Cambridge Specialist Mathematics Units 1 2 AC VCE Complete Book PDF

e ns ac l va a y l E W n b e o s ae la s re c h g L ip T i o u y id M D a a v K D

S p e M a t e c c i i t h a a he em l l i i ma s s t t t t 2 i i c c & s s 1 s n it U

C a m mb r r i i S e d d e n g n i io e e o r r M a t t h h e e m m Au s a s t t t i i tr a c c r a s s l l C u ia i n a n

rr r r i i c c u ul l u m u m / V VC E

Cambridge Senior Maths AC/VCE Specialist Mathematics 1&2

INCLUDES INTERACTIVE TEXTBOOK POWERED BY CAMBRIDGE HOTMATHS

Cambridge University Press ISBN 978-1-107-5 978-1-107-56765-8 6765-8 © Evans et al. 2016 Photocopying Photocopying is restricted under law and this material must not be transferred to another party.

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Contents

Introduction

ix

Acknowledgements

xi

An overvi overview ew of of the the Cambridg Cambridge e complet complete e teacher teacher and learn learning ing resou resource rce

1

Algebra I

2

Number systems and sets

xii 1

1A Indices . . . . . . . . . . . . . . . . . . . . . . . . . . . 1B Standard form . . . . . . . . . . . . . . . . . . . . . . . 1C Solving linear linear equations and simultaneous simultaneous linear linear equations 1D Solving problems problems with linear equations . . . . . . . . . . 1E Solving problems problems with simultaneous linear equations equations . . . 1F Substitution and transposition transposition of formulas . . . . . . . . . 1G Algebraic fractions . . . . . . . . . . . . . . . . . . . . . 1H Literal equations . . . . . . . . . . . . . . . . . . . . . . 1I Using a CAS calculator for algebra algebra . . . . . . . . . . . . .

2 5 . . . . . . 8 . . . 13 . . . 17 . . . 19 . . . 22 . . . 25 . . . 28 Review of Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . 33


2A 2B 2C 2D 2E 2F 2G 2H

Set notation . . . . . . . . . . Sets of numbers . . . . . . . . The modulus function . . . . Surds . . . . . . . . . . . . . Natural numbers . . . . . . . Linear Diophantin Diophantinee equations equations The Euclidean algorithm . . . Problems involving sets . . . . Review of Chapter 2 . . . . .

. . .

39

40 . . . . 43 . . . . 48 . . . . 51 . . . . 57 . . . . 62 . . . . 66 . . . . 70 . . . . 74

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


iv Contents

3

4

5

Variation 3A 3B 3C 3D 3E

82

Direct variation . . . Inverse variation . . Fitting data . . . . . Joint variation . . . . Part variation . . . . Review of Chapter 3

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

Sequences and series 4A 4B 4C 4D 4E 4F

110

Introduction to sequences . . . . . . . . . Arithmetic sequences . . . . . . . . . . . . Arithmetic series . . . . . . . . . . . . . . Geometric sequences . . . . . . . . . . . . Geometric series . . . . . . . . . . . . . . Zeno’ss paradox and Zeno’ and infinite geometric series series Review of Chapter 4 . . . . . . . . . . . .

. . . . . . . . . . . 111 . . . . . . . . . . . 118 . . . . . . . . . . . 122 . . . . . . . . . . . 127 . . . . . . . . . . . 133 . . . . . . . . . . . 137 . . . . . . . . . . . 140

Algebra II 5A 5B 5C 5D 5E

146

Polynomial identities . . . . . . . . . . . . . . Quadratic equations . . . . . . . . . . . . . . Applying quadratic quadratic equations equations to rate problems Partial fractions . . . . . . . . . . . . . . . . . Simultaneous equations . . . . . . . . . . . . Review of Chapter 5 . . . . . . . . . . . . . .

6

Revision of Chapters 1–5

7

Principles of counting

6A 6B 6C


83 87 91 98 101 104

. . . . . . . . . 147 . . . . . . . . . 151 . . . . . . . . . 157 . . . . . . . . . 162 . . . . . . . . . 169 . . . . . . . . . 173

177

Technologyechnology-free free questions . . . . . . . . . . . . . . . . . . . . 177 Multiple-choice Multiple-choi ce questions . . . . . . . . . . . . . . . . . . . . 179 Extended-response Extended-res ponse questions . . . . . . . . . . . . . . . . . . 182

7A Basic counting methods . . . . . . . 7B Factorial notation and permutations . 7C Permutations with restrictions . . . . 7D Permutations of like like objects objects . . . . . 7E Combinations . . . . . . . . . . . . . 7F Combinations Combinations with with restrictions restrictions . . . . 7G Pascal’s triangle . . . . . . . . . . . . 7H The pigeonhole principle . . . . . . . 7I The inclusion–exclu inclusion–exclusion sion principle . . . Review of Chapter 7 . . . . . . . . .

190 . . . . . . . . . . . . . . 191 . . . . . . . . . . . . . . 195 . . . . . . . . . . . . . . 201 . . . . . . . . . . . . . . 204 . . . . . . . . . . . . . . 207 . . . . . . . . . . . . . . 212 . . . . . . . . . . . . . . 216 . . . . . . . . . . . . . . 219 . . . . . . . . . . . . . . 223 . . . . . . . . . . . . . . 228


Contents

8

9

Number and proof 8A 8B 8C 8D 8E 8F

. . . . . . . . . . . . . . . . . . . . . 260

Geometry in the plane and proof

265

9B 9C 9D 9E 9F 9G 9H 9I

10

232

Direct proof . . . . . . . Proof by contrapositive . Proof by contradiction . Equivalent statements . Disproving Disprovin g statements . Mathematical induction Review of Chapter 8 . .

9A

. . . . . . . . . . . . . . . . . . . . . 233 . . . . . . . . . . . . . . . . . . . . . 238 . . . . . . . . . . . . . . . . . . . . . 242 . . . . . . . . . . . . . . . . . . . . . 246 . . . . . . . . . . . . . . . . . . . . . 249 . . . . . . . . . . . . . . . . . . . . . 251

Points, lines and angles angles . . . . Triangles and polygons . . . . Congruence and proofs . . . Pythagoras’ theorem . . . . . Ratios . . . . . . . . . . . . . An introduction introduction to similarity similarity . Proofs involving similarity . . Areas, volumes and similarity . The golden ratio . . . . . . . Review of Chapter 9 . . . . .

. . . . . . . . . . . . . . . . . . 266 . . . . . . . . . . . . . . . . . . 272 . . . . . . . . . . . . . . . . . . 277 . . . . . . . . . . . . . . . . . . 282 . . . . . . . . . . . . . . . . . . 286 . . . . . . . . . . . . . . . . . . 288 . . . . . . . . . . . . . . . . . . 295 . . . . . . . . . . . . . . . . . . 297 . . . . . . . . . . . . . . . . . . 304 . . . . . . . . . . . . . . . . . . 308

Circle geometry 10A Angle properties properties of the circle . 10B Tangents . . . . . . . . . . . 10C Chords in circles . . . . . . . Review of Chapter 10 . . . .

11


12

Sampling and sampling distributions

11A 11B 11C


12A 12B 12C

v

316 . . . . . . . . . . . . . . . . . . 317 . . . . . . . . . . . . . . . . . . 322 . . . . . . . . . . . . . . . . . . 326 . . . . . . . . . . . . . . . . . . 329

334


Populations and samples . . . . . . . . . . . . . . . . The distribution distribution of of the sample sample proportion proportion . . . . . . . Investigating the distribution distribution of the sample proportion proportion using simulation . . . . . . . . . . . . . . . . . . . . 12D Investigating the distribution distribution of the sample mean using simulation . . . . . . . . . . . . . . . . . . . . Review of Chapter 12 . . . . . . . . . . . . . . . . .

347 . . . . . 348 . . . . . 353 . . . . . 366 . . . . . 373 . . . . . 381


vi Contents

13

14

Trigonometric ratios and applications 13A 13B 13C 13D 13E

Reviewing trigonometry . . . . . . . . . . . . . . . . . The sine rule . . . . . . . . . . . . . . . . . . . . . . . The cosine rule . . . . . . . . . . . . . . . . . . . . . . The area of of a triangle . . . . . . . . . . . . . . . . . . . Circle mensuration . . . . . . . . . . . . . . . . . . . . 13F Angles of elevation, angles angles of depression depression and bearings . 13G Problems in three three dimensions dimensions . . . . . . . . . . . . . . 13H Angles between planes planes and more difficult difficult 3D problems problems . Review of Chapter 13 . . . . . . . . . . . . . . . . . . Further trigonometry 14A 14B 14C 14D 14E

15

15A 15B

Symmetry properties . . . . . . . . . . . . . . . . The tangent function . . . . . . . . . . . . . . . . Reciprocal functions functions and the Pythagorean Pythagorean identity Addition formulas formulas and double angle angle formulas . . . Simplifying a cos x + b sin x . . . . . . . . . . . . . Review of Chapter 14 . . . . . . . . . . . . . . .

Reciprocal functions . . . . . . . . . . . Locus of points . . . . . . . . . . . . . . Parabolas . . . . . . . . . . . . . . . . . Ellipses . . . . . . . . . . . . . . . . . . Hyperbolas . . . . . . . . . . . . . . . . Parametric equations . . . . . . . . . . . Polar coordinates . . . . . . . . . . . . . Graphing using polar coordinates coordinates . . . . Further graphing graphing using using polar coordinates Review of Chapter 15 . . . . . . . . . .


. . . . 393 . . . . 397 . . . . 400 . . . . 403 . . . . 408 . . . . 412 . . . . 416 . . . . 421

. . . . . . . 428 . . . . . . . 430 . . . . . . . 433 . . . . . . . 438 . . . . . . . 445 . . . . . . . 448

453 . . . . . . . . . . . . 454 . . . . . . . . . . . . 459 . . . . . . . . . . . . 462 . . . . . . . . . . . . 465 . . . . . . . . . . . . 469 . . . . . . . . . . . . 474 . . . . . . . . . . . . 483 . . . . . . . . . . . . 485 . . . . . . . . . . . . 488 . . . . . . . . . . . . 493

Complex numbers 16A 16B 16C 16D 16E

. . . . 388

427

Graphing techniques

15C 15D 15E 15F 15G 15H 15I

16

387

Starting to build the complex numbers numbers . . . . . Multiplication and division division of complex numbers Argand diagrams . . . . . . . . . . . . . . . . Solving equations equations over the complex numbers numbers . Polar form form of a complex complex number . . . . . . . . Review of Chapter 16 . . . . . . . . . . . . .

498 . . . . . . . . . 499 . . . . . . . . . 503 . . . . . . . . . 509 . . . . . . . . . 513 . . . . . . . . . 515 . . . . . . . . . 520


Contents

17


18

Matrices

19

20

21

17A 17B 17C

18A 18B 18C 18D 18E

524

Technologyechnology-free free questions . . . . . . . . . . . . . . . . . . . . 524 Multiple-choice Multiple-choi ce questions . . . . . . . . . . . . . . . . . . . . 526 Extended-response Extended-res ponse questions . . . . . . . . . . . . . . . . . . 531 535

Matrix notation . . . . . . . . . . . . . . . . . . . . . . . Addition, subtraction subtraction and multiplicatio multiplication n by a real real number Multiplication of matrices . . . . . . . . . . . . . . . . . Identities, inverses inverses and determinants for 2 × 2 matrices . . Solution of of simultaneous simultaneous equations using matrices matrices . . . . Review of Chapter 18 . . . . . . . . . . . . . . . . . . .

Transformations of the plane 19A 19B 19C 19D 19E 19F 19G 19H


. . . 540 . . . 544 . . . 547 . . . 552 . . . 555

. . . . . . . 561 . . . . . . . 565 . . . . . . . 571 . . . . . . . 574 . . . . . . . 577 . . . . . . . 581 . . . . . . . 585 . . . . . . . 590 . . . . . . . 593

598

Introduction to vectors . . . Components of vectors . . . Scalar product of vectors . . Vector projections . . . . . . Geometric proofs . . . . . . Vectors in three dimensions dimensions Review of Chapter 20 . . .

Revision of Chapters 18–20 21A 21B 21C

. . . 536

560

Linear transformation transformationss . . . . . . . . . . . . . . . Geometric transformation transformationss . . . . . . . . . . . . . Rotations and general general reflections . . . . . . . . . . Composition of transformations transformations . . . . . . . . . . Inverse transformation transformationss . . . . . . . . . . . . . . . Transformations of straight lines and other graphs Area and determinant . . . . . . . . . . . . . . . General transformation transformationss . . . . . . . . . . . . . . Review of Chapter 19 . . . . . . . . . . . . . . .

Vectors 20A 20B 20C 20D 20E 20F

vii

. . . . . . . . . . . . . . . . . . . 599 . . . . . . . . . . . . . . . . . . . 607 . . . . . . . . . . . . . . . . . . . 611 . . . . . . . . . . . . . . . . . . . 614 . . . . . . . . . . . . . . . . . . . 618 . . . . . . . . . . . . . . . . . . . 621 . . . . . . . . . . . . . . . . . . . 624

629



viii Contents ematics 22 Kin22A Position, velocity and acceleration . . . . . . . . . . . . . . . . 22B 22C 22D

Applications of antidifferentiation antidifferentiation to kinematics . Constant acceleration . . . . . . . . . . . . . . . Velocity–time graphs . . . . . . . . . . . . . . . Review of Chapter 22 . . . . . . . . . . . . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

tics of a particle 23 Sta23A Forces and triangle triangle of forces . . . . . . . . . . . . . . . . . . .

640

641 646 650 653 659 666

667 23B Resolution of forces . . . . . . . . . . . . . . . . . . . . . . . . 672 Review of Chapter 23 . . . . . . . . . . . . . . . . . . . . . . 676

vision of Chapters 22–23 679 24 Re24A Technologyechnology-free free questions . . . . . . . . . . . . . . . . . . . . 679 24B 24C

Multiple-choice questions . . . . . . . . . . . . . . . . . . . . 681 Multiple-choice Extended-response Extended-res ponse questions . . . . . . . . . . . . . . . . . . 683

Glossary

685

Answers

698

Included in the Interactive Textbook and PDF textbook only Chapter 25: Statistics 25A

Summarisi Summ arising ng univa univariat riatee data

25B

Displayin Displ aying g biva bivariat riatee data

25C

The corre correlatio lation n coef coefficien ficient t

25D

Liness on scatt Line scatterplo erplots ts

25E

The least squa squares res regr regressio ession n line

Review of Chapter 25 Chapter 26: Logic and algebra 26A

Sets Se ts and cir circui cuits ts

26B

Boolea Boo lean n alg algeb ebra ra

26C

Logical Logi cal conn connectiv ectives es and and trut truth h table tabless

26D

Logic Log ic cir circui cuits ts

26E 26 E

Karn Ka rnau augh gh ma maps ps

Review of Chapter 26 Chapter 27: Graph theory 27A

Graphs Grap hs and adja adjacency cency matr matrices ices

27B

Eulerr circu Eule circuits its and Hami Hamilton lton cycle cycless

27C

Matri Ma trix x pow powers ers and wal walks ks

27D

Complete Comp lete graph graphs, s, bipartit bipartitee graphs graphs and and trees trees

27E

Euler’ Eule r’ss formu formula la and the Plat Platonic onic solid solidss

27F

Appendix Appe ndix:: When ever everyy verte vertexx has has even degr degree ee

Review of Chapter 27 Appendix A: Guide to TI-Nspire CAS CX with OS4.0 Appendix B: Guide to Casio ClassPad II



Introduction

Cambridge Specialist Mathematics Australian Curriculum / VCE VCE Units 1 & 2 2 provides a

complete teaching and learning resource for the VCE Study Design to be implemented in 2016. It has been written with understanding as its chief aim and with ample practice o ff ered ered through the worked examples and exercises. All the work has been trialled in the classroom, and the approaches o ff ered ered are based on classroom experience and the responses of teachers to earlier versions of this book. Specialist Mathematics Units 1 & 2 o ff ers ers the material on topics from the Specialist Mathematics Study Design. The topics covered provide excellent background for a student proceeding to Specialist Mathematics Units 3 & 4. It also would be very useful for a student proceeding to Mathematical Methods Units 3 & 4. The book has been carefully prepared to reflect the prescribed course. New material has been included for many of the topics including geometry, proof, statistics, transformations, counting principles and algebra. The book contains five revision chapters. These provide technology-free, technology-free, multiple-choice and extended-response questions. The TI-Nspire calculator examples and instructions have been completed by Russell Brown and those for the Casio ClassPad have been completed by Maria Scha ff ner. ner.

Areas of Study The chapters in this book cover the diversity of topics that feature in the Specialist Mathematics Study Design. They are collected into Areas of Study. Topics from General Mathematics Units 1 & 2 2 are also available to be incorporated into a Specialist Mathematics course. The table opposite shows how courses can be constructed from Specialist Mathematics topics (indicated by SM , with prescribed topics marked as such) and General Mathematics topics (indicated by GM ). ‘ITB extra’ refers to a chapter that is accessed only in the Interactive Textbook. Cambridge Senior Maths AC/VCE Specialist Mathematics 1&2


x Introduction Refer to the Specialist Mathematics Study Design for a full list of conditions that a course must satisfy.

Area of study

Topic

Arithmetic and number

SM: Number Number systems and

Chapters

Prescribed

2, 4, 8, 16

recursion SM: Principles of counting

7

Geometry in the plane plane Geometry, measurement SM: Geometry and trigonometry

Prescribed

8, 9, 10, 13

and proof

Graphs of linear and non-linear relations

SM: Vectors in the plane

20, 22, 23

GM: Applications of trigonometry

13

SM: Graphs of non-linea non-linearr

Algebra and structure

Prescribed

15

relations SM: Kinematics

22

GM: Variation

3

SM: Transformations, trigonometry

14, 18, 19

and matrices GM: Linear relations and equations

1

GM: Number patterns and recursion

4

SM: Logic and algebra

ITB extra

Discrete mathematics

SM: Graph theory

ITB extra

Statistics

SM: Simulation, sampling and

12

sampling distributions GM:: In GM Inv ves esti tiga gati ting ng th thee re rela lati tion on be betw twee een n

ITB IT B ext xtra ra

two numerical variab variables les Other

SM: Chapter 5 provide providess more topics in algebra such as partial fractions.

The integration of the features of the textbook and the new digital components of the package, powered by Cambridge HOTmaths, are illustrated in the next two pages.

About Cambridge HOTmaths Cambridge HO Cambridge HOTmaths Tmaths is a compreh comprehensiv ensive, e, awar award-winning d-winning mathematics mathematics learning system – an interactive online maths learning, teaching and assessment resource for students and teachers, for individuals or whole classes, for school and at home. Its digital engine or platform is used to host and power the interactive textbook and the Online Teaching Suite, and selected topics from HOTmaths’ own Years 9 and 10 courses area are available for revision of prior knowledge. All this is included in the price of the textbook. Cambridge Senior Maths AC/VCE Specialist Mathematics 1&2


Acknowledge ments

The author and publisher wish to thank the following sources for permission to reproduce material: Cover: Used under license 2015 from Shutterstock.com / Tiago Tiago Ladeira.

korinoxe, p.1 / AlexVector, AlexVector, p.39 / ILeysen, ILeysen, pp.82, 232 / Images: Shutterstock.com / korinoxe, Sianapotam, p.110 / pzAxe, pzAxe, p.146 / tashechka, tashechka, pp.177, 316, 427, 453, 498 , 629 , 666 , 679 / Apostrophe, p.190 / Dinga, Dinga, p.265 / I_Mak, I_Mak, pp.334, 524 , 640 / Attitude, Attitude, pp.347, 387 , 560 , Melpomene, p.535 / 598 / Melpomene, Every eff ort ort has been made to trace and acknowledge copyright. The publisher apologises for any accidental infringement and welcomes information that would redress this situation.



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1

C h a p t e r 1

Algebra I Algebra

Objectives  To 

To solve problems with linear equations and simultaneous linear equations .

 To 

solve linear equations . use substitution and transposition with formulas.

To add and multiply algebraic fractions.

 To

solve literal equations.

 To

solve simultaneous literal equations.

Algebra is the language of mathematics. Algebra helps us to state ideas more simply. It also enables us to make general statements about mathematics, and to solve problems that would be difficult to solve otherwise. We know by basic arithmetic that 9 7 + 2 7 = 11 7. We could replace the number 7 in this statement by any other number we like, and so we could write down infinitely many such statements. statements. These can all be captured by the algebraic algebraic statement 9 x + 2 x = 11 x, for any number x . Thus algebra enables us to write down general statements.

×

×

×

Formulas enable mathematical ideas to be stated clearly and concisely. An example is the well-known formula for compound interest. Suppose that an initial amount P is invested at an interest rate R , with interest compounded annually. Then the amount, A n , that the investment is worth after n years is given by A n = P (1 + R)n . In this chapter we review some of the techniques which you have met in previous years. Algebra plays a central role in Specialist Mathematics at Years 11 and 12. It is important that you become fluent with the techni techniques ques introduced introduced in this chapter and in Chapter 5.



2

Chapter 1: Algebra I

1A Indices This section revises algebra involving indices.

Review of index laws For all non-zero real numbers a and b and all integers m and n : +



am





a

× an = a m n an

n

=

b

bn



am

÷ an = a m−n



a−n =

1

an





(am )n = a mn 1 a−n

= a n



(ab)n = a n bn



a0 = 1

Rational indices 1

If a a is a positive real number and n is a natural number, then a n is defined to be the n th root 1

1

of a a . That is, a n is the positive number whose n th power is a . For example: 9 2 = 3. 1

1

n is odd, then we can define a n when a is negative. If a a is negative and n is odd, define a n If n 1

to be the number whose n th power is a . For example: ( 8) 3 = 2.

−

−

In both cases we can write: 1

an =

√ n

a with

1 n

  an

= a

In general, the expression a x can be defined for rational rational indices, i.e. when x = and n are integers, by defining m an

m n

, where m

1 m

 

= an

To employ this definition, we will always first write the fractional power in simplest form.

Note: The index laws hold for rational indices m and n whenever both sides of the equation are defined (for example, if a a and b are positive real numbers).

Example 1 Simplify each of the following:

a x

2

× x

3

b

x4 x2

Solution

a x2 × x 3 = x2+3 = x5 b

x4

x2

=x

4 x 5

c

1 x2

d

1 3 2 ( x )

÷


4−2

=

=x

= 3 x2

c

1 x2

am an

1 4 − x2 5

=

3 − 10 x

d

1 3 2 ( x )

Explanation am

2

÷

4 x 5

am an

+

× an = a m n = a m−n = a m−n

(am )n = a mn


1A

1A Indices

3

Example 2 Evaluate:

a

1000 27

2 125 3

  b

Solution

2 3

2

1 2

Explanation

         b a 125 3 = 125 3 2 3

1000 27

1000 27

=

1 2 3

10 = 3

√

1

= 5 2 = 25

3

125 3 = 125 = 5 2

1 3

  

100 = 9

1000 27

=

3

1000 10 = 27 3

Example 3

 4

Simplify

x2 y3

1 2 x 2 y 3

.

Solution 1

 4

x2 y3

=

1 2 x 2 y 3

2

( x2 y3 ) 4

=

1 2 x 2 y 3

Explanation

3

x 4 y 4

(ab)n = a n bn

1 2 x 2 y 3

=

am

2 1 3 2 − − x 4 2 y 4 3

an

= a m−n

1

= x0 y 12 1

a0 = 1

= y 12

Section summary 

Index laws

•

am a

+

× an = a m n

  •

n

=

b



an bn

•

am

÷ an = a m−n

•

a−n =

Rational indices 1

• a n = √ a n

•

m an

1

an

• •

(am )n = a mn 1 a−n

= a n

•

(ab)n = a n bn

•

a0 = 1

1 m

 

= an

Exercise 1A Example 1

1 Simplify each of the following using the appropriate index laws:

x8

b a × a−3 p−5

x−4

p2

a x e


3

× x

4

5

f

2

c x × x −1 × x 2 1

2

g a 2 ÷ a 3

d

y3 y7

h (a−2 )4


4

1A


i ( y−2 )−7

j ( x )

1

1

m (n10 ) 5 2 5 5

s (ab3 )2 × a−2 b−4 ×

Example 2

x

o (a2 ) 2 × a−4

p

1

(43 n4 )

1 2

4

  − − l

5

n 2 x 2 × 4 x3

  − q ÷ 2n

3 20 5 − k (a )

5 3

1

x−4

3

r x3 × 2 x 2 × −4 x− 2

1

t 22 p−3 × 43 p5 ÷ (6 p−3 )



a2 b−3



0

2 Evaluate each of the following: a

1 25 2

49 36

  − e i

1 2

b

  c

f

1 27 3

1 144 2

81 16

3 92

1 2

16 9

1 64 3

g

  j

1 4

23 5

  k

1

d 16− 2 h

2 64 3

0

3

l 128 7

3 Use your calculator to evaluate evaluate each of the following, following, correct to two two decimal places: a 4.352 f

b 2.45

√ 4

c 1

d (0.02)−3

34.6921 2

g (345.64) − 3

2.3045

√

h (4.568) 5

e

√ 3

0.729

1

i

1

(0.064)− 3

4 Simplify each of the following, following, giving your answer with positive positive index: a2 b3

a

2a2 (2b)3 b (2a)−2 b−4

a−2 b−4 a2 b3

d

ab

× a−1b−1 a−2 b−4

2n 5 Write 2n 2

6

e

c

(2a)2 8b3 16a−2 b−4

×

f

a−2 b−3 a−2 b−4

2a2 b3 8a−2 b−4

÷ (2a16)−ab 1 b−1

× 8n in the form 2an b. × 16 Write 2− x × 3− x × 62 x × 32 x × 22 x as a power of 6. +

7 Simplify each of the following: following: 1

2

1

2

1

b a 4 × a 5 × a− 10

    d ×

  e ×

1 2 23

Example 3

1

a 2 3 × 2 6 × 2− 3 1 5 22

8 Simplify each of the following: a d g

1 2

23

1

23

2

2


3

2

× 2− 5

√ 3 2 √ 2 −1 √ √ b a3 b2 × a2 b−1 a b ÷ a b √ −4 2 √ 3 −1 √ 3 2 −3 √ 2 −1 −5 e abc × ab c a b × a b √ 3 2 √ −4 2 √ a b a b × × a3b−1 a3 b−1 a2 b−1 c−5 3

5

c 2 3 × 2 6 × 2− 3

√ 3 2 √ 2 −1 a b × a b √ 3 2 √ 2 −1 f a b ÷ a b

c

5

5

5

5


1B Standard form

5

1B Standard form Often when dealing with real-world problems, the numbers involved may be very small or very large. For example: 

The distance from Earth to the Sun is approximately 150 000 000 kilometres.



The mass of an oxygen atom is approximat approximately ely 0.000 000 000 000 000 000 000 026 grams.

To help deal with such numbers, numbers, we can use a more convenient convenient way to express them. This involv in volves es expressing expressing the number as a product of a number between between 1 and 10 and a powe powerr of 10 and is called standard form or scientific notation . These examples written in standard form are:  

× 108 kilometres 2.6 × 10−23 grams 1.5

Multiplication and division with very small or very large numbers can often be simplified by first converting the numbers into standard form. When simplifying algebraic expressions or manipulating numbers in standard form, a sound knowledge of the index laws is essential.

Example 4 Write each of the following in standard form:

a 3 453 000

b 0.00675

Solution a 3 453 000 = 3.453 × 10 6

b 0.00675 = 6.75 × 10−3

Example 5 32 000 000 0.000 004 Find the value of . 16 000

×

Solution

× 107 × 4 × 10−6 1.6 × 104 12.8 × 101 = 1.6 × 104 = 8 × 10−3

32 000 000 0.000 004 3.2 = 16 000

×

= 0.008



Significant figures When measurements are made, the result is recorded to a certain number of significant figures.. For example, figures example, if we say that the length of a piece of ribbon is 156 cm to the nearest centimetre, centim etre, this means that the length is between 155.5 cm and 156.5 cm. The number 156 is said to be correct to three significant figures. Similarly, we may record π as being 3.1416, correct to five significant figures.



6

Chapter 1: Algebra I When rounding off to a given number of significant figures, first identify the last significant digit and then: 

if the next digit digit is 0, 1, 2, 3 or 4, round down down



if the next digit digit is 5, 6, 7, 8 or 9, round up.

It can help with rounding rounding off if the original number is first written in scientific notation. So π = 3.141 592 653 . . . is rounded off to 3, 3.1, 3.1 3.14, 4, 3.14 3.142, 2, 3.14 3.1416, 16, 3.1 3.14159 4159,, etc. depending on the number of significant figures required. Writing a number in scientific notation makes it clear how many significant figures have been recorded. recor ded. For example, example, it is unclear whether 600 is recorded to one, two or three significant significant figures. However, when written in scientific notation as 6.00 102 , 6.0 102 or 6 102 , it is clear how many significant figures are recorded.

×

×

×

Example 6

√ a 5

Evaluate

b2

if a a = 1.34

× 10−10 and b = 2.7 × 10−8.

Solution

√ a √ 1.34 × 10−10 = (2.7 × 10−8 )2 b2 1 (1.34 × 10−10 ) 5 = 2.72 × (10−8 )2 = 1.45443 . . . × 1013 = 1.45 × 1013 to three significant figures 5

5

Many calculators can display numbers in scientific notation. The format will vary from calculator to calculator. For example, the number 3 245 000 = 3.245 106 may appear as 3.245e6 or 3.24506 .

×

Using the TI-Nspire on > Settings > Document Settings and Settings and change Insert a Calculator a Calculator page, page, then use Exponential Format field the Exponential the field to Scientific to Scientific . If you want this change to apply only to the current page, select OK to accept the change. Select Current Select Current to to return to the current page.

Using the Casio ClassPad The ClassPad calculator can be set to express decimal answers in various forms. To select a fixed number of decimal places, including specifying scientific notation with a fixed decimal accuracy, go to Settings to Settings O and in Basic in Basic format tap tap the arrow to select from the various Number formats available.



1B

1B Standard form

7

Section summary 

A number is is said to be be in scientific notation (or standard form ) when it is written as a product of a number between 1 and 10 and an integer power of 10. For example: 6547 = 6.457





× 103 and 0.789 = 7.89 × 10−1

Writing a number in scientific notation makes it clear how many many significant figures have been recorded. When rounding off to a given number of significant figures, first identify the last significant digit and then: if the next digit digit is 0, 1, 2, 3 or 4, round down down

• •

if the next digit digit is 5, 6, 7, 8 or 9, round up.

Exercise 1B Example 4

1 Express each of the following numbers in standard form: a 47.8 e 0.0023 i 23 000 000 000

b 6728 f 0.000 000 56 j 0.000 000 0013

c 79.23 g 12.000 34 k 165 thousand

d 43 580 h 50 million l 0.000 014 567

2 Express each of the following in scientific notation: a b c d e f

X-rays have a wavelength of 0.000 000 01 cm. The mass of a hydrogen hydrogen atom is 0.000 0.000 000 000 000 000 000 000 001 67 g. Visible Vi sible light has wavelength 0.000 05 cm. One nautical nautical mile is 1853.18 1853.18 m. A light year is is 9 461 000 000 000 km. The speed of light light is 29 980 000 000 cm / s. s.

3 Express each of the following as an ordinary number: a The star Sirius is is approximately 8.128 × 1013 km from Earth. b A single red blood cell cell contains 2.7 × 108 molecules of haemoglobin. c The radiu radiuss of an electron electron is 2.8 × 10−13 cm. 4 Write each of the following following in scientific notation, correct to the number of significant figures indicated in the brackets:

a 456.89 d 0.04536 Example 5

(2)

c 5 6 7 9 .0 8 7 f 4568.234

(2) (2)

(5) (5)

5 Find the value of: 324 000

a Example 6

b 3 4 5 6 7. 2 3 e 0 .0 9 0 4 5

(4)

× 0.000 0007

4000

b

5 240 000 0.8 42 000 000

×

6 Evaluate the following correct to three significant figures:

√ a 3

a


b4

√ a 4

if a a = 2

× 10

9

and b = 3.215

b

4b4

if a a = 2

× 1012 and b = 0.05


8


1C Solving Solving linear equations and simultaneous simultaneous linear equations Many problems may be solved by first translating them into mathematical equations and then solving the equations using algebraic techniques. An equation is solved by finding the value or values of the variables that would make the statement true. Linear equations are simple equations that can be written in the form a axx + b = 0. There are a number of standard techniques that can be used for solving linear equations.

Example 7 a Solve

x

5

−

2 =

x

3

.

b Solve

Solution a Multiply both sides of the equation

−

5 x

5

×

2 =

15 − 2 × 15 =

x

3 x

×

3

2

−

2 x − 4 = 5. 3

b Multiply both sides of the equation

by the lowest common multiple of 3 and 5: x

x−3

15

by the lowest common multiple of 2 and 3: x − 3 2 x − 4 − = 5 2 3 x − 3

2

×

6−

2 x − 4 3

6 = 5 × 6

×

3 x − 30 = 5 x

3( x − 3) − 2(2 x − 4) = 30

3 x − 5 x = 30

3 x − 9 − 4 x + 8 = 30

2 x = 30

−

x = ∴

x =

3 x − 4 x = 30 + 9 − 8

30 −2

x = 31

−

x =

15

−

x =

∴



Simultaneous linear equations The intersection point of two straight lines can be found graphically; however, the accuracy of the solution will depend on the accuracy of the graphs. Alternatively, the intersection point may be found algebraically by solving the pair of simultaneous equations. We shall consider two techniques for solving simultaneous equations.

31 −1 31

−

y

4

4 =

3

y

–

x

2

2

1 –3 –2 –1 O –1 –2

x

1

2

3 x

+ 2 y = – 3

–3 –4


(1, –2)


1C Solving linear equations and simultaneous linear equations

9

Example 8 Solve the equations 2 x y = 4 and x + 2 y =

−

Solution

−3. Explanation

Method 1: Substitution 2 x y = 4

(1)

x + 2 y =

(2)

−

−3

Using one of the two equations, express one variable in terms of the other variable.

From equation equation (2), we get x =

−3 − 2 y.

Substitute in equation (1):

Then substitute this expression into the other equation (reducing it to an equation in one variable, y ). Solve the equation equation for y .

2( 3

− − 2 y) − y = 4 −6 − 4 y − y = 4 −5 y = 10 y = −2

Substitute the value of y y into (2): x + 2( 2) =

−

−3

Substitute this value for y in one of the equations to find the other variable, x .

x = 1

Check in (1): LHS = 2(1) RHS = 4

− (−2) = 4

A check can be carried out with the other equation.

Method 2: Elimination 2 x y = 4

(1)

x + 2 y =

(2)

−

−3

To eliminate x , multiply equation equation (2) by 2 and subtract the result from equation (1). When we multiply equation equation (2) by 2, the pair of equations becomes: 2 x y = 4

−

2 x + 4 y =

(1)

If one of the variables has the same coefficient in the two equations, we can eliminate that variable by subtracting one equation from the other. It may be necess necessary ary to multiply one of the equations by a consta constant nt to make the coefficients of x x or y the same in the two equations.

(2 )

−6

Subtract (2 ) from (1):

−5 y = 10 y = −2 Now substitute for y in equation (2) to find x , and check as in the substitution method. intersection of the graphs of Note: This example shows that the point (1, −2) is the point of intersection the two linear relations. Cambridge Senior Maths AC/VCE Specialist Mathematics 1&2


10


Using the TI-Nspire Calculator application Simultaneous equations can be solved in a Calculator a Calculator application. application. 

Equations . Use menu > Algebra > Solve System of Equations > Solve System of Equations.



Complete the pop-up screen.



Enter the equations as shown to give give the solution to the simultaneous equations 2 x y = 4 and x + 2 y = 3.

−

−

Graphs application Simultaneous equations can also be solved graphically in a Graphs a Graphs application. application. 

Equations of the form a x + by = c can be entered directly using menu > Graph Entry/Edit > Equation > Line > Line Standard. Standard.



Alternatively, rearrange each equation to make y the subject, and enter as a standard Alternatively, function (e.g. f 1( x) = 2 x 4).

−

press tab . Press Pressing ing enter will hide the Entry Line. Note: If the Entry Line is not visible, press If you want to add more equations, equations, use  to add the next equation.  



The intersection point is found using menu > Analyze Graph > Intersection Intersection.. Move the cursor to the left of the intersection intersection point (lower bound), click, and move to the right of the intersection point (upper bound). Click to paste the coordinates coordinates to the screen.



1C Solving linear equations and simultaneous linear equations

11

Using the Casio ClassPad To solve the simultaneous equations algebraically: 

Open the M application and turn on the keyboard.



In Math1 , tap the simultaneous equations icon ~.



Enter the two equations as shown.





Type x , y in the bottom-right square to indicate the variables. Tap EXE .

There are two methods for solving simultaneous equations graphically.

Method 1 In the M application:   

Enter the the equation equation 2 x − y = 4 and tap EXE . Enter the equation x + 2 y =

3 and tap EXE .

−

Select $ from the toolbar to insert a graph window. An appropriate window can be set by Zoom > Quick > Quick Standard. Standard. selecting Zoom selecting



Highlight each equation and drag it into into the graph window window..



To find the point of intersection, go to Analysis to Analysis > G-Solve > Intersection Intersection..

Method 2 For this method, the equations need to be rearranged to make y the subject. In this form, the equations are y = 2 x − 4 and y = − 12 x − 32 . 

& Tabl able e Open the menu m; select Graph select Graph&



y 1 and enter 2 x − 4. Tap in the working line of y



y 2 and enter − 12 x − 32 . Tap in the working line of y



Tick the the boxes for y 1 and y 2.



Select $ from the toolbar.



Analysis > G-Solve > Intersection Intersection.. Go to to Analysis



.

If necessary, necessary, the view window settings settings can be adjusted by tapping 6 or by selecting Zoom selecting Zoom > Quick > Quick Standard. Standard.



12

1C


Section summary 

An equation is solved solved by finding the value or values of the variables variables that would make the statement true.



A linear equation is one in which the ‘unknown’ is to the first power. power.



There are often several diff erent erent ways to solve a linear equation. The following steps provide some suggestions:

1 Expand brackets and, if the equation involves fractions, multiply through by the lowest common denominator of the terms.

2 Group all of the terms containing containing a variable on one side of the equation and the terms without the variable on the other side. 

Methods for solving simultaneous linear equations in two variables variables by hand:

Substitution Make one of the variables variables the subject subject in one of the equations. equations.

• •

Substitute for that variable variable in the other equation.

Elimination

• •

Choose one of the two variables variables to eliminate.

•

Add or subtract the two two equations to eliminate the chosen variable. variable.

Obtain the the same or opposite opposite coefficients for this variable in the two equations. To do this, multiply both sides of one or both equations by a number.

Exercise 1C Example 7a

1 Solve the following linear equations: a 3 x + 7 = 15 d g j

Example 7b

2 x 3

− 15 = 27 3 x + 5 = 8 − 7 x x 6 x + 4 = − 3 3

b 8−

x

2

c 42 + 3 x = 22

−16

=

e 5(2 x + 4) = 13

f

h 2 + 3( x − 4) = 4(2 x + 5)

i

−3(4 − 5 x) = 24 2 x 3 − 4 = 5 x 5

2 Solve the following linear equations: a

x

2

3 x 4

− 2 + x = −8

d

5 x 4

− 4 + 2 x + 5 = 6

f

3 x

c e

b

+

2 x = 16 5

2 x

g


2

3 x 4

4

4

− − 2( x + 1) = −24 5

h

− x3 = 8

− 43 = 25 x 3 − 3 x 2( x + 5) − = 10

6

1 20

−2(5 − x) + 6 = 4( x − 2) 8

7

3


1C Example 8

1D Solving problems with linear equations

13

3 Solve each of the following pairs of simultaneous equations: a 3 x + 2 y = 2 2 x

b 5 x + 2 y = 4

− 3 y = 6

3 x y = 6

d x + 2 y = 12 x

e

− 3 y = 2

− 7 x − 3 y = −6 x + 5 y = 10

c

2 x y = 7

− 3 x − 2 y = 2

f 15 x + 2 y = 27 3 x + 7 y = 45

1D Solving problems with linear equations Many problems can be solved by translating them into mathematical language and using an appropriate mathematical technique to find the solution. By representing the unknown quantity in a problem with a symbol (called a pronumeral or a variable) and constructing an equation from the information, information, the value of the unkno unknown wn can be found by solvi solving ng the equation. Before constructing the equation, each variable and what it stands for (including the units) should shoul d be state stated. d. All the elements of the equation must be in units of the same system. system.

Example 9 For each of the following, form the relevant linear equation and solve it for x:

a The length of the the side of a square square is ( x − 6) cm. Its perimeter perimeter is 52 cm. b The perim perimeter eter of a square square is (2 x + 8) cm. Its area is 100 cm 2 . Solution a Perimeter = 4 × side length Therefore 4( x and so

− 6) = 52 x − 6 = 13 x = 19

b The perimeter perimeter of the square square is 2 x + 8. The length of one side is

2 x + 8 x+4 . = 4 2

Thus the area is x + 4 2 = 100 2

 

As the side length must be positive, this gives the linear equation x + 4

2

= 10

Therefore x = 16.



14


Example 10 An athlete trains for an event by gradually increasing the distance she runs each week over a five-week five-week period. If she runs an extr extraa 5 km each successive successive week and over the five weeks runs a total of 175 km, how far did she run in the first week?

Solution Let the distance run in the first week be x km. Then the distance run in the second week is x + 5 km, and the distance run in the third week is x + 10 km, and so on. The total distance run is x + ( x + 5) + ( x + 10) + ( x + 15) + ( x + 20) km. ∴

5 x + 50 = 175 5 x = 125 x = 25

The distance she ran in the first week was 25 km.

Example 11 A man bought 14 CDs at a sale. Some cost $15 each and the remain remainder der cost $12.50 each. In total he spent $190. How many $15 CDs and how many $12.50 CDs did he buy?

Solution Let n be the number of CDs costing $15. Then 14 n is the number of CDs costing $12.50.

−

∴

15n + 12.5(14 15n + 175

− n) = 190

− 12.5n = 190

2.5n + 175 = 190 2.5n = 15 n = 6

He bought 6 CDs costing $15 and 8 CDs costing $12.50.

Section summary Steps for solving a word problem with a linear equation 

Read the question carefully and write down the known information clearly. clearly.



Identify Identi fy the unknown quantity quantity that is to be found.



Assign a variable to this quantity. quantity.



 

Form an expression in terms of x x (or the variable being used) and use the other relevant information to form the equation. Solve the equation. Write a sentence answering the initial question.



1D

1D Solving problems with linear equations

15

Exercise 1D Skillsheet

Example 9

1 For each of the cases below below,, write down a relevant equation involvi involving ng the variables defined, and solve the equati equation on for parts a, b and c. a The lengt length h of the side of a square is ( x − 2) cm. Its perimeter perimeter is 60 cm. b The perimeter of a square is is (2 x + 7) cm. Its area is 49 cm 2 . c The length length of a rectangle is ( x − 5) cm. Its width is (12 − x ) cm. The rectangle is twice as long as it is wide.

d The length length of a rectangle is (2 (2 x + 1) cm. Its width is ( x − 3) cm. The perimeter perimeter of the rectangle is y cm.

e n people each have a meal costing costing $ p. The total cost of the meal is $ Q. f S people each have a meal costing $ p. A 10% service charge charge is added to the cost. The total cost of the meal is $ R.

g A machine working at a constant rate produces n bolts in 5 minutes. It produce producess 2400 bolts in 1 hour hour..

h The radius radius of a circle is ( x + 3) cm. A sector subtending subtending an angle of 60◦ at the centre is cut off . The arc length of the minor sector is a cm. Example 10

2 Bronwyn and Noel have a women’ women’ss clothing shop in Summerland. Bronwyn manages the shop and her sales are going up steadi steadily ly over a parti particular cular period of time. They are going up by $500 per week. If over a five-week five-week period her sales total total $17 500, how much did she earn in the first week?

Example 11

3 Bronwyn and Noel have a women’ women’ss clothing shop in Summerland. Sally, Sally, Adam and baby Lana came into the shop and Sally bought dresses dresses and handbags handbags.. The dresses cost $65 each and the handbags cost $26 each. Sally bought 11 items and in total she spent $598. How many dresses and how many handbags did she buy?

4 A rectangular rectangular courtyard courtyard is three times times as long as it is wide. If the perimeter perimeter of the courtyard court yard is 67 m, find the dimens dimensions ions of the courty courtyard. ard.

5 A wine merchant merchant buys 50 50 cases of wine. wine. He pays full full price for half half of them, but but gets a 40% discount on the remainder. remainder. If he paid a total of $2260, how much was the full price of a singl singlee case?

6 A real-estate real-estate agent sells sells 22 houses in six months. months. He makes a commissio commission n of $11 500 per house on some and $13 000 per house on the remainder remainder. If his total commission commission over the six months was $272 500, on how many houses did he make a commission commission of $11 500?

7 Three boys compare their marble collections. collections. The first boy has 14 fewer than the second boy, who has twice as many as the third. If between them they have 71 marbles, how many does each boy have?



16


1D

8 Three girls girls are playing playing Scrabble. Scrabble. At the end of the game, game, their three three scores scores add up to 504. Annie scored 10% more than Belinda, while Cassie Cassie scored 60% of the combined combin ed scores of the other two. What did each player score?

9 A biathlon event event involves involves running running and cycling. Kim can cycle 30 km / h faster than she can run. If Kim spends 48 minutes running running and a third as much time again cycling cycling in an event that covers a total distance of 60 km, how fast can she run?

10 The mass of a molecule molecule of a certain certain chemical chemical compound compound is 2.45 × 10−22 g. If each

molecule is made up of two carbon atoms and six oxygen atoms and the mass of an oxygen atom is one-third one-third that of a carbon atom, find the mass of an oxygen atom.

11 Mother’ Mother’ss pearl necklace fell to the floor. One-sixth of the pearls rolled under the fridge, one-third rolled under the couch, one-fifth of them behind the book shelf, and nine were found at her feet. How many pearls are there?

12 Parents say they don’t have have favourites, favourites, but everyone everyone knows that’s that’s a lie. A father distributes $96 to his three children according to the following instructions: The middle child receives $12 less than the oldest, and the youngest receives one-third as much as the middle child. How much does each receive?

13 Kavindi has achieved achieved an average mark of 88% on her first four maths tests. What mark would she need on her fifth test to increase her average to 90%?

14 In a particular class, 72% of the students have have black hair. hair. Five black-haired students leave the class, so that now 65% of the students have black hair. How many students were originally in the class?

15 Two tanks are being emptied. Tank Tank A contains 100 litres of water and tank B contains 120 litres of water. Water runs from Tank A at 2 litres per minute, and water runs from tank B at 3 litres per minute. After how how many minutes will the amount of water in the two tanks be the same?

16 Suppos Supposee that candle A is initial initially ly 10 cm tall and burns burns out after after 2 hours. Candle B is initially 8 cm tall and burns out after 4 hours initially hours.. Both candles are lit at the same time. Assuming ‘constant burning rates’:

a When is the height of candle candle A the same as the height of candle B? b When is the height of candle candle A half the height of candle B? c When is cand candle le A 1 cm taller than candle candle B? 10 hours.. He drove hours drove part of the way at an average speed speed 3 of 100 km / h and the rest of the way at an average speed of 90 km / h. h. What is the distance he travelled at 100 km / h? h?

17 A motorist motorist drove drove 320 km in



Cambridge Specialist Mathematics Units 1 2 AC VCE Complete Book PDF

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