ENGINEERING MATHEMATICS (For ESE & GATE Exam) (CE, ME, Pl, CH, EC, EE, IN, CS, IT) ----------- Salient Features : 1-----------• 282 topics under 32 chapters in 8 units • 1446 questions from last 26 years of GATE & ESE exams with detailed solutions • • • 642 Solved Examples for comprehensive understanding Only book having complete theory on ESE & GATE Pattern Comprising conceptual questions marked with '*' for quick revision Office: F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016 •Phone: 011-26522064 Mobile: 8130909220, 9711853908 • E-mail: info@iesmasterpublications.com, info@iesmaster.org Web: iesmasterpublications.com, iesmaster.org IES MASTER PUBLICATION F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016 Phone : 011-26522064, Mobile : 8130909220, 9711853908 E-mail : info@iesmasterpublications.com, info@iesmaster.org Web : iesmasterpublications.com, iesmaster.org All rights reserved. Copyright © 2018, by IES MASTER Publications. No part of this booklet may be reproduced, or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior permission of IES MASTER, New Delhi. Violates are liable to be legally prosecuted. First Edition : 2017 Second Edition : 2018 Typeset at : IES Master Publication, New Delhi-110016 PREFACE We, at IES M ASTER, have immense pleasure in placing the second edition of "Engineering Mathematics" before the aspirants of GATE & ESE exams. Dear Students, as we all know that in 2016 UPSC included Engineering Mathematics as a part of syllabus of common paper for ESE exam as well as of a technical paper for EC/EE branch, while l=ngineering Mathematics already has 15% weightage in GATE exam. We have observed that currently available books cover neither all the topics nor all previously asked questions in GATE & ESE exams. Since most of the books focus on only some selected main topics, students have not been able to answer more than 60-65% of 1446 questions that have been asked in GATE & ESE exams so far. Hence to overcome this problem, we have tried our best by covering more than 282 topics under 32 chapters in 8 units. (One should not be in dilemma that 282 topics are more than sufficient. These are the minimum topics from where GATE & ESE have already asked questions). Since we have covered every previous year questions from last 26 years of each topic, students can easily decide, how much time to allocate on each chapter based on the number of questions asked in that particular exam. Again, we have included only those proofs that are necessary for concept building of topics and we have stressed on providing elaborate solution to all the questions. It is the only book in the market that includes complete theory exactly on ESE & GATE Pattern. After each topic there are sufficient number of solved examples for concept building &_ easy learning. The book includes such types of 642 examples. It also covers all the previously asked questions in which conceptual questions are marked with '*' sign so that students can save their time, while revising. Having incorporated my teaching experience of more than 14 years, I believe this book will enable the students to excel in Engineering Mathematics. My source of inspiration is Mr. Kanchan Thakur Sir (Ex-lES). He has continuously motivated me while writing this book. My special thanks to the entire IES MASTER Team for their continuous support in bringing out the book. I strongly believe that this book will help students in their journey to success. I invite suggestions from students, teachers & educators for further improvement in the book. Dr. Puneet Sharma (M.Sc., Ph.D.) IES Master Publications New Delhi CONTENTS UNIT 1 : LINEAR ALGEBRA 1.1 Algebra of Matrices ...................................................................................... 01 - 32 (i) Definition of Matrix .......................... :................................................................ 01 (ii) Types of Matrices ............................................................................................. 01 (iii) Product of Matrix by a Scalar (or Constant) ....................................................... 03 (iv) Addition and Subtraction of Matrices ................................................................ 03 (v) Multiplication of Matrices .................................................................................. 03 (vi) Minors of Matrix ............................................................................................... 05 (vii) Cofactors of a Matrix ........................................................................................ 05 (viii) Properties of Determinants .............................................................................. 05 (ix) Adjoint of Matrix ............................................................................................... 06 (x) Inverse of a Matrix ............................................................................................ 06 (xi) Conjugate of a Matrix ....................................................................................... 07 (xii) Conjugate Transpose of a Matrix ...................................................................... 07 (xiii) Special Types of Matrices ................................................................................ 07 Previous Years GATE & ESE Questions ...................................................... 13 1.2 Rank of Matrices .......................................................................................... 33 - 51 (i) Definition of Rank ............................................................................................. 33 (ii) Elementary Transformations or E-Operation ..................................................... 34 (iii) Equivalent Matrices ............................................� ............................................. 34 (iv) Properties of Rank ........................................................................................... 34 (v) Echelon Form/ Triangular Form ....................................................................... 35 (vi) Normal Form of a Matrix (Canonical Form) ....................................................... 37 (vii) Elementary Matrices ........................................................................................ 40 (viii) Computation of the Inverse of Matrix by Elementary Transformation .................. 41 (ix) Row Rank and Column Rank of a Matrix ........................................................... 44 Previous Years GATE & ESE Questions ...................................................... 46 (vi) 1.3 System of Equations ................................................................................... 52-92 (i) Linear Dependence and Independence of Vectors ........................................... 52 (ii) System of Linear Equations ............................................................................. 55 (iii) Methods of Solving Non-Homogenous System ................................................. 56 (iv) Methods of Solving Homogenous System ........................................................ 63 (v) LU Decomposition (Factorization) Method ....................................................... 67 (vi) Dolittle's Method .............................................................................................. 67 (vii) Crout's Method ................................................................................................ 70 Previous Years GATE & ESE Questions ...................................................... 73 1.4 Eigen Values and Eigen Vectors ............................................................... 93-138 (i) Definition ......................................................................................................... 93 (ii) Algebraic & Geometric Multiplicity .................................................................... 94 (iii) Properties of Eigen Values & Eigen Vectors .................................................... 94 (iv) Similar Matrices & Diagonalisation ................................................................ 101 (v) Cayley-Hamilton Theorem .............................................................................. 102 Previous Years GATE & ESE Questions .................................................... 106 1.5 Pivoting & Vector Space .......................................................................... 139-147 (i) Normalisation ................................................................................................. 139 (ii) ILL-Conditioned System ......................................... , ....................................... 139 (iii) Diagonally Dominant System .......................-.................................................. 139 (iv) Gauss Elimination Method with Pivoting ......................................................... 139 (v) Gauss-Jordan Method .................................................................................... 142 (vi) Vector Space ................................................................................................. 143 Previous Years GATE & ESE Questions .................................................... 145 UNIT 2 : CALCULUS 2.1 Functions and their Graphs .................................................................... 148-162 (i) Cartesian Product .......................................................................................... 148 (ii) Relations ........................................................................................................ 148 (iii) Types of Relations .......................................................................................... 148 (iv) Functions ....................................................................................................... 149 (v) Types of Functions ......................................................................................... 149 (vii) (vi) Some Basic Graphs ....................................................................................... 150 (vii) Graph Transformation ..................................................................................... 156 Previous Years GATE & ESE Questions .................................................... 157 2.2 Infinite Series ............................................................................................ 163-176 (i) Expansion of Functions .................................................................................. 163 (ii) Maclaurin's Theorem (Series) ........................................................................ 163 (iii) Taylor's Theorem (Series) .............................................................................. 163 (iv) Convergence and Divergence of Infinite Series .............................................. 166 (v) Methods to Find Convergence of Infinite Series .............................................. 167 (vi) Power Series ................................................................................................. 168 Previous Years GATE & ESE Questions .................................................... 170 2.3 Limits, Continuity and Differentiability ................................................... 177 - 220 (i) Function ......................................................................................................... 177 (ii) Limit of a Function .......................................................................................... 177 (iii) Theorem on Limits ......................................................................................... 178 (iv) Indeterminate Forms ...................................................................................... 180 (v) L-Hospital Rule .............................................................................................. 181 (vi) Fundamentals of Continuity ............................................................................ 184 (vii) Kinds of Discontinuities .................................................................................. 186 (viii) Properties of Continuous Functions ................................................................ 191 (ix) Saltus of a Function ........................................................................................ 192 (x) Function of Two Variables .............................................................................. 193 (xi) Limit of a Function of Two Variables ............................................................... 193 (xii) Continuity of Function of Two Variables .......................................................... 198 (xiii) Differentiability ............................................................................................... 199 Previous Years GATE & ESE Questions .................................................... 205 2.4 Mean Value Theorems ............................................................................. 221-229 (i) Rolle's Theorem ............................................................................................. 221 (ii) Lagrange's First Mean Value Theorem ........................................................... 222 (iii) Cauchy's Mean Value Theorem ...................................................................... 222 (iv) Balzano Theorem ........................................................................................... 223 (viii) (v) Intermediate Value Theorem ........................................................................... 223 (vi) Darboux's Theorem ....................................................................................... 223 Previous Years GATE & ESE Questions .................................................... 225 2.5 Maxima and Minima ................................................................................. 230 - 259 (i) Increasing-Decreasing Function ..................................................................... 230 (ii) Maxima and Minima of Function of Single Variable ......................................... 231 (iii) Maxima-Minima of Functions of Two Variables ............................................... 234 (iv) Sufficient Condition (Lagrange's Conditions) .................................................. 235 (v) Lagrange's Method of Undetermined Multipliers ............................................. 238 Previous Years GATE & ESE Questions .................................................... 243 2.6 Partial Differentiation ............................................................................... 260-283 (i) Parial Derivatives ........................................................................................... 260 (ii) Homogeneous Functions ............................................................................... 265 (iii) Euler's Theorem on Homogenous Function .................................................... 266 (iv) Total Differential Coefficient ............................................................................ 268 (v) Change of Variables ...................................................................................... 268 (vi) Jacobian ........................................................................................................ 272 (vii) Chain Rule of Jacobians ................................................................................. 275 (viii) Functional Dependence ................................................................................. 276 Previous Years GATE & ESE Questions .................................................... 279 2.7 Multiple Integral ........................................................................................ 284-346 (i) Concept of Integration .................................................................................... 284 (ii) Some Standard Formulae .............................................................................. 284 (iii) Some Important Integration and their Hints...................................................... 285 (iv) Definite integrals ............................................................................................ 286 (v) Fundamental Properties of Definite Integrals .................................................. 286 (vi) Fundamental Theorem of Integral Calculus ..................................................... 288 (vii) Definite integral as the Limit of a Sum ............................................................ 288 (viii) Double Integrals ............................................................................................. 121 (ix) Leibnitz Rule of Differentiation Under the Sign of Integration ........................... 291 (x) Improper Integral ............................................................................................ 292 (ix) (xi) Beta and Gamma Functions ........................................................................... 293 (xii) Properties of Beta and Gamma Functions ...................................................... 293 (xiii) Double Integrals ............................................................................................. 300 (xiv) Double Integrals in Polar Coordinates ............................................................ 301 (xv) Triple Integrals ................................................................................................ 307 (xvi) Change of Order of Integration ....................................................................... 309 (xvii) Use of Jacobian in Multiple Integrals ...........................................................-.... 31 2 (xviii) Multiple Integral using Change of Variables Concept ...................................... 31 4 (xix) Area and Volume in Different Coordinates Systems ....................................... 31 5 (xx) Arc Length of Curves (Rectification) ............................................................... 31 8 (xxi) Intrinsic Equation of a Curve ........................................................................... 31 8 (xxii) Volumes of Solids of Revolution ..................................................................... 321 (xxiii) Surfaces of Solids of Revolution ..................................................................... 322 Previous Years GATE & ESE Questions .................................................... 324 UNIT 3: VECTOR CALCULUS 3.1 Vector & their Basic Properties ............................................................... 347 - 364 (i) Vector ............................................................................................................ 347 (ii) Dot Product .................................................................................................... 348 (iii) Cross Product ................................................................................................ 350 (iv) Scalar Triple Product ...................................................................................... 352 (v) Vector Triple Product ...................................................................................... 352 (vi) Direction Cosines & Direction Ratios ............................................................. 353 (vii) Straight Lines in 3D ........................................................................................ 354 Previous Years GATE & ESE Questions .................................................... 361 3.2 Gradient, Divergence and Curl ................................................................ 365 - 392 (i) Partial Derivatives of Vectors ......................................................................... 365 (ii) Point Functions .............................................................................................. 365 (iii) Del Operator .................................................................................................. 366 (iv) The Laplacian Operator ..................................... � ........................................... 366 (x) (v) Gradient ......................................................................................................... 367 (vi) Directional Derivative (D.D.) ........................................................................... 367 (vii) Level Surface ................................................................................................. 369 (viii) Divergence .................................................................................................... 373 (ix) Curl (Rotation) ................................................................................................ 374 (x) Vector Identities ............................................................................................. 375 Previous Years GATE & ESE Questions .................................................... 379 3.3 Vector Integration ...................................................................................... 393 - 411 (i) Line Integral ................................................................................................... 393 (ii) Surface Integral .............................................................................................. 395 (iii) Volume Integral .............................................................................................. 396 (iv) Gauss' Divergence Theorem .......................................................................... 397 (v) Stoke's Theorem ............................................................................................ 399 (vi) Green's Theorem ........................................................................................... 400 Previous Years GATE & ESE Questions .................................................... 402 UNIT 4: COMPLEX ANALYSIS 4.1 Complex Number ..................................................................................... 41 2 - 428 (i) IOTA .............................................................................................................. 412 (ii) Complex Numbers as Ordered Pairs .............................................................. 412 (iii) Euler Notation ................................................................................................ 413 (iv) Geometrical Representation ........................................................................... 413 (v) Algebraic Properties ............................................................. : ........................ 414 (vi) Complex Conjugate Numbers ......................................................................... 415 (vii) Modulus and Argument ................................................................................... 416 (viii) Equation of a Circle in Complex Form ............................................................ 419 (ix) Complex No. as a Rotating Arrow ................................................................... 421 Previous Years GATE & ESE Questions .................................................... 422 4.2 Analytic Functions ................................................................................... 429 - 453 (i) Neighbourhood of Complex Number z0 ................................................................................... 429 (ii) Function of a Complex Variable ...................................................................... 429 (xi) (iii) Types of Complex Function ............................................................................ 430 (iv) Complex Mapping/Transformation .................................................................. 430 (v) Continuity of Complex Function ...................................................................... 432 (vi) Differentiability of Complex Function .............................................................. 432 (vii) Analytic Functions/Regular Function/Holomorphic Function ............................. 433 (viii) Cauchy-Riemann Equations ........................................................................... 434 (ix) Polar Form of Cauchy-Riemann Equations ..................................................... 436 (x) Orthogonal System ......................................................................................... 438 (xi) Construction of an Analytic Function ............................................................... 442 Previous Years GATE & ESE Questions .................................................... 444 4.3 Complex Integration ................................................................................ 454-486 (i) Contour .......................................................................................................... 454 (ii) Elementary Properties of Complex Integrals ................................................... 455 (iii) Cauchy Integral Theorem ................................................................................ 458 (iv) Cauchy's Integral Formula ......................................................... ..................... 459 (v) Taylor Series of Complex Function ................................................................. 464 (vi) Zeroes of an Analytic Function ........................................................................ 465 (vii) Singularities of an Analytic Function ................................................................ 466 (viii) Types of Isolated Singular Points .................................................................... 466 (ix) Residue ......................................................................................................... 467 (x) Cauchy-Residue Theorem .............................................................................. 468 (xi) Methods of Evaluating Residues .................................................................... 468 Previous Years GATE & ESE Questions .................................................... 47 1 UNIT 5 : DIFFERENTIAL EQUATIONS 5.1 Ordinary Differential Equations ............................................................... 487-5 1 8 (i) Definition of Differential Equations ................................................................. 487 (ii) Order and Degree of Ordinary Differential Equation ....................................... 487 (iii) Non-Linear Differential Equation .................................................................... 488 (iv) Solution of Differential Equations ........................................... . ........................ 488 (xii) (v) Formation of Differential Equation .................................................................. 489 (vi) Wronskian ...................................................................................................... 490 (vii) Linearly Dependent (LO) & Linearly Independent (LI) Solutions ....................... 490 (vi ii) Methods of Solving Differential Equation ........................................................ 49 1 (ix) D ifferential Equations of F irst Order and First Degree .................................... 491 (x) Exact Differential Equation ............................................................................. 493 (xi) Equation Reducible to Exact Form ................................................................. 494 Previous Years GATE & ESE Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 499 5.2 Linear Differential Equations of Higher Order with Constant Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ...... . . . . . . . . . . . . . .. . . . . . . . . . . .. . . . . . . . . 51 9 - 562 (i) Complementary Function (C.F.) ...................................................................... 519 (i i) Rules for Finding the Complementary Function ............................................... 519 (iii) Particular Integral (P.1.) .................................................................................... 522 (iv) Methods of Evaluating (P. I) ............................................................................. 522 (v) Cauchy's Homogeneous Linear D ifferential Equation ..................................... 530 (vi) Legender 's Homogeneous Linear Differential Equations ................................ 530 (vii) Simultaneous Linear Differential Equation ...................................................... 534 (vi i i) Variation of Parameters .............................................................. , .................. 534 (ix) Differential Equations of First Order and Higher Degree ................................. 535 (x) Clairaut's Equation ......................................................................................... 538 Previous Years GATE & ESE Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 540 5.3 Partial Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563 - 585 (i) Order and Degree of Partial Differential Equation ........................................... 563 (i i) Linear P.O. Equation ...................................................................................... 563 (iii) Lagrange's Method of Solving Linear ............................................................. 564 (iv) Homogenous Linear P.D.E. with Constant Coefficient ..................................... 567 (v) Non Homogeneous Linear P.D.E. with Constant Coefficient ........................... 567 (vi) Classification of 2nd Order Linear P.D.E. in Two Variables ............................. 575 (vii) One Dimensional Wave Equation ................................................................... 576 (vi i i) One Dimensional Heat Equation .................................................................... 577 (xiii) (ix) Two Dimensional Heat Equation .................................................................... 578 (x) Laplace Equation ........................................................................................... 578 (xi) Clairaut Ordinary Differential Equation ............................................................ 580 (xii) Clairaut Partial Differential Equation ............................................................... 580 Previous Years GATE & ESE Questions .................................................... 581 UNIT 6: NUMERICAL METHODS 6.1 Numerical Solutions of Linear Equations .............................................. 586 - 594 (i) Gauss-Jacobi Method .................................................................................... 586 (ii) Gauss-Seidel Method of Iteration ................................................................... 589 Previous Years GATE & ESE Questions .................................................... 594 6.2 Numerical Solution of Algebraic and Transcendental Equations ........ 595 - 618 (i) Graphical Method ........................................................................................... 595 (ii) Bisection Method ........................................................................................... 596 (iii) Regula-Falsi Method ...................................................................................... 600 (iv) The Secant Method ........................................................................................ 602 (v) Newton-Raphson Method ............................................................................... 603 Previous Years GATE & ESE Questions .................................................... 607 6.3 Interpolation ............................................................................................. 619 - 627 (i) Interpolation and Extrapolation ....................................................................... 6 1 9 (ii) Various Methods of Interpolation .................................................................... 619 (iii) Notation of Finite Difference Calculus ............................................................. 6 1 9 (iv) Newton-Gregory Forward Interpolation Formula for Equal Intervals .................. 620 (v) Newton-Gregory Backward Interpolation Formula for Equal Intervals ............... 621 (vi) Newton's Divided Difference Interpolation Formula for Unequal Intervals ......... 623 (vii) Lagrange's Interpolation Formula for Unequal Intervals ................................... 624 Previous Years GATE & ESE Questions .................................................... 626 6.4 Numerical Integration (Quadrature) ........................................................ 628 - 643 (i) General Quadrature Formula .......................................................................... 628 (xiv) (ii) The Trapezoidal Rule ....... . . . ...... . . ........ ........ ................. . . .... . . ...... . ..... . . ..... . . . . ... 628 (iii) Simpson's One-Third Rule . ...................................................... ...... .............. . . 629 (iv) Simpsons 3/8 1h Rule ......... . . . . ..... . . . ..... . . ........... . . ................ ....... . ....... ............. . . 630 Previous Years GATE & ESE Questions .................................................... 636 6.5 Numerical Solution of a Differential Equation ................ ........................ 644-660 (i) Picard's Method of Successive Approximation ..... . .. . . . . ...... . . . . ..... . . ... . . . . ........... 644 (ii) Euler's Method/Forward Euler Method/Explicit Euler Method . ...... . . . .... . . . . . ..... . . 647 (iii) Backward Euler's Method/lmplict Euler Method .................................... . . . . . . . ... 647 (iv) Runge-Kutta Methods ............ .......... . ......... .................... . . ............. ... . . .......... ... 650 Previous Years GATE & ESE Questions .................................................... 656 UNIT 7 : TRANSFORM THEORY 7.1 Laplace Transform ................................................................................... 661 -690 (i) Integral Transform . ................... . ....... . . . .................... . . ......... ...... ..... .... . . .. . . . ....... 661 (ii) Definiti on of Laplace Transform ... . ...... ............................ .................... . . .... . . .... 661 (iii) Properties of Laplace Transform ..................... . . . .. . . . ........ . . .... . .............. ..... . .. . . 661 (iv) Laplace Transform of Periodic Function ............. ... . . ........ . . ... . . . .... ...... ... . . ... . . . . . 462 (v) Types of Laplace Transform ........... . ........... . . ... . . . . . . . . . ............ . . . ........ . . .. . . .... . . . . . 662 (vi) Evaluation of Integrals with the help of Laplace Transform .................. . .... ..... . . . 664 (vii) Inverse Laplace Transform ............... ...................... . . ......... . .. . ...... . ......... ... . . . . . . 665 (viii) Properties of Inverse Laplace Transform ................... ..................... ................ 665 (ix) Types of Inverse Laplace Transform ...... ....................... . ................ . . ....... . ........ 665 (x) Convolution of Two Functions (Falting of Two Functions) . ............... . ................ 666 (xi) Unit Step Function (Heaviside's Unit Step Function) ... . . ................. . ...... . . ........ 666 (xii) Unit Impulse Function (Dirac Delta Function) .............. .................. . . . . .............. 666 (xiii) Existence Theorem for Laplace Transform ... ........ . . . . ........ .................. . .. .......... 668 (xiv) Initial Value Theorem ......... . . .... ............... . .................................. . . . . .. . . ..... . . ...... 668 (xv) Final Value Theorem ........ . . ..................................... . . . . . . ........ . ... . . . .............. . . . . 668 (xvi) Application of Laplace Transform in Differential Equation . ............... ...... .... . . . .. 669 Previous Years GATE & ESE Questions .................................................... 67 1 (xv) 7.2 Fourier Series ........................................................................................... 691 - 703 (i) Periodic Function ........................................................................................... 691 (ii) Fourier Series : (Main Definition) .................................................................... 692 (iii) Euler's Formula of Fourier Series .................................................................. 692 (iv) Dirichlet's Conditions ..................................................................................... 694 (v) Fourier Expansion of Even and Odd Function ................................................. 695 (vi) Half Range Fourier Series .............................................................................. 697 Previous Years GATE & ESE Questions ............... ..................................... 699 7.3 Fourier and Z- Transforms ........................................................................ 704 - 71 4 (i) Fourier Integral Theorem ................................................................................ 704 (ii) Fourier Transforms ......................................................................................... 705 (iii) Z-Transform (Definition) .................................................................................. 709 Previous Years GATE & ESE Questions ................................. ................... 71 3 UNIT 8 : PROBABILIT Y & STATISTICS 8.1 Probability ................................................... ......................... .................... 71 5 - 768 (i) Some Basic Concepts ................................................................................... 715 (ii) Representation of Various Events Based on Algebraic Operations ................. 718 (iii) Probability (Definition) .................................................................................... 719 (iv) Addition Theorem of Probability ...................................................................... 723 (v) Odds in Favour and Odds Against .................................................................. 726 (vi) Independent Events ........................................................................................ 727 (vii) Conditional Probability ................................................................................... 731 (vii i) Binomial Theorem on Probability .................................................................... 731 (ix) Total Probability Theorem ............................................................................... 735 (x) Baye's Theorem (Inverse Probability Theorem) .............................................. 736 Pr.evious Years GATE & ESE Questions .................................................... 740 8.2 Statistics - 1 (Probability Distributions) .................................................. 769 - 81 9 (i) Random Experiment ...................................................................................... 769 (ii) Discrete Random Variable ............................................................................. 769 (xvi) (iii) The Bernoulli Random Variable . .......... . ...................... . . . . ....... . ...... . . . . . .... . ........ 771 (iv) The Binomial Random Variable . . ................. . ......... ....... . ..... . . .......... . ...... . . . . .... . 771 (v) The Geometric Random Variable . .......................... . . . . . . . . ....... . ......... . ......... . . . .. 773 (vi) The Poisson Random Variable . . . . . ..... . . ................. . . .... . . ......... . . ........ . . .... . ... . . . . . 773 (vii) Continuous Random Variable .. . . . ..... ... . . . . ....................... . . . . . ..... ..... . . ..... . ... . . ..... 774 (viii) Uniform Random Variable ..... ..... . . . ..... ............... . ........ ... . . . ...... ............... . . ..... . . 776 (ix) Exponential Random Variable ... ... . . . ...... . . . . . . ... ..... . . ........ . . ............. ......... . . ....... 777 (x) Normal Random Variable/Normal Distribution/Gaussian Distribution . . . . .......... 777 (xi) Standard Normal Distribution ... . ... . . . . . . . .. . . ...... . . . .... . ....... . . . ...... . ........ ..... . . ......... 778 (xii) Skewness in Normal Curve ... . . . . . ........ ..... . ... . . .................. . . .......... . ..... . . ........ . .. 779 (xiii) Expectation of Random Variable .. . . . . ........... . ... . .. . . .......... . . . . . ......... . .............. . .. 779 (xiv) Variance ................ . . . . . . . ... . . . ....... . . . . . . ..... . . . . . . . . . . .... . . . . .......... . . . . . ................. . . ... . . . 781 (xv) Standard Deviation . ... . . ... . . ........ . . . . . . ........... . . .... . .... . . ........ ......... . ............ . ........ 781 (xvi) Covariance . ........... . ... . . . ... . . ....... . . . . . . ....... . . . . . . ... . . . . . . . . ......... . ... . . ....... . . ....... ........ 781 (xvii) Cumulative Distribution Function (CDF) for Discrete Random Variable . ...... ... . 784 (xviii) Mean .. .... . . . . . .... . . . . .. . . ..... ... .... . . . . . . . . . ... . . ... . . . . . ... . . . . . . . . . . .... .......... . . ... . . .... . . . .... . . ..... 784 (xix) Median .......... . . .. . . ... . ... .... . . .......... . . ........... . ... ...... . . . . . .... ... .... ........ . . .... . . ............ 784 (xx) Mode .. . . . . . ..... . . ... . ... . ..... ... . . . ......... . . . . .................. . . . . . .... . . .......... .......... . . . .. . . ....... 785 Previous Years GATE & ESE Questions .................................................... 786 8.3 Statistics - 11 (Correlation & Regression) ................................................ 820 - 834 (i) Curve Fitting ....... ... ......... . . . . . . .......... . . . . . . . ....... . . . . . . . .......... ... . . . ......... . . ..... . . . ....... 820 (ii) Methods of Least Suqares . .......... . . . . . ... . . ................. . . ....... . ...... ......... . ..... . ....... 820 (iii) Correlation ................... . .... . . ...... ...... ... . ... . . . . . . . . . . . . . ... ......... .... . . ........ . . ....... ..... . .. 824 (iv) Methods of Estimating Correlation . ...... . ................... .................... .... . ....... ...... 825 (v) Regression Analysis ............... . . . . ....... . . . ... . . . . . . . ... . . . . . . ............. . ......... . . .... . . . ....... 727 (vi) Line of Regression of y on x ..... . . .............. . . . . . .... . . . . . . ... . ................... . . . .... ......... 827 (vii) Line of Regression of x on y . . . . . ......... . ... . . . ... ...... . . ... ...... . . ................. . . .... . . ....... 828 (viii) Properties of Regression Coefficients . . . . . . . ........ ........ . . . . ......... .... . ............. . . .... 829 (ix) Angle between Two Lines of Regression ... . . . .. . . ......... . .... . ........... . ..... . . ............ 829 Previous Years GATE & ESE Questions .................................................... 833 ■ DEFINITION OF MATRIX ) A set of m n objects or numbers (real or complex) arranged in a rectangular array of m rows and n columns, i.e. 81 1 a21 a1 2 a 22 a1n a 2n is called a matrix of order m x n matrix. The number a 11 , a 1 2, ... , amn are called the elements of the matrix. The vertical lines are columns or column vectors, and the horizontal lines are called rows or row vectors of the matrix. TYPES OF MATRICES 1. Row Matrix: A matrix having one row and any number of columns is called a row matrix, or a row vector, e.g., 2. Column Matrix: A matrix having one column and any number of rows is called a column matrix or a column vector. (a 1 1 ' a 1 2, .. . . a 1 nl or [a 1 , a2, ... , an] is a row matrix of order n or matrix of order 1 x n. e.g. a1 1 a1 1 a 1 is a column matrix of order m or matrix of order m x 1. � 1 or at am1 3. am 0 = · 4. l� � � �1 Null Matrix or Zero Matrix: Any matrix in which all the elements are zero is called a zero matrix or Null M atrix i.e. Square Matrix: A matrix in which the number of rows is equal to the number of columns is called a square matrix i.e. A = (aii)m x n is a square matrix if and only if m = n. A matrix A = [::: 83 1 ::: 832 ::: ] 8 33 3x 3 is a square matrix of order 3. The elements a 11 , a22, a33 of the above square matrix are called its diagonal elements and the diagonal containing these elements is called the principal diagonal or leading diagonal or main diagonal. Trace of Matrix: The sum of the diagonal elements of a square matrix is called trace of the matrix. 2 Algebra of Matrices 5. Diagonal Matrix: A square matrix is called diagonal matrix if all its non-diagonal elements are zero i.e. in general a matrix A = (aii) n • n is called a diagonal matrix if a ii = 0 for i A = For example 6. � �l is a diagonal matrix of order 3 '. Scalar Matrix: If all the elements of a diagonal matrix of order n are equal, i.e. if a ii = k V i , then the matrix is called a scalar matrix, i.e. A = 7. o o [0� * j; [ 5 o o � � l � �1 , is a scalar matrix of order 3. Unit or Identity Matrix: A square matrix is called a unit matrix or identity matrix if all the diagonal elements are unity and non-diagonal elements are zero. e.g. 1 3x3 = [ � � 0 0 1 1 2x2 = [� �l are identity matrices of order 3 x 3 and 2x2 respectively. 8. [H �1 Upper Triangular Matrix: A square matrix A = [aiil is called an upper triangular matrix if a ii = 0 whenever i > j. Thus in an upper triangular matrix all the elements below the pri ncipal diagonal are zero. For example, A = 0 0 0 : 6 1 a nd B a [� � �] are 4 x 4 and 3 x 3 upper triangular matrices respectively. 9. Lower Triangular Matrix : A square matrix of A = [a ii ] is called a lower triangular matrix if a ii = 0 whenever i < j. Thus i n a lower triangular matrix all the elements above the principal diagonal are zero. For example, A J� � 1 � � l5 1 and b a [ ! �] are 4 x 4 and 2x2 lower triangular m atrlces respectively. 4 3 4J rl� : ; �] �i 1 0. Sub matrix : A matrix obtained from a given matrix, say A = (aii)m• n by deleting some rows or columns or both is ca lled a su b m atrix of A. For example if A a of A. 1 then the matrix [ ; � 7 2 3 !] 7 8 0 and [: are sub matrices 7 2 . 1 1 . Equal matrices : Two matrices are said to be equal if (i) They are of the same order. l l (ii) The elements in the corresponding positions are equal. Thus if A = [ � : B = [� : Then A = B I n general if A = ( a ii txn and B = ( b ii txn are matrices each of order m x n and a ii = bii for all i and j then A = B. Engineering Mathematics 3 PRODUCT OF MATRIX BY A SCALAR (OR CONSTANT) Let A = [a ii· ]mxn b e a matrix of order m x n and k is a constant, then their product is matrix kA = [ka ii· ] ,... element of A is multiplied by k. For example, if A = [ 2 -1 0 4 5 -3 ] . Then we have 4A = [ 8 -4 0 1 6 20 -1 2 mxn i .e. every ] ADDITION AND SUBTRACTION OF MATRICES Let A and B be two matrices of the same order, then their sum A + B is defined as the matrix each element of which is the sum of the corresponding elements of A and B. I n general if A = (a ii ) mxn and B = ( bii txn then their sum is defined by the matrix. C = A + B = (Cij)m x n where c ii = a ii + b ii and i = 1 , 2, . . . . m and j = 1 , 2, 3, . . . . . . . ,n If A - [ 4 2 5 1 ] B-[ 1 1 1 3 -6 ' 0 2 3 1 4 ] 4+1 2+0 5+2 A+ B - [ Then 1 1 + 3 1 3 + 1 -6 + 4 ]-[ 5 2 7 1 4 1 4 -2 ] Similarly, if A and B are two matrices of the same order, then their difference is defined by A-B = [ 4-1 2-0 5-2 1 1 - 3 1 3 - 1 -6 - 4 ]=[ 3 2 3 ] 8 1 2 -1 0 Properties of Matrix Addition (i) Matrix addition is comm utative : Let A = [a ij txn and B = [ bii tx n b e matrices of the same order m x n then A + B = B + A. (ii) Matrix addition is associative : Let A, B, C can be the matrices of the same order, Then ( A + B) + C = A + (B + C) . (iii) Cancellation law for matrix addition : Let A, B, C be the matrices of the same order, then A + B = A + C holds if and only if B = C. M U LTIPLICATION OF MATRICES The product AB of two matrices A and B is possi ble only when the num ber of columns in A is equal to the num ber of rows i n B. Such matrices are said to be conformable for multiplication. Let A = [a ii ] mxn b e a matrix of order m x n and B = [b ik lxp b e a matrix of order nxp, then the product AB is defined as a matrix C = [ c ik ] mxp of order m x p where ci k = L>ii bik = au b1k + a i2 b2 k + . . . . + ain bn k j =1 i .e. (i, k)th element of AB = sum of the products of the elements of i th row of A with the corresponding elements of kth column of B, i .e. IES MASTER Publication 4 Algebra of Matrices [ 1 -2 3 B � [ : ; ] . Show that AB , BA ] -4 2 5 ' Exam ple 1 If A = Solution 1 .2 - 2.4 + 3.2 1 . 3 - 2 . 5 + 3. 1 O -4 AB = [ -4 .2 2 .4 + 5.2 -4. 3 + 2 . 5 + 5. 1 ] = [ 1 0 3 ] x + 2 2 Si nce number of col umns in B is equal to number of rows in A, so BA is also defined. 2 3 -2 BA = [ 4 5 ] [ _� 2 2 1 Hence AB !] * BA . - 1 0 2 27 = [ - 1 6 2 27 ] -2 -2 1 1 �i - Thus matrix m ultiplication is not com m utative in general . Example 2 : Solution �i[: �i [to �i If A = [� ; � ] • B = [ ; 2 3 1 AB = [ � ; 2 3 1 4 = 4 1 1 Find AB. Will BA exist? 15 5 Now, since number of columns of B is 2 and number of rows in A are 3. So BA does not exist. Therefore if AB is defined, it is not at all necessary that the product BA is also defined. Properties of Matrix M u ltiplication 1. Multipl ication of matrices is not commutative i.e. AB 3. Matrix m ultipl ication is distributive with respect to addition i .e. A(B + C) = AB + AC 2. 4. ,;c BA Multiplication of matrices is associative, i.e. A(BC) = (AB)C where A, B C are any three matrices of order m x n , nxp, n x p respectively. Positive integral power of square matrix : The product of AA is defined only when A is square matrix of order n. We shall denote it as A2 . If m and n are any positive integers, then we have A m An = A m +n 5. Zero Divisor : AB = 0 does not necessarily imply that at least one of the matrices A and B m ust be zero m atrix i .e. the product of two matrices can be zero matrix whi le neither of them is a zero matrix. Such matrices are said to be zero divisor. For example; if A= [ � �] and 8 = [� 1 �l then AB = [� �] [ �1 �] =[ � �] i .e. neither the m atrix A nor m atrix B is a zero matrix but their product matrix AB is zero matrix. 6. Multiplication with Identity Matrix : If A be n x n matrix and I n is a unit matrix of order n , then Al n = l n A = A i.e. a matrix remains unaltered when it is multiplied by a unit matrix of same order. Engineering Mathematics 5 M I NORS OF MATRIX The determinant value of the square matrix obtained from the original matrix of any order by the omission of the rows and columns is called a minor of a matrix. For example If A = [ 1 2 3 4] 0 3 7 8 2 1 2 3 is a matrix of order 3 x 4 . Then minors of A are 1 7 3 x 3 4 0 3 7; 8 2 7 I� :1 1; ;1 & etc. COFACTORS OF A MATRIX The cofactors of the element a ii is defined as A ) i = (- 1 ) +j [Mij l ij where ! M iil is the determinant obtained by deleting the ith row and jlh column from given matrix. EXPANSION BY COFACTORS We have row expansion of a determinant of a general 3x3 matrix, that is IAI = a 1 1 M1 1 - a 1 2M1 2 + a1 3M1 3 = a 1 1 A 1 1 + a 1 2 A 12 + a 1 3 A 1 3 Since b y the definition of cofactors, we have A 1+1 11 = ( -1 } 12 = ( - 1 )1+2 M1 2 A M1 1 = = M1 1 ' -M1 2 • and PROPERTIES OF DETERMINANTS 1. 2. 3. 4. I A '! = I A I If we interchange any two rows or columns then sign of determinants change. If any 2 rows or any 2 columns in a determinant are identical (or proportional) then the value of the determinant is zero. Multiplying a determinant by K means multiplying the elements of only one row (or one column) by K. e.g. then 5. IAI = I: 21AI = 2 !I = -2 2 2 4 = 8 - 1 2 = -4 = 2(-2) 1 3 3 41 4 1 1 1 If elements of a row in a determinant can be expressed as the sum of two or more elements then the given determinant can be expressed as the sum of 2 or more determinants. e.g. I A I = c d a b a+c b+d +l = · l le f f e fi l le IES MASTER Publication 6 Algebra of Matrices 6. 7. If we apply the operations like Ri ➔ Ri + KRi or C i ➔ C i + KCi , the value of the determinant remains unchanged. Thus, the determinant is the sum of products of elements of any row and their corresponding cofactors. i.e., I A I = 8i1 A i1 + 8i2 Ai2 + 8 i3 A i3 where i represents rows. This expansion is called the expansion by cofactors. Similarly, we can, expand the determinant as the sum of the product of elements of any column and the corresponding cofactors of the elements of the same column. That is, 8. I A I = a 1iA1 i + a 2 i A 2i + c13 i A 3 i This is expansion has a very important property that sum of products of elements of the ith row and the corresponding cofactors of the elements of the i /h row is zero, that is, a I-1 A11 1 + a I-2 A 11- 2 + a I.3 A11 3 = O , in the case of the determinant of general 3x3 matrix where i i = 1 , i1 = 3 I AI = = : :: : :: 8 31 8 32 * i 1 . Thus, if we choose a1 1 81 2 81 3 :: = a 1 1 A 31 + a 1 2 A 32 + a 1 3 A 33 = 8 1 1 1 8 8 1 + 8 1 2 (-1) I 8 21 22 2 3 8 : 8 1 1 ( 8 1 28 23 33 - 8 1 3 8 22 ) - 8 1 2 ( 8 1 18 23 - 81 3 82 1 ) + 8 1 3 ( 81 18 22 - 8 1 2 8 2 1 ) The same result holds good for the column expansion. That is a 1i A 1i1 + a 2i A 2i1 + a 3 i A 3 i1 = 0 where j *j 1 ADJOINT OF MATRIX Let A = [ a ii ] nx n be a square matrix and Aii is the cofactor of aii" Then the transpose of the matrix B of cofactors of the elements of A is known as the adjoint of A and is denoted by Adj A i .e. Adj A = B' where B = Aii = cofactor matrix of A. [a1 1 a, , a, , l Thus, if A = 82 1 8 31 8 22 8 23 8 32 8 33 A1 1 A 1 2 A,, l[ A , , A 21 A ,, ] adj A = transpose of [ A 21 A 22 A 23 = A 1 2 A 22 A 32 A 31 A 32 A 33 A 13 A 23 A 33 Then where Aii denotes the cofactors of the corresponding element a ii " RESULTS ON ADJOINT OF A MATRIX If A = [a-- ] IJ n x n be a square matrix of order n, then A.(Adj A ) = I A I .I n = (Adj A ) .A, where In is the identity matrix of the same order as A. INVERSE OF A MATRIX Definition : Let A be a square matrix of order n. If there is a m atrix B such that A.B = B .A. = I , then B is called the inverse of the m atrix A and denoted by A-1 . Thus, if A is square matrix of order n, then A-1 is also a square matrix of order n. Engineering Mathematics 7 A.A-1 = A- 1 A = I So But we know the relation between the matrix and its adj A. i.e. , A. (adj A) = adj A.A = IAI I n Dividing it by IAI , we get A. ( a�I J = ( a�I } A = r lAf Hence, by definition of inverse we have adj A A-1 = Therefore A-1 exists only if IAI *0. CONJUGATE OF A MATRIX Let A be any mxn matrix having complex numbers as its elements. The matrix of order m x n_which is obtained from A by replacing each element of A by its conjugate is called the conjugate of A denoted by A . Thus if A = [a .. ] IJ mxn ' A= [aIJ. ] mxn where ai1· is the conjugate of a.11.• �NOTE ½ I f A real matr;x, then A � A Example 3 : If A = [ 1 + 2i i 3 2 - 3i ] , then A=[ -i ] 2 + 3i 1 - 2i 3 J CONJUGATE TRANSPOSE OF A MATRIX The conjugate of the transpose of a matrix A is called the conjugate transpose of A and denoted by A*. Thus if A = [ a ii ] , then Clearly the conjugate of the transpose is the same as the transpose of the conjugate. (A*)* = A I f A is real matrix, then A* = A ' Example 4 : [ 1 + 2i 2 - 3i 3 + 4il 1 + 2i 4 - 5i [ If A 4 - 5i 5 + 6i 6 - ?i then A' = 2 - 3i 5 + 6i 8 7 + 8i 7 3 + 4i 6 - ?i SPECIAL TYPES OF MATRICES I D E M POTE N T M ATRIX A square matrix A is said to be idempotent if A 2 = A . [ 2 -2 -4 ] Example 5 : A = -1 3 4 1 -2 -3 IES MASTER Publication 8 Algebra of Matrices I NVOLUTORY MATRIX A square matrix A is said to be i nvolutory matrix if A2 = I (unit matrix). -5 -8 Example 6 : A = [3 1 5 � ] 2 -1 N I LPOTENT MATRIX A square matrix A such that Ak = 0 where k is the least positive integer, is called the nilpotent matrix of index k. A = [ _�2 �:b ] Example 7 = a a A2 = [_:2 �: ] [_:2 �: ] = [� b b �] = 0 A is nilpotent matrix of index 2. O RT H OGO NAL MATRIX A square matrix A is called an orthogonal matrix if the product of matrix A with its transpose matrix A' is an identity matrix, i.e . AA' = I Example 8 : Soluti on Show that the matrix A = i-[ ; � A= i[ ; � �2] i-[; � �2 ] ⇒ A' = -2 2 -1 i[ ; � � -2 2] -:i i[� � �i �i 2 -2 -1 2 [ 2] ; -2 2 -1 2 -2 -1 AA = ' is orthogonal. -2 2 -1 = 0 0 9 =[ � � 0 0 1 =I Hence A is an orthogonal matrix. NOTE 1. The value of determinant of an orthogonal matrix is either 1 or -1. Proof : Let A be any orthogonal matrix i.e. A' A = I ⇒ ⇒ ⇒ I AI - I A I = l I Al 2 = 1 I A' A I = I I I ⇒ ⇒ I A'I - I A I = 1 (·: I I l = I ) (·: I A ' I = A ) I A I = ±1 2. I f A and B are orthogonal matrices then A B and B A are also orthogonal. Engineering Mathematics 9 SYM M ETRIC MATRIX 1 2 3 A square matrix A = [ a ii ] is said to be symmetric if A = A' i . e. a ii = a i i for all values of i and j e.g. If A = [ 2 5 4 ] , �n 3 4 7 1 2 3 then A' = [ 2 5 4 ] = A , 3 4 7 :i , SKEW-SYM M ETRIC MATRIX A square matrix A = [ a ii lx n is said to be skew-symmetric if A = -A' i.e. aii = -ai i V i & j , e.g. , if A = [�3 � -4 -6 0 [o -3 - 4 ] then A' = 3 0 -6 = -A . 4 6 0 1. If A n• n such that A is skew symmetric and n is odd then IAI = 0 (always). 2. Every square matrix can be uniquely expressed as the sum of the symmetrical and skew­ A symmetrical matrices i.e. A = ( A ;A ' ) + ( A ; ' ) = P + Q where P is symmetric and Q is skew symmetric. 3. Symmetric matrix is symmtrical about leading diagonal. Example 9 : Solution : Show that every diagonal element of a skew-symmetric matrix is necessarily zero. Let A be any skew-symmetric matrix i .e. aij = -a i i v i and j . . . (i) For diagonal elements of a matrix, we can put i = j in (i) i.e., ⇒ every element in the principal diagonal is necessarily zero. Example 10 : Write the following matrix as the sum of a sym metric and a skew-symmetric m atrix. 1 2 3 [4 5 6 ] 7 0 0 Solution If A is any square matrix, then symmetric and skew-symmetric matrices are respectively and A can be written as 1 2 ( A + A') and 1 2 ( A - A') A = _.!_(A + A' ) + _.!_ (A - A' ) 2 Let !] A = [: ; 7 0 0 2 ⇒ 4 A' = 21 5 7] 0 [ 3 6 0 IES MASTER Publication 10 Algebra of Matrices 10 ] A + A ' = [: ; ! + [ ; ; : ] = [ ! 1� 6 l 7 0 0 10 6 0 3 6 0 ⇒ . . . (i) and . . . (ii) Adding (i) and (ii), we get 2A = ( A + A') + ( A + A') = [ ! 10 -4 2 l 1� 6 + [ � � 6] 4 -6 0 10 6 0 A = symmetric matrix + skew - symmetric matrix. H E R M I T IAN MAT RIX A square matrix X is called Herm itian if A = A*. Thus, a square matrix A = [a ;il is Hermitian if a;i = aii V i and j 1 2 + i 3 + 2i Example 1 1 : The matrix A = [ 2 - i 3 3 - 2i 3i 1 For A' = [ 2 +i 3 + 2i 2 -3i l is Hermitian. -2 - i 3 - 2il 1 and A* = A' = [ 3 3i -3i -2 2 + i 3 + 2il -i 3 -3i 3 - 2i 3i -2 2 =A . S KEW- H E RM ITIAN MATRIX A square matrix A is said t o be skew-Hermitian if A * = -A. Thus, a square matrix A = [a ;il i s skew-Hermitian if a;i = -ai; v i and j i Example 1 2 : The matrix A = [ -2 + i -3 + 2i 2 + i 3 + 2i 3i -3i -3i 0 l -2 + i -3 + 2il For A ' = [ 2 �i 3 + 2i NOTE 3i -3i -3i 0 [ -i is skew-Hermitian. and A* = A' = -2 - i -3 - 2il i 2 + i 3 + 2il -i -3i 3i = - [ -2 + i 3i -3i 3 - 2i 3i 0 -3 + 2i -3i 0 2 = -A 1. I f A and B are two matrices such that these are conformable for add ition then (A + B) = A + B 2. I f A and B are two matrices conformable for multiplicat ion, then ( A B) = A.B 3. If A and B are any two matrices conformable for add ition, then ( A + B)* = A * +B * 4. I f A and B are any two matrices conformable for multipl ication, then (A B)* = B* A * 5. The d iagonal elements of a Herm itian matrix are necessari ly rea l . Engineering Mathematics 11 6. The diagonal elements of a skew-Hermitian matrix are either purely imaginary or zero. 7. Every square matrix can be uniquely expressed as the sum of a Hermitian and a skew-Hermitian matrix. A = A +A*) A-A*) ( -- + ( -- = P + Q where P & Q are Hermitian and skew-Hermitian 2 2 matrices respectively. 8. Every square matrix can be uniquely expressed as P + iQ where P and Q are Hermitian. A = A + A * ·( A - A * ( -2- ) + I ---) 2i = p + iQ 9. Every Hermitian matrix A can be uniquely expressed as P + iQ form, where P and Q are real symmetric and real skew-symmetric. Let A be a Hermitian matrix i .e. A* = A. Now we can write A = i(A + A ) + {i(A - A) ] P = i ( A + A ) and O = � (A - A ) where i Here P and Q are real symmetric and real skew-symmetric matrices respectively. U N ITARY MATR IX A square matrix A is said to be unitary matrix if A* . A = I = AA * ===�- If the matrix A is real then A = A, A* = A' , so we can write A' A = T = AA' . Thus a unitary matrix NOTE on a field of real numbers is also an orthogonal matrix. 1 1+ i 1 i s unitary. . v 3 1 - 1 -1 ] . Example 1 3 : Show that the matrix A = r;:; [ Soluti on If A = �[ � v J3 1 ·1 1+i 1 1-i ] _ ] then A = � [ 1 v 3 1 + i -1 A* = ( A )' = So that 1 1+i 1 3[ ] . ✓ 1 - 1 -1 Hence A is unitary matrix. Theorem The modulus of the determinant of a un i tary matrix is unity. Prouf : Let A i s unitary matrix i.e. A*A= I ⇒ I A* I I A I = 1 ⇒ l( A)I I A * Al = I I I ⇒ I AI = 1 ⇒ ⇒ l( A Y I I AI = 1 I Al 2 = 1 [taking determinant on both s i des] [ · : I A 'l = I A I] [ ·: I A I = I A I J IES MASTER Publication 1 2 Algebra of Matrices ⇒ ⇒ I AI = 1 The modulus of the determinant of unitary matrix is unity. Example 14 : Express Solution : [ -2 + 3i 1-i 2+i 3 4 - 5i 5 1 1+i -2 + 2i l as the sum of a Hermitian and a Skew-Hermitian matrix. If A is any square matrix, then we can write A = _! ( A + A *) + _! ( A - A *) 2 n 2 where ½(A + A *) is a Hermitian matrix and ½(A - A *) is a Skew-Hermitian matrix. A = Let 2 i + i 1-i 3 n 4 - 5i -2: 2il 1+i 3 A = Then i 4 + 5i 1-i ( A)' [ -2 = A* = f 2 1+i -3 i - i l -2� 2i l 1 3 1+i 4 + 5i 1-i 2-i 5 -2 - 2i l 1. 3 1. 2 --1 - + -I 2 2 2 i 4-i 3+ 1. 1. -I 4 3 - -1 :i 8 6 - i = 2+ 2 2 3 - i 6 + i -4 1 . 1 . 3 - - -1 3 + -I -2 2 2 2 -2 Now _! (A 2 + A*) = which is Hermitian matrix. 1 1. -1- .! i - + - I 2 2 2 -2 - i 1 [ 6 +i i 1 . 1. _! (A - A*) = -½ 2 - i -5i -10i 4 + i = 1 - - 1 2 + -I Again 2 2 2 l -1+ i -4 + i 4i 1 1. 1. 2i - - + - I -2 + - I 2 2 2 which is Skew-Hermitian matrix. 3i 1. 3 1. 1 1. 3i 2 - -1 - + -I -1 - .! i - + - I 2 2 2 2 2 2 . 1. 1 1 . 1. 4 -5i 2 + -I 3 - -1 + 1 - -1 2 +-I 2 2 2 2 1 1. 1. 1. 3 1. - - -1 3 + -I - - + - I -2 + - I 2i -2 2 2 2 2 2 2 -2 Thus A = where the first matrix on the R.H.S. is Hermitian and the second matrix is skew-Hermitian. Engineering Mathematics 13 PREVIOUS YEARS GATE & ESE QUESTIONS Questions marked with aesterisk (*) are Conceptual Questions 1. (a) ri: 13 (c) 2.* r -1 1 ; 13 �:1 (b) 13 5 1 : ,:1 �:1 r� 1 -1 1 3l � ] is (c) [GATE-1994; 1 Mark] (d) (AB)T = STAT 0 1 1 1 0 [GATE-1994] -4 ] is an i nverse of the m atrix 1 -5 6 2 9 4 (d) 13 5 22 8.* (b) False [GATE-1994 (Pl)] 8 (b) 2 4 9 16 9 16 25 12 (d) -8 (c) -1 2 [GATE-1994 (Pl)] [i ; i] 0 (b) [ �1 [ � � �� (d) ] 5 1 o�l is �1 :] -1 2 2 2 -1 1 -1 2 2 2 0 0 1 [GATE-1995-EE; 2 Marks] If A a nd B are square m atrices of size n x n, then which of the fol lowi ng statement i s not true. (a) det(AB) = det(A) det(B) (c) det(A + 8 ) = det (A) + det (B) (d) 9.* det(N) = 1 /det (A-1 ) [GATE-1995; 1 Mark] If m atrix A is m x n and B is n x p , the number of m u l t i p l i cation operati o n s a n d a d d i t i o n operations needed to calcul ate the m atrix AB, respectively, are (a) m n2 p, mpn (c) (b) mpn, (n - 1 ) (d) m n 2 p , (m + p )n mpn, mp(n - 1 ) [GATE-1995; 1 Mark] 1 2 and M = [ � �] then 4 5 3 8 13 (b) det(kA) = k0 det(A) 9 L x M is (c) 1 Given m atrix L = 2 8 12 - is? The value of the following determi nant 4 (a) 22 5 1 9 (b) [GATE-1994 (CS)] The inverse of the m atrix - 1 is 2 The i nverse of the m atrix S = [ 1 -� 0 (a) (b) A = A-1 1 6. 7. 1 0 [51 -4] -1 13 5 6 1 If A and B a re real sym m etric m atrices of order n then whi ch of the fol l owi n g is true. The m atrix [ 8 [GATE-1994 (Pl)] 13 13 (a) True 5. (a) 13 1 (d) [ } (c) AB = BA 4. 3 � 13 (a) AN = I 3. [ The inverse of the m atrix A = 10 .* Let A be an i nvertible m atrix and suppose that 1 the i nverse of A is [ � � ] , the matrix A is 7 IES MASTER Publication 14 11.* Algebra of Matrices (a) 2/ 7 1 [ / 7 1/ 7 ] 4 (b) [ : � ] (c) 1 -4 / 7 ] [ -2 / 7 1 / 7 (d) [; � ] [GATE-1995; 1 Mark] (a) row is the same as its second row (b) row is the same as second row of A (c) column is the same as the second column of A (d) row is a zero row. r� (a) skew symmetric 17.' The determinant of the m atrix (b) symmetric about the secondary diagonal (c) always symmetric (d) another general matrix. 12. [GATE-1996; 1 Mark] (a) 11 (b) -48 (c) 0 (d) -24 [GATE-1997 (ME)] If A 1 = A-1, where A is a real matrix, then A is (c) 18.* (b) symmetric (a) normal Hermitian (d) orthogonal [GATE-1996; 1 Mark] 13.* If A and B are non-zero square matrices, then AB = 0 implies (a) A and B are orthogonal (b) A and B are singular (c) B is singular In A and B are two matrices and if AB exists, then BA exists (a) only if A has as many row as B has columns (b) only if both A and B are square matrices (c) only if A and B are skew symmetric matrices (d) only if both A and B are symmetric. 19. Inverse of matrix [GATE-1996; 1 Mark] cas e - sin e [ sin e cas e ] and [� �] [; If a = b or 8 = nrc, n is an integer (d) (b) always (c) never (d) If a cos 8 ""' b sin e !] ! �] � [; i� [GATE-1997-CE; 1 Mark] [GATE-1996 (CS)] . is M [� commute under multiplication. (a) � !] [GATE-1997-CE; 1 Mark] (d) A is singular . 14.* The m a t r i ce s I � �,J [GATE-1997 (CS)] Th_e matrix B = N, where A is any matrix is matrices such that AB = I. Let C = A [11 011, 20.' If the determinant of matrix [ then the determinant of the matrix and (a) -26 (c) CD = I. Express the elements of D in terms of the elements of B. [! � -:6] is (b) 26 0 (d) 52 [GATE-1997-CE; 1 Mark] [GATE-1996 (ME)] 16.* Let A n x n be matrix of order n and 1 1 2 be the matrix obtaind by interchanging the first and second rows of In. Then Al 1 2 is such that its first. �6 l is 26, 5 21 . 2 If A = 3 0] then A-1 = ° = [� 0 1 (a) [l Il [� ;] 0 0 (b) 1/3 0 -1 / 3 0 115 0 0 112 (d) 0 1 /3 O 1 / 3 [ ] [ � -1 / 2 0 1/2 0 115 (c) E ngineering Mathematics -ti 28. (c) 23. unsymmetric skew symmetric (b) always symmetric (d) sometimes symmetric [GATE-1998-CE; 1 Mark] In m atrix al gebra AS = AT (A, S , T, are m atrices of appropri ate order) i m plies S = T only if (a) A is symmetric (c) A is non singular (b) A is singular (d) A is skew sym metric [GATE-1998-CE; 1 Mark] (c) 25. If A = 2 0 then k = (a) -5 (c) -3 26.* 5 1 -2 and adj (A) = - 1 1 -9 (a) (c) 4 kn 10 k 32. (c) n - 3n + 2 (b) n k (d) kin (b) n ! (d) (n + 1 )2 15 (d) n (n + 1 ) [GATE-2000 (CS)] (b) 0 (d) 20 [GATE-1999-CE; 1 Mark] (c) 33. 8 1 +28 7 2 2 0 2 0 9 0 6 is 1 [GATE-2000 (CS)] Determ in�nt of the fol lowing m atrix is (a) -76 [GATE-1999 (ME)] 27.* N u m ber of terms in the expansion of general determ inant of order n is n2 n2 (b) n - 1 7 [GATE-1999; 1 Mark] (a) [GATE-1999-CE; 1 Mark] The determ inant of the matrix (c) -2 -3 If A is any n x n m atrix and k is a scalar, l kAI = alA I , where a is (a) 0 (a) 4 1 (b) 3 (d) 5 (1 , 2) , (2 , 1 ) , (0 , 0) i - j fo r all i, j, 1 ::::: i, j ::::: n then the suni of the elem ents of the array V is [GATE-1998 (CS)] 3 (1 , 1 ) , (0 , 0) , (2 , 2) 30.* An n x n array V is defi ned as follows v[i , j] = 31. (d) a + b + c -2 - 1 (c) (0 , 0) , (- 1 , 1 ) , (1 , 2) 2 0 0 0 (b) a - b abc (0 , 1 ) , (0 , 2) , (0 , - 1 ) [GATE-2000-CE; 1 Mark] 1 C ab a + b = 0 , represents a (d) A-1 c-1 s-1 24.* I f t,. = [1 b ca ] then wh i ch of the fol lowing is a (a) 1 -1 29.* IF A , B , C are square matrices of the same order, (ABC)- 1 is equal to (a) c-1A -1 s-1 (b) c-1 s- 1 A-1 1 a be factor of t,. . 1 1 (a) (d) 22.* I f A is a real square m atrix .then AN is 2 1 Y X2 X parabol a passing through the points (b) [GATE-1998-EE; 2 Marks] (a) The e q u ati on 15 (b) -28 [I � 1� ] (d) +72 [GATE-2001-CE; 1 Mark] [P] = [! !l[Q] = [: !] The product [P][Q] T of the fol lowing two m atrices [P] and [Q] is (a) (c) [!� !�] [ 32 24 ] 56. 46 (b) (d) [ 46 56 ] 24 32 [ 32 56 ] 24 46 [GATE-2001 -CE; 1 Mark] IES MASTER Publication 16 Algebra of Matrices 34.* Consider the following statements S 1 : The sum of two singular matrices may be singular. 38. a 1 -a + a - 1 1 - a = 0 then the i nverse of X is (b) S 1 & S 2 are both false (b) (a) (a) S 1 & S2 are both true (c) S 1 is true and S2 is false (d) S 1 is false and S2 is true �1 (c) (GATE-2001 (CS)] The determinant of the matrix (a) (c) 36. 2 S2 : The sum of two non-singular m atrices may be non singular. Which of the following statements is true. 35. If matrix X = [ 100 r, 0 0 1 00 1 0 1 00 200 1 00 200 300 is (b) 200 (d) � �i (c) (a) 37.* Real matrices [A] 3 [E] 5 • 5 and [F ] 5 • 1 are sym metric. [B] 3 • 3 , [C] 3 • 5 , [D] 5 • 3 , are given. Matrices [BJ and [EJ • 1, Following statements are made with respect to these matrices. 1. 2. Matrix product (FJ T (CF [BJ [CJ (FJ i s a scalar. Matrix product (FJT (FJ (DJ is always symmetric. 41 .' (d) Both the statements are false [GATE-2004-CE; 1 Mark] l [GATE-2004] (d) 2 0 - 1c - 1A- 1 If R = [ � ! (c) (2 0 -1] 2 [GATE-2004 (CS)] (b) CDA (d) does not exist [GATE-2004 (CS)] �l h•n top row of R-' is (b) (5 -3 1 ] (d) (2 -1 1 /2] [GATE-2005-EE; 2 Marks] • 3 > , Y< 4 • 3 > a n d P<2 • 3>. The order of [P(XTY)-1 P T]l wil l be 42.* C o n s i d e r t h e m a t rices X<4 (a) (2 (C) (4 (a) Statement 1 is true but 2 is false (c) Both the statements are true 2 2 (a) [5 6 4J With reference to above statements, which of the following applies? (b) Statement 1 is false but 2 is true [a2 - a + 1 a a 1-a (c) ABC (d) 1 2 (GATE-2004-ME; 1 Mark] 1 [ -a + a - 1 a - 1 l 2 zero determi nant and ABCD = I then B-1 = (b) 6 (c) 8 X + I 40.* Let A, B, C , D be n x n matrices, each with non 6 0 (a) 4 -a - with each elements being either O or 1 is [GATE-2002-EE; 1 Mark] For which value of x will the matrix given below become singular? 1 -a [ a2 � a + 1 � ] x2 39.* The num ber of different n x n symmetric matrices (d) 300 1 [ 1 :/ �1] ] and 43. If A = x X 2) 3) (b) (3 (d) (3 x x 3) 4) [GATE-2005; 1 Mark] 17 Engineering Mathematics 0 0 _! 0 2 0 0 0 _! 0 4 0 0 _! 0 2 0 _! 0 4 (a) 0 0 _! 0 0 0 (b) 0 _! 2 0 0 _! 0 2 0 0 0 _! 2 1 4 0 0 0 0 _! 0 4 0 (d) 0 0 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 [GATE-2006; 2 Marks] 47.* Multipl ication of matrices E and F is G. Matrices E and G are Ol cos 8 -sin8 E = sin8 cos8 0 [ 0 1 0 (a) cos 0 cos 0 oos 0 -si n 8 sine -cose [ sin8 cos8 ; ] (b) [ 0 0 0 0 (c) cos 0 sin e -s [ ne cose � 0 [2005-EC; 1 Mark] - 1 44.* Let, A = [ � � ] and k' = (a + b) = 45. (a) 20 (b) (c) 19 60 (d) Then 20 48. 11 20 [2005-EE; 2 Marks] The determ i nant of the matrix given below is 1 1 0 -1 0 1 0 2 ;] (a) [ i : �1] (c) ; [ �5 � ] [GATE-2005] 46.* Match the items in col umns I and I I . Column II 1 . Determinant is not defined The determinant 4. Eigenvalues are always real 5. Eigenvalues are not defined b b 1 1 + b 1 equals to 2b 1 (a) 0 (b) 2b(b-1 ) (c) 2(1-b)(1 +2b) (d) 3b(1 +b) [GATE-2007 (Pl)] 50. 2. Determinant is Q. Non-square always one matrix R. Real symmetric 3. Determinant is zero S. Orthogonal matrix ..![-7 -2] (d) 3 -5 -1 1+b 49. (d) 2 P. Singular matrix i [; � ] [GATE-2007-CE; 2 Marks] (b) 0 Column I [GATE-2006-ME; 1 Mark] (b) 2 0 0 1 -2 0 1 1 ;] sin e -oos e o : (d) [ cose sine O 0 1 0 The i nverse of the 2 x 2 matrix [ �] is ; 1 3 (a) -1 (c) 1 o ol and G = 0 1 0 [ 0 0 1 What is the matrix F? 0 _! 0 4 1 0 0 rn :l (a) The product of matrices (PQ)-1 P is (b) 0-1 (a) p-1 (c) p-1 a-1 p (d) pQp-1 [GATE-2008-CE; 1 Mark] 51 . · 0 1 0 The inverse of matrix 1 0 0 is 0 0 1 IES MASTER Publication 1 8 Algebra of Matrices (a) 0 1 0 1 0 0 (b) 0 0 (c) 52. 0 1 0 0 0 1 1 0 0 (d) For a m atrix M = [! il 0 -1 -1 0 0 0 0 -1 0 0 -1 0 0 0 -1 -1 0 0 (GATE-2008 (Pl)] the transpose of the 4 5 (c) 3 5 (b) 3 5 (d) 4 5 A square m atrix B is skew-sym metric if (a) BT = -B (c) B-1 = B I X I = O and !YI = o (d) I X I 70 o and !YI 70 o [GATE-201 0 (IN)] 57.* [AJ is a square m atrix which is neither sym metric nor skew-sym metric and [AF is its transpose. The sum and difference of these m atrices a re defi ned as [SJ = [AJ + [AJ T and [DJ = [AJ - [AF, respectively. W hich of the following statements is true? (a) Both [SJ and [DJ are symmetric (b) Both [SJ and [DJ are skew-symmetric (b) B T = B (d) B-1 = B T 1 3 2 The value of the determ inant 4 1 1 is 2 1 3 (a) -28 (b) -24 (c) -32 (d) 36 ;J and matrix B = [: :] the transpose of product of these two m atrices i .e . , (AB)l is equal to (a) [ 28 1 9 ] 34 47 (b) [ 1 9 34 ] 47 28 (c) [ 48 33 ] 28 1 9 (d) [ 28 1 9 ] 48 33 (GATE-1 1 (Pl)] 59.* W hat is the minimum number of multiplications involved i n computi ng the m atrix product PQR? Matrix P has 4 rows and 2 col umns, matrix Q has 2 rows and 4 col umns, and matrix R has 4 rows and 1 column.____ [GATE-201 3-CE; 1 Mark] [GATE-2009 (Pl)] 55 . I [GAT E-201 1 ; 1 Mark] [GATE-2009-CE ; 1 Mark] 54. I (d) [SJ is symmetric and [D] is skew-symmetric 58. If a m atrix A = [ � [GATE-2009-ME-Set-4 ; 1 Mark] 53. (c) I XI = O and !YI 70 o (b) I X I 70 o and !YI = o (c) [SJ is skew-symmetric and [DJ is symmetric matrix is equal to the i nverse of the matrix , [MF = [MJ-1 . The value of x is given by (a) (a) i . [3 + 2i . Is Th e .inverse of th e matrix ] -i 3 - 2i (a) _!_ 3 + 2i -i ] 12 [ i 3 - 2i _!_ 3 - 2i -i ] (b) 1 2 [ i 3 + 2i (c) _!_ 3 + 2i -i 3 - 2i ] 14 [ i _!_ 3 - 2i -i (d) 14 [ i 3 + 2i ] 60.* Let A be an m x n matrix and B an n x m matrix. It is given that determ inant (Im +AB) = determinant (1 0+BA), where l k is the k x k identity matrix. Using the above property, the determinant of the matrix given below is l l1 1 1 2 1 2 1 [GATE-2010-CE; 2 Marks] (a) 2 56.* X and Y are non-zero square matrices of size n x n . I f XY = 0 mxn (c) 8 11' (b) 5 (d) 1 6 (2013-EC; 2 Marks] Engi neering Mathematics 61 .* Which one of the following does NOT equal 1 66. x2 X 2 1 z z 2 (c) 1 z(z + 1) z + 1 2 1 z+1 z +1 (c) _ y 2 0 y - z y 2 _ 22 (d) 1 z2 z (M +N)l = M + N 67. 2 2 2 x+y x +y 2 2 y+z y +z 1 z z [i [! !] 2 [201 4-EE-SET-2; 1 Mark] 68.* 2 Consider the matrix 0 0 0 0 0 0 0 0 0 1 0 J6 = G i v e n t h a t t h e determ i n a n t of t h e m atrix 3 6 � is -1 2, the determinant of the matrix 0 6 12 0 (c) (d) 96 24 69.* The maximum value of the determinant among all 2 x 2 real sym m etric matrices with trace 1 4 is 63.* Which of the following equations is a correct [2014-EC-Set-2] 70.* With reference to the conventional Cartesian (x, y) coordinate system, the vertices of a triangle have the following coordinates identify for arbitrary 3 x 3 real matrices P, Q and R? ( x 1 , y 1 ) = ( 1, 0) ; (x 2 , y2 ) = (2, 2) and ( x3 , Y) = (a) P(Q + R) = PQ + RP (4 , 3). The area of the triangle is equal to (P - Q)2 = P2 - 2PQ + Q2 (c) det (P + Q) = det P + det Q (P + Q)2 = P2 + PQ + QP + Q2 r! � HJ (a) 3 2 (c) 4 5 [GATE 201 4; 1 Mark] The determinant of matrix 3 0 �i is 1 2 [GATE-201 4-CE-SET-11 ; 2 Marks] 65. Given J = [ ; T K J K is 0 0 0 0 [2014-EC-Set-1 ; 2 Marks] [GATE 201 4-ME-Set-1 ; 1 Mark] 64. 0 0 0 0 Let P = 1 6 +a J 6 , where a is a non-negative real num ber. The value of a for which det(P) = 0 is is (b) -24 (d) 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 which is obtained by reversing the order of the colu m ns of the identity m atrix 1 6 . ] (a) -96 (b) (d) M N = N M The determi nant of m atrix A is 5 and the determinant of matrix B is 40. The determinant of matrix AB is [GATE-201 3 (CS)] 62. T [201 4-EC-SET-1 ; 1 Mark] 2 1 y+1 y +1 2 (b) (cM)T = c(M)l T x+1 x +1 1 y ( y + 1) y + 1 (b) 0 x-y x (MT )l = M 2 1 x (x + 1) x + 1 (a) For matrices of same dimension M , N and scalar c, which one of these properties DOES NOT ALWAYS hold? (a) 1 y y ? 19 ! ;] 2 6 and K = [ -1 then product [GATE-2014-CE-SET-I; 2 Marks] 3 4 5 (d) 2 (b) [GATE-14 (CE-Set 1 )) 71 . 2 If the matrix A is such that A = - 4 [1 9 5] , 7 then the determi nant of A is equal to ___ [GATE-201 4 (CS-Set 2)) 72.* I f any 4 two 7 D = [: : :8l c ol u m n s of d e t er m i n a n t are interchanged, which one of the statement is correct IES MASTER Publication 20 Algebra of Matrices (a) absolute value remai ns unchanged but s ign will change (b) both value & sign will change (c) absolute value wi ll change but s ign wi ll not change (d) both absolute value and sign wi ll rema i n un­ changed [GATE 201 5-ME-SET-1 ; 1 Mark] 73. 4 For a g i ven m atri x P = [ the i nverse of matrix P is (a) (c) _ 1 4 - 3i 24 [ -i 4 + 3i 2� [ : 4 = 3i ] 4 3i i i ] (b) (d) -i ] i = 4 - 3i + 3i i i 1 25 [ 4 + 3i 4 2� [ : 4 ✓ -1 , - 3i _ - ] i 3i 4 = 3i ] [GATE 201 5-ME-Set-3 ; 2 Marks] is (a) sec2 x (c) 1 (d) 0 [201 5-EC-SET-3; 1 Mark] 45 4 7 9 1 05 1 3 2 1 95 (i) Add the th i rd row to the second row (ii ) Subtract the th i rd column from the fi rst col­ umn. The determ i nant of the resultant matrix is ___ [GATE-201 5 (CS-Set 2)] 76. A real square matri x A i s called skew-symmetr i c if (a) AT = A (c) Al = - A 4 * 79.* The m atr i x A = [GATE-2016-EC-SET-1 ; 1 Mark] l� � � 7 3 : ] has det(A) = 1 00 *I PQ * I (a) PQ = I but QP (b) QP = I but (c) PQ = I and QP = I (d) PQ - QP = I [GATE 2017 CE Session-I] 81 . ] [ If A = : : and B = (a) (c) [ 3 8 28] 3 2 56 [! :]. (b) [4 3 2 7 34 50] (d) [ ABT i s equal to 40 3 42 8 ] [ 38 32] 28 56 [GATE 201 7 CE Session-II] 82.* The figure shows a shape ABC and i ts m i rror i mage A 1 B 1 C 1 across the hori zontal ax i s (X -ax i s). The coord i nate transformat i on m atr i x that m aps ABC to A 1 B 1 C 1 i s y B original A� C --------x o <Jc, A, [GATE 2016-ME-SET-1 ; 1 Mark] * for n�2. 80.* The matr i x P is the inverse of a m atri x Q. If I denotes the i dent i ty matri x, wh i ch one of the followi ng options i s correct? (d) Al = A + A-1 * �J[�l and trace (A) = 1 4 . The value of l a-b l i s ___ [GATE-2016-EC-Set-2; 2 Marks] (b) AT = A-1 77.* Let M = I, (where I denotes the i dent i ty matr i x) and M I and M 2 I and M 3 I . Then , for any 1 natural number k, M- equals: (b) M4k+2 (a) M 4k+ 1 (d) M 4k (c) M4k+3 = ��11 ] [ � The in i t i al condit i ons are x[0] = 1 , x [ 1 ] = 1 and x[n] = 0 for n < 0. The value of x[1 2] is . . . . . . . . . . [GATE-201 6; 2 Marks] (b) cos4x 75.* Perform the followi ng operat i ons i n the matri x 3 [x I 1 tan x 74.* For A = [ ] , the determ i n a nt of - tan x 1 AT A-1 78.* A sequence x[n] i s specif i ed as mirror image B, (a) (c) [� �] 1 [� �] (b) [ �1 � ] (d) [ � �1] [GATE-201 7 (IN)] Engineering Mathematics 83. For the g iven orthogonal matrix Q . 3/7 2/7 84. 6/7 Q = [ -6 / 7 3 / 7 2 / 7 l 2 / 7 6 / 7 -3 / 7 The inverse is (a) (b) (c) (d) 3/7 2/7 [ -6 / 7 3 / 7 2/7 [ 6/7 2/7 6 / 7 -3 / 7 l l 85. 6 / 7 -3 / 7 -2 / 7 ] -2 / 7 -6 / 7 3 / 7 3 / 7 -6 1 7 [2 / 7 6/7 3/7 2/7 2/7 6/7 -3 / 7 Which one of the following m atrices is singular? (a) (c) -3 / 7 -2 / 7 -6 / 7 21 [ � !] [� :] (b) [ � !] (d) [ : �] [GATE 201 8 (CE-Morning Session)] 1 2 If A = [ 0 4 ; then det(A-' ) is _ (correct to ] 0 0 two decimal places) [GATE 201 8 (ME-Afternoon Session)] r3 / 7 -6 / 7 -2 / 7 -2 / 7 -3 / 7 -6 / 7 ] -6 / 7 -2 / 7 3 / 7 [GATE 201 8 (CE-Morning Session)] .. IES MASTER Publication 22 Algebra of Matrices 1. (c) 23. (c) 45. (a) 67. (200) 2. (d ) 24. (b) 46. (a) 68. (1) 3. (See Sol. ) 25. (a) 47. (c) 69. (49) 4. (a) 26. (c) 48. (a) 70. 5. (d) 27. (b) 49. (a) 71. (0) 6. (b) 28. (b) 50. (b) 72. (a) 7. (d) 29. (b) 51. (a) 73. (a) 8. (c) 30. (a) 52. (a) 74. (c) 9. (c) 31. (a) 53. (a) 75. (0) 10. (b) 32. (b) 54. (b) 76. (c) 11. (d) 33. (a) 55. (b) 77. (c) 12. (d) 34. (a) 56. (See Sol. ) 78. (233) 13. (b) 35. (c) 57. (d ) 79. (3) 14. (a) 36. (a) 58. (d) 80. (c) 37. (a) 59. (16) 81. (a) 15. (See Sol. ) (a) 16. (c) 38. (b) 60. (b) 82. (d) 17. (b) 39. (c) 61. (a) 83. (c) 18. (a) 40. (b) 62. (a) 84. (c) 19. (a) 41. (b) 63. (d) 20. (a) 42. (a) 64. (88) 21. (a) 43. (a) 65. (23) 22. (b) 44. (a) 66. (d) 85. (0.25) Engineering Mathematics 23 SOLUTIONS Sol-5: (d) Sol-1 : (c) IA A = [-} � ] J J Ai = , } � I = -3 - 1 0 = -1 3 ;c 0 Now so. Hence, Cof (A) = [ � = [ -� =!] ⇒ A-1 exists =;T _ [-� =�J A-1 = �; = - � [� � 1 = Sol-2: (d) r-1� 11 =�] 1 Cof (A) = [ � � -1 -2 = --8 Sol-6: (b) Sol-7: (d) s = � I Si = 1 (1 - 0) + 1 (1 - 0) ⇒ = 2 ;c 0 =�] NON, ⇒ 1 =;] & I A I = 1 (0 - 1 + 0 + 1 (-1 - 0) = -2 ;c 0 so A -1 exist. 1 So [i !] 1 Determinant of matrix S (AB)T = BT AT by using Reversal Law. Sol-3: 1 4 9 = 4 9 16 9 1 6 25 = 1 (225 - 256) - 4(100 - 144) + 9(64 - 81 ) adj = (Cof A)T = [� -1 � -1 1 1 1 -1] 1 � 0 -2 adj (A) _ __ A_1 = I A I - (-2) [_ 1 -1 1 adj (S) = [ [SJ-' = �1 �: H 1 � i �:] a��r Sol-8: (c) (a) It is true exist. col (S) = [ �1 �: ] � adj(S) = [ ⇒ s-1 = [ 1- 2 �� =��] I ABI = IAI I B I l kAI = k0 J Ai where n is order of matrix (b) Sol-4: (a) Given A = [ � �]=[: :] (c) (let) adj(A) _ _1_ d -b I A I ad - be [ -c a ] ⇒ A_1 = So, 1 -1 _ [1 A 1 = -5 + 4 [-1 4 5] = 1 -5 -4] I A + Bl = J Ai + I B I It is false (d) which is true. Hence (c) is the correct answer. So l-9: (c) IES MASTER Publication 24 Algebra of Matrices Sol-10: (b) A- 1 (A-1 t1 = So, =: =�] C = 1 = [4 } ] 7 � [ ( 1) ⇒ A = [ � �] where a1 1 81 2 A = [8 21 8 22 ] & B = [ CD = ·: Other three options are not necessarily true. N = A-1 AAT = AA- 1 T AA = I A is an orthogonal matrix. Sol-13: (b) ·: AB = 0 ⇒ I AB I = 0 B are si ngular Sol-14: (a) ⇒ I AI . I B I = 0 Given cos e Osine A = [ sin e cos e ] and B = [� �] cos e -sine a ] AB = [ sin e cose [O ⇒ acose -bsin e = [ asin0 bcose ] ⇒ A& 1 �] D = ⇒ D = ( A [� ⇒ D = [ 1 ⇒ AD = [� 1 �] ⇒ D = [� 1 c-1 A -1 (Using Reversal Law) 1 ]B � [using (3), ·: AB = I ⇒ B = A - 1 1 ::: ] We know that "Post multiplication in A with the elementary matrix E ii is equivalent to interchanging ith and jlh columns of matrix A" For example - Consider some matrix ac se -asine a c�s 8 -bsin 8 ] = [ � ] bsme bcose a s i n 8 bcos 8 A = [! !] and also unit matrix I = [� �] Then E 12 = [� �] a sin 8 = b sin 8 (a - b)sin 8 = 0 So either a = b or 8 = nn question) i n I ] (i.e. A and B com mute when a = b (or) 8 = nn, n is an integer) CD = 1 Let E 1 2 = Elementary Matrix obtained by R1 � R2 AB = BA Given or in I . On comparison Sol-15: 1 (by equation (2)) Sol-16: (c) acos e -a sin e ] = [ bsine bcose ⇒ [ �Jr ⇒ D = [ ] �1 � [ ::: a O cos e -sine J ] BA = [ O b [ sin e cose Now . . . (3) b1 1 b12 ] b21 b22 Pre multiply both sides by A ⇒ ⇒ ⇒ ... (2) From equation ( 1 ) , we have Sol-11: (d) Sol-12: (d) AB = I and ·: A -1 1 = 141 _2 1 = 7 - 8 = - 1 ] [ I 7 �] A [� . . .( 1 ) Now, A.E 1 2 = [i. e. R 1 B R2 (as given i n [! !] [� �] [! !] = Here the first column of AE12 is same as the 2nd column of A. : . Option (c) is correct. I 25 Engineering Mathematics Sol-1 7: (b) Given A = 1 3 2 = - 0 5 -6 = - ("26) 2 7 8 6 -8 1 1 0 2 4 6 0 0 4 81 0 0 0 -1 = -26 Sol-2 1 : (a) ⇒ It is an upper triangular matrix. ⇒ The determinant of an upper triangular matrix is equal to product of its diagonal elements i. e. IAI = TI(a; ; ) = = 6 2 X -VI X a 1 1 ,a22 ,a33 .a44 4 X (-1) = -4 8 Sol-1 8 : (a) If A and B are two matrices of order m x n and p x q respectively then BA will exist only if q = m. Sol-1 9: (a) A = �i ⇒ A-1 exists 0 1 Now, Cofactor A = 0 [ 1 � o�l adj A = A -' = ��� Hl Determinant of matrix [A] is I A I = (5 x 3) + (2 x -6) = 3 ⇒ A- 1 *0 so [! � �] Sol-20: (a) 2 7 8 1 3 2 0 5 -6 = 26 = 0 5 -6 1 3 2 2 7 8 �1 [� � -6 0 15 Co-factor matrix of A = ,', i.e., Adj [A] = [Cof (A)Jl = 6 ⇒ [AJ-1 = adj [A ] IAI Sol -22: (b) 1 [ � � -o l -6 0 15 T (AAT)T = (N)TN = AAT ⇒ AAT is always symmetric Sol -23: (c) AS = AT Multiply both sides by A-1 A- 1 AS = A-1 AT ⇒ ⇒ S = T *O Sol-24: (b) t,,. = R1 B R2 1 3 2 = 2 7 8 0 5 -6 � [� � -6 0 1 5 But A -1 will exist only if I A I 2 7 8 = -1 3 2 0 5 -6 0 6 = =[! � �] R2 B R3 * exist. 6 = [� � ff [� Adjoint of the matrix [A] is the transposed matrix of co­ factors of [A] . [�1 0� 0 0 1 0 I AI = 0 0 1 = 1 1 0 0 A = t,,. = a be b ca c ab 0 a - b be - ca b ca c ab IES MASTER Publication 26 Algebra of Matrices Take out common (a - b ) from R1 0 1 -c /'J. = (a - b ) 1 b ca c ab (a -b) is a factor of /'J. . Sol-25: (a) 1 adj. A = [�� � Sol-32: (b) +l a = kn Sol-27: (b) Max no. of terms is the expansion of I Al n x n = I.!} Sol-28: (b) ·: ⇒ 2 1 y 1 1 X 2 1 -1 = 0 X 3x 2 + x 2 By using options we can see that it is passing through the points (0, 0), (-1 , 1 ), (1 , 2). y = Hence, option (b) is correct. Sol-29: (b) (ABC)-1 = c-1 B-1 A-1 (Using Reversal Law) Sol-30: (a) Given that v [i, j] = i - j n = 3 Let then v�, = ![ �1 Here the given matrix is skew symmetric . . Sum of all the elements of V = 0. Sol-31 : (a) 2 0 0 0 1 8 1 7 2 CJ. = =2 0 2 0 2 0 0 0 6 1 = 2 X 2 X 2 = 1 Sol-33: (a) = -50 + 24 - 2 = -28 Sol-34: (a) 32 24 2 3 4 9 PQT = [4 5 ] [ 8 2 ] = [ 56 46] Both result are obviously true Sol-35: (c) Determinant of upper triangular matrix, [U] or lower triangular matrix, [L] is the product of leading diagonal elements. :. matrix. 7 2 2 0 6 1 Determinant of given matrix = 1 x 1 x 1 x 1 = 1 Sol-36: (a) Sol-37: (a) 8(0 - 1 2) - X (0 - 24) = 0 X = 4 Given real matrices [Ah x 1 • [Bh x 3 ,[Ch x 5 ,[D]5 x 3 ,[E] 5 x 5 ,[F] 5 x 1 • Also, [B]T = [BJ & [EjT = [E] (1) ⇒ ⇒ (2) ⇒ Order [F]T [C]T (BJ [C] (F] ( 1 >< 5) (5 >< 3) (3 >< 3) (3 >< 5) (5 >< 1 ) -L...t -L...t -L...t -L...t Scalar matrix of order (1 x 1 ) i .e. ( 1 ) is true. Order of [F]T [F] [ D] (1 x 5) (5 x 1 ) (5 x 3) Not Possible to multiply so (2) is false. Sol-38: (b) Given ⇒ ��] 1 � �I 5 3 2 1 2 6 = 5(20 -30) - 3 ( 1 0-1 8)+2 (5 -6) 3 5 10 �1 1O � � k cof A = ( adjA ? = so [ ] 1 -3 7 So K = Cof of a23 in A = C23 = (-1 )2+3M 23 -2 = ( 1) = -5 - 1� 5 1 Sol-26: (c) lkAI = alAI kn lAI = alAI where n is order of square matrix A ⇒ = (2)(2) X2 - X + I = 0 x-1 (X2 - X + I) = x-1 O (Pre multiplying x-1 both sides) X - l + X-1 = 0 x- 1 = 1 - X 1 x- = 1 - X = ] [� �] [-a2 : a - 1 1 � a -1 1-a ] = [ 2 a -a +1 a Engineering Mathematics Sol-39: (c) Let A = a1 1 a1 2 ··· a1n a21 a22 ··· azn then to form symmetric matrix we have to put aii = ajl ¥ i at j i.e. A = 81n 8 1 1 8 12 8 2n 8 12 822 8 1 3 823 a 33 83 n Now it is symmetric matrix. ( ·: it is symmetrical about leading diagonal) So, no. of different elements in A = n + (n - 1 ) + (n - 2) + . . ... + 3 + 2 + 1 n( n + 1) 2 Each element is filled by two ways (either O or 1 ) = n(n+1 ) 2so total no. of different symmetric Matrices = 2 - S0I-40: (b) ABCD = ➔ A(ABCD)o-1 = K1 .I.D-1 1 BC = A - . 0-1 (BC) c - 1 = A -1 0- 1 c- 1 So Sol-41 : (b) B = K10-1c-1 ⇒ B-1 (A-1 D-1 C-1 )~1 B-1 = (c-1 r1 (0-1 r 1 ( A- 1 t1 (using Reversal Law) = CDA R = [ � ! �:] Determinant of matrix [R] is = 1 at O Co-factor matrix of [R] ⇒ adjoint of the matrix [R] is the transposed matrix of co­ factor of [R], i.e., adj (R) = [Cof (R)JT R-1 exist. �1] Adj [R) = [ �6 � ⇒ :. [RJ-' = a ) ��� = [ �6 � �1] Top row of R- 1 is [5 -3 1 ] .· Sol-42: (a) Giv en, X<4 • 3) , Y(4 •3l , P <2•3l None of the given matrices is square, hence (AB)~1 = B-1 A-1 does not hold for the above given matrices. Now, Order of XTY i.e. ⇒ x[3x4 ) y4x3 = (3 X 3) Order of (XTY)~1 = (3 X 3) P(XTYy-1 = (2 X 3) Order of P(XTY)~1 pT = (2 X 2) ⇒ Order [P(XTY)~1 pT] = (2 X 2) Order of Sol-43: (a) If A is an orthogonal matrix then ⇒ AN = I (MTy-1 = (1)~1 = I r� �1 But given matrix is not orthogonal so we have to calculate M T first. 0 0 4 0 MT = i.e., 0 2 0 0 Since it is diagonal matrix. 1 0 4 So, I RI = 1 (2 + 3) - 1 (6 - 2) 27 (MTy-1 = 0 0 0 Sol-44: (a) 0 0 0 0 0 0 0 1 2 1 2 2 -0. 1 ] A = [0 3 ' A-1 = [i :] 0 IES MASTER Publication 28 Algebra of Matrices AA-1 = I ⇒ [� -� ] [i ⇒ [� ⇒ 1 :] = 2a - (0 . 1) b ] = 3b [ �] � [� �] 2a - (0. 1)b = 0 1 b = 3 So, Hence, Sol-45: (a) A = For A- 1 = & 3b = 1 & 1 a = 60 I AI = Sol-49: (a) 1 b 1+b 1 b +b I AI = 1 2b 1 1 b 1 2 + 2b = 2 + 2b 1+ b 1 IAI 2 + 2 b 2b 1 0 2 1 3 0 1 -2 0 1 Expanding along R3 1 1 0 0 1 0 = (-1) -1 1 1 1 -2 0 1 [ 7 -2 (1x 7 - 5 x 2) -5 1 ] -7 2 = � [ 5 -1] 3 1 7 21 1 = a + b = -+ - = , 60 60 20 3 0 -1 [; �] 1 b = 2(1 + b) 1 1 + b 2b =0 as C1 & C3 are identical. = (-1)(-1)(0 - 1) = -1 Sol-50: (b) (PQ j-1 p = (Q-1 p-1 ) p ; { ·. - ( AB f 1 = s-1 A - 1 } = 0-1 (P-1 P ) Sol-46: (a) By using standard definitions. = Sol-47: (c) a-1 (I) = a-1 Sol-51 : (a) Given, cos 8 - s i n 0 E = [ si 8 c0s 8s o l � 0 1 Let = [� ! ;] [! � ;]- E,, � A and EF = G = I Hence g iven matrix A is an elementary matrix so Now, 1 IF = E- 1 IE I = 1 A -1 = A itself because i nverse matrix corresponding to the elementary matrix is the m atrix itself. ⇒ cos e sin 8 E-1 = [ - s n 8 cos 8 � 0 Sol-48: (a) Formula, A = A-1 = [: :] d 1 [ (ad - be) -e ;] �] Sol-52: (a) [M] = [MJ-1 = Given, [! ?al 1 [¾ -Ys] � - 4X -X 25 5 = [MJ-1 [M]T Eng ineering Mathematics [¾½ ¾ x l - [¾ -Ys] ¾ 25 (9 - 20X) -X Comparing a 1 1 both sides ⇒ ⇒ 25 = 1 9 - 20X X = Sol-53: (a) 90 - 20X = 25 -4 -16 20 = 5 Sol-54: (b) 3 2 1 1 Let, A = [ : :] ⇒ N = [: ] : ⇒ [S] is symmetric. (using property of symmetri c matrices) Also, 1 3 = 1(3 - 1) - 3 (12 - 2) + 2(4 - 2) = -24 ⇒ [D] = I AI = (3 + 2i) ( 3 - 2i) - (- i) (i) - (9 + 4) - ( 1) = 12 A-1 = -i 3 - 2i 1 ] [ 1 2 -(-)i 3 + 2i = -i 3 - �i 1 12 [ + I 3 + 2i] Sol-56: · i.e. Q = [ IYI = R o For two matrices A& B of same order, we know that (A+B)T = AT + gT [S] = [A] + [A] T [S]T = ([A] + [A]T )T = [A]T + [Al{ . _. ([AF )T = [Al} = [A] + [A]:r = [S] [S] is symmetri c matrix [D] = [A] - [AF ; � = [A] - [A] Case I : m QR = [a e b f C g ] r�i d h] r s _ ap + bq + er + ds - [ ep + fq + gr + hs ] . -. Number of multipli cation from [Qb,4 [Rl4 x 1 is equal � tO (2 X 4 X 1 : 8). Similarly, number of multiplication from [Pl 4 x2 [0Rlzx 1 is � equal to (4 x 2 x 1 = 8). Hen ce, total number of multipli cations [D]T = ([A] - [AF)T T : � = Sol-57: (d) Also, ] Given, [Pl4 x2 , [0J i x4, [RJ4x 1 Multipli cation can be done only when number of column of the first matrix is equal to the number of rows of the second matrix. Let, I X I = 0 and So, � Sol-59: (16) 1 x 1.1 v 1 = 0 Now, c 28 1 9 5 2 1 = 9 ] [4 3 ] [48 33 ] (AB)T = BT AT = [ : I XY I = 1 0 1 or b [D] is skew -symmetric. (using prop of skew symmetric matri ces) XY = 0 or = [ o -(b - c ) We will use Reversal Law i 3 + 2i A = [ -i 3 - 2i] ⇒ [: =: :=:] Sol-58: (d) Sol-55: (b) ⇒ [D] is skew - symmetric matrix a+a b+c [S] = [ ] c+b d+d It is standard defi nition of skew-symmetri c matrix. Let = -([A] - [A]T ) = -[D] Note : {Tri ck for exami nation) Ix = -¼I IAI = 4 2 ⇒ 29 { . _. ([AF )T = [Al} = 8 + 8 = 16 IES MASTER Publication 30 Algebra of Matrices Case II: No. of multiplications required in PQ is 1 = -1 3 0 4 x 2 x 4 = 32 No. of multiplication required in [PQ] [R] is 4 X 4 X 1 = 16 Total number of multiplications = 32 + 1 6 = 48. :. Min. number of multiplications = 1 6. (using Case I ) Sol-60: (b) = - [-4 - 84] = 88 J = [ 1 x2 + X X+1 1 y2 + y y + 1 = 1 z2 + z z + 1 = 1 x2 + X 1 y2 + y y + 1 y2 + y 1 z2 + z z 1 x2 y2 1 z2 1 X X X X 1 z z Sol-62: (a) X 2 * 2 Sol-66: (d) 1 X x2 1 y y2 1 z z Sol-63: (d) X - 1 2 = - 96 Matrix Multiplication is not commutative in general so (P+Q)2 = (P + Q).(P + Q) = P 2 + PQ + QP + Q2 Sol-64: (88) 0 1 2 3 R 4 ➔ R 4 - 3R 2 = 1 0 3 0 2 3 0 1 3 0 1 2 0 1 1 0 2 3 0 0 3 0 -6 -8 = [6 + 16 + 1 ] = [23] Sol-67: (200) Sol-68: (1) 1 3 0 A = 2 6 4 = - 12 -1 0 2 = 8 a-{�] H-�l In general MN * NM , i.e., Commutative Law not holds. 2 2 6 0 4 1 2 8 = 23 I A I -2 0 4 R3 ➔ R 3 - 2R 2 = [6 1 z2 + z 1 y + 1 y y +O 1 z z z 2 = - 1 y y +0 Let K' JK = [1 2 - 11 [; � 1 x2 + X 1 X !l Ul k = ; � Sol-61 : (a) Let us take option (a) i.e. 3 1 2 = - [1 (-1 2 + 8) - 3 (4 + 24)] Sol-65: (23) Without using property just find the determinant by usual method. 2 -6 -8 IAI = 5, I B I = 40 AB I I = I AI I B I = 5 x 40 = 200 0 0 I I = P 0 0 a. 1 - a6 = 0 0 1 0 0 a 0 0 0 0 0 0 a a O a 1 0 0 0 1 0 0 0 a 0 0 = 0 (giv en) 0 0 1 a 6 = 1, a E R a = ±1 ⇒ a is non-negative real number. a = 1 ⇒ Sol-69: (49) Let A be any symmetric matrix. A = [: : ] 3 0 1 2 Trace (A) = a + c = 14 and IAI = ,a c - b2 IAI will be maximum when b = 0 and a = C. a = C= 7 ⇒ I Al max = 72 = 49 Sol-75: (0) Sol-70: (a) Area 1 0 1 = �2 2 2 Let 1 1 1 IAI = 1 = -� Sol-71 : (0) 2 18 10 9 5] == -4 -36 -20 l [ 7 63 35 3x3 IAI = 0 Sol-72: (a) If any two rows or columns of a determi nant are interchanged then the value of determinant will be equal to negative of the given determinant but absolute value remains same. Sol-73: (a) 4 3i -i ] 4 - 3i ' p = [ � 1 p- = ? We know that, So , [: :r 4 3i -i ] [ �1 4 - 3i 1 = 1 ct [ ( ad - be) -c i = ✓ -1 �] x = ------2 [( 4 + 3i) ( 4 - 3i ) + i ] i 4 - 3i ] [ -I. 4 + 3i i 4 - 3i ] = 2 [ -i 2 4 + 3i [ 1 6 - 9i + i ] 1 = Option (a) is correct. Sol-74: (c) i 1 4 - 3i _ ] [ 24 - i 4 + 3i jAj ic O ⇒ A-1 exists j ATA-1 1 j T -1 = A j jA 1 1 = IAI A I I = 1 31 3 4 45 7 9 1 05 1 3 2 1 95 3 4 3 = 1 5 7 9 7 == 1 5 x 0 == 0 1 3 2 13 2 4 3 1 Since area can not be -ve answer is ⇒ Engineering Mathematics Now we know that the value of the determinant remains same after performi ng the operations Ri + KRi and C i + KCi ' So.the determinant of resultant matrix = 0. Sol-76: (c) By definition Sol-77: (c) N = -A. As we know that MM-1 = Let M -1 = M4k+n M.M4k+n = ⇒ ⇒ ⇒ ⇒ M . M4k. M n = [as per options given] M.(M4 )k_ M n = M n+ 1 _1 = Now, if n = 3, then M n+ 1 = M4 = I Hence, ⇒ ⇒ M 4 .I = I.I = I Identity is satisfied Hence, (c) [ •: M4 = I - given] :. I n == 3 I Method II: i.e., M4 = M . M3 = (given) But we know that MM-1 = Hence on comparison M -1 = M 3 so for K = 1 , only option (c) satisfies above condition. Sol-78: (233) Given, [ x [ [� �\] = [� �T [�] n�2 x[O) = 1 x[1 ] = 1 IES MASTER Publication Algebra of Matrices 32 x[n] = 0 for n< 0 Let n = Post multiply by Q PQ = Q-1 Q (we know 0-1 Q = I) 2 PQ = I Again premultiply by Q QP = QQ-1 x[2] = 2 Let QP = I (QQ-1 = I) n = 3 PQ = I and QP = I So, Sol-81: (a) x[3] = 3 A = Similarly x[4] = 5, x[5] = 8, ..... Hence we obtain general sequence as 1, 1 , 2 , 3, 5, 8, 1 3, 2 1 , 34, 55, 89, 144 , 2 33, ..... x[1 2] = So, ABT = 233 Above sequence is also known as F IBONAC CI sequence. ⇒ .. . (i) a + b = 7 Also, det(A) = 100 ⇒ 38 2 8 = [ 3 2 56 ] Let A (x 1 , y 1 ) lies in 1 st quadrant then A1 (x 1 , - y 1 ) lies in IV quadrant. a + 7 + b = 14 or ⇒ Hence (d) is the required mirror. Sol-83: (c) = 100 A is orthogonal so ⇒ 10 ab = 100 ab = 10 ... (ii) - 2, 5 b = 5, 2 la - bl = 3 Sol-80: (c) Given, p = 0-1 = [!;; 6/7 IA I = So -6 3 /; 2/7 !�;] �3 1 7 =2X6-3X4=0 Sol-85: (0.25) (a - 5) -2 (a - 5) = 0 Hence, Q-1 2 4 I AI = 1 3 6 1 5a - 2 a + 1 0 = 0 a = Q-1 = QT Sol-84: (c) a2 - 7a + 1 0 = 0 a Q QT = I Hence 10 a+- = 7 a 2 a + 1 0 = 7a 2 � ] = [ x 1 , - Y1 ] · 1 ·: We have [x 1 , Y 1 ] [� Now, From (i) & (ii) ⇒ ⇒ ⇒ ⇒ [B] Sol-82: (d) trace (A) = 14 a 0 3 7 2 5 3 0 0 2 4 0 0 0 b :] 3 + 35 8 + 20 = [ 18 + 14 48 + 8 ] [ ·: sequence is starting from x(0)] Sol-79: (3) [; =[! :] [; :][� :] 1 2 3 0 4 5 = 1(4 - 0) = 4 0 0 1 1 1 det (A- 1 ) = -- = - = det(A) I A I 4 = 0. 25 ■ DEFINITION OF RANK A number r is said to be rank of a matrix A if (i) There exist at least one square submatrix of order r which is non singular. (ii) Every square submatrix of order (r + 1 ) is singular. OR The rank of a matrix is the order of any highest order non-vanishing minor of the matrix. We shall denote the rank of a matrix A by the symbol p (A) . [2 1 -1] Example 1 Find the rank of the matrix A = 0 3 -2 2 4 -3 Solution It is a square matrix of order 3. Therefore, the minor of it of largest order is the determinant I A I of order 2 1 -1 0 3 -2 = 2 (-9 + 8) - 0 (-3 + 4) + 2(-2 + 3) = -2 - 0 + 0 = 0 3. That is 2 4 -3 I� :1 Here the minor of order 3 zero. Therefore 2 and the first minor of order 2 is So, by the definition, the rank of A = 2. p ( A) * 3, p( A) < 3 . Now we consider the minors of A of order =6-0 =6 ! :] *0 (i.e., non-vanishing minor of order 2x2 exist). Example 2 : Find the rank of the matrix. A = [ � : 3 6 9 12 Solution The Matrix A is of order 3x4. So it has 4 square sub matrices of largest order 3 . l l 2 3 4 1 3 4 1 2 4 1 2 3 [ A 1 = [ 2 4 6 ] , A 2 = [ 2 4 8 A3 = 2 6 8 , A 4 = 4 6 8 [ 6 9 1 2] 3 6 12 3 6 9 3 6 12 We compute the minors of A 1 , A2 , A3 and A4 . 1 2 3 1 2 3 IA1 I = 2 4 6 = 2 1 2 3 = 0, 3 6 9 3 6 9 34 Ran k of Matrices 1�1 = 2 4 8 = 2 1 2 4 = o . 3 6 12 3 6 12 1 2 4 1 2 4 Similarly, jA3 j = 0 and jA 4 j = 0 . S o the rank of A cannot be 3 . I t will be less than 3 , p ( A) < 3 . Now we compute the minors of order 2. It is clear that every minor of order 2 of the elements of square submatrices A 1 , A2 , A3 and A4 is zero. S p( A) * 2, p (A) < 2 therefore p (A) = 1 . ELEMENTARY TRANSFORMATIONS OR E-OPERATION To find the rank of a matrix by the direct application of the definition would be very tedious except in the simplest cases. We therefore investigate some methods of altering a matrix in such way that the rank remains the same but is simpler to determine. These methods are based on the following type of three operations which are called elementary transformation of a matrix. (a) The interchange of any two ith and j th rows (columns) denoted by R i B Ri or C i B Ci (b) The multiplication of all elements of i th row (column) by the some nonzero constant k denoted by R i ➔ k R i or C i ➔ kCi (c) The addition of the i th row (column) of an arbitrary multiple of j th row (column) denoted by or These operations are called elementary row (column) operations, as the case may be. EQUIVALENT MATRICES Two matrices A and B are said to be equivalent, A~ B, if the matrix B can be obtained from the matrix A by applying the elementary transformation to A, and A can be obtained from B by applying the elementary transformation to B. I � The equivalent matrices have the same rank, i .e., the rank of matrix remains unchanged by NOTE . _ E-transformation. PROPERTIES OF RANK 1. If Amm then p (A) :S: min(m,n) 2. If A is non singular matrix of order n then p( A) = n 3. p(A) = p(A 8 ) = p (A T ) = p(A -1 ) = p (AA T ) = p (AA 8 ) 4. If A is non singular then p(A) = p (A-1 ) 5. Rank of a null matrix is not defined i .e. termed as zero 6. Elementary transformation do not alter the rank of a matrix i.e . equivalent matrices have same rank. 7. Rank of a matrix in echelon form is equal to no. of non zero rows. 8. Rank of a matrix in a normal form is equal to rank of an identity matrix used in it. 9. If A is column Matrix and B is row matrix then p( AB) = 1 1 0. Every non-singular matrix is a product of elementary matrices. Engineering Mathematics 35 11. The rank of a product of two matrices cannot exceed the rank of either matrix i.e. p ( AB) :,; p ( A ) and p ( AB) :,; p ( B ) 12. The rank of matrix does not alter b y pre m ultiplication (or post m ultiplication) with any non singular m atrix. 1 3 . p(A + B) * p(A ) + p(B) Methods of Finding Ra nk 1. Echelon form M ethod (Triangular form) 2. Normal form Method (Canonical form) ECHE LON FORM / TRIANGULAR FORM A matrix A is said to be in echelon form if (a) All the zero rows occurs below non zero rows i.e. all the zero rows occurs at the bottom of matrix. (optional) (b) No. of zeros before the first non-zero element in a row is less than no. of such zeros in the next row. (Compulsory) ri ;J 1 4 Eg ; 1 0 0 4 Eg ; 0 0 Eg ; ��i rI 0 0 Eg ; 0 4 � 1 2 1 4 1 2 0 1 2 2 0 0 4 0 0 0 0 0 4 0 0 0 0 0 0 0 0 0 0 0 (,) Not .in Echelon form 1J � : ; (') Not in Echelon form 0 4 2 Eg ; Eg ; Flow Chart to Convert a Matrix in an Echelon Form 1. Make a 11 unity 2. Make all the elements of C 1 zero using E-row transformation only 3. M ake a22 unity 4. M ake all the elements of C2 zero that lies below a22 . 5. Repeat the whole process until we get a matrix in an echelon form. INOTE ( (a) 0 11 , 0 2 2 , a 33 ..... are cal led PIVOTA L elements for C1 , C2 , C3 .... (b) Making PIVOTA L elements unity is optional for Echelon form. respectively. (c) Making PIVOTAL elements unity is compulsory (for identity matrix). (d) Rank of a matrix in an Echelon form is equal to no. of non zero rows. (e) A ll identity Matrices are in Echelon Form. IES MASTER Publication 36 Rank of Matrices Example 3 , f l 1 _ � � educe the matrix in echelon form and hence find its rank. A _ � 1 R Solution A- r1 0 2 O 0 0 0 fl� :�l 1 Find the rank of a matrix. 3 2 4 ! 7 Solution j A�r� � � �] -r� � ; Let =; -�] 1 2 Example 4 , 1 1 2 � J0 1 l � � 1 0 �] R , �t ➔ R, - R1, R3 ➔ R 3 - 2R 1 , R , ➔ R , - 3R 1 2 1 ➔ R, - R, which is the form of Echelon matrix. So the rank of matrix is the number of non-zero rows i.e. p ( A ) = 3 6 Example 5 : 3 81 4 2 6 -1 Find the rank of the following matrix. A= · 10 3 9 7 r 16 4 12 15 6 Solution 1 A = 4 3 81 2 6 -1 _ 10 3 9 7 16 4 12 15 81 6 1 3 4 2 6 -1 r 10 3 9 0 0 0 � 81 6 3 4 2 6 -1 r 10 3 9 10 3 9 6 R4 ➔ R4 - R 3 - R4 ➔ R4 - R1 ; 1 3 81 4 2 6 -1 6 -1 r� 2 0 0 0 R 3 ➔ R 3 - R1 -[� -[� Bl 1 3 2 6 -1 0 0 0 0 0 0 l [1 R 3 ➔ R 3 - R2 ~ 1/ 6 1/2 4/3 4/3 4 -19 / 3 0 0 0 0 0 0 Eng ineering Mathematics 1/6 1/2 4/3 6 -1 4 2 0 0 0 0 0 0 0 0 37 1 R 1 ➔ -R 1 6 R 2 ➔ R 2 - 4R 1 which is the form of Echelon matrix so the rank of matrix = number of non-zero rows i.e. p(A) = 2 Example 6 : Soluti o n [ Find the rank of the following matrix. A = ; A = [; : ! -3 -6 -9 l -[� � ! !] -3 -6 -9 0 0 which is the form of Echelon matrix so the rank of matrix A is one i.e. p ( A ) = 1 . NORMAL FORM OF A MATRIX (CANON ICAL FORM) B y performing elementary transformation, any non-zero matrix A can be reduced to one o f the following four forms, called the normal form of A : [ Ir O · J 0 0 ' Eg ; where Ir is the identity matrix of order r and O is zero matrix. The number r so obtained is called the rank of A and we write p ( A ) = r . Flow Chart to Convert a Matrix in a Normal Form Let A be any (m x n) matrix. 1. Make a 1 1 unity. 2. Make all the elements of C 1 zero using E-row operations only. 3. Make all the elements R1 zero using E-column operations only. 4. Make a22 unity. 5. Repeat step (2) and (3). 6. Repeat the whole process until we get a matrix in a normal form. IES MASTER Publication 38 - Ran k of Matrices 1 1 31 2 Example 7 : Find rank of matrix A = 2 1 3 -1 1 2 2 0 1 4 1 Solution A - ri �1 -; 1; 1 1 by reduci ng it to normal form. R 2 ➔ R 2 - 4R 1 , R 3 ➔ R3 - 3R 1 , R 4 ➔ R 4 - R 1 2 -1 3 -7 6 - 1 1 -7 4 -7 1 A~ A~ 1 0 0 0 -7 6 0 -7 4 0 0 1 1 0 0 0 -1 6 -2 -[ - 2 1 1 0 0 1 6 3 4 � -1 1 1 0 0 A~ C 2 ➔ -C3 -2 0 0 1 0 0 7 1 0 0 1 0 0 -1 1 26 -13 0 0 0 0 - 14 26 0 0 7 -13 1 0 0 1 0 0 0 1 0 R 3 ➔ R 3 - 3R 2 and R 4 ➔ R 4 + R2 -2 6 A~ 0 0 - 14 A~ C 2 ➔ C2 + C 3 -7 0 -3 4 0 A 0 C 2 ➔ C 2 - 2C1 , C3 ➔ C3 + C1 , C4 ➔ C4 - 3C1 1 0 0 26 0 0 - 1 / 2 -1 3 A 0 0 0 1 0 0 C 3 ➔ C 3 - 6C 2 and C 4 + 1 1C2 1 A ~ [; �1 [I' - 0 1 26 [� 0 0 1 C3 ➔ - C3 14 1 R4 ➔ R4 + - R3 2 C 4 ➔ C4 - 26C 3 0 0 0 1 0 ] ⇒ p ( A) = 3 ⇒A~ 0 � . 0 0 1 0 0 0 0 0 Engineering Mathematics 2 0 -1 Example 8 : Express the following matrix into normal form and find its rank : [ ; 1 2] 4 -2 3 2 5 Solution : A= [} 2 -[i -[i -[i - [� -[i -[i 39 2 0 1 1 �] 43 2 2 0 -1 -2 1 ] R 2 ➔ R2 - 3R 1 , R 3 ➔ R 3 + 2R 1 � 7 2 0 0 -2 1 7 2 �]c, ➔ C, - 2C, , c, ➔ C, + c, 0 0 1 -2 2 7 0 1 �] c, tt c, � ]R, 0 0 0 1 1 -7 ➔ R , - 2R , & C3 ➔ C3 + 2C2 , C4 ➔ C4 0 0 0 R ➔ ] , 0 1 -7 I 1 1 � - 5C2 it, 0 0: 7 1 0 : �]c, ➔ c , + c, 11 0 1: 0 ~ (1 3 ; O] which is the form of normal matrix, so the rank of a matrix is the order of identity matrix i.e p ( A) = 3 Example 9 : Solution Find the rank of the following matrix by reducing it to normal form. [ � Let A = 1 2 2 2 5 3 3 8 4 2 7 -[i 2 1 1 2 -1 2 -2 3 -3 �] 2 2 0 5 3 1 8 4 2 7 1 3 �J R, ➔ R , - 2R,, R, ➔ R, - 3R,, R , ➔ R , - 2R, IES MASTER Publication 40 Ran k of Matrices - li - li - li - li 0 0 - 1 �]c, ➔ C , - 2C , , c, ➔ c, - 2C , 2 -2 3 -3 0 0 0 -1 i]c, ➔ c, + C , , c, ➔ c, + C , 0 -2 0 -3 0 0 0 -1 0 0 0 0 0 0 i ] R, ➔ R, - 2R, , R , ➔ R , - 3R , -1 0 i] C, 0 0 0 - ri - [� �] 0 0 0 1 0 0 0 0 0 i] c , tt C , +C , which is the form of normal matrix, so the rank of a matrix is the order of identity matrix 1 2 i .e. p ( A) = 2 . ELEMENTARY MATRICES A matrix obtained from the unit matrix by applying to it any of the three elementary transformations is called an elementary matrix. We shall use the following notations about elementary matrices : (i) E ii denotes the matrix obtained by interchanging the i th row with the jth row of a unit matrix. It may be easily seen that the matrix obtained by interchanging i th and j th columns is the same. (ii) E i (k) denotes the matrix obtained by multiplying the elements of the i th row by a nonzero number k. This is the same as obtained by multiplying the i th column by k. (iii) Ei/ k) is a matrix obtained by adding to the elements of the i th row, k times the elements of the j lh row, where k *o. 0v) E1i ( k) denotes the transpose of Ei/ k) and can also be obtained by adding to the elements of the i th column, k times the corresponding elements of the j th column. Property of Elementary M atrices Every elementary row (column) transformation of a matrix can be brought about by pre-multiplying (post-multiplying) with the corresponding elementary matrix. Engineering Mathematics Example 1 0 = Let A 41 a1 1 a1 2 = a2 1 a 22 be any 3 x 2 matrix. [ a 3 1 a 32 ] a 31 a 32 By performing transformation R1 3 (R1 B R 3 ) on A , we have B = a2 1 a 22 and by performing operation ] [ a 1 1 a1 2 C 1 B C 2 on Solution A � �1 �1 , we have D = ::: ::: then show that E 1 3A = B and [ a 32 a 31 ] . E 1 2 = D. � 1 [: : A Then E1 3 = � � and E 1 3A = � � Take unit matrix 1, � � [ [ 0 1 [1 0 0 a 1 0 0 31 0 0 1 Take matrix 1 2 = � ] . Then E1 2 = [ ] and 1 1 O [ Thus we see that transformation of A a 1 1 a1 2 l [ [0 1 ] = D AE 1 2 = a 2 1 a 22 1 O a 31 a 32 by R 1 B R3 is the same matrix which is obtained by pre­ multiplication of A by E 1 3 and transformation of A by C 1 B C 2 is the matrix which is obtained by post­ multiplication of A by the corresponding elementary matrix E 1 2 . Row Equivalent Matrix A matrix B is said to be row equivalent to A , if it is obtained by performing elementary row operations on A. Colu mn Equ ivalent Matrix A matrix B is said to be column equivalent to A , if it is obtained by performing elementary column operations on A. · · · r� � � � 1 · · �r� � �1A 31 COMPU1ATION OF THE INVERSE OF MATRIX BY ELEMENTARY TRANSFORMATION Method Will be cleared in following examples Example 1 1 = Find the inverse of the matnx Solution We write A 2 2 2 3 2 3 3 3 by using E-transformat1 ons. A = 14A 0 1 2 2 1 1 2 3 = [2 2 2 31 2 3 3 3 Performi ng R1 B R 2 , we get 1 1 2 0 1 2 2 [ 2 2 2 3 \ = 2 3 3 3 = 0 0 1 0 0 0 0 1 r� � � �1A 0 0 1 0 0 0 0 1 IES MASTER Publication · 42 Rank of Matrices 1 1 2 3 0 1 2 2 0 0 -2 -3 [ 0 1 -1 -31 [ ' 1 1 0 0 0 1 2 2 0 0 -2 -3 0 0 -3 -5 1 0 0 1 0 1 0 -1 1 [ 0 0 1 3/2 0 0 0 0 1/2 1 [� � � �:1 · I : Performing R 4 ➔ -2R4 , we get r 0 1 - 1/20 3 3 -3 3 2 = 4 3 -3 4 -5 0 0 1 0 4 3 -2 -2 1 � 1A r-2 } [�0 �0 �0 �1 -3 3 -3 2 B = A-1 = r ⇒ I = BA {let) 3 -4 4 -2 -3 4 -5 3 2 -2 3 -21 -1 -3 3 1 -1 � Example 12 : Find the inverse of the matrix A = ; by using E-transformations. -5 2 -3 r -1 0 11 1 Eng i neering Mathematics Solution : 43 We write A = I 4 A 1 ⇒ r � �1 � � : � C 1 -1 1 0 Applying R 1 ➔ -R1 , we get r1 -1 0 0 0 3 1 -3 1 -1 0 2 -5 2 -3 -1 1 0 1 Applying R2 ➔ R 2 - R1 , 1 0 0 0 1 0 0 1 0 A 0 . 0 0 1 R 3 ➔ R 3 - 2R1 , R 4 ➔ R 4 ;+- R1 , we get -1 0 0 0 1 3 -3 0 -2 -2 -1 =� = 1 r0 -1 1 8 -5 0 4 -3 2 1 = r 0 0 1 1 2 0 1 0 -1 0 0 1 A 1 Applying R2 ➔ _ _!_ R2 , we get 2 � � 8 0 -1 1 1;2 -5 = 1�; 1;2 � � r2 -1 1 0 01 A 1 0 0 1 4 -3 2 Applying R 1 ➔ R1 - 3R2 , R 3 ➔ R 3 + 1 1R2 , R 4 ➔ R4 r0 0 0 0 1 -1 0 0 -3 1 0 0 1 2 -1 2 -1 2 0 1 2 - = 1 7 2 Applying R3 tt R 4 , we get 1 0 0 0 -1 0 0 1 0 0 -3 Applying R2 ➔ R 2 2 1 2 1 2 7 2 - 2 2 0 = 1 2 - 4R2 , we get 3 0 0 2 1 0 0 2 A 11 1 0 2 2 3 2 1 2 0 1 0 0 0 0 A 2 0 1 11 2 1 0 + R 3 , R4 ➔ R4 + 3R3 , we get 1 0 0 0 1 0 0 0 1 0 0 0 1 2 1 2 0 -1 2 = -1 2 1 2 1 1 2 3 0 0 2 3 0 A. 2 2 0 1 - 1 3 2 IES MASTER Publication 44 Rank of Matrices Applying R2 ➔ R2 - R4 , R1 ➔ R1 + R4 , we get, 1 0 0 [ 0 ] 0 1 0 0 0 0 1 0 0 0 0 1/2 = Applying R4 ➔ 2R4 , we get 0 2 1 3 1 1 -1 -2 1 2 0 1 A 1 1 1 2 2 [� �] [ t]A B = A-1 = [o 0 1 0 0 0 0 1 0 2 1 1 1 -1 1 2 0 -1 1 2 I = BA 2 1 1 -1 2 0 1 2 ROW RANK AND COLUMN RANK OF A MATRIX If A = [aiil m x n be any m x n matrix, then the number of maximum linearly independent rows (or row vectors) of matrix A is called its row rank. And the number of maximum linearly independent columns (or column vectors) of matrix A is called its columns rank. OR When matrix A is reduced in Echelon form, then the number of non-zero rows are said to be row rank of a matrix A. --�� and the row rank of A' (transpose of A) is called the column rank of matrix A. INOTE-, Row rank, column rank and rank of a matrix are same. 2 3 Example 13 , Find the row rank, column rank and rank of a matrix: A = [ � 3 5 3 4 Solution �l Reducing the given matrix in Echelon form 2 3 A - [ � -1 -1 1 1 -[� 2 3 -1 -1 0 0 :+, ;] R, ➔ R, - 2R,, R, ➔ R, - R, ➔ R, + R, 0 Since the row rank of A is same as the number of non-zero rows in A. ⇒ Row rank of A = 2 Since column rank of A = row rank of A' So we take il Engineering Mathematics 45 R, ➔ R, - 2R,, R, ➔ R, - 3R,, R4 ➔ R, - 2R, Thus column rank = 2 I NorE ( Instead of calculating row rank of A ', just use the follow ing result. Row rank = column rank = rank of matrix. IES MASTER Publication 46 Rank of Matrices 1 Questions marked with aesterisk (*) are Conceptual Questions 1 .* A 5 , 7 m at r i x h a s a l l i t s e n t r i e s e q u a l to -1 . Then the rank of a m atrix is (a) 7 (c) 1 2. (b) 5 8 .* (d) 0 [GATE-1994-EE; 1 Mark] ! !] is 3. (True/False) Rank of the matrix [ � -7 0 -4 (a) (b) Tm (c) Three (a) m (b) n (b) 1 (c) 2 [GATE-1994 (CS)] If for a matrix, rank equals both the number of rows and num ber of colum ns, then the matrix is called (b) singular (c) transpose (d) minor [ 1 2 - [GATE-1994 (Pl)] (c) 3 (b) 1 [GATE-1995; 1 Mark] (b) 2 1 4 0 0 4 2 8 3 3 7 0 1 3 1 2 24 2 (d) 4 [GATE-1998 (CS)] Rank o f t h e matrix given below i s : 1 3 2 [ -6 -4 1 8 1 2 8 -36 -91 (b) 2 (d) ✓2 [GATE-1999; 1 Mark] 1 2 3 The rank of matrix A = 3 4 5 i s (c) 2 1 2.* The rank of a 3 4 6 8 (b) 1 (d) 3 [GATE-2000 (IN)] x 3 matrix C ( = AB), found by m ultiplying a non-zero column matrix A of size 3 x 1 and a non-zero row matrix B of size 1 x 3 , is 1 (d) None of the above a" (b) 1 (a) 0 (a) 0 The rank of matrix 2 4 6 is 0 0 -8] (a) 2 a" [GATE-1996-EE; 1 Mark] (c) 3 1 1. (n + 1 ) m atrix, (d) depends on value of a (c) 2 (a) 5 1 (d) 3 1 a a2 x a" 2 The rank of the m atrix [1 994-EC; 1 Mark] 0 0 -3 (a) non-singular 7. 10. (d) None The rank of matrix 9 3 3 1 1 a a (a) 3 [GATE-1 994; 1 Mark] (a) 0 6.* 9. * (d) None of the above The rank of (m x n) matrix (where m < n ) cannot be more than (c) mn 5. 3 � The rank of the matrix A = (a) One 4. * [ 1 -!2 -�1] 1 1 a a2 (c) n [GATE-1994-ME; 1 Mark] 3. The rank of the following (n + 1 ) where 'a' is a real number is (c) 2 13. (b) 1 (d) 3 [GATE-2001 ; 2 Marks] The rank of the matrix [� �] is Engineering Mathematics (a) 4 (c) 1 4. (b) 2 1 (d} 0 2 1 Given m atrix [A] = [� 3 4 1 0 matrix is (a) 4 (b) 3 (c) 2 1 5. !] , (d) 1 [GATE-2001 (CE)] the rank of the i] (a) 0 (b) 1 (c) 2 (d) 3 (a) has rank Zero 1 7. (c) n - 1 is orthogonal AA'A = A AA'A = I [2006-EC; 1 Mark] (d) has rank n [GATE-2007-CE; 1 Mark] (b) (AA')2 = A (d) AA'A = A' [GATE-2008-EE; 2 Marks] If the rank of matrix A is N , then the rank of matrix B is 19. (b) N - 1 (d) 2N Rank of AT A is eq ual to 2. (c) Rank of AT A is greater than 2. (b) 22. The rank of the matrix 0 4 4 14 8 1 8 14 - 14 0 - 1 0 is [GATE-2014 (CE-Set 2)1 The rank of the m atrix M = (b) 1 (a) 0 (c) 2 [! ! ] I is (d) 3 [GATE-2017-EC Session-I] 1 -1 0 0 0 0 0 1 -1 0 23.* The rank of the matrix 0 1 -1 0 0 -1 0 0 0 1 0 0 0 1 -1 is : [GATE-2017-EC Session-II] 24. If v is a non-zero vector of dimension 3 x 1 , then the matrix A = vv T has a ran k [GATE-201 7 (IN)] 1 1 -1 -1 -2 - 1 25.* Let P = [ 2 -3 4 ] and Q = [ 6 1 2 1 6 ] be 5 10 5 3 -2 3 26. [GATE-2014-EE-Set-3; 1 Mark] 6 -2 [GATE-201 5-CE-Set-2; 1 Mark] (a) Rank of N A is less than 2. (b) has rank 1 p2 + q2 pr + q s A = [ P q] s = [ r s pr + q s r 2 + s2 l N 2 (c) N (d) n 21 .* Let A be a 4 x 3 real matrix with rank 2. Which one of the following statement is TRUE? 1 8 .* Two m atrices A and B are given below : (a) (b) 1 [GATE-2016-EE-Set-1 ; 2 Marks] identity matrix . Let matrix A' = (NA)-1 N. Then , which one of the fol l owing statement is TRUE? (c) (a) 0 (d) Rank of AT A can be any number between 1 and 3. A i s m x n ful l rank matrix with m > n and I is an (a) rank of A is is 1 6.* x = [x 1 x 2 • • • • x"] is an n-tuple non-zero vector. The n x n m atrix V = xxT (c) 20.* Let A = [ a ii ] 1 � i, j � n with n ;::: 3 and a ii = i.j the [GATE-2003-CE; 1 Mark] 1 1 The rank of the matrix [ 1 -1 1 1 47 two matrices. Then the rank of P + Q i s___ [GATE-201 7 PAPER-2 (CS)] The rank of the following matrix is 1 0 (a) 1 (c) 3 0 -2] 3 1 2 2 (b) 2 (d) 4 [GATE 2018 (CE-Afternoon Session)] IES MASTER Publication 48 27. Rank of Matrices The rank of the matrix � � [� = !] I (b) 2 (a) (c) 3 is (d) 4 [GATE 201 8 (ME-MorningSession)] I A N SWE R l{EY 1. (c) 8. (a) 1 5. (See Sol.) 2. (False) 9. (d) 1 6. (b) 3. (b) 10. (a) 1 7. (a) 4. (a) 11. 5. (c) 1 2. (b) 1 8. (c) 6. (a) 1 3. (See Sol.) 20 7. (a) 1 4. (c) 21 . (b) (c) Sol-1 : (c) If a m x n matrix has all its entries equal to k the rank of the matrix will always be 1 . * 0, then -7 0 -4l R ➔R 3 3 + R2 ) [ 0 2 2 0 4] � 0 0 !Rank of matrix = 21 R1 <-- R2 . [� 24. (1) 25. (2) 11 26. (b) (b) 27. (b) ' R2 ➔ R2 + 3 R1 R3 ➔ R3 - 2R1 Sol-2: IFalsel [� ! � 23. (4) 19. (2) Remember : If in any non-zero matrix all rows or all columns are same then the rank of that matrix will always be 1 . R2 ➔R 2-2R1 22. (c) II ⇒ - rn � �l Rank (A) = 2 Alternative Solution: 1 -2 -1 I AI = -3 3 0 2 2 4 I� = -1 (-6 -6) + 4 (3 - 6) = 1 2 - 12 = 0 Sol-3: (b) ⇒ M 11 = Rank (A) = 2 �1 = 12* 0 Engineering Mathematics Sol-4: (a) Sol-8: (a) If A is any matrix of order m x n then rank or A :::; min {m, n} :. rank of A � m Sol-5: (c) Given ( ·: m < n) 3 A = [� � � ] . 3 1 IAI = o⇒ 1 All the rows of the given matrix are same, hence the rank of this matrix should be 1 . Sol-9: (d) A = ri ,UJ l� � � Given p ( A) * 3 A ~ A R3 ➔2R3 + 3R2 ~ [� � i ] A - [� d p(A) = 2 = No. of Non zero rows. Given that both rows and columns are equal. So let us consider A nxn Also given that p(A) = number of rows of A = number of column of A. So ⇒ p(A) = n IAI Sol-7: (a) R2 ➔ R2 - 2R1 R3 ➔ R3 + R2 ⇒ * rn - rn 0 A= - [� Rank (A) = 2 2 1] 4 6 0 -8 2 1] 2 1] 0 6 0 -8 0 � 0 ,� -�4 -�9 = Echelon form ⇒ Sol-6: (a) 49 -1 9 -� p(A) = 4 (No. of Non zero rows) Sol-1 0: (a) 2 [ 3 +1 -9] -6 18 4 1 2 8 -36 R 2 ➔R 2 + 2R1 R3 --,R3 -4R1 jRan k of Matrix = 1j Sol-1 1 : (c) Given A R2 ➔ R2 - � ; R3 ➔ R3 R3 ➔ R3 - R2 A ~ [� 0 0 1 ! : :] [i �] [i ;1 =[ A ~ ) 2 - L!Ri 2 -2 -2 2 -2 0 So p (A) = No of non zero rows = 2 Sol-1 2: (b) Let A = [� ] & B = [a 2 b2 c2 ] rank of A = r(A) = 1 rank of B = r(B) = 1 Rank of (AB) � minimum (Rank (A), Rank (B)) r(AB) � 1 IES MASTER Publication 50 Rank of Matrices So, rank (AB) is either O or 1 . But AB t:- 0. ·: Rank (AB) = 1 Sol-1 8: (c) Sol-1 3: IAI = O Method 1 p(A) t:- 2 ⇒ so So only possibility is p(A) = 1 Sol-14: (c) [� �] [� l - [� �l 2 1 3 4 1 0 [A] = R1 ➔ R1 - 2R3 R2 ➔ R2 - 3R3 R 2 ➔ R2 - 4R1 ⇒ Sol-1 5: 0 1 0 4 1 0 Rank [A) = 2 A = 0 1 0 0 1 0 [i �] 1 -1 1 - [� -: �1] R2 ➔ R2 - R1 , R3 ➔ R3 - R1 1 AA' = AA- (AT f1 AT = I. I = I AA'A = IA = A I AI = I� � I IBI = l = ps - qr p2 + q2 pr + qs l pr + qs r 2 + s2 = (ps - qr)2 = I Al 2 :. So A & B becomes singular or non-singular simultaneously so r(B) = r(A) = N. Method 2 : A = [� � ] and B = [ :: : :: �� : :;] Now let us calculate MT = [� :] . [ :] � 2 = [ ⇒ Sol-19: (2) Let R1 B R 2 rank (A) = 2 Sol-1 6: (b) We know that p +q = B 2 rp + sq pr + qs r 2 + s2 p(A) = p(AAT ) = N p(A) = p(B) = N A = A ~ [ �2 0 14 4 ⇒ [-2 2 V = xxT A ~ [� Rank(V) = 1 0 So l l 14 8 18 6 0 4 1 4 -14 0 � 1 0 18 58 1 16 Rank (x) = 1 Rank (xT) = 1 Sol-1 7: (a) 4 8 18 1 4 -14 0 - 1 0 l Rank (AB) � min{Rank A, Rank B} & l :: �8 0 0 = Echelon form ::1 0 p(A) = No. of non-zero rows = 2 Engineering Mathematics Sol-20: (b) Rank of A = 1 Because each row will be scalar multiple of first row. So we will get only one non-zero row in row Echelon from of A. But now another row is not becoming zero, so at least one 4 x 4 matrix in A is not zero. So, Rank = 4 Sol-24: (1) Alternative : Rank of A = 1 Because all the minors of order greater than one will be zero. Sol-21 : (b) From the property of the rank of matrix. V = [ :: ] X 3 3x 1 10 1 0 i] 3 6 6 det {M} = 5(0-1 2)-1 (60-60)+3 (20-0) � = -60 - 0 + 60 = 0 The one of the minor of matrix M has non zero determinant value. e.g. M 11 = [ � !] and jM1 1 j =- 1 2 Hence rank of M i s 2. So l-23 : (4) r1 r2 r3 r4 rs 1 -1 0 0 0 0 0 1 -1 0 0 1 -1 0 0 -1 0 0 0 1 0 0 0 1 -1 r1 ➔ r 1 + r2 + r3 + r4 + rs We get, A = So, IAI = 0 0 0 0 -1 0 0 0 0 0 0 1 -1 0 1 -1 0 0 0 0 0 1 0 0 1 -1 5x5 ⇒ p(v) = 1 and we know that p (v) = p (vv T) = 1 , so p (A) = 1 also. Sol-25: (2) p + Q = Sol-22: (c) M = [ 51 [: I P + QI = 0 :1 �] ⇒ p(P + Q) ;t 3 ·: There exist non singular matrix of 2 x 2 So, p (P + Q) = 2 Sol-26: (b) A= [ ! 1 0 0 2 1 3 [� I] 1 0 -2 2 -3 3 -21 2. 1 R2 ➔R2 -2R1 R3 ➔R3 -4 R1 R3 ➔ R3 -�R2 2 So p(A) = 2. Sol-27: (b) -4 1 [ A = -1 -1 7 -3 [� 7] 1 0 -2 2 0 0 -1 -1 1 -1 -3 -1 ] 1 -1 -1] -3 1 1 I Il Hence p (A) = No. of non zero rows = 2 IES MASTER Publication ■ Ordered Pair : (a, b) is an ordered pair if (a, b) 1. * (b, a) i.e. position of a & b are fixed. Ordered Triplet : (a, b, c) is an ordered triplet where positions of a, b and c are fixed. 2. Ordered n-tuple: A set of n numbers (x 1 , x 2 , x 3 , . . . . . . . . . . . . . . x n ) in which position of each number is fixed is called 3. ordered n-tuple . Any ordered n tuple of numbers in column form is called n-vector. i.e . X = [x 1 x 2 . . . . . . . . . . x n]T here X is an n-vector. LINEAR DEPENDENCE AND INDEPENDENCE OF VECTORS A set of vectors are linearly dependent if at least one of them can be expressed as a linear combination of others and if there does not exist any linear combination between them, then they are called linearly independent. . . . (i) then vectors are called LO. ,?' r,= === =i<-� 1. LD means there exists a mathematical relation between the vectors in l i near form. �N OT E 2. Equation (1) is cal led l i near combination of vectors. Methods to Find L I and L D Vectors Let General Method: (i) If p(A) = No. of vectors then vectors are L I . (ii) I f p(A) < N o . of vectors then vectors are LO. Tricky Method: Let A be a square matrix (a) If IAI * ,?' NOT E Example 1 0 then vectors are LI. (b) 1. I n this chapter we are considering a homogenous system in w hich K1 , K2 , K 3 ,........•... are unknowns. ⇒ LD ⇒ ⇒ 2. LI Trivial sol ution Zero solution 3. Non Trivial solution ⇒ ⇒ Unique so lution. Non zero solution [-:l � ⇒ I nfinite so lutions. J � [i� Find the value of ').. for which the following vectors are LO. � = Solution If IA I = 0 then vectors are LO. Let A = = [ [ x, x, x, ] = = [ -: �1 l i�] Eng ineering Mathematics 53 I A I = (H + 25 ) - 2(-1 41c + 51c) + 3(-1 0 + ),) = 1 4 1c - 5 For vectors to be LO, we m ust have I A I = 0 ⇒ ⇒ 1 4 1c - 5 = 0 n 1c = 5 14 1 2 Example 2 : Show that row vectors of the matrix A are Linearly Independent where A = �1 } [ 2 Solution Xi Given, [iJ � A = [ X1 Let Since I A I Example 3 : = * X2 = [ X3 ] = 0, so vectors are LI [� 1 ]. = [ �1 -1 3 0 �2] Find linear dependence and independence of following vectors. Also find a relation between them . Solution x, 2 � Cofficient matrix A = [ X1 X2 X3 x, 1 � [ ! 3 R2 ➔ R2 - 2R1 , R3 ➔ R3 - 4R1 2 0 A [ i -5 1 1 3 -5 2 1 4 1 R3 ➔ R3 - R2 2 0 A [ i -5 1 0 1 1 - 3 13 1 2 0 -1 3 2 -n which is in echelon form So, p(A) = 3 & No. of vecotrs = 4 . Since p(A) < No. of vectors So, vectors are linearly dependent. Relation: Let required relation be K1 X1 + K2� + K3X3 + K4X4 = 0 i.e. AX = 0 ... (i) k 1 + 2k2 - 3k4 = 0 -5k2 + k3 + 1 3k4 = 0 k3 + k4 = 0 We have 3 equations in 4 unknowns so one variable should be assigned arbitrary value. Let k4 = t, then k3 = - t, k2 = 5t , 12 k1 = 5t -9 IES MASTER Publication 54 System of Equations By equ. ( i ), 5 -9 tx1 + 5 tx2 - tx3 + tx4 = 0 ⇒ 12 9x 1 - 1 2x2 + 5x3 Example 4 , Show that the following vectors are LI. X1 = [-� = [� Solution A = [� ] A- ; [ l Let � -4 -3 R2 ➔ R1 l X, � -4 -3 R, ➔ R, - 3R, ; R, ➔ R, + 4R, A -[i �l A -[i }4] 3 - 5x4 = 0 l wh i ch i s i n echelon form 0 p(A) = 2 & No. of vectors = 2 i.e., p(A) = No. of vectors So, they are l i nearly i ndependent. Example 5 = Show that following vectors are LI over real Field and LO over complex Field. X1 = Solution = [1 i ; ] , � = [ 1 : i] Consider a l i near comb i nation of X1 & � a s follows, . . . (i ) K1 X1 + K2 X2 = 0 1 i K1 [ ; ] + K 2 [ : ] = [ ] � 1 i (K1 (1 + i ) + K2 ) [ ] = 2i K1 + (1 + i )K2 [oo] K 1 ( 1 + i ) + K2 = 0 . . . ( ii ) K 1 2i + Ki 1 + i ) = 0 . . . (i i i ) By (ii) K1 (1 + i) = -K2 � = K2 = 'A . (1 + i) (- 1 ) . . . (Let) K 1 = -'A. & K2 = (1 + i )'A. These values also sati sfi es (i i i ). ⇒ Case I : If 'A. = 0 Case I I : If 'A. non zero all zero. K 1 = 0 & K2 = 0 ⇒ ⇒ Vectors are LI i n Real fi eld. ( · : No l i near comb i nat i on ex i sts.) K1 = -'A, & K2 = ( 1 +i) 'A, ⇒ Vectors are LO over complex fi eld. ·: K 1 & K2 are not Engineering Mathematics 55 SYSTEM OF LINEAR EQUATIONS Let us consider a system of m li near equations in n un knowns say x 1 , x2, . . . , xn as below: a1 1 X1 + a1 2 X2 + ....... + a1n Xn = b1 82 1 X1 + 822 X2 + . . . . . . . + 82 n Xn = b2 where a ii and bi are constants, may be real or complex. The above system of equations can be written in matrix form, as Or AX = B where A = are matrices of orders m x n, n x 1 and m x 1 respectively. The matrix A is k nown as the coefficient matrix, X is colum n matrix o f un k nowns and B i s the column matrix o f constants. If b1 = b2 = . . . bm = 0, the set of equations is said to be homogenous otherwise non-homogenous. Augmented M atrix 81 1 81 2 821 822 The matrix [A : BJ = is called the augmented matrix of the given system of equations . 8m1 8m2 Consistency of Non-Homogenous Linear Equation By the solution of the system of equations we mean that to find a set of values of the un knowns x 1 , x 2, . . . , xn which satisfy all the given m equations. But here we clear that linear equations do not always have a solution i . e. it is not always possible to find the values of the un k nowns. Consider the sets of equations. 2x + 3 y = 7 x-y=1 } ... (i) 2x + 3 y = 7 8x + 12y = 28 } ... (ii) 2x + 3 y = 7 4x + 6y = 20 } ... (iii) The system of equations (i) has a unique solution ; the second has an infinite num ber of solutions, and third has no solution. Consistent and Inconsistent A system of equations is said to be consistent if they have one or more solutions e.g. (i) and (ii) are consistent. When a particular system of equations has no solution then it is said to be inconsistent e . g . (iii) is i nconsistent. Important Result: The necessary and sufficient condition for a set of linear non-homogenous equations to be consistent is that the ran ks of the coefficient and the augmented matrices are equal. IES MASTER Publication 56 System of Equations METHODS OF SOLVING NON-HOMOGENEOUS SYSTE M : (GAUSSIAN ELI MINATION METHOD) AX B WHERE B * 0 = (1) If p(A) * p[A : B] then system is incosistent. (No solution) (2) If p(A) = p[A : B] = No. of unknows then; system is consistent with unique solution. (3) If p(A) = p[A : B] < No. of unknows; system is consistent with infinite solution & in that case (n-r) variables are assigned arbitrary values, where n = no. of unknowns & r = rank 1. 2. Example 6 Solution : 1 Try to convert augmented matrix into an echelon form using E-row operations only. Then find values of unknowns by using Backward substitution method. 9) 2x1 + 3x2 + x3 = Solve x1 + 2x2 + 3x3 = 6 3x 1 + Xz + 2x3 = 8 by Gaussian elimination method. Augmented Matrix is [A : B] 6 -3 - 1 0l � ] which i s in Echleon form . p(A) = 3; p[A : B] = 3 ⇒ p(A) = [A : B] = No. of unknown So, unique soln. exist. Eva luation of Solution Given system is (AX = B) which is equivalent to [ � X + 2y + 3z = 6 Solution is : ⇒ 29 5 35 ;X= ;Y= 18 18 18 35 29 5 (x, y, z) = ( ' ) 18 18 ' 18 - y - 5z = -3 ⇒ Z = ⇒ �1 _35] [�] [:3] = O 1 8z = 5 18 3x3 2 3x1 5 Engineering Mathematics Example 7 Solution 1 5x + 3y + 7z = 6 Solve: 3x + 26y + 2z = 9 7x + 2y + 1 0z = 5 Augmented Matrix is 1 [A : BJ = ~ 3 7 26 2 [ � 2 10 27 -1 26 2 2 10 [i -[� - [� 57 :] :] l! 27 - 1 . 8 -55 5 . - 1 5 -1 87 1 7 . -51 27 -1 -1 1 -1 -1 1 -1 27 -1 - 1 1 - [� � � 3] -3 ; ] which is in echelon form p(A) = 2 = p [A : BJ p(A) = p[A : BJ < No. of unknowns; so Infinitely many solution exist. Evaluation of Solution Given system is (AX = B) 27 which is equivalent to [ � -1 1 0 X + 27y - Z = 8 ⇒ -1 1 y - Z = 3 Since n - r = 1 , so one variable should be assigned arbitrary value. Let z = k, then y = ��� and x + 21 ( ��� ) - k = 8 71 + 2k _k = 8 ⇒ -1 1 1 3k + 1 59 ⇒ x = 11 x+ Sol is {x, y, z) = ( Example 8 1 -1 1 x + 71 + 2k + 1 1 k = 8 (-1 1 ) 1 3k + 1 59 -k - 3 ) k kER , 1 1- · 11 - Find the value of A for which given system have a solution x+y+z=1 x + 2y + 4z = 1c Solution 1 [A : BJ = [� � : 1 4 10 ;,l x + 2y + 1 Oz = A2 IES MASTER Publication 58 System of Equations R2 ➔ R2 - R1 ' R3 ➔ R3 - R1 1 1 1 3 B } - � [ 3 9 [A -[� R3 ➔ R3 - 3R3 1 11, - 1 (11, 2 - 1) l l 1 11, - 1 1 1 1 3 0 0 (11,2 - 311, + 2) System have a soln. when p(A) = p[A : BJ and p(A) = 2 so for p[A : BJ = 2 we have to take 11,2 - 311, + 2 = 0 ⇒ 11, = 1, 2 + 2x3 = 1 Example 9 : Solve the equations if consistent. 2x1 + 2x3 = 2 X1 - 3x2 + 4x3 = 2 X1 - X2 Solution ' The matrix representation of the sy&em is [ � �: AX = B Augmented matrix is: [A ⇒ ⇒ Example 10 1 BJ = • 1] [ -1 2 0 2 : 2 [ � -3 4 : 2 - [� ~ 1] 1 -1 2 0 2 -2 0 -2 2 -1 2 2 -2 : 0 0 0 : 1 ⇒ p[A : BJ = 3 and p[AJ = 2 !l [ : l �] = rn R2 ➔ R3 - 2R1 , R3 ➔ R3 - R1 R3 ➔ R3 + R2 p[A : BJ * p[AJ The given system of equation is inconsistent i.e. no solution. By the use of matrices, solve the equations: x+y+z=9 2x + 5y + 7z = 52 2x + y - z = 0 Solution The matrix representation of given system of equations is [ � i ] �] [ f ] = or AX = B The augmented matrix is 1 [A : BJ = 2 ; ; [ 2 1 -1 9 34 -1 8 l Engineering Mathematics 59 1 1 -3 : ~ 0 =1 -3 : [ 0. 3 5 1 ~ 0 �1 -� [ 0 0 -4 Above is the Echelon form of a matrix. ⇒ ⇒ p[A : BJ = p[AJ = 3 = number of variable The given equations will have a unique solution. Evaluation of Solution l =� l Now, to find the solution we see that the given system of equations is equivalent to the matrix equations, 1 [ � � �] [ � = [ ⇒ x + y+ z = 9 -y - 3z = -1 8 -4z = -20 Solving these equations, we get z = 5, y = 3, x = 1 Example 1 1 1 Test the consistency of the 4x - 2y + X + y1 5x - 3y + Solution 1 following equations, and, if possible, find the solution: 62 = 8 3z = -1 9z = 21 [ � -1 �3l[;l [�1 The matrix representation of given system of equations 2 1 5 -3 9 z = 21 or AX = B The augmented matrix is [A : BJ = [1� - r: -2 6 1 -3 -3 9 1 -3 -2 6 1 5 -3 9 ⇒ ⇒ - [� - [� 1 -3 -6 1 8 -1 8 54 �11 21 1 211 � R1 � R2 -1 1 2 R2 ➔ R2 - 4R1 , R3 ➔ R3 - 1 5R1 361 1 -3 : -1 -6 1 8 : 1 2 0 0 : 01 R3 ➔ R3 - 3R2 p[A : BJ = p[AJ = 2 < number of variables The given systems of equations are consistent and have an infinite solution. IES MASTER Publication 60 System of Equations l� � �l � l = [ � l The given system of equations is equivalent to the matrix equation. ⇒ X ⇒ + y - 3z = -1 and -6y + 1 8z = -12 y = 3z - 2, if z = k then y = 3k - 2 and x = 1 i.e. x = 1 , y = 3k - 2, z = k Thus for different values of k, the given system has infinite number of solutions. Example 12 1 I nvestigate for what values of a, b the equation X + 2y + 3z = 4 X + 3y + 4z = 5 x + 3y + az = b have (i) no-solution (ii) a unique soluiton (iii) an infinite solutions. Solution 1 [ � � :3][;] - [�1 2 The matrix representation of given system of equations 1 5 -3 9 or AX = B 21 z The augmented matrix is [A : B] = Case - I : When a = 4, b ;t 5 ⇒ p[A] = 2, and p[A : B] = 3 ⇒ p(A : B] ;t p[A] System is inconsistent, and have no solution. Case - II : When a ;t 4 ⇒ p(A) = 3 and p[A, B] = 3 = number of variables System is consistent and will have a unique solutions. for a ;t 4 & b e R Case - Ill : W hen a = 4 and b = 5 ⇒ p(A) = 2 = p[A, B] < number of variables System is consistent, but will have infinite solution. Example 13 1 Investigate the values of the ').., and µ so that the system : 2 x + 3y + 5z = 9 7x + 3y - 2z = 8 2x + 3y + ')..,z = µ has (i) a unique solution (ii) no solution. Solution I Here [� fl X=[�l B= [:] [� 9 ] 61 3 3 3 A = Augmented matrix Engineering Mathematics 3 5 3 -2 : 8 3 ')..,, : µ [A : B] = [A : BJ ~ 7R , R ➔ R -R By R2 ➔ R2 - 2 1 3 3 1 2 3 5 0 -1 5 -39 2 2 0 0 '}..,, - 5 9 -47 2 µ-9 Now following cases arise: Case (i) : There exist a unique solution if p([A : Bl) = 3 and p([A]) = 3 Then we have ')..,, - 5 *0 & µ can take any value. * Case (ii) : There exist no solution if p([A]) p([A : Bl) ⇒ ⇒ Example 14 Solution I 1 ')..,, - 5 = 0 and µ - 9 * 0 , ')..,, = 5 and µ * 9 then p([A]) = 2 , p([A : Bl) = 3 Find the solution of the system 5x 1 3x2 7x3 + x4 = 10 -x 1 + 2x2 + 6x3 - 3x4 = -3 X 1 + X2 + 4x3 - 5x4 = 0 - - The matrix representation of given system of equations: or -1 [ 51 - 2 13 AX = B The augmented matrix -3 -7 1 [A : B] = [ �1 2 6 -3 1 4 -5 - 1 4 -5 [ �1 2 6 -3 -3 -7 1 [� 1 3 4 10 -5 -8 -8 -27 26 (it is an underdetermined system) 10 -3] 0 �3 ] R1 tt R3 10 l �3 R2 ➔ R2 + R1 , R3 ➔ R3 - SR1 10 IES MASTER Publication 62 System of Equations 1 1 4 -5 0 1 10 -8 3 3 1 -1 R2 ➔ -R2 3 0 -8 -27 26 1 0 1 0 0 4 -5 1 0 -8 3 3 1 14 3 3 10 0 -1 R3 ➔ R3 + 8R 2 2 1 1 0 1 4 -5 10 8 3 3 0 0 1 -1 4 ⇒ ⇒ 0 -1 R3 ➔ -3R3 -6 p[A : BJ = p[AJ = 3 < number of variables the system is consistent and has infinite solutions. And we have 1 1 4 0 1 10 /3 [ 0 0 1 -��4{! which gives 1 l�;l = X 1 + Xz + 4X3 - 5x4 = 0 x3 - 1 4x4 = -6 ⇒ 10 8 Xz + - X3 - - X4 = -1 3 3 If x4 = k, then on solving these equations, we get x 1 = 8 1 k - 33, x2 = 57 - 1 32k, x 3 = 1 4k - 6, X4 = k. Since k is arbitrary so it has infinite number of solutions. Homogenous System (AX = 0) is called homogenous system where O is null matrix & Am xn · Solutions of Homogenous Equations The system AX = B of m linear equations in n unknowns is known as the system of homogenous linear equations if B = 0. Thus the above equation reduces to AX = 0 For such a system matrix A and augmented matrix [A : BJ are the same, so that the rank of coefficient matrix A and its augmented matrix are equal i.e. p[A : BJ = p[AJ . Hence a system of homogenous linear equation is always consistent. Trivial Solution : X = 0 i.e. x 1 = 0, x2 = 0, . . . . xn = 0 is always a solution of the giv�n system of equations. This solution is called a trivial solution or zero solution. Non-Trivial Solution : The system will have non-trivial solution if and only if the rank of the co-efficient matrix is less than the number of unknowns. Engineering Mathematics 63 M ETHODS O F SOLVI NG HOMOGENEOUS SYSTEM (1 ) If p(A) = No. of unknowns; then system is consistent with unique sol. or Trivial solution. (2) If p(A) < No. of unknowns; then system is consistent with infinite many solution i.e. Non trivial sol . Also exist. And in that case ( n-r) v a ri ab l es a re assig ned a r b i t ra ry v al ues. wh ere n = N o . of u n known & r = rank(A) 1. In case (1) no L I solution exist while in case (2) (n-r) L I sol. Exist. 2. Homogenous system is always consistent because in that case rank of coefficient matrix is always equal to rank of augmented matrix. 3. Nul l ity = Order - Rank & nullity of identity matrix = 0 4. Trivial solution means (x, y, z) = (0, 0, 0) "' zero solution. 5. Non Trivial sol means Non zero solution. 6. When the number of equations is less than the number of unknowns, the Homogeneous system w i l l always have an infinite number of solutions. Tricky Method for Homogeneous System When A is a square matrix of order n i.e. An • n * 0 then system is consistent with unique sol i.e.; Trivial sol . (i) If IA I (ii) I f IAI = 0 then system is consistent with infinite sol . i . e . ; Non Trivial soln. also exist. Example 15 Solution 1 1 Does the system of equ. have a solution, other than x = y = z = w = 0. If so, obtain them x + y + z + w = 0; x + 3y - 2z + w = 0; 2x - 3z + 2w = 0; x + y + w = 0 Coefficient matrix is A -- r 2�1 3 � -�3 -o 2 �1 1 So, p(A) = 3 < No. of unknown So, Infinitely many solution exist i.e. Non trivial solution also exist. IES MASTER Publication 64 System of Equations Evaluation of Solution Given system which is equivalent to r� mi r�1 [i mi r�1 1 1 3 -2 0 -3 0 AA = 0 = 1 1 2 -3 = 0 -8 0 0 x + y + 2 + w = 0 2y - 32 = 0 - 82 = 0 Here, n - r = 1 , So one variable should be assigned arbitrary value. ⇒ 2 = 0, y = 0 & X + W = 0 w = k then x = - k (x, y, 2, w) = (-k, 0, 0, k) k E R Let So solution is Example 16 Solution 1 (x, y, 2, w) = (-1, 0, 0, 1) . . . } Non-trivial solution (x, y, z, w) = (0, 0, 0, 0) ➔ Tr1v1al solution and = (-4, 0, 0, 4) Find the value of k for which (3k - 8)x + 3y + 32 = 0; 3x + (3k - 8)y + 3z = 0; 3x + 3y + (3k - 8)2 = 0 have Non Trivial solution. (3k - 8) 3 3 3 l (3k - 8) A= [ 3 (3k - 8) 3 3 For non trivial solution IA I = 0 C1 ➔ C1 + C2 + C3 3 3 (3k - 2) 3 (3k - 2) (3k - 8) = 0 3 (3k - 2) (3k - 8) (3k - 2) (3k - 1 1 ) (3k - 1 1 ) = 0 Example 17 : Solve the system of equations Solution 1 x+y+w =0 y+2=0 x+y+z+w = 0 x + y + 22 = 0 ⇒ ⇒ 1 3 3 3 (3k - 2) 1 (3k - 8 ) = 0 1 3 (3k - 8) K = 2 11 11 , , 3 3 3 I n matrix form , the given system is given by . . . (i) Engineering Mathematics 65 ⇒ A = where Thus By [i il AX = B A ~ R4 ➔ R4 - 2R3 ~ 1 1 1 1 ... (ii) 0 1 1 2 [� Il [� jJ 1 0 1 1 0 1 0 2 By R3 ➔ R3 - R1 , R4 ➔ R4 - R1 1 0 1 1 0 1 0 0 ⇒ p(A) = 4 Here number of variables = 4 So, p(A) = number of variables. Hence there exist only trivial solu't ion. E"xample 18 1 Solve the homogenous equations 3x + 4y - z - 6w = 0 2x + 3y + 2z - 3w = 0 2x + y - 1 4z - 9w = 0 x + 3y + 1 3z + 3w = 0 Solution I [; �El [�l [; -61 [1 The given equations are equivalent to AX = Now A = 4 -1 3 2 1 -14 3 13 = 4 -1 3 2 -3 1 - 1 4 -9 3 13 3 3 13 2 3 2 3 By R1 � R4 2 1 -1 4 -9 -6] 3 4 -1 : 3 13 3 -3 -24 -9 A - [� -5 -40 -1 5 -5 -40 -1 5 [ � � I �l l . . . (i) JC By R2 ➔ R2 - 2R1 , R3 ➔ R3 - 2R1 , R4 ➔ R4 - 3R1 A ~ ⇒ p(A) = 2 = r; and n = 4 (the number of variables) As n - r = 4 - 2 = 2, the equations have infinite number of solutions IES MASTER Publication 66 System of Equations ⇒ ,,. x + 3y + 1 3z + 3w = 0, y + 82 + 3w = 0 Now choosing z = k 1 , and w = k2 , we obtain x = 1 1 k 1 + 6k, y = - 8k 1 - 3k2 , z = k 1 , w = k2 where k 1 and k2 are arbitrary constants, at our choice. Hence the equations have an infinite number of solution. Example 19 1 Determine the value of A so that the equation 2x + y - 2z = 0 X + y + 3z = 0 4X + 3y + ').,, Z = 0 have non-zero solution & hence find the solution. Solutlon1 [ � l [ : ml [: �l [! �l [� .�J [� .�al [� �l The given system in the matrix form is given by 1 1 3 Now say AX = 0 = 1 1 3 A = 1 1 3 1 -1 -1 By R3 ➔ R3 ➔ R3 0 ⇒ 1 -1 0 m By R1 B R2 B y R2 ➔ R2 - 2R1 , R3 ➔ R3 - 4R1 1 -1 = . . . (i) if A = 8 p(A) = 2 < 3 i.e., rank is less than the number of variables l[�l ⇒ Non zero solution exists only when A = 8 . Now, the matrix form of the system reduces to [ � �1 � = - y - 4z = 0 0 = 0 ⇒ ⇒ X + y + 3z = 0 y = -4z, X = Z Choosing z = k, we get x = k, y = -4k, z = k This gives the general solution of the given system , where k is an arbitrary constant. Eng ineering Mathematics 67 LU DECOMPOSITION (FACTORIZATION) METHOD This method i s based on the fact that every square matrix can b e factorized into product of a lower a n d an upper triangular matrix. If all the principal minors of A are non-singular i.e. 811 * 8 0, 1 1 1 82 1 *0' 81 2 1 822 8 12 81 1 822 82 1 ... , And there are two types of LU Decomposition M ethod : (i) Dolittle's Method (ii) Crout's Method DOLITTLE'S M ETHOD Consider the system of equations. 81 1 X1 + 81 2 X2 + a1 3 X3 == b1 82 1 X1 + 822 X2 + 823 X3 == b2 831 X1 + 832 X2 + a33 X3 == b3 This system is equivalent to l . . . (i) " .(ii) AX = B where o o [�, l Now we will factorize A as the product of lower triangular matrix L = u and an upper triangular matrix So, = [ 1 0 "3 1 "32 1 u, , u, , U1 3 U 22 U 0 u33 AX = B ⇒ where 0 0 ⇒ l 23 i.e. A = LU ⇒ LUX = B . . . (iii) L(UX) = B LY = B ... (v) l...D( = y ...(iv) ;n . l ,l � l [ ,l Now first calculate Y using (v) and then calculate X using (iv) and L and U can be found from 0 i.e., �1 [ "3 1 u i . e., ,, [ � 1 U1 1 "3 1 1 1 U U 1 "32 12 � 1U1 2 + U22 "3 1U1 2 + "32 U22 LU = A U1 2 U 22 0 u1 3 u , U2 ,3 U 33 � 1 U1 3 + U23 "31 1 3 + "32 U23 + U33 U a" = [ 82 1 83 1 81 2 822 832 a1 1 821 81 2 822 831 832 = 823 833 82 3 833 Equating corresponding coefficients, we get nine equations in nine unknowns. From these nine equations, we can solve for 3fs and 6u's. IES MASTER Publication 68 I System of Equations ( To produce a unique solution it is convenient to choose either /ii = 1 or Uu = 1, i = 1 to n . NOTE _ When we choose . (i) /1; = 1 , the method is cal led the Doolittle's Method. (ii) uii = 1, the method is cal led the Crout's Method. Example 20 1 Solve the following system by the method of factorization, or by the method of triangularization or LU decomposition. x + 3y + 8z = 4, and x + 4y + 3z = -2, and x+ 3y + 4z = 1 Solution 1 The given system is AX = B, where Let LU = A where L = o o [�1 1 l:i 1 l:i2 0 l 1 and ⇒ U1 2 1 U ½ 1 2 + U22 l:J1 U 12 + l:J2 U22 u11 = 1 , u 1 2 = 3, . u 1 3 = 8 . 121 U 1 1 = 1 121 U 1 2 + U22 = 4 l21 U 1 3 + U23 = 3 ⇒ ⇒ ⇒ 121 = 1 ; l31 U 1 1 = 1 ⇒ 131 = 1 U22 = 4 - 12 1 U 1 2 = 4 - 1 (3) = 1 U23 = 3 - l21 U 13 = 3 - 1 (8) = -5 l3 1U 13 + l32 U23 + U33 = 4 U33 = 4 - 131 u 1 3 - l32 u23 = 4 - 1 .8 - 0.(-;-5) = -4 L = Let where then [i lJ)( = Y, y = 0 1 0 l�:l LY = B ;] 13 [ 8 and U = 0 1 -5 0 0 -4 l ... (i) . . . (ii) using (ii), and using (i), [i ! m:i - [�i Engineering Mathematics ⇒ 69 Y, • 4 . Y, . -2 . Y, . -3 [� � �l[�l · [f:l X + 3y + 8z = 4s y - 5z = -2 -4z = --3 By back substitution, z = y = and X 3 4 -4 + 5Z = -2 + 5 (¾ ) = i 2 = 4 - 3y - 8z = 4 - 3 (i) - 8(¾) = : 29 X = Example 21 1 y 4 ' 4' 7 = Solve the equations 2x + 3y + z = 9 X + 2y + 3z = 6 3x + y + 2z = 8 2 = 3 4 by the factorization method or LU decomposition. Solution I We have A • i] [� ; Now let A = LU, where L = [�1 l-:i 1 Thus we have U 11 l21 U 11 l21 U 1 2 + U 1 21 U 1 3 + U 22 = 2, U 1 2 = 1 = 2 = 3 l 3 1 U 1 2 + l 32U 22 = 1 l31 U 1 3 + l 32U23 + U33 = 2 23 ⇒ ⇒ ⇒ ⇒ ⇒ = 3, 1 21 = U 23 1 32 13 1 /2, 1 31 U 1 1 = 3 ⇒ 1 31 = 3/2 = 5/2 = U33 -7 = 18 Hence we can write A • LU . [ = 1 = 1 /2 1 · 22 U U ;;! : Further the given system can be written as m 3 1/2 0 IES MASTER Publication 70 l:'.! i m Jl�J . m l:'.! i nm . l:J . System of Equations AX = B ⇒ where ⇒ ⇒ L(UX) = B [m LY = B where Y = UX ⇒ 1 2 5 2 1;a � 5 1 = [ � + Now relation (i) gives y 1 = . ;J�] (i) (ii) 9, � + Y2 = 6 i.e. Y2 = ¾ and Now using (ii), solution of the original system is given by [ � ⇒ 1 2 5 2] = H 2] i8 � 2x + 3y + = z 9 y 5 3 - + -z = 2 2 2 ⇒ 1 8z = 5 X = 35 29 ·1 5 18 , Y = 18 , Z = 18 · CROUT'S METHOD In this case we use u ii = 1 in place of /ii = 1 . As before, we want to solve the system ... (i) AX = B a" where Suppose we decompose A = 82 1 [8 31 A LU then = L = ["' [ b' ] [� ] 81 2 � ' ] 822 823 , X = X2 and 8 = b 2 832 833 b3 X3 0 0.1 0.2 � ] "3 1 "32 "33 � a nd U = [0 and rest of the process is same as in Dolittle's method. U1 2 1 0 . . . (ii) �, l U23 1 Example 22 1 Solve by Grout's method, the following system of equations x + y + z = 3 2x - y + 3z = 16 3x + y - z = -3 Solution I We choose u ii = 1 and write A = . . . (i) LU � 1 U1 2 + ½_2 ½.1 U1 2 "3 1 U1 2 + "32 Engineering Mathematics 71 Equating, we get 111 = 1 , 121 = 2, 131 = 3 ⇒ U1 2 = 1 1 ⇒ U 13 = 1 -1 ⇒ 122 = -3 1 ⇒ /32 = -2 3 ⇒ U23 = -1 /3 -1 ⇒ 133 = - 1 4/3 111 U12 = 1 111 U13 = 121 U 1 2 + 122 = l31 U 1 2 + /32 = l31 U 13 + 122U23 = l31 U 1 3 + l32 U23 + 133 = �3 � ][� � 3] 1 -1 / A = LU = [ ; 1 3 -2 -14 / 3 0 0 Thus, we get The given system is AX = B ⇒ LUX = B ... (i) Let UX = Y so that (i) becomes LY = B [� _LJml [��] 0 -3 -2 = which gives Y1 = 3 2y1 - 3y2 = 1 6 3y1 - 2y2 = ⇒ 14 3 Y 1 = 3, Y2 = Now, lD( = y which gives, X + y + Z -3 , 10 Y3 = 4. [by forward substitution] = 3 1 10 y - -z = 3 3 z = 4 By back substitution x = 1 , y = -2, z = 4 Example 23 1 Show that the LU decomposition method fails to solve the system of the equations X1 + X2 - x3 = 2 2x 1 + 2x2 + 5x 3 = -3 3x 1 + 2x2 - 3x3 = 6 whose exact solution is (1 , 0, - 1 ). Solution 1 Case-I: If we write (taking 1 11 = 1 ) Dollttle's Method IES MASTER Publication 72 System of Equations We obtain, [� �1 ] 1 2 2 -3U 11 121 U 11 121 U 1 2 + U 22 = [�, ½1 0 1 ½2 ;n U 12 U 22 0 U1 3 U 23 U33 l = 1 , U 1 2 = 1 , u 13 = -1 = 2 ⇒ /21 = 2 = 2 ⇒ u 22 = 0 Hence LU decomposition method fails as the pivot u22 = 0. �1 ] H� l Case-II: If we write (taking u ii = 1) Grout's Method We obtain, [� 1 2 2 -3 121 U 1 2 1 11 = [� ' 0 ½1 ½2 ½1 ½2 = 1, '21 = 2, U 12 1 0 1 31 = 1 , ⇒ u1 2 = 1 1 U 13 11 = -1 , ⇒ u 1 3 = -1 + = 2, ⇒ '22 U 23 1 = 3 1 U 12 11 122 U1 3 = 0 Hence LU decomposition method fails as the pivot /22 = 0. Engineering Mathematics 73 PREVIOUS YEARS GATE & ESE QUESTIONS Questions marked with aesterisk (•) are Conceptual Questions 1 .* The number of linearly i ndependent solutions of the system of equations [� �][::] 7.* 0 -1 = 0 is equal to -2 (a) 1 (b) 2 (c) 3 {d) 0 [GATE-1994-EE; 2 Marks] 2. Solve the following system of equations X1 X1 + �+ = 8. � + X3 = 1 (a) Unique solution (b) No solution (c) I nfinite num ber of solutions Solve the system 2x + 3y + z X 3y - 7z = 6 = 9, 4x + y = 6.* (c) If m < n and B is a zero vector then the sys­ tem has i nfinitely many solutions (d) The system will have a trivial soution when m = n, B is the zero vector and rank of A is n. (a) -33 (b) 0 (c) (d) 9 For the followi ng set of sim ultaneous equations: 1.5x - 0.5y = 2 4x + 2y + 3z = 9 [3, 4, 7] , [3, 4, 7] [1, 0, OJ, [1, 1 , OJ (c) [1 , 0, 2J, [O, 5, OJ {d) [1, 1, 1J, [-1, -1, 1J the solution is unique (b) infinitely many solutions exist (d) finite number of m ultiple solutions exist [GATE-1997-ME; 1 Mark] 9.* 59 5 Among the followi ng, the pair of the v ector orthogonal to each other is (a) (b) (a) [GATE-1 995 (ME)] The solution(s) to the equations 2x + 3y = 1 , x - y = 4 4 x - y = a , will exits if a is eq ual to (a) diagonal matrix (b) lower triangular matrix upper triangular matrix (d) singualr matrix [GATE-1996-ME; 1 Mark] Express the given matrix A = � ; 1�] as a 6 27 3 1 product of triangular matrices L and U where the diagonal elements of the upper triangular matrices U are unity and L is an lower triangular matrix. [ [GATE-1997-EE; 2 Marks] 10. [GATE-1995 (ME)] I n the Gauss elimination m ethod fo r solving a system of linear algebraic equations, triangulari­ zation leads to (c) If m = n and B is non-zero vector then the system has a unique solution (c) the equations are incompatible 7, [GATE-1995; 1 Mark] 5.* (b) 7x + y + 5z = 10 [1 994-EC; 2 Marks] 4. The system has a solution, if p ( A ) = p (A/B) [GATE-1996 (CS)] 0 (d) Only one solution 3. (a) X3 = 3 X1 - � - Let AX = B be a system of linear equations where A is an m x n matrix B is an n x 1 column matrix which of the following is false? A set of linear equation is represented by the matrix equation AX = B. The necessary condition for the existence of a solution for this system is (a) A m ust be invertible (b) B must be linearly dependent of columns of A. (c) B m ust be linearly independent on the columns of A. (d) None 11. [GATE-1 998-EE; 1 Mark] Solve the followi ng set of simultaneous equations by Gauss elimination method. Z = -2y + z = X - 2y + 3 = 11 X + 3z 1 . .. (1 ) . . . (2) . . . (3) [GATE-1998; 2 Marks] IES MASTER Publication 74 System of Equations 12.* Consider the followi ng set of equations x + 2y = 5, 4x + 8y = 12, 3x + 6y + 3z = 1 5. This set 1 7 .* Consider the following system of linear equations (a) has unique solution (b) has no solution (c) has infinite num ber of solutions Notice that 2 nd and 3 rd columns of the coefficient matrix are li nearly dependent. For how many value of a , does systems of equations have i nfinitely m any solutions. (d) has 3 solutions 1 3. [GATE-1 998 (CS)] Consider the following two statements: I. II. The maxim um num ber of linearly independent column vectors of a matrix A is called the rank of A. If A is an n x n square matrix, it will be nonsingular if rank A = n. With reference to the above statements, which of the following applies? 1 4. (a) Both the statements are false (b) Both the statements are true (c) I is true but I I is false (d) I is false but I I is true [GATE-2000-CE; 1 Mark] 18. = 1 9. This system has 15. one solution (b) no solution (c) infinite solutions (d) four solutions [GATE-2001 ; 1 Mark] T h e following set o f equations has 3x + 2y + z = 4 x - y + z = 2 (d) infinitely many [GATE-2003 (CS)] A system of equations represented by AX = 0 where X is a column vector of unknown and A is a matrix contai ning coefficient has non-triv ial solution when A is 20. [l � :I�H�l What values of x, y, z satisfy the following system (a) X = 6, y = 3, Z = 2 {b) x = 12, y (c) X = (d) x = 6, y = = 12, y 3, z 6, = = Z = -3, z -4 -4 = 4 [GATE-2004] How many solutions does the following system of linear equations have -x + 5y = -1 X -y = 2 (b) a unique solution no solution (d) Hermitian [GATE-2003] - 2x + 2z = 5 (a) (b) singular non-singular of linear equations 5 (a) (c) 2 (c) symmetric + y = 2 2x + 2y (b) 1 (a) Consider the system of equations given below : X (a) O (c) m ultiple solutions (d) an i nconsistency [GATE-2002; 2 Marks] 1 6.* Consider the system of sim ultaneous equations X + 3y = 3 (a) infinitely many (c) unique (b) two distinct solutions (d) none [GATE-2004 (CS)] X + 2y + Z = 6 2x + y + 2z = 6 x + y + z = 5 21 .* A is a 3 x 4 matrix and AX = B is an i nconsistent system of equations. The highest possible rank of A is This system has (a) unique solution (b) infinite number of solutions (c) no solution (d) exactly two solutions [GATE-2003; 2 Marks] (a) 1 (b) 2 (c) 3 (d) 4 [GATE-2005-ME] 22.* A is a 3 x 4 real matrix and A x = b is an i nconsistent system of equations. The highest possible rank of A is Engineering Mathematics (a) 1 (b) 2 (c) 3 (d) 4 1 , 2 (d) Non-existent (c) [GATE 2005-ME; 1 Mark] 23.* Consider a non-homogeneous system of l i near equations, representing m athematically an over­ determ ined system . Such a system will be X = 1, y = Z = 2 [GATE-2006-CE; 1 Mark] 28.* A system of linear simultaneous equations is given 1 0 (a) consistent, having a unique solution (b) consistent, having m any solutions (c) inconsistent, having a unique solution (d) inconsistent, having no solution [GATE-2005; 1 Mark] 24.* In the m atrix equation Px = q , which of the following is necessary condition for the existence of at least one solution for the unknown vector x. (a) Augmented matrix [P : q] must have the same rank as matrix P (b) Vector q must have only non-zero elements (c) Matrix P must be singular (d) Matrix P must be square 25. [GATE-2005-EE] Consider the following system of equations in three real variable x 1 , x2 and x 3 : x2 + 3x3 = 1 3x 1 + 2x2 + 5x3 = 2 2x 1 - -X 1 + 4x2 + x 3 = 3 (b) a unique solution (c) more than one but a finite number of solutions (d) an infinite number of solutions [GATE-2005 (CE)] 26.* Let A be 3 x 3 matrix with rank 2. Then AX = 0 has 27. as AX = b , where A = 1 (b) 2 (c) 3 (d) 4 (a) 0 1 0 1 0 0 0 1 0 1 as Ax = b where A = and b = 0 1 1 0 0 1 0 0 0 1 which (a) x (b) x (c) x (d) x space and <x, y> denote their dot product. Then the determinant Solution for the system defined by the set of equations 4y + 3z = 8; 2x - z = 2 and 3x + 2y = 5 is (b) x = 0, y = 2' z 1 4 3 2 [< X , X > < X, Y > ] < y, x > < y, y > is zero when x and y are linearly independent. is positive when x and y are linearly indepen­ dent. is non-zero for all non zero x and y is zero only when either x and y is zero. det (a) (b) [GATE-2007-EE; 2 Marks] [GATE-2005 (IN)] = of the fol lowing statement is true? is null vector is unique does not exist is infinitely many values 30.* Let x and y be two vectors in a 3 dimensional (c) Z O ' [GATE-2006 (IN)] (d) = 0, y = 1 , 0 29.* A system of linear simultaneous equations is given (b) one independent solution (c) two independent solutions (d) three independent solutions X 1 0 0 and b = 0 0 0 then the rank of matrix A is (a) only the trivial solution X = 0 (a) 0 1 0 1 1 0 0 [GATE-2006 (IN)] This system of equation has (a) no solution 75 31.* For what v a l ues of a and b , t h e fol l owing simultaneous equations have an i nfinite number of solutions? X + y + Z = 5 '· X + 3y + 3z = g' x + 2y + az = b (a) 2, 7 (b) 3, 8 (c) 8, 3 (d) 7, 2 [GATE-2007-CE; 2 Marks] IES MASTER Publication 76 System of Equations 32.* q 1 , q 2, q 3 . . . qm are n-dimensional vecors with has m < n . This set of vectors is linearly dependent. Q is the matrix with q 1 , q2, q3 • . . qm as the columns. the rank of Q is (a) less than n (c) 33.* (a) a unique solution (b) (b) m (d) between m and n (d) n [GATE-2007 (EE)] Let A be an n x n real matrix such that A 2 = I and Y be an n-dimensional v ector. Then the li near system of equations Ax = Y has 38.* The following system of equations x 1 + x2 + 2x3 = 1 , x 1 + 2x2 + 3 x3 = 2, x 1 + 4x 2 + ax 3 = 4 has a uni que solution . The only possi ble value (s) for a is/are (a) 0 (b) unique solution (c) more than one but infinitely many dependent solutions 34.* [GATE 2007 (IN)] (a) (b) either O ( or) 1 (c) one of 0, 1 (or) -1 (d) any real number other than 5 [GATE-2008 (CS)] 39.** Let P be a 2 x 2 real orthogonal matrix and x is 1 Q will have four linearly independent rows and four li nearly independent columns a real vector [x1 , x2F with length Then, which one of the following statements is correct? (c) QQT will be invertible (a) QT Q will be invertible [GATE-2008-EE; 1 Mark] (b) For what value of a, if any, will the following system of equations i n x, y and z have a solution? (c) 2x + 3y = 4 x + y + z = 4 X + 2y - Z = a (a) (d) Any real number I Pxll �11xl where at least one vector satisfies I Pxll < l xl I Pxll = l xl for all vectors X I Pxll � l xl where at least one vector satisfies I Pxll > l xl No relationship can be established between and l xl I Pxll [GATE-2008-EE; 2 Marks] (b) 0 (c) 40. 1 (d) There is no such value 36. 37. l xl = (xf + x� )2 , (b) Q will have four linearly independent rows and five linearly independent columns (d) 35. infinitely many dependent solutions If the rank of a (5 x 6) matrix Q is 4 , then which one of the following statements i s correct? exactly two distinct solutions [2008-EC; 1 Mark] (a) no solution (d) no solution (c) an infinite number of solutions The value of x 3 obtai ned by solving the following system of li near equations is X1 [GATE 2008-ME; 2 Marks] 2x 1 + The following simultaneous equations x + y + z -X 1 = 3 X + 2y + 3 z = 4 x + 4y + kz = 6 will not have a unique solution for k equal to (a) zero (b) 5 (c) 6 (d) 7 [GATE-2008-CE; 2 Marks] The system of linear equations 4x + 2y = 7 2x + y = 6 + 2x2 (a) -1 2 (c) 41 . 0 + - X2 2x3 = 4 + X3 X2 - X3 = -2 = 2 (b) -2 (d) 12 For the set of equations [GATE-2009 (Pl)] X 1 + 2x2 + X 3 + 4 x4 = 2 x 3 1 + 6x2 + 3x3 + 1 2x4 = 6 The following statement is true: (a) Only the trivial solutions x 1 = X = X 2 3 = Engineering Mathematics (b) There are no solutions. (c) (c) A unique non-trivial solution exists. (d) M ultiple non-trivial solutions exist. [GATE-2010-EE; 2 Marks] 42. 43. The value of q for which the followi ng set of linear equation 2x + y = 0, 6x + qy = 0 can have non­ triv ial solution is 3 (d) 47. X + 2y + Z = 4 (b) 7 (c) (d) X - y + Z = 1 11 [GATE-2010 (Pl)] The system of algebraic equation given above has Consider the following system of equations 48. (c) infinite number of solutions (d) five solutions [GATE-2011-ME; 2 Marks] The system of equations x + y + z = 6 + 4y + 6z = 20 x + 4y + lz = m X * 20 I * 6, m * 20 (a) I =6, m = 20 (b) I = 6, m (c) I * 6, m = 20 (d) [2011 -EC; 2 Marks] 1 3 1) and [1 a a2) where a . = -- + J ✓ an d J. = vr-;1 are 2 2 (b) orthogonal (c) (d) collinear The matrix [A] = [! [GATE-2011 (EE)] �1] is decomposed i nto a product of a lower triang ular matrix x[L] and an u p p er tr i a n g u l a r m at r i x [ U ] . T h e p r o perly decom posed [L] and [U] matrices respectively are (a) (b) [ : �1 ] and [ � -�] [! �1 (b) only the two solutions of (x = 1 , y = 1 , z = 1 ) and (x = 2, y = , z = 0). (c) infinite number of solutions. (d) no feasible solution. 1 ] and [ � �] The equation [ � =�] [;:] =[�] ;J (a) no solution (b) only one solution [ (c) non-zero unique solution has = �] [ [GATE-2013-EE; 1 Mark] 49.* The dimension of the null space of the matrix 0 1 1 1 -1 0 -1 0 -1 is (a) 0 (b) 1 (c) (d) 2 3 [GATE-2013 (IN)] (a) orthonormal parallel a unique solution of x = 1, y = 1 and z = 1. (d) multiple solutions has NO solution for values of I and m given by 1 (a) [GATE-2012-ME-PI; 2 Marks] (a) a unique solution (b) no solution 45.* The two vectors [1 05 � ] ] and [ � (a) 2 This system has 46. [! �3 [GATE-2011-EE; 2 Marks] 2x 1 + x2 + x3 = 0 X2 - x3 = 0 X 1 + X2 = 0 44. 1 [ : �] and [ � � ] 2x + y + 2z = 5 9 77 50.* Given a system of' equations : x + 2y + 2z = b1 5x + y + 3z = b2 W h i c h of the followi ng i s true regardi ng its solution ? (a) The system has a unique solution for any given b 1 and b2 (b) The system will have infinitely many solutions for any given b 1 and b2 (c) Whether or not a solution exists depends on the given b 1 and b2 (d) The system would have no solution for any values of b 1 and b2 [GATE-2014-EE-Set-1 ; 1 Mark] IES MASTER Publication 78 System of Equations 51.* The matrix form of the linear system �: = 3x-5y dy and - = 4 x + 8y Is dt 5 :t t} = (b) :t t} = (c) :t t} =[ :t t} =[ (d) 52. [! -s [! (a) 56.* W hich o n e of t h e follow i ng descri bes t h e relationship among the three vectors, T + } + k, 2i + 3} + k and 5i + 6} + 4k ? (a) The vectors are mutually perpendicular (b) The vectors are linerly dependent (c) The vectors are linearly independent (d) The vectors are unit vectors ] t} �5] {;} [GATE-2014; 1 Mark] 57. -5 ] : 8 {;} 2x + 3y = 5 3x + PY = 10 : �5] {;} [GATE-2014-ME-Set 1 ; 1 Mark] The system of li near equations [GATE-201 5-CE-SET-I; 1 Mark] 58. a unique solution infinitely many solutions no solution exactly two solutions [2014-EC-Set-2; 2 Marks] 53.* W h i c h o n e of the following descri bes t h e relationsh i p among t h e three vectors, i + } + k, [201 5-EC-Set-1 ; 1 Mark] 59.* Let A be an n x n matrix with rank r(O < r < n). Then AX = 0 has p independent solutions, where p is 2i + 3} + k and 5i + 6} + 4k ? (a) (b) (c) (d) The vectors are mutually perpendicular The vectors are linearly dependent The vectors are linearly independent The vectors are unit vectors Consider a system of linear equations: X - 2y + 3z = -1 x - 3y + 4z = 1, and -2x + 4y - 62 = k The value of k for which the system has infinitely many solutio ns is_ _ _ has (a) (b) (c) (d) For what value of p the following set of equations will have no solution? (a) r (b) n (c) (d) n + r [GATE-2015 (EE-Set 2)] 60.* If the following system has non-trivial solution px + qy + rz rx + py + qz The system of equations, given below, has + 2y (a) (b) (a) a unique solution (c) no solution (d) more than two solutions 55. (c) (b) two solutions [GATE-201 4 (Pl-Set 1 )] Consider the following system of equations: 3x + 2y = 1 4x + 7z = 1 X + y +z = 3 X - 2y + 7z = 0 The n u m be r of solutions for this system is [GATE-2014 (CS-Set 1)] 0 = 0 then which one of the following opti ons is TRUE? + 4z = 2 4x + 3y + z = 5 3x + 2y + 3z = 1 X = qx + ry + pz = 0 [GATE-2014 (ME-Set 1)] 54. n -:- r (d) p - q + r = 0 or p = q = -r p + q - r = 0 or p = -q = r p + q + r = 0 or p = q = r p - q + r = 0 or p = -q = -r ! !] [GATE-2015 (CS-Set 3)] 61 . I n the LU decom position of the m atrix [ if the diagonal elements of U are both 1 , then the lower diagonal entry 1 22 of L is ___ [GATE-201 5 (CS-Set 1 )] 62. The solution to the system of equations is Engineering Mathematics (a) 6, 2 (b) -6, 2 (c) -6, -2 63. 68.* Let c 1 , (d) 6, -2 [GATE-201 6-ME-Set-1 ; 1 Mark] n i=1 Consider the set of l i near equations Ax = b. z:>i n where A = (a; , . . . . . . , a n] and b = . The set of i=1 equations has (a) a unique solution at x = J n where J n denotes a n-dimensional vector of all 1 (b) no solution (c) infinitely many solutions [GATE-201 6-CE-Set-ll; 2 Marks] (c) Gauss-Siedel (d) LU decomposition [GATE-2016 (EE-Set 1 )] (d) X1 46 2 5 5 2 10 0 0 2 5 3 X2 1 61 X3 61 5 X4 30 81 0 0 0 4 2 3 2 5 X5 69.* Let u and v be two vectors in R2 whose Euclidean norms satisfy ! l u l l = 2llvl l . What is the value of a such that w = u + av bisects the angle between u and v? The number of sol utions of the sim ultaneous algebraic equations y = 3x + 3 and y = 3x + 5 is (b) 1 (a) zero (c) 2 70. 2. if m > n, then none of these systems has a solution. 3. If m = n, then there exists a system which has a solution. W hich one of the following is CORRECT? (a) 1 , 2 and 3 are true (b) only 2 and 3 are true (c) only 3 is true (d) none of them is true [GATE-201 6 (CS-Set 2)] (c) (d) -1 /2 1 T h e s o l u t i o n of t h e system of e q u a t i o n s X + y + Z = 4, X - y + Z = 0, 2 X + y + Z = 5 is (a) X = 2, y = 2, Z = 2 (b) x = 1 , y = 4, z = 1 (c) X = 2, y = 4, Z = 3 Z = 1 [ESE 201 7 Common Paper] 71 . 67.* Consider the systems, each consisting of m linear equations in n variables If m < n, then all such systems have a solu­ tion. (b) 1 /2 (d) x = 1 , y = 2, (d) infinite [GATE-2016 (Pl)] 1. (a) 2 [GATE 2017] The value (rounded off to the nearest integer) of x3 i s _____ [GATE-2016 (CH, EE-Set 1 )] 66. Finitely many solutions [GATE 201 7] 65.* A set of simultaneous l inear algebraic equation s is represented i n a m atrix form as shown below: 0 0 0 4 13 en be scal ars, not all zero, such that z: Ci ai = 0 where a ; are column vectors i n R n . Consider the fol lowing l i near 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 (b) 3a + b - c = 0 (a) 7a - b - c = 0 (c) 3a - b + c = 0 (d) 7a - b + c = 0 64.* W h ich one of the fol lowi ng i s a n iterativ e technique for solving a system of simultaneous l inear algebraic equations? (a) Gauss elimination (b) Gauss-Jordan ......, 79 If the system 2x - y + 3z = 2 X + y + 2z = 2 5x - y + az = b has infinitely many solutions, then the val ues of a and b, respectively, are (a) -8 and 6 (b) 8 and 6 (d) 8 and -6 [ESE 2018(EE)] (c) -8 and -6 72. ::J Consider matrix A = x= [ [ k 2� k ::] and v ector The number of distinct real values of k for which the equation Ax = 0 has i nfinitely many solutions is ____ [GATE 201 8 (EC)] IES MASTER Publication 80 System of Equations : .'·- 1. (a) 2. (a) 20. (d) 22. 3. 4. 5. 6. 7. .· 19. (See Sol.) (c) (c) 23. (d) 25. (a) 26. 10. (b) 28. 11. 12. 1 3. 14. 1 5. 16. 17. 18. 27. (See Sol.) 29. (See Sol.) (b) 30. (b) 31. (b) 32. (b) 33. (c) 34. (b) 35. (b) 36. 37. (c) (b) 24. (b) (c) 21. 8. 9. ·· - - �· A ..N S· W E. R J<.EY 38. (b) (b) 41 . (d) 42. (b) (b) 45. (b) (d) 46. (b) 47. (b) 48. (1) 57. (4.5) 58. (c) (c) 55. 56. (b) 43. 44. (d) (d) 39. 40. (a) (b) (c) 61 . (5) 63. (b) 62. (d) (c) (d) (2) 59. 60. (b) (b) (c) (d) 64. (c) 65. (15) 66. (a) (a) 49. (b) (b) 51 . (a) 69. (a) 53. (b) 71. (b) (a) 72. (2) 50. (a) (a) 52. (d) 54. (b) 67. (b) 68. (b) 70. (c) (c) (d) Sol-1 : (a) or [i ;� �][::] where = 0 [i ;; �] tv<. = 0 M atrix A = R2 ➔ R1 -R1 R3 ➔R3 - 2R1 . . .( 1 ) A ~ [� �1 �l r� Rank of matrix A = 2 & order of A = 3 so Homogeneous system (1 ) have (3 - R = 1 ) linearly independent solutions. Sol-2: (a) [A : B] = �1 � 1 -1 1 �] 1 Engineering Mathematics ⇒ 1 1 -2 -1 -2 0 a = 3 1 : : -2 - 1 3 0 1 59 5 Sol-5: (c) The two vectors X1 & X2 are said to be orthogonal if 1l rank (A) = rank (A : B) = 3 : X1 .� = 0 mm: Let X1 '." (1 0 2]T and X2 = (0 5 O]T unique solution Alternative Sol. Method II = 1 (-1 - 0) - 1 ( 1 - 0) + 1 (-1 + 1 ) = -2 ct- 0 ⇒ Unique solution ⇒ [� : �I�l For orthogonal vectors : 1x0+°'5+2x 0= O a.b = 0 i.e. if b = X2i + y2i + 22k & Then: x 1 x2 + y1 y2 + 2 1 22 = 0 Option 'c' satisfies the condition. m AA = B = X1 .X2 = So, 1 1 1 1 -1 0 1 -1 1 IAI = Sol-3: Given 81 Sol-6: (c) In gauss elimination method for solving system of linear algebraicc eqation, the equation are reduce to an equivalent upper-triangular matrix. Sol-7: (b) . . .( 1 ) R3 [� �! ;; ][;] [-�7] ➔ 1 3R3 - 9R 2 0 0 -57 z = 1 14 According t o option (b) We can take m = n & B = 0 (Echelon form) ⇒ -57z = 1 14, 1 3y + 282 = -1 7 and x - 3y - 72 = 6 On solving, 2 = -2, y = 3 & x = 1 Sol-4: (d) 2x + 3y = 1 . . . (i) X-y = 4 . . . (ii) 4x - y = a ... (iii) If I A I * 0 , system have unique solution, if I A I = 0 system have infinite solutions. Hence, option (b) in wrong because condition of unique solution is not mentioned. Sol-8: (a) 1 .5 -0.5 0 IAI = From equation (i) and equation (ii) X = 13 5 1 y=- The solution of equations exists 7 5 4 2 7 1 3 5. = (1 .5)(1 0 - 3) + (0.5)(20-2 1 ) ⇒ = 1 0.5 - 0.5 = 1 0 ct- 0 Unique solution. IES MASTER Publication 82 m System of Equation s Sol-9: [A] = Matrix [: l [: ;] [� :1 1 8 1 � [� = [L] [ U] 27 3 1 0 [L] = Let -2 0 -2 � 1 A e [ U] = and [A g 1 0 ah ag bh + ci ( L] [UJ = [ : bg + c dg + e dh + ei + f On comparing with matrix A, i.e. l A = L. U. ⇒ a = 2, b = 4, d = 6 ag = 1 ⇒ ⇒ h = 5/2 bg + c = 8 ⇒ ... (i) 1 B X Augmented matrix [A C = [ 1:1 : BJ = 1 3 I 3 ! 11 [ � -2 1 i 1 -2 0 R2 ➔R2 -R1 R 3➔R3+R2 ·: (i) & (ii) are equivalent so g = 1/2, ah = 5 ⇒ l BJ [� [� -2 1 2 2 -2 1 -2 1 .2 2 0 3 ;1 ... (ii) :1 c = 6, bh + ci = 13 = 1/2 dg + e = 27 ⇒ e = 24, dh + ei + f = 31 ⇒ f = 4 . -. The matrix [L] and (U] are [L] = [U] = and Sol-1 0: (b) [: [� 0 6 24 1/ 2 1 0 5 ;1 / 21 Consider A3x3 then [A : Bbx4 , i.e. , Augmented M atrix have 4 vectors. But by property of rank, we have p ( A : B) � 3 which is definitely less than 4 so [A B] : has LD set of vectors so we can conclude that matrix [B] must be linearly dependent of columns of matrix [A] . Sol-1 1 : Gauss elimination method ⇒ Given ⇒ Ox - 2y + z = 1 X = 2, y = 1, Z = 3 l: : ml l:�l AA = B = Augmented M atrix [A / BJ = [ � ~ [� ~ [� � ➔ R2 - �; R3 ➔ R3 - � [A R2 B R3 X - 2y + 2 = 3 x + 0y + 3z = 11 ⇒ 2y + 2z = 8 ] 3z = 9 Sol-1 2: (b) 1/ 2 1 x - 2y + z = 3 BJ 2 [A : BJ 0 0 2 8 6 0 0 2 0 0 0 0 0 3 0 3 l �al 5 12 15 3 �8] ⇒ :. [� p(A) = 2, p(A : B ) = 3 p(A) * No solution exist. p(A : B) Sol-1 3: (b) By using standard definition of rank. ⇒ X + y = 2 . . . (1 ) 2x + 2y = 5 5 X + y = . . . (2) 2 In equation (1 ) and (2) sum of two variables gives two different values. r(A) * Hence, no solution. Sol-1 5: (b) Using R1 B R2 [A : BJ = R2 ➔ R2 - 3R 1 R3 ➔ R3 + 2R2 u - [i 2 R 3 ➔ R 3 + - R2 5 -1 1 5 -2 -2 4 1 -1 1 0 5 -2 16 0 0 5 :] Note: From third matrix, it can be easily seen that (y p(A) = pl[A : B) = 3 = Number of unknowns . - . Unique solution. Sol-1 7: (b) [� l -i I [ l Ri B R3 2 3 1 � :] -8 -1 2 - 4 Now, R2 ➔ R2 -4Ri ; R3 ➔ R3 -2Ri - [ 8 � :5 :20 0 -3 12 a -14 For infinitely many solutions, unknowns) ⇒ 1 - 5a = 0 ⇒ _; 3 l 7 -23 1 - 5al p(A) = p(A : B) < 3 (No of a =1/5 Hence for only one value of a = 1 / 5 , the system will have infinitely many solutions. Sol-1 8: (b) In case of homogeneous system AX = 0 Hence option (b) is correct. l 2 1 I 6 -3 0 i -6 -1 o -1 using � -4 -1 2 -8 If A is singular matrix AX = 0 wil l have non trivial solution. Consider the augmented matrix [A : bJ = [ -[ [A : BJ = 1 3 2 If A is non-singular AX = 0 will have trivial solution. Sol-1 6: (c) 2 1 6 1 2 : 6] 1 1 5 n = 2 & y = 1 ) which is impossible (Trick for examination) -�] 2 -2 41 5 l * r (A : B) -1 1 2 1 0 2 1 R ➔ - - R2 2 , 2 3 � (-1)R R ➔ 1 3 3 � rank (A) = 2 rank (A : b) = 3 Hence, no solution. Hence, system has no solution Alternative 6] 83 1 2 � : R ➔R -R 0 ~ [ 2 3 3 0 0 0 -1 Sol-14: (b) & Engineering Mathematics R2 ➔ R2 - 2R1 R 3 ➔ R 3 - R1 Sol-19: (c) [A : BJ = [� 2 3 2 3 4 3 1� l IES MASTER Publication 84 System of Equations R2 ➔ R2 - R1 ; R3 ➔ R3 - 2R, R3 ➔ R3 + 2R2 [i ~ [� 2 3 1 -2 -3 2 0 3 -1 �] !] Case II : Now if ir(A) = 21 , then it is possible that r(A : b) = 3 and hence system is inconsistent. Sol-23: (d) ·: Option (a) & (b) are partially correct and option (c) is totally wrong . Sol-24: (a) The matrix equation Px = q will have a unique solution in rank (P) = rank [P:q] = n where n, is order of matrix P & system will have infinite many solutions in rank (P) = rank [ P:q] < n . . .( 1 ) p(A) = 3 , p(A I B) = 3 & No. of unknown = 3 So unique solution exist. Sol-25: (b) Now given system is equivalent to (1 ) so Ax = B 2 -1 3 : 1 [A : B] = [ 3 2 -1 4 5 : 2l 1 : 3 3 2 R3➔(-R3 } & R1 � R2 On comparision z = -4, y + z = 2 & x + 2y + 3z = 6 or (z = -4, y = 6, x = 6) Sol-20: (c) [ : �� -1 -5 [0 43 0 [A ; BJ ~ [ � -1 5 [0 4 0 0 R3 ➔ R3 - 2R2 [A ; BJ ~ [ Here p(A) = p(A I B) = 2 No. of unknowns. :. [� R2 ➔ R2 - 2R1 ; R3 ➔ R3 - 3R1 1 [A BJ R1 �R3 p(A) U nique solution exists. = � -4 -1 7 5 14 8 -4 -1 7 5 0 -2 -4 -1 -1 3 2 5 3 �] - :i 11 -n �3 ] -3 3 = p(A I B ) = 3 = no. of unknowns. Unique solution exists. Sol-21 : (b) 5 2 -1 3 [1 -4 -1 Sol-26: (b) ·: A3 x4 then [A : Bhx 5 and p(A) :<:; 3 and p(A : B) :<:; 3 . B u t AX = B i s a n i n co n s i stent system p(A) * p(A : B) . ⇒ Hence, only possibility is that p(A) < p(A : B) ⇒ Maxp(A) = 2 & Maxp(A : B) Sol-22: (b) = 3 A3x real matrix & Ax = b is inconsistent If r (A) 4 r (A : b) * Case I : if r (A) = 3 (Max. possible) then r (A : b) 3 & system becomes consistent (which is not possible). = p(A) = 2 and order = 3 and in case of homogeneous system No. of LI solution = order - Rank. :. Number of independent solutions = 3 - 2 =1 Sol-27: (d) 4y + 3z = 8 2x - z = 2 3x + 2y = 5 Eliminating z from (i) & (ii) 4y + 3(2x - 2) = 8 ⇒ 3x + 2y = 7 ... (i) . . . (ii) . . . (iii) ... (iv) (iii) & (iv) imply that the given system of equations is inconsistent. Engineering Mathematics Sol-28: (d) A = Given R 3 ➔R 3 -R 1 li �1 85 2 2 2 2 - X 1Y 1 -X 2Y 2 -2X 1X2Y 1Y 2 0 1 1 0 0 0 0 2 2 2 2 = X1 Y2 +X2Y 1 - 2X1X2Y1Y2 I':,, = (X 1Y2 -X 2Y 1) 2 = ⇒ For linearly independence {O, when x & y are LO >O, when x & y are LI t:,, is positive. Sol-31: (a) The augumented matrix for the given system of equation is R 3 ➔R 3 -R 2 A ~ l� [A IB] = 0 1 0 (Echelon Form) 0 -1 ] 0 0 R 2 ➔R 2 -R 1 R 3 ➔R 3 -R 1 p(A) = 4 So, 1 R 2 ➔-R 2 ~ 2 Sol-29: (b) From the previous ·example p(A) = 4, p(A: B) = 4 and number of variables =4 p(A) = p(A: B) =No. of variables i.e. So, unique solution exist i.e. x must be unique. Sol-30: (b) For the sake of simplicity we will solve this question in 20. Let x.x x.y I':,, = l y.x y.yl Let X = X 1i +X 2 j I s 1 : 2 a- 1ib-5 j For infinite number of solutions a = 2, b = 7 (two linearly independent equations in three unknowns Sol-32: (a) Now ·: m < n so we can conclude that P(Q) < n also. Sol-33: (b) ⇒ ⇒ ⇒ y.y = yf +y� X 1Y 1 +X 2Y 2 1 p(Q) < m (No. of vectors) x.y = X 1Y 1 +X 2Y 2 IA I 1 But given that q 1 ,q 2 ,q 3....q m are dependent vectors 2 2 x.x = x1 +x 2 x 21 + x 22 1 1 2 2 1 a-1 b;S] (By property of Rank) we have p(Q) :-:; m y = Y1i + Y 2 i I = [! [� [� 1 1 J 5] 3 3:9 2 a1b X1Y 1 +X 2Y 2 1 yf +y� ⇒ A = A-1 A is invertible i.e. A-1 exists. A is non-singular Hence for the system AnxnXnx1 = Ynxi unique solution exist. 2 = (xf + x�)(y f +Y�)-(x 1Y 1 +X 2Y 2 ) 2 2 2 2 2 2 = x 1 y 1 +x 1 y 2 +x 2y21 + x 2 y22 AA = I Sol-34: (a) Rank 'A' of a matrix indicates maximum number (A) of linearly independent rows and columns. IES MASTER Publication 86 System of Equations Sol-35: (b) [A : BJ= R2 ➔R2 - 2R1 R3 ➔R3- R1 1 1 3 0:4 4 ] 2 -1:a -[i R3 ➔R3-R2 1 1 1 -2 1 -2 1 1 1 [0 1 -2 0 0 0 fI a= 0, then r(A)= 2 [� (1) ⇒ No solution .�] �] Hence, for infinite sol., Sol-37: (b) 1 1 0 1 2 K-7 p(A) = p(A:8)<3 ⇒ * rank No solution unk nown= 3 ⇒ * a 5 3�] �i : : 0 p(A) = p(A: 8) = No. of :. For all values of a except a=5, the system will have unique solution. Sol-39: (b) ⇒ K=7 Let an orthogonal matrix cose sine p= [-sine cose] So cose sine x 1 PX = [ -sine cose][x 2 ] x 1cose + x 2sine = [-x sine+x cose] 2 1 -[: : il 4 x + 2y = 7 2x + y = 6 � [� � � 0 0 a- 5 To have unique solution 2 II Pxll = J(x 1cose + x 2sin0) + (-x 1sin0 + x 2cos0) (A : 8) Alternative Sol : [! ! � !] 0 3 a-2 ~ 4 2 [A : BJ -- [ 2 1 rank (A) 2 ~ [� � 8) consistent. [! :] [ � !] - [� !] 7 AX.= 8 [AB] = 1 1 2 1 3 K-1 R, ➔ R, - 3R 2 Sol-38: (d) Given 1 1 2 3 4 K [ A : BJ= R2 ➔R2 -R1 R3 ➔R 3 - R1 2x + y = Not Possible ⇒ r(A: 8) = 2 ⇒ r(A) = r(A : Sol-36: (d) ⇒ = ... (1) ... (2) ⇒ xfcos 2e + x�sin 2e+ 2x 1x 2cosesin e +x�sin 2e + x�cos 2 0 - 2x 1x 2sin 0cos 0 = �= llxll IIPxll = llxll for any X 2 Engineering Mathematics Sol-40: (b) Sol-44: (b) 2 1 [A : BJ = [; -1 [A: B J = R 3 ➔R3+R2 [A : B J . = So, AX= B Sol-41: (d) � 2 -3 3 -2 5 -3 2 -3 0 -2 5 2 [A : = * rank (A: B) rank (A) = Let x3 =-2 As rank of matrix A =rank of matrix (A/ B) =1 < number of variables (4), so given set of equation have infinite solutions. Sol-42: (c) Pv(= 0 = [� ] For non-trivial solution, I A I= 0 q= 9 Sol-43: (c) Consider A = [L ⇒ I! �I= ⇒ 0 �1] IA I = 2( 0+1) + (-1- 1) = 0 146 µ- 6 l 1 1 3 5 0 11,-6 146 : µ- 2 0 A, =6, µ-ct-20 Sol-45: (b) -1 0 -2x 3 = -4 [! �][�] 1 3 5 3 A.- 1 A.- 6= 0 and µ- 2 0 -ct- 0 [A J [XJ = [ B J ⇒ . 6] 1 1 1 [1 4 6 : 2 0 1 4 "- . µ -[� - [� for No Solution ✓3 a = _ _.!.+i = w (Cube root of unity) 2 2 X 1.X 2 = 1x1+1xw+1xw 2 Now .. BJ_ = R3 ➔R3 -R2 ⇒ [� Tl[::1 [ �- ] � 2 -3 0 -3x 2+5x3 Hence, [ [ -:] -2 1 -1 87 ⇒ The homogeneous system of equations has infinite solutions. =1+w+w 2= 0 Hence they are orthogonal. Sol-46: (d) Matrix, [AJ = Let [LJ = and [UJ = ⇒ ⇒ [! [! a=2 b=4 ]= �1 ] = �1 [! [: �] [� �] [: �][� �] ] = [LJ [UJ �1 (using Crout's Method) [: ad=1 d= _.!. 2 ad ] bd +c bd+c = -1 2+·c =-1 C=3 IES MASTER Publication 88 System of Equations L = [: Sol-47: (c) u= Sol-51: (a) �3] dx = 3x - 5y dt 2 \] [� The augmented matrix for the given system of algebraic equation is given by ! 1 2 1 4 1 2: 5 [A : BJ= ] [� -1 1 l 1 dy = 4x + 8y dt matrix from is : Sol-52: (b) 2 1I 4 R ➔R3 -R1 : -3 0 1 -3 . 3 = ' [ -3 0: 3 l R2 ➔R2 - 2R1 1 :� = [ � =3 o R1 �!o\ o � 31; R3 ➔ R3 - R2 ⇒ = [; � 1 3 0 1 2 5 [�] Sol-49: (b) = Order -Rank = 3 -2 Sol-50: (b) R 2 ➔ R 2 - 5R1 ⇒ [A :BJ= [; 1 r(A) Let 2 2 1 3 2 rank (A) = rank (A:B) = 2 < 3 l l Infinite many solutions 2 : x 1= ( i + j +k ) = 1[ 1 1]' x 2 = (2T +3J +k) = [2 3 1]' X3 = (5T+ 6T +4k) = [5 6 4]' b1 b 2] b1 = [ 0 9 -7 : b 2 - 5b ] 1 = r [A : BJ -46 -23 Sol-53: (b) = 1 X+ 2y+ 2z= b 1 5x + y+3z= bz ⇒ 14 : 14 5 -14 : -46 0 : 0 Multiple solutions. Nullity = Dimension of Null space 14 -4 ] 5 � 2 5 0 1 1 3 [12 --12]= 0 �i 14 5 2 _ = [� 5 -14 0 -3 -7 Sol-48: (d) IA I = [! -:J{�} = [� ⇒ The system, is consistent and has infinite solutions. [� =�][::] = (A: B] :. Rank [A : BJ= Rank [ AJ= 2 < no. of unk nowns (3) of algebraic equation. = :t {�} = 2 < 3 ( No. of unk nowns) ⇒ Infinite many solutions for any given b 1 and b 2. To check linear dependency, letus consider the given vectors as columns of matrix (A) . i.e. A = [x, x, x,]�[i ! :] A I I= o So given vectors are linearly dependent. Engineering Mathematics Sol-54: (a) Let AX. = B [A: B] = [1 [A: B] - 2 4 : 2] : 5 4 3 1 3 2 3 : 1 -[: : R 4 ➔R 4 -R 3 4 -15 �31 -5 0 -4 -9 : 2 1 2 4 - [ 0 5 -15 : - 3 l 0 0 15 : -1 3 [ A : B] - [A·. BJ = R 1 BR 3 2 0 1 0 7 1 1 1 1 3 1 -2 7 0 [! 15 21 1 1 -1 - 3 8 0 1 5 21 0 0 0� l 1 1 1 IAI = 2 3 1 5 6 4 = 6 - 3 + (-3) = 0 ,I linearly dependent. Sol-57: (4.5) Given system of equations has no solution if 2 3 - ⇒ [A: B] = 3 R 2 ➔R 2 --R 1 2 when 3 5 *­ p 10 p= Method II : [A: B] = 0 Sol-56: (b) ⇒ Sol-55: (1) 1 3 1 -3 -8 0 15 21 Hence, unique solution exist. Rank A= Rank of [A: B] = 3 = number of variables. Hence unique solution exists. 1 :. p(A) = p(B) = 3 = no. of variables. = Echelon form ·: [� [� 89 C9l t..:lj' * - [! [: 3 : 5 p : 10] 5 3 (p-J) : 2.5 l p(A) = 1 & p(A: B) = 2 i.e., p(A) p(A: B). No solution. Sol-58: (2) : -1] 1 -2 3 [A: B] = [ 1 � 3 4 : -2 4 -6 : 1 K IES MASTER Publication 90 ⇒⇒ System of Equations The system has infinite many solutions. rank (A) �rank (A: B) <n K - 2 =0 K = 2 Sol-59: (c) The system AX = 0 has (n - r) linearly independent solutions; where'n' is number of variables and r' ' is rank of'/\. For homogeneous system AX= 0 , I p =(n - r) Nullity = No. of independent solution =order -rank So Sol-60: (c) ⇒ [; � p =n - r Sol-62: (d) ⇒ ⇒ �]x • o P q r q r P r p q t 1 q r (p +q+r) 1 r p 1 p q = o ⇒ (p +q +r) [p2 + q2 + r2 - pq -qr -pr] = 0 (p + q + r) [(p - q) 2 + (q - p)2 + (r - p)2] = 0 or p =q =r p + q + ( =0 Either Sol-61: (5) A = LU ⇒ [� !] = [::: l�J[� I 9 =4 X 1 + 1 22 1 -30 -5 2 ][ _! 0] [; ] = 2�[:4 _1 6+15 0 ] 2 6 [ 8 - 60 = 6 1 15 6 =[ ] [ 2 6 -5 2 -2 l X = 6, y =-2 [� :�][:] -3l [ l [ -3 R2➔2R1-R2 R3➔5R1-R 3 [� : a 9 2a -b - 9 5a -c 1· 2 + 0 1 -0 0 = a -9 : 2a-b 0 : 3a+b -c l for consistency p(A) = p(A: B) which is possible only ⇒ when ; I3a+b -c Sol-64: (c) =0 I Rest three methods are direct methods. Sol-65: (15) 0 0 0 4 13 4 6 2 1 0 1 61 [A : B] = 0 0 2 5 3 61 0 0 0 4 5 30 81 2 3 2 1 5 2 5 5 2 U ] ; Comparing the corresponding elements on both sides, 2 = 1 11 ⇒ -+: 122 [; ] = -4 l 3 � =0 u2 1 = 1 = Sol-63: (b) p+q+r q r p+q+r r p p+q+r P q 12 ⇒ [}4 !] [;] [-�O] [2 5]- [2] ⇒ =0 U 1 22 =5 = 111 or ⇒ 9 =l 21 · AX = 0 So for non-trivial sol. IA I =0 i.e. 2 = I 1 1'u 2 1 4 = 1 21 R 1 BR 2 Engineering Mathematics [A : B J ~ R 2 �R s 2 5 5 2 1 0 1 61 0 0 0 4 2 3 2 1 5 5 2 5 5 2 2 3 2 0 0 2 5 1 0 1 61 0 0 'Since the scalars are not all zero so linear combination exist. Hence, the column vectors a i for i= 1, 2, ... , n are linearly dependent. 30 81 ⇒ IA I [A : BJ= ( Echelon Form) AX= B So 2 Ax= b has infinitely many solutions. 2 5 5 2 1 0 1 61 0 0 0 0 0 4 0 5 8 0 -2 -3 -1 -5 - 81 0 0 2 5 3 61 0 0 0 0 0 0 0 2 0 0 ⇒ 5 4 0 3 5 8 X1 Xz X3 X4 X5 then we can assume u= (�) and v=(�) and let w 0 4 5 30 0 4 13 4 6 0 0 5 5 2 10 -2 -3 -1 -5 0 ·: llull= 2llvll 61 61 0 0 R 2 ➔R2 -R 1 ; R s ➔R s -R 4 30 16 1 61 - 81 = 61 30 16 On comparison x 5 = 2, x 4 = 5, x 3 = 15, x 2 = 1 0, X1 = 3 Sol-66: (a) Statement-2 is not correct because over determined system can also have unique solution or infinite solution. Statement-3 is correct because equally determined system have all three possibilities i.e. may have either unique solution, or infinite solution or no solution. ✓ 4 2 4+ a 2 ✓ a 1· 4+ a 2 ·: W is the angle bisector between u and v. ⇒ So, ✓ cos 0 1 = cos 0 2 4 2 4+ a Sol-70: (d) [A : B J = = 2 ✓ a = 2 [ ; ;] [� �] :1 [� 1 -1 1 1 -2 0 0 -1 tI is in Echelon Form. So, So ⇒ AX = B -4 x +y +z= 4 -2y = a 1- 4+ a 2 1 -2 0 -1 -1 Mathematically it is a contradiction. Statement-1 is not correct because under determined system may have no solution. cos 0 2 = and -3x + y = 5 Sol-67: (c) then u-w=4 and v-w = a lvllwlcos 0 2 = a COS 0 1 = -3x + y = 3 So no solution exist. (!) = lullwlcos 0 1 = 4 and Given simultaneous algebraic equations and ⇒ =0 Sol-69: (a) [A : B J = ⇒ Sol-68: (c) 0 0 4 13 4 6 0 2 5 3 61 5 3 91 R2 ➔R2 -R1 R3 ➔Rr2R2 R3 ➔R3 -_!R 2 2 -1 ⇒ [ � i � J [} [ � l -z = -1 X= 1, y= 2, Z= 1 IES MASTER Publication 92 System of Equations Which is in Echelon form. Sol-71: (b) [A Now for infinite solution -1 3 BJ = [ 1 2 ; -1 a p (A) = p (A : B) < 3 ·⇒ a =8, b =6 Sol-72: (2) k A = [ 2 k -k 2 1 1 -1 [A : B] ~ [0 -3 0 -6 a-10 R3 ➔ R3 - 2R2 , 2 1 1 0 -3 -1 [A: ] B ~ [ 0 0 a- 8 2 -2 b -1 0 2 l -2 b- 6 l :�] and given system is AX =0 We k now that homogeneous system have infinite solution if IA I = 0 ⇒ ⇒ k k 3 3 - 2k (k - 2k k 3 2 - k) + 2k 2(2 2 . = 0 = 0 -k) = 0 k = 0, 0, 2 Hence k have two distinct values. ■ DEFINITION Let A = [aiil nxn be a square matrix of order n. Consider the homogenous system of equation AX= AX or (A-Al)X=O ...(1) where I is an indentity matrix of order n and J, is a scalar. We know that homogenous system of equation (1) always has a trivial solution. Here our concern is to find the values of Jc, for which non-trivial (non-zero) solutions of the homogenous system exist. a11 -A a2 1 A -Al = is called a characteristc matrix of A, and the determinant a1 1 -A a21 det A = IA-All = a12 a22 - A a1 2 a22 -A a1n a2n a1n a2n is called the characteristic polynomial of A. Also the equation IA-AIJ = 0 is called the characteristic equation of A. In this chapter we are considering a homogenous system as follows AX = AX AX - AX = 0 (A-H)X Non zero solutiion X = Xl x2 = ... (1) 0 of (1) is called eigen vector or characteristic vector. 2. If X is a non-zero solution of (1) then KX is also a solution of (1) where K is any scalar. 3. Non zero solution ⇒ Non trivial solution equivalent statements. ⇒ Infinite solution ⇒ LD Vectors i.e. all are 94 Eigen Values and Eigen Vectors 4. (n-r) gives the value of the no. of LI solutions / LI Eigen vectors of above system (1). 5. If two eigen values are same and .:3 only one LI solution then there w i l l be same eigen vector for both the eigen values. 6. If two eigen values are same and 3 only two LI solution then there w i l l be different eigen vector for each eigen values. 7. Roots of a characteristics equation are known as eigen values or chacterstic values, or eigen roots or characteristic roots or Latent roots. 8. Set consisting eigen vectors corresponding to different eigen values is LI. 9. The set of eigen values of the matrix A is cal led the spectrum of the matrix A and the largest eigen value is cal led the spectral radius of A. ALGEBRAIC & GEOMETRIC MULTIPLICITY Algebraic Multiplicity : Number of times a particular eigen val ue repeats is known as its algebraic multiplicity. Geometric Multipl icity : Number of distinct LI eigen vectors f or a particular eigen value is known as its geometric multiplicity. PROPERTIES OF EIGEN VALUES & EIGEN VECTORS 1. Eigen val ues of a hermitian matrix are all real. 2. Eigen values of skew hermitian matrix are either 0 or purely imaginary. 3 . Eigen values of a orthogonal matrix are of unit modul us. 4. Eigen val ues of an ldempodent matrix are either zero or one. 5. Eigen values of a Unitary matrix are of unit modulus. 6. Eigen values of an lnvolutary matrix are either + 1 or -1. 7. If A and B are two matrices of same order then A and B have same eigen values where B = p-1 A P. OR Similar Matrices have same eigen val ues. 8. If 'A. is an eigen val ues of A then, (i) 'A. -1 is an eigen value of A-1 (ii) If 'A. 1 ' 'A. 2 , ;\,3 . . . 'A.n are Eigen val ues of A then Eigen val ues of A -1 are 0) IAI * 1 1 1 1 ( Pr ovided 9. Eigen value of adj A = ( ':' ) 1 0. The matrix A and its transpose A ' have the same characteristic roots. 11. If 'A. 1, 'A. 2 , . . ., 'A.n are the n-eigen val ues of square matrix A, then (i) Eigen val ue of kA are k'A.1 ' k'A. 2 , • • • , k'A. n 1 1 1 . (·1 ·1) E 1gen vaI ue of A-1 are i•�·· ··· � 2 (iii) Eigen value of A2 n are 'A.�, 'A.�, .. ., 'A.� (iv) Eigen val ue of A 3 are 'A. �, 'A.�, ..., 'A.� 1 2. Any two characteristic vectors corresp onding to two distinct characteristic roots of a unitary matrix I or real symmetric matrix A are orthog onal. Engineering Mathematics 95 (i) Product of eigen values of a matrix is equal to determinant of that matrix. Theorem 1 (ii) Sum of e igen values of a matrix is equal to trace of that matrix. OR Let A be a matrix of order n and let A- 1 , A-2, A 3 .. · "-n are the eigen values of A then = IA I (ii) "-1 + A2 + A 3 . . · "- n = Trace (A) (i) "- 1 , "- 2 , "- 3 .. , An Consider a 3x3 matrix A as follows : A = [ ::: ::: ::: ] a 31 a 32 a 33 Proof 1 Characteristic equation is ; IA - A.I I = 0 a12 (a 1 1 - A.) a2 1 (a 22 - 1c) a31 a 32 = 0 ... (i) I I So let 1c1 , 1c2 , 1c3 are the roots of equation (1 ), i.e . these are eigen values of A. d a b Sum of Roots = a Product of Roots = ⇒ i1 • A2 . 1c3 = _( - IA I ) = - IA I 1 a +a + a ) I ⇒ "-1 +A-2 + A-3 = ( 1 1 22 33 1 = a1 1 + a22 + a33 = Trace of (A) . So , in general we can say that. (i) Product of e igen values is equal to IAI. (ii) Sum of eigen value is equal to trace of A. INorE ( For the equation, ax 3 • bx2 • ex + d a O; Sum of roots a - � Product of roots taken two at a time = � and Pr.oduct of Roots = - a Theorem Proof a 1 d a The character ist ic roots of triangular matrix are just the diagonal eiements of the matrix. Let the triangular matrix be given as A = a 1 1 a 1 2 a 13 0 a 22 a 23 a 2n ,r a 1n 0 0 a 33 a 3n 0 0 0 a nn n Then its characteristic equation is I A- 1cl l = O⇒ a 1 1- A a 12 a 13 a 1n 0 a 22 - A a 23 0 0 a 33 - A. a 2n 0 0 0 a 3n a nn - A, =0 IES MASTER Publication 96 Eigen Values and Eigen Vectors (Expanding along first column) These a re the diagona l elements of the matrix A. Example 1 1 Find the cha racte ristic roots and the spectrum of the matrix [� �] 2 A= -4 0 Solution 1 2 3 1 - 1,., We have IA - 1,.,11 = 0 -4 - 1,., 2 = (1 - 1,.,) ( -4- 1,.,) (7 - A,) 7 - 1,., 0 0 Now the cha racte ristic equation of the matrix A is given by I A - 1,.,11 = (1 - 1,.,) ( -4 - 1,.,) ( 7 - 1,.,) = 0 1,., = 1 , -4, 7 Hence , the cha racte rstic roots of the matrix A a re 1 , -4, 7, while the spectrum of the matrix A is {1 , -4, 7) Example 2 1 z �] Determine the characteristic roots and the corresponding cha racte ristic vectors of the matrix. A • [� -6 7- A, �4 -4 3- 1,., Solution 1 ]=-A.3 2 +1 81,., - 451,., The cha racteristic equation of the matrix A is given by ⇒ ⇒ I A - 1,., II = -A.3 2 +1 8 1,., - 451,., = 0 -1,., (1,., - 3)(1,.,- 15) = 0 1,., = 0, 3 , 15 Hence, the characterstic roots of matrix A a re 0, 3 , 15. Now, the cha racte rstics vecto r of the matrix A corresponding to the cha racteristic root the non-zero solution of the equation (A - U) X = 0 ⇒ When 1,.,= 0, the corresponding cha racte ristic vector is given by 1,., is given by , Engineering Mathematics 97 On reduc ing to Echelon form using row transformation, we get : ⇒ [� TE] [� -�ol : l [� �l:l 10 -5 = 0 --4 By R 1 ➔R 1 - 4R 3 and R 2 ➔ R 2 + 3R 3 ⇒ --4 -1 10 1 By R 1 �R 3 , R 2 ➔ -R 2 5 ⇒ -4 1 = 0 = 0 0 By R 3 ➔R 3 -1 0R 2 ⇒ ⇒ 2x 1 = 4x 2 - 3x 3 = 4k - 3k = k Hence ⇒ x = , [Htl ⇒ This is characteristic vector of A corresponding to the characteristic root 'J.. = 0. When 'A,=3, s ubstituting in (i), the corresponding Eigen vecor is given by ⇒ [� �][::] [� ][::] -6 4 --4 = 0 Reducing Echelon form using row transformation , we get ⇒ Thus , we have 2 -8 0 = 0 � x 1 + 2x 2 + 2x 3 = 0, - 8x 2 - 4x 3 = 0 i.e. Let X 1 + 2x 2 + 2X3 = 0 & X3 = -2x 2 x2 = k , x 3 = -2k Hence, we get x 1 = -2x 2 - 2x 3 = - 2k + 4k = 2k so x, = [: Hil{�l IES MASTER Publication 98 Eige n Values an d Eige n Ve c t ors This is characteristic vector of A corresponding to the characteristic root 'A, = 3. When 'A, =15 , substituting in (i) , the corresponding characteristic vector is given by ⇒ [=; =: : ][::] = 0 2 - 4 -12 X 3 On reducing to Echelon form using row transformation, we get ⇒ X 1 - 22x 2 - 4 6x 3 = 0, - 2 0 X2 - 4 0x 3 = 0 Let x 3 = k , x 2 = -2k , Now x 1+4 4k - 4 6k = 0 Example 3 1 Solution 1 ⇒ This is the characteristic vector ⇒ X2 = -2 X3 x 1 = 2k A corresponding to the characteristic root 'A,= 15. Find all the eigen values and eigen vectors of the matrix. -2 2 -3 [ A= 2 1 - 6 ] - 1 -2 0 Characteristic equation of A is given by ⇒ ( 'A,+ 3)( 'A,+3)( 'A,- 5 ) = 0 ⇒ Hence eigen values are -3, -3, and 5. When i- 1 -3 -2- 'A, 2 2 [ 1- 'A, - 6 - 0 -2 0- 'A, -1 . ..(i) 'A,= -3, -3, 5 A, = -3 We have from (i) ⇒ x 1 + 2x 2 - 3x 3 = 0 Now let x 2 = k 1 and x 3 = k 2, so X= When 'A,= 5 we have from (i) 3 1<, - 2 k, ⇒ 3k ' X= [ t ' ] Engineering Mathematics 99 Solving, we get ⇒ X2 = 2x 1 , x 3 = -X 1 X2 = 2k3 , X3 = -k3 , where x 1 = K3 and K3 is arbitrary ⇒ X = Hence eigen vectors ara · Example 4 : Solution : , [: : l , ] and [ [ :: Find characteristic equation of A = · 3k , t] [1 2 2] 0 2 1 . Hence fi nd eigen values and eigen vectors. -1 2 2 Characteristic equation of A is I A - All = 0 (1 - tc) 0 -1 2 (2 - tc) A, 3 2 - 5tc 2 2 ,1 (2 - tc) = 0 + 8tc - 4 = 0 . . .(1) 3 2 2 A- - A- - 4A + 4tc + 4tc - 4 = 0 I t i s characteristic equation , 'A 2 ( 1c - 1) - 41c (1c - 1) + (tc - 1) = 0 ( 'A. - 1) l1c 2 - 4 'A. + 4J = 0 (1c - 1)('A. - 2) Eigen vectors for ( "- = 1) 2 = 0 "- = 1 , 2, 2 Eigen val ues. Consider a homogenous system (1 m I:l m (A- Al) X = 0 (A - I ) X = 0 [ t 2 � A) (2 � J::] ( [�1 ; = = IES MASTER Publication 1 00 Eige n Values an d Eige n Ve c t ors I • ... (2) which is in echelon form So, Rank of matrix is 2 =(r) & No. of unk nowns are 3 =(n) n - R =3 - 2 =1 So, one variable should be assigned arbitrary value. And also there will be only L I solution. {':Rank < No. of unk nowns Now by (2) , ⇒ Non trivial solutions ⇒ Non zero solutions ⇒ - X1+2 X 2+ X 3 = X2 + = [:} [=J f] X3 0} =0 X3 =k , X2 =-k, X 1 =....'.k Let So, Eigen vector X Eigen vectors for ie , fir at eigen vector is X , Infinite solutions exists) = [�: 1 (l = 2) ➔ Consider a homogeneous system m �Hl m (A - l i) X = 0 2 (1 2) � (2 - 2) [ -1 R3 ➔ R 3 - R 1 2 (A - 2 ) 1 X = 0 (2 � J::] [�1 2 0 -1 2 = = ... (3) which is in echelon form So, Rank =2, No. of unk nown =3 (n I r) =3 - 2 =1 So, one variable should be assigned arbitrary value. Engineering Mathematics { ' . · Rank < No. of unknowns , ⇒ Non trivial sol. And also there wil l be only one LI Sai n. ⇒ -x1 + 2x 2 + 2x 3 = 0 By (3) X3 Let x2 = k then x 1 = 2k Non zero sol . ⇒ 1 01 I nfinite sol . ) =0 So [J �. � [Note: Because two eigen values are same and there exist only one LI sol . so, there will be same eigen vector for both the repeating eigen values.] x, = i.e., M o_ f, 1' =__ on_ e_ ;_ G M_o_ f, 1'= -o n� el ; 1AM of '2' = two; GM of '2' = onel � l _ A_ where AM = Algebraic Multiplicity and GM = Geometric Multiplicity. SIMILAR MATRICES Let A and B are the two' matrices of same order then A is said to be similar to B if there exi st an i nvertible matrix P such that B = p-1 AP ,1 This transformation from A to B is known as similarity transformation. DIAGONALISATION A matrix A is said to be diagonaligable if it is similar to a diagonal matrix. OR Let A be a matrix of order n and if there exist n LI vectors then we can find an inverti ble matrix P such that p-1 AP is in a diagonal form i .e. p-1 AP = Diagonal Matrix l E.g. : Let A be a matrix of order 3 and let X 1 , �. X3 are non zero eigen vectors corresponding to eigen val ues A1 , A 2 , A3 then Diagonal Matrix = [A,1 D = p-1AP = 0 0 o o 0 A2 0 A3 where, P is Modal matrix defined as P = [X 1 �X3] � If there does not exist n LI eigen vectors then matrix is non diagonalizable where A n,n OR we � say that similarity transformation is not possible. Powers of A ➔ If D = p-1 AP then A" = p o n p-1 Example 5 1 Solution a . [2 Characteristic equation is IA-1cll = 0 Eigen val ues are ; = 3, 1, 1 .Elgen Vector are X 1 = !� ] . m, Reduce the m atrix A into a diagonal form A = ·. 1 . 0 Also fi nd A' 1J- X, = [� 1 X, = [� ] (Do yourself as practice) IES MASTER Publication 1 02 Eigen Values and Eigen Vectors 1 Now P = [X1 � X3] = [� � 0 0 1 �1[� � 1 �1 � ] . p-1 = ..!. [�1 � �11 3 1 0 0 2 1 So, D = p-1 AP = i- [ �1 � �1] [ � ; 0 0 2 0 0 1 � 1 0 0 1 = [� � 0 0 1 2nd Part: We know that if D = p-1 A P then A" = p on p-1 -1 - 1 8 1 0 A4 = 1 ' 1 0 ] 0 1 [ [ 0 0 1 0 0 82 8 0 80 = ..!_ [80 82 80'] 3 1 Eumplo 6 , 0 0 2 Find the eigen values and eigen vectors of tOe following matrix. A = [ � transformation exists? Solution = Char. Equn is; I A- ;\,II == 0 Eigen valus are A .= 1 , 2, 2 & Eig en vectors are X, = [ �;} X, = m, 1 � � . Is the simi larity ] X, - �] (Already done earlier) [ Similarity transformation is not possible because modal matrix P becomes sing ular and we cannot obtain p-1 . i.e P = [X1 � X3] "" [X1 � �] which is obvio usly singular. Theorem : Two similar matrices have the same charactristic roots. OR If A and P be square matrices of the same type and if P be invertible, show that the matrices A and p- 1 AP have the same characteristic roots. Proof 1 Let B = p-1 AP, then we will show that characteristic equations for both A and B are same and hence they have the same characteristic roots. ⇒ ⇒ ⇒ ⇒ ⇒ 1 = p- (A- ;\,l) P 1 1 1 1 IB - A.l l = I P- ( A- ;\,l) PI = I P - il (A- Al)I I PI = l(A- Al)I I P- I I PI = l (A - Al)I I P- PI = l ( A - Al) I P I IB - A.II = I A- All the matrices A and B have the same characteristic polynomial. the matrices A and B have the same characteristic equation. the matrices A and B have the same characteristic roots. CAYLEY-HAMILTON THEOREM "Every square matrix 11tl1fles its own ch1raderi1tic1 equation". Let A be a square matrix of order n then its char. Equn. is; I A- ;\,II == 0 a o + a 1;\, + a 2 "- 2 +... +a n-1 "--1 +a n ;\," = 0 ⇒ a ol + a 1A + a 2 A 2 + .. . + a n_1A n-1 +a n A " = 0 i.e. Caylay Hamilton theorem states that we can replace ;\, by A in Characteristic Equation. Example 7 1 Solution 1 [ !] 2 1 Engineering Mat hematics 1 03 Ver fi y C ayley- Ham li ton theorem for matr ix A = 2 4 an d hence f ni d A - an d also evaluate m atrix ' 3 5 Polynom ai l B . where, B =A 8 - 11A 7 - 4 A 6 + A 5 + A 4 - 11A 3 - 3A 2 + 2A + I Also evaluate A 4 Char. Equn. is IA - All= 0 ⇒ ⇒ ⇒ ⇒ (1- A) 2 2 (4 - A) 5 3 3 5 ( 6 - A) =O (1 - "-)[(4 - "-)( 6- A)- 25 ]- 2 [2( 6- A) -15 ] + 3 [1 0- 3(4 - "-)] =0 2 (1- A)[24 - 4 "-- 6A + A - 25 ]- 2 [1 2-2 A- 15 ] + 3 [1 0-12 + 3"-] =O (1- "-) [24 -1 0A + A2 - 25 ]- 24 ;4 A + 30 + 30-36 + 9A =0 ⇒ - A 3 + 11A2 + 4 ,._ -1 = 0 It is character si t ci Equat o i n. "-3 + 11A2 -4 ,._ + 1 = Accor d ni g to Cayley Ham ilton theorem ; A 3 Verification: !] 3 L.H.S. = [ ; : 3 5 6 = Evaluation of A-1 : =R. H. S. By (1) - 11A Pre multiply ni g both s ides by A A 3 (A ) -11A -1 - 11A 2 - 4 A ! !] 2 - 4 [ ! ] + [� � � 1 + I =0 ; : ...(1) [i:; ::� ;m-f fm : m-•l� : :H� ! ;1 A3 -1 - 1 1 [; 0 2 -1 , 3 5 - 6 1 6 0 0 1 = [� � �l 4 A + I =0 we get (A )- 4 A - (A ) + K1 I =A -1 o 2 3 5 A 2-11A - 4 I + K1 = 0 ⇒ Evaluation of Polynomial B A -1 = 11A + 4 1 - A2 8 4 6 5 2 7 3 B = A -11A - 4 A + A + A - 11A - 3A + 2A + 1 =A Evaluation of A4• We have A 3 5 A 3- 11A 2- 4 A + l ] + A [A 3 -11A 2- 4 A + l ] + A 2 + A + I [ 5 2 2 = A [0] + A + A + I =A + A + I - 11A 2 - 4 A + I =0 �: ;;]-[; 2 83 353 14 Mult iply ni g by A, A =11A + 4 A - A =11 ��: 51 0 636] + 4 [25 [ 34 1 636 793 2 9 5 6 70 4 3 2 14 4 1 3211 4 004 = 31 01 578 6 7215 ] [ 38 64 7215 8 997 3 IES MASTER Publication 1 04 Eigen Values and Eigen Vectors Prove that; A" = A n-2 + A2 Example 8 I If A = Solution Char. Eqn. is IA - AIi = 0 1 [ � � � 0 1 01 ⇒ (1 - A) ⇒ 0 0 1 0 1 (0 - A.) = 0 1 (0 - A.) 2 (1 - A.)IA. -11 = o ⇒ ⇒ I - (1 - A.) 0 0 1 -A. 1 = 0 1 -A. 0 "'2 -1 - A.3 + "' = o ⇒ "'3 _ "'2 _ "' +1 = o By Cayley Hamilton Theorem A3 OR, we can write, - A2 - A + I = 0 .. . (1) A3 - A2 = A - I 3 A4 - A = A2 - A 3 S 4 A - A = A - A2 6 S A - A = A4 - A3 2 2 A n-1 - A n- = A n- - A """" _ A n A n-1 = A n-1 _ A n-3 Adding Column wise. A" - A2 = A-2 - I or IA " = A "-2 + A 2 - II Example 9 , Find the characteristic equation of the matnx A, where A - [; Solution The cha racteristic equation of matrix A is given by I ! -:] · Hence find A-1. 1 � A. - l = O ⇒ A3 - 6A.2 +6A -11 = 0 2 1-�A By Cayley-Hamilton theorem , A must satisfy (i), i.e. 4 I A - All = [ A ... (i) ; 1 A 3 - 6A2 + 6A - 11 I = 0 Pre-multiplying both sides by A-1 , we g et 11A-1 = A 2 - 6A + 6 1 ⇒ 4 3 11A-1 [ - -6 = 2 1 - +6 [ [ [ � � 1 2� 1�l 1; �2 1�1 1 2 1�] ; 0� 0 1 1 1 2 Example 10 1 If A= [ ] , use Cayley-Hamilton theorem to express A 6 -1 3 polynomial in A. Solution 1 We have - [ 5 -1 -7] 1 \ = -4 3 1 0 A- = 1 3 -5 -2 4A 5 + 8A4 - 1 2A 3 + 1 4A 2 as a linear Engineering Mathematics -. . The cha racteristic equation of mat rix A is IA - Al l = A2 - 4 A +5 =0 By Cayley- Hamilton theorem , the matr ix A satisfy its character si t ci equation (i) , A 2 - 4 A + 5 1 = 0 or A 2 =4 A - 5 1 . - . We have Multip yl ing both sides of (ii) by A, A 2 , A 3 and A 4 respectively , we have A6 A 3 =4 A 4 - 4A 5 + 8A 4 A 5 =4 A 4 - 12A 3 - 5A , A 4 =4A 3 - 5A 3 , + 14 A 2 =(4 A 5 and A 6 =4 A 5 5 - 5A 4 ) - 4 A - 5A 2) . .. (i) ...( i) - 5A 2 , =3A 4 - 12A 3 + 14 A 2 =3(4 A 3 1 05 - 5A 4 • + 8A 4 - 12A 3 + 14 A 2 ( Substitut ing for A 6 ) - 12A 3 + 14 A 2 = - A 2 =- (4 A - 5 I) �i - ( Sub stitute for A 4 ) ( Substitute for A 2) = -4 A + 5 1, which is the required linear polynomial in A. Example 1 1 1 Find the characte ristic equation of the matrix A = [� � 1 1 2 hence evaluate the matrix equat o in Solution : 2 IA - All =f � l 1 A B - 5A 7 + 7A 6 A - 3A 5 +A4 - 5A 3 A 2 A' 3 A - 5A 2 = = [; 1 1 1 4 1 4 4] 1 4 A A =[ ; 1 0 0 1 0 ' 1 2I 4 4 5 5 4 [ + 7A - 3 I = 0 1 0 - 5 0 1 ] 13 13 14 4 4 [ .. (. )i m �]=[; i] 1 1 1 14 13 13 Hence , Cayley- Ham li ton theorem si ver fi ied . Now , + 8A 2 - 2A + I � l=0 1 �A 1 2- A By Cayley- Hamilton theorem , A must satisfy (i) , i.e. A 3 - 5A 2 + 7A - 3 1 = 0 Now Verify Cayley- Hamilton theorem and A B - 8A 7 + 7A 6 - 3A 5 +A4 1 4 13 13] [ = 0 1 0 13 13 14 ;]+ 7[; - 5A 3 1 0 1 -3 1 [� 1 �l 0 + 8A 2 - 2A + I =A 5 (A 3 -5A 2 + 7A - 3 1) + A (A 3 - 5A 2 0 0 ;] = [� �]= 0 0 + 7A - 3 ) 1 + A 2 + A + I =A 2 X O + A X O + A 2 + A + I =A 2 + A + I IES MASTER Publication 1 06 Eigen Values and Eigen Vectors PREVIOUS YEARS GATE & ESE QUESTIONS 1 .* Questions marked with aesterisk (*) are Conceptual Questions The eigen vector(s) of the matrix 7. 0 0 a 0 0 0 , a ""' 0 is (are) 0 0 0 (a) (0, 0, a) ( b) (a, 0, 0) (c) (0, 0, 1) (d) 8! (0, a, 0) [GATE-1993 (ME)] 0 0 2.* If A = 1 0 -1 0 -1 0 0 0 0 0 the matrix A 4, calculated by -i the use of Cayley-Hamilton theore m (or) otherwise is 3. For the following matrix [1 -11 2 3 [GATE-1993] the nu m be r of positive roots is (a) one ( b) two (c) four (d) cannot be found [GATE-1994 (Pl)] 4. 5. The eigen values of the matrix [ �] are : (a) (a + 1), 0 ( b) a, 0 (c) (a - 1), 0 (d) 0, 0 �1 Find out eigen values of the matrix A [ = ;0 �2 4 For any one of the eigen value s, find out the corresponding eigen vector. [GATE-1994-ME; 5 Marks] 6. The eigenvalues of 1 1 1 1 are [i ] 1 1 (a) 0, 0 0 ( b) 0, 0 1 (c) 0, 0, 3 (d) 1, 1 , 1 [GATE-1996-ME ; 2 Marks] The Eigen values of a square sym metric matrix are always ( b) real & i maginary (a) positive (c) negative 1 O. • The vector [-2 [ iJ (d) real [GATE-1996; 1 Mark] is an eigen vector of 2 A = 2 1 -6 then corresponding eigen value -1 -2 0 of A is (a) 1 ( b) 2 -31 (c) 5 (d) -1 [GATE-1998-E E ; 1 Mark] [GATE-1994-EE; 1 Mark] d 9. �· 1 0 Given the matrix A = � Its eigen values l [ -6 -11 -6 are [GATE-1995-E E ; 2 Marks] 1 The Eigen vectors of the matrix [ � � ] is/are 1 (a) (1, 0) ( b) (0, 1) (c) (1, 1) (d) (1, -1) [GATE-1994; 1 Mark] 0 1 11. 0 -1 0 the matrix A A•V 0 0 3 ; 0 is -11 The sum of the eigen value of (a) 10 ( b) -10 (c) 24 (d) 22 [GATE-1998-EE ; 1 Mark] 12. O btain the eigen values and eigen vectors of the matrix [� -4 2] [GATE-1998-CE ; 2 Marks] 13. The eigen values of mat rix A = [ � � ] are Engineering Mathematics (a) (c) 1, 1 ( b) - 1, - 1 J, - J (d) 1, - 1 [1998-EC; 1 Mark] 14. The eigen values of the matrix [! }3 ( a) 6 (c) ]are: ( b) 5 -3 (d) -4 [GATE-1999; 2 Marks] 15. Find the eigen values and eigen vectors of the ma trix 3 [- 1 3 -11 [GATE-1999] 16.* Show that the matrix [A] is orthogona l and determine its eigen values . [A] = 2 3 1 3 2 2 -3 1 3 - [GATE-1999; 2 Marks] (c) 2, - 2, 1, - 1 2, 3, 1, 4 ( b) (d) 1, 2, 2 0, 2, 3 1 [GATE-2000; 1 Mark] r� � 0 0 ( b) 2, 3, - 2, 4 (d) None 3 (c) ( 9.0 0 , 5 .0 0) 3 0 (a) (c) !] 3 ( b) ( . 85, 2. 9 ) 3 (d) ( 1 0. 1 6, .84) ( b) -3 and 5 3 and - 5 -3 and -5 (d) 3 and 5 [GATE-2002-CE; 1 Mark] 22.* O btain the eigen values of the matrix 1 2 34 3 49 o· o 1 04 0 2 4 0 0 -2 0 94 1, 2 , -2, -1 1, 2, 2, 1 -1 ( b) -1, -2, -1, -2 (d) none [GATE-2002 (CS)] 23. For the matrix [ � :] the eigen value are (a) 3 and -3 3 and 5 ( b) -3 and -5 (d) 5 and 0 [GATE-2003; 1 Mark] 24. The sum of the eigen values of the matrix given below is 01 0 0 -2 0 are -1 4 (a) (c) 5 9 [� ; !] ( b) 7 (d) 18 [GATE-2004-ME; 1 Mark] 2 [2000-EC; 1 Mark] 19. The eigen values of the matrix [ � (a) (5 . 1 , 9.4 2) [GATE-2001 (IN)] -1 21. Eigen values of the following matrix are: [ 4 (c) 18. The e gi en values of the matrix (a) i ts eigen vectors should be inde penden t its eigen values should be real the matrix is non -singular (a) (c) A = � � ! are [ ] 1, 2, 3 1 , 0, 0 ( b) (c) (d) its all eigen values should be dis tinct -1 17. The three characte ristic roots of the following matrix (a) (c) (a) 2 - 3 3 2 -2 3 3 0 0 2 20.* The necessa ry condition to diagonalize a matrix is tha t A = 3 1 07 are, [GATE-2001; 1 Mark] 25. The eigen values of the ma trix [ _: �] are (a) (c) 1 and 4 0 and 5 26.* For the matrix ( b) -1 and 2 (d) Cannot be determined [GATE-2004-CE; 2 Marks] -2 2 P = [� -2 1] 0 1 IES MASTER Publication Eigen Values and Eigen Vectors 1 08 one of the eigen values is equal to -2. Which of the following is an eigen vector? (a) (c) (b) [ ;] Hl (d) m [ _;] [GATE-2005-EE; 2 Marks] 27.* W hich one of the following is an eigen vector of the matrix [� �j [-ij [�j [J i1J 0 0 5 5 0 2 0 3 (a) (c) (b) (d) [ [GATE-2005-ME; 2 Marks] 28.* Consider the system of equation A(n • n) x (n •I) = "'<n xl) where 11, is a scalar, Let ( \· x i ) be an eigen pair of an eigen value and its corresponding eigen vector for real matrix A. Let I be a (n x n) unit m atrix. Which one of the following statements is not correct? (a) For a homogeneous n x n system of linear equations, (A- 11,l ) x = 0 havi ng a non-triv ial solution, the rank of ( A-Al) is less than n (b) For matrix Am , m being a positive integer, (Ar , xr ) will be the eigen pair for all i (c) If AT = A-1 then [Ai ] = 1 for all i (d) If AT = A, then Ai is real for all i [GATE-2005-CE; 2 Marks] 29.* Given the matrix -4 [4 !] , the eigen vector is (b) [ :] 1 (d) [ � ] [2005-EC; 2 Marks] 30. W h at are the eigen v a l u e s of t h e fol lowi ng 2 x 2 m atrix [ (a) -1 , 1 2 -4 -1 51 (b) 1 , 6 (d) 4, -1 (c) 2, 5 [GATE-2005 (CS)] 31 . The eigen values of the matrix M given below are 1 8 -6 2 -4 5, 3 and 0. M = -6 7 2 -4 , the value of the 3 determ i nant of a m atrix is (a) 20 (b) 1 0 (c) 0 1 (d) -1 0 [GATE-2005 (Pl)] 32. Identify which one of the followi ng is an eigen vector O of the matrix A = [ - 1 -2] (a) [-1 (c) [1 1 jT -W (b) [3 -1 ]T (d) [-2 1 ]1 33. * Eigen values of a matrix S = � [ 1 1 [GATE-2005 (M E)] !] 1 are 5 and . W hat are the eigenvalues of the matrix S2 = SS? (a) and 25 (c) 5 and 1 (b) 6 and 4 (d) 2 and 0 [GATE-2006-ME; 2 Marks] 1 2 -2 3 34. For a given matrix A= -2 - 6 , one of the eigen [ 1 2 0] values is 3. The other two eigen values are (a) 2, - 5 (c) 2, 5 (b) 3, -5 (d) 3, 5 [GATE-2006-CE; 2 Marks] Engineering Mathematics 35.* The eigen values and the corresponding eigen vectors of a 2 x 2 matrix are given by Eigen vector Eigen value 11 = 8 V1 12 = 4 V2 The matrix is (a) (c) 36. [ � !] [! :] =CJ = [ �1 For the matrix [ : !] ] the eigen value corresponding [2006-EC; 2 Marks] 1 1 A[ � ] = - [� ] and A[ � ] = -2 [ � ] then the matrix 2 2 A is 2 (a) A = [ - 1 (b) A = [ (c) 1 A= [ . - (d) A = [ 1 1 1 -1 1 - 1J l 0 1 1 1 1 l l -2 - 1 -2l 1 �1] 1 -1 -2J [ 1 -2 0 2 1 =; �], (a) A+ 31 + [GATE-2007 (Pl)] then A satisfies the relation 2A-1 = 0 (b) A2 + 2A + 21 = 0 (c) (A+ l )(A+ 3I) = O (d) eA = O [-3_1 02] , then A equals 9 1 (a) 51 A + 51 0I (c) 1 [GATE-2007-EE] 54A + (b) 300A + 1 041 1 551 (d) e9A [GATE-2007-EE ; 2 Marks] 42 . If a square matrix A is real and symmetric, then the eigen values (a) are always real (d) occur in complex conjugate pairs [GATE-2007-ME; 1 Mark] [� ;J 43.* The number of linearly independent eigen vectors of 1 [GATE-2006 (IN)] 38.* The linear operation L(x) is defined by the cross product L(x) = b x X, where b = [O O]T and T = [x are three dimensional vectors. The x x ] X 2 3 1 3 x 3 m atrix M of this operation satisfies 1 L(x) = { ] : [GATE-2007; 2 Marks] (c) are always real and non-negative ] -2 [ - 1 - l -31 (d) i, -i, 0 (b) are always real and positive o [ [ �2] � �] � 40.* If A = [ 41.* If A = (b) 4 37.* For a given 2 x 2 matrix A, it is observed that 1 1 , -1 , 1 (d) n pairs of complex conjugate num bers, not necessarily distinct [2006 -EC; 2 Marks] (d) 8 (b) (c) n distinct paris of complex conjugate numbers 1 01 (c) 6 (c) + (b) 2n real val ues not necessari ly distinct 1 01 to the eigen vector [ ] is (a) 2 1 , -1 i , -i , 1 (a) 0, (a) 2n distinct real values [: ! ] (d) then the eigen values of M are 39.* If A is square symm etric real values m atrix of dimensions 2n, then the eigen val ues of A are [ : !] . (b) 1 09 (a) 0 is (b) 1 (c) 2 ( d) infinite [GATE-2007-ME; 2 Marks] 44.* The minimum and the maximum eigen values of the 1 1 3 m atrix [ 1 5 1] 3 1 1 are -2 and 6, respectively. W hat is the other eigen val ue? IES MASTER Publication 110 Eigen Values and Eigen Vectors (a) (c) (a) (b) 3 5 1 (d) -1 [GATE-2007-CE; 1 Mark] 45.* All t he four entries of t he 2 x2 matrix P = [ ;:: ::: ]are nonzero, and one of its eigenva u l es is zero. W hic hof t he following statements is true ? (a) P1 1P22 - P 1 2 P 2 1 =1 (c) P1 1 P 22 - P 1 2 P 21 = 0 (b) (d) [GATE-2008-EC; 1 Mark] 46. T he eigen vector pair of t he matrix (c) [ � ] [�2] (b) [�] [ �2] 2 (d) [� ][;] 2 [! � 3] is [GATE-2008 (Pl)] 47.* How many of t he following matrices have an eigen value 1 ? (a) (c) one (b) two (d) four t hree [GATE-2008 (CS)] 48.* T he eigen vectors of t he matrix [ in t he form [ :] and [:] . W hat is a + b ? (a) (c) 0 1 49. T he matrix [ ; � :i 1 1 p (d) p - 3 2 [GATE-2008-ME; 1 Mark] 50. T he eigen value of t he matr ix [ P] (a) (c) = [: � ] are 5 (b) -6 and 5 -7 and 8 (d) 1 and 3 and 4 2 [GATE-2008-CE; 2 Marks] a( 'A.) =I 'A.1- P I = 'A.3+ 'A.2+ 2'A.+1 = 0 P 1 1 P 22 + P 1 2 P 21 = 0 [:][�2] (b) p - 1 51.* T he c haracteristic equat o i n of a (3 x 3) matrix P is defined as P 1 1 P 22 - P 1 2 P 21 =-1 (a) (c) P p - (b) � �] are written If I denotes identity matrix, t hen t he inverse of matrix P will be (a) ( P 2 + P + 2 I) (b) ( P 2 + P + 1) , (c) (d) -( P 2 + P + 2 ) 1 [GATE-2008-EE; 1 Mark] 52. T he trace and determ inant of a 2 x 2 matr ix are k nown to be -2 and -35 respectively its eigen values are (a) (c) -30 and -5 -7and 5 (b) -37 and -1 (d) 17.5 and -2 [GATE-2009-EE; 1 Mark] 53. T he eigen values of t he fol lowing matrix are (a) (c) [ 1 : � �] 3, 3+5 j, 6- j 3+j, 3 - j, 5 +j (b) - 6+5 j, 3+j, 3 -j (d) 3, -1 +3j, - 1 -3j [GATE-2009-EC; 2 Marks] 54.* T he eigen values of a 2 x 2 matrix X are -2 and -3. T he eigen values of matrix (X + ) 1 -1 (X + 5 ) 1 are (a) � -( P 2 + P + I) (c) -3, -4 -1, -3 (b) -1 , -2 (d) -2, -4 2 (d) 2 [GATE-2008-ME; 2 Marks] has one e gi en value equal to 3. T he sum of t he ot her two eigen values is 55. An eigen vector of P =[ (a) (c) [ -1 1 1]l [ 1 -1 2]T (b) (d) i � !] [1 2 [GATE-2009 (IN)] is 1]T [ 2 1 -1]l [GATE-2010-EE; 2 Marks] Engineering Mathematics 56.* A real n x n matrix A = [aii] is defined as follows { a ii = i, 'v'i = j = 0, otherwise The sum of all n eigen val ues of A is (a) (c) n (n + 1} 2 n (n + 1) (2n + 1) 2 n (n - 1) (b) (d) n 2 2 [GATE-2010 (IN)] 57. Consider the following matrix A= [: : - If the eigen ] values of A are 4 and 8 , then (a) X = 4, y = 1 0 (c) X = -3, y = 9 (b) x = 5, y = 8 (d) x = -4, y = 1 0 [GATE-2010 (CS)] 58. If {1 , 0, -1 }T is an eigen vector of the following matrix 1 -1 0 -1 -1 is 2 0 - 1 then the corresponding eigen value 1 (a) 1 (d) 5 [GATE-2010 (Pl)] !] is (b) (d) g} { �1 } [GATE-2010-ME; 2 Marks] 60. The eigen values of a skew symmetric matrix are (a) (b) (c) (d) always zero always pure imagi nary either zero or pure imaginary always real 61 . Eigen values of a real symmetric matrix are always (a) positive (c) real [GATE-2010-EC; 1 Mark] (b) negative (d) complex [GATE-2011-ME; 1 Mark] -3 2 1 6 has eigen val ues - 1 -2 0 3, -3, 5. An eigen vector corresponding to the eigen value 5 is [1 2 -1 ]1. O ne of the eigen vector of the m atrix M 3 is . 62.* The matrix M = (a) [1 - (c) -W 8 [1 � · - 1] 1 (b) [1 2-W (d) [1 1 -1)1 [GATE-2011 (IN)] 63. The eigen values of the following matrix [ � � are ..=-1:] (b) 6, -8 (a) 4, 9 (c) 4, 8 (d) -6, 8 [GATE-2011 (Pl)] 64. Consider the matrix as given below O 4 7 . Which 1 2 3 0 0 3 one of the following options provides the correct values of the eigen val ues of the m atrix? (b) 3, 7, 3 (c) 7, 3, 2 59. One of the eigen vectors of the matrix A = [� 2 (a) 1 , 4, 3 (b) 2 (c) 3 -2 111 (d) 1 , 2, 3 [GATE-2011 (CS)] 65. G iven that A= -5 -3 and I = � � ] the value A3 is [ [2 0] (a) 1 5A+ 12I (c) 1 7A + 1 51 (b) 1 9A+ 30I (d) 1 7A + 2 1 1 [GATE-2012-EC, EE, IN ; 2 Marks] 66.* For the matrix A = [� !] , ONE of the l1 l l� l normalized eigen vectors is given as (•) al (c) [ � ffo (b) ] (d) [GATE-2012-ME, Pl; 2 Marks] IES MASTER Publication Eigen Values and Eigen Vectors 112 [! !] 67. The eigen values of mat rix (a) (c) -2 .4 2 and 6.8 6 47 . 0 and 6.8 6 a re (b) 3.4 8 and 13.5 3 (d) 6.8 6 and 9.5 0 [GATE-2012-CE; 2 Marks] 68. A mat ri x has eigen values -1 and -2 . The co rresponding eigen vectors a re [�] and [ -�] 1 respectively . The mat rix is (b) [ � - :] (d) �3] [ �2 [GATE-2013-EE; 2 Marks] 69.* The pai r of eigen vecto rs co rresponding to the two 1 e_ igen values of the mat rix [ � � ] is (b) (c) [ 1 [�] (d) [�l • 1 [ � [1 [�] ] [GATE-2013 (IN)] 70. The eigen values of a symmet ric mat rix a re all (a) complex with non-ze ro positive imagina ry pa rt (c) real (b) (d) complex with non -ze ro negative imagina ry pu re imagina ry [GATE-2013; 1 Mark] 71 . The minimum eigen value of the fol o l wing mat rix is [� ❖ ;] (a) (c) 0 2 (b) 1 (d) 3 [2013-EC; 1 Mark] 72. Which one of the following statements is t rue fo r all real symmet ric mat rices ? (a) All the eigen values a re real (c) All the eigen values a re distinct (b) (d) All the eigen values a re positive Sum of all the eigen values is ze ro [GATE-2014-EE-SET-2; 1 Mark] 73.* A system mat rix is given as follows : 1 A = [ �6 -� 1 � ] -6 -11 5 The absolute value of the ratio of the maximum eigen value to the minimum eigen value is ___ [2014-EE-SET-1] 74.* Which one of the following statements is N O T t ru e fo r a squa re matrix A ? (a) fI A is uppe r t riangula ,r the eigen values of A a re the diagonal elements of it (c) fI A is real, the eigen values of A and A T a re always the same (b) (d) fI A is real symmet ric, the eigen values of A a re always real and positive fI all the p rincipal mino rs of A a re positive, all the eigen values of A a re also positive. [GATE-2014 (EC-Set 3)) 75.* Fo rthe mat rix A satis fying th eequation given below, the eigen values a re 1 2 3 1 2 3 [A] 7 8 9 = 4 5 6 4 5 6 7 8 9 (a) (1, -j, j) (c) (b } (1, 1 , 0) (1, 1, -1) (d } (1, 0, 0) [GATE-2014 (IN-Set 1)) 76.* The value of the dot p roduct of the eigen vecto rs co rresponding to any pai r of diffe rent eigen values of a 4 x 4 symmet ric positive definite mat rix is [GATE-2014 (CS-Set 1)) 77.* The product of the non-ze ro eigen values of the 1 0 0 0 1 0 1 1 1 0 mat rix 0 1 1 1 0 is 0 1 1 1 1 0 0 0 0 1 [GATE-2014 (CS-Set 2)) 78.* Which one of the following stat ements TRUE about eve ry n x n mat rix with only real eigen values ? Eng ineering Mathematics (a) If the trace of the m atrix is positive and the determ inant is negative, at least one of its eigen val ues is negative 85. The lowest eigen value of the (c) If the deteminant of the m atrix is positive, all its eigen values are positive. 86. * The two eigen value of the m atrix [ � (b) If the trace of the matrix is positive, all its eigen values are positive. (d) If the product of the trace and determ inant of the matrix is positive, all its eigen val ues are positive. [GATE-2014 (CS-Set 3)) 79. One of the eigen vectors of the m atrix (a) (c) {�1} 2 (b) {� } {:1} (d ) �m:l 2] -5 -9 6 [ IS { �} [GATE-2014-ME-SET-2; 1 Mark] 80.* Consider a 3 x 3 real symmetric m atrix S such that two of its eigen values are a * 0, b * 0 with respective eigen vectorn equals (a) a (c) ab [ If a, b then x,y, +x,Y,+x y , , (b) b (d) 0 [GATE-2014-ME-SET-3; 1 Mark] 81 . The sum of eigen value matrix (M) is W hen 2 1 5 650 795 [ M ] = [655 1 50 835] 485 355 550 (a) 91 5 (c) 1 640 (b) 1 355 (d) 2 1 80 82.* A real (4x4) matrix A satisfies the equation A2 = /, where / is the (4x4) identity matrix. The positive eigen value of A is ___ [GATE-2014-CE-SET-I ; 2 Marks] is ---- [GATE-2015-EE-Set-1; 1 Mark] 84.* At least one eigen value of a si ngular matrix is (a) positive (c) negative (b) zero (d) imaginary [GATE-2015-ME-SET-2; 1 Mark] x 2 m atrix [� !] [GATE-2015-ME-SET-3; 1 Mark] i] have a ratio of 3: 1 , for P = 2 . W hat is another value of P for which the Eigen val ues have the same ratio of 3: 1 ? (b} 1 (a) -2 (d} 1 4/3 (c) 7/3 [GATE-2015-CE-Set-2; 2 Marks] 87.**The maximum value of 'a' such that the m atrix -3 0 0 -2 a -2 -1 0 has three li nearly independent real eigen vectors is (a) (c) 2 3 ✓ 3 (b) 1 + 2✓3 3 1 ✓ 3 1 3 (d) + ✓ 3✓ 3✓ 3 3 [GATE-2015 (EE-Set 1)) 88. C o n s i d e r t h e fol l ow i n g 2 x 2 m at r i x A where two elements are unknown and are marked by a and b. The eigen values of this matrix are -1 and 7. W hat are the val ues of a and b? (a) a = 6, b = 4 A = (: :) (b) a = 4, b = 6 (d} a = 5, b = 3 (c) a = 3, b = 5 [GATE-2015 (CS-Set 1)] 89. The larger of the two eigen values of the m atrix [: � ] is ___ [2014-EC-Set-1; 1 Mark] 83.* I f t h e s u m of t h e d i a g o n a l e l e m e nts of a 2 x 2 matrix is -6, then the maximum possible value of determi nant of the matrix is 2 113 (GATE-2015 [CS-Set 21) 1 -1 2 1 2 90. I n the given matrix 0 1 0 , one of the eigen 1 value is 1 . The eigen vectors corresponding to the eigen value 1 are (a) (b) {a(4, 2, 1) l a ar 0, a E R} {a(-4, 2 , 1)la ar 0, a E R} IES MASTER Publication 114 Eigen Values and Eigen Vectors (c) {a(✓, 0, 1) la � 0, a ER} 2 95.* We have a set of 3 linear equations in 3 unknowns. ' X = Y ' means X and Y are equivalent statements and ' X ;/,. Y ' means X and Y are not equivalent statements. ✓ (d) {a(- , 0, 1) la � 0, a ER} 2 (GATE-201 5 [CS-Set 31) 91 . The smallest and largest Eigen value of the following matrix are [! � !] Which one of the following is TRUE? (b) 0.5 and 2.5 (c) 1.0 and 3 .0 (d) 1.0 and 2.0 H• [GATE-2015-CE-SET-1 ; 2 Marks] 92.• The value of P such that the vecto, [ Q : The equations are linearly independent. R : All eigen val ues of the coefficient matrix are nonzero. S : The determinant of the coefficient matrix is nonzero. 2 -3 5 (a) 1.5 and 2.5 P : There is a unique solution. (a) P= Q = R = S (b) P=R ;/:. Q = S (c) P= Q ;J:. R = S (d) P ;J:. Q ;J:.R ;J:. S [GATE-2015-EE-Set-2; 1 Mark] an e igen vector of the matrix [ ; � � ] is 1 4 -4 1 0 [2015-EC-SET-1 ; 1 Mark] 93.* The value of x for which all the e igen val ues of the matrix given below are real is 1 0 5+ j [ X 20 4 2 � �i 96.* The number of linearly independent eigen vectors of matrix A = [ � 0 3 is ____ [GATE-2016-ME-SET-2; 2 Marks] 97.* Consider a 3 x 3 matrix with every element being equal to 1. Its only non-zero eigen value is __. [GATE-201 6-EE-SET-1 ; 1 Mark] 98.* A 3 x 3 matrix P is such that, P3 = P. Then the eigen values of P are (a) 1 , 1 , -1 (a) 5 + j (b) 5-j (b) 1, 0.5 + j0. 866, 0.5 - j0. 866 (c) 1 -5j (d) 1 +5j (c) 1 , -0.5 + j0. 866, -0.-5 - 0. 866 [2015-EC-Set-2; 1 Mark] = 94.* We have a set of 3 linear equations in 3 unknowns. ' X Y ' means X and Y are equivalent statements and ' X ;/,. Y ' means X and Y are not equivalent statements. P : There is a unique solution. Q : The equations are linearly independent. R : All eigen values of the coefficient matrix are nonzero. S : The determinant of the coefficient matrix is nonzero. Which one of the following is TRUE? (a) P= O =R = S (b) P = R ;/,. Q = S (c) P Q R S (d) P Q R S [GATE-2015-EE-Set-2; 1 Mark] (d) 0, 1, -1 [GATE-201 6-EE-SET-2; 1 Mark] 99.* Let the e igen values of a 2 x 2 matrix A be 1 , -2 with eigen vectors x 1 and x 2 respectively. Then the eigen values and eigen vectors of the matrix A 2 - 3A + 41 would, respectively, be (a) 2, 1 4, X 1 , X2 (b) 2, 1 4, X1 + (c) 2, 0, X1 , X2 (d) 2, 0, X1 + X 2 , X 1 -X 2 X 2 , X 1 -X 2 [GATE-2016-EE-Set-1 ; 2 Marks] 100. * The condition for which the eigen values of the matrix are positive, is A = [ � �] Engineering Mathematics (a) k > 1/2 (c) (b) k > - 2 k > 0 (d) k < - 1/2 [GATE-2016-ME-SET-2; 1 Mark] 101 .* If the entries in each co lumn of a square matrix M add up to 1, then an eigen value of M is (a) (b) 3 4 (c) 2 (d) 1 [GATE-201 6-CE-Set-l; 1 Mark] ! ; is --- 4 13 has zero as an eigen value -9+x l [GATE-2016-EC-SET-2; 1 Mark] 103.* Consider a 2 x 2 square matrix A = [: :] where x is unk nown. fI the eigen values of the matrix A are ( cr+ joo) and ( cr- joo) , then x is equal to (a) (c) +joo +oo 104.** Let P= � [ ;J. (b) - joo (d) - oo [GATE-2016-EC-SET-1 ; 1 Mark] Consider the set S of all vectors (:) such that a 2+b 2= s is 1 where ✓-fa (:) = P(:) . Then (a) a circle of rad u is (c) an ellipse with ma jor axis along (�) (d) an ellipse with minor axis along (�) (b) a circle of radius � ...; 1 0 [GATE-201 6 (EE-Set 2)1 105.* Consider the matrix A = [ � ; : ] whose eigen -1 -1 -2 3 values are 1 , -1 and 3. then trace of (A -3A 2 ) is 106. The eigen values of the matrix [ � � ] are 1 (a) (c) (b) 1 and -1 i and -i 0 and 1 (d) 0 and -1 [GATE-2016 (Pl)] 107.* Two eigen value of a 3 x 3 real matrix P are (2 + ✓ --1) and 3. the determinant of P is ___ [GATE-2016 (CS-Set 1)1 108.* Suppose that the eigen values of matrix A are 1, 2, 102. The value of x for which the matrix A= [ - 6 -4 1 15 ----- [GATE-2016 (IN)] 1 4. The determinant of (A - f 109.* Consider the 5 x 5 matrix 1 5 A= 4 3 2 2 1 5 4 3 3 2 1 5 4 4 3 2 1 5 is ____ 5 4 3 2 1 It is given that A has only one real eigen value. Then the real eigen value of A is (a) (c) -25 (b) 15 1 10.* The matrix A = 0 (d) 25 [GATE-2017-EC Session-I] 3 2 o 1 2 0 -1 O 2 o 3 2 has three distinct 1 eigen values and one of its eigen vectors is [ ]. � Which one of the following can be another eigen vector of A ? (a) [ (c) [ �J iJ (b) (c) m [ �1 ] [GATE-2017-EE Session-I] 1 1 1 . The eigen values of the matrix given �elow are: [� i3 :] IES MASTER Publication E igen Values and E igen Vectors 116 1 (a) (0, - , -3) (b) (0, -2, -3) (c) (0, 2, 3) (d) (0, 1 , 3) [GATE-2017-EE; Session-II] [! �1 112.* Consider the matrix ] which one of the following statements is TRUE for the eigen values and eigen vectors of the matrix? (a) Eighen value 3 has a m ultipl icity of 2 and only one independent eigen vector exists (b) Eigen value 3 has a multiplicity of 2 and two independent eigen vectors exist (c) Eigen value 3 has a multiplicity of 2 and no i ndependent eigent vector exists (d) Eigen values are 3 and -3 and two independent eighen vectors exist. [GATE-2017-CE; Session-I] 113.* Consider the following simultaneous equation (with c1 and c2 beings constants): 3x 1 + 2x2 = c 1 4x 1 + X2 = c2 The characteristic equation for these simultaneous equations is (b) 11,2 -411, + 5 = 0 (a) ,,,2 - 411, - 5 = 0 (c) 11,2 + 411,-5 = 0 (d) 11,2 + 411, + 5 = 0 [GATE-2017-CE; Session-II] 114. The product of eign values of the matrix P is p =[: 0 �3 2 ; -1 l (a) -6 115. Consider the matrix (d) -2 �1 [GATE-2017-ME; Session-I] p = value of the matrix is is--- --=1 0 _.1_ ../2 � ../2 r� Which one of the following statements about P is INCORRECT? 1 (a) Determi nant of P is equal to . (b) P is orthogonal (c) I nverse of P is equal to its transpose. (d) Al l eigen values of P are real numbers [GATE-2017-ME; Session-I] 1 0, the other eigen value [GATE-2017-ME; Session-II] 50 70 1 1 7.*Consider the matrix A = [ whose eigen 70 80 ] vectors corresponding to eigen values 11,1 and 11,2 are x 1 = 70 11, - 80 . and x 2 = 2 ] , respectively. 11,1 - 50] 70 [ [ xT X The value of 2 is __ [GATE-2017-ME; Session-II] 1 1 8.*Let A be n x n real val ued square symmetric matrix LLA� = 50. n of rank 2 with n i=1 j=1 Consider the following statements. (I) One eigen value must be in [-5, 5] (11) The eigen value with the largest magnitude must be strictly greater than 5. Which of the above statements about eigen values of A is/are necessarily CORRECT? (a) Both ( I ) and ( I I ) (c) ( I I ) only ( b ) ( I ) only (d) Neither (I) nor ( I I ) [GATE-2017-CS/IT] 119. The eigen values of the matrix !] 1 are A = [� � 0 -6 5 1 , 5 .±. j6 (a) 1 , 5, 6 (b) 2 (c) 6 116. The determinant of a 2 x 2 m atrix is 50. If one eigen (c) (b) 1 , - 5 (d) 1 , 5, 5 .±. j6 [GATE-2017 (IN)] 120.* If the characteristics polynomial of 3 x 3 m atrix M over 3 R 2 (the set of real n u m bers) is 11, - 411, + a11, + 30, a E R, and one eigen value of M is 2, then the largest among the absolute values of the eighen values of M is __. [GATE-2017 PAPER-2 (CS/IT)] 1 1 121. * If a square matrix of order 00 has exactly 5 distinct eigen values, then the degree of the m inimal polynomial is 1 Always 1 5 (a) At least 5 (c) 1 (b) At most 5 1 (d) Exactly 00 [ESE 2017 (EE)] Engineering Mathematics 122. Let the Eigen vector of the matrix [ � !] be written in the form [:] and [ :]. What i s the value of (a + b) ? (a) 0 (b) (c) 1 123. 1 2 (d) 2 [ESE 2018 (Common Paper)] Eigen value s of the Matrix 3 -1 -1 -1 3 -1 are -1 -1 3 (a) 1, 1, 1 (c) 1, 4, 4 (b) 1, 1, 2 2 -4 124. The matrix ( )ha s 4 -2 (a) (b) (c) (d) 125. Let M be a real 4 x 4 matrix. Con sider the following statement s : S1: M ha s 4 linearly indendent eigen vector s. S2 : M ha s 4 di stinct eigen value s. S3 : M i s non- singular (invertible) . Which one among the following i s TRU E ? (b) S1 implie s S3 (a) S1 implie s S2 (c) real eigen value s and eigen vector s real eigen value s but complex eigen vector s complex eigen value s but real eigen vecto rs complex eigen value s and eigen vector s S2 implie s S1 (d) S3 implie s S2 [GATE 2018 (EC)] 126. Con sider a non -singular 2 x 2 square matrix A. fI trance (A) =4 and trace (A 2) =5 , the determinant of the matrix A i s _ _ _(up to 1 decimal place) . (d) 1, 2, 4 [ESE 2018(EE)] 117 127. 0 2 0 -1 -2] 0 [GATE 2018 (EE)] and B =A 3 -A 2 �4A + 5 ,1 where I i s the 3 x 3 identity matrix. The determinant of B i s___(up to 1 decimal place) [GATE 2018 (EE)] [GATE 2018 (CE-Afternoon Session)] IES MASTER Publication 118 Eigen Values and Eigen Vectors , 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 1 2. 1 3. 1 4. 1 5. 1 6. 17. 18. 1 9. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31 . 32. (b & d) (See Sol .) (d) 33. 34. 35. (a) 36. (c & d) 38. (c) 40. (See Sol.) (-1 , -2, -3) (d) 37. 39. 41 . (c) 42. (See Sol .) 44. (a) 43. (d) 45. (See Sol.) 47. (a & d) (See Sol.) (b) (b) 46. 48. 49. 50. (d) 51 . (a) 53. (d) 52. (1 , 2 , -2, -1 ) 54. (c) 55. (c) 57. (a) 59. (b) (d) (b) (c) 56. 58. 60. 61 . (b) 62. (b) 64. (c) 63. A N SWE R l{EY (a) (b) 65. 66. (a) 67. (c) 69. (b) 71 . (c) (d) 68. 70. (a) 72. (a) 74. (b) 76. (b) 78. (a) (b) (c) (a) 73. 75. 77. 79. (b) 80. (b) 82. (c) 84. (c) 86. (a) 88. (a) 90. (c) 92. (b) 94. (a) 96. (c) 81 . (d) 83. (d) 85. (b) 87. (d) 89. (a) 91 . (c) (b) 93. 95. (b) 97. (b) 99. (a) (b) 98. (d) 1 00. (c) 1 02. (1) (a) 1 04. (b) 1 06. (0) 1 08. (0.1 25) (a) 101. (d) (a) (a) (d) (a) 1 03. (0.31 to 0.35) 1 05. (-6) (d) (d) (a) (c) 1 07. (1 5) (6) 1 09. (a) 1 1 0. (d) 1 1 2. (1 ) 114. (d) 111. (a) 1 1 3. (9) 1 1 5. (c) (c) (a) (a) (a) (b) (d) (b) 1 1 6. (5) (d) 1 1 8. (d) 1 20. (5) (b) 1 22. (17) 1 24. (2) 1 1 7. (0) (b) 1 1 9. (6) 1 21 . (b) (c) (a) (b) (d) 1 23. (b) 1 25. (a) 1 27. (1 ) (2) (3) (c) (d) (c) 1 26. (5.5) 119 Engineering Mat hematics Sol-1: (b) & (d) Let ⇒ [H �] A = By C ayley H amilton theorem , we c an repl ace ;i., ;i.,x . .- - It is an upper tri angul ar m atr ix. Therefore di agon al elements are eigen v alues of A i.e. = 0 , 0,0 Now for Eigen Vectors we c an consider AX = or ⇒ [o � ;i., o � ;i., 0 0 Put S = 0 [ � ][::] (A - Al) X = 0 0 - A. x3 = � � � HJ = m �] [0 Case I: Put K 2 = 0 ., x , = Case II. Put � = 0 , x, A = [� m-m Sol-3: (d) Given l 0 C E is I A - Al = ⇒ or or = 2 ±i A Sol-4: (a) a 1 M atrix A -- [ ] a 1 A 2 a A l � IA - Al l = 0 1 � AI -( a+1) A+ _ a- a A = 0 = 0 = 0, a +1 sum of eigen v alues =tr ace of m atrix Hence option A is correct. Sol-5: A = IA - Al l = Since it is U. T.M. So eigen v alues of A are A= 1,-1, ,i -i so C.Eq. of A c an be t ak en as ⇒ A2-4 A+5 = 0 Alternative Solution : ( A- 1) ( A+1) ( A-i )( A+i) 1- A -1 = 0 I 2 3 - AI Hence both roots are complex. ⇒ ⇒ ij 1 A 4 =I ⇒ Hence (b) & (c) c an be obt ained by X 1 & X 2 resp. G vi en by ' /J\ Ch ar acteristic equ ation of m atrix A, = [:,] - [�] Sol-2: A - I = 0 So So , we h ave to assume X 1 = K 1 & X 2 = K 2 So Eigen vector is A 4 =0 p--2- 1)(11,2+ 1) = 0 [� � !] 1- A 0 2 32 0 A 0 1 4- A = 0 ( 1- A )[( 3- 11,)( 4 - A )- 2 ] = 0 2 3 A - 8 A +17 A-1 0 = 0 ( 1- A H A- 5 )( A- 2 ) = 0 A = 1 , 2, 5 IES MASTER Publication 1 20 Eigen Values and E igen Vectors mm l � ! !l m m So, eigen values are 1 , 2, 5 lT eigen vector corresponding to ')., = 2 3 2 � 1 = 4 ! 21 = -a = 0 a = 0 ⇒ -A, 0 1 -A, So, ⇒ -'}.,[')., ')., ⇒ ⇒ 3 2 + 6')., + 1 1]-6 = 0 2 + 6')., + 1 1')., +6 = 0 A, = -1 , -2 , -3. Sol-8: (c) Characteristic equation of m atrix is = I A- ').,I I = 0 1- ')., 1 1 1 = 0 1 - '}., 1 - ')., I 2 ( 1 -').,)[(1 - ').,) - 1] - [(1 - '}., ) - 1] + 1 [1 - (1 - ').,)] = 0 3 ( 1 - '}., ) - (1 - '}., ) + 2')., A,3 So eigen vector correspond ing to A = 2 Is [ _:] . Sol-6: (c) & (d) 1 -1 A = [ -1 1 ] For eigen val ues ⇒ 2 0 = 0 ')., ('}., - 3) = 0 1')., = o,o, 31 Sol-9: (d) Eigen val ues of a symmetric matrix are always real. If ').,i be the eigen value of matrix A and Xi the 1-1 A. = 0 I -1 1-'}.,I corresponding eigen vector then . ⇒ For ')., = 0 Ax 2 - 3')., = Sol-10: (c) IA- UI = 0 1 + ').,2 -2'}., - 1 = 0 A, = 0, 2 = 0 0] _ x-y= ⇒y - X -x + y = 0 Let = 0 -6 -1 1 -6 - ')., b + c = 0 b = -C 0 1 x = k ⇒ ⇒ -2 AXi = ').,i Xi 2 L.H.S. = [ 2 1 -1 -2 ').,i = 5 is the eigen value of A corresponding to the eigen vector [12 -W Sol-11: (a) y = k So, Hence option (c) is correct. By similar logic for ')., = 2, we get X = k � ] [ 1 So option ( d) is also correct. Sol-7: (-1, -2, -3) Characteristic equation of matrix [A] is I A- A.I I = 0 Sum of eigen value of any matrix [A] = sum of the diagonal elem ents of matrix = trace of matrix [A] Sol-12: LAi = 2 + 1 + 3 + 4 = 1 0 [A- ').,I ][X ] = 0 8-')., = 1· 2 (8 - ').,)(2 - '}.,)-8 = 0 4 2 - ').,1 = 0 Enginee ring Mat hematics ,..,2- 1 011,+24 Sol-15: = 0 C. E. is 11, = 4, 6 (Given values) Now, for A, = 4 [ A - Al] = [8 ; = [ [ ] = 0 [A - 11,l] X : 4 2 _\] =:] ⇒ 2x - 4y = 0 ⇒ x = 2y 3- A. -1 ] = [ - 1 3- A. x 5 "' IA - lI l = 0 ⇒ I _ / _ "' 1 ; (5- 1) ( 3-) 1 - 9 = 0 1 2 - 2 I - 24 = 0 (I - 6) ( 1 + 4) = 0 I"' _! ] = 0 6,-4 1 ⇒ x 1 = x 2 =k (let) X1 = [ �] 1 when y = k , x = 2k I an eigen value "' = 2. :. X 2 =k 1 [ 1 � ] is an eigen vecto rof A correspon ding to an eigen value 11, = 4 Sol-16: [A] = A.2 - 1 = 0 [: [oo] :. X = k[�] is an eigen vector of A co rrespon ding to - "' I = 0 11 -� A = ... (1) so I Sol-14: (a & d) ⇒ (A - 11,l) X o 1 - x [ 1 11 ] [ 1 ] = x2 - Cha racteristic equation is A - A.I = 0 ⇒ = 2&4 AX = 11,X Put 11, = 4 in (1) , Eigen vecto r = [ �] Sol-13: (d) I /\, so [ A - 11,][ l X] = 0 = 0 3- A. -1 I -1 3- A. = 0 Put 11,=2 , A, = 6 2x - 4y ⇒ ⇒ Eigen vecto r = [ ] � Again for ⇒ I A - A.I I = 0 Consi de r 4x - 4y = 0 ⇒ x = y =k 2x - 2y = 0 } ⇒ 1 21 A, = ±1 ⇒ 2 3 2 3 13 ..!_ 13 23 23 23 1 3 2 3 2 1 2 -2 2 1 [A] = 3[ l 1 2 -2 T ranspose of matrix [A '] = A ' = ..!_[� 3 2 -2 2 1 1l -2 2 AA ' IES MASTER Publication 1 22 Eigen Values and Eigen Vectors AA' = �[� � �1 ⇒ 53 = 0 2 A. 9- A I ⇒ 0 0 9 1 (5- A) (9- A) - 6 = 0 AA' = I = Identity matrix A 2 - 1 4A +39 = 0 He nce A is an orthogo nal matrix A = 2 1 2- A 1 1 =0 2A -2 I A- All = 3 1 2 -2 - A 1 4 ± 2 ✓10 = ± ✓10 7 2 A = -1 0. 1 6, 3. 84 A = -(2- A)(2- A)(2 + A) + 1- 8 - 2(2 - A) - 2(2 +A)- 2(2- A) = 0 (2- A)(A2 - 4)- 7- 1 + 2A- 4- 2A- 4 + 2A = 0 -A3 + 2A2 + 4 A - 8 - 7 - 4 + 2A - 8 = 0 -A3 + 2A 3 + 6A - 27 = 0 Sol-20: (d) Necessary co ndito n is I A l at O while sufficie nt co nditions is "A is diago nalizable if a nd o nly if it has n li nearly i ndependent eigen vectors." Sol-21: (a) For eige n values A = -3 , % + 1.66i , % - 1.66i Sol-17: (b) ⇒ Cha racte ristic equatio n of A is IA - A l l = 0 2 4 -1 A 1 4- -1 - AI = O A2 + 2A- 1 5 = 0 A = -5, 3 Alternative Solution = 0 Sum of eigen values = trace (A) This property is satisfied by o nly (a). 1A = 1. 2. 21 Sol-22: (1, 2, -2, -1) Sol-18: (b) Characteristic equatio n is I A- Ai l = 0 ⇒ IA- AIi = o 2 (-1- A) - 1 6 = 0 2 3 1-A 3 = 0 2 2- A 0 0 2-A (1- A) ( 2- A) ✓ 1 4 ± 196- 1 56 2 (1) 0 0 0 4- A = 0 0 3- A 0 -2 -A (2- A) 0 -1 0 0 4- A = 0 (2 - A) (3 - A) (-2 - A) ( 4- A) = 0 A = 2, 3, -2 , 4 For eigen values IA - All = 0 Sol-23: (c) 4 0 2 - A -1 0 0 3- A 0 0 -2- A -1 0 0 Sol-19: (d) Since it is U.T. M. so diago nal eleme nts are the eigen values. ⇒ ⇒ IA- All = 1 � A 4 � Al =0 1 6- 8A + A2 - 1 = 0 A2 - 8A + 1 5 = 0 A = 3, 5 Alternative Solution: Sum of eigen values = trace of matrix = 8 So only option (c) satisfies it. Sol-24: (b) Sum of eigen values = Trace of the matrix. = Sum of diagonal element = ( 1 + 5 + 1) = 7 Engineering Mathematics Sol-25: (c) Sol-29: (c) 2 [ A = _: � ] ⇒ I A- 11.II = Characteristic equation is I A- 11.I I ⇒ 0 1 -\-A, 4 = 0 :/ 1�\I (4- 11. ) (1 - 11, )- 4 = 0 = 0 11. = 0, 5 (A - 11.l)x = 0 for 11. = 5 [� ml m ⇒ -2 0 0 ⇒ X + 2y = 0 X = -2y X = = (A + 2I) X = 0 X = = ⇒ [=�] x = 2k K [�1] . Sol-30: (b) 0 &z= 0 2 = Y & 5 �] y = -k 11.1 + 11.2 SO, k' x [ = 5x - 2y + 2z = = Let "-1 Let y 0 [: !] [:] If "-i is the eigen value of matrix and X i the corresponding eigen vector then, "-i [ Xi] = (A + 5I)x Sol-26: (d) (A- 11.l) X = 0 = 0 for eigen vector Note: Here, sum of eigen values = trace (A) is satisfied by (a) & (c). Hence we have to solve the characteristic equation. = 0 11.2 + 11. - 20 = 0 A. = -5, 4 4 - 5 11, + 11.2 - 4 A[Xi] 3 � "' I = (-4 - 11.)(3 - 11.) - 8 = 0 1 ⇒ 1 23 Z = 0 = Trace (A) & + "-2 = 7 & "-1 =1, "-2 "-1"-2 "-1 · "-2 =I A I =6 =6 Sol-31: (c) IMI 2 -k ' z = 0 5 = The product of the eigen values : 15 X 3 X O : 0 Sol-32: (b) Eigen values are 1, -2 ( •: it is L.T. M.) so For 11. = 1, eigen vectors are given by (A - 11.l)x = 0 :. Eigen vector corresponding to eigen value -2 is [2 5 0]l . Sol-27: (a) ·: Only option (a) is satisfying AX = 11.X for 11. = 5. Sol-28: (b) If 11. is an eigen value of matrix A then eigen value of A" is 11." . If X is an e igen vector of matrix A then eigen vector of A" is X. (A -l)X = 0 or or or So let y x + 3y = 0 = k, then x = -3k So IES MASTER Publication Eigen Values and Eigen Vectors 1 24 Sol-33: (a) fI 'A. is an eigen value of mat rix A then the eigen value of N is 'A.n, n E N . Sol-34: (b) ⇒ Now, we For J.., = -\ X1 = [�1] 1 _ - (- 2) ] < ] [-2 [ -2 For l-2 = - 2, X2 = [-�J 1 know that square matr ix A nxn is said to be diagonali zable if there exists a non-singula rmatrix P such that p-1A P= D (diagonal matrix) whose diagonal elements are eigen values of A and co lumns of P are eigen vectors of A and P is known as M O DA L matrix Tric k : Let other eigen values are 'A.1, 'A. 2. i.e. Now, 3 + 'A.1 + 'A. 2 = 2 + (- 1) + 0 = 1 'A.1 + 'A.2 = - 2 Chec k options, only (b) satisfies (i) ... (i) Hence eigen values are 3, -5. Note : This tric k will fail when many options satisfy (i) . In that case, use the standard procedure. Sol-35: (a) Let A and A = P = [ X 1 X 2J . 0= So, p-1A P = 0 A = ⇒ [: :] [�1] [�1] =4 a - b = 4 & c - d = -4 ... (1) ...(2) Solving (1) and (2) a = 6, b = 2, c = 2, d = 6 A = [: Alternative Sol: Sol-36: (c) ⇒ [: Ax = 'A.x 1 01 1 01 . = 'A. [ [ 1 0 1] 4 ] 1 0 1] Sol-37: (c) Given A [�1] �2] [�1 �1] L (x ) = b xX so, L (x ) Again L(x ) 0 X1 k j 0 X 2 X3 = X3i+ O J - X1k = [x 3 0 -x l = {: ] (g v i en) where M is a 3 x 3 matrix !] Sum of e gi en values = trace (A) is satisfied by option (a) 2 1 O . 'A. 2 ] Sol-38: (d) = s [ �] 1 = P D P -1 Hence option (c) is correct. [: :] [ : :] [ ] = � a + b = 8 &c +d = 8 0 [ �1 -�] [� A x = 'A.x ⇒ ⇒A ['A, A, = 6 = (-1 { �1] ⇒ 81X1 + b 1 X2+ C 1 X 3 = X3 8 2 X1+b 2 X2+ C 2X 3 = 0 ⇒ 8 3 X1+b 3 X 2+ C 3X 3 = -X1 a 1 = 0, b 1 = 0, C 1 = 1 - 0, b 2 = 0, c 2 = 0 a2 - 83 = - 1,b 3 = 0, C 3 = 0 Engi neering Mathematics = -2 55 x (-3A - 2 I )-2 54A = 5 1 1 A + 5 1 01 ⇒ Sol-42: (a) Eigen val ues of a real and symmetric matrix are always real. For eigen values I M - 'A, II = 0 ⇒ ⇒ 0 1 A, 0 0 A, -1 0 A, Sol-43: (b) = 0 Eigen val ues : J A - 'A, IJ = 0 - 3+ = 0 A, A, ⇒ 2 ⇒ 'A, ( 11,2 + 1 ) = 0 ⇒ A, Sol-39: (b) = 0, ± i We know that the number of eigen val ues of Anxn is n and eigen values of real symmetric matrix are always real . Hence for real symmetric matrix A2n x 2n there m ust exist 2 n real eigen values which may or may not be repeated. Sol-40: (a) The characteristic equation of given matrix A is I A-'A,I I = 0 ⇒ 2 -3 = 0 A, 11, 1 1 � ⇒ 3 'A, + 3'A, + 2 = 0 A2 + 3A + 2 1 = 0 . . . (i) Multiplying by matrix [AJ -1 ⇒ ⇒ ⇒ A4 = (3A + 21 A2 2 21) = 0 = -(3A + 21) = 9A2 + 41 + 1 2A = -2 7A - 1 81 + 1 2A + 41 = -1 5A - 1 4 1 AB = (1 5A + 1 41)2 2 = 225A + 1 961 + 420A ⇒ = = � "'1 -675A + 42 0A + 1 961 - 4501 -2 55A - 2 541 Ag = -255A2 - 2 54A = 0 = I 2, 2 (Repeated) [� �] p ( A- 2 I) = 1 & = Null ity Order - Rank = 2 - 1 = 1 Hence, for 'A, = 2, only one LI Eigen vector. Sol-44: (b) We know that for any square matrix Sum of eigen values = trace of the matrix ⇒ (-2 ) + (6) + 'A, = 1 + 5 + 1 { ' 'A, ' is the required eigen value} ⇒ = 3 A, I P I =:= Product of eigen values = 0 P 1 1 P22 - P 1 2 P21 = O Sol-46: (b) ⇒ A + 31 + 2A- 1 = 0 A2 + 3A + 2 A- 2 1 = Now, C.E. is using (i) Sol-41: (a) I � 'A, Sol-45: (c) By Cayley-Hamilton theorem ⇒ 1 25 I A- 'A,I I = 0 Consider AX = 11,X ⇒ ⇒ For 'A, = 5, [ 3� (A- 'A,I) = 0 A, -3 � A, [ 2 � [ = ] ] ] - [� _ :][::] ] [ :: X1 = [ �2 k - [� ] k] IES MASTER Publication 1 26 Eigen Values and Eigen Vectors � Sol-47: (a) Let -x + 2y = 0 [- : : ] [ :: ] = [ �] For 1c =-5, [� �l A = 1 C = [� � l = [-k: 2] ~ [-�] B =[ � 1 D = [� �l �] 1 X = 2y Now, let y = k, x = 2k the n (given) 1 a + b = 0 + 1 /2 = 2 Alternative Solutio n So, 1 Eigen values of A are 1, 0. Eigen values of B are 0, 0. Eigen values of C are 1 +i,1 -i and the eige n values of D are -1,-1 . ⇒ I A- All = 1 � (1- A)(2- A) = 0 A A1 Given matrix, A2 ] � The characteristic equation is \A- 1cl \ = 0 -A = 0 � 1 �- A l ( A-1) ( A- 2) = 0 ⇒ ⇒ ⇒ A = 1, 2 Thus the eige n values are 1, 2. It x, y be the compo ne nt of a n eig e n ve ctor corresponding to the eigen value I, then (A -:- Al) X = 0 ][x] 1- A 2 [ 0 2- A y I � � I [� = [ ] � J y = 0 So, ⇒ X = k, k ER = 1, X 1 = [ :] = 1, X2 = [; ] Ax = AX [ � � 1 ] [: ] = 1 [:] + 2a = 1 a =0 [� �][;] 1 = 2[ ] � + 2b = 2 b = =O correspo nding to I = 1, we have Let ⇒ 2 -0 2- A1 = 1, 2 He nce only one matrix A has an eigen value 1. Sol-48: (b) A = [� /c Sol-49: (c) 1 a + b = O + __! = __! 2 2 For any matrix, sum of eigen values = sum of the pri ncip le diagona l eleme nts (or trace of the matrix) ⇒ 1 + 0 + p = 3 + sum of other two e ig e n values ⇒ Sum o f other two eige n values X 1 = [ �] - [� ] = [ : ] (give n) =p - 2 Sol-50: (b) .-. [ � ] eigen vector for eige n value I = 1 Correspo nding to I = 2, we have [P] = [ : � ] its eigen values are give n by 5 4 \ P- AI\ = I ; A _ / _ Al = 0 Engineering Mathematics ⇒ Eigen val ues of I are 1, 1 ( 4 - )..,) (-5 - )..,) - 1 0 = 0 Eigen values of + X I are (-2+1) & ( -3 +1) i.e. -1, -2 ( A+5 ) ( )..,- 4 ) - 1 0 = 0 )..,2+ )..,- 2 0 - 1 0 ( A+ 6) ( A- 5) = ⇒ Eigen values o f (X+l f1 me -1,- i 0 Given ( X+l f1(X+5 ) 1 = ( X+l f1( X+l+4 1) = 0 ).., = -6, 5 Note : Only (b) satisfies sum of eigen values of P . (t rick for Exam) Sol-51 : (d) Multiplying by matrix [ PJ-1 P 2 + p + 2 1 + p-1 = 0 ⇒ 2 P+2 ) 1 ⇒ p-1 = -( P + Sol-52: (c) Trace o f matrix = sum of eigen values Determinant of matrix ⇒ ⇒ ⇒ ⇒ ⇒ )..,1)..,2 A-1(-2 - A-1) = -2 =product of eigen val ues = -35 = = ( X+l f 1( X+ ) 1 +4( X+l f 1 = I+4 ( X+l f1 = trace According to Cayley- Hamilton theorem , every square matrix satisfies its own characteristic e quation P 3 + P 2 + 2 P+ I = 0 ⇒ A-1+ A-2 The eigen values of 1+4( X+l f1 = 1 + 4 ( -1) & 1+4 (- i) p= -7, 5 Sol-53: (d) ⇒ :. Eigen values o f matrix P are 1, 2 and 3 5 3 -1- A -3 -1- ).., 6 3- ).., 0 0 ( A- 3)( )..,2+2 A.+1 0) Eigen values of X are -2 , 3 A- 1 = 1, )..,2 = 2, ).., 3 = 3 say, Eigen vector for ).., = 3 ⇒ I A - Ai l Sol-54: (c) ] � � ; ⇒ A- 2 = 5, -7 : . Eigen values are ( -7, 5) . Characteristic e q is [ Remember : the eigen values of upper triangular matrix (U T M) or l ower triangularmatrix ( L TM) are the leading diagonal elements. [ P - )..,l] [ X] = 0 -35 ( A- 1+7) ( A1- 5) = 0 = = -3 & -1 Sol-55: (b) )..,f+2 )..,1- 35 = 0 A-1 1 27 ).., = 0 = 0 = 0 = 3, -1 ±3j Now let x 3 = k then x 2 = 2k & x 1 ⇒ [X 1 : X 2 : X 3] = [ 1 : 2 : 1] = k : . Eigen vector corresponding to eigen value 3 is [1 2 1 jT Sol-56: (a) A = 1 0 0 0 2 0 0 0 3 0 0 0 0 0 0 . . . ... ... n = diagonal Matix nxn IES MASTER Publication 1 28 Eigen Values and Eigen Vectors ⇒ Eigen val ues of A are the co rresponding diagonal elements i.e. 1, 2, 3, ....., n Sum of all e igen val ues = 1 + 2 + 3 + .... + n n(n + 1) = 2 Sol-57: (d) tc1 = 4 & tc2 = 8 Let Ia ii = 1c1 + 1c2 & I A I = 1c1.1c2 2 + y = 4 + 8 & 2y- 3x = (4)( 8) So y = 10 X =- 4 & Sol-58: (a) Corresponding to I = 4, we have X -y = 0 x = y = k (let) then, E igen Vector is [� ~ [ �] ] Sol-60: (c) Eigen val ues of a skew symmet ric matrix a re either purely imag inary or zero. Sol-61: (c) Eigen values of a real symmetric matrix are always real. A X = 1cX Sol-62: (b) Consider AX = 1cX where x is an e igen vector of A ⇒ 1c = 1 Sol-59: (a) Given , matrix !] A =[ � The cha racter istic eq uation is IA - lll = 0 ⇒ ⇒ -A =0 1� !- Al 1cm x ( where X is an Eigen vector of Am ) A m x = 1cm x where X is an eigen vector of A m . Hence A and A m have same eigen vecto rs. So [1 2- 1]T is one of the eigen vector of M3 also. Sol-63: (b) C. eq uation is 1 2 - 51 + 4 = 0 (I - 4) (I -1) = 0 = 1, 4 Thus the eigen values are 1, 4 If x, y be the component of an eigen vector I A- tclI = 0 1 0 - 1c -1 4 = O 1 1 8 -1 2- tcI 1c 2 - (-2)tc +(- 120 + 72) = 0 corresponding to the e igen val ue I, then (A- ?, I ) X = 0 ⇒ 1c = 6,- 8 Corresponding to I = 1, we have [ � �] [ : ] = [ �] X + 2y = 0 So, let y =k , x = -2k. So , Eigen Vector is Sol-64: (a) •: G iven matrix is an U.T.M. & We know that the diagonal elements of an uppe r triangular matrix are eigen values. So 1 , 4, 3 are the eigen values of g iven matrix. Sol-65: (b) Acco rding to Cay ley-Hamilton theo rem, every sq uare matr ix satisfies its own characteristic eq uation. The characteristic eq uation of g iven mat rix A is Engineering Mat hematic s IA- l I A ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ -5 , 2 A A 2 -3 - A, = 0 = 0 = +5 A + 6 Sol-67: (b) ⇒ ⇒ A 2 + 5A + 61 = 0 A -5A 3 = Sol-66: (b) 2 - 6A -5( -5A -6I ) - 6A A3 = 1 1 ; 8 � Al 9 3 = ] 3- A 2 - 8 A+ 12 A = A A + 30 1. A1 = 1 A = [ : :] = AX ⇒ ( A- 6 )( A - 2 ) = 0 I = 2, 6 -1 , X = [�1] A2 = -2, X2 = [ �2] Let 0 3.4 8, 1 3.53 Sol-68: (d) 9 0 = Prop. : ' Sum of ei gen values = trace (A )' is satisfied by only (b) The characteristic equation is IA - All= o 5 -A [1 = 0 ( - A )( 8 - A) = 25 Alternate method A 3 + 5A 2 + 6A = 0 A A 9 0 3 = AX Thus the ei gen values are 2 , 6 fI x,y be the component of ei gen vector correspondin g to the ei gen value I , then (A - Al)= 0 = Correspondin g to I x +y = [ � �][ :] = [ � ] [� 1 = =k ,x = -k 6, we have , :3] [ :] = [ � ] So , Ei gen Vector = and a - 2b=-2 c -2d =4 } & ei gen vector is So , norma il zed ei gen vector for ( A- 2) Correspondin g to I ...(1) 2 , we have 0 i.e. , when y 1 29 3k ⇒ a [�] X = 3y [ :]= [k ] - [ �] So , Normalized Ei gen Vector for ( A a 6) a [�] From ( ) 1 and (2 ) ⇒ = = a 1 , c = -2 , d= - 3 1 I 0 [ � �] so C.Equation is I A - Al = ( O A) I l � ( o�\) Now consider A X= AX ⇒ or 0,b = A = [ �2 -�] Sol-69: (a) Given A ...(2) O [ � A = 0 ⇒ A= ±i (A - ·A I ) = 0 0�\](::) = (� ...(1) ) IES MASTER Publication 1 30 E igen Va lues and E igen Vectors A. max. = -1 Case-I : Put 'A, = j i n equation ( 1 ) A.min. = -3 i.e. ⇒ Now let Sol-74: (b) X2 = K then X1 ·: Eigen values of real symmetric matrix m ay be -ve also. = jK Sol-75: (c) So The given m atrix equation can be written as W e get X2 = So X1 = j [� ] j [� ] X2 = and So option (a). Notice that, the m atrix 'C' can be obtai ned from B ' ' by i nterchanging R2 and R3 . [� ] x( -j) x( - 1) [ �j] [� 1] Sol-70: (c) Eigen values of a symmetric matrix are always real. Sol-71 : (a) Let then 5 ⇒ 5 2 2 7 12 - 'A, 7 5-'A, = O 'A,3 - 20'A,2 + 33'A, = 0 'A, = 0, 1 0 ± ./ITT Sol-72: (a) The eigen values of a real symmetric matrix are always real . Sol-73: (0.31 to 0.35) Characteristic equation is ⇒ ⇒ ⇒ , IA- 'A I1 = 0 1 -A. -1 6 = 0 -6 -1 1 - 'A, -6 -1 1 5 - 'A, 'A, = -1 , -2, -3 I E23 = [� _ _ [ 1 � Now E,,B = i .e. E 23B = C Characteristic eq is I A - 'A,l j = 0 3-'A, . . . (i) AB = C Similarly when we put 'A, = -j in ( 1 ) I �]ol 0 � [� � m:: � u si n g R 2 { HJ Compari ng (i) and (ii), A = E 23 = [� � 0 1 B R3 l �i . . . (ii) 0 Now, C. equation of A is jA-'A,lj = 0 ⇒ ⇒ ⇒ Sol-76: (0) 1 - 'A, 0 0 0 -A. 0 -A. 1 1 = 0 2 (1 - 'A,)( 'A, - 1) = 0 'A, = 1 , 1 , -1 Eigen vectors corresponding to distinct eigen val ues of real symmetric matrix are orthogonal to each other. So, Sol-77: (6) Let the given m atrix be 'A' Engineering Mathematics I A-All 0 1-A ⇒ 0 1- 0 1 0 1 0 0 1-A 1 1 A, 1 0 1 0 • (1 - A) 1-A 0 1-A 0 1 1-A 0 0 1 1 (1 - A) or 2 1 1-A 1 0 0 1 1 -(1) for A = 4 1 1 = 0 0 1-A 1 1 1 1-A 1 = 2 2 BI I [(1 - A) - 1] = = 0 0 0 . . . (1 ) [using ( 1 )] ⇒ So A (A -2)(A - 3) A = = Hence required product 0, 0, 0 , 2 , 3 Sol-78: (a) = 2 x 3 = 6 So, determ i nant is negative ⇒ there exists at least one eigen val ue, which is negative. I A- All = 0 = x [ ;Ht]- [!] k (let) 0 ⇒ [:: ] [�:l = x1 y1 + x2y2 + x,y, = 0 Summation of eigen values is equal to trace of matrix. Sol-82: (1) A2 = 1 = A = A = . . . (1 ) . . . (2) A-1 If 1c is an eigen val ue of matrix A then A A2 = 1 ·: Determinant of a matrix = product of its eigen values. Sol-79: (d) = T h e eigen vecto rs of a real sym metric m atrix corresponding to distinct eigen val ues are orthogona l . From ( 1 ) and (2) 0 = = We know that AK1 -A (A - 3) [A(A -2)] = 0 3 9x = 2y Sol-80: (d) Sol-81: (a) So (1 - A) B I l - (1) IB I ⇒ ¾k , so eigen vector y so, XV 2 = [ �] satisfies eigen vector. 1 Let ⇒ 0 so, eigen vector is [� ] ~ [� ] for A = - 3 , option 'd' 1 1-A [=: �] [�] ⇒ 1-A 1- A, 1-A = for A = -3 1 1 0 1c = 4 , -3 Let y = k, x = 1 1-A 1-A ⇒ A, 1 1 1 A2 - A - 12 ⇒ 0 1-A 0 0 0 0 0 0 1-A 1- = (-5 - A) (6 - A) + 1 8 0 1 1-A = 0 0 1 1 0 1 1 Expanding by 1 st row = 1 31 A ± 1 Positive eigen value of A = 1 . Sol-83: (9) Let A1 , A2 are the Eigen values, then IES MASTER Publication Eigen Values and E igen Vectors 1 32 11,1 + 11,2 = -6 {Trace) IAI = 11,1 .11,2 ( P roduct of Eigen Val ues) 2 1, 2 I AI max = (1 1 ) = ( -3) = 9 & So, Sol-84: (b) Alternative Solution A = [ � Characteristic equation is : For singular matrix, determ i nant = 0 P roduct 1 of eigen v alues = Value of determ i nant = 0 :. Atleast one eigen val ue should be zero Sol-85: (2) A = Characteristic equation. ⇒ A2 - n + 1 0 = o ⇒ 11, = 2,5 ⇒ ;J P = 2 sum of eigen value = 11,1 + 11,2 = 2 + P product of eigen values = 11,11c2 = 2P - 1 ⇒ (1 ) (2) ⇒ 311,z + 1c2 411,2 A2 ⇒ ⇒ ⇒ 3A.22 3( P:2 J p 1 = = 2 2 = = + p 4 P - 1 2 P -1 2, . . . (2 ) ⇒ and + p P+2 = -- = 2 ... ( 1 ) ⇒ At = 311,2 14 3 from options. P � = 3: 1 . 53 -3 - 11, 1 0 0 -2 a -2 - 1c -1 - 11, 0 = 0 (-3 - 11,)(-1 - 11,)(-2 -11,) = 0 a = \ 3 + 11,)(1 + 11,)(2 + 1c) 2 Now using maxima-minima concept for 'a' Let 11,1 and 11,2 be the eigen value of matrix A. "-2 A1 ( 2 - 11,) ( P - 11, ) - 1 = 0 The characteristic equation is \A - 11,I\ = 0 1 2 - 411, - 311, + 11, - 2 = 0 � = 3 = 0 Sol-87: (b) 2 "-1 3 = "'2 1 for I Hence ratio of two eigen val ues = ( 4 - 11,) (3 - 11,) - 2 = 0 [� P -11, Option (D) 1 4/3 above equations gives value 5, 5/3 (where I is the identity matrix) 4= 0 1 � 3 - �1 A = 1 By putti ng values of \A- 11,I\ = 0 Sol-86: (d) 1 \A- I11,\ = 0 11,2 - ( P + 2) A + (2 P - 1) = 0 [� �] Lowest eigen value is 2. 2 - 11, ;J At da = 0 d1c 2 \311, + 12A. + 1 1) 2 A, = = da2 d1c 2 A, da2 d11, 2 1 -2 ± J3 (-611, - 1 2 ) = ( = ✓3 (- 2 _k) = - ✓3 _k} 6 >0 : . We get maxi mum value at A, -2 + -6 = -<0 A, = da 2 2 d11, 0 ( -2 + _k) . . . (i) Engineering Mathematics So 'A. = (-2 + �) is the point of m axima and max 1 = value of 'a' = [- _!_ (3 + 'A.)(1 + ,,) (2 + 'A.)] 2 3-v,,;3 A=-2+� Sol-88: (d) Given that A = (: :J has eigen val ues -1 and 7. 'A. 1 · 'A.2 So Trace (A) = 'A.1 + 'A.2 and I AI = a + 1 = -1 + 7 For eigen val ues IA - 'A.I i = 0 ⇒ -2 3 - 'A. 4 4 - - 'A. 2 Sol-89: (6) ⇒ ⇒ 'A. +2 (20 - 4'A. - 12)+2(-12 + 8 + 2'A.) = 0 ⇒ 'A.3 - 4'A. 2 + 5'A. - 2 = 0 = 0 1 � 'A.I (4 - 'A.)(1 - 'A.) - 1 0 = 0 ⇒ ('A. 2 - 5'A. - 6) = 0 'A. = 6 and - 1 Larger eigen value = 6 ⇒ ⇒ Sol-93: .(b) G iven that Sol-90: (b) 1 Let ⇒ Consider A = [ � � AX = 'A.X -2 � ] (A- 'A.l)X = 0 For 'A. = 1, (A - l)X = 0 Largest eigen value = 2 AX = I A- 'A.II = 0 I ; ⇒ 'A, = 12 & 2'A. = p + 7 p = 17 10 5 + j A = X 20 [4 2 Ae = ⇒ 'A.X [A]T l [A] = [I x 10 = 5 j 20 [ 2 � · : Ei gen values of A are real :. A is Herm itian matrix ⇒ ⇒ x + 2y = 0 and -y + 2z = 0 Let z = k, then y = 2k and x = -4k ,.. So = 0 ( 3 - 'A.) (-20 + 4'A. - 5'A. + 'A.2 + 1 8 ) Sol-92: (17) 4 5 - 'A. Smallest Eigen Val ue = 1 b= 3 C.E. is -3 2 6 'A. = 1 , 1 , 2 a = 5 and ⇒ Sol-91: (d) a - 4b = (-1 )(-7) and 1 33 1 5 � j 2XO : [ 4 2 -10 _;o] 5-j 20 2 _;o] 5 j : = � 2� [ 2 -1 0 4 1 _u l x = 5 + j and x = 5 - j :. The value of x = 5 - j Sol-94: (2) A = [� Hl IES MASTER Publication 1 34 Eigen Val ues and Eigen Vectors So for eigen vectors IA- 11,II = 0 , 1 0 2 - 11 0 2 - 11, 0 0 3- 0 ⇒ ⇒ "' = O A. 2-11,)[(2 11,)(3-11,)-0]-1(0 - 0) + 0 = 0 ( = 3, 2, 2 F o r 11, = 2, we a re n ot s u re a bo u t n u m ber of independent eigen vectors as it is repeated eigen value so we wil l check p (A-2I) = ? A- 21 = [� � 0 0 -1 p (A-2I) = 2 � l So, number of LI Eigen vectors for (11, = 2) = O rder - Rank = 3 - 2 = 1 . So, overall we have two independent eigen vectors corresponding to "' = 3 & 'A = 2 respectively. Sol-95: (a) • If 3 l inear equations in 3 u nknowns are having a unique solution then equations are li nearly independent. • Equations are l inearly independent when determ inant of the coefficient matrix is non zero. p 3 -P = 0 , 3 A, - A = 0 ⇒ ⇒ ⇒ ⇒ 11, ( 11,2 -1 ) = 0 11, ( 11, - 1) ( 11, + 1 ) = 0 A, = 0, 1 , -1 Sol-99: (a) From the property of eigen values, if 11,1 , 11,2 , . . . . An a re the eigen values of a m atrix A, then Am has the eigen values Hence the eigen values of the m atrix A2 - 3A + 41 wil l be ( 1 )2 - 3 ( 1 ) + 4 = 2 2 ( -2 ) - 3 ( -2 ) + 4 = 4 & Sol-100: (a) ⇒ Sol-96: (3) ⇒ *o ⇒ Eigen values must be non A 3 x 3 matrix will have 3 eigen val ues and according to the question there is only one non-zero eigen value. Hence the eigen val ues will be 0, 0, 11, . Sum of the eigen values = Trace of the given matrix 0 + 0 + 'A = 1 + 1 + 1 A, = 3 ⇒ Sol-97: (a) • If 3 l inear equations in 3 unknowns are having a unique solution then equations are li nearly i ndependent. • Equations are l inearly i ndependent when determ inant of the coefficient matrix is non zero. • ·: Determinant of the matrix of the m atrix. So Determinant zero. Sol-98: (d) *o ⇒ = product of eigen val ues Eigen v alues m ust be non p3 = p According to the Cayley-Hamilton Theorem, every square matrix satisfies its own characteristic equation. 'A� , - - · ·"'� ( m being a positive integer) and the corresponding eigen vector is the same . Hence, by using this property, eigen values of any polynomial of A can be obtained by replacing A by 11, . • ·: Determinant of the matrix = product of eigen values of the matrix. So Determinant zero. 11,r , IA- A ll = 0 2 A = 0 l � k� I A. (2 - 'A)(k - 11,) - 1 = 0 ⇒ ⇒ 11,2 - 'A(k +2) + (2k - 1 ) = 0 ⇒ ⇒ For both values of ' "' ' to be positive, this implies to 11,1 + 'A2 > 0 2k - 1 > 0 ⇒ 2+k>0 k > 1 2 k > -2 and 11,1 .11,2 > 0 Now taking i ntersection of both inequalities, we get 1 k >2 Sol-101: (d) Assume any matrix in which sum of elements of any col umn is 1 and then proceed. For example, A = [ � ; -�] -4 0 1 Now we can fi nd eigen values of A by using I A-11,I I = 0 ⇒ "' = 1. Note : If sum· of the elements in each row/column of a square m atrix is equal to a then a is an eigen value of that matrix. Eng ineering Mathematics Sol-102: (1) ⇒ ⇒ Sol-105 : (-6) ·: A h as the eigen values -1 , 1 , ! � ⇒ (A 1: Matrix A = [-6 -4 -9 + x ] 3 [(1 )3 eigen value of mati rx A = 0 determ i nant [A] = 0 Hence, ⇒ ⇒ ⇒ ⇒ ⇒ + x) - (--4 )(1 3 )] -2[9 (-9 + x) - ( 1 3 )(-6)] + 4 [9 (--4 ) - (-6)(7)] = 0 3 [-6 3 + 7x+ 52] -2 [-81 + 9x + 78] + 4 [-36 + 42] = 0 -33 + 21 X + 6 - 1 8x + 24 = 0 -3 + 3x = 0 3x = = 1 X Sol-103 : (d) ⇒ 3 ⇒⇒ ⇒ "- 1 ·"-2 = IAI 2 (J2 + CO2 = cr - cox 2 CO = -cox (0 = -x X = -co Given, then , ⇒ ⇒ ⇒ Now, or - [ P - 1 1 3 = [� 1 0x2 + 1 0y2 + 12xy = 1 JA- 1clJ = 0 or ⇒ ⇒⇒ -1c I -1 - �I = 0 2 "- + 1 = 0 2 "- "- = -1 = i , -i A3 x 3 · : Com plex roots occurs in pai r only then 3 value can be given as 1c 3 = 2 - i . Now I P I = product of eigen value ⇒ ;J[:] ( 3x + y)2 + (x + 3y)2 = 1 C equation is = 5 X X (2 - i) 3 : 15 X rd eigen 3 G iven that 1 , 2, 4 are eigen val ue of A. So I AI = product of eigen values ] and IAI = 8 1 JA- I = 1 lAf = 1 1 J(A- ) I = A- = AI , f I [ ·: D eterm i n ant of m atrix = determ i nant o f it's transpose] Now, 3x + y = a and x + 3y = b a 2 + b2 = 1 Sol-106 : (a) Sol-108: (0.125) [:] = P [:] [:] = Sum of eigen val ues = - 4 - 2 + 0 = -6 = (2 + i) Sol-104 : (d) 3 - 3A2 ) Let 1c1 = 3 and 1c2 = 2 + i are the two eigen values of det [A] = cl - cox ·: Eigen values of m atrix A are (cr + jw) and (cr- jco) so 2 - 3 (-1 ) ] , = --4 , -2, 0 Sol-107: (15) A = [ : :] 3. ) has the ei gen values [(-1 )3 3 2 2 3 ( 1 ) ] a nd [( 3 ) - 3 ( 3 ) ] - 3A So, Trace (A3 IAI = 0 3 [7(-9 - 2 1 35 . . . (i) Now comparing it with ax2 + by2 + 2hxy = 1 , we get a = 1 0, b = 1 0, h = 6 h2 - ab = (6)2 - (1 0)(1 ) = -64 < 0 So (i) represents an ellipse with major axis is al ong y = -x ,., [ � ] and m i nor axis is along y = x ,., [ � ] 1 So, option (d) is correct. 1 T Sol-109 : ( c) Characteristic equation is IA- 1cll = 0 1-A 5 4 3 2 2 3 4 5 4 2 3 1-A 2 3 5 1-A = 0 5 4 2 1-A 5 1-A 3 4 IES MASTER Publication 1 36 E igen Val ues and E igen Vectors 4 5 2 3 1 5- A 2 3 4 1 5 - k (1 - A ) 2 3 (1 - A ) 5 15-A 2 4 5 15-A (1 - A ) 4 (1 - A ) 3 5 15-A 2 3 2 4 3 2 On solving it; A = 2 so given Eigen Vector corresponds to = 0 A Sol-110: ( c) 5 4 AX = AX A = 0 0 -1 1 2 0 a i � im m (A - l )X = 0 = 1 5 is the real eigen val ue. 3 2 =2 Thus for A a + 1 by laki ng X a [� we have l 1 (1 - A ) 3 (1 - A ) 5 = 0 or (1 5 - A ) 1 2 4 5 (1 - A ) 1 4 5 (1 - A ) 1 3 So only A ⇒ 2 2 1 1 - x + -z = 0 2 2 -2y = 0 1 2 & 3 - So Eigen Vector for 0 So y = 0 & x = -z. Let z = k then x = -k 2 =1, A To find the eigen val ues of A, det(A- A l3 ) = 0 i.e., 3 --A 2 0 1 2 0 -1 - A 0 1 2 0 3 --A 2 Hence option (c) satisfies. Sol-111: (a) = 0 A To fi nd eigen val ues, equate I A- Al I = 0 (% - A ) [(-1 - A {¾ - A )] + i[O - i (-1 - A)] = 0 (- 1- A {(¾ - A r-±] = 0 Now, let X be an eigen vector of A associated to then 3 2 AX = - 0 0 -1 -1 2 0 - AX 0 1 0 -3 (-4 - A ) = 0 On solving, ⇒ ⇒ -(1 + A)(A - 1)( A -2) = 0 Hence eigen val ues of A are -1 , 1 and 2. So, A 1 0 -A -(1 + A ) [ A 2 -3A + 2] = 0 or a [� }3 !] A , (-A) [ (-4 - A ) - (-3)] - 1[0] + 0 = 0 A ⇒ ⇒ 2 A 3 (4 + A ) + 3A = 0 + 4A2 + 3A = 0 A(A2 + 4A + 3) = 0 A Sol-112 : (a) For eigen value = (0, -1 , -3) 5 A IA- A ll = I 4 1 =\I = 0 ⇒ 1 37 Engineering Mat hematics (5 - 11,)(1- 11,) + 4 = 0 5 - 5 11,- 11, + 11,2 + 4 2 A - 611, + 9 = 0 = 0 2 = 0 (11,- 3) . 11, = 3, So it has multiplicity t' wo .' IPI = 1 (i) (ii) 3 p For eig en v ector ⇒ L et x = k , th en y = 2k Sol-113: (a) + 2x 2 = c 1 3 1 ; "' 1 � 11,1 11,2 - 4 11,- 3+ 8 11,2 - 4 11,- 5 Sol-114: (b) = O 2 ✓ 0 ✓ 12 2 ✓ 2 ✓ 2 ✓ 0 1 -12 0 ✓ = [ � � 0 1 �i 1 0 1 0 1 0 ✓2 -1 0 1 ✓2 = I 0 0 1 p-1 = (iii) Sol-117: (0) 12 ✓ 0 0 1 2 -21 0 ✓ 0 ✓ 0 12 = pl 11, = 5 70 8 0- 11, I = O (5 0- 11,) ( 8 0 - 11,) - 4 90 0 = 0 Solving quadratic egn. 0 ✓2 - 12 0 ✓ ✓ 5 0 - 11, 1 70 = 0 = 2 12 Charact eristic equation is = 0 2 0 1 _ 4 -3 3 0 2 -1 p = 0 0 1 1 0 x 11, = 5 0 = O = 2 (3 - 6) + 1 ( 8 - 0) Sol-115: (d) ✓ Sol-116: (5) Product of eig en valu es = D et erminant of matrix Product of eig en valu e = IPI u 1 2 ✓ H enc e (iv) is wrong. 2 = [A ] 1] I [A] - 11,(]I I 2 ✓ 4 x 1 + X2 = c 2 3 0 1 -12 0 H enc e P is orthogonal as p_ pl = I So, only on ei _nd ep end ent eig en v ector exists for "' = 3. [4 ✓ ✓ y = 2x. So 3x 1 = p_ pl = [A - 11,I] [ ] = 0 ; 2x -y = 0 l 12 1 0 1 2 ✓ For 11, = - 6.5 89, + 1 36.5 89 "'1 = - 6.5 89 1 37 70 X 1 = [ 11, - 5 0 ] = [-5 ;_� 89] = [- - � ] 1 For 11,2 =1 36.5 89 5 89 80 X 2 = [ "'\- ] = [ \� ] = [ : ] 1. 37 o xTx2 = [-1.2 37 1 1 [ _; ] = (Ol 1 37 IES MASTER Publication 1 38 Eige n Values an d Eige n Ve c t ors Sol-1 1 8: (b) ⇒ IA = I 0 one eigen value must be For n > 2, p(A) = 2 an d A n • n i.e., p(A) <n [ 5, 5] E (1 . . ) is true For n = 2 an d let A = [� ⇒ eigen values = 5, 5 '0' Sol-1 19: (c) 1- 11, -1 Character si tic equation is 0 5 - 11, 0 -6 5 6 =0 5 - 11, 2 11, =1, 11, - 1 011,+ 61= 0 ⇒ 11,=1 an d "' = 5 ± j6 Sol-120: (5) Since one eigen value of M is 2 2 - 4( 2)2 + a( 2) + 30 = 0 : . Characteristic polynomial is a = -11 2 11,- 4 11, -1111, + 30 ( 11,- 2)( 11,- 5)( 11,+3) = 0 = 0 "' = 2.5 - 3 Largest absolute value of "' is 5. Sol-121: (a) By property of "minimal polynomial ". If matrix has 15 distinct Eigen values, then its minimal polynomial must be of at least 15 degree. Sol-122: (b) A= [� �] ·: it is U T M so 11,1 =1 , 11,2= 2 Now let Eigen vectors are X 1= [ :] & X 2= [:] Accor ding to definition, we have, AX 1 = "'1X 1 & � = "'2X 2 [ � �] [ : ] = 1 [ : ] & [ � �] [ : ] = 2 [ :] [ (1 �:a )] 0 & b = 1 2 = Sol-1 23: (c) 1 2 ·: Sum of eigen values ⇒ "'1+ "'2+ A3 = 9 ( I)I is false. ⇒ = Hence a + b �] One eigen value must be in [ 5, 5] an d largest eigen value magnitu de is not greater than 5. ⇒ ⇒ a ( 1 �:b)] �] = [:] & [ = [ 2 = Trace (A) Only option (c) is satisfying it < Sol-124: (d) 2 -4 A = [ 4 -2 ] ⇒ = 11,1 an d i1 + "-, = Trace (A) = O "- 1 • "-2 = IAI = 1 2 - 11,2 an d ( 11,)(11,)1 = 1 2 1 11,f = -1 2 ⇒ 3 11,2 = -2 ✓i 3 "'1 = 2 ✓i Hence, Eigen values are complex an dconsequently Eigen Vectors will also be complex. Sol-125: (c) "Eigen vectors correspon ding to distinct eigen values are linearly in depen dent " Hence S 2 ⇒ S 1 Sol-1 26: (5.5) Let 11,1 an d 11,2 Eigen value of A then 11,1 + 11,2=4 an d 11,f an d "'� will be eigen value of A 2 then 11,f + 11,�=5 . 2 = 11,f + 11,� + 2 "' 1 "'2 (4) 2 = 5 + 2 11,1 11,2 IA I = Pro duct of values 11 = "'1 "'2 = 2 = 5.5 ( "'1 + 11,2) So, Sol-127: (1) C. Equation of A is IA - 11,II = 0 (1- A) -1 0 ⇒ ⇒ 0 ) (2 - 11, 0 -1 0 (- 2- 11,) ( - 2 - 11,) [(1- 11,) (2 - 11,) - 0 ] (11,+ 2) (A- 1) (11,- 2) = 0 = 0 = 0 3 2 "' - "' - 4 A.+ 4 = 0 So by Cay e l y -Hamilton theorem ; = Now, B Hence, B = A3 - A 2 - 4A + 5 1 = (A 3 - A 2- 4 A+4)1 + 1 ⇒ IBI = 1 = 0+ I N ORMALISATION Defi nition : The proce sso f mak ni g d iagonal element un ti y ( ni echelon form) i s called normal isat o i n. ILL-CONDITIONE D SYSTEM Con sider the sy stem X 1+ 2x 2 =1 0 1.1 X 1+2x 2= 1 0.4 in . Obv oi u sly, th is sy stem ha s the solut o i n x 1 =4 , x 2 =3. Suppo se we mod fi y the coeff ci ei nt of x 1 ni the second e quat o from 1 .1 to 1.05, then the so u l t oi n change s to x 1= 8, x 2= 1. Th is show s that somet m i e s re sult s g v i e h gi h va riat o in due to slight change s ni the coeff ci ei nt s of the var ai ble s or ni the r gi ht hand side. Such type of sy stem s are called un stable sy stem or li l -cond ti ioned or ill-po sed sy stem s. An li l -conditioned sy stem i s that where the small change s in the coeff ci ei nt s re sult s in the large change s ni the solut oi n. On the other hand, a well-cond ti oi ned or well-po sed sy stem is that where the small change s ni the coeff ci e i nt s re sult s ni a sim ilar small change ni the solut oi n. DIAGO NALLY DOM I NANT SYSTEM · fI for each row, d ai gonal element ha s la rge st numer ci al value among all the value s ni that row, then sy stem i s called d ai gonal dom ni ant sy stem. nI thi s sy stem there is no need of any type of P I VO T I N G and it can be solved by u sual method. GAUSS ELIMINATION M ETHOD WITH PIVOTING i j is k nown a s a p vi ot element. Each row is normal ized nI the ba sic Gau ss el imination method, the element a ii when = by d v i iding the coeff ci ient s of that row by ti s pivot element that is a k. a ki = -1 where j =1, 2,...n. a kk fI a kk =0, k th row cannot be normal ized . Therefore the procedure fa li s. One way to overcome th is problem is to ni terchange thi s row with another row below ti wh ci h doe s not have a zero element ni that po sition. tI is sugge sted that the row w ti h zero p vi ot element should be ni terchanged w ti h the row hav ni g the large st (ab solute value) coeffic e i nt ni that po sit oi n. The Steps Used in Reordering of Equations ( )i ( ii) Search and locate the large st ab solute value among the coeff ci ei nt s in the f ri st column. Interchange the f ri st row w ti h the row conta ni ni g that element. ( i )i Then el m i ni ate the f ri st unk nown (var ai ble) ni the second e quat oi n a s explained ea rl ei r, 1 40 Piv oting & Ve c t or Space (iv) When the second row becomes the pivot row, search for the coefficients in the second co u l mn from the second row to the nth row and locate the la rgest coefficient, Interchange the second row containing the largest coefficient. (v) Continue this procedure till (n - 1) unk nowns are eliminated. This process is referred to as partial pivoting. There is an alternative scheme also, which is k nown as complete pivoting in which at each stage, the largest element in any of the remaining rows is used as the pivot element. Complete pivoting requi res a lot of calculations and, therefo re, it is not generally used. Example 1 1 Solve the equations using Gauss elimination method 1 0x 1 -x 2 + 2x 3 x 1 + 1 0x 2 -x 3 - 4 =0 2x 1 + 3x 2 + 2 0x 3 =7 The given system is diagonally dominant and therefore no pivoting is necessary . Augment !,d matrix is Solution Operating R32 ( 1 0 -1 2 0 1 01 - 6 10 5 0 1 6 98 5 5 _E), 1 01 The equivalent system is 1 0x 1 - X2 4 13 5 31 5 2 1 0 -1 0 1 01 -6 5 10 0 0 2 01 8 0 1 01 0 ⇒ + 2x 3 =4 Using back substitution, we get X1 = 0.375, 4 13 5 54 30 1 01 0 13 1 01 6 - X1 - - X3 = 10 5 5 X2 =0.2 8 9, X3 I =; �1 5430 2 01 8 0 --x 3 = 1 01 0 1 01 0 =0.2 69. Example 2 : Solve the following equations by Gauss elimination method 3x - y + 2z =12 X+ 2y+ 3z =11 2x - 2y - z =2 Solution ⇒ Here, we will rearrange the equations so that the system becomes diagonal yl dominant and therefore no pivoting is necessary. . [A : B] = [! 2 2 �1 : 3 : 11 Engineering Mathematics Operating R 21 (-f), (-i), R 31 -1 4 3 7 3 2 7 3 7 3 12 2 7 3 7 4 12 0 -1 4 3 0 3x - y + 2z = 1 2 ⇒ 3 0 Operating R 32 ( i) , 0 3 0 The equivalent system is 4 - 6 7 - 6 7 -2 7 -- y -- z = -6 3 3 Using back substitution, we get z = 2, y = 1 and x = 3 Example 3 1 1 41 ⇒ 7 = -7 --z 2 4 Solve the following system of equations using Gauss elimination method with partial pivoting x + y + z = 7 3x + 3y + 4z = 24 2x + y + 3z = 1 6 and also find the pivots. Solution 1 In matrix form, the given system can be written as ... (i) To start with, we observe that the pivot element a 11 = 1(;e 0) . However, a glance at the first column reveals that the numerically large element is 3 which is in second row. Hence we interchange the first row with the second row and equation (i) becomes ... (ii) after partial pivoting . Eliminating x from the last two rows, we get ... (iii) The se cond row in equation (iii) cannot be used as the pivot row as a 22 = 0. Therefo re, interchanging second and third rows, we get IES MASTER Publication 1 42 Pivoting & Vector Space 3 3 o Bl 0 ... (iv) 0 3 which is a n upper triang ular form Using back substitutio n, we find z = 3 , y = 1, x = 3 & pivots for this system are 3 & -1. ""' ,-N-""'� In partial pivoting, we observe that the unknown column vector remains unaltered whi l e the right OT . E _ hand side column vector gets changed. GAUSS-JORDAN M ETHOD This method is a modificatio n of the Gauss elimi natio n method. I n this method, the coefficie nt matrix A of the system of equations AX = B is brought to a diago nal matrix or unit matrix rather than the upper triang ular matrix. there is no need to apply back substitution that is why it reduces the number of operations. This method can be used with or without pivoting. Thus, Gauss-Jordan elimination method red uces the system of equations AX = B in the followi ng form : 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 X1 Xz X3 = xn b'1 b'2 b 3' b'n Example 4 : Solve the system of equations 5x - 2y + z = 4 7x + y - 5z = 8 3x + 7y + 4z = 1 0 Using (i) Gauss eliminatio n method with partial pivoting, (ii) Gauss-Jorda n method. Solution (i ) Gauss elimi nation metho d with partial pivoting , The a ugme nted matrix of give n system of equations is [5 [A : B] = 7 3 -2 1 7 4 -5 : 8 [! :1 1 = 1 -2 7 7 0 0 1 19 7 46 7 ... (i) : 1 0l 4 -5 1 4 -5 32 7 43 7 ... (ii) R 1 B R2 10 8 12 7 46 7 5 R2 ➔ R2 - - R1 , R3 7 = 3 ➔ R3 - - R 1 7 7 0 6 ,� , 0 19 7 43 7 -5 8 46 7 32 7 12 -7 , R 2 B R3 Engineering Mathematics 7 1 0 46 7 0 0 = -5 43 7 3 27 46 8 46 7 1 19 R 3 ➔R3 +-R 2 46 1 43 ... (iii) 2 4 46 , from second row, y == 8 , and from first row of matrix equation 327 3 27 46 . . 122 and p1vots are 7 and ( ..111), x == 7. 1 09 Thus, the required solution is Thus, from last row, z == X= 1 .11 93, y = 0. 8685, z = 0.1 407. (ii) Gauss-Jordan Method: We form aug mented matrix and interchanging the first and the second row to get equation (ii) 1 above in part (i). Now, we perform R 1 ➔ R 1 to get 7 [A: BJ - [7 = = 1 1 / 7 -5 / 7 / 7 1 7 -5 / 7 '. 8 / 7 � 5 2 1 . 4 = [ 0 46 / 7 43 / 7 ] 3 7 4 : 10 0 -1 9 / 7 32 / 7 [� [� 1/7 -5 / 7 1 4 3 / 46 -1 9 / 7 32 / 7 0 -39 / 46 1 4 3 / 46 0 327 / 46 8 l / 1 7 , R 2 ➔ _I__R 2 -46 -1 2 / 7 l l 8/7 46 / 7 , R 2 �R 3 -1 2 / 7 1 46 , R 3 ➔ -R 3 1 327 46 / 327 1 0 0 : 1 22 /1 0 9 R 1 ➔R 1 + 39R 3, 46 = [ 0 1 0 : 284 / 327] ' 43 0 0 1 : 46 / 327 R 2 ➔ R 2 - R 3 46 Thus, the required solution is 1 22 46 2 84 y = X = = ' ' Z 1 09 327 327 VECTOR SPACE Vector space (V) : The space formed by the vectors ??? is known as vector space e.g. R 2, R 3, R etc. Subspace (W) : Any subset of vector space. If W s;;;; V st. W is itself a V-space then W is subspace of V. Linear span : Let S be any set of vectors then set of all l inear co mbinations of vectors in S is called linear span of s i.e . L(S) = {t ci x i ; X1 E s } . (;1 IES MASTER Publication 1 44 Pivoting & Vector Space INarE5" If W is a subspace spanned by S then it W = L(S). Basis : Let B be any subset of V then B is called basis of V if (i) B consist LI set of vectors (ii) B spans V, i.e. L(B) = V, i.e. any vector of V can be expressed as linear combination of vectors in B. e.g. Basis for R 3 in B = {( 1, 0, 0), (0, 1, 0), ( 0, 0, 1 )} = { E1, E2, E 3 , } Dimension of Vector Space : dim (V) = No. o f vectors in basis of V. ___.� INOTE , {l ) If dim (V) = n, then any set of n, LI vectors forms basis of V. (2) dim (V1 + V2 ) = dim(V1 ) + dim(V2 ) - dim(V1 n V2 ) e.g. Vectors X 1 = (1, 0, -1), X 2 = (1, 2 , 1), X3 = {O, -3, 2) from basis of R 3 . (Bcoz dim (R3 ) = 3 and and X 1 , X2 , X3 are also LI) Engineering Mathematics PREVIOUS YEARS GATE & ESE QUESTIONS • Questions marked with aesterisk (*) are Conceptual Questions 1 .** The real symmetric matric C coresponding to the quad ratic form Q = 4x 1 x 2 - 5x 2x 2 is 1 45 • 4.** It is given that X1 , �• . . .� are M non-zero, orthogonal vectors. The dimension of the vector space spanned by the 2M vectors X 1 , � • ... XM, -X1 , -� •... -XM is (a) 2M (b) M + 1 ( d) (c) M [ � }5 ] (d) dependent on the choice of X 1 , � • ... XM [GATE-1998 (CE)] [GATE 2007; 2 Marks] 2.** An orthogonal set of vectors having a span that 5.** In the solution of the following set of linear e quations by Gauss elimination usin g partial pivoting 5x + y + 2z = 3 4 ; 4y - 3z = . 1 2; and 1 0x - 2y + z = -4; the pivots for elimination of x and y are contains P, Q, R is ___, where P = (-1 0,- 1,3)' , Q = (-2,-5,9)' & R = (2,-7,1 2)' are three vectors. 6.* (a) 10 and 4 (b) 1 0 and 2 (c) 5 and 4 (d) 5 and -4 [GATE-2009-CE; 2 Marks] If V 1 and V2 a re 4-dimensional subspaces of a 6dimensional vector space V, then the smallest possible dimensions of V1 n V2 is _____ [GATE-20 14 (CS-Set 3)) 7.** If the vectors, e 1 = ( 1,0,2) ,e 2 = (0,1,0) and e 3 = ( -2,0, 1) from a n orthogonal basis o f the three dimensional real space JR3 , then the vector 3 u = ( 4, 3 - 3) E lR can be expressed as m m [GATE-2006; 2 Marks] 3.** The following vector is linearly dependent upon the solution to the previous problem (a) (c) (b) (d) [;�,] [�:] [GATE-2006; 2 Marks] 5 5 2 11 (d) u = - -e 1 + 3e 2 - -e 3 [GATE-2016; 2 Marks] IES MASTER Publication -------- 1 46 Pivoting & Vector Space . AN SWE R l<.EY 1. 2. 3. (d) 4. (a) 5. (b) 6. (c) · The given quadratic• form is in 2- variables x1 and x2 • So we can represent it as Q = 0X , 1X1 +2X1X 2 +2X 2X1 -5X 2X 2 So corresponding matrix C can be represented as (2) 4 -6 -3 6 -2 3 -2 -1 7 = O 30 Hence it is linearly dependent. Sol-4: (c) Since x 1 , x 2 , . . .x m are orthogonal they span a vector space of dimension M. Sol-2: (a) We know that two vectors (a1 , b1 , c1 ) & (a 2 , b 2 , c 2) are said to be orthogonal if a1 a 2 + b1 b 2 + c1 c 2 = 0 ... (i) We are looking for an o rthogonal set of vectors having a span that contain P, Q & R. Take option (a) l & [� = �6 x 4 + (-3) x (-2) + 6(3) ⇒ [:H�] The (d) SOLUTIONS Sol-1: (d) [=;] 7. (b) = -24 + 6 + 1 8 = 0 The span of [ =;] Sol-5 : (b) Method I In the process of converting given matrix into an Echelon form , the largest absolute value in a column, that is to be used for making Echelon form, are called PIVOT for that column. [A : B ] = 1 are o rthogonal. space spanned by Since -x 1 , -x 2 , -x 3 , . . . , -x m are linearly dependent on x 1 , x 2, . . . , x m , so the set (x 1 , x 2 , x 3 , . . . , x m) will span a vector space of d imension M only. these vectors is 2 34] 4 -3 : 1 2 R 1 [ 1� -2 1 : -4 R 3 ➔R 3 - JR, [ �R 3 1 0 -2 � t is linealy dependent The vector [ -2 -1 7 3 0 upon the solution of the previous q uestion. -2 1 4 -3 1 2 0 5 1 -4 -4 12 34 ] � -3 : 1 2 2 3 / 2 : 3 6] �1 30 & [ �] contains P,Q & R. So, P Q, R can be represented as l inear combination of these two vectors by solving K1 & K2 • Option (b), (c) & (d) are not orthogonal. Sol-3: (b) IT Echelon form Here 1 0 and 4 are the largest elements that are used in making of Echelon fom. So , they can be taken as PIVOTS for C 1 and C2 respectively Method II 5x + y + 2z = 34 4y - 3z = 12 Engineering Mathematics 1 Ox - 2y ⇒ Min. value of z = -4 + Elements of C 1 are 151, I OI, 11 01 so largest element can be taken as 1st Pivot for C 1 Hence , R 1 ⇒ ... (1 ) 4y - 3z = 12 ... (2) 2z = 34 ... (3) + y + = 4 R R3 - 1 �➔ 2 1 Ox - 2y + z = -4 2 Now elements of C2 : 14 1 , 121 so 4 can be taken as 2nd Pivot for C 2 Using R 3 ➔R 3 - JR 2 ⇒ + 4y - 3z = 12 Ox + 0y + 3z = 30 Now, we have reached at Echelon form , so pivots are : X = 1 0, y = 4 e 1 = ax where , ⇒ + + K'e 2 + K2e 3 2a z where a x = i , a y = j , a 2 = k A A e 3 = -2a x + a z ⇒ ⇒ Hence , 4a x + 4a x + u = 4a X + u = Ke 1 + 3a y - 3a Z K'e 2 + K2e 3 3a y - 3a 2 = K(a x + 2a 2)+ K'8y+K2(-2a x + a 2) 3a y - 3a 2 = (K - 2 K2)a x + K '8y+(2K+ K2)a 2 On comparing , K - 2K2 = 4 K' = 3 2K + 2 K = - 3 ... (i) .. :(ii) . .. (iii) On solving we get K = - 2/5 K' = 3 K2 = -11/5 = dim(V1 ) + dim (V2 ) - dim(V1 n V2 ) dim(V1 n V2 ) = d im V1 + dim V2 - dim(V1 + V2 ) max of dim (V1 + V2) [ ·: dim(V) = 6] e2 = a y Sol-6: (2) dim (V1 + V2 ) - u = Ke 1 1 Ox - 2y + z = -4 Ox d im (V2 ) We know that any vector can be expressed as linear combination of it 's basis vectors so let Ox + 4y - 3z = 12 3 Ox + 2y + z = 36 + 4-6 = 2 Sol-7: (d) 1 Ox - 2y + z = -4 5x ⇒ �R 3 dim(V1 n V2 ) = dim(V1 ) + 1 47 · Hence, IES MASTER Publication CARTESIAN PRODUCT Given two non -empty sets P and Q. The Cartesian product P x Q is the set of all ordered pairs of elements from P and Q, i.e. P x Q ={(p, q) : p e P, q e Q} A ={1, 2 } ; B ={a, b, c } e g. ., 1. 2 ordered pairs are e qual if and only if the corresponding first elements are e qual and the second elements are also e qual. 2. fI there are P elements in set A and q element in set B then there will be p q elements in A x B i.e., n(A x B) = n(A) x n( B) 3. 4. fI A and B are non -empty sets and either A or B is an infinite set, then so is A x B. A x A x A ={(a, b, c) : a, b, c e A } where (a, b, c) is called an ordered triplet. RELATIONS A relation R from a non -empty set'A' to non -empty set 'B' is a subset of A x B. Representation of Relation 1. 2. 3. R ={(a, b) : aRb, a e A, b e B } The set of all first elements of the ordered pairs in a relation R is called the domain of the relation R. The set of all second elements in a relation R from A ➔ B , is called the range of the relation R. The whole set B is called the co-domain of the relation R, where range is always a subset of co -domain i.e., Range s;;;; Co ­ domain. The total no. of relations that can be defined from a set A to B is the no. of possible subsets of (A x B) . fI n(A) = p and n( B) =q, then n(A x B) =n q and the total no. of relations 2 Pq_ TYPES OF RELATIONS 1. Reflexive : fI Va e Domain,(a,a) eR then R is ref e l xive ⇒ 2. Symmetric : fI (a,b) eR 4. Equivalence relation : A relation which is all reflexive, symmetric and transitive is called an e quivalence relation. 3. (b,a) eR where (a,b) eR then R is called symmetric. Transitive relation : fI (a,b) eR and (b,c) eR ⇒ (a,c) eR then R is transitive. Example 1 : A relation R defined on set A ' = {O,1, 2,...,12 }, R ={(a, b) : a -b is divisible by 4 and a, b e A }. Show that R is an e quivalence relation. Solution : Reflexive : Let a e A ⇒ Let (a, a) e R ⇒ (a - a) is div. by 4 0 is div. by 4, which is true. Hence R is reflexive. Engineering Mathematics Symmetric : Let a, b E A ⇒ If (a, b) E R ⇒ ⇒ a - b is div. by 4. (b - a) is divisible by 4. Transitive : a, b, c E A (b, a) E R ⇒ a- b · -- = p, p E l 4 b- c If (b, c) E R ⇒ -- = q, q E l 4 a- b b- c Adding (i) and (ii), -- + -- = p + q 4 4 a -c ⇒ - = p + q, p +q EI ⇒ a - c is div. by 4 4 Hence R is an equivalence re lation. If (a, b) E R 1 49 . ... (i) ... (ii) ⇒ (a, c) E R so R is ,transitive also. FUNCTIONS A relation R from set A to B is said to be a function (f) if every element of set A has one and only one image in set B. It is denoted by f : A ➔ B, where A = domain of function and B = the co-domain of function. The elements of set B which are the image of the elements of set A are called range. Let f : N ➔ N such that f (n) = 2x then domain = N, range =even natural no. = {2, 4, 6, 8, ...... }, co-domain = N TYPES OF FUNCTIONS 1. One-One function ( Injective) : A function f such that f : X ➔ Y is said to be one-one (injective) if the images ⇒ of distinct element of X unde r of are distinct. x 1 =x 2 then f is 'one-one'. Otherwise the function is 'many one '. i.e. if f (x 1 ) = f (x 2 ) 2. Onto function (Surjective) : A function f : X ➔ Y is said to be onto (surjective) if every e lement of Y is the image of some element of X under f. i.e., for every element y E Y , there exists an element x E X such that f (x) = y. Othe rwise the function is INTO. 3. Bijective function : A function which is both one-one and onto is called bijective function/one one correspondence. Example 2 : Show that the function f : R ➔R defined as f (x) = 3x + 4 is bijective. Solutio n : For One-One : Consider x 1 , x 2 E R (Domain) If f(x 1 ) = f(x2) 3x 1 + 4 = 3x 2 + 4 3x 1 = 3x 2 For Onto : Consider y ER then f (x) = 3x + 4 ⇒ ⇒ x 1 = x 2 so f is One-One. (Co-domain) y = 3x + 4 or x = - y- 4 3 ·: y E R then x E R So, for every e lement y ER (co-domain) there exist an e lement x ER such that f (x) =y. So, f is Onto. IES MASTER Publication 1 50 Functions and their Graphs SOME BASIC GRAPHS ( Polynomial ] Algebraic Rational ( Modulus ) Irrational [ Signum ) ( Greatest integer function ) ( Fractional part function ) ( Least integer function ) FUNCTIONS [ Trigonometric ) ( Exponential ) Transcendental ➔ Graph of f(x) = x y ( Logarithmic/Inverse of exponential ) ( Inverse trigonometric curves ) Graph of f(x) = x2 y 0 Graph of f(x) = x3 y Graph of f(x) = y 1 X y = -¼ 0 1 Engineering Mathematics Graph of f(x) = 1 x2 1 51 Graph of f(x) = x112 y y ,'!f y = X ----� y = ✓x y : .1, X 0 Graph of f(x) = x1 ,2 Graph of f(x) = x1 '3 Y = x'" Modulus Function Graph of l(x) = y = -x 1•1 = r� y Signum Function X<O X=O X>O i y= Graph of f(x) = sgn (x) = r n x' X ;c 0 0, X = O = X <O X>O X=O y-axis y y=x x-axis X -1 IES MASTER Publication 1 52 Functions and their Graphs (Greatest Integer Function) (Fractional Part Function) {-2 Graph of f(x) = [x] = , -2 � X < - 1 -1 � X < 0 -1 o: O�X<1 1�X<2 1, y-axis ---< 3 2 1 x + 2 -2 � X < -1 x + 1 -1 � X < 0 Graph of f(x) = {x} = x - [x] = { x 0�x<1 x-1 1�X<2 y-axis ---< 0 1 -3 -2 -1 � -1 2 3 4 x-axis -2 -3 -2 -1 -3 2 3 2 3 (Least Integer Function) y-axis 3 Graph of f(x) = Ix7 = { 2 -1 -2 < X :,::; -1 o: -1 < X :,::; 0 1, 0 < X :,::; 1 2, 1 < X :,::; 2 --4 -3 -2 -1 >-- 0 1 -1 -2 -3 Graph of Sine Function y y = sin x 1 (- ;, -1) - ----------- - - - - - - - - ► y- (3;, -1) 4 x-axis Engineering Mathematics 1 53 Graph of Cosine Function y (-2 1t, 1 ) (0, 1 ) (21t, 1 ) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ► y - COS X Graph of Tangent Function y y = tan x ... Graph of Cosecant Function y ... I I I I I - - -1- I :y = • Sin X ----+-------tf---+------1---r---t----'i,---+---+--------->;--► X y = cosec x y = cosec x y = cosec x Graph of Secant Function y IES MASTER Publication 1 54 Functions and their Graphs Graph of Cotangent Function y y = cot x 21t -2m I en Q) I I I .8 .8 I I (Exponential Function) Graph of f(x) = a x y x y=a , a>1 ... � y = ax y-axis --�--+-,------ x-axis 0 (Logarithmic Function) Graph of f(x) = log8 x if 0 < a < 1 y-axis Parabola (i) y2 = 4ax Vertex Focus Axis Directrix Eqn. of Latus Rectum Length of Latus Rectum : y if a > 1 y-axis (ii) y2 = -4ax (0, 0) (a, 0) x-axis or y = 0 X = -a X= a 4a Vertex Focus Axis Directrix Eqn . of Latus Rectum Length of Latus Rectum : y- = -4ax (0, 0) (-a, 0) x-axis or y = 0 X= a X = -a 4a y ------+---+-----♦ X V x I!! 0 X = -a y- = 4ax ... x=a Engineering Mathematics (iii) x2 = 4ay 1 55 (iv) x2 = -4ay Vertex (0 , 0) Focus (0 , a) Axis y-axis or x = 0 Directrix y = -a Eqn . of Latus Rectum y =a Length of Latus Rectum : 4a x2 = 4ay y Vertex Focus Axis Directrix Eqn. of Latus Rectum Length of Latus Rectum : y (0, 0) (0, -a) y-axis or x = 0 y =a y =a 4a - - - - - - - - - -. y = a X - - - - - - - - - -. y = -a Directrix x2 = -4ay a2 > b2 Ellipse (i) (ii) x2 y2 - += 1 (a2 > b 2) a2 b2 b Centre (0 , 0) Fo cus (±ae , 0) Vertex (±a, 0) Eccentricity b2 e = 1- 2 a Directrix X = ±� e ✓ x = -ale T x = ale T (0, b) (-a, 0) x2 y2 - + - = 1 (a2 < b 2) a 2 b2 0 (a, 0) (0, -b) - - - - - - - - - - - - - - - - - - -► y = -We directrix x2 y2 - 2 =1 2 b a Centre (0 , 0) Focus (±ae , 0) Vertex (±a, 0) Eccent ricity e= � f -t- � Directrix -b y 2 - - -b- ->-a'- - - ► y = ble Hyperbola (i) a -a y X = ±� e (-ae ,0) I I I I I I I x = -ale T x = ale T IES MASTER Publication 1 56 (ii) Fun ction s an d t heir Gr aphs x2 y2 = a 2 (Rectangular hyperbola) As asymptotes are perpendicular. Therefore y=x - it is called rectangular hyper bola . y =- x xy = c 2 (iii) Here, the asymptotes are x-axis and y-axis. Normal Curve - f(x) = e-x sin(x) Sine function - f(x) =-­ x 2 y y (0, 1 ) x' y = e- (0, 1 ) -==-------.----==+ x G RAPH TRANSFORMATION 1. (i) (ii) 2. (i) (ii) 3. (i) f(x) + a f(x) - a f(x + a) f(x - a) af(x) ➔ ➔ shift the graph o f f(x) upward by a units. ➔ shi ft the graph of f(x) downward by a units. ➔ shift the graph of f(x) leftward by a units. ➔ shi ft the graph of f(x) rightward by a units. stretch the graph of f(x), a times along y axis. 4. (i) _!_f(x) ➔ shrin k the graph of f(x), a times along y axis. a f (ax) ➔ stretch th e graph of f(x), a times along x axis. 5. (i) (ii) (iii) f (-x) ➔ Ta ke the mirror image o f f(x) a bout y axis. - f(x) ➔ Ta ke the mirror image of f(x) a bout x axis. - f( -x) ➔ First ta ke the mirror image a bout y axis and ta ke the mirror image o f new graph a bout x axis. (ii) t (\x \) ➔ First unit that portion of graph which lies in the left side of y axis, and then ta ke the mirror image a bout y axis of the remaining portion of the graph. (ii) (ii) 6. (i) (iii) t( �) ➔ stretch the graph of f(x), a times along x axis. it(x)I ➔ Ta ke the mirror image a bout x axis, of that portion of graph, which lies below x axis . While graph that lies a bove x axis remains as it is. It (lx l)I ➔ First form the graph o f it(x )I using part (i) and then from the graph of it(x )I using part (ii). t{ r� . o o , ,',_, l l o;.x l , *'oglxl , � l oglxl l Engineering Mathematics 1 57 PREVIOUS YEARS GATE & ESE QUESTIONS 1 .* Questions marked with aesterisk (*) are Conceptual Questions The curve given by the equation x 2 + y2 = 3axy is 6. (a) -x (a) symmetrical about x-axis 2. (d) tangential to x = y = a/3 The function f(x) = ex is [GATE-1 997] 7.* 4. [GATE-1 999; 1 Mark] The derivative of the symmetric function drawn in given figure wil l look like (c) A. A. k, 4 � • [GATE-2005; 2 Marks] Let x denote a real number. Find out the INCORRECT statement. (a) S = {x : x > 3} represents the set of all real numbers greater than 3 (b) S = {x : x < O} represents the em pty set. 2 (c) 5.* S = {x : x E A and x E B} represents the union of set A and set B (d) S = {x : a < x < b} represents the set of all real num bers between a and b, where a and b are real numbers. [GATE-2006] W hi ch one of the following functions is strictly bounded? (a) x2 (c) x 2 (b) e Given x ( t ) = 3 sin ( 1 000 nt ) and y(t) = 5 cos(1 0oont + i) . The x-y p lot will be (c) a hyperbola 8.* x (d) e- x 2 [GATE-2007; 1 Mark] (d) an el lipse [GATE-20 1 4 (IN-Set 1 )] Choose the most appropriate equation for the function drawn as a thick line, in the plot below. y (b) (d) [GATE-2008 (IN)] (b) a m ulti-loop closed curve (c) Neither even nor odd (d) None of the above (a) (d) -x-1 (a) a circle (a) Even (b) Odd 3.* (b) x (c) x-1 (b) symmetrical about y-axis (c) symmetrical about the l i ne y = x The expression e--lnx for x > 0 is equal to (a) x = y - I YI 9. (c) x = y + I Y I (b) x = - (y - I Y I ) (d) x = - (y + I Y I ) [GATE-201 5-CS-SET-3; 2 Marks] A function f(x) is linear and has a value of 29 at x = - 2 and 39 at x = 3. Find its val ue at x = 5. (a) 59 (c) 43 (b) 45 (d) 35 [GATE-201 5-CS-SET-3; 1 Mark] 1 0.* Consider an ant crasl ing along the curve (x - 2) 2 + y2 = 4, where x and y are i n meters. the ants starts at the point (4, 0) and moves cou nter C lockwise with a speed of 1 . 57 m eters per second. The time taken by the ant to reach the poi nt (2, 2) is (in seconds) _ ___ [GATE-201 5 (ME-Set 1 )] 1 1 . The fu nction f(x) = x 2 = x + x + x + . . . x tiems, is defined (a) at all real values of x (b) only at positive integer values of x (c) only at negative integer values of x (d) only at rational values of x [GATE-201 5 (P l)] IES MASTER Publication 1 58 Fun c tions an d t heir Gr aphs 1 2 . If g(x) = 1 - x and h(x) = h(x) g(x) (a) -1 x ( b) g(x) h(x) (c) (d) g(h(x) ) . X IS then h(g(x) ) x-1 X ( 1- x) 2 (a) ( b) (c) (d) * a then Jf(x) dx is 2� 2 b 4 b [a (ln 2- 25 ) + ; ] (c) -5 . . . . . . . . . : . - � [a (2 1n 2 -25 ) + ] 2 a -b [a (ln 2- 25 )- ; ] L [GATE-201 5 (CS-Set 3)1 ( b) � X f(x) (X) � X (d) . r ' X � [GATE-201 5-EC-Set-1 ; 2 Marks] 1 5.* Consider the plot of f(x) versus x as shown below: f(x) � +2 -5 :o ............,: -2 +5 !° :o [GATE-201 6-EC Set I] decreases exponential ly with an increase in load. At a load of 8 0 units, it tak es 1 0 0 cycles for failure. When the load is halved, ti tak es 1 0 0 0 0 cycles for failur e. The load for which the failure w li l happen in 5 0 0 0 cycles is ___ (a) (c) X 1 X 1 6. I n a process, the num ber of cycles to failure 4 b 2 ·► � (d) 47b b o . . . . - . �5 � f(x) function f(x) = e -x(x 2+x+1) ? (c) ( b) 47b � X f(x) � � [a (2 1n 2- 25 ) + ] 2 a -b a2 ► + :o f(x) � 1 1 4.* Which one of the following graphs descri bes the (a) -5 · [GATE-201 5 (CS-Set 1 )1 2 b f(x) � (a) 1 3 .* If for non-zero x, af (x) + bf (}) = }- 25 where a Suppose F(x) = Cf(y )dy. Which one of the following is a graph of F(x) ? ► X 40. 0 0 60. 01 ( b) 4 6. 02 (d) 92.02 [GATE-2016 Set-I] + 1 7.* A function f : N + _, N , defined on the set of positive + integers N , satisfies the following properties f(n) = f (%) if n is even f(n) = f(n + 5) if n is odd Let R = {i l :3j: f ( i =i)} be the set of distin ct value that f tak es. The maximum possi ble size of R is 1 8.* Let f(x) [GATE-20 1 6 (CS-Set 1 )1 be a polynomial and g(x) = f '(x ) be its derivative. If the degree of f(x ) + f(-x) is 1 0, then the degree of g(x) - g (-x) is [GATE-201 6 (CS-Set 2)1 Engineering Mathematics 19.* Let y 2 Jy - 2y + 1 = x and equals _ x+ decimal paces). ✓X + y = 5 . The value of ._ (Give the answer up to three [GATE-2017-EE Session-II] 20. Let x and y be integers satisfying the following equations 2x2 + y 2 = 34 X + 2y = 11 The value of (x + y) is [GATE-2017-EE Session-II] 21. What is the form of the f unction f (x) for the following data? X O 1 2 3 f (x) 3 6 1 1 1 8 (a) x2 + 2x + 3 (c) x2 + 2x - 3 (b) x2 - 2x + 3 2 (d) x - 2x - 3 [ESE 2018 (Common Paper)] 22. Which of the fo llowing f unct ion (s) is an acc urate description of the graph for the range (s) indicated? (i) 1 59 y = 2x + 4 for -3 ::; x ::; -1 (ii) y = Jx - 1 1 for -1 ::; x ::; 2 (iii) y = l l x l - 1 l for -1 :o; x ::; 2 (IV) y = 1 for 2 ::;X ::; 3 (a) (i), (ii) and (iii) only (b) (i), (ii) and (iv) only (c) (i) and (iv) only (d) (ii) and (iv) only [GATE 2018 (CE-MomingSession)] 23. Which of the following f unction describe the graph shown in the below fig ure? 1-+i! -+-!, --til 3 ]..!.... ..+·i · · -!--! -, f--i---i--t1 · �,·+ -- - 1; . ...... ...... y 3 (a) y = ll x l + 1 1 - 2 (b) y = ll x l - 1 1 - 1 (c) y = ll x l + 1 1 - 1 (d) y = l l x - 1 1 - 1 1 [GATE 2018 (ME-MorningSession)] -2 +---t--------,1--•----- ----- ------ ----- -3 IES MASTER Publication 1 60 Functions and their Graphs 1. 2. 3. 4. (c) 7. (c) 9. (d) 11. 8. (c) 1 0. (c) 5. 6. (b) 14. (2) 1 6. (c) 1 5. (b) 1 7. 12. (c) 1 3. (d) Sol-1 : (c) On i nterchanging x & y if equation of curve remains unaltered then curve is symmetric about the l i ne y = X Sol-2: (c) = f(x) ex f(-x) = e-x ⇒ *e x Neither even nor odd Sol-3: (c) = e-x ⇒ 2 = -2xe -x -�� 2(x - 1)(x + 1)e-x 2 • f ' (x) Let �(x) = 2 & f"(x) ⇒ f ' (x) then �'(x) = and decreasing in [-1 , 1 ] . Hence, only option (c) satisfies this property. Sol-4: (c) Sol-5: (d) 0 s x < oo 1 0 < 2 < <X) or X It is not bounded _) 0 (7) (a) (b) 23. (2) (9) (b) (5.732) (b) y = e' - <X) < X < <X) e--<X> < e x < e"' 0 0 < e x < oo Or (d) It is not bounded. - oo < X < <X) y = x' 0 s x2 <<X) It is not bounded or - oo < X < oo 0 s x2 < <X) 2 - <X) < - x s o 0 e- "' < e- x' s e0 0 < e- x ' s 1 or Sol-7: e- l nx (d) Let 8 = and ; = = 1 e- log0 x = elog8 (x - ) = X-1 = _.!_ X 1 000rct then � = sin 8 3 cos(s + i) = � (cos 8 - si n 8) Solving ( 1 ) and (2) y = � (cos s - 1 ) 5 It represents the i ntersection of A and B. 2 21 . It is bounded. Sol-6: ( c) x = ±1 are the critical poi nts of - <X) < X <<Xl 22. f"(x) �(x). So �(x) or f'(x) is increasing in (-<X), -1 ] u [1, <X)) (a) (b) = +1 -1 Sign of �'(x) Now, �'(x) = 0 20. (c) or - e-x Given g raph is equivalent to normal curve so we can take f(x) (b) (c) 18. (a) 1 9. (a) y=2 x So, ⇒ y✓ 2, X -- + 5 3 = cos8 sin2 8 + cos2 8 2. y -✓ + (� + J 9 5 3 � = = 1 . . .(� i . . . (2) . . . (3) Engineeri ng Mat hemat ics 2 2x 2 2 ✓ 2 -+ -xy+ -y 2 - 1 = 0 15 9 25 ...(4 ) On c omparision with ax 2 + 2hxy + by 2 + 2 9x + 2fy + C = 0 2 h2 a = � 'h = ✓ ' b 15 9 - ab -2 = -<0 225 = -3._ 25 So, (4 ) r epr es ents an Elli ps e Sol-8: (b) y = 0 at X y = - 1 at =2 = -(y - I Y I ) Sol-9: (c) X f( -2 ) =2 9 ...( i) 3a + b =39 ...(ii ) f(3) =39 From (i ) and (ii ) a =2, b =33 Sol-10: (2) g(x ) 1 -x ...(1 ) 1 Replacing x with X ⇒ ...(2 ) ⇒ (a 2 - b 2)f(x ) =a (�- 25 )- b(x -25 ) Int egrating f(x ) = � [a (�- 25 )- b(x -25 )] a2 b2 2 ° -5 0 )} - 5 = 2:2 [ {a(log 2 - 5 0 )- b ( 2 a (4, 0) -{a( 0 - 25 )- b (1 -25 )}] Sol-14: (b) f(x ) > 0 vx f( O) =1 f( oo) = 0 f (' x ) = e - x (2x+1)+ e - x (x 2+x+1 ) S peed =1.57 m/s ec. = m Is ec. 2 Dis tan c e = _n_ = 2 s ec . S peed n12 47 = +[a(log 2 - 25 )+ b ] a -b 2 2 f(x ) = e-x (x +x+1 ) 7t Tim e = C�1) = h(x ) a f(x )+bf (�) = �- 25 Th e l ength of arc of th ecircl e cov er ed by th e ant from 2 nr 2 nx2 (4, 0 ) to (2, 2 ) = = --= n 4 4 So .Distanc e cov er ed by an Ant. = 1t mtr So, = ff(x )dx = + {a(log x - 25x ) -b (�- 25 x )} a -b 2 1 f(5 ) =2(5 ) + 33 =4 3 (0, 0) -(6) 2 f(x ) =2x + 33 (0, 2) 5 times Sol-12: (a) Now [(1 ) x a - (2 ) x b] - 2a + b =2 9 ⇒ For exam pl e, l et x =5 th en 5 2 =5+5+5+5+5 '---------r---' a -f G)+b·f(x ) =x - 25 f(x ) is lin ear f(x ) =ax + b L et ⇒ x times Giv en that By o ptions w ecan can s ee that th es e points satisfi es th e only equation ⇒ Let X E N th er, =x 2 =x+x+x+.....+x is valid. Sol-13: (a) = 0 X Sol-11: (b) 1 1-h(x ) -X g(h(x ) = = = � ( 1- x ) h(g(x ) (1 -x )2 g(x ) -1 (1 -x )- 1 a h g 2 h b f =- - ,c 0 /',,. = 225 g f C and 1 61 Now , f '( x ) = e -x (-x 2+x ) f '( x ) = 0 ⇒ X =0, 1 IES MASTER Publication 1 62 Fu nctions and their Graphs 2 f"(x) = e- x (-2x + 1) - e -x (-x + x ) Now, g(x) = f'(x) is a poly of degree 9 ⇒ 2 ⇒ ⇒ = e-x ( x - 3x + 1 ) 1 f"(1) = e- (1-3 + 1) < 0 g(x) - g(-x) is also poly of degree 9. Sol-1 9: (5.732) Given : y2 f(x) has local maxima at x = 1 f"(0) = 1 > 0 f(x) has local minima at x = 0. So only option (b) satisfies these conditions. ⇒ ⇒ ⇒ and ⇒ Now, Sol-1 7: (2) Given that c2 c1 = c1 and ✓Y - 8y = 24 x = 4 = 4+ 2x2 + y2 = 34 ✓3 1 Oy = 4 + 1 . 732 = 5.732 . . . (i) . . . (ii) x + 2y = 1 1 2 11-x 2x 2 + ( --) = 34 2 8x2 + 121 + x 2 9x 2 - - 22x = 1 36 22x - 1 5 = 0 18 -10 as a root, as it is given that x and y are i ntegers. y = 4 if n is even We can observe that f(1 ) = f(6) = f(3) = f(8) = f(2) = f(1 ) . . . . ..... . Case I : Now, and Discarding t = 46.02 {%) y = 3 2y + 1 = 25 + y2 22 ± 32 -1 0 x = - = 3 ' 18 18 1 06 5000 = 1 0 6 e-txo. 1 1 s 1 f(n) = - 2y + 1 = (5 - y)2 Using sridharacharya form ula, e -aoxo. 1 1 s 1 f(n) = f(n + 5) if n is odd Case II ⇒ ⇒ . . . . (2) = 0. 1 1 51 1 00 = y2 - Solving (i) & (ii), Let N = No. of cycles; t = load; N = C 1e -C2 t . . . . (1 ) 1 00 = C 1 e-80c2 40C 2 = Zn1 00 y2 X+ Sol-20: (7) Now i ntegration of linear function should be parabolic in nature so only option (c) satisfies it. Sol-1 6: (b) ⇒ ⇒ ⇒ ⇒ . . . (ii) = 5 From (i) and (ii), we get, We know that odd function is sym metrical in opposite quadrants and even function is symmetrical about y­ axi s. 1 0000 = C1 e -4°c2 . . . (i) 2y + 1 = x ✓x + y and Sol-1 5: (c) Given function is an odd function and i ntegration of an odd function is an even function, so option (a) & (b) are wrong. - f(5) = f(1 0) = f(5) . . . . . . . . . . . f(7) = f(12) = f(6) f(9) = f( 1 4) = f(7) = f(12) = f(6) So f(7) and f(9) a re not distinct, they can be obtai ned by using case I . S o overall , range of f(x) wil l contain 2 distinct elements only, f(1 ) and f(5). Sol-18: (9) ·: f(x) is a polynom ial of degree 1 0 then f(x) + f(-x) is also polynom ial of deg ree 1 0. Hence value of (x + y) = 3 + 4 = 7. Sol-21 : (a) 0 1 2 3 X= f (x) = 3 6 1 1 1 8 So given points are (0,3), ( 1 ,6), (2, 1 1 ), (3, 1 8) 0 . -. O nl y option (a) i . e . jf ( x ) = x2 + 2x + 3j i s satistying given points. S o ( a ) is correct. Sol-22: (b) By standard definition of Graphs. Sol-23: (b) 2 I� I � I=� I� I� I�I 1 These points are satisfied by y = llxl-1 1-1 ) 1/ EXPANSION OF FUNCTIONS We can expand some functions of x in the ascending powers of x as infinite series. For this, Maclaurin's series and Taylor's series are used in general. MACLAURIN'S THEOREM (SERIES) Let f(x) be a function of x which can be expanded in powers of x and let it is differentiable term by term any number of times, then f(x) can be expanded in the neighboured of x =0 as follows 2 n f(x) = f(O)+xf'(O) + � f"(O) +...+� f (n)(O)+... 2! n! Important Maclaurin's Expansion 1. 3. 5. 7. x2 x3 x4 e x =1+x+-+-+-+ ...oo 2! 3! 4! x2 x4 x6 cosx = 1--+- - -+...oo 2! 4 ! 6! x . x3 xs e -e-x =X+-+-+...oo smh x= 3! 5! 2 x3 2 tanh x = x- -+-x s +...oo 3 15 2. x3 xs x7 . smx = x- -+-- -+...oo 3! 5! 7! 4. x2 2 tanx = x+-+-x 3 +...oo 3 15 6. cosh x= x x 2 x4 e x +e--= 1+-+-+...oo 2! 4! 2 TAYLOR'S THEOREM (SERIES) Let f(x+ h) be a function of x, which can be expanded in powers of h, and let this expansion is differentiable any number of times with respect to x, then f(x) can be expanded in the neighbourhood of x =a as follows: hn n h2 f(a + h) = f(a)+hf'(a)+-f"(a)+...+-f ( l (a)+... (where x =a+ h) 2! n! 2 X - a) f(x) =f(a) + (x - a) f'(a)+ ( 2 f"(a)+.... I.: Failure of Taylor's Theorem Taylor's theorem fails if any one of the following cases arises: (i) (ii) f(x) or any one of its derivativs, becomes infinite between the values of the given variable. f(x) or any one of its derivatives becomes discontinuous between the same values. (iii) The remainder �f n (a + 0nh) tends to non-zero value as n tends to infinite and in this case the series n! is divergent. L� f (x) n! n 164 Infinite Series Example 1 1 Solution : Expand log( 1 + x) by Maclaurin's theorem. Let f(x) = log( 1 + x) f(O) = 0 ⇒ ... (i) Differentiating successively, we get ⇒ 1 f'(x) = 1+x ⇒ f'(O)= 0 1 1 2 f"(x) = �(1+xf =-(1+xf =- - -2 dx (1+x) 2 2 f'"(x) = �[-(1+xr 1 = - -3 dx ( 1+x) ⇒ 6 3 f (iv ) (x) = �[2( 1+xf ) =- - -4 dx (1+x) ⇒ f"(O)=-1 f"'(O) = 2 ⇒ f (iv) (O) =-6, ... x3 x2 By Maclaurin's Theorem, we have f(x) = f(O)+xf'(O)+-f"(O)+- f 111(O)+... 2! . 3! 2 3 4 3! 4! ...(ii) Put all the values in equation (ii), we get f(x) = 0 +x - 1+ �(- 1)+ �( 2 )+�(-6)+... . ⇒ Example 2 Solution 1 3 2! 4 3! 4! 2 3 4 2 3 4 E xpand log cosx using Maclaurin's theorem upto the coefficient of x4 . Let f(x) = log cosx 1 2! x 2x 6x x x x f(x) = log( 1+x) =x--+---+... = x--+-- -+ ... 2 ⇒ f(O) = 0 ...(i) Differentiating above function successively, we get 1 f'(x) = - -(-sinx)=- tanx cosx d 2 f"(x) = -(-tanx) =-sec x dx ⇒ ⇒ f'(O)=O f"(O) =-1 f"'(x) = �(-sec 2 x)=-2 sec 2 x tanx dx ⇒ f"'(O)=O 4 2 2 2 ) f (iv)(x) = d:(-2 sec x tanx =(-4 sec xtan x- 2 sec x) By Maclaurin's theorem, we have 2 3 4 2! 3! 4! ⇒ f(ivl(Q ) = -2 f(x) = f(O)+xf'(O)+ � f"(O)+�f"'(O)+�f (iv) (O)+ ... Put·all the values, we get 4 2 f(x) = logcosx= O+x(0)+�(- 1)+�(0)+�(-2)+ ... = -(x + 2 x +...) 2! 3! 21 4! 41 Example 3 Solution : 1 Expand sin(¾+ 0) in powers of 0. Let f(x) = sinx ⇒- f'(x) = COSX ⇒ ⇒ f"(x) ⇒ f (¾)= sin¾= � (4n)=Cos4n = ✓1 2 n .n 1 . = -smx ⇒ f"( )= - sm =- 2 ⇒ f"'(x) = -COSX f' 4 ⇒ 4 ✓ f"(n)=-Cos n=- .fj_"' 1 4 4 f(x + h) 4 f (¾+e) sin (¾+e)=f (¾) +ef '(¾)+ �� f"(¾) + �� f "'( ¾) +... = ( ) . ) )+ � +e(� ��(-� +�� -� + .. = �(1+0-��-��+ ...) = Solution 1 Expand sin x in powers of ( x-¾) . f(x) = sin x=sin [¾ +(x-¾)] Let 1 165 7t and h =0, we get by Taylor's theorem: Put x = Example 4 3! 2! h2 h3 f(x)+hf'( x)+-f"(x) +-f "'(x)+... = Engineering Mathematics T he Taylor's theorem is given by f(a + h) Put a = = x-¾} we get ¾• h=( f(x) = f"(x) = ⇒ sin x ⇒ f '( x) = cos x f"'(x) 2! 3! h2 h3 f (a)+hf'(a)+- f"(a)+ - f "'(a)+... {¾)=sin (¾)=� f '(¾)=cos (¾)=� f"(47t)=-s.in(47t)=- 1'2. 1 = -cos x ⇒ f,, (7t)=-cos (7t )=- .J2 4 4 . ⇒ -smx ✓ Put all the values in equation (i), we get f(x) Example 5 Solution 1 1 ( � +(x-¾ )�+;i(x-¾r( -�) + ;,(x -¾r -�) +... = sinx= = 1+ x �[ ( -¾) -;,(x-¾r-;Jx-¾r +...] Expand esin-1 x in powers of x by Maclaurin's theorem and hence obtain the value of e8 where sin e = x Let y = f(x) ⇒ f '(x) = = . -1 e5'" ⇒ x f(O)=1 or (y0 )=1 1 e sin- x ·-1 ✓1 -x 2 Similarily (y2 )0 = 1 and so on ⇒ By Maclaurin's theorem, we have f(x) = 2! Y1 - y ✓1-x 2 ⇒ 2! ( Y 1 )o = 1 . -1 x2 x2 e5'" x =f(O)+xf'( O) + -f"(O)+...=1+x+-+... Taking sin-1x = 0 ⇒ . ..(i) . ..(ii) 2! • . • • ... 9 . sin2 0 x=sin0,put x=sin0 m equation (111), we get e =1+s In 0+--+... IES MASTER Publication 166 Infinite Series Example 6 : Solution : Express the polynomial 2x3 + 7x2 + x - 1 in powers of (x - 2). Let f(x) = 2x3 + 7x2 + x - 1. f'(x) = 6x2 +14 x+1, f"(x) =12x + 14, f "'(x) = 12, ... f(x) = f[(2+(x -2)] = f(2)+( x-2)f'(2)+ ( f(2) = 2(2)3 + 7(2)3 + 2 - 1 = 45 ; x -2)2 (x 2)3 f"(2)+ - f"'(2)+... 2! 3! f'(2) = 6(2) + 14(2) + 1 = 53; 2 ' ..(i) (by Taylor's Theorem) iv f"(2) = 12(2)+14 = 38; f"'(2) =12; f ( ) (2) =0, ... Put all the values n equation (i), we get (x- 2)2 (x 2)3 (38)+ - (12) 2! 3! 45 + 53(x - 2) + 19(x - 2)2 + 2(x - 2)3 2 3 f(x) = 2x +7x +x- 1= 45+(x-2)(53)+ = If f(x) = x3 + 8x2 + 15x - 24, calculate f(��) by taylor series. Example 7 : Solution : By Taylor series, we know that ( )2 ( )3 f(x) = f (a)+(x-a) f'(a)+ x-a f" (a)+ x-a f"'(a)+··· � @ x - a = h i.e. x = a + h Let f(a+h) Put a = 1, h = h2 � 1 10 11 f( ) 10 h3 = f (a)+hf' (a)+-f" (a) +-f"'(a) +··· ( = f ( 1)+_!_ f' (1)+ 1� 10 � @ r r ( f" ( 1)+ 1� @ f"' (1)+ o +o +... 1 1 1 = ( x3 +Bx2 + 15x-2y )x=1 + [ 3x 2 + 16x+15] x=1 + [6x+16 ]x=1 + 6 = 10 200 6000 [ ]x 1 = 0+ 6 34 22 +--=3.4+0.11+0.001 + 10 200 6000 = 3.5 11 CONVERGENCE AND DIVERGENCE OF INFINITE SERIES oo n 1. If lim Ian = finite then Ian is called Convergent. n➔oo n=1 n=1 oo n 2. If lim Ian = ± 00 then Ian is called Divergent. n➔ oo n=1 n=1 3. 4. ��[tan l = Neither finite nor infinite then tan is called Oscillatory. Necessary condition for the convergence of infinite series ➔ lim an = 0. (if we are not able to find sum of n➔ oo 1st n terms then necessary condition can be used to cross check for the divergence of the series). 2 Engineering Mathematics n 1 2 2) 2) Example 8 : T he series 1+ +( +....... is convergent because +.......+( 3 3 167 3 = 3 finite. Example 9 The series log 2 +1og¾+log¼+log¾+.......... is divergent because 1 = lim[fan ] = lim[log { 2 }.-±.� .......... n+1}] = lim[log(n+1)] = co n➔oo n➔oo n ➔oo 2 3 4 n lim Sn n ➔oo Example 10 : The series n =1 (-1 )n 1 1, even (-1 )n = 1-1+1-1+1-1+.......... = { 0, odd (-1)" .n = -1+ 2 - 3 +4 -5 +6 +.......... = different values for different n and Example 11 oscillates finitely and (-1 )n .n oscillates infinitely because ✓2(nn+1) +······· Series ff+�+·······+ 1-1 2 does not converge. Proof, We can not find I,an , so we will use necessary condition, i.e. n =1 lim(an ) = lim n ➔oo n ➔oo ✓2(nn+1) 1 1 = lim -2 [n➔oo ✓ 1+ .! n = � v2 *0 So, necessary condition n�t holds. Hence, given series is not convergent. It may be divergent or oscillatory. 00 .. 1 1 1 1 1 + -+ -+..........+ -+.......... converges when p > 1 and diverges when p � 1. = Note .· T he infinite series P nP 1P 2P 3P n =1 n L, - METHODS TO FIND CONVERGENCE OF INFINITE SERIES 1. Comparison Test I If I,an and I, bn are two series of +ve terms such that a n < bnVn then (i) if I, bn converges, then I,an also converges. (ii) if L,an diverges, then L,bn also diverges. . 1 1 1 1 Example 12 , Test the convergence of the sen es 1+2 +3 +4 +..... +-+ ..... 4 nn 2 3 1 1 "bn = Let "a n & let .LJ .LJ n = 2n n Solution, ·: L,bn is infinite GP, which can be added so it is convergent. Also we have an < bn V n > 2 . So by comparison test I,an is also convergent. 00 Example 13 1 Test for convergence of L,= 2 +1 3 n 1 n n ? IES MASTER Publication 168 Infinite Series Solution a Let an = 2n +3 n and bn = 3n (Infinite GP) an < bnVn � 1 and L bn is convergent so will Lan also. < 1, series La n converges an+1 2. Ratio Test I If lim ( = { > 1, series Lan diverges n➔«> an ) = 1, test fail < 1, series converges = l > 1, series diverges =1, test fail 1n 3. Cauchy's Root Test I If lim (an ) ' n ➔«> Example 14 1 Test the convergence of the series? Solution a r f n! 3,5, 7,.....(2n+1) n= 1 (n+1)! . 3,5,7, .....(2n+3) . a . n+1 1 hm -.!!±! = hm -- = - < 1 . So by ratio test given series is convergent. 1 = nhm 1 n n➔«> n ➔«> a ➔ · «> 2n+3 2 n · 3,5,7,.....(2n+1) Solution a a lim n+1 = lim• n➔«> n ➔«> a n 2n+1 / 2 /(n+1)! lim•( --) = O < 1 = n➔ «> n +1 2n I /n! So, by ratio test given series is convergent. Example 16 : 13 23 3 3 43 n3 ......· Test the convergence of the series - + 2 +3 +4 +......·+ -+ 3 3 3 3 3n 3n n° 1 1 ' = lim [n / ] = =- < lim (an }11n = lim [�] n n➔«> n ➔«> n ➔«> 3 3 3 3 1 n Solution: So by Cauchy's root test, given series is convergent. POWER SERIES (i) An infinite series of the type L=0 a x n n n is known as Power Series about x = 0. n (ii) An infinite series of the type Lan (x-x0 ) is general power series about x = x0. n =0 Radius of Convergence : The set of values of x for which given power series converges is called radius of convergence. (i) If the Power Series lxl > f a an xn is such that lim l n+1 i =_!_ then n ➔oo an R n =0 R and R is called radius of convergence. !xi> R. Example 17 La x 00 (ii) If the Power Series 1 n n=O n is such that lim I an n➔oo 1 11n = 1 R La x n Engineering Mathematics 169 is convergent for lxl < R and divergent for then series is convergent for lxl < R and divergent for n X 1 32 1 3 53 Find the radius of convergence of the power series -+- · -x +- · -· -x +......... 2 2 5 2 5 8 1.3.5.....(2n-1) an = -------''-------'- so series can be assumed as 2.5.8.....(3n -1) Solution, Now So R = 3 2 I: a x n n . 1.3.5.........(2n+1) 2.5.8..........(3n+2) = lim n➔oo 1.3.5..........(2n -1) 2.5.8 ..........(3n-1) = nlim ➔oo I 2n+1 3n+2 I =3 3 i.e., series converges for Ix I< ¾ and diverges for Ix I> ¾. IES MASTER Publication 170 -Infinite Series PREVIOUS YEARS GATE & ESE QUESTIONS • 1. 2.* Find the Taylor's series expansion of, sin x at x = 0. [GATE-1994) The third term in the Taylor series expansion of e� about x = a would be (a) (c) 3. 4. ,• Questions marked with aesterisk (*) are Conceptual Questions ea x2 2 7. ea 3 (d) -(x- a) 6 [GATE-1995; 1 Mark) x3 x5 T he series f(x) = x+ -+-+ .... is the expansion . 3! 5! d (a) cosh x (c) sinh x (b) sin x (d) cos x [GATE-1996; 1 Mark) For real values of x, cos (x) can be written in one of the forms of a convergent series given below: X = 7t 6' 1 (c) 8. sum the of · �)cos x)2 n = cos2 x +cos4 x+ .... Is (a) ea 2 (b) -(x - a) 2 (x - a) For the series (b) 3 7t (d) 1 00 . . 1 1 The mf'Irn'te series 1++-+.......... 2 (a) converges [GATE-1998) 3 (b) diverges (d) unstable (c) oscillates [GATE-1998; 1 Mark) 9.* A discontinuous real function can be expressed as (a) Taylor's series and Fourier's series (b) (c) (d) Taylor's series and not by Fourier's series neither Taylor's series nor Fourier's series not by Taylor's series, but by Fourier's series. [GATE-1998) 10.* The Taylor expansion of sin x about x = � is given by (d) 5.* cos(x)= x- x2 3 - x .......oo 11 21 31 + [GATE-1997; 1 Mark) The sum of the infinite series, 1 1 1 1+-+-+-+... is 2 3 4 (a) 7t (c) 4 6. x2 (a) 7t 2 4 [GATE-1997] The Taylor's series expansion of sin x is ; x4 (a) 1--+---2! 4! (c) x 3 x5 x+-+---3! 5! ; x4 (b) 1+-+---4! 4! (c) (d) (b) infinity (d) (b) x3 x5 (d) x--+--- 3! 5! [GATE-1998; 1 Mark) i+ !(x -i)-�(x-iJ-!(x-ir+... x3 x5 x7 x--+---+.... 3! 5! 7! x x x (x-�) ( -ir + ( -iJ ( -iJ + .... 6 3! 5! 7! 1 2 [GATE-2000; 1 Mark] 11. The limit of the following series as x approaches is (a) (c) 21t 3 7t 3 (b) 7t 2 7t 2 (d) 1 [GATE-2001-CE; 1 Mark] Engineering Mathematics 12.* T he summation of series S = 11 3 +...+ 00 is 2 (a) 4.50 5 8 2+ - + + 2 22 (b) 6.0 [GATE-2004; 1 Mark] 13.* For lxl « 1 , cot h{x) can be approximated as (a) (c) (b) x2 X 1 {d) X x2 [GATE-2007-EC; 1 Mark] 14. For the function e-x , the linear approximation around X = 2 is (a) (3 - x)e-2 2 2 (b) 1 - x {d) e-2 [GATE-2007] 15. Which of the following functions would have only odd powers of x in its Taylor series expansion about the point x = O? (a) sin(x3 ) {b) sin(x2 ) (d) cos(x2 ) (c) cos(x3) [GATE-2008) 16. In the Taylor series expansion of exp(x) + sin(x) about the point x = n, the coefficient of (x-n)2 is (c) exp(n)+1 (b) 0.5 exp(n) (d) exp{n)-1 (GATE-2008) 17. In t he Taylor series expansion of e x a bout x = 2, the coefficient of (x- 2)4 is (a) 1 /4! (c) e2 /4! {b) �/4! (d) e4 /4! (GATE 2008-ME] sin x 18.* The Taylor series expansion of - - at x X-n given by (a) (c) (x-n)2 1+ +... 3! 1 _(x-n)2 +... 3! = n is 2 (x-n) +... {b) -13! {d) -1+ (a) ( x-n)2 +... 3! [GATE-2009) x3 x5 x7 x--+-- -+.... 3! 5! 7! = (b) sin(x) cos(x) {d) e x sinh(x) (GATE-2010 (ME)] 20. A series expansion for the function sin 0 is (a) (c) 9 2 04 1 --+- 2! 4! 92 9 3 1+-+-2! 3! 9 3 95 (b) 0--+--.. . 3! 5! 9 3 95 (d) 0--+-+... 3! 5! [GATE-2011-ME; 1 Mark] . � 1 m 21.* T he series L, m (x -1)2 converges for ffi;O 4 (a) -2 < X < 2 (c) -3 < X < 1 (c) [3 + 2✓ -(1+✓ ) x] e -2 (a) exp(n) converges to · (c) (d) 10.0 (c) 6.75 19. T he infinite series f(x) 171 (b) -1 < X < 3 {d) X <3 [GATE-2011 (IN)] x2 x3 x4 x5 22. T he infinite series 1+x+-+ -+-+ -+... · 2! 3! 4! 5! corresponds to (b) ex (a) sec x · (d) 1 + sin2x (c) COS X (GATE-2012-CE; 1 Mark] 23.**The maximum value of q until which the approximation sinq"' q holds to within 10% (b) 18° (a) 10° (d) 90° (c) 50° [GATE-2013) 24. The Taylor series expansion of 3sinx + 2 cosx is x3 (a) 2 + 3x - x2 - - +... 2 x3 (b) 2 - 3x + x2 - - +... 2 (c) (d) 2 + 3x + x2 + x3 +... 2 x3 2 - 3x - x2 + - +... 2 1 25.* T he series " L, n;o n! converges to "' (a) (c) 2 In 2 2 [GATE�2014] (b) ✓2 (d) e [GATE-2014; EC-Set 4) IES MASTER Publication 1 72 Infinite Series 26. ( 1 )n oo The value of � n 2 is __ [GATE 201 5-EC Set Ill] 27.* The quadratic approximation of f(x) = x 3 _ 3x2 - 5 at the point x = 0 is (a) 3x2 -6x-5 (c) -3x2 +6x-5 28.* Let S= range L no. n n;Q (b) -3x2 -5 . (d) 3x2 -5 [GAT E-2016-CE-Set-2; 1 Mark] where lal < 1 . The value of a in the o <a<1, such that s = 2o. is ____ [GATE-2016] 29.* T he coeffi cient o f x 1 2 i n (x 3 + x 4 + x 5 + x6 ••••••) 3 is [GATE-201 6-CS-Set 1 ] 30." Let f(x) = e x+x2 for real x. From among the following, choose the Taylor series approximation of f(x) around 1. 2. 3. 4. 5. 6. 7. 8. Sol-1 : (See Sol.) · (b) 9. 10. 11. (c) 12. (b) 1 3. (b) 1 4. (d) 1 5. (b) 1 6. (b) f(x) = sinx Taylor's expansion (a) 1 + x + x2 + x3 3 (b) 1 + x + - x2 + x3 2 (c) 1 + x + (d) (a) (d) (d) (c) (a) (a) (b) 3 2 7 3 x2 + x 6 (d) 1 + x + 3x2 + 7x3 [GATE-2017-EC Session-I] 31 . Taylor series expansion of f(x) = J e (�) dt around 0 x = 0 has the form f (x) = a0 +a1 x +a2 x2 +..... The coefficient a2 (correct to two decimal places) is equal to ____ [GATE 201 8 (EC)] A N SW E R l<. E Y (x-a) (x - a)2 ( x - a)n n f'a+ f (x) = f (a)+ f"...+ f (a) � � � at x = 0 i.e. a = 0 x = 0, which includes all powers of x less than or equal to 3. 1 7. 18. 19. 20. 21 . 22. 23. 24. (c) (d) ·. , . · 25. 26. (b) 27. (b) 29. (b) (b) (b) (a) 28. 30. 31 . (d) (2) (b) (0.293) (10) (c) (0) f(x) = f(O)+xf( a)+� f"(a)+. . . . ; fn (a)... f(x) = sinx f(O) = sinO = 0 f(O) = cosO = 1 f"(O) = -sinO = 0 Engineering Mathematics f m( o) = -cosO =- 1 Hence, Sol-2: (b) sin x = O + x x 1 + sin x = x- f(X) x3 - 0 + � x(- 1) � x2 x 3 xs x7 - .... + � � l.Z = eX = eX- a+a = eaex-a 2 = e [ 1 + ( x- a) + ( x - a) + ...] 21 a Third term Sol-3: (c) Sol-4: (b) x 2 x4 x s cos (x) = 1- - + -- -- 2! 4 ! 6! Sol-5 : (b) Let we know that, S = 1+ 1 1 1 + 3 + 4 + ..... 2 log(1-x) = - [ x + x2 x3 x4 + + + .....] 2 3 4 x2 x3 -log ( 1 -x) = x + - + - + ..... 2 3 put X = 1 1+ 1 + Sol-6: (d) i +i + . . . Sol-7: (b) S = cos2 x + cos4 x + cos6 x + ....oo ·: Ratio of infinite G.P. = cos2 x < 1 s = At X = s = - 7t6 ' x2 x 3 = X + -+-+. . . . . 2 3 = log (J) x 2 x3 log(1-x) = -x---- -2 3 1 1 Put x = 1 ⇒ -00 = - 1 - - - - --2 3 ⇒ 1 + -1 + -1 + - - - - = 00 2 3 Sol-9: (d) Taylor's series exist only for continuous and differen­ ' tiables functions and Fourier's series exist even though the function have finite number of discontinuous points. f(x) = sin x, f (i ) = f " ( x) f'" . f" (67t) = -21 = -smx, ( x ) = - cosx, f" '(i ) = - � Sin X = ¾ + �(X Sol-11 : (d) r- r- - -i)-± ( -i ! ( -i X X x 3 xs x 7 = X- - + -- --3! 5! 7! f(x) = sin x f(X) Lim f(x) = Lim sin x = 1 1l 1l 00 x3 xs sin x = x - - + -- - - 3! 5! i f'( x) = cos x, f(i ) = + � = log oo 1 1 + +... = 2 3 cos2 x - = cot2 x sin2 x 3j js Sol-8: ( b) cos2 x 1-cos2 x cot2 (i ) s Sol-10: (a) x3 xs = x ++ +- -3! 5! 1 73 X➔ - 2 Sol-12: (d) Let, X➔- 2 5 8 1 1 .. + - + , oo S = 2 + -+2 22 2 3 IES MASTER Publication 1 74 Infinite Series 2 5 8 S = -+ -+ -+.. • oo ⇒ 2 2 2 2 2) ( 8 - 5) ( 1 1- 8) · -+ • • 00 ⇒ (s - ¾ ) = 2+ -( 5 -2-+--+2 2 2 3 2 3 + ···00 ) = 2+ � (1+ ..!. + __!_ 2 2 22 ⇒ =2 � 2 +%(�] S = 10 1-2 = 5 _ ex = e " e x-n = e" [1+(x-n)+ Cofficient of (x - 1t )2 = Sol-17: (c) Let y = ex then, y"" : 9X 4 ( ) ( x-2)3 ( ) x-2 ) + � y "' 2 + l1 y ""( 2 + ... ⇒ y (x) = e + (x - 2).e 2 (Neglect x2 and its high Powers) + 2 : e-2 e-(x - 2) for linear approximation neglect (x - 2)2 and its higher powers = e-2 (1 - X + 2) is e2 l1 . I (x -2)2 (x -2)3 (x -2)4 ] = e 1+(x- 2)+ - - +--+ --+... 2! 4 ! 3! [ 2 2 Coefficient of (x - 2)4 = : . , Sol-18: (d) sinx X- 7t sin(n+ x -n) X- 7t = -sin(x -n) (x - n) -1 (x- n )3 (x-n)- -'----'-+ . .. = -[ 3! (x- n) ( x3 )3 (x3 )s . 3 sin( x3) = x -�+� -......... . sin x contains odd powers of x only. So the cofficients of even parts will be zero in the expansion of sinx so only ex has the even powers which is as follows. (x-2)4 Alternative Solution = e-2 (3 - x) f(x) = ex+ sin x 2 ( x-2)2 2 ( x-2)3 2 ( x-2)4 2 l.f e + � e + l1 e +..... :. Coefficient of Sol-16: (b) y' = ex, y" = e x, y"' = ex, According to Taylor series expansion, + Sol-15: (a) n 2 cosh(x) e• + e-• cot h(x) = sinh(x) = e• -e-• e-x = e-2-x 2e ( x- 2) y(x) = y(2) + (x - 2) y'(2)+ -- y"(2) l.f Sol-13: (c) Sol-14: (a) 1 (X- 1t)2 + ...] 21 Sol-19: (b) (x-1t)2 = -1+-- .... . . 3! ] Engineering Mathematics Sol-20: (b) 93 95 sin 8 = 8-- + -3! 5! Sol-21: (b) mth Sol-24: (a) term, am ⇒ Ii� m ➔«> I aa +1 I m m = lim l (x-1) 4 m+1 r (x-1)2 Im -m➔ a 4 (x-1)2 4 am 1 + 1 < 1 am (x - 1)2 4 ⇒ or < 1 -1 < X < 3 Sol-22: (b) Sol-23: (b) s = s = s = x2 x3 ff" = 1+x+-+-+ ... 2! 3! 1 1 e = 1+1+-+-+ .... 2! 3! sin e E rror should 1 0% of e 93 ( ) :,;; 0. 1 8 , then higher order terms also going to 6 be less than (0. 1 ) 0 . If Hence, ⇒ x3 = sin e = e+ error So, sin Put x = 1 Sol-26: (2) -2 < X - 1 < 2 or = 2+ 3x - x 2 Sol-25: (d) 2m+ 2 We know that any power series converges if m ➔«> f(x) = 3sinx + 2cosx = 3 ( x- ;� ..... )+2 ( 1- ;� ....) 1 __ 2m = m (x-1) 4 am 1 = + 1 75 s� 8 3 approximation is valid 9 :,;; ( 0.1) 8 ⇒ 6 82 <0.6 ⇒ 8 < ../o.6 Smax . = ../o.6=0. 7746 radians 8 (in degrees) = 0.7746 X 1 80 7t Sol-27: (b) f(x) = x3 - 3x2 - 5 then f'(x) = 3x2 - 6x f"(x) = 6x - 6 - f"(O)= -6 Approximation at a =0 2 = f(O) + f'(O)x + f" (O)� 2 x2 = -5 +O x X+(-6)2 Smax = 45° (for error with in 1 0%) :. The maximum value of e amongst e = 1 0° and 1 8° (so that error is with in 1 0%) is 1 8°. ⇒ f'(O) =O (x -a)2 x- a) + f"(a ) f(x) = f(a)+ f'(a / 2! 1! 45° Hence from options 8 < Smax are 9 = 1 0° and 1 8° ⇒ Sol-28: (0.293) = -3x2 - 5 IES MASTER Publication 1 76 Infinite Series S = 0 + 1 .a1 + 2 .a 2 + 3.a3 + ... aS = 0 + ( 1-a)S = ⇒ For ⇒ ⇒ ( 1- a ) S = s = s = ( 1-a ) 2 1-a ⇒ a = = = Sol-29: (10) 1a2 + O + a + a2 a 1-a a 2 ( 1-a ) 2a= 1 2 2a 3 + ... + a 3 + .... ...(i) ...(ii) Sol-30: (c) Given, 2 x+x f(x) = e f '(x) = ex + x ( 1 + 2x) 2 f"(x) : ( 1 + 2X )2 eX+X + eX+X 2 2 X 2 a ( 1-a ) = ex+x (3 + 4x2 + 4x) f"'(x) = (3 + 4x + 4x2 ) 2 1 � 1 = 0.293 1� (x3 + x4 + x5 + xs ...)3 = x9(( 1 + x + x2 + ...)3 ( 1 + 2x) ex+x + ex+x (4 + 8x ) 2 Now, 2 2 f '(O) = 1 , f"(O) = 3, f "'(O) = 3 + 4 = 7 Taylor series of f(x) around x = 0 is f" ( O ) 2 f"' ( O ) 3 f' ( O ) f(x) = f(O) + --x + �x + �x + ........ 1I Sol-31: (0) = x9 ( 1 - x t3 = x9 [ 1 + (-3)(-x ) + (-3)(-3-1) (-x)z + � (-3(-3- 1)(-3-2) (-x)3 + ..... . � ] = x9 [ 1 + 3x + 6x2 + 1 Ox3 + ..... .] So, Coefficient of x 12 = 1 0 Then f '(x) = e x2 2 x2 2 and f"(x) = -x e Taylor series exp of f(x) about x = 0 is . 2 f(x) = f {O) + x f'(O) + � f"(O) + ..... Hence on comparison, f"(O) 0 a2 = --=.-= 0 � � FUNCTION Let A and B be two given sets and suppose that a relation is given which connects each member x E A to a member y E B . Then this relation is called a function from the set A into a set B. the function is normally denoted by the letter f, F, <j> etc. If we denote the function by f, then V x E A, there exist unique y E B such that f(x) = y The set A is called the domain of the function and the set B is called the range of the function. Thus, the range of the function f with domain A is the set {f(x) : x E A} . LIMIT OF A FUNCTION Let us consider a function f(x) defined in an interval I. If we see the behaviour of f(x) become closer and closer to a number l as x ➔ a then l is said to be limit of f(x) at x = a (e - o) Definition Let f be a function defined in the neighbourhood of a, then real value l, is said to be the limit of f as x ➔ a, if given E > 0 , however smal l , there exist o > 0 (depends on E ) such that If - ll < E whenever O < Ix - a l < o In other words, we can say that f approaches to a limit l as x ➔ a, and we can write The E-o lim f(x) = l definition does not give the value of l. It will help t o see whether a value l i s the l imit of f(x) o r not. X➔8 Left Hand Limit A function f(x) is said to approach l as x ➔ a from left if for an arbitrary positive small number number o (depends on E ) such that It can also be written as: if(x) - ll < E E, whenever a - o < x < a :I a small positive lim f(x) = lim f(x) = l f(a - 0) = x ➔a-0 X➔a- Right Hand Limit A function f(x) is said to approach l as x ➔ a from right if for an arbitrary positive small number number o (depends on E ) such that It can also be written as: if(x) - ZI < E whenever a < x < a + o E , :I small positive 1 78 Limits , Conti n u ity and Differentiability f(a + b) = lim f(x) = lim f(x) = l + If f(a + 0) = f(a - 0) = l as x ➔ a, then the finite definite value l is said to be limit of f(x) at x = a. X➔a+O X➔a Limit at Infinity (i) f(x) is said to have limit / as x ➔ oo if for given e > 0 :lo > 0 (depends on e ) s . t. lf(x) - ll < E whenever x 2 o . We can write lim f(x) = l . X ►- oo (ii) f(x) is said to have l i m it l as x ➔ - oo if for given Here we can write Infinite Limits lim f(x) = l . E > 0 :lo > 0 (depends on E) s.t. lf(x) - ZI < E whenever x :,; - o . X ➔ -oo (i) A function f : X ➔ R, (X c R) is said to have a l i m it oo as x ➔ a if for a given E > 0 :lo > 0 (depends on E ) s.t. x E X, 0 < Ix - al < o ⇒ f(x) > E. So, we have lim f(x) = oo . X➔a (ii) A function f : X ➔ R , (X c R) is said to have a l i m it -oo a s x ➔ a if for a given E > 0 :l o > 0 (depends on E ) s.t. x E X, 0 < Ix - al < o ⇒ f(x) > - E. So, we have lim f(x) = - oo . X ➔a Difference between the Limit and the Value of the Function We know that if z is the l i m it of f(x) at x = a, then . . . (i) lf(x) - ll < E for every x, other than ' a.' such that Ix - al < o . Hence in case of l i m it , x d oes not take up the v a l u e X = a, But we can obtain the value o f the function, when we put x = a i n the given function. Hence f(a) is called functional value of f(x) at x = a THEOREMS ON LIMITS Theorem 1 : Theorem 2 : Theorem 3 : Theorem 4 : The l i m it of a function if exists, is u nique. If lim f(x) = l1 , lim g(x) = l2 , then f(x) (i) lim [f(x) ± g(x)] = Z1 ± l2 (ii) lim [f(x) · g(x)] = Z1 • Z2 (iii) lim = (where l2 x ➔a x ➔a x ➔ a g(x) l2 X➔a X➔a 3. 5. 7. lim (1 +x) If lim f(x) = l, then lim e X ➔8 X➔8 11x . sinx I Im -- = 1 X➔O lim tan x = 1 X➔O X X➔O i, If lim f(x) = l, then lim l f(x)I = Ill x -► a X➔a Important Results on Limits 1. (Uniqueness Theorem). =e r< x > = e1 . 2. 4. X 6. lim ( 1 +! ) = e x X➔O lim cos x = 1 X X➔O n lim � = na -1 , n is integer. x ➔a x - a n n 1 . . 1 I I m sin - = 1 ·I m cos - = osci llatory value, so l i m it does not exist. X➔O X X➔O X * 0) Engineering Mathematics Important Expansions 1. 3 2 x x x e x = 1 + x + - + - + - + . . . oo 2 ! 3 ! 4! 2. . x3 x5 x7 sm x = x - - + - - - + . . . oo 3 ! 5! 7 ! 5. 5 3 . e x - e- x x x sm h x = - = x + - + - + . . . oo 2 3! 5! 6. 4 x x 2 x e + ex cosh x = --- = 1 + - + - + . . . oo 2 2 ! 4! 3. 7. 4 4 x x x COS X = 1 - - + - - - + . . . oo 2! 4! 6 ! 4. 6 2 3 2 5 x tanh x = x - - + - x + . . . oo 3 15 x3 2 5 tan x = x + - + - x + . . . oo 3 15 Example 1 : Exam ine the limit of the following function f(x) at x Solution : - LHL = 1 79 1 f(x) = - - X 2 1 = 2' 3 --x 2 = 1 O<X<2 1 X =2 1 -<X<1 2 1 , when f(x) is defined as: 2 f ( _:1_ - h ) = lim _:1_ _ ( � - h ) = lim h = O h➔O 2 h➔O 2 2 RHL = f (_:l_ + h ) = lim � - ( _:l_ + h ) = lim 1 - h = 1 h➔ O 2 h ➔O 2 2 Now, {½ - h ) Example 2 : Solution : * {½ + h ) , . tan x - si n x Eval uate l 1 m 3 x➔O X RHL = f(O + h ) = lim h ➔O = lim h➔O = so limit of f(x) at x = ; does not exist. ta n(O + h ) - sin(O + h) (O + h) 3 = lim �xample 3 : Solution : h ( h + � + � h5 + . . . ) - ( h - � + � - . . . ) h3 ! h 3 + (terms of higher powers of h ) � - ---lim 2--- 3 h➔O h ' tan (O - h ) - sin(O - h ) I LHL = f(O - h) = h� o (O _ h )3 ⇒ h➔ O tan h- sinh f(O + h ) = f(O - h ) = 1 2 = 1 2 lim h➔ O 3 - tanh+ sinh -h 3 (unique val ue) hence limit exists = lim h➔O tanh- sinh +h 3 = _:1_ 2 Eval uate lim f(x), where f(x) = 1 + x, x :,; 2 X➔2 = 5 - X, X>2 RHL = f(2 + h) = lim 5 - (2 + h ) = lim 3 - h = 3, h➔O h >- 0 IES MASTER Publication 180 Limits, Continuity and Differentiability ⇒ Example 4 : Solution : LHL = f(2 - h) = lim 1+(2 - h) = lim 3-h = 3, h➔O h➔O f(2 + h) = f(2 - h) = 3 n3 5n2 ) 2_ ( 1__ _ E valuate : lim [- 2 _+ - _�] n ➔ oo (2n + 3) (5n+1) l � )1 1 Put n = - as n ➔ oo, x ➔ 0, we have X lim X➔O 2 c 3 +3 ) + 1( � rn +1) 2 x2 - 5 + X➔O X(2+3X2 ) X(5+X) ] = lim ' [ 2- 13x + 3x 3 2 - 1 3 x O + 3 x 03 = _!_ lim = = x➔ D 10 +2x +15x 2 +3x 3 10+ 2 x O+15 x 0 2 + 3 x 03 5 or , ⇒ Example 5 : Solution : 2-13(0+h)+3(0+h)3 f(O+ h) = lim h➔O 10+2(0+h)+15(0+h)2 +3(0+h)3 2-13h+3h3 = lim h➔O 10+2h+15h 2 +3h3 3 2- 13(0- h)+3(0 - h) 3 2 f(O - h) = h1 • O � 10+2(0-h)+15(0 - h) +3(0 - h) f(O+ h) = f(O - h) = E valuate : lim X ➔O 5. x - lxl X RHL = f(O + h) = r (O+h)- IO+hl h-h . h- lh l = IIm - -= 1·Im - - =O h�o h➔O h h➔O h (O+h) lim LHL = f(O - h) = h➔O ⇒ 5 (0-h)-IO-hl (0- h) = IIm --= 2 . -h- h h➔O -h x- lx l . lim -- does not exists. X ➔O X (·: LHL * RHL) INDETERMINATE FORMS Let us consider a function lim f(x) f(x) x➔a f(x) • 1 en Im x = Im =--• F th ) 1 ( F(x) = x➔a x➔a g(x) lim g(x) g(x) X➔a f(x) When lim f(x) = lim g(x) = 0 or lim f(x) = lim g(x) = oo then the function F(x) = is said to have indeterminate g(X) x➔a x➔a x➔a X➔a 00 0 form of or O 00 respectively. The other important indeterminate forms are O x oo, oo - oo , 0°, 1"' and 00°. Engineering Mathematics 1 81 T he limiting value of indeterminate forms is known as true value. T he most standard form among all the indeterminate forms is 0 00 or . We can find the value of these two forms by using L-Hospital Rule. O 00 L-HOSPITAL RULE f (x) f (x) f(x) provided g'(x), .. ., g( n l (x) must not be = lim ' =...... = lim t: x ➔ a g n (x) g(x) x ➔ a g'(x) zero, where f (n) and g<n> are nth derivative of f(x)and g(x). When lim f(x)= lim g(x)= 0 then lim x ➔a x ➔a x ➔a L-Hospital Rule for the Form ( oo -oo, o x oo) For the evaluation of J� [f(x)-g(x)], if it is in the form (00 - 00) , we will convert it into the form (% ) by simplification. 00 The same process is also used in the form (0 x oo). Then we use the L -Hospital Rule. L-Hospital Rule for the Form (0° , 100 , 00°) I n the evaluation of lim [f(x)]9(x ) , we have to simplify by taking the log and convert it into the form X➔a we can use the L- Hospital Rule (2-) . After that 0 Remarks (i) log 1 =0 Example 6 : Solution : (iv) log1 x = oo (iii) log oo = oo, (ii) log O = -oo . tan x - sinx E valuate : IIm - - -x ➔O x3 We have (%) tanx-sinx I., m --x3 x ➔o Differentiating both numerator and denominator, we get sec 2 x-cosx I., m ---x ➔O 3x 2 (% ) Again this is in indeterminate form, so again using L- Hospital Rule, we get (%) +sinx 2 sec 2 x tanx I., m ----6x x➔o Again differentiating both numerator and denominator, we get lim X ➔O lim X➔O Example 7 : E valuate : · lim Solution : We have, X➔O lim X➔O 4 sec 2 xtan2 x+2 sec4 x + cosx 6 tanx- sinx = 2 x3 0+2+ 1 6 1 2 x-sinx x3 x- sinx x3 (%) = lim X➔O 1- cosx 3x 2 (%) IES MASTER Publication 1 82 Limits, Continu ity and Differentiability . sin x = I Im - ­ x ➔ o 6x = lim 6 X➔O Example 8 : Solution : F ind We have, lim X >- 0 X 2 . x cos x - log (1 + x) ---=-'-------'I Im x➔o x2 = lim Solution : Example 1 0 : Solution : We have, Evaluate : We have, Example 11 : Show that : Solution : Example 1 2 : Solution : We have, F ind : We have, . (%) COS X - X S i n X - - 2X = lim • m m . x -a hm --­ x➔a x - a (% ) . a -1 I Im -x x . a -1 I Im - x x➔o x lim x cot x = 1 X➔O 1 (%) . ✓ +x - � l1m ----- x➔o 1 . ✓ +x hm -- x ➔o x a x log a - 0 1 . = I Im --e�- = log8 a X➔O (% ) _! 1 _! 1 - (1+x) 2 +- (1 - x ) 2 2 = lim 2 x ➔O 1 1 1 1 - + -- ) = = lim � ( - Example 1 3 : Solution : Eval uate : We have, lim x➔o lim x➔o X ➔0 2 ✓ +x 1 (J_ - x 2 sin x 2 1 -) 2 sin x (J_ - x 2 ) 2 m . mx -1 = hm -- = ma m- 1 X ➔8 1 x -� - 1 (% ) -2 (1 +x) x (0 . 00 ) = lim __ x ➔ o tan x l i m x cot x X➔O I 2 m m . x -a hm --­ x➔a x - a x➔o 11+x - sIn x - sI n x - x cosx+, X -> 0 Eval uate : (%) x cos- log(1+x) X -> 0 Example 9 : 1 = 6 COS X � (oo - oo) form (%) 1 2 - = lim cos x = 1 = lim -2 X ➔ O sec X X ➔O Eng ineering Mathematics . sin x - x = hm - 2 - 2 x ---, o x sin x 2 = lim x >- D 2 2 si n x cos x - 2x (%) 2x sin x + 2x sin x cos x 2 2 sin2x -- 2x = lim ---x ---, o 2x sin 2 x + x 2 sin 2x = l im x ---> 0 = lim x ---> D = lim x ---, o = 1 83 (%) 2 cos 2x - 2 ------------2 2 2 sin x + 2x. sin2x + 2x sin 2x + 2x cos 2x cos 2x - 1 (2-) sin x + 2x sin 2x + x: cos 2x 0 2 2 -2 sin2x . 2 sin 2x + 2 sin2x + 4x cos2x + 2 x cos2x - 2x sin 2x -2 sin 2x lim -------- -2 3xin2x + 6x cos2x - 2x sin 2x (%) x >- D -4 cos 2x - ---�--- -= lim ----x ---, o 6 cos2x + 6 cos2x - 1 2x si n 2x - 4x sin2x - 4x 2 cos 2x 1 4 = -- = -3 12 Example 1 4 : Find : Solution : We have, Let lim ( X ---> 0 lim ( X ---> 0 tan x X tan x X ) ) tan x ) A = (- 1 /x 2 1 /x 2 1 /x (1 ) co 2 ⇒ x tan x 1 - - 1 og ( - - ) 1 og A _ 2 X x Taking limit as x ➔ 0 both the sides, we get tan x I og ( -- ) x lim log A = lim 2 X -> 0 = X ---> 0 lim - x ---> D ca: x ) - - x 2 x sec x - tan x --2x 2 . x sec x - tan x = I ,m ---� 3 x -> D 2x x 2 -- (%) ­ (%) [. , . 2 2 2 . x • 2 sec + sec- x -- sec- x - tan- x -- x= I 1m -2 x ---, o 6x 2 x . 2x tan x sec= I 1m - - -2 x ---, o 6x lim X ---> 0 tan x X = 1] ( %) IES MASTER Publication 184 Limits, Continuity and Differentiability (%) 2 . tan x sec x = I1m -- -x ➔o 3x . sec 2 x . sec 2 x + tan x • 2 sec 2 x tanx = I 1 m ------ -----x ➔o 3 ⇒ . sec 2 x+2tanx sec2 x = I1 m ------- x ➔o 3 1 A = e1 3 1/x 2 tan x ⇒ X1.1m ) (-➔O X 1 = e1 3 Example 1 5 : Find the values of a and b in order that lim X➔O Solution : We have, . x(1+ acosx)-b sin x - A = I1m ----3 x➔o x 1 3 x(1+acos x)-b sin x X (%) 3 may be equal to 1. b cos x . 1+ acos x + x(-a sin x) --- --'---A = I1 m ----�� 2 x ➔O 3X . 1+(a-b)cos x- ax sin x I1m --'--�- �-= x➔o 3x 2 = lim X ➔O (Qo) -(a -b)sin x -a sin x -ax cos x 6X . (-2a+b)sin x -ax cos x = I 1m -'-----'----x➔o 6x _ _ (-2a + b)- a =1 6 Solving (i) and (ii), we have a = (%) . (-2a+b)cos x-acos x + ax sin x _ ,m _ 1 ➔O 6 1 X ⇒ only when 1+ a - b = 0... (i) 5 -2, b ⇒ - 3a +b = 6 (given) ...(ii) 3 = -2 FUNDAMENTALS OF CONTINUITY Continuity A function y =f(x) is said to be continuous if the graph of the function is a continuous curve. On the other hand, if a curve is broken at some point say x =a, we say that the function is not continuous or discontinuous. Definition of Continuity A function f(x) is said to be continuous at x =a if and only if the following three conditions are satisfied: (i) f(x) exists; that is f(x) is defined at x =a (ii) lim f(x) exists X➔a (iii) lim f(x) = f(a) X➔a If the function is continuous at every point of a given interval [a, p] , then it is said to be continuous in that interval. 'Engineering Mathematics 1 85 Cauchy's Definition of Continuity A function f(x) is said to be continuous at a point x =a in (a, P) if for given E such that Right and Left Hand Continuity i f(x) - f(a)i < E E > 0 there exists o > 0 depending upon whenever O :,; Ix - al < o If a function f(x) is defined only for x :2: a, then we say that f(x) is continuous on the right at x = a if lim+ f(x) =f(a) or f(a + 0) =f(a) X➔a Similarly, for f(x) defined for x :,; a, we say that f(x) is continuous on the left at x = a if lim f(x) =f(a) or f(a-0) =f(a) X➔a Jump of a Function at a Point Let f(x) be a function for which the two limits f(a + 0) and f(a -0) at x = a both exists, where f(a+0)= lim f(a + h) h➔O and f(a - 0)= lim f(a- h) h➔O Then their non -negative difference i f(a +0)- f(a- 0)i is called the jump of the function at x = a . Finite Number of Jumps A function which has a finite number of jumps in a given interval is called piecewise continuous or sectionally continuous function. Geometrical Meaning of Continuity of a Function at a Point The continuity of a function f(x) at a point x = a geometrically means that there is no break in the graph of the curve y = f(x) at x = a or given e > 0 :3 o > 0, such that the graph of y = f(x) from x = a - o to x = a+ o lies between two lines y = f(a) - e and y = f(a)+E . Fundamental Theorems on Continuity Theorem : If f(x) and g(x) are continuous at x = a , then the functions (i) (iv) Theorem : f(x) + g(x) f(x) , g(x) g(x) * 0 and (ii) f(x) - g(x) (iii) f(x)g(x) (v) f(g(x)) are also continuous at x = a . The necessary and sufficient condition for a function f(x) to be continuous at x = a is that for a given E > 0 :3 0 > 0 S.t. Theorem : If a function f(x) is continuous at a point x =a, then lf(x)I is also continuous at x = a. Discontinuous Functions A function f(x) is said to be discontinuous at x = a if we have any of the following cases: (i) lim f(x) does not exist. X➔8 (ii) lim f(x) X➔8 * f(a) (iii) f(a) is undefined IES MASTER Publication 186 Limits, Continuity and Differentiability KINDS OF DISCONTINUITIES Removable Discontinuity A function f(x) is said to have a discontinuity of removable kind at x = a if lim f(x) exist but not equal to the value X➔a of function at x = a i.e., f(a + 0) =f(a -0) or lim f(x) X➔a ' 1 Example 1 6 : f(x) = x sin-, x X Solution : = 2, X *0 * f(a) * f(a) =0 1 f(O - h) = lim (O-h) sin -- = lim h sin ! =o h➔ O (0 - h) h ➔ O h 1 f(O + h) = lim (O+h) sin ---- = lim hsin ! =o h➔ O (0-h) h ➔ O h * f(O) = 2 . So, the discontinuity is of removable kind. f(O + h) = f(O - h) = 0 Discontinuity of First Kind/Jump Discontinuity A function f(x) is sai d to have a discontinuity of first kind at x = a if both f(a - 0) and f(a + 0) exist but are unequal. The point x = a is said the point of discontinuity from left or from right according to as follows * f(a) = f(a+0) f(a-0) I t is also known as ordinary discontinuity. Example 17 : f(x) = [x] , x 0 = 0, X = 0 * Solution : or f(a -0) = f(a) * f(a + 0) f(O - h) = lim [0 -h] = lim (-1) =-1 h➔O h➔O f(O + h) = lim [O+h] = lim O = 0 h➔O f(O-h) * f(O) =f(O+h) . h➔O The discontinuity is of first kind from left . Discontinuity of Second Kind A function f(x) is said to have a discontinuity of second kind at x = a if none of the limit f(a - 0) and f(a + 0) exist at x = a. The point x = a is the point of discontinuity of second kind from left or right accordingly f(a - 0) or f(a + 0) does not exists. . 1 Example 1 8 : f(x) = s in - , x 0 X = 0, X =0 Solution : * 1 f(O - h) = lim sin -- = lim (-sin !) = oscillatory value. h➔O (0-h) h ➔ O h . 1 . . 1 f(O + h) = I Im sin -- = 1·Im sin- = oscI·11 atory vaIue. + h ➔ O h➔ O h (0 h) So, f(O - h), f(O + h) does not exist, hence discontinuity is of second kind. Engineering Mathematics Mixed Discontinuity 187 A function f(x) is said to have a discontinuity of mixed kind at x = a if f(x) has a discontinuity of second kind on one side of a and on the other side it has discontinuity of first kind or may be continuous. Example 1 9 : Solution : 11 x f(x) = e sin _! , X f(O - h} = lim h ➔0 1 x*0 e ( O -h ) 1 1 sin-- = lim (- ;,h sin !) = 0 x (oscillatory value) = 0 (O-h) h ➔0 e h 1 1 1 . f(O + h) = I .Im eoti + sin-- = 1·Im eh - =oo ' h ➔0 (0 + h} h ➔ o sinh and f(O) is not defined. So, we have a discontinuity of first kind from left and second kind from right. Hence, f(x) has a discontinuity of mixed kind at x = 0. Infinite Discontinuity A function f(x) at x = a is said to have discontinuity of infinite kind if f(a + 0) or f(a - 0) is oo or -oo. Example 20 : Solution : 1 f(x) = - at x = 0. X 1· ( 1 1 . f(O - h) = I Im -- = Im --)= -oo h ➔0 (O-h) h ➔0 h 1 1 . f(O + h) = I Im -- = 1·Im - = oo h ➔ O (O + h} h➔ O h Example 21 : So, there is infinite discontinuity at x = 0. Find all points of discontinuity of the following functions: f(x) = (ii) f(x) = � x-a (iii) Solution : x2 +3x-10 x+5 (i) (i) 2 2 f(x) = -- if x x-a 2 x -a 2 *a and f(x) = 2a if x = a x2 + 3x-1O (x+5)(x- 2) - -- - ---�) (x+5) x+5 T his shows that the function is continuous at all points except possibly at x = -5 because f(-5) = Not defined so we will try to find its limit when x ➔ -5 (x+5)(x- 2) . . 1 , e. , X1•Im [----] - X1 Im [x - 21 -_ -7 ➔➔5) (X + 5 5 T hus, the function f(x) approaches a definite limit at x = -5 whereas the functional value of f(x) at x = -5 is not defined showing that x = -5 is a point of discontinuity. (ii) Hence, f(x) is continuous for all values of x except for x = -5. · (x-a)(x+a) . . which shows that f(a) is not defined and hm [ f(x)] = hm (x+a) = 2a f(x) = X➔8 X➔a (X - a ) Thus, f(x) is discontinuous at x = a while it is continuous for all the remaining values of x IES MASTER Publication 1 88 Limits, Continuity and Differentiability (iii) Here lim ( X➔B x 2 - a2 ) = 2a (as before), and f(x) = 2a. (as given) X- a so lim [f(x)]= f(a) X➔B Hence, the function f(x) is continuous at x = a. i.e., it is continuous for all values of x. Example 22 : Solution : * 1 Discuss the continuity of the function f(x) = (x -a)sin-- fo x a x-a =0 for x = a To test continuity at x = a we have f(a) = 0 f(a + h) = hsin ! :. f(a+0) = lim f(a+h) = lim (hsin ! ) = o h➔ O h➔ O h h . 1 . 1 Also, f(a - h) = -h sin = h sm -h 11 f(a + h) = f(a - h) = f(a) Taking limit as h ➔ 0 Hence, the function is continuous at x = a. To test the continuity at points other than x f(b + h) = (b + h - a) sin Also, and i.e., f(b - h) = (b - h - a)sin � b+ - a :. f(a - h) = 0 = a say x = b, we have 1 f(b+0) =(b-a)sinb-a 1 f(b- 0)= (b - a)sin-b-a b- h - a 1 f(b) = (b- a) sin -b- a f(b + h) = f(b - h) = f(b) for b a. * Hence, f(x) is continuous for all the real values of x. Example 23 : A function f(x) is defined as follows: -1 - x 2 when x < 0 when x= 0 0 f(x) = 2 l 1+x when x > 0 Find the points of discontinuity. Solution : LHL= lim f(x)= lim (-1-x 2 ) =-1 X➔O X➔O- RHL= lim f(x)= lim ( 1+x 2 )= +1 X➔O + Since, f(0 + 0) * X➔O f(0 - 0), the point x = 0 is the only point of discontinuity. Example 24 1 Test for continuity the function f(x) = e 11x at x = 0. Solution 1 RHL f(0 +h)= lim ( e11h )= = f(0 + h) = hlim ➔O h➔O 1/0 and f(0) = e = oo 00 (does not exist) Engineering Mathematics 189 Since RHL and functional value does not exists so function f(x) has an infinite discontinuity of second kind from right at x = 0 Example 25 1 Solution : Test for continuity the function f(x) Here f(0 + h) Also, = f(0 - h) and f(0) = e- e- 1 Ix at x = = 0. lim f(0+h)= lim (e-11h ) =0 (exist) h➔O = 110 h➔O lim f(0-h) = lim (e1Ih ) =oo (does not exists) h➔O =e ---«> h➔O = 0 ( exist) So, x =0 is a point of infinite discontinuity of llnd kind. Example 26 : Test for continuity the function Solution : Here f(x) = f( 1 + h) = f( 1 _ h) = f( 1 ) = 1 - at x = 1 (x 1)2 - lim f( 1 + h) = lim h ➔O h➔O lim f( 1-h) = lim h➔O h➔O = 00 ( 1-1)2 hence the function has an infinite discontinuity at x and Example 27 : A function f(x) is defined as follows : x2 -a f(x) = a =0, 1 = lim -;- =oo ( 1 + h - 1)2 h ➔ O h 1 = oo ( 1-h - 1)2 = 1, X<a X =a a2 =a- -, X >a X Prove that the function f(x) is continuous at x = a. Solution : ⇒ Example 28 : Solution : f(a + h) = f(a - h) = a2 +ah - a 2 lim (a- _L ) = lim = lim � = o a+h h ➔ o a+h (a+ h) h ➔D h➔D • I Im h➔O • (a- h)2 a 2 + h2 - 2ah-a2 (---- a J = 1 Im ------ = 0 a f(a+ h) = f(a - h) = f(a) = 0. Hence, f(x) is continuous at x = h➔O a a. Find the points and kinds of discontinuity of the function defined by 2 f(x) = -x , = 5x - 4, At x = 0 :o; 0 0 < X :,; 1 X = 4x 2 - 3x, 1 < x < 2 =3x + 4, x�2 2 2 f(0 - h) = lim -(0-h) = lim - h =0 h➔O h➔O IES MASTER Publication 1 90 Limits, Continuity and Differentiability f(O + h) = lim 5 (0+h) - 4= Iim 5h - 4 = -4 h➔O ⇒ h➔O f(O+h) ,d(O - h), so f(x) is discontinuous at x = 0 of first kind. f(1 - h) = lim [5(1 - h ) - 4] = lim (1- 5h) = 1 At x = 1 h➔O 2 f(1 + h) = lim [4(1+h) h➔O ⇒ - 3(1 + h)] h➔O 2 2 f(2 - h) = lim [4(2 - h ) - 3(2 - h)] = l i m (1 0+4h - 1 3h)=10 h➔O h➔O f(2 + h ) = lim (3(2+ h) +4] = lim (1 0+3h) = 1 0 h➔O Example 29 = lim [1+ 4h 2 +5h] = 1 f(1 - h ) = f(1 + h) = f(1 ), so f(x) is continuous at x = 1. At x = 2 ⇒ h➔O h➔O f(2) = 1 0 f(2 - h) = f(2 + h) = f(2) = 1 0. So, f(x) is continuous at x = 2 also Hence, f(x) is continuous at x = 1, 2 but disconti nuous at x = 0 of fi rst kind. 1 Discuss the continuity of the function f(x) at x = a : 1 1 f(x) = --cosec ( --) , X ;t a x-a x-a = 0, X=a . 1 1 -- cosec ( ) f(a - h) = I Im ➔ o (a - h ) - a h (a - h) - a Solution 1 = lim - !h cosec ( - !) h 1 . = I Im -�� = oo h➔O h➔ O ⇒ Example 30 : Solution f(a - 0) does not exists. Investigate the kind of discontinuity of the following function f(x) at x = 0: h➔O . tan f(O - h) = I Im h➔O Example 31 (-h1 ) So, f(x) i s discontinuous at x = a and the discontinuity is of infinite kind. f(O + h) = lim tan - 1 h s .in 1 ( 1 --) O+h 1 f(x) = tan - ( �) . f(O) = 0 at x = 0 = lim tan- 1 h➔O (!) h x ;t 0 = 2: 2 1 1 7t -1 (-) = I"Im - tan -I ( -) = - h 2 0-h h➔O Hence, f(O + h) and f(O - h) both exist but unequal so the discontinuity is of fi rst kind. 1 Examine the continuity of the function f(x) at x = 0 and x = 1, where f(x) = 3 if X < O = 3x +2, if O � x � 1 X x-1 if 1 < X Solution : Engi neering Mathematics f(0 + h) = lim 3(0+h )+2= lim 3h+2=2 At x = 0 f(0 - h) = lim 3 = 3 h➔O 1 91 h➔O So, f(x) is discontinuous at x = 0 of first kind. h➔O 1 +h . 1+h . f(1 + h) = I I m --- = 1 Im - = oo h➔O h ➔ O (1+h) - 1 h At x = 1 Example 32 : Solution : So, f(x) is discontinuous at x = 1 of i nfinite kind from right. Discuss the continuity of the functionat x = a: f(a - h) = hlim f(a - h ) = hlim ➔ O ➔O 1 1 . f(x) = -- s in - - ; x ;tc a x-a x-a = 0, X=a 1 1 sin (a - h) - a (a - h) - a 1 sin oo . 1 . 1 . -1 . ( -) = = I Im - sin ( --) = , Im - sin h➔O h h ➔O h h h 0 ⇒ Example 33 : Solution : = Finite but oscillatory value 0 = 00 f(a - h) does not exist, hence f(x) is discontinuous at x = a. The discontinuity is of infinite kind. 2 2 x2 x x + + +···00, show that f(x) is discontinuous at x = 0. If f(x) = -2 2 2 (1+ X 2 )3 1 +X (1 +X ) 0 0 + +. . . oo = 0 We have f(0) = -2 2 2 1+0 (1 + 0 ) 2 2 (-h) (-h) -+. . . oo ] f(0 - h) = f(0 - h) = lim f(-h ) = lim [---+--2 h ➔ O 1+( -h) h➔O (1+(-h)2 )2 2 2 ] h h . - 2 +. . . co = llm [-- 2 +- 2 h ➔ O 1+h (1+h ) 2 2 h h 2 ) (1+ ) (1+h � =1 (su m of infin ite G.P.) = lim = lim 1 h ➔0 h ➔O h 1-(1+h 2 ) (1+h 2 ) 2 h (h) . . +. . . oo ] = 1 -+ f(0 + h) = f(0 + h) = h m f(h) = h m [h ➔ O 1+h 2 h➔O (1+h 2 )2 2 ⇒ 2 2 2 f(0 + h) = f(0 - h ) ;tc f(0) . Hence, the function f(x) is discontinuous at x = 0. The discontinuity is of removable kind. PROPERTIES OF CONTINUOUS FUNCTIONS (i ) (i i ) f I A function which is continuous in a closed interval is also bounded in that i nterval . A continuous function which has opposite signs at two poi nts vanishes a t least once between these points and v anishing point is called root of the function. (iii) A continuous function f(x) in the closed interval [a, b] assumes at least once every value between f(a) and f(b), it being assumed that f(a) * f(b ). IES MASTER Publication 1 92 Limits , Continu ity and Differentiability INOTE.. ----.:f 1 2. If M and m are the upper and lower bounds of f(x) in {a, b) such that they are attained at x = a and x = p respectively, then by the property (iii) it fol lows that f(x) has any value between M and m at least one as x passes from a to p when a < a < p < b. T he converse o f the property (iii) i s not true, i.e., i t cannot b e adopted a s the definition of continuity since a function f(x) which takes all values between f(a) and f(b) is not necessarily continuous in the interval {a, b ). The above statement is evident from the following examples. Example 34 : Let f(x) be defined for the interval (0, 1) as follows: =0 for X = 0 f(x) = = 1- x for 0 < x < 1 =1 for X =1 Solution : Show that f(x) is not continuous at the points x =0 and x = 1 although f(x) assumes once and only once every value between f(0) and f(1). Here f{0) and =0 f(0 + h) lim f(0 + h) lim (1- h)= 1 = h➔O f(1 - h) = lim f(1- h) = lim {1- (1-h)}= 0 h➔O Since, f(0 + h) -ct- f{0) the function f(x) is discontinuous at x = 0 on the right. Also, f(1) = 1 and h➔O h➔O Since, f(1- h) -ct- f(1) , the function f(x) is discontinuous at x = 1 on the left. And f(x) assumes all values between 0 and 1 as x takes all values from x = 0 to x = 1. Yet the function si not continuous in the closed interval [0, 1]. Example 35 : Let f(x) = sin ! for x -ct- 0 and f(0) = 0. X Solution : ( -2 2) Test for continuity for all points in the interval --;- • ; Here f(0) =0. and f{h) f(h) = lim sin ! = sin oo = hlim ➔O h➔O h which oscillates between -1 and +1. Thus, f(x) is discontinuous at x = 0. But f(-2/n)= sin(-n/2)= -1, and f(2/n) = sin{n/2) =1 so, f(x) assumes all values between :....1 and +1 as x-axis varies from x =-2/n to x= 2/n including x = 0. still f(x) is discontinuous at x = 0 SALTUS OF A FUNCTION T he difference between upper and lower bound of a function in the small neighbourhood of a given point is known as saltus at that point s(f) = U(f) - L(f) If f is unbounded above or below, then the Saltus is given by s(f) = + oo Example 36 : Show that for the function, f(x)= x sin ! , X =0, x *o X= 0 l the oscillation at x = 0 is 2. Engineering Mathematics 1 93 For the given function f(x) the oscillation [in a small neighbourhood of origin (0 - h, 0 +h)], are ranging between - 1 and + 1 . Solution : i.e., U(f) = 1 , L(f) =- 1 Hence, the difference between the greatest and least =2. Example 37 : Solution : 1 For the function f(x) = 3 lim tan- ( 1 / x) show that the saltus at x = 0 is 2. 7t X ➔ O Take a small neighbourhood (0 - h, 0 + h) of the origin. Here· the oscillation ranges between the values of tan-1 (00) to tan-1 (- oo ) i.e. from % to(- % ). Thus U(f) = ; · % = 1 and L(f ) =� {- % ) =- 1 Hence, the difference between greatest and least is 2. FUNCTIONS OF TWO VARIABLES Let x and y are two independent variables. If a third variable z depends upon x and y, then z is called a function of two indep1:1ndent variables x, y, which is represented by the functional relation z =f(x, y) The set of points (x, y) for which f(x, y) is defined is called the d omain of the function. The collection of corresponding values of z is called the range of the function. Each point of the domain is an ordered pair and is mapped into the z-axis. The domain is sometimes called the reg ion and the points within it are called the argument points or simply, the argument. LIMIT OF A FUNCTION OF TWO VARIABLES We say that f(x, y) ➔ l as (x, y) ➔ (a, b) if and only if for every E I I > 0, there exist 8 > 0 such that f(x, y) - l < E * Whenever Ix -al < 8 and I Y - bl < 8 and (x, y) (a, b). In other words, a function f(x, y) is said to have limit l as (x, y) ➔ (a, b) if the value of f(x, y) can be made as close as a fixed number l by taking (x, y) close to the point (a, b) but not equal to (a, b). Symbolically, we write lim (x, y ) ➔ (a, b ) f(x, y) = l which is called the limit (the d o u b l e l i mit or the s i m u ltaneo us l i mit) of f(x, y) when (x, y) tends to (a, b). Remark 1 : It is not necessary that (a, b) be in the domain of f. T hat is, a limit may exist with f(a, b) being undefined. Remark 2: Since, the domain of f(x, y) consist region, therefore the concepts of the left hand limit and right hand limit of a function of single variable are not applicable. because (x, y) can approach to be (a, b) along infinite number of paths. Remark 3 : If limit depends upon these paths along which (x, y) ➔ (a, b) then we say that limit does not exist and if limit is same along infinite no. of paths then we say that limit exist. Remark 4: If lim (x, y) ➔ (a,b ) f(, y) = l and if y = $(x) is any function such that $(x) ➔ b, when x ➔ a, then lim f(x, $(x)) must exist and be equal to l. X ➔a I ES MASTER Publication 1 94 Limits, Continuity and Differentiability Thus, if we can determ ine two functions $.1 (x) and $ 2 (x) such that lim f(x, $ 1 (x)) x➔a * lim➔a f(x, $ then, we shall conclude that the simultaneous limit X 2 (x)) lim f(x, • Y ) does not exist . { x,y)➔ (a, b) Algebra of Limits Let f(x, y) and g(x, y) be two functions defined on some neighbourhood of a point (a, b) such that I i m f(x, y) = l, l i m g(x, y) = m , a s (x, y) ➔ (a, b), then (i) lim f(x, y) ± lim g(x, y) = l + m lim [f(x, y) ± g(x, y)] = ( x, y)➔ {a,b) (x, y)➔ {a, b) { x,y)➔ {a, b) .. ( .1 1 1) lim f(x, y) f(x, y) � = J_ provided m = {x, y)➔ {a, b ) lim lim g(x, y) m ' {x, y)➔ (a,b) L g(x, y) {x, y ) ➔ (a, b) (ii) f(x, y) • lim g(x, y) = l · m lim {f(x, y) • g(x, y)} = lim {x, y) ➔ (a,b) {x,y)➔ (a , b) (x, y)➔ (a, b) r } *0 x2 _ y , as (x, Y) ➔ (0, 0). Evaluate the limit of f(x, y) = -2-2 X +y 2 Example 38 : Solution : Let (x, y) ➔ (0, 0) along the path y = mx, then 2 . . . x - x2 m2 1 - m 2 = hm f(x, y) = hm f(x, mx) = hm 2 x ➔ 0 X +m 2 X 2 - 1+m 2 X➔0 (x,y)➔ {0,0) which depends upon m i.e. on path chosen. Hence, the limit does not exist. Example 39 : Solution : Example 40 : Solution : 2xy , (x, y) Eval uate the limit of f(x, y) = -2-X +y 2 * (0, 0) as (x, y) ➔ (0, 0). Let (x, y) ➔ (0, 0) along the path y = mx, then 2m x2 lim f(x, y) = lim f(x, mx) = lim / � 2 2 X ➔ 0 x +m x (x, y ) ➔ (0, 0) X ➔0 1+m which depends upon m i.e. on path chosen. Hence, the limit does not exist. Evaluate the limit of the following function as (x, y) ➔ (0, 0) (i) (i) f(x, y) = -2-2 X +y (ii) x3y3 f(x, y) = -2-2 X +y x 3 - y3 Let (x, y) ➔ (0, 0) along the path y = mx, we have lim f(x, y) = (x, y)➔ (0, 0) = ⇒ lim f(x, y) ➔ 0, y ➔ mX lim f(x, mx) X ➔O X The limit is i ndependent of m i.e. on path chosen. hence, limit exist. Engineering Mathematics (ii) 195 Let (x , y) ➔ (0 , 0) along the path y = mx , we have lim (x, y ) ➔ (0, 0) ⇒ f(x , y) = lim f(x , mx) X➔O (1 - m3 )x x3 -m3 x3 =O = lim = lim 2 x ➔ D x +m2 x2 x ➔ D 1 + m2 The limit is independent of m i.e. on path chosen. Hence, limit exist. Example 41 : E valuate the limit of the function Solution : Let (x , y) ➔ (0 , 0) along y = mx , then f(x , y) lim (x, y)➔(O, O ) x2 y X +y = -4--2 , when (x , y) ➔ (0 , 0). f(x , y) = lim f(x , mx) = lim X ➔O · approach gives · S o, any st raIg · ht 1me X➔Y X lim (x,y)➔ "(O, O ) 4 mx 3 =0 + m2 X 2 f(x y) = 0 ' But , if we approach (x , y) ➔ (0, 0) along y = mx2, then lim (x, y ) ➔ (O , O ) f(x , y) = lim f(x , mx 2 ) = � x ➔O 1+m which shows that the limit depends upon m i.e. on path chosen. Hence , exist. Example 42 : Solution : E valuate the limit of the function lim (x, y)➔( O, O ) f(x , y) does not 2 lim (x, y)➔ ( D,O ) 2xy x2 +y 4 Let (x , y) ➔ (0, 0) along the path x = my2 , we have 2xy 2 2xy2 = lim -�4 2 2 4 (x, y)➔( O , O ) x + y y ->D,x ➔ml X + y lim 2xy 2 . 4] [ 1 . Im Im = y ➔O 2 X ➔ m/ X + y I. = lim ➔O y 2my 4 (m2 +1) y 4 = 2m 1+m2 which shows that the limit depends upon m i.e. on path chosen. Hence , exist. x 3 + Y3 , ( x , y) x-y * (0 , 0) Example 43 : E valuate the limit for the functions f(x , y) = Solution : Let us take (x, y) ➔ (0, 0) along the path y =x - mx3 , we get lim (x,y)➔ "(O, O ) f(x , y) = lim x ➔ D, y ➔(x,-mx 3 ) [ = Xlim ➔D lim lim (x, y)➔ (D,O ) - 2xy2 - does not x2 + y 4 at origin. f(x , y) y➔ (x-mx 3 ) x 3 + Y3 ] x-y IES MASTER Publication . 1 96 Limits, Contin uity and Differe ntiability 2 - m 3 x 6 - 3mx 2 + 3m 2 x 4 ] 2 . = hm [--------- = x ➔o m m Si nce the limit depends on m i.e. on path chosen, thus it follows that limit does not exist: Repeated Limits The lim its lim lim f(x, y ) and lim lim f(x, y) are called repeated or iterated limits. X ➔a y ➔ b y➔b x➔a If lim lim f(x, y ) = l, a y➔ b then l is repeated limit of f(x, y) as y ➔ b, x ➔ a. If we change the order of taking the lim its, we get the other repreated limit X➔ i.e., lim lim f(x, y) = m Y ➔b X ➔ a when first x ➔ a and then y ➔ b. These two lim its may or may not be equal. Remark : I n case the double limit exists, the above two repeated lim its are necessarily equal but the converse is not true. However, if lim lim f(x, y) X ➔ a y ➔b the double limit does not exist. * ylimb xlima f(x, y) ➔ ➔ Example 44 = Show that the double l imit as (x, y) ➔ (0, 0) for the function Solution = Method-I : By repeated or interated method, we have (y - x)(1+x) f(x, y) = (y +x )(1+ y) does not exist. (y - x)(1 + x) [-(1+ x)] = - 1 lim lim f(x, y) = lim [ lim ] = Xlim ➔0 X ➔ 0 y ➔ 0 (y +X)(1+y ) X ➔0 y ➔ 0 1 (y - x)(1 +x) -- = 1 lim lim f(x, y) = lim [ lim ] = ylim ➔ D 1+y X ➔0 y ➔ 0 X ➔ 0 (y+x )(1+y) Y ➔D ⇒ lim lim f(x, y) x ➔0 y ➔0 * ylim0 xlim0 f(x, y) ➔ ➔ i.e. the two repeated limits exist but are unequal. Consequently, the double limit does not exist. Method-II : Let us make (x, y) ➔ (0, 0) along the line y = mx, we get (y - x)(1+x) lim lim f(x, y) = lim lim f(x, y) = lim [ lim X ➔ 0 y ➔ mX (y +x )(1+y) ] X ➔ 0 y ➔0 X ➔ 0 y ➔ mX _ (m - 1)(1+x) m-1 [ ]= - xlim m+1 ➔ o (m+1)(1+mx) Example 45 = Solution = which shows that the l i m it depends upon m i.e. on path chosen. Hence, double limit does not exist. Find the limit of the function f(x, y) = We have lim lim f(x, y) = lim 0= 0 X ➔0 y ➔0 x➔0 and X 2 xy +y 2 , (x, y) * (0, 0) as (x, y) ➔ (0, 0). lim lim f(x, y ) = lim (0) = 0 y ➔ 0 X ➔0 y➔0 Thus, the repeated limit exist and are equal. Engineering Mathematics 1 97 And the simultaneous limit is given by choosing y = mx. lim lim f(x, y) = lim X ➔ O y ➔ mx X ➔O lim [y ➔ mx X 2 xy + y2 ] 1+m2 which depends upon m i.e. on path chosen. Thus, simultaneous limit does not exist. Remark : From the above example, it is clear that even though repeated limits are equal, but it may happen that lim ( x, y ) ➔ (O, O ) Example 46 : If Solution : Repeated limits are f(x, y) does not exist. f(x, y) = x4 + y 4 _ x 2 y 2 Show that the repeated limits exist at origin but simultaneous limit does not exist. lim lim f(x, y) = lim lim X lim lim f(x, y) = lim [ lim y ➔ O x_➔ O X X➔O y➔O y➔O X➔O X➔O [y ➔ O 4 4 x>2 ] = lim [OJ = 0 + y _ X2 y2 X ➔ O + X:l y - X2 y2 ] =0 T hus, the repeated limits exist and are equal. And the simultaneous limit is given by if y = mx, then x2 y 2 x4 y 2 lim lim f(x, y) = lim [ lim 4 ] = lim x ➔ o y ➔ mx x + y 4 -x2 y 2 x ➔ O y ➔ mx x ➔ O [ x4 +x4 m4 -m2 x2 ] m2 1+m4 -m2 which depends upon m i.e. on path chosen. Thus, simultaneous limit does not exist. Hence, the repeated limits exist at origin but simultaneous limit does not exist. Example 47 : Evaluate the limit for the function Solution : Method-I : By using repeated method x2 _ y 2 , where (x, y) (0, 0). X2 +y2 Show that the two repeated limits of this function exist but simultaneous limit does not exist. * f(x, y) = x -y x lim lim f(x, y) = lim lim : : ] = lim [ : ] = 1 X➔O X X ➔ O [Y ➔ O X + y X➔Oy➔O 2 x -y lim f(x,y) = lim [ 1im : : ] = lim [ � ]=-1 and ylim y➔O ➔O X➔O y➔O X➔O X + y y T hus, two repeated limits exist but are unequal. Therefore, double limit also does not exist. Method II: Let (x, y) ➔ (0, 0) along the path y = mx, we get IES MASTER Publication 1 98 Li mits, Continu ity and Differentiability lim (x, y) ➔ (O, O ) ⇒ f( x , y) = lim X ➔ O, y ➔mx f( x, y) = lim [ X ➔O 2 2] lim � 2 X ➔ mX x +y 2 . 1-m - I 1m [ -] - X ➔O 1+m 2 2 1 - m2 1+m 2 the limit depends on m i.e. on path chosen, it follows that the double lim it does not ex ist. CONTINUITY OF FUNCTION OF TWO VARIABLES A function f(x , y) is said to be conti nuous at a point (a , b) if (i ) f(x, y) is defined at the point (a, b) (iii) (x, y ) ➔ (a,b) (i i) lim lim f( x , y ) exist, and f( x , y) = f(a, b). In otherwords, a function f(x , y) is said to be continuous at a point (a, b). (x, y) ➔ ( a ,b) If for every whenever > 0, there ex ists a number 8 > 0 depends on a such that !f( x , y) - f(a, b)I < E I x - a l < 8 and I Y - bl < 8 A function is said to be continuous in a region if it is continuous at every point in that region. E Example ,a : Solution : I nvestigate the continuity at (0, 0) of f(x , y) = Metho d 1 : To show that lim (x,y)➔(O, 0 ) lim ( lim f( x , Y )) = lim ( lim ; X ➔O y➔ O X➔O Y➔ O x 3 +y 3 = 0, 2 , (x, y) (0, 0) (x, y) = (0, 0) * f(x, y) exists. We have )=0 + y3 xy X 2xy 2 E and xy lim ( lim f(x, y) ) = lim ( lim ; Y ➔ O X ➔ O X +y y ➔ O X ➔O 2 But if ( x , y) ➔ (0, 0) simultaneously along the l i ne y = lim (x, y) ➔ (O, O) f( x, y) = lim f( x , x ) = lim ( x➔O X ➔O 2x :) = 1* O 2x x, then 3)=O Thus, we find that f(x , y) assume different l i mits for different paths. It does not tend to unique limit w h i c h i s i n d e p e n d e n t of t h e path. T h e refo r e , l i m i t does n o t e x i st a n d h e n ce f(x , y) is not continuous at (0, 0). Metho d I I : Let us make (x , y) ➔ (0, 0) along the path y = m x , we have lim (X, Y ) ➔ (O, O ) f(x, y ) = lim f( x , m x ) = lim X ➔O 2 2m x 3 X ➔ O x 3 (1+m 2 ) = 2m2 1+ m 2 Since, the l i m it depends upon m i . e. the path chosen, therefore the limit does not ex ist and the function cannot be continuous at (0, 0). Example '9 : Show that the function xy f(x , y) = ___ 2 2 ' Jx +y = 0, is conti nuous at the origin. ( x , y) 0 ( x, y) = (0, 0) * Solution : To show that and lim (x, y)-> (0, 0) f(x, y) exists. We have lim ( lim f(x, y) ) = lim l lim X -> 0 y -> 0 X -> 0 y -> 0 r lim ( lim f(x, y) ) = lim [ lim y -> 0 X -> 0 y -> 0 X -> 0 ,b J .V+;J X X 2 + 2 + y 2 2 lim f(x, y) = lim f(x, mx) = lim X -> 0 X -> 0 y =0 2 Q X 1 + m2 Also, let (x, y ) ➔ (0, 0) along y = mx2 , then 2 lim f(x, y) = lim f(x, mx ) = lim Solution : x -, o x -, o x ✓1 =0 mx3 +m 2 x2 =o Si nce the limit along any path is same, the limit exist and equal to zero, which is the value of the function at origin. Hence, f(x, y) is continuous at origin. � =! � Example 50 : x 1 99 =0 If (x, y) ➔ (0, 0) along the path y = mx, then y -> mX X -> 0 Eng ineering Mathematics Examine the continuity of the function f(x, y) at (0, 0). f(x, y) = x3 +y 3 x-y = 0, (x, y) (0, 0) (x, y) = (0, 0) * Let us take (x, y) ➔ (0, 0) along the path y = x - mx3 , we get lim (x,y)->(0, 0) f(x, y) = lim y-> x-mx X -> 0 f(x, y) = lim f(x, x - mx ) ·, 3 3 X -> 0 3 3 6 3 2 2 . x +(x - mx ) 2 - m x - 3mx +3m x4 . 2 = hm [-- -- -] = hm [--------- ] - _ 3 x x -, o o m -, -· m x - (x - mx ) 3 Since, the limit of a fucntion depends upon m i.e. on path chosen. So, the limit does not exist. Hence, the given function is not continuous. DIFFERENTIABILITY Differential Coefficients It is the rate of change of the dependent variable y with respect to another independent variable x. I n other words, if the variable y is a continuous function of x, then the change in y relative to x is called the derivative of y with respect to X. Differentiability f(a - h ) - f(a) f(a +h) - f(a) both exists and are equal then this limit is called the differential coefficient and lim h -> 0 h ( -h ) o r the derivative of f(x) at the point x = a. If lim h -> 0 Non-Differentiability A function f(x) is said to be non differentiable at x = a if either or both of the above two limits do not exist, or they are unequal IES MASTER Publication 200 Limits , Continu ity and Differentiability Progressive and Regressive Derivative If lim h➔ O+ [ f(a +h) - f(a) ] exists as h approaches to zero through positive val ues only, we say that f(x) has a progressive h or the right hand derivative at x = a and is denoted by Rf'(a). Thus, Rf'(a) = lim h➔O Again, if lim [ h ➔O [f(a+h) - f(a) ] where h is positive. h f(a - h) - f(a) ] exists as h approaches zero through negative values only then f(x) is said to be have -h a left hand or Regressive Derivative at x = a and is denoted by Lf'(a). Thus, Lf'(a) = lim h ➔O [ f(a - h) - f(a) ] where h is positive. -h The function f(x) has unique derivative at x = a if Rf'(a) = Lf'(a), and the common value is called derivative of f(x) at x = a. Chain Rule of Differentiability If <j>(x) = 'l'[f(x)], then <!>'(x) = 'l'' [f(x)] f'(x). Theorem : Let f and g be functions defined on an interval I and f, g are differentiable at x = a E I then (i) (ii) f ± g is differentiable and (f ± g)' (a)= f'(a) ± g'(a). cf is differentiable and (cf)'(a) = cf'(a) : c E R. (iii) f · g is differentiable and (fg)'(a) = f'(a) g(a) + f(a) g'(a). (iv) (v) - is differentiable at x = a and f 1 (!) f (a) = - f'(a) ; provided f(a) [f(a)] 2 * 0. f g(a) - f(a) g'(a) - is differentiable at x = a and ( -f ) ' (a) = - f'(a) - �-��� ; provided g(a) 2 g g [g(a)] A Necessary Condition for the Existence of a Finite Derivative Continuity is a necessary but not the sufficient conditions for the existence of a finite derivative. Example 51 : Show that the function f(x) Solution : f(x) = lx l = -x, x < 0 l 0, x = 0 x, x > 0 = !xi is conti nuous but not differentiable at x LHL = 0 ⇒ f(0) = 0 RHL=0 Hence, f(x) is continuous at x = 0. For differentiabil ity, we have Rf'(0) = lim h➔O f(O + h) - f(O) lhl - 0 = lim =1 h➔O h h = 0. * 0. Engineering Mathematics Lf'(0)= lim h➔O -hl -O f(O - h)- f(O) = lim (�) =- 1 = lim l h ➔ O -h h ➔ O -h -h Hence, f'(0) does not exist i.e. f(x) is not derivable at x = 0 altough it is continuous at x Example 52 : Show that the function f(x) defined by f(x) = x sin(1/x), is continuous but not differentiable. Solution � Rf'(0)= lim h➔ O = 0, 201 = 0. * x 0 X =0 (O + h)sin { 1t(O + h)} -O = lim sin (1/h) h ➔O h which does not exists as it oscillates between -1 and 1. Similarly, Lf'(0) does not exists. Hence, f is not differentiable at x For continuity f(0 + h) = lim f(0 + h)= lim (hsin ! )= 0 h➔O h➔O h I I and f(0- h) = lim f(0- h)= lim {-hsin(_!__ )} = 0 and f(0) h➔O h➔O -h = 0. = 0 (given) So, the function f(x) is continuous at x = 0 but is not diferentiable at x =0. Example 53 1 Solution : If f(X) = ✓ x ,, = h, h Here f(0 + h) = ✓ Also, f(0 - h) = and f(0) X�0 X <O h ✓- (-h) = ✓ = 0 by definition. Test for continuity and derivability at x = 0. h : . f(0 + h)= lim f(0 + h)= lim ( ✓ ) =O h ➔O h ➔O h :. f(0- h)= lim f(0 - h)= lim ✓ = 0 h➔O h➔O Since f(0 + h) = f(0 - h) = f(0) the function is continuous at x = 0. For derivability, we have h 1 f(O + h)-f(O) = lim ✓ -O = lim h = + co Rf'(0) = lim h➔O h➔O h h➔O ✓ h h f(O- h)- f(O) ✓ -O 1 Lf'(O)= lim [ h =-oo = lim (-] = lim h ➔ O -h h➔O ✓ ) h➔O -h Hence, f(x) is non -derivable at x = 0 although it is continuous at that point. 1 Example 54 : The function f(x) = - which is not continuous at x = 0 has no derivative at x = 0. X Solution : 1 The graph for the function y = f(x) = - is the curve xy = 1 which is a rectangular hyperbola whose asymptotes are the axes of coordinates. This is obviously discontinuous at x = 0 •: f(0) f(0 + 0)= + co and f(0- 0)=- co Also, Rf'(0) = lim [ h➔O 1th - 110 f(O + h)- f(O) ] = lim [ ] = co (Does not exist) h➔O h h = oo IES MASTER Publication 202 Limits, Continuity and Differentiability [ -1111 - 110 [ f(O - h ) - f(O) ] = lim ] = -w (Does not ex i st) h➔O h h Lf'(0) = lim h➔O Example 55 : Test f(x) = x 1I3 for its conti nuity and derivabil ity at x = 0. Solution : • f(0 + h ) - f(0) . 1 . h 1/ 3 - 0 H ere Rf'(0 ) = 1 1 m ---- = 1 1m - = 1 1m ()=+w h➔O h➔O h ➔ O h 2' 3 h h 1 f(O - h) - f(O) (-h ) 13 - f(O) ] ] = lim [ h -> 0 L h➔O h -h I Lf'( 0) = lim 1 1 . = 1·1m - = + w = I 1 m -11 ➔ 0 (-h) 2/ 3 h ➔ O h 2 /3 Si nce Rf'(O) = Lf'(0) = + w f'(0) = w ⇒ [Si nce [(-h)2] 1I3 = [(-h2 )] 1 I3 = h213] So, f(x) is not differentiable at x = 0 Again f(0) = 0 by definition f(0 + h ) = lim (h 1I3 ) = 0 f(0 + h) = hlim ➔O h➔O f (0 - h) = lim [(-h 1I3 )]= 0 f(0 _ h) = hlim ➔O h➔O Hence, f(x) is continuous a t x = 0. Example 56 : Test the function defined by f(x) = x 2 sin � , x f(0) = 0 * Ol with regard to (i) continuity (ii) differentiabi l ity and (iii) continuity of the derivative at x = 0 lim [f(x)] = lim ( x 2 sin ! ) = 0 = f(0) O X➔O X Solution : Hence the function f(x) is conti nuous at x = 0 X➔ Also, Rf'(0) = lim h➔D Si nce Lf'(0) = lim h >- 0 2 f(O + h) - f(O) [h sin 1/h - O] = lim = lim [h sin 1/h] = 0 h➔D h ➔D -h h h 2 sin(-1/h ) - O ] f(O - h ) - f(O ) = = lim [h sin 1/h] = 0 lim [ h -, 0 h➔D -h -·h f'(0) = 0 Hence f(x) is also differentiable at x = 0 Rf'(0) = Lf'(O) = 0 ⇒ Now, f'(x ) = 2xsin 2 + x 2 cos ( ! ) ( - � ) = 2x sin 2 - cos ! , ( x * 0) X X X X X . . . (i) But lim [f'( x )] = lim [2x sin 1/x - cos 1/x] = - cosoo which oscillates between - 1 and + 1 , X ➔D X ➔D Hence, lim [f'( x )] does not exist. Thus lim [f'(x )] X➔D ' X >- 0 * f'(0) So, f'( x) i s not contin uous a t x = 0 . Example 57 : Solution : X If f(x) = --1,x 1 + e 1X * 0, f( O ) == Eng ineering Mathematics 0 s h ow t h at f is continuous at x = 0, but f'( O ) does not exist. O +h Now, f( O + 0) = lim h ➔ 0 1 + e 1l( O+h) == 203 h == O lim __ -> h 0 1 + e1/ h -h f(O - h ) = lim == 0 h ---, 0 1 + e -1/h Similarly, f(O + h ) = f( O - 0) = f(O ) Since The function is continuous at x = 0. We now proceed to find the derivative of f(x) at x = 0. We have, Rf'( O ) == lim [ h >- 0 f( O + h ) - f( O ) h ] == lim [ 11 ---, 0 1 - 1111 == lim Lf'( O ) == lim 1 + e 0 11 ---, 0 -h 11 __, 1 + e -1 111 Si nce Rf'( O ) * Lf'( -h h / (1 + e h == O) 1111 )-0 1 1 == __ == 0 ] == lim __ h >- 0 1 + e 1111 1'+ e "' 1 1 == __ == 1 ---oo 1 + 0 1+e The derivative of f at x = 0 does not exist. Example 58 : Let f(x) Solution : = e~1 2 1x sin 1/x for x * 0, 1 e - 1l(O + h ) sin -- - 0 O + h Rf'( O ) == lim h ---, 0 h = = ��o 2 . 1 2 2 1/h (�/ h )- 2 . h [1 + - - + + . . .] 2! 1! Si n - a finite quantity 00 == f(O ) == = 0. Test th e differentiabil ity of the function at x = 0. sin! lim __h_ h ---, 0 h·e111i2 l . 1 e == ��o Sin- h + _]__ + � + . . . . h 2 h1 3 . 1 . . ) 0 ( ·: s1n osc1llates between -1 and +1 a nd hence finite. h 1 . 1 0 h/ . s1 n e -1 /( - sin - - - O h O == 0 - �_ == lim __h_ Similarly, Lf'( O ) == ----�2 -h 11---,0 he 1111 Si nce Rf'(O) == Lf'(O) == 0, t h e function f is differentiable at x = 0 and its derivative is zero. Exa mple 59 : Test t h e differentiabil ity of t h e function: Solution : 1 f(x ) = x tan~ (�} At x = 0 x*O and f(O) = 0 at x = 0 ' f( O - h ) - f (O ) _ . f(-h ) - O _ . l .f'( O ) - I 1m -'-----'----'-� - 1 1 m -'--'---- - 1 1m h---,0 h ---, 0 11 ---, 0 -h -h f(h) - O f( O + h ) - f(O) == lim Rf'(O ) == lim h --, 0 h---,0 h h ⇒ Lf'( O ) * Rf'( O ), == lim h ---, 0 (-h ) tan-1 -h _ 1 .1 m - t an -1 ! _ --�� ( ·- _ 2:_ h➔0 h 2 -h h tan- h (J_) 1 (2) h so f(x) is not differentiable at x = 0. == lim tan h >- 0 1 ( !) h == � 2 ) IES MASTER Publication 204 Limits, Continuity and Differentiability Example 60 : xe1/ x -If a function f(x) is defined as: f(x) = 1+ e1, x = 0 x *0 X=0 Show that f(x) i s continuous but not differentiable at x =0. Solution : - 1/ h -0 -11h 1 f(O - h)- f(O) 1 + e -= 0 = lim = lim Lf'(0) = lim 1 th h ➔ O h ➔ O -h h➔O -h e +1 h e1t h -O f(O +h)-f(O) 1 = lim � = lim - - =1 Rf'(0) = lim h➔O 1 h➔O h➔O h h +1 1/h e (-h) e ��- -he-1/h Now, f(0- h)= lim f(0- h) = lim h ➔ O 1+e - 11 h ➔O 1 h __ _ = lim = 0 = hlim ➔O e1/h +1 h ➔ O __! 2 + 2 + � +... ) h 2! h h( and ⇒ he1' h h f(0 + h) = lim f(0 + h)= lim--= lim --- = 0 h➔ O h ➔ O 1+e11h h ➔ O 1 +1 1/h e Lf'(0) etc Rf'(0) but f(0 - 0)= f(0 + 0)= f(0)= 0 Hence, f(x) is continuous but not differentiable at x =0. Engineering Mathematics 205 PREVIOUS YEARS GATE & ESE QUESTIONS Questions marked with aesterisk (*) are Conceptual Questions 1 .* 2. . x ( e x -1)+2( cos x-1) L Im�-�---= X➔O x (1-cos x ) 1 1- e -i5X Lim X➔0 1Q 1-e-jX (a) 0 X➔«> (b) 2 (c) 1 4. (d) 0 (a) 0 (b) 1 The limit of (a) 1 (b) 0 (c) (d) oo sin8 Lt - .Is 8 8➔0 7.* x3 - cos x . hm 2 2 equal x ➔"' x +( sin x) (a) may or may not exist [GATE-1994; 1 Mark] [GATE-1996; 1 Mark] 1 0. If y = lxl for x < 0 and y = x for x :2: 0, then (a) 11. �� is discontinuous 'at x = 0 (b) y is discontinuous at x = 0 (c) y is not defined at x = 0 (d) None of these [GATE-1997; 1 Mark] . sinm8 hm -- , where m is an integer, i s one of the 8➔0 8 following: (c) (b) 0 (d) does not exist [GATE-1995-CS; 2 Marks] T h e function f(x) = Ix + 1 1 o n the interv al [-2, OJ is (a) continuous and differentiable (b) continuous on the integers but not differentiable at all points (c) neither continuous nor differentiable (d) differentiable but not continuous [GATE-1995; 1 Mark] (a) oo (c) 2 If a function is continuous at a point its first derivative (a) m 00 (d) None of the above 6.* [GATE-1994; 1 Mark] [GATE-1995; 1 Mark] (d) has unique value 1 1 The value of [ 1im(. ----)] is : tan x x➔"' sin x (c) 2 5. [GATE-1994; 1 Mark] (d) Non-existant (c) will not exist X (a) oo (b) 0 (b) exists always [GATE-1993] sinx . The value of [ rIm -] Is: X (c) 1 9. (d) 1 lim x sin (� ) is X➔O (a) oo [GATE-1993 (ME)] (b) 1.1 (c) 0.5 3. 8. m8 (b) mn (d) 1 [GATE-1997; 1 Mark] 1 2.* Find the limiting value of the ratio of the square of the sum ·of a natural numbers to n times the sum of squares of the n natural number as, n approaches infinity. [GATE-1997] 13.* The profile of a cam in a particular zone is given by 3 x = ✓ cos8 and y = sin8 . The normal to the cam profile at 8= ¾ is at an angle (with respect to x axis): (a) (c) 7t 4 7t (b) 7t 2 (d) 0 [GATE-1998] IES MASTER Publication 206 Limits, Conti nuity and Differentiabil ity (a) (c) (c) (b) 0 2 (d) 1 00 [GATE-1 999; 1 Mark] 22. 1 5. Value of the function lim (x - a)(x - a) is (a) 1 (c) 1 6. (b) 0 X >- 8 (d) a co ) Lt ( x - 1 is (X - 1) (a) 00 (c) 2 2 (b) 0 (d) 1 X >- 1 1 7. The limit of the function f(x) given by (a) 1 (c) [GATE-1 999; 1 Mark] [GATE-2000; 1 Mark] = [ 1 - :: ] as x ➔ co is (a) 1 (c) 0 1 9. sin 2 ( x _ 2:_ ) 4 n n x ---> X-4 4 l im (b) -1 (d) Does not exist [GATE-2001 ; 1 Mark] = (b) .(a) 0 [GATE-2001 (IN)] x2 (d) f(x) (c) 24. = maximum(x, -x) [GATE-2002; 1 Mark] 21 .* The l i m it of the following sequence as n ➔ oo is: X - n1/ n n - (b) [GATE-2003; 1 Mark] = 7 x3 + x2 lim 3 is x- ,o 2x - 7x2 (d) infinite 7 [GATE-2004; 1 Mark] If x = a ( 0 + sin 0 ) a n d (a) sin (%) (b) cos (%) (c) tan (% ) {d) cot (% ) dy/dx will b e equal to y = a (1 - cos 0) , then [GATE-2004] 25.* W ith a 1 unit change in b, what is the change in x in the solution of the system of equations x + y = 2, 1 .0 1 x + 0.99 y = b? (a) zero (b) 2 units (c) 50 un its (d) 1 00 units (a) y = 3x - 5 20.* Which of the following functions is not differentiable in the domain [-1 , 1 ]? (a) f(x) = {b) f(x) = x - 1 (c) f(x) = 2 . 2 . Sin X . I Im -- Is equa I to X--->0 X (a) 0 (b) oo (c) 1 (d) -1 [GATE-2005] 26.* Equation of the l i ne normal to function f(x) (x - 8)213 + 1 at P(0, 5) is 1 2 (d) 2 (c) 1 {d) - oo [GATE-2002-CE; 1 Mark] (a) zero (d) Zero [GATE-2000; 1 Mark] 1 8. W h a t i s t h e d e r i v a t i v e of f ( x ) = l x l a t X = 0? oo 23. The value of the function f(x) (b) ex p[-a4] co (b) 1 (a) 0 n 1 4. Li mit of the function lim � is 2 n--->oo v n + n (c) 3y = 27. If f(x) = X -7 x + (c) 0 = (d) 3y = x - 5 3 5x2 - 1 2x - 9 (a) - 1 /3 28.* If, y (b) y = 3x + 5 + 15 2x 2 = [GATE-2006; 2 Marks] , then lim f(x) is x---,3 (b) 5/1 8 {d) 2/5 [GATE-2006; 2 Marks] x + x + x + .Jx + . . .oo , then y(2) ✓ ✓ = (b) 4 only (a) 4 or 1 (c) 1 only 29. lim ➔O (a) 0 X X3 (c) 1 /3 30. (d) undefined 2 -(1 +x + �2 'j ex Engi neering Mathematics [GATE-2007-ME; 2 Marks] = (b) 1/6 (d) 1 [GATE-2007-ME; 2 Marks] ' sin(0 / 2) I 1 m ---'---------'- .1 s 0---,0 0 (a) 0 . 5 (b) (c) 2 (d) not defined [GATE-2007-EC; 1 Mark] 31 .* Consider the functio n f ( x ) = lx 3 I , where x i s real . then the function f(x) at x :::; 0 i s (a) continuous but not differentiable (b) once differentiable but not twice (c) twice differentiable but not thrice (d) three differentiable 32. What i s the value of lim " X ➔4 (a) (c) ✓ 2 x -2 33. The value of lim -- X >- 8 ( X - 8 ) (c) 34. 1 16 (a) (c) 12 1 8 1 (d) 4 [GATE-2008-ME; 1 Mark] 1 (b) -1 (d) -oo x - sin x lim ➔ OO X +cos x X (b) 00 [GATE-2008 (CS)] sin x • i1m -- i s 1 (a) indeterminate (c) 1 X X➔O [GATE-2008 (IN)] (b) 0 (d) a. [GATE-2008 (IN)] sin( x ) lim [ J is The value of the expression X--->O ex X (a) 0 (b) (d) 38.* The function y = 12 - 3xl 1 2 1+e [GATE-2008 (Pl)] (a) is continuous Vx E R and differentiable vx E R (b) is continuous Vx E R and differentiable Vx E R except at x = 3/2 (c) is continuous Vx E R and differentiable Vx E R except at x = 2/3 (d) X-4 [GATE-2007 (Pl)] dy . Given y = x 2 +2x +1 O the value of is equal d X l X=1 to (a) 0 (b) 4 (c) 1 2 (d) 13 (c) 1 7t 113 (a) 37. cos x - sin x (d) l i m it does not exist -✓ 36. [GATE-2007 (IN)] (b) 0 2 35. 207 is c o n t i n u o u s Vx E R except x = 3 a n d differentiable V x E R 39. The lim X ➔D (a) 2/3 (c) 1 /4 40 . sin [� x ] 3 is X [GATE-2010-ME; 1 Mark] (b) 1 (d) 1 /2 [GATE-201 0-CE; 1 Mark] I f f(x ) = s i n lx l then the value of � at x = - � i s dx 4 (a) 0 1 (b) ✓2 (d) 1 [GATE-201 0 (Pl)] IES MASTER Publication 208 41. Limits, Con tinuity and Differentiability ) n 1 2 What is the value of li m ( 1-n➔oo n (b) e -2 (a) 0 (c) e -11 2 (d) 1 . si n e equal to? 42. What is hm 9➔0 e (a) e ? (GATE-2010 (CS)] (b) si n e (d) 1 [GATE-2011-ME; 1 Mark] (c) 0 43. What should be the value of 'A. such that the functi on defined below is continuous at x - 'A. cos x - I'f X 7' 7t 1 2 = n /2? f( x )=l % - x (a) zero (c) 1 1 if x =1t I 2 45. -cos x ) is llm c X2 X ➔O 1 (a) 4 (c) 1 (c) 30 (d) 35 (GATE-2013-EE; 1 Mark] 47.* Whi ch one of the followi ng functi on i s conti nuous at X = 3 ? f ( x )={ (c) f ( x) ={ 4, if=3 8 - X, if X 7' 3 x +3 , if x �3 x -4, if x > 3 f ( x ) = --- , i f x X 3 - 27 x�a (;�:;x�) (a) 0 (c) 1 1 is *3 equal to (GATE-2013 (CS)] (b) 0.5 (d) 2 [GATE-2014-ME-Set-2; 1 Mark] (b) the functi on must be deri vable at the point (c) the li mi t of the functi on at the point tends to i nfi ni ty [GATE-2011-CE; 2 Marks] 46. A functi on y = 5x2 + 10 x is defi ned over an open i nterval x = (1, 2). At least at one poi nt i n this i nterval, dy i s ex actly dx (b) 25 (a) 20 (b) x+3 3 -- I'f X < 3 (a) the l imit of the functi on may not exi st at the point (d) n /2 1 (b) 2 (d) 2 [GATE-2012-ME, Pl; 1 Mark] f ( x) 49. If a functi on is conti nuous at a poi nt, (b) 2/ 7t 44. Consi der the functi on f (x) = l xl i n the i nterval -1 < x < 1. At the poi nt x = 0, f(x ) is (a) continuous and d i fferentiable. (b) non-conti nuous and d i fferentiable. (c) conti nuous and non-di fferenti able. (d) nei ther conti nuous nor d i fferenti able. (GATE-2012-ME, Pl; 1 Mark] (a) (d) 48 · ={ 2, if x =3 x -1, if x > 3 (d) the l i mit must exi st at the point and the value of limit should be same as the value of the function at that point 50. x -sin x Lt is x➔o 1-cos x [GATE-2014-ME-Set-3; 1 Mark] (a) 0 (c) 3 (b) 1 (d) not defi ned 51. If y = 5x y = 3 2 [GATE-2014-ME-Set-1; 1 Mark] + 3 than the tangent at X = 0, (a) passes through x (b) has a slope + 1 (c) is parallel to x-axi s (d) has a slope of -1 52. = 0, y = 0 [GATE-2014 SET-I; 1 Mark] x+si n x . ) i s equal to hm ( X X ➔cx> (a) - 00 (c) 1 (b) 0 (d) oo [GATE-2014-CE-Set-1; 1 Mark] Engineering Mathematics 53 . . Xa - 1 . I Im - - ts equaI to a➔O 61 . a (a) log x 0 (b) (d) oo (c) X log X [GATE-2014-CE-SET-2; 2 Marks] x 54. T he value of lim (1+ ..!.) is X➔oo X (a) ln2 (d) oo 56. The value of lim (a) (c) X➔ O 0 1- cos( x 1 4 4 2X (b} 1 is 58 .* L im j1+ ..!. j X➔ oo X (a) e 2(c) 1 (a) 0 (c) 1 x lim x 1/ is (a) oo (c) 1 63. 2 [GATE-201 5-ME-Set-1 ; 1 Mark] sinx ) is -. + x cosx [GATE-2015-ME-Set-3 ; 1 Mark] (b) e (d) e2 [GATE-201 5-CE-Set-2; 1 Mark] 1 2 (d) oo [GATE-201 5 (Pl)] (b) 0 (d) not defined [GATE-201 5 (CS-Set 1 )] (d) - 4 and -1 [GATE-201 6-ME-Set-2; 1 Mark] loge 1+4x) ; is equal to X ➔O e X- 1 1 12 (a) 0 (b) (c) (d) 1 (a) 4 3 [GATE-201 6-ME-Set-1 ; 1 Mark] [.Jx 2 +x -1- x] 0 (c) 1/2 is (b) oo (d) --oo [GATE-2016; 2 Marks] ?. 65.* What is the value of lim __!'!__ 2 2 X➔ OX y ➔O +y (b) -1 (a) 1 (c) 0 (d) Limit does not exist is equal to (b) [GATE-201 5 (CS-Set 3)) Lt X ➔oo x 2 -xy 59.* T he value of lim r. c is (x,y)➔(O,O) vX - v Y 60. (c) -4 and 1 64.* lim x -,o 2 stnx 2x (d} oo x2 - 3x - 4 is NOT continuous are x2 +3x - 4 (a) 4 and - 1 (b} 4 and 1 (d) undefined 57.* T he value of lim ( 1 2 62. The values of x for which the function [GATE-201 4 (CS-Set 1 )) 2) (b) (c) 1 [GATE-201 4; 1 Mark] equation : T he value of t is __ is f(x) = 55.* T he function f(x) = xsin x satisfies the following f " (x )+ f (x )+t cosx = 0 0-x X➔ oo (a) 0 (b) 1.0 (c) e 2 The value of lim (1+ x ) 209 [GATE-201 6-CE-Set-2; 1 Mark] lim ( n2 + n - n2 + 1) is -----66. n->oo ✓ ✓ 67. [GATE-201 6 (IN)] s_ in_ (x__4�) = _ -----X ->4 X -4 lim 68.* At x = 2 0, [GATE-20 1 6 CS-Set 1 ] the function f(x) = l sin tl ,( - oo < X < oo, L > 0) is (a) (b) (c) (d} Continuous and differentiable Not continous and not differentiable Not continuous but differentiable Continuous but not differentiable [GATE-201 6 (Pl)] IES MASTER Publication 210 69. 70. Limits , Continuity and Differentiability (a) (1 , 0) 5x e - 1 )2 . [-hm X-+0 X is equal to ------ - [GATE-2016 (P l)] Given the following statements about a function f: R -----, R, select the right option. 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 not be differentiable at x = x0 • R: If f(x) is differentiable at x = x0 , then it is continuous at x = x0 (a) P is true, Q is false R is fase (b) P is false, Q is true, R is true (d) P is true, Q is false, R is true [GATE-2016 (EE, EC-Set 1)] 71. The angle of intersection of the curve x2 = 4y and y2 = 4x at point (0, O) __is (c) 45 72.* A (b) 30° ° ° function (d) 90 [GATE-2016-CE-Set-2; 2 Marks] f(x) is d efined (c) (1, 1) (d) ( 74. The value of lim X ➔O (a) 0 as f(x) = e x< 1 f \ } , where x E R Which one of lln x +ax + bx x ;;,: 1 the following statement is TRUE? (a) f(x) is NO T differentiable at x = 1 for any values of a and b. (b) f(x) is differentiable at x = 1 for the unique value 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 [GATE-2017 EE Session-I] 73. T he tangent to the curve represented by y = x /n x is required to have 45° inclination with the x-axis. The coordinates of the tangent point would be ✓2 ✓2 , ) [GATE-2017 CE Session-II] x3 -sin ( x) is X (b) 3 (c) 1 (d) -1 [GATE-2017 ME Session-I] 75. The value of lim X ➔1 (a) is 0 x7 -2x5 +1 3 2 X - 3X +2 (b) is -1 (c) is 1 (c) P is false, Q is true, R is false (a) 0° (b) (0, 1) (d) Does not exist . tan(x) . 76. The value of IIm -- Is X➔O [GATE-2017] X [GATE-2017 (CH)] 1-X, X � 0 1 . and f(x) = { 2 x+1, x ;;,: 1 x , X> O Consid er the composition of f and g. i.e. , (f o g)(x) = f(g(x)). The number of discontinuities in (f o g)(x) present in the interval (--oo, 0) is: 77.* Let g(x) = (a) 0 (c) 2 { -X, X� (b) 1 (d) 4 [GATE-2017 EE Session-II] 78. Let f be a real-valued function of a real variable defined as f(x) = x2 for x � 0, and f(x) = - x2 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. [GATE 2018 (EE)] Engineering Mathematics 21 1 1. (See Sol. ) 21 . (b) 41 . 3. (d) 23. (b) 43. 2. 4. 5. 6. 7. 8. 9. 1 0. 11. 12. 22. (c) 26. (a) 27. (b) 28. (b) (a) 29. (a) 31 . 30. (a) 32. (0.75) (c) ·33. 1 5. (a) 35. 1 6. 17. 1 8. 1 9. 20. 34. (d) 36. (c) (a) 38. (d) 39. (d ) 40. (d) (a) 47. 49. 51 . 52. 54. (c) 55. 56. 57. 58. 59. 60. (d) 66. 67. 68. 69. 70. 71 . 72. (a) (-2) (-0.33) (d) (25) (b) (d) (b) (c) 78. (a) (1) 75. 77. (d) (0.5) (a) 76. (c) (c) 73. 74. (d) (1) (a) (d) (c) II SOLUTIONS Sol-2: ( c) Using L' Hospital Rule (e x -1) +xex - 2 sin x O . . ��---_ 11m Req. L1m1· t (-) X ➔O (1-cos x) +x(sin x) 0 Again using L' Hospital rule x 65. 64. (d) (c) (c) 53. (c) (c) (c) (a) (c) 37. 50. 46. 63. (c) (c) (c) 62. (d) 48. 45. 61 . (b) (c) (c) (b) (b) (a) (b) 44. (c) (c) (b) (b) (b) (b) (a) (c) (c) (b) (a) (b) 25. (a) 42. (a) 24. (d) 1 3. 1 4. Sol-1 : AN SWE R l{ E Y x x . (e +e +xe - 2cos x o ----- (-) = I1m�x ➔O sinx +sinx+xcos x 0 Again using L' Hospital Rule . ex +e x +ex +xex + 2 sin x = hm -------­ x ➔o cos x +cos x+ cosx-x sinx 1+1+1+0+0 =1 = 1+1+1-0 Using L' Hospital rule 5e x 1-e-j5 x . lim J_ x i -js =�=0.5 hm · = 10 X ➔0 1Q je -JX X ➔O 10(1-e -JX ) Sol-3: (d) sin 1 / y . sin x I1m ­ = lim ➔ Y O 1 /y x ➔oo , X y sin(�) =0 = ylim ➔O y . sin x I Im - = 0 X X➔OO ( ·: lsinxl � 1) IES MASTER Publication 212 Limits, Continuity and Differentiability Sol-4: (d) 1 - os x 1 lim (- --) = lim c � ) x➔oo s in x tan x sin x Sol-8: (b) x➔oo lim = X➔oo L'Im8➔0 Sol-6: (a) lim x➔oo s in e e 2 s in� cos � 2 2 lim tan� = ONE 2 = Sol-5: (a) 2 sin2 � 2 X➔oo 3 x - cos x x2 +(sin x )2 X➔oo lim X➔OO 1- Sol-10: (a) 1- x X➔oo Sol-7: (b) x 2 3 - cos x + (s in x ) 2 f(x) = I x + 1 1 = - (x + 1) for X < - 1 = fo r X � - 1 (x + 1) Only concern is x = -1 Left limit = lim - (x + 1) = 0 lim (x + 1) = 0 --f lf(x ) is co ntinuo us inthelnterval [-2,0JI Right limit = X➔ LO = - 1 Right Derivative = + 1 !Hence f(x) is not differentiableat x=-1I Alternative Solution -1 It is clear from graph f(x) is co ntinuous every where but no t differentiate at x =. -1 becaus e there is sharp change in sl o pe at x = -1. = lim (x ) X➔O = Right limit x x <0 >0 0 Left derivative = -1 Right derivative = +1 Left derivative Right derivative 1 (s in x )2 -+ - 3 X x = 00 = dy -1, ={ 1, dx IY is continuo us at x = OI * 00 x x, for x < 0 {and x, for x > 0 Left limit 3 1 1- 0 = -- = 1 0 - +0 Y = lxl Right limit cos x x [-1 , 1] X➔O 3 2 (s in x )2 x + -3 3 x X left limit = lim (-x) = 0 cos x x = 0 = 0 If f (x) is co ntinuo us at a point x = a then it may o r may not be differentiable at x = a. = lim = X Sol-9: (a) - = 1 = lim . 1 I'Im x s in- = 0 · s inoo X➔O Sol-11: (a) dy is dis continuous at x =0 dx Apply LH Rule . s inm e o L im- - (-) 8➔0 e o mcos m e = m 8➔0 1 Alternative Solution: Lim s inme s inm e . L .im m- - = m(1) = m 8➔0 e = Lim 0➔0 me Sol-12: (0.75) Sum o f the squares o f natural numbers = S2 S2 = 1 2 + 22 + 32 + 42 + ..... + n2 n(n+1)( 2n+1) 6 Sum of natural numbers = S1 S 1 = 1 + 2 + 3 + 4 + ..... + n Hence, ⇒ s1 = n(n+1) 2 ⇒ ⇒ ⇒ Engineering Mathematics 1 n2 (n+ )2 4 lim n➔cxr n x n(n+ )(2n+ 1) 6 1 1 (x - a) Apply L'H Rule = Lim 1 X➔a (x- a>2 1 1 - +-1 .1 3 n Im n➔oo 2 [ 2+� ] ⇒ 3 ⇒ 4 = 0.75 Sol-13: (c) Sol-16: (c) X = ✓ cos0 y = sin e Slope of cam profile at an angle e is dy dy - x - = dx d0 m = -1 (�� ) = - 1 - (�; ) 3 ✓ tan(¾) 3 Sol-14: (d) 60° n Lim = Lim n➔oo n 2 +n n➔oo ✓ = Lim n➔oo Sol-15: (a) =�I 1 R 1 1 R 1 +­ n (x- 1)(x + 1) (x- ) 1 1 1 Lt (x + 1) = + X➔1 Sol-17: (a) Lim f(x) = um [ 1 - � ] X➔oo x➔oo x4 a = 1-0 = 1 1- 4 Sol-18: (d) = (oo) X, X > 0 f(x) = IXI = {-X, X < 0 1 Left derivative = Right derivative =+ 1 LO * RD The derivative does not exist at x = 0. Alternative Solution : n n +n Lt ➔1 X 1 x2 - 1 Lt ( ) = 2 X➔1 X- 1 3 ✓ tan e m = ✓ Angle of Normal with x -axis is 3 a = tan-1 (m) = tan-1 ( ✓ ) la 1 1 y = e0 = = -cot e cos e 3 3 = ✓ - ✓ sin0 Slope of Normal to cam profile at angle e is = log y = Lim- (x- a) X➔a log y = 0 x2 Lt ( ) = X➔1 X- 3 m = 1 log(x- a) ( ) x- a = L' x�� � (n+ ) lim ➔ oo n 2 (2n+ ) 1 213 =1 It is clear from graph f(x) is not differentiable at x = 0. Sol-19: (d) y = Lim(x - a)<x-a ) X➔a Taki n g logarithm on both sides. Let log y = Lim log (x - a)<x-a) X ➔a = Lim (x-a) log(x - a) X➔a X ➔� 1 = 2 IES MASTER Publication 214 Limits, Continu ity and Differentiabil ity Function f(x) = m aximum (x, -x) is Sol-20: (d) so & x, X > 0 f(x) = lx l = + { -x, X < 0 LO (Left derivative) = -1 RD (Right derivative) = +1 RD LO The function is not differentiable at x = 0 * Sol-2 1: (b) y = Lim n 1In n➔oo Taking logarithm on both sides. Let log n 1 I og y = L .1 m - 1 ogn = L"1m-n➔oo n n➔oo n 11n =0 Apply L'H Rule = Lim n➔co 1 0 y = e = 1 Sol-22: (a) Given, Sol-23: (b) . . 2 " Sin2 X Sin X I 1 m -- = I "1m x X ➔O X ➔O X X2 sinx = ( lim x) lim ( J X ➔O X ➔O X = 0 X 12 = 0 x3 + x2 f(x) = lim 2 3 x ➔o 2x - 7x x (x + 1) = xlim ➔ o x2 (2x - 7) 2 = Sol-24: (c) x+1 (- -) ➔O 2X - 7 r1m 1 0+1 = -- = -7 0-7 X Solving the given system of equation for x we get 0. 02 X = b - 1 .98 Differentiating both sides 0. 02dx = db dx = 50 db ⇒ db = 1 (given) dx = 50 units. ⇒ Sol-25: (c) dy = a (0 + sin s ) dS ( �� ) = = sin s 1 + cos S 2 s 2 = tan 2 2 2cos � 2 S . S 2 Sln - . COS - f '(x) = = 1 2 3 (x - 8 ) 11 3 1 2 3 (-8)1 /3 - x -- Slope of normal = 3 1 = Slope of tangent 3 . - . Equation of normal at (0, 5) is y - 5 = +3 (x - 0) y = 3x + 5 Sol-27: (b) 2x2 - 7x + 3 0 . . I 1m f(x ) = 1 1m -2-- - (-) x ➔ 3 5x - 12x - 9 0 x➔3 4x - 7 lim (L. H. rule) = X➔ 3 1 0x - 1 2 = Sol-28: (b) At x = 2 5 18 y = x + x + x + x + . ...oo ✓ ✓ ✓ y = 2 + 2 + 2 + ...oo ✓ ✓ y = 2 + Jy (y - 2)2 = y y - 4y + 4 = y y2 - 5y + 4 = 0 (y-1 ) (y-4) = 0 2 COS S ) COS S ) f'(x) = At (0, 5) X = a ( S + si n S ) dx = a (1 + dS y = a (1 - f(x) = (x - 8)213 + 1 Sol-26: (b) (Squaring on bath sides) y = 1 or y = 4 = But y 1 does not satisfy equation ( 1 ) y(2) = 4 Sol-29: (b) Let . . . (1) e x - (1 + x + �) 2 p = lim X ➔O X3 Engineering Mathematics 3 = P = Sol-30: (a) p = 7 s x x x ff + 1§ + 11 +.. · .. · [ ·.- = x3 (;, + o + ... ) = 1 6 o➔o e � cos0/2 1 2 - = Apply L'H Rule = Lo➔imo -2 Sol-31: (c) 3 f(x) = l xl = -:3 l f(x) is continuous at x =0 ( ·: LHS = RHS = f(O) = 0) X if X if X>O if X =O X >0 <0 =0 X<O Again ·: if if if >0 <0 X =0 X f "'(x) = { 6 if X >0 -6 if X =0 Does not if x = 0 LHD =-6 & RHO =6 ie LHD So f(x) is not thrice differentiable. Sol-32: (c) . cos x- sinx O I Im ---- ( - form) X➔n/ 4 X-7t / 4 0 * RHO 3 4 1 12 . sinx · cos x [·: I1m -- = 1 1m - -- = X➔oo X X➔oo X Sol-35: (b) Sol-36: (c) It is standard result. ·: Sol-37: (c) Sol-38: (c) o] y = x2 + 2x + 10 �� = 2x+2 ⇒ [ X LHD =0 = RHO so f(x) is twice differentiable Again 3 8 sinx 1_ . x - sin x x = 1 - O =1 I Im --­ = lim X ➔oo X+COSX COSX 1+0 X ➔oo 1+ -- LHD = 0 = RHO so f(x) is once differentiable 6x I f "(x) = l � x ✓2 --12 1 1 1x1 = = -x2/3 = - Sol-34: (a) if if if Now, ✓2 1 -1 - �-0 3 X = 1- 0 (%) x3 (using L-Hospital Rule) . ( x1/3 - 2) llm ----,---­ x ➔B (x-8) Applying L- Hospital rule , sin8/2 sin 0/2 1 L im = = LO ➔imO O ➔O 0 2 2.0/2 Alternative Solution: sin 0/2 L im - - sinx- cos x lim ---­ x ➔ rr/4 _ 1 __ 1 == _ Sol-33: (b) 6 21 5 dy ] =4 dx x=1 sinx 1 I.1 m -- x ­ X ex x ➔O to. 2) I �' It is clear from graph, at x = 2/3 function is continuous but not differentiable. Sol-39: (a) . [ 2x 3 lim X ➔O X Sin - ] 2 . [2x 3 = lim 3 X ➔O 2 (3 -Sin - ] x) = 3..(1) = 3. 3 3 IES MASTER Publication 21 6 Limits , Contin u ity and D ifferentiabil ity f(x) = sin l x l Sol-40: (c) df = cos l x l - � dx x = { cos(- x ) • (-1) , x < O COS (X) · (1) , X > 0 - cosx , x < 0 = { COS X , X > 0 -cos We know that Sol-41 : (b) lim (1+� ) n➔oo n 1 lim (1+-) n➔oo n Sol-42: (d) Sol-43: (c) Given, n 2n (4-7t) -1 =✓ 2 lim f( x ) = X➔½ I 7t A. COSX - ' I' f X * 7t 2 --x 2 1 X =� 2 = ;\, x sin (i) = ;\, For the function to be continuous, Sol-44: (c) limf( x ) = f ( �) 2 X➔ ½ f(x) = lx l lim f ( X) = 0 ➔O- lim f ( x ) = lim+ f ( x ) X ➔O x➔O= f(O ) = 0 at X ➔ Q+ . A. COSX 1 1� - ; {·: LHL = RHL} X➔ 1 J2 ( 2: _ X ) A, = 1 lim f ( x ) = 0 ➔O+ oX For differentiability at X ➔ Q- , • (-sin x) = " 1 1m -x➔½ (-1 ) {By L' Hospital rule} ⇒ at X ➔ o+ :. The function is continuous. . sine cos9 cos O I Im -- = = = 1 8➔0 9 1 f(x) = For continuous at X ➔ X = e8 - x �O f(x) = {:x X < O ⇒ d - (-x ) = -1 dx df = �(x) = 1 l d x x�o+ dx · : The g iven function is not differentiable. Alternative Solution: -1 1 It is clear from graph that f(x ) is continuous at x = 0 but not differentiate at x = 0 because there sharp change in slope at x = 0. Sol-45: (b) x 1- (1- : 1- cos x . [ L lim hm X➔O ( X 2 ) = X ➔O X2 = +O = 1 2 l.f 1 Alternative Solution (1) lim X➔O 1- cos x x Apply L'H Rule 2 = = o Q sin x = _!_ lim 2 ➔ O 2X X +X:J] L Engineering Mathematics Sol-51: (c) Alternative Solution (2) 2 sin2 � . 1- cosx 2 I, m -= xlim -➔o X➔O x2 X2 = lim 2 ( X➔O Sol-46: (b) 2 sin x/2 ) 1 = 2 X/2 2 y = 5x2 + 10x, x E dy = 10x + 10 dx (1, 2) ⇒ :. :� is exactly 25 at atleast one point in the open interval (1, 2). Sol-47: (a) If LHL = RHL = f(a), i.e., Lt f(x) = Lt + f(x) = f(a), X➔a x➔a- then f(x) is continuous at x = a Let us take option (a) LHL = RHL = Lt f (x) = Lt X➔3 X➔T x+3 =2 3 Lt f(x)= Lt (x-1) = 2 X➔3 - 1 (0) e2x Lt -- X ➔O sin(4x) 0 Sol-53: (a) 2x1 2e2x = X➔O 4 COS 4 X 4x 1 (Using L'Hospital rule} = Lt Sol-54: (c) x-sin x 1-�osx Q Lt = Lt ( ) x ➔o 1- cos x x ➔ o smx 0 sin x X➔O COS X x- sinx =0 Lt x ➔o 1-cos x _!_ . x f'(x) = x cos x + sinx So, f"(x) = -x sin x + cos x + cos x f"(x)+ f (x)+t cos x = 0 ⇒ -x sin x + 2 cos x + x sinx + t cosx = 0 t COS X = - 2 COS X t = -2 Sol-56: (c) 2 2 x /2 . 2sin 1- cos x 2 --hm lim = x ➔o 2x4 X➔O 2X4 2 Sol-49: (d) I f a function is continuous at a point 'a' then limit at the point should exist and should be equal to f(a). Sol-50: (a) Lt xa ogx = 1 og x X➔ O 1 l = I1 m -�- Lim (1+ ) = e x➔co X Sol-55 : (-2) f(x) = x sin x e2x -1 1 Lt -- = ­ x ➔O sin4x 2 = sinx ) x+sinx ) lim ( 1 + = X➔oo = 1 X X . xa - 1 0 I1m - x ➔O a 0 Apply L' Hospitals rule X➔3 + and f(3) = 2 so f(x) is continuous at x = 3 :. Option (a) is correct Sol-48: (b) · E quation of tangent at x = 0, y = 3 is y - 3 = 0(x - 0) y = 3 ⇒ A line parallel to x axis. Sol-52: (c) X➔oo 10 < 10x < 20 20 < 10x + 10 < 30 y = 5x2 + 3 dy = 10x dx lim ( 1 < X< 2 Given that, 217 0 = 1 sinx 2 /2 _!_ lim ( ) = X➔ O 2 x 2 /2 4 Alternative Solution: 1-cos x2 Q ( ) X➔O 0 2X4 Apply L'Hospital Rule, lim . = lim X➔0 smx 2 • 2x 8X 3 . 1 smx2 = = lim x ➔o 4 x 2 4 IES MASTER Publication Limits, Continuity and Differentiability 218 Sol-61: (c) Sol-57: (-0.33) -sin x ) I' ( 2sinx+x cos x x'.� It is 0/0 form. So apply L' hospital rule. -cos x I' ( J x�O 2 cos x +[-x sin x +cos x] Let Taking logarithms K = lim (1+x2 f X➔oo Sol-58: (d) -1 3 = Xlim ➔oo J�\(1 +�rr Sol-59: (a) = e2 = x2 - xy - ✓x ✓Y Hence limit emits. 2 . x2 -mx -=_____,= -- Xllm ➔0 x _ ✓rr1 X ✓ = I Im �-� . x2 (1-m) X ➔0 (1--.fm ✓x . x312 (1-m) - -----,� = Xhm ➔0 (1--.fm = 0 (free from m) Sol-60: (c) K = lim x x X ➔oo Taking logarithms on both sides Let log K = lim log[ x11x ] X ➔oo 00 log K = lim F-logx} ( form) X ➔oo X 00 log K = lim (�) X ➔oo 1 log K = 0 K = eO = 1 (applying L.H. Rule) 2 ( ( oo 2x ] - form x➔oo (1+ x2 )ex 00 2 lim[ ] 2 X ➔oo (1 + X )eX +2XeX = 0 00 K = e0 = 1 x = -4 and x = 1 Sol-63: (c) log0 (1+4x) O . form) I Im -�-(x➔O 0 e3x -1 Applying L' Hospital's rule _1_ · (4) 4 lim 1+4x = x ➔0 e3x - 3-0 3 Sol-64: (c) ✓ lim { x2 +x-1-x) X ➔ oo l[ (x2 +x-1)-x2 . ) = llm [ x ➔ 00 ✓ [ = Xlim ➔ oo X 2 +X-1+ X 1-_! x 1 +1 1+- -X x2 ffi 1 1 = -= 1+1 2 ) (By L Hospital's Rule) So, Sol-62: (c) For discontinuity x2 + 3x - 4 = O (x + 4)(x - 1 ) = 0 ⇒ 00 form) 00 (By L Hospital's Rule) . = I Im[ Along the path y = mx where different values of m gives different paths lim (x,y}➔(0,0 } log(1+x2 ) eX = -0.33 2x 1 lim I1+-I x➔oo X (cxi° Form) log K = lim e-x - log(1 + x2 ) (0 - oo form) x➔oo (-cosO ) -cos x . I Im [- ---- ] = (3 cosO°-O) x ➔0 3 cosx-xsin x = x ⇒ oo l X➔ -1 ➔ O X Engineering Mathematics Sol-65: (d) A limit should be same for every function. But here for :. y = x , limit = J and for y = 2x limit = ¾ exist Sol-66: (0.5) Doesn't ✓ ✓ ✓ ✓ n2 +n + n2 +1 I.Im \Jn� �1 + n \Jn+ 1 x --===-----=== ] x ➔oo [ n2 +n + n2 + 1 (After Ralionatization) n- 1 . n2 +n- n2 - 1 llm --===-----=== = lim 2 2 2 X ➔oo [ X ➔oo n +n + n + 1 n +n + n2 ✓ ✓ = lim X➔«> Sol-67: ( 1) • 1 Im X➔4 ✓ ✓ ( 1 - 1 / n) We have = 25 Sol-70: (b) Continuity is the necessary condition for differentiability but it is not a sufficient condition. Sol-71: (d) In x2 =4y, = 1] 1 1+- + 1+n R n2 1 2 cos(x- 4 ) sin(x -4) form ) = lim [ ] ( % X➔4 x- 4 1 Sol-68: (d) 5x 2 = lim · [� =(5)2 X ➔O 1 ] 21 9 = cos(O)= 1 t )\ 2 f(x) = \sin ( dy dx = � = 0 at (0 0) 2 y2 = 4x ⇒ dy dx = ⇒ 0=0 � = oo at (0, 0) 2v 4x :. L between them = 90° Sol-72: (b) f(x) = ex ' ⇒ 0= � 2 X < 1 {ln x +ax 2 +bx, x � 1 f(x)-f( 1) Left hand derivative = lim X➔1 X- 1 where f( 1) = I n ( 1)+ a( 1)2+ b ( 1) =a+b so, Left hand derivatives (LHD) ex - (a+b) = lim X ➔1 X- 1 this limit exists if e = a + b ...( 1) because with this condition expression comes in at At x = 0 , LHL =RHL =f(O) =0 so f(x) is continuous = 0 X r· 2 - COS - X<O ' L L f'(x) = 2n 2nx +- cos( - ) X > O L L Again ( "'] LHD = f'(0- ) = & * ·: LHD Sol-69: (25) 2n L 2n L RHO so f(x) is NOT differentiable at x = 0 RHO = f'(O+ )= + 2 e5 x - 1 . -= [ llm -] x ➔o x form and we can use L'Hospital Rule i.e. ex LHD = lim - = e = a + b X ➔1 1 Right hand derivative (RHO) = lim X➔1 0 0 f(x) - f( 1) X- 1 2 . I n x +ax + bx - ( a + b) = llm - - ---x ➔1 X- 1 = 0 0 Hence by using L'Hospital Rule !+2ax +b RHO = lim x = 1 + 2a+b X ➔1 1 LHD will be equal to RHO if a = - 1 and b = e+ 1 Hence , f(x) is differentiable at x = 1 for unique values of a and b. IES MASTER Publication Limits, Continuity and Differentiability 220 Sol-73: (a) Target is having inclination of 45 with x-ax is = dy = tan 45° dx ⇒ d(xln x ) =1 dx ⇒ ⇒ 1 - (-x) , 1 - ( x + 1), ° (- x )2 , (X ⇒ X + 1) , X X = -1 . 7x6 - 1 0x4 hm ---­ x ➔ 1 3x 2 -6 x = 1 Sol-77: (a) (Using L-Hospital's rule) 2 . tan x I 1m - ­ = lim sin x = 1 x ➔o x X➔O ( 1 J g(x) = f(x) = -x { x + �. 1 -:- x , x 2 X :::; X :::,: x 1 1 :::; 0 X :::,: 0 ' - g(x), g( x ) :::; 0 f o g( x ) = f(g(x)) = 2 r(g(x)) , g(x) :::,: 0 { 1 <0 x < 0 0 :::; X :::; 1 Sol-74: (d) [Using L'Hospital Rule) :::,: X> 1 = 1 . x 3 - sin x . 3x 2 - cos x Llm --­ = I 1 m -- x➔o x ➔o x 1 + 1> 0, X 0 :::; X :::; 1 .-. At x = 1 , y = 1 x ln' 1 = 0 so tangent point =( 1 , 0). Sol-76: (1 ) - x > 0, x :::; 1 2 x l n x +� = 1 Sol-75: (c) - x :::; o , x :::; 1 x + 1 :::; 0, x :::,: 1 >1 X ⇒ f o g( x ) is continuous at every point in (-oo , 0). Number of points of discontinuities =0. Overall we have two points of discontinuities at x =0 and x = 1 but these do not belong to the given interval (-co , 0). Sol-78: (d ) f(x) = -x2 , { X2 , x X <0 :2:: 0 LHL = RHL = 0 = f(0) So f( x) is continuous at x = 0 and f'(x) = -2x , { 2X, x X LHS = RHD =0 So , f( x) is diff at x = 0 Now, i.e., -2, f"( x ) = { 2, <O >0 <0 X> 0 X LHD = -2 and RHD = 2 LHD f'( x ) is not diff. at * x RHD so =0 2., ROLLE'S THEOREM Let f(x) be a function such that y (i) f(x) is continuous in the closed interval (a, b) i.e. a :,; x :,; b f(a) = f(b) . . . . (ii) f'(x) exists in the open interval [a, b] i .e . a < x < b (iii) f(a) = f(b) then there exists at least one point x0 between a and b at which f '(x0 ) = 0 i.e., tangent of f(x) at x = x0 becomes horizontal. (""\ :B :A -+- 0 -a-Xo--b-....X Corollary-1 : Converse of Rolle's theorem is not necessarily true. Corollary 2 : When f(x) is a polynomial and all the conditions of Rolle's theorem are fulfilled subject to the condition f(a) = f(b) = 0 then another useful result can be deducted that between any two roots of an equation f(x) = 0 there is at least one root of f'(x) = 0 . Example 1 = Solution : Verify Rolle's theorem for the function f(x) = 2x3 + x2 - 4x - 2. f(x) = 2x3 + x2 - 4x - 2 is a continuous function because it is a polynomial function of x. Also, f(x) is differentiable for all real values of x. Thus, the first two conditions of Rolle's theorem are satisfied in any interval. Here f(x) = 2 0 gives 2x3 + x2 - 4x - 2 = 0 or (x3 Thus, f(✓ ) =f(- ✓2 ) = f(-1/ 2) =0 2 - 2)( 2x + 1) = 2 O i.e. x = ± ✓ ,-1/ 2 2 If we take interval (-✓ , ✓ ), then all the three conditions of Rolle's theorem are satisfied in this interval. 2 2 Consequently, there exist at least one value of x in the open interval (-✓ , ✓ ) for which f '(x) =0. Now, f '(x) =O i.e. ⇒ 6x2 + 2x-4 = 0 f '(-1) = f'( 2 / 3) = 0 ⇒ (3x - 2)(x + 1 ) = 0 or x 2 = -1 , 2 /3 2 Since both the points x = -1 and x = 2/3 lie in the open interval (-✓ , ✓ ) . Rolle's theorem is verified . Example 2 = Verify conditions of Roll e's theorem for the function f(x) = x2 in [-1 , 1 ]. Solution = x2 is continuous and differentiable in the interval [-1 , 1] Also, f(1) = 1 and f(-1) = 1 = f(1) Thus, all the conditions of the Rolle's theorem are satisfied. The function f(x) = f '( x) = 0 for at least one value of x E (-1,1) i.e. 2x = 0 or x = 0 which obviously lies in (-1 , 1 ). Hence, Rolle's theorem is satisfied. 222 Mean Value Theorems LAGRANGE'S FIRST MEAN VALUE THEOREM Let f(x) b e a function where (i) (ii) f(x) is continuous in the closed i nterval a :s; x :s; b. f'(x) exists in the open interval a < x < b f(b) then there exists at least one point x0 between a and b such that f'(x0 ) f(b) - f(a) = b-a i.e., tangent at x0 , becomes parallel to the chord �--·11: ·····: ····:,·: B : .·· : :..·· :: --�A f(a) · · · · · · · · . : .. ' ' ' ' a x0 joining (a, f(a)) and (b, f(b)) :' ' ' b Example 3 : Verify Lagrange's form of mean value theorem in the interval [0, 4) for function Solution : f(x) = (x - 1)(x - 2) (x - 3). We have f(x) = (x - 1)(x - 2)(x - 3) = x3 f(a) = f(0) = -6 and f(b) = f(4) = 6 - f(b) - f(a) 6 - (-6) _g = = =3 b-a 4-0 4 6x2 + 1 1x - 6 Also, f'(x) = 3x2 - 1 2x+1 1 gives f'(c) = 3c 2 - 1 2c + 1 1 Now, from Lagrange's mean value theorem we have f(b ) - f(a) = f'(c ), (a < c < b), we get b-a 2 1 2 ± 144 - 95 = 2 ± .Jj .J 3 6 As both these values lie inthe interval (0, 4), hence both of these values are the required values of c. 3 = 3c2 - 12c + 11 or 3c2 - 1 2c + 8 = 0 ⇒ c = CAUCHY'S MEAN VALUE THEOREM Let two functions f(x) and $(x) satisfy the condition of the mean value theorem in (a, b) i .e. f(x) and $(x) be continuous in the closed interval [a, b] and differentiable in the open interval (a, b) and also suppose $'(x) 0 for any point in (a, b) then there exists at least one point x0 such that a < x0 < b for which Remark-1 : Remark-2 : Proof : Remaks-3 : f(b) - f(a) f'(x0 ) = $(b) - $(a) $'(x 0 ) * Cauchy's Mean Val ue theorem reduces to the first mean value theorem if $(x) = x . Cauchy's formula cannot be deduced from Lagrange's form of the mean value theorem. By Lagrange's formula for the mean value theorem , we get f(b) - f(a) = (b - a)f'(X 1 ), a < X 1 < b By division, we get and $(b) - $(a) = (b - a)$'(X 2 ), a < X2 < b f(b) - f(a) f'(X 1 ) = $(b) - $(a) $'(X 2 ) . . .(v) where the numbers x1 and x2 wil l not, necessarily be same and they are different from the num ber x0 of Cauchy's formula. f'(x) and $'(x) should not vanish for the same value of x. This im portant fact is illustrated by the following example. Example , : Solution : Engineering Mathematics 223 Let f(x) = x4 and �(x) = x5 and the domain of interval is (-1 , 1 ). Here the conditions of the mean value theorem are satisfied. Then, by Cauchy's form ula we m ust have But f(1) - f(-1) f'(x) �-� = - where - 1 < x < 1 �(1) - �(-1) �'(x) f(1 ) - f(- 1) = 1 4 (...:.1)4 = 0 and �(1) - �(-1) = 15 - (-1)5 = 2 f' (x) =0 �• (x) Now at x = 0, f'(x) = 0, and g'(x) = 0 ⇒ LHS of (ii) takes Hence, Cauchy's form ula of mean value theorem is not valid. - So, the L.H.S. of (i) is zero . . . (i) 7 ... (ii) form which is meaningless. BOLZANO THEOREM If f(x) is continuous and differentiable function in a given domain and let x = a & x = b are any two points in the domain of f(x) such that f(a) and f(b) have opposite sign then there exist at least one root 'C' of f(x) between a & b i.e there exist C E (a,b) such that f(c) = 0. INTERMEDIATE VALUE THEOREM If f(x) is continuous and differntiable function in [a,b] such that f' (a) * f' (b) then there exist at least one point C such that f' (c) = K where k is any number between f'(a) & f'(b) i.e K E DARBOUX'S THEOREM [f' (a) , f' (b)] . E (a,b) If f is finitely differentiable in a closed interval [a, b] and f'(a), f'(b) are of opposite signs, then there exists at least one E [a, b], such that f'(c) = 0 . · point c Example 5 : Solution : Example 6 : Solution : Verify whether Rolle's theorem is applicable to the function f(x) or not. At x ⇒ ⇒ = 1 = 2 + (x - 1 )213 in the interval [0, 2] 2 + (1 - h - 1)2/3 - 2 f(1 - h) - f(1) 1_ = -oo = lim = lim _ _113 h➔O h➔O h➔O -h -h h f(x) is not differentiable in (0, 2) Lf'(1) = lim Rolle's theorem is not applicable for f(x) in [0, 2]. Find 'c' of Lagrange's mean value theorem when f(x) = x3 - 3x - 1 in the interval [ - Si nce f(x) is polynomial function of x, so it is continuous and differentiable in [ val ues of x. By Lagrange's M.V.T., we have f(¥)-f(-¥) = 3C ( 1:)- (-�1) 2 -3 ⇒ f(b ) - f(a) . = f'(c) b-a 1 1 1 3] , . 7 7 1 1 1 3] , for all real 7 7 C = ± 1 E (-� �) which verifies Lagrange's M.V.T. 7 ' 7 IES MASTER Publication 224 Mean Value Theorems Example 7 : Solution : Discuss the application of Rolle's theorem to the function f(x) = x213 in (-1, 1). At x = 0 ⇒ ⇒ Example 8 � Solution : 1 ( h)2'3 -O f(O-h)- f(O) =-oo = lim - -1= lim h➔O h 13 h➔O -h -h lim Lf'(O) = h➔O Lf'(0) does not exist. f(x) = x213 is not differentiable at x = 0. Hence Rolle's theorem is not applicable. Discuss the application of Rolle's theorem to the function f(x) = x2 in (-1, 1). If applicable then verify it for f(x) in (-1, 1]. At x = 0 2 . (O-h) - 0 Lf'(0) = I1 m ---- = 11·m h = 0 h➔O h➔O -h Rf'(0) = lim h➔O (O+h)2 -O = lim h= 0 h➔O h ⇒ f'(x) exists in (-1, 1) i.e., f(x) is differentiable in (-1, 1), hence it will be continuous in (-1, 1]. ⇒ Rolle's theorem is applicable. Also, f(-1) = (-1)2 = 1 = f(1). Then there exists a point c s.t. ⇒ f'( c) = 0 2C = 0 ⇒ C = 0 E (-1, 1), which verifies Rolle's theorem. Example 9 1 Verify Lagrange's mean value theorem for the function: Solution f(x) = (x + 1)(x - 2)(x - 3) Vx E [0, 1]. We can see that f(x) = (x+ 1)(x - 2)(x - 3) is continuous as well as differentiable on R ( ·: it is a polynomial in x). Then by Lagrange's M.V.T., we have ⇒ ⇒ f(1)-f(0) = f'(c) 1- 0 ⇒ f'(c) =-2 2 3c - 8c+3= 0 1 Now, c = ⇒ 2(-1)(-2)- (6) = f'(c) 1 3c 2 -8c+1=-2 ⇒ c= 4 ? ±✓ 3 4✓ E (0, 1), which verfies Lagrange's M.V.T. 3 Example 10 : Verify Cauchy's mean value theorem -for the functions x2 and x3 in (1, 2]. Solution : Let f(x) = x2 , g(x) = x3 ; then both functions are continuous and differentiable in (1, 2] and (1, 2) respectively. So, by Cauchy's mean value theorem, we have ⇒ 4 - 1 2x _ = 8 1 3x 2 ⇒ f(2)-f(1) f'(c) = g(2) - g(1) g'(c) 9x2 = 14x ⇒ x(9x-14) =0 ⇒ x =0, 9. 14 1 c = : E (1, 2). Hence Cauchy's mean value theorem is verified. Engineering Mathematics 225 - PREVIOUS YEARS GATE & ESE QUESTIONS 1 ." Questions marked with aesterisk (*) are Conceptual Questions The value of I; in the mean value theorem of f(b) - f(a) = (b-a)f' (I;) for f(x) = Ax2 + Bx + C in (a, b) is (a) b + a (c) 2. " (b+a) 2 (b) b - a (b-a) (d) 2 [GATE-1 994; 1 Mark] , then the lower and If f(0) = 2.0 and f'(x) = 5-X2 upper bounds of f(1) estimated by the mean value theorem are (a) 1 .9, 2.2 3. (c) 2.25 , 2.5 (b) 2.2, 2.25 (d) none of the above [GATE-1 995] A rail engine accelerates from its stationary position for 8s and travels a distance of 280m. According to the mean value theorem, the speedo­ meter at a certain time during acceleration must read exactly (a) 0 (c) 75 km/ h (b) 8 km/ h (d) 126 km /h [GATE-2005; 2 Marks] 5 " 4. T he polynomial p(x) = x + x + 2 has 7." Roots of the algebraic equation x3 + x2 + x + 1 = 0 are (a) (+1 , +j , -1) (c) (0, 0, 0) (c) 1 real and 4 complex roots (d) all complex roots [GATE-2007 (IN)] 5." Given that one root of the equation x3 - 10x2 + 31x - 30 = 0 is 5 then other roots are (a) 2 and 3 6. " (c) 3 and 4 (b) 2 and 4 (d) - 2 and - 3 [GATE-2007 (CE)] It is known that two roots of the non-linear equation x3 - 6x2 + 11x - 6 = 0 are 1 and 3. T he third root will be (a) (c) 2 (b) - j (d) 4 [GATE-201 0 (Pl)] (d) (-1 , +j , -j ) [GATE-201 1 -EE; 1 Mark] 4 2 3 8. " A polynomial f(x) = a4 X + a3X +a2X +a1x-ao with all coefficients positive has (a) no real roots (b) no negative real root (c) odd number of real roots (d) at least one positive and one negative real root [GATE-2� Mark] 9.* A non-z ero polynomial f(x) of degree 3 has roots at x = 1, x = 2 and x = 3. Which one of the following must be TRUE? (a) f(0) f(4) < 0 (c) f(0) + f(4) > 0 1 0." Let the function f ( e) = (a) all real roots (b) 3 real and 2 complex roots (b) (+1 , -1 , +1) sin e (b) f(0) f(4) > 0 (d) f(0) + f(4) < 0 [GATE-1 4 (CS-Set 2)] cos 8 tan e sin (i) cos (i) tan (i) sin (�) cos (�) tan (�) where 8 E [i , � ] and f' (e ) denote the derivative of f with respect to e . Which of the following statements is / are TRUE? There exists 8 E [� . � ] such that f' (e) = 0 . I. II. There exists 8 E [�. �] such f' ( e ) (a) I only (c) Both I and II (b) I I only *0 (d) neither I nor I I [GATE-2014 (CS-Set 1 )] 1 1 ." A function f(x) is continuous in interval [0, 2) . It is known that f(0) = f(2) = -1 and f(1) = 1. Which one of the following statements must be true ? IES MASTER Publication 226 Mean Value Theorems (a) There exists a 'y' in the interval (0, 1) such that f(y) = f(y + 1) 7. (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). [GATE-2014 ( CS-Set 1 )] (a) 1 8. 12. A function f(x) = 1 -x + x3 is defined in the closed 2 interval [-1 , 1]. The value of x, in the open interval (-1 , 1) for which the mean value theorem is satifsied , is (b) -1/3 (a) -1/2 (d) 1/2 (c) 1/3 (c) f(a) . f(b) > 0 (b) f(a) . f(b) < 0 (d) f(a) I f(b) ::: 0 [GATE-201 5 ( EE-Set 1 )] 1 9. 20. (d) (x+1) [GATE-2016 SET-I; 2 Marks] in the range (c) f(a)(b-c,) 21 . 0 <x < 6 x ;;::- 60 30 0 -30 (c) 2 -60 22. 10 [GATE-201 7 ( CH)] (b) 4 (d) 13 [ESE 201 8 ( Common Paper)] (b) f(b)(c,- a) (d) 0 [GATE 201 8 (ME-MorningSession)] Consider p(s) = s3 + a2 s2 + a 1 s + a0 with all real coefficients. It is known that its derivative p'(s) has no real roots. T he number of real roots of p(s) is (a) 0 1 50 1 20 90 [ESE 201 8 (Common Paper)] According to the Mean value theorem, for a continuous function f(x) in the interval [a , b] , there exists a value c, in this interval such that (a) f(c,)(b- a) 1 6. The number of positive roots of the function f(x) shown bel ow is (d) -1.1 and 1.4 f f(x)dx = [GATE-201 6 ( EE-Set 2 ) & ( CH )] (b) (x-1) (b) -1.2 and 1.4 The equation, 8x2 + 37x - 50 = 0 is factored and it has (3 + 4i) as one of its roots. What is the real root of this equation? x3 - (c) 6.5 f(x) = x3 + 5x , in the interval [1, 4 ] at a value (rounded off to the second deciaml place) of x equal to _____ (a) (x3 + 8) (c) (2x-5) [GATE-20 1 7 PAPER-2 ( CS)] Given that 0.8 is one root of the equation, x3 0.6x2 - 1.84x + 1.344 = 0. The other roots of this equation will be (a) 2 1 4 . The Lagrange mean-value theorem is satisfied for 1 5. If f(x) =2x7 + 3x - 5 , which of the following is a factor of f(x)? (d) 3 (c) 1.2 and -1.4 1 3 .* If a continuous function f(x) does not have a root (a) f(a) . f(b) =0 (c) 2 (a) 1.1 and -1.4 [GATE-2015; 1 Mark] in the interval [a, b], then which one of the following statements is T R U E ? (b) 1 (b) 1 (d) 3 [GATE 201 8 ( EC)] T he three roots of the equation f(x) = 0 are x = {- 2 , 0 , 3}. What are the three value of x for which f(x - 3) = 0? (a) -5 , -3, 0 (c) 0, 6, 8 (b) -2 , 0, 3 (d) 1 , 3 , 6 [GATE 201 8 (EE)] Engineering Mathematics 1. 2. 3. 4. 5. 6. Sol-1 : (c) (c) 7. . (d) 9. (d) 8. (b) 1 0. (c) (a) 11. 12. (c) A N SW E R l{ E Y (d) (a) (c) (a & d) (b) f(x) = 2Ax + B 1 5. (b) 1 4. 16. 17. 1 8. ⇒ (c) 1 1 < f{ 1)- 2 < 4 5 ⇒ Increasing function & f"(x) = is also an increasing function. Hence we can take, 1 and Max. f'(c) = 5- 1 1 4 ⇒ 2.2 < f{ 1) < 2.25 18 = 35 x -kmph 5 = 126 kmph Given p(x) = x5 + x + 2. Now we will use Descartes rule of sign. C ase II :p(-x) = -x5 - x + 2 There is only one change of sign from - to + in p(-x) f'(c) = f(1)- 2 (5 (c) 280-0 ) m/s = 35 mis ( 8- 0 same. T hen p(x) = 0 does not have any positive root. :. p(x) = 0 has at most one negative root. 1 - > 0 . So f(x) is an 5- x 2 1 5 (d) C a se I : The signs of all the terms in p(x) are It is given that f(0) = 2 and we have to find f(1) so we can assume that f(x) is defined in [0, 1] By lagrange's means value theorem 1 5- 0 (b) 22. = Sol-4: Sol-2: (b) in [0, 1] , f'(x) = - 21 . (c) (Ab + Bb + C)- ( Aa + Ba + C) b-a f(1)- f(O) f'(c) = 1-0 (a) 20. (3) 2 2 2A I; · + B = A (b + a) + B min f'(c) = (2.645) Speedometer will read f ( b) - f(a) b-a 3c E (0,1) such that (a) 1 9. Sol-3: (d) f(I; ) = 2A I; + B 2A I; + B = (c) Hence, Min{f'( x)} < f'(c) < Max{f'(x)} f(x) = Ax2 + Bx + C = 1 3. 227 Again, the degree of p(x) = 0 is odd then the equation has at least one real root. �:2 2 > 0 so f'(x) ) Sol-5: Further, p(x) has real coefficients so complex roots can be in pair, so p(x) has 4 complex roots. Hence, we can conclude that option (c) is correct. (a) Metho d I : Let f(x) = x3 - 10x2 + 31x - 30 = 0 IES MASTER Publication 228 Mean Value Theorems f(2) = 23 - 1 0(2)2 + 31 (2) - 30 = 0 f(3) = 33 - 1 0(3)2 + 3 1 (3) - 30 = 0 and So, x = 2 & 3 are the other roots of given equation Method II: x3 - ⇒ 1 0x2 + 3 1x - 30 = 0 (x - 5)(x2 5x + 6) =0 - Sol-6: (c) x3 - x2 - - 6x2 + 1 1 x - 6 = 0 5x2 + 5x + 6x - 6 = 0 x2(x - 1 ) - 5x(x - 1) + 6(x - 1) = 0 (x - 1 )(x2 - 5x + 6) = 0 (x - 1 )(x - 3)(x - 2) = 0 X Sol-7: (d) Sol--8: (d) = f(x) has one change of sign So f(x) will have maximum one +ve real root f(-x) =·a x4 - a x3 + a x2 - a x - a ' Sol-9: 2 1 given function f(S) is not a constant function. Sol-11: (a) & (d) Let So f(x) has at most 3 -ve real root :. f(x) has atleast one positive and one negative real root. (a) Let us take poly as f(x) =k(x - 1 )(x - 2)(x - 3) where k constant. cas e ⇒ 7t sin3 7t cos - tan 2: 3 3 ( -_ 1) _ 3_ _ - f_ f (_ 1) _ _ 1 -2x + 3x2 = 1 1 (-1) - -2- = 1' I 3 1 X = -3 E (- 1, 1) X Sol-13: (c) If f(x) is continuous and does not have a root in [a, b] then the curve y = f(x) does not intersect co-ordinate axes. I\ f(x) tans 7t 7t sin - cos - tan 2: 6 6 6 f(x) = 1- x 2 +x3 . x E [- 1, 1) f'(x) = f(0). f(4) < 0 f( S) = g(0) = f(0) - f( 1) = - 1 - 1 = -ve g( 1) = f( 1 ) - f(2) = 1 + 1 = +ve g(0) and g( 1 ) have opposite signs so there must By L.M.V.T. Now f(0) =-6k (-ve) and f(4) = +6k (+ve) sin e g(y) = f(y) - f(y + 1 ) where y E (0, 1) then 0 f(-x) has 3 changes of sign Sol-10: (c) ⇒ tangent at, at least one point must be other than horizontal, which is obviously true as Sol-12: (b) a0 , a 1 , a2 , a3 , a4 > 0 3 *0 g(y) = 0 ⇒ f(y) = f(y + 1 ) (Here we have used Bolzano theorem) By similar logic, option (d) is also correct. - 1 , ± j. f(x) = a4x4 + a3x3 + a2x2 + a 1 x - a0 4 i) as it is composite trignometric exist y E [ 0, 1) such that (x + 1 )(x2 + 1) = 0 x Now, f'(S) = 1 , 2, 3 x3 + x2 + x + 1 = 0 ⇒ function i• value 8 E [i , i] such that f'(S) =0 i.e., (I} is true = 5, 3, 2 x3 differentiable in ( So, By Rolle's theorem, there exists at least one or (x - 5)[(x - 3)(x - 2)] = 0 .'. X function f(S} i s continuous i n [i , i] and is 0 a b Both f(a) & f(b) are +ve 0 a b f(x) Both f(a) & f(b) are -ve Engineering Mathematics Therefore the curve y = f(x) completely lie either above x-axis or below x -axis. Hence, f(a) and f(b) should have same sign. :. f(a).f(b) > 0 Sol-14: (2.645) f(x) = x3 + 5x, f'(x)= 3x2 , Method II : Let we have two curves y = f(x) = ex and y = g(x) = 0.5x2 y - 229 2 x E [1,4] By Lagrange's mean value theorem, there exists C E ( 1 , 4) such that f(4)- f( 1) 4- 1 69- 6 3c2 = 3 f'(C) = 3c2 = 21 c2 = 7 C = ± ✓7 ·: - ✓7 � [ 1,4] So Lagrange's mean value theorem satisfied only for C= ✓7 Sol-15: (b) ⇒ = 2.645 f(x) = 2x 7 + 3x - 5 f( 1) = 2 + 3 -5 =0 (x - 1 ) is a factor f(x). Sol-16: (3) T he graph of the function cuts x - axis thrice in 0 < X < 6. :. The no. of positive roots = 3. Sol-17: (c) f(x) = ex+ 0.5x2 - 2 f(-5) = 1 0.50, f(-4) = 6.0 1 , f(-2) = 0. 135, f(-1 ) = -1. 13, f(0) = - 1, f( 1) = 1 .2 1, f(2) = 7.38, f(3), f(4), f(5) also+ve. As there are 2 sign changes from+ve to -ve and -ve to +ve, two roots will be there in the range [-5, 5]. Sol-18: (c) Given equation in x3 - 0.6x2 - 1 .84x+ 1 .344 = 0 ·: Only option (c) is satisfying given equation so it is the right answer. Sol-19: (a) Given equation is x3 - 8x2+ 37x - 50 = 0 only option (a) i.e. x = 2 is satisfying above equation so it is the only real root. Sol-20: (a) · If � E (a, b) then mean value theorem for integrals is stated as follows: J: f(x)dx = (b - a)f(�) Sol-21: (b) By Rolle's theorem we know that. "Between any two real roots of f(x) there exist at least one real root of f'(x)" But in this question it is given that there exist no real root of P'(s) so P(s) can not have two real roots. Hence only possibility is that P(s) have one real root and two complex roots. [as coefficients of P(s) are real] Sol-22: (d) Root of f(x) = 0 are (-2, 0, 3) so f(x) = (x+ 2)(x)(x - 3) Now, f(x - 2) = (x - 1)(x - 3)(x - 6) Hence, f(x - 3) = 0 ⇒ x = 1 , 3, 6 IES MASTER Publication INCREASING-DECREASING FUNCTION Let f(x) be a real valued function defined in D then f(x) is Increasing : If v- x1 , x2 E D , whenever x1 < x2 ⇒ f(x1 ) .'.5. f(x) Strictly increasing : If v x1 , x2 E D, whenever x1 < x2 Decreasing : If v x1 , x2 E D, whenever x1 < x2 ⇒ f(x1 ) Strictly decreasing:lf v x1 , x2 E D, whenever x1 < x2 / Increasing ⇒ f(x1 ) < f(x2) ⇒ f(x1 ) or f'(x) 2 0 < f(x2) or f'(x) > 0 < f(x2) or f'(x) < 0 or f'(x) � 0 � � Decreasing S.I. S.D. e.g: f(x) = x2 is S.I. in (0,oo) , SD in (-oo,0) but neither increasing. Nor decreasing in (-oo, oo) . Condition for strictly increasing (51) and strictly decreasing (SD) function For SI functions, f' (x) > 0 v x E D , i.e. tangent, every point makes an acute angle e. with +ve x-axis For SD functions, f' (x) < 0 v x E D , i.e. tangent, every point makes an obtuse angle 0 . with + ve x-axis Monotonic function: f(x) is said to be monotonic in D if it is either SI or S.D. Note: (1) Monotonic functions are always one-one, onto hence always invertible i. e. Monotonicity + Continuity {::} Bizection (2) f is Monotonic then f' is also monotonic. (3) For solving question based on increasing-decreasing function, first write domain and then proceed. Example 1 Solution : Find the interval in which the function f(x) = sin4 x + cos4 x, x E [O,i] is increasing or decreasing. f'(x) = 4 sin3 x cos x + 4 cos3 x (-sin x) Put f'(x) = 0 For increasing function = 4 sinxcos x(sin2 x -cos2 x) = -2sin2x cos2x = -sin4x -sin 4x = 0, O � x � � 2 sin 4x = 0, 0 � 4x � 2n 4x = 0,n,2n f'(x) 2 0 ⇒ X = 0 n/4 sign of f'(x) o' � ' � 4 2 n/2 -sin 4x � 0 X E For decreasing function Example 2 : Solution : sin 4x s; 0 ⇒ sin 4x s; 0 [¾ · � ] f'(x) s; 0 -sin 4x s; 0 X E ⇒ [O, ¾] Engineering Mathematics Find the interval in which the function f(x) = sin x + cos x, o :,:; x :,:; 2n is increasing or decreasing. Put f'(x) = 0 f'(x) = cos x - sin x cos x - sin x = 0 Hence, f(x) increases in tan x = 1 X E or x [O,¾]u[ = 5n ' 44 0 1t 5n , 2n ] 4 & 57!/4 sign of f'(x) 27! E [ ¾· 54 ] n f(x) y 2. 71/4 f(x) decreases in, x MAXIMA AND MINIMA OF FUNCTION OF SINGLE VARIABLE 1. 231 ----x -0 --1x=_b______ -=-+--------------, x x=a -0 For maxima or minima, f'(x) = 0 or undefined i.e. tangent must be horizontal at that point or does not exist. The values of x at which maxima or minima occur are called critical points or stationary point or turning point. local maxima f(x) �� - - - - - - - - - - - - - - maxima f(d) f(b) absolute minima minima IES MASTER Publication 232 Maxima and Minima Critical points / Turning points Those values of x for which f'(x) = 0 or undefined are called critical points Extreme Points Those critical points where we can obtain extreme values. Extreme values / Extremas / Stationary values : Values of f(x) at extreme points are known as extreme values (i.e. maximum or minimum values) or extremum. Necessary condition for Extreme values f (c) is called extreme value of f(x) if f ' (c) = 0 or undefined. Sufficient Condition for Maxima - Minima 1st Derivative Test : To find out maxima/minima, check the symbol of f ' (x) according to the following diagram. a (Maxima) a (Minima) a a (N either max nor min.) (Above diagrams are also applicable for end points) 2nd derivative test : We can also find maxima/minima according to the following flowchart, f'(a) = 0 f"(a) > 0 (Minima) f"(a) < 0 (Maxima) f"(a) = O f'"(a) * 0 (Saddle point) Note : x = a is called point of inflexion if f" (a) = 0 and f0+ 1 (a) is being done. * 0 for n even, and n is number of times differentiation Local Maxima : f (x) < f(a) V x E (a-o, a +o), x * a then f(x) is said to have maxima at x = a & is given as f(a). Local Minima : f (x) > f(a) VxE (a - o, a+o), x *a then f (x) is said to have minima at x = a and is given as f(a). Concave upward curve : J = f(x) is called concave upon and if curve lies above the tangent at each point or we can say that lf"(x) > OI In either case f '( x) is increasing so f"(x) > 0 Engineering Mathematics Concave Downward Curve : If curve lies below the tangent at each point or we have lf" (x) < 233 OI I n either case f'(x) is decreasing so f"( x) < 0 Point of Inflexion / Saddle point : Point where neither maxima nor minima occurs is called point of inflexion. OR It is a point where curve is changing from concave upward to concave downward or vice-versa. e.g for y = x3 here x = 0 is a point of inflexion. x = 0 it is concave upward in LHS of x = 0 it is concave downward and in RHS of y = x' 0 INOTE-, ----"1 1 2 . I f f(x) i s constant then a t least one maxima or one minima must lie between two equal values of f(x). 3. 4. Example 3 Solution : Maxima/Minima occurs alternatively. 1 A n nth degree polynomial bends exactly (n - 1 ) times so can have maximum (n - 1 ) extremas. Corner points are always extreme points. Find all the points of local maxima and local minima of the function f(x) = x3 f'(x) = 3x2 - 3 For maxima or minima ⇒ = 3(x - 1 )(x + 1) f'(x) = 0 3(x+ 1 )(x - 1) = 0 or x Hence, x = 1 is point of Minima & Min. value = = ( = -1 is point of Maxima & max. value Example 4 a Find local maximum and local minimum values of the function f(x) = 3x4+ 4x3 - 12x2+ 12 f'(x) = 12x3+ 12x2 Solution : For Critical Points : - 3x+ 3. -1 sign of f'(x) 1, -1 are the critical points. f(1) = 1 & x - = ) f(-1) = 5. 24x f "(x) = 36x2 + 24x - 24 f'(x) = 0 12x(x2 + x - 2) = 0 2 12x(x + 2x - x - 2) = 0 ⇒ 12(x3 + x2 - 2x) = 0 IES MASTER Publication 234 Maxima and Minima at X = 0, +1, -2 f"(0) = -24 < 0 so x = 0 is Point of Maxim a = 0, f(0) = 12 & Max. value X f"(1) = 36 > 0 so x = 1 is Point of Minima at X = 1 f(1) = 7 & Min. value at f"(-2) = 72 > 0 so x = -2 is Point of Minima = -2, f(-2) = -20 & Min. Val ue X Example 5 : An open box with a square base is to be made out with a sheet of 'c' sq. units. Show that the max. Solution 1 volume of the box is · 6 ✓3 . c3 Length = x; breadth = x; height = h Given : Area of sheet x2 + 4hx = C2 i.e., Now, ...(1) x2 2 C Volume = (x) x (x) x (h) = x ( \� J C2 x x3 V = --- i.e., 4 dV = 0 dx (Max Vol.t=.£_ = 6 ✓3 ⇒ c3 . . . (2) 4 dV c2 3x3 = --dx 4 For max. or min. so using (2), A = C2 ✓3 C2 3x2 --- = 0 4 4 ⇒ X = C ✓3 MAXIMA - MINIMA OF FUNCTIONS OF TWO VARIABLES Let Z = f(x, y) be a given surface shown in figure; z A A = Absolute maxima, F = Absolute minima C & E = Local maxima, B & D Local minima Maxima : Let Z = f(x, y) be any surface and let P(a, b) be any point on it then f(x, y) is called maximum at P(a, b) if f(a, b) > f(a + h, b + k) \;/ +ve and -ve values of h and k. M inima : Let Z = f(x, y) be any surface and let P(a, b) be any point on it then f(x, y) is called minimum at P(a, b) if f(a, b) < f(a + h, b + k) v +ve and -ve val ues of h and k. Extremum : The m aximum or m inimum value of the function f(x, y) at any point x = a and y = b is called the extremum value and the point is called extremum poi nt. Engineering Mathematics 235 Saddle Point : It is a point where function is neither maxi mum nor minimum. At this point f is maximum in one direction while minimum in another direction. e.g. Consider Hyperbolic Paraboloid z maxima nor minima. So, origin is the saddle for Hyperbol ic Paraboloid. = xy; since at origin (0, 0) function has neither Condition tor the Existence of Maxima and Minima (Extremal If f(a + h, b + k) - f(a, b) remains of the same sign for all values (positive or negative) of h, k then f(a, b) is said to be extremum value of f(x, y) at (a, b). (i) (ii) If f(a + h, b + k) - f(a, b) < 0, then f(a, b) is maximum. If f(a + h, b + k) - f(a, b) > 0, then f(a, b) is minimum. Necessary Conditions tor Maximum or Minimum By Taylor's theorem f(a + h, b + k) ⇒ = af at f(a, b)+( h -+k -) ax f(a + h, b + k) - f(a, b) = oy at at ( h -+-) ax oy a2 f a f 1 2 a f +- [ h 2 ] +2h k --+k +... 2 2 2! axoy ax 8y (a , b) (a, b) 2 2 a 2f a f 1 2 a f + [ h2 +2hk -- + k -.] +... 2 1 2. ax oy ax oy 2 (a,b) (a , b) 2 2 . . . . (i) Neglecting higher order terms of h2 , hk, k2 etc. (Since h, k are small), then above expression reduces to f(a + h, b + k) - f(a, b) = (h af ax +k � ) oy (a,b) The sign of L.H.S. of (ii) is governed by h � + k af which may be +ve or -ve depending on h, k. ax ay Hence, the necessary condition for f(a, b) to be a maximum or minimum is that af af ) ( h -+k ax oy ... (ii) (a , b) =0 af = O � = O ⇒ . . . (iii) ' ay ax By solving the equations, we get point x = a, y = b and now f(a, b) may be maximum or minimum. SUFFICIENT CONDITION (LAGRANGE'S CONDITIONS) Function m ust satisfy Lagrange's conditions as follows, Using the conditions (iii) in equation (i) and neglecting the higher order term h2 , hk, k2 etc., we get 2 2 a f 1 2 a f 2 a f ] f(a + h, b + k) - f(a, b) = - [ h -2 +2hk --+k 2 2! axoy ax oy (a, b) 2 Putting a (-2f } ax a f = r, ( --) 2 2 (a , b ) axoy 2 (a , b) f(a + h, b + k) - f(a, b) ⇒ f(a a f = S, (-2 J oy (a,b) = t, then = � [h 2 r+2hks+k 2 t] = i [ h 1 r +2h krs+k tr ] r 2 2 (hr +ks) k (rt - s ) ] + h, b + k) - f(a , b) � i [ � 2 If rt - s2 > 0 then numerator in R . H . S . of (iv) is positive. Here sign of L. H.S. Thus, if rt - s2 > 0 and r < 0, then f(a + h, b + k) - f(a, b) < 0 if rt - s2 > 0 and r > 0, then f(a + h, b + k) - f(a, b) > 0 2 2 2 = sign of r. . . . (iv) I ES MASTER Publication 236 Maxima and Minima Therefore, the Lagrange's conditions for maximum or minimum are Consider a function z = f(x, y) and let P(a, b) be any point on it, and let (�! l = r; ( !2 ; l = s; ( (i) If rt - s2 > 0 and r < 0, then f(x, y) has maximum value at {a, b). (iii) If rt - s2 < 0, then f(x, y) has neither a minimum nor m i nimum i.e. (a, b) is saddle point. (ii) (IV} �! l = t If rt - s2 > 0 and r > 0, then f{x, y) has minimum value at (a, b). It rt - s2 = 0, then case fail and we need further investingations to cal culate maxima or m i nima. Flowchart to find Maxima and Minima 1. 2. Consider a given function Z = f(x, y) f Calculate the val ues of x & y by using � = O and o = 0. 3. Let we get x = a and y = b from step (2) then critical point is P(a, b) 5. Now, maximum or minimum value is given by f{a, b). 4. ax ay Check Lagrange's conditions for maxima/minima. Example 6 : Solution 1 If u = xy(a - x - y), show that the maximum value of u is � . 27 3 u = xy(a - x - y}, then Let au = ay - 2xy - y2 and au = ax - x 2 - 2xy ax ay �u a� a� = -2y, s = -- = a - 2x - 2y and t = - = -2x r = 2 2 ⇒ . . . (i) axay ax ay For a maximum or minimum of u, we have Thus, we have ay - 2xy - y2 = 0 or ay - x 2 - 2xy = 0 . . . (ii) On solving these equations, we get the following pairs of values of x and y which make the function stationary: (0, 0), (0, a), (a, 0), (ti) We discuss the following cases : (i) (ii) At (0, 0) : r 0, s = -2a, s = = 0, s -a, t = a, t = 0 ⇒ rt - s2 < 0 using (ii) ⇒ Neither a maxim um nor a minimum of u at (0, a). ⇒ Neither a maximum nor a minimum of u at (0, a). ⇒ Neither a maximum nor a minimum of u at (a, 0). At (0, a) : r (iii) At (a, 0) : r (IV) = At = -a, t ⇒ rt - s2 < 0 using (ii) = -2a ⇒ rt - s2 < 0 using (ii) (i,i) : 2 2 1 r = - a, s = - a, t = - a 3 3 3 ⇒ = 0 rt - s2 > 0 using (ii) ⇒ Example 7 Solution 1 1 Engineering Mathematics (i, i) u has an extreme val ue at and it will be a maximum or minimum according as r is negative or positive i.e. according as a, is positive or negative. The extreme val ue of u is given by a a a a a3 u = 3 . 3 ( a - 3 - 3 ) (by (i)) = 27 Find the extreme values of function x3 + y3 Here f(x, y) = x3 + y3 fX = 3x2 - Now, fx ⇒ and = 3ay fy = 0 and fy I = 3axy 3y2 - 0 x2 y 2 Solving (i) and (ii) when x Now, 3ax r = fxx = 3axy. I ay = 0 ax = 0 0, y O; when x = = a, y There are two stationary points (0, 0) and (a, a). rt - s2 = 36xy - 9a2 At (a, a) : rt - s2 = 36a2 ⇒ - 6x s = fxy = -3a , t = fyy = 6y = I At (0, 0) : rt - s2 = -9a2 < 0 ⇒ - = . . . (i) . . . (ii) a. There is no extreme value at (0, 0). 9a2 = 27a2 > 0 f(x, y) has extreme val ue at (a, a) Now, r = 6a If a > 0, r > 0 so that f(x, y) has a minimum value at (a, a). and minimum value a3 + a3 = - 3a3 = -a3 • If a < 0, r < 0 so that f(x, y) as a maximum value at (a, a) and Maxi mum value Exam ple 8 : Test function f(x, y) Solution 1 237 = r Now, and s = = x4y2 - 36x2y - 8x3y - 9x2y2 = - 4x3y2 fxx = 36xy2 fxy t = fw = = 12x3 fx = 0 fy -a3 - a3 + 3a3 x3y2 (6 - x - y) for maxima and minima for poi nts other than origin. Here f(x, y) = x3y2(6 - x - y) = 6x3y2 fx = 18x2 y2 = = 0 From (i) and (ii), 4x + 3y ⇒ = - - - ⇒ 3x2 y3 1 2x2y2 2x4 - - and fy - = 12x3y ..,. 2x4y - 3x3y2 6xy3 = 6xy2 (6 - 2x - y) 6x3y = x2y(36 - Bx - 9y) x3(12 - 2x - 6y) x3y(12 - 2x - 3y) 2x + 3y = a3 . x3y3. x2y2( 1 8 - 4x - 3y) = 0 18 = = 0 12 and x = 0 = y . . . (i) . ..(ii) Solving, we get x = 3, y = 2 and x = 0 = y. Leaving x = 0 = y, we get x = 3, y = 2. Hence, (3, 2) is the only stationary point under consideration. Now, rt - s2 = 6x4y2 (6 - 2x - y)(12 - 2x - 6y) - x4y2 (36 - 8x - 9y)2 At (3, 2) rt - s2 > 0 (positive) Also r = 6(3)(4)(6 - 6 - 4) = negative (< 0) ⇒ f(x, y) has a maximum value at (3, 2). IES MASTER Publication 238 Maxima a n d Minima Example 9 1 Solution 3 3 Show that minimum value of u = xy +� +� is 3a2 . X y 2 au = y - a3x-2 . au = x - a 3 y -2 . r = a2 u = 2a3x-3 · s = � a2 u = 2a 3 y -3 = 1· t = • ax;:i.., •� . ax ax2 � • ay 2 ;:i., • Now, for maxi mum or minimum we must have So, from au = 0, ax au = au = au 0, = 0 ax oY we get y - a3x-2 = O or x2y = a3 . . . (i) 0, we get x - a3 y-2 = 0 or xy2 = a3 . . .(ii) 8y Solving (i) and (ii), we get x2y = xy2 or xy(x - y) = 0 or x = 0, y = 0 and x = y and from From (i) and (ii), we find that x = 0 and y = 0 do no hold as it gives a = 0, which is against hypothesis. : . We have x = y and from (i) we get x3 = a3 or x = a and therefore, we have x = a = y. This satisfies (ii) also. Hence, it is a solution. At x = a = y, we have r = 2a3a-3 = 2, s = 1, t rt - s2 = (2)(2) - 1 2 = 3> 0 = 2 Also, r = 2 > 0. Hence, there is minima at x = a = y. The minimum value of u = xy+( : ) +( � ) at x . IS = a - a +( a ) = a +a +a aa )+( a 3 3 2 2 2 = 1 az oY For the maxi mum or m i nimum we must have From From y 2 - = 2X + y + 3; - = X + 2 az ax = = 3a . Hence proved. Example 10 : Examine the following surface for high and low points z Solution a . = x2 + xy + 3x + 3y + 5. z az = 0, o = O ay ax - = 0, we get 2x + y + 3 = 0 oz ax - = 0, we get x +2 = 0 8y . . .(i) . . . (ii) az From (ii), we get x = -2 and : . from (i) y = 3+ 4 = 1. Hence, the solution is x = -2, y = 1 and for these val ues we have r = 2, s rt - s2 = (2)(0) - 1 2 = -1 < 0 There is neither maximum nor minimum at x = -2, y = 1 . = 1 , t = 0. LAGRANGE'S METHOD OF UNDETERMINED MULTIPLIERS To find the maximum or minimum values of a function of three or more variables. We use Lagrange's method of undetermined m ultipliers. Working Rule Let u = f(x, y, z) be a function which is to be examined for maxima or minima . . . (i) Engineering Mathematics 239 and let $((x, y, z)) = c ...(ii) is the given relation F = f+1'.$ Then consider the Lagrange's function as follows For maxima or minima of F we must have dF = 0 . ..(A) ( 1'. ; Lagrange's multiplier) df + 1'.d$= 0 [ ( : ) dx+ (: ) dy + (: ) dz ]+ 1'. [ (: ) dx + (: ) dy+ ( :) dz ] = 0 ⇒ a$ a$ a$ ( � +A ) dx+ ( � + A ) dy+ ( af +A ) dz= 0 ax ax ay ay az az at a$ =0 + 1'. ax ax . .. (iii) at a$ =0 + 1'. az az . ..(v) � + 1'. a$ =0 ay ay ...(iv) T hese equations together with equation (ii) gives the values of x, y, z and 1'. . Here, P(x, y, z) is a stationary point and maximum or minimum value can be calculated by equation (i). 2. Example 1 1 Solution I 1 Lagrange's method does not enable us, to find, the nature of the stationary points (i.e. whether there is a maxima or minima) this fact can be calculated with the physical consideration of the problem. Keep the process of above article in mind. Find the minimum value of x2+ y2+ z2, given that ax+ by+ cz = p. u = x2 + y 2 + z 2 Let $ (x, y, z) = ax + by+ cz - p = 0 where Consider Lagrange's functions, F(x, y, z) = (x2+ y2+ z2) + For stationary values, dF = 0 ⇒ ⇒ ⇒ 1'. (ax+ by+ cz - p) ...(i) . ..(ii) (2x + 1'.a)dx+ (2y + 1'.b)dy + (2z + 1'.c)dz = 0 (2x+ 1'.a)dx+(2y+ 1'.b)dy+(2z+1'.c)dz =0 . ..(iii) 2x+1'.a =0 ...(iv) 2y+1'.b =0 2z+1'.c =0 Multiplying (iii) by x, (iv) by y, (v) by z and adding, we get 2(x 2 + y 2 + z 2 )+1'.(ax+by+cz)= 0 1'. = _ 2u From (iii), (iv) and (v), or 2u+1'.p= 0 au bu cu , X = -, y= -, Z =p p p ...(v) [using (i) and (ii)] IES MASTER Publication 240 Maxima and Minima From (i), This is the maximum or minimum value of u. Now, u is the square of the distance of any point P(x, y, z) on the plane (ii) from the origin. Also, the length of perpendicular from O on the plane is p )a +b2 +c 2 the plane. 2 . Clearly, OP is least when P coincides with M, the foot of the perpendicular from O on Hence, the minimum value of u = Example 1 2 Solution 1 a 2 + b 2 +c 2 Find the maxima and minima of u = x2 + lx + my + nz = 0. 2 y · + z2 subject to the conditions ax2 + by2 + cz2 = 1 and Let us define a function F such that 1 F = x 2 + y 2 + z2 + ;\,1 (ax 2 +by 2 +cz 2 - 1)+;\,2 (lx + my+nz) For maxima and minima of F, we have aF = 2x +2ax;\,1 + l;\,2= 0 ax aF = 2y +2by;\, 1 + m;\,2 = 0 ay ...(ii) aF - = 2z +2cz;\, 1 + n;\,2 = 0 az Multiplying equations (ii), (iii), (iv) by x, y, z respectively and adding, we get ⇒ . ..(i) ...(iii) ...(iv) 2(x 2 + y 2 + z 2 )+2;\,1 (ax 2 +by 2 + cz2 ) + ;\,2 (lx+my+nz)= 0 2u+2;\,1 =0 ⇒ ;\,1 =-u From (ii), 2x - 2axu+l;\, 2 = 0 Similarly, y ⇒ l;\,2 X = -�- 2(au- 1) m;\,2 n;\, 2 = 2(b - ' Z = 2(c -1) u u 1) Substituting these values of x, y, z in lx + my + nz = 0, we get z2 n2 m2 - -+- - + - -= 0 au- 1 bu- 1 cu- 1 which gives the required maximum and minimum values of u. A-2 - +n - -n;\, '"'2 - +m2 - m--=---=---0 l - -� 2(au- 1) 2(bu-1) 2(cu-1) Example 1 3 Solution 1 1 ⇒ If x and y satisfy the relation ax2 + by2 = ab, prove that the extreme values of the function u = x2 + xy + y2 are given by the roots of the equation 4(u - a)(u - b) = ab. Here u = x2 + xy + 2 y and Consider Lagrange's function For stationary values, x2 y 2 � = -+--1 = 0 a b F = u + ;\,� x 2 F(x, y) = x +xy+ y 2 +;\, ( : + y: - 1) dF = 0 ... (i) Engineering Mathematics ⇒ ⇒ 21cx 2x+y+- = 0 b 21cy x + 2y+- =0 a ...(ii) . . . (iii) Multiplying (ii) by x and (iii) by y and addi ng, we get 2(x2 +xy+y 2 )+n ( ⇒ 2u +2A(1) = O From (ii), 2x+ y - From (v), X =. From (iii), From (vi), 2 x +Y ) = 0 b a 2 [ ·.- � + Y: 2ux = 0 b 1] [using (iv)] -1 ⇒ 1c = -u ⇒ ⇒ 2( 1 - � ) Equating val ues of -"-, we get Y ⇒ = [using (iv)] 2uy x +2y - - = 0 a y 241 4(� - 1)(� - 1) = 1 ⇒ -1 2 ( 1-� ) . . . (iv) 2 ( 1 - � ) x+y = O ...(v) x+2 (1 - � ) y = O . . . (vi) • -2 ( 1 - � ) 4(u - a)(u - b) = ab a Example 1 4 = Find the volume of the largest rectangular parallelepiped that can be inscribed i n the ellipsoid Solution 1 x2 y 2 2 2 = -+-+1 a 2 b2 c 2 Let (x, y, 2) be a vertex of the parallelepiped then it l ies on the ellipsoid 2 x 2 = y -+-+ 1 2 2 a b c2 2 2 . . .(i) Let 2x, 2y, 22 be the length, width and height of the rectangular parallelopiped inscribed in the ellipsoid. Vol ume V = 8xy2 Consider Lagrange' s function F = V + 1c4> . . . (ii) 2 2 22 x y - 1J F(x, y, 2) = 8xy2+1c ( -+-+2 2 2 a b c For stationary val ues, dF = 0 ⇒ 8y2+1c ( !n=O 8 zx+ A,(�n =0 . . . (iii) . .. (iv) IES MASTER Publication 242 Maxima and Minima 8xy+A ( �n =0 Multiplying (ii), (iv), (v) by x, y, z respectively and adding, we get ⇒ From (iii), 8yz- 12xyz ( !� ) =o ⇒ 24 xyz+2A= 0 2 1 - 3 x2 a =0 ⇒ A =-12xyz ⇒ b c Similarly, from (iv) and (v), we get y = ✓ 3 , z = ✓3 Volume of the largest rectangular parallelopiped = 8xyz = Sabe --13 . 3 ...(v) Engineering Mathematics 243 PREVIOUS YEARS GATE & ESE QUESTIONS· 1. Questions marked with aesterisk (*) are Conceptual Questions The function f (x, y ) =x 2y -3xy+2y+x has (b) fx (a, b) = 0; fy(a,b) = 0; (b) One local maximum put no local minimum (c) fx (a , b) = 0 ; fy (a, b) = 0; (a) No local extremum (c) One local minimum but no local maximum fxx and fyy < 0 at (a, b) (d) One local minimum and one local maximum [GATE-1 993 (ME)] 2. (d) 1 [GATE-1 994; 1 Mark] Find the height of the right circular cylinder of maximum volume V which can be inscribed in a sphere of radius R. [GATE-1 994; 5 Marks] The function f(x) = x 3 - 6x2+ 9x+ 25 has (a) a maxima at x = 1 and a minima at x = 3 (b) a maxima at x = 3 and a minima at x = 1 (c) no maxima, but a minima at x = 3 5.* (d) a maxima at x = 1, but no minima [GATE-1995; 2 Marks] Find the absolute maxima and minima of the function f(x, y) = x2 - xy - y2 - 6x+ 2 on the rectangular plate 0 s; x s; 5, -3 s; y s; 0 6. 7.* Where the subscripts x, y etc. denote partial derivatives. [GATE-1 998 ; 2 Marks] (b) Minimum (c) Neither 4. f;y -fxx fyy = O at (a,b) 250 The function, y = x 2 +- at x = 5 attains X (a) Maximum 3.* (d) fx (a, b) = fy (a , b) = 0 ; [GATE-1995] What is the maximum value of the function f(x) = 2x2 - 2x+ 6 in the interval [0, 2]? (a) 64 (b) (c) 10 (d) 64 3 128 4 [GATE-1 997 (CS)] The continuous function f(x, y) is said to have saddle point at (a, b) if (a) fx (a, b) = fy(a, b) = 0 ; 8. 9.* Find the points of local maxima and minima if any of the following function defi ned in o s; x s; 6 , x 2 -6x 2 +9x +15 . [GATE-1 998 (CS)] N umber of inflect i o n poi nts for the curve y = x + 2x4 is (b) (a) 3 (d) (n + 1)2 (c) 0 [GATE-1 999; 1 Mark] 10. Find the maximum and minimum values ofthe function f(x) = sinx + cos 2x over the range 0 < x < 2 n . [GATE-1999; 5 Marks] 1 1 .* From an equilateral triangular plate of side 'a', a square plate of maximum size has to be cut. The side of such a square plate is: 3 (a) (c) ✓a 2 ( 1+ ✓3 ) a 8 3 (b) (d) (2 + ✓ ) a 4 3 ✓a3 (2+ ✓ ) [GATE-2000; 2 Marks] 12. The maxima and minima of the function f(x) = 2x3 15x2+ 36x+ 10 occur, respectively at (a) x = 3 and x = 2 (b) x = 1 and x = 3 (c) x = 2 and x = 3 (d) x = 3 and x = 4 1 3. The m i n i mum point f(x) = (x 3/3) - x is at [GATE-2000; 1 Mark] of the function IES MASTER Publication 244 Maxima and Min ima (a) (c) X X (b) x = -1 = 1 (a) [GATE-2001 ; 2 Marks] (c) f(x) = x 5 - x2 ✓ (b) 2 (d) JI . ✓5 -� [GATE-2002-CE ; 1 Mark] (a) only one stationary point at (0, 0) (b) two stationary points at (0, 0) and (i, - i) (c) two stationary points at (0, 0) and (1 , -1) (d) no stationary point [GATE-2002; 2 Marks] 1 6. The function f(x) = 2x3 - 3x2 - 36x + 2 has its m axima at (a) x = -2 only (b) x = 0 only (c) x = 3 only (d) Both x = -2 and x = 3 (d) decreases to a m i n i m um v a l ue and then increases [GATE-2006; 2 Marks] 20. The minimum value of function y = x2 in the interval [ 1 , 5] is (a) 0 1 5. The function f (x, y) = 2x2 + 2xy - y3 has 1 7. (c) i n creases to a maxi m u m value and then decreases jj 1 4. The following function has a local minima at which value of x? ✓5 (b) monotonically decreases 1 (d) x = = 0 (a) monotonically i ncreases [GATE-2004; 2 Marks] (c) 25 (a) 2 x (a) 18 (c) (d) -1 [GATE-2005-EE; 2 Marks] 1 8.* The right circular cone of largest volume that can be enclosed by a sphere of 1 m radius has a height of (a) (c) 1 -m 3 2 -m 3 ✓2 1 9.* As x i n creased from e f(x) = -- x 1+e x 2 (b) - m 3 4 (d) - m 3 [GATE-2005; 2 Marks] - OCJ to OCJ , the function [GATE-2007; 1 Mark] - 2.25 (b) 10 (d) indeterminate [GATE-2007-EC ; 2 Marks] 22.* For real x, the m axi m u m v alue of (a) 1 (c) e .fi. (b) e (d) a esin x is ecos x [GATE-2007 ( IN)] 23. For the function f (x, y) = x2 - y 2 defined on R2 , the point (0, 0) is (a) a local minimum (b) n e i t h e r a local m i n i m u m ( n o r) a l o c a l maximum (b) 1 (c) 0 (d) undefined 21 . Consider the function f(x) = x2 - x - 2. The maxi­ m u m v a l u e of f (x) i n t h e c l o sed i nt e rv a l [-4, 4] is For the function f(x) = x e- , the maximum occurs when x is equal to 2 (b) (c) a local maximum (d) both a local m i nimum and a local maximum [GATE-2007 ( Pl)] 24. Consider function f(x) = (x2 - 4)2 where x is a real num ber. Then the function has (a) only one m i nimum (b) only two minimum (c) three minimum 25. (d) three maximum [GATE-2008; 2 Marks] For real values of x, the minimum value of the function f(x) = exp(x) + exp( - x) is Engineering Mathematics (a) 2 (c) 0.5 (b) 1 (d) 0 [GATE-2008-EC ; 1 Mark] 26.* A point on the curve is said to be an extrem u m if it i s a l ocal minimum (or) a local maxi m u m . The num ber o f disti nct extrema for the curve 3x4 - 1 6x3 + 24x2 + 37 i s (a) 0 (b) 1 (c) 2 (d) 3 [GATE-2008 (CS)] 27. Consider the function y = x2 - 6x +9 . The m axi­ mum v a l ue of y obtai ned when x varies over- the i nterval 2 to 5 is (a) 1 (b) 3 (c) 4 (d) 9 [GATE-2008 (IN)] 28.* A cubic polynomial with real coefficients (a) Can possibly have no extrema and no zero crossings (b) May have up to three extrema and upto 2 zero crossings (c) Cannot have more than two extrema and more than three zero crossi ngs. (d) Will always have an equal number of extrema and zero crossi ngs [GATE-2009; 1 Mark] 29.* The distance between the origin and the point nearest it on the surface z2 = 1 + xy is (a) 1 (c) (b) ✓ 2 3 ✓ 3 (d) 2 30.* At t = 0, the function f(t) = (a) a minimum (b) a discontinuity (c) a point of inflection 31 . (d) a maximum sin t has [GATE-2009] Given a function f(x, y) = 4x2 + 6y2 The optimum value of f(x, y) is - (a) a minimum equal to 10/3 (b) a maximum equal to 1 0/3 (c) a minimum equal to 8/3 (d) a maximum equal to 8/3 [GATE-201 0-CE; 2 Marks] 32. * If eY = x 11x, then y has a (a) maxi mum at x = e (b) minimum at x = e (c) maximum at x = e-1 (d) minimum at x = e-1 8x - 4y + 8. [GATE-2010-EC; 2 Marks] 33. The function f(x) = 2x - x2 +3 has (a) a m ax i m u m at x = 1 a n d a m i n i m a at X = 5 (b) a m axima at x = 1 X = -5 (c) only a maxi ma at x = 1 and a m i n i m a at (d) only a minima at x = 1 [GATE-201 1 -EE; 2 Marks] 34. The maximum value of f(x) = x3 - 9x2 + 24x + 5 in the interval (1, 6] is (a) 21 (c) 41 (b) 25 (d) 46 [GATE-201 2-EC,EE-IN; 2 Marks] 35. 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 [GATE-201 2-ME, P l ; 1 Mark] 36. The maximum value of f(x) = x3 - 9x2 + 24x + 5 in the i nterval [ 1 , 6] (a) 21 (c) 41 (GATE-201 0-EE; 2 Marks] 245 (b) 25 (d) 46 [GATE-201 2-EC, EE, I N ; 2 Marks] 37.* A pol itical party orders an arch for the entrance to the ground i n which the annual convention is being held. The profi le of the arch follows the equation y = 2x - 0. 1 x2 where y is the height of the arch i n m eters. The m axi m u m possible height of the arch is IES MASTER Publication 246 Maxima and Minima (a) 8 meters (c) 1 2 meters (b) 1 0 meters (d) 14 meters [GATE-201 2 (ME, Pl)] 38.**A scalar v a l ue function i s defined as f(x) = x T Ax + bTx + c, where A is a symmetric positive definite m atrix with d i m ension n x n ; b and x are vectors of dimension nx 1 . the minimum value of f(x) will occur when x equals. (a) (c) 39. (A T A Ar B -[ � ) 1 B (b) - (A Ar B T A-1B (d) 2 [GATE-201 4 (IN -Set 1 )] Let f(x) = x e- x. The maximum value of the function in the i nterval (0, w ) is (a) e(c) 1 1 - e-1 (b) e (d) 1 + e-1 [GATE-2014-EE-SET-1 ; 1 Mark] 40.* M i n i m u m of t h e real v a l ued f(x) = ( x - 1 ) 213 occurs a t x equal to (a) (c) - oo 1 (b) 0 f u n ct i o n (d) 00 [GATE-2014-EE-SET-2; 1 Mark] 41 . The minimum value of the function f(x) = x3 - 3x2 24x + 1 00 i n the interval [-3, 3] is (a) 20 (c) 16 (b) 28 (d) 32 [GATE-2014-EE-SET-2; 2 Marks] 42.* The best approximation of the minimum value attained by e-x si n(1 00x) for x > 0 is_ _ __ 43. [GATE-2014; 1 Mark] For o � t < oo , the m aximum value of the function f(t) = e-1 - 2e-21 occurs at (a) t = loge4 (c) t = 0 (b) t = loge2 (d) t = log e8 [GATE 201 4-EC-Set-2; 1 Mark] 44. T h e m ax i m u m v a l u e of t h e f u n c t i o n f(x) = l n ( 1 + x ) - x (where x > -1 ) occurs a t x = [GATE-201 4-EC-Set-3; 1 Mark] 45. T h e m a x i m u m v a l u e of f(x) = 2 x 3 - 9x 2 1 2x - 3 in the i nterval o � x � 3 is___ + [GATE-201 4-EC-Set-3; 2 Marks] 46.* For a right angled triangle, if the sum of the length of the hypotenuse and a side is kept constant, in order to have maximum area of the triangle, the angle between the hypotensuse and the side is (a) 1 20° (b) 60° (c) 30° (d) 45° [2014; 2 Marks] 47. At x = 0, the function f(x) = lxl has (a) a minimum (b) a maximum (c) a point of inflexion (d) neither a maximum nor m i nimum [GATE-201 5-M E-SET-2; 1 Mark] 48. While minimizing the function f(x), necessary and sufficient condition for a point x 0 to be a m i nima are and f"( x 0 ) = 0 (a) f( x 0 ) > 0 (c) f(x0 ) = 0 and f"(x 0 ) < 0 (b) (d) f"( x 0 ) < 0 and f"(x 0 ) = 0 f( x 0 ) = 0 and f"( x 0 ) > 0 [GATE-201 5-Set-l ; 1 Mark] 49.* The maximum area (in square units) of a rectangle whose vertices l i e o n the e l l i pse x 2 + 4 y 2 = 1 is--. [GATE-201 5-EC-Set-1 ; 2 Marks] 50.* The cu rve y y = x 4 i s (a) concave up for all values of x (b) concave down for al l values of x (c) concave up onl y for positive values of x (d) concave up onl y for negative val ues of x [GATE-201 5 (Pl)] 51 . The maximum value attained by the function f(x) x(x - 1 )(x - 2) in the i nterval [1 , 2] is = [GATE-201 6-EE-Set-1 ; 1 Mark] 52. Consider the function f(x) = 2x3 - 3x2 i n the domain [-1 , 2]. The global minimum of f(x) is [GATE-201 6-ME-SET-1 ; 2 Marks] 53. The optimum value of the function f ( x) = x 2 - 4x + 2 is (a) 2 (maximum ) (c) -2 (maximum) (b) 2 (minimum) (d) -2 (minimum) [GATE-201 6-CE-Set-2; 1 Mark] Engineering Mathematics 54. As x varies from - 1 to + 3, which one of the following describes the behavior of the function f(x) = x3 - 3x2 + 1? (a) f(x) increases monotonically. x 58. The minimum value of the function f(x) = ( ; occurs at (a) (b) f(x) increases, then decreases and increases again. (c) f(x) decreases, then increases and· decreases again. (ct) f(x) increases and then decreases. [GATE-2016-EC-Set-3; 1 Mark] 55. Let t : [-1 , 1] --> R, where f(x) = 2x3 - x4 - 10. T he minimum value of f(x) is ____ (c) f (x) = (k2 -4 ) x2 + 6x3 + 8x4 has a local maxima at point x = 0 is (a) k < -2 or k > 2 (c) -2 < k < 2 in the interval -100 � x � 100 occurs at x = [GATE-2017-EC Session-II] (b) x = -1 = 1 (d) X = = 0 .J31 [ESE 2017 (Common Paper)] XY The optimal value of Cr, rounded to 1 decimal place, is __ 60. [GATE-2017 (CH)] At the point x = 0, the function f(x) = x3 has (a) local maximum (b) local minimum (c) both local maximum and minimum (ct) -2 :S k :S 2 2 57.* The minimum value of the function f(x) = -½ x ( x - 3 ) x 12000 CT = 2x +-- +y+5 (b) k :S -2 or k � 2 [GATE-2016 (Pl)] X J- 59. The total cost (CT ) of an equipment in terms of the operation variables x and y is [GATE-2016-ME-Set 1] 56.* T he range of values of k for which the function X 247 (d) neither local maximum nor local minimum 61. [GATE 2018 (CE-MorningSession)] Let f(x) = 3x3 - 7x2 + 5x + 6. The maximum value of f(x) over the interval [O , 2] is __ (up to 1 decimal place). [GATE 2018 (EE)] IES MASTER Publication 248 · - 1-.: Maxima and Minima .;---<...-.-.�� .,..: = ·, ,.. _.._ ;-�- �---..·· - -. --.. : •��• ��r :.,:• ., 1. - • f •♦_, ; • ~ •� • ' " • (a) 2. 17. (b) 1 8. (;) 3. 4. 1 9. 20. (a) 5. (See Sol.) 6. 21 . 22. (c) 7. 23. (b) (See Sol.) 8. 9. 1 0. 11. 1 2. 1 3. 14. 1 5. 1 6. 24. 25. (c) ( 1 . 1 25, -2) 26. 27. (d) 28. (c) 29. (a) 30. (d) 31 . (b) (a) if{ N�s..��l r:::I; · R .;l"·� E :"1·. •, �- )' ,? T,T�; 33. (a) 34. (d) 35. (a) 36. (b) 37. (a) 38. (c) 39. (b) 40. (b) 41 . (a) 42. (b) 43. (c) 44. (c) 45. (a) 46. (d) 47. (a) ax af af oY = 2xy - 3y + 1 = = x2 - 3x + 2 = 0 o, ... (1) ... (2) On solving (1) & (2) we get x = 1 , y = 1 & x = 2, y = -1 So P 1 ( 1 , 1) and Pi2, -1 ) are the critical points Now, At ( 1 , 1) and (2, - 1) ; value of rt - s2 < 0 ⇒ [t!. t!.]-[ ax.2 X 8y2 8 f ]<0 ax.y 50. (c) 51 . (d) · 52. (c) 53. (b) 54. (c) 55. (a) (c) (--0.954) (6) (1 ) (a) (0) (-5) (d) (b) (-13) 58. (a) 60. (0) (d) (c) 59. (a) � 56. 57. (b) 61 . (-1 00) (91.5) (d) (12) (b) (a) y = X 2 250 +­ X dy 250 = 2x - 2 dx x for critical poi nts �� = 0 ⇒ ⇒ 2 Both P 1 & P2 are saddle points • 49. (c) Sol-2: (b) Using necessary conditons for maxima-minima 48. (a) SOLUTIONS Sol-1 : (a) ie. 32. -: . · • •. • • ⇒ 250 2x - = 2 X o X = 5 500 d2y = 2+ 3 dx2 x d2 y dx2 l x=5 = 2+ 500 53 y has minima at x = 5 >O Engineering Mathematics Sol-3: ( f '(x) at x 2 3.J ) v; Radius of Cylinder Height of Cylinder = = Volume of Cyli�der = R cos0 2Rsin0 2 n(Rcos 0) ( 2R sin0) Sol-5: and ⇒ f(x, y) 2sin2 0) cos 0(3 cos2 0-2 ) ⇒ 0= 7t = • so, we must have 3 cos2 0-2 =0 cos e = � Hence sin 0 = � Height of cylinder Sol-4: (a) x2 i.e. 0 4 X - f(x) = 3x2 2 + 3 = 0 (x - 3) (x - 1 ) = 0 X f'(x) at x = •: f"(x) at x X = 3 f'(x) 12x + 9 = 0 = 3, 1 = 3 is : f"(3) 6x - 1 2 = 6 x 3 - 12 lf "( 3 ) = 6 > 01 = ✓ 6x2 + 9x + 2 5 - of - y - 6 and -= -x-2 y ay = 2x ay 2x -y -6 = 0 -x - 2 y 0 = 6 12 X = 5' Y = - 5 is a critical point. 0 1:j = 2 R sin0=� f(x) = x3 - = (Not possible) 2 of ax x2 - xy-y2 6x +2 = 12 Hence ( , - 5 5 cos 0[cos2 0- 2(1-cos2 0)] = 0 cos 0 = 0 lf"( 1) = -6 < ol 6) (cos2 0.cos 0- 2 sin2 0cos0) = O 0- 6 x 1 - 12 ax (cos2 0.cos0-2 sin2 0cos0) = 0 cos0(cos2 = For critical points �=0 and �=0 V = 2 nR3 sin 0 cos2 0 z To maximi e volume dV = O d0 2 nR 3 1 is : f"(1) ·: f'(x) at x = 1 is < 0, the function has a maxima at X = 1 . then, ⇒ = 249 3 > 0, the function has a minima at Now, r = = s = rt - S2 : a 2f 2 = ax > 0 2 a2 f = -2 ay z a2 f = -1 axay 2 X -2 - ( 2 - 1) = - 4 -1 = - 5 < 0 So, given point is a saddle point. f(0, 0) = 2 Now, on corner points; 2 f(0,-3) = 0-0 - (-3} -6 x 0+2 = -7 f(6, 0) : f(5,-3) = 2 5-50 -02 - 6 X 5 +2 =-3 2 2 5 - 5 x -3 - (-3) - 6 x 5 +2=3 IES MASTER Publication 250 Maxima and Minima Hence function has absolute maxima at (5, -3) and absolute minima at (0, -3) Sol-6: (c) For critical points put �� = 0 1 f'(x) = 4x - 2 = 0 ⇒ x = 2 f"(x) = 4 > 0 ie ⇒ f(x) is m i nimum at x = 1 2 So m ax. value occurs only at corner point f(0) = 6, f(2) = 1 0 Sol-7: Max value = 1 0 (b) ·: at saddle point rt - s < 0 .⇒ Local minima No point of i nflection. f(x) = x3 Let - 6x2 + 9x + 1 5 f(x) = 3x2 - 1 2x + 9 = 0 ⇒ x = 1 , 3 are the turning points f'(x) = cos x - 2 sin 2x = 0 cos x - 2 sin 2x = 0 cos x - 2 x 2 sin x cos x = 0 cosx(1 - 4sinx) = 0 Either cos x = 0 or ⇒ sinx = 4 X = (2n + 1)-, 2 x = nn + (-1f sin0 < 2 Sol-8: f(x) = sin x + cos 2x Sol-10: (1.125, -2) x = X < 21t 7t 1 ({) 1t 31t . -1 ( 1 1 . , , sm ) , n - sm-1 ( ) 4 4 2 2 are the critical points and val ues of f(x) at critical points are: f "(x) = 6x - 1 2 ⇒ f"(1) = - 6 < 0 f(x) has maximum at x = 1 f"(3) = 6 > 0 ⇒ f(x) has minimum at x = 3 So x = 1 is a point of local minima & x = 3 is a point of local minima Sol-9: y = X + 2x4 (c) dy = 1 + 8x3 = 0 dx 1 X = 2 d2 y d2 y = 24x2 dx2 dx 2 I x = ! -2 1 24 ( -2 2 ) =6>0 = 1 . 1 25 t (x = n - sin- 1 (± )) = -0.625 So, Maximum value f(x) = 1 . 1 25 Minimum value f(x) = -2 Sol-11: (d) B ---- a -----1 C Engineering Mathematics Let the side of the square plate of maximum size be x. For !l ACD 3 ✓ -a-x 2 = tan 60 x/2 3 ✓ a - 2x X Sol-12: (c) 3 = ✓ = f' (x ) = 3 X f(x) = 2x3 - 15x2+ 36x+ 10 r ( )% f'(x) = 6x2 - 30x + 36 6(x - 2)(x - 3) = 0 X = 2, 3 ⇒ f"(x) = 12x - 30 f(x) has local maxima at x = 2 f "(3) = 12(3) - 30 = 6 > 0 Sol-15: (b) f(x) has local minima at x = 3 x3 -x 3 f'(x) = x2 - 1 for minima or minima, f '(x) = 0 x2 - 1 = 0 X = ± 1 for minima, f"(x) > 0 f"(X) = 2 X f"(x) (at x = 1) is 2 > 0 of ax of ay ✓ 2 5-x 2 f(x) has local maxima. f(x) has local minima of putting, ax 2x - 3y2 = 4x + 2y = 2x - 3y2 = 0 and of = 0, we get ay = 0 ...(i) .. .(ii) we get, -1 1 , x= � \ IY = 0, x =OI and l Y _ _ _ _ 3 �__ -� So we have two stationary points at (0, 0) and Thus, maxima at x = -1 f' ( x ) = x( 2x2 -15) 3 2 ( 5-x2 ) ' Solving equation (1) and (2) simultaneously f"(x) (at x = - 1) is - 2 < 0 X(-2x ) ±l 4x + 2y = 0 Thus, lminima at x = 11 ✓ ✓ f(x, y) = 2x2 + 2xy - y3 f(x) = f(x) = x 5-x2 5-2x 2 5-x2 5 � (-10 ) 2 ff' < 0 ) = �2- l, -l, At x = At x = f " (2 ) = 24 - 30 < 0 Sol-14: (d) 2 and f" (-l ) > 0 For critical points f' (x ) = 0 Sol-13: (a) = f "(x ) = f' ( x ) = 6 (x -2) (x - 3) ⇒ ✓5-x 5 - 2x2 = 0 ⇒ = 6 (x2 -5x+6) ⇒ -x2 + 5 - x2 For critical points f'(x) = 0 ✓a 3 ( 2+ ✓ ) X 251 1 (¾, � ) ✓ + 5-x 2 Sol-16: (a) f(x) = 2x3 - 3x2 - 36x+ 2 f'(x) = -6x2 - 6x - 36 ... (i) IES MASTER Publication 252 Maxima and Minima Point of extremum are given by f'(x) = 0 ⇒ ⇒ 6x2 - 6x - 36 = 0 X2 - X - (X - ⇒ at X = 3, 3) ( X For maximum volume V dV = 0 dh 4 h = 0 or h = 3 But h = 0 (Not possible) ⇒ 6=0 + 2) = 0 X = 3, - 2 f"(x) = 1 2x - 6 so, required height, h = Sol-19: (a) f"(3) = 30 > 0 Hence x = 3 point of minima. at X Sol-17: (a) x = - 2 is point of maxima. f(x) = x2 e-x f'(x) = -x2.e-x + e-x. 2x For maxima or minima, f '(x) = 0 ⇒ X = 0, 2 f"(x) = x2e-x - 2xe-x + 2e-x - 2xe-x = e-x [x2 - 2x + 2 - 2x] = e- x [x2 - 4x + 2] f"(0) = 2 > 0 ⇒ ⇒ minima. ⇒ m ex 1 +ex (1 +ex ) ex -ex (e' ) (1 +ex )2 f(x) is monotonically increasing function. Sol-20: (b) Given, Y = x2 dy = 2x > 0, Vx E [1, 5] dx y is increasing in the interval. Sol-21: (a) 2 Ymin = ( 1 ) = 1 f(x) = x2 - X - 2, XE f'(x) = 2x - 1 = 0 X Maxima at x = 2 Sol-18: (d) = [-4, 4] 1 2 E (-4, 4) 1 - 4 2 9 f(x) 18 -- 10 4 X Let height (AM) of the cone is h meters. Volume of the cone 4 3 ex > 0 f '(x) = -(1 +ex )2 f"( 2) = e-2[+4 - 8 + 2] = - 2e-2 < 0 OA = OB = 1 m Radius of base of right circular cone is r = BM = (OB2 - QM2 )½ r = (1 - (h-1)2 )½ r = ( 2h - h2 )½ f(x) = f'(x) = = - 2, f"(-2) = -30 < 0 2:(4h-3h2 ) = 0 3 -4 Maximum value of f(x) = 18 Sol-22: (c) f(x) = sin x e_ _ = esin x-cos x cos e x f '(x) = esinx- cosx [cosx + sinx] ... (1) . 2 . & f"(x) = esin x -cos[(cos X +sin X ) +(cos X - sin X )] Engineering Mathematics f(x) has minima at x =0 For critical point f'(x) =0 ⇒ Minimum value of f(x) = f(O) = e0 + e---0 cos x + sin x = 0 = 1 +1 tan x = - 1 X ·: f"( n) < 0 3 4 ⇒ Maximum value = Sol-23: (b) =2 Alternative Solution: 3n = -� & 4 4 A - M- <'. G • M · ex +e-x 3 maximum at x = ; t( 3n)= [ esin -cos x 4 f(x, y) = x 2 x] 2 ✓ X= 3n = e 4 - y2 fX = 2x ' fy =-2y fxx = 2 • fxy = 0 ' fYY = -2 ⇒ 2x = 0 & 2y = 0 4).2x X = 0, -2, 2 f"(x) = 4[3x2 - f"(0) = - 16 < 0 Sol-25: (a) 4] ⇒ f(x) = + e x - f'(x) = ex - e-x = 0 ⇒ eX = e-X X = 0 f"(x) = ex + e-x f"(x) = 1 + 1 > 0 - 4 8x2 + 4 8x = 0 f"(0) = 4 8 > 0 =0 ⇒ f"(2) =0 & f "'(2) Sol-27: (c) y = x2 maximum f"(-2) = f"(2) =32 > 0 ex 16x3 + 24x2 + 37 · - - 4 8x3 - 4 8x2 + 4 8x 96x2 + 4 8x ⇒ x = 0, 2, 2 f(x) has local minimum at *0 so x = 2 is saddle point Hence we have only one extreme value / extrema. For maxima or minima, f '(x) =0 ⇒ ·: X ·: 4)2 - f'(x) = 12x3 12x3 So, origin (0, 0) in the saddle point f'(x) - Now for turning points f'(x) =0 At (0, 0), rt - s2 = (2)(-2) - (0)2 =-4 < 0 = 2(x2 f(x) = 3x4 f"'(X) = 72 X - 96 So P(O, 0) is stationary point - Sol-26: (b) f"(x) = 36x2 Now, for critical points fx = 0 & fy =0 f(x) = (x2 ex + e-x <'. 2 Minimum value =2 . .. (1) & Sol-24: (b) 253 ⇒ minimum Put and - d2 y dy 6x + 9, - = 2x - 6 , -2 = 2 dx dx dy =0 dx ⇒ x = 3 is the critical point f"(3) = 2 > 0 f(x) has a minimum at x =3 :. Hence Maximum value of y occur at corner points i.e. :. y(2) = 1, y(5) = 4 The maximum value of y =4 IES MASTER Publication 254 Maxima and Minima Sol-28: (c) For critical points, f'(t) = 0 ⇒ t = 0 where y 3 2 y = ax +bx +cx+d ⇒ 1 f"(0) = - - < 0 3 f(t) is maximum at t = 0. Sol-31: (a) Given Let a > 0 y = ax3 + bx2 + ex + d at ax at y (X➔ oo) ➔oo dy dx ay y (X➔ -oo ) ➔ -oo = 0, gives two roots, Hence A cubic polynomial with real coefficients can have maximum two extremum and three zero crossings. Sol-29: (a) Let the point on the surface z 2 = 1 + xy be P (x, y, z) such that the distance between P and origin is minimum : Then, the distance between origin to P, Now, R = ✓(x- 0) +{y- 0 ) +(z- 0) 2 2 2 u = R2 = x2 + y2 + z2 = x2 + y2+ xy+ 1 ( ·. - z2=xy+1) au = 2x + y ax ...(1) au = 2y + X ay ...(2) Solving equation (1) and (2), we get x = 0, y = 0. a2 u r = = 2 Also ax 2 a2 u a2 u = 1, t = 2 = 2 s = and axay ay 2 > 0 and r > 0 rt - s Hence, 'u' is minimum at x = 0, y = 0 and z = 1. Therefore, the minimum distance, Sol-30: (d) R = ✓0 2 2 +0 2 +(1) = 1 sint t cos t -sin t f(t) = --, f'(t) = t t = Bx - B = 12y - 4 a2 f = 12 ay2 so it can have two extremum. Let 2 2 - Bx - 4y+ B f(x, y) = 4x+6y a2f = 0 axay Points of extrema are given by at ax ⇒ X = 1 at - =0 ay y = 1/3 Now, (�)(a2fJ-( ax2 = 0 and ay 2 and a2 f J axay 2 = B x 12 - 02 = 96 > o a2 f = B> 0 ax2 So, by using Lagranges conditions : 1 .-. At x = 1 and y = 3 f(x, y) is minimum and its value is fmin = 4 X 12+ 6 2 = 4 -3 Sol-32: (a) = 10 3 1 eY = xx u-r - B (1) - 4(-¾)+ B Engineering Mathematics ⇒ Taking logarithm on both sides. y = dy dx dy dx ⇒ logx log X = - X X 1 x-- log x (1) X x 2 f"'(x) = 6 f'(x) and f"(x) is zero at x = 0 and f"'(x) is positive hence it is a point of inflection. Sol-36: (c) d2 = -ve < 0 ( ;) dx x=e y has local maxima at x =e f(x) = 2x - ⇒ x2 + f '(x) = 2 - 2x 3 ⇒ X X 1 2 4 6 f(x) 2 1 25 21 4 1 Maximum value of f(x) = 4 1 Sol-37: (b) dy d2 y y =2x - (O. 1)x2 , - = 2-O.2x , -2 = -0.2 dx dx and ( =1 f"(x)lx=1 = -2 as f"(x) is negative, the function f(x) will have maxima at X = 1. Sol-34: (c) f(x) = x3 - 9x2 + 24x + 5 f'(x) = 3x2 - 18x+ 24 For maxima or minima, f'(x) =0 x2 - 6x + 8 = 0 X = 2, 4 f( 1) = (1)3 - 9( 1)2+ 24(1)+ 5 =21 f(2) = (2)3 - 9(2)2+ 24(2) + 5 = 25 f(4) = (4)3 - 9(4)2+ 24(4)+ 5 =21 f(6) = (6)3 9(6)2+ 24(6)+ 5 =4 1 - The maximum value o f f(x) in interval [ 1, 6] is 4 1. Sol-35: (d) Given, ⇒ ⇒ Now, ⇒ f(x) = x3 + 1 f'(x) = 3x2 =0 x = 0 is a point of maxima or minima. f"(x) = 6x f"(0) = 0 9x2 + 24x + 5, x E[l, 6 ] X = 2, 4 E (1, 6) For maxima or minima f '(x) = 0 - x2 - 6x + 8 = 0 x =e is a critical point. Sol-33: (c) ⇒ ⇒ f(x) = x3 f'(x) = 3x2 - 1 8x+ 24 =O 1-logx = =0 x2 log x = 1 and ⇒ = 1 255 ⇒ y) d2 dx2 dy = 0, x = 1 0 is a critical point dx x=1 0 = -0.2 < 0 y is maximum at x = 10 & Maximum height = y =2( 10) - (0.1)(100) =1Om Sol-38: (c) For eg.: Let us take n = 2 and A = [ � � = ] symmetric and x = : ] and b = [ ;] and c = [ c ] 1x1 [ : So, XT Ax = [X1 X2 1 [ � � ] [ : ] : = [ix� +2mx 1 x2 + and So, rx�] x1 1 bT x = [ a �] : ] = [ ax 1 + �x 2 k1 [ : So f is a function of x 1 and x2 For minimum value of f(x) we must have at = 0 8x1 IES MASTER Publication 256 Maxima and Minima ⇒ 2I x1 + 2mx 2 + a = 0 of and ⇒ ... (ii) = 0 2mx 1 + 2rx2 +P = 0 Sol-39: (a) Hence, f(x) will be minimum when sin ( 1 00x) = -1 or -b 1 00 X f(x) = xe- x, x E (0, oo ) ⇒ f'(x) = 0 x = 1 is a critical point. f"(x) = (x - 2)e-x f(x) has maxima at x = 1 1 e 1 f'( t) = -e-1 +4e-2 For critical points f'(t) = 0 ⇒ is a critical point. and ⇒ x = 1 .. A e-t = t = log84 f "(t) = e-1 - 8e-2I (minima) f" {log0 4) = e-1o9, 4 _ 8e-21og,4 f(x) has minima at x = 1 . = f(x) = x3 - 3x2 - 24x + 1 00, x e [-3, 3) f'(x) = 3x2 - 6x - 24 = 0 x2 - 2x - 8 = 0 X f(x) X = 4 � [-3, 3] 118 -2 1 28 Minimum value of f(x) = 28. 3 28 i- C�) 8 f(t) has maximum value at Sol-44: (0) = 4, -2 -3 1 4 et = 4 --=-Y.±_ ►� f' W) > 0 X Either 4e-1 - 1 = 0 � f'(r1 ) < o Sol-41: (b) ⇒ = 00 (Not possible) 1 So, f'(x) = 0 3n = 200 Sol-43: (a) or f(x) = (x - 1 )213 f'(x) = � (x - 1 r '3 3 X e-t = 0 Maximum value = f( 1 ) = 1 .e-1 = - . Sol-40: (c) 3n 2 ⇒ e _, ( 4e-1 - 1) = 0 f"( 1) = -e-1 < 0 ⇒ = -1 .. 3n/200 minimum f( x) = e e= -0 . 954 f'(x) = -xe-x + e-x = ( 1 - x)e-x So, sin 1 00x ex e-x has always +ve value so product will be minimum only when sin( 1 00x) will be minimum. 1 -A - - b = 2 X f(x) = e-x sin ( 1 00 x), x � 0 = ...(iii) Matrix form of (ii) and (iii) will be 2Ax = Sol-42: (-0.954) t = log84 f(x) = ln( 1+x)-x, x >- 1 f'(x) = ⇒ 1 = -- < 0 4 _1_ _ __ x_ 1= 1+x 1+x f'(x) = 0 X 1+X X = 0 = 0 Engineering Mathematics f"(x ) = - [ ⇒ f(x ) has maxima at Sol-45: (6) Method II (1+x )(1)-x (1) ] 2 (1+x ) f "(0) = -1 < 0 x h + y = K, K is constant Let A be the area of triangle. A = 0 f(x ) = 2x3 - 9x2 + 12x - 3, x f '( x ) = 6x2 - 1 8x + 12 = 0 X = 1, 2 E (0, 3) 2 = k (constant) ⇒ y = ,Jk2 -2kx Hence, du . F or max imum u, = O dx ⇒ ⇒ X = tan e = 3 x k/3 3 X< X� -1 , X < O f '(x ) = { 1 , x � 0 ⇒ x 0 0 = 0 is a critical point f(x) has minima at Sol-48: (d) x = 0 f(x ) has a local minimum at f'(x0 ) > 0 Sol-49: (1) k 3 y_ = k / ✓ = ✓ ⇒ if if x f '(O - ) < 0 and f '(0+ ) > 0 0 kx -(k-3 x ) = 0 2 Now By (1 ), So 2k2 x 2 = � kx = 4 l X f(x ) is not differentiable at x = 0 ...(2) u = ⇒ :t1 :':,�:2) f(x) = Ix I= {� So so K 1+sin0 ? ° 0 = 30° so angle between h & y = 60 Sol-47: (a) f(x) = Ix! I n graph we can see that f(x ) = l xl f(x ) has minima at x = 0 Alternative Solution ...( 1 ) 1 Now area = - x x x y 2 Let h2 -sin20 4 K K2 (1+sin e)2 (2cos20) - sin 0.2cos0(1 + sin e) = ] 0 4 4[ (1 + sin e) ⇒ y 2 + ,Jx +y ..! xy = ..!_(hcos 0)(hsin0) 2 2 For max imum Area :: = 0 Method I x = = h = ⇒ X = A = Max imum value of f(x ) = 6. Sol-46: (b) ATQ h+h sin e ⇒ f(x ) -3 2 1 6 e A ⇒ E [0, 3] 0 1 2 3 X 257 0 = 60 ° x = x0 if f'(x0 ) = 0 and Let 2x, 2y be the length and breadth respectively of the rectangle inscribed in the ellipse x2 + 4y2 = 1 ⇒ Let Area of rectangle = (2x )(2y) = 4x y f = (Area)2 = 1 6x2y2 = 4x2 (1 - x 2 ) IES MASTER Publication 258 Maxima and Minima f'(x) = Bx - 16x3 = 0 ⇒ X ⇒ 1 = 0. ± 2 ✓ �) -2 (minimum) 2 f(x) = x -4x+2 = 8 - 12 < 0 2 ⇒ f is maximum at x = 1/✓ 1 1 x = 2, Y = 8 At ✓ ✓ d2 y The curve is concave up if -2 > 0 dx d2 y - = 12x2 > v x (valid) Here dx2 Hence the curve is concave up for all values of x. f(x} = x(x - 1)(x - 2) 3x2 + 2x Given 3x2 - 6x + 2 = 0 f(1) = f(2) = 0 f(1.577) : 1.577 X 0.577 = -0.385 X -0.423 The maximum value attained by f(x) in interval (1, 2] is 0. Sol-52: (-5) f(x) = 2x3 - 3x2 in x E (- 1, 2] ⇒ f'(x) = 6x2 - 6x = 0 (for maxima or minima) x = 0 or 1 So total points of maximum or minimum are x = 0, 1, -1 and 2. Now, f(0) = 0 f(1) = 2 - 3 = -1 f(- 1) = -2 - 3 = -5 = x3 Critical points are x - 3x2 + 1 = 0 and 2 while corner points are - 1 and 3 so sign of f'(x) is : � -1 0 f(x) = 2x3 - 2 x = 1.577, 0.423 are two critical nd points but 2 does not lie in the given interval so we will neglect this point. Now, f(x) f'(x) ,= 3x2 - 6x = 3x(x - 2) Sol-55: (-13) For maxima or minima, f(x) = 0 ⇒ ⇒ Sol-54: (b) 1 f(2)=- 2 I 2 3 If we look at f(x) at x E [-1,+3] ; it is clear that f(x) first increases then decreases and again increases. = x(x2 - 3x + 2) - x = 2 is critical points. and min value = f(2)= (2)2 - 4 (2) + (2)= 6 - 8 = -2 Sol-50: (a) = x3 f' (x) = 2x- 4= 0 Now, f"(2) = z > 0 so x = 2 is point of minima Max. area = 4 (�) (Js)= 1 Sol-51: (0) Global min value = (-5). Sol-53: (d) f"(x) = 8 - 4 8x2 f ( · vL.2 f(2) = 16 - 12 = 4 f'(x) = 6x - x4 - 4x3 , 10, f "(x) = 12x - 12x2 , For C.P; x = X f'(x) = 0 6x2 - 4x3 = 0 ⇒ ⇒ f "'(x) = 12 - 24x x = 0, x = 3 2 ;t (- 1,1] = 0 3 are stationary points 2 So we have only one C point i.e., * f"(0) = 0 and f m (0) 0 So x = 0 is a point of inflexion So Minimum value occur only at corner points i.e., So Min. f(-1) = -13 & f(1) = -9 f(x) = - 13 Engineering Mathematics 259 Sol-56: (c) = 2x(k2 f'(x) and ·: x = - 4) + 1 8x2+ 32x3 2x[k2 - 4 + 9x + 1 6x2] = 2k2 ⇒ f"(0) < 0 ⇒ 2k2 - 8 < 0 ⇒ (k - 2) (k+ 2) < 0 ⇒ - 2 < k < 2 Sol-57: (-100) 2 f(x) = _:l_x (x - 3) 3 Its graph will be f(x) df - = 3x2 - 3 = 0 dx So, local maxima & minima occurs at x = 1 and - 1 and gives y = 2/3 and - 2/3. As it is in free fall below x = - 1 , so minimum will occur at x = - 1 00. :v=v: f'(x) = 3(x - 1 )(x+ 1 ) or - -1 max 1 min � So, from x = - 1 00 to x = - 1, function is an increasing function. Hence, minimum value occurs at x Sol-58: (a) f(x) = 3 df x -x then dx 3 For Critical Points At x = At x = So, x = = - 100. 2 d f x2 - 1 • dx2 df =0 ⇒ x dx = = 2x ±1 d2 f = 2+ ( ve) i.e., Minima occurs 1, dx 2 d2 f = -2 (-ve) i.e. Maxima occurs. - 1, dx2 = 1 is Point of Minima. 12000 . . Cost funct I0n Is Cr = 2x+ -- + y+5 xy To find optimum value of Cr we have 96x2 - 8+ 36x+ f "(x) 0 is a point of local maxima = Sol-59: (91.5) f(x) = (k2 - 4)x2+ 6x2+ 8x4 8Cr 8�T = 0 and =0 x2y = 6000 ...( 1 ) ⇒ xy2 = 1 2000 .. .(2) and So, x : y = 1 : 2, i.e., we can take x = k & y = 2k. So by ( 1 ) (k2)(k) = 6000 ⇒ k = 1 4.4 ⇒ X = 1 4 .4 & y = 28.8 So optimum value of Cr = 9 1 .5 Sol-60: (d) Graph of f(x) =x3 is Hence at x = 0, f(x) has Neither local maxima nor minima i.e., it is point of inflection. Sol-61: (12) f(x) = 3x2 - 7x2 + 5x+ 6 in [0, 2] and f'(x) f"(x) = = 9x2 - 1 4x+ 5 1 8x - 1 4 Critical Points are f'(x) f"( 1) = = +ve so x 0 = ⇒ x = 1 and 5 9 1 in point of minimum is point of maximum f " ( ) = -ve so x = % 9 and value = f( ) = 3 rnr _ 7 rnr +5 [iJ+6 % = 7. 1 But maximum value can also occur at corner points. So f(2) = 3(2)3 - 7(2)2+ 5(2)+ 6 = 12 So absolute maximum value = 1 2 IES MASTER Publication 2.6 PARTIAL DERIVATIVES The partial differential coefficients of f(x, y) are defined as : �= fx = lim ax h➔ o � = f = lim Y k➔ o ay And f(x+h, y)-f(x, y) h when y =constant f(x +y + k)-f(x, y) k when x constant So, the partial differential coefficients of f(x, y) with respect to x and y are ordinary differential coefficients of f(x , y) when y and x are regarded as constant respectively. The other partial derivatives are as follows : -�(�) - t a2 f ay2 ay ay 2. az ) a z * (ax 3· ( 2 ax 2 ( 4. Solution + :� �r 2 = - yy , -�(�)-t a2 f axay ax ay yx or * 2 (az )2 a2 z &y ay ( :� r 2 + (�r 2 +2 2 (:�}(�J a a a a a ) = 2+ +2 2 ax ax + ay axay ay . . 5. . . au 1s d er1vat1ve w h I' I e a 1s an operator. 6. Normal multiplication rule i s not appl icab le i n case of operators. i .e. ax ( Example 1 2 _�( at ) - a2 f - fxy • ayax ay ax ax 2 � � � u � � � au au au ) + + = ( + + )( + + � fJy fu ) � fJy fu � fJy fu a2 t a2 f If f = tan -1 y_ , verify that -- · =--. ayax axay x We have; () f = tan-1 (t) Differentiating (i) partially w.r.t. x and y, we get at = ax 1 ...(i) y) )2 (--;! y_ 1+( X = x 2 + y2 ...(ii) m- : 1+(� )' Eng ineering Mathematics 2 2 ,i f = j_ (� ) = (x +y )(-1) - (- y )(2y ) = ayax ay ax (x 2 + 2 ) 2 y 2 2 2 j_ 8f = a f = (x +y )(1 ) - x( 2x) ( ) ax ay axay (x 2 + y 2 ) 2 From (iv) and (v), we have -- = -­ ayax axay a2 Example 2 : Solution If u = sin -1 ( � ) +tan - 1 y We have; au ay X Now, Solution : X = = 1 ✓1 - (x/y) 1 ✓1 - (x/y) au au = +y ax ay 1 ·Y 2 then show that x 2 ✓y ( 2 1 1+( y /x) ( 2 X x -x 2 _ 1 yx x +Y 2 ✓Y 1 =·x ✓ . . . (iv) (x 2 +y 2 ) 2 2 2 y -x 2 2 2 (x +y ) ...(v) 2 X 1 - x2 x - Y ✓l - x 2 y -x 2 (t) + 2 . . . (i) 2 2 x +y y + 2 2 x +y X Y x x +y 2 2 =Q = a2 If z = f(x +ay ) +�(x - ay ) , prove that . ay 2 ax2 z = f(x +a y ) +�(x - ay ) We have; a2 z - = f'(x +ay ) +�'(x - ay) ax az B y (ii) and (iii) ⇒ = a2 2 ay ax2 a2 z a 2z a z = f"(x+a y )+�"(x - ay ) ax 2 2 ⇒ az Solution : 2 2 - = af'(x+a y )+(-a)�'(x - ay ) ay Example 4 : y = ) --;z - 7 J+ 1 +(y/x) 2 2 y -x au au . =o +y y ax 8 u = sin -1 ( � ) +tan -1 + = f Differentiating (i) partiall y w.r.t. x and y, we get au ax Example 3 : (Y), a2 f . . . (iii) x' : y' a2 z = a2 f"(x+a y )+a 2 �"(x - ay ) 2 ay ⇒ . . . (i) . . . (ii) ...(Ill" ' ) a2 z a v a v a v If V = (x2 + y2 + z2 )-1 I2 , show that -+-+2 az2 = O . ax2 ay We have; 261 2 v 2 2 = (x2 + y2 + 22 )-112 Differentiating (i) partiall y w. r.t. 'x', two times, we have av = - � (x 2 +y 2 + z2 r3/2 (2x) = -x(x2 + y 2 +z 2 r3/2 ax 2 2 2 2 �� = - [x {- % (x + y +z r5/2 (2x)} +(x2 +y 2 +z 2 r3 /2 ] . . . (i) IES MASTER Publication 262 Partial Differentiation = 3x2(x2 + y2+ 22 )-s,2 _ (x2 + y2+ 22 )-312 = (x2 + y2 + 22 )-s,2 [3x2 _ (x2+ y2 + 22)) ... (ii) = (x2+ y2+ 22 )-s,2 (2x2 _ y2 _ 22 ) Similarly, we can find a v = (x2 + y 2 + 22 r5/2 (2y2 - 22 - x 2 ) _ 2 ay 2 . . .(iv) a 2 v a 2 v a2 v . +=0 +11 , ("') 111 and ('1v ), we get Addmg (") 2 2 2 ax Method II ...(iii) ay az Let r2 = x2+ y2 = 22 Differentiating (1) partially w.r.t 'x' 2r� = 2x + 0 + 0 ax ar i .e. ax v Now Similarly Solution I = (x2+ y2+ 22)-1,2 = (r2)-1,2 = _1 r -[J__ r 3 a2 v 3y 2 = 5 ] and r az2 - [J__ r 3 322 ] r5 3 ( x 2 + y 2 + 22 ) 3 3 3 � ---=---� + =0 =- -3 + r rs r3 r3 Now Example I I ar y ar 2 X = - . Similarly, - = - and - = r ay r az r au a2u 1 . y2, show that -2 + -2 = f"(r)+ -f'(r). If u = f(r), where r2 = x2+ r ax ay We have; r2 = x2 + y2 . . . (i) ar 2r-=2X ax Differentiating both the side w.r.t. 'x', we get ar y Similarly, we can find ay = u = f(r) Now, ⇒ au ax r = i! =f'(r) � ar ax r ⇒ 2 a2u 1 /f"(r)-Lf'(r)] -= rf'(r)+ [ r ay2 r2 Solution 2 2 2 2 a u a u J_ [2rf'(r)+(x2+y2)f"(r)- (x +y ) f'(r)] = + 2 r ay2 r2 ax 1 1 ar ax x r -=- ...(ii) ...(iv) ...(v) +r2f"(r)-rf'(r)] = -f'(r)+f"(r) = J_[2rf'(r) r r2 1 u 222 x = (1+3xy2+x Y 2 ) e yz If u = exyz· show that axayaz a 263 ...(iii) x2 1 a2u =rf'(r)+x2f"(r)- - f'(r)] 2 2[ r ax r ar Adding (iv) and (v), we get Example 6 ⇒ au r [f'(r)+xf"(r) i!] -xf'(r) a a2 ax ax ) f'(r)� �= = ( 2 2 r ax r ax Similarly, Engineering Mathematics I 3 ...(i) We have; a a au a2u -=xye xyz --=-[xyexyz ] =x-(y e xyz ) = ay az · ayaz ay 2 �=�-( a u )=�• [e xyz (x2y2+x)] = exyz(2xy2 + 1 ) + y2exyz (x2y2 + x) axayaz ax ayaz ax Example 7 Solution = exyz (2xy2 + 1 + x2y222+ xy2) = exyz (1 + 3xy2 + x2y222) 1 1 If u = log(x3 + y3 + 23 - 3xy2); show that We have; u = log(x3 + y3 + 23 ⇒ au ax - 3xy2) (a ax + a +a ay az ) 2 u= 9 (x+ Y+ 2)2 . 3 22-3xy " au 3x2- 3y2 3y2-3x2 ' au = 3 = ...(iii) ) ( ( ... 1) ... 11 ' az x ·+y3 +23 -3xyz ' ay x 3 +y 3 +23 - 3xy2 x 3 +y 3 +2 3 -3xyz Adding (i), (ii) and (iii), we get 3 3(x2+y2+22-xy-y2-zx) au au au 3(x2+y2+22-xy-yz-zx) �� -+-+-=-C..-.,.C--� = = 2 3 2 3 3 2 ax ay az (x+y+2) (x+y+2)(x +y +2 -xy-y2-zx) x +y +2 -3xyz 3 ) U=(�+�+�)(�+�+�) u = (�+�+�)( ) (�+�+� ay ax az ax ay az ax ay a 2 ax ay az x+y+2 2 = 3 + = + [!C+ :+2) :(x+ :+2) !(x+ :+2)] (x+ �:2)2 Example 8 Solution 1 1 a2 2 If xxyY22 = c, show that at x = y = 2 axay = -(xlog ext1• We have; xxyY22 = c, where 2 is the function of x and y. Taking log both the side, we get IES MASTER Publication 264 Partial Differentiation xlogx + ylogy + zlogz = loge Differentiating (i ) partially w.r.t. 'x', we get (1+Iogx) 8z. ⇒ -=ax (1 +logz ) ( x�+logx )+ ( zi+log z):= 0 8z. (1+Iogy) . . . =Sm 1 1larly, we can find ay (1 +logz) 2 ; Solution 1 ! ·[ ��:::���] ( )= : = -(1 +1ogy) ! (1+1og zr 1 (1+Iogx) (1+Iogy ) ] [ (1+Iogz ) z(1+1ogz) 2 2 a u a u a u 2 If u = (x + y + z ) 11 then prove that - + - +-=- 2 2 2 2 2' u = We have; ax2 (x2 + y 2 2 + ay2 22 )112 2 oz2 u· 2 32 ay 2 a 2u _ = (x + y + 22) -31 (x + y2) az2 Adding (ii), (iii ) and (iv ) , we have 2 2 2 2 a u _ a u _ a 2u _ + + = (x + y + 2 ) -31 (2x + 2y + 2z ax ay 8z. 2 Now and 2 ...(i) =x[-!(x + y + z r3/ .2x] + (x + y + z r 1/2 = -x (x + y + z ) -3/ + (x + y2 + z ) -1/2 2 ...(ii) = (x + y + z2) -3i [-x2 + (x + y + z )] = (x + y + z )-3/ (y + z ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 a Similarly, � = (x2 + y2 + z2)- 1 (x2 + z ) ax 2 Let ...(iii ) -1 a z (1+1ogx ) ---=-[xlog(ex )r1 At X = y = z, --== = xlog(ex) ax.B y x(1 + logx ) x(1 +logx)3 a 2u Method II ...(ii) = -(1+1ogy ) !(1+1ogzr1: = -(1+1ogy{-(1+1ogzr2 i:] = 2 Example 9 = = ! ! Now, ...(i) 2 2 2 2 2 2 2 2 r2 = x2 + y2 + z2 then - = ar ax 2 X ar r ' ay u = (x2 + y2 + 22) 112 = (r2) 1,2 = r au ax = -=ax r ar = X !(:)= !(�) = 2 2 2) = _____ = 2 2 2 2 1 / U (x + y + z ) y ar z = ' r az r 2 ...(iii) ...(iv) 2 Similarly 1 r uw = y2 1 and u zz = r r 3 Engineering Mathematics z2 r 265 3 2 2 ! 3_ 3_ � x +y + z2 ) = �- = = uxx + uw + uzz = -( r r r r r u Now 3 Example 10 1 If x = r cos0, y = r sin0 then prove that -- = -2 (logr) = --2 (logr)=-2cos20 axay ax r ay Solution 1 a20 a2 a2 We have; x = r cos0, y = r sin0 then r2 = x2 + y2 and 0 a0 ay = 1 1 = x2 1 = x · ; x x 2 +y 2 ·x x2 +y2 1+x2 a20 a a0 = tan -1 ⇒ axay = ax ( y) = a 1 (f) y 2 -x 2 1-(x2 +y 2 )-2x-x = (x2 +y 2 ) 2 (x 2 +y 2 )2 ...(i) a a a 1 x 1 12 2 1, 2 ·2X= Now, - (logr)=-log(x +y ) =-· -log(x 2 +y 2 )] = -·2 ax ax ax 2 2 x 2 +y 2 x +y2 [ 2 2 2 2 _ y -x _ }_ (}_ logr )- 1-(x +y )-2x-x � (logr) . 2 y2 )2 ax ax (x 2 +y2 )2 · (x ax 2 + ...(ii) 1_, Now, }_(logr) =}_[.!.1og(x 2 +y2 )] =.!.. _ 2y= 2 Y 2 ay 2 2 x2 + Y 2 ay x +Y Now, a2 a x2 - y 2 y 1-(x 2 +y 2 )-2y-y = - --- = �--,,--'---'-,---,--'------'-=-- --'--(logr) ay2 (x 2 +y2 )2 oY x2 +y 2 ) (x 2 +y 2 )2 ( r 2 cos2 0 -r 2 sin2 0 cos2 0 -sin2 0 = = (x 2 +y 2 )2 r 4 (cos2 0+sin 2 0)2 x2 -y 2 From equation (i), (ii), (iii) and (iv), we get I cos20 r2 a2 0 = a 2 a2 1 = (logr) (logr) =2 cos20 -, ;:,,.. 2 2 aXvy 8X r 8y . . , (iii) ...(iv) HOMOGENEOUS FUNCTIONS A function of the form a 0yn + a 1xyn - 1 + a2x2 yn-2 + ... + anxn is an expression in x and y of degree n. Such a function is known as homogeneous function of x and y of degree n. Now, f(x, y) = ao yn + a1xyn -1 + a2 x2yn -2 + ... + anxn 1 2 n = x [a 0 +a 1 +a 2 +...+an ] = xn f(y/x) (�J (fJ- (fJ- So, every homogeneous function of x and y of degree n can be represented as xn f(y/x). n Note : We can also write f(x, y) as follows; f(x, y) = y F(�). Tricky Method to Check, whether given function is Homogeneous? If f(Ax, ;\,y) =;\, n f(x, y) then f(x, y) is called hpmogeneous function of degree 'n'. where n ER (y) 2 2 3 · Example 11 : f(x, y) = x y +xy +x sin x IES MASTER Publication 266 Partial Differentiation 3 2 3 2 3 3 2 3 3 2 f(1vx, 1vy) = 1v x y+1v xy +1v x sinG�) = 1v [ x y+xy +x sin(t)] = 1v3 f( x,y) So, f( x , y) is a homogeneous function of degree n. y 3 1 Example 12 : f( x, y) = �- + dlogx-logy] Y f(1v x, 1vy) = 0 1 A 2 x2 - A �\ + 3 1v 2 y2 [ 3 y "'x log( )] = � [ � - 3 + 2 [log(�)]] ="' 2 f( x y)· 1vy 1v x x y Y ' ' So, f( x , y) is a Homogeneous function of degree -2. Example 13 1 f( x, y) = 1 x3 1 x4 x y3 1 .!.! .!.! .!.! .! .![ .! .!.!] +y 4 - 1 _1 1v 3 x 3 -1v 3 y 3 "' 3 x 3 -y 3 A .! =� .! f(1v x, 1vy)= = 12 f( x, y) 1v 4 x 4 +1v 4 y 4 Example 1, "' 4 x 4 +y 4 So, f( x , y) is a Homgeogeneous function of degree 1 f( x , y) = x 2y2 + x y3 + sin( x4) * "' 4 1 12 . · f( x, y); so it is non-homogeneous function EULER'S THEOREM ON HOMOGENEOUS FUNCTION Let f( x , y) be a homogeneous function of degree 'n' then, x -+y- af = nf ay af ax Corollary 11 If f is a homogeneous function of degree 'n' then 2 at a2 f a2 f =n(n - 1) f +2 xy--+y 2 axay ax ay 2 x 2 Corollary 21 If V = f(u); where V is a homogeneous function of degree 'n' then (a) (b) x x au au nf(u) = +Y ax ay f' (u) 2 2 a 2u a 2u 2 a u - +2 xy-+y - =<)>(u) [<)>'(u)-1 ] axay ay 2 ax2 where <)>(u) =�� �{and <)>'(u) means differentiation w.r. to 'u'. ( Note : Here 'u' is not a homogeneous function. Corollary 31 If f be a homogeneous function of n variables be Example 1& Solution 1 1 x 1 -+ x2 af ax1 · at af -+... +x n -=nf ax n ax2 3 3 x +y If u = log-2--2 X +y We have; , x 1, x 2 , ... , x n au au prove that x-+ y- =1 . ax By x 3 +y 3 u = log -2 --2 X +y Here z is homogeneous function of degree 1 in ⇒ x of degree n, th_en the Euler's theorem will e = u 3 +y 3 = X +y x 2 2 z (say) ...(i) and y, so by Euler's theorem, we get oz. az ⇒ Example 16 Solution : 1 X-+y -=1 ·Z ax ay ⇒ oz. Engineering Mathematics 267 oz. au au X-•-+y-•-= Z au ax au ay au au ⇒ x-+y=1 ay ax Verify the Euler's theorem for function u = ayz + bxz + cxy. We have; u = a y z + bxz + cx y i .e ., the homogeneous function of degree 2 in x, y and z au au au x-+y-+z-=x(bz+cy)+y(az+cx)+z(ay+bx) = 2(ayz + bxz + cxy) = 2u ax ay oz. So, the Euler's theorem is verified. 1 1 Example 17 : Verify Euler's theorem for the function u = sin- �+tan - y_ . Solution : y X 1 1 u = sin - �+tan - y_ We have; y X u(1cx, 1cy)=1c0u(x, y) so, u is the homogeneous function of degree zero Hence, by Euler's theorem will be au au x-+y -= 0 ax ay ...(i) ...(ii) Verification: Differentiating (i) partially w.r.t. 'x' and 'y', we get au 1 1 1 = + ( 2 2 Y ax /t 1 + x2 1 -x ty � = ✓ ✓1- -7 ) y - 2 + ( ) 2 y ( ; ) 1+y�/x � :2 t 2 From (iii) and (iv), we have au au + = ax Y ay Example 18 : If u = sin - 1 Solution : We have; ✓ 2 x 2 +y 2 x+y 2 y au ax ⇒ -=--===+--- au ay ✓y 2 -x 2 X 2 y /-x ✓ x ____.!'/___ _ +____.!'!___ =0 2 2 2 2 / x + x + / -x -x y Y So, the theorem is verified. x x 1 ⇒ x +y 2 2 X x +y 2 2 ...(iii) ...(iv) ✓ au au show that x -+y-=tanu. ay ax x2 +y2 sin u = --=f (say ) x+y ...(i) Here f is the homogeneous function of degree 1 in x and y, so by Euler's theorem, we have a . au a . au . a . a . at at . x-+y-=1 • f ⇒ x -stnu+y -sinu=stnu ⇒ x-sinu -+ y -sinu-=stnu au ax au ay ax ay ax ay au au . au au ⇒ xcosu-+y cosu-=stnu ⇒ x-+y-=tanu ay ax ax ay x3 + 3 x- y au ax au ay . 1 Example 19 : If u = tan --y- , prove that x-+y-=stn2u, hence evaluate _ a 2u a 2u a 2u X 2 - +2xy--+y 2 -2 axay ay ax2 IES MASTER Publication 268 Partial Differentiation x3 + y3 tan u = --=f x-y We have; Solution (say) ...(i) Here, f is the homogeneous function of degree 2 in x and y. So, by Euler's theorem, we have of of a a a au a au x- + y-=2f ⇒ x -tanu + y-tanu = 2 tanu ⇒ x-tanu- + y--tanu- = 2 tanu ax ay ax ay au ax au ay au au 2 au 2 au ·2u sinucosu =sin ⇒ xsec u- + y sec u-=2 tanu ⇒ x- +y-= 2· ax ay ax ay ...(ii) 2 2 2 a 2u au au a u u a u au =2cos2u --;:;- ⇒ x a x-2 + + y + y--=(2cos2u-1)�, 2 ax a OX XOy ax ax 3y ax 3y ...(iii) Differentiating (ii) partially w.r.t. 'x' , we get 2 a2u a u "u =(2cos2u-1) � Differentiating (ii) partially w.r.t. 'y', we get x-,,,- + y2 oxay oy ay Multiplying (iii) by x and (iv) by y and adding, we get 2 2 2 a u a u a u au au x2 -2 + 2xy-- + y2 =(2cos2u-1) x- +y- ) 2 y . ax ay axa ax ay ( ...(iv) = (2cos 2u - 1) sin 2u = sin 4u - sin 2u TOTAL DIFFERENTIAL COEFFICIENT If u = f(x, y) and x = <j,(t), y = '-V(t), then u can be written as the function of t by putting the values of x and y in u = f(x, y). So, we can find the ordinary differential coefficient du/dt, which is said to be total differential coefficient of u w.r.t. 't' . But sometimes it is difficult to express u in terms of t by direct substitution. Then we use other method to find du/ dt. NOTE 1. If u = f(x1 , x2 , 2. Again if f(x, y) is the function of x and y; and y is some function of as x so the total differential coefficient of f w.r.t. 'x' will be 3. ... , xn ) and x1 , x2 , ... , au dx du au dx au dxz - =- -l +--+ ... +- -n dt ax1 dt ax2 dt axn dt df of at dy - =-+- ·dx ax ay dx If xn are all functions of t, then generaliz ed form will be [·: f ➔ (x, y) -> x alon e ] = f(x, y) and x = x(t) and y = y(t) i.e. known as total derivative and is given by z x, then y can be taken z ➔ (x, y)-> t, then derivative of ' z' w.r. to 't' is dz a z dx + az dy az z = or dz =(a )dx+( ldy dt ax dt ay dt ax ay) CHANGE TO VARIABLES If z az az and - are defined as = f(x, y) and x = 0(s, t) and y = \J,(s, t), i.e., z ➔ (x, y) ➔ (s, t) then at as oz oz ax oz ay oz az ax oz ay + = and -=--+-as ax as ay as at ax at ay at Here derivative of z w.r. to s and t is not total but partial. I NOTEe 1. If u ➔ (x, y, z) ➔ (r, 0, 0) then 2. au au ax au ay au az -=--+--+-ar ax ar ay ar az ar au au ax au ay au az -=--+--+--80 ax a0 ay a0 az a0 au au ax au ay au az -=--+--+-80 ax a0 ay a0 az a0 If u -> (x, y, z) ➔ 't' If u 4. If u ➔ (x, y, z) ➔ (r, 0) then 6. Change of variable concept du au dx au dy au dz -=--+---+- 1 dt ax dt ay dt oz dt Total derivative concept ou au au du= (-J dx+ (-) dy+ (-J dz j az ay ax 3. 5. Engineering Mathematics 269 = f(x, y) and y = 0(x), i.e., u ➔ (x, y) ➔ x then du= au dx au dy + dx ax dx ay dx . . l Total der1vat1ve concept du au au dy -=-+-·dx ax ay dx au au ax au ay au az = + + . ar ax ar ay ar az ar ) Change of variable concept au au dx au dy au dz -=--+--+-88 ax de ay de az de 2 a2 a2 cl a2 a2 a2 a a a +2+2(-+-+- ) =-+-+-+2aydz azdx axdy ax ay az ax 2 ay 2 az2 If P(x, y) be any point, then x and y are known as Cartesian coordinates. Now, {r, 8) are called Polar coordinates of P if x = rcos0 and y= r sin0 then we have 82 u a 2 u a 2 u 1 au 1 8 2 u -+= -+--+-8 r2 r ar r2 80 2 ax2 ay 2 where RHS is Laplace equation in Polar coordinates and LHS is Laplace equation in Cartesian coordinates. 7. Consider an implict function f(x, y) C ⇒ df = 0 ... (1) Now using total derivative concept is LHS of (1) 8. af dx+af dy=O ⇒ ax ay If we are not able to separate x and y then function is called implicit. If is of the type f(x, y) = C IES MASTER Publication 270 Partial Differentiation Example 20 1 Find Solution : dy if xY + yx dx We have; = c. xY + y x = c Here f(x, y) is the implicit function of x and y. ⇒ ⇒ (axat)/(atay) dy =dx We have; ⇒ So, where dy yx Y-1 +l logy =- [ ] dx x Y logx+xy x -1 Example 21 1 If u = x log xy, where x3 Solution 1 f(x, y) du du dx = + y3 (:) dx+(:) dy au = at = y-1 x at x-1 y ax y x +y logy, ay = xy +x log x 3xy + u = = 1 , find du/dx. x log xy au dy = -+-- ...(i) ...(i) ...(ii) ax ay dx au 1 Now' -=x--y+logxy= (1 +1ogxy) ax xy ...(iii) We also have; x 3 ...(v) X au 1 =x - · x =and xy ay Y y3 3xy 1 Differentiating (v) w.r.t. 'x', we get 3x 2 +3y 2 + + = dy dy +3 ( x +y•1) =0 dx dx ⇒ ...(iv) dy dx (x2 +y) 2 (y +x) du x (x 2 +y) Put all the values in equation (ii), we get - = (1+log x y)+- [- 2 ] dX Y (y +X ) ...(vi) Example 22 1 If the curves f(x, y) = 0 and qi(x, y) = 0 touch, show that at the point of contact Solution 1 �a4i_ata4i= ax ay ay ax 0 f(x, y) = O qi(x, y) = O ⇒ ⇒ :� = -( :X) /(:) ...(i) :� = -( :) /(:) ...(ii) When the two curves touch each other then at the point of contact the values of (dy/dx) for the two curves must the same. So we have -(:)/(:)=-(:)/(:) ⇒ : :-: : Example 23 , If x = r cose, y = r sme, prove that ( . Solution 1 ar )2 + ( ay ar )2 ax =0 =1 . ...(i) Engineering Mathematics Differentiating both sides w.r.t. 'x', we have 2r -=2X ar ax ar X ⇒ =- ...(ii) r ax Again differentiating (i) partially both sides w.r.t. 'y', we get 2r -=2y ay ar ar y ⇒ =ay r Squaring and adding (ii) and (iii), we get Example 2, Solution I 2 Differentiating r2 = x2 + partially w.r.t. 'x' and 2 y , 2, (x) ar r - x a ar ) a x 1-r-x r ax = = ( )= = ( ax r r2 r2 ax 2 ax ax a2 r �: = :(:l :m = Adding (ii) and (iii), we get = 1 r 2 x=r cose, y=r sine 2 y 2 ...(iii) a 2r . a 2r 1 ar + = ( + y then prove that { ax ax 2 ay If X = r cose, y=r sin0 or x2 ) ar = x2 +y2 1 = 2 +( ay r2 We have r2 = x2 + 1 = (axar ) y 2 ) r respectively we get 2 a 2r r-- x2 -= 2 ax r3 ⇒ 7fy ar ⇒ Now, we take R.H.S. l From (iv) and (v), we get �� +�� = �{ Example 21 Solution I 1 (!: x2 r: y r +(: 2 ar } ...(i) ...(ii) 2 a 2r r 2 -= -3 y 2 r ay ...(iii) .. .(iv) 2 2 +( + = = ( ( � { : r (:r } � { � r f r } � ( )2 +( ay ar X ar y = and = ay r ax r a 2r a 2r r - x r2 1 1 -- =-[2r 2 - (x +y )] = -+-= + r ax 2 ay 2 r3 r3 r3 2 2 271 = ) � ...(v) r} If u = f(y - z, z - x, x - y), prove that -+-+-=0 . ax ay az au au au Let - z = u 1 , z - x, = u2, x - = u3 then u ➔ f(u 1 , u2 , u3) ➔ f(x, y, z) so by using change of variable concept y y au au1 au au au3 au au +- 2 +- = ax au1 ax au2 ax au3 ax au au au au au2 au au3 - =- -1 +- -+ay au1 ay au2 ay au 3 ay au au au au au2 au au3 - =- -1 +- -+au3 au2 au1 az az az az ...(ii) ...(iii) ... (iv) IES MASTER Publication 272 Partial Differentiation We have, au1 ax =0 au 1 ' ay =1 au 1 ' az = -1 au 2 · ax = -1 au 3 au 3 au au au 2 3 = -1 =1 = 0. =0 2 =1 ' ay ' az ' ay ' az ' ax Put all the values in equation (ii), (iii) and (iv), we get au au au au =+ ax au 2 au3 ' i3y = � au 1 - au au 3 au au au . Adding these, we get - + - + - = 0 ax ay az and au au au =+ az au 1 au 2 Example 26 : If z is a function of x and y and x = eu + e-v; y = e-u - e-v, prove that We have; ·: z ➔ f(x, y) ➔ f(u, v) Solution : So, az az ax az ay az az ax az ay - = - - + - - ., - = - - + - au ax au ay au av ax av ay av az az az az = x -y au av ax ay . . . (i) Also, we have; x = eu + e-v; y = e- u - ev ax -u ay u ax -v ay v - = e - = -e - = e - = -e au · au · av ' av From (i), we have az _ az u ) az -u ) - - - (e + - (-e au ax ay Subtracting ' we get - - - = (e + e au av az JACOBIAN (i) az u - = - (-e av ax az -v )- - (e az ax -u az -v ) + - (-e ) ay az . . . (iii) v + e )- = x - - y ax ay ay v az az az If u and v are the functions of two independent variables x and y, then the determinant Jacobian (or functional determinant) of u and v with respect to x and y. It is written as (ii) . . . (ii) a(u, v) or J(u, v). a(x, y) au ax ax If u, v and w are the functions of three i ndependent variables x, y and z, then the determinant is called the Jacobian of u , v and w with respect to x, y and z. a(u, v, w ) It is written as �-- or J(u, v, w ) . a(x, y, z) Example 27 Solution : 1 a(x, y ) a(r, 8) . . and If x = r cos 8, y = r sin 8, then find . a(x, y ) a(r, 8) We have au - ay is called the av ay au ax av ax au ay av ay au az av az aw aw aw ax ay az ax ax a(x, y) ar = ay a(r, 0) ar Again we have r = = l cos 0 sin e ae ay Jx 2 + y ae 2 -r sine = r(cos2 0 + sin 2 0) = r I r cos 0 and 0 = tan -1 y_ � = _!_ (x 2 + y 2 ax 2 r1' X 2 · 2x = cos 0 � = _!_ (x 2 + y 2 ay 2 and co s 0 _1 ( - ) = _ si�0 and :: = ___ ; :� = __ 2 r y 1+­ 2 x a(r, 0) = a(x, y) ar ay ar ax ae ae ay ax E ngineering Mathematics cos 0 sin e sin e cos 0 r1 ' 2 · 2y = sin 0 r(cos 0 + sin 0) 2 2 Example 28 : If x = cos u, y = cosv sinu , z = cosw sinv sinu, then show that Solution : We have a(x, y, z) a(u, v, w ) ax ay ay ay ax au ay au ax az ay az az az az ay az au - si n u cos v cosu cos w sin v cosu 273 - sin v sinu cos w cos v sin u 0 Now, expanding the determ inant along first row, we get r a(x, y, z) = -sin3 u sin 2 v sinw. a(u, v, w) 0 0 - sin w sin v sin u 2 3 2 2 = - sinu [sin v sin usin w - O] = sin u sin v si n w Solution W e have u1 U2 U3 J (u 1 ' U z , u 3) = x1 2X3 = X = X X = 1 z 3 ' then ' then au au X au 1 = - 2;3 ' 1 = � . 1 = � ax1 x 1 ax 2 x 1 ax 3 X 1 au 2 � au 2 - X 1 3 au 2 � = = = , � ' ax1 X z ax2 ax 3 X z Xz 2 ' then au 3 = � . au 3 = � . 8U 3 = - X 1�2 ax1 X z ax 2 X 3 ax 3 x3 x3 X 1 X X au 1 ax 1 au 2 ax 1 au 3 ax 1 au 1 ax 2 au 2 ax 2 8u 3 ax 2 au 1 ax 3 au 2 ax 3 8u 3 ax 3 - z X3 x 21 X � X 1 � X 3 - 1 � � X � 3 X1 x 22 X � X 3 X - X 1 z 1 x 23 X X 2 IES MASTER Publication 274 Partial Differentiation -X 2X 3 1 = 2 2 2 X 2X 3 x1 x2 x3 X 2X 3 X 3X 1 -X 3X 1 X 3X 1 = 4 X 1X 2 X 1X 2 X 1X 2 1 -1 1 1 -1 a(x, Y, z) 2 Example 30 : If x = r sine cos <j>, y=r sine sin<j>, z=r cose, then show that = r sine . Solution : We have ax a4> ay a4> ax ax ar ay a(x, y, z) = ar a(r, e, 4>) a(r,e, 4>) ae ay -r sine case ae az az az ar ae a4> -r sine sin 4> r sin0 cos <j> r cose cos<j> r cos0 sin<j> sine cos <j> sine sin<j> 0 Now, expand!ng the determinant along third row; we get 2 2 2 2 2 2 2 2 = cos 8(r sine cose cos 4>+r sine cos 8sin 4>)+r sine (r sin e cos 4>+r sin e sin 4>) a(u, v, w) Example 31 : If u = xyz, v = xy + yz + zx, w = x + y + z, then compute the J acob .a1 n -- ­ Solution a(u, v, w) = a(x, y, z) au au au ax - az ay av ay av - av az ax aw aw aw ax ay az a(x , y, z) yz = y+z 1 z(x -y) x-y yz y+z y(x - z) zx z+x 1 xy x+y By C2 ➔ C 2 -C1 and C 3 ➔ C3 -C1 1 z(x-y ) y(x - z) z x-z =I I=(x-y)(x-z) I x-y 1 x- z 0 0 1 = (x - y)(x - z)(z - y) = (x - y)(y - z)(z - x) = a(x, y, z) =u2 v. Example 32 : If x + y + z = u, y + z = uv, z = uvw, then show that Solution : x = u - uv = u(1 - v), a(x, y, z) = a(u, v, w) y = uv - uvw = uv(1 - w), ax ax ax au av aw au ay vw aw az az az aw au av 1 = v(1-w) = 1-v v(1- w) ay ay av vw u2v(1 u(1-w) - w) + 0 uw u2vw = 0 -UV UV u2 v a(u, v, w) z = uvw -u u(1-w) uw 0 -UV UV Applying R 1 ➔ R 1+(R2 +R 3 ) y 1 I Engineering Mathematics 275 CHAIN RULE OF JACOBIANS Theorem I If u1, u2, ..., un are functions of y 1 , y2, ..., Yn and y1, y2, ..., Yn are functions of x 1, x2, ...xn then a(u 1, U 2 , .. ,,un ) a(u1, U2 , , .., un ) a(y 1 , Y2 , ..·, Y n ) = a(x1,X 2, , ..,X n ) a( Y1, Y 2 , , .., y n ) ' a(x 1,X2 , , .., x n ) Example 33 : If x = r cos0, y=r sin0, r =au, 0=bv, then find Solution 1 Using the chain rule for Jacobian, we have a(x, y) a(x, y) . a(r, 0) = = a(u, v) a(r, 8) a(u, v) ax ar ay ar ax ae ay ae ar au ae au a(x, y) . a(u, v) ar - -r sin 8 · a I r cos0 I 0 av = cos0 ae l sin 8 av � I=rab Example 34 : If J is the Jacobian of the u, v with respect to x, y and J' is the Jacobian of x, y with respect to u, v then prove that JJ' =1 . Solution : Let u = f(x, y) and v = �(x, y) . Differentiating these with respect to u and v partially, we get 1 and O Now, au = au . ax + au . ay ; 0 au au . ax + au . ay = = au ax au ay au . av ax av ay av av av ax av ay _ av av ax av ay -=- · -+- ·- 1=-=- · - +- · -- au ax au ay au ' av ax av ay av au au ax ax ay a(u, v) . a(x, y) ax au av = JJ' = av ay ay a(x, y) a(u, v) av - av ay au ax au ax au ay -au ax + -au ay -- + - ­ ax au ay au ax av ay av = av ax av ay av ax + av ay = 1 o1 �, - - -+ ax au ay au ax av ay av _ Example 35 : If u = xyz, v = x2 + y2 + z2, w = x + y + z, find Solution 1 s·ince a(x, y, z) a(u, v, w) =1=JJ' a(u, v, w) a(x, y, z) So, here we calculate a(u, v, w) J' = J(u, v, w) = a(x, y, z) au - ax av ax aw ax au av ay aw ay ...(i) ...(ii) (by using (i) and (ii)) = 1 a(x, Y, z) or J(x, y, z) . a(u, v, w) au - az av az aw az yz = 2x 1 2y 1 ...(i) xy 2z 1 = yz(2y - 2z) - zx(2x - 2z) + xy(2x - 2y) = -2(x - y)(y - z)(z - x) -1 . a(x, y, z) So ' by ( )1 �-'----'--= 2(x - y)(y z)(z-x) a(u, v, w) - IES MASTER Publication 276 Partial Differentiation a(x, y, z) a(u, v, w) _ _ _ Example 36 •. If x - u , y - u tan v and z - w, show that a(u, v, w) . a(x, y, z) = 1 Solution : We have Then x = u ⇒ ⇒ oY = tan v, ay = u sec 2 v, ay y = u tanv z = w ax = ax � =0 1, = au ' av aw au ⇒ az = az = 0 az a(x, y, z) J = a(u, v, w ) au au ax au oY au az av ' aw ax av ax aw av az av az aw oY av =1 u sec 2 v 0 Solving for u , v, w in terms of x, y, z, we have 1 u = x, v = tan-1 2'. = tan - 2'. w = z u x' Thus, a(u, v, w) J' = .a(x, y, z) au av ax aw ax aw ay aw az Hence, from (i) and (ii) JJ' = u sec 2 v • FUNCTIONAL DEPENDENCE Let u 1 , u 2 , u 3 , . . ., 1 -y x +y 2 1 0 u sec 2 v =0 0 1 = tan v aw 0 oY ax aw 2 =1 0 0 = u sec 2 v 0 0 X 0 2 x +y 2 o 0 u n be functions of n independent variables x 1 , x 2 , x 3 , ..., X 1 = --- · . . (ii) -2-x + y2 u sec2 v x n . Then the necessary and sufficient condition for the existence of a relation of the form F(u 1 , u 2 , . . . , u n ) = 0 is that the Jacobian vanish identically. . . . (i) a(u 1 , u 2 , · · ·· u n ) should a(x 1 , X 2 , . . . , X n Example 37 : If u = x + 2y +z, v = x - 2y + 3z and w = 2xy - xz + 4yz - 2z2 , show that they are not i ndependent. Fi nd the relation between u, v and w. Solution . . . a(u, v, w) = u, v and w will not be i ndependent If a(X, y, Z ) 0 . a(u, v, w ) = a(x, y, z) = 1 1 2y - z 1 2y - z 2 -2 2x + 4z 1 3 -x + 4y - 4z 0 -4 2x - 4z + 6z 0 2 By C 2 ➔ C 2 - 2C 1 , C 3 ➔ C 3 - C1 -x + 2y - 3z = -4(-x + 2y - 3z) - 2(2x - 4y + 6z) = 4x - 8y + 1 2z - 4x + 8y - 1 2z = 0 Hence u , v and w are not independent i.e. they are functionally dependent Eng ineering Mathematics u + v = 2x + 4z Now, Multiplying these, we get 277 u - v = 4y - 2z and (u + v)(u - v) = (2x + 4z)(4y - 2z) u 2 - v2 = 4(2xy + 4yz - zx - 2z2 ) u 2 - v2 = 4w. ✓ ✓ 2 2 Example 38 : If u = sin-1 x + sin-1 y, v = x 1 - y + y 1 - x , find Solution = relationship. a (u, v ) . I s u , v functionally related? If so , find the 8( X , y ) u = sin-1 x + sin-1 y au ax Also, Now, v = 1 ✓1 - x x ✓1 - y = 2 2 and ax av ax = au -1 = ay R + y� au - 8(u, v) 8(x, y) . . . (i) (given) ✓1 - x 1 au ay av ay ✓(1 - x · . . . (ii) -xy 2 1 R - ✓1xy- x 2 )(1 - y ) ✓1 - y 2 +1-1+ ' ✓1 - y ✓(1 - x -xy - 2 xy 2 2 2 )( 1 - y ) (given) 2 1-x -+� =0 Hence, u and v are not i ndependent. They are functionally related. We have from ( i ) , ⇒ ✓ u = sin-1x + sin-1 y = sin-1 ( x 1 - y 2 + y �) = sin-1v v = sin u (From (ii)) 8(u, v) x+y . . . Example 39 = If u = -- and v = tan- 1 x + tan-1 y, find - . A re u and v functionally related? If so, find the 1-� ' 8( � y ) relationship. 8(u, v) = a( x, y) Solution au au av av ax ax ay ay 1 + y2 2 = ( 1 - xy) 1+x (1 - xy) 2 1+y 1+x Hence, u , v are functionally related. Example ,o Solution = 2 2 2 1 2 (1 - xy) 1 x+y tan-1x + tan-1y = tan - ( 1 - xy ) _ ⇒ 1 (1 - � ) 2 =O v = tan-1 u = Show that the function u = x + y + z, v = xy + yz + zx, w = x 3 + z3 Find a relation between them . We have ⇒ - u = tan v 3xyz are not i ndependent. IES MASTER Publication 278 Partial Differentiation a(u, v, w) J = a(x, y, z) au ax au av au oz. av oz. ax ay aw aw aw ax ay oz. av 1 1 y +z x+z 3y 2 -3xz 3x2 - 3yz 1 x+y 3 z2 -3xy Per forming C2 = C2 - C 1 , C 3 = C 3 - C 1 and expanding the determinant , we have J = 0 which shows that u , v, w are not independent, i.e. dependent. To find relation betwen u, v, w we know that x3 + y3 + z3 - 3xy = (x + y + z)(x2 + y2 + z2 w = u{(x + y + z)2 - 3(xy + yz + zx)} or w = u(u2 - 3v) or u3 = 3uv + w or - xy - yz - zx) Example ,1 : Prove that the function u = x + y - z, v = x - y + z, w = x2 + y2 + z2 of one another. Also, find the relation between them. Solution 1 We have au ax a(u, v, w) av = J = ax a(x, y, z) au ay av ay au - oz. av oz. aw aw aw ax ay oz. 1 1 2x 1 -1 2y :- 2 z -1 1 2z-2y Since, J = 0, the functions u, v, w are not independent. 1 2x Now, u2 + v2 = (x2 + y2 + z2 + 2xy - 2yz - 2zx) + (x2 + y2 + z2 or = 2(x2 + y2 + z2 u2 + v2 2yz) = 2w = 2w is the required relation. - - 2 0 2yz are not independent 2(x+ y - z) - 0 0 =0 0 2xy - 2yz + 2zx) Engi neering Mathematics 279 PREVIOUS YEARS GATE & ESE QUESTIONS Questions marked with aesterisk (*) are Conceptual Questions 1. 2.* If H(x, y) is a homogeneous function of degree n, aH aH then x - + y - = nH. (True/False) ay ax 7.* If Z = f(x, y), dZ is equal to [GATE 1 994-ME; 1 Mark] 8. * 3.* If t ( x, y,z) = ( x2 + y 2 + z2 ) is equal to a2t az 2 (a) zero (c) 2 4. 6. 2t 2 ax then 2 a + a2 t 2 ay + (b) 1 9. (d) -3 ( x2 + y 2 + z2 r2 [GATE-2000; 1 Marks] = 0 to n) a re constant then x - + y - is ax ay of f n (c) rt (d) nJf (a) 0 (c) 1 (b) (d) Zn 1 2 ln2 [GATE-2008) The total derivative of the function 'xy' i s (a) (c) x dy + y dx dx + dy (b) x dx + y dy (d) dxdy [GATE-2009 (Pl)] (b) y - = X 8x ay x - + y- = 0 ay ax oz ax az az az X- = yay az az (d) y - + x - = 0 ay ax az az The contour on the x-y plane, where the partial derivative of x2 + y2 with respect to y is equal to the partial derivative of 6y + 4x with respect to x, is [GATE-2014) (a) y = 2 (c) X (b) x = 2 + y = 4 For the two functions f (x,y) = x 3 - 3xy 2 (d) x - y = 0 [GATE-201 5) 2 and g (x, y) = 3x y - y 2 W hich one of the fol l owing options is correct? at ag =ax ax ag at = -ay ax (b) (d) at - ax at 8y ag = -ax ag ax 10. Consider a function f(x, y, z) given by [GATE-201 6 (Pl)] f(x, y, z) = (x2 + y 2 - 2z2 )(y2 + z2 ) The partial derivative of this function with respect to x at the point, x = 2, y = 1 and z = 3 is ____ [GATE-2005 (IN)] a 2t Let f = yx. What is -- at x = 2, y = 1 ? axay (c) (c) of n (b) f (a) (a) 1 f = a 0 x n + a 1 x n - y + . . . + a n_1 xy n- 1 + a n y n wh ere a; ( '1 (a) 5.* -� [GATE-2000) If z = xy l n(xy), then 11. Let w = f(x, y), where x and y are functions of t. Then dw t is equal to according to the chain rule d [GATE-2017 EE Session-II] (a) dw dx dxdt + dw dt dy dt [GATE-201 7 CE Session-II] IES MASTER Publication 280 1 2. Partial Differentiation If u = x-+ Y-? au ax (a) 0 (c) u 1 3. log ( au ay · x 2 + y2 ) , what i s the v a l u e of x+y (b) 1 (a) '2Z. (c) 2b [ESE 201 8 (Common Paper)] az ax az ay (b) 2a (d) 2abZ = ax2 + by2 xy of of constants. If - = - at x [ESE 201 8(EE)] where a and b are = relation between a and b is (d) eu ' Let f(x , y) ax ay + If Z = eax+by F (ax - by) · the value of b · - a is 1 4. 1 5. 1 and y = 2, then the b (b) a = 2 (d) a = 4b b (a) a = 4 (c) a = 2b Let r = x 2 + y - z and z3 - xy + yz + y3 = 1 . Assume that x and y are independent variables. At (x, y, z) = (2, -1 , 1 ), the value (correct to two [GATE 201 8 (EC)] decimal places) of � is ax - --- [GATE 201 8 (EC)] Engineering Mathematics 1. ( TRUE) 5. (c) 9. (c) 1 3. (d) 3. (a) 7. (c) 11. (c) 1 5. (4.5) 2. 6. (a) 4. 8. (c) 10. (a) 12. (a) - Euler's theorem : If Z is a homogeneous function of degree 'n' in x and y then Sol-2: (a) az az x -+y - =n-Z ldz = �-dx+� • dyl Sol-3: (a) -1 f(x, y, 2) = ( x2 +y 2 +22 )2 at ax at ax a2f ax2 :�: Similarly a 2f ay 2 a2 f a2 2 32 5 2 ( x2 +y 2 +22 )- / - 3 ( x 2 +y2 +22 )- / = 0 Method II : Let r2 = x2 + y2 = 22 Differentiating (1) partially w.r.t 'x' 32 1 = --( x2 +y2 +22 ) - / (2x) 2 = a ( 2) a r ay Z = f (x, y) then, (b) (d) SOLUTIONS Sol-1 : ITruel ax 14. (40) 281 i.e. Similarly, Now a ax (x2 ) + a ( y 2 )+ a ax 2r .0:. = 2x + 0 + 0 ax ar ax ar ay f = = x y_r and ar az = (22 ) ax 2 r 1 (x2 + y2 + 22)~112 = (r2)~1 12 = _ r 32 = -x ( x2 +y2 +22 )- / = -X ( _3 2 5 2 ) ( x2 + y 2 + 2 2 r / X (2x )- ( x2 +y 2 +22 ) -3/ 2 5 = 3x 2 ( x 2 +y 2 +2 2 r /2 -( x2 +y 2 +22 r312 Z = f (x, y) 312 -5 = 3y 2 ( x2 +y 2 +22 ) 12 -( x2 +y 2 +22 )- 312 5 = 322 ( x2 +y2 +22 )- 12 -( x2 +y 2 +22 )- Similarly and IES MASTER Publication 282 Partial Differentiation az = xy ( 1 + l nxy) xax az Sol-4: (c) y - = xy (1 + lnxy) ay · : f(ax, 8y ) = an f(x, y) ⇒ f is a homogeneous function i n x and y of degree 'n' . - . By Euler's theorem for homogeneous function, we have So l-5 : (c) at at ax x - + - = nf ay f = yx Differentiating eq uation (1) w.r.t. x, we get at ax ayax = At x = 2 and y = 1 a 2f Sol-6: ( a) Let ayax Ixy=1=2 f(x, y) x- 1 y [x /ny + 1 ] = = ( -: log 1 = 0) xy at at = xyln ( xy ) xy ( ; ) + yln ( xy) : = y Similarly, az ay To find : at ax ag ax at 6xy, ay = ag ax at = -6xy ay ag = 3x 2 - 3y2 ay of l f(x, y, z) = (x2 + y2 - 2z2 )(y2 + z2 ) ax Sol-1 1 : (c) = 2xy2 + 0 - 0 + 2xz2 + 0 - 0 = 2(2)(1 )2 + 2(2)(3)2 = 4 + 36 = 40 Using Total Derivative Concept w = f (x, y) ⇒ or Sol-1 2 : (b) = y ( 1 + l n xy) = x (1 + lnxy) = 3x 2 - 3y2 , oX X=2, y=1, 2=3 f(x, y, z) = x 2y2 + y4 - 2y2z2 + x2 z2 + y2z2 - 2z4 = 1 2-1 [2 /n 1 + 1 ] df = - dx + - dy = ydx + xdy ax ay z y = 2 ar Then using total derivative concept So l-7: (c) 2y = 4 Sol-9: ( c ) Sol-1 0 : (40) y a ax 2 ay So, Again differentiating w. r.t. y, we get = xyx- 1 /ny + yx. a· ay - ( x 2 + y ) = - ( 6y + 4x ) " .( 1 ) - = yx /ny a2 f Sol-8: (a ) az ax x- = y- Let aw aw ax ay aw dx aw dy - - + -- dw = - dx + - dy dw dt = ax dt ay x 2 + y2 u = l oge ( x+y V = e" = x2 + y 2 x+y dt J Engineering Mathematics V { ),x),y) = "A V(x ,y) so v is homogenous so by Euler ' s theorem av x-+y- = ax ay a( e0 ) + a x- Y ( e0 ) ax ay au au x� + yax ay az ax Now az ay = 1 .e" = 1 ax+by F(ax - by) = aeax+by F ( ax - by) + eax +by a F' ( ax - by) = beax+by .F ( ax - by) + eax+by (-b)F'(ax - by) az ay az = 2ab. z f(x, y) = (: at ax t. 2 ) = = ax 2 + by 2 xy xy (2ax ) - (ax 2 + by 2 ) ( y) 2a - 8 b 4 x2 y 2 at xy ( 2 by) - (ax 2 + by 2 ) ( x ) at ay 4b - a 4 According to the question, so at ax 2a - 8b = 4 b - a 3a = 1 2 b ⇒ b- + a - = 2 ab eax+ y _F ( ax - by) So l-1 4 : (d) so Ill z = e S o l-1 3 : (d) and function 283 a ⇒ = 4 b Sol-1 5 : (4.5) Given that z3 - xy + yz + y3 = 1 and x, y are independent variables so we can take z = f(x, y) i.e. Differential ( 1 ) purtially w.r.t 'x ' 3z 2 - - y + y - + O = 0 az ax az ax ⇒ y az ax = ar ax = 2x + O - - 3z + y r = x2 + y - z Now ar ( ax ) (2 . -1. 1) 2 = 2x - y 3 z2 + y = 2x2= az ax -1 3 (1 )2 - 1 1 4 + - = 4. 5 2 IES MASTER Publication ■ CONCEPTS OF INTEGRATION This is the inverse process of differentiation, if the differentiation of F(x) with respect to x be f(x) then the integration of f(x) with respect to x is F(x) i . e . , � F(x) = f(x) dx But the derivative of a constant term is zero then ⇒ f f ( x ) d x = F(x) - [F(x) + c] = f(x), so we have dx J f(x) dx = F(x) + c d The process of finding the integral of a function is said to be integration and the function which is to be integrated is known as integrand. SOME STANDARD FORMULAE 1. f (ax + btdx = (ax + br1 , n ctc - 1 n+1 2. 1 1 J -- dx = -log(ax + b) a ax + b 5. f sin(ax + b) dx = - � cos(ax + b) 6. f cos (ax + b) dx = 2a sin(ax + b) 3. 7. 9. 11 . 1 3. 1 5. 1 7. 1 9. 21 . f e ax+bdx = � eax+b 1 J tan(ax + b) dx = -log sec (ax + b) a f sec2 (ax + b) dx = � tan(ax + b) f sec(ax + b) tan(ax + b) dx = � sec(ax + b) f sec(ax + b) dx = � log(sec(ax + b ) + tan(ax + b)) J sinh(ax + b) dx = � cosh(ax + b) f tanh(ax + b) dx = 81 Iogcosh(ax + b) f sec h2 (ax + b) dx = � tanh(ax + b) f sec h(ax + b) tanh(ax + b) dx = - 81 sec h(ax + b) 4. 8. 1 0. 12. 14. 1 6. 1 8. 20. 22. f abx+c dx = i abx+c log a e J cot(ax + b) dx = 2 1ogsin(ax + b) a (ax + b) f cosec2 (ax + b) dx = _ _]._cot a (ax + b) f cosec(ax + b) cot(ax + b) dx = _2cosec a + b) - cot(ax + b)) f cosec (ax+b) dx = 21og(cosec(ax a f cosh(ax + b) dx = 2a sinh(ax + b) f coth(ax + b) dx = 2a 1og sinh(ax + b) f cosec2 (ax + b)dx = _ _].a_cot h(ax + b) J cosec h(ax + b) coth(ax + b) dx = _2cosec h(ax+b) a 2 3. 25. 27. 30. 31. 33. 34. 35. dx . x = sin -1 � 2 2 a va - x J J h J 24. = sin h - 1 � = 1og ( x + Jx + a ) a a +x 2 J � = tan - � 2 2 a a a +x 2 dx 1 x dx = sec - � a x v x- - a- 285 1 ! J h = cos h - 1 � = log x + Jx2 + a2 ) ( 26. 2 2 Engineering Mathematics � X � a . - X J va- - x- dx = - va- - x - + - sin 1 - 28. 2 2 2 � X � a - X X � a � J v x - - a - dx = - v x- - a- - - cos h 1 - = - v x - - a- - - log ( x + v x- - a- ) 2 2 a 2 2 2 a 2 dx 1 x-a 1 a-x J - - = - log -- , x > a = - log --, x < a 32. 2 2 2a x+a 2a a+x x -a Ie ax cos (bX + C ) d X = e ax I a cos(bx + c) + b sin(bx + c) a +b 2 e S i n (bX ax . + C ) d X = e ax a sin(bx + c) - b cos(bx + c) a +b 2 2 2 J f · g dx = f J g dx - J [f' · J g dx ] dx, where f, g are functions of x . m+1 n +1 r- -r- 2 2 J sin x cos x dx = m+n+2 o 2r 2 n/2 m n (Gam m a Fu nction) SOME IMPORTANT I NTEGRATION AND THEIR HINTS 1. 2. 3. 4. 5. 6. 7. 8. Hints Integrands -Ja 2 + x 2 put x = a tan 8 or a cot 8 put x = a sec 8 or a cosec 8 ra=x Vrn 2 Jax + bx + c where X, Y are both li near. 1 xfl ' where X is quadratic, Y is li near. 1 xfl ' where X is li near, Y is q uadratic. put x = a sin 8 or a cos 8 put x = a cos 28 by m aking perfect square put ✓ = y put Y ✓Y = y 1 put X = t IES MASTER Publication 286 9. 1 0. 11. Multiple Integral 1 Y X✓ ' 1 pu t X = ­ t where X, Y are both quadrat ic. 1 - - , (a, b, c ;tc 0) --- .a COS X + b Sin X + C a + b cos x + c sin x l + m cos x + n sin x pu t a = r cos a, b = r sin a. f g DEFINITE INTEGRALS pu t f = Ag + Bg' + C, A, B and C are constant s. The definite integral is denoted by J: f(x) dx and is read as 'the integral of the func ion f(x) w.r.t. 'x' from x = a o x = b." b d - F(x) = f(x) then f(x) dx = F(b ) - F(a); dx a where F(b) and F(a) are t he values of the functions F(x) at x = b and x = a respectively. Let f t t FUNDAMENTAL PROPERTIES OF DEFINITE INTEGRALS (I) (Ill) (V) J: f(x) dx =:, J: f(t) dt (11) (IV) J: f(x) dx = J: f(a - x ) dx or J: f(x) dx = J: f(x) dx + J: f(x) dx, a < C < b f f(x) dx a -a = { 0 J: f(x) dx = - f: f(x) dx when f(x) is odd (VI) J:' t(x) dx = { 2l f(x)dx, when f(x) is even 2J: 1 ) dx, �� J: f(x)dx = J: f(a + b - x)dx if f(2a - x) = f(x) e.g. let f(x) = log sin x then f(n - x) = log sin(n - x) = log sin x = f(x) so Jr logsin xdx = 2 Jr o 0 Example 1 : Solution Prove t hat Let f log sin x dx = n/2 0 = = = _ 2: log2 2 f log cos x dx 0 f;'2 log sin x dx = J;'2 1og sin (% - x) dx f0n/2 log cos x dx Adding (i) and (ii), we get 21 = n/2 n n/2 if f(2a - x) = - f(x) log sin xdx (using prop. vi) prop. IV] f n/2 log s1n. x dx + r n/2 log cos x dx = r n/2 log(s1n. x cosx) dx Jo Jo 2 2 = J; ' 1og [¾ (2 sin x cosx)] dx = J;' (1og ¾ + log sin2x ) dx 0 = - log 2 Jr 1 dx + o n/2 f n/2 log s1n. 2x dx = - % 1og 2 + ¾ J; 1og sin t dt ⇒ 0 [ put 2x = t, dx = � ] . in/2 n n log s1n t dt = -- log2 + 1 = - - log 2 + o 2 2 = - - log 2 2 7t t . . . (i) . . . (ii) Engineering Mathematics Example 2 / 7t Show that f rr log(1 + tan 0) d e = - log2 . 0 4 8 So lution Let [prop. I V] rr14 rr 4 1 - tan e 2 14 ' r rr 14 i = f i og [ 1 + ) d e = f0rr og 2d0 - 0 Iog( 1 + tan 0) d0 ] d e = J 1 og ( o 1 + ta n 0 1 + ta n 0 J, = log 2(0)�14 - I Example 3 Eval uate : Solution Let ° 21 = � 1og 2 4 ⇒ J rr a cos Xx +dxb sin 0 2 2 2 2 ⇒ 7t x · (n - x ) dx 2 2 2 2 a cos (n - x ) + b sin (n - x) /2 21 = 2n J rr Put, b ta n x = t ⇒ dx a cos x + b sin x 2 2 0 2 7t rr /Z Evaluate : J 0 2 b sec x dx = dt As x ➔ 0, t ➔ 0 and x ➔ - , t - > oo. · 2 Solution I = - log 2. 8 . . . (i) rr = f 0 Example 4 : 287 Jsin x Jsinx Jcosx dx sin x + cos x Let = Then, = "j2 0 ⇒ 2 rr n dx. = r -I Jo a 2 cos 2 x + b 2 sin 2 x = 2 n rr fO /2 2 sec X dx a + b ta n x 2 2 2 "' dt - � [tan- 1 _!_] I = � f __- a o b Jo a 2 + t 2 ab Jsinx Jsi n x + Jcos x "' dx 7t 2 2ab . . . (i) .Jsin(n / 2 - x) J ---=======--=== d X 0 .Jsin(n / 2 - x) + .Jcos(n / 2 - x) "1 2 ! [Using f(x)dx = [ f(a + b - x) dx ] . . . (ii) ⇒ Adding (i) and (ii), we get 21 = = ⇒ = "1 2 rr/2 .Jsin x .Jcos x J ----dx + J ---- dx .Jsin x .Jcos x .Jsi n x + .Jcos x + 0 O J .Jsin x + .Jcos x � \/ Sln X + \/COS X n /2 � 0 dx = n/2 J 1 .dx = [x] ,2 = � - 0 3 2 7t 4 IES MASTER Publication 288 Multiple I ntegral Example 5 : Solution Evaluate : Let I = = = = = f (2Iogsin x - log sin 2x)dx n/2 0 f (2 Iog si n x - log sin 2x)dx n/2 0 f {2 Iog sin x - log(2 sin x cos x)}dx { ·: sin 2A n/2 0 f {2 Iog sinx - log 2 - logsin x - log cos x}dx n/2 0 f log sin x dx - (log2) f 1 - dx - f log cos n/2 n/2 n/2 0 0 0 ( 12 0 n /2 n/2 0 0 2 �2 I = - - log 2 2 2 sin A cos A} f log sin x dx - f log 2dx - f log cos x dx n/2 2: _ x ) dx log sin x dx [ log sin x dx - (log 2) [x]� - [ �2 = = 7t = {log 2 { % - 0 ) FUNDAMENTAL TH EOREM OF INTEGRAL CALCULUS I f f{x) is integrable in (a, b) and F(x) is a function such that d F(x) dx = f{x) in (a, b) then we have J: f(x) dx = F(b) - F(a) DEFINITE INTEGRAL AS THE LIMIT OF A SUM The value of the definite i ntegral J: f(x) dx is given as : J>(x) dx = � � o [h{f(a) + f(a + h ) + . . . . . . + f(a + n - 1 h)}] = �� h [ � f(a + rh)] Here nh = b - a; as h ➔ 0, n ➔ oo, and f(x) is a continuous function of x in the given integral. Deduction The fundamental theorem of integral calculus may be written as : For a f: f(x) dx = �� h [� f{a + rh)l nh = b - a 1 n 00 = 0, b = 1 ⇒ h = -, so we have n-1 1 r ➔ in the above equation, - ➔ x, - ➔ dx and Example 6 i Solution : n Evaluate : We have n L i r=O 1 o J: x dx from the definition of a definite integral as the limit of sum . 2 Engi neering Mathematics 289 2 2 2 2 2 2 2 2 h [ a + (a + 2ah + h) + (a + 2 h + 2 · 2ah) + . . . + (a (n - 1} h + 2a(n - 1)h )] = hlim ➔O 2 2 2 2 2 h [ na + 2ah{1 + 2 + 3 + . . . + (n - 1)} + h {1 + 2 + . . . + (n - 1) }] = hlim ➔O . n(n - 1)(2n - 1) 2 n(n - 1 ) 2 + h3 • - --- ] = I Im [n h a + 2ah • -2 h➔O But nh = b - a as h ➔ 0, n ➔ CX), so we have Jab x 2 6 dx = lim [ (b - a) a 2 + a(b - a){(b - a) - h} + ..!_ (b - a)(b - a - h)(2b - 2a - h )] h➔O 6 1 1 2 2 3 2 2 = (b - a ) a + a(b - a) + - 2 - (b - a) = - (b - a)[3a + 3a(b - a) + (b - a) ] 3 6 3 2 2 = ! (b - a)[b + ba + a ] = ..!_ (b - a ) 3 3 3 Example 7 : Solution By the definition of definite integral as a limit of sum , show that Ja sin x dx = cos a - cosb b We have . h . nh sin [a + (n - 1) - ] sin 2 J sin x dx = lim h [sin a + sin(a + h) + . . . + sin a + n - 1 h] = lim h a h ➔O h➔O . / Sln 2 b - Example 8 : Solution [ J sin x dx = lim � a h➔O 2 b by formula : sina + sin(a + p) + . . . + sin(a + n - 1 p) = cos (a - �) - cos (a + nh - � ) 2 2 h Sln 2 = lim � [cos(a - �) - cos (b - �)] h➔O Sin h/2 2 2 . [·: Sin a + - .( n - 1- ,.,r:i. ) Sin. np ] 2 _ sin £ 2 2 nh = b - a ] = 1 - (cos a - cosb) = cos a - cos b. 1 1 1 1 . Show that the limit of the sum - + -- + - - + . . . + - , when n ➔ CX) Is log 3. n n+1 n+2 3n 1 1 1 1 . 1 1 · (- 1 - + --1 + . . . + ) lim (! + -- + -- + . . . + J_) = I Im ( - + -- + . . . + - ) + 1 Im 3n 2n n ➔ oo 2n + 1 2n + 2 n n+1 n+2 3n n ➔ oo n n + 1 1 1 1 1 1 1 . 1 = I Im - + 1 ·Im ( -- + -- + . . . + - ) + 1 ·Im ( -- + -- + . . . + - ) 3n 2n n ➔ oo 2n + 1 2n + 2 n ~;"' n n ➔ oo n + 1 n + 2 n ➔ «> n n 1 1 . - + hm L -L n + r n➔ "' r 2n + r n➔"' r = O + h. m =1 =1 . = hm n➔ 00 n 1- ·-1 n 1 1 . L -- · - + h m L n n ➔ "' r= n r =1 1 + (� ) 1 1 1 1 = f -- dx + -- dx = (log(x + 1))6 + (log(x + 2))6 Jo 1 + X O 2+X = log 2 - log 1 + log 3 - log 2 = log 3 J 1 2 + (� ) IES MASTER Publication 290 Multiple Integral Example 9 : Solution : Find the limit when n ➔ oo of the product: {( 1 + � ) ( 1 + � ) (1 + ¾) . . . (1 + �) r n 'n Let A = ��J( 1 + �)( 1 + � ) ( 1 + ¾ ) + . . . + log ( 1 + � ) J Taking log both the side, we get log A = ��� [ 1og(1 + � ) + 1 og ( 1 + � ) + . . . + log ( 1 + � ) ] l og A = ��"' � [ 1og ( 1 + � ) + 1og(1 + � ) + . . . + log ( 1 + � ) ] = lim ..!_ f log (1 + _r:_ ) = r log(1 + x) dx = (x log(1 + x))6 n->OO n r = 1 ⇒ O A = e lo g( 4/ e ) A = Let 1'n = O 1 - x dx 1+ X · i. [1 e n. I.1 m ] n ->oo n n 1 /n Taking log both the side, we get = 1.1 m [ n -,oo n(n - 1 )(n - 2 ) . . . 3 . 2 - 1 nn log A = lim ! 1og [! . 3- . � . . . . . . n ->oo n n n n ] 1 /n (n - 1 ) . � ] = lim ! f 1og ( _r:_ ) n n n ->oo n r = 1 n 1 1 = J 1og x dx = ( x log x)� - Jro J_ . x dx = -(x)6 = - 1 O X ⇒ A = e Example 1 1 : Find the value, as n ➔ oo of the series 2 1 [( 1 + n ) ( 1 + n ) 2 Solution r- 4 = log 2 - ( x - log(x + 1))� = log2 - 1 + 1og 2 = log 4 - log e = log ­ e � Example 1 0 : Find the value : lim [ n ] n ->oo n Solution n 1'2 ( 3 1 + n) 1' 3 ( n . . . . . . 1 + n) 1/ n ] n 3 Let A = ��J( 1 + � ) (1 + � r ( 1 + ¾r . . . . . . ( 1 + �r ] Taking log both the side, we get 2 log A = ,)�J 1og ( 1 + � ) + 1og (1 + � r + . . . log ( 1 + � ) = 1' n ] 1 lim f 2 1og ( 1 + _r:_ ) = lim f -- log(1 + _r:_ ) . ! = n ->oo r = 1 (r/n) n n n r =1 r n -,oo f log(1 + x) dx O X Engineering Mathematics = ⇒ = = (1 + + + + ( X ;: ;: :: " I 2: 3: 4� . . ·) A = e 291 7: (Value of trigonometric series)/or using fourier series concept. rr2 112 LEIBNITZ RULE OF DIFFERENTIATION UNDER THE SIGN OF INTEGRATION Let f(x , t) is i ntegrand which is function of two variables x & t then l( d lj x > IJl ( x > a d d f(x, t ) dt l = -[ - f(x, t ) dt + __y .f( X , \Jt ) - � .f(x, cp ) dx dx dx ax (j)(X ) (j)(X) f Note: 1. 2. f Take care, here \v(x) and cp(x) are repl aced i n place of t i n 2nd & 3rd term . I f integrand is function of 't' alone then lj/ d\Jf d cp � [ T f(t)dt l = - .f(\j/ ) - - .f(cp) d dx dx (j)(X) � (x3 ) ( 1-) - � (x 2 ) (1-) = for exam ple, Example 1 2 : · log x dx sin 2 x f sin 0 -1 t ✓ dt + 2 cos x f 0 cos-1 t ✓ f cp = Let · log x dx 2 - 3x 2 2x x2 - x -= 3 1og x 2 1og x log x at x = ; dt = ? sin2 x Solution: 3 0 t sin- 1 ✓ dt + cos2 x f 0 t cos-1 ✓ dt . . . (1 ) then so ⇒ = { 2 si n x. cos x (x)} + { 2 cos x(- sin x)(x)} = O 1/2 1/i 0 0 cp(x ) = constant k (let) f sin-1 ✓t dt + f cos-1 ✓t dt = k T ( � dt = k ) 0 cp(x ) = i .e . , stn2 x or f 0 or TC 4 cos2 x t t 1 cos-1 ✓ dt = sin- ✓ dt + f [using ( 1 )] k = 4 [ ·: cp is constant] TC 0 IES MASTER Publication· 292 Multiple Integral Example 1 3 : If cp = J sin(t2 )dt then cp'(1) = ? 1/x dcp dx Solution: ( ✓x ). [ sin ( ✓x ) ] - � (� sin (� }[ d d� r] 2 sin x cp'(x) = - - + 2✓ x . 1 Sin ( - ) x2 x2 sin 1 sin1 3 . = - Sin 1 cp'(1) = - + 2 2✓1 (1)2 so IMPROPER INTEGRAL An i ntegral J: f ( x ) dx is called im proper, if either range of integration is i nfinite or f(x) i s unbounded with in the range of integration. Improper integral of 1 st kind If f(x) is bounded and region of integration is infinite for e.g. J; f ( x) dx or J: f ( x ) dx ort f: f ( x ) dx. Improper integral of ll nd kind If both the lim its are finite but f(x) is unbounded at one or more poi nts with in the interval a s x s b. For eg. let C E (a,b) such that f(c) ➔ provided both the lim its exist fin itely. l!::::::==::::!I oo then process of evaluating integral will be as foll ows. Ja f ( x ) dx b cb = lim Ja e f ( x ) dx + lim Jfc +E f (x ) dx e➔O e➔O It is not necessary that value of improper integral always exist. J 1 dx Example 14 : Evaluate _1 2f3 = ? x Solution : f(x) = 2;3 and at x = 0, f(O) = oo (unbounded) So it is an im proper integral of l l nd kind. Hence X J 13 = 3 [0 - (-1)113 ] + 3 [(1) 1 - o] = 3 + 3 = 6 1 dx Example 15 : Show that J _ does not exist. 0 ( 1 x) Solution f(x) = So, [ / ]-E [ ,/ 0 -c dx . x 11 3 . x1 3 . f 1 dx . dx hm + lim J lim + lim = = & ➔0 1/3 c ➔0 -1 X 2/3 &➔ 0 Jo+t x 2/3 t➔ 0 1/3 _ -1 x 2/3 1 1 1 and at x = 1 , it is unbounded. 1_x ]1 & J01 1dx- x E ngineering Mathematics lim [ - 1og( i: ) - log 1 ] = lim fO-e � = lim [- 1og (1 - x) ]O-e = - e➔O 1 1 e➔O s➔O = - [log 0 - 0) = - ( --OJ) = + Hence the value of 1-X J 1dx_ x 293 oo does not exist. 1 O BETA AND GAM MA FUNCTIONS These functions are basically im proper integrals which are commonly known as Eulerian I ntegrals of the first and second kind respectively after the name of a Mathematician Euler who introduced them in 1 729. These functions are defined as A. Beta function (1 st Eulerian Integral) : m, n > 0 not necessarily a n int eger. 8. Gamma functions (2nd Eulerian Integral) : In = J;e-x x -1dx, n n > 0 and n may not be an i ntegral value. PROPERTIES OF BETA AND GAM MA FUNCTIONS Property 1 : Property 2: Beta function is symmetrical about m and n i .e. p(m, n) = p(n, m) f oo Another useful transformation of beta functions P (m, n) = Jo 1 x m -1 + x n -1 dx. Example 1 5 : Eval uate fo m +n ('/ + x ) Solution : n -1 (1 + x)m+n X dx We know that Consider 1 x m -1 00 1 x m-1 x mf oo dx -dx + J - dx = p(m, n) = J o + I0 1 (1 + x) m + n (1 + x r n (1 + x)m + n I1 = -J1 (1 + x r oo X m 1 +n dx Now, puttinn x = t ' we have from (ii) 1 11 = I1o ( ur- 1 1 + -) t 1 m+ n ( 1 t2 t = f0 -dt = (1 + t) n 1 1 m+n Putting the value of r.t(m n) :: 1-' ' J 00 1 x m -1 (1 + x) m -·1 m+n dt ) = n 0 ( n m+n n -1 Xf1 � Jo (1 + x r + n dx in (i), we get x J1 -- dx + J (1 + x r + 0 f m 1 1 - 1 ( ) t t2 1 O (t + 1) n 1 x --+- dx = (1 + x r n J 1 O m+n . . . (i) . . . (ii) dt 1 x m - + x n- 1 -- dx (1 + x)m + n IES MASTER Publication 294 Multiple Integral Property 3: m In Relation between beta and gamma function p(m, n) = / (m + n) Property 4: r(m + 1 ,( 2 ) where m > - 1 and n > -1 . 2 ) � 0 J o sin 8 cos d 8 = · m+ +2 ; 21 ( ) Property 5: Dupl ication formu la : 1 m r ( m + i = :_J (2m), m > 0 ) 2 1 f" 2 m n = m ay b e f h er re mar ed ha r_ t__ u_ r_ _)__oo t r_ t ______k___ t___ _____ co ...._____ ______________,______J Property 6: Property 7: "' In . . -lx n 1 Transformat1on of G amma F u nction Jro e x - d x = n I, n_ Reflection or Complement Form ulae I ni (1 - n) = ___ s1nnn Example 1 6 : Eval uate the val ues of Solution : 1 st fT, R, f=3 1 2 1 2 12 Part : Put n = 1 /2 in ( 1 ) rG ) r ( i ) = 2 nd Part : Put n = 3/2 in (1 ) n ( ( r i) = n r . n' Sin2 n 1 ¾ = r( ) r( � ) . 3rc ' Sin 2 3 rd Part : Put n = 5/2 in ( 1 ) r( % ) r( - } ) Property 8: = J n r G ✓n r( � ) = -n 1 ) = ✓n (neglecti ng -ve val ue) 1 , ( - 1 = -2 2 n (% J G ) ) ( - r r _23 ) = (1) . 5n ' Sin - 2 ✓n 3 ,(-} ) = i.J;, Product of Two Explicit Integrals J: t (x) dx x f: g( y) dy = J: J: t (x) g(y) dx dy Example 16 : Eval uate of r ( J ) using above theorem Solution : If we take n = 1 we get 2 1 r(¾ ) = J; e- c11 dt Putting t = x 2 or t 1 2 = x 2 1 ⇒ rG r112 dt = 2dx 2 ) = 2J; e -x dx . . . (i) Engineering Mathematics M ultiplying (i) and (ii), we get Transform i ng to polar, we get [r(i)T ⇒ Example 1 7 : Eval uate Solution : Let Putting J;if>< e -✓xdx . ✓x Solution : Let Putting . . . (ii) f0w e - 2 dx x Jro e-Y2 dy x w 2 w 2 rr/2 rr/ 2 w 2 rr/ 2 = 4f0 f0 e-r r d8 dr = 2 0 {e -r } d8 = 2 f0 d8 = 2(8); / = n r (i ) = f ✓n = f: t 112 e -1 2t dt = 2 J; t 312 e- 1dt = 2r (% ) = 2 - f r (%) = 2 - f - i r ( i ) = f J; ✓x e-rxdx . 'lfx = t or x = t3 ⇒ Then from (i), we have O . . . (i) = t or x = t2 we have dx = 2t dt Then from (i), we have Example 1 8 : Eval uate 4 295 (by definition) ✓n . . . (i) 2 dx = 3t dt 1 2 1 712 31 2 I = f: t e - 3t dt = 3 f: t e - dt = 3 r (¾ ) = 00 x Example 1 9 : Eval uate Jf dx. o ax a Solution : Let Putt i ng ax = e1 ⇒ I = t = x loga Then from (i), we have fo <£J ⇒ xa dx. ax dx = dUlog a . . . (i) w 1 t 1 t dt r w e- t t adt I - r ( f(a + 1) J eJ 1 J o o log a (log a r log a - (log a r1 Solution : Let a . . . (i) IES MASTER Publication 296 Multiple Integral 1 . Putting log - = t or x = e-1 we have dx = -e-1 dt Then from (i ) X n 1 f� x - [1og( )r \x � Putting nt = u Then from (ii) Example 21 : Evaluate Solution : f t=� or n ⇒ Then i"' du n /1 + x t- (1 - x)q- dx . 1 1 dx = -2 sin 2ede 1 f (1 + xt - (1 - x)q- dx - dt = 1 1 = = fn/ 0 frr/ 2 0 (1 + cos 2xt-1 (1 - cos 2x)q-\-2 sin 2ede) (1 + 2 cos2 e - 1 t-1 (1 - 1 + 2 sin 2 e)q-1 (-4 sin e cos e de) 2 2 2 1 1 = 4f;' 2 P- cos P-2 e2q- sin q- 2 e sin e cos e de 2 = 2p+q J; sin2 q-1 e cos2P -1 e de 1 Example 22 : Eval uate JO Solution : Let x n f0 1 dx n (1 - X ) 2 p+q-1 1,n · r(p) r(q) r(p + q) = sin 2 e or x = sin 21n e so that dx = 3_ sin<21n >-1 e ca s e de dx (1 - x n )1 / n . . . (ii) m 1 _ u U - du _ 1 J "' -u m-1 _ 1 f' e u du - (m ) I - o e - ( -) - - n n nm o nm Put x = cos2e, then 1 = f� (e - t f- 1 tm -1 (-e - t dt) = J; e -nt tm-1 dt t 21n 1 2 rrt2 sin< 21n >- 1 e cos e de _ "'2 (2/n) sin< >- e cos e de _ J - f0 2 O (cos 2 e)1/n (1 - sin e)1 / n n 12 1 < 21 n 21n 1 = 3. Jo" sin< >- ecos - > e de = n = = 3. r (� ) r (� ) n 7t n Sin n . 7t r(1) [ ;+ 1 ] r [ 1 t 1 ] 2 r �- n r 3. - 1 + 1 + 2 - 3. [n n] 2 Eng ineering Mathematics 2 1 611: 3 1 3 1 r Example 23 : Show that J o x(8 - x ) dx = -J?, . g Solution : ⇒ 3x2dx For the L.H.S. let u s take x 3 = 8y Now, for x = 0, y = 0 and for x = 2, y = 1 . = 8dy 1 13 113 L.H. S. = Jro\ B y) ( B - B y) Therefore 297 8d �13 3(8y) � (\ (3- i) � r 1 -11 3 11 3 (1 - y ) d y = y 3 Jo 3 t-' 3 ' 3 = 8 r(2/3) r(4/3) 8 r(1 - 1/3)1 / 3r(1 / 3) = =3 [(6/3) 3 r(2) = n/2 p 11: 2 � _n_ 1 611: = R.H.S. (Proved) = 9 sin n/3 g -J?, pn 2 Example 24 : Show that Jo tan 8 d8 = - se c - and indicate the restriction on the values of P. H e nce , find r::::::.,:; r n/2 � r n/2 \/cot 8 d8. v tan 8 d8 and Jo Jo f0n/2 tan p 8d8 Solution 12 = J; sinP 8 co s -P 8 d 8 ( 2r(1) - +1 = J r (1;P ) r ( P ) 2 1-p > p<1 OJ (1p+>p -> 1OJ (& - 1 < p < 1) 11: 11: 11: 1 1 pn = sec ( ) = 2 cos p 2: 2 2 s in P + \ 2 2 2 n/2 � n/2 � 71: vtan 8 d8 = ✓2 ' . and for p = -1/2 we hav e JO vcot 8 d8 O = For p = 1 /2 we have 12 f Example 25 : Show that J; .Jcot8 d8 = J r (±) r ( ¾} Solution : W e know that J;' Now, 2 s in P 8 co s q 8 d8 = 2 r' 0 ottee d8 ✓cco q 1 r ( P ;1) r ( ; ) + +2 2r( P � ) n/2 co s112 8 2 d8 = J; ' sin -112 8 co s11 2 8 d8 112 8 s in = Jo (using formula) IES MASTER Publication 298 Multiple I ntegral = Example 26 = Prove the following integral values: 2r(1) r(m)cos me m-1 oo -ax e COS bX X d X = ----'-,---'-�� 0 (a 2 + b 2 t 12 J (a) Jo e (b) Solution r ( ±) r ( ¾ ) oo -ax . Sin bX X m -1 _ r(m)sin me dX - 2 2 (a + b )m 12 r(n) oo -kx n -1 We know that o e x dx = kn J Let us take k = a - ib, then r(m) (a - ib t = where e = tan - 1 (bla) r(n)(a + ib r 2 m 2 (a + b ) (a + ib) = r(cose + i sin e). Now, let r = (a2 + b2 ) 1I2 and e = tan-1 (b/a) Then, 2 2 2 r(m)(a + b r' (cos e + i si n e r Therefore (a + b )m 2 2 r(m)(cosme + i sin me) (a 2 + b2 r1 2 Equating real and imagi nary parts, we get Jo e 00 Corollary : -ax cos bX X m -1 _ r(m )cos me dX (a 2 + b 2 )m/2 and If we put a = 0, so that e = % in the above results, we obtain r(m) mn J oo x m-1 cos bx dx = -m- cosa b 2 oo 1 / Example 27 : Show that Jfo e- x dx = vn. 2 Solution = 2 and r(m) sin me (a 2 + b2 )m /2 ) mn Jooo x -1 sin bx dx = r(m sin 2 bm m Putting x2 = y or x = y 1I2 so that dx = -½ y 1I2 dy . We get Proved � 3n rr 3/2 Example 28 = Prove that Jo (cos2e) cos e de = ---r:::· /4 Solution = (using prop (6) Putting 1 6v 2 .J2. sin e = sin� ⇒ d0 = cos (j> d(j> ✓ 2. cos e = J;/ (1 - sin <j>) = 1 2 ✓ 2 2 Engineering Mathematics 3I2 1 n/ 2 4 12 cos <j> d<j> = ✓ 2 J0 cos <I> d<j> ✓ 1 nt2 . ,i, 1 r(1/2)f(3/2) 4 ,i, A- _ r;:;, r sin o '!'cos 'f' d 'f' - r;:;, = _ i o J r(3) '-I L. v2 Example 29 : Evaluate Solution : J� ✓(1 - x ) dx. 4 ✓n • (3/2) r(3/2) r(3) 1 2 = ✓ ✓n • (3 t 2) · J ✓n 2-2·1 = 299 3 TC 2 1 6✓ 1 3/ . Putting x4 = t or x = t 1 /4 so that dx = - t- 4 dt 4 1 = r\ 1 - t)1 ' 2 . ..!r3' 4 dt = ..!. r t -3I4 (1 - t)1' 2 dt o Jo 4 4J 1 11 4 1 312 1 ..!_ r(1/4)r( 3/2) = ..!_ Jro t - ( 1 - t ) - dt = ..!. l3 ( ..!. �) = ] 4 4' 2 4[ 4 r (¾ + U - ..!. r ( ¾ } J l 4 � r (�) 4 4 n/2 d0 n/2 � Example 30 : S how that i0 � x J0 _i(sm 0 } d0 = TC. _i (sin 0} Solution : and n) - ✓ - ..!. r (¾) l 6 r (�) 4 ✓n j Pr oved. f{1/4} f{1/2) n12 . n12 d0 2 0 f �= = f (s1n 0r11 cos 0 d0 = ( 2f(3/4) ) o o .J(sin a ) r(3/4) r(1/2) 12 12 J; .J(sin 0) d0 = J; (sin 0)1I2 cos0 0 d0 = 2r(5/4) Multiplying (i) and (ii), we get n/2 d0 r(1/4) r(1/4)r(1 / 2) f{3/4)r(1/2) r n /2 x r .J(sin0} d0 = x = Jo .J(sin 0} J o 2r(3/4) 2r(5/4) 4r(5/4) 2 1 dx TC x dx r1 � . xr �= Example 31 : Prove that Jo r;:;, Jo 4 4 v(1 + x ) 4v2 v (1 - x ) Solution : Let Put x2 = sin 0 I1 -- 1o 1 . . . (ii) TCr(1/4) = TC 1 4 x - r(1 / 4 ) 4 2 x dx ✓(1 - x ) 2xdx = cos 0 d0 11 = ✓n ✓n . . . (i) 4 ..!_ r n/ 2 J[sin0} - cos 0 d0 = ..!_ r n' \ sin 0}1 / 2 d0 = 2 T siJ 0coso 0d0 2 o cos 0 2 Jo 2 Jo . IES MASTER Publication 300 Multiple I ntegral = Let � = � . r(3/4) r(1/2) 2 2r(5/4) r(3/4) r(1/2) 13 / 411 / 2 = 4r(5/4) 1114 Jo ✓(1dx+ x4 ) 1 . . . (i) 2 Put x 2 = tan $ so that 2x dx = sec $ d$ dx = 12 = Put 2$ = t' so that 2d$ = dt sec2 $ d$ 2 tan $ ✓ 1 r'4 2 d$ 1 2 (sin $cos$) = ✓ ° ✓ r'4 ✓(sind$2$) ° _ .!_ 2 1 r(1/4) r(1/2) 1 rr/ rr/ 2 112 2 · sin 2 tcos 0t dt = f 12 = � Io (sin t f dt = ✓ 2r(3/4) 22 0 2✓ 1 . r(1/4) r(1/2) 2 ✓ r(3/4) 4 12 = M ultiplying (i) and (ii), we get L.H.S. . . . (ii) 1 r(1/4) r(1/2) TC r(1/4) r(1/2) 2 · 2 x ----- = 2r(3/4) r(3/4 ) 4✓2 4✓ (·: ✓1!2 = ✓n ) DOUBLE INTEGRALS Theorem : If the region A is the area bounded by the curves y = f/x), y = fz(x) and the coordinates x = a and x = b, then JJ f(x, y) dA = A y x=a f x) f b f 2{ [f(x, a f1 { x ) y) dy] dx y = f2 (x) x=b A = f, (x) 0 - y-+- - ----+ X where the integration is carried w.r.t. y first and x is treated as constant i .e. we sum along a vertical strip. -;;0-;:; Remark : It can be shown in the same way that if the region is bounded by the curves x = g 1 (y), x = gz (y) and y = c and y = d, then Y ffA f(x, y) dA = Jcr d = J d f 9 2M c g {y) J 1 [f 92 <Y J g1{Y) \(x, y ) dx] dy f(x, Y ) dx dy if the brackets are omitted . � II X Multiple Integral : Consider a function y = f(x), then integration of f(x) w.r. to x is given as f: ydx = J: f(x) dx X C Q-+-----'----'---+ Here, the integration is fl rst carried w.r.t. x treating y as a constant i .e. we sum along a horizontal strip first. I = � II Now, consider a function Z = f(x , y) then concept of i ntegration can be extended to z also as follows; X b y2 b Y2 = J J zdydx=J J f(x, y) dy dx a Y1 a Y1 d x2 b x2 I = Jfc J zdx dy=J J f(x, y)dxdy or x1 a x1 This generalisation of definite integrals are known as multiple integrals. Engineering Mathematics 301 ........ (Type 1 ) ...... . .(Type 2) Note : Here, y1 = f 1 (x) and y2 = fix); x 1 = f 1 (y) and x2 = fiY) Double Integral Working Rule Consider a function z = f(x, y) 1. 2. 3. If integral is of the type (1 ) i.e. variable limits are of y and constant limits are of x then we first integrate dy keeping x constant and then dx as follows; I = ! a {Jf2 (X) f(x, y) dy} dx b f1 (X) If integral is of the type (2) i.e. variable limits are of x and constant limits are of y then we first integrate dx keeping y constant and then dy as follows; f {J 2 M f(x, y) dx} dy I = Jc f1 ( Y ) d 1 If both limits are constants then we can integrate dx and dy in any order i.e. Note : If f(x, y) has discontinuities within the region of intergration then above point (3) in not applicable always i.e. DOUBLE INTEGRAL IN POLAR COORDINATES If P(x, y) are Cartesian coordinates & P(r, 0) are Polar coordinates then x = r cos0 and y=r sin0. Because in general 0 1 and 0 2 are constant and r1 and r2 are variables, so we will first integrate dr keeping 0 constant, and then d 0 . Area in Cartesian Coordinates 1. Let y = f(x) and y = g(x) be two curves and we have to find area enclosed by these curves using Double integral concept then it will be as follows. Area = HR (1) dxdy = y ( ) Area=J x= b - g x (1) dy } dx x a y =f( x ) {f 2. Similarly Area bounded by the curves x = f(y) and x = g(y) is Area in Polar Coordinates : Area = IIR (1) dy dx=fY=C Area = y=d HR (1) dxdy {J X=g( ) dx} dy X=f( y ) y = Jt!JI drd0 IES MASTER Publication 302 Multiple Integral 2 Area =J: {( rdr } d0 1 (J = Jacobian) {Since I J I = r in polar coordinates} Now : Let 0 = oc and 0 = p be two radii vectors of the curve r = f(0) then area bounded by this curve between the radii vectors 0=oc and 0 =p is Area = B=P {f r=f(O) r dr} d0 f8=oc r=O Here r is the distance of strip from origin, and 0 is the angle in an anticlockwise direction. Some I mportant Formulae (for Double Integral) 1. = .! . tan- 1 ( �) f� a x 2 +a 2 a 5. J h = log ( x+ x2 -a2 ) x 2 -a 2 3. 7. 11. - J_ log ( a+x ) f� 2 2 2a a-x a -x f ✓x2 +a 2 dx=� ✓x 2 +a2 +� log (x +✓x 2 +a2 ) 6. dx . -1 ( X ) . -=== Sin 2 2 a Ja -x 8. 1 0. if f(2a-x) = f(x) if f(2a-x) =-f(x) y1x = log(x + ✓x 2 +a 2 ) fh 2 2 x +a f f ✓x 2 -a2 dx =�2 ✓x2 -a 2 - a2 log (x+ ✓x 2 -a 2 ) 2 e - - (a sin bx - bcos bx) f e ax sin bx dx = a 2 +b 2 ax J2J; f(x) dx ; 12. C f(x) dx= O; l eax ax - - (a cos bx + b sin bx ) J e cosbx dx=2 a +b 2 Example 32 : Evaluate : J� dx J; e - J_ log( x - a ) f� 2 2x+a 2a x a 4. ✓ 2 J f (x) dx ; 1 3. J: f(x) dx = f : . o .· l 2. dy. = (e-1) [_{] = .!. (e-1) 2 o 2 1 Solution 1 ✓1+ x 2 Exam ple 33 : Evaluate f o o Solution if f(-x)= f(x) i.e.f(x) is even if f(-x) = -f(x) i.e. f(x) is odd 1 ✓1 + x2 fO O f f dydx 1+x 2 +y2 dydx = 1+X 2 +y2 fO1 dx fOQ 1+Xdy 2 +y2 = fo dx 1 J1 + x 2 n 4 = 2: 4 1 1 Y_ = Jfo dx [---tan_1 __ ,---:-Z ,---:-Z v1 + xv1 + x- o f� -1-2 dx J1 + x ✓ ]Q 2 = 2: [1og {x+ x +1 }] 4 0 1 Example 34 Solution : Engineering Mathematics 2 J2 x - x Evaluate f J dx dy. Jo O 2 1 2 2 r 2 r hx-x dx dY = r2 [ Y ]hx- x dx o Jo Jo Jo = J: ,hx - x 2 dx = (on i ntegrating w.r.t. y fi rst, treating x as constant) f: ✓X(2 - x) dx Putting x = 2 sin 2 8, dx = 4 si n 8 cos8 d8, we have = J; Example 35 : Evaluate Solution = 12 ✓2 si n 2 2 8 - 2 cos 8 4 si n 0 cos 0 d0 = J; 12 ✓2 sin 2 2 0 - 2 cos 0 4 si n 0 cos 0 d0 ff (x + y) dx dy over the region in the positive quadrant for which x + y � 1 . x + y = 1 is a straight line making intercepts unity from the positive direction of the axes. Hence, i n this case. I f we take any strip parallel to y-axis then i t i s bounded b y x-axis a s one side for which y = 0 and by the line y = 1 - x. On the other side for which Limits of x are clearly O to 1 . Hence, JfA (x + y) d dy = f� J;-x (x + y ) dx dy = = Example 36 : Evaluate Solution = 303 = f� f� [ 2 1-x y ] dx = xy + 2 o (1 ) } dx fa1{ x(1 - x ) + + 2 2 2(x - x 2 ) + x 2 - 2x (1 - x ) ) dx : } dx = J; ( { 2 1 i[x - x; ]=i( -i)=i ff xy(x + y) dx dy over the area between y 2 (0, 1 ) 0 X (1 ,0 ) X "'"or--_......_ y = O_____,,_,........ = x and y = x . Let us first m ark the region of i ntegration which is bounded by and Solving (i) and (ii), we get y = x y = 2 . . . (ii) which is a parabola. X x2 = X . x = 0 . . . (i) which is a straight line or x = 1 which gives y = 0 and y = 1 respectively. Thus, the point is ( 1 , 1 ). Now, divide the region i nto strips parallel to y-axis. This strip is bounded by the line y = x on one side and y2 = x or y = on other side. .Jx Thus, the lim its for y will be y = x as lower limit and y = ./x as upper l i m it. Then the lim its for x will be O to 1 . Hence, the given integral = J; fx'fx xy(x + y ) dx dy 1 x 2 y 2 xy·3 = f [- + Jo 2 3 ]../x X dy IES MASTER Publication 304 Multiple I ntegral = f 1 (x o Example 37 : Ev a l uate Solution : y x 5 4) + - x dx 2 -3- 6 3 512 x4 2x 712 x 5 ] _ 1 2 1 _ 3 _ [- -+--- - + -- - 8 2 1 6 56 8 21 6 0 1 ff xy dx dy over the region in the positive quadrant for which x + y � 1 . x + y = 1 represents a line AB in the figure x + y < 1 represents the plane OAS. The region of integration is OAS as sh aded in the figure. By drawing PQ pa ra llel to y-axis, P l ies on the line AB (x + y = 1 ) a nd Q lies on x-axis. The lim its of y a re 1 - x a nd 0. Required integra l f� x dx ft y dy = f� y x dx [ ; r x r1 2 = ..!. J (x dx) (1 - x ) 2 o 1 x 2 2x 3 x 1 2 3 = - f (x - 2x + x ) dx = - [ - - - + -] 2 Jo 2 2 3 4 0 4 1 Example 38 : Ev a luate Solution : fJ xy R = 2: 1 p .Y "" Q II ''-1- y=0 0 A (1 , 0) dx dy where R is the qu adra nt of the circle x 2 + y2 = a2 where x :2'. 0 a nd y :2'. 0. Let us consider region of i ntegration be the first qu adra nt of the circle OAS. First we integrate w.r.t. y a nd then w.r.t. x . The limits for y a re 0 a nd f f xydxdy R Example 39 : Evaluate Solution : J[J- �+ ±] = B y fJ xy A ✓a 2 _ x 2 and for x , 0 t o a . y 0 -+--y.......... = 0-�---x A O) (a, (0,0) ff xy dx dy where A is the domain bounded by x-axis, ordinate x = 2a and the curve x = 4ay. 2 A dx dy The a rea of integration is OAS. Here we integrate fi rst w.r.t. y a nd then w.r.t. x . The lim its of integration a re O to 2 a for x a nd 0 to � for y. 4a 2 2a r / 4 a r x Thus, xy dx dy = xy dx dy Jo J o A 2 JI = /4 a 2a y2 x r x 2 /4 a r 2a y dy = J x dx [ ] x dx Jo Jo 0 2 0 2a __ 1 r 2a 5 x4 x dx dX = -_ = Jro x 32a 2 32a 2 J o y 2 B ----=o--=--+ --..-.x A (2a , ) (0,0) """"- O Engineering Mathematics 305 64a6 1 x6 ] 1 6 = = [( 2a) - O] = [ 1 9 2a 2 32a 2 6 o 1 92a 2 2a Second Method a4 3 Y . Now, we integrate first w.r.t. x and then w.r.t. y. The limits of integration are '14ay to 2 a for x and 0 to a for y. JJ xy dx dy=f: y dyf�x dx Thus, A = J; y dy [ = Example 40 : Evaluate Solution I = 1 Example 41 : Find 1 Find f1 J:' 2 1 Example 43 s a 2 � ] 2 a . 2 ,/4ay _ Y; I = f y( 2a2 -2a y) dy= 2a f (ay- y 2 ) dy O 0 =2a [� -�l=� 4 dx dy dz. <x+y +z ) e<x+y+z)dx dy dz = I:I: u:+� e V < x +y+z ) dz) dx dy 3 e 4x 3e 2x e1 2 � 6 e 12 3 +e x ] = - e +e3 _ ..!.+� - 1 = - --e 6 +e 3 -8 4 8 4 8 4 8 4 8 o z J: f: (x + 2 y 2 ) dy dx. f: foa (x2 +y2 ) dy dx Solution = Solution s: :+y e s:s: s:+y = [ Example 42 I: 2a[ x2 I f12 JfoY/2 y dy dx. 2 y dy dx = Jt ( f:' dx ) ( ydy) = 1 µ y Find fO fO Let = sine 2 ft [xy]::b' 2 = ft( y; ) dy = (i x y; )� = i( y dy dx. dy ,. : -1) 2 = Solution y ⇒ dy=cos e de and for y = 0, 0=0 and y � = 1 , 0=% IES MASTER Publication 306 Multiple Integral Therefore, from (i), the required integral is Example " Solution 2 12 " 2 (sin 30+sin0) = f " sin0 cos 2 e de = ..!_ fo"' sin20 cos e de = ..!_ f ' de J Jo 2 2 2 Jo " '2 1 cos 30 1 1 1 = - 4 [-- +cose ] = +4 [ +1] = + 3 3 3 o 1 2 Evaluate ff .jxy - y dy dx, where S, is a triangle with �ertices (0, 0), (1 0, 1 ) and ( 1 , 1 ). s y Let the vertices of a triangle OBA be (0, 0), (1 0, 1 ) and (1 , 1 ). Equation of OA is x = y. 1 Equation of OB is x = 1 0 y. The region of �OBA, given by the limits 7 y :,:; X :,:; 1 Oy and O :,:; y :,:; 1 = 6 0y 2 2 ffs .jxy � y dx dy = J� dy ( (xy - y ) 112 dx = Example ,1 Solution 1 /2 a 2 Evaluate 0n fa(1-c s8) r dr de . o 0 = 18 [ f a fOn12 d e fa(1-cos8) r2 dr de 1 21 l1 dy [-(xy - y ) 3y = 0 r' 2 2 312 y ] =�=6 3 o 3 3 1 ] 10 Y y = f01 -23 -y1 (9y2 )312dy = 1 8l0 y2dy 1 r3 ] 3 a(1-cos8) de [ a 3 3 "2 a 3 f '2 [ £ a (1 - cose ) ) " 1 - (1 - cose )3 ] de = = f ' de ( Jo 3 3 3 Jo a 3 12 = - f " [1 - (1 - 3 cose + 3 cos2 0 - cos3 0 )] de 3 Jo Example ,, Solution 1 1 Evaluate ffr de dr over the area of the circle r = a sine . r = a sine is a circle with its centre C on the line e = %· Initial line is tangent at the pole 0. Circle is symmetrical about the line e = !:. . Hence, the value of the integral 2 = 2 value of the integral of the half circle 307 a2 sin 20 n/Z a 2 n/2 ] [e= = - Jr (1-cos2e ) de 2 2 o 2 o TRIPLE I NTEGRALS It is denoted by JffR f (x, Y, z) dV �[%] r =a sine a2 = n 4 and is called the triple integral of f(x, y, z) over R. By taking dx dy dz as the element of volume, we can easily find the volume of a solid by means of triple integral Hf dx dy dz under suitable limits. 1 . Consider a function u = f(x, y, z) then Working Rule 2. If all the limits are constants then above order of integration is immaterial provided limit changes accordingly i.e. J:=J:=c (e (f) dz dy dx=(J:=c f:=a f dx dy dz=J:=J;= J dy dz dx Multiple I ntegrals We can further extend the triple integrals to finite integrals of higher order as log 2 x r x + log y x +y+z e dz dy dx Example 47 � Evaluate the integral Jfo Jfo J o Solution 1 ff .. .f dx 1, dx 2 . . .dx n . log 2 x+log y z x log 2 r x e ydy ( ez ) x+log y = Jr e xdx Jr e ydxJr e d z= Jro e xdx Jo 0 o o o X log 2 log 2 X = Jf e XdxJf e ydy (ye X -1)= Jf e xdx Jf (ye x -1) eydy o o o o log 2 e3Iog2 e 3Iog2 1 1 = -- e31og2 ------ +elog 2 +-+--1 3 9 3 9 3 8 8 8 1 1 8 19 = -log-2 - --+2 +-+--1=-log2-3 9 3 9 3 3 9 IES MASTER Publication 308 Multiple Integral Example '8 Solution 1 2 2 2 Evaluate HfR (x +y +z ) dx dy dz where R denoted the region bounded by x = 0, y = 0, z = 0 and x + y + z = a, (a > 0). fHR (x2 +y2 +z2 ) dx dy dz 1 x + y + z = a or z = a - x - y Upper limit of z = a - x - y On x-y plane, x + y + z = a becomes x + y = a [ •: z = 0 on xy plane] Upper limit of y = a - x Now on X-axis upper limit of x = a = J:=o dx = J;:: d y J:::-y (x +y +z ) dz = 2 2 2 dx f: f:-x 3 a- x - y z dy(x 2z+/z+ ) 3 0 f: dx f:-x dy [x 2 (a-x-y)+y2 (a-x-y)+(a-x3-y) 3 [ ·: on x axis y = OJ ] 4 a- x a y 3 y4 � x 2 y2 (a-x-y) 2 f � --+(a = x) dx x (a-x)y ] Jo [ 3 12 2 4 0 4 a (a-x)3 - (a-x)4 (a - x) ] x2 2 2 + = ro dx [x 2 (a-x, --(a - x) +(a-x) J 2 3 4 12 Example '9 Solution 1 1 dx dy dz if the region of integration is bounded by the coordinate planes and the (x+y+z+1)3 plane x + y + z = 1 . Compute Hf Let the given region be R, then R is expressed as o s i s 1-x-� 0 s y s 1 - � 0 s x s 1 dx dy dz iJ JrR (x+y+z+1) 3 = 1 1 dz fo dxfo-x dyJro1-x-y (x+y+z+1) 3 1-x-y 1 1-x 1 = fo dxfo dy [-2(x +y +z+1)2 ] o 1 x dy = -� f dxf~ 0 [ (x+y+1-x - y+1)2 2 ° 1 (x+y +1)2 ] Engineering Mathematics 309 1-x 1 1 1 ..!_ r1 dx Y+ 1 _ dy = ..!_ ro dx ro -x [ ..!_ ] [ J 4 (x + y +1)2 ] 4 x +y +1 0 2J 2 Jo 1 1 1 (1 - x)2 X 1 = -2 [--- + 2 - log(x+ 1)] = - 2 [2 - log2+ ] 8 8 o 1 Example 50 Solution 1 1 Evaluate = _ ..!_ [� - log2 ]= ..!_ log2 - � 2 8 2 16 fffs xyz dx dy dz where S = [(x, y, z) : x 2 +y 2 +z 2 s 1 , x � 0, y � 0, z � OJ S is the sphere x2 + y2 + z2 = 1 . Since, x, y, z are all positive, we have to consider only the positive octant of the sphere. Now, x2 + y2 + z2 = 1 so that z = ✓1-x 2 - y2 . The projection of the sphere on xy plane is the circle x2 + y2 = 1 . This circle is covered as y varies from O to = fff xyz dx dy dz s = Jfo x dx Jfo Q 1 ✓1-x 2 , and x varies from O to 1 . [ 2]� 1 1 1-x -y z dz= fo x dx f -x y dy � y dy fo� o J J J 2 r 1 X dx rQ y dy 1 - x = Jo [ Jo 2 2 l]=21 Jro1 -R X rQ (y - X 2 y -y 3 ) dy dx Jo 0 (1 -x--')'-l)Q =-1 ro x dx [1-x - x (1 - x ) - -'-] - 1 1 x2 y 2 = _ ro x dx - ----2J 2 2 4 (l 4J 1 2 2 2 4 4 = ..!_ ro\ dx (2 - 2x -2x + 2x - 1 - x +2x ) 0 2 2 2 2 2 2 aJ s 1 1 1 x2 x4 · x6 = - f (x -2x 3 + x ) dx=- ( - -- + - ) = 8 Jo ' 8 2 2 6 0 1 Example 51 Solution 1 2 ll ll/4 a 2 Evaluate fo fo ro r sine dr de d�. J J J s:ll s;'4 s: r sine dr de d� = 2 s:ll s;'4 des: d� sine r 2 dr=e d� ll s;'4 r sine de [ ; I n/4 a3 2 n a3 2 n d� f sine de= f d� [-cose]�I4 f o 3 o 3 o ll ll a = � s: d� ( - cos ¾+cos o)= : (1 - l) S: df = 2 1 2na 3 ✓ a3 2 { -1 ) = - ( 1--2) 2n=✓ ✓ 3 3 CHANGE OF ORDER OF INTEGRATION If f has no discontinuities within the region of integration and limits of x and y both are constant then the order of integration in fJR f(x, Y) dx dy is immaterial. IES MASTER Publication 31 0 Multiple Integral But if either of the limits is in variable form then by changing order of integration we can easily solve the double integration. Let us consider the integral fb J f2 (X ) a f1 (x) f(x, Y ) dx dy . This integral is a summation of strips parallel to y-axis. If we change the order of integration, then the summation will be that of strips parallel to x-axis. Flow chart to change order to Integration Type-I 1. 2. 3. If variable limits are of y; it means it is given that strip is parallel to y-axis. Now, make a strip parallel to x-axis. Now, put the limits according to the new strip. Type-II 1. 2. If variable limits are of x ; it means it is given that strip is W to x-axis. Now, make a new strip W to y-axis. 3. Now, put the limits according to new strip. Change of Order in Polar Coordinates Process is similar to that adopted in Cartesian coordinates. The only difference is that in order to evaluate an integral of the type HA f(r, 0) d0 dr The summation is completed by dividing the region into triangular strips ,(unlike the rectangular strips in Cartesian coordinates). Then we make strip-wise summation to cover the whole area for which the value 0 will be suitable chosen. When we change the order of integration, we get an integral. of the type HA f(r, 0) dr d0 Here, the integration is first performed with respect to 0 , r being kept constant which means that the summation is first made inside a circular strip. Now, the value of r is chosen so as to cover the whole region. Example 52 : Change the order of integration J: J: f(x, y) dx dy . Solution : The region of integration is bounded by y = 0, y = x, x = 0 and x = a. Thus OAB is the region of integration. Consider the elementary strip parallel to x-axis value of x at the extremities of this strip are x = y and x = a. These will be the lower and upper limits of x. Also, y varies from O to B, hence its limits are O to a . B (a, a) 0 a a a x r f(x, y) dx dy = r f f(x, y) dy dx Therefore • Io Jo Jo y Example 53 : Change the order of integration Solution : i a lx f 0 mx f(x, y) dy dx. Here the region of integration is bounded by the lines y = mx, y = lx, x = 0 and x = a. Thus, the region of integration is clearly OPQO. The region of integration is divided into two parts namely OPM and PQM. Now, in the region OPM any strip parallel to x axis has the limits for x as y/l and y/m. Also, the limits of y are from O to ma. y=O A Engineering Mathematics 31 1 Again in the region PQM any strip parallel to x axis has the limits for x as yll and a and limits of y are from m a to la. Hence, we have f0a ilxmx f(x, y) dy dx = fma f y/m O y/1 f(x, y) dx dy + fla fa ma y/1 f(x, y) dx dy Example 54 1 Changing the order of integration of J: J: e-xy sin nx dx dy, show that r oo sinnx dx = � Jo X 2 The region of integration is bounded by x = 0, x = oo , y = 0, y = oo , i.e. first quadrant. Solution c -xy oo oo e-{-y sin x - n cos nx} ] fo dy fo e-xy sin nx dx = Jfooo dy [2 n + y2 = 00 "' f dy [o + �] = f �dy = [tan - 1 yI = � o J Jo 2 x +y x +y On changing the order of integration, J: J: e-xy sin nxdxdy J: sin nx dxJ: e-xy • dy sin nx [ -1 ] e-x J sin nx dx [ - -J = fo --dx - = J:J: sin nx dx 00 o = r oo Jo = From equation (i) and (ii), Example H Solution 1 I f0 00 sin nx X x y 00 00 o X 00 sin nx sin nx dx [-0 + 11 = o dx J X I X r 00 0 . . . (i) 00 exy o dx = � 2 . . . (ii) 1 2-x Change the order of integration in I = f f 2 xy dydx and hence evaluate the same. 0 x 1 2 x I = fo J 2- xy dy dx J x The region of integration is shown by shaded portion in the figure bounded by parabola y = x2 and the line y = 2 - x. The point of intersection of the parabola y = x2 and the line y = 2 - x is 8(1 , 1 ). In the first figure , we have taken a strip parallel to y-axis and the order of integration is 1 2 -x xydydx 0 2 x I n the second figure, we have taken a strip called parallel to x-axis in the area OBC and second strip f f in the area ABC. The limits of x in the area OBC are O and are O and 2 - y. So, the given integral is I = f f xydxdy+ ff xydxdy Jy = o y dy o f 1 f OBC ABC [ ]✓Y 2 2 y 1 x2 x dx + f y dy Jo - x dx = Jo y dy 2 1 o + f y dy 2 1 ✓Y and the limits of x in the area ABC [ 2 2- y x 2 ]o IES MASTER Publication 31 2 Multiple Integral y=x 2 = ..!. 96 -128 +48-24 +1 6 -3 __!_ � � � + [ ]= + = = 12 6 2 6 24 24 8 __!_ Example 56 = Evaluate JO oo Solution : J"' ey- - dy Y . dx. II , Here the elementary strip PQ extends from y = x to y = oo and this vertical strip slides from x = 0 to x = oo . The shaded portion of the figure is, therefore, the region of integration. X On changing the order of integration, we first integrate w.r.t. x along a horizontal strip RS which extends from x = 0 to x = y. To cover the given region, we then integrate w.r.t. 'y' from y = 0 to y = oo . I = = J ' J �dydx = J J �dxdy 00 00 x;Qy;x -y 00 f y;Qx;Q Y "' e-y y - dy= 0 y = -[�-1 ]=1 Y f y Q -\- Y [r e- y 1 e dy= - - [ - ] -1 0 eY 0 "' _ y X -y R 00 -f USE OF JOCOBIAN IN MULTIPLE INTEGRALS Definition If u = u(x, y) and v = v(x, y) i.e. u and v are the functions of x and y then, Jocobian of (u, v) with respect to (x, y) au . . a(u, v) ax in defined as, J(u, v) = -= a( x, y ) av ax au ay av ay Similarly, if u = u(x, y, z), v = v(x, y, z), w = w(x, y, z) i.e. u, v, w are the function of x, y, z, then, Jocabion of a(u, v, w) (u, v, w) with respect to (x, y, z) is defined as, J(u, v, w) = a(x, y, z) Chain Rule for Jocobian au ax av au av ay ax ay aw aw ax ay au az av az az a(u, v) a(x, y) . . _ 11 1 . If u = u(x, y) and v = v(x, y) 1. e. u and v are the functions of x and y then a(x, y) · a(u, v) =1 or [J · J' - Engineering Mathematics 2. If u = u(x, y); v = v(x, y); x = x(r, s); y = y(r, s) 31 3 . i.e. (u, v) are the function of (x, y) and x, y) are the function of r(s). a (u, v) a(x, y) a(u, v) Then - - • - - =- a(x, y) a (r, s) a(r, s) ' Jacobian of Implicit Functions If u and v are the implicit functions of x and y given by the relations f 1(u, v, x, y) = 0 and fiu, v, x, y) = 0 then Jacobian of (u, v) with respect to (x, y) is a(f1. f2 ) a(u, v ) = (-1 2 ) a(x, y) a(x, y) a(f1, f2 ) a(u, v) Note : Similarly, if u, v, w are the implicit functions of x, y, z given by the relation f 1 (u, v, w, x, y, z) = 0 and fiu, v, w, x, y, z ) = 0 and fiu, v, w, x, y, z) = 0 then Jacobian of (u, v, w) with respect to (x, y, z) is Functional Dependence a(f1, f2 , f3 ) a(u, v, w ) 3 a(x, y, z ) = (-1) a(x, y, z ) . a(f1, f2 , f3 ) a(u, V, W) Two function u and v are called functionally dependent if their Jacobian is zero, i.e. if u = u(x, y) and v = v(x, y) then u and v are dependent if a(u, v) =0 a(x, y) INOTE( 2. Two functions u and v are not independent if their jacobian is 1. Functionally dependent means there exist a relation between them. zero (true). Transformation on Change of Variables · Sometimes, it is convenient to change the variables in the evaluation of multiple integrals. The process of changing the variables in a multiple integral is called the transformation of multiple integrals. Suppose we want to transform the multiple integral to another system of variables u, v where JfA f (x, y) dx dy x=�(u, v) and y =\Jf (U, v), say then the transformation would be completed in the following three steps: (i) Determine the function f(x, y) in terms of u, v this is done by algebraic substitutions and eliminations. Let F(u, v) be its new value. (ii) Assign new limits. (This is also an algebraic process and geometrical considerations.) for u & v . (iii) Determine the new element of integration i.e. J. From differential calculus, we know that dxdy = Jdudv, where J is the Jacobian of transformation and is defined as Thus, we have ax ax a(x, y) au = J = a (u, v ) av Jf f(x, y) dx dy=fJF(u, v) J du dv ay ay av au IES MASTER Publication 314 Multiple Integral a(x, y, z) In case of three variables, we have dx dy dz = u, v, w ) du dv, d w a( MULTIPLE INTEGRAL USING CHANGE OF VARIABLES CONCEPT Let x and y are the functions of u and v as follows: x = <l>(u, v) and y='I'(u, v) Let J = Then, �-�� a(x, y) and we know that !dxdy=!JI dudvl a(u, v) I = HR f(x, y) dx dy=HR' f (u, v)-IJ!du dv where, R and R' are the region of intergration in (x, y) coordinate system and in (u, v) coordinate system respectively. 1. Change Cartesian Coordinates into Polar Coordinates Let P(x, y) are the cartesian coordinates then P(r, 0) are called Polar coordinates if x = r cos0 and y=r sin0 J = Now, a(x, Y) =r so; a(u, v) jdx dy= !JI drd0 = rdr dej = JR f(x, y) dxdy=HR. f (r, 0) r dr d0 And, where R and R' are the region of integration in cartesian coordinate system and in Polar coordinate system respectively. """ Gene a l ly r varies from 'o' to 'r' and Nr::Jr _ E� ', '•I_ � question. 0 varies from o to 21t but it may be different according to 2. To change cartesian coordi nate into spherical coordinates Let P ( x , y, z) are cartesian coordinates then x = r sin0 cos $, y=r sin0sin$, z = r cos0 P(r, 0, <I>) are cal l ed spherical coordinates if a(x, Y, z) 2 2 J = 8(r, 0, <I>) =r sine and ldxdydz =!JI dr d0 d$=r sin 0drd0d<l>I Now, 2 I = HJR f (x, y, z) dx dy dz = Htf(r, 0, $) r sin0 drd0 d$ So, where R and R' are the region of integration in cartesian and spherical coordinate system respectively. � General ly r varies from 'o' to 'r'; 0 varies from o to 1t (on z-axis); and INr::JrE • - xy plane). <I> varies from O to 21t (on But it may be different according to question. 3. To change cartesian coordinates into cylindrical coordinates Let P(x, y, z) are cartesian coordinates the P(p, <I>, z) are called cylindrical coordinates if Now, x = p cos <1>, y=p sin <j>, z =z a(x, Y, z) J = 8(p, qi z)= p and jdxdydz =!JI dr d<I> dz=p dr d<I> dz.j Engineering Mathematics 31 5 I = HJR f(x, y, z) dx dy dz=Ht f(p, $, z) dr d$ dz where R and R' are region of integration in c artesi an and cylindric al coordinate system respectively. Generally P varies from o to P ; $ varies from O to 2n (on x y plane); and z varies from -z to _ INOTE ( • - z on z-ax,s. But it may be different according to question. AREA AND VOLUME IN DIFFERENT COORDINATES SYSTEMS 1 . In cartesian coordinate system 2. In polar coordinate system 3. In spherical coordinate system 4. In cylindrical coordinate system Area Area = = Volume HR (1) dxdy; Volume=HJR (1) dxdy dz HR (r) dr d0 = 2 HJR (r sin0)dr d0 d$ Volume = JHR (p)dp d�dz where limits can be rE;1pl aced according to the coordinate system. Example 57 1 Evaluate the double integra l Solution 1 fO fo a Ja 2 -x2 (x2 +y 2 ) dx dy by changing to pol ar coordinates. L aws of transformation from C artesian to pol ar is x = r cos0, y=r sin0 x2 + y2 = r 2 (cos 2 0 + sin 2 0)=r 2 J = ox ar ox dx dy = rd0 dr case ar -I ify -rsin0 ify ae sine =r r cos0 I Now, the region of integration is bounded by y = 0 (x-axis), y = a 2 _ x 2 (a circle with the centre at origin), x = 0 and x = a . This is clearly the positive quadra nt of the circle of radius a . In order to cover the area by means of polar coordinates, we find that the limits of r should be from O to a and those of e from O to 2 . Hence, we have 7t 4 rca4 = � [S]�/2 = 4 8 Example 58 1 Ev a luate the integra l Jo Jo . ✓ r1 r x 0=� 2 -�--�--0 = o 0 = (r O) x 3dxdy � by changing to polar coordinates. v x +y IES MASTER Publication 31 6 Multiple Integral Solution : Region of integration is bounded by the lines y = 0, y = x, x integration is the region OAS where B is the point (1 , 1 ). Equation of the line AB is x = 1 . r cos 0 = 1 or = 0 and x = 1 . Thus, the region of r = sec s Therefore, in any triangular strip OPQ into which the area OAS is divided, r varies from O to sec s and then for all such strips 0 varies from O to n/4 . Thus, the integral when transformed to polar i 8 = !.: 2 n/4 l sec e (rcos 0)3 rd0 dr _ l n/4 r4 secs -] cos3 e d e �-�-- o [4 0 o o r ' = i = n/4 o sec 0 -- d0 4 = -1 [log (sec 0 + tan 0)�/4 4 ✓ = }og ( sec ¾ + tan¾ ) = � log ( + 1 ) Solution 1 2 2 2 Changing the given integral into polars by putting x = r cos 0, y = sin 0 · ⇒ x 2 '+ y = r We have using Jacobian dx�y = r d 0 dr . 0=� 2 Since the limits of x and y are from O to oo i.e. the first Therefore, in polars the limits of r wil l be from O to oo and that of e wil l be from O to %. Now, the given integral transforms to f: f: e-(x +y ldx dy 2 2 = r'2 e-r r d0 dr = J;' 2 0 r 0 Solution f r1 o Jo 1 1 2 dx dy ✓(1 - x r 1 r1 2 2 )( 1 - y ) dx dy ✓(1 - x 2 )( 1 - y ) 2 - Jo - For the inner integral let y = sin e ⇒ and for y (J: e-r r dr) d0 2 2 = .! [0]�,2 = . 2 -r = r'2 ( e r d0 = J;'2 ( o + i ): de 0 2 - Example 60 : Find , , Jo J o 0 -,-....�------ 0=0 f 1 1 ✓1 - x dy = cos e de = 0, e = o , for y = 1 , 0 = � 2 2 0 4 [f o 1 1 ✓1 - y 2 dy ) dx . . . (i) r° ' ✓ 1 dy . = r0 -1-y 2 Therefore ✓ Engineering Mathematics r nt2 cos0 de = cose d0 = J o � cose 2 1-sin2 0 2 1 31 7 Substituting this in (i) we get the required integral as fo 1o 1 1 1 1 1t 1t n - fo � - dx = - x - ( by the same process as above) 2 2 2 2 v1 - x (1-x )(1-y ) ✓ dxdy 2 2 = 2 10 taken over the area of the triangle bounded by the lines x = 0, y = 0, x + y = 1 . 1I2 Example 61 : Using the transformation x + y = u, y = uv, show that ff [xy(1-x-y)] dx dy= � . Integration being Solution = 1I2 = Jf [xy(1-x-y)] dx dy Let x + y = u and y = uv or = X U - y = U - UV a(x, y) du dv = dxdy = a(u, v) u x + uv x o Upper limit of u = 1 U On putting x = u - uv, y = uv ax ax au au av du dv av 1 - v -u 1 dudv = u dudv 1 v u X + y = 1 u u(1 - v) y = UJ 0 = UJ u(1 - v) 0, V = 1 U = 0, dxdy = = = = = = ay ay dxdy = ududv in equation (i), we get I = fJ (u-uv) 1I2 r ( 9/2) 2 r(3/2) X = 7 5 3 . 3/2) r 2 '2 ( r( 3) 1 1 2 r (1/2) 2 r (1/2) 2 2 2 2 ✓2 Example 62 : Evaluate r r x-x (x 2 +y 2 ) dy dx . Jo o Solution : Let J Limits of y = ✓2x-x 2 V = (uv)1I2 (1-u)1I2 u dudv r( 3)r( 3/2) r( 3/2)r (3/2) = � ��� x ��� � 0 r(m) r (n) 1 1 r (._. Jo\ m- [1-xt- dx= ) r (m+n) 1 7 5 2 2 2 x = --3 � 2 ✓n ! ✓n 1 2 ✓2 I = r r x-x (x 2 + y2 ) dy dx Jo Jo or y2 =2x-x 2 ...(i) 2n 1 05 ...(i) 2 or x2 + y2 - 2x = 0 2 equation (ii) represents a circle whose centre is (1 , 0) and radius is 1 . ...(ii) Lower limit of y is O i.e. x-axis. Region of integration is upper half curve. Let us convert (i) into polar coordinates by putting IES MASTER Publication 318 Multiple Integral x = r cos 0, y = r sin 0 = 0 or r = 2 cos 0 r - 2r cos 0 2 Limits of r are O to 2 cos 0 & Limits of 0 are O to Then, I = cos9 2 2 r (r d0, dr) J;' : f ' = 4 f" Jo 2 COS4 2 7t = r d0 f: 2 2 r r dr = f:' d0 [ ; I cos9 3 3 x 1 x n _ 3n _ ed S - 4 4 X2X 2 4 cos0 0 ARC LENGTH OF CURVES (RECTIFICATION) This is the process of finding the length of the arc of a curve. Here we wil l discuss the arc length of the curves, whose equations are given in cartesian, polar or parametric forms. It is generally denoted by 's'. (i) W hen the equation of curve is given in cartesian form : s = · fx 1 x2 with the suitable limits of i ntegration. (ii) When the curve is given in polar form: 1+( (iii) When the curve is given in parametric form : s = f � ( I2 dy ) dx dx or 2 s= fY1Y2 ( dx 1+ dy )2 dy dx dy ) + ( ) dt' where t is the parameter. dt dt 2 2 I NTRINSIC EQUATION OF A CURVE Let s be the arc length of the curve measured from some suitable fixed point to any point P on the curve & 'I' be the angle that the tangent at P makes with x-axis, then the equation of the form "s = f('I' ) " i.e. , the equation that relates the arc length 's' and angle ''I'' is known as intrinsic equation of the curve. In cartesian form, let y = f(x) be the equation of the curve, then we have tan 'I' = s = and dy = y'(x) = f'(x) dx f: ✓1 + [f'(x) 2 . . . (i) dx = �(x) Now by elliminating x from (i), (ii), we will get relationship between s and 'I' . Example 63 : Find the length of the curve y Solution 1 We have, ⇒ = log[(ex - 1 )/(ex + 1 )] from x = 1 to x = 2. dy e x + 1 e x (e x + 1) - e x (e x - 1) · = x 2 x dx e -1 (e + 1) 2ex (e 2x _ 1) . . . (ii) . . . (i) Engineering Mathematics s = ⇒ f1 2 2 2 dy 4e 2x (e 2x +1)2 -- - dx = 1+--1 + 1+( -) dx= dx 1 1 dx (e2x _ 1)2 (e 2x _ 1)2 f 2 x 2x 2 e +e e2x +1 --dx= ----,---dx f 1 e2 x _ 1 1 e2x _ e -x x x ⇒ (e + e - )dx = dt, so we have x"'2 dt =(logt) �:� =[log(e x -e -x )J21 s = f = Now, put, ex - e -x = t f 31 9 f 2 X=1 t Example 64 : Find the entire length of the cardioid r = a(1+cos 8), and show that the arc of the upper half is bisected The given cardioid is symmetric about the initial line. Solution 0 = n/2 The shape is given in the figure. The required length of the curve is given by s = 2f; = 2a r2 +( �r d8=2f0\}a 2 (1+cos2 8+2 cos 8+sin2 8) d8 0=0 J; -J2+2cos 8 d8=2✓a J; ✓2 cos � d8 2 = 4a cos-d8=8a 0 2 i lt i Let the line 8= AB = Jf0" , 8=1t t3 8 2 i"'2 cos u du=8a 0 cuts the curve at B , then rr/3 rr/6 dr 8 r2 +(-) d8=2a Jf cos-d8=4a Jf cosu du o o d8 2 (as above) 2 . u )"O'6=2a= 1 (length of upper half of the curve). = 4a ( sin 2 Example 65 : Find the intrinsic equation of the parabola y2 = 4ax, origin being taken as fixed point. Solution : We have, On differentiating, we get ⇒ y2 = 4ax 4a ⇒ tan = \If dy 4a 2a = = dx 2y y y dx s = fa 1+( ) dy= dy 2 r )1+ O y2 4a 2 ...(ii) dy 2 2 2 2 • 4a2 1og(y+ 4a +y ) ] = _!_K.[Y 4a +y +_:1_ 2 o 2 � �- �-2 2 2 2 2 2 = _!_ · [Y 4a +y +2a log(y + 4a +y ) -2a log2a ] 2a 2 ✓ ✓ ...(i) ✓ y ✓ IES MASTER Publication 320 Multiple Integral 2 2 (y + 4a + y ) 1 = - · [ Y 4 a2 + y 2 + 4 a 2 log ] 4a 2a ✓ ✓ Now, elimin ating 'y' between ( ii) a nd ( iii), we get - ⇒ . . . ( iii) 1 2a cot 1J1 + 4a2 + 4a2 cot2 IJI 2 2 2 2 s = - · [2a cot IJIV/4a + 4a cot IJI + 4a 1o g ---�-- ---] 2a 4a ✓ s = a cot 'l' cosec \11 + a log (cot \11 + cosec \11) which is the required intrinsic equ ation of the curve. Example 66 : Find the intrinsic equ ation of the curve 3ay 2 = 2 x 3 . We h ave, Solution : ⇒ 3 ay2 dy 1I2 = tan \11 = [2x dx V3a . . . (i) x x 3X dY rx dx = b .J2a + 3x dx s = J 1 + ( ) dx = f 1 + Jo o Jo 2a dx v 2a 3/ 2 3 x + 2a ) 1 = __ [ ( 3 . 3/ 2 ✓2a Eliminating 'x ' between ( i ) a nd ( i i ) , we get Example 67 Solution : 1 ✓ 2 ]x 0 r;:::. 2 [ 2 "_ L. (3 x + 2a )3' 2 - (2a )3' ] =- g.Ja 3 /2 4a 2a 3 2 3 12 " L. s = - ( 3 - - - ta n 1J1 + 2a ) - ( 2a ) ] = - (sec \lf - 1) 3 9 g.Ja r;:::. [ . . . (ii) Show th at the length of the a rc of y2 = 4 ax cut off by the line 3y = 8x is a [log 2 + 1 5/1 6] y2 = 4 ax , 3y = 8x, on solving, we get ⇒ 3y 3a y2 = 4 a . 8 = 2 Y s = f: a12 ⇒ Y = 0, 2 3 1 + � r dy = J: ( � a12 3a t 2 2 2 = J_ f ( 2a ) + y dy Jo 2a ✓ a ✓1 + ( ;a r dy 4a = J_ [Y.)y 2 + 4 a2 + log ( y + 2 2a 2 2 ✓l 3a /2 + 4 a )] . o 2 3a 9a + 4a 2 + 4a log 3a + 9a + 4 a2 ) - 2a 2 log 2a = J_ [ ] (2 2a 4 4 2 4 ✓ 2 2 ✓ 1 1 5a2 15 = - [-- + 2a 2 log 2 ] = - a + a log 2 = 2a 8 16 2 Eng ineering Mathematics 321 Volumes and Surfaces of Revolution If any plane curve revolves about some straight line i n the plane of curve, then the solid thus generated i s said to be sol id of revol ution. The surface generated by the perimeter of curve is called as surface of the revolution and the volume generated by the area of the curve i s called as vol ume of the revol ution . I n this topic we will discuss about the curved surface areas and vol umes of the sol id generated by the revolution of the curves. VOLUMES OF SOLIDS OF REVOLUTION Theorem : The volume of the sol id generated by the revol ution of the area bounded by the curve y = f(x), the x-axis and the ordinates x = a and x = b about the x-axis if NOTE J: n ldx . ( i ) The vo lu m e generated by the revolution of the area lying between t he curve x y-axis and the l ines y a, y b a b out y axis is = V = = J: n x2 dy = �(y), the ( i i ) I f t he equat ion of curve i s given i n paramet r i c f or m i .e., x = f 1 (t), y = f 2 (t) then = f:ny2 dx ⇒ J: n y2dx ⇒ 2 2 V = J� n [ f2 (t) ] - : [fi (t) ] • dt t 2 2 2 V = J: nx dy ⇒ V = f� n [fi (t)] - : [f2 (t) ] • dt, and t where t 1 , t 2 are suita b l e l i m its of param eter t. V J: n (r sin e)2 · d� (r cos e) - de d (rsine) - de V = J: n (r cos e) de ( i i i ) When the equat ion o f curve i s given i n polar f or m then t h e f or m ulas w i l l b e V = V = J: n x dy 2 ⇒ V= 2 • where a, � are suita b le l i m its of e . These can be reduced to and 3 V = � nf\ s i n e de; 3 a (when rotation is a b out e = 0) nJ'3 r3 cos e de; V = � 3 a (when rotation is ab out e = n/2) Example 68 : Find the volume of a spherical cap of height h cut off from a sphere of radius a. Solution : Let us consider a circle of rad i us a with centre at origin so its eq uation is x 2 + y2 = a2 . W hen it i s rotated about x-axis, we get sphere of radius a. a 3 a a x ) 2 2 V = nJ y 2 dx = n f (a - x ) dx = n ( a 2 x - a -h X=a-h 3 a -h 3 3 h _ · (a - h) a 3 3 2 ] = nh ( a - ) n [( a - ) - a ( a - h) + 3 3 3 Example 69 : Find the volume of the paraboloid generated by the revolution of the parabola y2 = 4ax about the x-axis from x = 0 to x · = h . Solution : The required volume x 2 2 2 V = nJ:= y dx = nC (4ax} dx = n4a ( ; ): = n(2ah ) = 2n ah 0 IES MASTER Publication 322 Multiple Integral Example 70 : The cardioid r Solution : = a(1 + cos 8) is rotated about the initial line. Find vol ume of the solid generated. The curve r = a(1 + cas e) is symmetric about i nitial line i.e., 8 = 0. So, we have 2n V = 3 = l" r 0 3 2n . s1n 8 d8 = 3 2 3 O 3 - - na f t dt 3 l" a (1 + cos 8) sin 8 d8 3 0 3 [put (1 + cos 8) = t 2 ⇒ . 8 3 - sin 8 d8 = dt] = - na 3 SURFACES OF SOLIDS OF REVOLUTION Theorem : ,..INOTE:, --�� The surface of the sol id generated by the revolution of the area bounded by curve y = f(x), x-axis and the coordinates x = a and x = b about x-axis is given by S Here s = Arc l ength = f = 2nJ: y ds . 1 + ( :� r dx ( i ) The surface o f t h e sol id generated b y revolution of t h e area bounded b y x ·y = a, y = b; about y-axis is given by 5 = 211:J: x ds . ( i i ) If the equation of curve is in parametric form i .e., 5 and 5 = = 211:J: y ds ⇒ = �(y) , y-axis and x = f1 (t), y = fit) then the equation becomes 5 = 2nf�2 f2 (t) {�: } dt 211:J: x ds where t 1 , t 2 are suitable values of t 1 parameter t. ( i i i ) If the equation of curve is in polar form i .e. , r = f(8) then we have 5 5 and = = 211:J: y ds ⇒ 5 = 2nf: r si n e { �: } de 211:J: x ds ⇒ 5 = 2nf: r cose { �: } de, where a, � are suitable values of e . Example 71 : Find the surface generated by the revol ution of an arc of the catenary y = c cash ( � ) about the x-axis. Solution We have, ⇒ y dy = sinh ( �) dx c = ⇒ c cash ( �) ds dy = 1+( ) dx dx 2 or ds dx = S = 2nf: y (�: ) dx = 2nf: c cash ( � ) cash (� ) dx = 2 2ncf: cosh ( � ) dx = n c J:[ 1 + cash ( t)] 2 dx Engineering Mathematics 323 c . 2x c . 2x = n c[ x +-smh - ] =nc [x +- smh - ] 2 C o 2 C x Example 72 : Prove that the curved surface of the solid generated by revolution about the x -axis of the area bounded by the parabola y2 = 4a x , the x-a xis and the ordinate Solution 1 The surface area of revolution of solid is given by S.A. = 2nJ: =0 y ds=2nJ: ✓4a x ( �: ) dx = 4n h x +a) = 4n a r ✓x +a d x=4n a [ < Jo 3/2 ✓ ✓ 312 ] ✓a [ (a+h) = h is % n x ✓aC ✓x h o = 8n 3 312 -a 312 ] ✓aC ✓x ✓1+� d 1+(��r dx=4n ✓a [ (a + h) 1 -a 3 2 312 ] x Example 73 : Find the surface of the solid formed by the revolution of the cardioid r = a(1+cose) about the initial line. Solution 1 The curve is symmetric about the initial line, we have r = a(1+cose) ds = - de ⇒ ✓ �=-a sine de r 2 +( �� r = a2 +(1+cos2 e +2 cose)+a2 sin2 e ✓ = a./2 1+cose =a./2 s ✓ 2 cos 2 (e/ 2) = 2a cos � 2 = J; 2 ny (�: ) de =2 nJ;r sine-(2a cos� ) de = 2 n J; a (2cos �) 2 sin�cos�-(2a cos �) de 2 = 1 6na 2 r cos4 � sin � de = 1 6na 2 r cos4 <j>sin<j> (2d<j>) Jo Jo 2 2 n = 32na2 f'5/2f'1 = 3 2 na 2 2f'7/2 5 n Example 74 : Find the surface of the solid generated by the rotation of the curve r = a(1+cose), - � :-:; e :-:; � . about the axis of the y. Solution : The required surface area is S.A. = r 12 e ds 3 n/2 de f 2n x ds = 4n 8=0 x ( -) de = 8na io (1+cose) ·cose cos2 de 8 =0 nt2 2 n1 2 = 8na2 J 2cos2 � ( 1- 2 sin2 � ) cos� de o 2 2 2 ( put t=sin� ⇒ dt =_!_cos � de ) 2 2 2 2 2 1 4 11 2 s /✓ 48 � 2 = 1 6na2 1 ✓ (1-3t 2 +2t ) dt=32a 2 n[t-t 3 +-t ] =- v 2 na 0 . 5 5 0 IES MASTER Publication 324 Multiple Integral PREVIOUS YEARS GATE & ESE QUESTIONS Questions marked with aesterisk (*) are Conceptual Questions 1 .* 2.* 3.* The value of the double i ntegral The value of (c) 5. 6. * 7.* 1 0 3 [GATE-1 993] The volume generated by revolving the area bounded by the parabola y2 = 8x and the line x = 2 about y­ axis is (a) 4. 1 f y 2-e-Y dy is . . . . . . 00 8. [GATE-1 994-ME; 1 Mark] 1 28rc 5 5 1 28 rc (b) 1 27 5rc The integration of (d) None of above f log x.dx [GATE-1 994; 1 Mark] has the value (a) (x log x - 1 ) + C (b) log x - x + C (c) x (log x - 1 ) + C (d) None of the above I ntegrate f [GATE-1 994; 1 Mark] dx 2 - x - x2 [GATE-1994; 5 Marks] By reversing the order of integration. may be respresented as (a) f f f { x, y ) dy dx (c) 4 2 2x 0 x2 ✓Y f -f 0 y/2 ( x, y) dx dy (b) (d) f f f ( x, Y ) dydx 2 2x o x2 f f ( x, y) d x dy 2 ✓Y 0 y (c) oo x2 f cos t dt ' then 1 dy = dx -- Area bou nded by the curve y = x 2 and l i nes x = 4 and y = 0 is given by, (b) (a) 64 . (c) 1 28 3 (d) [GATE-1 995 (Pl)] 64 3 1 28 4 1 0 . A triangle ABC consists of vertex points A (0,0), B ( 1 , 0) and C (0, 1 ). The value of the i ntegral [GATE-1 997; 1 Mark] ff 2xdxdy over the triangle is (b) (a) 1 (c) 8 1 1 .* If (j>(x) = (a) 2x2 x, (d) f ✓tdt, 0 then 1 3 1 9 [GATE-1 997; 2 Marks] d(j> . IS : dx (b) Jx (d) 1 (c) 0 1 2.* The static moment of the area of a half circle of unit radius about y-axis (see Fig.) f 2xydx x=1 [GATE-1 998] is equal to f f (x, y ) dy dx 2x 2 x2 o The area bounded by the parabola 2y = x 2 and the line x = y - 4 is equal to (a) 6 9. Given Y = [GATE-1995-ME; 1 Mark] (b) 1 8 (d) None of the above [GATE-1 995; 1 Mark] (a) (c) 2 3 2 1C (b) (d) 8 1C 4 7t [GATE-1999; 2 Marks] Engineering Mathematics 1 3.* J J sin ( x+y ) dxdy is: 1 9. The-area enclosed between the parabola y = x and the straight line y = x is ll ll (a) 1/8 22 00 (a) 0 (c) 7t 2 [GATE-2000; 2 Marks] lim J x dx 1 4.* The following integral 8➔00 a -4 1 (b) Converges to V =i n/4 (c) 7t 8 7t 8 1 4 0 7t 1 (d) --+8 4 4 The value of the integral is (a) (c) (d) None of these [GATE-2002; 1 Mark] J xln x dx (b) 1 /4 (d) 1 i ntegral I= [GATE-2005-EE; 1 Mark] I =J J f ( x, y) dydx 0 x/4 J f f (x, y)dxdy . What is q? sq rP (a) 4 y (c) X 23. (b) 1 6y2 (d) 8 a -a (a) 2J sin 6 xdx co-ordinate planes and the plane x + y + z = 2. The density of the solid is P ( x, Y, z) =2x . Calculate the mass of the solid. [GATE-2002] [GATE-2005; 1 Mark] 0 (b) 2J sin7 xdx a a 0 1 8.* A solid region in the first octant is bounded by the to J ( sin 6 x + sin7 x ) dx is equal to a [GATE-2002; 1 Mark] leads 8 2 (c) 2 J (sin6 x+sin7x)dx (b) 0 (d) 1 7t 4 [GATE-2004; 2 Marks] 1 [GATE-2001] 1 (c) -1/4 (d) 7t 6 22. Changing the order of the integration in the double 1 7. The value of the following integral is (a) 1 /4 2n 3 (a) -1/3 (c) 1/2 f sin2x dx n 1+cos x 2 (b) 2 0 (b) 00 ll (c) 0 7t 3 0 21 . If S = J x·3 dx , then S has the value 1 6. The value of the following definite integral is: (a) -2 In 2 r sin �drd�d0 2 Jt In/3 i 1 2 0 ·0 3 1 5. The value of the integral is I = J cos2 xdx [GATE-2003; 2 Marks] ordinates is given by 1 [GATE-2000; 1 Mark] (b) (d) 1/2 20. The vloume of an obj ect expressed in spherical co­ 1 (c) Converges to - 3 (d) Converges to 0 a 7t 1 (a) -+8 4 (b) 1 /6 (c) 1/3 (b) n (d) 2 (a) Diverges 325 0 (d) zero 24.* By a change of variable x (u, v) = uv, y(u, v) = v/u in [GATE-2005-M E; 1 Mark] double integral, the integrand f(x, y) changes to f(uv, v/u) � (u, v). Then, � (u, v) is (a) 2 v/u (c) V 2 (b) 2 UV (d) 1 [GATE-2005; 2 Marks] IES MASTER Publication 326 Multiple Integral x 25. The value of the integral I = � 1 exp (- ; ) dx is (b) n (a) 1 (c) 2 (d) 2n [2005-EC; 2 Marks] 1 26 . The value of the integral f - dx is x2 -1 1 (b) does not exist (a) 2 (d) (c) ·-2 oo 1 31. The integral 2n (a) sin t cos t f sin(t - t)cos tdt 2rr (b) 0 0 equals (d) (1/2) sin t (c) (1/2) cos t [GATE-2007-EE; 2 Marks] 32.* The following plot shows a function y which varies linearly with x. The value of the integral I = y f Y dx is 2 1 [GATE-2005 (IN)] 2 2 27.* The expression V = nR (1-h / H) dh for the volume f of a cone is equal to (a) (b) (c) (d) 0 fnR2 (1 - h / H)2 dr R (a) 1 .0 (c) 4.0 1 nRH (1-k r dr 33.* The value of f 2nr H(1-r / R)dh 1 21tr H (1+� r dr (a) [GATE-2006-EE; 2 Marks] 28.* A surface S(x, y) = 2x + 5y - 3 is integrated once over a path consisting 2 of the points that satisfy (x +1 )2 + (y -1 )2 = ✓ . The intergral evaluates to (a) 1 7✓ (c) ✓ 17 (b) _!I_ 12 2 (d) 0 2 29. The integral (a) 1 /2 (c) 4/3 f sin 0 3 [GATE-2006; 2 Marks] 0d0 is given by (b) 2/3 (d) 8/3 (a) 0.524 (c) 1 .047 a2 (b) 0.614 a2 (d) 1 .228 a2 [GATE-2006; 2 Marks] 2 (c) n (a) rr/ 4 f f e- 00 00 00 x2 (d) 5.0 • [2007-EC; 1 Mark] e-Y dx dy is 2 (b) (d) ✓7t 4 7t [GATE-2007 (IN)] 34.* Which of the following integrals is unbounded? f tanxdx 0 1 (b) f � dx 2 0 x +1 00 1 dx (d) f 0 1-x 1 [GATE-2008-ME; 2 Marks] 35. Given the shaded triangular region P shown in the fig. what is [2006; 2 Marks] 30.* What is the area common to the circle r = a and r = 2a case ? a2 (b) 2.5 (a) (c) 1 6 7 16 ff xy dx dy ? :�, (b) 2 9 2 (d) 1 [GATE-2008-ME; 2 Marks] Engineering Mathematics 36. The value of J: J; (6 - x - y)dx dy is 42. The value of the integral (b) 27.0 (a) 1 3.5 (c) 40.5 (d) 54.0 [GATE-2008-CS; 2 Marks] 3 37.* The length of t h e curve y = ?_ x 12 b etwe e n 3 x = 0 and x = 1 is (b) 0.67 (a) 0.27 (d) (c) 1 1 .22 [GATE-2008; 2 Marks] 38.* T h e v a l u e of t h e i nt e g r a l of t h e f u n ct i o n g ( x, y) = 4 x3 + 1 0y 4 along the straight line segment from the point (0, 0) to the point ( 1 , 2) in the xy­ plane is (a) 33 (b) 35 (d) . 56 (c) 40 39. The value of the i ntegral rr/2 J ( x cos x ) d x i s -rr/2 (b) n - 2 (a) 0 (c) [GATE-2008 (EC)] (d) n + 2 n (a) (c) -n (b) 7t 2 327 f � is -1+x (d) n 7t 2 [GATE 201 0-ME; 1 Mark] ✓x , 1 :::; x :::; 2 is revolved around the x-ax is. The volume of the solid of revolution is 43. The parabolic arc y = (a) (c) (b) 7t 4 31t 4 7t 2 31t (d) 2 [GATE-201 0; 1 Marks] 44.* 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 2 equation of the parabola is y = 4h; , where x is the L horizontal coordinate and y is the vertical coordinate with the origin at the centre of the cable. The expression for the total length of the cable is [GATE-2008 (Pl)] 40.* f(x , y) is a continuous function defined over (x , y ) E [0, 1 ] x [0, 1 ]. Given the two constraints, and y > x2 , the volume under f(x, y) is (a) (b) (c) (d) 41 . y=1 X=,JY J J f( x, y) dxdy y=O x=y2 y=1 x > y2 X=1 f f 2 f( x,y)dxdy [GATE-201 0 ; 1 Marks] 1 00 45.* The integral -- J e-x2 12 d x is equal to Y =X2 x=y Y=1 X=1 J J f(x, y)dxdy y=O X=O y=.fx X=,JY f y=O f X=O (a) 1 2 . ..fi;c -00 [GATE-2009-EE; 2 Marks] The value of the quantity P, where x P = J xe d x , is equal to 0 (b) 1 (a) 0 (d) 1 /e (c) e [GATE-201 0-EE; 1 Mark] 1 ✓2 (d) a (c) 1 f( x, y)d xdy (b) [GATE-201 0 (Pl)] 46. If f(x) is an even function and a is a positive real a number, then J f (x) dx equals (a) 0 (c) 2a -a (b) a a (d) 2J f ( x ) d x [GATE-201 1 -ME; 1 Mark] IES MASTER Publication 328 Multiple Integral 47. What is the value of the definite integral, - ✓x d. X ? J 0 ✓x + .Ja - x (a) zero (c) a 1 (a) -(e - 1) 2 (b) a/2 (d) 2a [GATE-201 1 -CE; 2 Marks] 48. T h e area enclosed between the straight l i ne y = x and the parabola y = x 2 in the x-y plane is (a) 1 4 1 (d) 2 1 (b) 6 1 (c) 3 [GATE-201 2-ME, Pl; 1 Mark] 49. The value of definite integral J ✓x l n (x)dx is (a) (c) i9 ✓e3 + 3.9 (b) ✓e3 + i9 9 (d) 3. 1 3_9 ✓e3 _ i9 0 (a) 0 e v a l uate [GATE-201 3-M E; 2 Marks] 3 (b) 1 /1 5 double i ntegral 2x - y y u = ( -- ) and v = . The integral will reduce to 2 2 (c) J: { J: 2u du ) dv J: (J� u du ) dv (b) J: (J� 2u du ) dv (d) J: (J: u du ) dv [GATE-201 4-EE-Set-2; 2 Marks] 52. The value of the integral . f (x - 1/ sin ( x - 1) d X IS 2 (x 1) + cos ( x 1) 0 2 (a) 3 (c) -1 (b) 0 (d) -2 e- r � [GATE-201 4-ME-Set-1 ; 2 Marks] 54.* The volume under the surface z(x,y) = x + y and above the triangle in the x-y plane defined by { 0 :s; y :s; x and o :s; x :s; 1 2 } is . . . . . [2014; 2 Marks] 55.* If Jo lxsinxl dx = kn , then the val ues of k is equal to r 2n [GATE-201 4 (CS-Set 3)] (a) (c) -2n (b) n (d) 2n -n [GATE-20 1 4 (CS-Set 3)] sin(4nt) oo r s ( y12 )+1 2x - y . . - ) dx ) dy , we make the subst1tut1on Jo ( Jy/Z ( 2 (a) J( (d) is 57.* The value of the integral J-oo 1 2cos(2nt) ---dt 4nt [GATE-201 3-CE; 2 Marks] the 2 1 2 (b) - ( e - 1 ) 2 J; x 2 cos x dx i9 ✓e3 _ 39_ (d) 8/3 (c) 1 51 .* To 4 2 (c) I ( e - e ) 2 00 56. T h e v a l u e of t h e i nt e g r a l g i v en b e l ow i s 50.* The solution for J cos 38 sin 68 de is n /6 2x ff e x+y dydx 53. The value of the integral a [GATE-2014-ME-Set-4; 1 Mark] is [2015-EC-Set-2; 2 Marks] 58. The volume enclosed by the surface f(x, y) = ex over the triangle bounded by the lines x = y; x = O; y = 1 in the xy plane is �-- [GATE-201 5; 2 Marks] 59.* Consider a spatial curve in three dimensional space given in parametric form by, x(t) = cost, y(t) = sint, z(t) = ; t , 0 :s; t :s; % then the length of curve is __. 60. The double integral (a) (c) JJ a y 0 0 f(x,y)dx dy is equivalent to J: J: f ( x, y) dx dy (b) J: [ [GATE-201 5] Jo J: f ( x, y ) dx dy a f ( x, y) dy dx (d) Jo [ f ( x, y ) dy dy a [GATE-201 5 (IN)] 61 .* Let f(x) = x-< 113> and A denote the area of the region bounded by f(x) and the x-axis, when x varies from -1 to 1 . Which of the following statements is/are TRUE? I. Engineering Mathematics 68. The region specified by 329 "' ,,, ,,, n < "' < n < < {( p, 'I'• z ) : 3 .:::, p .:::, 5'- - 'I' - - ' 3 - z - 4 .5} 8 4 f is continuous is [-1 , 1] in cylindr ical coordinates has volume of Ill. A is non-zero and finite 1 69.* The integral - f'fJ (x+y +1 0) dx dy , where D 2n o denotes the disc: x 2 +y2 � 4 , ev aluates to II. f is not bounded in [-1 , 1] [GATE-201 6-EC-Set 1] (b) Ill only (a) II only (d) I , II and Ill (c) II and Ill only [GATE-20 1 5 (CS-Set 2)] 62. The area between the parabola x2 = By and the straight line y = 8 is __. • 63.* The value of (a) (c) 2 7t [GATE-201 6-CE-Set-2 ; 2 Marks] sinx dx f "' -1 - dx+f "'-o 1+X 2 o (b) n X is (d) 1 3n 2 [GATE-2016-Set-1 ; 2 Marks] 64. The area of the region bounded by the parabola y = x2 + 1 and the straight line x + y = 3 is (a) (c) 59 6 9 (b) 2 10 3 (d) 6 [GATE-201 6-CE-Set-1 ; 2 Marks] 1 dx f 65. The integral is equal to ___ O .J(1-x) [Gate-201 6-EC-Set-1 ; 1 Mark] 66. A triangle in the xy-plane is bounded by the straight lines 2x = 3y, y = 0 and x = 3. The volume above the triange and under the plane x + y + z = 6 is 67.* The integral J__ JJ ( 2n D [GATE-201 6 ; 2 Marks] x + y + 1 0 ) dxdy, where D denotes the disc: x 2 + y 2 s 4, ev aluates to [GATE-201 6 (EC-Set 1 )] [GATE-201 6; 2 Marks] 70.** A three dimensional region R of finite Volume is described by x 2 + y2 s z 3 ; 0 s z s 1 . where x , y, z are real. The volume of R (upto two decimal place) is____ [GATE-2017-EC; Session-I] 71 .* The values of integrals f (f x- y o o (x + y) 3 dy l dx and f (f x-y o o (x +y ) (a) same and equal to 0.5 3 dx ] dy are (b) same and equal to - 0.5 (c) 0.5 and -0.5 , respectively (d) -0.5 and 0.5 respectively 72. Let I = c [GATE-2017-EC; Session-II] J JR xy dxdy , where R is the region shown in 2 the figure and c = 6 x 1 0-4 . The value of I equals __. (Give the answer up to two decimal places). y 10 2 R 5 [GATE-2017-EE; Session-I] 73. Let x be a continuous variable defined over the interval e (-oo, CX)) and f(x) = e-x- -x . The integral g(x) = J f(x) dx is equal to (a) ee . -x (b) e -e (d) e -x -x [GATE-201 7-CE; Session-I] IES MASTER Publication 330 Multiple Integral 74. Consider the following definite integral: 1 (si n-1 x} 2 l=f � o 1-x The value of the integral is (a) (c) 7[ (b) 3 24 (d) 3 7[ 48 2 16 (a) - and 7[ 4 (c) - and 0 dx 7[ 12 64 (c) [GATE-2017-CE; Session-II] nu } y = sin (;) 2 in the range O � u � 1 is rotated about the X-axis by 360 degrees. Area of the surface generated is (c) (a) 3 75.* A parametric curve defined by x= cos( (a) 78. (b) n 7[ 2 2n (d) 4n [GATE-2017-ME; Session-I] n 76.* I f f ( x ) = R s i n ( ; ) + s & f ' G) = ✓2 2R , f 10 f(x)dx = - respectively. 7[ 2 (b) - and 0 7[ 4 16 (d) - and 7[ and then the constants R and S are, 2n ffo Mn (c) n2 / 2 79. J:"( 9 ) de is ! + n2 0 (b) 2Mn (d) 2n [ECE 2017 (Common Paper)] The value of the integral (a) n2 / 8 7[ [GATE-201 7-PAPER-2 (CS)] 77. The value of the integral 3 7[ 7[ 7[ f x cos2 x dx 0 (b) n2 / 4 is (d) n2 [GATE 2018 (CE-MomingSession)J Let f be a real valued function of a real variable defined as f(x) = x - [x] , where [x] denote the largest integer less than or equal to x. The value of 1 .25 f 0 . 25 f(x)dx is __ (up to 2 decim al places). [GATE 201 8 (EE)] Engineering Mathematics 331 1. 2. (See Sol.) 8. 9. 10. 11. 12. 1 3. 14. 1 5. 16. 17. 18. 19. 26. (See Sol.) 28. (d) (c) (d) (d) (b) (d) (d) 35. (c) 36. (2) 38. (a) (a) 37. (c) (d) (a) 39. (b) (85.33) (b) 34. (a) 62. 27. 33. (b) (d) (d) 32. (d) 42. (a) 31 . (a) (a) 25. 30. (a) (c) (a) 29. (b) 60. (a) (c) (b) (a) 23. 24. (b) 40. (c) (c) (a) x xdx I = J[f' �] X 1 + y2 0 = f� x[tan-1 y]� x J� (coC x - tan 1 _: 1 x ) xdx = f� x [% - 2 tan-1 x]dx 1 2f x tan-1 x dx 2: = 4 0 41 . 43. 44. 45. 46. 47. 48. 49. 50. 51 . 52. 53. 54. 55. 56. 57. 58. 59. Sol-2: dx 1 1 = f> [ tan- ( �) - tan- x] dx = --_- (a) 21 . 22. 5. 7. Sol-1 : (-;J (a) 6. 20. (See Sol.) 3. 4. A N SWE R l{EY Let 61 . (b) (a) 63. (d) (b) (d) 64. (d) 66. (10) 68. (4.71) 65. . (2) (c) 67. (b) (a) (20) (20) 69. (c) (b) 70. (b) 72. ( 2: ) 4 71. (b) (b) (c) (0.99) 73. (b) 75. (c) 74. (864) (4) (-21t) (a) 76. (c) 78. (b) 77. (3) (e - 2) (1 .86) (-;) (b) (a) 79. = I = (0.5) 'It [ 'It 1 4 - 2 4 - 2] -'It = 4+1 J y1 e-Y dy 12 0 3 y3 = t y = t1/3 13 dy = -�-T2 dt 3 I = f _!_ t116-213 e-tdt 0 3 IES MASTER Publication 332 Multiple Integral 1 -1 1 1 = J � t - 12 e - dt = � f e- . ti dt=� 3 \/2 30 3 0 Sol-3: (a) = � 3 y X' /I ✓re P(2, 4) y2 = 8x [Parabola] X y' Sol-4: (c) = Jlogxdx = log X ' X - J � . X dx Sol-5: Q(2, -4) f (2 - dx x - x2 ) x = 2 [line] 2 0 2 0 = 8 ✓nJ x 312 dx dx f (2-x-x 2) 2 2 0 = 8 ✓ X 7t X � [ 2] ' Method 2 : 1281t Volume = -- Sol-6: (c) 5 2 5 X = f-(x2 dx + x-2 ) = -dx f (x+2) (x - 1 ) = -dx f x2 +2x -x -2 ( x:2 - (x�1) ) dx dx J x+.2 x-1 dx = ln( x +2 )-ln( x-1)+C = zn ( x +2 ) +c x-1 I = J J f(x,y)dydx 2 2x To reverse the order of integration we will have to form horizontal strip y ➔ x2 to 2 x X ➔ 0 to 2 0 x2 Total volume of the cylindrical shaped solid formed by the revolution of the surface POQ about y-axis = nr 2h = 1t(2)2 (8)= 321t Now, volume of the solid formed by the revolution of the curve POQ about y-axis = 2 volume of solid formed by curve PO about y-axis y = 2x ' = 2 · J nx dy = 2 · f n ('L) dy 321t X = J = J 4nx.J'sx dx 5 log X - X + C (log X -1 ) + C X = J Volume = J (2nxdx) x 2y 2 = = 2 y=O 321t 1281t So , Required volume = 32n--= - - 5 5 64 I = J J f(x, y)dx dy 4 ✓Y 0 y/2 Hence (c) is the correct option Sol-7: (b) 2y = x2 ⇒ X = y-4 x2 = 2x + 8 ⇒ Engineering Mathematics x2 - 2x - 8 = 0 X = 4, - 2 = J J 2x dy dx = 1 = Shaded area is required area = = J J 4 X=-2 J( x+4 x2 Y= 2 (1) dy dx Sol-11 : (a) x+4- �) dx 2 2 - [ Sol-8: y = 3 2 J1 x 2 r y = 4 costdt= [+sint]� x2 f f X=0 y=0 (1) dy dx 4 . 64 x3 ] 4 = J x 2 dx = [- =- sq. units 3 0 3 0 Sol-1 0: (b) Equation of line CB is y = 1 -X =· Jf2xdxdy 0 dx J2x(1 - x) dx 1 0 f ✓t dt $(x) = x2 0 = f .ft dt x2 Apply Newton Leibnitz theorem Sol-12: (a) 0 d$ =O+2x - ✓x2 - O=2x2 dx Substituting y = I = J 2xy dx X=1 0 ✓f _ x 2 is above integral. put, 1 - x2 = t -2x dx = dt I = f -.ft dt 0 = , : t3/ 2 r I = o- ( C A 2 1 = 1 - -=3 3 $(x)= 2 = 4 = [Y]6- x Alternative Solution: and y = 0 Shaded area required area ___.::::,,..,���X2 , X J 2x � � dy = 2xcos x 2 So dx We can a l so use lei bnitz form u l a of Note "Differentiation under the sign of integration" (b) 0 0 3 $(x) = 3. x 3 = sinx2 - sin1 Sol-9: 1 1- x 2 3 2 = [x _ ; I x x - +4x - 2 6 -2 = 1 8 sq . units. = 333 -2 ) 3 B IES MASTER Publication 334 Multiple Integral Sol-1 3: (d) I = = n/2 n/2 J J sin (x + y) dxdy 0 n/2 J [-cos(x + y )]� 12 0 J (sin y + cos y) dy 1 0 1 1 x2 J - - - dx J = [ ln x - � 2 0 0X 2 dy n/2 J x lnx dx I = 0 = T [ - cos ( % + y ) + cos y] dy = Sol-17: (c) -[ �1 = -¾ = Sol-18: (2) 2 z 0 2 . = [ - cos y + sm yr0' Sol-14: (b) Sol-1 5: (a) 11 = -cos 2: + sin 2: _ (-cos0+sin0) 2 2 I = - 0 + 1 - (- 1 + 0) = 8 Jx -4 dx I = Lim 8 ➔«> 1 8 1 = · Lim [ 8 ➔«> -3 x3 1 I = 0 Mass = M = J pdv = = = 2 + c s 2x T (1 � ) dx 8 4 n/2 , sin 2 x = J dx 1 + cos x -n 2 sin 2x f(x) = 1 + cos x sin 2 ( -x ) = sin 2x 1 + cos x 1 + cos ( -x ) = - f(x) odd function X+y=2 Density p = 2x ] � sin 2x "'4 ] = [ + 2 4 0 7t 1 = -+- (2, 0, 0) Plane x + y + z = 2 J cos x dx n/4 f(-x) = ⇒ 'f(x) is an ⇒ I=0 X 1 + !J = ! = Lim [ - -8 ➔«> 3 3a 3 3 = Sol-16: (c) 21 = JJ J (2x)dx dydz ) J ( J-x (2x)(2 - x - y)dy dx 2 2 J ( -Jx (4x - 2x 0 0 2 2 0 0 Sol-19: (b) 2 - 2xy)dydx J ( 4xy - 2x y - xy )( -x) dx 2 0 2 = f[ o M = ) 4 3 2 2 0 4x(2 - x) - 2x2 (2 - x) -x(2 - x)2 y (0, 0) Area enclosed = JJ (1) dy dx = 1 X f f X=O y=x2 (1) dy dx ] dx 1 = J ( x-x2 ) dx 0 Sol-24: (a) ff f (x, y)dx.dy = Sol-20: (a) I = = 1 1 = 3 6 2 1t 3 0=0<!>=0 3 r J sin cj>.dcj>d0 f" f [3 O 0 =0<!>=0 2 3 2 1t 2: 1 = - f d0.[-cos cj> )3 3 e=O = Sol-22: (a) [�:r 8 2 I = fo f f(x, y) dy dx J x/4 ⇒ X = 0, X = 8, y = y = --[O -1) = 1 2 = r 4X I 6 { dx = I = U 1 = 2v u 2 t ✓2 .ft 1 dt � 1 = __ ✓n = sin6x is even function } sin7x is odd function V -V = X = 2✓ ✓ y = x/4 7 a (u, v) av -x2 y = 2 l:sin x dx + l: sin x dx a( X, y) 8y 1 I = __ Io"" eadx � x2 = 8t (Using vertical strip) f:f(x, y) dx dy I = Sol-25: (a) 2 s:=of:0 f(x, y) dx dy (Using Horizontal strip) ⇒ q = 4y Sol-23: (a) J = ⇒ 1 After changing the order of integration. I = a (x, y) au - J = a (u, v ) - 8y au x (u,v) = uv, y (u, v) = v/u Let = i x 2n x [1 -i] = Sol-21 : (c) f f f (uv, v / u)cj>(u, v ) du.dv ax ax av ( r 2 sincj> dr dcj> d0 0 f f = 335 where, cj> ( u, v ) = Jacobian function x3 = [� - J 2 3 o 1 Engineering Mathematics r r 0 0 -t ✓ e - dt 2 .ft e-lt -112 dt !_1 1 00 - J e -t , e 2 dt ✓n 0 1 � ✓n 2. I = � ✓n I = 1 Sol-26: (d) x = 0 is a point of dis continuity lies with in the [-1 , 1 ] S o it is a n improper integral of 2 nd kind 0 1 1 1 1 1 dx = -2 dx + 2 = oo I = 2 -1 x ox -1 x ·: Sol-27: (b) Method I Grven f 2 v = 1 nR f f -dx (1-�r dh IES MASTER Publication 336 Multiple Integral Put ⇒ So 1-� H dH = = x2 + y2 = ✓ 2 t -Hdt and h = 0, t = 1 , while at h = H , t v = ! 1t R2 (1 - � r dh 0 2 2 2 = f 1tR t (-Hdt) = -1tR H ( f ) 3 1 1 1tR H i.e. volume of cone i Similarly let us take option (b) = 2 0 1-� R and At r = 0, t So v = = dr ⇒ = = = ⇒ ⇒ = 1 1tRH ( 1 - � r dr H ! s f S (r, 0)rdrd0 = s = f [2 ( r cos 0) + s ( r sin 0)].rdrd0 s f r (2 cos0 + 5 sin 0) .rdrd0 = s f (2cos 0 + 5 sin 0)r2 dr d0 0 Sol-29: (c) h 2 1tR2 ( 1 - H ) dh s = 0 Hence option (b) is correct. Method II f S ( x, y) dxdy I = = i.e. volume of cone I f r2 dr. f ( 2 cos 0 + 5 sin0)d0 = 0. = r 2n 0 0=0 J; sin 0 d0 = 2 3 r H dh = -dr R 2 V = f 1tR ( 1 - � ) .!::!dr R R 0 ., Sol-28: (d) (x + 1 )2 I = + (y - 1 )2 = ✓ x + 1 =X y - 1 = Y 2 ⇒ ⇒ x = f sin3 0cos 0d0 0 Alternative Solution: = I 1tRH ( 1 - � r dr Equation of Circle: Let 2 nt2 . 3 n/2 r h H = H⇒ h = r R R. ⇒ sin e d e Jo = 2 From figure, X-1 y = Y + 1 2 = 2X + 5Y R, t = 0 1 2 H f 1tRHt2 (-Rdt) = -1tR 3 ✓ is = 2(X - 1 ) + 5(Y + 1 ) - 3 - Rdt 1 , while at r = Now, Equation of Surface: S(x, y) = 2x + 5y - 3 v = 1 1tRH ( 1 - � r dr Put V = = Let the parametric points on X2 + Y2 X = r cos e y = r sin e J; sin 0 d0 3 = ..!_ fo (3sin 0 - sin 30)d0 4J 71 1 cos 3 0 = - [-3 cos 0 + -3 ] 4 = = 71 ¾[ 4 3 3- i +3- i] o Sol-30: (d) Sol-32: (b) C (a,1t/3) s/ 1-0 (x + 1 ) 0+1 y = X + 1 r = 2a cos 0 I = Common area ⇒ [ 0 = -3 1 2 Shaded Area = 2 Area of upper portion cos 0 = 7t = a2 /3 rn 2 f 71/ 3 (1 + 2 cos 20)d0 Jo = a2 [ 0 + sin 20]� = a2 [ i+ �] 13 = 1 .91 32a2 So Common Area = na2 - 1 .91 32a 2 Sol-31 : (d) = (3. 1 428 - = 1 .9132)a2 = 1 f sin(t - -r)cos -rd-r 2n f 2 00 00 2 00 2 2 00 Put x = r cos 0, y = r si n 0, I JI = r ( J a co b i a n of polar coordinate system) then to cover whole 1 st Quadrant 0 varies from O to % and r varies from So, 0 to oo = = 2 f: f: e-( x +ll • dxdy 71/ 2 oo f f e-r 0=O r=O 2 • (rd rd0) 7t 7t -2 = - f r · e r · dr = - (Put r2 = t) 4 2 r=0 71 For 271 - f [sin t + sin(t - 2n)) d-r 71 = f f e-x e-Y dxdy =f f e-( x +y ldxdy 00 00 On integrating only option (d) does not have finite ans. 1 71 = - 2 • sin(t ,... .) cosi:d-r 4n 0 = � [sin t ( .)� + ( 4 = 2-5 2 Sol-34: (d) 0 2 0 Sol-33: (d) = 1 .228a2 271 1 = 4n ft (x + 1)dx 1 = 2 +2 - - - 1 2 (4cos 0 - 1)d0 Jo 71/3 f (2(1 + cos20) - 1)d0 = a2 Jo = a2 I = ft y dx y-0 = (2a, 0) COS S 337 The equation of line passing through (-1 , 0) and (0, 1 ) is given by 0=0 r = a = 2a Engineering Mathematics -cos - 2n) � ): ] 71 sint 1 = -(2n - 0) + - [cos(t - 4n) - cos t] 41t 81t 1 = !sin t + - [cost - cost] 81t 2 1 .i = -s n t 2 Sol-35: (a) = f0 1 dx 1 = [-log(1 - x)]0 1_x = [1 ogC�JI = log oo - log1 = oo The equation of straight line is ⇒ �+Y = 1 2 1 y = 1-� 2 y IES MASTER Publication 338 Multiple Integral Now, ff xy.dxdy = f f (xy)dydx 2 2 0 y;0 = f x[ 0 = L2 J; 0 dx 0 ✓ +x ) dx (·.1 3 '2 1 1+x ) 3/2 ] o dy =x½ ) dx 32 = 3. (2 / -1] == 1 .21 9 == 1 .22 3 Sol-38: (a) Equation of straight line from 2 (0, 0) to (1 , 2) is y = 2(x) x2 -x) dx f -2x (1+-· 4 2 J( = [< f -2x (1--2x r dx 0 = = 1- � = So 0 3 1 = - (x+�-x2 ) dx 20 4 f g(x, y)dx 2 f Sol-36: (a) = J[ J: J; (6-x-y) dx dy = J [4x3 +1 0 x 1 6(x)4 ]dx 0 Sol-39: (a) 2 +1- i ] = � = 3 X f f X;0 y;0 rr/2 f (6 - x-y) dy dx = f (6y -xy- y )�;0 dx o = x cos x dx = 0 ( ·: integral is an odd function) Sol-40: (a) 2 x f0 (6x-x2 --)dx 2 2 3 3x fo (6x--)dx 2 2 3 x = (6 �- r 2 2 o 3 Sol-37: (d) - rr/2 2 3 = = 33 27 54 27 = - -- = = 1 3 .5 2 2 2 2 32 . . Given, curve Is y = x / 3 We know that Length of a curve = f [ 1+(�� r ) dx Length of the given curve between x = 0 and X = 1 X1 The volume under f(x, y) = ff f(x, y) dxdy In case of horizontal strip, x is varying from x = y2 and y is varying from y = 0 to 1 , therefore, to x = the volume will be ✓Y Sol-41 : (b) = f f f(x, y)dxdy 1 ✓Y 0 y 2 P = J x ex dx = ( x.ex ]� -f e x dx 1 1 0 0 = e - [ ex ]� = e - (e - 1 ) = 1 Engineering Mathematics Sol-42: (d) 1 .dx f1+X2 ) = x2 = t 2 Put 00 ( ---«> 339 2f--1 2 .dx = 0 1+x 00 = 2 [tan -1 xJ : = 2 [tan-1 .X)-tan -1 0] Sol-43: (d) Sol-46: (d) Volume of the elemental disc of thickness (dx) dV = ny2 dx Total volume, V = f ny2 dx 2 = Sol-44: (d) = f dV Y. 2 f nxdx 2 ⇒ x2 L2 Supports (� . o ) u2 64h 2 = 2 r x dx 1+ Jo L4 ✓ = - -f g;, 0 = ✓x 2f Hx)dx ✓x 0 + -Ja - x dx f:(�7 a-x+ a-(a-x) (i) + (ii) ⇒ 0 ... (i) ) dx ... (ii) + x } -Ja - x� + r(✓x 21 = ✓x = J;1dx =(a-0) = a/2 Given, equation of straight line is y = X = r�;2 1+( ��r �x 00 = Ja 0 Sol-48: (a) 8h dy = ( 2 )x L dx 2 f f ( x ) dx -a {Using J: f(x)dx= J: f(a+b - x)dx} Total length of the cable Sol-45: (c) Sol-47: (b) f(-x) = f(x) I = 3n X = - (4- 1 ) = 2 2 y = 4h - A, B ⇒ X=1 8 ( -2L , 0 ) Given, f(x) is even function And equation of parabola is y = x2 ...(1) ...(2) 2 2 e-x 12 .dx (0 , 1 ) From equation (1 ) and (2) we get X2 = X ⇒ X 2 - = Q X = 0, 1 X IES MASTER Publication 340 Multiple Integral x = 0 and x = 1 Sol-51 : (b) y = 0 and y = 1 ⇒ (Using vertical strip) Enclosed Area = Jf (1) dy dx = 1 X (1) dy - dx X =0 y=X2 f f X ➔ "j_ to 1 + 1 2 2 1 = f ( x-x2 ) dx Sol-49: (c) f .Jx ln(x) dx=(33 1n x x '2 ) - J0 � (33 x ' ) dx 1 = = Sol-50: (b) 0 = 3 3 e3/2 _ 3 2 (Integration by parts) 1 I X 9 J = Jacobian matrix au av ay ay au av I 3 e3I2 _ _i e3I2 + i 3 9 9 9 9 3 ✓e3 + i Pu t 30 = t, 3d0=dt, d0 = ( � ) t = f;' cos4 t sin3 2 { � ) 2 t Sol-52: (b) put x-1 = 2 = Jfo ' cos4 t(2sin t cos t)3 � 3 71 71I2 3 = � x 8 fo cos7 t sin t dt J 3 = ax ax _i (x3'2 )8 71 6 I = fo ' cos4 30 sin3 60 d0 J = y 2 V = Y At x = - , u = 0 & at x = y_ + 1 u = 1 2 2 y ➔ 0 to 8 At y = 0, v = 0 & at y = 8, v = 4 From (1) and (2) X = U + V, y = 2V 1 1 1 = --- = 2 3 6 3 2x- y 2 u = 8 � � 8 14 12 = 3 7 + +2 2 ! � 8 � [.g 3 2 - l.§ = 1 15 3 216 (x-1 l sin(x -1 ) dx 1 = J 2 0 (x -1 ) +cos ( x - 1) t ⇒ 1 = f(t) = dx = dt j1 _1 t t 2 sin t dt +cos t 2 sint t +cos t t 2 2 f(-t) = -f(t) So the function is an odd function. Hence, Sol-53: (b) I = 2x ff ex+ydydx 00 ...(1) . . .(2) Engineering Mathematics = f ex · [ eY Jo dx 2 0 2 = - - e +- 2 e Sol-54: (864) Volume = = Sol-55: (4) Let = 2 1 4 Note: (1) (2) f f(x)dx = b a 0 f x lsinxl dx = = 2n 0 [ ·: Ix I = x for x > OJ f (2n-x) lsinxl dx [using Note (2) below] 0 f 2n lsin xi dx 21t 2n f = 2nflsinxl dx 7t 0 = 4n ⇒ I = 4n [using Note (1) below ] f lsinxl dx =4n [-cos x]�12 n/2 0 f lsinxl dx 0 if f(2a -x)=-f(x) f f(a+b-x)dx b a f x2 cosxdx=[x2 (sin x)- f 2x(sin x)dx 7t 0 = [ x2 sinx +2xcosx-sinx ] � = Jo Sol-57: (3) 7t J: 21tCOS 1t = -27t = = kn 4.n = k - n ⇒ lk =41 _g f 2 cos2ntsin4nt dt 4n O �[J sin 6nt t sin 2nt dt ] dt+f o t {Using 2sinA cosB = sin(A + B) + sin(A - B)} = n o t sin at 7t oo - dt=- for a > 0 will be discussed in f t 2 0 Laplace transform. 0 2n 0, if f{2a -x)=f(x) f 12 X = n lsinxl dx So, 0 = [ x2 sinx-2 { x(-cos x)- 1(-cos x)dx} f f f (1) dz dy dx f lx sinxl dx = I = X=O y=O z=O f f ( x+y) dy dx = 864 0 Sol-56: (-2n) 1 2 X X +y X =O y=O f f(x)dx 2a 2n 2n 21 = 2 f f(x}dx, X 341 Sol-58: (e - 2) = 3 Given z = f(x, y) = ex and surface lies on xy plane so z varies from O to ex. y=x ---=o------- - x Now, according to given strip y varies from x to 1 and x varies from O to 1 IES MASTER Publication 342 Multiple I ntegral So, Req. volume = HJ (1) dx dy dz = = = 1 ex 1 X=O Y=XZ=D f f f 1 1 x = O Y= X f f 1 X =O f (1) dz dy dx ex dy dx X = dx dt COS dy . = -smt, dt (1 - x) ex • dx =0 . Sin t, Z 2t = 7t Therefore, f(x) is not continuous in [-1 , 1] Ill : Area = = f sin2 t + cos2 t + � dt 7t Sol-62: (85.33) x = 0 to x = a & y = x to y = a ff f(x, y ) dx dy = ff f(x, y) dy dx ay 00 aa Ox -1 2/3 -1 1 r[ 0 -1 2 /3 J -� I 3 -1 ± ; / 3 o 2 1 2 3 3 -- + - = 0 2 2 Area = y Hence to change order of i ntegration we will from vertical stri p for which, variationof lim its wi l l be as follows : 1 f x-113 dx + f x- 113 dx 0 2 ✓ y has constant limits so strip used in given integral is Horizontal. 1 f f(x) dx = f x-1I3dx 1 /3 / 3 3 = -[ (-1)2 ] 3 + - [ (1)2 ] 4 = 2: 1 + t2 2 7 Region of integration is bounded by y = 0 to y = a and x = 0 to x = y. X=-1 y=O (1)dy dx = = - � (-1)2 /3 + � (1)2/3 dz 2 = cos t, = n dt n/2 1 f(x ) f f = [ = � = 1 .86 units Sol-60: (c) I = I : Here, f(x) is not continuous at x = 0 because f(O) = D.N.E. Q :::; t :::; 2: 2 1[ (�;)' +(�;)' +( ��)} . = So -1 = [ (2 - x) ex J :=o = (e - 2) t, y = Length of cu � e • . 1_ __ f(x) = X 3 1 X /3 I I : f(x) i s not bounded in [-1 , 1 ] , because f(O) is not finite = [(1 - x) ex + f ex dx J: Sol-59: (1 .86) Sol-61 : (a) Sol-63: (b) 1 8 X=-6 oo sin x 2 8 f f dx + f -- dx Iooo Jo x 1 + x- 2 Y= X a (1) dy dx 00 sin x "" = [ tan -1 x ]o + J -o x [Using Laplace transform method, we know that 2 Engineering Mathematics Sol-64: (b) Given, y = x2 +1 4x 2 2 x 3 2 x3 = [2 - 3 3 - g· ] 3 On solving 3 - x = x2 + 1 x2 + x - 2 = 0 x 2 + 2x - 2 = 0 ( x + 2) ( x - 1) = 0 ⇒ 1 = J (3 - x - x -2 2 x = -2 and 1 1 J X=-2 (1) dy dx - 1) dx y = r si n 0 then O :,; r :,; 2 & Sol-65: (2) r1 dx J o -J1 - x JJ ⇒ Sol-66: (1 0) 1 = - [r(sin 0 - cos0) + 1 00]� 71 r - dr 2n r=O 2 J 1 (1 - x )112 = -2 1 - x \ = 2 = 1 o 1/2 0 ✓ 1 Sol-68: (4.71) Vol ume in cylindrical coordi nations = The t ri a n g l e i n XY p l a n e bounded by l i nes y= 2 3 x, x = 3 and y = 0 is as shown. Equation of plane: x + y + z = 6 :. Z = (6 - y) 0 :,; Z :,; 6 - X - y SO, 2 Now, on xy plane, 0 :,; Y :,; - x and O :,; x :,; 3 3 so, X - V = V = = = HJ (1) dz dy dx 3 2 x/3 r, I J ( 6 - x - y ) dy dx X Oy O = 3 xL = [5y - xy - y2 2 o 3 dx 2x 2x 4x2 x ,3 - x x 3 dx [ 6 9x2 J xL 3 r = 0=0 7! 1 2 2 = [r(cos 0 + si n 0) + 1 0]r dr d0 2n r=0 8=0 9 2 = 0 I = J__ J J [(x + y) + 1 0] dx dy 2n 0 = C (2dx - xdx - x 2 dx) = 0 = n/2 Let x = r cos e , 0 :,; 0 :,; 2n & J 3 2 4x9 2 = -- - - x 27 - - x 27 27 9 2 = 10 Sol-67: (20) ff (1) dy dx = Area = so, f ( 4x - � x2 - � x 2 ) dx 3 9 . X= O = x+y=3 and 343 Sol-69: (20) Let = JJJ 4 . 5 n/4 5 2=3 <1>=2'_ p=3 8 4 . 5 n/4 2=3 <1>=2'. f f x2 + (p)dpd�d z 8 - d�dz = J (n)dz = 1 .5n = 4.71 4. 5 2=3 x = r cos e y = r sin 0 y2 :,; 4 r 2 cos2 0 + r 2 sin2 0 :,; 4 y dxdy = rdrd0 IES MASTER Publication 344 Multiple Integral ⇒ Now, I = I = ⇒ ⇒ Method I I : I n cyl i ndrical coordi nates (x, y, z) r::, (p, cp, z) where x 2 + y 2 :,; z3 & O :,; z :,; 1 � JJ ( x + y + 1 0) dxdy 2 � 2 ff ( r cos 0 + r sin 0 + 1 0) rd rd 0 ⇒ 0 :,; <I> :,; 2n & 0 :,; p :,; z 2 3 V = Jff (1) dx dy dz = 2 2n 1 r 2 cos0drd0 I = 2 7t r=0 0=0 f f = 2n 1 + r 2 sin 0drd0 2 7t r=0 0=0 2 f f ⇒ ⇒ = r=2 0=2n 1 1 0rdrd0 + 2 7t r=0 0=0 1 I = 21t 3 f f 2 [cj n r 2n . - cos 0lo + 33 10 I = 20 So l-70: ( � ) 4 0 . sin 0l� = 2 Sol-71 : (c) ·: 0 :,; z :,; 1 zm in = 0 and zmax = 1 Min(/ + /) = 0 so, x2 + y 2 :,; z3 ⇒ / ⇒ � Max(/ + /) = 1 min x = 0 and min y = 0 (i) 1 1 f [f = x- Let x + y = t SO, so , = = and max x = 1 = 0 :,; Z � 1 , 0 :,; y :,; 1 - X 2 , 0 :,; X :,; 1 ✓ Volume = JJf (1} dx dy dz = = = 1 1 M X=O Z=O Y =O f f L 1 L [L f (1) dy dx dx 2 .J1 - x d x}z 7 (4t ) dz = 47t f f f 1 2n p · dp · dcp dz 2 ) dcp dz (2 cj, L L z 2n 4 3 f �J dcp cj, ( 8 =0 0 !8 2j (1) dcp 0 = � 4 So after changing the l i m its we get, m i n x = 0 and m i n y = 0 2 z=O 0=0 r=O :. dy = dt y = (t - x) ✓ ✓1 - x 2n z312 1 \ dy) dx o o ( x + y) max y = 1 - x2 and max x = 1 max y = fff p dp dcp dz (ii) Let = 1 x +1 2x - t - dt fX f f (t 0 x + y = t ![ 1 f ( -: + -1t ) 0 l[ dx = dt X 3 1 x +1 1 2 ) dt dx [ ( t: ) (x - y � dx ) dy o � ( x + y) J[J X t t2 x+1 X 2] ( x �1) = (t - · y) dx dx = 0. 5 So after changing the limits we get, = J [T = = Sol-72: (0.99) = 1 l1+ 0 y ( t - y) : dt ] dy Y( l l 1 f O 1 t t2 - t3 y f -2 dy (y + 1) 1 -1 0 I = cff xy dx dy R = = c [ f J:\y dy dx ] 2 c [rtn[ dx l S u bstitude Sol-74: (a) g(x ) = x) = = c [n l g(x_) = f - :: g(x) = e-1 g(x ) = e-e- x dx = f -e-1 dt (sin- 1 x )2 = f -�d x a ,h - x 2 1 = cos ( � ) u x G i v e n 0 :s; u :,; 1 . S o , x dx f e- x-e- x dx = f e-X = t -e- x dx = dt x 2 f(x ) = e- x-e- x g(x ) = dt ✓1 - 2 y 4 e-x e e- x 0 :s; x :,; 1 , 0 :s; y :,; 1 i.e., 0 :,; 0 :,; � 2 So, we will get as quarter circle in x-y plane and by revolving it by 360 ° , we will get a hemisphere of unit radius. 8 5 ( 5 -1)] 15 f f( dx I t represents a circle in x -y plane. Area of hemisphere Sol-76: (c) II = 0.991 (u pto two decimal places) Sol-7 3 : (b) = t 345 nu y = sin ( ) 2 x 2 + y2 = 1 ; = - 0. 5 = c [ 8 x 5 \� ] = c [ 8 ( 5 5 - 1 ) ] 15 15 = 6 x 1 o-4 x [ x Sol-75: (c) 10 Region R is bounded by I l ) dt dy y+1 1 [=- + ; ] dy t t 2 y = 0 and y = 2x 2y ⇒ P ut sin- 1 Engineering Mathematics ⇒ dx f '( x ) = 2n(1)2 = 2n . : cos ( � ) n = f '(1 / 2 ) = ✓2 ⇒ x 2 R �=✓⇒R=± n 2v 2 1 1 2R -2R 1 7tX + S(x )0 = 2R/n Also f f( x )d x = - ⇒ [- cos] 2 7t 7t 0 X =O ⇒ S=0 So l-7 7: (a) 3 d0 de = 2 J 1 = J 2 2 9 + sin sin 0 0 + 9 0 0 2n = nt2 4f 0 = 12 f 3 d0 9 + sin2 0 n/2 0 n 3 sec 2 0 d0 9 sec2 e + tan 2 e IES MASTER Publication 346 Multiple Integral sec 2 8 d8 = 12 f 2 0 9 + 1 0 tan 8 T[ rr/ 2 2 2re cos x dx cos2 x dx = ef 21 = r 0 0 rr/2 1 0 ta n 0 = t Put v� So, Sol-78: (b) sec 2 8d8 - ⇒ oo dt/M = 1 2f 0 = I 9 + t2 f x cos T[ 0 2 _g_ fw = [ f � M I tan- 1 3 3 0 oo 0 2 xdx = f ( re - x ) cos (re - x ) dx T[ (By definite integral property) 0 2 2 I = ref cos xdx - x cos x dx T[ (_!_)J T[ f 0 I = re l l [� �] 2 2 = e r 0 + + 2 21 � f(x) = x - [x] Sol-79: (0 . 5) So, = = { 1 ✓n ✓ . - 2fj_ 0 :s; X < 1 X - 1, 1 ::,; X < 2 X, 1 1 . 25 ( x - 1 ) dx 1 0.2 5 1 1 . 25 = J- x dx + (x - 1) dx 0.25 1 f f ( x ) dx + J ( f f + � - x r = 0.5 = [� 2 1/4 2 2 2 4 n = re 2 3. 1 VECTOR A quantity which has magn itude as well as direction is called vector quantity. AB = j AB j AB ⇒ - � B Final or terminal point AB = I AB I AB A I nitial point A vector whose magnitude is one unit is a unit vector. Unit Vector i = unit vector along x-axis, ] Example 1 Solution : = IaI = 1 unit vector along y-axis, k = unit vector along z-axis. Find a vector whose magnitude is 8 units, lie along the vector (in the direction of vector) 2i + 3] + 61< a = 21 + 3] + 61< Let then, a = 1a1a Ad dition or Subtraction of Two Vectors ⇒ New vector = 2· 3. 6 · 8 x (- i + - j + - k ) 7 7 7 ➔ a ➔ R -b ➔ R ➔ ➔ ➔ R = a + (-b) Position Vector : A vector whose initial point is origi n is position vector of a point. Let A(x , y, z) be any point then, Vector Jointing Two Points : OA OA + AB = = Position vector of point A = x i + y] + zk OB ⇒ AB = OB - OA A B 348 Vector and their Basic Properties Modulus of Vector : OA = ai +b}+ck I OA I = e.g. 2 +b2 +c 2 a = 2i+3} +61< -Ja ⇒ Two vectors are said to be parallel (collinear) if Parallel Vectors l a I = J4g = 7 a = 1,,6 where e.g., a = 3i-2} +4k ; 6 = -6i +4}-8k then 6 = -2(3i+2} +41<) 6 = -2a so a & 6 are parallel. ⇒ Parallel and col l i near vectors are same vectors. DOT PRODUCT la.6 = I a 11 b I cos el 8 ( i ) a.6 = 6.a (iii) f.S = (ii) J.k=k.i=o a-b = I a 1 1 6 I cose then ⇒ Triangle I n equality Proof : cose = I a+6 i 2 I a+5 1 2 I a +5 i 2 I a+6 1 Example 2 : Solution : = JS = k.k=1 ( i v) a.a = I a 12 (most important) Angle between Two Vectors Let f.f a·b l a ll b l _ - a 1b 1 +a 2 b2 +a 3 b3 ✓af +a�+a� ✓bf+b� +b� = I a 1 2 +I 6 1 2 +2a.6 = I a 1 2 +I 6 1 2 +2 I a I I 6 I cos 0 s I a 12 + I 5 1 2 +2 I a 1 1 5 I s s ( I a I + I 6 1 )2 l a l +l b l If I a I = 3, I 5 I = 4 and a ..l 6 then find I a+5 I 2 2 (a+5).(a +6) = a.a+a.6+6.a +6.6 = 1 a 1 + 2a.b+ l 6 i 2 I +6 1 = 9 + a o + 1 6 = 25 Engineering Mathematics Example 3 = Solution = l a +5 I = so s If a + 5 +c = 0 and a, 5, c are 3 unit vectors. Find the value of a • 5+5 • c +c • a ? a+5+c = o 2 2 l (a+5+c) l = 0 ⇒ 1 + 1 + 1 + 2 (a.5+c.a +5.c) = o I a I2 +1 5 1 2 +I c I 2 +2(a-5 + 5-c+c - a) = o ⇒ 3 - - - - - (a - b+b-c+c-a) = -2 Example 4 : The scalar product of the vector i +} +k with a unit vector along the sum of vectors Solution Let (21+4}-51< ) and ( Ai+2} +31< ) is 1 . Find "' Now = = ✓(2+A,) +36+4 ✓(2+A-) +40 2 2 = J-Js2 +4+4A +40 A,2 +36 +12 A, = A-2 +4A +44 8A. = 8 A, = 1 A.+6 a, 5, c are mutually perpendicular vectors of equal magnitudes : show that the vector a +5 +c is equally inclined to vector a, 5, c Given a-5 ( a +5+c) • a so, cosa ( a+5+c )-5 Similarly, & and hence, = 0 5-c =0 c-a = I a+5 + c I I a I cos a 2 = la+5+cl lal cos a lal +o+o lal = la+5+cl cos a ⇒ Example 6 = = i+}+k, 5 = 2i +4}-5k and c = Ai+2} +3k , then +c = 1 5 + c = (2+A)i +6}-21< so according to question, a. � I b+c I ( 2+A )+6- 2 or Solution a (i +}+k ) · [( 2+A ) i +6}-21<] ⇒ Example 5 = 349 (a+5+c)-c = lal la + 5 +cl = la+5+cl l5l cos p =0 ⇒ = l a+5+c l l c l cosy ⇒ & cosp cos y lal = = = I 5 I = I c I so, cos a = cosp+cosy a = P=y IaI If a and 5 are two unit vectors and (i) 0 1 cos -=- 1a +b 1 2 2 e = l5l = lcl 15 1 la+5+cl lei la +5+cl ...(1) . . . (2) .. . (3) .. . (4) is the angle between them, then prove that (ii) . 0 1 sin- =- 1a +b 1 2 2 IES MASTER Publication 350 Vector and their Basic Properties Solution : We have I a + 5 I2 = la + 5I Example 7 : Solution : = I a 1 2 + I 5 1 2 +2 I a I I 5 I cos 8 = 1 + 1 + 2 X 1 x 1 x cos 8 = 2 + 2 cos 8 = 2 (1 + cos 8 ) I a-5 1 = 2 8 2 = 2 x 2 cos 2 - 8 cos- = 2 Similarly, Hence, 2 2 (a + 5) - (a + 5) = a - a + a - 5 + 5 - a + 5 - 5 = la-i2 + 151 + 2 a. 5 -1 2 Ia + bl - 2 2 (a - 5) · (a - 5) = 1 a 1 + 1 5 i = I a 1 2 + I 6 1 2 -2 I a 1 1 6 I cos 8 - 2a . 6 = 1 + 1 - 2 x 1 x 1 cos 8 = 2(1 - cos 8) = 4 sin2 � 2 sin - = - l a - b l 2 2 . 8 1 - P.T. r = ( r. i) i + (r. }) } + (r . k ) k where r is the position vector of any point in 3D space. Let j' = xi + y} + zk r. i = ( xi + x] + zk) . ( i + o} + O k ) = X then, Similarly, r.j = y & r.k = z RHS = (r . , ) 1 + ( r . ] )] + (r . 1<) 1< = xi + y} + zk = LHS So, CROSS PRODUCT la x b = I a 1 1 b I sin 0 x nl ,__._e____➔ -t \ 8 ' is the angle between the two vectors di rected outwards. n is a unit vector perpend icular to the vectors a and and NOTE: (i) (ii) (iii) (iv) (v) l a x b l = l a l l b l sin 8 l n l = l a l l b l s i n 0 1 1 1 = l a l l b l sin e axb A ,.. xj = -b X a = k A ,.. i xi = j X ,.. = kxk = 0 A A A a = a/ + azj + aJ j then a x b = A '1t A and b = A j � k A jXk = ; kxf = j bii + bzJ + bJ 03 = (a2 � - b2 a3 )i - J (a1 � - a3 b1 ) + k (eib2 ·- a2 6i ) bl bz b3 6. Engineering Mathematics Area of Triangle 351 BC = a , BA = 5 , I BC I = I a I . I BA I = I b I A AD = AB sin 0 = l b l sin e ti e B D C a 1 1 J. Area = - x BC x AD = . I a I I b I sin e = - j a x b I 2 2 2 Area of Parallelogram Let AB = a and AD = 5 , then AC = a + 5 and BO Area = I AB x AD I = a x 5 I I = in terms of sides 5-a C ! AC x B D I = ! (a + b ) x (b - a) ! in terms of diagonals 2 2 Lagrange's Identity & Area = 1 - - Example 8 : Solution : Example 9 : Solution 1 - L.H.S. - Iax51 2 j ax5 I 2 2 2 2 = I a 1 I 5 1 -(a.5) = I a I2 I 5 I 2 sin2 0 2 2 2 I a I I 5 ! (1 - cos 0) = I a !2 I 5 I 2 -(I a I I 5 I cos e)2 = = 2 2 2 I a I I 5 I - (a - 5) If a, 5, c are the position vectors of the vertices of L'lABC then find area of t,. BC = c - a We have We know that Let a • I + 4] + 2k , 5 Area = a and b and c.d = 1 5 . & - ! BC x BA ! 2 1 = - I (c - b) x (a - b) 1 2 = 1 - - A (a) BA = a - 5 B (b) e - 1 J. = - j cxa-cx b - b x a + b x b ! = . 1 cxa+ax5+5x c ! 2 2 3i - 2] + 7k , c = 2i - ] + 4k . Find a vector (a x b) = a a 1 3 11 a x 5 4 -2 = "- (a x 5 ) k 2 = 7 d = "- (32i - } - 1 41<) C (c ) a which is perpendicular to both 321 - J - 1 4k = 32Ai - ,c} - 1 4,-k . . . (1 ) IES MASTER Publication 352 Vector and their Basic Properties c.d = 1 5 Again (2i - } + 4k).(32tci - 1c] - 1 41c1<) = 1 5 641c + 1c - 561c = 1 5 91c - 1 5 Hence, by ( 1 ) cf SCALAR TRIPLE PRODUCT If we have three vectors a, b, c - ⇒ le = 3 32 X 5 o 5 o 1 4 X 5 • 1 60 0 5 " 70 . - k = - 1 - - J - -k - 1 --J 3 3 3 3 3 3 then their scalar triple product is defined as [ a, 5 , c ] = a.(5 x c) . Generally, it represents the volume of paral lelopiped and mathematically it can be calculated as follows : 81 b1 [ a, 5 , c ] = C1 82 bz Cz 83 b3 C3 =a T+a }+a 5 = b T+b J+ b = c i +c j +c where a 1 2 3 1 2 3 c 1 2 3 k k I< [ a, b, c] = [b, c, a] = [ c a b] , i . e . , i n cyc l i c order they are same. 2. Dot and c ross are i nterchangeable ⇒ [ a, b, c] 4. a, b, c a 3. 5. ' [ a b c] = - [ a c b] = a -b x c = a x b -c are coplanar vectors then, [ b c] = 0 In Scalar Tr i p l e Product if any two vectors are same then it must be zero , i .e. [a b a] = o . VECTOR TRIPLE PRODUCT If we have three vectors a, 5, c then their vector triple product is defined as la x (5 x c ) = C a . c ) b - ( a . 6 ) cl Example 1 0 : Find the vol ume of parallelopiped whose edges are represented by a Solution Required Volume = 2i - 3} + 41< ; = [a 6 c] = Example 1 1 : P.T. [a + b , b + c, c + a] = 2 [a 6 c] Solution : ( a + b) · [ (6 + c ) x ( c + a ) ] 5 = 2 -3 4 1 2 -1 3 -1 -2 T + 2} - 1< ; c = 3i - 1} - 21< = 2 x ( -5 ) + 3 x 1 + 4 ( 6 + 1 ) = 1 1 = ( a + b) {b x c + b x a + O + c x a] = (a + 6) . [b x c + 6 x a + c x a J = a. ( 6 x c ) + a. (6 x a ) + a. ( c x a) + 6. ( 6 x c ) + 6. (6 x a ) + 6. ( c x a) = a . (b x c) + o + o + o + o + 6 . (c x a) = [a 5 Engineering Mathematics c] + [5 c a] 353 = [a 5 c] + [a 5 c] = 2 [a 5 c] Example 1 2 : Prove that a, 5, c are coplanar if a + 5 , 5 + c , c + a are coplanar. Solution : or a + b , b + c , c + a are coplanar so thei r Scalar Triple Product will be 0, i.e., [a + 5 5 + c c + a] = 0 2 [a ti c] = 0 i.e., [a, 5, c] = 0 So, a 5 c are coplanar. {': [a + b b + c c + a] = 2[a b c]} DIRECTION COSINES & DIRECTION RATIOS Direction Cosine y A (a,b,c) z F rom figure it is clear that QA = a T + bJ + ck , then Cosines of the angles which a vector makes with +ve directions of coordinate axis are called di rection cosines of vector. If QA m akes a, �. y angle with X, Y, Z axis respectively, then direction cosines are ( l , m, n ) where l = cos a = -;==== · ✓a 2 + b2 + c2 ' and <a, b, c> are called O-Ratios of OA e.g . , b m = cos � = -;==== · 2 .Ja + b2 + c 2 n = cos y = -;==== .Ja2 + b2 + c 2 Direction Ratio of a Line Joining Two Points AB = O 8 - OA = ( X2 i + Y2 ] + Zzl< ) - (X 1 i + y1 ] + zi ) AB = (X 2 - X 1 )i + ( Y 2 - Y 1 )l + (Z2 - Z1 )k D.R.'s = ( x 2 - X 1 , Y2 - Y1 , 22 - 21 ) IES MASTER Publication 354 Vector a n d their Basic Properties m, Relation between D. cosines <l, n> and D ratios <a, b, c> are as f/ollows : l1 = l2 = cos 0 = So, a1 m1 = )r.a � a2 ✓I-a� b1 Jr. a� � , m2 = g· . a1 a 2 + Jr.a � n2 = b 1 b2 ✓I-a� ✓I-a� ✓I-a� ✓I-a� cos 0 = l 1 l2 + m 1 m 2 + n 1 n 2 Parallel Lines C1 , n1 = + C2 Jr.a� wh ere La 2 1 2 2 2 = a 1 + b1 + c 1 2.. wh ere ,,..., a 22 = a 22 + b22 + c 22 C 1C 2 Jr.a � �� l1 = cos a. = l2 m 1 = cosp = m2 n 1 = cosy = n 2 So, if li nes are parallel then their direction ratios are same. Perpendicu lar Lines cos 0 W hen l ines are perpendicular then 0 = 90 ° l1 li + m1 m2 + n 1 n2 = 0 a 1 a2 + b1 b2 + c 1 c 2 = 0 ⇒ Example 1 3 : Show that the vector i+}+k is equally inclined with the axis a = Solutio n : i+}+k; cos u = cos u = cos p = cos y Hence, this vector is equally inclined to the axis. ⇒ J31 , cos p cos y a = P=y STRAIGHT LINES IN 3D 1. Equation of line passing through a point (x1 , y 1 , 2 1 ) and having the ratios as <a, b, c> is ⇒ 2. l� = f = �I (Cartesian Form) Equation of line passing through two points. X - X 1 = Y - Y 1 = 2 - 21 X 2 - X 1 Y 2 - Y 1 22 - 21 a (x,, y,, z, ) Engineering Mathematics 355 " c=====i<-'""':r' 1 . Component of vecto r ii in t h e d i r ection o f 6 is = a · b NOTE Any I ine in 3-D ➔ 2. X - X1 01 3. = y - Y1 bl = z - Z1 . . . (i) ½ It p asses t h roug h t h e point (x 1 , y 1 , z 1 a nd p ara llel to t h e vecto r o,/ + ) 4. D i rection rat ios of l i ne (i) are < Oi , b1 , <; > w here Oi = Al, b1 = Am, c1 = ten and 2 10 = 2 0i 2 2 + b1 + c; . D i rection cos i ne of l i ne ( i ) are ( /, m , n) w here, ± bl .J0i2 + b12 + <;2 5. biJ + ei_k , n Here, signs to be taken a l l +ve o r a l l -ves . ± <; = .Ja� + biz + c; If a l i ne ( i ) ma kes an ang l e a., �,Y from x ,y and z axis respectively t h e n 2 d i rection cosi nes are l = c o s a., m = cos �, n = cos y f r o m w here l + m2 + n2 = 1 . x-4 y-1 z-0 x-1 y-2 z-3 Example 1 4 : Show that the l i nes -- = -- = -- . . . (L 1 ) and -- = -- = -- . . . (L2 ) intersect and find the 2 1 4 2 3 5 point of i ntersection. Solution The general point on line L 1 is x = n+1 , y = 31c + 2 , z = 41c + 3 i .e. , P(n + 1, 31c + 2, 4H3). If P is the point of i ntersection of two l ines then it will satisfy the equation of L2. ⇒ 2A. + 1 - 4 41c + 3 - 0 31c + 2 - 1 = = 5 2 1 2), - 3 1 31, + 41c + 3 = = - 5 2 1 = and 41c - 6 1 5,c + 5 31c + 1 = 81c + 6 1, = -1 A = -1 Since the value of A is same, hence two l ines intersect each other at P = (-1 , -1 , -1 ) x-1 y+1 z-1 x-2 y-1 z+1 Example 1 5 : Show that the l i nes -- = -- = -- and -- = - = - do not intersect. 2 3 5 3 4 -2 Solution Let us assume that . . .(1 ) Let and x-1 y+1 = 3 3 z-1 = A 5 So any point on line ( 1 ) can be taken as P(31c + 1, Putting this point in (2), we get ⇒ x-2 y-1 = 4 3 ⇒ z+1 -2 . . . (2) x = 31c + 1 , Y = 21c - 1 , Z = 51c + 1 n + 1, 57s + 1 ). 3/c + 1 - 2 2A. - 1 - 1 = - -3 4 51c + 1 + 1 -2 IES MASTER Publication 356 Vector and their Basic Properties 91c - 3 = 8 1c - 8 Taking I & I I Taking I I & I l l 1c = -5 ⇒ 1c = -2/1 9 ⇒ ·: 1c is not same so we can conclude that above two li nes do not intersect. Example 1 6 : Find the co-ordinates of foot of perpendicular from P(2, 3, -8) to Solution : P(2, 3, -8) The general point N 4 ; x = f- 1 z ; N X = -2A. + 4 , Y = 61c , Z = -3A. + 1 i.e., N(-2A + 4, 61c, - 31c + 1) & P(2, 3, -8) & D-Ratios of given line are (-2, 6,-3) ""' <b 1 , b2 , b3> ·: Both are perpendicular. ⇒ a 1 b 1 + a2 b2 + a3 b3 = 0 (-21c + 2)(-2) + (6 1c - 3)6 + (-3 1c + 9)(-3) = 0 41c - 4 + 361c - 1 8 + 9A. - 27 = 0 491c = 49 So, N(2, 6, -2) is required foot of perpendicular. Example 1 7 : F i nd the poi nts o n the l i ne Solution N(1 , 3, 3). ⇒ 1c = 1 x+2 _ y+1 _ _ z - 3 at a d i sta nce of 5 u n its from the point 3 2 2 Let the general point be P (31c - 2, 21c - 1, 21c + 3) then, ⇒ ✓(31c - 2 - 1) 2 PN = 5 + (21c - 1 - 3)2 + (21c + 3 - 3 )2 = 5 (31c - 3)2 + (2A - 4)2 + (2A)2 = 25 ,a 91c2 + 9 - 1 81c + 41c2 + 1 6 - 1 61c + 41c 2 = 25 1 n2 - 341c + 25 = 25 1 7A(1c - 2} = 0 1c = 2, 0 So, there exists two points : P(-2, -1 , 3) and 0(4, 3, 7). p - - - - - - - - - - - - - - -- - - - - - - -- - - - - -• N(1,3,3) Engineering Mathematics Example 18 : Find the coordinates of the foot of the perpendicular from (0, 2, 3) on P(O, 2, 3 ) 357 x 3 Y 1 z 4 ; = ; = ; Solution -------- L, N x+3 5 G iven line is = y-1 = z+4 = 'A. So, we can take point N as x = 5'A. - 3 , y = 2'A. + 1 , z = 3'A. - 4 2 3 D R of line are <5, 2, 3> = <a 1 , a2 , a 3> (let) DR of PN are < 5'A. - 3 - 0, 2"' + 1 - 2, 3'A. - 4 - 3 > Now using perpendicularity condition = a 1 b + a 2 b2 + a 3 b3 = 0 1 ( ( 5'A. - 3 ) x 5 + ( 2'A. - 1 ) x 2 + ( 3'A. - 7 ) x 3) = 0 ⇒ So, coordinates of N are 25'A. - 1 5 + 4'A. - 2 + 9'A. - 2 1 = 0 38 'A. = 38 Example 1 9 : Find the i mage of the point (1 , 6 , 3) in the line Solution : < 5'A. - 3, 2'A. - 1, 3'A. - 7 > = D R' s of line < 1 , 2, 3> = <a 1 a2 a3> (let) <b1 , b2 , b3> (let) A, = 1 ⇒ N = (2, 3, -1 ) y-1 z-2 = -- = -3 1 2 X General point N is x = 'A, , y = 2'A. + 1 , z = 3'A. + 2 , i.e. , N('A., 2'A. + 1, 3'A. + 2 ) 3 r· •. 1 I I I I I P'(a, 13, y) Direction Ratios of PN are < 'A. - 1, 2'A. + 1 - 6, 3'A. + 2 - 3 > = < 'A. - 1, 2'A. - 5, 3'A. - 1 > = <b 1 b2 b3> (let) Now using condition of perpendicularly we have a 1 b 1 + a2 b2 + a3b3 = 0 ⇒ So, Hence, ⇒ So, P' is ( 1 , 0, 7) ( 'A. - 1 ) 1 + ( 2'A, - 5 ) 6 + ( 3'A. - 1 ) 3 = 0 'A. - 1 + 1 2"' - 30 + 9'A. - 3 = 0 A, = 1 N = ('A., 2'A. + 1, 3'A. + 2) = ( 1 , 3, 5) 1 = a+1 y+3 P+6 5 = - 3 = 2- · -2 · 2 a = 1, p = 0, y = 7 IES MASTER Publication 358 Vector and their Basic Properties 3D PLANE 1. Let P(x, y, z) be a ny point a 3D-pl a ne, a nd < e , m , n> a re the direction cosines of norm a l vector on it i.e. 2. Simil a rly if Direction-Ratios of norm a l vector a re given i . e. < a , b, c)> a re the D-Ratios then equation of pl a ne is ax + by + cz = d where d is a ny sca lar. 3. 4. n= e i + mj + nk then equ ation of pl a ne is ex + m y + nz = 'p ' where p is dista nce of pl a ne from origin. Equ ation of pl a ne passing through point (x 1 , y 1 , z1 ) & perpendicul a r to vector AB = a i + b} + ck is given as a (x - x 1 ) + b(y - y 1 ) + c(z - z1 ) = 0 Here < a , b, c> a re ca l led D-ratios of pl a ne. Dista nce of Point from Plane : Let ax + by + cz + d = 0 be a ny pl a ne & let P(x, y, z) be a ny point from where dista nce is to be mea sured. then required dista nce is D - 1 - ax1 + bY 1 + cz1 + d ✓a 2 + b2 + c2 i x-2 y+2 z-3 Example 20 = Prove that line -- = � = -- i s p a ra l lel to x + 5y + z = 7 1 4 Solution = P (x,, y,, 2 1 ) D ax + by + CZ + d = 0 Direction Ratios of Pl a ne a re < 1 , 5, 1 > = < a 1 a2 a 3> i . e . , these a re the D-Ratios of norm a l vector the pl a ne a nd Direction R atios of line a re < 1 , -1 , 4> = <b 1 b2 b 3> a 1 b 1 + a2 b2 + a3 b3 = 1 x 1 + 5 (-1 ) + 1 4 = O So, l i ne and norm a l vector a re at 90 . Hence we ca n conclude th at pl a ne is pa ra llel to given l i ne. ° x x x+1 y-2 z+6 Example 21 : If -- = -- = -- is p a rallel to 2x - 3y + kz = 0 then find K. 3 4 5 Solution ·= D-Ratios of pl a ne = D-Ratios of Norm a l Vector t6 Plane = <2, -3, K> & D-Ratios of Line = <3, 4, 5> So by usi ng perpendicul a rity condition a 1 b 1 + a2b2 + a3 b3 = 0 (3) x 2 + 4(-3) + 5 (k) = 0 k = 6 5 Example 22 : If 2x + 3y + 4z = 7 a nd kx + 4y - 62 = 9, Find k such th at both pl a ns a re perpendicul ar. . . . (1 ) Let 2x + 3y + 4z = 7 Solution : and kx + 4y - 62 = 9 ⇒ 2k + 1 2 - 24 = 0 ·: D-Ratios of Plane ( 1 ) = <2, 3, 4> and D-Ratios of Pl a ne (2) = <K, 4, -6> So by perpendicul a rity condition a 1 b 1 + a2 b2 + a3 b 3 = 0 2k = 1 2 I< = 6 . . . (2) Exam ple 23 = F i nd the length a n d foot of perpendicul a r d ra wn from the point ( 1 , 1 , 2) to the p l a n e Solution T · ( 2 i - 2} + 4k ) + 5 = o Let T = xi + y} + zk Engineering Mathematics r - (2i - 2} + 4k) + 5 = o then 2x - 2y + 4z + 5 = 0 or ( xi + y} + zk) - (2i - 2} + 4k) + 5 = 0 ⇒ . . .( 1 ) ( 1 ) is given equation of plane so O-Ratios of Normal to Plane i . e . , of P N = <2, -2, 4> 359 [9, 2 ) S o equation of line P N having D-Ratios <2, -2, 4 > and passing through P(1 , 1 , 2 ) is x-1 2 ⇒ = y-1 -2 = X = 2'A, + 1 ; z-2 4 =A y = -2'A, + 1 ; Z i.e., we can take coordinates of Point N as (2'A, + 1, of plane. So by using ( 1 ) ( 2'A, + 1 ) X 2 - 2 ( -2'A, + 1 ) + 4 ( 4'A, + 2 ) + 5 = Q 4'A, + 2 + 4'A, - 2 + 1 6), + B + 5 = 0 24), + 1 3 = 0 I 13 A, = 24 or So, Point N is given as, d = PN = and � = 12 ✓ 6 = 4'A, + 2 n + 1, 4'A, + 2) (1 + 1 12 and this poi nt will satisfy equation ) + 12 ( ( 1 25 ) + 2+ ) 6 12 2 2 Example 2, : Find the distance of the point (1 , -2, 3 ) from the plane... (x - y + z = 5) measured parallel to the line · Solution y-3 z+2 x-1 - = - - = -2 3 -6 D-Ratios of given l i ne = <2, 3, 6> . 6 So, D-Ratios of l i ne PQ = <2, 3, 6> (as both are parallel). Now, equation of l i ne PQ, passes through P(1 , -2, 3) and having D-Ratios <2, 3 , 6> is General poi nt on Q ; x-1 2 So, y+2 3 = x = 2'A, + 1 , By using equation of plane x - y + z ⇒ = 5 ( 2'A, + 1 ) - ( 3'A, - 2 ) - 6'A, + 3 = 5 Hence = z-3 -6 z = -6'A, + 3, a n d y = 3), - 2 = -1 or 'A, = X = 9/7, PQ = (1 - y = 'A, = 7 -1 1 /7, Z = 1 5/7 i . e . , Q = ( 9 ) + ( -2 + 1 1 ) + ( 3 - 1 5 ) 7 8 7 2 2 2 = 1 1 -2, 3 ) / (1 , 3 ,-2) Q Given Line 9 -1 1 1 5 ) y ' y ' -7 IES MASTER Publication 360 Vector and their Basic Properties Example 25 : Find the distance of the point 2 i + Solution : Let r J-k from the plane r · = xi + y} + zk then equation of plane in terms of Cartesian form is or r . (i - 2} + 41<) and given point i s Hence, = 9 (xi + y} + zk ) - (i - 2} + 4k ) = 9 ⇒ x - 2y + 4z = 9 = ax + by + cz = d required distance = = ! ax1 + by1 + cz1 - d I ✓a 2 + b2 + c 2 2-2-4-9 ✓1 + 4 + 1 6 (i - 2J + 4k) = 9 = 13 ✓'21 = 11 X 2 - 2 X 1 + 4 X -1 - 91 units ✓f + (2 ) 2 + 42 Engineering Mathematics 361 PREVIOUS YEARS GATE & ESE QUESTIONS 1 .* Questions marked with aesterisk (*) are Conceptual Questions If P, Q and R are three points having coordinates (3, -2, -1), (1 , 3 ,4), (2,1 ,-2) in XYZ space (0 being the origin of the coordinate system) then distance of point P from plane OQR is (a) 3 2. (b) 7 (d) 9 (c) 5 [GATE-2003; 1 Marks] The angle between two unit-magnitude coplanar vectors P(0.866, 0.500, 0) and Q (0.259, 0.966, 0) will be (b) 30° (a) 0° (d) 60° (c) 45° [GATE-2005; 1 Marks] 3.* If a vector R(t) has a constant magnitude than (a) R· dR =0 dt dR =0 dt (b) R x - - dR (d) R x R=dt [GATE-2005; (IN)] 4.* The area of a triangle formed by the tips of vectors, (c) -·- dR R R=dt a,b and c is (a) (c) 5. -i ( a-6) .(a-c) 12 1a x b x c 1 7.* (d) -i ( a x 6 ) - c [GATE-2007-ME; 2 Marks] The angle (in degree) between two planar vectors 1 b = -✓ 1 + J. ·1 s a = ✓ 1 + J and 2 2 2 2 3 (a) 30 6. (b) -1- l (a - b) x (a-c) I . 1 . 3 . (c) 90 (b) 60 (d) 120 (a) zero (b) 30 [GATE-2007 (Pl)] The inner (dot) product of two vectors P and Q is zero. The angle (degree) between the two vectors is (c) 90 (d) 120 [GATE-2008; 2 Marks] If a and 5 are two arbitrary vectors with magnitudes a and b, respectively, Ja x 5{ will be equal to (a) a2 b2 -( a - 5 ) (c) a2 b2 +(a.5 ) 8. 2 2 (b) ab-a-5 (d) ab + a.5 [GATE-2011 -CE; 1 Marks] If A (0, 4, 3) B(0 , 0, 0) an C (3, 0, 4) are three points defined in x, y, z coordinate system, then which one of the following vectors is perpendicular to both the vectors AB and BC (a) 1 6T - 9T-12 K (c) 1 6 T-9T-12 K (b) 1 6 T - 91+12 K (d) -1 6 T-9T+12 K [GATE-2011 (Pl)] 9.* For the parallelogram OPQR shown in the sketch, OP=ai+b} and OR=ci+d} . The area of the parallelogram is . R 0 (a) ad - b e (c) ad + b e (b) a c + bd (d) a b - cd [GATE-2012-CE; 1 Marks] 10.* A particle, starting from origin at t = Os, is traveling along x-axis with velocity V=%cos (%t ) m/s At t = 3s, the difference between the distance covered by the particle and the magnitude of displacement from the origin is_ _ [GATE-2014; 1 Marks) 1 1 .* A particle move along a curve whose parametric equations are : x = t 3 + 2 t , y = -3e-21 and z = 2 sin (5t), where x, y and z show variations of the distance covered by the particle (in cm) with time t (in s). The magnitude of the acceleration of the particle (in cm/s2 ) at t = 0 is____ _ _ [GATE-201 4 (Pl-Set 1 )) 12.** Consider the time-varying vector I = x 1 5 cos (cot) + y5 sin (cot) in Cartesian coordinates, where. co > 0 is a constant. When the vector magnitude ll l is at its IES MASTER Publication 362 Vector a nd their Basic Properties 2. minimum value, the angle e that I makes with the x axis (in degrees, such that 0 < 0 < 1 80) is ___ 3. [GATE-201 6; 1 Mark] 1 3. The vector that is NOT perpendicular to the vectors (i + j + k) and (i + 2j + 3k) _____ (a) (i - 2j + k) (c) (Oi + oj + Ok) 1 4. 4. (b) (-i + 2j - k) (d) (4i + 3j + 5k) The smaller angle (in degrees) between the planes x + y + z = 1 and 2x - y + 2z = 0 is ____ (a) (c) [GATE-201 7-EC Session-II] x2 = [-1 2 8 1 6]' in radian is ___ 16. [GATE-201 7 (IN)] W h i ch of the following statements are correct regardi ng dot product of vectors? 1. (a) 3. (a) 2. (c) 6. (d) 9. (b) 4. 5. 7. (c) 8. (d) 1 0. (2) Si nce, Q and R lie on (i), [ESE 201 7 (EE)] (b) 73.7° (d) 53.7° 1 7. (d) . . . (i) [ESE 201 8(EC)] (b) (a) (54.73) (0.65 to 0.8) SOLUTIONS ax + �y + yz = 0 where a and b are constants. (d) 2, 3 and 4 only 1 6. (90) 1 5. (a) x + ay + bz = 0 (b) 1 , 3 and 4 only 1 , 2 and 4 only (12) 1 4. A pJane through origin (OQR) is given by ⇒ 1 , 2 and 3 only (c) 63.7° 1 3. Given poi nts P(3,- 2, -1 ), Q(1 , 3, 4), R(2, 1 , - 2) Sol-1 : Dot product is equal to the product of one vector and the projection of the vector on the first one. (a) 83.7° 12. (a) %. 3a x + 4a y + a 2 and B = 2a y - 5a2 is nearly 11. (a) vectors is less than or greater than The a n g l e 0A8 betwe e n the v ectors A = 17. Dot product is less than or equal to the prod­ uct of m agnitudes of two vectors. 1. Dot product of two vectors is positive or nega­ tive dependi ng whether the angle between the Select the correct answer using the codes given below: [GATE-201 6 (IN)] 1 5. The angle between two vectors x 1 = [2 6 1 4]' and When two vectors are perpendicular to each other, then their dot product is non-zero. 1 + 3a + 4b = 0 2 + a - 2b = 0 1 . On solving, a = - 1 and b = 2 : . Equation of the plane OQR is ⇒ z x - y + - = 0 2 2x - 2y + z = 0 : . Distance of P from (iv) is . . . (ii) . . . (iii) . . . (iv) Engineering Mathematics = 1 2( 3) - 2(-2) +(-1) = 3 Sol-2: (c) J2 +(-2) +1 2 2 2 1 a x 6 = ( la! l6lsin0 ) n Let, la X 61 The two vectors can be rewritten as cos0 = = = p.q IPll ql 0.866 0.259+0.5 0.966 1x 1 = 0.707 0 (a) :;e = X X dR n = = So 0 Sol-6: (c) ⇒ ⇒ ⇒ Sol-7: li5I (a) = = IJ(6- a) x (c-a)I IJ(a-6) x (a-c)I 3 1 ('JI +JJ } ( fI +{)= � +�= � 1 & 161 =1 ( 1 a·6 1 ° cos- 1 ( ) =cos- - ) =1 20 2 !al 1 6 1 13.o lol cos0 COS 0 0 la x 61 2 = 0 = 0 = = = 2 2 lal l6l\1-cos2 0) 2 0 goo BC ( a x 6 ) .( a x 6 ) Qi-4]- 31< = 3i = = A = = ⇒ Sol-1 0: (2) 0 3 + O]+41< k -4 -3 0 4 -1 61 - 9]+1 21< Sol-9: (a) Area of parallelogram = Area of the triangle ABC = ..!_IAB x Aci 2 !al AB AB x BC 45° - = l al l6l sin2 0(n,n) {!al=a; 151= b} Sol-8: (d) 1 2 ✓ ·: Angle between R & - is dt 2 Sol-4: (b) Sol-5: (d) {a, 6, n form a triad} o.259i+0.966]+01< = a-6 = p = 0.8661 + 0.5} +Ok q Sol-3: 2 363 OP x OR k a b O c d 0 IA I = ad - be = (ad - bc) k At t = 3 s, the distance covered by the particle is 1 + 1 + 1 = 3m and displacement from the origin is -1 . V For, Integrating At, t = 0, t = 1, t = 2, t = 3, = %cos ( % t) �� _ %cos (%t) \s =sin%t\ s=O �1 s = 1 Start End s=O s = -1 Difference between the distance covered by the particle and the magnitude of displacement from origin is 3 - 1-1 1 = 2. IES MASTER Publication 364 Vector and their Basic Properties cos 8 = Sol-1 1 : (12) Let r = xi + y} + zk be the position vector of any point in space then r = (t3 + 2t) i - 3e-2 1 }+ 2 sin 5tk Velocity = dt d2 r Acceleration = dt2 Sol-1 5: (0.65 to 0.8) = Sol-1 2: (90) Given, I = x1 5 cos (cot) + y5 sin (cot) where co = � so cot = T 2nt at t = T/4; cot = n/2; I = 5y 248 -J235 -J454 ✓ 8 = 0.72 radians 62 MAA Sol-1 6 : (b) Dot product is given by A - B = I A I I B l cos 8 where 8 is angle between the vector A & B. Also at t = 0; cot = 0; I = 1 5x (2)(-1 2) + (6)(8) + (14 )(1 6) -J4 + 36 + 1 96 144 + 64 + 256 = :::::: 0.75 -· - 1 2} : . The required magnitude = l -1 2JI = 1 2 3 3 x --/3 ' 8 = 54.73° cos 8 = = 6ti - 1 2e -2 t} - 50 sin 5tk :. At t = 0; -13 . Jg 8 = cos-1 (�) dr = (3t2 + 2)i + 6e-2 1 J + 1 0 cos 5tk ⇒ cos 8 = [2 - 1 + 2] B at t = T/2; cot = n; l = -1 5x at t = 3T /4; cot = 3n / 2; I = -5y at t = T; cot = 2n; I = 1 5x +5 t=T/4 lllmin = ±5 Angle making with x-axis = ± 90 ° -5 0 :,; 8 :,; 1 80° as given 8 = + 90° Sol-1 3: (d) t=3T/4 On verification, option (d) is not perpendicular to both the given vectors 2 ✓ 1 1 2 2 = + = 3 3✓ 3 ✓ Sol-14: (54.73) and x + y + z = 2x - y + 2z = 0 cos 8 = A-B I A II B I --+ So dot product is· product of one vector and the projection of the other vector on the first vector. B cos 0 From the figure ⇒ ��-----i� - - Sol-1 7 : (a) A = 3ax + 4ay + az s = 2ay - 5az A · B = IAI lslcos8 AB A-B = 4 x 2 - 5 = 3 IAI = -Jg + 1 6 - 1 = ✓26 Isl = -J25 + 4 = 3 = ✓26 m ✓'29 cos 8 3 A6 PARTIAL DERIVATIVES OF VECTORS Let r = T x, y, z ) i .e. r be a vector function of three scalar variables x, y and z . Then partial derivative of r with respect ( T ( x + 8 x,y, z) - T ( x, y, z) to x Is defined as - = hm -------� . Thus . ar ax . 8x ➔O OX i s nothing but the oridnary derivatives of r with respect to x provided the other variables y and z are regarded ar ax as constants. Sim ilarly we may define the partial derivatives ay ar r. and ai a If r has continuous partial derivatives of the second order at least, then _ is i m material. If r = T x, y, z ) , the total differential dr of r is given by dr = ( Example 1 a2 r = a2r axay ayax a r dx a r dy + a r dz . ax + ay az i .e. the order of differentiation 2 If <l>(x,y, z) = xy z and A = xzi - xy 2 } + yz 2 k find � ( <!>A) at (2, -1, 1) . ax az Solution ⇒ a 3 ( <l> A) ax az 2 = 4y2 z•i - 2y4 .j = 4 i - 2j at (2, -1, 1) • • POINT FUNCTIONS (i ) (i i) ( iii ) Scalar Point Functions: If to each point P in a region R, there corresponds a scalar <l>(P) , then we say that <j> is a scalar point function. Vector Point Function: If to each point P in a region R, there corresponds a vector T (P) , then we say that T is a vector point function. Level Surfaces: f (x, y, z) = C, where C is an arbitrary constant, represents a family of surfaces, known as level su rfaces in three di mensional spaces. 366 Gradient, Divergence and Curl DEL OPERATO R (NABLA OPERATOR/HAMI LTON IAN OPERATO R) The vector differential operator v (read as del or nabla) is defined as V = � i + � } + �k It is not a vector but combine both vectorial and differential properties ox ay az T H E LAPLACIAN OPERATOR It Is defined as v 2 . . a2 ox2 a2 ay 2 a2 = 'v - 'v · 2 82 = -+-+- a2 f a2 f a2 f If f is a scalar point function, then v 2f is also a scalar quantity defined as V 2 f =++2 2 2 lf ax v ay az a2- a2- a2is a vector quantity, then v 2v is also a vector quantity defined as v 2v =_y_ +_y_ + _y_ ox 2 . ay2 az2 Laplace's Equation : It is defined as V 2 <j> = 0 where <j> ➔ <j>(x, y, z) . Example 2 Solution 1 Show that V 2 ( x/r 3 )=0 1 Now v 2 (x/r 3 ) = a2 ax 2 r 3 (X) = = (! 2 2 + ! ! )C� ) 2 { ( ox ox ?'" �{_!_ ax a{ a a r3 x + )} = 2 { ax ?'" a 1 - 7 ox } 3x 2 � ( ·: r2 =x2 + y2 + z gives � =� ) ox r r4 r } 1 3x2 -3 ar = ox 3 _ 5 }= 4 r r r ox _ 6x 1 5x 2 ar + 6 r5 r ox -3 x 6x 1 5x 2 x 9x 1 5x 3 +-6- -=--5 +-7 = --4 5 r r r r r r r Again Similarly, 3x ar 3x 1 5xy2 --5 +-r7 r -3x 1 5xz2 a2 ( X ) -+-= az 2 r 3 r5 r7 Therefore, adding we get v2 c� ) = ( :2 + :2 + ;2 )C� ) = _ �: + 1 �t _ !: + 1 5 y2 _ � r Harmonic Function: Any function which satisfies Laplace equation is called Harmonic Function. Here Function <j> !: + 1 5 z2 � r is Harmonic Engineering Mathematics 367 GRADIENT (OR SLOPE) OR (NORMAL VECTOR) OF A SCALAR POINT FUNCTION If f(x,y,z) = c i s a scalar point function then (grad f) i s a vector function defined as grand f = �a �a .a Vf = ( 1 - + J - + k - ) f ax ay az = �at - a t " (_at 1 - + J - + k- ) = ( �at 1 -) ax ay az L.., grad f ax Geometrical lnterperation of grad f : Geometrically g radient of f shows a vector normal to the surface f(x,y,z) = c and it has a magnitude equals to the 0.0. in that direction. f(x, y, z) = C DIRECTIONAL DERIVATIVE (D. D.) Component of grad f i n the direction of vector a is known as directional derivative in that direction i.e. grad f \o.o. = (vf) · a\ . ➔ a / f(x, y, z) = C ----e INOTE:, Component of vector a in the dir:ction of = OAcos e = l a l cose = lal b a.b = a.b lal lb I = OM o�� �----------+ i. B • a-b Geometrical Interpretation o f Directional Derivative M b Si nce component of any vector in its own direction is maxi m u m . So, directional derivative wi ll be maximum in the d i rection of g rand f. A nd � INOTE 5 2 (gradf ) lgrad ti . ) f rad g ( . f ) f grad (grad = = = l gradf i Max 0 . 0 = ( _ ) l g radf l l g raf fj 1. Directional derivative is maximum in the direction of normal to the surface which is also the direction of grad f. 2. D irection of grad f = Direction of normal vector. · 3. A unit vector normal to the surface {f(x,y, z ) Theorem = = C} is = graf f Jgrad fJ I f f and g are two scalar point functions, then g rad (f + g) = g rad f + grad g or V (f + g) = Vf + Vg Theorem : The necessary and sufficient condition for scalar function to be constant is that vf = o Proof : If f (x,y,z) is constant, then at = 0, at = 0, � = 0 ax Therefore grad f = i at ax + }� + k ay ay at az Hence the condition i s necessary. az · = 0i + 0j + 0k = 0 IES MASTER Publication 368 Gradient, Divergence and Curl Conversely, let grad f = . af . af . af 0, then i -+ j-+k-= 0 af af af Therefore, -= 0, -= 0, -=0 ax ax oY fJz & oY :. f must be independent of x, y and 2 . Theorem :. f must be constant. Hence the condition is sufficient. Gradient of the product of two scalar point functions is defined as grad (fg) = f grad g + g grad f or V { fg)= fVg+gv'f Ir particular if c is a constant. Then V (ct)= cv'f +fVc = cv'f +0 = cv'f. LEVEL SURFACE Let f (x,y,2) be a scalar field over a region R. The point satisfying an equation of the type f(x,y,2) = C (arbitrary constant) constitute a family of surface in three dimensional space. The surfaces of this family are called level surfaces. Theorem Proof : The vector Vf at any point is normal to level surface f (x,y,2) = C which passes through that point. Let r= xi+y} +21< be the position vector of any point P(x,y, 2) on the level surface f (x,y,2) = C. And O (x+8x, y+8y, 2+82 ) be a neighbouring point on this surface. Then the position vector of O =r+8r=(x+8x) i +(y+8y) }+(2+82 ) 1< PQ = (r+8r)-r =8r =8xi +8y}+82k As Q ➔ P, the line PQ tends to tangent at P to the level surface. :. dr = dxi+dy}+d2k lies in the tangent plane to the surface at P. Now, so, Vf -dr =0 Vf.dr ⇒ = = � J · (dxi+dy}+d2k ) = � dx+� dy+� d2 (1 �+}�+k 8x 8y 82 8x 8y 82 d f = 0 since f (x,y,2) Vf is a vector perpendicular to dr = ⇒ C Vf is perpendicular to the tangent plane at P and hence Vf is normal to the level surface f (x,y, 2) = C. S O LVED EXAMPLES Example 3 : Solution : Find the angle between normals to the surface x 2 +y2 +22 =9 and 3 =x2 +y2 -2 at the point (2 , -1 , 2). Equations of the surfaces are cp x2 +y2 +22 = = = x2 + y2 - 2 = constant. constants & �2 Normal to two surfaces, at (2 , -1 , 2) are parallel to V�1 and V�2 respectively at that point. and � 8cp1 � 8cp1 • 8cp1 V cp1 = 1 +.J 8y +k Vcp2 �18 cp2 +� 8cp2 +k. 8cp2 J 8x 8y 82 8x = If 0 is angle between normals then 82 = = � •) 2 ( x�1+yJ+2k • 2 yj-k • • 2xi+ = = • • • at ( 2,-1, 2 ) 4i-2j+jk • 2•j-k• at ( 2,-1, 4i- 2) Example 4 : Solution : (Vcp1 ) · ( Vcp2 ) cas e - I'vcp1 II'vcp I 2 Engineering Mathematics ( 41 - 2] + 41<) ( 41 - 2] - k) 8_ � � - � - --'-====== • -'-==� - _� -...,2 1 �6-...,21 3v21 63 16+4+1 16+4+16 369 ✓ ✓ Find a unit normal vector to a level surface x 2 y + 2xz == 4 at the point (2, -2, 3) . Given that the equation of the level surface as f ( x , y, z) = x 2 y + 2xz == 4 T_he vector grad f l ies along the normal to the surface at the point (x,y,z) Therefore Vf = V ( x2 y + 2xz) = ( 2xy + 2z) I + x 2 ] + 2xk : . at the point ( 2, -2, 3) , grad f == -21 + 4] + 41< : . -2 1 + 4] + 41< is a vector along the normal to the given surface at the point (2, -2, 3) Hence unit normal vector to the surface at this point n = Example 5 : Solution : -21 + 4] + 41< == .:..21 + 4] + 41< -1 grad f 1 + � ] + � 1< == = 3 3 3 lgrad f ll 4+16+16 1-21 + 4} + 41<1 ✓ 1 � 2o 2 · 1 � 2o 2 · The vector - ( - 1 + J + k ) == 1 - J - k is also a unit normal to the given surface at the point ( 2, -2, 3) . 3 3 3 3 3 3 Find the directional derivatives of a scalar point function f in the direction of coordinate axes. We know that Vf = (atax I + �] + � k) ay az The directional derivative of f in the direction of I == grad f. I == ( : I + : } + : k ) = : Example 6 : Solution Similarly, the directional derivatives of f in the di rection of ] & k are : and : . Find the directional derivative of f (x,y,z) = x 2 yx + 4xz2 at the point (1 , -2, -1 ) in the direction of the vector 21 - } - 21< . We have f (x,y,z) = x 2 yz + 4xz2 2 2 : . grad f = ( 2xyz + 4z2 ) 1 + x z} + ( x y + 8xz)k = 81 - } - 1 01< at the point (1 , -2, -1 ) If a be the unit vector in the direction of the vector 21 - ] - 21< then 2 I - ] - 21< 2 • 1 0 2 • a = -----;====== = - 1 - - J - - k 3 3 4+1+4 3 Therefore, the required directional derivative is ✓ Example 7 : Solution : 1 6 1 20 37 � = grad f a == (81 - ] - 1 01<) - ( � 1 - J_ } - � k ) = - + - + - == ds 3 3 3 3 3 3 3 Si nce this is positive, f is increasing in this direction. Find the directional derivative of xyz2 in the direction of the vector 2 I + } - k at (2,3, 1 ). Equation of the surface is � = xyz2 �acp o 8<p • 8cp • • Vcp = 1 - + J - + k - = yz2 i + xz2 j + 2xyzk = 3i + 2} + 1 2k at (2, 3,1) ax ay az A IES MASTER Publication 370 Gradient, Diverg ence and Curl Unit vector i n the given direction is 2T + } - 1< 2i + J - k • ✓6 a = ---=== = ✓4 + 1 + 1 Therefore, required directional derivatives v cp.a• - (3"I + 2"J + 1 2k" ) · n Example 8 : Solution 2i + j - k _ 4 ✓ 6 6 ✓ ----==-- -- 2 2 2 Find the directional derivative of the function f = ( x - y + 2z ) at the point P ( 1, 2, 3) i n the direction A A A 2xi - 2yj + 4zk of the l i ne PQ where Q is the poi nt (5, 0, 4). of 7 of 7 of • Here grad f = - 1 + - J + - k ax A lso ay az = = 2i - 4} + 1 2k at the point (1 ,2,3) PO = ( sT + o} + 41<) - ( T + 2} + 3k) = 4i - 2} + k If a be the unit vector in the direction of the vector PO Then 4T - 2} + 1< 4i - 2} + k • a = ---==== = ffi ✓1 6 + 4 + 1 : . The req uired directional derivative Example 9 : Solution = A (� 0 ") (gra d f ) • a = 2 1 - 4 J + 1 2k · { 4i - 2} + k ffi } = 28 ffi I n what direction from the point ( 1 , 1 , -1 ) is the directional derivative of f = x 2 - 2y 2 + 4x2 maximum? Also find the value of this maximum directional derivative. We have grad f = 2xi - 4y} + 8xk = 2i - 4} - 8k at the point ( 1 , 1 ,-1 ) The directional derivative of f is a maxi mum in the direction of grad f = 2i - 4} - 8k . The maxi mum value of this directional derivative Example 10 : For the function f = y/( x 2 + y 2 ) , find the value of the directional derivative making an angle 30 ° with the positive x-axis at the poi nt (0, 1 ). Solution We have -2xy , x 2 - y2 o = . o at oJ = at 7I + 1+ 2 J - J at the point (0, 1 ) grad f = ( X + y2 ) ( X 2 + y2 ) 8x 8y If a is a unit vector along the l i ne which makes an angle 30 ° with the positive x-axis, then A a = A COS 30°'I . : . Th e re q uIre d d'Irect'Iona I d eriva ' t'Ive Is ' = A r;::; 3A 1• • 30° J' = '\j,j I• + - J• + Sin A ( 2 0) grad f • a = -J · 2 ( ✓3 C 1 0 = 1 I + JJ 2 2 2 Example 1 1 : W hat is the greatest rate of increase of u = xyz2 at the point ( 1 ,0, 3)? Solution : We have Vu = yz2 i + xz2 } + 2xyzk Engineering Mathematics 371 at the point (1 ,0,3) we have Vu = Oi + 9} + Ok = 9} The greatest rate of increase of u at the point (1 ,0,3) = the maximum value of �� at the point (1 ,0,3) = I Vul , at the point (1 ,0,3) = \9}\ = 9. Example 12 : Show that the directional derivative of a scalar point function at any point along any tangent line to the level surface at the point is zero. Solution Let f (x,y,z) be a scalar point function and let f (x,y,z) = c. a be a unit vector along a tangent line to the level surface We know that Vf is a normal vector at any point of the surface f (x,y,z) = c. Therefore the vectors Vf and a are perpendicular. Now the directional derivative of f in the direction of a = a · Vf = 0 Example 1 3 : Find the values of the constants a, b, c so that the directional derivative of f = axy 2 + byz + cz2 x3 at (1 , 2,-1 ) has maximum magnitude 64 in the direction parallel to z-axis. Solution 2 3 2 3 2 2 2 v f = ( 1 :x + ]; + k !J (axy + byz + cz x ) = (ay + 3x z c ) i + (2axy + bz) } + (by + 2czx ) k Vf = (4a + 3c) 1 + (4a - b) } + (2b - 2c) k at (1, 2, - 1) Since maximum value of directional derivative occurs in the direction of Vf itself which according to given condiion is parallel to z-axis, having direction cosines 0,0, 1 hence ⇒ ⇒ Thus, 4a + 3c 4a - b 2b - 2c -= -- = -0 0 1 4a + 3c = 0 4a - b = 0 . . . (i) .. . (ii) y'f = (2b - 2c )k Also maximum value of directional derivative is iVfi Therefore I Vfi = 64 ⇒ 2b - 2c = 64 . . . (iii) From (i), (ii) and (iii), we get a = 6, b = 24, c = -8. Example 14 : If f(x, y, z) = 3x2 y - y3z2 , find grad f at the point (1 , -2, - 1 ). Solution = We have grad f = Vf = ( 1 :x + } ; + k ! J (3x 2 y - y 3z2 ) = i ! (3x 2 y - y3z2 ) + } ; (3x2 y - y3 z2 ) + k ! (3x2 y - y3z2 ) = 1 (6xy) + } (3x 2 - 3y 2 z2 ) + 1< (-2y 3z) = 6xyl + (3x 2 - 3y 2z2 ) } - 2y3zk Putting x = 1 , y = -2, z = -1 , we get 2 2 v t = 6 (1)( -2) 1 + { 3 (1) - 3 ( -2) ( - 1) (.- 1) } 1 - 2 ( -2) ( - 1 ) 1< = -1 21 - 8} - 1 61< 2 2 2 IES MASTER Publication 372 Gradient, Divergence and Curl Example 1 5 : Find V� at (2,-1 , 1 ), where �=4xz3 - 3x2 y2 z Solution : 2 2 2 V� = (i ! + J ; + k !) (4xz3 -3x y z ) = (4z3 -6xy 2 z ) i +(-6x2 yz ) } + (12xz -3x 2y 2 ) k = Bi + 48] + 841< at (2,-1,1) Example 16 : Prove that : (i) vr m =mr m-27 , where r = lrl (i) X '-'�1 -r B m '-'�mr m-1 Cf '-' � m2 '-'�1mr m-1 Vr m = L.. - = L.. = L.. 1mr - x ax =L.. 1 ax r (ii) grad (a•r)=a , where a is a constant vector and r is 3D radial vector Solution . (•1 •1) grad (a - r) = " L..i a (a-r ) ax " � - ar " � aa - _ ar = L.. 1 ( x • r + a • x )=L.. 1 ( a • ) a a [· : rxi=xi+y}+zk=r ] ax Example 1 7 : Shot that grad f (r )= f ' (r) · r Solution : Let r = xi + y} + zk r2 = x 2 + y2 +z2 grad f(r) = v ( f (r ) ) = ar ar x & 2rar 2y ⇒ -=ar y & or _ _z 2 r- = 2x ⇒ -=ax oy r az r ax r ax = a k· a ) {t (r )} ( �1axa �J az + + 8y = f' (r) [i � + ]Y+k �J = f' (r ) i_ r r r r Example 1 8 : Evaluate v (loglrl ) Solution : So lution : = .J f'(r ) = f'(r) • f . (r ) ar ar .jf' (r) ar +kf' 8Y + ax r 1 • v (loglrl ) = v (logr) = -•r = 2 r r Let r = X i + yj + zk then r = \/ X • • · ✓ 2 2 and rn =( x + y + ( '2 + y2 + z2 ) z2 r Now deifferentiating with respect to x, nrn -1 - ar n -2 or nr n -1 - = nx (x 2 + y2 + z2 )2 ax ax az ... (i) Engineering Mathematics -2 - ar nr n 1 - = nx( x 2 + y 2 + z2 r ✓ ax nr n-1 - = ny ( x2 + y 2 + z2 r Similarly, ✓ ar ay n ( grad r ) = v (rn ) Now = n 1 2 & nr - - = nz ( x + y 2 + z2 r ✓ ar az 2 373 -2 ci a + -j a + k- a y rn ) ax ay az J� [ ar J + nr n -r ar ar = [ nr n-1 -l1 + [ nr n - 1 -] k ax ay az = nx ( x 2 + y2 + z2 r i + ny ( x2 + y2 + z2 r } + n z ( Jx2 + y2 + z2 r k ✓ 2 ✓ 2 2 Example 20 : grad (a • r ) = a i.e v (a • r) = a where a is constant vector & f is position vector of any point in 30. Solution : Let a = a1 i + a 2 } + ai and f = x.i + y} + zk then, v (a • r ) Now DIVERGENCE If f is a vector point function given by f = f1 i + f2 } + ti then d ivergence of f is defined as - - L (· axa'f J (· axa div f = v . f = NOTE 1•- = , i - + j - + k - · f1 i + f2 j + f3 k - a - a ) ( - - - ) = at + at + 2 at ay az _.1_ ___1_ ax ay az Divergence of f is a sc a l a r point function It is a measurement of rate of fl uid, originating at a point per unit volume. Physical I nterpretation Solenoidal Vector If div f is zero i.e. v . f = 0 then f is cal led solenoidal vector. Divergence of a Sum Let V and W be two vector point functions then V . Incompressible Fluid (v + W ) = v . V + V . W . Consider a fl uid motion hav i ng velocity v = v 1 i + v 2 } + vi W div v = 0 . . . (i) then fluid is called i ncompressible and (1 ) is known as equation of continuity for incom pressible fl uid. IES MASTER Publication 374 Gradient, Divergence and Curl CURL (ROTATION) If f is a vector point given by f = f1 i + f2 J + f3'< then, Here curl f is a vector point function i Curl f = V x f = I:( i x :) = : x f1 Physical Interpretation j 1< If a rigid body is in a motion then curl of its linear velocity at any point is twice of its angular velocity i.e. !curl v = 2wl Proof : then Now i wxr = W1 X W2 y k W3 z Curl v = curl ( w X r ) = V x ( w x r) i a = lrrotational Vector a ( w 2 Z - W 3 Y ) (w 3 x - w 1 z) ay = 2 (w 1 i + w 1 J + wJ) = 2w . If curl f = 6 then f is called irrotational vector. Hence Proved. 1. If f is an irrotational vector then we can always write f = V<jl w here potential. 2. If f is a fluid motion then <l> <l> is cal l ed velocity potential . Curl of a Sum of Vectors If V , W be two vector point functions possessing partial derivatives with respect to x, y, z then v x (v + w ) . = v x v + v x w Let, a = a 1 i + a 2 J + a J (constant vector) and Learn as Formulae 1. grad f ( r ) = v ( f (r )) = f' (r)f = f'(r ) I. r n eg. grad (rn ) = n r -1 f = nr n-2 r 2. grad (a · r) = a 4. div ( r ) = 3 and curl (r) = 3. div (a) = 0 and curl (a) = 5 r = xi + yJ + zk {i .e. for constant vector div and curl both are zero} (6) (position vector) is cal led scalar Eng ineering Mathematics div ( r x a) = 0 and curl (r x a) = (-2a) n n 2 div (grad rn ) = v' (r ) = n ( n + 1 ) r -z 5. 6. eg. div ( grad n = V 2 U) 375 =0 VECTOR IDENTITIES Learn the following vector identities as standard results; + v grad u 1. grad (u v) = u grad 5. div (a x 5) = 5 - curla - a - curl b 9. div Cu r l f = v - (v x 3. 7. 2. grad (a - 5) = a x curl b + b x curl a + (a - v ) 5 + (5 - v) a 6. Curl (a x 5) = a div 5 - 5div a + (5 - v) a + (a - v) 5 div (u a) = u diva + a.gradu 4. div grad f = V f 8. 2 11. V t) = 0 Curl Curl f = grad div f - V 2f NOTE 1. 2. 3. Curl (u a) = u curl a + (grad u) x a Curl grad f = v' x v'f = 6 10. grad div f = Curl curlf + V 2f Let f(x, y, z) be any scalar field then ,. af af ,. af grad f = x - + y,. - + z ax ay az ,. af ,. l af ,. af grad f = p - + <j> -- - + z ap p a<j> az ,. af ,. l af ,. 1 af grad f = r - + 0 - - + <J>--ar r ae r s i n e a<j> (in Cartesian Form ) (in cyl i ndrical form ( p, <j>, z )) (in spherical form (r, 0, <j>) ) Laplace equation i n various coordinate system v 2 f = 0 ⇒ a2 f + a2 f + a2 f = 0 (in Cartesian Coordinates) (i ) ax2 ay2 az 2 a2 f + -1 a2 f a2 f 1 at ++ -- = 0 (in Cyl i ndrical Coord inat�s ) (ii) 2 822 P ap ap2 P2 a<1> 1 a2 f 2 8f cot e af a2 f + 1 a 2 f + - - + -2- -+ -(iii) (in Spherical Coordinates) r ae ar2 r2 ae2 r2 sin 2 0 8<j>2 r ar ( I. ) L et A- = . - ax x,. + ay y,. + °zZ,. be any vector f 1e I d t hen v'•A = ,. ,. - 1 a(p · aP ) 1 aa<t> a°z (·1 1· ) i f A = aP p + a,. <J> + °zZ,. t h en v'• A = -- + -- + " p ap p a<j> dz aax a') aaz +-+ay � � 2 1 a(r - C\- ) + 1 1 aa.i, a = -- - (aO si. n 0) + -- -2 r s i n e B<J> r s i n e ae ar r S O LVED EXAM PLES 82 Example 21 : If A = x2 yzi - 2xz3 j + xz2 k B = 2zi + yj - x2k find value of --(A x B ) at (1 , 0, -2) axay Solution : We have IES MASTER Publication 376 Gradient, Divergence and Curl A X B = i x yz -2xz 2 2z y k xz2 3 -x2 82 3 2 -(A x B ) = � { � ( A x B ) } = -z i+2x zj+4xyz k Again axay ... (i) ax ay Putting x = 1 , y = 0 and z = -2 in (1 ), we get the required derivative at the point (1 , 0, -2) = -4i - 8 j . Example 22 : Find the divergence and curl of F=( x2 + yz ) i +( y2 +zx ) ] +( z2 +xy ) 1< at (1 , 2 , 3) Solution : div F = V . F = :x ( x2 +yz)+; ( y2 +zx)+ !( z2 +xy) = 2x+2y+2z = 2(1 + 2 + 3) = 12 at (1, 2 , 3) Curl F = a ax k j -a a az ay x2 +yz y2 +zx z2 +xy = 1 [ :y ( z2 +xy )- ! ( y - zx)]- } [ :j z + xy )-! (x +yz )] + k [! ( y +zx )-; (x +yz )] 2 2 = i [ x-x ] +} [y-y ] +k [z-z ] = O 2 2 2 Example 23 : Prove that (i) div r =3 (ii) curl r =6 r Proof : V -r V -r = xi+y} +zk = - (x)+-(y)+-(z) = 1+1+1=3 a ax = a ay k a az a a a = ax ay az X y z 1( az _ ay ay az ax )- 1 ( axaz _ az ay _ ax = ) ay )+k ( ax Example 24 : If f=x2 yi - 2xz} + 2yzk find (i) div f ; (ii) curl f ; (iii) curl curl f Solution : (i) We have 2 div f = v . f = ( 1 �+] �+k �)- (x yi-2xz}+2yzk ) ax (ii) We have Curl f = V x f = ay az 2 = �(x y ) +� (-2xz)+�(2yz) = 2xy+0+2y=2y(x+1) ax i a ax ay a ay k az x y -2xz 2yz 2 Dz 0 Engineering Mathematics = [; (2yz )- ! (-2xz)] i- [ 377 ! (2yz )-! (x y )] 1 +[ :x (-2xz) - :y (x z )}< 2 2 = (2z +2x) i - 0} + (-2z-x2 ) k = (2x +2z) i - ( x 2 +2z) k (iii) We have curl curl f = v x ( v x T) = V x [(2x +2z) i-( x2 +2z) 2 = [ ; (-x -2z )] i -[! k] = a ax (2x+2z) ! (-x2 -2z )- = Oi-(-2 x-2) J +(O+O ) k=(2x+2 ) J a ay 0 I< a az (-x 2 -2z) (2x +2z)] 1 + [0- ; (2x+2z )] k Example 2 5 : Determine the constant a so that the vector V = ( x + 3y) i + ( y-2 z ) J + ( x +a z) k i s solenoidal. Solution : A vector V is said to be solenoidal if div V= 6 . - - a div V = V-V =- (x+3y)+- ( y - 2z)+- (x +az) We have ax = 1 +1 +a = 2+a Now div V = 0 if 2 + a = 0 i.e. if a = -2 a ay a az Example 26 : Show that the vector V = ( sin y + z) i +( x cos y-z) J + ( x-y) k is irrotational. Solution : A vector V is said to be irrotational if curl V = 6 . We have curl V = VxV = a ax a ay I< a az sin y+ z xcos y- z x-y (x-y)-�(sin y + z)] J +[ � (xcos y-z)-� ( sin y + z)] k = [ � (x-y)- � az ( xcos y - z)] i-[ � ax az ax a a y y = (-1+1) i-(1-1) } +(cos y-cos y ) k =0 V is irrotational. Example 27 : If V is a constant vector, show that (i) div V = 0, (ii) Curl V = 0 Solution : (i) We have (ii) We have div V = Curl V = Example 28 : Prove that : f (r)r irrotational Solution : �1·-+J av � · -+k av . · av = i - 0+j • 0 +k - 0= 0 ay az ax �I X av + �J x -+k av . x av = i x O +j. x O+k x 0=0 az ay ax Curl [ f (r).r] = f ( r) (curl r ) + [grad f (r )] x i' {Since we have curl (u a)=ucurla +(gradu) x a } it is vector identity (4) = 6+ [ f (r)•r ] x r = 6+ f (r \r x i' ) = 6 r IES MASTER Publication 378 G radient, Divergence and Curl Example 29 : Prove that v x Solution : Curl c� ) =6 c�) C! . = curl r ; r ) = r; ( cu rl r ) + [ grad r ] X (using vector identity 4) -2 = _2_(0) + [ 3 - r ] x r r2 r -2 - - {since; grad f(r) = f ' (r) - f} - = 4(r x r) = o r Example 30 : Show that V 2f (r ) = f" (r) + -f' (r) . Solution : 2 r f' r) LHS = v' 2 f (r) = div (grad f (r)) = div (f' (r)f) = div [ � r ] [We know that; div (u,a) = u div a + (grad u ) a ] , it is vectory identity (3) = = f' r) � div r + ( gra /' �r) } r = 3fY) + [{ r f" (r) t (r)(1) r r } } � 3f ' (r) r +( r f " (r) - f ' (r) 3 . r J (r - r) r f " (r) - f ' (r) 3f' (r) 2f " (r) -+ -- = f (r) + -= " ( J r r r {sin ce r · r = r2 } Example 31 : If a is a constant vector, find (i) div (r x a) , (ii) Curl (r x a) Solution : r = xi + y} + zk (i) We have We have a r xa = i x 81 a 1< Y z 82 83 = (a 3 y - a2 z) i + (a 1 z - a 3 x ) } + (a2 x - a 1 y)k a (i) div (r x a) = 8 (a3 y - a2z) + a (a 1 Z - 8 3 X) + 8x (a 2 X - 81 Y ) = 0 + 0 + 0 = 0. y x (ii) curl (r x a) = v x (r x a) = i a ax 1< i a az a ay = -2a1i - 2a 2 } - 2ai :,, -2( a 1i + a 2 } + -----...... ai) = -2a Engi neering Mathematics 1 .* If 2 v e l o c i ty 1 - yz k is v given by then the angular velocity 2 at the point (1 , 1 , -1 ) is ___ [GATE-1 993] The derivative of f ( x, y) at point ( 1, 2) in the direction of vector i + j is 2 ✓2 and in the direction of the vector -2j is -3. Then the derivative of f ( x, y) in di rection -i - 2j is (a) 3. Questions marked with aesterisk (*) are Conceptual Questions l i near \I = x yT + xyz w 2.* the (c) ✓2 + 3/2 2 - 2✓ - 3/2 2 (b) -7/ (d) 1/✓ 5 (c) 5 ✓ 5 ✓ (b) (d) 4 4 - ✓5 5 ✓ 4 [GATE-1994; 1 Marks] 4.** If V is a differentiable vector function and f is a sufficient differentiable scalar function, then curl ( f v ) is equal to (a) (grad f) x ( v) + f (curl \/ ) (b) 0 (c) f curl (d) (v) (grad f) x (v) 5.** The expression curl (grad f), where f is a scalar function, is (a) equal to [GATE-1995-ME; 1 Marks] v 2f (b) equal to div (grad f) 6. (c) a scalar of zero magnitude (d) a vector of zero magnitude T h e d i rect i o n a l d e r i v a t i v e of t h e f u n ct i o n f (x, y,z) = x + y a t the point P(1 , 1 ,0) along the [GATE-1996-ME; 1 Marks] direction T + } is 7. (c) 1 (b) ✓2 2 -✓ 8.* The magnitude of the gradient of the function f = xyz3 at (1 ,0,2) is: [GATE-1 996; 1 Marks] (b) 3 (a) 0 (d) oo For the function $ = ax y - y to represent the velocity 2 3 [GATE-1 998; 1 Marks] potential of an ideal fluid, V 2 $ should be equal to zero. I n that case, the value of 'a' has to be (b) 1 (a) -1 9. ✓2 (d) 2 (c) 8 [GATE-1994; 1 Marks] point P ( 2, 1, 3 ) in the direction of the vector a = T - 2k is 4 (a) 5 ✓ The directional derivative f ( x, y, z) = 2x2 + 3y2 + z2 at (a) 379 (c) -3 (d) 3 If the velocity vector in a two dimensional flow field is given by curl (a) v v = 2xyi + ( 2y wi ll be 2y2 } (c) zero 2 [GATE-1 999; 1 Marks] - x ) }, 2 (b) 6yk the vorticity vector, (d) - 4xk [GATE-1 999; 1 Marks] 1 0 .* The m agn itude of the di rectional derivative of the function f(x , y) = x 2 + 3y2 i n a d i rection normal to the circle x 2 + y2 = 2, at the point (1 , 1 ), is (a) (c) 4 ✓2 2 7✓ (b) 5 (d) 1 1 . The divergence of vector (a) T+1+k (c) 0 ✓2 2 9✓ r = x T + y} + zk is (b) 3 (d) 1 [GATE-2000 (CE)] 12.* The maximum value of the directional derivative of the [GATE-2001 ; 1 Marks] function $ = 2x 2 + 3y 2 + 5z2 at a point (1 , 1 ,-1 ) is (a) 1 0 (c) ✓1 52 (b) - 4 (d) 1 52 [GATE-2002; 2 Marks] IES MASTER Publication 380 1 3. Gradient, Divergence and Curl The di rectional derivative of the following function at (1 ,2) in the direction of (4i + 3j ) is: f ( x, y) = x + y (a) (c) 4/5 2/5 2 (b) 4 (d) 1 2 2 2 1 4. A velocity vector is given as v = 5xyi + 2y } + 3yz k . The divergence of this velocity vector at (1 , 1 , 1 ) is (a) 9 (c) 14 [GATE-2002; 1 Marks] (b) 1 0 (d) 1 5 [GATE-2002-CE; 1 Marks] 1 5. The vector field F = xi - y} (where vector) is i and } are unit (a) divergence free, but not irrotational (b) irrotational, but not divergence free (d) neither divergence free nor irrorational [GATE-2003; 2 Marks] x2 y 2 For the scalar field u = - + - , the magnitude of 3 2 the gradient at the point (1 , 3) is (a) (c) 3 g {13 f9 ./5 (b) # 9 2 [GATE-2005-EE; 2 Marks] (d) 1 7 .* A scala: field is given by f = x 213 +y 213 , where x and y are the Cartesian coordinates. The derivative of 'f' along the l i ne y = x d i rected away from the origin at the point (8, 8) is (a) (c) ✓2 3 2 ✓3 (b) (d) ✓3 2 3 ✓2 1 8. The directional derivative of f ( x, y, z) = 2x + 3 y + z at the point P (2, 1 ,3) in the direction of the vector [GATE-2005 (IN)] 2 a = i - 21< is (a) - 2. 785 (c) - 1 . 789 2 2 (b) - 2 . 1 45 (d) 1 .000 (a) (c) P x V x P - V 2P v 2P + V x P [GATE-2006-CE; 2 Marks] (b) V 2 P + V (V x P) (d) V (V · P) - V 2 P [2006-EC; 1 Marks] 20.* D i v e rg e n c e of t h e v e c t o r f i e l d v(x, y, z) = -(x cos xy + y) + (y cosxy)j + [(sin z2 ) + x2 + y2 Jk (a) 2z cos z2 (b) sin xy + 2z cos z2 (c) x sin xy - cos z (d) none of these 21 . The divergence of the vector field 3xz i + 2xy} - yz2 k at a point (1 , 1 , 1 ) is equal to [GATE-2007 (Pl)] (a) 7 (c) 3 22. T h e (c) divergence free and irrotational 1 6. 1 9.** v x v x p (where P is a vector) is equal to (b) 4 d i v e rg e nce (d) 0 of the ( x - y) i + (y - x) j + (x + y + z) k is (a) 0 (c) 2 v ector field [GATE-2008-ME; 1 Marks) (b) 1 (d) 3 23. The directional derivative of the scalar function [GATE-2008-ME; 1 Marks] f ( x, y, z) = x 2 + 2y 2 + z at the point P = ( 1, 1,2) in the direction of the vector a = 3T - 41 is (a) - 4 (c) - 1 (b) - 2 (d) 1 [GATE-2008-M E; 2 Marks) 24. For a scalar function f ( x,y,z) = x2 + 3y 2 + 2z2 , the gradient at the point P (1, 2, -1) is (a) 21 + 6} + 41< (d) ✓56 (b) (c) 21 + 1 2} - 41< 2i + 1 2} + 41< [GATE-2009; 1 Marks) 25. For a scalar function f ( x,y,z) = x2 + 3y 2 + 2z2 the directional derivative at the point P(1 ,2 -1 ) in the direction of a vector i - } + 2k is (a) -1 8 (c) 2)6 (b) -3J6 (d) 1 8 [GATE-2009-CE; 2 Marks] Eng ineering Mathematics 26.* A sphere of u n i t rad ius is centred at the ori g i n . T h e u n it norm al a t a point (x , y , z ) on t h e surface of the sphere is the vector. (a) (x, y, z) (b) (c) (d) (.J3' F3' -f3J l ' 1 1 1 ✓3 , ✓3 , ✓3 J ()2' F2' .f2J [GATE-2009 (IN)] 27.* Divergence of the three dimensional radial vector field is r (a) 3 (c) (b) 1/r (d) 3 (T + } + T + ] + 1< k) [GATE-201 0-EE; 1 Marks] 28. The divergence of the vector field is (a) 0 A = xax + ya y + za 2 (d) 3 [20 1 0 ; 1 Marks and 201 3 EC] 29. If T (x, y, z) = x 2 + y2 + 2 z2 defines the tem perature at any location (x, y, z) then the magnitude of the tem peratu re g radient at point P( 1 , 1 , 1 ) is (a) (b) 4 2$, (d) (c) 24 30. 2 2 [GATE-201 1 (Pl)] 2 For the spherical surface x + y + z = 1 , the unit outward normal vector at the point ( � , � , 0 J is given by (a) 1 � 2J ✓2 I + ✓ (b) ✓2 (c) 1< (d) 1 1 7 1 � 2J I-✓ 1 � 1 · 1 7 ✓3 I + ✓3 J + ✓3 k [GATE-201 2-ME, Pl; 1 Marks] 31 .**The d i rection of vector A is radially outward from the orig i n , with I A I = Kr" where r2 = x 2 + y 2 + z2 and K is constant. The value of n for whi ch V• A = 0 is (a) 4xy a x + 6yz a y + 8zx a 2 (c) (4xy + 4z2 Ja x + (2x2 + 6yz)ay + (3y2 + 8zx)a 2 (b) 4a x + 6ay + 8a 2 (d) 0 [GATE-201 3; 1 Marks] 33.* For a vector E, which one of the following statement i s NOT TRUE? (a) I f V . E = O, E i s called solenoidal (b) If V x E = O, E is called conservative (c) If V x E = O, E is called irrotational If V • E = O, E is called irrotational [GATE-201 3 (IN)] 34.* Let V • ( f v) = x 2 y + y 2 z + z2 x, where f and v are scalar and vector fields respectively. If v = yT + zJ + xk, then v - Vf is (a) x 2 y + y 2 z + z2 x 35. Curl of vector (a) (b) 2xy + 2yz + 2zx (d) 0 [GATE-2014-EE-Set 3; 1 Marks] F = x2 z2 i - 2xy2 z} + 2y2 z3 k ( 4yz 3 + 2xy2 ) i + 2x 2 z} - 2y 2zk (b) 2 2 2 3 ( 4yz + 2xy ) i - 2x z} - 2y zk (d) 2xz2 i + 4xyz} + 6y 2 z2 k (c) 7 (d) 0 [GATE-201 2-EC, EE, IN] 32.* The curl of the gradient of the scalar field defined by V = 2x2 y + 3y 2 z + 4z2 x i s (c) x + y + z J6 (b) 2 (c) 1 (d) (b) 1/3 (c) (a) -2 381 is 2xz 2 i - 4xyz} + 6y 2z 2 k [GATE-201 4-ME-Set 2; 1 Marks] 36. Divergence of the vector field x 2 zi + xy} - yz2 k at ( 1 , -1 , 1 ) is (a) 0 (c) 5 (b) 3 (d) 6 [GATE-201 4-ME-Set 3; 1 Marks] 37. The magnitude of the g rad ient for the function f ( x , y, z) = x2 + 3y 2 + z3 at the point ( 1 , 1 , 1 ) i s [201 4-EC-sec 4 ; 1 Marks] IES MASTER Publication 382 G radient, Divergence and Curl 38. The directional derivative of f(x, y) = F2" ( x + y) at ( 1 , 1 ) i n the direction of the unit vector at an angle of with y-axis, is given by. . . . . . re 4 [2014-EC-sec 4 ; 1 Marks] 39.* l f r = xa x + ya y + Zaz and l r l = r, (r2 V (in r ) = ____ then + ( c1 x + c2 y + c 3z ) k and a 1 = 2 & c 3 = -4 . The value of b2 i s _ 45. T h e 40.* D i rectional derivative of � = 2xz - y 2 at the point (c) 2 i - 6j + 2k d i recti o n a l 2 (a) p is solenoidal , but not irrotational (c) p is neither solenoidal nor irrotational (b) (d) 4i - 6j - 2k (d) [GATE-201 4-PI-Set 1] If p is both solenoidal and i rrotational the v ector curl ( �V) = v (�Di�V ) (c) Div (curl V ) = O Div V = 0 �v) = �Div V [GATE-201 5-ME-Set 3; 1 Marks] 44.* The velocity field of an i ncomp ressible flow i s given by (b) 0.0, 3.0, 2 . 1 (d) 4.0, 3.0, 2.0 [GATE-2017 EC Session-II] 48.* The figures show diagramatic representations of vector fields, x , y and z , respectively. Which one of the following choices is true? [GATE-201 5-ME-Set 2; 1 Marks] (a) Div ( F = a: (3y - k 1 z) (d) 3i - 6k 43.* Let � be an arbitrary smooth real val ued scalar function and v be an arbitrary smooth vector valued function in a three-dimensional space. W hich one of the following is an identity? (d) f u n ct i o n (b) 3i (a) - 3i (b) [201 5-EC-Set 1 ; 2 Marks] +ay (k2 x - 2z) -a: ( k 3 y + z) is irrotational , then the values of the constants k 1 , k2 and k3 , respectively, are (c) 0.3, 0.33, 0.5 (c) 3i - 4j field p is irrotational , but not solenoidal (a) 0.3, -2 .5, 0.5 Curl of vector v ( x, y,z) = 2x 2 i + 3z2 } + y 31< at x = y = z = 1 is the 46.* A vector p is g iven by P = x3 yax - x 2 y2 a y - x2 yza2 • Which one of the following statements is TRUE? 47. 42. of [GATE-201 5-SET-I ; 1 Marks] (b) 4i - 6j + 2k [GATE-2014-PI-Set 1 ] derivative (i + } - 2k) at point (2, -1 , 4) is _. ( 1 , 3 , 2 ) becomes maximum i n the d i rection of (a) 4i + 2j - 3k [GATE-201 5-ME-Set 1 ; 2 Marks] u (x, y, z) = x - 3yz i n the direction of the vector div [GATE-201 4 (EC-Set-2)] ____ V = {aix+82y+3:iz) i + ( b1 x + b2 y + b 3z ) } (a) (b) (c) (d) * 0, V x Z = 0 v.x * 0, V x Y = 0, V x Z * 0 V.X * 0, v x Y * 0, V x Z * 0 V.X = 0, V x Y v.x = 0, V x Y = 0, V x Z = 0 [GATE-201 7-EE Session-II] 49. The divergence of the vector field V = x 2 i + 2y3 j + z4 k at x = 1 , y = 2, z = 3 is __ 50.* C o n s i d e r the [GATE-2017-CE Session-II] two-d i m e n s i o n a l v e l o c i ty field given by V = (5 + a1 x + b1 y ) i + ( 4 + a 2 x + b 2 y ) }. where a 1 , b 1 a2 and b2 are constants. Which one of the following conditins needs to be satisfied for the flow to be i ncom pressible? (a) a 1 + b 1 = 0 (c) a2 + b2 = 0 (b) a 1 + b2 = 0 56. (c) 2✓ (b) 1 5 52. For the vector (d) 0 [GATE-2017-ME Session-I] V = 2yz i + 3xz} + 4xy k, .'y . ( v x v ) is _. 57. 55. (c) 3x2 i + 2yj (d) 2yi - 3x2j The value of v . A; where A = 3xya x + xa y + xyza 2 at a point (2, -2, 2) is (a) -1 0 (c) 2 (b) -6 (d) 4 [GATE-2017 (CH)] field [GATE 2018 (ME-AfternoongSession)] For a position vector r = xi+ y} + zk the norm of the 2 2 2 vector can be defined as I r I = x + y + z . Given ✓ (a) r [GATE-2017-ME Session-II] (b) 6x2y v ector a function � = In I r I, its gradient 53. The divergence of the vector -yi + xj is__ . (a) 3x2 i - 2yj U = e x (cos yi + sin y } ) i s the (d) 2ex sin y [GATE-201 7-ME Session-I] gradient of the function i.e., VF is given by of (c) 2ex cos y the value of 54. Let i and j be the unit vectors in the x and y directions, respectively. For the function F(x, y) = x 3 + y2 , the d i v e rgence 383 (b) ex cos y + ex sin y [GATE-2017-ME Session-I] 51 .** F o r a ste a d y fl ow, the v e l ocity f i e l d i s (a) 2 The (a) 0 (d) a2 + b2 = 0 V = (-x 2 + 3y ) i + ( 2xy)]. The m agnitude o f the acceleration of a particle at (1 , -1 ) is Engineering Mathematics 58. r (c) r.r (b) (d) -r IrI v� is - Ir1 3 [GATE 201 8 (ME-AfternoongSession)] The value of the d i rectional derivative of the � function(x, y, z) = xy2 + yz2 + zx2 at the point (2, -1 , 1 ) in the direction of the vector p = i + 2j + 2k is (a) (c) 1 0.93 (b) 0.95 (d) 0.9 [GATE 2018 (EE)] [ESE 2018(EC)] IES MASTER Publication 384 Gradient, Divergence and Curl 1. 2. 3. 4. (See Sol.) 1 6. (c) (b) 1 8. (c) 20. (a) 22. (d) 24. (b) 26. (a) 28. (d) 30. (a) (b) (a) 5. 6. (d) 7. (c) 8. (d) 1 0. (a) 9. (b) (d) 11. (b) 1 3. (b) 12. (c) 1 4. (d) 1 5. (c) 17. (a) 1 9. (d) 21 . (c) 23. (b) 25. (b) 27. (a) 29. (a) 31 . (a) 32. (d) 34. (a) 36. (c) 33. (d) 35. (a) 37. (7) curl \/ = i j (3) 40. (b) a x a ay xyz 2 -yz2 = [T(-z2 - xy) - J(O - O) + k(yz - 0) ] (Curl \/) At (1 , 1 , - 1 ) = [-2T - k] = -(2 i + k ) So Sol-2: (b) Let w = } curl (v) = - i (2T + k) a 5 c = -i - 2} vt - a = 2 ✓ 2 v'f . t, = - 3 51 . (c) (b) 52. (0) 54. (0) (c) (a) 57. (c) (c) 56. (c) 43. (c) 58. (a) 45. (-5.71 5) (a) (2) (1) ⇒ (2) ⇒ w (3) (4) ⇒ _. 2 ( + ) v'f · T 1 = 2 ✓ ✓2 v'f · ( T + }) v'f · ⇒ ⇒ 4 . . . (3) v'f · J = 3 . . . (4) ( -2 1 ) 2 X = = -3 vt = x i + yj + y = 4 y = 3 X = 1 v'f = i + 3} Directional derivative in the direction of = i +j = -2 } (134) 53. (3) ⇒ oz 49. 41 . 44. a (c) (b) 55. 42. k 48. 50. SOLUTIONS Sol-1 : (a) 47. 38. 39. 46. . . .(1 ) . . . (2) Sol-3: c = v'f · c -1 - 6 _ -7 � � ( -T - 2} ) 5 = · ( t + 3J ) - �� = -5- - - (b) ✓5 ✓ f(x, y, z) = 2x2 + 3y2 + z2 vf = 4x i + 6y } + 2z k ✓ Sol-5: (d) Vfl( 2 ,1, 3 ) = 8i+6}+6k and a Engineering Mathematics = i -21< grad f = -1+-J+- k 8 -12 5 Sol-4: (a) Let ✓ 4 ✓ ✓ 5 = ( k = [ az wat ay +f aw .ay -v at az -f k [�(vf)-�(uf) ax ay ] i az ] av at au at aw . + [u-+f--w--f-]J az = t[( aw ay az ax) ax az ay lcurl(fV) = f(curlV) +(gradf)x Alternative Solution: vi axay ayax ax (3) div. (curl f) = v'.(v' x f) = 0 Sol-6: (b) f = X +y v'f = i + j (v0( 1,1.o) = i +j a = i +j Directional derivative = vt. a = ax ay curl ( fv) = f(curl V)+(�T+� }+ ax azax axaz lcurl(gradf) = oi +o} +okl au_ aw']+ av_ au av _ ) T+( ( ) k] az a 2 f _ _t_!_ a2 f _ a2 t )k ) j+( (2) div. (grad f) = v'.v'f = V 2 f ax az ayaz azay )i+( (1) Curl(grad f) = v' xVf = 0 vf wf [�(wf)-�(vf)]i+[�(uf)-�(wf) ]j+ ay a 2 f _ a2 f ax ay az Note : Important Repeated Operations by Operator ( v' ) a a a ax ay az uf k j at at at V = ui+v}+wk curl (fv) = = -- az a a a curl (grad f) = ax ay az -2 ) = (sT+6}+61<)·0 5 1< = ax ax vt .a at at. at. Directional Derivative : = 385 at az k) x(ui +v}+wk) Curl(fv) = V x(fV) = (vf) xv+ f(V x V) lt(curl V) = (gradf) XV+f (curlv)I It is a vector identity just learn it without paying attention on its proof. Sol-7: (c) +1) (�l+J�) . (i✓2 = 2-=✓2 ✓2 f = xyz3 at. at . at vf = -1+-J+- k ax at (1, 0, 2) Sol-8: (d) = yz3i + ay xz3 az j + 3xyz2k v' f = Oi + 1 X 23 j + Ok vf = 8j <I> = a x2y - y3 a<1> ay = ax 2 -3y 2 = - 6 y IES MASTER Publication 386 Gradient, Divergence and Curl a4> ay = 2 axy and 4> 4> v24> = -+2 = 0 2 a ax 2 2 ay - 6y = 0 Sol-9: (d) = 2 ay a ay 2 ⇒ a 2 4> ax2 (given) Curl V = v' x V i a ax The max imum directional derivatives of a function ' 4> ' at a given point 'p' is obtained in the same direction as that of the direction of grad 4> at p. a oz 2xy 2 y2 - x2 0 So, max imum directional derivative = v'(j> . az + [ -(2 xy)--(O)] j a az · a ax + [�(2 y2 -x2 )- �2 xy ]k ax lcurl V = -4 xkl al Sol-13: (b) v'f = 2xi+ 6yj a = 4i+3} f(x, y) = x2 + y2 v'f = 2xi+2 y} (Vf)(1, 2) = 2i+4} Directional derivative 8 + 12 = 4i 3} 4 = (2 i+4 })( ; ) = 5 Sol-14: (d) v = 5xyi+2y2 j + 3yz2 k (Vf)(1,1 l = 2 i+ 6j div v = v'.v Let 4> = x2 + y2 - 2 than grad 4> = 2 xi + 2 y] 11 = grad(j> 2 i+2 j i+j 2 = 2 = lgrad4> I 2✓ 2✓ The required directional derivative = (v'f)( 1.1l · 11 Sol-11: (b) ✓ = Vf -8 f(x, y)=x2 + 3y2 than grad f= 2 xi+ 6y] So, v'(j> V I 4>I = IV4>I = 4 2 +52 + 1 02 ay = of +0]+ (-2 x -2x)k ⇒ 4> = 2x2 + 3y2 + 5z2 v4> = 4 1+ 6 ]-101< k j a ay ay i.e., = 1 + 1 + 1=3 Sol-12: (c) at point (1, 1, -1) 2 2 = [� ( 0)- �(2 y -x )} Sol-10: (a) az ay v'(j> = 4 xi+6y}+10zk V = 2xyT +(2 y2 -x2 )1 llvfl ax gradient 4> la = v . r = (�i+� ]+�kJ•(xi+y]+zk) r = xT + y] +zk 2 = � (5xy)+�(2 y )+� (3yz2 ) ax ( div Sol-15: (c) v \ 1 1 1l Vector field, . . ay = 5y + 4y + 6yz az = 15 F = xi- y j (i) Divergence of the field v.F = -(x)+-(-y) =1 - 1 =o a ax a ay Engineering Mathematics 387 F is divergence free. VxF = (ii) i j k X -y 0 a a ax ay az F is an irrotational field. ✓ auA Let auA ay az 2 = x a x + y ay +O A 3 V Sol-17: (a) ,/1 2 f= x +22 = ✓ y =x P(8, 8) Let, 1 2 - - 2 �­ Vf = _ x 3 i +-y 3j 3 3 1. 1 . 'vf = -I+-J At (8, 8) 3 3 Directional derivative . . Sol-18: (c) f(x, y, z) = 2x2 + 3y2 + z2 v f = 1-+ J-+k- 7at �at ·at ax ay az = i(4 x)+ }(6y)+k(2z) = 8i+ 6}+6k; at P(2,1, 3) Directional derivative in the direction of a = i - 21< is a ay az = 2z cos z2 A = 3xzi +2xy}- yz2 k 2 V.A = _f_(3xz)+ _f_(2xy)+ _f_(- yz ) ay ax rr/4 ⇒ a + _f_[sinz2 +x2 +y2 ] Sol-21: (c) 0 y av1 + av2 + av3 ax ay az = -[ -x cosxy +y]+-[y cosxy] 5 Unit vector along (y = x) is 2/3 + 2l3 4 = - ✓5 ax OP = 8i +8} i+J i OP = ..f ✓ 1 +4 = V1 i +V2 } +Vi then div v = at (1,3) the value of v u is :. Magnitude l vu l = 5 (s.i +6.j +6k. ).(i.J- 2 1<) = V(V · P) - \7 2 P A Vu(l 1,3) = ax + 2ay 8- 12 = - 1.789 Sol-20: (a) Vu = -a x +-ay + -a z ax auA 1a1 = V X (V X P) = (V . P)V -(V.V)P (Using vector identity) lmax. dire ctinonal de rivative = -1521 Sol-16: (c) = Sol-19: (d) = i(O)- } (O)+k(O)=O Gradient of scal ar fiel d u is a = v f· = 3z + 2x - 2yz az Divergence at (1, 1, 1) V.A = 3 Sol-22: (d) Let, div. A = (x - y)i + (y - x) } + (x + y + z) k A Sol-23: (b) = -(x - y) +-(y - x) +- (x + y + z) a ax a ay . a az = 1+1+1 = 3 Given the scal ar function f(x, y, z) = x2 + 2y2 + z Grad f = 1-+ J-+k- = I- (x2 + 2xy2 + z) 7at �at •at ax ay az 7a ax + j -(x2 + 2 y2 + z) A a ay + k - (x2 + 2 y2 + z) A a ay IES MASTER Publication 388 Gradient, Divergence and Curl Vf = i(2x)+j(4y)+k(1) Vf = 2i+4J+k xi+ y} + zk At the point P (1 , 1, 2) The required directional derivative = (Vf) • (3i- 4D = (21+4}+k) · -h2 +42 = Sol-24: (b) a 6 16 = -2 � grad f = = v.r iJz ay Sol-28: (d) grad f at (1 , 2 , -1) A 2 i +12 }-41< grad f �of �of .of 1-+ J-+k- ax Sol-29: (a) iJz ay Vf = i(2 x)+J(6y)+k(4z) = 2i+12 }-4k;at P (1,2, -1) The required directional derivative = (Vf)· = ( vf ). = Sol-26: (a) = So (i- } +2k) Jf + (-1) +(-2) 2 2 (2i +12 }-41<).(i -} +21<) ✓6 2 -·12 -8 _ � - - 3v6 ✓6 2+ Let S : x y 2 z - 1 = 0 So, grads = lgradsl = -(x )+-(y)+-(z) = xax + ya y + zaz = ("I -+J-+ a ca l< a ax a a So, 2xi+2 y} + 2zk J4x2 +4y2 +4z2 a az ry = 1+1+1 = 3 ax ay = 1+1+1 = - • xax + Yay +za2 a iJz )(· .. ) 3 gradT = VT = 2xi+2 y}+4zk (gradT)p = 2i + 2 }+41< l gradTI P = ✓4 +4+16 = ffe = 2 Sol-30: (a) ✓6 Unit outward normal vector at the point P ( �. � , O) of the spherical surface = 2 grad S = 2 xi+2y}+2zk r = div (A) = v-A Given, f(x, y, z) = x2 + 3y2 + 2z2 = 1 1 a 2 • r) = 2.3r 2 =3 = --(r 2 r ar r r = xi+ y} + zk = i(2x)+ }(6y)+k(4z) Sol-25: (b) Let the radial vector is r . Divergence in spherical co-ordinate system vf �of �of .of = 1-+J -+k- = Sol-27: (a) Althernative: x2 + 3y2 + 2z2 ax xi+ y} + zk = (x, y, z) 'v.r Given scal ar function, f(x,y,z) = V� = 'v� IV�I IPIP (_Q_i+_Q_ } +_Q_ k) (x2 + y2 + z2 - 1) ax ay iJz v� = 2 xi +2y} +2zk at point P ( �. �. o ) Engineering Mathematics :. Unit normal vector, 2 = 2 i2 j ✓ i + ✓ } - -+2 2 +2 ✓ ✓ div f = ar r at (1, -1, 1):V-F = 2 x 1 x 1 + 1 - (2x-1x1) and Sol-37: (7) d(x,y,z) = x2 + 3y2 +z3 n = -2 V = 2x2 y + 3y2z + 4z2x grad V = Vv (using vectory identity 8) curl ( grad V)=V xVV = 5 •: by definition of sol enoidal vector Given, and V-E = 0 v,(f v) = x2 y + y2z + z2 x v = yi+ z}+ xk Now, using vector identity, ⇒ v, ( fv) = f(v,v )+ vf,v V,Vf Sol-35: (a) curl F = i a ax x z 2 2 a ay k a az -2 xy z 2 y z 2 2 3 (v0(1,1,1) = 2 i+ 6 }+3k IVfl = Sol-38: (3) Vf = v,v = o = O+V•Vf (as dot product is commutative) = V, (f v) = x2 y + y2 z + z2 x Vf = i2x + }6y + k3z2 f(x, y) = Sol-33: (d) Sol-34: (a) (2 y2 z3 )]} = 2 xz + x - 2yz K (n+ 2 ) n+1 .r = 0 r2 K(n+ 2)rn-1 = 0 ⇒ ! ax ay az (K. rn+2 ) = 0 __!_� r2 ar Sol-32: (d) (x2 z2 )- aF1 aF2 aF3 V-F = -+-+- (r2 I A 1 ) = 0 __!_� r2 ar ⇒ az Sol-36: (c) V·A = 0 ⇒ [! a (-2 xy2 Z )] 7I 3) -- I curl F = ( 4yz3 + 2 xy2 )i +(2x2 z ) } -(2y2 z)kl - 1 a2V · f = 2(-r I f I) I A I =krn (given), ⇒ + Z 2 2 2 + [� (-2 xy z)-� ( x z ) ]k ay ax If f--* f(r), then using standard result ⇒ y ✓ Sol-31: (a) so a( 2 [a 2y 389 ✓4+36+9 = 7 F2 (x+ y ) � x2 + 2 xy 2 xy + y2 + I 2 ] J [ ✓22 ] L ✓2 7 I 3 7 3 � 2 2 Vflat(1,1) = -1+-J ✓ ✓ a = (sin¾}+ ( cos¾ 1 7 1 � 2 1+-2 J =✓ ✓ )1 ..- Directional derivative=Vf. a Sol-39: (3) We know that IES MASTER Publication 390 Gradient, Diverg ence and Curl grad f(r) = f'(r) f = v'f(r) = f'(r)i r S o, v' logr = 1 i' rr axay ax . az ay . ax ( ·: ay = 2z i -2y}+ 2xk vector. Sol--41: (c) v'<j> = (6x2 y2 z4 )i + (4 x3 y2 z4 )} + (3x3 y2 z3 )k a2 <1> a2 <1> a2 <1> +-+ax2 ay 2 az2 v' X V = a ax 2x2 j a ay 3z2 [·: v'2 <1> = v'. v'<j>) k v,v 2+ b2 + (--4) ay V3 az ax az ax ay av3 _ av1 a v2 _ a v1 av av )+� = �l 3 _ 2 ) � ( az az ( ay ) ax ay az ay b2 Sol--45: (-5.715) u (x, y, z) ax ax ...11.. = o = = = 0 2 x2 - 3yz Vu = 2xi-3zj -3yk v'ulat (2 ,- 1 , 4) = 4i-1 2}+3k Directional derivative = Vu-8 1 + }-21< = (4 1-12}+3 1<)- = ⇒ 8V2 _ 8V1 aV3 _ 8V2 8V3 _ 8V1 ) ) +k( )- J( = 1( Div ( Curl V ) k V2 ayaz az . ax az . ay a1 + b2 + c3 = 0 Sol--46: (a) a a a ax ay az V1 a2 v1 The flow is incompressibl e, if y3 Sol--43: (c) Curl \/ = _ 'V x V is perpendicul ar to the pl ane of v and \/) a =(3y2 -6z) i +O} +Ok az At x = y = z =1 , v' x V = - 3 i i a2 v2 +(c1 x + c2 y + c3 z)k az direction of grad <I> so Normal vector is the required i + V = (a1 X + azy + a3 z)i + (b1 X + b2 Y+ b3 z)} ( grad<j>)( 1 , 3 , 2 > = (4 }-6}+21<) Sol--42: (a) a2 v1 Sol-44: (2) and we know that directional derivative is max in the & + vector identities) a a a grad <I> = V<I> = <1> 1+ <1> ] + <1> k = a2 v3 _ Div(Curl V)= v-(vx V )= O ( obvious result using ax S o, a2 v2 Alternative Solution div(r2 v' logr) = div(i')= 3 Sol--40: (b) _ = 0 r v' l ogr = i' 2 a2 v3 4 -12-6 ✓6 7✓6 =- 5.715 3 p is sol enoidal vx P = i a = -14 ✓6 ✓6 a ay k a az x3 y -x2 y2 -x2 yz = i -x z+O -j(-2xyz-O)+k -2xy -x .( 2 ) . .( = -x2 z i +2xyz}-(2xy 2 + x3 )1< 2 3) Engineering Mathematics vxP * o Sol-47: (b) Sol-51: (c) j I< ay oz u= -x 2 + 3y and v= 2xy For steady flow (3y-k 1 z) (k2 x - 2z) -(k 3 y+z) i[ -k 3 +2] - }[ 0+k 1 ] +k [k 2 -3] = 0 so, So, k= 2, k 1 = 0, k2 = 3 3 & ⇒ Sol-48: (c) From the given figures we can observe that: Fig (1): X is diverging field hence its divergence of X i.e. v. x * o. Fig (2) : Y is circularly rotating field hence its curl of Y i.e. (v'xY) 0. * Fig.(3): Z is also a circularly rotating field hence its curl of Z i.e., (V x Z) O . * Hence, option (c) satisfies the above three conditions. Sol-49: (134) Divergence ( V ) ay = 2x + 6y2 + 4z3 X 33 V= (5+a 1 x+b1 y)i+( 4 +a 2 x+b2 y)} = ui+ v} For, incompressible flow div V= 0 ⇒ au au a x= u-+v- ax ay iPI av aY = u-+vax i3y a x= (-x 2 +3y)(-2x) +( 2xy)(3) a x= 2x 3 - 6xy + 6xy Similarly, ay = (-x2 + 3y)(2y) + 2xy(2x)= 2x2y + 6y2 a/1, -1)= 4 Hence, Note for Q.51 : If Velocity anet= � a net = ✓4+16 anet = .J20=2✓ m/s 5 v = ui+ v}+wk then acceleration is given as a= ax [Dive rgence (V)l (x = 1, Y = 2, z = 3) = 2 X 1 + 6 X 22 + 4 =134 Sol-50: (b) au av =O = at at ax(1, -1)= +2 2 3 2 = �(x )+�(2y )+�(z ) ax a1 + b2 = 0 V = (-x 2 +3y)i+(2xy)}=ui+v} ⇒ a a a ax ax ay Given flow filed For irrotational vector F, curl xF =0 or au -+- = 0 or P is not irrotational. 391 or or if then d\/ dt av dx av dy av dz av = --+-- +--+ax dt ay dt oz dt at av av av av a = u-+v -+w-+ax ay oz at au au au au a = u-+v-+w-+oz at x ax ay iPI iPI iPI av a Y = u-+v-+w-+OX i3y OZ at 8z = u-+v-+w-+oz at ax a aw aw y aw aw IES MASTER Publication 392 Gradient, Divergence and Curl At (2, -2, 2) & net accel eration is Sol-52: (0) Given, 2 2 2 a = J aX +ay + aZ Sol-56: (c) a div u = (:) +(:) +(:) k a a ax f}y az 2 yz 3xz 4xy ax + �(-2 y)+ �(z) 'v,(VxV) = ax &y az = 1 - 2 +1 'v.(vx v) = 0 Alternatively : Divergence of a curl is alway s z ero i.e. div( curlV) = 0 ( using vector identity) so, Sol-54: (c) Sol-55: (a) = ex cosy + eY cosy+0 = 2 ex cosy Sol-57: (c) = xi -2 y} + zk Let f =xi+y} + zk and r = We know that grad f( r) = f'( r) So, = -(-y) +-(x) +-(0) ax az &y = O +O +O = O a a a Sol-58 : (a) r -aF -aF VF = i- +j - = (3x2 i+(2 y) j. ax f}y A = 3xya x + xa y + xyza z aAx aAy aAz v-A = - +- +aAy f}y az v-A = �(3xy) + ax v-A = 3y + xy a(x) a( xyz) + az &y i.e. lrl = ✓x 2 + y2 +z2 (f) grad <p = grad ( log r) f = -y i+xj div f = 'v. f = ( ex cosy) i+( ex siny) } u V = 2yzi+3xz} +4 xyk 'vx\/ Sol-53: (0) 'v·A = -(6+ 4) = - 10 Q = xy2 + yz2 + zx2 then Grad � = (y2 + 2xz)i + (2xy+ z 2 )I+ (2yz + x2 )k so (grad�)p = [C-1)2 -2 (2)(1)] 1<+[2 (2)(-1)+(1)2 ] } +[2 ]1< = 5i-3}+2 1< So required DD = 5 - 6+4 =1 3 3.3 LINE INTEGRAL Any integ ral which is to be eveal uated al ong a curve is cal l ed a l ine integ ral. Let C be a reg ul ar curve from A to B . Suppose r is position vector of a point P (x,y,z). i. e r(t) = xi+ y] + zk where x,y, z are scal ar function of t. Let F (x, y,z) = F1 i + F2} + Fi be a vect or point function, which i s differentiable and cont inuous al ong t he curve C, where F1 , F2 , F3 a re scalars wh ich are function of x,y, z . H ence l ine integral of F with respect to arc leng th along curve C, from A t o B i s defined as LI = f: F -dr = f: ( F1 dx+F2 dy+F3 dz) Work Done by a Fore� � A I f F be a fo rce field, then total work done by F as part icle moves from A to B along curve C is W= S imple Closed C111 rve B F - dr fA - A curve C is call ed a simpl e closed curve if it does not intersect its elf any where. Ci rcu lation The line integ ral of a vecto r point functi on F along a simple cl osed curve is called cir cul ation of F al ong C. It is sy mb olically denoted as f/ - dr O r t/ • dr = fJ F1 dx + Fzdy + F3 dz ) l rrotatioal If ci rculat io n of F along every closed curve in a reg ion vanishes then F is said to be irrotat ional in a reg ion. Thus f/ · dr = 0 ⇒ F is irrot ation al . Conservative Field The field F is conservative, if the work done in moving t he particle from one point to anot her is i ndependent of the pat h j oining the point s. The nece ssary and suf ficient condition that a fiel d be conservative is t hat there ex ists a sing l e valued function � such that F = v'� In other words, we can say that a fiel d F is conservative if curl F = 5 . 394 Vector Integration Example 1 = Solution 2 Ev al uate J F - dr where F = xz2 i+ y} - x2 k al ong the curv e x = sin8 cos8 , y = sin 8 , z = cos8 with 8 increasing from O to rc/2 . 2 2 2 2 F = xz2 i+y}-x2 k = sin8 cos 8i+ sin 8} - sin 8 cos 8 k r = x i + y }+ z k = si n8 cos8i+ sin2 e} + cos8 k and ⇒ dr = (cos2 8 - sin2 e ) i +2 sin8 sin8} - sin8k d8 - dr F= sin8 cos2 8 (cos2 8- sin2 8 ) +2 sin3 8 cos8 + sin3 8 cos2 8 = sin8 cos4 8 +2 sin3 8 cos8 d8 2 4 3 J l= -dr = Jt F ·�: d8 = J;t (s in8 cos 8 +2 sin 8 cos 8 ) de = Example 2 : Solution = Example 3 : Solution -( c os5 8 / cos4 8 / J J +2 ( 5 4 0 n 2 n2 0 = _!_ +_!_ = 2_ 5 2 10 · If F = 3xyi -y2 } ev aluate fc F · dr where C is giv en as, y = 2x2 from (0,0) to (1,2). The parametric equation of the parabol a y = 2 x2 can be taken as x = t, y = 2 t2 . At the point (0,0), x = 0 and so t = 0. Again at the point (1, 2), x = 1 and so t = 1 N ow Ev al uate f/ · dr - fc F•dr = fc ( 3xy i - y } ) · ( dxi+ dy}) (- : xi+ y} , so thatdr = dxi+ dy} ) 2 - y i - xj - & C is: where F = -x2 + y2 2 2 (i) Circul ar path x + y = 1 described counter cl ockwise. (ii) The squar e formed by the l ines x = ±1, y = ± 1 , counter clockwise. (i) We take parametric equation of giv en circl e as Then x = cos8, y = sine dx = - sin8 d8, dy = cos8 d8 Jfc, F -dr = = fc F1 dx+Jc F dy 2 ydx xdy - J -= Jc -2 2 2 c x +y x + y2 J:'' sin8 (- sin8 )d8 -J:'' cos8 cos8 d8 (On substi tuting v al ues of x, y, dx and dy) = J: - d8 = -2rc n (ii) . Jc F dr = JABCDA F · dr = J c = y dx - x dy x2 + y2 D � y=1 C X = -1 fAB F .dr+ JfB cF.dr+ fCD F. dr +JfoA F.dr X=1 A y = -1 B Eq uation of line AB is y = -1 ⇒ dy = 0, hence fAB F . dr = f \ dx = f + +1 -1 -1 Eng ineering Mathematics 395 +1 1 -1 dy y = (-tan- 1 x) = -rr/2 dy = + _ 2 2 2 1 X +y - 1+ y -1 Similarl y, since eq uation of BC is x = 1 ⇒ dx = 0 therefore f +1 +1 +1 � +1 -1dy ( -1 ) f J s cF - dr = f 1 F2 dy = f- 1 x2 + y2 dy = f 1 1 + y2 = -tan y _1 = -rr/2 . - Similarly, Eq uation of CA is y = 1 fCD F · dr ⇒ 1 . 1 dx _ = f F1 dx = f - -= ( tan 1 x)- = -rr/2 1 1 1 1+ 2 1 X Similarly, Eq uat ion of DA is x = -1 lDA Example 4 : Solution dy = 0, therefore ⇒ dx = 0, therefore _ - 1 - 1 -xdy - 1 dy - = rr/2 =f F - dr - f F2 dy = f -1 y2 +1 1 1 X2 + y2 Adding all the four line integrals, we get fc F•dr = -2rr F ind the work done in moving a particle once round the circle x2 + y2 = a2 , z = 0 under the for ce field given by F = (2 x - y + z)i + (x + y + z2 ) } +(3x -2y +4z )k . \ Since z = 0 ⇒ dz = 0, therefore P utting z = 0 in F 1 , F2 Using parametric eq uation of cir cl e x2 + y2 = a2 as x = a cos t, y = a sin t, so that dx = - a sin t dt, dy = a cos t dt and integrating within limits 0 to 2rr , we have fc F.dr = J: n (2 a cos t - a sin t)(-a sin t)dt +f:" (a cos t + a sin t )a cos t dt si = a' 1:· (3 sin t cos t+1)dt = •' H ;' SURFACE I NTERGRAL r n = a2 f: (-2 cos t sin t + sin2 t + cos2 t + sin t cos t) dt = a2 [ 0 +2rr ] = 2rr a2 +(t): " l Let dS be the surf ace area of an el ement of surface S surrounding the point P. I f n be the unit out ward normal vector to dS at P. Then, the normal surface integral of the vector point function F over S is given by sI = H/ •n ds =H/ - ds Flux The n ormal surface integral fis F · dS of a continuous vector point function F over a closed surf ace S is call ed fl ux of F over S. Scalar Ca rtesian Expression L et F =F1 i + F2 } + Fi then fJ/ · n ds = fis (F1 dydz + F2 dzdx + F3 dxdy ) IES MASTER Publication 396 Vector Integration VOLUME INTEGRAL Su ppos e V is a vol u me bou nded by surface S. Suppos e f(x , y, z) is a s ingl e valued function defined over V . Then t he volume integral of f (x,y,z) over V is defined as : Jff} (x, y, z )dv or fv f( x, y, z)dV or Hf} (x, y,z)dx dy dz If F (x, y, z ) is a vector function, then Hf/ dV is al so an ex ampl e of volume integral . 1. 2. 3. Example 5 : Solution ii is the unit outward drawn normal to the surface S. i . e . ,., ,., ,., grads n = cos a i +cos pj +cosyk = grads 1 - 1- ds = dxdy cosr ds = ds n = dydz = cos a. dxdy = dydz dzdx dzdx = = cos p ,, .f n -k n·S [·.- a = la! a] = ds (cos a f + cos PJ +cos yk ) = ds [ dydz ds dxd z s dxdy k] = dydz i +dxdzJ +dxdy k + ds ds r+ F ind the circul ation of F round the curve C where F = y i +z} + xk and C is th e circle x2 + y2 = 1, z = 0 B y definit ion, the circu lation of F al ong t he curve C is = j/ -dr, where f = y i+z} +xk = jJ yi+z} + xk) · ( dxi+ dy} + dzk) = fJ ydx+zdy +xdz ) [ · : on C, x = cos 0, y = sin0) = jc ydx =- f 2n 0 2n 1 1 s in20 n . si n2 0 d0 =-Jf (1- cos 20) d0 = - [ 0 ---] = -n o 2 2 0 2 2 Example 6 p E valuate fs F · n dS , where F =yzi + zx} + xyk and S is that part of the su rface of the sp here Solution A vector normal to the s urface S is given by f x2 + y2 + z2 = 1 which lies in the first octant. grad S = v (x2 +y2 +z2 ) = 2 xi+2 y}+2zk Therefore ii =unit normal to any point (x,y,z) of S = 2 x i+2y} +2zk � � • y(I 4x +4 y +4z ) = xI+YJ +zk 2 2 2 d We have Hs F · n dS = H F ·n ; :r , where R is the poj ection of s on the xy- pl ane. The region R is R , l bounded by x- ax is, y- ax is and the circl e x2 + y2 = 1, z = 0 . F .n = (yzi + zx} + xyk) . (xf + y} +zk) = 3xyz We have Als o H ence n.k = (xi+y}+zk ) ·k= Z H/ • n dS =H R 2 0 :. \n ·k l =z 3 2 '; dx dy =3HR xy dxdy = 3f��� ( (r cos0 )(r sin0 )r d0 dr J n/ [-r4 =3 ⇒ 4 ] 1 0 3( 1) 3 cos 0 s in0 d0 = - - =4 2 8 0 Engineering Mathematics Example 7 : Solution 397 E valuate ffs F · n dS, where F = z i + x} -3y2 zk and S is the surface of the cy l inder x2 + y 2 = 16 included in the first octant between z = 0 and z=5. A vect or normal to the surface S is given by grad S = v (x2 + y2 ) =2 xi + 2y} . . Therefore n = a uni t normal to- any poi nt of S = surface S. 2x i + 2yJ ✓(4 x 2 + 4y 2) X i + yj . = -- , si nce x2 + y 2 = 16 , in the 4 - dxdz We hav e ff s F - .ndS = fJrR F - .n. � , where R is the pro j ection of S on the x-z pl ane. It should be noted n-J 1 I that in this case we can not take the pro j ection of S on the x-y pl ane as the surface S is perpendicular to the x-y pl ane. Now 2 F -n = (z i + x} -3y zk) { n-] = ( xi y ] : } ]=i x i y} : J = ¾ (xz + xy ), xz + xy dx dz Therefore the required surface integ ral is = fJrR -- / y 4 4 = = Example 8 : Solution 5 = 4 = xz 2) J: (4 z + 8 )dz = 9 0 . 2 + x ] dx dz, ( since Y = f (1 6-x ) on SJ V E val uate f ( v x i=) - nds, for F = (2 x - y ) i -yz2 ]-y 2 zk where S is surface of sphere x2 + y 2 + z2 = a2 s above xy plane. v'x F = H ence, a ax a =k ffs (vx i= ) -n dS= Jf/•n dS [ where R is pro j ection of hemisphere on xy pl ane, which is circl e x2 + y 2 = a2 ] On changing to polar co- ordinates. fs C url F · n dS, a ay k 2 x -y -yz2 -y 2 z r = JJ R k ' 11 Remark : [ fz O f x O ✓( 1 _ 6 x dxdy r = JJ dxdy R I n . kl = r 2 rr r a 2 J e=O J r =O rd8 dr = n a can al so be evaluated very easily by apply ing stokes theorem. GAUSS' DIVERGENCE THEOREM Suppose V is the volume bounded by a closed piecewise smooth surface S. Suppose F (x,y, z) is a vector f unction which is continuous and has continuous partial derivatives in V. Then IES MASTER Publicatio·n 398 Vector Integration IJJ/ • n dS = JJL v - F dvl where n is the out ward drawn unit normal vector to S. or The surf ace integral of the normal component of a vector F taken over a closed surface is equal to the integral of the divergence of F taken over the vol ume enclosed by the surf ace. Example 9 : If S is c l osed surf ace, enclosing volume V, prove : Solution (i) Ap pl ying the divergence theorem f or r we have (i) !J r . n dS = V ' 3 s (ii) ff Vr 2 . dS = 6V fJ · n dS = L divrdv=L 3dV = 3V (·: (ii) Using �i'fm = mrm-2 r, we get Vr2 = 2r V- f = 3) div (vr2 ) = div(2r )= 2 divf = 2 x 3 = 6 H ence applying divergence theorem f or vector f unction Vr2 we have (ii) ff vr2 - dS = Jf vr2 - n dS= L ( divvr2 ) dv = fv 6dV = 6V f/- n dS= L divF dV = L ( divV<p )dV Example 10 : Show that S o lution = L V2<p dV, since V•V<p = V2<p = L --4 11:p dV = -411: p dV fs[ ( x -yz) i-2 x y} +2 1<] · n dS = a3 2 3 5 f where S denotes the surf ace of the cube bounded by pl anes x =0, x = a, y =0, y = a, z =0, z= a. By Divergence Theorem Ex,ample 1 1 : If F = ax i + by} + czk, prove that (i) fs j= . n dS = i3 n (a+ b+ c) (ii) E valuate Solution (i) ffs ( ax 2 where S is surf ace of unit sphere. + by2 + cz2 ) ds over sphere x2 + y2 +z2 F = axi+by}+ czk ,,:, 1 . V-F = -(ax )+-(by )+-(cz z- ) =(a + b + c) Theref ore, by Divergence theorem a ax a ay a· · _a Engineering Mathematics f/ - n dS (ii) B y Divergence theorem f/ - n dS = 399 L divF dV =(a+ b+ c) L dV = (a+ b+ c) v n = (a+ b+ c) � = L divF dV Given F•n = ax2 + by2 + cz2 , we have to find F . n = N ow, 'Vcp . where cp = x2 + y2 + z2 I Vcpl n = (2 xi +2y}+2zk )/ 4 (x2 + y2 +z2 ) Therefore ✓ = xi+ y}+zk (·: F -ri = (ax2 + by2 + cz2 ) = (ax i + by}+ czk )•(xi+ y}+zk ) x2 + y2 +z2 = 1) F = (ax i + by} + czk ) ⇒ N ow, this reduces to the part (i) that is already sol ved. H ence Example 1 2 : E valuate Solution : Let ffs ax2 + by2 + cz2 dS = ¼ n (a+ b+ c) JfJ x then N ow Simil arly 2 dy dz + y2 dzdx + z2 dxdy] where S is surface of cube ( O � x � 1, 0 � y � 1, o � z � 1 ) F = F1 i + F2 } + F3'<, = ffs{ F1 dy dz + F2 dzdx + F3 dxdy) = = ffL [: + a:: + a� ] dv = 2J� f� f� (x + y +z) dxdy dz f� f� f > dxdy dz f� f� f� y dx dy dz x = ( ; )> = ff/· n dS = L div F dV f� f� f� (2x + 2 y + 2z) dxdy dz y)� (z )� = i . .. (i) = J0 J0 f0 z dx dy dz = ! 1 1 1 2 Substituting these values in (i) we get I = 2 [! + ! +! ] = 3 2 2 2 STOKE'S THEOREM Let F (x, y, z) be a continuous vector function which has continuous first partial derivatives in a region of space which contains S in its interior. And S is an open surface bounded by a simple closed curve C. Then, If/ · dr = ff ⇒ (v x F ) • n dS = ffs ( curl F ) · dSI where C is traversed in the positive direction. or.. .. anti - cl o ckwise. sense. . . In o ther words, Stoke' s theorem may be stated as : IES MASTER Publication 400 Vector Integration "The line integral of the tangential component of vector F taken around a simple closed curve C is equal to the surface integral of the normal component of the curl F taken over any surface S having C as i ts boundary. GREEN'S TH EOREM . • 8F2 8F1 Let F- = F1 i+F2j then Green' s theorem is � -F- dr- = ff (-- ) dx dy C 8X S ⇒ ay ... (i) where s is any pl ane surface bounded by cl osed curve 'c' . r I ""' I NOTE � Stoke's theorem in plane is also referred as Green's t heorem. Corollary: Proof By using Gree n' s theorem show that area bounded by a simple cl osed curve C is; Area = _:l__ J (x dy -ydx) in xy plane 2 C a F1 a F2 a = -(x) = 1 and - = -1 L et F1 = -y, F2 = x, then - ax ax H ence, by Green' s theorem ay f (F dx + F dy) = ff (8FOX 1 2 2 fc (-ydx+ xdy) = fc (-ydx+xdy) C S - a F1 ay ) dx dy Thus the requi red area is: A = _:l__ f ( xdy - y dx) 2 C Example 13 : F ind the area of ell ipse by using Green' s theorem. Solution : We know that with the help of Green' s theorem area of any closed curve C is given as A = _:l__ f (xdy -y dx) 2 C At any point on the ellipse: x = a cos0, y = b sin0 therefore 2 2 A= i-fc (a cos0 bc os0 d0+b sin0 asin0 d0) = 1 s: ab ( cos 0 + sin 0) d0 = rr Example 14 : P rove that Proof : curv e C. fc r xdr = 2fs dS 1 2rr ab[ 0 ]0 =n ab 2 = twice the vector area of an open surface S bounded by the closed Appl y ing Stoke' s theorem to axr where a i s any constant vector, We have fc(a X r) . dr = H curl (a X r) · n dS = Jf 2a -n ds, Since [(axr)- dr = a -(rxdr)] ⇒ ⇒ fc a · (r xdr) = 2a -Jf n ds = 2a . Jf n ds ⇒ a • l fc (r xdr) -2 J n dsl =o s curl(a xr) = 2a (·: dS = n dS) Example 15 Engineering Mathematics 1 401 Use Stoke's theorem to evaluate f curl F • dS over open hemispherical surface x2 + y 2 + z2 = a 2 , z > O , where F = yi + zx] + yk . By Stoke's theorem Solution: fs curl F - dS = f/ - dr (where C is circumference of circle x2 + y2 = a2 in the plane z = 0) = fc{ yi + zxj+ yk) · (dxi + dy} + dzk) = f/ ydx+zxd y + ydz) = fc ydx,since z = O ⇒ dz = O = s: a sin8(-a sin 8)d8 x Example 16 Solution: 2 = -4a 1 f:' sin 2 2 (Using perametric co-ordinates; x = a cos 8, y = a sin 8 ) 8d8 = -na2 Verify stokes theorem for the function F = x2T + xy} in the square region in the xy-plane bounded by the lines x = 0, x = a; y = 0, y = a . Consider th F . dr = th i= . dr + J F . dr + r F . dr + r F . dr Jco 'foABCO 'foA AB Jee = OA ( x2 dx + xydy ) + ( x2dx + xydy) + ( x2 dx + xydy) + AB ec f f f fco ( x dx+ xydy) 2 Equation of OA is y = 0, equation of line AB is x = a, line BC is y = a, line CO is x = 0 For OA; y = 0, dy = 0, Q :-,; x :-,; a For OB; x = a, dx = 0, 0 :-,; y :<,; a X= For BC; y = 0, dy = 0, a :-,; x :-,; O For CO; x = 0, dx = 0, a :-,; y :-,; Q So Now, So, Here f F - dr = 0 VxF = C 0 ,, 0 y ,= a B ' x= a y=O A f: x2 dx+ f: aydy + J: x2 dx + J: Ody i j k a a a x2 ay az xy 0 a3 a3 a3 a 3 = -+- - -+O = 2 2 3 3 = T(O)+ ] (O)+k ( y - 0) = yk a ffs(vx F) • nds = ff yk - kds = f: f:y dxd y = f0 dxf f/ • dr = fffs(v x i=) - nds a 0 ydy = a a2 = a3 2 2 Hence Stoke's theorem verified. IES MASTER Publication 402 Vector Integration PREVIOUS YEARS GATE & ESE QUESTIONS 1.* Questions marked with aesterisk (*) are Conceptual Questions fJ p u n dA = fffv V ( pu) dV where p is a scalar, u A is a vector, A is surface area, n is unit normal vector to the surface and V is the volume. (True/False) [GATE-1 994-ME; 1 Mark] 2.* Match the following: 1. <Jf D.ds = Q 3. HJ (V · A ) dv = # A. ds P. Stoke's Theorem Q. Gauss's Theorem 2. R. Divergence Theorem S . Cauchy's Integral Theorem Codes: 3. 4. (a) (b) (c) (d) p 2 4 4 3 Q 1 1 3 4 4. R 4 3 1 2 t f(z) dz = 0 ff (V x A)·ds = j A · dl 3 2 2 1 surface integrals to volume integrals surface integrals to line integrals vector quantities to other vector quantities line integrals to volume integrals [GATE-2001 ; 1 Marks] Given a vector field F, the divergence theorem states that (a) t/ · dS = jv V • Fdv (b) f/ · dS = fv v x Fdv (c) j/ · dS = f)vi=)dv 5. (d) f/ · dS = j)v X i=)dv [GATE-2002-EE; 1 Marks] Stokes theorem connects (a) a line integral and a surface integral (b) a surface integral and a volume integral (c) a line integral and a volume integral (d) gradient of a function and its surface integral [GATE-2005-ME; 1 Marks] l i ne i n tegral f V • dr of t h e v ector \I = 2xyzi + x z} + x yk from the origin to the point (1 , 1 , 1 ) 2 2 (a) is 1 (b) is zero (c) is -1 (d) can not be determined without specifying the path [GATE-2005; 2 Marks] 2 7.* Value of the integral fj xy dy - y dx) , where C is the square cut from the first quadrant by the lines x = 1 and y = 1 will be (use Green's theorem to change the line integral into double integral) 1 2 3 (c) 2 (a) s [GATE-2000 (CE)] The Gauss divergence theorem relates certain (a) (b) (c) (d) 6.* The E 8. (b) 1 (d) 5 3 [GATE-2005; 2 Marks] Which one of the following is Not associated with vector calculus? (a) Stoke's theorem (b) Gauss Divergence theorem (c) Green's theorem (d) Kennedy's theorem 9. [GATE-2005 (Pl)] ff ( V x P) · ds , where P is a vector, (a) f P•dl (c) j v x P•dl is equal to (b) t V x V x P•dl (d) Hf v•Pdv [2006; 1 Marks] 10. Consider points P and Q in the x-y plane, with P = ( 1 , 0 ) a n d Q = (0 , 1 ) . T h e l i n e i ntegral 2J: (xdx+ ydy) along the semicircle with the line segment PQ as its diameter (a) is -1 (c) is 1 (b) is 0 (d) depends on the direction (clockwise or anticlock wise) of the semicircle) (2008-EC; 2 Marks] Engineering Mathematics 1 1 . If r is the position vector of any point o n a closed surface S that encloses the volume V then ff (r •ds) is equal to (a) 2 (c) 2V !v (b) V (d) 3V [GATE-2008 (Pl)] 2 2 1 2.* F(x, y) = ( x +xy)a x +(y +xy)ay , It's line integral over the straight line from (x,y)=(0,2) to (2, 0) evaluates to (b) 4 (a) - 8 (d) 0 (c) 8 [GATE-2009-EE; 2 Marks] 13.* A path AB in the form of one quarter of a circle of unit radius is shown in the figure. Integration of (x + y)2 on path AB traversed in a counter-clockwise sense is y A X (b) �+1 2 (a) � - 1 2 n (c) 2 (d) 1 14. I f a vector field [GATE-2009; 2 Marks] V is related to another field V= v x A, which of the following is true? Note: C and Sc refer to any closed contour and any surface whose boundary is C. (a) (b) (c) (d) D y 3 �--+---+----+ X Q 2/✓3 (a) 0 (b) 2 3 ✓ (d) 2✓ (c) 1 3 [201 0-EC; 2 Marks] 1 6. The line integral of the vector function F=2xi+ x2 ] along the x-axis from x = 1 to x = 2 is (a) 0 (c) 3 (b) 2.33 (d) 5.33 [GATE-2010 (EE)] 17. The line integral J (y dx + x dy) from P 1 (x 1 , y 1 ) to P2 B A through 403 f V· d l= ff A · ds f A· d l = ff V · ds f V· V · dl = ff v·A· ds f v· A·dl=ff V · ds C Sc C Sc C Sc C Sc y �----+ X (a) X 2 Y2 - X 1 Y 1 (b) (y� - y�)+(x� - xf) (c) (d) (x2 -x1 HY2 - Y1 ) ( Y2 - Yd +(x2 - xd [GATE-2011 (Pl)] 1 8 . Consider a closed surface 'S' surrounding a volume V. If [GATE-2009 (EC)] 1 5.* If A= xya x + x a y, t A . di over the path shown in c the figure is 2 P 2 (x 2, y 2) along u{J semi-circle P 1 P2 shown in the figure is with r n is the position vector of a point inside s the unit normal on 'S', the value of the integral <JP 5 f. n ds is (a) 3V (c) 10 V (b) 5V (d) 15 [GATE-2011 (EC)] IES MASTER Publication 404 Vector Integration 2 2 19. Given a vector field F = y xa x - yza y - x a 2 , the line integral f F · di evaluated along a segment on the x­ axis from x = 1 to x =2 is (a) - 2.33 (b) 0 (c) 2.33 (d) 7 [GATE-2013; 1 Marks] 20.* The following surface integral is to be evaluated over a sphere for the given steady velocity vector field F = xi + y} + zk defined with respect to a cartesian coordinate system having i, j and k as unit base vectors. ffs {(F • n)dA Where S is the sphere, x + y2 + z2 = 1 and n is the outward unit normal vector to the sphere. The value of the surface integral is 2 (a) n (b) 2n (c) 3n/4 (d) 4n [GATE-2013-ME; 2 Marks] 21 . Consider a vector field A (f) . The closed loop line integral � A.di can be expressed as (a) # ( v' x A) · a s over the closed surface bounded (b) (c) (d) by the loop # (v x A ) dv over the closed volume bounded by the loop (c) (b) 7t (d) 7t 2 7t 4 4 [GATE-201 4-ME-Set 1 ; 2 Marks] 24.**Given F = za x + X8 y + ya z , I f s represents the portion of the sphere x2 + y2 + 22 = 1 for z .::'.. 0, then J (V x F.)ds is ___ [GATE-2014 (EE-Set 1)] 25.* A vector i s defi ned as f = yi + xj + zk where i, } and k are unit vectors in Cartesian (x, y, z) coordinate system . The surface integral # f . ds over the closed surface S of a cube with vertices having the following coordinates: (0 0, 0), ( 1 , 0, 0), (0 0, 1 ), (1 , 0, 1 ), (1 , 1 , 1 ), (0, 1 , 1 ), (1 , 1 , 0) is 26.* T h e l i ne i ntegra l of [GATE-2014 (IN-Set 1)1 the v ector field F = 5xzi + (3x + 2y) } + x zk along a path from 2 2 2 (0, 0 , 0 ) to (1 , 1 , 1 ) parametrized by (t, t , t) i s 27. The surface integral [GATE-201 5; 2 Marks] ffa �(9xi -3yi) · nds over the sphere given by x + y + z = 9 is __. 2 2 2 [GATE-201 5-ME set 2; 2 Marks] Jff(v · A ) dv over the open volume bounded by 2 28. The value of Jc[(3x -sy )dx+(4y - 6xy ) dy] , where C ff ( v' x A) · a s over the open surface bounded by [GATE-201 5-ME set 3; 2 Marks] the loop the loop [2013-EC; 1 Marks] 22.* The line integral of function F = yzi , in the counter clockwise direction, along the circle x2 + y 2 = 1 at z = 1 is (a) -2n (b) -n (c) (a) 0 n (d) 2n 23. The integral [GATE-2014-EE-Set 1 ; 2 Marks] fc ( ydx - xdy) is evaluated along the 1 circle x + y = traversed in counter clockwise 4 direction. The integral is equal to 2 2 is the boundary of the region bounded by x = 0, y = 0 and x + y = 1 is ____ 2 29.** A vector field D = 2p aP + za2 exists inside a cylindri­ cal region enclosed by the surfaces P = 1 , z = 0 and z = 5. Let S be the surface bounding this cylindrical region. The surface integral of this field on s(# D • ds ) is _. s [GATE-2015-M E; 2 Marks] 30.* The value of the l i ne i ntegral 2 2 f ( 2xy dx + 2x ydy + dz) along a path joining the origin (0, 0, 0) and the point (1 , 1 , 1 ) (a) 0 (c) 4 (b) 2 (d) 6 [GATE-201 6-EE-Set-2; 1 Mark] Engineering Mathematics 405 + (x + z)j + (y + z)k and n is the unit outward su rface normal , yiel ds___ 31. A scalar potential � has one following gradient; v'� = yzi + xzi + xyk . Consider the integral l: : : fc v'�.dr on the c u rve r = xi + y} + zk . The curve C i s parameterized a s follows: 2 z = 3t and 1 :o; t :o; 3 2 2 i ntegral F(r). dx. for F(r) = x i + y } along C which is a straight line joining (0, 0) to (1 , 1 ) is__ fc [GATE 2018 (CE-Afternoon Session)] [GATE-2016 ; 2 Marks] 32.* The value of the line integral circle of radius ✓n units is _ jc F · dr, __ [GATE-2016; 2 Marks] Suppose C is the closed curve defined as the circle x 2 + y 2 = 1 with C oriented anti clockwise. The value 2 j( xy dx + x ydy) 2 ov er the curve C eq ual s The value of the integral # f · fidS over the closed s u rface S bou n d i n g a v o l u m e V, where r = xi + y} + zk is the position vector and n is the normal to the surface S, is (b) 2V (a) V (d) 4V (c) 3V 37. [GATE 2018 (ME-MorningSession)] As shown i n the figure, C is the arc from the point (3, 0) to the point (0, 3) on the circle x2 + y2 = 9 . The value of the integral J (y2 + 2yx) dx + (2xy + x2)dy is _ _ (up to 2 decimal places) y (0,3) C (GATE-2016) J J F• n dS over the surface S 34. The surface i ntegral s of the sphere x + y + z2 = 9 , where F = (x + y)i 2 36. where C is a Here, F ( x, y ) = yi + 2x] and r is the unit tangent vector on the curve C at an are lengths from a reference point on the curve. i and ] are the basic vectors in the x-y cartesian reference. In evaluating the line integral, the curve has to be traversed in the counter clockwise direction. of The value (up to two decimal places) of the line 2 The value of the integral is ____ 33. 35. [GATE 2017 M E Session -II] 2 -+-------'---.... x (3,0) [GATE 2018 (EE)] IES MASTER Publication 406 Vector Integration 1. (FALSE) (d) 21. (d) 31. (726) 3. (b) 11. (a) 13. (b) 23. 33. (0) 5. (a) 15. 16. (c) 24. (c) 25. (0.67) (d) 18. (a) (d) 27. (4.17) 35. 17. 26. (1) (216) (b) 20. 2. 14. (a) 4. 6. (a) 7. (c) 8. 9. 10. Sol-1: 12. 19. (a) 22. (d) (b) (c) (a) According to gauss divergence theorem: ffF - f)dA = f ff v • FdV while in the given statement A ff (p u ) rjdA Sol-2: (b) V = fff,,grad( p u) d\7' f Adi = Stokes theorem : 30. (b) ff ( V x A)ds # A · ds = HJ (V · A)dv Divergence theorem Cauchy's Integral theorem : f(Z)dz = 0 f Sol-3: (a) Gauss divergence theorem states that # (F • n) ds = 36. 37. fff (v - F) dv i.e. it connects surface integral to volume integral. V Sol-4: (a) The divergence theorem states that the total outward flux of a vector field F through the closed surface S is same as the volume integral of the divergence of F (16) (226.19) (c) (0) (78.53) ⇒ Sol-5: (a) � F.dS = � V.F d v V Strokes theorem connects a line integral and a surface integral. Sol-6: (a) f V.dr # D - ds = Q Gauss's theorem : S (1.67) SOLUTIONS IFalseI 34. (3.14) 28. 29. (b) 32. (b) = = f (2xyzdx+ x2zdy + x2 ydz ) (1,1,1) f ( 0,0,0 ) 1 1) d (x2 yz ) = (x2 yzf , (0, 0, 0 ) = 1 -0 = 1 Method-II x-O Y -O z -O = t (let) = Equation of St. line = . 1-0 1-0 1-0 ⇒ so, x = y = z = t so dx = d y = dz = dt and Sol-7: (c) fc \I • df = 0 :,:; t :,:; 1 fc (2xyzdx+ x zdy + x ydz) 2 = f� 4t3 dt = 1 X=1 � y=1 ' R C --+---+----t-----. X 0 2 Engineering Mathematics 407 By Green's theorem f (f dX + f d ) f aF2 - a F1 dxdy 2 Y =f 1 Jc JR ( � � ) or fc (Mdx+Ndy) = � (: - :) dxdy 2 fc (-y dx+xydy) = ff[(Y -'- (- 2y)] dx dy = = J 1J 1 X=Oy=O 1 f0 [�2 y ]y=Odx 2 f P - dl Jf (V x P).ds 2 f: xdx+ydy Sol-13: (b) Method I 2 0 0 2 f {x2 +x(2 - x)) dx+f {y 2 +(2-y)y) dy = [ x2 ]� +[ y2 J� x2 = 4-0 + y2 = 1 = 0 + = ff/ • ds = xi+ y]+zk then div f = 1 + 1 + 1 By G-D theorem Sol-12: (d) = 0. rd0 n 2 f (x +y)2rd0= J (cos 0+sin0)2d0 0 n f (1+sin20) d0 2 0 n 2 cos20 ' ] = �+1 2 0 We have equation for path AB x2 + y2 = 1 = fJfv (divf)dV = 3 fv (1)dV ff = 3V Equation of straight line X y -+- = 1 2 2 X + y = 2 = = [0 - Method I I 1 f 0-4 + il f/dr = = 2 L xdx+2f>dy Sol-11 : (d) Now, L = Now, = = -1 = 0 ⇒ L F = (x + y)2 and dr By Stokes Theorem = (x2 + xy) a x + (y2 + xy) a y f (x2 +xy)dx+(y2 +xy)dy = y = sin 0 , 0 E [ 0, Sol -9: (a) I f i=(x, y).d Z Let x = cos 0 , Kennedy's theorem = Line integral of F(x, y) (3y)dy dx Sol-8: (d) Sol-10: (b) F and dZ = dxa x + dy a y 3 2 Differentiating above equation w.r.t. y, we get ⇒ Let dx 2x-+2y dy ,�� = =- �, � = (x f ( x + y) dr = 2 0 AB = = + y)2 . Then f ( x+y)2 J(dx)2 +(dy)2 AB ls (x J8 2 2 + y + 2xy) 1+(�� J dy (x 2 +y2 +2xy ) . (1+ �: J dy ✓ IES MASTER Publication 408 Vector Integration = = = 1 dy f (1+2xy).-. X AB ( f (✓ � + 2y ) dy f X AB 1 1-y AB 2 2 It is the statement of stokes theorem. A - d i = (xy a x + x 2a y ).( dxa x + dya y ) A - di = xydx + x2d y d y 3 1 ✓3 2 3 ✓ X 1 8 3 2 = -+ - - --- = 1 Method-II : 3 Using Green's Theorem fc (Md x + Nd y)= ffs (: -:) dx dy J { xy dx + x d y)= 2 = Jf(2x - x ) dxd y 2/ ✓ 3 3 f f x dx dy 1 =1 X = .J3 y = 1 Sol-1 7: (a) f (2xdx) [·: along x a xis ; y = 0 ⇒ dy = OJ = 3 I = J (yd x+ xdy) = f d( xy) P2 P2 P1 P1 = [ xy]:� Sol-18: (d) ⇒ 3 2 2 3 /✓3 1/✓ = L,./3 x d x+ L ( 23 ) d y + f ,J3 3xdx + f 1 ( 13 ) d y 2 3 ✓ ✓ 2 = C # F - nds 2 3 = f (2xi + x2 }) - (dxi + dy}) f (2xdx + x2dy) C From Gauss Divergence Theorem f\ - d l = f: A - d l+J: A - d l + J: A - d l+f: A - d l 2 = J = �+1 Sol-15: (c) u= f t - dr + 2y dy = [sin-1 y + y2 J: Sol-14: (b) Sol-16: (c) #5r - nds = HJ d iv F dv = 5HJ div r dv V = 5HJ (3)dv V = 15 V V Sol-19: (b) F and di F . dl y 2 xa x - yzay - x 2a2 A = A dxa x + dya y A + dza2 = y2xdx - yz dy - on x-axis, y = 0, z = 0 F . dT= 0 ⇒ X=2 Sol-20: (a) J F. dl= o X=1 F= x i + y} + zk v - i== 3 By Guass Divergence Theorem �4 Jf (F• n)dA = s ± HJ 3 dv=¾(V) x 2d z Sol-21 : (d) Engineering Mathematics So, J ('v X F) · ds = J ('v X F) · n ds By stake's theorem f A - di s = H ('v X A).dS s = ' J (i +}+k) - kds (i.e closed loop line integral to open surface integral) Sol-22: (b) = J ds C : x2 + y2 = 1 , z = 1 = Area of the circle x2 + y2 = 1 F = yzi, dr = dxi+dy}+dzk x= cas e, y= sin e, e E [0, 21t] Line integral= J F . dr C = J yz dx C [ ·: for z = 0, sphere will convert into circle] 2 = 1t(1) = 3.14 Sol-25: (1) divf = -(y)+ -(x)+-(z) = 1 a ax 2n = J sine. 1(-sin e)de 1 1 1 _y1 - cos 2e de 000 2 0 si 2e " = -J[e - � J: = - 1t Sol-23: (c) Applying Green's theorem C F F = JJ(8 2 - 8 1 )dxdy R ax 8y = -2Jf dxdy = -2 x (area of circle) R f ydx - xdy = -21t X = H J1 dx dy dz Sol-26: (4.17) 000 F= 5xzi+(3x2 +2y)}+x 2 zk :. dx = dt, dy = 2tdt, dz = dt and at(O, 0, 0), t = 0 and at (1, 1, 1 ), t = 1 J F.dl= JJ5.t.t.dt+( 3(t 2 )+2t 2 )2tdt+t2 t.dt] ⇒ = J�( 5t2 + 5t 2 .2t+ t 3 ) dt 7t(J)2 = J�(5t 2 +11t3 ) dt 1 3 4 = [ � t +.!.!t ] 3 4 0 C Sol-24: (3.14) = 1 (Volume of the unit cube) Path is parametrizedy by (t, t2 , t) it means x = t, y = t2, Z = t R = -2 1 1 1 (using G-D theorem) F.d l= 5xzdx+ ( 3x2 +2y)dy+ x2zdz f ydx - xdy= JJ (-1 -1) dxdy C a az a ay = H J divf dv 0 = 409 i a Curl F= v x i== - j k X y a a ax ay az z = i+}+k & n = k [·: prozection of s will lie on xy plane] Sol-27: (216) = �+.!.! = 4.167 � 4. 1 7 3 4 By Gauss Divergence theorem, JJi=.n ds = Jff (v.F ) .dV JJ __!_( 9xi - 3y])• Ads = _§_ Jv dV , s 7t 7t IES MASTER Publication 41 0 Vector Integration = = Sol-28: (1 .67) 4 3 �xv = !x6x 7t (3) 3 7t 7t 216 = ._. <J>J( 3x - 8y 2 ) - dx+(4y-6xy) - dy ] Apply Green theorem, <f> F1dx + F2dY c Here, = n( 8 F2 Sol-30: (b) F - 8 1 , dx - dy = 4y - 6xy So, <J>J(3x-8y ) - dx+(4y-6xy)dy ] 2 = = = = u[( ⇒ l a a ( 4y - 6 xy)) - ( ( 3 x - 8y 2 )) dx dy ax ay ff[-6y + 16y] - dx - dy ⇒ 0 ff(10y)· dx · dy ff Sol-29: (78.53) D ! 1 0 - (-1)1 = = 5 = 1.67 3 f D - ds= f (V · D) dV where v'•D = So by (1) f (v' · D) dv 1 a 1 ao<I> aoz - - (pDP )+- +az p op p o� V = = = Put x = = ... (1) (Div in Cylindrical Coordinates) 1 a - - (p · 2p2 )+ 0+1 P op 6p + 1 1 2n 5 f f f (6p +1) pdp d� dz p=0 <J>=0 Z=0 = t, y = Sol-32: (16) fc f . df = z = 1 3 f�( 2'A.3 +2'A. +1)d'A. ⇒ z = v� = yzi + zx]+ xyk f/ yzdx + zxdy + xydz) t2, z = 3t2 1 :-s: t s: 3 f(t2 . 3t2 · dt+3t2 · t · 2tdt + t · t2 · 6tdt) = 1 3 f 15t4dt=l3t5 11 =3(343 -1)= 726 = 3 1 fc (ydx + 2xdy)= ffs (2 - 1) dx � 2n 4/✓n f rf 8 =0 = 0 Sol-33: (0) rdrd0 = Using Jacobian Concept 2n 8 -d0 = 16 8=0 7t f Using Green's theorem, = 'A.= 1 4 3 f�( 4'A. +1)d'A.=[ 'A. +'A.]�=2 Using Green's Theorem = = = fc ( yzi + zx} + xyk) •(idx+ jdy+kdz) = The value 2p2aP +za2 Using Gauss-Divergence theorem s 'A. =0 and when x = y ⇒ Now, f 0 s [<x/I = dx=dy=dz= d'A. and when x= y Given f= xi+ y] + zk and 0 0 1 ( 1 x) = J [5y 2 ] - - dx= 5 (1 - x)2 dx = X= Y =Z=A. Sol-31: (726) 1 1-x (10y) dy dx 0 78.53 x-0 = y - 0 z - 0 =A = 1-0 1-0 1-0 which is equation of straights line in 3D. F 1= 3x - 8y2 F2 2 + i) I = J (2xy 2dx +2x2 ydy + dz) c Path joining (0, 0, 0) and (1, 1, 1) can be written as ax a y ) R = 1on ( Eng ineeri ng Mathematics aN aM � Mdx+Ndy = f f (ax - ay ) dx dy J (xy dx + x ydy) = 2 2 Sol-34: (226.1 9) f J F -nds = f div.F dV V [Gauss-Diverg enceTheorem] = v{ (x+ y ) i+(x + z)] + ( y + z)k] dV f = (1+ 0+1) dV f 4 n.(3)3 = 36n 3 1 C f = x i+ y} + 3 1< Sol-37 : (0) J f (f · n )ds = ff f) 3) dv = 3v C f {f dx+f dy) 1 C 2 2 +2xy ) i+(x2 + 2xy ) } C url f is conservative, i.e path independent, so we can integ rate al ong easiest possib l e path i.e. along the straig ht line A(3, 0) to B (O, 3). AB : 1f + =1 ⇒ y 3- x 2 I=J [ (y +2xy ) dx+(x2 +2xy )] C 2 = [ (2 x dx) = � =0.67 Sol-36: (c) ⇒ so dy = -dx, 3 ::; x ::; 0 . So, LI = J 1.df = J (x2 dx + y2 dy) 1 f fs (t • n)ds = f f fv (div8 ) dv ti = (y y =X X :,; so by Gauss-Diverg ence theorem 1 = f1 i +f2 }+ Now, al ong (0, 0) to (1, 1) equati on of straig ht line 0 :,; a Surf ace is closed with volume V On comparison 2 2 1 = x i+y ] and dr = dx i+ dy ] So, f.df = (x2 dx + y2 dy) dy=dx a = 1 + 1 + 1=3 = 2 X 361t = 226.19 Sol-35 : (0.67) a 2 2 I = J [ (y +2xy ) dx + (2 xy + x ) dy] = 2f dV = 2V V = div (f ) = -(x)+-(y)+-(z) ax ay az R Jf (2xy -2xy) dxdy = 0 41 1 C then 2 2 = J [(3 - x) +2x (3-x)d x] +[x + 2x (3-x) (-dx)] = AB f (9 -6x) dx = (9 x-3x X=3 �=0 2) IES MASTER Publication IOTA The value of square root of - 1 i.e. � can not be evaluated in real number system . So Euler gave a symbol for it denoted by i and read as iota. i.e. Ii = �I So we can conclude, i2 = -1, i 3 = -i, i4 = 1, i5 = i, i6 = -1 and so on . . . . . hence cycle of iota consist four terms. �NOTE g 1 . lin = t I where r = Remai nder when n is d i vi ded by 4 i .e. r = Remainder ( % ) e.g., i 34 = i 2 = - 1 3 Rem ( : ) = 2 e.g., i 4 3 = i 3 = -1 ·: Re m ( : ) = 3 3 n+3 n n+t = ol , n E I i .e. sum of four consecutive power of iota is always zero 2. lin + i + i +z + i e.g., i 6 3 + i 64 + i65 + i66 = i 3 + iO + il + i 2 = -i + 1 + i - 1 = 0 3. If a, b > 0, t hen b b ✓a ✓ = ..Job . But if a, b < 0, then the product ✓a ✓ = -..Job , i .e., GM for two +ve nos = eg. geometric mean of -4 and Example 1 u�r 1-1 ⇒ [ 1 - + 2i : r is defi ned as . ..Job while for two -ve nos. GM = -..Job . -6 = - ✓(-4)(-6) = -9 1+i m If ( � ) = 1 find the least positive i ntegral value of m . Solution b ✓a ✓ = 1 ⇒ 1+i 1+i [ - x -r = 1 1-i 1+i 1 ⇒ lir = = 1 ⇒ m ⇒ = [ 1 + i + i + i2 1 - ·I2 r_ - 1 4 COMPLEX NUMBERS AS O RDERED PAIRS Any ordered pair of real numbers x, y written as (x , y) is called complex number, Usually a complex number z = x + iy is written as z = (x , y). E ng ineeri ng Mathematics 41 3 S uch a pl ane whose points are represented by compl ex numbers is cal l ed Argand pl ane or Argand di agram. I t is al so call ed compl ex pl ane or Gaussian pl ane. The real numbers x and y are call ed as the real and i magi nary parts of z and are denoted as• Re(z) = x, l m(z) = y where both x & y are real numbers itsel f. EULER NOTATION F ol l owing two notations are cal l ed E ul er Notati ons le;e = cos0 + isin0 e.g., H ence, we have foll owing resul ts. De-Moivere's theorem e2ni 1 = = cos2n + i sin 2n e2 ni or e2nni ' and = -1 1 = e- ie = cos0 -isinel eni (cos0 ± isinet = cosn0 ± i sin0 , n E I or Q ' i = e2 , - i IT. -I = e 2 1t . -- 1 ( cos0 +isiner = (eie r = eine = cos( n0) +isin( n0) (Whil e ( si n0 ± icoser ,;<c sin0 ± icos n0 and (cos � ± i sinet INOTE � ""' 1 * cos n� + isinn0 ) P ut x = 0 · 2 4 3 5 y +y y +y - ) ) +I ( y -= ( 1 -2 .1 4 .1 ""' 3 .I 5 I. "" eiY = cos y + i sin y o r w e can take e;e = cos 0 +i s in 0 e iz - e-iz sin z = sinz = -2i 2 . Circular function: COS Z = - e iz + e -iz 2 e z -e-z e z +e-z , cosh z = 2 2 4. Relation between hyperbo l ic and circu lar functions: 3 . Hyperbo lic function: sin hz = sin ix = i sin hx, cos ix GEOM ETRICAi:. REPRESENTATION = cos hx, tan ix = i tan hx Cartesian Representation We can represent every compl ex number by a point in the Argand pl ane. Let z = x + iy be a compl ex number. This z can be represented by a point P with Cartesian coordinates (x,y) referred to rectangul ar axes OX and OY usual l y call ed the real and imaginary ax es, respectivel y. P(Z) y I I 1 y 0 X X Polar Representation Obvi ousl y the pol ar coordi nates of P are ( r, 0) where r i s the mudul us and 0 is the argument of the compl ex number z, i.e., z = r(cos0 + i si n0) = re;e . IES MASTER Publication 41 4 Complex Number Vecto r Representation I f a complex numbe r 2 = x + iy is re pre se nte d by a point P(x, y) in the Argand plane the n the modulus and argume nt of 2 are re pre se nte d by the magnitude and dire ction of the ve ctor OP re spe ctive ly and vice ve rsa. ALGEBRAIC PROPERTIES F or any two complex numbe rs 21 = ( x1 , y1 ) and 22 = ( x2 , y2 ) we have the foll owing de finitions : 1 . Equality of Two Complex Numbers Two compl e x numbe rs 21 = ( x1 , y1 ) and 22 = ( x2 , y2 ) are said to be e qual if re al parts of both the numbe rs as we ll as the imaginary parts of both the numbe rs are e uqal i.e . x1 = x2 and y1 = y2 . 2. Order Relation: {> or <} relations are called order relations) Inequalitie s like ' gre ate r than' or 'le ss than' have no me aning in be twee n two complex numbe rs. Ine qualitie s may occur only in be twee n the mudulli of complex numbe rs. i.e . if 21 = 10 +8 i & 22 = 2+ 3 i the n 21 > 22 (F) while 121 1 > 1 22 1 .(T) Hence orde r re l ations are not foll owe d by compl e x numbe rs. i.e . Z 1 < Z2 or Z2 < Z 1 has no me aning but IZ 1 1 < IZ2 1 or IZ1 1 > IZ2 1 has got its me aning be co2 IZ 1 1 and IZ2 1 are re al nos. 3. Sum and Difference of Two Complex Numbers Represented by a Point Le t 21 = x1 + y1 , 22 = x2 + y2 be the two complex numbe rs. 21 + 22 = (X 1 +X2 ) + i(y1 + Y2 ) or (21 ± 22 ) = (x1 ± x2 , y1 ± y2 ) which is also a complex numbe r, He nce the complex numbe r (21 ± 22 ) is re pre se nte d by a point whose coordinate s are (x1 ± x2 , Y1 ± Y2 ) . Commutative law and associative law for addition of complex numbe rs hol ds good as in the case of re al numbe rs i.e . (Associative law) ( commutative law) 4. Multiplication of Complex Numbers 2122 = (x1 x2 - Y1 Y2 , 1 Y2 +X 2 Y1 ) Commutative law, associative law and distributivity for multiplication of complex numbe rs holds good as following: 21 2 2 = 22 2 1 X 21 (22 23 ) = (2122 ) 23 (Commutative law for multiplication) (Asso ciative law for multiplication) 21 (22 ± 23 ) = 21 22 ± 21 23 (Distributive law) The product of two non zero complex numbers is always non zero and if the product is zero then at least one of the factors z1 and z2 must be zero. 5. Division F or give n two complex numbe rs 21 � Hence if 1 22 1 *0 22 = x1 X 2 + + = ( x1 , y1 ) iy1 iy2 (x1 and 22 + = ( x2 , y2 ) iy1 )(x2 - iy2 ) (X� + the division 2d22 is we ll de fine d. y n = the division 2d22 is obtaine d as (x1 X2 + Y1Y 2 ) + i(X 2 Y1 - X 1 Y2 ) X� + Y� Engineering Mathematics 41 5 6. Identities The number 0 = (0,0) and 1 = (1,0) are the additive and multiplicative identities respectively. They cover the entire complex number system i.e. z+ 0 = z and z.1 = z or (a,b)+(0,0) = (a, b)-(1,0) = (a, b) 7. I nverses 1. The additive inverse of complex numbers z=(x + iy) is defined as: -z = (-x,-y) satisfying the equation z+ ( -z)=0 . The additive inverse of any complex number z is unique. 2. The multiplicative inverse of z(x, y) is z-1 such that zz-1 = 1 . z-1 = where Example 2 Solution: -y } X- -(x +y ' x +y 2 2 2 2 3 1 Find all values of z such that e 2 = 1+i✓ 3 1 + ✓i=rei9 = r cos 8+ ri sin 8 Let then ... (1) r = ..jx2 + y 2 = 1+ 3= 2 ✓ Therefore from (1) 8 = tan-1 ( t) = tan-1 (�) = � 1+ ✓i = 2eirr/3 3 Now from the given relation Taking log both the sides; log (e2) -i z = log 2+(2n+ }d, ⇒ COM P LEX CONJUGATE NUMBERS If z = x+iy is a complex number, then x -iy is said to be con jugate of z and is usually denoted as z . i.e. if z = ( x, y) then z=(x,-y) i .e. if the coordinates of the point z are (x, y) then coordinates of z are (x, -y) Again If z = r(cos 8+isin8) then z = r [cos(-S)+isin(-8)] So if the polar coordinates of z are (r, 8) then those of z are (r,-8) i.e. we have amp z = -amp z. Geometrically, if P and Q are the points represented by z and z then z is the reflection or image of z in the real axis. Properties of Conjugate Let z = X + iy then z = X - iy 1. Z + Z=2Re ( z)=2X & Z -Z=2i lm(z)=2iy 2. 4. lzl = z z = x2 + y2 2 Z1 Z2 = 21 22 IES MASTER Publication 41 6 Co mpl ex N umb er 6. 7 . R epr esentation of complex number z in var ious quadr ants Let z = x + iy :::: (x,y) Then and al so we have Example 3 : Solution 1 ( z" ) = (z)° -z = -x iy :::: (-x, -y), Z = X- iy :::: ( x,- y ), -Z = -X+iy :::: (-x, y) z1 z (-x+iy) = 1 -z 1 = lzl = 1-zl . lm(Y) -z (-x-iy) z(x+iy) z (x-i y) I f z1 and z2 ar e two non-zer o complex numbers such that lz1 1 = lz21 and ar g(z1 ) + ar g(z2) = n then Z1 = ? Let z1 lies in 1st quadrant & arg (z1 ) = 8 & l et z1 = (x + iy) :::: (x,y) It is given that ar g z1 + ar g z2 = n is given that IZ 1 I = IZ2 1 ⇒ ar g z2 = ( n -8 ) means z2 will lie in 2 nd quadr ant & also it So we can take z2 = - x + iy :::: (-x, +y) i.e. z2 = (-x, y) ⇒ z2 = (-x, -y) ⇒ z2 = (x, y)= z1 Second method : Let IZ 1 I = IZ2 1 = r and let ar g z1 = 8 then ar g z2 = n -8 z1 = r( cos8+isin8 ) and z2 = r [ cos( cosn - S )+isin(n - 8 )] so = r(- cos8 +isin8) or MODULUS AND ARGUM E NT -22 = r ( cos8 - isin8 ) ⇒ -z2 = r ( cos8 +isin8) = z1 Let z = x + iy be any complex number if x =r cos8 , y = r sin8 then r = .Jx2 +y2 is said to be modulus of z and 8 = tan- 1 (y/x) is cal led the ar gument or amplitude of z wr itten as ar g z. Geometri call y, lzl is the distance of the point z from or igin. Now, if 8 be a value of the ar gument, so also is 2 nn+0 , where n = o , ± 1, ± 2 ......i.e the ar gument of a complex number is not unique. Principal value of the ar gument is that val ue of argument 8 which lies between -n < 8 � n . __ 5 _.,.,,;:, INOTE 1. Z = 0 is a lso complex No . but it's Argument is not def ined . 2 . Arg Z = 0 or 3 . Arg Z = ± 2 TC n ⇒ ⇒ Z is purely real i.e . Z = K, KER . Z is purely imaginary i.e . Z = iK, K E R . Engineering Mathematics 41 7 Properties of Modulus and .Argu ment 1. The m o d u l u s of t h e sum of two complex n u mbers can never be g reater than the sum of their moduli, ⇒ l z1 + z2 I � l z1 I + l z2 I . This is also called the triangle inequality. Remark : Since l z1 + zJ = lzl + l zJ + 2Re l z1z2 I and 2. I21 - zJ = 1 21 ' + l zl - 2Re l z122 I 2 adding, we get 1 21 - 22 ' + 1 21 - zl = 2 i l zl + l zl l 2 The modulus of the difference of two complex numbers can never be less than the difference of their moduli. 1 21 - 22 1 � ( l 21 l - l 22 I) 2 2 l az1 - bzl + jbz1 + azl = (a + b ) {l zl + l zl ) 3. 4. To calculate Jzl in any question, always try to use JzJ 2 = zz 5. 6. Product of Two Complex Numbers Let us take two complex numbers: Multiplying, we have Let Z2 = X2 + iY2 z 1 z2 = (x1X 2 - Y1 Y2 ) + i(X1 Y2 + X2 Y1 ) Z 1 = X 1 + iY 1 , So that r1 , r2 are the modulii and 01 , 02 are the amplitudes of z1 and z2 , then z1z2 = rf2 [cos 01 cos 02 - sin 01 sin 02 + i. (cos 01 sin 02 + cos 02 sin 01 )] Hence = rf2 [cos(01 + 02 ) + i sin (01 + 02 )] . . . (1 ) The line representing z1z2 is of length equal to the product of the lengths of the vector z1 and z2 and amp is the sum of amp z1 and amp z2 . In general Thus we see that l z1z2 ......zn l = rf2 .... ,rn = l z1 l l z2 ' , .... l zn l arg (z1 z2 .... zn ) = 01 + 02 + ..... + 0n = arg z1 + arg z2 + ...... + arg zn Hence, the modulus of the product of complex numbers is equal to the product of the moduli of these numbers and argument of the product of complex numbers is equal to the sum of argument of these numbers. 5. Quotient of Two Complex Numbers : Let the two complex numbers be z1 and z2 then their quotient is � r1 (cos 01 + i si n 01 ) r1 (cos 01 + i si n 01 )(cos 02 - i sin02 ) = = r2 (cos 02 + i si n 02 )(cos02 - i sin 02 ) z2 r2 (cos 02 + i sin 02 ) [ ·: z1 = x1 + iy1 = r1 cos 01 + ir1 sin 01 & z2 = x2 + iy2 = r2 cos02 + ir2 sin02 ] [ cos (01 - 02 ) + i sin(01 - 02 )] = !:!._ r2 IES MASTER Publication 41 8 Complex Number Hence, the modulus of quotient is the quotient of the modulus of the numerator and the denominator, and amplitude of the quotient is the difference of the amplitudes of the numerator and the denominator. lm(Y) Method of calculating principal values of arguement Let z = x + iy then to calculate principal value of e first find a , by using the result a = tan -1 , �, where a is acute angle then which z l ies i .e. Example 4 : Solution and (-1 + i) is __? Real(X) 0 0 = -(1t-o.} 0 = -o. 7t 4 37t 1 But 0 2 = n - a = z = 1 + i a = tan � IYJ = 2: 2 X 4 ' 4 3n = argz2 4 and - argz1 = Find loci of the complex number z if arg ( W e know that z = x + iy Therefore, z, 1 z 1 = 1 + i then a = tan- JYI = 2: . So 01 = a = 2: ' 4 ' X 4 Let So angle between Solutio n : is according to quadrant, in e 0 = 0. I n the complex plane, the angle between the line (1 + i) Let Example 5 : 0 = (1t-o.) n 2 2-1 ) = 2: z+1 3 x +iy - 1 ( X - 1 + iy) ( X + 1 - iy) z-1 = �-� x �-� = X + iy + 1 z+1 ( X + 1 + iy) ( X + 1 - iy) = = ) (x - 1)(x +1) + iy (x +1) - iy ( x - 1) - i2 y 2 (x+1} - i2 y 2 x2 - 1 + 2iy + y 2 (x + 1) + y 2 2 2 = x2 + y 2 - 1 2 (x+1) + y 2 . +I ' 2y (x + 1) + y 2 2 2-1 2: z-1 2y ' . . )arg (-) = t an -1 2 . . . . (1 ) an d 1 t Is given th at arg ( 2 z+1 z 1 3 + X + y -1 7t 2y 7t ,,, = tan- = '13 ⇒ 2 2y2 ) = So we get, tan-1 ( 3 3 X +y -1 x2 + y2 - 1 ⇒ ⇒ Example 6 : Soluti o n : X 2 2 + y - 1 --y = 0 J3 2 ⇒ 2 x2 + y 2 - J3 y - 1 = 0 which is a circle. Find the locus of the complex number Z: for which arg ( W e have ⇒ arg ( z -3i) - arg(z -3) arg ( 7t 3 n z -3i ) = 3 z-3 2 - 3i z -3 ) = 2: 3 Engineering Mathematics ⇒ arg [x +(y -3)i] - arg[x - 3 + iy ] = ⇒ ⇒ y -3_ _ Y_ 1 x x 3 tan y -3 y• 1+ x x-3 ⇒ ⇒ ⇒ ; tan _1 yx -3x -3y + 9 -yx -'-------'---_.c..._ x2 - 3x + y2 - 3y n 3 = 9 -3x-3y x2 - 3x + y 2 - 3y ⇒ 3(3- x-y) = ✓ (x 2 - 3x + y 2 -3y) 3 3 3 3 x2 + y 2 - (3 - ✓ ) x - (3- ✓ ) y - 3✓ = 0 3 3 ⇒ = (·: = n tan 3 41 9 z = x + iy) n 3 3 3✓ - ✓ x-✓ y = x2 - 3x + y 2 -3y ⇒ 3 3 x 2 + y 2 -(3-✓ )(x + y)-3 ✓ = 0 This is the required locus of the complex number z = x + iy . EQUATION OF A CIRCLE IN COMPLEX FORM IZ - Z0 I = R is a circle with centre Z0 , radius R or zz + az + az + b = 0 with centre -a, , radius ✓la\ 2 -b Equation of a straig ht line in complex form az + az + b = 0, b E R D istance between two points Let P(Z 1 ) and Q(Z2) are two points then PQ Example 7 : Solution : = IZ 1 - Z2 1 Find k for which zz +(-3 + 4i)z -(3 + 4i)z + k = 0 represents drcle. On comparision with standard form. zz + az + zz + b We have a = - 3 + 4i and b So its radius = But Radius � 0 ✓lal ⇒ 2 = k -b = (9 + 16)- k = ✓ 25 - k � 0 ⇒ lk � 25I = 0, b E R ✓25 - k ROOTS O F A COMPLEX NUMBER Let z = a + ib be any complex number, then z 1 1n has n distinct values at n = 0, 1, 2, .... (n - 1). Cu be Roots of U nity : Consider ⇒ Hence, we have x = (1) x3 - 1 = 0 X 113 ⇒ 3 (x - 1)(x2 + x + 1) = 0 3 -1 + i✓ -1-i✓ - 1, , - 1, co , co2 (let) 2 2 . 2rc . r;:; I-1 h/ -'3 + =e 3 have three roots 1 , co, co where oo = 2 2 2 IES MASTER Publication 420 Complex Number Properties 1 . ro 30 = 1 , e.g. ro 3 = 1 2. 3. 4. 5. 1 + - 00 n + 00 2n 1 = 2 = ro = - and n is multiple of 3 0, n is not multiple of 3 {3, e.g. 1 + ro + ro 2 ro co2 = l zl = 1 = O I ro I � I ro 2 1 = 1 Cube roots of unity lie on the boundary of unit circle and divide its circumference into three equal parts. (0 Example 8 1 i Solution 1 If x + ..!. = 1 , then find the value of x2000 + 1 =1 X X +- X X = 1 ± .J=i=4 2 When x CHe I I ⇒ = -ro x2 + 1 = X = ⇒ x2 - X + 1 = 0 1 ± i✓ 1 + i✓ 1 - i✓ = -- ' -2 2 2 3 X 2�00 3 3 . = -ro2 , - ro respectively. 1 1 1- = = 002 + (-ro)2000 + -�� x 2000 + 2 2000 2000 00 (-ro ) x [because when 2000 is divided by 3 we get remainder 2] 0)4 +1 -0)2 = (02 = ro+1 (02 = 0)2 = -1 1 - = -1 Similarly for x = -ro2 · we get x 2000 + 2000 x CIH II 1 Cube Roots of -1 : Consider x 3 +1 = 0 ⇒ (x +1)(x 2 - x + 1) = 0 ⇒ x = -1 and J ± i "7 . r,; . 2rr I·1 hJJ3 So cube roots of -1 are -1, - ro, -ro2 where ro : - - + - = e 3 2 2 nth Roots of Unity : Let Z = (1)110 then we can take nth roots of unity as 1, a., a. 2 , a. 3 , . . . .a. n-1 where a. = ei2rr/n .2,r I5 For eg 5 roots of unity are z = (1 ) 115 "" 1, a., a. 2 , a.3 , a. 4 where a. = e th Properties 1. 2. nth roots of unity form a GP with common ratio a. Sum of nth roots of unity is always zero 3. 4. Engineering Mathematics Product of nth roots of unity is (-1)0-1 421 nth roots of unity lie on a unit circle 1 2 1 = 1 and divide its circumference into n equal parts. COMPLEX NO. AS A ROTATING ARROW 2 = x + iy = r(cos 0+isin 0) = r.e i9 Let i Thus multiplication by e a in 2, the vector Similarly for 2e -ia , vector Example 9 OP OP rotates in anticlockwise sense through an angle a. . is rotated clockwise sense by angle a. . Find the area of the triangle formed by 2, i2 and 2 + i2. 1 I I I I. I IZ I I I B iz i2 = e ¥ .2 and (1 + i)2 = i Solution ✓/j,.i 2 , 2 z it means i2 is obtained by rotating 2 through an angle � in 2 anticlockwise sense and (1 + i)2 is obtained by rotating 2 through ¾ and 121 = li2I. So it is right angle isosceles triangle. Example 1 0 area = Hence 1 J Base x height = Find the area of triangle formed by z, ro2 and 2+ ro2 . 1 l2l.li2I = 1 2} B { coz Solution J ro2 = e and 2n 3 0 l A z 2 , hence ro2 is obtained by rotating 2 through an angle 2 + ro2 = (1+ ro)2 = -ro2 2 = ( -1)e -I 4n. 3 . .2 = e n1 _ e -I 4n. 3 .2 = e - 7ni 3 . 2n in anti-clockwise sense 3 - -I 2 = e 2n1 .e 3 . 2=e 3 . 2 ni n. hence 2 + ro2 is obtained b y rotating 2 through a n angle � · i n anti-clockwise sense Now, Hence, 121 = lro2I =12+ro2I so equilateral triangle area = JJ l2l 2 4 IES MASTER Publication 422 Complex Number - PREVIOUS YEARS GATE & ESE QUESTIONS Questions marked with aesterisk (*) are Conceptual Questions 1. 2. The real part of the com plex number z = x + i y is given by 2-Z (b) Re (z) = (a) Re (z) = z - z 2 z+z (d) Re(z) = z + z (c) Re (z) = 2 cos qi can be represented as (a) (c) 3. e i<I> + e-i<I> (a) 0 ei<I> + e-i<I> (d) 2 7C 2 (a) 2n n [GATE-1 994 (IN)] (b) e-n/2 (d) 1 [GATE-1996-ME] (b) 2ni (d) in [GATE-1 997-CE] (a) a c i rc l e with ( 1 , 0) as t h e cehtre a n d radius 1 (b) a circle with (-1 , 0) as the centre and radius 1 (c) y-axis [GATE-1 997 (IN)] 6.* Which one of the following is not true for complex number 2 1 and 22? 21 Z1Z2 (a) 2 - z 12 2 I 2 (c) 5✓ - 2 2 ../34 (b) (d) 5✓ 2 [GATE-2005 (IN)] 8.* Let 23 = z, where z is a complex n u m ber not equal to zero. Then z is a solution of (a) z2 = 1 (c) z4 = 1 9. (b) z3 = 1 (d) z9 = 1 [GATE-2005 (IN)] Let j = � . Then one value of ji is (a) (c) (b) -1 ✓ 1 ✓ 3 (d) 2 e- ½ [GATE-2007 (IN)] 1 0.* I f a complex number z = � + i J then z4 is The complex number z = x + j y which satisfy the equation l z + 11 = 1 I ie on (d) x-axis Consider the circle l z - 5 - 5il = 2 i n the complex number plane (x, y ) with z = x + i y. The m inimum distance from the origin to the circle is (a) [GATE-1 994 (IN)] ei4> _ e-i4> (b) 2u e2 is a periodic with a period of (c) 5. ei4> _ e-i4> 2 i i , where i = ✓1- , is given by (c) 4.* . . . 7. (a) 2✓ + 2i 2 (c) ✓ .1 - - 12 2 3 1 .✓ (b) - - + 12 2 3 ✓ .1 (d) - - 1 8 8 3 1 1 . The value of the expression (a) 1-2i (c) 2 - i (b) (d) [GATE-2007 (Pl) -5 + i1 0 3 + 4i 1 2.* The equation sin (z) = 1 0 has 1 +2i 2+ i (a) no real (or) complex solution [GATE-2008 (Pl)] (b) exactly two distinct complex solution (b) lz 1 + 22 1 .:Slz 1 l+lz2 I (c) lz1 - z2 I .:S lz1 H z2 I (d) lz1 + 22 1 2 +lzc 22 1 2 = 21 21 1 2 + 21 22 1 2 [GATE-2005-CE] (c) a unique solution (d) an infinite number of complex solutions [GATE-2008 (ME)] 1 3.* If Z = x + j y where x, y are real when the value (b) e (a) 1 ✓x2 +y2 (d) e-Y (c) e Y 1 4.* O n e of the roots of equatio n positive square roots of -1 i s (a) (c) ✓32 j 2 (b) x3 [GATE-2009 (IN)] = j, where j i s the [GATE-2009 (IN)] (a) 1 + 6i (c) 9 + 8i (b) 9 - 8i (d) 17 + i6 [GATE-2009 (Pl)] 1 6. The modulu s of the com plex n umber (a) 5 (c) 1 ✓5 1 7. I f a com plex (b) (d) n umber co ✓5 1 5 (a) 0 (c) 2 (c) -3 -4i the equatio n (b) 1 (d) 4 1 8. The product of two complex 2 - 5i i s (a) 7 - 3i s [GATE-2010-CE] s ati sfie s 1 co3 = 1 then value of 1 + co + - i s (��ii) i [GATE-201 0 (Pl)] n umbers (b) 3 - 4i (d) 7 + 3i 1 + i and [GATE-2011-ME] n/2 cos x + , s , n x '. '. dx ? definite i n tegral f COS X ISm X 0 (b) 2 (a) 0 (d) [GATE-201 1 (CS)] 423 (b) e"a (d) 1 X [GATE-201 2-EC,EE,IN] 21 .* Square roots of -i, where i = � , are (a) i, -i ( 4n ) + .i .m (- 4n ) , cos (43n ) + .ism. (43n ) n .. n n .. n co ( ) + i m ( 3 ) , co ( 3 ) + i m ( ) 4 4 4 4 n . n n . . ( 3n ) cos ( 3 ) +i .m ( - 3 ) , cos ( - 3 ) +ism 4 4 4 4. (b) co s - (c) s s s s s s [GATE-2013-EE] 1+i 1-I 22. The argume nt of the complex n umber --. , where i = � . is (a) (c) -n n 2 (b) (d) n n 2 [GATE-2014-ME-SET-1] 23.* All the values of the multi-valued complex function 1 ;, where i = � , are (a) purely imagi n ary (b) real and non-negative (c) o n the u n it circle (d) equal in real a nd imagi n ary parts [GATE-2014EE-SET-2] 2-3i _5 + i , can be e x presed a s (a) -0.5 - 0.5i (b) -0.5 + 0.5i 24. z = (c) 0.5 - 0.5i (d) 0.5 + 0.5i [GATE-2014-CE-SET-2] az + b ' If f(z1 ) = f(�) for all 21 7' �. a=2, b=4 CZ + d a nd c=5, then d should be equal to___ 25.* Let f(z) = 1 9.* G ive n i = � , what will be the evaluatio n of the (c) --i = � , then the value of x' i s (a) e-n12 (d) ✓32 _ j2_ 1 5 . The product of complex num bers (3 - 2i) and (3 + i4) res ults i n x (c) ✓32 + l2 (d) - 20.* If Engineering Mathematics [GATE-2015-EC-Set-2] 26. Give n two com plex number z1 = 5 + ( 5 z2 = (a) 0 2 ✓3 (c) 60 + 2i, the argume nt of (b) 30 (d) 90 ✓3 ) i and 21 i n degree i s 22 [GATE-2015-ME-SET-1] IES MASTER Publication 424 Complex Number 27. G iven i = J::1, the value of the definite i ntegral, I = I (a) (c) ½ cos X + i sin X d X 0 . . COS X - I Sin X IS (b) -1 (d) ---i [GATE-201 5 (CE-Set 2)] 28.* Which one of the following options correctly describes the location of the roots of the equation s4 + s2 + 1 = 0 on the complex plane? (a) Four left half plane (LHP) roots (b) One right half plane (RHP) root, one LHP root and two roots on the imaginary axis (c) Two RHP roots and two LHP roots (d) All four roots are on the imaginary axis (GATE-2017 EC Session-I] 2 2 +1 29• For a complex number z, 2lim is ➔i z3 + 2z - i(z2 + 2) (a) -2i (c) (b) ---i (d) 2i [GATE-2017 EE Session-I] 30. Let z = x + jy where j = J::1. Then cos z = (a) cos z (c) sin z 31 . The real part of 6e (b) cos z (d) sin z 1s __ . in/3 . (GATE-2017 (IN)] [GATE-201 7 (CH)] E ng ineering Mathemat ics . 1. (c) 9. (d) 17. 3. (b) 11. (b) 19. 2. 4. I (c) 8. (c) Given (c) 15. (b) 16. x = R e(z) = (b) cos� = (b) cos �+isin� ⇒ e-i4> ei4> + e-i4> 2 = i log x = i l og i X = {i 1 og( 02 +12 )+ i tan-1 = cos � - isin� - 18. (a) 20. (a) 21. 22. 23. 24. ,• -• .,., 25. (10) 27. (c) 28. 29. (b) 30. (c) 31. (b) (b) < < (b) - 26. (d) · (a) (c) (d) (b) (3) "6'] Centre = Z0 "" ( x0 , y 0 ) ra dius r Now the equation of the given circl e is I z+ 1 1= 1 ( or) Centre Z0 -1 l z- (-1) I = 1 ra d 1 :. C entre = (-1, 0) and radius = 1 = = = = Sol-6: (c) Zl2 2 1 Z2 1 Z 1Z2 - Z1 . . -, ( gi ven) .1.e. ( a) I.s correct. = z22 2 2 2 Let lz 1 I =OA=BC & l z2I = OB = AC Now in the fol l owing fi gure, Im f(z) = ff f(z + T) = ez+T = ez .eT F or a periodic function ⇒ (a) where · [· 27t] = -27t f(z ,. � The general equation of circl e i s given by I z-z0 I= r log x = I 1 Sol-4: (b) • Sol-5: z +z 2 B y E uler Notation we know that Let (b) z = x + iy ⇒ z' = x - iy Sol-2: (d) Sol-3: (d) 14. (a) 7. (d) 13. (b) 6. (b) 12. (b) 5. Sol-1: 10. (d) ··_ : AN·SW.E R'-, ltE;Y · ·. ·:: ���-�--,. · ·-• -� . :.· :· • 425 + T) = f(z) for n = 1 eT = 1 = cos2nn+isi n2 nn eT = cos2n+ i sin2n = ei2rr :. P eriod of f(z) = i2n then, l z1 + z2I i.e. 1 2 1 1 + 1 22 1 .'.".. 1 2 1 + 22 1 = OC & l z 1 z2 1 = BA We know that in any triangl e sum of any two sides .'.".. the thi rd side - Al so, Difference of any two sides .'.5. the thi rd side IES MASTER Publication 426 i.e, Complex Number H ence, (c) is not true. Note that option (d) is true. Sol-7: (a) y ·.- I z -5 -5i I= 2 ⇒ Circl e of rad 2 & having centre at (5, 5) OC = So ✓2 5+ 25 = 5✓2 PC = 2 OP = OC -PC = 5✓2 -2 :. The minimum d istance from the origin to the circl e is 5✓ -2 . 2 Sol-8: (c) Let then Z = z Z = re i8 3 r3 e3ie = re-is ⇒ r2 e4ie = 1 or in Z = rei8 = 1.e2 = i So onl y option (c) satisfies it. We know that = So Sol-10: (b) So, Sol-11: (b) L et = cos2:+isin2: = .n ( /2 2 )i n = e-2 2 = -- +11 2 z = . --13 1 = 10 ⇒ eiz - = 2 0i 812 2i ⇒ a.2 -2 0ia -1 = 0, where a = eiz ⇒ ⇒ a, = -(- 2 0i)± -4 00+4 2 ✓ 2 0i± 6i✓1 2 0i± -J - 39 6 = = 2 2 1 eiz = 10 i± 3JIT. i or l oge (eiz ) = l og(1 Oi± 3✓1 i) ⇒ 1 iz = l og[i (10 ± 3✓11 )] ⇒ = l ogi+ l og(10 ± 3✓1 ) 1 ( -"-+2nn)i -"-i [·: l ogi = l oge e2 = l og1+ l oge e 2 - ) iz = l og1+{% ± 2 nTC) +109 (1 0 ± 3--./11) iz = {% ± 2 nTC) + l og(1 0 ± 3✓1) 1 z = (� ± 2 nTC) - i l og(10± 3✓1 ) ⇒ 1 where n = 0, 1 , 2 ... . :. sin z = 10 has infinite numb er of compl ex sol utions. Sol-13: (d) in e2 l eiz l = I ei( x+jy ) I = I ei x-y I = I e- Y I I ei x I = e-Y [· :I Z1 Z2 l=I Z1 1 - 1 Z 2 I] (·: I ei x I = I cosx + jsinx I = 1) (Using eul er notation) .n 4 2n . 2TC . . 2TC ( 1-6 ) -1 = 8 3 = C0S -+ISl nz = e 4 sin z = 10 ⇒ H ence, (d) Sol-12: (d) ⇒ 2ni r2 e4i8 = 1.e On comparison r = 1 & 0 = TC / 2 . Sol-9: = ⇒ X 0 0 (0, 0), C (5, 5) & PC = 2 (R ad iu s) So -15 +2 0i+ 30i+4 0 9 +16 -25 + 50i = 1+2 i = 25 I I 2 1 I - I 2 2 I I � 1 21 - 2 2 1 3 3 2 -5 + 1 0i _ (-5 + 1Oi)(3 -4 i) 3+4 i (- 3+4 i)(3 - 4 i) (After R ational ization) Sol-14: (b) x3 = j where j = � B y verification option (b) wil l satisfy the given equation x3 = j F or, Then x3 = H ence verified. TC . . TC , cos- + i sm - = 1 2 2 Engineering Mathematics Sol-1 5: (b) 1/2 X = [ cos( :n: )+isin(- %)] 2 (3 -2 i) (3+4 i) = 9 +12 i-6i-8 ( i) = 17 i+6i Sol-1 6: (b) Sol-1 7: (a) Given w = 1 (or) w = (1) unity. 113 1 X ⇒ w is the cub e root of 1+0) + - = 1+0) +00 = 0 0) Sol-18: (a) 2 (1 + i) (2 - 5i) = 2 - 5i + 2 i + 5 = 7 - 3i Sol-19: (d) We know that by E ul er Notation, So I= "12 X + i sinX e dx = f . dx f COS -, x cos x - isinx /2 rr/2 . 21x. 2ix e dx = [� r = 2i 0 0 0 0 e 1 . 1 -1 = 2' [ e"1 - 1]= 2'[-1-1] = --:-I -i =i = i2 I Sol-20: (a) L et 4 x = i 2 = ii 1 = {J 1 og(02+12 ) + itan- = ·I ·1 2 :n: [ ] Z = e n = 2 z ⇒ ⇒ Sol-23: (b) 2 4 4 Sol-22: (c) 1+i 1-i = 4 4 1+i 1+ i 1- i 1+i z = z = z z = -x- 1+2i+i2 (1+ i) = 2 f - ·I2 4 = ei rr/ 2 "" reie = 1° ii X = 1i = [ cos 2kn + i sin2k i Let :n: where k is an integer. i = [ ei2kn ] = e-2kn All values are real and non- negative. Sol-25: (10) (2 -3i) (2 -3i) (-5- i) = 2 2 (-5 + i) (-5) - i -1 0 -2 i+1 5i-3 = 25+1 -13+13i = -0.5+0.5i 26 a2+b c2+d f(21) = f(22 ), 21 etc 22 f(2) = a22+b a21 +b = C22 + d C21 + d 2 21+4 222 +4 = 52 1 + d 522 + d Method II x = � = (- i) 7t ¾] 4 3 . . 3 .. x = cos(- :n: ) +1si n(- :n: ) , cos( :n: ) +1si n( :n: ) Sol-24: (b) log 2 = i log i Sol-21: (b) I i x Taking log on b oth sides Let x = cos(4 n-1) �+isin(4 n- 1) � . n=0, 1 ix f 2 = [ cos(4 n-1) %+isin(4 n- 1)%J' , n = 0, 1 eie = cos0 +isin0 & e-i0 = cos0 - isin0 "12 2 X = [ cos(2n:n: - %)+isin(2n:n: -%)J' , n = 0, 1 5 3+4 i .Jg + 16 1 =✓ = 1-2 i + 4 ✓ l 1 3 So 2 427 11 2 (221 +4 )(522 + d) = (521 + d)(222 + 4 ) 1 02122+2 d21 + 2 022 +4 d = 102122 +2 021+2 d22 +4 d IES MASTER Publication 428 ⇒ ⇒ ⇒ Co mpl ex Numb er 2dz1 +20z2 = 20z1 +2dz2 Sol-28: (c) Given equation is, s4 + s2 + 1 = 0 2d (z1 -z2 ) = 20(z1 -z2 ) 2d = 20 d = 10 Sol-26: (a) Let R ational ize = 2 3 -+ 2'I ✓ 3 l- 2 � -10i+10i -10 ✓ i _± _ 4i2 3 3 Z2 ⇒ arg (::) = = 3 40 3 5✓ 3 x - =-2 ✓ 16 0 = 5+5,,J3 i = 2 3 +2·I ✓ arg z1 = tan- 1 ( Z2 f) 5 = 60° 3 arg (:: ) = arg (z1 ) - arg (z2) = 60° - 60 ° = 0 Sol-27: (c) We know that by E uler N otation, So eie = cos8+isin8 e-ie = cos8 - i sin8 . . n/ 2 n /2 i x cosx+1sm x d x = � dx = cosx - isinx -, x 0 O e f = = n /2 f 0 s2 = s = s & s s & /2 . 2i x e2, x dx = [� r 2i 0 0 3 ✓ = co jm co CO2 = ±co112 ±co = ni ± e3 = -{J) 2ni ±e3 = i.e., equation has two right half pl ane (RHP) roots and two l eft hal f plane (LH P) roots i. e., option (c). z2 +1 . I 1m----3 z➔i z +2 z - i(z2 +2 ) This is % form, so on d ifferentiating b oth numerator 2 1 ° 3 ) = tan- ( ✓ ) = 60 arg z2 = tan- 1 (-2/✓ & & s2 y -1± 1-4 -1 .✓ = -± J 2 2 2 2 ni co = e 3 Sol-29: (d) Alternative Solution: z1 = = = & CO2 where so 3 _!Q_3 +10✓ � is a real numb er 2i ) ( 1 +2 i) ( 1 - 2 i) = ✓ -+4 3 y ⇒ ( 5+5✓ i) ( = y2 + y + 1 then 3 � = 5+5✓ i Z2 s2 f -i 1 . 1 -1 =i z [ e"' - 1] = 2" [-1-1] = --:-- = I I I i2 and d enominator 2z . = I 1 m- 2 ➔i +2 -2zi 3z z Sol-30: (b) = _ _ = Sol-31: (3) _ 2i_ _ 3 (i2 ) +2 - 2 ( i) i [ = __2_ i_ = 2i -3+2+2 1-� +� -.. , -2 -4 = CO S (z ) l in re . . re ] = 6 cos3 re +1. ( 6 . re ) 6e3 = 6[ cos3 +1sm 3 sm 3 1 re R eal part = 6 cos- = 6 x - = 3 . 2 3 NEIGHBOURHOOD OF COMPLEX NUMBER z0 A set of all points of z such that lz - z0 1 < 8 where 8 is an arbitary small positive number, is called a neighbourhood of the point zO . The number 8 is called the radius of neighbourhood of zO . Deleted Neighbourhood of point z0 If the point z0 itself is excluded from the neighbourhood (nbd) then such a nbd is called the deleted nbd of z0 . Domain: A set of points is said to be domain (or region or connected) if any two of the points can be joined by a continuous curve which consists only the points of the set. If the boundary points of the set are also added to an open domain then it is called a closed domain. FUNCTIONS OF A COMPLEX VARIABLE Let S be a set of complex numbers. A function f is defined on S is a rule that assigns to each z in S a complex number w. The number w is called the value of f at z and is denoted by f(z) that is w = f(z). The set S is called the domain of f. If w = u + iv is the value of a function f at z = x + iy then w = f(z) ⇒ u + iv = f(x+iy) Here, u and v depends on the x and y and thus f(z) can be expressed in terms of a pair of real valued functions u and v as follows. in case of polar coordinates f(z) = u (x, y) + iv(x, y ) w = f(z) ⇒ u + iv = f (rei0 ) . . . (1 ) (where w = u + iv and z = rei8 ) so in this case f(z) can be expressed in terms of pair of real valued functions as follows. Example 1 : Solution : 2 If f(z) = z , then f(z) = u(r, 0) + iv (r, 0) So using cartesian coordinates Hence, f ( x + iy) = ( x + iy ) = x2 - y 2 + i2xy u (x, y) = x2 - y 2 and v (x, y) = 2xy When polar coordinates are used Consequently, 2 . . . (2) f (rei0 ) = (rei0 ) = r 2 ei2 8 = r2 cos20 + ir2 sin20 2 u (r, 0) = r2 cos20 and v (r, 0) = r 2 sin20 430 Analytic Functions Path of a Complex Variable In case of a function f(x) of a real variable, the independent variable x takes val ues corresponding to the points lying on the x-axis. But in the case of the function f(z) of a complex variable the i ndependent variable z takes val ues lying in a plane (called the z-plane) so path of th� complex variable z is either a straight l i ne or a curve on z plane or xy plane. TYPES OF COMPLEX FUNCTION If w = f(z) takes only one value for one value of z i n the region D , then f(z) is a single val ued function of z. If there corresponds two or more val ues of w for one value of z in the region D, then w is called as m ultival ued function of z. For example, we know that z112 has two values, while z2 has only one val ue. so w = z112 is m ulti val ued and w = z2 is single val ued function. Log of a complex number Let z = --�if 1 a + ib be any complex number then log z INOTE, = log re ie = log r + i0 co is said to be a l ogar ithm of z if one can write z = e"' l og z So, = co or log z = = log l zl + iAmp(z) ⇒ z = e• +2nni 1 (·: e2nni = 1) co + 2n1ti It means log of a complex number has i nfinite no. of values and is therefore a mult i-valued function. ,. The general value in written as log z and it's princ ipal value is obtained by taking n = 0 i n l og z. !log z = 2nni + logzl i.e. 2 . If we take y = 0 in above equation we have logx = 2n1ti + logx it means log of a real quantity is also a multivalued. log(-2) = log(2) + log(-1) = log 2 + Iog(e ni ) = log2 + i1t 3. COMPLEX MAPPING / TRANSFORMATION Let co = f(z) or u + iv = f(x + iy) be a complex function, then if a point P(x, y) moves along a curve C i n z plane, and poi nt P'(u, v) moves along a curve C' i n co pl ane, then we say that curve C in xy plane i s mapped i nto corresponding curve C ' in uv plane by m apping co = f(z). Hence to plot graph of com plex function we need two planes known as XY plane and UV plane V y -----,Oc+----- X z plane Example 2 = Solution = 1 Find the i mage of i nfinite strip under co = z Let z = x + iy, and co = u + iv, then -----,,o----- u 1 4 1 2 (a) - < y < - w plane 1 (b) 0 < y < 2 co ⇒ = z= co 1 z ⇒ 1 4 -v 1 ⇒ -2 2 > 1 431 1 u-iv -v u - 2 and y = -⇒x=+ iy= --=-2 2 2 2 u+iv u + v u +v u +v 2 X y > (a) again Engineering Mathematics y y < or 4 u +v -v 1 ⇒ -2 < 2 2 or u2 + (v + 1 )2 2 u +v u2 + (v + 2)2 < 4 (interior of circle) > 1 (exterior of circle) V ��..,...,....j'-,-,-,�� y=1 /2 ���.........��� y= 1 /4 --+ x ---�1---z- lane (b) when y > and when y 0 0 -v -- 0 ⇒ ⇒ u2 + v 2 > 1 2 < -, 4 p 1 -v -<⇒ 2 u +v 2 2 v ⇒ < w- plane 0 u2 + (v + 1)2 > 1 (exterior of circle) V /1/&ff,W11l/;1/ Y:"' y=O Example 3 1 Find the image of iz - 2il = y 1 2 under the mapping co = z V -+----- u ____0 ------ v=-1/4 - i/4 Solution 1 Given curve is jz - 2i)= 2 -v u2 - -- 2) = 4 +( 2 2 2 2 u + v2 (u + v ) 2 Example , 1 Solution 1 ⇒ Ix + iy - 2il= 2 ⇒ 1+ 4v -- = 0 ⇒ ⇒ 2 u v2 + 2 x2 + (y - 2)= 4 -1 V= 4 Detemine the region of co -plane into which region of z plane bounded by the st. lines x= 1 , y= 1 and x + y= 1 is mapped by transformation w= z2_ Case I : When x= 1, W= z2 => u + iv= (x + iy)2 => u= x2-y2 & v= 2xy U= 1 - y2 & V= 2y u= 1 - v2 4 => 2 - 4 (u - 1) v= ... (1 )(Parabola opening leftward) IES MASTER Publication 432 Analytic Functions Case II : When y= 1, u = x2 - 1 &v = 2x v2 . . . (2) (Parabola opening rightward) -1 v2 = 4(u + 1) 4 Case I l l : When x + y = 1 , u = x2 - y2 =(x - y) (x + y) =(x - y). 1 =x - y and v = 2xy ⇒ u = * (x - y)2 = (x + y)2 - 4xy and u2= -2 ⇒ u2 = 1 - 2v (v -J) . . . '(3) (Parabola opening downward) w- plane z-plane CONTINUITY OF COMPLEX FUNCTION A function f is continuous at a point z0 if all the following three conditions are satisfied. (i) lim f(z)exists (ii) f(z0)exists Z➔Zo (iii) lim f(z) = f(z0 ) Z➔Zo In other words we can say that the function f(z) of a complex variable z is said to be continuous at the point z0 if, given any positive number e , there exists a positive number o such that it(z)-f(zo)I < e whenever lz - z0 I < o A function of a complex variable is said to be continuous in a region R if it is continuous at each point in R. If two functions are continuous at a point, their sum and product are also continuous at that point. The quotient is continuous at any such point where the denominator is not zero. DIFFERENTIABLILITY OF COMPLEX FUNCTION Let f be a function whose domain contains a neighbourhood of a point z0 . The derivative of f at z0 is defined as, written f'(z0 )= lim 2➔2o f(z)- f(z0 ) Z - Zo . . . (1) provided this limit exists. The function f is said to be differentiable at z0 when its derivative at z0 exists. If we define a new variable t,;z as t,;z = z -z0 then definition (1) of differentiability can be written as f'(zo) = lim f(zo +ti.z)-f(zo) 62➔0 8.Z . . . (2) If we take z0 as general point z and introduce a number b.w= f( z + b.z)- f(z) and write �: =f'(z) then equation (2) becomes For e.g . , f(z) = z2 then f'(z)= lim b.w 62 ➔0 8.Z f (z + b.z) - f (z) dw . • b.w ] f'(z) = - = 1 1m - = 1 1m [ 62➔0 t,;z dz 62➔0 b.z = lim 62➔0 ( z+b.z )2 - z2 8.Z = lim (2z+ti.z.) = 2z 62➔0 . . . (3) Example 5 : Solution = Consider a function f(z) = lzl Here, 2 8 82 Engineering Mathematics . show that it is differentiable only at origin. lz+ 821 - Jzl 82 W = ---� = 2 8w Now, we have to find the lim 82➔0 82 2 433 (2+82)(2+82 ) - 22 82 = 2+82+2 82 82 Firstly let us take the case when 82 ➔ 0 through the point (�x.O) on the real axis. Then on the real 8w - = 82 . Hence 1 · m -= · 82 . z+ z on the rea I axis. axis I 82➔0 82 Secondly, let us take that 82 ➔ 0 through the point (0, 8y) on the imaginary axis. Then we have for imaginary axis 82 = -82 . . . . . 8W Hence II m -= z - z on the 1magmary axis. 82➔0 82 Since the limit are unique, we must have 2+ z=2 - z ⇒ !: !: z = o for the existence of Therefore, we conclude that of z = 0. exists only at the point z = 0 and nowhere else in the neighbourhood ANALYTIC FUNCTIONS / REGULAR FUNCTION / HOLOMORPHIC FUNCTION A function f of the complex variable z is analytic in an open set if it has derivative at each point in that set. If a function f is analytic in a set S which is not open, then f is analytic in an open set containing S. Analytic at a point z0 f is said to be analytic at z0 if it is not only differentiable at z0 but also differentiable in the neighbourhood of z0 . 2 For example, the function f(z )= 1/z is analytic at each non-zero point in a finite plane. But the function f( z) = lzl is not analytic at any point since its derivative exists only at z = 0 and not in the n bd of any point including z = 0. Entire Fundion An entire function is a function that is analytic at each point in the entire z-plane. For example, every polynomial is an entire function since the derivative of polynomial exists everywhere. Bounded Fundion A function f(z) analytic in a domain D is said to be bounded if there exists a number M > 0 such that lf(z)I � M V z e D. Liouville's Theorem If a function f(z) is analytic for all finite values of z and is bounded, then it is constant. or If f(z) is regular in the whole of the z-plane and if lf(z )I � M Vz, then f(z) must be constant. or If an entire function is bounded for all values of z, then it is a constant function. Singular Points If a function f fails to be analytic at a point z0 but is analytic at some point in every negihbourhood of z0 , then z0 is called a singular point or singularity of f. For example ( 1 ) The point z= 0 is a singular point of the function f(z) = 1/z (2) The function f(z ) = lzl 2 has no singular points since it is no where analytic. IES MASTER Publication 434 Analytic Functions INOTE5 1. If two functions are analytic in a domain D, their sum and product are both analytic in D. Simi larly, their quotient is analytic in D provided the function in the denominator does not vanish at any point in D. For instance, the quotient p(z)/Q(z) of two polynomials is analytic in any domain throughout which Q ( z ) *0. Again, a composition of two analytic functions is analytic. For example, consider a function f(z) which is analytic in a domain D. Then the composition g{f(z)} is analytic in D, with derivative. d g [ f ( z)] = g' [ f ( z )] f' ( z) dz 2. If f(z) & g(z) are analytic then f(z) + g(z), f(z). g(z) are analytic but f(z)/g(z) is not analytic. 3. If f' {z) = 0 everywhere in a domain D, then f(z) must be constant throughout D. CAUCHY RIEMANN EQUATIONS Let the function f(z) = u (x, y) + iv(x, y ) be defined throughout some neighbourhood of a point z0 = x0 + iy 0 . Suppose that the first order partial derivatives of the functions u and v with respect to x and y exist everywhere in that neighbourhood and they are continuous at { x0 , Yo ) . Then, these partial derivatives satisfies the Cauchy-Riemann equations at ( x0 , Yo ) as follows: ux = Vy , au = or ax av ay ' U y = -V x au ay ___.{f INOTE:, If f(z) is an analytic function then f ' (z) = =u x ax + ivx =v y - iuy Example 6 : Suppose that f ( z) = ex (cos y + i sin y ) where y is taken into radians. Show that f' ( z) exists everywhere Solution : From given value of f {z) we have and find its value. Now u(x, y) = e x cos y and v (x, y ) = e x sin y u x = ex cos y = v y and v x = ex sin y = -u y Therefore Cauchy Riemann equations are satisfied now since those derivatives are continuous everywhere therefore f' ( z) exists everywhere and Example 7 : Solution : f' ( z) = u x + iv x = ex {cos y + i sin y) = f { z) Given that function f ( z) = z2 = x2 - y 2 + i 2xy show that f(z) is differentiable everywhere and verify that Cauchy Riemann equations are also satisfied everywhere. For the given value of f ( z) we have u(x, y) = x2 - y 2 , v (x, y) = 2xy ⇒ ux = 2x = vy and Uy Hence Cauchy Riemann equations are satisfied. Also we know, that f' ( z) = u x + iv x = 2x+ i2y = 2z = -2y = -Vx Engineering Mathematics Example 8 = Solution = 435 Consider a function f(z)= lzl . Show that f(z) is not differentiable for z 0 . 2 * f(z)= lzl = x 2 + y2 Given that 2 = x2 + y2 and v(x, y)= 0 . If Cauchy Riemann equations are satisfied, then we must Here u(x, y) have -u y ⇒ 0 = -2y ⇒ y = 0 and v= x v y ⇒ 2x=0 ⇒ x=0 u= x Hence Cauchy Riemann equations are satisfied for x= y= 0 only and not for z 0 . Consequently, Example 9 = Solution : Example 1 0 : Solution = * f'(z) does not exist for z * 0 . The function f(z) = lzl has a derivative at z= 0 then find it 2 ·: u x= v y and v x= -u y at (0,0) ⇒ C-R equations are satisfied at origin. ⇒ f'(0) exists and f'(0) = u x + iv x = 0 + i0= 0 Show that the function 2x + ixy2 is nowhere analytic. Given that f(z)= 2x + ixy 2= u(x, y)+ iv(x, y)(say) Here u x *V u =- = 2 au ax X y and v x * -u av 2 V x= -= y , u =-=0 ay y au ' y ⇒ ax u(x, y)=2x, av = 2xy V Y=ay v(x, y) = xy2 i.e. Cauchy Riemann equation is not satisfied anywhere. Therefore, the given function is not analytic anywhere. Example 1 1 : With the help of Cauchy Riemann equations, show that each of the function is nowhere analytic Solution = (b) f(z)= 2xy + i ( x2 -y 2 ) (a) f(z)= xy + iy (a) Let f(z)= u(x, y)+ iv(x, y) Then u(x, y)= xy and v(x, y)= y Since u and v are polynomials in x and y, they are continuous at each point. Hence f(z) is everywhere continuous. Further, we have Hence : ( b) Here y, -= X ay ' au f(z)= 2xy + i(x2 - y2 ) u(x,y)= xy, v(x,y)= x2 - y2 Differentiating partially, we get Hence dv av =0 =1 ay ' dx * : and : * -: . So that Cauchy Riemann equations are not satisfied. Hence f(z) is not analytic. Given that au= ax = y, au ax ;= x, av -=2X ax , av -=-2y ay au av au av * , so that Cauchy Riemann equations are not satisfied. and * ax ay ay ax Hence f(z) is not analytic. IES MASTER Publication 436 Analytic Functions POLAR FORM OF CAUCHY-RIEMANN EQUATIONS We know that the coordinate transformation from ( x, y) to (r, 0) is given by x = r cose, y = r sin 0 , then C-R equation are defined as follows. au = 1 av r ae ar ... (1 ) au = av -r - and ae ...(2) ar Equations (1 ) and (2) are Cauchy-Riemann conditions in the polar form. Derivative of w in Polar Form We know that for w = f ( z) dw = dz ow az . av sin e aw ar aw ae aw cos + = 8r = 8 S a0 + l a0 ) -r- ( : . W = U + iv) r ax 80 ax (au ( av . au sin e aw = 8rcos 0 - -r + ir ) -- (from Cauchy-Riemann equation) r ar ar . . aw . av . aw = - cos 0 - 1 - + 1- ) sm 0 = -cos e - 1-sm e ar ·(au ar ar ar -i0 aro = e ·­ ar Another way to find the derivative is dw = dz ow ar = ( aw � + aw ae ar ax ae ax au ar +i av ar ) cos e - aw sine ae r -i ia aro . . i aw = -- ( cos 0 - i sm e) - = - • e - · r ae r ae This is also polar form of differentiation. Thus we have dw aw -i . . . . - = ( cos 0 - i sm e ) - = - ( cos 0 - i sm e) ae dz ar r Example 1 2 1 Consider the function f ( z) = ..!. = z re Solution 1 Given that Here f(z) = ow -ie- . Find f' ( z) if it exists. 1 1 cos 0 - i si n 0 r cose -sin e u,(r,0) = --, v ( r, e) _- -r r Since the first order partial derivatives of the function u and v with respect to any point (r, 0) exist so these partial derivatives satisfies the polar form of a Cau chy-Riemann equation. i.e e = -Va and vr = 2 s1n e = -Ua u r = -Cos r r r r2 -1 1 1 . -1 Engi neer ing Ma thema tic s 437 He nce gi ve n func ti on i s di ffe re nti ab le at any non ze ro poi nt z = rei0 and i ts de ri vati ve i s gi ve n as Example 1 3 : - ie f' (z) = (c os8 - isi n8): = e-i8 (ur +ivr ) = ;: e- ie = :� If n is re al, show that rn (c osn8 + i sinn8 ) i s anal y tic except possib l y whe n r = 0 and that i ts de rivati ve i s nr0-1 [c os(n-1) 0 +i si n(n-1)0] . av w = r" (cos n0+i si nn0) = f( z) = u+i v Let Solution 1 u = r" c osn0, So he re v = rn si nn0 au n . = nrn -1 c os n0 , - = nr -1 sm n0 ar ar au av n . = -nrn sm n0, - = nr c osn8 as ae the n av au 1 a v 1au T hus we see that - = -- and --= -as r ae r ae ar T hat i s, Cauc hy-Rie mann c ondi ti ons are sati sfied. dw He nce the func ti on w = r" (c osn0 +i si nn0) i s anal y ti c , i f dz exi sts for all fi ni te val ue s of z. aw dw = (c os8-i si n8) - nrn-1 (c osn8 +i si nn8) = (c os0-i si n0) dz ar = nr"-1 [ c os(n-1) 0 +i si n(n-1)0] We have Theorem Proof = T his exists for all fi nite value s of r i nc l udi ng ze ro, except whe n r = 0 and n � 1 . 1 Le t f(z) be analytic i n a d omai n D, i f f(z) i s also analytic i n D. P rove that f(z) red uce s to a c onstant i n D. Gi ve n that f(z) i n analytic. f(z) = u + i v Le t T he n, by Cauc hy Rie mann e quati ons ...(1) au av av au = - and - - ay ay ax ax f(z) = u-i v = U +i(-v) Also av} ... (2) T he n f(z) to be analytic , we must have by Cauc hy Rie mann e quati ons. au _ -- av ax au = ay Comb i ning (2) and (3), we have ⇒ av av ax - = 0 and -= 0 ay ay ⇒ = a(-v) _ ay - - -- ay a(-v) ax -av ay = av ax and ... (3) -av av -= ax ax v = c onstant. Si mil arly we c an show that, u = c onstant. T hus f(z) =u+i v is a c onstant func ti on i n D. IES MASTER Publication 438 Analytic Functions ORTHOGONAL SYSTEM Two families of curves u (x, y) = c1 & v (x, y) = c2 are said to form an orthogonal system if they intersect at right angles at each of their point of i ntersection. Theorem Real and imaginary parts of an Analytic function together constitute an Orthogonal system, i.e., if f(z) = u + iv is an Analytic function then u = u(x, y) & v = v(x, y) intersect at 90° . Proof : Let By total derivative concept, U(x,y) = c 1 and v(x,y) = c2 du = d(c 1 ) and dv = d(c2 ) au / ax dy - - au / ay dx i.e., u So we- can take m 1 = - _x and m 2 = -v /vy Uy and dy dx -- av I ax av I ay Now Hence u and v together constitute an orthogonal system. ----� INOTE :) Velocity Potential qi and stream function 'I' together constitute an Analyti c function known as complex potential co , i.e., co = qi+ i 'I' is an Analytic Function. Here, qi and 'I' are orthogonal to each other because they are the real and i maginary parts of analytic function ' co' . And the relationship between qi and 'I' is nothing but the C-R equations i.e., 84> = 8"' & 8$ = - 8"' ax ay ay ax HARMONIC FUNCTIONS A real valued function H of two real variable x and y is said to be harmonic if it has continuous partial derivatives of the first and second order and satisfies the partial differential equation. Hxx ( X, y) + H yy ( x, y) = 0 ... (1 ) known as Laplace's equation, i.e., those function that satisfies L. Equation are called Harmonic functions. Theorem : Proof : If f ( z) = u + iv is an analytic function then u and v both are Harmonic functions. Let f ( z) = u + iv be an analytic function then we have = au -av = ay ax : :} ... (1 ) Cauchy-Riemann Equations Also because u and v are the real and imaginary parts of an analytic function, therefore derivatives of u and v, of all orders exist and are contunuous functions of x and y so that we have Engineering Mathematics a 2v = a2 v . . . (2) axO'/ O'/ax Partially Differentiating eq u ation ( 1 ) w.r.t. x and y respectively. a2 v a2 u a2 u a2 v and 2 2 O'fax O'fax ay ax 2 2 au au Adding these, we have + - = 0 by (2) ax2 0'/2 - = -- 439 = Similarly -----� INOTE, Hence both u and v satisfy Laplace's eq uation 1. Converse of the above theorem is not necessarily true. 2. Such functions u and v are cal led conjugate harmonic function or simply harmonic functions. In other words we can define conjugate harmonic function as If two functions u and v are harmonic in a domain D and their first order partial derivatives satisfy the Cauchy Riemann equations throughout D, then v is said to be a harmonic conjugate of u. 3. If for a function f ( z) = u + iv, v is a harmonic conjugate of u then it is not necessary that u is a harmonic conjugate of v, As for example, consider the function f(z) = z 2 . f(x, y) = x 2 - y2 + i(2 xy) = u + iv ⇒ Here, u(x, y) = x 2 - y2 and v(x, y) = 2 xy Since, these are components of the entire function f(z) = z 2 therefore v is harmonic conjugate of u throughout the plane. But since 2 xy + i(x 2 - y 2 ) is not analytic anywhere, u can not be a harmonic conjugate of v. 4. Theorem : Proof : If u & v are harmonic functions such that u + iv is an analytic function then u & v are cal led harmonic. conjugate of each other. If f ( z) = u + iv is an analytic fu nction and u (x,y) is given then find the harmonic conj ugate v ( x, y) of the fu nction of u ( x, y) Since v is a fu nction of x,y therefore av av au au dv = -dx + -dy = --dx + -dy by Ca uchy Riemann eq u ations O'/ ax ay ax au au dv = --dx + - dy ay ax Th u s . . .(1 ) The right hand side of eq u ation ( 1 ) is of the form Mdx + N dy where So that . au au M = -- and N = O'/ ax : = :i ;u ) = �;:u and : = ! ( !� ) = �� a2u a2u is h armonic fu nction, therefore i t satisfies Laplace's equation, i . e . , + - = 0 or ax2 0'/2 a2 u -a2u aM aN = 2 which makes - . 2 O'/ ax ax 0'/ Si nce u IES MASTER Publication 440 Analytic Functions Example 1 4 : Solution : Hence equation (1) satisfies the conditions of exact differential equation. So equation (1) can be integrated and thus we can find v. Prove that u = y3 - 3x2 y is a harmonic function. Determine its harmonic con jugate and find the corresponding analytic function f(z) in terms of z. u = y3 - 3x2y We have So that au - = -6xy, ax au a 2u 2 2 =-6y - = 3y -3x & ' ax2 ay that is, u satisfies Laplace's equation, hence u is a harmonic function. Let v be the harmonic con jugate of u, then we have -au au av av dv = -dx+-dy = - dx+-dy (by Cauchy Riemann equation) ay ax ay ax 2 2 2 2 = -(3y - 3x ) dx - 6xydy = - (3y dx+6xydy)+3x dx Integrating we have v = 3xy2 + x3 + C This is the required harmonic con jugate of u. Example 1 5 : Solution : Now f {z) = u+iv ;= y 3 - 3x2 y+i(-3xy2 +x3 +C) = i(x+iy} +iC = iz3 +iC 3 Show that the function u=-ilog( x 2 + y2 ) is harmonic and find its harmonic con jugate. We have and so that u = -i1og (x2 +y 2 ) X au = 2 x + y2 ax au . Y_ = _ ay x2 +y 2 ⇒ ⇒ a 2 u a 2u y 2 - x2 += 2 ax2 ay 2 (x2 +y 2 ) Hence u is a harmonic function. y2 - x 2 x2 + y2 - 2x2 a 2u = = 2 ax2 (x2 + y2 ) ( x 2 + y2 ) x2 +y 2 _ 2y 2 a 2u X 2 -y2 = = 2 2 ay 2 (x2 +y 2 ) (x2 +y 2 ) (y 2 - x2 ) (x2 +y2 ) =0 Let v be the con jugate harmonic function. av av -au au -dy = - dx +-dy (by Cauchy Reimann equation) We have dv = -dx+ ay ax ax ay xdy - ydx -y X -- dy = -- dx += 2 2 X +y X 2 +y 2 X 2 +y2 Example 16 : Solution : Integrating, we get v = tan-1 1+ C where c is a real constant. Prove that, if u= x2 -y2 , v =-y/( x 2 + y2 ) both u and v satisfy the Laplace's equation, but u+iv is not an analytic function of z. We have u = x2 - y2 , v = � x2 +y 2 ⇒ au = 2x, ax au -=-2y ay av ay ⇒ - (x2 +y2 ) +2y2 = (x + y 2 2) (y2 - x2 ) 2 av = and 2 2xy (x +y2 ) 2 441 2 2 y (y2 -3x2 ) a2 v 2 y (x2 -3y2 ) a2 v a 2u = = -2, 3 3 ' a y2 = ay2 ax2 (x2 +y2 ) (x2 +y2 ) = 2, 2 a2 u ax = 2-2 = 0 So -+ax2 ay2 and B ut we notice that and a 2u 2) (x +y 2 E ngi neeri ng Mathemati cs a2 u 2 y (y2 - 3x2 ) - 2y (y2 - 3x2 ) a2 v a2 v -�--�-�-� = 0 + = 3 ax2 a y2 (x2 +y2 ) He nce b oth u and v sati sfy Lapl ace' s e quati on. au ax * av ay that i s Cauchy R ie m ann condi ti ons are not sati sfie d. -av * a y ax au He nce u+i v is not analyti c function. Example 1 7 : Show that u(x, y ) i s harm oni c i n some dom ain and fi nd the harm oni c conjugate v(x, y), i f U ( X, y )= Solution : y/( X 2 + y2 ) . He re u= y/x2 +y2 ax2 = -2y ( x +y ) 2 2 + 2 au = ax ⇒ 2 xy · 2 (2 x ) (x2 + y2 ) 3 = 2 -2 yx2 -2 y3 +8 x y (x2 +y2 ) 3 x2 +y2 _ 2y2 1 y•2 y au . . = S1m 11 ar1 y - = 2 2 2 2 ay x +y (x2 +y2 ) (x2 + y2 ) +- =0 Addi ng (1) and (2), we ge t ax2 ay2 a2 u a2 u 6x2 y - 2y 3 (x2 +y2 ) x2 _ y 2 (x2 +y 2 ) 3 2 and a2 u = 2y3 -6yx2 a y2 (x2 +y2 ) 3 ... (1) ....(2) He nce u is a harm oni c functi on. N ow, le t v be i ts harm oni c conjugate. au dy (by Cauchy Riem ann e quati ons) dy = -- dx +dv = - dx +ay ax ay ax The n av av (x2 - y2 ) dx = (x2 +y2 ) au 2 xydy (x2 + y2 ) 2 y2 dx -2 xy dy -x2 dx (x2 + y2 ) 2 v = x/ (x2 +y2 ) +c ⇒ 2 = (y2 - x2 ) dx -2xydy -x ) 2 x +Y2 = d( (x2 +y2 ) 2 Example 1 8 1 Show that the functi on u(x, y )= x3 - 3xy2 i s harm onic and fi nd the corre spondi ng analy tic functi on. Solution : Give n that ⇒ 3 2 u = x - 3xy au a2 u 6X = 3x2 -3y2 , -= ax ax 2 and =-6xy, ay2 = -6x ay au a2 u IES MASTER Publication 442 Analytic Functions a 2 u + a2 u ax2 ay 2 Now = 6x + (-6x) = o ⇒ u is a harmonic function. f' (z) = � (u + i v) = u x + i v x = u x - iu y (by Cauchy Riemann equation) Now, ax = (3x2 -3y 2 ) - i (-6xy ) = 3x2 -3y 2 + 6xyi = 3(x2 - y2 + 2ixy) = 3z2 f(z) = z3 + C ⇒ CONSTRUCTION OF AN ANALYTIC FUNCTION we have two methods Method I : (Using total derivative concept) in this method if u is given we will try to find its harmonic conjugate v by using total derivative concept and C - R equations as follows Let u is given then by C-R equation u x = v Y and uy = - vx v = v(x,y) so by using total derivative concept Now by integrating, we can find v. so analytic function can be given as f(z) = u + iv Method II : (Milne Thomson method) : With the help of this method we can directly construct f(z) without finding the other function when one of the part (either u or v) is given: Case I: (When real part 'u' is given) Step 1 : Find au <!>1 (z, 0 ), Step 2 : Find au Step 1 : Find 8v <!> (z, 0 ), Bx � 2 Step 2 : Find 8v � <!>1 (z, 0 ), By Bx � By Case II: (When only imaginary part 'v' is given Example 1 9 : Solution : Find an analytic function f(z) = u + iv if it's imaginary part is v = 3x2 y - y3 . I maginary part is given so by using case I I of MILNE THOMSON method. Step 1 . av = 6xy "' <P2 ( X, Y ) Step 2 : <!> 2 ( z, 0) = Step 4 : <!>1 ( z, 0 ) = 3z2 Step 3 : Method II : � <!> 2 (z, 0), Step 5 : av - 0 = 3x2 - 3y 2 "' <j> 1 (x, y ) f(z) = = J[ <!>1 (z, 0 ) + i <j>2 ( z, 0 )] dz + c J[3z 2 + i (0 ) ] dz + C = z3 + C f(z) = u + iv so f'(z) = u x + i v x Engineering Mathematics 443 i.e. So on integrating Method Ill f'( z) = V y 2 2 + iv x =(3x -3y )+i(6 xy) 2 2 2 = 3 ( x = y +2ixy) = 3z f(z) = 3z2 u = u(x,y) so by using total derivative concept. du = ( = + C au au ax ) dx + (ay ) dy ( av ) ( avax ) ay dx + - dy 2 2 = (3x - 3y )dx-6xydy 2 2 = 3x dx - 3 (y dx + 2xy dy) On integrating Hence 2 du = 3x dx -3d ( xy 2 ) u = x3 -3 xy 2 +c f(z) = u + iv 3 2 2 3 = ( x -3 xy )+i(3x y - y )+c = z3 + C IES MASTER Publication 444 Analytic Functions PREVIOUS YEARS GATE & ESE QUESTIONS Questions marked with aesterisk (*) are Conceptual Questions z -1 = - 1 . * The bilinear transformation w z +1 (a) maps the inside of the unit circle in the z-plane to the left half of the w-plane (b) maps the outside the unit circle in the z-plane to the left half of the w-plane (c) maps the inside of the unit circle in the z-plane to right half of the w-plane (d) maps the outside the unit circule in the z-plane to the right half of the w-plane = u + iv = � l;g( x2 2.* The function w is not analytic at the point. (a) (0, 0) 3. * (c) (1, 0) (b) (d) [GATE-2002 (IN)] + y 2 ) + itan-1 (t) (0, 1) (2, a) (b) set of concentric circles (c) set of confocal hyperbolas 4. (d) 2x2y2 (c) x2 - y2 [GATE-2007-CE] 6 . * A complex variable z= x + j(0.1) has its real part x varying in the range - oo to oo . Which one of the 1 following is the locus (shown in thic k lines) of z in the complex plane? (I) C (a) ·gE �X - Real axis (I) \ [GATE-2005 (Pl)] = In Z (where, For the function of a complex variable W W= u + jv and Z= x + jy), then u = constant lines get mapped in z-plane as (a) set of radial straight lines (d) set of confocal ellipses 5. * Potential function � is given as �= x2 - y2 • What will be the stream function ( 'I' ) with the condition 'I'= 0 at x= y= 0? (a) 2xy (b) x2 + y2 -j 1 0 Imaginary axis (b) 0 Real axis [GATE-2006-EC] If �( x,y) and 'l' ( x, y) are functions with con-tinuous second derivatives, then <I> (x, y) + i 'I' (x, y) can be expressed as a.n analytic function of x + i y(i = �) when (a) (b) a� O\Jf � O\Jf = = ax ax ' ay ay a� · O\Jf a� a'V = = ay - ax · ax ay a 2� a 2 'l' a 2 'l' a 2 'l' + - = - + -=1 (c) ax2 ay 2 ax 2 ay 2 (d) a� + O\Jf = O\Jf + O\Jf = 0 ax ay ax ay (c) � -­ C ·5i Cl) Real axis -1 0j (d) [GATE-2007-CE] [GATE-2008 (IN)] 7. 8. An analytic function of a complex var iable z = x + iy is expressed as f(x) = u(x,y) + i v(x,y) where i= � . If u = xy, the expression for v should be 2 x2_ y 2 (x+ y) +k (b) -(a) +k 2 2 y 2 - x2 +k (c) -2 (d) [GATE 2009-ME] J=1 and f(x + iy) is an analytic function then $(x, y) is (c) (b) 3x2 y - y 3 (d) xy - y2 x 4 - 4x 2 y [GATE-2010 (Pl)] For an analytic function, f(x + iy)= u(x, y) + iv(x, y), u is given by u = 3x2 - 3y2. The expression for v, considering K to be a constant is (a) 3y 2 - 3x2 + K (c) 6y - 6x + K (b) 6x - 6y + K (d) 6xy + K [GATE-2011-CE] 10.* A point z has been plotted in the complex plane, as shown in figure below Im Im Unit circle (c) Unit circle (d) [GATE-2011 -EE] 11. An analytic function of a complex varia ble z= x + iy is expressed as f(z)= u(x,y) + iv(x, y), whe r e i = I f u( x , y) = 2xy, then v(x, y) must be (a) x2 + y2 + constant n. (b) x2 - y2 + constant (c) -x2 + y2 + constant (d) -x2 - y2 + constant 1 · . 1 Ies m the curve Im (a) (b) Unit circle [GATE 2014-ME-Set-3] 12. An analytic function of a complex variable z = x + i y is expressed as f(z) = u(x, y) + iv(x, If u(x, y) = x2-y2, then expansion y), where i = for v(x,y) i n terms of x, y and a general constant c would be n. (a) xy + C Z 445 2 If f(x + iy) = x3 - 3xy 2 + i $( x, y) where i = (a) y 3 - 3x2 y 9. (x-y) +k 2 Engineering Mathematics (c) 2xy + C x2 +y 2 (b) --+c 2 - (d) (x-y ) +c 2 2 [GATE 2014-ME-Set-3] 13.* The real part of an analytic function f(z) where z = x + jy is given by e-Ycos(x). The imaginary part of f(z) is (a) evcos(x) (b) e-Ysin(x) (c) -eYsin(x) (d) -e-Ysin(x) [GATE-2014-EC-Set-2] 14.* Let S be the set of points in the complex plane corresponding to the unit circle. That is S = [z : l zl = 1]. Consider the function f(z) = zz' where z' denotes the complex con jugate of z. The f(z) maps S to which one of the following in the complex plane (a) unit circle (b) horizontal axis line segment from origin to (1 , 0) IES MASTER Publication 446 Analytic Functions (c) the point (1 , 0) (c) f(z) is not continuous but is analytic (d) the entire horizontal axis 15.* Given f (z) [GATE-2014-EE-SET-1] g (z) + h(z) where f, g, h are complex valued functions of a complex variable z. Which one of the following statements is TRUE? (a) = (d) f(z) is neither continuous nor analytic 20. If f(z) = (x2 + ay2 ) + ibxy is a complex analytic function of z (b) a (b) If g (z) and h(z) are differentiable at z0, then f(z) is also differentiable at z0 • (d) If f (z) is differentiable at z0, then so are its real and imaginary parts. 16.* C o n s i d e r (c) a [GATE-2016-EC-Set-2] 17. f(z) = u (x, y) + i v(x, y) is an analytic function of = 1, b [GATE-2016-ME-SET-2] z + z* where z is a com plex variabl e and z* denotes its complex conj ugate. W hich one of the fol lowing is TRUE? = (a) f(z) is both continuous and analytic (b) f(z) is a continuous but not analytic [GATE-2017 ME Session-II] (b) -1 22.* (d) -2 Given \If = x2 - y2 + �. then the function � is X +y Y- + c (a) -2xy+x2 + y2 23. [ECE 2017 (EE)] If W = � + i\11 represents the complex potential for an electric field. (c) -2xy +- is given as f(x, y) = u(x, y) + iv(x,y), where u(x, y) = 2kxy and v(x, y) = x2 - y2 • The value of k, for which the function is analytic, is __. 2 (c) 2 (c) x2 + y2 + constant 18. A function f of the com plex variable z = x + i y, = (a) 1 (b) x2 - y2 + constant [GATE-2016-ME-SET-1] �. then -1 , b = 2 is harmonic? (a) -x2 + y2 + constant 19.* Consider the function f(z) = = 21.* What is the value of the m for which 2x - x2 + my2 complex variable z = x + i y where i = �- If u(x, y) = 2 xy, then v(x, y) may be expressed as (d) - (x2 + y2 ) + constant x + iy, where i (d) a = 2, b = 2 [GATE-2015-EE-Set-2] t h e com pl ex v a l ued f u n c t i o n f(z) = 2z3 + b lzi 3 where z i s a complex variable. The value of b for which the function f(z) is analytic is___ = (a) a = -1 , b = -1 If f (z) is differentiable at z0, then g(z) and h(z) are also differentiable at z0 . (c) If f (z) is continuous at z0, then it is differen­ tiable at z0• [GATE-2016-EE-Set 2] x +y 2 X 2 -+C (b) 2xy + -2-+C 2 X +y X Y- +c (d) 2xy - 2 x + y2 [ECE 2017 (Com mon Paper)] F(Z) is a function of the complex variable z = x + iy given by F(Z) = iz + k Re(Z) + i Im (Z) For what value of k will F(Z) satisfy the Cauchy­ Riemann equations? (a) 0 (c) -1 (b) 1 (d) y [G�TE 2018 (ME-MorningSession)] Engineering Mathematics 1. (a & d) 3. (b) 2. (a) 4. 13. (b) 9. (d) 15. (b) 8. 11. (a) 6. (c) 10. (b) 5. 7. 12. (b) 14. (b) 16. (d) (c) 17. (c) 18. (c) 19. (b) 21. (a) 20. (0) (b) 22. (a) 23. (a) (k = -1) 447 (b) SOLUTIONS Sol-1: (a) & (d) Sol-3: (b) -w-1 z-1 Given co = - ⇒ Z = - - wher e co = f (z) = u + iv co - 1 z+ 1 ⇒ ⇒ ⇒ ⇒ ⇒ -w-1 < I w-1 I w x2 + y2 Sol-4: (b) I (u + 1 ) + iv 12' < I (u-1) + iv 12 ⇒ u 2 + f + 2 u+ v 2 < u2 + 1 - 2u + v 2 4u < 0 ⇒ u < 0 z 1 H ence, the function w = - maps the inter ior of unit z+ 1 cir cl e in the z pl ane, to the l eft hal f of the w pl ane. Sol-5. (a) = 1 2 1 2 e2u 1 log (x2 + y2 ) + itan- y_ = u + iv X l og (x2 + y2) = (eu )2 = K2 � � �o � � Similar ly, option (d) is al so correct. Sol-2: (a) We know that potential function � and str eam function \Jf together constitute an analytic function known as complex potential 'co ' i.e. which is obviousl y not defined at z = o "' (O, O) So w=l og z is not anal y tic at (0, 0) co = �+ i\jf is analy tic wher e � = x2 - y2 u Given function is the Cartesian repr esentation of l og z. We have w = log Z :: = : ; : = -Z (C-R equations) Metho d I V l zI < 1 In (x + iy) L et w = � (x, y) + i\Jf (x, y), then w is anal y tic when 2 (u + 1)2 + v < (u -1}2 + v 2 y = u = ⇒ I u + iv + 1 I < l u + iv - 1 I = = 1 n(rei0 ) = 1 n(r) + i0 Consider I z I < 1 which r epr esents interior of unit cir cle I zI = 1 W=I n Z Step 1 ·: R eal par t is given so by ca se I of M I LNE THOMSON method. Step 2 Step 3 : Step 4 : Step 5 a� ax = 2x � �1 (x, y) �1 (z, 0) = 2z a� ay - 2 y � �2 { x, y ) �2 (z,0) = 0 co = f(z) = J[ � (z, 0) - i� 1 2 (z, o )] dz + c IES MASTER Publication 448 A naly tic Fu nctio ns oo = J (2 z)dz+c = z2+c <l>+i\jl = (x 2 - y2 )+2 ix y+c ( ·: At X = y = 0, \JI = 0, c = 0,) H ence str eam function \JI = 2xy + c = 2xy oo = <I>+i\Jf is an analytic function Method II : By C - R equation, <l>x : \jl Y & <l>y = -\jl X \JI ➔ f(x , y) = (-: } x+(:} y (By C - R eq is S) = (2y) dx + (2x) dy Integrating b oth sides \JI = 2xy + C (By C.R. eq' s) <!>y = -\jl x ... (i) -2y = -[2 y+f' (x )] ⇒ B y using (i) , f' (x) = 0 f(x) = C \JI = 2 x y + C At x = 0, y = 0, \JI = 0 ⇒ C = 0 \JI = 2xy Note : I t will b e better to use MILNE-THOMSAN method, while sol vi ng Q uesti ons 3, 7 , 8 , 9, 1 0 & 11 . Sol-6: L et (b) u u? +v 2 -v & y=u2 +v 2 2 2 -- = ⇒ u +v +1 0v = 0 u2 +v 2 10 -v u2+ (v+5)2 = 25 Method I Using total derivative concept V = f(X , y) y dJ = (:) dx+(:} z = x + iy and w = u+ iv = (-:}x+(:) dy (By C - R equation) On integrati ng, H ence stream function \JI = 2xy. Again, by C.R . eq' s ⇒ u-iv 1 = 2 u+iv u +v 2 dJ = -x dx + y dy B ut at x = y = 0, \JI = 0 so c = 0 \JI = 2xy + f(x) Now y = 0.1 So d\Jf = 2 (x dy + y dx) = 2 d(xy) <l> x = 2x = \Jf y = 1 w whi ch is circle in w-plane having centre at (0,-5) & rad = 5. Sol-7: (c) d\Jf = (:} x+(:} y Method 111 : x+ iy = So ⇒ So by total derivative concept <I> = x 2 _ y2 z = ⇒ or Method II : and Using MILNE THOMSON method Step 1 Step 2 Step 3 : Step 4 : Step 5 x 2 y2 V = - -+-+C 2 2 2 2 Y -x - k V = - + 2 ( •: u = x y) f(z) = u+ iv (Analytic) u = x y (R eal part is g·iven) - = Y "" <l>1 (x, y ) au <1>1 (z, 0) = 0 - = au ay X <1>2 ( z, 0) = z :::< <p2 (x, y ) f(z) = J [<!>1 (Z, 0)- i<p2 ( Z, 0)] dz+c = J[o - i(z)] dz+c iz2 = --+c 2 Engineering Mathematics Again, by C . R. eq's = - �(x2 - y 2 +2ixy)+c y x u + iv= xy+ { ; )+c 2 2 Hence, Method Ill : ⇒ So by (i), Sol-10: (d) Let By Cauchy - Riemann equation 2 = L+f(x) 2 au ay ⇒ x2 f(x) = --+k 2 So by (i), where then V = - = 3x3 - 3y2 & 'V = - 6xy y �! } y = -'l' dx+ 'l'x dY (Using C-R eqns.) y 3 2 = -(-6xy)dx+ (3x - 3y )dy d$ = d(3x y - y ) 3 $ = 3x 2 y - y3 +C ⇒ u= 3x2 - 3y2 u x = 6x = vy (By C.R. Eq's.) v = 6xy + f(x) 1 = z 1 x - iy = 2 2 x+iy x +y ⇒ x _ 0 & ----=L. <0 _ x2 +y2 x 2 + y2 > = = >1 l x �iy l � (-.- 0 < ✓x2 + y 2 < 1) Method I ·: It is given that f(z) = u + iv is analytic & we have also given that u = 2xy i.e. real part is given so we can use case I of M ILNE THOMSON method i.e. Step 1 : . .. (i) au = 2y "" $ (x y) , 1 ax Step 2 $1 (z,0) = 0 Step 4 $2(z, 0) = 2z Step 3 : Step 5 : For an analytic function, f(x + iy) = u(x,y) + iv (x,y). where, ✓ 1 lies outside the unit circle in IV quadrant. Z Sol-1 1 : (c) \jl(X, y) = x 3 - 3xy2 x+( d$ = (� !} Sol-9: (d) 0 < x < 1, 0 < y < 1 and 0 < x2 + y2 < 1 ⇒ .. = \jl(X, y)+i$(X, y) Now using total derivative concept, ⇒ Z = X + jy Iii y2 - x2 -+ k 2 3 = 6xy + k 1 lies in IV quardant z f(x +iy) = (x3 - 3xy2 )+i$ (x, y) \jl X V Since x > 0, y > 0 ax X = -f'(x) ⇒ Sol-8: (b) . . . (i) av = -- Again, f(x) = k As Z lies in the first quadant in the unit circle av au = -= y ax ay V f'(x) = 0 ⇒ u(x, y)= xy ⇒ -6y = -6y - f'(x) ⇒ y2 - x2 V = -- + C 2 f(z) = u(x,y) + i v(x,y) 449 = 2x "" $2(x, y) au ay f(z) = 1 2 = J(0 - i (2z) dz)+c =-i z2 + c = So J[$ (z, 0) - i� (z, o)] dz+c -i[ x 2 - y 2 + 2ixy] +c u + iv = u+2 xy+i( y2 - x2 )+c V = y2 - x2 +c IES MASTER Publication 450 Analytic Functions Method I I : u = 2xy ⇒ LIX = 2y & U y = 2x So, by C-R equation, -2X Vy= 2y & V= X Method II : Using Total derivative concept v= V(x, y) Now, by Total Derivative Concept in 'v' V = V(x, y) dJ So, v ⇒ Method Ill : f(z) ,,; -x2 + y2 + C = u(x, y) + i v(x, y) au u(x, y) = 2xy ⇒ - = 2y ax Using Cauchy-Riemann condition - = V au ay = y2 + f(x) av ay ax 2x = -f'(x) f(x) ⇒ = -x2 + c . . . (ii) Method I Using MILNE THOMSON method Step 2: Step 3: Step 4: Step 5: Hence = u + iv u = x2 - y2 (Real part is given) f(z) · $1 (z,O) = 2 z au = -2y � $2 ( X, Y) 8y $2 (2, 0)= 0 f(z) = J [ $1 (z, 0 ) -i$2 (z,O)]dz+c = J(2 z - O) dz+c = z2 + C u + iv= x2 - y2 + 2 ixy + c V= 2xy + C dJ = 2 d(xy) on integrating V= 2xy + C Method Ill : u(x, y)= x2 - y2 Using Gauch-Riemann equation av au 2x= . . . (i) Sol-12: (c) Step 1 : (By CR equation) = (2y) dx + (2x) dy ( •: u = x2 - y2) ax 2 cl +_J y2_-_x_ so using (i) lLv____:_ _ where ⇒ av au ax Again, = (- �} x+(: ) dy = (:)dx+( :}y = J -2x dx+J 2y dy + C = (:} x+(:} y dV Again, So by (i), Sol-13: (b) ay av ay v = 2xy au ay = av + f(x) - ax -2y= -2y - . . . (i) at f = constant ax Iv = 2xy+cl Method I : Using M ILNE THOMSON method where Step 1 Step 2 : Step 3 : Step 4 : Step 5 f(z)= u + iv (Analytic) u = e -Y cos x (Real part) au ax = -e Y sin x � $1 (x, y ) $1 (x, O)= -sin z -Y ay = -e cos x � $2 (x, y) $2 (z,O) = - cos z au f(z) = J [ $1 (z, 0)-i$2 (z, o)] dz+c = J C-sinz+icos z) dz + c = (cos z + isinz) + c = e;2 +c { ·: e;9=cos 0+isin0 } = ei(x+iy) + C Sol-14: (c) = eix-y +C =e-Y (eix )+C u + iv = e-Y ( cosx + i sinx)+c H ence, v = e-Y sinx Method I I (for any c = 0) Using total d erivative concept u = e-Y cosx then and au ay = -e-Y COS X au = ax -e-Y sinx v = v(x, y) dJ = (: ) d x+(:} y = (- :} x+(:) d y (By C - R equation) = (+e-Y cosx )d x + (-e-Y sinx)d y dJ = d (e-Y sinx) On integrating, v = e-Y sin x +c I f we take c = 0 then v = e-Y sinx Method Ill : Let ⇒ f(z) = u + iv U = e- Y COS X ux = -e- Y sinX & Uy = -e-Y cos X B y C.R . eq ns. ux = vY v = -e- Y sinx y Integrati ng both sid es w.r.t. y v = e- Y sinx + <l>( x ) [ · : when w e integrate P.O .E ., w e get an Arbitrary function instead of a rbitrary constant] B y C.R. eqn.' s ⇒ so, vx = e- Y cosx + <J>' (x ) U y = -Vx - e- Y cosx = - e- Y cos- <j>' (x) <J>' (x ) = 0 4>(x) 1 = C v = e-y sin x +C Note : We can also use MI LNE THOMSON method to find v. E ng ineering Mathematics f(z) = 22· where z' = s = {z : l z l = 1} 451 z f(z) = zz = l zl 2 = 1 = 1 + 0 i = (1, 0) f(z) maps s to the point (1 , 0) in the compl ex plane. Sol-15: (b) a is incorrect as g and h may ind ivid ual l y be ind ifferentiable inspite of their sum being d ifferentiable at a point. b is correct because if both are ind ivid ual ly d iff erentiable then their sum is d efini tel y d ifferentiabl e. c is incorrect as continuity d oesn' t ensure d ifferen­ tiabil ity. d is incorrect as it is not correct to say that if f is d ifferentiabl e at Z0 then real and i maginary parts of f are d ifferentiabl e at real and imaginary parts of Z0 . Sol-16: (0) Let z = x+j y f(z) = 2 (x+j y)3 + bl (x+j y)l 3 ⇒ ⇒ ⇒ f(z) = 2[x3 + 3x2 j y - 3xy2 - j y3] + b(x2 + y2)3'2 f(z) = 2 (x3 - 3xy2) + b(x2 + y2)312 +j[6x2 y - 2y3] Let f(z) = u + j v u = 2 (x3-3xy2) + b(x2 + y2)3'2 au ⇒ ax and V av ⇒ ay = 6x2 - 6y2 + 3bx (x2 + y2)1 12 = 6x2 y - 2y3 = 6x2 - 6y2 Now, f(z) to be analy tic · au ax av ay and this is true onl y when, Sol-17: (a) Method I Step 1 b = 0 · : It is given that f(z) = u + iv is analy tic & we have also given that u = 2xy i.e. real part is given so we can use case I of MILNE THOMSON method i.e. Step 2 : ax 4>1 (z, 0) = 0 IES MASTER Publication 452 Analytic Functions Step 3 Step 4 : Step 5 : au oy = f(z) = $2(2,0) u + iv = 2z J [$ (z,0)- i$ (z,0)]dz + c 2 1 f(0 - i (2z) dz)+ c = -i z2 + c = -i [ x2 - y 2 + 2ixy J + c = V = So Method II = V(x, y) 2x � $2(x, y) u + 2 xy + i (y 2 -x2 ) + c y 2 - x2 + c W e have u(x, y) = 2xy and we know that au ax av oy = and av = - au ax oy (using C-R eqns.) au au = 2y & = - 2x Now = = ax ax - oy oy Now, using Total Derivative Concept in v av av ⇒ Sol-18: (k = -1) = = f -2xdx + f 2ydy -x2 + y2 + constant u(x, y) = 2kxy and v(x, y) = x2 - y2 au ax av au _ av = and = ax oy oy 2ky = -2y and 2kx = -2x Now, ⇒ ⇒ Sol-1 9: (b) k = Z = -1 and k = -1 X + iy & z* = X - iy f(z) = z + z· = 2x Then ·: v = f(z) is in polynomial form hence continuous Again f(z) = 2x = u + iv u = 2x & V = 0 ⇒ ·: C R equation are not satisfied throughout the domain of f(z) hence not analytic. Sol-20: (b) f(z) u( x, y) = = ( x2 + ay 2 ) + ib x y = u( x, y) + iv( x, y) x2 + ay 2 bxy au au = 2X, ax ax = av av = by, 2ay, ay ax Using Cauchy Reimann equations au av = ax oy and Hence, au oy = - ⇒ 2x = bx av a ax , = a = -1, b = 2 ⇒ = bx b=2 -b = -1 2 Sol-21 : (a) $ = 2x - x2 + my2 Let Any function is HARMONIC if it satisfies Laplace equation. $xx + $yy i.e., = 0 a 2$ a2$ +- = 0 ax2 0'/2 a a , - (2-2x) +- (2my) = 0 ax oy di = V = -2 + 2m = 0 · Sol-22: (a) Method I 'V ⇒ m =1 w = cp + i'V (Complex Potential) and = x x2 - y 2 + 2 x + Y2 Using Milne Thomson method, Step 1 Step 2 Step 3 Step 4 : y2 -x2 a'V = 2x + 2 � $2 (x, y) (let) ax ( X + y 2 )2 1 $2 (2, 0) = 2z - 22 -2xy a'V � $1 (x, y) (let) = -2y + 2 ay ( X + y 2 )2 $1 (2,0) = 0 Engineering Mathematics 1 = i (x2 - y2 + 2ixy + -. )+c x + Iy Method I I Y_+c )+i (x2 - y2 +_x_) = (-2xy+_ 2 x + y2 x2 +y2 w = � + i\jl Using Total derivative concept X 2 2 = X - y - -x2 + y2 then 8; = 2x + and mv = -2y y2 - x 2 (x +y ) 2 2 2 2xy = -2(xdy + ydx)+ � = Hx, y) so (x2 + y2 ) 2 dJ = - 2d(xy)+ d (�) X +y On integrating y+c V = - 2xy + 2 x + y2 Sol-23: (b) F(z) = iz + kRe(z) + i l m(z) = i(x+iy) + k(x) + i(y) 2 2 = (kx -y) + i(x+ y) d � = (: ) dx+(:)dy = (:) dx + (-: ) dy (By C - R equation) 2xy -2xydx-(y 2 -x 2)dy 2xydx + x2 d \- y2 dy = _ 2d(xy)+ (x2 + y2 ) (x + y ) 2 453 J [ y2 -x2 2x+ = (-2y x2 d 2 ] dy (x2 +y2) ( x2 +y2 ) So = u + iv u = kx - y and v = x + y By C-R equation: V X = Vy and uy = - VX (R = 1) = R(-1 = -1) Hence k = 1 IES MASTER Publication 4.3 In the case of real variables, a definite integral is regarded as the limit of a sum and an idefinite integral is regarded as a process inverse to differentiation. While in case of complex variables definite integrals are line integral and indefinite integrals are functions whose derivative is equal to a given analytic function in a region. Again, in the case of real variable, the path of integration of J: f (x)dx is always along the real axis from x = a to x = b. But in the case of a complex function f(z), the path of definite integral J: f ( z)dz may be along any curve joining the paths z = a and z = b. Thus, generally the value of the integral depends upon path of integration. Continuous Arcs An arc z = �(t) +hv (t) , (where the parameter t runs through the interval range a � t � b ) is said to be continuous arc. If �( t) and \J/ ( t) are real continuous functions of the real variable t. In other words, an arc is the set of all image points of a closed finite interval under a continuous mapping. Differentiable and Regular Arcs An arc z(t) is said to be differentiable if z' (t) exists and is continuous. If in addition z' (t) = 0 , then we say that the arc is regular or smooth. M u ltiple Point If the equation z(t) = x(t) + y(t) is satisfied by more than one value of t in the given range, then the point ( x, y) is called a multiple point of the arc. J o rdan Arc A continuous arc without multiple points is called Jordan Arc. Thus, for a Jordan arc Jordan arc is also called simple arc z(t1 ) = z(t2 ) only when t1 = t2 Closed Cu rve If the points corresponding to the value a and b coincide, the arc is said to be closed curve. CONTOUR Contuour is a Jordon curve consisting of continuous chain of a finite number o f regular arcs. If A is the starting point of the first arc and B the end point of the last arc, then integral along such a curve is written as f f ( z )dz. If the starting point A of the arc coincides with the end points B of the last arc, then the contour AB AB is said to be closed. Engineering Mathematics 455 Simply and Multiply Connected Domains A domain in which ev e ry closed curve can be shrinked to point without passing out the region is call ed a simply connected domain; oth e rwise it is said to be multiply conn ected. ELEMENTARY PROPERTIES OF CO MPLEX INTEGRALS · 1. Linearity : If f (z) and g(z) are two compl ex functions and a, p are constants, the n ( { a (z) + p g(z) } dz = a J f (z) dz+ pfc g (z)dz This property can be g en eralized fo r any finite number of functions. 2. Sense Reversal : 3. Partition of Path : fc f ( z) dz = ' fc, f ( z)dz+ where the curv e c consists of the curve c 1 and c 2 . 4. t In gene ral, if c = c 1 +c2 +c3 +...........c n , th en f f (z)dz = c If a is a compl ex constant, the n fc aF (z) dz Solution = Evaluate fc � z-a whe re c lz - al = r Solution Evaluate (i) Z c1 f (z)dz+ f f (z)dz+ .... + J f (z)dz . C the circl e l z - al = r. ⇒ z = a+ rei8 whe re O ::; 0 ::; 21t & dz = irei8 d0 · iG 2n ire 2n fc z dz f om z = 0 to z = 4 + 2i along th = t 2 + it Cn = Jo _1e d0 = i f d0 = i (0]2o n = 21ti Jo re H ence Example 2 = re pre se nts Param etric equation of the circl e f = a f/ ( z) dz . 5. An i mportant in equality l f/ (z) dzl ::; fc l f ( z) l l dzl . Example 1 f ( z) dz r e curve c d efin ed by (ii) The lin e from z = 0 to z = 2i and th e n from z = 2i to z = 4 + 2i (i) The curve c given by z = t2 + it it represents a parabola whose param etric equations a re x = t 2 , y = t y2 = X Furth er, th e point z = 0 corresponds to t = 0 and the point z = 4+2i co rresponds to t = 2. ⇒ Now, fc z dz = c: ( t 2 - it) (2t+i)dt, (since z = t2 +it giv es dz = (2t + i) dt & z = t2 - it ) 4 2 3 = J; [(2t3 +t) - it2 ] dt = [ ( i t + i t ) - (i it (ii) Let OA be th e straight line from z = 0 to z = 2i. The n )J: the point O is (0,0) and the point A is (0,2). on th e line OA : x = 0, dx = 0 and y moves from O to 2. Let AB b e th e straight lin e from z = 2i to z = 4 + 2i . Cl ea rly, AB is parall e l to th e real axis. Also, th e point B is (4, 2). On the line AB: y = 2, dy = 0 and x moves from O to 4. = [ ( s +2) - J ( Ls )] - o = 1 0 - ! i (0, 2) y 2__ 8 (4, 2) =.., y_ A .--_ x=O 0-------- x (0, 0) IES MASTER Publication 456 Complex Integration Hence = f JoA z dz + fAB z dz where z = x -iy = f: ( -iy)idy + Jo4 ( x - i2)d x= u- y 2 J: + [J ( x - 2i) I = 2 + J {( 4 - 2i}2 - ( -2i)2 } 2 = 2 + J {(16 - 4 -16i)- ( -4)} = 10 - 8i = 2 ( 5 -4i) Example 3 = Evaluate same. Solution : ft (z Since z = x 2 + z )dz , by choosing two different paths of integration, and show that the results are + iy then dz = dx + id y 2 2 and z2 + z = ( x + iy) + ( x + iy) = (x - y + x ) + i ( 2xy + y) so, we have 2 ft ( z 2 + z) dz= ft { ( x 2 - y2 + x ) + i ( 2xy + y)( d x + idy)} ... (1) (i) First we choose the path through straight line segment joining the points D(z = 0) and M ( z=1 + i) . On the line OM, we have y = x so that dy = dx and as z moves from 0 to M, x varies from 0 to 1. So from (1) we have ft (z2 + z) dz= f� {( x 2 - x 2 + X) + i ( 2x · X + X)} ( dx + idy) y=x M (1 , 1 ) 0 (0, 0) = f� { x + i ( 2x + x)} (1 + i)dx= f� { x - (2x 2 + x )} + i { x + (2x2 + x )} d x f� { -2 = x 2 2 3 . 1 3 1 2 + 2i( x2 + x )} d x= -3 x + 2{ 3 x + x ) 2 [ J 2 5. = {-� + 2{J + J)} - o= - - + -1. 3 3 o . . . (2) (ii) Now we choose the path OLM consisting of two straight line segments OL and LM, where L is the point z= 1. So from (1) we have ft ( z 2 2 2 + z) dz = foL { ( x - y + x ) + i ( 2xy + y)} ( d x + idy) + IM {( x 2 - y2 + x ) + i ( 2xy)} ( d x + idy) ... (3) On the line OL, we have y = 0 so that d y = 0 and x varies from 0 to 1 . Also on the line LM, we have x = 1 so that dx = 0 , and y varies from 0 to 1 . Hence from (3) we get 2 2 ft (z + z) dz = f� {( x -0 + x ) + i ( 2x - O+ o)} ( dx + i0) + f� { (12 - y 2 +1) + i ( 2 -1 - y + y)} (o + id y) = f� ( x 2 + x ) dx + f�1 ( 2 - y2 + 3i yidy) 2 = J� ( x + x ) dx + J� {-3 y + i(-2y 2 )} d y 2 5. -+ -I 3 3 . .. (4) Engineering Mathematics y 457 M(1 , 1 ) 0 y=0 x=1 L(1 , 0 ) -+------'-----+ X Example , = From (2) and (4) we find that the two results are same. Find the value of the integral ft ( x - y+ix2 )dz (i) along the straight line from z= 0 to z=1+i (ii) along the real axis from z=0 to z= 1 and then along a line parallel to the imaginary axis from Solution z = 1 to z=1+i . We have z= x+iy so dz = dx +idy Let A be the point of affix 1 and B be the point of affix 1+i in the argand plane. Join OB and AB. 1 y x = (i) OB is the straight line joining z = 0 to z = 1 +i. Obviously on OB we have y= x. So dy = dx. 2 f08 (x-y+ix )dz = f� (x-x+ix2 )(1+i)dx Now 1 O 3 = f i(1+i)x2 dx = (-1+i)[! x ] = !(-1+i) 1 Jo 3 0 3 (ii) OA is the line from z = 0 to z = 1 along the real axis and AB is the line z = 1 to z=1+i parallel to the imaginary axis. On the line OA, y = 0. Hence On the line AB, x = 1 , and so z=1+iy which gives dz= idy So, Hence Example 5 Solution So, 1 1 B (1 , 1 ) fA8 (x - y+ix2 )dz= f�(1+i - y)idy = {(1+i)y - Y; I =-1+i B (1, 1 ) x= 1 0 (0, 0 ) y=0 A (1 , 0 ) 1 5. 1 i i = -+--1+-= --+-I 2 6 2 2 3 Evaluate the integral I= Jiz =r Re z dz . l We know that Re z= !(z+z). Also on izl= r , 2 z = re ia, z= rei9 and dz = ire i9 d0 & O � 0 :::; 21t 2 Jiz =rRe z dz = J: J(rei9 +re -i9 )ire i9d0 = J ir f: J(cos 20+isin20+1)d0 1 n n IES MASTER Publication 458 Complex Integration _ _ ! •,r 2 [ sin 20_ .1 cos20 +e ] . 2 2 2 2n 0 1. 2 . 2 = -ir • 2n = mr 2 Example 6 = Prove that the value of the integral of i _ 1 ir 2 (0 -0)-i ( 1 - 1 )+( 2n - 0 )] [ _2 2 2 along a semi-circle lzl=1 from-1 to 1 is -ni or ni according as the arc lie s above or below the real axis . The given circle is lzl=1 . Parametric equation of the cir cle i s Solution: z = e ie where 0 � 0 � 2n . and B As z m oves from -1 to 1 al ong the semi-circular arc above the real axi s, 0 varie s from 1 1t ICBA -dz z C to 0. In this case, we have = 1 . ·e . 1 e, d0 fno e' e ·fno . = I dS = -17t Again when z moves from -1 to 1 along the semi-circular arc bel ow the real axis, So Example 7 Solution : 1 0 -1 A ----.--. C --,---0 -1 varie s from n to 2n . fCDA !z dz 2n 2n = f _!e_ iei9 d0 = if d0 = ni n n e' D Show that the integral of z al ong a semi-circular arc lzl =1 fr om -1 to 1 has the value -ni a ccording as the arc lie s above or below the real axis. The parametric equation of the c ircle l zl= 1 i s lzl= 1 z = e -ie ' ⇒ 0 0 if d0 = -ni fC A zdz= fn0 e-i0iei9d= n 0 So , B f z dz= JcoA CAUCHY INTEGRAL THEOREM If f(z) is an analytic function of z and if f'(z) i s fn2n e-·eie'·ed0= 1 continuous 2n . I·f d0 = 7tl n 0 varie s fr om n at each po int within and on a closed contour C, then fc f(z)dz = 0 In general, if a closed curve C c ontains non-interse cting closed curves C 1 , C 2, ........C n then, fC f(z)dz= fC f(z�z+Jcf 2 f(z)dz+. ......+fC f(z)dz 1 Cauchy Goursat Theorem ni varie s from n to 0. Again when z m ove s from -1 to 1 along the semi-circular arc below the real axis, to 2n . or z=em , where 0 � 0 � 2n , dz=iei9d0 and As z m oves from -1 to 1 along the semi-circular arc above the real axis, So , A If f(z) is an analytic function and single valued inside and on a simple closed 0 contour C , then Example 8 d Evaluate Jc � , where C is JzJ = 3 . z 5 1 Since the point z = 5 is outside the simple closed curve C. JzJ = 3 , the function Solution : and on C. Hence by Cauchy's theorem, We have Example 9 Solution Engineering Mathematics dz fC o z-5 - Z-5 _J_ 459 is analytic inside d Evaluate Jc � where C is the circle JzJ < 1 z 1 1 1 Let JzJ = r be the circle C, where r < 1 . Then the function - is analytic on and inside C. Hence by z-1 Cauchy's theorem, 1 We have f zdz- 1 - 0 CAUCHY'S INTEGRAL FORMULA If f ( z) is analytic within and on a closed curve C and 'a' is any point within C, then f (z) 1 f(a) = -. fc -- dz . 21tI ( z - a ) Extension o f Cauchy's Integral Formula For M ultiply Connected Domain If f(z) is analytic in the region bounded by two closed curves C1 and C2 , & 'a' is a point in the region between c 1 and c 2 then f (z) 1 1 dz f(a) = - J �dz - 2ni c, z - a 21ti c2 ( z - a) where C2 is the outer curve traversed in anticlockwise sense and C 1 is the inner curve traversed in clockwise sense. f Cauchy"s Integral Formu la for Higher Order Derivative of an Analytic Function If a function f(z) is analytic within and on closed contour c and 'a' is any point within c, then derivatives of all orders are analytic and are given by Example 1 0 Solution 1 f (z)dz _ f (n ) (a) 21ti C (z - a r1 �f 1 If C is a closed contour containing the origin inside it, and f(z) = eaz then, prove that : From the Cauchy's integral formulae for higher order derivatives of an analytic function, we have f " ( a) = Now choose f ( z) = e82 f" (O) = ⇒ f (z)dz n! 21ti c (z - a r1 f f (z)dz · �f 21ti c n+1 f " ( z) = a" e82 z ⇒ f " ( 0) = a" nl e a" 1 e 82 Therefore from (1 ), we have a" = . J n+2 dz or - = - Jf -dz · 27tl c z n! 21ti c z"+1 , 82 ... ( 1 ) IES MASTER Publication 460 Complex Integration Example 1 1 Solution 1 1 d Evaluate c � , where C is lzl= 2 2 1 f Cauchy's integral formula is: dz= Jc zf(z) -a 2 rcif(a) , where z = a is a point inside contour C and f(z) is analytic within and on C . Here C i s lzl = 2, which represents a circle centered a t z = 0 and having radius 2 . Also, a = 1, which lies inside C . Taking f(z) = 1 , it follows that Example 1 2 Solution 1 1 = 2 i[ since f(1) Jc -z 1-dz=2rcif(1) = 1] -1 rc 1' dz Evaluate 'f - - where C is any simple closed curve and z = a is (i) outside C (ii) inside C . cz-a 1 (i) when z = a is outside the simple closed curve C, the function -- is analytic inside and on C . z a 1' dz _ =0 Hence by Cauchy's theorem, we have ':J' -c z -a , (ii) When z = a is inside the simple closed curve C, and f(z) is analytic within on C , by Cauchy's integral formula, we have dz= 1c zf(z) -a Taking f(z) = 1 , it follows that i.e. 1 = 1c --dz z-a . Example 1 3 : Evaluate : (1) Solution 1 2 rc i f(a) 2rci X 1 , Since f(a)= 1 = 2rci f c z�a GOS 1Z =1 z 2-Z4)dz 1 1 ( e 2z + 22 (ii) f dz = J lzl 2 z -1 cos z is analytic within and on the circle lzl = 1 , and the singular point z = 0 z-4 lies inside this circle, Therefore, (i) The function f(z) = GOS Z f d2 = 2rc i f(O) = = 1 z J l l z(z- 4) ( GOS Z )y f � z-0 f dz = f (z) dz=2rci (f10) c z J -0 [Using C . I . F.] (ii) The function f(z) = e22 + z2 is analytic within and on the circle lzl= 2 , and the singular point z = 1 lies inside this circle. Therefore, f ( 2) e 2 +z2 f --< - dz=2rci (f1) dz = = Jlzl 2 z -1 (z lzl=2 -1) J- = 2 rci(e1 +f )=2(e+1)rci [Using C I F] Example 1, Solution Engineering Mathematics 1 Using Cauchy's integral formula, calculate the following integrals: (i) (ii) 461 dz f where C is lz + 3il = 1 J c Z(Z + 7t 'l) ' fc ez -dzi , where C is the ellipse lz - 2I + lz + 21 = 6 82 1t By Cauchy's integral formula, we have 1 fc zf(-z)a dz = 21ti f(a), where z= a is a point inside the contour C and f(z) is analytic within and on C. = I (i) Let fc z(Zdz+ ' . Take f(z)=J_ . Then 7t l) Z f dz - ' 1 - f, (z) ' - -2 - 27t1· ( -. ) 27tlf(-7t 'l) I J C Z -( -7t l ) - 7t l Here z= -1ti lies inside C and f(z) in analytic within C . (ii) Let I = e82 dz fc z - 1t i Here C is the ellipse lz - 21 + lz + 21 = 6 2 1/2 2 2 2 1/ I(x-2) + y 1 = 6 - {(x + 2} + y2 } i.e., Squaring, we get x 2 + y 2 + 4 - 4x = 36 + (x2 + y 2 + 4 + 4x)-12{(x + 2) + y 2 t 2 i.e, i.e, 1/2 1 2 ( x2 + y 2 +4 + 4x) = 36 + 8x 1/2 3 ( x2 + y 2 + 4 + 4x ) Again squaring, i.e., 2 = 9 + 2x 9 (x 2 + l .+ 4 + 4x) = 81 + 4x2 + 36 x 5x2 + 9y2= 45 ⇒ x2 y 2 -+- = 1 5 9 Comparing it with ; + Y2 = 1 we get a 2 = 9, b2 = 5 so that a = 3, b = ✓ , = 2.2 approx. a b 2 2 5 Evidently, z = 1ti=3.1 4 i lies outside C . e 32dz Hence --. is analytic within and on C. So, I = 0 by Cauchy's theorem Z - 7tl Example 1 & Solution 1 1 e2z dz , where C is circle lzl = 3 . Using C auchy's integral formula, evaluate � 4 C(z + 1) We know that IES MASTER Publication 462 Complex I ntegration n) � f f(z) dz f ( (a) 2ni Jc(z-a r1 Putting n= 3 and a= -1 this gives 3 f ( l(-1)= �f 2nl c f(z) (z +1) 4 dz .... (1) Let us take f( z) = e2 2 so that f ( 3)(z) = 8e22 , which gives 2 3 2 f ( l (-1)= a e- = 8/e . Here f 3 (-1) =f " (-1) Hence from (1 ), we find that i.e Example 1 6 1 (i) 6 e22 8 - dz f = 2ni c(z + 1)4 e2 8ni e2z dz= f 4 c(z .+1) 3e2 dz Evaluate J 2 , where C is the square having verties (0, 0), (- 2, 0), (-2, -2), (0, -2) C Z + 2z + 2 oriented in anti-clockwise direction. (ii) Evaluate Solution 1 1 1 r sin zdz z- n = 3 , where C is 4 2 Jc 1 (z - 4 n) I By Cauchy's integral formula, we have 1 f(z)dz f(a)= _ J 2ni c z - a This implies that Jc and Jc f(zz)dz -a f(z)dz ( z - a r1 and I · f(z)dz n! f" (a) - -I 2G c(z - a r1 ·J = 2n i f(a) = ... (1) 2ni (n ) f (a) n! ... (2) where z= a lies insdie C and f(z) is analytic within and on C. dz J 2 (·1) Let = I c z + 2z + 2 where C is the square OABC. Now z 2 + 2z + 2 = O gives ( z + 1) + 1 = 0 , 2 i.e. , (z + 1) = -1 = i2 , 2 i.e, z +1 = ±i z = -1 ± i = -1 + i, -1- i = a, p, say Here z = p lies inside C as shown in the figure. So, y • a. (-2, 0) O (0, 0) A � B (-2, 2 ) • p r f(z)dz r dz 1 = = I = Jc(z - a)(z - p) Jc z-13 ' where f (z) z - a = 2nif(P), using (1) (0, -2) X 2ni 2ni = -- = - = -n � - a -2i Take f ( z) = sinz so, Then 463 1 1 sin zdz I = r , where C is z - n = 3 Jc 4 2 1 (ii) Let I (z - 4 n) ⇒ 1 = n) -J2 I = /, = )r Jc z (z -4 1 Poles of integrand are I f'(z) = cosz , and f'' (z) = -sin z 1 f" (4 Example 17 : Evaluate the integral Solutio n : Engineering Mathematics (2 �¾� r n{- �) = ��i f"(¾•l, [Usi ng CIF for f' (a)] 1 dz , where C is the circle Jzl = � . 2 z - 2) z(z - 1)(z - 2) = 0 z = 0, 1, 2 ⇒ z = 0, 1 lie inside C while z = 2 lies outside C. So, 4 -32 4 -32 (z 1) (z 2) z 4 -3z d z + f (z - 2) dz dz = f J c2 z - 1 Jc1 z c z(z 1)(z - 2) J 4 -3z 4 -3z = 27t·l ----· ] + 27tl· ---] [ [ , (z - 1) ( z - 2) z;() z(z - 2) z=1 [by Cauchy's integral Formula] = 4ni - 2ni = 2ni Example 1 8 : Use Cauchy's Integral formula to evaluate Solution Poles are : z2 + z = 0 1 Solution I z(z + 1) = 0 ⇒ 2- z = 0, - 1 z = 0 lies inside C while z = -1 lies outside C. So, Example 19 ⇒ Jc -z2z ++ z1 where C is Jzj = -21 . · 2z + 1 Jc z2z ++ z1 dz = Jc z(z d2 + 1) 2 = 2ni 1 2ni( 2z + 1 ) z + 1 z=O (by Cauchy's integral theorem) dz where C is Jz + 3 iJ = 1 . Evaluate by Cauchy's integral forumula f Jc Z ( z + 7t'l ) Singularities are: z( z + ni) = 0 ⇒ z = 0, - ni z = -ni lies inside C while z = 0 lies outside C. IES MASTER Publication 464 Complex Integration fc z(zdz+ ni)= So, Example 20 Solution 1 1 Evaluate 21ti = -2 -ni = 2 + dz , where C is the circle f 322: -1 C Poles are : z2 - (by Cauchy's I ntegral Formula) lz -11 = 1 1= 0 or (z -1)(z+1) = 0 or z = 1, -1 z = 1 lies inside C while z= -1 lies outside C . 3z2 +z ( z +1 ) _ 3z2 +z _ 3z2 +z · -- dz 1t1 dz -) 2 ( z+1 c z2 -1 c z -1 Z=1 = 41ti f f (by Cauchy's integral Formula) TAYLOR SERI ES OF COMPLEX FUNCTION I f f(z) is analytic inside a circle C with centre at 'a' then for z inside C. f "(a) f" (a) f(z)= f (a ) + f'(a)(z - a) + - (z -a)2 + . . .+-- (z - a)" + . . . 2! n! f(z)= L an (z - a)" or, f (t) 1 dt an= _ 2ni (t - a)"+1 z= a + h h2 f(a + h) = f(a) + hf'(a) + f2(a) + . . . 21 n=O where or, put LAURENT'S SERIES f If f(z) is analytic in the ring shaped region R bounded by the two concentric circles C and C 1 of radii r and r1 (r > r1 ) and with centre at a, then for all, z in R. f(z)= a0 + a 1 (z - a) + aiz - a)2 + . . . + a_/z - a)-1 + a_iz - a)-2 + . . . 1 f(t) _ dt an= 2ni Jr (t-a)"+1 ' where r being any curve in R encircling C I f f(z) is analytic inside the curve C then a_= n 0 and Laurent series reduces to Taylor series. Example 21 Solution 1 1 Expand f(z)= (a) lz l < 1 (c) lzl > 2 (z -1)� z - 2) (a) By partial fractions in the region, , (b) 1 < lzl < 2 (d) 0 < lz - 1 1 < 1 Engineering Mathematics 1 2 -1 1 1 1 - - ) + (1 - z r 1 = __ _ __ = ( (z - 1)(z - 2) 2 z-2 z-1 2 . . . (i) 1 For lzl < 1 , both Iii and lzl are less than 1 . Hence (i) gives on expansion, = 465 1 3 7 2 15 3 + 2 + 2 + {/ + . . . which is a Taylor's series. 1 8 2 4 (b) For 1 < lzl < 2 we write (i) as, f(z) = -1 1 1 1 · z 2 ( 1 - - } z(1 - z- ) and notice that both Iii and jr1 1 are less than 1 . Hence expansion becomes f(z) = -J- {1 + i + : + : + . . - } - i (1 + z-1 + z-2 + z-3 + . . . ) 2 which is a Laurent series. 2 = ( � :2 ;2 (c) For jzj > 2 we write (i) as f(z) = = = 1 z(1 - 22-1 ) �2 1 3 1 . . . . . . . . .) 2(1 -2- 1 +( : 2: 2� .. . ....... ) ) r1 {1 + 2r1 + 4r2 + . . . } - r1 (1 + r1 + r2 . . . ) . . . + 7r4 + 3r3 + r2 + . . . (d) For O < j z - 1 1 < 1 we write (i) as f(z) = = 1 1 1 1 - -- = - [1 - (z - 1)r - (z - 1 r (z - 1) - 1 (z - 1) -(z - 1 t1 - [1 + (z - 1 ) + (z - 1 )2 + (z - 1 )3 + . . .] ZEROES OF AN ANALYTIC FUNCTION A zero of analytic function is defined as the value of t for which f(z) = 0. If f(z) is analytic in the neighbourhood of point z = a, then by Taylor's theorem. = If a0 = a 1 = a2 ... L a (z - a)" n=O n f" (a) where ' a n = -n! = an_1 = 0, then f(z) i s said t o have a zero of order n a t z = a. IES MASTER Publication 466 Complex Integration SINGULARITIES OF AN ANALYTIC FU NCTION A point at which the function ceases to be analytic, i.e., at which f(z) fails to exist, is called singular point or singularity of the function. If z= a is a singularity of f(z) such that f(z) is analytic at each point in its neighbourhood (i.e., there exists a circle with centre a which has no other singularity), then z= a is called an isolated singularity and if there exist so many singularities in the nbd of z= a then it is called Non Isolated Singularity. y y Isolated singularity Non-isolated singularity --------- x Example 22 : f(z)= co{� ) Solution : . It is not analytic where tan z =1, J. i . . . --------- x (2:) z = 0 i .e., at the points 2: z = nrc or, z = _!.n i.e., (n = 1, 2, 3, . . . ). Thus, are isolated singularities as there are no other singularities in their neighbourhood. But when n is large, z= 0 is such a singularity that there are infinite number of other singularity in the neighbourhood of z= 0. Thus z= 0 is the non-isolated singularity of f(z). TYPES O F ISOLATED SINGULAR POINTS (i) Removable Singularity A removable singularity of a function is a point at which the function is undefined, but it is possible to define the function at that point in such a way that the function is analytic in the neighbourhood of that point. sinz Example 23 : f(z)= z Solution (ii) Pole This singularity can be removed by defining f(0) = 1 which is the limit of f as z tends to 0. The resulting function is holomorphic. + If all negative power of (z - a) in Laurent's series around z= a, i.e., f(z)= a0 + a 1 (z - a) + ai(z - a)2 + a_1 (z - a)-1 + a_i(z - a)-2 + . . . after the nth term are missing, then the singularity at z= a is called pole of order 'n'. A pole of first order is simple pole. Qii) Essential Singularity If the number of negative powers of (z - a) in L'series is infinite, then z= a is called essential singularity. In this case lim f(z) does not exists. Z➔a Example 24 : Find the nature and location of singularities of the following function. (i) z -sin z 22 1 (ii) (z+1)sin (--) z-2 (111 " ') 1 cos z -sinz 1 (IV) 1 - ez Solution ' (1) (v) Engineering Mathematics e2z (2 - 1)4 1 2 (vi) 2e f 2 2 - sin 2 . . . , here 2 = O Is a smgu Ian·ty. 2 2 _!_{ 2 23 � 27 2 - sin 2 2-(2 + - + . . . )}= = 2 2 3! 3! 5! 7! 2 2 Also, 467 5! 3 2 ... Since there are no negativ e powers of 2 in the expansion so, 2= 0 is a remov able singularity. (ii) (2 + 1) sin ( � ) 2 2 Put t = 2 - 2 :. 2 + 1 = (t + 3) . 1 1 3 1 3 (2 + 1 )sin ( � ) = (t + 3) sm(1/t ) = ( 1 -+ - -. . . ) + (- -+ - - . . .) 2 2 t 2t 3 5 ! t 5 3 ! t2 5 ! t 4 3 1 1 1 3 1 ... = 1+---- + ---. . . = 1 + - ----t 6t 2 2t 3 1 2Ot 4 (2-2) 6(2 - 2)2 Since there are infinite number of terms in the (-v e) power for (2 - 2), 2 = 2 is an essential singularity. (iii) Poles of f(2) = ⇒ (iv) (v) where, 1 . , are cos 2 - sin 2 =0 cos 2 - sm 2 2 = 4 ⇒ is a simple pole of f(2) tan 2 =1 f(2) = , poles of f(2) ⇒ 1 - e2 = O 1 -e2 ff- = 1 ⇒ e2 =e20Pi ⇒ 2 =2npi n= (0, ±1 , ±2, ±3, . . . ) e2z f(2) = -put 2 - 1 = t ⇒ t + 1 = 2 (2-1)4 f(2) = 2 (2t)3 e2( 1+1 ) e21+2 e2 e21 � { 2t (2t) = . . ·} = 1 + + + = 1! 2! 3! t4 t4 t4 t4 = e2 {J.t4 +� + � + __! + � + � + . . .} t 3 t 2 3t 3 1 5 Since there are finite number of terms containing negativ e power 2= 1= pole of 4th order. 1 1 1 2-3 2 -5 lz2 �' -- 2 + 2-1 + 2 . ,v1) f(2) - 2e . . + } + - 2 {1 + 2 + + 6 + . . . oo 6 1!2 2!24 3!2 RESIDUE Infinite number of term s in the negative power. Hence, 2= 0 is an essential singularity. The coefficient of (2 - at1 in the expansion of f(2) around an isolated singularity 'a' is called the residue of f(2) at that point. Thus in the Laurent's series expansion of f(2) around 2=a i.e. , f(2) = a0 + a 1 (2-a) + ai2 - a)2 + . . . + a_1 (2 - at1 + a_2 (2 - at2 + . . . , the residue of f(2) at 2= a is a_1 IES MASTER Publication 468 Complex Integration Res f(a) = � f(z) dz 21tl C f f f(z) dz = 2ni Res f(a) C CAUCHY-RESIDUE THEOREM If f(z) i s analytic in closed curve C except at a finite number of singular points within C, f f(z) dz = 2ni then C ) x (Sum of the residue at the singular points within C) C2 ' Applying Cauchy theorem we have f f(z) dz C = • •• •• •• ® .la.:\ • • c_\:J n C f f(z) dz + f f(z) dz + ... + f f(z) dz C1 C2 = 2ni[Res [fa 1 ] + Res [fa2] + ... + Res f(an)l Cn METHODS OF EVALUATING RESIDUES (1) If f(z) has a simple pole at z = 0 then Res [f(a)] = Lt (z - a )f(z) (2) If f(z) = �(z) where ' ' (z) = (z - a)f(z) f(a) , I 'l'(z) "# 0 Z➔8 �(a) 'l''(a) Res f(a) = (3) If f(z) has a pole of order n at z = a, then Res f(a) = Example 25 Solution I 1 Find the sum of the residues of f(z) = [(z - a)" f(z)it (n �1) ! { d::�1 =a sin 2 at its poles inside the circle lzl = 2. zcosz . 2: - 31t , ... f(z) h as s1mpI e poI es at z - 0, ± , + 2 2 Only the points z = 0 and z = ± Res f(0) = n Res f (- ) _ = 2 = "i lies inside lzl = 2 Lt [zf(z)] = Lt ( z➔O Z➔O sin z GOSZ )=o Lt [ (z - 2: )f(z)] = Lt � 2 Z➔ z ➔ -� 2 2 ( z - � )cosz + sin z 2 Lt -'----'------➔ � cos z - z sinz Z 2 l( ---------------I---- z - � ) sin z Z CO S Z -3rt / 2 3rt / 2 ➔ } [Apply L'Hospital rule] Engineering Mathematics 1 -2 = -= -2 Re s t (-% ) = So on : . Sum of residue= 0. 469 2 Example 26 : Determine the poles of function : f(z)= and residue at each pole. Hence evaluate (z-1)2 (z + 2) fc f (z) dz , where C is the circle lz= l 2.5. Solution a Poles of F(z)= are z= -2 & 1. (z -1)2 (z + 2) So, function has simple pole at z = -2 and Res f(-2) = lim (z + 2)f(z) Z➔-2 Similarly, function has a pole of order two at z= 1. So, Res f (1) = ! [� {(z -1)2 f(z)} ] 1! dz z=1 (L)] = [� dz z + 2 Z=1 c: lzl = 2.5 5 9 Clearly f(z) is analytic on l zl = 2.5 and at all points inside except the poles z = -2 and z = 1. Hence by residue theorem. fc f(z)dz= 2ni [Res f(-2) + Res f (1)]= 2n{� + z -3 Example 27 a dz where C is the circle. C z2 + 2z + 5 %] = 2ni f (i) izl Solution a =1 The poles of f(z) = (iii) l z + 1 + ii= 2 (ii) iz + 1 - ii= 2 z -3 are given by z2 + 2z + 5 = 0 i.e., z = -1 ± 2i z + 2z + 5 2 (i) Both the poles -1 + 2i and -1 - 2i lies outside the circle l zl= 1. Therefore f is analytic everywhere within C. z -3 dz _ Hence by Cauchy's theorem c 2 -0 z + 2z + 5 f (ii) Here one pole z= -1 + 2i lies inside the circle : iz + 1 - ii= 2. Therefore f(z) is analytic within C except at this pole and C : lz -(-1 + i)I= 2, i.e., centre is (-1 + i) "' (-1, 1) Res(-1 + 2i) = = Lt z ➔-1+2i Lt [ z - (-1 + 2i)] { f(z)} = Z ➔ ( -1+2i) Lt z ➔ -1+2i -4 + 2i z -3 = =i + ! 2 4i Z + 1 + 2i (z +1- 2i)(z-3) z2 + 2z + 5 IES MASTER Publication 470 Complex Integration -1 +2i C Hence by Residue theorem, @ (-1-2 i) fc f(z) dz= 2niRes f(-1+2i) = 2n{i+J) = n(i - 2) (iii) Here only the pole z = - 1 - 2i lies inside the circle C : lz + 1 + ii = 2 with centre = -1 - i � (-1 ,-1) Therefore, f(z) is analytic within C except at this pole. Res (-1 - 2i)= z➔L-t -2i 1 = (z+i+2i)(z-3) z2 +2z+5 z-3 -4 - 2i __!_ _ = = i Lt 2 Z ➔ -1-2i Z + 1 - 2i -4i Hence by Residue theorem fc f(z) dz= 2ni Res f(-1 - 2i)=2ni (J - i)= n(2 + i) -1 +2i @ C Engineering Mathematics 471 PREVIOUS YEARS GATE & ESE QUESTIONS Questions marked with aesterisk (*) are Conceptual Questions 1.* Co n sider likely applicabil ity of Cauchy's i n tegral theorem to evaluate the followi ng i ntegral counter clockwise around the unit circle C, I = f sec jw z d z, z j2 being a complex variable. The value of I wil l be (a) I = 0, singularities set = $ (b) I = 0, singularities set = {± 2n +1 2 n; n = 0, 1, 2 . . .} (a) j n (c) - n (c) I = � , singularities set = {± nn; n =0, 1, 2 ...} (d) None of the above 2. positive sense is (c) -jn/2 lz-jl;2 z (b) - n/2 (d) n/2 is (c) ✓ .1 - + I- (b) i + { f) (d) 3 2 2 1 z in [GATE-2006-EC] Assum ing i = � a n d t is a real (a) 4. f -2-+-4 d The value of the contour integral (a) jn/2 3. [GATE-2005-CE] n um ber, n/3 i1 f e dt 6. i + {1 - f) 7 .* 8 .* (c) 5.* 4n - . 6 7tl 81 (d) 1 [GATE-2006-CE] If the semicircular contour D of radius 2 is as shown i n the figure, the n the value of the i n tegral f ( S2 - 1 ) D 1 ds is c 1 n z z 2 (b) n (d) ni tan- 1 z [GATE-2007-EC] sin z For the function -3- ,of a complex variable z , the z point z = 0 is The residue of the fu n ction f(z ) = (a) 3 (b) 2: - 6ni 8 z 2 (c) a pole of order 1 (d) not a si ngularity Using Cauchy's integral form ula, the value of the i ntegral ( i nteg ration bei n g taken i n cou n ter 2n . (a) - - 4 7tl 81 2 ni [GATE-2007-EC] (a) a pole of order 3 (b) a pole of order [GATE-2006] 5 clockwise di rectio n ) 'f J. Z - d z is within the u n it 3z - i circle = 1 is (b) -j n (d) n d i f-where C is the co tour - -I 2 ( + ) 1· (c) tan-1 z 3 2 The value of (a) 1 ✓ - .1 2 -j2 9. (c) ( z + 2 )2 ( z - 2 )2 1 1 32 The integral (b) (d) 16 1 at f unit cfrcle z [GATE-2007 (IN)] = 2 1 16 32 is [GATE-2008-EC] f(z)dz evaluated arou nd the u n it circle on the com plex plane for f ( z ) = (a) (c) 2 ni -2 ni z (b) 4ni (d) 0 cos z is z [GATE-2008-ME] 10.* G iven X(z ) = --2 with !z! > a, the residue of X(z ) z n -1 at z (z - a) = a for n ?'. O will be IES MASTER Publication 472 Complex Integration (b) a" (a) an-1 (d) n an-1 (c) n a" 1 1 .* If f(z) = c0+c1 z·1 , then (a) 2 1t C1 [GATE-2008 (EE)] . . 1 + f(z) dz 1s given b y t -- u n it circle Z (b) 2 1t (1+C0) (d) 2nj(1+C0) [GATE-2009-EC) (c) 2 1ti C1 z-1 1 2. The analytic function f(z) = -2- has singularities z +1 at (b) 1 and 1 (a) 1 and -1 (d) i and -i (c) 1 and -i [GATE-2009-CE] cos(2nz) 1 3. The value of the integral r J c (2z - 1)(z - 3) dz (where C i s a closed curve given by lz 1 = 1 ) is (a) 1ti (b) 5 21ti 5 (c) (d) 1ti 1 4. The val ue of £sin z - dz, j z [GATE-2009-CE] where the contour of the integration is a simple closed curve around the origin is (b) 21t j (a) 0 a (c) ( ) cl 21t j 1 1 5. T h e [GATE-2009 (IN)] res i d ues of a c o m p l e x f u n ct i o n 1 - 2z X(z) - ---- at its poles are z(z - 1)(z - 2) 1 1 1 1 (b) - - and -1 (a) - -- and 1 2' 2 2' 2 (c) 3 1 - 1 and - 2' 2 (d) 1 2, -1 and 3 2 [GATE-2010-EC] 16. The contour C in the adjoining figure is described b y x2 + y2 = 1 6 . T h e n t h e v a l u e of J C z +8 ---- dz (0.5) z - (1 .5 ) j 2 z = plane (a) (c) - 21t j (b) 21t j (d) -41t j 41t j [GATE-201 0 (IN)] ,r_ -3z + 4 . dz where c 1 7. The value of the integral 'f 2 c (z + 4z + 5) is the circle I Zl=1 is given by (b) 1/10 (a) 0 (d) 1 (c) 4/5 18.* The contour i ntegral J [GATE-2011 -EC] e½ dz with C as the counter clock wise unit circle in the z-plane i s equal to (b) 2 n (a) 0 (d) a (c) 2n� +- [GATE-2011 (IN)] 1 9.* The value of rf dz, using Cauchy's integral j z -1 2 formula, around the circle l z + 11 = 1 where z = x + i y is (a) 2ni (c) (b) -ni / 2 -3ni/2 [GATE-2011 (Pl)] 20. Given f(z) = - - -- . If C is a counterclockwise z+1 z+3 path i n the z-plane such that lz + 1 1 = 1 , the value 1 of �f f(z)dz is 21tj (a) -2 (c) 1 C 2 (b) -1 (d) 2 [GATE-2012-EC, EE, IN] 21 . 2 J- z - 4 evaluated anticlockwise around the circle 'f z2 + 4 lz - i i = 2 , where i = ✓-1 , is (a) (b) 0 -4n (d) (c) 2+ n 2 + 2i (a) -4 n (1 + j 2 ) (c) -4 n (3 + j2 ) [GATE-2013-EE] J- z - z + 4j ) dz is ( 're z+ 2J_ 2 (b) 4 n (3 - j 2 ) (d) 4 1t (1- j 2 ) [GATE-2014-EC-Set-1] . 3i dz is 5 z 23. If z is a complex variable, the value of (a) - 0.51 1 -1 .57i (c) 0.51 1 - 1 .57i 24. I nteg ra t i o n of (b) - 0.5 1 1 + 1 .57i (d) 0.51 1 + 1 .57i J- [GATE-2014-ME-Set-4] the com pl ex function z2 f(z) = � , in the counterclockwise direction, z -1 around lz - 1 1 = 1 , is (a) -ni (c) ni (b) 0 (d) 21ti [GATE-2014-EE-Set-3; 2 Marks] 25.* Let z = x + iy be a complex variable. Consider that contour integration is performed along the unit circle in anticlockwise direction. Which one of the following statements is NOT TRUE? z (a) The residue of 2 at z = 1 is 1 /2 z _1 2 (b) 1c z dz = 0 1 (c) - ,r,e 1z = 1 2ni 'r z (d) 2 n nj (b) 0 nj 2n (c) (d) 2 n n [GATE-2015-EC-Set-3] 27.* If C denotes the counterclockwise unit circle, the 22. C is a closed path in the z-plane given by l zl = 3. The value of the integral (a) Engineering Mathematics 473 z (complex conjugate of z) is an analytical function. [GATE-2015-EC-Set-1] 26. If C is a circle of radius r with centre z0, in the complex z-plane and if n is a non-zero integer, then dz J, 'fc (z - zo r+1 equals value of the contour integral � J, Re{z}dz is __ 21tJ 'fc [GATE-2015-EC-Set-2] 28. Consider the following complex function : f(z) = 9 (z - 1) (z+ 1) 2 Which of the followi ng is one of the residues of the above function ? 9 16 9 (d) 4 (a) -1 (c) (b) 2 29. The value of f-;.. dz z C [GATE-2015 SET-I] where the contour is the unit circle traversed clockwise is (a) (c) 30.* I n (b) 0 -2ni (d) 4ni 2 ni [GATE-1 5 (IN -Set 2)1 t h e fol l ow i n g i ntegra l , t h e C encloses the points 2 nj and -2 nj. - 1 ,r, sin z 2n 'fc (z- 2nj )3 dz The value of the integral is .............. . 1 31 . The values of the integral - f e [GATE-2016] 27tj. C -Z- 2 2 contour dz along a clOSed contour c i n anti-clockwise direction for (i) the point z0 = 2 i nside the contour c, and (ii) the point z0 = 2 outside the contour c, respectively, are (a) (i) 2.72 , (ii) 0 (b) (i) 7.39, (ii) 0 (d) (i) 0, (ii) 7.39 (c) (i) 0, (ii) 2 . 72 [GATE-2016-EC-Set-3] 3z - 5 _ dZ along a closed path 'fr (z _ 1)(z 2 ) 32.* The value of J, - 1. r is equal to (41ti), where z = x + i y and i = ✓ IES MASTER Publication 474 Complex Integration The correct path y (a) r is f 1 z2 +1 . 36. The value of the integral -. -2- dz where z 2TCJ C Z -1 is a complex number and C is a unit circle with center at 1 +Oj in the complex plane is ____ [GATE-2016 (IN)] z 2 +1 37. The function f(z) = -2-- is singular at z +4 (b) -1 -1 (a) - and 1�5 27 1 -1 (c) - and 5 27 33 .** The value of the integral [GATE-2016-ME-Set-2] sin x dx evaluated 2 f -<1J X + 2X + 2 using contour integration and the residue theorem is -TC sin (1 )/e (c) sin (1)/e (b) -TC cos (1 )/e (d) cos(1)/e [GATE-2016-ME-Set-3; 2 Marks] . The value of t he integral J c 2z+5 1 2 ( z - ) ( z -4z+5 ) 2 over the contour lzl = 1 , taken in the anti-clockwise direction, would be (a) (c) [GATE-2016 (Pl)] 38. The residues of a function f(z) = (d) 34. (d) z = ±2i (c) z = ±i (c) (a) (b) z = ±1 (a) z = ±2 24TCi 13 24 13 (b) sin (z) - , t h e r e si due of the pole 35. For f (z) = z2 z = 0 is _____? [GATE-2016-EC-SET-1 & 2; 1 Mark] (z-4)(z+1) 3 1 -1 (b) - and 125 125 1 -1 (d) - and5 125 [GATE-2017 EC Session-II] 39. An integral I over a counterclockwise circle C is z2 -1 given by I=th c -2-e 2 dz 'Y z +1 If C is defined as lzl= 3 , then the value of I is (a) (c) -TCi sin (1) -3TCisin(1) (b) -2TCisin(1) (d) -4TCi sin (1) [GATE-2017 EC Session-II] 40.* Consider the line integral I= J (x 2 + iy2 ) dz, where z = x + iy. The line c is shown in the figure below � l- - --- (1 . i) � 7 48TCi 13 12 13 [GATE-2016-EE-Set-1, 1 Marks] (d) 1 (0, 0) The value of I is 1 2 3. (c) - I 4 (a) -I X 1 2. 3 4, (d) - I 5 [GATE-2017 EE Session-I] (b) ·I 41 . The value of the contour integral in the complex- t z3 - 2z+3 plane --- dz along the contour lzl = 3, z- 2 taken counter-clockwise is: (a) -1 8ni (a) 1 (c) 0 43. f �. zsmz [GATE-2017 EE Session-II] where c is x2 + y 2 = 1 (b) 2 (d) -1 The sum of residues of f(z) = singular point is (a) --8 is (a) --8 (c) 0 45. [ECE 201 7 (EE)] [ESE 201 7 (EE)] 23 at z = 3 (z - 1) (z - 2)(z - 3) (b) (d) 3 2 (c) 3 4 50. 1 2 (d) -1 (b) 3 2 (d) 3 4 [ESE 201 8 (Common Paper)] [ESE 201 8(EE)] J. 1 --dz = Ani 'fc 5z - 4 The countour C given below is on the complex plane z = x + jy, :,vhere j = � . y 1 J. dz The value of the integral --;'f-2- is ____ 7tj c Z - 1 [GATE 2018 (EC)] J. z+1 The value of the integral 'f -2- dz in counter c Z -4 clockwise direction around a circle C of radius 1 with center at the point z = - 2 is. (a) (c) 51 . (b) 1 /2 (d) 4/5 [GATE 2018 (ME-AftemoongSession)] C (b) [ESE 201 8 (Common Paper)] 1 - e 2z What is the residue of the function -- at its z4 pole? (a) 49. [ESE 201 7 (Common Paper)] 1 (z _ 1) (z _ 2) (d) n(3 + 2i) Let z be a complex variable. For a counter-clockwise integration around a unit circle C. centred at origin. (a) 2/5 (c) 2 1 01 16 27 16 (b) 0 (c) - I 2 4 1 valid in the region 1 < l z l < 2 , the coefficient of 2 z is (c) 1 where c is the rectangular region defined by x = 0, x = 4, y= -1 and y = 1 the value of A is 2z In the Laurent expansion of f(z) = (a) 0 46. (d) 4 The residue of f(z) = 48. 475 1 z [ (z _ 1)3 _ (z _ 3) 7t . ) at its 2 (z - 1) (z - 2 (b) -4 Evaluate (a) 1 (d) 48ni (c) 0 44. 47. (b) 0 (c) 1 4ni 42.* Evaluate Engineering Mathematics 1ti 2 1ti 2 (b) 21ti (d) -21ti [GATE 2018 (EE)] ­ If C is a circle lzl = 4 and f(z) = --(z2 - 3z + 2)2 ' then p f(z)dz is 2 (a) 1 (c) -1 2 (b) 0 (d) -2 [GATE 2018 (EE)] IES MASTER Publication 476 Complex Integration 1. (a) 14. (a) 16. 2. (d) 4. (a) 3. 5. 7. (b) 9. (a) 8. 11. (d) 13. (c) 29. (b) 42. (c) (a) 30. (b) (d) 43. (c) 32. (b) 45. (d)- (a) 34. (b) (b) 36. (1) 31. 33. (c) (a) 46. (1) 48. (c) 24. (c) 37. (d) (b) 39. (d) 35. (d) 26. 38. I = (b) C 1} None of the poles lie within C : jzi= 1 so sing. set = $ Hence by Cauchy's integral theorem = I 0 Sol-2 : (d) t� lz -il=2 z +4 = f lz-ii=Z Poles are, z2 + 4 = 0 _dz . (z - 2J )(z+ 2J ) = Sol-4: (a) z3 -6 J.--dz = r (3z-i) C is jzi = 1 f(z) has a pole 3 0 0 3 49. (0) 51. (b) (a) (a) it f e _dt = f (cos t+i sin t)dt ✓32 + i(1-..!.2 ) = I3 ✓3 +..!. i 2 2 f f(z)dz C Residue of f(z) at (z = i/3) = ⇒ J. _ dz 1t 1 " )= 'Y z2 +4 = 2n j( 4J 2 lz -i1=2 Sol-3: (a) " (b) z = i/3 inside c. z = ± 2j z = - 2j lies outside the curve jz - jj = 2 " 47. I3 Pole of sec z are given by cos z = 0 ·: (b) = (sin t)o +i (-cost)o f sec z dz z = {(2n+1)% I n E (d) 50. SOLUTIONS Sol-1 : (a) 44. (b) 22. 25. (1 4ni) 41. (-133.87) (c) 23. (d) 12. (b) 17. (d) 21. (d) 10. 40. 28. 20. (a) (1/2) (c) 19. (b) 27. 15. 18. (a) 6. (a) = �� {( z - � ) t(z)} z 3 lim Z➔i/3 { (z 3 -6)} 3 = J [ ;� - 6] = -� - 2 1 By cauchy's residue theorem f f(z)dz C 21t . = 2n{ ;� -2 ) = - - 4 7tl 81 E ng ineering Math ematic s Sol-5: (a) rf. ds 1 2 R es { f(z) ; 2} = -- [� (z - 2 ) f(z )] 2 -1 dz 1 z =2 ds th = = ' 2 'f ( s-1)( s +1) f s -1 = 2 11:jx( sum of residues) = 2 11:{�) = 11:j [ where s = -1 is outside D and s = 1 is inside D, so R es { f( s), 1} = Sol-6: (b) 1 1 = ] 1+1 2 dz dz = p I = th 2 (z i) (z + i) ! (1+z ) c 1 f(z) = L et 1 +2 2 Countour C represents a cir cle, having radius 1 unit and centre at ( O,�) . The pole, z = i l ies inside the contour, whil e z = -i outside. Sol-9: (a) cosz z f(z) has a simpl e pole at z = 0 f(z) = f f(z) dz = 2rc i {R es. of f(z) at z = O} = 2rc i[ lim zf(z )] z➔O Sol-1 0: (d) Given = 2rc i { cos O} = 2rc i X(z) = z z a (z-a)2 ' l l > f(z) = X(z)zn-1 = L et = (z-a)2 :. R es (f(z) : z = a) = m ( -1)! 2 (z- a)2 zn-1 H ere z = a is a pole of f(z) of order 2 . Re (0,-i) Sol-7: (b) . 1 = 2 TCI X - = i+i TC (Using Meclaurin series ex pansion) 1 1 z2 z4 - + ---+ ... = 2 z 3! 5! 7! z = 0 is a pol e of order 2 . Sol-8: (a) z = 2 is a pole of order 2 1 where m = 2 ( order of pol e) � (z -a)2.f(z) dz [ dm - 1 = [� { (z-a)2 �}] dz (z-a)2 The value of integral = 2rc i [R es f(z) at z = ]i 477 = Sol-11: (d) It nzn-1 = nan-1 z➔a ] z=1 Z=a Method I : Using Cauchy' s R esidue theorem It is given that C1 , f(z) = C o +z and L et J..( z) 'I' c : l zl = 1 C1 1 + f(z ) _ 1 +C o +Z _ z + c0 z + c1 = z2 -2- Z H ence poles of � (z) are z2 = 0 IES MASTER Publication 478 Complex Integration (Double pole) z = 0, 0 ⇒ cos(2nz) = lim { } Z➔� 2(2-3) 2-1 1 R= Res $ (z) = _ [ d 2_ (z - 0)2 $ (z)] (z=O) I2 1 dZ 1 z=O z= 0 lies inside the given contour 'c' so $ (z) is non analytic inside 'c' so by cauchy's residue theorem = cos(2n. J ) 2 G -3 ) Cauchy's residue's theorem , Sol-1 4: (a) 1 =s 21ti . 1 I = ( 21t1 ) (S) = 5 Z= 0 is a simple pole of integrand and f(z) = f $(z)dz z = Method II 2ni [sum of residues]= 2ni(1 + c0 ) f(z) = + (z) = I f ! dz = so, f c1 +(1;c0 ) z dz z = 0 is a pole of order 2 where, Sol-12: (d) lzl=1 f'(z)= 1 + c0 z-1 f(z) = 2 z +1 Singularities are given by z2 +1 = 0 z2= -1 ⇒ ⇒ Sol-13: (c) z= ±-.J-1= ±i f cos (2nz) dz= f f( 2)d2 Jc (2z -1)(z - 3) Jc where, C : izl = Within C only one pole z = ½ lies. Res (f (z) at z= ½) z So Res f(z) ; (z= 0)= lim [zf(z)]= lim [sin(z)]= 0 Z➔O Z➔O •: Z= 0 lies with in the contour. Hence by Cauchy Residue theorem 2ni [Residue] Sol-15: (c) 2 (O) 2n{ = 2nj(1+c0 ) = I 1! f(z)= c 1 + (1 + c0) z sin z = 2ni [OJ = 0 X(z) = 1 - 22 z(z-1)(z-2) z= 0, 1, 2 ase simple poles. z(1-2z) Res {x(z); O}= Lim ---2➔0 z(z -1)(z - 2) = 1 1 = (-1) (-2) 2 = (-1) = 1 1(-1) ' (z -1) (1 - 22) Res {X (z); 1}= L Im 2➔1 z(z -1)(z - 2) ' (z - 2)(1- 22) Res {X(z); 2}= L Im 2➔2 z(z -1)(z-2) Sol-16: (d) = C : x 2 + y2 =16 ⇒ Now f(z) = 1-4 2(2 -1) = 3 2 l 4 I z= (0.5)Z - (1.5) j Pole of f(z) is Z= 3j (simple pole) which lies with in the given contour Engineering Mathematics Re s f(z) = lim (Z-3 j)f(z) z2 +8 ] = -2 Z j 1/ 2 S o by Cauchy residue the orem [ = lim ➔3 I Sol-1 7: (a) C Z➔3j (z=3) I = f (z)dz = 2n j [sum f C = -4n j 0 of residues] = lim ( J, -3z+4 c z + 4z+5 Z➔- 1 -dz , c : l z l = 1 = 'f 2 - Sol-1 8: (c) The Laurent Expansion in the neighbourhood of z = is given as 0 1 Z in Laurent expansion = 1 So by Cauchy residue the orem. 1 f e 2dz = 2ni(R 1 ) = 2ni(1) = 2n� Sol-1 9: (b) M ethod I Using Cauchy's residue the orem f(z) Let z2 z -1 = -;i-- so p oles of f(z) a re ; z4 - 1 = ⇒ i.e. the re exist simple p oles. four 0 z = ±1 and ± i poles 1 , -1, i, - i and all a re N ow given contour is c : lz + 11 = 1 (ci rcle) with centre at z0 = - 1 = (-1 , 0) and radius = 1 S o only z = - 1 lies inside 'c' R = Res f(z)= lim (z+1) f(z) Z➔ -1 (z=-1) (z - 1) 3 z2 + 2z ) = -� 4 4 z3 (Q0 fro m ) Hence by Cauchy's residue the orem. f f(z)dz = 2ni[ R ] = - Method II z2 f �z = Z -1 .!_ 1 1 1 ++ .... f(z) = e 2 = 1 + - + 2 z 2 ! z 3 ! z3 R1 = Res f(z); (z = 0) = Coefficient of z2 = lim [(z+1)-] 4 Z➔-1 z = -2 ± i a re simple p oles which lies outside c, so , I = 0 (By Cauchy integ ral theo rem) So 479 = ni 2 z dz f (Z -1)(Z +1) 1 2 2 1 2 1 - 2[ ( (Z -1/ (z +1)) dz 2 2 C is j Z + 1 j= 1 1 1 dz dz+� f = �rh _ 2 (Z+i)(z -i) 2 'J' (Z-1)(Z+1) = ( z�1) dz+0 �f 2 (Z+1) L Only z = -1 lies inside C.] integ ral becomes analytic and it's integ ral will be ze ro using cauchy integ ral the orem. S o 2nd C N ow by C auchy Integral fo rmula, 1 = � 2ni [-] 2 Z -1 Z=-1 ni = 2 N ote: We can also use cauchy Residue the orem to evaluate above question. Sol-20: (c) Method I : 1 2 f(z)= - - - and c : lz + 1 1 = 1 (ci rcle) z+1 z+3 -Poles of f(z) a re z = - 1 and - 3 but only z = 1 lies inside 'c' so we will calculate residue at z = - 1 only . i.e. IES MASTER Publication 480 Complex Integration R = Res f(z) = lim (z+1)f(z) Z➔-1 1 2 z+1 2+3 (z=-1) = lim (2+1)[-- - --] 2➔-1 -3 2(2+1) = lim (1) =1 Z➔-1 2+3 So pole lies inside 'c' C R= Res f(z)= lim _[(z+2 j)f(z)] (z=-2j) 0 lim ( z2 - z+4 j)=- 4+6 j = Z➔-2] By C - R - T, So by C - R - T Method II Jf 1 z+1 2 z+3 dz 1 <p � <p f (2)dz= 2n j C z+1 21tj C C : <p f(2)dz = 2n j (Sum of residues) f(z)dz= 2ni (R) = 1 f(z)= 12 + 11= 1 = 2n ( j -4+6 j)= -12n - 8n j 1 ,f, 2dz · 'Y 2+3 2nJ C z= -3 lies outside c and z= -1 lies inside c � ,f, f(2)dz = � x 2njg(-1)-0 21tJ 21tJ (Using Couchy Integral Formula) =1 (where g(z) =1) ! = - 4n(3+2 j) 22 - 2 4 j � dz, J2J=3 z+2J C Method II : I = <p 2 =-2j is singular point which lies inside Jzl = 3 . By cauchy integral formula 22 -z+4 i dz = 2n jf(-2 j) z+2j C <p Sol-21 : (a) z2 - 4 22 - 4 ,f, _ dz C · lz- ·11 _ dz = J. -2 2 2 +4 (z-2i)(2+2i) ' · ! Z➔-2J ! z= -2i, 2i are simple poles z= -2i lies outside the curve and z = 2i lies inside the curve. 2 = 2ni(4 j +2 j+4i) = 2nj(6 j - 4) Sol-23: (b) 2 = 2n(+6 j - 4 j) = -4n(3+2i) l l n zl3· s' z2 - 4 J. --dz = 2ni Res{f (z}, 2i} 2 'j' 2 +4 C = In 3i - ln5 = -41t 5 = 2ni ( Sol-22: (c) Method I f(z) = ( C : l zl= 3 22 ;:� 4 = In 3 + In 8 4i _ 4 )= 2ni(- ) 4 ·, 2i+2i 2 3 . 7t = l n-+1- Sol-24: (c) i) s o pole is 2 = - 2 j eint2 - In 5 2 - 0.511+1.571ij 22 2 z -1 z= -1, 1 are the simple poles of f(2) and z = -1 lies outside the curve C: 12-1 I = 1 f(z)= Engineering Mathematics and z = 1 lies inside the curve C : lz - 1 1 = 1 so, p t(z)dz = 2ni Res[f(z),1] Sol-28: (d) = 21t i c: ) = 1ti Sol-25: (d) 1 We will go through options z f(z) = 2 _ (a) 2 1 Res{f(z); z = -1 } = ,r, _ 1 f (z) dz = f(2tJ) 21ti 'j' C Z - 20 1 Let f(z) = 2 so poles are z = 0, 0 (Double pole) z and c : lzl = 1 so pole lies inside 'c' f(z) = Z = X - iy U = X , V = -y d2 R = Res f (z) = � [ :�1 ( z - o )2 f (z) ] I2 1 dz (z=O) z=O u x = 1 , vx = 0 Ux Vy -1 = [�(1)] = 0 dz z=0 f(z) does not satisfy C.R. eqn ⇒ # By C - R - T, f(z) is not analytic. Sol-26: (b) ,r, 'j' C " _ 2nif ( z0 ) _ 0 n+1 n! ( 2 - Zo ) (Using Cauchy integral formula) dz where f(z) = 1 Sol-27: (1) C : lzl = 1 0 � 0 � 2n f" (z0 ) = 0 ⇒ ⇒ z ei0 & dz = je i8 d0 where = . . 1 8 8 = -. Re(e1 ) j e1 d0 21tJ 0 2n f = /f J .. 21tJ 0 cos 0.i(cos 0 + j sin 0)d0 1 2 = - [ cos 0d0 - cos 0 sin 0 d0 2n o o 2" f 9 ( z - 1) --2 ] Method I : ,r,(- = ⇒ g'(z) = 9 4 = Sol-29: (b) 1 1 - ) dz = f(0) = 1 2ni ! z - 0 0, Vy g' ( -1) 1! 9 2 - - ( -1-1) (c) By Cauchy integral formula = ( z - 1) ( z + 1)2 9 9 [where g(z) = z-1 p z2dz = 0 Uy 1 1 (n - 0) = 2n 2 9 9 \ =Res{f(z); z = 1 } = -(1 + 1)2 4 (b) f(z) = z2 is analytic every where By cauchy's theorem (d) f(z) = Poles are z = 1 , -1 1 Res {f(z),1 } = lim (z - 1) f (z) = lim (-- ) = � Z➔1 Z➔1 2 + 1 2 ⇒ = 481 2n f l pt (z)dz = 2ni ( R) = 2ni ( 0) = 0 Method II : ·: Cauchy integral formula for 1 st derivative is f'(Zo ) = 1 J, f(z)dz 2ni 'Y (z - z0 )2 - rh 1 dz - 2nif'(0) = 0 - 'f ( Z - 0)2 C So C [·: f(z) = 1 ] Sol-30: (-1 33.87) Method I : f(z) = Let sin z ( z - 2nj)3 So poles are z = 2nj (poles of order 3) R = Res f (z) (z=2nj) IES MASTER Publication Co mpl ex I ntegratio n 482 = d3 3 (z -2nj) f(z)] [ 1 3� 1 dz;�1 2= 2 n 2J = � [�2 ( si nz)] � dz = 2=2 nj e-2 n - e2 n -4 j = -( (· : si nz ) 2 2=2 nj sin8 = eie - e-ie ) 2i P ole lies inside 'c' so b y C - R - T. --1 f 2n . st n2 (z - 2ajf 1 dz = -- cj> f(z)dz 2n c 1 = - - [ 2n i(R)] 2n · e-2 n - e2 n - e-2 n - e2 n 1 = - - [ 2 m ( ---)] - --2n 4 -4 j = - 133.8 7 1 rf- si nz dz Method II : I = - - 'f 3 2n c (z - j2n) Using cauchy ' s i ntegral formula; ( 2n) f" j I = _ _!_ x j2nx 2n 2! L et ⇒ f(z) = si n z f'(z) = cos z f"(z) = -sin z at z = j 2 n ; f"(z = j2 n ) = - si n (j2 n ) 1 . - sinj ( 2n) - = - - xJ2n x 2! 2n = -133.8 7 Sol-31 : (b) 1 e2 1 2 (i) - f - dz = -. 2nj f ( 2) = e "' 7 .39 2nJ 2nj c z-2 (ii) 1 th e2 2nj 'Y z-2 C = 0 Sol-32: (b) Method I L et 3z -5 f(z) = ( 2 _ 1)(2 _ 2) so poles are z = 1 & 2 R1 = R es f(z)= lim((z -1)f(z)) = 2 2➔1 (2=1) � = R es f(z)= li m((z - 2 )f(z)) = 1 2➔2 (2=2) L et us assume that z = 1 li es inside gi ven contour and z = li es outsi de i t. Then by C - R - T, f f(z)dz = 2ni (sum of residues at poles inside) = 2ni (2) = 4ni H ence verifi ed, So our assumption that z = 1 li es inside r and z = 2 lies outside r i s correct so corre ct path of r is defi ned by option (b) Method I I Using Cauchy' s formul a, 3z-5 3z -5 rfdz _- rf- z -2 dz 'f z -1 'f ( z-1)( z-2) . 3z - 5 = J 2Jt [-- ] z - 2 at z=1 = j2n [�= �] = j 2n [ =�] =j4n H ence, z = 1 li es inside the cl osed path r. and z = 2 l ies outsi de the closed path r . Sol-33: (a) Given: I = "' sinx -- dx J2 x +2 x +2 -«) Now; L et us assume: ei x = cosx + j sinx ⇒ Now; si n x = lmag [ei><] ; L et us assume a new function f(z) = l mag [ eiz ] z2 +2z+2 where z2 + 2z + 2 = 0 represents poles of f(z) ⇒ z = -2 ± 4 -4 ( 2) ✓ 2 Engineering Mathematics = -2 ± --1-4 2 j z = -1± 1 = -2 ± j2 2 From the z-plane it is clear that z = -1 + j1 is the only pole lying in f(z) > 0, i.e., in upper half plane. .a -R �· Re s f(z) = Z➔ -1+ j1 = R lim . [z - (-1+ j1)] -1 +J1 Z➔ X e iz [z -(-1+ 1j ) ][z - { -1- j1)] e iz lim Z➔-1+j1 Z+(1+ 1 j) e i (-1+i1) -1+ 1j +1+j1 e +1 j2 (using Residue theorem) i = Imag [ n:- ] = lmag [� [cos (1) - jsin (1)]] nsin (1) = - -e [taking imaginary parts] I = f(z) = sin z _!_ _ z3 z5 = 2 [z + ... ] 3! 5 ! · z2 z f(z) = sin(z) _!_ _ z3 z5 + = 2 [z 3 ! 5! z2 z 1 z z3 f(z) = z - ! + . . . 3 5! : . Residue o f the pole (at z = 0 ) = Coefficient of 1 . . . ( _ ) in power series expansion = 1 z 0 Method II tc 2Z+5 (z - i ) { z -4Z + 5 ) 2 dZ Only Pole Z = 1/2 lies inside the circle IZI = 1 I = 2ni x [ Residue at Z=i] 1 z z2 = - - -+ -! -.... z 3! 5 So Z = 0 is pole of order 1 (simple pole} _ . . . .] So, Res f(z);(z =0)= lim (Z - 0)f(z) Z➔ O -i = lmag [ j2n • � ] J 2e Sol-34: (b) 12ni 48ni = --= 13 1 -+3 ⇒ +1 lmag (eiz ) dz = lmag [ j2n · e ] 2 72 c z +2z+2 . J .. = Sol-35: (1) Method I Given: 483 Sol-36: (1 ) Given : . sin z = I1m (-) = 1 Z➔O Z 2 2 _1_ ,.f, z +1 dz = _1_ ,.f, z +1 dz 2 2n j 'f z -1 2n(f (z -1)(z+1) Poles are z=1,-1 (Both simple) C Given C is (x -1)2 + y 2 = 1 ⇒ C :I z -1 = 1 1 Clearly 1 lies inside of C and -1 outside C Then Res f(z) (at z = 1 ) = Lt (z -1) 2 ➔1 z2 +1 (z -1)(z+1) z2 +1 = lim ( ) =1 Z➔1 Z+1 :. By Cauchy's Residue theorem 1 1_ ,.f, Z2 + 1 dz = -- x 2njx1= 1 2n j 2n j 'j' z2 -1 IES MASTER Publication 484 Complex Integration Sol -37 : (d) substituting y = z2 + 1 z2 +4 For singular points i .e. for poles, z2 + 4 = 0 f(z) = Residue for I = f ( x2 + ix2 ) (dx + idx ) f x2 (1+i)(1+i)d x 1 = = ( 5)3 1 25 = x ( 1 - 1 + 2i) d x -1 1 2 = - X --=3 2! ( -5) 1 25 = 2{ ; I = 2 i { �] f(z) = 1 ( z - 4) = Residue for Sol-39: (d) Poles are z = As both z = Residue at Residue at ��4 1 [( z - 4) f ( z)] ±i ± i lie inside the circle lzl = 3 = Lt 2➔ i - ( z - i)(z - i) 2 (z - i) (z+i) -'--------'--- 8 2 -1 - 1 = -- e1 = iei zi (z + i) = (z+i)(z - 1) Lt - ��------'---e 2 z➔ -i (z - i)(z+i) 2 = -ie-i By Residue thorem I = 2n:i (iei - ie-i ) Sol-40: ( b ) Given i ntegral I = C 1 ( z - 4) ( z+1)3 -1 1 . and 1 25 1 25 (z - i) C y = 1 u 1 d2 [( 1 3 1) X = 1-3 ( z+1) z➔- 2 ! dz2 Z + (z - 4) (z + 1) ] So, residue is in given integral we get = z = ±2i Sol-38: ( b) x = - 2n:(cos1 +i sin 1 - cos1+i sin1) = -4ni sin1 f ( x2 + iy 2 ) dz, z= x + iy the line C is a straight l ine patsing through origin and its equation is given by y = x f C 2 X 1 f 2i x 2dx = 2if� x2 dx x 2i Sol-41 : ( 1 4n:i ) . f -z - 2z+3 -- dz z-2 Given contour I ntegral Is 3 where z3 - 2z+3 and the countour is lzl = 3 in counter f(z) = (z _ 2) 1 clockwise. The pole z = 2 lies inside the contour lzl = 3. Res(f(z)) = Lim(z - 2) [ 2➔2 z3 - 2z+3 ] (z - 2) = 8 - 2(2) + 3 = 7 By Cauchy residue theorem J, Z3 - 2z+3 dz = 2n:i[Res(f(z))] 'f (z - 2) Sol-42 : (c) = 2n:i(7) = 1 41ti f(z) 1 = zsinz ⇒ , c : lzl = 1 Poles of f(z) are z sin z = 0 z = 0 and sin z = 0 z = 0 and z = nn , n E I Engineering Mathematics 485 z= 0 and z= ..... -2n, - n, 0, n, 2n, ..... so z= 0 (double pole) lies inside ,c. R= Re sf(z)=-� i [�(z-0)2 f(z)] (2 1. ) dz {z ;Q) z;Q f(z)= 2ni [sum of residue inside c] 2z (z-1)2 (z-2)' lim(z-2)f(z) = lim [�] = 4 R1 = Resf(z)= 2➔2 2➔2 (z -1) {z;2) Rz = Re sf(z) = - � , [�(z-1)2 f(z)] (2 1. ) dz {z;1 ) z;1 --4 ) ( [� dz � z-2 ]z;l 23 are z = 2, 3 & 1 Poles of f(z) = (z-1)4(z-2)(z-3) in which 2 and 3 are simples poles and 1 is pole of order 4. Re s f(z) = lim(z-3)f(z) Z➔ 3 {z;3) 27 23 = lim = 4 16 z➔ 3 (z-1) (z-2) l l 1 1 1 1 = -z 1 112[ z[ 2 = - 1 [ -1 [ z 1-�] - - z-1 2 -1 1 1-z] 2 3 11 4 = -2(J_3)-2(J_2 )-i(!)_3_�( z)+... 3 z 3 120 2 z Which is Laurent expansion of f(z) about z = 0 and it is pole of order 3. So re·sidual of f(z) at (z= . 0)= coefficient of (�) 4 3 1 th dz I -- 'f ( 3 c z-1) ( z-3) Sol-44: (d) z-2 1-e 2 z z4 f(z) = Sol-47: (b) Hence r equired sum = R 1 + R 2 = 0 f(z) = ( z-1) ( z '..... 2) 1 1 1 )- - 2- 3 z z 2 1-(1+22+ (22) + (22) + (22) +···) � [_g = z4 poles are z = 2 (simple pole) and z = 1 (double pole) Sol-45: (d) z3 16 Sol-46: (b) = 2ni[0] = 0 Sol-43: (c) z2 8 So coefficient of ( :) is = - 1 2 so by Cauchy residue theorem, f f(z)dz = 1 z 2 4 = ( Poles of integrad are z= 3 (simple pole) z= 1 (pole of order 3) F(z)lim(z-3) F(z) =! R 1= Res (z; ) z➔3 8 3 1 d2 3 R2 = Res F (z) = I 'l 1 [ 2 (z -1) F ( z)] dz {z;1l � z;1 1 8 Now by cauchy-Residue theorem (':1<lzl<2) Sol-48: (a) 1 C : lzl = 1 and f(z) = -5z- y Pole of f(z) is z = ¼ (simple pole) IES MASTER Publication 486 Complex Integration @c Res f(z) <2 = 415 = �, ( z - �)t(z) ) 5 1 = lim (!)(5z - 4) (5z-4) 2➔ ± 5 ! = lim ( )= _! 5 2➔ ± 5 z = -2 lies inside C. Only R = Resf ( z) (Z=-2 ) = Lim ( z+2) f ( z) 2➔-2 5 Pole lies inside 'C'. So by Cauchy's Residue theorem = = j f(z)dz C 1 f(-)dz 5z-4 211:i = - = Ani (given) 5 C = Hence on comparison A = ¾ = 0.4 Sol-49: (0) Sol-51: (b) 1 Let f(z) = -2- then poles are z = ± 1 . z -1 Both are simple poles and lies inside the given contour R1 = Resf(z)lim(z-1)f(z) (2 =1) 2➔1 � = Res f (z)= lim (z +1)f (z) 2➔ -1 ( 2 =- 1 ) 1 lim (- ) =-_! 2 2-1 2➔ -1 But we will take R2= + 1 2 / because second contour is in clockwise direction. So by Cauchy's residue theorem 1 1- dz = �[211:j(sum of residues)] 11:j z2 -1 71:J I = 2[J+J]= 2 Sol-50: (a) C : lz + 21 = 1 and f ( z) = � z2 -4 z+ 1 -2 + 1 = _! )= Z-2 -2-2 4 f(z) = 11:i 2 [By Cauchy Residue Thoerem] (z2 -32+ 2/ 22 2 = 2 (z-1) (z-2) So poles are Z= 1 and 2 (Both are Double pole) i.e. m= 2 1 ! = lim(- )= 2➔1 z +1 2 = Lim ( 2➔-2 = f/( z) dx So, = 211:i (sum of residues) fa� � C : IZI = 4 . Hence both the poles lies inside 'C' 1 = Resf(z) = [�(z-1)2 f(z)] 1') 1 dz < 2=1> t£=-1 2 =1 = 2 [ :z (2� 2)2 1=1 =4 1 � = ResF(z) = [ (z-2)2 f(z)] 1 � 1') dz t£=-1 > <2=2 2= 2 = 2 d =-4 [ dz (z � 1)2 1= 2 So By Cauchy Residue Theorem, I= f f (z)dz = 2ni[R1 +R2 )= 0 C Indefinite Integration Collection of all the anti derivatives (primitive) is known as indefinite Integration. __![cp(x) +c] = f(x) or ff(x) dx=cp(x) +c dx DEFINITION OF DIFFERENTIAL EQUATIONS An equation which consist dependent variable, independent variable and differential coefficient of dependent variable with respect to independent v�riables is known as differential equation.· Ordinary Differential Equations (ODE) If there is only one independent variable in a differential equation then it is ordinary differential equation. Partial Differential Equations (PDE) If there is more than one independent variable, present in the differential equation then it is called partial differential equation. Examples : (1) (��r 3 2 +5xy=sinx (3) ( �� =( �) : r dY d y (2) (- 2)+( ) +logx=5 dx dx ri z & (4) -+-+5xy=0 axay ax ORDER AND DEGREE OF ORDINARY DIFFERENTIAL EQUATION Order = Degree = The order of highest derivative occurs in a differential eqn is known as its order. Degree of a differential equation is the exponent of highest order derivative when it is mad!=) free from fractional notations and Radical signs. OR It is the exponent of highest order derivative when it is written as a polynomial in differential coefficient. [ Example 1 Solution 2 1+ ( ) = Find the order and degree of following; p � d y dx First we will write above in form of polynomial i.e. 2 d2 y d = [1 +(_J_ ) ] p2 dx dx 3 2 ' 2 r2 488 Ordinary Differential Equations Squaring both side p 2 (::; r = [ 1 +(:�rr ⇒ ORDER➔ 2 & DEGREE➔ 2 2. ORDER➔ 1 & DEGREE➔ 3 3 dy d2 y + 1+( ) = 0 2 dx dx 3. ,. 3 dy d2 y ( = 1+( ) ) 2 dx dx 4 ( 5. ORDER➔ 2 & DEGREE➔ 2 ORDER ➔ 2 & DEGREE➔ 1 B dy d3 y ) -6x 2 ( ) = 0 3 dx dx ORDER➔ 3 & DEGREE➔ 4 NON-LINEAR DIFFERENTIAL EQUATION A differential equation is said to be non-linear if it satisfies any of the following four properties. 1. Degree is more than one. 2. Exponent of dependent variable (i.e. y) is more than one. 3. Exponent of any differential coefficient is more than one. 4. Product containing dependent variable and its any differential coefficient is present. Linear Differential Equation If a differential equation does not possess any of the above four properties then it is linear. Example a x(:�)+2 = 0 1. x 2 (::; r + 2. dy d2 y dy +x( ) +2y = 5 (Non linear) (exponent of is more than 1 ) 2 dx dx dx (Non linear) ·: degree more than 1 . 2 3. 4. 5. y'"+ yy'+2x 2 = 7 Non linear (Property 4) " y '+(y' ')2 +y 2x = sinx Non linear (Property 2 & 3) d3 d2 d --{ +---{+sin x --1. = logx (Linear) dx dx dx SOLUTION OF DIFFERENTIAL EQUATIONS Relation, between dependent variable and independent variable which satisfies the given differential equation is known as its solution. Types of Solutions (i) General solution (ii) Particular solution Engineering Mathematics 489 (i) General solution : If a solution contains same no. of arbitrary constant as the ord er of differential equation then it is known as general solution. (ii) Particular solution : If we assign particular value to arbitrary constant then it is known as particular solution. FORMATION OF DIFFERENTIAL EQUATION By eliminating the given number of arbitrary constants from general solution, we can easily form a d ifferential equation. Example 2 Solution 1 Find the d ifferential equation whose solution is, y = ex (Acosx + Bsinx) where A and B are arbitrary constant. y' = ex (A cosx +Bsinx)+ex (-A sin x +Bcosx) = y+ ex (-A sinx +Bcosx) 1 Again differentiating, ⇒ ⇒ Example 3 Solution 1 1 1 ⇒ y " = 2y'-2y ⇒ Find a differential equation of which xy = Aex + Be- ⇒ ⇒ Solution y " = y '+(y'- y)- y xy = Aex + Be- 1 Example 4 y'+e x (-A sinx +Bcosx)+ ex (-A cosx-Bsinx) x + x2 y+xy' = Aex -Be-x +2x 2y'+xy"-(xy-x 2 )-2 = 0 ⇒ y"-2y'+2y = 0 x + x 2 is the solution by eliminating A and B. y'+(y'+xy") = Ae x +Be -x +2 Find differential equation of all circles having radius r. We can take equation of circle as (x - a)2 + (y - b}2 = r2 where a & b are Arbitrary Constants. Differentiating w.r.t. x ...(1) 2(x- a)+2(y-b)�� = O Again differentiating 2+2 (y-b) 2 2 y dy +2 ( ) dx dx 2 d ...(2) = O -1 (y - b) = So, Put this value in (2), we get ...(3) -(��r d 2 y dx 2 1 - - (dx r dy (x- a)+ = 0 2 dy y 2 dx d dx dy dy +( J dx dx So, (x - a) = 2 d y 2 dx Put these values in (1 ), we get IES MASTER Publication 490 Ordinary Differential Equations = r2 Example 5 : Solution : (.d2 y ]2 dx 2 Find Differential Equations of all parabolas given by y2 = 4a(x + a), where a is an arbitary constant y2 = 4a(x + a) Here 'a' is Arbitrary Constant Differentiating w.r. to x ... (1) 2yy' = 4a ⇒ Put this value in (1), we get y2 (y')2 + 2xy a = yy' 2 dy -y2 = 0 dx 2 d d 2 y 2 ( _J_) +2xy _J_ - y = 0 dx dx i.e., Example 6 : Which one of the following differential equations has a solution given by the function (a) � � -¾cos(3x) = 0 y = 5sin ( 3x + i) d -f + 9y = 0 2 (c) Solution : (c) d X ·: �� = 15cos( 3x + �) and::; = -45sin( 3x + �) So only option (c) satisfying it. WRONSKIAN 2 If y1 , y2 are two solutions of second order differential equation a 0 as; W = I Y1 y; dy d ; + a1 + a 2 y = 0 then their wronskian is defined dx dx If, y1 , y2, y3 are three solutions of third order differential equation then their Wronskian is defined as; W = Y1 Y1 Y1 Y2 Y2 Y2 Y3 Y3 Y3 LINEARLY DEPENDENT (LD) AND LINEARLY INDEPENDENT (LI) SOLUTIONS Linearly dependent (LD) Let y1 , y2, y3 are solutions of given differential equation and if there exist a relation between y1 , y2, y3 such that c 1 y1 + c2y2 + c3y3 = 0 where c 1 , c2, c3 not all zero simultaneously then they are linearly dependent. Engineering Mathematics Linearly independent (LI) 491 If there does not exist any relation between y1 , y2 , y3 then they are linearly independent (LI) i.e. Relation c 1 y1 + c2 y2 + c3y3 = 0 exist only when c 1 = c2 = c3 = 0 * 0 ⇒ LI Solutions If wronskian of y1 , y2 , y3 is zero then they are linearly dependent, i.e., W = 0 ⇒ LD solutions and W 2. Example 7 : Solution : If y 1 , y 2, are two LI solution of a given differential equation then y = c 1y1 solution of that equation. + c 2y 2 is the general Determine the differential equation whose set of independent solutions are {eX, xeX, x2e'<} * Y1 = eX, Y2 = xeX, Y3 = x2ex Let ⇒ then general solution is, w 0 ⇒ y1 , y 2 , y3 are LI y = C 1 Y1 + C2Y2 + C3Y3 = c1 ex + c2xex + c3x2ex ...(1) where c1 , c2, c3 are arbitrary constant. X X 2 X 2 X X y' = e c1 + ( e c 2 + xe c 2 ) +( xe c 3 +x e c 3 ) = y +e X c 2 +2xe X c 3 Now, x x x x x y" = y' + e c 2 +(2e c 3 +2xe c 3 ) = y' + 2e c 3 +(y' -y) = 2y' -y +2e c 3 y"' = 2y" -y' +2e x c 3 = 2y" -y' + (y" -2y' + y) = 3y" - 3y' + y Example 8 : Solution y"' -3y" + 3y' -y = 0 Prove that 1, x, x2 are linearly independent and form the differential equation whose roots are {1, x, x 2} 1 W = 0 0 X 1 0 X 2 2x =2 i.e. W 2 Y = C 1 + XC2 + x2c3 y' = C 2 + 2XC3 ⇒ *0 So, given Solutions are L.I. Hence, general solution is y" = 2c 3 = C 2 +2XC 3 ⇒ ⇒ y" = 2c 3 METHODS OF SOLVING DIFFERENTIAL EQUATION 1. 2. 3. 4. 5. 6. y'" = 0 Observation Method Differential equations of first order and first degree (a) Variable separable method (c) Linear DE method Exact differential equation method (b) Homogeneous DE method (d) Bernoullie DE method Linear differential equation of nth order with constant coefficient (CF + Pl) method Methods of solving partial differential equation + Methods of solving non linear differential equation DIFFERENTIAL EQUATIONS OF FIRST ORDER AND FIRST DEGREE The general form of first order, first degree differential eqn. is given by : f ( x, y, �)=0 : Geometrical Interpretation : It establish a relationship between the co-ordinates of a point and slope of tangent at that point e.g. dy + 2x = y dx IES MASTER Publication 492 Ordinary Differential Equations Method of Solving First Order - First Degree Differential Equation (i) Variable separable method (ii) Homogeneous DE method (iii) Linear differential eqn. method Homogeneous Function : A function f(x, y) is said to be homogeneous if it can be represented in following form f(tx, ty) = t" f(x, y), i.e. the d egree of each term is equal where n is any real no. Homogeneous Differential Equation= If a given d ifferential equation of 1 st order and 1 st d egree can be represented in form. dy dx = f(x, y) where f(x, y) and g(x, y) are homogeneous functions of same d egree then it is known as g(x, y) homogeneous differential equations. Linear Differential Equation lot first order and first degree) dy dx + Py = Q where P & Q are function of x alone or constant IF = ef y(IF) = J Q(IF) dx +C p-dx Solution is; Integrating Factor UFJ A special term such that by multiplying it with the given equation it can be solved easily. Bernoullie Differential Equation A DE, which can be reduced to Linear Form by assuming suitable transformation is called Bernoullie Differential Equation dy For e.g., -+Py= Qy n., n > 0 it can be reduced to Linear Form by assuming y1 -n dx Equation. Example 9 = Solve: y-x Solution 1 dy dx = a(y2 + dy dx ) by using variable separable method y-ay2 ⇒ = (x+a) dy = ⇒ dx y J y(1-ay) d log(x +a) = logy+ ⇒ ⇒ J x+ax d log(x +a) alog(1-ay) +loge a = ⇒ = logy- log(1-ay)+loge ⇒ = v so it is Bernoullie Differential (x + a) = a J [�+y 1-ay ] dy log(x +a) cy 1-ay Example 10 1 Solve; (1 + exiY)dx + exiY (1 - x/y)dy = 0 by using the concept of homogeneous DE method Putting x = vy ⇒ dx/dy = v+yd v/dy ⇒ ev ⇒ (1 + v(1 )dx + e - v)dy = 0 dv -ev (1-v) -v y- = --dy (1+ev ) ⇒ ⇒ dx dy = ⇒ dv -ev (1- v) v+y­ = --dy (1 + eV) ⇒ dv y- = dy Engineering Mathematics ⇒ (1 +ev ) dv = (e v +v) J v+ev = ⇒ Example 11 Solution 1 1 Solve : + loge ⇒ f -dy y C ⇒ y 493 log(v+ev ) = -logy+ loge �+ex/y = y y C dy Y + logy = Y2 (logy)2 by using the concept of LOE method. dx x x Right now it is not LOE, so we will try to convert it into standard form of LOE. Dividing whole equation by y(logy)2 ...(1) 1 dy 1 1 1 -+-x- - = 2 ---x 2 dx x logy y(logy) x 1 Let Differentiating w.r.t. x logy = V dv d 1 dy dv d 1 -1 = ( ) ⇒ ( ) = dx ⇒ dy logy dx dx dx logy (logy)2 Putting this value in ( 1 ) 1 dy dv y dx = dx ...(2) -1 dv -1 +( ) v = 2 x dx x ⇒ It is LOE in V & X, so I.F. = e Hence solution of (2) is J�dx x = e- log x = 1 /x v-(IF) = fO(IF}dx+c V X = ⇒ -(2)+C ⇒ 2X 2 v-(�) = f ( :: ) (�) dx + c .1 _1_ +C = x logy 2x 2 EXACT DIFFERENTIAL EQUATION A differential Equation is said to be exact. If it can be obtained by directly differentiating its primitive (solution/ Antiderivative) without any further operation of elimination and reduction. Example = xdy + ydx = 0 is an, exact differential equation because it can be directly obtained by differentiating XY = C. Necessary and Sufficient Condition For a differential equation ...(1) Mdx + Ndy = 0 8M ay to be exact and solution of ( 1 ) is given as follows : · f yconstant = 8N ox f Mdx + (those terms of N which are free from x) dy = C IES MASTER Publication 494 Ordinary Differential Equations Example 12 : (eY + 1) cosx dx + eY sinxdy = 0 Solution : Comparing with Mdx + Ndy = 0 M = (eY + 1)cosx, N = eYsinx Hence, equation (1) is exact DE ⇒ . . .(1) · ax ax aM = e Y cos x aN=e Y cos x ⇒ aM aN = ay ay So, solution of (1) is : J (e Y + 1)cos xdx + J Ody= C, ( keeping y canst. in 1st ) (eY + 1)sinx = c Example 13 : (x2 + y2 - a2)x dx + (x2 - y2 - b2)y dy = O Solution : Comparing with Mdx + Ndy = 0 .. .(1) N = (x2 - y2 - b2 )y M = (x2 + y2 - a2)x; ax ax aM aM aN aN = 2yx, - = 2xy ⇒ -=ay ay So, it is exact eqn. Hence, Solution of ( 1) is 2 2 2 J x + y -a )xdx + J (-y 3 -b 2 st y) dy=c (Keeping y constant in 1 ) Example 1 ' : (1 + exiY)dx + exiY(1 - x/y)dy = Solution : o Comparing with Mdx + Ndy = 0 M = (1 + exiY), N = exiY(1 - x/y) : = e x/y ( n � ax aM aN = ay Since Hence, equation is in exact form. So, solution is J Mdx + J (those terms of N which are free from x) dy=C J y = co nstant X x + J e Y dx + J Ody = c C+ e ⇒. f) dx + J Odx = c x + J e t ydt = c EQUATION REDUCIBLE TO EXACT FORM Rule 1 Rule 2 : Rule 3 : ⇒ X x + yet = c If Mx + Ny * 0 and equation is homogeneous then IF = Mx + Ny ⇒ x + ye Y = c If Mx -Ny * O and Equation is of the form f 1 (xy)ydx + fixy)xdy = 0 then I F = ax aM aN J d _ If _!_( )=f( x) alone then IF=e t(xJ x N ay Mx - Ny 1 aN BM - ) = f(y) only then If - (- M ax ay Rule 4 : Rule 5 : If eqn. is of the form Engineering Mathematics If=e f f( y ) dy x ayb(pydx + qxdy) + x cyd (rydx + sxdy) = 0 Let its integrating Factor is IF = x y h k By multiplying it in given eqn. (1 ), it will become exact so it satisfies ax 495 . . . (1) aM aN = ay Now, we can compare the coefficients of x and y by which h and k can be evaluated easily. Example 15 : (y3 - 3xy2) dx + (2x2y - xy2) dy = 0 BM Here we have ay Solution : ...(1) aN *ax So, above is not exact eqn, so by using Rule-1 Mx + Ny = xy3 - 3x2y2 + 2x2y2 - xy3 = - x2y2 Now, IF = Multiplying it in given eqn. (1) 1 Mx + Ny -y 3 -2 1 +- ) dx +(- +- ) dy = 0 (2 X x Y X BM ay Now -1 x 2 y2 . . . (2) ax BN Hence (2) is an exact eqn so its soln is 1 Example 16 Solution 1 1 -y [ � ] +31ogx -2 1ogy = c 1 (x2y3 + 2y) dx + (2x - 2x 3y2)dy = O BM Here we have ay ⇒ (y constant in first) Y + 3 1ogx -21og y = c X 8N *ax So above eqn is not exact eqn so by using Rule 2. 1 IF = --Mx -Ny Multiplying it in given eqn (1) Now ax 8M BN = ay (x 2 y3 +2y)x -(x -2x 3 y2 )y . . .(1) 1 3x 3 y3 2 2 _ I ) dy] = o � [(� + ) dx + (3 x x 3 y2 x2 y3 y 2 2 - I ) dy = 0 + - dx +((� 2 3 x x 3 y2 ) y x y . . . (2) IES MASTER Publication 496 Ordinary Differential Equations Hence (2) is an exact eqn, so its solution is f Mdx +f (those items of N which are free from x) dy = c ax aN · aM Now - = ay -2 2 fy const(!x +) dx+f ( ) dy = log c ⇒ Y y X 2 1 logx+-2 Iogy = log c xy2 2 -2 logx+logy -loge+-;. = xy -1 X o ⇒ [keeping y constant in Is� - log2 cy 2 - xy ⇒ -1 2 xy X -1 =e cy2 2 xy2 x = cy e 3 Example 17 , (x +2 y ) Solution , dy 3 2 = y +2x y dx Given equation is (y + 2x 3y2)dx - (x + 2y3)dy = 0 Here we have, ax aM aN So, (1) is not exact -cf. ay Using Rule 4; -2(1+2x 3 y) -2 1 3 = f(y) alone = [-1-1-4x y] = y(1+2x 3y) y+2x 3 y 2 Y Jf y IF = e ( ) dy =e 2 y Jy -2 2 = e- 1og y = elog y = y - Multiplying the given eqn (1) by I.F. [ (x +2 y3 ) dy 1 x y) - = -+2x 3 (-+2 2 dx y y Now ...(1) d 2 1 7 dy = y +2x 3 y 2 ] � dx y ⇒ ...(2) ax aM ay aN So (2) is an exact eqn. so it' s soln will be 3 r (t+2x ) dx+ (-2 y) dy = c f 4 y2 X 2x -+ -+(-2) - = 4 2 y Example 18 , Solution 1 (x2 + y2 + x) dx + (xydy)= 0 Since aM 8y -cf. ax aN Using Rule 3; C ⇒ X (y constant in first) X 2 -+- - y = C y 2 4 Hence eqn (1) is not in exact form 1 1 1 -(2y- y) = -(y) = - = f(x) alone xy xy x ...(1) Engineering Mathematics Multiplying the given eqn. (1) by I . F. x[(x2 + y2 + x)dx + (xy)dy) = 0 (x3 + xy2 + x2) dx + (x2y) dy = 0 (2) is exact equation because 497 .. . (2) BM BN = ay ax Hence solution of (2) is JM · dx + J (those terms of N which are free from x) dy = c 3 J (x + xy 2 + x 2 ) dx + J O • dy = c y const Example 1 9 Solution 1 1 x 4 x2 y 2 x 3 - + -- + - = 4 2 3 C xy3(ydx + 2xdy) + (3ydx + 5xdy) = 0 Given equation is xy4dx + 2x2 y3dy + 3ydx + 5xdy = 0 (xy4 + 3y) dx + (2x2y3 + 5x) dy = O Compairing (1) with Mdx + Ndy = 0 we have . . . (1) a;; * : ⇒ (1) is not exact equation. Let x y be an h k I . F. , of (1 ). [Here we are using Rule 5) so we can multiply it in (1) + (x h+1 yk+4 + 3yk+1 xh) dx + (2x2+h y3+k + 5x1 h yk) dy = o Now it is an exact equation . . . (2) BM ay BN ax BM BN . . S mce, ( 2 ) 1s an exact eqn so we. h ave 8y = ax Comparing coefficients of x and y (k + 4) = 2 (h + 2) 3(k + 1) = 5(h + 1) By solving, we get h = 2 , k = 4 So, ⇒ ⇒ k - 2 h = OJ 3k - 5h = 0 I . F. = x2y4 Putting value of h and k in eqn. (2) (x3y8 + 3y5x2) dx + (2x4 y7 + 5x3y4) dy = 0 Now, eqn. (3) is an exact eqn. so its solution is . . . (3) J M • dx + J (those terms of N which are free from x) dy = c 3 8 5 2 J (x y + 3y x ) dx + J o - dy = c y const x4 x3 - y 8 + 3y 5 - = 4 3 C x4 - y a + x3 y 5 = c 4 IES MASTER Publication 498 Ordinary Differential Equations Example 20 1 (3xy + 8y5) dx + (2x2 + 24xy4) dy = Solution 1 8y "- ax aM aN o ...(1) . . Hence above eqn Is not m exact form Let I.F. = xhyk Multiplying given eqn. by I.F. [Using Rule 5] x hyk [(3xy + 8y5) dx + (2x2 + 24xy4)dy] = 0 (3xh + 1 yk+1 + axhy5+k) dx + (2xh+2yk + 24xh+1 yk+4) dy = 0 . ..(2) ax 8M 8N . . . . . Now (2) Is an exact equation, so It wIII satIsfy 8y = (k + 1 )yk x 3xh+1 + 8(5 + k)yk+4xh = (h + 2)2ykxh+1 + 24(h + 1 )xhyk+4 Comparing coefficient both sides, and 3(k + 1 ) = 2(h + 2) ⇒ 8(k + 5) = 24(h + 1) ⇒ 3k + 3 = 2h + 4 8k + 40 = 24h + 24 On solving, we will get k = 1 and h = 1 So, I. F. = xy Putting value of h and k in eqn. (2) (3x2y2 + 8xy6)dx + (2x3y + 24x2y5) dy = 0 . . .(3) Now, eqn. (3) is an exact eqn. so its solution is J M • dx +J those terms of N which are free from x) dy f (3x2 y 2 f = c 6 + 8xy ) dx + O dy = c 3x 3 y 2 8x 2 y6 -- +-- = C 3 . 2 x3y2 + 4x2y6 = c [y constant in 1 s� Engineering Mathematics ��J: PREVIOUS YEARS GATE & ESE QUESTIONS 499 .,.__,..-1 .� t•-:r.....- Questions marked with aesterisk (*) are Conceptual Questions 1 .* The differential (a) linear (c) homogeneous 2. (2) linear differential equation with constant coefficients (b) non-linear (3) linear homogeneous differential equation (d) of degree two (4) non-linear homogeneous differential equation [GATE-1 993 (ME)] (5) non-linear first order differential equation The necessary and sufficient for the differential equa­ tion of the from M(x,y) dx + N (x,y) dy = 0 to be exact is 8M 8N (b) -= (a) M = N ax 8M 8N (c) - = - ay 3. (1) non-linear differential equation dy d2 ;+ + sin y = O is dx dx ax (a) A-1 , B - 2, C - 3 (b) A - 3, B - 4, C - 2 (c) A - 2, B - 4, C - 3 ay 82M 82M = (d) ax2 ay 2 [GATE-1994) (d) A - 3, B - 1 , C - 2 7. (d) Linear and Fourth degree [GATE-1994) dy 3 3 = x +y dx [GATE-1994-ME] 4. Solve the differential equation, xy2 5. dy For the differential equation + 5y = 0 with y(O) dt = 1 , the general solution is (a) e5t (c) 5e-5t (b) e- 5t (d) e.J-51 [GATE-1 994 (ME)] 6. * Match each of the items A, B, C with an appropriate item from 1 , 2, 3,4 and 5 d2 y dy + a2y- + a3 y = a 4 (A) a 1 2 dx dx 3 d y + a2 y = a 3 (B) a1 dx 3 dy d3 y + a2x - + a3x 2 y = O (C) a1 2 dx dx y" +(y3 sinx) 5 y'+y = cosx 3 is (b) nonlinear (c) second order linear (d) nonhomogeneous with constant coefficients 8. (c) Non-Homogeneous The differential equation (a) homogeneous d2 d4 The differential equation El � +p ; + ky = O is dx dx (where, E, I, P, K are functions of x -only) (a) Linear of Fourth order (b) Non-linear of Fourth order [GATE-1 994 (EC)] [GATE-1995) If at every point of a certain curve, the slope of the tangent equals -2x/y , the curve is __. (a) a straight line (c) a circle (b) a parabola (d) an ellipse [GATE-1 995 (CS)] dy = f (x, y) is dx homogeneous if the function f(x, y) depends only 9.* A differential equation of the form (or) � on the ratio y X y (a) true (b) false [GATE-1 995 (ME)] 1 0.* For the differential equation, f(x, y)�� + g(x, y) = 0 to be exact, ax at ag (a) - = ag (b) �= (c) f= g (d) ay ax . ay a2 f a2 g = ax 2 ay 2 [GATE-1 997-CE) IES MASTER Publication 500 11. Ordinary Differential Equations The differential equation, :� + Py=Q , is a linear equation of first order only if, (a) P is a constant but Q is a function of y (b) P and Q are functions of y or constants (c) P is a function of y but Q is a constant 1 2. 1 3. (d) P and Q are functions of x or constants [GATE-1 997-CE] If c is constant, solution of the equation dy = 1 + y2 is dx (a) y = sin (x + c) (b) y = cos (x + c) (c) y = tan (x + c) (d) y = ex + c [GATE-1 999-CE] 2 d y dy + y = xa - 8. + ( x 2 + 4x ) 2 dx dx The above equation is a (a) partial differential equation (b) nonlinear differential equation (c) non-homogeneous differential equation (d) ordinary differential equation [GATE-1999) 1 4. The solution for the following differential equation with boundary conditions y(0) = 2 and y 1 ( 1 ) = -3 d2 y . - = 3x - 2 is? Where dx 2 (a) x x = 3x - 2 y= 3 2 3 2 x - 5x + 2 2 3 5x x 2 -x (c) y = +2 (b) y = 3x - 2 2 2 x 3 (d) y = x 3 - - + 5x + - 2 2 is 1 (a) Y=­ x +c -x 3 (b) Y=- + c 3 (c) cex (d) unsolvable as equation is non-linear [GATE-2003-ME] 1 7.* Biotransformation of an organic compound having concentration (x) can be modeled using an ordinary 2 different;al equation �: + kx =0, where k is the re­ action rate constant. If x = a at t = 0, the solution of the equation is (a) x = ae-kt (c) x 18. The = a (1 - e-kt) following 1 1 (b) -=- + kt X a (d) x = a + kt differential + y2 + 2=X + 4 (� ::n �r (a) degree = 2, order = 1 3( (c) degree = 4, order = 3 1 5.** The following equation is sometimes used to model the population growth: ��=aN (lnR") where, N is population at time t, a > 0 is the growth coefficient and K > 0 is a constant. Given, at t = 0, N = N 0 and 0 < N 0 < K. (a) Find an expression for the population with time. (b) What is the population as t ➔ oo (c) Find a constant c E (0, 1) such that the popula­ tion growth rate is maximum at N = cK. [GATE-2002) equation has [GATE-2005-EC] 1 9. The solution of the first order differential equation x'(t) = -3x(t), x(0) = x 0 is (a) x(t) = x 0e-31 (b) x(t) = x0e-3 (c) x(t) = x e-V3 (d) x(t) = x e-1 0 [GATE-2001 (CE)] [GATE-2004-CE) · (b) degree = 1 , order = 2 (d) degree = 2, order = 3 2 3 2 1 6. The solution of the differential equation :� + y = 0 [GATE-2005-EE] 0 20.* Transformation to linear form by substituting dy + p(t)y=q(t)yn ; v = y 1 -n of the equation dt n > 0 will be dv (a) dt +(1 -n)pv=(1 -n)q dv (b) dt +(1 -n)pv=(1+n)q dv (c) dt +(1 +n)pv=(1-n)q dv (d) dt + (1+ n)pv=(1 +n)q [GATE-2005-CE] Enginee ring Mathematics 21 . The differential equation 27.* The solution of dy/dx = y2 with initial value y(O) = 1 bounded in the interval (a) 2 nd order and 3rd degree (b) 3rd order and 2nd degree (c) 2 nd rd order and 2 nd rd degree (d) 3 order and 3 degree [GATE-2005 (P l)] 22. The solution of the differential equation 2 dy -+2xy=e -x with y(O) = 1 is dx 2 2 +x (a) ( 1 +x ) e (c) (1-x ) e+x (b) ( 1 +x ) e- x 2 2 x (d) ( 1 -x ) e[GATE-2006-ME] 23.* 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 (b) 9 months (a) 6 months (c) 12 months 24.* The solution (d) Infinite time of the [GATE-2006] differential equation dy +2xy -x + 1=Q given that at x = 1, y = 0 is dx 1 1 1 1 1 (b) - - - -(a) - - - +2 X 2x 2 2 X 2x 2 1 1 1 1 1 1 (c) - +- +(d) - - +- +2 2 X 2x 2 X 2x 2 x2 [GATE-2006-CE] 25. The degree ·of the differential equation is (a) zero (b) 1 (c) 2 (d) 3 2 d; +2x 3 = 0 dt [GATE-2007-CE] 26. The solution for the differential equation with the condition that y = 1 at x ; 0 is 1 (a) x3 y=e 2x (c) ln(y)= x2 2 501 (b) ln(y)=-+4 3 3 x (d) y=e 3 dy = x2 y dx [GATE-2007-CE] (a) -CL) < X < CL) (c) X < 1, X > 1 (b) -CX) < x $ 1 (d) -2 s; x s; 2 [GATE-2007-ME] (a) 35.2° C (c) 28.7°C (b) 31.5° C (d) 15°C 28.* A body originally at 60° C cools down to 40 ° C in 15 min when kept in air at a temperature of 25° C. What will be the temperature of the body at the end of 30 min? [GATE-2007-CE] 29. Which of the following is a solution to the differential dx(t) equation dt +3x(t)=0 ? (a) x(t) = 3e-1 (b) x(t) = 2e-31 (c) x(t) = -3 2 2t (d) x(t) = 3t2 [GATE-2008-EC] x dy . = 30. Solution of at x = 1 and y = dx (a) X - y2 = -2 (b) X + y2 = 4 -Y (c) x2 - y2 = -2 ,,/3 is (d) x2 + y2 = 4 [GATE-2008] 31 . Consider the differential equations :� = 1 + y 2 Which one of the following can be particular solution of this differential equations? (a) y = tan (x + 3) (c) x = tan (y + 3) 32. The o rder of (b) y = (tan x) +3 (d) x = (tan y) + 3 [GATE-2008 (I N)] the d (d ) _J_ + _J_ + y4 =e-t .IS dt dt2 3 2 differential (b) 2 (a) 1 [GATE-2009-EC] (d) 4 (c) 3 equation 4 33. The solution of x :� + y=x with the condition y (1)=_§_ is 5 (c) x4 y=-+1 5 4x 4 4 (b) y = - +5 5x (d) y = xs 5 +1 [GATE-2009-ME] IES MASTER Publication 502 Ordinary Differential Equations 34.* Match each differential equation in Group I to its family of solution curves from Group II Group I Group II A. 1 . Circles B. C. 0. dy y = dx X dy = -y X dx x (a) y = e (c) A-2, B-1 , C-3, 0-3 (GATE-2009-EC) differential 3y�� + 2x = 0 represents a family of (a) ellipses equation (b) circles (c) parabolas (d) hyperbolas (GATE-2009-CE] 2 d 36. The solution of the differential equation -{ = 0 dx with boundary conditions (i) dy = 1at X = 0 dx (a) y = 1 (ii) dy = 1at x = 1 is dx (b) y = X (c) y = x + c where c is an arbitrary constants (d) y = C1x + C2 where C1 , C2 are arbitary con­ stants 37.* The Blasius equation, ' (b) y = -Jx (d) y = tan (x + ¾) (GATE-2010 (Pl)] (d) A-3, B-2, C-1 , 0-2 the 2 (c) y = cot (x + ¾) (b) A-1 , B-3, C-2, 0-1 of [GATE-2010-CE] satisfying the conditions y(0) = 1 is (a) A-2, B-3, C-3, 0-1 35. Solution (d) 3 and 1 39. The solution of the differential equation �� - / = 1 3. Hyperbolas dy - x -= y dx d3 d 3 2 4 + ( ) +y = o are respectively _J__ _J_ dx dx 3 (b) 2 and 3 (a) 3 and 2 (c) 3 and 3 2. Straight lines dy = dx y 38.* The order and degree of the differential equation [GATE-2009 (Pl)] f d2 d3 : = 0 , is a �+ drt 2 drt (a) second order nonlinear ordinary differential equation (b) third order nonlinear ordinary differential equation (c) third order linear ordinary differential equation (d) mixed or der nonlinear ordinary differential equation [GATE-2010-ME] 40. Which one of the following differential equations has a solution given by the function y = 5 sin (3x + �) dy 5 (a) - - -cos (3x) = 0 dx 3 dy 5 (b) - + -cos(3x) = 0 dx 3 d2 y + 9y = 0 (c) d2x d2 y - 9y = 0 (d) d2 x (GATE-201 0(PI)] x 41 . Consider the differential equation �� + y = e with y(0) = 1 . Then the value of y(1 ) is (a) e + e-1 (b) 1 2 [e - e -1] [GATE-201 0 (IN)] 42. With K as a constant the possible for the first order dy . · = e-3x Is d'ff 1 erent Ia I equat·I0n dx 1 3x (a) -- e- + K 3 (c) -3e- 3x + K 3x (b) _..!. e + K 3 (d) -3e- x + K [GATE-2011 -EI] Engineering Mathematics 43. The solution of the differential equation �� = ky , y(0) = c is (a) x = ce-kY (b) x = kecy (d) y = ce-1oc (c) y = cekx 44. Consider the differential equation [GATE-2011 -EC] dy ( = 1 + y2 ) x . dx The general solution with constant c is 2 (a) x y = tan- + tanc (b) y = tan 2 ( � + c ) 2 (c) y = tan 2 ( �) + c (d) y = tan (� + c) [GATE-2011-ME] dy + y_ = x 45. The solution of the differential equation dx x with the condition that y = 1 at x = 1 , is (a) (c) 2 X +Y =3x2 3 2 X Y = -+ 3 3 (b) y = - + 2 2x X 1 2 x2 (d) Y = - + 3x 3 [GATE-2011 -CE] 46. The solution of the ordinary differential equation �� + 2y = O for the boundary condition, y = 5 at X = 1 is (a) y = e-2x (c) y = 10.95e-2x (b) y = 2e-2x (d) y = 36.95 e-2x [GATE-2012-CE] 47. With initial condition x(1) = 0.5, the solution of the dx differential equation, t + x = t is dt 1 1 2 (b) X = t - (a) X = t - 2 2 (c) t2 X=2 t 2 [GATE-2012-EC, EE, IN] (d) x = 48. ** A system described by a linear, constant coefficient ordi nary, first order differential equation has a� exact solution given by y(t) for y > 0, when the forcing functio is x(t) and the initial condition is 503 y(0). If one wishes to modify the system so that the solution becomes -2 y(t) for t > 0, we need to (a) change the initial condition to -y(0) and the forcing function to 2x(t) (b) change the initial condition to 2y(0) and the forcing function to (t) (c) change the initial condition to jv'2y (O) and the 2 forcing function to j✓ x (t) (d) change the initial condition to -2y(0) and the forcing function to -2x(t) [GATE-201 3 (EC)] 49.* Choose the CORRECT set of functions, which are l inearly dependent. (a) sin x, sin2x and cos2 x (b) cos x, sin x, and tan x (c) cos 2x, sin2 x and cos2 x (d) cos 2x, sin x and cos x 50. T h e s o l u t i o n of t h e i n i t i a l v a l u e problem �� = -2 xy; y(0) = 2 is (a) 1 + e- x 2 (c) 1 + e x 2 [GATE-201 3 (ME)] 2 (b) 2 e- x (d) 2 e x 2 [GATE-2014-ME-Set-4] 51 . Which ON E of the followi ng is a l i near non­ homogeneous differential equation, where x and y are the independent and dependent variables respectively? (a) (c) dy + xy = e- x dx dy - + xy = e- y dx (b) (d) dy + xy = 0 dx dy +e Y =0 dx [GATE-2014-EC-Set-3] 52. The integrating factor for the differential equation dp + K2P = K 1 Lo e-K1I is dt (b) e-K2t (d) eK 2t [GATE-2014-CE-Set-ll] 53.* The general solution of the different equation dy = cos(x + y) , with c as a constant, is dx (a) y + sin(x + y) = x + c x Y (b) tan ( ; ) = y + c IES MASTER Publication 504 Ordinary Differential Equations x y (c) cos ( ; ) = x + c x y (d) tan ( ; ) = x + c [GATE-2014-ME-Set-2) 54.* The figure shows the plot of y as a function of x. The function shown is the solution of the differential equation (assuming all initial conditions to be zero) is 5 3 , "' 0 -1 -3 -5 -4 (a) (c) d2 y =1 dx2 1v/ _I '"' ¾/ey X I dy - = -X dx ➔X 2 4 dy (b) - = -X dx (d) dy = x dx l l [GATE-2014 (IN-Set 1 )) 55.* Water i s flowing at a steady rate through a homogeneous and saturated horizontal soil strip of 1 0 m length. the strip is being subjected to a constant water head (H) of 5 m at the beginning and 1 m at the end. If the governing equation of d2 flow in the soil strip is � = 0 (where x is the dx distance along the soil strip), the value of H (in m) at the m iddle of the strip is ____ [GATE-201 4 (CE-Set 2)) 56. The general solution of the differential equation dy 1 + cos2y . = is dx 1- cos 2x (a) tan y - cot x = c (c is a constant) (b) tan x - cot y = c (c is a constant) (d) tan x + cot y = c (c is a constant) initial condition x(O)=1. The response x(t) for t > 0 is (b) 2 _ eo.21 (a) 2 _ e-0.2 1 (c) 5O - 49e-0·21 (d) 5O - 49e 0·2 1 [GATE-2015-EC-Set-2] 59.* Consider the following differential equation : X Which of the following is the solution of the above equation (c is an arbitrary constant) ? X . y y X (b) - S i n - = C (a) - COS - = C y X y X �/ 0 dx 58. Consider the differential equation dt = 10 - 0.2x with x ( ydx + xdy)cos Y = y ( xdy - ydx ) sin Y � // -2 The value of y at t = 3 is (b) 2e-1 0 (a) -se-1 0 1 5 (c) 2e(d). -15e2 [GATE-201 5-ME-Set-2] (c) tan y + cot x = c (c is a constant) [GATE-2015-EC-Set-2] 57. Consider the following differential equation: �� = -Sy ; initial condition: y = 2 at t = 0 (c) xycos Y = c X (d) xy sin Y = c X [GATE-2015-CE-SET-I] 60. The solution to 6yy' - 25x = 0 represents a (a) family of circles (b) family of ellipses (c) family of parabolas (d) family of hyperbolas 61 . A differential equation [GATE-201 5 (Pl)) di - O.2i = 0 is applicable dt over -10 < t < 1 0. If i(4) = 10, then i (-5) i s [GATE-201 5 (EE-Set 2)) 62.* Consider the following statements about the linear dependence of the real valued functions y1 = 1, y2 = x and y 3 = x2 , over the field of real numbers. I. 11. Ill. Yi , y2 and y3 are linearly independent on -1 ::; X ::; 0 y 1 , y 2 and y3 are linearly dependent on 0 ::; X ::; 1 y1 , y2 and y3 are linearly dependent on 0 :,; X ::; 1 IV. y1 , y2 and y3 are linearly independent on -1 ::; X ::; 1 Which on among the following is correct? Engineering Mathematics (a) Both I and IV are true (a) .J1- x 2 = c (b) Both I and I l l are true (c) Both I I and IV are true (b) (d) Both I l l and I V are true (c) [GATE-2017 EC Session-I] 63. 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 (d) 68. (d) y = 1 - x + tan (x + c), where c is a constant [GATE-2017 EC Session-I] 64.* Consider the differential equation dy (t2 - 81) + 5ty == sin(t) with y(1) = 2n . There dt exists a unique solution for this differential equation when t belongs to the interval (b) (-10, 10) (d) (0, 10) (c) (-10, 2) [GATE-2017 EE Session-I] 65. The solution of the equation �� + Q == 1 with Q = 0 at t = 0 is (a) Q(t) = e - t -1 (c) Q(t) = 1- e1 (b) Q(t) = 1+ e-t (d) Q(t) = 1- e-1 [GATE-2017 CE Session-I] 66. For the initial value problem dx == sin(t}, x (0) == 0 dt the value of x at t = 67. ½• is __. [GATE-2017 (CH)] The solution of the differential equation ✓ ✓ 2 y 1 - x2 dy + x 1- y dx == 0 is ✓1 - y 69. 2 = c ✓1 - x + ✓1 - y ✓1 + x 2 2 2 == c + F+"? == C The solution of the equation x through the point (1, 1) is (a) (c) (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 (a) (-2 , 2) 505 (b) x2 (d) x-2 x x-1 dy + y == 0 passing dx [GATE 2018 (CE-Afternoon Session)] If y is the solution of the differential equation y3 dy + x3 == O, y(0) == 1, the value of y(-1) is dx (b) -1 (a) -2 (d) 1 (c) 0 70. [ESE 2017 (EE)] [GATE 201 8 (ME-AfternoongSession)] A curve passes through the point (x = 1 , y = 0) and satisfies the differential equation dy x2 + y 2 Y. == + dx 2y x The equation that describes the curve is (a) In(1 + �: ) == x - 1 (b) I1n(1 + L ) == x - 1 2 x2 (c) (d) 1n (1 + � ) == x - 1 J 1n (1 + � ) == x - 1 [GATE 2018 (EC)) IES MASTER Publication 506 Ordinary D ifferential Equations 1. 2. 3. 4. 5. 6. 7. 8. 9. 1 0. 11. 12. 1 3. 14. 1 5. 1 6. 17. 1 8. AN SWE R l{EY ·j.,__ 19. (b) 20. (c) (a) 21. (b) 23. (See Sol.) 22. 24. (a) 25. (b) (d) (a) (b) (d) (c) (c & d) (c) (See Sol.) (a) (b) (b) 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. (a) 38. (a) 39. (c) 40. (b) 41 . (a) 42. 60. 45. (d) 47. (d) 49. (d) (c) 67. 51 . (a) 69. (c) 46. (b) (b) 48. (d) (a) 50. (b) (a) 52. (a) 53. (a) (c) 54. It is free from all the four properties of non-linear DE. du dy = u + xdx dx (a) 59. (c) 44. Sol-3: (a) Let 58. (d) ·: It is the standard result. dy = x3 + y 3 dx xy2 y = ux (c) 43. Sol-2: (c) dy = x3 + y 3 dx 57. (d) (b) Given equation is a non-li near differential equation because y has an exponent m ore than one. xy 2 56. (a) (a) Sol-1 : (b) Sol-4: 55. (b) ...(i) . 61 . (c) 62. 63. (d) (3) (c) (c) (c) (c) (d) (1 .652) (a) 64. (d) 66. (d) 65. (d) . - 68. (b) 70. (d) (a) (0.5) (c) (c) (c) (a) (d) (d) So, by eq. (i) x 3 + u 3x 3 du u + x- = x.u 2 x 2 dx 1 + u3 du u + x- = u2 dx du u + x- = +u dx u2 du 1 x- = 2 dx u J u 2 du = f u3 3 dx x = lnx + C y3 = In x + C 3x3 . Engineering Mathematics Sol-5: (b) dy ⇒ dt ⇒ ⇒ ⇒ +5y = 0 dy . . .(i) = -5dt y Integrating both sides tan-1 y = x + c ⇒ ⇒ log y = -5t + log c y = ce-5t . . .(ii) :. The general solution is y= e-51. Sol-6: (a) By using stand ard d efinitions. Sol-8: (d) -2x dy = dx Y d y = -2x dx y A homogeneous equation is of the form f(x, y) = g(x ) ,Y dx dy 2 d y 2 dx 2 on ir:itegrating Y = -x2 + c 2 ⇒ or ⇒ x 2 y2 -+= 1 C 2c which represents ellipse Sol-9: (a) ⇒ �� = f(x, y) � x"t (�) or +g(x, y ) = 0 g(x, y) dx + f(x, y) if dy = ax = Sol-11: (d) By stand ard definition of LOE of dy dx dy = 1 + 1+ y 2 = dx = 3x - 2 . ..(i) 3x 2 = --2x+C1 dx 2 dy x3 y = --x2 +C1x+ C2 2 ...(ii) ...(iii) Using y(O) = 2, C2 = 2 and using y' (1) = -3 , C1 = - It is an exact d ifferentiable equation Sol-12: (c) 8 A differential equation is said to be linear if the independent variable and its differential coefficient occur only in first d egree and not multiplied together. Sol-14: (c) Slope of tangent at point is 2x = y dx - where, f (x, y) and g(x, y) are homogeneous functions of the same d egree in x and y. An ord inary differential equation is that in which all differential coefficients are with respect to a single independent variable. A partial d ifferential equation has two or m0re independent variable. Sol-7: (b) ·: y has exponent other than 1 . dy tan (x + c) 2 d y dy (x 2 +4x)-+ 8 -+ y = x 2 dx dx log(t) = -5t f(x, y) y= Sol-13: (c) & (d) Using y(o)= 1 , we get c= 1 Sol-10: (b) 507 2 y 1 st y" t (�) So, solution is Sol-15: ag ay ord er. 5x x3 y = --x 2 --+2 2 2 dN = aNln(�) dt 0 N = KeP Let, So, by (i) KeP 5 2 ⇒ ...(i) In(�) = P dN dp = KeP dt dt dp dt = aKeP - P dp = ap dt lnP = at + C IES MASTER Publication 508 at t Ordinary Differential Equations = lnln(N / K) 0, N = = N0 lnln(�) C Using (ii) = at+ C a X ... (ii) 0+ C Differential equation for biotransformation of an organic compound is = lnln(�) In( � ) In(� ) = dN dt dt2 = N ⇒ dy + y2 dx = 1 X 3( a 3 d2 y dy + 4( ) + y 2 + 2 = x. 2 ) dt dt d2 ; dt Hence the power of the highest order derivative = 1 . Therefore; degree = 1 The derivative with highest order = Sol-19: (a) = 1 -+ Kt Given differential equation 0 ck, Inc = Sol-18: (b) 1 dN [ aN--..!.+aln(�)] (N / K) K K dt f Xo ⇒ ⇒ = 2. x' (t) = -3x(t) x(t) 1:: ' [l n x(t) 1 /nx(t) - /nx 0 ⇒ ⇒ dy + dx = 0 y2 order x (t) dx(t) -1 0 x(t) Xo = t -f3.dt 0 = -3 (t - 0) = = -3t e- 3t x(t) = Xae-3t Sol-20: (a) Integrating we get 1 -- + x = - C, y ⇒ a or 1 c = e-1 i.e., c = e Sol-16: (a) ⇒ = ... (ii) X aNln(�) In (�) = -1 ... (i) C = _ __!_ a Hence, using eq. (ii) 1 1 - -+ Kt = d2 N = 0 dx + Kdt = 0 x2 Integrating, we get ⇒ Using x(0) = a =0 For growth rate to be maximum dt2 a + aln(�) = X = eat = dx + Kx 2 dt __!_+ Kt = C at N = /at _rn(�) K (b) as t ➔ oo , N ➔ oo (c) x+ c Sol-17: (b) lnln(�) = at + lnln(�) if"In((�) �) l y = 'c' is a constant Given the differential equation dy + p(t)y dt = q(t)yn ; n > 0 Engineering Mathematics dy + p(t)y1 -n = q(t) y-� dt y(1-n) = V Putting ⇒ R ➔ radius ...(i) ⇒ dy dv (1- n)y-n cit = dt Putting these values in (i) m , we get, 2 dR - � n (3R ) = K ( 4 n R2 ) 3 dt ...(ii) 1 dv - -- + p(t)v = q(t) ( 1 - n) dt Given at 2 y.e x2 y ex ⇒ ⇒ Sol-23: (a) 2 = f e- x2 = X + C = ex R = 0.5 cm. 0.5 = - k(3) + 1 e dx + c y = (x + 1 ) e s; 2 -x x2 dV -- = KS dt 'K' constant of proportionality. ⇒ V = i nR3 3 S = 4nR2 t. R = -- + 1 6 R = 0 = 6 months dy + 2xy - x + 1 = 0 dx ⇒ ...(i) 1 dy 2 - + - y = ( \� ) dx x f�dx I .F. = e X = e2 1n x = x2 Solution of (i) is given by y (I.F.) yx2 = f (x - 1 ) dx + c {where V, S are insta-ntaneous volume & surface respectively} 6 Sol-24: (a) 2 = (0 + c) e---0 C = 1 Given ( - :� ) oc ⇒ x2 x y = (x + c) e y(0) = 1 1 K = Finally, 2 = -0 + C t = 3 month ·: It is free from fractions and radicals. So, order = 2 and degree = 2 Sol-22: (b) ( 2x )dx R = 1 cm R = - Kt + 1 at I .F. = e f General solution, t = 0 C = 1 ⇒ Sol-21 : (c) dy + 2xy = e -x ; y(0) = 1 dx Integrating factor, dR = - K dt R = - Kt + C 1 dv cit + (1 - n)pv = (1 - n)q 509 Given at x2 yx2 = - - x + c 2 X = 1 y = 0 1 0 = - -1 + c 2 ⇒ B] IES MASTER Publication 51 0 Ordinary Differential Equations 1 1 1 y = -- - + 2 X 2x 2 Method II Given (2xy - x +1)dx + x2dy = 0 On comparison with Given, ...(1) Mdx + Ndy = 0 M = 2xy - x + 1 we get N = x2 and x3 3 y(0) = 1 l n y = -+C Here MY = 2x = Nx so equation (1) is exact differential equation. l n1 = 0 + C ⇒ C = 0 ⇒ lny = y = e( x /3 ) Sol-27: (c) f (2xy-x + 1)dx + f (O)dy = C y =const [ ·: Standard Result is f ⇒ x dy = y2 (M dx) + J (those terms . =cons t y C 2 x y-- + x = C 2 C = _!_ 2 Using y(1 ) = 0, so solution is 2 x2 X y--+X = 2 1 2 1 1 1 + -- y = 2 2x 2 X i.e., Sol-25: (b) Given, d2x · 3 + 2X = 0 ... (i) dt2 ·: The given differential equation is free of radicals. : . Degree of (i) = exponent of highest order derivative = 1 Sol-26: (d) dy = x 2 y; {variables are separable} dx Given, dy = x2dx y ⇒ Integrating both sides, we get J dy y = J x 2dx + C f dx 1 X = -- + c y y(0) = 1 f of N which are from x) dy = C] 2 3 dy = y2 dx Solution of given equation is :. x3 3 = 1 1 1-x 0 1 E (-00,00 ) - {1} > 1, < 1 y = 1 -X But X X or X X * * S.ol-28: (b) Let, T : temperature at time 't' TO : temperature of the surrounding (air) T0 = 25°C Given, By Newton's law of cooling dT = -K (T - To) dt {K ➔ constant of proportionality} dT = -Kdt integrating, we get (T _ To ) l n (T - T0) = -Kt + C T - To = e- Kt ec ; {C ➔ constant of integration} Given at t = 0, T = 60° 60 - 25 = e-o ec ⇒ at t = 1 5 min eC : 35 Engineering Mathematics 51 1 T = 40°C 40 - 25 = e -K. lS e15K = 35 15 at t = 30 m i n T - 25 = e X 7 3 -K30. Sol-32: (b) 35 35 Given differential equation dy d2 y + ( ) + y4 = e-t dt2 dt 3 The derivative with highest order 2 = (e1 5K)(-2) x 35 = GJ x 35 9 T = 25 + - x 35 49 ⇒ ⇒ X = -3dt x y -+2 2 2 Given at At x dy = dx 2 y = 1, ⇒ 3 2 (✓ ) 2 +f 2 X y 3 y.(x) = f x3 . x dx + c = - + c . . . (i) ... (ii) = C 3 = ✓ Given ⇒ dy 1+ y2 y dy dx 1 C = -+-⇒ C = 1 5 1 x4 1 = +5 X = ·, y = dx X X Integrating both sides log y = log x + log c = 1 + y2 = dx ⇒ y(1) = 6/5 5 Sol-34: (a) XS � � dy y ...(1) tan-1 (y) =x+c y = tan(x + c) where c can ta ke any arbitrary value: ⇒ Given = C C = 2 dy dx = el nx Hence, solution of (1 ) is Le. x2 + y2 + 4 Sol-31 : (a) ...(1) J _!dx x = 1, y = ✓ ydy + xdx = 0 ⇒ dy + ( �) y = x3 dx x 31 x = c e- Sol-30: (d) dy + y = x4 dx I . F. = e I .F. = X Integration both sides log x = -3t + log c at Sol-33: (a) x dx(t) + 3x (t) = 0 dt dx Hence order = 2 . The differential equation is = 31.429°c Sol-29: (b) d2 y dt2 = ⇒ y = ex which is equation of family of straight lines. dy dx dy = y Integrating both sides y x dx X IES MASTER Publication 51 2 Ordinary Differential Equations X log y = - log xy = C + log C Sol-38: (a) Making the given differential equation free of radicals, we get (�:; r which is equation of family of hyperbolas. X dy - = dx Y ydy = x dx Integrating both sides. x 2 - y2 = C Given, ⇒ 3y Sol-39: (d) C Given C 3y2 +2� = 2 2 2 2 2 2x + 3y = C (C is a constant) ⇒ family of ellipses ⇒ d2 y dx2 dy dx ... (1) c1 = 1 so general solution is Sol-37: (b) Given, dy 1+ y2 ⇒ J 1+dyy 2 tan-1 (y) = 1 = 1 + y2 = dx = J dx+c = X+C Using y(0) = 1 , c = 7t 4 ...(ii) ¾) Sol-40: (c) y = 5 sin[3x+�] ⇒ y = C1x + C2 Using y'(0) = 1 , we get c1 = 16 [(�� r +y2 J :. solution is y = tan (x + = 0 = dy -- y2 dx ⇒ dy +2x = 0 dx Integrating, we get, Given 'J = 2. dy dx ⇒ 3ydy + 2xdx = 0 (variable separable form) Sol-36: (c) +y Degree = exponent of highest order derivative X y which is equation of family of circles. Sol-35: (a) (:�l' -4 = 3 xdx + ydy = 0 Integrating x 2 + y2 = ( Order = order of highest order derivative which is equation of family of hyperbolas. dy = dx = y = s [sin3x-cos � +cos3x-sin �] or !Y = x +C2 I y = C1 sin 3x + C2 cos 3x (let) then f d2 f 2 dr/ � cross product term = 0 Hence, non-linear differential equation obviously order is three. y = ( 5 cos �) sin(3x)+( 5 sin �) cos (3x) dy dx = 3C1 cos 3x - 3C2 sin 3x d2 y dx2 = - 9C1 sin 3x - 9C2 cos x d2 y dx2 = -9y ⇒ _J_2 +9y = 0 = -9 [C1 sin(3x)+C2 cos3xJ d2 dx Engineering Mathematics Sol-41: (c) dy + y = ex dx It is LOE in y & X. ...(1) dx x I.F = ef = e :. The general solution of equation (1) is y(I.F) = y.ex = f (I.F)ex dx + C f ex e xdx + C e2x y.ex = - + C 2 Now using y(O) = 1 Hence, solution is ⇒ C = 1 2 e2x 1 yex = - + - 2 Sol-42: (a) ⇒ Sol-43: (c) f dy = f e-3x dx e- 3x y = -+K 1 3x y = --e- + K 3 dy = ky, y(O) = c dx dy = kdx y Integrating both sides log y = kx + log G y = Gekx ⇒ Sol-44: (d) Given Integrating, we get . dy J = 1 + Y2 y(O) = C G = dy = dx C x dx f x2 = -+c 2 ⇒ ...(2) Sol-45: (d) dy f + ... (i) = x, y(1) = 1 dx x It is a linear differential equation of first order. I.F. = e Solution of (i) is e e - 1 e + e- 1 y(1) = - + - = -2 2 2 -3 ⇒ dy = xdx 1 + y2 2 dy = e-3x dx 51 3 y.(I .F. ) = yx = 1 ⇒ f(!\. x x f = e cnx = X f x:(I.F.)dx + C f x.xdx + C y(1) = 1 X 1 1 = -+C 3 C = ⇒ 2 ( ·: 3 X x2 2 y = -+3 3x Sol-46: (d) x3 = -+C 3 = 1 , y = 1) dy + 2y = 0 dx and at x = 1, y = 5 Given Differential equation is, ⇒ (D + 2)y = 0 It' s solution is At X = 1 , y = 5 ⇒ ⇒ D = -2 5 = c1 e-2 5e2 = c1 = 36.95 Hence required solution is Y = 36.95e- 2x IES MASTER Publication 51 4 Ordinary Differential Equations Sol-49: (c) Sol-47: (d ) Sol is ⇒ ⇒ ⇒ cos2x = (1 )cos2x + (-1 )sin2x (1)y1 - (1)y2 + (1)y3 = 0 C 1 Y1 + C2Y2 + C3Y3 = 0 So, linear combination exist between y1 , y2, h So, they are linearly dependent. x.(I.F.) = J1.(IF)dt + C Sol-50: (b) t = J t - 1dt + c X cos2x = cos2x - sin2x then, J � dt lo I.F. = e 1 = e g t = t X Y1 = cos2x, y2 = cos2x, y3 = sin2 x Let 1 dx t - + x = t, x(1) = 2 dt 1 dx - + -x = 1 t dt which is a linear differential equation. dy = -2xy ; y(O) = 2 dx t2 tx = - + C 2 ⇒ 1 x(1) = 2 {i) = 1 + 2 t2 tx = 2 So, Sol-48: (d ) C ⇒ ⇒ X C = I. F = e f a ctt y (I.F) = J x ( t ) . 1.F dt + c 1 C J x ( t ) - IF dt + IF I.F ·: C can be evaluated using initial condition so we can y(t) = J_ J x ( t) I.F dt + y(O) I.F Multiply both sides with (-2) or -2y(t) = -2 J x ( t ) I.F. dt + (-2)y (O ) I.F. -2y(t) = J__ J [ -2x ( t) ] 1.F. dt + [-2y (O)] I.F. Hence, for the solution to be -2y(t), we have to change x(t) by -2x(t) and the initial condition y(O) by -2y(O). ⇒ = � x2 c = /n 2 Sol-51: (a) (a) dy + xy = e-x is first order non-homogeneous linear dx eg. (b) � � + xy = 0 is first order homogeneous linear eg. (c) (d) are non linear eq's. Sol-52: (d) dp + K2 P = K 1 Lo e-k1t dt y = C take y(O) = lF So, ln(!) 2 dy +ay = x(t) dt then, integrating factor ⇒ Condition : y(O) = 2 =0 Consider first order differential equation is Linear form with constant coefficient & solution is dy = -2x dx y In y = -x2 + c _ I. F - ef +k2dt = ek2t Sol-53: (d ) put, ⇒ X dy = cos(x + y) dx + y = t 1 + dt dy = dx dx dt - - 1 = cos t dx dt dx 1 + cos t dt 2cos2 _!_ 2 = dx ... ( 1 ) [using (1)] 51 5 Engineering Mathematics � sec 2 _!_dt = dx 2 2 Integrating both sides f dyy l at t = 0, y = 2 tan(�) = x+c Sol-54: (d) From the graph y = :. y at t = 3 x ;; x�0 x ; -2 X :<, 0 2 Sol-58: (c) x; X > 0 dy = { - = lx l dx -x; x :-:; 0 (By definition of Mod) ⇒ Sol-55: (3) ln(y) = -5t + loge ⇒ t tan- = x + c 2 d2 H = 0 dx2 with H(0) = 5 & H(10) = 1 On integrating dH = C1 dx Again integrating dy 2 cos y = dx 2 sin2 x 2 sec ydy = cosec x dx tan y = -cotx+C dy = - y dt S dx = 10 - 0.2x dt log(10 -p.2x) = t + C (-0.2) x(0)= C = 1 log9.8 (-0.2) log(10- 0.2x) log9.8 (-0.2) = t (-0.2) 2 Sol-57: (c) y = e-sx3 x 2 = 2 e-1 s ⇒ Now using boundary conditions, we get 2 C1 = -- & C1 = 5 5 2 H = --x+5 so, 5 2 H(5) = - - x 5+ 5 = 3 ⇒ 5 Sol-56: (c) dy 1+cos2y = dx 1- cos2x tan y +cotx = C C = 2 dx = dt 1 0-0.2x Integrating both sides H = C1x+C2 Integrating both sides. y = ce-51 . .. (1) Given that, 2 = f-5dt log (10- 0.2x) = - 0.2t 9.8 x(t) = 50-49e-0 ·21 Sol-59: (c) x(ydx + xdy)cos Y = y (xdy - ydx) sin Y X Let X ydx +xdy Y Y xdy- ydx = X tan X y = v·x dy = V dx + xdv vxdx + vxdx +x 2dv = v tan v vxdx +x 2dv- vxdx xdv+2vdx = v tan v xdv 1 + 2 v dx = v tan v x dv 2 v dx = v tan v - 1 x dv IES MASTER Publication 51 6 Ordinary Differential Equations 2 dx I ntegrating both si d es 2 log x = log 1sec v ⇒ x2 = ⇒ x . y_ = ⇒ 2 X xy cos Sol-60: (d) 6y y X dy dx Case : 1 If x E [ 0, 1] (tan v - �) d v = x c sec v I - log v + log c c sec y/x y2 2 6y2 = = = di loge (i ) = = = 25 x 2 -·-+C 6 2 At Hence, solution is so Sol-62: (a) i(-5) = = = 0.2i ...(1) (0.2)dt (0.2)t + log C ceo.21 ...(2) 1 oe-0.a 1 0e-o. a . eo. 2 1 1 0e-1 ·8 = 1 .652 Y 1 = 1, Y2 = x, Y3 = x ⇒ a + bx + cx 2 = = 2 0, a, b, c E R 0 ...(2) 0 = b = = a + bx + cx2 0 ⇒ -1 ⇒ -b 1 2 X = -- c = = 0 ⇒ 0 = a = 0 + c = 0 ...(4) b c -- + - = 0 2 4 ⇒ a a,b,c, E R = b = c = ...(5) O y1 , y2 and y3 are linearly independent for -1 � x � 0 . Sol-63: (d) Given, Let, Then, Or dy dx x + y - 1 dy 1 + -- 0 dx dy dx From equation ( 1 ) �-1 dx dt 1 + t2 Or Linear combination is given by ay1 + by2 + cy3 0 = From equation (4) an d (5), we get b = c = 0 25x2 + K Now using i(4) = 1 0 1 0 = Ceo.a ⇒ C ....(1) y1 , y2 an d y3 are linearly independent for o � x � 1 . At x = The solution to given d ifferential equation represents hyperbolas. Sol-61 : (1 .652) = a At x 25 - - x dx 6 :. di dt b C -+2 4 b + C X 0 = Case II : If x E [-1, 0] 25x or - a ⇒ 1 = 2 At x = 1 , At ⇒ = X=O 0 = From equation ( 1 ) an d (2), we get b = c = 0 = C y dy so At a + bx + cx2 dt J1+t Or Or Or = = = = X + y- 1 y t dy -= ( x + dx y - 1)2 dt dx �-1 dx t2 = dx = tan-1 t = Or Or 2 = = = = J X dx + C tan (x+c) tan (x+c) 1 - x + tan (x+c) ... ( 1 ) Engineering Mathematics Q(t)e1 = e1 + C Sol-64: (a) Given differential equation is · At t= 0 Q= 0 (t2 -81) �� +5ty = sin t y( 1 ) initial condition = ...(1) 2n sin t t2 -8 1 = This is of the form :. x ✓ ✓1-y y.dy 21 ·: ln [ e x =X] J sin t(t2 -8 1)½ dt+ c 32 2 J sint - (t -81) ' dt 5 2 (t -81) ' C +----5 2 (t2 -8 1 ) ' and solving from the options by verifying initial condition we get unique solution. If t= ± 9 then solution is not unique hence range (-10, 1 0), (- 1 0, 2), (0, 1 0) can be eliminated, then left option is (-2, 2) Sol-65: (d) I.F. = 1 dt ef p t = ef1• d =e t Q(t)e1 = J e1 dt + C -✓1-x xdx = -J � = On integrating, 2 ✓1-x 2 i.e.,ydy= -udy and xdx= -vdv sin t (t2 - 8 1 )% dt+ c y(t2-8 1 )5/2 = J 2 (t - 8 1 ) 2 2 Put 1 - y2= u2 and 1 - x 2= v2 y(I.F)= J Q I F dt + c = ✓ (2 ) eh 1 -a1 and sol. of (2) is dQ +Q dt 1 Using variable separable, 1.F = ( t2 - 8 1)% y= = y 1-x 2 dy + x 1-y2 dx = 0 �1n(t2 -a1) = e2 = 0 ⇒ c - cost + 1 Sol-67: (c) f pdt (I.F)= e dt f� 2 = e 1 -01 y(t2-8 1)% = = 1t 1t 1 -1 at t= -, X =-COS -+ 1 =-+ 1 = - =0.5. 3 3 2 2 We know integrating factor = Sol-66: (0.5) Using x(0) -� p= � ° Qt2 - 81 t 2 - 81 5 Q(t)= 1 - e-1 �: =sin t ⇒ J dx =J sin t dt ⇒ x =-cos t+ C ...(2) dy + py = Q dt where 0= 1 + C C = -1 Converting the given equation into standard form dy 5t +( )y 2 dt t - 8 1 51 7 so, J - u�u ⇒ -u or ✓1 - x Sol-68 : (c) -✓1 -y + ✓1- y 2 2 2 vdv V = -J- = v+ c ✓1-x = 2 +c = c xdy - + Y = 0; y(1 )= 1 dx dy y dx X On integrating log y= -log x + log c Using y( 1 )= 1 ⇒ C = 1 so lxy =11 1 . 1.e., y= - = x-1 x IES MASTER Publication 51 8 Ordinary Differential Equations Sol-69: (c) dy y3 + x3 = o dx y3dy = -x3dx ⇒ . . y4 x4 On mtegratmg, = -- + C 4 4 1 Using y(O) = 1 ⇒ C = ...(i) 4 ⇒ lx 4 + y 4 = 11 Hence, y = (1 - x4) 1 14 4] 114 y(-1) = (1 - (-1) Sol-70 : (a) Put ⇒ ⇒ = 0 x2 + y 2 Y dy + = 2y X dx y = vx dv dy = v + x­ dx dx x vx dv x- = - + 2v 2 dx ...(ii) x4 1 . . Y4 S o, so I ut1on 1s - = -- + 4 4 4 ⇒ x2 + v 2 x 2 dv +v v + x- = 2 vx dx ...(1) Now ⇒ so ) dv = dx (� 1 + v2 log(1 + v 2) = X y(1) = 0 + C= - 1 C ...(1) A differential equation of the form where a0 , a 1 , a2 , . . . , an are cosntants are Q is a function of x only or constant is known as LOE of nth order with constant coefficients. Let - = D so dx _ ( aoDn +a1o n-1 +a 2o n -2 . . .+an ) y Q ⇒ lf(D) · y = 01 So complete solution of (2) is IY = CF+PII . . .(2) COMPLEMENTARY FUNCTION (C.F.) Consider an equation, f(D).y = 0 ...(3) Let y = emx be the solution of (3), its solution is known as CF for equation (2) Dy = memx, 02y = m2emx, ... , o ny = mnemx Put all these value in equation (3) emx *O So, a0mn +a 1 mn-1 +...+an _1 m+an - 0 1 It is known as Auxiliary Equation (A.E.) of (1 ), which is the nth degree polynomial in variable m. Let its roots are: m= m1 , m2 , m3, ... , mn Steps for Finding Auxiliary E quation Step 1 : Replace y by 1 Step 2 : Replace dy/dx by m Step 3 : Replace d2y/dx 2 by m2 and so on replace d nytdx n by mn . Step 4 : By doing so, we get an algebraic equation in m of degree n called auxiliary equation. RULES FOR FINDING THE COMPLEMENTARY FUNCTION Case 1: When all roots of the auxiliary equation are real and distinct : Let m = m 1 , m2 , m3, ... , mn be the roots of the A.E. Hence the general solution of (i) is ...(iv) (·: P.I. = 0) 520 Linear Differential Equations of Higher Order with Constant Coefficients d3 y dy 7--6y= 0 Example 1 1 Solve -dx dx 3 Solution : (03 - 70 - 6) y = 0 We have The auxiliary equation is ⇒ m3 - 7m - 6 = 0 ⇒ (m + 1 )(m + 2)(m - 3) = 0 ⇒ m = -1 , -2 , 3 the roots are real and distinct. Hence the complete solution is Case 2 : When the roots of auxiliary equation are equal. (a) When two roots of auxiliary equation are equal. m1 = m2 Let Hence the general solution of f(O)y = 0 is IY m X mX m X = (C 1 +C 2 X) e 1 +C 3e 3 +... +c n e n (b) If however, three roots of the auxiliary equation are equal say m1 = I (·: P.I.= 0 for Q = 0) m2 = m3 then general solution will be The above rule can be generalized for any number of repeated roots. Example 2 1 Solution 1 d3 y d 2y dy --3-+3-y =0 3 2 dx dx dx Equation is (03 - 302 + 30 - 1 )y = 0 A.E. is m3 - 3m2 + 3m - 1 =0 or (m - 1 )3 = 0 ⇒ Hence, the general solution is y = (c1 + c2x + c3x2)ex Example 3 1 (04 - 2 03 - 302 + 40 + 4)y = Solution : The auxiliary equation is o m4 - 2m3 - 3m2 + ,4m + 4 = 0 ⇒ (m + 1)(m3 - 3m2 + 4) = 0 ⇒ ⇒ m = 1, 1, 1 ⇒ ⇒ m4 + m3 - 3m2 - 3m2 + 4m + 4 = 0 (m + 1)(m3 + m2 - 4m2 + 4) = 0 (m + 1)(m + 1 )(m2 - 4m + 4) = 0 ⇒ (m + 1)2 (m - 2)2 = O m = -1 , -1 , 2 , 2 Hence the general solution is y = (c1 + c2x)e-x + (c3 + c4x)e2x Case 3: When two roots of auxiliary equation are imaginary. Let a * ip be the pair of imaginary roots of the auxiliary equation i.e. let m1 = a+ ip, m2 = a -ip then from (iv) m X x i x C. F. = C1 e(a + p) +C 2 e(a -ip ) +C 3 e 3 +. . .+c nemnX = eClX (C 1 ei px +C 2 eipx )+C 3 em3x +. . .+c n emnX = eax [c 1 (cospx +isinpx)+c 2 (cospx- isinpx)] +c3 em3 x +. . .+c n em"x = eax [(c 1 +c 2 )cospx +i(c 1 -c 2 )sinpx] +c 3 em3x + ... + c nem"x C.F. = eax [A cospx +Bsin px]+c 3 em3x + ...+c n em"x Hence the general solution of equation f(O)y = 0 is [taking A = c 1 + c2 & B = i(c1 - c2)] Engineering Mathematics 521 Example 4 1 Solution = Solve : (D4 - n4)y = O Complete solution of (1) is y = CF + Pl For CF : AE of (1) is m4 ⇒ (m 2)2 - (n2)2 = 0 m = ±n and - ⇒ n4 = O (m 2 n2)(m 2 + n2 ) = 0 - m = ±ni + c2e-nx + e0x[c3cosnx + c4sinnx] Now, CF = c 1 enx nx nx So, the complete solution of (1) is y = c 1 e + c 2 e- + [c3 cos nx + c 4 sin nx] + 0 Here Pl = 0 Example & 1 d4 y Solve + a4 y = 0 dx 4 (D4 + a4)y = 0 Equation is Solution = m4 + a4 = 0 The A.E. is ⇒ ⇒ m4 + 2m 2a2 + a4 2 2 m 2a 2 = 0 - 2 2 (m2 + a + ✓ ma)(m + a 2 ⇒ 2 - ✓ ma) 2 2 (m 2 + a2 )2 - ( ✓ ma/ = 0 = 0 (m 2 + a 2 + a✓ - m) = 0 Now, when 2 ✓ then -a✓ ± 2a 2 - 4a 2 -a ± ia 2 m = --- ---- = -2 ✓ then a✓ ± 2a 2 - 4a 2 m = 2 2 and when (m 2 + a 2 - 2✓ m) = 0 2 ⇒ -a . a 2 + 2 m=- I✓ ✓ ✓ 2 (m 2 + a 2 - 2✓m ) = 0 2 ✓ a✓ ± 2a 2 - 4a 2 then m = 2 Hence the general solution is Case 4 : When imaginary roots of auxiliary equation are repeated. Let m 1 = m 2 = a + ip and m3 = m 4 = a - ip, then by case-2 Example 6 Solution 1 1 Solve (D4 - 4D3 - 8D2 - 8D + 4)y = 0 The auxiliary equation is m 4 ⇒ (m 2 - 2 m + 2)2 = 0 - ⇒ 4m 3 - 8m 2 - 8m + 4 = 0 2 ± 2i . . . m = -- ( twice) = 1 ± 1, 1 ± 1 2 Hence the general solution is y = ex[(c 1 + c2x)cos x + (c3 + c4x)sinx] IES MASTER Publication 522 Linear Differential Equations of Higher Order with Constant Coefficients Case 5 : When roots of auxiliary equation are irrational. m1 = a+Jp, m2 = a - Jp and rest are distinct. C. F. = e0x l C 1 cosh Jpx+C 2 sinh Jp XJ + C 3 em3x +. . . c n emnx 2. d ( e x e-x J ; � (cos h x) = = sin h x = cos h x dx dx d f 3 . � (sin h x) = cos hx = f sin hx dx dx Example 7 : Solution = Solve (D2 - 4 D + 1 )y = 0. The auxiliary equation is m2 ⇒ m = 4± - 4m + 1 = 0 2 .Ji""6=4 = 4± ✓ 2 2 3 = 2 ± ✓3 3 Hence the general solution is y = e2x ( c 1 cosh ✓3x + c 2 sinh ✓ x ) Case 6 : When irrational roots of auxiliary equation are repeated. Let m1 = m2 = a+ Jp and m3 = m4 = a- ✓3 , then by case-2 C.F. = eax [(c 1 +c 2x)coshJpx + (c 3 + c 4 x)sinhJpx] + c 5 em5x + ... + c n em"x PARTICULAR INTEGRAL ( P.I.) Consider n th order linear differential equation with constant coefficient as follows f(D).y = Q ...(2) 1 If we take y = -- . Q then y will satisfy equation (2) it means it is solution of (2). It is known as particular integral f(D) for given equation. f, � is an operator. e.g . f; = 2 = ff etc. METHODS OF EVALUATING (P.I.) Case 1 : When Q = e3X , where a is any constant. INOTE !) * -1- eax = -1- eax , provided f(a) 0. f(a) f(D) (a) if f(a) = 0 then PI = (b) if f'(a) = 0 then PI = - � ·e f( ) 0x = ; -e f' a) 0x provided f' (a) 1 · 0x x2 - e = -- · e ax provided f"(a) f"(a) f(D) [Note : *0 * 0 and so on 2 f(D) = D -5D+ 7 ⇒ f'(D) = 2D- 5] Engineering Mathematics Example 8 : Solve (03 + 2 02 + O)y = e2x. A.E. is m 3 + 2m 2 + m = 0 Solution : m(m 2 + 2m + 1 ) = 0 ⇒ ⇒ P.I . = m(m + 1 )2 = 0 ⇒ 523 m = 0, -1 , -1 ⇒ 1 1 e 2x = 3 e 2x = J__ e 2x 2 2 18 2 +2 ·2 +2 0 + 20 + 0 3 Hence the general solution is given by y = C.F. + P.I. x 2x Y = c 1 + (c 2 + c 3x) e - + J__ e 18 Example 9 : Solve (02 + 0 + 1)y = (1 + ex)2 ; 0 = Auxiliary equation is m2 + m + 1 = O Solution : m = ⇒ = -1 ± d . dx 3 ✓i Jf=4 = _ _! ± 2 2 2 1 1 1 eox + 2 e 2x + 2 ex = (0)2 + (0) + 1 (2) + (2) + 1 (1) + (1) + 1 Hence the complete solution is -�2 3 1 2 ✓3 ✓ y = C.F. + P.I. = e ( C 1 COS 2 X + C 2 Sin 2 x ) + 1 + e2x + e x 7 3 where c 1 and c2 are arbitrary constants of integration . Case 2 : When Q is of the form sin ax or cos ax. * (a) Case of ---;- sinax or \- cos ax, when f(-a 2 ) 0. We can use following result f(O ) f(O ) * 1 . sin ax 2 sin ax = --, 'f1 f(-a ) 0 -2 f(O ) f(-a 2 ) and * 1 ---;-cos ax = --cosax, if f(-a 2 ) 0 f(-a 2 ) f(O ) Thus, we conclude that to find the P.I. corresponding to these forms, put -a2 in place of 02. But we cannot write anything for 0. Again 0 3 = 0 2 . = -a 2O, 0 4 = 0 2 . 0 2 = a 4 , 0 5 = 04 - 0 = -a4 O and so on. o by substitution f(O) is reduced to a linear factor and can be solved further by multiplying above and Hence, we see that below by the conjugate factor of the linear factor. In this way we will get the term 02 in the denominator for which we -----� put -a2. [ eg; 2 2 2 f(O 2 ) = 0 3 - 20 2 + 50 + 7)f(-a ) = -a O - 2(-a ) + 50 + 7 ] . 1 . X X . 2 - - sm ax - sm ax = - s m ax = -INOTE" (a ) if f(-a ) = 0 then PI = -2 2 2 ' ' f'(D ) f'(-a ) f(D ) 1 . x 2 . 2 s m ax and so on. · s m ax = (b) if f'(-a ) = 0, then, P.I. = -2 2 f(D ) f"(-a ) IES MASTER Publication 524 Linear Differential Equations of Higher Order with Constant Coefficients d3 y dy . 3 - + 2y = 1 0 s1n x. Example 1 0 : Solve 3 dx dx Solution : The equation is (0 2 A.E. is given by m 2 C . F. :. = 30 + 2)y = 1 0 sin x - - x c1e + c 2 e 3m + 2 = 0 P.I. = (m - 2)(m - 1 ) = 0 ⇒ 2x ⇒ 1 1 1 0 sinx = 1 0 ---- sin x (-1)0 + 30 + 2 D - 30 + 2 3 0-1 1 sin x = 1 0 --sin x = 5 (0 + 1)(0 - 1) 2(0 + 1) = 5 = -�[O sin x - sin x] = -�[cos x - sin x] 2 2 Therefore the general solution is y = C.. F + P..l = d3y dy . + a2 - = sin ax. Example 1 1 : Solve 3 dx dx Solution = C . F. = c 1 + c2 cos ax + c3 sin ax = (0 - 1)sin x 02 - 1 = 0, i.e., m = x 3 ( -a 2 ) + a2 dx dx = . x sinax c 1 + c 2 cos ax + c 3 s1n ax - - 2a 2 = -37 sin 3x or (02 + 20 + 1 0)y Auxiliary equation is m 2 + 2m + 1 0 C . F. = e-x(c1 cos3x + c2sin3x) P. I. = (0 - 1)sin x -1 - 1 (sinax) = -xsinax 2a 2 Given equation can be written as 2 5 0, ± ia . d2 y dy + 2- + 1 0y + 37 s1n3x = 0, and find the value of y when x Example 1 2 : Solve dx dx 2 dy = 0, when x = 0. dx d d __y + 2--1. + 1 0y 2 = 1 . X ( s1nax . s1n ax = 2 ) D 3 + a2 D 3D + a 2 [ ·: on replacing D2 ➔ -a2 , f(D) becomes zero] . . H ence, compI ete soI ut1on 1s y Solution (·: 0 2 = -1) c 1 e x + c2e 2x - � (cos x - sin x) 2 A.E. is m 3 + a2m = 0 or m(m 2 + a2) P.I . = m = 1, 2 ⇒ = m = -1 ± 3i = n -, being given that y = 3, 2 -37sin3x 1 1 (-37 sin 3x) = -37 · sin 3x -9 + 20 +10 D + 2D + 1 0 2 = -37 · (2D - 1) . (20 - 1) . s1nx = -37 • 2 _ _ 1 s,n 3x 4(-9) 40 1 = 20 sin 3x - sin 3x Hence the general solution is y = e-x(c1 cos3x + c2sin3x) + 6cos3x - sin3x Differentiate (i) w.r.t. x, we get = 6 cos 3x - sin 3x . ..(i) Engineering Mathematics dy = -e- x (c 1 cos3x + c 2 sin3x) + e -x (-3c 1 sin3x + 3c 2 cos3x)-18 sin3x -3 cos3x dx Putting x = 0, y = 3 in (i) we get 3 = c 1 + 6 ⇒ c 1 = -3 = 0, dy/dx = 0, in (ii) we get O = -c 1 + 3c2 - 3 ⇒ and putting x c2 525 . . . (ii) =0 Substituting t�e values c 1 and c2 in (i), we have y = e-x(-3cos3x) + 6cos3x - sin3x, Now ( Y) =n/2 = 1 x Case 3 : When Q = xm , m being a positive integer. P. I. = 1 m � Q=[f(D )r x f( ) In this case expand [f( D)J-1 x m in ascending powers of 1 -- xm = ( D -a) 2. 3. 4. Using binomial theorem as follows: m 1 x - m(m -1)x m-2 1 m m(m -1). . . 2.1 -+ ] = -- (X + m + ...+ 2 a a a am [All the differential coefficients of x m of an order higher than m are zero.) Steps : 1. D. Take out the lowest degree term from f( D ) to make the first term unity (so that Binomial Theorem for a negative index is applicable). The remaining factor will be of the form 1 +q>(D) or 1 -q>(D ). Take this factor in the numerator. It takes the form (1 + q>(D)r1 or (1-q>( D)r1 . Expand it in ascending powers of = 0 and so on. D Operate on x m _term by term. as far as the term containing o m, since om (x m ) = (m! ); om+1 (x m) = O; o m+2(xm) (where n is -ve integrer or fraction) 2 Example 1 3 : Solve (D 3 - D2 - 6D )y = 1 + x ; D = � ­ dx Solution : Auxiliary equation is m3 - m2 - 6m = 0 2x C.F. = C 1 + c2eP. I. = = D 3 + c3e 1 2 3x -D -6D ( 1 + X2 ) = ' (�-6D l} 6D 1 + - { 2 ⇒ m(m - 3)(m + 2) = 0 1 2 -6 D -D + D 3 (1 + X 2 ⇒ m = 0, -2, 3 ) 0 (1 + x' ) - ;� H / ]J\1 + x' ) 2 2 2 2 D -D 1 D D 1 ) + ( D - D ) - . . .] (1 + x 2 ) = _ _ ] (1 + x 2 ) [1-Q_ + + = _ _ [1-( 6D 6 6 6D 6 6 36 [ ·: D (x2) = 3 D4(x)2 = ... = OJ IES MASTER Publication 526 Linear Differential Equations of Higher Order with Constant Coefficients He nce the complete solution is y = C.F.+P.l. = c 1 +c2 ~2 +c 3 x e e 3 where c1 , c2 and c3 are arbitrary constants of inte gration. x -�(x 2 -�+ 18 2 25 6 ) 3 d2y dy dy 2 2 2 +x +x. + e +-= e Solv : 14 Example dx 3 dx 2 dx x Solution : A.E. is m3 + 2 m2 + m = 0 or m(m + 1)2 = 0, i.e ., m = 0, -1 , -1 C. F. = c1 + c2 + c3x) - . e ( P.I. = x 0(0�1)2 e 2 x + ( 0(0�1)2 x 2 : 2 e 2x +5 (1+ o r2 (x 2 +x) 2( 2 1) +x) = 2 2 2 2 � e2x + (1- 2O+3O ...)(x +x) = � e +5 [ ( x +x)- 2( 2 x+1)+6] 1 5 1 x2 x2 x2 x3 x 3 x2 1 1 = 18 e2 +[ 3+2-42- 2 x+6x] = w e2 +3-32-32+4x x = x x 1 x3 x2 3 2 He nce , complete solution is y = c 1 +(c2 +c 3 x ) e- +- e 2 +--3-+4x x x 18 Example 1 5 : Find the solution of y " +4y' +4y = x 2 , y(0) = 1, y(1) = 1 . Solution : The give n equation can be writte n as ( 02 + 40 + 4)y = x2 ⇒ A.E. is m2 C.F. + 4m + 4 = (c1 + c2x) ~ e 2 x ⇒ = 0 (m + 2)2 = 0, i.e . m = -2 , -2 2 3 0 - 2 1 ·2 1 1 -2 ) -3) 0 2 . 2 = ) 4 1+2 X = �[ 1- 2 9- + ( ( X ..] x2 = 4 [ X - 2X +X] ( P. I . = 2 4 4 2 2 0+ 2 ) ( He nce, the complete solution is given by y = C. F. + P.I. . . . (1) Putting x = 0 and y = 1 And x = 1 , y = 1 gives ( 7 2 C 1 +C 2 ) = -e y ⇒ 7 8 -e C2 = 2 5 -8 = (¾+i 2 x- ¾ x ) e~2x +±(x2 - 2x+%) e Case 4 : To find particular i ntegral, when Q is of the form of eaxv where V is a function of x. i .e. Q = eaxv. In these types of questions we use 1 _1_ ( e V ) = ea V f(O +a) f(O) x ax 1 1 i. e . to find P.I. = f ( e v ) we take out ea and put (0 + a) for O in f(O) and then operate V with operator f(O+a) O) ( which is one of the cases already discussed. ax x Example 1 6 : Obtain the ge neral solution of the differe ntial equation. Solution d2 y dy -+4--1 2y 2 dx dx = x -1) ( e 2 x The give n equation is (02 + 40 - 1 2 )y = (x - 1 )e2x. Engineering Mathematics 527 Auxiliary equation is m2 + 4m - 12 = 0 ⇒ (m - 2)(m + 6) = 0 C.F. = c1 e 2x + c2e -6x P.I. = ⇒ 1 2 0 + 40 -12 m = 2, -6 (x -1) e 2x = e 2x . = e 2x 1 1 1 2x (x -1)= e _ _ (1 + !2.) (x -1) 2 80 8 0 + 80 1 . -- (1-!2.)(x -1) (Leaving higher power terms) 80 8 2x = e _ 2x = e 1 (x -1) {(0 + 2) + 4(0 + 2) -12} 2 . � ( x -1-i)=e2x . � ( x 8 8 -¾) x x e = ; ( ;_ 9;) Hence, the complete solution is y = C.F. + P.1.=c 1 e2x + c 2e-6 x + e2x ( �: - :: ) where c1 and c2 are arbitrary constants of integration. d3 y dx d2 y dx dy dx x x -3+ 3 - - y=x e + e . Example 17 : Solve 2 3 Solution = A.E. is m3 - 3m2 + 3m - 1 = 0 or (m - 1)3 = 0, i.e. , m = 1, 1 , C. F. = (c 1 + C2X + C3X 2)ex. x 1 1 1 1 1 1 e - x4 ex - x3 x x -� x + e-� - 1 = e x - x + e x -1=- -+ -xex + __ ex = eP. l . _ __ 3 3 3 3 3 3 6 24 (0 + 1 -1) (0 +1-1) 0 0 (0 -1) (0 - 1) Hence, complete solution is d4 Example 18 : Solve ---{ - y=e x cosx. Solution = dx A. E. is m4 - 1 = 0 i.e. , m = ±1, ±i C. F. = c 1 ex + c2e-x + c3cosx + c4sinx 1 1 1 -ex cos x=e x · cos x = e x . 4 cos x P. I. =3 0 4 -1 (0 + 1)4 -1 0 + 40 + 60 2 + 40 x = e •- . Case 5 : To find the P.I. for 1 1 x -cos x [put O2 =-1) = --e cos x 1-40 -6 + 40 5 1 n . 1 n x smax or D x cos ax f(D) f( ) ·1ax ·iax 1 1 _ x" _ W nte · - x " (cos ax + isin ax)=- x "e =e f(D) f(D) f(D + ia) 1 1 1 iax x" cos ax=Real Part (RP) in eiax . x" x " and Now, -- x" sin ax=Imaginary Part (IP) in e f(O + 1a ) f( O) f(D) . f(D + ia) IES MASTER Publication 528 Linear Differential Equations of H igher Order with Constant Coefficients Example 1 9 1 Solve (D4 + 2D2 + 1 )y = x2 cos x. Solution : A.E. is m 4 + 2m 2 + 1 = 0 or (m 2 + 1 )2 = 0 i.e. m = ±i , ±i C.F. = (c1 + c2x) cosx + (c3 + c4x)sinx 1 1 x 2 cos x = Real part in x 2 eix = Real part in ei x \ 2 x2 2 2 (D2 + 1)2 (D + 1) [(D + i) + 1] P.I. = = Real part in ei x ei 0 1 2 : 2 (1 + . ) x2 Real part in = x 21 (D 2 + 2Di)2 4D i 2 D 3 D2 . 3 2 . eix 1 . ei x 1 2 2 = Real part m - - - 2 (1- 7 + - 2 ......) x = Real part m - - - 2 (1 + 1D - - D ......) x 4 4 D 1 4 4 D i 3 3 . ei x 1 . eix x 4 . x . 3 2 ( ) x + 21x = Real part in + 1- - x2 ) = Real part in -- · ( 2 4 12 3 4 4 D 2 1 4 1 . 3 2 x cos x + x 3 sm x + x cos x 48 12 16 4 ( ) ( ) . 1 3 . 3 2 1 4 Hence, compI ete so I ut.1on .1s y = c1 + c 2x cos x + c3 + c 4 x sm x - x cos x + x sm x + x cosx 16 48 12 x4 . x . . . 1 = Rea I pa rt in - (cos x + 1 s1n x) ( + 1 3 3 2) 12 3 4 x = - Case 6 : When Q = xV where V is any function of x 1 1 1 f'(D) • Pl = -- - 0 = -- - (x • V ) = x - -- - V (V) f{D) f(D) f(D) [f(D)]2 Case 7 : When Q is any other function of x (General Method) Resolve f(D) into linear factors. Let Then (Partial Fraction) mx = A 1e 1 Qe-m1xdx + A 2e m2X Qe -m2xdx + ...... + A nemnX Q e-m"x dx f f Remarks : Remember the following formulae : f 1 ax ax (i) __ Q = e J e- Qdx D-a 1 (ii) -- 0 = e -ax J e axQ dx D+a d2 Example 20 1 Solve -{ + y = cosec x. dx Solution 1 Given equation is (D2 + 1 )y = cosec x. A uxiliary equation is m2 + 1 = 0 C.F. = C 1 COSX + C2sinx P.I. = = ⇒ m = ± i -.! (- 1 -11 1 cosec x = (D + i)(D - i) cosec x = 2i D - i - D + i ) cosec x 2 0 +1 1 1 _.!_2i (- cosec x - -- cosec x ) D +i D-i [Partial Fractions] ...(i) Engineering Mathematics 529 1 Now -- cosec x = e i x f cosec x e-ix dx (D - i) = eix fcosec x (cos x - i sin x) dx = ei x f (cot x - i) dx = ei x (log sin x - ix) 1 Changing i to -i, we have --. cosecx = e-1x (logsin x + ix) D+I ix P.I. = � [ eix (logsin x - ix) - e - (log sin x + ix)] [Using equation (i)] i eix e-ix ) ( eix �e-ix ) -x = log sin x ( ; = (log sin x ) - sinx - x cos x i Hence the complete solution is y = C.F. + P.I . = c 1 cosx + c2sinx + sinxlogsinx - xcosx where c 1 and c2 are arbitrary constants of integration. 2 d y 2 + a Y = tan a x. Example 21 : Solve dx 2 Solution = Given equation is (D2 + a2)y = tan a x Auxiliary equation is m 2 + a2 = 0 C.F. = c 1 cos ax + c2sin ax P.I . = ⇒ m = ± ia -1 1 1 1 1 tan ax = ----- tanax = - ( -- - -- ) tan ax 2ia D - ia D + ia (D + ia)(D - ia) D 2 + a2 [Partial Fractions] 1 1 1 = -- [-- tan ax - --tan ax ] 2 ia D - ia D + ia Now, ...(i) 1 -- tan ax = eia x ftan ax • e-iax dx D - ia 2 = e iax f tan ax (cos ax - i sinax) dx = eiax f ( sin ax - i sin ax ) dx cosax 2 ax ) iax dx = e f [sin ax - i(sec ax - cosax)] d� cosax cos ax i . sin ax ·1 8x [ --- -log(sec ax + tan ax) + 1--] = e a a a = e iax f ( sin ax - i 1 - cos = _ _!_ eiax [(cosax - i sin ax) + i log(secax + tan ax)] a iax ax iax = _ _!e [e-i + i log(secax + tan ax)] = _ _![1 + ie log(secax + tan ax)] Changing i to -i , we have a a --. tanax = --[1 - e- ,ax log(sec ax + tan ax)] D + 1a a 1 P.I . = 2�a 1 · [-� {1 + ieiax log(secax + tan ax)} + �{1- ie-iax log(secax + tan ax)}] [Using (i)] iax 1 + e -iax ) 1 (-e ax + tan ax) = --log(sec = 2 Iog(sec ax + tanax) - cos ax 2 2 a a Hence the complete solution is y = C.F. + P. 1. = c 1 cos ax + c 2 sinax - --;. cosax log(secax + tan ax) a where c1 and c2 are arbitrary constants of integration. IES MASTER Publication 530 Linear Differential Equations of Higher Order with Constant Coefficients CAUCHY'S HOMOGENEOUS LINEAR DIFFERENTIAL EQUATION An equation of the form; 1 dn y dn -2 Y n 1 dn - Y - )-- +a2 (x n-2 )-. . . + an y = Q + a (x xn n n 1 n 2 1 dx dx dx - ...(i) ( X n Dn + 8 1 X n-1 on-1 +8 2X n-2 0n -2 . . . + a n ) y = Q or where (a 1 , a2, . . . , a n) are constants and Q is a function of x alone or constants is known as homogeneous LOE. To solve such type of equations we use following transformation. Put x= e 2 ⇒ d z = log x, let D' = - Then we have, dz 3 3 2 2 xD = D' and x D = D'(D' -1), x D = D'(D' -1)(0' -2) (1) is Cauchy's homogeneous L.D.E. ====! . . . . . xn on = D'(D' -1)(D' - 2). . . . . . (D' - n + 1) 2 2 Prove that : x D = D ' & x D = D '(D ' -1 ) we can write dy dy dz = · dx dz dx Again we can write ⇒ dy =_!_ · dy ⇒ xd =� ⇒ x D = D' z dx d y dx 2 = x dz dx dz -1 dy d dy d 1 dy ( )= ( ) = 2( ) dx dx dx dz x dz x 1 d dy +x dx ( dz ) 2 -1 dy 1 d dy dz -1 dy 1 d y - +- -(- ) · - = - - + - · = 2 2 2 2 x dz x dz dz dx x dz x dz 2 2 -dy d y 2 d y -+x = dz dz 2 dx 2 2 x 2 D 2 = -D' + (D ' ) =D '( D ' - 1) LEGENDER'S HOMOGENEOUS LINEAR DIFFERENTIAL EQUATIONS n 1 dn 1 d +. . . . . . +a n y=Q . Put, (ax + b) = e2 then z = log(ax + b}, and An equation of the form (ax + bt -1. + a 1 (ax + bt_ ____y n dx dx n -1 let d dz =D', then we have 3 d2 d 3 d 2 (ax + b)­ = aD', (ax + b }2 -= -= a 3 D'(D' -1)(0' -2) D'(D' -1), (ax + b ) a 2 3 dx dx dx By the above substitution given differential equation changes into a l inear differential equation with constant coefficients in y and z. d2 d Example 22 : Find the general solution of the equation: x 2 -1. + x _J_ _ y=0 dx dx 2 Solution Putting x=e2 ⇒ z=logx and x 2 dy 2 d y - Dy, x 2 dx dx ea D(D -1) y Engineering Mathematics The equation reduces to [0(0 - 1) + 0 - 1 ]y ⇒ ⇒ 2 [0 -0 + 0 -1]y = 0 = (0 2 -1)y = 0 531 d 0 where O = dz which is linear equation with constant coefficient. Its auxiliary equation is m -1 = 0 ⇒ m = ± 1 Hence general solution is y = c 1 ex + c2e- x 2 2 d dx d dx Example 23 : Solve x 2 ---{- -3x i + 4y = 2x 2 Solution : Putting x = e 2 ⇒ z = logx 2 2 d y = 0(0 -1)y, Then x 2 dx d where O = ­ dz Using these values in given differential equation [0(0 - 1) - 30 + 4] = ⇒ 2e22 [02 - 40 + 4]y = 2e22 which is the linear differential equation with constant coefficients. Its auxiliary equation is m2 - 4m + 4 = 0 ⇒ (m - 2)2 = 0 C.F. ⇒ (c1 + c2 z) e = 22 p. I. = m = 2, 2 1 (0 -2)2 2e 2 2 = 2e2 2 1 (0 + 2 -2) 2 -f_ · ( 1) [Using case (iv)] 1 = 2e 2 z . __ 1 = 2e 2 z . = z 2 e 2z 2 02 Hence complete solution is y = (c 1 + c2z) e22 + z2e22 = (c 1 + c2 Iogx)x 2 + (logx)2 .x 2 Example 24 : Solve the differential equation x 2 Solution : Putting x = e 2 2 ⇒ z= logx d 2 dy ; -x - 3y = x 2 log x dx dx d d d Then x 2 = 0(0 - 1), x - se O where O = ­ dz dx dx 2 By using these values in given differential equation [ 0(0 - 1 ) - 0 - 3]y = ze22 ⇒ [02 - 20 - 3]y = ze22 which is the linear differential equation with constant coefficient. Its auxiliary equation is m2 - 2m - 3 = 0 ⇒ (m - 3)(m + 1 ) C.F. = c1 e- 2 + c2 e32 P. I. = 2 1 0 -20 -3 ze 2 z = = ze ( 0 + 1): 0 - 3 ) 0 2z ⇒ = e2z m = -1, 3 1 1 .z . z= e2 z . (0 + 3)(0 -1) (0 + 2 + 1)(0 + 2 - 3) 1 2 2 . . 3 Hence comp Iete soI utIon Is y = c 1 e-z +c 2e z - e z ( z + ) 3 3 IES MASTER Publication 532 Linear Differential Equations of Higher Order with Constant Coefficients Example 25 : Solve x 3 . Solution : 2 3 d d ; + 2x 2 ; + 2y=1 o ( x + _!_) x dx dx = = logx and O = � then given differential equation reduces to d [0(0 - 1)(0 - 2) + 20(0 - 1) + 2]y = 10(e2 + e-2) or (03 - 02 + 2) y = 10(e2 + e-2) Putting x e2 so that z which is a linear equation with constant coefficients. Its Auxiliary equation is m3 - m2 + 2 = 0 or m = -1, C.F. = 2± � = 1, 1 ± i 2 c 1 e-z + e (c2cosz + c3sinz) 2 p. .I = 10 1 3 2 0 -0 -2 = 10 ( = (m + 1 )(m2 - 2m + 2) = 0 S + x [c 2 cos(logx)+ c3 sin(logx)] X (e z + e -z ) = 10 ( 1 3 2 0 -0 + 2 e z+ 3 1 0 - 02 + 2 e- z ) 1 1 1 e -z J e -2 )=10 (_!_ e2 + z • e2 + z • 2 2 2 2 30 -20 3(-1) - 2(-1) 1 -1 + 2 3 = 5e2 + 2ze -2 =5x + � logx Hence the complete solution is y = C.F. +P.I.=S + x[c 2 cos(log x) + c 3 sin(log x)] + 5x + �log x X X where c 1 and c2 are arbitrary constants of integration. Example 26 : Solve the following homogeneous differential equation Solution d2 y dy -3x - + 5y = x2 sin (log x) x2 2 dx dx Putting x = so that z = log x and let O = � then given differential equation reduce to dz [0(0 - 1) - 30+ 5]y = e22 sin z ⇒ (02 - 40 + 5)y = e22 sin z 2 e Auxiliary equation is m2 - 4m + 5 = 0 ⇒ m = C. F. = 4± e ./i"6=2o= 4 + 2i=2 ± i 2 2 (c 1 cosz + c2sin z) 22 P. I. = = 1 0 2 -40 + 5 e (e 22 sin z)=e22 - 1 (0 + 2)2 -4(0 + 2)+ 5 1 1 22 , (sin z)= e z - -- sin z= -� e 22 cosz 2 20 2 · + 1 0 22 Hence the complete solution is 22 22 y = C.F. + P.I.=e (c 1 cos z + c 2 sin z)-½ e cos z 2 x 2 y = x [c 1 cos(logx)+ c 2 sin(logx)] -- logx cos(logx) 2 where c 1 and c2 are arbitrary constants of int egration. d 2 d dx Example 27 : Solve (1+ x)2 ---{ + (1 + x)--1 + y=sin 2 {Iog(1 + x)} Solution : We have dx sinz Engineering Mathematics 533 2 d d (1 +x)2 ---1. +(1 +x)-1. +y = sin 2 {Iog(1 +x)} 2 dx dx 2 Put 1 + x = e ⇒ log(1 + x) = z dy (1 +x)- = Dy dx ⇒ ⇒ 2 d y d (1 +x)2 = D(D - 1)y, where D aa 2 dx dz log(1 + x) = z and Now, substitute these values in the given differential e quation, we obtain D(D - 1 )y + Dy + y = sin 2z ⇒ (D2 + 1)y = sin 2z which is the linear differential equation with constant coefficients. ⇒ m2 + 1 = 0 The A.E. is m = ±i C. F. = c 1 cosz + c2sinz P. I. = Also 1 1 . 3 1 - 4 +1 -- S ·i n 2 Z = -- Sl· n 2 Z = - - Sln 2 z 2 D +1 1 Complete solution is y = C. F. + P. I. = c 1 cos z +c 2 sinz --sin 2z 3 Thus, y = c 1 coslog(1 +x)+c 2 sinlog(1 +x)-..!.sin2 1og(1 +x) 3 2 d d Example 28 : Solve (3x + 2)2 -{- +3(3x +2)--1. -36y = 3x 2 + 4x +1 . dx dx Solution : Let 3x + 2 = 2 e ⇒ z = loge(3x + 2) . d From this, we get (3x +2) � = 3Dy and (3x +2)2 d !} 2 On putting these values in given e quation, we get ⇒ 2 2 2 9 D(D - 1) + 9 Dy - 36y = �(e -2) + i (e - 2) + 1 ⇒ ⇒ The A.E. is Now 9 d where D aa dz 3 2 2 9(02 -4)y = ..!.( e -2)( e - 2 + 4) +1 3 2 2 2 9(0 - 4)y = ..!.( e -2)( e +2) +1 3 22 e 1 1 2z 2 9(0 - 4)y = - (e -4) +1= - - 3 3 3 9(m2 - 4) = 0 ⇒ m2 -4= 0 1 1 ( 3 e 2z ) - C. F. = c 1 e 2 2 +c 2e -2z P. I. = = - Therefore, the complete solution is or =9D(D -1)y 2 9(0 -4) ⇒ 1 2 m=± 2 9(0 -4) 1 Q. • 3e z 22 1 z 1 z e 1 2z 2 - e z +-=- · - +- = __ ( ze +1 ) 1 08 27(2D) 1 08 54 2 108 1 2z 2 2 Y = c 1 e z +c 2e - z +__( ze +1) 1 08 1 C2 2 2 +- [loge (3x +2) · (3x + 2) +1] Y = c 1 (3x +2) + 2 (3x +2) 108 · IES MASTER Publication 534 Linear Differential Equations of Higher Order with Constant Coefficients SIMULTANEOUS LINEAR DIFFERENTIAL EQUATION If we have a set of differe n ti al equat i on in wh ich one indepe nd ent vari able a nd two or more tha n two d ependen t vari ables are present the n this set is k nown as system of s imultaneou s lin ear d ifferential equations . . f1 (D)x + f2 (D)y=T1 } � 1 (D) x + � 2 (D)y = T2 . . . (A ) where, T 1 and T2 are funct io ns of i ndependent vari able 't' a nd f 1 , f2 , <D 1 , <D 2 are function s of D with constant coefficien ts. NOTE: 1. In this topic we w i l l take � = D dt 2. No. of arbitrary constant in the complete solution of equation A is equal to the degree of D in the following determinant. I fi ( D) f2 (D ) � 2 ( D) � 1 (D ) VARIATION OF PARAMETERS I Consider a second ord er d iffere n t i al equatio n as follows : . . . (1) y" +Py' +Qy = R Let u a nd v are parts of C. F. then, complete solutio n of (1) is ... (2) or y = c 1u +c 2 v + Au + Bv y = CF + Pl where c 1 , c2 are arb i trary co nsta n ts & A and B are to be d etermi ned. A = -J R R dx � ,B=J dx where W=I �, � � I , First calcu late R by comparing the coeffic ient, after making coefficient of y" unity. Example 29 : U sing the method of variation of parameters, solve Solution y" + y = sec x On comparison wi th y" +Py' +Qy = R we get R = sec Now, Aux i lary equat i o n is So, y" + y = m2 + 1 = 0 CF = Hence, we ca n take u = cos so sec x x c1 ⇒ ⇒ cos X (D2 +1)y = sec and v = si n c 2 sin X + x as parts of CF, So, A = -J : dx = -f B = +f � ...(1) x m = ± i Wronskian Now, x. f dx= sec x sin x s ec � d x=-fta n x dx = log cos f co s xd x= (1)dx=x \ Pl = Au + Bv = (log cos x )cos x + (x ) sin x x Hence, general solution is Engineering Mathematics y = CF + Pl = C1 COS X + C2 sin X + COS X. log COS X + X sin 535 X E�ample 30 : Using the method of variation of parameters, solve Solution y" - 3y' + 2y = Given differential equation is (02 - 30 + 2)y = where A.E is Hence, parts of C. F. are u = ex, v = e2x Now A = B = -fw Rv ⇒ dx = w = I u� :, I = e e x .e2x -f (1 +e ).e x 3x 3x dx = e-dx =log(e-f e- + 1 x x = f [-·J dx = f f -dx w 1+ e e e (e Ru ex x ex 3x x 1 x + 1) x + 1) f 1 1 dx - -- ) dx - (ex ex + 1 so by variation of parameters, general solution is y = C F + Pl (C 1 u + C2v) + (Au + Bv) = C1 ex + C2 e2x + (log (e-x + 1)).ex + {-e-x + log(e- x + 1)}.e2x DIFFERENTIAL EQUATIONS OF FIRST ORDER AND HIGHER DEGREE In this chapter, we shall deal with some special types of differential equations which will involve (dy/dx) in higher degree and (dy/dx) will be denoted by p. Here, the differential equation will be of the form f(x, y, p) = 0 . . . (i) in which, the power of p is not equal to one. Obviously, the equation is non-linear. Since the equation (i) is of first order, its general solution will be of the form ...(ii) F(x, y, c) = 0 containing only one parameter c. There are mainly three types of equations IES MASTER Publication 536 Linear Differential Equations of H igher Order with Constant Coefficients Case (i) : Equations Solvable for p: If we can solve for p, the equation of the nth d egree in p can be resolved into n factors as [p - f1 (x, y)][p - f2 (X, y)] ......[p - fn (X, y)] = 0 p = f1 (x, y), p = fz (x, y)...... p = fn (x, y) ⇒ ...(iv) Solving these n equations, the general solution is F 1 (x, y, c 1 ) = 0, ...(iii) Fz (x, y, c2) = 0 ..... . F n (x, y, en) = 0 .. .(iv) We can replace the constants c1 , c2 , ...en by a single constant c, as the equation (iii) is of first order. Hence, the general solution of (iii) is ...(v) Example 31 : Solve x2 + 1 + p2 . x2 = 1 + p2 or p2 = x2 - 1 Solution : ✓ ± ✓x p = ± x2 - 1 ⇒ dy ⇒ ✓ = 2 - 1 dx ✓ 2 Integrating, we get y = ± � x - 1 + �1og ( x + x 2 - 1 ) + c 2 2 Example 32 : Solve xp2 + (y - x)p - y = 0. Solution : The equation is solvable for p, as it can be factorized to give (p - 1) (xp + y) = 0 ⇒ ⇒ (p - 1) = 0 or p = -y/x ⇒ y = x + c or xy = c dy = dx or On integration d y/y = -dx/x Case (ii) : Equations Solvable for y: ⇒ ⇒ d y/dx = 1 or d y/d x = -y/x (y - x - c)(xy - c) = 0 If equation (i) is solvable for y, then it can be written in the form y = cp(x, p) Differentiating it w.r.t. x, we get p = � ( X, p, �� ) ...(vi) ...(vii) which is a first ord er an d first degree d ifferential equation with p as d epen dent variable and x as in depend ent variable. General solution can be obtained by eliminating p from (vi) and (vii) or otherwise as Example 33 : Solve y = (x - a) p - p2 . Solution : F(x, y, c) = 0 y = (x - a)p - p2 Differentiating (i) w.r.t. 'x ' we obtain dy dp dp dp = p + (x - a) - 2p ⇒ p = p + (x - a) - 2p d p dx dx dx dx dx dp dp dp dp = 0 ⇒ O = (x - a) d x _ 2p d x ⇒ o = dx [x - a - 2p) ⇒ dx Integrating, we get p = c Putting the value of p in (i), we get y = (x - a) c - c2 . ..(i) Engineering Mathematics Example 34 : Solve y + xp = x 4p2. Solution : y = -xp + x4p2 We have Differentiating (i) w.r.t. 'x' we obtain ⇒ ⇒ 3 p = -x 3 2p(1 -2x p) + x �� (1-2x p) = dp dx . . . (i) 3 2 - p + 4x p + 2x 4p o ⇒ 537 3 dp dx (1- 2x p {2p + x �� ) = o For the first factor, we get a singular solution, which we are not considering. For the second factor, we have d p/p + 2dx/x = 0 Integrating, we get px2 = c . . .(ii) Eliminating p between (i) and (ii), we get the general solution of the given equation as xy = c2x - c Case (iii) : Equations Solvable for x: If the equation (i) can be solved for x, then it can be put in the form X Differentiating it w.r.t. y, we get . . . (viii) = cp(y, p) ; = � ( y, p, �� ) . ..(ix) y = 2px + yp2 . . .(i) Eliminating p from equation (viii) and equation (ix) gives the required general solution. Example 35 : Solve y = 2px + yp2. Solution : Given that y 2x = - -yp p Differentiating (ii), w.r.t. y, we get dp ! _ l'_ d p dp dx -p - y 2 = ⇒ !p + p = -y (_!_2 + 1) d y dy p p2 dy dy p or · 2 1 + p2 dp 1+ p -y-- ⇒ 1 = = ⇒ 2 dy p p Integrating we get -logy = logp + loge ⇒ log PY = log c ⇒ PY = c or Putting the value of p in (i), we get (C )2 ⇒ c y = 2yx + y y Example 36 : Solve y2 In y = xyp + p2 . Solution 1 We have y2 = 2cx + c2 p ⇒ -y d p dy dp - - ⇒ -- = p dy y dy p = c/y ⇒ y2 = c(2x· + c) x = Y... In y -£. 1 y dp 1 p 1 dp - In y -ln y - + - + -- ­ 2 p d y p y2 y dy p p 1 1 y dp In y ) - = (+ - In y) +(2 2 p dy y p y . . .(i) y p Differentiating w.r.t. y, we get ...(ii) ⇒ dp In y ) ( +l (! 2 dy y p _E_) y = o First factor gives the singular solution which we are not considering. Hence, the general solution is given by IES MASTER Publication 538 Li near Differential Equations of Hig her Order with Constant Coefficients dy dp = y p E_ = y ⇒ . . . (ii) C Eliminating p between (i) and (ii), we get the general solution as I n y = ex + c2 CLAIRAUT'S EQUATION Clairaut's equation is a first order and higher degree differential equation of the form y = This equation (i) is linear in y and x. Here <j> dy . = p, equation (i) takes the form I ntroducing dx X �� + q>( �� ) . . . (i) is a known function of �� . IY = xp + q>(p ll Differentiating both sides of (ii) with respect to x, we have �� = P + x �� + �: · �� . . . (ii) Since p itself is a function of x. On simpl ification x dq> dp = 0 [ + ] d p dx Thus and I ntegrating (iii), we get X+ dp = 0 dx . . . (iii) d q> = dp . . .(iv) p = constant = c Substituting p = c in (ii), the complete solution (or i ntegral) of the Clairaut's equation (i) is given by y = X · C + q>(C ) . . .(v) which is a one parameter fam ily of straight li nes; with c as the parameter. Thus the complete solution of Clairaut's equation is obtained simply by replacing �� (i.e. p) by constant c. Besides the complete i ntegral (v), one may obtain a singular solution of Clairaut's equation which satisfies the Clairaut's equation (i) but not obtained from the complete integral (v) for any value of c. Equation which is not in Clairaut's form can be reduced to Clairaut's form by suitable substitutions. Example 37 = Solve p = log(px - y) Solution = p = log(px - y) This equation can also be written as which is Clairaut's equation eP = px - y ⇒ y = px - eP The general solution (or complete i ntegral) is obtained by replaci ng p by c. Thus, y = ex - e c. This is the general solution of given differential equation. . . . (i) Example 38 : Solve Solution = p Engineering Mathematics = tan( px - y). The given d ifferential equation can be written as tan-1 p = px - y ⇒ y = px - tan-1 p which is the form of Clairaut's equation. Differentiate equation (i) with respect to x , we get dy dx p+ = x 1_ dp __ dx 1 + p 2 dx dp 1 dp ⇒ [x - ⇒ p = c or p = 1+p 2 ] dx ⇒ = 0 ⇒ ✓1 - � Substituting the value of p 1 1+p ] = p + [x - -2 p dp dx = 0 or x - dp dx 1 = 0 1 + p2 ...(i) dy [·: = c in (i), we get y = ex - tan-1 c dx = P] (neglecting 2 nd term) This is the general solution of given d ifferential equation. Example 39 : Solve p2(x2 - a2 ) - 2xyp + y2 - b2 = O. Solution = p 2x2 ⇒ - 2 2 p a - 2xyp + y2 - b2 = 0 (p2x2 - 2xyp + y2) - (p2a2 + b2 ) = 0 ⇒ (px - y)2 = p 2a 2 + b2 This is the form of Clairaut' s equation. Hence, the general solution is ✓ y = cx - c 2a 2 + b 2 ⇒ Example 40 : Solve y = 4x p - 1 6y3 p2 . Solution : ⇒ y= px - ✓ p 2 2 a + b2 (y - cx) 2 = c2a 2 + b2 This equation is not in Clairaut' s form. Multi p lying the given equation by y3 , we have y4 = 4xy3p - 1 6y6p2 , 3 dy Put yA = v, so 4y = dx Then (i) becomes 539 dv dx V = X �� ⇒ 4y 3 p = -(��r dv dx dy dx = p) . . . (i) dv which is a Clairaut's equation in v. Its general solution is obtained by re placing by a constant c. dx Thus the general solution is v = xc - c2 ⇒ y4 = ex - c2 (-,- V = y 4 ) . IES MASTER Publication 540 Linear Differential Equations of Higher Order with Constant Coefficients • . - }�REVIOUS)YEARS·GATE & ESE QUESTIONS -"'�--;-�-;�J;.. .<-...,.-�_;1'-"'!_ .... -�- __,_ _•<• • ••: 1. 2. 3. 4. ·t --::. ,--:�� equation has non-trivial solutions, the general value of "' is (c) 1.5 sin x [GATE-1 993 (ME)] y = e - 2x is a solution of the differential equation y" + y' -2y=0 (True/False) Solve the dy d2 y - 7 - +6y 2 dx dx Solve for y if e-t = differential [GATE-1 994-EC] equation, e2x [GATE-1 994] dy d2 + y = 0 with y(0) ; +2 dt dt = 1 and (b) (1 + t) e t (c) (1 + t) e- (d) (1 - t) e t t [GATE-1 994 (Pl)] The solution to the differential equation f"(x) + 4f '(x) + 4f(x) = 0 is (a) f1 (x) = e (c) f1 (x) = e- (b) (d) -2 x 2x , f2 ( x ) = xe-2x f1 (x) = e2X , f2 ( x ) = e-x d2 y - y = Stn t ' t > 0 dt 2 The s o l u t i o n of a (c) x C1 e + c 1 e 2x C 1 e- x + c 2 e - 2x [GATE-1 995-ME] [GATE-1 995] e q u ation (b) C 1 e - x + c 2 e 3x (d) C 1 e - 2x + c 2 rx [GATE-1 995] The particular solution for the differential equation dy d2 y +2y = 5 cos x is +3 2 dx dx (d) 0.5 cos x d4 v 4 9. * Solve + 4A v = 1 + x + x 2 4 dx [GATE-1996-ME] [GATE-1 996] 1 0. The general solution of the differential equation d2 y dy x2 - x - +y = 0 is: dx dx 2 (a) Ax +Bx 2 (A, B are constants) (b) Ax + Blogx (A, B are constants) (d) Ax + Bxlogx (A, B are constants) (c) Ax + Bx 2 logx (A, B are constants) [GATE-1998] 1 1 .* The radial displacement in a rotating disc is governed by the differential equation d2u 1 du u +- - -- = 8x dx 2 x dx x 2 i If u = 0 at x = 0, and u = 2 at x = 1, calculate the differential y" + 3y' + 2y = 0 is of the form (b) 1.5 cos x + 0. 5 sin x where u is the displacement and x is the radius. f1 (x) = e 2x , f2 ( x ) = xe-2x Find the general solution to the ordinary differential equation (a) 8. •• Questions marked with aesterisk (*) are Conceptual Questions the conditions y(0) = 0, y (A) = 0. In order that the (a) (1 - t) 7. - (a) 0.5 cos x + 1.5 sin x y (0) = -2 6. ··_- .:�-- The differential equation y" + y=0 is subjected to 1 5. _--::_•• displacement at x = d y - y = 15 cos 2x 12. Solve dx 4 [GATE-1998] 4 13. The solu tion 2 [GATE-1 998-CE] of the differential equation d y dy + +y = 0 is dx 2 dx (a) Ae x +Be- x (b) ex ( Ax +B ) (d) e� {A cos( f ) x +Bsin (f ) x } (c) e- x {A cos( f ) x +Bcos( f ) x} x [GATE-2000] 14.* Find the solution of the differential equation d2 y +1/ y=cos rot +K dt 2 dy (0)=0 . Here 1c, ro with initial conditions y(0) = 0, dt and k are constants. Use either the method of undetermined coefficients or the operator ( D= :t ) based [GATE-2000] method. 1 5. The solution for the following differential equation with boundary conditions y(0) = 2 and y'(1) = -3 is, (a) y= 3 2 d2 y -=3x -2 dx 2 3 - 2 +3x - 6 x x 3 x2 (d) y = x 3 - - +5x + - [GATE-2001] 2 1 6. Solve the differential equation, d2 y +y=x dx 2 with the following conditions : (i) at X = 0, y = 1 (ii) at X = (a) p (a) y=e2x +e-3x (c) 1 8. y = e-2x +e 3x = 541 (b) p = 3, q = 4 3 (d) p = 4, q = 4 [GATE-2005-ME] 20. Which of the following is a solution of the differential equation dy d2 y +p -+(q +1) y=0 ? 2 dx dx (a) e-3x (b) xe-x (c) xe-2x [GATE-2005-ME] 21 . For the equation x"(t) + 3x'(t) + 2x(t) = 5, the solution x(t) approaches to the following value as t ➔ oo (b) 5/2 {d) 10 [GATE-2005-EE] d2 y dy +2 - +17y= 0 · y(0) = 1 22. The solution of dx ' dx 2 �� (¾) = o in the range O < x < ¾ is given by x (a) e- ( cos4x +±sin 4x) 7t 2, y = x (b) e ( cos 4x -±sin 4x) 7t 2 --4x (c) e ( cos x -±sin x) [GATE-2001] (b) y=e2x +e3x --4 x {d) e ( cos 4x -±sin 4x) 4D The complete solution of the ordinary differential d y dy +p -+ qy=0 is y = c 1 e-x + c2e-3x equation dx dx 2 + 4 )y = 0 is of the ( given D = � and C1 , C2 are constants ) . d [GATE-2005-EC] Statement fo r L i n ked Answer Questions 1 9 and 20. [GATE-2005-CE] 23. The general solution of the differential equation (D2 (d) y=e-2x +e-3x 2l n ( x ) 2 dy If x - +2xy=- - and y(1) = 0 then what is x ' ' dx Y (e)? (b) 1 (a) e (d) 1/e2 (c) 1/e [GATE-2005-ME] 2 3, q (c) p = 4, q = 3 1 7. A solution of the following differential equation is given dy d2 y -5- +6y=0 by 2 dx dx = (c) 5 2 2 1 9.* Then, p and q are (a) 0 x2 (b) y=3X3 - -- 5x +2 2 5x x3 2 (c) y=- - X - -+2 2 Engineering Mathematics (a) C 1 e2x 24. (c) C 1 e2x + C2e2x For is form [GATE-05 (I N)] (b) C1 e2x + C2e-2x (d) C1 e2x + C2 xe2x dy d2 ; +4 +3y=3e2x , the particular integrals dx dx 1 2x (a) -e 15 (c) 3e2x 1 2x (b) - e 5 (d) C 1 e-X + C2e3-x [GATE-2006-ME] IES MASTER Publication Linear Differential Equations of Higher Order with Constant Coefficients 542 d2 y dx + k2Y 25.* For the differential eq uation 2 boundary condition s are (i) y = 0 for x = 0 and = 0 the (ii) y = 0 for x = a The form of non-zero solutions of y (where m varies over all integers) are " mrcx . mrcx " L, Am cos (a) y = L, A m sm (b) y = a a m m mrr 26.* F o r i n i ti al v a l u e prob l e m [GATE-2006-EC] y + 2 y + (101) y = (10.4)e , y(O) = 1 . 1 and y (O) = - 0.9. Variou s solution s are writen in the fol lowing group s . Match the type of solution with the correct expre ss ion. X Group -1 P. General sol ution of homogeneous equation s R. Total solution ·s ati sfying boundary condition s Q. Particular i ntegral Group-II (1) 0.1 ex (2) e-x [A co s 1 0x + B sin 1 Ox] (3) e -x cos 1 0x + o.1ex 27.* The s o l ut i o n [GATE-2006 (IN)] of the d i fferent i a l equation d2 y k2 = Y - Y2 under the boundary conditions dx2 (i) y = y1 at x = 0 and (ii) y = y2 at x = CJJ , where k, y1 and y2 are constants, is (a) y = (y1 - y2 ) exp ( - x/k2) + y2 (b) y = (y2 - Y1 ) exp ( - x/k) + Y1 (c) y = (y1 - y2 ) sinh (x/k) + y1 (d) y = (y1 - YJ exp ( - x/k) + y2 28. G iv e n that x (0) = 0, x +3x = what i s x(1 )? 0, (d) 0.99 d2 y dx cos x + Q s in x cos X s in x si n 2 x [GATE-2008-ME] + Y = 0 is 29. The general solution of 2 (a) (b) (c) (d) y = P y = P y = P y = P 30. Solve for y, if d2 y dy + = ; with +2 y O dt dt2 y(O) = 1 and y' (O) = -2 31 . It i s given that y" + 2y' + y What i s y(0.5)? (a) 0 32. The so l u t i o n s = 0, y(O) = 0, y(1) = 0. (b) 0.37 (c) 0.62 [GATE-2008] (d) 1.13 [GATE-2008-ME] of the d iffe re n t i a l e q u at i o n s d2y dy + 2 - + 2 y = 0 are 2 dx dx 1 ) ) ) (1+ (1+ ) (a) e - i x _ e -(1- i x (b) e i X , e( - i x 1 i) 1+ ) e -( i x , e( - x (d) e(1+i)x , e-(1-i)x [GAT E-2008 (Pl)] 33.* The homogeneou s part of the differential equation (b) P - 1, Q - 3, R - 2 (d) P - 3, Q - 2 , R - 1 (b) - 0.16 (c) 0.16 (c) (a) P - 2 , Q - 1 , R - 3 (c) P - 1, Q - 2, R - 3 (a) - 0.99 dy d2 . --{ + P + qy = r (p, q, r are con stants) ha s real dx dx di stinct roots if (a) (c) (b) p2 - 4q < 0 p2 - 4q > 0 (d) p2 - 4q = r p2 - 4q = O 34.* A · function n(x) s ati sfies [GATE-2009 (P l)] the differential equation d n(x) _ n(x) = o where L i s a con stant. The boundary dx2 L2 condition s are: n(O) = K and n( C/J ) = 0. The solution to thi s equation i s 2 (a) n(x) = K exp(x/L) [GATE-2007-EC] and x(O) = 1, (b) n(x) = K exp(-x!../L) (c) n(x) = K2 exp(-x/L) (d) n(x) = K exp(-x/L) [GATE-201 0-EC] Engineering Mathematics 35. For the differential equation d2x dx -+6-+ Bx = 0 with initial conditions x(O) = 1 2 dt dt = 0, the solution is �: 11�0 (a) x(t) = 2e--6I - e-2I (b) x(t) = 2e-2I - e--4I 40. The maximum value of the solution y(t) of the differential equation y(t) + y (t) = 0 with initial conditions y (O) = 1 and y(O) = 1, for t � O is (c) [GATE-201 0-EE] (b) y = C1 e3x + C2 e2x (d) y = C1 + C2 e2x + C2 e-2x [GATE-201 0-CE] (d) e2 2 [GATE-201 1 (IN)] 38. The solution of the differential equation dy d2 y + 6- + 9y = 9x + 6 2 dx dx constant is (a) y = (C1 x + C2 ) e-3x with c1 and c2 as (c) Y = (C1 X + C2 ) e-Jx + X (d) y = (C1 x + CJe3• + x 39. Consider the differential equation [GAT E-201 1 ( Pl)] 2 dy dy 0 + x -4y = dx dx 2 with the boundary condition of y(O) = 0 and y(1 ) = 1 . The complete solution o f the differential equation is x2 (a) x2 x (c) e sin ( n;) (b) u - u (c) (d) u = 1-e-kx _ U [-----=i<L uJ 1-e . nx (b) Stn ( ) 2 - . nx (d) e x Stn ( ) 2 [GATE-201 2-ME, Pl] 1- ekl +e u[11+e :: J [GATE-201 3-ME] d2 y dy -+2a -+y = 0 has two equal roots, then the 2 dx dx values of a are 43. (a) ± 1 (c) ± j (b) 0,0 (d) ± 1/2 [GATE-2014-EC-Set-2] If a and b are constants, the most general solution d2 dx : +2 +x =0 is dt dt (b) ae-I+bte-I (d) ae-21 of the differential equation (a) ae-1 (c) aeI + bte-I (b) b = C1 e3• + C2 e-3x - [�J (a) u= U� L 42.* If the characteristic equation of the differential equation 37. Consider the differential equation y + 2y + y = 0 with boundary conditions y(O) = 1 &y (1 ) = 0. The value of y(2) is (b) -e- 1 (a) -1 (c) -e- [GATE-20 1 3 (I N)] d2u du 0 where k is constant, -k = 2 dx dx subjected to the boundary conditions u(O) = 0 and u(L) = U , is 36. The solution to the ordinary differential equation d2 y dy -+- 6y= 0 is dx 2 dx (a) y = C1 e3x + C2 e-2x e-3x (d) n 41 . The solution to the differential equation (c) x(t) = -e--6I + 2e--4I (d) x(t) = e-2I + 2e--4I (c) y = C1 J2. (b) 2 (a) 1 and e-3x 543 [GATE-2014-EC-Set-4] 44. The solution for the differential equation with initial conditions x(O) = 1 · and (a) t2 + t + 1 dx = -9x dt2 2 dx 1 , is l = dt 1=0 (b) sin3t+ ! cos3t+� 3 ! sin3t+cos3t 3 (d) cos 3t + t (c) 3 [GATE-201 4-EE-Set-1] 45.* Consider the differential equation x 2 d2 Y dx 2 + 0 . Which of the following is a solution to x �� -y= this differential equation for x > 0 ? IES MASTER Publication 544 Linear Differential Equations of Higher Order with Constant Coefficients (b) x2 (a) e x (d) lnx (c) 1/x 46. If y = f(x) is the solution of d 2 [GATE-2014-EE-Set-2] ; = 0 with the bound­ dx ary con d itions y = 5 at x = 0, and �� = 2 at x = 10. �15) = ___ [GATE-2014-ME-Set-1] 47.* Consi der two solution x(t) = x 1 (t) and x(t) = xi{t) of the d ifferential equation d 2 x(t) -2 - + x(t) = 0' t > 0' dt dt I d y dy + 2 - + y = 0 with y(0)=y'(0)=1 is 2 dt dt (b) ( 1 + 2t)e- t (a) ( 2 - t) e1 (c) (d) ( 1 - 2t) e1 ( 2 + t) e- 1 [GATE-201 5-EC-Set-1] 52. Consid er the d ifferential equation d 2 x(t) dt 2 +3 d x(t) dt + 2x(t) = 0 Given x(0) = 20 and x(1 )=10/e, where e = 2.718, the value of x(2) is___ [GATE-201 5-EC-Set-3] d ifferential equation [GATE-2015-EE-Set-1] (a) 1 54. The solution to x2 y" + xy' - y = o is (b) -1 (d) rc/2 (c) 0 [GATE-2014-ME-Set-3] 48. With initial values y(0)=y'(0) = 1, the solution of the d ifferential equation d2 y dy + 4 - + 4y = 0 at x = 1 is--2 d x dx [GATE-2014-EC-Set-4] 49. Consi der the following second order linear d ifferential d 2 y dx 2 = -12x2 + 24x - 20 The bound ary cond ition are at x = 0, y = 5 and at X = 2 , y = 21 The value of y at x = 1 is __ 50. Fin d the solution of d y = ! ex - e-x 2 x x (c) y = ! (e - e- ) 2 [GATE-2015-CE-Set-ll] 2 ; = y which passes through dx origin an d point (1n 2, (a) equation d y dy + 5 - + 6y = 0 is such that y(0) = 2 an d y(1) = 2 dt dt 1 - 3e dy - ( 0) is --3- . The value of dt e The Wronskian equation d i fferential 2 =1 t=O of the 2 53. A solution of the ordinary such that x 1 (0) d x 2 (t) 51 . The s o l u t i o n ¾) x (b) y = ! (e + e-x ) 2 x x (d) y = ! e + e2 [GATE 201 5-ME-Set-1] [GATE-201 5 (Pl)] 55. A function y(t) , such that y(0) = 1 a n d y(1) = 3e-1 , is a solution of the d ifferential equation dt2 + 2 dt + y = 0. Then y(2) is d 2y dy (a) se-1 (c) ?e- 1 (b) se- 2 (d) ?e-2 [GATE-201 6-EE-Set-1] 56. If y = f(t) satisfies the boundary value problem 2 y" + 9y = 0, y(0) = 0, y(rc / 2) = ✓ , then y(rc / 4) is [GATE-201 6] 57. The particular solution of the initial value problem given below is y dy + 1 2 + 36y = 0 2 dx dx d 2 dy with y(0) = 3 an d dx = = -36 lx o Engineering Mathematics (b) (3 + 25x) e-6x (d) (3 - 1 2x) e-6x (a) (3 - 18x) e-6x (c) (3 + 2Ox) e -6x [GATE-2016) 58. Let y(x) be the solution of the differential equation dy d2 y - 4 - + 4y = 0 with initial conditions y(O) = 0 2 dx dx dy l = 1 . Then the value of y(1) is __. and dx x � [GATE-2016-EE-Set-2] x 2 2 fs x (c) K 1e(- +fs) + K 2e(- - ) x 2x 2 (d) K 1e(- + Fa) + K 2e(- Fa) [GATE-2017 EC Session-II] 62.* Consider the following second-order differential equation: y " - 4 y' + 3y = 2t - 3t 2 The particular sol ution of.the differential equation is (a) - 2 - 2t - t2 59.* The respective expressions for complimentary func­ tion and particular integral part of the solution of the differential equation -{ + 3 -{ = 108x2 are dx dx d (a) ct2 l c1 + c 2 x + c3 sin .fix + c4 cos .fix J and 2 L3x - 1 2 x + cJ 4 (b) 4 l c2 X + c3 sin .fix + c4 cos .fix J and 2 L5x -12x + cJ 4 (c) 2t - 3t2 2 L5x - 1 2x + cJ 4 [GATE-2016-CE-SET-I] 60. What is the solution for the second order differential 2 (c) exactly one solution (d) infinitely many solutions [GATE-2017 ME Session-I] 64. C o n s i d e r the d i ffe rent i a l equation 3y"(x) + 27y(x) = 0 with initial conditions y(O) = 0 and y'(O) = 2000. The value of y at x = 1 is __. d y equation + Y = 0, with the i nitial conditions dx2 65. Y l x =O (a) =5 dy and d X l X=0 (a) y = 5 + 0 sin x = 10? (b) (c) (b) y = 5 cos - 5 sin x (d) (C) y = 5 cos X + 1 Ox (d) y = 5 cos x + 1 0 sin x [GATE-16 (CH)] 61. The general solution of the differential equation dy d2 y + 2 - - 5y = O dx dx2 in terms of arbitrary constant K 1 and K 2 is (a) 1 fs x 1 x K 1e(- + fs) + K 2 e(- - ) 1 sx 1 x (b) K 1 e(- +fs ) + K 2e(- - f ) [GATE-2017 CE Session-II] (b) exactly two solutions 2 (d) lc1 + c 2 x + c3 sin .fix + c4 cos .fix J and (d) - 2 - 2t - 3t2 (a) no solution (c) lc1 + c3 sin ✓ x + c4 cos .fixJ and 4 (b) - 2t - t2 d2 y + 16 Y = 0 for y(x) with 63. The differential equation dx2 b o u n d ary two cond i t i o n s the dy l = dy l -1 has = 1 and dx x=! dx x=0 3 L3x -12x + cJ 545 66. [GATE-2017 ME Session-II] The general solution of the differential equation 3 dy d2 y d y d4y - 2 - + y = u,, 2 2 + dx 4 dx 3 dx dx 2 y = (c 1 - c2 x) e• + c3 cos x + c4 sin x y = (c1 + c2 x) e• - c3 cos x + c4 sin x y = (c 1 + c2 x) e• + c 3 cos x + c4 sin x y = (c1 + c2 x) e• + c3 cos x - c4 sin x [ESE 2017 (EE)] The solution of the differential equation 2 d y _ dy _ = 3e2x 2y dx 2 dx where, y(O) = 0 and y'(O) = -2 is (a) y = e-x - e2x + xe2x (b) y = ex - e-2x - xe2x (c) y = e-x + e2x + xe2x (d) y = ex - e-2x + xe2x [ESE 2018(EE)] IES MASTER Publication Linear Differential Equations of Higher Order with Constant Coefficients 546 67. If d2 dy ; + y = 0 under the conditions y= 1 , =0, dt dt when t = 0 then y is equal to d2 y dy -+--6y =0 dx2 dx dy (0)= 1, the value of y(1 ) is dx _ (correct to two decimal places). (d) cot t (c) tan t Given the ordinary differential equation with y(0) = 0 and (b) cos t (a) sin t [ESE 2018(EE)] 68. 69. The solution (up to three decimal place) at x = 1 of the d2 dy ; +2 + y +0 subject to dx dx dy ( 0)= - 1 is boundary conditons y(0) = 1 and dx differential equation [GATE 2018 (CE-MorningSession)] 70. [GATE 2018- (M E-AfternoongSession)] The position of a particle y(t) is described by the differential equation d2 y _ _ dy _ 5y dt2 - dt 4 The initial conditions are y(0) = 1 and dy l = 0. dt t=O The position (accurate to two decimal places) of the particle at t= 7t is ---[GATE 2018 (EC)] Engineering Mathematics 1. 2. (TRUE) 4. (a) 3. 5. (c) 7. (c) 8. 9. 1 0. 11. 12. 13. 14. 1 5. 1 6. 17. Given 24. 25. 26. (a) (See Sol. ) 27. (d) 28. (5/8) 29. (See Sol. ) 30. (d) 31 . ( See Sol. ) 32. (c) ( See Sol. ) ( nrc) , n (c) 56. (-1) (a) 40. (d) (b) 42. 37. (b) 39. 35. (b) . . . (i) 69. ⇒ ⇒ (b) (a) (7.39) (a) (d) (a) (a) (a) (94.08) (c) (a) (b) (0.368) (1 .47) 70. - (-0.21) 0 = C 1 cos A. + C2 sin A. C2 sin A. = 0 ( ·: C2 * sin A. = 0 0 for non-trivial solution) A. = nrc, n E z .-. Equation (i) has non-trivial solution for A . E . is m 2 + 1 = 0 m = ±i = 0 ± i cosx + C2sinx] Now using y(0) = 0 in (ii), we get C 1 = 0 Again using y(A) = 0 in (ii) A. * nrc, n * o , n E I and solution is given as y = C2 sin A Sol-2: . - . Solution of equation (i ) is given by i.e., 68. (-3) ⇒ (D2 + 1 )y = 0 1 67. SOLUTIONS From equation (i), we have y = 66. (0.8556) 53. y" + y = 0 e0x[C 65. (b) 52. with y(0) = 0 and y(A) = 0 ⇒ 64. (c) 51 . (d) 63. (18) 50. (a) 62. (0.54) 49. (a) (c) 61 . (a) 48. (a) 60. (35) 47. (a) (a) (c) 46. (b) 59. (b) 45. (d) 58. (b) 44. (a) 57. (a) 43. (a) 55. (c) 41 . (d) (a) ' * 0, n E I 38. (c) 33. 34. (b) (c) (c) 23. (See Sol. ) 54. 20. 22. _ (c) (d) 21 . � 36. 1 8. 19. ( See Sol. ) 6. Sol-1 : * ( n rc) , n 0 , n E I A N SW-E R ]{EY 547 . . . (ii) (True) y" + y' - 2y = 0 A.E is D2 + D - 2 = 0 D = -2 , 1 y = e-2x and y = ex are The solutions of the given differential eq . IES MASTER Publication Linear Differential Equations of H igher Order with Constant Coefficients 548 Sol-3: 1 1 . . sm t = - - sin t P.I . = -2 dy d2y - 7 - + 6y = e 2x dx dx 2 A.E. is 02 - 70 + 6 = 0 0 = 1, 6 C.F. = C1 Sol-7: ( c) ex + C2 e6x 1 1 ---e 2x = --e 2x P.l . = 4 02 - 70 + 6 So , general solution is 2x y = CF + Pl = C 1 ex + C2e6x _ _!_ e 4 Sol-4: (a) Given (02 + 2 0 + 1 )y = 0 So, AE is m2 + 2 m + 1 = 0 (m + 1 )2 AE is ⇒ m = -1 and -2 dy d2y + 3 + 2y = 5cosx 2 dx dx 1 (0 + 30 + 2) P. I. = -2--- - 5 cosx 1 1 -cos(ax + b) = --2- cos(ax + b) 2 f(O ) f(-a ) 1 P.I . = 5 ----cos x (-f) + 3O + 2 ...(ii) = --cos x (30 + 1) ...(iii) 30 -1 = 5 x -2- cos x 90 - 1 Hence the general solution is y = (1 - t)e-1 = 5x f2(x) + 4 f'(x) + 4 f(x) = 0 = AE is (02 + 40 + 4) f(x) = 0 Sol-6: A.E. is d y dt2 -y 2 02 - = sin t 1 = 0 0 = ± 1 C.F. = C 1 e1 + C 2e-t 5 30 - 1 COS X 9(-1) - 1 _10 x (3D -1)cos x 1 P.I. = - 2 x (3O - 1)cos x 02 + 40 + 4 = 0 f(x) = (c1 + c2x)e-2x f1 (x) = e-2x, f2(x) = xe-2x + 3m + 2 = 0 ...(i) y = c 1 e -x + c2e-2x is the general solution of eq. (i) Again using y'(0) = -2 in (iii), C2 = -1 Thus, the solution is m2 Sol-8: (a) :. m = -1 , -1 (real and equal) dy = (C 1 + C2t)(-e- 1) + C 2e-t dt Now using y(0) = 1 in (ii), C 1 = 1 y " + 3y' + 2 y = 0 (02 + 30 + 2)y = 0 Using: = 0 y = C 1e1 + C2 e-1 _ _!_ si n t 2 ⇒ ...(i) Solution is y = (C1 + C2t)e-t Sol-5: (c) 2 0 -1 0 = -2 , -2 Sol-9: IP.I. = _ _!_ x [(3(-sin x) - cos x] 2 1 .5 sin x + 0.5cosxl ...(i) ⇒ The auxiliary equation is ⇒ m4 + 4A4 = 0 2 (m2 + 2A. 2 }2 - 4m A2 = 0 (Using (a + b)2 - 2ab = a2 + b2) Engineering Mathematics ⇒ So, C.F. = Ae2 + Be- 2 = Ax + m = -A ± A,i , A ± Ai CF = e-),x [C1 COS AX + C2 sin Ax] + P. I. = 1x + e ' [C3 COS AX + C4 sin Ax] P. I. = ( 1 04 + 4A4 2 }1 + X + X ) O4 = -;- 1 + 4 (1 + x + x 2 ) 4A [ 4A r o 1 4 0 8 ..... ] (1 + x + x 2 ) 1-= + -8 4 4 [ 4A 4A 1 2 16A +x+x ) = -(1 4A4 [ ·: higher derivatives vanishes] Hence the complete solution of given equation is v = C.F. + P. I. x = e-1.. [C1 cos Ax + C 2 sin AX] + Sol-1 0 : (d) e1'x [C3 COS AX + C4 sin AX] + (1 + x + x 2 ) 4A4 It is Homogeneous Linear O.E. d2 y - x dy + y = 0 x 2 dx dx 2 Let z = logx [O'( O' - 1) - O' + 1] y = 0 0'2 - 20' + 1 = 0 O' = 1, 1 y = (A + Bz)e2 Sol-1 1 : (5/8) y = (A + B logx)x Put du d2 u x2 + x - - u = 8 x3 dx dx 2 X = ff [O'(O' -1) + O' - 1] u = 8e3z (0' - 1) u = 8e 2 A.E. is ⇒ m2 - 1 = 0 m = ±1 3z B 8 e3z ( 02 -1) 8 ( 3 - 1) = e3z = x 3 3z = --e 2 u = CF + Pl So, ⇒ lu=Ax + � +x 3 1 Thus solution is putting condition gives : B = 0, A = 1 So displacement at X ⇒ = 2 Sol-1 2: d4 y -Y = 15 cos 2x dx 4 A.E. is m4 - 1 = 0 m = 1, -1, i, -i General soluti o n : y(x) = C 1 ex + Particular solution Pl = 2 d u 1 du u + - - - - = 8x dx 2 x dx x 2 549 Sol-1 3: (d) 1 4 (0 -1) = 15 C3cosx + C4sinx (15cos 2x) 1 (-4)2 - 1 y(x) = C 1 ex + c2e-x + cos 2x= cos 2x c2e-x + C3cosx + C4sinx + cos2x d2y dy + -+ y = 0 dx 2 dx 02 + 0 + 1= 0 3 -1 ± ✓ i 0 = 2 x y = e~ 12 [ A cos � x + B sin � x ] IES MASTER Publication 550 Sol-14: Linear Differential Equations of Higher Order with Constant Coefficients d2 y 'A,2 + y = cos cot + k , y(0) = 0, dy (O) = O dt dt2 A. E. is 0 2 + 'A,2 = 0 2 D + 'A, 2 cos cot + 2 1 0 + 'A,2 K K coscot + = 2 2 2 A, A, -co ⇒ m2 + 1 = 0 y = -( K = 1, m = ±£ y = C 1 COS X + C2sin X Particular integral is : 1 X ( 0 2 +1) =(D2 + 1)-1 IP · I = (1 - 02 + 04 + ... ) X xi So, complete solution is Putting condition, ⇒ coscot K K + + 2 ) cos ').,.,t + 2 A,2 - co2 A, A, - co2 2 A, 1 y(0) = 1 and Y( % ) = C2 = 0 d2 y = 3x - 2 dx2 Integrating both sides w.r.t. x X ⇒ \y = c 1 cos x + c 2 sin x + x\ 0 = C2 'A, Sol-15: (c) ⇒ Auxiliary equation is P.I. = 1 ⇒ d2 y +y = X dx 2 It is a form of linear differential equation. K +2 0 = C1 + 2 A, - co2 A, dy (O) = 0 dt = 2 5 x3 2 y = - -x -- x + 2 2 2 y(0) = 0 ⇒ ⇒ c2 ⇒ Sol-1 6: 1 At x3 5 y = - - X2 -- X + C2 2 2 At x = 0, y = 2 ⇒ ⇒ 0 = ± 'A,i P. I. = Again, integrating both sides w.r.t. x dy 3 2 = -x -2x + C1 dx 2 dy = -3 dx 3 - 3 = - -2 + C1 2 5 C1 = dy 5 3 2 = -x -2x -dx 2 2 7t Solution is Sol-17: (b) 2 ⇒ = c1 + 0 + 0 n = 0 + C2 + ⇒ 2 IY = Cos x + x\ d2 y dy + 6y -5 dx dx2 = Given the differential equation· 2 log X dy + 2xy = X dx 2 I � 0 2, 3 = C 1 e2x + C2 e3x y x2 I�C_ = 0 0 Sol-18: (d) c-'1\ \._ 1 __, = 0 02 - 50 + 6 A. E. is % , y(1) = 0 Engineering Mathematics dy ⇒ dx + (I ) y = x Solutio n of (i) is 2 Iog x x J 3. d x 21nx = I.F = e X . = e y. x 2 = = yx2 = f x Sol-21: (b) d 2 y dx 2 +p dy dx ⇒ + ( 2 d y dx ...(1) Hence from equation (2), (-1)2 + p(-1) + q = 0 and (-3)2 + p(-3) + q = 0 Solving (3) and (4), we get p = 4 dx Putting the value of p and q, we get y dx 2 +4 +2 dy dx o· + 17y = Its auxiliary equation is m2 + 2m + 17 = 0 ... (i) -2 ± .J4 - 4 x 17 2 = - 1 ±. 4i m = y = e-x(C 1 cos 4x + C2 sin 4x) ...(ii) Given, y(0) = 1 ⇒ Als o, 1 = 1 (C1 + c1 = 1 �� (¾) = 0 Differentiating (ii) w.r.t. Using (iii), dy + p - + (q + 1) y = 0 2 2 2 c2 x x o) ... (iii) �� = e- x [-4C1 sin4x + 4C2 cos4x - C1 cos4x - C2 sin4x] q = 3 d 2 5 2 So lutio n of (i) is ... ( 4) 3p - q = 9 = C1 e-t + C2e-21 + � ➔ co) = ...(2) ...(3) p-q = 1 dx x (t x(t) 1 5 .5 = 2 D2 + 30 + 2 Given the differential equation m = -1, -3 2 d y P. I. = Sol-22: (a) 1 e2 Since, its soluti on is given as y = C 1 e- x + C2 e-3x Sol-20: (c) D = -1, -2 C.F. = C 1 e-t + C2e-2t 'og1 J + c=0 -1- + qy = O = 5 D2 + 3D + 2 = 0 lo g x � r+ 2 X x Its auxiliary equation is m2 + pm + q = 0 ⇒ x "(t)+ 3 x '(t)+ 2x(t) A.E is 'o x y(x) = ( ; r Sol-1 9: (c) y = (c 1 + C2X)e-2X Its solutio n is xe-2x 2 1og x -- dx + c C = 0 y(e) = m = -2, -2 2 2(Iog x )2 +c 2 y(1) = 0 Its auxiliary equation is (m2 + 4m + 4) = 0 ... (i) 2 Iog x -- 3 - x 2 d x + c X f y = ( ⇒ 3 551 dy dx + 4y = 0 ⇒ ⇒ ⇒ 0 = C 1 - 4C2 = 0 e-rr/4 [0 - 4C2 + C1 - 0] 1 - 4C2 = 0 C2 = 1 4 IES MASTER Publication 552 Linear Differential Equations of Higher Order with Constant Coefficients 1 4 Sol-23: (d) AE is (02 - 40 + 4)y = 0 So, solution is m = 2, 2 Sol-25: (a) so A.E. + e-x [ -c 1 10 sin10x +c2 1 0cos10x) 1 e2x (02 +40+3) 3 x 1 e2 x = - e2 15 2 +4.2+3 2 1 Pl.. = - e2 x 5 So, solution is Hence, P - 2, Q - 1, R - 3 is the correct option. Sol-27: (d) d 2y + K2y = 0 dx 2 (02 + K2)y = 0 is m2 + K2 = 0 m2 = -K2 A.E. is 1 +­ 0 = k P.I. = y = C2 sin Kx 2 ⇒ c2 sin ka = 0 At Ka = mn ⇒ mn K= a Given So AE is (02 + 2 0 + 101 )y = 1 0.4ex m2 + 2 m + 101 = 0 m = -1 ± 10i k2 ( - �� )= Y2 = 0, y = Y1 lc 1 +C 2 = Y1 - Y2 I mn c 1 = 0 and K = a for y= c 1 cosKx + c2 sinKx Sol-26: (a) X 1 1 O -- X = a, y = 0 At ...(3) Now using y(0)= 1 .1 & y'(0)= -0.9 in (2) & (3), ' we get c 1 = 1 & c 2 = o X = 0, y = 0 ⇒ ⇒ 1 0.1ex 10.4)ex = (1+2 +1 01/ �� = -e-x [ c1 cos10x+c2 sin10x)+0.1ex m = ±Ki y = C1 cos Kx + C2 sin Kx At (0 +20 +1 01) (10.4)ex y = e- x [c 1 cos(1 0)x+c 2 sin(1 0)x]+(0.1)ex ... (2) d2y d 4 _1_ +3y : 3e2X -+ 2 dx dx = 3 = 1 2 So great solution is y = CF + Pl Given the differential equation P.I. = 3 P.I = . . . (i) m2 - 4m + 4 = 0 Sol-24: (b) C. F. is e-x [c1 cos 10x + c2 sin 10x] y = e x- (cos 4x + sin 4x); At X = ex:, , y = Y2 ⇒ ...(1) ⇒ (1) ⇒ c1 = 0 C2 = Y1 - Y2 Y = ( Y1 - Y2 )e- x/k + Y 2 ...(1) Engineering Mathematics Sol-28: (b) x + 3x -2 = -1 + B B = -1 = 0 ......(i) Initial conditions are x(0) = 1 x(0) = 0 Auxiliary equation of (i) m2 + 3 = 0 Sol-31 : (a) Given the differential equation y2 + 2y¢ + y = 0 m = ±i General solution of (i) is Now, ⇒ and ✓ x (t ) = -A x(t) = Sol-29: (a) 3 Auxiliary equation of (i) m 2 + 2m + 1 = 0 3 ✓ t + B sin ✓ t 1 = A+ B .0 3 ⇒ 3 ⇒ A= 1 3 3 ✓ sin ✓ t + B✓ cos ✓ t ⇒ ⇒ ...(i) Given, Its auxiliary equation is m2 + 1 = 0 m = ±i General solution of (i) is y = P cos x + Q sin x Sol-30: (a) A.E. is Solution is d2 y dy + 2 - + y = 0; dt dt2 y(0) = 1 and y'(0) = -2 (D2 + 2 D + 1 ) y = 0 (D + 1 )2 = 0 D = -1 , -1 y = (A + Bt) e-1 ; y' = -(A + Bt) e-1 + B e-1 Putting conditions 1 = [A + B(0)] e-0 A = 1 Now, and x(1) = x(1) = - 0. 1 606 -2 = - [1 + B(0)] e-O + B e-0 ...(1) ...(2) m = -1 , -1 General solution of (i) is y(x) = (A + Bx) e-x ⇒ 3 ✓ ⇒B=0 3 cos ✓ t 3 cos ✓ 0 = 0 +B ⇒ Boundary conditions are y(0) = y(1 ) = 0 3 x(t) = A cos x(0) = 1 x(0) = 0 553 Hence, Sol-32: (a) y(0) = 0 0 = (A + Bo) e0 A = 0 y(1 ) = 0 0 = (0 + B . 1 )e-1 B = 0 IY ( X ) = 01 y(0.5) = 0 Given (D2 + 2D + 2)y = 0 ⇒ m = -1 ± i The auxiliary equation is m2 + 2 m + 2 = 0 So, y = e-x [C1 cos x + C2 sin x] 1 1 or C 1 e( - +i )x + C2 e( - -i)x = C1 e-( 1 -i)x + C2 e-( 1 +i)x 1 1 Here e-( -i)x and e-( +i)x are i ndependent solutions. Sol-33: (a) Given ⇒ dy d2 y + p + qy = r 2 dx dx (D2 + pD + q)y = r The auxiliary equation is m 2 + pm + q = 0 -p ± � 2 2 If (p - 4q) > 0 then the roots of AE are real and different. m = IES MASTER Publication 554 Linear Differential Equations of Higher Order with Constant Coefficients Sol-34: (d) d2n d x2 n - L2 Sol-37: (c) = 0 ⇒ (1) . . .(1) = K I "' Sol-38: (c) A.E is m d2 x dx + 8x = 0 + 6 dt dt 2 02 + 60 + 8 = 0 x(0) ⇒ So, l �: lt =O m = -3, -3 P. I = Now, = 0 = -2, -4 ⇒ = 0 --4 t x(t) = 2e- -e So l-3 9: (a) 21 d2 y dy + - - 6y = 0 d x 2 dx Its auxiliary equation is ⇒ ⇒ :. Solution of (i) is 1 ( 0 + 3) 1 9 [1 + 2 ( 9x + 6) o ]-2 (9 3 x + 6) 2 = ..!_ ( 9x + 6 ) -3_ = X + 3. _ 3_ = X 9 3 3 3 So, C1 = 2 , c2 = -1 Given, 1 ( (x )) f ( O) a = i [1- 2 % + 3 ( � ) -.... }9x + 6 ) C1 + 2C 2 = 0 Sol-36: (c) + 6m + 9 = 0 = = 1 ... (1) C. F. = (C 1 x + C2)e-3x So, dx dt -1 y(2) = (1 - 2)e-2 = 2 e So, 2 y = ke-x /L A. E. is y = (1 - x)e-x (02 + 60 + 9)y = 9x + 6 C2 = k Sol-35: (b) ...(2) So, solution of (1) is Given, c1 = 0 ⇒ y = (C1 + C2x)e-x ...(1) Using y(1) = 0, C2 = -1 0 = c1 e + 0 i.e. , m = -1, -1 Using y(0) = 1 , C 1 = 1 n(0) = K n(CX)) = 0 2 So, -L h + C2 m + 2m + 1 = 0 A. E is 1 0 = +­ ⇒ (02 + 20 + 1)y = 0 Given, 1 02 - 2 = 0 L A.E is y = C1 e-3x + c2 e2X m2 + m - 6 = 0 (m + 3)( m-2) = 0 m = -3, 2 y = y = CF + Pl (C1 x + C2)e-3x + x It is Homogeneous Linear 0. Equation x Let 2 dy d2 y + x - - 4y = 0 2 dx dx z = log :. (O'(O' -1)+ O' -4)y = 0 where D' = d dz (0'2 - 4)y = 0 x - m2 Aux. eqn. 4 = 0 Auxiliary equation becomes m (m - k) = 0 m = -2, 2 The solution is m = 0, k 0 Kx u(x) = c 1e · x + c 2e +2z 2z y = C1 e + C2 eY = c1 x2 + c2 x-2 For boundary condition y(0) = 0 ⇒ h 1 ⇒ ⇒ y = C1 x2 ⇒ ⇒ a ·= b y (t) + y ( t) = 0 ... (1) 2 m + 1 = 0 2 A. E is ⇒ m = ±i The general solution of equation (1) is y = c 1 cos t + c2 sin t Using y(0) = 1 & y (0) = 1 , we get C 1 = 1 and c2 = 1 . - . The solution of equation (1) is y = cos t + sin t ...(2) For maximum (or) minimum of equation (2), we have dy = 0 dt or ... (3) KL = C 1 + c 2e . Putting value of c2 in equation (2) we get IY = x2 I (0 + 1)y = 0 u u (1- e KL) C2 = - U -U _ C 1 = - C2 = __ eKL -1 The complete solution of given diff. equation is x2 . Given . . . (2) Solving equation (2) and (3) we get � Sol-40: (d) 0 = C1 + C2 u(L) = ⇒ = 0 From y(1) = 1 Kx = c1 + c2 e u(0) = 0 ⇒ 0 = 0 + � 0 ⇒ 555 Engineering Mathematics -sin t + cos t = 0 tan t = 1 ⇒ t = 7t 4 Hence, the maximum value of equation (2) is -U U + (� ) eKx u(x) = --KL 1- e 1- e 1-eKx U = U ( �J 1- e Sol-42: (a) dy d2 y + 2a - + y = 0 dx dx 2 Characteristic equation is 0 2 + 2aO + 1 = 0 For equal roots ⇒ 82 - 4AC = 0 2 (2a } -4 (1)(1) = 0 4a2 = 4 Sol-43: (b) a2 = 1 a = ±1 d2 x dx 2- + x = 0 -+ 2 dt dt Auxiliary equation is 02 + 20 + 1=0 Sol-41 : (b) Given differential equation is 2 du du -k = 0 dx dx 2 (02 - KO)u = 0 ...(1) Sol. is 0 = -1 , -1 Sol-44: (c) IES MASTER Publication 556 Linear Differential Equations of H igher Order with Constant Coefficients A.E is ⇒ d2 x + 9x = 0 dt 2 02 + 9 = 0 02 = -9 0 = ± 3i = C 1 cos3t + C2 sin 3t x(0) = 1 ---dt dx l =o = t dt ⇒ ⇒ c2 = 1 3 dy d2 y -y = 0 +x x 2 dx dx Let z A. E is = A.E. is 0' -1 = 0 O' = ± 1 y = C1 ez + C2e-z C2 X 1 y = x, y = - are the solution of the given differential X equation. 2 d y = 0 2 dx Cond ition : ⇒ = C1 y = c 1 X + c2 y(0) = 5 C2 = 5 dy dx �I y(0) = 1 c1 = 1 dy dx = 2 at X = 10 = -2 (C1 + C2 x ) e -2x + C2 e-2 x y' (0 ) = ⇒ 1 = -2C1 + C2 c2 = 1 + 2C 1 = 3 y = (1 +3x ) e~2x Sol-46: (35) dx 1�1 0 2 + 40 + 4 = 0 0 = -2, -2 ⇒ 2 1 sin t i -sin t COS t l= rr/2 y = (C1 + C2 x ) e~2x O'(O' -1) + 0' -1 = 0 dy = cos t 2 . d y + 4 d y +4y = 0 dx dx 2 log x y = C1x + 2x + 5 x 1 = cost and x2 = sin t satisfies the given condition. 3 2 = x = c 1 cost + c2 sin t Sol-48: (0.54) It is Homogeneous Linear O.E. of 2nd order y 0 = ± i 1 . X = cos3t + Sin 3t Sol-45: (c) ⇒ 2 x 15 + 5=351 W(t) = I = 3C2 2 d2 x ( t ) + x(t) = 0 dt 2 02 + 1 = 0 1 = c1 dx = -3C 1 sin3t + 3C2cos 3t ) 1y c15 Sol-47: (a) X ⇒ = C1 Sol-49: ( 1 8) y(1) = 4e -2 = �=0.54 e d2 y = -12x2 + 24x - 20 2 dx Integrating both sides w.r.t. x dy = - 4x3 + 12x2 - 20 x + c 1 dx Again Integrating both si des w.r.t. x y = -x4 + 4x3 - 10x2 + c 1 x + c2 ... (1) Engineering Mathematics At x ⇒ At X ⇒ = 0, y 15 5 = = C2 I Square both sides = 2, y = 21 2c1 = 21 + 1 6 - 32 + 40 - 5 c 1 = 20 y = -x + ⇒ Put X ⇒ Sol-50: ( c) = 1 4 4x3 - 3C� - 3 C1 = 0 c1 1 0x + 20x + 5 c1 y = 18 Square both sides, y2 + 1 = e2x + y2 - 2exy dy d2 y dy = 2y 2 d x dx dx d ( dy 2 r = �(y ) dx dx dx I ntegrating both sides, f ✓Y dy 2 r + Cf 1 X -X y = -(e - e ) 2 Alternative Solution By options we can obtai n that the curve y = = y2 + Cf = ✓ ! ( ex - e- x ) is passing through (0, 0) & (.Cn2, 3 /4) . 2 Sol-51 : (b) Y + Cf 2 f dx d2 y dy + 2 - + y = 0, y (0) = y'(0) = 1 dt dt2 z n [ y + .}y2 + cf ] = X + C2 It passes through (0, 0) ⇒ So, ⇒ log [C1 ] = c2 z n [y + .}y + cf ] = X + log c 1 A.E. ixs 0 2 + 20 + 1 = 0 0 = -1 , -1 y = ( C1 + C 2 t)e-t 2 zn[ Y+ ✓�: + Cf J It passes through ( e. n2, 3/4) ln 4 [�+ �i ifttl -n.,1 C1 ✓ � + g + Cz 4 16 1 .}y 2 + 1 = e x - y d2 y = y dx2 dy = dx = 1 o, Hence, by (i) y = -1 + 4 - 1 0 + 20 + 5 ( �� = C 1 = 0 is not possible 2 M ultiply 2 ( �� ) both sides, 2 = 2C1 - 3 / 4 9 9 3 C21 + - = 4C12 + - C1 16 16 21 = -1 6 + 32 - 40 + 2c1 + c2 2c1 = 40 Hs 557 = X = In 2 = 2C 1 . . . (i) ⇒ ⇒ y (0) = 1 1 = C1 y'(t) = - ( C1 + C2 t) e-t + C2 e-t y' (0) = 1 = -C1 + C2 c2 = 1 + C1 = 2 y = (1 + 2t ) e-1 IES MASTER Publication 558 Linear Differential Equations of Higher Order with Constant Coefficients B = -1 Sol-52: (0.8556) d 2 dt x 2 +3 dx dt A. E. is dy 10 +2x=0, x (0) = 20, x (1) = C dy dt D 2 +30 +2 = 0 D = -1 , -2 x = C1e-t + C2e- x(0) = 20 ⇒ 21 ... (1) 10 e C C2 10 = -1 + e e2 e ⇒ ⇒ From ( 1) and (2) C2 10 = C 1 +e c1 = ... (2) 10e -20 ' 10e C2 = e -1 e-1 10e _21 10e - 20 - t x(t) = ( )e )e +( e-1 e -1 x(2) = ( dt2 +dt + 6 y 2 d y 1 Oe -4 10e - 20 -2 ) e +( )e e-1 e -1 = 0.8556 Sol-53: (-3) 5dy ( D + 2) ( D + 3) y = 0 [ where D = n � Ae- 2 + Be-3 = -e-3 + 3e-2 = A = 3 -3 Put In x = z so that x = e2 and then x 2 d y x2 = dx 2 D(D - 1)y dy = Dy dx d . dz Given differential equation is x2y" + xy' - y = 0 where D = D(D - 1)y + Dy - y = 0 ⇒ A.E. is m2 - 1 = 0 The,solution is ⇒ (D2 - 1)y = 0 m = 1, - 1 Sol-55: (b) C2 X d2 y dy +2 +y = O 2 dt dt . . . (i) D = -1, -1 ➔ two same roots :. The complete solution of equation (i) is y = (C1 + C2 x) e-x Given, y(0) = 1 and y( 1) = 3e- 1 1 C1 = C1, = 1, C2 ⇒ 3e -2 -e-3 ... (1) y = C 1 e + C2e-z z y = C1 x + or ⇒ y(0) = 2 1- 3e ) e3 = -6 + 3 It is Homogeneous Linear D.E. of 2nd order so following assumptions are valid . ⇒ y(t) = Ae -21 + se-31 y(1) = -( = ⇒ D = -2, -3 A + B = 2 = -2(3) -3B Auxiliary equation (A.E) D2 + 2D + 1 = 0 = 0 ( D2 +5D + 6 ) y = 0 (0) = -2A - 3B Sol-54: ( c) C1 + C2 = 20 x(1) = 31 21 = -2Ae - -3Be- dt Sol-56: (-1 ) ⇒ 3e-1 = 2 = ( 1 + C2 ) e- 1 y(x) = (1+2x )e-x y(2) = (1 + 2 x 2)e-2 = 5e-2 y" + 9y = 0 y = A sin 3t + B cos 3t 0 ⇒ A.0 + B.1 = 0 ⇒ 0 y( ) = 2 ✓ ⇒ y( ) = % ⇒ 2 4 l) dy d y + 1 2- + 36y = 2 dx dx = -1 0; 0, ⇒ ⇒ 0 = - 6, -6 y = (C1 y = 3 + C2 x )e---6x 2 ...(i) Hence, Sol-58: (7.39) Method I = -18 ⇒ y(s) = 0 0 0 0 ⇒ = = +3 = 0 = ±i--.13 !0 ( 1 o sx 2 ) � 3 2 l7J :;�2 = 0 1 +3 0 d2 y +y = dx 2 0 0 With y( ) = 5 & y' ( ) = 1 ...(1) 0 Auxiliary equation is = 0 m = 0 ± i m2 + 1 General solution y = C1 cos X + C2 sin X • . . (2) 0 0 Using y( ) = 5 & y'( ) = 1 , we get c2 = 1 & c 1 = 5 Hence by (2), 0 0 s y(s) -sy( ) -y' ( ) -4 [sy (s) -y ( )] + 4y(s) = y(s) ( s2 - 4s + 4 ) -1 = 02 ( 0 2 + 3) 0 y = ( 3 -18x ) e---6x 0 0 4 + 302 = 0 Given that 0 dy d2 y --4 + 4y = 2 dx dx 0 For complementary solution Option (A) Sol-60: (d) Taking Laplace transform o n both side of above equation 2 d4 y d2 y + 3= 1 08x2 4 dx dx 2 P.I. = 3x 4 -12x 2 + C -36 = C 2 + (3 + 0) ( -6) c2 + Pl method. = dy = c 2e-6x + ( c 1 + c 2 x )e-6 x (-6) , dx [using (i)] -36 = c e0 + (c + c ( ) ) e0 (-6 ) 1 x = e2 = 7.39 Particular integral 0 2 2 1 C.F. = C 1 + C2x + C3 sinfix + C4 cos--.13x c1 = 3 ⇒ ⇒ ⇒ 1 .e So, Complementary solution : 3 = c 1 e0 + cz(0)e0 dy = -36 at x = dx y(1) = Method II : This question can also be solved by using CF 02 0 0 (0 + 6 ) (0 + 6 ) = y(t) = te2I 02 2 02 + 1 20 + 36 = ⇒ ⇒ 559 Sol-59: (a) dy y( ) = 3 & = -36 l dx x =O at X = ⇒ 2 -✓ si n 3n = -✓ ( 0 also, 2 0=✓ 2 -✓ sin 3t y (¾) = Sol-57: (a) A - (-1 )+ B · 2 y = So B =0 -✓ A = ⇒ ⇒ ⇒ ⇒ Engineering Mathematics (s -2) 2 0 Sol-61 : (a) 0 y = 5 cos x + 1 d2 y 2dy -Sy = + dx 2 dx 0 ⇒ Auxiliary Equation is m2 + 2m - 5 = 0 0 sin x (02 + 20 - 5)y = 0 6 m = -1± ✓ IES MASTER Publication 560 Linear Differential Equations of Higher Order with Constant Coefficients So, in terms of arbitrary constants k 1 and k2 we get, m2 x mx Y = k1 e 1 + k 2 e Sol-62: (a) Sol-64: (94.08) 3y" ( x ) +27y ( x) = 0 y" ( x) +9y ( x ) = 0 (02 + 9)y = 0 So characteristic equation is given by: y" - 4y' +3y = 2t - 3t2 m2 +9 = 0 f(D) = 02 - 40 + 3 m = ± 3i = 0 ± 3i 1 2 P I = --(2t -3t ) .. f(D) y = (C1 cos3x +C2 sin x ) e0x 1 (2t -3t2 ) 02 -40 +3 = y = C1 Cos3x +C2 sin 3x y' = -3C 1 sin 3x + 3C2 cos 3x y'(O) = 3C2 = 2000 2000 C2 = -3 y(O) = 0 = C1 ( 1) +C2 (0 ) 2 2 3 t 32 = ( t - � ) -i( t -� + � +... ) x ( t - � ) c1 = 0 y = 2 3t 3t 2 +� - 8 - � ) = (t +1-3t -3 ) -� (13 2 3 3 2 = - 2 - 2t - t2 Sol-63: (a) Sol-65: (c) d2 y +16y = 0 dx 2 Let ( 02 +16 ) Y = 0 m2 +16 = 0 (this is a complex equation) 0x y = ( C1 cos 4x +C2 sin 4x)e y = C1 cos 4x +C2 sin 4x y' = -4C1 sin 4x + 4C2 cos 4x y' ( O) = 4C2=1 y'(� ) 4 = -1 -1 = - 4C1 sin 2n + 4C2 cos 2n -1 = 4C2 c2 = 1 m4 A.E. is 02 = m2 C2 = 2000 sin3 = 94.08 3 Given DE is yiv(x) - 2yiii(x) + 2yii(x) - 2yi (x) + y = O ( 04 - 203 + 202 - 20 + 1]y = 0 . . .(i) or m = ± 4i = 0 ± 4i ⇒ ⇒ y(1) = 2000 sin3x 3 -4 So, we are not sure about C2 . Hence no solution. - 2m3 + 2m2 - 2m + 1 = 0 (m - 1 ) (m2 + 1) = 0 2 m = 1,1, ±i So, CF = (C1 + C2x)ex + C3cosx + C4sinx Pl = 0 Hence solutions is y = CF + Pl Sol-66: (a) = (C 1 + C2x)ex + C3cosx + C4sinx d2 y dy -- -2y = 3 e2x dx 2 dx (02 - D - 2)y = 3 e2x AE in m2 ⇒ ⇒ - m-2 = 0 (m + 1) (m - 2) = 0 m = - 1 and 2 So CF = c 1 e-X + c2 e2X ... (1) Engineering Mathematics y = CF + Pl 1 1 ( 3 e2x ) QNow Pl = f (D) 02 _ 0 _ 2 and x x - ( e2x ) ]=3 [ ( e2x )] = xe2x = 3 [2 D -1 2 x 2 -1 General solution of ( 1) is y dy dx = usi ng y(0) = 0 using y' (0) = ⇒ 0 ⇒ CF + Pl ⇒ -2 On solving C2 = - 1, C 1 = 1 So solution of (1), y Sol-67 : (b) d y dt2 +y 2 ⇒ = 0 Where D = = e-x - e2x + xe2x (D2 + 1) y = 0 dt ⇒ AE is m2 + 1 = 0 So CF = ...(1) Hence G. solution is y = CF + Pl = 1 ⇒ Using y' (0) = 0 ⇒ So y = cos t Sol-68: (0.368) dy d2 y +y = 0 +2 dx dx 2 (D2 + 2D + 1) y = 0 1 = C1 + ⇒ m = - 1, - 1 ⇒ y(0) = 1 C1 = 0 dy (0) = - 1 dx - 1 = - C 1 + C2 c2 = o 1 so y(1) = e-1 = e 1 = - = 0.368 e Sol-69: (1 .47) d2 y dy + - - 6y = 0 dx 2 dx (D2 + D - 6)y = 0 Auxilliary equation is m + m - 6 = 0 2 ...(1) So, CF = C 1 e-3x + C2e2x & Pl = 0 ... (3) C1 = 1 = ...(i) 0 = m = -3 & 2 0 C2 (m + 3)(m - 2) 0 Solution of (1) is y = CF + Pl = C 1 e~3x + C2e2x Using y(0) = 0 dy Now dx Using ( = ⇒ C 1 + C2 = 0 dy ) dX (0) By (2) y = = 1 ⇒ 1 1 5 -3 & C2 = 5 + 1 5 -3C 1 + 2C2 = 1 1 --e-3x +-e 2x 5 5 so, y(1) = - e Sol-70: (-0.21) ...(2) -3C 1 e-3x + 2C2e2x On solving, C 1 = So, CF = (C1 + C2 x) e-x and Pl = 0 So solution of (i) is = ...(2) o ⇒ 0 = 0 + C2 AE is m + 2m + 1 = 0 2 ... (iii) x By (ii), solution is \y=e-x \ ⇒ e01(C1 cos t + C2sin t] and Pl Using y(0) ...(ii) m2 + 3m - 2m - 6 = 0 m = ± dy . = - C1 sin t + C2 cos t dx - (C 1 + C2 x) e-x + C2 e- ⇒ ⇒ -2 = -C1 + 2C2 +0 + 1 or -C1 + 2C2 = - 3 = Using C1 + C2 = dy dx Using y'=-C 1 e~X + 2 C2 e2x + 2 xe2x + e2x = y = (C1 + C2 x) e-x 561 e2 = 1.4678 dy - Sy -dt 4 1 5 IES MASTER Publication 562 Linear Differential Equations of Higher Order with Constant Coefficients (02 + D +¾) y = AE is ⇒ dy = _ _!_e-112 [ c 1 cost + c 2 sin t] dt 2 0 m2 + m + - = 0 4 1 . m = -- ± 1 2 CF = e-112 [ c 1 cos t + c 2 sin t] Pl = 0 and Solution of (1) is y = CF + Pl = e-112 [ c 1 cos t + c 2 sin t] ... (2) using ⇒ using ⇒ +e-t /2 (c 1 sin t + c2 cos t] ...(3) y(O) = 1 C = 1 1 y'(O) = 0 C 1 2 = 2 Hence by (2) solution of given D E is Hence y = e-t1 2 [cos t + -isin t] - rr/ 2 [ cos n +O ] Y( n) = e = -e- "12 =-0.21 lillllS 5.3 ORDER AND DEGREE OF PARTIAL DIFFERENTIAL EQUATION The differential equation which contains one or more partial derivatives is called P.D.E. i.e. (P.D. E). contains two or more than two independent variables. az ax a 2z ' ax az ' &y a2z ' a y a2z ax&y - =p - =q =r = t and -- = s 2 2 Order : Degree = Example : (let) The highest order derivative occurring in PDE is called its order. The degree of the highest order derivative occurring in it after the differential equation is made free from radicals and fractions. (1) yz- + zx- = xy, order 1 , Degree 1 . (or yzp + zxq = xy) az ax az ay 8z a 2 z a2 z +(2) - + - = 1 order 2 , Degree 1 (or p + r + s = 1 ) x2 ax a axay LINEAR PARTIAL DIFFERENTIAL EQUATION P.D.E. is said to be linear if it satisfies following five conditions : a. Exponent of z should be one. b. Degree should be one. c. All the partial derivaties has exponent one. d. Product should not be present of z and partial derivatives. e. Product should not be present of partial derivatives. Non Linear Partial Differential Equation A differential equation which is not linear is said to be non linear P.D.E. Examples : Li near 1. 2. 2 2 a2z a z 8z x a z x+ (3x + 4y)-+ 5x- = x 2 + y 2 + e 2 2 ax axay ay a2 u a 2u a 2u = 3u x2 + z2 + y2 2 y 2 a ax 82 2 ax (Here u is dependent and x, y, z independent variables) 564 Partial Differential Equations Examples : Non Linear 1· 2. 3. 4. 2 82 z az az + 4 ( ) + 5 + 6z = 9 ax ay ax 2 or (r + 4p 2 + 5q + 6z = 9) az az a (x 2 y 2 ) az . z - xy ( + J - 1 = o ax &y ax &y (:r 2 (x2 - y2 ) 2 (p2 x + q y = z) or x + (: r Y = Z or a z + za2 z + (y + cos x) az = 5 x2 axay &y ax or (due to property 2) pq - xy(p + q) - 1 = O (due to property 5) (due to property 2) (xr + (y + cos x)s + z - q = 5) (due to property 4) LAGRANGE'S METHOD OF SOLVING LINEAR PARTIAL DIFFERENTIAL EQUATION OF 1ST ORDER The P.D.E. of the form !Pp + Qq = RI , is the standard form of a (where P, Q , R are the functions of x, y, z) linear P.D.E. of 1st order, and is called Lagrange's linear equation. Working R u le Step 1 : Put the given P.D.E. in standard form, i.e., in the form Pp + Qq = R ...(1) d Step 2 : Write down the Lagrange's Auxiliary equation as follows, : = � = : Step 3 : Now try to find two independent integrals of Auxiliary equation. Let it be u = a and v = b [where u & v are functions of x, y, z}. Step 4 : Lastly the general solution of (1) is given by $(u, v) = 0 {or u = $(v) or v = $(w)} where $ is an arbitrary function. Example 1 : Solution : Solve : x2 p + y2q = nxy Comparing it with Pp + Qq = R, we get P = x 2 , Q = y2 , R = nxy Lagrange's subsidiary equations are dx 7 = dy dz 2 = nxy y dx Taking 1 st two dy dy = y2 ⇒ y2 = nxy �z ( �) ⇒ ; Taking last two, ⇒ ; = 7 1 1 1 1 +- = a = -- + a ⇒ u = - X y Y X dz = �z (� _a J n Integrating, z = - log ( 1 - ay) + b a ⇒ z = ( (1-+� + �} + b i: r t l So, general solution of given equation is $ {..!. - ..!., z y X nxy y-X log (r)} = X o where dy ⇒ 1 - ay dz n ⇒ n dy dz = -1 - ay y y ⇒ z = ;� x log ( t) + b ⇒ v = z - ;� x log ( t ) = b $ is an arbitrary function Engineering Mathematics Example 2 : Solution : Solve : p + 3q = 5z + tan(y - 3x) ... (1) P = 1, Q = 3, R = 5z + tan(y - 3x) Lagrange's auxiliary equations are dz dy dx = = 5z + tan(y -3x) 3 1 dy dx . . Taking first two, 1= 3 So, u = y - 3x = a Taking first and third, ⇒ 1 Y + C1 3 x = dz dx = 5z + tan(y - 3x) 1 ⇒ ⇒ 3C1 = y - 3x ⇒ ... (2) a = y - 3x dz - - 5z = tan a dx Which is linear ordinary differential equation of 1st order and 1st degree, IF = e Solution of ( 4) is ze-5x = f tan a(e-5x ) dx + C ⇒ 2 z(IF) = J P dx 565 ... (3) ... ( 4) J 5d 5 = e - x = e- x f Q . (IF) dx + C 2 tan a 5 ze-5x = --=se - x + C2 5ze-5x + tan(y - 3x) . e-5x = 5C2 = b (let) ... (5) or V = [5z + tan(y - 3x)Je-5x = b So general solution of (1) is given by $(u, v) = 0 5 or $(y - 3x, e - x (5z + tan(y -3x))=0 where $ is an arbitrary function. Example 3 : Solve Solution : x 2(y - z)p + y2(z - x)q = z2(x - y) P = x2(y - z), Q = y2(z - x), R = z2(x - y) Lag range's subsidiary equations are dx x 2 (y -z) = dy y2 (z -x) dz = 1 1 1 -dx + -dy + -dz 2 2 x z2 y 1 1 1 Using multipliers, 2 , 2 , 2 , each fraction of (1) is = Q X y Z ⇒ ⇒ 1 1 1 -dx + -dy + -dz = x2 z2 y2 ... ( 1) o 1 1 1 Integrating -- - - -- = C1 X y Z 1 1 1 U =-+-+- = a X y Z or 1 1 1 -dx + -dy + -dz Z X y Now taking multipliers, _!, _! ).. each fraction of (1) is = X y Z 0 ⇒ 1 1 1 - dx + -dy + -dz = Q X y Z Integrating, logx + log y + log z = log b ⇒ xyz = b or v = xyz = b General solution of given equation is $( _! + _! + X y Example 4 : Solve Solution : (x - yz)p + (y - zx)q = z - xy 2 2 Lagrange's auxiliary equation is 2 -.!, xyz)=0 Z IES MASTER Publication 566 Partial Differential Equations dy dz dx = = 2 2 zx z - xy y x - yz Taking the multipliers, (1, -1 , 0); (0, 1 , -1 ); (1 , 0, 1 ), we get dx P = dy dz = Q R ⇒ 2 dz-dx dy-dz dx-dy = = (x-y)(x+y+z) (y-z)(x+y+z) (z-x)(x+y+z) dy-dz dx - dy Now taking first two, -- = y-z x-y Integrating, log(x - y) = log(y - z) + log a ⇒ dy-dz dz-dx Taking last two, we get y - z = z -x x-y u = y-z = a y-z log(z - x) + log b or v = -- = b z-x x - y y -z - , --) = 0 General solution is $(u, v) = 0 or $( y- z z-x Integrating log(y - z) Example 5 : Solve Solution : = y-z z-x _ x-y (--) p+(--)q --yz zx xy Given equation can be written as (xy - zx)p + (yz - xy)q = xz - yz dx dy Lagrange's auxiliary equations are -- = -xy - zx yz-xy = dz -xz-yz Now taking 1 , 1 , 1 as multipliers each fraction of (1) is = ⇒ dx + dy + dz = 0 or lx +y+z = al 1 1 1 -, - as multipliers each fraction of (1) is Again taking -, X y Z ... (1) dx +dy+dz 0 = ! dx + ! dx + ! dz X y Z 0 1 1 1 ⇒ x dx + y dx + 2 dz = O ⇒ log x + log y + log z = log b ⇒ lxyz = bl General solution is $(x +y+z, xyz) = 0 Example 6 : Solve : (y + z)p + (z + x)q = x + y Solution : dy dx Lagrange's auxiliary equations are- - = z+x y+z dx - dy dx+ dy+dz dy-dz = = 2(x+y+z) -(y- z) -(x -y) ⇒ Taking first two, Integrating u = dz x+y dx - dy dx+dy +dz = 2(x +y+z) -(x-y) 1 1 og(x +y+ z) = -log(x - y) 2 + logC 1 ⇒ (x + y + z)(x - y)2 = a Taking l ast two, dx-dy x-y dy-dz y-z log(x + y + z) = -log(x - y)2 + logC / Engineering Mathematics Integrating log(x - y) = log(y - z) + logC 1 [ 2 General solution is � (x + Y + z) , - X -yJ - = 0 y-z ⇒ 567 x-y v = y_ = b z HOMOGENOUS LINEAR P.D.E. WITH CONSTANT COEFFICIENT If all the Partial Derivatives appearing in the equation are of the same order then it is called Homogeneous LPQE.. It is of the form a"z a"z a"z -- + A 2 a"z + .....A n - + A1 = f(x y) n 1 n 2 2 ax" ax - ay ax - ay ay" ' where A1 , A2 . . . . . . A" are constants �{D, D')z = f(x, y) NON HOMOGENEOUS LINEAR P.D.E. WITH CONSTANT COEFFICIENT If all the partial derivatives appearing in the equation are not of same order then it is non homogeneous linear P.D.E. i.e. When � (D,D') is not a homogeneous function of D and D'. e.g. Homogeneous : 1. (202 + 5DD' + 2D'2 )Z = o Non Homogeneous : 1. (02 - 0'2 + D - D')Z = 0 Note : D = 2. (03 - 402 0' + 400' 2 )Z = 0 2. ( D - D')(D - D ' - 2)z = e2x + x a2 a a2 a _t__ , .... .D" _- __!._ 2 = , , , D' = ay ' 02 = ay2 DD' = D'D = axay ay" ax ax 2 0' Method of Solving Linear Homogeneous P.D.E. with Constant Coefficients Consider a"z ao - + 8 1 or ax" a"z a"z a"z + 81 n 2 2 + ..... + an - = f{x, y) ax"-1 ay ay" ax - ay �(D, D') = f(x, y) ...(1) C.F. : It must contains n arbitrary functions (n order of PDE) 1 P.l : Pl of (1) = cp(D D/{ x, y) , Z = CF + Pl General Sol. Rules for Finding CF AE of (1) is �{ m, 1) = 0 . ..(2) and Let m 1 , m 2 , 1. ..... m" are the roots of (2) If all are distinct, then CF = f1 (y + m 1 x) + fiY + m 2x)... + fn (Y + m nx) 2. If m 1 = m 2 and rest are distinct, then CF = f1 (y + m 1 x) + xf2(y + m 1 x) . .... + fn(Y + m"x) 3. If m 1 = m 2 = m 3 , and rest are distinct then CF = f1 (y + m 1x) + xf2(y + m 1 x) + x2f3{y + m 1 x) + f4{y + m 4x) + ..... + f"(y + m"x) IES MASTER Publication 568 Partial Differential Equations Rules for Finding Pl Case I : Case II 1 1_ ax+by ax + b _ e If f(x ' y) = e•x+by then Pl = -- e y = cp(D,D') cp(a,b) If f(x, y) = sin(ax + by) or cos(ax + by) then 1 1 1 sin(ax + by) = sin(ax + by) P l = -- sin(ax + by) = 2 2 2 cp(D, D') cp(D , DD', D' ) cp(-a , -ab, -b2 ) Similar is the case for cos(ax + by) NOTE 1. In Case I & II, if denominator is zero then use, PI = 2. Short Cut Method of Case degree n) of PI = cp{D, D') & D' and Case X a (cp{D, D')) aD and f{x, y) is of the form f (ax + by) then f (ax + by) = f f .....f f(v) dvd�.....dv cp( a' b) n times 1 When in ${D, D ') powers of then 4. PI = 1 (ax + by) = cp{D, D'/ If f(x, y) = x"y" then , P l = f(x, y) II : When $(0, D') is a homogeneous function (of where v = ax + by & $(a,b) -ct 0 . When ${a,b) = 0 then proceed same as Note 1 3. Case Ill : 1 D I 1 f{x , y) = cp{ D, D') 1 D' in highest degree term is greater than that of D & ${a,b) = 0 y f(ax + by)..... and so on � � {cp{D, D ')} a ' cp(D,D') ( Xm , y n ) = [cp(D,D- 1 if m > n, expand it in terms of D' / D r1 x y m n if m < n, expand it in the ascending powers of o / D'. Case IV : If f(x, y) = e•x•bY.V where V is a function of x & y. 1 1 ax b ax+b (V) Pl = __ (e + y .V) = e y . · cp(D, D') cp(D + a, D' + b) Case V (General Method) : If f(x, y) be any other fun ction of x & y then Now use the formula f(x y) = f(x, y) (D - m p'):D - m 2 D') cp(D� D') , Pl = f(x, y) = J f(x, c - mx) dx, where y + mx = c and c is repla ced by y + mx after integration. (D - m D') By repeated appli cation of above formula Pl can be evaluated. 1 Note : Logic behind the formula (D - m D') f(x, y) = f f(x, c - mx)dx 1 If we have to evaluate Pl = Let z = (x, y) (D -�o/ (D _�D') ⇒ f(x, Y) DZ - mD'Z = f(x, y) ⇒ az az - - m- = f(x, y) ay ax or p - mq = f(x, y) Engineering Mathematics Now subsidiary equations of ( 1 ) are dx 1 = dy -m dz = f(x, y) Taking 1 st two we will get one solution as y = -mx + (taking 1 st and 3rd) we will get another solution as So , Pl = 569 c f f(x, y)dx 1 (x, y) (D - m o/ = f dz ⇒ z= = J t( x, c - mx)dx f f(x, c - mx)dx Questions based on Linear Homogeneous Partial Differential Equations with Constant Coefficients Example 7 : Solve : Solution (i) a 2 z - 6a2 z - a2 z = 0 ax 2 axay ay2 (i) Let D = a a ax an d .D' = ay (D 2 - DD' - 60' 2 )z = 0 (ii) the equation redu ces to The A.E. is m 2 - m - 6 = 0 ⇒ (m - 3)(m + 2) = 0 So C.F. = f 1 (y + 3x) + f2 (y - 2x) & P.I. = 0 ⇒ m = 3, -2 ... (1 ) :. Solution is z = f 1 (y + 3x) + fiY - 2x) (ii) A.E. is m 4 - 1 = 0 ⇒ ( m 2 - 1 )(m 2 + 1 ) = 0 ⇒ m = ±1 , ±i :. C.F. = f 1 (y + x) + fiY - x) + fiY + ix ) + fiY - ix) and Pl = 0 So, complete solution is Z = CF + Pl = f 1 (y + x) + fiY - x) + f3(y + ix) + f4 (y - ix) Example 8 : Solve : (D 2 - 3OD' + 20' 2 )z = e2x- y + ex+y + cos(x + 2y) Solution : The A.E. is m 2 - 3m + 2 = 0 So, Now, P.I. = = Now, & & ⇒ ( m - 2)(m - 1 ) = 0 or m = 1 , 2 C.F. = f1 (y + x) + fiY + 2x) 1 ( e2x-y + ex + y + cos (x + 2y)) (D 2 - 3OD' + 20'2 ) 1 1 1 ( e2x-y ) + ( e x +y ) + cos(x + 2y) 2 2 2 2 (D - 3OD' + 20' ) (D - 3OD' + 20' 2 ) (D - 3OD' + 20' ) 2 1 1 ( e 2x-y ) = J_( e 2x-y ) ( e 2x -y ) = 2 2 12 4 3(2)(-1) + 2(-1) (D - 3OD' + 20' ) 2 1 1 ( e x+y ) = x.-1- e x+ y = ( e x+y ) = X. 2 2D 3D' (2 - 3) (D - 3OD' + 20' ) 2 1 1 cos(x + 2y) c os(x + 2y) = 2 2 -1 - 3(-2) + 2(-22 ) D - 3OD' + 2D' 2 1 1 cos(x + 2y) = - cos(x + 2y) = ---1 + 6 - 8 3 IES MASTER Publication 570 Partial Differential Equations 2x x+ p. I. = _!_ e - y -xe y _ J_cos( x +2y) 12 3 So, So general solution of given P.D.E. is z = CF + Pl 1 e 2x -y z = f1 (Y + x) +f2 (y + 2x) +-- -xe x+y -- cos(x +2y) 12 3 or Example 9 : Solve, (D2 - 60D' + 90'2) Z = 12x 2 + 36 xy Solution : The A.E. is, m2 - 6m + 9 = 0 => (m - 3)2 = 0 .- . C.F. = f 1 (y + 3x) + X f2 (y + 3x) P. I. = ⇒ m = 3, 3 1 1 1 2 (12x 2 + 36xy) (12x 2 + 36xy) = 2x + 36xy ) = 2 2 2 ( D2 - 60D' + 90'2t 3 ' ( 0 - 3D' ) 02 ( 1- � ) 2 D' (-2) (3) 90' 2 3D' 1 1 2 ) (12x + 36xy 1+ 6-+ -�- - -2-. . . ) (12x 2 + 36xy) 1= ( = ) 2( D D 1.2 D D D2 18 1 � x2 1 2 2 = 2 12x + 36xy + 6 · j = 2 [120x + 36xyl 0 0 19" x 4 6 xJ ,A'O · - + )-8 - y J j = 10x4 + 6x 3 y So complete solution is Z = f1 (y + 3x) + xf2 (y + 3x) + 10x 4 + 6x 3 y Example 1 0 : Solve: Solution 82 z 8 2z 82z + 6 = ycosx 2 2 ax axay ay � = D & � = D'. Let ax ay ( 0 2 + DD' - 6D' 2 ) z = ycos x A.E. is, m2 + m - 6 = 0 => (m + 3) ( m - 2) = 9 => m = 2, -3 :. . . . (1) C.F. = f1 (y + 2x) + f2 (y - 3x) Now, P.I. = 1 1 y (cos x) YCOS X = ( D - 2D ' ) ( D + 3D ') ( D 2 + DD' - 6D' 2 ) 1 1 1 (ycos x) } = ( D - 2D' ) = ( { ) D - 2D' D + 3D' {J(c + 3x) cos x dx} where y = C + 3x 1 1 = ( {(C + 3x) sinx + 3cos x } = (0 _ 20 , ) {y sinx + 3cos x} ) D 2D' = j {(b - 2x) sinx + 3 cos x} (where y = b - 2x) Engineering Mathematics 571 = -(b - 2x) cos x - (-2) (-sin x) + 3 sin x = -(b - 2x) cos x + sin x = -y cos x + sin x where b = y + 2x Hence, complete solution of P.D.E. is, Z = f1 (y + 2x) + f2 (y - 3x) - ycos x + sinx Example 1 1 Solution riz 82z - -- = sin x cos2y ax 2 axay (02 - D D') z = -½[sin(x + 2y) + sin(x - 2y)] A.E. is, m2 - m = 0 ⇒ m = 0, 1 C . F. = f 1 (y) + f2 (y + x) 1 1 sin (x + 2y) + 2 sin(x - 2y)j Pl. . - .![ 2 - 2 D - DD' D - DD' 1 1 1 1 sin (x - 2y)I = - [ sin(x + 2y) - - sin(x - 2y)j sin (x + 2y) + 2 = .! 2 I -12 - (-2) 3 2 -1 - (-2) Complete Solution is Z = CF + Pl Z = f1 ( y) + f2 (y + x) + .! sin (x + 2y) - .!sin(x - 2y) 2 6 Example 1 2 : Find a real function V of x, y, reducing to zero when y = 0 and satisfying, 8 2v 8 2v += -4n ( X 2 + y2 ) 2 2 Solution 8x ay 2 2 The given equation is (02 + 0'2) V = - 4n ( x + y ) A.E. is, m2 + 1 = 0 ⇒ m = ± i :. C. F. = f 1 (y + ix) + f2 (y - ix) 2 2 0' 0' ( 2 1 2 1 1 2 2 2 2 2 2 X + y )) = -4n - l X + y - - ( X + y )] { - 41t ( X + y )} = - 41t (1 + P. I. = --,------2 ---------,o2 o2 ( 0 + 0' 2 ) o2 o2 2 2 = - 4n �2 [ x + y - �2 (2)] = - 4n �2 [/ + y2 - /] = - 4n Y ;2 2 Since V is a real function. So, 2 = -2n x y 2 2 2 V = -2n x y which reduces to zero when y = 0. Exam ple 13 : Solve, r - 2s + t = sin (2x + 3y) where notations have their usual meaning Solution : ( 0 2 - 200 '+ D' 2 )z = sin(2x + 3y) AE is m2 - 2m + 1 = 0 ⇒ m = 1, 1 CF = � 1 (y + x) + x<h ( y + x) P. l . = 2 1 D - 2DD' + D' 2 sin(2x + 3y) = 2 1 2 ff sin v (dv} 2 - 2.2.3 + 3 = -sin v = -sin (2x + 3y) 2 where v = 2x + 3y Hence general sol of given equation is Z = CF + Pl z = �1 (y + x) + x �2 ( y + x) - sin(2x + 3y) IES MASTER Publication 572 Partial Differential Equations Example 14 : Solve : 4r - 4s + t = 16 log (x + 2y) Solution : 82z 82z 82z 4 - - 4 -- + = 16 Iog(x + 2y ) axay ay 2 ax 2 Now, P. l. = ⇒ 4 D2 - 4OD' + 0'2 = 16 log (x + 2y) 1 4D2 - 4OD' + D' 2 1 (16 Iog(x + 2y)) = -� 16 - log(x + 2y) (2D - D')2 ·: Denominator becomes zero when we put D = 1 and D' = 2. 1 , • f(ax + by) where cp(a,b)=0 so we have and it is of the form --$ ( D, D ) P. I = 2 (2D �D' ) (2 ) · 16 1og(x + 2y) Again, denominator becomes zero when we put D = 1 and D' = 2. So 2 P. I. = x · 1 6 Iog(x + 2y) = 2x 2 log (x + 2y) 2 ( 2) 2 83z 83 83 z 1 Example 1 5 : Solve : -- 2 -=2 3 2 + 2 x 8y 8x8y 8x 8y Solution ( 0 2 0' - 2DD' 2 ) z = � x 0' (0 2 - 2OO' + O' 2 ) z = � x Part of CF corresponding to D' is f 1 (x) AE of remaining factor m2 - 2m + 1 = 0 ⇒ CF = f1 (x ) + f2 (y + x ) + xt3 (y + x) Pl O = o'' + o'�' - 2DO' ' t1, l � 30' ' m = 1, 1 + ;, Y J_ dv dv 2 2 3 (0 ) + (1) 2 - 4 (1) (0 ) JJ v 1 - 400' l x ' l y f -dv v-1 = -ylogv = -ylogx = 1 i-. 1 x � (1x + 0y) ' l 1, where v = x = (1x+Oy) -1 z = CF + Pl = f1 (x) + f2 ( y + x ) + xf3 (y + x) - ylogx Methods of Solving Non-Homogeneous Linear P.D.E. with Constant Coefficients If, in the equation $(O, D')Z=f(x, y), the polynomial $(O, D') is not homogeneous, then it is called non-homogeneous L.P. D. equation. Its complete solution is Z = CF + Pl. Methods of Finding CF Consider the equation $(O, D')Z = 0 To find CF, we will resolve $(O, D') into linear factors as follows, (D -mp' -a1 )(D - m2 D' - a2 ). . . . . (D -mn D' -an )Z = 0 ...(3) Case I : Engineering Mathem.atics If all li near factors are disti nct, then 573 U X a x a X CF = e 1 f1 (y + m1X) + e 2 f2 (Y + m2 X) . . . . . +e n fn ( Y +mn x ) Case I I : I n case of repeated linear factors , for e.g. , (D -mD' -a)3 Z = 0 then, CF = eax f1 (y +m1 x ) +xeax f2 (y +m2 x ) + x 2 eax f3 (y +m3 x) Methods of Finding Pl These are very similar to those of homogen eou s L.P. D . EQuation with constant coefficie n ts . Q uestions based on Linear Non- Homogeneous Partial D ifferential Equations with Constant Coeffi cients 2 Example 1 6 : Solve: Solution Let or, So, 2 2 [) z [)z _ az [) z _ � +4 2+ 4 2 = e x+ y 2 [)x[)y [)x [) 2 8y [)x [)y [) - = D & - = D' Then [)x [)y ( 0 2 - 400' + 40' 2 + D - 2D' ) z = e x + y (D - 20') (D - 20' + 1 ) Z = ex+y C.F. = e0x f1 (y + 2x) + e - x f2 (y + 2x) P. I. = D 2 - 1 2 400' + 40' + D - 20' So Solution is , Z = f 1 (y + 2x) + e- x = x ⇒ { (0 - 20) + (D - 2D')} z = e 2 x +y 1 1 x e x + y = -x ex+y e +y = X · (2 - 4 + 1) 20 - 40' + 1 f 2 (y + 2x) - xex+y Example 1 7 : Solve, (D - D'-1) (D - D' - 2) Z = s in (2x + 3y) Solution : C. F. = ex f 1 (y + x ) + e2x f2 (y + x) P. I. = = = = = si n (2x + 3y) , , ( ( D - D + 1) D - D - 2) 1 sin (2x + 3y) ( 0 2 - DD' + D - DD' + 0' 2 + D' - 20 + 20' + 2 ) 1 s in (2x + 3y) ( 02 - 200' + D' 2 - 30 + 30' + 30' + 2) -41 DD' o 2 = -6 = -9 0'' � 1 sin (2x + 3y) , - 4 - 2 (-6) - 9 - 30 + 30 + 2 1 1 sin(2x + 3y) sin (2x + 3y) = , , -30 + 30 + 1 ( 30 - 30 -1) 30 - 30' + 1 (30 - 30' + 1) . s i n (2x + 3y) - 2 = -- - 22 1 sin (2x + 3y) = ) 90 + D' -1800' 1 (30 - 30' � = - 30 - 30' + 1 . sin (2 x + 3 Y) = 9(- 4 ) + 9(-9) - 18(-6 ) - 1 10 (30 -30 + 1) 1 J__[3co s(2x + 3y) • 2 -3cos(2x + 3y) • 3 + sin (2x + 3y)] = 10 = J__ [s in(2x + 3y) - 3cos (2x +3y)] 10 , s ·i n(2x + 3y) IES MASTER Publication 574 Partial Differential Equations So, general solution is x 2x z = e f1 ( y + x) + e f2 ( y + x ) + J_ {sin( 2x +3 y )-3 cos( 2x +3 y )} 10 2 2 Example 1 8 : Solve : (0 -0' -30 +3D')z = Solution : The equation can be written as {(D + D')(D -D')-3(0 -D')} z = xy xy + ex+2Y ⇒ + ex+2y (D -D')(D + D' -3)z C. F. = e0 x + f 1 (y + x) + e3x + fz(y - x) P.I. = Taking I, = Taking 1 1, 1 1 1 _ ( e x +2 y ) = I + II ( xy +e x+ 2y ) ( xy ) + (D -D')(D + D' -3) (D -D')(D +D' -3) (D -D')(D + D' -3) 1 D' D D' 200' -1 D' D D' 200' -1 (1- ) (1+ 3 + 3 +-- +... .) ( xy ) = ( 1 + + . ...) (1+ 3 + 3 + -- +....) ( xy ) 30 D 9 30 D 9 - e x+2 y - - - -(D2 -0' 2 + 30' -30) . . . P. I. = _ ! [ 3 2 x y 2 Z y 1 - - e( x +2 y ) = -yex+2y · (-20' +3) 2X x 2 + � + + £] -ye + y 3 3 9 9 xy + So, general solution is - 2 2 X X3 ] xy 2X 1 [X y 3x -+-+ - -yex+ 2 y + -+ = f1 (Y + X ) + e f2 (Y -X )-- 3 2 3 3 9 3 Example 19 : Solve : (D -3D' -2)3 z = 6e2xsin(3x + y) Solution : = x y + ex+2y C.F. = e2xf /y + 3x ) + P. I. = 1 , (D - 30 -2)3 2x = 6e 1 xe 2x + fz(y + 3x ) + • (3 x + )} { 6e 2x sin y 3 (D - 30') = x 6e e 2x ( sin(3 x + y)) = 6e 2x . x = f{e2 x ; sin(3 x + y ) 2 2xf i Y + 3x) 1 { (D + / )-30' -,2} 1 2 3(0 -30') 3 (sin(3 x + y)) 2x sin(3x + y ) · = 6e .x 2 1 sin(3x + y ) 6(0 - 30') General Solution is z = e2x f1 ( y + 3x) + xe2 x f2 (y + 3 x) + x 2 e2x f3 ( y + 3 x ) + X 3 e2x sin(3x + y ) Example 20 : Solve : (0 2 -DD' + D' -1)z = cos( x + 2 y) + eY Solution : { (D +1)(D - 1)-D'(D -1)} z = cos( x + 2 y) + eY ⇒ (D -1)(0 - D' +1)z . = cos(x + 2 y) + eY Engineering Mathematics P. l . = 1 2 (0 -DD' + D' -1) cos(x + 2y) + 1 2 (0 -DD' + D' -1 ) 575 (e Y ) 1 1 1 cos(x + 2y) + x e<Y> = cos(x + 2y) - xe Y = ..!_ sin(x + 2y) -xe Y (-1 - (-2) + 0' - 1 (-0' + 20 ) D' 2 = So General Solution is x x z = e f1 (y) + e - f2 (y + x ) + ..!_sin(x + 2y) - xe Y 2 CLASSIFICATION OF ORDER LINEAR P.D.E. IN TWO VARIABLES 2ND Consider the linear P. D.E in two independent variables x and y 2 a u 2 a u axay 2 a u ay =f(x, y, u, u x , u y ) + -- + CA2 B 2 ax where A, B, C may be functions of x and y and A is positive. At any point (x, y) the equation is said to be (a) Elliptic if B2 - 4AC < 0 (b) Hyperbolic if B2 (c) Parabolic if B 2 - - 4AC > 0 4AC = 0 Example 21 : Classify the following equation. 2 axay ax Solution 2 2 8z a z a z 8z a z -- + (1 - y 2 )- + x -+ 3x 2 y-- 2z=0 (1 -x 2 )-2xy 2 2 We can write, 2 2 a z 2 a z axay a z (1 -x 2 )- 2xy-+ (1 - y2 ) = 2 2 ax ay Here, A = 1 - x2 , B = -2xy, C = 1 - y2 B2 So, above eqn. will be - ( ay ax ay az az ) 2z -x --3x 2 Yay ax 4AC = (-2xy)2 - 4(1 - x2 )(1 - y2 ) = 4(-1 + x2 + y2 ) (a) Elliptic if, B2 - 4 AC < 0 ⇒ x2 + y2 < 1 (b) Hyperbolic if, B2 - 4 AC > 0 ⇒ x2 + y2 > 1 (c) Parabolic if, B 2 - 4 AC = 0 ⇒ x + y = 1 2 [if B2 [if B2 2 - 4AC < OJ 4AC > OJ [if B - 4AC = OJ 2 Hence, given equation is parabolic on the circle x + y = 1 , hyperbolic outside the circle and elliptic inside the circle. a 2u 2 2 a2u =C2 Example 22 : Show that the equation 2 2 is hyperbolic . at Solution 2a u a u -We can write, C 2 = 0 2 2 2 ax at ax A = c2 , B = 0 and 4AC = 0 - 4(c )(-1 ) = 4c So, given equation is hyperbolic. B2 - 2 C = -1 2 (+ve) Practice Questions 1. 2. Show that the given equation is parabolic uxx - 2uxy + uw = 0. Classify the following equations, (a) Zxx = Zw, (b) Zxx + Zw = 0, (c) Zxx = ZY [Ans : (a) hyperbolic, (b) elliptic, (c) parabolic] IES MASTER Publication 576 Partial Differential Equations ONE DIMENSIOANL WAVE EQUATION ...(1) y(O, t) = 0 and y ( e , t) = 0 Boundary conditions are y (x, 0) = f(x); Initial conditions are (ay ) at (x,O) =0 Solution of Wave E quation Y = Let xr ...(2) be a sol ution of (1 ), where X is a function of x alone and T is a function of t alone. Then by using (1) we can write XT" = c2X'T 1 T" x" = K (let) = -X c2 T Case 1 : When K = 0 .. . (3) X" 1 T" 0 = --= X c2 T X" = 0 and T " = 0 X = C 1 X + C2 & T = C3t + C4 y = (c 1 x + c2)(c3t + c4) by using (2) Case 2 : When k = p2 X" 1 T" = p2 = -X c2 T X" = p2X and (02 - p2)X = 0 m2 _ p2 = 0 and and m = ±p and CF = c 1 ePx + c2e-px (02 - c2 p2)T = 0 m2 _ c2 p2 = 0 m = ±cp CF = C3ecpt + c4e--<:pt and y = (c 1 ePX + C2e-PX)(c3ecpt + C4 e--<:PI) X" X = 1 T" c2 T X" = -p2X = -p2 (02 + p2)X = 0 m2 + p 2 = 0 m = ±ip CF = c 1 cos px + c2 sin px Pl = 0 by using (2) and X = c 1 ePx + c2e-px by using (2) AE is T" = c2p2T and Pl = 0 Case Ill : When k = -p2 ...( 4) X = (c 1 cos px + c2 sin px) and and and and and and and Pl = 0 · ... (5) T" = -c2p2T (02 + c2 p2 }T = 0 m2 + c2p2 = 0 m = ±icp CF = c3 cos cpt + c4 sin cpt Pl = 0 T = (c3 cos cpt + c4 sin cpt) y = (c 1 cos px + c2 sin px) (c3 cos cpt + c4 sin cpt) ...(6) Engineering Mathematics 577 Out of these three solutions (4), (5) and (6), we take only that solution which is suitable for physical nature of the problem. Since we are dealing with the problems of vibrations it means solution should contain periodic terms which is possible only in eqn . (6). So general solution of wave equation (1) is given by (6) Boundary conditions are : Initial conditions are y = (c 1 cos px + c2 sin px) (c3 cos cpt + c4 sin cpt) y(0, t) = 0 and y ( e , t) = 0 y (x, 0) = f(x); ( Z ) D-Alembert's Solution of Wave Equation ( x,O ) =0 a2 y a2y =c 2 such that y(x, 0) = f(x) and Consider the wave equation 2 ax2 at Then it's solution can be taken as (ay) at (x,O) = g(x) x+cl 1 1 y(x, t) = -[f(x +ct) +f(x -ct)] + - J g(x)dx 2 2c x-ct ONE DIMENSIONAL HEAT EQUATION Consider a flow of heat in a uniform rod, if temp. u depends only on the distance x and time t then one-D heat equation is given as au Boundary conditons are : & Initial temperature is: Solution of Heat Equation a 2u -=C 2 at ax 2 u(0, t) = o = u( u(x, 0) = f(x) u Let e, ... (1) t) (temperature at corner points) = xr ...(2) be a solution of (1 ), where X is a function of x alone & T is a function of t alone. Then by using (1) we can write Xf' Case 1 : When K by using (2) = c2X'T X" T" =k = X c2T 0 Case 2 : When K = p = T' X" = -=0 X c2T 2 X" = 0 X X = C1 X = C 1 X + c2 = (c 1 x + c2)c3 X" X T" c2T X" = p2 X ...(3) (7 and and and =P2 and T' = 0 c2 T T = C3 T = C3 T' c2 T = p2 ... (4) IES MASTER Publication 578 Partial Differential Equations AE is X " - p2 X = 0 and (D2 - p2)X = 0 and m2 _ p2 = 0 m = ±p CF = c1 epx + c2e-px Pl = 0 by using (2) Case Ill : When k = -p2 X = c1 epx + c2e-px J c2 p2 dt +logc 3 log T = c2p2t + log c3 and log T = c2 p2t + log c3 and and and log T = c2 p2t + log c 3 log T = c2 p2t + log c3 2 2 T = c 3 ec p t 2 2 u = (c 1 epx + c2e-Px) (c 3ec P t ) X" = --p2 X T' and c2T T' and c 2T m2 + p2 = 0 and m = ±ip Pl = 0 and = --p2 J -p2 c 2 .dt +logc 3 2 2 log T = -c p t +logc 3 and and ... (5) = --p2 T' and f r dt = (D2 + p2)X = 0 CF = c 1 cos px + c2 sin px . by using (2) dt = and X" + p2X = 0 AE is I� X = (c 1 cos px + c2 sin px) and 2 2 . u = ( c 1 cos px + c2 sin px) c 3 e-c p t 2 2t T = c 3 e-c p T= C 3 e-c Pl = c 3 e 2 2t p -c2p2 t c2 2 t T = c 3 e- p ... (6) Out of these three solutions (4), (5) and (6), we take only that solution which will consistent with the physical nature of the problem. Solution ( 4) is not possible because it gives a constant temperature at all times. Solution (5) also not possible because here u increases infinitely as t increases. So, our suitable solution is given by (6) because here u ➔ o as t ➔ oo Boundary conditons are & Initial temperature is: u(O, t) = o = u( e , t) u(x, 0) = f(x) Steady State : When temperature distribution in a body is independent from time then body is said to be in steady state condition, i.e., au = 0 at TWO DIMENSIONAL HEAT EQUATION and Consider a flow of heat in a metal plate. If temperature is free from z-coordinate and depends only on (x, y & t), then it is known as two dimensional heat equation. Given by, au - 2 a 2 u a 2 u - c ( 2 + ) at ax a y2 It is a temperature distribution in a metal plate in transient state. Engineering Mathematics LAPLACE EQUATION au For steady state, 579 = 0 at So, It is known as two dimensional Laplace equation. Hence, Laplace Equation represents temperature distribution in rectangular plate at steady state. Boundary conditions are Solution of Laplace Equation Let u(0, y) = u( a 2 u +a2 u = 0 2 ax ay 2 e , y) = u(x, 0) = 0 & u(x, a) = f(x) ...(1) u = XY ... (2) be a solution of (1), where X is a function of x alone & Y is a function of y alone. Then by using ('I ) we can write. Now, Case I : when k = 0 X"Y +XY" = 0 -Y" X" = -= k X y X" X -Y" y X" = 0 =0 X' = C 1 By using (2) Using. u(0, y) = 0 in (4) Using u( e , y) = 0 in ( 4) X = C 1 X +C2 ... (3) Y" = 0 and Y' = C3 and y = C 3 Y +C 4 and u = XY =r= ( c 1 x +c 2 ) ( c3 y +c4 ) 0 = (0 +c 2 ) ( c 3 y +c 4 ) ⇒ 0 = ( c 1 e +c 2 ) ( c3 y +c 4 ) By ( 4), u = 0, which is not possible. So this case is not acceptable. Case II : When k = p2 in (3) X" -Y" = -= p2 X y 2 X" = p X X" -p2 X = 0 (02 - P2 ) x = 0 AE is m2 -p2 = 0 m = ±p CF = C 1 epx +C 2e-px X = C 1 epx +C 2 e-px c2 = 0 ⇒ and and and 0 ⇒ .. . (4) c 1 e ( c 3 y +c4 ) ⇒ c1 = 0 2 -Y" = p Y Y " +p2 Y = 0 (02 +p2 ) y = 0 and AE is m2 +p2 = 0 and and and m = ±ip CF = (C 3 cos PY + C 4 sin py) y = c 3 cos py +c 4 sin py IES MASTER Publication 580 Partial Differential Equations u = XY = (C1 epx +c 2 e-px )(c3 cos py + c 4 sin py) By using (2), 0 = (c 1 +c 2 )(c 3 cos py +c 4 sin py) Using u(O, y) = 0 in (5) Using u( e , y) = 0 in (5) c 1 = -c2 0 = (C1 Cpr +c 2 e-p( )(C 3 COS py + C4 Sin py) 0 = C 1 (ePC _ e-PC )(C3 COs py +c 4 sin py) u = 0 Again this is not possible. By (5) Case Ill : When k = -p2 in (3) - X" X" 2 = - = -p y X 2 X" = -p X A.E. is ⇒ c 1 = 0 & C2 = 0 Y" - p2 Y = 0 and and A.E. is m2 - p 2 = 0 m2 +p2 = 0 m = ±ip m = ±p and CF = c 1 cospx +c 2 sinpx Pl = 0 and CF = C 3 ePY + C4 e- PY and y = C 3 ePY +C 4 e- PY Pl = 0 and X = c 1 cospx +c 2 sinpx By using (2), ⇒ 2 -Y" = -p Y and X" + p2 X = 0 . . . (5) u = XY = (c 1 cospx + c 2 sinpx)(c 3 ePY +c4 e-PY ) . . . (6) ·: Solution given by ( 4) & (5) are not possible, so (6) can be taken as solution of Laplace equation. CLAIRAUTS ORDINARY DIFFERENTIAL EQUATION where So, So p = dy dx dy = xp' +p + �'(p ).p' dx So, general sol ution of (1) is y = cp + � (c) For example, ".(1) y = xp +�(p) p' = 0 ⇒ = IP cl p = tan(px - y) ⇒ CLAIRAUTS PARTIAL DIFFERENTIAL EQUATION where Differential partially with respect to 'x', Differnetial partially with respect to 'y' , So, general solution of (1) is For example ⇒ [x + �'(p )] p' = 0 y = px - tan-1 p ⇒ " .(1) z = px + qy + f(p, q) p = j)z ax · q= j)z ay p = p + xp' +0 + f'(p,q).p'q q = O +q'y + q +f'.pq' z = C 1 x + C2y + f(C 1 , C2) z = px + qy + pq y = ex - tan-1 c ⇒ ⇒ ⇒ (x +f'.q) p'=0 (y +f'. p)q' = 0 ⇒ ⇒ IP = C1 I \q = C2 \ z = C 1 x + C2y + C 1 C2 Engi neering Mathematics 1. Questions marked with aesterisk (*) are Conceptual Questions The one dimensional heat conduction partial differential equation a2 T is ax z , (b) hyperbolic aT at (a) parabolic 2. (c) ell iptic (d) m ixed differential equation (c) ell iptic ax 2 is (b) hyperbolic 3.* The number of boundary conditions required to solve (a) 2 (b) 0 (c) 4 The 2 a � ax 2 ax + 2 al 2 is [GATE-2001 (CE)] d i ff e re n t i a l equation a� + a� = 0 ax ay (d) 1 p a rt i a l a �+ 2 ay (a) degree 1 and order 2 has (b) degree 1 and order 1 (c) degree 1 and order 1 5.* (d) degree 2 and order 2 ax (a) z = px + qy 6. (c) z = px + qy + pq [GATE-2007 (ME)] oz) ay (b) z = px + pq (d) z = qy + pq [GATE-2010-CE] The type of t h e parti a l d i ffere n t i a l equation 2f at 2 is a at = ax (a) parabolic 8. (c) hyperbolic (d) non-linear equation of order 2 T h e type (b) elliptic (d) nonlinear [GATE-201 3 (IN)] a2 P + axz + ay z a 2P (a) elliptic 9.* of p a r t i a l 3 a2 P axcty (c) hyperbolic +2 [GATE-201 3-ME] d i ffere n t i a l = o is aP - aP ay ax equation (b) parabolic (d) none of these [GATE-201 6-CE-Set-l] The solution of the partial differential equation 8t = a. aa au u . 1s of the form xz 2 (a) C cos( kt) l C 1 e( �x + C2e -( �x (b) Cekt l ci�x + Cze+. lkta )x J j (c) Cekt lC1 cos(�) x + C2 sin (�) x (d) The partial differential equation that can be formed from z = ax + by + ab h a s t h e form ( with p = az andq = 2 a u au (c) l i near equation of order 1 [GATE-1 996 (ME)] the differential equation is a - + u- = ax ax2 at (b) non-linear equation of order 1 (d) mixed a2 � + a 2� The partial differential equation au [GATE-1996] aT = a 2 T at 7. (a) l i near equation of order 2 The one d i m ensional heat co nduction part i a l (a) parabolic 4. 581 C cos ( � ) x C sin(kt) l 1 j +C 2 Sin ( -�) X j [GATE-201 6-CE-Set-l] 1 0.* Which one of the following is a property of the solutions to the Laplace equation: V 2f = O ? (a) The sol utions have neither maxima nor m i nima anywhere except at the boundaries. (b) The sol u t i o n s a re not sepa rab l e i n t h e coordinates. (c) The solutions are not continuous. (d) The solutions are not dependent on the boundary conditions. [GATE-201 6, 2 Marks] 1 1 .* Solution of Laplace's equation having continuous second-order partial derivatives are called (a) biharmonic functions (b) harmonic functions IES MASTER Publication 582 Partial Differential Equations (c) conjugate harmonic functions (d) error functions [GATE-201 6; 2 Marks] 1 2 . Consider the following partial differential equation 2 2 (a) 2 a � a � a � 3+B-- +3+ 4� = 0 2 axay ay2 ax (c) (a) � ( x 2 + y 2 , y2 - yz ) = 0 1 3. Consider the following partial differential equation u(x, y) with the constant c > 1 : (b) � ( x 2 - y2 , y 2 + yz) = 0 (c) � (xy, yz) = 0 au Solution of this equation is 1 7 .** (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) 1 4.* The complete integral of a2 u (a) 1 1 8. 2 C [ESE 2017 (Common Paper)] 1 5.* The solution of the following partial differential equaa2u a2u 9 - is tion ax 2 - ay2 The solution at x = of u(O) = 3x and (z -px -qy)3 = pq +2(p2 + q)2 is (d) z = ax + by + (d) � (x + y, ln x - z) = O [ESE 201 8(EE)] 1, t = 1 of the partial differential a2 u =25equation 2 subject to initial conditions 2 ax at [GATE-201 7 ME Session-I] (c) z=ax + by + � + �2(a 2 +b) [ESE 201 7 (Common Paper)] y2p - xyq = x (z - 2y) is [GATE-2017 CE Session-I] au (d) (3y2 - x 2 ) sin(3x - 3y) 1 6.** The general integral of the partial differential equation For the equation to be classified as parabolic, the value of B2 must be ___ -+ c - = 0 ax ay (b) 3x 2 + y2 sin(3x - y) (c) 4 81 au (0) = 3 is ____ (b) 2 (d) 6 [GATE 2018 (CE-MorningSession)] Consider a function it which depends on position x and time t . The partial differential equation is known as the au a 2u = at ax 2 (a) Wave equation (b) Heat equation (c) Laplace's equation (d) Elasticity equation [GATE 201 8 (ME-AfternoongSession)] Engineering Mathematics 1. 6. (a) 11. (b) 1 6. (c) 8. (c) 1 3. (b) 1 8. 2. (a) 4. (a) 3. 5. Sol-1 : (a) 7. 9. 1 0. (c) 14. (b) at . . . (i) ax 2 Comparing (i) with aT aT a2 T 8 2T + H=0 + C 2 + 2G + 2F A 2 +B ay ax axay ay ax 82 T A = 1, B = 0, C = 0 B2 - 4AC = 0 ⇒ Parabolic Partial Differential Equation (a) The general form of 2nd order partial differential equation is: a2u au au a2 u a 2u + f ( x, y,u, - , - ) + CA+ B 2 2 ax ay axay ay ax = 0 . . . ( 1) On comparison, A = 1, B = 0, C = 0 B2 - 4AC = 0 - 4(1 )(0) = 0 :. Given equation is parabolic. Sol-3 : (c) The general solution of Laplace equation contains 4 arbitrary constants. Therefore, we require 4 boundary conditions. i.e .. � = (C1 cos px + C2 sinpx ) (C3 ePY + C4e-PY ) where C 1 , C2, C3 , C4 are arbitrary constants. Sol-4: (a) Order = 2 (order is the highest derivative) Degree = 1 (exponent of highest derivative) (a) (d) (b) (a) Sol-5: = 1 7. (b) 1 5. (a) (a) Sol-2: (36) 1 2. (d) 583 Given, (c) z = ax + by + ab ... (i) Differentiating the above equation partially w.r.t. x & y, separately, we get az ax ay a = p (say) ... (ii) = b = q (say) . . . (iii) = :. z = px + qy + pq; the required differential equation Sol-6: (a) A 2nd order partial differential equation of the form 8 2u a2 u + B-a2 u- + CA+ f ( x, y,u, p, q) = 0 will be axay ay2 ax 2 (i) Parabolic if B2 -4AC=0 (ii) Hyperbolic if B2 - 4AC > 0 (iii) Elliptic if B2 - 4AC < 0 Given equation can be written as 1 2 2 _ a2 t . a t + 0 . a t + ( -at ) + 0 at axat ax2 at2 On comparison, A = 1 , B = 0, C Now, B2 - 4AC = 0 - 4( 1) = 0 = 0 = 0 :. The given equation is parabolic. Sol-7: (d) Sol-8: (c) The given equation is non-linear equation of order 2. - aP aP a2 p Aa2 P 82p _ 0 - + B- + C - + D- + E- _ axay ax 2 ay 2 ax ()If A = 1 8 = 3 C = 1 B2 - 4AC > 0 So hyperbola. IES MASTER Publication 584 Partial Differential Equations Sol-9: (b) au at Let u = au ...(1) xr is the solution of (1) then au - = X.T' and Now, ...(2) ⇒ a2 u = X"T ax 2 Using separation of variables concept in (1) at aX" ⇒ X XT' = aX ".T Taking aX" = KX we get (02 -�)x T' ⇒ - Taking T' = KT ⇒ ⇒ = 0 T T' = - = K = K Sol-1 0: (a) The solution of laplace's equation having second order partial derivatives are called harmonic functions. So l-1 2 : (36) 2 2 2 2 2 a � � + 3 2 + 4� = o 3 2 +B axay ax ay 2 � a � a a � � A 2 +B + C 2 + D� = 0 axay ay ax a Compare A = 3; B = ?; C = 3 Equation to be parabolic · B2 - 4AC = 0 B2 - Sol-1 3: (b) Let 4 X 3 X b = -ac a = b = -c u = f (1.x -c.y) = f (x -cy) Method II : Using hit and trial method we see that option (b) satisfies given P.D.E. z = px + qy + pq ...(1) az ... (2) and it's solution is z = ax + by + ab where p = constant. az ax and q = ay and a, b are arbitrary Using (2) we can easily get az ax = a and az ay = b. Hence to find general solution of Clairaut equation, we can replace p by a and q by b. Now, given equation is : (z -px -qy)3 = pq + 2(p2 + q)2 z = px + qy + [pq + 2(p2 + q)]113 so complete solution is Sol-1 5: (a) z = ax + by +[ab + 2(a 2 +b)]1I3 Given DE is ...(1) By hit and trial let us take option (a) is the solution, i.e., 3 = 0 B2 = 36 u = b + c x a = O au - x � a(ax -+ ----'by) _ 0 8(ax + by) ax General form of Clairaut's equation is Laplace equation: V 2 f = 0 can be used in any of the three co-ordinate system and its solutions are continuous. Moreover, the constants that occur while solving v 2 f=0 can be found only using boundary conditions. Sol-1 1 : (b) a 8(ax + by) ay Sol-1 4: (b) [;, + c,J• J ee" u = [ C1 e au -- x --� + C 8( ax + by) x X = C1 e-[ + C2 e[ x So by (2), solution is �+ C � = O ay ax T log T = Kt + log C ⇒ ⇒ ⇒ - = f' (ax + by) a (ax + by) f(ax + by) then u = sin(3x -y) au = 3 cos (3x - y) and ax Engineering Mathematics a 2u = -9 sin(3x -y) ax 2 au - Sol-17: (d) Standard form of wave equation is -cos(3x- y) a 2u = -sin(3x-y) ay 2 and Hence, option (a) satisfies (1) so it is the required solution. and we have given = 25� at2 2 Sol-16: (a) y2 p - xy q = x(z - 2y) On comparision with P.p + Qq = R p = y2 , Q = - xy, R = x(z - 2y) By Lagrange's Auxiliary equation dy = dz dx = P o R i.e. dy dx dz = = -xy x(z- 2y) y2 ⇒ xdx + ydy = 0 (II) (I) (Ill) xdx+ydy 0 (IV) dx xdx+ydy = Taking I and IV, y2 O i.e. x2 + y = c1 2 dz dy Again By II and 111, - = x(z-2y) -xy zdy - 2ydy = - ydz or ydz + zdy = 2ydy or d(yz) = d(y2) i.e. yz - y2 = C or y2 - yz = c 2 So General solution is ⇒ <I> <I> (C 1 , C2) = 0 (x2 + y2, y2 - yz) = 0 585 = 1 1 a 2u 2 so, C = 5 25 ax and f(x) = 3x and g(y) = 3 So by De-alemberts solution of wave equation. + 1 1 x ct u(x, t)=-[f(x+ ct)+f(x- ct)]+- J g(y)dy 2 2c x t -c 1 15 15 2 = -[6 x]+-[x+ ct-x+ ct]=3x+-(-t ) 2 5 2 2 u(x, t) = 3x + 2t So, u(1, 1) = 3 + 3 = 6 Sol-18: (b) Standard form of Heat equation is au at 2 = c a2 u . ax2 where u ➔ temperature. IES MASTER Publication 6.1 In this chapter, we will deal with the system of linear algebraic equations which has the form a 11X 1 +a12 X2 +· .. +a1n Xn = b1 a2 1X1 +a22 X 2 + ·· · +a2n Xn = b2 orAX = B n is the number of equations. For small number of equation (n � 3), the above system of linear equation can be solved by simple techniques. However, for four or more equations, techniques to solve linear system become tedious.After the availability of computer, we go to numerical method, which are suited for computer operations. These numerical methods are of two types namely: 1. Direct Methods (or Exact Methods) Direct methods are those in which the solution is obtained in a finite number of steps. The direct methods are also called elimination methods. The direct methods of solving linear equation reduce the given system to an equivalent set of equations, which are much easier to solve. The solution of direct methods do not contain any truncation errors, e.g. Gauss elimination methods, Gauss-Jordan method, Crouts method, Dolittle's method (or Factorization method). 2. Iterative Methods This gives a sequence of approximate solutions, which converges when the number of steps tends to infinity e.g. Gauss­ Jacobi method, Gauss-Seidel method. Iterative methods of solving system of linear equations: Iterative methods are such that in which we start the process by assuming a solution which is nearer to the true solution and we obtain the solution by successive repetition of the process. The number of iteration depends on the degree of accuracy required. This iterative method is not always successful to all systems of equations. If this method is to succeed, each equation of the system must possess one large coefficient and the large coefficient must be attached to a different unknown in that equation. This condition will be satisfied if the large coefficient are along the leading diagonal of the coefficient matrix. When this condition is satisfied, the system will be solvable by the iterative method and in that case system is called Diagonally Dominant. The system will be solvable if la 11 I > iad+lad GAUSS-JACOBI METHOD Consider the system of equations. Engineering Mathematics a1 x + b 1 y + c 1 z = d1 a2 x + b2y + c2z = d2 587 ... (1) a 3x + b3y + c3z = d3 lb2 I > la2 I + lc2 I , Let us assume Then, iterative method can be used for the system (1). Solve for x,y,z (whose coefficients are the larger values) in terms of the other variables. That is, 1 x = -( d1 -b1Y-C1z ) a1 z = -( d3 -a 3 x-b3 y ) C3 y = 1 -( d2 -a2 x-c2 z ) b2 1 If x( 0l, y(0l, z(O) are the initial values of x,y,z respectively, then x ( 1) = y( 1) ...(2) o __!_( d -b y O) - ci )) a 1 1 ( 1 ... (3) z(1) Again using these values xPl, y( 1 l, z( 1) in (2), we get x(2) = 1 1 ; (d1 -bi ) -ci )) 1 z(2) = __!_( d -a x C y(2) 3 3 3 (1) -b 3 y (1)) ...(4) Proceeding in the same way, if the rth iterates are x(r), y(r), z(r), the iteration scheme reduces to x(r+ 1) = y(r+1) = z(r+ 1) = __!_( d - bi') -ci'l) a 1 1 __!_( d -a x r )-c z r)) b 2 2 2 ( 2 ( __!_( d -a X r ) -b y r )) C 3 3 3 ( 3 ( ...(5) The procedure is continued till convergence is assured (correct to required decimals). In the absence of the initial values of x,y,z, we take usually (0, 0, 0) as the initial estimate. Example 1 Solution Solve by Jacobi's method 4x + y + 3z = 17; X + 5y + Z = 14; The above equations can be written as 17 y 3z 4 4 4 2x - y + 8z = 12 X=----- 14 X Z Y = 55-5 3 X y Z=----2 4 8 ... (1) IES MASTER Publication 588 Numerical Solution of Linear Equations On substituting x0 = y0 = z0 = 0 on the right hand side of (1), we get 14 -3 y1 -5• z1 2 Again substituting these values of x,y,z on R. H.S. of (1), we obtain X1 - 17 4' 14 17 3 3 17 7 63 z2 = --- + - = 2 16 20 80 33 Y2 -- -------· 5 20 10 20 ' Again putting these values on R.H.S. of (1), we get next approximations. X3 = 14_ 97 _ 63 863 17 _ 33_189 1039 = = 2.16 . = 3.25 . Y3 = = ' ' 5 200 400 400 4 80 320 320 Substituting, again, the value of x,y,z on R.H.S. of (1), we get 23 97 � 176 = 1.1 = + = �_ 2 160 160 160 !2__2.16_3(1. 1) = 2.885 . y = 14_ 3.25_.!J,_1_93 _ 2 = �- 3.25 + 2.16 = 0_96 X4 = ' 4 ' 4 2 4 8 5 5 5 4 4 4 Repeating the process for x = 2.885, y = 1.93, z = 0.96, we have _ 1·93 -� x0.9 + 6=4.25- 0.48-0.72=3.05 X5 = !2_ 4 4 4 2·885 0 96 - · = 2.8-0.577 -0.192 = 2.03 y 5 = 2_± _ 5 5 5 885 93 2 1 �- · + · =1.5-0.7 21+0.241=1.02 z5 = 2 8 4 After 5th iteration x = 3.05, y = 2.03, z = 1.02 Example 2 Solution I 1 & the actual values are x = 3, y = 2, z = 1. Solve the following equations by Jacobi method 20x + y - 2z = 17; 3x + 20y - z Given equations can be written as = -18; 2x - 3y + 20z = 25 X = �( 17- y +2z ) 2 1 y =- (-18-3x + z) 20 Z = �(28-2x + 3y) 2 Let us take initial approximation as x0 = 0, y0 = 0, z0 = 0 The First approximation using (1), we get 1 17 X1 = - ( 17- y 0 + 2z0 ) = - = 0.85 20 20 1 18 (-18-3x0 + z0) = -- = -0.9 Y1 = 20 20 25 1 Z 1 = - ( 25- 2x 0 + 3y 0) =- =1.25 20 20 Now, the second approximation is 1 1 (17-y1 + 2z1 ) = (17 + 0.9 + 2x1.25) = 1.02 20 20 1 1 y2 = (-18-3x1 + z1 ) = (-18 -3x0.85 + 1.25) = - 0.965 20 20 1 1 (25- 2X1 + 3y1 ) = (25 - 2x0.85 + 3x -0.9) = 1.03 z2 = 20 20 X2 = ...(1) Engineering Mathematics 589 Now, substitute these values into (1), we get third approximation x3 = z3 = 1 1 ( 17- y 2 + 2z2 ) = (17 + 0.965 + 2x1.03) = 1.0013 20 20 1 1 1 1 - ( -18 - 3x2 + z2 ) = - (-18-3 x 1.02 + 1.03) = -1. 0015 20 20 - ( 25-2x 2 + 3y2 ) = - (25-2x1.02 + 3 x-0.965) = 1. 0032 20 20 Substitute these values into (1), we get fourth approximation x4 = y4 = 1 1 (17-y 3 + 2z3 ) = (17 + 1.0015 + 2x1.0032) = 1.0004 20 20 1 20 1 1 (-18-3x3 + z3 )= 20 1 (-18- 3x1.0013 + 1.0032) = -1.00003 ( 25- 2x3 + 3y 3 ) = (25- 2x1.0013 + 3x1.0015) = 0.9996 20 20 Substituting these values into (1), we get fifth approximation z4 = x5 = 1 20 1 (17 -y 4 + 2z4 ) = 20 (17 + 1.00003 + 2x0.9996) = 0.99996 20.0016 = -1.00008 20 19.99911 1 1 (25-2x1.004 + 3x-1.00003) = (25 -2x 4 + 3y 4 ) = = 0.99995 20 20 20 The values in the fourth and fifth approximation are nearly same. Hence the solution is z5 = X = 1, y = - 1, Z = 1 GAUSS-SEIDEL METHOD OF ITERATION This method is modification of Jacobi's iteration method sometimes gives faster convergence. In the Jacobi's method, even though we have computed the new x<1l, we still do not use this in computing new x< 2>, while in Gauss-Seidel method whenever x is calculated, it is immediately used in the next equation to determine values of other variables. 1 X = -(d1 - b1y-c 1 z ) Consider 81 ...(6) We start with the initial values y< 0>, z< 0> for y and z and get x<1> from the first equation. That is, ( )_ o x( 1) = _!_ (d1 -b1Y O ci >) 81 While using the second equation, we use z< > for z and x( 1 l for x instead of x< 0> as in the Jacobi's method, we get 0 y(1) = 1 � (d2 -a 2 x (1) -c 2 z(o)) Now, having known x< 1> and y<1>, usages x<1> for x and y<1> for y in the third equation, we get z( 1l = 1 - (d3 -a 3 x ( 1) -b 3 Y( 1l ) C3 In finding the values of the unknowns, we use the latest available values on the right hand side. If x<r>, y(rl, z(rl are the rth iterates, then the iteration scheme will be IES MASTER Publication 590 Numerical Solution of Linear Equations 1 1 1 1 r (r+1) (d1 -b1 Y(r)_C1Z (rl) ; -b 3 y (r+ ) ) y(r+1) = _ (d2 -a 2 X (r + )_c 2 z< >) ; z(r+1 ) = _!_(d3 -a 3 X C3 a1 b2 This process is repeated until the successive approximations converges closely enough to the solution. This procedure is also called as the method of successive approximations. The rate of convergence of Gauss-Seidal method is more rapid than Jacobi's method but this is not always true. x(r+1 ) = 2. For all systems of equations, this method will not work. It converges only for special systems of equations. i.e., for diagonally dominant system Iteration method is self-correcting method. That is, any error made in computation is corrected in the subsequent iterations. Example 3 = Solve the system of equations by Gauss-Seidal iterative method 54x + y + z = 110 2x Solution + 15y + 6z = 72 -x + 6y + 27z = 85 The given equations can be written as 1 X = - [1 10- y -z] 54 1 y = -[72-2x-6z] 15 z = For first iipproximation: Putting y = 0, z = 0 in (1 ), we get Putting x = 2.037, z = �[85 +x -6y] 2 11 O = 2.037 54 0 in (2), we get x<1l = y(1 ) = _!_[72-2(2.037)-0] = 4.528 15 Putting x = 2.037, y = 4.528 in (3), we get 1 z(1 ) = - [ 85+ 2.037-6x4.528]=2.217 27 For second approximation: Putting y = 4.528, z = 2.217 in (1), we get 1 x(2) = - [110- 4.528- 2.217] = 1.912 54 Putting x = 1.912, z = 2.217 in (2), we gef 1 y(2) = -[72- 2x1.912-6x2.217]=3.658 15 Putting x = 1.912, y = 3.658 in (3), we get For third approximation: z(2) = 2 [85+ 1.912-6x3.658]=2.406 � Putting y = 3.658, z = 2.406 in (1 ), we get x<3> = y( 3 ) = _!_[ 110 - 3.658- 2.406] = 1.925 54 Putting x = 1.925, z = 2.406 in (2), we get _!_[72- 2x1.925-6x2.406] = 3.581 15 ...(1) ...(2) ...(3) Engineering Mathematics Putting x = 1.925, y = 3.581 in (3), we get For fourth approximation: Putting y = Putting x = Putting x = 591 3.581, z = 1.926, z = 1.926, y = z( 3) 1 -[85 + 1.925-6 X 3.581] = 2.424 27 = 2.424 in (1), we get x(4l = _2_[110- 6.581- 2.424] = 1.926 54 2.424 in (2), we get 1 y(4) = -[72- 2x1.926- 6x2.424]=3.574 15 3.574 in (3), we get 1 [85+1.926-6x3.574] = 2.425 z(4) = 27 For fifth approximation: Putting y = 3.574, z = 2.425 in (1), we get 1 x(5) = -[110- 3.574-2.425] = 1.926 54 Putting x = 1.926, z = 2.425 in (2), we get Putting x = 1.926, y = 1 y(S) = -[72-2 X 1.926-6 X 2.425] = 3.573 15 3.573 in (3), we get 1 z(5) = -[85 + 1.926-6 X 3.573] = 2.425 27 Thus, fourth and fifth iterations are practically same. Example 4 : Solution Hence the solution is x = 1.926, y = 3.573 and z = 2.425 Use Gauss-Seidal iterative method to solve the following equations as Perform four iterations. 9x + 4y + z = - 17 X - 2y - 6z = 14 X + 6y = 4 The given system is not diagonally dominant. Hence rearranging the equations as 9x + 4y + z = - 17 X + 6y = 4 X - 2y - 6z = 14 Now, Gauss-Seidal's iterative method can be applied From the above equations, we get First approximation: =- = 1 -(-17- 4y- z ) 9 1 ... (1) i(x-2y-14 ) . .. (3) y = -(4 - x) 6 Starting with y0 = 0 = z0, we obtain Now, putting x(1 l X z = 17 x(1 l = -- = -1.8888 9 1.8888 and z0 = 0 in equation (2), we get y(1) = 0.9815 ... (2) IES MASTER Publication 592 Numerical Solution of Linear Equations Again, putting x(1) = - 1.8888 and y<1) = 0 in equation (3), we get z<1 l = - 2.9753 Second approximation: x<2) = !{-17-4y < 1 l -z(1l } = -1.9945 z(2l = !{ x < 2) -2y < 2l -14} = -2.9988 y(2) x<3) Third approximation: x<4) Fourth approximation: Hence after four iterations, we obtain Example 5 : Solution X = -2, 9 = !{4- x< 2 l } = 0.9991 6 6 = - 1.9997; y<3) = 0.9999; z(3) = - 2.9999 = -1.9999 ""-2 ; y(4) = 0.9999 ""1 ; z<4) = -2.9999 se-3 y= 1, Z =- 3 Solve the following system of equations using Gauss-Seidal iterative method 2x + 1Oy + z = 51 1 Ox + y + 2z = 44 x + 2y + 1 Oz = 61 The given system is not diagonally dominant. Hence rearranging the equations as 1Ox + y + 2z = 44 2x + 10y + z = 51 x + 2y + 1Oz = 61 Now, Gauss-Seidal's iterative method can be applied. From the above equations, we get X = -(44-y-2z) 1 10 ...(1) (61-X -2y) 1� ...(3) y = _!__(51-2x-z) 10 z = First approximation: Starting with y0 = 0 = z0, we obtain from (1), x<1) = 4.4 Now, putting x(1) = 4.4 and z0 = 0 in equation (2), we get y<1) = 4.22 ...(2) = 4.4 and y<1l =