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2ND EDITION MATHS QUEST 11 TI-NSPIRE CAS CALCULATOR COMPANION Advanced General Mathematics



2ND EDITION VCE MATHEMATICS UNITS 1 & 2 PATRICK SCOBLE MARK BARNES RUTH BAKOGIANIS KYLIE BOUCHER MARK DUNCAN TRACEY HERFT ROBYN WILLIAMS JENNIFER NOLAN GEOFF PHILLIPS MATHS QUEST 11 TI-NSPIRE CAS CALCULATOR COMPANION Advanced General Mathematics

First published 2013 by John Wiley & Sons Australia, Ltd 42 McDougall Street, Milton, Qld 4064 Typeset in 10/12 pt Times LT Std © John Wiley & Sons Australia, Ltd 2013 The moral rights of the authors have been asserted. ISBN: 978 1 118 31771 6 978 1 118 31768 6 (flexisaver) Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL). Reproduction and communication for other purposes Except as permitted under the Act (for example, a fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher. Cover and internal design images: © vic&dd/Shutterstock.com Typeset in India by Aptara Printed in Singapore by Craft Print International Ltd 10 9 8 7 6 5 4 3 2 1 Acknowledgements The authors and publisher would like to thank the following copyright holders, organisations and individuals for their permission to reproduce copyright material in this book. Images: Texas Instruments: Screenshots from TI-Nspire reproduced with permission of Texas Instruments Every effort has been made to trace the ownership of copyright material. Information that will enable the publisher to rectify any error or omission in subsequent editions will be welcome. In such cases, please contact th\ e Permissions Section of John Wiley & Sons Australia, Ltd.

Contents Introduction vi CHAPTER 1 Number systems: real and complex 1 CHAPTER 2 Transformations 7 CHAPTER 3 Relations and functions 9 CHAPTER 4 Algebra 13 CHAPTER 5 Trigonometric ratios and their applications 23 CHAPTER 6 Sequences and series 27 CHAPTER 7 Variation 37 CHAPTER 8 Further algebra 45 CHAPTER 10 Linear and non-linear graphs 51 CHAPTER 11 Linear programming 61 CHAPTER 12 Coordinate geometry 65 CHAPTER 13 Vectors 67 CHAPTER 14 Statics of a particle 69 CHAPTER 15 Kinematics 71 CHAPTER 16 Geometry in two and three dimensions 75

Introduction This booklet is designed as a companion to Maths Quest 11 Advanced General Mathematics Second Edition. It contains worked examples from the student text that have been re-worked using the TI-Nspire CX CAS calculator with Operating System v3. The content of this booklet will be updated online as new operating systems are released by Texas Instruments. The companion is designed to assist students and teachers in making deci\ sions about the judicious use of CAS technology in answering mathematical questions. The calculator companion booklet is also available as a PDF fi le on the eBookPLUS under the preliminary section of Maths Quest 11 Advanced General Mathematics Second Edition. vi Introduction

Chapter 1 • Number systems: real and complex 1 Chapter 1 Number systems: real and complex Worked example 4 Express each of the following in the form a b , where a ∈ Z and b ∈ Z \{0}. a 0.6  b 0.23 c 0.41  d 2.1234   think Write a 1 Write 0.6 in expanded form. a 0.6  = 0.666666… [1] 2 Multiply [1] by 10. 10 × 0.6 = 6.66666… [2] 3 Subtract [1] from [2]. 9 × 0.6  = 6 0.6 = 6 9 4 State the simplest answer. 2 3 = b 1 Write 0.23 in the expanded form. b 0.23 = 0.232323… [1] 2 Multiply [1] by 100. 100 × 0.23  = 23.232323… [2] 3 Subtract [1] from [2]. 99 × 0.23 = 23 4 State the simplest answer. 0.23 23 99 =  c 1 Write 0.41   in the expanded form. c 0.41  = 0.41111… [1] 2 Multiply [1] by 10. 10 × 0.41  = 4.11111… [2] 3 Subtract [1] from [2]. 9 × 0.41  = 3.7 4 State the simplest answer. 0.41 3.7 9 37 90== d 1 Write 2.1234  in the expanded form. d 2.1234  = 2.1234234… [1] 2 Multiply [1] by 1000. 1000 × 2.1234  = 2123.423423… [2] 3 Subtract [1] from [2]. 999 × 2.1234  = 2121.3 4 State the simplest answer. 2.1234 2121.3 999 21213 9999 2357 1111 = = =  Contents Number systems: real and complex 1

2 Maths Quest 11 Advanced General Mathematics Note: The CAS calculator can perform some of these calculations for you. On a Calculator page, press: • MENU b • 2: Number 2 • 2: Approximate to fraction 2 Complete the entry line as: 0.6666666666666666666666¢ approxFraction(5E-14) Then press ENTER ·. Worked example 13 Expand and simplify the following where possible. a 7( 18 3) − b 23 (105 3) −− c (5 36)(23 2) +− think Write a , b & c 1 On a Calculator page, complete the entry lines as: 7( 18 3) − 23 (105 3)−− (536)(23 2) +− Press ENTER · after each entry. Note: The calculator must be set in either Real or Auto modes. 2 Write the answers. a b c 3(21 )7 − 30 230 − (2 32 )563 182 −− +

Chapter 1 • Number systems: real and complex 3 Worked example 16 If z 1 = 2 − 3i and z 2 = −3 + 4i, find a z1 + z 2 b z1 − z 2 c 3z 1 − 4z 2. think Write a , b & c 1 On a Calculator page, press: • MENU b • 1: Actions 1 • 1: Define 1 Complete the entry lines as: Define z1 = 2 − 3i Define z2 = 3 + 4i z1 + z2 z1 − z2 3z1 − 4z2 Press ENTER · after each entry. 2 Write the answers. a b c z1 + z 2 = −1 + i z 1 − z 2 = 5 − 7i 3z 1 − 4z 2 = 18 − 25i Worked example 17 Simplify a 2i(2 − 3i) b (2 − 3i)( −3 + 4i). think Write a Expand the brackets. a 2i(2 − 3i) = 4i − 6i 2 = 6 + 4i b Expand the brackets as for binomial expansion and simplify. b (2 − 3i)( −3 + 4i) = −6 + 8i + 9i − 12i 2 = 6 + 17i Alternatively, on a Calculator page, complete the entry lines as: 2i × (2 − 3i) (2 − 3i) × ( −3 + 4i) Press ENTER · after each entry.

4 Maths Quest 11 Advanced General Mathematics Worked example 18 If z 1 = 2 + 3i and z 2 = −4 − 5i, find a zz12+ b zz12+ c zz12 d zz12 . think Write On a Calculator page, complete the entry lines as: Define z1 = 2 + 3i Define z2 = −4 − 5i conj(z1) + conj(z2) conj(z1 + z2) conj(z1) × conj(z2) conj(z1 × z2) Press ENTER · after each entry. Note: ‘conj’ can be typed directly onto the screen or can be found by pressing: • MENU b • 2: Number 2 • 9: Complex Number Tools 9 • 1: Complex Conjugate 1 Worked example 19 Express each of the following in a + bi form. a i 4 2− b i i 3 3 −4 c (3 − 2i) −1 d i i 23 2 − + think Write a , b , c & d On a Calculator page, complete the entry lines as: i 4 2− i i 34 3 − (3 − 2i) − 1 i i 23 2 − + Press ENTER · after each entry.

Chapter 1 • Number systems: real and complex 5 Worked example 21 Factorise each of the following quadratic expressions over C. a 2z 2 + 6z b 2z 2 − 6 c 2z 2 + 3 think Write a, b & c Ensure the calculator is set in Rectangular mode. On a Calculator page, press: • MENU b • 3: Algebra 3 • C: Complex C • 2: Factor 2 Complete the entry lines as: cFactor(2z 2 + 6z, z) cFactor(2z2 − 6, z) cFactor(2z2 + 3, z) Press ENTER · after each entry. Worked example 23 Solve the following using the formula for the solution of a quadratic equation. a 2z 2 + 4z + 5 = 0 b 2iz 2 + 4z − 5i = 0 think Write a & b Ensure the calculator is set in Rectangular mode. On a Calculator page, press: • MENU b • 3: Algebra 3 • C: Complex C • 1: Solve 1 Complete the entry lines as: cSolve(2z 2 + 4z + 5 = 0, z) cSolve(2iz2 + 4z − 5i = 0, z) Press ENTER · after each entry.



CHAPTER 2 • Transformations 7 CHAPTER 2 Transformations WORKED EXAMPLE 4 Find the image rule for each of the following, given the original rule and translation. a y = x, T − 2, − 3 b y = 2x 2, T −   4, 5 c y = f (x), T   h, k THINK WRITE a 1 State the image equations. a x = x − 2 y = y − 3 2 Find x and y in terms of x and y. x = x + 2 y = y + 3 3 Substitute into y = x.y = x ⇒ y + 3 = x + 2 ⇒ y = x − 1 4 Express the answer without using the  primes.Given y = x under translation T − 2,  − 3 , the equation  of the image (or image rule) is y = x − 1. 5 Note: The effect of the  transformation can be illustrated on a  CAS calculator. To do this, open a Graphs page. Complete the entry lines as: f 1(x) = x f 2(x) = x − 1 Press ENTER · after each entry. b 1 State the image equations. b x = x − 4 y = y + 5 2 Find x and y in terms of x and y. x = x + 4 y = y − 5 3 Substitute into y = 2x 2.y = 2x 2 ⇒ y − 5 = 2(x + 4) 2 ⇒ y = 2(x + 4) 2 + 5 4 Express the answer without using the  primes. Note: In f rst form of the answer,  the turning point is ( −4, 5), which  was the answer expected as  (0, 0)  ( −4, 5). Given y = 2x 2 under translation T − 4, 5 , the equation  of the image (or image rule) is y = 2(x  + 4) 2 + 5 or y = 2x 2 + 16x + 37 Contents Transformations 7

8 Maths Quest 11 Advanced General Mathematics 5 Note: The effect of the transformation  can be illustrated on a CAS calculator. To do this, open a Graphs page. Complete the entry lines as: f 1(x) = 2x 2 f 2(x) = 2(x + 4) 2 + 5 Press ENTER · after each entry. c 1 State the image equations. c x = x + h y = y + k 2 Find x and y in terms of x and y. x = x − h y = y − k 3 Substitute into y = f (x). y − k = f (x − h) ⇒ y = f (x − h) + k 4 Express the answer without using the  primes. Given y = f (x) under translation T h, k, the equation  of the image (or image rule) is y = f (x − h) + k.

ChapTer 3 • Relations and functions 9 ChapTer 3 Relations and functions Worked example 3 Find the range for the following functions. a f: R + → R, f (x) = 4x − 1 b f: R → R, f (x) = −x2 − 4x + 5 c f: R → R, f (x) = 2 x − 1 Think WriTe a 1 f (x) = 4x − 1 is linear. The domain is x ∈ R + or x ∈ (0, ∞). a When x = 0, f (0) = 4(0) − 1 = −1 2 f (0) = −1, but (0, −1) is not included and therefore this lower end of the range must be represented using a round bracket. State the range. The range: y ∈ ( −1, ∞). b 1 f (x) = −x2 − 4x + 5 is an inverted parabola over the set of real numbers. Use x b a 2 =− to determine the x-value of the turning point, as this can be used to indicate the maximum y-value of the graph. b = = − − − − x (4 ) 2 2 2 Substitute this x-value into f (x) to determine the maximum y-value. f ( −2) = −4 + 8 + 5 = 9 3 State the range. The range: y ∈ ( −∞, 9]. c 1 f (x) = 2 x − 1 is an exponential graph over the set of real numbers. As x → −∞, 2 x → −1, y = −1 is an asymptote. Use a CAS calculator to draw this graph. c 2 Use the graph and the information described above to state the range. The range is: y ∈ ( −1, ∞). Contents Relations and functions 9

10 Maths Quest 11 Advanced General Mathematics Worked example 4 If f: R → R, f (x) = 2x 2 − 4x + 1, find a f (x 2) b f (2x + 1) Think WriTe a To find f (x 2), substitute x 2 for x and simplify. a f (x) = 2x 2 − 4x + 1 f (x 2) = 2(x 2)2 − 4(x 2) + 1 = 2x 4 − 4x 2 + 1 b To find f (2x + 1), substitute 2x + 1 for x and simplify. b f (x) = 2x 2 − 4x + 1 f (2x + 1) = 2(2x + 1) 2 − 4(2x + 1) + 1 = 2[4x 2 + 4x + 1] − 8x − 4 + 1 = 8x 2 + 8x + 2 − 8x − 4 + 1 = 8x2 − 1 Alternatively, on a Calculator page, complete the entry lines as: Define f (x) = 2x 2 − 4x + 1 f (x 2) f (2x + 1) Press ENTER · after each entry. Worked example 6 If f: ( −∞, −1] → R, f (x) = x 2 + 2x + 2, find the domain, range and rule of f − 1(x), and sketch the graphs of f and f − 1 on the same set of axes. Think WriTe 1 First, determine if the inverse function, f −1(x), exists. Since f (x) = x 2 + 2x + 2 is an upright parabola, it is necessary to locate the turning point using = − x b a 2 . ==− − x2 2 1 Since the turning point occurs at x = −1, and the domain is x ∈ ( −∞, −1], f (x) is a 1–1 function and f − 1(x) exists. 2 Determine the range of f (x). The domain is x ∈ ( −∞, −1], so substitute the end value of x to determine the range (this value is at the turning point). x = −1 ⇒ f ( −1) = 1 − 2 + 2 = 1 The point ( −1, 1) is the minimum point on the graph. 3 The domain of f (x) = the range of f − 1(x). The range of f (x) = the domain of f − 1(x). State the domain and range of f − 1(x). Domain f (x): x ∈ ( −∞, −1] ⇒ range of f − 1(x) is y ∈ ( −∞, −1] Range of f (x): y ∈ [1, ∞) ⇒ Domain of f − 1(x) is [1, ∞)

ChapTer 3 • Relations and functions 11 4 To determine the rule of f − 1(x), let f (x) = y and interchange x and y. Then make y the subject. Let y = x 2 + 2x + 2 ⇒ y = (x + 1) 2 + 1 Interchange x and y ⇒ x = (y + 1) 2 + 1 ⇒ x − 1 = (y + 1) 2 ⇒= ±− − yx 11 5 Since f (x) = x 2 + 2x + 2, and x ∈ ( −∞, −1] (left side of the parabola), then f − 1 should be =− − − yx 11 . Fully define the rule for the inverse function. ∞→ =−− −− −fR fxx :[ 1, ) ,() 1111 6 Sketch the graphs over the required domains, showing the line y = x. −1y x −2 −3 −4 −5 1 2 3 5 4 −2 −3 −4 −5 1 0 2 3 4 5 y = x −1 f−1(x) =  −1− f(x) = x 2 + 2x + 2 √x − 1 7 To view the graph and its inverse on a CAS calculator, open a new Graphs page. Complete the function entry lines as: f 1(x) = x 2 + 2x + 2 | x ≤ −1 =− −− fx x 2( )1 1 | x ≥ −1 f 3(x) = x Press ENTER · after each entry.



Chapter 4 • Algebra 13 Chapter 4 Algebra Worked example 9 Transpose each of the following formulas to make the pronumerals indicated in brackets the subject. a A = 4 3π r2 (r) b = − P ab ac d a ( ) c =− mp qr ss() think Write a 1 Write the equation. a A = 4 3 πr2 2 Multiply both sides of the equation by 3. 3 × A = 4 3 πr2 × 3 3A = 4 πr2 3 Divide both sides by 4 π. π π π π = =Ar A r 3 4 4 4 3 4 2 2 4 Take the square root of both sides. Note: From an algebraic point of view we should write ± in front of the root. However, since r represents a physical quantity (radius of a sphere in this case), it can take only positive values. π π = = A r r A 3 4 34 2 b 1 On a Calculator page, press: • MENU b • 3: Algebra 3 • 1: Solve 1 Complete the entry line as: solve(p = (a × b − a × c) ÷ d, a) Then press ENTER ·. b 2 Write the answer. = − a dP bc Note: Capital P should be used in the answer. Contents Algebra 13

14 Maths Quest 11 Advanced General Mathematics c 1 Write the equation. c =−mp qr s 2 The inverse of x is x 2 so square both sides. m 2 = pq − rs 3 Subtract pq from both sides. m 2 − pq = pq − rs − pq m2 − pq = −rs 4 Divide both sides by −r. − = = − − − − − mp q r rs r s mp q r 2 2 5 Multiply the numerator and denominator by −1 (optional). = − s pq m r 2 Worked example 12 Solve for x in 2(x + 5) = 3(2x − 6). think Write 1 On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(2(x + 5) = 3(2x − 6),x) Then press ENTER ·. 2 Write the answer. Solving 2(x + 5) = 3(2x − 6) for x, gives x = 7.

Chapter 4 • Algebra 15 Worked example 13 Find the value of x that will make the following a true statement: + =− xx 2 3 5 2. think Write 1 On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve + =−      xx x 2 3 5 2, Then press ENTER ·. 2 Write the answer. Solving + =−xx 2 3 5 2 for x, gives = x 26 5 or 5 1 5. Worked example 14 Solve the following equation for x: += − xx x 23 2 1 1. think Write 1 On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve += −       xx xx 23 2 1 1, Then press ENTER ·. 2 Write the answer. Solving += − xx x 23 2 1 1 for x, gives = x 7 5 or 1 2 5.

16 Maths Quest 11 Advanced General Mathematics Worked example 15 Solve the following pair of simultaneous equations graphically: a x + 2y = 4 b y + 3x = 17 x − y = 1 2x − 3y = 4 think Write a 1 Rule up a set of axes. Label the origin and the x and y axes. a (See graph at step 7 on page 107.) 2 Find the x-intercept for the equation x + 2y = 4, by making y = 0. x-intercept: when y = 0, x + 2y = 4 x + 2 × 0 = 4 x = 4 The x-intercept is at (4, 0). 3 Find the y-intercept for the equation x + 2y = 4, by making x = 0. Divide both sides of the equation by 2. y-intercept: when x = 0, x + 2y = 4 0 + 2y = 4 2y = 4 y = 2 The y-intercept is at (0, 2). 4 Plot the points on graph paper and join them with the straight line. Label the graph. (Refer to the graph at step 7.) 5 Find the x-intercept for the equation x – y = 1, by making y = 0. x-intercept: when y = 0, x − y = 1 x − 0 = 1 x = 1 The x-intercept is at (1, 0). 6 Find the y-intercept for the equation x − y = 1, by making x = 0. Multiply both sides of the equation by −1. y-intercept: when x = 0, x − y = 1 0 − y = 1 −y = 1 −y × −1 = 1 × −1 y = −1 The y-intercept is at (0, −1). 7 Plot the points on graph paper and join them with the straight line. Label the graph. 1 1 −1 2 24 (2, 1) y x x – y = 1 x + 2 y = 4 0 8 From the graph, read the coordinates of the point of intersection. The point of intersection between the two graphs is (2, 1). 9 Verify the answer by substituting the point of intersection into the original equations.Substitute x = 2 and y = 1 into x + 2y = 4. LHS = 2 + 2 × 1 RHS = 4 = 2 + 2 = 4 LHS = RHS Substitute x = 2 and y = 1 into x − y = 1 LHS = 2 − 1 RHS = 1 = 1 LHS = RHS In both cases LHS = RHS; therefore, the solution set (2, 1) is correct.

Chapter 4 • Algebra 17 b 1 Rearrange both equations to make y the subject. To do this, on a Calculator page, complete the entry lines as: solve(y + 3x = 17, y) solve(2x − 3y = 4, y) Press ENTER · after each entry. b 2 On a Graphs page, complete the function entry lines as: = − fxx 1( ) 2( 2) 3 f 2(x) = 17 − 3x Press ENTER · after each entry. 3 To find the point of intersection, complete the following steps. Press: • MENU b • 8:Geometry 8 • 1:Points & Lines 1 • 3:Intersection Point(s) 3. Click on each line. The point of intersection will appear. ENTER ·. 4 Write the answer. The point of intersection between the two graphs is (5, 2).

18 Maths Quest 11 Advanced General Mathematics Worked example 18 Solve the following simultaneous equations. 2x + 3y = 4 3x + 2y = 10 think Write 1 On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Then complete the entry line as: solve(2x + 3y = 4 and 3x + 2y = 10, x). Then press ENTER · Note: The term ‘and’ can be found in the catalogue k or can be typed. 2 Answer the question. Solution: x = 22 5, y = − 8 5 or −       ,22 5  8 5 .

Chapter 4 • Algebra 19 Worked example 21 A train (denoted as train 1) leaves station A and moves in the direction of station B with an average speed of 60 km/h. Half an hour later another train (denoted as tr\ ain 2) leaves station A and moves in the direction of the first train with an average speed of 70 km/h. Find: a the time needed for the second train to catch up with the first train b the distance of both trains from station A at that time. think Write 1 Define the variables. Note: Since the first train left half an hour earlier, the time taken for it to reach the meeting point will be x + 0.5. Let x = the time taken for train 2 to reach train 1. Therefore, the travelling time, t, for each train is: Train 1: t 1 = x + 0.5 Train 2: t 2 = x 2 Write the speed of each train. Train 1: v 1 = 60 Train 2: v 2 = 70 3 Write the distance travelled by each of the trains from station A to the point of the meeting. (Distance = speed × time.) Train 1: d 1 = 60(x + 0.5) Train 2: d 2 = 70x 4 Equate the two expressions for distance. Note: When the second train catches up with the first train, they are the same distance from station A — that is, d 1 = d 2.When the second train catches up with the first train, d 1 = d 2. 5 Solve the equation. On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Then complete the entry line as: solve(60(x + 0.5) = 70x,x). Then press ENTER ·. 6 Substitute 3 in place of x into either of the two expressions for distance, say into d 2. Substitute x = 3 into d 2 = 70x d 2 = 70 × 3 7 Evaluate. = 210 8 Answer the questions. a The second train will catch up with the first train 3 hours after leaving station A. b Both trains will be 210 km from station A.

20 Maths Quest 11 Advanced General Mathematics Worked example 23 Two hamburgers and a packet of chips cost $8.20, while one hamburger and two packets of chips cost $5.90. Find the cost of a packet of chips and a hamburger. think Write 1 Define the two variables. Let x = the cost of one hamburger. Let y = the cost of a packet of chips. 2 Formulate an equation from the first sentence and call it [1]. Note: One hamburger costs $x, two hamburgers cost $2x. Thus, the total cost of cost of two hamburgers and one packet of chips is 2x + y and it is equal to $8.20. 2x + y = 8.20 [1] 3 Formulate an equation from the second sentence and call it [2]. Note: One packet of chips costs $y, two packets cost $2y. Thus, the total cost of two packets of chips and one hamburger is x + 2y and it is equal to $5.90. x + 2y = 5.90 [2] 4 Solve for the simultaneous equations. On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Then complete the entry line as: solve(x + 2y = 5.90 and 2x + y = 8.20,x). Then press ENTER ·. 5 Answer the question and include appropriate units. A hamburger costs $3.50 and a packet of chips costs $1.20.

Chapter 4 • Algebra 21 Worked example 25 Simplify a − − xx32 1 b + − − axx 2 3 2 3. think Write a & b 1 On a Calculator page, press: • MENU b • 2:Number 2 • 7:Fraction Tools 7 • 4:Common Denominator 4 Complete the entry lines as: comDenom − −       xx32 1 comDenom + − −       axx 2 3 2 3 Press ENTER · after each entry. a & b 2 Write the answer. a b − −= − − xx x xx 32 1 3 2 + − −= −− − − a xx ax xa x 2 3 2 3 22 66 9 2 Worked example 26 Simplify a × x x 3 4 20 9 2 b + × + x y y x 4 6 2 52 0 2 22 . think Write a & b 1 On a Calculator page, complete the entry lines as: × x x 3 4 20 9 2 + × + x y y x 4 6 2 52 0 2 22 Press ENTER · after each entry. a & b 2 Write the answers. a b ×= x xx 3 4 20 9 5 3 2 + × += x y y xy 4 6 2 52 01 15 2 22



Chapter 5 • Trigonometric ratios and their applications 23 Chapter 5 Trigonometric ratios and their applications Worked example 1 Determine the value of the pronumerals, correct to 2 decimal places. a x4 50° b 7 h 24° 25 think Write a 1 Label the sides, relative to the marked angles. a x O 4 H 50° 2 Write what is given. Have: angle and hypotenuse (H) 3 Write what is needed.Need: opposite (O) side 4 Determine which of the trigonometric ratios is required, using SOH–CAH–TOA. sin ( θ ) = O H 5 Substitute the given values into the appropriate ratio. sin (50°) = x 4 6 Transpose the equation and solve for x. 4 × sin (50°) = x x = 4 × sin (50°) 7 Round the answer to 2 decimal places. = 3.06 b 1 Label the sides, relative to the marked angle. b 7A H 24° 25h 2 Write what is given. Have: angle and adjacent (A) side 3 Write what is needed.Need: hypotenuse (H) 4 Determine which of the trigonometric ratios is required, using SOH–CAH–TOA. cos ( θ ) = A H 5 Substitute the given values into the appropriate ratio. cos (24°25′ ) = h7 Contents Trigonometric ratios and their applications 23

24 Maths Quest 11 Advanced General Mathematics 6 Solve for h. On a Calculator page, complete the entry line as: solve  ′=     h h cos (242 5) 7 , Then press ENTER ·. 7 Round the answer to 2 decimal places. h ≈ 7.69 Worked example 2 Find the angle θ, giving the answer in degrees and minutes. think Write 1 Label the sides, relative to the marked angles. 12A O 18 θ 2 Write what is given. Have: opposite (O) and adjacent (A) sides Need: angle 3 Write what is needed. 4 Determine which of the trigonometric ratios is required, using SOH–CAH–TOA.tan ( θ ) = O A 5 Substitute the given values into the appropriate ratio.tan ( θ ° ) = 18 12 6 Transpose the equation and solve for θ, using the inverse tan function. To calculate tan − 1, on a Calculator page, complete the entry line as:          − ta n 18 12 1 ¢ DMS Then press ENTER ·. Note: ¢ DMS is located in the catalogue. 7 Write the answer to the nearest minute. θ ° =       − ta n 18 12 1 = 56° 19′ 12 18 θ

Chapter 5 • Trigonometric ratios and their applications 25 Worked example 8 In the triangle ABC, a = 10 m, c = 6 m and C = 30°. a Show that the ambiguous case exists. b Find two possible values of A, and hence two possible values of B and b. think Write a & b 1 Draw a labelled diagram of the triangle ABC and fill in the given information. a = 10 c = 6 A 30° C B 2 In part a it was shown that the ambiguous case of the sine rule exists. Therefore, on a Calculator page, complete the entry line as: solve =       |≤ ≤ a aa 10 si n () 6 si n (3 0) ,0 180 Then press ENTER ·. 3 Convert the angles to degrees and minutes. A = 56°27′ or A = 123°33′ 4 Calculate the size of the angle B given each angle A. If A = 56°27′, B = 180 − (30 + 56°27′) = 93°33′ If A = 123°33′, B = 180 − (30 + 123°33′) = 26°27′ 5 To find the side length b, on a Calculator page, complete the entry lines as: solve  ′=       b b si n (9 33 3) 6 si n (3 0) , solve  ′=       b b si n (2 62 7) 6 si n (3 0) , Press ENTER · after each entry. 6 Write the answers. If B = 93°33′, b = 11.98 m If B = 26°27′, b = 5.35 m

26 Maths Quest 11 Advanced General Mathematics Worked example 9 Find the third side of triangle ABC given a = 6, c = 10 and B = 76°, correct to 2 decimal places. think Write 1 Draw a labelled diagram of the triangle ABC and fill in the given information. b a = 6 c = 10 A C B 76° 2 Write the appropriate cosine rule to find side b. b 2 = a 2 + c 2 − 2ac cos (B) 3 On a Calculator page, complete the entry line as: solve(b 2 = 6 2 + 10 2 − 2 × 6 × 10 × cos (76),b) Then press ENTER ·. 4 Since b represents the side length of a triangle, then b > 0. b = 10.34, correct to 2 decimal places. Worked example 10 Find the smallest angle in the triangle with sides 4 cm, 7 cm and 9 cm. think Write 1 Draw a labelled diagram of the triangle ABC and fill in the given information. a = 4 c = 7 b = 9 A C B 2 Write the appropriate cosine rule to find the angle A. a 2 = b 2 + c 2 − 2bc cos (A) 3 On a Calculator page, complete the entry line as: solve(4 2 = 9 2 + 7 2 − 4 × 9 × 7 × cos (a),a) |0 ≤ a ≤ 180 Then press ENTER ·. 4 Round the answer to degrees and minutes. A = 25.2088° = 25°13′

Chapter 6 • Sequences and series 27 Chapter 6 Sequences and series Worked example 1 a Find the next three terms in the sequence, b: {14, 7, 7 2, . . .}. b Find the 4th, 8th and 12th terms in the following sequence: e n = n 2 − 3n, n ∈ {1, 2, 3, . . .}. c Find the 2nd, 3rd and 5th terms for the following sequence: k n + 1 = 2k n + 1, k 1 = −0.50. think Write a 1 In this example the sequence is listed and a simple pattern is evident. From inspection, the next term is half the previous term and so the sequence would be 14, 7, 7 2, 7 4, 7 8, 7 16. a The next three terms are 7 4, 7 8, 7 16. 2 On a Calculator page, complete the entry lines as: 14 Ans × 0.5 Press ENTER · repeatedly to generate the sequence. b 1 This is an example of a functional defi nition. The nth term of the sequence is found simply by substitution into the expression e n = n 2 − 3n. b en = n 2 − 3n 2 Find the 4th term by substituting n = 4. e 4 = 4 2 − 3 × 4 = 4 3 Find the 8th term by substituting n = 8. e 8 = 8 2 − 3 × 8 = 40 4 Find the 12th term by substituting n = 12. e 12 = 12 2 − 3 × 12 = 108 Contents Sequences and series 27

28 Maths Quest 11 Advanced General Mathematics 5 On a Calculator page, press • MENU b • 6: Statistics 6 • 4: List Operations 4 • 5: Sequence 5 Complete the entry line as: seq(e(n) = n 2 − 3n,n,4,12,4) Then press ENTER ·. Note: The fi rst 4 and the 12 are the highest and lowest values required, and the last 4 is the step value. An alternative method to the one above for generating a sequence is shown below. Insert a new Lists and Spreadsheets page. Go to the title cell of column A and press: • MENU b • 3: Data 3 • 1: Generate Sequence 1 Complete the sequence fi elds as shown. Then press OK. 6 Scroll down to fi nd the required terms. c 1 This is an example of an iterative defi nition. We can fi nd the 2nd, 3rd and 5th terms for the sequence k n + 1 = 2k n +1, k 1 = −0.50 by iteration. c kn + 1 = 2k n + 1, k 1 = −0.50 2 Substitute k 1 = −0.50 into the formula to fi nd k 2. k 2 = 2 × −0.50 + 1 = 0 3 Continue the process until the value of k 5 is found. k 3 = 2 × 0 + 1 = 1 k 4 = 2 × 1 + 1 = 3 k 5 = 2 × 3 + 1 = 7

Chapter 6 • Sequences and series 29 4 Write the answer. Thus k 2 = 0, k 3 = 1 and k 5 = 7. 5 Follow the same steps as for the alternative method in part b; however, because the formula is defined as u(n), the rule needs to be entered as: u(n) = 2 × u(n − 1) + 1 with the initial term as: −0.5 Press OK to view the sequence.

30 Maths Quest 11 Advanced General Mathematics Worked example 2 Given that a = 2 and t 0 = 0.7, use the logistic equation to generate a sequence of 6 terms, and state whether the sequence is convergent, divergent or oscillating. If the sequence is convergent, state its limit. think Write 1 Insert a new Lists & Spreadsheet page. Go to the title cell of column A and press: • MENU b • 3: Data 3 • 1: Generate Sequence 1 Complete the sequence fields as shown. Then press OK. 2 Scroll down to find the required terms. Note: Here the initial term corresponds to cell A1; however, we need to remember that this is actually t 0. Hence it is the sixth term when the sequence first converges to 0.5, not the fifth term.

Chapter 6 • Sequences and series 31 Worked example 6 Find the 16th and nth terms in an arithmetic sequence with the 4th term 15 and 8th term 37.\ think Write 1 Write the two equations that represent t 4 and t 8. t 4: a + 3d = 15 [1] t 8: a + 7d = 37 [2] 2 To solve equations [1] and [2] simultaneously, open a new Calculator page and press: • MENU b • 3: Algebra 3 • 1: Solve 1 Complete the entry line as: solve(a + 3d = 15 and a + 7d = 37,a) Then press ENTER ·. 3 Write the answer. If =− a3 2 and = d 11 2 , = − t n 11 14 2n t16 = 81

32 Maths Quest 11 Advanced General Mathematics Worked example 7 Find the sum of the first 20 terms in the sequence t n: {12, 25, 38, . . .}. think Write 1 On a Lists & Spreadsheet page, place the cursor in the grey header cell of column A. Then press: • MENU b • 3: Data 3 • 1: Generate sequence 1 Complete the sequence fields as shown. 2 Scroll down to the 20th term. 3 Write the answer. S 20 = 2710

Chapter 6 • Sequences and series 33 Worked example 10 The fi fth term in a geometric sequence is 14 and the seventh term is 0.56. Find the common ratio, r, the fi rst term, a, and the nth term for the sequence. think Write 1 Write the general rule for the nth term of the geometric sequence. t n = ar n − 1 2 Use the information about the 5th term to form an equation. Label it [1].When n = 5, t n = 14 14 = a × r 5 − 1 14 = a × r 4 [1] 3 Similarly, use information about the 7th term to form an equation. Label it [2]. When n = 7, t n = 0.56 0.56 = a × r 7 − 1 0.56 = a × r 6 [2] 4 Solve equations simultaneously: Divide equation [2] by equation [1] to eliminate a. [2 ] [1] gives = ar ar 0.56 14 6 4 5 Solve for r. r 2 = 0.04 r0.04 =± = ± 0.2 6 As there are two solutions, we must perform two sets of computations. Consider the positive value of r fi rst. Substitute the value of r into either of the two equations, e.g. [1], and solve for a. If r = 0.2 Substitute r into [1]: a × (0.2) 4 = 14 0.0016a = 14 a = 8750 7 Substitute the values of r and a into the general equation to fi nd the expression for the nth term. The nth term is: t n = 8750 × (0.2) n − 1 8 Now consider the negative value of r. If r = −0.2 9 Substitute the value of r into either of the two equations, e.g. [1], and solve for a. (Note that the value of a is the same for both values of r.)Substitute r into [1] a = ( −0.2) 4 = 14 0.0016a = 14 a = 8750 10 Substitute the values of r and a into the general formula to fi nd the second expression for the nth term. The nth term is: t n = 8750 × ( −0.2) n − 1 11 Write the two equations that represent t 5 and t 7. t 5: 14 = a × r 4 [1] t 7: 0.56 = a × r 6 [2]

34 Maths Quest 11 Advanced General Mathematics 12 To solve equations [1] and [2] simultaneously, open a new Calculator page and press: • MENU b • 3: Algebra 3 • 1: Solve 1 Complete the entry line as: solve(14 = a × r 4 and 0.56 = a × r 6,r) Then press ENTER ·. 13 Write the answer. When r = −0.2 and a = 8750, t n = 8750 × ( −0.2) n − 1 When r = 0.2 and a = 8750, t n = 8750 × (0.2) n − 1

Chapter 6 • Sequences and series 35 Worked example 11 Find the sum of the fi rst 5 terms (S 5) of these geometric sequences. a tn: {1, 4, 16, . . .} b tn = 2(2) n − 1 , n ∈ {1, 2, 3, . . .} c tn + 1 = 1 4 tn, t1 = 1 2 − think Write a 1 On a Lists & Spreadsheet page, place the cursor in the grey header cell of column A. Then press: • MENU b • 3: Data 3 • 1: Generate sequence 1 Complete the sequence fi elds as shown. a 2 To sum the terms of any sequence, place the cursor in cell B1 and type = 3 Then press Catalogue and scroll down to sum(. Then press ENTER ·.

36 Maths Quest 11 Advanced General Mathematics 4 To sum the fi rst 5 terms, complete the entry line as: sum(a1:a5) Then press ENTER ·. 5 The answer will appear in cell B1. 6 Write the answer. If t n: {1, 4, 16, . . .} then, S 5 = 341. b 1 Write the sequence. b tn = 2(2) n − 1 , n ∈ {1, 2, 3, . . .} 2 Compare the given rule with the general formula for the nth term of the geometric sequence t n = ar n − 1 and identify values of a and r; the value of n is known from the question. a = 2; r = 2; n = 5 3 Substitute values of a, r and n into the general formula for the sum and evaluate. = − − = − = S 2( 21 ) 21 2(32 1) 1 62 5 5 c 1 Write the sequence. c tn + 1 = 14tn, t1 = 1 2− 2 This is an iterative formula, so the coeffi cient of t n is our r; a = t 1; n is known from the question. r = 1 4; a = −1 2; n = 5 3 Substitute values of a, r and n into the general formula for the sum and evaluate. = − − = ×− = −            −       − − S 1 1 15 1 2 1 4 1 4 1 2 1 1024 3 4 341 512 5

Chapter 7 • Variation 37 Chapter 7 Variation Worked example 1 For the given data, establish whether direct variation exists between x and y using: a a numerical approach (clearly specify k, the constant of variation, if applicable) and b a graphical approach. c Confi rm your result using a CAS calculator. x47810 y 5.2 9.110.4 13 y x think Write/draW a 1 Find the ratio y x for each of the four pairs of values. One variable varies directly as the other if the ratio between any two corresponding values is constant. a Ratio = y x First pair: = 5.2 4 1.3 Second pair: = 9.1 7 1.3 Third pair: = 10.4 8 1.3 Fourth pair: = 13 10 1.3 2 Compare each of the four ratios and answer the question. Since all four ratios are the same (that is, 1.3), y varies directly as x. 3 Copy and complete the table. x 47810 y 5.2 9.110.4 13 y x1.3 1.31.31.3 b 1 2 Plot the information from the table onto a set of axes. Join the given points and see if a straight line is obtained. Note: For a direct variation to exist between two variables x and y, a straight line passing through the origin (0, 0) must be obtained. b 13 10.4 0 10 9.1 5.2 87 4 x y The given points fi t perfectly on a straight line. If the straight line is extended it will pass through the origin. Therefore y varies directly as x. Contents Variation 37

38 Maths Quest 11 Advanced General Mathematics 3 Calculate the gradient using any two points on the straight line. Answer the question. Note: The gradient of the straight line will equal k, the constant of variation, if direct variation exists between the variables x and y. Let (x 1, y1) = (4, 5.2) and let (x 2, y2) = (10, 13) = − − m yy xx21 21 = − − 13 5.2 10 4 =7.8 6 = 1.3 The gradient of the straight line is equal to k, the constant of variation. c 1 On a Lists & Spreadsheet page enter the data into the spreadsheet. Label the columns x and y. c 2 To draw the scatterplot of the data, on a Data & Statistics page: Tab to each axis to select ‘Click to add variable’. Place x on the horizontal axis and y on the vertical axis. 3 To check there is a linear relationship, press: • MENU b • 4: Analyze 4 • 6: Regression 6 • 1: Show linear (mx + b) 1 In X List select x; in Y List select y. Interpret the graph; if all points are on the line, then it confi rms that the relationship is one of direct variation. The relationship is one of direct variation.

Chapter 7 • Variation 39 Worked example 5 For the given data, establish the rule relating the variables x and y then graph the relationship using a CAS calculator. x 36710 y 28.8 115.2 156.8 320 think Write/display 1 On a Calculator page, complete the entry lines as: {3,6,7,10} → x {28.8,115.2,156.8,320} → y Press ENTER · after each entry. 2 Find the rule that fits the data. To do this press: MENU b • 6: Statistics 6 • 1: Stat Calculations 1 • 9: Power Regression 9 Enter x in X List and y in Y List Then press ENTER ·. 3 Write down the rule for x and y. y = 3.2x 2 4 Graph the rule. On a Graphs page, use the NavPad and the up arrow ` to display f 1(x), which has already been defined in this document. Then press ENTER · to display the graph.

40 Maths Quest 11 Advanced General Mathematics Worked example 8 For the data represented in the table below, establish whether an inverse variation exists between x and y using: a a numerical approach (clearly specify k, the constant of variation, if applicable) b a graphical approach. x 12345 y 20 10 62 3 54 xy think Write/display a 1 Find the product of xy for each of the 5 pairs of values. Note: One variable varies inversely as the other if the product between any 2 corresponding values is constant. a Product = xy First pair: 1 × 20 = 20 Second pair: 2 × 10 = 20 Third pair: 3 × 623 = 20 Fourth pair: 4 × 5 = 20 Fifth pair: 5 × 4 = 20 2 Compare each of the five products and answer the question. Since the product of the corresponding values is the same in each case (that is 20), y varies inversely as x. 3 Copy and complete the table. x 12345 y 20 10 623 54 xy 20 20202020 b 1 Calculate the values of x 1. Place these values into a table. b x 12345 x 1 1 1 2 13 14 15 y 20 10 623 54 2 On a Calculator page, enter the x 1 and y values by completing the entry lines as: {1, 1 2, 13, 1 4, 15 } → x {20, 10, 62 3, 5, 4} → y Press ENTER · after each entry.

Chapter 7 • Variation 41 3 To find the rule that fits the data, press: • MENU b • 6: Statistics 6 • 1: Stat Calculations 1 • 3: Linear Regression (mx + b) 3 Enter x in X List and y in Y List Then press ENTER ·. 4 To graph the rule, open a Graphs page. Complete the entry line as f 2(x) = f 1(x) Then press ENTER · to view the graph. Adjust the window if necessary. 5 Answer the question. The graph of y against x 1 is a straight line directed from, but not passing through, the origin, hence an open circle at the point (0, 0). Therefore y ∝ x 1.

42 Maths Quest 11 Advanced General Mathematics Worked example 16 My telephone bill consists of 2 parts: a fixed charge of $32 (paid whether any calls are made or not) and a charge proportional to the number of calls made. Last quarter I made 296 calls an\ d my bill was $106. a Find the equation of variation. b Find the amount to be paid when 300 calls are made. think Write/display a 1 Define each variable to be used. a Let A = the total amount to be paid, in dollars Let n = the number of calls 2 Write the equation of variation. A = k n + 32 3 Substitute the values for A and n into the equation.When n = 296 and A = 106, 106 = 296k + 32 4 To solve the equation for k, on a Calculator page, press: • MENU b • 3: Algebra 3 • 1: Solve 1 Complete the entry line as: solve(106 = 296 × k + 32,k) Then press ENTER ·. 5 Rewrite the equation substituting 1 4 in place of k. So A = 1 4n + 32 b 1 Substitute n = 300 into the given equation. b When n = 300, A = 1 4(300) + 32 2 Evaluate. = 1 4 × 300 + 32 = 75 + 32 = 107 3 Answer the question and include the appropriate unit. The amount to be paid when 300 calls are made is $107.

Chapter 7 • Variation 43 Worked example 18 The following table shows the values of the total surface area, TSA, of spheres and their corresponding radii, r. Radius (r) (cm) 12345 TSA (cm 2) 12.5750.27113.1201.06 314.16 Graph the values given in the table and comment on the shape of the graph. Using the graph, \ or otherwise, fi nd the equation which relates total surface area of the sphere, TSA, and its radius r. think Write/display 1 On a Lists & Spreadsheet page, enter values and label in column A: radius: 1, 2, 3, 4, 5 and in column B: tsa 12.57, 50.27, 113.1, 201.06, 314.16 2 Draw the scatterplot of the data on a Data & Statistics page. Tab to each axis to select ‘Click to add variable’. Place radius on the horizontal axis and tsa on the vertical axis. To check the power relationship, press: • MENU b • 4: Analyse 4 • 6: Regression 6 • 7: Show power 7 In X List select x; in Y List select y 3 Comment on the graph obtained. The graph is not a straight line, passing through the origin, so direct variation does not exist between the two variables. Hence, there is no direct variation between the radius and the total surface area of the sphere. 4 Make assumptions about the graph obtained.The graph resembles a parabola, so it is reasonable to assume that area is directly proportional to the square of the radius. 5 Write the variation statement for the assumption made. TSA ∝ r 2 6 Write the variation equation. TSA = kr 2 7 Transpose the equation to make k the subject. k = r TS A2

44 Maths Quest 11 Advanced General Mathematics 8 Test the assumption by fi nding the values of r 2 and check that the ratio r TS A2 is constant. On the Lists & Spreadsheet page, label column C: ratio In the header (grey) cell, complete the entry line as: =tsa/(radius)^2 Then press ENTER ·. 9 Comment on the result obtained. The ratio is constant for each corresponding pair (when rounded to 2 decimal places). Hence, TSA ∝ r 2 TSA = kr 2 TSA = 12.57r 2 10 Alternatively, once the values of r 2 have been calculated, rule up a table of values and plot TSA versus r 2. r2 1 4916 25 TSA 12.57 50.27113.1201.06 314.16 11 Comment on the graph obtained. 300 200 100 01 49 16 25 r2 TSA The graph is a straight line, passing through the origin. 12 Establish the value of k by substituting any pair of values from the table into the equation of variation and write the equation relating the two variables. TSA = kr 2 When r = 1, TSA = 12.57, k = ? 12.57 = k × 1 12.57 = k k = 12.57 TSA = 12.57r 2

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