëam1

a12 ...

a1n ù æ x1 ö ç ÷ ú a2n ú x , X = ç 2 ÷. ç .... ÷ .... .... .... ú çç ÷÷ ú am2 .... amn úû è xn ø a22 ...

Here, A is called the coefficient matrix. It is easy to verify that x1 = x2 = x3 = ... = xn = 0 is a solution of (i) [called the trivial solution]. Hence, the system (i) is consistent. Further if rank (A) < n(= number of unknowns), then the system (i) has an infinite number of solutions. Remember: If a system has “n” homogeneous equations in “n” unknowns, then

Hence, rank (A) = 2 < 3 (= number of unknowns). Thus, the given system of equations has an infinite number of solutions. Alternative method: Here det(A) = (15 - 16) - 2(10 - 12) + 3(8 - 9) = 0. Hence, the system has an infinite number of solutions.

1.9 EIGENVALUES AND EIGENVECTORS 1.9.1 Characteristic Roots (Eigenvalues) of a Matrix If A be a square matrix of order n and I be an identity matrix of order n, then A - lI is called the characteristic matrix of A, l being a scalar. The determinant |A - lI| is called the characteristic polynomial of A, which is basically a polynomial in l of n th degree. The equation |A - lI| = 0, i.e., det (A - pI) = 0 is called characteristic equation of the matrix A and the roots of the characteristic equation are called the eigenvalues or characteristic roots of the matrix A. Example: é1 3 ù Let A = ê ú. ë1 -1û

System has an infinite number of solutions ⇔ determinant of the coefficient matrix is zero.



é1 3 ù é1 0 ù Then A - lI = ê ú -lê ú ë1 -1û ë0 1 û 3 ù é1 - l =ê ú. -1 - l û ë 1



Then the characteristic equation is given by

Example: Let us consider the system of equations: x + 2y + 3z = 0, 2x + 3y + 4z = 0, 3x + 4y + 5z = 0. æ1 2 3 ö ç ÷ Here A = ç 2 3 4 ÷ ç3 4 5÷ è ø æ1 2 3 ö ç ÷  ç 0 -1 -2 ÷ ç 0 -2 -4 ÷ è ø

|A - lI| = 0 1-l 1

or, 

( using R2 ® R2 - 2R1 , R3 ® R3 - 3R1 )



æ1 2 3 ö ç ÷  ç 0 -1 -2 ÷ ç0 0 0 ÷ è ø



which has “2” non-zero rows.

EMEP.CH01_3PP.indd 15

( using R3 ® R3 - 2R2 )

3 =0 -1 - l

or, (1 - l) (- 1 - l) - 3 = 0 or,  -(1 - l2) = 3 or,  l2 = 4 or,  l=±2

Therefore, eigenvalues of A are 2 and - 2.

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16 • Engineering Mathematics Exam Prep Properties:

Application: T

(i) If A be a square matrix, then A and A both have the same eigenvalues. (ii) If A be a square matrix of order n, then det(A) = product of all eigenvalues of A. (iii) If the matrix A is invertible and l is an eigenvalues of A, then l–1 will be an eigenvalue of A–1 (iv) If a square matrix A have eigenvalues l1, l2, ..., lm then the matrix kA have eigenvalues kl1, kl2, ..., klm. (v) If l1, l2, ..., lk are eigenvalues of a square matrix A, then l1m, l2m, ..., lkm will be the eigenvalues of Am, m being a positive integer. (vi) If both A and P are square matrices and P is non-singular, then both the matrices A and P–1 AP have the same eigenvalues.

If l2 + l + 3 = 0 be the characteristic equation of a square matrix A of order 3, then we can write A2 + A + 3I = O. 1.9.2.  Trace of a Matrix Let A = [aij]n×n be a square matrix of order n. Then the trace of the matrix A is denoted by tr(A) and is tr( A ) =

defined by

n

åa r =1

rr

Thus, the trace of a given square matrix is equal to the sum of the diagonal elements of the matrix. Example: æ2 3 5 ö ç ÷ Let A = ç 3 7 1 ÷ ç0 -1 4 ÷ è ø



Then tr(A)

(vii) If both A and B are square invertible matrices of same order, then AB and BA will have the same eigenvalues.



= Sum of the diagonal elements = 2 + 7 + 4 =13.

(viii) “0” is an eigenvalue of a square matrix A if and only if A is singular.

If A and B be two square matrices of same order. Then

(ix) All the eigenvalues of an identity matrix is “1.”

(i) tr (kA) = k tr(A), k being a scalar

(x) The eigenvalues of a lower (or upper) triangular matrix are the diagonal elements of the matrix.

(ii) tr (A + B) = tr (A) + tr (B)

(xi) The eigenvalues of a diagonal matrix are the diagonal elements of the matrix. (xii) The eigenvalues of a unitary matrix are of unit modulus, i.e., if l be an eigenvalue of an unitary matrix A, then |l| = 1. (xiii) The eigenvalues of a Hermitian matrix are all real. (xiv) The eigenvalues of a real symmetric matrix are real. (xv) The eigenvalues of a skew-symmetric matrix are either purely imaginary or zero. (xvi) The eigenvalues of an orthogonal matrix are of unit modulus. (xvii) The eigenvalues of an idempotent matrix are 0 and 1. (xviii) (Cayley-Hamilton theorem). Every square matrix satisfies its own characteristic equation.

EMEP.CH01_3PP.indd 16

Properties:

(iii) tr (AB) = tr (BA) (iv) tr (A) = sum of all eigenvalues of A. Remember: (i) If l2 + al + b be the characteristic polynomial of a 2 × 2 matrix A, then tr(A) = -a and det(A) = b. (ii) If l3 + al2 + bl + c be the characteristic polynomial of a 3 × 3 matrix A, then tr(A) = -a and det(A) = -c. 1.9.3. Eigenvectors or Characteristic Vectors Let A be a square matrix of order n and be l an eigenvalue of A. If there exists a column matrix X (≠ 0) of order n × 1 such that AX = l X holds, then X is called an eigenvector or characteristic vector corresponding to the eigenvalue l. Remember: If A is a square matrix of order 2, then we assume X in the form X = æç x ö÷ and if A is a square è yø

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Linear Algebra • 17

matrix of order 3, then we assume X in the form æxö ç ÷ X = ç y÷ çz ÷ è ø

(i) (x1, x2, x3, ..., xn) is also termed as n-tuple vector. (ii) Every row matrix as well as every column matrix is also termed as a vector. Thus,

Example: æ1 Let A = ç è1

é1 ù

[2 5 1] and êê5 úú are vectors.

3ö ÷. -1ø

êë6 úû

Then as discussed earlier, the eigenvalue of A are 2 and - 2 (see Section 1.9.1). For l = 2: AX = lX 

Remember:

æ1 3 öæ x ö æxö é æ x öù ÷ç ÷ = 2 ç ÷ ê where X = ç ÷ ú è1 - 1 øè y ø è yø ë è y øû æ x + 3 y ö æ 2x ö ⇒  ç ÷=ç ÷ è x - y ø è 2y ø

⇒  ç

Equating the corresponding elements we get, x + 3y = 2x and x - y = 2y  i.e., x - 3y = 0 æxö è yø

æ 3y ö ÷= èy ø

æ3ö yç ÷ è1 ø



\  X = ç ÷ = ç



Which is the eigenvector for l = 2

For l = -2.

(iii) A vector whose components belong to a field F is said to be a vector over the field F. The set of all n-tuple vectors over a field F (denoted by Vn(F) is called a vector space over F. The elements of F are called scalars. (iv)  A subset S of a vector space Vn ( F ) is called a vector subspace if aX + bY ∈ S  " X, Y ∈ S and a, b ∈ F. Example:

{



⇒ç



⇒ç

we have aX + bY = a( x , y, z ) + b( x ¢, y¢, z ¢) = ( ax , ay, az ) + (bx ¢,by¢,bz ¢) = ( ax + bx ¢, ay + bx ¢, az + bz ¢)

æ1 3 öæ x ö æxö é æ x öù ÷ç ÷ = -2 ç ÷ ê where X = ç ÷ ú è1 - 1 øè y ø è yø ë è y øû



(where R3 is theset of all points in three dimensional space)

Then for X = (x, y, z), Y = (x′, y′, z′) ∈ S and a, b ∈ F,

AX = lX

}

Let S = X = ( x , y, z ) Î R3 : x + y + z = 0

æ x + 3 y ö æ -2x ö ÷=ç ÷ è x - y ø è -2 y ø

Now ( ax + bx ¢ ) + ( ay + bx ¢ ) + ( az + bz ¢ ) = a( x + y + z ) + b( x ¢ + y¢ + z ¢)

Equating the corresponding elements we get, x + 3y = -2x i.e., x + y = 0 i.e., y = - x

= a ´0 + b´0 =0

and x - y = -2y i.e., x + y = 0 i.e., y = -x.

 X ,Y Î S æ ö ç ÷ ¢ ¢ ¢ è Þ x + y + z = 0, x + y + z = 0 ø So aX + bY Î S.

æxö è yø

æx ö æ1 ö ÷ = xç ÷ è -x ø è -1 ø



\  X = ç ÷ = ç



which is the eigenvector for l = -2.

1.10 Vectors 1.10.1 Introduction An ordered n-tuple X = (x1, x2, x3, ...., xn) is called a n-vector. x1, x2, x3, ..., xn are called components of the vector X.

EMEP.CH01_3PP.indd 17



Hence, S is a vector sub-space.

1.10.2 Linear Dependence and Linear Independence (i) The vectors X1, X2, X3, ..., Xk are said to be linearly dependent if there exist scalars a1, a2, a3, ..., ak (not all zero) such that a1X1 + a2X2 + a3X3 + ... + akXk = O

8/9/2023 8:28:58 PM

18 • Engineering Mathematics Exam Prep Example: Let X1 = (1, 0, - 2) and X2 = (-2, 0, 4). Then,

(vi) Three vectors X1 = (x1, x2, x3), X2 = (y1, y2, y3) and X3 = (z1, z2, z3) are said to be linearly

2 X1 + X 2 = 2(1,0, -2) + ( -2,0,4)

independent if

= (2 - 2,0 + 0, -4 + 4) = (0,0,0) = O

Therefore, the vectors X1 and X2 are linearly dependent.

Then a1 X1 + a2 X 2 = O

Þ a1 (1,0, -2) + a2 (0,3,0) = O

Þ ( a1 ,0, -2a1 ) + (0,3a2 ,0) = O



Þ ( a1 + 0,0 + 3a2 , -2a1 + 0) = (0,0,0)



Þ a1 = a2 = 0

Þ a1 = 0,3a2 = 0, - 2a1 = 0

Therefore, the vectors X1 and X2 are linearly independent. Remember: (i) Rank of a matrix = number of linearly independent eigenvectors of the matrix.

1 -1 2  3 0 5 = 0, 2 -2 4



the vectors (1, -1,2),(3,0,5)and (2, -2,4) are linearly dependent.

1.10.3 Inner Product and Norm of Vectors (i) The inner product of two vectors X = (x1, x2, x3, ..., xn) and Y = (y1, y2, y3, ..., yn) over the field R of real numbers is denoted by and is defined by = x1y1 + x2y2 + x3y3 + ... + xnyn Example: Let X1 = (1, 0, -2) and X2 = (3, 3, 4). Then X ,Y = 1 ´ 3 + 0 ´ 3 + ( -2) ´ 4 = -5. (ii)  The norm of a vector X = (x1, x2, x3, ... xn) over the field R of real numbers is denoted by X and is defined by X = x12 + x22 + x32 + ...... + xn2

(ii) Rank of a matrix = number of linearly independent columns (or rows) of the matrix.

Example:

(iii) If A be a matrix of order n×n such that A has “n” distinct eigenvalues, then A has “n” linearly independent eigenvectors.

Remember:

(iv) Two functions f(x) and g(x) are said to be lin-

If q be the angle between the two vectors X and Y, then cos q =

wise they called linearly dependent.

Example:

f ( x ) g ( x ) h( x ) f ¢( x ) g ¢( x ) h¢( x ) ¹ 0 ; otherwise they are f ¢¢( x ) g ¢¢( x ) h¢¢( x )

called linearly dependent.

EMEP.CH01_3PP.indd 18

Let X = (1, 3, -2). Then X = 12 + 32 + ( -2)2 = 14.

f ( x ) g( x ) ¹ 0 ; otherearly independent if f ¢( x ) g¢( x )

(v) Three functions f(x), g(x) and h(x) are said to be linearly independent if

x3 y3 ¹ 0 ; otherwise z3

Example:

a1 X1 + a2 X 2 + a3 X 3 + .... + ak X k = O

Let X1 = (1, 0 - 2) and X2 = (0, 3, 0).

x2 y2 z2

they are said to be linearly dependent.

(ii) The vectors X1, X2, X3, ...., Xk are said to be linearly independent if there exist scalars a1, a2, a3, ..., ak such that Þ a1 = a2 = a3 = ..... = ak = 0 Example:

x1 y1 z1

X ,Y

X Y

.

Let X1 = (2, 1, - 2) and X2 = (3, 0, 4). Then X ,Y = 2 ´ 3 + 0 ´ 1 + ( -2) ´ 4 = -2. Also X = 22 + 12 + ( -2)2 = 3,

Y = 32 + 02 + 42 = 5.

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Linear Algebra • 19 X ,Y

\cos q =



X Y

=

1.10.5  Basis and Dimension A collection of n-tuple vectors X1, X2, X3, ... Xk is said to form a basis of the vector space Vn(F) if

-2 2 =- . 3´5 15

æ 2 ö Now, So q = cos-1 ç - ÷ . è 15 ø

(i) X1, X2, X3, ..., Xk are linearly independent (ii) any arbitrary n-tuple vector X can be expressed as

2

Remember : x = < x , x > .

1.10.4 Orthogonal and Orthonormal Vectors (i) Two vectors X and Y are said to be orthogonal if 〈X, Y〉 = 0. Example: Let X1 = (2, 1, -2) and X2 = (-3, 0, -3). Then X ,Y = 2 ´ ( -3) + 1 ´ 0 + ( -2) ´ ( -3) = 0.

Therefore, the vectors X and Y are orthogonal.

(ii) A set of n-tuple vectors X1, X2, X3, ..., Xk is called an orthonormal set of vectors if (i) X i = 1 for i = 1,2,3,..., k (ii) X i ,Y j = 0 for i ¹ j; i, j = 1,2,3,..., k Example:

X = b1 X1 + b2 X 2 + b3 X 3 + .... + bk X k where b1 ,b2 ,b3 ,....,bk Î F .

If the n-tuple vectors X1 , X 2 , X 3 ,....., X k forms a basis of Vn(F), then we say that dimension of the vector space Vn(F) is “k.” Example: Let X1 = (1,0,0), X 2 = (0,1,0)and X 3 = (0,0,1). 1 0 0 Then 0 1 0 = 1 ¹ 0. 0 0 1

Therefore, the vectors X1 , X 2 , and X 3 are linearly independent. Now let X = ( a,b, c ) be any arbitrary 3-tuple vector.

Let X1 = 1 (1,1,1)and X 2 = 1 ( -1,0,1) .



Then

Hence, the vectors X1 , X 2 , and X 3 , form a basis of V3(R) and dimension of V3(R) is “3.”

3

2

æ 1 1 (1,1,1) = ç , , 3 è 3 3 æ -1 1 ( -1,0,1) = ç ,0, X2 = 2 è 2

X1 =

1

1 ö ÷ and 3ø 1 ö ÷. 2ø



1 æ -1 ö 1 1 \ X1 , X 2 = ´ç ´ ÷+0+ 3 è 2ø 3 2 =0



Therefore, the vectors X1 and X2 are orthogonal.

Also X1

2

æ



2

2

æ 1 ö æ 1 ö æ 1 ö = ç ÷ +ç ÷ +ç ÷ = 1, è 3ø è 3ø è 3ø

X2 = ç è

2

2

1 ö 2 æ 1 ö ÷ +0 +ç ÷ =1 2ø è 2ø

Hence, the vectors X1 and X2 form an orthonormal set of vectors.

EMEP.CH01_3PP.indd 19

Then we can write X = ( a,b, c ) = a(1,0,0) + b(0,1,0) + c(0,0,1) = aX1 + bX 2 + cX 3

Fully Solved MCQs (Level-I)

1. For the matrices A3×1, B1×3, C3×5, D5×3 which of the followings is possible? (a) AB only (b) CD only (c) AB and CD only (d) AB, BA, BC, CD, DA, DC only. 2. A square matrix “A” of order 2 which commutes with every 2 × 2 matrix is of the form æ0

(a) ç èa æa

(c) ç è0

aö ÷ (b) 0ø

æa ç è0 0ö æ0 ÷ (d) ç bø èb æ0

0ö ÷ aø aö ÷ 0ø

aö

3 3. If A = ç ÷ , then A + A = 0, whenever èb 0 ø (a) ab = 2 (b) ab = 1 (c) ab ≠ 0 (d) ab = -1

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20 • Engineering Mathematics Exam Prep





4. A square matrix becomes a diagonal matrix if and only if (a) it is upper triangular (b) it is lower triangular (c) both lower and upper triangular (d) none of these 5. If a matrix A is symmetric as well as skewsymmetric, then A is (a) Diagonal matrix (b) Null matrix (c) Unit matrix (d) None of these æ6

xö

6. Let A = ç ÷ and A = A′. Then 0ø èy (a) x = 0, y = 6 (b) x + y = 6 (c) x = y (d) None of these æ1

7. If A = ç è -4

-2 2

æ2 3ö ç ÷ , B = ç4 7ø ç2 è

3ö ÷ 5 ÷ , then 0 ÷ø

(a) AB, BA exist but not equal (b) AB exists but BA does not exist (c) AB does not exist but BA exist (d) AB, BA exist and both are equal 8. If A and B are square matrices of order 3 such that det (A) = - 1, det (B) = 3; then det (3AB) = ? (a) 9 (b) 81 (c) 27 (d) -81 æ4

9. If A = ç è1

5ö ÷ , then A. adj A = ? 2ø

æ3 0ö

æ -3

(a) ç ÷ (b) ç è0 è0 3ø æ10 0 ö (c) ç ÷ (d) è 0 10 ø

0ö ÷ -3 ø

æ 1 3ö ç ÷ è -3 0 ø

10. If A, B, C are square matrix of the same order, then AB = AC ⇒ B = C if (a) |A| ≠ 0 (b) |A| = 0 (c) A = I (d) A = 0 11. If w is a complex cube root of unity, then the æ1 ç matrix A = ç w2 çç èw



(a) Singular (c) Symmetric æ 2x

12. If A = ç è x

EMEP.CH01_3PP.indd 20

(a) 1 (c) 1/2

w w 1

w2 ö ÷ 1 ÷ is ÷ w2 ÷ø

(b) Non-singular (d) None of these

æ1 0ö -1 ÷ and A = ç -1 xø è

(b) 2 (d) - 1

0ö ÷ , then x = ? 2ø

13. If A is a square matrix, then adj(A′) - (adj A)′is equal to (a) 2|A| (b) 2|A|I (c) null matrix (d) I 14. If A is an idempotent matrix and A + B = I, then B is (a) idempotent (b) involutary (c) null matrix (d) none of these 15. If A is a real skew-symmetric matrix such that A2 + I = 0, then A is (a) idempotent (b) involutary (c) null matrix (d) orthogonal æ ab

16. The matrix A = çç

è -a



(a) Idempotent (c) Nilpotent æ ç ç ç 17. The matrix ç ç ç ç è



1 3 1 3 1 3

(a) Orthogonal (c) Symmetric 18. The matrix A =



2

(a) Unitary (c) Symmetric

b2 ö ÷ is - ab ÷ø

(b) Orthogonal (d) None of these 1 6 -2 6 1 6

-1 ö ÷ 2 ÷ ÷ 0 ÷ is ÷ 1 ÷ ÷ 2ø

(b) Idempotent (d) None of these 1 æ 1 ç 2 è -i

iö ÷ is -1ø

(b) Idempotent (d) None of these

é 7 1+i 2ù ê ú 19. For the matrix M = ê1 - i 4 3i ú which of êë 2 -3i 0 úû

the following is correct? (a) M is skew-Hermitian and iM is Hermitian (b) M is Hermitian and iM is Skew-Hermitian (c) M and iM both are Hermitian (d) M and iM both are Skew-Hermitian æ2 ç 20. If the rank of the matrix A = ç 3 ç1 è

is “3,” then the value of x is?

4 2ö ÷ 1 2÷ 0 x ÷ø

4 3 (b) ¹ 5 5 2 (c) ¹ (d) none of these 5

(a) ¹

8/9/2023 8:29:04 PM

Linear Algebra • 21

21. The rank of the unit matrix I of order n is (a) n - 1 (b) n (c) n2 (d) n+1 22. If A is a non-singular matrix of order n, then the rank of A is (a) n - 1 (b) n (c) 2 (d) n+1 æ1 ç

1

1ö ÷

ç4 è

6

8 ÷ø

23. The rank of the matrix ç 2 3 4 ÷ is



(a) 2 (c) 1 æ0 ç 0 24. If A = ç ç0 ç è0

(a) 2 (c) 1

(b) 0 (d) 3 2 0 0 0

0 3 0 0

0ö ÷ 0÷ , then rank (A) =? 4÷ ÷ 0ø

(b) 0 (d) 3

æ1 ç 25. The rank of the matrix ç 0 ç0 è



2 4 0

3 1 5

4ö ÷ 2 ÷ is 0 ÷ø

(a) 1 (b) 2 (c) 3 (d) 4 26. The system of given equations x + y + z = 6, x - y + z = 2, 2x + y - z = 1 has (a) unique solution (b) two solutions (c) an infinite number of solutions (d) no solution 27. A system of “m” homogeneous linear equations AX = 0 in “n” unknowns has only trivial solution if (a) m ≠ n (b) m = n (c) rank (A) = m (d) rank (A) = n 28. The system of equations 4x + 6y = 5, 6x + 9y = 7 has (a) a unique solution (b) an infinite number of solutions (c) no solution (d) finite number of solutions 29. The equations 2x - 3y + 6z = 4, 5x + 7y 14z = 1, 3x + 2y - 4z = 0 has (a) a unique solution (b) no solution (c) infinitely many solutions (d) none of these

EMEP.CH01_3PP.indd 21

30. The values of “a” for which the system of equations ax + y + z = 0, x +ay + z = 0, x + y + z = 0 posses non-zero solutions are given by (a) 1, 2 (b) 1, 1 (c) 1, - 1 (d) - 1, - 2 31. The system of equations x + 2y + 3z = 1, 2x + y + 3z = 2, 5x + 5y + 9z = 4 has (a) a unique solution (b) no solution (c) infinitely many solutions (d) none of these 32. The system of equations x + y + z = 0, 2x + y - z = 0, 3x + 2y = 0 has (a) a unique solution (b) no solution (c) infinitely many solutions (d) none of these 33. The system of equations x + 2y + 3z = 0, 2x + 3y + 4z = 0, 3x + 4y + 5z = 0 has (a) unique solution (b) an infinite number of solutions (c) trivial solution (d) none of these 34. The system of equations x - 2y + z = 0, x - 2y - z = 0, 2x - 4y - 5z = 0 has (a) unique solution (b) an infinite number of solutions (c) trivial solution (d) a none of these 35. Consider the system of equations 5x + 2y - z = 1, 2x + 3y + 4z = 7, 4x - 5y + lz = l - 5 It will have a unique solution if (a) l = 14 (b) l ≠ 14 (c) l ≠ -14 (d) l = -14 36. The system of equations x1 + x2 + x3 + x4 = 0,

x1 + 3x2 + 2x3 + 4 x4 = 0,



2x1 + x3 - x4 = 0.

has

(a) unique solution (b) an infinite number of solutions (c) trivial solution (d) none of these 37. The system of equations x + 2y + 3z = 0, 2x + 3y + z = 0, 3x + y + 2z = 0 has

8/9/2023 8:29:04 PM

22 • Engineering Mathematics Exam Prep

det (P + I) = 0, where I is the identity matrix, then the eigenvalues of P are? (a) -1, -2, 3 (b) 4, -4, 0 (c) 1, 2, 3 (d) 1, 1, 6

(a) two solutions (b) an infinite number of solutions (c) trivial solution (d) none of these æ 1 2ö 4 ÷ , , then A = ? 1 3 è ø

46. If the vectors (0, 1, a), (1, a, 1) and (a, 1, 0) are linearly dependent, then the value of “a” will be (a) 0, 1 (b) 1, -1

38. If A = ç

(a) 24 A - 55I (b) -4A + 55I (c) A + I (d) A - I



æ2 1 0ö ç ÷ 39. The eigenvalues of A = ç1 2 0 ÷ are ç1 1 1 ÷ è ø

(a) 1, 2, 3 (c) 1, 1, 3

(a) 1, 3, 1 (c) 0, 2, 3

0 1 0

1ö ÷ 0 ÷ are 2 ÷ø

48. If the vectors (1, 0, 1), (1, 1, 0) and (p, q, r) are linearly dependent, then which of the following is true? (a) p + q + r = 0 (b) p = q = r (c) p = q + r (d) r = p + q

(b) 1, 2, 3 (d) 1, 1, 1

æ 2 -2 0 ö ç ÷ 41. The eigenvalue of A = ç -2 1 - 2 ÷ are ç 0 -2 0 ÷ è ø



Answer key

(a) 1, 2, 4 (b) 1, -2, 4 (c) 1, -2, -4 (d) -1, 2, 4

æ1 ç 42. If A = ç1 ç0 è

0 0 1

0ö ÷ 1 ÷ , then A–1 = ? 0 ÷ø

(a) A2 + A - I (b) A2 - A - I (c) A2 + A + I (d) None of these 43. The eigenvalue and eigenvector æ0 A=ç è0

1ö ÷ are, respectively 0ø

of

æ0ö ækö ç ÷ (b) 1, ç k ÷ è ø è0ø ækö (c) 0, æç k ö÷ (d) 1, ç ÷ èkø èkø

(a) 0,

é1

é1 ù

(a) 2 (c) 3

2. (b) 7. (a) 12. (c) 17. (a) 22. (b) 27. (b) 32. (c) 37. (c) 42. (d) 47. (d)

3. (d) 8. (d) 13. (c) 18. (a) 23. (a) 28. (c) 33. (b) 38. (a) 43. (a) 48. (c)

4. (c) 9. (a) 14. (a) 19. (b) 24. (d) 29. (b) 34. (b) 39. (c) 44. (b)

5. (b) 10. (a) 15. (d) 20. (a) 25. (c) 30. (b) 35. (c) 40. (a) 45. (a)

1. (d) The number of column of A is 1 and the number of rows of B is 1. So AB is possible. -n ù

(b) -2 (d) 1

45. Let P be a 3 × 3 matrix with real entries such that det (P) = 6 and trace (P) = 0. If

EMEP.CH01_3PP.indd 22

1. (d) 6. (c) 11. (a) 16. (c) 21. (b) 26. (a) 31. (a) 36. (b) 41. (b) 46. (c)

Explanation

44. If ê ú is an eigenvector of ê ú , then ë -3 2n û “n”ë=-1?û

(d) 0

47. If the vectors (a, b) and (c, d) are linearly dependent, then which of the following is true? (a) ab = ac (b) a + c = b + d (c) a = d = 0 (d) ad - bc = 0

(b) 1, 3, 3 (d) 1, 1, 1

æ2 ç 40. The eigenvalue of A = ç 0 ç1 è



(c) 0, ± 2

The number of column of B is 3 and the number of rows of both A and C are 3. So BA and BC both are defined. The number of column of C is 5 and the number of rows of D is 5. So CD is possible. The number of column of D is 3 and the number of rows of both A and C are 3. So DA and DC are possible.

8/9/2023 8:29:05 PM

Linear Algebra • 23 æx yö 2. (b) Let, B = ç ÷ be any arbitrary 2 × 2 èz t ø æa 0ö matrix and A = ç ÷ . Then è0 aø æ a 0 öæ x y ö æ ax ay ö AB = ç ÷ç ÷=ç ÷, è 0 a øè z t ø è az at ø æx

y öæ a 0 ö

æ ax

ay ö

BA = ç ÷ç ÷=ç ÷ è z t øè 0 a ø è az at ø Hence,  AB = BA which means A commutes with B.

æ0

a öæ 0 ÷ç 0 øè b

3. (d) A2 = AA = ç èb

æ ab A =AA= ç è 0 3

2

0 öæ 0 ÷ç ab øè b

Then,

a ö æ ab ÷=ç 0ø è 0

0 ö ÷ ab ø

aö æ0 ÷=ç 0 ø çè ab2

a2b ö ÷ 0 ÷ø

a2b ö æ 0 a ö æ 0 0 ö ÷+ç ÷=ç ÷ 0 ÷ø è b 0 ø è 0 0 ø



Adding (i) and (ii) we get 2AT = O



⇒ (AT)T = OT⇒ A = O. æ6



⇒ç

Then



⇒ a2b + a = 0, ab2 + b = 0





⇒ a(ab + 1) = 0, b(ab + 1) = 0



⇒ ab + 1 = 0  (assuming a ≠ 0, b ≠ 0)



⇒ ab = -1 æa ç 4. (c) Consider A = ç d çg è

b e h

c ö ÷ f ÷ k ÷ø

clearly a diagonal matrix.

0 e 0

EMEP.CH01_3PP.indd 23

4 1

5 =4×2-5×1=3 2 æ1

0ö

æ3

0ö ÷ 3ø

A. adj A = |A|I = 3 ç ÷=ç è 0 1ø è 0

11. (a)

0ö ÷ 0 ÷ , which is k ÷ø

5. (b) A is symmetric ⇒  AT = A…(i)

9. (a) |A| =

10. (a) |A| ≠ 0 ⇒ A–1 exist. \  AB = AC ⇒ A–1(AB) = A–1(AC) ⇒ (A–1 A)B = (A–1 A)C ⇒  IB = IC ⇒  B = C

If A is upper as well as lower triangular, then the elements lying below and above of the principal diagonal will be all zero. So b = c = d = f = g = h = 0. æa Hence, A becomes çç 0 ç0 è

yö

8. (d) Det (3AB) = 33 det(A) det(B) = 27 × (-1) × 3 = -81 (since AB is a square matrix of order “3”).



æ 0 a2b + a ö æ 0 0 ö ÷=ç ÷ ç 2 0 ø÷ è 0 0 ø è ab + b

æ6

7. (a) A is a matrix of order 2 × 3 and B is a matrix of order 3 × 2. Therefore, AB and BA are defined. Also AB is a matrix of order 2 × 2 and BA is a matrix of order 3 × 3. So obviously they are not equal.

æ 0 A + A = 0 ⇒ çç 2 è ab 3

xö

6. (c) A = A′ ⇒ ç ÷=ç ÷ 0ø èx 0ø èy Then equating the corresponding elements we get x = y.

A is skew-symmetric ⇒  AT = -A

…(ii)

1

w w2

A = w2 w w

1 + w + w2 w w2

1 = 1 + w + w2 w

1 w2

1

1 + w + w2 1 w2

( by C1 ® C1 + C2 + C3 ) 0 w w2 = 0 w 1 =0

(

0 1 w2

 1 + w + w2 = 0 Hence, A is singular.

)

8/9/2023 8:29:06 PM

24 • Engineering Mathematics Exam Prep 12. (c)

2x A = x

0 = 2x 2 x

\ A -1 =

1 adj( A ) A T

1 æx = 2ç 2x è 0 =

-x ö ÷ 2x ø

0 ö ÷ 2x ø

1 æ x ç 2x 2 è - x

æ 1 ç 2x =ç ç- 1 ç è 2x



æ1 1 1 ç3 + 6 + 2 ç 1 2 = çç - + 0 3 6 ç 1 ç +1 -1 ç3 6 2 è

18. (a) A=





è -a



æ a 2b2 - a 2b2

= çç

2 è -a b + ab

2

b2 ö æ ab b2 ö ÷´ç ÷ - ab ÷ø çè -a 2 - ab ÷ø

ab3 - ab3

ö ÷=O 2 2 2 2÷ -a b +a b ø

Hence, A is a nilpotent matrix of index 2. 17. (a)



EMEP.CH01_3PP.indd 24

æ ç ç ç T AA = ç ç ç ç è

1 3 1 3 1 3

1 6 -2 6 1 6

-1 ö æ ÷ç 2 ÷ç ÷ç 0 ÷ç ÷ç 1 ÷ç ÷ç 2 øè

1 3 1 6 -1 2

1 3 -2 6 0

1 ö ÷ 3÷ 1 ÷ ÷ 6÷ 1 ÷ ÷ 2ø

T

1 æ1 -i ö 1 æ1 i ö ç ÷ = ç ÷ i 1 2è 2 è -i -1 ø ø 1 æ1 i ö 1 æ1 i ö \ AA q = ç ÷´ ç ÷ 2 è -i -1 ø 2 è -i -1 ø 2 i -i ö 1 æ1 - i = ç ÷ ç 2 è -i + i -i2 + 1 ÷ø 1 æ 2 0 ö æ1 0 ö = ç ÷=ç ÷=I 2 è0 2ø è0 1 ø

Similarly AAq = I. Hence, AqA = AAq = I. So A is unitary.

15. (d) A is skew-symmetric ⇒ AT = -A ⇒ AAT = -A A = -A2 = I(using A2 + I = 0). Hence, A is orthogonal. æ ab

1 æ1 i ö ç ÷ 2 è -i -1 ø

Þ Aq =

14. (a) A + B = I   ⇒  B = I - A ⇒  B2 = (I - A)2 = I - 2IA + A2 ⇒  B2 = I - 2A + A (since A is idempotent, so A2 = A) ⇒  B2 = I - A = B Hence, B is idempotent.

1 1 1ö + 3 6 2÷ ÷ 1 2 - + 0 ÷÷ 3 6 ÷ 1 1 1÷ + + 3 6 2 ÷ø

So A is orthogonal.

13. (c) adj(A′) - (adj A′) = adj (A′) - adj(A′) = O = null matrix.

16. (c) A2 = A × A = çç

æ1 0 0 ö ç ÷ = ç0 1 0÷ = I ç0 0 1÷ è ø



ö 0÷ æ 1 0ö ÷=ç ÷ (given) 1 ÷ è -1 2 ø ÷ xø 1 1 Þ =1 Þ x = . 2x 2

1 2 - +0 3 6 1 4 + +0 3 6 1 2 - +0 3 6

19. (b)

é 7 1-i 2 ù ê ú M = ê1 + i 4 -3i ú and so êë 2 3i 0 úû

é 7 1+i 2ù ê ú q M = M = ê1 - i 4 3i ú êë 2 -3i 0 úû 1-i 2ù é7 ê ú = ê1 + i 4 - 3i ú = M ê 3i 0 úû   ë2

T

( )

T

Hence, M is Hermitian matrix and so iM is skew-Hermitian. 20. (a) Rank (A) = 3

⇒  A has non-zero determinant



⇒ det(A) ≠ 0.

8/9/2023 8:29:07 PM

Linear Algebra • 25



2 4 2 Þ 3 1 2 ¹0 1 0 x

27. (b) Any system of “n” homogeneous linear equations in “n” unknowns has a trivial solution.

Þ 2( x - 0) - 4(3x - 2) + 2(0 - 1) ¹ 0 Þ -10x + 6 ¹ 0

28. (c)

Þx¹

3 . 5

21. (b) Since I is an unit matrix of order “n” and det (I) = 1 ≠ 0. So rank (I) = order (I) = n. 22. (b) Since A is non-singular, so det (A) ≠ 0. Hence, rank (A) = order of A = n. 23. (a)



D=



D1 =

2 -3 6 Here D = 5 7 -14 = 0, 3 2 -4 4 -3 6 D1 = 1 7 -14 = 0, 0 2 -4

24. (d) A has only three non-zero rows. Hence, rank (A) = 3. 2 4 0

3 1 5

4ö ÷ 2÷ 0 ÷ø

Then A is a matrix of order 3 × 4. Therefore, rank (A) ≤ min {3, 4} = 3.  Now consider a sub matrix B of order 3, where æ1 ç B = ç0 ç0 è

2 4 0

3ö ÷ 1 ÷ . Then det (B) = 1 × 4 × 5 = 20 ≠ 0 5 ÷ø

Therefore, by Cramer’s rule, the system has no solution. 30. (b) The homogeneous system has a non-zero solution ⇒ rank of the co-efficient matrix < 3 (here number of unknowns = 3) Þ A =0 a 1 1 Þ 1 a 1 =0 1 1 1

Alternative method:

26. (a)



1 1 1 Here D = 1 -1 1 = 6 ¹ 0. 2 1 -1

Therefore by Cramer’s rule, the given system of equations has a unique solution.

EMEP.CH01_3PP.indd 25

2 4 6 D2 = 5 1 -14 ¹ 0 3 0 -4



Therefore, rank (A) = order of the square sub-matrix B = 3. The given matrix has three non-zero rows. Hence, its rank = 3.

6 = 45 - 42 = 3 ¹ 0 9

29. (b)

which has two non-zero rows. Hence, rank = 2.

æ1 ç Let, A = ç 0 ç0 è

5 7

6 = 36 - 36 = 0 9

Therefore, by Cramer’s rule, the system has no solution.

æ1 1 1 ö æ1 1 1 ö ç ÷ ç ÷ ç 2 3 4 ÷  ç 2 3 4 ÷ (by R3 ® R3 - 2R2 ) ç4 6 8÷ ç0 0 0÷ è ø è ø

25. (c)

4 6

Þ a 2 - 2a + 1 = 0 Þ a = 1,1

31. (a)



1 2 3 Here D = 2 1 3 = 3 ¹ 0. 5 5 9

Therefore, by Cramer’s rule, the system has a unique solution.

8/9/2023 8:29:10 PM

26 • Engineering Mathematics Exam Prep 32. (c)



34. (b) æ1 ç A = ç1 ç2 è æ1 ç  ç0 ç0 è

1 1 1 Here A = 2 1 -1 = 0. 3 2 0

Therefore, rank of the coefficient matrix < 3 (here number of variables = 3). Hence, the system has an infinite number of solutions.



(by R2 ® R2 - R1 , R3 ® R3 - 2R1 )



æ 1 -2 1 ö ç ÷  ç 0 0 -2 ÷ ç0 0 0 ÷ è ø 7 (by R3 ® R3 - R2 ) 2



which has two non-zero rows.

33. (b)



æ1 2 3 ö ç ÷ A = ç2 3 4÷ ç3 4 5÷ è ø æ1 2 3 ö ç ÷  ç 0 -1 -2 ÷ ç 0 -2 -4 ÷ è ø (by R2 ® R2 - 2R1 , R3 ® R3 - 3R1 )



æ1 2 3 ö ç ÷  ç 0 -1 -2 ÷ ç0 0 0 ÷ è ø (by R3 ® R3 - 2R2 )



which has two non-zero rows.

Therefore, rank (A) = 2 < 3 (= number of unknowns) Hence, the system has an infinite number of solutions. Alternative method: Here, co-efficient determinant 1 2 3 = 2 3 4 3 4 5 = (15 - 16) - 2(10 - 12) + 3(8 - 9) = 0.

Therefore, the rank of the coefficient matrix < 3 (here number of variables = 3). Hence, the system has an infinite number of solutions.

EMEP.CH01_3PP.indd 26

-2 1 ö ÷ -2 -1 ÷ -4 -5 ÷ø -2 1 ö ÷ 0 -2 ÷ 0 7 ÷ø

Therefore rank (A) = 2 < 3 (= number of unknowns) Hence, the system has an infinite number of solutions. Alternative method:

Here, co-efficient determinant 1 -2 1 = 1 -2 -1 2 -4 -5 = (10 - 4) + 2( -5 + 2) + 1( -4 + 4) = 0.

Therefore, rank of the co-efficient matrix < 3 (here number of variables = 3). Hence, the system has an infinite number of solutions. 35. (c)

5 | A |= 2 4

2 3 -5

-1 4 l



= 5(3l + 20) - 2(2l - 16) -1(-10 - 12)



= 11l + 154



System has a unique solution



⇒ |A| ≠ 0 ⇒ 11l + 154 ≠ 0 ⇒ l ≠ -14.

8/9/2023 8:29:11 PM

Linear Algebra • 27

36. (b)

38. (a)

æ1 1 1 1 ö ç ÷ A = ç1 3 2 4 ÷ ç 2 0 1 -1 ÷ è ø

æ1 1 ç  ç0 2 ç 0 -2 è (by R2 ® R2

1 1 ö ÷ 1 3÷ -1 -3 ÷ø - R1 , R3 ® R3 - 2R1 )

æ1 1 1 1 ö ç ÷  ç0 2 1 3÷ ç0 0 0 0÷ è ø (by R3 ® R3 + R2 )



which has two non-zero rows



Therefore rank (A) = 2 < 4.



(here number of unknowns = 4)

Hence, the system has an infinite number of solutions. 37. (c) æ1 ç A = ç2 ç3 è æ1 ç  ç0 ç0 è

2 3ö ÷ 3 1÷ 1 2 ÷ø 2 3ö ÷ -1 -5 ÷ -5 -7 ÷ø

(by R2 ® R2 - 2R1 , R3 ® R3 - 3R1 )



which has three non-zero rows



Therefore, rank (A) = 3 = number of unknowns

Hence, the system has a unique solution. So, the only solution is x = y = z =0 (trivial solution). Alternative method:

|A - lI| = 0 ⇒

1-l 2 =0 -1 3-l



⇒  (1 - l) (3 - l) + 2 = 0



⇒  l2 - 4l + 5 = 0



Therefore, by the Cayley Hamilton theorem, A2 - 4 A + 5I = O Þ A2 = 4 A - 5I

( )

Þ A2



2

2

= ( 4 A - 5 I ) = 16 A 2 - 40 A + 25 I

Þ A 4 = 16 ( 4 A - 5 I ) - 40 A + 25 I = 24 A - 55 I .

39. (c) 2 1 | A |= 1 2 1 1



0 0 = 2(2 - 0) + 1(0 - 1) = 3 = 1 × 1 × 3 1

= product of the eigenvalues. 40. (a)

2 | A |= 0 1



0 1 0

1 0 = 2(2 - 0) + 1(0 - 1) = 3 = 1 × 1 × 3 2

= product of the eigenvalues. 41. (b) Trace of the matrix = 2 + 1 + 0 = 3 = sum of eigenvalues = 1 + 4 + (-2). 42. (d)

1-l |A - lI| = 0 ⇒ 1 0

0 0 0-l 1 =0 1 0-l

⇒  (1 - l) × (l2 - 1) = 0 ⇒  l2 - 1 - l3 + l = 0

Here, co-efficient determinat

⇒  l3 - l2 - l + 1 = 0

1 2 3 = 2 3 1 3 1 2

Therefore, by the Cayley Hamilton theorem,

= (6 - 1) - 2(4 - 3) + 3(2 - 9) = -18 ¹ 0.

Hence, the system has a unique solution which is the trivial solution.

EMEP.CH01_3PP.indd 27

A3 - A2 - A + I = 0 ⇒  A–1(A3 - A2 - A + I) = A–10 ⇒  A–1A3 - A–1A2 - A–1A + A–1I = 0 ⇒  A2 - A - I + A–1 = 0 ⇒  A–1 = A + I - A2

8/9/2023 8:29:12 PM

28 • Engineering Mathematics Exam Prep 47. (d) The vectors (a, b) and (c, d) are linearly dependent

43. (a) æ0 è0

A - lI = ç

1ö æ1 ÷ - lç 0ø è0

0 ö æ -l 1 ö ÷=ç ÷ 1ø è 0 - l ø

-l 1 =0 0 -l



\ |A - lI| = 0 ⇒



⇒  l2 = 0 ⇒ l = 0, 0



Therefore, 0 is the only eigenvalue.

Þ





1 0 1 Þ 1 1 0 =0 p q r



Fully Solved MCQs (Level-II)



(by replacing x by k)



which is the eigenvector for eigenvalue “0.” 44. (b)

AX = lX

é 1 -n ù é 1 ù é1 ù Þê úê ú = lê ú 3 2 n 1 ë ûë û ë -1û é 1+n ù é l ù Þê ú=ê ú ë -3 - 2n û ë -l û Þ 1 + n = l, - 3 - 2n = -l Þ -3 - 2n = -(1 + n) Þ n = -2.

45. (a) Det(P) = 6 = product of the eigenvalues = (-1) × (-2) × 3 Trace(P) = 0 = sum of the eigenvalues = (-1) + (-2) + 3. 46. (c) The vectors (0, 1, a), (1, a, 1) and (a, 1, 0) are linearly dependent 0 1 a Þ 1 a 1 =0 a 1 0

1. If A and B are square matrices of the same order such that (A + B)2 = A2 + B2 + 2AB, then (a) AB = BA (b) A = B (c) A + B = 0 (d) A = -BT 2. Let A be a 3 × 5 matrix and B be a matrix such that ATB and BAT are both defined. Then B is of the type: (a) 3 × 5 (b) 5 × 3 (c) 3 × 3 (d) 5 × 5 æ cos a

4. If A and B are square matrices of same order so that AB = A, BA = B, then (a) Both A and B are singular. (b) Both A and B are non-singular. (c) Both A and B are unit matrix. (d) A2 = A, B2 = B 5. If the matrices A and B commute, then (a) (AB)n = AnBn (b) (AB)n = AB n n (c) (AB) = B (d) none of these æ 1

6. If A = ç è -1

Þ 0 - (0 - a ) + a(1 - a ) = 0 Þ 2a - a3 = 0

EMEP.CH01_3PP.indd 28

sin a ö

3. If Aa = ç ÷ , then è - sin a cos a ø (a) Aan = Aa (b) Aan = nAa (c) Aan = Ana (d) Aan = 0

2



Þ (r - 0) - 0 + (q - p) = 0 Þ p =r+q

Þ y=0

æx ö ækö Hence, X = ç ÷ = ç ÷ è0ø è0ø



Þ ad - bc = 0

48. (c) The vectors (1, 0, 1), (1, 1, 0) and (p, q, r) are linearly dependent

\ AX = lX æ 0 1 öæ x ö æxö Þç ÷ç ÷ = p ç ÷ è 0 0 øè y ø è yø æ yö æ0ö Þç ÷=ç ÷ è0ø è0ø

a b =0 c d

0ö n ÷ , then A = ? 1ø

æn

0ö

æn

0ö

æ1

0ö

æ1

0ö

Þ a(2 - a 2 ) = 0

(a) ç ÷ ÷ (b) ç è -n n ø è -1 n ø

Þ a = 0, ± 2



(c) ç ÷ (d) ç ÷ è -n 0 ø è -n 1 ø

8/9/2023 8:29:15 PM

Linear Algebra • 29 æ i

7. If A = ç è -i 8 then A = ?

æ 1 - iö ÷ and B = ç -1 iø è

-1ö ÷ and 1ø

(a) 128B (c) 116B

(b) 130B (d) 8B écos q - sin q ù

éa 0 ù

9. Find P50 if P is given by

é1 50 100 ù ê0 1 50 ú ê ú êë0 0 1 úû é50 100 150 ù é1 50 1275 ù ê ú (c) ê 0 50 100 ú (d) êê0 1 50 úú 0 50 úû êë0 0 1 úû ëê 0 é 0 wù

10. Let P = ê ú , where w is the complex ëw 0 û cube root of unity . Then P24 is equal to (a) P2 (b) P (c) Identity matrix (d) 0 11. Let A = (aij)3×3 be a matrix with aij ∈ R. Let B be a matrix obtained by interchanging two columns of A. Then det (A + B) = ? (a) 2det (A)+ det (B) (b) 0 (c) 2 det (A) (d) det (A) - det (B) 12. What is the value of the following determinant of order “n”? 1 0 1 1 2 1 1 2 .. .. .. ..

EMEP.CH01_3PP.indd 29

1 2

0 ... ... 0 0 ... ... 0 1 ... ... 0 3 .. .. .. .. .. .. .. .. 1 3

..

..

n

(c) 0

(d)

æ cos x 13. If A = çç - sin x ç 0 è

(a) f(x) (c) -f(x)

sin x cos x 0

1 n(n + 1)

0 ö ÷ 0 ÷ = f(x), then A–1 = ? 1 ÷ø

(b) f(-x) (d) -f(-x)

14. If Ak = O, then I + A + A2 + ... + Ak–1 is equal to (a) Null matrix (b) (I + A)k (c) I (d) (I - A)–1 15. If the matrix A, B, A + B are non-singular, then [A(A + B)–1B]–1? (a) A + B (b) A–1 + B–1 (c) (A + B)–1 (d) AB

é1 1 1ù ê ú P = ê0 1 1ú êë0 0 1úû é1 100 500 ù ê ú (a) ê0 1 100 ú (b) êë0 0 1 úû

1

n!



8. Two matrices ê ú and ê 0 b ú ësin q cos q û ë û commute under multiplication. Then (a) a = b or q = nπ (n is an integer) (b) always (c) a = 4b (d) q = π/3



(a) 1 (b) 1

1 n

16. Let A, B be two 3 × 3 invertible matrices with A + B = AB. Then (a) A–1 + B–1 = 0 (b) A–1 + B–1 = B–1 A–1 (c) I - A–1 is invertible (d) I + B–1 is invertible 17. Let A and B be two non-zero 2 × 2 matrices such that AB = 0. Then (a) both A and B are non-singular (b) exactly one of A and B is singular (c) both A and B are singular (d) A + B is singular æ0 18. If A = çç1 ç3 è

1 2 x

2ö ÷ 3 ÷ , A–1 = 1 ÷ø

1 æ1 ç2 - 2 ç 3 ç -4 ç5 3 ç 2 è2

1ö 2÷ ÷ y ÷ , then 1÷ ÷ 2ø

(a) x = 1, y = -1 (b) x = -1, y = 1 (c) x = y =

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

19. Let A and B be symmetric and skew-symmetric matrices, respectively, of same order. Then AmBnAm is (a) Skew-symmetric for all m, n (b) Symmetric for all m, n (c) Skew-symmetric if n is odd and symmetric if n is even. (d) Symmetric if m is even and skew-symmetric if m is odd.

8/9/2023 8:29:18 PM

30 • Engineering Mathematics Exam Prep 20. If A and B are symmetric and commute, then (a) A–1B is symmetric only (b) AB–1 is symmetric only (c) A–1 B–1 is symmetric only (d) A–1B, AB–1, A–1B–1 are all symmetric 21. Let A (a square matrix of order n) be a nilpotent matrix of index “p.” Then (a) In - A is invertible (b) In + A is a zero matrix (c) In + A is nilpotent (d) In - A is a zero matrix 22. If A, a square matrix of order “n,” be nilpotent, then (a) In - A is singular and In + A is non-singular (b) In + A is singular and In - A is non-singular (c) Both In - A and In + A are singular. (d) Both In + A and In - A are non-singular. æ 1 ç 23. The matrix A = ç 5 ç -2 è





(a) idempotent (b) Nilpotent of index 3 (c) skew-symmetric (d) none of these 24. The values of α, β, and g so that the matrix æ0 A = çç a ça è

2b b -b

æ1 2 3 1 ö

29. The rank of the matrix A = çç 2 4 6 4 ÷÷ is ç1 2 3 2 ÷ è ø



1 ,b = ± 2 1 ,b = ± (b) a = ± 2 (c) a = ± 1 , b = ± 2

(d) none of these

1 ,g =± 6 1 ,g =± 2 1 ,g=± 6

1 3 1 3 1 2

25. If A is a square matrix and A - I and A + I 2 2

are both orthogonal, then (i) A is (a) idempotent (b) involutary (c) skew-symmetric (d) symmetric (ii) A2 =



EMEP.CH01_3PP.indd 30

(b) 1

(c) 2

(a) 0

(b) 1

(d) 3 1 3 8 ö ÷ 2 6 -1 ÷ is 3 9 7 ÷ ÷ 4 12 15 ø

(c) 2

(d) 3

æ1 1 31. The rank of the matrix A = çç1 - 1 ç3 1 è



(a) 0

(b) 1

(c) 2

-1 1 ö ÷ 2 - 1 ÷ is 0 1 ÷ø

(d) 3

æ2 3 -1 ç 32. The rank of the matrix A = ç1 - 1 - 2 ç3 1 3 ç 6 3 0 è



(a) 4

(b) 1

(c) 2



(a) 4

(b) 1

(c) 2

æ0 ç 34. The rank of the matrix A = ç1 ç3 è

3 2



(a) 4

(b) 1

(c) 2

-1 ö ÷ -4 ÷ is - 2÷ ÷ - 7ø

(d) 3

æ0 1 - 3 ç 33. The rank of the matrix A = ç1 0 1 ç3 1 0 ç è1 1 - 2

(a) - 3 I (b) - I 4 3 (c) - I (d) I 4

(a) 0

æ6 ç 4 30. The rank of the matrix A = ç ç10 ç è16

gö ÷ - g ÷ is orthogonal are g ÷ø

(a) a = ±



1 3ö ÷ 2 6 ÷ is - 1 - 1 ÷ø

26. Let A be a n × n matrix with integral entries and B = A + (1/2)I, where I is the n × n identity matrix. Then B is (a) idempotent (b) nilpotent (c) invertible (d) none of these 27. Suppose A and B are two orthogonal matrices such that det (A) + det (B) = 0, then (a) A + B = -I (b) A + B = I (c) det (A + B) = 0 (d) A + B = null matrix 28. If A and B are 3 × 3 matrices such that rank (AB) = 1, then rank (BA) cannot be (a) 0 (b) 1 (c) 2 (d) 3

-1 ö ÷ 1 ÷ is 2÷ ÷ 0ø

(d) 3 1 2 1

2 3 1

1ö ÷ 2 ÷ is 3 ÷ø

(d) 3

8/9/2023 8:29:20 PM

Linear Algebra • 31

35. The rank of the matrix æ1 0 1 ç ç1 1 0 A = ç0 1 1 ç ç0 0 1 ç0 1 0 è



(a) 4

0 0 0 1 1

42. If a, b, c are in A.P. with common difference

0ö ÷ 0÷ 0 ÷ is ÷ 0÷ 0 ÷ø

(b) 1

é4 5 a ù

“d” and the rank of the matrix êê5 6 b úú is êë6 l

(c) 2

(d) 3

1ö æ1 1 ç ÷ -1 1 - 1 ÷ ç 36. The rank of the matrix A = is ç 1 1 5÷ ç ÷ è -1 0 - 3 ø



(a) 4

(b) 1

(c) 2

(d) 3



-1 2 0 1

(a) 4

-3 3 1 1

-1 ö ÷ -1 ÷ is 1÷ ÷ -1ø

(b) 1

(c) 2



(a) 4

0 1 1 1

2 0 0 1 0 2 0 0

(b) 1

(a) 3

(b) 4

(d) 3

11 11 , m = 22 (b) l ¹ - , m ¹ 22 3 3 11 11 (c) l ¹ - , m = 22 (d) l ¹ - , m ¹ 22 3 3

2ö ÷ 0÷ is 1÷ ÷ 0ø

(a) l = -

(c) 2 3 7 0 0

5 9ö ÷ 2 -1 ÷ is 1 1÷ ÷ 0 2ø

(c) 1

(d) 2

the value of l is (a) 1

(b) - 1

(c) 0

(d) 4

æ 1 1 0ö ç ÷ -1 1 2 ÷ 41. Let A = ç , then rank (A) = ? ç 2 2 0÷ ç ÷ è -1 0 1 ø



EMEP.CH01_3PP.indd 31

(a) 1

(b) 2

(c) 3

44. Consider the system of equations x + y + z =

(d) 3

0ö æ l -1 0 ç ÷ 0 l -1 0÷ 40. The rank A = ç is 3. Then ç 0 0 l -1 ÷ ç ÷ è -6 11 - 6 1 ø



11 , m ¹ 22 3 3 11 (c) l ¹ - , m ¹ 22 (d) l ¹ - 11 , m ¹ 22 3 3

(a) l = - 11 , m = 22 (b) l = -

(iii) Infinite solution

æ1 ç 0 39. The rank of the matrix ç ç0 ç è0



11 11 (b) l = 3 3

(ii) No solution for

38. The rank of the matrix æ2 ç 0 A=ç ç2 ç è0

(a) l ¹ -

(c) l ≠ -1 (d) l = 0

37. The rank of the matrix æ -2 ç 1 A=ç ç1 ç è 0

c úû

“2.” Then the value of d and l are given by (a) d = 0, l = any arbitrary number (b) d = any arbitrary number, l = 7 (c) d = l = 7 (d) none of these 43. The system of given equations x + 2y + z = 8, 2x + y + 3z = 13, 3x + 4y - lz = m (i) has an unique solution if

(d) 4

6, x + 2y + 3z = 10, x + 2y + lz = m Then the system has (i) An unique solution if (a) l = 3 (b) l ≠ 3 (c) l ≠ 1 (d) l = 1 (ii) No solution for (a) l = 3, m ≠ 10 (c) l = 3, m = 10 (iii) Infinite solution

(b) l ≠ 3, m ≠ 10 (d) l ≠ 3, m = 10

(a) l = 3, m ≠ 10 (b) l ≠ 3, m ≠ 10 (c) l = 3, m = 10 (d) l ≠ 3, m = 10 45. Consider the system of equations

x + 4 y + 2z = 1, 2x + 7 y + 5z = 2m, It has 4 x + ly + 10z = 2m + 1.

(i) no solution for

(a) l = 14, m ¹ 1 (b) l = 14, m = (c) m ≠ 14

2

1 2

(d) None of these

8/9/2023 8:29:22 PM

32 • Engineering Mathematics Exam Prep (ii) a unique solution if (a) l = 14



(b) l ≠ 14

(c) l ¹ 1 (d) l = 1 2

2

(iii) infinite number of solutions for (a) l = 14, m = 1 (b) l ¹ 14, m = 1 2

2

(c) l = m = 14 (d) None of these 46. Consider the system of equations x1 + x2 + x3 = 1,



x1 + 2x2 - x3 = m,



x + 2 y + 3z = l x , 3x + y + 2z = ly, 2x + 3 y + z = l z ;

has a non-trivial soluation, is (a) l = 4 (b) l = 5 (c) l = 6 (d) l = 1 48. The system of equations given below is x1 + 2x2 + 3x3 + x4 = 1,

x1 - x2 + 2x3 - x4 = -3,



3x1 + 3x2 + 8x3 + x4 = -3

(a) consistent (b) inconsistent (c) non-trivial solution (d) none of these 49. The system of equations given below is 2x1 + x2 + 4 x3 = 4, x1 - 3x2 - x3 = -5,

-3x1 + 2x2 - 2x3 = 1,



EMEP.CH01_3PP.indd 32

8x1 - 3x2 + 8x3 = 2

(a) 1, 1, 4 (c) 4, 4, 4

(b) 1, 4, 4 (d) 1, 1, 1

æ 6 -2 2 ö ç ÷ A = ç -2 3 -1 ÷ are ç 2 -1 3 ÷ è ø

5x1 + 7x2 + lx3 = m2

(a) l = 1, m = -1 or 3 (b) l ≠ 1, m ≠ -1 or 3 (c) l = 1, m ≠ -1 or 3 (d) None of these 47. The value of l for which the system

æ 2 -1 1 ö ç ÷ 50. The characteristic values of ç -1 2 -1 ÷ are ç 1 -1 2 ÷ è ø

51. The eigenvectors of the matrix

It has

(i) no solution for (a) l = 1, m ≠ -1, 3 (b) l ≠ 1, m = -1, 3 (c) l ≠ 1, m = -1, 3 (d) none of these (ii) a unique solution if (a) l ≠ -1 (b) l ≠ 3 (c) l ≠ 1 (d) l = 1 (iii) infinite number of solutions

(a) consistent (b) inconsistent (c) non-trivial solution (d) none of these

æ1 ö æ 0 ö æ 2 ö

æ1 ö æ 0 ö æ 2ö

ç 0 ÷ ç1 ÷ ç 1 ÷ è ø è ø è ø

ç1 ÷ ç 0 ÷ ç1 ÷ è ø è ø è ø

ç ÷ ç ÷ ç ÷ (a) çç 2 ÷÷ , çç 1 ÷÷ , çç -1 ÷÷ , (b) ç 0 ÷ , ç 1 ÷ , ç 1 ÷ æ 1 ö æ1 ö æ 2 ö (c) çç 0 ÷÷ , çç1 ÷÷ , çç 0 ÷÷ ç 1 ÷ ç1 ÷ ç 1 ÷ è ø è ø è ø

(d) None of these

æ1 0 2ö ç ÷ 52. If A = ç 0 -1 1 ÷ , then A9 = ? ç0 1 0÷ è ø



(a) 12 A2 - 22A - 9I (b) -12A2 + 22A - 9I (c) 12A2 + 22A + 9I (d) None of these 53. Which of the following can be an eigenvector

æ1 1 1 ö ç ÷ of A = ç1 1 1 ÷ ? ç1 1 1 ÷ è ø 2 æ ö (a) çç 2 ÷÷ (b) ç9÷ è ø

æ1 ö ç ÷ ç1 ÷ ç2÷ è ø

æ1 ö (c) çç1 ÷÷ (d) ç1 ÷ è ø

æ0ö ç ÷ ç0÷ ç1 ÷ è ø

54. The possible set of eigenvalues of a 4 × 4 skew symmetric orthogonal real matrix is (a) ± i (b) ± 1, ± i (c) ± 1 (d) 0, ± i 55. Let A be a 2 × 2 complex matrix such that tr(A) = 1 and det(A) = - 6. Then tr(A4 - A3) = ? (a) 55 (b) 78 (c) 70 (d) 88

8/9/2023 8:29:24 PM

Linear Algebra • 33

56. Let a =



2 pi e 5

é1 ê ê0 ê and M = ê0 ê ê0 ê0 ë

a a2

a3

a a2

a3

0 a2

a3

0 0

a3

0 0

0

a4 ù ú a4 ú ú a4 ú ú a4 ú a4 úû

then trace of I + M + M 2 is

(a) 5 (b) - 5 (c) 1 (d) 0 57. If a 3 × 3 real skew symmetric matrix has an eigenvalue 3i, then one of the remaining eigenvalues is (a) 0 (b) 1/3i (c) - 1/3i (d) 1 58. Let H be a 3 × 3 complex Hermitian matrix which is unitary. Then the distinct eigenvalues of H are (a) ± 1 (b) ± i, ± i (c) 1 ± i (d) (1 ± i )/2

é ê1 ê 59. If A = êê i ê ê êë0

0

0

-1 + i 3 2

0

1 + 2i

-1 - i 2

ù ú ú ú ú , then ú 3ú úû

tr(A102) = ?

(a) 0

(b) 1

(c) 2

(d) 3

60. Suppose that the matrix é 40 -29 -11 ù ê ú A = ê -18 30 -12 ú has a certain complex êë 26 24 -50 úû

number l ≠ 0 as an eigenvalue. Then which of the following must also be an eigenvalue of A? (a) l + 20 (b) l - 20 (c) 20 - l (d) None of these é0 ê 3 61. Consider the matrix M = ê ê2 ê ë0

3 0 1 0

2 1 0 2

0ù ú 0ú . Then 2ú ú 0û

(a) M has no real eigenvalues (b) all eigenvalues of M are positive (c) all eigenvalues of M are negative (d) M  has both are positive and negative real eigenvalues

EMEP.CH01_3PP.indd 33

62. The linear operation L(X) is defined by the cross product L(X) = b × X, where b = [0 1 0]T and X = [x1 x2 x3]T are three dimensional vec-

é x1 ù tors. Also, given that L(X) = A êê x2 úú . Then the eigenvalues of A are êë x3 úû

(a) 0, ± 1 (b) 0, ± i (c) 1, ± i (d) -1, ± i 63. Let A be a 3 × 3 matrix with eigenvalues 0, 1 and -1. Then det (I + A100) = ? (a) 3 (b) 4 (c) 5 (d) 6 64. If w is a non-real cube roots of unity, then the eigenvalues of the following matrix are

é1 1 ê ê1 w ê 2 ë1 w



é ê 1 ù é1 0 0 ù ê1 1 ú ê 1 ú ê w2 ú ´ ê0 -1 0 ú ´ ê1 w ú w4 û ëê0 0 0 ûú ê ê1 1 êë w2

ù 1 ú ú 1 ú w2 ú ú 1 ú w4 úû

(a) 1, 1, -1 (b) 1/3, 1/3, 0 (c) 1, -w, -w2 (d) 3, -3, 0 65. Let P, M, N be square matrices of order “n” such that the matrices M and N are non-singular. If X be an eigenvector of P corresponding to the eigenvalue l, then an eigenvector of N–1MPM–1N corresponding to the eigenvalue l is (a) MN–1X (b) M–1NX (c) M–1 N–1X (d) N–1MX

éa -1 4 ù ú b 7 ú be a matrix with real enêë 0 0 3 úû

66. Let A = êê 0

tries. If the sum and product of all the eigenvalues of A are 10 and 30, respectively, then a2 + b2 equals (a) 29 (b) 45 (c) 58 (d) 60 67. If A be a 3 × 3 non-zero matrix such that A2 = O, then the number of non-zero eigenvalues of A is (a) 0 (b) 1 (c) 2 (d) 4 68. Let A be a 3 × 3 matrix with eigenvalues 1, -1 and 3. Then (a) A2 + A is non-singular (b) A2 - A is non-singular (c) A2 + 3A is non-singular (d) A2 - 3A is non-singular

8/9/2023 8:29:26 PM

34 • Engineering Mathematics Exam Prep 69. If aij = 1 for 1 ≤ i, j ≤ m, then the characteristic equation of the matrix A = {aij}m×m is (a) lm - 2lm–1 + 2 = 0 (b) lm - m = 0 (c) lm - mlm–1 = 0 (d) lm + m = 0 æ4 0ö ÷ . Then è0 4ø

70. Consider the 2 × 2 matrix ç

which of the following vectors is not a valid eigenvector: é1 ù é -2ù (a) ê ú (b) ê ú (c) 0 ë û ë1 û

é4ù ê ú (d) ë -3û

é0 ù ê ú ë0 û

71. One of the eigenvectors of the matrix é 1 -1 0 ù æ1 ö ç ÷ ê ú A = ê 0 1 -1ú is V = ç1 ÷ . ç1 ÷ êë -1 0 1 úû è ø

Then

the

corresponding eigenvalue is? (a) -1

(b) 2

(c) 0

(d) 4

72. Let f(x) = x - 1, g(x) = x + 1, h(x) = x2 - 1 and q(x) = x2 + 1. Then these functions are (a) linearly independent (b) linearly dependent because f(x) × g(x) = h(x) (c) linearly dependent because f(x) - g(x) h(x) = q(x) (d) none of the above 73. If Vn(R) be the vector space over the field of real ì

ü

numbers and W = ïí( a1 , a2 ,....) : lim an = b ïý ïî

n ®¥

be a subspace of Vn(R). Then (a) b is real (b) 0 < b < 1 (c) b = 0 (d) b = 1 74. The set {1, x, x(1 - x)} is (a) linearly independent for all x (b) linearly independent for x = 0 only (c) linearly independent for x = 1 only (d) linearly dependent

ïþ

Answer key 1. (a) 10. (c) 19. (c)

2. (a) 11. (b) 20. (d)

3. (c) 12. (a) 21. (a)

28. (d) 29. (c) 30. (c) 37. (c) 38. (a) 39. (b) 44. (i) (b), (ii) (a), (iii) (c) 47. (c) 48. (b) 49. (a) 56. (a) 57. (a) 58. (a) 65. (d) 66. (a) 67. (a) 74. (a)

4. (d) 13. (b) 22. (d)

5. (a) 14. (d) 23. (b)

31. (c) 32. (d) 33. (c) 40. (a) 41. (b) 42. (b) 45. (i) (a), (ii) (b), (iii) (a) 50. (a) 51. (a) 52. (b) 59. (d) 60. (c) 61. (d) 68. (c) 69. (c) 70. (d)

Explanation

1. (a) (A + B)2 = A2 + 2AB + B2 ⇒ (A + B) (A + B) = A2 + 2AB + B2

⇒  A2 + AB + BA + B2 = A2 + AB + AB + B2



⇒  AB = BA

2. (a) Order of A is 3 × 5 ⇒ Order of AT is 5 × 3 ATB is defined if number of columns of AT = Number of rows of B. Therefore number of rows of B = 3.  BAT is defined if number of columns of B = number of rows of AT.

EMEP.CH01_3PP.indd 34

6. (d) 15. (b) 24. (a)

7. (a) 8. (a) 9. (d) 16. (c) 17. (c) 18. (a) 25. (i)-(c), 26. (c) 27. (c) (ii)-(a) 34. (d) 35. (a) 36. (d) 43. (i) (a), (ii) (b), (iii) (a) 46. (i) (a), (ii) (c), (iii) (a) 53. (c) 54. (a) 55. (b) 62. (b) 63. (b) 64. (d) 71. (c) 72. (c) 73. (c)

Therefore, the number of columns of B = 5. Hence, B is a matrix of order 3 × 5. 3. (c) A2a = Aa × Aa æ cos a sin a ö æ cos a sin a ö =ç ÷´ç ÷ è - sin a cos a ø è - sin a cos a ø æ cos2 a - sin 2 a sin a cos a + sin a cos a ö =ç ÷÷ ç - sin a cos a - sin a cos a - sin 2 a + cos2 a è ø a a cos2 sin 2 æ ö =ç ÷ a a sin 2 cos2 è ø



\  Ana = Ana is satisfied for n = 2.

8/9/2023 8:29:27 PM

æ 1 + 1 -1 - 1 ö =ç ÷ è -1 - 1 1 + 1 ø æ 2 -2 ö =ç ÷ è -2 2 ø

4. (d) AB = A ⇒  A(BA) = A[  BA = B]

⇒  (AB)A = A



⇒  AA = A[  AB = A]



⇒  A2 = A

5. (a) (AB)2 = (AB) (AB) = A(BA)B

æ 1 -1 ö = 2ç ÷ = 2B è -1 1 ø 2

[by associative law]

Similarly considering BA = B it can be shown that B2 = B.

Þ B 4 = ( 2B ) = 4 B 2 = 4 ´ 2B = 8 B



écos q - sin q ù éa 0 ù ú and B = ê ú q q sin cos ë û ë0 bû

Then, BA = AB

éa 0 ù écos q - sin q ù Þê úê ú ë 0 b û ësin q cos q û écos q - sin q ù éa 0 ù =ê úê ú ësin q cos q û ë 0 b û

[  A and B commutative, so AB = BA] = A2B2,



which satisfies (AB) = A B for n = 2. n

æ1

n

0ö æ 1

éa cos q -a sin q ù é a cos q -b sin q ù Þê ú=ê ú ëb sin q b cos q û ë a sin q b cos q û Þ -a sin q = -b sin q Þ ( a - b)sin q = 0 Þ a = b or sin q = 0 Þ a = b or q = np

n

0ö

æ 1-0

0ö

6. (d) A2 = ç ÷´ç ÷=ç ÷ è -1 1 ø è -1 1 ø è -1 - 1 1 ø æ1



\ 

0ö

=ç ÷ è -2 1 ø



æ1

0ö



7. (a) æ i -i ö æ i -i ö A2 = ç ÷´ç ÷ è -i i ø è -i i ø æ i2 + i2 -i2 - i2 ö =ç ÷ ç -i2 - i2 i2 + i2 ÷ è ø 2 2 æ ö =ç ÷ 2 2 è ø



A8 = (-2B)4 = 16B4…(i)



æ 1 -1 ö B=ç ÷ è -1 1 ø æ 1 -1 ö æ 1 -1 ö Þ B2 = ç ÷´ç ÷ è -1 1 ø è -1 1 ø æ 1 + 1 -1 - 1 ö =ç ÷ è -1 - 1 1 + 1 ø æ 2 -2 ö =ç ÷ è -2 2 ø

ésin q = 0 = sin 0 Þ q = np + ( -1)n 0 ù ë û é1 1 1ù ê ú 9. (d) P = ê0 1 1ú êë0 0 é11úû 1 1ù é1 1 1ù 2 P = P ´ P = êê0 1 1úú êê0 1 1úú êë0 0 1úû êë0 0 1úû é1 2 3 ù = êê0 1 2úú êë0 0 1 úû

An = ç ÷ is true for n = 2. è -n 1 ø

æ 1 -1 ö = -2 ç ÷ = -2B è -1 1 ø

\ (i ) Þ A 8 = 16 ´ 8 B = 128 B.

8. (a) Let A = ê

= A(AB)B



Linear Algebra • 35

 





é1 2 1 + 2ù = êê0 1 2 úú 1 úû ëê0 0

é1 P 3 = P 2 ´ P = êê0 êë0 é1 = êê0 êë0

2 3 ù é1 1 1ù 1 2úú êê0 1 1úú 0 1 úû êë0 0 1úû 3 6ù 1 3 úú 0 1 úû

é1 3 1 + 2 + 3ù = êê0 1 3 úú êë0 0 úû 1

æ 1 -1 ö = 2ç ÷ = 2B è -1 1 ø 2

Þ B 4 = ( 2B ) = 4 B 2 = 4 ´ 2B = 8 B \ (i ) Þ A 8 = 16 ´ 8 B = 128 B. EMEP.CH01_3PP.indd 35

8/9/2023 8:29:28 PM

36 • Engineering Mathematics Exam Prep é1 3 6 ù é1 1 1ù P = P ´ P = êê0 1 3 úú êê0 1 1úú êë0 0 1 úû êë0 0 1úû é1 4 10 ù = êê0 1 4 úú ëê0 0 1 ûú 4

é1 50 1 + 2 + 3 + .......... + 50 ù ú = êê0 1 50 ú êë0 0 úû 1

é 0 wù é 0 wù

ëê 0



2

éw

0 ù éw2 úê w2 úû êë 0

0ù ú w2 úû

0ù ú w2 úû

0ù ú w4 úû

0 ù é1 0 ù ú=ê ú= w24 ûú ë0 1 û æ1 2 ç 11. (b) Let A = ç 4 1 ç6 7 è

0 ù éw6 ú=ê w2 ûú êë 0

0ù ú w6 ûú

I (since w24 = ( w3 )8 = 1) . 3ö ÷ 2÷ 5 ÷ø

Let B be a matrix obtained by interchanging two columns C2 and C3 of A.

EMEP.CH01_3PP.indd 36



= 2(36 - 36) - 5(96 - 36) + 5(96 - 36) = 0

1 1 = . n n!

æ cos x sin x ç = ç - sin x cos x ç 0 0 è æ cos x - sin x ç = ç sin x cos x ç 0 0 è

T

0ö ÷ 0÷ 1 ÷ø

0ö ÷ 0÷ 1 ÷ø

æ cos( -x ) sin( -x ) 0 ö ç ÷ = ç - sin( -x ) cos( -x ) 0 ÷ ç 0 0 1 ÷ø è = f ( -x ).

14. (d)  I - A k = ( I - A )( I + A + A 2 + ... + A k -1 )

Continuing like this we get, P24

éw24 = ê ëê 0

\  det (A + B)

= cos2 x + sin 2 x = 1 1 A -1 = adj( A ) A

P2 = P × P = ê úê ú=ê ëw 0 û ëw 0 û êë 0

0 ù éw2 úê w4 ûú ëê 0



cos x sin x 0 13. (b)   A = - sin x cos x 0 0 0 1

é 0 wù

éw4

and

1 1 2 3

10. (c)   P = ê ú ëw 0 û

P 6=P 4×P 2= ê



æ2 5 5ö ç ÷ A+B= ç8 3 3÷ ç12 12 12 ÷ è ø

ç6 è

= 1 ´ ´ ´ ..... ´

50(50 + 1) ù é ê1 50 ú 2 ê ú 50 50 P = ê0 1 ú ê0 0 ú 1 ê ú ë û 1 50 1275 é ù 50 úú = êê0 1 êë0 0 1 úû

éw4 ê êë 0

2ö ÷ 1÷ 7 ÷ø

ç B = ç4

12. (a) Since the associated matrix is lower triangular, so value of the determinant = product of the elements lying in the principal diagonal

Continuing like this we get,

éw2 P4 = P2 × P2 = ê êë 0

3 2 5

Then,



é1 4 1 + 2 + 3 + 4 ù ú = êê0 1 4 ú êë0 0 úû 1



æ1

3

Þ I = ( I - A )( I + A + A 2 + ... + A k -1 ) [ A k = O ]..........(1) Þ ( I - A )( I + A + A 2 + ... + A k -1 ) = I Þ I - A I + A + A 2 + ... + A k -1 = 1 (by AB = A B ) Þ I-A ¹0

Þ I - A is invertible Þ ( I - A )-1 exist

Then multiplying both sides of (1) by (I - A)–1, we get

8/9/2023 8:29:30 PM

Linear Algebra • 37

(I - A)–1 I = (I - A)–1 (I - A) (I + A + A2 + ... + Ak–1)

⇒ (I - A)–1 = I(I + A + A2 + ... + Ak–1)



= I + A + A2 + ... + Ak–1 15. (b) 

0 1 2

18. (a)  A = 1 2 3 = 0 - (1 - 9) + 2( x - 6) 3 x 1

é A( A + B )-1 B ù ë û

= 2x - 4.

-1

= B -1 éë( A + B )-1 ùû

-1

é 2 ê ê x ê 1 adj( A ) = ê ê x ê ê 1 ê 2 ë

A -1

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

( ) ( I + BA ) I + (B B) A

= B -1 AA -1 + BA -1 = B -1 = B -1

-1

-1

16. (c)  A + B = AB Þ A -1 ( A + B ) = A -1 ( AB ) Þ A -1 A + A -1 B = ( A -1 A )B Þ A -1 A + A -1 B = ( A -1 A )B

Now,

Þ I + A -1 B = IB = B Þ I = B - A -1 B = ( I - A -1 )B

A -1 =

1 æ1 ç 2 -2 ç Þ ç -4 3 ç 5 3 çç 2 2 è

Þ I = ( I - A -1 )B Þ I - A -1 B = 1 Þ I - A -1 ¹ 0

0 2 3 1

2 3

-

0 2 1 3

T

1 adj( A ) A 1ö æ 2 - 3x 2x - 1 -1 ö 2 ÷÷ 1 ç ÷ y÷ = -6 2÷ ç 8 x 2 4 ç x -6 1÷ 3 -1 ÷ø è ÷÷ 2ø

æ 2 - 3x 1ö ç 2x - 4 2 ÷÷ ç 8 ç y÷ = ç 2 x -4 1÷ ç ÷÷ ç x - 6 2ø ç è 2x - 4 2 1 2 - 3x , y= Þ = 2x - 4 2 2x - 4 1 æ1 ç 2 -2 ç Þ ç -4 3 ç 5 3 çç 2 è 2

Þ I - A -1 is invertible. æ1 0 ö æ0 0ö ÷,B =ç ÷. 0 0 è ø è0 1ø æ 1 0 öæ 0 0 ö æ 0 0 ö Then AB = ç ÷ç ÷=ç ÷. è 0 0 øè 0 1 ø è 0 0 ø Here A = 0, B = 0 and so both are singular.

17. (c)  Let A = ç

0ö æ0 0ö ÷+ç ÷ 0ø è0 1ø 0ö ÷. 1ø

1 0 = 1 ¹ 0. 0 1

2 1

1 2 ù ú 3 x ú 0 1ú ú 3 xú ú 0 1 ú 1 2 úû

é 2 - 3x 8 x - 6 ù 3 úú = êê 2x - 1 -6 êë -1 2 -1 úû é2 - 3x 2x - 1 -1ù 2 úú = êê 8 -6 êë x - 6 -1úû 3

= B -1 + A -1

Hence, A + B =

1 3 3 1

T

-1

= B -1 + IA -1

æ1 Now A + B = ç è0 1 æ =ç è0

3 1

Þ 2x - 4 = 4 - 6 x , y =

2x - 1 2x - 4 -6 2x - 4 3 2x - 4

-1

ö 2x - 4 ÷ ÷ 2 ÷ 2x - 4 ÷ ÷ -1 ÷ ÷ 2x - 4 ø

2 2x - 4

2 2x - 4 2 Þ x = 1, y = = -1. 2-4 Þ 8x = 8, y =



So A + B is non-singular.



Here, |A| = 0, |B| = 0 and so both are singular

NOTE If A and B be two non-zero n × n matrices such that AB = 0. Then both of them are singular.

EMEP.CH01_3PP.indd 37

8/9/2023 8:29:31 PM

38 • Engineering Mathematics Exam Prep 19. (c) A is symmetric ⇒ AT = A. Also, B is skewsymmetric ⇒ BT = -B. m

n

m T

(A B A )

T n

2

Þ 1 = I n - A I n + A + A + ... + A

T m

Thus( A m B n A m )T = A m B n A m if n is even Þ A m B n A m is symmetric if n is even n

m

m T

m

n

m

Also ( A B A ) = - A B A if n is odd



Þ A B A is skew-symmetric if n is odd

20. (d) As A and B are symmetric, so AT = A,BT = B. Also A and B commute, so AB = BA. Then

( A B) -1

T

=B

(A ) -1

T

( )

T

=B A

T

-1

-1

= BA ..(1)

Now AB = BA Þ A -1 AB = A -1 BA

(

)

(

\ (1) Þ A -1 B

)

( ) = ( A B) I = A

Þ I n = ( I n + A )( I n - A + A 2 - ... - A p -1 ) Þ I n = ( I n + A )( I n - A + A 2 - ... - A p -1 ) Þ 1 = I n + A I n - A + A 2 - ... - A p -1 (by AB = A B ) Þ In + A ¹ 0

Þ I n + A is non-singular

-1

-1

B

= A -1 B

Þ I n = ( I n - A )( I n + A + A 2 + ... + A p-1 ) Þ 1 = I n - A I n + A + A 2 + ... + A p-1 (by AB = A B ) Þ In - A ¹ 0

Þ I n - A is non-singular.

Similarly, it can be shown that AB–1 is symmetric.

(

-1

) = (B ) ( A ) = (B ) ( A )

Now A B

-1

T

-1

-1

T

T

T

-1

23. (b) A2 = A × A

T

-1



3ö æ 1 1 3ö æ 1 1 ç ÷ ç ÷ 5 2 6 ´ 5 2 6÷ =ç ÷ ç ç -2 -1 -3 ÷ ç -2 -1 -3 ÷ è ø è ø



æ1 + 5 - 6 ç = ç 5 + 10 - 12 ç -2 - 5 + 6 è

1+2-3 5+4-6 -2-2+3



æ0 = çç 3 ç -1 è

0ö ÷ 9÷ , - 3 ÷ø



A3 = A2 × A

= B -1 A -1 = ( AB )



= ( BA )

-1

-1

-1

=A B

-1

Hence, A–1 B–1 is symmetric. 21. (a) Since A is a nilpotent matrix of index p, so AP = 0. Then,

EMEP.CH01_3PP.indd 38

-1

Þ I n = ( I n - A )( I n + A + A 2 + ... + A p-1 )

So A–1 B is symmetric.

2

n p I n + A = ( I n + A )( I n - A + A - ... - A )

n p I n - A = ( I n - A )( I n + A + A 2 + ... + A -1 )

Þ BA -1 = A -1 B AA -1 Þ BA -1

22. (d) Let A be a nilpotent matrix of index p. So, AP = 0. Then,

Again,

Þ IB = A -1 B A

T

p -1

Þ I n - A is invertible

ìï A m B n A m if n is even =í m n m ïî- A B A if n is odd

m

)

Þ In - A ¹ 0

= ( -1)n A m B n A m

n

)

p -1

( AB = A B )

= A m ( - B )n A m

m

p -1

Þ I n = ( I n - A )( I n + A + A + ... + A

= ( A ) (B ) ( A )



Þ I n = ( I n - A )( I n + A + A 2 + ... + A 2

= ( A m )T ( B n )T ( A m )T T m

I n - A p = ( I n - A )( I n + A + A 2 + ... + A p -1 )

0 3 -1

3+6-9 ö ÷ 15 + 12 - 18 ÷ - 6 - 6 + 9 ÷ø

8/9/2023 8:29:33 PM

Linear Algebra • 39



æ0 = çç 3 ç -1 è

0 3 -1



æ0 + 0 + 0 ç = ç 3 + 15 - 18 ç -3 - 5 + 6 è



æ0 = çç 0 ç0 è

0 0 0



0 ö æ1 ÷ ç 9÷ ´ ç5 - 3 ÷ø çè -2

1 2 -1

3 ö ÷ 6÷ - 3 ÷ø

0+0+0 3+6-9 -1 - 2 + 3

0+0+0 ö ÷ 9 + 18 - 27 ÷ - 3 - 6 + 9 ÷ø





-b

æ 4b2 + g 2 ç ⇒ ç 2b2 - g 2 ç ç -2b2 + g 2 è

æ 1 -1 ö 1 æ 1 0 ö Then B = ç ÷+ ç ÷ è2 3 ø 2 è0 1ø æ1 ö æ3 0÷ ç æ 1 -1 ö ç 2 2 =ç ÷=ç ÷+ç è2 3 ø ç 0 1 ÷ ç 2 ç ÷ ç 2ø è è 21 \ det( B ) = + 2 ¹ 0. 4

-g

2b2 - g 2 a2 + b2 + g 2 a2 - b2 - g 2

aö ÷ -b ÷ = I g ÷ø

-2b2 + g 2 ö ÷ a2 - b2 - g 2 ÷ ÷ a2 + b2 + g 2 ÷ø

27. (c) Since A and B are two orthogonal matrices, so ATA = A AT = I and BTB = B BT = I. Then



⇒ 4b2 + g2 = 1 

…(i)

  2b2 - g2 = 0

…(ii)

  a2 + b2 + g2 = 1

…(iii)

0ö ÷ 0÷ 1 ÷ø

(



A-

Þ A A + B B = BT + A T I = BT + A T T

Þ A A + B B = (B + A) Þ A+B =0

è

2ø

2 øè

2ø



⇒ AAT - AI - IA + I = I



T ⇒ AAT - A - A + I = I …(i) 2 2 4

2

2

æ A B - 1 = 0 ç ç Þ A = B = -1 or A = B = -1 çç è Þ A + B ¹ 0, a contradiction

ö ÷ ÷ ÷÷ ø

28. (d)   rank( AB ) = 1 ¹ 3 = order of AB

4

Similarly, A + I is orthogonal

EMEP.CH01_3PP.indd 39



T

2

= A+B

Þ A + B ( A B - 1) = 0

I is orthogonal 2 2 øè

)

Þ A A + B B = BT + AT BBT

1 1 1 ,b = ± ,g=± . 2 6 3

T I öæ T I ö æ Iö ⇒ æç A - I öæ A ÷ç ÷ = I ⇒ ç A - ÷ç A - ÷ = I

è

(

Þ A T A + B BT = I + AT B BT

25. (i)-(c); (ii)-(a)

)

AT ( A + B ) BT = AT A + AT B BT

Solving the above three equations, we get a=±

ö -1 ÷ ÷ 7÷ ÷ 2ø

So, B is invertible.

æ1 = çç 0 ç0 è



0 1 0

a b

4

2 3 I 3I which implies AAT = and so A( - A ) = 4 4 3 i.e., A2 = - I . 4 æ 1 -1 ö 26. (c)   Let A = ç ÷ è2 3 ø

24. (a) A is orthogonal ⇒  A × AT = I g ö æ0 ÷ ç - g ÷ ´ ç 2b g ÷ø çè g

2

Adding (i) and (ii), we get, 2 AAT + I = 2 I

0ö ÷ 0 ÷ = 0. 0 ÷ø

2b b

2

Subtracting (i) from (ii), we get, A + AT = O which gives AT = -A. Thus, A is skew-symmetric.

Hence, A is a nilpotent matrix of index 3.

æ0 ç ⇒ ça ça è

T ⇒ AAT + A + A + I = I …(ii)

Þ AB = 0 Þ A B = 0.



So, at least either |A| or |B| should be zero



Hence, |BA| = |B||A| = 0. Therefore, BA is singular.



Hence, rank (BA) cannot be 3.

8/9/2023 8:29:36 PM

40 • Engineering Mathematics Exam Prep æ1 29. (c)   A = çç 2 ç1 è æ1 ç  ç0 ç1 è

2 3 1ö ÷ 4 6 4÷ 2 3 2 ÷ø 2 3 1ö ÷ 0 0 0 ÷ (by R2 ® R2 - 2R3 ) 2 3 2 ÷ø



which has two non-zero rows.



Hence, rank (A) = 2. æ6 ç 4 30. (c)   A = ç ç10 ç è16 æ6 ç 4 ç ç10 ç è0 æ6 ç 4 ç ç0 ç è0



which has three non-zero rows.



Hence, rank (A) = 3. æ0 ç 1 33. (c)  A = ç ç3 ç è1

1 -3 -1 ö ÷ 0 1 1 ÷ 1 0 2÷ ÷ 1 -2 0 ø æ1 0 1 1 ö ç ÷ 0 1 -3 -1 ÷ ç ç3 1 0 2 ÷ ç ÷ è 1 1 -2 0 ø (by R1 « R2 )

1 3 8ö ÷ 2 6 -1 ÷ 3 9 7÷ ÷ 4 12 15 ø 1 3 8ö ÷ 2 6 -1 ÷ [ by R4 ® R4 - ( R1 + R3 )] 3 9 7÷ ÷ 0 0 0ø 1 3 8ö ÷ 2 6 -1 ÷ [ by R3 ® R3 - ( R1 + R2 )] 0 0 0÷ ÷ 0 0 0ø



which has two non-zero rows.



Hence, rank (A) = 2.

0 1 1 ö ÷ 1 -3 -1 ÷ 1 -3 -1 ÷ ÷ 1 -3 -1 ø (by R3 ® R3 - 3R1 , R4 ® R4 - R1 ) æ1 ç 0 ç ç0 ç è0

 

æ1 31. (c)   A = çç 1 ç3 è æ1 ç  ç1 ç0 è

1 -1 1 ö ÷ -1 2 -1 ÷ 1 0 1 ÷ø 1 -1 1 ö ÷ -1 2 -1 ÷ 0 0 0 ÷ø ( by R3 ® R3 - (2R1 + R2 ) )



Which has two non-zero rows.



Hence, rank (A) = 2. é2 3 -1 ê 1 -1 -2 32. (d)  A = ê ê3 1 3 ê 6 3 0 ë

æ1 ç 0 ç ç0 ç è0

-1 ù ú -4 ú -2 ú ú -7 û

é2 3 -1 -1 ù ê ú 1 -1 -2 -4 ú ê ê3 1 3 -2 ú ê ú ë0 0 0 0 û ( by R4 ® R4 - ( R1 + R2 + R3 ) )

0 1 1 ö ÷ 1 -3 -1 ÷ 0 0 0÷ ÷ 0 0 0ø (by R3 ® R3 - R2 , R4 ® R4 - R2 )



which has two non-zero rows.



Hence, rank (A) = 2. 34. (d) Since order of A is 3 × 4, so rank (A) ≤ min {3, 4} = 3. Consider a sub-matrix B, where æ0 1 2ö ç ÷ B = ç1 2 3 ÷. ç3 1 1÷ è ø Then B = 0 - (1 - 9) + 2(1 - 6) = -2 ¹ 0.



Thus, there exist a square sub-matrix B of order “3” whose determinant value is non-zero.

Hence, rank (A) = 3.

35. (a) 

æ1 ç ç1 A = ç0 ç ç0 ç0 è

0 1 1 0 1

æ 0 -2 ç ç1 1  ç0 1 ç ç0 0 ç0 1 è

EMEP.CH01_3PP.indd 40

1 0 1 1 0

0 0 0 1 1

0 0

0 1 1 0

0 0 1 1

0ö ÷ 0÷ 0÷ ÷ 0÷ 0 ÷ø 0ö ÷ 0÷ 0 ÷ [ by R1 ® R1 - ( R2 + R3 )] ÷ 0÷ 0 ÷ø

æ0 1 0 0 0ö ç ÷

8/9/2023 8:29:37 PM

A= æ1 ç ç1 A = ç0 ç ç0 ç0 è



æ0 ç ç1  ç0 ç ç0 ç0 è æ0 ç ç1  ç0 ç ç0 ç0 è

æ0 ç ç1  ç0 ç ç0 ç0 è æ0 ç ç1  ç0 ç ç0 ç0 è



1 5÷ ÷ 0 -3 ø



æ0 ç 0 ç ç0 ç è -1

1 0ö ÷ 1 0÷ (by R1 ® R1 + 2R3 ) 0 1 ÷ ÷ 0 -3 ø



æ0 ç 0 ç ç0 ç è -1

0 1 0 0ö ÷ 1 0 0 0÷ 1 1 0 0÷ ÷ 0 1 1 0÷ 1 0 1 0 ÷ø -2 0 0 0 ö ÷ 1 0 0 0÷ 1 1 0 0 ÷ [ by R1 ® R1 - ( R2 + R3 )] ÷ 0 1 1 0÷ 1 0 1 0 ÷ø 1 0 0 0ö ÷ 1 0 0 0÷ 1 ù é 1 1 0 0 ÷ ê by R1 ® - R1 ú 2 û ÷ ë 0 1 1 0÷ 1 0 1 0 ÷ø

1 1 1 0 0

0 0 1 1 0

1 0 1 0 0

( by R2

0ö ÷ 0÷ 0 ÷ ( by R5 ® R5 - R1 ) ÷ 0÷ 0 ÷ø 0 0 0ö ÷ 0 0 0÷ 1 0 0÷ ÷ 1 0 0÷ 0 1 0 ÷ø ® R2 - R1 , R4 ® R4 - R5 )



1 0 0 0ö ÷ 0 0 0 0÷ 0 1 0 0÷ ÷ 0 1 0 0÷ 0 0 1 0 ÷ø



æ0 ç ç1  ç0 ç ç0 ç0 è

1 0 0 0 0



which has four non-zero rows.



Hence, rank (A) = 4.

0 0 1 0 0

0 0 0 0 1

æ1 ç -1 36. (d)  A = ç ç1 ç è -1

æ1 ç 0 ç ç1 ç è -1 æ1 ç 0 ç ç0 ç è -1 æ0

1 1 ö ÷ 2 0÷ (by R2 ® R2 + R1 ) 1 5÷ Linear Algebra • 41 ÷ 0 -3 ø æ1 1 1 ö ç ÷ 0 2 0÷ ç (by R3 ® R3 - R1 ) ç0 0 4÷ ç ÷ è -1 0 -3 ø æ 0 1 -2 ö ç ÷ 0 1 0÷ ç ç0 0 1 ÷ ç ÷ è -1 0 -3 ø é by R ® R + R , R ® 1 R ,ù 1 1 4 2 2 ú ê 2 ê ú 1 ê ú R3 ® R3 ëê ûú 4

0 0 0 1 1

æ0 ç ç1  ç0 ç ç0 ç0 è

EMEP.CH01_3PP.indd 41

ç1 ç è -1 æ1 ç 0 ç ç1 ç è -1

0ö ÷ 0÷ 0÷ ÷ 0÷ 0 ÷ø

( by R3 ® R3 - R1 )



0 0ö ÷ 1 0÷ (by R1 ® R1 - R2 ) 0 1÷ ÷ 0 -3 ø

which has three non-zero rows. Hence, rank (A) = 3.

Alternative method: Rank(A) ≤ min {number of rows, number of colums} = min {4, 3}. So rank (A) ≤ 3.

( by R4 ® R4 - R3 )

1 1 ö ÷ 1 -1 ÷ 1 5÷ ÷ 0 -3 ø 1 1 ö ÷ 2 0÷ (by R2 ® R2 + R1 ) 1 5÷ ÷ 0 -3 ø 1 1 ö ÷ 2 0÷ (by R3 ® R3 - R1 ) 0 4÷ ÷ 0 -3 ø 1 -2 ö



Let us consider a third order square sub-matrix



æ1 1 1 ö ç ÷ B = ç -1 1 -1 ÷ ç1 1 5÷ è ø



Then det(B) = 1(5 + 1) - ( - 5 + 1) + 1(-1-1) ≠ 0.

Thus, there exist a square sub-matrix B of order 3 whose determinant value is non-zero. Hence, rank (A)=3. æ -2 -1 -3 -1 ö ç ÷ 1 2 3 -1 ÷ 37. (c)   A = ç ç1 0 1 1 ÷ ç ÷ è 0 1 1 -1 ø æ 0 -1 -1 1 ö ç ÷ 0 2 2 -2 ÷ ç  ç1 0 1 1 ÷ ç ÷ è 0 1 1 -1 ø

(by R1 ® R1 + 2R3 , R2 ® R2 - R3 ) æ 0 -1 -1 1 ö ç ÷

8/9/2023 8:29:39 PM

æ -2 -1 -3 -1 ö ç ÷ 1 2 3 -1 ÷ A=ç ç1 0 1 1 ÷ 42 •çEngineering Mathematics Exam Prep ÷ è 0 1 1 -1 ø æ 0 -1 -1 1 ö ç ÷ 0 2 2 -2 ÷ ç ç1 0 1 1 ÷ ç ÷ è 0 1 1 -1 ø (by R1 ® R1 + 2R3 , R2 ® R2 - R3 )

40. (a)  rank ( A ) = 3

Þ rank ( A ) < 4 ( = order of the square matrix) Þ A =0 -1 0 0 0 l -1 0 Þ =0 0 0 l -1 -6 11 -6 1 l

æ 0 -1 -1 ç 0 0 0 ç ç1 0 1 ç è0 0 0

1ö ÷ 0÷ 1÷ ÷ 0ø (by R2 ® R2 + 2R1 , R4 ® R4 + R1 )



which has two non-zero rows.



Hence, rank (A) = 2.

0 l -1 0 l -1 Þ -6 11 l - 6 -6 11 -6

(by R 3 ® R 3 + R 4 )

l -1 0 Þ 0 l -1 = 0 -6 11 l - 6

38. (a) rank(A) ≤ min {number of rows of A, number of columns of A} = min {4, 5}. So rank(A) ≤ 4. Let us consider a 4th order square sub-matrix



æ0 ç 1 B= ç ç1 ç è1

2 0 0 0

0 1 det( B ) = 1 1

0 1 2 0 2 0 0 0

Þ l[ l( l - 6) + 11] - ( -1)(0 - 6) = 0 Þ l3 - 6l2 + 11l - 6 = 0

2ö ÷ 0÷ 1÷ ÷ 0ø 0 1 2 0

Þ l2 ( l - 1) - 5l( l - 1) + 6( l - 1) = 0 Þ ( l - 1)( l2 - 5l + 6) = 0 Þ ( l - 1)( l - 2)( l - 3) = 0

2 0 1 0

So, l = 1, 2, 3. 41. (b)

39. (b) Let A be the given matrix. Since A is upper triangular, so det(A) = 1 × 7 × 1 × 2 = 14 ≠ 0.



0ù ú 2ú ú 0úú 1úû 0ù ú 2ú ú 0úú 1úû æby R2 ® R2 + R1 , R3 ® R3 - 2 R1 ,ö÷ çç ÷÷ çè R4 ® R4 + R1 ø÷ é 1 1 0ù ê ú ê 0 0 0ú ú ê ® R2 - 2 R4 ) ê 0 0 0ú (by R2 ê ú ê 0 1 1ú ë û



which has two non-zero rows.

So rank (A) = order of the square matrix A = 4.



Therefore, rank (A) = 2.

1 1 = 1 0

0 0 0 2

0 1 2 0

0 0 1 2

é1 1 ê ê-1 1 A= ê ê2 2 ê ê-1 0 ë é1 1 ê ê0 2 ê ê0 0 ê ê0 1 ë

(by R1 « R4 )

= 1 ´ cofactor of 1

(since 1 is the only non-zero element in first row)



0 1 0 = 0 2 1 = 2. ¹ 0 2 0 2

Thus, there exist a square sub-matrix B of order 4 whose determinant value is non-zero. Hence, rank (A) = 4.



EMEP.CH01_3PP.indd 42

0 0 =0 0 1

8/9/2023 8:29:40 PM

Linear Algebra • 43 é4 5 ê ê5 6 êë6 l é4 ê  ê10 êë 6

42. (b) 

Hence, the system has no solution for l = - 11 3 and m ≠ 22

aù ú bú c úû

5 aù ú 12 2bú l c úû

(by R2 ® 2R2 )

5 a é4 ù ê ú  ê0 7 - l 2b - ( a + c ) ú êë6 úû l c

[by R2 ® R2 - ( R1 + R3 )]

(iii) The system has an infinite number of solutions D = D1 = D2 = D3 = 0 (by Cramer's rule)

Þ D = D3 = 0



Þl=-

Alternative method: ëé A : B ûù é1 2 1 8 ù ê ú = ê2 1 3 13 ú êë3 4 -l m úû 1 8 ù é1 2 ê ú 1  ê0 -3 -3 ú êë0 -2 -l - 3 m - 24 úû

5 aù é4 ê ú  ê0 7 - l 0 ú êë6 l c úû [ a, b, c are in A.P, so 2b = a + c ] é4 5 a ù ê ú  ê0 0 0 ú ê6 l c ûú ë

(if 7 - l = 0)

(by R2 ® R2 - 2R1 , R3 ® R3 - 3R1 )

Thus, for l = 7, we get two non-zero rows in the final equivalent matrix and so rank becomes “2.” But since the rank is independent of “d,” so we have d = any arbitrary number. 43. (i) - (a), (ii) - (b), (iii) - (a) (i) By Cramer’s rule the system has a unique solution ÛD¹0 1 2

Û 2 1

1

3 ¹0

3 4 -l



Û 3l + 11 ¹ 0 11 Ûl¹- . 3

(ii) Now if the system has no solution, then D = 0.

D = 0 Þ= 1 D3 = 2 3

11 3

2 8 1 13 = -3m + 66 4 m

If D3 ≠ 0, then -3m + 66 ≠ 0 i.e; m ≠ 22 and vice versa. But D = 0, D3 ≠ 0 ⇒ system has no solutions (by Cramer’s rule).

EMEP.CH01_3PP.indd 43

11 , m = 22 3



é ù ê1 2 1 8 ú ê ú 1 -3 ú  ê0 -3 ê ú 11 m - 22ú ê0 0 -l 3 ë û 2 (by R3 ® R3 - R2 ) 3

Case-I: -l - 11 ¹ 0 i.e; l ¹ - 11 . 3

3

In this case, rank (A) = 3 = number of variables. So the system has a unique solution. Case-II: -l -

11 11 = 0 and m - 22 ¹ 0 i.e; l = and m ¹ 22. 3 3

In this case, rank (A) = 2 ≠ 3 = rank ([A:B]). So the system has no solution. Case-III: -l -

11 11 = 0 and m - 22 = 0, i.e; l = and m = 22. 3 3

In this case, rank (A) = rank([A:B]) = 2 < 3 = number of variables. Hence, the system has an infinite number of solutions. 44. (i) - (b), (ii) - (a), (iii) - (c) (i) The system has a unique solution

8/9/2023 8:29:41 PM

44 • Engineering Mathematics Exam Prep ÞD¹0

Case-III:

1 1 1 Þ1 2 3 ¹0 1 2 l

l - 3 = 0 and m - 10 = 0 i.e., l = 3 and m = 10.

Þ l -3 ¹ 0 Þ l ¹ 3.

(ii) Now if the system has no solution, then D = 0. But D = 0 Þ l = 3



1 1 6 D3 = 1 2 10 = m - 10 1 2 m

In this case, rank (A) = rank([A:B]) = 2 < 3 = number of variables. Hence, the system has an infinite number of solutions. 45. (i) - (a), (ii) - (b), (iii) - (a) 1 4 2 (i) D = 2 7 5 4 l 10 1 4 2 =0 -1 1 0 l - 16 2

If D3 ≠ 0, then m ≠ 10 and vice versa.

(by R2 ® R2 - 2R1 , R3 ® R3 - 4 R1 )

But D = 0, D3 ≠ 0 ⇒ System has no solution.

= 1( -2 - l + 16) = -l + 14

Hence, the system has no solution for l = 3 and m ≠ 10

1 1 2 D2 = 2 2m 5 4 2m + 1 10

(iii) The system has an infinite number of solutions ⇔  D = D1 = D2 = D3 = 0 ⇔ D = D3 = 0 ⇔ p = 3, m = 10

1 1 2 = 0 2m - 2 1 0 2m - 3 2

(by Cramer’s rule).

Alternative method éë A : B ùû é1 1 1 6 ù ê ú = ê1 2 3 10 ú êë1 2 l m úû 1 6 ù é1 1 ê ú  ê0 1 2 4 ú êë0 1 l - 1 m - 6 úû

(by R2 ® R2 - R1 , R3 ® R3 - R1 ) 1 6 ù é1 1 ê ú  ê0 1 2 4 ú êë0 0 l - 3 m - 10 úû

(by R2 ® R2 - 2R1 , R3 ® R3 - 4 R1 )



= 1(4m - 4 - 2m + 3) = 2m - 1

Then D = 0 ⇒ -l + 14 = 0, i.e., l = 14 and,

D2 ¹ 0 Þ 2m - 1 ¹ 0 i.e; m ¹

1 . 2

But D = 0, D2 ≠ 0 ⇒ System has no solution (by Cramer’s rule).

Hence, the system has no solution for l = 14

and m ¹

1 2

Case-I: l - 3 ≠ 0 i.e., l ≠ 3.

(ii) Now if the system has unique solution, then D ≠ 0. D ≠ 0 ⇔ l ≠ 14. Thus, system has a unique solution for p ≠ 14

In this case, rank (A) = 3 = number of variables. So the system has a unique solution.

(iii) The system has an infinite number of solutions



(by R3 ® R3 - R2 )

Û D = D1 = D2 = D3

Case-II: l - 3 = 0 and m - 10 ≠ 0 i.e; l = 3 and m ≠ 10. In this case, rank (A) = 2 ≠ 3 = rank([A:B]). So the system has no solution.

EMEP.CH01_3PP.indd 44



Û D = D2 = 0 Û D = 0, D2 = 0 Û p = 14, m =

1 2

8/9/2023 8:29:42 PM

Linear Algebra • 45 D3 ¹ 0

Alternative method: éë A : B ùû 1 ù é1 4 2 ê ú = ê2 7 5 2m ú êë4 l 10 2m + 1úû

1 1 Þ1 2 1 1 Þ0 1

Þ 1(m2 - 5 - 2m + 2) ¹ 0 Þ (m + 1)(m - 3) ¹ 0 Þ m ¹ -1 and m ¹ 3.



But D = 0, D3 ≠ 0 ⇒ system has no solution (by Cramer’s rule)

Case-I: l - 14 ≠ 0 i.e., l ≠ 14 In this case, rank (A) = 3 = number of variables. So the system has a unique solution. Case-II: l - 14 = 0 and 1 - 2m ≠ 0 i.e., l = 14 and m ≠ 1 . 2

In this case, rank (A) = 2 ≠ 3 = rank([A:B]). So the system has no solution.

1 1 1 Þ 1 2 -1 = 0 5 7 l 1 1 1 Þ 0 1 -2 = 0 0 2 l -5 (by R2 ® R2 - R1 , R3 ® R3 - 5R1 )

Þ 1( l - 5 + 4) = 0 Þ l =1

EMEP.CH01_3PP.indd 45

Hence, the system has no solution for l = 1 and m ≠ -1 or 3. (ii) If the system has a unique solution, then D ≠ 0 (by Cramer’s rule) and vice versa. But D ≠ 0 ⇒ l ≠ 1. Hence, the system has a unique solution for l ≠ 1. (iii) The system has an infinite number of solutions Û D = D1 = D2 = D3 = 0 (by Cramer's rule) Û D = D3 = 0

1 . 2

In this case, rank (A) = rank ([A:B]) = 2 < 3 = number of variables. Hence, the system has an infinite number of solutions. 46. (i) - (a), (ii) - (c), (iii) - (a) (i) D = 0

m -1 ¹ 0

(by R2 ® R2 - R1 , R3 ® R3 - 5R1 )

(by R3 ® R3 - 2R2 )

l - 14 = 0 and 1 - 2m = 0 i.e., l = 14 and m =

1

0 2 m2 - 5

4 2 1 ù é1 ê ú  ê0 -1 1 2m - 2 ú êë0 l - 14 0 1 - 2m úû

Case-III:

m ¹0

5 7 m2

4 2 1 ù é1 ê ú 1 2m - 2ú  ê0 -1 êë0 l - 16 2 2m - 3 úû (by R2 ® R2 - 2R1 , R3 ® R3 - 3R1 )



1



Û l = 1, m = -1 or 3

Alternative method: ëé A : B ûù é1 1 1 1 ù ê ú = ê1 2 -1 m ú ê5 7 l m2 ú û ë 1 4 2 D= 2 7 5 4 l 10 1 4 2 =0 -1 1 0 l - 16 2 (by R2 ® R2 - 2R1 , R3 ® R3 - 4 R1 )

= 1( -2 - l + 16) = -l + 14

8/9/2023 8:29:42 PM

é1 2 3 1 1 ù ê ú  ê0 -3 -1 -2 -4 ú êë0 -3 -1 -2 -6 úû (by R2 ® R2 - R1 , R3 ® R3 - 3R1 )

46 • Engineering Mathematics Exam Prep Case-I: l - 1 ≠ 0 i.e., l ≠ 1.

é1 2 3 1 1 ù ê ú  ê0 -3 -1 -2 -4 ú êë0 0 0 0 -2 úû (by R3 ® R3 - R2 )

In this case, rank (A) = 3 = number of variables. So the system has a unique solution. Case-II: l - 1 = 0 and m2 - 2m - 3 ≠ 0 i.e., l = 1 and m ≠ -1, 3 In this case, rank (A)=2≠3=rank([A:B]). So the system has no solution. Case-III: l - 1 = 0 and m2 - 2m - 3 = 0 i.e., l = 1 and m = -1 or 3. In this case, rank (A) = rank ([A:B]) = 2 < 3 = number of variables. Hence, the system has an infinite number of solutions.



Therefore, rank ([A: B]) = 3. Now ignoring the last column in the final equivalent matrix, we see that there are only two non-zero rows. Hence, rank (A) = 2. Thus, rank (A) ≠ rank ([A: B]) and so the system is inconsistent. (i.e., no solution exist) 49. (a)

47. (c) The given system of equations can be written as:

(1 - l )x + 2 y + 3z = 0 3x + (1 - l ) y + 2z = 0 2x + 3 y + (1 - l )z = 0

The system of equation has a non-trivial solution ⇒  Rank of the coefficient matrix < 3 (here Number of variables = 3)



⇒  co-efficient determinant = 0 1-l Þ

3 2

2 1-l 3

3

2 =0 1-l

6-l 2 3 Þ 6-l 1-l 2 = 0 (by C1 ® C1 + C2 + C3 ) 6-l 3 1-l 1 2 3 2 =0 Þ (6 - l ) 1 1 - l 1 3 1-l Þ6-l =0 Þ l = 6.

48. (b)

é1 2 3 1 1 ù ê ú éë A : B ùû = ê1 -1 2 -1 -3ú êë3 3 8 1 -3úû



EMEP.CH01_3PP.indd 46

é1 2 3 1 1 ù ê ú  ê0 -3 -1 -2 -4 ú êë0 -3 -1 -2 -6 úû (by R2 ® R2 - R1 , R3 ® R3 - 3R1 ) é1 2 3 1 1 ù ê ú  ê0 -3 -1 -2 -4 ú êë0 0 0 0 -2 úû (by R3 ® R3 - R2 )

which has three non-zero rows.



4 4ù é2 1 ê ú 1 -3 -1 -5ú éë A : B ùû = ê ê -3 2 -2 1 ú ê ú ë 8 -3 8 2 û é 1 -3 -1 -5ù ê ú 2 1 4 4ú ê ê -3 2 -2 1 ú ê ú ë 8 -3 8 2 û (by R1 « R2 ) é1 -3 -1 -5 ù ê ú 0 7 6 14 ú ê ê0 -7 -5 -14 ú ê ú ë0 21 16 42 û

[by R2 → R2 - 2R1,

R3 → R3 + 3R1,

R4 → R4 - 8R1] é1 -3 -1 -5ù ê ú 0 7 6 14 ú ê ê0 0 1 0ú ê ú ë0 0 -2 0 û

[by R3 → R3 + R2

R4 → R4 - 3R2]



é1 -3 -1 -5ù ê ú 0 7 6 14 ú ê [by R4 ® R4 + 2R3 ] ê0 0 1 0ú ê ú ë0 0 0 0 û



which has three non-zero rows.

8/9/2023 8:29:44 PM

AX = lX æ 6 -2 2 öæ x ö æxö ç ÷ç ÷ ç ÷ Þ ç -2 3 -1 ÷ç y ÷ = 2 ç y ÷ Linear Algebra • 47 ç 2 -1 3 ÷ç z ÷ çz÷ è øè ø è ø æ 6 x - 2 y + 2z ö æ 2x ö ç ÷ ç ÷ Þ ç -2x + 3 y - z ÷ = ç 2 y ÷ ç 2x - y + 3z ÷ ç 2z ÷ è ø è ø

Therefore, rank ([A: B]) = 3. Now ignoring the last column in the final equivalent matrix, we see that there are only three non-zero rows. Hence, rank(A) = 3. Thus, rank (A) = rank ([A: B]) and so the system is consistent. 50. (a) 

2 -1 1 A = -1 2 -1 = 2(4 - 1) + ( -2 + 1) + (1 - 2) = 4 1 -1 2



Þ 2x - y + z = 0, 2x - y + z = 0, 2x - y + z = 0 Þ y = 2x + z

Using |A|= product of eigenvalue, we get l = 1, 1, 4. 51. (a)  A - lI æ 6 -2 2 ö æ1 0 0 ö ç ÷ ç ÷ = ç -2 3 -1 ÷ - l ç 0 1 0 ÷ ç 2 -1 3 ÷ ç0 0 1÷ è ø è ø 2 ö æ 6 - l -2 ç ÷ = ç -2 3 - l -1 ÷ ç 2 -1 3 - l ÷ø è

\ A - lI = 0

6 - l -2 2 Þ -2 3 - l -1 = 0 2 -1 3 - l Þ (6 - l ){(3 - l )2 - ( -1) ´ ( -1)} - ( -2){ -2(3 - l ) - ( -1) ´ 2} + 2{( -1) ´ ( -2) - 2 ´ (3 - l )} = 0



æx ö æ x ö æ1 ö æ0ö ç ÷ ç ÷ ç ÷ ç ÷ \ X = ç y ÷ = ç 2x + z ÷ = x ç 2 ÷ + z ç 1 ÷ çz÷ ç z ÷ ç0÷ ç1 ÷ è ø è ø è ø è ø



For l = 8:

æ1 ö ç ÷ Therefore, x = 1, z = 0 gives eigenvector = ç 2 ÷ ç0÷ and è ø æ0ö ç ÷ x = 0, z = 1 gives eigenvector = ç 1 ÷ ç1 ÷ è ø AX = lX



Þ (6 - l )( l2 - 6l + 8) + 8l - 16 = 0 Þ 6l 2 - 36l + 48 - l3 + 6l 2 - 8l + 8l - 16 = 0 Þ l3 - 12l2 + 36l - 32 = 0 Þ l2 ( l - 2) - 10l( l - 2) + 16( l - 2) = 0 Þ ( l - 2)( l2 - 10l + 16) = 0 Þ ( l - 2)( l - 2)( l - 8) = 0 Þ l = 2, 2, 8

For l = 2 : AX = lX



EMEP.CH01_3PP.indd 47

Þ 6 x - 2 y + 2 z = 2 x , - 2 x + 3 y - z = 2 y, 2x - y + 3z = 2z

æ 6 -2 2 öæ x ö æxö ç ÷ç ÷ ç ÷ Þ ç -2 3 -1 ÷ç y ÷ = 8 ç y ÷ ç 2 -1 3 ÷ç z ÷ çz÷ è øè ø è ø æ 6 x - 2 y + 2z ö æ 8 x ö ç ÷ ç ÷ Þ ç -2x + 3 y - z ÷ = ç 8 y ÷ ç 2x - y + 3z ÷ ç 8 z ÷ è ø è ø Þ 6 x - 2 y + 2 z = 8 x , - 2 x + 3 y - z = 8 y, 2x - y + 3z = 8 z Þ -2x - 2 y + 2z = 0, - 2x - 5 y - z = 0, 2x - y - 5z = 0 Þ x + y - z = 0 i.e., z = x + y............(1) 2x + 5 y + z = 0..........(2) 2x - y - 5z = 0..........(3) (3) & (1) Þ 2x - y - 5( x + y ) = 0 Þ x = -2 y........(4) (4) & (1) Þ z = -2 y + y = - y

Þ (6 - l ){9 - 6l + l2 - 1} - (12 - 4l - 4) + ( -8 + 4l ) = 0

æ 6 -2 2 öæ x ö æxö ç ÷ç ÷ ç ÷ Þ ç -2 3 -1 ÷ç y ÷ = 2 ç y ÷ ç ÷ç ÷ ç ÷ è 2 -1 3 øè z ø èzø æ 6 x - 2 y + 2z ö æ 2x ö ç ÷ ç ÷ Þ ç -2x + 3 y - z ÷ = ç 2 y ÷ ç 2x - y + 3z ÷ ç 2z ÷ è ø è ø

Þ 6x - 2 y + 2z = 2x , - 2x + 3 y - z = 2 y, 2x - y + 3z = 2z Þ 4 x - 2 y + 2z = 0, - 2x + y - z = 0, 2x - y + z = 0



æ x ö æ -2 y ö æ 2 ö ç ÷ ç ÷ ç ÷ \ X = ç y ÷ = ç y ÷ = ç -1 ÷ (for y = -1) ç z ÷ ç -y ÷ ç 1 ÷ è ø è ø è ø æ2ö ç ÷ Therefore, y = -1 gives eigenvector = ç -1 ÷ ç1 ÷ è ø

8/9/2023 8:29:47 PM

48 • Engineering Mathematics Exam Prep \ A - lI = 0

Þx+ y+z =0 Þ x = -y - z

1-l 0 2 Þ 0 -1 - l 1 = 0 -l 0 1

æ -y - z ö ç ÷ \X = ç y ÷ = ç z ÷ è ø

52. (b)

Þ (1 - l ){( -l )( -1 - l ) - 1} - 0 + 0 = 0 Þ (1 - l )( l2 + l - 1) = 0



2

3

Þ l + l -1 - l - l + l = 0 Þ l3 - 2l + 1 = 0



⇒ l3 - 2l + 1 = 0



Therefore, by the Cayley Hamilton theorem, A3 - 2 A + I = 0

AX = lX

\ A 9 = ( A 3 )3 = (2 A - I )3 = 8 A 3 - 12 A 2 + 6 A - I [Using (1)]

= -12 A 2 + 22 A - 9 I

53. (c) A - lI = 0 1-l 1 1 Þ 1 1-l 1 =0 1 1 1-l Þ (1 - l ) ´ {(1 - l )2 - 1} - 1 ´ (1 - l - 1) + 1 ´ (1 - 1 + l ) = 0 Þ (1 - l )( l2 - 2l ) + 2l = 0 2

3

æ -1 ö ç ÷ and y = 0, z = 1 gives eigenvector = ç 0 ÷ ç1 ÷ è ø

For l = 3:

or, A 3 = 2 A - I ...........(1)



Therefore, y = 1, z = 0 gives eigenvector æ -1 ö ç ÷ =ç1÷ ç0÷ è ø

2

= 8 ( 2 A - I ) - 12 A 2 + 6 A - I



2

Þ l - 2l - l + 2l + 2l = 0 Þ l3 - 3l2 = 0



Þ l2 ( l - 3) = 0 Þ l = 0, 0, 3



Therefore, the eigenvalues are 0, 0, 3

æ1 1 1 öæ x ö æxö ç ÷ç ÷ ç ÷ Þ ç1 1 1 ÷ç y ÷ = 3 ç y ÷ ç1 1 1 ÷ç z ÷ çz÷ è øè ø è ø + + x y z 3 x æ ö æ ö ç ÷ ç ÷ Þ ç x + y + z ÷ = ç3y ÷ ç x + y + z ÷ ç 3z ÷ è ø è ø

Þ x + y + z = 3x , x + y + z = 3 y, x + y + z = 3 y ì-2x + y + z = 0 .............(1) ï Þ í x - 2 y + z = 0.............(2) ï x + y - 2z = 0.............(3) î

(1) - (2) Þ -3x + 3 y = 0 Þ x = y \ (3) Þ x + x - 2z = 0 Þ z = x æ -y - z ö æ x ö æ1 ö ç ÷ ç ÷ ç ÷ \ X = ç y ÷ = ç x ÷ = x ç1 ÷ ç z ÷ çx÷ ç1 ÷ è ø è ø è ø æ1 ö ç ÷ Therefore, x = 1 gives eigenvector = ç1 ÷ . ç1 ÷ è ø

54. (a) We know that the eigenvalues of a skew symmetric matrix are either imaginary number or zero. But since the eigenvalues of an orthogonal matrix are of unit modulus, so the possible eigenvalues are ± i.

For l = 0 AX = lX



EMEP.CH01_3PP.indd 48

æ1 1 1 öæ x ö æxö ç ÷ç ÷ ç ÷ Þ ç1 1 1 ÷ç y ÷ = 0 ç y ÷ ç1 1 1 ÷ç z ÷ çz÷ è øè ø è ø æx + y + z ö æ0ö ç ÷ ç ÷ Þ çx + y + z ÷ = ç0÷ çx + y + z ÷ ç0÷ è ø è ø

æ -1 ö æ -1 ö ç ÷ ç ÷ + yç 1 ÷ z ç 0 ÷ ç0÷ ç1 ÷ è ø è ø

55. (b)



æ0 3ö Let A = ç ÷ è2 1ø Then, tr( A ) = 0 + 1 = 1; det(A) = 0 - 6 = -6

æ0 3öæ0 3ö æ6 3ö Now, A 2 = ç ÷ç ÷=ç ÷ è2 1øè2 1ø è2 7ø æ6 3öæ0 3ö \ A3 = A2´ A = ç ÷ç ÷ è2 7øè2 1ø

4

3

æ 6 21 ö =ç ÷ è14 13 ø æ 6 21 öæ 0 3 ö

8/9/2023 8:29:48 PM

Now M = MM é1 ê ê0 ê = ê0 ê ê0 ê0 ë

æ0 3ö Let A = ç ÷ è2 1ø Then, tr( A ) = 0 + 1 = 1; det(A) = 0 - 6 = -6 æ0 3öæ0 3ö æ6 3ö Now, A 2 = ç ÷ç ÷=ç ÷ è2 1øè2 1ø è2 7ø æ6 3öæ0 3ö \ A3 = A2´ A = ç ÷ç ÷ è2 7øè2 1ø æ6 =ç è14 æ6 A4 = A3´ A = ç è14

æ 42 A4 - A3 = ç è 26 æ 36 =ç è 12

21 ö ÷ 13 ø

a=

2 pi e 5

=



1 2 ù5

2 5

= ( -1) = é( -1) úû ëê \ a is a fifth root of 1.



EMEP.CH01_3PP.indd 49

a3

0 0

a3 0

é1 ê 2 ê- a = êê - ê- ê êë - 2

0 0 -

-

-

-

4

-

-

a6

-

-

a

2

4

-ù ú -ú ú -ú -ú ú a8 úû

6

-

a6

-

2

1 - a10 1-a

2

=

5

1 -1

1 - a2

=0

Hence, tr( I + M + M 2 ) = tr( I ) + tr(M) + tr(M2 ) = 5 + 0 + 0 = 5.

2

Also, ω = 1 and 1 + ω + ω = 0

2

é1 0 0ù ê ú A = êi w 0 ú which is the lower trianguê0 1 + 2i w2 ú ë û

lar matrix with 1, ω, ω2 as the diagonal elements.

a a2

a3

a a2

a3

0 a2

a3

0 0

a3 0

8

0 0



Therefore, eigenvalue of A are 1, ω, ω2.



So tr(A) = Sum of eigenvalues = 1 + ω + ω2 = 0

Now 1, ω, ω2 are eigenvalues of A ⇒ 1102, ω102, (ω2)102 are eigenvalues of A102.

Now M2 = MM

0 a2

-

a

2

Also tr(M) = 1 + a + a2 + a3 + a4 = 0

a3

-

4

3

0ù ú 0ú 0ú ú 0ú 1 úû \ tr(I) = 1 + 1 + 1 + 1 + 1 = 5

a a2

-

-ù ú -ú ú -ú -ú ú a8 úû

1, where w = -1 + i 3 , w2 = -1 - i 3

0 0 0 1 0

a4 ù é1 úê a 4 ú ê0 úê a 4 ú ê0 úê a 4 ú ê0 a4 ûú ëê0

-

59. (d) We know that 1, ω and ω2 are cube root of

1 = 15

1.a.a2 .a3 .a4 = 1 i.e., a10 = 1

a3

-

a4 ù ú a a 2 a3 a 4 ú ú 0 a 2 a3 a 4 ú ú 0 0 a3 a 4 ú Linear Algebra • 49 0 0 0 a4 úû

58. (a) Since the eigenvalue of Hermitian matrix are all real and the eigenvalues of an unitary matrix are of unit modulus. So ± 1 are the only possible eigenvalues for H.

So, 1 + a + a2 + a3 + a4 = 0 and

a a2

a3 0

0 0

a3

57. (a) We know that the eigenvalue of a 3 × 3 real skew symmetric matrix are either purely imaginary number or zero and since complex eigenvalues occurs in conjugate pairs, so the other eigenvalues are 0 and - 3i.

2

é1 ê ê0 ê = ê0 ê ê0 ê ë0

0 0

=

( )

0 0 1 0 0

a3

a a2

( ) ùúû

2 e pi 5

0 1 0 0 0

0 a2

é 1 ´ ê1 - a2 ë = 1 - a2

39 ö æ 6 21 ö ÷-ç ÷ 55 ø è14 13 ø 18 ö ÷ 42 ø

é1 ê ê0 Here I = ê0 ê ê0 ê0 ë

a3

a4 ù é1 úê a 4 ú ê0 úê a 4 ú ê0 úê a 4 ú ê0 a4 úû êë0

tr( M ) = 1 + a + a4 + a6 + a8

= ( cos p + i sin p ) 5



a a2

2

So tr(A 4 - A 3 ) = 36 + 42 = 78.

56. (a)

a3

é1 ê 2 ê- a = êê - ê- ê êë - -

21 ö æ 0 3 ö ÷ç ÷ 13 ø è 2 1 ø æ 42 39 ö =ç ÷ è 26 55 ø

Hence,

a a2

a4 ù ú a4 ú ú a4 ú ú a4 ú a4 úû

Therefore, tr(A102) = 1102 + ω102 + (ω2)102 = 1+(ω3)34 + (ω3)68 = 1 + (1)34 + (1)68 = 1 + 1 + 1 = 3. 40 -29 -11 0 -29 -11 60. (c) det( A ) = -18 30 -12 = 0 30 -12 26 24 -50 0 24 -50 [ By C1 ® C1 + C2 + C3 ]

=0

8/9/2023 8:29:49 PM

50 • Engineering Mathematics Exam Prep Therefore, A is singular and so “0” is an eigenvalue of A. Let l1 and l2 be other two eigenvalue.

Now the sum of eigenvalues = trace of the matrix



⇒ 0 + l1 + l2 = 40 + 30 -50 ⇒ l2 = 20 - l1.

Then, A - lI = 0 0-l 0 1 Þ 0 0-l 0 =0 -1 0 0-l

Thus, if l(≠ 0) be the eigenvalue of A, then 20 - l will be another eigenvalue. 61. (d) Since M = M, so M is a real symmetric matrix with real eigenvalues. But since tr(M) = 0, so M has some positive and some negative eigenvalues.

Þ l3 + l = 0 Þ l(l 2 + 1) = 0

T



é1 0 0 ù 63. (b)  Let A = êê0 -1 0 úú . Then, A has eigenêë0 0 0 úû

62. (b) L(X) = b × X = cross product of b and X

iˆ = 0 x1

jˆ 1 x2

values 0, 1 and -1 (since A is a diagonal matrix). Then,

kˆ 0 = x3iˆ - x1kˆ = [ x3 0 - x1 ]T x3

æ a1 b1 ç Let A = ç a2 b2 ça b è 3 3

A2 = A ´ A é1 0 0 ù é1 0 0 ù ê ú ê ú = ê0 -1 0 ú ´ ê0 -1 0 ú êë0 0 0 úû êë0 0 0 úû é1 0 0 ù ê ú = ê0 1 0 ú , êë0 0 0 úû

c1 ö ÷ c2 ÷ c3 ÷ø

é x1 ù ê ú Then, L( X ) = A ê x2 ú êë x3 úû é x3 ù æ a1 b1 c1 ö é x1 ù ÷ê ú ê ú ç Þ ê 0 ú = ç a2 b2 c2 ÷ ê x2 ú ç ÷ ëê -x1 ûú è a3 b3 c3 ø ëê x3 ûú



é a1 x1 + b1 x2 + c1 x3 ù é x3 ù ê ú ê ú Þ êa2 x1 + b2 x2 + c2 x3 ú = ê 0 ú êë a3 x1 + b3 x2 + c3 x3 úû êë -x1 úû



Comparing both sides we get,



a1x1 + b1x2 + c1x3 = x3 = 0x1 + 0x2 + 1x3



a2 x1 + b2 x2 + c2 x3 = 0 = 0x1 + 0x2 + 0x3 ,



a3 x1 + b3 x2 + c3 x3 = -x1 = ( -1)x1 + 0x2 + 0x3 .



Therefore, we have,

A4 = A2 ´ A2





EMEP.CH01_3PP.indd 50

é1 0 0 ù \A = êê0 1 0 úú and so êë0 0 0 úû é1 0 0 ù é1 0 0 ù ê ú ê ú 100 I+A = ê0 1 0 ú + ê0 1 0 ú êë0 0 1 úû êë0 0 0 úû é2 0 0 ù ê ú = ê0 2 0 ú . êë0 0 1 úû

a3 = -1,b3 = c3 = 0.

é 0 0 1ù ê ú Therefore, A = ê 0 0 0 ú êë -1 0 0 úû

é1 0 0 ù é1 0 0 ù ê ú ê ú = ê0 1 0 ú ´ ê0 1 0 ú êë0 0 0 úû êë0 0 0 úû é1 0 0 ù ê ú = ê0 1 0 ú and so on. êë0 0 0 úû 100

a1 = b1 = 0, c1 = 1, a2 = b2 = c2 = 0,



Þ l = 0, ± i ( l 2 + 1 = 0 Þ l 2 = -1 = i2 )



Hence, det( I + A100 ) = 2 ´ 2 ´ 1 = 4.

8/9/2023 8:29:51 PM

Linear Algebra • 51

64. (d) é1 ê A = ê1 ê ë1

é1 ê = ê1 ê1 ë



é ù ê1 1 ú 1 1 1 ù é1 0 0 ù ê ú 1 1 ú ú ê 2ú ê w w ú ´ ê0 -1 0 ú ´ ê1 w w2 ú ú ú w2 w4 û ëê0 0 0 ûú ê 1 ú ê1 1 êë w2 w4 úû é 1 w3 ù = w2 ,ú ê = ú 1 ùê w w -1 0 ù é1 1 ú úê 1 ú ê w3 2 -w 0 ú ´ ê1 w w úê 2 = 2 = w ú w ú ê ú êw -w2 0 úû ë1 w w2 û ê ú w6 1 ê 4 = 4 = w2 ú êë w w úû

é 1 - 1 1 - w2 1 - w ù ê ú = ê 1 - w 1 - w3 1 - w2 ú ê ú 1 - w2 1 - w4 1 - w3 ú ëê û

[ w3 =1and w4 =(w3 )w = w ]

Therefore, tr(A) = 0 = sum of eigenvalues = 3 + (-3) + 0. 65. (d) Since X is the eigenvector of P corresponding to the eigenvalue l, so PX = lX. Now (N–1MPM –1N)(N–1MX)

= N–1MPM–1(NN –1)MX



= N–1MP(M –1M)X



= (N–1M)PX



= (N –1M)lX



= l(N –1MX) [since NN –1 = I= M –1M]

By (i) and (ii)



a 2 + b2 = ( a + b)2 - 2ab = 49 - 20 = 29

67. (a) The eigenvalues of A2 are square of eigenvalues of A. Since A2 is a zero matrix. So, all its eigenvalues will be zero. Hence, all the ­eigenvalues of A will also be zero. 68. (c) We knot that the characteristic equation for A is |A - lI| = 0 …(i)

Since “1” is an eigenvalue, so (i) gives

|A - I| = 0 ⇒ |A2 - A| = 0. Hence, A2 - A is singular. “-1” is eigenvalue

é 0 1 - w2 1 - w ù ê ú = ê1-w 0 1 - w2 ú ê ú 1 - w2 1 - w 0 ú ëê û







⇒ |A + I| = 0.



⇒ |A2 + A| = 0. Hence, A2 + A is singular



Since “3” is an eigenvalue, so |A - 3I| = 0(by (i))



\ |A2 - 3A| = 0. Hence, A2 - 3A is singular



Since “0” and “-3” are not eigenvalues,



so, |A| ≠ 0 and |A + 3I| ≠ 0.

Hence, A 2 + 3 A = A ´ A + 3 I ¹ 0. Therefore, A2 + 3A is non-singular. 69. (c)

é1 1ù ú ë1 1û

Let m = 2. Then A = ê

The characteristic equation of A is given by det( A - lI ) = 0 or,



–1



This shows that N MX is an eigenvector of



N–1MPM–1N corresponding to the eigenvalue l.

1-l 1 = (1 - l )2 - 1 = 0 1 1-l

or, l2 - 2l = 0

This satisfies the equation lm - mlm -1 = 0 for m = 2.

66. (a) Trace of a matrix = sum of eigenvalues a + b + 3 = 10 ⇒ a + b = 7 …(i)

70. (d) Let A be the given matrix. Then A = 4I and the eigenvalues of A are 4 and 4. Then AX = lX ⇒ 4IX = 4I, which is true for any

determinant of a matrix = product of eigenvalues

X≠O= ê ú. ë0 û

3ab = 30 ⇒ ab = 10

EMEP.CH01_3PP.indd 51

é0 ù

…(ii)

8/9/2023 8:29:53 PM

52 • Engineering Mathematics Exam Prep 71. (c) é 1 -1 0 ù æ1 ö æ1 ö ç ÷ ê ú ç ÷ AV = lV Þ ê 0 1 -1ú ç1 ÷ = l ç1 ÷ ç1 ÷ êë -1 0 1 úû çè1 ÷ø è ø æ 1 -1 ö æ l ö ç ÷ ç ÷ Þ ç 1 -1 ÷ = ç l ÷ ç -1 + 1 ÷ ç l ÷ è ø è ø Þl=0



72. (c) f(x) - g(x) - h(x) = (x - 1) - (x + 1) - (x2- 1) = x2 + 1 = q(x)

⇒  f(x) - g(x) - h(x) - q(x) = 0



⇒  the functions are linearly dependent. 73. (c)

Let X = ( a1 , a2 ,....) and Y = (b1 ,b2 ,....) Î W .

PREVIOUS YEARS SOLVED PAPERS (2000-2018)

1. The eigenvalues of the matrix



(b) n-1 (d) n(n + 1) (CS GATE 2000) 0ù ú 2ú 0ú ú 1û

\ lim ( aan + bbn ) = a lim an + lim bn

4. If A, B, C are square matrices of the same order, then (ABC)–1 is equal to (a) C–1 A–1B–1 (b) C–1B–1A–1 (c) A–1B–1C–1 (d) A–1C–1B–1  (CE GATE 2000)



n ®¥

= ab + bb = b (if b = 0)

n ®¥

Thus, W is a subspace of Vn(R) if b = 0. Let f(x) = 1, g(x) = x, h(x) = x(1 - x) = x - x2. f ( x ) g ( x ) h( x ) f ¢( x ) g ¢( x ) h¢( x ) f ¢¢( x ) g ¢¢( x ) h¢¢( x ) 1 x x - x2 = 0 1 1 - 2x 0 0 -2 = -2(1 - 0) = -2 ¹ 0

Therefore, the given set of vectors is linearly independent for all x.

(c) 15 (d) 20 (CS GATE 2000)

é1 2 3 ù ê ú 5. The rank of the matrix A = ê3 4 5 ú is êë4 6 8 úû

74. (a) Then

(b) 0

0 7 2 6



n ®¥

(a) 4

0 1 0 0

= a ( a1 , a2 ,....) + b (b1 ,b2 ,....)

= ( aa1 + bb1 , aa2 + bb2 ,...., aan + bbn ,....)

EMEP.CH01_3PP.indd 52

(b) 2,3,-2,4 (d) none of these (EC GATE 2000)

é2 ê 8 3. The determinant of the matrix ê ê2 is ê ë9

Now,



(a) 2, -2, 1, -1 (c) 1, 2, 3, 4

(a) 0 (c) n2 - 3n + 2 

n ®¥

aX + bY

0ù ú 0ú are 0ú ú 4û

2. An n×n array V is defined as follows: V[i, j] = i - j for all i, j, where 1 ≤ i, j ≤ n. Then the sum of the elements of the array V is

Then lim an = b and lim bn = b . n®¥

é2 -1 0 ê ê0 3 0 ê0 0 -2 ê ë0 0 -1

(a) 0 

(b) 1

(c) 2 (d) 3 (IN GATE 2000)

6. The product [P][Q]T of the following two matrices [P] and [Q] is

é2 3ù é4 8 ù [P ] = ê ú , [Q ] = ê ú ë4 5 û ë9 2 û é32 24 ù

é46 56 ù

é35 22ù

é32 56 ù

(a) ê ú (b) ê24 32 ú ë û ë56 46 û (c) ê ú (d) ê24 46 ú ë û ë61 42 û  (CE GATE 2001)

8/9/2023 8:29:54 PM

Linear Algebra • 53

7. The rank of a 3 × 3 matrix C = BA, found by multiplying a non-zero column matrix A of size 3 × 1 and a non-zero row matrix B of size 1 × 3 is (a) 0 b) 1 (c) 2 (d) 3  (GATE 2001) æ1 1 ö ÷ is è0 0ø

8. The rank of the matrix ç 

(a) 4

(b) 1

(c) 2 (d) 0 (CE GATE 2001)

14. Obtain the eigenvalues of the matrix é1 ê ê0 ê0 ê ë0



(a) (5.13, 9.42) (c) (9, 5)

(b) (3.85, 2.93) (d) (10.16, 3.84) (CE GATE 2001)

10. Consider the system of equations given below:

x+y=2 2 x + 2 y = 5



of

the

matrix

(EE GATE 2002)

1 ù 2 ú and X - X + I = 0, êë -a + a - 1 1 - a úû

12. The following set of equations has 3x + 2y + z = x - y + z = 2, -2x + 2z = 5 (b) a unique solution (d) an inconsistency (GATE 2001)

13. Consider the following statements:  S1 : The sum of two singular matrices may be singular

a



é1 - a -1ù ú (b) 2 a ûú ëê a

(a) ê

-1ù é 1-a ê 2 ú êëa - a + 1 a úû

1 ù ú 2 ëê -a + a - 1 a - 1ûú é

(c) ê

-a

éa2 - a + 1 a ù ú 1 1 - a ûú ëê

(d) ê 

(GATE 2002)

17. Consider the system of simultaneous equations x + 2y + z = 6, 2x + y + 2z = 6, x + y + z =5 This system has 

(a) a unique solution (b) an infinite number of solutions (c) no solution (d) exactly two solutions (ME GATE 2003) 18. Consider the following system of linear equations:

Which of the following statements is true? (a) S1 and S2 are both true (b) S1 and S2 are both false (c) S1 is true and S2 is false (d) S1 is false and S2 is true  (CS GATE 2001)

2

then the inverse of X =?

 S2 : The sum of two non-singular matrices may be non-singular

EMEP.CH01_3PP.indd 53

0ù ú 0ú is 0ú ú 1û

(b) 200 (d) 300

é

(a) one solution (b) no solution (c) infinite solution (d) four solutions (GATE 2001)

(a) no solution (c) multiple solutions

(a) 100 (c) 1

16. If X = ê

11. The necessary condition to diagonalize a matrix is that (a) its all eigenvalues should be distinct (b) its eigenvalues should be independent (c) its eigenvalues should be real (d) the matrix is non-singular  (IN GATE 2001)



determinant

0 0 é 1 ê 0 ê100 1 ê100 200 1 ê ë100 200 300

This system has 

(a) 1, 2, -2, -1 (b) -1, -2, -1, -2 (c) 1, 2, 2, 1 (d) none (CS GATE 2002) 15. The

æ5 3ö 9. The eigenvalues of the matrix ç ÷ are è2 9ø



2 34 49 ù ú 2 43 94 ú 0 -2 104 ú ú 0 0 -1 û



æ 2 1 -4 ö é x ù éa ù ç ÷ê ú ê ú ç 4 3 -12 ÷ ê y ú = ê 5 ú ç 1 2 -8 ÷ ê z ú ê 7 ú è øë û ë û

Notice that the second and third columns of the coefficient matrix are linearly dependent. For

8/9/2023 8:29:55 PM

54 • Engineering Mathematics Exam Prep how many values of α, does this system of equations have infinitely many solutions? 

(a) 0 (c) 2

(b) 1 (d) infinitely many (CS GATE 2003)

19. A system of equation represented by AX = 0, when X is a column vector of unknowns and A is a matrix containing coefficients has a non-trivial solution when A is (a) non-singular (b) singular (c) symmetric (d) Hermitian  (GATE 2003) æ4 ç 20. Given matrix A = ç 6 ç2 è



of the matrix is (a) 4 (b) 3

2 3 1

1 3ö ÷ 4 7 ÷ . The rank 0 1 ÷ø

(c) 2 (d) 1 (CE GATE 2003)

æ4 1 ö ÷ , the eigenvalues are è1 4 ø

21. For the matrix ç 

(a) 3 and -3 (b) -3 and -5 (c) 3 and 5 (d) 5 and 0 (ME GATE 2003) æ4

- 2ö

22. The eigenvalues of the matrix A = ç ÷ è -2 1 ø are 

(a) 1, 4 (b) -1, 2 (c) 0, 5 (d) cannot be determined (CE GATE 2004) 23. The sum of the eigenvalues of the matrix given below is



é1 2 3ù ê ú ê1 5 1 ú êë3 1 1 úû

(a) 5

(a) 2n

(b) 7



(a) 4

0ö ÷ 2 ÷ become singular? 0 ÷ø

(b) 6

(c) 12 (d) 8 (ME GATE 2004)

26. Let A, B, C, and D be n×n matrices, each with non-zero determinant. If ABCD = I, then B–1 = ? (a) D–1C–1A–1 (b) CDA (c) ADC (d) does not necessarily exist.  (CS GATE 2004) 27. Real matrices A3×1, B3×3, C3×5, D5×3, E5×5, F5×1 are given, where the matrices B and E are symmetric. Following statements are made with respect to these matrices: (i) The matrix product FTCTBCF is a scalar (ii) The matrix product DTFD is always symmetric With reference to above statements, which of the following is true? 

(a) Statement (i) is true but (ii) is false. (b) Statement (i) false but (ii) is true. (c) Both the statements are true. (d) Both the statements are false. (CE GATE 2004)

28. How many solutions does the following system of linear equations have? -x + 5y = -1, x - y = 2, x + 3y = 3 

(c) 9 (d) 18 (ME GATE 2004)

(b) 2n2 (c)

n2 + n 2 2

(d)

n2 - n 2 2

(CS GATE 2004)

EMEP.CH01_3PP.indd 54

æ8 x ç ç4 0 ç12 6 è

(a) infinitely many (b) two distinct solutions (c) unique (d) none (CS GATE 2004) 29. What values of x, y, z satisfy the following system of linear equations

24. The number of n×n symmetric matrices with each elements being either 0 or 1 is

25. For which value of “x” will the matrix



é1 2 3 ù é x ù é 6 ù ê úê ú ê ú ê1 3 4 ú ê y ú = ê 8 ú êë2 2 3 úû êë z úû êë12úû

(a) x = 6, y = 3, z = 2 (b) x = 12, y = 3, z = -4 (c) x = 6, y = 6, z = -4 (d) x = 12, y = -3, z = 4 

(GATE 2005)

8/9/2023 8:29:56 PM

Linear Algebra • 55

30. Which of the following is an eigenvector of é5 ê 0 the matrix êê 0 ê ë0

0 5 0 0

é1 ù ê ú -2 (a) ê ú (b) ê0ú ê ú ë0û

0 5 2 3

0ù ú 5ú ? 1ú ú 1û



é0 ù ê ú ê0 ú (c) ê1 ú ê ú ë0 û

é1 ù ê ú ê 0 ú (d) ê0ú ê ú ë -2û

é1 ù ê ú ê -1ú ê2ú ê ú ë1 û

(ME GATE 2005)





(b) For matrix Am, m be a positive integer, (lmi , Xim) will be eigenpair for all i.

(c) If AT = A–1, then |li| = 1 for all i.



(d) If AT = A, then li is real for all i.

æ1 4ö ÷ are èa 2ø

æ 3 -2 32. For the matrix P = çç 0 -1 ç0 0 è

0ö ÷ 1 ÷ , one of the 1 ÷ø

eigenvalue is equal to -2. Then which of the following is an eigenvector? é -3ù ê ú ê2 ú (c) êë -1úû

é3 ù (a) êê -2úú (b) êë1 úû

é1 ù ê ú ê -2ú (d) êë3 úû

é2 ù ê ú ê5 ú êë0 úû

(EE GATE 2005)



33. What are the eigenvalues of the following 2 × 2 matrix: é 2 -1ù ê ú ë -4 5 û

(a) -1 and 1 (c) 2 and 5 

(b) 1 and 6 (d) 4 and -1 (CS GATE 2005)

36. The eigenvalue of the matrix M given below are 15, 3, 0.





é4 ù

é2ù

é -1ù

38. A is a 3 × 4 real matrix and AX = B is an inconsistent system of equations. The highest possible rank of A is (a) 1 (b) 2 (c) 3 (d) 4  (ME GATE 2005) 39. Consider the following system of equations in three real variables x1, x2, and x3: 2x1 - x2 + 3x3 = 1,

EMEP.CH01_3PP.indd 55

3x1 - 2x2 + 5x3 = 2,



the

system

of

An´n X n´1 = lX n´1 , where l is a scalar. Let (l1,

(ME GATE 2005)



Xi) be an eigenpair of an eigenvalue and its corresponding eigenvector for real matrix A.

35. Consider

(c) 0 (d) -10 (PI GATE 2005)

0ù ú 1 2û ë é -1ù é3ù é1 ù é -2ù (a) ê ú (b) ê ú (c) ê ú (d) ê ú 1 1 1 ë û ë û ë û ë1 û

equations

(EC GATE 2005)

(b) 10

genvector of the matrix A = éê 1





(a) 20

37. Identify which one of the followings is are ei-

é -4 2ù ú . The eigenvector is 3û

é3 ù

é 8 -6 2 ù ê ú M = ê -6 7 -4 ú êë 2 -4 3 úû

Then the value of the determinant of the matrix is

34. Given the matrix ê ë4

(a) ê ú (b) ê ú (c) ê ú (d) ê ú ë2û ë2 û ë3 û ë -1û

(CE GATE 2005)



31. The eigenvalues of the matrix A = ç

real and non-negative for the condition (a) - 1/16 ≤ a ≤ 1/16 (b) - 1/2 ≤ a ≤ 1/2 (c) - 1/2 ≤ a ≤ 1/16 (d) - 1/16 ≤ a ≤ 1/2  (CS GATE 2005 )

Let I be a n×n unit matrix. Then which of the following statements is not correct? (a) For a homogeneous n×n system of linear equations (A - lI)X = 0, having a non-trivial solution, the rank of A - lI is less than n.

-x1 - 4 x2 + x3 = 3.

This system of equations has (a) no solution (b) a unique solution (c) more than one but finite number of solution (d) an infinite number of solutions (CS GATE 2005)

8/9/2023 8:29:58 PM

56 • Engineering Mathematics Exam Prep 40. Consider a non-homogeneous system of ­linear equations representing mathematically an over determined system. Such a system will be (a) consistent having a unique solution (b) consistent having many solutions (c) inconsistent having a unique solution (d) inconsistent having no solution  (CE GATE 2005) 41. Let A be a 3 × 3 matrix with rank 2. Then AX = 0 has (a) only the trivial solution X = 0 (b) one independent solution (c) two independent solutions (d) three independent solutions  (GATE 2005) 42. In the matrix equation PX = Q, which of the following is a necessary condition for the existence of at least one solution for the unknown vector X: (a) The augmented matrix [P:Q] must have the same rank as matrix P (b) Vector Q must have only one-zero elements (c) Matrix P must be singular (d) Matrix P must be square  (EE GATE 2005) 43. The determinant of the matrix given below is 0 1 -1 1 0 0 1 -2

(a) - 1 

(b) 0

0 1 0 0

(c) 1

EMEP.CH01_3PP.indd 56

0 1 4

0 0

0

1 2

0

0

ö 0÷ ÷ 0 ÷÷ ÷ (b) 0 ÷ ÷ 1 ÷÷ 2ø

æ1 ç2 ç ç0 ç ç ç0 ç ç ç0 è

0 1 0 0

0 0 1 0

 æ2



0

0

1 4

0

0

1 4

0

0

ö 0÷ ÷ 0 ÷÷ ÷ 0÷ ÷ 1 ÷÷ 4ø

(EC GATE 2005) 45. Let A = ç è0 b=? (a) 7/20

æ1 ö - 0.1 ö -1 ç a÷ , A , = 2 ÷ çç ÷÷ then a + 3 ø 0 b è ø

(b) 3/20

(c) 19/60 (d) 11/20 (EC GATE 2005)

46. Consider the matrices X(4×3), Y(4×3), and P(2×3). Then the order of [P(XTY)– 1PT]T will be (a) 2 × 2 (b) 3 × 3 (c) 4 × 3 (d) 3 × 4  (CE GATE 2005) æ 1 0 -1 ö 47. If R = çç 2 1 -1 ÷÷ then top row R–1 is ç2 3 2 ÷ è ø

(b) [5 -3 1]



(a) [5 6 4]



(c) [2 0 -1] (d) ê2

é ë

-1

1ù 2 úû

(EE GATE 2005)



2 3 1 1

48. Multiplication of matrices E and F is G. Matrices E and G are as follows

(d) 2 (GATE 2005)

44. For an orthogonal matrix A of order 4 × 4, (AA′)–1 is æ1 ç4 ç ç0 ç (a) ç ç0 ç ç ç0 è

æ1 ç 0 (c) çç 0 ç è0

æ1 ç4 ç 0ö ç0 ÷ 0÷ ç (d) ç ÷ 0 ç0 ÷ ç 1ø ç ç0 è

0

0

1 2

0

0

1 2

0

0

ö 0÷ ÷ 0 ÷÷ ÷ 0÷ ÷ 1 ÷÷ 2ø



æ cos q ç E = ç - sin q ç 0 è



What is the matrix F ? æ cos q

(a) çç sin q

ç 0 è æ cos q ç (b) ç - cos q ç 0 è æ cos q ç (c) ç - sin q ç 0 è

sin q cos q 0

- sin q cos q 0 cos q sin q 0 sin q cos q 0

0 ö æ1 ÷ ç 0 ÷, G = ç0 ç0 1 ÷ø è

0 1 0

0ö ÷ 0 ÷. 1 ÷ø

0ö ÷ 0÷ 1 ÷ø

0 ö ÷ 0 ÷ 1 ÷ø

0 ö ÷ 0 ÷ 1 ÷ø

8/9/2023 8:29:59 PM

Linear Algebra • 57 æ - cos q ç (d) ç sin q ç 0 è

sin q cos q 0



0 ö ÷ 0 ÷ 1 ÷ø

æ3 è2

54. The eigenvalues of a matrix S = ç (ME GATE 2006)

49. A system of linear simultaneous equations is given as AX = B, where æ1 0 1 0 ö æ0ö ç ÷ ç ÷ 0 1 0 1÷ 0 ç , B=ç ÷ A= ç1 1 0 0 ÷ ç0÷ ç ÷ ç ÷ è 0 0 0 1ø è1 ø



(a) 1

(b) 2

(c) 3 (d) 4 (IN GATE 2006)

æ1 1 1 ö ç ÷ 50. The rank of the matrix ç1 - 1 0 ÷ is ç1 1 1 ÷ è ø



(a) 0

(b) 1

(c) 2

(d) 3

(EC GATE 2006)

 é4 2 ù

51. For the matrix ê ú , the eigenvalue correë2 4 û é101ù sponding to the eigenvector ê ú is ë101û



(a) 2



(b) 4

(c) 6

(d) 8

(EC GATE 2006) 52. Match the items in Columns I and II

Column-I P. Singular matrix Q. Non square matrix R. Real symmetric S. Orthogonal matrix

Column-II 1. Determinant is not defined 2. Determinant is always one. 3. Determinant is zero 4. Eigenvalues are always real 5. Eigenvalues are not defined

(a) P → 3, Q → 1, R → 4, S → 2 (b) P → 2, Q → 3, R → 4, S → 1 (c) P → 3, Q → 2, R → 5, S → 4 (d) P → 3, Q → 4, R → 2, S → 1  (ME GATE 2006) æ 2 -2 3 ö 53. For a given matrix, A = çç -2 -1 6 ÷÷ one of ç1 2 0 ÷ø è



EMEP.CH01_3PP.indd 57

and 1. What are the eigenvalues of S2 ? (a) 1, 25 (b) 6, 4 (c) 1, 5 (d) 2, 10 (ME GATE 2006) 55. The eigenvalues and the corresponding eigenvectors of a 2 × 2 matrix are given by

Eigenvalue

Then the rank of the matrix A is





the eigenvalue is 3. The other eigenvalues are (a) 2, -5 (b) 3, -5 (c) 2, 5 (d) 3, 5 (CE GATE 2006)

2ö ÷ are 5 3ø



l1 = 8



l2 = 4



Eigenvector é1ù v1 = ê ú ë1û é1 ù v2 = ê ú ë -1û

Then the matrix is

æ6 2ö ÷ (b) è2 6ø æ2 4ö (c) ç ÷ (d) è4 2ø

(a) ç



æ4 ç è6 æ4 ç è8

6ö ÷ 4ø 8ö ÷ 4ø

(EC GATE 2006)

56. For a given 2 × 2 matrix A, it is observed that

é1 ù é1 ù é1 ù é1 ù A ê ú = - ê ú , A ê ú = -2 ê ú . 1 1 2 ë û ë û ë û ë -2û



Then the matrix A is é2

1 ù é -1

é1

1 ù é1 0 ù é 2

é1

1 ù é -1

(a) ê ú´ê ë -1 -1û ë 0

0ù é1 1 ù ú´ê ú -2û ë -1 -2û 1ù

(b) ê ú´ê ú´ê ú ë -1 -2û ë0 2 û ë -1 -1û (c) ê ú´ê ë -1 -2û ë 0

0ù é2 1ù ú´ê ú -2û ë -1 -1û

é1 -2ù

(d) ê ú ë1 -3û 

(IN GATE 2006)

57. If a square matrix A is real and symmetric, then the eigenvalues (a) are always real (b) are always real and positive (c) are always real and non-positive (d) occur in complex conjugate pairs  (ME GATE 2007)

8/9/2023 8:30:00 PM

58 • Engineering Mathematics Exam Prep 58. The minimum and maximum eigenvalues of

63. The inverse of the 2 × 2 matrix A = æç1

é1 1 3ù ê ú the matrix ê1 5 1 ú are -2 and 6, respecêë3 1 1 úû

è5

tively. Then what is the other eigenvalue?

(a) 5

(b) 3

(d) -1

(c) 1

(CE GATE 2007)



59. Consider the set of column vectors defined by X = {x Î R3 : x1 + x2 + x3 = 0, where x T = [ x1 , x2 , x3 ]T.} Then which of the following is true? (a) {[1, - 1, 0]T, [1, 0, - 1]T} is the basis for the subspace X (b) {[1, - 1, 0]T, [1, 0, - 1]T} is a linearly independent set, but it does not span X and therefore not a basis of X (c) X is not a subspace of R3 (d) none of the above  (CS GATE 2007) 60. (common data linked question) The Cayley Hamilton theorem states that a square matrix satisfies its own characteristic é -3 2 ù ú ë -1 0 û

equation. Consider a matrix A = ê

(i) A satisfies the relation (a) A + 3I + 2A–1 = O (b) A2 + 2A + 2I = O (c) (A + I) (A + 2I) = I (d) exp(A) = O (ii) A9 Equal to (a) 511A + 510I (b) 309 A + 104I (c) 154A + 155I (d) exp(9A)  (EE GATE 2007) 1+b b 1 61. The determinant b b + 1 1 equals to 1 2b 1



(a) 0 (c) 2(1 - b)(2b + 1)

(b) 2b(b - 1) (d) 3b(1 + b) (PI GATE 2007)

62. X = ëéx1 x 2 . . . . x n ûùT 

EMEP.CH01_3PP.indd 58

is a n-tuple

non-zero vector. Then the n×n matrix V = XX′ (a) has rank zero (b) has rank 1 (c) is orthogonal (d) has rank n. (EE GATE 2007)

1 æ -7 2 ö ç ÷ 3 è 5 -1 ø 1 æ 7 -2 ö (c) ç ÷ 3 è -5 1 ø (a)



2ö ÷ is 7ø

1 æ7 2ö ç ÷ 3 è5 1 ø 1 æ -7 2 ö (d) ç ÷ 3 è -5 1 ø (b)

(CE GATE 2007)

64. The solution for the system defined by the set of equations: 4y + 3z = 8, 2x - z = 2 and 3x + 2y = 5 is (a) x = 0, y = 1, z = 4 , (b) x = 0, y = 1 ,z = 2 3 2 1 (c) x = 1, y = , z = 2 (d) non- existent 2

(GATE 2007)



65. For what value of α and β, the following simultaneous equations have an infinite number of solutions? x + y + z = 5, x + 3y + 3z = 9, x + 2y + az = b 

(a) 2, 7

(b) 3, 8

(c) 8, 3 (d) 7, 2 (CE GATE 2007)

66. Let A be an n×n real matrix such that A2 = I and Y be an n-dimensional vector. Then, the linear system of equations AX = Y has (a) no solution (b) unique solution (c) more than one but finitely many dependent solutions (d) infinitely many dependent solution.  (IN GATE 2007) 67. The number of linearly independent eigenæ2 1ö ÷ is è0 2ø

vectors of ç 

(a) 0 (c) 2

(b) 1 (d) infinite (ME GATE 2007)

68. It is given that X1, X2, ..., XM are M non-zero orthogonal vectors. Then the dimension of the vector space spanned by the 2M vectors X1, X2, ..., XM, - X1, - X2, .... -XM is (a) 2M (b) M + 1 (c) M (d) dependent on the choices of X1, X2, ..., XM  (EC GATE 2007)

8/9/2023 8:30:02 PM

Linear Algebra • 59

69. The following simultaneous equations x + y + z = 3, x + 2y + 3z = 4, x + 4y + kz = 6 will not have a unique solution for k is equal to (a) 0 (b) 5 (c) 6 (d) 7  (CE GATE 2008) 70. For what value of “a” if any, the following system of equations is x, y, z have a solution? 2x + 3y = 4, x + y + z = 4, x + 2y - z = a 

(a) 0 (c) 1

(b) any real number (d) there is no such value (ME GATE 2008) 71. The following system of equations

(a) P11P22 - P12P21 = 1 (b) P11P22 - P12P21 = -1 (c) P11P22 - P12P21 = 0 (d) P11P22 + P12P21 = 0 

to 3. Then the sum of other two eigenvalues is (a) p (b) p-1 (c) p - 2 (d) p - 3  (ME GATE 2008) 77. How many of the following matrices have eigenvalue “1”?

x1 + x2 + 2x3 = 1,

x1 + 2x2 + 3x3 = 2,

x1 + 4 x2 + ax3 = 4. has a unique solution. Then the only possible values of “a” is / are ? (a) 0 (b) either 0 or 1 (c) one of 0, 1 and - 1 (d) any real number other than 5  (CS GATE 2008) 72. The system of linear equations 4x + 2y = 7, 2x + y = 6 has (a) a unique solution (b) an infinite number of solutions (c) no solution (d) exactly two distinct solutions  (EC GATE 2008) 73. The product of matrices (PQ)–1P is (a) P–1 (b) Q–1 (c) P–1Q–1P (d) PQP–1  (CE GATE 2008) 74. A is m×n full rank matrix with m > n and I is an identity matrix. Let A1 = (ATA)–1AT. Then which of the following statement is true? (a) AA1A = A (b) (AA1)2 = A 1 (c) AA A = I (d) AA1A = A1  (EE GATE 2008)

75. All the four entries of the 2×2 matrix æP P = ç 11 è P21

P12 ö ÷ are non-zero and one of its P22 ø

eigenvalues is zero. Then which of the following is true?

EMEP.CH01_3PP.indd 59

(EC GATE 2008)

é1 2 4 ù 76. The matrix êê3 0 6 úú has eigenvalue equal êë1 1 P úû



æ 1 0 ö æ 0 1 ö æ1 -1 ö æ -1 0 ö ç ÷, ç ÷, ç ÷, ç ÷ è 0 0 ø è 0 0 ø è1 1 ø è 0 -1 ø

(a) one

(b) two

(c) three (d) four (CS GATE 2008)

78. The eigenvalues of the matrix P = æç 4 (a) -7, 8

(b) -6, 5

(c) 3, 4

5ö ÷ are è 2 -5 ø

(d) 1, 2

(CE GATE 2008)



æ1 2ö ÷ are writè0 2ø

79. The eigenvectors of the matrix ç é1 ù

é1 ù

ten in the form ê ú and ê ú . Then a + b = ? ëa û ëbû 

(a) 0

(b) 1/2

(c) 1

(d) 2

(ME GATE 2008)

80. The characteristic equation of a 3 × 3 matrix P is defined as P(l) = |P - lI| = p3 + p2 + 2p + 1 = 0. If I denotes identify matrix, then the inverse of the matrix P will be (a) P2 + P + 2I (b) P2 + P + I (c) -(P2 - P + 2I) (d) -(P2 + P + 2I)  (EE GATE 2008) 81. The trace and determinant of a 2 × 2 a matrix are known to be -2 and -35, respectively. Its eigenvalues are? (a) -30, -5 (b) -37, -1 (c) -7, 5 (d) 17.5, -2  (EE GATE 2009)

8/9/2023 8:30:03 PM

60 • Engineering Mathematics Exam Prep 82. The eigenvalues of the following matrix æ -1 3 5 ö ç ÷ ç -3 -1 6 ÷ are ç 0 0 3÷ è ø

(a) 3, 3 + 5j, 6 - j (b) -6 + 5j, 3 + j, 3 - j (c) 3 + j, 3 - j, 5 + j (d) 3, - 1 + 3j, -1 -3j 

(EC GATE 2009)

4ö 5 ÷÷ the transpose of , 3÷ ÷ 5ø

the matrix is equal to the inverse of the matrix. Then, the value of x = ? (a) - 4/5 (b) - 3/5 (c) 3/5 (d) 4/5  (ME GATE 2009) 85. A square matrix B is skew-symmetric if (a) BT = -B (b) BT = B (c) B–1 = B (d) B–1 = BT  (CE GATE 2009)

86. The value of x3 obtained by solving the followings systems of linear equation is x1 + 2x2 - 2x3 = 4 2x1 + x2 + x3 = -2, -x1 + x2 - x3 = 2 (a) - 12 (b) -2 (c) 0 (d) 12  (GATE 2009) 87. The value of “q” for which the following set of linear equations 2x + 3y = 0, 6x + qy = 0 can have non-trivial solution is (a) 2 (b) 7 (c) 9 (d) 11  (GATE 2009) æ 3 + 2i i ö 88. The inverse of the matrix ç ÷ is 3 - 2i ø è -i

(a)

EMEP.CH01_3PP.indd 60

-i ö -i ö 1 æ 3 + 2i 1 æ 3 + 2i ç ÷ (d) ç ÷ 14 è i 14 è i 3 - 2i ø 3 + 2i ø

(CE GATE 2010)



83. The eigenvalues of a 2 × 2 matrix X are - 2 and - 3. Then, the eigenvalues of the matrix (X + I)–1 (X + 5I) are (a) - 3, - 4 (b) - 1, - 2 (c) - 1, - 3 (d) - 2, - 4  (IN GATE 2009) æ3 ç 84. For a matrix M = ç 5 çx ç è

(c)

1 æ 3 - 2i - i ö -i ö 1 æ 3 + 2i ç ÷ (b) 12 ç i 3 + 2i ÷ 12 è i 3 - 2i ø è ø

89. For the set of equations x1 + 2x2 + x3 + 4x4 = 2, 3x1 + 6x2 + 3x3 + 12x4 = 6 which of the following statement is true? (a) Only the trivial solution x1 = x2 = x3 = x4 = 0 exist (b) There are no solutions (c) A unique non-trivial solution exists (d) Multiple non-trivial solution exists  (EE GATE 2010) 90. A real n×n matrix A = [aij]n×n is defined as follows:



ìi , for i = j aij = í î0, otherwise

Then, the sum of all n eigenvalues of A is

(a) n(n + 1) (b) n(n - 1) 2 n(n + 1)(2n + 1) (c) (d) n2 2

(IN GATE 2010)



91. If (1, 0, - 1)T is an eigenvector of the following matrix é1

-1

0ù ú -1ú , then the corresponding ei-1 1 úû

ê  ê -1 2 êë 0



genvalue is (a) 1 (b) 2

(c) 3 (d) 5 (PI GATE 2010) æ2 3ö ÷ . If the eièx yø

92. Consider the matrix A = ç

genvalues are 4 and 8, then (a) x = 4, y = 10 (b) x = 5, y = 8 (c) x = -3, y = 9 (d) x = - 4, y = 10  (CS GATE 2010) 93. One of the eigenvector of the matrix

æ2 2ö A=ç ÷ is è1 3 ø é2 ù é4 ù é1 ù é2 ù (a) ê ú (b) ê ú (c) ê ú (d) ê ú ë1 û ë1 û ë -1û ë -1û



(ME GATE 2010)

8/9/2023 8:30:06 PM

Linear Algebra • 61 é2

é1 1 0 ù 94. An eigenvector of P = êê0 2 2 úú is êë0 0 3 úû



(a) [-1 1 1]T (c) [1 -1 2]T

(b) [1 2 1]T (d) [2 1 -1]T (EE GATE 2010)

é -2 2 -3ù 95. The matrix M = êê 2 1 6 úú has eigenvalêë -1 -2 0 úû



ues 3, -3, and 5. An eigenvector corresponding to the eigenvalue 5 is [1 2 -1]T. One of the eigenvector of the matrix M3 is (a) [1 8 -1]T (b) [1 2 -1]T

(c) éë1 

3

2

T

-1ù û

(d) [1 1 -1] (IN GATE 2011) T

96. Consider the matrix as given below: é1 2 3ù ê ú ê0 4 7 ú êë0 0 3úû

1ù

99. The matrix ëé A ûù = ê ú is decomposed ë4 -1û into a product of lower triangular matrix [L] and an upper triangular matrix [U]. The properly decomposed [L] and [U] matrices, respectively, are é1

0ù

é1

1ù

(a) ê ú and ê ú ë4 -1û ë0 -2û é1 0 ù

é2

1ù

é1 0 ù

é2

1ù

(b) ê ú and ê ú ë0 -3û ë2 1 û (c) ê ú and ê ú ë0 -1û ë4 1 û é2

0ù

é1 0.5ù ú 1 û

(d) ê ú and ê ë4 -3û ë0  æ -5 è 2

100. If A = ç

-3 ö æ1 ÷, I = ç 0ø è0

(EE GATE 2011) 0ö 3 ÷ , then A = ? 1ø



(a) 15A + 12I

Which one of the following options provides the correct values of the eigenvalues of the matrix?



(c) 17A + 15I (d) 17A + 21I



101. Consider the following system of equations: 2x1 + x2 + x3 = 0, x2 - x3 = 0, x1 + x2 = 0 This system has



(a) 1,4, 3

(b) 3, 7, 3 (c) 7, 3, 2 (d) 1, 2, 3 (CS GATE 2011)

97. The system of equations x + y + z = 6, x + 4y + 6z = 20, x + 4y + lz = m has no solution for values of l and m and given by (a) l = 6, m = 20 (b) l = 6, m ≠ 20 (c) l ≠ 6, m = 20 (d) l = 6, m ≠ 20  (EC GATE 2011) 98. [A] is a square matrix which is neither symmetric nor skew symmetric and [A]T is its transpose. The sum and difference of these matrices are defined as [S] = [A] + [A]T and [D] = [A] - [A]T, respectively. Then which of the following statement is true? (a) both [S] and [D] are symmetric. (b) both [S] and [D] are skew-symmetric. (c) [S] is skew-symmetric and [D] is symmetric (d) [S] is symmetric and [D] is skew-symmetric  (CS GATE 2011)

EMEP.CH01_3PP.indd 61

(b) 19A + 30I

(EC GATE 2012)





(a) a unique solution



(b) no solution



(c) an infinite number of solutions



(d) five solutions (GATE 2012)



102. Consider x + 2y + z = 4, 2x + y + 2z = 5, x - y + z = 1. Then the system of algebraic equations given above has

(a) a unique solution: x = 1, y = 1, z = 1



(b) only two solutions: x = 1, y = 1, z = 1 and x = 2, y = 1, z = 0



(c) an infinite number of solutions



(d) no feasible solution



(ME GATE 2012)

8/9/2023 8:30:07 PM

62 • Engineering Mathematics Exam Prep é9 5 ù

103. The eigenvalues of the matrix ê ú are ë5 8 û (a) -2.42 and 6.86 (b) 3.48 and 13.52 (c) 4.70 and 6.86 (d) 6.86 and 9.50  (CE GATE 2012) 104. Let A be the 2 × 2 matrix with elements a11 = a12 = a21 = 1 and a22 = -1. Then, the eigenvalues of the matrix A19 are (a) 1024, -1024 (b) 1024 2 , -1024 2 (c) 4 2 , -4 2 (d) 512 2 , -512 2  (CS GATE 2012) 105. One pair of eigenvectors corresponding to é0 -1ù ú is 0û ë

the eigenvalues of the matrix ê 1

é0 ù é -1ù

é1 ù é j ù

(a) ê ú , ê ú (b) ê ú , ê ú ë - j û ë -1û ë1 û ë 0 û

(c) cos2x , sin2 x , and cos2 x (d) cos2x , sin x , and cos x  (ME GATE 2013)



(IN GATE 2013)

106. A matrix has eigenvalues -1 and -2. The coré1 ù responding eigenvectors are éê 1 ùú and ê ú , ë -2û ë -1û respectively. The matrix is

2ù é1 1 ù é1 (a) ê ú (b) ê ú ë -1 -2û ë -2 -4 û é0 1ù é -1 0 ù (c) ê ú (d) ê -2 -3ú ë û ë 0 -2û



é3 5 2 ù ê ú matrix ê5 12 7 ú is êë2 7 5 úû



(a) 0

(b) 1

(c) 2 (d) 3 (EC GATE 2013)

108. Choose the correct set of functions that are linearly dependent 2

2

(a) sin x , sin x , and cos x (b) cos x , sin x , and tan x

EMEP.CH01_3PP.indd 62

ë x2 û

ë0 û

(GATE 2013)

110. There are three matrices P4×2, Q2×4, and R4×1. Then, the minimum number of multiplication required to compute the matrix PQR is______________________ 

(CE GATE 2013)

111. Which of the followings does not equal to



1

x

x2

1

y

y2 ?

1

z

z2

1 x ( x + 1) x + 1 (a) 1 y( y + 1) y + 1 1 z( z + 1) z + 1 1

x +1

x2 + 1

(b) 1

y +1

y2 + 1

1

z +1

z2 + 1

0

x-y

x 2 - y2

(c) 0

y-z

y2 - z 2

z

z2

(EE GATE 2013)

107.  The minimum eigenvalue of the following

é0 ù

(c) non zero unique solution (d) multiple solutions

é1 ù é0 ù é1 ù é j ù (c) ê ú , ê ú (d) ê ú , ê ú ë j û ë1 û ë j û ë1 û



é2 -2ù é x ù

1 109. The equation ê ú ê x ú = ê ú has 1 1 ë û ë 2 û ë0 û (a) no solution é x ù é0 ù (b) only one solution ê 1 ú = ê ú

1

2 x+ y (d) 2 y + z 1 z 

x2 + y 2 y2 + z2 z2 (CS GATE 2013)

112. Let A be a m×n matrix and B be a n×m matrix. It is given that det(Im + AB) = det(In + BA), where Ik denote the identity matrix of

8/9/2023 8:30:11 PM

Linear Algebra • 63

order k. Use the above property, the determinant of the matrix given below is é2 ê ê1 ê1 ê ë1



(a) 2

1 1 2 1

0 0 0 0 0

0 0 0 1 0

0 0 1 0 0

118. With reference to the conventional Cartesian (x, y) co-ordinate system, the vertices of a triangle have the following co-ordinates: (x1, y1) = (1, 0), (x2, y2) = (2, 2) and (x3, y3) = (4, 3).

(c) 8 (d) 16 (CE GATE 2013)





the

0 1 0 0 0

1ù ú 0ú 0ú ú 0ú 0 úû

matrix

which is obtained by reversing the order of the columns of the identity matrix I6. Let P = I6 + αJ6, where a is non-negative real number. Then the value of α for which det(P) = 0 is_____________ 

(ECE GATE 2014)

114. Which one of the following equations is a correct identity for arbitrary 3 × 3 real matrices P, Q, and R? (a) P(Q + R) = PQ + RP (b) (P - Q)2 = P2 - 2PQ + Q2 (c) det(P + Q) = det (P) + det (Q) (d) (P + Q)2 = P2 + PQ + QP+ Q2  (ME GATE 2014) é2ù 115. If the matrix A is such that A = êê -4 úú ëé1 9 5ùû , êë 7 úû

then the det(A) =?

(CS GATE 2014)

é6

0 4 4 ù ú 14 8 18 ú is êë14 -14 0 -10 úû

116. The rank of the matrix êê -2

117. Two matrices A and B are given below:

EMEP.CH01_3PP.indd 63

æ p2 + q 2 qö = , B ç ÷ ç pr + qs sø è

pr + qs ö ÷. r 2 + s2 ÷ø

If the rank of A is “N,” then the rank of matrix B is

(a) 3/2

(b) 3/4

(c) 4/5 (d) 5/2 (CE GATE 2014)

dx = 3x - 5 y and dy = 4 x + 8 y is dt dt d é x ù é3 -5ù é x ù (a) ê ú=ê úê ú dt ë y û ë4 8 û ë y û



é x ù é3 8 ù é x ù (b) d ê ú = ê úê ú dt ë y û ë4 -5û ë y û é x ù é4 -5ù é x ù (c) d ê ú = ê úê ú dt ë y û ë3 8 û ë y û d éx ù

é4

8 ù éx ù

(d) ê ú=ê úê ú dt ë y û ë3 -5û ë y û 

(GATE 2014)

120. Consider the following system of equations: 3x + 2y = 1, 4x + 7z = 1, x + y + z = 3, x - 2y + 7z = 0.  The number of solutions for this system is________________. 

(CS GATE 2014)

121. The system of equations given below has x + 2y + 4z = 2, 4x + 3y + z = 5, 3x + 2y + 3z = 1 (a) a unique solution (b) two solutions (c) no solution (d) more than two solutions  (GATE 2014) 122. The system of linear equations has

(CE GATE 2014)



The area of the triangle is equal to

119. The matrix form of the linear system

_________________.

æp A=ç èr

(c) N (d) 2N (EE GATE 2014)

1ù ú 1ú 1ú ú 2û

(b) 5

113.  Consider é0 ê ê0 J 6 = ê0 ê ê0 ê1 ë

1 2 1 1

(a) N/2 (b) N - 1 



æ2 ç ç3 ç1 è

1 0 2

3 öæ a ö æ 5 ö ÷ç ÷ ç ÷ 1 ÷ç b ÷ = ç -4 ÷ ÷ ç ÷ 5 ÷ç øè c ø è14 ø

(a) a unique solution (b) infinitely many solutions (c) no solution (d) exactly two solutions (EC GATE 2014)

8/9/2023 8:30:12 PM

64 • Engineering Mathematics Exam Prep 123. Given a system of equations: x + 2y + 2z = b1, 5x + y + 3z = b2 which of the following is true regarding it’s solution? (a) The system has a unique solution for any given b1 and b2 (b) The system will have infinitely many solutions for any given b1 and b2 (c) Whether or not a solution exists depends on the given b1 and b2 (d) The system would have no solution for any values of b1 and b2  (EE GATE 2014) 124. The determinant of matrix A is 5 and the determinant of matrix B is 40. Then determinant of matrix AB is____?  (EC GATE 2014) 125. For the matrix A satisfying the equation given below, the eigenvalues are 

é1 2 3 ù é1 2 3 ù ê ú ê ú A ê7 8 9 ú = ê4 5 6 ú is êë4 5 6 úû êë7 8 9 úû

(a) (1, -1, 1) (c) (1, 1, -1)

(b) (1, 1, 0) (d) (1, 0, 0) (IN GATE 2014)

126. Which of the following statements is true for all symmetric matrices? (a) all eigenvalues are real (b) all eigenvalues are positive (c) all eigenvalues are distinct (d) sum of the eigenvalues is zero  (EE GATE 2014) 127. Which of the following statements is not true for a square matrix A? (a) If A is upper triangular, the eigenvalues of A are the diagonal elements of it (b) If A is real symmetric, the eigenvalues of A are always real and positive (c) If A is real, the eigenvalues of A and AT are always the same (d) If all the principal minors of A are positive, all the eigenvalues are all positive.  (EC GATE 2014)

EMEP.CH01_3PP.indd 64

128. The sum of eigenvalues of matrix M is, where é215 650 795 ù ê ú M = ê655 150 835ú êë485 355 550 úû



(a) 915 (c) 1640

(b) 1355 (d) 2180

(CE GATE 2014)

129. The maximum value of the determinant among all 2 × 2 real symmetric matrices with the trace of 14 is_______________  (EC GATE 2014) 130. One of the eigenvectors of the matrix é -5 2 ù ê ú is ë -9 6 û é -1ù

é -2ù

é2ù

é1ù

(a) ê ú (b) ê ú ë1 û ë9û (c) ê ú (d) ê ú ë -1û ë1û 

(ME GATE 2014)

131. Which one of the following statements is true about every n × n matrix with only real eigenvalues? (a) If the trace of the matrix is positive and the determinant of the matrix is negative, at least one of its eigenvalues is negative (b) If the trace of the matrix is positive, all its eigenvalues are positive (c) If the determinant of the matrix is positive, all its eigenvalues are positive (d) If the product of the trace and determinant of the matrix is positive, all its eigenvalues are positive  (CS GATE 2014) 132. Consider a 3 × 3 real symmetric matrix A such that two of its eigenvalues are a ≠ 0, b ≠ 0 with é x1 ù

é y1 ù

êë x3 úû

êë y3 úû

respective eigenvectors êê x2 úú and êê y2 úú . If a ≠ b, then x1y1 + x2y2 + x3y3 equals 

(a) a

(b) b

(c) ab (d) 0 (ME GATE 2014)

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Linear Algebra • 65

133. The product of the non-zero eigenvalues of é1 ê ê0 the matrix ê0 ê ê0 êë1

0 1 1 1 0

0 1 1 1 0

0 1 1 1 0

1ù ú 0ú 0 ú  is____________. ú 0ú 1 úû

(CS GATE 2014)



134. A real 4 × 4 matrix A satisfies the equation A2 = I, where I is the identity matrix of order 4. Then, the positive eigenvalue of A is____________.  (EC GATE 2014) 135. A system matrix is given as follows:



1 -1 ö æ0 ç ÷ A = ç -6 -11 6 ÷ ç -6 -11 5 ÷ è ø

{ (d) {a( -



(EE GATE 2014)

é3 -2 2 ù ê ú ê4 -4 6 ú : êë2 -3 5 úû

(a) 1.5 and 2.5 (c) 1 and 3

æ2 1 ö ÷ è1 p ø

139. The two eigenvalues of the matrix ç

have a ratio 3 : 1 for p = 2. What is the another value of “p” for which the eigenvalues have the same ratio of 3:1 ? (a) -2 (b) 1 (c) 7/ 3 (d) 14/3  (CE GATE 2015) 140. At least one eigenvalue of a singular matrix is (a) positive (b) zero (c) negative (d) imaginary  (ME GATE 2015) é -3 0 -2ù ê ú trix ê 1 -1 0 ú has three linearly indepenêë 0 a -2úû



(b) 0.5 and 2.5 (d) 1 and 2 (CE GATE 2015)

é4 1 2ù ê ú ê p 2 1 ú is _________ êë14 -4 10 úû

(EC GATE 2015)

é1 -1 2 ù 138. In the given matrix êê0 1 0 úú , one of êë1 2 1 úû



EMEP.CH01_3PP.indd 65

dent real eigenvectors is

2 (b) 1 3 3 3 3 1+2 3 1+ 3 (c) (d) 3 3 3 3

(a)

(EE GATE 2015)



é1 ù 137. The value of “p” such that the vector êê2úú êë3 úû is an eigenvector of the matrix



(CS GATE 2015)

141. The maximum value of “a” such that the ma-

136. The smallest and the largest eigenvalues of the following matrix are

}

2,0,1) : a ¹ 0, a Î R



The absolute value of the ratio of the maximum eigenvalue to the minimum eigenvalue is_______________. 

}

(c) a( 2,0,1) : a ¹ 0, a Î R

the eigenvalues is “1.” The eigenvectors corresponding to the eigenvalue “1” are (a) {a(4, 2, 1): a ≠ 0, a ∈ R} (b) {a(-4, 2, 1) : a ≠ 0, a ∈ R}

142. Consider the following 2 × 2 matrix, A where two elements are unknown and are marked by “a” and “b.” The eigenvalues of this matrix are -1 and 7. What are the values of “a” and “b”? é1 4 ù ê ú ëb a û

(a) a = 6, b = 4 (c) a = 3, b = 5 

(b) a = 4, b = 6 (d) a = 5, b = 3 (CS GATE 2015)

143. The value of “x” for which all the eigenvalues of the matrix given below are real is



é10 5 + j 4 ù ê ú 20 2 ú êx êë 4 2 -10 úû

(a) 5 + j

(b) 5 - j

(c) 1 - 5j (d) 1 + 5j (EC GATE 2015)

8/9/2023 8:30:15 PM

66 • Engineering Mathematics Exam Prep é

1

144. For A = ê ATA–1 is ë - tan x 

(a) sec2x

tan x ù ú , the determinant of 1 û

(b) cos4x

(c) 1 (d) 0 (EC GATE 2015)

145. If any two columns of a determinant é4 7 8 ù ê ú ê3 1 5 ú are interchanged, which one of êë9 6 2 úû



the statement is correct? (a)  absolute value remains unchanged but sign will change. (b) both value and sign will change. (c) absolute value will change but sign will not change. (d) both absolute value and sign will remain unchanged. (ME GATE 2015)

146. Let A = [aij], 1 ≤ i, j ≤ n with n ≥ 3 and aij = i × j. Then the rank of A is (a) 0 (b) 1 (c) n-1 (d) n  (CE GATE 2015) 147. Consider a system of linear equations: x - 2y + 3z = -1, x - 3y + 4z = 1, -2x + 4y - 6z = k The value of “k” for which the system has infinitely many solutions is_____________  (EC GATE 2015) 148. For what value of “p,” the following set of equations will have no solution? 2x + 3y = 5, 3x + py = 10  (CE GATE 2015) 149. If the following system has non-trivial solution: px + qy + rz = 0, qx + ry + pz = 0, rx + py + qz = 0; then which of the following options is true? (a) p - q + r = 0 or p = q = -r (b) p + q - r = 0 or p = -q = r (c) p + q + r = 0 or p = q = r (d) p - q + r = 0 or p = -q = -r  (CS GATE 2015)

EMEP.CH01_3PP.indd 66

éa ê 2 150. The matrix A = ê ê0 ê ë0



0 5 0 0

3 1 2 0

7ù ú 3ú has det(A) 4ú ú bû

= 100 and trace(A) = 14. The value of |a-b| is_____________ (EC GATE 2016) é3 1 ù

151. Let P = ê ú . Consider the set S of all ë1 3û éx ù

vectors ê ú such that a2 + b2 = 1 where ë yû éa ù éx ù ê ú = P ê ú . Then S is ëb û ë yû



(a) a circle of radius 10 (b) a circle of radius 1



(c) an ellipse with major axis along ê ú 1



(d) an ellipse with minor axis along ê ú 1

10

é1ù ë û é1ù ë û

[EE GATE 2016]



152. If the entries in each column of a square matrix M add up to “1,” then an eigenvalue of M is (a) 4 (b) 3 (c) 2 (d) 1  [CE GATE 2016] 153. Consider a 3 × 3 matrix with every element being equal to “1.” Its only non-zero eigenvalue is _____________.  (EE GATE 2016] 154. Let M 4 = I ( where I denote the identity matrix) and M ≠ I, M 2≠ I, M 3 ≠ I. Then for any natural number k, M–1equals (a) M4k+1 (b) M4k+2 (c) M 4k+3 (d) M 4k  (EC GATE 2016) 155. The condition for which the eigenvalues of é2 1 ù

the matrix ê ú are positive is ë1 k û (a) k > 1/2 (b) k > -2 (c) k > 0 (d) k < -1/2  (ME GATE 2016)

8/9/2023 8:30:16 PM

Linear Algebra • 67

156. Two eigenvalues of a 3 × 3 matrix real matrix P are 2 + -1 and 3. Then, the determinant of P is_________  (CS GATE 2016) 157. Consider the following linear system:  x + 2y - 3z = a, 2x + 3y + 3z = b, 5x + 9y - 6z = c. The system is consistent if a, b, and c satisfy the equation (a) 7a - b - c = 0 (b) 3a + b - c = 0 (c) 3a - b + c = 0 (d) 7a - b + c = 0  (CE GATE 2016)

(a) i, - i  163. Consider



(b) -6, 2 (c) -6, -2 (d) 6, -2 (ME GATE 2016)

159. The number of solutions of the simultaneous algebraic equations y = 3x + 3 and y = 3x + 5 is? (a) zero (b) 1 (c) 2 (d) infinite  (PI GATE 2016) 160. Consider the systems, each consisting of “m” linear equations in “n” variables. I. If m < n, then all such systems have a solution II. If m > n, then none of the system has a solution III. If m = n, there exist a system which has a solution. Then which of the following is correct? 

(a) I, II, and III are true (b) only II and III are true (c) only III is true (d) none of them is true (CSE GATE 2016) és x ù

161. Consider a 2 × 2 square matrix P = ê ú ëw s û where “x,” is unknown. If the eigenvalues of the matrix A are (s + jw) and (s - jw), then “x” is equal to (a) jw (b) - jw (c) w (d) -w  (EC GATE 2016)

EMEP.CH01_3PP.indd 67

the

matrix

(c) 0, 1 (d) 0, - 1 (PI GATE 2016) æ2 1 1 ö ç ÷ A=ç 2 3 4 ÷ ç -1 -1 -2 ÷ è ø

whose eigenvalues are 1, - 1, and 3. Then the trace of A3 - A2is_______ (IN GATE 2016) 2 4 ù é3 ê ú A=ê9 7 13 ú has zero as an eigenêë -6 -4 -9 + x úû

é 2 5ù é x ù é 2 ù ê úê ú = ê ú is ë -4 3û ë y û ë -30 û

(a) 6, 2

(b) 1, - 1

164. The value of “x” for which the matrix

158. The solution to the system of equations



æ 0 1ö ÷ are è -1 0 ø

162. The eigenvalue of the matrix A = ç

value, is _____ 

(EC GATE 2016)

165. Suppose that the eigenvalues of a matrix A are 1,2,4. Then the determinant of (A–1)T is ________________  (CS GATE 2016) 166. A real squire matrix A is called skew-symmetric if (a) AT = A (b) AT = A–1 (c) AT = - A (d) AT = A + A–1  (ME GATE 2016) 167. A 3 × 3 matrix P is such that P3 = P. Then the eigenvalues of P are (a) 1, 1, -1 (b) 1, 0.5 + 0.866j, 0.5-0.866j (c) 1, -0.5 + 0.866j, -0.5 - 0.866j (d) 0, 1, -1  (EE GATE 2016) 168. A sequence x[n] is defined as n



é x[n ] ù æ1 1 ö é1 ù ÷ = ê ú for n ³ 2. ê úç ë x[n - 1]û è1 0 ø ë0 û

The initial conditions are : x[0] = 1, x[1] = 1 and x[n] = 0 for n < 0. The the value of x[12] is___________  (EC GATE 2016)

8/9/2023 8:30:17 PM

68 • Engineering Mathematics Exam Prep 169. Let A be a 4 × 3 real matrix with rank 2. Then which of the following statements is true? (a) Rank of ATA is less than 2 (b) Rank of ATA is equal to 2 (c) Rank of ATA is greater than 2 (d) Rank of ATA can be any number between 1 and 3  (EE GATE 2016)

Which of the above statements about eigenvalues of A is/are necessarily correct?

170. If the vectors e1 = (1, 0, 2), e2 = (0, 1, 0) and e3 = (- 2, 0, 1) form an orthogonal basis of the three-dimensional real space R3, then the vector u = (4, 3, - 3) ∈ R3 can be expressed as

175. Consider the matrix A = ç

(a) u = - 2 e1 - 3 e2 - 11 e3

5 5 2 11 (b) u = - e1 - 3 e2 + e3 5 5 2 11 (c) u = - e1 + 3 e2 + e3 5 5 2 11 (d) u = - e1 + 3 e2 - e3 5 5



(a) 0

(EC GATE 2016)

(c) 2 (d) 3 (EC GATE 2017)

172. If the determinant of a 2 × 2 matrix is 50 and one eigenvalue is 10, then the other eigenvalue is_____________________  (ME GATE 2017) 173. The product of the eigenvalues of the matrix P is



æ2 0 1 ö ç ÷ ç 4 -3 3 ÷ ç 0 2 -1 ÷ è ø

(a) -6 

(c) 6 (d) -2 (ME GATE 2017)

(b) 2

matrix of rank 2 with

n

n

åå A i =1 j =1



EMEP.CH01_3PP.indd 68

70

æ

ö

æ l - 80 ö

2 and l2 are X1 = ç ÷ and X 2 = ç ÷, l1 - 50 ø 70 ø è è respectively. The value of X1TX2 is____________ (ME GATE 2017)



é1 -1 5 ù ê ú A = ê0 5 6 ú  are êë0 -6 5 úû

(a) 1, 5, 6 (c) 1, 5 ± 6j

(b) 1, -5 ± 6j (d) 1, 5, 5 (IN GATE 2017) é5 -1ù

177. Consider the matrix ê ú . Then which ë4 1 û one of the following statements is true for the eigenvalues and eigenvectors of this matrix? (a) Eigenvalue 3 has a multiplicity of 2 and only one independent eigenvector exists (b) Eigenvalue 3 has a multiplicity of 2 and two independent eigenvector exist (c) Eigenvalue 3 has a multiplicity of 2 and no independent eigenvector exists (d) Eigenvalues are 3 and -3 and two independent eigenvector exist  (CE GATE 2017) 178. Consider the 5 × 5 matrix

174. Let A be a n×n real valued square symmetric



genvectors corresponding to eigenvalues l1

176. The eigenvalues of the matrix

é5 10 10 ù ú 0 2 ú is êë3 6 6 úû

(b) 1

(CS GATE 2017) æ 50 70 ö ÷ whose eiè 70 80 ø



171. The rank of the matrix M = ê1 ê 

(a) both I and II (b) I only (c) II only (d) neither I nor II 

ij

2

=50. Consid-

er the following statements: I. one eigenvalue must be in [-5, 5] II. The eigenvalue with the largest magnitude must be strictly greater than “5.”

é1 ê ê5 A = ê4 ê ê3 êë2

2 1 5 4 3

3 2 1 5 4

4 3 2 1 5

5ù ú 4ú 3ú ú 2ú 1 úû

It is given that A has only one real eigenvalue. Then, the real eigenvalue of A is (a) -25 (b) 0 (c) 15 (d) 25  (EC GATE 2017)

8/9/2023 8:30:18 PM

Linear Algebra • 69 é 1 ê ê 2 179. Consider the matrix P = ê 0 ê ê- 1 êë 2

0 1 0

1 ù ú 2ú 0 ú. ú 1 ú 2 úû

Which one of the following statements about P is incorrect? (a) Det(P) = 1 (b) P is orthogonal –1 T (c) P = P (d) All eigenvalues of P are real numbers  (ME GATE 2017) 180. Consider the following simultaneous equations (c1 and c2 being constants): 3x1 + 2x2 = c1



4 x1 + x2 = c2

é 1 -1 0 0 0 ù ê ú ê 0 0 1 -1 0 ú 183. The rank of the matrix ê 0 1 -1 0 0 ú ê ú ê -1 0 0 0 1 ú êë 0 0 0 1 -1úû

is________________

 é1

-1ù é -1 -2 -1ù ú ê ú 4 ú and Q = ê 6 12 6 ú êë3 -2 3 úû êë 5 10 5 úû



be two martices, then the rank of P + Q is_______ (CS GATE 2017)

0

tinct eigenvalues and one of the eigenvector

186. If V is a non-zero vector of dimension 3  × 1, then the matrix A =VVT has a rank _____________.  (IN GATE 2017)

é3 ê2 ê 181. The matrix A = ê 0 ê1 ê ë2

1ù 2ú ú -1 0 ú has three dis3ú ú 0 2û

é1 ù is êê0 úú . Which of the following can be another êë1 úû

é1 5 ù é3 7 ù T ú and B = ê ú , then AB is 8 4 ë6 2 û ë û

187. If A = ê equal to

eigenvector of A? é0ù (a) êê 0 úú (b) êë -1úû

é -1ù ê ú ê 0 ú (c) êë 0 úû

é1 ù ê ú (d) ê0ú êë -1úû

é1 ù ê ú ê -1ú êë 1 úû

(EE GATE 2017)

182. If the characteristic polynomial of a 3 × 3 matrix M over R (where R is the set of real numbers) is l3- 4l2 + al + 30 for a∈ R and one eigenvalue of M is 2, then the largest among the absolute values of the eigenvalue of M is ____________.  (CS GATE 2017)

EMEP.CH01_3PP.indd 69

1

184. Let P = ê2 -3 ê

185. The matrix P is the inverse of a matrix Q. If I denotes the identity matrix, which one of the following option is correct? (a) PQ = I but QP ≠ I (b) QP = I but PQ ≠ I (c) PQ = I and QP = I (d) PQ - QP = I  (CE GATE 2017)

The characteristic equation for these simultaneous equations is (a) l2 - 4l - 5 = 0 (b) l2 - 4l + 5 = 0 2 (c) l + 4l - 5 = 0 (d) l2 + 4l + 5 = 0  (CE GATE 2017)



(EC GATE 2017)

é38 28 ù

é3

é43 27 ù

é38 32 ù

(a) ê ú (b) ê ë32 56 û ë42

40 ù ú 8û

(c) ê ú (d) ê28 56 ú ë û ë34 50 û  (CE GATE 2017) 188. Which of the following matrix is singular? æ2 5ö æ3 ÷ (b) ç 1 3 è ø è2 æ2 4ö æ4 (c) ç ÷ (d) ç 3 6 è ø è6

(a) ç



2ö ÷ 3ø

3ö ÷ 2ø

(CE GATE 2018)

8/9/2023 8:30:20 PM

70 • Engineering Mathematics Exam Prep 193. Let M be a 4 × 4 matrix. Consider the following statements: S1: M has 4 linearly independent eigenvectors.

189. For the given orthogonal matrix Q, æ 3 ç 7 ç 6 Q = çç 7 ç ç 2 ç 7 è



2 7 3 7 6 7

6 ö 7 ÷÷ 2 ÷ 7 ÷ ÷ 3 - ÷÷ 7ø

The inverse is

æ 3 ç 7 ç (a) Q = ç -6 ç 7 ç ç 2 ç 7 è

2 7 3 7 6 7

æ -3 6 ö ç 7 ÷ 7 ÷ ç ç 6 2 ÷ (b) Q = ç 7 7 ÷ ç ÷ -3 ÷ ç 2 ç 7 ÷ è 7 ø

-2 7 -3 7 -6 7

6 ö 7 ÷÷ -2 ÷ 7 ÷ ÷ 3 ÷ 7 ÷ø

æ3 ç7 ç (c) Q = ç 2 ç7 ç ç6 ç7 è

-6 7 3 7 2 7

æ -3 2 ö ç 7 ÷ 7 ÷ ç -2 6 ÷ (d) Q = çç ÷ 7 7 ç ÷ -3 ÷ ç 6 ç 7 ÷ 7 ø è

6 7 -3 7 -2 7

-2 ö 7 ÷÷ -6 ÷ 7 ÷ ÷ 3 ÷ 7 ÷ø

(CE GATE 2018)



190. The rank of the following matrix is



æ 1 1 0 -2 ö ç ÷ ç2 0 2 2 ÷ ç4 1 3 1 ÷ è ø

(a) 1

(b) 2

(c) 3 (d) 4 (CE GATE 2018)

æ -4 1 -1 ö ç ÷ 191. The rank of the matrix ç -1 -1 -1 ÷ is ç 7 -3 1 ÷ è ø



(a) 1

(b) 2

(c) 3 (d) 4 [ME GATE 2018]

æ 2 -4 ö ÷ has è 4 -2 ø

192. The matrix ç



(a) Real eigenvalues and eigenvectors. (b) Real eigenvalues but complex eigenvectors. (c) Complex eigenvalues but real eigenvectors. (d) Complex eigenvalues and eigenvectors.



(CS/IT-GATE-2018)



EMEP.CH01_3PP.indd 70



S2: M has distinct eigenvalues.



S3: M is non-singular (invertible).



Which one among the following is TRUE?



(a) S1 implies S2 (c) S2 implies S1

(b) S1 implies S3 (d) S3 implies S2 [EC GATE 2018]

æ 1 0 -1 ö ç ÷ 194. Let A = ç -1 2 0 ÷ and B = A3 - A2 - 4A ç 0 0 -2 ÷ è ø



+ 5I, where I is the 3 × 3 identity matrix. The determinant of B is _________? (Up to 1 decimal places). [EC GATE 2018]

195. Consider the following system of linear equation 3x + 2ky = - 2, kx + 6y = 2 Here x and y are the unknowns and k is a real constant. The value of k for which there are an infinite number of solutions is (a) 3 (b) 1 (c) - 3 (d) - 6  [IN GATE 2018] æ

k

2k ö

196. Consider matrix A = ç 2 and vector ç k - k k2 ÷÷ è ø éx ù X = ê 1 ú . The number of distinct real values ë x2 û

of k for which the equation AX = O has an infinitely many solutions is _________?  [EC GATE 2018] 197. The diagonal elements of a 3 × 3 matrix are -10, 5, and 0, respectively. If two of its eigenvalues are -15 each, the third eigenvalues is _______?  [PI GATE 2018] é1 ù

198. Consider a matrix A = UVT, where U = ê ú , ë2 û é1ù T V = ê ú Note that V denotes the transpose ë1û

of  V. The largest eigenvalue of A is _____?  [CS/IT GATE 2018]

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Linear Algebra • 71

199. Consider a non-singular 2 × 2 square matrix A. If trace (A) = 4 and trace (A2) = 5, the determinant of the matrix A is __________ ?  [EC GATE 2018] 200. Let N be a 3 × 3 matrix with real numbers entries. The matrix N is such that N2 = 0. The eigenvalues of N are (a) 0, 0, 0 (b) 0, 0, 1 (c) 0, 1, 1 (d) 1, 1, 1

Answer key 177. (a) 182. 5. 187. (a) 192. (d)

197. 25. 198. 3.

EMEP.CH01_3PP.indd 71

179. (d) 184. 2. 189. (c) 194. 1. 199.

180. (a) 185. (c) 190. (b) 195. (c)

11 . 2

181. (c) 186. 1 191. (b) 196. 2. 200. (a)

Explanation

Answer key 1. (b) 2. (a) 3. (a) 4. (b) 5. (c) 6. (a) 7. (b) 8. (b) 9. (d) 10. (b) 11. (d) 12. (b) 13. (a) 14. (a) 15. (c) 16. (b) 17. (c) 18. (b) 19. (b) 20. (c) 21. (c) 22. (c) 24. (c) 25. (a) 26. (b) 27. (d) 28. (c) 29. (c) 30. (a) 31. (d) 32. (d) 33. (b) 34. (c) 35. (b) 36. (c) 37. (b) 38. (b) 39. (b) 40. (a), (b), (d) 41. (b) 42. (a) 43. (a) 44. (c) 45. (a) 46. (a) 47. (b) 48. (a) 49. (d) 50. (c) 51. (c) 52. (a) 53. (b) 54. (a) 55. (a) 56. (c) 57. (a) 58. (b) 59. (a) 62. (b) 61. (a) 60. (i) - (a); (ii) - (a) 63. (a) 64. (d) 65. (a) 66. (b) 67. (b) 68. (c) 69. (d) 70. (a) 71. (d) 72. (c) 73. (b) 74. (a) 75. (c) 76. (c) 77. (a) 78. (b) 79. (b) 80. (d) 81. (c) 82. (d) 83. (c) 84. (a) 85. (a) 86. (b) 87. (c) 88. (b) 89. (d) 90. (a) 91. (a) 92. (d) 93. (a) 94. (b) 95. (b) 96. (a) 97. (b) 98. (d) 99. (b) 100. (b) 101. (c) 102. (c) 103. (b) 104. (d) 105. (a),(d) 106. (d) 107. (a) 108. (c) 109. (d) 110. 16 111. (a) 112. (b) 113. 1,-1 114. (d) 115. 0. 116. 2. 117. (c) 118. (a) 119. (a) 120. one 121. (a) 122. (b) 123. (b) 124. 200 125. (c) 126. (a) 127. (b) 128. (a) 129. 49 130. (d) 131. (a) 132. (d) 133. 6. 134. 1. 135. 3. 136. (d) 137. 17 138. (b) 139. (d) 140. (b) 141. (b) 142. (d) 143. (a) 144. (c) 145. (a) 146. (b) 147. 2 148. 4.5 149. (c) 150. 3 151. (d) 152. (d) 153. 3 154. (c) 155. (a) 156. 15. 157. (b) 158. (d) 159. (a) 160. (c) 161. (d) 162. (a) 163. -6. 164. 1. 165. 1/8. 166. (c) 167. (d) 168. 233 169. (b) 170. (d) 171. (c) 172. 5. 173. (b) 174. (b) 175. [0] 176. (c)

178. (c) 183. 4 188. (c) 193. (c)

1. (b) The trace of the matrix = 2 + 3 + (-2) + 4 = 7 = sum of eigenvalues = 2 + 3 + (-2) + 4. 2. (a) Here V[i, j] denotes the element lying in ith row and jth column.



æ V[1,1] V[1,2] V[1,3] ö ç ÷ For n = 3, V = ç V[2,1] V[2,2] V[2,3] ÷ ç V[3,1] V[3,2] V[3,3] ÷ è ø æ1 - 1 1 - 2 ç = ç2 -1 2 - 2 ç3 -1 3 - 2 è æ 0 -1 -2 ö ç ÷ = ç 1 0 -1 ÷ ç2 1 0 ÷ø è



1-3ö ÷ 2 - 3÷ 3 - 3 ÷ø ( V [i, j ] = i - j )

Therefore, the sum of the elements in the array V is zero. 3. (a) 2 8 2 9

0 1 0 0

0 7 2 6

0 2 0 1

= 2 ´ cofactor of “2” (since “2” is the only non-zero element in first row) 1 7 2 =20 2 0 0 6 1



= 2´2 = 4.

4. (b) (ABC)–1 = [(A(BC)]–1 = (BC)–1 A–1 = C–1B– 1 –1 A .

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72 • Engineering Mathematics Exam Prep 5. (c)

13. (a) é1



A = êê3 êë4 é1 ê  3 ê êë0

2 4 6 2 4 0

3ù

ú ú 8 úû 3ù ú 5 [by R3 ® R3 - ( R1 + R2 )] ú 0 úû 5

which has 2 non-zero rows Therefore, rank (A) = 2. 6. (a) [P][Q]T =

é 2 3 ù é4 9 ù é 2 ´ 4 + 3 ´ 8 2 ´ 9 + 3 ´ 2 ù ê ú´ê ú=ê ú ë4 5 û ë 8 2 û ë4 ´ 4 + 5 ´ 8 4 ´ 9 + 5 ´ 2 û é32 24 ù =ê ú. ë56 46 û

é0 1 ù é0 0 ù ú ,B = ê ú .Then det(A) 0 0 ë û ë0 1 û

Case-I: Let A = ê

= 0, det(B ) = 0 and det(A + B) =

In this case, each of A and B is a singular matrix and A+B is also singular. Thus, the statement S1 is true. é -1 0 ù é3 0 ù ú ,B = ê ú . Then, ë 0 2û ë0 2 û

Case-II: Let A = ê

det(A) ≠ 0, det(B) ≠ 0 and det(A + B) =

14. (a) The trace of the matrix = 1 + 2 + (-2) + (-1) = 0 = sum of eigenvalues = 1 + 2 + (-2) + (-1). 15. (c) Since the determinant value of an upper triangular matrix is the product of the diagonal elements, so the determinant value of the given matrix =1 × 1 × 1 × 1 = 1. 16. (b) X2 – X + I = O Þ I = X - X2

9. (d) The trace of the matrix = 5 + 9 = 14 = sum of eigenvalues = 10.16 + 3.84.



1 1 D= = 0, 2 2

    

(

Þ X -1 = XX -1 - X 2 X -1 = I - X



EMEP.CH01_3PP.indd 72

= 3( -2 - 0) - 2(2 + 2) + 1(0 - 2) ¹ 0

Thus, the system has a unique solution.

1 ù a é1 0 ù é =ê ú ú-ê 2 ë0 1 û êë -a + a - 1 1 - a ûú -1ù é 1-a =ê 2 ú ëêa - a + 1 a úû

17. (c)

1 2 1

D= 2 1 2 =0 1 1 1

12. (b) 3 2 1 D = 1 -1 1 -2 0 2

)

Þ IX -1 = X - X 2 X -1

2 1 D1 = ¹0 5 2

 Therefore, the system has no solution (by Cramer’s rule). 11. (d) The necessary condition to diagonalize a matrix is that the matrix is non-singular, whereas the sufficient condition is that the matrix has n linearly independent eigenvectors (where “n” is the order of the square matrix).

2 0 ¹ 0. 0 4

In this case, each of A and B is non-singular matrix and A + B is also non-singular. Thus, the statement S2 is true.

7. (b) C  = BA = B1×3 A3×1. Then, C is a matrix of order 1×1. Therefore, rank(A) ≤ min {1, 1} = 1. So the possible cases are rank(C) = 0 or rank(C) = 1. But since both A and B are non-zero matrices, so the product AB (=C) contains at least one non-zero element. Hence, rank(C) = 1. 8. (b) Since the given matrix has only one nonzero row, so rank = 1.

10. (b)

0 1 = 0. 0 1

6 2 1



D1 = 6 1 2 = 3 ¹ 0 5 1 1

   D = 0, D1 ≠ 0, so the system has no solutions (by Cramer’s rule).

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Linear Algebra • 73

Alternative method:

æ4 ç  ç0 ç2 è æ0 ç  ç0 ç2 è

é1 2 1 6 ù ê ú [A : B] = ê2 1 2 6 ú êë1 1 1 5 úû é1 2 1 6 ù ê ú  ê0 -3 0 -6 ú êë0 -1 0 -1úû

(by R2 ® R2 - 2R1 , R3 ® R3 - R1 )

é1 2 1 6 ù ê ú  ê0 0 0 -3ú by ( R2 ® R2 - 3R3 ) êë0 -1 0 -1úû



é1 2 1 6 ù ê ú  ê0 -1 0 -1ú êë0 0 0 -3úû (by R2 « R3 )





Which has two non-zero rows



Therefore, rank (A) = 2.

22. (c)

18. (b)  The given system of equations can be written as 2x + y - 4z = α, 4x + 3y - 12z = 5, x + 2y - 8z = 7. The system has infinitely many solutions

|A - lI| = O ⇒



⇒ (4 - l) (1 - l) - 4 = 0



⇒ l2 - 5l = 0



⇒ l = 0, 5

Alternative Method If we take the eigenvalues 0 and 5, then product of the eigenvalues = 0, which is equal to det(A). 23. (b) The sum of eigenvalues = trace of the matrix = 1 + 5 + 1 = 7.

2 1 a Þ 4 3 5 =0 1 2 7

24. (c) For n = 2, the symmetric matrices with each element equal to either 0 or 1 are

Þ 2(21 - 10) - (28 - 5) + a(8 - 3) = 0 1 Þa= . 5



So a has only one value. 19. (b) A homogeneous system of equations has a non-trivial solution if the coefficient matrix is singular. æ4 2 1 æ4 2 1 3ö ç 5 ç ÷ A = ç6 3 4 7÷  ç0 0 ç 2 ç2 1 0 1÷ ç è ø 2 1 0 è

EMEP.CH01_3PP.indd 73

(by R2 ® R2 -

3ö ÷ 5÷ 2÷ 1 ÷ø

3 R1 ) 2

é0 ê ë0 é1 ê ë0

0 ù é1 0 ù é0 ú ,ê ú ,ê 0 û ë0 1 û ë1 0 ù é0 0 ù é1 ú ,ê ú ,ê 0 û ë0 1 û ë1

1 ù é1 1ù ú,ê ú, 0 û ë1 1û 1 ù é0 1 ù ú,ê ú. 0 û ë1 1û

Thus, for n = 2, there exist eight such symmetric matrices. Now, 8 =

20. (c)



4-l -2 =0 -2 1-l



Þ D3 = 0



æ0 0 0 0ö ç ÷  ç 0 0 1 1 ÷ (by R1 ® R1 - R2 ) ç2 1 0 1÷ è ø

21. (c) The trace of the matrix = 4 + 4 = sum of eigenvalues = 3 + 5.

Hence, rank ([A:B]) = 3 and rank(A) = 2. Since they are unequal, the system has no solution.





2 1 3ö 2 ÷ 0 1 1 ÷ (by R2 ® R2 ) 5 1 0 1 ÷ø 0 1 1ö ÷ 0 1 1 ÷ (by R1 ® R1 - 2R3 ) 1 0 1 ÷ø

22 + 2 2 2

Hence, option (c) is correct. 25. (a)



æ 8 x 0ö ç ÷ A = ç 4 0 2÷ ç12 6 0 ÷ è ø

Then, A is singular ⇒ det(A) = 0 ⇒ x = 4.

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74 • Engineering Mathematics Exam Prep 26. (b) A, B, C, D has non-zero determinants A–1, B–1, C–1, D–1 exist Then, ABCD = I ⇒ A–1(ABCD) = A–1I = A–1 ⇒  (A–1 A)(BCD)D–1 = A–1D–1 ⇒  IBC(DD–1) = A–1D–1 ⇒  BCI = A–1D–1 ⇒  BC = A–1D–1 ⇒  (BC)C–1 = A–1D–1C–1 ⇒  B(CC–1) = A–1D–1C–1 ⇒  BI = A–1D–1C–1 ⇒  B = A–1D–1C–1 = (CDA)–1 ⇒  B–1 = ((CDA)–1)–1 = CDA 27. (d) Since product of matrices (if defined) is again matrix. Hence, the product FTCTBCF cannot be a scalar. Hence, the statement (i) is false. Again, F is a matrix of order 5×1 and D is a matrix of order 5×3 ⇒ FD is not defined (since number of columns of F ≠ number of rows of D and so DTFD is not defined. Therefore, the statement (ii) is false. 28. (c) Solving the first two equations we get, x = 9/4 and y = 1/4; which satisfies the third equation. So the system has a unique solution given by x = 9/4 and y = 1/4.

1 2 6 D3 = 1 3 8 2 2 12 1 2 6 =0 1 2 0 -2 0 (by R2 ® R2 - R1 , R3 ® R3 - 2R1 ) =4 Thus,



x + 2y + 3z = 6, x + 3y + 4z = 8, 2x + 2y + 3z = 12. These equations are satisfied for x = y = 6 and z = -4. 30. (a) A - lI = 0

5-l 0 0 0 0 5-l 5 0 Þ =0 0 0 2-l 1 0 0 3 1-l 5-l 5 0 2-l 1 =0 Þ (5 - l ) 0 0 3 1-l

1 2 3 D = 1 3 4 = -1 2 2 3

6 2 3 6 2 3 D1 = 8 3 4 = 2 1 1 = -6 12 2 3 6 0 0



(by R2 ® R2 - R1 , R3 ® R3 - R1 )

Þ l = 5 is an eigenvalue. Then AX = lX



1 6 3 =0 2 1 0 0 -3 (by R2 ® R2 - R1 , R3 ® R3 - 2R1 )



= 1(-6 - 0) = -6

EMEP.CH01_3PP.indd 74

é5 ê 0 Þê ê0 ê ë0

0 5 0 0

0 5 2 3

0ù éx ù éx ù úê ú ê ú 0ú ê yú y = 5ê ú ú ê ú ê 1 z zú úê ú ê ú 1û ë t û ët û

é 5x ù é5x ù ê ú ê ú 5 y + 5z ú ê5 y ú Þê = ê 2z + t ú ê 5z ú ê ú ê ú ë 3z + t û ë 5t û Þ 5x = 5x ,5 y + 5z = 5 y,2z + t = 5z ,3z + t = 5t Þ z = 0, t = 0 Then eigenvector,

1 6 3 D2 = 1 8 4 2 12 3



D D1 D = 6, y = 2 = 6, z = 3 = -4. D D D

Alternative Method The given system of equations can be written as

29. (c) Using Cramer’s rule we have



x=

éx ù éx ù é 1 ù ê ú ê ú ê ú y y -2 X =ê ú=ê ú=ê ú ê z ú ê0 ú ê 0 ú ê ú ê ú ê ú ë t û ë0 û ë 0 û

(taking x = 1, y = -2)

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Linear Algebra • 75

31. (d)

34. (c)

A - lI = 0

Þ

A - lI = 0

1-l 4 =0 2-l a

Þ

Þ (1 - l )(2 - l ) - 4a = 0

Þ ( l - 3)(4 + l ) - 8 = 0

2

Þ l2 + l - 20 = 0 Þ l = -5,4 For l = -5,

Þ l - 3l + (2 - 4a ) = 0 3 ± ( -3)2 - 4(2 - 4a ) 2 3 ± 16a + 1 , Þl= 2 which is real if 1 16a + 1 ³ 0, i.e; a ³ 16 Þl=



AX = lX é -4 2ù é x ù éx ù Þê ú ê ú = -5 ê ú y 4 3 ë ûë û ë yû é -4 x + 2 y ù é -5x ù Þê ú=ê ú ë 4 x + 3 y û ë -5 y û

Also, l ³ 0





Þ

Þ -4 x + 2 y = -5x , 4 x + 3 y = -5 y

3 ± 16a + 1 ³0 2



3 16a + 1 Þ ³ 2 2 9 16a + 1 Þ ³ 4 4 1 Þa£ 2 1 1 \£a£ . 16 2



PX = lX

æ 3 -2 2 ö é x ù éx ù ç ÷ê ú ê ú Þ ç 0 -2 1 ÷ ê y ú = - 2 ê y ú ç 0 0 1 ÷ êz ú è øë û ëê z ûú é3x - 2 y + 2z ù é -2x ù ê ú ê ú Þ ê -2 y + z ú = ê -2 y ú êë úû êë -2z úû z

é x ù é -2 y ù é 2 ù \ Eigenvector, X = ê ú = ê ú=ê ú ë y û ë y û ë -1û (taking y = -1)



Thus, (a) is correct.



AT = A–1 ⇒ A is orthogonal



⇒ |li|= 1, (c) is correct



AT = A ⇒ A is symmetric



⇒ eigenvalues li are real



Thus, (d) is correct.



Therefore, (b) is not a correct statement.

36. (c) Value of the determinant = product of eigenvalues

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



Þ 5x - 2 y + 2z = 0, z = 0 Þ x =

37. (b) Since A is a lower triangular, so the eigenvalues are the diagonal elements, i.e. 1 and -2. For l = 1

2 y, z = 0 5

é2 ù ê 5 y ú é2 ù ê ú ê ú \ eigenvector, X = ê y ú = ê5 ú (for y = 5) ê z ú ê0 ú ê ú ë û ëê ûú

33. (b) Determinant of the matrix = 10 - 4 = 6 = product of eigenvalues = 1 × 6.

EMEP.CH01_3PP.indd 75

Þ x = -2 y

35. (b) System has a non-trivial solution ⇒ rank (A - lI) < n.

32. (d)



-4 - l 2 =0 4 3-l

= 15 × 3 × 0 = 0

AX = lX



0 ù éx ù é1 éx ù Þê ú ê ú =1ê ú 1 2 y ë ûë û ë yû é x ù éx ù Þê ú=ê ú ë -x - 2 y û ë y û

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76 • Engineering Mathematics Exam Prep

Þ x = x , -x - 2 y = y Þ x = -3 y

Thus, eigenvector,

é x ù é -3 y ù é 3 ù X =ê ú=ê ú=ê ú ë y û ë y û ë -1û

(for y = -1)

44. (c) A is orthogonal ⇒ AA′ = I æ1 0 0 ç 0 1 0 -1 -1 Þ ( AA¢) = I = I = ç ç0 0 1 ç è0 0 0 45. (a)

38. (b) A has order 3 × 4

A =

⇒  rank (A) ≤ min{3, 4} ⇒  rank (A) ≤ 3 Also rank ([A : B]) ≤ min {3, 5} = 3 If rank (A) = rank ([A : B]) = 3, then the system will be consistent. Hence, for the system to be inconsistent, the highest possible rank of A will be 2. 2 -1 3 D = 3 -2 5 = 2 ¹ 0 -1 -4 1



[ P ( X T Y )-1 P T ]T = ( P T )T [( X T Y )-1 ]T P T = P[( X T Y )-1 ]T P T = P2´3[{( X T )3´4 Y4´3 }-1 ]T ( P T )3´2 = P2´3[{( X T Y )3´3 }-1 ]T ( P T )3´2 = P2´3[{( X T Y )-1 }3´3 ]T ( P T )3´2 = P2´3[{( X T Y )-1 }T ]3´3 ( P T )3´2 = P2´3[{( X T Y )-1 }T P T ]3´2



42. (a) Rank (P) = rank ([P : Q]) ⇔ at least one solution exists.

0 1 0 -1 = 0 0 1 -2

0 1 0 0 0 1 0 0

= [ P {( X T Y )-1 }T P T ]2´2

= P2×3[{(XTY)–1}T]3×3(PT)3×2 = P2×3[{(XTY)–1}TPT]3×2 = [P{(XTY)–1}TPT]2×2 47. (b) 1 0 -1

R = 2 1 -1 = 1(2 + 3) - 0 - 1(6 - 2) = 5 - 4 = 1 2 3 2

2 3 1 1 2 4 1 1

1 é3 0.1ù ê ú 6 ë0 2 û é 1 0.1 ù é1 ù ê2 aú 6 ú ê =ê (given) ú= 2 ê 0 1 ú êê 0 b úú ë û êë 3 úû 0.1 1 1 1 7 ,b = . So a + b = . \a = + = 6 3 60 3 20

46. (a)

40. (a), (b), (d) In case of non-homogeneous system of equations, a consistent system will have either unique solution or infinitely many solutions, where as an inconsistent system will have no solution.

0 1 -1 1 0 0 1 -2

é 1 ê ê 3 ê 0 adj( R ) = ê ê 3 ê ê 0 ê 1 ë

[by R2 ® R2 + R4 ]



EMEP.CH01_3PP.indd 76

-1 2 -1 2 -1 -1

-

2 -1 2 2

1 -1 2 2 -

1 -1 2 -1 T

= 1 ´ cofactor of “1” [  “1” lies in 1st column] 1 0 2 1 0 = -1 1 4 = = -1 -1 1 0 0 1

1 adj( A ) A

=

Therefore, the system has a unique solution (by Cramer’s rule).

41. (b) No. of independent solution  = order of the square matrix - rank of the coefficient matrix = 3 - 2 = 1.

2 -0.1 =6 0 3

A -1 =

39. (b)

43. (a)

0ö ÷ 0÷ 0÷ ÷ 1ø

R -1



é 5 -6 4 ù é5 ê ú ê = ê -3 4 -3ú = ê -6 êë 1 -1 1 úû êë 4 é 5 -3 1 1ê = adj( R ) = ê -6 4 1 R êë 4 -3

2 1 ù ú 2 3 ú 1 0ú ú 2 3ú ú 1 0 ú 2 1 úû

T

-3 1 ù ú 4 -1ú -3 1 úû 1ù ú -1ú 1 úû

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Linear Algebra • 77

ÞR

-1



= 1 ´ cofactor of “1”

é 5 -3 1 ù ê ú = ê -6 4 -1ú êë 4 -3 1 úû

(since “1” is the only non-zero element of first column)

Therefore, top row of R–1 = [5 -3 1] 48. (a)



cos q sin q 0 E = - sin q cos q 0 0 0 1

Thus, A is a non-singular matrix of order 4. Therefore, rank (A) = 4 50. (c)

= cos2 q + sin 2 q = 1 ¹ 0. \ E -1 exist. adj( E ) é cos q ê ê 0 ê sin q = êê 0 ê ê sin q ê cos q ë

0 1

-

0 1

- sin q 0 0 1 cos q 0 0 1

0 0

-

cos q 0 - sin q 0

- sin q cos q ù ú 0 0 ú cos q sin q ú ú 0 0 ú ú cos q sin q ú - sin q cos q úû

T

T



écos q - sin q 0 ù é cos q sin q 0 ù ú ê ú ê = ê - sin q cos q 0 ú = êsin q cos q 0 ú êë 0 êë 0 0 1 úû 0 1 úû

écos q - sin q 0 ù 1 ê ú -1 E = adj( E ) = êsin q cos q 0 ú R êë 0 0 1 úû Now EF = G Þ E -1 ( EF ) = E -1G Þ ( E -1E ) F = E -1 I (G = I,the identity matrix) Þ IF = E -1 Þ F = E -1

49. (d)

1 0 A = 1 0



EMEP.CH01_3PP.indd 77

1 0 = 0 0

écos q - sin q 0 ù ê ú = êsin q cos q 0 ú . 0 1 ûú ëê 0 0 1 1 0

1 0 0 0

0 1 1 0 1 -1 0 0

1 0 1 = 1 -1 0 = -1 ¹ 0. 0 0 1



æ 1 1 1ö æ 1 1 1ö ç ÷ ç ÷ ç 1 -1 0 ÷  ç 1 -1 0 ÷ (by R3 ® R3 - R1 ) ç 1 1 1 ÷ ç 0 0 0÷ è ø è ø



which has 2 non-zero rows



Therefore, rank = 2. 51. (c)

AX = lX



é4 2 ù é101ù é101ù Þê úê ú = lê ú ë2 4 û ë101û ë101û



é4 ´ 101 + 2 ´ 101ù é101l ù Þê ú=ê ú ë2 ´ 101 + 4 ´ 101û ë101l û



Þ 101l = 4 ´ 101 + 2 ´ 101



⇒ l = 4 + 2 = 6. 52. (a)

53. (b) Let l1 and l2 be the other two eigenvalues. Then, the sum of eigenvalues = trace of the matrix

⇒ 3 + l1 + l2 = 2 - 1 + 0 = 1



⇒ l2 = -2 - l1

If l1 = 3, then l2 = - 2 - 3 = -5. If l1 = 2, then l2 = -2 - 2 = -4. 54. (a) The eigenvalues of S2 = square of the eigenvalues of S = 1, 25.

0 1 0 1 0 1 (by R3 ® R3 - R1 ) 0 1

55. (a)

æa bö Let A = ç ÷ è c dø Then, Av1 = 8v1



æ a b ö é1ù é1ù Þç ÷ê ú =8ê ú è c d ø ë1û ë1û é a + b ù é8 ù Þê ú =ê ú ë c + d û ë8 û

ì a + b = 8............(1) Þí îc + d = 8............(2) Also,

8/9/2023 8:30:30 PM

æa bö Let A = ç ÷ è c dø Then, Av1 = 8v1

æ a b ö é1ù é1ù 78 ç• Engineering Exam Prep Þ ÷ ê ú = 8M ê athematics ú è c d ø ë1û ë1û é a + b ù é8 ù Þê ú =ê ú ë c + d û ë8 û

é 1 1 ù é -1 0 ù é 2 1 ù ê ú´ê ú´ê ú ë -1 -2û ë 0 -2û ë -1 -1û é 1 1 ù é -2 -1ù =ê ú´ê ú ë -1 -2û ë 2 2 û

ì a + b = 8............(1) Þí îc + d = 8............(2) Also, Av2 = 4v2

æa bö é 1 ù é1 ù Þç ÷ê ú =4 ê ú è c d ø ë -1û ë -1û éa - bù é 4 ù Þê ú =ê ú ëc - d û ë -4 û

ì a - b = 4............(3) Þí îc - d = -4............(4)



Solving (1) and (3) we get, a = 6, b = 2.



Solving (2) and (4) we get, c = 2, d = 6. æ6 2ö ÷. è2 6ø

Hence, A = ç 56. (c) éa b ù A=ê ú ë c dû Then

é1 ù é1 ù Aê ú = -ê ú ë -1û ë -1û éa b ù é 1 ù é1 ù Þê úê ú = -ê ú ë c d û ë -1û ë -1û éa - bù é -1ù Þê ú=ê ú ëc - d û ë 1 û

Þ a - b = -1......(1), c - d = 1.....(2)



é1 ù é1 ù Again, A ê ú = -2 ê ú ë -2û ë -2û

éa b ù é 1 ù é1 ù Þê ú ê ú = -2 ê ú ë c d û ë -2û ë -2û éa - 2bù é -2ù Þê ú=ê ú ëc - 2d û ë 4 û



Þ a - 2b = -2......(3), c - 2d = 4.....(4)



Solving (1) and (3) we get b = 1, a = 0



Solving (2) and (4) we get d = -3 , c = -2.

é0 1ù Thus, A = ê ú . Now, ë -2 -3û

EMEP.CH01_3PP.indd 78



é -2 + 2 -1 + 2ù é 0 1 ù =ê ú=ê ú=A ë 2 - 4 1 - 4 û ë -2 -3û

57. (a) A is symmetric ⇒ eigenvalues are all real. 58. (b) Let l be the other eigenvalue. Then, the sum of eigenvalues = trace of the matrix ⇒ l - 2 + 6 = 5 + 1 + 1 ⇒ l = 3. 59. (a) Clearly the set X forms a subspace of R3. a[1, -1,0]T + b[1,0, -1]T = O



Þ [ a, -a,0]T + [b,0, -b]T = O



Þ [ a + b, -a + 0,0 - b]T = [0,0,0]T Þ a + b = 0, - a = 0, - b = 0 Þa =b=0

Hence, the vectors [1, -1,0]T and [1,0, -1]T are linearly independent. Now x T = [ x1 , x2 , x3 ]T = a[1, -1,0]T + b[1,0, -1]T Þ [ x1 , x2 , x3 ]T = [ a, -a,0]T + [b,0, -b]T Þ [ x1 , x2 , x3 ]T = [ a + b, -a + 0,0 - b]T Þ x1 = a + b, x2 = -a = 0, x3 = -b



\x1 + x2 + x3 = a + b - a - b = 0, a = x2 ,b = -x3 .

so  xT = [x1, x2, x3]T

= (-x2)[1, -1, 0]T + (-x3) [1, 0, -1]T

Thus, the vector “x” can be expressed as a linear combination of the vectors [1, -1,0]T and [1,0, -1]T

{

}

Consequently, [1, -1,0]T ,[1,0, -1]T is the basis for the subspace X. 60. (i) - (a) ; (ii) - (a) A - lI = 0



Þ

-3 - l 2 =0 -1 0-l

Þ l( l + 3) + 2 = 0



⇒  2l(l+3) + 2 = 0



2 ⇒  Þ l l= -+1,3l -2+ 2 = 0



⇒  l = -1, -2

Þ l + 3l + 2 = 0.........(*)

…(*)

8/9/2023 8:30:31 PM

Linear Algebra • 79

Using the Cayley Hamilton theorem, we get from (*),

é x12 ê êx x =ê 2 1 ê .. êx x ë n 1 éx1 ê x ê 1 ê .. ê êëx1

A 2 + 3 A + 2 I = O........(**) Þ A -1 ( A 2 + 3 A + 2 I ) = A -1O Þ A + 3 A -1 A + 2 A -1 = O (since, A -1 A 2 = ( A -1 A ) A = IA = A )



Þ A + 3 I + 2 A -1 = O



Again (**) Þ A 2 = -3 A - 2 I

\  A 4 = A 2 . A 2 = (3 A + 2 I )2 = 9 A 2 + 12 A + 4 I



= 9( -3 A - 2 I ) + 12 A + 4 I = -15 A - 14 I A 8 = A 4 . A 4 = (15 A + 14 I )2



= 225 A 2 + 420 A + 196 I = 225( -3 A - 2 I ) + 420 A + 196 I = -255 A - 254 I Hence, A 9 = A 8 ´ A = ( -255 A - 254 I ) A = -255 A 2 - 254 A

= -255( -3 A - 2 I ) - 254 A = 511 A + 510 I .



x2 x2 .. x2

.. .. .. ..

®

which has one non-zero row



Therefore, rank (A) = 1. 63. (a)

æ1 è5

Let, A = ç

æ 7 è -2

adj (A) = ç

 \  A–1 =

1+b b 1 = -1 1 0 -b b 0

64. (d) Here,

)

2ö ÷ . Then, det (A) = 7 - 10 = - 3 7ø T

- 5ö - 2ö æ7 ÷ =ç ÷ 1ø 1ø è -5

2ö 1 1 æ 7 - 2 ö 1 æ -7 adj( A ) = ç ÷= ç ÷ | A| ( -3) è -5 1 ø 3 è 5 -1ø

= -b + b = 0



62. (b)

EMEP.CH01_3PP.indd 79

xnx2



[by R2 ® R2 - R1 ,R3 ® R3 - R1 ]



..

.. .. .. .. x1x n ù ú .. .. .. .. x 2x n ú ú .. .. .. .. .. ú .. .. .. .. x 2n úû .. .. .. x n ù ú .. .. .. x n ú .. .. .. .. ú ú .. .. .. x n úû

ö 1 1 R2 , ...., Rn ® Rn ÷ x2 xn ø .. .. .. x n ù ú .. .. .. 0 ú .. .. .. .. ú ú .. .. .. 0 û (by R2 ® R2 - R1 , R3 ® R3 - R1 ,...., Rn ® Rn - R1

b 1 1+b b b +1 1 1 2b 1

é x1 ù ê ú ê x2 ú ê . ú ê ú V = XX ¢ = ê . ú ´ éëx1 ê . ú ê ú ê . ú êx ú ë nû

x 22

æ 1 R1 , R2 ç by R1 ® x 1 è éx1 x 2 .. ê 0 0 .. ê ê .. .. .. ê ë 0 0 ..

61. (a)



x1x 2

0 4 3 D = 2 0 -1 = 0 - 4(0 + 3) + 3(4 - 0) = 0, 3 2 0 8 4 3 D1 = 2 0 -1 = 8(0 + 2) - 4(0 + 5) + 3(4 - 0) 5 2 0

x 2 . . . . x n ùû



¹0

Since D = 0 and D1 ≠ 0, so the system is inconsistent (by Cramer’s rule) and hence the solution in non-existent.

8/9/2023 8:30:33 PM

80 • Engineering Mathematics Exam Prep 65. (a) By Cramer’s rule, the system has an infinite number of solutions. Û D = 0, D3 = 0

1 1 1 1 1 5 Û 1 3 3 = 0, 1 3 9 = 0 1 2 a 1 2 b Û 2a - 4 = 0, 2b - 14 = 0



Þ a = 2, b = 7

Alternative Method: [A : B]

é1 1 1 5 ù ê ú = ê1 3 3 9 ú êë1 2 a b úû 1 5 ù é1 1 ê ú 2 4 ú  ê0 2 êë0 1 a - 1 b - 5úû

(by R2 ® R2 - R1 , R3 ® R3 - R1 ) 1 5 ù é1 1 ê ú  ê0 2 2 4 ú êë0 0 a - 2 b - 7 úû



(by R3 ® R3 -

1 R2 ) 2

Hence, if a-2 = 0 and b - 7 = 0 i.e; if a = 2 and b = 7, then rank([A:B]) = rank(A) = 2 0)

é 1 æ 1 öù 1 1 = - lim ê - çç- ÷÷÷ú = - lim - , ç 2 øú e® 2- ê e - 2 è e® 2- e - 2 2 ë û

Here

1-e1

1

1 dx ( x - 2) 2 e® 2-

0

2

1 1 dx + ò dx 2 ( x -1) ( x -1) 2 1

2.9.3 Beta Function The beta function is denoted by b (m, n) and is de-

Example:

= lim

dx

Remember:

Case-V: f has a discontinuity at x = b

e

2

é 1 ù é 1ù = - lim ê - (-1)ú - lim ê1- ú ú e2 ® 0+ ê e2 ú e1 ® 0+ ê -e ë 1 û ë û 1 1 = lim + lim - 2, which does not exist. e1 ® 0+ e e2 ® 0+ e 2 1

= log 2 - lim log e,

0

f ( x) dx

é é 1 ù 1 ù ú + lim êú = lim êe1 ® 0+ ê x -1ú e2 ® 0+ ê x -1ú ë û0 ë û1+e2

e® 0+

ò

ò

c+e2

1-e1

= lim [log 2 - log e ]

2

1

e1 ® 0+

1 1 2 dx = lim ò dx = lim [log x ]e e ® 0 + e ® 0 + x x e

Here

e2 ® 0+

0

= lim

e

Example: 2

f ( x) dx + lim

a

Example:

= lim éêë tan -1 X - tan -1 (- X )ùúû X ®¥ -1

b

Properties:

(i) b (m, n) = b (n, m) 1

(ii) b (m, n) =

ò 0

x m-1 + x n-1 dx (1 + x) m+n

2.9.4 Gamma Function The gamma function is denoted by G(n) and is defined by G(n) =

¥

òe

-x

x n-1 dx (n > 0)

0

Sometimes G(n) is denoted also by n .

8/9/2023 8:38:26 PM

128 • Engineering Mathematics Exam Prep Properties:

G( m ) G( n ) (I)  b (m, n) = G(m+ n) p 2

(II) 

ò 0

æ m + 1ö÷ æ n + 1ö÷ G çç Gç çè 2 ÷÷ø ççè 2 ÷÷ø m n sin q cos q d q = æ m + n + 2 ÷ö 2 G çç ÷÷ çè ø 2

(III)  G(n) G(1- n) =



such that lim j ( x) = b = lim y ( x) , but



lim f ( x, j( x) ) ¹ lim f ( x, y ( x) ) , then we say

that

x® a

x ®a

x® a

x ®a

lim

( x , y ) ® ( a ,b )

2.  lim

f ( x, y ) does not exist. f ( x, y ) is called double or simultane-

( x , y ) ® ( a ,b )

ous limit. 3.  lim lim f ( x, y ) and lim lim f ( x, y ) are called

p sin np

x ® a y ®b

ìï nG(n) , if n > 0 ïïîn ! , if n is a positive integer

(IV)  G(n + 1) = ï í

æ ö çè 2 ÷ø

1 (V)  G çç ÷÷ = p

2.10 FUNCTIONS OF SEVERAL VARIABLES AND PARTIAL DERIVATIVES

y ®b x ® a

repeated or iterated limits. These two limits may or may not exist. If the double limits exists, then the repeated limits will exist but the converse is not true in general. ( x + 2)( y - x) Consider, f ( x, y ) = ( y + 2)( y + x) Then,

lim

( x , y )® (0,0)

= lim

f ( x, y ) =

( x + 2)( y - x) x)

lim

( x , y )® (0,0) ( y + 2)( y +

( x + 2)(mx - x) x)

x ®0 ( mx + 2)( mx +

2.10.1  Functions of Two Variables Let x, y be two independent variables and z be a variable which takes a value corresponding to a pair of values (x, y). Then we say that z depends on x & y and z is a function of x and y. We write z = f (x, y).

(Taking the curve y = mx, so that when y → 0, x → 0)

For example, z = x2 + y2 + xy,





2

y

z = f (x,y) = x + e sinx etc.

Remember: z = f (x, y) represents a surface in three dimensional space. 2

2

For example, z = 1- x - y represents a sphere with center (0, 0, 0) and radius “1” unit in three dimensional space. 2.10.2  Limit of Functions of Two Variables Let z = f (x, y) be a function of two variables. If “l” be a real number and for a given ε > 0, there exist δ > 0 (δ depends on ε) such that | f (x, y) – l | < ε whenever 0 < | x – a | < δ and 0< | y – b | < δ,

then we say that

lim

( x , y )®( a ,b )

f ( x, y ) = l .

Remember: 1. If there exist two functions y = ϕ(x) and y = ψ(x)

EMEP.CH02_3PP.indd 128



= lim

( x + 2)(m -1)

x®0 ( mx + 2)( m +1)

=

(0 + 2)(m -1) m -1 = , (0 + 2)(m +1) m +1

which depends on “m” and so is not unique.



Hence, the double limit does not exist. ( x + 2)( y - x) Now, limlim f ( x, y ) = lim lim x®0 y®0 x®0 y®0 ( y + 2)( y + x )

{

}

ì-( x + 2)ü = limí x®0î 2 ýþ (0 + 2) ==-1 2 ( x + 2)( y - x) Again, limlim f ( x, y ) = lim lim y®0 x®0 y®0 x®0 ( y + 2)( y + x ) = lim

( x + 2)(0 - x)

x®0 (0 + 2)(0 + x )

{

= lim

(0 + 2)( y - 0)

}

= lim 2 y®0 y + 2 = 2 =1 0+ 2

y®0 ( y + 2)( y + 0)



Thus, limlim f ( x, y ) ¹ limlim f ( x, y ) . x®0 y®0

y®0 x®0

2.10.3 Continuity of Functions of Two Variables A function z = f (x, y) is called continuous at (a, b) if

8/9/2023 8:38:29 PM

Calculus • 129

(i) lim

f ( x, y ) exists and

∴  f (x, y) is a homogeneous function of ­ egree “2.” d

(ii) lim

f ( x, y ) = f (a,b) , provided f (a, b) is

2.10.6  Euler’s Theorem If u = f (x, y) be a homogeneous function of degree “n,” then x ¶u + y ¶u = nu , i.e ¶x ¶y

( x , y )®( a ,b ) ( x , y )®( a ,b )

defined. Alternatively, f (x, y) is said to be continuous at (a, b), if for a a given ε > 0, there exist δ > 0 (δ depends on ε) such that | f (x, y) – f (a, b) | < ε whenever 0 f (a, b) for all positive or negative small values of h and k . 3. A saddle point is a point where the function f ( x, y ) is neither maximum nor minimum. 2.11.2 Working Rule to Find the Maximum and Minimum Values of f ( x, y ) : Step-I:

¶f ¶f ¶ 2 f ¶ 2 f ¶2 f Find , , , and . ¶x ¶y ¶x 2 ¶x ¶y ¶y 2 ¶f ¶f Step-II: Consider = 0 and = 0 and then ¶x ¶y solve these equations for x and y . Let (a, b) be one of the values of ( x, y ) . Step-III: Calculate the followings:

A=

¶2 f



¶x 2

, B= (a,b)

¶2 f ¶x ¶y ( a , b )

&C=

¶2 f ¶y 2

. (a,b)

Step-IV: (i) If AC - B 2 > 0 and A < 0 , then f ( x, y ) has a maximum value at (a, b) . (ii) If AC - B 2 > 0 and A > 0 , then f ( x, y ) has a minimum value at (a, b) .

(iii) If AC - B 2 < 0 , then f ( x, y ) has neither a maximum value nor a minimum value at

Let f ( x, y ) = x 22 + y 22 - 2 x - 2 y -1 Let f ( x, y ) = x + y - 2 x - 2 y -1 ¶f ¶f Then ¶f = 2 x - 2, ¶f = 2 y - 2, Then ¶x = 2 x - 2, ¶y = 2 y - 2, ¶x ¶y ¶ 22 f ¶ 22 f ¶2 f ¶ f2 = 2, ¶ f = 0 and ¶ 2 f2 = 2. ¶x 2 = 2, ¶x ¶y = 0 and ¶y 2 = 2. ¶x ¶x ¶y ¶y ¶f ¶f \ ¶f = 0, ¶f = 0 \ ¶x = 0, ¶y = 0 ¶x ¶y Þ 2 x - 2 = 0, 2 y - 2 = 0 Þ 2 x - 2 = 0, 2 y - 2 = 0 Þ x = 1, y = 1. Þ x = 1, y = 1. ¶ 22 f ¶ 22 f = 2, B = ¶ f So, A = ¶ f2 =0 = 2, B = ¶x ¶y So, A = ¶x 2 =0 (1,1) ¶x (1,1) ¶ ¶ x y (1,1) (1,1) ¶ 22 f = 2. and C = ¶ f2 = 2. and C = ¶y 2 ¶y (1,1) (1,1) \ AC- B22 = 4 > 0 and A = 2 > 0 \ AC- B = 4 > 0 and A = 2 > 0 Hence, f ( x, y ) has a minimum value at (1,1) and minimum value of f ( x, y ) is

f (1,1), i.e, 1 + 1- 2 - 2 -1 = -3. 2.11.3 Lagrange’s Method for Undetermined Multipliers Let u = f ( x, y, z ) be a function of three variables x, y, z which are connected by the relation

f( x, y, z ) = 0 . Suppose we need to find the ex-

treme values (maximum or minimum values) of the function u = f ( x, y, z ) . Then we need to follow the steps given below: Step-I: Take F ( x, y, z ) = f ( x, y, z ) + lf( x, y, z ) Step-II: Consider

¶F ¶F ¶F = 0, = 0 and = 0 and ¶x ¶y ¶z

then solve these equations for x, y, and z . Let

(a, b) . In this case, the point (a,b) is called a

(a, b, c) be one of the values of ( x, y, z ) . Then, (iv) If AC - B 2 = 0 , then no conclusion can be u = f ( x, y, z ) will have a stationary value (maxiwithdrawn and further investigation is needed. mum/minimum value) at (a, b, c) . saddle point.

EMEP.CH02_3PP.indd 131

8/9/2023 8:38:44 PM

132 • Engineering Mathematics Exam Prep Application: value of x y 2 z 3 . Here, f( x, y, z ) = x + y + z -1,

f ( x, y, z ) = xy 2 z 3 .

\ F ( x, y, z ) = f ( x, y, z ) + lf( x, y, z ) = xy 2 z 3 + l ( x + y + z -1) Then,

¶F ¶F ¶F = 0, = 0 and =0 ¶x ¶y ¶z

Þ y 2 z 3 + l = 0, 2 xyz 3 + l = 0 and 3 xy 2 z 2 + l = 0 2 3

2 2

( x + y + z = 1) \ F ( x, y, z ) = f ( x, y, z ) + lf( x, y, z ) = xy 2 z 3 + l ( x + y + z -1) ¶F ¶F ¶F Then, = 0, = 0 and =0 ¶x ¶y ¶z Þ y 2 z 3 + l = 0, 2 xyz 3 + l = 0 and 3 xy 2 z 2 + l = 0 Þ -l = y 2 z 3 = 2 xyz 3 = 3xy 2 z 2 1 1 1 Þ 2 3= = 3 2 xyz 3xy 2 z 2 y z x y z Þ 2 3= = 2 3 xy z 2 xy z 3xy 2 z 3 x+ y+z = 2 3 xy z + 2 x y 2 z 3 + 3xy 2 z 3 x y z 1 Þ 2 3= = = 2 3 2 3 xy z 2 xy z 3xy z 6 xy 2 z 3



EMEP.CH02_3PP.indd 132



2 3 Therefore, f ( x, y, z ) = xy z has a maximum

æ ö çè 6 3 2 ø÷

1 1 1 value at çç , , ÷÷ . Hence, the maximum value of 2

3

Þ -l = y z = 2 xyz = 3 xy z 1 1 1 Þ 2 3= = 3 2 xyz 3xy 2 z 2 y z x y z Þ 2 3= = 2 3 xy z 2 xy z 3xy 2 z 3 x+ y+z = 2 3 xy z + 2 x y 2 z 3 + 3xy 2 z 3 x y z 1 Þ 2 3= = = 2 3 2 3 xy z 2 xy z 3xy z 6 xy 2 z 3

1 1 x y = = , , 2 3 2 3 2 3 6 xy z 2 xy z 6 xy 2 z 3 xy z z 1 = 2 3 3 xy z 6 xy 2 z 3 1 1 1 Þ x= ,y= ,z= . 6 3 2 Þ

Given that x + y + z = 1 . Let us find the maximum

( x + y + z = 1)

æ öæ ö x y 2 z 3 is çç 1 ÷÷çç 1 ÷÷ èç 6 ÷øèç 3 ÷ø

3

æ ö çç 1 ÷÷ . çè 2 ÷ø

2.12 CHANGE OF ORDER OF INTEGRATION Let us consider a double integral

ò ò f (x, y)dxdy, R

where R is the region. Now assume that the region R lies between the lines x = x1 and x = x2; y = y1 and y = y2. Consider any value of x in x1  x  x2 and keep the value constant. y varies between ϕ1(x) and ϕ2(x) where ϕ1(x) and ϕ2(x) are the ordinate of the points at which the boundary of R is intersected by the line through (x,y) and parallel to y-axis. Now first integrate f (x,y) w.r.t y and then integrate the result w.r.t. “x.” Thus,

x2 ìj2 ( x )

üï ï f ( x , y ) dxdy = f ( x , y ) dy í ý dx . òò ò ò ïj1 ( x ) R x1 î þï

Again consider any y in the interval y1 ≤ y ≤ y2 and keep the value constant. x varies between ψ1(y) and ψ2(y) where ψ1(y) and ψ2(y) are the abscissa of the points at which the boundary of R is intersected by the line through (x, y) and parallel to the x-axis. Now first integrate f (x,y) w.r.t x and then integrate the result w.r.t. “y.” Thus,

ò ò f ( x , y)dxdy = R

ìïy2 ( y ) üï òy íï y ò( y ) f ( x , y)dx ýï dy. þ 1 î 1

y2

Hence, we can conclude that in a region, the order of integration when evaluating a double integral can always be changed.

8/9/2023 8:38:46 PM

Calculus • 133

y

Remember: The easiest way to find the limits of integration is to sketch the region of integration and determine the limits from the sketch. Example:

1 x

ò ò (x

Let us consider

2

y =y ( x) x= b

x= a

)

+ y 2 dydx

0 x2

y =j ( x)

Hence, the region of integration is described by 0 ≤ x ≤ 1, x2  y  x .

x Example: 3 ì x dy üï ï 1 dxdy = ò0 ( x 2 + y 2 ) òx =1 íï yò=0 x 2 + y 2 ýï dx î þ

3 x

1



(1, 1)

(0, 1)

ò

3

é1 -1 òx =1 êë x tan

ò { x tan

x

=

y

=



x

(1, 0)

(0, 0)

1ìx 1ì y üï ï ï ïü Hence, ò í ò x2 + y 2 dyý dx = ò í ò x2 + y 2 dxý dy . ïþ 0ï 0ï îx2 îy þï

(

)

-1

}

x - 1 tan -1 0 dx x x x

2.13.1  Double Integrals Let the region R be divided into rectangular partitions and dx be the length of a subrectangular partition and dy be its width so that dxdy represents an





=



EMEP.CH02_3PP.indd 133

R

)

pö æ -1 ç tan 1 = 4 ÷ è ø

òò

f ( x, y )dxdy =

R

ìï y ( x ) üï f ( x , y ) dy í ý dx ò ò ïþ x=a ï y =j ( x ) î

ìï y ( y ) üï òy =c íï x =jò( y ) f ( x, y)dx ýï dy î þ d

y y= d

R

1. Let the region R be bounded by the continuous curves y = ϕ(x), y = ψ(x) and the ordinates x = a and x = b. Then we write

òò

1= p 4

p log e 3 ( log e 1 = 0) 4

x =j( y)

x =y( y) y= c

Evaluation of double integral:

b

-1

3 = p [ log e x ]1 = p ( log e 3 - log e 1) 4 4 p = log e 3 ( log e 1 = 0 ) 4

òò f ( x, y)dxdy

is called the double integral of the single valued function f (x,y).

f ( x, y )dxdy =

(tan

3

2. Let the region R be bounded by the continuous curves x = ϕ(y), x = ψ(y) and the abscissas y = c and y = d. Then we write

2.13 DOUBLE AND TRIPLE INTEGRALS

elementary area. Then, the integral

1

= ò 1 tan -1 1dx = p ò dx 41 x x 1

y x y

)

x =1 3

From the figure above, it is clear that the region of integration can be described as 0 ≤ x ≤ 1 and

(

3

=

x= x

x

yù dx x úû y = 0

x 2.13.2  Triple Integrals If the region of integration R is given by

R = {( x, y, x) : axb, cyd , ezf } ,



then we define

8/9/2023 8:38:49 PM

134 • Engineering Mathematics Exam Prep

òòò

b

R

f

d

ò ò ò

f ( x, y, z )dxdydz =

x =a y =c z =e

Remember: If the region of integration R is given by

y1 ( x, y )zy 2 ( x, y )}



Then we define



é j2 ( x ) æ ìï y2 ( x , y ) üï ö ù = ò ê ò ç í ò f ( x, y, z )dz ý ÷ dy ú dx ç ï z =y ( x , y ) ÷ x = a ê y =j1 ( x ) è î 1 þï ø ûú ë

òòò



Let us evaluate

1 1- x

ò ò

0z 1 - x 2 - y 2 2 2 1 1- x 2 1- x - y

ò ò 0

ì ï = ò í x =0 ï î 1 æ = òç ç x =0 ç è 1 æ = òç ç x =0 è 1 æ = òç ç x =0 è

=

ò

0

1

EMEP.CH02_3PP.indd 134

1- x - y

0

ò

2

xyzdxdydz

0

{

1



2

R = ( x, y, z ) : 0x1, 0y 1 - x 2 ,





2

Here region of integration,

\

0

}

xyzdxdydz

0

1- x 2

ò

y =0

æ ç ç è

1- x 2 - y 2

ò

z =0

ö ü ï xyzdz ÷ dy ý dx ÷ ï ø þ

ö ÷ dx dy ò ÷ ÷ y =0 ø 2 1- x é1 - x 2 - y 2 ù ö÷ xy ò ëê 2 ûú dy ÷ dx y =0 ø 1- x 2

1- x

ò

2

y =0

1- x 2 - y 2

é 2ù xy ê z ú ë 2 û z =0

(

)

2 ü ö ïì x 1 - x x y 3 ï dy ÷ dx y í 2 2 ýï ÷ ïî þ ø

2 2 2 ö æ çç x(1- x ) é y 2 ù 1-x - x é y 4 ù 1-x ÷÷dx ò çè 4 ëê ûú 0 8 êë ûú 0 ÷ø÷ x=0

)

2

)

(

)

´ x dx 8



(

)

1

(

)

=1 1-1+1 = 1 . 8 2 2 6 48

2.14  ARC LENGTH OF A CURVE (i) The arc length of a curve y = f (x) between the points where x = a and x = b is given by 2

b

0



(

(

4 6 é 2 ù = 1 ò x - 2 x 3 + x 5 dx = 1 ê x - x + x ú 80 8ë 2 2 6 û0

R

Example:





f ( x, y, z )dxdydz

where f (x,y,z) is a continuous function in R.



1

)

= ò 1 - x2 1

b

)

(

where f (x,y,z) is a continuous function in R.

R = {( x, y, z ) : axb, j1 ( x)yj2 ( x),

(

æ x 1 - x2 1- x 2 1- x 2 ö ÷ dx éë y 2 ùû = òç - x éë y 4 ùû 0 0 4 8 ç ÷ x =0 è ø 1 ì x 1 - x2 ü 2ï ï 1 - x 2 - x 1 - x 2 ý dx = òí 4 8 0î ï þï 1

f ( x, y , z )dxdydz

æ dy ö ò 1+ çè dx÷ø dx a

(ii) The arc length of a curve x = f (y) between the points where y = c and y = d is given by 2

d

æ dx ö ò 1+ çè dy ÷ø dy c

(iii) The arc length of the parametric curve x = f (t), y = g(t) between the points where t = α and t = β is given by b

( )

2

2 dx + æ dy ö dt ò dt çè dt ÷ø a

(iv) The arc length of the curve θ = f (r) between the points where r = a and r = b is given by b

ò a

( ) dr .

1+ r 2 d q dr

2

2.15 VOLUMES OF SOLIDS OF REVOLUTION 2.15.1 Working Formulas (i) The volume V of a solid formed by the revolution of a plane area bounded by the continuous curve y = f (x), the ordinates x = a, x = b and the x-axis about the x-axis is given by b

V = p ò y 2 dx . a

8/9/2023 8:38:52 PM

Calculus • 135

(ii) The volume V of a solid formed by the revolution of a plane area bounded by the curve x = f (y), the abscissas y = c, y = d and y-axis about y-axis is given by d

c

(iii) The volume V of a solid formed by the revolution of a plane area bounded by the parametric curve x = f (t), y = g(t), x-axis and the ordinates where t = a, t = b, about x-axis is given by

( )

b

2 V =p ò y 2 dx dt =p ò{g (t )} f ¢(t )dt dt a a

(iv) The volume V of a solid formed by the revolution of a plane area bounded by the parametric curve x = f (t), y = g(t), y-axis and the abscissas where t = c, t = d, about y-axis is given by d

d

2 æ dy ö

(v) The volume V of a solid formed by the revolution of a plane area bounded by the polar curve r = f (θ), the radii vectors θ = θ1 and θ = θ2, about the initial line i.e; θ = 0 (x axis) is given by

S = 2p ò r sin q´ r 2 + dr dq q1

2

dq

Fully Solved MCQs (Level-I)

ì x-2

ï 1. If f ( x) = í x - 2 , x ¹ 2 , then ïî 0, x = 2

(a) lim f ( x) exist x® 2



(b) lim f ( x) does not exist x® 2

(c) lim f ( x) = –1 x®2+

(d) lim f ( x) =1 x®2-

ìx - x ï , x ¹ 0 , then x ïî 1, x = 0

(a) lim f ( x) does not exist

2.16 SURFACE AREAS OF SOLIDS OF REVOLUTION (i) The surface area S of a solid generated by the revolution of the curve y = f (x) about x-axis between the ordinates x = a and x = b

x ®0 +

(b) lim f ( x) does not exist x®0

(c) lim f ( x) does not exist x ®0 -

(d) lim f ( x) = f(1) x®0

2

is given by S = 2p y 1 + æç dy ö÷ dx ò è dx ø

3. lim

a

(ii) The surface area S of a solid generated by the revolution of the curve x = f (y) about y-axis between the abscissas y = a and y = b b

( )

q2

3

b

( )

2. Given that f ( x) = í

V = 2 p ò { f (q)} sin qd q 3 q1 .

x ®0

sin x 0 =? x

(a) π  (b) 0  (c) π/90  (d) π/180 4. lim

2

is given by S = 2p ò x 1 + æç dx ö÷ dy è dy ø a (iii)  The surface area S of a solid generated by the revolution of the parametric curve

EMEP.CH02_3PP.indd 135

t1

2

2 dx + æ dy ö dt çè dt ÷ø dt

(iv) The surface area S of a solid generated by the revolution of the polar curve r = f (θ) about x-axis between the limits θ = θ1 and θ = θ2 is given by

2

V =p ò x ç ÷ dt =p ò{ f (t )} g ¢(t )dt. dt c c è ø

q2

t2

S = 2p ò y



V = p ò x2 dy .

b

x = f(t), y = g(t) about x-axis between the limits t = t1 and t = t2 is given by

x ®0



3x - 1 1+ x -1

=?

(a) 9 (b) 4 (d) none of these

(c) 1

8/9/2023 8:38:56 PM

136 • Engineering Mathematics Exam Prep 5. lim (1 + x )cos ecx = ? x ®0

(c) e

(d) e2

sin 4 x - sin 2 x + 1 6. lim =? x ®¥ cos 4 x - cos 2 x + 1 (a) 1/2 (b) –1 (c) 0

(d) 1



(a) 0

(b) 1

x®a

(a)

2 (b) a

1 (c) 1 (d) 1 a 2 2a

tan x - sin x =? x ®0 x3 (a) 0 (b) 1/2

(c) 1

(d) –1

tan 2 x - x =? x ®0 3 x - sin x (a) 1/2 (b) 1/3

(c) 2/3

(d) 0

9. lim

14. lim

8. lim

(a)

(a) f is bounded (b) f is differentiable (c) f may be unbounded (d) none of these

=?

(b) 1 (c) 2p (d) p

(d) 1

ìï 1 + cos 2 x ïüï 17. lim ï í ý=? p (p - 2 x) 2 ïþï x® ï ï î 2 (a)

1 1 (b) - 2 2

(c) 0

(d) 1

ìï 43 x - 34 x üïï 18. lim ï í ý=? x®0 ï ïþï x îï æ 64 ö æ ö (a) log çç 4 ÷÷  (b) log çç ÷÷   (c) 0  (d) 1 çè 81 ÷ø çè 31÷ø

3

ìï 1 + cos3 x üïï 19. lim ï í ý=? 2 x®p ï ïþï sin x îï

x 4

Then, f is differentiable (a) everywhere except at x = 4 (b) nowhere (c) everywhere (d) only at x = 4 12. If f ( x) = k sin x + 2 x and f ¢(0) exist, then

the value of “k” is? (a) 4 (b) 3 (c) 1 (d) 0 2 ìï x - ax + b, x < 1 ïï 13. Let f ( x) = ï í a + b, x = 1 . ïï 2 ïïî4 x - bx + a, x > 1

EMEP.CH02_3PP.indd 136

(a) 0

x2

(d) 0, 2

ìï 2 - 3 cos x - sin x üï ïý = ? 16. lim ï í 2 p ïï (6 ) p x x® ï ï þ 6 î 1 1 (a) 0 (b) 1 (c) (d) 18 36

f(x)



sin (p cos 2 x)

(c) 1, 1

log(4 + x) - log(4 - x) =? x®0 x 1 1 (a) (b) - (c) 0 2 2

11. The graph of a function f is given below:

0

(b) 0, 3

15. lim

10. If f : R ® R be a continuous function, then

1 , 2 2

x® 0

3x - a - x + a =? x-a

7. lim

If f is continuous at x = 1, then the values of “a” and “b” are, respectively

(a)



3 4 (b) 2 9

(c) 0

(d) 1

20. Which of the following is true? æbö b-a b-a (a) < log çç ÷÷÷ < ç èaø b a

(b)

æbö b-a b-a < log çç ÷÷÷ < çè a ø a b

(c)

æbö 1 1 < log çç ÷÷÷ < çè a ø b a

(d)

æbö 1 1 < log çç ÷÷÷ < çè a ø a b

8/9/2023 8:39:02 PM

Calculus • 137 2 21. The points on the curve y = ( x + 1) at which the tangent is parallel to x-axis are given by (a) 2, 4 (b) –1 (c) 1,1 (d) 1,2

22. The point on the curve y = ( x -1)(x- 2) at which the tangent is parallel to the chord joining the points (1, 0) and (2, 0) is given by (a) x = 1.7 (b) x = 1.5 (c) x = 1.3 (d) x = 1.2 23. Let f , g be two differentiable functions on R and g ¢( x) ¹ 0 for x Î R . Also let f (1) = 6, g (1) = 0, f (2) = 8 and g (2) = 4.Then ,

æ 1 (a) çç-¥, çè e

÷÷ö (b) (-¥, ¥) ÷ø

æ1 ö (c) çç , ¥÷÷ (d) (0,¥) ÷ø çè e intervals on which the function x are strictly increasing and f ( x) = log x strictly decreasing are given by, respectively

29. The

(a) (e, ¥); (0, e)

(b) (0, e); (e, ¥)



(a) $ c Î (2, 4) such that g ¢(c) = 2 f ¢(c)

(c) (e, ¥); (0, e) -{1}



(b) $ c Î (2, 4) such that g ¢(c) = 4 f ¢(c)

(d) (e, ¥); (0, e) -{-1}



(c) $ c Î (1, 2) such that g ¢(c) = 2 f ¢(c)

30. The value/values of l for which the function



(d) $ c Î (2, 4) such that g ¢(c) = 3 f ¢(c)

f ( x) = l x 3 - 9l x 2 + 9 x + 5 is strictly increasing on R is/are given by

f ¢ exist " x Î [1, 4].

24. Let

"xÎR

&

f ¢( x ) ³ 3

If f (1) = -3 , then

(a) f (4) ³ 3



(b) f (4) ³ 4

(c) f (4) ³ 5

(d) f (4) ³ 6 x-2 25. The function f ( x) = (where x ¹ -1) is x +1 (a) strictly increasing (b) strictly decreasing (c) increasing (d) decreasing

26. The function f ( x) = log

(

)

x 2 + 1 - x is

(a) strictly increasing for all x (b) strictly decreasing for all x (c) increasing for all x (d) decreasing for all x



27. Which of the following functions are æ pö ­decreasing on çç0, ÷÷ ? èç 2 ÷ø

9 (a) cos x (b) 3 x 3 - 5 (c) tan x (d) x

28. The interval on which the function f ( x) = x x ( x > 0) is strictly increasing is given by

EMEP.CH02_3PP.indd 137

(a) l Î (0, ¥) (b) l Î (0,1)

æ 1ö æ 1ö (c) l Î çç1, ÷÷ (d) l Î çç0, ÷÷ çè 3 ÷ø çè 3 ø÷ 31. The intervals on which the function f ( x) = 6 - 9 x - x 2 are strictly increasing and strictly decreasing are given by, respectively

æ ö 9ö æ 9 (a) çç- ¥, - ÷÷ ; çç- , ¥÷÷ ÷ ÷ø èç 2 ø èç 2 æ ö 9ö æ9 (b) çç- ¥, ÷÷÷ ; çç , ¥÷÷÷ çè ç ø 2ø è 2 (c) (- ¥, 0); (0, ¥) (d) (1, ¥); (0, 1) 32. Which of the following functions has a local minima or local maxima? (a) e x (b) log x (c) cos x in (0, p ) (d) none of these

log x

33. The function f ( x) = has a local maxix mum value at (a) x = e (b) x = 1 (c) x = 0 (d) none of these

8/9/2023 8:39:13 PM

138 • Engineering Mathematics Exam Prep x- p) (x- p) ( 2 2 (a) 1+ -.........¥ 2

34. The function f ( x) = x + 1- x , x £ 1 has a local maximum value (a) 0 (b) 1 (c)

5 1 (d) 4 4

2!

2 1 2 1 , (b) - , 3 3 3 6

( ) ( ) ( ) (c) 1+ 1 ( x - p ) + 1 ( x - p ) + ......¥ 2! 2 4! 2 3



1 1 1 = + R R1 R2

If R1 + R2 = k (constant), then R is maximum when



Wx Wx3 M= - 2 3 3L The M is maximum at x = ? (d) 2 L

(d) 2

40. The expansion of sinx in powers of x - p is 2 given by

EMEP.CH02_3PP.indd 138

n=0

be the Taylor series expansion of

(a)

¥

å an x k +n n=0 ¥

(c) å an x

2n

(b)



¥

å an x 2k n=0



(d) none of these

43. The Taylors series expansion of  x 3 + 5 x 2 + 2 x - 5 in a powers of x - 1 is given by

2 3 39. The value of log3 + 1 (log3) + 1 (log3) + ...¥ is 2! 3!

(c) 1

å an x n

n=0

38. The series 1+ 3 + 9 + 27 + .......¥ is 1! 2! 3! (a) convergent (b) divergent (c) p-series (d) neither convergent nor divergent

(b) 0

) ) ) )

the function f (x) at x = 0, then at x = 0, what will be the Taylor series expansion of xkf(x) ?

(d) none of these

at the two ends. The bending moments M at a distance x from one end is given by

(a) –1

( ( ( (

3 5 æ ö (a) log 1- x = 2ç x + x - x + ......¥ ÷ 1+ x 3 5 è ø 3 5 æ ö (b) log 1+ x = 2ç x + x + x + .......¥ ÷ 1- x 3 5 è ø 3 5 æ ö (c) log 1+ x = 2ç x - x + x -.......¥ ÷ 1- x 3 5 è ø 3 5 æ ö (d) log 1- x = 2ç x - x + x -.......¥ ÷ 1+ x 3 5 è ø ¥

37. A beam is uniformly loaded and is supported



(d) none of these

42. If

(a) R1 = 3R2 (b) 2R1 = R2

L L L (b) (c) 2 3 2

4

41. Which of the followings is true?

1 1 2 1 , (d) - , 2 3 3 6 36. The combined resistance R of two resistors R1 and R2 is given as

(a)

5

2

(c)

(c) R1 = R2

4!

(b) x - p - 1 x - p + 1 x - p -.......¥ 2 3! 2 5! 2

35. If f ( x) = A log x + Bx 2 + x has extreme values at x = 1 and x = 2, then the values of “A” and “B” are, respectively (a)

4

(a) 2( x - 1)3 - 10( x - 1) 2 + 8( x - 1) + 3

3 ( x - 1) 2 - 5( x - 1) + 2 2 (c) ( x - 1)3 - 5( x - 1) 2 + 8( x - 1) + 10 (b) ( x - 1)3 +

(d) ( x - 1)3 + 8( x - 1) 2 + 15( x - 1) + 3 44. Let f ( x) =

tan x and F ( x) be its ansin x cos x

æpö

ti-derivative. If F çç ÷÷ = 4 , then F ( x) =? çè 4 ø÷ (a) 2 tan x + 2 (b) 2 cot x + 1 (c) 2 tan x + 1 (d) 2 cot x + 2

8/9/2023 8:39:21 PM

Calculus • 139

45.

ò

e tan

-1

x

tan (a) ( x -1) e

-1

x



-1

x

(d)

-1

x

p 2

+C

ò

+C

p 2

dq = log f (q ) , then f (q ) =? sin q cos q

f ( x) dx = f( x), then ò x f ( x ) dx =? 5

3

1 3 3 x f( x ) - ò x 2f( x 3 ) dx 3 1 (b) x 2f ( x 2 ) - ò x 2f ( x 3 ) dx 2 1 (c) x 3f ( x 3 ) - x 2f ( x 2 ) dx ò 3 1 (d) x 4f ( x 3 ) - ò x 3f ( x 4 ) dx 4 1

ò

(a)

p p 1 + log 3 (b) - log 2 2 4 2

(c)

p 1 p 1 - log 2 (d) - log 2 3 3 4 2 1

52. If



¥

0

(a)

1 (b) 1 5 2

x

is

(c) 0

(d) 1

ò x f ( x) dx = ? a

a +b 2 òa b

(c)

1

EMEP.CH02_3PP.indd 139

54.

ò 0

(c) 2

(d) 3

2

e- x x 2 dx = ?

p p p (b) (c) 2 4 4 1 1- x 6

(d) 0

dx = ?

æ1ö æ1ö G çç ÷÷÷ G çç ÷÷÷ ç p çè 6 ø (a) p è 6 ø (b) 6 æç 2 ö÷ 2 æç 1 ÷ö G ç ÷÷ G ç ÷÷ çè 3 ø çè 3 ø

(d) None of these

¥

b

a -b f ( x) dx f ( x) dx (b) 2 òa

a +b 3 f ( x) dx 2 òa

(b) 1

æ1ö G çç ÷÷÷ 1 çè 6 ø (c) 6 æç 1 ÷ö G ç ÷÷ çè 3 ø

b

(a)

ò

(a)

50. If f ( x) = f (a + b - x) , then

b

0

0

cos (t 2 ) dt

x®0

(1- x + x ) dx = k ò tan -1 x dx,

(a) 0 53.

1 1 1 (a)  (b)  (c) 0 (d) 2550 1550 15

ò

1

2

then the value of k is

0

49. The value of lim

ò cot

-1

0

x (1- x) 49 dx is

x

is

0

(a)

48. The value of

sin x dx

ò 1 + sin x + cos x

­value of

(c) 2sin q (d) 2 cos q

ò

then the

0

(a) cot q (b) tan q

47. If

dx

ò 1 + sin x + cos x = log 2 ,

51. If

(d) none of these 46. If

3a + 4b f ( x) dx 2 òa

+C

-1 tan (b) tan x ´ e

(c) xe tan

b

æ1 + x + x 2 ö÷ çç ÷ dx =? çè 1 + x 2 ÷÷ø

55. The integral

ò 1

1 dx x2

(a) exist and converges to “0” (b) exist and converges to “1” (c) doesn’t exist (d) converges to “2”

8/9/2023 8:39:29 PM

140 • Engineering Mathematics Exam Prep 1

56. The value of the integral

ò

-1



(a) 6

(b) 4

1 dx is x 2/3

(c) 0

(a)

- x2

òe



(d) 2

57. Which of the following is NOT correct? ¥

¥ y

dx has a finite value



(b)

òe

dx is convergent

¥

- x2

òe

dx is convergent

0

(d)

òe

x

dx is divergent

58. The function f ( x, y ) = x 3 + y 3 - 6 xy has maximum/minimum value at (a) (0,0) (b) (1, 2) (c) (2, 2) (d) (1, 1) 59. The minimum value of the function 1 1 u ( x, y ) = xy + + is x y (a) 3 (b) 10 (c) 2 (d) 4 60. The minimum value of x 2 + y 2 + z 2 , given that xyz = 8 , is (a) 12 (b) 10 (c) 15 (d) 8 61. By changing the order of integration 4 4 x- x2

ò ò (a)

(b)

(c)





0

2 2 2+ 4+ y

ò ò

f ( x, y )dxdy

ò ò

4+ y2

0

0

ò ò

¥ ¥ -y

ò

0 0

y

0

a

a2 - y 2

ò ò

f ( x, y )dxdy

ò ò

f ( x, y )dxdy

a

a2 - y 2

0

0

ò ò

f ( x, y )dxdy

(d) none of these 4 2 x

ò ò



(a) 16 3

dydx is equal to

x2 4

0

(b) 4 3

(c) 2 3

(d) 0

65. What will be the value of





ò ò(x

2

)

+ y 2 dxdy,

R

where R is the region bounded by x = 0, y = 0 and x + y = 1 (a) 1 (b) 1 (c) 1 (d) 1 2 3 6 12 p 2 p

ò ò sin( x + y)dydx

dxdy; I = 0

is

0

(a) 0

(b) 1

(c) 2

(d) 3

67. Let I = òò 4 x2 - y 2 dxdy, where R is the R

¥ ¥

(a) I = ò

-a

0 - a2 - y 2

0

f ( x, y )dxdy

e

a2 - y 2

66. The value of the double integral f ( x, y )dxdy

4

a

value of

4+ y2

2

f ( x, y )dydx is equal to

64. By changing the order of integration, the

-y 62. Consider I = ò ò e dydx. Then which of the y 0 x following is true?

EMEP.CH02_3PP.indd 140

0

(c)

(d) none of these



a 2 - x2



0

0 - 4- y2



a

(b)

f ( x, y )dydx can be expressed as

0 2- 4- y2



(d) none of these

(a)

-¥

0

0



0

e- y dxdy; I = 1 ò y 2 -¥

ò ò

0

(c)

¥ 0

(c) I = ò

63. By changing the order of integration,

0

¥ -2 x

-y (b) I = ò ò e dxdy; I =1 y 0 0



t­riangle formed by the straight lines y = 0, x = 1 and y = x. Then which of the following is true? (a) I = 1 3 3 + 2p (b) I = 1 3 + p 18 6 1 (c) I = 6 3 + 2p (d) I = 2 2 3 + 4p 3 3

(

(

)

)

( (

)

)

8/9/2023 8:39:34 PM

Calculus • 141

68. The value of (a) 3 p 2



2

4- y2

0

0

ò ò (x

2

(b) 2p

)

+ y dydx is

òò



(d) p 2

(c) 4p 1 1- x

69. Find the value of

2

y

e x+ y dydx

(



R



(d) 16 945

- x2 + y 2 ) 71. The integral = ò ò e ( dxdy is equal to ¥ ¥ 0 0

(a) p 2



(b) p 4

72. The value

òò x R

(c) p

(d) 2p

4 3

y dxdy, of where

R = {( x, y ): x 0, y 0, x + y 1} is (a) 1 (b) 3 (c) 1 (d) 5 504 208 302 12



2

1. (b) 6. (d) 11. (a) 16. (c) 21. (b) 26. (d) 31. (a) 36. (c) 41. (b) 46. (b) 51. (c) 56. (a) 61. (a) 66. (c) 71. (b) 76. (a)

ed by x = 0, y = 0 and x + y = 1 (c) 4 235

2. (b) 7. (b) 12. (d) 17. (a) 22. (b) 27. (a) 32. (d) 37. (a) 42. (a) 47. (a) 52. (c) 57. (d) 62. (b) 67. (a) 72. (a)

3. (d) 8. (b) 13. (a) 18. (b) 23. (c) 28. (c) 33. (a) 38. (a) 43. (d) 48. (d) 53. (b) 58. (c) 63. (c) 68. (b) 73. (d)

73. The length of the arc of the parabola y = 8x measured from the vertex to an extremity of the latus rectum is (a) 12

æ ö (b) log ç 4 + 3 2 ÷ 2 è ø

(c) 10 2

ì æ öü (d) 2 í6 2 + log ç 4 + 3 2 ÷ ý è 2 øþ î



x®a

x ®0 +

æ ö (a) 2 ç e -1÷ è ø

æ ö (b) 2 ç e +1÷ è ø

(c) 2 2

(d) e p +1

p 2

75. The arc length of the Cartesian curve y = log(secx) between the points where x = 0 and x = p is 6

EMEP.CH02_3PP.indd 141

1. (b) We know that lim 2. (b) lim f ( x) = lim

74. The arc length of the curve x = et sint, y = et cost between the points where t = 0 and t = p is given by 2 p 2

2

2

4. (d) 9. (a) 14. (d) 19. (a) 24. (d) 29. (c) 34. (c) 39. (d) 44. (a) 49. (d) 54. (b) 59. (a) 64. (a) 69. (a) 74. (a)

5. (c) 10. (c) 15. (a) 20. (a) 25. (a) 30. (d) 35. (b) 40. (a) 45. (c) 50. (a) 55. (b) 60. (a) 65. (c) 70. (d) 75. (b)

Explanation 2



)

Answer key

 ò ò xy 1 - x - y , where R is the region bound(b) 1 235

(

)

)

(b) 1 e2 +1 2 (d) e + 1 2

70. Find the value of the double integral

(a) 2 745

(

(b) 1 log3 2 (d) log 2 + 3

76. The total length of the astroid x 3 + y 3 = a 3 is given by (a) 6a units (b) 12a units (c) 8a units (d) 4a units

0 0

(a) 1 (e -1) 2 (c) e 2

(a) 1 log2 2 (c) log 2 + 2

x ®0 +

x-a x-a

x- x x

does not exist.

= lim

x ®0 +

x-x = 0, x

x- x x - (-x) = lim 0 x® 0x ® x x 2x = lim =2 x®0- x

lim f ( x) = lim

x® 0-



 lim f ( x) ¹ lim f ( x), so lim f ( x) does not  x ®0 +

x ®0 -

exist. 3. (d) lim x®0

x®0

sin x 0 x

æ p x ö÷ sin çç çè180 ÷ø÷ = lim x®0 x

c ö æ çç 10 = p ÷÷ çè 180 ÷÷ø

8/9/2023 8:39:37 PM

142 • Engineering Mathematics Exam Prep é æ px ö ù ÷÷ ê sin çç ú ê èç180 ø÷ p ú p = lim ê ´ = ´ lim ú px px 180 ú 180 p x ®0 ®0 ê 180 180 ê ú êë 180 úû ö p p æç sin x = ´1 = = 1÷÷÷ ççè lim 0 x ® ø x 180 180

é æ p x öù ÷÷ ú ê sin çç ê èç180 ÷ø ú ê ú ê px ú ê ú êë 180 úû

7. (b) lim x®a

= lim

x®a

4. (d)

= lim

3 x -1 lim x®0 1 + x -1

x®a



é é ù ê ê 3 x -1 ú 3 x -1 ê ê ú ê ê ú x x ê = lim ê ú = lim 1 1 1 x® 0 x® 0 ê ê ú 2 2 2 ê + + x x (1 ) 1 (1 ) 1 ê ú ê ê ú x êë (1 + x) -1 ë û æ 3x -1ö÷ ÷ lim çç x®0 ç log e 3 è x ÷÷ø = = 1 1 1ü ì -1 ï ï (1 + x) 2 -12 ï ï 1 ´12 ï ï lim í ý 2 x® 0 ï (1 + x ) -1 ï ï ï ï ï ï ï î þ



= 2 log e 3 = log e 9.

ù ú ú ú ú ú ú ú úû

= lim x®a

8. (b) 

x®0

æç x ö÷ ÷

ç ÷ 1 üxlim 1 x ü ì ì ´ ï ï ï ï ®0 èç sin x ø÷ x sin x x = lim í(1 + x ) = lim í(1 + x) ý ý x®0 ï ïþ ïþï ï ï ï x®0 ïî î

x ®¥

sin 4 x - sin 2 x + 1 cos 4 x - cos 2 x + 1

(1 - cos x ) - (1 - cos x ) + 1 = lim 2

2

4

2

2

cos x - cos x + 1 cos x - cos 2 x + 1 = lim x ®¥ cos 4 x - cos 2 x + 1 = lim 1 = 1 x ®¥

4

x ®¥

EMEP.CH02_3PP.indd 142

(

3x - a + x + a

3x - a + x + a

( 3x - a ) - ( x + a ) ( x - a) ( x - a)

(

( (

3x - a + x + a 2( x - a) 3x - a + x + a 2

3x - a + x + a

)

)

) )

)

tan x - sin x x3 sin x - sin x = lim cos x 3 x®0 x sin x (1 - cos x) = lim ´ lim 2 x®0 x ® 0 x cos x x (1 - cos x) ù é sin x = lim ê ´ lim 2 x®0 ë x x ® 0 x cos x ú û x 2 sin 2 2 = 1 ´ lim 2 x ® 0 x cos x ì ü x ï 2 sin 2 ï ï 2 ´ 1ï = lim í ý 2 x®0 ï æ x ö cos x 4 ï ïî çè 2 ÷ø ïþ 2 ìæ ü xö ïïç sin ÷ 1 1 ïï 2 = lim íç ÷ ´ ý 2 x®0 ïç x ÷ cos x ï ç ÷ îïè 2 ø þï x ®0

cos ecx

6. (d)  lim

( x - a)

)(

lim

lim (1 + x )

= e1 = e.

3x - a - x + a

2 2 1 . = = = 2a    3a - a + a + a 2 2a

5. (c)

=e

(

x®a

= lim

æç ö÷ çç ÷÷÷ 1 çç ÷÷÷ çç ÷÷ sin x ÷ çç lim çè x®0 x ø÷÷

3x - a - x + a x-a

2

x öü ì æ ç sin 2 ÷ ïï 1 ïï 1 = ´ ílim ç ÷ý ´ x x 2 ï ®0 ç cos x ÷÷ ï lim 2 x®0 ç è 2 ø þï îï 1 1 1 = ´ 12 ´ = . 2 1 2

8/9/2023 8:39:40 PM

Calculus • 143

13. (a) f is continuous at x = 1 Þ lim f ( x) = lim f ( x) = f (1)

tan 2 x - x 3 x - sin x tan 2 x tan 2 x - 1 lim -1 x ®0 x x = lim = x ®0 sin x sin x 33 - lim x ® 0 x x tan 2 x ö æ 2 ´ ç lim ÷ - 1 2 ´1 - 1 1 è 2 x ®0 2 x ø = = = . sin x 3 -1 2 3 - lim x ®0 x  

9. (a)  lim x ®0

x®1-

Þ lim ( x - ax + b) = lim (4 x 2 - bx + a) = a + b x ®1-

Þ 1 - a + b = a + b, 4 - b + a = a + b Þ 2a = 1, 2b = 4 Þ a =

lim

= lim dx x®0

f ( x) - f (0) x k sin x + 2 x - 0

x ® 0+

x k sin x + 2 x = lim x ® 0+ x sin x x = k lim + 2 lim x ® 0+ x 0 ® + x x = k ´1 + 2 = k + 2 x ® 0+

Lf ¢(0) = lim

x ® 0-

= lim

f ( x) - f (0) x k sin x + 2 x - 0

x k (- sin x) + 2 x = lim x ® 0x sin x x = -k lim + 2 lim x ® 0x 0 ® x x = -k ´1 + 2 = 2 - k x ® 0-



Now, f ¢(0) exist

EMEP.CH02_3PP.indd 143

Þ Rf ¢(0) = Lf ¢(0) Þ k + 2 = 2 - k Þ k = 0 .

ççform çè

2



= lim

0 ö÷

÷

0 ÷ø

{sin (p cos x)} 2

d dx

f ( x) £ K "x Î R.



x

x®0

A function f : R ® R is said to be bounded if there exists a positive number “K” such that

= lim

sin (p cos 2 x) æ d

Rf ¢(0) = lim

1 , b=2 2

14. (d)

Remember:

12. (d) Clearly, f (0) = k sin 0 + 0 = 0 .

x ®1+

Þ 1- a + b = 4 - b + a = a + b

10. (c)  Let f ( x) = x 2 "x Î R. Then clearly f is unbounded.

11. (a) A function is differentiable at only those points where its graph is smooth, i.e; it has no edges. Here, the function has an edge only at x = 4 and so it is differentiable everywhere except at x = 4.

x®1+

2

{x2 }

(by the L'Hospital Rule)

cos (p cos 2 x)´ p ´ 2 cos x ´ (- sin x) 2x

x®0

æ sin x ÷ö çè x ÷÷ø

= -p lim (cos (p cos 2 x)´ cos x )´ lim çç x®0

x®0

= -p (cos (p cos 2 0)´ cos 0)´1



= -p cos p = -p (-1) = p.

15. (a)  lim éê log(4 + x) - log(4 - x) úù çæform x®0 ê úû ççè x ë d {log(4 + x) - log(4 - x)} = lim dx x®0 d ( x) dx (by the L'Hospital Rule) é 1 (-1) ù ê ú ê (4 + x) (4 - x) ú = lim ê ú x®0 ê ú 1 ê ú ëê ûú é 1 ù 1 ú = lim ê + x® 0 ê (4 + x ) (4 ) x ë ûú =

0 ö÷ ÷ 0 ÷ø

1 1 1 + = . (4 + 0) (4 - 0) 2

ìï 2 - 3 cos x - sin x üï æ ïý çform 16. (c) limp ïí 2 ïï èçç (6 x ) p x® ï ï þ 6 î d {2 - 3 cos x - sin x} = lim dx p d x® 6 {(6 x - p)2 } dx (by the L'Hospital Rule)

0 ö÷ ÷ 0 ø÷

8/9/2023 8:39:43 PM

144 • Engineering Mathematics Exam Prep ìï 0 + 3 sin x - cos x üï æ 0ö = lim íï ýï çççform ÷÷÷ p ï ï 12(6 x - p ) 0ø x® î þï è 6 ï d 3 sin x - cos x dx (by the L'Hospital Rule) = lim p d x® {12(6 x - p )} 6 dx

{

ìï 3´ 43 x ´ log 4 - 4´34 x ´ log 3üïï = lim ïí ý x® 0 ï ïþï 1 îï = lim { 43 x log 43 - 34 x ´ log 34 }

}

x® 0

= 40 log 43 - 30 ´ log 34 æ 64 ö = log 64 - log 81 = log çç ÷÷÷. çè 81 ø

ìï 3 cos x + sin x üï ïý = lim ïí p ïï ´ 12 6 x® ï ï þ 6 î



p p 3 1 3 cos + sin + 3´ 6 6= 2 2= 1. = 72 72 36

ìï 1 + cos3 x üïï æ ö ççform 0 ÷÷ lim ïí ý 2 ç x®p ï 0 ÷ø îï sin x ïþï è d {1 + cos3 x} dx (by the L'Hospital Rule) = lim x®p d 2 sin x dx ìï 0 + 3cos 2 x (- sin x) üïï 3 = lim ïí { cos x} ý = - lim x®p ï x®p ï 2sin cos 2 x x îï þï

19. (a)

17. (a) ü ïì 1 + cos 2 x ï ï lim ïí ý p ï (p - 2 x ) 2 ï x® î ï þ 2ï

æ ççform çè

0 ÷ö ÷ 0 ÷ø

d {1 + cos 2 x} (by the L'Hospital Rule) = lim dx p d x® (p - 2 x ) 2 } 2 { dx ì ï -2sin 2 x ü ï ï = lim ï í ý p ï -2(p - 2 x ) ´ 2 ï x® î ï ï þ

3 3 3 = - ´ cos p = - ´ (-1) = . 2 2 2

20. (a)  Let f ( x) = log x . Then clearly f satisfies all the conditions of Lagrange’s Mean Value Theorem. Hence, there exist a point f (b) - f (a ) c Î (a, b) such that f ¢(c) = . b-a

2

ì ï sin 2 x ü ï 1 ï = lim ï í ý p ï 2 x® ï ( 2 ) x p ï î þï 2



æ ççform çè

0 ö÷ ÷ 0 ÷ø

d {sin 2 x} 1 dx (by the L'Hospital Rule) = lim 2 x ® p d (p - 2 x ) { } 2 dx ì ï 2 cos 2 x ü ï 1ì ï cos p ü ï 1 1 ï ï = lim ï = ï í ý í ý= . p ï ï ï 2 x® î ï (-2) þ ï 2î ï (-1) ï ï 2 þ



1 Here, f (a ) = log a, f (b) = log b, f ¢( x) = . x \ f ¢ (c ) =



2

f (b) - f (a ) b-a

æbö log çç ÷÷÷ ç 1 log b - log a èaø Þ = = c b-a b-a Now, c Î (a, b)

Þ a 0 and A = 2 > 0

60. (a) Here,

f ( x, y, z ) = x 2 + y 2 + z 2 , f( x, y, z ) = xyz - 8. \ F ( x, y, z ) = f ( x, y, z ) + lf( x, y, z )

= 12. (2, 2)

\ AC- B2 = 144 - 36 > 0 and A = 12 > 0.

= 2, B =

Thus, the given function has a minimum value at (1, 1). Thus, the minimum value of u ( x, y ) is 1+1+1, i.e, 3.

¶2 f = 12, B = = -6 ¶ x ¶ y (2, 2) (2,2) 2

¶2 f ¶x 2



= x 2 + y 2 + z 2 + l ( xyz - 8) ¶F ¶F ¶F = 0, = 0 and =0 Then, ¶x ¶y ¶z Þ 2 x + l yz = 0, 2 y + l xz = 0, 2 z + l xy = 0

8/9/2023 8:40:09 PM

Calculus • 153

Hence,

2x 2 y 2z = = yz xz xy

Þ -l =



Therefore, the minimum value of x 2 + y 2 + z 2 is k + k + k i.e; 4 + 4 + 4 = 12 . 61. (a)

ò ò 0



f ( x, y )dydx = ò

f ( x, y )dydx …(i)

Here the limits of integration are given by x = 0, x = 4; y = 0, y = 4 x - x

The region of integration is OAB, which is shown in the following figure: Clearly, y varies from 0 to ∞ and x varies from 0 to y.



2

¥ ¥ -y ¥ì - y y ü Now, I = ò ò e dydx = ò ïíe ò dxïýdy y ï y 0 þï 0 x 0î

y

2

Þ ( x - 2) + ( y - 0) = 2 Þ x - 2 = ± 4 - y Þ x = 2± 4- y

B

2

A x=¥

x=0

x

O y=0

y

¥

yü ì -y = ò íe ´ [ x]0 ý dy y þ 0î

B

(0,2)

O

(2,0)

x=4 A

y=0 x

¥ -y







=ò e 0

y

¥

´ ydy = ò e- y dy = éë- e- y ùû 0

Clearly, y varies from 0 to 2 and x varies from 2 - 4 - y to 2 + 4 - y 2

0

æ ö = - éëe-¥ - e0 ùû =1 ç  e-¥ = 1¥ = 0÷ è ø e

The given integral is

a

a 2 - x2

0

0

ò ò



f ( x, y )dydx.

The limits of integration are

x = 0, x = a; y = 0, y = 2

¥

63. (c)

EMEP.CH02_3PP.indd 153

y= x

y =¥

2

Here the region of integration is bounded by y = 0, x = 0 and the circle x2 + y2 – 4x = 0. In the following figure, OABO is the region of integration.



¥ y

-y -y Thus, I = ò ò e dydx = ò ò e dxdy y y 0 x 0 0

Now y = 4 x - x2 Þ x2 + y 2 - 4 x = 0 2

f ( x, y )dxdy.

Here the limits of integration are

¥ ¥

0

2

ò

0 2- 4- y2

x = 0, x = ∞; y = x, y = ∞.

Þk=4

Let, I =

0

ò

-y 62. (b) Given that I = ò ò e dydx. y 0 x

\ xyz = 8 Þ x 2 y 2 z 2 = 64 Þ k ´ k ´ k = 64



0

2 2 2+ 4- y

¥ ¥

Þ x 2 = y 2 = z 2 = k (say)

4 4 x- x2

4 x- x2

I=ò

x y z x2 y2 z2 Þ = = Þ = = yz xz xy xyz xyz xyz



4

a2 - x2 . Now,

y = a2 - x2 Þ x2 + y2 = a2 Þ x = a2 - y2 .

8/9/2023 8:40:13 PM

154 • Engineering Mathematics Exam Prep The region of integration is OAB, which is shown in the following figure: y ) x=0 ,a B

(0

A (a

, 0)

y=0

O x2 + y2 = a2

x

4



65. (c) Here, R = {( x, y ) : 0 x 1, 0 y 1 - x} . Thus, x varies from 0 to 1 and y varies from 0 to (1 – x)

x= a

é 32 ù 3 3 2y y3 =ê - ú = 4 4 2 - 4 = 4 ´8 - 16 = 16 . ê 3 12 ú 3 12 3 3 3 êë 2 úû0

(

R

=

Hence,

a

a -x

0

0

ò ò

a

a -y

0

0

f ( x, y )dydx = ò

ò



4

Here the region of integration is OABCO, which is shown in the following figure: y



x=0

4x y =

C

B (4



,4 ) x



x=4

Clearly, y varies from 0 to 4 and x varies from y to 2 y 4 æ 2 ö y2 2 ç  y = 4 x Þ x = 4 and x = 4 y Þ x = 2 y ÷ è ø

Hence, by changing order of integration we have,

ò ò

EMEP.CH02_3PP.indd 154

0

x2 4

dydx = ò 0

ò

)

{

}

R = ( x, y ) : 0  x  p , 0  y  p 2 \ ò ò sin( x + y )dxdy = 0 0 p 2

ò [ - cos( x + y)]

p

x =0

y =0

p 2

p ïì ïü í òx =0 ï yò=0 sin( x + y)dy ýï dx î þ



p 2

dx = - ò {cos(p + x) - cos x} dx

p 2

2

4 2 y

(

= 1 - 1 - 1 (1 - 1) - - 1 ´ (0 - 1) 4 3 4 12 12 = 1 + 1 = 1. 12 12 6

=



4 2 x

4 é 3 ù = ê x - x - 1 ( x - 1) 4 ú 3 4 12 ë û0

p 2 p

A

O (0,0)

3

66. (c) Here, the region

x2 = 4 y 2

3

1

2

Now y = x Þ x2 = 4 y and y = 2 x Þ y 2 = 4 x .

2

0



2 x = 0, x = 4; y = x and y = 2 x . 4

x =0 y =0

1

f ( x, y )dxdy.

64. (a) The limits of integration are:

+ y 2 dy dx

= ò x 2 (1 - x) + 1 (1 - x)3 dx 3 0

2

ò

) }

2

é 2 y3 ù ê x y + 3 ú dx ë û y =0

1

2

ò ò {( x

{ } = ò { x - x - 1 ( x - 1) } dx 3

Clearly, y varies from 0 to a and x varies from 0 to a2 - y2 . 2

1 1- x

1- x

1

x =0

2

)

\ ò ò x 2 + y 2 dxdy =

0

p 2

= - ò {- cos x - cos x} dx = 2 ò cos xdx 0

0

p 2 0

p æ ö = 2 [ sin x ] = 2 ç sin - sin 0 ÷ = 2(1 - 0) = 2. 2 è ø

67. (a)

y

y= x

dxdy

y2 4

4 4 æ y2 ö 2 y = ò [ x] y2 dy = ò ç 2 y - ÷ dy 4ø è 0 0 4

R O

y=0

x x=1

8/9/2023 8:40:18 PM

Calculus • 155



Here, the region of integration,





R = {( x, y ):0  x 1, 0  y  x} .

So, y = uv, x = u – y = u – uv = u(1 – v).

òò

\I =



\ x = 0 Þ u (1- v) = 0 Þ u = 0, v =1; y =1- x Þ x + y =1 Þu =1; y = 0 Þuv = 0 Þu = 0, v = 0.



Hence, new region of integration,



R ¢ ={(u ,v):0  u 1, 0  v 1}.



¶x ¶u Jacobian, J = ¶y ¶u

4 x2 - y 2 dxdy

R

=



ìï

1

x =0 ï î y =0

éy 4 x2 -1 y ù 2 2 ò êë 2 4 x - y + 2 sin 2 x úû y=0 dx x =0

æ çè using 1

ò

=



x =0

ò

{(

x

1

=



üï 4 x2 - y 2 dyý dx ïþ

x

ò íò

4a2 - y 2 dy =

2 y yö 4a2 - y 2 + a sin -1 ÷ 2 2 2a ø

(sin

1

æ ö = ò ç 3 x2 + 2 x2 ´ p ÷ dx 2 6ø 0è



}

)

x 4 x2 - x2 + 2 x2 sin -1 x - (0 + 0) dx 2 2x -1 1

=p 2 6

)





{



}

∴ Given integral =

1

1

uv

ò ò e u u dvdu

u =0 v=0

(  dxdy = J × dudv) =



1

1

ò udu ´ vò=0 e dv u =0 v

1

68. (b) Here, the region of integration, R = ( x, y ):0  y  2, 0  x < 4 - y 2

¶x ¶v 1- v -u = ¶y v u ¶v

= u(1 – v) – v(–u) = u.

1

3 3ù é = ê 3 ´ x + p ´ x ú = 3 + p = 3 3 + 2p . 18 ë 2 3 3 3 û0 6 9

Let us put x + y = u and y = uv.



1 é 2ù = ê u ú ´ éëev ùû = 1 e1 - e0 = 1 (e - 1). 0 2 2 ë 2 û0

(

)

But, x  4 - y 2 Þ x2 + y 2  22............(i)

70. (d) Let us put x + y = u and y = uv Then, y = uv, x = u – y = u – uv = u(1 – v)



Let us put x = rcosθ, y = rsinθ.





Then Jacobian, J = r.

Also, x + y = 1 ⇒ u = 1 and



2

2

\ x = 0 Þ u (1 - v) = 0 Þ u = 0, v = 1.

(i) gives, r  2 and so 0  r  2 .



y = 0 Þ uv = 0 Þ u = 0, v = 0.

Since, x  0, y  0; so 0  q  p .



∴  New region of integration is



Hence, the new region of integration is

R ¢ = {(u , v) :0  u  1, 0  v 1}



R ¢ = (r ,q): 0 r £ 2, 0  q  p 2





¶x ¶u Jacobian, J = ¶y ¶u

2

{

}

Then, =



2

òò ( x

p 2

R

ò òr

2

2

)

(

)

+ y 2 dydx = òò x2 + y 2 dydx

´ r drd q

R¢

( dxdy = Jdrd q)

r =0 q=0



ìï 2 üï ìïp 2 üï é 4 ù2 p 2 = í ò r 3 dr ý ´ í ò d qý = ê r ú ´ [q]0 4 ë û 0 îïr =0 þï îïq=0 þï



= 16 ´ p = 2p 4 2 .

69. (a) Here, region of integration,

EMEP.CH02_3PP.indd 155

R ={( x, y ):0  x 1, 0  y 1- x}.

¶x ¶v 1- v -u = ¶y v u ¶v

= u(1 – v) – v(–u) = u.

Hence, given integral =

1

1

ò ò u (1 - v)uv

1 - u ududv

u =0 v =0

=

1

òu

3

1 - u du ´

u =0 1



= ò u 4-1 (1 - u ) 0

1

ò v(1 - v)dv

v =0 3 -1 2

1

du ´ ò v 2-1 (1 - v) 2-1 dv 0

8/9/2023 8:40:24 PM

156 • Engineering Mathematics Exam Prep 1 æ ö æ 3ö = b ç 4, ÷ ´ b(2,2) ç b(m, n) = ò x m-1 (1 - x )n-1 dx ÷ è 2ø 0 è ø



æ3ö æ3ö G(4)G ç ÷ G ç ÷ ´ (1!)2 2 ø G(2)G(2) 2 è = ´ = è ø ( G(2) = 1!) 3 ö G(2 + 2) æ æ 11 ö Gç4 + ÷ Gç ÷ 2ø è è 2ø

=





71. (b) Here, region of integration,

Let us put x = rcosθ and y = rsinθ



Then Jacobian, J = r



  x  0, y  0, so we must have 0  q  p .



Also, 0 ≤ x, y < ∞



⇒0≤x +y 0 , then f is ï ï ï î 0, x £ 0



(a) discontinuous at x = 0 (b) differentiable at x = 0 (c) continuous but not differentiable at x = 0 (d) none of these ì æç 1x - 1x ö÷ ïï x2 çç e 1 -e 1 ÷÷, x >0 26. If f ( x) = í çè e x +e- x ÷ø ï ïî 0, x £ 0

Then, f ( x) is (a) discontinuous at x = 0 (b) continuous at x = 0 (c) f ( x) is not defined at x = 0 (d) none of these

EMEP.CH02_3PP.indd 159

kx x 27. If lim e - e - x = 3 , then k = ? 2 x® 0 x 2

1 3 (b) 2 2 (c) 0 (d) 1

(a)

ïì x - 3 ïüï 28. lim ïí ý=? x®3 ï log ( x - 2) ï a îï þï (a) log e a (b) log a e



(c) 0

(d) 1 üï x 2 + x +1 - ax - bïý = 4, then ïþï x +1

ïì 29. If lim ïí x®¥ ï îï (a) a=1, b=4 (c) a = 1, b = –4

(b) a = 4, b = 1 (d) a = 4, b = –1

30. Let f : R ® R be such that f (1) = 3 and 1

ì üx f ¢(1) = 6 . Then, lim ïïí f ( x + 1) ïïý = ? x®0 ï îï f (1) ïþï (a) e (b) e 2 (c) 0 (d) 1

31. If q1, q2, and q satisfies the condition p  0 < q1 < q < q2 < , then which of the follow2 ing is true? (a) cot q =

sin q1 - sin q2 cos q1 - cos q2

(b) tan q =

sin q1 - sin q2 cos q1 - cos q2

(c) cot q =

sin q2 - sin q1 cos q1 - cos q2

(d) tan q =

sin q2 - sin q1 cos q1 - cos q2



32. Let f , g : [ a, b ] ® R be continuous on [ a, b ] ,

differentiable on (a, b) with g ( x) ¹ 0 in [ a, b ].

Then, g (a ) f (b) - f (a ) g (b) = ? g (c ) f ¢ (c ) - f (c ) g ¢ (c ) (b - a ) g (a ) g (b) (a) g (a ) g (b) (b) 2 2 ( g (c ) ) ( g (c ) )

(b - a ) f (a ) g (b) (c) (b - a ) f (a ) f (b) (d) 2 2 ( f (c ) ) ( g (c ) )

8/9/2023 8:40:53 PM

160 • Engineering Mathematics Exam Prep 39. If f ( x) be a real valued function defined for all x ³ 0 such that f (0) = 0 and f ¢¢( x) > 0 " x ,

33. Let 2 p + 3q + 6r = 0 then at least one root of the equation px 2 + qx + r = 0 lies in the interval (a) (0, 1) (b) (0, 2) (c) (1, 2) (d) (2, 4) 34. If p ( x) be a polynomial function with real coefficients and a, b be two consecutive roots of the equation p ( x) = 0 (where a < b), then $ c Î (a, b) such that

(a) 50 p ¢(c) + p (c) = 0



p (c ) (b) p ¢(c) + =0 50 (c) p ¢(c) + 50 p(c) = 0



(d) none of these

35. If a function f : [ a, b ] ® R is continuous on [ a, b ] and differentiable on (a, b) , then for bf (a ) - af (b) =? c Î ( a, b ) , b-a (a) f (c) - cf ¢(c) (b) f (c) - f ¢(c) (c) cf (c) - f ¢(c) (d) cf ¢(c)

then the function g ( x) =

f ( x) is x



(a) decreasing on (0, ¥)



(b) increasing on (0, ¥)



(c) increasing on (0, 1) & decreasing on (1, ¥) (d) none of these



40. Which of the following points is the point of a local minima or local maxima of the function æ pö f ( x) = sin 4 x + cos 4 x in çç0, ÷÷÷ ? çè 2 ø p p 3p (a) (b) (c) p (d) 2 4 4 41. The maximum value of the function x æ 1 ÷ö ç f ( x) = ç ÷÷ is çè x ø 1

(a) e e (b) ee (c) 10 (d)

1 +e e

36. Let f be twice differentiable function for

42. The positive numbers a and b are such that

all x and f (1) = 1, f (2) = 4, f (4) = 16. Then which of the following is true? (a) f ¢¢( x) = 2 " x Î (1, 4)

3 a + b = 60 and ab is maximum, are, respec-

(b) f ¢¢( x) = f ¢( x) = 5 " x Î (2, 4) (c) f ¢¢( x) = 4 " x Î (2, 4) (d) f ¢¢( x) = 2 for some x Î (1, 4) 37. The function f ( x) = sin x + cos x ( 0 £ x £ 2p ) is strictly increasing on æ 5p ö æ 5p ö æ pö (a) çç , 2p÷÷ (b) çç , 2p÷÷ È çç0, ÷÷ ÷ çè 4 çè 4 ø ø÷ èç 4 ø÷



æ 5p ö æ pö æ pö (c) ççç0, ÷÷÷ (d) ççç , 2p÷÷÷ Ç ççç0, ÷÷÷ è4 ø è 4ø è 4ø

38. The intervals on which the function f ( x) = 2 x 3 -15 x 2 + 36 x + 1 are strictly increasing and strictly decreasing are given by, respectively (a) (- ¥, 2) È (3, ¥); (2, 3)

tively (a) 15, 45 (b) 10, 50 (c) 20, 40 (d) 25, 35 43. An open box with a square base is to be made out of a given quantity of card board of area c 2 square units. Then the maximum volume of the box is c3 c3 (a) cubic units cubic units (b) 2 3 (c)

44. A wire of length 36 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the lengths of the two pieces so that the combined area of the circle and square is minimum? (a)

36p 144 28 112p ; ; (b) p+4 p+4 p+4 p+4

(c)

12p 16 28 112 ; (d) ; p+4 p+4 p p

(b) (- ¥, 2); (2, 3) (c) (3, ¥); (2, 3) (d) (- ¥, 2) È (2, ¥); (1, 2)

EMEP.CH02_3PP.indd 160

c3 c3 cubic units (d) cubic units 6 3 2 3

8/9/2023 8:41:04 PM

Calculus • 161 2

3n 2 - 2 is n! n =1 ¥

45. The value of

å

51. If

(a) 4e (b) 4e + 1 (c) 4e - 1 (d) 4e + 2 46. Consider the Taylors series expansion of the x 2

function f ( x) = e about x = 3. Then the co-efficient of ( x - 3)5 in the expansion is



3

5

1 e2 1 e2 (a) ´ (b) ´ 5! 32 3! 25

(c) -1

2

h f ¢¢(q h), where 2!

x < log(1 + x) < x if x > 0 1+ x



(d) all the above

x2 if x > 0 2

1

49. If I = x n tan -1 x dx , then the value of n ò 0

(n + 1) I n + (n -1) I n-2 is p p 2 (a) p + 1 (b) p - 1 (c) - (d) 2 2 n 2 n 2 50. Let f be a function such that f ¢, f ¢¢, f ¢¢¢ all

exist and u = f ¢( x) cos x - f ¢¢( x) sin x and v = f ¢( x) sin x + f ¢¢( x) cos x . Then,



ò

2

2

æ du ö÷ æ dv ö÷ çç ÷ + çç ÷ dx = ? çè dx ø÷ èç dx ø÷

p

p 2

ò

f (cos x) dx (b)

ò

f (cos x) dx

0

p 2

(c) p ò f (cos x) dx (d) O 0

2

53. ò x3 - x dx is equal to -1

(a)

2 1 1 11 (b) (c) (d) 21 30 2 4 1

54. Let I1 = ò 0

then

1

ex x2 dx & I 2 = ò x 3 dx , 3 1+ x 0 e (2 - x )

I1 =? I2

(a) 3e (b) 2e (c) 3e + 2 (d) 2e -1 é 1 1 1ù 55. lim ê + + ..... + ú =? 2 n®¥ ê n úû 4n - 2 2 ë 2 n -1 3p p (a) (b) (c) p -1 (d) 0 2 2 é1 1 1 1ù 56. lim ê + + + ..... + ú = ? n®¥ ê n n +1 n + 2 3n úû ë

(a) 3 + log 5 (b) 1- log 2 (c) log 3

(d) 0

p 4

57. ò log(1 + tan q ) d q = ? 0

(a) f(x) + f″(x) + C (b) f ¢( x) + f ¢¢¢( x) + C

(a)

p p log 2 (b) log 2 2 8

(c) f ¢( x) + f ¢¢( x) + C (d) f ¢¢( x) + f ¢¢¢( x) + C

(c)

p log 3 3

EMEP.CH02_3PP.indd 161

(d) 0

52. ò x f (sin x) dx is equal to

0

1 5 9 (a) 0 (b) (c) (d) 2 9 25 48. Which of the following is true? x (a) 0 < 1 log æ e - 1 ö < 1 ç ÷ x è x ø

(c) x > log(1 + x) > x -

(a) 3p (b) p (c) 4p 2 2

(a)

q Î (0,1); then for h = 1and f ( x) = (1 - x) , the value of q will be ?

(b)

-1

k , then the value of “ k ” is p2

p

5 2



?

x sin p x dx =

0

(d) 0

47. If f (h) = f (0) + hf ¢(0) +

ò

(d) 0

8/9/2023 8:41:14 PM

162 • Engineering Mathematics Exam Prep f ( x)

58. If

ò

t dt = x cos p x, then f ¢(4) = ?

0



æ1ö æ 2ö æ 3ö æ 4ö æ8ö 65. G çç ÷÷÷´G çç ÷÷÷´G çç ÷÷÷´G çç ÷÷÷´.......´G çç ÷÷÷ = ? èç 9 ø èç 9 ø èç 9 ø èç 9 ø èç 9 ø

2

(a) 1

1

(b)

5

(c)

2 3

1 12

2 3



(d) 0

(a)

66. If u = log ( x3 + y3 + z3 - 3xyz ) , then which of the

1

59. lim æç n ! ö÷n = ? çè n ø÷÷ n®¥ ç n

(a) 1

following is not true?

(b)

1 (c) 2e e

(d) –1



60. If g (x) = cosx- ò ( x - t ) g (t ) dt , then 0



g ( x) + g ¢¢( x) = ?

(b) sin x (c) - cos x (d) 0

(a) 1 1

61. ò 0

1 1 n n

(1- x )



62. ò 0

(a)

(c)

x2 1- x 4

dx = ?

æ 3ö æ 3ö G çç ÷÷÷ G çç ÷÷÷ çè 2 ø p çè 4 ø (b) p æ1ö 4 æ1ö G çç ÷÷÷ G çç ÷÷÷ çè 4 ø èç 2 ø

æ 3ö G çç ÷÷÷ çè 4 ø p æ1ö G çç ÷÷÷ èç 4 ø

p 2

p 2

0

0

63. ò sin q d q ´ ò (a) p (b) ¥

64. ò 0

4

¥

(d) None of these

æ ö 9 (b) ç ¶ + ¶ + ¶ ÷ u = è ¶x ¶y ¶z ø ( x + y + z)2 2 2 2 3 (c) ¶ u2 + ¶ u2 + ¶ u2 = 2 ¶x ¶y ¶z x + y + z) (

(d) none of these

2 2 2 -2 (a) ¶ V +¶ V +¶ V = 2 2 2 ¶x ¶y ¶z 2 x + y2 + z2

(

)

(

)

2 ¶2 V + ¶2 V = 2 (b) ¶ V + ¶x2 ¶y 2 ¶z 2 x2 + y 2 + z 2 2



2

2

2

1 (c) ¶ V +¶ V+¶ V =2 ¶x2 ¶y 2 ¶z 2 x + y2 + z2

(

)

2

(d) none of these 68. If u = f (r), x = rcosθ and y = rsinθ, then uxx + uyy = ?



(a) 1 f ¢¢(r ) + f ¢(r ) r

(b) f ¢¢(r ) + f ¢(r )

(c) f ¢¢(r ) + 1 f ¢(r ) r

(d) none of these

æ 2 2ö 69. If u = sin -1 ç x + y ÷, then x ¶u + y ¶u = ? è x+ y ø

1 dq = ? sin q p p (c) (d) 2p 2 3



0

(a) tanu (c) secu

¶x

¶y

(b) cotu

(d) cosecu

70. If u = f (x,y) and x = ef + e–ψ, y = e–f –eψ, then ¶u - ¶u = ? ¶j ¶y

4

e- x x 2 dx´ ò e- x dx = ?

p p p p (a) (b) (c) (d) 8 2 4 2 3

EMEP.CH02_3PP.indd 162

3 x+ y+ z

67. If V = log(x2 + y2 + z2), then which of the following is true?

dx = ?

p p p np (a) sin (b) sin 2 n n 2 p p (c) cos ec (d) None of these n n 1

(a) ux + u y + uz =

2

x



1 16 4 8 4 p (b) 3p (c) p 4 (d) p 4 3 3 3



(a) x ¶u + y ¶u (b) x ¶u - y ¶u ¶x ¶y ¶x ¶y



(c) y ¶u - x ¶u (d) 0 ¶x ¶y

8/9/2023 8:41:22 PM

Calculus • 163

71. If u = f (2x – 3y, 3y – 4z, 4z – 2x), then,

78. The minimum and maximum distance of the point (1, 2, 3) from the sphere x 2 + y 2 + z 2 = 1 are, respectively

1 u + 1u + 1 u =? 2 x 3 y 4 z



(a) –1

(b) 1

(c) 0

(d) 2u

72. If u = x + y and v = tan -1 x + tan -1 y , then which 1- xy

(a) 14 -1, 14 + 1 (b) 14, 2 14 (c) 15 (d) 8



(a) u and v are functionally related



¶ (u ,v) (b) =0 ¶ ( x, y )



(d) all of these è xø

æ yö (a) xf ç ÷ è xø æ yö (c) f ç ÷ è xø





¶y

æ yö (b) - yf ç ÷ è xø

(a)



(c)



2

x +y x+ y

2

(d) -

x +y x+ y



2

y x

(c) log xy -1+ x y



(b) log xy +1+ x y (d) log xy +1- x y

EMEP.CH02_3PP.indd 163

(a) 8/7

(b) 1/7

(c) 15

(a) 1 840

(d) 4 21

ò ò ò ( x + y + z ) xyz,

(b) 2 525

òòòx

3

(c) 1 420

taken over the

(d) 1 2

y 4 z 2 dxdydz , where V is the

1 2 11! ( )

(b) 3!4!5! (11)!

(c) 1!2!3! (11)!

(d) 2!3!4! (11)!

(a)



2 l ( x) m ( x, y )

ò ò ò 0



77. The minimum value of x 2 + y 2 + z 2 , given that x + 2 y + 3 z = 4 , is

(c) 1 90

83. If the triple integral bounded by the planes 2x + y + z = 4, x = 0, y = 0 and z = 0 is given by

æ1 1ö (d) maximum value at çç , ÷÷÷ çè 3 3 ø



y 2 z dzdydx is

­region given by x0, y0, z0 and x + y + z1

(a) minimum value at (0, 1) (b) maximum value at (1, 0) æ1 1ö (c) minimum value at çç , ÷÷÷ çè 3 3 ø



3

V

dx

76. The function f ( x, y ) = xy (1- x - y ) has

òòòx

(b) 1 25

82. Evaluate

75. If u = xlog(xy) and x2 + y2 + xy = 1, then du = ? (a) log xy -

1 x y

volume V bounded by x = 0, y = 0, z = 0 and x + y + z = 1 is

x +y x + y2 2

(d) 1 16

(c) 8

V

2

(b) -

(a) 1 10

81. The value of

3

x +y x + y2

(b) 32

0 0 0

(d) 0

2

(a) 16

80. The value of

dy 74. If x + y + 3xy -1= 0 , then dx = ? 3

)

+ y 2 + z 2 dxdydz ,

where V is the volume of a cube bounded by the co-ordinate planes x = 0, y = 0, z = 0 and the plane x = y = z = 2

(c) u = tan v

¶x

2

V

73. If u = xf æç y ö÷ + g æç y ö÷, then x ¶u + y ¶u = ? è xø

ò ò ò(x

79. Find the value of

of the following is true?

(d) 8

0

dzdydx, then l ( x) -m ( x, y ) = ?

0

(a) x + y

(b) x - y

84. The value of

1

x2

(c) x

(d) y

y

ò ò ò ( y + 2 z ) dzdydx

is

x =0 y =0 z =0



(a) 4 21

(b) 2 21

(c) 5 21

(d) 8 21

8/9/2023 8:41:28 PM

164 • Engineering Mathematics Exam Prep 85. The volume of the solid generated by the

31. (c) 36. (d) 41. (a) 46. (a) 49. (b) 54. (a) 59. (b) 64. (d) 69. (a) 74. (b) 79. (b) 84. (b) 88. (c)

revolution of the curve y = 8 2 about x-axis 4+ x from x = –∞ to x = ∞ is given by (a) 2p2



(b) 4p2

(c) 8p2

(d) 16p2

86. The volume of a solid formed by the revolution of the loop of the curve y2(2 + x) = x2(2 – x) about x-axis is

(

(

)

(a) 8p log 2 + 2 (b) 16p log 2 - 2 3 3

(

)

(

(c) 16p log 2 + 2 (d) 8p log 2 - 2 3

3

)

)

87. The volume of the solid generated by the revolution of the cardiod r = a(1 + cosθ) about the initial line is (a) 8 pa3 3 (c) 2 pa3 3



88. What will be the surface area of a solid generated by revolving the arc of the parabola y2 = 4ax about x-axis bounded by its latus rectum?

)

(



2 (b) 16pa 1 + 2 2 3

)

)

(d) none of these

89. The surface area of the solid generated by the 2

2

2

revolution of the astroid x 3 + y 3 = a 3 about x-axis is 2 (a) 12pa 5



2 (b) 3pa 5

2 (c) 8pa 5

Hence, by the Sandwich theorem, we get,



lim

EMEP.CH02_3PP.indd 164

2. (a) 7. (a) 12. (d) 17. (a) 22. (a) 27. (b)

3. (d) 8. (b) 13. (d) 18. (b) 23. (a) 28. (a)

4. (b) 9. (a) 14. (c) 19. (a) 24. (a) 29. (c)

5. (a) 10. (a) 15. (c) 20. (a) 25. (b) 30. (b)

x + 3 sin x = 1. x-5

x ®¥





1-

1 x = 1 - 0 = 1 and

x -1 = lim x ®¥ x 1 lim 1 = 1 , so by the Sandwich theorem we get,

Since, lim

Answer key 1. (b) 6. (a) 11. (c) 16. (a) 21. (a) 26. (b)

x ®¥

2. (a) x - 1 < [x ] £ x x - 1 [x ] Þ < £1 x x

90. What will be the surface area of the solid formed by the revolution of the cardioid r = a(1 + cosθ) about the initial line?

35. (a) 40. (d) 45. (d) 48. (d) 53. (d) 58. (c) 63. (a) 68. (c) 73. (a) 78. (a) 83. (d) 87. (a)



(d) 2pa2

2 2 2 2 (a) 4pa (b) 8pa (c) 16pa (d) 32pa 5 5 5 5

34. (c) 39. (b) 44. (a) 47. (d) 52. (c) 57. (a) 62. (c) 67. (b) 72. (d) 77. (a) 82. (d) 86. (b)

1. (b) -1 £ sin x £ 1 Þ -3 £ 3sin x £ 3 Þ x - 3 £ x + 3sin x £ x + 3 x - 3 x + 3sin x x + 3 Þ £ £ x -5 x -5 x -5 3 1x -3 x = 1 - 0 = 1, Now lim = lim x ®¥ x - 5 x ®¥ 5 1- 0 1x 3 1+ x+3 x = 1 + 0 = 1. = lim lim x ®¥ x - 5 x ®¥ 5 1- 0 1x

(d) 4pa3

( (

33. (a) 38. (a) 43. (c) 48. (d) 51. (c) 56. (c) 61. (c) 66. (d) 71. (c) 76. (d) 81. (a) 85. (b) 90. (d) Explanation

(b) 4 pa3 3

2 (a) 4pa 2 +1 3 2 (c) 8pa 2 2 - 1 3

32. (b) 37. (b) 42. (a) 47. (d) 50. (b) 55. (a) 60. (c) 65. (a) 70. (b) 75. (d) 80. (c) 85. (b) 89. (a)

x ®¥

lim x ®¥

[x] = 1. x

x3 x5 + - ........¥ 3! 5! 3 5 Therefore, sin x - x = - x + x - ..........¥ 3! 5!

3. (d) We know that sin x = x -

8/9/2023 8:41:32 PM

{

} = lim ( -1) cos {p ( n - n + n )} ì n- n +n n+ n +n ü )( ) ïý ï ( = lim ( -1) cos íp C • 165 ï ( n + n + n ) ïþ î ì æ öü ïï ç n - ( n + n ) ÷ ïï ø = lim ( -1) cos íp è ý n+ n +n) ï ï ( ïî ïþ = lim ( -1) cos np - p n 2 + n n

n ®¥

n

2

n ®¥

2

2

n

n ®¥



sin x - x x ®0 x3 é x3 x5 ù ê - 3! + 5! - .......¥ ú = lim ê ú x ®0 x3 ê ú êë úû

\ lim



(

)

é x + x 2 + x3 + ..... + x50 - 50 ù ú = lim ê ú x -1 x®1 ê ëê ûú



é x - 1 x 2 - 1 x3 - 1 x50 - 1 ù = lim ê + + + ....... + ú x -1 x -1 x - 1 úû x®1 êë x - 1 é1 + ( x + 1) + ( x 2 + x + 1) + .........ù ê ú = lim ê ú 49 48 x®1 ê ú + ( x + x + .... + 1) ë û = 1 + (1 + 1) + (1 + 1 + 1) + .... + (1 + 1 + ...... + 1)



50 (50 + 1) = 1 + 2 + 3 + ....... + 50 = = 1275 2 æ n (n + 1)ö÷ ÷ ççç 1 + 2 + 3 + ........ + n = çè 2 ÷÷ø



ì -p n ï = lim ( -1) ´ cos í n ®¥ ïî 1 + 1 + 0

(



{

{

}

} = lim ( -1) cos {p ( n - n + n )} ì n- n +n n+ n +n ü )( ) ïý ï ( = lim ( -1) cos íp ï ( n + n + n ) ïþ î ì æ öü ïï ç n - ( n + n ) ÷ ïï ø = lim ( -1) cos íp è ý n+ n +n) ï ï ( ïî ïþ n

n

2

n ®¥

2

2

n

n ®¥

2

2

2

2

æ ö n æ pö æ pö = lim ( -1) ´ cos ç - ÷ = 0 çcos ç - ÷ = 0 ÷ n ®¥ è 2ø è 2ø è ø

æp ö æp ö lim n cos ç ÷ sin ç ÷ n ®¥ è 8n ø è 8n ø 1 1ö æpx ö æpx ö æ = lim cos ç sin ç putting n = ÷ ÷ ÷ ç x ®0 x xø è 8 ø è 8 ø è é æp x öù sin ç ÷ú ê æpx ö è 8 øú ê = lim cos ç ´ lim ÷ x ®0 x è 8 ø x ®0 ê ú êë úû é ù æpx ö ê sin ç 8 ÷ p ú è ø´ ú = cos 0 ´ lim ê x ®0 8ú ê px êë úû 8

= lim ( -1) cos np - p n 2 + n n ®¥

)

ü ï ý ïþ

6. (a)

5. (a) n ®¥

ì ü ïï ïï -p = lim(-1) ´ lim í ý x®¥ x®¥ æ ï 1+ 1+ 1 ö ï ç ÷ n ø ïþ ïî è n

é ( x - 1) + ( x 2 - 1) + ( x3 - 1) + ....( x50 - 1) ù = lim ê ú x -1 x®1 ê úû ë

lim cos p n 2 + n

)

(

å xk - 50 x -1

2

ì ü -np n ï ï = lim ( -1) cos í ý n ®¥ 2 ï n+ n +n ï î þ ì ü ï ï -np n ï ï = lim ( -1) cos í ý n ®¥ ï n æ1 + 1 + 1 ö ï ÷ ï ç n ø ïþ î è ì ü ï ï -p n ï ï = lim ( -1) ´ lim cos í ý n ®¥ n ®¥ ï æ1 + 1 + 1 ö ï ÷ ïç n ø ïþ îè

50

k =1

2

2

n

4. (b)

lim

2

n ®¥

æ 1 x2 ö = lim ç - + - .........¥ ÷ x ®0 è 3! 5! ø 1 0 1 1 = - + - .......¥ = - = - . 3! 5! 3! 6

x®1

alculus

2



é æpx ö sin ç ÷ ê p è 8 ø = 1´ ´ lim ê 8 p x ®0 ê p x 8 êë 8

ù ú p p ú = ´1 = . 8 8 ú úû

n

n ®¥

EMEP.CH02_3PP.indd 165

ì n ï = lim ( -1) cos í n ®¥

2

- np 2

ü ï ý

8/9/2023 8:41:34 PM

166 • Engineering Mathematics Exam Prep x m sin n x x® 0 sin x k k ìæ sin x ön ï ïü ÷÷ ´ x ´ x m+n-k ïý ç = lim ï í ç k ÷ x® 0 ïç ïï ïè x ø sin x î þ

7. (a) lim

n

æ sin x ö÷ 1 = ççlim ´ ´lim ( x m+n-k ) kö çè x®0 x ø÷÷ æ x®0 çç lim sin x ÷÷ k ÷ k çè x ®0 x ÷ø





æ ö æ x ö = logex çlim ç a x- 1 ÷ = log e a ÷ ÷ ÷ ç x ®0 ç è ø è ø So, f(xy) = loge(xy) = loge(x) + loge(y) = f(x) + f(y).



10. (a)

( x ® 0 Þ x k ® 0)

= lim x ®1

( = lim

(provided m + n - k = 0 i.e; k = m + n)

æ n2 - n + 1 ö lim ç 2 ÷ n ®¥ n - n - 1 è ø

( x - 1) (

)(

24 + 25 - x 2

24 + 25 - x 2

( 24 - 25 + x )

)

)

2

= lim x ®1

n ( n -1)

= lim x ®1

n ( n -1)

1 ù é ê1 + n (n - 1) ú = lim ê ú n ( n -1) ®¥ ê1 - 1 ú êë n (n - 1) úû (n ® ¥ Þ n (n - 1) ® ¥ )

= lim x ®1

=

x

é 1ù ê1 + ú = lim ê x ú (putting n (n - 1) = x) x ®¥ 1 ê1 - ú ë xû x

æ 1ö lim ç1 + ÷ x ®¥ è x ø = e = e2 . = x e -1 æ 1ö lim ç1 - ÷ x ®¥ è xø

9. (a)

EMEP.CH02_3PP.indd 166

24 - 25 - x 2

n ( n -1)

é n ( n - 1) + 1 ù = lim ê ú n ®¥ n ( n - 1) - 1 ë û

x -1

24 - 25 - x 2 x -1

x ®1

)

(

- 25 - x 2 - - 25 - 12

x ®1

8. (b)



f ( x) - f (1) x -1

= lim

x®0



lim x ®1

1 = 1n ´ ´lim x m+n-k = lim x m+n-k x®0 1 x®0 0 = lim x = 1, a finite non-zero value



æ xm -1 ö æ 1ö = lim ç ÷ ç putting m= ÷ m ®0 nø è m øè

æ 1n ö æ 1n ö x -1 f ( x) = lim n ç x - 1÷ = lim ç 1 ÷ ç ÷ n ®¥ è ø n®¥ ç n ÷ è ø m æ x -1 ö æ 1ö = lim ç m ÷ ç putting m = ÷ m ®0 xø è ø è



=

(

( x - 1) (

( x - 1) (

(

24 + 25 - x 2

(x

2

- 1)

24 + 25 - x 2

( x + 1)

24 + 25 - x 2

(1 + 1) 2

24 + 25 - 1

) )

) 2

=

) (

24 + 24

)

2 1 . = 2´ 2 6 2 6

11. (c) f (1,1) = f (1 - 1, f (1,1 - 1)) = f (0, f (1, 0)) = f (0, f (1 - 1,1)) = f (0, f (0,1)) = f (0, 2) = 2 + 1 = 3. 12. (d) 1ö æ f çx+ ÷ xø è 2



= x2 +

1 1ö 1 æ + 3 = ç x + ÷ - 2´ x´ + 3 x2 x x è ø

8/9/2023 8:41:38 PM

Calculus • 167

In a similar manner, it can be shown that lim j ( x) = lim j ( x) = j (3). As a result j is conx ®3+ x ®3-

2

1ö æ = ç x + ÷ +1 xø è Þ f ( x) = x 2 + 1 (by replacing x +



1 by x) x

tinuous at x = 3 . 15. (c) f ( x) = x sin x = x sin x

13. (d) Case-I: x > 0

ì(- x)(- sin x), - p < x £ 0 =í 0£ x £p î x sin x, = x sin x, - p < x < p

Here x = x.

2

x -3 x + 2 = 0 Þ x 2 - 3 x + 2 = 0 Þ ( x - 1)( x - 2 ) = 0 Þ x = 1, 2



Then,

Case-II: x < 0 Here x = - x.



x -3 x + 2 = 0



Þ x 2 + 3 x + 2 = 0 Þ ( x + 1)( x + 2 ) = 0



Since each of x , sin x , cos x is

Þ x = -1, - 2



differentiable, so f ¢ is also differentiable. \  f″ (x)



Thus, four solutions exist. 14. (c)

\ f ¢¢( x)

lim j ( x)



d d d ( f ¢( x)) = ( x cos x) + (sin x) dx dx dx = - x sin x + cos x + cos x = - x sin x + 2 cos x, which exists "x Î R.



Thus, we can say that f is twice differentiable.



\ f ¢¢ exist, so it is continuous.

=

x®2+

1

= lim

x®2+

1+ e

1 x-2

=

-1

+ e ( x -3) 1

2

1

1 + lim e x - 2 + lim e x®2+



-1 ( x -3)2

=0

x®2+

( lim e

1 x-2

x®2+

16. (a) We know that ì x, x ³ 0 x =í î - x, x £ 0

= ¥)

lim j ( x)

x®2-

1

= lim

x®2-

1+ e

1 x-2

= 1 + lim e

+e 1

1 x-2

x®2-

1+ 0 + e

and sin(- x) = - sin x, cos(- x) = cos x. Now, Rf ¢(0)

-1 ( x -3)2

+ lim e

-1 (2 -3)2

=

x ®0 +

1 1 + e -1

1

( lim e x - 2 = 0) x®2-

 lim j ( x) ¹ lim j ( x), x®2+

at x = 2

EMEP.CH02_3PP.indd 167

x®2-

so j

f ( x) - f (0) x q cos x + q1 sin x + q2 x 3 - q0 æ 0 ö = lim 0 ç form ÷ x ®0 + x è0 ø 2 -q sin x + q1 cos x + 3 x q2 = lim 0 x ®0 + 1 (by the L'Hospital rule) = -q0 ´ 0 + q1 + 0 = q1. = lim

-1 ( x -3)2

x®2-

1

=



d d (sin x) + sin x ´ ( x) dx dx = x cos x + sin x

f ¢( x) = x ´

2



Clearly, f ¢( x) exists, i.e, f is differentiable (since each of x and sin x is differentiable).

is discontinuous



8/9/2023 8:41:43 PM

168 • Engineering Mathematics Exam Prep Again, Lf ¢(0) f ( x) - f (0) = lim x ®0 x q cos x - q1 sin x - q2 x 3 - q0 æ 0 ö = lim 0 ç form ÷ x ®0 x è0 ø 2 -q sin x - q1 cos x - 3 x q2 = lim 0 x ®0 1 (by the L'Hospital rule)



= -q0 ´ 0 - q1 ´ 1 - 0 = - q1



 f is differentiable at x = 0,

Þ 6 f (- x) + 9 f ( x) - 4 f ( x) - 6 f (- x) = 3x 2 + 3x + 3 - 2 x 2 + 2 x - 2 Þ 5 f ( x) = x 2 + 5 x + 1 Þ 5 f ¢( x) = 2 x + 5



Þ f ¢( x) =

2x + 5 7 Þ f ¢(1) = . 5 5

19. (a) lim f ( x) = lim

x ®0 +

x ®0 +

x 1+ e

1 x

=

0 =0 1+ ¥ 1

1 ® ¥ Þ e x ® ¥) x 0 = =0 1+ 0

( x ® 0+ Þ

So Rf ¢(0) = Lf ¢(0) i.e; q1 = - q1 Þ q1 = 0.

lim f ( x) = lim

17. (a) f ( x + y ) = f ( x) f ( y ) Þ f ( x) = f ( x) ´ f (0) (for y = 0)

x ®0 -

x ®0 -

x 1

1+ ex

1

( x ® 0- Þ e x ® 0) Hence, lim f ( x) = lim f ( x) = 0 = f (0).



⇒  f(x) {f(0) – 1} = 0





⇒  f(0) – 1 = 0 (assuming f(x) ≠ 0)

Consequently, f is continuous at x = 0 .



⇒  f(0) = 0





f '5 = lim h ®0

18. (b) 2 f ( x) + 3 f (- x) = x 2 - x + 1.......(i ) Replacing x by - x , we get from (i),

2 f (- x) + 3 f ( x) = x 2 + x + 1.......(ii )



Then, 3 ´ Eqn (ii) - 2 ´ Eqn (i)

x ®0





= lim x ®0

1 - cos kx x sin x

0ö æ ç form ÷ 0ø è

d (1 - cos kx) = lim dx x ®0 d ( x sin x) dx k sin kx = lim x ®0 x cos x + sin x

(by the L'Hospital rule) 0ö æ ç form ÷ 0ø è

d (k sin kx) dx

0ö æ form ÷ ç d 0ø ( x cos x + sin x) è dx k 2 cos kx = lim x ®0 - x sin x + cos x + cos x k 2 ´1 k2 = = 0 +1+1 2

= lim x ®0



Now f is continuous at x = 0 Þ lim f ( x) = f (0) x ®0



EMEP.CH02_3PP.indd 168

x ®0 -

20. (a) lim f ( x)

f (5 + h) - f (5) h

Þ f ( x){ f (0) - 1} = 0 Þ f (0) - 1 = 0 (assuming f ( x) ¹ 0) Þ f (0) = 1 f (5 + h) - f (5) f ¢(5) = lim h ®0 h f (5) f (h) - f (5) = lim h ®0 h 2{ f (h) - 1} = lim ( f (5) = 2) h ®0 h f (h) - f (0) = 2 lim ( f (0) = 1) h ®0 h = 2 f ¢(0) = 2 ´ 3 = 6.

x ®0 +

Þ

k2 1 = Þ k 2 = 1 Þ k = ±1. 2 2

8/9/2023 8:41:49 PM

Calculus • 169

21. (a) lim f ( x)

Lf ¢(0) = lim

x®2

x 3 + x 2 - 16 x + 20 = lim x®2 ( x - 2) 2

x ®0 -

0ö æ ç form ÷ 0 è ø

1

xe x 1

3 x 2 + 2 x - 16 = lim (by the L'Hospital rule) x®2 2( x - 2) 1 3 x 2 + 2 x - 16 æ 0ö = lim ç form ÷ x ® 2 2 ( x - 2) 0ø è 1 6x + 2 (by the L'Hospital rule) = lim x ® 2 2 1 6´ 2 + 2 = = 7. 2



-0

x = lim 1 + e x ®0 x

Rf ¢(0) = lim

x ®0 +

1

= lim

x ®0 -

f ( x) - f (0) x

e

-

1 x

= 0,

+1

1

xe x 1 x



Now f is continuous at x = 2

-0

= lim 1 + e x ®0 + x

1

lim

x ®0 +

e

1 x

= 1.

+1

Since Lf ¢(0) ¹ Rf ¢(0) , so f ¢(0) does not exist.

Þ lim f ( x) = f (2) Þ 7 = k .



f ( x) - f (0) x

x®2

24. (a) Let k = 0 22. (a) RHL at x = 0 = lim f ( x) = lim x ®0 +

xe

Rf ¢(0) = lim

1 x 1 x

x ®0 +

= lim

x ®0 +

-

x ®0 +

0 = =0 0 +1

x 1 x

e +1 1+ e 1 1 ( x ® 0+ Þ - ® -¥ Þ e x ® e -¥ = 0) x



= lim = lim

= lim f ( x) = lim

x ®0 -

xe x 1 x

= lim

x ®0 -

x -

1 x

=

ö ÷ ÷ ø

-0

x xe

2 x

-0

= lim e

x ®0 + x f ( x) - f (0) Lf ¢(0) = lim x ®0 x x ®0 +

1

0 =0 ¥ +1

e +1 1+ e 1 1 ( x ® 0- Þ - ® ¥ Þ e x ® e¥ = ¥) x



xe

æ1 1 -ç + çx x è

x ®0 +

LHL at x = 0

x ®0 -

f ( x) - f (0) x

= lim

xe

æ1 1 -ç + çx x è

ö ÷ ÷ ø

-

2 x

= 0,

-0

x xe - 0 = lim = lim 1 = 1. x ®0 x ®0 x x ®0 -

-0

Hence, f is continuous at x = 0 .



Again

 Rf ¢(0) ¹ Lf ¢(0), so f is not differentiable for k = 0. It is easy to verify that f ( x) is not continuous for k ¹ 0 and hence not differentiable for k ¹ 0.

23. (a) f ( x) £ x 2

Þ - x 2 £ f ( x) £ x 2 Þ - x £

f ( x) £x x



f ( x ) - f (0) Þ lim( -x ) £ lim £ lim x x ®0 x ®¥ x ®0 x



⇒ 0 ≤ f’(0) ≤ 0 [|f(0)| ≤ 02 ⇒ f(0) = 0]



⇒ f’(0) = 0 [by the Sandwich theorem]

25. (b) Rf ¢(0) = lim

x ®0 +



EMEP.CH02_3PP.indd 169

f ( x) - f (0) x

é - x12 ù ì ü e -0ú ï 1 ï ê lim = lim ê = í ý x ®0 + x ú x ®0+ ï x12 ï êë úû î xe þ ì ü ïï ïï 1 = lim í ý x ®0 + ï x çæ1 + 1 + 1 ´ 1 + ........ ÷ö ï 2 2! x 4 ø þï îï è x ì ï

ü ï

8/9/2023 8:41:52 PM

é - x12 ù ì ü e -0ú ï 1 ï ê = lim ê = lim í 1 ý 0+ x ®0 + x Múathematics 170 • xE®ngineering 2 ï P rep ïî xeExxam þ ëê ûú

28. (a) ìï x - 3 ïüï lim ïí ý x®3 ï log ( x - 2) ï a îï þï ì ü ï ï x -3 ï = lim ï í ý x®3 ï log e ´ log ( x - 2) ï a e ï ï î þ

ì ü ïï ïï 1 = lim í ý x ®0 + ï x çæ1 + 1 + 1 ´ 1 + ........ ÷ö ï ïî è x 2 2! x 4 ø ïþ ì ü ï ï 1 1 = lim í = 0. ý= x ®0 + 1 1 ï x + + 3 + ........ ï 0 + ¥ x 2x î þ Similarly Lf ¢(0) = 0. \Rf ¢(0) = Lf ¢(0).



26. (b)

lim f ( x )

x ®0 +

1

= lim

x ®0 +

-

1

1 x

e +e

-

1 x

-

= lim

2

x 2 (1 - e x )

x ®0 +

-

2 x

(1 + e )

=

0(1 - 0) =0 (1 + 0)

2 2 2 ( x ® 0+ Þ - ® -¥ Þ e x ® e -¥ Þ e x ® 0) x lim f ( x )



x ®0 -

1

= lim

x ®0 -

-

1 x

e +e

-

( x ® 0- Þ



1

x 2 (e x - e x ) 1 x

2

= lim

x 2 ( e x - 1)

x ®0 -

2 x

( e + 1)

2 2 2 ® -¥ Þ e x ® e -¥ Þ e x ® 0) x

x ®0 +



1

sin (p cos x) æ 2

x

x®0

d = lim dx x®0



ççform çè

2

÷

0 ÷ø

{x }

1



2

d

ì f ( x + 1) ü x Lety = í ý . Then î f (1) þ

0 ö÷

{sin (p cos x)} dx

= lim

(by the L'Hospital Rule) 1

ìï f ( x + 1) üïï x Let y = ïí ý .Then ïîï f (1) ïþï

cos (p cos 2 x)´ p ´ 2 cos x ´ (- sin x)

x®0

\ k = 2.

ì f (x + 1) ü x y= í ý î f (1) þ

2

1

2x

ìï f ( x + 1) üïï x y = ïí ý ïîï f (1) ïþï

ì 2e 2 x - e x - 1ï ü ï ïý çæform 0 ÷÷ö = lim ïí çç x®0 ï ï 2x 0 ÷ø ï ïè î þ d (2e2 x - e x -1) dx (by the L'Hospital rule) = lim x®0 d (2 x) dx ì ï 4e 2 x - e x ü ï 4e 0 - e 0 4 -1 3 ï = lim ï = = = í ý x® 0 ï ï 2 2 2 2 ï ï î þ



ìï x 2 + x + 1 üï lim ïí - ax - bïý x®¥ ï ïþï îï x + 1 ïì x 2 + x + 1- ax 2 - ax - bx - b ïüï = lim ïí ý x®¥ ï ïþï x +1 îï

30. (b)

27. (b) lim

1 1 lim ( x - 2) = ´1 = log e a. x ® 3 log a e log a e

x ®0 -

continuous at x = 0 .



=

29. (c)

0(0 - 1) = =0 (0 + 1)

Thus, lim f ( x) = lim f ( x) = 0 = f (0) , so f is

Þ log y = Þ log y =

ïì f ( x + 1) ïüï 1 log íï ý ïîï f (1) ïþï x log [ f ( x + 1) ] - log[ f (1)]

Þ lim (log y )

x

x® 0



ìï log [ f ( x + 1) ] - log [ f (1) ]üïæ ïýçform = lim ï í x® 0 ï ïïççè x ï î þ ïìï f ¢( x + 1) ïü - 0 ïï ïï ïï f ( x + 1) Þ log lim y = lim ïí ý x®0 x®0 ï ïï 1 ïï ïï ïïî ïþ

(

EMEP.CH02_3PP.indd 170

0 ÷ö ÷ 0 ø÷

d ( x - 3) 1 dx = ´ lim log a e x®3 d (log e ( x - 2)) dx (by the L'Hospital rule) ìï üï ïï ï ïï 1 ïïï 1 = ´ lim í ý log a e x®3 ïï 1 ïï ïï ï îï ( x - 2) ïþï

Hence, f is differentiable at x = 0.

x 2 (e x - e x )

æ ççform èç

0 ÷ö ÷ 0 ÷ø

)

8/9/2023 8:41:54 PM

Þ log y =

îï þï log [ f ( x + 1) ] - log[ f (1)]

Þ lim (log y )

x

x®0

ì ï log [ f ( x + 1) ] - log [ f (1) ]ïüïæ = lim ï í ýçççform x®0 ï ï x ï ïè î þ ì ü ¢ ï ïï f ( x + 1) - 0 ïïï ï f ( x + 1) ïï Þ log lim y = lim ï í ý x®0 x®0 ï ïï 1 ï ï ïï ï ï ïþ î

(

0 ÷ö ÷ 0 ÷ø

1



ì f ( x + 1) ï ü ï ï x = e2 . Þ lim y = e 2 Þ lim íï ý x®0 x®0 ï îï f (1) ïþï



ì ï (1- a ) x 2 + (1- a - b) x + (1- b) ïü ï = lim ï í ý x®¥ ï ï x +1 ï ï î þ ì ü a b b (1 ) (1 ) ï ïï ï + (1- a ) + 2 ï ïï ï x x = lim í ý x®¥ ï 1 1 ïï ï + 2 ï ïï ï x x î þ æ ö÷ 0 çform if 1- a = 0 i.e; if a = 1÷ èçç ø÷ 0 ì (-b) (1- b) ü ï ï ï ï + 2 ï ï ï ï çæform 0 ÷ö x x = lim í ÷÷ ý x®¥ ï 1 1 ï èçç ø 0 ï ï + 2 ï ï ï ï x x î þ ì ü (1- b) ï ï ï ï -b + ï ï ï x ï = lim í ý x®¥ ï 1 ï ï ï 1+ ï ï ï ï x î þ -b + 0 = = -b = 4 (given) 1+ 0 \ b = -4; a = 1.

31. (c) Let f ( x) = sin x and g ( x) = cos x . Then ­clearly the functions f and g satisfy the Cauchy’s Mean Value Theorem on [q1 , q2 ] . Hence, $ q Î (q1 , q2 ) such that f ¢(q ) = f (q2 ) - f (q1 ) . g ¢(q ) g (q2 ) - g (q1 )

Now, f ¢( x) = cos x and g ¢( x) = - sin x .



EMEP.CH02_3PP.indd 171

Þ - cot q =

)

(by the L'Hospital rule) f ¢(0 + 1) f ¢(1) 6 = = = =2 3 f (0 + 1) f (1)



Calculus • 171

\

f (q2 ) - f (q1 ) f ¢(q ) = g ¢(q ) g (q2 ) - g (q1 )

Þ

sin q2 - sin q1 cos q = - sin q cos q2 - cos q1

Þ cot q =



sin q2 - sin q1 cos q2 - cos q1

sin q2 - sin q1 . cos q1 - cos q2

32. (b)

Let f( x) = f ( x) . Then clearly f satisfies all g ( x) the conditions of the Lagrange’s Mean Value Theorem on [a, b]. Then there exist a point c Î (a, b) f(b) - f(a ) such that f ¢(c) = . b-a f(b) - f (a ) Now, f ¢(c) = b-a f (b) f (a ) g (c ) f ¢ (c ) - f (c ) g ¢ (c ) g (b) g (a ) Þ = 2 b-a ( g (c ) )



é ù ê  f ¢( x) = g ( x ) f ¢( x ) - f ( x ) g ¢( x ) ú ê ú 2 ( g ( x )) êë úû Þ Þ



g (c ) f ¢ (c ) - f (c ) g ¢ (c ) 2

( g (c ))

=

f (b) g ( a ) - f ( a ) g (b) (b - a ) g ( a ) g ( b )

g (a ) f (b) - f (a ) g (b) (b - a ) g (a ) g (b) = . 2 g (c ) f ¢ (c ) - f (c ) g ¢ (c ) ( g (c ) )

33. (a)

Let f ( x) = 1 px3 + 1 qx 2 + rx + s. 3 2 Then f (0) = s and 1 1 2 p + 3q + 6r +s p+ q+r +s = 3 2 6 0 = +s= s 6

f (1) =



\ f (0) = f (1).

Now f being a polynomial function, is continuous on [0, 1] and differentiable on (0, 1). Hence, f satisfies all the conditions of Rolle’s theorem. Therefore, there exists a point c Î (0,1) such that f ¢(c) = 0. But f ¢( x) = px 2 + qx + r. Consequently, at least one root of the equation px 2 + qx + r = 0 lies in the interval (0, 1). 34. (c) Since a, b are two consecutive roots of the equation p ( x) = 0 , so p (a ) = 0 = p (b).

8/9/2023 8:41:58 PM

172 • Engineering Mathematics Exam Prep Let f ( x) = e50 x p ( x) and then

f ¢( x) = 50e50 x p ( x) + e50 x p ¢( x). Therefore, f is continuous on [a, b] and differentiable on (a, b). Also f (a ) = e50 a p (a ) = 0 ( p(a) = 0), f (b) = e50b p (b) = 0 ( p(b) = 0).



Thus, f satisfies all the conditions of Rolle’s theorem on [a, b]. Hence, there exists a point c Î (a, b) such that f ¢(c) = 0. Now f ¢(c) = 0 Þ 50e50 c p (c) + e50 c p ¢(c) = 0 Þ 50 p (c) + p ¢(c) = 0.



f ( x) 1 and h( x) = . Then the x x functions g , h : [ a, b ] ® R are continuous on [ a, b ] and differentiable on (a, b) . Hence, by the Cauchy’s Mean Value Theorem on [a, b], there exists a point c Î (a, b) such that g ¢(c) g (b) - g (a ) = ................(i) h ¢(c) h(b) - h(a ) xf ¢( x) - f ( x) 1 Here, g ¢( x) = and h ¢( x) = - 2 . 2 x x ¢ g ( c ) g ( b ) g ( a ) Therefore, (i) Þ = h ¢(c) h(b) - h(a )

Now consider the function g ¢ . Then clearly g ¢ satisfies all the conditions of the Lagrange’s Mean Value Theorem on [c1 , c2 ] . Hence, there exists a point c Î [c1 , c2 ] such that g ¢(c2 ) - g ¢(c1 ) g ¢¢(c) = . c2 - c1 But since g ¢(c1 ) = 0 =g ¢(c2 ), so g ¢¢(c) = 0 for some c Î [c1 , c2 ] . Now g ¢¢(c) = 0 Þ f ¢¢(c) - 2 = 0 Þ f ¢¢(c) = 2 . Hence, we can conclude that f ¢¢( x) = 2 for some x Î (1, 4) . 37. (b) f ( x) = sin x + cos x æ 1 ö 1 = 2 çç sin x + cos x÷÷÷ çè 2 ø 2 æ pö = 2 cos çç x - ÷÷÷ çè 4ø

35. (a) Let g ( x) =



cf ¢(c) - f (c) f (b) f (a ) 2 c a Þ = b 1 1 1 - 2 c b a af (b) - bf (a ) Þ - éë cf ¢(c) - f (c)ùû = a -b



bf (a ) - af (b) Þ = f (c) - cf ¢(c). b-a



Û f ¢( x) > 0 æ pö Û - 2 sin çç x - ÷÷÷ > 0 çè 4ø



2

g (2) = f (2) - 2 = 4 - 4 = 0,



g (4) = f (4) - 42 = 16 -16 = 0.

æ pö p Û sin çç x - ÷÷÷ < 0 Û p < x - < 2p çè 4ø 4 p p 5p 9p < x < 2p + Û 0

Hence, g satisfies all the conditions of Rolle’s theorem on [1, 2] and [2, 4]. Therefore, there exists points c1 Î (1, 2) and c2 Î (2, 4) such that

Û 6 x 2 - 30 x + 36 > 0

g ¢(c1 ) = 0 and g ¢(c2 ) = 0.



Û x 2 - 5 x + 6 > 0 Û ( x - 2)( x - 3) > 0 Û ( x - 2) > 0, ( x - 3) > 0 or ( x - 2) < 0, ( x - 3) < 0 Û x > 2, x > 3 or x < 2, x < 3 Û x > 3 or x < 2 Û x Î (- ¥, 2) È (3, ¥)

EMEP.CH02_3PP.indd 172

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Û f ¢( x ) > 0 Û 6 x 2 - 30 x + 36 > 0

f ( x) = sin 4 x + cos 4 x

Û x 2 - 5 x + 6 > 0 Û ( x - 2)( x - 3) > 0 Û ( x - 2) > 0, ( x - 3) > 0

Þ f ¢( x) = 4sin 3 x cos x + 4 cos3 x (- sinx) 2 x) = -4sin x cos x (cos 2 x -Csin alculus • 173 = -2sin 2 x cos 2 x = - sin 4 x

or ( x - 2) < 0, ( x - 3) < 0

\ f ¢( x) = 0 Þ - sin 4 x = 0 Þ sin 4 x = sin p p æ pö Þ 4 x = p Þ x = Î çç0, ÷÷÷ 4 çè 2 ø Also, f ¢¢( x) = -4 cos 4 x and so

Û x > 2, x > 3 or x < 2, x < 3 Û x > 3 or x < 2 Û x Î (- ¥, 2) È (3, ¥)



Again, f is strictly decreasing Û f ¢( x ) < 0 Û 6 x 2 - 30 x + 36 < 0 Û x 2 - 5 x + 6 < 0 Û ( x - 2)( x - 3) < 0 Û ( x - 2) > 0, ( x - 3) < 0



Therefore, x = the given function.

or

æ1ö f ( x) = çç ÷÷÷ çè x ø

Û x > 2, x < 3 Û x Î (2,3)

39. (b) f ¢¢( x) > 0 Þ f ¢( x) is strictly increasing. f ( x) Now, g ( x) = x ¢ xf ( x) - f ( x) Þ g ¢( x) = x2 f ( x) f ¢( x) x ........(i) = x



This implies f ¢(c) = f ( x) = g ( x) < f ¢( x) x é  f ¢( x) is strictly increasing and ù ê ú ê ú ¢ ¢ 0 < c < x Þ f ( c ) < f ( x ) ë û



Now g ( x) < f ¢( x) f ( x) f ( x) Þ < f ¢( x ) Þ f ¢( x ) >0 x x Þ g ¢( x) > 0 (using (i))

Consequently, g ( x) is increasing on (0, ¥) . 40. (d) f ( x) = sin 4 x + cos 4 x Þ f ¢( x) = 4sin 3 x cos x + 4 cos3 x (- sinx)



EMEP.CH02_3PP.indd 173

= -4sin x cos x (cos 2 x - sin 2 x) = -2sin 2 x cos 2 x = - sin 4 x

x

x

æ1ö æ1ö Þ log f ( x) = log çç ÷÷÷ = x log çç ÷÷÷ çè x ø çè x ø Þ log f ( x) = x {log1- log x} = -x log x





Þ

d d [log f ( x)] = - [ x log x ] dx dx

Þ

f '( x ) 1 = - log x - x ´ = - log x - 1 f (x ) x

Þ f '( x ) = - f ( x ) ´ {log x + 1}

1 f ¢( x) = - log x - x ´ = - log x -1 f ( x) x Þ f ¢( x) = - f ( x)´{log x + 1} Þ

By the Lagrange’s Mean Value Theorem on [0, x], $ f ( x) - f (0) f ( x) c Î (0, x) such that f ¢(c) = = . x-0 x

p is a point of local minima of 4

41. (a)

( x - 2) < 0, ( x - 3) > 0 Û x > 2, x < 3 or x < 2, x > 3



æpö f ¢¢ çç ÷÷÷ = -4 cos p = -4 ´ (-1) > 0 çè 4 ø

\ f ¢( x ) = 0 Þ log x + 1 = 0 æ1ö Þ log x = -1 = - log e = log çç ÷÷÷ çè e ø Þx=



1 e

Now f ¢¢( x) = - f ¢( x)´{log x + 1} +

1 f ( x). x

æ1 ö æ1 ö ì æ1 ö ü 1 æ1 ö \ f '' ç ÷ = - f ' ç ÷ ´ ílog ç ÷ + 1ý fç ÷ e è ø è e ø î è e ø þ æ1 ö è e ø çe÷ è ø



1 æ ö æ1 ö æ1 ö = -e ´ f ç ÷ = -e ´ e e < 0 ç f ' ç ÷ = 0 ÷ e èeø è ø è ø

Hence, the given function has a local maxi-

mum value at

1 x = . Therefore, the local e 1 æ 1 ÷ö e

maximum value of the function= f çç ÷ = e . ç ÷

èeø

\ f ¢( x) = 0 Þ - sin 4 x = 0 Þ sin 4 x = sin p p æ pö Þ 4 x = p Þ x = Î çç0, ÷÷÷ 4 çè 2 ø Also, f ¢¢( x) = -4 cos 4 x and so æ p ö÷

8/9/2023 8:42:09 PM

174 • Engineering Mathematics Exam Prep 44. (a) Let r be the radius of the circle and x be the length of each side of the square. Then ATQ,

42. (a) 3

ab = (60 - b)b 3

3

( a + b = 60)

4

= 60b - b = f (b) (say)

18 - 2 x p 2 \ Combined area, A = p r + x 2 2p r + 4 x = 36 Þ r =

\ f ¢(b) = 0 Þ 180b 2 - 4b3 = 0 Þ b = 45 \ a = 60 - 45 = 15.

2

2

Now f ¢¢(b) = 360b -12b and so f ¢¢(45) = 360´ 45 -12´ 452 < 0



f (b) i.e; ab3 is maximum when a = 15 and b = 45 .

Hence,

43. (c) Let the length, breadth, and height of the box be, respectively, x, x and y units. Then ATQ,



\

x 2 + xy + xy + xy + xy = c 2

or, x 2 + 4 xy = c 2 or, y =

c2 - x2 4x

Thenvolume volume box, The ofof thethe box, V V



3 ææcc22-x 2xö2 ö÷ c 2 xc-2 xx x3 = xx ççç = ÷ = = ÷ x x ø ÷÷ø 4 4 èçè 44 22

dV c 2 - 3x 2 c =0Þ =0Þ x= dx 4 3 d 2V 0 - 6x = Also, and so 2 4 dx d 2V -6 c = ´ < 0. 2 4 dx x= c 3

Thus,

V

is

maximum

when

36 p +4

> 0.

36 . p+4

Now x =

3

volume

d2A 8 d2A = + 2;so p dx 2 dx 2 x=

Hence, the combined area, A is minimum

when x =

\



æ 72p - 8 x ÷ö dA = 0 Þ -çç ÷÷ + 2 x = 0 çè ø dx p 36 Þx= p+4

Now,

= xx×´xx×´yy==x2xy2 y =

æ18 - 2 x ö÷ = p çç + x2 çè p ø÷÷ æ18 - 2 x ö÷ dA Then, (-2) + 2 x = 2p çç çè p 2 ÷÷ø dx æ 72p - 8 x ÷ö = -çç ÷÷ + 2 x çè ø p

36 p+4

æ 36 ÷ö 18 - 2´çç çè p + 4 ÷÷ø 18 . Þr= = p p+4 18 36p and \ 2p r = 2p ´ = (p + 4) p + 4 36 144 4 x = 4´ . = p+4 p+4

c . Hence, maximum volume, V = x= 3 Hence, the combined area, A is minimum 3 c æç c ö÷ 2 36p c ´ - ç ÷÷ when the lengths of the two pieces are and 3 ç c è ø 3 3 p+4 144 = cubic units. . 4 6 3 p+4

EMEP.CH02_3PP.indd 174

8/9/2023 8:42:12 PM

Calculus • 175

45. (d) ¥ 3n 2 - 2 å n! n =1

3

1 1 e2 = ´ f v (3) = ´ . 5! 5! 32

¥ 1 n2 - 2å n =1 n ! n =1 n !

47. (d)

¥

= 3å

¥ é12 ¥ n 2 ù 1 = 3 ê + å ú - 2å n =1 n ! ë 1! n = 2 n ! û ¥ ì ¥ ææ é 1 ö öü n ù = 3 ê1 + å - 2 íå ç ç 1 + ÷ - 1÷ ý ú ë n = 2 (n - 1)!û î n =1 è è n ! ø ø þ ¥ é (n - 1) + 1 ù = 3 ê1 + å ú - 2(e - 1) ë n = 2 (n - 1)! û

¥ ¥ é 1 1 ù +å = 3 ê1 + å ú - 2(e - 1) ë n = 2 (n - 2)! n = 2 (n - 1)!û





é æ1 1 1 öù ê1 + ç 0! + 1! + 2! + ...... + ¥ ÷ ú è øú = 3ê ê æ1 1 ú ö ê + ç + + ...... + ¥ ÷ ú ø ë è 1! 2! û - 2(e- 1)

= 3 [1 + e + (e - 1) ] - 2e + 2 = 4e + 2.



46. (a) By the Taylor series expansion about x = 3, ( x - 3) 2 f ( x) = f (3) + ( x - 3) f ¢(3) + f ¢¢(3) 2! ( x - 3)3 ( x - 3) 4 iv + f ¢¢¢(3) + f (3) 3! 4! ( x - 3)5 v + f (3) 5! Here, x

x 2

1 1 1 1 1 x e2 f ( x) = e ,so f v ( x) = ´ ´ ´ ´ e 2 = 2 2 2 2 2 32 \ f v (3) =

3 2

e . 32

Hence, coefficient of ( x - 3)5 in the expansion

EMEP.CH02_3PP.indd 175

h2 f ¢¢(q h) 2! 5 1 2 æ 5 ö h 15 Þ (1 - h) 2 = 1 + h ´ ç - ÷ + ´ (1 - q h) 2 è 2ø 2 4 Then f (h) = f (0) + hf ¢(0) +

¥ é 1ù 1 1 ê e = 1 + 1 ! + 2! + .......¥ = 1 + å n !ú n =1 ë û





5 3 5 f ( x) = (1 - x) 2 Þ f ¢( x) = - (1 - x) 2 , 2 1 15 f ¢¢( x) = (1 - x) 2 . 4 1 5 15 \ f ¢(0) = - and f ¢¢(q h) = (1 - q h) 2 . 2 4



5 æ ö 2 ç f (0) = (1 - 0) = 1÷ è ø

5 1 5 15 Þ (1 - 1) 2 = 1 - + (1 - q ) 2 2 8 15 5 3 1-q = -1 = Þ 8 2 2 3 8 4 Þ 1-q = ´ = 2 15 5 16 9 Þ q = 1= . 25 25

( for h = 1)

48. (d) (a) Let, f ( x) = e x . Then by Taylor's theorem, f ( x) = f (0) + x f ¢(q x) Þ e x = e0 + x eq x = 1 + xeq x 1 Þ eq x = (e x - 1) x æ ex -1 ö Þ log eq x = log ç ÷ è x ø æ ex -1 ö æ ex -1 ö 1 Þ q x = log ç Þ = log q ÷ ç ÷ x è x ø è x ø Now 0 < q < 1Þ 0
0) 1+ x 1+q x x Þ < log(1 + x) < x 1+ x \ (b) is correct.

1 x . = 1 + qx 1 + qx \ q > 0 Þ qx > 0 Þ 1 + qx > 1 Þ log(1 + x ) = log1 + x ´



1 x 0) 1+q x 1+q x Þ log(1 + x) < x..........(ii) Þ



Combining (i) and (ii) , we get x > log(1 + x) > x -



\ (c) is correct.

x2 2

49. (b) 1



1

I n = ò x n tan -1 x dx = ò x n-1 ( x tan -1 x) dx

(c) Let f ( x) = log(1 + x).

0

Then f ¢( x) =

é ù = êê x n-1 ò x tan -1 x dxúú ëê ûú 0 1 é ù ìd ü - ò êêïí x n-1 ïý´ ò x tan -1 x dxúú dx.........(i) ïï dx ïþï úû 0 êëî

1 1 , f ¢( x) = 2 1+ x (1 + x )

\ By Taylor's theorem, x2 f ¢¢(q x) 2! ü x2 ì 1 Þ log(1 + x) = 0 + x ´ 1 + ´ í2 ý 2 î (1 + q x) þ f ( x) = f (0) + xf ¢(0) +



But

Then, q > 0 Þ q x > 0 Þ (1 + q x) 2 > 1 1 Þ> -1 (1 + q x) 2 Þ-

éìï 1 üï x 2 êí êïîï1 + x 2 ýïþï 2 ë

ù ú dx ú û

=

x2 tan -1 x - ò 2

=

2 x2 1 ( x + 1) -1 dx tan -1 x - ò 2 2 1+ x2

=

x2 1 ì 1 üï tan -1 x - ò ïí1ý dx 2 2 ïîï 1 + x 2 ïþï

1

Þ log(1 + x) > x -



x dx

x2 1 tan -1 x - { x - tan -1 x} 2 2 \ (i) Þ

x2 x2 > x2 2 2(1 + q x)

x2 ...........(i) 2 Again, by Taylor's theorem, f ( x) = f (0) + xf ¢(q x)

-1

=

2

x x >2 2(1 + q x) 2

Þ x-

ò x tan

éïì d ù ü ï = tan -1 x ò x dx - ò êêí tan -1 xý ò x dx úú dx ï ï ï þ ëêïî dx ûú

1 æ ö = 1÷ ç f (0) = log1 = 0, f ¢(0) = 1+ 0 è ø 2 x = x2 (1 + q x) 2

2

0

1



é æ x2 öù 1 I n = êê x n-1 çç tan -1 x - { x - tan -1 x}÷÷÷úú ÷ø çè 2 2 ë û0 1 é 2 æ x x tan -1 x ö÷ùú ÷ dx - ò êê (n -1) x n-2 çç tan -1 x - + çè 2 2 2 ÷÷øúû 0 ë

1 x = . 1+q x 1+q x \q > 0 Þ q x > 0 Þ 1 + q x > 1 Þ log(1 + x) = log1 + x ´

EMEP.CH02_3PP.indd 176

8/9/2023 8:42:16 PM

Calculus • 177 2

1 -1 1 tan 1- {1- tan -1 1} 2 2

Þ In =

ò

1

-1

1

(n -1) (n -1) x n tan -1 x dx + x n-1dx ò 2 0 2 ò0 1

1

p

1

2

4

2

Þ In = ´ -

{

{ } 1-

p 4

-

2

2

\ ò x sin p x dx In +

( n -1) 2

´

n

( n -1) 2

}

In =

p 4

1

- + 2 p 2

( n -1) 2n -1 +

-

( n -1) 2

2

æ du ö æ dv ö \ç ÷ +ç ÷ è dx ø è dx ø

( n -1) n

=

2

= { f ¢( x ) + f ¢¢¢( x )} sin x + { f ¢( x ) + f ¢¢¢( x )} cos x 2

2

2

= { f ¢( x ) + f ¢¢¢( x )}



ò

2

2

k

ò x sin px dx = p

-1



EMEP.CH02_3PP.indd 177

2

ìx sin px , for - 1 £ x £ 1 x sin px = í îx ( - sin px ), for 1 £ x £ 2

-1

1

-x cos p x sin p x + p p2 Then (i) Þ

= 2

ò

x sin p x dx

-1

1

2

-1

1

= ò x sin p x dx - ò x sin p x dx



1

2

é -x cos p x sin p x ù é -x cos p x sin p x ù ú -ê ú + + 2 êë p p úû -1 êë p p 2 úû1

=ê

é - cos p - cos(-p ) ù é -2 cos 2p cos p ù ú-ê ú + + êë p úû êë p p p úû

=ê

2

51. (c)

2

éd ù = x ò sin p x dx - ò ê ( x) ´ ò sin p x dx ú dx êë dx úû é æ - cos p x öù é - cos p x ù ú - ò ê1´çç = xê ÷÷ú dx êë çè p ÷øúû êë p úû

=

æ du ö æ dv ö ç dx ÷ + ç dx ÷ dx = f ¢( x ) + f ¢¢¢( x ) + C è ø è ø

1

1

Now, ò x sin p x dx

1 - . 2 n

p

2

2

2

= ò x sin p x dx + ò x (- sin p x) dx........(i)

I n-2

50. (b) u = f ¢( x) cos x - f ¢¢( x) sin x du Þ = f ¢¢( x) cos x - f ¢( x) sin x - f ¢¢( x) cos x dx - f ¢¢¢( x) sin x du Þ = -{ f ¢( x) + f ¢¢¢( x)} sin x dx Again, v = f ¢( x) sin x + f ¢¢( x) cos x dv Þ = f ¢¢( x) sin x + f ¢( x) cos x - f ¢¢( x) sin x dx + f ¢¢¢( x) cos x dv Þ = { f ¢( x) + f ¢¢¢( x)} cos x dx

Hence,

1

= ò x sin p x dx + ò x sin p x dx -1

Þ ( n + 1) I n + (n -1) I n-2 =



-1

1

(n -1) I n-2 2

Þ 1+



( n -1)

k p2

ïì x sin p x, for - 1 £ x £ 1 x sin p x = ïí ïïî x (- sin p x), for 1 £ x £ 2

(n -1) x n-2 tan -1 x dx 2 ò0

-

x sin p x dx =

1 1 2 1 4 + + + = . p p p p p



( sin p = sin(-p ) = 0, cos p = cos(-p ) = -1, cos 2p = 1) 2

Hence, ò x sin p x dx = -1

k k 4 Þ = 2 2 p p p Þ k = 4p

8/9/2023 8:42:18 PM

\ x 3 - x = 0 Þ x( x + 1)( x -1) = 0 Þ x = 0, -1,1 Moreover we have, x3 - x

178 • Engineering Mathematics Exam Prep 52. (c)

= x( x + 1)( x -1) = x x + 1 x -1

p p

òò

Let ) dxx. ) Then, Let II == xf (sin x f x(sin dx.Then pp

0 0

òò

II == xf (sin x ) dxx) dx x f (sin



0

0

p

Þ I = ò (p - x) f {sin(p - x)} dx 0



ìï(-x) ( x + 1){-(x-1)} for -1 £ x £ 0 ïï = í x ( x + 1){-(x-1)} for 0 £ x £ 1 ïï x ( x + 1) (x-1) for 1 £ x £ 2 ïïî 3 ìï x - x for -1 £ x £ 0 ïï = ïí-( x3 - x) for 0 £ x £ 1 ïï ï 3 îï x - x for 1 £ x £ 2 2

p

ò

\

= ò (p - x) f (sin x) dx

-1

0

p

p

0

0

0

2

p 0

0

ö÷ ÷÷ ì ü æ ö p ï ï f íïsin çç2´ - x÷÷÷ýï = f {sin(p - x)}÷÷÷ ç ï øï ÷÷ ï è 2 ï î þ ÷÷ ÷÷ = f (sin x), ÷÷ p ÷÷ p 2 ÷÷ ÷ (sin ) 2 (sin ) = f x dx f x dx ò ò ÷÷÷ø 0 0



=

0

4

2

x2 3

e x (2 - x 3 )

dx

1

1 1 dt = ò t 3 0 e (2 - t ) æ putting x 3= t so that 3 x 2 dx = dt ; ÷ö çç ÷ ççèalso x = 0 Þ t = 0 & x = 1 Þ t = 1 ÷÷ø

ì öü ï çæ p ï = pò f ï ísin çç - x÷÷÷ï ý dx ï ï è ø 2 ï ï î þ 0

=

p 2

= p ò f (cos x) dx

1 3e

=-

0

1

e1-t

ò {1 + (1- t )} dt 0

1 3e

0

ey

ò 1 + y dy 1

æ ö çç putting 1- t = y so that - dt = dy;÷÷ ÷ çèalso t = 0 Þ y = 1 & t = 1 Þ y = 0 ÷ø

53. (d) x3 - x = x( x 2 -1) = x( x + 1)( x -1)

ì(-x) ( x + 1){-(x-1)} for -1 £ x £ 0 ï ï ï = í x ( x + 1){-(x-1)} for 0 £ x £ 1 ï ï x ( x + 1) (x-1) for 1 £ x £ 2 ï ï î ì ï x3 - x for -1 £ x £ 0 ï ï 3 =ï í-( x - x) for 0 £ x £1 ï

1

I2 = ò

0

= x( x + 1)( x -1) = x x + 1 x -1

4

11 . 4

54. (a)

p 2

x3 - x

1

2 1

éx éx éx x ù x ù x2 ù = ê - ú -ê - ú +ê - ú ê4 2 úû -1 êë 4 2 úû 0 êë 4 2 úû1 ë é 1 1 ù é 1 1 ù éæ16 4 ö æ 1 1 öù = ê- + ú - ê - ú + êçç - ÷÷÷ - çç - ÷÷÷ú ëê 4 2 ûú ëê 4 2 ûú êëçè 4 2 ø çè 4 2 øúû

Þ I = p ò f (sin x) dx

\ x 3 - x = 0 Þ x( x + 1)( x -1) = 0 Þ x = 0, -1,1 Moreover we have,

3

0

2 0

4

p 2

EMEP.CH02_3PP.indd 178

3

-1

p 2

Þ 2 I = p ò f (sin x) dx = p ´ 2 ò f (sin x) dx



1

1

= ò ( x - x) dx - ò ( x - x) dx + ò ( x 3 - x) dx

0



2

3

0

0

= p ò f (sin x) dx - I



3

-1

p

æ çç ççç çç çç çç çç çç ç çççso è

1

= ò x - x dx + ò x - x dx + ò x 3 - x dx

= p ò f (sin x) dx - ò x f (sin x) dx



x 3 - x dx

=



1 3e

1

ò 0

1 ey dy = 1+ y 3e

1

ò 0

ex dx = I1 1+ x

(using the properties of definite integral) \

I1 = 3e. I2

8/9/2023 8:42:19 PM

Calculus • 179

55. (a) é 1 1 1ù lim ê + + ..... + ú 2 n®¥ ê n úû 4n - 2 2 ë 2 n -1 é 1 1 1 ù ú = lim ê + + ..... + 2 n®¥ ê 4n - 2 2 n 2 úû ë 2n -1 n n 1é = lim ê + + ... 2 n®¥ n ê 2´ 2n - 2 2 ë 1´ 2n -1 ù n ú ........ + n ´ 2n - n 2 ûú



é ê ê 1ê 1 1 = lim ê + + 2 2 n®¥ n ê æ 1 ÷ö æ 1 ÷ö æ 2 ÷ö æ 2 ÷ö ê 2´çç ÷ - çç ÷ 2´çç ÷÷ - çç ÷÷ ê èç n ÷ø èç n ÷ø èç n ø èç n ø ë ù ú ú 1 ú ........ + ú 2 æ n ö÷ æ n ö÷ ú 2´çç ÷÷ - çç ÷÷ ú èç n ø èç n ø úû 1 n 1 = lim å 2 n®¥ n r =1 ærö ærö 2 ´çç ÷÷÷ - çç ÷÷÷ ènø ènø 1

=ò 0

1

=ò 0

1 2x - x2

é ù ê ú ê ú æ ö 1 ê f ç r ÷ = ú ççè ÷ø÷ ê 2 ú n ê ærö ærö ú 2 ´çç ÷÷÷ - ç ÷÷÷ ú ê è n ø èç n ø úû êë

1 1

0



p . 2

56. (c) é1 1 1 1ù + + ..... + ú lim ê + n®¥ ê n 3n úû n +1 n + 2 ë



EMEP.CH02_3PP.indd 179

n ®¥



+ lim n ®¥

1é n n n ù + + ... + n êë n + 1 n + 2 n + n úû

n n ù 1é n + + ... + n êë 2n + 1 2n + 2 2n + n úû

é ù ê ú 1 1 1 1 ú = lim êê + + ..... + n®¥ n n úú ê1 + 1 1 + 2 + 1 êë n n n úû é ù ê ú 1 1 1 1 ú + lim êê + + ..... + n®¥ n n úú ê2+ 1 2+ 2 + 2 êë n n n úû 1 n 1 1 n 1 = lim å + lim å n®¥ n n®¥ n æ ö æ r rö r =1 ç r =1 ç 1+ ÷ 2 + ÷÷÷ ççè n ÷÷ø çèç nø 1

=ò 0

1

1 1 dx + ò dx 1+ x 2 x + 0 1

1

= [log(1 + x) ]0 + [log(2 + x) ]0



= log 2 - log1 + log 3 - log 2 = log 3. p 4

57. (a) Let I = log(1 + tan q ) d q. ò 0

p 4

Then, I = ò log(1 + tan q ) d q 0

p 4

é æp öù Þ I = ò log ê1 + tan çç - q ÷÷÷ú d q êë èç 4 øúû 0 p 4

1 - ( x -1) 2

= éëêsin -1 ( x -1)ùûú = 0 - sin -1 (-1) =

= 0 + lim

é 1 1 1 1 ù ú = lim + lim ê + + ..... + n®¥ n n®¥ ê n + 1 n+2 n + n úû ë é 1 1 1 ù ú + lim ê + + ..... + n®¥ ê 2n + 1 2n + 2 2n + n úû ë 1é n n n ù ú = 0 + lim ê + + ..... + n®¥ n ê n + 1 n+2 n + n ûú ë 1é n n n ù ú + lim ê + + ..... + n®¥ n ê 2n + 1 2n + 2 2n + n ûú ë

é 1 - tan q ù ú dq = ò log ê1 + ê úû 1 tan q + ë 0 p 4

é 2 ù ú dq = ò log ê ê ë1 + tan q úû 0 p 4

= ò [log 2 - log(1 + tan q ) ] d q 0

p 4

p 4

0

0

= log 2 ò d q - ò log(1 + tan q ) d q = Þ 2I =

p log 2 - I 4

p p log 2 Þ I = log 2. 4 8

8/9/2023 8:42:22 PM

= lim

n®¥

ê log + log + ........ + log ú n êë n n n úû

r 1 n log å n®¥ n n r =1

= lim 1

0

58. (c) f ( x)

ò

t 2 dt = x cos p x

0

f ( x)

é t3 ù Þê ú ê3ú ë û0

= x cos p x 3

Þ

{ f ( x )} 3

= x cos p x 1

Þ f ( x) = (3 x cos p x)3 Þ f ¢( x )



Þ f ¢(4) 2 1 = (3´ 4 cos 4p ) 3 [3cos 4p + 3´ 4 (- sin 4p )] 3 2 1 1 ¢ Þ f (4) = ´12 3 [3 + 0]= 2 . 3 12 3 ( cos 4p = 1,sin 4p = 0)

1



1

0

x ù d d éê Þ g (x) = êcosx- ò ( x - t ) g (t ) dt úú dx dx ëê 0 ûú

d Þ g ¢(x) = - sinx- ò ( x - t ) g (t ) dt dx 0

1

Þ lim log y x ®¥

x ®¥

1 n -1 n ù é1 2 log ê ´ ´ ...... ´ ´ ú n n nû ën n

1é 1 2 nù log + log + ... + log ú x ®¥ n ê n n nû ë

= lim

1 n r log å x ®¥ n n r =1

= lim 1



EMEP.CH02_3PP.indd 180

= ò log x dx = [ x (log x - 1)]0 = -1 0

x

g (x) = cosx- ò ( x - t ) g (t ) dt

x

æ n ! ön æ n ! ön Let y = çç n ÷÷÷ .Then y = çç n ÷÷÷ çè n ø çè n ø æ n!ö 1 Þ log y = log çç n ÷÷÷ çè n ø n é n ´ (n -1)´ (n - 2)´......´ 2´1ù 1 ú = log ê n n ´ n ´ n ´.........´ n ´ n ëê ûú

= lim

æ n ! ön 1 Þ lim çç n ÷÷÷ = . n®¥ ç èn ø e 60. (c)

59. (b)



é  x(log x -1) 1 = 1(log1-1) - lim x log x ]0 êë [ x ® 0+ log x æç ¥ö form ÷÷÷ = -1- lim ç çè x ® 0+ 1 ¥ø x æ 1 ö÷ çç ÷÷ = -1- lim ççç x ÷÷÷ (by the L'Hospital rule) x ® 0+ ç 1 çç - 2 ÷÷÷ è x ø = -1 + lim x = -1 + 0 = -1ùú x ® 0+ û Þ log éê lim y ùú = -1 Þ lim y = e-1 n®¥ ë n®¥ û

2 1 = (3 x cos p x) 3 [3cos p x + 3 x (-p sin p x )] 3



1

= ò log x dx = [ x(log x -1)]0 = -1

180 • Engineering Mathematics Exam Prep

1



éd x ù = - sinx- êê ò h( x, t ) dt úú êë dx 0 úû (assuming h( x, t ) = ( x - t ) g (t )) éx ¶ ù = - sinx- êê ò h( x, t ) dt + h( x, x) - 0úú êë 0 ¶x úû x

= - sinx- ò g (t ) dt 0

x ù d d éê ¢ g (x) = ê- sinx- ò g (t ) dt úú Þ dx dx ëê 0 ûú

Þ g ¢¢( x) = -cosx - {0 + g ( x)} Þ g ( x) + g ¢¢( x) = -cosx.

8/9/2023 8:42:23 PM

Calculus • 181



61. (c)

p 2

Let x n = sin 2 q.

2 Þ x = sin 2/ n q and dx = sin q cos q d q n Also x = 0 Þ sin q = 0 Þ q = 0 & p x = 1 Þ sin q = 1 Þ q = 2



=ò 0

1

(1- sin 2 q) p 2

2

1 n

2 ´ sin n

2 -1 n

p 2

2 2 -1 12 n n q q dq sin cos n ò0





EMEP.CH02_3PP.indd 181

p 2

æ1 ö çç + 1÷÷ æ 0 + 1ö÷ ÷÷ ç G çç 2 ÷÷ ´G ççç ÷ è 2 ÷ø çç 2 ÷÷ ÷ø 1 çè = æ1 ö 2 çç + 0 + 2 ÷÷ ÷÷ 2 G ççç 2 ÷÷ 2 çç ÷÷ èç ø÷ æ 3ö æ 3ö G ç ÷÷ ´ p G ç ÷÷´ p 1 èçç 4 ø÷ 1 èçç 4 ø÷ = = æ 5 ö÷ 4 4 æç 1 ö÷ ç G ç ÷÷ G ç + 1÷÷ çè 4 ø çè 4 ø æ 3ö æ 3ö G çç ÷÷÷ G çç ÷÷÷ ç p è 4ø èç 4 ø = = p . æ1ö 4 1 æç 1 ö÷ G ç ÷÷ G çç ÷÷÷ èç 4 ø 4 èç 4 ø p 2



p 2

ò

1 dq sin q

sin q d q ´ ò

0

0

p 2

1 2

p 2

-1

= ò sin q cos qdq ´ò sin 2 q cos0 qdq 0

0

0



æ1 ö æ 1 ö ç 2 +1 ÷ ç - 2 +1 ÷ æ 0 +1 ö æ 0 +1 ö Gç ÷´ Gç ÷ Gç 2 ÷´ Gç 2 ÷ 2 2 è ø è ø çç ÷÷ çç ÷÷ ø ø = è ´ è æ1 ö æ 1 ö ç 2 +0+2÷ ç-2 +0+2÷ 2G ç 2G ç ÷ ÷ 2 2 çç ÷÷ çç ÷÷ è ø è ø



æ3ö æ1 ö æ1 ö æ1 ö Gç ÷´ Gç ÷ Gç ÷´ Gç ÷ 4ø 2ø 4ø è è è è2ø = ´ æ5ö æ3ö 2G ç ÷ 2G ç ÷ è4ø è4ø

Then x 2 = sin q

x = 1 Þ sin q = 1 Þ q =

sin q d q

63. (a)

62. (c) Let x 2 = sin q.

1 cos q d q 2 sin q Also x = 0 Þ sin q = 0 Þ q = 0 &

1 cos q d q 2 sin q

p 2

-1

Þ x = sin q and dx =

´

1 1 = ò sin 2 q cos 0 q d q 2 0

q cos q d q

æ2 ö æ ö çç -1 + 1÷÷ çç1- 2 + 1÷÷ ÷÷ ÷÷ ç G ççç n ÷÷ ´G çç n ÷ çç çç 2 ÷÷ 2 ÷÷÷ ÷ø çè 2 çè ø÷ = æ2 ö 2 n ççç -1 + 1- + 2 ÷÷÷ n 2 G çç n ÷÷÷ çç 2 ÷÷ çè ø÷ æ 1ö æ1ö G çç ÷÷÷ ´G ççç1- ÷÷÷ p 2 çè n ø è nø 1 ( G(1) = 0! = 1) = ´ = n 2 G(1) n sin p n p p = cos ec n n

1 2 ò0

=

2 sin n q cos q = ò dq 2 n 0 n cos q =

1- sin q p 2

Therefore, given integral p 2

2

0

2 -1 n



sin q

=ò

Then x n = sin 2 q



Therefore, given integral

8/9/2023 8:42:24 PM

2 G çç ÷÷ 2 çç ÷÷ çè ø÷ æ1ö æ1ö æ1ö æ 3ö G çç ÷÷÷ ´G ççç ÷÷÷ G ççç ÷÷÷ ´G ççç ÷÷÷ çè 4 ø è 2ø è 4ø è 2ø ´ = æ 5 ö÷ æ 3 ö÷ 182 • Engineering 2 G çç ÷÷ Mathematics 2 G çç Exam ÷ Prep èç 4 ø èç 4 ø÷ 2 æ1ö p G çç ÷÷÷ çè 4 ø = =p 1 æç 1 ö÷ 4´ G ç ÷÷ 4 çè 4 ø

æ 1 ö æ 2 ö æ 3ö æ 4 ö æ8ö G çç ÷÷÷´G çç ÷÷÷´G çç ÷÷÷´G çç ÷÷÷´.........´G çç ÷÷÷ èç 9 ø èç 9 ø èç 9 ø èç 9 ø èç 9 ø æ æ 1 ö æ 8 öö æ æ 2 ö æ 7 öö æ æ 3 ö æ 6 öö = ççG çç ÷÷÷ G çç ÷÷÷÷÷÷´ççG çç ÷÷÷ G çç ÷÷÷÷÷÷´ççG çç ÷÷÷ G çç ÷÷÷÷÷÷ èç èç 9 ø èç 9 øø èç èç 9 ø èç 9 øø èç èç 9 ø èç 9 øø

( )

æ æ1ö ö çç G çç ÷÷ = p & G çæç 5 ÷ö÷ = G æçç 1 + 1÷ö÷ = 1 G æçç 1 ö÷÷÷÷ çè 4 ø÷ çè 4 ø÷ 4 èç 4 ø÷ø÷ çè èç 2 ø÷



æ æ 4 ö æ 5 öö ´ççG çç ÷÷÷ G çç ÷÷÷÷÷÷ èç èç 9 ø èç 9 øø æ æ 1 ö æ 1 öö æ æ 2 ö æ 2 öö = ççG ççç ÷÷÷ G ççç1- ÷÷÷÷÷÷´ççG ççç ÷÷÷ G ççç1- ÷÷÷÷÷÷ çè è 9 ø è 9 øø èç è 9 ø è 9 øø

64. (d) ¥

- x4

òe 0

¥

æ æ 3 ö æ 3 öö æ æ 4 ö æ 4 öö ´ççG çç ÷÷÷ G çç1- ÷÷÷÷÷´ççG çç ÷÷÷ G çç1- ÷÷÷÷÷ çè çè 9 ø çè 9 øø÷ èç çè 9 ø çè 9 øø÷

0

¥

1

= ò e- y y 2



4

x 2 dx´ ò e- x dx

0

1 - 34 y dy´ 4

¥

ò 0

e- y

1 - 34 y dy 4

=

1 3 é ù ê putting x = y 4 so that dx = 1 y 4 dy; ú ê ú 4 ê ú êë Also x = 0 Þ y = 0 & x = ¥ Þ y = ¥úû ¥

¥

1 3 1 = ò e- y y 4 dy´ ò e- y y 4 dy 16 0 0 ¥

=

¥

3 1 -1 -1 1 -y 4 -y 4 e y dy ´ e y dy ò ò 16 0 0

1 çæ 3 ÷ö æç 1 ö÷ 1 æç 1 ö÷ æç 1 ö÷ G ç ÷´G ç ÷ = G ç1- ÷´G ç ÷ 16 èç 4 ø÷ èç 4 ø÷ 16 èç 4 ø÷ èç 4 ø÷ 1 p p = ´ = p 16 sin 8 2 4 æ ö ççusing G(n) G(1- n) = p ÷÷ çè sin np ø÷ =



=



=

æ æ 4 ö æ 5 öö ´ç G ç ÷ G ç ÷ ÷ è è 9 ø è 9 øø

EMEP.CH02_3PP.indd 182

4p 4

æ p 2p ö 4p 3 çç2sin sin ÷÷÷ sin çè 9 9ø 9 4p 4

æ p 3p ö 4p 3 ççcos - cos ÷÷÷ sin çè 9 9ø 9

4p 4 = æ p 1ö 4p 3 ççcos - ÷÷÷ sin çè ø 9 2 9

æ æ 1 ö æ 8 öö æ æ 2 ö æ 7 öö æ æ 3 ö æ 6 öö = çGç ÷Gç ÷÷ ´ çGç ÷Gç ÷÷ ´ çGç ÷Gç ÷÷ è è 9 ø è 9 øø è è 9 ø è 9 øø è è 9 ø è 9 øø



p4

[ 2sin A sin B = cos ( A - B) - cos ( A + B)]

æ1 ö æ2ö æ3ö æ4ö æ8ö G ç ÷ ´ G ç ÷ ´ G ç ÷ ´ G ç ÷ ´ .... ´ G ç ÷ è9ø è9ø è9ø è9ø è9ø

æ æ3ö æ 3 öö æ æ 4 ö æ 4 öö ´ ç G ç ÷ G ç1 - ÷ ÷ ´ ç G ç ÷ G ç1 - ÷ ÷ 9 9 9 9 øø øø è è ø è è è ø è p p p p = ´ ´ ´ p 2p 3p 4p sin sin sin sin 9 9 9 9

´

3 2p 4p p sin sin sin 2 9 9 9 æ ö çç sin 3p = sin p = 3 ÷÷ ÷ çè 9 3 2 ÷ø

=

65. (a)

æ æ1 ö æ 1 öö æ æ 2 ö æ 2 öö = ç G ç ÷ G ç1 - ÷ ÷ ´ ç G ç ÷ G ç1 - ÷ ÷ 9 øø è è 9 ø è 9 øø è è9ø è

p p p ´ ´ 2p 3p 4p p sin sin sin sin 9 9 9 9 æ ö ççusing G(n) G(1- n) = p ÷÷ çè sin np ø÷ p

=

=



8p 4

æ p 4p 4p ö 3 ççç2sin cos - sin ÷÷÷ è 9 9 9ø 8p 4 æ 5p 3p 4p ö + sin - sin ÷÷÷ 3 ççsin çè 9 9 9ø

[ 2sin A cos B = sin ( A + B) + sin ( A - B)]

8/9/2023 8:42:26 PM

Calculus • 183

8p 4 8p 4 16p 4 = = 3p 3 3 3 sin 3´ 9 2

=

é æ 5p 4p ö 4p ê  sin = sin ççp - ÷÷÷ = sin ç êë è 9 9ø 9







(

3

3

u = log x + y + z - 3 xyz

ù ú úû

æ ö 1 w2 w + ¶ç + + 2 2 ÷ ¶y è x + y + z x + yw + zw x + yw + zw ø

)



{

}

= log ( x + y + z )( x + yw2 + zw )( x + yw + zw2 )

æ ö 1 w w2 + ¶ç + + 2 2 ÷ ¶z è x + y + z x + yw + zw x + yw + zw ø =-

1 1 2 ( x + y + z) x + yw2 + zw

-

1 w4 ( x + y + z )2 x + yw2 + zw

-

1 w2 ( x + y + z )2 x + yw2 + zw

(where ω is the cube root of unity) 2



= log( x + y + z ) + log( x + yw + zw ) + log( x + yw + zw2 ) 1 1 1 \ ¶u = + + , ¶x x + y + z x + yw2 + zw x + yw + zw2 ¶u = w2 w 1 + + , ¶y x + y + z x + yw2 + zw x + yw + zw2 ¶u = w w2 1 + + . ¶z x + y + z x + yw2 + zw x + yw + zw2



Hence, ¶u + ¶u + ¶u ¶x

¶y

¶z

2 2 3 + 1+ w 2 + w + 1+ w + w 2 x + y + z x + yw + zw x + yw + zw 3 é 1 + w + w2 = 0 ù = ë û x y z + +

=



∴ (a) is true. 2

Now, æç ¶ + ¶ + ¶ ö÷ u è ¶x ¶y ¶z ø æ öæ ö = ç ¶ + ¶ + ¶ ÷ ç ¶u + ¶u + ¶u ÷ è ¶x ¶y ¶z ø è ¶x ¶y ¶z ø



æ öæ 3 ö =ç ¶ + ¶ + ¶ ÷ç è ¶x ¶y ¶z ø è x + y + z ÷ø



æ ö ¶æ ö ¶æ ö 3 3 3 = ¶ç + + ¶x è x + y + z ÷ø ¶x çè x + y + z ÷ø ¶z çè x + y + z ÷ø



9 =( x + y + z)2



So option (b) is true.

Again,

EMEP.CH02_3PP.indd 183

¶2 u + ¶2 u + ¶2 u ¶x 2 ¶y 2 ¶z 2

( )

æ ö 1 1 1 = ¶ç + + 2 2 ¶x è x + y + z x + yw + zw x + yw + zw ÷ø

66. (d) 3

( )

æ ö = ¶ ¶u + ¶ ç ¶u ÷ + ¶ ¶u ¶x ¶x ¶y è ¶y ø ¶z ¶z



(

(

=-

(

( w

3 ( x + y + z )2

4

-

1

-

w2

-

w4

) ( x + yw + zw ) 2

2 2

) ( x + yw + zw ) 2

2 2

) ( x + yw + zw ) = w, 1 + w + w = 0 ). 2

2 2

2

So option (c) is true.

67. (b) V = log(x2 + y2 + z2) (given) \¶V = 2 2 x2 2 , ¶x x + y + z 2y ¶V = , 2 ¶y x + y 2 + z 2 2z ¶V = . ¶z x 2 + y 2 + z 2

( )

2 æ ö Now ¶ V = ¶ ¶V = ¶ ç 2 2x2 2 ÷ , 2 x x x ¶ ¶ ¶ ¶x è x + y +z ø

(x = =





2

)

+ y 2 + z 2 ´ 2 - 2 x´ 2 x

(x

2

+ y2 + z2

2 y 2 + 2z 2 - 2x2

(x

2

+ y2 + z2

)

2

)

2

.

Similarly we can get, 2 2 2 2 2 2 ¶2V = 2x + 2z - 2 y , ¶2V = 2x + 2 y - 2z . 2 2 2 2 2 2 ¶y ¶z x + y +z x2 + y 2 + z 2

(

)

2 2 2 Hence, ¶ V +¶ V +¶ V 2 2 2

¶x

¶y

¶z

8/9/2023 8:42:31 PM

184 • Engineering Mathematics Exam Prep = =



(

) (

4 x2 + y 2 + z 2 - 2 x2 + y 2 + z 2

(

(

x2 + y 2 + z 2

)

2

)

69. (a)

2 . 2 x + y2 + z2

)

68. (c) Given x = rcosθ, y = rsinθ, x2 + y2 = r2.

y \2 x = 2r ¶r , 2 y = 2r ¶r i.e; x = ¶r , = ¶r . ¶x ¶y r ¶x r ¶y



¶ 2r So 2 = ¶x 2

¶r So 2 = ¶x



r ´1 - x ´ r2 r ´1 - x ´



¶r r - x 2 ¶x = r æ ¶r = x ö , r 2 çè ¶x r ÷ø ¶r r - x 2 ¶x r 2 r2

r2 ¶r y2 r ´1 - y ´ r2 ¶r ¶y r = = r2 r2 ¶y2



æ ¶r y ö ç ¶y = r ÷ . è ø

¶u

¶r

Similarly, it can be shown that ¶y = f ¢(r )´ ¶y . Then,

( ) {

}

{ }

æ yö = x1 f ç ÷ (say) è xø



x ¶ (sin u ) + y ¶ (sin u ) =1´sin u ¶x ¶y



or, x é ¶ (sin u )ù ¶u + y é ¶ (sin u )ù ¶u = sin u ëê ¶u ûú ¶x ëê ¶u ûú ¶y



or, x cosu ¶u + y cosu ¶u = sin u ¶x ¶y ¶ u ¶ u or, x + y = tan u. ¶x ¶y

Þ ¶u = ¶ f (r ) = ¶ { f (r )}´ ¶r = f ¢(r )´ ¶r . ¶x ¶x ¶r ¶x ¶x



2 ìï æ y ö 2 üï ì ü x 2 í1+ ç ÷ ý ï1+ æ y ö ï ç ÷ x è ø ï þï = x ï è x ø ï = î í ý y ï 1+ y ï x 1+ x ï x ïî þ

∴ sinu is a homogeneous function of degree “1.” So by Euler’s theorem,

æ ¶r x ö ç ¶x = r ÷ , è ø

Now u = f (r)

æ x2 + y 2 ö x2 + y 2 u = sin -1 ç Þ sin u = ÷ x+ y è x+ y ø

70. (b)



¶ 2u = ¶ ¶u = ¶ f ¢(r ) ¶r ¶x ¶x 2 ¶x ¶x ¶x

x = eϕ + e–ψ Þ ¶x = ej , ¶x = -e -y ,



2 = f ¢(r )´ ¶ r2 + ¶r ´ ¶ { f ¢(r )}´ ¶r ¶x ¶x ¶x ¶r

y = e–ϕ – eψ Þ ¶y = -e -j , ¶y = -ey .



= f '(r ) ´

¶j

2

¶ 2r æ ¶r ö + f ''(r ) ´ ç ÷ . ¶x 2 è ¶x ø

¶j

¶j ¶x ¶j ¶y ¶j

(

¶ 2u ¶ 2r æ ¶r ö + f '(r ) ´ 2 f ''(r ) ´ ç ÷ . 2 ¶y ¶y è ¶y ø



¶u = ¶u ´ ¶x + ¶u ´ ¶y ¶y ¶x ¶y ¶y ¶y



= ¶u ´ -e -y + ¶u ´ -ey .........(ii ) ¶x ¶y

2 2 2 ü ìï ì 2 æ ö üï = f ¢(r )´ í ¶ r2 + ¶ r2 ý + f ¢¢(r )´ í ¶r + ç ¶r ÷ ý ¶x è ¶y ø þï î ¶x ¶y þ îï



∴ (i) – (ii) ⇒

1 2r - ( x 2 + y2 ) ì x 2 y2 ü r + f ''(r ) ´ í 2 + 2 ý f '(r ) ´ 2 r r þ îr



¶u - ¶u = ¶u ej + e -y + ¶u ey - e -j ¶j ¶y ¶x ¶y



= x ¶u - y ¶u ¶x ¶y

2

2

¶x

¶y

So, ¶ u + ¶ u 2 2



EMEP.CH02_3PP.indd 184

)

¶u j ¶u -j = ¶x ´ e + ¶y ´ -e .........(i ) 2



¶y

Then ¶u = ¶u ´ ¶x + ¶u ´ ¶y

Similarly,



¶y

( )

1 = f '(r ) + f ''(r )  x 2 + y2 = r 2 r

(

)

(

)

( )

(

)

(

)

[ x = eϕ + e–ψ, y = e–ϕ – eψ]

8/9/2023 8:42:36 PM

Calculus • 185

71. (c) u = f (2x – 3y, 3y – 4z, 4z – 2x) (given) Let, p = 2x – 3y, q = 3y – 4z, r = 4z – 2x Then ¶p = 2, ¶p = -3, ¶p = 0; ¶x ¶y ¶z

¶q ¶q ¶q = 0, = 3, = -4; ¶x ¶y ¶z



¶r = -2, ¶r = 0, ¶r = 4. ¶x ¶y ¶z

¶u = ¶u ´ ¶p + ¶u ´ ¶q + ¶u ´ ¶r ¶x ¶p ¶x ¶q ¶x ¶r ¶x = ¶u ´ 2 + ¶u ´ 0 + ¶u ´ (-2) ¶p ¶q ¶r æ ö = 2ç ¶u - ¶u ÷ è ¶p ¶r ø ¶u = ¶u ´ ¶p + ¶u ´ ¶q + ¶u ´ ¶r ¶y ¶p ¶y ¶q ¶y ¶r ¶y = ¶u ´ (-3) + ¶u ´3 + ¶u ´ 0 ¶p ¶q ¶r æ ö = 3ç ¶u - ¶u ÷ è ¶q ¶p ø



¶u = ¶u ´ ¶p + ¶u ´ ¶q + ¶u ´ ¶r ¶z ¶p ¶z ¶q ¶z ¶r ¶z = ¶u ´ 0 + ¶u ´ (-4) + ¶u ´ 4 ¶p ¶q ¶r æ ö = 4ç ¶u - ¶u ÷ r q ¶ ¶ è ø



\ 1 ux + 1 u y + 1 uz 2 3 4 æ ¶u ¶u ö æ ¶u ¶u ö æ ¶u ¶u ö = ç - ÷ + ç - ÷ + ç - ÷ = 0. è ¶p ¶r ø è ¶q ¶p ø è ¶r ¶q ø





¶u ¶y ¶v ¶y

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

EMEP.CH02_3PP.indd 185

(1+ y ) ´ = (1- xy )

2

(

)

2 1 - 1+ x ´ 1 = 0. 2 1+ y 2 (1- xy ) 1+ x 2

(

)

(

)

∴  u and v are functionally related.

Now, v = tan -1 x + tan -1 y = tan -1 æç x + y ö÷ = tan -1 u è 1- xy ø \u = tan v . 73. (a) y y Let v = xf çæ ÷ö and w = g çæ ÷ö . è xø è xø Then clearly v is a homogeneous function of degree “1” and w is a homogeneous function of degree “0.” So by Euler’s theorem, we have x ¶v + y ¶v =1´ v .................(i ) ¶x ¶y and x ¶w + y ¶w = 0´ w = 0...............(ii ) ¶x ¶y



Now (i) + (ii) ⇒



æ ö x ¶v + ¶w + y ç ¶v + ¶w ÷ = v + 0 ¶x ¶x ¶ y ¶ y è ø

(

)

æ yö or , x ¶u + y ¶u = xf ç ÷ ¶x ¶y è xø

  u = v + w ⇒ é ¶u ¶v ¶w ¶u ¶v ¶w ù êëu = v + w Þ ¶x = ¶x + ¶x and ¶y = ¶y + ¶y úû 74. (b) Let f ( x, y ) = x3 + y 3 + 3xy -1 . Then ¶f = 3x 2 + 3 y and ¶f = 3 y 2 + 3x . ¶x ¶y

(

72. (d) ¶u ¶ (u ,v) ¶x = ¶ ( x, y ) ¶v ¶x

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

2



Again, u = f (p, q, r) and so





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



) ( (

) )

2 2 ¶f ¶x - 3 x + 3 y - x + y dy \ == = ¶f ¶y dx 3 y 2 + 3x x + y2

75. (d) Let f ( x, y ) = x 2 + y 2 + 2 xy -1 . (1- xy )´1- ( x + y )´ (- x) (1- xy ) 2 1 1+ y 2

Then ¶f = 2 x + 2 y and ¶f = 2 y + 2 x . ¶x ¶y 2 x + 2 y ( ) = -1. ¶f ¶x dy \ == dx ¶ f ¶ y 2 y + 2 x

8/9/2023 8:42:41 PM

186 • Engineering Mathematics Exam Prep Hence, du = ¶u + ¶u ´ dy dx ¶x ¶y dx ì ü = ílog( xy ) + x ´ 1 ´ yý+ x ´ 1 ´ x ´ (-1) xy xy î þ = log( xy ) +1- x . y 76. (d)



At (0,0):

A=

¶f = y (1- x - y ) + xy ´ (-1) ¶x = y - 2 xy - y 2 , ¶f = x(1- x - y ) + xy ´ (-1) ¶y

æ ö çè 3 3 ø÷

A=

¶ f ¶2 f = = 1- 2 y - 2 x y 2 , ¶x 2 ¶x ¶y

¶f ¶f = 0, =0 ¶x ¶y x - 2 xy - x 2 = 0...............(ii)

çè 3 3 ÷ø

Therefore, f ( x, y ) has a maximum value at

æ 1 1 ö÷ çç , ÷ . çè 3 3 ø÷

Case-II: x + y = 1 2 Equation (i) ⇒ y - 2 y (1- y ) - y = 0

or, y 2 - y = 0

or, either x = y or x + y = 1

2 2 Equation (i)⇒ y - 2 y - y = 0

or, y ( y -1) = 0 or, y = 0,1

y = 0 Þ x = 1 & y = 1 Þ x = 0 ( x + y = 1)

Therefore, in this case the points are (1, 0) and



At (1,0):

(0,1)

A=

1 3 1 1 y = 0 Þ x = 0 & y = Þ x = ( x = y ) 3 3

or, y = 0,

Therefore, in this case the points are (0,0) and

æ 1 1 ö÷ çç , ÷ çè 3 3 ø÷

EMEP.CH02_3PP.indd 186

2 =- . æç 1 1 ö÷ 3 ç , ÷

4 1 \ AC- B2 = - > 0 and A < 0 9 9

or, y (3 y -1) = 0



¶2 f ¶y 2

çè 3 3 ÷ø

x - y - ( x2 - y 2 ) = 0

or, 3y 2 = y



çè 3 3 ÷ø



Case-I: x = y

2 ¶2 f 1 ==- , B = 3 3 ¶x ¶y ççæ1 , 1÷÷ö ççæ 1 , 1 ÷÷ö

Subtracting (i) from (ii) we get,

or, ( x - y ) (1- x - y ) = 0

¶2 f ¶x 2

and C =

¶2 f = -2 x. ¶y 2

Þ y - 2 xy - y 2 = 0...............(i)



= 0. (0,0)

1 1 At çç , ÷÷ :

= x - 2 xy - x ,



¶2 f =1 ¶x ¶y (0,0)

\ AC- B = -1 < 0. Therefore, f ( x, y ) has neither a maximum nor a minimum at (0, 0).

2

\

(0,0)

¶2 f and C = 2 ¶y

2



= 0, B =

2

Here,

and

¶2 f ¶x 2

¶2 f ¶x 2

and C =

= 0, B = (1,0)

¶2 f ¶y 2

¶2 f = -1 ¶x ¶y (1,0)

= -2. (1,0)

2

\ AC- B = -1 < 0.

Therefore, f ( x, y ) has neither a maximum nor a minimum at (1, 0).

8/9/2023 8:42:45 PM

Calculus • 187



At (0, 1):

A=

¶2 f ¶x 2

= -2, B = (0,1)

¶2 f and C = 2 ¶y

78. (a) Let P( x, y, z ) be any point on the given sphere and A(1, 2,3) be the given point. Then,

¶2 f = -1 ¶x ¶y (0,1)

AP = (x-1) 2 + ( y - 2) 2 + ( z - 3) 2 . 2

= 0.

\ AP = (x-1) 2 + ( y - 2) 2 + ( z - 3) 2 = f ( x, y, z ) ( say )

(0,1)

2

\ AC- B = -1 < 0.

Let f( x, y, z ) = x 2 + y 2 + z 2 -1. \ F ( x, y , z ) = f ( x, y, z ) + lf( x, y, z )

Therefore, f ( x, y ) has neither a maximum nor a minimum at (0, 1). æ ö Hence, f ( x, y ) has a maximum at çç 1 , 1 ÷÷ only çè 3 3 ø÷ but it does not have any minimum value.

= (x-1) 2 + ( y - 2) 2 + ( z - 3) 2 + l ( x 2 + y 2 + z 2 -1)

77. (a) Here f ( x, y, z ) = x 2 + y 2 + z 2 ,

f( x, y, z ) = x + 2 y + 3 z - 4. \ F ( x, y , z ) = f ( x, y, z ) + lf( x, y, z ) 2



2

¶F ¶F ¶F = 0, = 0 and =0 ¶x ¶y ¶z Þ 2( x -1) + 2l x = 0, 2( y - 2) + 2l y = 0, 2( z - 3) + 2l z = 0 Then,



Þ -l =

2

= x + y + z + l ( x + 2 y + 3 z - 4) ¶F ¶F ¶F = 0, = 0 and =0 Then, ¶x ¶y ¶z

=

Þ 2 x + l = 0, 2 y + 2l = 0, 2 z + 3l = 0



2x 4 y 6z 2x + 4 y + 6z Þ = = = 1 4 9 1+ 4 + 9 2( x + 2 y + 3 z ) = 14 8 4 = = 14 7 [ x + 2 y + 3 z = 4]

2 4 6 Þ x = , y = and z = . 7 7 7



2 2 2 Therefore, minimum value of x + y + z is

2

2

Þx= z=

(x- 1) 2 + ( y - 2) 2 + ( z - 3) 2 x2 + y 2 + z 2 f ( x, y , z ) = 1 1

1+

f ( x, y , z ) 3

1+

f ( x, y , z )

f ( x, y , z ) ,y=

2 1+

f ( x, y , z )

,

.

Then, x 2 + y 2 + z 2 = 1 14 Þ =1 2 1 + f ( x, y , z )

(



Þ 1+

)

f ( x, y, z ) = ± 14

Then, x2 + y2 + z2 = 1 Þ

2

æ 2 ö÷ æ 4 ö÷ æ ö çç ÷ + çç ÷ + çç 6 ÷÷ i.e; 8 . çè 7 ÷ø çè 7 ÷ø çè 7 ÷ø 7

EMEP.CH02_3PP.indd 187

=

2x y 2z Þ -l = = = 1 1 3 2x y 2z 2x 4 y 6z Þ = = Þ = = 1 1 3 1 4 9

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



(

14

1 + f ( x , y, z )

)

2

=1

Þ 1 + f ( x , y, z ) = ± 14

8/9/2023 8:42:48 PM

188 • Engineering Mathematics Exam Prep



Þ

f ( x, y, z ) = ± 14 -1

Þ

f ( x, y, z ) = 14 -1, 14 + 1

Hence, the minimum and maximum values of AP are, respectively, 14 -1 and 14 + 1 . 79. (b)

ò ò ò(x

2



x + y + z = 1}.



Let us put, x + y + z = u, x + y = uv, y = uvw.

Then, y = uvw, x = uv – y = uv – uvw = uv(1 – w).

v

é 2 ìï 2 2 üï ù 2 2 ê òx =0 ê yò=0 íîï zò=0 x + y + z dz ýþï dy úú dx ë û 2 2 æ 2 3 ïì é ù ïü ö = ò ç ò í ê x 2 z + y 2 z + z ú ý dy ÷ dx ç ïë 3 û z =0 ï ÷ x =0 è y =0 î þ ø =

= =

(

)

(

)

ìï 2 8 dy üï dx 2 2 + + 2 2 x y í ý ò ï yò=0 3 ïþ x =0 î 2

2

é

ò êë 2 x

2

y+

x =0 2

{

)

1 x y

òòòx

3

= u 2 v - u 2 vw + u 2 v 2 w + u 2 vw - u 2 v 2 w

ìï x æ y ö üï = ò í ò ç ò x 3 y 2 z dz ÷ dy ý dx ç ÷ x =0 ï ø ïþ î y =0 è z =0 1

ìï x é 3 2 z 2 ù y üï òx =0 íï yò=0 êë x y 2 úû z =0 dy ýï dx î þ 1 ì x 2 y ï ïü = ò í ò x3 y 2 ´ dy ý dx 2 ï ï y =0 x =0 î þ =



1

1





3 = ò x 2 x=0

x

1 3 é y5 ù x ´ x 5 dx dx = ê5ú ò 10 ë û0 0

EMEP.CH02_3PP.indd 188

é 9ù = 1 êx ú = 1 . 10 ë 9 û 0 90

Þ u = 1,

x = 0 Þ uv(1 - w) = 0 Þ u = 0, v = 0, w = 1 y = 0 Þ uvw = 0 Þ u = 0, v = 0, w = 0



z = 0 Þ u (1 - v) = 0 Þ u = 0, v = 1



Thus, the new region of integration is given by R ¢= {(u , v, w) : 0u1, 0v1, 0w1}.



Hence, the given integral =



1

1

1

ò ò ò u ´uv(1- w)´uvw´u (1- v)´u vdudvdw 2

u =0 v=0 w=0

( ( dxdydz = Jdudvdw = u 2 v dudvdw )

1



= u 2v

Now, x + y + z = 1

y 2 z dzdydx

0 0 0

2

+ u 2 v 2 w + u 2 vw(1 - v)



80. (c)

- uv {-uvw - uw(1 - v)}

= u v (1 - w) + u v(1 - w) - u 2 v 2 (1 - w)

2



}

2 2

= ò 4 x 2 + 16 + 16 dx 3 3 0 é 3 ù = ê 4 x + 16 x + 16 x ú 3 3 3 ë û0 4 2 16 ´ = ´8 + ´ 2 = 32. 3 3

¶u ¶w ¶y ¶w ¶z ¶w

= v(1 - w) 0 + u 2 v - u (1 - w) {0 - uv(1 - v)}

2y 8 ù + y ú dx 3 3 û y =0

(

¶x ¶v ¶y ¶v ¶z ¶v

v(1 - w) u (1 - w) -uv uw uv = vw -u 1- v 0



2

3

z = u – x – y = u(1 – v). ¶x ¶u ¶y \ Jacobian, J = ¶u ¶z ¶u

)

+ y 2 + z 2 dxdydz

2

81. (a) Here, V = {(x,y,z) : x = 0, y = 0, z = 0,

=



1

ò

u =0

u 6 du ´

1

ò

v =0

v 3 (1 - v)dv ´

1

ò

w(1 - w)dw

w=0

8/9/2023 8:42:52 PM

Calculus • 189 1

1

85. (b) Required volume of the solid of revolution

1

5 3 é 7ù é 4 ù é 2 ù = êu ú ´ êv - v ú ´ ê w - w ú 3 û0 ë 7 û0 ë 4 5 û0 ë 2



(

¥

)( )

= 1´ 1 -1 ´ 1 -1 = 1´ 1 ´1 = 1 7 4 5 2 3 7 20 6 840 .



¥

= p ò y 2 dx = p ò



-¥

-¥

¥

= 2p ò

82. (d) We know that m -1 n -1 l -1 ò ò ò x y z dxdydz =



V

G(m)G(n)G(l ) , G(m + n + l )



where V = {( x, y, z ) : x0, y0, z0, x + y + z1}



∴  Given integral

=

òòòx

4 -1 5 -1 3-1

y

z

= =



dxdydz

p 2

= 2p ò

G(4)G(5)G(3) (here, m = 4, n = 5, l = 3) G(4 + 5 + 3) 3! ´ 4! ´ 2!  n = (n - 1)! for n Î Z + 11!

(

)







ò ò 0

dzdydx =

0

2 4-2 x 4-2 x- y

ò ò

x =0 y =0

ò

dzdydx

z =0

84. (b)

ìï x üï 2 y é ù + yz z dy í ò òë û z=0 ý dx ï y =0 x =0 î þï 1

Given integral =

=

1



EMEP.CH02_3PP.indd 189

2

2

´ 2sec2 q d q

x = 0 ⇒ θ = 0; x = ∞ ⇒ θ = π/2 ] p 2

2 = 256p ò sec qd4 q =16p ò cos2 qd q 16sec q 0 0 p 2

= 8p ò 2cos q d q= 8p ò (1+ cos2q) d q 2

0

0

= 8p éq+ sin2q ù 2 ûú0 ëê = 8p é p + sin p - 0 - 0ù = 4p2 êë 2 úû

(



2

( sin p = 0)

0 2

0

x2 (2 - x) dx 2+ x

x2 { (2 + x) - 2 x} = pò dx 2+ x 0

= pò 0

2

3 x2 (2 + x) dx - 2p ò x dx (2 + x) x+2 0

(

)

x2

2

2

ìï x 2 2 üï ò í ò y + y dyýdx ïþ x=0 ï îy=0 1

(4tan q+ 4) 2

= p ò y 2 dx = p ò

So, l ( x) -m ( x, y ) = y .

2

64

Hence, required volume of the solid of revolution

Hence, l ( x) = 4 - 2 x, m( x, y ) = 4 - 2 x - y .



dx

86. (b) Clearly x varies from 0 to 2.

2 l ( x) m ( x, y )

0

dx

p 2

⇒x=2



0

p 2



2x + y + z = 4⇒ 2x = 4 (for y = 0, z = 0)

\ò

2

p 2

Again, 2x + y + z = 4⇒ y = 4 – 2x (for z= 0)



)

x +4

2

[ the integrand is an even function]

83. (d) 2x + y + z = 4 ⇒ z = 4 – 2x – y

(

64 2

(x + 4)

[putting x = 2tanθ so that dx = 2sec2θdθ;

2! ´ 3! ´ 4! = . (11)!







V



0

64

2

1

1 é 2 y3 ù 2 x6 dx = 2 é x7 ù =òê = dx ú ò 3 û 3 3 êë 7 úû0 x=0 ë 0 y=0 = 2´1 = 2 . 3 7 21



)

2 x3 + 23 - 8 ìï 2 üï 2 = p íò x dx - 2ò dx ý ( x + 2) ïî 0 ïþ 0

(

)

2 2 ( x + 2) x2 - 2 x + 4 2 ïìé 3 ù ïü = p í ê x ú - 2ò dx + 16ò dx ý x + 2ï ( x + 2) ïîë 3 û0 0 0 þ

8/9/2023 8:42:57 PM

190 • Engineering Mathematics Exam Prep 2

(

)

= 8p - 2p ò x2 - 2 x + 4 dx + 16p [ log( x + 2)]0 3 0

2

2

é 3 ù = 8p - 2p ê x - x2 + 4 x ú + 16p ( log 4 - log 2) 3 3 ë û0 = 8p - 2p é 8 - 4 + 8ù + ( 32log 2 - 16log 2) p êë 3 úû 3 = - 32p + 16p log 2 3 = 16p log 2 - 2 3

(



Here the end points of the latus rectum arc (a, 2a) and (a, –2a). So x varies from “0” to “a.” Hence, the required surface area of the solid of revolution



0

a

= 2p ò y 1 + a dx x 0

)

y

2

a

æ dy ö = 2p ò y 1 + ç ÷ dx è dx ø



æ dy aö 2 ç  y = 4ax Þ y = 2 a x Þ dx = ÷ xø è



= 2p ò 2 a x a + x dx x 0

a

x

(2, 0)

a

a

x = –2

= 4p a ò

87. (a) Required volume of solid of revolution = 2p 3



= 2p 3



p

ò

3 2pa 3 ò {a (1 + cos q)} sin qd q = - 3 ò z dz 2

q= 0

3 ù é 2 ( x + a ) ú = 4p a ê ê 3 ú ëê 2 ûú0



3 3 ì ü = 4p a ´ 2 í(2a ) 2 - a 2 ý 3î þ



2 = 8p a 2 2 a a - a a = 8pa 2 2 - 1 . 3 3

2 0

p

0



r 3 sin qd q

q= 0

1 ( x + a ) 2 dx

(putting 1 + cosθ = z, so that –sinθdθ = dz and θ = 0 ⇒ z = 1 + 1 = 2, θ = π ⇒ z = 1 – 1 = 0)

(

)

89. (a)

0

(

(0,a)

)

( –a,0)

y

2

a

y

O

x

Clearly x varies from 0 to a in first quadrant. Then the surface area of the solid of revolution



(a ,2a) B A (a,0) C (a, – 2 a)

EMEP.CH02_3PP.indd 190

(a, 0) (0,–a)

x

88. (c)

)

y

2 é 4ù = - 2pa ê z ú 3 ë 4 û2 2 2 = - 2pa ´ 0 - 16 = 8pa . 3 4 3



(

æ dy ö = 2 ´ 2p ò y 1 + ç ÷ dx è dx ø 0

a

x





3

2

2 2 æ 2 ö y3 = 4p ò ç a 3 - x 3 ÷ 1 + 2 dx ø 0è x3 2 2 2 -1 - 1 dy æ 3 + y3 = a3 Þ 2 x 3 + 2 y 3 x  =0 çè dx 3 3

8/9/2023 8:43:01 PM

Calculus • 191

Þ

a



p

dy y1 3 ö =- 13÷ dx x ø æ 23 a

= 4p ò ç 0è

-

2 x3

= 2p ò a (1 + cos q)sin q a2 (1 + cos q)2 + a2 sin 2 qd q 0

3 ö2

÷ ø



2 y3

+

1 x3 3

a

2 x3

p

dx



1

3 2

1 3



2 2 é 2 3 3 3 êputting x = a sin q so that x = a sin q ê 2 êdx = 3a sin q cos qd q, x = 0 Þ q = 0; ê p êx = a Þ q = 2 ë

(

p 2

(

)

(

)

)

ù ;ú ú ú ú ú û

)

p 2

= 12pa2 ò cos3 q ´ sin q cos q d q 0 0



= 12pa2 ò t 4 (- dt ) 1



(putting cos q = t so that - sin qd q = dt and q = 0 Þ t = 1, q = p Þ t = 0) 2



2 é 5ù = -12pa ê t ú = 12pa . 5 ë 5 û1 2





EMEP.CH02_3PP.indd 191

p

ò r sin q

q= 0

( )

2

2



0



2

= 32pa . 5 Previous Years Solved Papers (2000-2018)

1. Limit of the function 1 - a4 f ( x) = 4 as x ® ¥ is given by x 4 (a) 1 (b) e - a (c) ¥ (d) 0  [GATE 2000] 2. Limit of the following series as x approaches p is 2

r 2 + dr d q dq

3

= 2pa2 ò 2 t 2 (-dt )

é 5ù 5 2 êt 2 ú = -2 2pa = 2 2pa2 ´ 2 ´ 2 2 ê5ú 5 êë 2 úû2

x3 x5 x7 f ( x) = x - + - + ........¥ 3! 5! 7!

p (a) 2p (b) (c) p (d) 1 2 3 3  [CE GATE 2001]

é p öù æ ê sin 2 ç x - 4 ÷ ú è øú = ? 3. lim ê p p x® ê ú x4 êë úû 4

Also, r = a (1 + cosθ) Þ dr = - a sin q . dq ∴  Required surface area of the solid of revolution = 2p

0



0

90. (d) Here, x-axis is the initial line and θ varies from 0 to π.

3

ò 2 (1 + cos q) 2 sin qd q

(putting 1 + cos q = t so that - sin qd q = dt and q = 0 Þ t = 2, q = p Þ t = 0)

3

ì 2 ü2 = 12pa ò ía 3 1 - sin 2 q ý sin q cos qd q þ 0î

(

p

0



2 3

2



æ ö ´ 3a sin 2 q cos qd q = 4pò ç a - a sin 2 q ÷ 1 a è ø 0 a 3 sin q 2 3

0

= 2pa

2 2 3 2 2 2 æ 2 ö æ ö = 4p ò ç a 3 - x 3 ÷ a 1 dx ç  x 3 + y 3 = a 3 ÷ ø x3 è ø 0è p 2

= 2pa2 ò (1 + cos q) sin q 1 + 1 + 2cos q d q



(a) 0

(b)

1 2

(c) 1

(d) 2

[IN GATE 2001]

8/9/2023 8:43:07 PM

192 • Engineering Mathematics Exam Prep 12. The expression e- ln x for x > 0 is equal to

sin 2 x =? x ®0 x (a) 0 (b) ¥

4. lim 

(c) 1

(d) -1 [GATE 2003]

(a) 0



(b)

æq ö sin ç ÷ è 2 ø is 6. lim q ®0 q



(a) 

æ x2 ö e - ç1 + x + ÷ 2 ø è 7. lim =? 3 x ®0 x

(b)



1 1 (c) (d) 1 6 3 [ME GATE 2007]

8. If y = x + x + x + x + .....¥ then y (2) = ? 

(a) 4 or 1 (c) 1 only

(b) 4 only (d) Undefined [ME GATE 2007]

9. What is the value of Lt cos x - sin x ? p p x® x4 4 (a) 2 (b) 0 (c) - 2 (d) limit doesn’t exist  [PI GATE 2007]



EMEP.CH02_3PP.indd 192

2 3

3 (d) ¥ 2 [CE GATE 2010]

(b) 1

(c)

(a) 0

(b) e -2

(c) e

-

1 2

2n

?

(d)1

[CS GATE 2010]

16. What is Lt sin q equal to? q ®0 q (a) q (b) sinq (c) 0 (d) 1  [ME GATE 2011] x 17. The value of Lt æç x + 1 ö÷ is x ®¥ xø è (a) ln2 (b) 1.0 (c) e (d) ¥  [EC GATE 2014] 18. Lt æç x + sin x ö÷ equals to x ®¥ x è ø -¥ (a) (b) 0 (c) 1 (d) ¥  [CE GATE 2014] 19. Given x(t ) = 3sin(1000 p t ) and

(b) -1 (c) ¥ (d) -¥ [CS GATE 2008]

11. Lt sin x is x ®0 x (a) indeterminate (c) 1 

[ME GATE 2008]



x - sin x =? x ®¥ x + cos x

(a) 1

1 1 1 1 (b) (c) (d) 4 16 12 8

15. What is the value of Lt æç1 - 1 ö÷ n ®¥ è nø

10. Lt 

x -2 is x -8



[EC GATE 2007]

(d) - x -1

[IN GATE 2008]

1 3

æ2 ö sin ç x ÷ è 3 ø is 14. The Lt x ®0 x

(a) (c) 2

(c) x -1



x

(a) 0

(b) x

x ®8

-1 1 (c) (d) ¥ 7 7 [GATE 2004]

(a) 0.5 (b) 1 (d) not defined



(a) -x

13. The value of Lt

5. The value of the function, x3 + x is f ( x) = lim 3 x ®0 2 x - 7 x 2



(b) 0 (d) ¥ [IN GATE 2008]



pö æ y (t ) = 5cos ç1000 p t + ÷ 4ø è The x-y plot will be

(a) a circle (b) a multiloop closed curve (c) a hyperbola (d) an ellipse [IN GATE 2014]

8/9/2023 8:43:17 PM

Calculus • 193 2x

æ 1ö 20. Lt ç1 + ÷ is equal to x ®¥ è xø -2 2 (a) e (b) e (c) 1 (d) e  [CE GATE 2015]

21. A function f ( x) is linear and has a value of 29 at x = -2 and 39 at x = 3. Find its value x = 5. (a) 59 (b) 45 (c) 43 (d) 35  [CE GATE 2015] 22. Choose the most appropriate equation for the function drawn as a thick line in the plot below:

log e (1 + 4 x) is equal to? x ®0 e3 x - 1 1 4 (a) 0 (b) (c) (d) 1 12 3 [ME GATE 2016]

28. Lt 

3 29. The value of lim æç x - sin x ö÷ is x ®0 x è ø (a) 0 (b) 3 (c) 1 (d) -1  [EE GATE 2017] x ö 30. lim æç tan ÷ is equal to ___? 2 x ®0 x - x è ø  [CE GATE 2017]

31. Which of the following function(s) is an accurate description of the graph for the range(s) indicated?

(a) x = y - y (b) x = -( y - y ) (c) x = y + y (d) x = -( y + y )  23. Lt

n ®¥

 24. Lt

x ®¥



( (

(a) 0



)

[CS GATE 2015]

n 2 + n - n 2 + 1 is ____?

)

2

[IN GATE 2016]

x + x - 1 - x is 1 (b) ¥ (c) (d) -¥ 2 [ME GATE 2016]

2

5x 25. Lt æç e - 1 ö÷ is equal to ____? x ®0 è x ø  [PI GATE 2016]



26. Lt sin( x - 4) = ? x®4 x-4

[CS GATE 2016]

27. How many distinct values of x satisfy the x equation sin x = , where x is in radians? 2 (a) 1 (b) 2 (c) 3 (d) 4 or more  [EC GATE 2016]

EMEP.CH02_3PP.indd 193

(i) y = 2 x + 4 for -3 £ x £ -1 y = x - 1 for -1 £ x £ 2 (ii) y = x - 1 for -1 £ x £ 2 (iii) y = 1 for 2 £ x £ 3 (iv)



(a) (i), (ii) and (iii) only



(b) (i), (ii) and (iv) only



(c) (i) and (iv) only



(d) (ii) and (iv) only



[CE GATE 2018] 32. Consider two functions f ( x) = ( x - 2) 2 and g ( x) = 2 x - 1 , where x is real. The smaller



value of x for which f ( x) = g ( x) is ____? [IN GATE 2018]

8/9/2023 8:43:24 PM

194 • Engineering Mathematics Exam Prep 33. Which of the following functions describe the graph shown in the below figure?

df 38. If f ( x) = sin x , then the value of at dx -p x= is 4 (a) 0 (b) 1 (c) - 1 (d) 1 2 2  [PI GATE 2010] 39. The function y = 2 - 3 x



(a) is continuous "x Î R and differentiable "x Î R

(b) is continuous "x Î R and differentiable 3 "x Î R except at x = 2 (c) is continuous "x Î R and differentiable

2 "x Î R except at x = 3 (d) is continuous "x Î R and except at x = 3 and differentiable "x Î R

(a) y = x + 1 - 2 (b) y = x - 1 - 1 (c) y = x + 1 - 1 (d) y = x - 1 - 1  [PI GATE 2018] 34. Which of the following functions is not differentiable in the domain [-1,1] ? (a) f ( x) = x 2 (b) f ( x) = x - 1 (c) f ( x) = 2 (d) f ( x) = maximum ( x, - x)  [GATE 2002] 35. If x = a (q + sin q ) and y = a (1 - cos q ) , then



[ME GATE 2010] 40. At t = 0 , the function f (t ) = sin t has t (a) a minimum (b) a discontinuity (c) a point of inflection (d) a maximum  [EE GATE 2010] 

41. What should be the value of l such that the function defined below is continuous at x p ?

dy =? dx q q (b) cos 2 2 q (c) tan q (d) cot 2 2  [GATE 2004]

(a) sin

36. Consider the function f ( x ) = x 3 , where x is real. Then the function f ( x ) at x = 0 is (a) continuous but not differentiable (b) once differentiable but not twice (c) twice differentiable but not thrice (d) thrice differentiable  [IN GATE 2007] 37. Given y = x 2 + 2 x + 10 the value of dy is dx x =1 (a) 0 (b) 4 (c) 12 (d) 13  [IN GATE 2008]

EMEP.CH02_3PP.indd 194

2

p x= ? 2





p ì l cos x , if x ¹ ïp 2 ï -x f ( x) = í 2 ï p , if x = ï1 2 î

(a) 0

(b) 2p

(d) p 2 [CE GATE 2011]

(c) 1

42. Consider the function f ( x) = x in the inter 

val -1 £ x £ 1 . At the point x = 0, f ( x) is (a) continuous and differentiable (b) non- continuous and differentiable (c) continuous and non-differentiable (d) neither continuous nor differentiable [ME, PI GATE 2012]

8/9/2023 8:43:33 PM

Calculus • 195

48. Given the following statements about a function f : R ® R, select the right option:

43. A function y = 5 x 2 + 10 x is defined over an open interval x = (1, 2). At least at one point dy exactly dx (b) 25 (c) 30 (d) 35 [EE GATE 2013]



P: If f ( x) is continuous at x = x0 , then it is also differentiable at x = x0

Q: If f ( x) is continuous at x = x0 , then it may

44. Which of the following functions is continuous at x = 3?



in this interval, 

(a) 20

ì ï2 if x = 3 ï (a) f ( x) = í x - 1 if x > 3 ïx+3 ï if x < 3 î 3 ì4 (b) f ( x) = í î8 - x ìx + 3 (c) f ( x) = í îx - 4 1 (d) f ( x) = 3 x - 27 

if x £ 3 if x > 3 x¹3

[CS GATE 2013]

45. If a function is continuous at a point, (a) the limit of the function may not exist at the point (b)  the function must be derivable at the point (c) the limit of the function at the point tends to infinity (d) the limit must exist the point and the value of limit should be same as the value of function at the point  [ME GATE 2014] 46. The function f ( x) = x sin x satisfies the following equation: f ¢¢( x) + f ( x) + t cos x = 0. The value of t is ___? [CS GATE 2014]



47. The values of for which the function x 2 - 3x - 4 is NOT continuous are x 2 + 3x - 4 (a) 4 and -1 (b) 4 and 1 (c) -4 and 1 (d) -1 and -4  [ME GATE 2016]



EMEP.CH02_3PP.indd 195

R: If f ( x) is differentiable at x = x0 , then it is also continuous at x = x0 

if x = 3 if x ¹ 3

if

not be differentiable at x = x0

(a) P is true, Q is false, R is false (b) P is false, Q is true, R is true (c) P is false, Q is true, R is false (d) P is true, Q is false, R is true [EC GATE 2016]

49. At x = 0 , the function is 2p x f ( x) = sin (-¥ < x < ¥, L > 0) L (a) continuous and differentiable (b) not continuous and not differentiable (c) not continuous but differentiable (d) continuous but not differentiable  [PI GATE 2016] 50. A function f ( x) is defined as



x ïìe , x < 1 f ( x) = í 2 ïîln x + ax + bx , x ³ 1

Which one of the following statements is TRUE? (a)  f ( x) is not differentiable at x = 1 for any values of a and b. (b)  f ( x) is differentiable at x = 1 for the unique values of a and b. (c)  f ( x) is differentiable at x = 1 for all values of a and b such that a + b = e. (d)  f ( x) is differentiable at x = 1 for all values of a and b.  [EE GATE 2017] x £1 ì - x, 51. Let g ( x) = í and î x + 1, x ³ 1

f ( x) =



x£0 ì1 - x, f ( x) = í 2 îx , x > 0

Consider the composition of f and g , i.e., ( f  g ) ( x) = f ( g ( x) ) . Then the number of



8/9/2023 8:43:41 PM

196 • Engineering Mathematics Exam Prep ­ iscontinuities in ( f  g ) ( x) present in the interval d (-¥, 0) is (a) 0 (b) 1 (c) 2 (d) 4  [EE GATE 2017] 52. The tangent to the curve represented by y = x ln x is required to have inclination 45 with the x-axis. The coordinates of the tangent point would be (a) (1,0) (b) (0,1) (c) (1, 1) (d) 2, 2

(



)

57. The value of the expression Lt sin x is x ® 0 xe x 1 1 (a) 0 (b) (c) 1 (d) 2 1+ e  [PI GATE 2008] xö 58. Lt æç 1 - cos ÷ is 2 x ®0 x è ø 1 1 (a) (b) (c) 1 (d) 2 4 2  [ME, PI GATE 2012]

53. A real-valued function y of real variable x is such that y = 5 x . At x = 0 , the function is (a) discontinuous but differentiable (b) both continuous and differentiable (c) discontinuous and not differentiable (d) discontinuous but not differentiable  [PI GATE 2018] 54. Let f be a real valued function of a real variable defined as f ( x) = x 2 for x ³ 0, and f ( x) = - x 2 for x < 0. Which one of the following statements is true? (a) f ( x) is discontinuous at x = 0

(b)  f ( x) is continuous but not differentiable



at x = 0 (c)  f ( x) is differentiable but its first derivative is not continuous at x = 0 (d)  f ( x) is differentiable but its first derivative is not differentiable at x = 0 [EE GATE 2018]



55. Limit of the following sequence as n ® ¥ is __? 1



xn = n n



(a) 0



(b) 1

(c) ¥ (d) -¥ [CE GATE 2002]

56. What is the value of Lim cos x - sin x p p x® x4 4 (a) 2 (b) 0 (c) - 2 (d) Limit does not exist  [PI GATE 2007]

EMEP.CH02_3PP.indd 196

x - sin x is x ®0 1 - cos x (a) 0 (c) 3

59. Lt

[CE GATE 2017] 

æ e2 x - 1 ö 60. Lt ç ÷ is x ®0 sin 4 x è ø (a) 0 (b) 0.5 

(b) 1 (d) not defined [ME GATE 2014]

(c) 1 (d) 2 [ME GATE 2014] a x -1 61. The expression Lt is equal to a ®0 a (a) log x (b) 0 (c) x log x (d) ¥  [CE GATE 2014] 62. The value of - sin x æ ö Lt ç ÷ is _____? x ® 0 2sin x + x cos x è ø  [ME GATE 2015] 2

63. The value of Lt 1 - cos x is ___? x ®0 2x4 1 (a) 0 (b) 2 1 (c) (d) undefined 4  [ME GATE 2015] 64. Lt log(1 + 4 x) is equal to x ®0 e3 x - 1 1 4 (a) 0 (b) (c) (d) 1 12 3  [ME GATE 2016] 7 5 x - 2x +1 65. The value of lim 3 is x ®1 x - 3 x 2 + 2 (a) 0 (b) -1 (c) 1 (d) does not exist  [CS/IT GATE 2017]

8/9/2023 8:43:49 PM

Calculus • 197

66. Let the function sin q

69. According to the Mean Value Theorem, for a continuous function f ( x) in the interval cos q

tan q

[a, b] , there exists a value x in this interval

æp ö æp ö æp ö f (q ) = sin ç ÷ cos ç ÷ tan ç ÷ 6 6 è ø è ø è6ø æp ö æp ö æp ö sin ç ÷ cos ç ÷ tan ç ÷ 3 3 è ø è ø è3ø



such that

(I) There exists q Î æç p , p ö÷ such that è6 3ø f ¢(q ) = 0



(II) There exists q Î æç p , p ö÷ such that è6 3ø f ¢(q ) ¹ 0



(a) I only (c) Both I and II

(b) II only (d) neither I nor II [CS GATE 2014]

67. A function f(x) is continuous in the interval [0, 2]. It is given that f(0)=-1=f(2) and f(1)=1. Which one of the following statements must be true? (a) there exist a “y” in the interval (0,1) such that f(y)=f(y+1)

(b) for every “y” in the interval (0, 1), f(y)=f(2-y)



(c) the maximum value of the function in the interval (0, 2) is 1



(d) there exists a “y” in the interval (0, 1) such that f(y)=-f(2+y) [CS GATE 2014]



68. A function f ( x) = 1 - x + x is defined in the closed interval [-1,1]. The value of x , in the 2

3

open interval (-1,1). for which the mean value theorem is satisfied, is 1 1 -1 (a) -1 (b) (c) (d) 3 2 3 2



EMEP.CH02_3PP.indd 197

[CE GATE 2015]

ò f ( x)dx = a

(a) f (x )(b - a ) (b) f (b)(x - a ) (c) f (a )(b - x ) 

where q Î é p , p ù and f ¢(q ) denotes the deêë 6 3 úû rivative of f with respect to q . Which of the following statement/statements is/are true?

b

(d) 0 [ME GATE 2018]

70. As x varies from -1 to 3, which of the following describes the behavior of the function f ( x) = x 3 - 3 x 2 + 1?

(a)  f ( x) Increases monotonically

(b) f ( x) Increases, then decreases and increases again

(c)  f ( x) decreases, then increases and decreases again

(d) f ( x) Increases and then decreases  [EC GATE 2016] 71. The following function has local minima at which value of x ,

f ( x) = x 5 - x 2



(a) - 5 (b) 2



5 (c)

5 (d) 5 2 2

[CE GATE 2002] 72. The function f ( x) = 2 x3 - 3x 2 - 36 x + 2 has its maxima at

(a) x = -2 only (b) x = 0 only (c) x = 3 only (d) both x = -2 and x = 3  [GATE 2004] 73. For the function f ( x) = x 2 e - x , the maximum occurs when x is equal to (a) 2 (b) 1 (c) 0 (d) -1  [EE GATE 2005] 74. For real x the maximum value of 

(a) 1

esin x is ecos x

2 (b) e (c) e (d) ¥ [IN GATE 2007]

8/9/2023 8:43:58 PM

198 • Engineering Mathematics Exam Prep 75. The minimum value of function y = x 2 in the interval [1, 5] is (a) 0 (b) 1 (c) 25 (d) undefined  [ME GATE 2007] 76. Consider the function f ( x) = x 2 - x - 2. The maximum value of f ( x) in the closed interval [-4, 4] is

(a) 18 (c) -2.25 

(b) 10 (d) indeterminate [EC GATE 2007]

77. Consider the function f ( x) = ( x 2 - 4) 2 where x is a real number. Then the function has (a) Only one minimum (b) Only two minima (c) Three minima (d) Three maxima  [EE GATE 2007] 78. Consider the function f ( x) = x 2 - x - 2. Then the maximum value of f ( x) in the closed interval [-4, 4] is (a) 18 (c) -2.25



(b) 10 (d) indeterminate [EC GATE 2007]

79. A point on the curve is said to be an extremum if it is a local minimum (or) a local maximum. The number of distinct extrema for the curve 3 x 4 - 16 x3 + 24 x 2 + 37 is 

(a) 0

(b) 1



84. The maximum value of  f ( x) = x 3 - 9 x 2 + 24 x + 5 in the interval [1, 6] is 

(a) 1

(b) 3

(c) 4 (d) 9 [IN GATE 2008]

81. For the real value x , the minimum value of the function f ( x) = e x + e - x is 

(a) 2

(b) 1 1 x

y 82. If e = x , then y has a (a) maximum at x = e (b) minimum at x = e

EMEP.CH02_3PP.indd 198

(c) 0.5 (d) 0 [EC GATE 2008]

(a) 21

(b) 25 (c) 41 (d) 46 [EE, EC, IN GATE 2012]

85. At x = 0, the function f ( x) = x3 + 1 has 

(a) a maximum value (b) a minimum value (c) a singularity (d) a point of inflection [ME, PI GATE 2012]

86. A political party orders an arch for the entrance to the ground in which the annual convention is being held. The profile of the arch follows the equation y = 2 x - 0.1 x 2 where y is the height of the arch in meters. The maximum possible height of the arch is (a) 8 meters (b) 10 meters (c) 12 meters (d) 14 meters  [PI, ME GATE 2012] 87. For 0 £ t < ¥, the maximum value of the function f (t ) = e - t - 2e -2t occurs at

2



[EC GATE 2010]

83. The function f ( x) = 2 x - x 2 + 3 has (a) a maxima at x = 1 and minima at x = 5 (b) a maxima at x = 1 and minima at x = -5 (c) only a maxima at x = 1 (d) only a minima at x = 5  [EE GATE 2011]

(c) 2 (d) 3 [CS GATE 2008]

80. Consider the function y = x - 6 x + 9. The maximum value of y obtained when x varies over the interval 2 to 5 is

(c) maximum at x = e -1 (d) minimum at x = e -1

(a) t = log e 4 (b) t = log e 2 (c) t = 0 (d) t = log e 8  [EC GATE 2014] 88. The maximum value of the function f ( x) = ln(1 + x) - x (where x > -1) occurs at x = ? 

[EC GATE 2014]

89. The maximum value of f ( x) = 2 x3 - 9 x 2 + 12 x - 3 in the interval 0 £ x £ 3 is ____? 

[EC GATE2014]

8/9/2023 8:44:07 PM

Calculus • 199

90. For a right-angled triangle, if the sum of the lengths of the hypotenuse and a side is kept constant, in order to have maximum area of the triangle, the angle between the hypotenuse and the side is

96. The maximum value attained by the function f ( x) = x( x - 1)( x - 2) in the interval [1, 2] is_____?  [EE GATE 2016] 97. Let f :[-1,1] ® R ,

(a) 12 (b) 36 (c) 60 (d) 45  [EC GATE 2014] 91. Let f ( x) = xe - x . The maximum value of the function in the interval (0, ¥) is -1 (a) e (b) e

f ( x) = 2 x 3 - x 4 - 10 . The minimum value of f ( x) is _______? [IN GATE 2016]



98. Consider the function f ( x) = 2 x3 - 3 x 2 in the domain [-1, 2]. The global minimum of f ( x) is ____________? [ME GATE 2016]

(c) 1 - e -1 (d) 1 + e -1 [EE GATE 2014]



92. Minimum of the real valued function 2 3

f ( x) = ( x - 1) occurs at x equal to

(a) -¥ 

(b) 0

(c) 1 (d) ¥ [EE GATE 2014]

93. The minimum value of the function f ( x) = x 3 - 3 x 2 - 24 x + 100 in the interval [-3,3] is 

(a) 20

(b) 28

(c) 16 (d) 32 [EE GATE 2014]

94. The maximum area (in square units) of a rectangle whose vertices lie on the ellipse x 2 + 4 y 2 = 1 is ____?  [EC GATE 2015] 95. Which one of the following graphs describes the function? f ( x) = e - x ( x 2 + x + 1) ?

where



99. The range of values of k for which the function f ( x) = (k 2 - 4) x 2 +6 x3 + 8 x 4 has local maxima at point x = 0 is (a) k < -2 or k >2 (b) k £ -2 or k ³ 2 (c) -2 < k < 2 (d) -2 £ k £ 2  [PI GATE 2016] 100. At the point x = 0 , the function f ( x) = x3 has (a) local maximum (b) local minimum (c) both local maximum and minimum (d) neither local maximum local minimum  [CE GATE 2018] 101. Let f ( x) = 3 x3 - 7 x 2 + 5 x + 6 . The maximum value of f ( x) over the interval [0, 2] is ____ ? (up to 1 decimal places).  [EE GATE 2018] 102. The Taylor series expansion of sin x about p x = is given by 6 (a) 1 2

+

3æ 2

çxè

2

3

ö - 1 æ x - p ö - 3 æ x - p ö + ....¥ ÷ ç ÷ ç ÷ 6 ø 4è 6ø 12 è 6ø

p

(b) x -

x3 x5 x7 + - + ......... 3! 5! 7! 3

(c) (d) 

EMEP.CH02_3PP.indd 199

[EC GATE 2015]



5

7

æx - p ö æx - p ö æx - p ö æx - p ö ç ÷ ç ÷ ç ÷ ç ÷ 6ø è 6ø 6ø 6ø è è è + + .....¥ 1!

1 2

3!

5!

7!

[CE GATE 2000]

8/9/2023 8:44:16 PM

200 • Engineering Mathematics Exam Prep 103. For the function e - x , the linear approximation around x = 2 is (a) (3 - x) e -2 (b) 1 - x (c) é3 + 2 2 - (1 + 2 x) ù e -2 ë û (d) e 

-2

[EC GATE 2007]

103. (a) Let f ( x) = e - x . Then f ¢( x) = -e - x , f ¢¢( x) = e - x ,....... So by the Taylor series expansion around x = 2 , we have, 2 f ( x) = f (2) + ( x - 2) f ¢(2) + ( x - 2) f ¢¢(2) + .....¥ 2! Therefore, around x = 2 , the linear approxi-x mation of e = f (2) + ( x - 2) f ¢(2)

= e -2 + ( x - 2)(-e -2 ) = (3 - x)e -2 .

104. For x y 2 and y >x 2 , the volume under f ( x, y ) is

line x = y, x = 0, y = 1 in the xy plane is ___?

183. If f ( x, y ) is continuous function defined over

(a)

y =1 x = y

ò ò

f ( x, y ) dx dy

188. The integral 1 ( x + y + 10) dx dy where D denotes the 2 p òò D

y =0 x = y 2

y =1 x =1

(b)

ò ò

f ( x, y ) dx dy

y = x2 x = y 2

disc: x 2 + y 2 £ 4, evaluates to ______?

y =1 x =1

(c)

ò xò=0 f ( x, y) dx dy

y= x x= y

ò ò

f ( x, y ) dx dy

y =0 x =0

[EE GATE 2009]

184. The volume under the surface Z ( x, y ) = x + y and above the triangle in the xy plane defined by {0 £ y £ x and 0 £ x £ 12} is __________ ? 

4 1 4 2 æ ö (a) æç 2udu ö÷ dv (b) ç 2udu ÷ dv ò0 è ò0 ò0 è ò0 ø ø

186. The value of the integral

æ 21 ö ò0 çè ò0 2udu ÷ø dv [EE GATE 2014] 4

2 x

òòe

x+ y

dy dx is

0 0

1 (a) 1 (e - 1) (b) (e 2 - 1) 2 2 2 2

(c) 1 (e 2 - e) (d) 1 æç e - 1 ö÷ 2è eø 2  [ME GATE 2014]

EMEP.CH02_3PP.indd 207

189. The region specified by p p ì ü í( r , j , z ) : 3 £ r £ 5, £ j £ ,3 £ z £ 4.5ý 8 4 î þ  in cylindrical coordinates has volume of _______ ? [EC GATE 2016]



190. A triangle in the xy -plane is bounded by

[EC GATE 2014]

185. To evaluate the double integral y æ +1 ö 8 2 ç æ 2x - y ö ÷ ò0 ç òy çè 2 ÷ø dx ÷ dy we make the substituç ÷ è 2 ø y tion u = æç 2 x - y ö÷ and v = . The integral 2 è 2 ø will reduce to

4 1 æ ö (c) ò ç ò udu ÷ dv (b) 0è0 ø 

[EC GATE 2016]



y = x2

(d) 

[EE GATE 2015]



the straight lines 2 x = 3 y, y = 0, and x = 3. The volume above the triangle and the under the plane x + y + z = 6 is________? [EC GATE 2016]

 191. The values of the integrals



æ1 x- y ö ò0 çè ò0 ( x + y)3 dy ÷ø dx and 1 1 æ x- y ö ò0 çè ò0 ( x + y)3 dx ÷ø dy are



(a) and equal to 0.5



(b) same and equal to - 0.5

1

(c) 0.5 and - 0.5, respectively (d) - 0.5 and - 0.5, respectively [EC GATE 2017]



192. Let I = c xy 2 dxdy, where R is the region òò R

shown in the figure and c = 6 ´10-4

The value of I equals _______ ?

8/9/2023 8:45:46 PM

208 • Engineering Mathematics Exam Prep 197. A three-dimensional region R of finite ­volume is described by x 2 + y 2 £ z 3 , 0 £ z £ 1 , where x, y, z are real. The volume of R correct to two decimal places is _____________?  [EC GATE 2017] 198. A parametric curve defined by æ pu ö æ pu ö x = cos ç ÷ , y = sin ç ÷ in the range 0 £ u £ 1 è 2 ø è 2 ø [EE GATE 2017]



2 3 193. The length of the curve y = x 2 between 3 x = 0 & x = 1 is

is rotated about x-axis by 360 degrees. Area of the surface generated is (a) p (b) p (c) 2p (d) 4p 2



(a) 0.27

(b) 0.67





(c) 1

(d) 1.22 [MEGATE 2008]



194. A parabolic cable is held between two supports at the same level. The horizontal span between the supports is L. The sag at the mid – span is h. The equation of the parabola is x2 y = 4h 2 , where x is the horizontal co-­ L ordinate and y is the vertical coordinate with the origin at the center of the cable. The expression for the total length of the cable is (a)

(c)

L

[EE GATE 2017] Answer key 1. (d)

2. (d)

3. (d)

4. (a)

5. (b)

6. (a)

7. (b)

8. (b)

9. (c)

10. (a)

11. (c)

12. (c)

13. (b)

14. (a)

15. (b)

16. (d)

17. (c)

18. (c)

19. (d)

20. (d)

21. (c)

22. (b)

1 23. 2 .

24. (c)

25. 25

26. 1.

27. (c)

28. (c)

29. (d)

30. –1

31. (b)

32. 1.

33. (b)

34. (d)

35. (c)

36. (c)

37. (b)

38. (c)

39. (c)

40. (b)

L 2

41. (c)

42. (c)

43. (a)

44. (a)

45. (d)

46. -2

47. (c)

48. (b)

49. (d)

50. (b).

53. (d)

54. (d)

55. (b)

58. (b) 62. - 0.333

59. (a)

60. (b)

61. (a)

63. (c)

64. (c)

0

h 2 x 2 (b) h3 x 2 dx 4 2 1 + 64 dx ò L L4 0

51. (a)

52. (a)

L 2

L 2

56. (c)

57. (c)

ò ò

1 + 64

2 2 h 2 x 2 (d) 2 1 + 64 h x dx 1 + 64 4 dx ò0 L4 L

65. (c)

66. (c)

67. (a)

68. (b)

69. (a)

[CE GATE 2010]

70. (b)

70. (b)

71. (d)

72. (a)

73. (a)

195. Consider a spatial curve in three-dimensional space given in parametric form by 2 p x(t ) = cos t , y (t ) = sin t , z (t ) = t , 0 £ t £ . p 2 The length of the curve is __________?  [ME GATE 2015]

74. (c)

75. (b)

76. (a)

77. (b)

78. (a)

79. (b)

80. (c)

81. (a)

82. (a)

83. (c)

84. (c)

85. (d)

86. (b)

87. (a)

88. 0

89. 6

90. (c)

91. (a)

92. (c)

93. (b)

94. 1.

95. (b)

96. 0

97. -13

98. -5

99. (c)

100. (d) 101. 12

102. (a) 103. (a)

196. The parabolic arc y = x is revolved around the x -axis . The volume of the solid of revolution is p 3p 3p p (a) (b) (c) (d) 2 4 2 4  [ME GATE 2010]

104. (c)

105. (a) 106. (b)

107. (c) 108. (d)

109. (b)

110. (b) 111. (b)

112. (b) 113. (a)

114. (d)

115. 2

119. 0

120. (a) 121. (c)

122. (b) 123. (a)

124. (d)

125. (b) 126. (d)

127. (a) 128. (a)

129. (b)

130. (b) 131. (d)

132. (d) 133. (a)

0



EMEP.CH02_3PP.indd 208

116. 0.293 117. (b) 118. (c)

8/9/2023 8:45:51 PM

Calculus • 209 134. (c)

135. 1.

136. (b)

137. 4.

139. (c)

140. (a) 141. 2

144. (c)

145. (b) 146. 0.005. 147. 0.5 148. (b)

149. (c)

150. (a) 151. (c)

152. (d) 153. (c)

154. (d)

155. (b) 156. 3

157. (d) 158. (a)

159. (c)

160. (a) 161. (a)

162. (c) 163. (a)

164. (c)

165. 40. 166. (c)

167. (d) 168. (d)

169. 4.5

170. (b) 171. (b)

172. (c) 173. (a)

174. (a)

175. (a) 176. (c)

177. (d) 178. (a)

179. (a)

180. (d) 181. (a)

182. (a) 183. (a)

184. 864

185. (a) 186. (b)

187. e-2. 188. 20

142. (a) 143. (b)

189. 4.71 190. 10 191. (c)

192. 1666.13

193. (d)

196. (d) 197. π/4

194. (d) 195. 1.86

4. (a)

138. (a)

198. (c)



5. (b) f ( x)

Lim x ®¥



6. (a)



1 - a4 1 = 1 - a 4 Lim 4 = 0 4 x ®¥ x x 1 æ ö =0 ÷ çLim x ®¥ x è ø

(

)



2. (d) x3 x5 x7 + - + ........¥ 3! 5! 7! Þ f ( x) = sin x p \ lim f ( x) = lim sin x = sin = 1. p p 2 x® x® 2

2

3. (d)



EMEP.CH02_3PP.indd 209

pö æ sin 2 ç x - ÷ 4ø è lim p p x® x4 4 pö æ sin 2 ç x - ÷ sin 2t 4ø è = lim = lim p t ® 0 p t x - ®0 x4 4 p pö æ ç putting x - = t so that t ® 0 when x ® ÷ 4 4ø è sin 2t é sin 2t ù = lim ê ´ 2 ú = 2 lim = 2 ´1 = 2. t ®0 2 t ®0 2t ë 2t û

æq ö sin ç ÷ è2ø Lt q ®0 q é æq öù sin ç ÷ ú ê è 2 ø ú ´ 1 = 1´ 1 = 1 æ lim sin x = 1ö = ê Lt ç ÷ q q 2 2 è x ®0 x ø ê 2 ®0 ú 2 êë 2 úû

7. (b)

f ( x) = x -



x3 + x x 2 ( x + 1) lim = x ®0 2 x 3 - 7 x 2 x ®0 x 2 (2 x - 7) x +1 0 +1 1 = lim = =- . x ®0 2 x - 7 2´ 0 - 7 7 = lim

Explanation

1. (d)

sin 2 x x ®0 x æ sin x ö = lim ç ´ sin x ÷ x ®0 è x ø sin x = lim ´ lim sin x = 1´ sin 0 = 0. x ®0 x ®0 x

lim



æ x2 ö e x ç1 + x + ÷ 2 ø lim è x ®0 x3 æ ö æ x2 x3 x4 x2 ö + + + ...¥ ÷ - ç1 + x + ÷ ç1 + x + 2! 3! 4! 2 ø ø è = lim è x ®0 x3

æ x2 ö e x - ç1 + x + ÷ 2 ø è lim 3 x ®0 x æ ö æ x 2 x3 x 4 x2 ö ç 1 + x + 2! + 3! + 4! + .....¥ ÷ - ç 1 + x + 2 ÷ ø è ø = lim è 3 x ®0 x x3

= lim 3!



+

x4 4!

+ .....¥

1

+

x

+ .....¥

= lim 3! 4! x ®0 x ®0 x3 1 1 0 1 = + + ........¥ = . 3! 4! 6

8/9/2023 8:45:55 PM

210 • Engineering Mathematics Exam Prep 8. (b)

-1 £ sin x £ 1 and - 1 £ cos x £ 1 Þ 1 ³ - sin x ³ -1 and - 1 £ cos x £ 1 Þ x - 1 £ x - sin x £ x + 1 and x - 1 £ x + cos x £ x + 1 Þ x - 1 £ x - sin x £ x + 1 and

y = x + x + x + x + .....¥ Þ y = x+ y Þ ( y - x) 2 = y Þ y 2 - 2 xy + x 2 = y





1 1 £ £ x + 1 x + cos x x - 1 x - sin x Þ £ £ x + 1 x + cos x



1 x -1 x = 1- 0 = 1 Now Lt = Lt x ®¥ x + 1 x ®¥ 1 1+ 0 1+ x 1 1+ x +1 x = 1 + 0 = 1. = Lt and Lt x ®¥ x - 1 x ®¥ 1 1- 0 1x



Therefore, by the Sandwich theorem, we get

At x = 2,



y2 - 4 y + 4 = y



or, y 2 - 5 y + 4 = 0



or, y = 1 , 4



But at x = 2,



y = 2 + 2 + 2 + ....¥ > 1



\ y = 4 only.

9. (c)

1-



cos x - sin x p x4 é 1 ù ê 2 ( cos x - sin x ) ú ú = 2 Lt ê p p x® ê ú x4 êë úû 4 1 é 1 ù ê 2 cos x - 2 sin x ú ú = 2 Lt ê p p x® ê ú x4 4 ëê ûú p 4

é é p öù æp öù æ ê sin ç 4 - x ÷ ú ê sin ç x - 4 ÷ ú è ø è øú ú = - 2 Lt ê = 2 Lt ê p p p p ú ú x® ê x - ®0 ê 4 4 ê x-4 ú ê x-4 ú ë û ë û sin x æ ö = - 2 ´ 1 = - 2 ç lim = 1÷. è x ®0 x ø



Lt

x ®¥

x - sin x = 1. x + cos x

11. (b) Result follows from the fundamental ­formulas. -1 1 12. (c) e - ln x = e- loge x = eloge x = x -1 = . x 13. (b)

Lt

x®

1 x -1 x +1 x -1

1

x3 - 2 Lt x ®8 x - 8 1

1

x 3 - 8 3 1 13 -1 = Lt = (8) x ®8 x -8 3 n æ ö x - an using Lt = n a n -1 ÷ ç x®a x-a è ø



1 -2 1 = (8) 3 = . 3 12

14. (a) 10. (a)



–1 sin x ≤ 1 and –1 ≤ cos x ≤ 1



⇒ 1 ≥ – sin x ≥ –1 and –1 ≤ cos x ≤ 1



⇒ x – 1 ≤ x – sin x ≤ x + 1 and



x – 1 ≤ x + cos x ≤ x + 1



⇒ x – 1 ≤ x – sin x ≤ x + 1 and

EMEP.CH02_3PP.indd 210



æ2 ö sin ç x ÷ è3 ø Lt x ®0 x é æ 2 öù ê sin ç 3 x ÷ ú 2 ø ú ´ = 1´ 2 = 2 æ Lt sin x = 1ö = Lt ê è ç ÷ 2 2 3 3 è x ®0 x x ®0 ê ø x ú 3 3 êë 3 úû

8/9/2023 8:46:00 PM

Calculus • 211

b n 2 2n é æ 1ö æ 1ö ù Lt ç1 - ÷ = ê Lt ç1 - ÷ ú = e -1 n ®¥ n ®¥ è nø è n ø ûú ëê

15. (

( )

) 2

-2

=e .



n é ù æ 1ö -1 ê nLt ç1 - ÷ = e ú ®¥ è nø ëê ûú

16. (d) Follows from fundamental formulas. 17. (c) x

1 æ 1ö Lt ç1 + ÷ = Lt (1 + x ) x = e . x ®¥ è x ø x ®0



18. (c) æ x + sin x ö Lt ç ÷ x ®¥ x è ø sin x æ sin x ö = Lt ç1 + = 1 + 0 = 1. ÷ = 1 + xLt ®¥ x ®¥ x ø x è 1 é ù = 0ú êë sin x is bounded and xLt ®¥ x û



ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 , we get



2 2 2 2 2 , b = , g = 0, a = , 2h = i.e; h = 9 15 15 25 f = 0 and c = -1.

Therefore, ab - h 2 = æç 4 - 2 ö÷ = 2 > 0, è 225 225 ø 225 2 9 2 D= 15 0

x ®¥

19. (d) Let q = 1000 p t . Then x x = 3sin q i.e; = sin q ..........(1) 3 p é ù y = 5cos êq + ú i.e; 4 ë û



pù y é = cos êq + ú 5 4û ë p p = cos q cos - sin q sin 4 4 1 = ( cos q - sin q ) .............(2) 2



Using (1) in (2), we get



y 1 æ xö = ç cos q - ÷ 5 3ø 2è æ y 2 xö or, çç + ÷÷ = cos q ..........(3) 3ø è 5

Squaring and adding equations (1) and (3), we get,

EMEP.CH02_3PP.indd 211

2x2 2 2 2 y2 + xy + - 1 = 0..........(4) 9 15 25

Now comparing equation (4) with

If f ( x) is bounded and Lt g ( x) = 0 , then Lt [ f ( x) g ( x) ] = 0.

i.e.,



Remember:

x ®¥

x2 æ 2 2 2 2 x2 ö + çç y + xy + ÷÷ 9 è 25 15 9 ø = sin 2 q + cos 2 q = 1

2 15 2 25 0

0 0 =-1

2 ¹0 225



Hence, equation (4) represents an ellipse.

Remember: ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 represents a an ellipse if ab - h > 0 and D = h g

h b f

2



g f ¹ 0. c

2

2x é æ 1 öx ù æ 1ö 2 20. (d) xLt ç1 + ÷ = ê xLt ç1 + ÷ ú = e . ®¥ ®¥ x è xø è ø êë ûú

21. (c) Let f ( x) = a x + b.............(i) Then, f (-2) = 29 Þ -2 a + b = 29........(i) and f (3) = 39 Þ 3a + b = 39.........(ii) Solving equations (i) and (ii), we get, a=2 and b=33. Hence f(x) f ( x)==2x2 x++3333and andsoso (5)2 =× 43. Hence, f(5)f = 5 +33 = 43.

22. (b) The point (2, -1) lies on the curve (b) only. 1 23. . 2

8/9/2023 8:46:06 PM

212 • Engineering Mathematics Exam Prep Lt

n ®¥

(

= Lt

n2 + n - n2 + 1

(

n2 + n - n2 + 1

(

n ®¥

= Lt

n ®¥



(

n2 + n + n2 + 1

24. (c)

(

x2 + x -1 - x

( = Lt (x = Lt

)

x2 + x -1 - x

(

x ®¥

x ®¥

2

(n 2 + n) - (n 2 + 1)

(n - 1) æ 1 1 çç 1 + + 1 + 2 n n è 1ö æ ç1 - n ÷ è ø = Lt x ®¥ æ 1 1 çç 1 + + 1 + 2 n n è

Lt

)(

n2 + n + n2 + 1

n + n + n +1

x ®¥

x ®¥

26. 1.

2

= Lt



)

)

)

)



27. (c) Let f ( x) = sin x & g ( x) =

ö ÷÷ ø

ö ÷÷ ø



=

)(

1-0 1 = . 1+0 + 1+0 2

x2 + x -1 + x

x2 + x -1 + x

)



æ e5 x - 1 ö Lt ç ÷ x ®0 è x ø

 From the figure above it is clear that x intersects at 3 f ( x) = sin x and g ( x) = 2 points. Hence, three solutions exist. 28. (c) log e (1 + 4 x) e3 x - 1 ì log e (1 + 4 x) ü log e (1 + 4 x) Lt ïï ï ï 0 ® x x x = Lt í ý= 3x 3x x ®0 1 -1 e e ï ï Lt ® x 0 ïî ï x x þ + log (1 4 ) x ì ü e í 4Lt ý´ 4 x ®0 1´ 4 4 4 x î þ = = = . 3x 1´ 3 3 ì e - 1ü í3Lt ý´3 î x ®0 3 x þ Lt

x ®0

2

2

ìïæ e5 x - 1 ö üï ìï æ e5 x - 1 ö üï = Lt íç ´ 5ý = 25 ´ í Lt ç ÷ ÷ý 5 x ®0 x ®0 è 5 x ø þï îïè 5 x ø þï îï

EMEP.CH02_3PP.indd 212

x £ 1 i.e; - 2 £ x £ 2 2 Let us draw both curves for x Î [-2, 2].

+ x -1 - x2 )

25. 25



We know that -1 £ sin x £ 1 .

)

æ 1ö ç1 - ÷ 1- 0 1 è xø = Lt = = . x ®¥ æ ö 1+ 0 - 0 +1 2 1 1 ç 1 + - 2 + 1÷ x x è ø



x . 2

\-1 £

æ 1ö x ç1 - ÷ è xø = Lt 2 x ®¥ æ ö 1 1 x + x -1 + x x ç 1 + - 2 + 1÷ x x è ø

2

sin( x - 4) x®4 x-4 sin( x - 4) sin y = Lt = Lt =1 x - 4®0 y ® 0 x-4 y (taking x - 4 = y ). Lt

æ æ ex -1 ö ö = 25 ´12 = 25. ç Lt ç ÷ = 1÷ x ®0 è x ø ø è

2



8/9/2023 8:46:10 PM

Calculus • 213

29. (d) æ x3 - sin x ö lim ç ÷ x ®0 x è ø sin x ö sin x æ = lim ç x 2 x 2 - lim ÷ = lim ® ® x ®0 x x 0 0 x ø x è = 0 - 1 = -1. 30. –1. æ tan x ö lim ç 2 ÷ x ®0 x - x è ø æ tan x ö tan x ç ÷ lim 0 x ® x = 1 = -1. = lim ç 2 x ÷ = x ®0 ( x - 1) 0 - 1 ç x - x ÷ lim ç ÷ x ®0 x è ø 31. (b) The equation of the line joining the points (-3, -2) and (-1, 2) is given by: y+2 2 - (-2) = x - (-3) -1 - (-3) or, y + 2 = 2( x + 3) = 2 x + 6 or, y = 2 x + 4 for -3 £ x £ -1



 The equation of a line passing through the points ( x1 , y1 ) and ( x2 , y2 ) is given by y - y1 y 2 - y1 . = x - x1 x2 - x1 32. 1. f ( x) = g( x) Þ x2 - 4x + 4 = 2x -1 Þ x2 - 6x + 5 = 0



y-2 0-2 = x + 1 1 - (-1) or, y - 2 = -( x + 1) or, y = 1 - x = -( x - 1) for -1 £ x £ 1.........(1)

33. (b) From the given graph we have the following points: (0, 0), (±1, -1), (±2, 0). It is easy to verify that all these points lie on the curve (b) only. 34. (d) Since all polynomial functions and all constant functions are differentiable, so the functions f ( x) = x 2 , f ( x) = x - 1 and f ( x) = 2 are all differentiable in [-1,1].



Combining (1) and (2), we get,



y = x - 1 for -1 £ x £ 2 \  (ii) is correct.



 Again from the graph, we have, y = 1 for 2 £ x £ 3. So (iv) is also correct.

EMEP.CH02_3PP.indd 213

Consequently, option (b) is correct.

ì x, x > 0 ü ï ï max = ( x , x ) f ( x) = í - x, x < 0 ý = x ï0, x = 0 ï î þ which is not differentiable at x = 0∈[-1, 1].

35. (c) Given that



dx = a (1 + cos q ), dq dy y = a (1 - cos q ) Þ = a sin q . dq



dy dy dq a sin q \ = = dx dx a (1 + cos q ) dq q q 2sin cos 2 2 = tan q = q 2 2 cos 2 2

x = a (q + sin q ) Þ

The equation of the line joining the points (1, 0) and (2,1) is given by: y - 0 1- 0 = x -1 2 -1 or, y = x - 1for1 £ x £ 2..........(2)

Þ ( x - 1)( x - 5) = 0 Þ x = 1,5.

 Hence, the smaller value of x for which f ( x) = g ( x) is 1.

\ (i) is correct.

The equation of the line joining the points (-1, 2) and (1, 0) is given by:



Remember:

ì x3 , x > 0 36. (c) f ( x ) = x 3 = ïí- x3 , x < 0 ï0, x = 0 î

8/9/2023 8:46:20 PM

214 • Engineering Mathematics Exam Prep R.H.L at x = 0

39. (c) We know that if y = ax + b , then y is con3

3

= lim f ( x) = lim x = 0 = 0. x ®0 +

tinuous for all “x” and dy = a ax + b , which dx (ax + b) exist for x ¹ -b . a

x ®0 +

L.H.L at x = 0

(

)

= lim f ( x) = lim - x 3 = -03 = 0 x ®0 -

x ®0 -

Thus lim f ( x) = lim f ( x) = 0 = f (0).



x ®0 +

x ®0 -

\ f ( x ) is continuous at x = 0.





(





sin t = 1, t sin t L.H .L = Lt f (t ) = Lt = 1. t ®0 t ®0 - t Thus R.H .L = L.H .L, but f (0) is not given. t ®0 +

41. (c) f ( x) is continuous at x =

Hence f ( x ) is twice differentiable only. dy = 2x + 2 dx

= 2 ´1 + 2 = 4.

42. (c) We know that if f ( x) = ax + b , then f ( x)

f ( x) = sin x Þ



EMEP.CH02_3PP.indd 214



æ p öö æ ç -l sin ç x - 2 ÷ ÷ è ø ÷ =1 Þ Lt ç p p æ ö ÷ x - ®0 ç 2 ç -ç x - 2 ÷ ÷ è ø ø è æ p öö æ ç sin ç x - 2 ÷ ÷ è ø ÷ = 1 Þ l ´ 1 = 1 Þ l = 1. Þ l Lt ç p pö ÷ æ x - ®0 ç 2 ç çx- 2 ÷ ÷ ø ø è è

x =1

38. (c)

df \ dx

p 2

æ ö ç l cos x ÷ æp ö Þ Lt f ( x) = f ç ÷ Þ Lt ç ÷ =1 p x® è 2 ø x ® p2 ç p - x ÷ 2 è 2 ø

ì6, x > 0 and f ¢¢¢ ( x ) = í î-6, x < 0 But f ¢¢¢ ( x ) does not exist if x = 0 .

dy dx

t ®0 +

Hence, the function f (t ) has a discontinuity at t =0

ì6 x, x > 0 Similarly, f ¢¢ ( x ) = ïí-6 x, x < 0 ï0, x = 0 î

\

sin t t

\ R.H .L = Lt f (t ) = Lt

ì3 x 2 , x > 0 Thus f ¢ ( x ) = ïí-3 x 2 , x < 0 ï0, x = 0 î

37. (b) y = x 2 + 2 x + 10 Þ

2 differentiable "x Î R except at x = . 3

40. (b) Given that f (t ) =

)

Hence, Rf ¢(0) = Lf ¢(0) and so f ( x ) is differentiable at x = 0.

2 - 3x dy = -3 . dx (2 - 3 x)

Thus, y = 2 - 3 x is continuous "x Î R and

Rf ¢(0) f ( x) - f (0) = lim x ®0 + x 3 x -0 = lim = lim x 2 = 0. x ®0 + x ®0 + x Lf ¢(0) f ( x) - f (0) = lim x ®0 x - x3 - 0 = lim = lim - x 2 = 0. x ®0 x ®0 x

\ y = 2 - 3x Þ

p p p 1 4 = cos - = - cos = . p 4 4 2 4 -

x =-

p 4

x df = cos x dx x

ax + b is continuous for all “x” and f ¢( x) = a , (ax + b) -b which exist for x ¹ . a

Here a = 1 and b = 0. So f ( x) = x is continuous and non-differentiable at x = 0.

8/9/2023 8:46:29 PM

Calculus • 215

49. (d) Since any modulus function is continuous, so f ( x) = sin 2p x is continuous for all valL ues of x.

43. (a) y = 5 x 2 + 10 x Þ \



dy dx

dy = 10 x + 10. dx

= 10 ´1 + 10 = 20.

Now, Rf ¢(0) f ( x) - f (0) = Lt x ®0 + x æ 2p x ö æ 2p x ö sin ç sin ç ÷ ÷ è L ø è L ø = Lt = Lt x ®0 + x ®0 + x x ì æ 2p x ö ü sin ç ÷ï ïï è L ø ï ´ 2p = 1 ´ 2p = 2p . = í Lt ý 2p x L L ï L ®0+ 2p x ï L L ïî ïþ

x =1

44. (a) Consider the function f(x) defined by: ì ï2 if x = 3 ï f ( x) = í x - 1 if x > 3 ïx+3 ï if x < 3 î 3



Then, æ x +3ö 3+3 L.H .L = Lt f ( x) = Lt ç = 2, ÷= x ®3x ®33 è 3 ø R.H .L = Lt f ( x) = Lt ( x - 1) = 3 - 1 = 2 x ®3+

Lf ¢(0)

and f (3) = 2. \L.H .L = R.H .L = f (3).



Hence, f(x) is continuous at x=3. 45. (d) If a function f ( x) is continuous at x = a ⇒ Lt f ( x) = f (a ) i.e; limit exists and is equal x®a

to the value of the function at that point. 46. -2 Given f ( x) = x sin x .



Since Rf ¢(0) ¹ Lf ¢(0) , so f ( x) is not differentiable at x = 0 .

\ f ¢( x) = x cos x + sin x, f ¢¢( x) = - x sin x + 2 cos x. Then, f ¢¢( x) + f ( x) + t cos x = 0 Þ - x sin x + 2 cos x + x sin x + t cos x = 0 Þ (2 + t ) cos x = 0 Þ t + 2 = 0 \ t = -2.



x 2 - 3x - 4 is not dex 2 + 3x - 4 fined at x = 1 and x = -4 .

47. (c) The function f ( x) =

50. (b).



48. (b) A continuous function may not be differentiable where as a differentiable function is always continuous.

EMEP.CH02_3PP.indd 215

x ïìe , x < 1 f ( x) = í 2 ïîln x + ax + bx , x ³ 1 ìe x , x < 1 ï Þ f ¢( x) = í 1 ï + 2ax + b , x ³ 1 îx

Now, if f ( x) is differentiable at x=1, then Lt f ¢( x) = Lt f ¢( x) x ®1+

\ The function f ( x) is not continuous at x = -4,1.



f ( x) - f (0) x ®0 x æ 2p x ö æ 2p x ö sin ç - sin ç ÷ ÷ è L ø è L ø = Lt = Lt x ®0 x ®0 x x ì æ 2p x ö ü sin ç ÷ï ïï è L ø ï ´ 2p = -1 ´ 2p = - 2p . = - í Lt ý 2p x L L ï L ®0- 2p x ï L L îï þï = Lt

x ®3+

x ®1-



1 or, + 2a + b = e 1 or, 2a + b = e - 1............(1)



Again, if f ( x) is differentiable at x=1, then



f ( x) is continuous at x=1. This implies,

8/9/2023 8:46:36 PM

216 • Engineering Mathematics Exam Prep Lt f ( x) = Lt f ( x)

x ®1+

x ®1-



or, ln1 + a + b = e



or, a + b = e............(2)



Solving (1) and (2) we get, a = -1, b = e + 1 .

Thus, we can conclude that f ( x) is differentiable at x = 1 for the unique values of a and b. 51. (a) In (-¥, 0) , g ( x) = - x. Then, ( f  g ) ( x) = f ( g ( x) ) = f (- x) = (- x) 2 = x 2

2

Thus,

( f  g ) ( x)

is a polynomial function and

so it has no points of discontinuity in (-¥, 0) . 52. (a) y = x ln x dy 1 Þ = x ´ + ln x = 1 + ln x dx x dy æ ö Þ tan 45 = 1 + ln x ç slope = = tan 45 ÷ dx è ø Þ ln x = 0 = ln1 Þ x = 1. At x = 1, y = 1´ ln1 = 0.

Hence, the required point is (1, 0). 53. (d) Since any modulus function is continuous, so y = 5 x . is continuous at x = 0 .





EMEP.CH02_3PP.indd 216

x ®0 +

Taking logarithm on both sides we get, 1



log y = Lim log n n n ®¥

Þ log y = Lim n ®¥



1 ¥ log n æç form ö÷ n è¥ ø

Using the L’Hospital rule, we get, d 1 log n log y = Lim dn = Lim n = 0 = log1 n ®¥ n ®¥ 1 d n dn 1



\ y = 1 i.e; Lim n n = 1. n ®¥

56. (c)

cos x - sin x æ 0 ö ç form ÷ p 0 è ø x4 d (cos x - sin x) (by the L'Hospital Rule) = Lim dx p d æ pö x® 4 çx- ÷ dx è 4ø - sin x - cos x = Lim p 1 x® Lim p x® 4

54. (d) ì 2 Given that f ( x) = ïí x , x ³ 0 2 ïî- x , x < 0 f ( x) - f (0) Lf ¢(0) = Lt x ®0 x

4

2

-x - 0 = Lt (- x) = 0, x ®0 - x - 0 x ®0

Rg ¢(0) = Lt

n ®¥

Hence, f ( x) is not differentiable at x = 0 .

= Lt

-2 x - 0 g ( x) - g (0) = Lt = -2, x ® 0 x x

1

\ Lf ¢(0) ¹ Rf ¢(0).



x ®0 -

55. (b) Let y = Lim n n .

x ®0 +



Lg ¢(0) = Lt

Hence, g ( x) i.e; f ¢( x) is not differentiable at x =0.

f ( x) - f (0) -5 x - 0 = Lt = -5 x ® 0 x x ( x = - x for x < 0 ) ,



f ( x) is differentiable at

2x - 0 g ( x) - g (0) = Lt = 2. x ® 0 + x x Thus, Lg ¢(0) ¹ Rg ¢(0) .



x ®0 -

5x - 0 f ( x) - f (0) = Lt =5 x ® 0 + x x ( x = x for x > 0 )

x2 - 0 = Lt ( x) = 0. x ®0 + x ®0 x

= Lt

ì 2 x, x ³ 0 Then, f ¢( x) = í = g ( x) (say) -2 x, x < 0 î Now

Lf ¢(0) = Lt

Rf ¢(0) = Lt

x ®0 +

f ( x) - f (0) x

  Lf ¢(0) = Rf ¢(0), x =0.



( x Î (-¥, 0) Þ - x Î (0, ¥) Þ f (- x) = (- x) )





Rf ¢(0) = Lt



= - sin

p p 1 1 - cos = = - 2. 4 4 2 2

8/9/2023 8:46:45 PM

Calculus • 217

57. (c) sin x æ 0ö Lt ç form ÷ x ®0 e x x 0ø è d (sin x) = Lt dx (by the L'Hospital rule) x ®0 d (e x x ) dx cos x cos 0 = Lt x = = 1. x ® 0 x xe + e 0 + e0 58. (b) æ 1 - cos x ö æ 0 ö Lt ç ÷ ç form ÷ 2 x ®0 è x ø è0 ø



æ d ö ç dx {1 - cos x} ÷ = Lt ç ÷ (by the L'Hospital rule) x ®0 d 2 çç {x } ÷÷ è dx ø sin x 1 sin x 1 é sin x ù = Lt = ´ Lt = ê Lt = 1ú x ®0 2 x x 0 x 0 ® ® x x 2 2 ë û

59. (a) x - sin x Lt x ®0 1 - cos x



d ( x - sin x) (by the L'Hospital rule) = Lt dx x ®0 d (1 - cos x) dx 1 - cos x æ 0 ö = Lt ç form ÷ x ®0 sin x è 0 ø d (1 - cos x) (by the L'Hospital rule) = Lt dx x ®0 d (sin x) dx sin x sin 0 = Lt = = 0 (sin 0 = 0). x ®0 cos x cos 0

xa -1 æ 0 ö ç form ÷ a ®0 a è0 ø d a ( x - 1) = Lt da a ®0 d (a) da Lt

x a log x [using the L'Hospital rule] a ®0 1 = log x ´ Lt x a = log x ´ 1 = log x. = Lt



a ®0

62. - 0.333



- sin x æ ö Lt ç ÷ 2sin cos x + x x è ø

x ®0

0ö æ ç form ÷ 0 è ø

d (- sin x) dx

= Lt

(by the L'Hospital rule) d (2sin x + x cos x) dx - cos x = Lt x ® 0 2 cos x + cos x - x sin x - cos x - cos 0 -1 = Lt = = = -0.333. x ® 0 3cos x - x sin x 3cos 0 - 0 3 x ®0

æ0 ö ç form ÷ 0 è ø

60. (b) e2 x - 1 æ 0 ö Lt ç form ÷ x ®0 sin 4 x è0 ø d 2x (e - 1) (by the L'Hospital rule) = Lt dx x ®0 d (sin 4 x) dx 2e 2 x 2e 0 2 = Lt = = = 0.5. x ®0 4 cos 4 x 4 cos 0 4

EMEP.CH02_3PP.indd 217

61. (a)

1 - cos ( x 2 ) æ form 0 ö ç ÷ 0ø è x ®0 2x4

63. (c) Lt



d (1 - cos x 2 ) dx (by the L'Hospital rule) = Lt x ®0 d 2x4 ) ( dx 2 x sin x 2 1 sin( x 2 ) 1 1 Lt = Lt = ´ = ´1 = . 3 2 2 x ®0 0 x ® 4 4 4 8x x + 4 x) 64. (c) Lt log(1 3 x x ®0 e -1

æ0 ö ç form ÷ 0 è ø

ìd ü (log(1 + 4 x)) ï ïï dx ï (by the L'Hospital rule) = Lt í ý x ®0 ï d ( e3 x - 1) ï îï dx þï



4 ì 1 ü ïï 1 + 4 x ´ 4 ïï 1 + 0 4 = Lt í = . ý= 3x 0 x ®0 3 ï 3e ï 3e ïî ïþ

8/9/2023 8:46:50 PM

218 • Engineering Mathematics Exam Prep 65. : x 7 - 2 x5 + 1 æ 0ö lim 3 ç form ÷ x ®1 x - 3 x 2 + 2 0ø è d 7 x - 2 x5 + 1) ( = lim dx (by the L'Hospital rule) x ®1 d 3 2 x x + 3 2 ( ) dx 7 x 6 - 10 x 4 7 x 5 - 10 x 3 = = lim lim x ®1 x ®1 3x - 6 3x 2 - 6 x 7 - 10 = = 1. 3-6 66. (c) It is easy to verify that f (q ) is continuous on é p , p ù and differentiable on æç p , p ö÷ . êë 6 3 úû è6 3ø

f (1) - f (-1) 1 - (-1) (1 - 1 + 1) - (1 - 1 - 1) or, - 2c + 3c 2 = =1 2 or, 3c 2 - 2c - 1 = 0 f ¢(c) =

or, c = 1, -



b

ò f ( x) dx = f (x )(b - a) . a

70. (b)



\ By Rolle’s theorem, there exists at least one æp p ö value of q Î ç , ÷ such that f ¢(q ) = 0 è6 3ø



67. () Let g(y)= f(y)–f(y+1) for y∈[0, 1]. Then we have, g(0)= f(0)–f(0+1)=–1–1=–20. Thus, g(0)× g(1) 0. Therefore, f ( x) has a minimum at x=0.  \The function f ( x) increases from - 1 to 0 and decreases from 0 to 2 and then increases from 2 to 3. 71. (d)

Remember: If f :[a, b] ® R be a function such that f is con-

1 \ c = - Î (-1,1) [ c = 1Ï (-1,1) ] 3

69. (a) According to the Mean Value Theorem of ­Integral Calculus, we have, If a function f ( x) is continuous on [a, b] , then a point such that x Î ( a, b) $

Also f çæ p ÷ö = 0 = f çæ p ÷ö . è6ø è3ø

Thus, ( I ) is true. Again f ¢(q ) is always zero if f (q ) is a constant function. Since f (q ) is not a constant function, so ( II ) is also true.

1 3

f ( x) = x 5 - x 2 gives,

f ¢( x) =

f ¢¢( x) =



- x2 5 - x2

+ 5 - x2 =

(5 x - 2 x3 ) 2

(5 - x ) 5 - x

2

-

5 - 2 x2 5 - x2 4x 5 - x2

,

.

Now, f ¢( x) = 0 Þ 5 - 2 x 2 = 0 Þ x 2 = Þx=

5 2

5 5 ,2 2

8/9/2023 8:46:57 PM

Calculus • 219

73. (a) f ( x) = x 2 e - x

æ 5ö f ¢¢ çç ÷÷ è 2ø 5

=



5 2

æ 5ö ÷ è 2ø

Þ f ¢( x) = 2 xe - x - x 2 e - x = (2 x - x 2 )e - x ,

3

- 2ç

4 -

5



2

2 é æ 5ö ù æ 5ö æ 5ö ê5 - ç 5-ç ÷ ú 5-ç ÷ ÷ êë è 2 ø úû è 2ø è 2ø 2 æ 5 ö üï 5ì ï 5 5 ÷ ý í5 - 2 ç 4 2ï 2ø ï 4 è 2 = 2 î þ= 0 < 0. 5 5 é5 - 5 ù 5 - 5 55êë 2 úû 2 2 2 2

2

\ f ( x) has a local maxima at x = 5 . 2

5

=



è

5ö

3

÷ 2ø

2

4

2

-



\ f ( x) has a minimum value at x = 0.



f ¢¢(2) = (22 - 4 ´ 2 + 2)e -2 = -2e -2 < 0.

So, f ( x) has a maximum value at x = 2.

æ 5ö ç -4 ÷ 2ø è æ

5ö

è

2ø

5-ç-

Þ esin x -cos x (sin x + cos x) = 0 Þ sin x + cos x = 0 Þ sin x = - cos x

2

÷

2 > 0. 5 52

\ f ( x) has local minimum at x = -

5 2

Þ f ¢( x) = 6 x 2 - 6 x - 36, f ¢¢( x) = 12 x - 6.





\ x = 3, -2 are stationary points.

So, f ( x) has a maximum value at x = -2.

So, the stationary points are x = - p , 3p . 4 4 Now, we have f ¢¢( x) + (sin x + cos x)esin x -cos x (cos x + sin x)

{

}

= esin x -cos x (sin x + cos x) 2 + (cos x - sin x)

Now, f ¢¢(3) = 12 ´ 3 - 6 = 6 > 0 . Thus, f ( x) has a minimum value at x = 3. Also, f ¢¢(-2) = 12 ´ (-2) - 6 < 0.

ì æ pö ï tan ç - 4 ÷ p ï è ø Þ tan x = -1 = - tan = í 4 ï æ pö tan p - ÷ ïî çè 4ø p 3p Þx=- , 4 4

= esin x -cos x (cos x - sin x)

Þ 6 x 2 - 6 x - 36 = 0 Þ x2 - x - 6 = 0 Þ ( x - 3)( x + 2) = 0 Þ x = -2,3

esin x = esin x -cos x . ecos x

So f ¢( x) = esin x -cos x (sin x + cos x). \ f ¢( x) = 0

Then, f ¢( x) = 0

EMEP.CH02_3PP.indd 219

x = 0, 2 are the stationary points.

Now, f ¢¢(0) = (02 - 4 ´ 0 + 2)e -0 = 2 > 0.

5

72. (a) f ( x) = 2 x3 - 3 x 2 - 36 x + 2

Þ x(2 - x)e - x = 0



Given f ( x) =

é æ 5ö ù æ 5ö ê5 - ç ÷ ú 5-ç÷ êë è 2 ø úû è 2ø

= 0+



2

æ

- 2ç -

Now, f ¢( x) = 0 Þ (2 x - x 2 )e - x = 0

74. (c)

æ 5ö Again f ¢¢ çç ÷÷ è 2ø -5

f ¢¢( x) = ( x 2 - 4 x + 2)e - x .



æ pö \ f ¢¢ ç - ÷ è 4ø 2 1 1 ìïæ 1 - 1 ö æ 1 1 ö üï = e 2 2 íç + + ÷ +ç ÷ý > 0 2 2ø è 2 2 ø ïþ ïîè æ 1 æ pö æ pö 1 ö , cos ç - ÷ = ç sin ç - ÷ = ÷ 4 2 2ø è ø è 4ø è

8/9/2023 8:47:05 PM

220 • Engineering Mathematics Exam Prep So f ( x) has a minimum value at - p . 4 Since, sin æç 3p ö÷ = sin æç p - p ö÷ = sin p = 1 4ø 4 2 è 4 ø è and cos æç 3p è 4 we have,

pö p 1 ö æ , ÷ = cos ç p - ÷ = - cos = 4ø 4 2 ø è

æ 3p ö f ¢¢ ç ÷ è 4 ø =e

1 1 + 2 2



2 ìïæ 1 1 ö æ 1 1 ö üï íç ÷ +ç÷ý < 0 2ø è 2 2 ø þï îïè 2

3p So, f ( x) has a maximum at x = . 4 Hence, maximum value of f ( x) is sin æ 3p ö f ç ÷ i.e; e è 4 ø 3p 3p sin - cos 4 e 4



3p 3p - cos 4 4

=e

1 1 + 2 2

\ f ( x) has a maximum value at x = 0 and  minimum values at x = 2 and x = -2 .

78. (a) f ( x) = x 2 - x - 2 Þ f ¢( x) = 2 x - 1, f ¢¢( x) = 2. 1 \ f ¢( x) = 0 Þ x = . 2 1 At x = , f ¢¢( x) = 2 > 0. 2 \ x = 1 is a point of local minimum. 2   As f (4) = 42 - 4 - 2 = 10, f (-4) = 42 + 4 - 2 = 18 , so the maximum value of f ( x) in the closed interval [-4, 4] is max{10, 18} i.e; 18.



79. (b) Let f ( x) = 3 x 4 - 16 x 3 + 24 x 2 + 37.

. =e

2 2

\ f ¢( x) = 12 x3 - 48 x 2 + 48 x, f ¢¢( x) = 36 x 2 - 96 x + 48. Now f ¢( x) = 0

= e 2,

so the maximum value of f ( x) is e 2 .



Þ 12 x 3 - 48 x 2 + 48 x = 0

75. (b)

 Given y = x 2 in [1 ,5]. So dy = 2 x > 0 in [1,5]. dx This shows that y is an increasing function in [1,5] . Hence, y is minimum when x is minimum in [1, 5]. Consequently, minimum value of y is (1) 2 , i.e, 1. 76. (a) f ( x) = x 2 - x - 2 Þ f ¢( x) = 2 x - 1 < 0 for x Î [-4, 4] \ f ( x) is a decreasing function on [ -4, 4].

Þ x( x 2 - 4 x + 4) = 0



Þ x( x - 2) 2 = 0 Þ x = 0, 2.



\ x = 0, 2 are the stationary points.

Now f ¢¢(0) = 0 - 0 + 48 > 0 and so



80. (c) y = x2 - 6x + 9 Þ

Hence, the maximum value of f ( x) occurs at x = -4.

\

77. (b) f ( x) = ( x 2 - 4) 2 Þ f ¢( x) = 4 x( x - 4), f ¢¢( x) = 4( x 2 - 4) + 4 x ´ 2 x = 4(3 x 2 - 4). \ f ¢( x) = 0



EMEP.CH02_3PP.indd 220

Þ 4 x( x 2 - 4) = 0 Þ x = 0, 2, -2. Now f ¢¢(0) = -16 < 0, f ¢¢(2) = 4(12 - 4) > 0 and f ¢¢(-2) = 4(12 - 4) > 0.

dy d2 y = 2 x - 6, 2 = 2. dx dx

dy = 0 Þ 2 x - 6 = 0 Þ x = 3. dx So, x = 3 is the only stationary point.

\ the maximum value of f ( x) is f (-4) i.e; 18.

2

f ( x) has a minimum at x = 0. f ¢¢(2) = 36 ´ 4 - 96 ´ 2 + 48 = 0 and so f ( x) has no extremum at x = 2.





d2 y dx 2

= 2 > 0, f ( x) has a minimum at x = 3. x =3

Since y (2) = 22 - 6 ´ 2 + 9 = 1, y (5) = 52 - 6 ´ 5 + 9 = 4,

\ the maximum value of y is max{ f (2), f (5)} i.e; 4.

8/9/2023 8:47:14 PM

Calculus • 221

81. (a) f ( x) = e x + e - x

f ¢( x) = 0 Þ 2 - 2x = 0

Þ f ¢( x) = e x - e - x and f ¢¢( x) = e x + e - x . f ¢( x) = 0 Þ e - e x

-x

Þ x = 1 is a stationary point. Now f ¢¢(1) = -2 < 0

2x

= 0 Þ e =1

Þ e 2 x = e0 Þ 2 x = 0 Þx=0 \ x = 0 is a stationary point.



84. (c) f ( x) = x3 - 9 x 2 + 24 x + 5 Þ f ¢( x) = 3 x 2 - 18 x + 24, f ¢¢( x) = 6 x - 18. Now, f ¢( x) = 0

0 0 Then, f ¢¢(0) = e + e = 1 + 1 = 2 > 0.



\ f ( x) has a minimum value at x = 0.



Minimum value = f (0) = e0 + e -0 = 2.

Þ 3 x 2 - 18 x + 24 = 0 Þ x 2 - 6 x + 8 = 0 Þ x = 2, 4 are the stationary points. f ¢¢(2) = 12 - 18 < 0, f ¢¢(4) = 24 - 18 > 0. \ x = 2 is a local maxima and x = 4 is a local minima. Again, f (1) = 1 - 9 + 24 + 5 = 21,

82. (a) 1

1

e y = x x Þ log e y = log x x Þ y =

1 log x. x

1 x ´ - logx dy 1 - logx x \ = = and 2 dx x x2 d d x 2 (1 - logx ) - (1 - logx ) ( x 2 ) d2y dx = dx 2 2 dx 2 x

{ }

1ö æ x 2 ç 0 - ÷ - (1 - logx ) 2 x xø = è x4 -1 - 2 (1 - logx ) -3 + 2 log x = = x3 x3



dy =0 dx 1 - log x Þ = 0 Þ 1 - log x = 0 x2 Þ log x = 1 = log e Þ x = e. \ x = e is the stationary point Now,

2

d y -3 + 2 log e -3 + 2 = = < 0. dx 2 e3 e3 \ y has a maximum value at x = e. At x = e,



83. (c)



EMEP.CH02_3PP.indd 221

\ f ( x) has a (local) maximum value at x = 1.

f ( x) = 2 x - x 2 + 3 Þ f ¢( x) = 2 - 2 x and f ¢¢( x) = -2.

f (6) = 63 - 9 ´ 62 + 24 ´ 6 + 5 = 41, f (2) = 23 - 9 ´ 22 + 24 ´ 2 + 5 = 25, and f (4) = 43 - 9 ´ 42 + 24 ´ 4 + 5 = 21. \ Maximum value of f ( x) in[1, 6] is



max { f (2), f (1), f (6)} i.e; max {25, 21, 41} i.e; 41.

85. (d) Given f ( x) = x3 + 1 . 2 \ f ¢( x) = 3 x , f ¢¢( x) = 6 x, f ¢¢¢( x) = 6 . f ¢( x) = 0



Þ 3x 2 = 0 Þ x = 0 is the stationary point.



Now, f ¢¢(0) = 0 & f ¢¢¢(0) ¹ 0 \ f ( x) has point of inflection at x = 0.

86. (b) Given y = 2 x - 0.1 x 2 . dy d2y \ = 2 - 2(0.1) x, 2 = -2 ´ 0.1 = -0.2 dx dx .

dy = 0 Þ 2 - 2(0.1) x = 0 Þ x = 10 dx .



\ x = 10 is the stationary point.

Since,

d2 y dx 2

= -0.2 < 0, x =10

so y is maximum at x = 10. \ Max. height = y (10) = 2 ´ 10 - (0.1) ´ 102 = 10 m.

8/9/2023 8:47:21 PM

222 • Engineering Mathematics Exam Prep 87. (a) f (t ) = e - t - 2e -2t Þ f ¢(t ) = -e - t + 4e -2t . f ¢(t ) = 0 1 Þ e - t (-1 + 4e - t ) = 0 Þ e - t = 4 æ1ö Þ -t = log ç ÷ = - log 4 Þ t = log 4. è4ø Now, f ¢¢(t ) = e - t - 8e -2t . At t = log 4, log

1 4

log

1 16



f ¢¢(t ) = e - log 4 - 8e -2 log 4 = e - 8e 1 1 1 = - 8 ´ = - < 0. 4 16 4



\ f (t ) has a maximum value at t = log 4.

88. 0 f ( x) = ln(1 + x) - x Þ f ¢( x) =

1 - 1. 1+ x

1 = 1 Þ 1 + x = 1 Þ x = 0. 1+ x Thus, x = 0 is a stationary point. -1 = -1 < 0 at x = 0 Now, f ¢¢( x) = (1 + x) 2 \ f ( x) is maximum at x = 0 and maximum So f ¢( x) = 0 Þ



value of the function= ln(1 + 0) - 0 = 0.

89. 6 f ( x) = 2 x 3 - 9 x 2 + 12 x - 3



so x + x 2 + y 2 = c , where c is constant or,

x2 + y 2 = c - x

or, x 2 + y 2 = (c - x) 2



or, y 2 = c 2 - 2cx..........(1)



Let the area of the triangle be



1 xy. 2 x2 y 2 x2 2 \ A2 = = (c - 2cx) 4 4 1 2 2 = ( c x - 2cx3 ) = f ( x) (say). 4



1 \ f ¢( x) = (2c 2 x - 6cx 2 ), 4



f″(x) =



f″(x) = 0



⇒ 2c2x – 6cx2 = 0



⇒ 2cx (c – 3x) = 0 ⇒ x = 0, c .



\ x = 0,

A. Then, A =

1 4

c 3

(2c2 – 12cx).

3

are stationary points.

1 (2c 2 - 12cx). 4 f ¢( x) = 0

Þ f ¢( x) = 6 x 2 - 18 x + 12, f ¢¢( x) = 12 x - 18. f ¢( x) = 0

f ¢¢( x) =

Þ 6 x 2 - 18 x + 12 = 0 Þ x = 1, 2 Î [0,3]. At x = 1 f ¢¢( x) = 12 ´1 - 18 = -6 < 0.

Þ 2c 2 x - 6cx 2 = 0



\ x = 1 is a point of local maximum.



At x = 2 f ¢¢( x) = 12 ´ 2 - 18 = 6 > 0 \ x = 2 is a point of local minimum.



Here, f (0) = 0 - 0 + 0 - 3 = -3, f (1) = 2 - 9 + 12 - 3 = 2 and and f (3) = 2 ´ 27 - 9 ´ 9 + 36 - 3 = 6. Hence, maximum of f ( x) in[0,3] = max { f (0), f (1), f (3)} = max {-3, 2, 6} = 6.

EMEP.CH02_3PP.indd 222

90. (c) Let base=x and height=y. Then, hypotenuse= x 2 + y 2 . Given that sum of the lengths of the hypotenuse and a side is constant

c Þ 2cx(c - 3x) = 0 Þ x = 0, . 3 c \ x = 0, are stationary points. 3 1 c2 > 0. Clearly, f ¢¢(0) = (2c 2 - 0) = 4 2 \ f ( x) has a minimum value at x = 0.



cü c2 æcö 1ì Again, f ¢¢ ç ÷ = í2c 2 - 12c ´ ý = - < 0. 3þ 2 è3ø 4 î c \ f ( x) has a maximum value at x = . 3

8/9/2023 8:47:25 PM

Calculus • 223

c c2 Now y 2 = c 2 - 2cx = c 2 - 2c ´ = . 3 3 c \y= . 3

If q be the angle between the hypotenuse and the side, then we have c x 1 3 = = = cos 60 cos q = 2 2 2 2 2 x +y c c + 9 3  \q = 60 . 91. (a) Given f ( x) = xe - x . Þ f ¢( x) = - xe - x + e - x = (1 - x)e - x ,



\ f ( x) has a minimum value at x = 4. Now, f (4) = 43 - 3 ´ 42 - 24 ´ 4 + 100 = 20, f (-3) = -33 - 3 ´ 32 + 24 ´ 3 + 100 = 118,



Since 4 ∈ [– 3, 3]. Therefore, required minimum value of f(x) = min f(-3), f(3)}=min{118, 28} = 28. 94. 1. Let 2x and 2 y be the length and breadth of the rectangle.

f ¢¢( x) = -(1 - x)e - x - e - x = ( x - 2)e - x .



If A denotes the area of the rectangle, then

Then, f ¢( x) = 0 Þ (1 - x)e - x = 0 Þ x = 1. \ x = 1 is the only stationary point.



A = 2 x ´ 2 y = 4 xy.

Now, f ¢¢(1) = (1 - 2)e -1 = -e -1 < 0 \ f ( x) has a maximum value at x = 1.

Then, A2 = 16 x 2 y 2 = 4 x 2 (1 - x 2 ) = 4 x 2 - 4 x 4

Hence, the maximum value of f ( x) = f (1) = e -1.



2 1 2 f ( x) = ( x - 1) 3 Þ f ¢( x) = ( x - 1) 3 . 3 \ f ¢( x) = 0 1 2 Þ ( x - 1) 3 = 0 Þ 3



1 ( x - 1)

1 3

Þ 8 x(1 - 2 x 2 ) = 0 1 1 ,. 2 2 1 æ 1 ö Here f ¢¢ ç ÷ = 8 - 48 ´ 2 < 0. è 2ø

Þ x = 0, 2

1 ì ü Now f ( x) = ( x - 1) = í( x - 1) 3 ý ³ 0. î þ

Thus, the minimum value of f ( x) is “0” which occurs at x = 1 . 93. (b) f ( x) = x3 - 3x 2 - 24 x + 100



Thus, f ( x) is maximum at x = 1 and so area 2 is maximum at x = 1 . 2

Þ f ¢( x) = 3 x 2 - 6 x - 24, f ¢¢( x) = 6 x - 6. f ¢( x) = 0 2

EMEP.CH02_3PP.indd 223

)

+ 4 y2 = 1

f ¢¢( x) = 8 - 48 x 2 . Now f ¢( x) = 0

= 0,

2 3



2

Then, f ¢( x) = 8 x - 16 x3 and

which is never possible for any value of x.



( x

Let f ( x) = 4 x 2 - 4 x 4 .

92. (c)



f (3) = 33 - 3 ´ 32 - 24 ´ 3 + 100 = 28.

Þ 3 x - 6 x - 24 = 0 Þ x = -2, 4. \ x = -2 and x = 4 are stationary points. f ¢¢(-2) = 6 ´ (-2) - 6 = -18 < 0 \ f ( x) has a maximum value at x = 2. –2. f ¢¢(4) = 6 ´ 4 - 6 = 18 > 0.

\ Maximum Area





= 4 xy = 4 x ´

= 2´

1 - x2 = 2x 1 - x2 2

1 1 ´ 1 - = 1. 2 2

8/9/2023 8:47:33 PM

224 • Engineering Mathematics Exam Prep 95. (b) Given f ( x) = e - x ( x 2 + x + 1) Þ f ¢( x) = (2 x + 1)e - x - e - x ( x 2 + x + 1)

98. -5 f ( x) = 2 x3 - 3x 2

Then, f ¢( x) = 0

Þ f ¢( x) = 6 x 2 - 6 x, f ¢¢( x) = 12 x - 6. Then, f ¢( x) = 0

Þ (- x 2 + x)e - x = 0 Þ x(1 - x) = 0 \ x = 0,1are stationary points. Now, f ¢¢( x)

Þ 6 x 2 - 6 x = 0 Þ 6 x( x - 1) = 0 Þ x = 0,1 are stationary points. Now, f ¢¢(1) = 12 - 6 > 0.

= (-2 x + 1)e - x - e - x (- x 2 + x) = e - x ( x 2 - 3x + 1).

\ f ( x) has a minimum at x = 1and f (1) = 2 - 3 = -1.

= ( - x 2 + x )e - x .

\ f ¢¢(0) = 1 > 0 and f ¢¢(1) = -e - 0 < 0. Hence, f ( x) has minimum at x = 0 and maximum -1

Again, f (-1) = -2 - 3 = -5, f (2) = 16 - 12 = 4.

at x = 1which is true for the graph (b) only.



\ Global minimum = min{ f (1), f (-1), f (2)} = min{-1, -5, 4} = -5.

96. 0 f ( x) = x( x - 1)( x - 2) in [1, 2] Þ f ¢( x) = 3 x 2 - 6 x + 2. Then, f ¢( x) = 0

99. (c) f ( x) = (k 2 - 4) x 2 +6 x3 + 8 x 4 Þ f ¢( x) = 32 x 3 + 18 x 2 +2 x(k 2 - 4)

6 ± 36 - 4 ´ 3 ´ 2 3 ± 3 Þx= = . 2´3 3



Þ f ¢¢( x) = 96 x 2 + 36 x + 2(k 2 - 4). Now, f ( x) has a local maxima at x = 0 Þ f ¢¢(0) < 0

These two stationary points lie outside the given interval [1, 2]. Therefore, the maximum value of f ( x) exists at one of the end points of the interval. We have f (1) = 1(1 - 1)(1 - 2) = 0 and



f (2) = 2(2 - 1)(2 - 2) = 0 .

\maximum value of f ( x) = max{ f (1), f (2)} = 0. 97. -13 f ( x) = 2 x 3 - x 4 - 10 Þ f ¢( x) = 6 x 2 - 4 x 3 , f ¢¢( x) = 12 x - 12 x 2 . \ f ¢( x) = 0 Þ 6 x 2 - 4 x 3 = 0 Þ 2 x 2 (3 - 2 x) Þ x = 0,

3 are stationary points. 2

3 lies outside the interval [-1,1]. 2 Now f (-1) = -2 - 1 - 10 = -13, f (1) = 2 - 1 - 10 = -9, f (0) = 0 - 0 - 10 = -10.

But x =



EMEP.CH02_3PP.indd 224

\ Minimum value of f ( x) in [-1,1] =min{f (-1), f (1), f (0)}= - 13.

Þ 2(k 2 - 4) < 0 Þ k 2 < 4



Þ k < 2 Þ -2 < k < 2.

100. (d) f ( x) = x 3 Þ f ¢( x) = 3 x 2 f ¢¢( x) = 6 x , f″′(x) = 6 Then, f ¢( x) = 0 Þ 3 x 2 = 0 Þ x = 0.

\ x = 0 is a stationary point. Now, f ¢¢(0) = 0 and f ¢¢¢(0) = 6 ¹ 0 .

\ x = 0 is a point of inflection (i.e. neither a maximum nor a minimum). 101. 12 f ( x) = 3x3 - 7 x 2 + 5 x + 6

Þ f ¢( x) = 9 x 2 - 14 x + 5 , f ¢¢( x) = 18 x - 14.

Then, f ¢( x) = 0

Þ 9 x 2 - 14 x + 5 = 0 Þ ( x - 1)(9 x - 5) = 0



Þ x = 1,

5 are stationary points. 9

8/9/2023 8:47:40 PM

Calculus • 225

Now, f ¢¢(1) = 18 - 14 > 0 and

Therefore, around x = 2 the linear approxima-x tion of e = f (2) + ( x - 2) f ¢(2)

æ5ö æ5ö f ¢¢ ç ÷ = 18 ç ÷ - 14 < 0 . è9ø è9ø



Thus, f ( x) has a local minimum at x = 1 and local maximum x = 5 . 9 Now, f (0) = 6, f (2) = 3 ´ 8 - 7 ´ 4 + 5 ´ 2 + 6 = 12 and

3



104. (c) e x + e- x cosh x e x + e- x coth x = = x 2 -x = x -x sinh x e - e e -e 2 2 4 é x ù x 2 ê1 + + + .......¥ ú 2! 4! û = ë 3 5 é ù x x 2 ê x + + + .......¥ ú 3! 5! ë û

2

5 æ5ö æ5ö æ5ö f ç ÷ = 3 ´ ç ÷ - 7 ´ ç ÷ + 5 ´ + 6 = 7.13. 9 è9ø è9ø è9ø \ The maximum value of f ( x) in [0, 2]



= max ìí f (0), f (2), f æç 5 ö÷ üý è9ø þ î = max {6,12, 7.13} = 12 .



é x2 x4 ù ê1 + 2! + 4! + .......¥ ú û = ë é x2 x4 ù x ê1 + + + .......¥ ú ë 3! 5! û 1 = x [by neglecting x 2 and higher power of x as

102. (a) Let f ( x) = sin x. Then f ¢( x) = cos x f ¢¢( x) = - sin x , f ¢¢¢( x) = - cos x . Therefore,



3 1 æp ö 1 æp ö æp ö , f ¢¢ ç ÷ = - , f ç ÷ = , f ¢ç ÷ = 2 è6ø 2 è6ø 2 è6ø 3 æp ö . f ¢¢¢ ç ÷ = 2 è6ø



By the Taylor series expansion of f ( x) about p x = , we have, 6



3

æ -p ö çx ÷ 6 ø ¢¢¢ æ p ö +è f ç ÷ + .............¥ 3! è6ø 2

=

( )=x

sin x

(x ) + (x ) 3

3

= x3 -

3!

3

3

5

5!

- ........¥

x9 x15 + - ........¥ 3! 5!

106. (b) f ( x) = e x + sin x \ f (p ) = ep + sin p = ep ,

3

103. (a) Let f ( x) = e - x .

f ¢(p ) = ep + cos p = ep - 1,



f ¢¢(p ) = ep - sin p = ep .

Then using the Taylor series expansion of f ( x) about x = p , we can write,

Then, f ¢( x) = -e - x , f ¢¢( x) = e - x ,....... So by the Taylor series expansion around x = 2 , we have

EMEP.CH02_3PP.indd 225



3

Þ f ¢( x) = e x + cos x, f ¢¢( x) = e x - sin x.

1 3æ p ö 1æ pö 3æ pö + çx- ÷- çx- ÷ ç x - ÷ + ......¥ 2 2 è 6 ø 4è 6 ø 12 è 6ø

f ( x) = f (2) + ( x - 2) f ¢(2) +

x 0

4. If f ( x) = x - 1 , then

æ dx ö æ dx ö = ò 2p y ç ÷ + ç ÷ du (a = 0, b = 1) è du ø è du ø 0



defined

(b) f ( x) = log e x

b

æ ds ö = ò 2p y ç ÷ du è du ø a 1

be

f ( x) be continuous. If satisfies æ xö f ç ÷ = f ( x) - f ( y ) "x, y and f ( e) = 1 , è yø then which of the following is true? (a) f ( x) is bounded

198. (c)

f ( x)

3. Let

é z4 ù p = ò p r dz = p ò z dz = p ê ú = . ë 4 û0 4 0 0 2



ù 1 é æ xö of f ( x) f ( y ) - ê f ç ÷ + f ( xy ) ú is 2 ë è yø û 1 (a) 0 (b) -1 (c) 1 (d) 2 2 n ù é 1 6. lim ê + + ..... + =? 2 n ®¥ 1 - n 2 1- n 1 - n 2 úû ë

1 (a) -1 (b) - 2

7. lim (1 - x) tan x ®1

(a)

8. lim x ®¥

(c) 1

(d) 0

px =? 2

p 1 2 (b) (c) (d) p 2 p p

x - sin x =? x + cos 2 x

(a) ¥ (c) 1

(b) 0 (d) -1

8/9/2023 8:49:40 PM

242 • Engineering Mathematics Exam Prep 9. lim

1 - cos 2 ( x - 1)

15. If lim

x -1

x ®1

1 - cos 2 x 10. lim =? x ®0 2x (a) 1 (c) -1





(a) continuous as well as differentiable for all x (b) continuous for all x but not differentiable at x = 0 (c) neither continuous nor differentiable at x=0 (d) discontinuous everywhere

(b)  f is differentiable at x = 0 but not at x =1 (c)  f is differentiable at x = 1 but not at x=0 (d)  f is differentiable neither at x = 0 nor at x =1

13. Suppose f is differentiable at x = 1 and 1 lim f (1 + h) = 5 , then f ¢ (1) is equal to h® 0 h (a) -5 (b) 5 (c) 4 (d) -4 14. The set of points where f ( x) = x is dif1+ x ferentiable is (a) ( -¥, 0)  ( 0, ¥) (b) ( -¥, - 1)  ( -1, ¥) (c) ( -¥, ¥) (d) ( 0, ¥)

EMEP.CH02_3PP.indd 242

(a) 2

=?

f (1) = 1 and f ¢ (1) = 2, then

(b) 1

(c) 0

(d) -1

16. If f ( x) = x - [ x ] , then f ¢ æç 1 ö÷ is equal to è2ø

(b) 0 (d) does not exist

ì æ 1 ö x - 1) sin ç , x ¹1 12. Let f ( x) = ïí( . Then è x - 1÷ø ï x =1 0, î which of the following is true? (a) f is differentiable at x = 0 and x = 1





(a)

ì -æç 1 + 1 ö÷ ç ÷ 11. If f ( x ) = ïí xe è x x ø , x ¹ 0 ; then f ( x) is ï x=0 0, î



x -1

x ®1

(a) exists and equals to 2 (b) exists and equals to - 2 (c) exists and equals to 1 (d) does not exist



f ( x) -1

1 2

(b) 1

17. If f ( x ) = 1 + x +

(a) 1000

(c)

(c) 0

(d) -

1 2

x2 x1000 + .... + , then f ¢ (1) = ? 2 1000 (b) 5000

101 2

(d) 1

ì 1 + kx - 1 - kx , -1 £ x < 0 ï x 18. Let f ( x ) = ïí 2x +1 ï , 0 £ x £1 x-2 îï

If f ( x) is continuous "x , then k = ? 1 (a) 1 (b) 2 2



(c) 1

(d) 0

if ì 5, ï 19. If f ( x ) = íax + b, if ï 21, if î

x£2 2 < x < 10 x ³ 10

and f ( x) is continuous "x Î R , then the values of a and b are, respectively:

(a) 2, 1

(b) 1, 2

ìl x 2 + m , ï 20. Let f ( x ) = í 1 ï x, î

(c) -1, 2 (d) 2, -1 x n ï n! ï x n -m , if n > m ïî ( n - m ) !

(x )

2 ì ü 1 ï æ D 2 - 2D ö æ D 2 - 2D ö ï 1 + ........¥ ý ( x ) ç ÷ ç ÷ í ç ÷ ç ÷ 2ï è 2 2 ïþ ø è ø î 1ì 1 ü = í1 - D 2 + D ý ( x ) 2î 2 þ

can be re-written as

d y dy + a1 + a2 y = X 2 dx dx

ïì æ D 2 - 2D ö ïü 2 í1 + ç ÷÷ ý ç 2 ø þï îï è

=

D2y + 3Dy + 4y = x2 or (D2 + 3D + 4)y = x2 4.4.3 Particular Integral (P.I) Let us consider the differential equation

1

-1

2 and D2 for d 2 .

2

2

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

=

=

1 e4 x D - 2D + 3 2

e4 x 4 - 2´4 + 3 2

é f ( D ) = D 2 - 2D + 3 ë

and f (4) = 42 - 2 ´ 4 + 3 ¹ 0 ù û

e4 x 11

Case-III: Let X = sin(ax + b) or cos(ax + b). Assume that f (D) contains only even powers of D. Then f (D) = f (D2) In this case, yPI =

1 sin( ax + b) f -a 2

(

)

or

1 cos( ax + b) f -a 2

(

)

[provided f (–a2) ≠ 0]

8/3/2023 5:23:31 PM

Ordinary Differential Equations • 301

Example: (i) Let us find the P.I. of y′′ + y = sin2x. Here, the given differential equation can be re-written as: (D2 + 1)y = sin 2x

\ yPI = =

1 sin 2x D2 + 1

1

1 sin 2x = - sin 2x 3 -22 + 1

1 cos x D2 + D + 1



=

1 cos x -1 + D + 1

=

1 cos x = cos xdx = sin x D

2

ò

[provided f ′(–a2) ≠ 0] (ii) If f (–a2) = 0, then x cos ax 1 cos ax = f ( D) f ¢( -a 2 )

[provided f ′(–a2) ≠ 0] Case-IV: Let X = eaxV, where V is a function of x. Example:

1 1 e ax V = e ax ´ (V) f ( D) f ( D + a)

(

)

dx

dx

1 1 xe x = xe x ( D - 1)2 D - 2D + 1

( )

2

1

x

ò

ò

Remember: If X = xV, where V is any function of x, then

EMEP.CH04_3PP.indd 301

yPI =

Let y = emx be a trial solution of (ii) Then (ii) gives m2emx – 4memx + 4emx = 0 or, m2 - 4m + 4 = 0

é emn ¹ 0 ù ë û

2

or, (m - 2) = 0 or, m = 2, 2 Here, the values of m are real and equal. So, yCF = (c1 + c2x)e2x Now, equation (i) can be re-written as (D2 – 4D + 4)y = ex \ yPI =



1 1 =e (x ) = e (x ) = ex xdx 2 2 D D éë{( D + 1) - 1}ùû 1 æ x2 ö x2 x3 ex x 3 = ex ç = ex = . dx = e x ´ ÷ 2 2´3 6 D çè 2 ÷ø x

d2 y dy +4 + 4 = 0 …(ii) dx dx 2

(D

1 2

- 4D + 4

)

1

ex =

é d ì 1 üù 1 1 ( xV) = x V+ê í ýú (V) f ( D) f ( D) ë dD î f ( D ) þû

2

( D - 2) =

This equation can be written as (D2 – 2D + 1)y = xex

dx

2

Let us consider, d y2 - 2 dy + y = xe x

\ yPI =

…(i)

where a0, a1, a2, and X are functions of x or constants. For the equation (i), the complete (general) solution is given by y = yCF + yPI, which contains two arbitrary constants in yCF part. Example:

Consider

x sin ax 1 sin ax = f ( D) f ¢( -a 2 )

Here, yPI =

d2 y dy + a1 + a2 y = X dx dx 2

dx

Remember: (i) If f (–a2) = 0, then

yPI =

a0

2 Let us solve d y2 + 4 dy + 4 = e x …(i)

æ 1 ö ç D stands for integration ÷ è ø

yPI =

1 f ¢( D ) V(V) 2 f ( D) ëé f ( D ) ûù

4.4.4 Complete (General) Solution Let us consider the second order linear differential equation

(ii) Let us find the P.I of (D2 + D + 1)y = cosx Here yPI =

=x

ex

ex = ex (1 - 2)2

Hence, the complete solution is given by y = yCF + yPI = (c1 + c2x)e2x + ex.

4.4.5 Homogeneous Linear Differential Equations of Order Two An equation of the form

x2

d2 y dy + a1x + a2 y = X , 2 dx dx

where a1, a2 are con-

stants and X is a function of x, is called a linear homogeneous differential equation of order two. To solve this type of equation, put logx = z. 2 2 Then dy = x dy and x 2 d y2 = d y2 - dy

dz

dx

dx

dz

dz

8/3/2023 5:23:33 PM

302 • Engineering Mathematics Exam Prep Using these results the given equation reduces to a linear differential equations with constant coefficients which can be solved by finding out its CF and PI. Example:

where P1 and P2 are constants and X is a function of x alone, can be reduced to a linear homogeneous differential equation of order 2 by substituting ax + b = u. Fully Solved MCQs

2 Let us solve x 2 d y2 - x dy + y = 2 log x …(i)

dx

dx

1. The differential equation of all straight lines passing through the origin is given by

The given differential equation is a linear homogeneous equation of order 2. So put z = logx

(a) dy = y (b) dy = x

Then dy = x dy

(c) xdy + ydx = 0



dz

∴ (i) gives

dx

d 2 y dy d2 y = x2 2 2 dz dz dx

and

dx

d y dy dy + y = 2z dz 2 dz dz

4

2 or d y2 - 2 dy + y = 2z …(ii)

dz

dz

Consider

d2 y dy -2 + y = 0 …(iii) 2 dz dz

Let y = emz be a trial solution of (iii)

Then, (iii) ⇒ m2emz – 2memz + emz = 0



⇒ m2 – 2m + 1 = 0 ( emz ≠ 0)



⇒ m = 1, 1

\  yCF = (c1 + c2z)e1z = (c1 + c2 log x)x ( z = log x)

Now equation (ii) can be written as:

(D – 2D + 1)y = 2z, where D º d

dz



-1 1 = 2 (2z ) = éê1 + D 2 - 2D ùú (2z ) ë û D - 2D + 1

(

(

) (

)

)

2 é ù = ê1 - D 2 - 2D + D 2 - 2D .......ú (2z ) ë û

( using (1 + x ) = 1 - x + x = éê1 - ( D - 2D ) ùú (2z ) ë û -1

2

- x 3 + .........¥

)

2

= 2z - D 2 (2z ) + 2D(2z ) = 2z - 0 + 4 = 2(log x + 2)

Hence, the complete solution of (i) is given by y = yCF + yPI = x(c1 + c2 log x) + 2(log x + 2) Remember: An equation of the form

EMEP.CH04_3PP.indd 302

( ax + b)2

d2 y dy + ( ax + b)P1 + P2 y = X , 2 dx dx



é æ dy ö2 ù 5 d2 y ê1 + ç ÷ ú = dx 2 êë è dx ø úû

(a) 2,1

(b) 2,2

(c) 2,5

(d) 5,2

3. The general solution of a differential equation is given as y = a cos (bx + c), here “b” is the only parameter. Then the differential equation is giver by: (a) y2 = b2y (b) y2 + b2y = 0 (c) y2 + y = 0 (d) y2 – y = 0 4. y = a +

b x

is a general solution of a differential

equation given by 2 d2 y dy (a) x d y2 + 2 dy = 0 (b) x 2 - 2 = 0 dx dx dx dx

2

\ yPI

y

(d) none of these

2. The order and degree of the following differential equation are, respectively:

2



dx

x

(c) x

d2 y dy =2 dx dx 2



(d) none of these

5. xy = aex + be–x is a general solution of a differential equation given by (a) x2y2 + 2xy1 = y (b) xy2 + 2y1 = xy (c) y2 + y1 = y (d) none of these 6. xy = Aex + Be–x + x2 is a general solution of a differential equation given by (a) x2y2 + xy1 – xy + y2 – 2 = 0 (b) x2y2 + xy1 + y2 – 2 = 0 (c) xy2 + 2y1 – xy + x2 – 2 = 0 (d) None of these 7. The solution (1 + x2)tan–1 xdy + y2dx = 0 is (a) log ( tan -1 x ) - 1 = C y

8/3/2023 5:23:35 PM

Ordinary Differential Equations • 303

(b) log (tan–1x) + y = C



(c) log ( cot -1 x ) - 12 = C y



(d) none of these

8. (xy2 + x)dx + (yx2 + y)dy = 0 has the solution 2

(a) log çç x 2 + 1 ÷÷ = C è y +1 ø æ

ö

2



2

(b) log(x + 1) = C log (y + 1)

{

}

(c) 1 log ( x 2 + 1)( y2 + 1) = C 2



(d) none the these

9. The solution of log æç dy ö÷ = 2x + 3 y is given by è dx ø –3y 2x (a) 2e + 3e = C (b) 3e–3y + 2e2x = C 2x+3y (c) 2 + e = Cxy (d) none of these y x

10. Solve: x ( ydx + xdy) = y tan ( xdy - ydx )

(c) sec y = cy x



dy + y = 0, y(0) = 2 dx



(a) a unique solution (b) no solution (c) finite number of solutions (d) infinite number of solutions 2 19. Solve: d y2 + dy - 2 y = 0, when x = 0, y = 2 and

dx dx dy 4 x 2 -2x = 0, y = e + e dx 3 3

(c)

(d) none of these

x

(a) xe (c)

xe

x2 y

= C (b) ye

dx

x2 2 y2

1 9

(a) y = - sin 3x + c1x + c2 =C

(b) y =

y2

= C (d) ye x = C

(c) y =

12. The solution of x dy + y = y2 log x is given by dx



dx

3 2 (a) y ( x 2 + 1) = x + c (b) y ( x 2 + 1) = x + c

2

(c) y(x2 – 1) = xy + c (d) none of these 14. The solution of (x + y + 1)dy – dx = 0 is given by (a) x – y = ce–y (b) x + y + 2 = cey (c) 2x + y = cey (d) 2x – y = ce–y 15. The solution of

dy x + sin 2 y = x 2 cos2 y dx

is

2 4 (a) x 2 tan y = x + c (b) x 2 tan y = x + c

4

2

4

4 (c) x 2 cot y = x + c (d) x 2 sec y = - x + c

4

(d) none of these dx 2

13. Solve: ( x 2 + 1) dy + 2xy = x 2 3



1 sin 3x + c1x + c2 4 1 - sin 3x + c1x + c2 4

2 21. Solve: d y + 2 dy + y = e x - e -2x

(a) y (1 + log x) = 1 + Cxy (b) x (1 + log x) = 1 + Cxy (c) y log x = Cx (d) x log x = Cy

EMEP.CH04_3PP.indd 303

1 1 3 3 3 3 4 2 4 2 y = e - x + e2x (d) y = e x + e -2x . 3 3 3 3

2 20. Solve: d y2 = sin 3x

11. Solve: xydx =(x2 – y2)dy x2 2 y2

has

(a) y = 1 e x + 2 e -2x (b) y = e x + e -2x

y cx y = cxy (d) cos = x y x

(a) sec

16. If y(x) satisfies the initial value problem (x2 + y)dx = xdy with y(1) = 2, then y(2) = ? (a) 4 (b) 5 (c) 6 (d) 8 17. The homogeneous equation (ax + by) dx + (cx + fy) dy is exact if and only if (a) a = c (b) b = f (c) b = c (d) a = f 18. The differential equation

4

dx

(a) (c1 + c2x)ex + ex + e–2x 1 4

(b) ( c1 + c2 x ) e - x + e x - e -2x 1

(c) ( c1 + c2 x ) e - x + e x - e -2x 2 (d) none of these 22. Solve: ( D2 + 1) y = sin 2x ; y = 0 & dy = 0 at x = 0 dx

(a) y = 2 sin x - 1 sin 2x 3

(b)

3

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

(c) y = 1 sin x + 2 sin 2x 3

3

1 3

4 3

(d) y = sin x - sin 2x

8/3/2023 5:23:37 PM

304 • Engineering Mathematics Exam Prep 23. If y = 3e2x + e–2x –ax is the solution of the initial value problem

d2 y + b y = 4a x , y(0) = 4 dx 2



(a) a = 3, b = – 4 (c) a = –3, b = – 4 24. Solve: y′′ + 4y = x2

and dy dx

x =0

= 1,

then

(b) a = 3, b = 4 (d) a = – 3, b = 4

30. Let W(y1, y2) be Wronskian of two linearly independent solutions y1 and y2 of the equation y′′ + P(x)y′ + Q(x)y = 0. Then the product W(y1, y2) × P(x) is equal to_____? (a) y1′′ y2 – y1y2′′ (b) y1y2′ – y2y1′ (c) y1′y2′′ – y1′′y′2 (d) y2′y1′′ – y1′y2′′ Answer key

1æ 1ö (a) c1 cos2x + c2 sin 2x + ç x 2 - ÷ 4è

2ø

1

1

æ 2 ö (b) c1 cos2x + c2 sin 2x + 2 ç x + 2 ÷ è

ø

1æ 1ö (c) c1 cos2x + c2 sin 2x - ç x 2 - ÷



2è

(d) none of these 25. Solve: y′′ – 4y = 2 sin 3x + ex

(a) (b) (c) (d)

2ø

–4x

3. (b)

4. (a)

5. (b)

8. (c)

9. (a)

10. (a)

11. (b)

12. (a)

13. (a)

14. (b)

15. (a)

16. (c)

17. (c)

18. (b)

19. (d)

20. (a)

21. (b)

22. (a)

23. (a)

24. (a)

25. (b)

26. (d)

27. (b)

28. (b)

29. (d)

30. (a)

1. (a) Let, y = mx be the straight line passing through the origin. dy Then, dy = m and so y = ´ x dx

dx



i.e; dy = y dx

x

4

2. (c) 4x

26. The Wronskian of e , e and e is (a) 48e–2x (b) 12e–2x (c) 24e2x (d) –96e2x 27. For three solutions x2 – 1, x2 + x + 1, 2x2 + x; which of the following is true? (a) W(x) ≠ 0 (b) The solutions are linearly dependent (c) The solutions are linearly independent (d) W(x) = 2x2 – x 28. For the solutions 1, sin2x, cos2x which of the following is true? (a) W(x) ≠ 0 (b) The solutions are linearly dependent (c) The solutions are linearly independent (d) None of these. 29. Let y1(x) and y2(x) be two linearly independent solutions of y′′ + y sin x = 0, 0 ≤ x ≤ 1. Let g(x) = W(y1, y2). Then which of the following is true? (a) g′ (x) > 0 on [0, 1] (b) g′ (x) < 0 on [0, 1] (c) g′(x) vanishes at only one point of [0,1] (d) g′(x) vanishes at only all points of [0,1]

EMEP.CH04_3PP.indd 304

2. (c) 7. (a)

Explanations

1 1 c1e2x + c2e -2x + sin 3x - e x 13 3 2 1 2x -2 x c1e + c2e - sin 3x - e x 13 3 1 1 ( c1 + c2x ) e2x + 13 cos3x + 2 e x 2 1 ( c1 + c2x ) e -2x - 13 cos3x - 2 e x 2x

1. (a) 6. (c)



é æ dy ö2 ù 5 d2 y ê1 + ç ÷ ú = dx 2 êë è dx ø úû

5 é æ dy ö2 ù æ d2 y ö ú Þç = ê1 + ç ÷ ç dx ÷ dx ÷ø ú è ø ëê è û

4

Here highest order derivative occurring in the

2 equation is d y2 . So order = 2.

dx



Again, power of highest order derivative i.e;

power of

d2 y dx 2

is 5. Hence, degree = 5.

3. (b) y = a cos (bx + c) ⇒ y1 = –ab sin (b + c)

⇒ y2 = –ab2 cos (bx + c)



⇒ y2 = –b2y



⇒ y2 + b2y = 0. 4. (a) y = a +



⇒

b x

⇒ dy = d æç a + b ö÷ dx dx x è

ø

dy b =0- 2 dx x

8/3/2023 5:23:40 PM

Ordinary Differential Equations • 305

8. (c) (xy2 + x)dx + (yx2 + y)dy = 0

dy = -b dx



⇒ x2



⇒ d æç x 2 dy ö÷ = d ( -b )



2 ⇒ 2x dy + x 2 d y2 = 0



dx è

dx ø

dx

⇒

(

dx

dx

d y dy x 2 +2 = 0. dx dx

d d ( xy ) = dx ae x + be - x dx

(



⇒ x dy + y = ae x - be - x



)

⇒ d æç x dy + y ö÷ = d ( ae x - be - x )



⇒ xy2 + 2y1 = aex + be–x = xy.

dx è dx

ø

dx

6. (c) xy = Aex + Be–x + x2 d d ( xy ) = Ae x + Be - x + x 2 dx dx

(



⇒



⇒ xy1 + y = Ae – Be + 2x

)(

)

dy = e 2 x + 3 y = e 2 x ´ e3 y dx dy Þ 3 y = e2x dx Þ e -3 ydy = e2x dx e

)

–x

x



Integrating we get,



òe

xy2 + y1 + y1 = Aex + Be–x + 2



or, xy2 + 2y1 = (Aex + Be–x + x2) + (2 – x2)



or, xy2 + 2y1 = xy + 2 – x2

or, or,



or, xy2 + 2y1 – xy + x2 – 2 = 0 +

dy =0 y2



(1 + x ) tan x dx dy Þò +ò =C (1 + x ) tan x y -1

2

-

-1

e2x e -3 y C + = 2 3 6

y ( xdy - ydx ) x x y æ xdy - ydx ö Þ 2 d( xy ) = y tan ç ÷ xè x x2 ø x ( ydx + xdy ) = y tan

d( xy ) y æ yö = tan d ç ÷ - log C xy x èxø

( where

1 =C y

é putting tan -1x = t so that 1 dx = dt ù ú ë 1 + x2 û 1 1 Þ log t - = C Þ log tan -1 x - = C. y y

)

y æ yö 1 d( xy ) = tan d ç ÷ xy x èxø

Integrating we get,

ò

2

(

e -3 y e2x C = -3 2 6

Þ

7. (a) (1 + x2)tan–1 xdy + y2dx = 0 dx

C 6

or, 3e2x + 2e -3 y = C.

10. (a)

(\  xy = Aex + Be–x + x2)

2

ò

dy = e2x dx 6

…(i)



-3 y

(where - C is the constant of integration)

Again, differentiating both sides of (i) w.r.t “x” we get,

EMEP.CH04_3PP.indd 305

(

Þ

( )

Þ





)

9. (a) log æç dy ö÷ = 2x + 3 y è dx ø

dx

dt

ò

(

Þ

⇒

òt

)

ò

ò

Þ



Þ

ò

Þ

2

Þ

(

xdx ydy + =0 x 2 + 1 y2 + 1 xdx ydy + =C 2 x +1 y2 + 1 1 2xdx 1 2 ydy + =C 2 x 2 + 1 2 y2 + 1 1 1 log x 2 + 1 + log y2 + 1 = C 2 2 1é 2 log x + 1 y2 + 1 úù = C. û 2 êë

Þ

5. (b) xy = aex + be–x



)

Þ x y2 + 1 dx + y x 2 + 1 dy = 0

ò

- log C is the constant of integration )

yö æ or, log( xy ) = log ç sec ÷ - log C x è ø yö æ or, log ç sec ÷ = log xy + log C = log(Cxy ) xø è



or, sec

y = Cxy x

8/3/2023 5:23:41 PM

306 • Engineering Mathematics Exam Prep 11. (b) xydx =(x2 – y2)dy y dy xy æ yö x Þ = = = f ç ÷ ( say ) …(i) 2 dx x 2 - y2 èxø æ yö 1-ç ÷ x è ø

or,

∴  the given differential equation is homogeneous. So let us put y = vx.

or, or,



Then,

dy dv =v+x . dx dx

∴ (i) gives

dv v v+x = dx 1 - v2

dv v v - v + v3 or, x = -v = 2 dx 1 - v 1 - v2 æ 1 - v2 ö dx or, ç 3 ÷ dv = ç v ÷ x è ø dx æ 1 1ö or, ç 3 - ÷ dv = vø x èv



1 1 æ 1ö æ 1ö = log x ´ ç - ÷ ´ dx + C xy x x çè x ÷ø è ø 1 log x 1 =- +C xy x x 1 log x 1 = + -C xy x x 1 = y(1 + log x ) - Cxy y(1 + log x ) = 1 + Cxy

or,

Alternative Method:

x

dy + y = y2 log x dx

Þ

1 dy y log x + = x y2 dx xy2



[on dividing both sides by xy2]



Þ

1 dy 1 log x + = x y2 dx xy

1 - log v = log x - log C 2v2 ( where - log C is the constant of integration ) or, log C - log v - log x = or, log C - log xv = 1 æC ö or, log ç ÷ = 2 è xv ø 2v or,

y



or, ye



x2 2 y2

yö æ ç v = x ÷ è ø

x



EMEP.CH04_3PP.indd 306

ò

ò

log x +C x2

…(ii)

æ 1ö ù ì æç - 1 ö÷dx ü é æ - log x ö ò çè - x ÷ødx ú ï òè x ø ï ê ´e z ´ íe dx - C ý= ç êè x ø÷ ú ïî ïþ ë û æ log x ö - log x or, ze - log x = ç dx - C ´e x ÷ø è 1 log x 1 or, z ´ = ´ dx - C x x x -1 1ö æ - log x = e log x = x -1 = ÷ ç e xø è

ò

ò

ò



z log x dx - C =x x2 é 1 1 æ 1ö 1 ù = - ê - log x - ç - ÷ ´ dx ú - C or, xy è xø x û ë x

ò

or,

ò

1 1 1 = log x + - C xy x x or, y(1 + log x ) = 1 + Cxy. or,



13. (a)

Integrating we get, d( xy ) = x 2 y2

dz æ 1 ö log x + z=dx çè x ÷ø x

which is a linear equation in z. So the solution of (ii) is given by

Þ xdy + ydx = y2 log x dx



dz z log x + = dx x x



dy + y = y2 log x dx

Þ d( xy ) = y2 log x dx d( xy ) log x Þ 2 2 = 2 dx x y x

or,

=C

12. (a)

dx



1

2 C = e 2y y

y dx

Then (i) becomes,



C 2 = e 2v xv x2

or,

1 2v2

1 2v2

…(i)

Let 1 = z , then - 12 dy = dz

Integrating we get, -

ò

or, -

(x or,



2

+1

dy + 2xy = x ) dx

2

dy æ 2x ö x2 +ç 2 y= 2 ÷ dx è x + 1 ø x +1

which is linear in y.

8/3/2023 5:23:42 PM

Ordinary Differential Equations • 307 \ eò



Pdx

=e

2x

æ 1ö

ò x 2 +1 dx = elog ( x 2 +1) = x 2 + 1.

Then, the solution is given by

æ x2 ö y ´ x +1 = ç 2 ´ x 2 + 1 dx + c ç x + 1 ÷÷ è ø

(

(

2

) ò

(

) ò

or, y x 2 + 1 = x 2dx + c =



)



Here, I.F = e



\ the



x3 +c 3



dx = x + y +1 dy

Þ

dx + ( -1)x = y + 1, dy

\ eò



Pdy

= eò

( -1)dy

which is linear in x

= -e - y ( y + 1) +

ò (e

-y

)

´ 1 dy + c

So, ¶M = b, ¶N = c. ¶y

¶M ¶N = i.e; if b = c. ¶y ¶x

y

18. (b) dy + y = 0 Þ dy = 0 and y = 0

dy x + sin 2 y = x 2 cos2 y dx dy 2sin y cos y Þ sec2 y + =x dx x cos2 y dy 2 Þ sec2 y + tan y = x dx x

dz 2z + = x, dx x

which is linear in z.

z ´ x 2 = x ´ x 2dx + c



x4 + c. 4

16. (c) ( x 2 + y ) dx = xdy dy x 2 + y y = = +x dx x x dy æ 1 ö Þ + y = x, dx çè x ÷ø

Þ



EMEP.CH04_3PP.indd 307

Which is lirear in y.

dx

⇒ y = c and y = 0 (c is a constant) But this contradicts the fact that y(0) = 2. Hence, the given differential equation has no solution. 19. (d) Since the R.H.S. of the given equation is

dy dz sec y = . dx dx

2 Pdx = e2log x = e log x = x 2 Then e ò = e ò \  the required solution is,

or, x 2 tan y =

dx

2

2 dx x

ò

¶x

 ∴ the given differential equation is exact if

or, x + y + 2 = ce y .



ò

Here, M = ax + by and N = cx + fy.

= -( y + 2)e - y + c

Let tan y = z. So Then (i) gives

1 1 = x ´ dx + c x x

¶M ¶N = ¶y ¶x

= -e - y ( y + 1) - e - y + c



ò ( x ´ I .F ) dx + c

y ´ ( I .F ) =

17. (c) The equation Mdx + Ndy = 0 is exact if

ò

15. (a)

1 . x

Now, y(1) = 2 ⇒ 2 = 1 + c ⇒ c = 1 \  y(x) = x2 + x Hence, y(2) = 22 + 2 = 6.

= e- y

or, x = -( y + 2) + ce

= x -1 =

or, y = x 2 + cx

x ´ e - y = ( y + 1) ´ e - ydy + c



-1

y or, = x + c x

So, the solution is given by



= e - log x = e log x

solution is given by

or, y ´

14. (b) (x + y + 1)dy – dx = 0 Þ

ò çè - x ÷ødx

“0,” so let us assume that y = emx be a solution of the equation.

2 Then d y2 + dy - 2 y = 0

dx



dx

⇒ m e + memx – 2emx = 0 ⇒ m2 + m – 2 = ( emx ≠ 0) ⇒ m2 + 2m – m – 2 = 0 ⇒ m(m + 2) – 1(m + 2) = 0 ⇒ (m + 2)(m – 1) = 0  ⇒  m = 1, – 2 ∴  the required solution is given by y = c1e1x + c2e–2x = c1ex + c2e–2x…(i) 2 mx

So, dy = c1e x - 2c2e -2x …(ii) dx

When x = 0, y = 2; So (i) ⇒ c1 + c2 = 2  When x = 0, dy = 0 ;

…(iii)

dx

8/3/2023 5:23:44 PM

308 • Engineering Mathematics Exam Prep

So, (iii) ⇒ c1 – 2c2 = 0 



Solving (iii) and (iv) we get,



∴  The required solution is given by y=





Hence, the complete solution is given by y = yCF + yPI = ( c1 + c2 x ) e - x +

Þ D 2 y + y = Sin 2x Þ

2

d y = sin 3x dx 2

Consider

Integrating both sides w.r.t “x” we get, d2 y dx = sin 3xdx + c1 dx 2 d æ dy ö 1 or, dx = - cos3x + c1 dx çè dx ÷ø 3

ò

ò

mx

¹0

)

⇒ m = – 1 = i ⇒ m = ± i = 0 + i, 0 – i  \  yCF = e0x (c1 cos x + c2 sin x) = c1 cos x + c2 sin x.

Again, integrating both sides w.r.t “x” we get



So, the complete solution is given by



y = yCF + yPI = c1 cos x + c2 sin x -



\

1

dy

1

2

d 1 ( y )dx = - sin 3x + c1x + c2 or, 9 dx 1 or, dy = - sin 3x + c1x + c2 9 1 or, y = - sin 3x + c1x + c2 . 9

ò ò

21. (b) Given equation is 2

d y dy +2 + y = e x - e -2x …(i) dx dx 2

Consider

d2 y dy +2 + y = 0 …(ii) dx dx 2

Let y = emx be trial solution of (ii) Then, (ii) ⇒ m2emx + 2memx + emx = 0 ⇒ m2 + 2m + 1 = 0 (  emx ≠ 0) ⇒ m –1, –1 \  yCF = (c1 + c2x)e–x. Now (i) can be expressed as: (D2 + 2D + 1)y = ex – e–2x 1

(e

x

- e -2x

\ yPI =



1 1 (ex ) - 2 ( e -2x ) 2 D + 2D + 1 D + 2D + 1 ex e -2x = 2 1 + 2 ´ 1 + 1 ( -2) + 2 ´ ( -2) + 1 1 = e x - e -2x . 4

(

D 2 + 2D + 1

)

yPI =

2

1 sin 2x 1 sin 2x = 2 = - sin 2x 3 -2 + 1 D2 + 1

1 sin 2x 3

2 dy = -c1 sin x + c2 cos x - cos2x dx 3

When x = 0, y = 0;

…(ii)

…(iii)

1 3



So, (ii) gives 0 = c1 + c2 ´ 0 - ´ 0 or c1 = 0.



Again for x = 0,



0 = 0 + c2 -



dy =0; dx

so (iii) gives

2 2 i.e; c2 = . 3 3

Hence, the required solution is given by y = 0 ´ cos x + =

2 1 sin x - sin 2x 3 3

2 1 sin x - sin 2x . 3 3

23. (a) y = 3e2x + e–2x – ax Þ Þ

)



dy = 6e2x - 2e -2x - a ................(i) dx d2 y = 12e2x + 4e -2x dx 2

(

)

= 4 3e2x + e -2x - a x + 4a x = 4 y + 4a x

=

EMEP.CH04_3PP.indd 308

(\ e

1 æ dy ö = - cos3x + c1 or, d ç ÷ 3 è dx ø dy 1 = - cos3x + c1 or, dx 3

2

ò dx dx = - 3 ò cos3xdx + ò c dx + c



d2 y + y = 0...........(1) dx 2

Þ m2emx + emx = o Þ m2 + 1 = 0

ò



d2 y + y = sin 2x dx 2

Let y = emx be a trial solution of (1) Then(1)

ò



1 x e - e -2x . 4

22. (a) ( D2 + 1) y = Sin 2x

4 x 2 -2x e + e . 3 3

20. (a) Given that

…(iv)

4 2 and c2 = c1 = 3 3



Þ

2

d y - 4 y = 4a x ...................(ii) dx 2 d2 y

Comparing (ii) with 2 + b y = 4a x , dx we get β = –4.

8/3/2023 5:23:47 PM

Ordinary Differential Equations • 309 Now

dy dx

x =0

=1

Þ1 = 6 - 2 -a

éëusing (i) ùû

Þ a = 3.



24. (a) The given equation is y′′ + 4y = x2 …(i) Consider y′′ + 4y = 0 …(ii) mx Let y = e be a trial solution of (ii) Then (ii) gives m2emx + 4emx = 0 or, m2 + 4 = 0 or, m2 = – 4 = i2 × 22 or, m = 2i = 0 + 2i, 0 – 2i \  yCF = c1 cos2x + c2 sin 2x. Now (i) can be written as: (D2 + 4)y = x2 \ yPI

1 = 2 (x 2 ) = D +4

2

-1

1

æ D2 ö 4 ç1 + ÷ ç 4 ÷ø è

( )

ù 1é D2 2 = êx 2 x + 0 - 0 + ........ú 4 ëê 4 ûú

( )



1é 2 1 1ö ù 1æ x - ´ 2!ú = ç x 2 - ÷ 4 êë 4 2ø û 4è

Hence, the complete solution is given by y = c1 cos2x + c2 sin 2x +

1æ 2 1ö x - ÷. 4 çè 2ø

25. (b) Given y′′ – 4y = 2 sin 3x + ex Consider y′′ – 4y = 0 …(i) mx Let y = e be a trial solution of (i) Then (i) gives, m2emx – 4emx = 0 or, m2 = 4 or, m = ± 2 \  yCF = c1e2x + c2e–2x Now, the given differential equation can be rewritten as: (D2 – 4)y = 2 sin 3x + ex

\ yPI =

1 2sin 3x + e x 2 D -4

(



EMEP.CH04_3PP.indd 309

=-

)

2 1 sin 3x - e x 13 3

(

)

2 1 sin 3x - e x . 13 3

\ W( y1 , y2 , y3 ) = 2e





e -4 x

2x

4e2 x

-4e

e4 x

-4 x

4e4 x

16e -4 x

16e4 x

1 1 1 = e2x ´ e -4 x ´ e4 x 2 -4 4 = -96e2x . 4 16 16

27. (b) Let y1(x) = x2 – 1, y2(x) = x2 + x + 1 and y3(x) = 2x2 + x.

Then 1 × y1(x) + 1 × y2(x) + (–1) × y3(x) = 0.



So clearly the solutions are linearly dependent.

Here, y′1 (x) = 2x, y′′1 (x) = 2, y′2 (x) = 2x + 1,



y′′2(x) = 2, y′3 (x) = 4x + 1, y3′′(x) = 4. x 2 - 1 x 2 + x + 1 2x 2 + x \ W( x ) = W( y1 , y2 , y3 ) = 2x 2x + 1 4x + 1 2 2 4

(

)

= x 2 - 1 {4(2x + 1) - 2(4 x + 1)}

( + ( 2x

)

- x 2 + x + 1 {4 ´ 2x - 2(4 x + 1)}



2

)

+ x {2 ´ 2x - 2 ´ (2x + 1)}

= 2(x2 – 1) + 2(x2 + x + 1) – 2(2x2 + x) = 0. 28. (b)

Let f(x) = 1, g(x) = sin2x, h(x) = cos2x f (x )

g( x )

h( x )

Then, W( x ) = f ¢( x ) g ¢( x ) h¢( x )

f ¢¢( x ) g ¢¢( x ) h¢¢( x )

=

(

= c1e2x + c2e -2x -

e2x

)

1 1 (2sin 3x ) + 2 (ex ) D2 - 4 D -4 1 1 = ex 2sin 3x + 2 2 -3 - 4 1 -4

Hence, the complete solution is given by y = yCF + yPI

26. (d) Let y1(x) = e2x, y2(x) = e–4x, y3(x) = e4x Then, y′1 (x) = 2e2x, y′2(x) = –4e–4x, y′3(x) = 4e4x, y′′1(x) = 4e2x, y′′2(x) = 16e–4x and y′′3(x) = 16e4x

(x )

( )

=



2

1æ D ö 2 ç1 + ÷ x 4 çè 4 ÷ø 2 é ù 1 D2 æ D2 ö = ê1 +ç - .........¥ ú x 2 ÷ ú 4ê 4 èç 4 ø÷ ë û =





1 sin 2 x = 0 2sin x cos x 0 2cos2x

cos2 x -2cos x sin x -2cos2x

= sin 2x × (–2 cos 2x) – (– sin 2x) × 2 cos 2x = 0

∴  the functions f(x), g(x) and h(x) are linearly dependent.

8/3/2023 5:23:48 PM

310 • Engineering Mathematics Exam Prep 4. The differential equation of all circles touching both the axes is given by

29. (d)

   y1(x) and y2(x) are solutions of y′′ + y sin  x = 0

So, y1′′ + y1sin x = 0 and y2′′ + y2 sin x = 0 i.e; y1′′ = – y1 sin x and y2′′ = – y2 sin x



Then, g ( x ) = W( y1 , y2 ) =

y1

y2

y1¢

y2¢



W( y1 , y2 ) =

y2

y¢

y¢

6. The solution of (3x + 2y – 6)dx + (2x + 3y – 6) dy = 0 is (a) x2 + y2 + xy – 6x – 6y = c (b) x2 + y2 + 6xy – 3x – 3y = c

= y1 y2¢ - y1¢ y2 .

…(i) …(ii)

or, ( y1¢ y2 - y2¢ y1 )P( x ) = y2¢¢ y1 - y1¢¢ y2 or, - W( y1 , y2 ) ´ P( x ) = y1 y2¢¢ - y2 y1¢¢ or, W( y1 , y2 ) ´ P( x ) = y1¢¢ y2 - y1 y2¢¢ .

Fully Solved MCQs

1. The differential equation corresponding to the solution y = e–2x (A cos x + B sin x) is given by (a) y′′ + 4y′ + 3y = 0 (b) y′′ + 4y′ + 5y = 0 (c) y′′ – 4y′ + 3y = 0 (d) y′′ – 4y′ + 5y = 0 2. The differential equation of the system of circles having a constant radius “a” and having their centers on the x-axis is given by (a) (1 + y12)y2 = a2 (b) yy12 = a2 (c) 1 – y12 = a2y2 (d) none of these

EMEP.CH04_3PP.indd 310

given that

y1

(c) 1 ( x 2 + y2 ) + 3xy - 2x - 3 y = c (d)

y1¢¢ y2 + P( x ) y1¢ y2 - y2¢¢ y1 - P( x ) y2¢ y1 = 0



dy dy = 1 + x2 ; dx dx

y = 2 when x = 1 (a) xy + x + y + 1 = 0 (b) xy – y + 2x = 1 (c) xy + y – 3x – 1 = 0 (d) none of these

1 2  y1 and y2 are solutions of y′′ + P(x)y′ + Q(x)y = 0. So y1′′ + P(x)y1′ + Q(x)y1 = 0 y2′′ + P(x)y2′ + Q(x)y2 = 0 (i) × y2 – (ii) × y1 ⇒



2

= y1 ( - y2 sin x ) - ( - y1 sin x ) y2 = 0 " x Î [0,1]

30. (a)

)

dy æ dy ö + =0 dx çè dx ÷ø (d) none of these

5. Solve: y - x

= y1 y2¢¢ - y1¢¢ y2



(

(c) x 2 + y2 - 2xy

= y1 y2¢ - y1¢ y2 .

\ g ¢( x ) = y1 y2¢¢ + y1¢ y2¢ - y1¢¢ y2 - y1¢ y2¢

2

dy æ dy ö + x 2 - 2xy ç ÷ =0 dx è dx ø dy =0 (b) x 2 + y2 - 2xy dx

(a) y2 - 2xy - 2xy

3. The differential equation of all circles touching y-axis at origin is given by (a) (x2 + y2)dy – 2xydx = 0 (b) (y2 – x2)dx + 2xydy = 0 (c) (x2 + y2)dy + 2xydx = 0 (d) (y2 – x2)dx – 2xydy = 0

2 3 2 x + y2 + 2xy - 6x - 6 y = c 2

(

)

7. Consider the differential equation p

dy + qy = re - kx dx

where p, q, r are positive constants and k is a non-negative constant. Then every solution of the given differential equation approaches to r as x → ∞ when q

(b) k > 0 (c) k = p (d) k = q

(a) k = 0

r

q

8. The solution of the differential equation yy′ + y2 – x = 0 is given by (a) y2 = x + ce–2x (b) y2 = x + ce -2x - 1 (c) y2 = x + ce2x

1 + 2

2

(d) y2 = 2x + ce2x + 1

9. The general solution of yy′′ – (y′)2 = 0 is given by cx (a) y = c2ec1x (b) y = ( c2 - x ) e 1 2

2

2

cx

(c) y = ( c2 + x ) ec1x (d) y = c2e 1 10. Consider the differential equation

dy - y = - y2 . dx

Then lim y( x ) = ? y ®¥

8/3/2023 5:23:50 PM

Ordinary Differential Equations • 311



(a) 0 (b) 2 (c) 1 (d) –1 11. If g(x, y)dx + (x + y)dy = 0 is an exact differential equation and if g(x, 0) = x2, then the general solution of the differential equation is given by (a) x3 + 2xy + y3 = c (b) 2x3 + 6xy + 3y3 = c (c) 2x3 + 4xy + y3 = c (d) x2 + 6xy + 3y3 = c 12. If λ be a constant such that xy + k = isfies the differential equation dy x = x 2 - x - 1 y + x - 1, dx

(



)

( x -1)2 e 2

then “k” is equal to

(a) 0 (b) –1 (c) 1 (d) 2 k 13. If y is an integrating factor of the differential equation 2xydx – (3x2 – y2)dy = 0, then k = ? (a) –4 (b) 4 (c) –1 (d) 1 14. Consider the differential equation dy  = ay - by2 , where a, b > 0 and y(0) = y0. dx

When x → ∞, y(x) tends to– (a) 1 (b) 0

(c)

a b

(d)

b a

15. Let α and β be two real numbers such that 2 every solution of d y2 + 2a dy + b y = 0

satisfies lim y( x ) = 0 . Then,

(c)

(a) 3α2 + β < 0 and α > 0 (b) α2 + β > 0 and α < 0 (c) α2 – β > 0 and α > 0 (d) α2 – β > 0, α > 0 and β > 0 16. Let y(x) be the solution of the differential equation d æç x dy ö÷ = x ; y(1) = 0 and dy dx

dx è dx ø





(d) 0

x2

d2 y dy + lx + y = 0, x > 0; then l=? dx dx 2

(a) 3 (b) 3 (c) 2 (d) 1 19. Consider the initial value problem x2y′′ – 6y = 0, y(1) = k and y′(1) = 6 If y(x) → 0 as x → 0 +, then “k” is equal to– (a) 1 (b) 2 (c) 4 (d) 0 2

dy

d y =0 20. Solve: cos2 y 2 = tan y; given that dx dx when y = 0 (a) (sin y + cex)(sin y + ce–x) = 0 (b) (sin y + c)(cos y – ex) = 0 (c) (cos x + cey)(sin y + ce–x) = 0 (d) none of these 21. Solve: (x2D2 + xD + 1) y = sin (2 log x) 1 2

(a) y = c1 cos(log x ) + c2 sin(log x ) - sin(log x 2 ) 2 3

1 3

x =1

1 3

(d) y = c1 cos(log x ) + c2 sin(log x ) - sin(2log x ) Answer key

1. (b) 6. (d) 11. (b) 16. (a) 21. (d)

=0

(a) 3 - 1 log e 2 (b) 3 + 1 log e 2 4

2. (a) 7. (a) 12. (c) 17. (c)

2

3. (d) 8. (b) 13. (a) 18. (a)

4. (a) 9. (d) 14. (c) 19. (b)

5. (c) 10. (c) 15. (d) 20. (a)

Explanations

(d) none of these

–2x

17. Let y1(x) and y2(x) be two linearly indepen-



dent solution of the differential equation x2y′′ – 2xy′ – 4y = 0 for x Î éë1, 9 ùû . If W(x) = y1(x)



EMEP.CH04_3PP.indd 311

16

18. If 1 ( A + B log e x ) is the general solution of the x differential equation

Then, y(2) = ?

(c)

5 16

(c) y = c1 cos(log 2x ) + c2 sin(log 2x ) - sin(log x )

x ®¥

4 2 1 + log e 2 2

(b) 1

(a) 16

(b) y = c1 cos(log x ) + c2 sin(log x ) + sin(log x )

dx

dx



sat-





y′2(x) – y2(x)y′(x) and W(1) = 5 Then what will be the value of W(2) = ?

1. (b) y = e (A cos x + B sin x) ye2x = A cos x + B sin x…(i) Differentiating both sides of (i) w.r.t “x,” we get e2x

dy + 2 ye2x = - A sin x + B cos x …(ii) dx

8/3/2023 5:23:51 PM

312 • Engineering Mathematics Exam Prep Now differentiating both sides of (ii) w.r.t “x,” we get

2e2x

2 or e2x d y2 + 4e2x dy + 4 ye2x = - ye2x [using (i)]

dx

dx



2. (a) Let the equation of the circle be (x – a)2 + y2 = a2…(i) Here, radius of the circle = α and center of the circle = (a, 0) In equation (i), “a” is an arbitrary constant and α is a parameter. Differentiating (i) w.r.t “x” we get,



dy =0 dx

i.e; x - a = - y

dy dx

Then equation (i) becomes,

or





2

dy ö 2 2 y ç ÷ +y =a è dx ø

or y2 (1 + y12 ) = a2



3. (d) Let the equation of the circle be (x – a)2 + y2 = a2 i.e.; x2 – 2ax + y2 = 0 Now differentiating (i) w.r.t “x” we get,





…(i)

dy 2x - 2a + 2 y =0 dx dy or, a = x + y dx



\ (i) gives



dy ö æ + y2 = 0 x - 2x ç x + y dx ÷ø è dy or, y2 - x 2 - 2xy =0 dx 2



(



dy æ ç x + y dx x + y - 2x ç çç 1 + dy dx è

dy ö æ ÷ ç x + y dx ÷ - 2y ç ÷÷ çç 1 + dy dx ø è

2

ö ÷ ÷ ÷÷ ø

dy æ ç x + y dx +ç çç 1 + dy dx è

2

ö ÷ ÷ =0 ÷÷ ø

2

dy ö dy öæ dy ö æ æ or, x 2 + y2 ç1 + - 2( x + y ) ç x + y 1+ dx ÷ø dx ÷ç dx ÷ø è è øè

(

)

2

dy ö æ +çx + y ÷ =0 dx è ø 2 é dy æ dy ö ù ú or, x 2 + y2 ê1 + 2 +ç dx è dx ÷ø ú êë û

(

)

2 é dy æ dy ö ù -2( x + y ) ê x + ( x + y ) + yç ÷ ú dx è dx ø úû êë 2 é dy æ dy ö ù ú=0 + ê x 2 + 2xy + y2 ç ÷ dx è dx ø ûú ëê

(

or, x 2 + y2 + 2 x 2 + y2

dy + (x ) dx

2

æ dy ö + y2 ç ÷ è dx ø

)

2

2

æ dy ö 2 + x 2 + y2 ç ÷ - 2x - 2xy è dx ø

(

)

(

-2 x 2 + 2xy + y2 + x 2 + 2xy

)

dy æ dy ö - 2xy + 2 y2 ç ÷ dx è dx ø

(

)

2

2

dy æ dy ö + y2 ç ÷ =0 dx è dx ø

or, y2 - 2xy - 2xy

2

dy æ dy ö + x 2 - 2xy ç ÷ = 0. dx è dx ø

(

)

5. (c) y - x dy = 1 + x 2 dy dx

)

or, y2 - x 2 dx - 2xydy = 0

4. (a) Let the equation of the circle be (x – a)2 + (y – a)2 = a2 i.e.; x2 + y2 – 2ax – 2ay + a2 = 0

EMEP.CH04_3PP.indd 312

Then, (i) gives





2

dy ö æ 2 2 ç - y dx ÷ + y = a è ø 2æ

dy dy - 2a - 2a +0 =0 dx dx dy x+y dx or, a = dy 1+ dx

2

or y’’ + 4y′ + 5y = 0





d2 y dy +4 + 4y = -y dx dx 2

2( x - a ) + 2 y

Differentiating (i) w.r.t “x” we get, 2x + 2 y

dy d2 y dy é ù + e2x + 2 êe2x + 2 ye x ú dx dx dx 2 ë û

= – A cos x – B sin x

or



…(i)



dx

dy Þ y - 1 = x (1 + x ) dx dx dy Þ = x (1 + x ) y - 1

Integrating we get,

8/3/2023 5:23:54 PM

Ordinary Differential Equations • 313

ò x(1 + x ) = ò y - 1 + C



Now from (ii) we have

( x + 1) - x dx = log( y - 1) + C x (1 + x ) dx dx Þ = log( y - 1) + C x x +1 Þ log x - log( x + 1) = log( y - 1) + C.............(i)



dy¢ -(3x ¢ + 2 y¢) = = dx ¢ 2 x ¢ + 3 y¢



which is a homogeneous equation. Let us put y′= vx′.

dx



Þ

dy

ò ò

ò

Putting y = 2 and x = 1, we get from (i), log1 – log(1 + 1) = log(2 – 1) + C or, C = log1 – log2 – log1 = –log2 ∴ (i) gives logx – log(x + 1) = log(y – 1) – log2

So

6. (d) (3x + 2y – 6)dx + (2x + 3y – 6)dy = 0  ⇒ 3xdx + 3ydy + 2ydx + 2xdy – 6dx – 6dy = 0  ⇒ 3(xdx + ydy) + 2(ydx + xdy) – 6(dx + dy) =0 Integrating we get,

ò

ò



or,

)

ò





Let us put x = x′ + h and y = y′ + k Then dx = dx′ and dy = dy′ ∴ (i) gives,



EMEP.CH04_3PP.indd 313

-2

} + log {(c¢) } 2

é where (c¢)2 = c1 ù ë û ì ü ï 2 ï ï 3v + 4v + 3 ï or, log í ý = log c1 2 1 ï ï ïî ïþ ( x ¢) 2 4 y¢ ïì 3( y¢) ïü + + 3ý = c1 or, ( x ¢)2 í 2 ¢ x ¢ ( x ) îï þï

or, 3( y¢)2 + 4 x ¢y¢ + 3( x ¢)2 = c1

- {3x ¢ + 2 y¢ + (3h + 2k - 6)} = .........(ii ) {2x ¢ + 3 y¢ + (2h + 3k - 6)}

or, 3( y - k )2 + 4( x - h )( y - k ) + 3( x - h )2 = c1 2

…(iii) and (iv)

Solving (iii) and (iv) we get, h = 6 & k = 6 . 5

{

or, log(3v2 + 4v + 3) = log ( x ¢ )

or, ( x ¢)2 (3v2 + 4v + 3) = c1

dy¢ - {3( x ¢ + h) + 2( y¢ + k ) - 6} = dx ¢ {2( x ¢ + h) + 3( y¢ + k) - 6}



ò



…(i)

Let us consider 3h + 2k – 6 = 0 2h + 3k – 6 = 0 

ò

1 (6v + 4)dv = - log x ¢ + log c¢ 2 3v2 + 4v + 3 1 or, log(3v2 + 4v + 3) = - log x ¢ + log c¢ 2 or, log(3v2 + 4v + 3) = -2log x ¢ + 2log c¢ or,

)

dy -(3x + 2 y - 6) = (2x + 3 y - 6) dx

dx ¢ (2 + 3v)dv =+ log c¢ 2 x¢ + 4v + 3

ì 1 ü ï ï = log c1 or, log(3v2 + 4v + 3) - log í 2ý ïî ( x ¢ ) ïþ

Alternative Method: The given equation can be re-written as

dv -3 - 4v - 3v2 = 2 + 3v dx ¢ (2 + 3v)dv dx ¢ = or, 2 x¢ 3v + 4v + 3

ò 3v

3 2 x + y2 + 2xy - 6x - 6 y = c 2

(

dv -3 - 2v = -v 2 + 3v dx ¢

Integrating we, get,



ì1 ü or, 3 d í x 2 + y2 ý + 2 d( xy ) - 6( x + y ) = c î2 þ

(

2 + 3v

dx ¢



3 ( xdx + ydy ) + 2 ( ydx + xdy ) - 6 d( x + y ) = c

ò

Then (v) ⇒ v + x ¢ dv = -(3 + 2v)

or, x ¢

or, 2x = (y – 1)(x + 1) or, xy + y – 3x – 1 = 0

ò

…(v)

dy¢ dv = v + x¢ . dx ¢ dx ¢

or, x ¢

x y -1 = log or, log x +1 2 x y -1 = or, x +1 2

y¢ ö æ -ç3 + 2 ÷ ¢ø x è y¢ 2+3 x¢

5



2

6ö 6 öæ 6ö 6ö æ æ æ or, 3 ç y - ÷ + 4 ç x - ÷ç y - ÷ + 3 ç x - ÷ = c1 5ø 5 øè 5ø 5ø è è è 36 y 108 24 x 24 y 2 or, 3 y + + 4 xy 5 25 5 5 144 36x 108 2 + + 3x + = c1 25 5 25

8/3/2023 5:23:55 PM

314 • Engineering Mathematics Exam Prep or, 3( x 2 + y2 ) + 4 xy - 12x - 12 y = c1 3 2 ( x + y2 ) + 2xy - 6x - 6 y = c, 2 1æ 216 ö where c = ç c1 . 2è 25 ø÷

216 25



or,





dx

è pø

or, ye

or, ye

qx p

qx p

qx p

p

qx

qx p

ò

= e æq

´

=



r - kx e dx + C p



æ y¢ ö or, log ç ÷ = log c1 è yø or, dy = c1dx y



r ´ +C p æq ö ç - k÷ èp ø

to

r q

qx p

r - ox e + Ce q

=

r + Ce q

…(i)



10. (c) dy

,

Þ

which approaches

e

ø

2

8. (b) Let us put y = z. Then, 2 y

dy dz = dx dx

i.e; yy¢ =





1 dz +z-x =0 2 dx

Þ

dz + 2z = 2x , dx

\ eò

Pdx

= eò

2dx

which is linear in z. = e2x .



z ´ e2x = e2x ´ 2xdx + c Þ y 2 e 2 x = 2x ´ Þ y2e2x = xe2x



EMEP.CH04_3PP.indd 314

Þ y2 = x -

e2x e2x - 2´ dx + c 2 2 e2x +c 2

1 + ce -2x . 2

ò

- y = - y2 dy = dx y(1 - y )

dy = y(1 - y ) dx {(1 - y) + y} dy = dx Þ y(1 - y )

Þ

Integrating we get, ò æç 1 + 1 ö÷ dy = ò dx + log c1 y 1- y è

ø

æ y ö or, log ç ÷ - log c1 = x è1 - y ø ì ü y or, log í ý=x c y (1 ) î 1 þ or,

y = ex c1 (1 - y )

y = c1e x 1- y y or, + 1 = c1e x + 1 1- y 1 or, = 1 + c1e x 1- y 1 or, 1 - y = 1 + c1e x

or,

Hence, the solution is given by

ò

dy = c1dx + log c2 y

or, log y - log(1 - y) = x + log c1

1 dz 2 dx

Then, yy′ + y2 – x = 0 Þ

ò

æ1 1 ö Þç + ÷ dy = dx è y 1- yø

when x ® ¥ æç e -¥ = 1¥ = 0 ö÷ . è

y¢ = c1 y

or, y = c2ec1x dx

qx p

ù

æ yö or, log ç ÷ = c1x è c2 ø y = ec1x or, c2

When k = 0, (i) gives, y=

f ¢( x )

ò f ( x ) dx = log ( f ( x )) + c úû

log y = c1x + log c2

æq ö ç -k ÷ x eè p ø

qx p

or,

Integrating we get,

ö

æ r ö - kx or, y = ç + Ce ÷e è q - pk ø

1

é êusing ë or, log y¢ - log y = log c1

r çè p -k ÷ø x e dx + C = p

ò

y¢

or, log( y¢) = log( y) + log c1

dx Pdx Here, e ò = e ò p = e p . ∴  the solution is given by,

y´e

y¢¢ y¢ = y¢ y

Integrating we get, y¢¢

which is linear in y. q

Þ

ò y¢ = ò y + log c

7. (a) p dy + qy = re -kx Þ dy + æç q ö÷ y = r e -kx dx

9. (d) yy¢¢ - ( y¢)2 = 0

ò



or, y = 1 -

c ex c 1 = 1 x = -x 1 x 1 + c1e 1 + c1e e + c1

8/3/2023 5:23:58 PM

Ordinary Differential Equations • 315

Now x→∞, y( x ) =

e

c1

-¥

+ c1

=

c1 = 1. 0 + c1

∴  lim y( x ) = 1. 11. (b) g(x, y)dx + (x + y)dy = 0 is exact

¶ ¶g ( x + y) = ¶x ¶y

Þ

¶g =1 ¶y

Given that g(x, 0) = x2. So we can take g(x, y) = x2 + y, 2



which satisfies g(x, 0) = x and



¶g =1 ¶y

y2 2

ò ò (taking x as constant) Hence, the general solution is given by 3

12. (c) x dy = ( x 2 - x - 1) y + x - 1



1

dy



\ I.F = e ò

=e

loge x -

which is linear

ø

in y. æ

Pdx

=e

( x -1)2

1

ö

ò èç1+ x -x ø÷dx

=e

é1

2



∴  The solution is given by



ì - ( x -1)2 ü ï ï y ´ íxe 2 ý = ïî ïþ



EMEP.CH04_3PP.indd 315

or, xye =

-

( x -1)2 2

ù

ò ëê x - ( x -1)ûúdx

- ( x -1) 2 = eloge x e

2

ò



=

( x -1)2 xe 2



ò

= ( x - 1)e

( x -1)2 2 dx

+c

ò ( e ) dz + c -z

= -e

-z

+ c = -e

( x -1)2 2

+c

)} = ¶¶y (2xy ) k +1

or, - 6xyk + 0 = 2x ( k + 1) yk or, - 3 = k + 1 or, k = -4.

dy = ay - by2 dx 1 dy a Þ 2 - = -b .............(i) y dx y y dx

dx

Then (i) gives, -

dz - az = -b dx

or,

dz + az = b, dx

\ I.F = e ò

Pdx

which is linear in z.

= eò

adx

= e ax

So, the solution is given by

ò



b 1 or, e ax = e ax + c y a 1 b or, = + ce - ax ................(ii) y a b 1 Now, y(0) = y0 Þ = +c y0 a

Þ c=

1 b y0 a

Then from (ii) we get, 1 b æ 1 b ö - ax = +ç - ÷´e y a è y0 a ø

æ ö ( x - 1)2 = z so that ( x - 1)dx = dz ÷ çç by putting ÷ 2 è ø -

{(

¶ - 3x 2 y k - y k + 2 ¶x

z ´ e ax = e ax ´ bdx + c

( x -1)2 ù é êæç1 - 1 ö÷ ´ xe 2 ú dx + c êè ú xø ë û -

we get,

Put, 1 = z , so that - 12 dy = dz .



1

è

,

14. (c)



dy æ 1ö 1ö æ = x - 1 - ÷ y + ç1 - ÷ dx çè xø xø è

æ ö Þ + 1+ - x÷ y =1- ,  dx ç x x

( x -1)2 e 2

)

y

dx



............(i)

Comparing (i) with xy + k = k = 1 and c = 1.

\

2

or, 2x 3 + 6xy + 3 y2 = 6c1 = c (taking c = 6c1 ).

Þ

( x -1) 2

(

x y + xy + = c1 3 2



or, xy + 1 = ce

Þ 2xy ´ ykdx - 3x 2 - y2 ´ yk dy = 0 ´ yk

x3 + xy 3

Ndy = ( x + y )dy = xy +

=c

 ⇒ 2xyk+1 dx – (3x2yk – yk+2)dy = 0, which is exact.

(taking y as constant)

( x -1)2 2

13. (a) 2xydx – (3x2 – y2)dy = 0

Now, ò Mdx = ò g ( x , y)dx = ò ( x 2 + y ) dx =

-

2



y ®¥

Þ

or, ( xy + 1)e

when x ® ¥ , e - ax ® e -¥ = 0 \



1 b æ 1 bö = +ç - ÷´0 y a è y0 a ø

i.e;

y=

a . b

8/3/2023 5:24:00 PM

316 • Engineering Mathematics Exam Prep 15. (d) Let y = emx be a solution of

d2 y dy + 2a +by=0 2 dx dx …(i)



Then (i) ⇒ m2emx + 2amemx + bemx = 0 2

Þ m + 2a m + b = 0 ( e

mx

When, x = 1, dy = 0; dx



So, (ii) Þ 0 = 1 + c1 Þ c1 = - 1



2 Thus, we have, y = x - 1 log e x - 1

¹ 0)

2

-2a ± (2a ) - 4 ´ 1 ´ b Þm= 2 ´1 = -a ± a 2 - b

Let α2 – β > 0 In this case, values of m are distinct real numbers. So, the solution is given by ( -a+ a2 -b )x + c e( -a- a2 -b )x y( x ) = c e 1

2

Now, lim y( x ) = 0 ( given) ,

( -a+

)

a2 -b x

®0

and e

( -a-

)

a2 -b x

®0

as x → ∞



i.e; if -a + a2 - b < 0 and - a - a2 - b < 0



i.e; if α > 0 and β > 0



∴ option (d) is correct. dx è dx ø



d æ dy ö

ò dx çè x dx ÷ø dx = ò xdx + c

1

æ dy ö x 2 Þ dçx + c1 ÷= è dx ø 2

ò

2



…….(i) which is a linear homogeneous differential equation of order “2.”



Þ y=



ò

x2 + c1 log e x + c2 ..................(i) 4 dy x c1 \ = + ...................(ii) dx 2 x



EMEP.CH04_3PP.indd 316

So, (i) Þ 0 = 1 + c1 log e 1 + c2 4

Þ c2 = -

1 4

(log e 1 = 0)

dx

d2 y dy d2 y = x 2 2 = x 2 y¢¢ 2 dz dz dx d2 y dy dy -2 - 4y = 0 dz dz 2 dz

d2 y dy -3 - 4 y = 0............(ii) dz dz 2

Let y = emz be a trial solution of (ii) Then, (ii) ⇒ m2emz – 3memz – 4emz = 0 or,  m2 – 3m – 4 = 0  ( emz ≠ 0) or,  m = – 4, 1 ∴ complete solution is given by y( x ) = c1e -4 z + c2e z =

c1

x4

+ c2 x

1ö æ z -z ç e = x and e = x ÷ è ø

Let, y1 ( x ) =

When, x =1, y = 0;

Let us put log x = z.

\ (i) Þ



ò

4

17. (c) The given equation is x 2 y¢¢ - 2xy¢ - 4 y = 0

dy x = + c1 dx 2 æx c ö Þ dy = ç + 1 ÷ dx è2 x ø æx c ö Þ dy = ç + 1 ÷ dx + c2 è2 x ø Þx

2

22 1 1 3 1 - log e 2 - = - log e 2. 4 2 4 4 2

\ y(2) =

or,

16. (a) d æç x dy ö÷ = x Þ

4

dz

which is possible

if e

2

Then, dy = x dy = xy¢ and

x ®¥





2



1 and y2 ( x ) = x . x4

Then clearly y1(x)

and y2(x) are an independent solution of (i) Now, W(x) 1 x y1 ( x ) y2 ( x ) x4 = = y1¢ ( x ) y2¢¢ ( x ) -4 1 x5 1 æ -4 ö 5 = 4 ´ 1 - x ´ ç 5 ÷ = 4 , which satisfies W (1 ) = 5. x èx ø x \W(2) =

5 5 = 4 16 2

8/3/2023 5:24:02 PM

Ordinary Differential Equations • 317 2 18. (a) The equation x 2 d y2 + lx dy + y = 0 is a

dx



dx

linear homogeneous differential equation of order “2.” So, let us put z = logex 2

2

Then, dy = x dy , d y2 - dy = x 2 d y2 dz



dx

dz

dz





or,

x ®0 +

d2 y dy + ( l - 1) + y = 0 …(i) 2 dz dz 1



Þ

1 ( A + Bz ) ez

19. (b) The given differential equation is a linear homogeneous differential equation of order 2. So, let us put log x = z Then,



d2 y dy d2 y = x2 2 2 dz dz dx

dy dy =x , dz dx



∴ the given differential equation becomes



d y dy - 6 y = 0 …(i) dz 2 dz

cos2 y



y = c1e -3z + c2e2z =

c1

x3

+ c2 x 2 ................(ii)



EMEP.CH04_3PP.indd 317

Also (ii) gives,

3c dy = - 41 + 2c2 x dx x

d2 y = tan y dx 2

d2 y = sec2 y ´ tan y dx 2 dy d2 y dy = 2sec2 y ´ tan y 2 dx dx dx 2



Þ

d æ dy ö dy = 2tan y sec2 y dx çè dx ÷ø dx

Now integrating both sides w.r.t “x” we get 2

æ dy ö 2 ç dx ÷ = 2 tan y ´ sec ydy + c1 è ø = 2´



ò

( tan y )2 + c

æ ç using ç è

2

ò

1

2

= ( tan y ) + c1

2 é f ( y ) ûù ö ÷ f ( y ) ´ f ¢( y )dy = ë ÷ 2 ø

2

æ dy ö 2 or, ç ÷ = tan y + c1 ..................(i) è dx ø

Now when y = 0, dy = 0 dx

∴ (i) gives, 0 = tan20 + c1 or, c1 = 0

æ z 1 -z ö ç log x = z Þ x = e , x = e ÷ è ø

Then, y(1) = k ⇒ k = c1 + c2 [from (ii)]…(iii)

Þ

Þ2

2

Let y = emz be a trial solution of (i) Then (i) becomes, m2emz – memz – 6emz = 0 or  m2 – m – 6 = 0 or  m = – 3, 2 Hence, the solution of (i) is given by

é 6 - 2k ù Þ lim ê 3 + (3k - 6)x 2 ú = 0 x ®0 + ë x û 6 - 2k Þ lim + (3k - 6) ´ 0 = 0 x ®0 + x 3 æ 6 - 2k ö Þ lim ç ÷=0 x ®0 + è x 3 ø Þ 6 - 2k = 0 Þ k = 3.

20. (a)

is the general solution of (i)

Þ ( A + Bz ) e - z is the general solution of (i) ⇒ m = –1, –1 are the roots of the auxiliary equation. ⇒ y = e–z is a trial solution of (i) Then, (i) gives e–z – (l – 1) × (e–z) + e–z = 0 or, 1 + (1 – l) + 1 = 0  \ l = 3

x

Þ lim y( x ) = 0

dx

Now, ( A + B log e x ) is the general solution x of (i)

…(iv)

Then (ii) becomes, y( x ) = 6 - 32k + (3k - 6)x 2

Now, y(x) → 0 as x→ 0 +

∴ the given differential equation becomes d 2 y dy dy +l +y=0 dz dz 2 dz

∴ y′(1) = 6 ⇒ 6 = –3c1 + 2c2  Solving (iii) and (iv), we get, c1 = 6 – 2k, c2 = 3k – 6

2

Thus, we have æç dy ö÷ = tan2 y è dx ø

dy = ± tan y dx or, cot ydy = ±dx or,



Integrating we get,

8/3/2023 5:24:04 PM

318 • Engineering Mathematics Exam Prep

ò cot ydy = ± ò dx + log c¢

or, log(sin y ) = ± x + log c¢ or, log(sin y ) - log c¢ = ± x æ sin y ö or, log ç ÷ = ±x è c¢ ø sin y = e±x or, c¢ or, sin y = c¢e x x

or, sin y - c¢e = 0 or, sin y + ce x = 0



Hence, the complete solution is given by y = yCF + yPI = c1 cos(log x ) + c2 sin(log x ) -

PREVIOUS YEARS QUESTIONS (2000-18) and and

1. The solution of the differential equation

sin y = c¢e - x sin y - c¢e

-x

d2 y = 3x - 2 with dx 2

=0

and sin y + ce - x = 0 (taking c = -c¢)

(a) y =

21. (d) (x2D2 + xD + 1)y = sin (2 log x) ⇒ x2D2y + xDy + y = sin (2 log x)

(c) y =

Þ x2

2



∴ (i) gives,

dz

dz

dx

d2 y dy dy + + y = sin 2z dz 2 dz dz



or,

d2 y + y = sin 2z ...................(ii) dz 2

Consider

m2emz + emz = 0

or, m2 + 1 = 0 or, m2 = -1 = i2 or, m = ±i = 0 + i, 0 - i

 \  yCF = e0z(c1 cos z + c2 sin z) = c1 cos z + c2 sin z = c1 cos(log x) + c2 sin(log z) Now equation (ii) can be re-written as:

(D

EMEP.CH04_3PP.indd 318

)

+ 1 y = sin 2z

\ yPI



2. The differential equation 

d where D º dz

1 1 1 = 2 (sin 2z ) = 2 sin 2z = - sin 2z 3 D +1 -2 + 1 1 = - sin(2 log x ) 3

2 ìï æ dy ö2 üï æ 2 ö 2 d y 1 + = C is of ç ÷ í ç ý ç dt 2 ÷ dt ÷ø ï è ø îï è þ

(a) Second order and third degree (b) Third order and Second degree (c) Second order and Second degree (d) Third order and Third degree [PI GATE-2005] 3. The following differential equation has 3 æ d2 y ö æ dy ö 3ç 2 ÷ + 4 ç + y2 + 2 = t ÷ ç dt ÷ dt è ø è ø

d2 y + y = 0 …(iii) dz 2

Let y = emz be a trial solution of (iii) Then (iii) gives

2

[CE GATE 2001]

2

3

2 2 Then, dy = x dy and d y2 - dy = x 2 d y2

dx

5x x3 - x2 +2 2 2 2

tial equation of order “2” So let us put logx = z dz

x2 - 5x + 2 2

(d) y = x 3 - x + 5x + 3 

Equation (i) is a linear homogeneous differen

x3 x2 - 3x - 2 3 2

(b) y = 3x 3 -

d2 y dy +x + y = sin(2 log x ) …(i) dx dx 2

boundary conditions y(0)

= 2 and y′(1) = –3is

Hence, the required solution is (sin y + cex)(sin y + ce–x) = 0



1 sin(2log x ). 3



(a) degree = 2, order = 1 (b) degree = 1, order = 2 (c) degree = 4, order = 3 (d) degree = 2, order = 3 [EC GATE 2005] 4. The degree of the differential equation d2 x + 2x 3 = 0 dt 2



(a) 0

is

(b) 1



(c) 2

(d) 3

[CE GATE 2007] 5. The order of the differential equation 3



d2 y æ dy ö 4 -t +ç ÷ +y =e dt 2 è dt ø

(a) 1

(b) 2

(c) 3 (d) 4 [EC GATE-2009]

8/3/2023 5:24:06 PM

Ordinary Differential Equations • 319

6. The order and degree of the differential equation d y3 + 4 æç dy ö÷ + y2 = 0 are, respecdx è dx ø tively 

7. Which of the following differential equations has a solution given by the function?

(c)

(a) (c)

d2 y d2 y + = - 9y = 0 y 9 0 (d) dx 2 dx 2



[PI GATE 2010] 3 2 8. The Blasius equation, d f3 + f d f2 = 0, is a

dn



2 dn

(a) second-order nonlinear ordinary differential equation (b) third-order nonlinear ordinary differential equation (c)  third-order linear ordinary differential equation (d) mixed order nonlinear ordinary differential equation [ME GATE-2010] 9. The solution of the differential equation dy + y2 = 0 is dx

(a) y = 

1 -x 3 (b) y = + c (c) x +c 3

cex

(d) unsolvable as equation is non-liner [ME GATE 2003] 10. Biotransformation of an organic compound having concentration “x” can be modeled using an ordinary differential equation dx + kx 2 = 0 , dt

where k is the reaction rate con-

stant. If x = a at t = 0, the solution of the equation is (a) x = ae–kt (b)

1 1 = + kt x a

(c) x = a(1 – e–kt) (d) x = a + kt  [CE GATE 2004]

EMEP.CH04_3PP.indd 319

(b)

(d) 

dy 5 dy 5 - cos3x = 0 (b) - cos3x = 0 dx 3 dx 3



dy + p(t ) y = q(t ) yn ; n > 0 dt

(a)

pö æ y = 5sin ç 3x + ÷ 3ø è





(b) 2 and 3 (d) 3 and 1 [CE GATE 2010]

(a) 3 and 2 (c) 3 and 3



11. Transformation to linear form by substituting v = y1–n of the equation

3

3

will be

dv + (1 - n) pv = (1 - n )q dt dv + (1 + n ) pv = (1 + n )q dt dv + (1 + n ) pv = (1 - n )q dt dv + (1 - n) pv = (1 + n )q dt

[CE GATE 2005]

12. The solution of the first-order differential equation x′(t) = – 3x(t), x(0) = x0 is (a) x(t) = x0e–3t (b) x(t) = x0e–3 (c) x(t) = x0e–1/3 (d) x(t) = x0e–1  [EE GATE 2005] 13. A spherical naphthalene ball exposed to the atmosphere loses volume at a rate proportional to its instantaneous surface area due to evaporation. If the initial diameter of the ball is 2 cm and the diameter reduces to 1 cm after 3 months, the ball completely evaporates in (a) 6 months (b) 9 months (c) 12 months (d) infinite time  [CE GATE2006] 14. The solution of the differential equation 2 dy + 2xy = e - x with y(0) = 1 is dx

(a) (1 + x ) e + x 2 (b) (1 + x ) e - x

2

(c) (1 - x ) e + x 2 (d) (1 - x ) e - x

2

[ME GATE 2006]



15. The solution of dy/dx = y2 with initial value y(0) = 1 is bounded in the interval (a) – ∞ ≤ x ≤ ∞ (b) – ∞ ≤ x ≤ 1 (c) x < 1, x > 1 (d) –2 ≤ x ≤ 2  [ME GATE 2007] 16. The solution for the differential equation dy = x2y dx

with the condition that y = 1 at x =

0 is

1

x3 (a) y = e 2x (b) In( y) = + 4 3

x3

x

3

(c) In( y) = (d) y = e 3 2  [CE GATE 2007]

8/3/2023 5:24:08 PM

320 • Engineering Mathematics Exam Prep 17. A body originally at 60°C cools down to 40°C in 15 minutes when kept in air at a temperature of 25°C. What will be the temperature of the body at the end of 30 minutes? (a) 35.2°C (b) 31.5°C (c) 28.7°C (d) 15°C  [CE GATE 2007] dy x =dx y

18. Solution of

at x = 1 and y = 3 is

2

2

(a) x – y = 2 (c) x2 – y2 = –2 

(b) x + y = 4 (d) x2 + y2 = 4 [CE GATE 2008]



tion

(c) y =

3

21. Match List-I with List-II and select the correct answer using the codes given below: List-I List-II

A. dy = y



B.

dy y =dx x



C. dy = x



D. dy = - x

dx

dx

dx



y

2. Straight lines 3. Hyperbolas

[EC GATE 2009] 22. Solution of the differential equation

EMEP.CH04_3PP.indd 320

x + 1 (d) y = x + 1 5 5

[ME GATE 2009] with y(0) = 1. Then the value of

y(1) is (a) e + e–1 (b) (c) 1 ée + e -1 ù ë û

(d) 2[e – e–1]

2



1é e - e -1 ù û 2ë

[IN GATE 2010]

25. The solution of the differential equation dy - y2 = 1 satisfying the conditions y(0) = 1 is dx

2

(a) y = e x (b) y = x (c) y = cot æç x + p ö÷ (d) y = tan æç x + p ö÷ 4 è

ø



4ø

è

[PI GATE 2010] the

26. Consider

dy = 1 + y2 x . The dx

(

)

differential

equation

general solution with con-

(a) y = tan

ö 2æx x2 + tan c (b) y = tan ç + c ÷ 2 è ø 2 æ

2

ö

(c) y = tan2 æç x ö÷ + c (d) y = tan çç x + c ÷÷ 2 è2ø è

ø

[ME GATE 2011]

27. The solution of the differential equation

A B C D (a) 2 3 3 1 (b) 1 3 2 1 (c) 2 1 3 3 (d) 3 2 1 2



5x

5



Codes:

dy 3y + 2x = 0 dx

5

stant c is

y



4 (b) y = 4 x + 4

4

dy + y = ex dx

1. Circles

x

with the condi-

24. Consider the differential equation

dt

(c) x (t ) = t 2 (d) x(t) = 3t2 2  [EC GATE 2008]

x4 1 + 5 x

is

dy + y = x4 dx



which one of the following can be particular

(a) x(t) = 3e–t (b) x(t) = 2e–3t

6 y(1) = 5

(a) y =

dx

20. Which of the following is a solution to the differential equation dx (t ) + 3x (t ) = 0?

(b) circles (d) hyperbolas [CE GATE 2009]

23. The solution of x

19. Consider the differential equations dy = 1 + y2 solution of this differentials equation? (a) y = tan(x + 3) (b) y = tan x + 3 (c) y = tan(y + 3) (d) x = tan y + 3  [IN GATE 2008]

(a) ellipses (c) parabolas

represents a family of

dy = ky, y(0) = c dx

is

(a) x = ce–ky (b) x = kecy (c) y = cekx (d) y = ce–kx  [EC GATE 2011] 28. The solution of the differential equation dy y + = x, dx x

with the condition that y = 1 at x

= 1, is

8/3/2023 5:24:11 PM

Ordinary Differential Equations • 321

(a) y =

2 x + 3x 2 3

(b) y =

2 x (c) y = + (d) y = 3

3

x 1 + 2 2x

æ (b) tan ç

2 x2 + 3x 3

æx+ yö (c) cos ç ÷ = x +c è 2 ø

x+ yö ÷ = y+c è 2 ø

[CE GATE 2011]



29. With K as a constant, the possible solution for dy

the first-order differential equation = e -3x dx is



3

3

(c) –3e–3x + K

(d) –3e–x + K [EE GATE 2011]

30. With initial condition x(1) = 0.5, the solution the differential equation, t dx + x = t is dt

(a)

1 2 1 x = t - (b) x = t 2 2

(c)

t2 x= 2



(d)

t x= 2

dy + 2y = 0 dx

(a)

for the boundary condi-

dy (c) dy + xy = e - y (d) + e- y = 0



32. The matrix form of the linear system dx dy = 3x - 5 y and = 4 x + 8 y is dt

(c)

d ìx ü é4 -5ù ìx ü í ý=ê úí ý dt î y þ ë3 8 û î y þ

(d)

d ì x ü é4 8 ù ì x ü í ý=ê úí ý dt î y þ ë3 -5û î y þ



2

(c) 1 + e x (d) 2e x

dy = cos ( x + y ) , dx

EMEP.CH04_3PP.indd 321

2

36. Consider the following difference equation x ( ydx + xdy ) cos

y y = y ( xdy - ydx ) sin x x

Which

of the following is the solution of the above equation (C is an arbitrary constant)? (a) x cos y = C (b) x sin y = C y



y

x

y x

x

y x

[CE GATE 2015]

37. The solution to 6yy′ – 25x = 0 represents a (a) family of circles (b) family of ellipses (c) family of parabolas (d) family of hyperbolas  [PI GATE 2015] 38. Consider the following differential equation:

[ME GATE 2014]

with “c” as a constant, is

(a) y + sin(x + y) = x + c

2

[ME GATE 2014]



dy = -5 y ; dt

initial condition: y = 2 at t = 0 The

value of y at t = 3 is

33. The general solution of the differential equation

is

(c) xy cos = C (d) xy sin = C

ìx ü é3 -5ù ìx ü (a) d í ý = ê úí ý dt î y þ ë4 8 û î y þ

(b)

dy = -2xy; y(0) = 2 dx

(a) 1 + e - x (b) 2e - x

–2x

d ì x ü é3 8 ù ì x ü í ý=ê úí ý dt î y þ ë4 -5û î y þ

[EC GATE 2014] 35. The solution of the initial value problem

(a) y = e (b) y = 2e (c) y = 10.95e–2x (d) y = 36.95e–2x  [CE GATE 2012]

dt

dx

dx

tion. y = 5 at x = 1 is –2x

dy dy + y = e - x (b) + xy = 0 dx dx

2

[EC, EE, IN GATE2012]

31. The solution of the ordinary differential equation

[ME GATE 2014]

34. Which ONE of the following is a linear nonhomogeneous differential equation, where x and y are the independent and dependent variables, respectively?

(a) - 1 e -3x + K (b) - 1 e3x + K 

x+ yö ÷ = y+c è 2 ø

æ (d) tan ç



(a) –5e–10 (c) 2e–15

(b) 2e–102e -10 (d) –15e2 [ME GATE 2015]

8/3/2023 5:24:14 PM

322 • Engineering Mathematics Exam Prep 39. The general solution of the differential equation 

dy 1 + cos2 y = dx 1 - cos2x

is

(a) tan y – cot x = c (c is a constant) (b) tan x – cot y = c (c is a constant) (c) tan y + cot x = c (c is a constant) (d) tan x + cot y = c (c is a constant) [EC GATE 2015] 40. Consider the differential equation



of the particle immediately after the removal of the force is _______. [CE GATE 2017] 45. The solution of the equation dQ + Q = 1 with dt

Q = 0 at t = 0 is (a) Q(t) = e–t–1 (b) Q(t) = 1 + e–t (c) Q(t) = 1 – et (d) Q(t) = 1 – e–t  [CE GATE 2017]



dx = 10 - 0.2x dt



x(0) = 1. The response x(t) for t > 0 is (a) 2 – e–0.2t (b) 2 – e0.2t –0.2t (c) 50 – 49e (d) 50 – 49e0.2t [EC GATE 2015]

ing through the point (1, 1) is (a) x (b) x2 (c) x–1 (d) x–2  [CE GATE 2018]

41. A differential equation di - 0.2i = 0 is appli-

47. A curve passes through the point (1, 0) and satisfies the differential equation

with initial condition

46. The solution of the equation x dy + y = 0 passdx

dt

cable over –10 < t < 10. If i(4) = 10, then i(–5)

dy x 2 + y2 y = + dx 2y x

is _______.

the curve is

[EE GATE 2015]



42. Which one of the following is the general solution of the first-order differential equation dy 2 = ( x + y - 1) dx

where x, y are real? (a) y = 1 + x + tan–1(x + c), where c is a constant. (b) y = 1 + x + tan(x + c), where c is a constant. (c) y = 1 – x + tan–1(x + c), where c is a constant. (d) y = 1 – x + tan(x + c), where c is a constant.  [EC GATE 2017]

2 ö 2 ö æ æ (a) ln ç1 + y2 ÷ = x - 1 (b) 1 ln ç1 + y2 ÷ = x - 1 ç 2 ç x ÷ x ÷

è

dy + 5ty = sin(t ) dt



equation when t belongs to the interval (a) (–2, 2) (b) (–10, 10) (c) (–10, 2) (d) (0, 10) [EE GATE 2017]



44. A particle of mass 2 kg is traveling at a ­velocity of 1.5 m/s. A force f(t) = 3t2 (in N) is applied to it in the direction of motion for a duration of 2 seconds, where t denotes time in seconds. The velocity (in m/s, up to one decimal place)

EMEP.CH04_3PP.indd 322

è

ø

è

ø

[EC GATE 2018]

48. If y is a solution of the differential equation dy y3 + x 3 = 0, y(0) = 1 , the value of y(–1) is dx ______? 

(a) –2 (c) 0

(b) –1 (d) 1 [ME GATE 2018]

49. The solution for the following differential equation with boundary conditions y(0) = 2 and y′(1) = –3 is

with y(1) = 2p. There

exists a unique solution for this differential

ø

(c) ln çæ1 + y ÷ö = x - 1 (d) 1 ln çæ1 + y ÷ö = x - 1 2 è xø x

43. Consider the differential equation

 (t2 - 81)

. The equation that describes



d2 y = 3x - 2 dx 2

3 2 (a) y = x - x + 3x - 6

3

2

2

(b) y = 3x 3 - x - 5x + 2 2

3

(c) y = x - x 2 - 5x + 2 2

2

2

(d) y = x 3 - x - 5x + 3 2 2 

[GATE 2001]

8/7/2023 6:26:13 PM

Ordinary Differential Equations • 323

50. Solve the differential equation,

55. Which of the following is a solution of the dif2 ferential equation d y2 + p dy + (q + 1) y = 0?

2

d y +y=x dx 2

(ii) at x = p , y = p 2 2 

(a) e–3x (b) xe–x (c) xe–2x (d) x2e–2x  [ME GATE 2005] 56. A solution of the following differential equa[GATE 2001]

d2 y dy +2 + 17 y = 0; dx dx 2

51. The solution of dy æ p ö

y(0) = 1, = 0 in the range 0 < x < p  dx ç 4 ÷ è

dx

dx

With the following conditions: (i) at x = 0, y = 1

4

ø

is

tion is given by

d2 y dy -5 + 6y = 0 2 dx dx

(a) y = e2x + e–3x (b) y = e2x + e3x (c) y = e–2x + e3x (d) y = e–2x + e–3x  [EC GATE 2005] 2

57. For d y2 + 4 dy + 3 y = 3e2x , the particular intedx gral isdx

given by (a) e - x æç cos4 x + 1 sin 4 x ö÷

(a)

(b) e x çæ cos4 x - 1 sin 4 x ÷ö 4



(c) e -4 x æç cos x - 1 sin x ö÷

58. For the initial value problem y′′ + 2y′ + 101y = 10.4ex, y(0) = 1.1 and y′(0) = –0.9; various solutions are written in the following groups. Match the type of solution with the correct expression. Group -1 P. General solution of homogeneous equations Q. Particular integral R. Total solution satisfying boundary conditions Group-II (1) 0.1ex (2) e–x[A cos 10x + B sin 10x] (3) e–x cos 10x + 0.1ex (a) P – 2, Q – 1, R – 3 (b) P – 1, Q – 3, R – 2 (c) P – 1, Q – 2, R – 3 (d) P – 3, Q – 2, R – 1  [IN GATE 2006]

4

è

ø

è

ø

è

4

ø

1

æ ö (d) e -4 x ç cos4 x - sin 4 x ÷ 4 è ø  [CE GATE 2005]

52. The complete solution of the ordinary differential equation d2 y dy +p + qy = 0 dx dx 2



Then, p and q are (a) p = 3, q = 3 (c) p = 4, q = 3 

is y = c1e–x + c2e–3x. (b) p = 3, q = 4 (d) p = 4, q = 4 [ME GATE 2005]

53. The general solution of the differential equation (D2 – 4D + 4)y = 0 is of the form d æ ö ç given D = dx and C1 ,C2 are constants ÷ è ø

(a) C1e2x (b) C1e2x + C2e–2x 2x –2x (c) C1e + C2e (d) C1e2x + C2xe2x  [IN GATE 2005] 54. For the equation,  x′′ (t) + 3x′(t) + 2x(t) = 5, the solution x(t) approaches to the following value as t → ∞ 5 2



(a) 0

(b)



(c) 5

(d) 10 [EE GATE 2005]

EMEP.CH04_3PP.indd 323

1 2x 1 e (b) e2x 15 5

(c) 3e2x (d) C1e–x + C2e–3x [ME GATE 2006]

2 59. The general solution d y2 + y = 0 is

dx

(a) y = P cos x + Q sin x (b) y = P cos x (c) y = P sin x (d) y = P sin2x 

[EC GATE 2007]

8/3/2023 5:24:17 PM

324 • Engineering Mathematics Exam Prep 60. Given that x + 3x = 0 , and x(0) = 1, x (0) = 0 , what is x(1)? (a) –0.99 (b) –0.16 (c) 0.16 (d) 0.99  [ME GATE 2008] 61. It is given that y′′ + 2y′ + y = 0, y(0) = 0, y(1) = 0 what is y? (a) 0 (b) 0.37 (c) 0.62 (d) 1.13  [ME GATE 2008]

(c) x(t) = –e–6t + 2e–4t (d) x(t) = e–2t + 2e–4t 

67. Consider the differential equation



62. The solutions of the differential equations

-(1+i ) x (1-i ) x (1+i ) x -(1-i ) x ,e ,e (c) e (d) e  [PI GATE 2008]

63. The homogeneous part of the differential equation 2

(a) y = (C1x + C2)e–3x (b) y = C1e3x + C2e–3x (c) y = (C1x + C2)e–3x + x (d) y = (C1x + C2)e3x + x 

d y dy +P + qy = 0 (p and q, r are contants) has dx dx 2

(b) P2 – 4q < 0 (d) P2 – 4q = r [PI GATE 2009]

64. The solution to the ordinary differential equation

d2 y dy + - 6y = 0 dx 2 dx 3x

–2x

3x

2 2x

(a) y = c1e + c2e (b) y = c1e + c e (c) y = c1e–3x + c2e2x (d) y = c1e–3x + c2e–2x  [EE GATE 2010] 65. The function n(x) satisfies the differential equad2n( x ) n( x ) - 2 =0 dx 2 L

(a) –1 (b) –e–1 (c) –e–2 (d) e2 [IN GATE 2011] d2 y dy +6 + 9 y = 9x + 6 with C1 and C2 as dx dx 2 constant is

(a) e -(1+i )x , e -(1-i )x (b) e(1+i )x , e(1-i )x

real distinct roots if (a) P2 – 4q > 0 (c) P2 – 4q = 0 

d2 y dy +2 + y = 0 with boundary conditions 2 dx dx y(0)=1 & y(1)=0. The value of y(2) is

68. The solution of the differential equation

d2 y dy +2 + 2 y = 0 are dx dx 2



[EE GATE 2010]

[PI GATE 2011]

69. Consider the differential equation x2

d2 y dy +x - 4y = 0 2 dx dx

With the boundary condition of  y(0) = 0 and y(1) = 1. The complete solution of the differential equation is æ px ö (a) x2 (b) sin ç ÷ è 2 ø

(c) ex sin æç px ö÷ (d) e- x sin æç px ö÷ è 2 ø

è 2 ø

[ME, PI GATE 2012]



where L is a constant.

70. The solution to the differential equation

The boundary conditions are: n(0) = K and n(∞) = 0. The solution to this equation is (a) n(x) = K exp (x/L) (b) n( x ) = K exp( -x / L ) (c) n(x) = K2 exp (–x/L) (d) n(x) = K exp (–x/L)  [EC GATE 2010]

where k is constant, subjected

tion

66. For

the

d2 x dx +6 + 8x = 0 dt dt 2 = 1 and dx = 0, dt t =0

differential

with initial conditions x(0) the solution is

(a) x(t) = 2e–6t – e–2t (b) x(t) = 2e–2t – e–4t

EMEP.CH04_3PP.indd 324

equation

d2u du -k =0 2 dx dx

to the boundary conditions u(0) = 0 and u(L) = U is kx

æ1-e ö (a) u = U x (b) u = U çç kL ÷ ÷ è1 - e

L

- kx

ø

(c) u = U çç 1 - e-kL ÷÷ (d) u = U ç 1 + ekL ÷ ç ÷ æ

è1 - e



ö

æ

ø

è1 + e

kx

ö ø

[ME GATE 2013] 71. The maximum value of the solution y(t) of the differential equation y(t ) + y(t ) = 0 with initial conditions y (0) = 1 and y(0) = 1 , for t ≥ 0 is

8/3/2023 5:24:20 PM

Ordinary Differential Equations • 325



(a) 1

(b) 2

77. If the characteristic equation of the differen-

(c) p (d) 2  [IN GATE 2013] 72. The solution for the differential equation d2 x + 9x = 0 dt 2

and dx dt

t =0

with initial conditions x(0) = 1

= 1,

2 differential equation d y2 = -12x 2 + 24 x - 20.

dx

(b) sin 3t + 1 cos3t + 2 3

The boundary conditions are at x = 0, y = 5

3

(d) cos 3t + t [EE GATE 2014 ]

73. Consider two solutions x(t) = x1(t) and x(t) = x2(t) of the differential equad 2 x (t ) dt 2

such

+ x (t ) = 0,t > 0,

t = 0, x2 = 0, x1 (t ) W (t ) = dx1 (t ) dt



dx1 (t ) dt

x 2 (t )

dx2 (t )

=0.

that

and x = 2, y = 21. The value of y at x = 1 is _______. 79. A solution of the ordinary differential equation

The Wronskian

dx

conditions

dy = 2at x = 10, f (15) = ? dx

[ME GATE 2014]



75. If a and b are constants, the most general solution of the differential equation d2 x dx +2 +x =0 dt dt 2

dt

[EE GATE 2015 ]

is

80. The solution of the differential equation d2 y dy +2 +y=0 2 dt dt



76. Consider the differential equation

d2 y dy  x +x - y = 0 . Which of the following is a dx dx 2

(a) (2 – t)et (c) (2 + t)e–t

(b) (1 + 2t)e–t (d) (1 – 2t)et [EC GATE 2015]

81. The solution to x2y′′ + xy′ – y = 0 is (a) y = C1x2 + C2x–3 (b) y = C1 + C2x–2 (c) y = C1x +

C2 x

(d) y = C1x + C2x4



[PI GATE 2015] 82. Find the solution of

2

d y =y dx 2

which passes

through the origin and the point



3ö æ ç ln2, 4 ÷ è ø

2

solution to this differential equation for x = 0? (a) ex (b) x2 (c) 1/x (d) In x  [EE GATE 2014]

with y(0) = y′(0) = 1



(a) ae–t (b) ae–t + bte–t (c) aet + bte–t (d) ae–2t  [EC GATE 2014]

EMEP.CH04_3PP.indd 325

. The value of dy (0) is



2 74. If y = f(x) is the solution of d y2 = 0

y = 5 at x = 0 and

1 - 3e e3

is such that y(0) =

_______.

at t = p/2 is

boundary

d2 y dy +5 + 6y = 0 dt dt 2

2 and y(1) = -

(b) –1 (d) p/2 [ME GATE 2014]

the

[CE GATE 2015]



at

dt

(a) 1 (c) 0

with

roots, then the values of a are (a) ±1 (b) 0, 0 (c) ±j (d) ±1/2  [EC GATE 2014] 78. Consider the following second-order linear

1 sin 3t + cos3t 3

tion

dx

dx

is

(a) t2 + t + 1 (c) 

2 tial equation d y + 2a dy + y = 0 has two equal 2



(c) 

1 2 1 x y = e - e-x 2

(a) y = e x - e - x (b) y =

(

1 x e + e-x 2

(

) (d) y = 12 e

x

)

+ e-x

[ME GATE 2015]

8/3/2023 5:24:22 PM

326 • Engineering Mathematics Exam Prep 83. Consider the differential equation

d2 x (t ) dx (t ) 2 + 3 dt + 2x (t ) = 0. dt

Given x(0) = 20 and

x(1) = 10/e, where e = 2.718, then the value of x(2) is ______? [EC GATE 2015]



88. A function y(t), such that y(0) = 1 and y(1) = 3e–1, is a solution of the differential equation

84. The particular solution of the initial value problem given below is

d2 y dy +2 + y = 0 . Then y(2) is 2 dt dt



with y(0) = 3 and (a) (3 – 18x)e–6x (c) (3 + 20x)e–6x



dx

with the two boundary conditions dy

dy dx

x =0 =

and

-36

(b) (3 + 25x)e–6x (d) (3 – 12x)e–6x [EC GATE 2016]

85. If y = f(x) satisfies the boundary value probp

æ ö lem y¢¢ + 9 y = 0, y(0) = 0, y ç ÷ = 2, then y æç p ö÷ è2ø è4ø is ______.

[ME GATE 2016]



86. Let y(x) be the solution of the differential equation

d2 y dy -4 + 4y = 0 2 dx dx

tions y(0) = 0 and dy of y(1) is _____. dx

x =0

with initial condi-

=1 .



Then the value

are

(a) éëc1 + c2 x + c3 sin( 3x ) + c4 cos( 3x )ùû and

é3x 4 - 12x 2 + c ù ë û

(b) éëc2 + c3 sin( 3x ) + c4 cos( 3x )ùû and éë5x 4 - 12x 2 + c ùû (c) éëc1 + c3 sin( 3x ) + c4 cos( 3x )ùû 4

has

d2 y dy +2 - 5y = 0 2 dx dx

(a) K1e(

)

-1+ 6 x

(b) K1e(

( (c) K1e

)

-1+ 8 x

)

-2 + 6 x

(d) K1e( 

)

-2 + 8 x

+ K 2e

( -1- 6 )x

+ K 2e + K 2e

( -1- 8 )x

( -2- 6 )x

+ K 2e

( -2- 8 )x [EC GATE-2017]

92. Consider the differential equation 3y′′(x) + 27y(x) = 0 with initial conditions y(0) = 0 and y’(0) = 2000. The value of y at x = 1 is ______.  [ME GATE 2017] 93. The solution (up to three decimal places) at x = 1 of the differential equation d2 y dy +2 +y=0 dx dx 2

(d) éëc1 + c2 x + c3 sin( 3x ) + c4 cos( 3x )ùû

tions y(0) = 0 and [CE GATE 2016]

in terms of arbitrary

constants K1 and K2 is

and éë3x - 12x + c ùû

EMEP.CH04_3PP.indd 326

-1

p= 2

x =0 = 1

91. The general solution of the differential equa-

2

and éë5x 4 - 12x 2 + c ùû 

x=

dx

(a) no solution (b) exactly two solution (c) exactly one solution (d) infinitely many solutions [ME GATE 2017]

tion

87. The respective expressions for complimentary function and particular integral part of the solution of the differential equation

dy dx

90. Consider the following second-order differential equation:  y′′ – 4y′ + 3y = 2t – 3t2. The particular solution of the differential equation is (a) –2 – 2t –t2 (b) –2t – t2 2 (c) 2t – t (d) –2 – 2t – 3t2  [CE GATE 2017]

[EE GATE 2016]



d4 y d2 y + 3 2 = 108x 2 4 dx dx

(b) 5e–2 (d) 7e–2 [EE GATE 2016]

2 89. The differential equation d y2 + 16 y = 0 for y(x)

d2 y dy + 12 + 36 y = 0 dx dx 2



(a) 5e–1 (c) 7e–1



subject to boundary condidy dx

x =0

= -1 is ______.

[CE GATE 2018]

8/7/2023 6:29:19 PM

Ordinary Differential Equations • 327

94. The position of a particle y(t) is described by the differential equation

d y dy 5 y . =+ dt 4 dt 2

The initial conditions are y(0) = 1 and dy dt

t =0

= 0.

The position (accurate to two decimal places) of the particle at t = p is ______. [EC GATE 2018]



the

95. Consider 2

2

d y + 8y = 0 dt 2

and dy dt

t =0

differential

equation

with initial conditions y(0) = 0

= 10 .

dy 3x 2 = - 2x + C.............(i) dx 2 3 Now y¢(1) = -3 Þ -3 = - 2 + C ( from (i )) 2 5 ÞC = 2 5 dy 3x 2 \ (i) gives, = - 2x dx 2 2 Þ

2

Integrating both sides with respect to “x,” we get,

ò

Then the value of y (correct

1. (c) 6. (a) 11. (a) 16. (d) 21. (a) 26. (d) 31. (d) 36. (c) 41. 1.652 45. (d)

50.   cosx +x

2. (c) 7. (c) 12. (a) 17. (b) 22. (a) 27. (c) 32. (a) 37. (d) 46. (c) 51. (a)

55. (c) 56. (b) 60. (d) 61. (a) 65. (d) 66. (b) 70. (b) 71. (d) 75. (b) 76. (c) 80. (b) 81. (c) 84. (a) 85. (–1) 89. (a) 90. (a) 93. (–0.37)

Answer Key 3. (b) 8. (b) 13. (a) 18. (d) 23. (a) 28. (d) 33. (d) 38. (c) 42. (d) 47. (a) 52. (c) 57. (b) 62. (a) 67. (c) 72. (c) 77. (a) 82. (c) 86. (7.38) 91. (a) 94. (–0.21) Explanations



x3 5x - x2 +K 2 2 x3 5x or, y = - x2 + K .............(ii ) 2 2 Now y(0) = 2 gives, 2 = 0 - 0 - 0 + K , i.e, k = 2.



So from (ii) we have, y =

ò

d2 y = 3x - 2 dx 2

Þ



EMEP.CH04_3PP.indd 327

ò ò

58. (a) 59. (a) 63. (a) 64. (c) 68. (c) 69. (a) 73. (a) 74. 35 78. 18 79. –3 83. (0.8566) 87. (a) 88. (b) 92. (94.08) 95. 4.54

5x x3 - x2 +2. 2 2

2. (c) The highest order derivative occurring in

2 the equation is d y2 . So order of the equation

dy

= 2. Also the power of the highest order ded2 y

rivative, i.e, the power of 2 is “2.” Hence, dy degree of the equation = 2. 3. (b) The highest order derivative occurring in 2 the equation is d y2 . So order of the equation

dy

= 2. Also the power of the highest order de2

rivative, i.e, the power of d y2 is “1.” Hence, degree of the equation = 1.dy 4. (b) The highest order derivative occurring in dt

d2 y dx = (3x - 2)dx + C dx 2

ò

2

d æ dy ö x dx = 3 - 2x + C ç ÷ dx è dx ø 2

x2 æ dy ö Þ dç =3 - 2x + C ÷ 2 è dx ø

ò

5. (b) 10. (b) 15. (c) 20. (b) 25. (d) 30. (d) 35. (b) 40. (c) 44. 5.5 49. (c) 54. (b)

2 the equation is d x2 . So degree of the equa-

1. (c)

Þ

4. (b) 9. (a) 14. (b) 19. (a) 24. (c) 29. (a) 34. (a) 39. (c) 43. (a) 48. (c) 53. (d)

ò

or, dy =

upto two decimal places) at t = 1 is ______. [PI GATE 2018]



ìï 3x 2 dy 5 ïü dx = í - 2x - ý dx + K dx 2 2 þï ïî

tion = power of the highest order derivative = power of

d2 x dt 2

= 1.

5. (b) The highest order derivative occurring in 2 the equation is d 2y . So order of the equation dt = 2.

8/3/2023 5:24:26 PM

328 • Engineering Mathematics Exam Prep

6. (a) 3

3

d y æ dy ö 2 +4 ç ÷ +y =0 dx 3 è dx ø



ì ü ⇒ d y3 + 16 ïíæç dy ö÷ + y2 ïý , which is radical free. ïîè dx ø

ïþ

 \ order of the equation = 3 (since highest order 3 derivative occurring in the equation is d y3 ).

dx

The degree of the equation is “2” (since power of highest order derivative is “2”).







d2 y d2 y = 9 y Þ + 9y = 0 dx 2 dx 2



8. (b)

d3 f f d 2 f = 0, 3 + 2 dn2 dn

is third-order ordinary differ-

ential equation (since highest order derivative occurring in the equation is

d3 f dn3

) and it is non-

linear (since the product f ´

d2 f dn2

is not allowed

9. (a)

dy dy + y2 = 0 Þ 2 = -dx dx y (by separation of variables)

Integrating, we get

dy = - dx + K y2 1 or, - = -x + K y 1 or, = x - K = x + C (where C = - K ) y 1 or, y = . x +C

ò



ò

10. (b) dx + kx 2 = 0 Þ dx2 = -k dt

dt x (by separation of variables)

EMEP.CH04_3PP.indd 328

Thus, we get,

1 1 = -kt ( from (i )) x a 1 1 or, = kt + x a

dy + p(t ) y = q(t ) yn dt 1 dy 1 Þ n + n -1 p(t ) = q(t ) dt y y - n dy Þy + y1-n p(t ) = q(t )........(i) dt

Let us put v = y1–n. Then we have dv dy = (1 - n) y -n dt dt 1 dv - n dy or, y = dt (1 - n ) dt



Substituting this in (i), we get,



1 dv + vp(t ) = q(t ) (1 - n ) dt dv or, + (1 - n)vp(t ) = (1 - n)q(t ) dt

in case of linear differential equation).



1 = -kt + C........(i ) x At x = a,t = 0. So (i ) gives, 1 - = -k ´ 0 + C a 1 or, C = a

11. (a)

pö d2 y æ Þ = -45sin ç 3x + ÷ 2 3ø dx è Þ

ò

= -k dt + C

-

7. (c) pö æ y = 5sin ç 3x + ÷ 3ø è pö dy æ Þ = 15 cos ç 3x + ÷ dx 3ø è

2

or, -

3

dx

dx

òx



3

Integrating, we get

12. (a)



x ¢(t ) = -3x (t ) dx dx Þ = -3x Þ = -3dt dt x

Integrating, we get, dx = -3 dt + ln C x (where ln C is the constant of integration) or, ln x = -3t + ln C or, ln x - ln C = -3t x or,ln = -3t i.e; x = Ce -3t ..........(i ) C Given that x (0) = x 0 . so (i) gives,

ò

ò

x = Ce0 i.e; C = x

0 0 Hence, x = x0e–3t..

8/3/2023 5:24:28 PM

Ordinary Differential Equations • 329

13. (a) If V be the volume and A be the surface area of the ball, then ATQ,



dV = -kA …(i) dt





4 3

Here V = pr3 , A = 4 pr 2 .



dV dr \ = 4 pr 2 dt dt



dr 4 pr = -k ´ 4 pr 2 dt dr or, = -k dt or, dr = -kdt 2



1 0.5 = -k ´ 3 + 1 i.e; k = . 6 (iii),

1 r = - t + 1 …(iv) 6

When the ball completely evaporates, we have r = 0. In this case, we get from (iv), 1 0 = - t + 1, i.e, t = 6 months 6

14. (b) The given equation is 2 dy + 2xy = e - x , dx

Integrating factor, I.F = e ò



Hence, the solution is given by

Pdx

= eò

2 xdx

…(i)

1=-

1 i.e; C = -1 0+C 1 x -1





dy = x2y dx

Þ

dy = x 2dx y

Þ

ò

dy = x 2dx + C1 y

ò

Þ log e y =

x3 + C1 3

x 3 +C 1

Þ y=e3

x3

= eC1 ´ e 3

x3

Þ y = C ´ e 3 (where C = eC1 )…(i) Given x = 0, y = 1; so (i) gives 0

1 = C ´ e3



i.e; C = 1



\  y = e 3 is the required solution.

x3

17. (b) By Newton’s law of cooling, we have = ex .



dq = -k( q - q0 ) dt

or, ye x = x + C..........(i )

or,

ò q - q = ò -kdt + C

Given that at x = 0, y = 1. So (i) gives,

or, ln (q – q0) = –kt + C1 or, q – q0 = Ce–kt (where C = eC1) or, q = q0 + Ce–kt Given that, q0 = 25°C Now at t = 0, q = 60° ; so from (i) we get 60 = 25 + C.e0 ⇒ C = 35 \  q = 25 + 35e–kt  …(ii)

ò

= e

x2

´e

-x2

dx + C

1e = 0 + C i.e., C = 1 Hence, the required particlular solution is 2

2

ye x = x + 1 i.e; y = e - x ( x + 1).

15. (c)

EMEP.CH04_3PP.indd 329



2

0



1 x +C

dq = -kdt q - q0

x2

2



\y = -

or,

y´e



1 = x +C y

16. (d)

which is linear in y.





Þ-

+C

Now y is bounded when x – 1 ≠ 0, i.e, when x ≠ 1, i.e, when x < 1  or  x > 1.



Now substituting this value of r in equation (iii) we get,



2

Hence, y = -

Integrating, we get, r = –kt + C…(ii) At t = 0, r = 1. So (ii) gives, 1 = –k × 0 + C  i.e., C = 1. So r = –kt + 1 …(iii) Now at t = 3 months, r = 0.5 cm; we get from



dy

ò y =ò dx

When x = 0, y = 1; So (i) gives

Substituting this in (i) we get,



⇒

dy = y2 dx

dq

0

1

8/3/2023 5:24:31 PM

330 • Engineering Mathematics Exam Prep Again at t = 15 minutes, q = 40°C. So from (ii) we have \  40 = 25 + 35e(–k×15) Þe



-15k

3 = 7

Now at t = 30 minutes, we get q = 25 + 35e -30k = 25 + 35( e -15k )2



2

æ3ö = 25 + 35 ´ ç ÷ = 31.4 o C  31.5o C è7ø



18. (d)

dy -x = dx y



⇒ y dy = – x dx

ò

ò

Þ y dy = - x dx + C



Þ

y2 x2 =+C 2 2 …(i)

Given x = 1, y = 3 ; so (i) given

( 3)

2



2



=

-12 +C 2

i.e; C = 2

The required solution is i.e.,  x2 + y2 = 4. 19. (a)



dy = 1 + y2 dx



⇒

dy = dx 1 + y2

Integrating, we get, tan–1(y) = x + c or, y = tan(x + c), where c can take any arbitrary value. 20. (b)



Þ



Þ

dx = -3dt + C x

ò

⇒x=t+c ⇒ x = e–3t+c ⇒ x = ec.e–3t = c1e–3t(c1 = ec) For C1 = 2, x = 2e–3t. 21. (a)

A. dy = y dx

EMEP.CH04_3PP.indd 330

ò

dy = y

ò

dx x +log c



B. dy = - y



⇒

dx x dy -dx = Þ y x

ò

dy dx =y x

ò

+ log c

⇒ log y = –log x + log c ⇒ log y + log x = log c ⇒ log yx = log c ⇒ yx = c ⇒ y = c/x, which represents a family of hyperbolas.

C. dy = x



⇒ ydy = xdx

dx

y

ò

ò

Þ y dy = x dx + Þ

C2 2

y2 x 2 C 2 = + 2 2 2



⇒ y2 – x2 = e2



⇒

y2 x 2 =1 c2 c2

which represents a family of hy-

perbolas.

D. dy = -x dx

y

ò

ò

Þ y dy = - x dx +

⇒

c2 2

y2 x 2 c 2 + = 2 2 2

⇒ x2 + y2 = c2 which represents a family of a circle. 22. (a)

dx = -3dt x

ò

Þ

⇒ log y = log x + log c = log cx ⇒ y = cx, which represents a family of straightlines.

dx = -3x dt



dy dx = y x



y2 - x 2 = +2, 2 2

Þ

x



3y

dy + 2x = 0 dx

⇒ 3 ydy = –2x dy

ò

ò

Þ 3 y dy = - 2x dx + Þ

C 2

3 2 x2 C y = -2 + 2 2 2

⇒ 3 y2 + 2x2 = C Þ

x2

æ Cö çç ÷÷ è 2ø

2

+

y2

æ Cö çç ÷÷ è 3 ø

2

=1,

which represents a

family of ellipse.

8/3/2023 5:24:34 PM

Ordinary Differential Equations • 331 dy + y = x4 dx

23. (a) x

dy æ y ö + = x 3 …(i) dx çè x ÷ø



Þ



which is a liner equation of the form



dy + Py = Q dx



\  integrating factor (I.F.) = e ò 1 dx Inx x = e ò = e = x

Pdx



ò

y ´ I .F . = Q ´ ( I .F .) dx + C

ò



3

or, y ´ x = x ´ x dx + C



or, yx =



x +c 5 …(ii)

6 1 ´1 = + C 5 5



So we get from (ii)

x5 +1 5

4 i.e; y = x + 1 .

5

x

dy + y = ex , dx



Given equation is



which is linear in y. dx x ∴  I. F = e ò = e

Hence, the general solution of the equation is

ò = ò e e dx + C = ò e

y ´ I.F = (I.F) ex dx + C or, ye

x

x x

e2 x or, y e = +C 2 x



2x

dx + C

Thus, the solution is given by y ex =



e 1 + 2 2

\ y(1) =



25. (d)

EMEP.CH04_3PP.indd 331

2x

.

e e -1 e + e -1 + = . 2 2 2

dy - y2 = 1 dx

2

x2 +C 2

Þ tan -1 y =

æ x2 ö Þ y = tan ç + C ÷. ç 2 ÷ è ø



dy = ky dx

Þ

dy = kdx y

Þ

ò

dy = k dx + logC y

ò

⇒ log y = kx + log C Which satisfies y(0) = C Now, (i) ⇒ log y – log C = kx or, log y = kx

…(i)

C

...(i )

Now using y(0) = 1,we get from (i ) C =



dy

ò 1 + y = ò xdx + C

27. (c)

24. (c)





Hence, we have yx =



dy = (1 + y2 )x dx



6 1 - =1 5 5

ÞC =



6 . 5

pö æ \ required solution is y = tan ç x + ÷ 4ø è

Þ

5

Given that y(1) =

⇒ tan–1(y) = x + c…(i)

26. (d)

So the solution is given



ò

4

x



ò

Given y(0) = 1, so (i) gives, c = p .

Here, P = 1 and Q = x3

dy = 1 + y2 dx dy Þ = dx 1 + y2 dy Þ = dx + c 1 + y2 Þ

or, 1 . 2

y = e kx C

or, y = Cekx 28. (d)

The equation dy + y = x is a liner differential dx

x

dy

1 equation of the form and + Py = Q with P = dx x Q = x..



Here, integrating factor (I. F) = eò

Pdx

= eò x

1 dx

= elnx = x

8/3/2023 5:24:37 PM

332 • Engineering Mathematics Exam Prep

Therefore, the solution is given by

31. (d)

ò

y ´ ( I .F .) = Q ´ ( I .F .) dx + C



ò Þ yx = ò x dx + C

Þ y.x = ( x .x )dx + c



C



2



x2 +C 3

Þ yx =



Þy=

3

x C + 3 x

…(i)

For y(1) = 1, we get from (i),



12 C + =1 3 1



i.e; C = 2/3



So, the required is solution is 29. (a)

2

x 2 y= . + 3 3x

dy = e -3x dx



ò

ò

Þy=



-3x

dx + k K



-3 x

e 1 + K = - e -3x + K -3 3

30. (d)

dt



t

\ Integrating factor (I. F.) 1 ò dt Pdt = e t

=e

log e t

= t.

ò

or, x.t = ò1.t.dt + C



ò

⇒ log y = –2x + log C ⇒ log y – log C = –2x Þ log Þ

y = -2x C

y = e -2x C

⇒ y = Ce–2x \  y(1) = 5 ⇒ 5 = Ce–2 i.e., C = 5e2 = 36.95 Hence, (i) gives y = 36.95e–2x.

2



t 2

dx = 3x - 5 y dt

…(i)

and dx = 4 x + 8 y dt

C t

…(i)

Þ

dz - 1 = cosz dx

Þ

ò 1 + cos z = ò dx + C

Þ

1 æzö sec2 ç ÷ dz = x + c 2 è2ø

dz

ò

æzö Þ tan ç ÷ = x + c è2ø æx+ yö Þ tan ç ÷ = x +c. è 2 ø

dx

2



where P and Q are functions of x or constants. Here only option (a) is in the above form. 35. (b)

⇒C=0

t x= 2

dx

34. (a) We know that the general form of linear differential equation in “y” is dy + Py = Q,

Given x (1) = 1 ; so (i) gives 1 C 1 + = 2 1 2

dx

\  dy = cos( x + y)



or, x = +

EMEP.CH04_3PP.indd 332

dy = -2dx + log C y

dx

2

or, xt = t + C

Hence,

ò

ìx ü é3 - 5ù ìx ü So d í ý = ê úí ý. dt î y þ ë4 8 û î yþ

Hence, the solution is given by x ´ ( I .F .) = Q ´ ( I .F .) dt + C



Þ

Then, dz = 1 + dy .

dx x Þ + =1 dt t

= eò

dy = -2 y dx

33. (d) Let us put z = x + y

dx t +x =t dt

which is a liner differential equation of the dx form + Px = Q where P = 1 and Q = 1.



Þ

32. (a)

Þ dy = e





dy + 2y = 0 dx

dy

is the required solution.



dx

= -2xy Þ

dy dx

+ 2xy = 0.

8/3/2023 5:24:41 PM

Ordinary Differential Equations • 333



2 2xdx = ex I.F. = e ò So, the solution is given by



38. (c) dy

ò

= -5 y ⇒ dt (i)

y ´ I .F = I .F ´ 0 dx + C 2

or, ye x = 0 + C = C Now y(0) = 2 Þ 2 ´ e0 = C Þ C = 2 Hence, ye



x2

= 2 i.e; y = 2e

-x2

.

y x

36. (c) x ( ydx + xdy) cos = y( xdy - ydx )sin

y x

⇒ ydx + xdy = y tan y …(i)



xdy - ydx

x

x







or, xdv + 2vdx = v tan v xdv

or,



2v dx 1+ = v tan v x dv

or, 2v dx = v tan v - 1 x dv

or, 2 dx = æç tan v - 1 ö÷ dv x v è

ø

or, 2 dx = æç tan v - 1 ö÷ dv + log c òx ò v è





Þ x2 =

cx æ yö = sec ç ÷  (since y èxø

Þ xy cos



ø

c sec v v

v = y/x)

y =c x

Þ

y

2

=

2

6

x dx Þ 2

C 25 x . 6 2 12

2

2

2

2

2

2

Þ 25x - 6 y = C Þ



EMEP.CH04_3PP.indd 333

2

æC ö ç ÷ è5ø

2

-

y

25 C2 xdx 6 12

ò

2

Þ 6 y = 25x - C x

ò

ydy =

2

æ C ö ç ÷ è 6ø

2

2

dy 1 + cos2 y = dx 1 - cos2x

Þ

dy dx = 1 + cos2 y 1 - cos2x

Þ

dy dx = 2cos2 y 2sin 2 x

ò

æ 10 - 0.2x ö Þ log ç ÷ = -0.2t è 9.8 ø 10 - 0.2x Þ = e -0.2t 9.8 5 - 0.1x 50 - x Þ = e -0.2t Þ = e -0.2t 4.9 49

dy Þ 6y = 25x dx 25

or, y = e -5t or, y = 2e–5t.

Hence, at t = 3, y = 2e–15.

ò

37. (d) 6 yy¢ - 25x = 0

Þ y dy =

y = -5t 2

dx = 10 - 0.2x dt dx Þ = dt 10 - 0.2x dx Þ = dt + c 10 - 0.2x log(10 - 0.2x ) Þ =t+c ( -0.2) log 9.8 \ x (0) = 1 Þ c = ( -0.2) log(10 - 0.2x ) log 9.8 Hence, =t ( -0.2) ( -0.2)

or, log x 2 = log æç c sec v ö÷ Þ x2 =

⇒ ln y = –5t + C …

40. (c)

or, 2 log x = log | sec v| – log v + log c v

ò

⇒ sec2 ydy = cos ec2xdx Integrating, we get tan y = – cot x + c or, tan y + cot x = c.

ø

è

dy = - 5dt y

39. (c)

2

vxdx + vxdx + x dv = v tan v vxdx + x 2dv - vxdx

ò

Given that at t = 0,y = 2; so (i) gives, ln 2 = C. Thus, we get, ln y = –5t + ln 2

or, ln

Let y = v . x Then dy = vdx + xdv. So (i) gives,

which represents the family of hyperbolas.



Þ x = x (t ) = 50 - 49 e -0.2t .

41. 1.652 = 1,



di - 0.2i = 0 dt

Þ

di = 0.2dt i

8/3/2023 5:24:44 PM

334 • Engineering Mathematics Exam Prep Þ





di

ò i = ò 0.2dt + log C



⇒ i = Ce

0.2t

…(i)





dx

ò

or, tan–1 t = x + c or, tan–1(x + y – 1) = x + c or, x + y – 1 = tan(x + c) or, y = 1 – x + tan(x + c)



(



which is of the form



EMEP.CH04_3PP.indd 334

5t t 2 - 81



=e

⇒ (8 – 0) = 2(V – 1.5) ⇒ V = 5.5 m/s 45. (d)

The given equation is dQ + Q = 1 dt 1dt \  I.F. = e ò = et



2

dy + Py = Q. dt

and Q = 2sin t . t - 81

= (t - 81)

5 2

So the solution is given by Q.et = ò 1.et dt + C = et + C

Integrating both sides, we get, xy = C …(i) Given that the solution passes through the point (1, 1). So (i) gives C = 1. Hence, the required solution is xy = 1 i.e; y=

1 = x -1 . x

47. (a) The given differential equation is dy x 2 + y2 y = + dx 2y x

So I.F 5t 5 2t dt 2 Pdt 5 2 = eò = e ò t -81 dt = e 2 ò t2 -81 = e 2 log(t -81) In(t 2 -81)5 / 2

2

Þ ét3 ù = 2(V - 1.5) ë û0

dy +y=0 dx Þ xdy + ydx = 0 Þ d( xy ) = 0.

dy 5t sin t …(i) + y= 2 dt t 2 - 81 t - 81

Þ



0

x

)

Here, P =

ò



46. (c)

dy t - 81 + 5ty = sin t dt



,

or, Q = 1 + Ce–t…(i) When t = 0, Q = 0; so equation (i) ⇒ 0 = 1 + C ⇒ C = –1 Hence, Q(t) = 1 – e–t.

1 dt = dx + C 2 t +1

2

C (t 2 - 81)5 / 2

which exists for t ≠ –9, 9. Hence, option (a) is correct.





dt = dx t2 + 1

43. (a)

+

2

Integrating, we get,

ò

(t 2 - 81)5 / 2

Þ 3t 2dt = 2(V - 1.5)

dt - 1 = t2 dx

or,

- 81)3 / 2dt

0

Now using (ii) and (iii) , we get from (i),



2

0

dy dt = - 1 …(iii) dx dx



ò (sin t)(t

+C

ò f (t )dt = m(V - V )

…(ii)

Then, 1 + dy = dt and so

5 sin t .(t 2 - 81) 2 dt - 81

2

t

dy = ( x + y - 1)2 …(i) dx

Let x + y – 1 = t

òt

44. 5.5 Here m = 2 kg, V0 = 1.5 m/sec. Then,

42. (d)

dx

y=

or,

Given i(4) = 10 ; so (i) gives, 10 = Ce(0.2)4 or, 10 = C(2.225) \ C = 4.493 Hence, i = (4.493)e0.2t and so at t = –5 , i = 4.493e–10 = 1.652.

5

y(t 2 - 81) 2 =



æi ö Þ log ç ÷ = 0.2t èC ø i Þ = e0.2t C

ò

y ´ I .F = I .F ´ Q dt + C



⇒ log i = 0.2t + log C ⇒ log i – log C = 0.2t



Hence, the solution is given by



or, 2 y

dy æ 2ö - ç1 + ÷ y2 = x 2 ..........(i ) dx è xø

8/3/2023 5:24:47 PM

Ordinary Differential Equations • 335

Let us put y2 = z. So 2 y dy = dz . Then (i) bedx dx comes,

2ö dz æ - 1 + ÷ z = x 2 …(ii) dx çè xø



Clearly, equation (ii) is linear and it is integrat-



ing factor is given by

I. F =

æ 2ö - ç1+ ÷dx e è xø

ò

= e - x ´ e log x



-2

=

=e

- x -2log x

=e

-x

´e

-2 log x

ò

e-x . x2

z ´ I.F = (I.F)x 2dx + C or, z

-x



-x

æ y2 ö y2 + 1 ÷ e - x = e -1 i.e; 2 + 1 = e x -1 çç x 2 ÷ x è ø æ y2 ö i.e; ln ç 2 + 1 ÷ = x - 1. çx ÷ è ø

dy + x3 = O dx

ò

Þ y3dy = -x 3dx Þ y3dy = - x 3dx +

y P.I. =

C 4

Again, y çæ p ÷ö = p Þ 0 + c2 + p = p Þ c2 = 0 2 2 è2ø 2 \  y = cos x + x. 51. (a) d2 y dy +2 + 17 y = 0 dx dx 2

1

Þ m2emx + 2memx + 17emx = 0 (taking y = emx as a trial solution)

1 4

\ y( -1) = (1 - 1) = 0.



EMEP.CH04_3PP.indd 335

d2 y = 3x - 2 dx 2

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

Hence, the complete solution is y = c1 cos x + c2 sin x + x Now, y(0) = 1 ⇒ 1 = c1 + 0 + 0 ⇒ c1 = 1

y4 x4 C =+ Þ y4 = -x 4 + C........(i ) 4 4 4

49. (c) Given

dx 2

Let y = emx be a trial solution of (ii). Then (ii) gives, m2emx + emx = 0 ⇒ m2 + 1 = 0 ⇒m=0±i Then, yC.F = c1 cos x + c2 sin x.

Hence, (i) becomes, y4 = -x 4 + 1 i.e; y = (1 - x 4 ) 4 .



2 The given equation is d y + y = x …(i)

= x - D 2 x + D 4 x - ..... = x - 0 + 0 - ......... =x

Now given that y(0) = 1. So (i) gives, 14 = –04 + C i.e., C = 1.

x3 5x - x2 + 2. 2 2

= (1 - D 2 + D 4 - .....) x

48. (c)

Þ

Thus, y =

dx

Consequently, the required solution i.e; the curve is

ò

x3 5 - x 2 - x + B.........(ii ) 2 2 At x = 0, y = 2; So (ii ) Þ = 2 = 0 - 0 - 0 + B i.e., B = 2

Consider, d y2 + y = 0 …(ii)

æ0 ö -1 -1 ç 2 + 1 ÷ e = C i.e; C = e . è1 ø

y3

Again, integrating both sides w.r.t x we get,

2

Now since the curve passes through (1, 0), so y(1) = 0 and hence from (iii) we get,



5 dy 3 2 = x - 2x dx 2 2

50. y = cos x + x

ò

e = -e - x + C x2 æ y2 ö or, ç 2 + 1 ÷ e - x = C..........(iii ) çx ÷ è ø or, y2



e e = ´ x 2dx + C = e- x dx + C x2 x2

ò

\

y=

Hence, the general solution of the equation (ii) is -x



dy 3 2 = x - 2x + A........(i ) dx 2 dy At x = 1, = -3; dx 3 5 so (i ) Þ - 3 = - 2 + A Þ A = 2 2

Integrating both sides w.r.t x, we get,

Þ m2 + 2m + 17 = 0



Þm=

-2 ± 4 - 68 = -1 ± 4i 2

Hence, the solution is given by

8/3/2023 5:24:50 PM

336 • Engineering Mathematics Exam Prep y = e–x(c1 cos 4x + c2 sin 4x)…(i)



\

dy = -e - x (c1 cos4 x + c2 sin 4 x ) dx

Then, x = x(t) = c1e–t + c2e–2t 1 5 (D2 + 3D + 2) 5 1 = (1) 2 æ D2 + 3D ö çç1 + ÷÷ 2 è ø

x P .I . =

+ e - x ( -4c sin 4 x + 4c cos4 x ).........(ii)

1 2 Since, y(0) = 1, so (i) gives, 1 = e0(c1 cos 0 + c2 sin 0) = c1 i.e., c1 = 1.

Since, dy æç p ö÷ = 0 , so (ii) gives, dx è 4 ø -p

0 = -e 4 (c1 cos p + c2 sin p)

5æ D2 + 3D ö çç1 + ÷÷ 2è 2 ø

=

5æ D2 + 3D ö 5æ 0+0ö 5 = çç1 ÷÷ (1) = ç1 2è 2 2 2 ÷ø 2 è ø

-p

+ e 4 ( -4c1 sin p + 4c2 cos p)



\ x (t ) = xC . F + x P . I . = c1e -t + c2e -2t +

or, c2 =

Þ x (t ) t ®¥ = c1 ´ 0 + c2 ´ 0 +

c1 1 = . 4 4



52. (c) y = c1e - x + c2 e -3 x is the solution Þ (m + 1)(m + 3)=0 is the auxiliary equation 2

Þ m +4m+3=0 is the auxiliary equation 2 mx

Þm e 2

Þ



d y dx

2

+4me

+4

dy dx

mx

+3e

mx

2

d y dy +p + qy = 0 , dx dx 2

2

dx

2

+4

dy dx

+ 3y = 0

with

53. (d) Consider (D2 – 4D + 4)y = 0 …(i) mx Let y = e be a trial solution of (i). Then (i) gives, m2emx - 4memx + 4emx = 0 Þ m2 - 4m + 4 = 0 Þ m = 2,2.

Then, y = y(x) = (c1 + c2x)e2x 54. (b) Given x′′(t) + 3x′(t) + 2x(t) = 5 …(i) Consider x′′(t) + 3x′(t) + 2x(t) = 0 …(ii) Let x = emt be a trial solution of (ii). Then (ii) gives,



EMEP.CH04_3PP.indd 336

Þ m2 + 3m + 2 = 0 Þ m = -1, - 2.



Let us take p = 4 and q = 3. Then (i) becomes,



d2 y dy +4 + 4 y = 0 …(ii) 2 dx dx

m2emx + 4memx + 4emx = 0

we get p = 4 and q = 3.

m2emt + 3memt + 2emt = 0

d2 y dy +p + (q + 1) = 0 …(i) 2 dx dx

gives,

mx

d y



Let y = emx be a trial solution of (ii). Then (ii)

=0

+ 3 y = 0 (taking y = e )

Now comparing

5 5 = . 2 2

5 2

55. (c) The given equation is

1 æ ö y = e - x ç cos4 x + sin 4 x ÷ 4 è ø



(1)

or, - c1 cos p + 4c2 cos p = 0

Hence, the required solution is,



-1

=

Þ m2 + 4m + 4 = 0 Þ m = -2, -2.

Then the solution is given by, y = (c1 + c2x)e–2x Therefore, e–2x and xe–2x are two solutions of (ii). The solution y = xe–2x satisfies option (c).

56. (b) The given equation is

d2 y dy -5 + 6 y = 0 …(i) dx dx 2

Let y = emx be a trial solution of (i). Then (i) gives,

m2emx - 5memx + 6emx = 0



Þ m2 - 5m + 6 = 0 Þ m = 2,3.

Then the solution is given by, y = c1e2x + c2e3x = e2x + e3x for C1 = C2 = 1 57. (b)



d2 y dy +4 + 3 y = 3e2x  (given) 2 dx dx

8/3/2023 5:24:52 PM

Ordinary Differential Equations • 337

(D2 + 4D + 3)y = 3e2x Hence, P .I =

1 1 3e2x = 3 2 e2x ( D 2 + 4 D + 3) ( D + 4 D + 3)

=3



1 ax 1 ax e = e f ( D) f (a)

]

58. (a) The given equation is y′′ + 2y′ + 101y = x

10.4e …(i) Consider y′′ + 2y′ + 101 y = 0 …(ii) Let y = emx be a trial solution of (ii). Then (ii) gives, m2emx + 2memx + 101emx = 0 Þ m2 + 2m + 101 = 0 Þm=

-2 ± 4 - 404 -2 ± 20i = = -1 ± 10i 2 2

Then yC.F = e–x(c1 cos10x + c2 sin 10x). 1 (10.4 e x ) (D + 2D + 101) 10.4 = 2 e x = 0.1 e x (1 + 2 + 101)



2

So, the complete solution is y = yC . F + yP . I

= e - x [c1 cos10x + c2 sin10x ] + 0.1 e x .......(iii )



\ 

dy = -e - x [c1 cos10x + c2 sin10x ] + 0.1 e x dx

+e–x[–10c1 sin 10x + 10c2 cos10x]…(iv) Now using y(0) = 1.1 and y′(0) = –0.9 we get from (iii) and (iv), c1 =1 and c2 = 0 Hence, y = e–x cos 10x + 0.1ex So P – 2, Q – 1, R – 3 is the correct option. 2 59. (a) The given equation is d y2 + y = 0 …(i)

d2 x 60. (d) Given x + 3x = 0 i.e; 2 + 3x = 0 …(i)



EMEP.CH04_3PP.indd 337

dt

Then taking x = emt, we get from (i),

Hence, the solution is given by

x = e0 ( A cos 3t + B sin 3t )

( 3t ) + B sin ( 3t ) ..............(ii) dx \ = x = - A 3 sin ( 3t ) + B 3 cos ( 3t ) dt

..............(iii ) At t = 0, x = 1; so from (ii) we get, A = 1. .

At t = 0, x = 0; so from (ii) we get, B = 0. Thus, from (ii), we have x = cos 3t .

Therefore, x (1) = cos 3 = 0.99.

61. (a) Given y′′ + 2y′ + y = 0 …(i) mx Let y = e be a trial solution of (i). Then (i) gives, m2emx + 2memx + emx = 0 Þ m2 + 2m + 1 = 0 Þ m = -1, -1.

Then, the solution is given by y = (C1 + C2x)e–x. Then, y(0) = 0 ⇒ 0 = (C1 + C2(0))e–0 ⇒ C1 = 0 and y(1) = 0 ⇒ 0 = (C1 + C2)e–1 ⇒ C1 + C2 = 0 ⇒ C2 = 0 \ y = (0 + 0x)e–x = 0 is the solution. Consequently y(0.5) = 0. 62. (a)

2 Given equation is d y2 + 2 dy + 2 y = 0 …(i)

mx

dx

dx

Let y = e be a trial solution of (i). Then, (i) Þ m2emx + 2memx + 2emx = 0

dx

Let y = emx be a trial solution of (i). Then, (i) ⇒ m2emx + emx = 0 ⇒ m2 + 1 = 0 ⇒ m = 0 ± i Hence, the solution is given by y = e0(P cos x + Q sin x) = P cos x + Q sin x,where P and Q are constants.

Þ m2emt + 3emt = 0 Þ m2 + 3 = 0 Þ m = 0 ± i 3

= A cos

1 e2x . e2x = 2 5 (2 + 4 ´ 2 + 3)

[Using

yP . I . =



Þ m2 + 2m + 2 = 0 Þm=

-2 ± 4 - 8 = -1 ± i 2



Hence, the general solution is given by



So two solutions are e( -1+i )x and e( -1-i )x .

y = C1e( -1+i ) x + C2e( -1-i ) x .

63. (a)

The given equation is

d2 y dy +P + qy = 0 dx dx 2

…(i)

8/3/2023 5:24:55 PM

338 • Engineering Mathematics Exam Prep Now x(0) = 1 ⇒ A + B = 1 

Let y = emx be a trial solution of (i). Then, (i)

Also

Þ m2emx + Pmemx + qemx = 0



Þ m2 + Pm + q = 0 Þm=

- P ± P 2 - 4q 2

Thus if P2 – 4q > 0, then we have real and distinct roots. 64. (c)

2 Given d y2 + dy - 6 y = 0 …(i)

dx

dx

mx

Let y = e be a trial solution of (i). Then (i) gives, m2emx + memx - 6emx = 0 Þ m2 + m - 6 = 0 Þ m = -3,2.



\ the required solution is y = c1e–3x + c2e2x. 65. (d)

2 Given d n(2x ) - n( x2 ) = 0 …(i)

dx

L

mx

Let n = e gives,

be a trial solution of (i). Then (i)



⇒ –2A – 4B = 0

…(iii)

Solving (ii) and (iii) we get, A = 2 and B =–1. Hence, the solution is x(t) = 2e–2t – e–4t. 67. (c)



The given equation is

d2 y dy +2 + y = 0 …(i) dx dx 2

Let y = emx be a trial solution of (i). Then, (i) ⇒ m2emx + 2memx + emx = 0 ⇒ m2 + 2m + 1 = 0 ⇒ m = – 1, – 1 Hence the general solution is given by y = (C1 + C2x)e–x. Then, y(0) = 1 ⇒ C1 = 1. Again, y(1) = 0 ⇒ C2 = –1. Hence, y = y(x) = (1 – x)e–x and so y(2) = –e–2. 68. (c)

The given equation is

dx

1

Then, n(0) = K ⇒ C1 + C2 = K  Also, n(∞) = 0 ⇒ C2e∞ = 0 ⇒ C2 = 0 \  C1 = K (using (ii)) -1 x Hence, the solution is n(x) = Ke L .

d2 y dx

2

+6

dy + 9 y = 9x + 6…(i) dx

dx

1

Let y = e be a trial solution of (ii). Then, (ii) ⇒ m2emx + 6memx + 9emx = 0 ⇒ m2 + 6m + 9 = 0 ⇒ m = –3, – 3 Hence, yC.F = (C1x + C2)e–3x 1 (9x + 6) D2 + 6D + 9 1 = (9x + 6) æ D2 + 6D ö 9 ç1 + ÷÷ ç 9 è ø

yP . I =

…(ii)

-1

66. (b)

2 Given, d x2 + 6 dx + 8x = 0 

=

…(i)

dt

dt

2 mt

m e

+ 6me

mt

+ 8e

mt

=

=0

2

Þ m + 6m + 8 = 0 Þ m = -4, -2. –2t

So the solution is x = Ae

Therefore,

–4t

+ Be

dx = -2 Ae -2t - 4 Be -4t dt

D2 + 6D ö 1æ (9x + 6) ç1 + ç ÷÷ 9è 9 ø

D2 + 6D ö 1æ = ç1 ÷÷ (9x + 6) ç 9è 9 ø

Let x = emt be a trial solution of (i). Then (i) gives,

EMEP.CH04_3PP.indd 338

= 0,

mx

∴ The solution is given by n(x) = C1e - L x + C2e L x .



t =0

…(ii)

2 Consider d y2 + 6 dy + 9 y = 0…(ii)

mx

e =0 L2 1 Þ m2 - 2 = 0 L 1 Þm=± . L

m2emx -

dx dt



=

D 2 (9x + 6) + 6 D(9x + 6) ö 1æ çç 9x + 6 ÷÷ 9è 9 ø

1æ 0 + 54 + 0) ö 9x + 6 ÷=x 9 çè 9 ø

Hence, the complete solution is y = yC.F + yP.I = (C1x + C2)e–3x + x.

8/3/2023 5:24:57 PM

Ordinary Differential Equations • 339

69. (a)

d2 y dy x2 2 + x - 4y = 0 dx dx

Given equation is

…(i)

which is a linear homogeneous differential equation of order “2.” Let us put logx = z.

2 2 Then dy = x dy and d y2 - dy = x 2 d y2

dz

\ (i) Þ or,

dx

dz

dz

dx

d2 y dy dy + - 4y = 0 dz 2 dz dz

d2 y - 4 y = 0............(ii) dz 2

Let u = emz be a trial solution of (ii) Then, (ii) ⇒ m2emz – 4emz = 0 or  m2 – 4 = 0( emz ⇒ 0) or  m = –2, 2 ∴ the complete solution is given by

y( x ) = c1e -2z + c2e2z =

c1

x2

t=

Now, y(0) = 0 ⇒ c1 = 0. Again, y(1) = 1 ⇒ 0 + c2 = 1 ⇒ c2 = 1 Thus, y = y(x) = x2. 70. (b) The given differential equation is …(i) d2u du -k

dx

=0

Let u = emx be a trial solution of (i). Then (i) gives, m2emx – kmemx = 0 ⇒ m2 – km = 0 ⇒ m = 0, k. So the solution is given by u = u(x) = Aeox + Bekx = A + Bekx Then, u(0) = 0 ⇒ A + B = 0 …(ii) and u(L) = U

EMEP.CH04_3PP.indd 339

Þ A + Be kx = U Þ A - Ae kL = U Þ A =

p = 45 4

Since y = - cos x - sin x , so at t = 45°, y = -2 2 < 0.

+ c2 x 2

1ö æ z -z ç e = x and e = x ÷ è ø

dx 2

71. (d) Given, y(t) + y(t) = 0 …(i) Let y = emt be a trial solution of (i). Then (i) gives, m2emt + emt = 0 ⇒ m2 + 1 = 0 ⇒m=0±i So the solution is y = A cost + B sin t and y = - A sin t + B cos t. Since y(0) = 1,  \  1 = A × 1 + B × 0 i.e; A = 1. Since y(0) = 1, \  1 = – A × 0 + B × 1 i.e; B = 1. Hence, y = cos t + sin t and y = – sin t + cos t. For y to be maximum, we must have  y = 0 or –sin t + cos t = 0 or tan t = 1 or

U 1 - e kL

So (ii) gives B = - U kL . 1-e kx ù U U é u= e kx i.e; u = U ê 1 - e ú . kL kL kL 1-e 1-e êë1 - e ûú

Hence, y is maximum at t = 45°. So the maximum value of y is cos 45° + sin45°; 2. 72. (c)

2 Given, d x2 + 9x = 0 …(i)

dt

Let x = emt be a trial solution of (i). Then (i) gives, m2emt + 9emt = 0 ⇒ m2 + 9 = 0 ⇒ m = ±3i  ∴ x = C1 cos 3t + C2 sin 3t and dx = -3C1 sin 3t + 3C2 cos3t. dt

Now, x(0) = 1 ⇒ C1 = 1. Also, dx dt

t =0

=1

⇒ 3C2 = 1 Þ C2 = 1 . 3

1 3

Hence, x = cos3t + sin 3t

73. (a) The given differential equation is d 2 x (t ) dt 2

+ x (t ) = 0 …(i)

Let x = emt be a trial solution of (i). Then (i)

8/3/2023 5:25:00 PM

340 • Engineering Mathematics Exam Prep gives, m2emt + emt = 0 Þ m2 + 1 = 0 Þm = 0±i

So x = x(t) = C1 cos t + C2 sin t and dx = -C1 sin t + C2 cos t. dt

Let x1(t) = cos t and x2(t) = sin t. Then the conditions x2 (0) = 0,

dx1 (t ) dt

x1 (t )

cos t dx2 (t ) = - sin t dt

dt



sin t cos t

2

= cos t + sin t = 1 for all “t.” 74. 35



2 Given equation is d y2 = 0



Integrating both sides we get,



Again, integrating both sides we get,

dx

dy =A dx

dy

75. (b)

2 Given d x2 + 2 dx + x = 0 …(i)

dt

dt

Let x = emt be a trial solution of (i). Then (i) gives, m2emt + 2memt + emt = 0 Þ m2 + 2m + 1 = 0 Þ m = -1, -1 \ x = x (t ) = ( a + bt )e -t = ae -t + bte -t .



Given

d2 y dy x2 2 + x - y = 0 …(i) dx dx

Let us put z = log x

Then, dy = x dy dz





EMEP.CH04_3PP.indd 340

dx

and

∴ (i) gives

d2 y - y = 0.........(ii ) dz 2

2

2

d y dy d y = x2 2 dz 2 dz dx

d2 y dy dy + -y=0 dz 2 dz dz

or,

and “x” are two solutions.

dx

Let y = emx be a trial solution of (i). Then the auxiliary equation is given by (m2 + 2am + 1) = 0 -2a ± 4a2 - 4 2



⇒ m=



Now both roots i.e; values of “m” are equal ⇒ 4a2 – 4 = 0 ⇒a=±1 d2 y = -12x 2 + 24 x - 20. dx 2

Integrating both sides w.r.t. x, we get dy = -4 x 3 + 12x 2 - 20x + c1 dx

Again, integrating both sides w.r.t. x, we get y = –x4 + 4x3 – 10x2 + c1x + c2…(i) At x = 0, y = 5; so (i) gives, c2 = 5. Again at x = 2, y = 21; so (i) ⇒ 2c1 + c2 = 45 …(ii) Solving (i) and (ii) we get, c2 = 5 and c1 = 20. So, (i) ⇒ y = –x4 + 4x3 – 10x2 + 20x + 5 So for x = 1, (i) ⇒ y = –1 + 4 – 10 + 20 + 5 = 18. 79. –3

76. (c)



1 x

c2 x

2 Given d y2 + 2a dy + y = 0 …(i)



= 2Þ A = 2 x =10

\ y( x ) = 2x + 5 and so y(15) = 30 + 5 = 35.



77. (a)

Given

\ y(0) = 5 Þ B = 5. dx

Hence,

78. 18

y = y( x ) = Ax + B

Again



\ y = c1e z + c2 e z = c1 x +

dx

t =0

x2 (t )

Then, W (t ) = dx (t ) 1 2

= 0 are satisfied.

Let y = emz be a trial solution of (ii) Then, (ii) ⇒ m2emz – emz = 0 ⇒ m2 – 1 = 0 ( emz ≠ 0) ⇒ m = –1, 1

2 Given d 2y + 5 dy + 6 y = 0 …(i)

dt

dt

mt

Let y = e gives,

be a trial solution of (i). Then (i)

m2emt + 5memt + 6emt = 0 Þ m2 + 5m + 6 = 0 Þ m = -2, -3 \ y = y(t ) = c e -2t + c e -3t

1 2 Then, y(0) = 2 ⇒ c1 + c2 = 2

…(ii)

8/3/2023 5:25:03 PM

Ordinary Differential Equations • 341

Again, y(1) = - çæ 1 - 3e ÷ö Þ c1 + c2 = - çæ 1 - 3e ÷ö…(iii) 3 2 3 3 è e

e

ø

e

è e

Now, dy = -6e -2t + 3e -3t dt

è dt øt = 0

80. (b) Given

= -6 + 3 = 3

–3.

d y dy +2 + y = 0 …(i) dt dt 2

Let y = emt be a trial solution of (i). Then (i) gives, m2emt + 2memt + emt = 0

+c e

-t

d2 y dy d2 y Then, dy = x dy and 2 - = x 2 2 dz

dz

dx

2

d y dy dy \ (i) Þ + -y=0 dz 2 dz dz

d2 y or, - y = 0............(ii) dz 2

Let y = emz be a trial solution of (ii) Then, (ii) ⇒ m2emz – emz = 0 or,  m2 – 1 = 0 (  emz ≠ 0) or,  m = –1, 1 ∴  the complete solution is given by y( x ) = c2e

82. (c) Given

1ö æ z -z c + c1e = 2 + c1x . ç e = x and e = x ÷ è ø x z

d2 y = y …(i) dx 2

Let y = emx be a trial solution of (i). Then (i) gives,

EMEP.CH04_3PP.indd 341

3 4



⇒ C2 + 4C1 = 1.5…(iv) Solving (iii) and (iv) we get, C1=0.5, C2 =-0.5.

=

x -x Hence, from (ii), we get, y = e - e .

2

dt

dt

-t

81. (c) The given equation is x2y′′ + xy′ – y = 0… (i) which is a linear homogeneous differential equation of order “2.” Let us put log x = z.



C 3 2C1 + 2 4 2

⇒

2 Given d x2(t ) + 3 dx (t ) + 2x (t ) = 0. …(i)

-t

and y¢ = y¢(t ) = -(c + c t )e

-z

3 4

83. 0.8566

1 2 2 Then, y(0) = 1 ⇒ c1 = 1. Again, y′(0) = 1 ⇒ –c1 + c2 = 1 ⇒ c2 = 2 Hence, y(t) = (1 + 2t)e–t.

dx

C1e ln 2 + C2e - ln 2 =





Þ m2 + 2m + 1 = 0 Þ m = -1, -1

dz

So y = C1ex + C2e–x …(ii) Since (ii) passes through origin, so C1 + C2 = 0 …(iii) Again, (ii) passes through (ln 2, 3/4); so

2

\ y = y(t ) = (c1 + c2t )e

Þ m2 - 1 = 0 Þ m = -1,1

ø

Now solving equation (ii) and (iii), we get c1 = 3 and c2 = –1. So y(t) = 3e–2t – e–3t.

Therefore, æç dy ö÷

m2 emx - emx = 0

mt

Let x = e gives,

be a trial solution of (i). Then (i)

m2emt + 3memt + 2emt = 0 Þ m2 + 3m + 2 = 0 Þ m = -1, -2 \ x = x (t ) = C e -t + C e -2t

1 2 Then, x(0) = 20 ⇒ C1 + C2 = 20 x(1) = 10/e ⇒ C1 + C2e–1 = 10 Solving (ii) and (iii), we get



C1 =

…(ii) …(iii)

10e - 20 æ 10e ö ; C2 = ç ÷. e -1 è e -1 ø

10e - 20 ü -t ì 10e ü -2t ýe + í ýe . î e -1 þ îe -1þ

ì So x = x (t ) = í

Hence, x (2) = æç 10e - 20 ö÷ e -2 + æç 10e ö÷ e -4 è



e -1

ø

è e -1 ø

= 0.8566 84. (a)

2 Given d y2 + 12 dy + 36 y = 0 …(i)

dx

dx

Let y = emx be a trial solution of (i). Then (i) gives,

m2emx + 12memx + 36emx = 0



⇒ m2 + 12m + 36 = 0



⇒ m = –6, –6

So y = (C1 + C2x)e–6x. Then y(0) = 3 ⇒ C1 = 3.

8/3/2023 5:25:06 PM

342 • Engineering Mathematics Exam Prep y′ = –6C1e–6x + C2e–6x – 6C2xe–6x

87. (a)

So y′(0) = –36 ⇒ –36 = –6C1 + C2

Given



⇒ –36 = –18 + C2



⇒ C2 = –18 –6x



\  y = 3e

–6x

– 18xe

–6x

i.e., y = (3 – 18x)e

mx

Let y = e gives,

…(i)

be a trial solution of (i). Then (i)



m2emx + 9emx = 0



⇒ m2 + 9 = 0



⇒ m = ±3i

4 2 Consider d y4 + 3 d y2 = 0 

dx



m4emx + 3m2emx = 0



⇒ m4 + 3m2 = 0



⇒ m2(m2 + 3) = 0 Þ m = 0,0,0 ± 3i ( 3x )



Then, y(0) = 0 ⇒ C1 = 0 2

è ø

2

-1

3p 1 æpö y ç ÷ = - 2 sin =- 2´ = -1 4 2 è4ø æ 3p pö p 1 ö æ = sin ç p - ÷ = sin = ç sin ÷ 4 4ø 4 2ø è è

ò

86. 7.38

Let y = emx be a trial solution of (i). Then (i) gives,

m2emx – 4memx + 4emx = 0



⇒ m2 – 4m + 4 = 0 ⇒ m = 2, 2



∴  y = (C1 + C2x)e2x

2 Given d 2y + 2 dy + y = 0 …(i)

= C2 e 2 x + 2C2 xe2 x

x =0

Let y = e gives,

=1

.

⇒ C2 = 1

Thus, y = xe2x and so y(1) = e2 = 7.38.

EMEP.CH04_3PP.indd 342

dt

dt

So, y = C2xe2x.

dx



88. (b)

Now, y(0) = 0 ⇒ C1 = 0

Then, dy

ò

mt



dx

æ x3 2 ö 1 æ x3 2 ö - x ÷ = 36 ç - x dx çç ÷ ç 3 3 ÷÷ Dè 3 3 ø è ø æ x4 1 2 ö - x ÷ = 3x 4 - 12x 2 = 36 ç ç ÷ è 12 3 ø = 36

d2 y dy -4 + 4 y = 0 …(i) dx dx 2

dy

1 x2 2 ö æ D 3 D 2 ç1 + ÷ ç 3 ÷ø è

D2 ö 1 æ 2 1+ ÷ x 2 ç ç 3 ÷ø D è ö D2 D4 1 æ = 36 2 ç1 + - ......... ÷ x 2 ç ÷ 3 9 D è ø 2 2 4 2 ö D x D x 1 æ = 36 2 ç x 2 + - ......... ÷ ÷ 3 9 D çè ø 1 æ 2 2ö 1 æ 2 2ö = 36 2 ç x - ÷ = 36 x - ÷ dx 3ø 3ø D çè D è

Therefore, y = - 2 sin 3x and so

Now,

1 (108x 2 ) D 4 + 3D 2

= 36

⇒ C2 = - 2

Given

Now, P.I = = 108

Again y æç p ö÷ = 2 ⇒ C1 cos 3p + C2 sin 3p = 2 2



…(ii)

 \  CF = (C1 + C2x) + C3 sin ( 3x ) + C4 cos

So y = C1 cos 3x + C2 sin 3x…(i)



dx

Let y = emx be a trial solution of (ii). Then (ii) gives,

85. –1. Given y′′ + 9y = 0

d4 y d2 y + 3 2 = 108x 2 …(i) 4 dx dx

be a trial solution of (i). Then (i)

m2emt + 2memt + emt = 0 ⇒ m2 + 2m + 1 = 0 ⇒ m = –1, –1 ∴  y(c1 + c2t)e–t Now, y(0) = 1 ⇒ c1 = 1 So, y = (1 + c2t)e–t Again, y(1) = 3e–1 ⇒ (1 + C2)e–1 = 3e–1 ⇒ c2 = 2 Hence, y = (1 + 2t)e–t and so y(2) = 5e–2.

8/3/2023 5:25:08 PM

Ordinary Differential Equations • 343

89. (a) Given

91. (a)

d2 y + 16 y = 0  dx 2

…(i)

Let y = emx be a trial solution of (i) So the auxiliary equation is m2 + 16 = 0 which gives m = ± 4i Therefore, y = c1 cos 4x + c2 sin 4x and y′ = –4c1 sin 4x + 4c2 cos 4x. Now, y′(0) = 1 ⇒ 1 = 4c2 ⇒ c2 = 1/4 Again, y′(p/2) = –1 ⇒ –1 = –4c1 sin 2p + 4c2 cos 2p ⇒ –1 = 04c2  ⇒ c2 = –1/4, which contradicts the fact that c2 = 1/4. Hence, the given differential equation has no solution. 90. (a) The equation y′′ – 4y′ + 3y = 2t – 3t2 can be re-written as: (D2 – 4D + 3)y = 2t – 3t2 So P.I. =

1 (2t - 3t 2 ) D2 - 4 D + 3

= = =

2(3 - D )(1 - D ) 1 2(1 - D ) 1 2(1 - D )

= =

=



1 2(3 - D ) 1

æ è

6 ç1 -

Dö 3 ÷ø

2

(2t - 3t )

-1

1 1æ Dö (1 - D )-1 (2t - 3t 2 ) - ç1 - ÷ (2t - 3t 2 ) 2 6è 3ø 1 1æ D D2 ö 2 (1 + D + D 2 )(2t - 3t 2 ) - ç1 + + ÷ (2t - 3t ) 2 6 çè 3 9 ÷ø 1 2 2 2 2 (2t - 3t ) + D(2t - 3t ) + D (2t - 3t ) 2 2 ü D D 1 ìï 2 ï - í(2t - 3t 2 ) + (2t - 3t 2 ) + (2t - 3t ) ý 6 ïî 3 9 ïþ

{

}

= =

1 1 ( -4 - 4t - 3t 2 ) - ( -3t 2 ) 2 6

EMEP.CH04_3PP.indd 343

= –2 – 2t – t2



m=

-2 ± 4 + 20 -2 ± 2 6 = = -1 ± 6. 2 2

Hence, y = K1e( -1+

6 )x

+ K 2e( -1-

6 )x

.

92. 94.08 Given 3y′′(x) + 27y(x) = 0 Let y = emx be a trial solution of (i) Then the auxiliary equation is 3m2 + 27 = 0

…(i)

or, m2 + 9 = 0 or, m = ±3i 

So the solution is y(x) = y = c1 cos3x + c2 sin 3x …(ii)

Given y(0) = 0. Therefore (ii) gives, C1=0. So, y = C2 Sin 3x

\ y=

2000 sin 3x . 3

Hence, y(1) =

(2t - 3t 2 )

1é 1é 2 - 6t -6 ù 2t - 3t 2 + 2 - 6t - 6 ù - ê2t - 3t 2 + + û 6ë 2ë 3 9 úû



Let y = e be a trial solution of (i) So the auxiliary equation is m2 + 2m–5 = 0 which gives,



2

2

(2t - 3t ) -

mx

So y′(0) = 2000 ⇒ 2000 = 0 + 3c2 ⇒ c2 =

(2t - 3t )

(2t - 3t 2 ) -

dx

dx

Now y′ = 3c2 cos 3x.

1 = (2t - 3t 2 ) (1 - D )(3 - D ) (3 - D ) - (1 - D )

2 Given d y2 + 2 dy - 5 y = 0 …(i)

2000 3

2000 sin 3 = 94.08. 3

93. –0.37. The given differential equation is:

d2 y dy +2 + y = 0 …(i) 2 dx dx

Let y = emx be a trial solution of (i) So the auxiliary equation is m2 + 2m + 1 = 0 which gives, (m + 1)2 = 1 i.e; m = –1, –1. So the general solution of (i) is given by y = (A + Bx)e–x…(ii) \  y(0) = 0 ⇒ 0 = (A + 0)e–0 ⇒ A = 0. Again, dy dx

x =0

= -1 Þ é -( A + Bx )e - x + Be - x ù = -1 ë û x =0

⇒ –(0 + 0)e–0 + Be–0 = –1 ⇒ B = –1. Now putting the values of A and B we get from (ii), y = –xe–x.

\  y(1) = –e–1 = – 0.37.

8/3/2023 5:25:10 PM

344 • Engineering Mathematics Exam Prep 94. –0.21

So

The given equation is 2

d y dy 5 y …(i) =+ dt 4 dt 2

Let y = emt be a trial solution of (i) So the auxiliary equation is



m2 = -m -



m=



So the general solution of (i) is given by



which gives,

-4 ± 16 - 80 -4 ± 8i 1 = = - ± i. 2´4 8 2

y = ( A cos t + B sin t )e

-

t 2

dy = ( - A sin t + B cos t )e dt dy So =0 dx t =0

-

…(ii)

t 2

-

1 ( A cos t + B sin t )e 2

-

t 2



Now putting the values of A and B we get from

(ii),

y=e

-

p 2

p

1 æ ö 2 ç cos p + 2 sin p ÷ = -e = -0.21. è ø

95. 4.54. The given equation is

d2 y 2 2 + 8 y = 0 …(i) dt

Let y = emt be a trial solution of (i) So the auxiliary equation is 2m2 + 8 = 0 which gives, m = 0 ± 2i. So the general solution of (i) is given by y = (A cos 2t + B sin 2t)e0 = A cos 2t + B sin 2t…(ii)  \  y(0) = 0 ⇒ (A cos 0 + B sin 0)e–0 = 0 ⇒ A = 0. Also, dy = -2 A sin 2t + 2B cos 2t = 2B cos 2t. dt

EMEP.CH04_3PP.indd 344

(c)

e 1-e



(d) 0

2. Let y(x) be the solution of the differential equation

dy = ( y - 1)( y - 3) dx

satisfying y(0) = 2.

Then which of the following is not true? (c) y(x) is bounded (d) y(x) is not bounded above

1 æ ö ç cos t + 2 sin t ÷ . è ø

\ y( p) = e

2e e (b) e -1 e -1

x ®¥

1 Þ ( -0 + B )e - (1 + B ´ 0)e0 = 0 2 1 ÞB= . 2 t 2

(a)

(a) lim y( x ) = 1 (b) lim y( x ) = 3 x ®-¥

0

-

1. Let y(x) be the solution of the differential equation  ( xy + y + e-x ) dx + ( x + e-x ) dy = 0 satisfying y(0) = 1. Then y(–1) = ?

 \  y(0) = 1 ⇒ (A cos 0 + B sin 0)e–0 = 1 ⇒ A = 1. Also

Now putting the values of A and B we get from (ii), y = 10 sin 2t. \  y(1) = 10 sin 2 = 10 × 0.9092 = 4.54. Questions For Practice



5 i.e;4m2 + 4m + 5 = 0 4

dy = 10 Þ 2B cos0 = 10 Þ B = 5. dt t =0

3. The nonzero value of n for which the differential equation (3xy2 + n2x2y)dx + (nx3 + 3x2y)dy = 0 becomes exact is (a) –3 (b) –2 (c) 2 (d) 3 4. An integrating factor of the differential equation (a)

dy 2xy2 + y = dx x - 2 y3

1 y

is

(b) 12 (c) y (d) y2 y

5. If y(x) is a solution of the differential equation y′′ + 4y = 2ex, then lim e - x y( x ) = ? x ®¥

(a)

1 5

(b)

2 1 (c) 5 7

(d)

2 7

6. The differential equation (1 + x2y3 + lx2y2)dx + (5 + x3y2 + x3y)dy = 0 is exact if λ = ? 1 5 2 (a) 3 (b) (c) (d) 2

2

3

5

8/3/2023 5:25:13 PM

Ordinary Differential Equations • 345

7. If y(x) is a solution of

d2 y dy +4 + 4y = 0 2 dx dx

sat-





æ

p

ö

Then the value of y ç e 2 ÷ is

–2x

(a) 4e (c) 4e–2x + 16x

(b) 2a – b = 0 (d) 2a = b + 1

14. Let y(x) be the solution of the initial value problem x2y′′ + xy′ + y = x, y(1) = y′(1) = 1.

isfying y(0) = 4 and dy = 8 ; then y(x) is dx x =0 given by 2x

(a) 3a – 5b = 0 (c) 3a + 5b = 0

(b) (16x + 4)e (d) 4e–2x + 16e2x

ç è

÷ ø

8. Consider the following differential equation (x + y +1)dx + (2x +2 y + 1)dy = 0.

(a)

pö pö 1æ 1æ ç1 - e 2 ÷ (b) ç1 + e 2 ÷ ÷ 2ç ÷ 2ç è ø è ø

 Then which of the following statements is true? (a) The differential equation is homogeneous (b) The differential equation is exact (c) ex+y is an integrating factor (d) A suitable substitution transforms the differential equation to the variable seperable form.

(c)

1 p 1 p + (d) 2 4 2 4

15. If the integrating factor of (x7 y2 + 3y)dx + (3x8y – x)dy = 0 is xmyn, then (a) m = –7, n = 1 (b) m = 1, n = –7 (c) m = n = 0 (d) m = n = 1 16. The particular solution of the equation y sin x = y p

æ ö ln y satisfying the initial condition y ç ÷ = e is è2ø

9. Consider the differential equation 2 cos(y2)dx – xysin(y2) dy = 0. Then which of the following is true? (a) ex is an integrating factor. (b) e–x is an integrating factor. (c) x2 is an integrating factor. (d) x3 is an integrating factor. 10. Let y1(x) and y2(x) be two linearly independent solutions of y′′ + P(x)y′ + Q(x)y = 0.

(a)

are linearly independent if (a) ad = bc (b) ad ≠ bc (c) ac = bd (d) ab ≠ cd 11. An integrating factor of

dy x + (3x + 1) y = e -2x dx

ì î

EMEP.CH04_3PP.indd 345

ü

è

øþ

î

è

øþ

17. The differential equation dy dx

= k( a - y)(b - y) when solved with the condition y(0) = 0, yields the result b( a - x ) = e(b-a )ky (a) b( a - y) = e( a -b)kx (b) a(b - x )

a(b - y )

(c) a(b - y) = e( a -b)kx (d) xy = ke b( a - y )

18. Solve: y’’ – 4y′ + 4y = x2; when x = 0, y =

is

(a) xe3x (b) 3xex (c) xex (d) x3ex [Hint: Express the equation in the form Pdx ù dy + Py = Q , Then I.F = e ò ú dx û 2 12. The general solution of x y′ – 5xy′ + 9y = 0 is given by (a) (c1 + c2x)e3x (b) (c1 + c2 ln x)x3 3 (c) (c1 + c2x)x (d) (c1 + c2 ln x)x2 13. If x3y2 is an integrating factor of (6y2 + axy)dx + (6xy + bx2)dy,

(b)

æxö cot ç ÷ e è2ø

(c) ln ítan æç x ö÷ ý (d) ln ìícot æç x ö÷ üý 2 2

Then, y3(x) = ay1(x) + by2(x) and y4(x) = cy1(x) + dy2(x)

æxö tan ç ÷ e è2ø

dy =1 dx

2

(a) xe2x + x + x + 1 (b) 2

(c)

8

3 8

and

xe2x x 2 1 + +x+ 2 2 4

xe2x x 2 x 3 y= + + + 2 4 2 8

(d) none of these

19. Solve: (D2 – 1)y = 2; given that dy = 3 when y dx

= 1; and x = 2 when y = – 1 (a) y + 4 = e2x (b) y + 2 = ex–2 (c) y = e2x + 2 (d) y – 2 = ex+2

8/3/2023 5:25:16 PM

346 • Engineering Mathematics Exam Prep 20. The equation of a curve passing through the point (–2, 3) given that the slope of the tangent to the curve at any point (x, y) is 2x2 , is (a) y = ( 3x 2 + 15 ) (b) y2 = 2x2 + 15 3 2

(c) y = ( x + 5 ) (d)

(

y 2 = 3x 2 - 4

)

1 3

è2ø

2

2

(a) y cos x = x 2 + p (b) y cos x = x 3 - p 2

6

2

2

(c) y sin x = 2x 2 - p (d) y sin x = x 3 - 2p 2

22. Solve:

3

dy + 2 y tan x = sin x ; y = 0 dx

p when x = 3

2

2

(a) y = cos x – 2 cos x (b) y = cos x + sin x (c) y = sin x – 2 sin2 x (d) y = sin x + cos2x 23. Solve:

dy + x cot y = 2 y + y2 cot y ; dx

0, when

given that x =

p y= 2 p2 4

(b) x sin y = y2 sin y -

p2 4

(d) none of these

dy y - = 2x 2 dx x

(a) y = x3 + cx (b) y = x2 + 2cx (c) y =

2

3

x x + cx + cx 2 (d) y = 4 3

æ dy ö log ç ÷ = 3x + 4 y è dx ø

= 0, is given by (a) 4e–3x + 3e3y = 5 (c) 3e3x + 4e4y = 5



27. Solve: ( x + 1)

given that y = 0 when x (b) 4e–3x + 3e4y = 7 (d) 4e3x + 3e–4y = 7

dy = 2e - y - 1; y(0) = 0 dx

1ö æ (a) y = log æç 2 - 1 ö÷ (b) y = log ç1 + ÷ è

EMEP.CH04_3PP.indd 346

î2

x +1 ø

2

2

þ

(b) y = cot ìí p + 1 + 1 (1 + log x )2 üý î4

2

2

þ

(c) y = tan ìí p + 1 - 1 (1 + log x )2 üý î2

2

2

þ

(d) none of these 31. The solution of the differential equation K2

d2 y = y - y2 dx 2

under the boundary conditions

è

(a) y = ( y1 - y2 ) exp æç - x2 ö÷ + y2 k è

ø

(b) y = ( y2 - y1 ) exp æç - x ö÷ + y1 è kø

(c) y = ( y1 - y2 )sinh çæ x ÷ö + y1

25. Solve: (x + y + 1)2dy = dx; y(–1) = 0 (a) x + y = tan y + 2 (b) (x + y)2 = cot y + 1 (c) x + y + 1 = tan y (d) x + y – 1 = tan y 26. A particular solution of the differential equation

(a) y = tan ìí p - 1 + 1 (1 + log x )2 üý

(i) y = y1 and x = x0 and (ii) y = y2 and x = ∞, where K, y1 and y2 are constant, is

4

24. What will be the solution of

cos2 x

(a) y = e (b) y = e sinx+cosx (c) y = e (b) y = ecosx–sinx 29. Solve: (1 + e2x)dy + (1 + y2)exdx = 0; y = 1 when x = 0 (a) y = ex (b) y = e–x (c) xy = ex (d) xy = e–x 30. Solve: (1 + y2) (1 + log x)dx + xdy = 0; y = 1 when x = 1



2 (c) (c) x (sin y + cos y) = p



dx



(a) x cos y = y2 sin y +

(d) none of these

sin2 x

21. Solve: dy + y cot x = 4 xcosecx ; y çæ p ÷ö = 0 dx

1 = c + ey x

28. Solve: dy = y sin 2x ; y(0) = 1

y

1 3

2

(c)

xø

èkø

æ xö (d) y = ( y1 - y2 ) exp ç - ÷ + y2 è kø

Answer key

1. (b) 6. (a) 11. (a) 16. (a) 21. (c) 26. (d) 31. (d)

2. (d) 7. (b) 12. (b) 17. (a) 22. (a) 27. (a)

3. (d) 8. (d) 13. (a) 18. (c) 23. (b) 28. (a)

4. (b) 9. (d) 14. (b) 19. (b) 24. (a) 29. (b)

5. (b) 10. (b) 15. (a) 20. (a) 25. (c) 30. (c)

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Ordinary Differential Equations • 347

10. (b) y1 and y2 are linearly independent solu-

Hints

1. (b) Express the differential equation in the

tions

form dy + Py = Q. Here I .F . = xe x + 1]

Þ W ( y1 , y2 ) ¹ 0

dx

Þ

3. (d) The given equation is exact Þ



(

)

(

¶ ¶ nx 3 + 3x 2 y = 3xy2 + n2 x 2 y ¶x ¶y



)

and then express the equation in the form ¶y

)

(

¶ ¶ 5 + x 3 y2 + x 3 y = 1 + x 2 y3 + l x 2 y 2 Þ ¶x ¶y

)

8. (d) Use the transformation x + y = u so that dy du 1+ = and then proceed. dx

EMEP.CH04_3PP.indd 347

dx

y3 y3¢

y4 y4¢

ay1 + by2 ay1¢ + by2¢

cy1 + dy2 cy1¢ + dy2¢

= ( ay1 + by2 )(cy1¢ + dy2¢ ) - ( ay1¢ + by2¢ )(cy1 + dy2 ) = acy1 y1¢ + bcy1¢ y2 + ady1 y2¢ + bdy2 y2¢

¶x

6. (a) The given differential equation is exact

(

Þ y1 y2¢ - y1¢ y2 ¹ 0 ..........(i)

=

Mdx + Ndy = 0. Then show that ¶M = ¶N



y2 ¹0 y2¢

Now, W( y3 , y4 ) =

4. (b) Multiply numerator and denominator by 1 y2

y1 y1¢

- acy1 y1¢ - ady1¢ y2 - bcy2¢ y1 - bdy2 y2¢



= (bc - ad )( y1¢ y2 - y2¢ y1 )



But from (i) we get y′1 y2 – y′2y1 ≠ 0

 \  W(y3, y4) ≠ 0 if bc – ad ≠ 0 i.e; if ad ≠ bc. Thus, y3 and y4 are L.I. solutions if ad ≠ bc

8/3/2023 5:25:23 PM

EMEP.CH04_3PP.indd 348

8/3/2023 5:25:23 PM

CHAPTER

5

Partial Differential Equations 5.1  BASIC CONCEPTS 5.1.1 Introduction Partial differential equations (PDE for short) have applications in many practical situations such as Brownian motion, population growth, traffic flow along a highway, stochastic process, etc. A partial differential equation is an equation involving partial differential coefficients of a function of two or more variables. Let us, consider that x and y are independent variables and z = f(x, y) [or u = f(x, y)] be a function of two variables x and y. We shall use the following notations throughout this chapter:

p=

¶z æ ¶u ö or p = ç ¶x è ¶x ÷ø , 

r=

¶2z ¶x 2

q=

æ ¶ 2u ö çç or r = 2 ÷÷ , ¶x ø è  

¶z æ ¶u ö ç or q = ÷ ¶y è ¶y ø

s=

¶2z ¶x ¶y

æ ¶ 2u ö çç or s = ÷ ¶x ¶y ø÷ è

2 2 ö æ and t = ¶ z2 ç or t = ¶ u2 ÷ . ç ÷

¶y

è

¶y ø

A few examples of partial differential equations are ¶ 2u ¶ 2u ¶ 2u ¶u + xy + y2 2 - (2x + 1) = 3y (i ) x 2 ¶x ¶y ¶x ¶x ¶y 2



¶z ¶z (ii ) x +y = 3z ¶x ¶y

5.1.2  Order and Degree The highest order derivative occurring in the partial differential equation is called the order of the equation.

EMEP.CH05_3PP.indd 349

The degree of a partial differential equation is the greatest exponent of the highest order when the equation is made free from all radicals. Examples: (i) The order and degree of the PDE 2 ¶z + 5x ¶z = xy2 are respectively 1 and 1. ¶x

¶y

(ii) The order and degree of the PDE 2 2 ¶ u2 + ¶ u2 + x ¶u + y ¶u = 3u are respectively 2

¶x

¶x

¶y

¶y

and 1. 5.1.3 Linear and No-Linear Partial Differential Equations A partial differential equation is called linear if it is of first degree in the dependent variable and its partial derivatives, i.e., if the powers or the products of the dependent variable and its partial derivatives remain absent in the equation. A PDE which is not linear is called a nonlinear partial differential equation. Examples: (i) x ¶u + y ¶u = 2u is a linear PDE. ¶x

¶z

2

2

(ii) çæ ¶z ÷ö + æç ¶z ö÷ = 1 is a non-linear PDE. è ¶x ø

è ¶y ø

2 2 (iii) x 2 ¶ z + 2 ¶ z + ¶z - e y ¶z = x + y is a non-­ 2

¶x

¶x ¶y

¶x

¶y

linear PDE.

8/3/2023 5:52:40 PM

350 • Engineering Mathematics Exam Prep Here, u = u( x , y ) = X ( x )Y ( y) = XY (say) ¶u ¶u = X ¢Y and = XY ¢ Then, ¶x ¶y

5.1.4 Formation of Partial Differential Equations Let us consider an equation

F(x,y,z,a,b) = 0 

…(i)

where a & b are arbitrary constants and z is a function of two independent variables x and y. Now differentiating (i) partially w.r.t x and y, respectively, we get,

¶F ¶F +p = 0 …(ii) ¶x ¶z

and



or,log X - log c1 = kx or,log

Then eliminating the arbitrary constants a and b from equation (i), (ii), and (iii), we get a partial differential equation of the form f(x, y, z, p, q) = 0.







5.3 HEAT, WAVE, AND LAPLACE EQUATIONS 5.3.1  Solution by Separation of Variables In this method, we assume that the dependent variable is the product of two functions each of which involves only one of the independent variables. Example:

Integrating we get,

ky

or,Y = c2 e 4

ky



\ u( x , y ) = XY = c1c2 e kx e 4 = c1c2 e



yö æ kç x + ÷ è 4ø

\ c1c2 = 8,

EMEP.CH05_3PP.indd 350

¶u ¶u , =4 ¶x ¶y

given that u(0, y) = 8e–3y

= 8e -3 y

k = -3 i.e., c1c2 = 8, k = -12 4

Hence, u( x , y) = 8e the required solution.

yö æ -12ç x + ÷ è 4ø

= 8e -12 x -3 y ,

which is

5.3.2 One-Dimensional Heat (Diffusion) Equation and Its Solution The one dimensional heat equation is given by ¶u ¶ 2u = c2 2 , ¶t ¶x

0 0 ¶t 2 ¶x ¶u ¶t

a

2 æ npx ö f ( x )sin ç dx . a ÷ø æ npb ö è a sinh ç ÷0 è a ø

Fully Solved MCQs

x + ct

1 1 + g ( x)dx. u( x ,t ) = éë f ( x + ct ) + f ( x - ct ) ùû 2c x -ct 2

u(0,y) = 0, u(a,y) = 0, u(x,b) = 0 and u(x,0) = f(x).

EMEP.CH05_3PP.indd 351

where, an =

is given by

5.3.4  The Laplace Equation and its Solution The Laplace equation is given by

n sin ç

n =1

¶x

with initial condition u(x,0) = f(x), and ¶u ¶t

æ npx ö ì np(b - y ) ü , ý ÷ sinh í a è a ø î þ

åa

¶x

¶ 2u ¶ 2u = c 2 2 , - ¥ < x < ¥, t > 0 2 ¶t ¶x

2



¥

(c)  z = ¶z - ¶z (d)  none of these

2. The D’Alembert’s solution of the following problem:

u( x , y ) =

¶x

0

1. The D’Alembert’s solution of the following problem:





(a)  z = ¶z ´ ¶z (b)  z = ¶z + ¶z

l

2 npx g ( x )sin dx . npc l

Remember:



The general solution is given by

2 ¶ 2u 2 ¶ u = c , 0 0, – ∞ < x < ∞ satisfying the conditions u(x, 0) = x, ut (x, 0) = 0 is (a)  x (b)  1

11. Let u(x, t) be the solution of

¶u ¶u = a and =b ¶x ¶y

2

¶u ¶u æ ¶u ö æ ¶x ö +y + +ç ÷ ¶x ¶y çè ¶x ÷ø è ¶y ø

i.e. u = px + qy + p2 + q2. 2. (a) z = (x + a)(y + b)…(i) ¶z ¶z = y + b, =x+a ¶x ¶y

Þ

Substituting these values of (x + a) and (y + b) in (i), we get z = ¶z ´ ¶z . ¶x

¶y

3. (c) u = (x2 + a)(y2 + b)…(i) Then differentiating (i) partially w.r.t x and y, we get,

2

1 ¶u ¶u = 2x ( y2 + b) i.e. y2 + b = ¶x 2x ¶x

2 y ¶y

¶y

2

Substituting these values of x + a and y2 + b in (i), we get u = 1 ¶u ´ 1 ¶u i.e; ¶u ¶u = 4 xyu. 2 y ¶y

2x ¶x

Þ

¶y = (- a ) f ¢ ( x - at ) + ag ¢ ( x + at ) ¶t

(a)  p (b)  1 - p



(c) 1 (d)  1 + p 13. Let u = Y(x, t) be the solution to the initial value problem utt = uxx for – ∞ < x < ∞, t > 0 with u(x, 0) = sin x, ut(x, 0) = cos x

Also y = f(x – at) + g(x + at)



2

Then the value of

(a) 

3 2

(b) 

is

and



1 2



(c)  1 (d) 1

Þ

¶y = f ¢( x - at ) + g ¢( x + at ) ¶x

and \

¶2 y = af ¢¢( x - at ) + a 2 g ¢¢( x + at ) ¶t

¶2 y = f ¢¢( x - at ) + g ¢¢( x + at ) ¶x 2

¶2 y ¶2 y = a2 2 2 ¶t ¶x

.

5. (b) Comparing the given equation with

2

Answer key 1. (a)

2. (a)

3. (c)

4. (c)

5. (b)

6. (b)

7. (a)

8. (a)

9. (c)

10. (a)

11. (b)

12. (a)

13. (a)



Explanation

1. (a) u = ax + by + a2 + b2 …(i) Then differentiating (i) partially w.r.t x and y, we get

EMEP.CH05_3PP.indd 352

¶x ¶y

4. (c) y = f(x – at) + g(x + at)

Then u ( p, p 2 ) = ?

æp pö yç , ÷ è2 6ø

…(ii)

and ¶u = 2 y( x 2 + a ) i.e. x 2 + a = 1 ¶u …(iii)

¶ 2u ¶ 2u ¶u = 0, u( x ,0) = sin x , ( x ,0) = 1 ¶t ¶t 2 ¶x 2

2

2



A

¶ 2u ¶ 2u ¶ 2u ¶u ¶u +B +C 2 + D +E + Fu = G , 2 ¶x ¶y ¶x ¶y ¶x ¶y

we get A = 1, B = –2 sin x, C = – cos2x \  B2 – 4AC = 4 sin2x – 4 × 1 × (– cos2 x) = 4(sin2x + cos2x) = 4 > 0 Hence, the given equation is hyperbolic. 6. (b) Comparing the given equation with 2 2 2 A ¶ u2 + B ¶ u + C ¶ u2 + D ¶u + E ¶u + Fu = G , ¶x¶y ¶x ¶y ¶x ¶y

we get A = y, B = x – y, C = –x. So B2 – 4AC = (x – y)2 – 4y(–x)

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Partial Differential Equations • 353









= (x – y)2 + 4xy = (x + y) 2 > 0 Hence, the given equation is hyperbolic. 7. (a) Comparing the given equation with



¶ 2u ¶ 2u ¶ 2u ¶u ¶u A 2 +B +C 2 + D +E + Fu = G , ¶x ¶y ¶x ¶y ¶x ¶y 2

2

we get A = y , B = –2xy, C = x . B2 – 4AC = (–2xy)2 – 4y2x2 = 4x2y2 – 4x2y2 = 0 Hence, the given equation is parabolic. 8. (a) Comparing the given equation with A

we get A = sin2x, B = – sin2x, C = cos2x. ∴  B2 – 4AC = (–sin2x)2 – 4 sin2x cos2x = 0. Hence, the given equation is parabolic. 9. (c) Comparing the given equation with ¶ 2u ¶ 2u ¶ 2u ¶u ¶u A 2 +B +C 2 + D +E + Fu = G , ¶x ¶y ¶x ¶y ¶x ¶y

2

=

2

1 1 éë f ( x + t ) + f ( x - t ) ùû + 2 2

{

x + ct

ò

(

0dx

x -ct

ò 0dx

x -t

}

)

(

)

12. (a) Here, c = 1, f(x) = sin x, g(x) = 1. Hence, u( x ,t ) =

=



EMEP.CH05_3PP.indd 353

3 pö p æ = sin ç p - ÷ = cos = . 3 3 2 è ø



Fully Solved MCQs

1. The PDE corresponding to ax + by + cz = f(x2 + y2 + z2) is ¶y

¶y

(c)  (by - cz ) ¶z + (cz - ax ) ¶z = ax - by ¶x

¶y

(d)  none of these 2. The PDE corresponding to u = xy + g(x2 + y2) is (a)  px – qy = x2 – y2 (b)  py + qx = 2y2 – 2 x2 (c)  py – qx = x2 – y2 (d)  py + qx = y2 + x2 3. The PDE corresponding to u = F(xy) + G(x/y) is 2 2 (a)  y2 ¶ u2 + x 2 ¶ u2 + y ¶u - x ¶u = 0

¶x

¶x

¶y

¶x

¶x

¶y

¶x

¶x

¶y

x -t

2 (a)  xy ¶ u = x ¶u + y ¶u - u

x -t

ò 1dx

1 1 = ´ ( 2sin x cos t ) + {( x + t ) - ( x - t )} 2 2 = sin x cos t + t

¶y

2 2 (c)  y2 ¶ u2 + x 2 ¶ u2 + x ¶u - y ¶u = 0

x +t

ò g(x)dx

¶y

2 2 (b)  x 2 ¶ u2 - y2 ¶ u2 + x ¶u - y ¶u = 0

(d)  none of these 4. The PDE corresponding to u = yj(x) + xY(y) is

1 1 é f ( x - t ) + f ( x + t ) ûù + 2ë 2

1 1 ésin( x + t ) + sin( x - t ) ùû + 2ë 2

x +t

2p æp pö æp pö \ y ç , ÷ = sin ç + ÷ = sin 3 è2 6ø è2 6ø

¶x

x +t

} {

ò cos xdx

x -t

(b)  (cy + bz ) ¶z + ( az + cx ) ¶z = bx + ay

1é ( x + t ) - ( x + t )2 + ( x - t ) - ( x - t )2 ùú û 2 êë 1é 2 2 ù 2 2 = ê 2x - 2 x + t ú = x - x + t û 2ë \ u(1,1) = 1 - (1 + 1) = -1. =

x +t

1 1 x +t ésin( x - t ) + sin( x + t ) ûù + ëésin x ûù x -t 2ë 2 1 1 = ëésin( x - t ) + sin( x + t ) ûù + ëésin( x + t ) - sin( x - t ) ûù 2 2 = sin( x + t )

¶x

1 é( x + t ) + ( x - t ) ùû = x . 2ë

u( x ,t ) =





1 1 ésin( x - t ) + sin( x + t ) ûù + 2ë 2

(a)  (cy - bz ) ¶z + ( az - cx ) ¶z = bx - ay

11. (b) Here, c = 1, f(x) = x – x2, g(x) = 0. Hence,



y( x , t ) = =

we get A = 1 + x2, B = 1 + y2, C = 0. ∴  B2 – 4AC = 0 – (1 + x2) (1 + y2) < 0. Hence, the given equation is elliptic. 10. (a) Here, c = 2, f(x) = x, g(x) = 0.



13. (a) Here, c = 1, f(x) = sinx, g(x) = cosx. Hence,

¶ 2u ¶ 2u ¶ 2u ¶u ¶u +B +C 2 + D +E + Fu = G , 2 ¶x ¶y ¶x ¶y ¶x ¶y

Hence, u( x ,t ) = 1 ëé f ( x + ct ) + f ( x - ct )ûù + 1

p p p æ pö \ u ç p, ÷ = sin p cos + = . 2 2 2 è 2ø .

¶x ¶y

¶x

¶y

¶y

2 (b)  ¶ u = y ¶u + x ¶u - u

¶x ¶y

¶x

¶y

8/3/2023 5:52:44 PM

354 • Engineering Mathematics Exam Prep 2 (c)  ( x + y) ¶ u = x ¶u - y ¶u - u

(c)  1 é32 - (2 - 2t )2 - (2 + 2t )2 ù + 1

2 (d)  xy ¶ u = ¶u - xy ¶u + 2u

(d)  1 é16 - (2 - 2t )2 ù + 1 é1 - max {1 - t, -1}ù ë û

¶x ¶y

¶x ¶y

5. The PDE + y( y - 1)2

¶x

¶x

2ë

¶y

2ë

¶y

x



2

¶ 2u ¶ 2u = , ¶t 2 ¶x 2

è2

¶t

t =0 =

0, otherwise

ì and ut ( x ,0) = ïí1, if x  2

4

7

5

Answer key

0,

1. (a)

2. (c)

3. (b)

4. (a)

5. (c)

6. (c)

7. (a)

8. (d)

9. (b)

10. (b)

11. (b) Explanation

1. (a) Given that ax + by + cz = f(x2 + y2 + z2)…(i) Differentiating (i) partially w.r.t x and y, we get

a+c

¶z ¶z ù é = f ¢( x 2 + y2 + z 2 ) ´ ê2x + 2z ú …(ii) xû ¶x ¶ ë

And é ¶z ¶z ù = f ¢( x 2 + y2 + z 2 ) ´ ê2 y + 2z ú …(iii) ¶y ¶y û ë



b+c



Dividing (ii) by (iii) we get, ¶z ¶z x+z ¶x = ¶x ¶z ¶z b+c y+z ¶y ¶y

a+c



îï 0, otherwise

(b)  5

(c)  4 (d)  4

2

ïî

and u æç 0, 1 ö÷ = 1 è 2ø 2

ø

4

then u(1,0) = ? (a) 0 (b) 5 (c) 4 (d) 1 9. Consider the diffusion problem: uxx = ut, 0 < x < p, t > 0 with u(0, t) = 0 = u(p, t) and u(x, 0) = 3 sin 2x. Then its solution is given by (a) 3e–t sin 2x (b) 3e–4t sin 2x (c) 3e–9t sin2x (d) 3e–2t sin 2x 10. Consider the one dimensional wave equation: utt = 4uxx, – ∞ < x < ∞,t > 0 ì with u( x ,0) = ïí16 - x , if x £ 4

¶x

æ 1ö u ç1, ÷ = 1 è 2ø

(a)  7  

is

with u(x,0) = x2, ¶u

¶t

Then, u æç 1 ,1 ö÷ = ?

(a)  elliptic in the region x < 0, y < 0, xy < 1 (b)  elliptic in the region x > 0, y > 0, xy > 1 (c)  parabolic in the region x < 0, y < 0, xy < 1 (d)  hyperbolic in the region x < 0, y < 0, xy < 1 8. If u(x, y) be a solution of the initial value problem

2

2 2 wave equation ¶ u = ¶ u , 0 < x < 1, t > 0 2 2

If u æç 1 ,0 ö÷ = 1 , è2 ø 4

¶ u ¶u ¶u +x +y =0 ¶x ¶y ¶y2

¶ 2u ¶ 2u ¶ 2u ¶u ¶u + 2xy +y 2 + + =0 2 ¶x ¶y ¶y ¶x ¶x ¶y

û

11. A function u(x, t) satisfies the

¶ 2u ¶ 2u x - ( y2 - 1)x 2 ¶x ¶y ¶x 2

is hyperbolic in a region in the XY-plane if (a)  x ≠ 0 and y = 1 (b)  x = 0 and y ≠ 1 (b)  x ≠ 0 and y ≠ 1 (d)  x = 0 and y = 1 6. In the region x > 0, y > 0, the PDE (x2 – y2)zxx + 2(x2 + y2)zxy + (x2 – y2)zyy = 0 (a)  is elliptic (b)  is parabolic (c)  is hyperbolic (d)  changes its type 7. The PDE

û

¶z ö æ ¶z ö æ ¶z ö æ ¶z ö æ or, ç a + c ÷ ç y + z ÷ = ç b + c ÷ ç x + z ¶x ø è ¶y ø è ¶y ø è ¶x ÷ø è or, ( cy - bz )

¶z ¶z + ( az - cx ) = bx - ay. ¶x ¶y

(a)  1 é16 - (2 - 2t )2 ù + 1 é1 - min {1, t - 1}ù ë û ë û

2. (c) u = xy + g(x2 + y2)…(i) Differentiating (i) partially w.r.t x and y, we get

(b)  1 é32 - (2 - 2t )2 - (2 + 2t )2 ù + t





Then for 1 < t < 3, u(2, t) is equal to 2

2ë

EMEP.CH05_3PP.indd 354

2

û

¶u = y + 2xg ¢( x 2 + y2 ) …(ii) ¶x

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Partial Differential Equations • 355

and ¶u = x + 2 yg ¢( x 2 + y2 ) …(iii) ¶y







From (ii) and (iii) we get, ¶u ¶u -x -y ¶y ¶x = 2x 2y or, y

¶u ¶u - y2 = x - x2 ¶x ¶y

or, x

¶u ¶u -y = y2 - x 2 ¶y ¶x

or,  py – qx = x2 – y2. 3. (b) u = F(xy) + G(x/y)…(i)

So ¶u = yF ¢( xy) + 1 G¢( x y) …(ii) ¶x



y

4. (a) Given that u = yj(x) + xY(y)…(i) \



¶u = j( x ) + x y¢( y ) …(iii) ¶y



Differentiating (iii) w.r.t. x, we have



¶ 2u = j ¢ ( x) + y ¢ ( y ) …(iv) ¶x ¶y



From (ii) and (iii) we get,





é x2 ù ¶ 2u 2x 2 ¢¢ = ( ) + x F xy ê 4 G ¢¢ ( x y ) + 3 G ¢ ( x y ) ú 2 ¶y y úû ëê y \ x2

2 ¶ 2u 2 ¶ u y ¶x 2 ¶y2

x2 = x 2 y2 F ¢¢( xy ) + 2 G ¢¢ ( x y ) y -x 2 y2 F ¢¢( xy ) =-

2x G ¢ ( x y ) …(iv) y

Also,



EMEP.CH05_3PP.indd 355

x2 2x G ¢¢ ( x y ) G¢ ( x y ) 2 y y

¶u ¶u x -y ¶x ¶y

= xyF ¢( xy ) +

x x G ¢ ( x y ) - xyF ¢( xy ) + G ¢ ( x y ) y y

2x G ¢ ( x y ) …(v) = y

Using (v), we get from (iv), x2

é ¶u ¶ 2u ¶ 2u ¶u ù - y2 2 = - ê x -y ú 2 ¶y û ¶x ¶y ë ¶x

or, x 2

¶ 2u ¶ 2u ¶u ¶u - y2 2 + x -y = 0. 2 ¶x ¶y ¶x ¶y

ù ¶ 2u 1 é ¶u ù 1 é ¶u = ê - y( x ) ú + ê - j( x ) ú ¶x ¶y y ë ¶x û x ë ¶y û

or, xy

¶ 2u ¶u ¶u =x - x y( y ) + y - yj( x ) ¶x ¶y ¶x ¶y

¶x ¶y

¶x

¶y

xY(y)] 5. (c) Comparing the given equation with

Differentiating (ii) partially w.r.t x partially we get, Again, Differentiating (iii) partially w.r.t y we get,

and

¶ 2u ¶u ¶u  =x +y - u [  u = yj(x) + or xy

¶u x = xF ¢( xy ) - 2 G ¢( x y ) …(iii) ¶y y

¶ 2u 1 = y2 F ¢¢( xy ) + 2 G ¢¢( x y) 2 y ¶x

¶u = yj( x ) + x y( y ) …(ii) ¶x



A

¶ 2u ¶ 2u ¶ 2u ¶u ¶u + B + C +D +E + Fu = G , 2 2 x y x ¶ ¶ ¶ ¶y ¶x ¶y

A

¶ 2u ¶ 2u ¶ 2u ¶u ¶u +B +C 2 + D +E + Fu = G , 2 ¶x ¶y ¶x ¶y ¶x ¶y

we get A = x2, B = –(y2 – 1)x, C = y(y – 1)2. ∴  B2 – 4AC = {–x(y2 – 1)} – 4x2y(y – 1)2 = (y – 1)2x2{(y + 1)2 – 4y} = (y – 1)2x2(y – 1)2 = x2(y – 1)4 > 0  (if  x ≠ 0 and y ≠ 1) 6. (c) Comparing the given equation with

we get A = x2 – y2, B = 2(x2 + y2), C = x2 – y2. \  B2 – 4AC = {2(x2 + y2)}2 – 4(x2 – y2) × (x2 – y2) = 4(x2 + y2)2 – 4(x2 – y2)2

(

ì = 4 í x 2 + y2 î 2 2

) - (x 2

2

2 2

- y2

) üýþ 2

= 4 ´ 2x y = 8x y > 0 (since x > 0, y > 0) Hence, the given PDE is hyperbolic in the given region x > 0, y > 0. 7. (a) Comparing the given equation with A

¶ 2u ¶ 2u ¶ 2u ¶u ¶u + B + C +D +E + Fu = G , 2 2 x y x ¶ ¶ ¶ ¶y ¶x ¶y

we get A = x, B = 2xy, C = y. \  B2 – 4AC = (2xy)2 – 4xy = 4xy (xy – 1) …(i) Now if x < 0, y < 0 and xy < 1, then xy > 0 and xy – 1 < 0 i.e; xy(xy – 1) < 0. So from (i), B2 – 4AC = 4xy (xy – 1) < 0. Hence, the given PDE is elliptic.

8/3/2023 5:52:47 PM

356 • Engineering Mathematics Exam Prep 8. (d) Here, u( x ,t ) = j1 ( x + ct ) + j2 ( x - ct )

Þ 3sin 2x = b1 sin x + b2 sin 2x + b3 sin 3x + .......¥

= j1 ( x + t ) + j2 ( x - t )

Þ 3sin 2x = 0 + b2 sin 2x + 0 + 0 + ......¥

( bn = 0 " n( ¹ 2) Î N )

2

Then, u(x, 0) = x ⇒ j1(x) + j2(x) = x2…(i) ¶u = j1 ( x + t ) - j2 ( x - t ) ¶t

Here,



¶u \ ¶t

= 0 Þ j1 ( x ) - j2 ( x ) = 0 …(ii)

t =0

(i) + (ii) Þ 2j1 ( x ) = x 2 Þ j1 ( x ) = x

2

2

2 (i) – (ii) Þ 2j2 ( x ) = x 2 Þ j2 ( x ) = x

2

( x + t )2 ( x - t )2 + = x 2 + t2 . 2 2 \ u(1,0) = 12 + 02 = 1.



Þ b2 = 3

Hence, (i) gives = 3e



u( x ,t ) =

åb n =1

=

¥

å n =1

=

bn e

- n2p2t p2

æ npx ö sin ç ÷ è l ø

x + 2t

So, u(2,t ) = 1 éë f (2 - 2t ) + f (2 + 2t )ùû + 1 2

2 + 2t

4



æ npx ö sin ç ÷ è p ø éë here c = 1, l = p ùû

¥

åb e n =1



n

- c2n2p2t 2 e l

n

- n2t

sin nx ..........(i)



ò



p

2 æ npx ö 3sin 2x ´ sin ç ÷ dx p è p ø

ò 0

=

+

1 é(2 + 2t ) - (2 - 2t ) ùû 4ë

=

1é 32 - (2 - 2t )2 - (2 + 2t )2 ù + t. û 2ë

}

ò

æ1 ö 1 æ1 ö æ1 ö 1 \ u ç ,0 ÷ = Þ j ç ÷ + y ç ÷ = …(i) 2 4 2 è ø è ø è2ø 4 1ö 1ö æ 1ö æ æ u ç1, ÷ = 1 Þ j ç1 + ÷ + y ç1 - ÷ = 1 2 2 2ø è ø è ø è



æ 1ö 1 æ1 ö æ 1ö 1 u ç 0, ÷ = Þ j ç ÷ + y ç - ÷ = è 2ø 2 è2ø è 2ø 2



Eqn (ii) + Eqn (iii) – Eqn (i) ⇒

3 écos(nx - 2x ) - cos(nx + 2x ) ùû dx p ë

ò

3 écos {(n - 2)x} - cos {(n + 2)x}ùû dx p ë

ò

p

3 é sin{(n - 2)p} sin{(n + 2)p} ù = ê ú n-2 n+2 pë û 3 ´ (0 - 0) = 0 p



[ for n ¹ 2] ¥

(i) Þ u( x ,0) = å bn sin nx n =1

EMEP.CH05_3PP.indd 356

} {

æ3ö æ1 ö Þ j ç ÷ + y ç ÷ = 1 …(ii) è2ø è2ø

3 é sin{(n - 2)x } sin{(n + 2)x } ù = ê ú n-2 n+2 pë û0



{



0

=

2 + 2t

1é 16 - (2 - 2t )2 + 16 - (2 + 2t )2 úù + 1 1dx û 2 ëê 4 2 -2t 1 = é32 - (2 - 2t )2 - (2 + 2t )2 ù û 2ë =

0

ò p



g ( x)dx



3 2sin 2x sin nxdx p

0

=

ò

2 -2t

p

p

=

g ( x)dx

u( x ,t ) = j( x + t ) + y( x - t )

0



ò

x -2t

11. (b)

l

2 æ npx ö bn = f ( x )sin ç ÷ dx l è l ø =

sin 2x

1 é f ( x - 2t ) + f ( x + 2t ) ûù + 1 2ë 2´2

\ u( x ,t ) =

9. (b) The solution is given by ¥

-4t

10. (b) Comparing utt = 4uxx with utt = c2uxx, we get c = 2, Here u(x, 0) = f (x).

Hence, u( x ,t ) =



(

u( x ,t ) = b2e -4t sin 2x  b1 = b3 = b4 = b5 = .... = 0 )



…(iii)

æ3ö æ1 ö æ1 ö æ 1ö æ1 ö æ1 ö jç ÷ + y ç ÷ + jç ÷ + y ç - ÷ - jç ÷ - y ç ÷ 2 2 2 2 2 è ø è ø è ø è ø è ø è2ø =1+

1 1 2 4

æ3ö æ 1ö 5 Þ jç ÷ + y ç - ÷ = è2ø è 2ø 4 æ1 ö æ1 ö 5 æ1 ö 5 Þ jç + 1÷ + y ç -1÷ = Þ u ç ,1 ÷ = . è2 ø è2 ø 4 è2 ø 4

8/3/2023 5:52:49 PM

Partial Differential Equations • 357 PREVIOUS YEARS SOLVED PAPERS (2000-2018)

1. The number of boundary conditions required 2

2

to solve the differential equation ¶ j + ¶ j = 0 ¶x 2 ¶y2 is (a) 2 (b) 0 (c) 4 (d) 1  [CE GATE 2001] 2. The partial differential equation

¶ 2j ¶ 2j ¶j ¶j + + + =0 ¶x 2 ¶y2 ¶x ¶y

has

(a)  degree 1 and order 2 (b)  degree 1 and order 1 (c)  degree 2 and order 1 (d)  degree 2 and order 2  [ME GATE 2007] 3. The partial differential equation that can be formed from z = ax + by + ab has the form æ ¶z ¶z ö ,q= ç with p = ÷ ¶ x ¶y ø è

7. Which one of the followings a property of the solutions to the Laplace equation Ñ2 f = 0 ? (a) The solutions have neither maxima nor minima anywhere except at the boundaries (b) The solutions are not separable in the coordinates (c)  The solutions are not continuous (d)  The solutions are not dependent on boundary conditions  [ME GATE 2016]

is

(a)  z = px + qy (b)  z = px + pq (c)  z = px + qy + pq (d)  z = qy + py  [CE GATE 2010]

8. The solution of the partial differential equation

¶u ¶u ¶ u +u = ¶t ¶x ¶x 2

is a

(a)  linear equation of order 2 (b)  non-linear equation of order 1 (c)  linear equation of order 1 (d)  non-linear equation of order 2  [ME GATE 2013] 5. The type of partial differential equation ¶f ¶ 2 f = ¶t ¶x 2

is

(a) parabolic (c) hyperbolic 

(b) elliptic (d) non-linear [IN GATE 2013]

k/a x

¶2P ¶2P ¶2P ¶P ¶P + 2 +3 +2 = 0 is 2 ¶x ¶y ¶x ¶y ¶x ¶y

(a) elliptic (c) hyperbolic 

EMEP.CH05_3PP.indd 357

(b) parabolic (d)  none of these [CE GATE 2016]

)

(b)  Cekt éêC1e(

)

ë

k/a x

ë

{(

+ C2e

+ C2e

-

(

-

(

)

k/a x ù

)

ú û

k/a x ù

ú û

{ ( k / a ) x}ùúû (d)  C sin kt éC cos {( k / a ) x} + C sin {- ( k / a ) x}ù êë úû (c)  Cekt éC1 cos êë

)}

k / a x + C2 sin -

1

2

[CE GATE 2016]



9. Consider the following partial differential equation

¶u ¶u +c = 0. ¶y ¶x

Solution of this equation is (a)  u(x, y) = f(x + cy) (b)  u(x, y) = f(x – cy) (c)  u(x, y) = f(cx + y) (d)  u(x, y) = f(cx – y)  [ME GATE 2017] 10. Consider the following partial differential 2 2 2 equation 3 ¶ j2 + B ¶ j + 3 ¶ j2 + 4j = 0.

¶x

¶x ¶y

¶y

For the equation to be classified as parabolic, the value of B2 must be________?  [CE GATE 2017] 11. Consider a function u which depends on position “x” and time “t.” The partial differential

6. The type of the partial differential equation

is of the form

(a)  C cos kt éêC1e(

4. The partial differential equation 2

¶u ¶ 2u =a 2 ¶t ¶x



2 equation ¶u = ¶ u is known as the ¶t ¶x 2 [ME GATE 2018]

(a)  Wave equation (b)  Heat equation (c)  Laplace’s equation (d)  Elasticity equation

8/3/2023 5:52:50 PM

358 • Engineering Mathematics Exam Prep 12. The solution at x = 1, t = 1 of the partial differential equation

¶ 2u ¶ 2u = 25 ¶x 2 ¶t 2

subject to ini-

tial conditions of u(0) = 3x and ¶u ¶t

_______

t =0

=3

is

[CE GATE 2018]

 (a) 1 (c) 4

6. (c) Comparing the given equation with

Answer key 2. (a)

3. (c)

4. (d)

5. (a)

6. (c)

7. (a)

8. (b)

9. (b)

10. 36

11. (b)

12. (d)

¶2P ¶2P ¶2P ¶P ¶P + B + C +D +E + FP = G , 2 2 ¶x ¶y ¶x ¶y ¶x ¶y

we get A = 1, B = 3, C = 1. \  B2 – 4AC = 9 – 4 > 0. Hence, the given PDE is hyperbolic. 7. (a) For a Laplace equation, solutions are always continuous and are separable in any coordinate system. The arbitrary constants occurring in the solution can be determined using the boundary conditions.

(b) 2 (d) 6

1. (c)

A

8. (b) Here, u = u( x ,t ) = X ( x )T (t ) = XT (say) Then,

Explanation

dX dY ö æ ç Here X ¢ = dx , T ¢ = dt ÷ è ø

1. (c) We know that the general solution of the 2 2 Laplace equation ¶ j + ¶ j = 0 contains four

¶x

2

¶y

arbitrary constants. Hence, “4” boundary conditions are needed in this case.

p = ¶z = a and q = ¶z = b. ¶x ¶y

Substituting the values of “a” and “b” we get from (i), z = px + qy + pq. 4. (d) Order of the equation = order of highest order derivative in the equation = 2. Again, the product term u ¶u is present in ¶x the equation.  Hence, the given equation is a non-linear equation of order 2. 5. (a) Comparing the given equation with

EMEP.CH05_3PP.indd 358

¶ 2u ¶ 2u ¶ 2u ¶u ¶u A 2 +B +C 2 + D +E + Fu = G , ¶x ¶y ¶x ¶y ¶x ¶y

we get A = 1, B = 0, C = 0. \  B2 – 4AC = 0 – 0 = 0 Hence, the given PDE is parabolic.

¶u ¶ 2u = a 2 Þ XT ¢ = aX ¢¢T ¶t ¶x aX ¢¢ T ¢ Þ = = k(say) X T k aX ¢¢ = k Þ X ¢¢ - X = 0........(i ) Then, X a

\

2

2. (a) The order of the equation = order of highest order derivative in the equation = 2; The degree of the equation = exponent of the highest order derivative in the equation = 1. 3. (c) The given equation is z = ax + by + ab …(i) Differentiating (i) partially w.r.t x and y, we get

¶u ¶ 2u ¶u = X ¢T , 2 = X ¢¢T and = XT ¢ ¶x ¶t ¶x



Auxiliary equation of (i) is given by: m2 -

k k = 0 i.e; m = ± . a a

Hence, the solution of (i) is given by X = X ( x ) = C1

æ kö x çç a ÷÷ eè ø

+ C2

æ kö -ç x ç a ÷÷ e è ø .

T¢ 1 dT dT =kÞ =kÞ = k dt T T dt T Integrating we get, log T = kt + log C or,log T - log C = kt T or,log = kt C or,T = Ce kt



ì æç ï ç \ u( x ,t ) = XT = íC1eè ï î

kö ÷x a ÷ø

+ C2

æ kö ü -ç x ç a ÷÷ ï e è ø Ce kt .

ý ï þ

9. (b) Let u(x, y) = f(x – cy).

∴  ¶u = f ¢( x - cy) and ¶u = -cf ¢( x - cy) . ¶x

¶y

Then, ¶u + c ¶u = -cf ¢( x - cy) + cf ¢( x - cy) = 0. ¶y

¶x

8/3/2023 5:52:51 PM

Partial Differential Equations • 359

10. 36 Comparing the given equation with

Questions for Practice

¶ 2u ¶ 2u ¶ 2u ¶u ¶u A 2 +B +C 2 + D +E + Fu = G , ¶x ¶y ¶x ¶y ¶x ¶y



we get A = 3, B = B, C = 3. Then, equation is parabolic

1. Solve by the method of separation of ­variables,

2. Solve: 3 ¶u + 2 ¶u = 0; u( x ,o) = 4 e- x

Þ B2 - 4 ´ 3 ´ 3 = 0

¶x

Þ B 2 = 36.

2 mensional heat flow equation is ¶u = c2 ¶ u . 2

¶t

For c = 1, the equation becomes ¶u = ¶ u2 . ¶t

¶x

Hence, the given partial differential equation is a heat equation. 12. (d) We know that the D’Alembert’s solution of the following problem: ¶ 2 u = c 2 ¶ 2 u , - ¥ < x < ¥, t > 0 ¶t 2 ¶x 2

with initial condition u(x,0) = f(x), and ¶u ¶t

t =0 =

g( x )

is given by



1 1 + g ( x)dx. u( x ,t ) = ëé f ( x + ct ) + f ( x - ct ) ûù 2c x -ct 2

ò

Here, c2 = 1 i.e; c = 1 , f ( x ) = 3x and g ( x ) = 3. 25

\ u(1,1) =

5

1 1 éë f (1 + c ) + f (1 - c ) ùû + 2 2c

=

1+ c

ò g(x)dx

1- c

1é æ 1ö 1 öù 5 æ ê f 1 + ÷ + f ç1 - ÷ ú + 2 ë çè 5ø 5 øû 2 è

1+

1 5

ò

1é æ 1ö 1 öù 5 æ = ê3 ´ ç1 + ÷ + 3 ´ ç1 - ÷ ú + 2ë è 5ø 5 øû 2 è 1

1+ 1 é18 12 ù 15 = ê + ú+ ´ éëx ùû 15 2ë 5 5û 2 15



EMEP.CH05_3PP.indd 359

g ( x)dx

1 15

15 éæ 1ö æ 1 öù =3+ ´ ê 1 + ÷ - ç1 - ÷ ú = 6. 2 ëçè 5ø è 5 øû

1+

1 5

ò 3 dx

1-

1 5

-x +

3y 2

2 2 2 3. The equation t ¶ u + 3 ¶ u + x ¶ u + ¶u = 0 is 2 2

¶x ¶t

¶t

¶x

¶y

(a)  Hyperbolic if 4xt > 9 (b)  Parabolic if 4xt ≤ 9 (c)  Elliptic if 4xt > 9 (d)  Parabolic if 2xt = 3 2 4. The equation ¶ u = ¶u is 2

¶x

¶y

(a) Parabolic (b) Hyperbolic (c) Elliptic (d)  None of these 5. Classify the following equation in second quadrant:

x + ct

u = 4e

(c)  u = 4e–x+2y (d)  none of these.

¶x

2



¶y

(a)  u = 6ex–2y (b) 

11. (b) We know that the general form of one di-



where u(x, 0) = 6e–3x

(a)  u(x, 0) = 6e–3x (b)  u = 6e3x–4t (c)  u = 6e3x+4t (d)  u = 8e–(2x+3t)

Þ B 2 - 4 AC = 0



¶u ¶u =2 + u, ¶x ¶t

x 2 + y2 +x

¶ 2u ¶ 2u ¶ 2u + 2( x - y ) + y2 + x 2 2 2 ¶x ¶y ¶x ¶y

¶u ¶u +y =0 ¶y ¶y

(a)  Elliptic for x > 0 (b)  Parabolic for y < 0 (c) Hyperbolic (d)  Parabolic for x = 0 6. The partial differential equation that can be formed from z = ax + by + ab is given by (a)  z = xzx + yzy + zxzy (b)  z = yzx + xzy + zxzy (c)  z = zx + zy + xyzxzy (d)  None of these. 7. The PDE corresponding to 2z = (ax + y)2 + b is given by (a)  x ¶z + y ¶z = 0 ¶y

¶y

(b)  x ¶z + y ¶z = çæ ¶z ÷ö ¶x

¶y

2

è ¶y ø

(c)  y ¶z + x ¶z = ¶z ¶z ¶x

¶y

¶x ¶y

(d)  None of these

8/3/2023 5:52:52 PM

360 • Engineering Mathematics Exam Prep 8. Let u(x, t) be the D’Alembert’s solution of the initial value problem for the wave equation  utt - c2uxx = 0; u( x ,0) = f ( x ), ut ( x ,0) = g ( x ); where c is a positive real number and f(x), g(x) are add functions. Then, u(0,1) is equal to (a) –1 (b) 1 (c) 0 (d) 2 9. Let u(x, t) be the solution of the initial value problem: utt = uxx , u( x ,0) = x , ut ( x ,0) = 1, Then, u(2, 2).is equal to (a) 4 (b) 6 (c) –2 (d) 0

EMEP.CH05_3PP.indd 360

10. Let u(x, y) = 2cos (x – 2y) × f(y) be a solution of the initial value problem 2 ux + uy = u; u( x ,0) = cos x Then, f(1) = (a)  1 (b) e 2

(c) 

e (d) 1 2

1. (a)

2. (b)

6. (a)

7. (b)

Answer key 3. (c) 4. (a) 8. (c)

9. (a)

5. (c) 10. (c)

8/3/2023 5:52:52 PM

CHAPTER

6

Laplace Transforms

6.1 BASICS OF LAPLACE TRANSFORMS

Example: 1 L(t 2 )

6.1.1  Definition of the Laplace Transform Let f(t) be defined for all real t ≥ 0. Then the ­Laplace transform of f(t), denoted by L[f(t)] or ¥

=

æ1 ö æ3ö G ç + 1÷ G ç ÷ 2 è ø = è2ø 1 3 s2



s2 2! 2 = . s2+1 s3

and L(t 2 ) =

F(s) or f ( s) , is defined by L[ f (t )] = F ( s) = e f (t )dt, ò

(iii) L( eat ) = 1

provided the integral on the right-hand side exists (here s is a real or complex variable).

Example:

- st

0

6.1.2 Linear Property of the Laplace Transform Let f(t) and g(t) be two functions of t and a, b be two constants. Then,

s-a

L( e5t ) =



L(sin 3t ) =

Example:

(i) L(1) = 1 (provided s > 0)



(ii)

EMEP.CH06_3PP.indd 361

ì G(n + 1) if s > 0 (in general) ïï n +1 s L(t n ) = í n! ï if s > 0 and n Î Z + îï sn +1

a s2 + a 2

provided s > 0

3 3 = 2 2 s +3 s +9 2

(v) L(cos at ) =

6.1.3 Fundamental Formulas of the Laplace Transform s

s>a

Example:

L(2et – 3 sin t) = 2L(et) – 3L(sin t)

provided

1 1 , L( e -2t ) = . s-5 s+2

(iv) L(sin at ) =

L{af(t) + bg(t)} = a L{f(t)} + bL {g(t)} Example:

+1

L(cos 2t ) =

s s2 + a 2

s s = s2 + 22 s2 + 4

(vi) L(sinh at ) = é

a s2 - a 2

êRemember : sinh x = ëê

provided s > 0

provided s > a

ex - e -x ù ú 2 ûú

8/3/2023 6:13:59 PM

362 • Engineering Mathematics Exam Prep Example: L(sinh t ) =



Example: 1 s2 - 1

(vii) L(cosh at ) =

L( e 3t sinh 2t) = s s - a2 2

provided s > a

é ex + e -x ù êRemember : cosh x = ú 2 ëê ûú

(v)

L( e t t 3 ) =

Example: s s -9 2

6.1.4  First Shifting Theorem If L{f(t)} = F(s), then L{eat f(t)} = F(s – a), “a” being a real or complex constant. Application: t

Application: We know that L( e2t ) = 1 .

1

s2 + 1

.

L{ e t sin t } = L{ e at f (t )} = F (s- a) 1 = F ( s - 1) = . 1 s ( )2 + 1

6.1.5 Some Advanced Formulas of the Laplace Transform s-a (i) L( e at cos bt) = 2 2 (s - a) + b

Example: s -1 L( e cos 2 t) = ( s - 1)2 + 4 t

(ii) L( e at sin bt) =

b ( s - a )2 + b2



Let L{f(t)} = F(s). Also let f (t ), f ¢(t ), f ¢¢(t ),..., f (n -1) (t ) are all continuous for t ≥ 0. Then the Laplace transform of nth order derivative of f(t) is denoted by L{f (n)(t)} and is defined by L{ f ( n ) (t )} = sn F ( s) - sn -1 f (0) - sn -2 f ¢(0)



3 ( s + 1)2 + 9



1 . s2 + 1 Here f (0) = sin 0 = 0. f ¢(t ) = cos t and so f ¢(0) = cos0 = 1.Therefore, L{ f ¢¢(t )} = s2 F ( s) - sf (0) - f ¢(0) 1 = s2 ´ 2 - (s ´ 0) - 1 s +1

s+2 cosh t) = ( s + 2)2 - 1 b ( s - a )2 - b2

L{ f ¢¢¢(t )} = s3 F ( s) - s2 f (0) - sf ¢(0) - f ¢¢(0). Let f (t ) = sin t.Then L{ f (t )} = F ( s) =

s-a cosh bt) = ( s - a )2 - b2

(iv) L( e at sinh bt) =

EMEP.CH06_3PP.indd 362

In particular, L{ f ¢(t )} = s F ( s) - f (0),

Example: L( e

- ........... - f ( n -1) (0)



L{ f ¢¢(t )} = s2 F ( s) - sf (0) - f ¢(0),

L( e -t sin 3 t) =

-2t

1 . s-2 1 1 1 \ L{f(3t )} = F ( s 3) = ´ 3 3 æs ö ç 3 - 2÷ è ø 1 = s-6

6.2 LAPLACE TRANSFORM ON DERIVATIVES

Example:

(iii) L( e

s-2

Take f (t ) = e . Then F ( s) =

Hence, by1st shifting theorem,

at

3 ( s - 1)4

6.1.6  Change of Scale Property If L{f(t)} = F(s), then L{ f ( at )} = 1 F ( s a ), “a” being a a real or complex constant.

at

\F(s) = L{ f (t )} = L{sin t } =



n , n ÎZ+ ( s - a )n +1

2t

L{ e sin t } = L{ e f (t )} Here, a = 1and f (t ) = sin t.



L( e at t n ) =

Example:



L(cosh 3t ) =

2 ( s - 3)2 - 4



=

-1 s2 -1 = 2 . 2 s +1 s +1

8/3/2023 6:14:01 PM

Laplace Transforms • 363

Some important results: If L{f(t)} = F(s), then (i) L{t f (t )} = - d F ( s ) n ds (ii) L{t n f (t )} = ( -1)n d F ( s ) ds n (for n = 0,1,2,3......)

find the value of the integral. This technique actually reduces the number of steps for obtaining the value of integration. Example: ¥

òe

If L{f(t)} = F(s), then Remember:

ò

If L{f(t)} = F(s), then

¥

é f (t ) ù Lê ú = F ( s)ds, ë t û s

ò

provided the integral is e­ xists.

6.4 LAPLACE TRANSFORM ON PERIODIC FUNCTIONS If L{f(t)} = F(s) and f(t) be a periodic function with period “T,” then T



L{ f (t )} = F ( s) =

òe 0

- sT

¥

ò

1 - e - sT

.

Remember: (i) f(t) is periodic with period T ⇔ f(t + T) = f(t). (ii) Let f(t) be defined as follows: ì f (t ), 0 < t < k1 f (t ) = í 1 î f2 (t ), k1 < t < k 2 OR ì f (t ), 0 £ t < k1 f (t ) = í 1 î f2 (t ), k1 < t £ k 2 OR ì f (t ), 0 < t £ k1 f (t ) = í 1 î f2 (t ), k1 < t < k 2 OR



ì f (t ), 0 £ t £ k1 f (t ) = í 1 î f2 (t ), k1 £ t £ k 2



Then, the period of f(t) is k2.

6.5 EVALUATION OF INTEGRALS USING LAPLACE TRANSFORMS If the range of integration are “0” and “∞,” then the definition of Laplace transform can be used to

EMEP.CH06_3PP.indd 363

= e - st f (t ) dt

[where s = 3, f (t ) = sin t ]

0

= F ( s) = L ( f (t ) ) = L(sin t )



=

1 1 1 = = [ s = 3] s2 + 1 32 + 1 10

6.6 INITIAL AND FINAL VALUE THEOREMS 6.6.1  Initial Value Theorem

If L ( f (t ) ) = F ( s),then lim f (t ) = lim sF ( s) t ®0

s ®¥

(provided the limit exists) 6.6.2  Final Value Theorem

If L ( f (t ) ) = F ( s),then lim f (t ) = lim sF ( s) t ®¥



f (t ) dt

sin t dt

0

6.3 LAPLACE TRANSFORM ON INTEGRALS ét ù F ( s) L ê f(t)dt ú = s ê ú ë0 û

-3t

s ®0

(provided the limit exists)

6.7 FUNDAMENTALS OF INVERSE LAPLACE TRANSFORM 6.7.1 Definition of Inverse Laplace Transform If L[f(t)] = F(s), then f(t) is called inverse Laplace transform of F(s), and is written as f(t) = L–1[F(s)].

( )

L e2t =

1 æ 1 ö 2t Þ L-1 ç ÷=e s-2 ès-2ø

Remember: (Linearity property) If L[f1(t)] = F1(s), L[f2(t)] = F2(s) and c1, c2 be two constants, then L–1[c1f1(s) + c2f2(s)] = c1L–1[F1(s)] + c2L–1[F2(s)] = c1f1(t) + c2f2(t). 6.7.2 Useful formulas on Inverse Laplace Transforms (i) L-1 æ 1 ö = 1, s > 0 çs÷ è ø

(ii)

ì t n -1 , provided s > 0, n > 0 ï æ 1 ö ï G(n) L-1 ç n ÷ = í ès ø ï t n -1 , if n Î Z + ï (n - 1)! î

8/3/2023 6:14:02 PM

364 • Engineering Mathematics Exam Prep Example:

Example: 3 -1

2

2

t t æ1 ö t = = , L-1 ç 3 ÷ = 3 1 2 2 ès ø

ö 1 ö t3 / 2-1 t1/ 2 æ 1 = L ç 3/2 ÷ = ç n + 1 = n n and = p ÷ 2 3 1 1 è ès ø ø 2 2 2 2t1/ 2 2t1/ 2 = = 1 p 2 -1 æ



(iii) L-1 æ 1 ö = eat , provided s > a çs-a÷ è

2

1 ö 1 æ 1 ö -1 æ L-1 ç 2 = sin 2t ÷=L ç 2 2 ÷ s 4 s + è ø è +2 ø 2 s ö = cos at, provided s > 0 2 ÷ ès +a ø 2

Example:

æ 1 è s2 - a 2

L-1 ç

s ö 1 -1 æ ÷ = sinh at; L ç 2 ø a è s - a2

ö ÷ = cosh at ø

1 ö 1 æ 1 ö -1 æ L-1 ç 2 = sinh 5t ÷=L ç 2 2 ÷ è s - 25 ø è s -5 ø 5 n -1 ù 1 at t , = e ú n n -1 ë (s - a) û

(vii) L-1 éê

if n Î Z +

Example: é 1 ù L-1 ê = 3ú ë ( s - 2) û

6.8 IMPORTANT THEOREMS ON INVERSE LAPLACE TRANSFORMS (i) If L–1[F(s)] = f(t), then, î ds

Example:



é ù é ù 1 1 L-1 ê 2 = L-1 ê 2 2ú 2 2ú ë ( s + 4) û ë (s + 2 ) û 1 (sin 2t - 2t cos2t ) = 2 ´ 23 1 = (sin 2t - 2t cos2t ) 16

tf (t ) = -L-1 ìí d F ( s)üý

s ö æ s ö -1 æ L-1 ç 2 = cos3t ÷=L ç 2 2 ÷ ès +9ø ès +3 ø

(vi)

2

Example:

Example:





é ù é ù 1 s s L-1 ê 2 = L-1 ê 2 = t sin 2t 2ú 2 2ú + + s s ( 4) ( 2 ) ë û ë û 4 ù 1 1 = 3 (sin at - at cos at ) 2 2ú ë ( s + a ) û 2a

1 ö 1 = sin at, provided s > 0 2 ÷ ès +a ø a

(v) L-1 æ ç

2

Example:

æ 1 ö 5t L-1 ç ÷=e s 5 è ø

(iv) L-1 æ ç



ù 1 s t sin at = 2 2ú ë ( s + a ) û 2a

(x) L-1 éê

(xi) L-1 éê

ø

Example:



é ù s-2 = e2t cos t L-1 ê 2 2ú ë ( s - 2) + 1 û

t3-1 t2 t2 = e2t = e2t 3 -1 2 2

þ

(ii) (Convolution theorem) if L–1[F(s)] = f(t) and L–1[G(s)] = g(t), then t



ò

f * g = L-1 ëé F ( s)G( s) ûù = f (u ) g (t - u )du 0

6.9 UNIT STEP FUNCTION AND UNIT IMPULSE FUNCTION 6.9.1 Unit Step Function The unit step function (Heaviside’s unit step function) u(t – a) is defined by

ì0, for t < a u(t - a ) = í î1, for t ³ a

(viii) L-1 éê



[a being a positive real number]

Example:

(i) L {u(t - a )} = 1 e -a s



e2t

ù 1 at 1 = e sin bt 2 2ú ë (s - a) + b û b



é ù 1 t 1 = e sin 7t L-1 ê 2 2ú ë ( s - 1) + 7 û 7 ù s-a = e at cos bt 2 2ú + s a b ( ) ë û

(ix) L-1 éê

EMEP.CH06_3PP.indd 364

Remember: s

(ii)

ì f (t ), a < t < c If f (t ) = í 1 , then î f2 (t ), c < t < b

 f(t) = f1(t) {u(t – a) – u(t – c)} + f2(t) {u(t – c) –u(t – b)}

8/3/2023 6:14:04 PM

Laplace Transforms • 365 ì f (t ), a < t < b , then î f2 (t ), t > b

(iii) If f (t ) = í 1

f(t) = f1(t){u(t – a) – u(t – b)} + f2(t) u(t – b)} ì f1 (t ), a < t < c

(iv) If f (t ) = ïí f2 (t ), c < t < b , then ï f (t ), t > b î 3

 f(t) = f1(t){u(t – a) – u(t – c)} + f2(t){u(t – c) – u(t – b)} + f3(t)u(t – b)

Suppose a given ordinary differential equation contains the variable y = y(t) and its derivatives. Assume that L[y(t)] = Y(s). Then to solve the equation (i.e., to get y(t)) the following steps can be ­followed: Step-1: Take the Laplace transform on both sides of the given differential equation and apply the given conditions. Step-2: Express Y(s) as a function of “s” and then resolve the R.H.S. part as a sum of partial fractions. Step-3: Take inverse Laplace transform on both sides and put L–1(Y(s)) = y(t).

Example: ì0, 0 < t < 1 ï If f (t ) = í1, 1 < t < 2 , then ï 2, t>2 î

Fully Solved MCQs (Level-I) at 1. The Laplace transform of f (t ) = e - 1 is

f (t ) = 0 ´ {u(t - 0) - u(t - 1)}

+ 1 ´ {u(t - 1) - u(t - 2)} + 2 ´ u(t - 2)



6.9.2 Second Shifting Theorem If L–1[F(s)] = f(t), then L[f(t – a) × u(t – a] = e–as F(s) = e–as L(f(t))

(a)

1 s-a

(c)

1 (d) 1 s+a s( s - a )

{

}

( here f (t - 1) = t - 1; so f (t ) = t )



= e -1s L ( f (t ) ) = e - s L (t ) = e - s

1 s2

6.9.3 Unit Impulse Function The unit impulse function (Dirac Delta function) d(t – a) is defined by ì 0, for t ¹ a d(t - a ) = í î¥, for t = a



æ ç provided ç è

ö d(t - a ) dt = 1 ÷ ÷ -¥ ø ¥

ò

Remember: (i) L{d(t – a)} = e–as (ii)

¥

ò

f (t ) d(t - a ) dt = f ( a )

0

6.10 SOLVING ORDINARY DIFFERENTIAL EQUATIONS The Laplace transform method can be used to solve an ordinary differential equation without finding the general solution.

EMEP.CH06_3PP.indd 365

a

2. The Laplace transform of (sin t + cos t)2 is

Example: L (t - 1) u(t - 1) = L { f (t - 1) u(t - 1)}

(b)

1 s( s + a )



(a) 1 + s

(c)

2 s +4 2

(b) 1 - 2 1

1 1 + 2s s2 + 4

s

(d)

3. L(sin 3t cos t) = ? (a)

1 2 + s2 + 16 s2 + 4

(b)

(c)

2 1 + s2 + 16 s2 + 4



2

( s + 1)( s + 25)

(c)

1 ( s2 + 1)( s2 + 5)

(c)

1 s2 + 5s + 6



2s2 + 6 ( s + 1)( s2 + 5) 2

(d) None of these

5. L(e–2t – e–3t) = ? (a) 2 5 (b) s + 5s + 6

1 1 + s2 + 16 s2 + 4

(d) None of these

4. L(sin 3t sin 2t) = ? 12s (a) (b) 2

s +4

1 2 + 2 2 s s +4

s s + 5s + 6 2

(d) None of these

6. L (sin2 2t) = ?

(a) 1 æç 1 - s ö÷ (b) 1 æç 1 + 2 s ö÷ 2 è s s + 16 ø 2 è s s2 + 16 ø (c) 1 æç 1 - 2 4 ö÷ (d) 1 æç 1 + 2 4 ö÷ 2 è s s + 16 ø 2 è s s + 16 ø

8/3/2023 6:14:06 PM

366 • Engineering Mathematics Exam Prep 7. L(sin3 t) = ? (a)

4s ( s2 + 1)( s2 + 9)

(b)

6 ( s2 + 1)( s2 + 9)

(c)

6s 2 ( s + 1)( s2 + 9)

(d) None of these

s +1

(a) 1 (b) 1

s+2

s -1

(c) 1 (d) 1 s-3

s+4

14. L(t sin 2t) = ?

8. L (1 pt ) = ? (a) 1 ps (b) 1 s (c) p

13. If L { f (t )} = 1 , then L { f (4t )} = ?

s (d)

1 2 s

2 –2t

4 ( s2 + 4)2

(a)

(c) 22

1 ( s + 2)3

(b)

s ( s + 2)3

(a)

1 (s- 2)2

(c)

2 ( s + 2)3

(d)

3 ( s + 2)3

(c)

-1 ( s - 2)2

(a)

b ( s + a )2 - b 2

(c)

s s 2 2 (d) (s + a ) - b (s + a ) 2 + b 2

(b)

b ( s + a )2 + b 2

11. Find L(f(t)) if

1- s 1- s (a) 1 - e (b) e - 1

s

1-s

(d) None of this

12. The Laplace transform of sin(2t + 5) is

(a)

(b) (c) (d)

EMEP.CH06_3PP.indd 366

(s (

1 2

-4

1

(s

2

(s

-4

)

)

( s cos5 - 5sin 5 )

)

( s cos5 + 5sin 5 )

1 2

( 2 cos5 - s sin 5 )

( 2 cos5 + s sin 5 )

s2 + 4 1

)

+4



(d)

1 (s+ 2)2 -1 (s+ 2)2

÷ ø

t

(a) log æç s- 1 ö÷ (b) log æç s ö÷ è s- 1 ø è s ø (c) log æç s ö÷ (d) log æç s+ 1 ö÷ s+ 1 s è

æ ç è

ïìe , 0 < t < 1 f (t ) = í ïî 0, t > 1

-1

ç è

17. L ç e

t

1-s

(b)

t 16. L æç 1 - e ö÷ = ?

10. L(e–at sin h bt) = ?

1- s

s +4

15. L(te ) = ?

(a)

(c) e

(d) 24



s +4

4s ( s + 4)2 2

2t

9. L(t e ) = ?



(b)

- at

ø

- e -bt t

è

ø

ö ÷÷ = ? ø

(a) log æç s+ b ö÷ (b) - log æç s+ b ö÷ è s+ a ø

è s+ a ø

(c) 2log æç s+ a ö÷ (d) None of these è s+ b ø

18. The Laplace transform of sint is t

–1

(a) cot s (b) tan–1s (c) cot -1 1 (d) tan -1 12 2 s

s

19. The Laplace transform of sinht is t

(a) log s + 1 (b) log s - 1 s -1

s +1

(c) log æç s + 1 ö÷ (d) log æç s - 1 ö÷ è s +1 ø è s -1 ø

8/3/2023 6:14:09 PM

Laplace Transforms • 367

20. Find the Laplace transform of the periodic function f(t) which is defined as follows: f (t ) =



ct for 0 < t < T and f (t + T ) = f (t ). T

(a) 1 -

(

sT

s 1-e

(c) c 2 sT

ce - sT

(

- sT

)

ce - sT

s 1-e

(b) c sT

)

- sT

(

3è

- sT

)

2 1- s p e ( ) -1

(1 - s ) (1 - e -2ps )

(c) e

2(1- s ) p

(b)

- 1 (d)

(1 - e ) -2 ps

e 2 sp - 1

(1 - s ) (1 - e -2ps )

(a) æç1 - t ö÷ sin t + t cos t (b)

(b) 1 (c) 1 (d) 0 2

3

è

13

13

13

2

t

(a) p (b) p (c) p (d) - p 3

4

2

sin 4t dt = ? t 4

4

3 ü ý=? î ( s + 2)(s+ 5) þ

26. L-1 ìí

(a) e2t – e5t (b) e–2t – e–5t (c) e–2t + e–5t (d) None of these

EMEP.CH06_3PP.indd 367

2

(c) e–2tt3

(d) e–2tt2

31. L-1 ìí s + 4 3 üý = ? î ( s + 2) þ

(a) te2t(1 + t) (b) te–2t(1 + t)

p (a) p (b) - p (c) p (d) 2

1 ü ý=? + (s 2)3 þ î

30. L-1 ìí

(a) 1 e -2tt 2 (b) 1 e2tt 2

¥

24. e -t sin t dt = ? ò 2

2ø

(d) None of these

(a) 1 (b) 3 (c) 4 (d) 5 13

2ø è t sin t + t cos t 2

(c) æç1 + t ö÷ sin t - t cos t

0

0

1 ( sin 3t - 3cos3t ) 27

î (s + 1) þ

23. e -3t sin t cos t dt = ? ò

ò

(b) 1 ( sin 3t - t sin 3t )

ì ü 29. L-1 í s2 + 2 2 ý = ?

1/s

¥

25.

54

(d) None of these

0

¥

(a) 1 ( sin 3t - 3t cos3t )

(c)

¥

0

2

27

22. te -t dt = ? ò (a) 1

ø

ü 1 =? 2ý î (s + 9) þ

28. L-1 ìí

(d) 0

f(t) = et for 0 < t < 2p

(a)

2

(b) 1 æç 1 sin t - sin 2t ö÷ 3è2 ø (c) sin t + sin 2t (d) None of these

21. Find the Laplace transform of the periodic function f(t) where:

2

(a) 1 çæ sin t - 1 sin 2t ÷ö

e - sT

s 1+e

ü 1 ý=? 2 î ( s + 1)(s + 4) þ

27. L-1 ìí

2

(c) e–2t (2 + t) (d) None of these 32. L-1 ìílog æç s + 1 ö÷ üý = ? î

è s - 1 øþ

(a) 2 sinht (b) 2 cosht t

(c) t sin t 2

t

(d) t cos t 2

8/3/2023 6:14:12 PM

368 • Engineering Mathematics Exam Prep 1 ù ú=? ë s( s + 1) û

33. L-1 éê

(a) 1 – e–t (b) 1 + e–t

40. Solve y′ + y = 1; given y(0) = 0 (a) et (b) e–t (c) 1 + e–t (d) 1 – e–t Answer key

(c) 1+et (d) 1–et ¥

34. ò e 2t d (t - 2) dt = ? 0

(a) e–3 (b) e4 –4

(c) e (d) None of this 35.

é e -2s ù L ê 2 ú=? ëê s + 1 ûú -1

1. (c)

2. (a)

3. (c)

4. (a)

5. (c)

6. (a)

7. (b)

8. (b)

9. (c)

10. (a)

11. (b)

12. (b)

13. (d)

14. (b)

15. (a)

16. (a)

17. (a)

18. (a)

19. (a)

20. (c)

21. (a)

22. (a)

23. (a)

24. (c)

25. (a)

26. (b)

27. (a)

28. (a)

29. (c)

30. (a)

31. (b)

32. (a)

33. (a)

34. (b)

35. (c)

36. (a)

37. (b)

38. (c)

39. (a)

40. (d)

(a) u(t – 2) × sin t

Explanation

(b) e–2t × sin t

1. (c)

(c) u(t – 2) × sin (t – 2) (d) e2t × sin (t – 2) 36. L{et–2 u(t – 2)} = ? (a)

1 1 e -2s (b) e -2s ( s - 1) ( s + 1)

(c)

1 1 e2s (d) e 2s ( s - 1) ( s + 1)



2. (a) L{ f (t )}

ì t -2 37. Find L(f(t)) if f (t ) = ïíe , t > 2 is ïî 0, t < 2

-2s -s (a) e (b) e

s -1

s -1

(c) e

-3s

s -1

-s (d) e

38. Find L(f(t)) if

(c)

-p s e 3 (d) e s2 + 1 s2 + 1

ps

39. Solve y′ = 1; given y(0) = 0 (a) t (b) t + 1 (c) 2t (d) t2

EMEP.CH06_3PP.indd 368

sin 3t cos t 1 = (2sin 3t cos t ) 2 1 = {sin(3t + t ) + sin(3t - t )} 2 1 = (sin 4 t + sin 2t ) 2 \ L ( sin 3t cos t )

ps

(a)

= L{(sin t + cos t )2 } = L(1 + sin 2t ) = L(1) + L(sin 2t ) 1 2 = + 2 s s +4

3. (c)

ì pö p æ ïsin ç t - ÷ , t > ï 3 3 è ø f (t ) = í p ï 0, t< ïî 3 -

-



s +1

e -p s e 3 (b) 2 2 s -1 s -1

æ e at - 1 ö L{ f (t )} = L ç ç a ÷÷ è ø 1 at = L( e - 1) a 1 = { L( e at ) - L(1)} a 1æ 1 1ö 1 = ç - ÷= a è s - a s ø s( s - a )



ì1 ü = L í (sin 4 t + sin 2t ) ý î2 þ 1 = { L(sin 4 t) + L(sin 2t)} 2 1æ 4 2 ö 2 1 = ç 2 + = + 2 è s + 16 s2 + 4 ÷ø s2 + 16 s2 + 4

8/3/2023 6:14:13 PM

Laplace Transforms • 369

4. (a)

8. (b)

sin 3t sin 2t 1 = (2sin 3t sin 2t ) 2 1 = {cos(3t - 2t ) - cos(3t + 2t )} 2 1 = (cos t - cos 5t ) 2 \ L(sin 3t cos t )



ì1 ü = L í (cos t - cos 5t ) ý 2 î þ 1 = { L(cos t) - L(cos 5t )} 2 1æ s 12s s ö = ç 2 = 2 è s + 1 s2 + 52 ÷ø ( s2 + 1)( s2 + 25)

(

pt =

9. (c)

(

{(

7. (b)

=



EMEP.CH06_3PP.indd 369

=



=

)

=

)

2

s3

2

)

=

) (

)}

1 æ s+a +b-s-a +b ö ç ÷ 2 è ( s + a - b)(s+ a + b) ø b (s+ a )2 - b 2

Alternative Method b s2 - b2

L ( sinh bt ) =

By 1st shifting theorem,

1 æ3 ö = L ç sin t - sin 3t ÷ 4 è4 ø

=

(

L t -1/ 2

é1 ù = L ê { e(b-a )t - e -(b+ a )t } ú ë2 û 1 L e(b-a )t - L e -(b+ a )t = 2 1æ 1 1 ö = ç ÷ 2 è s - (b - a ) s + (b + a ) ø

1 1 L(2sin2 2t ) = L(1 - cos4t ) 2 2

3 1 L(sin t) - L(sin 3t ) 4 4 3 1 1 3 ´ - ´ 4 s2 + 1 4 s2 + 9 3æ 1 1 ö 4 çè s2 + 1 s2 + 9 ÷ø 6 2 ( s + 1)( s2 + 9)

p

{s- ( -2)}3 [by first shifting theorem] 2 = (s+ 2)3

(

1 1 (4 sin3 t ) = {3sin t - sin 3t } 4 4 [Using sin 3x = 3sin x - 4 sin3 x ] 3 1 = sin t - sin 3t 4 4 \ L(sin3 t )

1

æ ebt - e -bt ö e - at sinh bt = e - at ç ÷÷ ç 2 è ø 1 (b-a )t -(b + a )t = {e -e } 2

1 { L(1) - L(cos4t )} 2 1 æ1 s ö = ç - 2 2 è s s + 16 ø÷

=

s3

L e - at sinh bt

sin3 t =

)

t =

10. (a)

=



2

( )

L t2 =

\ L t 2 e -2t =

6. (a) L(sin 2 2t ) =

(

L 1

æ 1 ö æ1 ö G ç - + 1÷ Gç ÷ è 2 ø = 1 è2ø 1 1 ( - +1) p s 2 s2 ù 1 p 1 é æ1 ö = = êsince G ç ÷ = p ú 2 s ë p s è ø û



L(e-2t - e-3t ) = L(e-2t ) - L(e-3t ) 1 1 = s+2 s+3 1 1 = = ( s + 2)( s + 3) s2 + 5s + 6



p

1 = p

5. (c)



1

)

L 1



(

)

L e - at sin bt =

b ( s + a )2 - b2

11. (b)

L{ f (t )} ¥

ò

= e - st f (t )dt 0

1

¥

0

1

ò

ò

= e - st et dt + e - st 0dt



8/3/2023 6:14:14 PM

370 • Engineering Mathematics Exam Prep 1

ò

= e

(1- s )t

0

=



1

é e(1-s )t ù dt = ê ú ëê 1 - s ûú 0

1- s

0



1- s

e e e -1 = 1-s 1-s 1-s

12. (b)

{

æ 1 - et ö Lç ç t ÷÷ è ø é f (t ) ù = Lê ú, ë t û

}

= L ( cos5 ´ sin 2t + sin 5 ´ cos2t )

= cos5 ´ L ( sin 2t ) + sin 5 ´ L ( cos2t )

=



(s

1 2

+4

(s )

2

2

+4

)

+ sin 5 ´

(s

s 2

+4

)

1

s 1+ 4

=



n

ò

¥

¥

1 s+4

n

1 æ ö æ 1 ö = log ç ÷ - log ç 1 - (1/ s) ÷ 1 0 è ø è ø 1 [ s ® ¥ Þ ® 0] s æ s ö æ s -1 ö = - log ç ÷ = log ç s ÷ (log1 = 0 ) 1 s è ø è ø

17. (a)

(

)

( ) ( )

L e - at - e -bt = L e - at - L e -bt

1 1 = F (s) s+a s+b æ e - at - e -bt ö \Lç ÷÷ ç t è ø é f (t ) ù é - at -bt ù = Lê ú , where f (t ) = e - e û ë t û ë =

¥

¥

1 ö æ 1 ds = F ( s)ds = ç s + a s + b ÷ø è s s

ò

15. (a)

ò

¥

( ) = L (t

EMEP.CH06_3PP.indd 370

1 ö ¥ æ1 = ç ÷ ds = ëé log s - log( s - 1) ûù s s s 1 è ø s é é æ 1 öù æ s öù = êlog ç ÷ ú = êlog ç 1 - (1/ s) ÷ ú 1 s è ø è øûs ë ûs ë

d F ( s) dsn [w here n = 1, f (t ) = sin 2t ] 1 d L{ f (t )} = ( -1) ds d = - L(sin 2t ) ds d æ 2 ö =- ç 2 ds è s + 4 ÷ø ( -1) ´ 2s 4s = ( -2) ´ 2 = ( s + 4)2 ( s2 + 4)2



s

ò

ò



L (t sin 2t ) = L t n f (t ) = ( -1)

L te

s

ò

= { L(1) - L( et )}ds [Using linear property]

14. (b)



s

s

1 æsö 1 F = ´ 4 çè 4 ÷ø 4

)

¥

¥

1 æsö L { f ( at )} = F ç ÷ . a èaø

(

¥

¥

We know that if L { f (t )} = F ( s),then,



¥

ò

13. (d)

1 æ ö ç F ( s) = s + 1 (given) ÷ è ø

é where f (t ) = 1 - et ù ë û

= F ( s)ds = L[ f (t )]ds = L(1 - et )ds

( 2 cos5 + s sin 5 )

\ L { f (4t )} =

( -1) 1 = 2 ( s - 2) ( s - 2)2

16. (a)

L sin ( 2t + 5 )

= cos5 ´

= ( -1) ´

2t

n

)

n

f (t ) = ( -1)

= éëlog( s + a ) - log( s + b) ùû s

dn F ( s) dsn

[w here n = 1, f (t ) = e2t ] 1 d = ( -1) L{ f (t )} ds d = - L( e2t ) ds d æ 1 ö =- ç ds è s - 2 ÷ø

¥

é æ s + a öù = êlog ç ÷ú è s + b øûs ë

¥

é æ 1 + ( a / s) ö ù = êlog ç ÷ú è 1 + (b / s) ø û s ë



æ 1 + ( a / s) ö æ1 + 0 ö = log ç ÷ - log ç 1 + (b / s) ÷ è1 + 0 ø è ø 1 [ s ® ¥ Þ ® 0] s

8/3/2023 6:14:15 PM

Laplace Transforms • 371 æs+aö = log1 - log ç ÷ è s+b ø æ s+b ö = log ç ÷ ès+aø



=

=

18. (a) ¥

æ sin t ö Lç ÷ = L ( sin t ) ds = è t ø s

ò

¥

ò s

1 ds s2 + 1

=

¥

= étan -1 s ù = tan -1 ¥ - tan -1 s ë ûs p -1 = - tan s = cot -1 s 2



=

=

19. (a) æ sinh t ö Lç ÷ è t ø

=

¥

ò

= L ( sinh t ) ds s

¥

=

òs s

1 ds -1



= =

=

=



=

òe 0

- st

)

c

)

é æ e - st êt çç ëê è -s

(

c

)

é ïì æ e - sT ê íT ç êë îï çè -s

(

T 1 - e - sT c

(

T 1-e

- sT

)

é -Te - sT e - sT 1ù - 2 + 2ú ê s s úû êë s

c

)

é -Te - sT ù 1 + 2 1 - e - sT ú ê s ëê s ûú

(

T 1 - e - sT

(

ce - sT

s 1 - e - sT

ò 0

T

=

)

+

ò

ö ù ÷÷ dt ú ø úû 0

T

ö e - st ù ÷÷ - 2 ú ø s ûú 0

ö e - sT ïü ïì e -0 üïù ÷÷ - 2 ý - í0 - 2 ýú s ïþ ïî s ïþúû ø

)

c s2T

2p

=

1 - e - sT 1

2p

(1 - e ) ò -2 ps

1

(1 - e ) -2 ps

1

(1 - e ) -2 ps

òe 0

- st

´ et dt

1 - e -2ps

1- s t e( ) dt =

0

1

(1 - e ) -2 ps

2p

é e(1-s ) t ù ê ú êë (1 - s ) úû 0

é e(1-s ) 2p e0 ù ê ú 1 - s ) (1 - s ) ú ëê ( û é e(1-s ) 2p 1 ù ê ú êë (1 - s ) (1 - s ) úû

2 1- s p e ( ) -1

(1 - s ) (1 - e -2ps )

¥

ò te

-t

dt

0

¥

1 - e - sT

ö æ e - st ÷÷ - 1 ´ çç ø è -s

(

e - st f (t ) dt

ò

= e - st f (t ) dt f (t ) dt

T

é æ e - st êt çç êë è -s

22. (a)

L{ f (t )} T

EMEP.CH06_3PP.indd 371

=

1 s +1 æ s +1 ö log ç ÷ = log s - 1 . 2 è s -1 ø

20. (c)

t dt

0

(integrating by parts)

T 1 - e - sT

T

é 1 öù æ ç 1 - s ÷ú 1ê = ê log1 - log ç ÷ú 2ê çç 1 + 1 ÷÷ ú êë s ø úû è

=

(

- st

L{ f (t )}

é 1 öù æ ç 1 - s ÷ú 1ê æ1 - 0 ö = êlog ç ÷ú ÷ - log ç 2ê è1 + 0 ø çç 1 + 1 ÷÷ ú s ø ûú è ëê

é 1 öù æ ç 1 - s ÷ú 1ê = ê0 - log ç ÷ú 2ê çç 1 + 1 ÷÷ ú êë s ø úû è 1 æ s -1 ö = - log ç ÷ 2 è s +1 ø

c

e )ò

21. (a) Here, f(t) is a periodic function with ­period 2π.

¥

1 æ ö çs ® ¥ Þ s ® 0 ÷ è ø

- sT

T 1 - e - sT



é 1 öù æ ¥ ç 1 - s ÷ú 1é 1ê æ s - 1 öù = ê log ç ÷ú ÷ ú = êlog ç 2ë è s + 1 øûs 2 ê çç 1 + 1 ÷÷ ú êë s ø úû s è



(

T 1-e

=-

2

T

c

òe 0

- st

´

ct dt T

1 - e - sT

[where s = 1, f (t ) = t ]

0



= F ( s) = L ( f (t ) ) = L(t ) =

1 = 1 [ s = 1] s2

8/3/2023 6:14:16 PM

372 • Engineering Mathematics Exam Prep ¥

¥

23. (a) e -3t sin t cos t dt ò

=

0

¥

=

ò

= e - st f (t ) dt 0

( here s = 3 and f (t ) = sin t cos t )



3 ì ü L-1 í ý s ( + 2)(s + 5) î þ ì + + 2) ü s s ( 5) ( = L-1 í ý s ( + 2)(s + 5) þ î

= L éë f (t ) ùû

1 1 = L éësin t cos t ùû = L éë2sin t cos t ùû = L éësin 2t ùû 2 2 1 2 1 1 = ´ 2 = = ( s = 3 ) 2 s + 22 s2 + 4 13

1 ö æ 1 = L-1 ç s 2 s 5 ÷ø + + è æ 1 ö -1 æ 1 ö = L-1 ç ÷ - L ç s+ 5 ÷ ès+2ø è ø

24. (c) ¥

òe

-t

0

sin t dt t

p s p 0 p - tan -1 = - tan -1 = 2 4 2 4 2 (s = 0 )

26. (b)

= F ( s) (by definition of Laplace transform)



4é sù s tan -1 ú = tan -1 ¥ - tan -1 4 êë 4 ûs 4



¥

= e -2t - e -5t

27. (a)

ò

= e - st f (t ) dt 0

sin t ö æ ç here s = 1 and f (t ) = t ÷ è ø é sin t ù = F ( s) = L ëé f (t ) ûù = L ê ú ë t û ¥

ò

= L(sin t ) ds = s

¥

òs s

2

=

1 ds +1

=

¥

= é tan -1 s ù = tan -1 ¥ - tan -1 s ë ûs p p -1 = - tan s = cot -1 s = cot -1 1 = 2 4 (s = 1)



25. (a)

¥

ò 0



òe 0

ü ì ü 1 1 = L-1 í 2 2 2 ý 2ý î (s + 9) þ î (s + 3 ) þ

-0t

1 (sin 3t - 3t cos3t ) 2 ´ 33 1 = (sin 3t - 3t cos3t ) 54

sin 4t dt t

29. (c)

ò 0

sin 4t ö æ ç here s = 0 and f (t ) = t ÷ è ø é sin 4t ù = F ( s) = L éë f (t ) ùû = L ê ú ë t û = L(sin 4t ) ds =



EMEP.CH06_3PP.indd 372

s

2

=

= e - st f (t ) dt

ò

1æ 1 ö sin t - sin 2t ÷ 3 çè 2 ø

28. (a) L-1 ì í

¥

¥

=

1 -1 æ (s2 + 4) - ( s2 + 1) ö L ç 2 ç ( s + 1)(s2 + 4) ÷÷ 3 è ø

1 é -1 æ 1 ö öù -1 æ 1 êL ç 2 ÷-L ç 2 ÷ú 3ë è s +1 ø è s + 4 øû

sin 4t dt t ¥

=

ì ü 1 L-1 í 2 ý 2 + + ( 1)(s 4) s î þ ü 1 -1 ì 3 = L í 2 ý 2 3 + + ( 1)(s 4) s î þ

¥

òs s

2

4 ds + 42



ì s+2 ü L-1 í 2 2ý î (s + 1) þ ì ü s 1 = L-1 í 2 2 2 + 2 2 2 2 ý (s 1 ) (s 1 ) + + î þ ì ü ì ü s 1 = L-1 í 2 2 2 ý + 2L-1 í 2 2 2 ý î (s + 1 ) þ î (s + 1 ) þ 2 t (sin t - t cos t ) = sin t + 2 2 ´ 13 tö æ = ç1 + ÷ sin t - t cos t 2ø è

8/3/2023 6:14:17 PM

Laplace Transforms • 373

30. (a)

.................(1)

é ù ì 1 ü 1 -1 ê ú L-1 í = L ý 3 ê {s- ( -2)}3 ú î (s+ 2) þ ë û =

e -2tt3-1 1 -2t 2 = e t 3 -1 2

é 1 ù -t g (t ) = L-1 éëG( s) ùû = L-1 ê ú=e . ës +1û

ì s+4 ü ì ( s + 2) + 2 ü L-1 í = L-1 í 3ý 3 ý î ( s + 2) þ î ( s + 2) þ ì 1 1 ü = L-1 í + 2´ ý 2 ( s + 2)3 þ î ( s + 2)

\ f (u) = 1, g (t - u ) = e -(t -u ) = eu -t So (1)gives t

= e -2t (t + t 2 )

ò

t

= e -t é e u ù = e -t é e t - e 0 ù ë û0 ë û

¥

ò

\ e2t d(t - 2) dt = f (2) = e2´2 = e4 .

æ s +1 ö Let F(s) = log ç ÷. è s -1 ø



æ s +1 ö Then f (t ) = L-1 ( F(s) ) = L-1 ç ÷ è s -1 ø tf (t ) d = -L-1 { F ( s)} ds d ì æ s + 1 öü = -L-1 ílog ý ds î çè s - 1 ÷ø þ



éd ù log ( s + 1) - log ( s - 1 ) ú = -L-1 ê ds ë û 1 ö -1 æ 1 = -L ç ÷ è s +1 s -1 ø



1 ö æ -2 ö -1 æ = -L-1 ç 2 ÷ = 2L ç 2 ÷ = 2sinh t è s -1 ø è s -1 ø 2 Þ f (t ) = sinh t t



ì æ s + 1 öü 2 Þ L-1 ílog ç ÷ ý = sinh t î è s - 1 øþ t

{

}

0

35. (c) L éëu(t - a ) f (t - a ) ùû = e - as F ( s)



Þ u(t - a ) f (t - a ) = L-1 ée - as F ( s) ù ë û



(by second shifting theorem)



-2s Now comparing e–as F(s) with e , we get



1 a = 2, F ( s) = 2 s +1

s2 + 1

é 1 ù \ f (t ) = L-1 ëé F ( s) ûù = L-1 ê 2 ú = sin t ës +1û é e -2s ù -1 - as Hence, L-1 ê 2 ú = L éëe F ( s) ùû êë s + 1 úû = u(t - a ) f (t - a ) = u(t - 2) f (t - 2) = u(t - 2) ´ sin(t - 2)

36. (a)

{

é 1 ù L-1 ê ú ë s( s + 1) û é 1 1ù = L-1 ê ´ ú ëê ( s + 1) s ûú

}

( )

L et -2u(t - 2) = e -2s L et =

t

ò

= L-1 ëé F ( s) ´ G( s) ûù = f (u ) ´ g (t - u ) du 0

EMEP.CH06_3PP.indd 373

= e -t é e t - 1 ù = 1 - e -t ë û



32. (a)

33. (a)

ò

34. (b) Here, f(t) = e2t and a = 2.

= te -2t (1 + t )



t

é 1 ù u -t -t u L-1 ê ú = 1 ´ e du = e e du ë s( s + 1) û 0 0

ì 1 ü ì 1 ü = L-1 í + 2L-1 í 2ý 3ý + ( 2) s î þ î ( s + 2) þ e -2tt 2-1 e -2tt3-1 +2 (2 - 1)! (3 - 1)!

the Convolution theorem )

1 1 ù é ê where, F ( s) = s , G( s) = s + 1 ú ë û é1 ù Here, f (t ) = L-1 éë F ( s) ùû = L-1 ê ú = 1, ësû

31. (b)

=

( by

( by

1

( s - 1)

e -2s .

the second shifting theorem )

37. (b) We know that if f (t ) = ìí g (t - a ), t > a , then, L[f(t)] = e–as × L[g(t)].

î

0, t < a

8/3/2023 6:14:19 PM

374 • Engineering Mathematics Exam Prep Here, a = 2 and g (t - 2) = et -2 . So g (t ) = et . \ L ëé f (t ) ûù = e =e

-2s

-2s



´ L ëé g (t ) ûù = e

-2s

1 . ´ ( s - 1)

38. (c)

ì g (t - a ), t > a , î 0, t < a

–as

L[f(t)] = e

1. Find L(f(t))

We know that if f (t ) = í



s +1

then,

(c)

× L[g(t)].

-

ps 3

=e

-

ps 3

´ L ëé g (t ) ûù = e ´



(s

1 2

+1

)

-

ps 3

´ L ëésin t ûù

.

39. (a) y¢ = 1 Þ y¢(t ) = 1 Þ L[ y¢(t )] = L(1) Þ sY ( s) - y(0) = 1 s 1 Þ Y ( s) = 2 s Þ sY ( s) =

1 s

( y(0) = 0)



Þ sY ( s) - y(0) + Y(s) =



1 s

( y(0) = 0)

1 ( s + 1) - s 1 1 = = s( s + 1) s( s + 1) s s +1

1 ö æ1 Þ L-1 éëY ( s) ùû = L-1 ç ÷ è s s +1 ø æ1 ö æ 1 ö = L-1 ç ÷ - L-1 ç ÷ èsø è s +1 ø



EMEP.CH06_3PP.indd 374

Þ y(t ) = 1 - e -t

s +1 s2 + 2s + 5

(a)

1 ( s + 1)(s + 2s + 5)

(b)

(c)

2 ( s + 1)(s2 + 2s + 5)

(d) None of these

2

4. L{(1 + te–t)3} = ? (a) 1 +

3 6 6 + + (s+ 1)2 (s+ 2)3 (s+ 3)4

(b) 1 +

2 3 4 + + ( s + 1)2 ( s + 2)3 ( s + 3)4

s

(c) - 1 + s

y¢ + y = 1 Þ y¢(t ) + y(t ) = 1 Þ L[ y¢(t )] + L[ y(t )] = L(1)

Þ Y ( s) =

1 s

s -s (a) 2e (b) 2e s3 s3 2e - s 2e s (c) 2 (d) 2 s s –t 3. L(e sin2 t) = ?



40. (d)

1 s

(d)

ìï 0, 0 < t < 1 if f (t ) = í 2 ïî(t - 1) , t > 1

s

æ 1 ö t 2-1 Þ L-1 ëéY ( s) ûù = L-1 ç 2 ÷ = =t è s ø 2 -1 Þ y(t ) = t

Þ ( s + 1)Y ( s) =

1 + e - sp s2 + 1

1 s2 + 1

2. Find L(f(t))

So g (t ) = sin t. \ L ëé f (t ) ûù = e

ìsin t, 0 < t < p if f (t ) = í î 0, t > p

- sp (a) 1 -2 e (b)

p p pö æ Here, a = and g (t - ) = sin ç t - ÷ . 3 3 3 è ø



Fully Solved MCQs (Level-II)

´ L é et ù ë û

2 6 6 + + ( s + 1)2 ( s + 2)3 ( s + 3)4

(d) None of this 5. The Laplace transform of e–t(4 sin 2t – 5 cosh 2t) is

(a)

5 ( s + 1) 8 s2 + 2s + 5 s2 + 2s - 3

(b)

5 ( s + 1) 8 s2 - 2s + 5 s2 + 2s - 3

(c)

10 ( s + 1) 4 - 2 s - 2s + 5 s + 2s - 3



2

(d) none of these 6. The Laplace transform of cosh2 2t is

(a) 1 + 2s

(

s 2

2 s - 16

)

(b) 1 2s

(

s 2

2 s + 16

)

8/3/2023 6:14:21 PM

Laplace Transforms • 375

(c) 1 + s

s

(

2

2 s -4

(d) 1 -

)

s

(s

s 2

-4

t

æ ö 13. L ç e -t sin t dt ÷ = ? ò

)

ç è

(b) (c) (d)

3

4

+

+

2 ( s - 1)

2

+

3 s ( - 1)

2

+

2

+

( s - 1)3 ( s - 1)2 2

3

+

3

+

3

+

( s - 1) 1

( s + 1) 2

( s + 1)

4

( s - 1) 3

( s + 1) 4

( s + 1)

s +1

(c)

2

(c)

æ s -1 ö ÷ 3 ø

-ç 1 e è ( s - 1)

(d)

ç è0

÷ ø

s ( s + 2)

æ s -1 ö ÷ 3 ø

ç 1 eè ( s + 1)



æ s +1 ö ÷ 3 ø

-ç 1 e è ( s + 1)

p ì ïïsin kt, 0 < t < k f (t ) = í ï 0, p < t < 2p ïî k k

(a)

2 (a) s2 - 3 + 2s 2 (b)

(s + 2s + 5)



1 (s2 + 2s + 5)2

(b)

æ t ö 1 æ 1 10. If L çç 2 ÷÷ = 3 ,then L ç è pt è pø s2 (a) 1 (b) s

(c)

ö ÷=? ø 1 s

(d)

11. If L ( f (t ) ) = tan -1 æç 1 ö÷ ,then L (tf (t ) ) = ? è

(c)

ø

s 2 s +1

(c)

EMEP.CH06_3PP.indd 375

(

)

then L[f(t)] = ?

(d) None of this

(a)

æ at ö 12. L ç e - cos bt ÷ = ? ç ÷ t è ø

æ s- a ö log ç 2 2 ÷ è s +b ø

k p ö æ - s ç 1 - e k ÷ s2 + k 2 ç ÷ è ø

)

ì1, 0 < t < p f (t ) = í î-1, p < t < 2p

1 s +1 2

æ ö æ (a) log ç s- a ÷ (b) - log ç s- a ç s2 + b2 ÷ ç s2 + b2 è ø è

k 2p ö æ s ç 1 - e k ÷ s2 + k 2 ç ÷ è ø

)

16. If f(t) is a periodic function with period 2p and f(t) is given by

èsø

(a) tan -1 æç 12 ö÷ (b) s

1 2p ö æ s ç 1 + e k ÷ s2 + k 2 ç ÷ è ø

(

(d) None of this

s

)

(

(d) None of this

(c) -1

1 p æ - sö ç 1 + e k ÷ s2 + k 2 ç ÷ è ø

(

9. L(te cos 2t) = ?

5 - s2 (s2 + 2s + 5)2

s ( s - 2)

15. Find the Laplace transform of the periodic function f(t) where

–t

(c)

s ( s - 2)

(c) ( s - 1) (d) ( s + 1)

4 s ( + 1)

(b)

ö

s2 ( s + 2 )

s

1 s ( - 1)

æt

(a) ( s + 1) (b) 2( s - 1)

( s + 1)

-s

(a)

s +1

1 1 -1 tan -1 s (d) cot s s s

14. L ç e -t cosh t dt ÷ = ? ò

8. If L { f (t )} = e , L {e -t f (3t )} = ? æ s +1 ö ç ÷ eè 3 ø

÷ ø

(a) 1 tan -1 ( s + 1) (b) 1 cot -1 ( s + 1)

7. The Laplace transform of e–t(t + 2)2 is (a)

t

0

ö ÷ ÷ ø

(d) None of these

(c) (d)

1

s( e -2ps - 1) s( e

1

-2 ps

- 1)

(b)

e -ps

s( e -2ps - 1)

{2e -ps - e -2ps - 1}

e -ps s

8/3/2023 6:14:24 PM

376 • Engineering Mathematics Exam Prep ¥

17. ò e

- at

0

(b) 1 éêt + 1 sin(2t )ùú

- e -bt dt = ? t

2ë

(a) log æç b ö÷ (b) log æç a ö÷ èaø



èbø

(c) log (a + b) ¥

18. e ò

-2t

0

(d) log (ab)

2

2

(d) 0

4 ,then lim f (t ) = ? t ®¥ s (s + 4)



(a) 0

(b) 1



(c) –1

(d) 4

2 20. If L ( f (t ) ) = 4s - 3s + 8 ,then lim f (t ) = ?

(

s s2 + 4

)

t ®¥



(a) 0

(b) 1



(c) –1

(d) 2

s+2 ö ÷=? 2 è s - 4s + 13 ø

21. L-1 æç

(a) e2t (cos3t + 4 sin 3t ) 3

(b) e -2t (cos3t + 4 sin 3t ) 3

(c)

4 et (cos2t - sin 3t ) 3

(d) None of these 22. L-1 æç 3s + 4 ö÷ = ? 2 è s + 2s + 5 ø

(a) e -t cos2t + 1 e -t sin 2t 3

(b)

1 3e cos2t + e -t sin 2t 2

(c)

1 e cos2t + et sin 2t 2



-t

t

(d) None of these ü 1 ý=? 2 î s ( s + 2) þ

23. Find L-1 ìí (a)

EMEP.CH06_3PP.indd 376

(c) 1 é1 - 1 sin(2t )ù ê ú 2ë

2

û

ë

(c) 1 19. If L ( f (t ) ) =

û

(d) 1 éêt + 1 sin( 2t )ùú 2 2

sinh t dt = ? t

(a) 1 log 2 (b) 1 log 3

2

2

ù 1é 1 tsin( 2t ) ú 2 êë 2 û

ì

û

ü ï ý=? 2 îï s ( s + 1) þï 1

24. L-1 ïí

2

(a) t – 2 + (t + 2)e–t (b) t + 2 + (t – 2e–t) (c) t + 2 + (t – 2)et (d) t – 2 + (t + 2)e–2t ù s ú=? 2 ë ( s + 1)( s + 1) û

25. L-1 éê

(a) 1 {- cos t + sin t + et } 2

(b) 1 {cos t - sin t - e -t } 2

(c) 1 {cos t - sin t + e -t } 2

(d) 1 {cos t + sin t - e -t } 2

ü s +1 =? 2ý î ( s + 2s + 5) þ

26. L-1 ìí

2

(a) 1 tet sin 2t (b) 1 te -t sin 2t 4

2

(c) 1 tet sin t (d) 1 te -t sin 2t 2

4

é ì 2 üù 27. L-1 êlog íï s + 1 ýïú = ? ëê

îï s( s + 1) þïúû

(a) 1 (1 + et - 2cos t ) (b) 1 (1 - e -t + 2cos t ) t

t

(c) 1 (1 + e -t - 2cos t ) (d) None of these t

28. The Laplace transform of f(t), where z ì0, 0 < t < 1 ï f (t ) = í1, 1 < t < 2 ï 2, t>2 î

is

(a) 1 ( e -s - e -2s ) (b) 1 ( e -s + e -2s ) s

s

(c) 1 ( e -2s - e -s ) (d) 1 ( e -3s + e -s ) s

s

8/3/2023 6:14:29 PM

Laplace Transforms • 377

29. The Laplace transform of f(t) , where 1 ì ï t, 0 < t < 2 ï 1 ï f (t ) = ít - 1, < t < 1 2 ï t >1 ï 0, ï î



(a) (b) (c)

34. L{t2u(t – 1)} = ?

is

(a) æç 13 - 22 + 1 ö÷ e -s (b) æç 1 - 1 + 1 ö÷ e -s sø s ès s3 s2 s è

ès

s 1æ 1 1ö - ç e 2 + e- s - ÷ sç s s÷ è ø

(a) (c)

(

2 1-e

s2 + 4

1 - e -ps s2 + 4

ès

s

sø



(a) 1 – e–t(1 + t)

(b) 1 + et (t – 1)



(c) 1 – te–t (d) tet 36. Solve: y′′ – 4y′ = et; given y(0) = 0 = y′(0)

(a) 1 e -2t - 1 et + 1 e2t 4

30. The Laplace transform of f(t), where

-ps

sø

s

35. Solve: y′′ + 2y′ + y′ = 1; given y(0) = 0 = y′(0)

s 1æ 1 1ö - ç e 2 + e- s + ÷ ç s s s÷ è ø

ìsin 2t, 0 < t < p f (t ) = í t>p î 0,

ø

(c) æç 23 + 22 + 1 ö÷ e -s (d) æç 23 - 22 - 1 ö÷ e -s

s 1 æ - 2 1 -s 1 ö çe - e + ÷ sç s s÷ è ø

(d) None of these



(c) 22 + æç 12 + 1 ö÷ e -s (d) -22 - æç 12 + 1 ö÷ e -s sø s sø ès s ès

4

) (b) 2 (1 + e ) -ps

s2

12

(b) 1 e2t - 1 et + 1 e -2t

is

(d)

3

3

12

(c) 1 e -t - 1 et + 1 e -2t 4

3

12

(d) 1 e2t - 1 et + 1 e -2t 12

1 + e -ps s2 + 4

3

12

Answers key 1. (c)

31. The Laplace transform of f(t), where

2. (a)

3. (c)

4. (a)

5. (a)

6. (a)

7. (d)

8. (d)

9. (a)

10. (b)

11. (b)

12. (b)

13. (b)

14. (a)

15. (d)

16. (c)

17. (a)

18. (b)

19. (b)

20. (d)

s s (a) 6e (b) -6e

21. (a)

22. (b)

23. (a)

24. (a)

25. (d)

26. (b)

27. (c)

28. (b)

29. (c)

30. (a)

-s (c) 6e (d)

31. (c)

32. (b)

33. (a)

34. (c)

35. (a)

0 < t 1 ïî(t - 1) ,



is

s4

s4

s3 2e s s3

36. (b)

32. The Laplace transform of f(t), where ì1, f (t ) = í î t,



(a)

01 î t,

p

ò

= e - st f (t )dt + e - st f (t )dt

is

2 æ 1 1 ö - s (b) 2 æ 1 1 ö - s + ç - ÷e - ç + ÷e s2 è s2 s ø s2 è s2 s ø

p

¥

0

p

ò

ò

= e - st sin t dt + e - st ´ 0 dt p

ò

= e - st sin t dt



0

8/3/2023 6:14:31 PM

378 • Engineering Mathematics Exam Prep p

é e - st ù =ê 2 ( -s sin t - cos t )ú ëê s + 1 ûú 0

=



ò

[Using e ax sin bx dx

4. (a)

e ax = 2 ( a sin b x - b cos b x )] a + b2 =

e

- sp

s2 + 1

( -s sin p - cos p ) -

e

0

s2 + 1

( s + 1) ( s

-t

(

ò

= e

- st

Now, L (t ) =

( )

L t2 =

f (t )dt

ò

f (t )dt + e

0

¥

( )

ò

= e

L t3 =

- st

f (t )dt

ò

- s(1+ q ) 2

q dq = e

0

-s

¥

òe

- sq 3 -1

q

dq

0

= e -s = e -s = e -s

2 s3 2 s3

é ¥ G(n) ù ê e - aq q n -1 dq = ú ê 0 an ú ë û

ò

)

(

)

(

2

s3 3

=

1

s2 2

s3 6

=

1

s2

,

,

1

2

1

=

) ùúû = 1s + (s+31) 3

+

2

( since L(1) = 1 / s )

5. (a)

{

= 4´

)

1 L e -t (2sin2 t ) 2 1 = L e -t (1 - cos2 t) 2 1 1 = L( e -t ) - L( e -t cos2 t) 2 2 ö 1æ 1 s +1 = ç ÷ 2 è s + 1 ( s + 1)2 + 4 ø

EMEP.CH06_3PP.indd 378

=

=

6

+

3

(s + 2)

(

)

-t

}

(

-t

= 4 L e sin 2t - 5L e cosh 2t

(



s2

3 -3t

6 (s + 3)4

L e -t ( 4 sin 2t - 5cosh 2t )

L e -t sin2 t

=

(

é L ê 1 + te -t ë

é G(n) = n - 1 if n Î Z+ ù ë û

3. (c) =

s G(4)

G(2)

) ) + L (t e )

, {s- ( -1)} (s+ 1)2 2 2 = , L t 2 e -2t = 3 {s- ( -2)} (s+ 2)3 6 6 = L t3 e -3t = {s- ( -3)}4 (s+ 3)4 Hence, from (1) we get,

(t - 1) dt

G(3) s3

3

( )

1

= e

G(3)

L te -t =

2

æ Putting t - 1 = q so that dt = dq ö ç ÷ è and t = 1 Þ q = 0; t ® ¥ Þ q ® ¥ ø ¥

3

= 4 = 4 s4 s s So by the first shifting theorem,

1

- st

( ) + 3L ( t e

2 -2t

-t

......................(1)

¥

-t

3

= L (1) + 3L te

0

1

) ùúû

= L 1 + 3te -t + 3t 2e -2t + t3e -3t

L{ f (t )}

ò

2

= 1 + 3te -t + 3 te -t

é L ê 1 + te -t ë



= e

( ) + (te )

3

(

( -s sin 0 - cos 0 )

2. (a)

- st

)

= 1 + 3te -t + 3t 2e -2t + t3e -3t

1 e - sp e - sp + 1 + 2 = 2 2 s +1 s +1 s +1 ( sin p = 0, cos0 = 1, cos p = -1)

¥

+ 2s + 5

(1 + te )

=



2 2

{ {

}



}

(

2 æ 2 1 ç s + 2s + 5 - s + 2s + 1 2 çç ( s + 1) s2 + 2s + 5 è

(

)

=

)

( s + 1) ( s + 1) + 2 ( s + 1)2 - 22 5 ( s + 1) 8 2

2

s2 + 2s + 5

2

- 5´

s2 + 2s - 3

6. (a) æ e2t + e -2t 2 cosh2 2t = ( cosh 2t ) = ç ç 2 è x æ e +e x ö çç cosh x = ÷÷ 2 è ø

) ö÷ ÷÷ ø



=

ö ÷÷ ø

2

e4t + 2 + e -4t e4t 1 e -4t = + + 4 4 2 4

8/3/2023 6:14:32 PM

Laplace Transforms • 379

(

\ L cosh2 2t



(

L te -t cos2t

æ e 4t ö æ e -4t ö æ1 ö = Lç + Lç ÷ + Lç ç 4 ÷÷ ç 4 ÷÷ è2ø è ø è ø 1 1 1 4t = L e + L (1) + L e -4t 4 2 4

( )

=



)

1

= L{t( e -t cos2t )} = ( -1 )

( )

==-

1 1 1 1 1 1 ´ + ´ + ´ 4 (s - 4) 2 s 4 (s + 4)

s 1 1 2s 1 = + ´ = + 2 2 2s 4 s - 16 2s 2 s - 16

(

7. (d)

(

L e -t ( t + 2 )

(

)

2

(

)



)

=



2!

3

)

( )

( )

3

1!

+4´

( s + 1)

+

2

+4´

1 s ( + 1)

4

( s + 1)

2

+

4

( s + 1)

8. (d) We know that

3

1

d L ( f (t ) ) ds d æ1 ö = - tan -1 ç ÷ ds èsø

è3ø

æ s +1 ö

}

=

9. (a) L ( cos2t ) =

EMEP.CH06_3PP.indd 379

æ s +1 ö -ç ÷ e è 3 ø

1 1 = 3 æ s + 1 ö ( s + 1) ç 3 ÷ è ø

s s2 + 4

\ By first shifting theorem, s +1 s +1 L e -t cos2t = = 2 2 ( s + 1) + 4 s + 2s + 5

(

)

=-

-s

e . s 1 æ s +1 ö \ L e -t f (3t ) = F ç 3 è 3 ÷ø

æ s +1 ö -ç ÷ e è 3 ø

æ 1 ö 1 1 Þ Lç ÷ = s´ 3 -0 = 1 . è pt ø s2 s2

Þ L (tf (t ) ) = -

L {e -t f (3t )} = F ç (by the first shifting 3 è 3 ÷ø theorem) Given L { f (t )} = F ( s) =

t .Then, p

èsø

Therefore, L { f (3t )} = 1 F æç s ö÷ and so



s2 + 2s - 3 (s2 + 2s + 5)2

11. (b) L ( f (t ) ) = tan -1 æç 1 ö÷

1 æsö L { f ( at )} = F ç ÷ if L { f (t )} = F ( s ) . a èaø

{

)

æ t ö 1 = 3 (given) F ( s) = L ( f (t ) ) = L ç 2 ç p ÷÷ è ø s2 2 1 1 ´ = Also, f ¢(t ) = and f (0) = 0. p 2 t pt Hence, L ( f ¢(t ) ) = sF ( s) - f (0)

( s + 1) ( s + 1) ( by the first shifting theorem ) 2

(

{( s2 + 2s + 5) ´ 1 - ( s + 1)(2s + 2)} ( s2 + 2s + 5)2

Let f (t ) = 2

)

= L e -t t 2 + 4 L e -t t + 4 L e -t =

=

d æ s +1 ö ds çè s2 + 2s + 5 ÷ø

d L e -t cos2t ds

10. (b)

= L éêe -t t 2 + 4t + 4 ùú ë û

(

)

=

1 æ -1 ö ´ ìï æ 1 ö2 üï çè s2 ÷ø í1 + ç ÷ ý s îï è ø þï

1 s2 + 1

12. (b) L ( eat - cos bt ) = L ( eat ) - L ( cos bt )

1 s = F (s) s - a s2 + b2 æ e at - cos bt ö \Lç ÷÷ ç t è ø é f (t ) ù at = Lê ú , where f (t ) = e - cos bt ë t û =

8/3/2023 6:14:34 PM

380 • Engineering Mathematics Exam Prep ¥

¥

s ö æ 1 = F ( s)ds = ç - 2 ÷ ds s a s + b2 ø è s s

ò

ò

ò

¥

1 é ù = êlog( s - a ) - log( s2 + b2 ) ú 2 ë ûs é æ s-a = êlog ç ç 2 2 è s +b ëê

¥ é æ 1 - ( a / s) öù ÷ ú = êlog ç ÷ ç 1 + (b / s)2 ø ûú s êë è

æ 1 - ( a / s) æ 1-0 ö = log ç ÷ - log ç ç 1 + (b / s)2 è 1+0 ø è

¥

öù ÷ú ÷ú øûs

ö ÷ ÷ ø

1 ® 0] s æ s-a s-a ö ÷ = - log ç ç s2 + b2 s2 + b2 ÷ø è





Here, f(t) is a periodic function with period 2π/k. T

L[ f (t )] = ö ÷ ÷ ø

=

æ t sin t ö Lç dt ÷ ç ÷ t è0 ø æt ö é sin t ù = L ç f (t )dt ÷ ê where f (t ) = ç ÷ ë t úû è0 ø 1 1 1 é sin t ù = F ( s) = L{ f (t )} = L ê s s s ë t úû ¥

¥

ò

ò

¥ 1é 1 tan -1 s ù = étan -1 ¥ - tan -1 s ù ûs û së së 1 ép 1 ù = ê - tan -1 s ú = cot -1 s s ë2 û s

=



cot -1 ( s + 1) = s +1



(

F ( s) = L ( f (t ) ) = L e -t cosh t = =



EMEP.CH06_3PP.indd 380

s - ( -1)

s +1

=

òe

2p s 0 k

- st

f (t )dt

ò

ò

p k

1 e - st ´ sin kt dt 2p ö æ s ç1 - e k ÷ 0 ç ÷ è ø

ò

p



é - ps ù ê e k ( -s sin p - k cos p ) - ( -k ) ú = ê ú 2p ö æ s2 + k 2 s ê ú ç1 - e k ÷ ëê ûú ç ÷ è ø



é - ps ù ê ke k + k ú sin p = 0,cos p = -1) = 2 2 ú ( 2p ö ê æ s ê s +k ú k ç1 - e ÷ë û ç ÷ è ø é - ps ù êe k +1ú k = ê 2 2 ú p p æ - s öæ - sö s +k ú ç1 - e k ÷ç1 + e k ÷ êë û ç ÷ç ÷ è øè ø k = p ö æ - s ç 1 - e k ÷ s2 + k 2 ç ÷ è ø

1

1

)

{s - ( -1)}2 - 1 s2 + 2s

2p k



14. (a) Here, f (t ) = e -t cosh t and

1 - e - sT

é e - st ( -s sin kt - k cos kt ) ù k = ê ú 2p ö æ s2 + k 2 s ê ûú 0 ç1 - e k ÷ ë ç ÷ è ø

By first shifting theorem,

ò

f (t ) dt

1

=

t æ sin t ö÷ 1 L ç e -t dt = cot -1 { s - ( -1)} ç ÷ s - ( -1) t è 0 ø

0

- st

2p ép ù k êk ú ê e - st ´ sin kt dt + e - st ´ 0 dt ú = 2p ö æ ú s ê p ç1 - e k ÷ ê 0 úû k ç ÷ë è ø

ò

1 1 1 L(sin t ) ds = ds s s s2 + 1 s s

-

òe

1

ò

=

1 1-e



13. (b)

ò

15. (d)

[ s ® ¥ Þ æ = log1 - log ç ç è

æt ö æt ö \L ç e -t cosh t dt ÷ = L ç f (t ) dt ÷ ç ÷ ç ÷ è0 ø è0 ø F ( s) s +1 . = = 2 s s ( s + 2)

s +1 s ( s + 2)

(

)

8/3/2023 6:14:35 PM

Laplace Transforms • 381

16. (c) We know that if f(t) is a periodic function with period T, then L[ f (t )] =

=

=



(1 - e ) ò - sT

2p

1

(1 - e

T

1

-2 ps

e )ò

- st

e

- st

f (t )dt

0

f (t )dt

( here T = 2p )

0

(1 - e ) ò -2 ps

[ e - st ´ 1dt +

2p



0

òe

- st

( -1)dt ]

òe

p

- st

\

ì é - st ù p é - st ù 2p ü 1 e ï e ï ú -ê ú ý íê (1 - e -2ps ) ï êë -s úû 0 êë -s úû p ï î þ



1 = { e -2ps - 2e -ps + 1} s(1 - e -2ps )



s( e

- 1)

g (t ) dt we get, s = 0. ¥

e - at - e -bt dt = e - st g (t ) dt = G( s) = G(0) t

ò 0

æ 0+b ö æbö = log ç ÷ = log ç a ÷ è0+aø è ø

18. (b) ¥

1

ò 0



-2 ps

ò

e -2t

0

¥

sinh t dt = e - st f (t ) dt t

ò 0

sinh t ö æ ç here s = 2 and f (t ) = t ÷ è ø é sinh t ù = F ( s) = L éë f (t ) ùû = L ê ú ë t û

{2e -ps - e -2ps - 1}

¥

ò

= L(sinh t ) ds =

17. (a)

(

)

s

( ) ( )

L e - at - e -bt = L e - at - L e -bt



¥

ò

é 1 öù æ ç 1 - s ÷ú 1é 1ê æ s - 1 öù = ê log ç ÷ú ÷ ú = ê log ç 2ë è s + 1 øûs 2 ê çç 1 + 1 ÷÷ ú êë s ø úû s è

1é æ s - 1 öù 1 é æ s - 1 öù ê log1 - log ç ÷ ú = ê0 - log ç s + 1 ÷ ú 2ë è s + 1 øû 2 ë è øû 1 1 æ 2 -1 ö æ1 ö 1 = - log ç ÷ = - 2 log ç 3 ÷ = 2 log 3 ( s = 2 ) 2 è 2 +1 ø è ø =

¥

= ëélog( s + a ) - log( s + b) ûù s

¥



é é æ 1 + ( a / s) ö ù æ s + a öù = êlog ç ú = êlog ç ÷ú ÷ è s + b øûs ë è 1 + (b / s) ø û s ë



æ 1 + ( a / s) ö æ1 + 0 ö = log ç - log ç ÷ ÷ è1 + 0 ø è 1 + (b / s) ø 1 [ s ® ¥ Þ ® 0] s s + a æ ö æ s+b ö = log1 - log ç = log ç ÷ ÷ è s+b ø ès+aø

EMEP.CH06_3PP.indd 381

s

1 ds -1

2

é 1 öù æ ç 1 - s ÷ú 1ê æ1 - 0 ö = êlog ç ÷ú ÷ - log ç 2ê è1 + 0 ø çç 1 + 1 ÷÷ ú s ø ûú è ëê

1 ö æ 1 ds = F ( s)ds = ç s a s + + b ÷ø è s s ¥

òs

¥

1 1 = F (s) s+a s+b æ e - at - e -bt ö \Lç ÷÷ ç t è ø é f (t ) ù - at -bt = Lê ú , where f (t ) = e - e ë t û

ò

¥

¥

=

¥

e - at - e -bt dt with t

0

1 1 ì 1 é -ps ü = - e - e0 ù + ée -2ps - e -ps ù ý -2 ps í ë û ë û s s (1 - e )î þ

=

ò 0

¥

¥

=

¥

Now comparing

p

1

e - at - e -bt .Then t æ e - at - e -bt ö æ s+b ö L{ g (t )} = L ç ÷÷ = log ç ÷ ç t ès+aø è ø æ s+b ö Þ G( s) = log ç ÷ ( if L{ g (t )} = G( s) ) ès+aø Let g (t) =



19. (b) lim f (t ) = lim sF ( s) = lim sL ( f (t ) )

t ®¥

s ®0

s ®0

4 ïì ïü = lim ís ´ ý s ®0 ï 4 s s + ( ) î þï



4 ïì 4 ïü = lim í = 1. ý= s ®0 ï ( s + 4 ) ï 0 + 4) ( î þ

8/3/2023 6:14:36 PM

382 • Engineering Mathematics Exam Prep 20. (d)

=

lim f (t ) = lim sF ( s) = lim sL ( f (t ) )

t ®¥

s ®0

(

s ®0

ì 4s2 - 3s + 8 ï = lim ís ´ s ®0 s s2 + 4 ïî



(

)

) üï

(



)





ì ü s-2 1 = L-1 í +4 2 2 2 2ý ( s - 2) + 3 þ î ( s - 2) + 3 ì ü ì ü s-2 1 = L-1 í + 4 L-1 í 2 2ý 2 2ý + + ( s 2) 3 ( s 2) 3 î þ î þ 2t 2t 4 = e cos3t + e sin 3t 3 4 = e2t (cos3t + sin 3t ) 3





ì ( s + 1) - s ( s + 1) - s ü ï ï = L-1 í 2 2 ý 1 s s + ( ) s ( s + 1) ïþ ïî ìï ( s + 1 ) ( s + 1) + s üï s = L-1 í 2 - 2 2 2 ý s ( s + 1) þ ï îï s ( s + 1 ) s ( s + 1 ) s ( s + 1 )



ì1 1 1 1 üï ï = L-1 í 2 + ý s ( s + 1) s ( s + 1) ( s + 1)2 ï ïî s þ



ì 1 ( s + 1) - s ( s + 1) - s 1 üï ï = L-1 í 2 + ý s ( s + 1) s ( s + 1) ( s + 1)2 ïþ ïî s ì1 1 1 1 1 1 üï ï = L-1 í 2 - + - + + ý s s + 1 s s + 1 ( s + 1 )2 ï ïî s þ ì1 2 ü 2 1 ï ï = L-1 í 2 - + + ý s s + 1 ( s + 1 )2 ï ïî s þ

1 -t e sin 2t 2

23. (a) ì ü 1 L-1 í 2 2 ý î s ( s + 2) þ é1 ù 2 = L-1 ê ´ 2 2 ú ë 2 s ( s + 2) û =



EMEP.CH06_3PP.indd 382

=

ì1 ü ì1 ü ì 1 ü = L-1 í 2 ý - 2L-1 í ý + 2L-1 í ý îs þ îsþ îs +1þ ì 1 ü ï ï + L-1 í 2ý ïî ( s + 1) ïþ

1 -1 é ( s2 + 2) - s2 ù L ê 2 2 ú 2 ëê s ( s + 2) ûú 1 -1 é 1 1 ù L ê 2 - 2 ú 2 ( s + 2) û ës

ì ü ï ( s + 1) - s ï = L-1 í ý 2 2 îï s ( s + 1) þï

ì ü 1 1 ï ï = L-1 í 2 2ý 1 + s s ( ) s ( s + 1) ïþ ïî

ì 3( s + 1) ü 1 = L-1 í + 2 2 2 2ý ( 1) 2 ( 1) 2 s s + + + + î þ ì ü ì ü ( 1) 1 s + = 3L-1 í + L-1 í 2 2ý 2 2ý î ( s + 1) + 2 þ î ( s + 1) + 2 þ = 3e -t cos2t +

2

ì ( s + 1) ü s ï ï = L-1 í ý 2 2 2 2 s ( s + 1) þï îï s ( s + 1)

22. (b) æ 3s + 4 ö L-1 ç 2 ÷ è s + 2s + 5 ø ì 3( s + 1) + 1 ü = L-1 í 2 2ý î ( s + 1) + 2 þ

( )

öù ÷ú ÷ú ÷ú øû

ì ü 1 ï ï L-1 í 2ý 2 îï s ( s + 1) þï

s+2 ö L ç 2 ÷ è s - 4s + 13 ø -1 æ



é æ 1 ê -1 æ 1 ö 1 -1 ç L L ê ç 2÷ ç 2 2ê ès ø çs + 2 è ë ù 1é 1 sin( 2t ) ú = êt 2ë 2 û

24. (a)

21. (a)

ì ( s - 2) + 4 ü = L-1 í 2 2ý î ( s - 2) + 3 þ

1 öù ö -1 æ ÷-L ç 2 ÷ú ø è s + 2 øû

=

ý ïþ

ì 2 ü ï 4s - 3s + 8 ï 0 - 0 + 8 = lim í = 2. ý= 2 s ®0 ïî s + 4 þï ( 0 + 4 )

1 é -1 æ 1 êL ç 2 2ë ès

= t - 2 + 2e -t + te -t



= t - 2 + (t + 2 ) e -t

8/3/2023 6:14:38 PM

Laplace Transforms • 383

25. (d)

\ tf (t )

é ù s L-1 ê ú 2 ë ( s + 1)( s + 1) û é1 ù 2s = L-1 ê ´ ú 2 2 ( s 1)( s 1) + + ë û ù 1 -1 é 2s = L ê ú 2 2 ë ( s + 1)( s + 1) û



=

1 -1 é ( s + 1)2 - ( s2 + 1) ù L ê ú 2 2 êë ( s + 1)( s + 1) úû

=

1 -1 é ( s + 1)2 ( s2 + 1) ù L ê ú 2 2 2 êë ( s + 1)( s + 1) ( s + 1)( s + 1) úû

=

1 -1 é s + 1 1 ù L ê 2 ú 2 ës +1 s +1û

=

1 ì -1 é s ù ù -1 é 1 -1 é 1 ù ü íL ê 2 ú+L ê 2 ú - L ê s + 1 úý 2î ës +1û ës +1û ë ûþ

=



1 cos t + sin t - e -t 2

{

d {F (s)} ds ìï s2 + 1 üïù d é = -L-1 êlog í ýú ds êë îï s( s + 1) þïúû éd ù log s2 + 1 - log s- log ( s + 1 ) ú = -L-1 ê ë ds û = -L-1



{ (

}

)

1 1 ö æ 2s = -L-1 ç 2 - ÷ è s +1 s s +1 ø æ s ö -1 æ 1 ö -1 æ 1 ö = -2L-1 ç 2 ÷ + L ç s ÷ + L ç s +1 ÷ è s +1 ø è ø è ø = -2cos t + 1 + e -t 1 Þ f (t ) = -2cos t + 1 + e -t t 1 -1 Þ L { F ( s)} = -2cos t + 1 + e -t t

(



}

)

(

)

28. (b) Using the concept of the unit step function, we can write: f (t ) = 0 ´ {u(t - 0) - u(t - 1)}

26. (b)



\ L { f (t )}

ì üï 1 1 æ ö -1 ï L-1 ç 2 ÷=L í 2 2ý s 2 s 5 ( s 1) 2 + + + + è ø îï þï

= L ëé0 ´ {u(t - 0) - u(t - 1)}ûù

1 Þ f (t ) = L { F ( s)} = e -t sin 2t 2 1 æ ö ç where F ( s) = 2 ÷ s + 2s + 5 ø è ìd æ 1 öü Then, tf (t ) = -L-1 í ç 2 ý ÷ î ds è s + 2s + 5 ø þ -1

ïì ( -1)(2s + 2) ïü = -L-1 í 2 ý ïî ( s + 2s + 5)2 ïþ



ì ü s +1 = 2L-1 í 2 2ý î ( s + 2s + 5) þ



ì ü 1 s +1 Þ L-1 í 2 = tf (t ) 2ý î ( s + 2s + 5) þ 2 1 1 = t ´ e -t sin 2t 2 2 1 -t = te sin 2t 4

+ L éë1 ´ {u(t - 1) - u(t - 2)}ùû + L ëé2 ´ u(t - 2) ûù

= L(0) + L {u(t - 1) - u(t - 2)} + 2L {u(t - 2)}

= 0 + L {u(t - 1)} - L {u(t - 2)} + 2L {u(t - 2)}

EMEP.CH06_3PP.indd 383

1 - s 1 -2s e + e s s 1 - as ö æ ç using L {u(t - a )} = s e ÷ è ø

= L {u(t - 1)} + L {u(t - 2)} =



29. (c) Using the concept of the unit step function, we can write: 1 ü ì f (t ) = t ´ íu(t - 0) - u(t - ) ý 2 þ î 1 ì ü + (t - 1) ´ íu(t - ) - u(t - 1) ý + 0 ´ u(t - 1) 2 î þ

\ L { f (t )}

é ì 1 üù = L êt ´ íu(t - 0) - u(t - ) ýú 2 þû ë î é 1 ì üù + L ê(t - 1) ´ íu(t - ) - u(t - 1) ýú + L ëé0 ´ u(t - 1) ûù 2 î þû ë

ìï s2 + 1 üï 27. (c) Let F(s) = log í ý îï s( s + 1) þï é ìï s2 + 1 üïù Then, f (t ) = L-1 { F ( s)} = L-1 êlog í ýú îï s( s + 1) þïûú ëê

+ 1 ´ {u(t - 1) - u(t - 2)} + 2 ´ u(t - 2)

ì æ 1 öü ì æ 1 öü = L {tu(t - 0)} - L ítu ç t - ÷ ý + L í(t - 1) u ç t - ÷ ý è 2 øþ î è 2 øþ î



{

}

- L (t - 1) u(t - 1) + L(0)

8/3/2023 6:14:39 PM

384 • Engineering Mathematics Exam Prep ì æ 1 öü = L (t - 0 ) u(t - 0) - L íu ç t - ÷ ý - L (t - 1) u(t - 1) î è 2 øþ

{

}

{

}

= L[u(t – 0)] + L [(t – 3) × u(t – 3)] + 2L [u(t – 3)] f (t ) = 1´ {u (t - 0) - u (t - 3)} + t ´ u (t - 3)

s

1 = e -0s L(t ) - e 2 - e -1s L(t ) s 1 - as ö æ ç using the second shifting theorem & L {u(t - a )} = s e ÷ è ø s s æ ö 1 1 1 1 1 1 = 2 - e 2 - 2 e -s = - ç e 2 + e -s - ÷ . s sç s s÷ s s è ø

= u (t - 0) + ( t - 1) ´ u (t - 3)

= u (t - 0) + ( t - 3) ´ u (t - 3) + 2u (t - 3)

\ L { f (t )}

= L [u (t - 0) ] + L éë( t - 3) ´ u (t - 3) ùû + 2 L [u (t - 3) ] 1 1 = e -0 s + e -3 s ´ L ( t ) + 2 ´ e -3 s s s æ by second shifting theorem and using ö ç ÷ 1 çç ÷÷ L ( u (t - a ) ) = e - as s è ø -3 s 2 1 e = + 2 + e -3 s s s s

30. (a) Using the concept of the unit step function, we can write: f (t ) = sin 2t ´ {u(t - 0) - u(t - p)} + 0 ´ u(t - p) = sin 2t ´ u(t - 0) - sin 2t ´ u(t - p) + 0

\ L { f (t )}

= L ëésin 2t ´ u(t - 0) ûù - L ëésin 2t ´ u(t - p) ûù



= L éë2sin t cos t ´ u(t - 0) ùû - L éë2sin t cos t ´ u(t - p) ûù = L ëé2sin(t - 0) cos(t - 0) ´ u(t - 0) ûù

- L ëé2sin(t - p) cos(t - p) ´ u(t - p) ûù æsin(t - p) = - sin( p - t ) = - sin t, ö ç ÷ è cos(t - p) = cos( p - t ) = - cos t ø

= e -0s L ( 2sin t cos t ) - e -ps L ( 2sin t cos t )

(by the second shifting theorem)

= L ( sin 2t ) - e =



2

s2 + 22

-e

-ps

-ps

´

L ( sin 2t ) 2

s2 + 22

=

(

2 1-e

-ps

s2 + 4



33. (a) f(t) = 2t × {u(t – 0) – u(t – 1)} + t × u(t – 1)

{

3

3! s

4

=

= 2 ´ e -0s L (t ) - e -1s L (t ) -



1 1 e -1s - e -s ´ 2 2 s s s

2 æ 1 1 ö -s - ç + ÷e s2 è s2 s ø

{

}

2 = L é (t - 1) + 1 u(t - 1) ù êë úû

{

6e - s 4

}

{

}

2 é ù 2 = L ê (t - 1) + 2(t - 1) + 1 u(t - 1ú ë û 2 = L é(t - 1) u(t - 1) ù + 2L ëé(t - 1) u(t - 1) ûù + L ëéu(t - 1) ûù ëê ûú

1 = e -1s L t 2 + 2e -1s L (t ) + e - s s 1 -s - s 2! -s 1 =e + 2e + e s3 s2 s 2 1ö æ 2 = ç 3 + 2 + ÷ e -s sø s ès

( )

= u(t – 0) + (t – 3) × u(t – 3) + 2u(t – 3)

EMEP.CH06_3PP.indd 384

e -1s s

L t 2u(t - 1)

= u(t – 0) + (t – 1) × u(t – 3) \  L{f(t)}

=

34. (c)

32. (b) Using the concept of the unit step function, we can write: f(t) = 1 × {u(t – 0) – u(t – 3)} + t × u(t – 3)



= 2´



( )

s

}

- L éëu (t - 1) ùû

3 \ L { f (t )} = L é(t - 1) ´ u(t - 1) ù = e -1s ´ L t3 êë úû ( by the second shifting theorem of Laplace transform )

= e -s ´

{

\ L éë f (t ) ùû = 2L (t - 0 ) ´ u (t - 0 ) - L éë(t - 1) ´ u (t - 1) ùû

3

f (t ) = 0 ´ {u(t - 0) - u(t - 1)} + (t - 1) ´ u(t - 1)

}

= 2 (t - 0 ) ´ u (t - 0 ) - (t - 1 ) ´ u (t - 1 ) - u (t - 1 )

)

31. (c) Using the concept of the unit step function, we can write: = (t - 1) ´ u(t - 1)

= 2t u(t) – t × u(t – 1)



8/3/2023 6:14:41 PM

Laplace Transforms • 385

35. (a)

=

y¢¢ + 2 y¢ + y = 1 Þ y¢¢(t ) + 2 y¢(t ) + y(t) = 1

Þ s2Y ( s) - sy(0) - y¢(0) + 2{ sY ( s) - y(0)} 1 s

Þ s2Y ( s) - 0 - 0 + 2sY ( s) + Y ( s) =



Þ ( s2 + 2s + 1)Y ( s) =

1 s

Þ y(t ) = 1 - e -t -

e ´t 2 -1

= 1 - e -t - te -t

= 1 - e -t (1 + t )



1. If L{f(t)} = s2+ 2 , L{ g (t)} = s +1

2 1 (a) s + 1 (b)

(c)

1

s -1

Þ s Y ( s ) - 4Y ( s ) =



(

2

)

Þ s - 4 Y ( s) =

Þ Y ( s) =

EMEP.CH06_3PP.indd 385

[ y(0) = 0 = y¢(0)]

s -1 1

s -1

1

(

( s - 1) s2 - 4

(d) None of these (EC GATE 2000)

æt

ö

è

ø

(a) L æç df ö÷ = 1 F ( s); L ç f (t ) dt ÷ = sF ( s) - f (0) çò ÷ è dt ø s 0 æt

ö

è

ø

æt

ö

æt

ö

è

ø

(b) L æç df ö÷ = sF ( s) - F(0); L ç f (t ) dt ÷ = - dF çò ÷ ds è dt ø 0

)

=

1 ( s - 1)( s - 2)( s + 2)

( s + 2) - ( s - 2) 4( s - 1)( s - 2)( s + 2) 1 1 = 4( s - 1)( s - 2) 4( s - 1)( s + 2) ( s - 1) - ( s - 2) [( s + 2) - ( s - 1)] = 4( s - 1)( s - 2) 12( s - 1)( s + 2) =



1

s2 + 1 ( s + 2) + ( s + 3)( s + 2) s2 + 1

2. Let F(s) = L(f(t)) denotes the Laplace transform of the function f(t). Which of the ­following statements is true?

Þ L [ y¢¢(t )] - 4 L [ y¢(t )] = L é et ù

ë û 2 é ù ¢ Þ s Y ( s ) - sy(0) - y (0) - 4 [ sY ( s ) - y(0)] ë û

s+3

s+3

t

2

s2 + 1 ( s + 3)( s + 2)

0

Þ y¢¢(t ) - 4 y¢(t ) = et

=

1 12( s + 2)

and h(t ) = f (T ) g (t - T )dT , then L{h(t)} is ò



36. (b) y¢¢ - 4 y¢ = e

+

t

æ1 1 1 ö Þ L-1 éëY ( s) ùû = L-1 ç ÷ 1 + s s ( 1)2 ø + s è æ1 ö æ 1 ö -1 ì 1 ü = L-1 ç ÷ - L-1 ç ÷-L í 2 ý èsø è s +1 ø î ( s + 1) þ 2 -1

1 12( s - 1)

Previous Years Questions (2000-2018)

1 1 = 2 s( s + 2s + 1) s( s + 1)2 ( s + 1) - s 1 1 = = s( s + 1) ( s + 1)2 s( s + 1)2

-t

-

1 2t 1 t 1 t 1 -2t e - e e + e 4 4 12 12 1 1 1 -2t e = e2t - et + 4 3 12



1 s

1 1 1 s s + 1 ( s + 1)2

1 4( s - 1)

Þ y(t ) =

Þ Y ( s) =

=

-

1 é 1 ù 1 -1 é 1 ù Þ L-1 éëY ( s) ùû = L-1 ê ú - L ê ( s - 1) ú 4 ë ( s - 2) û 4 ë û 1 -1 é 1 ù 1 -1 é 1 ù L ê L ê ú+ ú 12 ë ( s - 1) û 12 ë ( s + 2) û

Þ L ëé y¢¢(t ) ûù + 2L ëé y¢(t ) ûù + L ëé y(t) ûù = L (1) + Y ( s) =

1 4( s - 2)

(c) L çæ df ÷ö = sF ( s) - F(0); L ç ò f (t ) dt ÷ = F ( s - a ) ç ÷ è dt ø è0 ø (d) L çæ df ÷ö = sF ( s) - f (0); L ç f (t ) dt ÷ = 1 F ( s) çò ÷ s è dt ø 0 

(GATE 2000)

8/3/2023 6:14:44 PM

386 • Engineering Mathematics Exam Prep 3. The Laplace transform of the function L(sin22t) is (a) 1 2s

(c)

(

s

2 s2 + 16

)

(b)

(b) d(t ) = ìí¥,t = 0

î0, otherwise

1 s (d) 2 s s s2 + 4 s +4

(GATE 2002) 4. The Laplace transform of the following function is



ìsin t, 0 £ t £ p f (t ) = í î 0, t > p

(a)

(s

1 2

+1

)

+1

(c) d(t ) = ìí1, t = 0

î0, otherwise

(d) d(t ) = ìí¥,t = 0

î0, otherwise

for s > 0 (b)

)

(s

1 2

+1

e -ps

(s

2

+1



)

for s < p

)

for s > 0

(a)

1 a 2 (b) 2 s -a s + a2 2

s s (c) 2 2 (d) 2 s +a s - a2

(CE GATE 2003)



6. The Laplace transform of a function sin wt is

ò d(t) dt = 1

-¥

and

¥

ò d(t) dt = 1

-¥

(EC GATE 2005)

9. A solution for the differential equation x′(t) + 2x(t) = d(t) with initial condition x(0) = 0 is (a) e–2t u(t) (b) e2t u(t) (c) e–t u(t) (d) et u(t) (u(t) denotes the unit step function) (GATE 2006)



(GATE 2002) 5. If L defines the Laplace transform of a function, then L(sin at) will be equal to

¥

and



-ps (c) 1 + e for s > 0 (d) 2

(s

(a) d(t ) = ìí1,t = 0

î0, otherwise

s s2 + 16



8. The Dirac delta d(t) is defined as

¥

10. Evaluate sint dt ò 0

t

p (a) p (b) p (c) (d) p 4

2



8

(GATE 2007) 11. The Laplace transform of the function 8t3 is

16 (a) 84 (b) 4 (c) 244 (d) 484 s

s



s

s

(PI GATE 2008) 12. The Laplace transform for the function f(x) = cos h (ax) is

(a)

s 2 s + w2

(b)

w 2 s + w2

(c)

s 2 s - w2

(d)

w 2 s - w2

(a)

a s - a2

(b)

s s - a2

(ME GATE 2003)

(c)

a s2 + a 2

(d)

s s2 + a 2



7. A delayed unit step function is defined as

ì0, for t < a u(t - a ) = í î1, for t ³ a



Its Laplace transform is

13. If F(s) is the Laplace transform of the f(t), t

then the Laplace transform of f ( t) dt is ò s

(c) 1 eas (d) 1 eas 

EMEP.CH06_3PP.indd 386

2

(CE GATE 2009)



,

(a) ae–as (b) 1 e -as s

2

0

(a)

1 1 F ( s) (b) F ( s) - f (0) s s

a

(c) sF(s) – f(0) (d) ò F ( s) ds

(ME GATE 2004)



(ME GATE 2007, EC GATE 2009)

8/3/2023 6:14:49 PM

Laplace Transforms • 387

14. The inverse Laplace transform of 21 is s +s (a) 1 + et (b) 1 – et –t –t (c) 1 – e (d) 1 + e  (ME GATE 2009) 15. The Laplace transform of a function f(t) is 1 . s2 ( s + 1)



Then f(t) is



(a) g(t) = f(2t – 3) è



If lim f (t ) = 1, t ®¥



(a) 1

then the value of “k” is (b) 2

(c) 3 (d) 4 (EE GATE 2010)

17. If u(t) represents the unit step function, then the Laplace transform of u(t – t) is - st (a) 1 (b) 1 (c) e (d) e–st

s

s-t



18. The integral

¥

ò

-¥



(a) 6

pö æ d ç t - ÷ ´ 6 sin t dt 6ø è

(b) 3

19. Given f(t) and g(t) shown below: f(t)

(EE GATE 2011) then the initial

and final values of f(t) are, respectively (a) 0, 2 (b) 2, 0 (c) 0, 2 (d) 2 , 0 7

7

(GATE 2011)

22. The unilateral Laplace transform of y(t) is 1 . The unilateral Laplace transform of s2 + s + 1

evalutes to

(c) 1.5 (d) 0 (IN GATE 2010)

ø

21. Given two continuous time signal x(t) = e–t and y(t) = e–2t which exist for t > 0 then the convolution z(t) = x(t)*y(t) is (a) e–t – e–2t (b) e–2t (c) e–t (d) e–t + e–3t  (GATE 2011)

(IN GATE 2010)



ø

è

s + 4s + 7

2

st

ø

20. If F ( s) = L{ f (t)} = 22( s + 1)

ù ú = f (t). ë s + 4s + ( k - 3)s û 3

è



3s + 1

é

L-1 ê

(b) g (t ) = f æç t - 3 ö÷ 2

(c) g (t ) = f æç 2t - 3 ö÷ (d) g (t ) = f æç t - 3 ö÷ 2 2 2

(a) t – 1 + e–t (b) t + 1 + e–t (c) –1 + e–t (d) 2t + et  (ME GATE 2010) 16. Given

g(t) can be expressed as

tf(t) is (a) (c)

(s

(s

s 2

+ s +1 s

2

+ s +1

)

)

2

2

(b) - ( 2s + 1)

(s

(d)

(s

2

+ s +1

( 2s + 1) 2

+ s +1

)

)

2

2

(EE, EC GATE 2012)



23. The inverse Laplace transform of the function F ( s) =

1

t

1

24. Consider the differential equation

d2 y(t ) dy(t ) +2 + y(t ) = d(t ) 2 dt dt

dy |t =0 = 0. dt

1

EMEP.CH06_3PP.indd 387

is given by

(a) f(t) = sin t (b) f(t) = e–t sin t (c) f(t) = e–t (d) f(t) = 1 – e–t  (ME GATE 2012)

g(t)

0

1 s ( s + 1)

3

5

t



with y(t )|t =0 = -2 and

Then the numerical value of dy |t =0 is

(a) –2 (c) 0

dt

(b) –1 (d) 1 (EE, IN GATE 2012)

8/3/2023 6:14:52 PM

388 • Engineering Mathematics Exam Prep 25. The function f(t) satisfies the differential equation

d2 f +f =0 dt 2

tions are

f (0) = 0,

30. The Laplace transform of the function f(t) is given by

and the auxiliary condidf (0) = 4 . dt

The Laplace

0

The Laplace transform of the function shown below is given by

transform of f(t) is given by (a) 2 (b) 4 s +1

(c)

¥

F(s) = L{ f (t )} = f (t ) e -st dt . ò



f(t)

s +1

4 2 (d) 4 s2 + 1 s +1



(ME GATE 2013)



3s - 15 s2 + 10s + 21

be the Laplace transform of

2

a signal x(t). Then x(0) is (a) 0 (b) 3 (c) 5 (d) 21 (EE GATE 2014)

0

26. Let

27. The Laplace transform of cos wt is

s 2 s + w2

.

The Laplace transform of e–2t cos 4t is (a) (c)

s-2

( s + 2 )2 + 16 s-2

( s + 2 )2 + 16

(b) (d)



s+2

28. The Laplace transform of ei5t, where i = -1 is (a) s2 - 5i (b) s2 + 5i

-s -s (c) 2 - 2e (d) 1 - 2e

s



ì1, if a £ t £ b f (t ) = í î0, otherwise

es a - b) (a) a - b (b) ( s

s

s( a -b ) - as -bs (c) e - e (d) e

s



EMEP.CH06_3PP.indd 388

s

(EC GATE 2015)

¥

1

ò (1 + x ) 2

0

¥

dx +

ò 0

sinx dx x

is

(a) p (b) p 2

(c) 3p 2 

(d) 1 (CE GATE 2016) ¥

32. The value of the integral 2 sin 2pt dt is ò -¥

s + 25

29. The bilateral Laplace transform of a function f(t), where

(GATE 2015) 31. The value of

s + 25

(ME GATE 2015)



s



(c) s + 5i (d) s2 - 5i s2 - 25

2s

s

s+2

(ME GATE 2014)

s - 25

-s -2s (a) 1 - e (b) 1 - e

( s - 2 )2 + 16 ( s + 2 )2 + 16

t

1



(a) 0 (c) 1

pt

(b) 0.5 (d) 2 (EE GATE 2016)



33. The Laplace transform of the causal periodic square wave of period T shown in the figure below is (a) F(s) = (c) F(s) =

1 1+e

- sT 2

(b) F(s) =

1 s(1 + e

- sT 2

)

(d) F(s) =

1 s(1 - e - sT ) 1 1 - e sT

8/3/2023 6:14:55 PM

Laplace Transforms • 389

Where u(t) denotes the unit step function.

Then the value of lim x 2 (t ) + x 2 (t ) 1 2



is ________________

1

t ®¥

(EC GATE 2017)



38. If the Laplace transform of a function is s + 1 , then the initial and final values of f(t) s ( s + 2)

0

T/2

T

(GATE 2016)



34. Let y(x) be the solution of the differential equation

d2 y dy -4 + 4y = 0 2 dx dx



are, respectively (a) 0, 1

(b) 1, 1/2



(c) 1/2, 1

(d) 1 , 0

t

3T/2 2T

with initial condi-

dy

(GATE 2016)



35. The solution of the differential equation, for t > 0, y”(t) + 2y’(t) + y(t) = 0 with initial conditions y(0) = 0 and y’ (0) = 1, is (u(t) denotes the unit step function). (a) te–t u(t) (b) (e–t – te–t)u(t) –t –t (c) (–e + te )u(t) (d) e–tu(t)  (GATE 2016) 36. Let a causal LTI system be characterized by the following differential equation, with initial rest condition

s+6



Where x(t) and y(t) are the input an output, respectively. The impulse response of the system is (u(t) is the unit step function) (a) 2e–2tu(t) – 7e–5tu(t) (b) –2e–2tu(t) + 7e–5tu(t) (c) 7e–2tu(t) – 2e–5tu(t) (d) –7e–2tu(t) + 2e–5tu(t) (EE GATE 2017)



37. Consider the state space realization

EMEP.CH06_3PP.indd 389

é x1 (t ) ù é0 0 ù é x1 (t ) ù é 0 ù ê ú=ê ú + ê ú u(t ) úê ë x2 (t ) û ë0 -9 û ë x2 (t ) û ë45 û

With the initial condition

é x1 (t ) ù é0 ù ê ú=ê ú ë x2 (t ) û ë0 û

y(t) at t = 0.1 is __________ unit. (a) ∞ (b) –2.19 (c) 1 (d) non-existant (IN GATE 2017) 40. The Laplace transform F(s) of the exponential function f(t) = eat when t ≥ 0 where “a” is a constant and s – a > 0 is

(a)

1 s+a

(b)

(c)

1 a-s





1 s-a

(d) ∞ (CE GATE 2018)

41. If F(s) is the Laplace transform of the function f(t) = 2t2e–t, then the value of F(1) is _______?

d2 y(t ) dy(t ) dx (t ) +7 + 10 y(t ) = 4 x (t ) + 5 dt dt dt 2



(GATE 2017)

39. The Laplace transform of a casual signal y(t) is Y ( s) = s + 2 . Then the value of the signal

= 1. Then the value tions y(0) = 0 and dx x =0

of y(1) is ___________.

2



(ME GATE 2018)

 1. (b) 6. (b) 11. (d) 16. (d) 21. (a) 26. (b) 31. (b) 35. (a)

2. (d) 7. (b) 12. (b) 17. (c) 22. (d) 27. (d) 32. (d) 36. (b)

40. (b)

41.

Answers key 3. (a) 4. (c) 8. (d) 9. (a) 13. (a) 14. (c) 18. (b) 19. (d) 23. (d) 24. (d) 28. (b) 29. (c) 33. (b) 34. 7.389 37. 5 38. (b)

5. (b) 10. (b) 15. (a) 20. (b) 25. (c) 30. (c) 39. (b)

1 . 2

8/3/2023 6:14:57 PM

390 • Engineering Mathematics Exam Prep 5. (b) By fundamental formulas of Laplace transform.

Explanations

1. (b)

L{f(t)} =

and

6. (b) Use the fundamental formula.

s+2 = F ( s) s2 + 1

7. (b) L éu(t - a )ù = 1 e -as ë û s

s2 + 1 L{ g (t)} = = G( s ) ( s + 3)( s + 2)

8. (d) Use the definition of the Dirac Delta function.

By the convolution theorem,

9. (a) x′(t) + 2x(t) = d(t)

t

ò

L-1[ F ( s)G( s)] = f (T ) g (t - T )dT



Þ L ëé x ¢(t ) ûù + 2L ëé x (t ) ûù = L ëéd(t ) ûù

0

Þ sX ( s) - x (0) + 2 X ( s) = 1

⇒ L–1[F(s)G(s)] = h(t)

( assuming L( x (t ) = X ( s) )

Þ L{h(t )} = F(s)G(s)

Þ X (s)(s+ 2) = 1 1 Þ X ( s) = s+2

ö s2 + 1 æ s + 2 öæ =ç 2 ÷ çç ( s + 3)( s + 2) ÷÷ è s +1 øè ø =



1 ( s + 3)

2. (d) Using the Laplace transform on derivative and integral.

é 1 ù Þ L-1 ëé X ( s) ûù = L-1 ê ú ës + 2û



10. (b)

3. (a)

¥

1 1 L(sin 2t ) = L(2sin2 2t ) = L(1 - cos4t ) 2 2 1 = { L(1) - L(cos4t )} 2 1 æ1 s ö = ç - 2 2 è s s + 16 ø÷

ò

2



Þ x (t ) =   e -2t u(t )

s

¥

ò



0

p

¥

0

p

ò

ò

= e - st ´ sin t dt

14. (c)

p



EMEP.CH06_3PP.indd 390

1 ds +1

é e - st ( -s sin t - cos t ) ù =ê ú s2 + 1 êë úû 0

( -s sin p - cos p ) - e ( -s sin 0 - cos0 ) -0

s2 + 1

1 é - sp 0 - ( -1) + 0 + 1ù e û s2 + 1 ë

{

e +1 s2 + 1

3! 48 . = s3+1 s4

13. (a)

0

=

2

12. (b),

p

- sp

S

11. (d) L ( 8t3 ) = 8L (t3 ) = 8 ´

ò

= e - st ´ sin t dt + e - st ´ 0 dt

=

òs

¥

ò

e

¥

= étan -1 s ù = tan -1 ¥ - tan -1 s ë ûs p p = - tan -1 s = ( s = 0) 2 2

¥

=

æ sin t ö ÷ è t ø

S

= e - st f (t ) dt

- sp

s

sin t ö æ ç here s = 0, f (t ) = ÷ t ø è

= L ( sin t ) ds =

L ( f (t ) )



ò

= L ( f (t ) ) = L ç

4. (c)



¥

sin t dt = e - st f (t ) dt t

}

s2 + 1

1 ü æ 1 ö -1 ì L-1 ç 2 ÷ = L í s( s + 1) ý ès +sø î þ ì ( s + 1) - s ü = L-1 í ý î s( s + 1) þ 1 ö æ1 = L-1 ç ÷ è s s +1 ø

æ1 ö æ 1 ö = L-1 ç ÷ - L-1 ç ÷ s è ø è s +1 ø = 1 - e -t

8/3/2023 6:14:59 PM

Laplace Transforms • 391

15. (a)

t -3 ì ï 0, 2 < 0 ï t -3 æ t 3ö æ t -3ö ï \fç - ÷= fç = í1, 0 < 0 î

f (t )

é ù 1 = L-1 ê 2 ú + ( 1) s s ë û ù 1 -1 é =L ê 2 ú ( 1) + s s ë û



é ( s + 1) - s ù = L-1 ê 2 ú ë s ( s + 1) û é1 1 ù = L-1 ê 2 s( s + 1) úû ës

20. (b)

é 1 ( s + 1) - s ù =L ê 2 s( s + 1) úû ës -1

æ1 = L-1 ç 2 ès æ1 = L-1 ç 2 ès

Initial value of f (t ) = lim f (t ) t ®0

1 1 ö + s s + 1 ÷ø ö -1 æ 1 ö -1 æ 1 ö ÷ - L ç s ÷ + L ç s +1 ÷ ø è ø è ø

= lim sF ( s )

-

s ®¥

16. (d)

s ®0

( using final value theorem )

é ù 3s + 1 Þ 1 = lim s ê 3 ú s ®0 ë s + 4 s2 + ( k - 3)s û é ù 3s + 1 Þ lim ê 2 ú =1 s ®0 ë s + 4 s + ( k - 3) û 0 +1 Þ =1 0 + 0 + ( k - 3) Þk=4



17. (c) Use L{u(t - a )} = e 18. (b)

- as

s

¥

ò f (t ) d(t - a) dt = f (a)

t ®¥

s ®0

s ®0



21. (a) Here, x(t) = e–t and y(t) = e–2t t

ò

\ x * y = x (u) ´ y(t - u ) du 0

pö p æ Þ d ç t - ÷ ´ 6 sint dt =6 sin = 3 6ø 6 è -¥

ò

ì 0, t < 0 ï Here, f (t ) = í1, 0 < t < 1 and ï 0, t > 0 î



EMEP.CH06_3PP.indd 391

ì0, 0 < t < 3 ï g (t ) = í1, 3 < t < 5 ï 0, t > 5 î

é 2( s + 1) ù ú 2 ë s + 4s + 7 û é 2( s2 + s) ù = lim ê ú s ® 0 ê s2 + 4 s + 7 ú ë û 2 (0 + 0) = = 0. 0+0+7 = lim s ê

¥

19. (d)

ù ú ú ú ú û

= lim sF ( s )

-¥



s ®¥

é æ 1ö ê 2 ç1 + ÷ sø è = lim ê 4 7 s ®¥ ê ê1 + s + 2 s ë 2 (1 + 0 ) = =2 1+0+0 Final value of f (t ) = lim f (t )

lim f (t ) = lim sF ( s ) t ®¥

é 2( s + 1) ù ú 2 ë s + 4s + 7 û

= lim s ê



= t - 1 + e -t



ì0, 0 < t < 3 ï = í1, 3 < t < 5 ï 0, t > 5 î = g (t )

t

t

ò

ò

Þ z(t ) = e -u e -2(t -u ) du = e -2t eu du 0

=e



0

-2t

[ e - 1] = e

22. (d) Let, F(s) = Here,



t

-t

-e

-2t

1

s2 + s + 1

L-1 { F ( s)} = f (t )

Then, tf (t )

8/3/2023 6:15:01 PM

392 • Engineering Mathematics Exam Prep = -L-1

25. (c)

d d 1 ö {F ( s)} = -L-1 ds æç 2 ÷ ds è s + s +1 ø

d2 f +f =0 dt 2 Þ f ¢¢(t ) + f (t ) = 0

é ù ê ( -1)(2s + 1) ú = -L ê 2ú 2 ê s + s +1 ú ë û -1



(

)

é ù ê (2s + 1) ú = L-1 ê 2ú 2 ê s + s +1 ú ë û 2s + 1 \ L éëtf (t ) ùû = s2 + s + 1

(

Þ és2 F( s) - sf (0) - f ¢(0) ù + F ( s) = 0 ë û Þ s2 F( s) - 0 - 4 + F( s) = 0

)

(



Þ L éë f ¢¢(t ) ùû + L éë f (t ) ùû = 0

Þ ( s2 + 1) F( s) = 4

)

2



Þ F ( s) =

26. (b)

23. (d) ïì 1 ïü L-1 { F ( s)} = L-1 í ý îï s ( s + 1) þï



L ( x (t ) ) =

é ù 3 ( s - 5) = L-1 ê 2 ú êë s + 10s + 25 - 4 úû é 3 ( s + 5 ) - 20 ù ú = L-1 ê ê ( s + 5 )2 - 22 ú ë û

1 ü ì1 = L-1 í ý +1þ s s î ì1 ü ì 1 ü = L-1 í ý - L-1 í ý s î þ îs +1þ Þ f (t ) = 1 - e -t

-1

d2 y(t ) dy(t ) +2 + y(t ) = d(t ) dt 2 dt é d 2 y(t ) ù é dy(t ) ù Þ Lê + 2L ê ú + L ëé y(t ) ûù = 1 2 ú dt ë dt û ëê ûú

27. (d)

( s + 2s + 1) 2 1 =s + 1 ( s + 1)2

{2 ( s + 1) + 1}





Þ y(t ) = -2e -t - te -t dy Þ = 2e -t + te -t - e -t dt



EMEP.CH06_3PP.indd 392

dy Þ = 2 + 0 -1 = 1 dt t = 0

L ( cos wt ) =

s

(

)

Þ L e -2t cos4t =

( s + 1)2

1 ù é 1 ù -1 é Þ L-1 éëY ( s) ùû = -2L-1 ê ú-L ê 2ú 1 + s ë û ë ( s + 1) û

cosh 2t -

s2 + w2 s Þ L ( cos4t ) = 2 s + 16

Þ ( s2 + 2s + 1)Y ( s) = -3 - 2s =-

-5t

( cosh 0 = 1;sinh 0 = 0 )



Þ és2Y ( s) + 2s - 0 ù + 2 éësY ( s) + 2ùû + Y ( s) = 1 ë û

( 2s + 3 )

ù ( s + 5 ) ù - 20L-1 é 1 ú ê ú 2 2 2 2 êë ( s + 5 ) - 2 úû êë ( s + 5 ) - 2 úû

20 -5t e sinh 2t 2 \ x (0) = 3e0 cosh 0 - 10e0 sinh 0 = 3 = 3e

Þ éë s2Y ( s ) - sy(0) - y¢(0) ùû + 2 [ sY ( s ) - y(0)] + Y ( s ) = 1

2

é

= 3L ê

24. (d)

Þ Y ( s) = -

3s - 15 s + 10s + 21 2

æ 3s - 15 ö Þ x (t ) = L-1 ç 2 ÷ è s + 10s + 21 ø

ïì ( s + 1) - s ïü = L-1 í ý îï s ( s + 1 ) þï



4 s2 + 1

28. (b)

( )

L ei5t =

1 s + 5i = s - 5i ( s - 5i )( s + 5i ) =



s - ( -2)

[ s - ( -2)]2 + 16 [by first shifting property] s+2 = ( s + 2 )2 + 16

s + 5i

2

s - ( 5i ) 2

=

s + 5i . s2 + 25

8/3/2023 6:15:03 PM

Laplace Transforms • 393

29. (c)

¥

\ L[ f (t )]

=

ò

= e - st f (t ) dt 0

a

b

¥

ò

ò

ò

b

- st

= e - st ´ 0 dt + e - st ´ 1 dt + e - st ´ 0 dt 0

a

b

b

ée ù e = e - st dt = ê ú = s úû a ëê a

ò

- as

- e -bs s



30. (c)

ò

Here, f (t ) = ìí1 0 < 1 £ T / 2

¥

\ L ( f (t ) ) = e - st f (t ) dt

ò

0 1

¥

0

1

ò

sin x æ ö ç f ( x ) = x ÷ ç ÷ ç Þ f ( - x ) = sin( - x ) = - sin x = f ( x ) ÷ ç ÷ ( -x ) -x ç ÷ and so f(x) is even ç ÷ ç ÷ è ø ¥ æ 4 p sin x pö dx = ÷ = ´ = 2 ç ç x p 2 2÷ è 0 ø

33. (b)

ì 0, t < 0 ï Here, f (t ) = í2, 0 < t < 1 ï 0, t > 1 î



F(s)



=

ò

= e - st ´ 2 dt + e - st ´ 0 dt é e - st ù = 2 e - st dt = 2 ê ú êë -s úû 0 0

ò

(

-2 e - s - 1

31. (b)

s

) = 2 - 2e

1

2

0

-1

-1

¥

xù û0

= tan ¥ - tan p p = -0 = 2 2 ¥

sinx dx = x

ò 0

¥

\

1

ò (1 + x ) 2

0



¥

ò 0

0

sin t p dt = t 2 ¥

dx +

ò 0

=

=



sinx p p dx = + = p. x 2 2

=



2

ò

-¥

=4

sin 2pt dt pt ¥

ò

-¥

= 4´



EMEP.CH06_3PP.indd 393

- st

dt

0

T 2

ò

1.e - st dt +

0

1 1 - e sT

T

ò 0.e

- st

dt

T 2

-1

(1 - e ) - sT

(1 - e

- sT

s(1 - e

1 . .( e - sT / 2 - 1) s

/2

- sT 2

)

(1 - e s(1 - e

(

) - sT 2

)

- sT 2

)(1 + e - sT / 2 )

- sT 2

)

1

s 1+e

34. 7.389

32. (d) ¥

ò f (t )e

é e - st ù 1 = ê ú 1 - e - sT ëê -s úû 0 =

-1

T /2 < t £ T

T /2

s

ò (1 + x ) dx = éëtan

T

1 1 - e - sT



-s



¥

î0

1 = 1 - e sT

1

1

=

ò 0

¥



2 sin x dx ´2 x p

sin 2pt dt 2pt 1 2p

¥

ò

-¥

sin x dx ( putting x = 2pt ) x



d2 y dy -4 + 4y = 0 2 dx dx



é d2 y ù é dy ù Þ L ê 2 ú - 4 L ê ú + 4 L[ y ] = 0 ë dx û êë dx úû

⇒  s2Y(s) – sy(0) – y’(o) – 4(sY(s) – y(0)) + 4Y(s) = 0 

(taking L(y(x)) = Y(s))

8/3/2023 6:15:05 PM

394 • Engineering Mathematics Exam Prep

⇒  s2Y(s) – 0 – 1 – 4sY(s) + 4Y(s) = 0



⇒  (s2 – 4s + 4)Y(s) = 1 1 s2 - 4 s + 4



⇒  Y(s) =



⇒  L–1 [Y(s)] = L–1 éê



⇒  y(x) = xe2x



So y(1) = e2 = 7.389

=

é Y (s) ù = L-1 ê ú ëê X ( s ) ûú

1 ( s - 2)2

1 ù 2ú ë ( s - 2) û

35. (a) L[y′′] + 2L[y′] + L[y] = 0



é 1 ù é 1 ù -1 = 7L-1 ê ú - 2L ê ú s 5 + ( ) ëê ûú ëê ( s + 2 ) ûú

⇒  s2Y(s) – sy(0) – y′(0) + 2(sY(s) – y(0))



+ Y(s) = 0  (Where Y(s) = L[y(x)])

⇒  s2Y(s) – 0 – 1 + 2sY(s) – 0 + Y(s) = 0



⇒  Y(s) =



⇒  L–1[Y(s)] = L–1 ê



1 1 = ( s2 + 2s + 1) ( s + 1)2

é x1 (t ) ù é0 0 ù é x1 (t ) ù é 0 ù ê ú=ê ú + ê ú u(t ) úê ë x2 (t ) û ë0 -9 û ë x2 (t ) û ë45 û 0 é x (t ) ù é ù Þê 1 ú=ê ú  + x ( t ) 9 x ( t ) 45 u ( t ) 2 ë 2 û ë û

1 ù ú ( s + 1)2 û ë

Þ x1 (t ) = 0, x 2 (t ) = -9x2 (t ) + 45 u(t ) Þ L éë x1 (t ) ùû = L éë0 ùû = 0,

⇒  y(t) = te–t = te–tu(t) 36. (b) Let L(x(t)) = X(s) and L(y(t)) = Y(s). Then the initial rest conditions are



x(0) = 0 = y(0); x′(0) = 0 = y′(0). then



d2 y(t ) dy(t ) dx (t ) +7 + 10 y(t ) = 4 x (t ) + 2 dt dt dt

= 7e -5t u(t ) - 2e -2t u(t )

37. 5 The initial conditions are x1(0) = 0, x2(0) = 0. Now

é



é ù 4 + 5s = L-1 ê ú êë ( s + 2 )( s + 5 ) úû é 5( s + 2) - 6 ù = L-1 ê ú êë ( s + 2 )( s + 5 ) úû é 1 ù é ù 1 -1 = 5L-1 ê ú - 6L ê ú êë ( s + 5 ) úû êë ( s + 2 )( s + 5 ) úû é 1 ù é ù -1 ( s + 5 ) - ( s + 2 ) = 5L-1 ê ú - 2L ê ú êë ( s + 5 ) úû êë ( s + 2 )( s + 5 ) úû é 1 ù é 1 ù é 1 ù -1 -1 = 5L-1 ê ú - 2L ê ú + 2L ê ú êë ( s + 5 ) úû êë ( s + 2 ) úû êë ( s + 5 ) úû

L éë x 2 (t ) ùû = -9L éë x2 (t ) ùû + 45L éëu(t ) ùû



Þ sX1 ( s) - x1 (0) = 0,

45 s 45 Þ X1 ( s) = 0, ( s + 9 ) X 2 ( s) = s sX 2 ( s) - x2 (0) = -9 X 2 ( s) +

Þ L-1 éë X1 ( s) ùû = L-1 éë0 ùû = 0,

é 45 ù L-1 éë X 2 ( s) ùû = L-1 ê ú ëê s ( s + 9 ) ûú

2

é d y(t ) ù é dy(t ) ù Þ Lê + 7L ê + 10L[ y(t )] 2 ú dt úû dt ë ëê ûú

é (s + 9) - s ù = 5L-1 ê ú êë s ( s + 9 ) úû é 1 ù é1 ù = 5L-1 ê ú - 5L-1 ê ú ësû êë ( s + 9 ) úû



é dx (t ) ù = 4 L[ x (t )] + 5L ê ú ë dt û



⇒ [s2Y(s) – sy(0) – y′(0)] + 7[sY(s) – y(0)]



+ 10Y(s) = 4X(s) + 5[sX(s) – x(0)]

Þ x1 (t ) = 0, x2 (t ) = 5 - 5e -9t



⇒  s2Y(s) + 7sY(s) + 10Y(s) = 4X(s) + 5sX(s)

\ lim



EMEP.CH06_3PP.indd 394

Þ

= 5 - 5e -9t

t ®¥

Y ( s) 4 + 5s 4 + 5s = = X ( s) s2 + 7s + 10 ( s + 2)( s + 5)

Then, the impulse response

=



=

x12 (t ) + x22 (t )

lim x12 (t ) + lim x22 (t )

t ®¥

t ®¥

2

0 + (5 - 0) = 5

8/3/2023 6:15:07 PM

Laplace Transforms • 395

38. (b)

Questions for Practice

Initial value of f (t ) = lim f (t )

1. Find the Laplace transform of (sin 2t – cos 2t)2

t ®0

= lim sF ( s ) s ®¥

é s +1 ù = lim s ê ú s ®¥ ëê s ( s + 2 ) ûú

(a)

1ù é ê1 + s ú = lim ê ú s ®¥ ê1 + 2 ú s ûú ëê

(c) 1 s

s cos7 + sin7 (c) 2cos72 - s sin7 (d) 2

= lim sF ( s ) s ®0

3. Find the Laplace transform of f(t), where

és +1ù ú ës + 2û

= lim ê



0 +1 1 = . 0+2 2



39. (b) y(t )

è

és + 2ù = L-1 ê ú ës + 6û

è

(a)

= d(t ) - 4e -6t

(c)

\ y(0.1) = d(0.1) - 4e -6 ´ 0.1 = 0 - 4e -0.6 = -2.19

- 2p s 3

ps

(b) æç 2 s ö÷ e 3 s +1 è



ø

(d) none of these

4

2

( s + 1)( s - 2s + 1) 4 ( s - 1)( s2 - 2s + 5)

(b)

1 s( s + 1) 2

4 ( s + 1)( s - 2s + 5) 2

(d) none of these

5. Find the Laplace transform of f(t), if s

40. (b)



41. 1 . 2

ì4, 0 £ t £ 1 f (t ) = í î2, t > 1

(a) 2 (2 - e s ) (b) 2 (2 + e s ) s

F ( s)

(

2

= L( f (t )) = L 2t e 2

= 2 ( -1)

EMEP.CH06_4PP.indd 395

ø

- 2p s 3

4. Find L[e–t f(2t)] given that L[ f (t )] =

é 1 ù = L-1 éë1ùû - 4 L-1 ê ú ës + 6û



ø

(c) æç s ö÷ e s2 + 1

é (s + 6) - 4 ù = L-1 ê ú êë s + 6 úû 4 ù é = L-1 ê1 ú ë s + 6û



2p 2p ì ïïcos(t - 3 ), if t > 3 f (t ) = í ï0, if t < 2p ïî 3

(a) æç 21 ö÷ e s -1



s +4

s +4

é s +1 ù ú ëê s ( s + 2 ) ûú

= lim s ê

=

s +4

s +4

t ®¥

s ®0

1 2 2 (d) + 2 s s +4 s2 + 4

s cos7 - 2sin7 (a) s cos72 + sin7 (b) 2

Final value of f (t ) = lim f (t )

s ®0

(b)

1 4 + s s2 + 16

2. L[cos (2t + 7)] = ?

1+0 = =1 1+0



1 4 s s2 + 16

-t

) = 2L{t

2

-t

g (t )}[where g (t )=e ]

d2 d2 d2 G( s) = 2 2 L { g (t )} = 2 2 L e -t 2 ds ds ds

{ }

d2 æ 1 ö 1 =2 2ç =4 . ÷ ds è s + 1 ø ( s + 1)3 \ F (1) = 4

1 1 = . (1 + 1)3 2

s

(c) 1 (1 - e s ) (d) 1 (1 + e s ) s

s

6. L{e–2t(2 sin 3t – 5 cos 2t)} = ? (a)

5 6( s + 2) + s2 + 4s + 13 s2 + 4s + 8

(b)

2 3( s + 2) - 2 s + 4s + 13 s + 4s + 8 2

8/14/2023 2:32:10 PM

396 • Engineering Mathematics Exam Prep (c)

5 6( s + 2) - 2 2 s + 4s + 13 s + 4s + 8

(d)

6 5( s + 2) s2 + 4s + 13 s2 + 4s + 8

¥

13. e2t cos3t dt = ? ò 0

(a)

7. Find the Laplace transform of e–2t sin 5t (a) (c)

5 s2 + 4s + 26

(b)

5 s + 2s + 25 2

5 s2 + 4s + 29

(d) none of these

8. L(cos3t) = ?

s ö ÷=? è s - 2s + 5 ø

14. L-1 æç

15. L-1 éêlog æç s + 2 ö÷ ùú = ? (a) 2 sinh 2t (b) 2 sinh t. (c) 1 sinh 2t. (d) 1 sinht t

3 3t

9. L(t e ) =

t

16. Find the Laplace transform of f(t) where

(a)

6 ( s - 3)4

(b)

(c)

6 ( s - 3)3



5 ( s + 3)4



(d) none of these

îï0, t > 2

4 + 2s + 1 (b) e (a) e

4 -2s

2-s

4 -2s +1 (c) e

2+s

(a) 22 + æç 1 + 2p - 32 ö÷ e - ps s s s è

-1 2-s

(d) 22 + æç 1 - 2p - 22 ö÷ e - ps s s s è

17. If f(t) is defined by

2

2

( s + 5)

2 2 (c) s2 - 9 2 (d) s2 - 9 2

( s + 9)

ø

(d) none of these

2 2 (a) s + 9 (b) s + 9

( s - 9)

ø

(c) 22 + æç 2p - 22 ö÷ e - ps s è s s ø

11. L(t cos 3t) = ? 2

ì2t, 0 < t < p f (t ) = í î1, t > p

(b) 22 + æç 1 + 22 ö÷ e - ps s ès s ø

ì 2t 10. Find L{f(t)} if f (t) = ïíe , 0 < t < 2

( s + 9)

¥



ìt,0 < t < p f (t ) = í îp - t, p < t < 2 p,



then find the Laplace transform of f(t)

-ps -ps (a) 1 + e ( ps - 1) (b) e ( p-pss+ 1)2

(1 + e

(1 + e -ps )s

-ps ( ps + 1) (c) 1 - e -p s 2

12. ò e2t cos3t dt = ?

(1 + e

0

(a) -2 (b) 2

13

(c) 5 (d) -5

EMEP.CH06_3PP.indd 396

t

t

(d) none of these

13

è s - 2 øû

ë

(c) 1 êé 2 s + 23s úù 4 ës + 9 s +1û

13

2

3 et cos2t - et sin 2t 2

(d) et cos 2t + 2et sin 2t

(b) 1 êé 2 s - 22s úù 4 ës + 9 s +1û

2

2

(a) et cos2t – et sin 2t (b) et cos2t + 1 et sin 2t (c)

(a) 1 êé 23s + 2 s úù 4 ës + 9 s +1û



(c)

-2 (b) 2 13s 13s 2 (d) 1 12s 13

13

)s

)s

(d) none of these

18. Find L{t2u(t – 3)} -3s -3s -3s (a) e 3 + 5e 2 + e

s

s

s

8/3/2023 6:15:13 PM

Laplace Transforms • 397

(b) (c)

26. Find L-1 æç s2+ 2 ö÷ = ?

e -3s 4e -3s 6e -3s + 2 + s s3 s -3s

2e s3

+

-3s

6e s2

+

è 2s + 8 ø

-3s

9e s

(a) 1 (cos2t - sin t ) (b) 1 (cos2t + sin 2t ) 2

(d) none of these 19. If

1 L{ g (t )} = , s-2

(a)

(c) 1 (cos t + sin t ) 2

then L{g(2t)} = ?

s-a

1 (b) 1 s-4 s+4 s-3

to _______ ?

s+3

20. Find the Laplace transform of f(t), if ì0, t < 4 f (t ) = í î1, t ³ 4

(a)

e -4s s

(c)

e -2s s

(d) none of these

27. If L ( eat ) = 1 , then L(3e5t sin h 5t)is equal

(c) 1 (d) 1



2

(a)

5 (b) 5 2 s - 5s s - 2s

(c)

15 (d) 2 10 s2 - 10s s - 15s

2

28. If a is a constant, then the value of the

(b)

e 4s s

(d)

e 2s s

¥



integral a2 ò xe -ax dx is ? 0 (a) 4 (b) 3 (c) 2 (d) 1 29. The Laplace transform of f (t ) = 1 is ______ ?

21. L[cos (t – 2)u(t – 2)] = ?

t

(a) e -t æç s ö÷ (b) e -t æç 22s ö÷ 2 è s +1 ø è s +1 ø

(a)

(c) e -2t æç 2 s ö÷ (d) e2t æç 2 s ö÷ è s +1 ø è s +1 ø

(c) 1 (d)

5 (b) 2 5 s - 4s + 25 s - 4s + 29

(c)

5 s2 - 2s + 10

p 2s

(b)

p s

s

30. The Laplace transform of the function shown in the figure below is ______ ?

22. The Laplace transform of is (a)

2p s

2

(d) none of these

V

23. If L-1 æç 13 ö÷ = x 3 , then x = ? ès ø

(a) 4

(b) 3

(c) 2

ü s = f (t ) * g (t ), 2 2ý ( s 4) + î þ ì

24. If L-1 í

(d) 1

then find f(t) and

a

t

b

g(t), respectively (by the convolution theorem)

(a) V ( e -as - e -bs ) (b) V ( eas - ebs )



(a) cos 2t and sin 2t (b) cos t and sin 2t

(c) V ( e -as - ebs ) (d) V ( eas - e -bs )



(c) cos 2t and

2

sin t 2

2



(d) none of these

1 ü ý=? î s( s + 4) þ

25. L-1 ìí

2

(c) 1 (1 - e -4t ) (d) 1 (1 + e -4t ) 4

EMEP.CH06_3PP.indd 397

s

s

s

31. For the time-domain function f(t) = t, the t

Laplace transform of f (t ) dt is _____ ? ò 0

(a) 1 (1 - e -2t ) (b) 1 (1 + e -2t ) 2

s

4

(a) 1 (b) 1 s

s2

(c) 13 (d) 13 s

2s

8/3/2023 6:15:18 PM

398 • Engineering Mathematics Exam Prep 32. Given that the Laplace transform of the function below a single period 0 < t < 2 is 2 1 (1 - e-s ) . Then the Laplace transform of

Explanation

1. (a)

s2

(sin 2t - cos2t )2

the periodic function over 0 < t < ∞ is

2. (b) cos(x + y) = cos x cos y – sin x sin y 3. (c), 4. (c), 5. (a), 6. (d), 7. (b), 8. (c)

1 f(t)

0

2

t

4

s è1 - e

s è1 + e

ø

ø

-s ö æ -s ö æ (c) 12 ç e -s ÷ (d) 12 ç 1 - es ÷ ç ÷ ç ÷

s è1 + e

s è

ø

e



ø

33. Consider the differential equation y′′ + 2 y′ + y′ = 0 with boundary condition y(0) = 1, y(1) = 0. The value of y(2) is (a) e2 (b) –e2 –2 (c) e (d) –e–2 1

34. The inverse Laplace transform of 2 2s + 3s + 1 is _______ ? (a)

t e 2

(c)

-

e

t 2

+ et -e

(b)

(d) -t

t

e

t 2

+ e -t

îï ( s - 1) þïúû

(a)

2 t e - cos t t

(c)

2 -t e - cos t t

( (

2 t e + cos t t

) (b) (

ì ü ( s - 1) ü 1 + L-1 í 2 2ý 2 2ý î ( s - 1) + 2 þ î ( s - 1) + 2 þ ì

= L-1 í

15. (a), 16. (d) 17. (c) 18. (c) t 2u(t - 3) = {(t - 3) + 3}2 u(t - 3)



= (t – 3)2 u(t – 3) + 6(t – 3)u(t – 3) + 9u(t – 3)



= L{(t - 3)2 u(t - 3)} + 6L{(t - 3)u(t - 3)} + 9L{u(t - 3)}



= e -3s L(t 2 ) + 6e -3s L(t ) + 9L{u(t - 3)}

19. (a), 20. (a), 21. (c), 22. (b), 23. (d)

)

2 -t e + cos t t

) (d) (

s æ ö L-1 ç 2 ÷ è s - 2s + 5 ø ì ( s - 1) + 1 ü = L-1 í 2 2ý î ( s - 1) + 2 þ

L[t 2u(t - 3)]

é ì 2 üù 35. L-1 êlog íï s + 12 ýïú = ? ëê

1 1 (4 cos3 t ) = (cos3t + 3cos t ) 4 4

Thus,

e 2 - e -t -

cos3 t =

9. (a), 10. (b), 11. (d), 12. (a), 13. (c) 14. (b)

-s ö -s ö æ æ (a) 12 ç 1 + e -s ÷ (b) 12 ç 1 - e -s ÷ ç ÷ ç ÷

24. (a) ìï üï s f (t ) * g (t ) = L-1 í 2 2ý îï ( s + 4) þï

)

1 ïü ïì s = L-1 í 2 ´ 2 ý + + 4) þï ( 4) ( s s îï

Answers key 1. (a)

2. (b)

3. (c)

4. (c)

5. (a)

6. (d)

7. (b)

8. (c)

9. (a)

10. (b)

11. (d)

12. (a)

13. (c)

14. (b)

15. (a)

16. (d)

17. (c)

18. (c)

19. (a)

20. (a)

21. (c)

22. (b)

23. (d)

24. (a)

25. (c)

26. (b)

27. (c)

28. (d)

29. (d)

30. (a)

31. (c)

32. (b)

33. (d)

34. (c)

35. (a)

EMEP.CH06_3PP.indd 398

= sin 2 2t - cos2 2t - 2sin 2t cos2t = 1 - sin 4t

= L-1 { F ( s) ´ G( s)}



æ ö 1 s , G( s ) = 2 çç where, F ( s) = 2 ÷ ( s + 4) ( s + 4) ÷ø è æ ö s \ f (t ) = L-1 ( F ( s) ) = L-1 çç 2 ÷÷ = cos2t, è ( s + 4) ø æ 1 ö 1 g (t ) = L-1 (G( s) ) = L-1 çç 2 ÷ = sin 2t ÷ è ( s + 4) ø 2

8/3/2023 6:15:21 PM

Laplace Transforms • 399

25. (c) 26.  (b) 27.  (c) =

28. (d) ¥

ò xe

- ax

¥

¥

ò

ò f (t )e

dx = te - at dt =

0

0

- at

0

= L[ f (t )] = L(t ) =

1 s2

1

=

a2

2



¥ 0

Þ L ëé y¢¢(t ) ûù + 2L ëé y¢(t ) ûù + L ëé y(t ) ûù = L ëé0 ûù

æ -1 ö æ 1 ö 2 L[ f (t )] = L ç ÷ = L çç t ÷÷ è tø è ø 1 1 - +1 2 = = 21 = 1 - +1 s 2

Þ éë s2Y ( s ) - sy(0) - y¢(0) ùû + 2 [ sY ( s ) - y(0)] + Y ( s ) =0

pù ú s úû

s2

Þ s2Y ( s) - s - k + 2sY ( s) - 2 + Y ( s) = O [assume y¢(0) = k ] Þ ( s2 + 2s + 1)Y ( s) = s + 2 + k



30. (a) Here,

ì 0, t < a ï f (t ) = íV , a < t < b ï 0, t > b î

( s + 1) + ( k + 1) = ( s + 1)2 ( s + 1)2 1 1 = + ( k + 1) ´ s +1 ( s + 1)2

Þ Y ( s) =

¥

ò

\ L[ f (t )] = e - st f (t ) dt b

ù = e - st ´ 0 dt + e - st ´ V dt + e - st ´ 0 dt ú ú 0 a b û

ò



¥

ò

ò

Þ y(t ) = e -t + ( k + 1)e -tt Now,

31. (c)

ét ù 1 1 L ê f (t ) dt ú = F ( s) = L[ f (t )] s ê0 ú s ë û 1 1 1 1 = L(t ) = ´ 2 = 3 s s s s

ò



L[ f (t )] =

1 1 - e - sT = =

=



EMEP.CH06_3PP.indd 399

òe 0

1

1 - e -2s 1 1-e

2

òe 0

´

-2s

1

- st

f (t ) dt ( here T = 2 )

1

(1 - e )

s

2

Hence, y(2) = -e -2



34. (c) 1 æ ö L-1 ç 2 ÷ 2 + 3 + 1 s s è ø 1 ö -1 æ =L ç 2 ÷ è 2s + 2s + s + 1 ø

f (t ) dt.....(i )

2

(1 - e ) ´ s

2

ì 1 ü = L-1 í ý î 2s( s + 1) + (s+ 1) þ ì 1 ü = L-1 í ý î ( s + 1)(2s + 1) þ

2

-s

-s

( )

1 - e -s

- st

y(1) = 0 Þ 0 = e -1 + ( k + 1)e -1 Þ k +1 +1 = 0 Þ k = -2

\ y(t ) = e -t + ( -2 + 1)e -tt = e -t (1 - t )

32. (b) We know that if f(t) be periodic with period T, where 0 < t < ∞ then, T

s+2+k

1 ù é 1 ù -1 é Þ L-1 [Y ( s )] = L-1 ê ú + ( k + 1)L ê 2 ú ( s + 1) û ës +1û ë

0

a

-s

y¢¢ + 2 y¢ + y = 0 Þ y¢¢(t ) + 2 y¢(t ) + y(t ) = 0

ò

29. (d)

s2

33. (d)

\ a 2 xe - ax dx = 1



-s

2

-s

dt

[where s = a, f (t ) = t ]



(1 - e )(1 + e ) (1 - e ) ùú = s (1 + e ) úú û -s

(1 - e ) ´ -s

1

2



ì 1 ü = 2L-1 í ý î (2s + 2)(2s + 1) þ

8/3/2023 6:15:23 PM

400 • Engineering Mathematics Exam Prep ì (2s + 2) - (2s + 1) ü = 2L-1 í ý î (2s + 2)(2s + 1) þ æ 1 ö æ 1 ö = 2L-1 ç - 2L-1 ç ÷ ÷ è 2s + 1 ø è 2s + 2 ø



35. (a) ìï s2 + 1 üï Let, F(s) = log í 2ý îï ( s - 1) þï Then,

ì ü ï ï ì 1 1 ï ï ïü -1 ï = 2L-1 í ý - 2L í ý ïî 2 ( s + 1) ïþ ï2æ s + 1 ö ï ç 2 ÷ø þï îï è æ ö ç 1 ÷ -1 æ 1 ö =L ç ÷-L ç ÷ è s +1 ø çç s + 1 ÷÷ 2ø è

\tf (t )

ìï s2 + 1 üïù d d é F ( s)} = -L-1 êlog í ú { 2ý ds ds êë ïî ( s - 1) ïþúû éd ù = -L-1 ê log s2 + 1 - 2log ( s - 1) ú ë ds û

= -L-1

{ (

-1



EMEP.CH06_3PP.indd 400

=e

-

t 2

L-1 { F ( s)} = f (t )



)

}

2 ö æ 2s = -L-1 ç 2 s + 1 ÷ø è s +1

- e -t

8/3/2023 6:15:23 PM

CHAPTER

7

Numerical Analysis 7.1  ERRORS AND APPROXIMATIONS 7.1.1  Rounding off For computation purpose, we generally cut off some unwanted digits from given numbers. This process of dropping unnecessary digits is called rounding off. The general rules for rounding off a number to n significant figures are as follows: Discard all digits to the right of the nth digit and if among these discarded digits the digit in the (n + 1)th place is (i) greater than 5 then the digit in the nth place is increased by 1. (ii) less than 5 then the digit in the nth place is left unchanged. (iii) exactly 5 then leave the nth digit unaltered if it is even and increase the nth digit by 1 if it is odd. For example, the following numbers are rounded off to four significant figures: 7.029886 becomes 7.029 3.5634 becomes 3.563 89.3999 becomes 89.40 8.52854 becomes 8.528 8.42757 becomes 8.428 7.1.2  Errors and Their Computation Let xT and xA denote the true and approximate value, respectively, of a number. Then the absolute error, Ea involved in xA is defined by Ea = |xT – xA|. The relative error, Er is defined by Er = xT - x A , xT provided xT ≠ 0.

EMEP.CH07_3PP.indd 401



The percentage error (or relative percentage

error), E = E ´ 100 = xT - x A ´ 100 . p r xT

7.2 CALCULUS OF FINITE DIFFERENCES Let y = f(x) be a real-valued function of x defined in an interval [a, b] and its values are known for (n + 1) equally spaced points xi(i = 0, 1, 2, ..., n) such that xi = x0 + ih(i = 0, 1, 2, ..., n) where x0 = a, xn = b and h is the spacing (space length). Then xi (i = 0, 1, 2, ..., n) are called nodes and the corresponding values yi are termed as entries. 7.2.1  Forward Difference Operator The differences y1 – y0, y2 – y1, y3 – y2,..., yn – yn–1 are denoted by Dy0 , Dy1 , Dy2 ,..., Dyn -1 , respectively.

Thus, we have Dyi = yi +1 - yi ( i = 0,1,2,...., n - 1)

where D is called forward difference operator. In general, the forward difference operator is defined by Df ( x ) = f ( x + h ) - f ( x ) . Similarly, the higher order forward differences are defined as D2 yi = Dyi +1 - Dyi , D3 yi = D2 yi +1 - D2 yi ,

........ ........... .............. r

r -1

r -1

D yi = D yi +1 - D yi . where i = 0, 1, 2, …, n – 1 and r (1 £ r £ n ) is a positive integer.

8/3/2023 6:23:14 PM

402 • Engineering Mathematics Exam Prep Properties of D: If a and b be any two constants, then (i) Da = 0 (ii) D {af ( x )} = aDf ( x ) (iii) D {af ( x ) ± bg ( x )} = aDf ( x ) ± bDg ( x ) (iv) D[f(x)g(x)] = f ( x + h ) Dg ( x ) + Df ( x ) g ( x + h ) ì f ( x ) ïü g ( x ) Df ( x ) - f ( x ) Dg ( x ) ý= g (x ) g (x + h) îï g ( x ) þï

(v) D ïí

7.2.2  Backward Difference Operator The differences y1 - y0 , y2 - y1 , y3 - y2 ,...., yn - yn -1

are denoted by Ñy1 , Ñy2 , Ñy3 ,..., Ñyn , respectively.

Thus, we have Ñyi +1 = yi +1 - yi ( i = 0,1,2,...., n - 1) where ∇ is called backward difference operator. In general, the backward difference operator is defined by Ñf ( x ) = f ( x ) - f ( x - h ) .

Similarly, the higher order backward differences are defined as Ñ2 yi = Ñyi - Ñyi -1 , Ñ3 yi = Ñ2 yi - Ñ2 yi -1 ,

Suppose the function y = f (x) is not known explicitly, but the values of f(x) are known for (n + 1) distinct values of x, say x0, x1, ...,xn called arguments or nodes. Let y0 = f(x0), y1 = f(x1), ......, yn = f(xn) be the corresponding entries and there is no other information available about the function. The problem of interpolation is to compute the value of f(x), at least approximately, for a given argument x, not found in the table. When x lies slightly outside the interval [xo, xn], then the process is called extrapolation. Our target is to find a polynomial function j(x) such that f(xi) = j(xi) at i = 0, 1, 2, ..., n. The polynomial j(x) is called interpolating polynomial. 7.3.1 Newton’s Forward Difference Interpolation Formula Let y = f(x) be a real valued function of x defined in [a, b] and is known for the corresponding (n + 1) equi-spaced arguments xi(i = 0, 1, 2, ..., n) such that xi = x0 + ih(i = 0, 1, 2, ...,n) with x0 = a, xn = b and h is the space length. Let yi = f(xi) (i = 0, 1, 2, ..., n) Then, y = f (x )  j(x )

........ ........... .............. r

r -1

r -1

Ñ yi = Ñ yi - Ñ yi -1 . where i = 1, 2, …, n and r (1 £ r £ n ) is a positive integer.

7.2.3  Shift Operator The shift operator is denoted by E and is defined as E { f ( x ) } = f ( x + h ) , h being the spacing.

\ E 2 f ( x ) = Ef ( x + h ) = f ( x + 2h ) , E 3 f ( x ) = Ef ( x + 2h ) = f ( x + 3h )

and so on.

Thus, in general, E n f ( x ) = f ( x + nh ) The inverse shift operator, E–1 is defined by –n E -1 f ( x ) = f ( x - h ) and in general, E f(x) =

f(x – nh). Remember: (i) E º D + 1

(ii)



= yn + sÑyn +

7.3 INTERPOLATION

EMEP.CH07_3PP.indd 402

u ( u - 1)

7.3.2 Newton’s Backward Difference Interpolation Formula Let y = f(x) be a real valued function of x defined in [a, b] and is known for the corresponding (n + 1) equi-spaced arguments xi (i = 0, 1, 2, ..., n) such that xi = x0 + ih(i = 0, 1, 2, ..., n) with x0 = a, xn = b and h is the space length. Let yi = f(xi) (i = 0, 1, 2, ..., n). Then, y = f(x) ≈ j(x)

E -1 º 1 - Ñ

Let f(x) be a function of x defined in the interval [a, b] in which it is assumed to be continuous and continuously differentiable.

u ( u - 1) (u - 2) 3 D2 y0 + D y0 2! 3! u ( u - 1) ... ( u - n + 1) n + ....... + D y0 n! x - x0 where u = . h = y0 + uDy0 +



+ ........... + where s =

s ( s + 1)

Ñ2 yn +

s ( s + 1) ( s + 2)

2! 3! s ( s + 1)( s + 2 ) ... ( s + n - 1) n!

Ñ3 yn

Ñn yn

x - xn . h

8/3/2023 6:23:15 PM

Numerical Analysis • 403

7.3.3  Lagrange’s Interpolation Formula Let y = f (x) be a function of x, continuous and (n + 1) times continuously differentiable in [a, b]. Let us divide the interval [a, b] by (n + l) points a = x0, x1, ...xn = b which are not necessarily equispaced and the corresponding entries are yi = f(xi) (i = 0, 1, 2, ..., n). Then, y = f(x) ≈ j(x) =

( x - x1 )( x - x2 )( x - x3 ) ....... ( x - xn ) y ( x0 - x1 )( x0 - x2 )( x0 - x3 ) ....... ( x0 - xn ) 0 ( x - x0 )( x - x2 )( x - x3 ) ....... ( x - xn ) y + ( x1 - x0 )( x1 - x2 )( x1 - x3 ) ....... ( x1 - xn ) 1 ( x - x0 )( x - x1 )( x - x3 ) ....... ( x - xn ) y + ( x2 - x0 )( x2 - x1 )( x2 - x3 ) ....... ( x2 - xn ) 2

7.4.2 Differentiation Formula Based on Newton’s Backward Difference Formula (i) f ¢ ( xn ) 

1é 1 1 1 ù Ñyn + Ñ2 yn + Ñ3 yn + Ñ4 yn + .......ú h êë 2 3 4 û



(ii) f ¢¢ ( x0 ) 

1 h2

11 4 é 2 ù 3 êÑ yn + Ñ yn + 12 Ñ yn + ..........ú ë û

7.5  NUMERICAL INTEGRATION b

Consider the definite integral I = f ( x ) dx ò a

7.3.4  Error in Interpolation

and divide the interval [a, b] of integration into n equal subintervals such that a = x0 < x1 < x2 < ... < xn = b and xi = x0 + ih (i = 0, 1, 2, ..., n), h is the space length. Also, let the function f(x) be known at the nodes xi, i.e., the values yi = f(xi) (i = 0, 1, 2, ..., n) are given.

The error in approximating f(x) by j(x) is given by

7.5.1 Trapezoidal Rule



+ ................... ..................... ........................ +

( x - x0 )( x - x1 )( x - x2 ) ....... ( x - xn-1 ) y ( xn - x0 )( xn - x1 )( xn - x2 ) ....... ( xn - xn -1 ) n

Rn +1

( x - x0 )( x - x1 )( x - x2 ) ....... ( x - xn ) f ( n+1) (t ) =

(n + 1)! where a = x0 < t < xn = b



xn



h

òx f ( x ) dx  2 éë y

Let f(x) be a function of x defined in the interval [a, b] in which it is assumed to be continuous and continuously differentiable. Suppose the function y = f (x) is not known explicitly, but the values of f(x) are known for (n + 1) distinct values of x, say x0, x1, ..., xn. Let y0 = f(x0), y1 = f(x1) ..., yn = f(xn) be the corresponding entries. Our target is to find a polynomial function j(x) such that f(xi) = j(xi) at i = 0, 1, 2, ..., n and then differentiate it as many times as we desire. 7.4.1 Differentiation Formula Based on Newton’s Forward Difference Formula

In particular,

0

x2

h

ò f ( x ) dx  2 éë y x

0

+ y2 + 2 y1 ùû .

0 Remember: (i) The error in computing the value of the integral using trapezoidal rule is given by

ET = -

nh3 f ¢¢(t ) where a = x0 < t < xn = b 12

(ii) The trapezoidal rule geometrically interprets that the curve f(x) in [x0, xn] is replaced by n straight lines joining the points (x0, y0), (x1, y1), (x2, y2),…….., (xn, yn). 7.5.2  Simpson’s 1/3rd Rule

ò f ( x ) dx

1é 1 1 1 ù Dy0 - D2 y0 + D3 y0 - D4 y0 + ......ú h êë 2 3 4 û

x0





(ii) f ¢¢ ( x0 )  12 éê D2 y0 - D3 y0 + 11 D4 y0 - ..........ùú h ë

EMEP.CH07_3PP.indd 403

h

ò f ( x ) dx  2 éë y0 + y1 ùû , x

xn

(i) f ¢ ( x0 ) 

+ yn + 2 ( y1 + y2 + ... + yn -1 ) ùû

0

x1

7.4  NUMERICAL DIFFERENTIATION

0

12

û

h é y0 + yn + 4 ( y1 + y3 + ...... + yn -1 ) 3ë

+2(y2 + y4 + ... + yn–2)

8/3/2023 6:23:16 PM

404 • Engineering Mathematics Exam Prep Remember: (i) The error in computing the value of the integral using Simpson’s 1/3rd rule is given by ES = -

nh5 iv f (t ) where a = x0 < t < xn = b 180

(ii) Simpson’s 1/3rd rule gives us exact result for a polynomial of degree less than or equal to three. (iii) Simpson’s 1/3rd rule geometrically interprets that the curve f(x) in [x0, x2] is replaced by a second degree parabola through the points (x0, y0), (x1, y1), (x2, y2). 7.5.3 Weddle’s Rule xn

ò f ( x ) dx

x0



3h  é y0 + yn + ( y2 + y4 + y8 + ... + yn -4 ) 10 ë

+5 ( y1 + y5 + y7 ... + yn -1 ) + 6 ( y3 + y9 + y15 ... + yn -3 ) +2 ( y6 + y12 + y18 ... + yn -6 ) ùû

In particular, x6

3h

ò f ( x ) dx  10 éë y0 + 5 y

1



ò x

0

(b - a ) f ( x ) dx  é y0 + 3 y1 + 3 y2 + y3 ûù 8 ë

7.6 SYSTEM OF LINEAR ALGEBRAIC EQUATIONS 7.6.1 Gauss Elimination Method In this method, the given system of equations is reduced to an equivalent upper triangular system by a systematic elimination procedure from which the unknowns are found by back substitution. Let us consider the system of three equations in three unknowns given by

(1) (1) (1) a11 x1 + a12 x2 + a13 x3 = b1(1) ü ïï (1) (1) (1) a21 x1 + a22 x2 + a23 x3 = b2(1) ý …(i) ï (1) (1) (1) a31 x1 + a32 x2 + a33 x3 = b3(1) ï þ

First suppose that a11 ≠ 0, a22 ≠ 0, a33 ≠ 0. ­Using the Gauss elimination method, the above system (i) reduces to:

EMEP.CH07_4PP.indd 404



(1) (1) (1) a11 x1 + a12 x2 + a13 x3 = b1(1) ü ïï (2) (2) a22 x2 + a23 x3 = b2(2) ý …(ii) ï (3) a33 x3 = b3(3) ï þ

1 1 1 1 b( ) a ( ) b( ) a ( ) 2 1 2 1 b2( ) = b2( ) - 1 1 21 , b3( ) = b3( ) - 1 131 , () () a11 a11 (1) (1) (1) (1) ( 2) (1) a21 a12 ( 2) (1) a21 a13 a22 = a22 , a23 = a23 , (1) (1) a11 a11

(1) (1) (1) (1) ( 2) (1) a31 a12 ( 2) (1) a31 a13 a32 a a = a32 , = , 33 33 (1) (1) a11 a11 2 2 ( 2) ( 2) b( ) a ( ) (3) 3 2 ( 2) a32 a23 b3( ) = b3( ) - 2 232 , a33 = a23 , ( ) ( 2) a22 a22

The elements a11, a22, a33 are called pivots. The solutions of the (i) are then obtained from (ii) by back substitutions. 7.6.2  LU Decomposition Method This method is also known as triangular decomposition method. This method is based on the fact that every square matrix can be expressed as the product of a lower and an upper triangular matrix. Let us consider a system of three equations with three unknowns given below: a11x1 + a12 x2 + a13 x3 = b1

+ y2 + 6 y3 + y4 + 5 y5 + y6 ]

x0 7.5.4  Simpson’s 3/8th’s Rule

x3

where

a21x1 + a22 x2 + a23 x3 = b2



a31x1 + a32 x2 + a33 x3 = b3

Here, the coefficient matrix é a11 ê A = êa21 êë a31

a12 a22 a32 é l11

a13 ù ú a23 ú a33 úû 0

where L = êl21 l22 ê êël31 l32



\  A = LU gives



é l11 0 0 ù ê ú êl21 l22 0 ú êël31 l32 l33 úû



can be written as A = LU 0ù é1 u12 u13 ù ú, 0 ú U = ê0 1 u23 ú ê ú l33 úû êë0 0 1 úû

é1 u12 u13 ù ê ú ê0 1 u23 ú êë0 0 1 úû

Then we have, l11 = a11, l21 = a21, l31 = a31, l11u12 = a12 Þ u12 =

é a11

a12

a13 ù

êë a31

a32

a33 úû

= êêa21 a22 a23 úú

a a12 , l11u13 = a13 Þ u13 = 13 , l11 l11

l21u12 + l22 = a22 Þ l22 = a22 - l21u12 = a22 - a21a12 , a11



l31u12 + l32 = a32 Þ l32 = a32 - l31u12 = a32 l21u13 + l22u23 = a23 Þ u23 =

a31a12 , a11

a23 - l21u13 , l22

and l31u13 + l32u23 + l33 = a33 Þ l33 = a33 - l31u13 - l32u23 .

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Numerical Analysis • 405

Solving all these equations, we obtain the values of l11, l21,..... and u12, u13...., and thus, we get the matrices L and U. 7.6.3  Gauss–Seidel Iteration Method The sufficient condition for the convergence of the Gauss–Seidel iteration method is that the system of equations must be strictly diagonally dominant, i.e; for the system of equations AX = b where A = (aij)n×n, we must have aii >

n

aij . å j =1, j ¹i

Let us consider the system of three equations in three unknowns given by a11x1 + a12 x2 + a13 x3 = b1

a21x1 + a22 x2 + a23 x3 = b2

a31x1 + a32 x2 + a33 x3 = b3 Let us consider the initial approximations: x1 = x1(0), x2 = x2(0), x3 = x3(0). Then the iteration formula of the Gauss–­Seidel method are as follows:

1 é x1 = b - a x ( k ) - a13 x3( k ) ù , û a11 ë 1 12 2 1 é x2( k +1) = b - a21x1( k ) - a23 x3( k ) ù , û a22 ë 2 1 é x3( k +1) = b - a31x1( k ) - a32 x2( k ) ù . û a33 ë 3 ( k +1)



7.7 SOLUTION OF ALGEBRAIC AND TRANSCENDAL EQUATIONS 7.7.1  Method of Bisection This method is based on the well-known theorem which states that if f(x) be a continuous function in [a, b] and f(a) f(b) < 0, then there exist at least one real root of the equation f(x) = 0 in (a, b). In this method, we first find out a sufficiently small interval [a0, b0] containing the required root a of the equation f(x) = 0. Then f(a0)f(b0) < 0. To generate the sequence of iterates {xn}, each member of which is a successively better approximation of the exact root, say α. We put x0 = a0 and



1 2

Next, set x2 = ( a1 + b1 ) and repeat the above

process till we obtain xn +1 = 1 ( an + bn ) with desired 2

accuracy with xn ® a as n ® ¥ . 7.7.2  Regula Falsi Method The Regula Falsi method or method of false position is the oldest method for computing real roots of an equation f(x) = 0. The general iteration formula for the Regula Falsi method is xn +1 = xn -

f ( xn )

f ( xn ) - f ( xn -1 )

( xn - xn-1 ) ,

where n = 0,1,2,3,....... 7.7.3 Newton–Raphson Method Let x0 be an initial approximation of the desired root a of the equation f(x) = 0. The iteration formula for the Newton–Raphson method is



xn +1 = xn -

f ( xn ) f ¢ ( xn )

(provided f ¢ ( xn ) ¹ 0 )

where n = 0, 1, 2, 3.....

7.8 NUMERICAL SOLUTION OF ORDINARY DIFFERENTIAL EQUATIONS 7.8.1  Euler’s Method It is a simple and single step method for solving an ordinary initial value problem differential equation, where the solution will be obtained as a set of tabulated values of variables x and y. Let us consider a first-order and first-degree differential equation as

dy = f ( x , y ), dx

With y(xo) = yo.

find f(x1). If f(a0) f(x1) < 0, then

We divide the range [x0, xn] into n-equal subintervals by the points x0, x1, x2, ...., xr–1, xr, xr+1, ...,xn where xr = x0 + rh (r = 1,2,3,..., n) and h = length of each subinterval. Then, Euler’s iteration formula is given by yn +1 = yn + hf ( xn , yn ) = y( x n +1 ) for n = 0,1,2,3,…..

set a1 = a0, b1 = x1 so that [a1, b1] = [a1, x1]. On the other hand, if f(x1)f(b0) < 0 a, then put a1 = x1, b1 = b0, i.e; [a1, b1] = [x1, b0]. Thus, we see that [a1, b1] contains the root a in either case.

7.8.2  Modified Euler’s Method This method is also known as the Euler–Cauchy method. This method gives us a rapid and moderately accurate result up to a desired degree of accuracy.

x1 =

EMEP.CH07_3PP.indd 405

1 ( a0 + b0 ) and 2

8/3/2023 6:23:19 PM

406 • Engineering Mathematics Exam Prep

The modified Euler’s iteration formula is yr( n ) = yr -1 +

hé f ( xr -1 , yr -1 ) + f ( xr , yr( n -1) ) ù û 2ë

where yr( n ) is the nth approximation of yr and yr(0) = yr -1 +

h é f ( xr -1 , yr -1 ) + f ( xr , yr ) ûù . 2ë

7.8.3  Runge–Kutta Method The Runge–Kutta methods for the numerical solution of an ordinary differential equation give us a greater accuracy. I.  Second-Order Runge–Kutta Method Consider the differential equation:

y¢ =

dy = f ( x , y) dx

with y( x0 ) = y0 .

The computational formulae for the Runge– Kutta method of order two is as follows: y1 = y0 + k where 1 ( k1 + k2 ), 2 k1 = hf ( x0 , y0 ), k=

k2 = hf ( x0 + h, y0 + k1 ). II.  Fourth-Order Runge–Kutta Method

The computational formulae for the Runge–Kutta method of order four is as follows: y1 = y(x0 + h) = y0 + k where 1 ( k1 + 2k2 + 2k3 + k4 ) 6 and k1 = hf ( x0 , y0 ),

Fully Solved MCQs

1. If p = 3.14 is used in place of 3.14156, the absolute and relative errors are, respectively (a)  0.00156, 4.966 × 10–4 (b)  0.0018, 4.25 × 10–4 (c)  0.06, 4.2 × 10–4 (d)  0.05, 4.06 × 10–4 2. Find the percentage error in approximating 4 to 1.3333 is 3

(a) 0.45 (b) 0.0036 (c) 0.0025 (d) 0.05 2 3. If y = x + 4 for y = 1, 3, 5, 7, 9; find the value of D3f(9) (a) 0 (b) 1 (c) 10 (d) 15 3. (a) x 1 3 5 7 9 y 5 13 29 53 85 = x2 + 4 (= y0) (= y1) (= y2) (= y3) (= y4) Ñy4 = y4 - y3 = 85 - 53 = 32, Ñ2 y4 = Ñy4 - Ñy3 = Ñy4 - ( y3 - y2 ) = 32 - (53 - 29) = 8,

Ñ f ( 9 ) = Ñ3 y4 = Ñ2 y4 - Ñ2 y3 3

= Ñ2 y4 - (Ñy3 - Ñy2 ) = Ñ2 y4 - {( y3 - y2 ) - (( y2 - y1 ))}

k=



k ö h æ k2 = hf ç x0 + , y0 + 1 ÷ , 2 2ø è k ö h æ k3 = hf ç x0 + , y0 + 1 ÷ , 2 2ø è

k4 = hf ( x0 + h, y0 + k3 ) . 7.8.4  Predictor-Corrector Method

Let us consider the differential equation

dy = f ( x , y) dx

with y( x0 ) = y0 .

In order to solve the above differential equation, we first obtain the approximate value of yn+1 by predictor formula and then improve this value by means of a corrector formula. The predictor-corrector formulas are given by ynP+1 = yn + hf ( xn , yn ),

hé f ( xn , yn ) + f ( xn +1 , ynP+1 ) ù , û 2ë where n = 0,1,2,3.......

ynC+1 = yn +



EMEP.CH07_3PP.indd 406

= 8 - {(53 - 29) - (29 - 13)} = 0.

   4. Which of the following relation is true? (a)  D.Ñ º 2D - Ñ (b)  D.Ñ º D - Ñ (c)  D.Ñ º D + Ñ (d)  none of these 5. D log f ( x ) = ? é Df ( x ) ù é Df ( x ) ù ú (b)  log ê1 ú f ( x ) úû êë ëê f ( x ) ûú

(a)  log ê

é ù (c)  log ê1 + f ( x ) ú (d)  log éê1 + Df ( x ) ùú ëê

Df ( x ) ûú

ëê

f ( x ) ûú

2 x 6. The simplest value of æç D e x ö÷ Ee is ç E ÷ D2e x è ø

(a)  ex (b)  e2x (c)  e–x (d)  e–2x 7. Which of the following is true? (a)  f(4)= f(3) + Df(2) + D2f(1) + D3f(1) (b)  f(4) = f(2) + Df(3) + D2 f(2) + D3f(2) (c)  f(4) = f(1) + Df(2) + D2f(3) + D3f(1) (d)  none of these

8/3/2023 6:23:21 PM

Numerical Analysis • 407

8. The missing terms in the following table are, respectively: x 0 1 y 0 ___ (a)  1, 8 (c)  3, 24 9. Given the data:

2 8

3 4 15 ___ (b)  2, 12 (d)  5, 25

5 35

3



y¢ ( x k ) =

(c)  a = 1 , b = 3 (d) 

then the value of y¢ ( 2 ) is (a) 4 (b) 8 (c) 12 (d) 16 x2

ò1 + x 0

3

using

dx

Simpson’s 1/3rd rule with h = 0.25 is



(a) 0.23108 (c) 0.10234 11. The values of



(b) 0.25251 (d) 0.20501 0.7

ò (e

x

+ 2x ) dx

4

4 4 2 3 a= ,b= 3 4

15. The root of the equation xex = 1 between 0 and 1, obtained by using two iterations of bisection method, is (a) 0.25 (b) 0.50 (c) 0.75 (d) 0.65 16. While solving the equation x2 – 3x + 1 = 0 ­using the Newton–Raphson method with the initial guess of a root as 1, the value of the root after one iteration is (a) 1.5 (b) 0.1 (c) 0.5 (d) 0 17. Consider the system of equations

1æ 1 1 ö Dyk + D2 yk - D3 yk ÷ , 2 4 h çè ø

1

4

4

x 1 2 3 4 5 y –1 2 –3 4 –5 If the derivative of y(x) is approximated as

10. The value of log 21/3 from

is exact for polynomials of degree as high as possible are (a)  a = 2 , b = 1 (b)  a = 3 , b = 1

é 5 2 1 ù é x1 ù é 13 ù ê úê ú ê ú ê -2 5 2 ú ê x2 ú = ê -22ú êë -1 2 8 úû êë x3 úû êë 14 úû

with the initial guess of the solution T

by the trapezoidal

0.1

é x ( 0 ) , x ( 0 ) , x ( 0 ) ù = é1,1,1ùT,  2 3 ú ë û ê 1 ë

û

the

approximate T

value of the solution é x1(1) , x2(1) , x3(1) ù after one ê ú ë

û

rule and Simpson’s 1/3rd Rule (taking h = 0.1, iteration by the Gauss–Seidel method is correct to 5-decimal places) are, respectively (a)  [2, – 4.4, 1.625]T (b)  [2, –4, – 3]T (a)  2.32445, 2.38541 (b)  1.25034, 1.00350 (c)  [2, 4.4,1.625]T (d)  [2, – 4, 3]T (c)  1.38934, 1.38858 (d)  none of these 18. The following system of equations by the LU-­ 12. If a quadrature formula 3 f æç - 1 ö÷ + K f æç 1 ö÷ + 1 f (1) factorization method has a solution: 3 æ 1ö æ1 ö 1 f ç - ÷ + K f ç ÷ + f (1) 2 è 3ø è3ø 2

2 è 3ø

that approximates

ò

1

-1

f ( x ) dx

è3ø

2

is found

to be exact for quadric polynomials, then the value of K is (a) 2 (b) 1 (c) 0 (d)  – 1 13. Consider the quadrature formula ò

1

-1

x f ( x ) dx =

1 é f ( x0 ) + f ( x1 ) ùû 2ë

where x0 and x1 are quadrature points. Then, the highest degree of the polynomial, for which the above formula of exact, equals (a) 1 (b) 2 (c) 3 (d) 4 14. The values of the constants a, b for which the quadratic formula

EMEP.CH07_3PP.indd 407

ò

1

0

f ( x ) dx = af ( 0 ) + bf ( x1 )

8x1 – 3x2 + 2x3 = 20 4x1 + 11x2 – 2x3 = 33 6x1 + 3x2 + 12x3 = 36 (a)  3, 3, 2 (b)  2, 1, 3 (c)  3, 2, 1 (d)  1, 2, 3 19. Using the Euler’s method taking step size = 0.1, the approximate value of y obtained corresponding to x = 0.2 for the initial value problem dy = x 2 + y2 and y(0) = 1, is dx

(a) 1.322 (b) 1.122 (c) 1.222 (d) 1.110 dy y x 20. Given with initial condition y = 1 = dx

y+x

at x = 0, then the value of y for x = 0.1 by the Euler’s method (correct up to 4 decimal places, taking step length h = 0.02) is (a) 2.0540 (b) 1.0528 (c) 1.0928 (d) 2.0345

8/3/2023 6:23:22 PM

408 • Engineering Mathematics Exam Prep 21. Given dy + y = 1 , y (1) = 1 . Then the value of 2 dx

x

x

y(1.2) by the modified Euler’s method correct up to 4 decimal places, is (a) 0.9858 (b) 1.0528 (c) 1.8776 (d) 0.0333 22. The value of y(1.1) using the Runge– Kutta method of fourth order, given that dy = y2 + xy, y(1) = 1 is



1. (a) 6. (a) 11. (c) 16. (d) 21. (a)

2. (c) 7. (a) 12. (c) 17. (a) 22. (d)

(b) 1.0528 (d) 1.2415 Answer key 3. (a) 8. (c) 13. (a) 18. (c)



4. (b) 9. (b) 14. (c) 19. (c)

5. (d) 10. (a) 15. (c) 20. (c)



= 3.14156 - 3.14



Relative error, E = xT - x A = 0.00156 r



3.14156

= 4.966 × 10 2. (c) True value, xT = 4 , approximate value, 3 xA = 1.333. Therefore, percentage error

=

xT - x A xT

4 - 1.3333 3 ´ 100 = ´ 100 = 4 3

0.0025.

4. (b) We have Df ( x ) = f ( x + h ) - f ( x ) and Ñf ( x ) = f ( x ) - f ( x - h ) .

\ D.Ñf ( x )

= D éë f ( x ) - f ( x - h ) ùû = Df ( x ) - Df ( x - h )



= Df ( x ) - éë f ( x ) - ( x - h ) ùû



\ D.Ñ º D - Ñ

= Df ( x ) - Ñf ( x ) = ( D - Ñ ) f ( x )

5. (d) We have, D log f ( x ) = log f ( x + h ) - log f ( x ) (h being the  space length)

EMEP.CH07_3PP.indd 408

Df ( x ) ïü ïì = log í1 + ý f ( x ) ïþ îï

æ D2 x ö Ee x e ÷ 2 x çç ÷D e è E ø x +h

x +h

) De e = ( D e ) De e 2 x -h

2 x

2 x

= e -h D 2e x

= e–h.ex.eh = ex,h being the space length.

e x .eh D2e x

7. (a) Df(3) = f(4) – f(3)

= 0.00156

ìï Df ( x ) + f ( x ) üï = log í ý f (x ) îï þï

éë\ Df ( x ) = f ( x + h ) - f ( x ) ùû

(



1. (a) Here, true value, xT = 3.14156 and approximate value, xA = 3.14. \  Absolute error, Ea = xT - x A

xT

f (x )

= D2 E -1e x

Explanation

–4

f (x + h)

6. (a) We have,

dx

(a) 1.4561 (c) 1.8776

= log

Þ f ( 4 ) = f ( 3 ) + Df ( 3 )

{

}

= f ( 3 ) + D f ( 2 ) + Df ( 2 ) ëé Df ( 2 ) = f ( 3 ) - f ( 2 ) ûù = f ( 3 ) + Df ( 2 ) + D2 f ( 2 )

{

}

= f ( 3 ) + Df ( 2 ) + D2 f (1) + Df (1)

= f ( 3 ) + Df ( 2 ) + D f (1) + D f (1) Thus, f ( 4 ) = f ( 3 ) + Df ( 2 ) + D2 f (1) + D3 f (1) 2

3

8. (c) Since we are given four values, therefore, we take f (x) to be a polynomial of degree 3 in x so that D4f(x) = 0 for all values of x. Now D4f(x) = 0 ⇒ (E – 1)4 f(x) = 0 Þ E 4 f ( x ) - 4 E 3 f ( x ) + 6E 2 f ( x ) - 4 Ef ( x ) + f ( x ) = 0  ⇒  f(x + 4) – 4f(x + 3) + 6f(x + 2) – 4f(x + 1) + f(x) = 0 …(i) Putting x = 0 in (i), we get f ( 4 ) - 4 f ( 3 ) + 6 f ( 2 ) - 4 f (1) + f ( 0 ) = 0 or  f ( 4 ) - 4 ´ 15 + 6 ´ 8 - 4 f (1) + 0 = 0

or,  f ( 4 ) - 4 f (1) = 12 …(ii) Again, putting x = l in (i), we get f ( 5 ) - 4 f ( 4 ) + 6 f ( 3 ) - 4 f ( 2 ) + f (1) = 0 or,  35 - 4 f ( 4 ) + 6 ´ 15 - 4 ´ 8 + f (1) = 0

or,  4 f ( 4 ) - f (1) = 93 …(iii)

Solving (ii) and (iii) we get, f (1) = 3, f ( 4 ) = 24 .

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Numerical Analysis • 409

9. (b) x 1 2 3 4 5 y –1 (y0) 2(y1) –3(y2) 4 (y3) –5(y4) Here, Dy1 = y2 - y1 = -3 - 2 = -5,



= (4 - ( -3)) - ( -3 - 2) = 12,

3

D y1 = D2 y2 - D2 y1 = ( Dy3 - Dy2 ) - D 2 y1 2

= ( y4 - y3 ) - ( y3 - y2 ) - D y1



= ( -5 - 4) - (4 + 3) - 12 = -28. Now, x1 = x0 + h = 1 + 1 = 2.



\ y¢ ( x1 ) =

1æ 1 1 ö Dy1 + D2 y1 - D3 y1 ÷ 2 4 h çè ø

Þ y¢ ( 2 ) =

1é 1 1 ù -5 + ´ 12 - ( -28 ) ú = 8. 1 êë 2 4 û

f (x ) x2 = 1 + x3

1

x2 dx 1 + x3 0



= 0.23108







ò

(

)

= log21/3.

x x0 = 0.1 x1 = 0.2 x2 = 0.3 x3 = 0.4 x4 = 0.5 x5 = 0.6 x6 = 0.7

f(x) y0 = 1.30517 y1 = 1.62140 y2 = 1.94986 y3 = 2.29182 y4 = 2.64872 y5 = 3.02212 y6 = 3.41375

x

+ 2x ) dx

= 1.3885813  1.38858.

ò

1

-1

3 æ 1ö +K f 2 çè 3 ÷ø

f ( x ) dx =

æ1 ö 1 f ç ÷ + f (1) …(i) è3ø 2

3 1 ´1 + K ´1 + ´1 2 2 3 1 Þ2= +K + 2 2 Þk=0

ò

1

1 dx =

-1

ò

1

-1

x f ( x ) dx =

Þ-

-

1 1 = éêlog 1 + x 3 ùú = log 2 ë û 3 3 0

ò (e

ò

0

-1

1 é f ( x0 ) + f ( x1 ) ùû 2ë

ò

x f ( x ) dx +

1

0

x f ( x ) dx =

1 é f ( x0 ) + f ( x1 ) ùû 2ë

…(i) For f(x) = 1, (i) gives,

x2 dx 1 + x3 0

1

0.7

For f(x) = 1, (i) gives,

1

Therefore, log 21/ 3  0.23108 11. (c) Here, f ( x ) = e x + 2x , a = 0.1,b = 0.7, h = 0.1

EMEP.CH07_3PP.indd 409

Again, by Simpson’s 1/3rd rule,

13. (a)

0.25 é0 + 0.5 + 4 ( 0.06154 + 0.39560 ) + 2 ´ 0.22222 ûù = 3 ë

Now,

= 1.389338  1.38934.

h é(y0 + y6 ) + 4(y1 + y3 + y5 ) + 2(y2 + y4 ) ùû 3ë 0.1 = é(1.30517 + 3.41375) + 4(1.62140 + 2.29182 3 ë +3.02212) + 2(1.94986 + 2.64872) ùû

ò

h  éë y0 + y4 + 4 ( y1 + y3 ) +2 y2 ùû 3

+ 2x ) dx

=



Here, h = 0.25. \  By Simpson’s 1/3rd rule we get,

x

h = éë(y0 + y6 ) + 2(y1 + y2 + y3 + y4 + y5 ) ùû 2 0.1 = é(1.30517 + 3.41375) + 2(1.62140 + 1.94986 2 ë +2.29182 + 2.64872 + 3.02212) ùû

12. (c) We have, 0 0.25 0.50 0.75 1 0 0.06154 0.22222 0.3956 0.5 (y0) (y1) (y2) (y3) (y4)

ò (e

0.1

10. (a) x

0.7

0.1

D2 y1 = Dy2 - Dy1 = (y3 - y2 ) - (y2 - y1 )



Then by the trapezoidal rule,

ò

0

-1

x dx + 0

ò

1

0

x dx = 1

1 é1 + 1ùû 2ë

é x2 ù é x2 ù 1 1 Þ -ê ú + ê ú =1 Þ + =1 Þ1 =1 2 2 2 2 ëê ûú -1 ëê ûú 0

For f(x) = x, (i) gives, -

ò

0

-1

x 2dx +

é x3 ù Þ -ê ú ëê 3 ûú

0

ò

1

0

x 2 dx = 1

1 é -1 + 1ùû 2ë

é x3 ù 1 1 + ê ú = 0 Þ - + = 0 Þ 0 = 0. 3 3 ëê 3 ûú

0 -1 2 For f(x) = x , (i) gives,

-

ò

0

-1

x 3dx + 0

ò

1

0

x 3 dx = 1

1 é1 + 1ûù 2ë

é x4 ù é x4 ù 1 1 1 Þ - ê ú + ê ú = 1 Þ + = 1 Þ = 1, 4 4 2 ëê 4 ûú -1 ëê 4 ûú 0



a contradiction.

\  Highest degree of f(x) is 1.

8/3/2023 6:23:27 PM

410 • Engineering Mathematics Exam Prep

14. (c) Given that,

ò

1

0

Þ 5x1 + 2x2 + x3 = 13,

f ( x ) dx = a f ( 0 ) + b f ( x1 ) …(i)

When f(x) = 1, we get from (i),

- 2x1 + 5x2 + 2x3 = -22,



1

ò0 Þ 1 = a + b …(ii) When f(x) = x, we get from (i), 1dx = a ´ 1 + b ´ 1

ò

1

0



x dx = a ´ 0 + b ´ x1 1

é x2 ù 1 Þ ê ú = bx1 Þ = bx1 …(iii) 2 2 ëê ûú 0



2

When f(x) = x , we get from (i), Þ



ò

1

0

x 2dx = a ´ 0 + b ´ x12 1



é x3 ù 1 Þ ê ú = bx12 Þ = bx12 …(iv) 3 3 ëê ûú 0



Now, dividing Eq. (iv) by Eq. (iii), we get, bx12 1 3 2 = i.e; x1 = . bx1 1 2 3

From Eq. (iii), we get, 1 = b ´ 2 i.e; b = 3 .



\  From Eq. (ii), we get,



a =1-b =1-

2

Thus

3

4

3 1 = . 4 4 1 3 a= ,b= . 4 4

15. (c) Let f(x) = xex – 1. Now, f ( 0 ) = -1 < 0 and f (1) = e - 1 > 0. Thus, the exact root lies in (0, 1).  Hence, first approximation of the root, 0 +1 x1 = = 0.5 .



0.5 + 1 1.5 = 2 2

f ( x0 ) f (1) ( -1) = 0. x1 = x0 =1=1f ¢ ( x0 ) f ¢ (1) ( -1)





x2( ) = -



EMEP.CH07_3PP.indd 410

1

x3( ) = 1



22 2 1 + ´ 1 - ´ 1 = 4.4, = 5 5 5

– 4.4

14 1 2 + ´ 1 - ´ 1 = 1.625. 8 8 8

T 1 1 1 \ é x1( ) x2( ) x3( ) ù = éë2.0, - 4.4, 1.625 ùû . úû ëê

18. (c) The given system of equations can be ­written as AX = b é8 -3

2ù

é20 ù

é x1 ù

12 úû

êë36 úû

êë x3 úû

where A = ê4 11 -1ú , b = ê33 ú , X = ê x ú ê ú ê ú ê 2ú êë6

3

Then, A = LU é8 -3 2 ù é l11 0 ê 11 -1ú = êl ê ú ê 21 l22 êë6 3 12 úû êël31 l32

0 ù é1

ú ú l33 úû

Þ 4

ê ê êë0

0 ´ 0



l11u12 é8 -3 2 ù é l11 ê ú ê Þ 4 11 -1 = l21 l21u12 + l22 ê ú ê êë6 3 12 úû êël31 l31u12 + l32



3 \ l11 = 8, l11u12 = -3 Þ u12 = - , 8



l11u13 = 2 Þ u13 =

2 1 = , l 8 4 21

u12

u13 ù

ú ú 1 úû

1

u23

0

l11u13

ù ú l21u13 + l22u23 ú l31u13 + l32u23 + l33 úû

= 4,

l21u12 + l22 = 11

17. (a)

é 5 2 1 ù é x1 ù é 13 ù ê úê ú ê ú ê -2 5 2 ú ê x2 ú = ê -22ú êë -1 2 8 úû êë x3 úû êë 14 úû é 5x1 + 2x2 + x3 ù é 13 ù ê ú ê ú Þ ê -2x1 + 5x2 + 2x3 ú = ê -22 ú êë -x1 + 2x2 + 8x3 úû êë 14 úû

û

(0)



= 0.75.

16. (d) We have, x0 = 1, f(x) = x2 – 3x + 1, f′(x) = 2x –3. \  f(1) = 1 – 3 + 1 = – 1, f′(1) = 2 – 3 = –1. Then by the Newton–Raphson method,

(0)

13 2 1 1 \ x1( ) = - ´ 1 - ´ 1 = 2, 5 5 5

So second approximation of the root, x2 =

ë

(0)

\  x1 = x2 = x3 = 1.

Then, f ( x1 ) = ( 0.5 ) e0.5 - 1 = -0.175 < 0.



(0) (0) (0) Now, the initial guess is éê x1 , y2 , z3 ùú = éë1 1 1ùûT .



2



13 2 1 - x 2 - x3 , 5 5 5 22 2 2 + x1 - x3 , x2 = 5 5 5 14 1 2 + x1 - x2 . x3 = 8 8 8

Þ x1 =





- x1 + 2x2 + 8x3 = 14.



æ 3 ö 25 Þ l22 = 11 - l21u12 Þ l22 = 11 - 4 ç - ÷ = , è 8ø 2

l21u13 + l22u23 = -1 Þ 4 ´



1 4

Þ u23 = -

+

25 2

u23 = -1

4 , 25

l31 = 6, l31 u12 + l32 = 3

8/3/2023 6:23:31 PM

Numerical Analysis • 411 21 æ 3ö Þ 6 ç - ÷ + l32 = 3 Þ l32 = , 4 è 8ø



l31u13 + l22u23 + l33 = 12



567 1 21 æ -4 ö . Þ 6´ + ´ + l33 = 12 Þ l33 = 50 4 4 çè 25 ÷ø

Hence,

é ê8 ê L = ê4 ê ê ê6 êë

0 25 2 21 4

3 é ù ê1 - 8 0 ú ê ú 0 ú , U = ê0 1 ê ú ê 567 úú ê0 0 êë 50 úû



1 ù 4 ú ú -4 ú 25 ú ú 1 ú úû

Now, AX = b

⇒  LUX = b ⇒  LY = b 

…(i)

where UX = Y 3 é ê1 - 8 ê i.e; ê0 1 ê ê ê0 0 êë



1 ù 4 ú éx ù é y ù ú 1 1 -4 ú ê ú ê ú x = y 2 2 25 ú ê ú ê ú ú êë x3 úû êë y3 úû 1 ú úû

3 1 ù é ê x1 - 8 x2 + 4 x3 ú ê ú é y1 ù 4 ê ú i.e; ê x2 x3 ú = ê y2 ú ê ú 25 ê ú êë y3 úû x3 ê ú ëê ûú

Þ y3 =

50 é 21 ´ 23 ù = 1. 36 - 15 567 êë 50 úû

Then from (ii), we have



This gives x3 = 1,



3 1 ù é5 ù é ê x1 - 8 x2 + 4 x3 ú ê 2 ú ê ú ê ú ê x - 4 x ú = ê 46 ú . 2 3 ê ú ê 25 ú 25 ê ú ê ú x3 ê ú ê1 ú êë úû ëê ûú

x2 -

4 46 46 4 x3 = Þ x2 = + = 2, 25 25 25 25

x1 -

x 3 5 5 3 1 x2 + 3 = Þ x1 = + ´ 2 - = 3. 8 4 2 2 8 4

19. (c) We have, dy = x 2 + y2 , x0 = 0, y0 = 1, h = 0.1 dx

…(ii)

From (i) we get,



é ê8 ê ê4 ê ê ê6 êë



é ù ê ú 8 y1 ê ú é20 ù 25 ê ú = ê33 ú or, 4 y1 + y2 ê ú ê ú 2 ê ú êë36 úû 21 567 ê6 y1 + y2 + y3 ú êë úû 4 50 25 21 567 or, 8 y1 = 20,4 y1 + y2 ,6 y1 + y2 + y3 = 36. 2 4 50



5 Now, 8 y1 = 20 Þ y1 = , 2



25 4 y1 + y2 = 33 Þ y2 = æç 33 - 4 ´ 5 ö÷ 2 = 46 , 2 2 ø 25 25 è

EMEP.CH07_3PP.indd 411

21 567 y2 + y3 = 36 4 50







0 25 2 21 4

6 y1 +

ù 0 ú ú é y1 ù é20 ù ê ú ê ú 0 ú ê y2 ú = ê33 ú ú ê y ú ê36 ú 567 úú ë 3 û ë û 50 úû



\  f(x, y) = x2 + y2. By the Euler’s method, y1 = y(x1) = y0 + hf(x0, y0) = 1 + 0.1(02 + 12) = 1.1, y2 = y(x2) = y(0.2) = y1 + h f(x1, y1) = 1.1 + 0.1 [(0.1)2 + (1.1)2] = 1.1 + 0.1(0.01 + 1.21) = 1.222. 20. (c)

Here, f ( x , y ) = y - x , x0 = 0, y0 = 1 and h = 0.02. y+x



\  By the Euler’s method, we get



y1 = y0 + h f(x0, y0) = 1 + 0.02 æç 1 - 0 ö÷ = 1.02.



y(0.04) = y2 = y1 + h f(x1, y1)



æ 1.02 - 0.02 ö = 1.02 + 0.02 ç ÷ è 1.02 + 0.02 ø

è1 + 0 ø

= 1.039231.

Similarly,

y ( 0.06 ) = y3 = y2 + h f ( x2 , y2 )





æ 1.039231 - 0.04 ö = 1.039231 + 0.02 ç ÷ è 1.039231 + 0.04 ø

= 1.057748

y ( 0.08 ) = y4 = y3 + h f ( x3 , y3 ) æ 1.057748 - 0.06 ö = 1.057748 + 0.02 ç ÷ è 1.057748 + 0.06 ø = 1.075601

8/3/2023 6:23:33 PM

412 • Engineering Mathematics Exam Prep .

y ( 0.10 ) = y5 = y4 + h f ( x4 , y4 ) æ 1.075601 - 0.08 ö = 1.075601 + 0.02 ç ÷ è 1.075601 + 0.08 ø = 1.092832  1.0928

21. (a) Here,

1 y f ( x , y ) = 2 - , x0 = 1, y0 = 1. x x

0 \ y1( ) = y0 + h f ( x0 , y0 ) = 1 + 0.1 ´ (1 - 1) = 1.

The modified Euler’s iteration formula is yr( n ) = yr -1 +

hé f ( xr -1 , yr -1 ) + f ( xr , yr( n -1) ) ù û 2ë

)

(

h 1 0 \ y1( ) = y0 + é f ( x0 , y0 ) + f x1 , y1( ) ù úû 2 êë é ù ì 1 0.1 ê 1 üïú ï 1 - 1) + í ( ý 2 2 ê 1.1 ïú ïî (1.1) þû ë

= 1 + 0.05(–0.08264) = 0.99587,



h 2 1 y1( ) = y0 + é f ( x0 , y0 ) + f x0 , y1( ) ù ûú 2 ëê

(

)

é ù æ 1 0.1 ê 0.99587 ö÷ ú 1 - 1) + ç =1+ ( ç (1.1)2 2 ê 1.1 ÷ ú è øû ë

= 1 + 0.05 (–0.078888) = 0.99606,

(

)

h 3 1 y1( ) = y0 + é f ( x0 , y0 ) + f x0 , y2( ) ù úû 2 êë é ù æ 1 0.99606 ö÷ ú = 1 + 0.05 ê(1 - 1) + ç ê ç (1.1)2 1.1 ÷ ú è øû ë

= 1 + 0.05(–0.079063) = 0.99607. Hence, y1 = y(1.1) = 0.9961. Thus, x1 = 1.1, y1 = 0.9961. \ f ( x1 , y1 ) =

1

2

-

0.9961 = -0.079. 1.1

( ) Now, y2(0) = y1 + h(x1, y1) = 0.9961 + 0.1 × (–0.079) = 0.98819 Then we have, y2( ) = y1 + 1



EMEP.CH07_3PP.indd 412

1.1

(

)

hé 0 f ( x1 , y1 ) + f x2 , y2( ) ù úû 2 êë

é 1 0.98819 ùú = 0.9961 + 0.05 ê -0.079 + 1.2 ú ê (1.2 )2 ë û

= 0.98569,

)

(

é 1 0.98569 ùú = 0.9961 + 0.05 ê -0.079 + 1.2 ú ê (1.2 )2 ë û

= 0.98580,

)

(

h 3 2 y2( ) = y1 + é f ( x1 , y1 ) + f x2 , y2( ) ù ûú 2 ëê é 1 0.98580 ùú = 0.9961 + 0.05 ê -0.079 + 2 1.2 ú ê (1.2 ) ë û

= 0.985797 Thus, y2  0.9858 , correct up to four decimal places. \ y (1.2 ) = y2  0.9858 . 22. (d) Here, f(x, y) = y2 + xy, x0 = 1, y0 = 1. Let us take h = 0.1. Then iterative formula of the Runge–Kutta method of order 4 gives: y1 = y0 +







Let us take h = 0.1 so that x1 = 1 + 0.1 = 1.1.

=1+





h 2 1 y2( ) = y1 + é f ( x1 , y1 ) + f x2 , y2( ) ù ûú 2 ëê

1

( k1 + 2k2 + 2k3 + k4 )

6 where k1 = hf(x0, y0) = 0.1(12 + 1 × 1) = 0.2,



1 1 ö æ k3 = hf ç x0 + h, y0 + k2 ÷ 2 2 ø è



= 0.1 × {(1.1)2 + 1.05 × 1.1} = 0.2365,





= 0.1 × f(1.05, 1.1)

1 1 ö æ k3 = hf ç x0 + h, y0 + k2 ÷ 2 2 ø è

= 0.1 × f(1.05, 1.11825) = 0.1 × {(1.11825)2 + 1.11825 × 1.05} = 0.2425, k4 = hf(x0 + h, y0 + k3) = 0.1 × f(1.1, 1.2425) = 0.1 × {(1.2425)2 + 1.1 × 1.2425} = 0.2910556. \ y (1.1) = y1 = y0 + =1+

1 ( k1 + 2k2 + 2k3 + k4 ) 6

1 ( 0.2 + 2 ´ 0.2365 + 2 ´ 0.2425 + 0.2910556 ) 6

 1.2415

(correct up to four decimal places)

PREVIOUS YEARS SOLVED PAPERS (2000-2018)

1. The trapezoidal rule for integration gives exact result when the integrand is a polynomial of degree (a)  but not 1 (b)  1 but not 0 (c)  0 or 1 (d) 2  [CS GATE 2002] 2. The accuracy of Simpson’s rule quadrature for a step size h is

8/3/2023 6:23:40 PM

Numerical Analysis • 413

(a) O(h2) (b) O(h3) (c) O(h4) (d) O(h5)  [ME GATE 2003] 3. Given a > 0, we wish to calculate the recipro1 cal value by the Newton–Raphson method a

for f(x) = 0. (i) The Newton–Raphson algorithm for the function will be æ ö (a) x k +1 = 1 ç x k + a ÷ 2è

xk ø

a 2

æ ö (b) x k +1 = ç xk + xk2 ÷ è

ø

(c) xk+1 = 2xk – axk2 (d) x k +1 = x k - a x k2

(a) Coulomb’s theorem (b) Newton–Raphson method (c) Euler method (d) Stokes’ theorem  [PI GATE-2005] 7. The real root of the equation xex = 2 is evaluated using the Newton–Raphson method. If the first approximation of the value of “x” is 0.8679, the second approximation of the value of “x” correct to three decimal places is (a) 0.865 (b) 0.853 (c) 0.849 (d) 0.838  [PI GATE-2005] 8. A second degree polynomial, f(x) has values of 1, 4, and 15 at x = 0, 1, and 2, respectively. 2

2

(ii) For a = 7 and starting with x0 = 0.2, the first two iterations will be (a) 0.11, 0.1299 (b) 0.12, 0.1392 (c) 0.12, 0.1416 (d) 0.13, 0.1428  [CE GATE 2005] 4. Match List-l with List-ll and select the correct combination: List-l A. Newton–Raphson method B. Runge–Kutta method equations C. Simpson’s 1/3rd rule D. Gauss elimination List-ll 1. Solving nonlinear equations 2. Solving linear simultaneous equations 3. Solving ordinary differential equations 4. Numerical integration method 5. Interpolation 6. Calculation of eigenvalues Codes: (a) A-6, B-1, C-5, D-3 (b) A-1, B-6, C-4, D-3 (c) A-1, B-3, C-4, D-2 (d) A-5, B-3, C-4, D-1  [EC GATE-2005] 5. Starting from x0 = 1, one step of the Newton– Raphson method in solving the equation x3 + 3x – 7 = 0 gives the next value x1 as (a) x1 = 0.5 (b) x1 = 1.406 (c) x1 = 1.5 (d) x1 = 2  [ME GATE 2005] 6. For solving algebraic and transcendental equations which one of the following is used?

EMEP.CH07_3PP.indd 413

The integral f ( x )dx is to be estimated by apò 0

plying the trapezoidal rule to this data. What is the error in the estimate? (a) - 4 (b) - 2 3

(c) 0

(d)

2 3

3

[CE GATE 2006]



9. The differential equation dy = 0.25 y2 is to be dx

solved using the backward (implicit) Euler’s method with the boundary condition y = 1 at x = 0 and with a step size of 1. What would be the value of y at x = 1? (a) 1.33 (b) 1.67 (c) 2.00 (d) 2.33  [CE GATE 2006] 10. The following equation needs to be numerically solved using the Newton–Raphson method: x3 + 4x – 9 = 0 The iterative equation for this purpose is (k indicates the iteration level) 3

2

(a) x k +1 = 2x k2 + 9 (b) x k +1 = 3x k2 + 4 3x k + 4

3x k + 9

2

(c) xk+1 = xk – 3x k + 4 2

(d) x k +1 = 4 x k + 3 2 9x k + 2

[CE GATE 2007] 11. Given that one root of the equation  x3 – 10x2 + 31x – 30 = 0 is 5, the other two roots are (a) 2 and 3 (b) 2 and 4 (c) 3 and 4 (d) –2 and –3  [CE GATE 2007] 

8/3/2023 6:23:42 PM

414 • Engineering Mathematics Exam Prep 12. The equation x3 – x2 + 4x – 4 = 0 is to be solved using the Newton–Raphson method. If x = 2 is taken as the initial approximation of the ­solution, then the next approximation using this method will be (a) 2 (b) 4 3



(c) 1

(d)

3 3 2

[EC GATE 2007]



13. Consider the series xn +1

x 9 = n + , x0 = 0.5 2 8xn

obtained from the Newton–Raphson method. The series converges to (a) 1.5 (b) 2 (c) 1.6 (d) 1.4  [CS GATE 2007] 14. Identify the Newton–Raphson iteration scheme for finding the square root of 2 (a) xn +1

1æ 2 ö 1æ 2 ö (b) xn +1 = ç xn = ç xn + ÷ ÷ 2 x 2è xn ø n ø è

(c) xn +1 = 2 (d) xn +1 = 2 + xn 

18. It is known that two roots of the non-linear equation x3 – 6x2 + 11x – 6 = 0 are 1 and 3. The third root will be (a) j (b) –j (c) 2 (d) 4  [IN GATE 2008] 19. The Newton–Raphson iteration xn +1 =

imal place. The value of

2p

ò sin xdx 0

when evalu-

ated using this calculator by trapezoidal method with 8 equal intervals, to 5 significant digits is (a) 0.00000 (b) 1.0000 (c) 0.00500 (d) 0.00025  [ME GATE 2007] 16. Match List-l with List-ll and select the correct combination: List-l P. Second order differential equations Q. Non-linear algebraic equations R. Linear algebraic equations S. Numerical integration List-ll 1. Runge–Kutta method 2. Newton–Raphson method 3. Gauss elimination 4. Simpson’s rule Codes: (a) P-3, Q-2, R-4, S-1 (b) P-2, Q-4, R-3, S-1 (c) P-1, Q-2, R-3, S-4 (d) P-1, Q-3, R-2, S-4  [PI GATE 2007]

1æ Rö ç xn + ÷ 2è xn ø

can be used to compute the

(a) square of R (b) reciprocal of R (c) square root of R (d) logarithm of R  [CS GATE 2008] 20. The recursion relation to solve x = e–x using the Newton–Raphson method is

xn

[IN GATE 2007] 15. A calculator has accuracy up to 8 digits after dec-

EMEP.CH07_3PP.indd 414

17. Equation ex – 1 = 0 is required to be solved using Newton’s method with an initial guess x0 = – 1. Then, after one step of Newton’s method, estimate x1 of the solution will be given by (a) 0.71828 (b) 0.36784 (c) 0.20587 (d) 0.00000  [EE GATE 2008]

- xn (a) xn +1 = e (b) xn +1 = xn - e - xn

- xn (c) xn +1 = (1 + xn ) e - xn

1+e

(d) xn +1 =

xn2



-e

- xn

(1 + xn ) - 1

x n - e - xn

[EC GATE 2008]

21. The minimum number of equal length subintervals needed to approximate

2

ò xe dx x

to

1

an accuracy of at least 1 ´ 10-6 using the trap3 ezoidal rule is (a) 1000e (c) 100e 

(b) 1000 (d) 100 [CS GATE 2008]

22. A differential equation dx = e -2tu(t ) has to be dt

solved using trapezoidal rule of integration with a step size h = 0.01 sec. Function u(t) indicates a unit step function. If x(0) = 0, then the value of “x” at t = 0.01 sec will be given by (a) 0.00099 (b) 0.00495 (c) 0.0099 (d) 0.0198  [EE GATE 2008]

8/3/2023 6:23:44 PM

Numerical Analysis • 415

23. Let x2 – 117 = 0. The iterative steps for the solution using the Newton–Raphson method is given by

(a) x k +1 = 1 æç x k + 117 ö÷ 2è

(a) 542 (c) 1444 

29. The square root of a number N is to be ­obtained by applying the Newton–­Raphson iterations to the equation x2 – N = 0. If i ­denotes the iteration index, the correct iterative scheme will be

xk ø

(b) x k +1 = x k - 117 xk

(c) x k +1 = x k - x k

117

æ ö (a) xi +1 = 1 æç xi + N ö÷ (b) xi +1 = 1 çç xi2 + N2 ÷÷

(d) x k +1 = x k - 1 æç x k + 117 ö÷ 2è

2è

xk ø

[EE GATE-2009] 24. During the numerical solution of a first-order differential equation using the Euler (also known as Euler-Cauchy) method with step size h, the local truncation error is of the order of (a) h2 (b) h3 (c) h4 (d) h5  [PI GATE 2009] 25. The table below gives values of a function F(x) obtained for values of x at intervals of 0.25: x 0 0.25 0.50 0.75 1 F(x) 1 0.9412 0.8 0.64 0.5 The value of the integral of the function between the times 0 to 1 using Simpson’s rule is (a) 0.7854 (b) 2.3562 (c) 3.1416 (d) 7.5000  [CE GATE 2010] 26. The Newton–Raphson method is used to compute a root of the equation x2 – 13 = 0 with 3.5 as the initial value. The approximation after one iteration is (a) 3.575 (b) 3.677 (c) 3.667 (d) 3.607  [CS GATE 2010] 27. Consider a differential equation dy - y = x 

dx

with the initial condition y(0) = 0. Using Euler’s first-order method with a step size of 0.1, the value of y(0.3) is (a) 0.01 (b) 0.031 (c) 0.0631 (d) 0.1  [EC GATE-2010] 28. Torque exerted on a flywheel over a cycle is listed in the table. Flywheel energy (in J per unit cycle) using Simpson’s 1/3rd rule is Angle 0 60 120 180 240 300 360 (degree) Torque 0 1066 –323 0 323 –355 0 (Nm)

EMEP.CH07_3PP.indd 415

(b) 993 (d) 1986 [ME GATE 2010]

2è

xi ø 2

xi ø

(c) xi +1 = 1 ç xi + N ÷ (d) xi +1 = 1 æç xi + N ö÷ ç ÷ æ

2è

ö

2è

xi ø



xi ø

[CE GATE 2011]

30. Solution of the variables x1 and x2 for the following equations is to be obtained by employing the Newton–Raphson iterative method: Equation (i) 10x2 sin x1 – 0.8 = 0 Equation (ii) 10x22 – 10x2 cos x1 – 0.6 = 0 Assuming the initial values x1 = 0.0 and x2 = 1.0, the Jacobian matrix is (a) éê10 - 0.8 ùú (b) éê10 ë0

0ù ú 10 û

(c) éê 0

é10 ê ë10

0ù ú - 10 û

ë0



- 0.6 û

- 0.8 ù ú (d) - 0.6 û

ë10

[EE GATE 2011]

31. A numerical solution of the equation f(x) = x + x - 3 = 0 can be obtained using the Newton–Raphson method. If the starting value is x = 2 for the iteration, the value of x that is to be used in the next step is (a) 0.306 (b) 0.739 (c) 1.694 (d) 2.306  [EC GATE 2011] 32. The matrix ëé A ûù = éê2 1ùú is decomposed into ë4 - 1 û a product of a lower triangular matrix L and an upper triangular matrix U. The properly decomposed L and U matrices, respectively, are

(a) éê1

0ù ú 4 1 ë û

and éê1

(b) éê2

and

0ù ú 4 1 ë û

1ù ú 0 2û ë 1ù é1 ê ú 0 1û ë

1ù é2 0ù é1 (c) ê ú and ê0 - 1ú 4 1 ë û ë û é2

(d) ê ë4 

0ù ú - 3û

é1

and ê ë0

0.5ù ú 1û

[EE GATE 2011]

8/3/2023 6:23:46 PM

416 • Engineering Mathematics Exam Prep 33. The integral

ò

31

1 rd

x

dx

, when evaluated by using

Simpson’s 1/3 rule on two equal subintervals each of length 1, equals (a) 1.000 (b) 1.098 (c) 1.111 (d) 1.120  [ME GATE 2011] 1.5 dx 34. The estimate of ò obtained using Simp0.5

x

son’s rule with three-point function evaluation exceeds the exact value by (a) 0.235 (b) 0.068 (c) 0.024 (d) 0.012  [CE GATE 2012] 35. The bisection method is applied to compute a zero of the function f(x) = x4 – x3 – x2 – 4 in the interval [1, 9]. The method converges to a solution after ______ iterations. (a) 1 (b) 3 (c) 5 (d) 7  [CS GATE 2012] 36. Match the correct pairs: Numerical Integration Order of Fitting scheme polynomial P. Simpson’s 3/8 Rule 1. First Q. Trapezoidal Rule 2. Second R. Simpson’s 1/3 Rule 3. Third (a) P-2, Q-1, R-3 (b) P-3, Q-2, R-1 (c) P-1, Q-2, R-3 (d) P-3, Q-1, R-2  [ME GATE 2013] 37. When the Newton–Raphson method is applied to solve the equation f(x) = x3 + 2x – 1 = 0, the solution at the end of the first iteration with the initial guess value as x0 = 1.2 is (a) –0.82 (b) 0.49 (c) 0.705 (d) 1.69  [EE GATE 2013] 38. Find the magnitude of the error (correct to two decimal places) in the estimation of the Integral

4

ò (x 0

4

)

+ 10 dx

using Simpson 1/3rd

rule [Take the step length as 1]  [CE GATE 2013] 39. While numerically solving the differential equation dy + 2xy2 = 0, y(0) = 1 using Euler’s dx

predictor-corrector (improved Euler-Cauchy) with a step size of 0.2, the value of y after the first step is

EMEP.CH07_3PP.indd 416

(a) 1.00 (b) 1.03 (c) 0.97 (d) 0.96  [IN GATE 2013] 40. Function f is known at the following points: x 0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 3.0 The value of

f(x) 0 0.09 0.36 0.81 1.44 2.25 3.24 4.41 5.76 7.29 9.00 3

ò f ( x ) dx

computed using the

0

trapezoidal rule is (a) 8.983 (b) 9.003 (c) 9.017 (d) 9.045  [CS GATE 2013] 41. Match the application to appropriate numerical method. Application: P1: Numerical integration P2: Solution to a transcendental equation P3: Solution to a system of linear equations P4: Solution to a differential equation Numerical method: M1: Newton–Raphson Method M2: Runge–Kutta Method M3: Simpson’s 1/3rd rule M4: Gauss Elimination Method (a) P1— M3, P2—M2, P3—M4, P4—M1 (b) P1— M3, P2—M1, P3—M4, P4—M2 (c) P1— M4, P2—M1, P3—M3, P4—M2 (d) P1— M2, P2—M1, P3—M3, P4—M4  [EC GATE 2014] 3 1 42. The definite integral ò dx is evaluated using 1

x

trapezoidal rule with a step size of 1. The correct answer is ______.  [ME GATE 2014] x 43. The function f(x) = e – 1 is to be solved using the Newton–Raphson method. If the initial value of x0 is taken as 1.0, then the absolute error observed at second iteration is ______.  [EE GATE 2014]

8/3/2023 6:23:47 PM

Numerical Analysis • 417

44. The real root of the equation 5x - 2cosx -1 = 0 (up to two decimal accuracy) is _______.  [ME GATE 2014] 4 45. The value of In( x )dx calculated using

ò

2.5

the trapezoidal rule with five subintervals is ______.  [ME GATE 2014] 46. In the Newton–Raphson method, an initial guess of x0 = 2 is made and the sequence x0, x1, x2..... is obtained for the function 0.75x3 – 2x2 – 2x + 4 = 0. Consider the statements (i) x3 = 0 (ii) The method converges to a solution in a finite number of iterations. Which of the following is TRUE? (a) Only (i) (b) Only (ii) (c) Both (i) and (ii) (d) Neither (i) nor (ii)  [CS GATE 2014] 47. Consider an ordinary differential equation dx = 4t + 4 . If x = x0 at t = 0, the increment dt

in x calculated using the Runge–Kutta fourthorder multi-step method with a step size of Dt = 0.2 is (a) 0.22 (b) 0.44 (c) 0.66 (d) 0.88  [ME GATE 2014] 48. Using the trapezoidal rule, and dividing the interval of integration into three equal sub+1 intervals, the definite integral |x |dx is ò 1 _______.  [ME GATE 2014] 49. With respect to the numerical evaluation of b

the definite integral K = x 2dx , where a and ò a

b are given, which of the following statements is/are TRUE? (i) The value of K obtained using the trapezoidal rule is always greater than or equal to the exact value of the definite integral. (ii) The value of K obtained using the Simpson’s rule is always equal to the exact value of the definite integral (a) (i) only (b) (ii) only (c) Both (i) and (ii) (d) Neither (i) nor (ii)  [CS GATE 2014] 50. A non-zero polynomial f(x) of degree 3 has roots x = 1, x = 2, and x = 3. Which one of the following must be true?

EMEP.CH07_3PP.indd 417

(a) f(0)f(4) < 0 (b) f(0) f(4) > 0 (c) f(0) + f(4) > 0 (d) f(0) + f(4) < 0  [CS GATE 2014] 2 51. If the equation sin x = x is solved by Newton–Raphson’s method with the initial guess of x = 1, then the value of x after second iteration would be ___?  [PI GATE 2014] 52. The iteration step in order to solve for the cube roots of a given number “N” using the Newton–Raphson method is (a) xk +1 = xk + 1 ( N - xk3 ) 3

æ ö (b) xk +1 = 1 ç 2xk + N2 ÷ ç ÷ 3è

xk ø

(c) xk +1 = xk - 1 ( N - xk3 ) 3

æ

ö

(d) xk +1 = 1 ç 2xk - N2 ÷ 3 çè x k ÷ø 

[IN GATE 2014]

53. In the LU decomposition of the matrix éê2 2 ùú , ë4

9û

if the diagonal elements of U are both 1, then the lower diagonal entry l22 of L is ______.  [CS GATE 2015] 54. The Newton–Raphson method is used to find the roots of the equation, x3 + 2x2 + 3x – 1 = 0. If the initial guess is x0 = 1, then the value of x after second iteration is _______.  [ME GATE 2015] 2 55. The quadratic equation x – 4x + 4 = 0 is to be solved numerically, starting with the initial guess x0 = 3. The Newton–Raphson method is applied once to get a new estimate and then the Secant method is applied once using the initial guess and this new estimate. The estimated value of the root after the application of the Secant method is _______.  [CE GATE 2015] 56. The Newton–Raphson method is used to solve the equation x3 – 5x2 + 6x – 8 = 0. Taking the initial guess as x = 5, the solution obtained at the end of the first iteration is_______.  [EC GATE 2015] 57. In the Newton–Raphson iterative method, the initial guess value (xini) is considered as zero while finding the roots of the equation: f(x) = – 2 + 6x – 4x2 + 0.5x3. The correction, Dx, to be added to xini in the first iteration is _______.  [CE GATE 2015]

8/3/2023 6:23:48 PM

418 • Engineering Mathematics Exam Prep 58. The values of function f(x) at 5 discrete points are given below: x 0 0.1 0.2 0.3 0.4 y = f(x) 0 10 40 90 160 Using trapezoidal rule step size of 0.1, the value of 

0.4

ò f ( x )dx

is _______.

0

[ME GATE 2015] 59. Using a unit step size, the value of integral 2

ò x ln x dx

by trapezoidal rule is _______.  [ME GATE 2015] rd 60. Simpson’s 1/3 rule is used to integrate the 1

function f ( x ) = 3 x 2 + 9 between x = 0 and x 5

5

= 1 using the least number of equal subintervals. The value of the integral is ______.  [ME GATE 2015] 61. For step-size, Dx = 0.4 , the value of following integral using Simpson’s 1/3rd rule is ______ 0.8



ò (0.2 + 25x - 200x

2

0

)

+ 675x 3 - 900x 4 + 400x 5 dx

[CE GATE 2015] 62. In numerical integration using Simpson’s 1/3rd rule, the approximating function in the interval is a (a) constant (b) straight line (c) cubic B-spline (d) parabola  [PI GATE 2015] 63. The velocity v in kilometer/minute) of a motorbike which starts from rest is given at fixed intervals of time t (in minutes) as follows: t 2 4 6 6 10 12 14 16 18 20

v 10 18 25 29 32 20 11 5 2 0

 The approximate distance (in kilometers) rounded to two places of decimals covered in 20 minutes using Simpson’s 1/3rd rule is _______.  [CS GATE 2015]

EMEP.CH07_3PP.indd 418

64. Solve the equation x = 10 cos(x) using the Newton–Raphson method. The initial guess is p x = . The value of the predicted root after the 4

first iteration, up to second decimal, is ______.  [ME GATE 2016] 65. The ordinary differential equation dx  = -3x + 2 , with x(0) = 1 is to be solved usdt

ing the forward Euler method. The largest time step that can be used to solve the equation without making the numerical solution unstable is ______.  [EC GATE 2016]. 66. Numerical integration using trapezoidal rule gives the best result for a single variable function, which is (a) linear (b) parabolic (c) logarithmic (d) hyperbolic  [ME GATE 2016] 67. Consider the first-order initial value problem dy

= y + 2x - x 2 , y ( 0 ) = 1(0 £ x < ¥ )

dx with exact solution y(x) = x2 + ex. For x = 0.1, the percentage difference between the exact solution and the solution obtained using a ­single iteration of the second-order Runge–Kutta method with step-size h = 0.1 is _______.  [EC GATE 2016]

68. The root of the function f(x) = x3 + x – 1 obtained after first iteration on application of the Newton–Raphson scheme using an initial guess of x0 = 1 is (a) 0.682 (b) 0.686 (c) 0.750 (d) 1.000  [ME GATE 2016] 69. The error in numerically computing the integral

p

ò (sin x + cos x )dx using

the trapezoidal

0



rule with three intevals of equal length between 0 and p is ______. [ME GATE 2016]

70. The Gauss Seidel method is used to solve the following equations (as per the given order): x1 + 2x2 + 3x3 = 5 2x1 + 3x2 + x3 = 1 3x1 + 2x2 + x3 = 3

8/3/2023 6:23:50 PM

Numerical Analysis • 419

Assuming initial guess as x1 = x2 = x3 = 0, the value of x3 after the first iteration is ______ .  [ME GATE 2016]

(a) 0.000171 (c) 0.000579 

71. The Newton–Raphson method is to be used to find foot of equation 3x – ex + sin x = 0. If the initial trial value of the roots is taken as 0.333, the next approximation for the root would be ______.  [CE GATE 2016]

Answer key 1. (c) 2. (c) 3. (i)-(c); (ii)-(b) 5. (c) 6. (b) 7. (b) 8. (a) 10. (a) 11. (a) 12. (b) 13. (a) 14. (a) 15. (a) 16. (c) 17. (a) 19. (c) 20. (c) 21. (a) 22. (c) 24. (a) 25. (a) 26. (d) 27. (b) 28. (b) 29. (a) 30. (b) 31. (c) 33. (c) 34. (d) 35. (b) 36. (d) 38. 0.53 39. (d) 40. (d) 41. (b) 43. 0.06 44. 0.54 45. 1.753. 46. (a) 48. 1.11 49. (c) 50. (a) 51. 0.88 53. 5 54. 0.3043 55. 2.33 56. 4.29 58. 22 59. 0.693 60. 2 61. 1.367 62. (d) 63. 309.33 64. 1.56 65. 0.66 67. 0.05 68. (c) 69. 0.186 70. –6 71. 0.3601 72. 10 73. (a) 74. 0.686 75. 8 76. 1 77. (c)

72. Only one of the real roots of f(x) = x6 – x – 1 lies in the interval 1 ≤ x ≤ 2 and bisection method is used to find its value. For achieving an accuracy of 0.001, the required minimum number of iterations is ______. (Give the answer up to two decimal places.)  [EE GATE 2017] 73. P(0, 3), Q(0.5, 4), and R(1, 5) are three points on the curve defined by f(x). Numerical integration is carried out using the both trapezoidal rule and Simpson’s rule within limits x = 0 and x = 1 for the curve. The difference between the two results will be (a) 0 (b) 0.25 (c) 0.5 (d) 1  [ME GATE 2017] 74. Starting with x = 1, the solution of the equation x3 + x = 1, after two iterations the Newton–Raphson’s method (up to two decimal places) is _______.  [EC GATE 2017] 75. Consider the equation du = 3t 2 + 1 with u = 0 dt

at t = 0. This is numerically solved by ­using the forward Euler method with a step size Dt = 2. The absolute error in the solution in the end of the first time step is _______. [CE GATE 2017]



76. The quadratic equation 2x2 – 3x + 3 = 0 is to be solved numerically starting with an initial guess as x0 = 2. The new estimate of x after the first iteration using the Newton–Raphson method is ___________.  [CE GATE 2017] 77. In order to evaluate the integral

1

ò e dx 0

x

with

Simpson’s 1/3rd rule, values of the function ex are used at x = 0, 0.5 and 1.0. The absolute value of the error of numerical integration is

EMEP.CH07_3PP.indd 419

(b) 0.000440 (d) 0.002718 [ME GATE 2018] 4. (c) 9. (c) 18. (c) 23. (a) 32. (d) 37. (c) 42. 1.166 47. (d) 52. (b) 57. 0.33 66. (a)

Explanation

1. (c) The Trapezoidal rule gives exact result for a polynomial of degree ≤ 1. 2. (c) 3. (i)-(c); (ii)-(b) (i) Let x = 1 . Then 1 = a i.e; 1 - a = 0. Take

a 1 f ( x ) = - a. x

1 \ f ( xk ) = -a xk

x

x

Then f ¢ ( x ) = - 1 . 2

and

x 1 f ¢ ( xk ) = - 2 . xk

 Now by the Newton–Raphson iteration ­formula,

x k +1 = x k -

f ( xk )

f ¢ ( xk )

or, x k +1 = x k -

(1/ xk - a ) -

1 x k2

or, x k +1 = 2x k - ax k2 (ii) For a = 7 and x0 = 0.2, the iteration formula, becomes x k +1 = 2x k - 7x k2 \  x1 = 2x0 – 7x02 = 2 × 0.2 – 7 (0.2)2 = 0.12 and x2 = 2x1 – 7x12 = 2 × 0.12 – 7(0.12)2 = 0.1392.

8/3/2023 6:23:51 PM

420 • Engineering Mathematics Exam Prep 4. (c) 3

5. (c) Given f(x) = x + 3x – 7 and x0 = 1. \  f′(x) = 3x2 + 3. Then, f(x0) = f(1) = 13 + 3 × 1 – 7 = –3 and f′(x0) = f′(1) = 3 × 12 + 3 = 6. Now the Newton–Raphson iteration formula gives,

x1 = x0 -

f ( x0 ) f ¢ ( x0 )

æ -3 ö or,  x1 = 1 - ç ÷ ´ 1 = 1.5. è 6 ø 6. (b)

7. (b) Take f(x) = xex – 2. Then f′(x) = xex + ex. Given that x1 = 0.8679. Now the Newton–Raphson iteration formula gives,

x1 = x0 -

f ( x0 )

f ¢ ( x0 ) æ 0.8679 ´ e0.8679 - 2 ç 0.8679 ´ e0.8679 + e0.8679 è

or,  x1 = 0.8679 - ç or,

ö ÷÷ ø

æ 0.8679 ´ 2.3819 - 2 ö x1 = 0.8679 - ç 0.8679 ÷ è 0.8679 ´ 2.3819 + e ø

or x1 = 0.853. 8. (a) Given that f(0) = 1, f(1) = 4, f(2) = 15. Here, h = 1. So by the trapezoidal formula, 2

ò



Since f(x) is second-degree polynomial, let f(x) = a0 + a1x + a2x2. Then, f(0) = 1  ⇒  a0 + 0 + 0 = 1  ⇒  a0 = 1. f(1) = 4  ⇒  a0 + a1 + a2 = 4 ⇒  1 + a1 + a2 = 4  ⇒  a1 + a2 = 3 …(i) f(2) = 15  ⇒  a0 + a1 + 4a2 = 15 ⇒  1 + 2a1 + 4a2 = 15 ⇒  a1 + 2a2 = 7 …(ii) Solving (i) and (ii), we get, a1 = –1 and a2 = 4. Thus, f(x) = 1 – x + 4x2. 2

Now, ò f ( x ) dx 0

2

=



EMEP.CH07_3PP.indd 420

ò( 0

2

é x 2 4x 3 ù 32 . + 1 - x + 4 x 2 dx = ê x ú = 2 3 úû 3 êë 0

)

3

dx

Iterative equation for the backward (implicit) Euler methods is given by yk +1 = yk + hf ( x k +1 , yk +1 ) , where dy = f ( x , y ). dx Then we have, yk +1 = yk + 1 ´ 0.25 y2



Now putting k = 0 in (i), we get,



0.25 y12 - y1 + y0 = 0



k +1

or,

0.25 yk2+1

- yk +1 + yk = 0 …(i)

or, 0.25 y12 - y1 + 1 = 0 or, y1 =

1 ± 1 - 4 ´ 0.25 = 2. 2 ´ 0.25

⇒  y1 = 2 10. (a) Here f(x) = x3 + 4x – 9. \ f ¢ ( x ) = 3x 2 + 4 , f ( x k ) = x k3 + 4 x k - 9 and f ( x k ) = 3x k + 4. Now iteration formula for the Newton–Raphson method gives, 2

¢



0

1 h  ( f (0) + 2 f (1) + f (2) ) = (1 + 8 + 15 ) = 12. 2 2

3

9. (c) Given, dy = 0.25 y2 , y0 = y(0)= 1 and h = 1.



f ( x ) dx

Hence, error = Exact – Approximate value = 32 - 12 = - 4 .



x k +1 = x k -

f ( xk ) f ¢ ( xk )

or, x k +1 = x k -

( 3x =

3 k

x k3 + 4 x k - 9 3x k2 + 4

) (

+ 4 x k - x k3 + 4 x k - 9 3x k2

+4

) = 2x

3 k 3x k2

+9 +4

.

11. (a) Since x = 5 is a root, so x – 5 is a factor. \ x 3 - 10x 2 + 31x - 30 = 0



Þ x 2 ( x - 5) - 5x ( x - 5) + 6( x - 5) = 0

(

)

Þ ( x - 5 ) x 2 - 5x + 6 = 0 Þ ( x - 5)( x - 2)( x - 3) = 0

Þ x = 5,2,3. Hence, the other roots of x2 – 5x + 6 = 0 are 2 and 3. 12. (b) Here, x0 = 2, f(x) = x3 – x2 + 4x – 4. \  f′(x) = 3x2 – 2x + 4, f(x0) = f(2) = 23 – 22 + 4 × 2 – 4 = 8, f′(x0) = f′(2) = 3 × 22 – 2 × 2 + 4 = 12.

 Now, the Newton–Raphson iteration formula

8/3/2023 6:23:54 PM

Numerical Analysis • 421 gives x1 = x0 or, x1 = 2 -



f ( x0 ) f ¢ ( x0 )

7

2

8

8xn

Then, xn +1 = xn + 9 2

8xn

2π

p 1 == -0.7071 4 2

y8 = sin 2p = 0

By the trapezoidal rule, we have, x8

ò f ( x ) dx

x0



a 9 Þ 8a 2 = 4 a 2 + 9 Þa= + 2 8a 9 3 Þ a2 = Þ a = = 1.5. 4 2



7p pö æ = sin ç 2p - ÷ 4 4 è ø

= - sin

When the series converges, we have, xn +1 = xn = a = root of equation.



y7 = sin

8 4 = . 12 3

13. (a) Given, xn +1 = xn + 9 , x0 = 0.5

7p 4

=

hé ( y + y8 + 2( y1 + y2 + ... + y7 )ùû 2ë 0

or,

2p



ò sin x dx 0

p 8

 = [2(0.70710 + 1 + 0.70710 – 0.70710 – 1 –

2

14. (a) Let x = 2 . Then x – 2 = 0. 0.70710)] Now let f(x) = x2 – 2. Then f′(x) = 2x. = 0.00000  Now iteration formula for the Newton–­ 16. (c) Raphson method gives, xn +1 = xn -

f ( xn ) f ¢ ( xn )

or, xn +1 = xn -



xn2 - 2 1 ïì 2 ïü = íxn + ý. 2x n 2 ïî xn ïþ

15. (a) Here, h = 2p - 0 = p and y = f(x) = sin x. 8

4

i 0 1

x 0

y = sin x y0 = sin 0 = 0

p 4

1 p y1 = sin = = 0.7071 4 2

2

p 2

3

3p 4

y2 = sin y3 = sin = sin

4 5

π 5p 4

EMEP.CH07_3PP.indd 421

6p 4

pö 3p æ = sin ç p - ÷ 4 4 è ø

p 1 = = 0.7071 4 2

y4 = sin p = 0 y5 = sin

5p pö æ = sin ç p + ÷ 4 4ø è

= - sin

6

p =1 2

p 1 == -0.7071 4 2

6p 3p y6 = sin = sin 4 2 pö p æ = sin ç p + ÷ = - sin = -1 2ø 2 è

17. (a) Here, f(x) = ex – 1. So f′(x) = ex. \  f(x0) = f(–1) = e–1–1, f′(x0) = f′(–1) = e–1. Now, the Newton–Raphson iteration formula gives,

x1 = x0 -

f ( x0 )

f ¢ ( x0 )

or, x1 = -1 -

e -1 - 1 -2e -1 + 1 = = -2 + e e -1 e -1

= -2 + 5.436563 = 0.71828



18. (c) Here product of the roots = –(constant term)/coefficient of x3 = 6/1= 6. If α be the third root, then we have, α × 1 × 3 = 6 i.e; α = 2. 19. (c) Given that the Newton–Raphson iteration formula is

xn +1 =

1æ R ö …(i) ç xn + ÷ 2è xn ø

If the above formula converges to the root after n iterations, then xn+1 = xn = a(say). Then (i) gives,

a=

1æ Rö a+ ÷ 2 çè aø

or, 2a =

a2 + R a

or, 2a2 = a2 + R or,  a2 = R,  i.e., a = R .

8/3/2023 6:23:57 PM

422 • Engineering Mathematics Exam Prep 20. (c) Here f(x) = x – e–x. So f′(x) = 1 + e–x. 23. (a) Here f(x) = x2 – 117. So f′(x) = 2x.  Now iteration formula for the Newton–­ Now, the iteration formula for the Newton– Raphson method gives, Raphson method gives, xn +1 = xn -

f ( xn ) f ¢ ( xn )

or, xn +1 = xn -



x k +1 = x k -

x n - e - xn 1+e

- xn =

e - xn (1 + xn ) . 1 + e - xn

21. (a) Here, f(x) = xex, f′(x) = xex + ex = ex(x + 1), f′′(x) = xex + ex + ex = ex(x + 2). Since, both ex and x are increasing functions of x, maximum value of f′′(t) in interval 1 ≤ t ≤ 2, occurs at t = 2. So, max f ¢¢ (t ) = e2 ( 2 + 2 ) = 4e2 . Now the truncation error for the trapezoidal rule =

h2 b-aù é = f ¢¢ (t ) ê n = (b - a )1max £t £2 12 h úû ë =



h2 h2 2 2 - 1) 4e2 = e ( 12 3

( )

Now ATQ, h e2 = 1 ´ 10-6 3

Þ h2 =

So minimum number of subintervals of equal length h

22. (c)

2 -1

(10

-3

/e

)

Þ dx = e -2tu(t ) dt Þ x = Þx= =



EMEP.CH07_3PP.indd 422

ò F ( x ) dx 0



0.01

=

h é( y0 + y4 ) + 4( y1 + y3 ) + 2 y4 ûù 3ë

0.25 [(1 + 0.5) + 4 ´ (0.9412 + 0.64) + 2 ´ 0.8] 3 0.25 = ´ 9.4248 = 0.7854 3 =



= 0.7854

26. (d) Here, f ( x ) = x 2 - 13 and so f ¢ ( x ) = 2x.

( 0) ( )  The Newton–Raphson iteration formula gives

= f ¢ 3.5 = 2 ´ 3.5 = 7.

x1 = x0 -

f ( x0 )

f ¢ ( x0 )

é -0.75 ù or, x1 = 3.5 - ê ú = 3.607 ë 7 û

27. (b) Comparing dy = f ( x , y ) with dy = x + y , dx

0.01

òe t =0

-2t

f (t ) dt,where f (t ) = e ò t =0

u(t ) dt

-2t

u(t )

h é f (0) + f (0.01) ùû (by trapezoidal rule) 2ë

0.01 é 0 e u(0) + e -2´0.01u(0) ù û 2 ë é ì0, t < 0 ù = 0.005 ´ é1 + e -0.02 ù ê u(t ) = í ú ë û î1, t ³ 0 û ë = 0.0099

=

1

= 1000e.

dx = e -2tu(t ) dt



25. (a) Using Simpson’s 1/3rd formula for n = 4, we have,

f¢ x

3

10-6 10-3 Þh= 2 e e

= n = b - a =

x k2 - 117 x k2 + 117 1 æ 117 ö = = ç xk + ÷. 2x k 2x k 2è xk ø

\ f ( x0 ) = f ( 3.5 ) = 3.52 - 13 = -0.75,

2



or, x k +1 = x k -

24. (a)

nh3 max f ¢¢ (t ) 12 1£t £2

(where n is number of subintervals of equal length)



f ( xk ) f ¢ ( xk )

dx



we get, f(x, y) = x + y. Here, step size, h = 0.1, x0 = 0, y0 = 0, x1 = x0 + h = 0 + 0.1 = 0.1 x2 = x0 + 2h = 0 + 2 × 0.1 = 0.2, x3 = x0 + 3h = 0 + 3 × 0.1 = 0.3. Euler’s first-order formula is given by y = yi + hf(xi yi)



\ y1 = y0 + hf ( x0 , y0 ) = 0 + 0.1 ´ (0 + 0) = 0,

y2 = y1 + hf ( x1 , y1 ) = 0 + 0.1 ´ (0.1 + 0) = 0.01,  y3 = y2 + hf(x2, y2) = 0.01 + 0.1 × (0.2 + 0.01) = 0.031.

8/3/2023 6:23:59 PM

Numerical Analysis • 423

28. (b) If T(q) represents the torque exerted, then flywheel energy

or, x1 = 2 -

2p

=



ò T ( q) dq 0

=

32. (d) Let us take u11 = 1 and u22 = 1.

(using Simpson's 1/3rd rule) p 3 = éë( 0 + 0) + 4(1066 + 0 - 355) + 2( -323 + 323 ) ùû 3 [since h = 60 degrees = p radians] 3

= 993.14 = 993 (approx) 29. (a) Here, f(x) = xi2 – N and so f′(x) = 2xi.  The Newton–Raphson iteration formula gives

é2 1 ù Then, ê ú = LU ë4 -1û é2 1 ù é l11 0 ù é1 u12 ù Þê úê ú=ê ú ë4 -1û ël21 l22 û ë0 1 û l11u12 ù é2 1 ù é l11 Þê ú ú = êl l u 4 1 ë û ë 21 21 12 + l22 û



xi +1 = xi -

f ( xi )

f ¢ ( xi )

æ x2 - N or, xi +1 = xi - ç i ç 2x i è



v ( x1 , x2 )

= 10x22

- 10x2 cos x1 - 0.6 .

¶u ù ¶x2 ú ú ¶v ú ú ¶x2 û





1 . 2 2

Now, the Newton–Raphson iteration formula gives f ( x0 )

0.5(= y1) 0.3333(= y2)

h ( y0 + 4 y1 + y2 ) 3 1 = (1 + 4 ´ 0.5 + 0.3333 ) = 1.1111 3



ò

0.5

dx 1.5 = éëlog x ùû 0.5 = log (1.5 ) - log ( 0.5 ) = 1.0986 . x

Thus, exact value of the integral = 1.0986. x 0.5 1 1.5 y=

\ f ( x0 ) = f (2) = 2 + 2 - 3 = 2 - 1,

f ¢ ( x0 )

1(= y0)

3

By Simpson’s 1/3rd rule, given integral

1.5

2 x

x1 = x0 -

1 x

2

34. (d)

31. (c) Here, f ( x ) = x + x - 3 and so f ¢ ( x ) = 1 + 1 .

EMEP.CH07_3PP.indd 423

y=

10 sin 0 ù é10 0 ù é10 cos0 =ê ú=ê ú. ë10 sin 0 20 - 10 cos0 û ë 0 10 û

f ¢ ( x0 ) = f ¢ ( 2 ) = 1 +

1

x

=

Therefore, at x1 = 0.0 and x2 = 1.0, the Jacobian matrix is



Þ l11 = 2, u12 = 0.5, l21 = 4, l22 = -3.

Thus, éê2 1 ùú = éê2 0 ùú éê1 0.5ùú ë4 -1û ë4 -3û ë0 1 û 33. (c)



10 sin x1 é10x2 cos x1 ù =ê ú ë10x2 sin x1 20x2 - 10 cos x1 û





Then the Jacobian matrix is é ¶u ê ¶x ê 1 ê ¶v ê ë ¶x1



Þ l11 = 2, l11u12 = 1, l21 = 4, l21u12 + l22 = -1

ö x2 + N 1 é Nù = ê xi + ú . ÷÷ = i 2x i 2ë xi û ø

30. (b) Let u ( x1 , x2 ) = 10x2 sin x1 - 0.8 and

6 2 -2 = 1.694. 2 2 +1

=



h é( y0 + y6 ) + 4( y1 + y3 + y5 ) + 2( y2 + y4 ) ùû 3ë

2 -1 2 2 ( 2 - 1) =21 2 2 +1 1+ 2 2

1 x

2(= y0)

1(= y1)

0.666(= y2)

Now using Simpson’s 1/3rd rule, we have the approximate value of the given integral h ( y0 + 4 y1 + y2 ) 3 0.5 = ( 2 + 4 ´ 1 + 0.666 ) = 1.111. 3 =

Therefore, approximate value – Exact value = 1.1111 – 1.0986 = 0.012499 » 0.012 So the estimate exceeds the exact value by 0.012.

8/3/2023 6:24:01 PM

424 • Engineering Mathematics Exam Prep

35. (b)

0 1 2

9 5 3

y1P = y0 + hf ( x0 , y0 )

xn +1

bn an (+ve) (–ve)

n

1 1 1

f(xn+1)

an + bn 2

=

x1 = 5 x2 = 3 x3 = 2

471(+ve) 41(+ve) 0

Thus, the method converges exactly to the root in three iterations. 36. (d)



40. (d) Clearly h = 0.3. Using the trapezoidal rule, the value of the 3

integral ò f ( x ) dx 0

=

So, f ( x0 ) = f (1.2) = (1.2 ) + 2 ´ 1.2 - 1 = 3.128,

=

2

f ¢ ( x0 ) = f ¢ (1.2 ) = 3 (1.2 ) + 2 = 6.32.

Now the Newton–Raphson iteration formula gives



hé f ( x0 , y0 ) + f ( x0 , y1P ) ù û 2ë 0.2 é or, y1C = 1 + ´ 0 - 2 ´ 0.2 ´ 12 ù = 0.96. û 2 ë y1C = y0 +

3

h 3

éë( y0 + y10 ) + 2 ( y1 + y2 + y3 + y4 + y5 + y6 + y7 + y8 + y9 ) ùû

0.3 é( 0 + 9 ) + 2 ( 0.09 + 0.36 + 0.81 + 1.44 + 2.25 3 ë +3.24 + 4.41 + 5.76 + 7.29 ) ùû

= 9.045

f ( x0 )

41. (b)

f ¢ ( x0 )

42. 1.166

3.128 = 0.705 6.32

or, x1 = 1.2 -

or, y1P = 1 + 0.2 é -2 ´ 0 ´ 12 ù = 1. ë û



37. (c) Here, f ( x ) = x 3 + 2x - 1. \ f ¢ ( x ) = 3x 2 + 2.

x1 = x0 -

After one iteration,

38. 0.53

y=

0 10 (y0)

x y = x 4 + 10

1 11 (y1)

2 26 (y2)

3 91 (y3)

4 266 (y4)



4

ò (x 0

4



h é( y0 + y4 ) + 4 ( y1 + y3 ) + 2 y2 ùû 3ë 1 = éë(10 + 266 ) + 4 ´ (11 + 91 ) + 2 ´ 26 ùû = 245.33. 3



The exact value of integral

=

4

ò (x

4

\  Magnitude of error = |exact value – estimated value| = 244.8 - 245.33 = 0.53



39. (d) dy dy + 2xy2 = 0 Þ = -2xy2 . dx dx



EMEP.CH07_3PP.indd 424

\ f ( x , y ) = -2xy2 . y(0) = 1 Þ x0 = 0, y0 = 1.

)

+ 10 dx

0

é x5 ù 45 =ê + 10x ú = + 10 ´ 4 = 244.8 5 êë 5 úû 0



3

1 x

1(= y0)

0.5(= y1)

0.3333(= y2)

31

dx x h = éë( y0 + y2 ) + 2 y1 ùû 2 1



)

+ 10 dx

4

2

By the trapezoidal rule,

ò

Using Simpson’s rule, the estimated value of the integral

1

x



=

1 2.333 = 1.166 éë1 + 0.333 + 2 ´ 0.5 ùû = 2 2

43. 0.06 Here, f ( x ) = e x - 1 and so f ¢ ( x ) = e x . Given that x0 = 1. In the Newton–Raphson method, iteration formula is

x k +1 = x k -

f ( xk ) f ¢ ( xk )

\ x1 = x0 -

f ( x0 ) e1 - 1 = 1 - 1 = e -1 = 0.367. f ¢ ( x0 ) e

Now, x2 = x1 =

f ( x1 ) e0.367 - 1 = 0.367 - 0.367 f ¢ ( x1 ) e

0.5297 - 1.443 + 1 = 0.06 1.443

Therefore, the absolute error observed at the second iteration = 0.06.

8/3/2023 6:24:03 PM

Numerical Analysis • 425

44. 0.54 Here, f(x) = 5x – 2cos x – 1, f ¢ ( x ) = 5 + 2sin x . Let us take x0 = 1 (in radian). The Newton–Raphson’s iteration formula is xn +1



f ( xn ) = xn f ¢ ( xn )

Again,

4 - 2 ´ 0.5403 = 0.5632, 5 + 2 ´ 0.8414

f ( x1 ) x2 = x1 f ¢ ( x1 )

\ x3 = x 2 -



f ( x2 ) f ¢ ( x2 )

5 ´ 0.5426 - 2cos(0.5426) - 1 = 0.5426 5 + 2sin(0.5426) 2.713 - 2 ´ 0.856369 - 1 = 0.5426 = 0.5425. 5 + 2 ´ 0.53389

\ Real root (up to two decimal accuracy) = 0.54. 45. 1.753.

Here, h = b - a = 4 - 2.5 = 0.3 n



I=



EMEP.CH07_3PP.indd 425

5

x y = f(x) = ln(x) 2.5 0.916 (= y0) 2.8 1.030 (= y1) 3.1 1.131 (= y2) 3.4 1.224 (= y3) 3.7 1.308 (= y4) 4.0 1.386 (= y5) By the trapezoidal rule,





5 ´ 0.5632 - 2cos(0.5632) - 1 = 0.5632 5 + 2sin(0.5632) 2.816 - 2 ´ 0.84555 - 1 = 0.5632 = 0.5426, 5 + 2 ´ 0.53389





f ( x0 ) 5 ´ 1 - 2cos1 - 1 \ x1 = x0 =1f ¢ ( x0 ) 5 + 2sin1 =1-

46. (a) Here, f(x) = 0.75x3 – 2x2 – 2x + 4, f’(x) = 2.25x2 – 4x – 2. The Newton–Raphson’s iteration formula is



+1.224 + 1.308 =

0.3 ´ 11.688 = 1.753 2

f '(2)

f ( x1 )

f ¢ ( x1 )

0.75 ´ 0 - 2 ´ 0 - 0 + 4 = 2, 2.25 ´ 0 - 0 - 2

x3 = x 2 =2-

= 2 – f (2)

0.75 ´ 8 - 2 ´ 4 - 4 + 4 = 0, 2.25 ´ 4 - 8 - 2

x2 = x1 =0-

f ( x0 ) f ¢ ( x0 )

f ( x2 ) f ¢ ( x2 )

0.75 ´ 8 - 2 ´ 4 - 4 + 4 = 0, 2.25 ´ 4 - 8 - 2

x 4 = x3 -

f ( x3 ) f ¢ ( x3 )

0.75 ´ 0 - 2 ´ 0 - 0 + 4 = 2. 2.25 ´ 0 - 0 - 2

dt

So, f(t, x) = 4t + 4. By the Runge–Kutta fourth order multistep method, 1 ( k1 + 2k2 + 2k3 + k4 ) , 6



x1 = x0 + k = x0 +



k1 = hf(t0, x0) = 0.2(0 + 4) = 0.8,



k ö 1 0.2 0.8 ö æ æ , x0 + k2 = hf ç t0 + h, x0 + 1 ÷ = 0.2 f ç 0 + 2 2ø 2 2 ÷ø è è = 0.2 f (0.1, x0 + 0.4) = 0.2 ´ (4 ´ 0.1 + 4) = 0.88,

ò In ( x ) dx

0.3 [( 0.916 + 1.386 ) + 2(1.0296 + 1.131 = 2

=2-

f ( xn )

f ¢ ( xn )

Thus, x0 = 2, x1 = 0, x2 = 2, x3 = 0, x4 = 2, ..... So, x3 = 0 is correct but it converges in an infinite number of iterations. 47. (d) Given, dx = 4t + 4, at t = 0, x = x0.

4

h é( y0 + y5 ) + 2( y1 + y2 + y3 + y4 ) ùû 2ë

\ x1 = x0 -

=0-

2.5

=

xn +1 = xn -



where,

k ö 1 æ k3 = hf ç t0 + h, x 0 + 2 ÷ 2 2 ø è 0.2 0.88 ö æ , x0 + = 0.2 f ç 0 + = 0.2 f ( 0.1, x0 + 0.44 ) 2 2 ÷ø è = 0.2 ( 4 ´ 0.1 + 4 ) = 0.88,

k4 = hf(t0 + h, x0 + k3) = 0.2 f(0.2, x0 + 0.88) = 0.2(4 × 0.2 + 4) = 0.96. \k =

1 ( k1 + 2k2 + 2k3 + k4 ) 6

8/3/2023 6:24:05 PM

426 • Engineering Mathematics Exam Prep 1 ( 0.8 + 2 ´ 0.88 + 2 ´ 0.88 + 0.96 ) = 6



=

0.88.



Hence, the increment in x = x1 – x0 = k = 0.88





b - a 1 - ( -1) 2 = = . n 3 3 -

–1

x

1 3

1 3

1

1 1 y = |x| 1 (= y0) 3 (= y1) 3 (= y2) 1 (= y3)



h é( y0 + y3 ) + 2 ( y1 + y2 ) ùû 2ë 2 é æ 1 1 ö ù 10 3 = ê(1 + 1) + 2 ç + ÷ ú = = 1.11 2ë è 3 3 øû 9



x k +1 = x k -



f ( xn ) f ¢ ( xn )

or, x k +1 = x k -

x k3 - N

=

3x k2

1æ N ö ç 2x + ÷. 3 çè k x k2 ÷ø

53. 5 é l11 0 ù é1 u12 ù é 2 2 ù ê úê ú=ê ú ël21 l22 û ë0 1 û ë4 9 û l11u12 ù é2 2 ù él Þ ê 11 ú=ê ú l l u ë 21 21 12 + l22 û ë4 9 û

49. (c) Case-I: When we use the trapezoidal rule: 3

Error = - nh f ¢¢ ( x ) .

Þ l11 = 2, l11u12 = 2, l21 = 4, l21u12 + l22 = 9

12

2

0.891392 - sin(0.89139) 2 ´ 0.89139 - cos(0.89139) 0.79457 - 0.77794 = 0.89139 = 0.87698. 1.78278 - 0.62833

52. (b) Here, f(x) = x3 – N, f′(x) = 3x2.  The Newton Raphson’s iteration formula gives:

By the trapezoidal rule, given integral =

f ( x1 ) f ¢ ( x1 )

= 0.89139 -

48. 1.11 h=

x2 = x1 -

Þ l11 = 2, u12 = 1, l21 = 4, l22 = 5.

Here, f(x) = x , so f ¢¢ ( x ) = 2 which is always



positive. So the sign of the error is always negative. Hence, approximate value always greater than or equal the exact value of the integral. Case-II: When we use Simpson’s 1/3rd rule:

54. 0.3043 Here, f(x) = x3 + 2x2 + 3x – 1, f′(x) = 3x2 + 4x + 3, x0 = 1.  The Newton–Raphson iteration formula gives:

5

Error = - nh f (iv ) ( x ) . 90

Here, f(x) = x2 , so f (iv)(x) = 0. So the error is always “0” which means that the approximate value is always equal to the exact value of the integral. 50. (a) Let f(x) = k(x – 1)(x – 2)(x – 3), k ≠ 0. Then f(0) f(4) = k(0 – 1) (0 – 2) (0 – 3) × k(4 – 1) (4 – 2) (4 – 3) = –36K2 < 0 51. 0.88(approx) Here, f(x) = x2 – sin x, f′(x) = 2x – cos x. Given x0 = 1. The Newton Raphson’s iteration formula is

xn +1 = xn \ x1 = x0 -

f ( xn )

f ¢ ( xn ) f ( x0 ) f ¢ ( x0 )





EMEP.CH07_3PP.indd 426

\ x1 = x0 x2 = x1 -



= 0.5 -

f ( xn )

f ¢ ( xn ) f ( x0 )

f ¢ ( x0 )

=1-

1 + 2 + 3 -1 = 0.5, 3+4+3

f ( x1 )

f ¢ ( x1 ) 0.53 + 2 ´ 0.52 + 3 ´ 0.5 - 1 = 0.3043 3 ´ 0.52 + 4 ´ 0.5 + 3

55. 2.33 Here, f(x) = x2 – 4x + 4, f′(x) = 2x – 4, x0 = 3. \  f(3) = 1 – 4 + 4 = 1, f′(3) = 2 × 3 – 4 = 2. By The Newton–Raphson method, we have

2

1 - sin1 1 - 0.84147 =1=1= 0.89139, 2 ´ 1 - cos1 2 - 0.54030

xn +1 = xn -



x1 = x0 -

f ( x0 )

f ¢ ( x0 )

=3-

f (3)

f ¢ (3)

=3-

1 5 = . 2 2

2

5 1 æ5ö æ5ö Then, f ( x1 ) = f ç ÷ = ç ÷ - 4 ´ + 4 = . 2 4 è2ø è2ø

8/3/2023 6:24:07 PM

Numerical Analysis • 427



Now, by the Secant method,



( x1 - x0 ) f ( x1 ) f ( x1 ) - f ( x0 )

x2 = x1 -





æ5 ö 1 - 3÷ ´ 5 çè 2 ø 4 = 7 = 2.33 = 1 2 3 -1 4



By Simpson’s 1/3rd rule, =

h 0.5 é 9 12 39 ù é y0 + y2 + 4 y1 ùû = + +4´ =2 3ë 3 êë 5 5 20 úû

= 0.2 + 25x - 200x 2 + 675x 3 4

- 900x + 400x

x0 = 5.



\ f ( 5 ) = 5 - 5 ´ 25 + 6 ´ 5 - 8 = 22, 3

f ¢ ( 5 ) = 3 ´ 25 - 10 ´ 5 + 6 = 31.

Now the Newton–Raphson iteration formula gives, x1 = x0 -



f ( x0 ) f (5) 22 =5= 4.29 =531 f ¢ ( x0 ) f ¢ (5)

0

h é y0 + 4 y1 + y2 ûù 3ë

=

0.4 ( 0.2 + 4 ´ 2.456 + 0.232 ) 3

f ( x ) dx

h é y0 + y4 + 2 ( y1 + y2 + y3 ) ùû 2ë 0.1 = é0 + 160 + 2(10 + 40 + 90) ùû = 22. 2 ë

59. 0.693 1 0

2 2 ln 2



1



h 1 é y0 + yn ùû = éë0 + 2 In2ùû = In2 = 0.693 2ë 2

EMEP.CH07_3PP.indd 427

0

0 ( = y0)

2

10(= y1)

4

18 ( = y2)

6

25 (= y3)

6

29 (= y4)

10

32 (= y5)

12

20 (= y6)

14

11 (= y7)

16

5 (= y8)

18

2 (= y9)

20

0 (= y10)

=

x

1

0.5

1

3 2 9 x + 5 5

9 5

39 20

12 5

=



ò vdt 0

60. 2

f (x ) =

v

20

=

By the trapezoidal rule, x ln x dx ò =

t

By Simpson’s 1/3rd rule, distance (in kilometers)

2



= 1.367

63. 309.33 km

f ( xini ) ( -2 + 0 - 0 + 0) =0= 0.33. (6 - 0 - 0) f ¢ ( xini )

x f(x) = x In x

0.2 2.456 0.232

5

=

=



0.8

By Simpson’s 1/3rd Rule, given integral

\  Dx = x1 – xini = 0.33 – 0 = 0.33 58. 22

ò

0.4

62. (d)

57. 0.33 Here, f(x) = – 2 + 6x – 4x2 + 0.5x3, f′(x) = 6 – 8x + 1.5x2, xini = 0. By the Newton Raphson method,

0.4

0

f (x )

f ¢ ( x ) = 3x 2 - 10x + 6,

x1 = xini

f (x )

x

f ( x ) = x 3 - 5x 2 + 6x - 8,



1

0

61. 1.367

56. 4.29 Here,



ò

h é( y + y ) + 4( y1 + y3 + y5 + y7 + y9 ) 3 ë 0 10 +2( y2 + y4 + y6 + y8 )ùû

2 é(0 + 0) + 4(10 + 25 + 32 + 11 + 2) 3ë +2(18 + 29 + 20 + 5)ùû

= 309.33 km

8/3/2023 6:24:10 PM

428 • Engineering Mathematics Exam Prep

64. 1.56 Here, f ( x ) = x - 10 cos x , p . 4

 Now, by the Newton–Raphson iteration ­formula, x1 = x0 -

f ( x0 ) p æ -6.2856 ö = f ¢ ( x0 ) 4 çè 8.0711 ÷ø

= 0.7854 + 0.7788 = 1.5642 » 1.56

65. 0.66 Given,

dx = -3x + 2, dt

68. (c) Here, f ( x ) = x3 + x - 1, f ¢ ( x ) = 3 x 2 + 1, x0 = 1. \ f ( x0 ) = f (1) = 1 + 1 - 1 = 1,

f ¢ ( x0 ) = f ¢ (1) = 3 + 1 = 4.  Then by the Newton–Raphson iteration ­formula,



y (0) = 1 .

i.e; if 0 < h < 2 . 3

= sin x + cos x



1 ( k1 + k2 ) 2

)



k2 = hf ( x0 + h, y0 + k1 ) = hf (0 + 0.1,1 + 0.1)



= 0.1 (1 + 0.1 ) + 2 ´ 0.1 - ( 0.1 )



EMEP.CH07_3PP.indd 428

\ y1 = y0 + =1+

2

} = 0.129.

1 ( k1 + k2 ) 2



p

By the trapezoidal rule, ( sin x + cos x ) dx ò h = { y0 + y3 + 2(y1 + y2 )} 2 æ 3 +1 p / 3 ìï 3 - 1 ö üï = + ÷ý í1 + ( -1) + 2 çç 2 ïî 2 ÷ø ïþ è 2 = 1.813799 p

Again, ( sin x + cos x ) dx ò

= éë( - cos x + sin x ) ùû = - cos p + cos0 = 2. 0

Then error = Exact value – approx. value = 2 – 1.813799 = 0.186 70. –6 ATQ, (x1)0 = (x2)0 = (x3)0 = 0. According to the Gauss Seidel method, the first iterations are given by

( x1 )1 = 5 - 2 ( x2 )0 - 3 ( x3 )0 = 5 - 0 - 0 = 5,

1 ( 0.1 + 0.129 ) = 1.1145 2

1

( x2 )1 = 3 éë1 - 2 ( x1 )1 - ( x3 )0 ùû

Again, the exact solution 2

= y ( 0.1) = ( 0.1) + e0.1 = 1.1151

Hence, error

–1(= y3)

p

= 0.1 1 + 2 ´ 0 - 02 = 0.1,

{

3 +1 3 -1 2 2 ( = y1 ) ( = y2 )

p

0

where

k1 = hf ( x0 , y0 ) = hf (0,1)

(

1(= y0)

2p 3

0

66. (a)The trapezoidal rule gives the best result when the function is linear, i.e, of degree 1. 67. 0.05 Here, f(x, y) = y + 2x – x2, x0 = 0, y0 = 1, h = 0.1. By the second-order Runge–Kutta method, we have, y1 = y0 +

p 3

0

f (x )

3



f ( x0 ) 1 = 1 - = 1 - 0.25 = 0.75. f ¢ ( x0 ) 4

x



Hence, hmax = 2 = 0.66

x1 = x0 -

69. 0.186

The solution of differential equations is stable if 1 - 3h < 1 i.e; if –1 < 1 – 3h < 1 i.e; if –2 < – 3h < 0

error ´ 100% exact value 0.0006 = ´ 100% = 0.05% 1.1151



p æpö p \ f ( x0 ) = f ç ÷ = - 10 cos = -6.2856, 4 4 4 è ø p f ¢ ( x ) = 1 + 10 sin = 8.0711. 4



= 1.1151 - 1.1145 = 0.0006.

\ % error =

f ¢ ( x ) = 1 + 10 sin x ,

x0 =

= exact value-approximate value

1 é1 - 2 ´ 5 - 0 ùû = -3, 3ë = 3 - 3 ( x1 )1 - 2 ( x2 )1 = 3 - 3 ´ 5 + 2 ´ 3 = -6.

=



( x3 )1

8/3/2023 6:24:13 PM

Numerical Analysis • 429

71. 0.3601 Here, f(x) = 3x – ex + sin x, f′(x) = 3 – ex + cos x, x0 = 0.333. Now, the Newton–Raphson iteration formula gives: x1 = x0 -

3 ´ 0.333 - e0.333 + sin(0.333) 3 - e0.333 + cos(0.333) 0.0693 = 0.333 + = 0.3601 2.5499

72. 10 Here, f(x) = x6 – x – 1, a = 1, b = 2, e = 0.001 = 10–3. If n be the number of iterations by the Bisection method then, b-a 2n





73. (a) Here, h = 0.5 and y0 = 3, y1 = 4, y2 = 5 1

By the trapezoidal rule, f ( x ) dx ò 0

0.5 h é( y0 + y2 ) + 2 y1 ùû = é(3 + 5) + 2 ´ 4 ùû = 4. 2ë 2 ë 1

é( y0 + y2 ) + 4 y1 ù =

0.5

Thus after the first iteration u = 2 when t = 2.

Again, t1 = t0 + h = 0 + 2 = 2.

Now, du = 3t 2 + 1

74. 0.686 Here, f(x) = x3 + x – 1, f′(x) = 3x2 + 1, x0 = 1. \  f(1) = 1 + 1 – 1 = 1, f′(1) = 3 + 1 = 4. By the Newton–Raphson method,

(

)

Þ du = 3t 2 + 1 dt Þ du = Þ u = 3´

ò

ò (3t

2

)

+ 1 dt

t3 + t + C = t3 + t + C …(i) 3

Putting u = 0 and t = 0 in (i), we get, C = 0. So, at t = 2, u = 8 + 2 = 10. Hence, Absolute error = Exact value – approx. value = 10 – 2 = 8. 76. 1 Here, f ( x ) = 2x 2 - 3x + 3, f ¢ ( x ) = 4 x - 3, x0 = 2.

\ f ( 2 ) = 8 - 6 + 3 = 5, f ¢ (1) = 8 - 3 = 5.

By the Newton–Raphson method, x1 = x0 -

f ( x0 )

f ¢ ( x0 )

=2-

5 = 1. 5

77. (c) x f(x)

éë(3 + 5) + 4 ´ 4 ùû = 4.

û 3ë 3 Thus, the difference between these two results will be zero.

EMEP.CH07_3PP.indd 429

u1 = u0 + hf ( u0 ,t0 )



0

h

By the Euler’s method = 0 + 2 f ( 0,0 ) = 2 ´ ( 3 ´ 0 + 1) = 2.

By the Simpson’s rule, f ( x ) dx ò =

\ f ( u,t ) = 3t + 1. 2





or, n > 9.96 Hence, the minimum number of iterations = 10.



0.753 + 0.75 - 1 3 ´ 0.752 + 1 0.171875 = 0.75 = 0.686 2.6875

dt



3ln10 3 ´ 2.30258 or, n > = ln 2 0.693147



f ( x1 ) f ( 0.75 ) = 0.75 ¢ f ( x1 ) f ¢ ( 0.75 )

dt

or, 2n > 103 or, ln 2n > ln 103

=

x2 = x1 -

75. 8 Given that, du = 3t 2 + 1; u0 = 0,t0 = 0; h = Dt = 2.

0 6. To solve the equation 2 sin x = x by the ­Newton–Raphson method, the initial guess was chosen to be x = 2.0. Consider x in radian only. The value of x (in radian) obtained after one iteration will be closest to (a) –8.101 (b) 1.901 (c) 2.099 (d) 12.101 7. The integral

x2

òx x

2

dx

with x2 > x1 > 0 is evalu-

analytically and J is the approximate value obtained using the trapezoidal rule, which of the following statements is correct about their relationship? (a) J > I (b) J < I (c) J = I (d) data is insufficient 8. An explicit forward Euler method is used to numerically integrate the differential equation dy = y using a time step of 0.1. With dx

the initial condition y(0) = 1 , the value of y(1) computed by this method is ________ (correct to two decimal places) (a) 2.11 (b) 4.34 (c) 2.59 (d) 4.56 9. The Newton–Raphson method is used to find the root of the equation x2 – 2 = 0. If the iterations are started from –1, then the iteration will (a) converge to –1 (b) converge to 2 (c) converge to – 2 (d) not converge 10. Simpson’s 1/3rd rule for integration gives exact result when f(x) is a polynomial function of degree less than or equal to (a) 1 (b) 2 (c) 3 (d) 4 11. The Newton–Raphson iteration formula for finding 3 c , where c > 0 is 3 3 3 3 (a) xn +1 = 2xn +2 c (b) xn +1 = 2xn -2 c

EMEP.CH07_4PP.indd 430

3x n

3

3

3x n

3x n

(c) xn +1 = 2xn +2 c (d) xn +1 = 2xn 2- c 12. Given the differential equation y¢ = x - y with initial condition y(0) = 0. The value of y(0.1) calculated numerically up to the third place of decimal by the second order the Runge– Kutta method with step size h = 0.1 is (a) 1/200 (b) 1/50 (c) 4/35 (d) 1/20 rd 13. Using Simpson’s 1/3 rule, the value of the

1

ated analytically as well as numerically using a single application of the trapezoidal rule. If I is the exact value of the integral obtained

3x n

integral

6

dx

ò (1 + x ) 0

2

taking six equal subintervals

of [0, 6] is

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Numerical Analysis • 431

(a) 0.751 (b) 0.849 (c) 0.904 (d) 0.356 rd 14. Using Simpson’s 1/3 rule, if we want to find the approximate value of

24

ò f ( x ) dx , then the

12

error committed is (a)

1 iv 1 iv f ( x) (b) - f ( x) 15 90

(c)

2 1 - f iii ( x) (d) - f iii ( x) 15 15

15. A system of equations AX = b where is said to be diagonally dominant if A = ( aij ) n´n

(a) aii > (b) aii
å aij (d) aii < å aij 16. The condition of convergence of the ­Newton– Raphson method when applied to an equation f(x) = 0 in an interval [a, b] is (a) f ¢( x ) ¹ 0 (b) f ¢( x ) < 1

2

EMEP.CH07_4PP.indd 431

4 x1 + 2x2 + 13x3 = 24



4 x1 - 2x2 + x3 = -8

(a) x1 = 1.005, x2 = 3.879, x3 = -2.667 (b) x1 = -1.240, x2 = 2.002, x3 = 4.123 (c) x1 = 1.205, x2 = 4.111, x3 = 3.533 (d) x1 = -1.423, x2 = 2.131, x3 = 1.956 21. If the third order differences of f(x) be constant and f ( -1) = -1, f ( 0 ) = 0, f (1) = 1, f ( 2 ) = 8 , (a) 15 (b) 64 (c) 45 (d) 45 22. The smallest degree of the polynomial that interpolates the data is x f(x)

–2 –58

–1 –21

0 –12

1 –13

2 –6

3 27

(a) 3 (b) 4 (c) 5 (d) 6 23. An iterative method of find the n th root ( n Î N ) of a positive number a is given by

(c) { f ¢( x )}2 > f ¢¢( x ) f ( x ) (d) none of these 17. The Runge–Kutta method is used to solve (a) a first order ordinary differential ­equation (b) a first order partial differential equation (c)  a second order ordinary differential equation (d) a second order partial differential ­equation 18. Using the Newton–Raphson method, an approximate real root of 3x – cos x – 1 = 0 is (up to three decimal places) (a) 0.607 (b) 0.809 (c) 0.004 (d) 1.560 19. Using Weddle’s rule and taking 12 subinter-

decimal places is

3x1 + 9x2 - 2x3 = 11

and f ( 3 ) = 27 , then the value of f ( 4 ) is

n

vals, the value of

(a) 3.0234 (b) 3.0211 (c) 1.1071 (d) 1.2075 20. Using the Gauss–Seidel iteration method, the solution of the following system is (correct up to four significant figures)

1

ò0 1 + x

2

dx

correct to four

x k +1 =

1é a ù ê x k + n -1 ú . A value of n for which 2 ëê x k ûú

this iterative method fails to converge is (a) 1 (b) 2 (c) 3 (d) 8 24. Solution the equation x 3 - 9x + 1 = 0 for the root lying between 2 and 3, correct to ­2-significant figures, is (a) 0.45 (b) 0.94 (c) 0.60 (d) 0.50 25. The following system of linear equations by the Gauss-elimination method has a solution x - 2 y + 9z = 8, 3x + y - z = 3, 2x - 8 y + z = -5.

(a) x = 1, y = 1, z = 1 (b) x = 1, y = 2, z = 3 (c) x = -1, y = 1, z = 0 (d) x = -2, y = -1, z = -3

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432 • Engineering Mathematics Exam Prep 26. The following system of equations by LU factorization method has a solution

2x - 6 y + 8z = 24 5x + 4 y - 8 z = 2 3x + y + 2z = 16 (a) x = 1, y = -1, z = 2 (b) x = 3, y = 2, z = 1 (c) x = 1, y = 1, z = 2 (d) x = 1, y = 3, z = 5

27. Given dy = x 3 + y, y(0) = 1. Then y(0.02) , by dx

Euler’s method correct up to two decimal places, taking step length h = 0.01 is

EMEP.CH07_4PP.indd 432

(a) 2.05 (c) 1.25

(b) 1.02 (d) 2.35 Answer Key

1. (c)

2. (c)

3. (b)

4. (d)

5. (d)

6. (b)

7. (a)

8. (c)

9. (c)

10. (c)

11. (c)

12. (a)

13. (b)

14. (b)

15. (a)

16. (c)

17. (a)

18. (a)

19. (c)

20. (d)

21. (b)

22. (a)

23. (a)

24. (b)

25. (a)

26. (d)

27. (b)

8/14/2023 2:51:24 PM

CHAPTER

8

Complex Analysis

8.1  BASICS OF COMPLEX ANALYSIS 8.1.1  Complex Number A number of the form x + iy, where i = -1 , x and y being real numbers, is called a complex number and is denoted by “z.” The real numbers x and y are called the real part of z and imaginary part of z, respectively. Symbolically, we write z = x + iy, Re(z) = x, Img(z) = y. 1

Example: 2 + 3i, - 2 + i, 5 i , etc., all are com3 plex numbers. Remember: (i) Consider z = x + iy.

(2)  z = – 1 – i represents a point (–1, –1) in the argand plane. 8.1.2 Modulus and Amplitude of a Complex Number Let the complex number z = x + iy corresponds the point P(x, y). Join O(origin) to P. Let the line segment makes an angle q with the positive direction of x-axis OP and OP = p. Then, x = r cos q and y = r sin q and so tan q = y/x. Therefore, z = x + iy = r(cos q+ i sin q), which is called the polar form or modulus amplitude form of the complex number “z.” y

If x = 0, then z = iy and so z is purely imaginary.

P(x,y)

If y = 0, then z = x. In this case, z becomes a real number. r

(ii) The form z = x + iy is called the Cartesian form of the complex number “z.” The complex number z can be regarded as an order pair (x, y). Thus, every complex number z = x + iy corresponds a point P(x, y) in the xy-plane and vice versa. In the theory of complex numbers, xy-plane is called complex plane or argand plane or Gaussian plane. Example: (1)  z = 1 + 2i represent a point (1, 2) in the argand plane.

EMEP.CH08_3PP.indd 433

q

x

Here OP = r = x 2 + y2 is called the modulus of z and is denoted by |z|.

Thus, |z| = r = x 2 + y2 .

q is called the amplitude (argument) of z and is denoted by amp (z) or arg (z).

Thus, amp(z) = q = tan–1 (y/x).

8/4/2023 5:13:40 PM

434 • Engineering Mathematics Exam Prep Remember:



Hence, the polar form of z is

(a) Let z = r (cos q + i sin q)........(1) be the polar form of z = x + iy



z = 2 æç cos 3p + i sin 3p ö÷ .

Then amp (z) (= q) is not unique because of equation (1) remains unaltered if we replace q by q + 2p. Thus, q can take infinite number of values which differ from each other by 2p.

Case-III:

If a value of q satisfies (i) and lies between –p and; p, then that value of q is called the principle value of the amplitude. In the polar form of the complex number, we need to consider the principle value of the amplitude. Case-I: The point (x, y) lies in the first quadrant. In this case, the polar form of z is

z = r (cos q + i sin q), where



r = x 2 + y2 and q = tan -1 y .

4

è

The point (x, y) lies in the third quadrant. In this case, the polar form of z is

z = r (cos q + i sin q), where



r = x 2 + y2 and q = -p + tan -1 y . x

Example: Consider z = –1–i which represents the point (–1, –1) that lies in third quadrant. Comparing z = –1–i with z = x + iy, we get x = –1, y = –1. Then, r = x 2 + y2 = ( -1)2 + ( -1)2 = 2 and q = -p + tan -1

x

Example: Consider z = 1 + i which represents the point (1, 1) that lies in first quadrant.

4 ø



y -1 = -p + tan -1 -1 x

= -p + tan -1 1 = -p +

3p p =4 4



Hence, the polar form of z is

Comparing z = 1 + i with z = x + iy, we get x =1, y = 1.



z = ìícos æç - 3p ö÷ + i sin æç - 3p ö÷ üý .

Then, r = x 2 + y2 = 12 + 12 = 2

Case-IV: The point (x, y) lies in the fourth quadrant. In this case, the polar form of z is

and q = tan -1 y = tan -1 1 = tan -1 1 = p . x



1

4

Hence, the polar form of z is z=

p pö æ 2 ç cos + i sin ÷ . 4 4ø è



z = r (cos q + i sin q), where



r = x 2 + y2 and q = p - tan -1 y .

Example: Consider z = –1 + i which represents the point (–1, 1) that lies in second quadrant. Comparing z = –1 + i with z = x + iy, we get x = –1, y = 1. Then, r = x 2 + y2 = ( -1)2 + 12 = 2 and



EMEP.CH08_3PP.indd 434

y 1 = p - tan -1 = p - tan -1 1 -1 x

p 3p = 4 4

4 øþ

è



z = r (cos q + i sin q), where



r = x 2 + y2 and q = - tan -1 y . x

Comparing z = 1 + i with z = x + iy, we get x = 1, y = –1. Then, r = x 2 + y2 = 12 + ( -1)2 = 2

x

= p-

4 ø

Example: Consider z = 1–i which represents the point (1, –1) that lies in fourth quadrant.

Case-II: The point (x, y) lies in the second quadrant. In this case, the polar form of z is

q = p - tan -1

è

î

and -1 p y = - tan -1 = - tan -1 1 = - . x 1 4



q = - tan -1



Hence, the polar form of z is



z = 2 ícos æç - p ö÷ + i sin æç - p ö÷ ý . 4 4

ì î

ü

è

ø

è

øþ

(b) z = r(cos q + i sin q) can be written as

z = reiq[  eiq = cos q + i sin q] and



z = r(cos q – i sin q) can be written as



z = pe–iq[  e–iq = cos q – i sin q]

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Complex Analysis • 435

8.1.3 Conjugate of a Complex Number The conjugate of the complex number z = x + iy is denoted by z and is defined by z = x - iy .

Remember: z z2

1. 1 =

Example:

=

The conjugate of z = 1 – i is z = 1 + i Remember: The conjugate of z = reiq is z = re -iq . 8.1.4 Properties of Modulus, Argument, and Conjugate (i) z = 0 Û z = 0 (ii) z1z2 = z1 z2 for any two complex numbers z1 and z2 (iii) z1 = z1 z2

z2



è z2 ø

(ix) arg(0) is not defined. (x) Arguments of any positive real number is 0. (xi) Arguments of any negative real number is ± p (xii) arg(z) – arg(–z) = ± p (xiii) z = z if and only if z is real. (xiv) zz = ( x + iy)( x - iy) = x 2 + y2 = z 2 (xv)

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

(xvi) z1 ± z2 = z1 ± z2 8.1.5 Sum, Difference, and Product of Two Complex Numbers If z1 = x1 + iy1, z2 = x2 + iy2 be two complex numbers, then (i) The sum of z1 and z2 is denoted by z1 + z2 and is defined by z1 + z2 = (x1 + x2) + i(y1 + y2).

x22 + y22

.

z1z2 = r[cos(q1 + q2) + i sin (q1 + q2)]

8.1.6 Cube Roots of Unity The cube roots of unity (i.e; 1) are 1, w, w2. where w, w2 =

-1 ± i 3 . 2

Remember: (i) If we take w = -1 + i 3 ,then w2 = -1 - i 3 and 2

2

if we take w = -1 - i 3 ,then w2 = -1 + i 3 . 2

(v) z1 - z2 £ z1 - z2

(viii) arg æç z1 ö÷ = arg( z1 ) - arg( z2 )

( x1x2 + y1 y2 ) - i( x1 y2 - y1x2 )

2. If z1 = r(cos q1 + i sin q1) and z2 = r(cos q2 + i sin q2) be two complex numbers, then

(iv) |z1 + z2| ≤ |z1|+|z2| (vi) z1 £ z1 - z2 + z2 (vii) arg(z1z2) = arg(z1) + arg(z2)

x1 + iy1 ( x1 + iy1 )( x2 - iy2 ) = x2 + iy2 ( x2 + iy2 )( x2 - iy2 )

2

(ii) 1 + w + w = 0 2

(iii)  w3 = 1, w4 = w3 × w = w, w5 = w3 × w2 = w2, etc. (iv) w3n = 1 where n ∈ Z+. 8.1.7  De Moivre’s Theorem If n ∈ Z, then (cos q + i sin q)n = cos nq + i sin nq and if n is a fraction of the form p (where p ∈ Z, q

q(≠ 0) ∈ Z), then one of the values of (cos q + i sinq)n will be cos nq + i sin nq. Application: Let x + 1 = 2cos p . 5

x

Then, x + 1 = 2cos p Þ x 2 - 2x cos p + 1 = 0 x



Þx=

2cos

5

5

p p ± 4 cos2 - 4 5 5 2

= cos

p pö p p æ ± - ç1 - cos2 ÷ = cos ± - sin 2 5 5ø 5 5 è

= cos

p p ± i sin 5 5

(ii) The difference of z1 from z2 is denoted by z1 – z2 and is defined by z1 – z2 = (x1 – x2) + i(y1 – y2).



(iii) The product of z1 and z2 is denoted by z1z2 and is defined by z1z2 = (x1x2 – y1y2) + i(x1y2 + y1x2)



5p 5p p pö æ \ x 5 = ç cos ± i sin ÷ = cos ± i sin 5 5ø 5 5 è



= cos p ± i sin p = –1  ( cos p = –1, sin p = 0)

EMEP.CH08_3PP.indd 435

5

8/4/2023 5:13:43 PM

436 • Engineering Mathematics Exam Prep 8.1.8  Hyperbolic Functions Hyperbolic functions are defined as follows

sinh x =

x

-x

x

-x

e -e 2

non-zero complex numbers (iii) Logz1z2 = z2 Logz1 + 2npi, where z1(≠ 0) and z2 are two complex numbers and n ∈ Z.

e +e 2

tanh x =

e x - e - x sinh x , = e x + e - x cosh x

sech x =

2 1 = , e x + e - x cosh x

coth x =

,

2 1 = , -x sinh x e -e x

ex + e-x 1 = . -x x tanh x e -e

Remember: (i) sin (ix) = i sinh x (ii) cos (ix) = cosh x (iii) tan (ix) = i tanh x 3

5

3!

5!

2

4

2!

4!

(iv) sinh x = x + x + x ........¥ (v) coshx = 1 + x + x ........¥ (vi) sinh 0 = 0 = tanh 0; cosh 0 = 1 (vii) cosh2x – sinh2 x = 1 (viii) sech2x + tanh2x = 1 (ix) coth2x – cosech2 x = 1 (x) sinh(2x) = 2 sinh x cosh x 8.1.9  Logarithm of a Complex Number Let us consider the complex number z in polar form given by z = r(cos q + isin q) = reiq, where r = z and q = amp(z).

Then we define



Log (z) = log r + i(2np + q), n ∈ Z



= log|z| + i (2np + amp(z))

When n = 0, we get the principal value of the logarithm denoted by log z which is given by log z = log |z| + i amp (z) Remember: (i) Log(z1 z2) = Logz1 + Logz2, where z1 and z2 are non-zero complex numbers

EMEP.CH08_3PP.indd 436

è z2 ø

,

cosh x =

co sech x =

(ii) Log æç z1 ö÷ = Logz1 - Logz2 , where z1 and z2 are

(iv) Log (–x) = log x + pi, where x is a real number (v) If a(≠ 0) and z be two complex numbers, then az = e Loga = ezlog a = ez(loga+2npi) z

8.2 CALCULUS OF COMPLEX VALUED FUNCTIONS 8.2.1  Function of a Complex Variable If for each value of a complex variable z(= x + iy), there corresponds one or more than one values of a complex variable w, then we say that w is a function of complex variable z and is denoted by w = f(z). Example: z3 + 2z + 2, sin z, ez + z, etc. are functions of complex variable z. We express w = f(z) by w = f(z) = u(x, y) + iv(x, y) or w = f(z) = u + iv. 8.2.2  Limit of a Complex Valued Function Let D be the domain of the complex valued function f(z) and z0 ∈ D. Then we say that l (a complex number) is a limit of f(z) as z tends to z0 if for a given ε > 0(however small), there exist a number d > 0 (depends on ε) such that

f (z) - l < e

whenever z - z0 < d .

Symbolically, we write lim f ( z ) = l. z ®z0

Remember: (i) The limit of a function, if exist, is unique. (ii) If lim f ( z ) = l1 and lim g ( z ) = l2 then, z ®z0

z ®z0

(a) lim éë f ( z ) ± g ( z )ùû = l1 ± l2 z ®z0 (b) lim éë f ( z ) g ( z )ùû = l1l2 z ®z0 (c) lim éê f ( z ) ùú = l1 provided l2 ¹ 0 z ®z0 ë g ( z ) û l2

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Complex Analysis • 437

8.2.3 Continuity of a Complex Valued Function Let D be a domain of the complex valued function f(z) and z0 ∈ D. Then f(z) is said to be continuous at z = z0 if for a given ε > 0, there exist a number d > 0 (depends on ε) such that f ( z ) - f ( z0 ) < e



whenever z - z0 < d .

Thus, f(z) is continuous at z = z0 if and only if lim f ( z ) = f ( z0 ).

z ®z0

f(z) is said to be continuous on D if f(z) is continuous at every point of D.

Remember: (i) The term regular and holomorphic are also used as synonyms for analytic. (ii) If f(z) is analytic at every point of a finite complex plane, then f(z) is called an entire function. (iii) A point z = a is said to be singular point of f(z) if f ′(a) doesn’t exist. (iv) If f(z) = u + iv be analytic and |z| = constant, then f(z) becomes constant.

Remember:

(v) If f(z) = u + iv be analytic and u = constant, then f(z) becomes constant.

If f(z) and g(z) are defined on a domain D and both are continuous on D, then

8.2.6  Cauchy Riemann Equations If f(z) = u(x, y) + iv(x, y) (= u + iv) be analytic, then

(a) f(z) ± g(z) is continuous on D.



¶u ¶v = ¶x ¶y



ux = vy and vx = –uy

(b) f(z) g(z) is continuous on D. (c) f ( z ) is continuous on D provided g(z) ≠ 0 g( z )

for any z ∈ D 8.2.4 Derivative of a Complex Valued Function Let the complex valued function f(z) is defined on D and z0 ∈ D. Then the derivative of w = f(z) at z = z0 is denoted by f′(z) or dw and is defined by dz



f ¢( z0 ) = lim

h®0

f ( z0 + h ) - f ( z0 ) h

, provided the limit

exists and is independent of the path. Remember: (i) If f(z) is differentiable at z = z0, then f(z) is continuous at z = z0 (ii)

f (z) - f (0) f ¢(0) = lim z ®0 z

8.2.5 Analytic Function A complex valued function f(z) defined on a ­domain  D is said to be analytic at z = z0 if there exist a neighborhood z - z0 < d of z0 at all points of which the function f(z) is differentiable i.e; f′(z) exist. If f′(z) exist at every point of the domain D, then f(z) is said to be analytic in D. Example: The function f(z) = ez is analytic in every finite region of the complex plane.

EMEP.CH08_3PP.indd 437

and ¶v = - ¶u i.e; ¶x

¶y

The above two equations are called the ­Cauchy-Riemann equations (CR Equations). Remember: (i) If ux, vx, uy, vy are all continuous and u, v are differentiable for a function f(z) = u + iv, and if ux = vy, vx = –uy; then f(z) becomes analytic. (ii) If z = r(cos q + i sin q) = reiq, then the polar form of the CR equations are: ¶u 1 ¶v and ¶v = - 1 ¶u . = ¶r

r ¶q

¶r

r ¶q

8.2.7 Conjugate Function If f(z) = u + iv is analytic and if u and v both

2 2 ­satisfy the Laplace equation, i.e., ¶ u2 + ¶ u2 = 0 and

¶x

¶ 2v ¶ 2v + =0. ¶x 2 ¶y2

¶y

Then u and v are called conjugate functions (or conjugate to each other). 8.2.8  Harmonic Function A function g(x, y) is called a harmonic func2 2 tion if ¶g , ¶g , ¶ g2 , ¶ g2 are all continuous and

¶x ¶y ¶x

¶y

¶2 g ¶2 g + = 0. ¶x 2 ¶y2

Example: 1 2

Let g ( x , y) = log( x 2 + y2 ) .

8/7/2023 6:20:35 PM

438 • Engineering Mathematics Exam Prep Then, ¶g = 1 ´ 2

¶x



(C being the constant of integration).

)

x 2 + y 2 ´ 1 - x ´ 2x ¶2 g y2 - x 2 = = , 2 2 ¶x 2 x 2 + y2 x 2 + y2 2



Then, v = M1 dx + N1 dy + C ò ò

¶g 1 2y y = ´ = . ¶y 2 x 2 + y2 x 2 + y2

(



2x x = 2 , 2 x +y x + y2 2

¶ g = ¶y2

(

(

)

)

(

x 2 + y2 ´ 1 - y ´ 2 y

(

2

x +y

2

=

)

2

)

2

(

x -y

2

x 2 + y2

)

2

.

Remember: The common terms in M1 dx and N1 dy should apò ò pear once while calculating v = M1 dx + N1 dy + C . ò ò Case-II: Suppose f(z) = u + iv is analytic and v is given. We need to find u. du =

2 2 Hence, ¶ g + ¶ g = 0. 2 2

¶x

¶y

2 2 Clearly ¶g , ¶g , ¶ g2 , ¶ g2 are all continuous.

¶x ¶y ¶x

¶y

¶y

Remember: If f(z) = u + iv is analytic, then u and v are both harmonic functions. 8.2.9 Construction of an Analytic Function (by Milne Thomson’s method) Case-I: Let u = u(x, y) be given. ì

¶x îï ( z ,0)



æ ¶v ö æ ¶v ö = ç ÷ dx + ç - ÷ dy è ¶x ø è ¶y ø = M1 dx + N1 dy

where M = ¶v , N = - ¶v . 1 1

Hence, g is harmonic.

Then f ( z ) = ïí ¶u ò



¶u ¶u dx + dy ¶x ¶y

-i

üï ¶u ý+C ¶y ( z ,0) ï þ

C being the constant of integration.

¶x

Then u = ò M1 dx + ò N1 dy + C (C being the constant of integration). Remember: The common terms in M1 dx and N1 dy should apò ò pear once while calculating u = ò M1 dx + ò N1 dy + C .

8.3  COMPLEX INTEGRATION

C being the constant of integration.

8.3.1 Curves Let z = x + iy, where x, y ∈ R. The “z” is a ­complex variable and z represents a point (x, y) which moves on the complex plane and makes a curve, say C. Then we say that the curve C is represented by the complex variable “z.”

8.2.10 Construction of Harmonic Conjugate Case-I: Suppose f(z) = u + iv is analytic and u is given. We need to find v.

real variable, then z = j (t ) + iy (t ) represents the curve C.

Case-II: Let v = v(x, y) be given. ì

Then f ( z ) = ïí ¶v ò

¶y îï ( z ,0)

üï ¶v ý+C ¶x ( z ,0) ï þ

Further if x = j (t ) and y = y (t ) , where “t” is a

For example, if z = x + iy and x = a cost, y = a sin t.

æ ¶u ö ¶u = çdy ÷ dx + ¶ y ¶x è ø = M1dx + N1dy

Then z forms a curve x2 + y2 = a2 cos2t + a2 sin2t = a2, which is a circle. Thus, z = a cos t + i(a sin t) represents a circle in the complex plane.

where M1 = - ¶u , N1 = ¶u . ¶y

EMEP.CH08_3PP.indd 438



¶v ¶v dx + dy ¶x ¶y

dv =



+i

¶x

A curve C is called a simple curve if does not intersect itself.

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Complex Analysis • 439 2

=



Curve – 1 (Fig: 1)  

Curve – 2 (Fig: 2)

Here, curve 1 is simple, where as curve 2 is not simple. A simple curve is called closed if the two end points of the curve co-inside. The followings are few examples of closed curves:



2

ò éëê3x (1 - x ) + i (1 - x ) ùûú (1 - i )dx

x =0

2 2ü ì é 2 é x3 ù x3 ù ï ï x 2 = (1 - i ) í3 ê ú + i êx - x + ú ý 3 úû 3 úû ï êë ïî ëê 2 0 0þ

ìï é 4 8 ù éæ ù üï 8ö = (1 - i ) í3 ê - ú + i êç 22 - 4 + ÷ - 0 ú ý 3ø û þï îï ë 2 3 û ëè



8i æ ö = (1 - i ) ç -2 + - 2i ÷ 3 è ø

= -2 + 2i +

2i 2 4 8i + =- + 3 3 3 3

8.3.3 Cauchy-Goursat Theorem Statement: A curve is called contour (piecewise smooth) if it is comprised of a finite number of curves each of which has an unique tangent at every point of it. For example, a circle is a contour.

If f(z) be analytic within and on a simple closed curve C, then, ò f ( z ) dz = 0 . C

Example: Let us find the value of

5

ò z - 3 dz, where C : z = 2.

8.3.2  Complex Line Integral Let f(z) be a complex valued function where z ­varies over a piecewise smooth and simple curve C. Then

Here, f ( z ) = 5

ò

does not exist only for z = 3, i.e., at the point (3, 0).

C

f ( z ) dz

is called the complex line integral along

C

z -3

and so, f ¢ ( z ) = -

5

( z - 3 )2

,

Now, z = 2 Þ x + iy = 2

the curve C. Example: Let us find the value of the integral

2-i

òi (3xy + iy ) dz 2

along the straight line joining z = i and z = 2 – i. Here z = i represents the point A(0, 1) and



Þ x 2 + y2 = 2

⇒ (x – 0)2 + (y – 0)2 = 22,

which represents a circle with center (0, 0) and radius “2” unit. Clearly z = 3 lies outside the circle C.



z = 2 – i represents the point B(2, – 1).



\  x varies from 0 to 2 and y varies from 1 to –1.

Hence, f(z) is analytic within and on C.



Now the equation of AB: y - y1 = x - x1





Þ

y2 - y1

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

x2 - x1



⇒  y – 1 = – x  ⇒  y = 1 – x



\  dy = –dx and so



dz = dx + idy = dx – idx = (1 – i)dx

Let C represents the straight line AB Then, f ( z ) dz = ò C

EMEP.CH08_3PP.indd 439

which

z =2-i

ò (3xy + iy ) dz z =i 2

ò

\ f ( z )dz = C

5

ò z - 3dz = 0 .

C



(by the Cauchy-Goursat theorem)

8.3.4  Cauchy’s Integral Formula If f(z) is analytic with and on a simple closed curve C and “z = a” be any point lying within C, then

ò

C

f ( z ) dz z-a

= 2pi ´ f ( a ) .

Example:

Let us find the value of 2z + 3dz , where z = 2 . ò C

z -1

8/4/2023 5:36:24 PM

440 • Engineering Mathematics Exam Prep 2z + 3 dz , z -1

ous example, we see that C represents a circle with center (0, 0) & radius “2” unit and z = 1 lies within C.

we get f(z) = 2z + 3 and a = 1 (which represents the point (1, 0).

\ By Cauchy’s integral formula on higher order derivatives, we get,



ò

Comparing the given integral with

C



z = 2 Þ x + iy = 2 Þ x 2 + y2 = 2

⇒ (x – 0)2 + (y – 0)2 = 22, which represents a circle with center (0, 0) and radius “2” unit.

f ( z ) dz

2z + 1

Clearly the point z = 1 lies within the circle C. y

ò ( z - 1) dz = ò ( z - a ) 2

C

C

2pi n ´ f ( ) (a ) n!

=

n +1



2pi 1 = ´ f ( ) (1) 1!



= 2pi × f′ (1) = 2pi × 2 = 4pi

[  f(z) = 2z + 1  ⇒  f′(z) = 2 ⇒ f′(1) = 2]

8.4 Taylor and Laurent Series (1, 0) 0, 0

a=1

8.4.1  The Taylor Series Let f(z) be a complex valued function which is analytic within a circle C : z - a = r . Then at every

x (2, 0)

¥



point “z” lying within C, f ( z ) = å an ( z - a )n where n =0 1 (n) an = f (a) .

Hence, by Cauchy’s integral formula,

ò

f ( z ) dz z-a

C

Þ



ò

= 2pi ´ f ( a )

( 2z + 3 ) dz = 2pi ´ f

C

n!

z -1

8.4.2  The Laurent Series Let f(z) be analytic in the closed ring shaped region bounded by the concentric circles C1 : z - a = R1

(1) = 2pi ´ 5 = 10i

and

C2 : z - a = R2

(where

R1 > R2 .

Then at

(  f(1) = 2 × 1 + 3 = 5).

­every point “z” lying in the ring-shaped region,

8.3.5 Cauchy’s Integral Formula on Higher Order Derivatives Let f(z) be analytic within and on a simple closed curve C and z = a be any point lying within C. If the nth order derivative of f(z) at z = a is denoted by f (n)(a), then

f (z) =

f ( z ) dz



ò (z - a)

n +1

C

=

2pi n ´ f ( ) ( a ) ; for n = 0,1,2,.... n!

f ( z ) dz

ò (z - a)

C

2

1 = 2pi ´ f ( ) ( a ) = 2pi ´ f ¢ ( a )

Example:

Let us find the value of

C : z = 2.



(For n = 1)

2z + 1

ò ( z - 1)

C

2

Comparing the given integral with

where 1 bn = 2pi

ò

C2

f ( z ) dz ( z - a )-n +1

¥

å b (z - a) n =1

an =

n

1 2pi

-n

f ( z ) dz

ò (z - a)

n +1

C1

and

8.5.1 Singular Point A point z = a is called a singular point of a complex valued function f (z) if f (z) is not analytic at z = a, i.e; if f′(a) does not exist. Example:

dz ,

where

f ( z ) dz

ò (z - a)

C

n +1

we get f(z) = 2z + 1, a = 1, n = 1. Also in the previ-

EMEP.CH08_4PP.indd 440

å n =0

an ( z - a )n +

8.5 Singularities

In particular,



¥

If f ( z ) =

1 , z -1

then f ¢( z ) =

-1

2

( z - 1)

,

which does not

exist at z = 1. Hence, z = 1 is singular point of f (z) =

1 . z -1

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Complex Analysis • 441

Remark: z = a is a singular point of f (z) ⇔ f (z) has a singularity at z = a. 8.5.2  Types of Singularities 8.5.2.1  Isolated singularity

f (z) =

¥

å a (z - a) n =0



Consider the Laurent series expansion of f(z) about z = a given by

(a)  f ′(a) does not exist, i.e; f (z) is not analytic at z = a.



(b)  f (z) is analytic in the deleted neighborhood of a i.e; f (z) is analytic in (a – ε, a + ε) – {a}, for any pre-assigned ε > 0.



Example: Consider f ( z ) = 1 . This function is analytic z -1

everywhere except at z = 1. Hence, z = 1 is an iso-

f (z) =

¥

å n =0

Example:



A singularity z = a is said to be a removable singularity of f (z) if lim f ( z ) = a finite quantity.



¥

åb (z - a) n

-n

Here,

2

3

1 æ 1 ö 1 æ 1 ö 1 + ç ÷ ´ + ç ÷ ´ + ....¥, z 2 è z 2 ø 2! è z 2 ø 3!

so clearly z = 0 is a singularity of f (z). ¥

å b ( z - 0) n =1

n

-n

=

1 1 1 + 4 + 6 + ....¥, 2 2z 6z z

which contains an infinite number of terms.

Hence, z = 0 is an essential singularity of f (z).

)

ze z - e z - 1 ´ 1 z2

…(i)

1

 ez =1 +

Example:

(

-n

Consider the function f ( z ) = e z2 .

8.5.2.2  Removable singularity

z

n =1

n

contains an infinite number of terms, then z = a is called an essential singularity of f (z).

2

z Suppose f ( z ) = e - 1 .

åb (z - a)

If the principal part of (i) i.e;

1

z ®a

¥

an ( z - a )n +

n =1

lated singularity of f (z).

Then, f ¢( z ) =

n

8.5.2.3  Essential singularity:

A point z = a is said to be an isolated singularity of f (z) if the following conditions are satisfied:



n

=

e z ( z - 1) + 1 z2

\  f′(z) does not exist at z = 0 and so f (z) is

not analytic at z = 0. Hence, z = 0 is a singularity of f (z). Again, lim f ( z ) z ®0



ìæ ü ö z2 z3 + + ....¥ ÷ - 1 ï ï çç1 + z + ÷ z 2! 3! e -1 ï ï ø = lim = lim í è ý z ®0 z ® 0 z z ï ï ï ï î þ



æ ö z z2 = lim ç1 + + + ....¥ ÷ ç ÷ z ®0 2! 3! è ø = 1 + (0 + 0 + ....¥ ) = 1, a finite quantity

8.5.3  Zeros and Poles (A) Zeros of an analytic function: z = a is called a zero of an analytic function f (z) if f (a) = 0. An analytic function f (z) is said to have a zero of order “n” if we can express f (z) in the following form:

f ( z ) = ( z - a )n j( z ) ,

where j(z) is analytic and j(a) ≠ 0 Example: Consider the function f ( z ) = ( z - 1)2

1 . z2 + 4

Then, f ( z ) = 0 Þ ( z - 1)2 ´ 21

=0

z +4

Þ ( z - 1)2 = 0

Þ z = 1, 1

Hence, z = 0 is a removable singularity of f (z)



Remember:

Remember:

If z = a is a removable singularity of f (z), then f (z) can be expressed as:

(i) z = a is called a simple zero of f (z) if z = a is a zero of order 1.

EMEP.CH08_3PP.indd 441

∴  z = 1 is a zero of order “2” of f (z).

8/4/2023 5:13:54 PM

442 • Engineering Mathematics Exam Prep (ii)  If a function f(z) has infinite number of “­ zeros” and these zeros approaches to a number say “α” (finite or infinite), then “α” is called the limit point of zeros of f(z).

Example: Consider the function f ( z ) = e z cos ec

(iii) The limit point of zeros is an isolated essential singularity.

Example: Consider the function



Then zeros of f (z) are given by æ 1 ö or, sin ç ÷=0 èz +2ø

f (z) = 0



æ 1 ö f ( z ) = sin ç ÷. èz +2ø

1 or, = np z+2 or, z = -2 +

1

( sin x = 0

Þ x = np, n Î Z +

)

, n ÎZ+

np So clearly z = –2 is a limit point of zeros, since the zeros -2 + 1 , - 2 + 1 , - 2 + 1 , etc. approaches p 2p 3p to “–2.”

Consequently, z = –2 is an isolated essential singularity of f (z). B. Poles A function f (z) is said to have a pole of order “n” if we can express f (z) in the following way: j( z ) , f (z) = ( z - a )n

lytic.

where ϕ(a) ≠ 0 and ϕ(z) is ana-

ez

Then, f ( z ) =



sin

1 z

.

∴  The poles of f (z) are given by sin 1 = 0 z

or,

1 = np, n Î Z +  sin x = 0 Þ x = np, n Î Z + z

(

or, z =

1 , n ÎZ+ np

So, clearly z = 0 is a limit point of the poles 1 ,

1 1 1 , , , etc. 2p 3p 4 p

p

Consequently, z = 0 is a non-isolated essential singularity of f (z). (V)  f (z) has a pole of order “n” at z = a ⇔ 1 has a zero of order “n” at z = a.

8.6.1  Residue at a Simple Pole The residue of f (z) at a simple pole z = a is defined by Res (z =a) = lim( z - a ) f ( z ) . z ®a

Example: z -1

2 Consider the function f ( z ) = z + 5 3



∴  Res (z = 1) = lim( z - 1) f ( z )

Then z = 1 is a pole of order “3.”



ze z = lim( z - 1) z ®1 z -1

( z - 1)

Remember: (i) If f (z) has a pole of order “k” say at z = a, then f (z) can be express as

å n =0

an ( z - a )n +

k

å n =1

bn ( z - a )-n

(ii) A pole of order “1” is called a simple pole. (iii) If a function f (z) has infinite number of poles and these poles approaches to a number say “β,” then β is called the limit point of poles of f (z). (iv) Limit point of poles is a non-isolated essential singularity.

EMEP.CH08_3PP.indd 442

f (z)

8.6 Residues

pole.

¥

)

z Consider f ( z ) = ze . Then clearly z = 1 is a simple

Example:

f (z) =

1 z



z ®1

= lim ze z = 1 ´ e1 = e. z ®1

8.6.2 Residue at a Pole of Order “n” Let f ( z ) = j( z ) , where j(a ) ¹ 0. n (z - a)

Then z = a is a pole of order “n” and the residue of

f (z) at z = a is defined by



Res (z = a) =

1 ´ j( n -1) ( a ), (n - 1)!

where j( n -1) ( a ) represents the (n – 1)th derivative of ϕ(z) at z = a.

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Complex Analysis • 443

Example: 3

Consider f ( z ) = z + 42 .



( z - 1)

Then clearly z = 1 is a pole of order “2.” Here, ϕ (z) = z3 + 4. Then ϕ(1)(z) = ϕ′(z) = 3z2 and so ϕ(1) (1) = 3 × 12 = 3.

1 j(2-1) (1) = j(1) (1) = 3. (2 - 1)!

∴  Res (z = 1) =

=-

1 = -1. (1 - 0)(1 - 0)

8.6.4 Cauchy’s Residue Theorem Suppose f (z) is analytic within and on a closed curve C except at a finite number of poles, namely, z1, z2,….., zm that lies entirely within C. Then

ò f ( z )dz = 2pi ´ (the sum of residues at the poles)

C

m

8.6.3  Residue at Infinity If an analytic function f (z) has a pole at z = ∞, then Res(z = ∞) = – coefficient of 1 in the expansion z of f (z).



Example:

Let us evaluate the integral ò , where C ( z - 1)( z - 2) C is the circle |z|=4.

z2 ( z - 2)( z - 3)

Consider the function f (z) =

Clearly z = ∞ is a pole of f (z). Now, f ( z ) =

z ( z - 2)( z - 3)



2ö æ 3ö z2 æ = ç1 - ÷ ç1 - ÷ 2 3 z zø æ ö æ ö ø è z 2 ç1 - ÷ ç1 - ÷ è z z è øè ø

-1

2 4 3 9 æ ö æ ö = ç1 + + 2 + ......¥ ÷ ´ ç1 + + 2 + ......¥ ÷ z z z z è ø è ø 4 6 ö æ2 3ö æ 9 = 1 + ç + ÷ + ç 2 + 2 + 2 ÷ + ......¥ z z ø èz zø èz 1

Hence, Res(z = ∞) = –(5) = – 5.







Res(z = ∞) = lim {-z f ( z )} .



z ®¥



Example: z . ( z - 1)( z - 2)

Then clearly f (z) in analytic at z = ∞.

Res(z = ∞) = lim {-z ´ f ( z )} z ®¥

= - lim



EMEP.CH08_3PP.indd 443

z ®¥

1 . ( z - 1)( z - 2)

The poles of f (z) are given by

The poles z = 1 and z = 2 are simple poles and lies within the circle C.

If f (z) is analytic at z = ∞, then



dz

Here, f ( z ) =

Remember:

Consider the function f ( z ) =

i

i =1

(z – 1)(z – 2) = 0, i.e; z = 1, 2.

 ∴  The coefficient of in the expansion of f (z) z is 5.

å Res( z = z ).

Application:



2

-1

=

= 2pi ´

z2 1 = - lim z ®¥ æ 1 öæ 2ö ( z - 1)( z - 2) ç1 - z ÷ ç1 - z ÷ è øè ø

Now Res (z = 1) = lim ( z - 1) f ( z ) z ®1

= lim ( z - 1) ´ z ®1

= lim z ®1

1 ( z - 1)( z - 2)

1 1 = = -1 ( z - 2) 1 - 2

Res (z = 2) = lim( z - 2) f ( z ) z ®2

= lim( z - 2) ´ z ®2

= lim

z ®2 ( z

1 ( z - 1)( z - 2)

1 1 = = 1. - 1) 2 - 1

Hence, by Cauchy’s residue theorem,

ò f ( z )dz = 2pi ´ (sum of residues at the poles)

C



= 2pi × (–1 + 1) = 0. Fully Solved MCQ’s

1. Let z1 and z2 be two complex numbers such that z1 = z2 and arg(z1) + arg(z2) = p. Then, z1 = ? (a) z2 (b) -z2 (c) z2

(d) –z2

8/4/2023 5:13:57 PM

444 • Engineering Mathematics Exam Prep 2. Let z1 and z2 be two complex numbers such that |z1| = |z2| = 1 and arg(z1) + arg(z2) = 0. Then, z1z2 = ? (a) 0 (b) –1 (c) 1 (d) 2 3. If z be a complex number such that, |z – 1| = |z – 3| = |z – i| then z = ? (a) 2 + 2i (b) –2 + 2i (c) –2 – 2i (d) 2 – 2i 4. If (z1, z2) and (z3, z4) be the two pair of conjugate complex numbers, then, æz arg ç 1 è z3

ö æ z2 ÷ + arg ç è z4 ø

ö ÷=? ø

(a) p (b) p 2

(c) 2p

(d) 0

5. The amplitude of the complex number z=

(1 + i 3 )2 is 4i(1 - i 3 )

(a) p (b) p 6

4

(c) p (d) p 3

2

6. If z1 and z2 be two complex numbers such

12. The function f(z) = xy + iy (a) is continuous every where (b) is analytic every where (c) may be analytic (d) satisfies the CR-equations 13. The function f(z) = x2 + iy2 is (a) nowhere analytic. (b) everywhere analytic. (c) continuous nowhere. (d) analytic at finite number of points 14. The function f ( z ) = z is (a) nowhere continuous. (b) nowhere analytic. (c) analytic everywhere. (d) none of these. 15. The function f(z) = log z is (a) analytic everywhere except z = 0 (b) nowhere analytic (c) analytic everywhere (d) none of these

that z1 + z2 = z1 + z2 , then which of the following is true? (a) arg(z1) + arg(z2) = 0 (b) arg(z1) = arg(z2) (c) arg(z1z2) = 0 (d) z1 = -z2 7. For any complex number z1 and z2, given that 2 2 2 z1 + z2 = z1 + z2 . Then

(a) 10 - 8 i (b) -10 + 8 i



(c) -10 - 8 i

(a) Re(z1/z2) = 0 (b) Im(z1/z2) = 0 (c) Re(z1z2) = 0 (d) Im(z1z2) = 0 8. The number of solutions of the equation z 2 = z is (a) 4 (b) 3 (c) 2 (d) 1 9. Let z1 and z2 be two complex numbers such that z1 = 12 and z2 - 3 - 4i = 5 . Then the minimum value of z1 - z2 is (a) 1 (b) 3 (c) 2



(d) 5

10. If x2 + y2 = 1, then 1 + x + iy = ? 1 + x - iy

(a) x + iy (b) x – iy (c) y + ix (d) y – ix 11. If z + 1 is purely imaginary, then z lies on

EMEP.CH08_3PP.indd 444

z +i

(a) a circle (c) an ellipse

(b) a parabola (d) a straight line

16. The values of the integral ò z dz from z = 0 to C

z = 4 + 2i along the curve C given by z = t2 + it is 3

3

(d) none of these

3

17. The value of

ò ( x - y + ix ) dz 2

along a straight

line from z = 0 to z = 1 + i is

(a) 1 ( i - 1) (b) 1 ( i + 1) 3 3

(c) - 1 ( i + 1) (d) i 3

18. The value of

3

1+ i

ò0 ( x

2

)

+ iy dz

along a straight

line joining z = 0 and z = 1 + i is (a) 1 ( i - 5 ) (b) 1 (1 - 5i ) 6

6

(c) 1 ( 5i - 1) (d) 1 ( i - 1) 6 6

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Complex Analysis • 445

19. Let C be any circle enclosing the origin and oriented counterclockwise. Then the value of the integral ò cos2 z dz is

z

C

(a) 2pi (c) –2pi

z

æ z ö , valid in z > 1 , is ç z -1 ÷ è ø

(a) –1

(b) 1

21. Let f ( z ) =

1 z 2 - 3z + 2

(c) - 1 (d) 1 2

2

z

series expansion of f(z) for z > 2 , is (a) 0

(b) 1

(c) 3

(d) 5

22. The Laurent series expansion of the function f (z) =

(a)

1 z 2 (1 - z )

about z = 0 is given by

n =1

¥

(c) f ( z )= 22 - 1 + 1 - å z n z

z

(d) none the these

å

( -1)n

n =0

(c)

z 2n +1 4n +1

z 2n + 3 ( -1)n 2n 4 n =0 ¥

å

(b)

¥

n =0

n +1

¥

(a) f ( z ) = å ( -1)n ( z - 1)n n =0

(b) f ( z ) = å ( -1)n ( z - 1)n +1

EMEP.CH08_3PP.indd 445

(c) 2 - å ( -1)n z n

(d) none of these

n =0



(z

e2z 2

+1

)

2

are given as

(a) –i, i, – i, i (b) –1, 1, i, –i (c) 0, 1, – 1, i (d) none of these 27. Consider f (z) = cosz – sinz. Then, z = ∞ is (a) a pole of order 1 (b) a zero of order 1 (c) a non-isolated essential singularity (d) an isolated essential singularity z

3

( z - 2 ) ( z + 1)4

,

which of the follow-

ì 29. The function f ( z ) = ïí z , z ¹ 0 has



ï î

0,

z=0

(a) a pole of order at z = 0 (b) an essential singularity at z = 0 (c) a zero of order 2 at z = 0 (d) a removable singularity at z = 0 30. Consider the function f ( z ) = 1 - e z

(d) none of these

z = 1 is given by

n =0

z 2 z +4

z 2n -1

å4

24. The Taylor series expansion of f ( z ) =

¥

n =0

sin z

about z = 0 is given by ¥

n =0

¥

n =1

23. The Taylor series expansion of f ( z ) =

(a)

¥

ing is not true? (a) f (0) = 0 (b) z = 2 is a pole of, order 3 (c) z = –1 is a pole of order 4 (d) z = 0 is a zero of order 2

(b) f ( z ) = 1 + 1 + 1 + å z n 2 z

¥

(a) -1 + 2å z n (b) 1 - 2å z n

28. For f ( z ) =

å

z

(d) none of these 25. The Taylor series expansion of f ( z ) = z + 1 z -1 about z = 0 is given by

¥

1 2 f ( z )= - 2 - - 1 + zn z z n =1 ¥





26. The poles of

Then, the coefficient of 13 in the Laurent

2 n =0

(b) 0 (d) undefined

20. The coefficient of 1 in the expansion of log



¥

(c) f ( z ) = 1 å ( -1)n ( z - 1)n

1 z

about

-z

. Then

which of the following is true? (a) z = 0 is a simple pole (b) z = 0 is an essential singularity (c) z = 0 is a removable singularity (d) none of these.

31. Consider the function f ( z ) = z - sin z . Then z z3 = 0 is (a) a pole of order 1 (b) a pole of order 2 (c) a removable singularity (d) an essential singularity

8/4/2023 5:14:01 PM

446 • Engineering Mathematics Exam Prep 32. Consider f ( z ) = cos 1 . Then z = 0 is

Answer key

z



1. (b) 6. (b) 11. (a) 16. (a) 21. (c) 26. (a) 31. (c) 36. (c)

(a) an isolated singularity (b) a non-isolated singularity (c) an isolated essential singularity (d) a non-isolated essential singularity 1

33. For the function f ( z ) = z 2e z2 , z = 0 is (a) a removable singularity (b) a pole of order 2 (c) a simple pole (d) an essential singularity 34. The value of

ò

C

2

(a) 2pie (c) –2pie2

35. The residue of

(a) 1

e2z dz z( z - 1)

Let

, where C: |z| = 2, is

z3 f (z) = 1 + z2

(b) –1 (c)



(a)

(

ez

z z2 + 4

)



(a) 0

\ z1 = eiq1 and z2 = eiq2

Now arg( z1 ) + arg( z2 ) = 0 Þ q1 + q2 = 0 Þ q2 = -q1 .



Let z = x + iy Then z - 1 = z - 3

e -2z dz z2 |z|=1

is

(c) 4pi

(d) 2pi

ò

39. Let f :C ® C be analytic except for a simple

Þ x + iy - 1 = x + iy - 3 Þ ( x - 1) + iy = ( x - 3) + iy Þ ( x - 1)2 + y2 = ( x - 3)2 + y2



Then, the value of

z ®0

{

}

Re s f ( z )

Þ x + iy - 1 = x + iy - i

is

Þ ( x - 1) + iy = x + i( y - 1) Þ 1 + iy = 2 + i( y - 1)

z ®0

(a) g(0) (b) g′(0) (c) lim z f ( z ) z ®0

EMEP.CH08_3PP.indd 446

(d) lim z f ( z ) g ( z ) z ®0

Þ x 2 - 2x + 1 + y 2 = x 2 - 6 x + 9 + y 2 Þ 4x = 8 Þ x = 2 Also z - 1 = z - i

pole at z = 0 and let g :C ® C be analytic. Re s f ( z ) g ( z )

\ z1z2 = eiq1 e -iq1 = e0 = 1

3. (a)

4

(b) –4pi

z1 = r1eiq1 and z2 = r2eiq2 .

Then z1 = z2 = 1 Þ r1 = r2 = 1

(c) 1 e2i (d) - 1 e -2i 38. The value of the integral

é eip = cos p + i sin p = -1 + i ´ 0 = -1ù ë û

Let

, then Res(z = –2i) = ?

2

Now arg( z1 ) + arg( z2 ) = p

2. (c)

2

3 2i (b) 1 -2i e - e 4 8

\ z1 = reiq1 and z2 = reiq2 .



(a) 1 sin 3 (b) - 1 sin 3 (c) 9 sin 3 (d) 0 37. If f ( z ) =

z1 = r1eiq1 and z2 = r2eiq2 .

\ z1 = reiq1 = rei( p-q2 ) = reipe -iq2 = ( -1)re -iq2 = -z2

( z + 1)

2

5. (d) 10. (a) 15. (a) 20. (a) 25. (b) 30. (c) 35. (a)

Þ q1 + q2 = p Þ q1 = p - q2

36. If f ( z ) = sin 3z3 , then Res(z = –1) = ? 4

4. (d) 9. (c) 14. (b) 19. (b) 24. (a) 29. (d) 34. (d) 39. (a)

Then z1 = z2 Þ r1 = r2 = r (say)

at z = ∞ is given by 1 (d) 1 2 2

3. (a) 8. (a) 13. (a) 18. (c) 23. (a) 28. (d) 33. (d) 38. (b) Explanation

1. (b) (b) 2pi(e2 + 1) (d) 2pi(e2–1)

2. (c) 7. (a) 12. (a) 17. (a) 22. (b) 27. (d) 32. (c) 37. (b)

Þ 12 + y2 = 22 + ( y - 1)2



Þ 1 + y2 = 4 + y2 - 2 y + 1 Þ 2 y = 4 Þ y = 2. \ z = x + iy = 2 + 2i.

8/4/2023 5:14:03 PM

Complex Analysis • 447

4. (d) Since z1 and z2 be two conjugate complex numbers, so let z1 = r1eiq and z2 = r1-iq . Since z3 and z4 be be two conjugate complex numbers, so let z3 = r2ei q and z4 = r2e -i q . 1

2

\



(1 + i 3 )2

4i(1 - i 3 )

=

1 + 2i 3 - 3 4i + 4 3



(

=

=

1

3

= p-

)

p p p - = . 3 6 2



Comparing the real and imaginary parts from both sides we get, x 2 - y2 = x .........................(1) 2xy = - y i.e; y(2x + 1) = 0...........................(2) (2) Þ 2x = -1 or y = 0 Þ x = -1/ 2 or y = 0. Case - I : y = 0

= (r1 cos q1 + r2 cos q2 ) + i(r1 sin q1 + r2 sin q2 ) 2

= ( r1 cos q1 + r2 cos q2 ) + ( r1 sin q1 + r2 sin q2 )

\ (1) Þ x 2 - 02 = x Þ x ( x - 1) = 0 Þ x = 0, x = 1 Case - II : x = -1/ 2

2

= r12 + r22 + 2r1 r2 (cos q1 cos q2 + sin q1 sin q2 )

\ (1) Þ y2 = x 2 - x = (1/ 4) + (1/ 2) = 3 / 4

= r12 + r22 + 2r1 r2 cos( q1 - q2 ) éëif cos( q1 - q2 ) = 1ùû

2



( r1 + r2 ) = r1 + r2 = z1 + z2 ,

Now cos( q1 - q2 ) = 1 = cos 0o Þ q1 - q2 = 0

Þy=±

3 . 2

Hence, the solutions are

which satisfies the given condition.

EMEP.CH08_3PP.indd 447

r1 æ r1 p pö ç cos + i sin ÷ = 0 + i 2 2ø r2 è r2 Hence, Re( z1 / z2 ) = 0.

Þ x 2 - y2 + 2ixy = x - iy

\ z1 + z2

Hence, arg( z1 ) = arg( z2 ).

p [Assuming r1 , r2 ¹ 0] 2

8. (a) z 2 = z Þ ( x + iy )2 = x - iy

= r1 (cos q1 + i sin q1 ) + r2 (cos q2 + i sin q2 )



2

=

Then z1 + z2 = r1 eiq1 + r2 eiq2

Þ q1 = q2 .

+ z2

p p Þ q1 = + q2 2 2 æ ö p i çç p +q2 ÷÷ i p 2 è ø z1 r1 e r1 e 2 eiq2 r i = 1 e2 \ = = z2 r2 r2 eiq2 r2 eiq2

z1 = r1 eiq1 and z2 = r2 eiq2

= r12 + r22 + 2r1 r2

2

Þ q1 - q2 =

3 2 - tan -1 -1 2 3

= p - tan -1 3 - tan -1

= z1

Þ cos( q1 - q2 ) = 0 = cos

2 3 + 2i

(

2

Þ r12 + r22 + 2r1 r2 cos( q1 - q2 ) = r12 + r22

é the complex number - 1 + i 3 represents ù ê ú ê the point ( -1, 3 ) that lies on second quadrant ú ê ú êand the complex number 2 3 + 2i represents ú ê ú ë the point (2 3,2) that lies on first quadrant û

=

( r1 cos q1 + r2 cos q2 )2 + ( r1 sin q1 + r2 sin q2 )2

\ z1 + z2

-1 + i 3

)

= p - tan -1

Let

= ( r1 cos q1 + r2 cos q2 ) + i( r1 sin q1 + r2 sin q2 )

\ z1 + z2

4i(1 - i 3 )

= arg -1 + i 3 - arg 2 3 + 2i

6. (b)

Then z1 + z2 = r1eiq1 + r2eiq2

1 + 2i 3 + (i 3 )2

æ -1 + i 3 ö \ arg( z ) = arg ç ç 2 3 + 2i ÷÷ è ø



= r22

= r12 + r22 + 2r1 r2 cos( q1 - q2 )

5. (d)



2

= r1 (cos q1 + i sin q1 ) + r2 (cos q2 + i sin q2 )

æz ö æz ö Then arg ç 1 ÷ - arg ç 2 ÷ z è z4 ø è 3ø = ( q1 - q2 ) - { -( q1 - q2 )} = 0.

=

2

Then z1 = r12 and z2

2

z1 r1 i( q1 -q2 ) z2 r1 -i( q1 -q2 ) = e , = e z3 r2 z4 r2

z=

7. (a) Let z1 = r1eiq1 and z2 = r2eiq2 .

1

x = 0, y = 0; x = 1, y = 0; x = -1 / 2, y = 3 / 2;



x = -1 / 2, y = - 3 / 2.



Thus, the total number of solutions = 4.

8/4/2023 5:14:05 PM

448 • Engineering Mathematics Exam Prep 9. (c)

æ z +1ö Re ç ÷ = 0. èz +iø æ z +1ö Now Re ç ÷=0 èz +iø

z1 = ( z1 - z2 ) + ( z2 - 3 - 4i ) + ( 3 + 4i ) Þ z1 £ z1 - z2 + z2 - 3 - 4i + 3 + 4i Þ 12 £ z1 - z2 + 5 + 32 + 42

Þ

Þ z1 - z2 ³ 12 - 5 - 5 = 2 Þ z1 - z2 ³ 2.



(1 + x + iy )(1 + x + iy ) = (1 + x + iy )2 (1 + x - iy )(1 + x + iy ) (1 + x )2 + y2 2

(1 + x ) + 2iy (1 + x ) - y (1 + x )2 + y2 1 + 2x + x 2 + 2iy (1 + x ) - y2 = 2

=



which is an equation of a circle.

12. (a) f(z) = xy + iy = u + iv. Therefore, u = xy, v = y

[ x 2 + y2 = 1]

Hence, ¶u ¹ ¶v & ¶v ¹ - ¶u .

=

¶x

2x + 2x + 2iy (1 + x ) 2

2 (1 + x )

¶x

¶y

¶y

Now ¶u = 2x , ¶u = 0, ¶v = 0, ¶v = 2 y. ¶x

x + ( y + 1)

¶x

2

2

x + x - ix - iy + y2 + y - i 2

2

x 2 + ( y + 1)

-i

x + y +1

¶y

Moreover, u = x2, v = y2 being both polynomials, are continuous and so f(z) is continuous everywhere. 14. (b)

2

x 2 + ( y + 1)

Now if z + 1 is purely imaginary, then z +i

¶y

the function is not analytic.

x 2 + ( y + 1) x 2 + y2 + x + y

¶x

Thus, CR-equations are not satisfied and so

x 2 + x + ixy - ix ( y + 1 ) - i ( y + 1 ) + y ( y + 1 ) 2

¶y

Hence, ¶u ¹ ¶v .

( x + 1 + iy ) {x - i ( y + 1)} = x { + i ( y + 1)}{x - i ( y + 1)}

=

¶y

¶x

13. (a) f(z) = x2 + iy2 = u + iv ⇒ u = x2, v = y2

z +1 z +i x + iy + 1 ( x + 1) + iy = = x + iy + i x + i( y + 1)

=

¶y

Thus, CR-equations are not satisfied and so the function is not analytic.

= x + iy.

11. (a)

=

¶x

1 + 2x + 1

2 + 2x 2x (1 + x ) + 2iy (1 + x )

2

2

2

æ 1 ö 1ö 1ö æ æ Þ çx + ÷ + ç y + ÷ = ç ÷ , 2ø 2ø è è è 2ø

Now, ¶u = y, ¶u = x , ¶v = 0, ¶v = 1.

=

EMEP.CH08_3PP.indd 448

2

1 æ1 ö æ1 ö =ç ÷ +ç ÷ = 2 2 2 è ø è ø

x 2 + y2 + 2x + x 2 + 2iy (1 + x ) - y2

=



2

1 æ1 ö æ1 ö 2 ÷ + y + 2´ ´ y + ç ÷ 2 è2ø è2ø

Since u and v are both polynomials in x and y, so they are continuous, and hence f(z) is continuous everywhere.

1 + 2x + x 2 + y 2



2

= 0 Þ x 2 + y2 + x + y = 0

´x +ç

2

1 + x + iy 1 + x - iy



1

2

10. (a)

=

2

x 2 + ( y + 1)

Þ x2 + 2 ´

Therefore, the minimum value of z1 - z2 is 2.



x 2 + y2 + x + y



f ( z ) = z = x + iy = x - iy = u + iv Þ u = x ,v = - y.

Now, ¶u = 1, ¶u = 0, ¶v = 0, ¶v = -1. ¶x

¶y

¶x

¶y

8/4/2023 5:14:06 PM

Complex Analysis • 449 \



18. (c) z = 0 = 0 + i × 0 represents the point A(0,0) &

¶u ¶v ¹ . ¶x ¶y

Thus, CR-equations are not satisfied and so the function is nowhere analytic. 15. (a) f ( z ) = log z Þ f ¢( z ) =



z = 0.

1 z

which is not defined at

So f′(z) exist except at z = 0. Thus, f(z) is analytic everywhere except at z = 0.



z = 1 + i represents the point B(1, 1).



\  x varies from 0 to 1:



Now, the equation of AB is: y - 0 = x - 0



i.e., y = x and so dy = dx.



\  dz = dx + idy = dx + idx = (1 + i)dx



\  the given integral 1

16. (a)

=

z = t 2 + it



2



ìï t + it = 0 for z=0 Þí 2 ïît + it = 4 + 2i for z = 4 + 2i



z=0 ìt = 0, for Þí ît = 2, for z = 4 + 2i



ò



C 2

=



ò (t

ò (t

2

C 2

)

0



)

2

ò (2t

3

)

- 2it 2 + t 2i + t dt

0



Now, the equation of AB is: y - 0 = x - 0

i.e., y = x and so dy = dx.

1-0

1-0



\  dz = dx + idy = dx + idx = (1 + i)dx



Hence, the given integral

ò ( x - x + ix ) (1 + i ) dx 2

EMEP.CH08_3PP.indd 449

n +1

we get f(z) = cos z, a = 0, n = 1.

f ( z ) dz cos z 2pi n = ´ f ( ) (a ) dz = 2 n +1 n ! z z a ) C C(

ò

=

2pi 1 ´ f ( ) (0) 1!

= 2pi × f′(0) = 2pi × 0 = 0.

[  f(z) = cos z  ⇒  f′(z) = – sin z  ⇒  f′ (0) = 0]. 20. (a)

1ö æ z ö æ log ç ÷ = - log ç1 - z ÷ è z -1 ø è ø



1 1 é1 ù = - ê + 2 + 3 + .....ú 3z ë z 2z û



é ù x2 x3 + + .......ú êusing log(1 - x ) = x + 2 3 ëê ûú

 So, coefficient of 1/z in the expansion of

0

3

1

éx ù = (1 + i ) x 2dx = ( i - 1) ê ú = 1 ( i - 1) ëê 3 úû 0 3 0

ò

)

Clearly, z = 1 lies within C.  \  By Cauchy’s integral formula on higher order derivatives, we get,



\  x varies from 0 to 1:

1

(



1 16i 8i 8i ´ 16 + + 2 = 10 - . 2 3 3 3

1

1 1 2 + 2i + 3i + 3i2 = ( 5i - 1) . 6 6

ò





=

C

17. (a) z = 0 = 0 + i × 0 represents the point A(0,0) & z = 1 + i represents the point B(1, 1).

=

)

f ( z ) dz

é t4 t3 t3 t 2 ù = ê2 - 2i ´ + i + ú 3 3 2 úû êë 4 0 =

1

é x 3 ix 2 ù + ix (1 + i ) dx = (1 + i ) ê + ú 2 ûú ëê 3 0

æ1 i ö 1 = (1 + i ) ç + ÷ = (1 + i )( 2 + 3i ) è3 2ø 6

- it dz

- it ( 2t + i ) dt =

2

ò (z - a)

2



0

\  the given integral = z dz =

ò (x

1-0

19. (b) Comparing the given integral with

Also dz = (2t + i)dt

1-0

æ z ö . log ç ÷ = -1, z > 1 è z -1 ø

8/4/2023 5:14:08 PM

450 • Engineering Mathematics Exam Prep 21. (c)

1 1 f (z) = 2 = z - 3z + 2 ( z - 2 )( z - 1)



=





( z - 1) - ( z - 2) 1 1 . =+ ( z - 2 )( z - 1) ( z - 1) ( z - 2 )

f (z) 1 1 -1 = = = {1 + ( z - 1)} z 1 + ( z - 1)

⇒ z > 2 and z > 1 Þ



=-



=







1 1 + z (1 - 1 / z ) z (1 - 2 / z ) -1

1ö æ - ç1 - ÷ zø è

-1 ù

ú úû





1 3 1 é1 3 ù = ê + 2 + ...ú = 2 + 3 + ...... z z z ëz z û



So, the coefficient of 1/z3 = 3.



(

(

23. (a)



EMEP.CH08_3PP.indd 450

zæ z2 ö = ç1 + ÷ 4 ÷ø æ z 2 ö 4 çè 4 ç1 + ÷ ç 4 ÷ø è

å

z

n

å

¥ ö z 2n +1 ( -1)n n +1 ÷÷ = 4 ø n =0

å

(z

e2z 2

+1

)

2

.

So the poles of f (z) are given by (z2 + 1)2 = 0

or, (z + i)2 (z – i)2 = 0

)

)

æ z2 z = ( -1)n ç ç 4 4 n =0 è

¥ æ ö xn ÷ ç using (1 - x )-1 = ç ÷ n =0 è ø

f (z) = 0 or, cos z - sin z = 0 p or, tan z = 1 = tan 4 p or, z = np + , n Î Z + 4

( tan x = tan a Þ x = np + a,



n ÎZ+

)

 ∴  The zeros of f (z) are p + p , 2p + p , 3p + p , 4 4 4 and so on. Clearly z = ∞ is a limit point of these zeros. Hence, z = ∞ is an isolated essential singularity.

å

¥

å

27. (d) The zeros of f (z) are given by

¥



¥

2 = 1 - 2(1 - z )-1 = 1 - 2 z n 1-z n =0

or,  z = i, i, – i, – i.

1 1 = 2 + +1 + zn . z z n =1

z f (z) = 2 = z +4

å

or, {(z + i)(z – i)}2 = 0

1 1 = (1 - z )-1 z 2 (1 - z ) z 2

1 1 + z + z 2 + z 3 + z 4 + .......¥ z2 1 1 = 2 + + 1 + z + z 2 + ........¥ z z



¥ é ù -1 ( -1)n x n ú êusing (1 + x ) = n =0 ëê ûú

2 z + 1 ( z - 1) + 2 = =1+ z -1 z -1 z -1

Here, f ( z ) =

éusing (1 - x )-1 = 1 + x + x 2 + x 3 + ....... for x < 1ù ë û

=

n

26. (a)





f (z) =



1 1 æ öù - ç1 + + 2 + ... ÷ ú z z è øû

f (z) =

n

n =0

=1-

1 éæ 2 4 ö êç1 - + 2 + ... ÷ z ëè z z ø

22. (b)

¥

å ( -1) ( z - 1)

25. (b)

1 1 + 1 z z ( ) ( - 2)

1 éæ 2ö = êç1 - ÷ z êè zø ë



=

2 1 < 1and < 1 z z

f (z) = -

å

24. (a)

Now, z > 2

¥ æ ö ( -1)n x n ÷ ç using (1 + x )-1 = ç ÷ n =0 è ø

-1

28. (d) The zeros of f (z) are given by f (z) = 0 or,



z

3

( z - 2 ) ( z + 1)4

=0

or, z = 0

8/4/2023 5:14:10 PM

Complex Analysis • 451



∴  z = 0 is a zero of order “1.”



The poles of f (z) are given by:

or,

1 p = 2np + z 2

(z – 2)3 (z + 1)4 = 0

or, z =

or  (z – 2)3 = 0, (z + 1)4 = 0 or z = 2, 2, 2 and z = –1, –1, –1, –1.  ∴  z = 2 is a pole of order “3” and z = –1 is a pole of order “4” 29. (d)

sin z lim f ( z ) = lim = 1, z ®0 z ®0 z



a finite quantity,



f (z) =



(

Now, lim 1 - e z ®0

-z

0ö æ ç form 0 ÷ è ø

z

0 + e-z z ®0 1 = lim e - z = e -0 = 1, a finite quantity z ®0

Hence, z = 0 is a removable singularity of f (z). 31. (c) Clearly, z = 0 is a singularity of f (z). Now,

z - sin z lim f ( z ) = lim z ®0 z ®0 z3

æ ö z3 z5 z7 + + .....¥ ÷ z - çz ç ÷ 3! 5! 7! è ø = lim z ®0 z3 ö 1 æ z 3 z 5 z7 = lim 3 ç + - .....¥ ÷ ÷ z ®0 z ç 3! 5! 7! è ø 2 4 æ1 z ö z = lim ç + - .......¥ ÷ ÷ z ®0 ç 3! 5! 7! è ø 1 1 = - 0 + 0 - ......¥ = , a finite quantity 3! 6





EMEP.CH08_3PP.indd 451

z

2

2 3 é ù 1 æ 1 ö 1 æ 1 ö 1 = z 2 ê1 + 2 + ç 2 ÷ ´ + ç 2 ÷ ´ + .....¥ ú z è z ø 2! è z ø 3! êë úû 1 1 = z 2 + 1 + 2 + 4 + ......¥ 2z 6z

Since infinite number of items containing the negative power of “z – 0” exist, so z = 0 is an essential singularity. 2z 34. (d) Here, f ( z ) = e dz . So the poles of f (z) ò C

z( z - 1)

are given by z(z – 1) = 0 i.e; z = 0, 1. The poles z = 0 and z = 1 are simple poles and lies within C.

e2z e0 = = -1 . z ®0 z - 1 0 -1

Now, Res(z = 0) = lim zf ( z ) = lim z ®0

Res(z = 1) = lim( z - 1) f ( z ) = lim e z ®1

z ®1

2z

=

z

e2 = e2 . 1



∴  By the Cauchy’s residual theorem,



e dz ò z( z - 1) = 2pi ´ ( -1 + e ) = 2pi ( e

2z

2

C

2

)

-1 .

35. (a) f (z) =



z3 = 1 + z2

z3

1 ö æ = z ç1 + 2 ÷ 1 ö z ø 2æ è z ç1 + 2 ÷ z ø è

-1

2 1 æ 1 ö ïì ïü = z í1 - 2 + ç 2 ÷ - .......¥ ý èz ø îï z þï

32. (c) The zeros of f (z) are given by f (z) = 0 or, cos 1 = 0

2

1

Hence, z = 0 is a removable singularity of f (z).

2

p 2

f ( z ) = z 2e z

= lim



, n ÎZ+

33. (d)

)

[  f ′(0) does not exist]

p 2np + 2

So clearly, z = 0 is a limit point of zeros. Hence, z = 0 is an isolated essential singularity.

ze - z - 1 - e - z ´ 1 1 - e-z Þ f ¢( z ) = z z2

⇒  z = 0 is a singularity of f (z)



1

1 1 Thus, the zeros of f (z) are , , p p 2p + 4p + , and so on.

6p +

So z = 0 is a removable singularity of f (z). 30. (c)

1

p æ +ö ç cos x = 0 Þ x = 2np + 2 , n Î Z ÷ è ø



=z-

1 1 + - .......¥ z z3

8/4/2023 5:14:12 PM

452 • Engineering Mathematics Exam Prep Res(z = ∞)

39. (a) Let f :C ® C be analytic except for a sim-



= – coefficient of 1 in the expansion of f (z)

ple pole at z = 0 and g :C ® C be analytic.



= – (–1) = 1.

Then, the value of

36. (c) f (z) =



z

sin 3z j( z ) = ( z + 1)3 ( z + 1)3

{

}

Res f ( z ) ´ g ( z ) z ®0

(say)

Res f ( z ) z ®0

lim( z - 0) f ( z ) ´ g ( z ) z ®0

(where j(z) = sin 3z)

=

Clearly, z = –1 is a pole of order “3.”

lim z f ( z ) = g ( 0 ) ´ z ®0 = g (0). lim z f ( z )

Here, j′(z) = 3 cos 3z, j′′(z) = –9sin 3z So, j(2) ( -1) = j¢¢( -1) = -9 sin( -3) = 9 sin 3 ∴ Res(z = –1) =



z ®0

1. The general values of ii is given by (a) e

(

ez

z z2 + 4

=

)

ez z( z + 2i )( z - 2i )

(b) e (c) e

ez e -2i = z ®-2i z( z - 2i ) -2i( -2i - 2i )



ö

-çç 4n - 1 ÷÷ p 2ø è æ

ö

è2

î

-2 z

z

2

=

e -2z j( z ) = (say) ( z - 0)2 ( z - 0)2

where j( z ) = e -2z .  ∴  z = 0 is a pole of order 2 and lies within the circle |z| =1. Now, j¢( z ) = -2e

è2

ø

øþ

3p (b) e - 4 ìícos æ 1 log 2 ö + i sin æ 1 log 2 ö üý ç ÷ ç ÷

î

Here, f ( z ) = e

(n Î Z )

p

38. (b)

-2 z

0

and so j¢(0) = -2e = -2 .

1 j(2-1) (0) = j(1) (0) = -2 . (2 - 1)!



∴ Res(z = 0) =



Hence, by Cauchy’s residue theorem,



e -2z dz = 2pi ´ Re s( z = 0) = 2pi ´ ( -2) = -4 pi. 2 z |z|=1

EMEP.CH08_3PP.indd 452

-çç 2n - 1 ÷÷ p 2ø è

(n Î Z )

(a) e - 4 ìícos æç 1 log 2 ö÷ + i sin æç 1 log 2 ö÷ üý

1 e -2i = - e -2i . 8 8i2

ò

ö

(d) e (n Î Z ) 2. The principal value of ( -1 + i )i is given by

ez z( z + 2i )( z - 2i )

= lim =

(n Î Z )

-çç 4n + 1 ÷÷ p 2ø è æ

z ®-2i

z ®-2i

ö

æ

∴ Res(z = –2i) = lim {z - ( -2i )} f ( z ) = lim ( z + 2i ) ´

-çç 2n + 1 ÷÷ p 2ø è æ

Clearly, z = –2i is a simple pole of f (z).



Fully Solved MCQ’s (Level-II)

37. (b) f (z) =

z ®0

1 j(3-1) ( -1) (3 - 1)!

1 9 = j(2) ( -1) = sin 3 2! 2



lim( z - 0) f ( z )

(c)

p e4

è2

è2

ø

øþ

ì æ1 ö æ1 öü ícos ç log 2 ÷ + i sin ç log 2 ÷ ý ø è2 øþ î è2

3p (d) e 4 ìícos æç 1 log 2 ö÷ + i sin æç 1 log 2 ö÷ üý

è2

î

ø

è2

øþ

3. sin æç i log a - ib ö÷ = ? a + ib ø

è

(a)

2ab a - b2

(b)

ab a 2 - b2

(c)

2ab a + b2

(d)

ab a + b2

2

2

2

8/4/2023 5:14:14 PM

Complex Analysis • 453

10. If a and b be two complex numbers such that

4. The equation tan æç i log x - iy ö÷ = 2 represents è



x + iy ø

(a) an ellipse (b) a hyperbola (c) a circle (d) a rectangular hyperbola 5. If sin (a + ib) = x + iy, then which of the following is true?

(a)

x2 y2 =1 2 sin a cos2 a

(c)

x2 y2 + =1 2 cosh b sinh 2 b

(d)

2

(a) –2

2



(c) 0

(d) 1

2

(a) 2sin p æ cos p + i sin p ö 10 ç 10 10 ÷ ø

(b) 2sin p çæ cos p + i sin p ÷ö 5 10 10 è

(c) 2 cos p çæ cos p + i sin p ÷ö 10 10 10 ø

(d) 2 cos p çæ cos p + i sin p ÷ö 10 10 10 è

ø

9. Let z be a complex such that 1 - iz = 1 . Then, z -i

(a) z is any complex number (b) z = 0 (c) z is purely real (d) z = ± i

EMEP.CH08_3PP.indd 453

) (d) ±i 3

(a) 1

(b) 4

(c) 3

(d) 8

0

,z = 0

14. For the function

ì xy2 ( x + iy ) ,z ¹ 0 ï f ( z ) = í x 2 + y4 ï 0 ,z = 0 î

which of the followings is true? (a) f(z) is analytic at z = 0 (b) f(z) is not analytic at z = 0 (c) f′(0) does not exist (d) both (b) and (c) are correct 15. Consider f ( z ) = xy . Then,



ø

è

3 +i

(a) CR equations are satisfied at z = 0 (b) CR equations are not satisfied at z = 0 (c) f(z) is analytic at z = 0. (d) none of these 16. The function f ( z ) = z is

5ø

è

(

(a) is analytic at (0, 0) (b) not analytic at (0, 0) (c) f′(0) exist (d) none of these.

p æ pö sin + i ç1 - cos ÷ is è

2

2

ï î

(a) 1 (b) –1 (c) 0 (d) 4 8. The polar form of the complex number 5

2

ì x 2 y5 ( x + iy )

7. If zn = cos pn + i sin pn , then z1z2z3..... ∞= ?

(d) 1

(c) 4

13. The function f ( z ) = ïí x 4 + y10 , z ¹ 0

x

(b) –1

is

(a) ± 1 (1 + i ) (b) ± 1 (1 - i )



6. If x + 1 = 2 cos p ,then x 5 + 15 = ?

(b) 1

b-a 1 - ab

12. The product of all the values of (1 + i 3 ) 4 is

x y =1 cosh 2 b sinh 2 b 5

(a) 2 11. i = ?

2

2

x



(c) ± 1

x2 y2 + =1 2 sin a cos2 a

(b)

a ¹ b and b = 1 . Then the value of

(a) differentiable for all z (b)  differentiable except finite number of points (c) differentiable except at z = 0 (d) not continuous 17. The harmonic conjugate of u(x, y) = x3 – 3xy2 is (a) u = y3 + 3x2y (b) u = y3 – 3x2y (c) u = –y3 + 3x2y (d) u = –y3 – 3x2y 18. Let f(z) = u + iu be analytic and u = x2 – y2. Then v = ? (a) 2xy (b) –2xy (c) x2y2 (d) x2 + 2y2

8/4/2023 5:14:16 PM

454 • Engineering Mathematics Exam Prep 19. Find an analytic function whose real part is u(x, y) = ex(x cos y – y sin y) (a) –ze–z + C (b) ze–z + C (c) –zez + c (d) zez + C 2

2

20. The harmonic conjugate of v = x – y – 2xy + 2x – 3y is ? (a) x2 + y2 – 2x + 3y (b) x2 + y2 – 2xy + 3x + 3y (c) x2 – 2x + y2 (d) –x2 + y2 – 2xy – 3x – 2y 21. The analytic function f(z) whose imaginary part v is given by v = e–x(x cos y + y sin y) is (a) ize–z + c (b) izez + c (c) ze–z + c (d) zez + c 22. If f(z) = u + iu be analytic and v – u = ex(cos y – sin y), then f(z) = ? (a) ze–z + c (b) e–z + c (c) ez + c (d) zez + c 23. Let us consider a function f(z) defined by ì 2 ï(z) f (z) = í z , z ¹ 0 ï z=0 î 0,

25. The value of

ò (z

C

zdz

2

)

+ 9 (z - i)

26. The value of the integral is the circle z + 3i = 1 , is

EMEP.CH08_3PP.indd 454

, where c : z = 2 is

p (b) p (c) p (d) p 4 4 2 2

(a) –2

(b) 2pie2p (d) 1

(a) –2p (c) 0 28. The value of integral

(b) –3

ò

C

dz , z ( z + pi )

(c) –4

where C

(d) 0

òz

C

2

dz + 2z + 2

where C is

the square having vertices at (0, 0), (–2, 0), (2, – 2), and (0, – 2), is given by (c) –p (d) 2p

(a) p (b) p

3

2

29. The value of

sin z dz

òæ

, where C : z -

3

pö çz - 4 ÷ è ø

C

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

2

2

p 1 = , is 4 2

(d) 0

-z 30. The value of e 2dz , where c : z = 1 , is ò

z

C

(c) 2pi (d) -pi

(b) –2pi

ò

z = eiq ,0 £ q £ p,

(a) C  R equations are not satisfied at (0, 0) and f ′(0)exist (b) CR equations are not satisfied at (0, 0) (c) f′(0) does not exist (d) f (z) is not analytic at z = 0 although CR equations are satisfied at z = 0 24. Let u(x, y) = 2x(1 – y) for all real x and y. Then, a function v(x, y), so that f(z) = u(x, y) + iv(x, y) is analytic, is (a) x2 – (y – 1)2 (b) (x – 1)2 – y2 (c) (x – 1)2 + y2 (d) x2 + (y – 1)2





31. The value of



(a)

C

z - 2 + z + 2 = 6 , is

(a) pi

Then,

given by

e2z dz

ò z - pi , where C is the ellipse

27. The value of

C

2

1 dz , z

where C is the circle

is

(a) - pi (b) pi 2

(c) –pi (d) pi

2

32. The coefficient of (z – p)2 in the Taylor series expansion of ì sin z , if ï f (z) = í z - p ï -1, if î



z¹p z=p

around p is (a) 1 (b) - 1 (c) 1 (d) - 1 3

2

6

6

33. The Taylor series expansion of the function f ( z ) =

z2 + 1 ( z + 2)( z + 3)

in the region |z| < 2 is

given by ¥ ü n (a) 1 + å ( -1)n ìí n5+1 - 10 z n +1 ý n =0

î2

3

þ

¥ ü n (b) 1 - å ( -1)n ìí n5+1 - 10 z n +1 ý n =0

î2

3

þ

8/4/2023 5:14:18 PM

Complex Analysis • 455 ¥ (c) 1 + å ( -1)n ìí 3n - 2n üý z n

2



î2

n =0

(c)

3 þ

Laurent

f (z) =

(a)

1 z 2 - 5z + 6

series

expansion

n

¥

1 1 æzö æzö ( -1)n ç ÷ + ( -1)n ç ÷ 3 n =0 2 n =0 è3ø è2ø

å

å

n

¥

¥

(b) 1 å ( -1)n æç z ö÷ - 1 å ( -1)n æç z ö÷ 3 n =0

è3ø

n

¥

2 n =0

¥

(c) - 1 å æç z ö÷ - 1 å æç 2 ö÷

3 n =0 è 3 ø

n

f (z) =

C

where C is the circle |z - 2|= 1 , is

pö æ z- ÷ 4ø æ p np ö çè cos ç + ÷ n! è4 2 ø n =0

å

(b)

pö æ z- ÷ ¥ ç p p n 4 æ öè ø sin ç + ÷ 4 2 ! n è ø n =0

pö æ np p ö æ sin ç - ÷ç z - ÷ 4 4 4 è øè ø n =0

å

n =0

EMEP.CH08_3PP.indd 455

(a) 2pi(1 + 15i)

(b) 2pi(1 – 15i)



(d) 2pi

(c) 4pi(1 + 15i)

2

n

n

¥

1 ö 2n ç1 - n +1 ÷ z 2 è ø

næ

å ( -1)

n =0

z +4

2

2

2

Answer key

)

1 ö 2n (b) ÷z ø

n +1

(z - i)

p (a) - (b) p (c) - ip (d) ip

(d) none of these 37. The Taylor series expansion of 1 is powers of z when |z| f (z) =

æ

(d) –2pi

C

å ¥

(c) 2pi

42. Let C be the positively oriented circle given by z - 3i = 2 . Then, the value of ò 2dz is

n

(a)

¥

(b) pi

(a) 0

C

¥

å çè1 + 2

,

f ( z ) dz is equal to 15 ò

4

(a)

2

n =0

about z = p is given by

)(

zdz

ò ( z - 1)( z - 2)

15

(d) none of these 36. The Taylor series expansion of f(z) = cosz

< 1, is given by

21

41. Let f ( z ) = å z n for z Î C. If C : z - i = 2 , then

n =0

+ 1 z2 + 2

16i

ø

40. The value of the integral

n =0

2

16

24

24



¥ (c) - 1 + 2å ( -1)n ( z + 1)n

(z

16

è

¥ (b) 1 + å ( -1)n ( z + 1)n



3

2

2 z =0

(c)

)

39. The residue of z 3 cos æç 1 ö÷ at z = 2 is g­ iven by z -2

about z = –1 is given by

(a) 1 å ( z + 1)n



+1

11 (a) - 143 (b) (c) - 143 (d) 0

¥

z +1

1 2

16i

(d) none of these 35. The Laurent series expansion of

z +1

(d) none of these

(a) 3 (b) 3i (c) 5i (d) 5

è2ø

z n =0 è z ø

z f (z) = ( z + 1)( z + 2)

(z



n

n



38. Find the residue of f (z) at z = i if

of

in 2 < |z| < 3 is given by

¥

1 ö n z n ÷ ø

æ

n =0

(d) none of these 34. The

¥

å çè1 - 2

1. (a) 6.(a) 11.(a) 16.(c) 21.(a) 26.(a) 31.(d) 36.(a) 41.(a)

2.(b) 7.(b) 12.(d) 17.(c) 22.(c) 27.(c) 32.(c) 37.(b) 42. (b)

3.(c) 8.(a) 13.(b) 18.(a) 23.(d) 28.(c) 33.(a) 38.(a)

4.(d) 9.(c) 14.(d) 19.(d) 24.(a) 29.(a) 34.(c) 39.(a)

5. (c) 10. (b) 15. (a) 20. (d) 25. (d) 30. (b) 35. (c) 40. (d)

8/4/2023 5:14:22 PM

456 • Engineering Mathematics Exam Prep 3. (c) Let a = r cos q and b = r sin q.

Explanation

1. (a)



ì i ïílog îï

=e

ü



( log1 = 0 )

-ç 2n+ 1 ÷ p ç 2 ÷ø è



ö



2

( -1)

Let z = -1 + i.Then r = z = amp ( z ) = q = p - tan

amp ( z ) = p - tan -1

2

+ 1 = 2.

1 p 3p = p- = . -1 4 4

-1

1 p 3p = p- = . -1 4 4



3p 3p ö æ \ z = r ( cos q + i sin q ) = 2 ç cos + i sin . 4 4 ÷ø è



Now, (–1 + i)i i

Log æç -1+i ö÷



= eiLogz = e



=e

è

ø

è

{

}

{

(taking z = – 1 + i)

}

i log z +iamp( z )+2npi

i ïílog 2 + i 34p + 2npiïý ì

ü

ïî

ïþ

ilog

2 - 3p -2np 4 ,



Then the principal value is obtained by

=e

i log 2 - 3p 4

putting n = 0 and is given by e



EMEP.CH08_3PP.indd 456

i.e., e



Þ tan i ´ ( -2iq ) = 2 Þ tan ( 2q ) = 2

{

(

- 3p ïícosæç log 2 ö÷ +i sinæç log 2 ö÷ ïý 4 îï è ø è ø þï ü

ì

ü

î

þ

æ ö æ ö - 3p ïícosçç 1 log2 ÷÷+i sinçç 1 log2 ÷÷ïý 4 ï è2 2 ø è øï

}

y x =2 2 æ yö 1-ç ÷ èxø 2

which is a rectangular hyperbola.

⇒ sin a cos (ib) + cos a sin (ib) = x + iy Þ sin a cos hb + i cos a sin hb = x + iy

n∈z

.



ëé cos(ib) = cosh b and sin(ib) = i sinh bûù



⇒  x = sin a cos hb, y = cos a sin hb



.

Þ xy = x 2 - y2 ,

)

ö -2iq =2 ÷÷ = 2 Þ tan i log e ø

5. (c) sin (a + ib) = x + iy

=e

i.e., e



æ e - iq Þ tan ç i log iq ç e è



ø







é ïì r ( cos q - isin q ) ïüù Þ tan êi log í ýú = 2 êë îï r ( cos q + i sin q ) þïúû

2tanq Þ =2Þ 1 - tan2 q

iLog æç -1+i ö÷

=e

x

æ x - iy ö tan ç i log ÷=2 x + iy ø è



i log z +2npi

ì

b a = 2ab . 2 a 2 + b2 æbö 1+ç ÷ èaø 2´

æ b r sin q ö ç a = r cos q = tan q ÷ è ø

Then,

=e

)

Let x = r cos q, y = r sin q. Then, tan q = y .

  z = –1 + i represents a point (–1, 1) that lies in second quadrant, so

reiq

4. (d) , n Î Z.

2. (b)

= sin {i × (–2iq)} = sin 2q 2tanq = = 1 + tan2 q

=e

=e

ý þï

- p -2np =e 2

(



a - ib ö æ \ sin ç i log = sin i log e -2iq a + ib ÷ø è

ü 02 +12 +i tan-1 ( ¥ )+2npi ï

i ïílog1+i p +2npi ïý 2 îï þï

æ



}

ì ü i ïílog i +iampæç i ö÷+2npi ïý è ø ïî ïþ e

ì



{

r ( cos q + i sin q )

a + ib

i log i +2npi

i

ii = e Logi = eiLogi = e =

- iq Then, a - ib = r ( cos q - i sin q ) = re = e–2iq



\

x2 y2 + cos h2b sin h2b

=

sin2 a cos h2b cos2 a sin h2b + cos h2b sin h2b

= sin2 a + cos2 a = 1.

8/4/2023 5:14:24 PM

Complex Analysis • 457

6. (a)

1 p x + = 2 cos x 5



p Þ x - 2x cos + 1 = 0 5 2





Þx=

=

2cos

2 cos

5

5

= sin2



\  x 5 = æç cos p ± i sin p ö÷ = cos 5p ± i sin 5p



(by De Moivre’s theorem)



= cos p ± i sin p = –1 ± 0 = –1

5

7. (b)

5ø

5



5

1 1 x + 5 = -1 + = -2. ( -1) x 5

z1 = cos

p p p p + i sin , z2 = cos 2 + i sin 2 , 2 2 2 2

z3 = cos

p p p p + i sin 3 , z4 = cos 4 + i sin 4 , 23 2 2 2

Then, tan q =





p pö æ p p ö æ = ç cos + i sin ÷ ´ ç cos 2 + i sin 2 ÷ 2 2ø è 2 2 ø è



p p ö æ p p ö æ ´ ç cos 3 + i sin 3 ÷ ´ ç cos 4 + i sin 4 ÷ ´ ....¥ 2 2 ø è 2 2 ø è



p p æp p ö = cos ç + 2 + 3 + 4 + .....¥ ÷ 2 2 2 2 è ø



= cos (q1 + q2 + ..... ∞) + i sin (q1 + q2 + .... ∞)]



æ p ö æ p ö ç 2 ÷ ç 2 ÷ = cos ç ÷ + i sin ç ÷ 1 çç 1 - ÷÷ çç 1 - 1 ÷÷ 2ø 2ø è è

EMEP.CH08_3PP.indd 457

= tan

p 10

\q =

p 10

1 - cos sin

p 5

p 5 =

2sin 2 2sin

p 10

p p cos 10 10

Hence, the required polar form is given by: z = r ( cos q + i sin q )



= 2sin

p æ p p ö cos + i sin ÷ . 10 çè 10 10 ø

9. (c) Let z = x + iy. Then, 1 - iz = 1 z -i

p p æp p ö +i sin ç + 2 + 3 + 4 + .....¥ ÷ 2 2 è2 2 ø

× (cos q3 + i sin q3) × .....∞

p 5

Let amp(z) = q.

\  z1z2z3z4... up to ∞

[using (cos q1 + i sin q1) × (cos q2 + i sin q2)

= 2 - 2cos







p p p + 1 - 2cos + cos2 5 5 5



and so on.

2

5ø

è

pö p æ = 2 ç1 - cos ÷ = 2 ´ 2sin 2 5ø 10 è p = 2sin 10





5ø

è

Then, r = z = sin2 p + çæ1 - cos p ÷ö

p p = cos ± i sin 5 5

Hence,

= cos p + i sin p = –1 + i × 0 = –1.

Let z = sin p + i æç1 - cos p ö÷

p p ± 4 cos2 - 4 5 5 2

5



8. (a)

p p ± 2 -1 1 - cos2 5 5 2

è



a ù é 2 3 ê a + ar + ar + ar + ...¥ = 1 - r ú ë û

Þ



1 - i ( x + iy ) x + iy - i

=1 Þ

(1 + y ) - ix x + ( y - 1) i

=1

Þ (1 + y ) - ix = x + ( y - 1) i Þ

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

2

= x 2 + ( y - 1)



⇒  y2 + 2y + 1 + x2 = x2 + y2 – 2y + 1



or, 4y = 0  or  y = 0

8/4/2023 5:14:27 PM

458 • Engineering Mathematics Exam Prep \  z = x + iy = x + i × 0 = x.





Hence, z is purely real. 10. (b) b-a

=

1 - ab



=

=



b-a bb - ab

b-a

(

b b-a

=

)

b-a

=

b b-a

( b = 1 Þ b

b-a

(

b b-a b-a

b b-a

2

= bb = 1



)



)

= 23 4 e

1 =1  z = z =1 b

(

)

= 0,1,2,3 3

So the values of (1 + i 3 ) 4 are: 3 ip

3 3ip

3 5ip

3 7ip

.

2 4 e 4 ,2 4 e 4 ,2 4 e 4 and 2 4 e 4



Hence, the product of these values



= 24

3

=

i ( 2n +1) p 4 ,n



( e

´4

ip

´e4

2npi

(1+3+5+7 )

= 1for n Î N

= 23 ´ e4 pi = 8

)

Remember: 11. (a) i

=



1

1 1 ´ 2i = ´ (1 + 2i - 1) 2 2

1



1 1 2 = 12 + 2i + i2 = (1 + i ) 2 2

{



z =1+i 3

12. (d) Let



. Then z represents the

= lim

1

z ®0

è

3ø

34

3

ì æ p p öü \ 1 + i 3 4 = z 4 = í2 ç cos + i sin ÷ ý 3 3 øþ î è =2



(using De Moivre’s theorem)



= 23/4(cos p + i sin p)1/4



= 23/4 {cos(2np + p) + i sin(2np + p)}1/4 (n ∈ Z) 34

=2



EMEP.CH08_3PP.indd 458

ìï ( 2n + 1) p + i sin ( 2n + 1) p üï , ícos ý 4 4 îï þï

n = 0,1,2,3



Let us consider a curve x2 = my5

(m = arbitrary constant). 5

3

æ

ö

Then, z = x + iy = m y 2 + iy = ç m y 2 + i ÷ y . ç è

14

3p 3p ö ç cos 3 + i sin 3 ÷ è ø

34æ



3

f ( z ) - f (0) z

x 2 y5 ( x + iy ) -0 x 2 y5 x 4 + y10 = lim = lim 4 .........(1) z ®0 z ®0 x + y10 x + iy

3

Hence, z = r ( cos q + i sin q ) = 2 æç cos p + i sin p ö÷

)

1

ì æ 2np + q ö æ 2np + q ö ü = r k ícos ç + i sin ç ÷ ÷ý , k k è ø è øþ î

f ¢(0)

If amp(z) = q, then q = tan -1 3 = p .

(

}

13. (b)

Now, r = z = 12 + ( 3 ) = 4 = 2 .



{

= r k cos ( 2np + q ) + i sin ( 2np + q ) k , n Î Z

where n = 0, 1, 2, 3, ...., k –1.

2

3

1

1

point (1, 3 ) that lies in first quadrant.



1

}k

= r k ( cos q + i sin q ) k 1

}

1 \ i =± (1 + i ) . 2



{

z k = r ( cos q + i sin q )



\  z → 0 ⇒ y → 0. Then, (1) Þ f ¢(0) = lim

my5 y5

y ®0 m2 y10



÷ ø

= lim

+ y10

m

y ®0 m2

+1

=

m

m2 + 1

which is not unique as it depends on m.

So f′(0) does not exist, and hence, f(z) is not analytic at (0, 0), i.e., at z = 0.

8/4/2023 5:14:29 PM

Complex Analysis • 459

14. (d) 2

f ( z ) - f (0) f ¢(0) = lim = lim z ®0 z ®0 z

xy ( x + iy ) -0 x 2 + y4 x + iy

\  f′(z) does not exist and so f(z) is not analytic. 16. (c) f ¢(0) = lim

xy2 ............(1) 2 z ®0 x + y4

= lim



z ®0

Let us consider a curve x = my2 (m = arbitrary constant). So, z = x + iy = my2 + iy = (my + i)y. \  z → 0 ⇒ y → 0 Then, (1) Þ f ¢(0) = lim

my2 y2

y ®0 m2 y4

+ y4 m m = lim 2 = 2 y ®0 m + 1 m +1

which is not unique it depends on m. So f′(0) does not exist and hence f(z) is not analytic at z = 0. 15. (a) f (z) =

xy + i0 = u( x , y ) + iv( x , y ).

xy =

\ u( x , y ) =

xy , v( x , y ) = 0 and so

u( x ,0) = 0, u(0, y ) = 0, v( x ,0) = 0, v(0, y ) = 0. Moreover, f (0) = 0 Þ u(0,0) = 0, v(0,0) = 0. Then, 0-0 ¶u u( x ,0) - u(0,0) = lim = lim = 0, x ®0 x ¶x (0,0) x ®0 x ¶u u(0, y) - u(0,0) 0-0 = lim = lim = 0, y ®0 y ¶y (0,0) y®0 y

= lim

z ®0 x

Thus,



¶u = ¶x (0,0)

¶v & ¶y (0,0)

¶u ¶v . =¶y (0,0) ¶x (0,0)

Hence, CR equations are satisfied at (0, 0), i.e., at z =0. xy - 0 f ( z ) - f (0) f ¢(0) = lim = lim z ®0 z ®0 x + iy z = lim

x ´ mx

x + imx ëéTaking y = mx so that z ® 0 Þ x ® 0 ûù x ®0

= lim

m

x ®0 1 + im



EMEP.CH08_3PP.indd 459

=

m 1 + im

( f (0) = 0 = 0)

x 2 + y2 z ®0 x + iy

Let us consider a curve y = mx (m = arbitrary constant). Then, z = x + iy = x + imx = (1 + im)x.

\  z → 0 ⇒ x → 0 x 2 + m2 x 2 x ®0 x + imx

Then, (1) Þ f ¢(0) = lim

1 + m2 1 + m2 = , x ®0 1 + im 1 + im

= lim



which is not unique as it depends on m.

So f′(0) does not exist and hence f(z) is not differentiable at z = 0. 17. (c) Let v be the harmonic conjugate of u. Then, dv = ¶v dx + ¶v dy ¶x



¶y

æ ¶u ö æ ¶u ö = -ç ÷ dx + ç ÷ dy è ¶x ø è ¶y ø

(

(by CR equations)

)

= - ( -6xy ) dx + 3x 2 - 3 y2 dy



(

)

= 6xydx + 3x 2 - 3 y2 dy

= Mdx + Ndy (say) 2

Now, Mdx = 6xy dx = 6 y xdx = 6 y ´ x = 3x 2 y, ò ò ò 2



ò Ndy = ò (3x



= 3x2y – y3



2

ò

)

ò

ò

- 3 y2 dy = 3x 2 dy - 3 y2dy

ò

\ v = Mdx + Ndy = 3x 2 y - y3 .

18. (a)

,

which depends on m and so is not unique.

+ iy

= lim



0-0 v( x ,0) - v(0,0) ¶v = lim = lim = 0, x ®0 x ¶x (0,0) x ®0 x 0-0 ¶v v(0, y ) - v(0,0) = lim = lim = 0. 0 0 ® ® y y ¶y (0,0) y y

f ( z ) - f (0) z z -0



u = x 2 - y2 Þ

¶u ¶u = 2x , = -2 y. ¶x ¶y

8/4/2023 5:14:31 PM

460 • Engineering Mathematics Exam Prep \ du =



= Mdx + Ndy (say)

¶u ¶u dx + dy ¶x ¶y

Then,

–(–(–2y))dx + (2x)dy – (–24)



ò Mdx = -ò (2x + 2 y + 3) dx = -2ò xdx - 2 y ò dx - 3ò dx



= 2ydx + 2xdy



= –x2 – 2xy – 3x,



= 2(ydx + xdy) = 2d(xy)



Integrating we get, v = 2xy.



ò Ndx =ò (2 y - 2x - 2) = 2ò ydy - 2x ò dy - 2ò dy

æ ¶u ö æ ¶u ö = ç÷ dx + ç ÷ dy è ¶x ø è ¶y ø



(by the CR equations)

19. (d) ¶u = e x ( x cos y - y sin y ) + e x ( cos y - 0 ) ¶x



¶u \ = e z ( z cos0 - 0 ) + e z ( cos0 ) ¶x ( z ,0 ) = ze z + e z = ( z + 1)e z .



¶u = e x ( -x sin y - sin y - y cos y ) . ¶y



¶u \ = e z ( - z´ 0 - 0 - 0 ) = 0. ¶y ( z ,0 )



By the Milne Thomson method,



ì ü ¶u ï ¶u ï f (z) = í -i ýdz + C x y ¶ ¶ ïî ( z ,0 ) ( z ,0 ) ïþ

ò

=



ò ( z + 1) e dz + C ò

ò

ò

= (zH)ez – ez + C = zez + C.



20. (d)

2

2



v = x - y - 2xy + 2x - 3 y ¶v ¶v Þ = 2x - 2 y + 2, = -2 y - 2x - 3. ¶x ¶y



¶u ¶u du = dx + dy ¶x ¶y



=

¶v æ ¶v ö dx + ç - ÷ dy ¶y è ¶x ø

= (–2y – 2x – 3)dx + {–(2x – 2y + 2)}dy

–(2x + 2y + 3)dx + (2y – 2x – 2)dy

EMEP.CH08_3PP.indd 460



ò

\ u = Mdx + Ndy

= –x2 – 2xy – 3x + y2 – 2y

21. (a) Here, v = e–x(x cos y + y sin y)

\

¶v = -e - x ( x cos y + y sin y ) + e - x cos y ¶x



and so ¶v



Again,



and so



Hence, by the Milne Thomson method,



ì ü ¶v ï ¶v ï +i f (z) = í ý dz + c y x ¶ ¶ ( z ,0 ) þï îï ( z ,0 )

ò

= ( z + 1) e z - e z dz + C





¶x ( z ,0 )



= -e - z ( z + 0 ) + e - z = (1 - z ) e- z .

¶v = e - x ( -x sin y + y cos y + sin y ) ¶y

z

é d = ( z + 1) e z dz - ê ( z + 1) e z dz ùú dz + C ë dz û



= y2 – 2xy – 2y

¶v ¶y ( z ,0 )

= 0.

ò

=

-z ò {0 + i (1 - z ) e } dz + c

é ìd ü ù = i ê(1 - z ) e - z dz - í (1 - z ) e - z dz ý dz ú + c î dz þ û ë

ò

{

(

ò

) ò ( -1)( -1) e

= i (1 - z ) -e - z -

{

ò

-z

}

dz + c

}

= i ( z - 1) e - z + e - z + c

= ize–z + c 22. (c) f(z) = u + iv ⇒ if(z) = iu + i2v = iu – v \ f ( z ) + if ( z ) = ( u + iv ) + ( iu - v )



= (u - v ) + i (u + v )

⇒  (1 + i)f(z) = (u – v) + i(u + v)

8/4/2023 5:14:34 PM

Complex Analysis • 461

⇒  F(z) = U + iV (say), where,



F(z) = (1 + i)f(z), which is analytic, U = u – v = ex (cos y – sin y)(given)

Now, f ( z ) = (

V = u + v.

Now, U = ex(cos y – sin y)



¶ ¶ Þ = e x ( cos y - sin y ) , = e x ( - sin y - cos y ) ¶ ¶y x ¶  ¶ Þ = e z (1 - 0 ) , = e z ( -0 - 1) ¶x ( z ,0 ) ¶y ( z ,0 )



Þ u + iv =

=

z

)

2

=

z

( x - iy )2 ( x + iy )

= u + iv

(for z ≠ 0)

( x - iy )3 ( x + iy )( x - iy )

x 3 - 3x 2iy + 3xi2 y2 - i3 y3 x 2 - i 2 y2

(x Þ u + iv =

3

) (

- 3xy2 + i -3x 2 y + y3 2

x +y

2

)

x 3 - 3xy2 y3 - 3 x 2 y , v . = x 2 + y2 x 2 + y2

¶  ¶ Þ = ez , = -e z . ¶x ( z ,0 ) ¶y ( z ,0 )





Then by the Milne Thomson method,

At z = 0, f(z) = u(x, y) + iv(x, y) = 0.



ì ü ¶U ï ¶U ï F (z) = í -i ý dz + C ¢ x y ¶ ¶ ïî ( z ,0 ) ( z ,0 ) ïþ

ò

=



ò

= (1 + i ) e dz + (1 + i ) C



z

ò {e

z

Þ (1 + i ) f ( z ) = (1 + i ) e z + (1 + i ) C



Þ f ( z ) = e z + C.

( )} dz + C¢

- i -e z

f ( z ) - f (0) z

z ®0

2

( x - iy )

z ®0

x + iy ( x + iy )



( x - imx )2 = lim z ®0 x + imx 2 ( ) (1 - im )2 z ®0 1 + im 2 ( )

= lim z z ®0

-0

2

, which depends on

“m” and so is not unique. \  f′(0) does not exist and hence, f(z) is not analytic at z = 0.

EMEP.CH08_3PP.indd 461



z

étaking y = mx so that x ® 0 ù ê ú ë whenever z ® 0 û

æ 1 - im ö =ç ÷ è 1 + im ø

x3 0-0 = x , u ( 0, y ) = = 0, x2 0 + y2

Þ u ( x ,0 ) =

Þ v ( x ,0 ) = \





= lim



2

( z = x + iy,so z = x - iy )

( x - iy )2 = lim z ®0 x + iy 2 ( )



x +y



f ¢ ( 0 ) = lim

= lim

3 2 Now, u = u ( x , y ) = x 2- 3xy2

v = v ( x, y) =

23. (d)

(z)

\  u(0, 0) + iv(0, 0) = 0 + i0 at z = 0.

So, u(0, 0) = 0, v(0, 0) = 0.

[taking (1 + i ) C = C ¢]



Þu=



y3 - 3 x 2 y x 2 + y2

0-0 y3 - 0 0, 0, = = = y. v y ( ) 0 + x2 y2 + 0

u ( x ,0 ) - u ( 0,0 ) ¶u æx -0ö = lim = lim ç ÷ = 1, x ®0 è x ø ¶x ( z ,0 ) x ®0 x

u ( 0,y ) - u ( 0,0 ) ¶u 0-0 = lim = lim = 0, y ®0 y ¶y ( z ,0 ) x ®0 y v ( x ,0 ) - v ( 0,0 ) ¶v æ0-0ö = lim = lim ç ÷ = 0, x ®0 è x ø ¶x ( z ,0 ) x ®0 x

v ( 0, y ) - v ( 0,0 ) ¶v y-0 = lim = lim = 1. y ®0 y y ¶y ( z ,0 ) y ®0

Hence, ¶u

¶x ( z ,0 )



=

¶v ¶u ¶v . and =¶y ( z ,0 ) ¶y ( z ,0 ) ¶x ( z ,0 )

So the CR equations are satisfied at z = 0. 24. (a) f(z) = u(x, y) + iv(x, y) is analytic



⇒ ¶u = ¶v and ¶u = - ¶v . ¶x

¶y

¶y

¶x

Given, u(x, y) = 2x(1 – y).

8/4/2023 5:14:36 PM

462 • Engineering Mathematics Exam Prep \

z = 0 represents the point(0, 0) which lies outside C.

¶v ¶u = = 2 (1 - y ) ¶y ¶x

æ y2 ö Þ v = 2 (1 - y ) dy + f ( x ) = 2 ç y ÷ + f (x ) ç 2 ÷ø è ¶v ¶u Þ = f ¢( x ) = = 2x ¶x ¶y

ò

ò

Þ f ( x ) = 2 x dx + C = x 2 + C æ y2 ö 2 2 2 Hence, v = 2 ç y ÷÷ + x + C = 2 y - y + x + C ç 2 è ø

z = –pi represents the point (0, –p) which lies inside C.



dz \ = z ( z + pi )

ò

C

=

= x 2 - ( y - 1)2 .

25. (d) (z2 + 9)(z – i) = 0  ⇒ (z + 3i)(z – 3i) (z – i) = 0 ⇒  z = i, – 3i, 3i. z = 2 Þ x + iy = 2 Þ ( x - 0)2 + ( y - 0)2 = 22 ,



which represents the circle with center (0, 0) and radius “2” unit. z = i represents the point (0, 1) which lies within C. z ± 3i represents the point (0, 3) and (0, –3) which lie outside C. \

ò (z

C



=

zdz

2

)

+ 9 (z - i)

f (z)

ò z - a dz

=

z z + 9 dz z -i 2

ò

C

C

(where f ( z ) =

= 2pi × f(a) (by Cauchy’s integral formula)



= 2pi ´ f ( i ) = 2pi ´

(i

2

+9

)

=

-2p p =- . -1 + 9 4

26. (a) z(z + pi) = 0  ⇒  z = 0, z = –pi.

z + 3i = 1 Þ x + iy + 3i = 1

Þ x + i ( y + 3) = 1

2

}

Þ ( x - 0 ) + y - ( -3 )

= 12 ,

which represents the circle with the center (0, – 3) and radius “1” unit.

EMEP.CH08_3PP.indd 462

= 2pi × f(a) (by Cauchy’s integral formula)



= 2pi × f(– pi) = 2pi ´ 1

( -pi )

= -2 .

27. (c)

z -2 + z +2 = 6

Þ x + iy - 2 + x + iy + 2 = 6 Þ

( x - 2 )2 + y2 + ( x + 2 )2 + y2

Þ

( x - 2 )2 + y2

=6-

=6

( x + 2 )2 + y2

Squaring both sides we get,

( x - 2 )2 + y2 = 36 + ( x + 2 )2 + y2 - 12 ( x + 2 )2 + y2

or,  x2 + 4 – 4x + y2 = 36 + x2 + 4 + 4 x + y2 - 12 x 2 + 4 x + 4 + y2

or, 3 x 2 + 4 x + 4 + y2 = 2x + 9

Again, squaring both sides we get,

9(x2 + y2 + 4x + 4) = 4x2 + 36x + 81

or, 5x2 + 9y2 = 45

2 2 2 2 or, x + y = 1 or x + y = 1 , which repre2 2

9

5

3

( 5)

Now, z – pi = 0 ⇒ z = pi which represents the point (0, p) that lies outside C.

2

{

z

sents the ellipse C.

Þ x 2 + ( y + 3) = 1 2

(where f ( z ) = 1 , a = -pi )

or, 12 x 2 + 4 x + 4 + y2 = 8x + 36

z ,a = i ) z2 + 9 i

z-a





(say)

f ( z ) dz

C

= 2 y - y2 + x 2 - 1 (taking C = -1)



ò

ò

C

1 z dz z - ( -pi )



So by the Cauchy-Goursat theorem,

ò

C

e2z dz = 0. z - pi

8/4/2023 5:14:38 PM

Complex Analysis • 463

higher order derivatives]

28. (c) z 2 + 2z + 2 = 0 Þ z =



-2 ± 4 - 4 ´ 2 = -1 ± i 2

=

2pi ( 2) æ p ö f ç ÷ 2! è4ø

–1 + i represents the point (–1, 1) which lies outside the square C.



–1 – i represents the point (–1, –1) which lies inside the square C.



æpö = pi ´ f ¢¢ ç ÷ è4ø



æ 1 ö = pi ´ ç ÷ 2 ø ëé f ( z ) = sin z Þ f ¢¢ ( z ) = - sin z ûù è

\



=

dz dz = 2 + z i z - ( -1 - i ) 1 + + z 2 z 2 ( ) C C

ò

ò

ò{

f ( z ) dz

}

(say)

z-a

C

}{

(where f ( z ) =



1 , a = -1 - i ) z +1 - i

= 2pi × f(a) (by Cauchy’s integral formula)



= 2pi × f(– 1 – i)



1 = 2pi ´ = -p . éë( -1 - i ) + 1 - i ùû



2

z = 0 represents the point O (0, 0) which lies within the circle C.

p 1 p 1 = Þ x + iy - = 4 2 4 2



\

pö 1 pö 1 æ æ Þ ç x - ÷ + iy = Þ ç x - ÷ + y2 = 4 2 4 2 è ø è ø 2



pö 2 æ æ1 ö Þ ç x - ÷ + ( y - 0) = ç ÷ 4ø è è2ø

,

and radius = 1 unit.

z=

p 4

2

represents the point A æç p ,0 ö÷ which lies è4

inside the circle C. \



=

sin z dz

òæ

3

C z- pö ç 4 ÷ø è

f ( z ) dz

ò (z - a)

C

[where

EMEP.CH08_3PP.indd 463

=

æp ö ç 4 ,0 ÷ è ø

n +1

=

òæ

sin z

C z- pö ç 4 ÷ø è

dz

2 +1

ø

dz

2pi n ´ f ( ) (a) n!

[by Cauchy’s integral formula on

dz =

e - z dz

ò ( z - 0)

1+1

C

=

f ( z ) dz

ò (z - a)

n +1

C

(where f ( z ) = e - z , a = 0, n = 1 )



=

2pi n ´ f ( ) (a) n!

(say)

[by Cauchy’s integral formula on

higher order derivatives]

=

2pi 1 ´ f ( ) ( 0 ) = 2pi ´ f ¢ ( 0 ) 1!



= 2pi ´ ( -1) = -2pi.



é f ( z ) = e - z Þ f ¢ ( z ) = -e - z Þ f ¢ ( 0 ) = -e -0 = -1ù ë û

31. (d) z = eiq Þ dz = ieiq`dq Þ

(say)

p f ( z ) = sin z , a = , n = 2 ] 4

2



2

which represents a circle with the center

e-z

òz

C

2



2

Þ ( x - 0 ) + ( y - 0 ) = 12 ,

which represents a circle with the center (0, 0) and radius “1” unit.

29. (a)

pi . 2

30. (b) z = 1 Þ x + iy = 1 Þ x 2 + y2 = 1



z-

=-



Þ

dz = idq e iq

dz = idq z p

\  given integral = idq = i éëqùû p = pi . 0 ò 0

8/4/2023 5:14:41 PM

464 • Engineering Mathematics Exam Prep 32. (c)

sin z z-p sin { p + (z - p)} - sin( z - p) = = z-p z-p ì üï 1 ï ( z - p)3 ( z - p)5 =+ + .......¥ ý í( z - p) 3! 5! z - p ïî ïþ = -1 +



¥

å ( -1)



which is the required Taylor series expansion

n =0

34. (c) f (z) =

Hence, the coefficient of (z – p)2 in the Taylor series expansion of f(z) is 1 . 6

f (z) =

2

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



=

2



( z + 2)( z + 3)

( z + 2)( z - 2) 5( z + 3) 5( z + 2) + ( z + 2)( z + 3) ( z + 2)( z + 3) ( z + 2)( z + 3)



z -2 5 5 = + z +3 z +2 z +3 ( z + 3) - 5 5 5 = + z +3 z +2 z +3 5 5 5 =1+ z +3 z +2 z +3 5 10 =1+ .......................(i) z +2 z +3



z 0) of this tree = 3i. 29. (d) The maximum number of nodes in a binary tree of depth 5 = 25 – 1 = 31. 30. (d) A full binary tree with height k has 2k+1 – 1 number of nodes and 2k number of leaves.

Therefore, the number of internal nodes



= (2k+1 – 1) – 2k = 2k–1.

31. (b) The number of different trees with “n” nodes = nn–2. Here, n = 5 and so the number of different trees with “5” nodes is 125. 32. (a) Consider a graph G with three isolated vertices. Then G has no cycles and G is not a tree. 33. (b) We know that any tree with n number of vertices contains n–1 edges and no cycles. 34. (d) The number of binary trees formed with n nodes

24. (c) Let “k” be the number of connected components, then we have

14 – k ≤ 7 + 2 i.e; k ≥ 5.



2n

8 Cn C4 = (here n = 4) n +1 4 +1 8´7´6´5 = = 14. 5 ´ 4 ´ 3 ´ 2 ´1

=

25. (a) Let G be a planar graph with “n” vertices, “m” edges, and “k” regions.

35. (d) It follows from the definition of a tree.



Then, by the Euler’s formula, n – m + k = 2…(i)

37. (c) Using Kruskal’s algorithm, we can find out the minimum spanning tree which does not necessarily give the shortest path.



2m Again, 2m ³ 3k Þ k £ 3 \(i )gives, 2m 2 = n-m+k £ n-m+ 3 m Þn³ 2 Þ m £ 3n - 6 = 3(n - 2) 3

26. (b) If we delete the edges BD, CF, and AB, then the graph becomes disconnected.

36. (c) Result follows from the four color’s theorem.

38. (d) In case of strongly connected graph, each vertex must be reachable from any other vertex. 39. (c) The total number of nodes

= total number of leaves + number of nonleaves

= n + (n – 1) = 2n–1.

27. (d) We know that a connected graph with “n” vertices and n–1 edges edges is always a tree.

40.  (b) The number of edges in spanning tree = n – 1.

Hence, T is a tree.



Hence, the number of edges to be removed =





m – (n – 1) = m – n + 1.

EMEP.CH11_4PP.indd 594

Since a tree contains no cycles, so T has no cycles.

8/14/2023 4:44:27 PM

Graph Theory • 595 = (V , E1 Ç E2 )

Previous Years Solved Paper (2000-2018)

the graph G1 È G2 = (V , E1 È E2 )

1. The minimum number of colors required to color the vertices of a cycle with “n” nodes (n = odd) in such a way that no two adjacent nodes have the same color is



(a) 2 (c) 4



(b) 3 (d) n – 2[n/2] + 2 (CS GATE 2002)



7. What is the maximum number of edges in an acyclic undirected graph with “n” vertices? (a) n – 1 (b) n (c) n + 1 (d) 2n – 1  (IT GATE 2004) 8. What is the number of vertices in an undirected connected graph with 27 edges, 6 vertices of degree 2, 3 vertices of degree 4 and remaining of degree 3? (a) 10 (b) 11 (c) 18 (d) 19  (IT GATE 2004)

(a) k and n (b) k – 1 and k + 1 (c) k – 1 and n – 1 (d) k + 1 and n – k  (CS GATE 2003) 3. How many perfect matching are there in a complete graph of six vertices? 

(a) 15

(b) 24

(c) 30 (d) 60 (CS GATE 2003)

9. An undirected graph G has “n” nodes. Its adjacency matrix is given by an n × n square matrix whose

4. The minimum number of colors required to the color the following graph, such that no two adjacent vertices are assigned same color, is

(i) diagonal elements are 0’s and (ii) nondiagonal elements are 1’s.



(a) 2

(b) 3

(c) 4 (d) 5 (CS GATE 2004)

5. How many graph on “n” labeled vertices

2

exist which have at least n - 3n edges? 2

(a)

( n2 -n )/ 2

(c)

( n2 -n )/ 2

C( n2 -3n )/ 2 (b) Cn (d)

( n2 -3n )/ 2

å k =0

n

( n2 -n )/ 2

Ck

å (n -n)/ 2Ck 2

k =0



(CS GATE 2004)



6. Let G1 = (V, E1) and G2 = (V, E2) be two connected graphs on the same vertex set V with more than two vertices. If G1 Ç G2

EMEP.CH11_4PP.indd 595

(a) Cannot have a cut vertex (b) Must have a cycle (c) Must have a cut-edge (bridge) (d)  Has chromatic number strictly greater than those of G1 and G2. (CS GATE 2004)



2. Let G be an arbitrary graph with “n” nodes and “k” components. If a vertex is removed from G, the number of components in the resultant graph must necessary lie between

is not a connected graph, then



Which one of the following is TRUE? (a) Graph G has no minimal spanning tree (MST) (b) Graph G has a unique MST of cost n – 1. (c) Graph G has multiple distinct MSTs, each of cost n – 1. (d) Graph G has multiple spanning trees of different costs. (CS GATE 2005)

10. Let G be a simple connected planar graph with 13 vertices and 19 edges. Then, the number of faces in the planar embedding of the graph is: 

(a) 6

(b) 8

(c) 9 (d) 13 (CS GATE 2005)

11. Let G be a simple graph with 20 vertices and 100 edges. The size of the minimum vertex cover of G is 8. Then, the size of the maximum independent set of G is: 

(a) 12 (c) Less than 8

(b) 8 (d) More than 12. (CS GATE 2005)

8/14/2023 4:44:27 PM

596 • Engineering Mathematics Exam Prep 12. Which of the following graphs is not planar? G2

15. Which of the following graphs has an Eulerian circuit?

G1



16. Consider a weighted graph with positive edge weights and let (u, v) be an edge in the graph. It is known that the shortest path from the source vertex “s” to “u” has weight 53 and the shortest path from “s” to “v” has weight 65. Which one of the following statements is always true?

G3

G4

(a) G1 (b) G2 (c) G3 (d) G4  (CS GATE 2005) 13. The 2n vertices of a graph G corresponds to all subsets of a set of size “n,” for n ≥ 6. Two vertices of G are adjacent if and only if the corresponding sets intersect in exactly two elements. (i) The number of vertices of degree “0” in G is (a) 1 (b) n (c) n + 1 (d) 2n (ii) The maximum degree of a vertex G is (a) n/ 2C2 ´ 2n (b) 2n–2 n–3 (c) 2 × 3 (d) 2n–1 (iii) The number of connected components in G is (a) n (b) n + 2 



EMEP.CH11_4PP.indd 596



(a) weight of (u, v) < 12 (b) weight of (u, v) = 12 (c) weight of (u, v) ≥ 12 (d) weight of (u, v) > 12

(IT GATE 2007)

17. What is the largest integer “m” such that every simple connected graph with “n” vertices and “n” edges contains at least “m” different spanning trees? 

(a) 1

(b) 2

(c) 3 (d) n (IT GATE 2007)

18. What is the chromatic number of the following graph?

(c) 2n/2 (d) 2n/n (CS GATE 2006)

14. Let T be a depth first search tree in an undirected graph G. Vertices “u” and “v” are leaves of this tree T. The degree of both “u” and “v” in G are at least 2. Then which of the following statements is true?

(a) Any k-regular graph where k is an even number. (b) A complete graph on 90 vertices. (c) The complement of a cycle on 25 vertices. (d) None of these. (CS GATE 2007)

(a) There must be a vertex “w” adjacent to both “u” and “v” in G (b) There must be a vertex “w” whose ­removal disconnects “u” and “v” in G (c) There must exist a cycle in G containing “u” and “v” (d) There must exist a cycle in G containing “u” and all its neighbors in G (CS GATE 2006)



(a) 2

(b) 3

(c) 4 (d) 5 (IT GATE 2008)

19. Let G be a simple undirected graph. Some vertices of G are of an odd degree. Add a node “v” to G and make it adjacent to each odd d ­ egree vertex of G. The resultant graph will be 

(a) regular (c) Hamiltonian

(b) complete (d) Euler (IT GATE 2008)

20. Let G be a graph with “n” vertices and 2n-2 edges. The edges of G can be partitioned

8/14/2023 4:44:28 PM

Graph Theory • 597

into two edge-disjoint spanning trees. Which of the following is NOT true for G? 

(a) For every subset of “k” vertices, the induced sub-graph has at most 2k-2 edges (b) The minimum cut in G has at least two edges (c)  There are two edge-disjoint paths between every pair of vertices (d) There are two vertex-disjoint paths between every pair of vertices (CS GATE 2008)

21. What is the chromatic number of an n-vertex simple connected graph which does not contain any odd length cycle? (assume n ≥ 2) 

(a) 2

(b) 3

(c) n–1 (d) n (CS GATE 2009)

22. Which one of the following is TRUE for any simple connected undirected graph with more than two vertices? 

(a) No two vertices have the same degree (b) At least two vertices have the same degree (c) At least three vertices have the same degree (d) All vertices have the same degree (CS GATE 2009)

23. Let G = (V, E) be a graph. Define x(G ) = å id ´ d , where id is the number of d

vertices of degree “d” in G. If S and T are two different trees with x(S ) = x(T ) , then 

(a) |S| = 2|T| (c) |S| = |T|

(b) |S| = |T| – 1 (d) |S| = |T| + 1 (CS GATE 2010)

24. K4 and Q3 are graphs with the following structures:

25. An undirected graph G (V, E) contains “n” (n > 2) nodes namely v1, v2, ...., vn. Two nodes vi, vj are connected if and only if 0 < |i – j| ≤ 2. Each edge (vi, vj) is assigned a weight i + j. A sample graph with n = 4 is shown below: v3 4

5

3

v1

v4

7

6

v2

(i) What will be the cost of the minimum spanning tree (MST) of such a graph with “n” nodes? (a) 1 (11n2 - 5n) (b) n2 – n + 1 12



(c) 6n – 11

(d) 2n + 1 (CS GATE 2011)

(ii) The length of the path from v5 to v6 in the MST with n = 10 is 

(a) 11

(b) 25

(c) 31 (d) 41 (CS GATE 2011)

26. Let G be a simple undirected planar graph on 10 vertices with 15 edges. If G is a connected graph, then the number of bounded faces in any embedding of G on the plane is equal to 

(a) 3

(b) 4

(c) 5 (d) 6 (CS GATE 2012)

27. The maximum number of edges in a bipartite graph on 12 vertices is __________ 

(CS GATE 2014)

28. Let G be a graph with n vertices and m ­edges. What is the tightest upper bound on the running time Depth First Search on G, when G is represented as an adjacency matrix? (a) Q(n) (b) Q(n + m) (c) Q(n2) (d) Q(m2)  (CS GATE 2014)

Which of the following statements is TRUE in relation to these graphs? (a) K4 is planar while Q3 is not (b) Both K4 and Q3 are planar (c) Q3 is planar while K4 is not (d) Neither Q3 nor K4 is planar  (CS GATE 2011)

EMEP.CH11_4PP.indd 597

29. Consider the tree arcs of a BFS traversal from a source node W in an unweighted, connected, undirected graph. The tree T formed by the three arcs is a data structure for computing

(a) the shortest path between every pair of vertices.

8/14/2023 4:44:29 PM

598 • Engineering Mathematics Exam Prep 

(b) the shortest path from W to every vertex in the graph. (c) the shortest path from W to only those nodes that are levels of T. (d) the longest path in the graph. (CS GATE 2014)



number of faces is at most n + 2 2



(CS GATE 2014)

34. Let G be a connected planar graph with ten vertices. If the number of edges on each face is three, then the number of edges in G is _____?  (CS GATE 2015)

30. Consider the directed graph given below. Q

P

(d)  There is a planar embedding in which

35. The minimum number of colors that is sufficient to vertex-color any planar graph is _____?  (CS GATE 2016) S

36. Consider the weighted undirected graph with four vertices, where the weight of edge {i, j} is given by the entry Wij in the matrix W.

R



Which one of the following is TRUE?

(a) The graph does not have any topological ordering. (b) Both PQRS and SRQP are topological orderings. (c) Both PSRQ and SPRQ are topological orderings. (d) PSRQ is the only topological ordering.  (CS GATE 2014)

é0 ê 2 W =ê ê8 ê ë5





8 5 0 x

5ù ú 8ú xú ú 0û

 The largest possible integer value of x, for which at least one shortest path between some pair of vertices will contain the edge with weight x is __________?

31. A cycle on n vertices is isomorphic to its complement. The value of n is ________?

(CS GATE 2016)



(CS GATE 2014)



2 0 5 8

32. If G is a forest with n vertices and k connected components, how many edges does G have?

37. Let G be a complete undirected graph on 4 vertices, having six edges with weights b ­ eing 1,2,3,4,5, and 6. The maximum possible weight that a minimum weight spanning tree of G can have is _______?

(a) [n/k] (b) [n/k] + 1 (c) n – k (d) n – k + 1  (CS GATE 2014)

(CS/IT GATE 2016)



33. Let d denote the minimum degree of a v­ ertex in a graph. For all planar graphs on n vertices with d ≥ 3, which one of the following is TRUE?

38. Let G = (V, E) is an undirected simple graph in which each edge has a distinct weight, and e is a particular edge of G. Which of the following statements about the minimum spanning trees (MSTs) of G is/ are TRUE?





(a) In any planar embedding, the number of faces is at least



n +2 2

(b) In any planar embedding, the number of faces is less than n + 2



2

(c)  There is a planar embedding in which number of faces is less than n + 2 2

EMEP.CH11_4PP.indd 598

I. If e is the lightest edge of some cycle in G, then every MST of G includes e. II. If e is the heaviest edge of some cycle in G, then every MST of G excludes e. 

(a) I only (c) Both I and II

(b) II only (d) Neither I nor II (CS/IT GATE 2016)

8/14/2023 4:44:29 PM

Graph Theory • 599

39. Let T be a tree with 10 vertices. The sum of the degrees of all the vertices in T is ___

19. (d)

20. (d)

21. (a)

22. (b)

23. (c)

24. (b)

25. (i)-(b); (ii)-(c)

26. (d)

27. 36

(CS (IT) GATE 2017)

28. (c)

29. (b)

30. (c)

31. 5

32. (c)

40. G is undirected graph with n vertices and 25 edges such that each vertex of G has degree at least 3. Then the maximum possible value of n is _________

33. (a)

34. 24

35. 4

36. 12

37. 7

38. (b)

39. 18

40. 16

41. 3

42. 4



(CS (IT) GATE, 2017)



Explanation



41. The chromatic number of the following graph is a

e

Now let n = 5. Then the minimum number of ­colors required to color the vertices of a cycle with “5” nodes is “3” which satisfies n – 2[n/2] + 2 (since 5 – 2[5/2] + 2 = 5 – 2 × 2 + 2 = 3).

c d

2. (a) There are two possibilities:

b

(i) If the removed vertex is an isolated vertex (it self a component), then the minimum number of components in the resultant ­ graph will be “k.”

f

(CS (IT) GATE, 2018)



42. The post order traversal of a binary tree is 8, 9, 6, 7, 4, 5, 2, 3, 1. The in-order traversal of the same tree is 8, 6, 9, 4, 7, 2, 5, 1, 3. The height of a tree is the length of the longest path from the root to any leaf. The height of the binary tree above is_______

(ii) If the removed vertex disconnects it is all components, then the maximum number of components in the resultant graph will be “n.”

(CS (IT) GATE, 2018)



1. (d) Let n = 3. Then the minimum number of ­colors required to color the vertices of a cycle with “3” nodes is “3” which satisfies n – 2[n/2] + 2 (since 3 – 2[3/2] + 2 = 3 – 2 × 1 + 2 = 3).

3. (a) The number of perfect matching in a graph with “n” vertices is n(n - 1) . Here, n = 6.

43. Consider the following undirected graph G:



2

So, the required number of perfect matching 4

x 1

=

3

6(6 - 1) = 15. 2

4. (c)

G 4

5 W

R

4

Choose a value for x that will maximize the number of minimum weight (MWSTs) of G. The number of MWSTs of G for this value of x is _______ (CS (IT) GATE, 2018)



Answer Key 1. (d)

2. (a)

3. (a)

4. (c)

5. (d)

6. (b)

7. (a)

8. (d)

9. (c)

10. (b)

11. (a)

12. (a)

13. (i)-(c); (ii)-(c); (iii)-(b)

14. (b)

15. (c)

16. (c)

EMEP.CH11_4PP.indd 599

17. (c)

18. (b)

B

B G

W

R

Here, “G” stands for the color “green,” “B” stands for the color “black,” “W” stands for the color “white” and “R” stands for the color “red.” Clearly “4” colors are required to the color the graph, such that no two adjacent vertices are assigned same color.

8/14/2023 4:44:30 PM

600 • Engineering Mathematics Exam Prep 5. (d) The maximum number of edges in a ­simple graph

=

n(n - 1) 2

= a (say).

Let b = (n2−3n)/2. Then, a – b = n and so number of graphs with minimum b edges = aCb + aCb+1 + aCb+2 + ......... + aCa

= aCa -b + aCa -b-1 + aCa -b-2 + ......... + aCa -a a

a

a

a

= Cn + Cn -1 + Cn -2 + ......... + C0 = =



n2 -n n2 -n n2 -n 2 2 Cn -1 + Cn -2 + ....... + 2 C0 n+

n2 -n 2 C n

å (n -n)/ 2Ck 2



8. (d) Let “x” be the total number of vertices. Then, the sum of degrees of the vertices = 2 × total number of edges

⇒ 6 × 2 + 3 × 4 + (x – 9) × 3 = 2 × 27



⇒ 24 + 3x – 27 = 54 ⇒ x = 19.



Thus, the total number of vertices = 19. 9. (c) If all nondiagonal elements are 1, then every vertex is connected to every other vertex in the graph with an edge of weight 1. Such a graph has multiple distinct MSTs with cost n–1.

10. (b) For a planar embedding of a simple connected graph G with “n” vertices, “m” edges, and “r” regions (faces), we have by Euler’s formula: n – m + r = 2.

Here, n = 13 and m = 19. So r = 8. 11. (a) We know that if a graph has “n” vertices, “m” edges, and “x” be the minimum vertex cover of G, then the size of the maximum independent set of G is n – x. Here, n = 20, m = 100, and x = 8. So the size of the maximum independent set of G is 20 – 8, i.e., 12.

EMEP.CH11_4PP.indd 600

(ii) The maximum degree of a vertex G

(

)

max k kC2 ´ 2n -k = 3C2 ´ 2n -3 = 3 ´ 2n -k

(iii) According to (i), n + 1 nodes of the graph are not connected to anyone while others are connected. Hence, the total number of connected components = n + 2 (n + 1 connected ­components by each of the n + 1 vertices +1 connected component by remaining vertices).

6. (b) Since G1 and G2 are two connected graphs on the same vertex set V with more than two vertices, so G1 È G2 must have a cycle. 7. (a) In a tree number of edges = no. of vertices–1 = n–1.



13. (i)-(c); (ii)-(c); (iii)-(b) (i) No. of vertices with degree zero = no. of subsets with size (≤1) = n + 1.

k =0





12. (a) We know that a graph is planar if it can be drawn in a plane without any edge crossing. Among the given graphs, only G1 can be redrawn in a plane without any crossing edges.

14. (b) Since the vertices “u” and “v” are leaves of this depth first search tree T, so there must be a vertex “w” whose removal disconnects “u” and “v” in G. 15. (c) We know that a graph has an Eulerian circuit if following conditions are true. (i) All vertices with nonzero degree are connected. (ii) All vertices have even degree.

(a) Any k-regular graph (where k is an even number) is not Eulerian as a k regular graph may not be connected. (b) A complete graph on 90 vertices is not Eulerian because all vertices have degree as 89. (c) The complement of a cycle on 25 vertices is Eulerian. The reason is in a cycle of 25 vertices, all vertices have degree as 2 and in the complement graph, all vertices would have degree as 22 and so the graph would be connected. 16. (c) weight(s, u) + weight (u, v) ≥ weight(s, v)

⇒ 53 + weight (u, v) ≥ 65



⇒ weight (u, v) ≥ 12

17. (c) We know that a graph is connected if and only if all nodes can be traversed from each node.

8/14/2023 4:44:30 PM

Graph Theory • 601

For a simple connected graph with n nodes, there will be n-1 minimum number of edges. Given that there are n edges which implies that a cycle is there in the graph. Now we can get a different spanning tree by removing one edge from the cycle, one at a time. Since the minimum cycle length can be 3, so, there must be at least 3 spanning trees in any such graph. 18. (b) R

B

B B W

R

B

R

R

Here, “B” stands for the color “black,” “W” stands for the color “white” and “R” stands for the color “red.” Clearly “3” colors are required to the color the graph, such that no two adjacent vertices are assigned same color. Hence, the chromatic number of the graph is “3.” 19. (d) When we add a node “v” to the simple undirected graph G and make it adjacent to each odd degree vertex of G, the graph will contain vertices each of even degree. Hence, the resultant graph will become an Euler graph since we know that a connected graph G is Eulerian if and only if all vertices of G are of even degree of v is also even since number of odd degree vertices in a graph is even. 20. (d) Let us take two copies of K4 (which is a complete graph on 4 vertices), say G1 and G2. Let V(G1) = {a, b, c, d} and V(G2) = {m, n, p, q}. Now let us construct a new graph G3 by using these two graphs G1 and G2 by merging at a vertex, say (d, m). The resultant graph is two edge connected, and of minimum degree 2 but there exists a cut vertex, the merged vertex. 21. (a) We know that (i) A simple graph with no odd cycles is a bipartite graph, (ii) A bipartite graph can be colored with two colors.

EMEP.CH11_4PP.indd 601

Hence, the chromatic number of an n-vertex simple connected graph which does not contain any odd length cycle will be “2.” 22. (b) The graph is simple implies neither loops nor parallel edges exist in the graph. Again the graph is connected implies that the degree of any vertex cannot be “0.” As a result, the degree of any vertex will lie between 1 and n-1. Hence, at least two vertices will have the same degree. 23. (c) Here, x(G ) denotes the sum of all degrees in G. Since x(S ) and x(T ) are same for the two trees S and T, so both of them have same number of vertices. 24. (b) Both of K4 and Q3 can be drawn in a plane without any edge crossing. 25. (i)-(b); (ii)-(c) To calculate MST, first we take the edges (v1, v2)and (v1, v3) as the two least cost edges. Next, we add the set of edges (v2, v4), (v4, v6), (v6, v8),... and (v3, v5), (v5, v7), (v7, v9), .... Thus, we have two chains-one with odd labeled vertices and other with all even numbered vertices. These two chains merge at v1. Hence, the cost of the minimum spanning tree (MST) of such a graph with “n” nodes = [weight (v1, v2) + weight (v2, v4) + weight (v4, v6) + weight (v6, v8) +………] + [weight (v1, v3) + weight (v3, v5) + weight (v5, v7) + weight (v7, v9) + .....]

= [(1 + 2) + (2 + 4) + (4 + 6) + (6 + 8) +…….]+



[(1 + 3) + (3 + 5) + (5 + 7) + (7 + 9) +………]



= 1 × 2 + [2 × 2 + 4 × 2 + 6 × 2 + 8 × 2 +…….]+



[3 × 2 + 5 × 2 + 7 × 2 + 9 × 2 +……….]

= n + (n–1) + 2 × [1 + 2 + 3 + 4 + ………..+(n – 2)] = n + (n - 1) + 2 ´



= n2 - n + 1.

(n - 2) {1 + (n - 2)} 2

(ii) Length of the path from (v5 to v6) in the MST with n = 10 = [weight (v1, v2) + weight (v2, v4) + weight (v4, v6)] + [weight (v1, v3) + weight (v3, v5)] = [(1 + 2) + (2 + 4) + (4 + 6)] + [(1 + 3) + (3 + 5)] = 31.

8/14/2023 4:44:31 PM

602 • Engineering Mathematics Exam Prep 26. (d) We know that,

no. of vertices – no. of edges + no. of faces = 2.



Here no. of vertices = 10, number of edges = 15.



So the number of faces = 2 – 10 + 15 = 7.

Out of these seven faces, one is an unbounded face. Hence, total number of bounded faces = 6. 27. 36 We know that, the maximum number of edges 2 in a bipartite graph = n .

4



Here, n = 12. So themaximum number of edges 12 ´ 12 in a bipartite  graph with 12 vertices is 4 i.e;36 edges.

28. (c) In adjacency matrix representation, graph is represented as an n × n matrix. To do DFS, for every vertex, we traverse the row corresponding to that vertex to find all ­adjacent vertices. (In adjacency list representation we traverse only the adjacent vertices of the vertex). Therefore, time complexity becomes Q(n2.) 29. (b) BFS is applied to find shortest path from the vertex u to the vertex v. According to the question, W is the source node. Therefore, the tree T formed by the three arcs is a data structure for computing the shortest path from W to every vertex in the graph. 30. (c) The graph does not contain any cycle, so there exist topological ordering. Q depends on P, R, and S. R depends on P and S. Hence, both PSRQ and SPRQ are topological orderings. 31. 5 It can be easily verified that a cycle graph on n = 1, 2, 3, 4 vertices is not isomorphic to its complement. Now consider n = 5. Then, we have the following cycle:



Clearly, these two graphs are isomorphic.



So, the value of n is 5.

32. (c) A tree has n vertices ⇒ it has n–1 edges. Therefore, to make it forest with k connected components, we have to delete k−1 edges from the tree. Hence, G has total (n − 1) − (k − 1) i.e; (n − k) edges. 33. (a) d ≥ 3 Þ 3 V £ 2 E Þ 3n £ 2 E Þ E ³ 3n . 2

Then, E = V + R - 2 (Euler's formula)



Þ n+ R -2 ³

3n n Þ R ³ +2 2 2

34. 24

By Euler’s formula, V + R = E + 2.

Given that, V = 10 and number of edges on each face = 3. Thus, 3 R = 2 E Þ R = 2 E . 3



Putting all the values in Euler’s formula, we get, 10 +

2 E = E + 2 Þ E = 24. 3

35. 4 By the 4-color theorem, every planar graph is 4- colorable. 36. 12 Let V = {a, b, c, d} be the set of vertices.

Then, the graph is given by 2

a

b

8

8 5



EMEP.CH11_4PP.indd 602

Its complement is given by

c

5

x

d

8/14/2023 4:44:31 PM

Graph Theory • 603

 The paths between the vertices a and d are s follows:

c-d with weight “x,”



c-a-d with weight = 8 + 5 = 13,



c-b-d with weight = 5 + 8 = 13,



c-a-b-d with weight = 8 + 2 + 8 = 18.

41. 3

Blue a

Blue e

Green c

d Blue

b Red

f Green

Thus, “3” colors (Red, Blue, and Green) are sufficient to the color the graph, such that no two adjacent vertices are assigned same color. Hence, the chromatic number of the graph is “3.”

Hence, the largest possible integer value of “x” will be 12. 37. 7

42. 4

4



The tree is given below:

3

1 5

1

6

2 4

5

4

1

7

6

2



3

2

b

The maximum possible weight = 1 + 2 + 4 = 7.

38. (b) Since MST asked so if e any heaviest edge in cycle, it obviously will not be involved. Therefore, only statement (II) is true.



39. 18

43. 4



The sum of degrees of all vertices



= 2 × Number of edges = 2 × 9 = 18.

For x = 1, 2, 3, 4 only 2 spanning trees are possible but for x = 5 total 4 spanning trees possible.



éin a tree,total no.of edges ù ê ú = total no.of vertices 1 = 10 1 = 9 ë û

The sum of degree of n vertices =



2 × Number of edges



Þ 2 ´ 25 ³ 3 ´ n (Since each vertex has degree at least 3)  50 Þn£ = 16.66 Þ n £ 16. 3

EMEP.CH11_4PP.indd 603

The height of the above tree = 1 + 1 + 1 + 1 = 4.

Questions for Practice



1. How many vertices does a regular graph of degree 4 with 10 edges have?

(a) 3 (b) 4 (c) 5 (d) 6 2. A simple graph with n vertices is connected if the graph has

40. 16

9

8



(a) (n - 1)(n - 2) 2



(b) more than (n - 1)(n - 2) edges 2

8/14/2023 4:44:32 PM

604 • Engineering Mathematics Exam Prep

(c) less than (n - 1)(n - 2) edges



(d) none of these

2

3. Which of the following graph is not regular?

(a) Kn (b) Cn (c) Wn (d) Qn 4. The number of distinct graph of order 5 and size 4 is (a) 6 (b) 5 (c) 4 (d) 3

5. The number of distinct graph of order n and size e is (a) n (b) e (c) min {n, e} (d) none of these



6. An undirected graph possesses an Eulerian circuit if and only if it is connected and its vertices are (a) all of even degree (b) all of odd degree (c) of any degree (d) even in number



7. For the given graphs G1, G2, and G3,





G1:



G2:



G3:



(a) G  1 is isomorphic to G2, and G2 is isomorphic to G3 (b) G  1 is not isomorphic to G2 and G2 is not isomorphic to G3 (c) G  1 is isomorphic to G2, but G2 is not isomorphic to G3 (d) G  1 is not isomorphic to G2, but G2 is isomorphic to G3 8. Which of the following statements is true regarding connected graphs?

EMEP.CH11_4PP.indd 604



(a) No Eulerian graph is a Hamiltonian graph (b)  Every Hamiltonian graph is an Euler graph (c) A graph can be neither Eulerian nor Hamiltonian (d) A Hamiltonian graph cannot be an Eulerian graph 9. Let G be a simple graph of size “e.” Let A(G), C(G), and K(G) be the incidence matrix, the circuit matrix, and cut-set matrix of G. Which of the following statement is false:

(a) rank(A) + rank(C) = e (b) rank(C) + rank(K) = e (c) rank(A) = rank(C) (d) rank(A) + rank(C) + rank(K) = e 10. A graph with n vertices and n-1 edges that is not a tree is (a) Euler (b) Circuit (c) Disconnected (d) Connected 11. The length of a Hamiltonian path (if exists) in a connected graph of n vertices is (a) n (b) n–1 (c) n + 1 (d) n/2 12. A path in graph G, which contains every vertex of G once and only once is (a) Euler tour (b) Hamiltonian path (c) Euler trail (d) Hamiltonian tour 13. Every cut-set of a connected Euler graph has (a) an odd number of edges (b) an even number of edges (c) at least three edges (d) none of these 14. If G is a connected graph, then (a) G  is unicursal if it has exactly one vertex of even degree (b) G  is unicursal if it has exactly one vertex of odd degree (c) G is never unicersal (d) G  is unicursal if it has no vertex of odd degree 15. Which of the following statement is true: (a) both K5 and K3,3 are planar (b) both K5 and K3,3 are nonplanar (c) K5 is planar but K3,3 is nonplanar (d) K5 is nonplanar but K3,3 is planar 16. Determine which of the following statements is not true (a) The Peterson graph is nonplanar (b) K5 is nonplanar

8/14/2023 4:44:33 PM

Graph Theory • 605

(c) K3,3 is nonplanar (d) Dual graph of a planar graph is nonplanar 17. Which of the following statement is true (a) K4 is a self-dual (b) if G** is the dual of a dual of G, G, and G** are always isomorphic (c) G** is always non-isomorphic to G (d) if G1 and G2 are isomorphic, so are their duals 18. Determine which of the following statements is not true. (a) The chromatic number of Kn is “n” (b) The chromatic number of any cycle is 2 (c) The chromatic number of any nontrivial tree is 2 (d) Every planar graph is 4 colorable 19. The graph whose chromatic polynomial is λ(λ–1)5 is (a) a tree having 6 vertices (b) K6 (c) K5 (d) A tree having 5 vertices 20. Let χ(G) be the (vertex) chromatic number of a graph G and δ, ∆, respectively, denote the minimum and maximum degree of a vertex of G, then (a) x > 1 + ∆ (b) x > 1 + δ (c) x ≤ 1 + ∆ (d) x ≤ 1 + δ 21. Let X(G) be the (vertex) chromatic number of a graph G having maximum degree ∆, then (a) if X ≤ ∆, G is either a complete graph or has an odd cycle (b) X  > ∆, G neither a complete graph nor has an odd cycle (c) if G is either a complete graph or has an odd cycle, then X > ∆ (d) if G neither a complete graph nor has an odd cycle, then X ≤ ∆ 22. Let x(G) be the (vertex) chromatic number of a graph G of order n, then (a) n < x (G) +x (G ) < n +1 (b) 2 n < c(G ) + c(G ) < n +1 (c) n < c(G ) + c(G ) < 2n (d) none of the above 23. Every complete graph is (a) regular (b) connected (c) simple (d) circuit

EMEP.CH11_4PP.indd 605

24. What is the minimum number of edges in a connected cyclic graph containing “n” vertices? (a) n (b) n + 1 (c) n–1 (d) 2n 25. Let G be a planar graph. Then G does not contain (a) subgraph isomorphic to K3 or K3,3 (b) subgraph homomorphic to K5 or K3,3 (c) subgraph homomorphic to K2 or K2,2 (d) subgraph isomorphic to K4,4 26. The adjacency matrix of the following graph is

(a)

é1 ê ê1 ê0 ê ë0

0 0 1 0

1 1 0 0

0ù ú 0 ú (b) 1ú ú 0û

(c)

é1 ê ê0 ê0 ê ë1

0 0 1 1

1 1 0 0

1ù ú 0ú 0ú ú 0û

é1 ê ê1 ê0 ê ë0

0 0 1 1

0 0 0 1

1ù ú 0ú 1ú ú 0û

(d) none of these

27. Consider a simple connected graph G with “n” vertices and “n”-edges (n > 2). Then, which of the following statement are true? (a) G has no cycles. (b) The graph obtained by removing any edge from G is not connected. (c) G has at least two cycle. (d) The graph obtained by removing any two edges from G is not connected. 28. Let G be a graph without any loop. If G has six vertices and ten edges, then the size of its incidence matrix is

(a) 10 × 6 (b) 6 × 10 (c) 10 × 10 (d) 6 × 6 29. If A = (aij)n×n be the adjacency matrix of a graph having “n” number of vertices. If a33 = 1, then which of the following is true? (a) there exist a self loop at vertex v3 (b) the vertex v3 is connected with each vertex of the graph (c) each edge is incident to v3 (d) there exists at least one parallel edge joining v2 and v3

8/14/2023 4:44:33 PM

606 • Engineering Mathematics Exam Prep 30. If A = (aij)5×5 be the adjacency matrix of a graph. If a24 = 0, then which of the following is true? (a) no edge connects the vertices v2 and v4 (b) v2 and v4 are adjacent vertices (c) each of v2 and v4 is an isolated vertex (d) none of the above 31. Which of the following graph is connected? (a)

(c)

34. The number of spanning trees in a complete graph G with “n” vertices is (a) nn–2 (b) nn (c) n2 (d) nn–1 35. A complete graph with “n” vertices is (a) 2-chromatic (c) n-1 chromatic



(d) none of these

(b) n/2 chromatic (d) n chromatic

Answer Key

(b)

32. The number of distinct simple graphs with up to four nodes is: (a) 25–1 (b) 24–1 (c) 27–1 (d) 24 + 1 33.  Let G be a simple graph all of whose vertices have degree 3 and E = 2 V - 3 . What can be said about G? (a) G has 6 vertices and 9 edges (b) G has 6 vertices and 12 edges

EMEP.CH11_4PP.indd 606

(c) G has 9 vertices and 6 edges (d) None of these

1. (c)

2. (b)

3. (a)

4. (a)

5. (d)

6. (a)

7. (b)

8. (c)

9. (d)

10. (c)

11. (b)

12. (b)

13. (b)

14. (d)

15. (b)

16. (d)

17. (a)

18. (b)

19. (b)

20. (c)

21. (d)

22. (b)

23. (b)

24. (a)

25. (b)

26. (a)

27. (d)

28. (b)

29. (a)

30. (a)

31. (a)

32. (b)

33. (a)

34. (a)

35. (d)

Hints

26. (a) Use the definition of adjacency matrix. 32. (b) The number of distinct simple graphs with up to “n” nodes is equal to 2n–1.

8/14/2023 4:44:34 PM

APPENDIX

A

Gate 2019 Solved Papers TOPIC: LINEAR ALGEBRA



é1 1 0 ù 1. Consider the matrix P = ê0 1 1 ú ê ú êë0 0 1 úû The number of distinct eigenvalues of P is (a) 0 (b) 1 (c) 3 (d) 2 [ME GATE 2019]

2. The set of equations x + y + z =1 ax - ay + 3 z = 5 5 x - 3 y + az = 6 has infinite solutions, if a =? (a) 4 (b) -4 (c) (d) 3 -3  [ME GATE 2019]





EMEP.App-A_3PP.indd 607

3. In matrix equation AX = R, 4ù é4 8 ì2ü ì32 ü ï ï ï ï ê ú A = ê8 16 -4 ú , X = í1 ý and R = í16 ý ï ï ï ï êë 4 -4 15 úû î4 þ î64 þ One of the eigenvalues of the matrix A is (a) 8 (b) 16 (c) 15 (d) 4 [ME GATE 2019] 4. Let X be a square matrix. Consider the following two statements on X. I. X is invertible II. Determinant of X is non zero. Which one of the following is true? (a) I implies II; II does not imply I (b) I does not implies II; II does not imply I



(c) I and II are equivalent statements (d) II implies I; I does not imply II [CS GATE 2019] 5. Consider the following matrix 8 ù é1 2 4 ê1 3 9 27 ú ú R=ê ê1 4 16 64 ú ê ú ë1 5 25 125û  the absolute value of the product of eigenvalues of R is _____?  [CS GATE 2019] 6. A 3×3 matrix has eignvalues 1, 2, and 5. The determinant of the matrix is ______?  [CS GATE 2019] 7. The number of distinct eigenvalues of 2 2 3 3 the matrix A = 0 1 1 1 is equals to 0 0 3 3 ________ 0 0 0 2  [EC GATE 2019] 8. M is a 2 ´ 2 matrix with eigenvalues 4 and 9. The eigenvalues of M 2 are (a) 16 and 81 (b) 2 and 3 (c) –2 and –3 (d) 4 and 9  [EE GATE 2019] é0 1 1ù 9. The rank of the matrix, M = ê1 0 1ú , is ê ú ____? êë1 1 0 úû 

[EE GATE 2019] 10. Consider a 2 ´ 2 matrix M = év1 v2 ù , where ë û v1 and v2 are the column vectors. Suppose

8/9/2023 8:25:29 PM

608 • Engineering Mathematics Exam Prep éu1t ù t t M -1 = ê t ú where u1 and u2 are the row vecëêu2 ûú tors. Consider the following statements: Statement 1: u1t v1 = [1] and u2t v2 = [1]



Statement 2: u1t v2 = [0] and u2t v1 = [0] . Which of the following options is correct? (a) statement 2 is true and statement 1 is false (b) statement 1 is true and statement 2 is false (c) both statements are false (d) both the statements are true [EE GATE 2019] 1. (b) 6. 10

Answer key 2. (a) 3. (b) 4. (c) 7. 3.

8. (a)

9. 3

Thus, for a = 4, D = D1 = D 2 = D 3 = 0 and so by Cramer’s rule, the system has an infinite number of solutions. 3. (b) AX = R ì2 ü é 4 8 4 ù ì2 ü ì32 ü ï ï ï ï ê ú Þ 8 16 -4 í1 ý = í16 ý = 16 ïí1 ïý , ê ú ï4 ï êë 4 -4 15 úû ïî4 ïþ ïî64 ïþ î þ which is of the form AX = l X ; where l is an eigenvalue. \ One of the eigenvalues of the matrix must  be 16 4. (c) A ¹ 0 Û A is invertible matrix



5. 12 10. (d)

\ I and II are equivalent statements. \ R 5. 12 1 2 4 8

= Explanations

1. (b) Since P is a upper triangular matrix, so its diagonal elements 1, 1, 1 are the eigenvalues. Thus the number of distinct eigenvalues of P =1.

1 2 4 0 1 5 = 0 2 12

8 by R2 ® R2 - R1 , 19 æ ö ç ÷ 56 è R3 ® R3 - R1 , R4 ® R4 - R1 ø 0 3 21 117

2. (a) Now the system has an infinite number of ­solutions

1 5 19 1 5 19 æ by R2 ® R2 - 2 R1 , ö = 2 12 56 = 0 2 18 ç ÷ R3 ® R3 - 3R1 ø 3 21 117 0 6 60 è

1 1 1 if a -a 3 = 0 5 -3 a i.e., if ( -a 2 + 9) - ( a 2 - 15) + 1( -3a + 5a ) = 0 i.e., if - a + a + 12 = 0 i.e., if ( a - 4)( a + 3) = 0 i.e., if a = 4 , -3

=

2









EMEP.App-A_3PP.indd 608

Now for a = 4, 1 1 1 1 1 1 D1 = 5 -a 3 = 5 -4 3 6 -3 a 6 -3 4 = (-16 + 9) - (20 - 18) + (-15 + 24) = 0, 1 1 1 1 1 1 D2 = a 5 3 = 4 5 3 5 6 a 5 6 4 = (20 - 18) - (16 - 15) + (24 - 25) = 0, 1 1 1 1 1 1 D 3 = a -a 5 = 4 -4 5 5 -3 6 5 -3 6 = (-24 + 15) - (24 - 25) + (-12 + 20) = 0.

1 3 9 27 1 4 16 64 1 5 25 125



2 18 = 120 - 108 = 12. 6 60

\ The absolute value of the product of eigenvalues = det( R) = 12.

6. 10 Since determinant of a matrix is the product of its eigenvalues, so determinant of the given matrix = 1´ 2 ´ 5 = 10. 7. 3. Clearly the given matrix A is upper triangular. So its eigenvalues are its diagonal elements i.e; 2, 1, 3, 2. \ The number of distinct eigenvalues = 3. 8. (a) We know that if l is an eigenvalue of a matrix M then l 2 is an eigenvalues of the matrix M 2 . Given, M is a 2 ´ 2 matrix with eigenvalues 4 and 9. So, the eigenvalues of M 2 are 42 ,92 i.e; 16, 81.

8/9/2023 8:25:31 PM

Gate 2019 Solved Papers • 609

2. The value of the following definite integral

9. 3

é0 1 1ù The matrix M = ê1 0 1ú is a non-singular ê ú êë1 1 0 úû square matrix of order 3. So, its rank = 3. 10. (d) éa a ù éa ù Let M 2´2 = ê 11 12 ú , where v1 = ê 11 ú , ë a21 a22 û ë a21 û



é a12 ù v2 = ê ú . ë a22 û



\ M -1 =

1 a11a22 - a21a12

é a22 - a12 ù . Then ê -a ú ë 21 a11 û

u1t =

1 [ a22 a11a22 - a21a12

u2t =

1 [ -a21 a11 ]. a11a22 - a21a12

- a12 ] ,

e

ò ( x ln x)dx is _____? (round off to three deci1

mal places) [ME GATE 2019] x 4 - 81 3. Compute lim 2 x ®3 2 x - 5 x - 3 (a) 108/7 (b) 1 (c) 53/12 (d) Limit does not exist  [CS GATE 2019] 

4. The curve y = f(x) is such that the tangent to the curve at every point (x, y) has a y-axis intercept c, given by c = -y. Then f(x) is proportional to x -1 (b) x2 (a) (c) x3 (d) x4  [CS GATE 2019] p p sin x 5. The value of the integral dx dy, is ò ò x equal to_________? 0 y

é ù é a11 ù a22 -a12 \ u1t v1 = ê  [EC GATE 2019] úê ú ë a11a22 - a21a12 a11a22 - a21a12 û ë a21 û 6. Consider a differentiable function f(x) on the é æ ö æ öù a22 -a12 set of real numbers such that f(–1) = 0 and = ê a11 ç ÷ + a21 ç ÷ú a a a a a a a a f ¢( x) £ 2. Given these conditions, which 11 22 21 12 11 22 21 12 è ø è ø ë û    one of the following inequalities is necessarily = [1] true for all x Î [ -2, 2] ? Similarly, u2t v2 = [1], u1t v2 = [0], u2t v1 = [0]. f ( x) £ 2 x f ( x) £ 2 x + 1 (b) (a) So, both statements are true. 1 1 (c) f ( x) £ x + 1 (d) f ( x) £ x TOPIC: CALCULUS 2 2  [EC GATE 2019] 1. A parabola x = y 2 with 0 £ x £ 1 is shown in 7. For a small value of h, the Taylor series the figure. The volume of the solid of rotation ­expansion of f(x + h) is obtained by rotating the shaded area by 360 h2 h3 (a) f ( x) - hf ¢( x) + f ¢¢( x) f ¢¢¢( x) + .........¥ around the x axis is 2 3 2 h h3 f ( x) - hf ¢( x) + f ¢¢( x) f ¢¢¢( x) + .......¥ (b) 2! 3! h2 h3 f ( x) + hf ¢( x) + f ¢¢( x) + f ¢¢¢( x) + .........¥ (c) 2 3 h2 h3 f ( x) + hf ¢( x) + f ¢¢( x) + f ¢¢¢( x) + .........¥ (d) 2! 3!



p p (b) (a) 2 4 (c) p 2p (d)  [ME GATE 2019]

EMEP.App-A_3PP.indd 609

[CE GATE 2019]

8. Which of the following is correct? sin 4 x tan x (a) lim = 1 and lim =1 x ®0 sin 2 x x ®0 x sin 4 x tan x (b) lim = ¥ and lim =1 x ®0 sin 2 x x ®0 x

8/9/2023 8:25:32 PM

610 • Engineering Mathematics Exam Prep sin 4 x tan x = 2 and lim =¥ x ®0 sin 2 x x ®0 x sin 4 x tan x (d) lim = 2 and lim =1 x ®0 sin 2 x x ®0 x  [CE GATE 2019]

1

(c) lim

0

x=0

(b) The function x x (x > 0) has the global minima at x = e (c) The function x x has the global maxima at x = e (d) The function x3 has neither global ­minima nor global maxima [CE GATE 2019]



10. The following inequality is true for all x close to zero. x2 x sin x 2- < 5% (c) 0.1 % and 1% and 1 and 1 £ x < ¥ and p(x) = 0 for -¥ < x < 1 . For p(x) to be probability density function, the value of A should be equal to 1 1 (a) m - 1 (b) m + 1 (c) (d) m -1 m +1 [CS GATE 2019]

8. If X and Y are random variables such that E (2 X + Y ) = 0 and E ( X + 2Y ) = 33 , then E ( X ) + E (Y ) = ___?  [EC GATE 2019] 9. Let Z be an exponential random variable with mean 1. That is, the cumulative distribution function of Z is given by ì1 - e - x if x ³ 0 then prob. (Z >2 | Z Fz ( x) = í î 0 if x < 0 >1), rounded off two decimal places, is equal to _____?  [EC GATE 2019] 10. The probability of a resistor being defective is 0.02. There are 50 such resistors in a circuit. The probability of two or more defective resistor in the circuit (round off to two decimal places) is _____ ?  [EE GATE 2019] 11. The probability density function of a continuous random variable distributed uniformly between x and y ( y > x) is 1 x- y (a) (b) x- y 1 y - x (d) (c) y-x  [CE GATE 2019] Answer key 1. (d) 6. (a)

2. 3 7. (a)

3. 0.17

4. 0.08

8. 11

1 9. e

5. 0.8 10. 0.26

11. (d) Explanation

1. (d) Here mean ( m ) = 440 mm and SD (s ) =1mm Let X be the random variable denoting the lengths of rods. Here, we need to find P [ 438 < X < 441].  Since the standard normal variable, X -m so Z= s 438 - 440 X = 438 Þ Z = = -2 and 1 441 - 440 X = 441 Þ Z = = 1. 1

8/9/2023 8:25:50 PM

620 • Engineering Mathematics Exam Prep



\ P [ 438 < X < 441]

\ Prob.( 3 x 2 + 3 xy + 3 y + 6 has only real roots)

= P [ -2 < Z < 1] = P [ -2 < Z < 0] + P [ 0 < Z < 1]

= P (Y £ -1) + P (Y ³ 2 )

æ 95.44 ö æ 68.26 ö =ç ÷% + ç ÷ % = 81.85% è 2 ø è 2 ø

= =



=





EMEP.App-A_3PP.indd 620

C1 8C1 1 8 4 ´ 15 = ´ = . 16 C1 C1 2 15 15

Þ

¥

ò

p ( x)dx = 1 Þ

-¥

1

¥

-¥

1

ò 0 dx + ò

¥

A dx = 1 xm ¥

é ù 1 Þ Aò x - m dx = 1 Þ A ê =1 m -1 ú ë (1 - m ) x û1 1 A Þ [0 - 1] = 1 Þ A = m - 1. 1- m

= 1 - {(0.05) + 15(0.05) ´ 0.05} » 0.17

i.e;if (6 y)2 - 4 ´ 3 ´ (3 y + 6) ³ 0 i.e;if ( y - 2)( y + 1) ³ 0 i.e;if ( y - 2) ³ 0, ( y + 1) ³ 0 or ( y - 2) £ 0, ( y + 1) £ 0 i.e;if y ³ 2, y ³ -1 or y £ 2, y £ -1 i.e;if y ³ 2 or y £ -1

8

7. (a) p(x) is a p.d.f

14

4. 0.08 Clearly, either first place or the last place is the worst possible location for the pivot ­element. 2 Hence, the required probability = = 0.08 25 5. 0.8 Since Y is distributed uniformly in the open interval (1, 6), so the p.m.f of Y is: ì1 ï , 1£ y £ 6 f ( y) = í 5 ïî0, otherwise The polynomial 3 x 2 + 6 xy + 3 y + 6 has only real roots if 3 x 2 + 6 xy + 3 y + 6 = 0 has a real root

1 6 6-2 4 = = 0.8. [ y] = 5 2 5 5

= P( A Ç B ) = P ( A) ´ P ( B / A)

= 1 - { 15C0 p15 q 0 + 15C1 p14 q} =1 - { p15 + 15 p14 q}



6

¥

1 f ( y ) dy + ò f ( y ) dy = 0 + ò dy 5 2 2

6. (a) The total number of balls = 8 + 8 = 16. Let A = the event of drawing a red ball in 1st withdraw & B = event of drawing a green ball in 2nd withdraw. Since the balls are drawn without replacement, so the required probability

= 1 - { P( X = 0) + P( X = 1)} 15

ò

-¥

2. 3

Mean = 3 + x + 2 + 4 Þ x = 9 + x Þ x = 3. 4 4  \ The data values are 3, 3, 2 and 4 among which ‘3’ is most frequently used. Therefore, Mode = 3. 3. 0.17 Let X be the random variable denoting the number of defective parts. Then X follows a Binomial distribution with number of trials (n) = 15 and probability of success (p) = 0.05. Therefore, q = 1–0.05 = 0.95. So, P [ X ³ 2] = 1 - P [ X < 2]

-1

1 æ ö çm > 1 Þ m -1 ® 0 when x ® ¥ ÷ x è ø



8. 11 E (2 X + Y ) = 0 Þ 2 E ( X ) + E (Y ) = 0.........(i) E (X + 2 Y) = 33 Þ E ( X ) + 2 E (Y ) = 33...........(ii)

Solving (i) and (ii) ; we have, E (X) = -11 and E (Y ) = 22. \ E ( X ) + E (Y ) = -11 + 22 = 11. 9.

1 e

-x Given, F ( x) = íì1 - e if x ³ 0 z î 0 if x < 0



ìe x if x ³ 0 \ Fz¢ ( x) = f ( x) = í î 0 if x < 0



Now Prob( z > 2 | z > 1)

8/9/2023 8:25:51 PM

Gate 2019 Solved Papers • 621 =

Prob( z > 2 Ç z > 1) Prob( z > 1)

Answer key

¥

=

Prob( z > 2) = Prob [ z > 1]



ò f ( x )dx

2 ¥

=

ò f ( x )dx 1

¥

=



éë -e - x ùû 2

¥

éë -e - x ùû 1

=

1. (b), (c) 2. 4.25

¥

-x ò e dx

2 ¥

Explanation

-x ò e dx 1

0 + e -2 e -2 1 = = . 0 + e -1 e -1 e

10. 0.26 The probability being defective, p = 0.02 and the number of resistor, n = 50. 2 \ l = np = 50 ´ 0.02 = 50 ´ = 1. 100 Let “X” be the random variable denoting the number of defective resistors. Clearly X follows the Poisson distribution with mean l =1. So, the required probability = P [ x ³ 2] = 1 - P [ x < 2]

= 1 - [ P( x = 0) + P( x = 1) ] é e- l l 0 e- l l ù = 1- ê + 1! úû ë 0!



= 1 - e - l [1 + l ] = 1 - e -1 [1 + 1] = 1 -



2 » 0.26 e

11. (d) The result follows from the definition of uniform probability distribution.

1. (b), (c) In case of labeled nodes, for an undirected complete graph G, we have, number ( n - 1)! . of Hamiltonian cycles = 2 In case of unlabeled nodes, for an undirected complete graph G, every Hamiltonian cycles will be similar and so number of Hamiltonian cycles =1. 2. 4.25 For a full binary tree T with eight leaves, the number of pairs in which two levels can be chosen independently= 8 × 8 = 64. Among them 8 pairs will at distance 0 unit from each others, 8 pairs will be at distance 2 unit from each other (Children of the same parent), 16 pairs will be distance 4 unit from each other and 32 pairs will be at distance 6 unit from each others. Hence, the total distance = 8 × 0 + 8 × 2 + 16 × 4 + 32 × 6 = 272 So, the expected distance = 272/64 = 4.25.

TOPIC: LAPLACE TRANSFORMS 1. Let Y(s) be the unit-step response of a casual system having a transfer function 3- s That is, Y ( s ) = G ( s ) . G (s) = ( s + 1)( s + 3) s

TOPICS: GRAPH THEORY 1. Let G be an undirected complete graph on n vertices, where n > 2. Then, the number of different Hamiltonian cycles in G is equal to ( n - 1)! (a) n ! (b) 2 (c) 1 (d) ( n - 1) !  [CS GATE 2019] 2. Let T be a full binary tree with eight leaves (A full binary tree has every level full.). Suppose two leaves a and b of T are chosen uniformly and independently at random. The ­expected value of the distance between a and b in T (i.e., the number of edges in the unique path between a and b) is (rounded off to two ­decimal places)_____  [CS GATE 2019]

EMEP.App-A_3PP.indd 621



The forced response of the system is

u (t ) - 2e - t u (t ) + e -3t u (t ) (a) (b) 2u (t) (c) u (t)

(d) 2u (t ) - 2e - t u (t ) + e -3t u (t )  [EC GATE 2019] 2. The inverse Laplace transform of s+3 H(s) = 2 for t ³ 0 is s + 2s + 1 (a) 3 te - t + e - t (b) 3e - t (c) 4te - t + e - t (d) 2te - t + e - t  [EE GATE 2019]

8/9/2023 8:25:53 PM



4 4 2 2 1 1 + - + 3s 3( s + 3) s + 1 s + 3 3s 3( s + 3) 1 2 1 = + s s +1 s + 3



\ y (t ) = L-1{Y ( s )}



é1 ù é 1 ù -1 é 1 ù = L-1 ê ú - 2 L-1 ê +L ê ësû ë s + 1 úû ë s + 3 úû = u (t ) - 2e - t u (t ) + e -3t u (t ).

622 • Engineering Mathematics Exam Prep 3. The output response of a system is denoted as y(t), and its Laplace transform is given by 10 . The steady state Y (s) = 2 s s + s + 100 2

(

)

value of y(t) is 1 (a) (b) 10 2 100 2 1 (c) (d) 100 2 10 2  [EE GATE 2019]

=

2. (d) L-1 [ H ( s ) ] é s+3 ù é s+3 ù = L-1 ê 2 = L-1 ê 2ú ú + + s 2 s 1 ë û ë ( s + 1) û

Answer key 1. (a)

2. (d)

3. (c) Explanation

1. (a) G ( s) 3- s = s s ( s + 1)( s + 3) 4 - (1 + s ) 4 1 = = s ( s + 1)( s + 3) s ( s + 1)( s + 3) s ( s + 3) ( s + 1) - s ( s + 3) - s =4 s ( s + 1)( s + 3) 3s ( s + 3) 4 4 1 1 = - + s ( s + 3) ( s + 1)( s + 3) 3s 3( s + 3) 4 ( s + 3) - s ( s + 3) - ( s + 1) 1 1 = -2 - + 3 s ( s + 3) ( s + 1)( s + 3) 3s 3( s + 3)



é 1 ù é 1 ù = L-1 ê + 2 L-1 ê 2ú ú s 1 + ë û ë ( s + 1) û

Y (s) =



EMEP.App-A_3PP.indd 622

4 4 2 2 1 1 = + - + 3s 3( s + 3) s + 1 s + 3 3s 3( s + 3) 1 2 1 = + s s +1 s + 3

é s +1+ 2 ù é 1 2 ù = L-1 ê = L-1 ê + 2 ú 2ú êë ( s + 1) úû êë s + 1 ( s + 1) úû



= e - t + 2te - t .

3. (c) We need to find steady state values of y(t)

i.e, y (¥). By the final value theorem, y (¥ ) 10 = lim sY ( s ) = lim s 2 s ®0 s ®0 s s + s + 100 2

(

= lim s ®0



(s

10 2

+ s + 100 2

)

=

)

10 1 = . 0 + 0 + 100 2 10 2

8/9/2023 8:25:54 PM

APPENDIX

B

Gate 2020 Solved Papers TOPIC: LINEAR ALGEBRA

4. Consider the following system of linear ­equations:

é1 -1 0 ù

1. Consider the matrix M = ê1 -2 1 ú . One of ê ú êë0 -1 1 úû

the eigenvectors of M is _____?

x1 + 2 x2 = b1 ,2 x1 + 4 x2 = b2 ,3 x1 + 7 x2 = b3 ,3 x1 + 9 x2 = b4

Which of the following conditions ensures that a solution exists for the above system? (a) b3 = 2 b1 ,3 b1 - 6 b3 + b4 = 0

é -1ù

é1ù

êë -1úû

êë 1 úû

(b) b3 = 2 b1 ,6 b1 - 3 b3 + b4 = 0

ê ú (d) (c) 1

é1ù

é1ù ê ú ê1ú êë -1úû

(c) b2 = 2 b1 ,3 b1 - 6 b3 + b4 = 0



[IN GATE 2020]

(a) ê 1 ú (b) ê -1ú ê ú ê ú

ê ú êë1úû

(d) b2 = 2 b1 ,6 b1 - 3 b3 + b4 = 0 [EC GATE 2020] 5. The number of purely real elements in a lower triangular representation of the given 3 × 3 matrix, obtained through the given decomposition is ______?

2. Let A and B be two n × n matrices over real numbers. Let rank (M) and det (M) denote the rank and determinant of a matrix M, respectively. Consider the following statements: I. rank( AB) = rank(A) ´ rank(B) II. det( AB) = det(A) ´ det(B) III. rank( A + B) £ rank(A) + rank(B) IV. det( A + B) £ det(A) + det(B) Which of the above statements are true? (a) III and IV only (b) II and III only (c) I and II only (d) I and IV only  [CSE/IT GATE 2020] 4

3. If V1 , V2 ,........, V6 are six vectors in R , which of the following statements is false? (a) If {V1 , V3 , V5 , V6 } spans R4, then it forms a basis for R4 (b) Any four of these vectors form a basis for R4 (c) These vectors are not linearly independent (d) It is not necessary that these vectors span R4  [EC GATE 2020]

EMEP.App-B_3PP.indd 623

é2 3 3 ù é a11 ê ú ê ê3 2 1 ú = ê a12 êë3 1 7 úû êë a13

(a) 6 

0 a 22 a23

(b) 5

0 ù é a11 úê 0 úê 0 a33 úû êë 0

a12 a 22 0

a13 ù ú a23 ú a33 úû

(c) 8 (d) 9 [EE GATE 2020]

6. A matrix P is given below:

é0 ê -2 P=ê ê0 ê ë0

1 3 0 0

3 0 6 1

(a) 1, 2, 3, 4 (c) 3, 4, 5, 7 

0ù 4ú ú The 1ú ú 6û

eigenvalues of P are (b) 0, 3, 6, 6 (d) 1, 2, 5, 7 [CE GATE 2020]

7. Consider the system of equations The value of x3 (round off to nearest integer), is __?

8/4/2023 8:42:46 PM

624 • Engineering Mathematics Exam Prep



é1 ê2 ê ê4 ê ë2

é ù 3 2ù ê ú éx ù 1 2 -3 ú ê 1 ú ê ú ú x = ê1 ú 4 -6 ú ê 2 ú ê ú 2 ú êx ú 5 2û ë 3 û ê ú ê1 ú ë û

[CE GATE 2020]



8. A matrix P is decomposed into its symmetric part S and skew-symmetric part V. If

é ê -4 ê S=ê4 ê ê ê2 ë

(a)

ù é 2ú ê 0 -2 ú ê 7ú ,V = ê 2 0 ê 2ú ú ê 7 2ú ê-3 2 û ë

4 3 7 2

ù 3ú ú 7ú , 2ú ú 2ú û é ê -2 ê ê -1 ê ê ê -2 ëê

é -4 2 5 ù ê ú (b) ê 6 3 7ú êë -1 0 2 úû -6

é4

1ù ú 0 ú (d) êë -5 -7 -2 úû

9 2 81 4 45 2

ù -1 ú ú 11 ú ú 73 ú ú 4 úû

é -4 6 -1ù ê ú ê2 3 0ú êë 5 7 2 úû

(c) êê -2 -3 

then P is

[ME GATE 2020]

9. Let I be a 100 dimensional identity matrix and E be the set of all its distinct real eigenvalues. The number of elements in E is _______ ?  [ME GATE 2020] 1. (c)

Answers key 2. (b) 3. (b) 4. (d)

6. (d)

7. (3)

8. (a)

5. (c)

1. (c) det(M - l I) = 0

-l Þ -l -l

Þ x - y = 0, x - 2 y + z = 0, - y + z = 0 Þ x = y=z

Therefore, the eigenvector corresponding to l = 0 is



-1 0 -2 - l 1 =0 -1 1-l

1 0ö ÷ è0 0ø

æ0 0ö ÷ =O. è0 0ø

So rank( AB) = 0 but rank( A) = 1 = rank(B) . Therefore, rank( AB) ¹ rank(A) ´ rank(B) .

EMEP.App-B_3PP.indd 624

Hence, the statement I is not correct. 1 0ö ÷. è0 1ø

Now, A + B = æç

So det( A + B) = 1 but det( A) = 0 = det(B) . Therefore, det( A + B) > det(A) + det(B) . Hence, the statement IV is not correct. 3. (b) Let V1 = (1,0,1,0), V2 = (-1,0, -1,0), V3 = (0,0,1,1), V4 = (0,0, -1, -1). Then, V1 + V2 + V3 + V4 = (0,0,0,0) = O , which shows that the set {V1 , V2 , V3 , V4 } is not linearly independent and hence does not form a basis of R4.

4. (d) The given system of equations can be expressed as AX = B,

1 -1 0 Þ -l 1 -2 - l 1 =0 1 -1 1-l

Now, MX = lX

and B = æç 0 0 ö÷ . è0 1ø

Then, AB = ç

-1 0 -2 - l 1 = 0 (by C1 ® C1 + C2 + C3 ) -1 1-l

Þ l = 0 is an eigenvalue

é x ù é x ù é1ù ê ú ê ú ê ú X = ê yú = ê x ú = ê1ú (for x = 1) ëê z ûú ëê x ûú ëê1ûú

2. (b) Let A = æç

9. (1)

Explanations 1-l Þ 1 0



é1 -1 0 ù é x ù é xù ê úê ú ê ú Þ ê1 -2 1 ú ê yú = 0 ê yú êë0 -1 1 úû êë z úû êë z úû é x - y ù é0 ù ê ú ê ú Þ ê x - 2 y + zú = ê0 ú êë - y + z úû êë0 úû



é1 ê2 where A = ê ê3 ê ë3

2ù é b1 ù ú êb ú 4 x ú , X = é 1 ù =, B = ê 2 ú ê ú ê b3 ú 7ú ë x2 û ú ê ú 9û ë b4 û

So, the augmented matrix

8/4/2023 8:42:47 PM

Gate 2020 Solved Papers • 625 é1 ê2 = [ A : B] = ê ê3 ê ë3







2 0

2 b1 ù 4 b2 ú ú 7 b3 ú ú 9 b4 û b1 ù æ by R2 ® R2 - 2 R1 , ö b2 - 2 b1 ú ç ÷ ú R3 ® R3 - 3 R1 , ÷ b3 - 3 b1 ú çç R4 ® R4 - 3 R1 ÷ø ú b4 - 3 b1 û è b1 ù b4 - 3 b1 ú ú ( by R « R ) 2 4 b3 - 3 b1 ú ú b2 - 2 b1 û

é1 ê0 ê ê0 ê ë0

1 3

é1 ê0 ê ê0 ê ë0

2 3 1 0

é1 ê0 ê ê0 ê ë0

2 b1 ù 0 6 b1 - 3 b3 + b4 ú ú 1 b3 - 3 b1 ú ú 0 b2 - 2 b1 û

( by R2 ® R2 - 3R3 )

é1 ê0 ê ê0 ê ë0

2 b1 ù 0 b3 - 3 b1 ú ú 1 6 b1 - 3 b3 + b4 ú ú 0 b2 - 2 b1 û

( by R2 « R3 )

Hence, rank(a) = 2. Now, the given system has a solution Û rank([ A : B]) = rank( A) Û rank([ A : B]) = 2 Û b2 - 2 b1 = 0,6 b1 - 3 b3 + b4 = 0 Û b2 = 2 b1 ,6 b1 - 3 b3 + b4 = 0

5. (c) 0 ù é a11 a12 a13 ù é2 3 3 ù é a11 0 ê ú ê ú ê ú = 3 2 1 a a 0 ú ´ ê 0 a 22 a23 ú 22 ê ú ê 12 êë3 1 7 úû êë a13 a23 a33 úû êë 0 0 a33 úû 2 é a11 ù a11 a12 a11 a13 ê ú 2 2 a12 a12 a13 + a 22 a23 ú Þ ê a21 a11 + a22 2 2 2 ê a13 a11 a13 a12 + a23 a 22 a13 + a23 + a33 ú ë û

Therefore, the number of purely real elements in the lower triangular representation of the given 3 × 3 matrix = 8. 6. (d) 0 -2 P = 0 0





2 2 2 + a23 + a33 =7 a13



2

EMEP.App-B_3PP.indd 625

2

= 2 ëé( 36 - 1 ) - 3 ( 0 - 0 ) + 0 ûù = 70

= Product of eigenvalues = 1 × 2 × 5 × 7 7. (3) é ù é ù é ù 3 2ù ê ú ê ú ê ú é x1 ù ê1 ú ê x1 + 3 x2 + 2 x3 ú ê1 ú ú 2 -3 ê ú ú x = ê1 ú Þ ê 2 x + 2 x - 3 x ú = ê1 ú 2 3 4 -6 ú ê 2 ú ê ú ê 1 ú ê ú ú êë x3 úû ê2 ú ê 4 x1 + 4 x2 - 6 x3 ú ê2 ú 5 2û ê1 ú ê2 x + 5 x + 2 x ú ê1 ú 2 3û ë û ë 1 ë û Þ x1 + 3 x2 + 2 x3 = 1, 2 x1 + 2 x2 - 3 x3 = 1, 4 x1 + 4 x2 - 6 x3 = 2,

2 x1 = 1 - 2 x2 + 3 x3 = 1 - 2 x2 Þ x1 =

(2),

9 x2 5

1 19 x2 ...........(5) 2 10



Using (4) and (5) we get from (1),



6x 1 19 x2 + 3 x2 - 2 = 1 2 10 5 -31 x2 + 30 x2 1 = or, 10 2 or, x2 = -5 3 \ x3 = - ´ (-5) = 3 5

8. (a)

5 5 1 i.e; a22 = i, a23 = (1 - a12 a13 ) , 2 2 a22

a33 = 7 - a13 - a23  Clearly, each of a22, a23, a33 is a complex ­number.

0 1 3 0 4 = -(-2) 0 6 1 1 0 1 6 6

2 x1 + 5 x2 + 2 x3 = 1 Þ x1 + 3 x2 + 2 x3 = 1...........(1) 2 x1 + 2 x2 - 3 x3 = 1...........(2) 2 x1 + 5 x2 + 2 x3 = 1...........(3) (3) - (2) Þ 3 x2 + 5 x3 = 0 3x Þ x3 = - 2 ..............(4) 5 3 x2 Putting x3 = we get from 5

3 3 3 Þ a11 = ± 2, a12 = ± , a13 = ± , a21 = ± , 2 2 2 2 a22 =-

3 0 6 1

é1 ê2 ê ê4 ê ë2

é2 3 3 ù ê ú = ê3 2 1 ú êë3 1 7 úû 2 Þ a11 = 2, a11 a12 = 3, a11 a13 = 3, a21 a11 = 3, 2 2 + a22 = 2, a12 a13 + a 22 a23 = 1, a12

1 3 0 0

2



1 1 P = (P + P T ) + (P - P T ) = S + V (given) 2 2 é ù ê -4 4 2 ú ê ú 1 7ú \ (P + P T ) = S = ê 4 3 ..............(i) ê 2 2ú ê ú 7 2ú ê2 2 ë û

8/4/2023 8:42:48 PM

626 • Engineering Mathematics Exam Prep and



é ê 0 -2 ê 1 (P - P T ) = V = ê 2 0 ê 2 ê 7 ê-3 2 ë (i) + (ii) Þ P = S + V

ù 3ú ú 7ú ..............(ii) 2ú ú 2ú û

é ê -4 ê =ê 4 ê ê ê2 ë

ù 3ú ú é -4 2 5 ù 7ú ê ú = 6 3 7ú 2ú ê ú ëê -1 0 2 ûú 2ú û

4 3 7 2

ù é 2 ú ê 0 -2 ú ê 7ú ê + 2 0 2ú ê ú ê 7 2 ú ê-3 2 û ë

9. (1) 1 is the only eigenvalue of an identity matrix of any order.

(a) 1

[CE GATE 2020]

(a)

1 + y2

, w.r.t x at the point (1,

0, e) is 1

(a)  (b) 1 (c) –1 (d) 0 e  [EC GATE 2020] 4. Consider the function f ( x, y) = x2 + y2 . The minimum value of the function attains on the line x + y = 1 (rounded off to one decimal place) is _______ ?  [IN GATE 2020] 5. The value of lim x ®¥ (a)

9 x2 + 2020 x+7

7 (b) 1 9

(c) indeterminate 

EMEP.App-B_3PP.indd 626

is

(d) 3 [CE GATE 2020]

2 x2 y + 2 =1 2 a b

equation

+1

is

(a) p ab (b) 4p ab



-

c +1 (b) c c c (d) c c +1

is

(c)  [ME GATE 2020] 8. The area of an ellipse represented by an

1. Consider the functions: I. e–x II. x2 - sin x III. x3 + 1 Which one of the functions above is/are increasing everywhere in [0, 1]? (a) I and III only (b) III only (c) II and III only (d) II only  [CSE/IT GATE 2020]

 f ( x, y, z) = e1- x cos y + xze

æ 1 - e- c(1- x) ö lim ç ÷ x ®1 1 - xe - c (1 - x ) è ø

7. The value of

(c)

1

(b) 



TOPIC: CALCULUS:

2. For real numbers x and y, with y = 3 x2 + 3 x + 1, the maximum and minimum value of y for x Î [-2,0] are, respectively ______ ? (a) 1 and 1/4 (b) 7 and 1 (c) 7 and 1/4 (d) –2 and –1/2  [EE GATE 2020] 3. The partial derivative of the function

x2 - 5 x + 4 is 4 x2 + 2 x 1 (c)  1 (d) 0 2 4

6. The value of lim x ®¥

3 4 p ab (d) pab 2

[CE GATE 2020] 9. If C represents a line segment between (0, 0, 0) and (1, 1, 1) in the Cartesian coordinate system, the value (expressed as integer) of the line integral ò éë( y + z ) dx + ( x + z ) dy + ( x + y) dzùû is C



[CE GATE 2020] 10. Let I = ò ò xy dydx. Then, I may also be x = 0 y= 0 1

x2

2

­expressed as 1

1

y= 0

x= y

(a)

ò ò

(c)

ò ò



1

y=0

y x=0

yx2 dxdy

1

1

y= 0

x= y

(b)

ò ò

xy2 dxdy (d)

ò ò

1

y=0

y x=0

xy2 dxdy yx2 dxdy

[ME GATE 2020]

11. Define [x] as the smallest integer less than or equal to x for each x Î (-¥, ¥). If y = [ x] , then area under y for x Î [1, 4] is _______? (a) 6 (b) 3 (c) 4 (d) 1  [ME GATE 2020] 1. (b)

Answer key 2. (c) 3. (d) 4. (0.5)

6. (c)

7. (c)

8. (d)

9. (3)

5. (d) 10. (a)

11. (a) Explanation

1. (b) Let f ( x) = x3 + 1. Then, f ¢( x) = Hence,

3 x2 2 x3 + 1

x3 + 1

³ 0 " x Î [0,1] .

is increasing everywhere

8/4/2023 8:42:49 PM

Gate 2020 Solved Papers • 627

in [0, 1] 2. (c) y = 3 x2 + 3 x + 1

d 2 ( x - 5x + 4) dx = lim ëéby the L'Hospital ruleûù x ®¥ d 4 x2 + 2 x ) ( dx 2x - 5 é ¥ù =lim êëform ¥ úû x ®¥ 8 x + 2

dy d2 y = 6 x + 3 and 2 = 6 dx dx dy Now, = 0 Þ 6 x + 3 = 0 Þ x = -0.5 dx Þ

d2 y At x = -0.5, = 6 > 0. dx2

Therefore, x = -0.5 is a point of local minima of y. Hence, the maximum value of y in [-2, 0]



7. (c)

= max{y(-2), y(0)}



d (2 x - 5) =lim dx éëby the L'Hospital ruleûù x ®¥ d (8x + 2) dx 2 1 = lim = . x ®¥ 8 4

= max{3 ´ (-2)2 + 3 ´ (-2) + 1,0 + 0 + 1} = 7

Also, the minimum value of y in [–2, 0] = min{y(-2), y(0), y(-0.5)} = min{7,1,3 ´ (-0.5)2 + 3 ´ (-0.5) + 1}



1ü 1 ì = min í7,1, ý = . 4þ 4 î

3. (d) f ( x, y, z) = e1- x cos y + xze

-

= lim x ®1

1 1 + y2



a



x + y = 1 Þ y = 1 - x. \ f ( x, y) = x + y = x + (1 - x) = g( x) (say) 2

2

2

2

a

=

4b a2 - x2 dx a ò0

(

So g¢( x) = 0 Þ 4 x - 2 = 0 Þ x = 0.5

=

a xù 4b é x 2 a - x2 + sin -1 ú ê a ë2 a û0 2

=

ù a2 a 4b é a 2 2 a a sin -1 - ( 0 - 0 ) ú + ê 2 a ë2 a û

=

4 b a2 p ´ sin -1 1 = 2 ab ´ = p ab a 2 2

\ g¢¢( x)

=4 > 0



9 x + 2020 x+7 2020 9+ 2 x = 9+0 = 3 = lim x ®¥ 7 1 1+ x x ®¥

6. (c)

x2 - 5 x + 4 é ¥ù lim êëform ¥ úû x ®¥ 4 x 2 + 2 x

2

)

a

C

2

lim

a2 - x2 is an even function

9. (3) The given line integral = ò éë( y + z ) dx + ( x + z ) dy + ( x + y ) dz ùû

= f (0.5,0.5) = 0.52 + 0.52 = 0.5

5. (d)

EMEP.App-B_3PP.indd 627

b 2 a - x2 dx a -a

= 2ò

Now g( x) = x2 + (1 - x)2 Þ g¢( x) = 2 x - 2(1 - x) = 4 x - 2, g¢¢( x) = 4

x = 0.5 Thus, x = 0.5 is a point of local minima of g(x). At x = 0.5, y = 1 - 0.5 = 0.5 . Therefore, (0.5, 0.5) is a point of minimum of f(x, y). Hence, the required minimum value of f(x, y)



In the given ellipse, x varies from – a to a. Hence, the required area = 2 ´ (area of upper half of the ellipse)

4. (0.5)



-c c c = = -(1 + cx) 1 + c ´ 1 c + 1

2 y2 x2 y x2 b 2 + = 1 Þ =1- 2 Þ y = a - x2 2 2 2 a b b a a

8. (d)

1

2 ¶f Þ = e1- x cos y ´ (- cos y) + ze 1+ y ¶x 1 ¶f 2 Þ = - e1-1cos 0 + e ´ e 1+ 0 = 0. ¶x (1,0, e)



æ 1 - e- c(1- x) ö æ 0ö lim ç ÷ ç form ÷ x ®1 1 - xe - c (1 - x ) 0 ø è ø è d - c (1 - x ) (1 - e ) = lim dx ( by the L'Hospital rule ) x ®1 d 1 - xe- c(1- x) ) ( dx - ce- c(1- x) = lim x ®1 - e - c (1 - x ) + cxe - c (1 - x ) { }



= ò éë( y dx + x dy ) + ( y dz + z dy ) + ( x dz + z dx ) ùû C

=

(1,1,1)

ò

(0,0,0)

éëd( xy) + d(yz) + d( xz)ùû (1,1,1)

= ëé xy + yz + zx ûù(0,0,0)



= (1 + 1 + 1) - (0 + 0 + 0) =3

8/4/2023 8:42:50 PM

628 • Engineering Mathematics Exam Prep 10. (a) The limits of integration are as follows: 0 £ x £ 1, 0 £ y £ x2 . In other words, 0 £ y £ 1, y £ x £ 1. Thus, changing the order of integration, we have, I=ò



1

ò

x2

x = 0 y= 0

11. (a) ì1, ï2, [ x] = ïí ï3, ïî4,

xy2 dydx = ò

1

y=0

ò

1 x= y

5. An ordinary differential equation is given ­below:

x

d2 x dt

 -5 2

4

2

3

4

1

1

1

2

3

0 and

= ò y dx = ò [ x ] dx = ò 1 dx + ò 2 dx + ò 3 dx 4

= ( 2 - 1 ) + 2 ( 3 - 2 ) + 3 ( 4 - 3 ) = 6.

dy = 2 x - y, y(0) = 1  dx

2. The general solution of

[EE GATE 2020] d2 y dy - 6 + 9y = 0 2 dx dx

3. Which of the following options contain two solutions of the differential equation dy = x(y - 1) ? dx

-x

x

with initial conditions x(0) = the solution is

Answer key 2. (b) 3. (c) 4. (d)

1. (0.886) The given equation is

5. (a)

dy + y = 2 x …(i) dx

Integrating factor (I.F.) of (1) = eò = e x Therefore, the solution of the linear equation (1) is dx

y x = ò (I.F.) ´ e x dx + C or, ye x = ò 2 xe x dx + C or, ye x = 2 xe x - 2 e x + C -x

or, y = 2 x - 2 + Ce Now y(0) = 1 gives 0–2 + C = 1 i.e; C = 3.  Therefore, the particular solution of (i) is ­given by y = 2x – 2 + 3e–x \ y(ln 2)

ln(y - 1) = 2 x2 + C, y = 1 ln(y - 1) = 2 x2 + C, y = -1



ln(y - 1) = 0.5 x2 + C, y = 1 ln(y - 1) = 0.5 x + C, y = -1

[EC GATE 2020] d2 u - 2 x2 u + sin x = 0 dx2

= 2 ln 2 - 2 + 3 e- ln 2 = 1.3862 - 2 + 3 ´

1  0.886 2

2. (b) Let y = emx be a trial solution of

2

is

(a) linear and homogeneous (b) non-linear and homogeneous (c) non-linear and non -homogeneous (d) linear and non-homogeneous  [CE GATE 2020]

EMEP.App-B_3PP.indd 628

x

Explanation

is

3x (a) y = c1 e3 x + c2 e-3 x (b) y = (c1 + c2 x)e 3x (c) y = (c1 + c2 x)e-3 x (d) y = c1 e  [EC GATE 2020]

4. The ODE

x

dx + 6x = 0 dt dx (0) = 10 , dt

1. (0.886) 6. (a)

1. Consider the initial value problem below. The value of y at x = ln2 (rounded off to 3 decimal places) is ______

(a) (b) (c) (d) 

-x

(b) y( x) = c1 xe 3 + c2 e 2

(a) -10 e2 t + 10 e3 t (b) 5 e2 t + 6 e3 t (c) 10 e2 t + 10 e3 t (d) -5 e2 t + 6 e3 t  [CE GATE 2020]

TOPIC: ORDINARY DIFFERENTIAL EQUATIONS



x 2

(c) y( x) = c1 e 3 + c2 e 2 (d) y( x) = c e 3 + c xe 2 1 2  [CE GATE 2020]

= [ x ]1 + 2 [ x ]2 + 3 [ x ]3



-

6. For the ordinary differential equation

4

3

Its solution is

-x

xy2 dxdy

1£ x