maths specialist cas

2ND

ED

ITIO

N

MATHS QUEST 11TI-NSPIRE CAS CALCULATOR COMPANION

Advanced General Mathematics

2ND

ED

ITIO

N

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 11TI-NSPIRE CAS CALCULATOR COMPANION

Advanced General Mathematics

First published 2013 byJohn Wiley & Sons Australia, Ltd42 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)

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10 9 8 7 6 5 4 3 2 1

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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 decisions 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 ab

, 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. 0.2323

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. 0.413.7

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. 2.12342121.3

99921213

99992357

1111

=

=

=

ContentsNumber 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 2Complete the entry line as: 0.6666666666666666666666approxFraction(5E-14)Then press ENTER .

Worked example 13

Expand and simplify the following where possible.

a 7( 18 3) b 2 3( 10 5 3) c ( 5 3 6)(2 3 2)+

think Write

a ,b& c

1 On a Calculator page, complete the entry lines as:

7( 18 3)

2 3( 10 5 3)

( 5 3 6)(2 3 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( 2 1) 7

30 2 30

(2 3 2) 5 6 3 18 2 +


Worked example 16

If z1 = 2 3i and z2 = 3 + 4i, finda z1 + z2 b z1 z2 c 3z1 4z2.

think Write

a ,b&c

1 On a Calculator page, press: MENU b 1: Actions 1 1: Define 1Complete the entry lines as:Define z1 = 2 3iDefine z2 = 3 + 4iz1 + z2z1 z23z1 4z2Press ENTER after each entry.

2 Write the answers. abc

z1 + z2 = 1 + iz1 z2 = 5 7i3z1 4z2 = 18 25i

Worked example 17

Simplifya 2i(2 3i) b (2 3i)(3 + 4i).

think Write

a Expand the brackets. a 2i(2 3i) = 4i 6i2= 6 + 4i

b Expand the brackets as for binomial expansion and simplify.

b (2 3i)(3 + 4i) = 6 + 8i + 9i 12i2= 6 + 17i

Alternatively, on a Calculator page, complete the entry lines as:2i (2 3i)(2 3i) (3 + 4i)Press ENTER after each entry.


Worked example 18

If z1 = 2 + 3i and z2 = 4 5i, find

a z z1 2+ b z z1 2+ c z z1 2 d z z1 2 .

think Write

On a Calculator page, complete the entry lines as: Define z1 = 2 + 3iDefine z2 = 4 5iconj(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 i4

2

b i

i3

3 4

c (3 2i)1 d ii

2 32

+

think Write

a , b ,c & d

On a Calculator page, complete the entry lines as:

i4

2

i

i

3 4

3

(3 2i)1i

i

2 3

2

+

Press ENTER after each entry.


Worked example 21

Factorise each of the following quadratic expressions over C.a 2z2 + 6z b 2z2 6 c 2z2 + 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 2Complete the entry lines as:

cFactor(2z2 + 6z, z)cFactor(2z2 6, z)cFactor(2z2 + 3, z)


Worked example 23

Solve the following using the formula for the solution of a quadratic equation.a 2z2 + 4z + 5 = 0 b 2iz2 + 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 1Complete the entry lines as:

cSolve(2z2 + 4z + 5 = 0, z)cSolve(2iz2 + 4z 5i = 0, z)


CHAPTER 2 Transformations 7

CHAPTER 2

TransformationsWORKED EXAMPLE 4

Find the image rule for each of the following, given the original rule and translation.a y = x, T2, 3 b y = 2x2, T 4, 5 c y = f (x), T h, k

THINK WRITE

a 1 State the image equations. a x = x 2y = y 3

2 Find x and y in terms of x and y. x = x + 2y = 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 T2, 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) = xf 2(x) = x 1Press ENTER after each entry.

b 1 State the image equations. b x = x 4y = y + 5

2 Find x and y in terms of x and y. x = x + 4y = y 5

3 Substitute into y = 2x2. y = 2x2 y 5 = 2(x + 4)2 y = 2(x + 4)2 + 5

4 Express the answer without using the primes.Note: In rst form of the answer, the turning point is (4, 5), which was the answer expected as (0, 0) (4, 5).

Given y = 2x2 under translation T4, 5, the equation of the image (or image rule) isy = 2(x + 4)2 + 5or y = 2x2 + 16x + 37

ContentsTransformations 7


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) = 2x2f 2(x) = 2(x + 4)2 + 5Press ENTER after each entry.

c 1 State the image equations. c x = x + hy = y + k

2 Find x and y in terms of x and y. x = x hy = 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 Th, 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 functionsWorked example 3

Find the range for the following functions.a f: R+ R, f (x) = 4x 1b f: R R, f (x) = x2 4x + 5c f: R R, f (x) = 2x 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 xb

a2=

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)

22

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) = 2x 1 is an exponential

graph over the set of real numbers. As x , 2x 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, ).

ContentsRelations and functions 9


Worked example 4

If f: R R, f (x) = 2x2 4x + 1, finda f (x2) b f (2x + 1)

Think WriTe

a To find f (x2), substitute x2 for x and simplify.

a f (x) = 2x2 4x + 1f (x2) = 2(x2)2 4(x2) + 1

= 2x4 4x2 + 1b To find f (2x + 1), substitute 2x + 1 for

x and simplify.b f (x) = 2x2 4x + 1

f (2x + 1) = 2(2x + 1)2 4(2x + 1) + 1= 2[4x2 + 4x + 1] 8x 4 + 1= 8x2 + 8x + 2 8x 4 + 1= 8x2 1

Alternatively, on a Calculator page, complete the entry lines as:Define f (x) = 2x2 4x + 1f (x2)f (2x + 1)Press ENTER after each entry.

Worked example 6

If f: (, 1] R, f (x) = x2 + 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) = x2 + 2x + 2 is an upright parabola, it is necessary to

locate the turning point using =

xb

a2.

= =

x2

21

Since the turning point occurs at x = 1, and the domain is x (, 1], f (x) is a 11 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

= 1The 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 = x2 + 2x + 2 y = (x + 1)2 + 1Interchange x and y x = (y + 1)2 + 1 x 1 = (y + 1)2

= y x 1 1

5 Since f (x) = x2 + 2x + 2, and x (, 1] (left side of the parabola), then f 1 should be =

y x 1 1. Fully define the rule for the inverse function.

=

f R f x x: [1, ) , ( ) 1 11 1

6 Sketch the graphs over the required domains, showing the line y = x.

1

y

x

2345

123

54

2345 10 2 3 4 5

y = x

1f1(x) = 1

f(x) = x2 + 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) = x2 + 2x + 2 | x 1

= f x x2( ) 1 1 | x 1

f 3(x) = xPress ENTER after each entry.

Chapter 4 Algebra 13

Chapter 4

AlgebraWorked example 9

Transpose each of the following formulas to make the pronumerals indicated in brackets the subject.

a A = 43 r2 (r) b =

Pab ac

da( ) c = m pq rs s( )

think Write

a 1 Write the equation. a A = 43 r2

2 Multiply both sides of the equation by 3.

3 A = 43 r2 3

3A = 4r2

3 Divide both sides by 4.pi

pi

pi

pi

=

=

A r

Ar

3

4

4

43

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.

pi

pi

=

=

Ar

rA

3

4

3

4

2

b 1 On a Calculator page, press: MENU b 3: Algebra 3 1: Solve 1Complete the entry line as:solve(p = (a b a c) d, a)Then press ENTER .

b

2 Write the answer. =

adP

b c

Note: Capital P should be used in the answer.

ContentsAlgebra 13


c 1 Write the equation. c = m pq rs

2 The inverse of x is x2 so square both sides.

m2 = pq rs

3 Subtract pq from both sides. m2 pq = pq rs pqm2 pq = rs

4 Divide both sides by r. =

=

m pq

r

rs

r

sm pq

r

2

2

5 Multiply the numerator and denominator by 1 (optional).

=

spq 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 1Complete 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.


Worked example 13

Find the value of x that will make the following a true statement: +

=

x x23

52

.

think Write

1 On a Calculator page, press: MENU b 3:Algebra 3 1:Solve 1Complete the entry line as:

solve+

=

x xx

2

35

2,

Then press ENTER .

2 Write the answer. Solving +

=

x x2

35

2 for x, gives =x

26

5 or 5

1

5.

Worked example 14

Solve the following equation for x: + =x x x

2 32

11

.

think Write

1 On a Calculator page, press: MENU b 3:Algebra 3 1:Solve 1Complete the entry line as:

solve + =

x x x x2 3

2

1

1,

Then press ENTER .

2 Write the answer. Solving + =x x x

2 3

2

1

1 for x, gives =x

7

5 or 1

2

5.


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 = 4The 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 = 2The 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 = 1The 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 = 1The 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

2 4

(2, 1)

y

x

x y = 1

x + 2y = 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 = 4LHS = RHSSubstitute x = 2 and y = 1 into x y = 1 LHS = 2 1 RHS = 1 = 1LHS = RHSIn both cases LHS = RHS; therefore, the solution set (2, 1) is correct.


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:

=

f xx

1( )2( 2)

3f 2(x) = 17 3xPress 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).


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 1Then 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 = 225 , y = 8

5 or ,225 8

5 .


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 train 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 trainb 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: t1 = x + 0.5Train 2: t2 = x

2 Write the speed of each train. Train 1: v1 = 60Train 2: v2 = 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: d1 = 60(x + 0.5)Train 2: d2 = 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, d1 = d2.

When the second train catches up with the first train, d1 = d2.

5 Solve the equation. On a Calculator page, press: MENU b 3:Algebra 3 1:Solve 1Then 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 d2.

Substitute x = 3 into d2 = 70x d2 = 70 3

7 Evaluate. = 2108 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.


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 1Then 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.


Worked example 25

Simplify

a x x

3 21

b +

ax x2

323

.

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

x x3 2

1

comDenom+

a

x x

2

3

2

3


a &b

2 Write the answer. a

b

=

x x

x

x x

3 2

1

32

+

=

a

x x

ax x a

x

2

3

2

3

2 2 6 6

92

Worked example 26

Simplify

a x

x34

209

2

b +

+

xy

yx

46

25 20

2

2 2 .

think Write

a&b


x

x

3

4

20

9

2

+

+

x

y

y

x

4

6

2

5 20

2

2 2


a&b


b

=x

x

x3

4

20

9

5

3

2

+

+=

x

y

y

x y

4

6

2

5 20

1

15

2

2 2

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

4H

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 SOHCAHTOA.

sin ( ) = OH

5 Substitute the given values into the appropriate ratio.

sin (50) = x4

6 Transpose the equation and solve for x.

4 sin (50) = xx = 4 sin (50)

7 Round the answer to 2 decimal places. = 3.06b 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)


cos ( ) = AH


cos (2425 ) = h

7

ContentsTrigonometric ratios and their applications 23


6 Solve for h. On a Calculator page, complete the entry line as:

solve = h hcos(24 25 )7,

Then press ENTER .

7 Round the answer to 2 decimal places.

h 7.69

Worked example 2

Find the angle , giving the answer in degrees and minutes.

think Write

1 Label the sides, relative to the marked angles.

12

A

O 18

2 Write what is given. Have: opposite (O) and adjacent (A) sidesNeed: angle3 Write what is needed.


tan ( ) = OA


tan ( ) = 18

12

6 Transpose the equation and solve for , using the inverse tan function.To calculate tan1, on a Calculator page, complete the entry line as:

tan

18

121 DMS

Then press ENTER .Note: DMS is located in the catalogue.

7 Write the answer to the nearest minute. = tan 18

121

= 56 19

12

18

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

A30

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 a a

10

sin( )

6

sin(30), 0 180

Then press ENTER .

3 Convert the angles to degrees and minutes.

A = 5627 or A = 12333

4 Calculate the size of the angle B given each angle A.

If A = 5627, B = 180 (30 + 5627) = 9333If A = 12333, B = 180 (30 + 12333) = 2627

5 To find the side length b, on a Calculator page, complete the entry lines as:

solve

=

bb

sin(93 33 )

6

sin(30),

solve

=

bb

sin(26 27 )

6

sin(30),


6 Write the answers. If B = 9333, b = 11.98 mIf B = 2627, b = 5.35 m


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 = 6c = 10

A C

B

76

2 Write the appropriate cosine rule to find side b.

b2 = a2 + c2 2ac cos (B)

3 On a Calculator page, complete the entry line as:solve(b2 = 62 + 102 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 = 4c = 7

b = 9A C

B

2 Write the appropriate cosine rule to find the angle A.

a2 = b2 + c2 2bc cos (A)

3 On a Calculator page, complete the entry line as:solve(42 = 92 + 72 4 9 7 cos (a),a) |0 a 180Then press ENTER .

4 Round the answer to degrees and minutes.

A = 25.2088= 2513

Chapter 6 Sequences and series 27

Chapter 6

Sequences and seriesWorked example 1

a Find the next three terms in the sequence, b: {14, 7, 72, . . .}.

b Find the 4th, 8th and 12th terms in the following sequence: en = n2 3n, n {1, 2, 3, . . .}.c Find the 2nd, 3rd and 5th terms for the following sequence: kn + 1 = 2kn + 1, k1 = 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:14Ans 0.5Press 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 en = n2 3n.

b en = n2 3n

2 Find the 4th term by substituting n = 4.

e4 = 42 3 4= 4


e8 = 82 3 8 = 40


e12 = 122 3 12= 108

ContentsSequences and series 27


5 On a Calculator page, press MENU b 6: Statistics 6 4: List Operations 4 5: Sequence 5Complete the entry line as:seq(e(n) = n2 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 sequencekn + 1 = 2kn +1, k1 = 0.50 by iteration.

c kn + 1 = 2kn + 1, k1 = 0.50

2 Substitute k1 = 0.50 into the formula to fi nd k2.

k2 = 2 0.50 + 1 = 0

3 Continue the process until the value of k5 is found.

k3 = 2 0 + 1 = 1 k4 = 2 1 + 1 = 3 k5 = 2 3 + 1 = 7


4 Write the answer. Thus k2 = 0, k3 = 1 and k5 = 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) + 1with the initial term as:0.5Press OK to view the sequence.


Worked example 2

Given that a = 2 and t0 = 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 1Complete 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 t0. Hence it is the sixth term when the sequence first converges to 0.5, not the fifth term.


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 t4 and t8.

t4: a + 3d = 15 [1]t8: 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 1Complete 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,

=

tn11 14

2n

t16 = 81


Worked example 7

Find the sum of the first 20 terms in the sequence tn: {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 1Complete the sequence fields as shown.

2 Scroll down to the 20th term.

3 Write the answer. S20 = 2710


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.

tn = arn 1

2 Use the information about the 5th term to form an equation. Label it [1].

When n = 5, tn = 1414 = a r 5 114 = a r4 [1]

3 Similarly, use information about the 7th term to form an equation. Label it [2].

When n = 7, tn = 0.560.56 = a r 7 10.56 = a r6 [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. r2 = 0.04r 0.04=

= 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 = 0.2Substitute 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: tn = 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:tn = 8750 (0.2)n 1

11 Write the two equations that representt5 and t7.

t5: 14 = a r4 [1]t7: 0.56 = a r6 [2]


12 To solve equations [1] and [2] simultaneously, open a new Calculator page and press: MENU b 3: Algebra 3 1: Solve 1Complete the entry line as:solve(14 = a r4 and 0.56 = a r6,r)Then press ENTER .

13 Write the answer. When r = 0.2 and a = 8750,tn = 8750 (0.2)n 1

When r = 0.2 and a = 8750,tn = 8750 (0.2)n 1


Worked example 11

Find the sum of the fi rst 5 terms (S5) of these geometric sequences.

a tn: {1, 4, 16, . . .} b tn = 2(2)n 1, n {1, 2, 3, . . .} c tn + 1 = 14 tn, t1 =

12

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 1Complete 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 .


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 tn: {1, 4, 16, . . .} then, S5 = 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 tn = 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.

=

=

=

S2(2 1)

2 12(32 1)

162

5

5

c 1 Write the sequence. c tn + 1 = 1

4tn, t1 =

12

2 This is an iterative formula, so the coeffi cient of tn is our r; a = t1; n is known from the question.

r = 14 ; a = 1

2 ; n = 5

3 Substitute values of a, r and n into the general formula for the sum and evaluate.

=

=

=

S1

1

1

5

1

2

1

41

41

2

1

10243

4341512

5

Chapter 7 Variation 37

Chapter 7

VariationWorked 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) andb a graphical approach.c Confi rm your result using a CAS calculator.

x 4 7 8 10

y 5.2 9.1 10.4 13yx

think Write/draW

a 1 Find the ratio yx 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 = yx

First pair: =5.2

41.3

Second pair: =9.1

71.3

Third pair: =10.4

81.3

Fourth pair: =13

101.3

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 4 7 8 10

y 5.2 9.1 10.4 13yx 1.3 1.3 1.3 1.3

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.

b13

10.4

0 10

9.1

5.2

874 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.

ContentsVariation 37


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 (x1, y1) = (4, 5.2) and let (x2, y2) = (10, 13)

=

my y

x x2 1

2 1

=

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) 1In 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.


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 3 6 7 10

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} yPress 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 9Enter x in X List and y in Y ListThen press ENTER .

3 Write down the rule for x and y. y = 3.2x2

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.


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 1 2 3 4 5

y 20 10 623

5 4

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 = xyFirst pair: 1 20 = 20Second pair: 2 10 = 20Third pair: 3 6 23 = 20Fourth pair: 4 5 = 20Fifth 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 1 2 3 4 5

y 20 10 6 23 5 4

xy 20 20 20 20 20

b 1 Calculate the values of x

1.

Place these values into a table.

b x 1 2 3 4 5

x1

1 12

1

31

4

1

5

y 20 10 62

35 4

2 On a Calculator page, enter the x

1

and y values by completing the entry lines as:

{1, 12, 13, 1

4, 1

5 } x

{20, 10, 623 , 5, 4} y



3 To find the rule that fits the data, press: MENU b 6: Statistics 6 1: Stat Calculations 1 3: Linear Regression (mx + b) 3Enter x in X List and y in Y ListThen 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 x1

.


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 and 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 dollarsLet n = the number of calls

2 Write the equation of variation. A = k n + 323 Substitute the values for A and n into

the equation.When n = 296 and A = 106,

106 = 296k + 324 To solve the equation for k, on a

Calculator page, press: MENU b 3: Algebra 3 1: Solve 1Complete 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 = 14n + 32

b 1 Substitute n = 300 into the given equation.

b When n = 300,A = 14(300) + 32

2 Evaluate. = 14 300 + 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.


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) 1 2 3 4 5

TSA (cm2) 12.57 50.27 113.1 201.06 314.16

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 7In 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 r2

6 Write the variation equation. TSA = kr2

7 Transpose the equation to make k the subject.

k = r

TSA2


8 Test the assumption by fi nding the values

of r2 and check that the ratio r

TSA2

is

constant. On the Lists & Spreadsheet page, label column C: ratioIn the header (grey) cell, complete the entry line as:=tsa/(radius)^2Then press ENTER .

9 Comment on the result obtained. The ratio is constant for each corresponding pair (when rounded to 2 decimal places).Hence, TSA r2 TSA = kr2 TSA = 12.57r2

10 Alternatively, once the values of r2 have been calculated, rule up a table of values and plot TSA versus r2.

r2 1 4 9 16 25

TSA 12.57 50.27 113.1 201.06 314.16

11 Comment on the graph obtained.

300

200

100

014 9 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 = kr2When r = 1, TSA = 12.57, k = ?

12.57 = k 112.57 = k

k = 12.57TSA = 12.57r2

ChapTer 8 Further algebra 45

ChapTer 8

Further algebraWorked example 2

If 5x3 + 2x2 7x + 1 = (2 a + b)x3 ax2 (b c) x + 1, then fi nd the values of a, b and c.Think WriTe

1 On a Calculator page, press: MENU b 3: Algebra 3 1: Solve 1Complete the entry line as:

solvea b

a a

b c

5

= ,

= 7

2 + = 2

Then press ENTER . Note: The template used here can be found in the Maths expression template.

2 Write the answer. a = 2, b = 9 and c = 2

Worked example 4

If x 4 is a factor of x3 6x2 + 2x + 24, fi nd the other factor.Think WriTe

1 On a Calculator page, press: MENU b 3: Algebra 3 2: Factor 2Complete the entry line as:factor(x3 6x2 + 2x + 24)Then press ENTER .

2 Write the answer. The quadratic factor of x3 6x2 + 2x + 24 is x2 2x 6.

ContentsFurther algebra 45


Worked example 5

Express +

xx4 5

3 in the form +

Ab

x 3.

Think WriTe

1 On a Calculator page, press: MENU b 3: Algebra 3 9: Fraction Tools 9 1: Proper Fraction 1Complete the entry line as:

propFrac +

x

x

4 5

3Then press ENTER .

2 Write the answer in the form

+

Ab

x 3.

+

= +

x

x x

4 5

34

17

3

Worked example 6

Express +

xx x

33 402

in partial fraction form.

Think WriTe

1 On a Calculator page, press: MENU b 3: Algebra 3 3: Expand 3Complete the entry line as:

expand+

x

x x

3

3 402

Then press ENTER .


++

A

x

B

x( 8) ( 5).

+

=

++

x

x x x x

3

3 40

11

13( 8)

2

13( 5),

2

x R\{5, 8}


Worked example 7

Express +

xx x

2 1( 2)( 1)2

in partial fractions.

Think WriTe


expand

+

x

x x

2 1

( 2)( 1)2

Then press ENTER .


++

++

A

x

B

x

C

x( 2) ( 1) ( 1)2.

x

x x x x x

2 1

( 2)( 1)

1

3( 2)

1

3( 1)

1

( 1),

2 2

+=

++

+

x R\{1, 2}

Worked example 8

Express + +

x xx

5 9 108

2

3 in partial fractions.

Think WriTe


expand + +

x x

x

5 9 10

8

2

3

Then press ENTER .


++

+ +

A

x

Bx C

x x2 2 4.

2

+ +

=

++

+ +

x x

x x

x

x xx R

5 9 10

8

4

2

3

2 4, \{2}

2

3 2


Worked example 9

Express +

x xx5 21

2 as a partial fraction.

Think WriTe


expand +

x x

x

5 2

1

2

Then press ENTER .

2 Write the answer.+

=

+ +x x

x xx

5 2

1

4

16,

2

x R\{1}

Worked example 10

Solve simultaneously: y = x and y = x2 + 3x + 1.Think WriTe

1 On a Calculator page, press: MENUb 3:Algebra3 1:Solve1Complete the entry line as:solve(y = x and y = x2 + 3x + 1, x)

Then press ENTER .

2 Write the answer. Solving y = x and y = x2 + 3x + 1 for x and y gives x = 1 and y = 1.That is, (1, 1).


Worked example 11

Solve simultaneously: y = x + 1 and x2 + y2 = 4.Think WriTe

1 On a Calculator page, press: MENUb 3:Algebra3 1:Solve1Complete the entry line as:solve(y = x + 1 and x2 + y2 = 4, x)

Then press ENTER .

2 Write the answer. Solving y = x + 1 and x2 + y2 = 4 for x and y gives

=

+x

( 7 1)

2 and =

y( 7 1)

2 or

=

x7 1

2 and =

+y

7 1

2

That is,

+

( 7 1)

2,

( 7 1)

2 or

+

7 1

2,

7 1

2.


Worked example 12

Solve simultaneously: y = 2x 1 and =

yx23.

Think WriTe

1 On a Calculator page, press: MENUb 3:Algebra3 1:Solve1Complete the entry line as:

solve y x yx

x2 1and2

3,= =

Then press ENTER .

2 Write the answer.Solving y x y

x2 1and

23

= =

for x and y gives

x y

x y

( 41 7)

4and

( 41 5)

4

or41 7

4and

41 5

4

=

=

=

+=

+

That is,

( 41 7)

4,

( 41 5)

4or

41 7

4,

41 5

4.

+ +

ChapTer 10 Linear and non-linear graphs 51

ChapTer 10

Linear and non-linear graphsWorked example 11

Convert [2, 23pi

] to Cartesian coordinates.

Think WriTe

1 Find the x-coordinate. x = r cos ( )

= 2 cos 23

pi = 2

1

2x = 1

2 Find the y-coordinate. y = r sin ( )

= 2 sin 23

pi = 2 3

2y = 3

3 State the Cartesian coordinates. Hence, the Cartesian coordinates are (1, 3).

4 Alternatively, open a Calculator page.Make sure the setting is on AUTO.Complete the entry lines as:

P Rx 2,2

3

pi P Ry 2,

2

3

pi Press ENTER after each entry. Note: P Rx and P Ry can be found in the catalogue.

[2, ]

x

y

23

23

23


Worked example 16

Sketch the graph of r = for 0 < < 4 using a CAS calculator.Think WriTe

Open a new Graphs page.Ensure the angle setting is on radians.Then press MENUb 3:GraphEntry/Edit3 4:Polar4Complete the entry line as:

pi

=

=

r

step

1( )

0 4 0.13

Then press ENTER .

The increment step is 6

pi (0.13) because this was

used in the introductory table.Compare the graphs of worked examples 15 and 16.


Worked example 17

Sketch the graph of r = 8 for 0 2.Think WriTe/display

1 Construct a table of values for 0 2 and find the corresponding r values to 2decimal places.

06

pi

3

pi

2

pi 2

3

pi 5

6

pi

r 8 8 8 8 8 8

7

6

pi 4

3

pi 3

2

pi 5

3

pi 11

6

pi2

r 8 8 8 8 8 8 8

2 Sketch the graph using a protractor and ruler to plot each of the points from the table. Remember r is the distance from the centre (the origin).

8

8

88

32

2

3 On a Graphs page, press MENUb 3:GraphEntry/Edit3 4:Polar4Complete the entry line as:

pi

=

=

r

step

1( ) 8

0 2 0.13

Then press ENTER .

Zoom out to see the whole circle.


Worked example 18

Sketch the graph of r = 2 sin ( ) for 0 2 using a calculator.Think WriTe

On a Graphs page, press MENUb 3:GraphEntry/Edit3 4:Polar4Complete the entry line as:

pi pi=

=

r

step

1( ) 2sin ( )

0 212

Then press ENTER .

Zoom out to see the whole circle.

Worked example 21

Express the following complex numbers in polar form.

a z = 1 + i b z i3= +

Think WriTe

a&b

1 Open a Calculator page.Enter the real and imaginary parts using the [ ] parentheses. Complete the entry lines as:[1,1] Polar

3,1 Polar

Press ENTER after each entry.Note: Polar can be found in the catalogue.

a&b


b

z = 1 + i in polar form is z 2cis4

pi=

. z i3= +

in polar form is

z 2cis5

6

pi=

.


Worked example 22

Express 3cis56pi in Cartesian form.

Think WriTe


P Rx pi 3,

5

6

P Ry pi 3,

5

6


2 Write the answer. i3cis5

6

3

2

3

2

pi = +


Worked example 24

Sketch the graph that results from the addition

of ordinates of the functions y = x and y = x1

, whose graphs are shown at right.

Think WriTe

1 Add the ordinates at the LHS end points of the graph: A small negative value and a large negative value should give a slightly larger negative value. Mark this point on the set ofaxes.

x

y3

2

1

1

2

3

1123 2 30

y = xy = 1x

2 Add the ordinates at the first point of intersection. Because they are the same we simply double the y-value of this point. Mark the point on the set of axes.

x

y

3

2

1

1

2

3

1123 2 30

y = xy = 1x

3 At its x-intercept, the straight line graph y = x has an ordinate of zero while the non-linear graph has an undefined ordinate. This indicates that our graph will also be undefined at the point where x = 0, giving a vertical asymptote.

Vertical asymptote at x = 0

4 Add the ordinates at the second point of intersection. Mark the point on the set ofaxes.

x

y

3

2

1

1

2

3

1123 2 30

y = xy = 1x

x

y

y = xy =

3

2

1

1

2

3

1123 2 30

1x


5 Add the ordinates at the RHS end points of the graph: A small positive value and a large positive value should give a slightly larger positive value. Mark this point on the set ofaxes.

x

y

3

2

1

1

2

3

1123 2 30

y = xy = 1x

6 Join the points with a smooth curve noting that the graph will asymptote toward the y-axis as x approaches zero, and toward the line y = x as x approaches positive or negative infinity.

0

xy = x + 1

x

y

3

2

1

1

2

3

1123 2 30

7 To perform the addition of ordinates on the CAS calculator, open a new Graphs page.Complete the function entry lines as:

f x x

f xx

f x f x f x

1( )

2( ) 1

3( ) 1( ) 2( )

=

=

= +



Worked example 25

Sketch the reciprocal of the graph of the function y = x2 shown at right.

Think WriTe

1 Sketch the given function and draw the horizontal line y = 1 on the same axes.

y

3

2

x

2

1

1112 20

3 3

2 2

11 21

3

3

21

2 Mark any points where the ordinates of the graph are equal to 1.

3 Select several points on the original graph with ordinates greater than 1 (above the line). Estimate their value and mark their reciprocals below the line.

4 Select several points on the original graph with ordinates between 0 and 1 (between the line and the x-axis). Estimate their value and mark their reciprocals above the line.

x

y

333 3 3

222 2 2

1

1112 20

21

31

331

311 2

121

21

5 Join the points with a smooth curve noting that the graph will asymptote toward the y-axis as x approaches zero, and toward the x-axis as x approaches positive or negative infinity.

x

y

333 3 3

222 2 2

1

1112 20

21

31

331

311 2

121

21

x21y =

6 Alternatively, on a Calculator page, complete the entry line as:Define f(x) = x2 Then press ENTER . On a Graphs page, complete the function entry lines as:f x f x

f xf x

1( ) ( )

2( )1

( )

=

=


x

y

3

2

1

1112 20

y = x2


Worked example 26

Sketch the square of the graph of the function y = x2

shown below.

Think WriTe

1 Sketch the given function and draw the horizontal lines y = 1 and y = 1 on the sameaxes.

x

y

3

2

1

1

2

3

1

1 1

123 2 3

2 Mark any points where the ordinates of the graph are equal to 1 and change any ordinates of 1 to 1. (Squaring 1 results in1.)

3 Select, estimate, and square several ordinates between the horizontal lines y = 1 and y = 1 marking these positive values closer to the x-axis.

x

y

3

4

2

1

1

2

3

4

1

1 1

1234 2 3 41

2

1214

14

4 Select, estimate, and square several ordinates outside the horizontal lines y = 1 and y = 1 marking these positive values further away from the x-axis.

5

6

7

8

9 99

44 4

3 32 21

1 1

122334

12 134 2 3 4 x

y

0

121

414

1 2

3

2

1

123

12 13 2 3 x

y

0

y = 2x


5 Join the points with a smooth curve noting that the graph will asymptote toward the y-axis as x approaches zero.

5

6

7

8

9 99

44 4

3 32 211 1

12233

4

12 134 2 3 4x

y

0

1214

x2y = 4

14

1 2

6 Alternatively, on a Calculator page, complete the entry line as:

Defi ne f xx

( )2

=

Then press ENTER . On a Graphs page, complete the function entry lines as:f1(x) = f(x)f2(x) = (f(x))2Press ENTER after each entry.

ChapTer 11 Linear programming 61

ChapTer 11

Linear programmingWorked example 3

Sketch the graph of the inequation y 4x 8 and indicate the required region.Think WriTe/draW

1 Make y the subject of the inequation by adding 4x to both sides.

y 4x 8 y 8 + 4x

2 On a Graphs page, delete = and complete the entry line as:y 8 + 4xThen press ENTER .

3 Indicate the required region.Note: The calculator shades the required region.

The required region.

ContentsLinear programming 61


Worked example 5

Sketch the following pair of simultaneous linear inequations, determine the point of intersection and indicate the required region.

2x + 3y 6x y 3

Think WriTe/draW

1 Make y the subject of the inequations. 2x + 3y 63y 6 2x

yx6 2

3 [1]

x y 3y 3 xy x 3 [2]

2 On a Graphs page, delete = and complete the entry lines as:

yx6 2

3y x 3Press ENTER after each entry.To change the colour of the inequations, highlight the line and press: Ctrl / MENU b 4: Colour 4 2: Fill colour 2Select a colour and press ENTER .

3 Indicate the required region. The required region.

4 To determine the point of intersection between the lines, press: MENU b 8: Geometry 8 1: Points & Lines 1 3: Intersection Point(s) 3Move the flashing cursor to each line, press ENTER to select each line and press ENTER again to lock in the point of intersection.

State the point of intersection. The point of intersection between

yx6 2

3 and

y x 3 is (3, 0).

ChapTer 11 Linear programming 63

Worked example 8

a Sketch the following system of linear inequations and indicate the required region.

x + y 10, y x 4, y 2x + 1, x 0, y 0b Determine the coordinates of the vertices of the feasible region.c Determine the maximum and minimum values of z = 3x y subject to the above constraints,

using the corner-point method.

Think WriTe

a 1 Make y the subject of the fi rst inequation and write the others.

a x + y 10y 10 x [1] x 0 [4]y x 4 [2] y 0 [5]y 2x + 1 [3]

2 On a Graphs page, complete the entry lines as:y 10 xy x 4y 2x + 1y 0Press ENTER after each entry. Note: x 0 cannot be drawn on the calculator.Change the colours as required.

3 Indicate the required region. The required region.

b 1 To determine the points of intersection between the lines, press: MENU b 8: Geometry 8 1: Points & Lines 1 3: Intersection Point(s) 3Move the fl ashing cursor to each line, press ENTER to select each line and press ENTER again to lock in the point of intersection.Repeat for each pair of lines and state the points of intersection.Zoom out to see all the coordinates.

b

The points of intersection are: (3, 7), (7, 3), (4, 0).

2 To determine the points of intersection between the line x = 0 and y = 2x + 1, substitute the value x = 0 into the equation y = 2x + 1.

y = 2x + 1x = 0 y = 2(0) + 1

y = 1That is the point (0, 1).From the graph, (0, 0) is also a corner point.


c 1 To determine the maximum and minimum values, on a Calculator page, complete the entry line as:Defi ne z = 3x yz | x = 0 and y = 0z | x = 0 and y = 1z | x = 3 and y = 7z | x = 7 and y = 3z | x = 4 and y = 0Press ENTER after each entry.

c

2 Write the maximum and minimum values of z.

zmin = 1 at (0, 1)zmax = 18 at (7, 3)

CHAPTER 12 Coordinate geometry 65

CHAPTER 12

Coordinate geometryWORKED EXAMPLE 20

ABCD is a parallelogram. The coordinates of A, B and C are (1, 5), (4, 2) and (2, 2) respectively. Find:a the equation of AD b the equation of DC c the coordinates of D.

THINK WRITE/DRAW

a 1 Draw the parallelogram ABCD.Note: The order of the lettering of the geometric shape determines the links in the diagram. For example: ABCD means join A to B to C to D to A. This avoids any ambiguity.

a

x

y

A

B

C

D

2141 4

5

2

42

2 Find the gradient of BC. m2 2

2 4BC =

4

2=

= 2

3 State the gradient of AD. Since mBC = 2and AD||BCthen mAD = 2

4 Using the given coordinates of A and the gradient of AD nd the equation of AD.

y = 2x + cLet (x, y) = (1, 5):

5 = 2(1) + cc = 3

Hence, the equation of AD is y = 2x + 3.

b 1 Find the gradient of AB. b m2 5

4 1AB =

3

3=

= 1

2 State the gradient of DC. Since mAB = 1and DC||ABthen mDC = 1

3 Using the given coordinates of C and the gradient of DC nd the equation of DC.

y = x + cLet (x, y) = (2, 2):

2 = (2) + cc = 0

Hence, the equation of DC is y = x.c 1 Solve simultaneously to nd D,

the point of intersection of the equations AD and DC.

c Equation of AD: y = 2x + 3 [1]Equation of DC: y = x [2]


2 On a Calculator page, complete the entry line as:solve( y = 2x + 3 and y = x, x)Then press ENTER .

3 Write the solution. Hence, the coordinates of D are (1, 1).

CHAPTER 13 Vectors 67

CHAPTER 13

VectorsWORKED EXAMPLE 14

a Draw a vector to represent a i j3 .= b Find the magnitude and direction of the vector a

.

THINK DRAW/WRITE

a 1 Draw axes with i and j

as unit

vectors in the x- and y-directions respectively.

a

x

y

321

1

2

21O1

j~

i~

2 Represent i j3

as a vector from 0 that is 3 units in the positive x-direction and 1 unit in the negative y-direction, and mark the angle between a

and the x-axis as. x

y

3

1

2

1O12

j~

i~

a~

b 1 The magnitude of a (that is,

a ) may

be found using Pythagoras theorem.b

a 3 ( 1)

10

2 2= +

=

2 On a Calculator page, complete the

entry line as: 3 ( 1)2 2+

Then press ENTER .

3 Find the value of angle using the tangent ratio.

tan () = 13

ContentsVectors 67


4 On a Calculator page, complete the

entry line as: tan1

31

Then press ENTER .Note: Ensure your calculator is in Degree mode.

5 Give the direction of vector a relative

to the positive x-axis.Vector a

makes an angle of 18.4 from the

positive x-axis.

CHAPTER 14 Statics of a particle 69

CHAPTER 14

Statics of a particleWORKED EXAMPLE 8

The forces acting on two bobs suspended from light inextensible strings are shown in the diagram at right. If the connecting string is at an angle of 10 to the horizontal nd T and m.

THINK WRITE/DRAW

1 Separate the two bobs. Draw a diagram representing the forces applied to the right bob.

5g

T1 sin (30)

T1 cos (30)T cos (10)

T sin (10)

2 Calculate the force, T. Set up two equations.To solve these equations simultaneously, using a CAS calculator, let T1 = x and T = y.On a Calculator page, complete the entry line as:solve(x sin(30) = 5g + y sin(10) and y cos(10) = x cos(30), x) | g = 9.8Then press ENTER .

T1 sin (30) = 5g + T sin (10) [1]T cos (10) = T1 cos (30) [2]

T = 124.1 N

10

T2 NT N

T N

T1 N

5g Nmg N

60 30

ContentsStatics of a particle 69


3 Draw a diagram representing the forces applied to the left bob.

mg

T sin (10)T2 sin (60)

T2 cos (60) T cos (10)

4 Calculate the value of m given T = 124.1 N.

T cos (10) = T2 cos (60)Given T = 124.1 N

=T124.1cos(10)

cos(60)2

T2 = 244.4 NT2 sin (60) + T sin (10) = mg

=+

m244.4sin(60) 124.1sin(10)

9.8 m = 23.80 kg

5 State the answers. T = 124.1 Nm = 23.80 kg

ChapTer 15 Kinematics 71

ChapTer 15

KinematicsWorked example 11

A particle is travelling in a straight line with its velocity, v cm/s, at any time, t seconds, given as

=

+v t

tt( )

81, 0.

Find the acceleration of the particle after 1 second.

Think WriTe/display

1 Given the expression, =+

v tt

( )8

1 we

want a(1).=

+v t

t( )

8

1

2 Find the acceleration equation by differentiating velocity with respect to time (a(t) = v(t)). To do this, on a Calculator page, complete the entry lines as:

Defi ne =+

v tt

( )8

1

d

dtv t( ( ))

Defi ne =+

a tt

( )8

( 1)2


3 Substitute t = 1 seconds into the formula for a(t).To do this, complete the entry line as:a (1)Then press ENTER .

4 State the solution. The acceleration of the particle at t = 1 seconds is 2 cm/s2.

ContentsKinematics 71


Worked example 13

A particle is travelling in a straight line with its velocity, v (in m/s), at any time, t seconds, given as:

v(t) = t2 + t, t 0Calculate the exact distance travelled during the first 4 seconds of its motion.

Think WriTe/display

1 On a Calculator page, press: MENU b 4: Calculus 4 3: Integral 3Complete the entry line as:

+t t dt( )204Then press ENTER .

2 State the exact distance travelled. The exact distance travelled during the first

4seconds of its motion, is 291

3 metres.

ChapTer 15 Kinematics 73

Worked example 14

A car accelerates from rest at 2 m/s2 for 5 seconds.a Write an equation for the acceleration.b Write the equation for the velocity.c Calculate the distance covered in the first 5 seconds.

Think WriTe

a 1 The acceleration is 2 m/s2. a a(t) = 2b 1 To determine the equation for

velocity, given a(t) = 2, on a Calculator page, complete the entry line as:

(2)dt + cThen press ENTER .

b

2 Write the equation for velocity. v(t) = 2t + c3 It is given that v = 0 when t = 0.

Calculate the constant, c, using this information.

0 = 2(0) + cc = 0 v(t) = 2t, 0 t 5

c 1 To calculate the distance, d(t), covered in the first 5 seconds, on a Calculator page, complete the entry line as:

t dt(2 )05Then press ENTER .

c

2 State the distance covered in the first 5 seconds.

The distance travelled in the first 5 seconds is 25 metres.

Chapter 16 Geometry in two and three dimensions 75

Chapter 16

Geometry in two and three dimensions

Worked example 4

a Use a ruler and a pair of compasses to bisect a line, AB.b Use a CAS calculator to bisect a line segment.

think draW/display

a 1 (a) Draw a line AB.(b) Place the compass point at A,

with any radius (more than half the length of the line).

(c) Draw a circle.

a

A B

2 With the same radius as in step 1, repeat for point B.

3 The two circles will intersect at two points. Join these points with a straight line.

b 1 Open a new Graphs page. To access the Geometry view, press: MENU b 2: View 2 2: Plane Geometry 2

b

2 To draw a line segment, complete the following steps.Press: MENU b 4: Points and Lines 4 5: Segment 5Move the pencil to the position for the end of the line segment and press CLICK a.Repeat the process for the other end.

ContentsGeometry in two and three dimensions 75


3 To bisect the line, press: MENU b 7: Constructions 7 3: Perpendicular bisector 3Place the cursor over each of the end points of the line segment and press CLICK a to view the bisected line.

Worked example 5

a Use a ruler and compasses to construct a line parallel to a given line.b Use a CAS calculator to construct a line parallel to a given line.

think draW/display

a 1 Pick any two points, A and B, on the given line.

a

A B

2 From point A, draw a circle of any radius (more than half the distance from A to B).

3 With the same radius repeat step 1 at point B.

4 Join the highest points (or lowest points) of the circles with a straight line. This will be parallel to AB.

b 1 Construct a straight line segment in the same manner as for worked example 4.To construct a parallel line, press: MENU b 7: Constructions 7 2: ParallelPress e until the word segment appears.

b

2 Then press CLICK a and use the NavPad to place the parallel line in the required position.


Worked example 6

a Use a ruler and compasses to bisect any angle. b Use a CAS calculator to bisect any angle.

think draW/display

a 1 With any radius, and the point of the compass at the vertex V, draw an arc of a circle which crosses both arms of the angle. The crossings are labelled A and B.

a

VB

A

2 With any radius and the point of the compass at A, draw an arc inside the angle. This arc should be long enough so that the line representing half the angle would cross it. V

B

A

3 With the same radius, repeat step 2, putting the point of the compass at B. The two arcs will cross at point C.

4 Join the vertex V to C. This line bisects the angle, namely:

AVC = AVB2

VB

A

C

b 1 Open a new Geometry page in a Plane Geometry view.Press: MENU b 7: Construction 7 4: Angle bisector 4Move the cursor to the end point of the first arm of the angle and press CLICK a.Using the navigation pad move the cursor to where the vertex of the angle will be and press CLICK a.Move the cursor to the end of the other arm and again press CLICK a.The bisector of the angle will automatically appear.

b

2 Pressing ENTER will show the three points and the bisecting line.


Worked example 8

Construct the perpendicular bisectors and median bisectors of each side of any triangle and investigate their properties.

think draW/display

1 Draw any triangle and add circles centred at each vertex. The radius should be large enough so that perpendicular bisectors can be drawn.

2 Use the construction circles to draw perpendicular bisectors. The black lines join pairs of intersecting arcs. It should be clear that the bisectors all meet at a point. This point is called the circumcentre.

3 Use this point as a centre and draw a circle which just touches each vertex. To do this, put the compass point at the point where the bisectors met and the pencil point at any vertex. You should observe that the resultant circumcircle (or outcircle) just touches each vertex.

4 Alternatively, open a new Geometry page.Construct a triangle by pressing: MENU b 5: Shapes 5 2: Triangle 2Press CLICK a and use the NavPad to generate the 3 vertices of the triangle.

5 To construct the perpendicular bisectors, press: MENU b 7: Construction 7 3: Perpendicular Bisector 3Move the cursor to the end points of each side of the triangle (vertices) and press CLICK a.


6 To draw the circumcircle, press: MENU b 5: Shapes 5 1: Circle 1Move the cursor over the intersection point (circumcentre) and press CLICK a.Move the cursor until it is over one of the vertices of the triangle and press CLICK a.

7 Furthermore, from step 2, we can determine the midpoint of each side from the perpendicular bisectors. Note the use of short bars to indicate the bisection of the sides, at points P, Q and R. P

R Q

8 Join each midpoint to its opposite vertex. These lines also meet at a single point, called the centroid, which has applications in physics and engineering, as it is effectively the point of symmetry of the triangle.

P

R Q


Worked example 9

Construct the incentre of the triangle shown at right:a by handb using a CAS calculator.

think draW/display

a 1 (a) Construct the angle bisectors by drawing arcs centred at each vertex (A, B and C).

(b) From the intersection of these arcs and the sides of the triangles, draw intersecting arcs between pairs of sides. In the figure at right this has been done to vertex A only, to keep the drawing uncluttered.

a

C

B

A

2 Complete the construction of the angle bisectors and observe that they, too, meet at a point the incentre.

C

B

A

3 By placing the compass point at the incentre and carefully drawing a circle, it is possible to construct the incircle, which just touches each side of the triangle.

C

B

A

b 1 Construct a triangle using the method set out in worked example8.Then press: MENU b 7: Construction 7 4: Angle Bisector 4Move the cursor over each vertex and press CLICK a.This will bisect the angle at the second vertex in the order in which they were clicked.

b

2 Repeat this process for each of the other angles of the triangle.

C

B

A


3 To draw the incircle, press: MENU b 5: Shapes 5 1: Circle 1Move the cursor over the intersection point (incentre) and press CLICK a.Move the cursor until the circle fits inside the triangle, touching each side and press CLICK a.


Worked example 18

Construct, with a straight edge and compasses, a tangent to a circle at any point.This task is relatively easy, relying on the earlier construction of a perpendicular bisector.

think display

1 To construct a circle, open a new Geometry page, and then press: MENU b 5: Shapes 5 1: Circle 1Press CLICK a to locate the centre of the circle. Use the navigation pad to increase the size of the circle then press CLICK a.

2 To draw a tangent to the circle, press: MENU b 4: Points and Lines 4 7: Tangent 7Press CLICK a to set the position of the tangent.

3 To add the line OB, press: MENU b 4: Points and Lines 4 4: Line 4Place the cursor over the centre of the circle and press CLICK a then move the cursor to the point of intersection of the tangent and the circle and press CLICK a.

ContentsIntroductionChapter 1 Number systems: real and complexChapter 2 TransformationsChapter 3 Relations and functionsChapter 4 AlgebraChapter 5 Trigonometric ratios and their applicationsChapter 6 Sequences and seriesChapter 7 VariationChapter 8 Further algebraChapter 10 Linear and non-linear graphsChapter 11 Linear programmingChapter 12 Coordinate geometryChapter 13 VectorsChapter 14 Statics of a particleChapter 15 KinematicsChapter 16 Geometry in two and three dimensions

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