1. HIGHER ENGINEERING MATHEMATICS

2. In memory of Elizabeth 3. Higher Engineering MathematicsFifth EditionJohn Bird, BSc(Hons), CMath, FIMA, FIET, CEng, MIEE, CSci, FCollP, FIIEAMSTERDAM BOSTON HEIDELBERG LONDON NEW YORK OXFORDPARIS SAN DIEGO SAN FRANCISCO SINGAPORE SYDNEY TOKYONewnes is an imprint of Elsevier 4. NewnesAn imprint of ElsevierLinacre House, Jordan Hill, Oxford OX2 8DP30 Corporate Drive, Suite 400, Burlington, MA01803, USAFirst published 1993Second edition 1995Third edition 1999Reprinted 2000 (twice), 2001, 2002, 2003Fourth edition 2004Fifth edition 2006Copyright c 2006, John Bird. Published by Elsevier Ltd. 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Because of rapid advances in the medicalsciences, in particular, independent verification of diagnoses and drug dosagesshould be madeBritish Library Cataloguing in Publication DataA catalogue record for this book is available from the British LibraryLibrary of Congress Cataloging-in-Publication DataA catalog record for this book is available from the Library of CongressISBN 13: 9-78-0-75-068152-0ISBN 10: 0-75-068152-7For information on all Newnes publicationsvisit our website at books.elsevier.comTypeset by Charon Tec Ltd, Chennai, Indiawww.charontec.comPrinted and bound in Great Britain06 07 08 09 10 10 9 8 7 6 5 4 3 2 1 5. ContentsPreface xvSyllabus guidance xviiSection A: Number and Algebra 11 Algebra 11.1 Introduction 11.2 Revision of basic laws 11.3 Revision of equations 31.4 Polynomial division 61.5 The factor theorem 81.6 The remainder theorem 102 Inequalities 122.1 Introduction to inequalities 122.2 Simple inequalities 122.3 Inequalities involving a modulus 132.4 Inequalities involving quotients 142.5 Inequalities involving squarefunctions 152.6 Quadratic inequalities 163 Partial fractions 183.1 Introduction to partial fractions 183.2 Worked problems on partial fractionswith linear factors 183.3 Worked problems on partial fractionswith repeated linear factors 213.4 Worked problems on partial fractionswith quadratic factors 224 Logarithms and exponential functions 244.1 Introduction to logarithms 244.2 Laws of logarithms 244.3 Indicial equations 264.4 Graphs of logarithmic functions 274.5 The exponential function 284.6 The power series for ex 294.7 Graphs of exponential functions 314.8 Napierian logarithms 334.9 Laws of growth and decay 354.10 Reduction of exponential laws tolinear form 385 Hyperbolic functions 415.1 Introduction to hyperbolic functions 415.2 Graphs of hyperbolic functions 435.3 Hyperbolic identities 445.4 Solving equations involvinghyperbolic functions 475.5 Series expansions for cosh x andsinh x 48Assignment 1 506 Arithmetic and geometric progressions 516.1 Arithmetic progressions 516.2 Worked problems on arithmeticprogressions 516.3 Further worked problems onarithmetic progressions 526.4 Geometric progressions 546.5 Worked problems on geometricprogressions 556.6 Further worked problems ongeometric progressions 567 The binomial series 587.1 Pascals triangle 587.2 The binomial series 597.3 Worked problems on the binomialseries 597.4 Further worked problems on thebinomial series 617.5 Practical problems involving thebinomial theorem 648 Maclaurins series 678.1 Introduction 678.2 Derivation of Maclaurins theorem 678.3 Conditions of Maclaurins series 678.4 Worked problems on Maclaurinsseries 688.5 Numerical integration usingMaclaurins series 718.6 Limiting values 72Assignment 2 75 6. vi CONTENTS9 Solving equations by iterative methods 769.1 Introduction to iterative methods 769.2 The bisection method 769.3 An algebraic method of successiveapproximations 809.4 The Newton-Raphson method 8310 Computer numbering systems 8610.1 Binary numbers 8610.2 Conversion of binary to denary 8610.3 Conversion of denary to binary 8710.4 Conversion of denary to binaryvia octal 8810.5 Hexadecimal numbers 9011 Boolean algebra and logic circuits 9411.1 Boolean algebra and switchingcircuits 9411.2 Simplifying Boolean expressions 9911.3 Laws and rules of Boolean algebra 9911.4 De Morgans laws 10111.5 Karnaugh maps 10211.6 Logic circuits 10611.7 Universal logic gates 110Assignment 3 114Section B: Geometry andtrigonometry 11512 Introduction to trigonometry 11512.1 Trigonometry 11512.2 The theorem of Pythagoras 11512.3 Trigonometric ratios of acuteangles 11612.4 Solution of right-angled triangles 11812.5 Angles of elevation and depression 11912.6 Evaluating trigonometric ratios 12112.7 Sine and cosine rules 12412.8 Area of any triangle 12512.9 Worked problems on the solutionof triangles and finding their areas 12512.10 Further worked problems onsolving triangles and findingtheir areas 12612.11 Practical situations involvingtrigonometry 12812.12 Further practical situationsinvolving trigonometry 13013 Cartesian and polar co-ordinates 13313.1 Introduction 13313.2 Changing from Cartesian into polarco-ordinates 13313.3 Changing from polar into Cartesianco-ordinates 13513.4 Use of RP and PR functionson calculators 13614 The circle and its properties 13714.1 Introduction 13714.2 Properties of circles 13714.3 Arc length and area of a sector 13814.4 Worked problems on arc length andsector of a circle 13914.5 The equation of a circle 14014.6 Linear and angular velocity 14214.7 Centripetal force 144Assignment 4 14615 Trigonometric waveforms 14815.1 Graphs of trigonometric functions 14815.2 Angles of any magnitude 14815.3 The production of a sine andcosine wave 15115.4 Sine and cosine curves 15215.5 Sinusoidal form A sin (t ) 15715.6 Harmonic synthesis with complexwaveforms 16016 Trigonometric identities and equations 16616.1 Trigonometric identities 16616.2 Worked problems on trigonometricidentities 16616.3 Trigonometric equations 16716.4 Worked problems (i) ontrigonometric equations 16816.5 Worked problems (ii) ontrigonometric equations 16916.6 Worked problems (iii) ontrigonometric equations 17016.7 Worked problems (iv) ontrigonometric equations 17117 The relationship between trigonometric andhyperbolic functions 17317.1 The relationship betweentrigonometric and hyperbolicfunctions 17317.2 Hyperbolic identities 174 7. CONTENTS vii18 Compound angles 17618.1 Compound angle formulae 17618.2 Conversion of a sin t +b cos tinto R sin(t +) 17818.3 Double angles 18218.4 Changing products of sines andcosines into sums or differences 18318.5 Changing sums or differences ofsines and cosines into products 18418.6 Power waveforms in a.c. circuits 185Assignment 5 189Section C: Graphs 19119 Functions and their curves 19119.1 Standard curves 19119.2 Simple transformations 19419.3 Periodic functions 19919.4 Continuous and discontinuousfunctions 19919.5 Even and odd functions 19919.6 Inverse functions 20119.7 Asymptotes 20319.8 Brief guide to curve sketching 20919.9 Worked problems on curvesketching 21020 Irregular areas, volumes and mean values ofwaveforms 21620.1 Areas of irregular figures 21620.2 Volumes of irregular solids 21820.3 The mean or average value ofa waveform 219Section D: Vector geometry 22521 Vectors, phasors and the combination ofwaveforms 22521.1 Introduction 22521.2 Vector addition 22521.3 Resolution of vectors 22721.4 Vector subtraction 22921.5 Relative velocity 23121.6 Combination of two periodicfunctions 23222 Scalar and vector products 23722.1 The unit triad 23722.2 The scalar product of two vectors 23822.3 Vector products 24122.4 Vector equation of a line 245Assignment 6 247Section E: Complex numbers 24923 Complex numbers 24923.1 Cartesian complex numbers 24923.2 The Argand diagram 25023.3 Addition and subtraction of complexnumbers 25023.4 Multiplication and division ofcomplex numbers 25123.5 Complex equations 25323.6 The polar form of a complexnumber 25423.7 Multiplication and division in polarform 25623.8 Applications of complex numbers 25724 De Moivres theorem 26124.1 Introduction 26124.2 Powers of complex numbers 26124.3 Roots of complex numbers 26224.4 The exponential form of a complexnumber 264Section F: Matrices andDeterminants 26725 The theory of matrices anddeterminants 26725.1 Matrix notation 26725.2 Addition, subtraction andmultiplication of matrices 26725.3 The unit matrix 27125.4 The determinant of a 2 by 2 matrix 27125.5 The inverse or reciprocal of a 2 by2 matrix 27225.6 The determinant of a 3 by 3 matrix 27325.7 The inverse or reciprocal of a 3 by3 matrix 27426 The solution of simultaneous equations bymatrices and determinants 27726.1 Solution of simultaneous equationsby matrices 27726.2 Solution of simultaneous equationsby determinants 279 8. viii CONTENTS26.3 Solution of simultaneous equationsusing Cramers rule 28326.4 Solution of simultaneous equationsusing the Gaussian eliminationmethod 284Assignment 7 286Section G: Differential calculus 28727 Methods of differentiation 28727.1 The gradient of a curve 28727.2 Differentiation from first principles 28827.3 Differentiation of commonfunctions 28827.4 Differentiation of a product 29227.5 Differentiation of a quotient 29327.6 Function of a function 29527.7 Successive differentiation 29628 Some applications of differentiation 29828.1 Rates of change 29828.2 Velocity and acceleration 29928.3 Turning points 30228.4 Practical problems involvingmaximum and minimum values 30628.5 Tangents and normals 31028.6 Small changes 31129 Differentiation of parametricequations 31429.1 Introduction to parametricequations 31429.2 Some common parametricequations 31429.3 Differentiation in parameters 31429.4 Further worked problems ondifferentiation of parametricequations 31630 Differentiation of implicit functions 31930.1 Implicit functions 31930.2 Differentiating implicit functions 31930.3 Differentiating implicit functionscontaining products and quotients 32030.4 Further implicit differentiation 32131 Logarithmic differentiation 32431.1 Introduction to logarithmicdifferentiation 32431.2 Laws of logarithms 32431.3 Differentiation of logarithmicfunctions 32431.4 Differentiation of [ f (x)]x 327Assignment 8 32932 Differentiation of hyperbolic functions 33032.1 Standard differential coefficients ofhyperbolic functions 33032.2 Further worked problems ondifferentiation of hyperbolicfunctions 33133 Differentiation of inverse trigonometric andhyperbolic functions 33233.1 Inverse functions 33233.2 Differentiation of inversetrigonometric functions 33233.3 Logarithmic forms of the inversehyperbolic functions 33733.4 Differentiation of inverse hyperbolicfunctions 33834 Partial differentiation 34334.1 Introduction to partialderivaties 34334.2 First order partial derivatives 34334.3 Second order partial derivatives 34635 Total differential, rates of change andsmall changes 34935.1 Total differential 34935.2 Rates of change 35035.3 Small changes 35236 Maxima, minima and saddle points forfunctions of two variables 35536.1 Functions of two independentvariables 35536.2 Maxima, minima and saddle points 35536.3 Procedure to determine maxima,minima and saddle points forfunctions of two variables 35636.4 Worked problems on maxima,minima and saddle points forfunctions of two variables 35736.5 Further worked problems onmaxima, minima and saddle pointsfor functions of two variables 359Assignment 9 365 9. CONTENTS ixSection H: Integral calculus 36737 Standard integration 36737.1 The process of integration 36737.2 The general solution of integrals ofthe form axn 36737.3 Standard integrals 36737.4 Definite integrals 37138 Some applications of integration 37438.1 Introduction 37438.2 Areas under and between curves 37438.3 Mean and r.m.s. values 37638.4 Volumes of solids of revolution 37738.5 Centroids 37838.6 Theorem of Pappus 38038.7 Second moments of area of regularsections 38239 Integration using algebraicsubstitutions 39139.1 Introduction 39139.2 Algebraic substitutions 39139.3 Worked problems on integrationusing algebraic substitutions 39139.4 Further worked problems onintegration using algebraicsubstitutions 39339.5 Change of limits 393Assignment 10 39640 Integration using trigonometric andhyperbolic substitutions 39740.1 Introduction 39740.2 Worked problems on integration ofsin2 x, cos2 x, tan2 x and cot2 x 39740.3 Worked problems on powers ofsines and cosines 39940.4 Worked problems on integration ofproducts of sines and cosines 40040.5 Worked problems on integrationusing the sin substitution 40140.6 Worked problems on integrationusing tan substitution 40340.7 Worked problems on integrationusing the sinh substitution 40340.8 Worked problems on integrationusing the cosh substitution 40541 Integration using partial fractions 40841.1 Introduction 40841.2 Worked problems on integration usingpartial fractions with linear factors 40841.3 Worked problems on integrationusing partial fractions with repeatedlinear factors 40941.4 Worked problems on integrationusing partial fractions with quadraticfactors 41042 The t =tan2substitution 41342.1 Introduction 41342.2 Worked problems on the t =tan2substitution 41342.3 Further worked problems on thet = tan2substitution 415Assignment 11 41743 Integration by parts 41843.1 Introduction 41843.2 Worked problems on integrationby parts 41843.3 Further worked problems onintegration by parts 42044 Reduction formulae 42444.1 Introduction 42444.2 Using reduction formulae forintegrals of the formxnexdx 42444.3 Using reduction formulae forintegrals of the formxncos x dx and xn sin x dx 42544.4 Using reduction formulae forintegrals of the formsinn x dx and cosn x dx 42744.5 Further reduction formulae 43045 Numerical integration 43345.1 Introduction 43345.2 The trapezoidal rule 43345.3 The mid-ordinate rule 43545.4 Simpsons rule 437Assignment 12 441 10. x CONTENTSSection I: Differentialequations 44346 Solution of first order differential equationsby separation of variables 44346.1 Family of curves 44346.2 Differential equations 44446.3 The solution of equations of the formdydx= f (x) 44446.4 The solution of equations of the formdydx= f (y) 44646.5 The solution of equations of the formdydx= f (x) f (y) 44847 Homogeneous first order differentialequations 45147.1 Introduction 45147.2 Procedure to solve differentialequations of the form Pdydx= Q 45147.3 Worked problems on homogeneousfirst order differential equations 45147.4 Further worked problems onhomogeneous first order differentialequations 45248 Linear first order differentialequations 45548.1 Introduction 45548.2 Procedure to solve differentialequations of the formdydx+Py=Q 45548.3 Worked problems on linear firstorder differential equations 45648.4 Further worked problems on linearfirst order differential equations 45749 Numerical methods for first orderdifferential equations 46049.1 Introduction 46049.2 Eulers method 46049.3 Worked problems on Eulersmethod 46149.4 An improved Euler method 46549.5 The Runge-Kutta method 469Assignment 13 47450 Second order differential equations of theform ad2ydx2+bdydx+cy=0 47550.1 Introduction 47550.2 Procedure to solve differentialequations of the formad2ydx2+bdydx+cy=0 47550.3 Worked problems on differentialequations of the formad2ydx2+bdydx+cy=0 47650.4 Further worked problems on practicaldifferential equations of the formad2ydx2+bdydx+cy=0 47851 Second order differential equations of theform ad2ydx2+bdydx+cy=f (x) 48151.1 Complementary function andparticular integral 48151.2 Procedure to solve differentialequations of the formad2ydx2+bdydx+cy=f (x) 48151.3 Worked problems on differentialequations of the form ad2ydx2+bdydx+cy=f (x) where f (x) is a constant orpolynomial 48251.4 Worked problems on differentialequations of the form ad2ydx2+bdydx+cy=f (x) where f (x) is anexponential function 48451.5 Worked problems on differentialequations of the form ad2ydx2+bdydx+cy=f (x) where f (x) is a sine orcosine function 48651.6 Worked problems on differentialequations of the form ad2ydx2+bdydx+cy=f (x) where f (x) is a sum ora product 488 11. CONTENTS xi52 Power series methods of solving ordinarydifferential equations 49152.1 Introduction 49152.2 Higher order differential coefficientsas series 49152.3 Leibnizs theorem 49352.4 Power series solution by theLeibnizMaclaurin method 49552.5 Power series solution by theFrobenius method 49852.6 Bessels equation and Besselsfunctions 50452.7 Legendres equation and Legendrepolynomials 50953 An introduction to partial differentialequations 51253.1 Introduction 51253.2 Partial integration 51253.3 Solution of partial differentialequations by direct partialintegration 51353.4 Some important engineering partialdifferential equations 51553.5 Separating the variables 51553.6 The wave equation 51653.7 The heat conduction equation 52053.8 Laplaces equation 522Assignment 14 525Section J: Statistics andprobability 52754 Presentation of statistical data 52754.1 Some statistical terminology 52754.2 Presentation of ungrouped data 52854.3 Presentation of grouped data 53255 Measures of central tendency anddispersion 53855.1 Measures of central tendency 53855.2 Mean, median and mode fordiscrete data 53855.3 Mean, median and mode forgrouped data 53955.4 Standard deviation 54155.5 Quartiles, deciles and percentiles 54356 Probability 54556.1 Introduction to probability 54556.2 Laws of probability 54556.3 Worked problems on probability 54656.4 Further worked problems onprobability 548Assignment 15 55157 The binomial and Poisson distributions 55357.1 The binomial distribution 55357.2 The Poisson distribution 55658 The normal distribution 55958.1 Introduction to the normaldistribution 55958.2 Testing for a normal distribution 56359 Linear correlation 56759.1 Introduction to linear correlation 56759.2 The product-moment formula fordetermining the linear correlationcoefficient 56759.3 The significance of a coefficient ofcorrelation 56859.4 Worked problems on linearcorrelation 56860 Linear regression 57160.1 Introduction to linear regression 57160.2 The least-squares regression lines 57160.3 Worked problems on linearregression 572Assignment 16 57661 Sampling and estimation theories 57761.1 Introduction 57761.2 Sampling distributions 57761.3 The sampling distribution of themeans 57761.4 The estimation of populationparameters based on a largesample size 58161.5 Estimating the mean of apopulation based on a smallsample size 58662 Significance testing 59062.1 Hypotheses 59062.2 Type I and Type II errors 590 12. xii CONTENTS62.3 Significance tests for populationmeans 59762.4 Comparing two sample means 60263 Chi-square and distribution-free tests 60763.1 Chi-square values 60763.2 Fitting data to theoreticaldistributions 60863.3 Introduction to distribution-freetests 61363.4 The sign test 61463.5 Wilcoxon signed-rank test 61663.6 The Mann-Whitney test 620Assignment 17 625Section K: Laplace transforms 62764 Introduction to Laplace transforms 62764.1 Introduction 62764.2 Definition of a Laplace transform 62764.3 Linearity property of the Laplacetransform 62764.4 Laplace transforms of elementaryfunctions 62764.5 Worked problems on standardLaplace transforms 62965 Properties of Laplace transforms 63265.1 The Laplace transform of eat f (t) 63265.2 Laplace transforms of the formeat f (t) 63265.3 The Laplace transforms ofderivatives 63465.4 The initial and final value theorems 63666 Inverse Laplace transforms 63866.1 Definition of the inverse Laplacetransform 63866.2 Inverse Laplace transforms ofsimple functions 63866.3 Inverse Laplace transforms usingpartial fractions 64066.4 Poles and zeros 64267 The solution of differential equations usingLaplace transforms 64567.1 Introduction 64567.2 Procedure to solve differential equationsby using Laplace transforms 64567.3 Worked problems on solvingdifferential equations using Laplacetransforms 64568 The solution of simultaneous differentialequations using Laplace transforms 65068.1 Introduction 65068.2 Procedure to solve simultaneousdifferential equations using Laplacetransforms 65068.3 Worked problems on solvingsimultaneous differential equationsby using Laplace transforms 650Assignment 18 655Section L: Fourier series 65769 Fourier series for periodic functionsof period 2 65769.1 Introduction 65769.2 Periodic functions 65769.3 Fourier series 65769.4 Worked problems on Fourier seriesof periodic functions ofperiod 2 65870 Fourier series for a non-periodic functionover range 2 66370.1 Expansion of non-periodicfunctions 66370.2 Worked problems on Fourier seriesof non-periodic functions over arange of 2 66371 Even and odd functions and half-rangeFourier series 66971.1 Even and odd functions 66971.2 Fourier cosine and Fourier sineseries 66971.3 Half-range Fourier series 67272 Fourier series over any range 67672.1 Expansion of a periodic function ofperiod L 67672.2 Half-range Fourier series forfunctions defined over range L 680 13. CONTENTS xiii73 A numerical method of harmonicanalysis 68373.1 Introduction 68373.2 Harmonic analysis on data given intabular or graphical form 68373.3 Complex waveform considerations 68674 The complex or exponential form of aFourier series 69074.1 Introduction 69074.2 Exponential or complex notation 69074.3 Complex coefficients 69174.4 Symmetry relationships 69574.5 The frequency spectrum 69874.6 Phasors 699Assignment 19 704Essential formulae 705Index 721 14. This page intentionally left blank 15. PrefaceThis fifth edition of Higher Engineering Math-ematicscovers essential mathematical materialsuitable for students studying Degrees, Founda-tionDegrees, Higher National Certificate andDiploma courses in Engineering disciplines.In this edition the material has been re-orderedinto the following twelve convenient categories:number and algebra, geometry and trigonometry,graphs, vector geometry, complex numbers, matri-cesand determinants, differential calculus, integralcalculus, differential equations, statistics and proba-bility,Laplace transforms and Fourier series. Newmaterial has been added on inequalities, differ-entiationof parametric equations, the t =tan /2substitution and homogeneous first order differen-tialequations. Another new feature is that a freeInternet download is available to lecturers of a sam-pleof solutions (over 1000) of the further problemscontained in the book.The primary aim of the material in this text isto provide the fundamental analytical and underpin-ningknowledge and techniques needed to success-fullycomplete scientific and engineering principlesmodules of Degree, Foundation Degree and HigherNational Engineering programmes. The material hasbeen designed to enable students to use techniqueslearned for the analysis, modelling and solution ofrealistic engineering problems at Degree and HigherNational level. It also aims to provide some ofthe more advanced knowledge required for thosewishing to pursue careers in mechanical engineer-ing,aeronautical engineering, electronics, commu-nicationsengineering, systems engineering and allvariants of control engineering.In Higher Engineering Mathematics 5th Edi-tion,theory is introduced in each chapter by a fulloutline of essential definitions, formulae, laws, pro-ceduresetc. The theory is kept to a minimum, forproblem solving is extensively used to establish andexemplify the theory. It is intended that readers willgain real understanding through seeing problemssolved and then through solving similar problemsthemselves.Access to software packages such as Maple, Math-ematicaand Derive, or a graphics calculator, willenhance understanding of some of the topics inthis text.Each topic considered in the text is presented in away that assumes in the reader only the knowledgeattained in BTEC National Certificate/Diploma inan Engineering discipline and Advanced GNVQ inEngineering/Manufacture.Higher Engineering Mathematics provides afollow-up to Engineering Mathematics.This textbook contains some 1000 worked prob-lems,followed by over 1750 further problems(with answers), arranged within 250 Exercises.Some 460 line diagrams further enhance under-standing.A sample of worked solutions to over 1000 ofthe further problems has been prepared and can beaccessed by lecturers free via the Internet (seebelow).At the end of the text, a list of Essential Formulaeis included for convenience of reference.At intervals throughout the text are some 19Assignments to check understanding. For example,Assignment 1 covers the material in chapters 1 to 5,Assignment 2 covers the material in chapters 6 to8, Assignment 3 covers the material in chapters 9 to11, and so on. An Instructors Manual, containingfull solutions to the Assignments, is available free tolecturers adopting this text (see below).Learning by exampleis at the heart of HigherEngineering Mathematics 5th Edition.JOHN BIRDRoyal Naval School of Marine Engineering, HMSSultan,formerly University of Portsmouthand Highbury College, PortsmouthFree web downloadsExtra material available on the InternetIt is recognised that the level of understand-ingof algebra on entry to higher courses isoften inadequate. Since algebra provides thebasis of so much of higher engineering studies,it is a situation that often needs urgent atten-tion.Lack of space has prevented the inclusionof more basic algebra topics in this textbook; 16. xvi PREFACEit is for this reason that some algebra topics solution of simple, simultaneous and quadraticequations and transposition of formulae havebeen made available to all via the Internet. Alsoincluded is a Remedial Algebra Assignment totest understanding.To access the Algebra material visit: http://books.elsevier.com/companions/0750681527Sample ofWorked Solutions to ExercisesWithin the text are some 1750 further problemsarranged within 250 Exercises. A sample ofover 1000 worked solutions has been preparedand is available for lecturers only at http://www.textbooks.elsevier.comInstructors manualThis provides full worked solutions and markscheme for all 19 Assignments in this book,together with solutions to the Remedial Alge-braAssignment mentioned above. The materialis available to lecturers only and is available athttp://www.textbooks.elsevier.comTo access the lecturer material on the text-bookwebsite please go to http://www.textbooks.elsevier.com and search for the book and click onthe manual link. If you do not have an accounton textbooks.elsevier.com already, you will needto register and request access to the books sub-jectarea. If you already have an account ontextbooks, but do not have access to the rightsubject area, please follow the request accesslink at the top of the subject area homepage. 17. Syllabus guidanceThis textbook is written for undergraduate engineering degree and foundation degree courses;however, it is also most appropriate for HNC/D studies and three syllabuses are covered.The appropriate chapters for these three syllabuses are shown in the table below.Chapter Analytical Further EngineeringMethods Analytical Mathematicsfor Methods forEngineers Engineers1. Algebra 2. Inequalities3. Partial fractions 4. Logarithms and exponential functions 5. Hyperbolic functions 6. Arithmetic and geometric progressions 7. The binomial series 8. Maclaurins series 9. Solving equations by iterative methods 10. Computer numbering systems 11. Boolean algebra and logic circuits 12. Introduction to trigonometry 13. Cartesian and polar co-ordinates 14. The circle and its properties 15. Trigonometric waveforms 16. Trigonometric identities and equations 17. The relationship between trigonometric and hyperbolic functions 18. Compound angles 19. Functions and their curves 20. Irregular areas, volumes and mean value of waveforms 21. Vectors, phasors and the combination of waveforms 22. Scalar and vector products 23. Complex numbers 24. De Moivres theorem 25. The theory of matrices and determinants 26. The solution of simultaneous equations by matrices and determinants27. Methods of differentiation 28. Some applications of differentiation 29. Differentiation of parametric equations30. Differentiation of implicit functions 31. Logarithmic differentiation 32. Differentiation of hyperbolic functions 33. Differentiation of inverse trigonometric and hyperbolic functions 34. Partial differentiation (Continued) 18. xviii SYLLABUS GUIDANCEChapter Analytical Further EngineeringMethods Analytical Mathematicsfor Methods forEngineers Engineers35. Total differential, rates of change and small changes 36. Maxima, minima and saddle points for functions of two variables 37. Standard integration 38. Some applications of integration 39. Integration using algebraic substitutions 40. Integration using trigonometric and hyperbolic substitutions 41. Integration using partial fractions 42. The t = tan /2 substitution43. Integration by parts 44. Reduction formulae 45. Numerical integration 46. Solution of first order differential equations by separation of variables47. Homogeneous first order differential equations48. Linear first order differential equations 49. Numerical methods for first order differential equations 50. Second order differential equations of the form ad2ydx2+ bdydx+ cy = 051. Second order differential equations of the form ad2ydx2+ bdydx+ cy = f (x)52. Power series methods of solving ordinary differential equations53. An introduction to partial differential equations 54. Presentation of statistical data 55. Measures of central tendency and dispersion 56. Probability 57. The binomial and Poisson distributions 58. The normal distribution 59. Linear correlation 60. Linear regression 61. Sampling and estimation theories 62. Significance testing 63. Chi-square and distribution-free tests 64. Introduction to Laplace transforms 65. Properties of Laplace transforms 66. Inverse Laplace transforms 67. Solution of differential equations using Laplace transforms 68. The solution of simultaneous differential equations using Laplace transforms69. Fourier series for periodic functions of period 2 70. Fourier series for non-periodic functions over range 2 71. Even and odd functions and half-range Fourier series 72. Fourier series over any range 73. A numerical method of harmonic analysis 74. The complex or exponential form of a Fourier series 19. Number and AlgebraA1Algebra1.1 IntroductionIn this chapter, polynomial division and the fac-torand remainder theorems are explained (in Sec-tions1.4 to 1.6).However, before this, some essentialalgebra revision on basic laws and equations isincluded.For further Algebra revision, go to website:http://books.elsevier.com/companions/07506815271.2 Revision of basic laws(a) Basic operations and laws of indicesThe laws of indices are:(i) am an = am+n (ii)aman= amn(iii) (am)n = amn (iv) amn= n am(v) an = 1an (vi) a0 = 1Problem 1. Evaluate 4a2bc3 2ac whena=2, b = 12 and c = 1124a2bc3 2ac = 4(2)2 12323 2(2)32= 4 2 2 3 3 32 2 2 2 122= 27 6 = 21Problem 2. Multiply 3x + 2y by x y.3x + 2yx yMultiply by x 3x2 + 2xyMultiply by y 3xy 2y2Adding gives: 3x2 xy 2y2Alternatively,(3x + 2y)(x y) = 3x2 3xy + 2xy 2y2= 3x2 xy 2y2Problem 3. Simplifya3b2c4abc2 and evaluatewhen a = 3, b = 18and c = 2.a3b2c4abc2= a31b21c4(2) = a2bc6When a = 3, b = 18and c = 2,a2bc6 = (3)2 18 (2)6 = (9) 18 (64) = 72Problem 4. Simplifyx2y3 + xy2xyx2y3 + xy2xy= x2y3xy+ xy2xy= x21y31 + x11y21= xy2 + y or y(xy + 1)Problem 5. Simplify(x2y)(x 3 y2)(x5y3) 12(x2y)(x3 y2)(x5y3) 12= x2y12 x12 y23x52 y32= x2+1252 y12+2332= x0y13= y13or1y 13or13 y 20. 2 NUMBER AND ALGEBRANow try the following exercise.Exercise 1 Revision of basic operations andlaws of indices1. Evaluate 2ab + 3bc abc when a = 2,b = 2 and c = 4. [16]2. Find the value of 5pq2r3 when p = 25 ,q = 2 and r = 1. [8]3. From 4x 3y + 2z subtract x + 2y 3z.[3x 5y + 5z]4. Multiply 2a 5b + c by 3a + b.[6a2 13ab + 3ac 5b2 + bc]5. Simplify (x2y3z)(x3yz2) and evaluate whenx = 12 , y = 2 and z = 3. [x5y4z3, 1312 ]6. Evaluate (a32 bc3)(a12 b12 c) when a=3,b = 4 and c = 2. [412 ]7. Simplifya2b + a3ba2b21 + ab8. Simplify(a3b12 c12 )(ab)13(a3b c)a116 b13c32 or6 a11 3 b c3 (b) Brackets, factorization and precedenceProblem 6. Simplifya2 (2a ab) a(3b + a).a2 (2a ab) a(3b + a)= a2 2a + ab 3ab a2= 2a 2ab or 2a(1 + b)Problem 7. Remove the brackets and simplifythe expression:2a [3{2(4a b) 5(a + 2b)} + 4a].Removing the innermost brackets gives:2a [3{8a 2b 5a 10b} + 4a]Collecting together similar terms gives:2a [3{3a 12b} + 4a]Removing the curly brackets gives:2a [9a 36b + 4a]Collecting together similar terms gives:2a [13a 36b]Removing the square brackets gives:2a 13a + 36b = 11a+36b or36b 11aProblem 8. Factorize (a) xy 3xz(b) 4a2 + 16ab3 (c) 3a2b 6ab2 + 15ab.(a) xy 3xz = x(y 3z)(b) 4a2 + 16ab3 = 4a(a + 4b3)(c) 3a2b 6ab2 + 15ab = 3ab(a 2b + 5)Problem 9. Simplify 3c+2c4c+c5c8c.The order of precedence is division, multiplication,addition and subtraction (sometimes remembered byBODMAS). Hence3c + 2c 4c + c 5c 8c= 3c + 2c 4c +c5c 21. 8c= 3c + 8c2 + 15 8c= 8c2 5c + 15or c(8c 5)+ 15Problem 10. Simplify(2a 3)4a+5 63a.(2a 3)4a + 5 6 3a= 2a 34a+ 5 6 3a= 2a 34a+ 30 3a= 2a4a 34a+ 30 3a= 12 34a+ 30 3a = 3012 34a 3a 22. ALGEBRA 3Now try the following exercise. AExercise 2 Further problems on brackets,factorization and precedence1. Simplify 2(p+3q r)4(r q +2p)+p.[5p + 10q 6r]2. Expand and simplify (x + y)(x 2y).[x2 xy 2y2]3. Remove the brackets and simplify:24p [2{3(5p q) 2(p + 2q)} + 3q].[11q 2p]4. Factorize 21a2b2 28ab [7ab(3ab 4)].5. Factorize 2xy2 + 6x2y + 8x3y.[2xy(y + 3x + 4x2)]6. Simplify 2y + 4 6y + 3 4 5y. 23y 3y + 127. Simplify 3 y + 2 y 1. 5y 18. Simplify a2 3ab 2a 6b + ab. [ab]1.3 Revision of equations(a) Simple equationsProblem 11. Solve 4 3x = 2x 11.Since 4 3x = 2x 11 then 4 + 11 = 2x + 3xi.e. 15 = 5x from which, x = 155= 3Problem 12. Solve4(2a 3) 2(a 4) = 3(a 3) 1.Removing the brackets gives:8a 12 2a + 8 = 3a 9 1Rearranging gives:8a 2a 3a= 9 1 + 12 8i.e. 3a= 6and a=63= 2Problem 13. Solve3x 2= 43x + 4.By cross-multiplying: 3(3x + 4)= 4(x 2)Removing brackets gives: 9x + 12= 4x 8Rearranging gives: 9x 4x= 8 12i.e. 5x= 20and x=205= 4Problem 14. Solve t + 3 t = 2.t t + 3 t= 2ti.e.t + 3= 2tand 3= 2t ti.e. 3=tand 9= t(b) Transposition of formulaeProblem 15. Transpose the formulav = u + f tmto make f the subject.u+ f tm= v from which,f tm= v uand mf tm= m(v u)i.e. f t = m(v u)and f = mt(v u)Problem 16. The impedance of an a.c. circuitis given by Z =R2 + X2. Make the reactanceX the subject. 23. 4 NUMBER AND ALGEBRAR2 + X2 = Z and squaring both sides givesR2 + X2 = Z2, from which,X2 = Z2 R2 and reactance X =Z2 R2Problem 17. Given thatDd= f + pf p,express p in terms of D, d and f .Rearranging gives: f + pf p = DdSquaring both sides gives:f + pf p= D2d2Cross-multiplying gives:d2( f + p)= D2( f p)Removing brackets gives:d2f + d2p= D2f D2pRearranging gives: d2p + D2p= D2f d2fFactorizing gives: p(d2 + D2)= f (D2 d2)and p= f (D2 d2)(d2 +D2)Now try the following exercise.Exercise 3 Further problems on simpleequations and transposition of formulaeIn problems 1 to 4 solve the equations1. 3x 2 5x = 2x 412 2. 8 + 4(x 1) 5(x 3) = 2(5 2x)[3]3.13a 2+ 15a + 3= 0 184.3t1 t= 6 [4]5. Transpose y = 3(F f )Lfor f .f = 3F yL3or f = F yL36. Make l the subject of t = 21gl = t2g427. Transpose m = LL + rCRfor L.L = mrCR m8. Make r the subject of the formulaxy= 1 + r21 r2r = x yx + y (c) Simultaneous equationsProblem 18. Solve the simultaneousequations:7x 2y = 26 (1)6x + 5y = 29 (2)5equation (1) gives:35x 10y = 130 (3)2equation (2) gives:12x + 10y = 58 (4)equation (3)+equation (4) gives:47x + 0 = 188from which, x= 18847= 4Substituting x = 4 in equation (1) gives:28 2y = 26from which, 28 26 = 2y and y=1Problem 19. Solvex8+ 52= y (1)11 + y3= 3x (2)8equation (1) gives: x + 20 = 8y (3)3equation (2) gives: 33 + y = 9x (4)i.e. x 8y = 20 (5) 24. ALGEBRA 5Aand 9x y= 33 (6)8equation (6) gives: 72x 8y= 264 (7)Equation (7) equation (5) gives:71x = 284from which, x= 28471= 4Substituting x = 4 in equation (5) gives:4 8y= 20from which, 4 + 20= 8y and y = 3(d) Quadratic equationsProblem 20. Solve the following equations byfactorization:(a) 3x2 11x 4 = 0(b) 4x2 + 8x + 3 = 0(a) The factors of 3x2 are 3x and x and these areplaced in brackets thus:(3x )(x )The factors of 4 are +1 and 4 or 1 and +4,or 2 and +2. Remembering that the productof the two inner terms added to the product ofthe two outer terms must equal 11x, the onlycombination to give this is +1 and 4, i.e.,3x2 11x 4= (3x + 1)(x 4)Thus (3x + 1)(x 4)= 0 henceeither (3x + 1)= 0 i.e. x = 13or (x 4)= 0 i.e. x = 4(b) 4x2 + 8x + 3 = (2x + 3)(2x + 1)Thus (2x + 3)(2x + 1)= 0 henceeither (2x + 3)= 0 i.e. x=32or (2x + 1)= 0 i.e. x = 12Problem 21. The roots of a quadratic equationare and 2. Determine the equation in x.13If 13and 2 are the roots of a quadratic equationthen,(x 13)(x + 2)= 0i.e. x2 + 2x 13x 23= 0i.e. x2 + 53x 23= 0or 3x2 + 5x2= 0Problem 22. Solve 4x2 + 7x + 2 = 0 givingthe answer correct to 2 decimal places.From the quadratic formula if ax2+bx+c = 0 then,x =b b2 4ac2aHence if 4x2 + 7x + 2 = 0then x =7 72 4(4)(2)2(4)=1787 =7 4.1238=7 + 4.1238or7 4.1238i.e. x= 0.36 or 1.39Now try the following exercise.Exercise 4 Further problems on simultan-eousand quadratic equationsIn problems 1 to 3, solve the simultaneousequations1. 8x 3y = 513x + 4y = 14 [x = 6, y = 1]2. 5a = 1 3b2b + a + 4 =0 [a = 2, b = 3]3.x5+ 2y3= 49153x7 y2+ 57=0 [x = 3, y = 4]4. Solve the following quadratic equations byfactorization:(a) x2 + 4x 32 = 0(b) 8x2 + 2x 15 = 0[(a) 4, 8 (b) 54 , 32 ] 25. 6 NUMBER AND ALGEBRA5. Determine the quadratic equation in x whoseroots are 2 and 5.[x2 + 3x 10 = 0]6. Solve the following quadratic equations, cor-rectto 3 decimal places:(a) 2x2 + 5x 4 = 0(b) 4t2 11t + 3 = 0(a) 0.637,3.137(b) 2.443, 0.3071.4 Polynomial divisionBefore looking at long division in algebra let usrevise long division with numbers (we may haveforgotten, since calculators do the job for us!)For example,20816is achieved as follows:1316208164848 (1) 16 divided into 2 wont go(2) 16 divided into 20 goes 1(3) Put 1 above the zero(4) Multiply 16 by 1 giving 16(5) Subtract 16 from 20 giving 4(6) Bring down the 8(7) 16 divided into 48 goes 3 times(8) Put the 3 above the 8(9) 3 16 = 48(10) 48 48 = 0Hence20816= 13 exactlySimilarly,17215is laid out as follows:11151721522157Hence17215= 11 remainder 7 or 11 + 715= 11715Below are some examples of division in algebra,which in some respects, is similar to long divisionwith numbers.(Note that a polynomial is an expression of theformf (x) = a + bx + cx2 + dx3 + and polynomial division is sometimes requiredwhen resolving into partial fractionsseeChapter 3)Problem 23. Divide 2x2 + x 3 by x 1.2x2 + x 3 is called the dividend and x 1 thedivisor. The usual layout is shown below with thedividend and divisor both arranged in descendingpowers of the symbols.2x + 3x 12x2 + x 32x2 2x3x 33x 3 Dividing the first term of the dividend by the firstterm of the divisor, i.e.2x2xgives 2x, which is putabove the first term of the dividend as shown. Thedivisor is then multiplied by 2x, i.e. 2x(x 1)=2x2 2x, which is placed under the dividend asshown. Subtracting gives 3x 3. The process isthen repeated, i.e. the first term of the divisor,x, is divided into 3x, giving +3, which is placedabove the dividend as shown.Then 3(x 1)=3x 3which is placed under the 3x 3. The remain-der,on subtraction, is zero, which completes theprocess.Thus (2x2 +x3) (x 1) = (2x + 3)[A check can be made on this answer by multiplying(2x + 3) by (x 1) which equals 2x2 + x 3]Problem 24. Divide 3x3 + x2 + 3x + 5 byx + 1. 26. ALGEBRA 7A(1) (4) (7)3x2 2x + 5x + 13x3 + x2 + 3x + 53x3 + 3x22x2 + 3x + 52x2 2x5x + 55x + 5 (1) x into 3x3 goes 3x2. Put 3x2 above 3x3(2) 3x2(x + 1) = 3x3 + 3x2(3) Subtract(4) x into 2x2 goes 2x. Put 2x above thedividend(5) 2x(x + 1) = 2x2 2x(6) Subtract(7) x into 5x goes 5. Put 5 above the dividend(8) 5(x + 1) = 5x + 5(9) SubtractThus3x3 + x2 + 3x + 5x + 1= 3x2 2x + 5Problem 25. Simplifyx3 + y3x + y(1) (4) (7)x2 xy + y2x + yx3 + 0 + 0 + y3x3 + x2yx2y + y3x2y xy2xy2 + y3xy2 + y3 (1) x into x3 goes x2. Put x2 above x3 of dividend(2) x2(x + y) = x3 + x2y(3) Subtract(4) x into x2y goes xy. Put xy above dividend(5) xy(x + y) = x2y xy2(6) Subtract(7) x into xy2 goes y2. Put y2 above dividend(8) y2(x + y) = xy2 + y3(9) SubtractThusx3 + y3x + y= x2 xy + y2The zeros shown in the dividend are not normallyshown, but are included to clarify the subtractionprocess and to keep similar terms in their respectivecolumns.Problem 26. Divide (x2 + 3x 2) by (x 2).x + 5x 2x2 + 3x 2x2 2x5x 25x 108Hencex2 + 3x 2x 2= x + 5 + 8x 2Problem 27. Divide 4a3 6a2b + 5b3 by2a b.2a2 2ab b22a b4a3 6a2b + 5b34a3 2a2b4a2b + 5b34a2b + 2ab22ab2 + 5b32ab2 + b34b3Thus4a3 6a2b + 5b32a b=2a2 2ab b2 + 4b32a b 27. 8 NUMBER AND ALGEBRANow try the following exercise.Exercise 5 Further problems on polynomialdivision1. Divide (2x2 + xy y2) by (x + y).[2x y]2. Divide (3x2 + 5x 2) by (x + 2).[3x 1]3. Determine (10x2 + 11x 6) (2x + 3).[5x 2]4. Find14x2 19x 32x 3. [7x + 1]5. Divide (x3 + 3x2y + 3xy2 + y3) by (x + y).[x2 + 2xy + y2]6. Find (5x2 x + 4) (x 1).5x + 4 + 8x 17. Divide (3x3 + 2x2 5x + 4) by (x + 2).3x2 4x + 3 2x + 28. Determine (5x4 + 3x3 2x + 1)/(x 3).5x3 + 18x2 + 54x + 160 + 481x 31.5 The factor theoremThere is a simple relationship between the factors ofa quadratic expression and the roots of the equationobtained by equating the expression to zero.For example, consider the quadratic equationx2 + 2x 8 = 0.To solve this we may factorize the quadratic expres-sionx2 + 2x 8 giving (x 2)(x + 4).Hence (x 2)(x + 4) = 0.Then, if the product of two numbers is zero, one orboth of those numbers must equal zero. Therefore,either (x 2) = 0, from which, x = 2or (x + 4) = 0, from which, x = 4It is clear then that a factor of (x 2) indicates aroot of +2, while a factor of (x + 4) indicates a rootof 4.In general, we can therefore say that:a factor of (x a) corresponds to aroot of x = aIn practice, we always deduce the roots of a simplequadratic equation from the factors of the quadraticexpression, as in the above example. However, wecould reverse this process. If, by trial and error, wecould determine that x = 2 is a root of the equationx2+2x8 = 0 we could deduce at once that (x2)is a factor of the expression x2+2x8.We wouldntnormally solve quadratic equations this way butsuppose we have to factorize a cubic expression (i.e.one in which the highest power of the variable is3). A cubic equation might have three simple linearfactors and the difficulty of discovering all these fac-torsby trial and error would be considerable. It is todeal with this kind of case that we use the factortheorem. This is just a generalized version of whatwe established above for the quadratic expression.The factor theorem provides a method of factorizingany polynomial, f (x), which has simple factors.A statement of the factor theorem says:if x = a is a root of the equationf (x) = 0, then (x a) is a factor of f (x)The following worked problems show the use of thefactor theorem.Problem 28. Factorize x3 7x 6 and use itto solve the cubic equation x3 7x 6 = 0.Let f (x) = x3 7x 6If x = 1, then f (1) = 13 7(1) 6 = 12If x = 2, then f (2) = 23 7(2) 6 = 12If x = 3, then f (3) = 33 7(3) 6 = 0If f (3) = 0, then (x3) is a factorfrom the factortheorem.We have a choice now.We can divide x3 7x 6 by(x 3) or we could continue our trial and errorby substituting further values for x in the givenexpressionand hope to arrive at f (x)=0.Let us do both ways. Firstly, dividing out gives:x2 + 3x + 2x 3x3 0 7x 6x3 3x23x2 7x 63x2 9x2x 62x 6 28. ALGEBRA 9AHencex3 7x 6x 3= x2 + 3x + 2i.e. x3 7x 6 = (x 3)(x2 + 3x + 2)x2 + 3x + 2 factorizes on sight as (x + 1)(x + 2).Thereforex3 7x 6 = (x 3)(x + 1)(x + 2)A second method is to continue to substitute valuesof x into f (x).Our expression for f (3) was 33 7(3) 6. Wecan see that if we continue with positive values of xthe first term will predominate such that f (x) will notbe zero.Therefore let us try some negative values forx. Therefore f (1) = (1)3 7(1) 6 = 0;hence (x + 1) is a factor (as shown above). Alsof (2) = (2)3 7(2) 6 = 0; hence (x + 2) isa factor (also as shown above).To solve x3 7x 6 = 0, we substitute thefactors, i.e.,(x 3)(x + 1)(x + 2) = 0from which, x = 3, x = 1 and x = 2.Note that the values of x, i.e. 3, 1 and 2, areall factors of the constant term, i.e. the 6. This cangive us a clue as to what values of x we shouldconsider.Problem 29. Solve the cubic equationx3 2x2 5x + 6=0 by using the factortheorem.Let f (x) = x3 2x2 5x + 6 and let us substitutesimple values of x like 1, 2, 3, 1, 2, and so on.f (1) = 13 2(1)2 5(1) + 6 = 0,hence (x 1) is a factorf (2) = 23 2(2)2 5(2) + 6= 0f (3) = 33 2(3)2 5(3) + 6 = 0,hence (x 3) is a factorf (1) = (1)3 2(1)2 5(1) + 6= 0f (2) = (2)3 2(2)2 5(2) + 6 = 0,hence (x + 2) is a factorHence x3 2x2 5x + 6 = (x 1)(x 3)(x + 2)Therefore if x3 2x2 5x + 6 = 0then (x 1)(x 3)(x + 2) = 0from which, x = 1, x = 3 and x = 2Alternatively, having obtained one factor, i.e.(x1) we could divide this into (x3 2x2 5x+6)as follows:x2 x 6x 1x3 2x2 5x + 6x3 x2 x2 5x + 6 x2 + x 6x + 6 6x + 6 Hence x3 2x2 5x + 6= (x 1)(x2 x 6)= (x 1)(x 3)(x + 2)Summarizing, the factor theorem provides us witha method of factorizing simple expressions, and analternative, in certain circumstances, to polynomialdivision.Now try the following exercise.Exercise 6 Further problems on the factortheoremUse the factor theorem to factorize the expres-sionsgiven in problems 1 to 4.1. x2 + 2x 3 [(x 1)(x + 3)]2. x3 + x2 4x 4[(x + 1)(x + 2)(x 2)]3. 2x3 + 5x2 4x 7[(x + 1)(2x2 + 3x 7)]4. 2x3 x2 16x + 15[(x 1)(x + 3)(2x 5)]5. Use the factor theorem to factorizex3 + 4x2 + x 6 and hence solve the cubicequation x3 + 4x2 + x 6 = 0.x3 + 4x2 + x 6= (x 1)(x + 3)(x + 2)x = 1, x = 3 and x = 2 6. Solve the equation x3 2x2 x + 2 = 0.[x = 1, x = 2 and x = 1] 29. 10 NUMBER AND ALGEBRA1.6 The remainder theoremDividing a general quadratic expression(ax2 + bx + c) by (x p), where p is any wholenumber, by long division (see section 1.3) gives:ax + (b + ap)x pax2 + bx + cax2 apx(b + ap)x + c(b + ap)x (b + ap)pc + (b + ap)pThe remainder, c + (b + ap)p = c + bp + ap2 orap2 + bp + c. This is, in fact, what the remaindertheorem states, i.e.,if (ax2 + bx + c) is divided by (x p),the remainder will be ap2 + bp + cIf, in the dividend (ax2 +bx +c), we substitute pfor x we get the remainder ap2 + bp + c.For example, when (3x2 4x + 5) is divided by(x 2) the remainder is ap2 +bp+c (where a = 3,b = 4, c = 5 and p = 2),i.e. the remainder is3(2)2 + (4)(2) + 5 = 12 8 + 5 = 9We can check this by dividing (3x2 4x + 5) by(x 2) by long division:3x + 2x 23x2 4x + 53x2 6x2x + 52x 49Similarly, when (4x27x+9) is divided by (x +3),the remainder is ap2+bp+c, (where a = 4, b = 7,c = 9 and p = 3) i.e. the remainder is4(3)2 + (7)(3) + 9 = 36 + 21 + 9 = 66.Also, when (x2 + 3x 2) is divided by (x 1),the remainder is 1(1)2 + 3(1) 2 = 2.It is not particularly useful, on its own, to knowthe remainder of an algebraic division. However, ifthe remainder should be zero then (x p) is a fac-tor.This is very useful therefore when factorizingexpressions.For example, when (2x2 + x 3) is divided by(x 1), the remainder is 2(1)2 + 1(1) 3 = 0,which means that (x1) is a factor of (2x2+x3).In this case the other factor is (2x + 3), i.e.,(2x2 + x 3) = (x 1)(2x 3)The remainder theorem may also be stated for acubic equation as:if (ax3 + bx2 + cx + d) is divided by(x p), the remainder will beap3 + bp2 + cp + dAs before, the remainder may be obtained by substi-tutingp for x in the dividend.For example, when (3x3 +2x2 x+4) is dividedby (x 1), the remainder is ap3 + bp2 + cp + d(where a = 3, b = 2, c = 1, d = 4 and p = 1),i.e. the remainder is 3(1)3 + 2(1)2 + (1)(1) + 4 =3 + 2 1 + 4 = 8.Similarly, when (x37x6) is divided by (x3),the remainder is 1(3)3+0(3)27(3)6 = 0, whichmeans that (x 3) is a factor of (x3 7x 6).Here are some more examples on the remaindertheorem.Problem 30. Without dividing out, find theremainder when 2x2 3x + 4 is divided by(x 2).By the remainder theorem, the remainder is givenby ap2 + bp + c, where a = 2, b = 3, c = 4 andp = 2.Hence the remainder is:2(2)2 + (3)(2) + 4 = 8 6 + 4 = 6Problem 31. Use the remainder theorem todetermine the remainder when(3x3 2x2 + x 5) is divided by (x + 2).By the remainder theorem, the remainder is given byap3 + bp2 + cp + d, where a = 3, b = 2, c = 1,d = 5 and p = 2.Hence the remainder is:3(2)3 + (2)(2)2 + (1)(2) + (5)= 24 8 2 5= 39 30. ALGEBRA 11AProblem 32. Determine the remainder when(x3 2x2 5x+6) is divided by (a) (x1) and(b) (x+2). Hence factorize the cubic expression.(a) When (x3 2x2 5x + 6) is divided by (x 1),the remainder is given by ap3 + bp2 + cp + d,where a = 1, b = 2, c = 5, d = 6 and p = 1,i.e. the remainder = (1)(1)3 + (2)(1)2+(5)(1) + 6= 1 2 5 + 6 = 0Hence (x 1) is a factor of (x3 2x2 5x + 6).(b) When (x3 2x2 5x + 6) is divided by (x + 2),the remainder is given by(1)(2)3 + (2)(2)2 + (5)(2) + 6= 8 8 + 10 + 6 = 0Hence (x+2) is also a factor of (x32x25x+6).Therefore (x1)(x+2)(x ) = x32x25x+6.To determine the third factor (shown blank) wecould(i) divide (x3 2x2 5x +6) by(x 1)(x + 2).or (ii) use the factor theorem where f (x) =x3 2x2 5x +6 and hoping to choosea value of x which makes f (x) = 0.or (iii) use the remainder theorem, again hopingto choose a factor (x p) which makesthe remainder zero.(i) Dividing (x3 2x2 5x + 6) by(x2 + x 2) gives:x 3x2 + x 2x3 2x2 5x + 6x3 + x2 2x3x2 3x + 63x2 3x + 6 Thus (x3 2x2 5x + 6)= (x 1)(x + 2)(x 3)(ii) Using the factor theorem, we letf (x) = x3 2x2 5x + 6Then f (3) = 33 2(3)2 5(3) + 6= 27 18 15 + 6 = 0Hence (x 3) is a factor.(iii) Using the remainder theorem, when(x32x25x+6) is divided by (x3), theremainder is given by ap3+bp2+cp+d,where a = 1, b = 2, c = 5, d = 6and p = 3.Hence the remainder is:1(3)3 + (2)(3)2 + (5)(3) + 6= 27 18 15 + 6 = 0Hence (x 3) is a factor.Thus (x3 2x2 5x + 6)= (x 1)(x + 2)(x 3)Now try the following exercise.Exercise 7 Further problems on the remain-dertheorem1. Find the remainder when 3x2 4x + 2 isdivided by(a) (x 2) (b) (x + 1) [(a) 6 (b) 9]2. Determine the remainder whenx3 6x2 + x 5 is divided by(a) (x + 2) (b) (x 3)[(a) 39 (b) 29]3. Use the remainder theorem to find the factorsof x3 6x2 + 11x 6.[(x 1)(x 2)(x 3)]4. Determine the factors of x3 + 7x2 + 14x + 8and hence solve the cubic equationx3 +7x2 + 14x + 8 = 0.[x = 1, x = 2 and x = 4]5. Determine the value of a if (x+2) is a factorof (x3 ax2 + 7x + 10).[a = 3]6. Using the remainder theorem, solve theequation 2x3 x2 7x + 6 = 0.[x = 1, x = 2 and x = 1.5] 31. Number and Algebra2Inequalities2.1 Introduction to inequalitiesAn inequality is any expression involving one of thesymbols , , or pq means p is less than qpq means p is greater than qpq means p is less than or equal to qpq means p is greater than or equal to qSome simple rules(i) When a quantity is added or subtracted toboth sides of an inequality, the inequality stillremains.For example, if p3then p+23+2 (adding 2 to bothsides)and p232 (subtracting 2from both sides)(ii) When multiplying or dividing both sides ofan inequality by a positive quantity, say 5, theinequality remains the same. For example,if p4 then 5p20 andp545(iii) When multiplying or dividing both sides of aninequality by a negative quantity, say 3, theinequality is reversed. For example,if p1 then 3p3 andp313(Notehas changed toin each example.)To solve an inequality means finding all the valuesof the variable for which the inequality is true.Knowledge of simple equations and quadratic equa-tionsare needed in this chapter.2.2 Simple inequalitiesThe solution of some simple inequalities, using onlythe rules given in section 2.1, is demonstrated in thefollowing worked problems.Problem 1. Solve the following inequalities:(a) 3+x 7 (b) 3t 6(c) z25 (d)p32(a) Subtracting 3 from both sides of the inequality:3+x 7 gives:3 + x 37 3, i.e. x4Hence, all values of x greater than 4 satisfy theinequality.(b) Dividing both sides of the inequality: 3t 6 by3 gives:3t363, i.e. t2Hence, all values of t less than 2 satisfy theinequality.(c) Adding 2 to both sides of the inequality z25gives:z 2 + 2 5 + 2, i.e. z 7Hence, all values of z greater than or equal to7 satisfy the inequality.(d) Multiplying both sides of the inequalityp32by 3 gives:(3)p3 (3)2, i.e. p 6Hence, all values of p less than or equal to 6satisfy the inequality.Problem 2. Solve the inequality: 4x+1x+5Subtracting 1 from both sides of the inequality:4x +1x +5 gives:4xx + 4Subtracting x from both sides of the inequality:4x x +4 gives:3x4 32. INEQUALITIES 13ADividing both sides of the inequality: 3x 4 by 3gives:x 43Hence all values of x greater than43satisfy theinequality:4x + 1x + 5Problem 3. Solve the inequality: 34t 8+tSubtracting 3 from both sides of the inequality:34t 8+t gives:4t 5 + tSubtracting t from both sides of the inequality:4t 5+t gives:5t 5Dividing both sides of the inequality 5t 5 by 5gives:t 1 (remembering to reverse theinequality)Hence, all values of t greater than or equal to 1satisfy the inequality.Now try the following exercise.Exercise 8 Further problems on simpleinequalitiesSolve the following inequalities:1. (a) 3t 6 (b) 2x 10[(a) t 2 (b) x 5]2. (a)x21.5 (b) x +25[(a) x 3 (b) x 3]3. (a) 4t 13 (b) 5x1[(a) t 1 (b) x 6]4. (a)72k41 (b) 3z+2z+3(a) k 32(b) z125. (a) 52y9+y (b) 16x 5+2x(a) y43(b) x122.3 Inequalities involving a modulusThe modulus of a number is the size of the num-ber,regardless of sign. Vertical lines enclosing thenumber denote a modulus.For example, | 4 |=4 and |4 |=4 (the modulus ofa number is never negative),The inequality: | t |1 means that all numberswhose actual size, regardless of sign, is less than1, i.e. any value between 1 and +1.Thus | t |1 means 1t 1.Similarly, | x |3 means all numbers whose actualsize, regardless of sign, is greater than 3, i.e. anyvalue greater than 3 and any value less than 3.Thus | x |3 means x3 and x3.Inequalities involving a modulus are demonstratedin the following worked problems.Problem 4. Solve the following inequality:| 3x +1 |4Since | 3x + 1 |4 then 43x +14Now 43x +1 becomes 53x,i.e. 53x and 3x +14 becomes 3x 3,i.e. x1Hence, these two results together become53x1and mean that the inequality | 3x +1 |4 is satisfiedfor any value of x greater than 53but less than 1.Problem 5. Solve the inequality: | 1 + 2t |5Since | 1+2t |5 then 51+2t 5Now 51+2t becomes 62t, i.e. 3tand 1+2t 5 becomes 2t 4 i.e. t 2Hence, these two results together become:3t 2Problem 6. Solve the inequality: | 3z 4 |2 33. 14 NUMBER AND ALGEBRA| 3z 4 |2 means 3z42 and 3z42,i.e. 3z6 and 3z2,i.e. the inequality: | 3z 4 |2 is satisfied whenz2 and z23Now try the following exercise.Exercise 9 Further problems on inequalitiesinvolving a modulusSolve the following inequalities:1. | t +1 |4 [5t 3]2. | y+3 |2 [5y1]3. | 2x 1 |4 32x 524. | 3t 5 |4 [t 3 and t 135. | 1k |3 [k 4 and k2]2.4 Inequalities involving quotientsIfpq0 thenpqmust be a positive value.Forpqto be positive, either p is positive and q ispositive or p is negative and q is negative.i.e.++=+ and=+Ifpq0 thenpqmust be a negative value.Forpqto be negative, either p is positive and q isnegative or p is negative and q is positive.i.e.+= and+=This reasoning is used when solving inequalitiesinvolving quotients, as demonstrated in the follow-ingworked problems.Problem 7. Solve the inequality:t +13t 60Sincet +13t 60 thent +13t 6must be positive.Fort +13t 6to be positive,either (i) t +10 and 3t 60or (ii) t +10 and 3t 60(i) If t +10 then t 1 and if 3t 60 then3t 6 and t 2Both of the inequalities t 1 and t 2 areonly true when t 2,i.e. the fractiont +13t 6is positive when t 2(ii) If t +10 then t 1 and if 3t 60 then3t 6 and t 2Both of the inequalities t 1 and t 2 areonly true when t 1,i.e. the fractiont +13t 6is positive when t 1Summarizing,t +13t 60 when t 2 or t 1Problem 8. Solve the inequality:2x +3x +21Since2x +3x +21 then2x +3x +210i.e.2x +3x +2 x +2x +20,i.e.2x +3(x +2)x +20 orx +1x +20Forx +1x +2to be negative or zero,either (i) x +10 and x +20or (ii) x +10 and x +20(i) If x +10 then x1 and if x +20 thenx 2(Note thatis used for the denominator, not ;a zero denominator gives a value for the fractionwhich is impossible to evaluate.)Hence, the inequalityx +1x +20 is true when x isgreater than 2 and less than or equal to 1,which may be written as 2x1(ii) If x +10 then x1 and if x +20 thenx 2 34. INEQUALITIES 15AIt is not possible to satisfy both x1 andx 2 thus no values of x satisfies (ii).Summarizing,2x +3x +21 when 2x1Now try the following exercise.Exercise 10 Further problems on inequali-tiesinvolving quotientsSolve the following inequalities:1.x +462x0 [4x 3]2.2t +4t 51 [t 5 or t 9]3.3z4z+52 [5z14]4.2xx +34 [3x2]2.5 Inequalities involving squarefunctionsThe following two general rules apply when inequal-itiesinvolve square functions:(i) if x2 k then xk or xk (1)(ii) if x2 k then kxk (2)These rules are demonstrated in the followingworked problems.Problem 9. Solve the inequality: t2 9Since t2 9 then t2 90, i.e. (t +3)(t 3)0byfactorizingFor (t +3)(t 3) to be positive,either (i) (t +3)0 and (t 3)0or (ii) (t +3)0 and (t 3)0(i) If (t +3)0 then t 3 and if (t 3)0 thent 3Both of these are true only when t 3(ii) If (t +3)0 then t 3 and if (t 3)0 thent 3Both of these are true only when t 3Summarizing, t2 9 when t 3 or t 3This demonstrates the general rule:if x2 k then xk or xk (1)Problem 10. Solve the inequality: x2 4From the general rule stated above in equation (1):if x2 4 then x 4 or x 4i.e. the inequality: x2 4 is satisfied when x2 orx2Problem 11. Solve the inequality:(2z+1)2 9From equation (1), if (2z+1)2 9 then2z+19 or 2z+19i.e. 2z+13 or 2z+13i.e. 2z2 or 2z4,i.e. z1 or z2Problem 12. Solve the inequality: t2 9Since t2 9 then t2 90, i.e. (t +3)(t 3)0byfactorizing. For (t +3)(t 3) to be negative,either (i) (t +3)0 and (t 3)0or (ii) (t +3)0 and (t 3)0(i) If (t +3)0 then t 3 and if (t 3)0 thent 3Hence (i) is satisfied when t 3 and t 3which may be written as: 3t 3(ii) If (t +3)0 then t 3 and if (t 3)0 thent 3It is not possible to satisfy both t 3 and t 3,thus no values of t satisfies (ii).Summarizing, t2 9 when 3t 3 which meansthat all values of t between 3 and +3 will satisfythe inequality.This demonstrates the general rule:if x2 k then kxk (2)Problem 13. Solve the inequality: x2 4 35. 16 NUMBER AND ALGEBRAFrom the general rule stated above in equation (2):if x2 4 then 4x 4i.e. the inequality: x2 4 is satisfied when:2x2Problem 14. Solve the inequality:(y3)2 16From equation (2), 16(y3) 16i.e. 4(y3) 4from which, 34y 4+3,i.e. 1y 7Now try the following exercise.Exercise 11 Further problems on inequali-tiesinvolving square functionsSolve the following inequalities:1. z2 16 [z4 or z4]2. z2 16 [4z4]3. 2x2 6 [x 3 or x3]4. 3k2 210 [2k 2]5. (t 1)2 36 [5t 7]6. (t 1)2 36 [t 7 or t5]7. 73y25 [y2 or y2]8. (4k +5)2 9 k 12or k 22.6 Quadratic inequalitiesInequalities involving quadratic expressions aresolved using either factorization or completing thesquare. For example,x2 2x 3 is factorized as (x +1)(x 3)and 6x2 +7x 5 is factorized as (2x 1)(3x +5)If a quadratic expression does not factorize, thenthe technique of completing the square is used. Ingeneral, the procedure for x2 +bx +c is:x2 + bx + c x + b22+c b22For example, x2 +4x 7 does not factorize; com-pletingthe square gives:x2 + 4x 7(x + 2)2 7 22 (x + 2)2 11Similarly,x2 6x 5 (x 3)2 5 32 (x 3)2 14Solving quadratic inequalities is demonstrated in thefollowing worked problems.Problem 15. Solve the inequality:x2 +2x 30Since x2 +2x 30 then (x 1)(x +3)0 byfactorizing. For the product (x 1)(x +3) to bepositive,either (i) (x 1)0 and (x +3)0or (ii) (x 1)0 and (x +3)0(i) Since (x 1)0 then x 1 and since (x +3)0then x 3Both of these inequalities are satisfied only whenx1(ii) Since (x 1)0 then x 1 and since (x +3)0then x 3Both of these inequalities are satisfied only whenx3Summarizing, x2 +2x 30 is satisfied wheneither x1 or x3Problem 16. Solve the inequality:t2 2t 80Since t2 2t 80 then (t 4)(t +2)0 byfactorizing.For the product (t 4)(t +2) to be negative,either (i) (t 4)0 and (t +2)0or (ii) (t 4)0 and (t +2)0(i) Since (t 4)0 then t 4 and since (t +2)0then t 2It is not possible to satisfy both t 4 and t 2,thus no values of t satisfies the inequality (i)(ii) Since (t 4)0 then t 4 and since (t +2)0then t 2Hence, (ii) is satisfied when 2t 4 36. INEQUALITIES 17ASummarizing, t2 2t 80 is satisfied when2t 4Problem 17. Solve the inequality:x2 +6x + 30x2 +6x +3 does not factorize; completing thesquare gives:x2 + 6x + 3 (x + 3)2 + 3 32 (x +3)2 6The inequality thus becomes: (x +3)2 60 or(x +3)2 6From equation (2), 6(x +3)6from which, (63)x (63)Hence, x2 +6x +30 is satisfied when5.45x 0.55 correct to 2 decimal places.Problem 18. Solve the inequality:y2 8y 100y2 8y10 does not factorize; completing thesquare gives:y2 8y 10 (y 4)2 10 42 (y 4)2 26The inequality thus becomes: (y4)2 260 or(y4)2 26From equation (1), (y4)26 or (y4)26from which, y4+26 or y426Hence, y2 8y100 is satisfied when y9.10or y1.10 correct to 2 decimal places.Now try the following exercise.Exercise 12 Further problems on quadraticinequalitiesSolve the following inequalities:1. x2 x 60 [x 3 or x 2]2. t2 +2t 80 [4t 2]3. 2x2 +3x 20 2x 124. y2 y200 [y5 or y4]5. z2 +4z+44 [4z0]6. x2 +6x +60[(33)x (33)]7. t2 4t 70[t (11+2) or t (211)]8. k2 +k 30k 134 12or k 134 12 37. Number and Algebra3Partial fractions3.1 Introduction to partial fractionsBy algebraic addition,1x 2+ 3x + 1= (x + 1) + 3(x 2)(x 2)(x + 1)= 4x 5x2 x 2The reverse process of moving from4x 5x2 x 2to1x 2+ 3x + 1is called resolving into partialfractions.In order to resolve an algebraic expression intopartial fractions:(i) the denominator must factorize (in the aboveexample, x2 x 2 factorizes as (x 2)(x +1)), and(ii) the numerator must be at least one degree lessthan the denominator (in the above example(4x 5) is of degree 1 since the highest poweredx term is x1 and (x2 x 2) is of degree 2).When the degree of the numerator is equal to orhigher than the degree of the denominator, thenumerator must be divided by the denominator untilthe remainder is of less degree than the denominator(see Problems 3 and 4).There are basically three types of partial fractionand the form of partial fraction used is summarizedTable 3.1Type Denominator containing Expression Form of partial fraction1 Linear factorsf (x)(x + a)(x b)(x + c)A(x + a)+ B(x b)+ C(x + c)(see Problems 1 to 4)2 Repeated linear factorsf (x)(x + a)3A(x + a)+ B(x + a)2+ C(x + a)3(see Problems 5 to 7)3 Quadratic factorsf (x)(ax2 + bx + c)(x + d)Ax + B(ax2 + bx + c)+ C(x + d)(see Problems 8 and 9)in Table 3.1, where f (x) is assumed to be of lessdegree than the relevant denominator and A, B andC are constants to be determined.(In the latter type in Table 3.1, ax2 +bx +c isa quadratic expression which does not factorizewithout containing surds or imaginary terms.)Resolving an algebraic expression into partialfractions is used as a preliminary to integrating cer-tainfunctions (see Chapter 41) and in determininginverse Laplace transforms (see Chapter 66).3.2 Worked problems on partialfractions with linear factorsProblem 1. Resolve11 3xx2 + 2x 3into partialfractions.The denominator factorizes as (x 1) (x +3) andthe numerator is of less degree than the denomina-tor.Thus11 3xx2 + 2x 3may be resolved into partialfractions.Let11 3xx2 + 2x 3 11 3x(x 1)(x + 3) A(x 1)+ B(x + 3) 38. PARTIAL FRACTIONS 19Awhere A and B are constants to be determined,i.e.11 3x(x 1)(x + 3) A(x + 3) + B(x 1)(x 1)(x + 3),by algebraic addition.Since the denominators are the same on each sideof the identity then the numerators are equal to eachother.Thus, 113x A(x +3)+B(x 1)To determine constants A and B, values of x arechosen to make the term in A or B equal to zero.When x =1, then113(1) A(1+3)+B(0)i.e. 8 =4Ai.e. A =2When x= 3, then113(3) A(0)+B(31)i.e. 20 =4Bi.e. B =5Thus11 3xx2 + 2x 3 2(x 1)+5(x + 3) 2(x 1) 5(x + 3)Check:2(x 1) 5(x + 3)= 2(x + 3) 5(x 1)(x 1)(x + 3)= 11 3xx2 + 2x 3Problem 2. Convert2x2 9x 35(x + 1)(x 2)(x + 3)into the sum of three partial fractions.Let2x2 9x 35(x + 1)(x 2)(x + 3) A(x + 1)+ B(x 2)+ C(x + 3)A(x 2)(x + 3) + B(x + 1)(x + 3)+C(x + 1)(x 2)(x + 1)(x 2)(x + 3)by algebraic addition.Equating the numerators gives:2x2 9x 35 A(x 2)(x + 3)+B(x + 1)(x + 3) + C(x + 1)(x 2)Let x= 1. Then2(1)2 9(1)35 A(3)(2)+B(0)(2)+C(0)(3)i.e. 24 =6Ai.e. A =246=4Let x =2. Then2(2)2 9(2)35 A(0)(5)+B(3)(5)+C(3)(0)i.e. 45 =15Bi.e. B =4515=3Let x= 3. Then2(3)2 9(3)35 A(5)(0)+B(2)(0)+C(2)(5)i.e. 10 =10Ci.e. C =1Thus2x2 9x 35(x +1)(x 2)(x +3) 4(x + 1) 3(x 2)+ 1(x + 3)Problem 3. Resolvex2 + 1x2 3x + 2into partialfractions.The denominator is of the same degree as thenumerator. Thus dividing out gives:1x2 3x +2x2 + 1x2 3x +23x 1For more on polynomial division, see Section 1.4,page 6. 39. 20 NUMBER AND ALGEBRAHencex2 + 1x2 3x + 2 1 + 3x 1x2 3x + 2 1 + 3x 1(x 1)(x 2)Let3x 1(x 1)(x 2) A(x 1)+ B(x 2) A(x 2) + B(x 1)(x 1)(x 2)Equating numerators gives:3x 1 A(x 2) + B(x 1)Let x = 1. Then 2 = Ai.e. A = 2Let x = 2. Then 5 = BHence3x 1(x 1)(x 2)2(x 1)+ 5(x 2)Thusx2 + 1x2 3x + 21 2(x1)+ 5(x2)Problem 4. Expressx3 2x2 4x 4x2 + x 2inpartial fractions.The numerator is of higher degree than the denom-inator.Thus dividing out gives:x 3x2 + x 2x3 2x2 4x 4x3 + x2 2x 3x2 2x 4 3x2 3x + 6x 10Thusx3 2x2 4x 4x2 +x 2 x 3+ x 10x2 +x 2 x 3+ x 10(x +2)(x 1)Letx 10(x +2)(x 1) A(x +2)+ B(x 1) A(x 1)+B(x +2)(x +2)(x 1)Equating the numerators gives:x 10 A(x 1)+B(x +2)Let x=2. Then 12 = 3Ai.e. A= 4Let x =1. Then 9 = 3Bi.e. B= 3Hencex 10(x +2)(x 1) 4(x +2) 3(x 1)Thusx3 2x2 4x4x2 +x2x3+ 4(x+2) 3(x1)Now try the following exercise.Exercise 13 Further problems on partialfractions with linear factorsResolve the following into partial fractions.1.12x2 9 2(x 3) 2(x + 3)2.4(x 4)x2 2x 3 5(x + 1) 1(x 3)3.x2 3x + 6x(x 2)(x 1)3x+ 2(x 2) 4(x 1)4.3(2x2 8x 1)(x + 4)(x + 1)(2x 1) 7(x + 4) 3(x + 1) 2(2x 1)5.x2 + 9x + 8x2 + x 61 + 2(x + 3)+ 6(x 2)6.x2 x 14x2 2x 31 2(x 3)+ 3(x + 1)7.3x3 2x2 16x + 20(x 2)(x + 2)3x 2 + 1(x 2) 5(x + 2) 40. PARTIAL FRACTIONS 21A3.3 Worked problems on partialfractions with repeated linearfactorsProblem 5. Resolve2x + 3(x 2)2 into partialfractions.The denominator contains a repeated linear factor,(x 2)2.Let2x + 3(x 2)2 A(x 2)+ B(x 2)2 A(x 2) + B(x 2)2Equating the numerators gives:2x + 3 A(x 2) + BLet x = 2. Then 7 =A(0) + Bi.e. B =72x + 3 A(x 2) + B Ax 2A + BSince an identity is true for all values of theunknown, the coefficients of similar terms may beequated.Hence, equating the coefficients of x gives: 2 = A.[Also, as a check, equating the constant terms gives:3 = 2A + BWhen A=2 and B=7,R.H.S. = 2(2) + 7 = 3 = L.H.S.]Hence2x + 3(x 2)2 2(x 2)+ 7(x 2)2Problem 6. Express5x2 2x 19(x + 3)(x 1)2 as thesum of three partial fractions.The denominator is a combination of a linear factorand a repeated linear factor.Let5x2 2x 19(x + 3)(x 1)2 A(x + 3)+ B(x 1)+ C(x 1)2 A(x 1)2 + B(x + 3)(x 1) + C(x + 3)(x + 3)(x 1)2by algebraic addition.Equating the numerators gives:5x2 2x 19 A(x 1)2 + B(x + 3)(x 1)+C(x + 3) (1)Let x=3. Then5(3)2 2(3) 19 A(4)2 + B(0)(4)+C(0)i.e. 32 = 16Ai.e. A = 2Let x =1. Then5(1)2 2(1) 19 A(0)2 + B(4)(0) + C(4)i.e. 16 =4Ci.e. C =4Without expanding the RHS of equation (1) it canbe seen that equating the coefficients of x2 gives:5=A+B, and since A=2, B=3.[Check: Identity (1) may be expressed as:5x2 2x 19 A(x2 2x + 1)+B(x2 + 2x 3) + C(x + 3)i.e. 5x2 2x 19 Ax2 2Ax + A + Bx2 + 2Bx3B + Cx + 3CEquating the x term coefficients gives:2 2A + 2B + CWhen A=2, B=3 and C=4 then2A + 2B + C = 2(2) + 2(3) 4= 2 = LHSEquating the constant term gives:19 A 3B + 3CRHS = 2 3(3) + 3(4) = 2 9 12= 19 = LHS] 41. 22 NUMBER AND ALGEBRAHence5x2 2x 19(x + 3)(x 1)2 2(x + 3)+ 3(x 1) 4(x 1)2Problem 7. Resolve3x2 + 16x + 15(x + 3)3 intopartial fractions.Let3x2 + 16x + 15(x + 3)3 A(x + 3)+ B(x + 3)2+ C(x + 3)3 A(x + 3)2 + B(x + 3) + C(x + 3)3Equating the numerators gives:3x2 + 16x + 15 A(x + 3)2 + B(x + 3) + C (1)Let x=3. Then3(3)2 + 16(3) + 15A(0)2 + B(0) + Ci.e. 6=CIdentity (1) may be expanded as:3x2 + 16x + 15 A(x2 + 6x + 9)+B(x + 3) + Ci.e. 3x2 + 16x + 15 Ax2 + 6Ax + 9A+Bx + 3B + CEquating the coefficients of x2 terms gives: 3 = AEquating the coefficients of x terms gives:16 = 6A + BSince A = 3,B = 2[Check: equating the constant terms gives:15 = 9A + 3B + CWhen A=3, B=2 and C=6,9A + 3B + C = 9(3) + 3(2) + (6)= 27 6 6 = 15 = LHS]Thus3x2 + 16x + 15(x + 3)3 3(x + 3) 2(x + 3)2 6(x + 3)3Now try the following exercise.Exercise 14 Further problems on partialfractions with repeated linear factors1.4x 3(x + 1)2 4(x + 1) 7(x + 1)22.x2 + 7x + 3x2(x + 3) 1x2+ 2x 1(x + 3)3.5x2 30x + 44(x 2)3 5(x 2) 10(x 2)2+ 4(x 2)34.18 + 21x x2(x 5)(x + 2)2 2(x 5) 3(x + 2)+ 4(x + 2)23.4 Worked problems on partialfractions with quadratic factorsProblem 8. Express7x2 + 5x + 13(x2 + 2)(x + 1)in partialfractions.The denominator is a combination of a quadraticfactor, (x2 +2), which does not factorize withoutintroducing imaginary surd terms, and a linear factor,(x +1). Let,7x2 + 5x + 13Ax + B (x2 + 2)(x + 1)(x2 + 2)+ C(x + 1) (Ax + B)(x + 1) + C(x2 + 2)(x2 + 2)(x + 1)Equating numerators gives:7x2 + 5x + 13 (Ax + B)(x + 1) + C(x2 + 2) (1)Let x=1. Then7(1)2 + 5(1) + 13(Ax + B)(0) + C(1 + 2)i.e. 15= 3Ci.e. C = 5 42. PARTIAL FRACTIONS 23AIdentity (1) may be expanded as:7x2 + 5x + 13 Ax2 + Ax + Bx + B + Cx2 + 2CEquating the coefficients of x2 terms gives:7 = A + C, and since C = 5,A = 2Equating the coefficients of x terms gives:5 = A + B, and since A = 2,B = 3[Check: equating the constant terms gives:13 = B + 2CWhen B=3 and C =5,B + 2C = 3 + 10 = 13 = LHS]Hence7x2 + 5x + 13(x2 + 2)(x + 1) 2x + 3(x2 + 2)+ 5(x + 1)Problem 9. Resolve3 + 6x + 4x2 2x3x2(x2 + 3)intopartial fractions.Terms such as x2 may be treated as (x +0)2, i.e. theyare repeated linear factors.Let3 + 6x + 4x2 2x3x2(x2 + 3) Ax+ Bx2+ Cx + D(x2 + 3) Ax(x2 + 3) + B(x2 + 3) + (Cx + D)x2x2(x2 + 3)Equating the numerators gives:3 + 6x + 4x2 2x3 Ax(x2 + 3) + B(x2 + 3)+(Cx + D)x2 Ax3 + 3Ax + Bx2 + 3B+Cx3 + Dx2Let x =0. Then 3=3Bi.e. B = 1Equating the coefficients of x3 terms gives:2 = A + C (1)Equating the coefficients of x2 terms gives:4 = B + DSince B = 1, D = 3Equating the coefficients of x terms gives:6 = 3Ai.e. A = 2From equation (1), since A=2, C = 4Hence3 + 6x + 4x2 2x3x2(x2 + 3) 2x+ 1x2+4x + 3x2 + 3 2x+ 1x2+ 3 4xx2 + 3Now try the following exercise.Exercise 15 Further problems on partialfractions with quadratic factors1.x2 x 13(x2 + 7)(x 2) 2x + 3(x2 + 7) 1(x 2)2.6x 5(x 4)(x2 + 3) 1(x 4)+ 2 x(x2 + 3)3.15 + 5x + 5x2 4x3x2(x2 + 5)1x+ 3x2+ 2 5x(x2 + 5)4.x3 + 4x2 + 20x 7(x 1)2(x2 + 8) 3(x 1)+ 2(x 1)2+ 1 2x(x2 + 8)5. When solving the differential equationd2dt2 6ddt10 =20e2t by Laplacetransforms, for given boundary conditions,the following expression for L{} results:L{} =4s3 392s2 + 42s 40s(s 2)(s2 6s + 10)Show that the expression can be resolved intopartial fractions to give:L{} = 2s 12(s 2)+ 5s 32(s2 6s + 10) 43. Number and Algebra4Logarithms and exponential functions4.1 Introduction to logarithmsWith the use of calculators firmly established, loga-rithmictables are now rarely used for calculation.However, the theory of logarithms is important, forthere are several scientific and engineering laws thatinvolve the rules of logarithms.If a number y can be written in the form ax, thenthe index x is called the logarithm of y to the baseof a,i.e. if y=ax then x=loga yThus, since 1000=103, then 3= log10 1000.Check this using the log button on yourcalculator.(a) Logarithms having a base of 10 are called com-monlogarithms and log10 is usually abbrevi-atedto lg. The following values may be checkedby using a calculator:lg 17.9=1.2528 . . ., lg 462.7=2.6652 . . . andlg 0.0173=1.7619 . . .(b) Logarithms having a base of e (where e is amathematical constant approximately equal to2.7183) are called hyperbolic, Napierian ornatural logarithms, and loge is usually abbrevi-atedto ln. The following values may be checkedby using a calculator:ln 3.15=1.1474 . . ., ln 362.7=5.8935 . . . andln 0.156=1.8578 . . ..4.2 Laws of logarithmsThere are three laws of logarithms, which apply toany base:(i) To multiply two numbers:log (AB)=logA+logBThe following may be checked by using acalculator:lg 10 = 1, also lg 5 + lg 2= 0.69897 . . . + 0.301029 . . . = 1Hence lg (52)= lg 10= lg 5+ lg 2(ii) To divide two numbers:log AB =log Alog BThe following may be checked using acalculator:ln52=ln 2.5 = 0.91629 . . .Also ln 5 ln 2 =1.60943 . . . 0.69314 . . .=0.91629 . . .Hence ln52 =ln 5 ln 2(iii) To raise a number to a power:lg An =n log AThe following may be checked using acalculator:lg 52 =lg 25 = 1.39794 . . .Also 2 lg 5 =2 0.69897 . . .=1.39794 . . .Hence lg 52 =2 lg 5Problem 1. Evaluate (a) log3 9 (b) log10 10(c) log16 8.(a) Let x = log3 9 then 3x =9 from the definitionof a logarithm, i.e. 3x =32, from which x =2Hence log3 9=2 44. LOGARITHMS AND EXPONENTIAL FUNCTIONS 25A(b) Let x = log10 10 then 10x =10 from the defin-itionof a logarithm, i.e. 10x =101, from whichx =1Hence log10 10=1 (which may be checked bya calculator)(c) Let x = log168 then 16x =8, from the defini-tionof a logarithm, i.e. (24)x =23, i.e. 24x =23from the laws of indices, from which, 4x =3 andx = 34Hence log16 8= 34Problem 2. Evaluate (a) lg 0.001 (b) ln e(c) log3181.(a) Let x = lg 0.001= log10 0.001 then 10x =0.001,i.e. 10x =103, from which x=3Hence lg 0.001=3 (which may be checkedby a calculator)(b) Let x = ln e= loge e then ex =e, i.e. ex =e1from which x =1. Hence ln e=1 (which maybe checked by a calculator)(c) Let x = log3181then 3x = 181= 134=34, fromwhich x=4Hence log3181=4Problem 3. Solve the following equations:(a) lg x =3 (b) log2 x =3 (c) log5 x=2.(a) If lg x =3 then log10 x =3 and x =103, i.e.x=1000(b) If log2 x =3 then x=23 =8(c) If log5 x=2 then x =52 = 152= 125Problem 4. Write (a) log 30 (b) log 450 interms of log 2, log 3 and log 5 to any base.(a) log 30 =log (2 15) = log (2 3 5)=log 2+log 3+log 5,by the first law of logarithms(b) log 450 =log (2 225) = log (2 3 75)=log (2 3 3 25)=log (2 32 52)=log 2 + log 32 + log 52,by the first law of logarithmsi.e. log 450 =log 2+2 log 3+2 log 5,by the third law of logarithmsProblem 5. Write log8 4 581in terms oflog 2, log 3 and log 5 to any base.log8 4 581=log 8 + log 4 5 log 81,by the first and secondlaws of logarithms= log 23 + log 514 log 34,by the laws of indices,i.e.log8 4 581=3 log 2+ 14 log 54 log 3,by the third law of logarithmsProblem 6. Evaluatelog 25 log 125 + 12 log 6253 log 5.log 25 log 125 + 12 log 6253 log 5=log 52 log 53 + 12 log 543 log 5=2 log 5 3 log 5 + 42 log 53 log 5= 1 log 53 log 5= 13Problem 7. Solve the equation:log (x 1) + log (x +1)=2 log (x +2). 45. 26 NUMBER AND ALGEBRAlog (x 1)+ log (x + 1) =log (x 1)(x + 1),from the first law oflogarithms=log (x2 1)2 log (x + 2) =log (x + 2)2=log (x2 + 4x + 4)Hence if log (x2 1) =log (x2 + 4x + 4)then x2 1 =x2 + 4x + 4i.e. 1 =4x + 4i.e. 5 =4xi.e. x=54 or 114Now try the following exercise.Exercise 16 Further problems on the lawsof logarithmsIn Problems 1 to 8, evaluate the givenexpression:1. log10 10000 [4] 2. log2 16 [4]3. log5 125 [3] 4. log218 [3]5. log8 2 13 6. lg 100 [2]7. log4 8 112 8. ln e2 [2]In Problems 9 to 14 solve the equations:9. log10 x =4 [10000]10. log3 x =2 [9]11. log4 x=212 13212. lg x=2 [0.01]13. log8 x=43 11614. ln x =3 [e3]In Problems 15 to 17 write the given expressionsin terms of log 2, log 3 and log 5 to any base:15. log 60 [2 log 2+ log 3+ log 5]16. log16 4 5274 log 2+ 14 log 53 log 317. log125 4 164 813[log 23 log 3+3 log 5]Simplify the expressions given in Problems 18and 19:18. log 27 log 9+ log 81 [5 log 3]19. log 64+ log 32 log 128 [4 log 2]20. Evaluate12log 16 13log 8log 412Solve the equations given in Problems 21and 22:21. log x4 log x3 = log 5x log 2xx =21222. log 2t3 log t = log 16+ log t[t =8]4.3 Indicial equationsThe laws of logarithms may be used to solve cer-tainequations involving powerscalled indicialequations. For example, to solve, say, 3x =27, log-arithmsto a base of 10 are taken of both sides,i.e. log10 3x =log10 27and x log10 3 =log10 27,by the third law of logarithmsRearranging givesx = log10 27log10 3= 1.43136 . . .0.4771 . . .= 3which may be readily checkedNote,log 8log 2is not equal to lg82 46. LOGARITHMS AND EXPONENTIAL FUNCTIONS 27AProblem 8. Solve the equation 2x =3, correctto 4 significant figures.Taking logarithms to base 10 of both sides of 2x =3gives:log10 2x =log10 3i.e. x log10 2 =log10 3Rearranging gives:x = log10 3log10 2= 0.47712125 . . .0.30102999 . . .= 1.585, correct to 4 significant figuresProblem 9. Solve the equation 2x+1 =32x5correct to 2 decimal places.Taking logarithms to base 10 of both sides gives:log10 2x+1 =log10 32x5i.e. (x + 1) log10 2 =(2x 5) log10 3x log10 2 + log10 2 =2x log10 3 5 log10 3x(0.3010) + (0.3010) =2x(0.4771) 5(0.4771)i.e. 0.3010x + 0.3010 =0.9542x 2.3855Hence2.3855 + 0.3010 =0.9542x 0.3010x2.6865 =0.6532xfrom which x = 2.68650.6532=4.11, correct to2 decimal placesProblem 10. Solve the equation x3.2 =41.15,correct to 4 significant figures.Taking logarithms to base 10 of both sides gives:log10 x3.2 =log10 41.153.2 log10 x =log10 41.15Hence log10 x = log10 41.153.2= 0.50449Thus x=antilog 0.50449=100.50449 =3.195 cor-rectto 4 significant figures.Now try the following exercise.Exercise 17 Indicial equationsSolve the following indicial equations for x, eachcorrect to 4 significant figures:1. 3x =6.4 [1.690]2. 2x =9 [3.170]3. 2x1 =32x1 [0.2696]4. x1.5 =14.91 [6.058]5. 25.28=4.2x [2.251]6. 42x1 =5x+2 [3.959]7. x0.25 =0.792 [2.542]8. 0.027x =3.26 [0.3272]9. The decibel gain n of an amplifier is given by:n = 10 log10P2P1where P1 is the power input and P2 is thepower output. Find the power gainP2P1whenn=25 decibels.[316.2]4.4 Graphs of logarithmic functionsA graph of y= log10 x is shown in Fig. 4.1 and agraph of y= loge x is shown in Fig. 4.2. Both areseen to be of similar shape; in fact, the same generalshape occurs for a logarithm to any base.Figure 4.1 47. 28 NUMBER AND ALGEBRAFigure 4.2In general, with a logarithm to any base a, it is notedthat:(i) loga1=0Let loga=x, then ax =1 from the definition ofthe logarithm.If ax =1 then x =0 from the laws of indices.Hence loga1=0. In the above graphs it is seenthat log101=0 and loge 1=0(ii) logaa=1Let loga a=x then ax =a from the definition ofa logarithm.If ax =a then x =1.Hence loga a=1. (Check with a calculator thatlog10 10=1 and loge e=1)(iii) loga0Let loga 0=x then ax =0 from the definition ofa logarithm.If ax =0, and a is a positive real number,then x must approach minus infinity. (Forexample, check with a calculator, 22 =0.25,220 =9.54107, 2200 =6.221061, andso on)Hence loga 04.5 The exponential functionAn exponential function is one which contains ex, ebeing a constant called the exponent and having anapproximate value of 2.7183. The exponent arisesfrom the natural laws of growth and decay and isused as a base for natural or Napierian logarithms.The value of ex may be determined by using:(a) a calculator, or(b) the power series for ex (see Section 4.6), or(c) tables of exponential functions.The most common method of evaluating an expo-nentialfunction is by using a scientific notationcalculator, this now having replaced the use oftables. Most scientific notation calculators containan ex function which enables all practical values of exand ex to be determined, correct to 8 or 9 significantfigures. For example,e1 =2.7182818 e2.4 = 11.023176e1.618 =0.19829489 correct to 8 significantfiguresIn practical situations the degree of accuracy givenby a calculator is often far greater than is appropriate.The accepted convention is that the final result isstated to one significant figure greater than the leastsignificant measured value. Use your calculator tocheck the following values:e0.12 =1.1275, correct to 5 significant figurese0.431 =0.6499, correct to 4 decimal placese9.32 =11159, correct to 5 significant figuresProblem 11. Use a calculator to determine thefollowing, each correct to 4 significant figures:(a) 3.72 e0.18 (b) 53.2 e1.4 (c)5122e7.(a) 3.72 e0.18 =(3.72)(1.197217 . . . )=4.454,correct to 4 significant figures(b) 53.2 e1.4 =(53.2)(0.246596 . . . )=13.12,correct to 4 significant figures(c)5122e7 = 5122(1096.6331 . . . )=44.94,correct to 4 significant figuresProblem 12. Evaluate the following correct to4 decimal places, using a calculator:(a) 0.0256(e5.21 e2.49)(b) 5e0.25 e0.25e0.25 +e0.25(a) 0.0256(e5.21 e2.49)= 0.0256(183.094058 . . . 12.0612761 . . . )= 4.3784, correct to 4 decimal places 48. LOGARITHMS AND EXPONENTIAL FUNCTIONS 29A(b) 5e0.25 e0.25e0.25 + e0.25= 51.28402541 . . . 0.77880078 . . .1.28402541 . . . + 0.77880078 . . .= 50.5052246 . . .2.0628261 . . .= 1.2246, correct to 4 decimal placesProblem 13. The instantaneous voltage v ina capacitive circuit is related to time t bytthe equation v = V eCR where V, C and Rare constants. Determine v, correct to 4 sig-nificantfigures, when t =30103 seconds,C =10106 farads, R=47103 ohms andV =200V.v = V etCR = 200 e(30103)(1010647103)Using a calculator,v = 200 e0.0638297... = 200(0.9381646 . . . )= 187.6VNow try the following exercise.Exercise 18 Further problems on evaluat-ingexponential functions1. Evaluate, correct to 5 significant figures:(a) 3.5 e2.8 (b) 65e1.5 (c) 2.16 e5.7(a) 57.556(b) 0.26776(c) 645.55In Problems 2 and 3, evaluate correct to 5decimal places.2. (a)17e3.4629 (b) 8.52 e1.2651(c)5 e2.69213 e1.1171(a) 4.55848(b) 2.40444(c) 8.051243. (a)5.6823e2.1347 (b)e2.1127 e2.11272(c)4(e1.7295 1)e3.6817 (a) 48.04106(b) 4.07482(c) 0.082864. The length of a bar, l, at a temperature is given by l =l0 e, where l0 and areconstants. Evaluate l, correct to 4 signifi-cantfigures, when l0 =2.587, =321.7 and=1.771104. [2.739]4.6 The power series for exThe value of ex can be calculated to any requireddegree of accuracy since it is defined in terms of thefollowing power series:ex = 1 + x + x22!+ x33!+ x44!+ (where 3!=321 and is called factorial 3)The series is valid for all values of x.The series is said to converge, i.e. if all the termsare added, an actual value for ex (where x is a realnumber) is obtained. The more terms that are taken,the closer will be the value of ex to its actual value.The value of the exponent e, correct to say 4 decimalplaces, may be determined by substituting x =1 inthe power series of equation (1). Thus,e1 =1 + 1 + (1)22!+ (1)33!+ (1)44!+ (1)55!+ (1)66!+ (1)77!+ (1)88!+ =1 + 1 + 0.5 + 0.16667 + 0.04167+0.00833 + 0.00139 + 0.00020+0.00002+ i.e. e =2.71828 = 2.7183,correct to 4 decimal placesThe value of e0.05, correct to say 8 significant figures,is found by substituting x =0.05 in the power series 49. 30 NUMBER AND ALGEBRAfor ex. Thuse0.05 = 1 + 0.05 + (0.05)22!+ (0.05)33!+ (0.05)44!+ (0.05)55!+ = 1 + 0.05 + 0.00125 + 0.000020833+0.000000260 + 0.000000003and by adding,e0.05 = 1.0512711, correct to 8 significant figuresIn this example, successive terms in the series growsmaller very rapidly and it is relatively easy to deter-minethe value of e0.05 to a high degree of accuracy.However, when x is nearer to unity or larger thanunity, a very large number of terms are required foran accurate result.If in the series of equation (1), x is replaced by x,then,ex =1 + (x) + (x)22!+ (x)33!+ i.e. ex =1 x + x22! x33!+ In a similar manner the power series for ex may beused to evaluate any exponential function of the forma ekx, where a and k are constants. In the series ofequation (1), let x be replaced by kx. Then,a ekx =a 1 + (kx) + (kx)22!+ (kx)33!+ Thus 5 e2x =5 1 + (2x) + (2x)22!+ (2x)33!+ =5 1 + 2x + 4x22+ 8x36+ i.e. 5 e2x =5 1 + 2x + 2x2 + 43x3 + Problem 14. Determine the value of 5 e0.5, cor-rectto 5 significant figures by using the powerseries for ex.ex = 1 + x + x22!+ x33!+ x44!+ Hence e0.5 =1 + 0.5 + (0.5)2(2)(1)+ (0.5)3(3)(2)(1)+ (0.5)4(4)(3)(2)(1)+ (0.5)5(5)(4)(3)(2)(1)+ (0.5)6(6)(5)(4)(3)(2)(1)=1 + 0.5 + 0.125 + 0.020833+0.0026042 + 0.0002604+0.0000217i.e. e0.5 =1.64872,correct to 6 significant figuresHence 5e0.5 =5(1.64872) = 8.2436,correct to 5 significant figuresProblem 15. Expand ex(x2 1) as far as theterm in x5.The power series for ex is,ex = 1 + x + x22!+ x33!+ x44!+ x55!+ Hence ex(x2 1)=1 + x + x22!+ x33!+ x44!+ x55!+ (x2 1)=x2 + x3 + x42!+ x53!+ 1 + x + x22!+ x33!+ x44!+ x55!+ Grouping like terms gives:ex(x2 1)= 1 x + x2 x22! + x3 x33!+x42! x44! +x53! x55!+ =1x+ 12x2 + 56x3 + 1124x4 + 19120x5when expanded as far as the term in x5. 50. LOGARITHMS AND EXPONENTIAL FUNCTIONS 31ANow try the following exercise.Exercise 19 Further problems on the powerseries for ex1. Evaluate 5.6 e1, correct to 4 decimal places,using the power series for ex. [2.0601]2. Use the power series for ex to determine, cor-rectto 4 significant figures, (a) e2 (b) e0.3and check your result by using a calculator.[(a) 7.389 (b) 0.7408]3. Expand (12x) e2x as far as the term in x4.12x2 8x33 2x412 ) to six terms.4. Expand (2 ex2 )(x2x12 + 2x52 + x92 + 13x132+ 112x172 + 160x2124.7 Graphs of exponential functionsValues of ex and ex obtained from a calculator,correct to 2 decimal places, over a range x=3to x =3, are shown in the following table.x 3.0 2.5 2.0 1.5 1.0 0.5 0ex 0.05 0.08 0.14 0.22 0.37 0.61 1.00ex 20.09 12.18 7.39 4.48 2.72 1.65 1.00x 0.5 1.0 1.5 2.0 2.5 3.0ex 1.65 2.72 4.48 7.39 12.18 20.09ex 0.61 0.37 0.22 0.14 0.08 0.05Figure 4.3 shows graphs of y=ex and y=exProblem 16. Plot a graph of y=2 e0.3x over arange of x=2 to x =3. Hence determine thevalue of y when x =2.2 and the value of x wheny=1.6.Figure 4.3Figure 4.4A table of values is drawn up as shown below.x 3 2 1 0 1 2 30.3x 0.9 0.6 0.3 0 0.3 0.6 0.9e0.3x 0.407 0.549 0.741 1.000 1.350 1.822 2.4602 e0.3x 0.81 1.10 1.48 2.00 2.70 3.64 4.92A graph of y=2 e0.3x is shown plotted in Fig. 4.4. 51. 32 NUMBER AND ALGEBRAFrom the graph, when x=2.2, y=3.87 and wheny=1.6, x=0.74.Problem 17. Plot a graph of y= 13e2x overthe range x=1.5 to x =1.5. Determine fromthe graph the value of y when x=1.2 and thevalue of x when y=1.4.A table of values is drawn up as shown below.x 1.5 1.0 0.5 0 0.5 1.0 1.52x 3 2 1 0 1 2 3e2x 20.086 7.389 2.718 1.00 0.368 0.135 0.0501e2x 6.70 2.46 0.91 0.33 0.12 0.05 0.023A graph of 13e2x is shown in Fig. 4.5.Figure 4.5From the graph, when x=1.2, y=3.67 andwhen y=1.4, x=0.72.Problem 18. The decay of voltage, v volts,across a capacitor at time t seconds is given byv=250 et3 . Draw a graph showing the naturaldecay curve over the first 6 seconds. From thegraph, find (a) the voltage after 3.4 s, and (b) thetime when the voltage is 150V.A table of values is drawn up as shown below.t 0 1 2 3et3 1.00 0.7165 0.5134 0.3679v=250 et3 250.0 179.1 128.4 91.97t 4 5 6et3 0.2636 0.1889 0.1353v=250 et3 65.90 47.22 33.83The natural decay curve of v=250 et3 is shown inFig. 4.6.Figure 4.6From the graph:(a) when time t =3.4 s, voltage v=80V and(b) when voltage v=150V, time t =1.5 s.Now try the following exercise.Exercise 20 Further problems on exponen-tialgraphs1. Plot a graph of y=3 e0.2x over the rangex=3 to x =3. Hence determine the valueof y when x =1.4 and the value of x wheny=4.5. [3.95, 2.05]2 e1.5x over a range2. Plot a graph of y= 1x=1.5 to x =1.5 and hence determine thevalue of y when x=0.8 and the value of xwhen y=3.5. [1.65, 1.30] 52. LOGARITHMS AND EXPONENTIAL FUNCTIONS 33A3. In a chemical reaction the amount of start-ingmaterial C cm3 left after t minutes isgiven by C =40 e0.006t . Plot a graph of Cagainst t and determine (a) the concentrationC after 1 hour, and (b) the time taken for theconcentration to decrease by half.[(a) 28 cm3 (b) 116 min]4. The rate at which a body cools is given by =250 e0.05t where the excess of tempera-tureof a body above its surroundings at timet minutes is C. Plot a graph showing thenatural decay curve for the first hour of cool-ing.Hence determine (a) the temperatureafter 25 minutes, and (b) the time when thetemperature is 195C.[(a) 70C (b) 5 min]4.8 Napierian logarithmsLogarithms having a base of e are called hyper-bolic,Napierian or natural logarithms and theNapierian logarithm of x is written as loge x, or morecommonly, ln x.The value of a Napierian logarithm may bedetermined by using:(a) a calculator, or(b) a relationship between common and Napierianlogarithms, or(c) Napierian logarithm tablesThe most common method of evaluating a Napierianlogarithm is by a scientific notation calculator, thisnow having replaced the use of four-figure tables,and also the relationship between common andNapierian logarithms,loge y = 2.3026 log10 yMost scientific notation calculators contain a ln xfunction which displays the value of the Napierianlogarithm of a number when the appropriate key ispressed.Using a calculator,ln 4.692 =1.5458589=1.5459, correct to 4