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MATHS QUEST 12 TI-NSPIRE CAS CALCULATOR COMPANION Specialist Mathematics 4TH EDITION



4TH EDITION VCE MATHEMATICS UNITS 3 & 4 RAYMOND ROZEN | PAULINE HOLLAND | BRIAN HODGSON HOWARD LISTON | JENNIFER NOLAN | GEOFF PHILLIPS MATHS QUEST 12 TI-NSPIRE CAS CALCULATOR COMPANION Specialist 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 31811 9 978 1 118 31809 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 Illustrated by Aptara and Wiley Composition Services 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 Coordinate geometry 1 CHAPTER 2 Circular functions 15 CHAPTER 3 Complex numbers 21 CHAPTER 4 Relations and regions of the complex plane 27 CHAPTER 5 Differential calculus 29 CHAPTER 6 Integral calculus 37 CHAPTER 7 Differential equations 47 CHAPTER 8 Kinematics 53 CHAPTER 9 Vectors 63 CHAPTER 10 Vector calculus 71 CHAPTER 11 Mechanics 79

Introduction This booklet is designed as a companion to Maths Quest 12 Specialist Mathematics Fourth 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 12 Specialist Mathematics Fourth Edition. vi Introduction

Chapter 1 • Coordinate geometry 1 Chapter 1 Coordinate geometry Worked example 4 Sketch the graph of yx x2 3 2 = + including all asymptotes and intercepts. think Write/display 1 On a Graphs page, complete the entry line as: fx x x 1( ) 2 3 2 = + then press ENTER ·. 2 To determine the equation of the oblique asymptote, divide y x x2 3 2 = + into two functions. To do this on a Calculator screen press: • MENU b • 2: Number 2 • 7: Fraction Tools 7 • 1: Proper Fraction 1 Complete the entry line as: propfrac(f1(x)) then press ENTER ·. 3 Write the equations of the asymptotes. The equations of the asymptotes are: x = 0 and yx 1 3 = .

2 Maths Quest 12 Specialist Mathematics 4 To determine the turning points, press: • MENU b • 3: Algebra 3 • 1: Solve 1 • MENU b • 4: Calculus 4 • 1: Derivative 1 Complete the entry line as: solve d dx fx x (1 ())0 , =       then press ENTER ·. 5 To fi nd the y -coordinates of the stationary points by substitution, complete the entry lines as: f 1 2 ) ( f12 ) (− Press ENTER · after each entry. Describe the nature and coordinates of the stationary points, as deduced from the graph. Solving d dx x x2 3 0 2+       = for x gives xx 2 or 2 . == − The coordinates of the stationary points are: Local minimum       2, 22 3 Local maximum 2,  22 3       − − 6 Sketch the graph of y x x2 32 = + . y x 0 −5 5 10 x = 0 (Asymptote) −5 510 −10 −10 y = x (Asymptote)1 – 3

Chapter 1 • Coordinate geometry 3 Worked example 12 Sketch the graph of xy (1 ) 25 (2 ) 9 1 22 − + − = . think Write/display 1 Compare xy (1 ) 25 (2 ) 9 1 22 − + − = with xh a yk b () () 1 2 2 2 2 − +− =. h = 1, k = 2 and so the centre is (1, 2). a 2 = 25 b 2 = 9 a = 5 b = 3 2 The major axis is parallel to the x-axis as a> b. 3 The extreme points (vertices) parallel to the x-axis for the ellipse are: ( −a + h, k) (a + h, k) Vertices are: ( −5 + 1, 2) (5 + 1, 2) = (−4, 2) = (6, 2) 4 The extreme points (vertices) parallel to the y-axis for the ellipse are: (h, −b + k) (h, b + k) and (1, −3 + 2) (1, 3 + 2) = (1, −1) = (1, 5) 5 Find the x- and y-intercepts. On a Calculator page, complete the entry lines as: solve − +− =      = xy xy (1 ) 25 (2 ) 9 1, 0 22 solve − +− =      = xy yx (1 ) 25 (2 ) 9 1, 0 22 Press ENTER · after each entry. The x-intercepts are: xx 355 3 , 35 5 3 = − =+ The y-intercepts are: yy 10 66 5 , 10 66 5 = − =+ 6 To sketch the graph of the ellipse, on a Graphs page press: • MENU b • 3: Graph Entry/Edit 3 • 2: Equation 2 • 4: Ellipse 4 • 1: xh a yk b 1 2 2 2 2 () () − +− = 1 Complete as shown: x y 1 5 2 3 1 2 2 2 2 ) ( ) ( − +− = Press ENTER · after the entry, the graph is shown.

4 Maths Quest 12 Specialist Mathematics 7 Sketch the graph of the ellipse. y x 0 − 2 2 4 6 −4 −6 −2 6 2 4 (6, 2) (1, 2) ( −4, 2) (1, 5) (1, −1) = 1 + (x − 1) 2 –––––– 25 (y − 2) 2 –––––– 9 10 + 6√6––––––– 50, 10 − 6√6––––––– 50, 3 − 5√5–––––– 3 , 0 3 + 5√5–––––– 3 , 0

Chapter 1 • Coordinate geometry 5 Worked example 13 Sketch the graph of xy (2 ) 9 (4 ) 16 1 22 − + + = . think Write/display 1 Compare xy (2 ) 9 (4 ) 16 1 22 − + + = with xh a yk b () () 1 2 2 2 2 − + − =. h = 2, k = −4 So the centre is (2, −4). a 2 = 9 b 2 = 16 a = 3 b = 4 2 The major axis is parallel to the y-axis as b> a. 3 The extreme points (vertices) parallel to the x-axis for the ellipse are: ( −a + h, k) (a + h, k) Vertices are: ( −3 + 2, −4) (3 + 2, −4) = ( −1, −4) = (5, −4) 4 The extreme points (vertices) parallel to the y-axis for the ellipse are: (h, −b + k) (h, b + k) and (2, −4 − 4) (2, 4 − 4) = (2, −8) = (2, 0) 5 Find the x- and y-intercepts. On a Calculator page, complete the entry lines as: solve xy xy (2 ) 9 (4 ) 16 1, 0 22 − + + =       = solve xy yx (2 ) 9 (4 ) 16 1, 0 22 − ++ =      = Press ENTER · after each entry. The x-intercept is x = 2. The y-intercepts are: yy 12 45 3 , 12 5 3 . = − =+−− 6 To sketch the graph of the ellipse, on a Graphs page press: • MENU b • 3: Graph Entry/Edit 3 • 2: Equation 2 • 4: Ellipse 4 • 1: Complete the entry line as: x y 2 3 4 4 1 2 2 2 2 () () − +− = − then press ENTER ·. Note that the viewing window has been changed.

6 Maths Quest 12 Specialist Mathematics 7 Sketch the graph of the ellipse. x 0 −1 6 2 1 35 y −4 −8 −102 (2, −4) = 1 + (x − 2) 2 –––––– 9 (y + 4) 2 –––––– 16 (−1, −4) (5, −4) (2, −8) 4 (2, 0) −2 −6 −12 + 4√5 –––––––– 30, −12 − 4√5–––––––– 30,

Chapter 1 • Coordinate geometry 7 Worked example 14 Sketch the graph of 5 x 2 + 9(y − 2) 2 = 45. think Write/display 1 Rearrange and simplify by dividing both sides by 45 to make the RHS = 1. 5x 2 + 9(y − 2) 2 = 45 2 Simplify by cancelling. xy 5 45 9( 2) 45 45 4522+ − = xy 9 (2 ) 5 1 22+ − = 3 Compare xy 9 (2 ) 5 1 22+ − = with xh a yk b () () 1 2 2 2 2 − +− =. h = 0, k = 2 and so the centre is (0, 2). a 2 = 9 b 2 = 5 as a, b > 0 a = 3 b = 5 4 Major axis is parallel to the x-axis as a > b. 5 The extreme points (vertices) parallel to the x-axis for the ellipse are: ( −a + h, k) (a + h, k) Vertices are: ( −3 + 0, 2) (3 + 0, 2) = (−3, 2) = (3, 2) 6 The extreme points (vertices) parallel to the y-axis for the ellipse are: (h, −b + k) (h, b + k) and (0, 5− + 2) (0, 5 + 2) or (0, 2 − 5 ) (0, 2 + 5) ≈ (0, −0.24) ≈ (0, 4.24) 7 Find the x-intercepts. On a Calculator page, complete the entry line as: solve +−== xy xy (5 9(2) 45,) |0 22 then press ENTER ·. xx 35 5 , 35 5 ==− 8 To sketch the graph of the ellipse, on a Graphs page press: • MENU b • 3: Graph Entry/Edit 3 • 2: Equation 2 • 6: Conic 6 • 1: Complete the entry line as: xx yy xy 50 90 36 90 22++++ += −− then press ENTER ·.

8 Maths Quest 12 Specialist Mathematics 9 Sketch the graph of the ellipse. (0, 2 − ) 5 y x 0 − 2 2 4 6 −2 −3 −4 −1 34 1 2 (3, 2) (0, 2) ( −3, 2) 5x2 + 9(y − 2) 2 = 45 (0, 2 + ) 5 −3√5 –––– 5 , 0 3√5 –––– 5 , 0

Chapter 1 • Coordinate geometry 9 Worked example 15 Sketch the graph of the relation described by the rule: 25x 2 + 150x + 4y 2 − 8y + 129 = 0. think Write/display 1 To locate the intercepts, on a Calculator page, complete the entry lines as: solve xxyy x y 25 150 48 1290, |0 22 () ++ −+= = solve xx yy y x 25150 48 1290, |0 22 () ++ −+= = Make a record of the intercepts. 2 To sketch the graph of the ellipse, on a Graphs page press: • MENU b • 3: Graph Entry/Edit 3 • 2: Equation 2 • 6: Conic 6 • 1: Complete the entry line as: xx yy xy 25 04 081290 22++ +++= − then press ENTER ·. 3 Write the x-intercepts. x15 46 5 = − − x 15 46 5 = + − 4 Sketch the graph of the ellipse. 0 6 −3 −1 −5 (−5, 1) (−3, 1) (−3, 6) (−3, −4) (−1, 1) y x −4 −15 + 4√6 –––––––– 5 , 0 −15 − 4√6 –––––––– 5 , 0

10 Maths Quest 12 Specialist Mathematics Worked example 16 Determine the Cartesian equation of the curve with parametric equations x = 2 + 3 sin (t) and y = 1 − 2 cos (t ) where t ∈ R. Describe the graph and state its domain and range. think Write/display 1 Use a CAS calculator to sketch the graph in a Graphs page, in parametric mode, by completing the entry line as: =+ =−     xt t yt t 1( )2 3sin( ) 1( )1 2cos( ) Then press ENTER ·. 2 Rewrite the parameters by isolating cos (t) and sin (t). x 2 3 − = sin (t) and y1 2 − − = cos (t) 3 Square both sides of each equation then add. xy (2 ) 9 (1) 4 22 − + − = sin 2 (t) + cos 2 (t) = 1 4 Describe the relation. This represents an ellipse with centre (2, 1). 5 The domain is the range of the parametric equation x = 2 + 3 sin (t).Domain is [2 − 3, 2 + 3] = [ −1, 5] 6 The range is the range of the parametric equation y = 1 – 2 cos (t).Range is [1 − 2, 1 + 2] = [ −1, 3]

Chapter 1 • Coordinate geometry 11 Worked example 25 Express each of the following as partial fractions. a xx xx 51 05 2 (2 )(4) 2+− −+ b xx x xx 25 37 2 32 2−+ + −− think Write/display a & b 1 On a Calculator page, press: • MENU b • 3: Algebra 3 • 3: Expand 3 Complete the entry lines as: xx xx xx x xx expand 51 052 (2 )( 4) expand 25 37 2 2 32 2+− −× +       −+ + −−       Press ENTER · after each entry. 2 Write the answers. a b xx xx xx xx x xx xxx 51 052 (2 )( 4)2 4 2 2 5 25 37 2 1 1 3 2 23 2 32 2+− −× += +− −+ −+ + −− = ++ − +−

12 Maths Quest 12 Specialist Mathematics Worked example 27 Sketch the graph of the function y xx x56 42 == −+ − . think Write/display 1 Use a CAS calculator to express the rational function as partial fractions by completing the following steps. Press: • MENU b • 2: Number 2 • 7: Fraction Tools 7 • 1: Proper Fraction 1 Complete the entry line as: xx x pr opF rac 56 4 2−+ −       then press ENTER ·. 2 Express the function as partial fractions. y x x 2 4 1 = − +− 3 Sketch the graphs of y 1 = x − 1 (asymptote) and y 2 = x 2 4 − on the same axes. y x 0 3 −1 x = 4 4 2 y 1 = x − 1 y 2 = 2 —— x − 4 1 4 Determine any x-intercepts. y = 0, x 2 − 5x + 6 = 0 (x − 2)(x − 3) = 0 ⇒ x = 2 and x = 3 5 Determine the y-intercept. x = 0, y 6 4 =− y 3 2 =− 6 Add the two graphs by addition of ordinates to obtain the graph of y xx x 56 4 2 = −+ − . 3 – 2 + y 1 = x − 1 y = x − 1 y 2 = 2 —— x − 4 2 —— x − 4 x = 4 y x (2, 0) 1 −1 2 3 4 (3, 0) (0, − ) 0

Chapter 1 • Coordinate geometry 13 7 Open a Graphs page, and complete the entry lines as; fxx fx x fx fx fx 11 2 2 4 31 2() () () () () =− = − =+ then press ENTER ·.



ChapTer 2 • Circular functions 15 ChapTer 2 Circular functions Worked example 2 If cosec (x) = 43 and, 0 ≤ x ≤ 90°, fi nd x (to the nearest tenth of a degree). Think WriTe 1 express the equation cosec (x) = 4 3 in terms of sin (x). x x co sec() 1 sin () 4 3 == 2 On a Calculator page, press: • Menu b • 3: Algebra 3 • 1: Solve 1 Complete the entry line as x xx so lve 1 si n( ) 4 3 ,| 09 0 =       ≤≤ Then press enTeR ·. Alternatively, the three reciprocal functions are built into the TI-nspire. They can be accessed by the µ key, or through the catalogue, or you can simply use the letter keys and enter csc, sec or cot as needed. 3 Write the solution. Solving cosec (x) = 43 for xx[0,90] , 48.5904 ∈° =° 4 Round off the answer to 1 decimal place. x = 48.6°

16 Maths Q uest 12 Specialist Mathematics Worked example 12 Solve cosec (x) = 1.8 over the interval 0 ≤ x ≤ 4 π. Give your answer(s) correct to 2 decimal places. Think WriTe 1 On a Calculator page, press: • Menu b • 3: Algebra 3 • 1: Solve 1 Complete the entry line as solve (csc(x) = 1.8, x) | 0 ≤ x ≤ 4 π Then press enTeR ·. 2 Write the solution. Solving cosec (x) = 1.8 over the interval 0 ≤ x ≤ 4 π gives 3 Round the answers to 2 decimal places.x = 0.59, 2.55, 6.87, 8.84

ChapTer 2 • Circular functions 17 Worked example 13 a Expand, and simplify where possible, each of the following. i sin (x − 2y) ii cos (x + 30°) b Simplify the expression sin (2x) cos (y) + cos (2x) sin (y). Think WriTe a i & ii On a Calculator page, press: • Menu b • 3: Algebra 3 • B: Trigonometry B • 1: expand 1 Complete then entry lines as shown, then press enTeR ·. a i & ii b 1 Write the appropriate compound- angle formula. b sin (A) cos (B) + cos (A) sin (B) = sin (A + B) 2 Substitute A = 2x and B = y to reveal the answer. sin (2x) cos (y) + cos (2x) sin (y) = sin (2x + y) 3 Alternatively, on a Calculator page, press: • Menu b • 3: Algebra 3 • B: Trigonometry B • 2: Collect 2 Complete the entry line as tCollect(sin(2x)cos(y) + cos(2x)sin(y)) Then press enTeR ·. Write the solution. sin (2x) cos (y) + cos (2x) sin (y) = sin (2x + y)

18 Maths Q uest 12 Specialist Mathematics Worked example 15 Simplify: a sin (270 − C °) b sec 2 π θ−    . Think WriTe a & b 1 On a Calculator page, press: • Menu b • 3: Algebra 3 • 3: expand 3 Complete the entry lines as: expand(sin (270 − c), c) expand se c 2 , π θθ−           Press enTeR · after each entry. Note: The calculator should be in degree mode for the first expansion above, and radian mode for the second. a & b 2 express the answer as a reciprocal function. s ec 2 cose c() π θθ−    =

ChapTer 2 • Circular functions 19 Worked example 16 Find the exact value of π      cot 5 12. Think WriTe 1 express 5 12π as the sum of 4 π and 6 π. cot 5 12 cot 46 ππ π      =+     2 express cot in terms of its reciprocal, 1 ta n. 1 ta n 46 ππ = +     3 use the appropriate compound-angle formula to expand the denominator. 1 ta n 4 ta n 6 1t an 4 t an  ππ ππ =    +    −    6               4 express in simplest fraction form. 1t an  4 ta n  6 ta n 4 ta n 6 ππ ππ = −           +    5 Simplify. 1( 1) 1 1 3 1 3 = − +       1 1 1 3 1 3 = − + 31 3 31 3 = −       +       31 31 = − + 6 Rationalise the denominator. (3 1)( 31 ) (3 1)( 31 ) = −− +− 7 Simplify. 32 31 31 = −+ − 42 3 2 =− 23 =−

20 Maths Q uest 12 Specialist Mathematics Note: It is possible to check the answer using a calculator. On a Calculator page, complete the entry line as: π      co t 5 12 then press enTeR ·.

ChapTer 3 • Complex numbers 21 ChapTer 3 Complex numbers Worked example 1 Using the imaginary number i, write down an expression for: a 16− b 5− . Think WriTe a & b 1 Change the document settings to Rectangular mode. To do this, press: • HOME c • 5: Settings 5 • 2: Settings 2 • 2: Document Settings 2 Tab down to Real or Complex and select Rectangular. a & b 2 On a Calculator page, complete the entry lines as: 16− 5− Press ENTER · after each entry.

22 Maths Q uest 12 Specialist Mathematics Worked example 4 Simplify z = i 4 − 2i 2 + 1 and w = i 6 − 3i 4 + 3i 2 − 1. Think WriTe 1 On a Calculator page, complete the entry lines as: i 4 − 2i 2 + 1 i 6 − 3i 4 + 3i 2 − 1 Press ENTER · after each entry. 2 Write the answer. z = 4 w = −8 Worked example 5 Evaluate each of the following. a Re(7 + 6i) b Im(10) c Re(2 + i − 3i 3) d iii Im 13 2 23 −− −       Think WriTe a ,b , c & d 1 On a Calculator page, press: • MENu b • 2: Number 2 • 9: Complex Number Tools 9 • 2: Real Part 2 or • 3: Imaginary Part 3 Complete the entry lines as: real(7 + 6i) imag(10) real(2 + i − 3i 2) ii i ima g13 2 23 −− −       Press ENTER · after each entry . 2 Write the answers. a Re(7 + 6i) = 7 b Im(10) = 0 c Re(2 + i − 3i 3) = 2 d ii i Im 13 2 1 23 −− −       = −

ChapTer 3 • Complex numbers 23 Worked example 10 Determine Re(z 2w) + Im(zw 2) for z = 4 + i and w = 3 − i. Think WriTe 1 On a Calculator page, complete the entry lines as: Define z = 4 + i Define w = 3 – i Press ENTER · after each entry. Then press: • MENu b • 2: Number 2 • 9: Complex Number Tools 9 • 2: Real Part 2 Complete the entry line as: real(z 2 × w) + imag(z × w 2) Then press ENTER ·. Note: The imaginary part can be found in the same menu as the real part. 2 Write the answer. Re(z 2 w) + Im(zw 2) = 37 Worked example 12 Write down the conjugate of each of the following complex numbers. a 8 + 5i b −2 − 3i Think WriTe a & b 1 On a Calculator page, press: • MENu b • 2: Number 2 • 9: Complex Number Tools 9 • 1: Complex Conjugate 1 Complete the entry lines as: conj (8 + 5i) conj ( −2 − 3i) Press ENTER · after each entry. 2 Write the answers. a 8 − 5i b −2 + 3i

24 Maths Q uest 12 Specialist Mathematics Worked example 16 If z = a + bi, find a and b such that − −=− z z i 51 5 1 43 . Think WriTe 1 On a Calculator page, press: • MENu b • 3: Algebra 3 • C: Complex C • 1: Solve 1 Complete the entry line as: cSolve z z iz51 5 143 , − −=−       Then press ENTER ·. 2 a is the real part of z, b is the imaginary part. a = 2, b = −3 Worked example 17 Find the modulus of the complex number z = 8 − 6i. Think WriTe 1 On a Calculator page, press: • MENu b • 2: Number 2 • 9: Complex Number Tools 9 • 5: Magnitude 5 Complete the entry line as: |8 − 6i| Then press ENTER ·. 2 Write the answer. | z | = |8 − 6i| = 10

ChapTer 3 • Complex numbers 25 Worked example 23 Express each of the following in polar form, r cis ( θ ), where θ = arg(z), −π < θ ≤ π. a z = 1 + i b =−zi 13 Think WriTe a & b 1 On a Calculator page, complete the entry line as: 1 + i Then press: • MENu b • 2: Number 2 • 9: Complex Number Tools 9 • 6: Convert to Polar 6 Then press ENTER ·. a & b 2 Write the answer. For a, i 12 cis 4 π +=      3 u se the relationship re iθ = r cos ( θ ) + ir sin ( θ ) to express the answer in the required form. The calculator always gives θ in principle valued form. Key in i 13 − and repeat the above procedure. 4 Write the answer. For b, ie 13 2 i 3 −= π− 2 cis 3 π =     −

26 Maths Q uest 12 Specialist Mathematics Worked example 36 a If f (z) = z 3 + 7z 2 + 16z + 10, find all factors of f (z) over C. b Factorise P(z) = z 3 − (3 −i)z 2 + 2z − 6 + 2i. Think WriTe a & b 1 On a Calculator page, press: • MENu b • 3: Algebra 3 • C: Complex C • 2: Factor 2 Complete the entry lines as cFactor (z 3 + 7z 2 + 16z + 10, z) cFactor (z 3 − (3 − i)z 2 + 2z − 6 + 2i, z) Press ENTER · after each entry. a & b 2 Write the answers in the required form. For a, the three factors of P(z) are (z + 1), (z + 3 − i) and (z + 3 + i) For b, =− ++ − Pz zi zi zi ()(3 )(2) (2 )

CHAPTER 4 • Relations and regions of the complex plane 27 CHAPTER 4 Relations and regions of the complex plane WORKED EXAMPLE 16 Express each of the following expressions in Cartesian form. a Re(z + 5) b Im(z − 2 − 3i) c | z − 4 + 2i | THINK WRITE a, b & c 1 On a Calculator page, complete the  entry lines as: Def ne  z = x + yi Then press ENTER ·. To answer part  a press: • MENU b • 2: Number 2 • 9: Complex Number Tools 9 • 2: Real Part 2 or • 3: Imaginary Part 3 or • 5: Magnitude 5 Complete the entry line as: real(z + 5) imag(z − 2 − 3i) | z − 4 + 2i  | Press ENTER · after each entry. a , b & c 2 Write the answers. For  a, Re(z + 5) = x + 5. For  b, Im(z − 2 − 3i) = y − 3. For  c, |− +|=− +++ zi xxyy 42 8 42 0 22 .



ChapTer 5 • Differential calculus 29 ChapTer 5 Differential calculus Worked example 1 Differentiate the following expressions with respect to x. a y = tan (6 x) b y x 2tan 4 3 =       Think WriTe a & b 1 On a Calculator page, press: • Menu b • 4: Calculus 4 • 1: Derivative 1 Complete the entry lines as: d dx x (t an ( 6) ) d dx x 2 tan 4 3            Press enTeR · after each entry. a & b 2 Write the solutions. For a, d dxx x [ ta n(6 )] 6 [cos (6) ] 2 = For b, d dx x x 2 tan 4 3 8 3c os 4 3 2             =         

30 Maths Q uest 12 Specialist Mathematics Worked example 4 Find the equation of the tangent to the curve y = 3x + cos (2x) + tan (x) where x 4.π= Think WriTe 1 On a Calculator page, press: • Menu b • 4: Calculus 4 • 9: Tangent Line 9 Complete the entry lines as: tangentLine xx xx 3c os(2) tan () ,, 4 π ++     Then press enTeR ·. 2 Write the solution. equation of the tangent is y = 3x + 1

ChapTer 5 • Differential calculus 31 Worked example 6 Find, using calculus, f ″( x) if f ( x) is equal to: a ecos (2x) + log e (x) b x x si n() . Think WriTe a & b 1 On a Calculator page, complete the entry line as: f (x) : = e cos (2x) + ln(x) Then press enTeR ·. Note: The syntax used here is another way of defi ning a function or variable. You can use the Defi ne or Store methods if you prefer. a & b 2 Complete the entry line as: d dx fx (( ))2 2 Then press enTeR ·. 3 Write the solution. The second derivative, f''x xxe x ()[4sin (2 )4 cos(2) ] 1 x 2 co s( 2) 2 =− − 4 On a Calculator page, complete the entry line as: fxx x () :si n () = Complete the entry line as: d dx fx (( )) 2 2 Press enTeR · after each entry.

32 Maths Q uest 12 Specialist Mathematics 5 Write the solution. The second derivative, fx xx x x x () 3 4 1 sin ( ) co s( ) 5 2 3 2 ′′ =−        − 6 You may rearrange the answer to a form similar to that given in the solution obtained manually as follows. Press: • Menu b • 3: Algebra 3 • 2: Factor 2 Then select and paste the previous answer to obtain the entry line: factor x x x x x 3 4 1 sin ( ) cos () 5 2 3 2−         −         Then press enTeR ·. 7 Write the solution. The second derivative: fxxx xx x () (3 4) sin( )4 cos( ) 4 2 5 2 ′′ = −− Worked example 16 Find the equation of the normal to the curve with equation: y x 2 co s 2 1 =     − at the point where x 3 = . Think WriTe 1 On a calculator page, press: • Menu b • 4: Calculus 4 • A: normalLine A Complete the entry line as: normalLine x x 2c os 2,, 3 1         − Then press enTeR ·. 2 Write your solution in an appropriate form. The equation of the normal is y x 2 3 23 .π =− +

ChapTer 5 • Differential calculus 33 Worked example 18 Find the antiderivative for each of the following expressions: a x 1 25 2 − b x 3 49 2 − − c x 20 16 2 + . Think WriTe a , b & c 1 On a Calculator page, press: • Menu b • 4: Calculus 4 • 3: Integral 3 Complete the entry lines as: x dx 1 25 2 ∫ −       x dx 3 49 2 ∫ −       − x dx 20 16 2 ∫ +       Press enTeR · after each entry. Note: The calculator finds the second form of the antiderivative in part b. Also, it does not include the constant. You will have to do that yourself. 2 Write your solutions, remembering to include the constant of integration. a b c x dx x c 1 25 s in 52 1 ∫ − =    + − x dx x c 3 49 3 sin 72 1 ∫ − =    + − −− x dx x c 20 16 5 tan 4 2 1 ∫ + =    + −

34 Maths Q uest 12 Specialist Mathematics Worked example 21 Differentiate the equation y 2 + 3x 2 = 4 to find dy dx in terms of x. Think WriTe 1 On a Calculator page, press: • Menu b • 4: Calculus 4 • e: Implicit Differentiation E Complete the entry line as: impDif (y 2 + 3x 2 = 4, x, y) Then press enTeR ·. 2 Substitute for y as in part a (which is preferable in this straightforward equation) or continue to use the calculator to make y the subject in the equation. Press: • Menu b • 3: Algebra 3 • 1: Solve 1 Complete the entry line as: Solve (y 2 + 3x 2 = 4, y) Then press enTeR ·. 3 express the domain, 3x 2 − 4 ≤ 0 shown in the screen in a more appropriate form. Take care to change ≤ to < as y is in the denominator in the solution. x 23 3 23 3

ChapTer 5 • Differential calculus 35 Worked example 23 For x 2y2 + y = 2, find the gradient of the tangent, dy dx , at the point (1, −2). Hence, determine the equation of the tangent at this point. Think WriTe 1 The gradient of the implicit function can be found using a CAS calculator. To do this, on a Calculator page, press: • Menu b • 4: Calculus 4 • e: Implicit Differentiation E Complete the entry line as: impDif (x 2 × y 2 + y = 2, x,y) | x = 1 and y = −2 Then press enTeR ·. 2 The gradient of the tangent at (1, −2) is 8 3. use this information and the general equation of a straight line to determine the equation of the tangent. yy mxx yx y x y x () 28 3 (1 ) 2 8 3 8 3 8 3 8 3 2 11 −= − −= − += − =− −− yx 8 3 14 3=−



ChapTer 6 • Integral calculus 37 ChapTer 6 Integral calculus Worked example 1 Find the antiderivative of the following expressions. a (x + 3) 7 b 4x(2 x 2 + 1) 4 c x xx 312 3 + + Think WriTe a & b 1 On a Calculator page, press: • Menu b • 4: Calculus 4 • 3: Integral 3 Complete the entry lines as: xd x xx dx (3 ) (4 (2 1)) 7 24 ∫ ∫ +×+ x xx dx 312 3 ∫ + +       Press enTeR · after each entry. Note: The calculator cannot determine the solution for part c. It simply returns the input you entered. You will have to do this problem manually. Also, it does not include the constant for any of the antiderivatives. ensure you include the constant with your answers. a & b 2 Write your solutions, remembering to include the constant of integration. xd xx c xx dxx c (3 ) (3 ) 8 4( 21 ) (2 1) 5 7 8 24 25 ∫ ∫ += + + += + + c 1 Recognise that 3x 2 + 1 is the derivative of x 3 + x. Let u = x 3 + x. c Let u = x 3 + x. 2 Find du dx . dudx x31 2 =+

38 Maths Q uest 12 Specialist Mathematics Worked example 2 Antidifferentiate the following functions with respect to x. a fx x xx () () 3 6 23 = + + b fx xx x () () cos () 13 23 =− − Think WriTe a 1 On a Calculator page, press: • Menu b • 4: Calculus 4 • 3: Integral 3 Complete the entry line as: x xx dx 3 (6 ) 23 ∫ + + Then press enTeR ·. The answer is in an equivalent but more complex form than the solution found manually. a 2 Collect the terms in factorised form over a common denominator as follows: Press: • Menu b • 3: Algebra 3 • 2: Factor 2 Complete the entry line as factor x xx x (9 ) 432 (6 ) 1 432 1 144 22 + + +−       − Then press enTeR ·. 3 Write your solution, remembering to include the constant of integration. x xx dx xx c 3 (6 ) 1 4( 6) 23 22 ∫ + + =− + + b 1 express in integral notation. b xx xd x (1 )cos( 3) 23 ∫ −− 2 Recognise that x 2 − 1 is a multiple of the derivative of 3x − x 3. 3 Let u = 3x − x 3. Let u = 3x − x 3. 4 Find du dx . dudx x 33 2 =− = 3(1 − x 2) = −3(x 2 − 1)

ChapTer 6 • Integral calculus 39 5 Substitute u for 3x − x 3 and xdu dx(1 ) 1 3 2−= − . So xx xd x (1 )cos( 3) 22 ∫ −− u du dx dx ud u dx dx co s 1 3 co s 3 ∫ ∫=× = − − = udu cos 3 ∫ − = ud u 1 3 co s∫ − 6 Antidifferentiate with respect to u. = u c si n 3 + − 7 Replace u with 3x − x 3. = xx c si n(3) 3 3 − + −

40 Maths Q uest 12 Specialist Mathematics Worked example 7 Find the antiderivative of the following expressions. a sin 2 x 2     b 2 cos 2 x 4     Think WriTe a 1 express in integral notation. a x dx si n 2 2 ∫     2 use identity 1 to change sin 2 x 2    . = xd x [1 cos() ] 1 2∫ − 3 Take the factor of 1 2 to the front of the integral. = xd x [1 cos()] 1 2∫ − 4 Antidifferentiate by rule. = xx c [s in( )] 1 2 −+ 5 Simplify the answer. = x xc 2 1 2 si n () −+ 6 If you use a calculator for problems such as these, you may find the answer expressed in a form that is different from the ones above. The screen dump shows the result, in the first line, of part a done using a CAS calculator. The compact form shown in the second line can be obtained as follows. Press: • Menu b • 3: Algebra 3 • B: Trigonometry B • 2: Collect 2 Complete the entry line as: tCollect xx x 2 si n 2co s 2 −            Then press enTeR ·. 7 Write your solution, remembering to include the constant of integration. x dx xx c si n 2 si n( ) 2 2 ∫     = − + b 1 express in integral notation. b x dx 2 co s 4 2 ∫     2 use identity 2 to change cos 2 x 4    . = x dx 21 cos 2 1 2 ∫ () +          3 Simplify the integral. = x dx 1c os 2 ∫ +          4 Antidifferentiate by rule. = x x c 2 sin 2 +    +

ChapTer 6 • Integral calculus 41 Worked example 12 For each of the following rational expressions: i express as partial fractions ii antidifferentiate the result. a x xx 7 23 () () + +− b x xx 23 342 − −− Think WriTe a i 1 express the rational expression as two separate fractions with denominators (x + 2) and (x − 3) respectively. a i x xx a x b x 7 (2 )(3)(2 )( 3) + +− = + + − 2 express the partial fractions with the original common denominator. = ax bx xx (3)( 2) (2)(3) −+ + +− 3 equate the numerator on the left-hand side with the right- hand side. so x + 7 = a(x − 3) + b(x + 2) 4 Let x = −2 so that a can be evaluated. Let x = −2, and thus 5 = −5a 5 Solve for a. a = −1 6 Let x = 3 so that b can be evaluated.Let x = 3, and thus 10 = 5b 7 Solve for b. b= 2 8 Rewrite the rational expression as partial fractions. Therefore x xx xx 7 (2 )(3) 1 2 2 3 + +− = + + − − 9 A CAS calculator can convert expressions in partial fraction form as follows. Press: • Menu b • 3: Algebra 3 • 3: expand 3 Complete the entry line as: expand x xx 7 (2 )(3) ++−       Then press enTeR ·. 10 Write the answer. x xx xx 7 (2 )(3) 2 3 1 2 + +− = −− + ii 1 express the integral in partial fraction form. ii x xx dx 7 (2 )(3)∫ + +− = xx dx 1 2 2 3 ∫ + + −       − 2 Antidifferentiate by rule. = −log e (| x + 2 |) + 2 log e (| x − 3 |) + c; x ∈R\{ −2, 3}

42 Maths Q uest 12 Specialist Mathematics 3 Simplify using log laws. = log e x x(3 ) 22 − |+ |       + c; x ∈R\{ −2, 3} b i 1 Factorise the denominator. b i x xx 23 34 2 − −− = x xx 23 (4 )(1) − −+ 2 express the partial fractions with denominators (x − 4) and (x + 1) respectively. = a x b x 41− + + 3 express the right-hand side with the original common denominator. = ax bx xx (1 )( 4) (4 )(1) ++ − −+ 4 equate the numerators. So 2x − 3 = a(x + 1) + b(x − 4) 5 Let x = 4 to evaluate a.Let x = 4, 5 = 5a 6 Solve for a. a = 1 7 Let x = −1 to evaluate b.Let x = −1, −5 = −5b 8 Solve for b. b = 1 9 Rewrite the rational expression as partial fractions.Therefore x xx 23 34 2 − −− = xx 1 4 1 1 − + + ii 1 express the integral in its partial fraction form. ii x xx dx 23 34 2 ∫ − −− = xx dx 1 4 1 1 ∫ − + +       2 Antidifferentiate by rule. = log e (| x − 4 |) + log e (| x + 1|) + c 3 Simplify using log laws. = log e (| x − 4 |)(| x + 1|) + c; x ∈R\{ −1, 4} or log e (| x 2 − 3x − 4 |) + c; x ∈R\{ −1, 4}

ChapTer 6 • Integral calculus 43 Worked example 16 Evaluate the following defi nite integrals. a x xx dx 2 54 2 0 2∫ − ++ b xx dx cos () 1sin( ) 0 2 ∫ + π Think WriTe a 1 Write the integral. a x xx dx 2 54 2 0 2 ∫ − ++ 2 Factorise the denominator of the integrand. Consider: x xx 2 54 2 − ++ = x xx 2 (1 )( 4) − ++ 3 express in partial fraction form with denominators x + 1 and x + 4. = a x b x 14 ++ + 4 express the partial fractions with the original common denominator. = ax bx xx (4)( 1) 54 2 ++ + ++ 5 equate the numerators. x − 2 = a(x + 4) + b(x + 1) 6 Let x = −1 to fi nd a.Let x = −1, −3 = 3a a = −1 7 Let x = −4 to fi nd b. Let x = −4, −6 = −3b b = 2 8 Rewrite the integral in partial fraction form. So x xx dx 2 54 2 0 2∫ − ++ = xx dx 1 1 2 402∫ + + + − 9 Antidifferentiate the integrand. = xx [log (1 )2log (4 )] ee 0 2|+|+ |+| − 10 evaluate the integral. = [ −log e (3) + 2 log e (6)] − [ −log e (1) + 2 log e (4)] = −log e (3) + 2 log e (6) − 2 log e (4) 11 Simplify using log laws. = 2 log e (1.5) − log e (3) = log e (2.25) − log e (3) = log e (0.75) (or approx. −2.88) 12 You may attempt this problem using a CAS calculator. On a Calculator page, press • Menu b • 4: Calculus 4 • 3: Integral 3 Complete the entry line as: ∫ − ++       x xx dx 2 (5 4)2 0 2 Then press enTeR ·. Pressing Ctrl / enTeR · will give an approximate answer. 13 Write the solution. ∫ − ++ =      ≈ − x xx dx 2 (5 4)lo g 3 4 2.88 e 2 0 2

44 Maths Q uest 12 Specialist Mathematics b 1 Write the integral. b xx dx cos () 1sin( ) 02∫ + π 2 Let u = 1 + sin (x) to antidifferentiate. Let u = 1 + sin (x) 3 Find du dx . du dx x cos () = 4 xd x du dx dx cos () = xdx du dxdx cos () = 5 Change terminals by fi nding u when x = 0 and x = 2 π. When x = 0, u = 1 + sin (0) = 1 When x 2 π=, u = 1 + sin 2 π    = 1 + 1 = 2 6 Simplify the integrand. So xx dx cos () 1sin( ) 0 2 ∫ + π = udu dx dx 1 2 1 2 ∫ = ud u 12 1 2 ∫ 7 Antidifferentiate the integrand. = u2 33 2 1 2       8 evaluate the integral. = 21 23 3 22 3 3 2 ×− × = 42 3 2 3 − or 42 2 3 − 9 using CAS, press • Menu b • 4: Calculus 4 • 3: Integral 3 Complete the entry line as: xx dx cos () 1sin( ) 02∫ + π Then press enTeR ·.

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