# The Free High School Science Texts: Textbooks for High School Students Mathematics

FHSST Authors The Free High School Science Texts: Textbooks for High School Students Studying the Sciences Mathematics Grades 10 - 12 Version 0 September 17, 2008 ii iii Copyright 2007 “Free High School Science Texts” Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no FrontCover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled “GNU Free Documentation License”. STOP!!!! Did you notice the FREEDOMS we’ve granted you? Our copyright license is different! 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Thousands of hours went into making them and they are a gift to everyone in the education community. iv FHSST Core Team Mark Horner ; Samuel Halliday ; Sarah Blyth ; Rory Adams ; Spencer Wheaton FHSST Editors Jaynie Padayachee ; Joanne Boulle ; Diana Mulcahy ; Annette Nell ; René Toerien ; Donovan Whitfield FHSST Contributors Rory Adams ; Prashant Arora ; Richard Baxter ; Dr. Sarah Blyth ; Sebastian Bodenstein ; Graeme Broster ; Richard Case ; Brett Cocks ; Tim Crombie ; Dr. Anne Dabrowski ; Laura Daniels ; Sean Dobbs ; Fernando Durrell ; Dr. Dan Dwyer ; Frans van Eeden ; Giovanni Franzoni ; Ingrid von Glehn ; Tamara von Glehn ; Lindsay Glesener ; Dr. Vanessa Godfrey ; Dr. Johan Gonzalez ; Hemant Gopal ; Umeshree Govender ; Heather Gray ; Lynn Greeff ; Dr. Tom Gutierrez ; Brooke Haag ; Kate Hadley ; Dr. Sam Halliday ; Asheena Hanuman ; Neil Hart ; Nicholas Hatcher ; Dr. Mark Horner ; Mfandaidza Hove ; Robert Hovden ; Jennifer Hsieh ; Clare Johnson ; Luke Jordan ; Tana Joseph ; Dr. Jennifer Klay ; Lara Kruger ; Sihle Kubheka ; Andrew Kubik ; Dr. Marco van Leeuwen ; Dr. Anton Machacek ; Dr. Komal Maheshwari ; Kosma von Maltitz ; Nicole Masureik ; John Mathew ; JoEllen McBride ; Nikolai Meures ; Riana Meyer ; Jenny Miller ; Abdul Mirza ; Asogan Moodaly ; Jothi Moodley ; Nolene Naidu ; Tyrone Negus ; Thomas O’Donnell ; Dr. Markus Oldenburg ; Dr. Jaynie Padayachee ; Nicolette Pekeur ; Sirika Pillay ; Jacques Plaut ; Andrea Prinsloo ; Joseph Raimondo ; Sanya Rajani ; Prof. Sergey Rakityansky ; Alastair Ramlakan ; Razvan Remsing ; Max Richter ; Sean Riddle ; Evan Robinson ; Dr. Andrew Rose ; Bianca Ruddy ; Katie Russell ; Duncan Scott ; Helen Seals ; Ian Sherratt ; Roger Sieloff ; Bradley Smith ; Greg Solomon ; Mike Stringer ; Shen Tian ; Robert Torregrosa ; Jimmy Tseng ; Helen Waugh ; Dr. Dawn Webber ; Michelle Wen ; Dr. Alexander Wetzler ; Dr. Spencer Wheaton ; Vivian White ; Dr. Gerald Wigger ; Harry Wiggins ; Wendy Williams ; Julie Wilson ; Andrew Wood ; Emma Wormauld ; Sahal Yacoob ; Jean Youssef Contributors and editors have made a sincere effort to produce an accurate and useful resource. Should you have suggestions, find mistakes or be prepared to donate material for inclusion, please don’t hesitate to contact us. We intend to work with all who are willing to help make this a continuously evolving resource! www.fhsst.org v vi Contents I Basics 1 1 Introduction to Book 1.1 II 3 The Language of Mathematics . . . . . . . . . . . . . . . . . . . . . . . . . . . Grade 10 3 5 2 Review of Past Work 7 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.2 What is a number? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.3 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.4 Letters and Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2.5 Addition and Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.6 Multiplication and Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.7 Brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.8 Negative Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.9 2.8.1 What is a negative number? . . . . . . . . . . . . . . . . . . . . . . . . 10 2.8.2 Working with Negative Numbers . . . . . . . . . . . . . . . . . . . . . . 11 2.8.3 Living Without the Number Line . . . . . . . . . . . . . . . . . . . . . . 12 Rearranging Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.10 Fractions and Decimal Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.11 Scientific Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.12 Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.12.1 Natural Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.12.2 Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.12.3 Rational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.12.4 Irrational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.13 Mathematical Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.14 Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.15 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3 Rational Numbers - Grade 10 23 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 3.2 The Big Picture of Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 3.3 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 vii CONTENTS CONTENTS 3.4 Forms of Rational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 3.5 Converting Terminating Decimals into Rational Numbers . . . . . . . . . . . . . 25 3.6 Converting Repeating Decimals into Rational Numbers . . . . . . . . . . . . . . 25 3.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 3.8 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 4 Exponentials - Grade 10 29 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 4.2 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 4.3 Laws of Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 4.3.1 Exponential Law 1: a0 = 1 . . . . . . . . . . . . . . . . . . . . . . . . . 30 4.3.2 Exponential Law 2: am × an = am+n . . . . . . . . . . . . . . . . . . . 30 4.3.3 Exponential Law 3: a−n = 4.3.4 4.4 m 1 an , a n 6= 0 . . . . . . . . . . . . . . . . . . . . 31 Exponential Law 4: a ÷ a = am−n . . . . . . . . . . . . . . . . . . . 32 4.3.5 Exponential Law 5: (ab)n = an bn . . . . . . . . . . . . . . . . . . . . . 32 4.3.6 Exponential Law 6: (am )n = amn . . . . . . . . . . . . . . . . . . . . . 33 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 5 Estimating Surds - Grade 10 37 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 5.2 Drawing Surds on the Number Line (Optional) . . . . . . . . . . . . . . . . . . 38 5.3 End of Chapter Excercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 6 Irrational Numbers and Rounding Off - Grade 10 41 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 6.2 Irrational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 6.3 Rounding Off . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 6.4 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 7 Number Patterns - Grade 10 7.1 45 Common Number Patterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 7.1.1 Special Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 7.2 Make your own Number Patterns . . . . . . . . . . . . . . . . . . . . . . . . . . 46 7.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 7.3.1 7.4 Patterns and Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . 49 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 8 Finance - Grade 10 53 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 8.2 Foreign Exchange Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 8.3 8.2.1 How much is R1 really worth? . . . . . . . . . . . . . . . . . . . . . . . 53 8.2.2 Cross Currency Exchange Rates 8.2.3 Enrichment: Fluctuating exchange rates . . . . . . . . . . . . . . . . . . 57 . . . . . . . . . . . . . . . . . . . . . . 56 Being Interested in Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 viii CONTENTS 8.4 Simple Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 8.4.1 8.5 8.6 8.7 CONTENTS Other Applications of the Simple Interest Formula . . . . . . . . . . . . . 61 Compound Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 8.5.1 Fractions add up to the Whole . . . . . . . . . . . . . . . . . . . . . . . 65 8.5.2 The Power of Compound Interest . . . . . . . . . . . . . . . . . . . . . . 65 8.5.3 Other Applications of Compound Growth . . . . . . . . . . . . . . . . . 67 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 8.6.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 8.6.2 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 9 Products and Factors - Grade 10 71 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 9.2 Recap of Earlier Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 9.2.1 Parts of an Expression . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 9.2.2 Product of Two Binomials . . . . . . . . . . . . . . . . . . . . . . . . . 71 9.2.3 Factorisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 9.3 More Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 9.4 Factorising a Quadratic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 9.5 Factorisation by Grouping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 9.6 Simplification of Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 9.7 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 10 Equations and Inequalities - Grade 10 83 10.1 Strategy for Solving Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 10.2 Solving Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 10.3 Solving Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 10.4 Exponential Equations of the form ka(x+p) = m . . . . . . . . . . . . . . . . . . 93 10.4.1 Algebraic Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 10.5 Linear Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 10.6 Linear Simultaneous Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 10.6.1 Finding solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 10.6.2 Graphical Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 10.6.3 Solution by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . 101 10.7 Mathematical Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 10.7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 10.7.2 Problem Solving Strategy . . . . . . . . . . . . . . . . . . . . . . . . . . 104 10.7.3 Application of Mathematical Modelling . . . . . . . . . . . . . . . . . . 104 10.7.4 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 106 10.8 Introduction to Functions and Graphs . . . . . . . . . . . . . . . . . . . . . . . 107 10.9 Functions and Graphs in the Real-World . . . . . . . . . . . . . . . . . . . . . . 107 10.10Recap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 ix CONTENTS CONTENTS 10.10.1 Variables and Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 10.10.2 Relations and Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 10.10.3 The Cartesian Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 10.10.4 Drawing Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 10.10.5 Notation used for Functions . . . . . . . . . . . . . . . . . . . . . . . . 110 10.11Characteristics of Functions - All Grades . . . . . . . . . . . . . . . . . . . . . . 112 10.11.1 Dependent and Independent Variables . . . . . . . . . . . . . . . . . . . 112 10.11.2 Domain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 10.11.3 Intercepts with the Axes . . . . . . . . . . . . . . . . . . . . . . . . . . 113 10.11.4 Turning Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 10.11.5 Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 10.11.6 Lines of Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 10.11.7 Intervals on which the Function Increases/Decreases . . . . . . . . . . . 114 10.11.8 Discrete or Continuous Nature of the Graph . . . . . . . . . . . . . . . . 114 10.12Graphs of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 10.12.1 Functions of the form y = ax + q . . . . . . . . . . . . . . . . . . . . . 116 10.12.2 Functions of the Form y = ax2 + q . . . . . . . . . . . . . . . . . . . . . 120 10.12.3 Functions of the Form y = a x + q . . . . . . . . . . . . . . . . . . . . . . 125 10.12.4 Functions of the Form y = ab(x) + q . . . . . . . . . . . . . . . . . . . . 129 10.13End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 11 Average Gradient - Grade 10 Extension 135 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 11.2 Straight-Line Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 11.3 Parabolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 11.4 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 12 Geometry Basics 139 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 12.2 Points and Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 12.3 Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 12.3.1 Measuring angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 12.3.2 Special Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 12.3.3 Special Angle Pairs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 12.3.4 Parallel Lines intersected by Transversal Lines . . . . . . . . . . . . . . . 143 12.4 Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 12.4.1 Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 12.4.2 Quadrilaterals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 12.4.3 Other polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 12.4.4 Extra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 12.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 12.5.1 Challenge Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 x CONTENTS 13 Geometry - Grade 10 CONTENTS 161 13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 13.2 Right Prisms and Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 13.2.1 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 13.2.2 Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 13.3 Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 13.3.1 Similarity of Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 13.4 Co-ordinate Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 13.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 13.4.2 Distance between Two Points . . . . . . . . . . . . . . . . . . . . . . . . 172 13.4.3 Calculation of the Gradient of a Line . . . . . . . . . . . . . . . . . . . . 173 13.4.4 Midpoint of a Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 13.5 Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 13.5.1 Translation of a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 13.5.2 Reflection of a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 13.6 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 14 Trigonometry - Grade 10 189 14.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 14.2 Where Trigonometry is Used . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 14.3 Similarity of Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 14.4 Definition of the Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . 191 14.5 Simple Applications of Trigonometric Functions . . . . . . . . . . . . . . . . . . 195 14.5.1 Height and Depth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 14.5.2 Maps and Plans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 14.6 Graphs of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . 199 14.6.1 Graph of sin θ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 14.6.2 Functions of the form y = a sin(x) + q . . . . . . . . . . . . . . . . . . . 200 14.6.3 Graph of cos θ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 14.6.4 Functions of the form y = a cos(x) + q . . . . . . . . . . . . . . . . . . 202 14.6.5 Comparison of Graphs of sin θ and cos θ . . . . . . . . . . . . . . . . . . 204 14.6.6 Graph of tan θ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 14.6.7 Functions of the form y = a tan(x) + q . . . . . . . . . . . . . . . . . . 205 14.7 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 15 Statistics - Grade 10 211 15.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 15.2 Recap of Earlier Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 15.2.1 Data and Data Collection . . . . . . . . . . . . . . . . . . . . . . . . . . 211 15.2.2 Methods of Data Collection . . . . . . . . . . . . . . . . . . . . . . . . . 212 15.2.3 Samples and Populations . . . . . . . . . . . . . . . . . . . . . . . . . . 213 15.3 Example Data Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 xi CONTENTS CONTENTS 15.3.1 Data Set 1: Tossing a Coin . . . . . . . . . . . . . . . . . . . . . . . . . 213 15.3.2 Data Set 2: Casting a die . . . . . . . . . . . . . . . . . . . . . . . . . . 213 15.3.3 Data Set 3: Mass of a Loaf of Bread . . . . . . . . . . . . . . . . . . . . 214 15.3.4 Data Set 4: Global Temperature . . . . . . . . . . . . . . . . . . . . . . 214 15.3.5 Data Set 5: Price of Petrol . . . . . . . . . . . . . . . . . . . . . . . . . 215 15.4 Grouping Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 15.4.1 Exercises - Grouping Data . . . . . . . . . . . . . . . . . . . . . . . . . 216 15.5 Graphical Representation of Data . . . . . . . . . . . . . . . . . . . . . . . . . . 217 15.5.1 Bar and Compound Bar Graphs . . . . . . . . . . . . . . . . . . . . . . . 217 15.5.2 Histograms and Frequency Polygons . . . . . . . . . . . . . . . . . . . . 217 15.5.3 Pie Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 15.5.4 Line and Broken Line Graphs . . . . . . . . . . . . . . . . . . . . . . . . 220 15.5.5 Exercises - Graphical Representation of Data . . . . . . . . . . . . . . . 221 15.6 Summarising Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 15.6.1 Measures of Central Tendency . . . . . . . . . . . . . . . . . . . . . . . 222 15.6.2 Measures of Dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 15.6.3 Exercises - Summarising Data . . . . . . . . . . . . . . . . . . . . . . . 228 15.7 Misuse of Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229 15.7.1 Exercises - Misuse of Statistics . . . . . . . . . . . . . . . . . . . . . . . 230 15.8 Summary of Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 15.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 16 Probability - Grade 10 235 16.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 16.2 Random Experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 16.2.1 Sample Space of a Random Experiment . . . . . . . . . . . . . . . . . . 235 16.3 Probability Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238 16.3.1 Classical Theory of Probability . . . . . . . . . . . . . . . . . . . . . . . 239 16.4 Relative Frequency vs. Probability . . . . . . . . . . . . . . . . . . . . . . . . . 240 16.5 Project Idea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242 16.6 Probability Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242 16.7 Mutually Exclusive Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 16.8 Complementary Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244 16.9 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 III Grade 11 17 Exponents - Grade 11 249 251 17.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251 17.2 Laws of Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251 √ m 17.2.1 Exponential Law 7: a n = n am . . . . . . . . . . . . . . . . . . . . . . 251 17.3 Exponentials in the Real-World . . . . . . . . . . . . . . . . . . . . . . . . . . . 253 17.4 End of chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254 xii CONTENTS CONTENTS 18 Surds - Grade 11 18.1 Surd Calculations . . . . . . . . . . √ √ √ 18.1.1 Surd Law 1: n a n b = n ab √ p n a 18.1.2 Surd Law 2: n ab = √ . . n b √ m 18.1.3 Surd Law 3: n am = a n . . 255 . . . . . . . . . . . . . . . . . . . . . . . . 255 . . . . . . . . . . . . . . . . . . . . . . . . 255 . . . . . . . . . . . . . . . . . . . . . . . . 255 . . . . . . . . . . . . . . . . . . . . . . . . 256 18.1.4 Like and Unlike Surds . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256 18.1.5 Simplest Surd form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257 18.1.6 Rationalising Denominators . . . . . . . . . . . . . . . . . . . . . . . . . 258 18.2 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 19 Error Margins - Grade 11 261 20 Quadratic Sequences - Grade 11 265 20.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 20.2 What is a quadratic sequence? . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 20.3 End of chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 21 Finance - Grade 11 271 21.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 21.2 Depreciation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 21.3 Simple Depreciation (it really is simple!) . . . . . . . . . . . . . . . . . . . . . . 271 21.4 Compound Depreciation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274 21.5 Present Values or Future Values of an Investment or Loan . . . . . . . . . . . . 276 21.5.1 Now or Later . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 21.6 Finding i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278 21.7 Finding n - Trial and Error . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279 21.8 Nominal and Effective Interest Rates . . . . . . . . . . . . . . . . . . . . . . . . 280 21.8.1 The General Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 21.8.2 De-coding the Terminology . . . . . . . . . . . . . . . . . . . . . . . . . 282 21.9 Formulae Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284 21.9.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284 21.9.2 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 21.10End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 22 Solving Quadratic Equations - Grade 11 287 22.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 22.2 Solution by Factorisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 22.3 Solution by Completing the Square . . . . . . . . . . . . . . . . . . . . . . . . . 290 22.4 Solution by the Quadratic Formula . . . . . . . . . . . . . . . . . . . . . . . . . 293 22.5 Finding an equation when you know its roots . . . . . . . . . . . . . . . . . . . 296 22.6 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299 xiii CONTENTS CONTENTS 23 Solving Quadratic Inequalities - Grade 11 301 23.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301 23.2 Quadratic Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301 23.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304 24 Solving Simultaneous Equations - Grade 11 307 24.1 Graphical Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 24.2 Algebraic Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309 25 Mathematical Models - Grade 11 313 25.1 Real-World Applications: Mathematical Models . . . . . . . . . . . . . . . . . . 313 25.2 End of Chatpter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 26 Quadratic Functions and Graphs - Grade 11 321 26.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 26.2 Functions of the Form y = a(x + p)2 + q . . . . . . . . . . . . . . . . . . . . . 321 26.2.1 Domain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322 26.2.2 Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323 26.2.3 Turning Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324 26.2.4 Axes of Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325 26.2.5 Sketching Graphs of the Form f (x) = a(x + p)2 + q . . . . . . . . . . . 325 26.2.6 Writing an equation of a shifted parabola . . . . . . . . . . . . . . . . . 327 26.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327 27 Hyperbolic Functions and Graphs - Grade 11 329 27.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329 27.2 Functions of the Form y = a x+p +q . . . . . . . . . . . . . . . . . . . . . . . . 329 27.2.1 Domain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330 27.2.2 Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331 27.2.3 Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332 27.2.4 Sketching Graphs of the Form f (x) = a x+p + q . . . . . . . . . . . . . . 333 27.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333 28 Exponential Functions and Graphs - Grade 11 335 28.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335 28.2 Functions of the Form y = ab(x+p) + q . . . . . . . . . . . . . . . . . . . . . . . 335 28.2.1 Domain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336 28.2.2 Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337 28.2.3 Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338 28.2.4 Sketching Graphs of the Form f (x) = ab(x+p) + q . . . . . . . . . . . . . 338 28.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339 29 Gradient at a Point - Grade 11 341 29.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341 29.2 Average Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341 29.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344 xiv CONTENTS 30 Linear Programming - Grade 11 CONTENTS 345 30.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 30.2 Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 30.2.1 Decision Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 30.2.2 Objective Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 30.2.3 Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346 30.2.4 Feasible Region and Points . . . . . . . . . . . . . . . . . . . . . . . . . 346 30.2.5 The Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346 30.3 Example of a Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 30.4 Method of Linear Programming . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 30.5 Skills you will need . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 30.5.1 Writing Constraint Equations . . . . . . . . . . . . . . . . . . . . . . . . 347 30.5.2 Writing the Objective Function . . . . . . . . . . . . . . . . . . . . . . . 348 30.5.3 Solving the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350 30.6 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352 31 Geometry - Grade 11 357 31.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357 31.2 Right Pyramids, Right Cones and Spheres . . . . . . . . . . . . . . . . . . . . . 357 31.3 Similarity of Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360 31.4 Triangle Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361 31.4.1 Proportion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361 31.5 Co-ordinate Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368 31.5.1 Equation of a Line between Two Points . . . . . . . . . . . . . . . . . . 368 31.5.2 Equation of a Line through One Point and Parallel or Perpendicular to Another Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371 31.5.3 Inclination of a Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371 31.6 Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373 31.6.1 Rotation of a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373 31.6.2 Enlargement of a Polygon 1 . . . . . . . . . . . . . . . . . . . . . . . . . 376 32 Trigonometry - Grade 11 381 32.1 History of Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381 32.2 Graphs of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . 381 32.2.1 Functions of the form y = sin(kθ) . . . . . . . . . . . . . . . . . . . . . 381 32.2.2 Functions of the form y = cos(kθ) . . . . . . . . . . . . . . . . . . . . . 383 32.2.3 Functions of the form y = tan(kθ) . . . . . . . . . . . . . . . . . . . . . 384 32.2.4 Functions of the form y = sin(θ + p) . . . . . . . . . . . . . . . . . . . . 385 32.2.5 Functions of the form y = cos(θ + p) . . . . . . . . . . . . . . . . . . . 386 32.2.6 Functions of the form y = tan(θ + p) . . . . . . . . . . . . . . . . . . . 387 32.3 Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389 32.3.1 Deriving Values of Trigonometric Functions for 30◦ , 45◦ and 60◦ . . . . . 389 32.3.2 Alternate Definition for tan θ . . . . . . . . . . . . . . . . . . . . . . . . 391 xv CONTENTS CONTENTS 32.3.3 A Trigonometric Identity . . . . . . . . . . . . . . . . . . . . . . . . . . 392 32.3.4 Reduction Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394 32.4 Solving Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 399 32.4.1 Graphical Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399 32.4.2 Algebraic Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401 32.4.3 Solution using CAST diagrams . . . . . . . . . . . . . . . . . . . . . . . 403 32.4.4 General Solution Using Periodicity . . . . . . . . . . . . . . . . . . . . . 405 32.4.5 Linear Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . 406 32.4.6 Quadratic and Higher Order Trigonometric Equations . . . . . . . . . . . 406 32.4.7 More Complex Trigonometric Equations . . . . . . . . . . . . . . . . . . 407 32.5 Sine and Cosine Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 32.5.1 The Sine Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 32.5.2 The Cosine Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412 32.5.3 The Area Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414 32.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 33 Statistics - Grade 11 419 33.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419 33.2 Standard Deviation and Variance . . . . . . . . . . . . . . . . . . . . . . . . . . 419 33.2.1 Variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419 33.2.2 Standard Deviation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 421 33.2.3 Interpretation and Application . . . . . . . . . . . . . . . . . . . . . . . 423 33.2.4 Relationship between Standard Deviation and the Mean . . . . . . . . . . 424 33.3 Graphical Representation of Measures of Central Tendency and Dispersion . . . . 424 33.3.1 Five Number Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . 424 33.3.2 Box and Whisker Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . 425 33.3.3 Cumulative Histograms . . . . . . . . . . . . . . . . . . . . . . . . . . . 426 33.4 Distribution of Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428 33.4.1 Symmetric and Skewed Data . . . . . . . . . . . . . . . . . . . . . . . . 428 33.4.2 Relationship of the Mean, Median, and Mode . . . . . . . . . . . . . . . 428 33.5 Scatter Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429 33.6 Misuse of Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432 33.7 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435 34 Independent and Dependent Events - Grade 11 437 34.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437 34.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437 34.2.1 Identification of Independent and Dependent Events . . . . . . . . . . . 438 34.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441 IV Grade 12 35 Logarithms - Grade 12 443 445 35.1 Definition of Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445 xvi CONTENTS CONTENTS 35.2 Logarithm Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 446 35.3 Laws of Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447 35.4 Logarithm Law 1: loga 1 = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447 35.5 Logarithm Law 2: loga (a) = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 448 35.6 Logarithm Law 3: loga (x · y) = loga (x) + loga (y) . . . . . . . . . . . . . . . . . 448 35.7 Logarithm Law 4: loga xy = loga (x) − loga (y) . . . . . . . . . . . . . . . . . 449 35.8 Logarithm Law 5: loga (xb ) = b loga (x) . . . . . . . . . . . . . . . . . . . . . . . 450 √ 35.9 Logarithm Law 6: loga ( b x) = logab(x) . . . . . . . . . . . . . . . . . . . . . . . 450 35.10Solving simple log equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452 35.10.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454 35.11Logarithmic applications in the Real World . . . . . . . . . . . . . . . . . . . . . 454 35.11.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455 35.12End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455 36 Sequences and Series - Grade 12 457 36.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457 36.2 Arithmetic Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457 36.2.1 General Equation for the nth -term of an Arithmetic Sequence . . . . . . 458 36.3 Geometric Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459 36.3.1 Example - A Flu Epidemic . . . . . . . . . . . . . . . . . . . . . . . . . 459 36.3.2 General Equation for the nth -term of a Geometric Sequence . . . . . . . 461 36.3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 461 36.4 Recursive Formulae for Sequences . . . . . . . . . . . . . . . . . . . . . . . . . 462 36.5 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463 36.5.1 Some Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463 36.5.2 Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463 36.6 Finite Arithmetic Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465 36.6.1 General Formula for a Finite Arithmetic Series . . . . . . . . . . . . . . . 466 36.6.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467 36.7 Finite Squared Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468 36.8 Finite Geometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 469 36.8.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470 36.9 Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471 36.9.1 Infinite Geometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . 471 36.9.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472 36.10End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472 37 Finance - Grade 12 477 37.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477 37.2 Finding the Length of the Investment or Loan . . . . . . . . . . . . . . . . . . . 477 37.3 A Series of Payments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 478 37.3.1 Sequences and Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479 xvii CONTENTS CONTENTS 37.3.2 Present Values of a series of Payments . . . . . . . . . . . . . . . . . . . 479 37.3.3 Future Value of a series of Payments . . . . . . . . . . . . . . . . . . . . 484 37.3.4 Exercises - Present and Future Values . . . . . . . . . . . . . . . . . . . 485 37.4 Investments and Loans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485 37.4.1 Loan Schedules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485 37.4.2 Exercises - Investments and Loans . . . . . . . . . . . . . . . . . . . . . 489 37.4.3 Calculating Capital Outstanding . . . . . . . . . . . . . . . . . . . . . . 489 37.5 Formulae Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489 37.5.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 490 37.5.2 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 490 37.6 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 490 38 Factorising Cubic Polynomials - Grade 12 493 38.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493 38.2 The Factor Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493 38.3 Factorisation of Cubic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . 494 38.4 Exercises - Using Factor Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 496 38.5 Solving Cubic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496 38.5.1 Exercises - Solving of Cubic Equations . . . . . . . . . . . . . . . . . . . 498 38.6 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 498 39 Functions and Graphs - Grade 12 501 39.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501 39.2 Definition of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501 39.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501 39.3 Notation used for Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502 39.4 Graphs of Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502 39.4.1 Inverse Function of y = ax + q . . . . . . . . . . . . . . . . . . . . . . . 503 39.4.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504 39.4.3 Inverse Function of y = ax2 . . . . . . . . . . . . . . . . . . . . . . . . 504 39.4.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504 39.4.5 Inverse Function of y = ax . . . . . . . . . . . . . . . . . . . . . . . . . 506 39.4.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 506 39.5 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507 40 Differential Calculus - Grade 12 509 40.1 Why do I have to learn this stuff? . . . . . . . . . . . . . . . . . . . . . . . . . 509 40.2 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 510 40.2.1 A Tale of Achilles and the Tortoise . . . . . . . . . . . . . . . . . . . . . 510 40.2.2 Sequences, Series and Functions . . . . . . . . . . . . . . . . . . . . . . 511 40.2.3 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512 40.2.4 Average Gradient and Gradient at a Point . . . . . . . . . . . . . . . . . 516 40.3 Differentiation from First Principles . . . . . . . . . . . . . . . . . . . . . . . . . 519 xviii CONTENTS CONTENTS 40.4 Rules of Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521 40.4.1 Summary of Differentiation Rules . . . . . . . . . . . . . . . . . . . . . . 522 40.5 Applying Differentiation to Draw Graphs . . . . . . . . . . . . . . . . . . . . . . 523 40.5.1 Finding Equations of Tangents to Curves . . . . . . . . . . . . . . . . . 523 40.5.2 Curve Sketching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524 40.5.3 Local minimum, Local maximum and Point of Inflextion . . . . . . . . . 529 40.6 Using Differential Calculus to Solve Problems . . . . . . . . . . . . . . . . . . . 530 40.6.1 Rate of Change problems . . . . . . . . . . . . . . . . . . . . . . . . . . 534 40.7 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 535 41 Linear Programming - Grade 12 539 41.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 539 41.2 Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 539 41.2.1 Feasible Region and Points . . . . . . . . . . . . . . . . . . . . . . . . . 539 41.3 Linear Programming and the Feasible Region . . . . . . . . . . . . . . . . . . . 540 41.4 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546 42 Geometry - Grade 12 549 42.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549 42.2 Circle Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549 42.2.1 Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549 42.2.2 Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 550 42.2.3 Theorems of the Geometry of Circles . . . . . . . . . . . . . . . . . . . . 550 42.3 Co-ordinate Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 566 42.3.1 Equation of a Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 566 42.3.2 Equation of a Tangent to a Circle at a Point on the Circle . . . . . . . . 569 42.4 Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 571 42.4.1 Rotation of a Point about an angle θ . . . . . . . . . . . . . . . . . . . . 571 42.4.2 Characteristics of Transformations . . . . . . . . . . . . . . . . . . . . . 573 42.4.3 Characteristics of Transformations . . . . . . . . . . . . . . . . . . . . . 573 42.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574 43 Trigonometry - Grade 12 577 43.1 Compound Angle Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 577 43.1.1 Derivation of sin(α + β) . . . . . . . . . . . . . . . . . . . . . . . . . . 577 43.1.2 Derivation of sin(α − β) . . . . . . . . . . . . . . . . . . . . . . . . . . 578 43.1.3 Derivation of cos(α + β) . . . . . . . . . . . . . . . . . . . . . . . . . . 578 43.1.4 Derivation of cos(α − β) . . . . . . . . . . . . . . . . . . . . . . . . . . 579 43.1.5 Derivation of sin 2α . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 579 43.1.6 Derivation of cos 2α . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 579 43.1.7 Problem-solving Strategy for Identities . . . . . . . . . . . . . . . . . . . 580 43.2 Applications of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . 582 43.2.1 Problems in Two Dimensions . . . . . . . . . . . . . . . . . . . . . . . . 582 xix CONTENTS CONTENTS 43.2.2 Problems in 3 dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . 584 43.3 Other Geometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586 43.3.1 Taxicab Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586 43.3.2 Manhattan distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586 43.3.3 Spherical Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 587 43.3.4 Fractal Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 588 43.4 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589 44 Statistics - Grade 12 591 44.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591 44.2 A Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591 44.3 Extracting a Sample Population . . . . . . . . . . . . . . . . . . . . . . . . . . . 593 44.4 Function Fitting and Regression Analysis . . . . . . . . . . . . . . . . . . . . . . 594 44.4.1 The Method of Least Squares . . . . . . . . . . . . . . . . . . . . . . . 596 44.4.2 Using a calculator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 597 44.4.3 Correlation coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . 599 44.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 600 45 Combinations and Permutations - Grade 12 603 45.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603 45.2 Counting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603 45.2.1 Making a List . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603 45.2.2 Tree Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604 45.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604 45.3.1 The Factorial Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . 604 45.4 The Fundamental Counting Principle . . . . . . . . . . . . . . . . . . . . . . . . 604 45.5 Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 605 45.5.1 Counting Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . 605 45.5.2 Combinatorics and Probability . . . . . . . . . . . . . . . . . . . . . . . 606 45.6 Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 606 45.6.1 Counting Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . 607 45.7 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 608 45.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 610 V Exercises 613 46 General Exercises 615 47 Exercises - Not covered in Syllabus 617 A GNU Free Documentation License 619 xx Part III Grade 11 249 Chapter 17 Exponents - Grade 11 17.1 Introduction In Grade 10 we studied exponential numbers and learnt that there were six laws that made working with exponential numbers easier. There is one law that we did not study in Grade 10. This will be described here. 17.2 Laws of Exponents In Grade 10, we worked only with indices that were integers. What happens when the index is not an integer, but is a rational number? This leads us to the final law of exponents, m an = 17.2.1 m Exponential Law 7: a n = √ n √ n am (17.1) am We say that x is an nth root of b if xn = b. For example, (−1)4 = 1, so −1 is a 4th root of 1. Using law 6, we notice that m m (a n )n = a n ×n = am (17.2) m therefore a n must be an nth root of am . We can therefore say √ m a n = n am √ where n am is the nth root of am (if it exists). For example, 2 23 = (17.3) √ 3 22 A number may not always have a real nth root. For example, if n = 2 and a = −1, then there is no real number such that x2 = −1 because x2 can never be a negative number. Extension: Complex Numbers There are numbers which can solve problems like x2 = −1, but they are beyond the scope of this book. They are called complex numbers. It is also possible for more than one nth root of a number to exist. For example, (−2)2 = 4 and 22 = 4, so both -2 and 2 are 2nd (square) roots of 4. Usually if there is more than one root, we choose the positive real solution and move on. 251 17.2 CHAPTER 17. EXPONENTS - GRADE 11 Worked Example 81: Rational Exponents Question: Simplify without using a calculator: 5 4−1 − 9−1 21 Answer Step 1 : Rewrite negative exponents as numbers with postive indices = 1 4 5 − 1 9 21 Step 2 : Simplify inside brackets 5 9−4 ÷ 1 36 1 5 36 2 × 1 5 = = 21 1 (62 ) 2 = Step 3 : Apply exponential law 6 = 6 Worked Example 82: More rational Exponents Question: Simplify: 3 (16x4 ) 4 Answer Step 1 : Covert the number co-efficient to index-form with a prime base 3 = (24 x4 ) 4 Step 2 : Apply exponential laws 3 3 = 24× 4 .x4× 4 = 23 .x3 = 8x3 252 CHAPTER 17. EXPONENTS - GRADE 11 17.3 Exercise: Applying laws Use all the laws to: 1. Simplify: 2 (a) (x0 ) + 5x0 − (0,25)−0,5 + 8 3 (c) 7 12m 9 11 8m− 9 1 1 (b) s 2 ÷ s 3 2 (d) (64m6 ) 3 2. Re-write the expression as a power of x: s r q √ x x x x x 17.3 Exponentials in the Real-World In Chapter 8, you used exponentials to calculate different types of interest, for example on a savings account or on a loan and compound growth. Worked Example 83: Exponentials in the Real world Question: A type of bacteria has a very high exponential growth rate at 80% every hour. If there are 10 bacteria, determine how many there will be in 5 hours, in 1 day and in 1 week? Answer Step 1 : P opulation = Initial population×(1+growth percentage)time period in hours Therefore, in this case: P opulation = 10(1,8)n , where n = number of hours Step 2 : In 5 hours P opulation = 10(1,8)5 = 188 Step 3 : In 1 day = 24 hours P opulation = 10(1,8)24 = 13 382 588 Step 4 : in 1 week = 168 hours P opulation = 10(1,8)168 = 7,687 × 1043 Note this answer is given in scientific notation as it is a very big number. Worked Example 84: More Exponentials in the Real world Question: A species of extremely rare, deep water fish has an extremely long lifespan and rarely have children. If there are a total 821 of this type of fish and their growth rate is 2% each month, how many will there be in half of a year? What will be the population be in 10 years and in 100 years ? Answer Step 1 : P opulation = Initial population×(1+growth percentage)time period in months 253 17.4 CHAPTER 17. EXPONENTS - GRADE 11 Therefore, in this case: P opulation = 821(1,02)n, where n = number of months Step 2 : In half a year = 6 months P opulation = 821(1,02)6 = 924 Step 3 : In 10 years = 120 months P opulation = 821(1,02)120 = 8 838 Step 4 : in 100 years = 1 200 months P opulation = 821(1,02)1 200 = 1,716 × 1013 Note this answer is also given in scientific notation as it is a very big number. 17.4 End of chapter Exercises 1. Simplify as far as possible: 2 A 8− 3 √ 2 B 16 + 8− 3 2. Simplify: 1 4 (b) (s2 ) 2 4 (d) (−m2 ) 3 4 (f) (3y 3 )4 (a) (x3 ) 3 5 (c) (m5 ) 3 4 (e) −(m2 ) 3 3. Simplify as much as you can: 3a−2 b15 c−5 (a−4 b3 c) −5 2 4. Simplify as much as you can: 9a6 b4 5. Simplify as much as you can: 6. Simplify: 7. Simplify: 3 3 a2 b4 12 16 √ x3 x √ 3 x4 b5 8. Re-write the expression as a power of x: r q p √ x x x x x √ 3 x 254 Chapter 18 Surds - Grade 11 18.1 Surd Calculations There are several laws that make working with surds easier. We will list them all and then explain where each rule comes from in detail. √ √ √ n n n a b = ab (18.1) r √ n a a n = √ (18.2) n b b √ m n am = a n (18.3) 18.1.1 Surd Law 1: √ √ √ n a n b = n ab It is often useful to look at a surd in exponential notation us to use the exponential √ as it1 allows√ 1 laws we learnt in section ??. In exponential notation, n a = a n and n b = b n . Then, √ √ 1 1 n n a b = an bn (18.4) = = 1 (ab) n √ n ab Some examples using this law: √ √ 16 × 3 4 = 3 64 = 4 √ √ √ 2. 2 × 32 = 64 = 8 √ √ √ 3. a2 b3 × b5 c4 = a2 b8 c4 = b4 c2 1. √ 3 18.1.2 Surd Law 2: If we look at p n a b p n a b = √ na √ n b in exponential notation and applying the exponential laws then, r n a b = a n1 b 1 = = Some examples using this law: 255 an 1 b√n n a √ n b (18.5) 18.1 1. CHAPTER 18. SURDS - GRADE 11 √ √ √ 12 ÷ 3 = 4 = 2 2. √ 3 3. √ √ √ a2 b13 ÷ b5 = a2 b8 = ab4 18.1.3 24 ÷ √ √ 3 3= 38=2 Surd Law 3: If we look at √ n m am = a n √ n am in exponential notation and applying the exponential laws then, √ 1 n am = (am ) n m = an (18.6) For example, √ 6 23 26 = 22 √ 2 = 18.1.4 3 = 1 Like and Unlike Surds √ √ n m m = n, otherwise they are called unlike surds. For Two surds √ b are called like surds if √ √ √ a and example 2 and 3 are like surds, however 2 and 3 2 are unlike surds. An important thing to realise about the surd laws we have just learnt is that the surds in the laws are all like surds. If we wish to use the surd laws on unlike surds, then we must first convert them into like surds. In order to do this we use the formula √ √ bn n (18.7) am = abm to rewrite the unlike surds so that bn is the same for all the surds. Worked Example 85: Like and Unlike Surds Question: Simplify to like surds as far as possible, showing all steps: Answer Step 1 : Find the common root √ √ 15 35 × 53 = 15 = 15 Step 2 : Use surd law 1 = = √ 35 .53 √ 243 × 125 √ 15 30375 15 256 √ √ 3 3× 55 CHAPTER 18. SURDS - GRADE 11 18.1.5 18.1 Simplest Surd form In most cases, when working with surds, answers are given in simplest surd form. For example, √ √ 50 = 25 × 2 √ √ 25 × 2 = √ = 5 2 √ √ 5 2 is the simplest surd form of 50. Worked Example 86: Simplest surd form √ Question: Rewrite 18 in the simplest surd form: Answer Step 1 : Break the number 18 into its lowest factors √ 18 = = √ 2×9 √ 2×3×3 √ √ 2× 3×3 √ √ 2 × 32 √ 3 2 = = = Worked Example 87: Simplest surd form √ √ Question: Simplify: 147 + 108 Answer Step 1 : Simplify each square root seperately √ √ √ √ 147 + 108 = 49 × 3 + 36 × 3 p p = 72 × 3 + 62 × 3 Step 2 : Take the values that have square root sign = 2 under the surd to the outside of the √ √ 7 3+6 3 Step 3 : The exact same surds can be treated as ”like terms” and may be added = √ 13 3 257 18.1 CHAPTER 18. SURDS - GRADE 11 18.1.6 Rationalising Denominators It is useful to work with fractions, which have rational denominators instead of surd denominators. It is possible to rewrite any fraction, which has a surd in the denominator as a fraction which has a rational denominator. We will now see how this can be achieved. √ √ Any expression of the form √ a + √b (where a and √ rational) can be changed into a rational √ b are number by multiplying by a − b (similarly a − b can be rationalised by multiplying by √ √ a + b). This is because √ √ √ √ ( a + b)( a − b) = a − b (18.8) which is rational (since a and b are rational). √ √ If we have a fraction which has a √ denominator which looks like a + b, then we can simply √ multiply both top and bottom by a − b achieving a rational denominator. √ √ c a− b c √ √ ×√ √ = √ (18.9) √ a+ b a− b a+ b √ √ c a−c b = a−b or similarly c √ √ a− b = = √ √ c a+ b √ ×√ √ √ a+ b a− b √ √ c a+c b a−b (18.10) Worked Example 88: Rationalising the Denominator √ Question: Rationalise the denominator of: 5x−16 x Answer Step 1 : Get√rid of the square root sign in the denominator √ To get rid of x in the denominator, you can multiply it out by another x. This √ ”rationalises” the surd in the denominator. Note that √xx = 1, thus the equation becomes rationalised by multiplying by 1 and thus still says the same thing. √ x 5x − 16 √ ×√ x x Step 2 : There is no longer a surd in the denominator. The surd is expressed in the numerator which is the prefered way to write expressions. (That’s why denominators get rationalised.) √ √ 5x x − 16 x x √ ( x)(5x − 16) x Worked Example 89: Rationalising the Denominator Question: Rationalise the following: 5x−16 √ y−10 258 CHAPTER 18. SURDS - GRADE 11 18.2 Answer Step 1 : Rationalise this denominator by using a clever form of ”1” √ y + 10 5x − 16 ×√ √ y − 10 y + 10 Step 2 : Multiply out the numerators and denominators √ √ 5x y − 16 y + 50x − 160 y − 100 Step 3 : There is no next step in this case. All the terms in the numerator are different and cannot be simplified and the denominator does not have any surds in it anymore. Worked Example 90: Rationalise the denominator y−25 Question: Simplify the following: √ y+5 Answer Step 1 : Multiply this equations by a clever form of ”1” that would rationalise this denominator √ y−5 y − 25 ×√ √ y+5 y−5 Step 2 : Multiply out the numerators and denominators √ √ y y − 25 y − 5y + 125 y − 25 18.2 √ y(y − 25) − 5(y − 25) = (y − 25) √ (y − 25)( y − 25) = (y − 25) √ y − 25 = End of Chapter Exercises 1. Expand: 2. Rationalise the denominator: 3. Write as a single fraction: √ √ √ √ ( x − 2)( x + 2) 10 √ x− 1 x √ 3 √ + x 2 x 4. Write in simplest surd form: 259 18.2 CHAPTER 18. SURDS - GRADE 11 (a) (c) (e) √ 72 √ (b) (d) (f) √48 12 √ 4√ ( 8÷ 2) 5. Expand and simplify: (2 + 6. Expand and simplify: (2 + 7. Expand and simplify: (1 + 8. Rationalise the denominator: √ √ 45 + 80 √ √ 18÷ √ 72 8 √ 16√ ( 20÷ 12) √ 2 2) √ √ 2)(1 + 8) √ √ √ 3)(1 + 8 + 3) y−4 √ y−2 9. Rationalise the denominator: 2x − 20 √ √ y − 10 10. Proof(without the use of a calculator) that: r r r r 13 2 8 5 1 +5 − = 3 3 6 2 3 11. Simplify, without use of a calculator: √ √ 98 − 8 √ 50 12. Simplify, without use of a calculator: √ √ √ 5( 45 + 2 80) 13. Write the following with a rational denominator: √ 5+2 √ 5 14. Simplify: √ √ 98x6 + 128x6 √ 1 √ 1 7 2 7 2 . 2+ 15. Evaluate without using a calculator: 2 − 2 2 16. The use a calculator permissible is not q in this question. Simplify completely by showing 1 √ √ 3 −2 12 + (3 3) all your steps: 3 17. Fill√ in the blank √ surd-form √ numberwhich will make the following equation a true statement: −3 6 × −2 24 = − 18 × ........... 260 Chapter 19 Error Margins - Grade 11 We have seen that numbers are either rational or irrational and we have see how to round-off numbers. However, in a calculation that has many steps, it is best to leave the rounding off right until the end. For example, if you were asked to write √ √ 3 3 + 12 as a decimal number correct to two decimal places, there are two ways of doing this as described in Table 19.1. √ √ Table 19.1: Two methods of writing 3 3 + 12 as , Method 1 / √ √ √ √ √ √ 3 3 + 12 = 3√3 + √ 4·3 3 3 + 12 = 3√3 + 2 3 = 5 3 = 5 × 1,732050808 . . . = 8,660254038 . . . = 8,66 a decimal number. Method 2 = = = 3 × 1,73 + 3,46 5,19 + 3,46 8,65 In the example we see that Method 1 gives 8,66 as an answer while Method 2 gives 8,65 as an answer. The answer of Method 1 is more accurate because the expression was simplified as much as possible before the answer was rounded-off. In general, it is best to simplify any expression as much as possible, before using your calculator to work out the answer in decimal notation. Important: Simplification and Accuracy It is best to simplify all expressions as much as possible before rounding-off answers. This maintains the accuracy of your answer. Worked Example 91: Simplification and Accuracy √ √ Question: Calculate 3 54 + 3 16. Write the answer to three decimal places. Answer Step 1 : Simplify the expression 261 CHAPTER 19. ERROR MARGINS - GRADE 11 √ √ 3 3 54 + 16 = = = = = √ √ 3 3 27 · 2 + 8 · 2 √ √ √ √ 3 3 3 3 27 · 2 + 8 · 2 √ √ 3 3 3 2+2 2 √ 3 5 2 5 × 1,25992105 . . . Step 2 : Convert any irrational numbers to decimal numbers √ 3 5 2 = 5 × 1,25992105 . . . = 6,299605249 . . . = 6,300 Step 3 : Write the final answer to the required number of decimal places. ∴ 6,299605249 . . . = 6,300 to three decimal places √ √ 3 3 54 + 16 = 6,300 to three decimal places. Worked Example 92: Simplification and Accuracy 2 p √ Question: Calculate x + 1 + 13 (2x + 2) − (x + 1) if x = 3,6. Write the answer to two decimal places. Answer Step 1 : Simplify the expression √ 1p x+1+ (2x + 2) − (x + 1) = 3 = = √ 1√ x+1+ 2x + 2 − x − 1 3 √ 1√ x+1+ x+1 3 4√ x+1 3 Step 2 : Substitute the value of x into the simplified expression 4√ x+1 3 = = = = 4p 3,6 + 1 3 4p 4,6 3 2,144761059 . . . × 4 ÷ 3 2,859681412 . . . Step 3 : Write the final answer to the required number of decimal places. 2,859681412 . . . = 2,86 To two decimal places p √ 1 ∴ x + 1 + 3 (2x + 2) − (x + 1) = 2,86 (to two decimal places) if x = 3,6. 262 CHAPTER 19. ERROR MARGINS - GRADE 11 Extension: Significant Figures In a number, each non-zero digit is a significant figure. Zeroes are only counted if they are between two non-zero digits or are at the end of the decimal part. For example, the number 2000 has 1 significant figure (the 2), but 2000,0 has 5 significant figures. Estimating a number works by removing significant figures from your number (starting from the right) until you have the desired number of significant figures, rounding as you go. For example 6,827 has 4 significant figures, but if you wish to write it to 3 significant figures it would mean removing the 7 and rounding up, so it would be 6,83. It is important to know when to estimate a number and when not to. It is usually good practise to only estimate numbers when it is absolutely necessary, and to instead use symbols to represent certain irrational numbers (such as π); approximating them only at the very end of a calculation. If it is necessary to approximate a number in the middle of a calculation, then it is often good enough to approximate to a few decimal places. 263 CHAPTER 19. ERROR MARGINS - GRADE 11 264 Chapter 20 Quadratic Sequences - Grade 11 20.1 Introduction In Grade 10, you learned about arithmetic sequences, where the difference between consecutive terms was constant. In this chapter we learn about quadratic sequences. 20.2 What is a quadratic sequence? Definition: Quadratic Sequence A quadratic sequence is a sequence of numbers in which the second differences between each consecutive term differ by the same amount, called a common second difference. For example, 1; 2; 4; 7; 11; . . . (20.1) is a quadratic sequence. Let us see why ... If we take the difference between consecutive terms, then: a2 − a1 a3 − a2 a4 − a3 a5 − a4 =2−1 =4−2 =1 =2 =7−4 =3 = 11 − 7 = 4 We then work out the second differences, which is simply obtained by taking the difference between the consecutive differences {1; 2; 3; 4; . . .} obtained above: 2−1 = 3−2 = 4−3 = 1 1 1 ... We then see that the second differences are equal to 1. Thus, (20.1) is a quadratic sequence. Note that the differences between consecutive terms (that is, the first differences) of a quadratic sequence form a sequence where there is a constant difference between consecutive terms. In the above example, the sequence of {1; 2; 3; 4; . . .}, which is formed by taking the differences between consecutive terms of (20.1), has a linear formula of the kind ax + b. 265 20.2 CHAPTER 20. QUADRATIC SEQUENCES - GRADE 11 Exercise: Quadratic Sequences The following are also examples of quadratic sequences: 3; 6; 10; 15; 21; . . . 4; 9; 16; 25; 36; . . . 7; 17; 31; 49; 71; . . . 2; 10; 26; 50; 82; . . . 31; 30; 27; 22; 15; . . . Can you calculate the common second difference for each of the above examples? Worked Example 93: Quadratic sequence Question: Write down the next two terms and find a formula for the nth term of the sequence 5, 12, 23, 38,..., ..., Answer Step 1 : Find the first differences between the terms. i.e. 7, 11, 15 Step 2 : Find the 2nd differences between the terms. the second difference is 4. So continuing the sequence, the differences between each term will be: 15 + 4 = 19 19 + 4 = 23 Step 3 : Finding the next two terms. So the next two terms in the sequence willl be: 38 + 19 = 57 57 + 23 = 80 So the sequence will be: 5, 12, 23, 38, 57, 80 Step 4 : We now need to find the formula for this sequence. We know that the first difference is 4. The start of the formula will therefore be 2n2 . Step 5 : We now need to work out the next part of the sequence. If n = 1, you have to get the value of term 1, which is 5 in this particular sequence. The difference between 2n2 = 2 and original number (5) is 3, which leads to n + 2. Check is it works for the second term, i.e. when n = 2. Then 2n2 = 8. The difference between term 2( 12) and 8 is 4, which is can be written as n + 2. So for the sequence 5, 12, 23, 38,... the formula for the nt h term is 2n2 + n + 2. General Case If the sequence is quadratic, the nt h term should be Tn = an2 + bn + c TERMS 1st difference 2nd difference a+b+c 4a + 2b + c 3a + b 9a + 3b + c 5a + b 2a 7a + b 2a In each case, the 2nd difference is 2a. This fact can be used to find a, then b then c. 266 CHAPTER 20. QUADRATIC SEQUENCES - GRADE 11 20.2 Worked Example 94: Quadratic Sequence Question: The following sequence is quadratic: 8, 22, 42, 68, ... Find the rule. Answer Step 1 : Assume that the rule is an2 + bn + c TERMS 1st difference 2nd difference 8 22 14 42 20 6 68 26 6 6 Step 2 : Determine values for a, b and c T hen 2a = 6 which gives a = 3 And And 3a + b = 14 → 9 + b = 14 → b = 5 a+b+c=8→3+5+c=8→c=0 Step 3 : Find the rule The rule is therefore: nt h term = 3n2 + 5n Step 4 : Check answer For n = 1, 1st term = 3(1)2 + 5(1) = 8 n = 2, 2nd term = 3(2)2 + 5(2) = 22 n = 3, 3rd term = 3(3)2 + 5(3) = 42 Extension: Derivation of the nth -term of a Quadratic Sequence Let the nth -term for a quadratic sequence be given by an = A · n 2 + B · n + C (20.2) where A, B and C are some constants to be determined. an = a1 a2 = = a3 = A · n2 + B · n + C (20.3) A(1) + B(1) + C = A + B + C A(2)2 + B(2) + C = 4A + 2B + C (20.4) (20.5) A(3)2 + B(3) + C = 9A + 3B + C (20.6) 2 Let d ∴d ≡ a2 − a1 = 3A + B ⇒ B = d − 3A (20.7) The common second difference is obtained from D = (a3 − a2 ) − (a2 − a1 ) = (5A + B) − (3A + B) = 2A ⇒A= Therefore, from (20.7), B = d− 267 D 2 3 ·D 2 (20.8) (20.9) 20.2 CHAPTER 20. QUADRATIC SEQUENCES - GRADE 11 From (20.4), C = a1 − (A + B) = a1 − D 3 −d+ ·D 2 2 ∴ C = a1 + D − d (20.10) th Finally, the general equation for the n -term of a quadratic sequence is given by an = 3 D 2 · n + (d − D) · n + (a1 − d + D) 2 2 (20.11) Worked Example 95: Using a set of equations Question: Study the following pattern: 1; 7; 19; 37; 61; ... 1. What is the next number in the sequence ? 2. Use variables to write an algebraic statement to generalise the pattern. 3. What will the 100th term of the sequence be ? Answer Step 1 : The next number in the sequence The numbers go up in multiples of 6 1 + 6 = 7, then 7 + 12 = 19 Therefore 61 + 6 × 6 = 97 The next number in the sequence is 97. Step 2 : Generalising the pattern TERMS 1st difference 2nd difference 1 7 6 19 12 6 37 18 6 61 24 6 6 The pattern will yield a quadratic equation since second difference is constant Therefore an2 + bn + c = y For the first term: n = 1, then y = 1 For the second term: n = 2, then y = 7 For the third term: n = 3, then y = 19 etc.... Step 3 : Setting up sets of equations a+b+c = 1 (20.12) 4a + 2b + c 9a + 3b + c = = 7 19 (20.13) (20.14) Step 4 : Solve the sets of equations eqn(2) − eqn(1) : 3a + b = 6 eqn(3) − eqn(2) : 5a + b = 12 eqn(5) − eqn(4) : 2a = 6 T heref ore a = 3, b = −3 and c = 1 Step 5 : Final answer The general formula for the pattern is 3n2 − 3n + 1 Step 6 : Term 100 Substitude n with 100: 3(100)2 − 3(100) + 1 = 29 701 The value for term 100 is 29 701. 268 (20.15) (20.16) (20.17) (20.18) CHAPTER 20. QUADRATIC SEQUENCES - GRADE 11 20.3 Extension: Plotting a graph of terms of a quadratic sequence Plotting an vs. n for a quadratic sequence yields a parabolic graph. Given the quadratic sequence, 3; 6; 10; 15; 21; . . . If we plot each of the terms vs. the corresponding index, we obtain a graph of a parabola. b a10 b Term, an a9 b a8 b a7 b a6 b a5 b a4 a3 a2 a1 b b b y-intercept, 1 2 a1 3 4 5 6 7 8 9 10 Index, n 20.3 End of chapter Exercises 1. Find the first 5 terms of the quadratic sequence defined by: an = n2 + 2n + 1 2. Determine which of the following sequences is a quadratic sequence by calculating the common second difference: A 6, 9, 14, 21, 30, . . . B 1, 7, 17, 31, 49, . . . C 8, 17, 32, 53, 80, . . . D 9, 26, 51, 84, 125, . . . E 2, 20, 50, 92, 146, . . . F 5, 19, 41, 71, 109, . . . G 2, 6, 10, 14, 18, . . . H 3, 9, 15, 21, 27, . . . I 10, 24, 44, 70, 102, . . . J 1, 2.5, 5, 8.5, 13, . . . K 2.5, 6, 10.5, 16, 22.5, . . . 269 20.3 CHAPTER 20. QUADRATIC SEQUENCES - GRADE 11 L 0.5, 9, 20.5, 35, 52.5, . . . 3. Given an = 2n2 , find for which value of n, an = 242? 4. Given an = (n − 4)2 , find for which value of n, an = 36? 5. Given an = n2 + 4, find for which value of n, an = 85? 6. Given an = 3n2 , find a11 ? 7. Given an = 7n2 + 4n, find a9 ? 8. Given an = 4n2 + 3n − 1, find a5 ? 9. Given an = 1,5n2 , find a10 ? 10. For each of the quadratic sequences, find the common second difference, the formula for the general term and then use the formula to find a100 . A 4,7,12,19,28, . . . B 2,8,18,32,50, . . . C 7,13,23,37,55, . . . D 5,14,29,50,77, . . . E 7,22,47,82,127, . . . F 3,10,21,36,55, . . . G 3,7,13,21,31, . . . H 1,8,27,64,125, . . . I 6,13,32,69,130, . . . J 2,9,17,27,39, . . . 270 Chapter 21 Finance - Grade 11 21.1 Introduction In Grade 10, the ideas of simple and compound interest was introduced. In this chapter we will be extending those ideas, so it is a good idea to go back to Chapter 8 and revise what you learnt in Grade 10. If you master the techniques in this chapter, you will understand about depreciation and will learn how to determine which bank is offering the better interest rate. 21.2 Depreciation It is said that when you drive a new car out of the dealership, it loses 20% of its value, because it is now “second-hand”. And from there on the value keeps falling, or depreciating. Second hand cars are cheaper than new cars, and the older the car, usually the cheaper it is. If you buy a second hand (or should we say pre-owned!) car from a dealership, they will base the price on something called book value. The book value of the car is the value of the car taking into account the loss in value due to wear, age and use. We call this loss in value depreciation, and in this section we will look at two ways of how this is calculated. Just like interest rates, the two methods of calculating depreciation are simple and compound methods. The terminology used for simple depreciation is straight-line depreciation and for compound depreciation is reducing-balance depreciation. In the straight-line method the value of the asset is reduced by the same constant amount each year. In the compound depreciation method the value of the asset is reduced by the same percentage each year. This means that the value of an asset does not decrease by a constant amount each year, but the decrease is most in the first year, then by a smaller amount in the second year and by even a smaller amount in the third year, and so on. Extension: Depreciation You may be wondering why we need to calculate depreciation. Determining the value of assets (as in the example of the second hand cars) is one reason, but there is also a more financial reason for calculating depreciation - tax! Companies can take depreciation into account as an expense, and thereby reduce their taxable income. A lower taxable income means that the company will pay less income tax to the Revenue Service. 21.3 Simple Depreciation (it really is simple!) Let us go back to the second hand cars. One way of calculating a depreciation amount would be to assume that the car has a limited useful life. Simple depreciation assumes that the value of 271 21.3 CHAPTER 21. FINANCE - GRADE 11 the car decreases by an equal amount each year. For example, let us say the limited useful life of a car is 5 years, and the cost of the car today is R60 000. What we are saying is that after 5 years you will have to buy a new car, which means that the old one will be valueless at that point in time. Therefore, the amount of depreciation is calculated: R60 000 = R12 000 per year. 5 years The value of the car is then: End End End End End of of of of of Year Year Year Year Year 1 2 3 4 5 R60 R60 R60 R60 R60 000 000 000 000 000 - 1×(R12 2×(R12 3×(R12 4×(R12 5×(R12 000) 000) 000) 000) 000) = = = = = R48 R36 R24 R12 R0 000 000 000 000 This looks similar to the formula for simple interest: Total Interest after n years = n × (P × i) where i is the annual percentage interest rate and P is the principal amount. If we replace the word interest with the word depreciation and the word principal with the words initial value we can use the same formula: Total depreciation after n years = n × (P × i) Then the book value of the asset after n years is: Initial Value - Total depreciation after n years = = P − n × (P × i) P (1 − n × i) For example, the book value of the car after two years can be simply calculated as follows: Book Value after 2 years = = = = = P (1 − n × i) R60 000(1 − 2 × 20%) R60 000(1 − 0,4) R60 000(0,6) R36 000 as expected. Note that the difference between the simple interest calculations and the simple depreciation calculations is that while the interest adds value to the principal amount, the depreciation amount reduces value! Worked Example 96: Simple Depreciation method Question: A car is worth R240 000 now. If it depreciates at a rate of 15% p.a. on a staight-line depreciation, what is it worth in 5 years’ time ? Answer Step 1 : Determine what has been provided and what is required P i = = n A = is R240 000 0,15 5 required 272 CHAPTER 21. FINANCE - GRADE 11 21.3 Step 2 : Determine how to approach the problem A = 240 000(1 − 0,15 × 5) Step 3 : Solve the problem A = = = 240 000(1 − 0,75) 240 000 × 0,25 60 000 Step 4 : Write the final answer In 5 years’ time the car is worth R60 000 Worked Example 97: Simple Depreciation Question: A small business buys a photocopier for R 12 000. For the tax return the owner depreciates this asset over 3 years using a straight-line depreciation method. What amount will he fill in on his tax form after 1 year, after 2 years and then after 3 years ? Answer Step 1 : Understanding the question The owner of the business wants the photocopier to depreciates to R0 after 3 years. Thus, the value of the photocopier will go down by 12 000 ÷ 3 = R4 000 per year. Step 2 : Value of the photocopier after 1 year 12 000 − 4 000 = R8 000 Step 3 : Value of the machine after 2 years 8 000 − 4 000 = R4 000 Step 4 : Write the final answer 4 000 − 4 000 = 0 After 3 years the photocopier is worth nothing Extension: Salvage Value Looking at the same example of our car with an initial value of R60 000, what if we suppose that we think we would be able to sell the car at the end of the 5 year period for R10 000? We call this amount the “Salvage Value” We are still assuming simple depreciation over a useful life of 5 years, but now instead of depreciating the full value of the asset, we will take into account the salvage value, and will only apply the depreciation to the value of the asset that we expect not to recoup, i.e. R60 000 - R10 000 = R50 000. The annual depreciation amount is then calculated as (R60 000 - R10 000) / 5 = R10 000 In general, the for simple (straight line) depreciation: Annual Depreciation = Initial Value - Salvage Value Useful Life 273 21.4 CHAPTER 21. FINANCE - GRADE 11 Exercise: Simple Depreciation 1. A business buys a truck for R560 000. Over a period of 10 years the value of the truck depreciates to R0 (using the straight-line method). What is the value of the truck after 8 years ? 2. Shrek wants to buy his grandpa’s donkey for R800. His grandpa is quite pleased with the offer, seeing that it only depreciated at a rate of 3% per year using the straight-line method. Grandpa bought the donkey 5 years ago. What did grandpa pay for the donkey then ? 3. Seven years ago, Rocco’s drum kit cost him R 12 500. It has now been valued at R2 300. What rate of simple depreciation does this represent ? 4. Fiona buys a DsTV satellite dish for R3 000. Due to weathering, its value depreciates simply at 15% per annum. After how long will the satellite dish be worth nothing ? 21.4 Compound Depreciation The second method of calculating depreciation is to assume that the value of the asset decreases at a certain annual rate, but that the initial value of the asset this year, is the book value of the asset at the end of last year. For example, if our second hand car has a limited useful life of 5 years and it has an initial value of R60 000, then the interest rate of depreciation is 20% (100%/5 years). After 1 year, the car is worth: Book Value after first year = P (1 − n × i) = R60 000(1 − 1 × 20%) = R60 000(1 − 0,2) = R60 000(0,8) = R48 000 At the beginning of the second year, the car is now worth R48 000, so after two years, the car is worth: Book Value after second year = P (1 − n × i) = R48 000(1 − 1 × 20%) = R48 000(1 − 0,2) = R48 000(0,8) = R38 400 We can tabulate these values. End End End End End of of of of of first year second year third year fourth year fifth year R60 R48 R38 R30 R24 000(1 − 1 × 20%)=R60 000(1 − 1 × 20%)=R60 400(1 − 1 × 20%)=R60 720(1 − 1 × 20%)=R60 576(1 − 1 × 20%)=R60 000(1 − 1 × 20%)1 000(1 − 1 × 20%)2 000(1 − 1 × 20%)3 000(1 − 1 × 20%)4 000(1 − 1 × 20%)5 = = = = = R48 R38 R30 R24 R19 000,00 400,00 720,00 576,00 608,80 We can now write a general formula for the book value of an asset if the depreciation is compounded. Initial Value - Total depreciation after n years = P (1 − i)n (21.1) 274 CHAPTER 21. FINANCE - GRADE 11 21.4 For example, the book value of the car after two years can be simply calculated as follows: Book Value after 2 years = P (1 − i)n = R60 000(1 − 20%)2 = R60 000(1 − 0,2)2 = R60 000(0,8)2 = R38 400 as expected. Note that the difference between the compound interest calculations and the compound depreciation calculations is that while the interest adds value to the principal amount, the depreciation amount reduces value! Worked Example 98: Compound Depreciation Question: The Flamingo population of the Bergriver mouth is depreciating on a reducing balance at a rate of 12% p.a. If there is now 3 200 flamingos in the wetlands of the Bergriver mouth, how many will there be in 5 years’ time ? Answer to three significant numbers. Answer Step 1 : Determine what has been provided and what is required P = R3 200 i n = = 0,12 5 is required A Step 2 : Determine how to approach the problem A = 3 200(1 − 0,12)5 Step 3 : Solve the problem A = = = 3 200(0,88)5 3 200 × 0,527731916 1688,742134 Step 4 : Write the final answer There would be approximately 1 690 flamingos in 5 years’ time. Worked Example 99: Compound Depreciation Question: Farmer Brown buys a tractor for R250 000 and depreciates it by 20% per year using the compound depreciation method. What is the depreciated value of the tractor after 5 years ? Answer Step 1 : Determine what has been provided and what is required 275 21.5 CHAPTER 21. FINANCE - GRADE 11 P = R250 000 i n = = 0,2 5 is required A Step 2 : Determine how to approach the problem A = 250 000(1 − 0,2)5 Step 3 : Solve the problem A = 250 000(0,8)5 = 250 000 × 0,32768 = 81 920 Step 4 : Write the final answer Depreciated value after 5 years is R 81 920 Exercise: Compound Depreciation 1. On January 1, 2008 the value of my Kia Sorento is R320 000. Each year after that, the cars value will decrease 20% of the previous years value. What is the value of the car on January 1, 2012. 2. The population of Bonduel decreases at a rate of 9,5% per annum as people migrate to the cities. Calculate the decrease in population over a period of 5 years if the initial population was 2 178 000. 3. A 20 kg watermelon consists of 98% water. If it is left outside in the sun it loses 3% of its water each day. How much does in weigh after a month of 31 days ? 4. A computer depreciates at x% per annum using the reducing-balance method. Four years ago the value of the computer was R10 000 and is now worth R4 520. Calculate the value of x correct to two decimal places. 21.5 Present Values or Future Values of an Investment or Loan 21.5.1 Now or Later When we studied simple and compound interest we looked at having a sum of money now, and calculating what it will be worth in the future. Whether the money was borrowed or invested, the calculations examined what the total money would be at some future date. We call these future values. 276 CHAPTER 21. FINANCE - GRADE 11 21.5 It is also possible, however, to look at a sum of money in the future, and work out what it is worth now. This is called a present value. For example, if R1 000 is deposited into a bank account now, the future value is what that amount will accrue to by some specified future date. However, if R1 000 is needed at some future time, then the present value can be found by working backwards - in other words, how much must be invested to ensure the money grows to R1 000 at that future date? The equation we have been using so far in compound interest, which relates the open balance (P ), the closing balance (A), the interest rate (i as a rate per annum) and the term (n in years) is: A = P · (1 + i)n (21.2) Using simple algebra, we can solve for P instead of A, and come up with: P = A · (1 + i)−n (21.3) This can also be written as follows, but the first approach is usually preferred. P = A/(1 + i)n (21.4) Now think about what is happening here. In Equation 21.2, we start off with a sum of money and we let it grow for n years. In Equation 21.3 we have a sum of money which we know in n years time, and we “unwind” the interest - in other words we take off interest for n years, until we see what it is worth right now. We can test this as follows. If I have R1 000 now and I invest it at 10% for 5 years, I will have: A = = = P · (1 + i)n R1 000(1 + 10%)5 R1 610,51 at the end. BUT, if I know I have to have R1 610,51 in 5 years time, I need to invest: P = = A · (1 + i)−n R1 610,51(1 + 10%)−5 = R1 000 We end up with R1 000 which - if you think about it for a moment - is what we started off with. Do you see that? Of course we could apply the same techniques to calculate a present value amount under simple interest rate assumptions - we just need to solve for the opening balance using the equations for simple interest. A = P (1 + i × n) (21.5) P = A/(1 + i × n) (21.6) Solving for P gives: Let us say you need to accumulate an amount of R1 210 in 3 years time, and a bank account pays Simple Interest of 7%. How much would you need to invest in this bank account today? P = = = A 1+n·i R1 210 1 + 3 × 7% R1 000 Does this look familiar? Look back to the simple interest worked example in Grade 10. There we started with an amount of R1 000 and looked at what it would grow 277 21.6 CHAPTER 21. FINANCE - GRADE 11 to in 3 years’ time using simple interest rates. Now we have worked backwards to see what amount we need as an opening balance in order to achieve the closing balance of R1 210. In practice, however, present values are usually always calculated assuming compound interest. So unless you are explicitly asked to calculate a present value (or opening balance) using simple interest rates, make sure you use the compound interest rate formula! Exercise: Present and Future Values 1. After a 20-year period Josh’s lump sum investment matures to an amount of R313 550. How much did he invest if his money earned interest at a rate of 13,65% p.a.compounded half yearly for the first 10 years, 8,4% p.a. compounded quarterly for the next five years and 7,2% p.a. compounded monthly for the remaining period ? 2. A loan has to be returned in two equal semi-annual instalments. If the rate of interest is 16% per annum, compounded semi-annually and each instalment is R1 458, find the sum borrowed. 21.6 Finding i By this stage in your studies of the mathematics of finance, you have always known what interest rate to use in the calculations, and how long the investment or loan will last. You have then either taken a known starting point and calculated a future value, or taken a known future value and calculated a present value. But here are other questions you might ask: 1. I want to borrow R2 500 from my neighbour, who said I could pay back R3 000 in 8 months time. What interest is she charging me? 2. I will need R450 for some university textbooks in 1,5 years time. I currently have R400. What interest rate do I need to earn to meet this goal? Each time that you see something different from what you have seen before, start off with the basic equation that you should recognise very well: A = P · (1 + i)n If this were an algebra problem, and you were told to “solve for i”, you should be able to show that: A/P = (1 + i)n (1 + i) = (A/P )1/n i = (A/P )1/n − 1 You do not need to memorise this equation, it is easy to derive any time you need it! So let us look at the two examples mentioned above. 1. Check that you agree that P =R2 500, A=R3 000, n=8/12=0,666667. This means that: i = = (R3 000/R2 500)1/0,666667 − 1 31,45% Ouch! That is not a very generous neighbour you have. 278 CHAPTER 21. FINANCE - GRADE 11 21.7 2. Check that P =R400, A=R450, n=1,5 i = = (R450/R400)1/1,5 − 1 8,17% This means that as long as you can find a bank which pays more than 8,17% interest, you should have the money you need! 8 Note that in both examples, we expressed n as a number of years ( 12 years, not 8 because that is the number of months) which means i is the annual interest rate. Always keep this in mind keep years with years to avoid making silly mistakes. Exercise: Finding i 1. A machine costs R45 000 and has a scrap value of R9 000 after 10 years. Determine the annual rate of depreciation if it is calculated on the reducing balance method. 2. After 5 years an investment doubled in value. At what annual rate was interest compounded ? 21.7 Finding n - Trial and Error By this stage you should be seeing a pattern. We have our standard formula, which has a number of variables: A = P · (1 + i)n We have solved for A (in section 8.5), P (in section 21.5) and i (in section 21.6). This time we are going to solve for n. In other words, if we know what the starting sum of money is and what it grows to, and if we know what interest rate applies - then we can work out how long the money needs to be invested for all those other numbers to tie up. This section will calculate n by trial and error and by using a calculator. The proper algebraic solution will be learnt in Grade 12. Solving for n, we can write: A = A = P P (1 + i)n (1 + i)n Now we have to examine the numbers involved to try to determine what a possible value of n is. Refer to Table 5.1 (on page 38) for some ideas as to how to go about finding n. Worked Example 100: Term of Investment - Trial and Error Question: If we invest R3 500 into a savings account which pays 7,5% compound interest for an unknown period of time, at the end of which our account is worth R4 044,69. How long did we invest the money? Answer Step 1 : Determine what is given and what is required 279 21.8 CHAPTER 21. FINANCE - GRADE 11 • P =R3 500 • i=7,5% • A=R4 044,69 We are required to find n. Step 2 : Determine how to approach the problem We know that: A = A = P P (1 + i)n (1 + i)n Step 3 : Solve the problem R4 044,69 = R3 500 1,156 = (1 + 7,5%)n (1,075)n We now use our calculator and try a few values for n. Possible n 1,0 1,5 2,0 2,5 1,075n 1,075 1,115 1,156 1,198 We see that n is close to 2. Step 4 : Write final answer The R3 500 was invested for about 2 years. Exercise: Finding n - Trial and Error 1. A company buys two typs of motor cars: The Acura costs R80 600 and the Brata R101 700 VAT included. The Acura depreciates at a rate, compunded annually of 15,3% per year and the Brata at 19,7&, also compunded annually, per year. After how many years will the book value of the two models be the same ? 2. The fuel in the tank of a truck decreases every minute by 5,5% of the amount in the tank at that point in time. Calculate after how many minutes there will be less than 30l in the tank if it originally held 200l. 21.8 Nominal and Effective Interest Rates So far we have discussed annual interest rates, where the interest is quoted as a per annum amount. Although it has not been explicitly stated, we have assumed that when the interest is quoted as a per annum amount it means that the interest is once a year. Interest however, may be paid more than just once a year, for example we could receive interest on a monthly basis, i.e. 12 times per year. So how do we compare a monthly interest rate, say, to an annual interest rate? This brings us to the concept of the effective annual interest rate. 280 CHAPTER 21. FINANCE - GRADE 11 21.8 One way to compare different rates and methods of interest payments would be to compare the Closing Balances under the different options, for a given Opening Balance. Another, more widely used, way is to calculate and compare the “effective annual interest rate” on each option. This way, regardless of the differences in how frequently the interest is paid, we can compare apples-with-apples. For example, a savings account with an opening balance of R1 000 offers a compound interest rate of 1% per month which is paid at the end of every month. We can calculate the accumulated balance at the end of the year using the formulae from the previous section. But be careful our interest rate has been given as a monthly rate, so we need to use the same units (months) for our time period of measurement. So we can calculate the amount that would be accumulated by the end of 1-year as follows: Closing Balance after 12 months = = = P × (1 + i)n R1 000 × (1 + 1%)12 R1 126,83 Note that because we are using a monthly time period, we have used n = 12 months to calculate the balance at the end of one year. The effective annual interest rate is an annual interest rate which represents the equivalent per annum interest rate assuming compounding. It is the annual interest rate in our Compound Interest equation that equates to the same accumulated balance after one year. So we need to solve for the effective annual interest rate so that the accumulated balance is equal to our calculated amount of R1 126,83. We use i12 to denote the monthly interest rate. We have introduced this notation here to distinguish between the annual interest rate, i. Specifically, we need to solve for i in the following equation: P × (1 + i)1 = P × (1 + i12)12 (1 + i) = (1 + i12)12 divide both sides by P i = (1 + i12)12 − 1 subtract 1 from both sides For the example, this means that the effective annual rate for a monthly rate i12 = 1% is: i = = = = (1 + i12)12 − 1 (1 + 1%)12 − 1 0,12683 12,683% If we recalculate the closing balance using this annual rate we get: Closing Balance after 1 year = = = P × (1 + i)n R1 000 × (1 + 12,683%)1 R1 126,83 which is the same as the answer obtained for 12 months. Note that this is greater than simply multiplying the monthly rate by 12 (12 × 1% = 12%) due to the effects of compounding. The difference is due to interest on interest. We have seen this before, but it is an important point! 21.8.1 The General Formula So we know how to convert a monthly interest rate into an effective annual interest. Similarly, we can convert a quarterly interest, or a semi-annual interest rate or an interest rate of any frequency for that matter into an effective annual interest rate. 281 Remember, the trick to using the formulae is to define the time period, and use the interest rate relevant to the time period. 21.8 CHAPTER 21. FINANCE - GRADE 11 For a quarterly interest rate of say 3% per quarter, the interest will be paid four times per year (every three month). We can calculate the effective annual interest rate by solving for i: P (1 + i) = P (1 + i4)4 where i4 is the quarterly interest rate. So (1 + i) = (1,03)4 , and so i = 12,55%. This is the effective annual interest rate. In general, for interest paid at a frequency of T times per annum, the follow equation holds: P (1 + i) = P (1 + iT )T (21.7) where iT is the interest rate paid T times per annum. 21.8.2 De-coding the Terminology Market convention however, is not to state the interest rate as say 1% per month, but rather to express this amount as an annual amount which in this example would be paid monthly. This annual amount is called the nominal amount. The market convention is to quote a nominal interest rate of “12% per annum paid monthly” instead of saying (an effective) 1% per month. We know from a previous example, that a nominal interest rate of 12% per annum paid monthly, equates to an effective annual interest rate of 12,68%, and the difference is due to the effects of interest-on-interest. So if you are given an interest rate expressed as an annual rate but paid more frequently than annual, we first need to calculate the actual interest paid per period in order to calculate the effective annual interest rate. monthly interest rate = Nominal interest Rate per annum number of periods per year (21.8) For example, the monthly interest rate on 12% interest per annum paid monthly, is: monthly interest rate = = = Nominal interest Rate per annum number of periods per year 12% 12 months 1% per month The same principle apply to other frequencies of payment. Worked Example 101: Nominal Interest Rate Question: Consider a savings account which pays a nominal interest at 8% per annum, paid quarterly. Calculate (a) the interest amount that is paid each quarter, and (b) the effective annual interest rate. Answer Step 1 : Determine what is given and what is required We are given that a savings account has a nominal interest rate of 8% paid quarterly. We are required to find: • the quarterly interest rate, i4 • the effective annual interest rate, i Step 2 : Determine how to approach the problem We know that: quarterly interest rate = Nominal interest Rate per annum number of quarters per year 282 CHAPTER 21. FINANCE - GRADE 11 21.8 and P (1 + i) = P (1 + iT )T where T is 4 because there are 4 payments each year. Step 3 : Calculate the monthly interest rate quarterly interest rate = = = Nominal interest Rate per annum number of periods per year 8% 4 quarters 2% per quarter Step 4 : Calculate the effective annual interest rate The effective annual interest rate (i) is calculated as: (1 + i) = (1 + i4)4 (1 + i) = (1 + 2%)4 i = (1 + 2%)4 − 1 = 8,24% Step 5 : Write the final answer The quarterly interest rate is 2% and the effective annual interest rate is 8,24%, for a nominal interest rate of 8% paid quarterly. Worked Example 102: Nominal Interest Rate Question: On their saving accounts, Echo Bank offers an interest rate of 18% nominal, paid monthly. If you save R100 in such an account now, how much would the amount have accumulated to in 3 years’ time? Answer Step 1 : Determine what is given and what is required Interest rate is 18% nominal paid monthly. There are 12 months in a year. We are working with a yearly time period, so n = 3. The amount we have saved is R100, so P = 100. We need the accumulated value, A. Step 2 : Recall relevant formulae We know that monthly interest rate = Nominal interest Rate per annum number of periods per year for converting from nominal interest rate to effective interest rate, we have 1 + i = (1 + iT )T and for cacluating accumulated value, we have A = P × (1 + i)n Step 3 : Calculate the effective interest rate There are 12 month in a year, so i12 = = = Nominal annual interest rate 12 18% 12 1,5% per month 283 21.9 CHAPTER 21. FINANCE - GRADE 11 and then, we have 1+i = i = = = = (1 + i12)1 2 (1 + i12)1 2 − 1 (1 + 1,5%)1 2 − 1 (1,015)1 2 − 1 19,56% Step 4 : Reach the final answer A = P × (1 + i)n = 100 × (1 + 19,56%)3 = 100 × 1,7091 = 170,91 Step 5 : Write the final answer The accumulated value is R170,91. (Remember to round off to the the nearest cent.) Exercise: Nominal and Effect Interest Rates 1. Calculate the effective rate equivalent to a nominal interest rate of 8,75% p.a. compounded monthly. 2. Cebela is quoted a nominal interest rate of 9,15% per annum compounded every four months on her investment of R 85 000. Calculate the effective rate per annum. 21.9 Formulae Sheet As an easy reference, here are the key formulae that we derived and used during this chapter. While memorising them is nice (there are not many), it is the application that is useful. Financial experts are not paid a salary in order to recite formulae, they are paid a salary to use the right methods to solve financial problems. 21.9.1 P i n iT Definitions Principal (the amount of money at the starting point of the calculation) interest rate, normally the effective rate per annum period for which the investment is made Rate the interest rate paid T times per annum, i.e. iT = Nominal Interest T 284 CHAPTER 21. FINANCE - GRADE 11 21.9.2 21.10 Equations Simple Increase : A = Compound Increase : A = P (1 + i × n) P (1 + i)n Simple Decrease : A = Compound Decrease : A = P (1 − i × n) P (1 − i)n Ef f ective Annual Interest Rate(i) : (1 + i) = 21.10 (1 + iT )T End of Chapter Exercises 1. Shrek buys a Mercedes worth R385 000 in 2007. What will the value of the Mercedes be at the end of 2013 if A the car depreciates at 6% p.a. straight-line depreciation B the car depreciates at 12% p.a. reducing-balance depreciation. 2. Greg enters into a 5-year hire-purchase agreement to buy a computer for R8 900. The interest rate is quoted as 11% per annum based on simple interest. Calculate the required monthly payment for this contract. 3. A computer is purchased for R16 000. It depreciates at 15% per annum. A Determine the book value of the computer after 3 years if depreciation is calculated according to the straight-line method. B Find the rate, according to the reducing-balance method, that would yield the same book value as in 3a after 3 years. 4. Maggie invests R12 500,00 for 5 years at 12% per annum compounded monthly for the first 2 years and 14% per annum compounded semi-annually for the next 3 years. How much will Maggie receive in total after 5 years? 5. Tintin invests R120 000. He is quoted a nominal interest rate of 7,2% per annum compounded monthly. A Calculate the effective rate per annum correct to THREE decimal places. B Use the effective rate to calculate the value of Tintin’s investment if he invested the money for 3 years. C Suppose Tintin invests his money for a total period of 4 years, but after 18 months makes a withdrawal of R20 000, how much will he receive at the end of the 4 years? 6. Paris opens accounts at a number of clothing stores and spends freely. She gets heself into terrible debt and she cannot pay off her accounts. She owes Hilton Fashion world R5 000 and the shop agrees to let Paris pay the bill at a nominal interest rate of 24% compounded monthly. A How much money will she owe Hilton Fashion World after two years ? B What is the effective rate of interest that Hilton Fashion World is charging her ? 285 21.10 CHAPTER 21. FINANCE - GRADE 11 286 Chapter 22 Solving Quadratic Equations Grade 11 22.1 Introduction In grade 10, the basics of solving linear equations, quadratic equations, exponential equations and linear inequalities were studied. This chapter extends on that work. We look at different methods of solving quadratic equations. 22.2 Solution by Factorisation The solving of quadratic equations by factorisation was discussed in Grade 10. Here is an example to remind you of what is involved. Worked Example 103: Solution of Quadratic Equations Question: Solve the equation 2x2 − 5x − 12 = 0. Answer Step 1 : Determine whether the equation has common factors This equation has no common factors. Step 2 : Determine if the equation is in the form ax2 + bx + c with a > 0 The equation is in the required form, with a = 2, b = −5 and c = −12. Step 3 : Factorise the quadratic 2x2 − 5x − 12 has factors of the form: (2x + s)(x + v) with s and v constants to be determined. This multiplies out to 2x2 + (s + 2v)x + sv We see that sv = −12 and s + 2v = −5. This is a set of simultaneous equations in s and v, but it is easy to solve numerically. All the options for s and v are considered below. 287 22.2 CHAPTER 22. SOLVING QUADRATIC EQUATIONS - GRADE 11 s 2 -2 3 -3 4 -4 6 -6 v -6 6 -4 4 -3 3 -2 2 s + 2v -10 10 -5 5 -2 2 2 -2 We see that the combination s = 3 and v = −4 gives s + 2v = −5. Step 4 : Write the equation with factors (2x + 3)(x − 4) = 0 Step 5 : Solve the equation If two brackets are multiplied together and give 0, then one of the brackets must be 0, therefore 2x + 3 = 0 or x−4=0 Therefore, x = − 23 or x = 4 Step 6 : Write the final answer The solutions to 2x2 − 5x − 12 = 0 are x = − 23 or x = 4. It is important to remember that a quadratic equation has to be in the form ax2 + bx + c = 0 before one can solve it using these methods. Worked Example 104: Solving quadratic equation by factorisation Question: Solve for a: a(a − 3) = 10 Answer Step 1 : Rewrite the equation in the form ax2 + bx + c = 0 Remove the brackets and move all terms to one side. a2 − 3a − 10 = 0 Step 2 : Factorise the trinomial (a + 2)(a − 5) = 0 Step 3 : Solve the equation a+2=0 or a−5=0 Solve the two linear equations and check the solutions in the original equation. Step 4 : Write the final answer Therefore, a = −2 or a = 5 288 CHAPTER 22. SOLVING QUADRATIC EQUATIONS - GRADE 11 Worked Example 105: Solving fractions that lead to a quadratic equation 4 3b + 1 = b+1 Question: Solve for b: b+2 Answer Step 1 : Put both sides over the LCM 3b(b + 1) + (b + 2)(b + 1) 4(b + 2) = (b + 2)(b + 1) (b + 2)(b + 1) Step 2 : Determine the restrictions The denominators are the same, therefore the numerators must be the same. However, b 6= −2 and b 6= −1 Step 3 : Simplify equation to the standard form 3b2 + 3b + b2 + 3b + 2 = 4b2 + 2b − 6 = 2b2 + b − 3 = 4b + 8 0 0 Step 4 : Factorise the trinomial and solve the equation (2b + 3)(b − 1) = 2b + 3 = 0 or −3 or b= 2 0 b−1=0 b=1 Step 5 : Check solutions in original equation Both solutions are valid Therefore, b = −3 2 or b = 1 Exercise: Solution by Factorisation Solve the following quadratic equations by factorisation. Some answers may be left in surd form. 1. 2y 2 − 61 = 101 2. 2y 2 − 10 = 0 3. y 2 − 4 = 10 4. 2y 2 − 8 = 28 5. 7y 2 = 28 6. y 2 + 28 = 100 7. 7y 2 + 14y = 0 8. 12y 2 + 24y + 12 = 0 9. 16y 2 − 400 = 0 10. y 2 − 5y + 6 = 0 11. y 2 + 5y − 36 = 0 12. y 2 + 2y = 8 13. −y 2 − 11y − 24 = 0 14. 13y − 42 = y 2 289 22.2 22.3 CHAPTER 22. SOLVING QUADRATIC EQUATIONS - GRADE 11 15. y 2 + 9y + 14 = 0 16. y 2 − 5ky + 4k 2 = 0 17. y(2y + 1) = 15 22.3 18. 5y y−2 + 3 y 19. y−2 y+1 = 2y+1 y−7 +2= −6 y 2 −2y Solution by Completing the Square We have seen that expressions of the form: a2 x2 − b2 are known as differences of squares and can be factorised as follows: (ax − b)(ax + b). This simple factorisation leads to another technique to solve quadratic equations known as completing the square. We demonstrate with a simple example, by trying to solve for x in: x2 − 2x − 1 = 0. (22.1) We cannot easily find factors of this term, but the first two terms look similar to the first two terms of the perfect square: (x − 1)2 = x2 − 2x + 1. However, we can cheat and create a perfect square by adding 2 to both sides of the equation in (22.1) as: 2 x2 − 2x − 1 = 0 (x − 1)2 = (x − 1)2 − 2 = 2 0 x − 2x − 1 + 2 = x2 − 2x + 1 = Now we know that: 0+2 2 √ 2 = ( 2)2 which means that: (x − 1)2 − 2 is a difference of squares. Therefore we can write: √ √ (x − 1)2 − 2 = [(x − 1) − 2][(x − 1) + 2] = 0. The solution to x2 − 2x − 1 = 0 is then: (x − 1) − or √ 2=0 √ (x − 1) + 2 = 0. √ √ This means x = 1 + 2 or x = 1 − 2. This example demonstrates the use of completing the square to solve a quadratic equation. 290 CHAPTER 22. SOLVING QUADRATIC EQUATIONS - GRADE 11 22.3 Method: Solving Quadratic Equations by Completing the Square 1. Write the equation in the form ax2 + bx + c = 0. e.g. x2 + 2x − 3 = 0 2. Take the constant over to the right hand side of the equation. e.g. x2 + 2x = 3 3. If necessary, make the coefficient of the x2 term = 1, by dividing through by the existing coefficient. 4. Take half the coefficient of the x term, square it and add it to both sides of the equation. e.g. in x2 + 2x = 3, half of the x term is 1. 11 = 1. Therefore we add 1 to both sides to get: x2 + 2x + 1 = 3 + 1. 5. Write the left hand side as a perfect square: (x + 1)2 − 4 = 0 6. You should then be able to factorise the equation in terms of difference of squares and then solve for x: (x + 1 − 2)(x + 1 + 2) = 0 Worked Example 106: Solving Quadratic Equations by Completing the Square Question: Solve: x2 − 10x − 11 = 0 by completing the square Answer Step 1 : Write the equation in the form ax2 + bx + c = 0 x2 − 10x − 11 = 0 Step 2 : Take the constant over to the right hand side of the equation x2 − 10x = 11 Step 3 : Check that the coefficient of the x2 term is 1. The coefficient of the x2 term is 1. Step 4 : Take half the coefficient of the x term, square it and add it to both sides The coefficient of the x term is -10. (−10) = −5. (−5)2 = 25. Therefore: 2 x2 − 10x + 25 = 11 + 25 Step 5 : Write the left hand side as a perfect square (x − 5)2 − 36 = 0 Step 6 : Factorise equation as difference of squares (x − 5)2 − 36 = 0 [(x − 5) + 6][(x − 5) − 6] = 0 Step 7 : Solve for the unknown value [x + 1][x − 11] ∴ x = −1 = or 291 0 x = 11 22.3 CHAPTER 22. SOLVING QUADRATIC EQUATIONS - GRADE 11 Worked Example 107: Solving Quadratic Equations by Completing the Square Question: Solve: 2x2 − 8x − 16 = 0 by completing the square Answer Step 1 : Write the equation in the form ax2 + bx + c = 0 2x2 − 8x − 16 = 0 Step 2 : Take the constant over to the right hand side of the equation 2x2 − 8x = 16 Step 3 : Check that the coefficient of the x2 term is 1. The coefficient of the x2 term is 2. Therefore, divide both sides by 2: x2 − 4x = 8 Step 4 : Take half the coefficient of the x term, square it and add it to both sides 2 The coefficient of the x term is -4. (−4) 2 = −2. (−2) = 4. Therefore: x2 − 4x + 4 = 8 + 4 Step 5 : Write the left hand side as a perfect square (x − 2)2 − 12 = 0 Step 6 : Factorise equation as difference of squares [(x − 2) + √ √ 12][(x − 2) − 12] = 0 Step 7 : Solve for the unknown value [x − 2 + √ √ 12][x − 2 − 12] = 0 √ √ ∴ x = 2 − 12 or x = 2 + 12 Step 8 : The last three steps can also be done in a different the way Leave left hand side written as a perfect square (x − 2)2 = 12 Step 9 : Take the square root on both sides of the equation √ x − 2 = ± 12 Step 10 : Solve for √x √ Therefore x = 2 − 12 or x = 2 + 12 Compare to answer in step 7. Exercise: Solution by Completing the Square Solve the following equations by completing the square: 292 CHAPTER 22. SOLVING QUADRATIC EQUATIONS - GRADE 11 22.4 1. x2 + 10x − 2 = 0 2. x2 + 4x + 3 = 0 3. x2 + 8x − 5 = 0 4. 2x2 + 12x + 4 = 0 5. x2 + 5x + 9 = 0 6. x2 + 16x + 10 = 0 7. 3x2 + 6x − 2 = 0 8. z 2 + 8z − 6 = 0 9. 2z 2 − 11z = 0 10. 5 + 4z − z 2 = 0 22.4 Solution by the Quadratic Formula It is not always possible to solve a quadratic equation by factorising and it is lengthy and tedious to solve a quadratic equations by completing the square. In these situations, you can use the quadratic formula that gives the solutions to any quadratic equation. Consider the general form of the quadratic function: f (x) = ax2 + bx + c. Factor out the a to get: c b (22.2) f (x) = a(x2 + x + ). a a Now we need to do some detective work to figure out how to turn (22.2) into a perfect square plus some extra terms. We know that for a perfect square: (m + n)2 = m2 + 2mn + n2 and (m − n)2 = m2 − 2mn + n2 The key is the middle term, which is 2× the first term × the second term. In (22.2), we know b that the first term is x so 2× the second term is ab . This means that the second term is 2a . So, (x + b b b 2 ) = x2 + 2 x + ( )2 . 2a 2a 2a In general if you add a quantity and subtract the same quantity, nothing has changed. This b 2 means if we add and subtract 2a from the right hand side of (22.2) we will get: f (x) c b = a(x2 + x + ) a a ! 2 2 b c b b 2 = a x + x+ − + a 2a 2a a ! 2 2 b c b = a x+ − + 2a 2a a 2 ! b2 b +c− = a x+ 2a 4a (22.3) (22.4) (22.5) (22.6) We set f (x) = 0 to find its roots, which yields: a(x + b 2 b2 ) = −c 2a 4a 293 (22.7) 22.4 CHAPTER 22. SOLVING QUADRATIC EQUATIONS - GRADE 11 Now dividing by a and taking the square root of both sides gives the expression r c b b2 − =± x+ 2 2a 4a a (22.8) Finally, solving for x implies that x r c b b2 − = − ± 2 2a 4a a r b b2 − 4ac = − ± 2a 4a2 which can be further simplified to: x= −b ± √ b2 − 4ac 2a (22.9) These are the solutions to the quadratic equation. Notice that there are two solutions in general, but these may not always exists (depending on the sign of the expression b2 − 4ac under the square root). These solutions are also called the roots of the quadratic equation. Worked Example 108: Using the quadratic formula Question: Solve for the roots of the function f (x) = 2x2 + 3x − 7. Answer Step 1 : Determine whether the equation can be factorised The expression cannot be factorised. Therefore, the general quadratic formula must be used. Step 2 : Identify the coefficients in the equation for use in the formula From the equation: a=2 b=3 c = −7 Step 3 : Apply the quadratic formula Always write down the formula first and then substitute the values of a, b and c. √ −b ± b2 − 4ac x = (22.10) 2a p −(3) ± (3)2 − 4(2)(−7) = (22.11) 2(2) √ −3 ± 56 (22.12) = 4 √ −3 ± 2 14 = (22.13) 4 Step 4 : Write the final answer The two roots of f (x) = 2x2 + 3x − 7 are x = √ −3+2 14 4 and √ −3−2 14 . 4 Worked Example 109: Using the quadratic formula but no solution Question: Solve for the solutions to the quadratic equation x2 − 5x + 8 = 0. 294 CHAPTER 22. SOLVING QUADRATIC EQUATIONS - GRADE 11 22.4 Answer Step 1 : Determine whether the equation can be factorised The expression cannot be factorised. Therefore, the general quadratic formula must be used. Step 2 : Identify the coefficients in the equation for use in the formula From the equation: a=1 b = −5 c=8 Step 3 : Apply the quadratic formula x = = = √ b2 − 4ac 2a p −(−5) ± (−5)2 − 4(1)(8) 2(1) √ 5 ± −7 2 −b ± (22.14) (22.15) (22.16) (22.17) Step 4 : Write the final answer Since √ the expression under the square root is negative these are not real solutions ( −7 is not a real number). Therefore there are no real solutions to the quadratic equation x2 − 5x + 8 = 0. This means that the graph of the quadratic function f (x) = x2 − 5x + 8 has no x-intercepts, but that the entire graph lies above the x-axis. Exercise: Solution by the Quadratic Formula Solve for t using the quadratic formula. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 3t2 + t − 4 = 0 t2 − 5t + 9 = 0 2t2 + 6t + 5 = 0 4t2 + 2t + 2 = 0 −3t2 + 5t − 8 = 0 −5t2 + 3t − 3 = 0 t2 − 4t + 2 = 0 9t2 − 7t − 9 = 0 2t2 + 3t + 2 = 0 t2 + t + 1 = 0 Important: • In all the examples done so far, the solutions were left in surd form. Answers can also be given in decimal form, using the calculator. Read the instructions when answering questions in a test or exam whether to leave answers in surd form, or in decimal form to an appropriate number of decimal places. • Completing the square as a method to solve a quadratic equation is only done when specifically asked. 295 22.5 CHAPTER 22. SOLVING QUADRATIC EQUATIONS - GRADE 11 Exercise: Mixed Exercises Solve the quadratic equations by either factorisation, completing the square or by using the quadratic formula: • Always try to factorise first, then use the formula if the trinomial cannot be factorised. • Do some of them by completing the square and then compare answers to those done using the other methods. 1. 24y 2 + 61y − 8 = 0 4. −5y 2 + 0y + 5 = 0 7. −12y 2 + 66y − 72 = 0 10. 6y 2 + 7y − 24 = 0 13. −25y 2 + 25y − 4 = 0 16. 35y 2 − 8y − 3 = 0 19. −4y 2 − 41y − 45 = 0 22. 9y 2 − 76y + 32 = 0 25. 64y 2 + 96y + 36 = 0 28. 3y 2 + 10y − 48 = 0 31. x2 − 70 = 11 34. 2y 2 − 98 = 0 22.5 2. −8y 2 − 16y + 42 = 0 5. −3y 2 + 15y − 12 = 0 8. −40y 2 + 58y − 12 = 0 11. 2y 2 − 5y − 3 = 0 14. −32y 2 + 24y + 8 = 0 17. −81y 2 − 99y − 18 = 0 20. 16y 2 + 20y − 36 = 0 23. −54y 2 + 21y + 3 = 0 26. 12y 2 − 22y − 14 = 0 29. −4y 2 + 8y − 3 = 0 32. 2x2 − 30 = 2 35. 5y 2 − 10 = 115 3. −9y 2 + 24y − 12 = 0 6. 49y 2 + 0y − 25 = 0 9. −24y 2 + 37y + 72 = 0 12. −18y 2 − 55y − 25 = 0 15. 9y 2 − 13y − 10 = 0 18. 14y 2 − 81y + 81 = 0 21. 42y 2 + 104y + 64 = 0 24. 36y 2 + 44y + 8 = 0 27. 16y 2 + 0y − 81 = 0 30. −5y 2 − 26y + 63 = 0 33. x2 − 16 = 2 − x2 36. 5y 2 − 5 = 19 − y 2 Finding an equation when you know its roots We have mentioned before that the roots of a quadratic equation are the solutions or answers you get from solving the quadatic equation. Working back from the answers, will take you to an equation. Worked Example 110: Find an equation when roots are given Question: Find an equation with roots 13 and -5 Answer Step 1 : Write down as the product of two brackets The step before giving the solutions would be: (x − 13)(x + 5) = 0 Notice that the signs in the brackets are opposite of the given roots. Step 2 : Remove brackets x2 − 8x − 65 = 0 Of course, there would be other possibilities as well when each term on each side of the equal to sign is multiplied by a constant. 296 CHAPTER 22. SOLVING QUADRATIC EQUATIONS - GRADE 11 22.5 Worked Example 111: Fraction roots Question: Find an equation with roots − 23 and 4 Answer Step 1 : Product of two brackets Notice that if x = − 32 then 2x + 3 = 0 Therefore the two brackets will be: (2x + 3)(x − 4) = 0 Step 2 : Remove brackets The equation is: 2x2 − 5x − 12 = 0 Extension: Theory of Quadratic Equations - Advanced This section is not in the syllabus, but it gives one a good understanding about some of the solutions of the quadratic equations. What is the Discriminant of a Quadratic Equation? Consider a general quadratic function of the form f (x) = ax2 + bx + c. The discriminant is defined as: ∆ = b2 − 4ac. (22.18) This is the expression under the square root in the formula for the roots of this function. We have already seen that whether the roots exist or not depends on whether this factor ∆ is negative or positive. The Nature of the Roots Real Roots (∆ ≥ 0) Consider ∆ ≥ 0 for some quadratic function f (x) = ax2 + bx + c. In this case there are solutions to the equation f (x) = 0 given by the formula √ √ −b ± b2 − 4ac −b ± ∆ x= = (22.19) 2a 2a Since the square roots exists (the expression under the square root is non-negative.) These are the roots of the function f (x). There various possibilities are summarised in the figure below. ∆ ∆ < 0 - imaginary roots ∆ ≥ 0 - real roots ∆>0 unequal roots ∆ a perfect square - rational roots 297 ∆=0 equal roots ∆ not a perfect square irrational roots 22.5 CHAPTER 22. SOLVING QUADRATIC EQUATIONS - GRADE 11 Equal Roots (∆ = 0) If ∆ = 0, then the roots are equal and, from the formula, these are given by x=− b 2a (22.20) Unequal Roots (∆ > 0) There will be 2 unequal roots if ∆ > 0. The roots of f (x) are rational if ∆ is a perfect√square (a number which is the square of a rational number), since, in this case, ∆ is rational. Otherwise, if ∆ is not a perfect square, then the roots are irrational. Imaginary Roots (∆ < 0) If ∆ < 0, then the solution to f (x) = ax2 + bx + c = 0 contains the square root of a negative number and therefore there are no real solutions. We therefore say that the roots of f (x) are imaginary (the graph of the function f (x) does not intersect the x-axis). Extension: Theory of Quadratics - advanced exercises Exercise: From past papers 1. [IEB, Nov. 2001, HG] Given: x2 + bx − 2 + k(x2 + 3x + 2) = 0 (k 6= −1) A Show that the discriminant is given by: ∆ = k 2 + 6bk + b2 + 8 2. 3. 4. 5. B If b = 0, discuss the nature of the roots of the equation. C If b = 2, find the value(s) of k for which the roots are equal. [IEB, Nov. 2002, HG] Show that k 2 x2 + 2 = kx − x2 has non-real roots for all real values for k. [IEB, Nov. 2003, HG] The equation x2 + 12x = 3kx2 + 2 has real roots. A Find the largest integral value of k. B Find one rational value of k, for which the above equation has rational roots. [IEB, Nov. 2003, HG] In the quadratic equation px2 + qx + r = 0, p, q and r are positive real numbers and form a geometric sequence. Discuss the nature of the roots. [IEB, Nov. 2004, HG] Consider the equation: k= x2 − 4 2x − 5 where x 6= 5 2 A Find a value of k for which the roots are equal. B Find an integer k for which the roots of the equation will be rational and unequal. 6. [IEB, Nov. 2005, HG] A Prove that the roots of the equation x2 − (a + b)x + ab − p2 = 0 are real for all real values of a, b and p. B When will the roots of the equation be equal? 7. [IEB, Nov. 2005, HG] If b and c can take on only the values 1, 2 or 3, determine all pairs (b; c) such that x2 + bx + c = 0 has real roots. 298 CHAPTER 22. SOLVING QUADRATIC EQUATIONS - GRADE 11 22.6 22.6 End of Chapter Exercises 1. Solve: x2 − x − 1 = 0 (Give your answer correct to two decimal places.) 2 2. Solve: 16(x + 1) = x (x + 1) 3. Solve: y 2 + 3 + 12 =7 +3 y2 (Hint: Let y 2 + 3 = k and solve for k first and use the answer to solve y.) 4. Solve for x: 2x4 − 5x2 − 12 = 0 5. Solve for x: A x(x − 9) + 14 = 0 B x2 − x = 3 (Show your answer correct to ONE decimal place.) 6 C x+2= (correct to 2 decimal places) x 2x 1 + =1 D x+1 x−1 6. Solve for x by completing the square: x2 − px − 4 = 0 7. The equation ax2 + bx + c = 0 has roots x = values for a, b and c. 2 3 and x = −4. Find one set of possible 8. The two roots of the equation 4x2 + px − 9 = 0 differ by 5. Calculate the value of p. 9. An equation of the form x2 + bx + c = 0 is written on the board. Saskia and Sven copy it down incorrectly. Saskia has a mistake in the constant term and obtains the solutions -4 and 2. Sven has a mistake in the coefficient of x and obtains the solutions 1 and -15. Determine the correct equation that was on the board. 10. Bjorn stumbled across the following formula to solve the quadratic equation ax2 +bx+c = 0 in a foreign textbook. 2c √ x= −b ± b2 − 4ac A Use this formula to solve the equation: 2x2 + x − 3 = 0 B Solve the equation again, using factorisation, to see if the formula works for this equation. C Trying to derive this formula to prove that it always works, Bjorn got stuck along the way. His attempt his shown below: ax2 + bx + c c b a+ + 2 x x b c + +a x2 x b a 1 + + x2 cx c b 1 + x2 cx 1 b ∴ 2+ x cx = 0 = 0 Divided by x2 where x 6= 0 = 0 Rearranged = 0 Divided by c where c 6= 0 = − + ... a c Complete his derivation. 299 Subtracted Got stuck a from both sides c 22.6 CHAPTER 22. SOLVING QUADRATIC EQUATIONS - GRADE 11 300 Chapter 23 Solving Quadratic Inequalities Grade 11 23.1 Introduction Now that you know how to solve quadratic equations, you are ready to learn how to solve quadratic inequalities. 23.2 Quadratic Inequalities A quadratic inequality is an inequality of the form ax2 + bx + c > 0 ax2 + bx + c ≥ 0 ax2 + bx + c < 0 ax2 + bx + c ≤ 0 Solving a quadratic inequality corresponds to working out in what region the graph of a quadratic function lies above or below the x-axis. Worked Example 112: Quadratic Inequality Question: Solve the inequality 4x2 −4x+1 ≤ 0 and interpret the solution graphically. Answer Step 1 : Factorise the quadratic Let f (x) = 4x2 − 4x + 1. Factorising this quadratic function gives f (x) = (2x − 1)2 . Step 2 : Re-write the original equation with factors (2x − 1)2 ≤ 0 Step 3 : Solve the equation which shows that f (x) = 0 only when x = 12 . Step 4 : Write the final answer This means that the graph of f (x) = 4x2 − 4x + 1 touches the x-axis at x = 21 , but there are no regions where the graph is below the x-axis. Step 5 : Graphical interpretation of solution x = 12 b -2 -1 0 301 1 2 23.2 CHAPTER 23. SOLVING QUADRATIC INEQUALITIES - GRADE 11 Worked Example 113: Solving Quadratic Inequalities Question: Find all the solutions to the inequality x2 − 5x + 6 ≥ 0. Answer Step 1 : Factorise the quadratic The factors of x2 − 5x + 6 are (x − 3)(x − 2). Step 2 : Write the inequality with the factors x2 − 5x + 6 ≥ (x − 3)(x − 2) ≥ 0 0 Step 3 : Determine which ranges correspond to the inequality We need to figure out which values of x satisfy the inequality. From the answers we have five regions to consider. A 1 Bb C Db 2 3 E 4 Step 4 : Determine whether the function is negative or positive in each of the regions Let f (x) = x2 −5x+6. For each region, choose any point in the region and evaluate the function. Region A Region B Region C Region D Region E f (x) f (1) = 2 f (2) = 0 f (2,5) = −2,5 f (3) = 0 f (4) = 2 x<2 x=2 2<x<3 x=3 x>3 sign of f (x) + + + + We see that the function is positive for x ≤ 2 and x ≥ 3. Step 5 : Write the final answer and represent on a number line We see that x2 − 5x + 6 ≥ 0 is true for x ≤ 2 and x ≥ 3. 1 b 2 b 3 4 Worked Example 114: Solving Quadratic Inequalities Question: Solve the quadratic inequality −x2 − 3x + 5 > 0. Answer Step 1 : Determine how to approach the problem Let f (x) = −x2 − 3x + 5. f (x) cannot be factorised so, use the quadratic formula to determine the roots of f (x). The x-intercepts are solutions to the quadratic 302 CHAPTER 23. SOLVING QUADRATIC INEQUALITIES - GRADE 11 23.2 equation −x2 − 3x + 5 = 2 x + 3x − 5 = ∴x = = x1 = x2 = 0 0 p −3 ± (3)2 − 4(1)(−5) 2(1) √ −3 ± 29 2√ −3 − 29 2√ −3 + 29 2 Step 2 : Determine which ranges correspond to the inequality We need to figure out which values of x satisfy the inequality. From the answers we have five regions to consider. A B C D b b x1 x2 E Step 3 : Determine whether the function is negative or positive in each of the regions We can use another method to determine the sign of the function over different regions, by drawing a rough sketch of the graph of the function. We know that the roots of the function correspond to the x-intercepts of the graph. Let g(x) = −x2 − 3x + 5. We can see that this is a parabola with a maximum turning point that intersects the x-axis at x1 and x2 . 7 6 5 4 3 2 1 x1 x2 −4 −3 −2 −1 −1 1 It is clear that g(x) > 0 for x1 Step 4 : Write the final answer and represent the solution graphically −x2 − 3x + 5 > 0 for x1 x1 x2 When working with an inequality where the variable is in the denominator, a different approach is needed. 303 23.3 CHAPTER 23. SOLVING QUADRATIC INEQUALITIES - GRADE 11 Worked Example 115: nominator Question: Solve Non-linear inequality with the variable in the de- 1 2 ≤ x+3 x−3 Answer Step 1 : Subtract 1 x−3 from both sides 2 1 − ≤0 x+3 x−3 Step 2 : Simplify the fraction by finding LCD 2(x − 3) − (x + 3) ≤0 (x + 3)(x − 3) x−9 ≤0 (x + 3)(x − 3) Step 3 : Draw a number line for the inequality - undef + -3 - undef 9 3 We see that the expression is negative for x < −3 or 3 < x ≤ 9. Step 4 : Write the final answer x < −3 or 23.3 3<x≤9 End of Chapter Exercises Solve the following inequalities and show your answer on a number line. 1. Solve: x2 − x < 12. 2. Solve: 3x2 > −x + 4 3. Solve: y 2 < −y − 2 4. Solve: −t2 + 2t > −3 5. Solve: s2 − 4s > −6 6. Solve: 0 ≥ 7x2 − x + 8 7. Solve: 0 ≥ −4x2 − x 8. Solve: 0 ≥ 6x2 9. Solve: 2x2 2 + x + 6 ≤ 0 10. Solve for x if: 11. Solve for x if: x < 2 and x 6= 3. x−3 4 ≤ 1. x−3 0 304 + CHAPTER 23. SOLVING QUADRATIC INEQUALITIES - GRADE 11 12. Solve for x if: 13. Solve for x: 14. Solve for x: 4 < 1. (x − 3)2 2x − 2 >3 x−3 −3 <0 (x − 3)(x + 1) 15. Solve: (2x − 3)2 < 4 16. Solve: 2x ≤ 15 − x x x2 + 3 ≤0 3x − 2 17. Solve for x: 18. Solve: x − 2 ≥ 19. Solve for x: 3 x x2 + 3x − 4 ≤0 5 + x4 20. Determine all real solutions: x−2 ≥1 3−x 305 23.3 23.3 CHAPTER 23. SOLVING QUADRATIC INEQUALITIES - GRADE 11 306 Chapter 24 Solving Simultaneous Equations Grade 11 In grade 10, you learnt how to solve sets of simultaneous equations where both equations were linear (i.e. had the highest power equal to 1). In this chapter, you will learn how to solve sets of simultaneous equations where one is linear and one is a quadratic. As in Grade 10, the solution will be found both algebraically and graphically. The only difference between a system of linear simultaneous equations and a system of simultaneous equations with one linear and one quadratic equation, is that the second system will have at most two solutions. An example of a system of simultaneous equations with one linear equation and one quadratic equation is: y − 2x = −4 (24.1) 2 x +y =4 24.1 Graphical Solution The method of graphically finding the solution to one linear and one quadratic equation is identical to systems of linear simultaneous equations. Method: Graphical solution to a system of simultaneous equations with one linear and one quadratic equation 1. Make y the subject of each equation. 2. Draw the graphs of each equation as defined above. 3. The solution of the set of simultaneous equations is given by the intersection points of the two graphs. For the example, making y the subject of each equation, gives: y = 2x − 4 y = 4 − x2 Plotting the graph of each equation, gives a straight line for the first equation and a parabola for the second equation. The parabola and the straight line intersect at two points: (2,0) and (-4,-12). Therefore, the solutions to the system of equations in (24.1) is x = 2, y = 0 and x = −4, y = 12 307 24.1 CHAPTER 24. SOLVING SIMULTANEOUS EQUATIONS - GRADE 11 4 x 6 4 2 = 2 − (1,0) y b −2 −2 − x2 −4 (-4,-12) 2 4 6 −4 −6 −8 −10 −12 −14 y= 4 −6 b Worked Example 116: Simultaneous Equations Question: Solve graphically: y − x2 + 9 y + 3x − 9 = 0 = 0 Answer Step 1 : Make y the subject of the equation For the first equation: y − x2 + 9 = y = 0 x2 − 9 and for the second equation: y + 3x − 9 = y = 0 −3x + 9 Step 2 : Draw the graphs corresponding to each equation. 40 −3 x+ 9 20 −9 (-6,27) 30 y= x2 b y= 10 b −8 −6 −4 −2 2 (3,0) 4 6 8 Step 3 : Find the intersection of the graphs. The graphs intersect at (−6,27) and at (3,0). Step 4 : Write the solution of the system of simultaneous equations as given by the intersection of the graphs. The first solution is x = −6 and y = 27. The second solution is x = 3 and y = 0. 308 CHAPTER 24. SOLVING SIMULTANEOUS EQUATIONS - GRADE 11 24.2 Exercise: Graphical Solution Solve the following systems of equations algebraically. Leave your answer in surd form, where appropriate. 1. b2 − 1 − a = 0, a + b − 5 = 0 2. x + y − 10 = 0, x2 − 2 − y = 0 3. 6 − 4x − y = 0, 12 − 2x2 − y = 0 4. x + 2y − 14 = 0, x2 + 2 − y = 0 5. 2x + 1 − y = 0, 25 − 3x − x2 − y = 0 24.2 Algebraic Solution The algebraic method of solving simultaneous equations is by substitution. For example the solution of y − 2x = −4 x2 + y = 4 is: y = x + (2x − 4) = 2 x2 + 2x − 8 = Factorise to get: (x + 4)(x − 2) = 2x − 4 4 into second equation 0 0 ∴ the 2 solutions for x are: x = −4 and x = 2 The corresponding solutions for y are obtained by substitution of the x-values into the first equation y = 2(−4) − 4 and: y = 2(2) − 4 = −12 for x = −4 = 0 for x = 2 As expected, these solutions are identical to those obtained by the graphical solution. Worked Example 117: Simultaneous Equations Question: Solve algebraically: y − x2 + 9 y + 3x − 9 = 0 = 0 Answer Step 1 : Make y the subject of the linear equation 309 24.2 CHAPTER 24. SOLVING SIMULTANEOUS EQUATIONS - GRADE 11 y + 3x − 9 = y = 0 −3x + 9 Step 2 : Substitute into the quadratic equation (−3x + 9) − x2 + 9 Factorise to get: = 0 x2 + 3x − 18 = 0 (x + 6)(x − 3) = 0 ∴ the 2 solutions for x are: x = −6 and x = 3 Step 3 : Substitute the values for x into the first equation to calculate the corresponding y-values. y = −3(−6) + 9 = and: y = −3(3) + 9 = 27 for x = −6 0 for x = 3 Step 4 : Write the solution of the system of simultaneous equations. The first solution is x = −6 and y = 27. The second solution is x = 3 and y = 0. Exercise: Algebraic Solution Solve the following systems of equations algebraically. Leave your answer in surd form, where appropriate. 310 CHAPTER 24. SOLVING SIMULTANEOUS EQUATIONS - GRADE 11 a − b2 + 3b − 5 = 0 1. a + b = 5 2. a − b + 1 = 0 3. a − (2b+2) 4 =0 4. a + 2b − 4 = 0 5. a − 2 + 3b = 0 6. a − b − 5 = 0 7. a − b − 4 = 0 8. a + b − 9 = 0 9. a − 3b + 5 = 0 10. a + b − 5 = 0 11. a − 2b − 3 = 0 12. a − 2b = 0 13. a − 3b = 0 14. a − 2b − 10 = 0 15. a − 3b − 1 = 0 16. a − 3b + 1 17. a + 6b − 5 = 0 18. a − 2b + 1 = 0 19. 2a + b − 2 = 0 20. a + 4b − 19 = 0 21. a + 4b − 18 = 0 a − b2 + 5b − 6 = 0 a − 2b2 + 3b + 5 = 0 a − 2b2 − 5b + 3 = 0 a − 9 + b2 = 0 a − b2 = 0 a + 2b2 − 12 = 0 a + b2 − 18 = 0 a + b2 − 4b = 0 a − b2 + 1 = 0 a − 3b2 + 4 = 0 a − b2 − 2b + 3 = 0 a − b2 + 4 = 0 a − b2 − 5b = 0 a − 2b2 − b + 3 = 0 a − b2 = 0 a − b2 − 8 = 0 a − 2b2 − 12b + 4 8a + b2 − 8 = 0 8a + 5b2 − 101 = 0 2a + 5b2 − 57 311 24.2 24.2 CHAPTER 24. SOLVING SIMULTANEOUS EQUATIONS - GRADE 11 312 Chapter 25 Mathematical Models - Grade 11 Up until now, you have only learnt how to solve equations and inequalities, but there has not been much application of what you have learnt. This chapter builds introduces you to the idea of a mathematical model which uses mathematical concepts to solve real-world problems. Definition: Mathematical Model A mathematical model is a method of using the mathematical language to describe the behaviour of a physical system. Mathematical models are used particularly in the natural sciences and engineering disciplines (such as physics, biology, and electrical engineering) but also in the social sciences (such as economics, sociology and political science); physicists, engineers, computer scientists, and economists use mathematical models most extensively. A mathematical model is an equation (or a set of equations for the more difficult problems) that describes are particular situation. For example, if Anna receives R3 for each time she helps her mother wash the dishes and R5 for each time she helps her father cut the grass, how much money will Anna earn if she helps her mother 5 times to wash the dishes and helps her father 2 times to wash the car. The first step to modelling is to write the equation, that describes the situation. To calculate how much Anna will earn we see that she will earn : 5 + 2 ×R3 for washing the dishes ×R5 for cutting the grass = R15 +R10 = R25 If however, we say, what is the equation if Anna helps her mother x times and her father y times. Then we have: Total earned = x × R3 + y × R5 25.1 Real-World Applications: Mathematical Models Some examples of where mathematical models are used in the real-world are: 1. To model population growth 2. To model effects of air pollution 3. To model effects of global warming 4. In computer games 313 25.1 CHAPTER 25. MATHEMATICAL MODELS - GRADE 11 5. In the sciences (e.g. physics, chemistry, biology) to understand how the natural world works 6. In simulators that are used to train people in certain jobs, like pilots, doctors and soldiers 7. In medicine to track the progress of a disease Activity :: Investigation : Simple Models In order to get used to the idea of mathematical models, try the following simple models. Write an equation that describes the following real-world situations, mathematically: 1. Jack and Jill both have colds. Jack sneezes twice for each sneeze of Jill’s. If Jill sneezes x times, write an equation describing how many times they both sneezed? 2. It rains half as much in July as it does in December. If it rains y mm in July, write an expression relating the rainfall in July and December. 3. Zane can paint a room in 4 hours. Billy can paint a room in 2 hours. How long will it take both of them to paint a room together? 4. 25 years ago, Arthur was 5 more than 31 as old as Lee was. Today, Lee is 26 less than twice Arthur’s age. How old is Lee? 5. Kevin has played a few games of ten-pin bowling. In the third game, Kevin scored 80 more than in the second game. In the first game Kevin scored 110 less than the third game. His total score for the first two games was 208. If he wants an average score of 146, what must he score on the fourth game? 6. Erica has decided to treat her friends to coffee at the Corner Coffee House. Erica paid R54,00 for four cups of cappuccino and three cups of filter coffee. If a cup of cappuccino costs R3,00 more than a cup of filter coffee, calculate how much each type of coffee costs? 7. The product of two integers is 95. Find the integers if their total is 24. Worked Example 118: Mathematical Modelling of Falling Objects Question: When an object is dropped or thrown downward, the distance, d, that it falls in time, t is described by the following equation: s = 5t2 + v0 t In this equation, v0 is the initial velocity, in m·s−1 . Distance is measured in meters and time is measured in seconds. Use the equation to find how far an object will fall in 2 s if it is thrown downward at an initial velocity of 10 m·s−1 ? Answer Step 1 : Identify what is given for each problem We are given an expression to calculate distance travelled by a falling object in terms of initial velocity and time. We are also given the initial velocity and time and are required to calculate the distance travelled. Step 2 : List all known and unknown information • v0 = 10 m · s−1 • t = 2s 314 CHAPTER 25. MATHEMATICAL MODELS - GRADE 11 • s =? m Step 3 : Substitute values into expression s = = 5t2 + v0 t 5(2)2 + (10)(2) = = 5(4) + 20 20 + 20 = 40 Step 4 : Write the final answer The object will fall 40 m in 2 s if it is thrown downward at an initial velocity of 10 m·s−1 . Worked Example 119: Another Mathematical Modelling of Falling Objects Question: When an object is dropped or thrown downward, the distance, d, that it falls in time, t is described by the following equation: s = 5t2 + v0 t In this equation, v0 is the initial velocity, in m·s−1 . Distance is measured in meters and time is measured in seconds. Use the equation find how long it takes for the object to reach the ground if it is dropped from a height of 2000 m. The initial velocity is 0 m·s−1 ? Answer Step 1 : Identify what is given for each problem We are given an expression to calculate distance travelled by a falling object in terms of initial velocity and time. We are also given the initial velocity and time and are required to calculate the distance travelled. Step 2 : List all known and unknown information • v0 = 0 m · s−1 • t =? s • s = 2000 m Step 3 : Substitute values into expression s = 2000 = 5t2 + v0 t 5t2 + (0)(2) 2000 = = 5t2 2000 5 400 = 20 s t2 ∴ t = Step 4 : Write the final answer The object will take 20 s to reach the ground if it is dropped from a height of 2000 m. 315 25.1 25.1 CHAPTER 25. MATHEMATICAL MODELS - GRADE 11 Activity :: Investigation : Mathematical Modelling The graph below shows the how the distance travelled by a car depends on time. Use the graph to answer the following questions. Distance (m) 400 300 200 100 0 0 10 20 30 Time (s) 40 1. How far does the car travel in 20 s? 2. How long does it take the car to travel 300 m? Worked Example 120: More Mathematical Modelling Question: A researcher is investigating the number of trees in a forest over a period of n years. After investigating numerous data, the following data model emerged: Year 1 2 3 4 Number of trees in hundreds 1 3 9 27 1. How many trees, in hundreds, are there in the SIXTH year if this pattern is continued? 2. Determine an algebraic expression that describes the number of trees in the nt h year in the forest. 3. Do you think this model, which determines the number of trees in the forest, will continue indefinitely? Give a reason for your answer. Answer Step 1 : Find the pattern The pattern is 30 ; 31 ; 32 ; 33 ; ... Therefore, three to the power one less than the year. Step 2 : Trees in year 6 year6 = hundreds = 243hundreds = 24300 Step 3 : Algebraic expression for year n number of trees = 3n−1 hundreds Step 4 : Conclusion No The number of trees will increase without bound to very large numbers, thus the forestry authorities will if necessary cut down some of the trees from time to time. 316 CHAPTER 25. MATHEMATICAL MODELS - GRADE 11 25.2 Worked Example 121: Setting up an equation Question: Currently the subsription to a gym for a single member is R1 000 annually while family membership is R1 500. The gym is considering raising all membershipfees by the same amount. If this is done then the single membership will cost 75 of the family membership. Determine the proposed increase. Answer Step 1 : Summarise the information in a table Let the proposed increase be x. Single Family Now 1 000 1 500 After increase 1 000+x 1 500+x Step 2 : Set up an equation 1 000 + x = 5 (1 500 + x) 7 Step 3 : Solve the equation 7 000 + 7x = 2x = x = 7 500 + 5x 500 250 Step 4 : Write down the answer Therefore the increase is R250. 25.2 End of Chatpter Exercises 1. When an object is dropped or thrown downward, the distance, d, that it falls in time, t is described by the following equation: s = 5t2 + v0 t In this equation, v0 is the initial velocity, in m·s−1 . Distance is measured in meters and time is measured in seconds. Use the equation to find how long it takes a tennis ball to reach the ground if it is thrown downward from a hot-air balloon that is 500 m high. The tennis ball is thrown at an initial velocity of 5 m·s−1 . 2. The table below lists the times that Sheila takes to walk the given distances. Time (minutes) Distance (km) 5 1 10 2 15 3 20 4 25 5 30 6 Plot the points. If the relationship between the distances and times are linear, find the equation of the straight line, using any two points. Then use the equation to answer the following questions: 317 25.2 CHAPTER 25. MATHEMATICAL MODELS - GRADE 11 A How long will it take Sheila to walk 21 km? B How far will Sheila walk in 7 minutes? If Sheila were to walk half as fast as she is currently walking, what would the graph of her distances and times look like? 3. The power P (in watts) supplied to a circuit by a 12 volt battery is given by the formula P = 12I − 0,5I 2 where I is the current in amperes. A Since both power and current must be greater than 0, find the limits of the current that can be drawn by the circuit. B Draw a graph of P = 12I − 0,5I 2 and use your answer to the first question, to define the extent of the graph. C What is the maximum current that can be drawn? D From your graph, read off how much power is supplied to the circuit when the current is 10 amperes? Use the equation to confirm your answer. E At what value of current will the power supplied be a maximum? 4. You are in the lobby of a business building waiting for the lift. You are late for a meeting and wonder if it will be quicker to take the stairs. There is a fascinating relationship between the number of floors in the building, the number of people in the lift and how often it will stop: If N people get into a lift at the lobby and the number of floors in the building is F , then the lift can be expected to stop F −1 F −F F N times. A If the building has 16 floors and there are 9 people who get into the lift, how many times is the lift expected to stop? B How many people would you expect in a lift, if it stopped 12 times and there are 17 floors? 5. A wooden block is made as shown in the diagram. The ends are right-angled triangles having sides 3x, 4x and 5x. The length of the block is y. The total surface area of the block is 3 600 cm2 . 3x 4x y Show that y= 300 − x2 x 318 CHAPTER 25. MATHEMATICAL MODELS - GRADE 11 25.2 6. A stone is thrown vertically upwards and its height (in metres) above the ground at time t (in seconds) is given by: h(t) = 35 − 5t2 + 30t Find its initial height above the ground. 7. After doing some research, a transport company has determined that the rate at which petrol is consumed by one of its large carriers, travelling at an average speed of x km per hour, is given by: x 55 + litres per kilometre P (x) = 2x 200 Assume that the petrol costs R4,00 per litre and the driver earns R18,00 per hour (travelling time). Now deduce that the total cost, C, in Rands, for a 2 000 km trip is given by: C(x) = 256000 + 40x x 8. During an experiment the temperature T (in degrees Celsius), varies with time t (in hours), according to the formula: 1 T (t) = 30 + 4t − t2 2 t ∈ [1; 10] A Determine an expression for the rate of change of temperature with time. B During which time interval was the temperature dropping? 9. In order to reduce the temperature in a room from 28◦ C, a cooling system is allowed to operate for 10 minutes. The room temperature, T after t minutes is given in ◦ C by the formula: T = 28 − 0,008t3 − 0,16t where t ∈ [0; 10] A At what rate (rounded off to TWO decimal places) is the temperature falling when t = 4 minutes? B Find the lowest room temperature reached during the 10 minutes for which the cooling system operates, by drawing a graph. 10. A washing powder box has the shape of a rectangular prism as shown in the diagram below. The box has a volume of 480 cm3 , a breadth of 4 cm and a length of x cm. Washing powder Show that the total surface area of the box (in cm2 ) is given by: A = 8x + 960x−1 + 240 Extension: Simulations A simulation is an attempt to model a real-life situation on a computer so that it 319 25.2 CHAPTER 25. MATHEMATICAL MODELS - GRADE 11 can be studied to see how the system works. By changing variables, predictions may be made about the behaviour of the system. Simulation is used in many contexts, including the modeling of natural systems or human systems in order to gain insight into their functioning. Other contexts include simulation of technology for performance optimization, safety engineering, testing, training and education. Simulation can be used to show the eventual real effects of alternative conditions and courses of action. Simulation in education Simulations in education are somewhat like training simulations. They focus on specific tasks. In the past, video has been used for teachers and education students to observe, problem solve and role play; however, a more recent use of simulations in education include animated narrative vignettes (ANV). ANVs are cartoon-like video narratives of hypothetical and realitybased stories involving classroom teaching and learning. ANVs have been used to assess knowledge, problem solving skills and dispositions of children, and pre-service and in-service teachers. 320 Chapter 26 Quadratic Functions and Graphs Grade 11 26.1 Introduction In Grade 10, you studied graphs of many different forms. In this chapter, you will learn a little more about the graphs of functions. 26.2 Functions of the Form y = a(x + p)2 + q This form of the quadratic function is slightly more complex than the form studied in Grade 10, y = ax2 + q. The general shape and position of the graph of the function of the form f (x) = a(x + p)2 + q is shown in Figure 26.1. 3 2 1 −5 −4 −3 −2 −1 1 2 3 4 5 −1 −2 −3 Figure 26.1: Graph of f (x) = 12 (x + 2)2 − 1 Activity :: Investigation : Functions of the Form y = a(x + p)2 + q 1. On the same set of axes, plot the following graphs: A B C D a(x) = (x − 2)2 b(x) = (x − 1)2 c(x) = x2 d(x) = (x + 1)2 321 26.2 CHAPTER 26. QUADRATIC FUNCTIONS AND GRAPHS - GRADE 11 E e(x) = (x + 2)2 Use your results to deduce the effect of p. 2. On the same set of axes, plot the following graphs: A f (x) = (x − 2)2 + 1 B g(x) = (x − 1)2 + 1 C h(x) = x2 + 1 D j(x) = (x + 1)2 + 1 E k(x) = (x + 2)2 + 1 Use your results to deduce the effect of q. 3. Following the general method of the above activities, choose your own values of p and q to plot 5 different graphs (on the same set of axes) of y = a(x+ p)2 + q to deduce the effect of a. From your graphs, you should have found that a affects whether the graph makes a smile or a frown. If a < 0, the graph makes a frown and if a > 0 then the graph makes a smile. This is shown in Figure 10.9. You should have also found that the value of p affects whether the turning point of the graph is above the x-axis (p < 0) or below the x-axis (p > 0). You should have also found that the value of q affects whether the turning point is to the left of the y-axis (q > 0) or to the right of the y-axis (q < 0). These different properties are summarised in Table 26.1. The axes of symmetry for each graph is shown as a dashed line. Table 26.1: Table summarising general shapes and positions of functions of the form y = a(x + p)2 + q. The axes of symmetry are shown as dashed lines. p<0 p>0 a>0 a<0 a>0 a<0 q≥0 q≤0 26.2.1 Domain and Range For f (x) = a(x + p)2 + q, the domain is {x : x ∈ R} because there is no value of x ∈ R for which f (x) is undefined. The range of f (x) = a(x + p)2 + q depends on whether the value for a is positive or negative. We will consider these two cases separately. If a > 0 then we have: (x + p)2 a(x + p)2 a(x + p)2 + q f (x) ≥ 0 ≥ 0 ≥ q ≥ q (The square of an expression is always positive) (Multiplication by a positive number maintains the nature of the inequality) 322 CHAPTER 26. QUADRATIC FUNCTIONS AND GRAPHS - GRADE 11 26.2 This tells us that for all values of x, f (x) is always greater than q. Therefore if a > 0, the range of f (x) = a(x + p)2 + q is {f (x) : f (x) ∈ [q,∞)}. Similarly, it can be shown that if a < 0 that the range of f (x) = a(x + p)2 + q is {f (x) : f (x) ∈ (−∞,q]}. This is left as an exercise. For example, the domain of g(x) = (x − 1)2 + 2 is {x : x ∈ R} because there is no value of x ∈ R for which g(x) is undefined. The range of g(x) can be calculated as follows: (x − p)2 2 (x + p) + 2 g(x) ≥ ≥ ≥ 0 2 2 Therefore the range is {g(x) : g(x) ∈ [2,∞)}. Exercise: Domain and Range 1. Given the function f (x) = (x − 4)2 − 1. Give the range of f (x). 2. What is the domain the equation y = 2x2 − 5x − 18 ? 26.2.2 Intercepts For functions of the form, y = a(x + p)2 + q, the details of calculating the intercepts with the x and y axis is given. The y-intercept is calculated as follows: y yint = = a(x + p)2 + q a(0 + p)2 + q (26.1) (26.2) = ap2 + q (26.3) If p = 0, then yint = q. For example, the y-intercept of g(x) = (x − 1)2 + 2 is given by setting x = 0 to get: = (x − 1)2 + 2 = (0 − 1)2 + 2 g(x) yint = (−1)2 + 2 = 1+2 = 3 The x-intercepts are calculated as follows: y = 0 = a(xint + p)2 xint + p xint = a(x + p)2 + q a(xint + p)2 + q −q r q − = a r q = ± − −p a 323 (26.4) (26.5) (26.6) (26.7) (26.8) 26.2 CHAPTER 26. QUADRATIC FUNCTIONS AND GRAPHS - GRADE 11 However, (26.8) is only valid if − aq > 0 which means that either q < 0 or a < 0. This is consistent with what we expect, since if q > 0 and a > 0 then − aq is negative and in this case the graph lies above the x-axis and therefore does not intersect the x-axis. If however, q > 0 and a < 0, then − aq is positive and the graph is hat shaped and should have two x-intercepts. Similarly, if q < 0 and a > 0 then − aq is also positive, and the graph should intersect with the x-axis. For example, the x-intercepts of g(x) = (x − 1)2 + 2 is given by setting y = 0 to get: g(x) = (x − 1)2 + 2 0 = (xint − 1)2 + 2 −2 = (xint − 1)2 which is not real. Therefore, the graph of g(x) = (x − 1)2 + 2 does not have any x-intercepts. Exercise: Intercepts 1. Find the x- and y-intercepts of the function f (x) = (x − 4)2 − 1. 2. Find the intercepts with both axes of the graph of f (x) = x2 − 6x + 8. 3. Given: f (x) = −x2 + 4x − 3. Calculate the x- and y-intercepts of the graph of f . 26.2.3 Turning Points The turning point of the function of the form f (x) = a(x + p)2 + q is given by examining the range of the function. We know that if a > 0 then the range of f (x) = a(x + p)2 + q is {f (x) : f (x) ∈ [q,∞)} and if a < 0 then the range of f (x) = a(x + p)2 + q is {f (x) : f (x) ∈ (−∞,q]}. So, if a > 0, then the lowest value that f (x) can take on is q. Solving for the value of x at which f (x) = q gives: q = a(x + p)2 + q 0 0 = a(x + p)2 = (x + p)2 0 x = x+p = −p ∴ x = −p at f (x) = q. The co-ordinates of the (minimal) turning point is therefore (−p,q). Similarly, if a < 0, then the highest value that f (x) can take on is q and the co-ordinates of the (maximal) turning point is (−p,q). Exercise: Turning Points 1. Determine the turning point of the graph of f (x) = x2 − 6x + 8 . 2. Given: f (x) = −x2 + 4x − 3. Calculate the co-ordinates of the turning point of f . 3. Find the turning point of the following function by completing the square: y = 12 (x + 2)2 − 1. 324 CHAPTER 26. QUADRATIC FUNCTIONS AND GRAPHS - GRADE 11 26.2.4 26.2 Axes of Symmetry There is one axis of symmetry for the function of the form f (x) = a(x + p)2 + q that passes through the turning point and is parallel to the y-axis. Since the x-coordinate of the turning point is x = −p, this is the axis of symmetry. Exercise: Axes of Symmetry 1. Find the equation of the axis of symmetry of the graph y = 2x2 − 5x − 18 2. Write down the equation of the axis of symmetry of the graph of y = 3(x − 2)2 + 1 3. Write down an example of an equation of a parabola where the y-axis is the axis of symmetry. 26.2.5 Sketching Graphs of the Form f (x) = a(x + p)2 + q In order to sketch graphs of the form, f (x) = a(x + p)2 + q, we need to calculate determine four characteristics: 1. sign of a 2. domain and range 3. turning point 4. y-intercept 5. x-intercept For example, sketch the graph of g(x) = − 12 (x + 1)2 − 3. Mark the intercepts, turning point and axis of symmetry. Firstly, we determine that a < 0. This means that the graph will have a maximal turning point. The domain of the graph is {x : x ∈ R} because f (x) is defined for all x ∈ R. The range of the graph is determined as follows: (x + 1)2 1 − (x + 1)2 2 ≥ 0 ≤ 0 1 − (x + 1)2 − 3 ≤ 2 ∴ f (x) ≤ −3 −3 Therefore the range of the graph is {f (x) : f (x) ∈ (−∞, − 3]}. Using the fact that the maximum value that f (x) achieves is -3, then the y-coordinate of the turning point is -3. The x-coordinate is determined as follows: 1 − (x + 1)2 − 3 2 1 − (x + 1)2 − 3 + 3 2 1 − (x + 1)2 2 Divide both sides by − 21 : (x + 1)2 Take square root of both sides: x + 1 ∴ x 325 = −3 = 0 = 0 = 0 = 0 = −1 26.2 CHAPTER 26. QUADRATIC FUNCTIONS AND GRAPHS - GRADE 11 The coordinates of the turning point are: (−1, − 3). The y-intercept is obtained by setting x = 0. This gives: yint 1 = − (0 + 1)2 − 3 2 1 = − (1) − 3 2 1 = − −3 2 1 = − −3 2 7 = − 2 The x-intercept is obtained by setting y = 0. This gives: 0 = 3 = −3 · 2 = −6 = 1 − (xint + 1)2 − 3 2 1 − (xint + 1)2 2 (xint + 1)2 (xint + 1)2 which is not real. Therefore, there are no x-intercepts. We also know that the axis of symmetry is parallel to the y-axis and passes through the turning point. −4 −3 −2 −1 −1 (-1,-3) 1 2 3 4 −2 b−3 b (0,-3.5) −4 −5 −6 −7 Figure 26.2: Graphs of the function f (x) = − 21 (xint + 1)2 − 3 Exercise: Sketching the Parabola 1. Draw the graph of y = 3(x − 2)2 + 1 showing all the relative intercepts with the axes as well as the coordinates of the turning point. 2. Draw a neat sketch graph of the function defined by y = ax2 + bx + c if a > 0; b < 0; b2 = 4ac. 326 CHAPTER 26. QUADRATIC FUNCTIONS AND GRAPHS - GRADE 11 26.2.6 26.3 Writing an equation of a shifted parabola Given a parabola with equation y = x2 − 2x − 3. The graph of the parabola is shifted one unit to the right. Or else the y-axis shifts one unit to the left. Therefore the new equation will become: y = (x − 1)2 − 2(x − 1) − 3 = x2 − 2x + 1 − 2x + 2 − 3 = x2 − 4x If the given parabola is shifted 3 units down, the new equation will become: (Notice the x-axis then moves 3 units upwards) y+3 = y 26.3 = x2 − 2x − 3 x2 − 2x − 6 End of Chapter Exercises 1. Show that if a < 0, then the range of f (x) = a(x + p)2 + q is {f (x) : f (x) ∈ (−∞,q]}. 2. If (2;7) is the turning point of f (x) = −2x2 − 4ax + k, find the values of the constants a and k. 3. The graph in the figure is represented by the equation f (x) = ax2 + bx. The coordinates of the turning point are (3;9). Show that a = −1 and b = 6. b (3,9) 4. Given: f : x = x2 − 2x3. Give the equation of the new graph originating if: A The graph of f is moved three units to the left. B The x - axis is moved down three. 5. A parabola with turning point (-1; -4) is shifted vertically by 4 units upwards. What are the coordinates of the turning point of the shifted parabola ? 327 26.3 CHAPTER 26. QUADRATIC FUNCTIONS AND GRAPHS - GRADE 11 328 Chapter 27 Hyperbolic Functions and Graphs Grade 11 27.1 Introduction In Grade 10, you studied graphs of many different forms. In this chapter, you will learn a little more about the graphs of functions. 27.2 Functions of the Form y = a x+p +q This form of the hyperbolic function is slightly more complex than the form studied in Grade 10. 5 4 3 2 1 −5 −4 −3 −2 −1 −1 1 2 3 4 5 −2 −3 −4 −5 Figure 27.1: General shape and position of the graph of a function of the form f (x) = The asymptotes are shown as dashed lines. Activity :: Investigation : Functions of the Form y = 1. On the same set of axes, plot the following graphs: 329 a x+p +q a x+p + q. 27.2 CHAPTER 27. HYPERBOLIC FUNCTIONS AND GRAPHS - GRADE 11 A a(x) = B b(x) = C c(x) = D d(x) = E e(x) = −2 x+1 −1 x+1 0 x+1 +1 x+1 +2 x+1 +1 +1 +1 +1 +1 Use your results to deduce the effect of a. 2. On the same set of axes, plot the following graphs: A f (x) = B g(x) = C h(x) = D j(x) = E k(x) = 1 x−2 + 1 1 x−1 + 1 1 x+0 + 1 1 x+1 + 1 1 x+2 + 1 Use your results to deduce the effect of p. 3. Following the general method of the above activities, choose your own values a + q to deduce the effect of q. of a and p to plot 5 different graphs of y = x+p You should have found that the value of a affects whether the graph is located in the first and third quadrants of Cartesian plane. You should have also found that the value of p affects whether the x-intercept is negative (p > 0) or positive (p < 0). You should have also found that the value of q affects whether the graph lies above the x-axis (q > 0) or below the x-axis (q < 0). These different properties are summarised in Table 27.1. The axes of symmetry for each graph is shown as a dashed line. Table 27.1: Table summarising general shapes and positions of functions of the form y = The axes of symmetry are shown as dashed lines. p<0 p>0 a>0 a<0 a>0 a<0 a x+p +q. q>0 q<0 27.2.1 For y = −p}. Domain and Range a x+p + q, the function is undefined for x = −p. The domain is therefore {x : x ∈ R,x 6= 330 CHAPTER 27. HYPERBOLIC FUNCTIONS AND GRAPHS - GRADE 11 We see that y = a x+p 27.2 + q can be re-written as: a +q x+p a y−q = x+p If x 6= −p then: (y − q)(x + p) = a a x+p = y−q y = This shows that the function is undefined at y = q. Therefore the range of f (x) = {f (x) : f (x) ∈ (−∞,q) ∪ (q,∞)}. For example, the domain of g(x) = at x = −1. 2 x+1 a x+p + q is + 2 is {x : x ∈ R, x 6= −1} because g(x) is undefined y = (y − 2) = (y − 2)(x + 1) = (x + 1) = 2 +2 x+1 2 x+1 2 2 y−2 We see that g(x) is undefined at y = 2. Therefore the range is {g(x) : g(x) ∈ (−∞,2) ∪ (2,∞)}. Exercise: Domain and Range Determine the range of y = Given:f (x) = 8 x−8 1 x + 1. + 4. Write down the domain of f . 8 +3 Determine the domain of y = − x+1 27.2.2 Intercepts a + q, the intercepts with the x and y axis is calculated by For functions of the form, y = x+p setting x = 0 for the y-intercept and by setting y = 0 for the x-intercept. The y-intercept is calculated as follows: y yint a +q x+p a = +q 0+p a +q = p 331 = (27.1) (27.2) (27.3) 27.2 CHAPTER 27. HYPERBOLIC FUNCTIONS AND GRAPHS - GRADE 11 For example, the y-intercept of g(x) = 2 x+1 + 2 is given by setting x = 0 to get: = 2 +2 x+1 2 +2 0+1 2 +2 1 2+2 = 4 y = yint = = The x-intercepts are calculated by setting y = 0 as follows: y= a +q x+p (27.4) a +q xint + p (27.5) = −q (27.6) = −q(xint + p) a −q a −p −q (27.7) 0 = a xint + p a xint + p = xint For example, the x-intercept of g(x) = 2 x+1 = (27.8) (27.9) + 2 is given by setting x = 0 to get: y = 0 = −2 = −2(xint + 1) = xint + 1 = 2 +2 x+1 2 +2 xint + 1 2 xint + 1 2 2 −2 −1 − 1 xint = xint = −2 Exercise: Intercepts 1 Given:h(x) = x+4 − 2. Determine the coordinates of the intercepts of h with the x- and y-axes. Determine the x-intercept of the graph of y = no y-intercept for this function. 27.2.3 5 x + 2. Give a reason why there is Asymptotes There are two asymptotes for functions of the form y = examining the domain and range. 332 a x+p + q. They are determined by CHAPTER 27. HYPERBOLIC FUNCTIONS AND GRAPHS - GRADE 11 27.3 We saw that the function was undefined at x = −p and for y = q. Therefore the asymptotes are x = −p and y = q. 2 + 2 is {x : x ∈ R, x 6= −1} because g(x) is undefined For example, the domain of g(x) = x+1 at x = −1. We also see that g(x) is undefined at y = 2. Therefore the range is {g(x) : g(x) ∈ (−∞,2) ∪ (2,∞)}. From this we deduce that the asymptotes are at x = −1 and y = 2. Exercise: Asymptotes Given:h(x) = 1 x+4 − 2.Determine the equations of the asymptotes of h. Write down the equation of the vertical asymptote of the graph y = 27.2.4 Sketching Graphs of the Form f (x) = a x+p In order to sketch graphs of functions of the form, f (x) = determine four characteristics: 1 x−1 . +q a x+p + q, we need to calculate 1. domain and range 2. asymptotes 3. y-intercept 4. x-intercept For example, sketch the graph of g(x) = 2 x+1 + 2. Mark the intercepts and asymptotes. We have determined the domain to be {x : x ∈ R, x 6= −1} and the range to be {g(x) : g(x) ∈ (−∞,2) ∪ (2,∞)}. Therefore the asymptotes are at x = −1 and y = 2. The y-intercept is yint = 4 and the x-intercept is xint = −2. Exercise: Graphs 1. Draw the graph of y = x1 + 2. Indicate the new horizontal asymptote. 1 − 2. Sketch the graph of h showing clearly the asymptotes 2. Given:h(x) = x+4 and ALL intercepts with the axes. 8 + 3 on the same system of axes. 3. Draw the graph of y = x1 and y = − x+1 4. Draw the graph of y = 5 x−2,5 + 2. Explain your method. 5. Draw the graph of the function defined by y = and intercepts with the axes. 27.3 8 x−8 +4. Indicate the asymptotes End of Chapter Exercises 1. Plot the graph of the hyperbola defined by y = x2 for −4 ≤ x ≤ 4. Suppose the hyperbola is shifted 3 units to the right and 1 unit down. What is the new equation then ? 2. Based on the graph of y = x1 , determine the equation of the graph with asymptotes y = 2 and x = 1 and passing through the point (2; 3). 333 27.3 CHAPTER 27. HYPERBOLIC FUNCTIONS AND GRAPHS - GRADE 11 6 5 4 3 2 1 −4 −3 −2 −1 −1 1 2 3 −2 −3 Figure 27.2: Graph of g(x) = 334 2 x+1 + 2. 4 Chapter 28 Exponential Functions and Graphs - Grade 11 28.1 Introduction In Grade 10, you studied graphs of many different forms. In this chapter, you will learn a little more about the graphs of exponential functions. 28.2 Functions of the Form y = ab(x+p) + q This form of the exponential function is slightly more complex than the form studied in Grade 10. 4 3 2 1 −4 −3 −2 −1 1 2 3 4 Figure 28.1: General shape and position of the graph of a function of the form f (x) = ab(x+p) +q. Activity :: Investigation : Functions of the Form y = ab(x+p) + q 1. On the same set of axes, plot the following graphs: A B C D E a(x) = −2 · b(x+1) + 1 b(x) = −1 · b(x+1) + 1 c(x) = −0 · b(x+1) + 1 d(x) = −1 · b(x+1) + 1 e(x) = −2 · b(x+1) + 1 Use your results to deduce the effect of a. 2. On the same set of axes, plot the following graphs: 335 28.2 CHAPTER 28. EXPONENTIAL FUNCTIONS AND GRAPHS - GRADE 11 A B C D E f (x) = 1 · b(x+1) − 2 g(x) = 1 · b(x+1) − 1 h(x) = 1 · b(x+1) 0 j(x) = 1 · b(x+1) + 1 k(x) = 1 · b(x+1) + 2 Use your results to deduce the effect of q. 3. Following the general method of the above activities, choose your own values of a and q to plot 5 different graphs of y = ab(x+p) + q to deduce the effect of p. You should have found that the value of a affects whether the graph curves upwards (a > 0) or curves downwards (a < 0). You should have also found that the value of p affects the position of the x-intercept. You should have also found that the value of q affects the position of the y-intercept. These different properties are summarised in Table 28.1. The axes of symmetry for each graph is shown as a dashed line. Table 28.1: Table summarising general shapes and positions of functions of the form y = ab(x+p) + q. p<0 a>0 p>0 a<0 a>0 a<0 q>0 q<0 28.2.1 Domain and Range For y = ab(x+p) + q, the function is defined for all real values of x. Therefore, the domain is {x : x ∈ R}. The range of y = ab(x+p) + q is dependent on the sign of a. If a > 0 then: b(x+p) a · b(x+p) a · b(x+p) + q f (x) ≥ ≥ ≥ ≥ 0 0 q q Therefore, if a > 0, then the range is {f (x) : f (x) ∈ [q,∞)}. 336 CHAPTER 28. EXPONENTIAL FUNCTIONS AND GRAPHS - GRADE 11 28.2 If a < 0 then: b(x+p) a·b a·b ≤ (x+p) (x+p) 0 ≤ ≤ +q f (x) 0 q ≤ q Therefore, if a < 0, then the range is {f (x) : f (x) ∈ (−∞,q]}. For example, the domain of g(x) = 3 · 2x+1 + 2 is {x : x ∈ R}. For the range, 2x+1 3 · 2x+1 ≥ ≥ 0 0 3 · 2x+1 + 2 ≥ 2 Therefore the range is {g(x) : g(x) ∈ [2,∞)}. Exercise: Domain and Range 1. Give the domain of y = 3x . 2. What is the domain and range of f (x) = 2x ? 3. Determine the domain and range of y = (1,5)x+3 . 28.2.2 Intercepts For functions of the form, y = ab(x+p) + q, the intercepts with the x and y axis is calulated by setting x = 0 for the y-intercept and by setting y = 0 for the x-intercept. The y-intercept is calculated as follows: y = yint ab(x+p) + q (0+p) = ab = abp + q (28.1) +q (28.2) (28.3) For example, the y-intercept of g(x) = 3 · 2x+1 + 2 is given by setting x = 0 to get: y 3 · 2x+1 + 2 3 · 20+1 + 2 = = yint 3 · 21 + 2 = = = 3·2+2 8 The x-intercepts are calculated by setting y = 0 as follows: y ab = 0 = = (xint +p) b(xint +p) ab(x+p) + q (28.4) (xint +p) (28.5) (28.6) ab −q q = − a 337 +q (28.7) 28.2 CHAPTER 28. EXPONENTIAL FUNCTIONS AND GRAPHS - GRADE 11 Which only has a real solution if either a < 0 or Q < 0. Otherwise, the graph of the function of form y = ab(x+p) + q does not have any x-intercepts. For example, the x-intercept of g(x) = 3 · 2x+1 + 2 is given by setting x = 0 to get: y 0 = 3 · 2x+1 + 2 = 3 · 2xint +1 + 2 −2 = 3 · 2xint +1 −2 2xint +1 = 2 which has no real solution. Therefore, the graph of g(x) = 3 · 2x+1 + 2 does not have any x-intercepts. Exercise: Intercepts 1. Give the y-intercept of the graph of y = bx + 2. 2. Give the x- and y-intercepts of the graph of y = 12 (1,5)x+3 − 0,75. 28.2.3 Asymptotes There are two asymptotes for functions of the form y = ab(x+p) + q. They are determined by examining the domain and range. We saw that the function was undefined at x = −p and for y = q. Therefore the asymptotes are x = −p and y = q. For example, the domain of g(x) = 3 · 2x+1 + 2 is {x : x ∈ R, x 6= −1} because g(x) is undefined at x = −1. We also see that g(x) is undefined at y = 2. Therefore the range is {g(x) : g(x) ∈ (−∞,2) ∪ (2,∞)}. From this we deduce that the asymptotes are at x = −1 and y = 2. Exercise: Asymptotes 1. Give the equation of the asymptote of the graph of y = 3x − 2. 2. What is the equation of the horizontal asymptote of the graph of y = 3(0,8)x−1 − 3 ? 28.2.4 Sketching Graphs of the Form f (x) = ab(x+p) + q In order to sketch graphs of functions of the form, f (x) = ab(x+p) + q, we need to calculate determine four characteristics: 1. domain and range 2. y-intercept 338 CHAPTER 28. EXPONENTIAL FUNCTIONS AND GRAPHS - GRADE 11 28.3 3. x-intercept For example, sketch the graph of g(x) = 3 · 2x+1 + 2. Mark the intercepts. We have determined the domain to be {x : x ∈ R} and the range to be {g(x) : g(x) ∈ [5,∞)}. The y-intercept is yint = 8 and there are no x-intercepts. 11 10 9 8 7 6 5 4 3 2 1 −4 −3 −2 −1 1 2 3 4 Figure 28.2: Graph of g(x) = 3 · 2x+1 + 2. Exercise: Sketching Graphs 1. Draw the graphs of the following on the same set of axes. Label the horizontal aymptotes and y-intercepts clearly. A B C D y y y y = bx + 2 = bx+2 = 2bx = 2bx+2 + 2 A Draw the graph of f (x) = 3x . B Explain whre a solution of 3x = 5 can be read off the graph. 28.3 End of Chapter Exercises 1. The following table of values has columns giving the y-values for the graph y = ax , y = ax+1 and y = ax + 1. Match a graph to a column. 339 28.3 CHAPTER 28. EXPONENTIAL FUNCTIONS AND GRAPHS - GRADE 11 x -2 -1 0 1 2 A 7,25 3,5 2 1,4 1,16 B 6,25 2,5 1 0,4 0,16 C 2,5 1 0,4 0,16 0,064 2. The graph of f (x) = 1 + a.2x (a is a constant) passes through the origin. A Determine the value of a. B Determine the value of f (−15) correct to FIVE decimal places. C Determine the value of x, if P (x; 0,5) lies on the graph of f . D If the graph of f is shifted 2 units to the right to give the function h, write down the equation of h. 3. The graph of f (x) = a.bx (a 6= 0) has the point P(2;144) on f . A If b = 0,75, calculate the value of a. B Hence write down the equation of f . C Determine, correct to TWO decimal places, the value of f (13). D Describe the transformation of the curve of f to h if h(x) = f (−x). 340 Chapter 29 Gradient at a Point - Grade 11 29.1 Introduction In Grade 10, we investigated the idea of average gradient and saw that the gradient of some functions varied over different intervals. In Grade 11, we further look at the idea of average gradient, and are introduced to the idea of a gradient of a curve at a point. 29.2 Average Gradient We saw that the average gradient between two points on a curve is the gradient of the straight line passing through the two points. b A(-3,7) C(-1,-1) y x b Figure 29.1: The average gradient between two points on a curve is the gradient of the straight line that passes through the points. What happens to the gradient if we fix the position of one point and move the second point closer to the fixed point? Activity :: Investigation : Gradient at a Single Point on a Curve The curve shown is defined by y = −2x2 − 5. Point B is fixed at co-ordinates (0,-5). The position of point A varies. Complete the table below by calculating the y-coordinates of point A for the given x-coordinates and then calculate the average gradient between points A and B. 341 29.2 CHAPTER 29. GRADIENT AT A POINT - GRADE 11 xA -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 yA y average gradient B A x b b What happens to the average gradient as A moves towards B? What happens to the average gradient as A away from B? What is the average gradient when A overlaps with B? In Figure 29.2, the gradient of the straight line that passes through points A and C changes as A moves closer to C. At the point when A and C overlap, the straight line only passes through one point on the curve. Such a line is known as a tangent to the curve. bA (a) (b) y y bA C b x C b x (d) (c) y C bA y x C bb A x Figure 29.2: The gradient of the straight line between A and C changes as the point A moves along the curve towards C. There comes a point when A and C overlap (as shown in (c)). At this point the line is a tangent to the curve. We therefore introduce the idea of a gradient at a single point on a curve. The gradient at a point on a curve is simply the gradient of the tangent to the curve at the given point. Worked Example 122: Average Gradient Question: Find the average gradient between two points P(a; g(a)) and Q(a + h; g(a+h)) on a curve g(x) = x2 . Then find the average gradient between P(2; g(2)) 342 CHAPTER 29. GRADIENT AT A POINT - GRADE 11 29.2 and Q(4; g(4)). Finally, explain what happens to the average gradient if P moves closer to Q. Answer Step 1 : Label x points x1 = a x2 = a + h Step 2 : Determine y coordinates Using the function g(x) = x2 , we can determine: y1 = g(a) = a2 y2 = g(a + h) = (a + h)2 = a2 + 2ah + h2 Step 3 : Calculate average gradient y2 − y1 x2 − x1 = = = = = (a2 + 2ah + h2 ) − (a2 ) (a + h) − (a) 2 a + 2ah + h2 − a2 a+h−a 2ah + h2 h h(2a + h) h 2a + h (29.1) The average gradient between P(a; g(a)) and Q(a+ h; g(a+ h)) on the curve g(x) = x2 is 2a + h. Step 4 : Calculate the average gradient between P(2; g(2)) and Q(4; g(4)) We can use the result in (29.1), but we have to determine what is a and h. We do this by looking at the definitions of P and Q. The x coordinate of P is a and the x coordinate of Q is a + h therefore if we assume that a = 2 then if a + h = 4, which gives h = 2. Then the average gradient is: 2a + h = 2(2) + (2) = 6 Step 5 : When P moves closer to Q... When point P moves closer to point Q, h gets smaller. This means that the average gradient also gets smaller. When the point Q overlaps with the point P h = 0 and the average gradient is given by 2a. We now see that we can write the equation to calculate average gradient in a slightly different manner. If we have a curve defined by f (x) then for two points P and Q with P(a; f (a)) and Q(a + h; f (a + h)), then the average gradient between P and Q on f (x) is: y2 − y1 x2 − x1 = = f (a + h) − f (a) (a + h) − (a) f (a + h) − f (a) h This result is important for calculating the gradient at a point on a curve and will be explored in greater detail in Grade 12. 343 29.3 29.3 1. CHAPTER 29. GRADIENT AT A POINT - GRADE 11 End of Chapter Exercises A Determine the average gradient of the curve f (x) = x(x + 3) between x = 5 and x = 3. B Hence, state what you can deduce about the function f between x = 5 and x = 3. 2. A(1;3) is a point on f (x) = 3x2 . A Determine the gradient of the curve at point A. B Hence, determine the equation of the tangent line at A. 3. Given: f (x) = 2x2 . A Determine the average gradient of the curve between x = −2 and x = 1. B Determine the gradient of the curve of f where x = 2. 344 Chapter 30 Linear Programming - Grade 11 30.1 Introduction In everyday life people are interested in knowing the most efficient way of carrying out a task or achieving a goal. For example, a farmer might want to know how many crops to plant during a season in order to maximise yield (produce) or a stock broker might want to know how much to invest in stocks in order to maximise profit. These are examples of optimisation problems, where by optimising we mean finding the maxima or minima of a function. We have seen optimisation problems of one variable in Chapter 40, where there were no restrictions to the answer. You were then required to find the highest (maximum) or lowest (minimum) possible value of some function. In this chapter we look at optimisation problems with two variables and where the possible solutions are restricted. 30.2 Terminology There are some basic terms which you need to become familiar with for the linear programming chapters. 30.2.1 Decision Variables The aim of an optimisation problem is to find the values of the decision variables. These values are unknown at the beginning of the problem. Decision variables usually represent things that can be changed, for example the rate at which water is consumed or the number of birds living in a certain park. 30.2.2 Objective Function The objective function is a mathematical combination of the decision variables and represents the function that we want to optimise (i.e. maximise or minimise) is called the objective function. We will only be looking at objective functions which are functions of two variables. For example, in the case of the farmer, the objective function is the yield and it is dependent on the amount of crops planted. If the farmer has two crops then the objective function f (x,y) is the yield, where x represents the amount of the first crop planted and y represents the amount of the second crop planted. For the stock broker, assuming that there are two stocks to invest in, the objective function f (x,y) is the amount of profit earned by investing x rand in the first stock and y rand in the second. 345 30.2 30.2.3 CHAPTER 30. LINEAR PROGRAMMING - GRADE 11 Constraints Constraints, or restrictions, are often placed on the variables being optimised. For the example of the farmer, he cannot plant a negative number of crops, therefore the constraints would be: x≥0 y ≥ 0. Other constraints might be that the farmer cannot plant more of the second crop than the first crop and that no more than 20 units of the first crop can be planted. These constraints can be written as: x≥y x ≤ 20 Constraints that have the form ax + by ≤ c or ax + by = c are called linear constraints. Examples of linear constraints are: x+y ≤0 −2x = 7 √ y≤ 2 30.2.4 Feasible Region and Points Constraints mean that we cannot just take any x and y when looking for the x and y that optimise our objective function. If we think of the variables x and y as a point (x,y) in the xyplane then we call the set of all points in the xy-plane that satisfy our constraints the feasible region. Any point in the feasible region is called a feasible point. For example, the constraints x≥0 y ≥ 0. mean that only values of x and y that are positive are allowed. Similarly, the constraint x≥y means that only values of x that are greater than or equal to the y values are allowed. x ≤ 20 means that only x values which are less than or equal to 20 are allowed. Important: The constraints are used to create bounds of the solution. 30.2.5 The Solution Important: Points that satisfy the constraints are called feasible solutions. Once we have determined the feasible region the solution of our problem will be the feasible point where the objective function is a maximum / minimum. Sometimes there will be more 346 CHAPTER 30. LINEAR PROGRAMMING - GRADE 11 30.3 than one feasible point where the objective function is a maximum/minimum — in this case we have more than one solution. 30.3 Example of a Problem A simple problem that can be solved with linear programming involves Mrs. Nkosi and her farm. Mrs Nkosi grows mielies and potatoes on a farm of 100 m2 . She has accepted orders that will need her to grow at least 40 m2 of mielies and at least 30 m2 of potatoes. Market research shows that the demand this year will be at least twice as much for mielies as for potatoes and so she wants to use at least twice as much area for mielies as for potatoes. She expects to make a profit of R650 per m2 for her mielies and R1 500 per m2 on her sorgum. How should she divide her land so that she can earn the most profit? Let m represent the area of mielies grown and let p be the area of potatoes grown. We shall see how we can solve this problem. 30.4 Method of Linear Programming Method: Linear Programming 1. Identify the decision variables in the problem. 2. Write constraint equations 3. Write objective function as an equation 4. Solve the problem 30.5 Skills you will need 30.5.1 Writing Constraint Equations You will need to be comfortable with converting a word description to a mathematical description for linear programming. Some of the words that are used is summarised in Table 30.1. Table 30.1: Phrases and mathematical equivalents. Words Mathematical description x equals a x=a x is greater than a x>a x is greater than or equal to a x≥a x is less than a x<a x is less than or equal to a x≤a x must be at least a x≥a x must be at most b x≤a Worked Example 123: Writing constraints as equations Question: Mrs Nkosi grows mielies and potatoes on a farm of 100 m2 . She has accepted orders that will need her to grow at least 40 m2 of mielies and at least 347 30.5 CHAPTER 30. LINEAR PROGRAMMING - GRADE 11 30 m2 of potatoes. Market research shows that the demand this year will be at least twice as much for mielies as for potatoes and so she wants to use at least twice as much area for mielies as for potatoes. Answer Step 1 : Identify the decision variables There are two decision variables: the area used to plant mielies (m) and the area used to plant potatoes (p). Step 2 : Identify the phrases that constrain the decision variables • grow at least 40 m2 of mielies • grow at least 30 m2 of potatoes • area of farm is 100 m2 • demand is twice as much for mielies as for potatoes Step 3 : For each phrase, write a constraint • m ≥ 40 • p ≥ 30 • m + p ≤ 100 • m ≥ 2p Exercise: constraints as equation Write the following constraints as equations: 1. Michael is registering for courses at university. Michael needs to register for at least 4 courses. 2. Joyce is also registering for courses at university. She cannot register for more than 7 courses. 3. In a geography test, Simon is allowed to choose 4 questions from each section. 4. A baker can bake at most 50 chocolate cakes in 1 day. 5. Megan and Katja can carry at most 400 koeksisters. 30.5.2 Writing the Objective Function If the objective function is not given to you as an equation, you will need to be able to convert a word description to an equation to get the objective function. You will need to look for words like: • most profit • least cost • largest area 348 CHAPTER 30. LINEAR PROGRAMMING - GRADE 11 Worked Example 124: Writing the objective function Question: The cost of hiring a small trailer is R500 per day and the cost of hiring a big trailer is R800 per day. Write down the objective function that can be used to find the cheapest cost for hiring trailers for 1 day. Answer Step 1 : Identify the decision variables There are two decision variables: the number of big trailers (nb ) and the number of small trailers (ns ). Step 2 : Write the purpose of the objective function The purpose of the objective function is to minimise cost. Step 3 : Write the objective function The cost of hiring ns small trailers for 1 day is: 500 × ns The cost of hiring nb big trailers for 1 day is: 800 × nb Therefore the objective function, which is the total cost of hiring ns small trailers and nb big trailers for 1 day is: 500 × ns + 800 × nb Worked Example 125: Writing the objective function Question: Mrs Nkosi expects to make a profit of R650 per m2 for her mielies and R1 500 per m2 on her potatoes. How should she divide her land so that she can earn the most profit? Answer Step 1 : Identify the decision variables There are two decision variables: the area used to plant mielies (m) and the area used to plant potatoes (p). Step 2 : Write the purpose of the objective function The purpose of the objective function is to maximise profit. Step 3 : Write the objective function The profit of planting m m2 of mielies is: 650 × m The profit of planting p m2 of potatoes is: 1500 × p Therefore the objective function, which is the total profit of planting mielies and potatoes is: 650 × m + 1500 × p 349 30.5 30.5 CHAPTER 30. LINEAR PROGRAMMING - GRADE 11 Exercise: Writing the objective function 1. The EduFurn furniture factory manufactures school chairs and school desks. They make a profit of R50 on each chair sold and of R60 on each desk sold. Write an equation that will show how much profit they will make by selling the chairs and desks? 2. A manufacturer makes small screen GPS’s and wide screen GPS’s. If the profit on small screen GPS’s is R500 and the profit on wide screen GPS’s is R250, write an equation that will show the possible maximum profit. 30.5.3 Solving the Problem The numerical method involves using the points along the boundary of the feasible region, and determining which point has the optimises the objective function. Activity :: Investigation : Numerical Method Use the objective function 650 × m + 1500 × p to calculate Mrs. Nkosi’s profit for the following feasible solutions: m 60 65 70 66 32 p 30 30 30 33 31 Profit The question is How do you find the feasible region? We will use the graphical method of solving a system of linear equations to determine the feasible. We draw all constraints as graphs and mark the area that satisfies all constraints. This is shown in Figure 30.1 for Mrs. Nkosi’s farm. Now we can use the methods we learnt previously to find the points at the vertices of the feasible region. In Figure 30.1, vertex A is at the intersection of p = 30 and m = 2p. Therefore, the coordinates of A are (30,60). Similarly vertex B is at the intersection of p = 30 and m = 100−p. Therefore the coordinates of B are (30,70). Vertex C is at the intersection of m = 100 − p and m = 2p, which gives (33 31 ,66 32 ) for the coordinates of C. If we now substitute these points into the objective function, we get the following: m 60 70 66 32 p 30 30 33 31 Profit 81 000 87 000 89 997 Therefore Mrs. Nkosi makes the most profit if she plants 66 32 m2 of mielies and 66 32 m2 of potatoes. Her profit is R89 997. 350 CHAPTER 30. LINEAR PROGRAMMING - GRADE 11 30.5 m 100 90 80 70 B 60 A C 50 40 30 20 10 p 10 20 30 40 50 60 70 80 90 100 Figure 30.1: Graph of the feasible region Worked Example 126: Prizes! Question: As part of their opening specials, a furniture store has promised to give away at least 40 prizes with a total value of at least R2 000. The prizes are kettles and toasters. 1. If the company decides that there will be at least 10 of each prize, write down two more inequalities from these constraints. 2. If the cost of manufacturing a kettle is R60 and a toaster is R50, write down an objective function C which can be used to determine the cost to the company of both kettles and toasters. 3. Sketch the graph of the feasibility region that can be used to determine all the possible combinations of kettles and toasters that honour the promises of the company. 4. How many of each prize will represent the cheapest option for the company? 5. How much will this combination of kettles and toasters cost? Answer Step 1 : Identify the decision variables Let the number of kettles be xk and the number of toasters be yt and write down two constraints apart from xk ≥ 0 and yt ≥ 0 that must be adhered to. Step 2 : Write constraint equations Since there will be at least 10 of each prize we can write: xk ≥ 10 and yt ≥ 10 Also the store has promised to give away at least 40 prizes in total. Therefore: xk + yt ≥ 40 Step 3 : Write the objective function 351 30.6 CHAPTER 30. LINEAR PROGRAMMING - GRADE 11 The cost of manufacturing a kettle is R60 and a toaster is R50. Therefore the cost the total cost C is: C = 60xk + 50yt Step 4 : Sketch the graph of the feasible region yt 100 90 80 70 60 50 40 30 B 20 A 10 xk 10 20 30 40 50 60 70 80 90 100 Step 5 : Determine vertices of feasible region From the graph, the coordinates of vertex A is (3,1) and the coordinates of vertex B are (1,3). Step 6 : Calculate cost at each vertex At vertex A, the cost is: C = 60xk + 50yt = = 60(30) + 50(10) 1800 + 500 = 2300 = 60xk + 50yt = = 60(10) + 50(30) 600 + 1500 = 2100 At vertex B, the cost is: C Step 7 : Write the final answer The cheapest combination of prizes is 10 kettles and 30 toasters, costing the company R2 100. 30.6 End of Chapter Exercises 1. You are given a test consisting of two sections. The first section is on Algebra and the second section is on Geometry. You are not allowed to answer more than 10 questions from any section, but you have to answer at least 4 Algebra questions. The time allowed 352 CHAPTER 30. LINEAR PROGRAMMING - GRADE 11 30.6 is not more than 30 minutes. An Algebra problem will take 2 minutes and a Geometry problem will take 3 minutes each to solve. If you answer xA Algebra questions and yG Geometry questions, A Formulate the constraints which satisfy the above constraints. B The Algebra questions carry 5 marks each and the Geometry questions carry 10 marks each. If T is the total marks, write down an expression for T . 2. A local clinic wants to produce a guide to healthy living. The clinic intends to produce the guide in two formats: a short video and a printed book. The clinic needs to decide how many of each format to produce for sale. Estimates show that no more than 10 000 copies of both items together will be sold. At least 4 000 copies of the video and at least 2 000 copies of the book could be sold, although sales of the book are not expected to exceed 4 000 copies. Let xv be the number of videos sold, and yb the number of printed books sold. A Write down the constraint inequalities that can be deduced from the given information. B Represent these inequalities graphically and indicate the feasible region clearly. C The clinic is seeking to maximise the income, I, earned from the sales of the two products. Each video will sell for R50 and each book for R30. Write down the objective function for the income. D Determine graphically, by using a search line, the number of videos and books that ought to be sold to maximise the income. E What maximum income will be generated by the two guides? 3. A patient in a hospital needs at least 18 grams of protein, 0,006 grams of vitamin C and 0,005 grams of iron per meal, which consists of two types of food, A and B. Type A contains 9 grams of protein, 0,002 grams of vitamin C and no iron per serving. Type B contains 3 grams of protein, 0,002 grams of vitamin C and 0,005 grams of iron per serving. The energy value of A is 800 kilojoules and the of B 400 kilojoules per mass unit. A patient is not allowed to have more than 4 servings of A and 5 servings of B. There are xA servings of A and yB servings of B on the patients plate. A Write down in terms of xA and yB i. The mathematical constraints which must be satisfied. ii. The kilojoule intake per meal. B Represent the constraints graphically on graph paper. Use the scale 1 unit = 20mm on both axes. Shade the feasible region. C Deduce from the graphs, the values of xA and yB which will give the minimum kilojoule intake per meal for the patient. 4. A certain motorcycle manufacturer produces two basic models, the ’Super X’ and the ’Super Y’. These motorcycles are sold to dealers at a profit of R20 000 per ’Super X’ and R10 000 per ’Super Y’. A ’Super X’ requires 150 hours for assembly, 50 hours for painting and finishing and 10 hours for checking and testing. The ’Super Y’ requires 60 hours for assembly, 40 hours for painting and finishing and 20 hours for checking and testing. The total number of hours available per month is: 30 000 in the assembly department, 13 000 in the painting and finishing department and 5 000 in the checking and testing department. The above information can be summarised by the following table: Department Assembley Painting and Finishing Checking and Testing Hours for ‘Super X’ Hours for Super ‘Y’ 150 50 10 60 40 20 Maximum hours available per month 30 000 13 000 5 000 Let x be the number of ’Super X’ and y be the number of ’Super Y’ models manufactured per month. 353 30.6 CHAPTER 30. LINEAR PROGRAMMING - GRADE 11 A Write down the set of constraint inequalities. B Use the graph paper provided to represent the constraint inequalities. C Shade the feasible region on the graph paper. D Write down the profit generated in terms of x and y. E How many motorcycles of each model must be produced in order to maximise the monthly profit? F What is the maximum monthly profit? 5. A group of students plan to sell x hamburgers and y chicken burgers at a rugby match. They have meat for at most 300 hamburgers and at most 400 chicken burgers. Each burger of both types is sold in a packet. There are 500 packets available. The demand is likely to be such that the number of chicken burgers sold is at least half the number of hamburgers sold. A Write the constraint inequalities. B Two constraint inequalities are shown on the graph paper provided. Represent the remaining constraint inequalities on the graph paper. C Shade the feasible region on the graph paper. D A profit of R3 is made on each hamburger sold and R2 on each chicken burger sold. Write the equation which represents the total profit, P, in terms of x and y. E The objective is to maximise profit. How many, of each type of burger, should be sold to maximise profit? 6. Fashion-cards is a small company that makes two types of cards, type X and type Y. With the available labour and material, the company can make not more than 150 cards of type X and not more than 120 cards of type Y per week. Altogether they cannot make more than 200 cards per week. There is an order for at least 40 type X cards and 10 type Y cards per week. Fashion-cards makes a profit of R5 for each type X card sold and R10 for each type Y card. Let the number of type X cards be x and the number of type Y cards be y, manufactured per week. A One of the constraint inequalities which represents the restrictions above is x ≤ 150. Write the other constraint inequalities. B Represent the constraints graphically and shade the feasible region. C Write the equation that represents the profit P (the objective function), in terms of x and y. D Calculate the maximum weekly profit. 7. To meet the requirements of a specialised diet a meal is prepared by mixing two types of cereal, Vuka and Molo. The mixture must contain x packets of Vuka cereal and y packets of Molo cereal. The meal requires at least 15 g of protein and at least 72 g of carbohydrates. Each packet of Vuka cereal contains 4 g of protein and 16 g of carbohydrates. Each packet of Molo cereal contains 3 g of protein and 24 g of carbohydrates. There are at most 5 packets of cereal available. The feasible region is shaded on the attached graph paper. A Write down the constraint inequalities. B If Vuka cereal costs R6 per packet and Molo cereal also costs R6 per packet, use the graph to determine how many packets of each cereal must be used for the mixture to satisfy the above constraints in each of the following cases: i. The total cost is a minimum. ii. The total cost is a maximum (give all possibilities). 354 CHAPTER 30. LINEAR PROGRAMMING - GRADE 11 30.6 Number of packets of Molo 6 6 5 5 4 4 3 3 2 2 1 1 0 0 00 11 22 33 44 55 Number of packets of Vuka 66 8. A bicycle manufacturer makes two different models of bicycles, namely mountain bikes and speed bikes. The bicycle manufacturer works under the following constraints: No more than 5 mountain bicycles can be assembled daily. No more than 3 speed bicycles can be assembled daily. It takes one man to assemble a mountain bicycle, two men to assemble a speed bicycle and there are 8 men working at the bicycle manufacturer. Let x represent the number of mountain bicycles and let y represent the number of speed bicycles. A Determine algebraically the constraints that apply to this problem. B Represent the constraints graphically on the graph paper. C By means of shading, clearly indicate the feasible region on the graph. D The profit on a mountain bicycle is R200 and the profit on a speed bicycle is R600. Write down an expression to represent the profit on the bicycles. E Determine the number of each model bicycle that would maximise the profit to the manufacturer. 355 30.6 CHAPTER 30. LINEAR PROGRAMMING - GRADE 11 356 Chapter 31 Geometry - Grade 11 31.1 Introduction Activity :: Extension : History of Geometry Work in pairs or groups and investigate the history of the development of geometry in the last 1500 years. Describe the various stages of development and how different cultures used geometry to improve their lives. The works of the following people or cultures can be investigated: 1. Islamic geometry (c. 700 - 1500) A Thabit ibn Qurra B Omar Khayyam C Sharafeddin Tusi 2. Geometry in the 17th - 20th centuries (c. 700 - 1500) 31.2 Right Pyramids, Right Cones and Spheres A pyramid is a geometric solid that has a polygon base and the base is joined to an apex. Examples of pyramids are shown in Figure 31.1. Figure 31.1: Examples of a square pyramid, a triangular pyramid and a cone. Method: Surface Area of a Pyramid The surface area of a pyramid is calculated by adding the area of each face together. 357 31.2 CHAPTER 31. GEOMETRY - GRADE 11 Worked Example 127: Surface Area Question: If a cone √ has a height of h and a base of radius r, show that the surface area is πr2 + πr r2 + h2 . Answer Step 1 : Draw a picture a h r h r Step 2 : Identify the faces that make up the cone The cone has two faces: the base and the walls. The base is a circle of radius r and the walls can be opened out to a sector of a circle. a 2πr = circumference This curved surface can be cut into many thin triangles with height close to a (a is called a slant height). The area of these triangles will add up to 12 ×base×height which is 12 × 2πr × a = πra Step 3 : Calculate a a can be calculated by using the Theorem of Pythagoras. Therefore: p a = r 2 + h2 Step 4 : Calculate the area of the circular base Ab = πr2 Step 5 : Calculate the area of the curved walls Aw = πra p = πr r2 + h2 Step 6 : Calculate surface area A A = Ab + Aw = πr2 + πr p Method: Volume of a Pyramid The volume of a pyramid is found by: V = 1 A·h 3 where A is the area of the base and h is the height. 358 r 2 + h2 CHAPTER 31. GEOMETRY - GRADE 11 31.2 A cone is a pyramid, so the volume of a cone is given by V = 1 2 πr h. 3 A square pyramid has volume V = 1 2 a h 3 where a is the side length. Worked Example 128: Volume of a Pyramid Question: What is the volume of a square pyramid, 3cm high with a side length of 2cm? Answer Step 1 : Determine the correct formula The volume of a pyramid is 1 V = A · h, 3 which for a square base means V = 1 a · a · h. 3 3cm b 2cm 2cm Step 2 : Substitute the given values = = = 1 ·2·2·3 3 1 · 12 3 4 cm3 We accept the following formulae for volume and surface area of a sphere (ball). Surface area = Volume = Exercise: Surface Area and Volume 359 4πr2 4 3 πr 3 31.3 CHAPTER 31. GEOMETRY - GRADE 11 1. Calculate the volumes and surface areas of the following solids: *Hint for (e): find the perpendicular height using Pythagoras. c) b) d) e) 3 b a) 6b b 4 13 5 14 24 b 24 7 a hemisphere a sphere a cone a pyramid with a square base a hemisphere on top of a cone 2. Water covers approximately 71% of the Earth’s surface. Taking the radius of the Earth to be 6378 km, what is the total area of land (area not covered by water)? 3. A right triangular pyramid is placed on top of a right triangular prism. The prism has an equilateral triangle of side length 20 cm as a base, and has a height of 42 cm. The pyramid has a height of 12 cm. A Find the total volume of the object. B Find the area of each face of the pyramid. C Find the total surface area of the object. 31.3 Similarity of Polygons In order for two polygons to be similar the following must be true: 1. All corresponding angles must be congruent. 2. All corresponding sides must be in the same proportion to each other. A If P E T Q B 1. Â = P̂ ; B̂ = Q̂; Ĉ = R̂; D̂ = Ŝ; Ê = T̂ and 2. S D AB PQ = BC QR = CD RS then the polygons PQRST are similar. R C Worked Example 129: Similarity of Polygons Question: 360 = DE ST = ABCDE EA TP and CHAPTER 31. GEOMETRY - GRADE 11 31.4 R Q x 3- x Polygons PQTU and PRSU are similar. Find the value of x. P T 3 U 1 S Answer Step 1 : Identify corresponding sides Since the polygons are similar, PQ PR ∴ TU SU 3 1 = x = x + (3 − x) x = ∴ 3 ∴x = 31.4 Triangle Geometry 31.4.1 Proportion 3 9 Two line segments are divided in the same proportion if the ratios between their parts are equal. AB x kx DE = = = BC y ky EF ∴ the line segments are in the same proportion C B x A y D kx E ky F If the line segments are proportional, the following also hold 1. AC · F E = CB · DF 2. CB AC = FE DF 3. AB BC = DE FE and BC AB = FE DE 4. AB AC = DE DF and AC AB = DF DE • Triangles with equal heights have areas which are in the same proportion to each other as the bases of the triangles. 361 31.4 CHAPTER 31. GEOMETRY - GRADE 11 h1 area △ABC ∴ area △DEF = h2 1 BC × h1 BC = = 21 EF EF × h 2 2 A D h1 h2 B E C F • A special case of this happens when the bases of the triangles are equal: Triangles with equal bases between the same parallel lines have the same area. area △ABC = A 1 · h · BC = area △DBC 2 D h B C • Triangles on the same side of the same base, with equal areas, lie between parallel lines. If area △ ABC = area △ BDC, then AD k BC. A D B C Theorem 1. Proportion Theorem:A line drawn parallel to one side of a triangle divides the other two sides proportionally. E A D A A h1 h2 E D B D C B C B E 362 C CHAPTER 31. GEOMETRY - GRADE 11 31.4 Given:△ABC with line DE k BC R.T.P.: AD AE = DB EC Proof: Draw h1 from E perpendicular to AD, and h2 from D perpendicular to AE. Draw BE and CD. area △ADE area △BDE = area △ADE area △CED = but area △BDE = area △ADE ∴ = area △BDE AD = ∴ DB ∴ DE divides AB and AC proportionally. 1 2 AD · h1 1 2 DB · h1 1 2 AE · h2 1 2 EC · h2 = AD DB = AE EC area △CED (equal base and height) area △ADE area △CED AE EC Similarly, AD AB AB BD = = AE AC AC CE Following from Theorem 1, we can prove the midpoint theorem.: Theorem 2. Midpoint Theorem: A line joining the midpoints of two sides of a triangle is parallel to the third side and equal to half the length of the third side. Proof: This is a special case of the Proportionality Theorem (Theorem 1). A If AB = BD and AC = AE, then DE k BC and BC = 2DE. B C E D Theorem 3. Similarity Theorem 1:Equiangular triangles have their sides in proportion and are therefore similar. A b D b G H E B C Given:△ABC and △DEF with Â = D̂; B̂ = Ê; Ĉ = F̂ 363 F 31.4 CHAPTER 31. GEOMETRY - GRADE 11 R.T.P.: AB AC = DE DF Construct: G on AB, so that AG = DE H on AC, so that AH = DF Proof: In △’s AGH and DEF ∴ ∴ ∴ ∴ ∴ ∴ AG = DE; AH = DF (const.) Â = D̂ △AGH ≡ △DEF (given) (SAS) AĜH = Ê = B̂ GH k BC AH AG = AB AC DF DE = AB AC △ABC ||| △DEF (corres. ∠’s equal) (proportion theorem) (AG = DE; AH = DF) Important: ||| means “is similar to” Theorem 4. Similarity Theorem 2:Triangles with sides in proportion are equiangular and therefore similar. A h1 h2 D B E C Given:△ABC with line DE such that AD AE = DB EC R.T.P.:DE k BC; △ADE ||| △ABC Proof: Draw h1 from E perpendicular to AD, and h2 from D perpendicular to AE. Draw BE and CD. 364 CHAPTER 31. GEOMETRY - GRADE 11 area △ADE area △BDE area △ADE area △CED AD but DB area △ADE ∴ area △BDE ∴ area △BDE ∴ DE k BC ∴ AD̂E and AÊD = = = = = 31.4 1 2 AD · h1 1 2 DB · h1 1 2 AE · h2 1 2 EC · h2 = AD DB = AE EC AE (given) EC area △ADE area △CED area △CED (same side of equal base DE, same area) = AB̂C (corres ∠’s) = AĈB ∴ △ADE and △ABC are equiangular ∴ △ADE ||| △ABC (AAA) Theorem 5. Pythagoras’ Theorem:The square on the hypotenuse of a right angled triangle is equal to the sum of the squares on the other two sides. Given:△ ABC with Â = 90◦ A 1 2 1 2 B C R.T.P.:BC 2 = AB 2 + AC 2 Proof: Let Ĉ = x ∴ Â2 ∴ Â1 = 90◦ − x (∠ ’s of a △ ) = x B̂ D̂1 = 90◦ − x (∠ ’s of a △ ) = D̂2 = Â = 90◦ ∴ △ABD |||△CBA and △CAD |||△ CBA (AAA) AB CA BD CD AD AD ∴ and = = = = CB BA CA CB CA BA ∴ AB 2 = CB × BD and AC 2 = CB × CD ∴ AB 2 + AC 2 i.e. BC 2 = = CB(BD + CD) CB(CB) = = CB 2 AB 2 + AC 2 Worked Example 130: Triangle Geometry 1 Question: In △ GHI, GH k LJ; GJ k LK and 365 JK KI = 35 . Determine HJ KI . 31.4 CHAPTER 31. GEOMETRY - GRADE 11 G L I K J H Answer Step 1 : Identify similar triangles ˆ LIJ J L̂I ∴ △LIJ ˆ LIK K L̂I ∴ △LIK = ˆ GIH = H ĜI ||| △GIH (Corres. ∠s) (Equiangular △s) ˆ GIJ = = J ĜI ||| △GIJ (Corres. ∠s) (Equiangular △s) Step 2 : Use proportional sides HJ JI GL and LI GL LI JK KI 5 3 5 3 = = = ∴ HJ JI = (△LIJ ||| △GIH) (△LIK ||| △GIJ) Step 3 : Rearrange to find the required ratio HJ KI HJ JI × JI KI 5 8 × 3 3 40 9 = = = Worked Example 131: Triangle Geometry 2 Question: PQRS is a trapezium, with PQ k RS. Prove that PT · TR = ST · TQ. P Q 1 2 1 2 T 2 2 1 1 R 366 S CHAPTER 31. GEOMETRY - GRADE 11 31.4 Answer Step 1 : Identify similar triangles Pˆ1 Sˆ1 = (Alt. ∠s) Q̂1 = R̂1 ∴ △P T Q ||| △ST R (Alt. ∠s) (Equiangular △s) Step 2 : Use proportional sides PT = TQ ∴ PT · TR = ST TR ST · T Q (△ PTQ ||| △ STR) Exercise: Triangle Geometry 1. Calculate SV S V 10 b b 20 U 35 T 2. CB YB = 23 . Find DS SB . D A S Z X C Y B 3. Given the following figure with the following lengths, find AE, EC and BE. BC = 15 cm, AB = 4 cm, CD = 18 cm, and ED = 9 cm. C A E D B 4. Using the following figure and lengths, find IJ and KJ. HI = 26 m, KL = 13 m, JL = 9 m and HJ = 32 m. 367 31.5 CHAPTER 31. GEOMETRY - GRADE 11 J K H L I 5. Find FH in the following figure. E 36 42 D G 21 b b F H 6. BF = 25 m, AB = 13 m, AD = 9 m, DF = 18m. Calculate the lengths of BC, CF, CD, CE and EF, and find the ratio DE AC . A D B C E F 7. If LM k JK, calculate y. J 2 L K y y-2 M 7 I 31.5 Co-ordinate Geometry 31.5.1 Equation of a Line between Two Points There are many different methods of specifying the requirements for determining the equation of a straight line. One option is to find the equation of a straight line, when two points are given. Assume that the two points are (x1 ; y1 ) and (x2 ; y2 ), and we know that the general form of the equation for a straight line is: 368 CHAPTER 31. GEOMETRY - GRADE 11 31.5 y = mx + c (31.1) So, to determine the equation of the line passing through our two points, we need to determine values for m (the gradient of the line) and c (the y-intercept of the line). The resulting equation is y − y1 = m(x − x1 ) (31.2) where (x1 ; y1 ) are the co-ordinates of either given point. Extension: Finding the second equation for a straight line This is an example of a set of simultaneous equations, because we can write: y1 y2 = = mx1 + c mx2 + c (31.3) (31.4) We now have two equations, with two unknowns, m and c. Subtract (31.3) from (31.4) y2 − y1 ∴ m Re-arrange (31.3) to obtain c y1 c = mx2 − mx1 y2 − y1 = x2 − x1 = mx1 + c = y1 − mx1 (31.5) (31.6) (31.7) (31.8) Now, to make things a bit easier to remember, substitute (31.7) into (31.1): y which can be re-arranged to: y − y1 = mx + c = = mx + (y1 − mx1 ) m(x − x1 ) (31.9) (31.10) (31.11) Important: If you are asked to calculate the equation of a line passing through two points, use: y2 − y1 m= x2 − x1 to calculate m and then use: y − y1 = m(x − x1 ) to determine the equation. For example, the equation of the straight line passing through (−1; 1) and (2; 2) is given by first calculating m m y2 − y1 x2 − x1 2−1 2 − (−1) 1 3 = = = and then substituting this value into y − y1 = m(x − x1 ) to obtain y − y1 = 1 (x − x1 ). 3 369 31.5 CHAPTER 31. GEOMETRY - GRADE 11 Then substitute (−1; 1) to obtain y − (1) = So, y = 13 x + 4 3 y−1 = y = y = 1 (x − (−1)) 3 1 1 x+ 3 3 1 1 x+ +1 3 3 1 4 x+ 3 3 passes through (−1; 1) and (2; 2). 3 (2;2) b 2 y = 31 x + (-1;1) b −3 −2 4 3 1 −1 1 2 3 Figure 31.2: The equation of the line passing through (−1; 1) and (2; 2) is y = 13 x + 43 . Worked Example 132: Equation of Straight Line Question: Find the equation of the straight line passing through (−3; 2) and (5; 8). Answer Step 1 : Label the points (x1 ; y1 ) = (x2 ; y2 ) = (−3; 2) (5; 8) Step 2 : Calculate the gradient m = = = = = y2 − y1 x2 − x1 8−2 5 − (−3) 6 5+3 6 8 3 4 Step 3 : Determine the equation of the line 370 CHAPTER 31. GEOMETRY - GRADE 11 y − y1 31.5 = y − (2) = y = = = = m(x − x1 ) 3 (x − (−3)) 4 3 (x + 3) + 2 4 3 3 x+ ·3+2 4 4 9 8 3 x+ + 4 4 4 17 3 x+ 4 4 Step 4 : Write the final answer The equation of the straight line that passes through (−3; 2) and (5; 8) is y = 3 17 4x + 4 . 31.5.2 Equation of a Line through One Point and Parallel or Perpendicular to Another Line Another method of determining the equation of a straight-line is to be given one point, (x1 ; y1 ), and to be told that the line is parallel or perpendicular to another line. If the equation of the unknown line is y = mx + c and the equation of the second line is y = m0 x + c0 , then we know the following: If the lines are parallel, then m If the lines are perpendicular, then m × m0 = m0 = −1 (31.12) (31.13) Once we have determined a value for m, we can then use the given point together with: y − y1 = m(x − x1 ) to determine the equation of the line. For example, find the equation of the line that is parallel to y = 2x − 1 and that passes through (−1; 1). First we determine m. Since the line we are looking for is parallel to y = 2x − 1, m=2 The equation is found by substituting m and (−1; 1) into: y − y1 y−1 = m(x − x1 ) = 2(x − (−1) y y = 2x + 2 + 1 = 2x + 3 y−1 y−1 31.5.3 = 2(x + 1) = 2x + 2 Inclination of a Line In Figure 31.4(a), we see that the line makes an angle θ with the x-axis. This angle is known as the inclination of the line and it is sometimes interesting to know what the value of θ is. 371 31.5 CHAPTER 31. GEOMETRY - GRADE 11 3 2 (-1;1) −3 −2 y = 2x + 3 b −1 y = 2x − 1 1 1 2 3 −1 −2 Figure 31.3: The equation of the line passing through (−1; 1) and parallel to y = 2x − 1 is y = 2x + 3. It can be seen that the lines are parallel to each other. You can test this by using your ruler and measuring the distance between the lines at different points. f (x) = 4x − 4 3 3 ∆y 2 1 g(x) = 2x − 2 2 1 ∆x θg θ 1 2 3 1 (a) θf 2 3 4 (b) Figure 31.4: (a) A line makes an angle θ with the x-axis. (b) The angle is dependent on the gradient. If the gradient of f is mf and the gradient of g is mg then mf > mg and θf > θg . 372 CHAPTER 31. GEOMETRY - GRADE 11 31.6 Firstly, we note that if the gradient changes, then the value of θ changes (Figure 31.4(b)), so we suspect that the inclination of a line is related to the gradient. We know that the gradient is a ratio of a change in the y-direction to a change in the x-direction. m= ∆y ∆x But, in Figure 31.4(a) we see that tan θ = ∴m = ∆y ∆x tan θ For example, to find the inclination of the line y = x, we know m = 1 ∴ tan θ ∴θ = = 1 45◦ Exercise: Co-ordinate Geometry 1. Find the equations of the following lines A B C D E through points (−1; 3) and (1; 4) through points (7; −3) and (0; 4) parallel to y = 21 x + 3 passing through (−1; 3) perpendicular to y = − 21 x + 3 passing through (−1; 2) perpendicular to 2y + x = 6 passing through the origin 2. Find the inclination of the following lines A B C D E y = 2x − 3 y = 13 x − 7 4y = 3x + 8 y = − 23 x + 3 (Hint: if m is negative θ must be in the second quadrant) 3y + x − 3 = 0 3. Show that the line y = k for any constant k is parallel to the x-axis. (Hint: Show that the inclination of this line is 0◦ .) 4. Show that the line x = k for any constant k is parallel to the y-axis. (Hint: Show that the inclination of this line is 90◦ .) 31.6 Transformations 31.6.1 Rotation of a Point When something is moved around a fixed point, we say that it is rotated. What happens to the coordinates of a point that is rotated by 90◦ or 180◦ around the origin? Activity :: Investigation : Rotation of a Point by 90◦ 373 31.6 CHAPTER 31. GEOMETRY - GRADE 11 Complete the table, by filling in the coordinates of the points shown in the figure. Point x-coordinate y-coordinate A B C D E F G H What do you notice about the x-coordinates? What do you notice about the y-coordinates? What would happen to the coordinates of point A, if it was rotated to the position of point C? What about point B rotated to the position of D? bC bB Db Eb F b b G bA b H Activity :: Investigation : Rotation of a Point by 180◦ Complete the table, by filling in the coordinates of the points shown in the figure. Point x-coordinate y-coordinate A B C D E F G H What do you notice about the x-coordinates? What do you notice about the y-coordinates? What would happen to the coordinates of point A, if it was rotated to the position of point E? What about point F rotated to the position of B? Db Eb bC bB bF b G From these activities you should have come to the following conclusions: 374 bA b H CHAPTER 31. GEOMETRY - GRADE 11 31.6 y P(x; y) b P’(y; -x) • 90◦ clockwise rotation: The image of a point P(x; y) rotated clockwise through 90◦ around the origin is P’(y; −x). We write the rotation as (x; y) → (y; −x). b x y b P(x; y) x P”(-y; x) b • 90◦ anticlockwise rotation: The image of a point P(x; y) rotated anticlockwise through 90◦ around the origin is P’(−y; x). We write the rotation as (x; y) → (−y; x). y b P(x; y) • 180◦ rotation: The image of a point P(x; y) rotated through 180◦ around the origin is P’(−x; −y). We write the rotation as (x; y) → (−x; −y). x b P”’(-x; -y) Exercise: Rotation 1. For each of the following rotations about the origin: (i) Write down the rule. (ii) Draw a diagram showing the direction of rotation. A OA is rotated to OA′ with A(4;2) and A′ (-2;4) B OB is rotated to OB′ with B(-2;5) and B′ (5;2) C OC is rotated to OC′ with C(-1;-4) and C′ (1;4) 2. Copy ∆XYZ onto squared paper. The co-ordinates are given on the picture. A Rotate ∆XYZ anti-clockwise through an angle of 90◦ about the origin to give ∆X′ Y′ Z′ . Give the co-ordinates of X′ , Y′ and Z′ . B Rotate ∆XYZ through 180◦ about the origin to give ∆X′′ Y′′ Z′′ . Give the co-ordinates of X′′ , Y′′ and Z′′ . 375 31.6 CHAPTER 31. GEOMETRY - GRADE 11 X(4;4) Z(-4;-1) Y(-1;-4) 31.6.2 Enlargement of a Polygon 1 When something is made larger, we say that it is enlarged. What happens to the coordinates of a polygon that is enlarged by a factor k? Activity :: Investigation : Enlargement of a Polygon Complete the table, by filling in the coordinates of the points shown in the figure. Point x-coordinate y-coordinate A B C D E F G H What do you notice about the x-coordinates? What do you notice about the y-coordinates? What would happen to the coordinates of point A, if the square ABCD was enlarged by a factor 2? b b F b1 B −1b −1C bG Activity :: Investigation : Enlargement of a Polygon 2 376 b E A b1D bH CHAPTER 31. GEOMETRY - GRADE 11 31.6 7 I’ 6 5 H’ 4 I 3 H 2 J’ K’ 1 K J 0 1 0 2 3 4 5 6 7 8 9 In the figure quadrilateral HIJK has been enlarged by a factor of 2 through the origin to become H’I’J’K’. Complete the following table. Co-ordinate H = (;) I = (;) J = (;) K = (;) Co-ordinate’ H’ = (;) I’ = (;) J’ = (;) K’ + (;) Length OH = OI = OJ = OK = Length’ OH’ = OI’ = OJ’ = OK’ = What conclusions can you draw about 1. the co-ordinates 2. the lengths when we enlarge by a factor of 2? We conclude as follows: Let the vertices of a triangle have co-ordinates S(x1 ; y1 ), T(x2 ; y2 ), U(x3 ; y3 ). △S’T’U’ is an enlargement through the origin of △STU by a factor of c (c > 0). • △STU is a reduction of △S’T’U’ by a factor of c. • △S’T’U’ can alternatively be seen as an reduction through the origin of △STU by a factor of 1c . (Note that a reduction by 1c is the same as an enlargement by c). • The vertices of △S’T’U’ are S’(cx1 ; cy1 ), T’(cx2 ,cy2 ), U’(cx3 ,cy3 ). • The distances from the origin are OS’ = cOS, OT’ = cOT and OU’ = cOU. 9 8 T’ 7 6 5 S’ 4 T 3 U’ 2 S U 1 0 0 1 2 3 4 5 377 6 7 8 9 10 11 31.6 CHAPTER 31. GEOMETRY - GRADE 11 Exercise: Transformations 1. 1) Copy polygon STUV onto squared paper and then answer the following questions. 3 S 2 T 1 0 -3 -2 -1 0 1 2 3 4 5 -1 V -2 U -3 A What are the co-ordinates of polygon STUV? B Enlarge the polygon through the origin by a constant factor of c = 2. Draw this on the same grid. Label it S’T’U’V’. C What are the co-ordinates of the vertices of S’T’U’V’ ? 2. △ABC is an enlargement of △A’B’C’ by a constant factor of k through the origin. A What are the co-ordinates of the vertices of △ABC and △A’B’C’ ? B Giving reasons, calculate the value of k. C If the area of △ABC is m times the area of △A’B’C’, what is m? 5 A 4 3 A’ B 2 B’ 1 0 -4 -5 -3 -2 -1 0 -1 -2 C’ -3 -4 C -5 378 1 2 3 4 5 CHAPTER 31. GEOMETRY - GRADE 11 31.6 5 M 4 3 2 N P 1 Q 0 -2 -1 0 1 2 3 4 5 -1 3. -2 A What are the co-ordinates of the vertices of polygon MNPQ? B Enlarge the polygon through the origin by using a constant factor of c = 3, obtaining polygon M’N’P’Q’. Draw this on the same set of axes. C What are the co-ordinates of the new vertices? D Now draw M”N”P”Q” which is an anticlockwise rotation of MNPQ by 90◦ around the origin. E Find the inclination of OM”. 379 31.6 CHAPTER 31. GEOMETRY - GRADE 11 380 Chapter 32 Trigonometry - Grade 11 32.1 History of Trigonometry Work in pairs or groups and investigate the history of the development of trigonometry. Describe the various stages of development and how different cultures used trigonometry to improve their lives. The works of the following people or cultures can be investigated: 1. Cultures A Ancient Egyptians B Mesopotamians C Ancient Indians of the Indus Valley 2. People A Lagadha (circa 1350-1200 BC) B Hipparchus (circa 150 BC) C Ptolemy (circa 100) D Aryabhata (circa 499) E Omar Khayyam (1048-1131) F Bhaskara (circa 1150) G Nasir al-Din (13th century) H al-Kashi and Ulugh Beg (14th century) I Bartholemaeus Pitiscus (1595) 32.2 Graphs of Trigonometric Functions 32.2.1 Functions of the form y = sin(kθ) In the equation, y = sin(kθ), k is a constant and has different effects on the graph of the function. The general shape of the graph of functions of this form is shown in Figure 32.1 for the function f (θ) = sin(2θ). Exercise: Functions of the Form y = sin(kθ) On the same set of axes, plot the following graphs: 1. a(θ) = sin 0.5θ 381 32.2 CHAPTER 32. TRIGONOMETRY - GRADE 11 1 −270 −180 −90 90 180 270 −1 Figure 32.1: Graph of f (θ) = sin(2θ) with the graph of g(θ) = sin(θ) superimposed in gray. 2. b(θ) = sin 1θ 3. c(θ) = sin 1.5θ 4. d(θ) = sin 2θ 5. e(θ) = sin 2.5θ Use your results to deduce the effect of k. You should have found that the value of k affects the periodicity of the graph. Notice that in ◦ the case of the sine graph, the period (length of one wave) is given by 360 k . These different properties are summarised in Table 32.1. Table 32.1: Table summarising general shapes and positions of graphs of functions of the form y = sin(kx). The curve y = sin(x) is shown in gray. k>0 k<0 Domain and Range For f (θ) = sin(kθ), the domain is {θ : θ ∈ R} because there is no value of θ ∈ R for which f (θ) is undefined. The range of f (θ) = sin(kθ) is {f (θ) : f (θ) ∈ [−1,1]}. Intercepts For functions of the form, y = sin(kθ), the details of calculating the intercepts with the y axis are given. There are many x-intercepts. The y-intercept is calculated by setting θ = 0: y yint = sin(kθ) = sin(0) = 0 382 CHAPTER 32. TRIGONOMETRY - GRADE 11 32.2.2 32.2 Functions of the form y = cos(kθ) In the equation, y = cos(kθ), k is a constant and has different effects on the graph of the function. The general shape of the graph of functions of this form is shown in Figure 32.2 for the function f (θ) = cos(2θ). 1 −270 −180 −90 90 180 270 −1 Figure 32.2: Graph of f (θ) = cos(2θ) with the graph of g(θ) = cos(θ) superimposed in gray. Exercise: Functions of the Form y = cos(kθ) On the same set of axes, plot the following graphs: 1. a(θ) = cos 0.5θ 2. b(θ) = cos 1θ 3. c(θ) = cos 1.5θ 4. d(θ) = cos 2θ 5. e(θ) = cos 2.5θ Use your results to deduce the effect of k. You should have found that the value of k affects the periodicity of the graph. The period of ◦ the cosine graph is given by 360 k . These different properties are summarised in Table 32.2. Table 32.2: Table summarising general shapes and positions of graphs of functions of the form y = cos(kx). The curve y = cos(x) is shown in gray. k>0 k<0 Domain and Range For f (θ) = cos(kθ), the domain is {θ : θ ∈ R} because there is no value of θ ∈ R for which f (θ) is undefined. The range of f (θ) = cos(kθ) is {f (θ) : f (θ) ∈ [−1,1]}. 383 32.2 CHAPTER 32. TRIGONOMETRY - GRADE 11 Intercepts For functions of the form, y = cos(kθ), the details of calculating the intercepts with the y axis are given. The y-intercept is calculated as follows: y yint = cos(kθ) = cos(0) = 1 32.2.3 Functions of the form y = tan(kθ) In the equation, y = tan(kθ), k is a constant and has different effects on the graph of the function. The general shape of the graph of functions of this form is shown in Figure 32.3 for the function f (θ) = tan(2θ). 5 −360 −270 −180 −90 90 180 270 360 −5 Figure 32.3: The graph of tan(2θ) superimposed on the graph of g(θ) = tan(θ) (in gray). The asymptotes are shown as dashed lines. Exercise: Functions of the Form y = tan(kθ) On the same set of axes, plot the following graphs: 1. a(θ) = tan 0.5θ 2. b(θ) = tan 1θ 3. c(θ) = tan 1.5θ 4. d(θ) = tan 2θ 5. e(θ) = tan 2.5θ Use your results to deduce the effect of k. You should have found that, once again, the value of k affects the periodicity of the graph. As k increases, the graph is more tightly packed. As k decreases, the graph is more spread out. The ◦ period of the tan graph is given by 180 k . These different properties are summarised in Table 32.3. Domain and Range ◦ ◦ For f (θ) = tan(kθ), the domain of one branch is {θ : θ ∈ (− 90k , 90k )} because the function is ◦ ◦ undefined for θ = 90k and θ = 90k . The range of f (θ) = tan(kθ) is {f (θ) : f (θ) ∈ (−∞,∞)}. 384 CHAPTER 32. TRIGONOMETRY - GRADE 11 32.2 Table 32.3: Table summarising general shapes and positions of graphs of functions of the form y = tan(kθ). k>0 k<0 Intercepts For functions of the form, y = tan(kθ), the details of calculating the intercepts with the x and y axis are given. There are many x-intercepts; each one is halfway between the asymptotes. The y-intercept is calculated as follows: y = tan(kθ) yint = tan(0) = 0 Asymptotes The graph of tan kθ has asymptotes because as kθ approaches 90◦ , tan kθ approaches infinity. In other words, there is no defined value of the function at the asymptote values. 32.2.4 Functions of the form y = sin(θ + p) In the equation, y = sin(θ + p), p is a constant and has different effects on the graph of the function. The general shape of the graph of functions of this form is shown in Figure 32.4 for the function f (θ) = sin(θ + 30◦ ). 1 −270 −180 −90 90 180 270 −1 Figure 32.4: Graph of f (θ) = sin(θ + 30◦ ) with the graph of g(θ) = sin(θ) in gray. Exercise: Functions of the Form y = sin(θ + p) On the same set of axes, plot the following graphs: 1. a(θ) = sin(θ − 90◦ ) 2. b(θ) = sin(θ − 60◦ ) 3. c(θ) = sin θ 4. d(θ) = sin(θ + 90◦ ) 5. e(θ) = sin(θ + 180◦ ) Use your results to deduce the effect of p. 385 32.2 CHAPTER 32. TRIGONOMETRY - GRADE 11 You should have found that the value of p affects the y-intercept and phase shift of the graph. The p value shifts the graph horizontally. If p is positive, the graph shifts left and if p is negative tha graph shifts right. These different properties are summarised in Table 32.4. Table 32.4: Table summarising general shapes and positions of graphs of functions of the form y = sin(θ + p). p>0 p<0 Domain and Range For f (θ) = sin(θ + p), the domain is {θ : θ ∈ R} because there is no value of θ ∈ R for which f (θ) is undefined. The range of f (θ) = sin(θ + p) is {f (θ) : f (θ) ∈ [−1,1]}. Intercepts For functions of the form, y = sin(θ + p), the details of calculating the intercept with the y axis are given. The y-intercept is calculated as follows: set θ = 0◦ y yint = sin(θ + p) = sin(0 + p) = sin(p) 32.2.5 Functions of the form y = cos(θ + p) In the equation, y = cos(θ + p), p is a constant and has different effects on the graph of the function. The general shape of the graph of functions of this form is shown in Figure 32.5 for the function f (θ) = cos(θ + 30◦ ). 1 −270 −180 −90 90 180 270 −1 Figure 32.5: Graph of f (θ) = cos(θ + 30◦ ) with the graph of g(θ) = cos(θ) shown in gray. Exercise: Functions of the Form y = cos(θ + p) On the same set of axes, plot the following graphs: 386 CHAPTER 32. TRIGONOMETRY - GRADE 11 32.2 1. a(θ) = cos(θ − 90◦ ) 2. b(θ) = cos(θ − 60◦ ) 3. c(θ) = cos θ 4. d(θ) = cos(θ + 90◦ ) 5. e(θ) = cos(θ + 180◦ ) Use your results to deduce the effect of p. You should have found that the value of p affects the y-intercept and phase shift of the graph. As in the case of the sine graph, positive values of p shift the cosine graph left while negative p values shift the graph right. These different properties are summarised in Table 32.5. Table 32.5: Table summarising general shapes and positions of graphs of functions of the form y = cos(θ + p). The curve y = cos θ is shown in gray. p>0 p<0 Domain and Range For f (θ) = cos(θ + p), the domain is {θ : θ ∈ R} because there is no value of θ ∈ R for which f (θ) is undefined. The range of f (θ) = cos(θ + p) is {f (θ) : f (θ) ∈ [−1,1]}. Intercepts For functions of the form, y = cos(θ + p), the details of calculating the intercept with the y axis are given. The y-intercept is calculated as follows: set θ = 0◦ 32.2.6 y = cos(θ + p) yint = = cos(0 + p) cos(p) Functions of the form y = tan(θ + p) In the equation, y = tan(θ + p), p is a constant and has different effects on the graph of the function. The general shape of the graph of functions of this form is shown in Figure 32.6 for the function f (θ) = tan(θ + 30◦ ). Exercise: Functions of the Form y = tan(θ + p) On the same set of axes, plot the following graphs: 1. a(θ) = tan(θ − 90◦ ) 387 32.2 CHAPTER 32. TRIGONOMETRY - GRADE 11 5 −360 −270 −180 −90 90 180 270 360 −5 Figure 32.6: The graph of tan(θ + 30◦ ) with the graph of g(θ) = tan(θ) shown in gray. 2. b(θ) = tan(θ − 60◦ ) 3. c(θ) = tan θ 4. d(θ) = tan(θ + 60◦ ) 5. e(θ) = tan(θ + 180◦ ) Use your results to deduce the effect of p. You should have found that the value of p once again affects the y-intercept and phase shift of the graph. There is a horizontal shift to the left if p is positive and to the right if p is negative. These different properties are summarised in Table 32.6. Table 32.6: Table summarising general shapes and positions of graphs of functions of the form y = tan(θ + p). k>0 k<0 Domain and Range For f (θ) = tan(θ + p), the domain for one branch is {θ : θ ∈ (−90◦ − p,90◦ − p} because the function is undefined for θ = −90◦ − p and θ = 90◦ − p. The range of f (θ) = tan(θ + p) is {f (θ) : f (θ) ∈ (−∞,∞)}. Intercepts For functions of the form, y = tan(θ + p), the details of calculating the intercepts with the y axis are given. The y-intercept is calculated as follows: set θ = 0◦ y yint = tan(θ + p) = tan(p) Asymptotes The graph of tan(θ + p) has asymptotes because as θ + p approaches 90◦ , tan(θ + p) approaches infinity. Thus, there is no defined value of the function at the asymptote values. 388 CHAPTER 32. TRIGONOMETRY - GRADE 11 32.3 Exercise: Functions of various form Using your knowledge of the effects of p and k draw a rough sketch of the following graphs without a table of values. 1. y = sin 3x 2. y = − cos 2x 3. y = tan 12 x 4. y = sin(x − 45◦ ) 5. y = cos(x + 45◦ ) 6. y = tan(x − 45◦ ) 7. y = 2 sin 2x 8. y = sin(x + 30◦ ) + 1 32.3 Trigonometric Identities 32.3.1 Deriving Values of Trigonometric Functions for 30◦ , 45◦ and 60◦ Keeping in mind that trigonometric functions apply only to right-angled triangles, we can derive values of trigonometric functions for 30◦ , 45◦ and 60◦ . We shall start with 45◦ as this is the easiest. Take any right-angled triangle with one angle 45◦ . Then, because one angle is 90◦ , the third angle is also 45◦ . So we have an isosceles right-angled triangle as shown in Figure 32.7. C b 45◦ b A b B Figure 32.7: An isosceles right angled triangle. If the two equal sides are of length a, then the hypotenuse, h, can be calculated as: h2 ∴ = a2 + a2 = 2a2 √ = 2a h So, we have: sin(45◦ ) opposite(45◦ ) hypotenuse a = √ 2a 1 = √ 2 389 = 32.3 CHAPTER 32. TRIGONOMETRY - GRADE 11 adjacent(45◦ ) hypotenuse a √ 2a 1 √ 2 cos(45◦ ) = = = tan(45◦ ) opposite(45◦ ) adjacent(45◦ ) a = a = 1 = We can try something similar for 30◦ and 60◦ . We start with an equilateral triangle and we bisect one angle as shown in Figure 32.8. This gives us the right-angled triangle that we need, with one angle of 30◦ and one angle of 60◦ . B 30◦ b b 60◦ b b C A D Figure 32.8: An equilateral triangle with one angle bisected. If the equal sides are of length a, then the base is 21 a and the length of the vertical side, v, can be calculated as: v2 = = = ∴ v = 1 a2 − ( a)2 2 1 a2 − a2 4 3 2 a 4√ 3 a 2 So, we have: 390 CHAPTER 32. TRIGONOMETRY - GRADE 11 sin(30◦ ) = = = cos(30◦ ) = = = tan(30◦ ) opposite(30◦ ) hypotenuse 32.3 sin(60◦ ) = a 2 a 1 2 = = adjacent(30◦ ) hypotenuse cos(60◦ ) √ 3 2 a √a 3 2 = = = = opposite(30◦ ) adjacent(30◦ ) = a √2 3 2 a = 1 √ 3 tan(60◦ ) = = = opposite(60◦ ) hypotenuse √ 3 2 a √a 3 2 adjacent(60◦ ) hypotenuse a 2 a 1 2 opposite(60◦ ) adjacent(60◦ ) √ 3 2 a a 2 √ 3 You do not have to memorise these identities if you know how to work them out. Important: Two useful triangles to remember 2 30 32.3.2 ◦ √ 2 60◦ 1 45 √ 3 45◦ 1 ◦ 1 Alternate Definition for tan θ We know that tan θ is defined as: tan θ = opposite adjacent This can be written as: tan θ = = opposite hypotenuse × adjacent hypotenuse hypotenuse opposite × hypotenuse adjacent But, we also know that sin θ is defined as: sin θ = and that cos θ is defined as: cos θ = opposite hypotenuse adjacent hypotenuse 391 32.3 CHAPTER 32. TRIGONOMETRY - GRADE 11 Therefore, we can write tan θ opposite hypotenuse × hypotenuse adjacent 1 sin θ × cos θ sin θ cos θ = = = Important: tan θ can also be defined as: tan θ = 32.3.3 sin θ cos θ A Trigonometric Identity One of the most useful results of the trigonometric functions is that they are related to each other. We have seen that tan θ can be written in terms of sin θ and cos θ. Similarly, we shall show that: sin2 θ + cos2 θ = 1 We shall start by considering △ABC, C b θ b b A We see that: sin θ = AC BC cos θ = AB . BC and B We also know from the Theorem of Pythagoras that: AB 2 + AC 2 = BC 2 . So we can write: 2 2 sin θ + cos θ = = = = = 2 2 AB AC + BC BC 2 2 AB AC + BC 2 BC 2 AC 2 + AB 2 BC 2 BC 2 (from Pythagoras) BC 2 1 392 CHAPTER 32. TRIGONOMETRY - GRADE 11 Worked Example 133: Trigonometric Identities A Question: Simplify using identities: 1. tan2 θ · cos2 θ 2. 1 cos2 θ − tan2 θ Answer Step 1 : Use known identities to replace tan θ = = = tan2 θ · cos2 θ sin2 θ · cos2 θ cos2 θ sin2 θ Step 2 : Use known identities to replace tan θ = = = = 1 − tan2 θ cos2 θ sin2 θ 1 − cos2 θ cos2 θ 1 − sin2 θ cos2 θ cos2 θ =1 cos2 θ Worked Example 134: Trigonometric Identities B Question: Prove: Answer 1−sin x cos x = cos x 1+sin x LHS = = = = = 1 − sin x cos x 1 − sin x 1 + sin x × cos x 1 + sin x 2 1 − sin x cos x(1 + sin x) cos2 x cos x(1 + sin x) cos x = RHS 1 + sin x Exercise: Trigonometric identities 393 32.3 32.3 CHAPTER 32. TRIGONOMETRY - GRADE 11 1. Simplify the following using the fundamental trigonometric identities: A cos θ tan θ 2 B cos θ. tan2 θ + tan2 θ. sin2 θ C 1 − tan2 θ. sin2 θ D 1 − sin θ. cos θ. tan θ E 1 − sin2 θ 2 θ F 1−cos − cos2 θ 2 cos θ 2. Prove the following: A 1+sin θ cos θ 2 = cos θ 1−sin θ B sin θ + (cos θ − tan θ)(cos θ + tan θ) = 1 − tan2 θ C D E F 32.3.4 (2 cos2 θ−1) 1 + 1 (1+tan2 θ) = 1−tan2 θ 1+tan2 θ 1 cos θ tan2 θ =1 cos θ − 1 2 sin θ cos θ 1 sin θ+cos θ = sin θ + cos θ − sin θ+cos θ 1 cos θ sin θ + tan θ · cos θ = sin θ Reduction Formula Any trigonometric function whose argument is 90◦ ± θ, 180◦ ± θ, 270◦ ± θ and 360◦ ± θ (hence −θ) can be written simply in terms of θ. For example, you may have noticed that the cosine graph is identical to the sine graph except for a phase shift of 90◦ . From this we may expect that sin(90◦ + θ) = cos θ. Function Values of 180◦ ± θ Activity :: Investigation : Reduction Formulae for Function Values of 180◦ ± θ 1. Function Values of (180◦ − θ) A In the figure P and P’ lie on the circle with radius 2. OP makes an angle θ = 30◦ with √ the x-axis. P thus has co-ordinates ( 3; 1). If P’ is the reflection of P about the y-axis (or the line x = 0), use symmetry to write down the co-ordinates of P’. y P’ b 2 180◦ − θ bP 2 θ θ 0 B Write down values for sin θ, cos θ and tan θ. C Using the co-ordinates for P’ determine sin(180◦ − θ), cos(180◦ − θ) and tan(180◦ − θ). (d) From your results try and determine a relationship between the function values of (180◦ − θ) and θ. 2. Function values of (180◦ + θ) 394 x CHAPTER 32. TRIGONOMETRY - GRADE 11 32.3 A In the figure P and P’ lie on the circle with radius 2. OP makes an angle θ = 30◦ with √ the x-axis. P thus has coordinates ( 3; 1). P’ is the inversion of P through the origin (reflection about both the x- and y-axes) and lies at an angle of 180◦ + θ with the x-axis. Write down the co-ordinates of P’. y 180◦ + θ θ P’ b x 0 θ B Using the co-ordinates for P’ determine sin(180◦ + θ), cos(180◦ + θ) and tan(180◦ + θ). bP 2 2 C From your results try and determine a relationship between the function values of (180◦ + θ) and θ. Activity :: Investigation : Reduction Formulae for Function Values of 360◦ ± θ 1. Function values of (360◦ − θ) A In the figure P and P’ lie on the circle with radius 2. OP makes an angle θ = 30◦ with √ the x-axis. P thus has co-ordinates ( 3; 1). P’ is the reflection of P about the x-axis or the line y = 0. Using symmetry, write down the co-ordinates of P’. y bP 360◦ − θ 2 θ 0 x θ 2 b B Using the co-ordinates for P’ determine sin(360◦ − θ), cos(360◦ − θ) and tan(360◦ − θ). P’ C From your results try and determine a relationship between the function values of (360◦ − θ) and θ. It is possible to have an angle which is larger than 360◦ . The angle completes one revolution to give 360◦ and then continues to give the required angle. We get the following results: sin(360◦ + θ) = ◦ cos(360 + θ) = tan(360◦ + θ) = sin θ cos θ tan θ Note also, that if k is any integer, then sin(360◦ · k + θ) = ◦ cos(360 · k + θ) = tan(360◦ · k + θ) = 395 sin θ cos θ tan θ 32.3 CHAPTER 32. TRIGONOMETRY - GRADE 11 Worked Example 135: Basic use of a reduction formula Question: Write sin 293◦ as the function of an acute angle. Answer We note that 293◦ = 360◦ − 67◦ thus sin 293◦ = = sin(360◦ − 67◦ ) − sin 67◦ where we used the fact that sin(360◦ − θ) = − sin θ. Check, using your calculator, that these values are in fact equal: sin 293◦ − sin 67◦ = = −0,92 · · · −0,92 · · · Worked Example 136: More complicated... Question: Evaluate without using a calculator: tan2 210◦ − (1 + cos 120◦) sin2 225◦ Answer tan2 210◦ − (1 + cos 120◦ ) sin2 225◦ = [tan(180◦ + 30◦ )]2 − [1 + cos(180◦ − 60◦ )] · [sin(180◦ + 45◦ )]2 = (tan 30◦ )2 − [1 + (− cos 60◦ )] · (− sin 45◦ )2 2 2 1 1 1 √ − 1− = · −√ 2 3 2 1 1 1 − = 3 2 2 1 1 1 = − = 3 4 12 Exercise: Reduction Formulae 1. Write these equations as a function of θ only: A B C D E F sin(180◦ − θ) cos(180◦ − θ) cos(360◦ − θ) cos(360◦ + θ) tan(180◦ − θ) cos(360◦ + θ) 2. Write the following trig functions as a function of an acute angle, then find the actual angle with your calculator: A sin 163◦ 396 CHAPTER 32. TRIGONOMETRY - GRADE 11 32.3 B cos 327◦ C tan 248◦ D cos 213◦ 3. Determine the following without the use of a calculator: A B C D tan 150◦. sin 30◦ + cos 330◦ tan 300◦. cos 120◦ (1 − cos 30◦ )(1 − sin 210◦ ) cos 780◦ + sin 315◦. tan 420◦ 4. Determine the following by reducing to an acute angle and using special angles. Do not use a calculator: A B C D E F G H I J K L cos 300◦ sin 135◦ cos 150◦ tan 330◦ sin 120◦ tan2 225◦ cos 315◦ sin2 420◦ tan 405◦ cos 1020◦ tan2 135◦ 1 − sin2 210◦ Function Values of (−θ) When the argument of a trigonometric function is (−θ) we can add 360◦ without changing the result. Thus for sine and cosine sin(−θ) = sin(360◦ − θ) = − sin θ cos(−θ) = cos(360◦ − θ) = cos θ Function Values of 90◦ ± θ Activity :: Investigation : Reduction Formulae for Function Values of 90◦ ± θ 1. Function values of (90◦ − θ) y A In the figure P and P’ lie on the circle with radius 2. OP makes an angle θ = 30◦ with √ the x-axis. P thus has co-ordinates ( 3; 1). P’ is the reflection of P about the line y = x. Using symmetry, write down the co-ordinates of P’. B Using the co-ordinates for P’ determine sin(90◦ −θ), cos(90◦ −θ) and tan(90◦ −θ). C From your results try and determine a relationship between the function values of (90◦ − θ) and θ. 397 b P’ θ 90◦ − θ 2 bP 2 θ 0 x 32.3 CHAPTER 32. TRIGONOMETRY - GRADE 11 2. Function values of (90◦ + θ) A In the figure P and P’ lie on the circle with radius 2. OP makes an angle θ = 30◦ with √ the x-axis. P thus has co-ordinates ( 3; 1). P’ is the rotation of P through 90◦ . Using symmetry, write down the coordinates of P’. (Hint: consider P’ as the reflection of P about the line y = x followed by a reflection about the y-axis) y P’b θ 2 90◦ + θ 2 P b θ 0 B Using the co-ordinates for P’ determine sin(90◦ +θ), cos(90◦ +θ) and tan(90◦ +θ). C From your results try and determine a relationship between the function values of (90◦ + θ) and θ. Complementary angles are positive acute angles that add up to 90◦ . e.g. 20◦ and 70◦ are complementary angles. Sine and cosine are known as co-functions. The other co-functions are secant and cosecant, and tangent and cotangent. The function value of an angle is equal to the co-function of its complement (the co-co rule). Thus for sine and cosine we have sin(90◦ − θ) cos(90◦ − θ) = cos θ = sin θ Worked Example 137: Co-co rule Question: Write each of the following in terms of 40◦ using sin(90◦ − θ) = cos θ and cos(90◦ − θ) = sin θ. 1. cos 50◦ 2. sin 320◦ 3. cos 230◦ Answer 1. cos 50◦ = cos(90◦ − 40◦ ) = sin 40◦ 2. sin 320◦ = sin(360◦ − 40◦ ) = − sin 40◦ 3. cos 230◦ = cos(180◦ + 50◦ ) = − cos 50◦ = − cos(90◦ − 40◦ ) = − sin 40◦ Function Values of (θ − 90◦ ) sin(θ − 90◦ ) = − cos θ and cos(θ − 90◦ ) = sin θ. These results may be proved as follows: sin(θ − 90◦ ) = sin[−(90◦ − θ)] = − sin(90◦ − θ) = − cos θ and likewise for cos(θ − 90◦ ) = sin θ 398 x CHAPTER 32. TRIGONOMETRY - GRADE 11 32.4 Summary The following summary may be made second quadrant (180◦ − θ) or (90◦ + θ) sin(180◦ − θ) = + sin θ cos(180◦ − θ) = − cos θ tan(180◦ − θ) = − tan θ sin(90◦ + θ) = + cos θ cos(90◦ + θ) = − sin θ third quadrant (180◦ + θ) sin(180◦ + θ) = − sin θ cos(180◦ + θ) = − cos θ tan(180◦ + θ) = + tan θ first quadrant (θ) or (90◦ − θ) all trig functions are positive sin(360◦ + θ) = sin θ cos(360◦ + θ) = cos θ tan(360◦ + θ) = tan θ sin(90◦ − θ) = sin θ cos(90◦ − θ) = cos θ fourth quadrant (360◦ − θ) sin(360◦ − θ) = − sin θ cos(360◦ − θ) = + cos θ tan(360◦ − θ) = − tan θ Important: 1. These reduction formulae hold for any angle θ. For convenience, we usually work with θ as if it is acute, i.e. 0◦ < θ < 90◦ . 2. When determining function values of 180◦ ± θ, 360◦ ± θ and −θ the functions never change. 3. When determining function values of 90◦ ± θ and θ − 90◦ the functions changes to its co-function (co-co rule). Extension: Function Values of (270◦ ± θ) Angles in the third and fourth quadrants may be written as 270◦ ± θ with θ an acute angle. Similar rules to the above apply. We get third quadrant (270◦ − θ) sin(270◦ − θ) = − cos θ cos(270◦ − θ) = − sin θ 32.4 fourth quadrant (270◦ + θ) sin(270◦ + θ) = − cos θ cos(270◦ + θ) = + sin θ Solving Trigonometric Equations Chapters ?? and ?? focussed on the solution of algebraic equations and excluded equations that dealt with trigonometric functions (i.e. sin and cos). In this section, the solution of trigonometric equations will be discussed. The methods described in Chapters ?? and ?? also apply here. In most cases, trigonometric identities will be used to simplify equations, before finding the final solution. The final solution can be found either graphically or using inverse trigonometric functions. 32.4.1 Graphical Solution As an example, to introduce the methods of solving trigonometric equations, consider sin θ = 0,5 (32.1) As explained in Chapters ?? and ??, the solution of Equation 32.1 is obtained by examining the intersecting points of the graphs of: y y = sin θ = 0,5 399 32.4 CHAPTER 32. TRIGONOMETRY - GRADE 11 Both graphs, for −720◦ < θ < 720◦, are shown in Figure 32.9 and the intersection points of the graphs are shown by the dots. y = sin θ 1 b b b b b −720 −630 −540 −450 −360 −270 −180 −90 b 90 180 b 270 360 b 450 540 y = 0,5 630 −1 Figure 32.9: Plot of y = sin θ and y = 0,5 showing the points of intersection, hence the solutions to the equation sin θ = 0,5. In the domain for θ of −720◦ < θ < 720◦, there are 8 possible solutions for the equation sin θ = 0,5. These are θ = [−690◦, −570◦, −330◦, −210◦, 30◦ , 150◦, 390◦ , 510◦] Worked Example 138: Question: Find θ, if tan θ + 0,5 = 1,5, with 0◦ < θ < 90◦ . Determine the solution graphically. Answer Step 1 : Write the equation so that all the terms with the unknown quantity (i.e. θ) are on one side of the equation. tan θ + 0,5 = tan θ = 1,5 1 Step 2 : Identify the two functions which are intersecting. y = tan θ y = 1 Step 3 : Draw graphs of both functions, over the required domain and identify the intersection point. y = tan θ 1 0 b 45 90 −1 The graphs intersect at θ = 45◦ . 400 y=1 720 CHAPTER 32. TRIGONOMETRY - GRADE 11 32.4.2 32.4 Algebraic Solution The inverse trigonometric functions arcsin, arccos and arctan can also be used to solve trigonometric equations. These are shown as second functions on most calculators: sin−1 , cos−1 and tan−1 . Using inverse trigonometric functions, the equation sin θ = 0,5 is solved as sin θ ∴ θ = 0,5 = arcsin 0,5 = 30◦ Worked Example 139: Question: Find θ, if tan θ + 0,5 = 1,5, with 0◦ < θ < 90◦ . Determine the solution using inverse trigonometric functions. Answer Step 1 : Write the equation so that all the terms with the unknown quantity (i.e. θ) are on one side of the equation. Then solve for the angle using the inverse function. tan θ + 0,5 = tan θ = ∴ θ = = 1,5 1 arctan 1 45◦ Trigonometric equations often look very simple. Consider solving the equation sin θ = 0,7. We can take the inverse sine of both sides to find that θ = sin−1 (0,7). If we put this into a calculator we find that sin−1 (0,7) = 44,42◦ . This is true, however, it does not tell the whole story. y 1 −360 −180 180 360 x −1 Figure 32.10: The sine graph. The dotted line represents y = 0,7. There are four points of intersection on this interval, thus four solutions to sin θ = 0,7. As you can see from figure 32.10, there are four possible angles with a sine of 0.7 between −360◦ and 360◦. If we were to extend the range of the sine graph to infinity we would in fact see that there are an infinite number of solutions to this equation! This difficulty (which is caused by the periodicity of the sine function) makes solving trigonometric equations much harder than they may seem to be. 401 32.4 CHAPTER 32. TRIGONOMETRY - GRADE 11 Any problem on trigonometric equations will require two pieces of information to solve. The first is the equation itself and the second is the range in which your answers must lie. The hard part is making sure you find all of the possible answers within the range. Your calculator will always give you the smallest answer (i.e. the one that lies between −90◦ and 90◦ for tangent and sine and one between 0◦ and 180◦ for cosine). Bearing this in mind we can already solve trigonometric equations within these ranges. Worked Example 140: Question: Find the values of x for which sin x2 = 0,5 if it is given that x < 90◦ . Answer Because we are told that x is an acute angle, we can simply apply an inverse trigonometric function to both sides. sin x2 = 0.5 (32.2) ⇒ ⇒ x 2 x 2 = = ∴x = arcsin 0.5 30◦ (32.3) (32.4) 60◦ (32.5) We can, of course, solve trigonometric equations in any range by drawing the graph. Worked Example 141: Question: For what values of x does sin x = 0,5, when −360◦ ≤ x ≤ 360◦ ? Answer Step 1 : Draw the graph We take a look at the graph of sin x = 0,5 on the interval [−360◦, 360◦ ]. We want to know when the y value of the graph is 0,5, so we draw in a line at y = 0,5. y 1 −360 −180 180 360 x −1 Step 2 : Notice that this line touches the graph four times. This means that there are four solutions to the equation. Step 3 : Read off the x values of those intercepts from the graph as x = −330◦, −210◦, 30◦ and 150◦. 402 CHAPTER 32. TRIGONOMETRY - GRADE 11 1st 1 2nd 3rd 32.4 4th 90◦ 2nd +VE 0 90 ◦ 180 ◦ 270 ◦ 360 ◦ 1st +VE 0◦ /360◦ 180◦ 3rd -VE 4th -VE 270◦ −1 +VE +VE -VE -VE Figure 32.11: The graph and unit circle showing the sign of the sine function. y 1 −360 −270 −180 −90 90 180 270 360 x −1 This method can be time consuming and inexact. We shall now look at how to solve these problems algebraically. 32.4.3 Solution using CAST diagrams The Sign of the Trigonometric Function The first step to finding the trigonometry of any angle is to determine the sign of the ratio for a given angle. We shall do this for the sine function first and do the same for the cosine and tangent. In figure 32.11 we have split the sine graph into four quadrants, each 90◦ wide. We call them quadrants because they correspond to the four quadrants of the unit circle. We notice from figure 32.11 that the sine graph is positive in the 1st and 2nd quadrants and negative in the 3rd and 4th . Figure 32.12 shows similar graphs for cosine and tangent. All of this can be summed up in two ways. Table 32.7 shows which trigonometric functions are positive and which are negative in each quadrant. sin cos tan 1st +VE +VE +VE 2nd +VE -VE -VE 3rd -VE -VE +VE 4th -VE +VE -VE Table 32.7: The signs of the three basic trigonometric functions in each quadrant. A more convenient way of writing this is to note that all functions are positive in the 1st quadrant, only sine is positive in the 2nd , only tangent in the 3rd and only cosine in the 4th . We express 403 32.4 CHAPTER 32. TRIGONOMETRY - GRADE 11 1 0 −1 1st 2nd 3rd 4th 8 6 4 2 0 90◦ 180◦ 270◦ 360◦ −2 −4 −6 −8 +VE +VE -VE -VE 1st 2nd 90 ◦ +VE 3rd 180 ◦ -VE 4th 270 ◦ +VE 360 ◦ -VE Figure 32.12: Graphs showing the sign of the cosine and tangent functions. this using the CAST diagram (figure 32.13). This diagram is known as a CAST diagram as the letters, taken anticlockwise from the bottom right, read C-A-S-T. The letter in each quadrant tells us which trigonometric functions are positive in that quadrant. The ‘A’ in the 1st quadrant stands for all (meaning sine, cosine and tangent are all positive in this quadrant). ‘S’, ‘C’ and ‘T’ ,of course, stand for sine, cosine and tangent. The diagram is shown in two forms. The version on the left shows the CAST diagram including the unit circle. This version is useful for equations which lie in large or negative ranges. The simpler version on the right is useful for ranges between 0◦ and 360◦ . Another useful diagram shown in figure 32.13 gives the formulae to use in each quadrant when solving a trigonometric equation. 90◦ S 180 A ◦ ◦ 0 /360 T S A 180◦ − θ θ T C 180◦ + θ 360◦ − θ ◦ C 270◦ Figure 32.13: The two forms of the CAST diagram and the formulae in each quadrant. Magnitude of the trigonometric functions Now that we know in which quadrants our solutions lie, we need to know which angles in these quadrants satisfy our equation. Calculators give us the smallest possible answer (sometimes negative) which satisfies the equation. For example, if we wish to solve sin θ = 0,3 we can apply the inverse sine function to both sides of the equation to find– θ = = arcsin 0,3 17,46◦ However, we know that this is just one of infinitely many possible answers. We get the rest of the answers by finding relationships between this small angle, θ, and answers in other quadrants. To do this we use our small angle θ as a reference angle. We then look at the sign of the trigonometric function in order to decide in which quadrants we need to work (using the CAST diagram) and add multiples of the period to each, remembering that sine, cosine and tangent are periodic (repeating) functions. To add multiples of the period we use 360◦ · n (where n is an integer) for sine and cosine and 180◦ · n, n ∈ Z, for the tangent. 404 CHAPTER 32. TRIGONOMETRY - GRADE 11 32.4 Worked Example 142: Question: Solve for θ: sin θ = 0,3 Answer Step 1 : Determine in which quadrants the solution lies We look at the sign of the trigonometric function. sin θ is given as a positive amount (0,3). Reference to the CAST diagram shows that sine is positive in the first and second quadrants. Step 2 : Determine the reference angle The small angle θ is the angle returned by the calculator: sin θ = 0,3 ⇒θ ⇒θ = = arcsin 0,3 17,46◦ Step 3 : Determine the general solution Our solution lies in quadrants I and II. We therefore use θ and 180◦ − θ, and add the 360◦ · n for the periodicity of sine. I: θ II : θ S T 180◦ − θ 180◦ + θ A C θ 360◦ − θ = 17,46◦ + 360◦ · n, n ∈ Z = 180◦ − 17,46◦ + 360◦ · n, n ∈ Z = 162,54◦ + 360◦ · n, n ∈ Z This is called the general solution. Step 4 : Find the specific solutions We can then find all the values of θ by substituting n = . . . , − 1,0, 1, 2, . . .etc. For example, If n = 0, θ = 17,46◦ ; 162,54◦ If n = 1, θ = 377,46◦; 522,54◦ If n = −1, θ = −342,54◦; −197,46◦ We can find as many as we like or find specific solutions in a given interval by choosing more values for n. 32.4.4 General Solution Using Periodicity Up until now we have only solved trigonometric equations where the argument (the bit after the function, e.g. the θ in cos θ or the (2x − 7) in tan(2x − 7)), has been θ. If there is anything more complicated than this we need to be a little more careful. Let us try to solve tan(2x − 10◦ ) = 2,5 in the range −360◦ ≤ x ≤ 360◦ . We want solutions for positive tangent so using our CAST diagram we know to look in the 1st and 3rd quadrants. Our calculator tells us that arctan(2,5) = 68,2◦. This is our reference angle. So to find the general solution we proceed as follows: tan(2x − 10◦ ) = I : 2x − 10◦ = 2x = x = 2,5 [68,2◦ ] 68,2◦ + 180◦ · n 78,2◦ + 180◦ · n 39,1◦ + 90◦ · n, n ∈ Z This is the general solution. Notice that we added the 10◦ and divided by 2 only at the end. Notice that we added 180◦ · n because the tangent has a period of 180◦ . This is also divided by 2 in the last step to keep the equation balanced. We chose quadrants I and III because tan 405 32.4 CHAPTER 32. TRIGONOMETRY - GRADE 11 was positive and we used the formulae θ in quadrant I and (180◦ + θ) in quadrant III. To find solutions where −360◦ < x < 360◦ we substitue integers for n: • n = 0; x = 39,1◦ ; 129,1◦ • n = 1; x = 129,1◦; 219,1◦ • n = 2; x = 219,1◦; 309,1◦ • n = 3; x = 309,1◦; 399,1◦ (too big!) • n = −1; x = −50,9◦; 39,1◦ • n = −2; x = −140,1◦; −50,9◦ • n = −3; x = −230,9◦; −140,9◦ • n = −4; x = −320,9◦; −230,9◦ Solution: x = −320,9◦; −230◦; −140,9◦; −50,9◦; 39,1◦ ; 129,1◦; 219,1◦ and 309,1◦ 32.4.5 Linear Trigonometric Equations Just like with regular equations without trigonometric functions, solving trigonometric equations can become a lot more complicated. You should solve these just like normal equations to isolate a single trigonometric ratio. Then you follow the strategy outlined in the previous section. Worked Example 143: Question: Write down the general solution isf 3 cos(θ − 15◦ ) − 1 = −2,583 Answer 3 cos(θ − 15◦ ) − 1 3 cos(θ − 15◦ ) cos(θ − 15◦ ) II : θ − 15◦ θ III : θ − 15◦ θ 32.4.6 = −2,583 = −1,583 = −0,5276... [58,2◦ ] = 180◦ − 58,2◦ + 360◦ · n, n ∈ Z = 136,8◦ + 360◦ · n, n ∈ Z = 180◦ + 58,2◦ + 360◦ · n, n ∈ Z = 253,2◦ + 360◦ · n, n ∈ Z Quadratic and Higher Order Trigonometric Equations The simplest quadratic trigonometric equation is of the form sin2 x − 2 = −1.5 This type of equation can be easily solved by rearranging to get a more familiar linear equation sin2 x ⇒ sin x = 0.5 √ = ± 0.5 406 CHAPTER 32. TRIGONOMETRY - GRADE 11 32.4 This gives two linear trigonometric equations. The solutions to either of these equations will satisfy the original quadratic. The next level of complexity comes when we need to solve a trinomial which contains trigonometric functions. It is much easier in this case to use temporary variables. Consider solving tan2 (2x + 1) + 3 tan (2x + 1) + 2 = 0 Here we notice that tan(2x + 1) occurs twice in the equation, hence we let y = tan(2x + 1) and rewrite: y 2 + 3y + 2 = 0 That should look rather more familiar. We can immediately write down the factorised form and the solutions: (y + 1)(y + 2) = 0 ⇒ y = −1 OR y = −2 Next we just substitute back for the temporary variable: tan (2x + 1) = −1 or tan (2x + 1) = −2 And then we are left with two linear trigonometric equations. Be careful: sometimes one of the two solutions will be outside the range of the trigonometric function. In that case you need to discard that solution. For example consider the same equation with cosines instead of tangents cos2 (2x + 1) + 3 cos (2x + 1) + 2 = 0 Using the same method we find that cos (2x + 1) = −1 or cos (2x + 1) = −2 The second solution cannot be valid as cosine must lie between −1 and 1. We must, therefore, reject the second equation. Only solutions to the first equation will be valid. 32.4.7 More Complex Trigonometric Equations Here are two examples on the level of the hardest trigonometric equations you are likely to encounter. They require using everything that you have learnt in this chapter. If you can solve these, you should be able to solve anything! Worked Example 144: Question: Solve 2 cos2 x − cos x − 1 = 0 for x ∈ [−180◦; 360◦ ] Answer Step 1 : Use a temporary variable We note that cos x occurs twice in the equation. So, let y = cos x. Then we have 2y 2 − y − 1 = 0 Note that with practice you may be able to leave out this step. Step 2 : Solve the quadratic equation Factorising yields (2y + 1)(y − 1) = 0 ∴ y = −0,5 or y=1 Step 1 : Substitute back and solve the two resulting equations We thus get cos x = −0,5 or cos x = 1 Both equations are valid (i.e. lie in the range of cosine). General solution: 407 32.4 CHAPTER 32. TRIGONOMETRY - GRADE 11 cos x II : x = −0,5 [60◦ ] = 180◦ − 60◦ + 360◦ · n, n ∈ Z = 120◦ + 360◦ · n, n ∈ Z cos x I; IV : x 1 [90◦ ] 0◦ + 360◦ · n, n ∈ Z = = = 180◦ + 60◦ + 360◦ · n, n ∈ Z = 360◦ · n, n ∈ Z = 240◦ + 360◦ · n, n ∈ Z Now we find the specific solutions in the interval [−180◦; 360◦ ]. Appropriate values of n yield x = −120◦; 0◦ ; 120◦; 240◦ ; 360◦ III : x Worked Example 145: Question: Solve for x in the interval [−360◦; 360◦ ]: 2 sin2 − sin x cos x = 0 Answer Step 2 : Factorise Factorising yields sin x(2 sin x − cos x) = 0 which gives two equations sin x = 0 2 sin x 2 sin x cos x 2 tan x tan x = cos x cos x = cos x = 1 = 1 2 Step 3 : Solve the two trigonometric equations General solution: sin x = 0 [0◦ ] tan x = ∴ x = 180◦ · n, n ∈ Z Specific solution in the interval [−360◦; 360◦ ]: I; III : x = 1 2 [26,57◦] 26,57◦ + 180◦ · n, n ∈ Z x = −360◦ ; −206,57◦; −180◦; −26,57◦; 0◦ ; 26,57◦; 180◦ ; 206,25◦; 360◦ Exercise: Solving Trigonometric Equations 1. A Find the general solution of each of the following equations. Give answers to one decimal place. B Find all solutions in the interval θ ∈ [−180◦; 360◦ ]. i. sin θ = −0,327 ii. cos θ = 0,231 iii. tan θ = −1,375 iv. sin θ = 2,439 408 CHAPTER 32. TRIGONOMETRY - GRADE 11 2. A Find the general solution of each of the following equations. Give answers to one decimal place. B Find all solutions in the interval θ ∈ [0◦ ; 360◦]. i. cos θ = 0√ ii. sin θ = 23 √ iii. 2 cos θ − 3 = 0 iv. tan θ = −1 v. 5 cos θ = −2 vi. 3 sin θ = −1,5 vii. 2 cos θ + 1,3 = 0 viii. 0,5 tan θ + 2,5 = 1,7 3. A Write down the general solution for x if tan x = −1,12. B Hence determine values of x ∈ [−180◦ ; 180◦]. 4. A Write down the general solution for θ if sin θ = −0,61. B Hence determine values of θ ∈ [0◦ ; 720◦]. 5. A Solve for A if sin(A + 20◦ ) = 0,53 B Write down the values of A∈ [0◦ ; 360◦ ] 6. A Solve for x if cos(x + 30◦ ) = 0,32 B Write down the values of x ∈ [−180◦; 360◦ ] 7. 32.5 32.5 A Solve for θ if sin2 (θ) + 0,5 sin θ = 0 B Write down the values of θ ∈ [0◦ ; 360◦ ] Sine and Cosine Identities There are a few identities relating to the trigonometric functions that make working with triangles easier. These are: 1. the sine rule 2. the cosine rule 3. the area rule and will be described and applied in this section. 32.5.1 The Sine Rule Definition: The Sine Rule The sine rule applies to any triangle: sin B̂ sin Ĉ sin Â = = a b c where a is the side opposite Â, b is the side opposite B̂ and c is the side opposite Ĉ. Consider △ABC. 409 32.5 CHAPTER 32. TRIGONOMETRY - GRADE 11 C b b A h b b a b c B The area of △ABC can be written as: area △ABC = 1 c · h. 2 However, h can be calculated in terms of Â or B̂ as: sin Â ∴ h h b = b · sin Â = and sin B̂ = h = ∴ h a a · sin B̂ Therefore the area of △ABC is: 1 1 1 c · h = c · b · sin Â = c · a · sin B̂ 2 2 2 Similarly, by drawing the perpendicular between point B and line AC we can show that: 1 1 c · b · sin Â = a · b · sin Ĉ 2 2 Therefore the area of △ABC is: 1 1 1 c · b · sin Â = c · a · sin B̂ = a · b · sin Ĉ 2 2 2 If we divide through by 12 a · b · c, we get: sin B̂ sin Ĉ sin Â = = a b c This is known as the sine rule and applies to any triangle. Worked Example 146: Lighthouses Question: There is a coastline with two lighthouses, one on either side of a beach. The two lighthouses are 0,67 km apart and one is exactly due east of the other. The lighthouses tell how close a boat is by taking bearings to the boat (remember – a bearing is an angle measured clockwise from north). These bearings are shown. Use the sine rule to calculate how far the boat is from each lighthouse. 410 CHAPTER 32. TRIGONOMETRY - GRADE 11 A 32.5 127◦ b bB 255◦ b C Answer We can see that the two lighthouses and the boat form a triangle. Since we know the distance between the lighthouses and we have two angles we can use trigonometry to find the remaining two sides of the triangle, the distance of the boat from the two lighthouses. A 0,67 km b bB 15 37◦ ◦ 128◦ b C We need to know the lengths of the two sides AC and BC. We can use the sine rule to find our missing lengths. BC sin Â BC AC sin B̂ AC = = AB sin Ĉ AB · sin Â sin Ĉ (0,67km) sin(37◦ ) = sin(128◦ ) = 0,51 km = = = = AB sin Ĉ AB · sin B̂ sin Ĉ (0,67km) sin(15◦ ) sin(128◦ ) 0,22 km Exercise: Sine Rule 1. Show that sin Â sin B̂ sin Ĉ = = a b c 411 32.5 CHAPTER 32. TRIGONOMETRY - GRADE 11 is equivalent to: a b = = c sin Â sin B̂ sin Ĉ Note: either of these two forms can be used. . 2. Find all the unknown sides and angles of the following triangles: A B C D △PQR in which Q̂ = 64◦ ; R̂ = 24◦ and r = 3. △KLM in which K̂ = 43◦ ; M̂ = 50◦ and m = 1 △ABC in which Â = 32,7◦; Ĉ = 70,5◦ and a = 52,3 △XYZ in which X̂ = 56◦ ; Ẑ = 40◦ and x = 50 3. In △ABC, Â = 116◦; Ĉ = 32◦ and AC = 23 m. Find the length of the side AB. 4. In △RST, R̂ = 19◦ ; Ŝ = 30◦ and RT = 120 km. Find the length of the side ST. 5. In △KMS, K̂ = 20◦ ; M̂ = 100◦ and s = 23 cm. Find the length of the side m. 32.5.2 The Cosine Rule Definition: The Cosine Rule The cosine rule applies to any triangle and states that: a2 = b2 c2 = = b2 + c2 − 2bc cos Â c2 + a2 − 2ca cos B̂ a2 + b2 − 2ab cos Ĉ where a is the side opposite Â, b is the side opposite B̂ and c is the side opposite Ĉ. The cosine rule relates the length of a side of a triangle to the angle opposite it and the lengths of the other two sides. Consider △ABC which we will use to show that: a2 = b2 + c2 − 2bc cos Â. C b b A a h b b d D b c-d B c In △DCB: a2 = (c − d)2 + h2 (32.6) b2 = d2 + h2 (32.7) from the theorem of Pythagoras. In △ACD: from the theorem of Pythagoras. 412 CHAPTER 32. TRIGONOMETRY - GRADE 11 32.5 We can eliminate h2 from (32.6) and (32.7) to get: b2 − d2 = a2 = = = a2 − (c − d)2 b2 + (c2 − 2cd + d2 ) − d2 b2 + c2 − 2cd + d2 − d2 b2 + c2 − 2cd (32.8) In order to eliminate d we look at △ACD, where we have: cos Â = d . b So, d = b cos Â. Substituting this into (32.8), we get: a2 = b2 + c2 − 2bc cos Â (32.9) The other cases can be proved in an identical manner. Worked Example 147: Question: Find Â: A 8 5 7 B C Answer Applying the cosine rule: a2 ∴ = cos Â = = ∴ = Â = b2 + c2 − 2bd cos Â b 2 + c2 − a 2 2bc 82 + 52 − 72 2·8·5 0,5 arccos 0,5 = 60◦ Exercise: The Cosine Rule 1. Solve the following triangles i.e. find all unknown sides and angles A B C D △ABC in which Â = 70◦ ; b = 4 and c = 9 △XYZ in which Ŷ = 112◦ ; x = 2 and y = 3 △RST in which RS= 2; ST= 3 and RT= 5 △KLM in which KL= 5; LM= 10 and KM= 7 413 32.5 CHAPTER 32. TRIGONOMETRY - GRADE 11 E △JHK in which Ĥ = 130◦; JH= 13 and HK= 8 F △DEF in which d = 4; e = 5 and f = 7 2. Find the length of the third side of the △XYZ where: A X̂ = 71,4◦ ; y = 3,42 km and z = 4,03 km B ; x = 103,2 cm; Ŷ = 20,8◦ and z = 44,59 cm 3. Determine the largest angle in: A △JHK in which JH= 6; HK= 4 and JK= 3 B △PQR where p = 50; q = 70 and r = 60 32.5.3 The Area Rule Definition: The Area Rule The area rule applies to any triangle and states that the area of a triangle is given by half the product of any two sides with the sine of the angle between them. That means that in the △DEF , the area is given by: A= 1 1 1 DE · EF sin Ê = EF · F D sin F̂ = F D · DE sin D̂ 2 2 2 F b b D b E In order show that this is true for all triangles, consider △ABC. C b b A h b a b c b B The area of any triangle is half the product of the base and the perpendicular height. For △ABC, this is: 1 A = c · h. 2 However, h can be written in terms of Â as: h = b sin Â So, the area of △ABC is: A= 1 c · b sin Â. 2 Using an identical method, the area rule can be shown for the other two angles. 414 CHAPTER 32. TRIGONOMETRY - GRADE 11 32.5 Worked Example 148: The Area Rule Question: Find the area of △ABC: A 50◦ C B Answer △ABC is isosceles, therefore AB=AC= 7 and Ĉ = B̂ = 50◦ . Hence Â = 180◦ − 50◦ − 50◦ = 80◦ . Now we can use the area rule to find the area: A = = = 1 cb sin Â 2 1 · 7 · 7 · sin 80◦ 2 24,13 Exercise: The Area Rule Draw sketches of the figures you use in this exercise. 1. Find the area of △PQR in which: A P̂ = 40◦ ; q = 9 and r = 25 B Q̂ = 30◦ ; r = 10 and p = 7 C R̂ = 110◦ ; p = 8 and q = 9 2. Find the area of: A △XYZ with XY= 6 cm; XZ= 7 cm and Ẑ = 28◦ B △PQR with PR= 52 cm; PQ= 29 cm and P̂ = 58,9◦ C △EFG with FG= 2,5 cm; EG= 7,9 cm and Ĝ = 125◦ 3. Determine the area of a parallelogram in which two adjacent sides are 10 cm and 13 cm and the angle between them is 55◦ . 4. If the area of △ABC is 5000 m2 with a = 150 m and b = 70 m, what are the two possible sizes of Ĉ? 415 32.6 CHAPTER 32. TRIGONOMETRY - GRADE 11 Summary of the Trigonometric Rules and Identities Pythagorean Identity cos2 θ + sin2 θ = 1 tan θ = sin θ cos θ Odd/Even Identities Periodicity Identities Cofunction Identities sin(−θ) = − sin θ cos(−θ) = cos θ sin(θ ± 360◦ ) = sin θ cos(θ ± 360◦ ) = cos θ sin(90◦ − θ) = cos θ cos(90◦ − θ) = sin θ Sine Rule Area Rule Cosine Rule Area = 21 bc cos A Area = 21 ac cos B Area = 21 ab cos C a2 = b2 + c2 − 2bc cos A b2 = a2 + c2 − 2ac cos B c2 = a2 + b2 − 2ab cos C sin A a 32.6 Ratio Identity = sin B b = sin C c Exercises P 1. Q is a ship at a point 10 km due South of another ship P. R is a lighthouse on the coast such that P̂ = Q̂ = 50◦ . Determine: 50 ◦ R A the distance QR 10 km 50◦ B the shortest distance from the lighthouse to the line joining the two ships (PQ). Q W 30◦ 2. WXYZ is a trapezium (WXkXZ) with WX= 3 m; YZ= 1,5 m;Ẑ = 120◦ and Ŵ = 30◦ Z 120◦ A Determine the distances XZ and XY B Find the angle Ĉ 3m 1,5 m Y X 3. On a flight from Johannesburg to Cape Town, the pilot discovers that he has been flying 3◦ off course. At this point the plane is 500 km from Johannesburg. The direct distance between Cape Town and Johannesburg airports is 1 552 km. Determine, to the nearest km: A The distance the plane has to travel to get to Cape Town and hence the extra distance that the plane has had to travel due to the pilot’s error. B The correction, to one hundredth of a degree, to the plane’s heading (or direction). A 4. ABCD is a trapezium (i.e. ABkCD). AB= x; BÂD = a; BĈD = b and BD̂C = c. Find an expression for the length of CD in terms of x, a, b and c. x c D 416 B a b C CHAPTER 32. TRIGONOMETRY - GRADE 11 32.6 5. A surveyor is trying to determine the distance between points X and Z. However the distance cannot be determined directly as a ridge lies between the two points. From a point Y which is equidistant from X and Z, he measures the angle XŶZ x A IfpXY= x and XŶZ = θ, show that XZ= X x 2(1 − cos θ) Z B Calculate XZ (to the nearest kilometre) if x = 240 km and θ = 132◦ 6. Find the area of WXYZ (to two decimal places): X 3,5 W 120◦ Z 3 Y 4 7. Find the area of the shaded triangle in terms of x, α, β, θ and φ: x A φ β θ E B α C D 417 Y θ 32.6 CHAPTER 32. TRIGONOMETRY - GRADE 11 418 Chapter 33 Statistics - Grade 11 33.1 Introduction This chapter gives you an opportunity to build on what you have learned in previous Grades about data handling and probility. The work done will be mostly of a practical nature. Through problem solving and activities, you will end up mastering further methods of collecting, organising, displaying and analysing data. You will also learn how to interpret data, and not always to accept the data at face value, because data are sometimes unscrupulously misused and abused in order to try to prove or support a viewpoint. Measures of central tendency (mean, median and mode) and dispersion (range, percentiles, quartiles, inter-quartile, semi-inter-quartile range, variance and standard deviation) will be investigated. Of course, the activities involving probability will be familiar to most of you - for example, you have played dice games or card games even before you came to school. Your basic understanding of probability and chance gained so far will be deepened to enable you to come to a better understanding of how chance and uncertainty can be measured and understood. 33.2 Standard Deviation and Variance The measures of central tendency (mean, median and mode) and measures of dispersion (quartiles, percentiles, ranges) provide information on the data values at the centre of the data set and provide information on the spread of the data. The information on the spread of the data is however based on data values at specific points in the data set, e.g. the end points for range and data points that divide the data set into 4 equal groups for the quartiles. The behaviour of the entire data set is therefore not examined. A method of determining the spread of data is by calculating a measure of the possible distances between the data and the mean. The two important measures that are used are called the variance and the standard deviation of the data set. 33.2.1 Variance The variance of a data set is the average squared distance between the mean of the data set and each data value. An example of what this means is shown in Figure 33.1. The graph represents the results of 100 tosses of a fair coin, which resulted in 45 heads and 55 tails. The mean of the results is 50. The squared distance between the heads value and the mean is (45 − 50)2 = 25 and the squared distance between the tails value and the mean is (55 − 50)2 = 25. The average of these two squared distances gives the variance, which is 12 (25 + 25) = 25. Population Variance Let the population consist of n elements {x1 ,x2 , . . . ,xn }. with mean x̄ (read as ”x bar”). The variance of the population, denoted by σ 2 , is the average of the square of the distance of each data value from the mean value. 419 CHAPTER 33. STATISTICS - GRADE 11 Frequency (%) 33.2 60 55 50 45 40 35 30 25 20 15 10 5 0 Tails-Mean Heads-Mean Heads Tails Face of Coin Figure 33.1: The graph shows the results of 100 tosses of a fair coin, with 45 heads and 55 tails. The mean value of the tosses is shown as a vertical dotted line. The difference between the mean value and each data value is shown. P ( (x − x̄))2 σ = . n 2 (33.1) Since the population variance is squared, it is not directly comparable with the mean and the data themselves. Sample Variance Let the sample consist of the n elements {x1 ,x2 , . . . ,xn }, taken from the population, with mean x̄. The variance of the sample, denoted by s2 , is the average of the squared deviations from the sample mean: s2 = P (x − x̄)2 . n−1 (33.2) Since the sample variance is squared, it is also not directly comparable with the mean and the data themselves. A common question at this point is ”Why is the numerator squared?” One answer is: to get rid of the negative signs. Numbers are going to fall above and below the mean and, since the variance is looking for distance, it would be counterproductive if those distances factored each other out. Difference between Population Variance and Sample Variance As seen a distinction is made between the variance, σ 2 , of a whole population and the variance, s2 of a sample extracted from the population. When dealing with the complete population the (population) variance is a constant, a parameter which helps to describe the population. When dealing with a sample from the population the (sample) variance varies from sample to sample. Its value is only of interest as an estimate for the population variance. Properties of Variance If the variance is defined, we can conclude that it is never negative because the squares are positive or zero. The unit of variance is the square of the unit of observation. For example, the 420 CHAPTER 33. STATISTICS - GRADE 11 33.2 variance of a set of heights measured in centimeters will be given in square centimeters. This fact is inconvenient and has motivated many statisticians to instead use the square root of the variance, known as the standard deviation, as a summary of dispersion. 33.2.2 Standard Deviation Since the variance is a squared quantity, it cannot be directly compared to the data values or the mean value of a data set. It is therefore more useful to have a quantity which is the square root of the variance. This quantity is known as the standard deviation. In statistics, the standard deviation is the most common measure of statistical dispersion. Standard deviation measures how spread out the values in a data set are. More precisely, it is a measure of the average distance between the values of the data in the set. If the data values are all similar, then the standard deviation will be low (closer to zero). If the data values are highly variable, then the standard variation is high (further from zero). The standard deviation is always a positive number and is always measured in the same units as the original data. For example, if the data are distance measurements in metres, the standard deviation will also be measured in metres. Population Standard Deviation Let the population consist of n elements {x1 ,x2 , . . . ,xn }. with mean x̄. The standard deviation of the population, denoted by σ, is the square root of the average of the square of the distance of each data value from the mean value. σ= rP (x − x̄)2 n (33.3) Sample Standard Deviation Let the sample consist of n elements {x1 ,x2 , . . . ,xn }, taken from the population, with mean x̄. The standard deviation of the sample, denoted by s, is the square root of the average of the squared deviations from the sample mean: rP (x − x̄)2 s= n−1 (33.4) It is often useful to set your data out in a table so that you can apply the formulae easily. For example to calculate the standard deviation of 57; 53; 58; 65; 48; 50; 66; 51, you could set it out in the following way: mean = = = = sum of items number of items P x n 448 6 56 Note: To get the deviations, subtract each number from the mean. 421 33.2 CHAPTER 33. STATISTICS - GRADE 11 X̄ 57 53 58 65 48 50 66 51 P X = 448 Deviation squared (X − X̄)2 1 9 4 81 64 36 100 25 P (X − X̄)2 = 320 Deviation (X − X̄) 1 -3 2 9 -8 -6 10 -5 P x=0 Note: The sum of the deviations of scores about their mean is zero. This always happens; that is (X − X̄) = 0, for any set of data. Why is this? Find out. Calculate the variance (add the squared results together and divide this total by the number of items). P (X − X̄)2 Variance = n 320 = 8 = 40 = √ variance rP (X − X̄)2 n r 320 8 √ 40 = 6.32 Standard deviation = = = Difference between Population Variance and Sample Variance As with variance, there is a distinction between the standard deviation,σ, of a whole population and the standard deviation, s, of sample extracted from the population. When dealing with the complete population the (population) standard deviation is a constant, a parameter which helps to describe the population. When dealing with a sample from the population the (sample) standard deviation varies from sample to sample. In other words, the standard deviation can be calculated as follows: 1. Calculate the mean value x̄. 2. For each data value xi calculate the difference xi − x̄ between xi and the mean value x̄. 3. Calculate the squares of these differences. 4. Find the average of the squared differences. This quantity is the variance, σ 2 . 5. Take the square root of the variance to obtain the standard deviation, σ. Worked Example 149: Variance and Standard Deviation 422 CHAPTER 33. STATISTICS - GRADE 11 33.2 Question: What is the variance and standard deviation of the population of possibilities associated with rolling a fair die? Answer Step 1 : Determine how many outcomes make up the population When rolling a fair die, the population consists of 6 possible outcomes. The data set is therefore x = {1,2,3,4,5,6}. and n=6. Step 2 : Calculate the population mean The population mean is calculated by: x̄ = = 1 (1 + 2 + 3 + 4 + 5 + 6) 6 3,5 Step 3 : Calculate the population variance The population variance is calculated by: P (x − x̄)2 σ2 = n 1 (6,25 + 2,25 + 0,25 + 0,25 + 2,25 + 6,25) = 6 = 2,917 Step 4 : Alternately the population variance is calculated by: X̄ 1 2 3 4 5 6 P X = 21 (X − X̄) -2.5 -1.5 -0.5 0.5 1.5 2.5 P x=0 (X − X̄)2 6.25 2.25 0.25 0.25 2.25 6.25 P (X − X̄)2 = 17.5 Step 5 : Calculate the standard deviation The (population) standard deviation is calculated by: σ = = p 2,917 1,708. Notice how this standard deviation is somewhere in between the possible deviations. 33.2.3 Interpretation and Application A large standard deviation indicates that the data values are far from the mean and a small standard deviation indicates that they are clustered closely around the mean. For example, each of the three samples (0, 0, 14, 14), (0, 6, 8, 14), and (6, 6, 8, 8) has a mean of 7. Their standard deviations are 7, 5 and 1, respectively. The third set has a much smaller standard deviation than the other two because its values are all close to 7. The value of the standard deviation can be considered ’large’ or ’small’ only in relation to the sample that is being measured. In this case, a standard deviation of 7 may be considered large. Given a different sample, a standard deviation of 7 might be considered small. Standard deviation may be thought of as a measure of uncertainty. In physical science for example, the reported standard deviation of a group of repeated measurements should give the precision of those measurements. When deciding whether measurements agree with a theoretical prediction, the standard deviation of those measurements is of crucial importance: if the mean of the measurements is too far away from the prediction (with the distance measured in standard 423 33.3 CHAPTER 33. STATISTICS - GRADE 11 deviations), then we consider the measurements as contradicting the prediction. This makes sense since they fall outside the range of values that could reasonably be expected to occur if the prediction were correct and the standard deviation appropriately quantified. See prediction interval. 33.2.4 Relationship between Standard Deviation and the Mean The mean and the standard deviation of a set of data are usually reported together. In a certain sense, the standard deviation is a ”natural” measure of statistical dispersion if the center of the data is measured about the mean. This is because the standard deviation from the mean is smaller than from any other point. Exercise: Means and standard deviations 1. Bridget surveyed the price of petrol at petrol stations in Cape Town and Durban. The raw data, in rands per litre, are given below: Cape Town Durban 3.96 3.97 3.76 3.81 4.00 3.52 3.91 4.08 3.69 3.88 3.72 3.68 A Find the mean price in each city and then state which city has the lowest mean. B Assuming that the data is a population find the standard deviation of each city’s prices. C Assuming the data is a sample find the standard deviation of each city’s prices. D Giving reasons which city has the more consistently priced petrol? 2. The following data represents the pocket money of a sample of teenagers. 150; 300; 250; 270; 130; 80; 700; 500; 200; 220; 110; 320; 420; 140. What is the standard deviation? 3. Consider a set of data that gives the weights of 50 cats at a cat show. A When is the data seen as a population? B When is the data seen as a sample? 4. Consider a set of data that gives the results of 20 pupils in a class. A When is the data seen as a population? B When is the data seen as a sample? 33.3 Graphical Representation of Measures of Central Tendency and Dispersion The measures of central tendency (mean, median, mode) and the measures of dispersion (range, semi-inter-quartile range, quartiles, percentiles, inter-quartile range) are numerical methods of summarising data. This section presents methods of representing the summarised data using graphs. 33.3.1 Five Number Summary One method of summarising a data set is to present a five number summary. The five numbers are: minimum, first quartile, median, third quartile and maximum. 424 CHAPTER 33. STATISTICS - GRADE 11 33.3.2 33.3 Box and Whisker Diagrams A box and whisker diagram is a method of depicting the five number summary, graphically. The main features of the box and whisker diagram are shown in Figure 33.2. The box can lie horizontally (as shown) or vertically. For a horizonatal diagram, the left edge of the box is placed at the first quartile and the right edge of the box is placed at the third quartile. The height of the box is arbitrary, as there is no y-axis. Inside the box there is some representation of central tendency, with the median shown with a vertical line dividing the box into two. Additionally, a star or asterix is placed at the mean value, centered in the box in the vertical direction. The whiskers which extend to the sides reach the minimum and maximum values. median first quartile third quartile minimum data value -4 maximum data value -2 0 Data Values 2 4 Figure 33.2: Main features of a box and whisker diagram Worked Example 150: Box and Whisker Diagram Question: Draw a box and whisker diagram for the data set x = {1,25; 1,5; 2,5; 2,5; 3,1; 3,2; 4,1; 4,25; 4,75; 4,8; 4,95; 5,1}. Answer Step 1 : Determine the five number summary Minimum = 1,25 Maximum = 4,95 Position of first quartile = between 3 and 4 Position of second quartile = between 6 and 7 Position of third quartile = between 9 and 10 Data value between 3 and 4 = 12 (2,5 + 2,5) = 2,5 Data value between 6 and 7 = 12 (3,2 + 4,1) = 3,65 Data value between 9 and 10 = 12 (4,75 + 4,8) = 4,775 The five number summary is therefore: 1,25; 2,5; 3,65; 4,775; 4,95. Step 2 : Draw a box and whisker diagram and mark the positions of the minimum, maximum and quartiles. first quartile third quartile median minimum 1 2 maximum 3 4 Data Values 425 5 33.3 CHAPTER 33. STATISTICS - GRADE 11 Exercise: Box and whisker plots 1. Lisa works as a telesales person. She keeps a record of the number of sales she makes each month. The data below show how much she sells each month. 49; 12; 22; 35; 2; 45; 60; 48; 19; 1; 43; 12 Give a five number summary and a box and whisker plot of her sales. 2. Jason is working in a computer store. He sells the following number of computers each month: 27; 39; 3; 15; 43; 27; 19; 54; 65; 23; 45; 16 Give a five number summary and a box and whisker plot of his sales, 3. The number of rugby matches attended by 36 season ticket holders is as follows: 15; 11; 7; 34; 24; 22; 31; 12; 9 12; 9; 1; 3; 15; 5; 8; 11; 2 25; 2; 6; 18; 16; 17; 20; 13; 17 14; 13; 11; 5; 3; 2; 23; 26; 40 A B C D E F G Sum the data. Using an appropriate graphical method (give reasons) represent the data. Find the median, mode and mean. Calculate the five number summary and make a box and whisker plot. What is the variance and standard deviation? Comment on the data’s spread. Where are 95% of the results expected to lie? 4. Rose has worked in a florists shop for nine months. She sold the following number of wedding bouquets: 16; 14; 8; 12; 6; 5; 3; 5; 7 A What is the five-number summary of the data? B Since there is an odd number of data points what do you observe when calculating the five-numbers? 33.3.3 Cumulative Histograms Cumulative histograms, also known as ogives, are a plot of cumulative frequency and are used to determine how many data values lie above or below a particular value in a data set. The cumulative frequency is calculated from a frequency table, by adding each frequency to the total of the frequencies of all data values before it in the data set. The last value for the cumulative frequency will always be equal to the total number of data values, since all frequencies will already have been added to the previous total. The cumulative frequency is plotted at the upper limit of the interval. For example, the cumulative frequencies for Data Set 2 are shown in Table 33.2 and is drawn in Figure 33.3. Notice the frequencies plotted at the upper limit of the intervals, so the points (30;1) (62;2) (97;3), etc have been plotted. This is different from the frequency polygon where we plot frequencies at the midpoints of the intervals. Exercise: Intervals 426 CHAPTER 33. STATISTICS - GRADE 11 Intervals Frequency Cumulative Frequency 33.3 0 < n≤1 30 30 1 < n≤2 32 30 + 32 2 < n≤3 35 30 + 32 + 35 3 < n≤4 34 30 + 32 + 35 + 34 4 < n≤5 37 30 + 32 + 35 + 34 + 37 30 62 97 131 168 5 < n≤6 32 30 + 32 + 35 + 34 + 37 + 32 200 Table 33.1: Cumulative Frequencies for Data Set 2. b b 160 b 120 b f 80 b 40 b 0 0 1 2 3 4 5 Intervals Figure 33.3: Example of a cumulative histogram for Data Set 2. 1. Use the following data of peoples ages to answer the questions. 2; 5; 1; 76; 34; 23; 65; 22; 63; 45; 53; 38 4; 28; 5; 73; 80; 17; 15; 5; 34; 37; 45; 56 A B C D Using an interval width of 8 construct a cumulative frequency distribution How many are below 30? How many are below 60? Giving an explanation state below what value the bottom 50% of the ages fall E Below what value do the bottom 40% fall? F Construct a frequency polygon and an ogive. G Compare these two plots 2. The weights of bags of sand in grams is given below (rounded to the nearest tenth): 50.1; 40.4; 48.5; 29.4; 50.2; 55.3; 58.1; 35.3; 54.2; 43.5 60.1; 43.9; 45.3; 49.2; 36.6; 31.5; 63.1; 49.3; 43.4; 54.1 A B C D E F G Decide on an interval width and state what you observe about your choice. Give your lowest interval. Give your highest interval. Construct a cumultative frequency graph and a frequency polygon. Compare the cumulative frequency graph and frequency polygon. Below what value do 53% of the cases fall? Below what value fo 60% of the cases fall? 427 33.4 CHAPTER 33. STATISTICS - GRADE 11 33.4 Distribution of Data 33.4.1 Symmetric and Skewed Data The shape of a data set is important to know. Definition: Shape of a data set This describes how the data is distributed relative to the mean and median. • Symmetrical data sets are balanced on either side of the median. It does not have to be exactly equal to be symmetric • Skewed data is spread out on one side more than on the other. It can be skewed right or skewed left. skewed right skewed left 33.4.2 Relationship of the Mean, Median, and Mode The relationship of the mean, median, and mode to each other can provide some information about the relative shape of the data distribution. If the mean, median, and mode are approximately equal to each other, the distribution can be assumed to be approximately symmetrical. With both the mean and median known the following can be concluded: • (mean - median) ≈ 0 then the data is symmetrical • (mean - median) > 0 then the data is positively skewed (skewed to the right). This means that the median is close to the start of the data set. • (mean - median) < 0 then the data is negatively skewed (skewed to the left). This means that the median is close to the end of the data set. Exercise: Distribution of Data 1. Three sets of 12 pupils each had test score recorded. The test was out of 50. Use the given data to answer the following questions. A Make a stem and leaf plot for each set. B For each of the sets calculate the mean and the five number summary. C For each of the classes find the difference between the mean and the median and then use that to make box and whisker plots on the same set of axes. 428 CHAPTER 33. STATISTICS - GRADE 11 Set 1 25 47 15 17 16 26 24 27 22 24 12 31 33.5 Set 2 32 34 35 32 25 16 38 47 43 29 18 25 Set 3 43 47 16 43 38 44 42 50 50 44 43 42 Table 33.2: Cumulative Frequencies for Data Set 2. D State which of the three are skewed (either right or left). E Is set A skewed or symmetrical? F Is set C symmetrical? Why or why not? 2. Two data sets have the same range and interquartile range, but one is skewed right and the other is skewed left. Sketch the box and whisker plots and then invent data (6 points in each set) that meets the requirements. 33.5 Scatter Plots A scatter-plot is a graph that shows the relationship between two variables. We say this is bivariate data and we plot the data from two different sets using ordered pairs. For example, we could have mass on the horizontal axis (first variable) and height on the second axis (second variable), or we could have current on the horizontal axis and voltage on the vertical axis. Ohm’s Law is an important relationship in physics. Ohm’s law describes the relationship between current and voltage in a conductor, like a piece of wire. When we measure the voltage (dependent variable) that results from a certain current (independent variable) in a wire, we get the data points as shown in Table 33.3. Table 33.3: Values of current and voltage measured in a wire. Current Voltage Current Voltage 0 0.4 2.4 1.4 0.2 0.3 2.6 1.6 0.4 0.6 2.8 1.9 0.6 0.6 3 1.9 0.8 0.4 3.2 2 1 1 3.4 1.9 1.2 0.9 3.6 2.1 1.4 0.7 3.8 2.1 1.6 1 4 2.4 1.8 1.1 4.2 2.4 2 1.3 4.4 2.5 2.2 1.1 4.6 2.5 When we plot this data as points, we get the scatter plot shown in Figure 33.4. If we are to come up with a function that best describes the data, we would have to say that a straight line best describes this data. 429 33.5 CHAPTER 33. STATISTICS - GRADE 11 Voltage htb 2 1 bb bb bb b b 1 bb b b b b bb bbbb 2 3 Current bbbbb 4 Figure 33.4: Example of a scatter plot Extension: Ohm’s Law Ohm’s Law describes the relationship between current and voltage in a conductor. The gradient of the graph of voltage vs. current is known as the resistance of the conductor. Activity :: Research Project : Scatter Plot The function that best describes a set of data can take any form. We will restrict ourselves to the forms already studied, that is, linear, quadratic or exponential. Plot the following sets of data as scatter plots and deduce the type of function that best describes the data. The type of function can either be quadratic or exponential. 1. x -5 -4.5 -4 -3.5 -3 y 9.8 4.4 7.6 7.9 7.5 2. x -5 -4.5 -4 -3.5 -3 y 75 63.5 53 43.5 35 Height (cm) 3. Weight (kg) x 0 0.5 1 1.5 2 x 0 0.5 1 1.5 2 147 168 52 63 y 14.2 22.5 21.5 27.5 41.9 x -2.5 -2 -1.5 -1 -0.5 y 11.9 6.9 8.2 7.8 14.4 x 2.5 3 3.5 4 4.5 y 49.3 68.9 88.4 117.2 151.4 y 5 3.5 3 3.5 5 x -2.5 -2 -1.5 -1 -0.5 y 27.5 21 15.5 11 7.5 x 2.5 3 3.5 4 4.5 y 7.5 11 15.5 21 27.5 157 178 57 70 160 180 59 72 150 170 53 64 152 173 54 66 155 175 56 68 163 183 60 74 165 61 Definition: outlier A point on a scatter plot which is widely separated from the other points or a result differing greatly from others in the same sample is called an outlier. 430 CHAPTER 33. STATISTICS - GRADE 11 33.5 Exercise: Scatter Plots 1. A class’s results for a test were recorded along with the amount of time spent studying for it. The results are given below. Score (percent) 67 55 70 90 45 75 50 60 84 30 66 96 Time spent studying (minutes) 100 85 150 180 70 160 80 90 110 60 96 200 A Draw a diagram labelling horizontal and vertical axes. B State with reasons, the cause or independent variable and the effect or dependent variable. C Plot the data pairs D What do you observe about the plot? E Is there any pattern emerging? 2. The rankings of eight tennis players is given along with the time they spend practising. Practice time (min) 154 390 130 70 240 280 175 103 Ranking 5 1 6 8 3 2 4 7 A Construct a scatter plot and explain how you chose the dependent (cause) and independent (effect) variables. B What pattern or trend do you observe? 3. Eight childrens sweet consumption and sleep habits were recorded. The data is given in the following table. Number of sweets (per week) 15 12 5 3 18 23 11 4 A B C D Average sleeping time (per day) 4 4.5 8 8.5 3 2 5 8 What is the dependent (cause) variable? What is the independent (effect) variable? Construct a scatter plot of the data. What trend do you observe? 431 33.6 33.6 CHAPTER 33. STATISTICS - GRADE 11 Misuse of Statistics Statistics can be manipulated in many ways that can be misleading. Graphs need to be carefully analysed and questions must always be asked about ’the story behind the figures.’ Common manipulations are: 1. Changing the scale to change the appearence of a graph 2. Omissions and biased selection of data 3. Focus on particular research questions 4. Selection of groups Activity :: Investigation : Misuse of statistics 1. Examine the following graphs and comment on the effects of changing scale. 16 14 b 12 10 earnings 8 6 b 4 2 b 0 2002 2003 432 2004 CHAPTER 33. STATISTICS - GRADE 11 33.6 80 70 60 50 earnings 40 30 20 10 0 2002 2003 2004 year 2. Examine the following three plots and comment on omission, selection and bias. Hint: What is wrong with the data and what is missing from the bar and pie charts? Activity Sleep Sports School Visit friend Watch TV Studying 433 Hours 8 2 7 1 2 3 33.6 CHAPTER 33. STATISTICS - GRADE 11 10 9 8 7 6 5 4 3 2 1 watch TV visit friend sports studying sleep school 0 sleep school stu d TV ch wat visit friend sports yin g Exercise: Misuse of Statistics The bar graph below shows the results of a study that looked at the cost of food compared to the income of a household in 1994. 434 CHAPTER 33. STATISTICS - GRADE 11 33.7 Food bill (in thousands of rands) 12 10 8 6 4 2 ¿50 40-50 30-40 20-30 15-20 10-15 ¡5 5-10 0 Income in 1994(in thousands of rands) Income (thousands of rands) <5 5-10 10-15 15-20 20-30 30-40 40-50 >50 Food bill (thousands of rands) 2 2 4 4 8 6 10 12 1. What is the dependent variable? Why? 2. What conclusion can you make about this variable? Why? Does this make sense? 3. What would happen if the graph was changed from food bill in thousands of rands to percentage of income? 4. Construct this bar graph using a table. What conclusions can be drawn? 5. Why do the two graphs differ despite showing the same information? 6. What else is observed? Does this affect the fairness of the results? 33.7 End of Chapter Exercises 1. Many accidents occur during the holidays between Durban and Johannesburg. A study was done to see if speeding was a factor in the high accident rate. Use the results given to answer the following questions. Speed (km/h) 60 < x ≤ 70 70 < x ≤ 80 80 < x ≤ 90 90 < x ≤ 100 100 < x ≤ 110 110 < x ≤ 120 120 < x ≤ 130 130 < x ≤ 140 140 < x ≤ 150 150 < x ≤ 160 435 Frequency 3 2 6 40 50 30 15 12 3 2 33.7 CHAPTER 33. STATISTICS - GRADE 11 A Draw a graph to illustrate this information. B Use your graph to find the median speed and the interquartile range. C What percent of cars travel more than 120km/h on this road? D Do cars generally exceed the speed limit? 2. The following two diagrams (showing two schools contribution to charity) have been exaggerated. Explain how they are misleading and redraw them so that they are not misleading. R200.00 R100 R200.00 R100 R100 3. The monthly income of eight teachers are given as follows: R10 050; R14 300; R9 800; R15 000; R12 140; R13 800; R11 990; R12 900. A What is the mean income and the standard deviation? B How many of the salaries are within one standard deviation of the mean? C If each teacher gets a bonus of R500 added to their pay what is the new mean and standard deviation? D If each teacher gets a bonus of 10% on their salary what is the new mean and standard deviation? E Determine for both of the above, how many salaries are within one standard deviation of the mean. F Using the above information work out which bonus is more beneficial for the teachers. 436 Chapter 34 Independent and Dependent Events - Grade 11 34.1 Introduction In probability theory an event is either independent or dependent. This chapter describes the differences and how each type of event is worked with. 34.2 Definitions Two events are independent if knowing something about the value of one event does not give any information about the value of the second event. For example, the event of getting a ”1” when a die is rolled and the event of getting a ”1” the second time it is thrown are independent. Definition: Independent Events Two events A and B are independent if when one of them happens, it doesn’t affect the other one happening or not. The probability of two independent events occurring, P (A ∩ B), is given by: P (A ∩ B) = P (A) × P (B) (34.1) Worked Example 151: Independent Events Question: What is the probability of rolling a 1 and then rolling a 6 on a fair die? Answer Step 1 : Identify the two events and determine whether the events are independent or not Event A is rolling a 1 and event B is rolling a 6. Since the outcome of the first event does not affect the outcome of the second event, the events are independent. Step 2 : Determine the probability of the specific outcomes occurring, for each event The probability of rolling a 1 is 16 and the probability of rolling a 6 is 16 . Therefore, P (A) = 16 and P (B) = 16 . Step 3 : Use equation 34.1 to determine the probability of the two events occurring together. 437 34.2 CHAPTER 34. INDEPENDENT AND DEPENDENT EVENTS - GRADE 11 P (A ∩ B) = P (A) × P (B) 1 1 = × 6 6 1 = 36 The probability of rolling a 1 and then rolling a 6 on a fair die is 1 36 . Consequently, two events are dependent if the outcome of the first event affects the outcome of the second event. Worked Example 152: Dependent Events Question: A cloth bag has 4 coins, 1 R1 coin, 2 R2 coins and 1 R5 coin. What is the probability of first selecting a R1 coin followed by selecting a R2 coin? Answer Step 1 : Identify the two events and determine whether the events are independent or not Event A is selecting a R1 coin and event B is next selecting a R2. Since the outcome of the first event affects the outcome of the second event (because there are less coins to choose from after the first coin has been selected), the events are dependent. Step 2 : Determine the probability of the specific outcomes occurring, for each event The probability of first selecting a R1 coin is 14 and the probability of next selecting a R2 coin is 23 (because after the R1 coin has been selected, there are only three coins to choose from). Therefore, P (A) = 41 and P (B) = 23 . Step 3 : Use equation 34.1 to determine the probability of the two events occurring together. The same equation as for independent events are used, but the probabilities are calculated differently. P (A ∩ B) = P (A) × P (B) 1 2 = × 4 3 2 = 12 1 = 6 The probability of first selecting a R1 coin followed by selecting a R2 coin is 61 . 34.2.1 Identification of Independent and Dependent Events Use of a Contingency Table A two-way contingency table (studied in an earlier grade) can be used to determine whether events are independent or dependent. 438 CHAPTER 34. INDEPENDENT AND DEPENDENT EVENTS - GRADE 11 34.2 Definition: two-way contingency table A two-way contingency table is used to represent possible outcomes when two events are combined in a statistical analysis. For example we can draw and analyse a two-way contingency table to solve the following problem. Worked Example 153: Contingency Tables Question: A medical trial into the effectiveness of a new medication was carried out. 120 males and 90 females responded. Out of these 50 males and 40 females responded positively to the medication. 1. Was the medication’s succes independent of gender? Explain. 2. Give a table for the independent of gender results. Answer Step 1 : Draw a contingency table Positive result No Positive result Totals Male 50 70 120 Female 40 50 90 Totals 90 120 210 Step 2 : Work out probabilities 120 P(male).P(positive result)= 210 = 0.57 90 = 0.43 P(female).P(positive result)= 210 50 = 0.24 P(male and positive result)= 210 Step 3 : Draw conclusion P(male and positive result) is the observed probability and P(male).P(positive result) is the expected probability. These two are quite different. So there is no evidence that the medications success is independent of gender. Step 4 : Gender-independent results To get gender independence we need the positve results in the same ratio as the gender. The gender ratio is: 120:90, or 4:3, so the number in the male and positive column would have to be 47 of the total number of patients responding positively which gives 22. This leads to the following table: Positive result No Positive result Totals Male 22 98 120 Female 68 22 90 Totals 90 120 210 Use of a Venn Diagram We can also use Venn diagrams to check whether events are dependent or independent. Definition: Independent events Events are said to be independent if the result or outcome of the event does not affect the result or outcome of another event. So P(A/C)=P(A), where P(A/C) represents the probability of event A after event C has occured. 439 34.2 CHAPTER 34. INDEPENDENT AND DEPENDENT EVENTS - GRADE 11 Definition: Dependent events If the outcome of one event is affected by the outcome of another event such that P (A/C) 6= P (A) Also note that P (A/C) = P P(A∩C) (C) For example, we can draw a Venn diagram and a contingency table to illustrate and analyse the following example. Worked Example 154: Venn diagrams and events Question: A school decided that it’s uniform needed upgrading. The colours on offer were beige or blue or beige and blue. 40% of the school wanted beige, 55% wanted blue and 15% said a combination would be fine. Are the two events independent? Answer Step 1 : Draw a Venn diagram S Beige 0.25 Blue 0.4 0.15 0.2 Step 2 : Draw up a contingency table Blue Not Blue Totals Beige 0.15 0.25 0.40 Not Beige 0.4 0.2 0.6 Totals 0.55 0.35 1 Step 3 : Work out the probabilities P(Blue)=0.4, P(Beige)=0.55, P(Both)=0.15, P(Neither)=0.20 Probability of choosing beige after blue is: P (Beige ∩ Blue) P (Blue) 0.15 = 0.55 = 0.27 P (Beige/Blue) = Step 4 : Solve the problem Since P (Beige/Blue) 6= P (Beige) the events are statistically independent. 440 CHAPTER 34. INDEPENDENT AND DEPENDENT EVENTS - GRADE 11 34.3 Extension: Applications of Probability Theory Two major applications of probability theory in everyday life are in risk assessment and in trade on commodity markets. Governments typically apply probability methods in environmental regulation where it is called ”pathway analysis”, and are often measuring well-being using methods that are stochastic in nature, and choosing projects to undertake based on statistical analyses of their probable effect on the population as a whole. It is not correct to say that statistics are involved in the modelling itself, as typically the assessments of risk are one-time and thus require more fundamental probability models, e.g. ”the probability of another 9/11”. A law of small numbers tends to apply to all such choices and perception of the effect of such choices, which makes probability measures a political matter. A good example is the effect of the perceived probability of any widespread Middle East conflict on oil prices - which have ripple effects in the economy as a whole. An assessment by a commodity trade that a war is more likely vs. less likely sends prices up or down, and signals other traders of that opinion. Accordingly, the probabilities are not assessed independently nor necessarily very rationally. The theory of behavioral finance emerged to describe the effect of such groupthink on pricing, on policy, and on peace and conflict. It can reasonably be said that the discovery of rigorous methods to assess and combine probability assessments has had a profound effect on modern society. A good example is the application of game theory, itself based strictly on probability, to the Cold War and the mutual assured destruction doctrine. Accordingly, it may be of some importance to most citizens to understand how odds and probability assessments are made, and how they contribute to reputations and to decisions, especially in a democracy. Another significant application of probability theory in everyday life is reliability. Many consumer products, such as automobiles and consumer electronics, utilize reliability theory in the design of the product in order to reduce the probability of failure. The probability of failure is also closely associated with the product’s warranty. 34.3 End of Chapter Exercises 1. In each of the following contingency tables give the expected numbers for the events to be perfectly independent and decide if the events are independent or dependent. A D Brown eyes 50 70 120 Not Brown eyes 30 80 110 Totals 80 150 230 Busses left late Busses left on time Totals Point A 15 25 40 Point B 40 20 60 Liked living there Did not like living there Totals Durban 130 140 270 Bloemfontein 30 200 230 B C Black hair Red hair Totals Improvement in health No improvement in health Totals Multivitamin A 400 140 540 Totals 55 35 100 Totals 160 340 500 Multivitamin B 300 120 420 Totals 700 260 960 2. A company has a probability of 0.4 of meeting their quota on time and a probability of 0.25 of meeting their quota late. Also there is a 0.10 chance of not meeting their quota on time. Use a Venn diagram and a contingency table to show the information and decide if the events are independent. 441 34.3 CHAPTER 34. INDEPENDENT AND DEPENDENT EVENTS - GRADE 11 3. A study was undertaken to see how many people in Port Elizabeth owned either a Volkswagen or a Toyota. 3% owned both, 25% owned a Toyota and 60% owned a Volkswagen. Draw a contingency table to show all events and decide if car ownership is independent. 4. Jane invested in the stock market. The probability that she will not lose all her money is 1.32. What is the probability that she will lose all her money? Explain. 5. If D and F are mutually exclusive events, with P(D’)=0.3 and P(D or F)=0.94, find P(F). 6. A car sales person has pink, lime-green and purple models of car A and purple, orange and multicolour models of car B. One dark night a thief steals a car. A What is the experiment and sample space? B Draw a Venn diagram to show this. C What is the probability of stealing either model A or model B? D What is the probability of stealing both model A and model B? 7. Event X’s probability is 0.43, Event Y’s probability is 0.24. The probability of both occuring together is 0.10. What is the probability that X or Y will occur (this inculdes X and Y occuring simultaneously)? 8. P(H)=0.62, P(J)=0.39 and P(H and J)=0.31. Calculate: A P(H’) B P(H or J) C P(H’ or J’) D P(H’ or J) E P(H’ and J’) 9. The last ten letters of the alphabet were placed in a hat and people were asked to pick one of them. Event D is picking a vowel, Event E is picking a consonant and Evetn F is picking the last four letters. Calculate the following probabilities: A P(F’) B P(F or D) C P(neither E nor F) D P(D and E) E P(E and F) F P(E and D’) 10. At Dawnview High there are 400 Grade 12’s. 270 do Computer Science, 300 do English and 50 do Typing. All those doing Computer Science do English, 20 take Computer Science and Typing and 35 take English and Typing. Using a Venn diagram calculate the probability that a pupil drawn at random will take: A English, but not Typing or Computer Science B English but not Typing C English and Typing but not Computer Science D English or Typing 442 Appendix A GNU Free Documentation License Version 1.2, November 2002 c 2000,2001,2002 Free Software Foundation, Inc. Copyright 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA Everyone is permitted to copy and distribute verbatim copies of this license document, but changing it is not allowed. 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