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 = 4.22 in equation (2), we get z<1) = 4.816;
Second approximation: x<2l = 3.0148;
z<2) = 4.995432
y<2) = 4.01544;
z<3l = 4.999946464
Third approximation: x<3l = 2.9993696;
y <3l = 4.00058288;
Fourth approximation: x<4) = 2.9999 = 3;
z<4l = 5.0000 = 5
y<4l = 4.0000 = 4;
Hence after four iteractions, we obtain x = 3, y = 4, z = 5.
Again, putting x(1)
Example 6 :
Solution :
Solve by Gauss-Seidal method, the following system
28x + 4y - z
= 32;
x + 3y + 10z
= 24;
2x + 17y + 4z = 35
Since the diagonal elements in the coefficient matrix are not dominant, we rearrange the equations as
follows, such that the elements in the coefficient matrix are dominant.
Engineering Mathematics
28x + 4y - z = 32;
1
[ 32-4y+ z];
28
Setting y = 0, z = 0, we get
Hence,
X =
First iteration:
Second iteration:
Third iteration:
Fourth iteration:
Fifth iteration:
X + 3y + 10z = 24;
1
Y = -[35- 2x-4z]·
17
593
2x + 17y + 4z = 35
z =
1
x( 1 l = - [32-4(0)+ 0] = 1.1429
28
1
�[24- x - 3y]
y(1) = _:!_ [35 -2(1.1429)- 4(0)] = 1.9244
17
z( 1 ) =
�[24-1.1429 - 3(1.9244)] = 1.8084
1
1
x C2) = - [32-4( 1.9244)+1.8084] = 0.9325
28
y(2) = _:!_[35- 2(0.9325) - 4(1.8084)] = 1.5236
17
z(2) = _:!_ [24 -0.9325- 3(1.5236)] = 1.8497
10
x C 3) =
� [32- 4(1.5236) + 1.8497] = 0.9913
2
y( 3) = _:!_ [35-2 ( 0.9913)- 4(1.8497)] = 1.5070
17
z( 3) =
�[24-0.9913- 3(1.5070)] = 1.8488
1
1
x<4l = - [32- 4 ( 1.5070) + 1.8488] = 0.9936
28
y(4) = _:!_[35- 2(0.9936) - 4(1.8488)] = 1.5069
17
z(4 l =
1
�[24-0.9936- 3(1.5069)] = 1.8486
1
x<Sl = - [32-4(1.5069)+ 1.8486] = 0.9936
28
y(5) = _:!_ [35- 2(0.9936)- 4(1.8486)] = 1.5069
17
� [24 -0.9936- 3(1.5069)] = 1.8486
1
Since the values of x,y,z in the 4th and 5th iterations are same, we stop the process here.
z(5) =
Hence, x = 0.9936, y = 1.5069, z = 1.8486
IES MASTER Publication
594
Numerical Solutions of Linear Equations
- PREVIOUS YEARS GATE & ESE QUESTIONS
Question marked with aesterisk (*) is Conceptual Question
1.*
Gauss-Seidel method is used to solve the following
equations (as per the given order):
x 1 + 2x2 + 3x3 = 5
Sol-1:
Assuming initial guess as x 1 = x2 = x3 = 0, the value
of x3 after the first iteration is ___
(GATE-2016)
2x 1 + 3x2 + x3 = 1
I
3x 1 + 2x2 + x3 = 3
ANSWER l{�Y
1.
I
.
(-6)
(-6)
⇒
Given that
Hence,
X1
3x2 = 1 - 2(5) = - 9
X2
= - 3
Now for x3, we will take x4 = 5 & x2 = -3
+ 2 (0) + 3(0) = 5
X1
= 5
Now for x2, we have x 1 = 5 & x3 = 0
i.e.,2x 1 + 3x2 + 0= 1
⇒
: -
3x2 = 1 - 2x 1
i.e.,3x 1 + 2x2 + x3 = 3
x 3 = 3 - 3x 1
X3
-
2x2
= 3 - 3(5) - 2(-3) = -6
Hence, - 6 is correct.
Note: Here we have not changed given system into
diagonally dominant system as it is mentioned that
we have to solve this system "As per the given order''.
The equation f(x) = 0 is said to be algebraic if f(x) is purely a polynomial in x. If f(x) contain some other functions,
namely, Trigonometric, Logarithm, Exponential etc., then the f(x) = 0 is called a Transcendental Equation.
For example, the equations
While, the equations,
x2 + 5x - 7 = 0 and x4 + 3x3 + 5x2
-
4x + 7 = 0 are algebraic equations
3 sin 4x = 2x + 7 and ex - cos x + 3x - 5 = 0 are transcendental equations.
Root of an equation
Algebraically, the real number u is called the real root (or zero of the function f(x)) of the equation f(x) = 0 if and only
if f(u) = 0 and geometrically, the real root of an equation f(x) = 0 is the value of x where the graph of f(x) meets the
x-axis in rectangular coordinate system.
There are two types of methods that can be used to find the roots of the equation f(x) = 0.
1. Direct Methods: These methods give the exact value of the root in a finite number of steps. These methods
determine all the roots at the same time.
2. Iterative Methods: These methods are based on the idea of successive approximations. Starting with one or
more initial approximations to the root, we obtain a sequence of iterates {xk} which in the limiting form converges
to the root. These methods determine only one root at a time.
Methods for finding the approximate value of the root
GRAPHIC.AL M ETHOD
In order t o solve the equation f(x) = 0, draw the graph o f function y = f(x). Then roots may be interpreted as the x­
coordinates of the point of intersection of the curve y = f(x) and the x-axis.
If the equation f(x) = 0 is written as f1 (x) = fix) where f 1 (x) and fix) both are the function of x. Then its real roots are
given by the points of intersection of the curves y = f1 (x) and y = f2(x). The x-coordinate of their points of intersection
will give the required roots of the given equation.
Example 1
Solution :
Find the approximate value of the root of x - sin x - 1 = 0.
The given equation can be written as
x - 1 = sin x
Now, we draw the graphs of two functions y = x - 1 and y = sin x
it is obvious that the x-coordinate of the point of intersection is near about 1.9. Hence
the approximate value of the root is 1.9.
Example 2 :
Solution :
Solve the equation xlog 1 0 x = 1, graphically.
The given equation is xlog 1 0 x = 1
y
596
N u merical Solution of Algebraic and Tra nscendental Equations
can be written as
1
log10 x = X
y=
y = log x
x
0
Now, constructing the curves
1
Now the x-coordinate of the point of intersection of the two curves is x
X
Hence the approximate value of the root of xlog 1 0 x = 1 is x = 2.5.
y
Example 3 :
Solution :
=
log 1 0x and y
= -,
=
2.5.
Find the approximation of the smallest real root of e-x = sin x.
Let us draw two curves for the functions y = e-x and y = sin x. The abscissa of the first point of
intersection of the two curves is appearing nearly about x = 0.6.
y = sinx
Hence x = 0.6 is an approximate value of the smallest root.
BISECTION METHOD OR ANALYTIC METHOD OR BOLZANO M ETHOD OR HALF
INTERVAL METHOD
I t i s based on the Rolle's theorem
"If f(x) is a real valued continuous funct_ion in the interval x = a to x = b and f(a) and f(b) have opposite signs such that
f(a). f(b) < 0, then the graph of the function y = f(x) crosses the x-axis is least once".
Bisection method of finding the roots is based on repeated application of the above theorem. The method is as follows­
Let us find a root of the equation f(x) = 0 in the interval [a, b]. We bisect the interval (a, b) and take the point of bisection,
1
as x1 = (a +b) , as the first approximation to the root of the equation f(x) = 0. where we have assumed that f(a) <
2
0 and f(b) > 0
Now there are three possibilities-
(i)
If f (x1 ) = 0, then x 1 is the required root and no need to investigate further.
(ii) If f(x1 ) > 0, then the root lies in the first half interval (a, x1 ) .
(iii) if f(x1 ) < 0, then the root lies in the second half interval (x 1 , b).
As a next step, we bisect the interval (a, x1 ) or (x1 , b), whichever contains the roots, and take the point of bisection,
say x2 , as the second approximation to the root. If f(x2 ) = 0, then x2 is the required root. Otherwise, we bisect the half
interval which contains the root.
In this way, we form sequence of approximate roots x1 , x2 , x3,
...
whose limit of convergence is the exact root.
We continue this process of bisection until the root is obtained to the desired accuracy.
Engineering Mathematics
597
Number of iterations required to achieve the accuracy
In this process, the interval width is reduced by a factor of half of each step and at the end of the nth step, the new
lb - al
interval will be (an , b nl of length � ·
lb -a l
We then have - �E where
2n
which gives on simplification
E
is very small +ve number
>
n -
0 (�)
log
log 2
Inequality (1) gives the number of iterations required to achieve the accuracy
...(1)
E.
Order of Convergence of Bisection Method
In bisection method, the original interval is divided into half interval and in each interation, we take midpoints of
successive intervals to be the approximation of the root.
Hence, error in successive iterations can be related as
e
.lim \ i+1 I
�A
Hoo ek
I
Comparing with
we get
= 0.5.
or
k = 1 and A = 0.5
Here k = order of convergence andA = asymptotic error.
The Bisection method is first order convergent or linearly convergent.
Example 4 :
Solution :
Find the real root of the equation f(x) = x 3
Here
-
f(x) = x 3
x - 1 = 0 by bisection method.
-
x-1 = 0
f(1) = -1 (-ve)
f(2) = 8 - 2 - 1 = 5 (+ve)
So, at least one real root will lie between 1 and 2.
Using Bisection method, we can take first approximation.
1+ 2 �
= = 1.5
2
2
f(1.5) = (1.5)3 - (1.5) - 1 = 3.375 - 1.5 - 1 = 0.875 (+ve)
X1
=
Hence the root lies between 1 and 1.5.
Now, second approximation
then
X2
=
f(1.25) = (1.25)3
Thus, the root lies between 1.25 and 1.5.
So, the third approximation
then
1 + 1.5
=1.25
2
-
(1.25) - 1 = 1.953 - 2.25 = - 0.297 (-ve)
1.25 + 1.5
= 1.375
2
f(1.375) = (1. 375)3 - (1.375) - 1 = 0.2246 (+ve)
X3
=
Thus, the root lies between 1.25 and 1.375.
IES MASTER Publication
598
Numerical Solution of Algebraic and Transcendental Equations
So, the fourth approximation is given by
1.25 + 1.375
= 1.313
2
f(1.313) = (1.313)3 - (1.313) - 1 = 0.0494 (-ve)
X4
Now,
=
Therefore f(1. 313) is negative and f(1.375) is positive. Thus root lies between 1.313 and 1.375.
Now, fifth approximation is
then
X5
1.313 + 1.375
= 1.344
2
=
f(1.344) = (1.344)3 - (1.344) - 1 = 0.0837 (+ve)
Therefore root lies between 1.313 and 1.344.
So, the sixth approximation is given by
1.313 +1.344
= 1.329
2
x6 =
f(1.329) = (1.329)3 - (1.329) - 1 = - 0.0158 (-ve)
The root lies between 1.321 and 1.329
Example 5 :
Solution :
Thus, upto two decimal places the root of f(x) = x3
-
x - 1 = 0 is 1.32.
Find a positive real root of x - cos x = 0 by bisection method, correct upto 4 decimal place
between 0 and 1.
Suppose
f(x) =
X - COS X
f(0.73) = -ve
f(0.74) = +ve
Hence the root lies between 0.73 and 0.74
First approximation to the root is
X1
=
0.73 + 0.74 =
0.735
2
f(0.735) = -ve
So, the root lies between 0.735 and 0.74.
Second approximation to the root is
0.735 + 0.74 =
0.7375
2
f(0.7375) = -ve
X2
=
Hence root lies between 0.7375 and 0.74.
Third approximation to the root is
X3
=
0.7375 + 0.74
= 0.73875
2
f(0.73875) = -ve
So, the root lies between 0.73875 and 0.74.
Fourth approximation to the root is
0.73875 + 0.74 =
0.739375
2
f(x4) = f(0.739375) = +ve
X4
=
Hence the root lies between 0.73875 and 0.739375.
Eng i neering Mathematics
Fifth approximation to the root is
X5 =
0.73875 + 739375
= 0_7390625
t
x6 =
0.739062 + 0.739375
= 0_7392 1875
\
599
f(x5) = -ve
Now,
Hence the root lies between 0.7390625 and 0.739375.
Sixth approximation to the root is
f(0.73921875) = + ve
Hence the root lies between 0.7390625 and 0.73921875.
Seventh approximation to the root is
0.7390625 +0.73921875
= 0.73914
2
f(0.73914) = + ve
X7 =
Hence the root lies between 0.7390625 and 0.73914.
Eighth approximation to the root is
Example 6 :
Solution :
0.7390625 + 0.73914
= 0.73910
2
Hence the approximate real root correct upto 4 decimal places is 0.73910.
X8 =
Find a real root of the equation xlog 1 0 x = 1.2 by bisection method.
f(x) = xlog1 0 x - 1.2
f(1) = 1log 1 0 1 - 1.2 = - 1.2 < 0
f(2) = 2Iog 1 0 2 - 1.2 = 0.602 - 1.2 = - 0.598 < 0
f(3) = 3Iog 1 0 3 - 1.2 : 3
X
0.4771 - 1.2 : 1.4313 - 1.2 = 0.2313 > 0
Therefore, root will lie between 2 and 3.
First approximation to the root is
X1 =
2 3
; = 2.5
f(2.5) = 2.5Iog1 0 2.5 - 1.2 = 2.5 x 0.3979 - 1.2 = 0.9948 - 1.2 (-ve)
Therefore root lies between 2.5 and 3.
Second approximation to the root is
X2 =
2.5 3
/ = 2.75
f(2.75) = 2.75Iog 1 0 2.75 - 1.2 = 2.75 x 0.4393 - 1.2 = 1.2081 - 1.2 = 0.0081 (+ve)
So, the root lies between 2.5 and 2.75.
Third approximation to the root is
X3 =
2.5 2.75
= 2.625
�
f(2.625) = 2.625Iog 1 0 2.625 - 1.2 = 2.625 x 0.4191 - 1.2 = 1. 1001 - 1.2 (-ve)
So, the root lies between 2.625 and 2.75.
Fourth approximation to the root is
f(2.6875) = 2.6875Iog 1 0
2.625 + 2. 75
= 2 _ 6875
2
2.6875 - 1.2 = 2.6875 x 0.4293 - 1.2 = 1.1537 - 1.2 = - 0.0463 (-ve)
X4 =
Therefore, the root lies between 2.6875 and 2.75.
IES MASTER Publication
600
Numerical Solution of Algebraic a nd Transcendental Equations
Fifth approximation to the root is
2.6875 + 2.75
= 2.71 88
2
f(2.71 88) = 2.71 88Iog1 0 2.71 88 - 1 .2 = 2.7188 x 0.4344 - 1 .2 = 1 . 1 81 0 - 1 . 2 = - 0 . 019 (-ve)
X
5
=
So, the root lies between 2.7 1 88 and 2.75.
Sixth approximation to the root is
2.71 88 + 2.75
= 2.7344
2
f(2.734) = 2.734Iog1 0 2.734 - 1 .2 = 2.734 x 0.4368 - 1 .2 = 1 . 1 942 - 1 .2 = - 0.0058 (-ve)
X
6
=
So, the root lies between 2.734 and 2.75.
Seventh approximation to the root is
2.734 + 2.75
= 2 .742
2
f(2.742) = 2.742 log1 0 2.742 - 1.2 = 2.742 x 0.4381 - 1 .2 = 1 .201 2 - 1 .2 = 0.0012 (+ve)
X7 =
So, the root lies between 2.734 and 2.742.
Eighth approximation to the root is
2.734 + 2.742
= 2.738
2
f(2.738) = (-ve)
X8 =
Hence from the approximation value of the root, the root is 2.74.
REGULA-FALSI METHOD OR THE M ETHOD OF FALSE POSITION
(METHOD OF CHORDS)
This is the oldest numerical method for finding the real root of an equation f(x) = 0. It closely remsebles the bisection
method and is based on better logic. It is sometimes known as method of Linear Interpolation.
I n this method, we choose a sufficiently small interval x = a to x = b in which the root of f(x) = 0 lies.
Now, in this interval (a, b), the graph of the function y = f(x) must cross the x-axis between x = a and x = b, and hence
f(a) and f(b) are of opposite signs.
Now, the equation of the chord j oining the points A(a, f(a)) and B(b,f(b)) is
_ f (b) - f ( a) ( _ )
x a
y - f(a) b-a
This chord will pass through x-axis.
So, putting y = 0 in (1 ), the point of intersection of the chord with x axis is given by
f (b) - f (a)
0 - f(a) = --- ( x - a)
b-a
b-a
i.e.
f ( a)
x = af ( b) - f ( a)
Hence, the first approximation to the root is
Now, there are two possibilities-
b-a
( a)
X1 = a f (b) - f (a/
(i) When f(a) and f(x 1 ) are of opposite signs.
In this case, the root lies between a and x 1 .
So, we replace b by x 1 in (2) and compute the next approximation x2 .
...(1)
,
... (2)
Engineering Mathematics
(ii) When f(x 1 ) and f(b) are of opposite signs.
601
In this case, the root lies between x 1 and b. So, we replace a by x 1 in (2)' and compute the next approximation
X 2.
Similarly using formula'(2) iteratively, we can obtain a sequence (x 1 ,x2,x3, ... ) of successive approximations of the
root.
Order of Convergence of Regula-Falsi Method
It is also linearly convergent and order of convergence is 1.
Example 7 :
Solution :
Find a real root of the equation 3x + sin x - ex = 0 by the method of false position correct to four
decimal places.
f(x) = 3x + sin x - ex
f(0.3) = -0.154(- ve) {
f(0.4) =
0.0975
T he root lies between 0 .3 and 0 .4.
Now, the first approximation,
x2 = x 0 - (
x1 -x o
) f (x 0 ) =
f ( x 1 )-f (x0 )
0
f(x2) =
·
3 -(
(+ve)
Xo = 0.3
x1 = 0.4
0 .4- 0 .3
0 . 00975+0.154
0.0 0 19
(+ve)
) <-o · 154) =
Hence the root lies between 0 .3 and 0.3612
T he second approximation,
x3 = x 0 -(
x 2 - xo
f ( x0 ) =
f ( x2 )-f (x0 ) )
0 3612- 0 3
· ) <-0.154)
·
0 .3-(
0 .00 19+0.154
f(x3) = - 0.00005 (-ve)
Hence the root lies between 0 .3604 and 0.3612.
:. T he third approximation,
X2 - X 3
) f ( x3 ) =
X 4 = X 3 - ( f (x
2 )- f (x 3 )
0 .3+
0
· 1 x 0 · 154 = 0.3612
0.2515
0 0612
= 0 .3+( ·
) <o.154) = o · 3604
0.1559
0 .3612-0.3604
0 .3604-(
) (-0 .00005)
0 .00 19+0. 0000 5
= o.3604+( 0 - 0008 ) < 0.00005) = o.36042
f(x4) =
Example 8 :
Solution :
0 . 00 19
0 .0 0 195
(+ve)
Since x3 and x4 are approximately the same hence the required root is 0.3604.
Find a real root of the equation x log 1 0 x = 1.2 by Regula-Falsi method correct to four decimal places.
f(x) = log 10 x - 1.2
Let
f(2.74) = - 0.0 005634 and f(2.741) = 0.0003087
Hence the root lies between 2.74 and 2.741.
Using method of Regula-Falsi
x2
= x0 - (
x 1 -x0
f (x 0 )
f ( x 1 )- f (x 0 ) )
= 2.74(
2 - 741- 2 - 74
0.0003 0 87+0.000 5634
= 2.74+
0 .00 1
0.00 0 8721
) <-0.0 005634)
x 0 .0005634 = 2.740646027
f(�) = - 0.000000060 16 (-ve)
Hence the root lies between 2.740646027 and 2.741.
IES MASTER Publication
602
N u merical Solution of Algebraic and Tra nsce ndental Eq uations
o
)x(- .00000006016 ) = 2.740646096
Now x3 = 2.740646027-( o oo�;i;;::�������:
.
.
016
Example 9 :
Solution :
Hence, the required root correct to four decimal places is 2.7406.
Solve x3
-
5x + 3 = 0 by Regula-Falsi method.
f(x) = x3 - 5x + 3
f(0.65) = 0.024625 (+ve)
f(0.66) = - 0.012504 (-ve)
Hence the root lies between 0.65 and 0.66.
Let Xo = 0.65, X1 = 0.66
Using method of False position
x2 = x0 -(
x 0·66 - 0·65
f( x 0 ) = 0.65 ;
(-0.012504- 0.024625 ) <o.024625) = 0.656632282
f( X1)- (Xo ) )
f(x2) = - 0.00004392 (-ve)
Hence the root lies between 0.65 and 0.656632282.
Now
0 ·656632282-0 ·65
x(0.024625) = 0.656620474
x3 = 0.65 -(
-0.00004392- 0.024625 )
Hence the root is 0.6566 correct upto four decimal places.
Example 1 0 : Find the root of the equation xex = cos x in the interval (0, 1) using Regula-Falsi method correct to
four decimal places.
Solution :
f(x)
= cos x - xex = 0
f(0) = 1 (+ve)
f(1) = - 2.17798 (-ve)
So, the root lies between 0 and 1.
By Regula-Falsi method
X2 - Xo -
1-0
X1 -Xo
(1) - 0.31467
f ( xo ) = 0-3.17798
f( x1 )- f( xo )
f(�) = 0.51987 (+ve)
i.e. the root lies between 0.31487 and 1.
Now, again using Regula-Falsi method
X3 = 0.31487 -[
and
1-0 31487
·
](0.51987) = 0.44673
(-2.17798-0.51987)
f(x3) = 0.20356 (+ve)
The root of the equation corrected upto 4 decimal places is 0.5177. (after performing some more
iterations)
THE SECANT M ETHOD
This method is a n improvement over Regula-Falsi method. Here w e omit the condition f(a).f(b) < 0. Thus in this method
the graph of the function y = f(x) is approximated by a secant line (chord) in the neighbourhood of the root but the interval
at each iteration need not contain the root. If a and b are the initial limits of interval, then the first approximation is given
by
X 1 = a -(
(b -a)
) f(a)
f(b)-f(a)
Now we can directly assume that a E (x1, b) where a is exact root.
Engineering Mathematics
So, 2nd iteration can be given as
X2
�
603
X1 -( �=;\ ) f( X1 )
( 1)
f(
Now for 3rd iteration we can assume that a E (x2 , b) and so on.......
Obviously, the secant method fails at any iteration if we have f(x n) = f(xn_1 ). So, the method does not necessarily
converge. This is a drawback of secant method over the method of false position (Regula-Falsi method) which always
converges. But if the secant method once converges, its rate of convergence is 1 .62 times faster than that of the method
of false position.
Order of Convergence of Secant Method
Thus the secant method converges with order 1 .62 approximately.
If we have to perform only one iteration then Regula-Falsi method and secant method has no
difference.
While for more than one iterations process should be as follows.
for 2nd iteration a E (x1 , b) ,
For 3rd iteration a E (x2 , b) and so
For 1st iteration a E (a, b) ,
on......
Example 1 1 1 Compute the root of equation x2
Solution :
Suppose
a = 1 .42
-
e-)l)Z = 1 in the interval (0, 2) using the secant method.
and b = 1 .43 and f(x) = x2
f(a) = - 0.0086 f(b) = 0.00034
So, by secant method
-
e- )l)Z - 1
1 .43 - 1 .42
x 1 = a-[ ( � ; ] f ( a ) = 1 .42- [
] x (-0.0086) = 1 .4296
f b - ( a)
0.00034 + 0.0086
so f(x2 ) = -0.0000 1 1
Now x2 = x 1 -[
1 .43 -1 .4296
-x
�
] <-0.00001 1 ) = 1 .42961 25
] f ( x 1 ) = 1 .4296 - [
0.00034+0.00001 1
f ( b -f ( x1 )
Hence the required root upto three decimal places is 1 .429.
NEWTON-RAPHSON METHOD1 (METHOD OF TANGENTS)
This method can be derived from Taylor's series in the following way. Let x0 is the initial approximation of a real root
of the equation f(x) = 0. Let (x0 + h) be the exact value or the better approximation of the corresponding root, h being
a small quantity.
Thus
Expanding by Taylor's series (in LHS)
f(x0 + h) = 0.
f ( x0 )+% f' ( x0 ) + �� f " ( x0 )+... = O
Since h is small, neglecting h2 and higher powers of h, we get
f ( Xo ) + hf' ( Xo ) = 0
⇒
h = -
f (xo )
provided f'(x o ) ,1c O
f(xo )
Hence i f x0 b e the initial approximation, then the next (or first) approximation x 1 i s given by
_
f ( x0 )
X 1 = Xo + h - Xo -f'(x )
0
...( 1 )
...(2)
IES MASTER Publication
604
Numerical Solution of Algebraic and Transcendental Equations
Now, using x 1 in place of x0 , the next approximation is given by
In general, the (n + 1 )th iteration, is
f(x )
( X1 )
1
X 2 = X 1 - -f'
...(3)
This formula is known as Newton-Raphson's Formulae or iterative equation for Newton-Raphson Method.
Order of Convergence of Newton-Raphson M ethod
The error is roughly proportional to the square of the error in the previous iteration. Therefore, the Newton - Raphson
method is said to have quadratic convergence, i.e., order of convergence is 2.
----if
INOTE., 1.
2.
3.
4.
5.
6.
If f ' ( x ) is. large, i.e. the graph of y = f(x) is nearly vertical to the x-axis, then Newton's method
i� very u_seful.
This method is also used to obtain complex roots.
The convergence of Newton's method for double root is l inear.
This method will fail if the initial guess is too far away from the required root and then process
may converge to some other root.
Regula - Falsi is surely convergent while Newton's method is conditionally convergent. But once,
Newton's method converges, it converges faster .
When f'(x) is large, near the root, correction to be applied is smal ler in case of Newton's
method and this method is preferred.
When f'(x) is small, near the root, correction to be applied is large in case of Newton's method
and hence in this case Regula - Falsi method should be applied.
\
Example 1 2 : By Newton-Raphson method, find the root of x4 - x - 1 0 = 0 which is nearer to x = 2.
7.
Solution :
Here f(x) = x4
-
x - 10, so that f'(x) = 4x3 - 1 . Therefore
=
i.e.,
x n+ 1 =
3x� + 1 0
4x 3n - 1
X n ( 4x� - 1) - ( x� - X n - 1 0)
4xn3 - 1
...(1)
An approximate value of the root is given to be 2. Taking x0
So,
3Xri +
)
= 2, from (1) we get
X1
1 0 3(2 + 1 0
= 1 · 87 1
= 4x � - =
'
1
4 (23 ) - 1
X2
=
4
3x1 + 1 0 3 ( 1 .871 4 ) + 1 0
= 1 .856,
=
4x f - 1
4(1.8713 ) - 1
3Xi + 1 0 3 ( 1 .85 64 ) + 1 0
= 1 .856 _
=
and
4x� - 1
4(1 .8563 ) - 1
Since x2 = x3 , the required root is 1 .856 .
X3 =
Example 13 : Find the real root of 3x - cos x - 1 = 0 by Newton-Raphson method.
Solution :
Here f(x)
Now
= 3x - cos x - 1 , so that f'(x) = 3 + sin x.
f ( xn )
3x n - cos x n - 1
= X n - ��-�
X n - -f' ( x n )
3 + sin x n
Taking x0
=
0.6, we get
Engi neerin g Mathematics
X 1 = 0.6-
3(0.6 ) -cos0.6 - 1
=0.6071
3 + sin0.6
x2 = 0 _ 6071
Also,
Since
605
_ 3(0.6071) -cos 0.6071-1
=0_6071
. 3 + sin0.6071
x1 = x2 = 0.6071, it is the required real root.
Example 14 : Find a positive value of (17) 113 correct to four decimal places by Ne�on-Raphson method.
Solution :
=
Assuming f(x)
x3 - 17
-;-J
The iterative formula is xn+1 = .!. ( 2xn +
3
Xn
Here a = 17
x0
=
2.5 because
X1
=
V8 =2 and 'ifn = 3
�J
.!. ( 2 X 2.5 +
2
3
(2.5)
= .!. ( 5 +____!2__ ) = 2.5733
3
6.25
x2 = J ( 2 X 2.5733 + _ �: 2
(2 5 3)
X3
=
X4 =
17
.!. ( 2x 2.5713 +
2
3
(2.5713 )
J
J
1
17
- (2x2.57128+
2
3
(2.57128)
=
17
.!. 5. 1466 + ---) =2.5713
3(
6.6220
17
= .!. ( 5. 1426 +
) = 2.57128
6.61158
3
J
=
17
1
- ( 5.14256 +---) = 2.57128
3
6.61148
Thus the required root is 2.5712 correct to four decimal places.
Example 1 5 : Evaluate ✓12 to four decimal places by Newton's Raphson method.
Solution :
Assuming f(x)
=
x2
-
12,
The iterative formula is
�J
- .!. ( x +
xn+1 2 n Xn
Now since f(3) = - 1(-ve) and f(4) = 4(+ve)
Root lies between 3 and 4.
Now take
X0 =
X1
X2
3.5
Here a = 12
_g_)
= 3.4643
= .!. ( 3.5 +
2
3.5
12
= .!. ( 3.4643 + ---) = 3.4641
2
3.4673
12
.!. 3.4641+ -- =3.4641
2(
3.4641)
X3 =
So ✓12 = 3.4641 correct to 4 decimal places.
Example 1 6 : Find a real root of the equation x
Solution :
or
x
xex
f'(x)
=
=
=
=
e-x using Newton's method.
e-x
1 so f(x) = xex - 1
(1 + x)ex
IES MASTER Publication
606
Numerical Solution of Algebraic and Transcendental Equations
Let
x0 = 1
X2
Similarly
e
1
x1 = 1-( � ) = ( 1+ �) = 0.6839397
2
i
X3
X4
So, the required root is 0.5671.
0 3553424
= 0.5774545
= 0.6839397 - ·
3.337012
= 0.5672297
= 0.5671 433
Engineerin g Mathematics
607
Questions marked with aesterisk (*) are Conceptual Questions
1.
2.
Let f(x) = x - cos x. Using Newton-Raphson method
at the (n + 1 )lh iteration , the point xn+1 is computed
from xn as?
(b) only if f is a polynom ial
(c) only if f(x0) < 0
[GATE-1 995)
The iteration form ula to· find the square root of a
positive real n u m ber b by using the N ewto n­
Raphson m ethod is
(a) X k+ 1 =
3.*
(a) always
(c) X k+1 =
3(xk + b)
2 Xk
(b) X k+1 =
xk - 2xk-1
(d) none
2� + b
x� + b
2Xk
[GATE-1 995 (CS)]
I n the i nte rv al [0, n ] the equation x = cos x has
(a) No solution
(b) exactly one solution
(c) exactly two solution
(d) none of the above
8.*
[GATE-1 999 (CS)]
E st i m ate
the
root
of
the
equation
(e-x - x) = 0 to four decimal accuracy by employing
the Newton-Raphson method starti ng with an initial
guess of x0 = 0.
[GATE-2000; 5 Marks]
9.* A piecewise l i near function f(x) is plotted using
thick sol id l i nes i n the figure (the plot is drawn to
scale).
f(x)
1 .0
(2.05, 1 .0)
a
(d) an infinite num ber of solutions
4.
[GATE-1 995 (CS)]
T h e N ewto n - R a p hson i t e rat i o n form u l a f o r
finding 'efc, where c > 0 is,
(a) x n+ 1 =
2x� + 'efc
(b) X n+ 1 = 2x� + 'efc
3·X 2n
3xn2
(c) Xn+1 = 2 X � +
3 x 2n
2x n3 - c
(d) X n+1 = �
C
[GATE-1 996 (CS)]
5.* The Newton-Raphson method is used to find the
root of the equation x 2 - 2. If the iterations are
started from - 1 , then the iteration will
(a) converge to - 1
(c) converge to 6.
✓2
(d) not converge
[GATE-1997 (CS)]
W e wish to solve x - 2 = 0 b y Newton Raphson
t ec h n i q u e .
Let
the
i n itial
g u ess
be
x0 = 1 .0 subsequent estimate of x (i.e. x 1 ) will be:
(c) 2.0
7.*
✓2
(b) converge to
2
(a) 1 .414
2.05
(b) 1.5
(d) None of the above
[GATE-1 999)
The Newton-Raphson method is to be used to find
the root of the equation and f'(x) is the derivative
of f. The m ethod converges
If we use the Newton-Raphson m ethod to find the
roots of f(x) = 0 using x0, x 1 and x2 respectively
as initial guesses, the mots obtai ned woul d be
(a) 1 . 3 , 0.6 and 0. 6 respectively
(b) 0.6, 0.6 and 1 . 3 respectively
(c) 1 . 3 , 1 . 3 and 0.6 respectively
(d) 1 . 3 , 0.6 and 1 . 3 respectively
[CS, GATE-2003, 2 Marks]
10. Starti ng from x0 = 1 , one step of Newton-Raphson
m et h o d
in
so l v i n g
the
equation
x3 + 3x - 7 = 0 gives the next value (x 1 ) as
(a) x 1 = 0.5
(c) x 1 = 1 .5
(b) x1 = 1 .406
(d) x1 = 2
[GATE-2005-ME]
IES MASTER Publication
Numerical Solution of Algebraic and Transcendental Equations
608
Statements for Linked Answer Questions 11 and 12
Given a > 0, we wish to calculate its reciprocal value 1 /
a by u sing Newton Raphson method for
f(x) = 0.
11. The Newton-Raphson algorithm for the function will
X +
x = x
x�
=_!_(
2 k Xk ) (b) k + 1 ( k + % )
_3_
(a)
Xk + 1
(c)
X k + 1 = 2X k
- ax�
(d)
a
2
X k + 1 = X k - 2 Xk
[GATE-2005-CE]
12. For a = 7 and starting with x0 = 0.2, the first two
iterations will be
(a) 0. 1 1 , 0. 1 299
(c) 0.12, 0. 1 4 1 6
(b) 0.1 2, 0. 1 392
(d) 0. 1 3, 0. 1 428
[GATE-2005-CE]
13. For solving algebraic and transcendental equation
which one of the following is used?
(a) Coulomb's theorem
(b) Newton-Raphson method
(c) Euler's method
(d) Stoke's theorem
[GATE-2005 (Pl)]
14. Newton-Raphson formula to find the roots of an
equation f(x) = 0 is· given by
(a) xn+1 =
(b)
X n+1
=
(c)
X n+ 1
=
(a)
2x� + 9
X k+1 _
3X� + 4
2x� + 4
(b) Xk+1 _
3X� + 9
(c)
Xk +1 = Xk - 3Xk2 + 4
(d) Xk+1 -
17. Identify the Newton-Raphson iteration scheme for
finding the square root of 2
( xn + X )
(a) x n+ 1 = _!_
2
n
_3_
(b)
X n+1
xn _ _3_)
= !
Xn
2(
(c)
X n+ 1
=
. (d)
15. The equation x3 - x2 + 4x - 4 = 0 is to be solved
u sing the Newton-Raphson method. I f
x = 2 is taken a s the initial approximation of the
solution, then the next approximation using this
method will be
2
3
4
3
3
(c) 1
[GATE-2007-EC]
(d)
2
16. The following equation needs to be numerically solved
using the Newton-Raphson method
x3 + 4x - 9 = 0
(b)
X n+ 1
2
Xn
✓
[GATE-2007 (IN)]
- 2 + xn
Xn
9
18. Consider the series x n.1 = 2 + B x , x 0 = 0.5
obtained from the Newton-Raphson method. The
series converges to
(b)
(a) 1 .5
.J2
(d) 1 .4
(c) 1 .6
[CS, GATE-2007, 2 Marks]
19. The recursion relation to solve x = e - x using Newton­
Raphson method is
(b)
f ( xn )
f ' ( Xn )
4x + 3
�
9X� + 2
[GATE-2007-CE]
(a)
[GATE-2005 (Pl)]
(a)
The I terative equ ation for this purpose is
(k indicates the iteration level)
(d)
X n+ 1
X n+1
= e-Xn
= X n - e- xn
2
_ X - e-Xn (1 + X n ) - 1
X n+1 - n
-x
Xn - e n
[GATE-2008-EC]
20.* Equation
1 = 0 is required to be solved using
Newton's method with an initial guess x0 = -1 . Then,
after one step of Newton's method, estimate x 1 of
the solution will be given by
ex -
(a) 0.71 828
(c) 0.20587
(b) 0.36784
(d) 0.00000
21. The Newton-Raphson iteration
x0•1 =
J(
[GATE-2008-EE]
xn + : ) can be used to compute the
Engineering Mathematics
(a) square of R
(b) reciprocal of R
(c) square root of R (d) logarithm of R
[GATE-2008 (CS)]
22. Let x - 1 1 7 = 0. The iterative steps for the solution
using Newton-Raphson's method is given by
2
(a)
(b)
(c)
(d)
Xk +1
X k +1
117
=!
xk +
2(
Xk J
=X
k
xk +1 = Xk -
X k +1
=X
117
-X
Xk
1 17
_ x +
__!_ ( k
k
2
Xk ]
(a) 0.865
(b) '0.853
(c) 0.849
(d) 0.838
x
[GATE-2009-EE]
[GATE-2009 (IN)]
24. Newton-Raphson method is used to compute a
root of the equation x 2 - 1 3 = 0 with 3.5 as the
initial value. The approximation after one iteration
is
(a) 3.575
(b) 3.677
(c) 3.667
(d) 3.607
[GATE-201 0 (CS)]
25.* The square root of a number N is to be obtained by
applying the Newton-Raphson iterations to the
equation x2 - N = 0. If i denotes the iteration index,
the correct iterative scheme will be
(c)
�J
xi+1 = _:l_ ( xi +
2
�
xi+1 = _:l_ ( xF +
2
N2
�
J
N
1
(b) xi+1 = _ ( xF + 2
2
�
J
_ti_]
(d) xi+1 = _:l_ ( xi _
2
�
[GATE-2011 -CE]
26. A num erical solution of the equation f(x) =
x +./x -3 = 0 can be obtained using Newton­
Raphson method. If the starting value is x = 2 for the
iteration, the value of x that is to be used in the next
step is
(a) 0.306
(c)
1.694
equation (i) 1 0x2 sinx 1
(b) 0.739
(d) 2.306
[GATE-2011 -EC]
-
0.8 = 0
equation (ii) 10x� - 1 0x2 cos x1 - 0.6 = 0
=�::]
Assum ing the initial values x 1 = 0 . 0 and
x2 = 1 .0, the Jacobian matrix is
[
1
1 17
23.* The real root of the equation xe = 2 is evaluated
using Newton- Rap hson's method. If the first
approximation of the value of x is 0.867 6, the 2nd
approximation of the value of x correct to three
decimal places is
(a)
27.** Solution of the variables x 1 and x2 for the following
equations is to be obtained by employing the Newton­
Raphson iterative method
(a)
k
609
0
0
1
(b) [
i
-0.8
]
-0.6
(d) [� �
1�]
-� o]
[GATE-2011 EE]
28. The bisection method is applied to compute a zero
of the function f(x) = x4 - x3 - x2 - 4 in the interval
[ 1 , 9]. The method converges to a solution after
___ _ iterations.
(a)
(b) 3
(c) 5
(d)
7
[CS, GATE-201 2, 2 Marks]
29. When the Newton-Raphson methodis applied to solve
the equation f(x) = x3 + 2x - 1 = 0, the solution at
the end of the first iteration with the initial guess
value as x0 = 1 .2 is
(a) - 0.82
(b) 0.49·
(c) 0.705
(d)
1 .69
[GATE-2013]
30.* T h e
real
root
of
the
equation
5x - 2cosx -1 = 0 (up to two decimal accuracy) is
[GATE-2014-ME-Set-3]
31. The function f(x) = ex - 1 is to be solved using
Newton-Raphson method. If the initial value of x0 is
taken as 1 .0, then the absolute error observed at 2nd
iteration is ----
[GATE-2014-EE-SET -3]
32. The iteration step in order to solve for the cube
roots of a given number 'N' using the Newton­
Raphson's method is
1
3
(a) X k+ 1 = xk + (N - xk )
3
(b)
__!_ 2xk + � )
3(
Xk
1
3
(c) Xk+1 = X k -- (N- xk )
3
(d) xk+ 1 = { ( 2xk - �)
[GATE-201 4 (IN-set 1 )1
IES MASTER Publication
Nu merical Solution of Algebraic and Tran scendental Equations
61 0
33. If the equation sin (x) = x2 is solved by Newton­
Raphson's m ethod with the i nitial guess of x = 1 ,
then the value of x after 2 iterations would be
[GATE-1 4 (Pl-Set 1 )]
34.* In the Newton-Raphson method, an initial g uess of
x 0 = 2 is made and the seq uence x0 , x 1 , x2 , . . . . i s
obtai ned for t h e function
0.75x
3
-
2x
2
-
2x + 4 = 0
Consider the statements
x3 = 0
(II) The method converges to a solution i n a finite
number of iterations
(I)
W h ich of the fol l owi ng is TRUE?
(a) only 1
(c) Both I and I I
(b) only I I
(d) Neither I nor I I
[GATE-1 4 (CS-Set 2)]
35. The Newton-Raphson method is used to solve the
equation f(x) = x3 - 5x2 + 6x - 8 = 0. Taki ng the
initial guess as x = 5, the solution obtained at the
end of the fi rst iteration is----
36.
[GATE-201 5-ME-Set-2]
Newton-Rapshon method is used to find the roots of
the equation, x3 + 2x2 + 3x - 1 = 0. If the initial
guess is x0 = 1 , then the value of x after 2 nd iteration
is ___
[GATE-2015-ME-Set-3]
37.* In Newton Raphson iterative method, the initial guess
value (Xini ) is considered as zero, whi le finding the
roots of the equation f(x) = -2 + 6x - 4x2 + 0.5x 3 .
The correction rue , to be added to Xini in the first
iteration is ____
[GATE-201 5-CE-SET-11]
38.* The quadratic equation x - 4x + 4 = 0 is the solved
numerical ly, starting with the initial guess x0 = 3.
The Newton-Raphson method is applied once to get
a new estimate and then the Secant method is applied
once using the initial guess and this new estimate.
The estimated value of the root after the appl ication
of the Secant method is ---2
[GATE-2015-CE-SET-I]
39. The Newton-Raphson method is used to solve the
equation f(x) = x 3 - 5x2 + 6x - 8 = 0. Taking the
i n itial g uess as x = 5 , the solution obtained at the
end of the fi rst iterations i s___
[GATE-201 5 (EC-Set 3)]
40.* How many distinct values of x satisfy the equation
sin(x) = x/2, where x is in radians?
(a)
1
(c) 3
(b) 2
(d) 4 or more
[GATE-2016; 1 Mark]
41 . Solve the equation x = 1 0 cos(x) using the Newton­
Raphson method. The i nitial guess is x = n/4 . The
value of the predicted root after the first iteration, up
to second deci mal , is _ _ _
[GATE 2016-ME-Set-1]
42. The root of the function f(x) = x3 + x - 1 obtai ned
after first iteration on appl ication of Newton Raphson
scheme using an initial guess of x0 = 1 is
(a) 0.682
(c) 0.750
(b) 0.686
(d) 1 .000
[GATE-2016-ME-Set-1]
43. Newton-Raphson method is to be used to find root of
equation 3x - ex + sin x = 0. If the initial trial value
for the root is taken as 0.333, the next approximation
for the root would be____ (note : answer up to
three decimal)
[GATE-2016-CE-Set-I]
44. To solve the equation 2 sin x = x by Newton­
Raphson m ethod , the i n itial guess was chosen to
be x = 2 . 0 . Consider x in rad ian only. The value
of x (in rad ian) obtai ned after one iteration wi ll be
closest to
(a) -8. 1 01
(c) 2.099
(b) 1 .901
(d) 1 2. 1 01
[GATE-2016 (Pl)]
45.* Only one of the real roots of f(x) = x6 - x - 1 lies i n
the i nterval 1 :-::: x :-::: 2 a n d bisection method is used
to find its val ue. For achieving an accuracy of 0.001 ,
the required minimum number of iterations