questions & answers ncea level 2 mathematics

NCEA

LEVEL 2MATHEMATICS

QUESTIONS AND ANSWERS

P J Kane

Published by Mahobe Resources (NZ) Ltd

ACKNOWLEDGEMENTS

NCEA Level 2 MathematicsQuestions & AnswersP. J. Kane This eBook was published in 2010.

Mahobe Resources (NZ) LtdP.O. Box 109-760Newmarket, AucklandNew Zealand www.mahobe.co.nz

© Mahobe Resources (NZ) LtdISBN 9781877216824 This eBook has been provided by Mahobe Resources (NZ) Ltd to The New Zealand Centre of Mathematics. Schoolteachers, University lecturers, and their students are able to freely download this book from The New Zealand Centre of Mathematics website www.mathscentre.co.nz. Electronic copies of the complete eBook may not be copied or distributed. Students have permission to print one copy for their personal use. Any photocopying by teachers must be for training or educational purposes and must be recorded and carried out in accordance with Copyright Licensing Ltd guidelines. The content presented within the book represents the views of the publisher and his contributors as at the date of publication. Because of the rate with which conditions change, the publisher and his contributors reserve the right to alter and update the contents of the book at any time based on the new conditions. This eBook is for informational purposes only and the publisher and his contributors do not accept any responsibilities for any liabilities resulting from the use of the information within. While every attempt has been made to verify the content provided, neither the publisher nor his contributors and partners assume any responsibility for errors, inaccuracies oromissions. All rights reserved. All the views expressed in this book are those of the author. The questions and suggested answers are the responsibility of the author and have not been moderated for use in NCEA examinations. Thats all the legal stuff over. We hope that the book is helpful!

3

YEAR 12 MATHEMATICS

CONTENTS

2.1 Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2 Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.3 Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.4 Co-ordinate Geometry . . . . . . . . . . . . . . . . . . 55

2.5 Sample Statistics . . . . . . . . . . . . . . . . . . . . . 73

2.6 Probability & Normal Distribution . . . . . . . . . . 87

2.7 Sequences . . . . . . . . . . . . . . . . . . . . . . . . 105

2.8 Trigonometry Problems . . . . . . . . . . . . . . . . 117

2.9 Trigonometric Equations . . . . . . . . . . . . . . . 130

The Answers . . . . . . . . . . . . . . . . . . . . . . . 144

Areas Under the Normal Curve . . . . . . . . . . . 171

Formulae Sheet . . . . . . . . . . . . . . . . . . . . . 172

Pages for Extra Notes . . . . . . . . . . . . . . . . . 173

4

YEAR 12 MATHEMATICS

STUDYING NCEA LEVEL 2 MATHS

Ø This book has been written for you to practise NCEA Level Two- type assessments. Nine chapters have

been designed to match the nine achievement standards at this level.

Ù Each chapter begins with a schedule of the requirements for that achievement standard. As you read

down each schedule, you will see that the challenges become more complex.

Ú In most chapters a preliminary set of exercises has been provided to set in motion the set of skills

required for the achievement objectives. Once you think that you have mastered the skill set progress

onto the first model assessment.

For external achievement standards allow 45 - 60 minutes.

For internal achievement standards allow 4-5 hours as these are more project orientated.

Check the solutions, and if yours do not quite match these, rework your calculations, or check with

friends or teachers until you are satisfied.

Û Attempt the second model assessment 1-2 weeks later to see if the themes you covered still ‘click’.

Again, check the solutions at the back of the book with yours. Remember, you can still learn from your

mistakes .... this side of the final exams.

Ü It is worth recognising that in Year 12 your mathematics may appear to have begun at a roundabout.

It introduces new themes which seem to go down different roads. One of the strengths of this subject,

however, is that these themes or roads are connected, though this may not be evident just yet.

Therefore as you are being assessed in discrete themes or standards, try to develop an eye for the

bigger picture. As always mathematics is about solving problems and finding patterns and reasons.

Hopefully your experiences this year will provide you with confidence and judgement for future

challenges.

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Algebra 5

YEAR 12 MATHEMATICS

MATHEMATICS 2.1

ACHIEVEMENT STANDARD 90284

Manipulate algebraic expressions and solve equations

Below are the requirements of this Achievement Standard.

C expand brackets (up to 3 bracket sets)

C factorise expressions including quadratics

C simplify and use fractional exponents

C simplify and use integer exponents

C interchange between exponential and

logarithmic statements

C solve linear equations or inequations using at

least 2 steps

C solve quadratic equations which can be

factorised

C solve simple logarithmic equations

C form then solve pairs of linear simultaneous

equations

C solve quadratics using the quadratic formula

C solve a pair of simultaneous equations, with

one being linear and one being non-linear

C solve exponential equations, which may involve

logarithmic methods

C complete algebraic challenges such as proving

an algebraic statement

C explore the nature of the roots of a quadratic

equation.

From the

straightforward

(less steps)

cases

to

situations

involving

more depth,

more steps

and

growing complexities

with

sensible interpretations

of the solutions(s).

6 Algebra

YEAR 12 MATHEMATICS

ALGEBRA - Revision Summary

The exercises in this section have been specifically chosen to underpin the Achievement Criteria that will be

tested in this Achievement Standard. Write your answers on the opposite page.

1. ASSUMED KNOWLEDGE

Rearrange (change the subject of) these formulae:

a. x, in y = 3x - 7

b. r, in A = 4πr2

Solve in factored form:

c. (2x - 5)(x + 11) = 0

d. 5x2(1 - 3x)2(x + 8) = 0

2. EXPAND BRACKETS AND SIMPLIFY

a. 3x(2x - 3) - 6(x2 - 2)

b. 2x(3x + 1)(3x - 5)

c. (x + 3)(2x - 7)2

3. FACTORISE ALGEBRAIC EXPRESSIONS

Factorise completely:

a. x2 - 19x + 84

b. 10x2 - x - 21

c. a2 - av + aw - vw

4. FRACTIONAL AND NEGATIVE INDICES

a. Write with positive indices.

b. Simplify .

c. Simplify .

5. ELEMENTARY PROPERTIES OF LOGARITHMS

a. Write log3 243 = 5 in exponential form.

b. Express as the log of a single number:

i. log 15 - log 3

ii. 3 log 2 + 2 log 3

iii. 2 log 6 - log 36 + log 5

c. Simplify

Algebra 7

YEAR 12 MATHEMATICS

8 Algebra

YEAR 12 MATHEMATICS

6. SIMPLIFY RATIONAL EXPRESSIONS

a.

b.

c.

d. ×

7. SOLVE LINEAR EQUATIONS AND INEQUATIONS

a. x - 3 = x + 7

b. 2(n + 3) = 5(n - 1) - 7(2n - 3)

c. 3(2x + 1) < 2x + 9

d. 2(x + 3) >

8. SOLVE QUADRATIC EQUATIONS

By factorising

a. x2 - x - 42 = 0

b. 3x2 - 7x + 2 = 0

c. 5x = 2x3 + 3x2

By quadratic formula (to 2dp)

d. 4x2 - 2x - 3 = 0

e. x2 + 4x - 2 = 0

f. 8 - x - x2 = 0

9. SOLVE LOGARITHMIC AND EXPONENTIAL EQUATIONS

Find the value of x in:

a. log3 x = 7

b. logx 343 = 3

c. 7x-1 = 26

Algebra 9

YEAR 12 MATHEMATICS

10 Algebra

YEAR 12 MATHEMATICS

10. SOLVE SIMULTANEOUS EQUATIONS

Linear pairs

a. x + 3y = 5 b. y = x + 5

2x + y = -5 3y + 4x = 1

Line and curve

c. y = x2 - 3x d. xy = 2

y = 2x - 6 y - x = 1

11. QUADRATIC ROOTS

a. Find the nature of the roots of:

i. 4x2 - 13x + 7 = 0

ii. 25x2 - 30x + 9 = 0

b. Use the discriminant of 2x2 - 2nx + 5 = 0 to find the values of n for which

there will be no real roots (i.e. imaginary).

Algebra 11

YEAR 12 MATHEMATICS

ALGEBRA - PRACTICE TEST 1

QUESTION ONE

1. Simplify:

2. Simplify fully:

3. Write as the log of a single number: log 112 - log 14

4. Solve the following equations:

a. =

b. logx 44 = 5

c. 3x2 - x = 4

12 Algebra

YEAR 12 MATHEMATICS

QUESTION TWO

A suburb in a major city has been infected by a foreign moth which could have devastating effects on

neighbouring farms and forests. An aerial spray campaign is launched where an aeroplane flies over the area

and spreads an insecticide, which though fatal to the moth is harmless to humans and other creatures.

The formula M = M0(0.85)t gives the number of moths (M) in the spray zone t days after the plane has sprayed.

M0 is the initial number of moths that the Ministry officials believe were in the zone. If they believe that 800

moths were present in the zone, how many days after spraying would it take the population to fall to 500

moths?

QUESTION THREE

A circular traffic island in the middle of an intersection is planned. The circle is represented by x2 + y2 = 36.

Also in the plan is a path of an electrical cable which runs underneath the traffic island. The cable path may

be shown by y = 2x + 6.

a. Find the x ordinates of the points where the cable meets the perimeter of the traffic island.

b. Find the y ordinates and hence write the points of intersection.

Algebra 13

YEAR 12 MATHEMATICS

QUESTION FOUR

A team of netballers and their supporters are fundraising in order to attend a Golden Oldies tournament in the

Cook Islands. One of their activities is a social at a local hall. The team has two options for pricing tickets to

this event.

Option Price A Option Price B

PA = PB =

Where x is the number of tickets sold Where x is the number of tickets sold

and the 9 best ticket sellers get free and the 12 best ticket sellers get free

tickets to the social. tickets to the social.

Solve PA = PB and find the minimum number of tickets which need to be sold so that the price of Option B

tickets would be less than the price of Option A tickets.

14 Algebra

YEAR 12 MATHEMATICS

QUESTION FIVE

The quadratic equation x2 - (k + 1)x + 4k = 0 has 2 roots.

If the difference between the roots is 1, find the value of k.

Algebra 15

YEAR 12 MATHEMATICS

ALGEBRA - PRACTICE TEST 2

QUESTION ONE

1. Expand and simplify: (x + 8)(3x - 1)(4x + 3)

2. Write this expression in positive index form:

3. Write as a single number:

4. Solve the equations:

a. logx 243 = 5 b. 3x2 - 8x = -4 c. 6x = 31

5. At the movies during the weekend, Moira served ice creams to a group of children from a birthday party.

Of the 9 she served, 7 wanted chocolate dipped while the other 2 wanted plain. If it cost a total of

$19.25 with a chocolate dipped ice cream being 50 cents more than a plain one, calculate the cost of

a plain ice cream.

16 Algebra

YEAR 12 MATHEMATICS

QUESTION TWO

Show that there is only one point of intersection between: x2 + y2 + 2x - 7 = 0 and y = x - 3.

QUESTION THREE

The height of a door is 1 metre longer than its width. The area of the door is 1.7 m2.

What are the dimensions of the door? (Give your answer to 1 dp.)

Algebra 17

YEAR 12 MATHEMATICS

QUESTION FIVE

After t hours of use, the value (V) of a certain brand of jetski (which was purchased new for $19 995) may be

estimated by: V = P(0.993)t where P is the retail price.

After how many hours of use would the jetski be worth of its original retail price?

18 Algebra

YEAR 12 MATHEMATICS

QUESTION SIX

A certain aeroplane can cover a distance of 5000 km travelling over a time, t hours, at a velocity v = .

If the same aeroplane flew the 5000 km again, this time increasing its speed by 250 km/h (i.e. v + 250), and

cutting the travelling time by an hour (i.e. t - 1), what would its speed have been in both instances?

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Graphs 19

YEAR 12 MATHEMATICS

MATHEMATICS 2.2


Draw straightforward non-linear graphs


C draw a quadratic graph whose equation

may be factorised, y = (cx - d)(x + e)

or expressed as y = ±(x - a)2 + b

C draw a polynomial graph whose equation could

be factorised (leading term (±1)xn )

C draw a rectangular hyperbola from an equation

such as y = or

C draw a circle from a given equation whose

centre is at (0, 0)

C draw an exponential function from a given

equation, y = ax

C draw a logarithmic function from a given

equation y = loga x

C show relevant features including: intercepts,

maxima or minima, asymptotes, symmetry

C draw rectangular hyperbola y =

C draw circles such as (x - a)2 + (y - b)2 = r2

C draw exponential functions y = ax - b + c

C draw log functions y = loga(x - b) + c

C write equations and interpret features of any

graphs of the above

C model any of the above (or combinations of) to

describe a situation, find points of intersection

and to solve related problems

From more

straightforward

relations

and their

graphs

to

graphs

with more

complexity

and more features.

Relations

given may have

coefficients

constraints and exponents

other than ± 1.

20 Graphs

YEAR 12 MATHEMATICS

GRAPHS - Revision Summary


tested in this Achievement Standard.

1. NON LINEAR GRAPHS

Draw graphs for each of the following. Insure that you include any intercepts, asymptotes, symmetry

and maxima or minima for quadratic curves.

a. i. y = x2 + 3x - 4

ii. y = (x + 3)2 - 4

iii. (x + 1)(x - 2)(x + 3)

b. i. y = ii. y =

Draw each of the following pairs of graphs on the same axes.

c. i. x2 + y2 = 49 ii. (x + 2)2 + (y - 1)2 = 49

d. i. y = 5x ii. y = 5x - 3

e. i. y = log8 x ii. y = log8 *x - 3*

Graphs 21

YEAR 12 MATHEMATICS

22 Graphs

YEAR 12 MATHEMATICS

Graphs 23

YEAR 12 MATHEMATICS

2. For each of the graphs sketched below, write the equations.

a. b.

c. d.

24 Graphs

YEAR 12 MATHEMATICS

3. Draw the curve xy = -1 and y = x - 2 on the same axes.

Label any points of intersection with the correct co-ordinates.

Graphs 25

YEAR 12 MATHEMATICS

GRAPHS - PRACTICE TEST 1

QUESTION ONE

Draw the graphs of the three equations below.

a. y = x2 - 4x -5 b. y = c. y = 4x

26 Graphs

YEAR 12 MATHEMATICS

QUESTION TWO

1. Identify THREE features of the graphs y = x2 - 3x + 5 and y + x = 4.

2. Millie bought a car for $11995 some years ago. She knows that the current value of her car may be

modelled by the equation: V = $11995(0.82)t, where V is the current value of her car and t is the number

of years since she bought her car.

a. Plot the graph of this equation of the car’s current value over the six years since she bought it.

b. During which year did the value of the car fall under half of the purchase price?

1.

2.

V (Value in $)

t(years)

Graphs 27

YEAR 12 MATHEMATICS

QUESTION THREE

Draw graphs of these equations:

a. (x - 2)2 + (y + 3)2 = 9

b. y = = 3 +

28 Graphs

YEAR 12 MATHEMATICS

QUESTION FOUR

Another car depreciation model that Millie discovered is given as: V = .

V is the value of the car (V) in dollars over t years.

The graph of the equation for the current value of the car is shown below.

a. What does the graph tell us about the rate at which the value of the car decreased?

b. What does the graph indicate about the value of the car after many years?

c. What does the y intercept tell us about the purchase price of the car?

V (Value in $)

t(years)

Graphs 29

YEAR 12 MATHEMATICS

QUESTION FIVE

For each of the graphs, write the equation.

y y

a. b.

x x

c.

30 Graphs

YEAR 12 MATHEMATICS

QUESTION SIX

Millie’s geology class has been studying volcanic crater lakes of the central North Island. One crater lake that

she studied had suddenly filled then burst one of its walls sending a torrent of water, mud and rock down the

mountain side. The data from the seismic monitoring station at the lake gave these figures:

Time Lake Depth Number of hours later

4pm, 10 Feb 6.6m (initial) 0

7am, 11 Feb 15.2m (burst) 15

10am, 13 Feb 8.0m 66

Millie models this situation with two hyperbolae (see graph below).

After the first 15 hours, the depth of the lake could be modelled by this hyperbole:

D is the depth of the crater lake (in metres) and t is the time (in hours) since the lake began to fill.

a. Write the equation for the (second) hyperbola which models the lakes depth after 15 hours.

b. Use your model equation above to estimate the time and the date when the crater lake returns to its

initial depth of 6.6 metres.

Graphs 31

YEAR 12 MATHEMATICS

GRAPHS - PRACTICE TEST 2

QUESTION ONE

a. Draw the graph of y = x(x - 1)(x + 3), showing all intercepts.

b. Draw the graph of y = 4 - (x + 1)2 showing key features.

32 Graphs

YEAR 12 MATHEMATICS

c. Draw the graph of y = log10 x.

d. Write three features of the circular graph illustrated below.

y

x

Graphs 33

YEAR 12 MATHEMATICS

QUESTION TWO

Write the equation of each of the following graphs.

a. b.

c.

34 Graphs

YEAR 12 MATHEMATICS

QUESTION THREE

Draw graphs of EACH of the following:

a. y = 2x2 - 3x - 5

b. y = -x3 + 1

Graphs 35

YEAR 12 MATHEMATICS

QUESTION FOUR

Draw the graph of y = 4 on the axes below for -5 # x #4.

y

x

36 Graphs

YEAR 12 MATHEMATICS

QUESTION FIVE

Helen and Don invest a sum of money into an education fund which compounds at 8% annually. The amount

in the account after t years may be given by the equation y = 45(1.08)t, where y the amount of money is in

hundreds of dollars. Below, a graph is given for the first 11 years.

a. What sum did Helen and Don invest initially?

b. If interest is calculated and added on at the end of every year, during which year would you expect

their original sum to have doubled?

c. 12 years after the fund began, Helen and Don need to withdraw $3000 for a family emergency. How

would this be represented on the graph?

Graphs 37

YEAR 12 MATHEMATICS

38 Graphs

YEAR 12 MATHEMATICS

QUESTION SIX

In the year of a general election a certain government department has been ordered to trim its spending (S)

according to this model equation: S = A - B log10 (x + 0.5), where S is in dollars, and x is the number of weeks

since the order was given.

By the end of Week 1, the Department has spent $107 449 for that week, but by the end of the tenth week,

their weekly spending was $64 011.

a. Find A and B (to the nearest $10), then rewrite the model equation with these values.

b. If the election was held seven months (30 weeks) after the order was given to the government

department, use your model equation to estimate how much had been spent by them in that election

week.

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Calculus 39

YEAR 12 MATHEMATICS

MATHEMATICS 2.3


Find and use straightforward derivatives and integrals


C find derivatives of polynomial expressions

such as 5x9 - 7x3 + 4

C find integrals of polynomial expressions

C relate the derived function to the gradient

of a curve

C relate the integral to the area under a curve

C use the derivative to find the gradient at a point

and locate the point given a gradient value

C use the integral to find a straight forward

area under a curve, and to extract an equation

given the gradient function

C use differentiation techniques to locate turning

points where f!(x) = 0, then determine their

nature(s), find the equation of a tangent to

a curve and solve rate of change problems

such as kinematics

C using integration techniques to find areas

(including compound) under polynomials

C use various calculus techniques to form

equations, to interpret results, to optimise

situations, to solve rates of change cases

(including kinematics) and to find relevant

areas

From more

straightforward

uses of

calculus techniques

and applications

and familiarity with

, f!(x)

and Idx notations

to

wider ranging

applications and contexts

involving those

techniques requiring

interpretation of the

solutions.

40 Calculus

YEAR 12 MATHEMATICS

CALCULUS - Revision Summary



1. Differentiate these functions with respect to x:

a. y = 3x9 - 5x2 + 7

b. f(x) = (3x - 5)(2x2 + 7) (Hint: expand, then find f !(x).)

2. Find the indefinite integrals for:

a. f ! (x) = 3x2 + 8x - 11

b. = 18x5 + + 1

3. a. Find f!(2) if f(x) = x4 - 5x2 + x.

b. Find the gradient of (the tangent to) the curve, y = x2 - 2x, at x = -1.

Calculus 41

YEAR 12 MATHEMATICS

4. For the curve y = x2 + 5x:

a. Find the equation of the tangent to this curve at (1, 6).

b. Find the equation of the tangent to this curve at x = 0.

5. Find the point on the curve y = x2 - 3x + 2 where the gradient is 1.

6. a. Evaluate the definite integral 3x2 . dx.

b. Find the area between the curve, y = 2x - x2 and the x axis between x = 0 and x = 2.

c. Find the total area between y = x(x - 1)(x + 3) and the x axis.

42 Calculus

YEAR 12 MATHEMATICS

7. Consider the curve, y = x3 - 3x = x(x2 - 3).

a. Find an expression for the gradient function .

b. Determine any turning points on y = x3 - 3x.

c. Along which values of x is the curve increasing and decreasing?

Calculus 43

YEAR 12 MATHEMATICS

8. A large model rocket is fired vertically into the air with an initial velocity of 245 m/s. After t seconds

the height of the rocket (h metres) is given by: h = 245t - 4.9t2.

a. Find an expression for the instantaneous velocity, v, of the rocket after t seconds.

b. What is the velocity of the rocket after 5 seconds?

c. What is the height of the rocket at the same time?

d. Show that the acceleration of the rocket is constant.

e. When does the rocket reach its maximum height above the ground, and what is this height?

44 Calculus

YEAR 12 MATHEMATICS

9. Optimisation situations require the use of calculus to find the maximum or minimum solution.

For example, in a new subdivision the developers wish to create rectangular sections, each having a

total boundary (or perimeter) of 108 m. What are the dimensions of such a rectangle, so that its area

could be a maximum?

Calculus 45

YEAR 12 MATHEMATICS

CALCULUS - PRACTICE TEST 1

Show ALL working.

QUESTION ONE

Find the gradient of the curve y = x3 - 6x - 5 at the point where x = 5.

QUESTION TWO

The graph shown below has the equation y = 3x2 + 1.

Calculate the shaded area.

46 Calculus

YEAR 12 MATHEMATICS

QUESTION THREE

The gradient function of a curve is f!(x) = 6x2 - 4x + 5.

The curve passes through the point (2, 11).

Find the equation of the curve.

QUESTION FOUR

Find the x co-ordinates of the two points on the graph of y = 2x3 - 6x + 8 where the gradient is parallel to the

x-axis.

Calculus 47

YEAR 12 MATHEMATICS

QUESTION FIVE

Find the equation of the tangent to the curve y = x3 - 3x2 - 7x + 1 at the point (-1, 4).

QUESTION SIX

Graeme returns to his car at the end of work and realises that he left the lights on, draining the battery.

Fortunately he parked on a slight slope earlier in the day, so he can roll the manual geared vehicle to push

start it.

As the vehicle slowly rolls forward, its velocity is given by v = 0.75t (m/s) where v = velocity in metres per

second and t = time in seconds from when the car begins to roll.

How far has the car rolled over the first 8 seconds?

48 Calculus

YEAR 12 MATHEMATICS

QUESTION SEVEN

Graeme designs rest areas along the edges of major highways. One of his more recent designs was the

computer designed area (part of which is shown as the shaded region on the graph below). As edges for the

area, he used these three equations:

y = 12 - 3x2

y = -36

and x = 1, where x and y are in metres.

Calculate the shaded (rest area) region .

Calculus 49

YEAR 12 MATHEMATICS

QUESTION EIGHT

To prevent flooding near a rural school the local council asked Graeme to design a drain along the rear

boundary. Graeme saves costs by designing a concrete structure to fit into an existing ditch.

Together the floor (width) and the heights of the two walls have a total

length of 5.6m.

Find the width of the floor which will allow the greatest flow of

stormwater through the cross-sectional area shown. Also, give this

maximum cross-sectional area.

50 Calculus

YEAR 12 MATHEMATICS

CALCULUS - PRACTICE TEST 2

Show ALL working.

QUESTION ONE

a. Find the gradient of the curve y = x4 - 3x2 + 5 at the point where x = 2.

b. Find the equation of the function which passes through the point (-1, 1) and whose gradient function

is = 8x3 + 6x2 - 4x - 1.

Calculus 51

YEAR 12 MATHEMATICS

c. Find the area under the curve, y = x3 + 2 for the values of x between 0 and 2.

d. Find the co ordinates of the point on the curve y = where the gradient is .

52 Calculus

YEAR 12 MATHEMATICS

QUESTION TWO

Find the area between the x - axis and the

curve y = (x + 1)(x - 4)

= x2 - 3x - 4

for values of x between 0 and 5.

Calculus 53

YEAR 12 MATHEMATICS

QUESTION THREE

3. An electronic powered model boat is being sailed on a small lagoon. Its velocity, in cm/s is given by:

v = 18 + 15t - 3t2 for 0 # t # 6

where t is the time in seconds after the boat is started.

a. After 2 seconds the boat is 65 cm from its owner who is controlling it from shore. How far was the

boat from the owner at the start?

b. Use calculus to find the maximum velocity of the boat.

54 Calculus

YEAR 12 MATHEMATICS

QUESTION FOUR

A manufacturer produces car polish in tin cans which have a volume of 335 cm3. Find the radius of the tin can

which requires the least amount of metal.

Note - for a cylinder V = πr2h and SA = 2πr2 + 2πrh.

planned orbit

actual orbit

C2

P

C1

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Coordinate Geometry 55

YEAR 12 MATHEMATICS

MATHEMATICS 2.4


Use coordinate geometry methods


C find the mid-point between 2 points

C find the distance between 2 points

C find the equation of a line

C find the equation of a parallel line

C find the equation of a perpendicular line

C find the coordinates of the point of intersection

of 2 lines

C find equations of medians, perpendicular

bisectors and altitudes

C formulate a proof (i.e. geometric cases relying

on the above techniques

C prove points are collinear

C prove more challenging situations

C solve more challenging contextual problems

From

straightforward contexts

in two dimensions

to

situations with

more complexity

including

three dimensional

possibilities

and

extended chains of

reasoning.

56 Coordinate Geometry

YEAR 12 MATHEMATICS

COORDINATE GEOMETRY - Revision Summary



1. Find the midpoint between:

a. (0, 5) and (4, 11)

b. (-1.2, 6) and (4.4, -7.6)

2. Find the distances between the points in Question 1.

3. What is the gradient of the line 2y - 3x = 7 ?


YEAR 12 MATHEMATICS

4. Find the equation of the line which passes through (-3, 2) and (1, 5).

5. A line has the equation 2x + 5y - 8 = 0.

Write the equation of a line (in the form ax + by + c = 0) which is:

a. Parallel to and passing through (1, -2).

b. Perpendicular to and passing through (5, 6).


YEAR 12 MATHEMATICS

6. Find the points of intersection of these pairs of lines:

a. 3x + 4y = 10

5x + 3y = 13

b. 0.1x - 0.2y - 0.7 = 0

0.4x + 0.3y - 0.6 = 0


YEAR 12 MATHEMATICS

7. P(1, 5), Q(3, 2) and R(-3, -1) are the vertices of a triangle.

a. Find the equation of the median drawn from R to the midpoint of .

b. Find the equation of the altitude from P to .

c. Find the equation of the perpendicular bisector of .


YEAR 12 MATHEMATICS

8. Prove that the triangle ªOPQ is isosceles.


YEAR 12 MATHEMATICS

COORDINATE GEOMETRY - PRACTICE TEST 1

A training track for horses has the shape shown. Use coordinate geometry techniques to solve all questions.

Use the axes on the grid below to help answer these questions.

Note the grid lines are 20 m apart.


YEAR 12 MATHEMATICS

QUESTION ONE

a. Calculate the distance along the length of the back straight between N(4, 11) and S(-10, -2).

b. Find the equation of the line along this back straight.

c. A fence line passes through the point (-2, 8) and follows a path parallel to the line y = x + 5.

Find the equation of this fence line.


YEAR 12 MATHEMATICS

QUESTION TWO

A drain runs along a straight line equidistant between the points (1, -2) and (3, -4).

Find the equation of the line which the drain follows.


YEAR 12 MATHEMATICS

QUESTION THREE

The ends of the back straight N(4, 11) and S(-10, -2) form a triangle with a trough at T(2, 0).

Find the equation of the median of this triangle through N(4, 11).


YEAR 12 MATHEMATICS

QUESTION FOUR

The equation of the road between the ends S(-10, -2) on the back straight, and P(-2, -10) on the front straight

is x + y + 12 = 0.

The altitude of the triangle SPT, through the horse trough (2, 0) is given by the equation: x - y - 2 = 0.

Calculate the length of the altitude of the triangle SPT through vertex T(2, 0).


YEAR 12 MATHEMATICS

QUESTION FIVE

The farmer who owns the property wishes to move the back straight of the track so that it now runs along the

line y = x + 8.

Calculate the closest distance this new piece of track comes to a new trough planned at the position (6, 0).


YEAR 12 MATHEMATICS

COORDINATE GEOMETRY - PRACTICE TEST 2

Part of a mini golf course is shown, with the first five tees (T, symbol !) and holes (H, symbol F). A plan is

on the office wall, set to a grid system. The office has co-ordinates (0, 0) and some of the other tees and holes

have been given coordinates. Two electrical cables run under the pond, from T3 to T2 and from L3 to L4 and

are shown by dashed lines.

Every unit represents one metre.

The diagram is not drawn to scale.


YEAR 12 MATHEMATICS

QUESTION ONE

a. A solar light, L2, is located halfway between the second tee T2 (25, 15) and the second

hole H2 (11, 28). Find the coordinates of L2.

b. What is the equation of the line from T2 (25, 15) to T3 (1, 24)?

c. The equation of the path from hole two at H2 to the third tee, T3 is y = x - .

Write the equation of a line which is parallel to at and which passes through the bend

point, B(10, 26).


YEAR 12 MATHEMATICS

QUESTION TWO

An old cable running from the bend at C(19, 12), under the pond, meets the service path at the midpoint

between the two tees, T1(11, -2) and T3(1, 24). What is the length of this underground cable?


YEAR 12 MATHEMATICS

QUESTION THREE

Consider a triangle formed by the three points T3, H2 and T2.

Show how the equation of the altitude of the triangle T3, H2, T2 which passes through the vertex at T2 is

5x + 2y - 155 = 0.


YEAR 12 MATHEMATICS

QUESTION FOUR

The path for the fifth hole has two ‘legs’. The first leg starts at T5(9, 0) and runs perpendicular to the line T1

H1 until it gets to the bend at D. From D, the second leg runs along a path which is perpendicular to the line

BH3, ending at the hole H5 (3, 15).

What are the co-ordinates of the bend at D?


YEAR 12 MATHEMATICS

QUESTION FIVE

Another light is going to be located halfway between T3 and B so that the area of the north side of the

pond can be lit up. A cable will run from this point and be connected to the existing cable which runs

between T3 and T2.

What will be the shortest distance between the new light and the line T3 T2?

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Sample Statistics 73

YEAR 12 MATHEMATICS

MATHEMATICS 2.5


Select a sample and use this to make an inference about the population


C select a sample from a population

(possibly supplied)

C provide evidence of the method

C use appropriate sample statistics such as: mean

median, quartiles, standard deviation, and

proportions from the selected sample

C describe the sampling method so that another

person could repeat the process

C comment on whether the sample is truly

representative, or shows bias

C refer to sample statistics (and suitable graphs)

to help justify the above position

C when evaluating the sampling process, consider

limitations of, and possible improvements

to reliability

C evaluate the accuracy of the results, considering

an improved interval for the question

C refer to how the data is distributed

From

more straightforward

inferences

taken from a sample

about the population

to

fuller justification

of the method(s) used

leading to

a more critical

evaluation

of the whole sampling

process

and

the results.

74 Sample Statistics

YEAR 12 MATHEMATICS

SAMPLE STATISTICS - Revision Summary

The exercises in this section have been specifically chosen to underpin the Achievement Criteria that will

be tested in this Achievement Standard.

1. Find the mode and the mean of this data set:

109, 59, 130, 40, 42, 33, 32, 40, 71, 138, 203, 162, 84,

136, 154, 39, 55, 36, 30, 40, 33, 36, 38, 87, 119, 36.

2. The number of nose-to-tail accidents during Labour weekend along a particular stretch of motorway

is given below for each direction. The results have been recorded over the last 2 decades.

North Bound Lanes: 14, 11, 25, 19, 10, 8, 23, 34, 21, 15, 20, 12, 25, 11, 13, 31, 10, 5, 33, 22

South Bound Lanes: 16, 18, 8, 14, 24, 2, 18, 13, 11, 21, 29, 17, 5, 16, 14, 20, 10, 9, 16, 30

a. Create a back to back stem and leaf plot for each data set.

b. List the five point summary for each data set (high, median, quartiles, low).

c. Form box and whisker plots for each on the same grid.

d. Calculate the mean number for each direction.


YEAR 12 MATHEMATICS

Price Midpoint

of Interval

Frequency

$50 - $55 1

$60 - $65 3

$70 - $75 9

$80 - $85 4

$90 - $95 2

$100 - $105 2

$110 - $115 3

$120 - 130 $125 1

3. For the frequency table:

x 1 2 3 4 5

y 13 9 4 2 2

a. Calculate the mean.

b. Calculate the standard deviation of the data.

4. The cost of a single person to stay one night at a motel in an east coast city varies considerably.

This frequency table shows the tariffs which were charged in 2005.

a. In which interval does the median price lie?

b. Estimate the mean price for a single person to stay

in a motel.

c. Draw a cumulative frequency curve of the data.


YEAR 12 MATHEMATICS

Each table below has columns to assist the manual calculation of the sample mean and sample standard

deviation.

a. Complete each table, then use the formulae to find each sample statistic.

b. Verify these answers, by using your calculator to find the values.

5. 6.

x x - (x - )2 x f x.f (x - )2 f.(x - )2

3 -5 25 11 2

6 -2 14 8

7 15 13

7 16 16

9 21 9

10 2 22 7

14 6 36 25 5

Totals 56 - Totals

= =

= =

S = S =

= =


YEAR 12 MATHEMATICS

SAMPLE STATISTICS - PRACTICE TEST 1

Select a sample and use this to make an inference about the population.

THE SITUATION

A newspaper claims that recent significant earthquakes seemed to strike more often in the early hours of

the morning rather than at any other time of the day.

YOUR TASK

Your local newspaper editor has asked you to check this claim and estimate which times of the day of an

arbitrarily chosen year had the most quakes.

YOU NEED TO:

a. Choose a sampling method and use it to obtain a representative sample of at least 30 significant

earthquakes.

b. Describe your sampling process clearly so that someone else can follow it.

c. Justify your choice of sampling method by describing the decisions made and the reason for these

decisions.

d. List the data for your sample that you gathered from the data sheet.

e. Explain whether your sample is actually representative of your population or not. (You do not have

to select another sample if it is not.)

f. Check appropriate statistics for your sample and use this to estimate when in the day there were

significant earthquakes.

g. Write a short paragraph outlining what you have found (from your estimates), and what you could

conclude about when a significant earthquake occurred. Comment on the reliability of your estimate

and therefore your conclusion.

h. Evaluate the sampling and statistical processes you have used. Comment on things such as:

C reliability of your sampling process

C limitations of your sampling process

C the accuracy of your estimate - when an earthquake of this size was most likely

C distribution of the data

You need at least three valid statements.


YEAR 12 MATHEMATICS

DATA SHEET - Significant Earthquakes of the World - 2005

(of magnitude 6.5 or greater and/or causing fatalities, injuries or substantial damage)

Data from US Geological Survey, Earthquake Hazards Program: http://earthquake.usgs.gov/eqcenter/eqarchives/significant/sig_2005.php

Downloaded 13 Jan 2006.

Date and Time Magnitude Location

Jan 01, 0625 6.7 off West coast of N. Sumatra

Jan 10, 1847 5.4 N. Iran

Jan 10, 2348 5.5 W. Turkey

Jan 12, 0840 6.8 Central-mid Atlantic Ridge

Jan 16, 2017 6.6 Yap, Micronesia

Jan 19, 0611 6.6 off E. Honshu, Japan

Jan 23, 2010 6.3 Sulawesi, Indonesia

Jan 25, 1630 4.8 Yunnan, China

Jan 25, 1644 5.9 Turkey - Iraq border

Feb 02, 0555 4.8 Java, Indonesia

Feb 05,0334 6.6 Anatchan, N.Mariana Is

Feb 05, 1223 7.1 Celebos Sea

Feb 08, 1448 6.8 Vanuatu

Feb 14, 2338 6.1 S. Xinjiang China

Feb 15, 1442 6.6 Kepulauan, Indonesia

Feb 15, 1946 5.5 off S. Honshu, Japan

Feb 16, 2027 6.6 S. mid-Atlantic Ridge

Feb 19, 0004 6.5 Sulawesi, Indonesia

Feb 22, 0225 6.4 Central Iran

Feb 26, 1256 6.8 Simeulue, Indonesia

Mar 02, 1042 7.1 Banda Sea

Mar 02, 1112 4.9 Pakistan

Mar 05, 1906 5.8 Taiwan

Mar 09, 1015 5.0 S.Africa

Mar 12, 0736 5.7 E.Turkey

Mar 14, 0155 5.8 E. Turkey

Mar 14, 0943 4.9 Maharashtra, India

Mar 20, 0153 6.6 Kyushu, Japan

Mar 21, 1223 6.9 Salta, Argentina

Mar 28, 1609 8.7 N.Sumatra, Indonesia

Apr 10, 1029 6.7 Kepulauan, Indonesia

Apr 10, 1114 6.5 Kepulauan, Indonesia

Apr 11, 1220 6.7 N. Coast of N.Guinea

Apr 11, 1708 6.8 S.E. of Loyalty Islands

Apr 19, 2111 5.5 Kyushu, Japan

May 01, 1623 4.5 Kyushu, Japan

May 03, 0721 4.9 W. Iran

May 05, 1912 6.5 S. of Panama

May 12, 1115 6.5 Pacific-Antartic Ridge

May 14, 0505 6.8 Nias Region, Indonesia

May 16, 0354 6.6 S. of Kermadec Islands

May 19, 0154 6.9 Nias Region, Indonesia

May 23, 0609 4.3 S. Africa

Date and Time Magnitude Location

Jun 04, 1450 6.1 E. New Guinea, PNG

Jun 06, 0741 5.7 E. Turkey

Jun 13, 2244 7.8 Tarapaca, Chile

Jun 14, 1710 6.8 Aleutian Is, Alaska

Jun 15, 0250 7.2 off N. Californian Coast

Jun 15, 1952 6.5 off Aisen coast, Chile

Jun 16, 2053 4.9 greater L.A. area, California

Jun 17, 0621 6.7 off N. Californian coast

Jun 20, 0403 4.7 off W. Honshu, Japan

Jul 02, 0216 6.6 off Nicaragua Coast

Jul 05, 0152 6.7 Nias Region Indonesia

Jul 05, 1653 2.7 S. Africa

Jul 23, 0734 6.0 off S. Honshu, Japan

Jul 24, 1542 7.3 Nicobar Is, India

Jul 25, 1543 5.0 Heilongjiang, China

Aug 05, 1414 5.2 Yunnan, China

Aug 13, 0458 4.8 Yunnan, China

Aug 16, 0246 7.2 off E. Honshu, Japan

Aug 21, 0229 5.1 off W. Honshu, Japan

Sep 09, 0726 7.7 New Ireland region, PNG

Sep 24, 1924 5.6 Ethiopa

Sep 26, 0155 7.5 N. Peru

Sep 29, 1550 6.7 New Britain region, PNG

Oct 01, 2219 5.3 S. Peru

Oct 08, 0350 7.6 Pakistan

Oct 15, 0424 5.2 SW Kashmir

Oct 15, 1551 6.5 NE of Taiwan

Oct 16, 0705 5.1 E. Honshu, Japan

Oct 20, 2140 5.9 off W. Turkey

Oct 27, 1118 4.2 Guangxi, China

Oct 29, 0405 6.5 S.E. Indian Ridge

Nov 06, 0211 5.2 Pakistan

Nov 08, 0754 5.1 S.China Sea

Nov 14, 2138 7.0 off E.Honshu, Japan

Nov 17, 1926 6.9 Potosi, Bolivia

Nov 19, 1410 6.5 Simeulue, Indonesia

Nov 26, 0049 5.2 Huber-Jiangxi, China

Nov 27, 1022 6.0 S. Iran

Dec 02, 1313 6.5 off E. Honshu, Japan

Dec 05, 1219 6.8 L. Tanganyila Region, Tanzania

Dec 11, 1420 6.6 N. Britain Region, PNG

Dec 12, 2147 6.6 Hindu Kush region, Afghanistan

Dec 13, 0316 6.7 Fiji region

Dec 24, 0201 4.5 W. Honshu, Japan


YEAR 12 MATHEMATICS


YEAR 12 MATHEMATICS

SAMPLE STATISTICS - PRACTICE TEST 2

THE SITUATION

A marine farm, about 800m offshore, has ropes anchored to the sea floor attached to floating longlines at

the surface. Shellfish grow along each rope and these are shown as numbered spaces on the map (of the

farm layout) below.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

30 29 28 27 26 25 24 23 22 21 20 19 18 17 16

31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

214 213 186 185 158 157 130 129 102 101 74 73 46

215 212 187 184 159 156 131 128 103 100 75 72 47

216 211 188 183 160 155 132 127 104 99 76 71 48

217 210 189 182 161 154 133 126 105 98 77 70 49

218 209 190 181 162 153 134 125 106 97 78 69 50

219 208 191 180 163 152 135 124 107 96 79 68 51

220 207 192 179 164 151 136 123 108 95 80 67 52

221 206 193 178 165 150 137 122 109 94 81 66 53

222 205 194 177 166 149 138 121 110 93 82 65 54

223 204 195 176 167 148 139 120 111 92 83 64 55

224 203 196 175 168 147 140 119 112 91 84 63 56

225 202 197 174 169 146 141 118 113 90 85 62 57

226 201 198 173 170 145 142 117 114 89 86 61 58

227 200 199 172 171 144 143 116 115 88 87 60 59

228 229 230 231 232 233 234 235 236 237 238 239 240 241

255 254 253 252 251 250 249 248 247 246 245 244 243 242


YEAR 12 MATHEMATICS

YOUR TASK

The owners of the marine farm have asked you to help estimate the average mass of shell fish per rope in

the farm.

By selecting a sample of the ropes:

1. Choose a sampling method and use it to design a sampling process to obtain a representative sample

of at least 30 shellfish ropes. Explain your process clearly so that someone else could follow it.

2. Justify your choice of sampling method taking into account the population and considerations of

bias. You may like to describe any decisions you made and the reasons for these decisions.

3. Use your sampling process to select a sample of shellfish ropes.

4. List the data for your sample which you gathered from the data sheet.

5. Explain whether your sample is representative of your population or not. You do not have to select

another sample if it is not.

6. Calculate appropriate statistics for your whole sample.

7. Use your sample and the statistics you have calculated to:

a. Estimate the average mass of shellfish per rope for the whole farm.

b. Estimate a suitable measure of spread for the mass of shellfish per rope for the whole farm.

8. Use your estimate to inform the marine farm owners of your conclusion as to the average mass of

shellfish per rope in the farm. Justify your estimates and therefore your conclusion.

9. Evaluate the sampling process you used. Comment on things like:

C the reliability of your sampling process

C limitations of your sampling process and ways to improve it

C accuracy or appropriateness of your estimate

C the distribution of your data

You need at least 3 valid statements.


YEAR 12 MATHEMATICS

Table showing mass (in kg) of shellfish on each rope.

1 404 38 488 75 165 112 358 149 159 186 370 223 429

2 413 39 187 76 168 113 269 150 480 187 208 224 446

3 160 40 449 77 508 114 504 151 394 188 139 225 331

4 451 41 397 78 227 115 291 152 257 189 520 226 314

5 337 42 309 79 207 116 189 153 187 190 357 227 180

6 477 43 357 80 186 117 164 154 248 191 499 228 401

7 499 44 255 81 395 118 310 155 378 192 174 229 360

8 186 45 185 82 459 119 195 156 389 193 411 230 219

9 445 46 509 83 191 120 479 157 135 194 504 231 208

10 108 47 185 84 247 121 178 158 407 195 463 232 352

11 216 48 307 85 279 122 484 159 220 196 458 233 267

12 173 49 345 86 468 123 207 160 249 197 285 234 409

13 286 50 175 87 247 124 161 161 281 198 308 235 190

14 191 51 143 88 366 125 252 162 192 199 348 236 167

15 458 52 309 89 336 126 264 163 257 200 240 237 480

16 145 53 365 90 207 127 518 164 294 201 348 238 375

17 425 54 506 91 225 128 210 165 485 202 257 239 385

18 447 55 361 92 275 129 158 166 139 203 381 240 251

19 306 56 490 93 177 130 195 167 460 204 190 241 310

20 327 57 440 94 164 131 338 168 501 205 507 242 428

21 240 58 336 95 458 132 406 169 409 206 214 243 228

22 385 59 359 96 493 133 311 170 350 207 460 244 177

23 438 60 159 97 456 134 411 171 167 208 338 245 201

24 447 61 498 98 168 135 403 172 506 209 464 246 355

25 447 62 291 99 501 136 367 173 298 210 208 247 196

26 405 63 155 100 305 137 490 174 373 211 447 248 265

27 248 64 158 101 413 138 362 175 149 212 191 249 408

28 318 65 429 102 307 139 176 176 334 213 268 250 506

29 165 66 370 103 361 140 195 177 381 214 452 251 177

30 316 67 476 104 489 141 461 178 393 215 295 252 236

31 379 68 257 105 472 142 486 179 232 216 208 253 325

32 354 69 187 106 468 143 348 180 407 217 465 254 297

33 445 70 184 107 175 144 278 181 446 218 260 255 434

34 245 71 344 108 343 145 411 182 160 219 162

35 202 72 440 109 399 146 287 183 155 220 259

36 407 73 187 110 516 147 396 184 274 221 416

37 500 74 410 111 479 148 404 185 362 222 382


YEAR 12 MATHEMATICS

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5

Probability & Normal Distribution

87

YEAR 12 MATHEMATICS

From

a straightforward

simulation process

and

normal distribution

calculations

to

using theoretical

probability

and

the normal distribution

in contexts

requiring several steps

and

further interpretation of

the model and the results.

MATHEMATICS 2.6


Simulate probability situations and apply the normal distribution


C design a simulation process using, for example,

coins, dice or random numbers

C use z = (the standard normal

transformation)

C find expected numbers

C apply theoretical techniques using: probability

trees, tables, informal conditional probability

C make predictions from simulations

C apply the normal distribution using more than

one z-value

C find expected numbers from theoretical

probability and normal distribution cases

C inverse normal distribution problems

C interpret results from normal distribution cases

and make recommendations

C relate the results of one simulation to a second

simulation

C combine theoretical and experimental

probabilities

C discuss any limitations of the model (or

process) used

88 Probability & Normal Distribution

YEAR 12 MATHEMATICS

PROBABILITY & NORMAL DISTRIBUTION

- Revision Summary



1. Find the probabilities of each event:

a. Choosing a vowel out of all the letters in the name “WAIKAREMOANA”.

b. Rolling a die and not selecting a prime number.

c. Picking a Jack or a Queen or a King from a shuffled deck of cards (Jokers removed).

2. A wheel is divided into 5 sections labelled A, B, C, D and E.

The wheel is spun around its centre, then a dart is thrown at it.

i. What is the probability that the dart lands in:

a. Region A?

b. Regions C or E?

c. Not in region B?

d. In regions A, B or D?

ii. If the wheel was spun 150 times, how many hits would you expect in region

B?


89

YEAR 12 MATHEMATICS

3. Two dice are thrown and the sums of the top two faces are recorded.

a. What is the chance that the sum is an even number?

b. What is the probability that the sum is 7?

c. What is the probability that the sum has double digits?

d. What is the probability that the sum is a square number?

e. Find the probability that the sum is a multiple of three.

(First draw a table of possibilities.)

4. Niko is a promising young school athlete who has been training on these throwing events - discus,

shot put and hammer throw for the local championships. He knows that he has a 70% chance of

winning the discus, an 80% chance of winning the shot put and a 75% chance of winning the hammer

throw. The order of the events is as above. Assume the result of each event is independent of the

others.

a. Draw a probability tree of the possible outcomes.

b. What is the probability that Niko wins all three events?

c. Find the probability that Niko wins any two of the three events.

d. Given that Niko wins the first event (discus) what is the chance that he loses the next two?


YEAR 12 MATHEMATICS

5. A paint manufacturer supplies tins of paint which are normally distributed with a mean volume of

4120 ml and a standard deviation of 60ml.

a. What is the probability that a randomly chosen tin of paint has between 4000 ml and 4180 ml?

b. If the manufacturer claims that each tin contains 4 l, what is the chance that a randomly

selected tin has less than 4 l.


91

YEAR 12 MATHEMATICS

6. Use the Normal Distribution table (page 171) to help answer these questions.

a. Find: b. Calculate:

c. A normal distribution has a mean of 55 and a standard deviation of 8.

Use the conversion formula to find:

i. Pr( x > 61 ).

ii. Pr( 50 < x < 66 ).


YEAR 12 MATHEMATICS


PRACTICE TEST 1

Show ALL working.

There are two parts to this activity:

Section A Requires you to design and carry out a simulation to answer questions.

Section B Tests your knowledge of the Normal Distribution. For each question you should write

correct probability statements and show working to support your answer.

These formulae may be useful:

z = or z =

The Good Garden Bag Company provides large bags for people to put their garden clippings into. These

large bags are then collected once a month and replaced by an empty bag. The clippings are tipped into a

truck which compacts then transports the clippings to a composting site.

The collectors on each truck monitor the clippings they pick up - they suspect that 10% of the bags will

have bamboo, while 25% will contain flax. When either bamboo or flax is discovered the homeowner is

given a warning letter which advises them to refrain from placing either of these (forbidden) items in their

bag.

The Good Garden Bag company asks you to investigate the situation. They believe that the presence of

bamboo and flax are independent of each other. (Both bamboo and flax are not wanted since these two

plants do not break down very readily and they may jam the auger at the composting plant).

SECTION A

Design a way to simulate the bag collection of a randomly selected truck, to find out how many of the next

80 homeowners will need to be given the written warning letter.

You need to:

1. a. Describe a method you use in sufficient detail so that another person could repeat it again with

your help.

b. Carry out at least 80 trials of the simulation.

c. Record the result of each trial of the simulation, e.g. in a table.

d. Use the results of your simulation to find the number of homeowners who will receive a letter of

warning for placing these materials in their bags.


93

YEAR 12 MATHEMATICS


YEAR 12 MATHEMATICS

2. In a normal day a driving team could collect 200 bags.

a. Use theoretical probability to calculate the expected number of households who would receive a

warning letter.

b. Use theoretical probability to calculate how many of the 200 households would be expected to

have both bamboo and flax in their garden bags.

c. Use the results of your simulation to find the expected number of homeowners who should get

the letter if 200 households have their bags collected.

d. Use the results of your simulation to find the expected number of households of the next 200 that

would get a letter for having both flax and bamboo in their bags.


95

YEAR 12 MATHEMATICS

3. Compare the results of your simulation with the theoretical probability. Make at least one comment

about your simulation.

You could comment on any similarities or differences between the simulation results and the

theoretical probability, or you could comment on ways in which your simulation could be improved

so that it is a better model.


YEAR 12 MATHEMATICS

SECTION B

The operators also gather evidence on the masses of the bags they collect. The maximum limit for each

bag should be 100 kg because manoeuvring these down driveways and paths to a truck can be very

challenging. Analysis of the results showed that the masses of bags were normally distributed with a mean

of 84 kg and a standard deviation of 8.5 kg.

Note: A suitably shaded design or use of proper notation is the minimum working expected.

4. Find the probability that a randomly chosen bag:

a. weighs between 84 kg and 90 kg

b. weighs under 94 kg

c. weighs under the maximum.


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YEAR 12 MATHEMATICS

5. a. Find the percentage of bags:

i. which weighed between 100 kg and 105 kg

ii. which weighed over 105 kg, so that the homeowner receives a letter reminding them of the

weight limit.

b. Bags which weigh over 104 kg are monitored for statistical purposes. Out of a 4 day collection of

850 bags, how many would be expected to be over 104 kg?


YEAR 12 MATHEMATICS

6. Very light bags are considered by some drivers as a waste of time, since they still require one person

to collect, replace and load the bag. If 8 bags out of a sample of 175 were found to be “too light”,

use the Normal Distribution and this information to describe a “very light bag” in terms of its mass.


99

YEAR 12 MATHEMATICS


PRACTICE TEST 2

Show ALL working.

Vai is collecting a set of 5 movie character wrist bands which can be found inside packets of Revita cordial

drinks. The manufacturer of Revita experienced some packing machine malfunctions with 30% of the packs

containing 2 wrist bands instead of 1 band.

1. Design a simulation to predict the number of packets of Revita that are required to obtain a full set

of movie character wrist bands. Describe this simulation in sufficient detail so that another person

could repeat it without your help.


YEAR 12 MATHEMATICS

2. Carry out the simulation and record your results.

3. Use your simulation to work out how many packets of Revita are needed on average to get a

complete set.


101

YEAR 12 MATHEMATICS

4. In Vai’s class there are 28 students altogether (including Vai) trying to collect the 5 wrist bands.

Use the results of your simulation to estimate:

a. How many students will have 4 different wristbands in the set after collecting 5 wristbands.

b. How many students would be expected to have the full set of 5 wrist bands once they have

collected less than or equal to 10 wristbands?

5. After a while, 25 of the 28 students in Vai’s class have only one wrist band to collect to make the

complete set.

a. Use theoretical probability to predict how many of these 25 students will complete their set with

the next packet of Revita.

b. Use theoretical probability to predict how many of these 25 students will complete their set

given that the next packet contains two wrist bands.


YEAR 12 MATHEMATICS

The weights of Revita Packs of cordial are normally distributed with a mean of 315 grams and a standard

deviation of 6 grams.

6. What is the probability that a pack of Revita cordial weighs between 315 grams and 323 grams?

7. What proportion of Revita packets weigh less than 323 grams?

A large carton delivered to a local supermarket contains 950 packets of Revita cordial.

8. How many of these packets would you expect to weigh less than 310 grams?


103

YEAR 12 MATHEMATICS

9. How many of these packets would you expect to weigh between 310 grams and 321 grams?

10. What is the chance that a packet of Revita cordial will weigh between 321 and 324.5 grams?

11. The manufacturer of Revita regularly checks the machines which weigh the packets of cordial.

Packets under 300 grams or over 325 grams are rejected. What proportion of the packets will the

manufacturer reject?


YEAR 12 MATHEMATICS

12. One day 599 packets are checked and 3 are found to be very light. If the machine is functioning

normally, what is the maximum weight of a very light packet according to this sample?

13. Calculate the mean weight that the machine needs to be set at so that 85% of Revita cordial packets

exceed 308 grams? (Assume the same standard deviation.)

planned orbit

actual orbit

C2

P

C1

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Sequences

105

YEAR 12 MATHEMATICS

Work

with straightforward cases

where a, d and r

are evident

to

situations

where other techniques

are used

to solve problems

which

require further

interpretation.

MATHEMATICS 2.7


Solve straightforward problems involving arithmetic and geometric

sequences


C find general terms of an arithmetic progression

(AP)

C find general terms of a geometric progression

(GP)

C find partial sums of an AP

C find partial sums of a GP

C find the sum to infinity, S4 of a GP

C use sigma (3) notation

C manipulate formulae to find a, d or r

C apply the techniques in contexts such as

radio active decay, increasing/decreasing %

and experiments which create sequences

C use logarithmic equations to find n in GP’s

C compare sequences

C discuss long term effects from the results

(process) used

106 Sequences

YEAR 12 MATHEMATICS

SEQUENCES - Revision Summary



1. Write the first four terms of the sequence whose nth term is defined by tn = n + .

2. Write down the next two terms of 0, 3, 8, 15, ......, .......

3. Find the first term, a, and the common difference, d, for the arithmetic progression which has

t4 = 4, t5 = 7 and t6 = 10.

4. Find the eighth term and the sum of the first sixteen terms for the arithmetic progression

8, 14, 20, ......, .......

Sequences 107

YEAR 12 MATHEMATICS

5. An Arithmetic Progression has t1 = 8 and t12 = 41.

Find the general term, and the sum of the first 13 terms.

6. How many terms of the series, 7+9+11+13+15+ ...... = 352 (i.e. find n when Sn = 352).

7. For the Geometric Progression 3, 9, 27, ...... find the next two terms, and the eleventh term.

8. A Geometric Progression has a fifth term of -80 and a sixth term of 160. Find an expression for the

general term, tn.

108 Sequences

YEAR 12 MATHEMATICS

9. a. Find the sum to twelve terms of the series 3200 + 1600 + 800 + ......

(Leave your answer as a fraction.)

b. Find the sum to infinity for the same series.

10. Evaluate:

a. 3k + 1

b. (2n - n2)

Sequences 109

YEAR 12 MATHEMATICS

SEQUENCES - PRACTICE TEST 1

Show ALL working.

QUESTION ONE

Carol spends 15 minutes texting her friends on the first day she bought her new mobile phone.

She spends 19 minutes texting on Day 2.

She spends 23 minutes texting on Day 3.

She continues texting daily at the same rate.

a. How long will she spend texting on the sixteenth day?

b. How much time will she spend texting on her new mobile phone over the first sixteen days?

110 Sequences

YEAR 12 MATHEMATICS

QUESTION TWO

Carol’s friend, Tony, spends 24 minutes texting his friends and family on the first day he got his mobile

phone. Each day he increases the time spent texting by 5% from the day before (i.e. he spends 1.05 times

as many minutes as he did the previous day).

What is the total time Tony has spent on his phone in the first 20 days?

QUESTION THREE

The Hong family lease a home security system and make payments every month. The payments reduce

each month by the same percentage. They paid $68 in the second month. They paid $49.13 in the fourth

month and $35.50 in the sixth month and so on.

How much did they pay in the first month?

Sequences 111

YEAR 12 MATHEMATICS

QUESTION FOUR

Tony buys a DVD for $495.

The DVD depreciates at the rate of 26% per year.

How many years to the nearest year, will it take for the DVD to reduce to one-sixth of its original value?

QUESTION FIVE

One of Tony’s uncles helps him to save money by employing him after school. In week 1 he pays Tony $96.

Each week, the uncle reduces the hours and the pay by 17.5% so that Tony can return slowly to full time

study. If this continued indefinitely, how much would Tony’s uncle pay him in total?

112 Sequences

YEAR 12 MATHEMATICS

QUESTION SIX

After several months of using her new mobile phone, Carol decides to reduce her texting by the same

number of minutes each week.

In Week 7, she texted for 595 minutes and by the end of Week 21, her total texting had amounted to

11,235 minutes.

How many minutes did Carol text in Week 1 of her texting reduction plan?

Sequences 113

YEAR 12 MATHEMATICS

SEQUENCES - PRACTICE TEST 2

Show ALL working.

QUESTION ONE

The rungs of a triangular painting trestle decrease uniformly in length. The bottom

rung is 88 cm and each successive rung is 2.75 cm shorter than the previous (lower)

rung. If there are 13 rungs on each trestle, what is the length of the top rung?

QUESTION TWO

A painter, Moe, has just bought some roller blades and is keen to practise as much as he can. On the first

evening after work he roller blades 6 km. Each evening he increases this by 1.25 km more than the

previous evening. If his first day of roller blading was on October 1, what was the total distance Moe would

have roller bladed by the end of October (31 days)?

114 Sequences

YEAR 12 MATHEMATICS

QUESTION THREE

On one of the painting jobs he had, Moe noticed a Yucca plant on the sundeck of the house. The height of

the Yucca was 95 cm and each week he noticed that it grew upwards by another 4% (i.e. 1.04 times taller).

Calculate the height of the Yucca at the end of week 10 (the nearest cm).

QUESTION FOUR

On another job, Moe saw the owner planting bamboo along a border. In week 1 it grew 48 cm after initially

being at ground level. Each week the bamboo’s extra growth length is 12.5% less than the previous week.

What is the maximum height the bamboo will ever grow?

Sequences 115

YEAR 12 MATHEMATICS

QUESTION FIVE

In a park a fountain and statue are surrounded by circular concrete rings.

Moe and his team must repaint the statue (S) with anti-graffiti paint. The concrete rings are centred

around the fountain and have different concrete textures. The percentage increase in area from one ring

to the next is constant. The first ring has an area of 201 m2 while the third ring has an area of 547.22 m2.

What is the area of the outside ring (Ring 4)?

QUESTION SIX

In the same park as the statue and fountain, a retaining wall has been constructed to hold in an earth

bank. Twelve horizontal timber pieces have been used and each is 150 mm shorter than the one below.

If 25.5 metres of timber was used altogether what must the length of the lowest piece be?

116 Sequences

YEAR 12 MATHEMATICS

QUESTION SEVEN

Moe plans to set up his own painting company so he has been looking for a tidy, recent model, second hand

van to transport his equipment to jobs. He sees one that will be perfect and notes that there are two

possible payment regimes.

PLAN A pay a deposit of $4950

first month pay $300

each successive month pay $20 more than the previous month

PLAN B pay a deposit of $7995

first month pay $615

each successive month pay $10 less than the previous month

Calculate how many months it will take before the total paid into Plan A would be the same amount paid

into Plan B.

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5

Trigonometry Problems 117

YEAR 12 MATHEMATICS

From

measurements taken

lead to

straightforward

calculations

to

more complex

trigonometry situations

where

suitable models

and rules are selected

resulting

in sensibly rounded

solutions (in context)

with appropriate units.

MATHEMATICS 2.8


Solve trigonometry problems requiring modelling of practical situations


C take measurements in a practical situation

with suitable calculations to follow

C calculations could include use of the Sine Rule,

Cosine Rule and/or areas of triangles

C contexts to explore could be bearings,

relative velocity, etc

C find length and angles using: Sine Rule

and/or Cosine Rule

C find circular measures using: arc length and/or

sector areas

C find triangular areas

C use 2 dimensional representations of

3 dimensional situations

C combine any of the above techniques

to solve more integrated contextual problems

118 Trigonometry Problems

YEAR 12 MATHEMATICS

TRIGONOMETRY PROBLEMS

- Revision Summary


be tested in this Achievement Standard. (Note, diagrams have not been drawn to scale.)

1. Calculate the lengths or angles that have been marked.

a. b. c.

2. Use the sine rule or cosine rule to find the missing lengths.

a. b.

3. Use the sine rule or cosine rule to find the missing angles.

a. b.

4. a. Convert these to radians. Give your answer to 2 decimal places.

i. 30E ii. 165E iii. 238E

b. Change these to degrees. Give your answer to 1 decimal place.

i. 1.06 rad ii. rad iii. 5.89 rad


YEAR 12 MATHEMATICS


YEAR 12 MATHEMATICS

5. Calculate the marked dimensions in these sectors.

a. b. c.

(Find θ in radians and degrees)

6. Find the areas of these figures.

a. b. c. d.

Find the shaded area.

7. Three fisherman are angling along the banks of a river. Two of them, F and G, are on the same bank

and are 50 metres apart. The third, H, is on the opposite bank of the river. It is known that pHFG is

62E and angle pFGH is 66E. How wide is the river estuary?


YEAR 12 MATHEMATICS

PRACTICAL TRIGONOMETRY PROBLEMS

PRACTICE TEST 1

Show ALL working.

This model assessment is in two parts.

PART A

1. You and some other students go to a local park near your school. On a level piece of ground your

teacher has marked out a large four sided area. You have been given tape measures, magnetic

compasses, trundle wheel and alidade.

2. From a reference point, X, inside this quadrilateral, you must begin your measurements. Note that

the point X is not allowed on any sides nor any diagonals of the quadrilateral.

3. You draw the quadrilateral and include point X.

4. Using the equipment, from point X, complete a radial survey. Indicate the measurements you will

need to take to calculate the area of the quadrilateral and mark these on your diagram.

NOTE

Usually you need to draw your diagram from scratch. If you use the diagram given, the highest grade

you will be awarded is ‘Achievement’. For the purpose of continuing PART A, assume that you have

drawn the shape below (to scale) and it is 1: 1500.


YEAR 12 MATHEMATICS


YEAR 12 MATHEMATICS

PART B

Use the sketch and your measurements from PART A to complete this section of the task. You do not need

any other measurements to complete this.

1. Council drainage contractors are to check the old pipe which lies diagonally across the quadrilateral

in Part A. Calculate the shortest distance between the pairs of opposite corners (i.e. the lengths of

both diagonals of the quadrilateral).

2. A memorial rose garden is going to be set up in the North-West corner of this quadrilateral. To assist

with the planning, you are asked to find the size of the interior angle in that corner of the

quadrilateral.


YEAR 12 MATHEMATICS

3. As the contractors explore the old drainage pipe, they find that there is a thick layer of sludge and

sediment which covers the bottom of the pipe.

The radius of the pipe (OP or OQ) is known to be 1.25 m.

a. If the distance across the top of the sediment (i.e. chord ) is

2.24 m, then calculate the area of the cross-section of the

sediment (shown by the shaded area on the diagram).

b. Assume this measurement is the average width of the sediment in

the pipe. Also assume that the pipe runs along the longer of the

two diagonals you calculated in #1.

What volume of sediment is the pipe holding?

(Round this number to the nearest 10 units.)


YEAR 12 MATHEMATICS

PRACTICAL TRIGONOMETRY PROBLEMS

PRACTICE TEST 2

Show ALL working.

This model assessment is in two parts.

PART A

1. As part of a measuring task, a class of Year 12 students met their teacher at the local ice skating

rink. There they were paired up and given the measuring task, with a measuring tape and, of course,

ice skates!

An arbitrary point, Q, on the blue line was given to them by the teacher. They had to measure and

record the lengths needed to find the angle pLQR. (Every student pair had a different location for

Q.)

A triangle is set with Q at one corner, on the blue line, and the other two corners, L and R, as the

goal posts.

NOTE: If a pair of students couldn’t identify the measurements which they needed to take, they

were supplied with a help sheet. This usually has the diagram of the triangle ªLQR,

supplied with labels L, R and Q.

If students use the help sheet, then the highest grade which they can be awarded for

this task is ‘Achievement’.

At the end of the session the measurements are usually handed in to the teacher.


YEAR 12 MATHEMATICS

PART B

1. For the position given (to one pair of students) and the measurements provided, calculate the angle

in the triangle on the blue line, pLQR.

2. If a student was standing at point Q, looking towards the goal mouth, what would the “apparent

width” of the posts be to her? i.e. find the length on the diagram.

(Note: ªRXQ is not right angled but isosceles.)


YEAR 12 MATHEMATICS

3. a. A student, Wayne (W) is 4.29 m from the right goal post, R. The distance between the goal posts

(LR) is 1.83 m. The bearing of the left post, L, from Wayne is 093E

and the bearing of the right post, R, from Wayne is 118E.

Another student, Sheena (S) is standing further along the goal line,

leaning on the rink wall.

What is the bearing of Sheena (S) from the left goal post?

Note - round all angles to a nearest degree.

b. How far is Wayne from the left goal post?


YEAR 12 MATHEMATICS

4. Near each goal line there are two “face off” spots. Each “face off” spot is the centre of a circle

with a 4.57 metre radius.

A sponsor usually paints their logo or one of their products inside these

circles before big games are played. One sponsor who imports sports

gear has painted a sports helmet inside the circle, and centred it at

the “face off” spot, O.

The angle (below) pAOC = 2.17 radians, while AO = OC = 3.36 metres.

Calculate the area inside the circle which has not been painted.


YEAR 12 MATHEMATICS

5. Just as the students are completing their measurements, an ice hockey team arrives for a training

session. The class decide to watch them practise. During one of the drills a player attempts to flick

the puck into the back of the net. (The height of the goal is 1.22 m.)

The player stands to the side of the goal mouth, so that the puck is:

3.5 metres from the goal line.

4.9 metres from the near post.

5.3 metres from the far post.

The player aims the puck at a point, T, which is 30 cm

below the top of the centre of the crossbar.

Find the angle of elevation pTPC.

Note - in any triangle ªABC then AB2 + AC2 = 2AD2 + 2BD2

planned orbit

actual orbit

C2

P

C1

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YEAR 12 MATHEMATICS

From

straightforward

(simple step)

equations

to

equations requiring

several steps

in contexts

or

situations involving

more challenging

manipulations.

MATHEMATICS 2.9


Solve straightforward trigonometry equations


C solve trigonometric equations using

trigonometric graphs in degrees within the

range 0E # θ # 360E and in radians within

the range π # θ # 2π

C solve trig equations such as:

5 sin 3x = 4 on [0E, 360E]

cos (x - 180E) = 0.5 on [-180E, 180E]

2 sin x - 1.5 = -1 on [0, 2π]

using degrees or radians as directed

C solve trigonometric equations in contexts

such as tidal motion or wheel revolutions

C manipulate more complex trigonometric

equations

C solve problems in contexts, including

mathematical situations

Trigonometric Equations 131

YEAR 12 MATHEMATICS

TRIGONOMETRIC EQUATIONS

- Revision Summary



1. a. Change these degree measures to radians (to 2 dp).

i. 30E ii. 114E iii. 268E

b. Write each degree measure as a fraction of π.

i. 30E ii. 105E iii. 320E

c. Change these radian measures to degrees (to 1 dp where necessary).

i. 0.92 radii. 2.63 rad iii. rad


YEAR 12 MATHEMATICS

2. a. Draw i. y = 2 sin x.

and ii. y = cos x - 1, on separate graphs.

On graph i. use 0 # x # 360E and on graph ii. use 0 # x # 2π.

b. On the first graph, draw the line y = 1.6 and find x where 2sin x = 1.6.

On the second graph, draw the line y = and find x where cos x - 1 = .


YEAR 12 MATHEMATICS

3. Solve these equations with a suitable method.

a. 2 sin x = 1, on 0E # x # 360E

b. 5 cos x - 2 = 1 on 0E # x # 2π

c. cos 2x = 0.85 on 0 # x # 2π

d. 2.5 sin 3x = -1.5 on 0E # x # 360E

e. sin (x - 45E) = 0.4 on -270E # x # 270E


YEAR 12 MATHEMATICS

4. On the interval 0 # x # 4π, which values of x would provide the greatest and least values

of y = 5 sin ?


YEAR 12 MATHEMATICS


PRACTICE TEST 1

Show ALL working.

QUESTION ONE

Solve the following trigonometric equations:

a. cos x = 0.4, 0E # x # 360E

b. sin x + 4 = 3.09 on 0E # x # 360E


YEAR 12 MATHEMATICS

c. 3 tan x = 4.8 on 0 # x # 2π

QUESTION TWO

Solve cos 2x = 0.78, 0E # x # 360E.


YEAR 12 MATHEMATICS

QUESTION THREE

Millie hears a “click - clack” sound while riding her bike. She dismounts

and finds a small tack has embedded itself into the rubber tread.

She knows that her bike wheel has a 66 cm diameter and once she is

home, she turns it and observes how the tack on the tyre rotates around

the central wheel hub.

The height (H) of the tack relative to the centre hub may be given by:

H = 33 sin (45t)E

H = height in centimetres.

t = time in seconds after the wheel begins to rotate.

a. After how many seconds will the tack first be 30 cm above the hub?

b. How long will it take the tack to return to its starting position?


YEAR 12 MATHEMATICS

QUESTION FOUR

While watching her piano being tuned, Millie notices a tuning fork for the note Middle C. The pure tone of

the tuning fork when struck, vibrates at 256 times per second.

Millie believes these vibrations may be modelled by the trigonometric function y = 0.6 sin (512 π t), where

the amplitude (loudness) is 0.6.

The speed of the vibrations suggests the number of cycles per second cannot be detected by us.

What is the least amount of time taken for the tuning fork to reach a loudness of 0.45?


YEAR 12 MATHEMATICS

QUESTION FIVE

Over a long period of time, Millie has observed the depth (d) of water at Flint’s Stone, a large rock near

the entrance to the local harbour. She believes that the tidal cycle is about 12½ hours. At low tide the

water depth beside the rock is only 1.5 metres, but at high tide it has reached 6.1 m.

Millie believes this formula, d = 2.3 cos + 3.8, could model the depth around Flint’s Stone,

where: d = depth of the water in metres.

t = time in hours after high tide.

Fishing boats may only enter the harbour when the water there is at least 3 metres deep. How long either

side of high tide, can boats safely travel in and out of the harbour?


YEAR 12 MATHEMATICS


PRACTICE TEST 2

Show ALL working.


QUESTION ONE

a. cos θ = 0.4, 0E # x # 360E

b. 2.5 sin θ + 1 = 1.5, 0E # x # 360E

c. tan θ + 2.7 = 1.1, 0 # θ # 2π


YEAR 12 MATHEMATICS

QUESTION TWO


a. cos (θ - 30E) = -0.7, 0E # x # 360E

b. sin 2θ = 0.61, 0 # θ # π


YEAR 12 MATHEMATICS

QUESTION THREE

A patient with a fever is admitted to hospital for further observation. Her temperature varies from a low

of 37E to a high of 40.2E Celcius.

A model of the temperature has been suggested as: T = 38.6 + 1.6 sin where t is the time in days.

During which day, following her admission to hospital, does the patient’s temperature first reach 40E C?


YEAR 12 MATHEMATICS

QUESTION FOUR

The eleven weeks from the middle of July until early October is a profitable time for a local winter sports

retailer to sell snow boards. Sales of snow boards have been monitored over the same period of time for

some years. They have developed this trig formula to model the numbers of snowboards that they sell over

the 11 weeks.

S = 30 - 33 cos , where week 1 # t # week 11.

According to the formula over how many weeks would they sell at least 45 snowboards?

144 Answers

YEAR 12 MATHEMATICS

THE ANSWERS

Note: the answers have been checked and rechecked. If your answer differs from the one listed then check with your

teacher, friends or write to Mahobe Resources. Due to space restrictions, we have not been able to set out all answers

as fully as we would have liked. Remember - in the exam you should put each step of your answer on a separate line.

Pages 6-10, ALGEBRA - Revision Summary

1. a. x = b. r = =

c. x = or x = -11 d. x = 0, , -8

2. a. 12 - 9x b. 18x3 - 24x2 - 10x, c. 4x3 - 16x2 - 35x + 147

3. a. (x - 12)(x - 7) b. (5x + 7)(2x - 3) c. (a - v)(a + w)

4. a. b. 3x3y5 c. m6

5. a. 35 = 243b. i. log 5, ii. log 72, iii. log 30 c. 2

6. a. 3x - 1 b.

c. a + b d. 2n+10

7. a. x = -50 b. n =

c. x < 1 d. x >-

8. a. x = 7, x = -6 b. x = , x = 2

c. x = - , x = 0, x = 1 d. x = 1.15, x = -0.65

e. x = 0.45, x = -4.45 f. x = -3.37, x = 2.37

9. a. x = 37 = 2187 b. x = = 7

c. x = 2.6710. a. (-4, 3) b. (-2, 3)

c. (2, -2) and (3, 0) d. (-2, -1) and (1, 2)11. a. i. ª = 57 > 0 ˆ 2 real distinct roots

ii. ª = 0 ˆ 1 real (repeated) root

b. (2n)2 - 4(2)(5) < 0 Y n2 < 10 ˆ - < n <

Pages 11-14, ALGEBRA PRACTICE TEST 1

QUESTION ONE

1. x - 52. 3x4

3. log 8

4. a. x = b. x = 2.13 (2 dp) c. x = or x = -1

QUESTION TWO

M = M0(0.85)t

500 = 800(0.85)t

= 0.85t

0.625 = 0.85t

log 0.625 = t log (0.85)

t =

t = 2.89 (2 dp) days= 3 days

QUESTION THREE

a. By substituting, equations becomex2 + (2x + 6)2 = 36x2 + 4x2 + 24x + 36 = 36 5x2 + 24x = 0

x(5x + 24) = 0

x = 0 or x = -

b. If x = 0, y = 6, ˆ first is (0, 6)

If x = - , y = -3.6, ˆ second is (-4.8, -3.6)

QUESTION FOUR

=

(x - 12)(1350 + 24x)= (1800 + 20x)(x - 9)24x2 + 1062x - 16200 = 20x2 + 1620x - 16200 4x2 - 558x = 0 2x(2x - 279) = 0ˆ x = 0 or x = 139.5Check: x = 139 tickets, PA < PB

x = 139.5 tickets, PA = PB

x = 140 tickets, PA > PB

ˆ sell 140 tickets or more, price B is cheaper.

QUESTION FIVE

Rule: α + β = , αβ =

Let roots be α and α + 1ˆ sum of roots, k + 1 = α + α + 1 i.e. k + 1 = 2α + 1

and k = 2α (A)Product of roots, 4k = α (α + 1)

4k = α2 + α (B)

ˆ from equation (A) α =

substituting this into equation (B)

= 4k

= 4k

k2 + 2k = 16k k2 - 14k = 0 k(k - 14) = 0

ˆ possible values: k = 0 or k = 14Check: if k = 0, x2 - x = 0 Y x(x - 1) = 0k = 14, x2 - 15x + 56 = 0

Y (x - 7)(x - 8) = 0

Answers 145

YEAR 12 MATHEMATICS

Pages 15-18, ALGEBRA PRACTICE TEST 2

QUESTION ONE

1. 12x3 + 101x2 + 37x - 24

2.

3. or 1.2

4. a. x = 3, b. x = or x = 2, c. x = 1.92

5. 7c + 2p = 1925 c = p + 50by substitution, 7(p + 50) + 2p = 1925

9p + 350 = 1925 9p = 1575 p = 175

ˆ plain ice creams cost $1.75

QUESTION TWO

By substitutionx2 + (x - 3)2 + 2x - 7 = 0x2 + x2 - 6x + 9 + 2x - 7= 0 2x2 - 4x + 2 = 0 2(x2 - 2x + 1) = 0

(x - 1)2 = 0 x = 1

By back substitution, y = -2there is only one co-ordinate set (1, -2)ˆ only 1 point of intersection

QUESTION THREE

Set up 2 equations: h × w = 1.7 h = w + 1

(w + 1)(w) = 1.7 w2 + w - 1.7 = 0

By quadratic formula

w =

w = or

w = 0.896 or -1.896Ignoring the negative resultwidth = 0.9 m (1dpf) and height = 1.9m (1dp)

QUESTION FIVE

of 19995 = 13330

ˆ solve 13330 = 19995(0.993)t with =

ˆ = (0.993)t

log = log (0.993)t

log = t log (0.993)

= t

t = 57.7 h or t = 58h (2 sf)

QUESTION SIX

First Instance (A) V =

Second Instance (B) V + 250 =

Substituting (A) into (B)

+ 250 =

multiply by t(t - 1) (t - 1)5000 + 250 t(t - 1) = 5000t

5000t - 5000 + 250t2 - 250t = 5000t 250t2 - 250t - 5000 = 0

t2 - t - 20 = 0 (t - 5)(t + 4) = 0

t = 5h (positive choice)

ˆ in first instance, t= 5h, V = ,

= 1000 km/hr

in second instance, t = 4h, V =

= 1250 km/hr

Pages 20 - 24, GRAPHS - Revision Summary

1. a. i. y = x2 + 3x - 4

ii. y = (x + 3)2 - 4

146 Answers

YEAR 12 MATHEMATICS

Pages 20 - 24 cont1. a. iii. (x + 1)(x - 2)(x + 3)

b. i. y =

ii. y =

c. x2 + y2 = 49(x + 2)2 + (y - 1)2 = 49

d. y = 5x y = 5x - 3

e. y = log8 xy = log8 |x - 3 |

2. a. y = (x + 5)(x + 1)(x - 2)b. y = (x - 1)2 - 3

= x2 - 2x - 2c. y = 5x - 2d. (x - 3)2 + (y + 2)2 = 25

Answers 147

YEAR 12 MATHEMATICS

Pages 20 - 24 cont3.

Pages 25 - 30, GRAPHS - PRACTICE TEST 1

1. a. y = x2 - 4x - 5= (x - 5)(x + 1)x intercept (5, 0), (-1, 0), y intercept (0, -5)Vertex x = (5 + -1) ÷ 2

y = (2 - 5)(2 + 1)ˆ (x, y) = (2, -9)

b. y = Using a table of values

x -4 -2 -1 1 2 4

y -1 -2 4 4 2 1

c. y = 4x

Using a table of values

x y

-3 0.0156

-2 0.0625

-1 0.25

0 1

1 4

2 16

3 64

QUESTION TWO

1. Features can include:< The two graphs meet at one point, (1, 3)< y + x = 4 is a straight line< y = x2 - 3x + 5 is a parabola< y + x = 4 is a tangent to the curve y = x2 - 3x + 5< y = x2 - 3x + 5 does not meet the x axis but does

intersect the y axis at (0, 5)< Vertex of the curve y = x2 - 3x + 5 is (1.5, 2.75)

2. a.

t V

0 $11 955

1 9 836

2 8 065

3 6 614

4 5 423

5 4 447

6 3 647

b. During the fourth year (t . 3.5 y)

148 Answers

YEAR 12 MATHEMATICS

Pages 25 - 30 cont

QUESTION THREE

a.

b. y =

= 3 +

x = 0, y = -2 (0, -2)y = 0, 3x + 2 = 0 (-2/3, 0)VA: x = 1 HA: as x 6 0, y 6 3

QUESTION FOUR

a. In the early years, the rate decreases the car value thequickest i.e. it loses half its value, one third its value,one quarter its value etc. In later years the value changesless rapidly.

b. The car loses its value less rapidly in later years with thevalue only changing marginally. The remaining value staysabove $1000.

c. The y intercept tells us that the car was bought for justunder $12,000 (i.e. $11, 995)

QUESTION FIVE

a. y = (x + 2)(x - 1)(x - 4) or y = x3 - 3x2 - 6x + 8

b. x2 + y2 = 25c. HA: y = 2, VA: x = -3

Intercepts (1.5, 0) and (0, -1)

ˆ y = or y = 2 -

QUESTION SIX

a. Use the 2nd and 3rd lines of the data table with theformula to attain two simultaneous equations.

15.2 = 8.0 =

15.2(15 - b)= a 8.0(66 - b) = a 228 - 15.2b = a 528 - 8.0b = a

ˆ 228 - 15.2b = 528 - 8.0b 8.0b - 15.2b = 528 - 228

- 7.2b = 300 b = -41.7 (1 dp)

or b = -42 (2 sf)

If using b = -41.7, then a = 528 - 8(-41.67)= 861.6

ˆ D =

If using b = -42, then a = 864

ˆ D =

b. Using D =

6.6 =

6.6(t + 41.7) = 861.6 6.6t + 275.2 = 861.6

6.6t = 586.4 t = 88.8h

ˆ 89 hours after the initial time (after 4pm, 10 Feb)i.e. 3 days, 17 hrs later, 9am on 14 February.

Pages 31- 38, GRAPHS - PRACTICE TEST 2

QUESTION ONE

a. intercepts at x = 0, x = 1, x = -3 or use a table

x -4 -3 -2 -1 0 1 2

y -20 0 6 4 0 0 10

b. vertex at (-1, 4)intercepts (0, 3), (-3, 0) and (1, 0) or use a table

x -4 -3 -2 -1 0 1 2

y -5 0 3 4 3 0 -5

Answers 149

YEAR 12 MATHEMATICS

Pages 31- 38, (cont)

QUESTION ONE

b.

c.

d. Features to include:< diameter = 6 units ˆ radius = 3 units< circle centre = (1, 0)< equation is (x - 1)2 + y2 = 32

< x intercepts at y = 0, (-2, 0) and (4, 0)< y intercepts at x = 0

(-1)2 + y2 = 32

y2 = 9 - 1

y = ˆ at (0, ) (0, )

QUESTION TWO

a. Equation y = (x - 2)2 or y = x2 - 4x + 4b. y = 2x

c. xy = 6 or y =

QUESTION THREE

a. Solve the equation first2x2 - 3x - 5 = 0(2x - 5)(x + 1) = 0ˆ x = 2.5, x = -1Intercepts (-1, 0), (2.5, 0) and (0, -5)

Vertex: x =

ˆ x =

ˆ y = 2( )2 - 3( ) - 5

= -6

ˆ vertex = ( , -6 )

y = 2x2 - 3x - 5

b. y = -x3 + 1

QUESTION FOUR

Using a table of values:

x -5 -4 -3 -2 -1 0 1 2 3 4

y 128 64 32 16 8 4 2 1

QUESTION FIVE

a. Initial investment was $4 500b. Since 4500(1.08)9 = $8995.52 is very close to double the

initial $4 500 then the end of year 9 is close.ˆ during year 10 the sum would double.

c. At the end of year 12 there is $11 332 (nearest dollar) inthe fund. If $3000 was withdrawn and assuming no penaltyclause (early withdrawal) with the same interest rate asbefore, the graph would plunge by $3 000, beforecontinuing to compound from the new balance of $8 332

150 Answers

YEAR 12 MATHEMATICS

Pages 31- 38, (cont)

QUESTION SIX

a. Using the two pieces of information, then equating witheach other.Week 1: 107449 = A - B log(1 + 0.5)Week 10: 64011 = A - B log(10 + 0.5)Making A the subject of each equation:107 449 + B log(1.5) = 64 011 + B log (10.5)

107 449 - 64 011 = B log (10.5) - B log (1.5)

43 438 = B log

= B log 7

= B

ˆ B = 51 400 (nearest 10)Using Week 1 A = 107 449 + B log (1.5)

= 107 449 + 51 400 log (1.5)= 116 500 (nearest 10)

Check A in other equation A = 64 011 + 51 400 log(10.5)

= 116 500 Tˆ Model equation for spending is:

S = 116 500 - 51 400 log(x + 0.5)b. If x = 30 then S = 116 500 - 51 400 log(30.5)

= $40 207 (nearest $)

Pages 40 - 44 CALCULUS - Revision Summary

1. a. = 27x8 - 10x

b. f(x) = 6x3 - 10x2 + 21x - 35ˆ f!(x) = 18x2 - 20x + 21

2. a. f(x) = x3 + 4x2 - 11x + cb. y! = 18x5 + 6x-4 + 1

ˆ y = 3x6 - 2x-3 + x + c

or y = 3x6 - + x + c

3. a. f!(x) = 4x3 - 10x + 1ˆ f!(2) = 32 - 20 + 1

= 13

b. = 2x - 2, ˆ gradient = -2 - 2

= -4

4. = 2x + 5

a. gradient 2+5 = 7equation at (1, 6) is y - 6 = 7(x - 1)

or y = 7x - 1 or 7x - y - 1 = 0

b. when x = 0, y = 0equation at (0, 0) is y = 5x

5. = 2x - 3

gradient = 1 Y 2x - 3 = 1ˆ x = 2

substituting x = 2 Y y = 4 - 6 + 2= 0

ˆ at (2, 0), gradient = 1

6. a. = 125 - 27

= 98b. A quick sketch helps

(2x - x2) dx =

= [4 - ] - [0]

= units2

c. y = x3 + 2x2 - 3x

ˆ Area = | y . dx | + | y . dx |

= | (x3 + 2x2 - 3x)| + | (x3 + 2x2 - 3x)|dx

= x4 + x3 - x2 + x4 + x3 - x2

= + - + 0 - ( - 18 - )

= +

= 11 units2

Answers 151

YEAR 12 MATHEMATICS

Pages 40 - 44 cont

7. a. = 3x2 - 3

b. turning point is when = 0

ˆ 3x2 - 3 = 0 3(x2 - 1) = 0

3(x + 1)(x - 1) = 0x = ± 1

substituting back into original equation y = x(x2 - 3)x = 1, y = -2 and x = -1, y = 2 and using cubic shape,

ˆ (-1, 2) is a local maximum(1, -2) is a local minimum

c. y = x3 - 3x increases before the max i.e. x < -1and after the minimum i.e. x > 1decreases between the max and min i.e. -1 < x < 1

8. h = 245t - 4.9t2

a. v =

= 245 - 9.8tb. v(5) = 245 - 9.8(5)

= 196 m/sc. h(5) = 245(5) - 4.9(5)2

= 1102.5 m

d. = - 9.8 m/s2

(since no t variable is present, a is constant at -9.8 m/s2)

e. h is a maximum when it turns, i.e. the turning point iswhen v = 0 ˆ solve 245 - 9.8t = 0

t =

when t = 25s it is at maximum height,and h(25) = 245(25) - 4.9(25)2

= 3062.5 m above the ground.

9. 1. First identify the two formulae

y

x

Perimeter is 2x + 2y = 108Area is xy (we need to optimise this)

2. Use the perimeter formula and make one of thevariables the subject2y = 108 - 2xy = 54 - x

3. Substititute this whole expression into the Area.This leaves just one variable to contend with.A = xy

= x (54 - x)= 54x - x2

4. DifferentiateA! = 54 - 2x

5. Solve A! = 0 to find maximum or minimum54 - 2x = 0

x = 27m6. Substitute back into original equation

y = 54 - 27= 27m

Since x = y = 27m, the rectangle is a square.Maximum area is 272 = 729 m2

Pages 45 - 49, CALCULUS - PRACTICE TEST 1

QUESTION ONE

= 3x2 - 6, ˆ gradient = 3(5)2 - 6

= 69

QUESTION TWO

A = (3x2 + 1) dx

= x3 + x + c

= [ 23 + 2 ] - [ 13 + 1] c cancels out= 10 - 2= 8 units2

QUESTION THREE

If f!(x) = 6x2 - 4x + 5

then f(x) = (6x2 - 4x + 5) dx

= 2x2 - 2x2 + 5x + cAt (2, 11), f(x) = 11 and x = 2ˆ 11 = 2(2)3 - 2(2)2 + 5(2) + c

11 = 16 - 8 + 10 + cc = -7

ˆ curve is f(x) = 2x3 - 2x2 + 5x - 7

QUESTION FOUR

Parallel to the x-axis implies the gradient = 0

ˆ = 6x2 - 6

gradient = 0 when 6x2 - 6 = 0 6(x2 - 1) = 06(x + 1)(x - 1) = 0ˆ x = -1 or x = 1 are the x coordinates

QUESTION FIVE

The gradient of the function is = 3x2 - 6x - 7

at x = -1, gradient m = 3(-1)2 - 6(-1) - 7= 2

equation at (-1, 4) is y - 4 = 2(x - -1) y - 4 = 2x + 2 y = 2x + 6

or 2x - y + 6 = 0

QUESTION SIX

d = v . dt

= 0.75 t . dt

d = 0.75 + c

since t = 0 and d= 0 (i.e. vehicle hasn’t moved from thestarting point) then c = 0

ˆ d = 0.75 or d = 0.375t2

after 8 seconds d = 0.375(8)2

d = 24m (moved from start of roll)

152 Answers

YEAR 12 MATHEMATICS

Pages 45 - 49 (cont)

QUESTION SEVEN

The curve would be translated 36 metres upwards

from to

ˆ Area = (12 - 3x2 + 36) dx

= (48 - 3x2)dx

= 48x - x3 + c

= [128] - [47]= 81 m2 (Area of the parking bay )

QUESTION EIGHT

First find some labels for dimensions

y y

x

x + 2y = 5.6 or x = 5.6 - 2yTo maximise the cross sectional area, A = xy

= (5.6 - 2y)y= 5.6y - 2y2

= 5.6 - 4y

= 0 when 5.6 - 4y = 0

ˆ y =

y = 1.4 msubstituting back into original equationsx = 5.6 - 2 × 1.4x = 2.8 mA = xy

= 2.8 × 1.4= 3.92 m2

Pages 50 - 54, CALCULUS - PRACTICE TEST 2

QUESTION ONE

a. = 4x3 - 6x

ˆ gradient = 4(2)3 - 6(2)= 32 - 12= 20

b. y = (8x3 + 6x2 - 4x - 1) dx

= 2x4 + 2x3 - 2x2 - x + cAt (-1, 1) y = 1 when x = -1ˆ 1 = 2 - 2 - 2 + 1 + cˆ c = 2ˆ y = 2x4 + 2x3 - 2x2 - x + 2

c. A = (x3 + 2) dx

= + 2x + c

= + 4 - 0

= 8 units2

d. If y = x-2, = -2x-3 or

ˆ =

x3 =

x3 = -8

x =

x = -2

ˆ y =

= ˆ Coordinates are (-2, )

QUESTION TWO

A1 + A2 = ( x2 - 3x - 4) + ( x2 - 3x - 4) dx

= - x2 - 4x + - x2 - 4x

= - 24 - 16 - 0

+ - - 20 - - 24 - 16

= -18 + -15 - -18

= 21 units2

Answers 153

YEAR 12 MATHEMATICS

Pages 50 - 54 (cont)QUESTION THREE

a. d = (18 + 15t - 3t2) dt

= 18t + t2 - t3 + c

when t = 2 s, d = 65 cm

65 = 18(2) + (2)2 - 23 + c

65 = 36 + 30 - 8 + c 7 = c

ˆ d = 18t + t2 - t3 + 7

when t = 0 (stationery), d = 7 cm from owner

b. If v = 18 + 15t - 3t2

the = 15 - 6t (acceleration)

maximum velocity is when a = 0i.e. 15 - 6t = 0

t =

t = 2.5sˆ maximum velocity

v = 18 + 15(2.5) - 3(2.5)2

= 36.75 cm/s

QUESTION FOUR

V = 335, V = πr2h

ˆ h =

S = 2 πr2 + 2 πrh

= 2 πr2 + 2 πr .

S = 2 πr2 + 670

To optimise, differentiate the surface area function.

ˆ = 4 πr - 670

0 = 4πr -

= 4πr

= r3

= r

r = 7.30 cm (3 sf), (h = 2.00 cm)

Pages 55 - 60, CO-ORDINATE GEOMETRY

1. a. Mid point = (2, 8)

b. = (1.6, -0.8)

2. a. d =

=

= 7.2 (s 2f)

b. d =

=

= 14.7 (1 dp)

3. Make y the subject2y - 3x = 7

2y = 3x + 7

y = x + ˆ m =

4. Using the points (-3, 2) and (1, 5)

ˆ m =

=

ˆ equation is y - 5 = (x - 1)

4(y - 5) = 3(x - 1) 4y - 20 = 3x - 3

3x - 4y + 17 = 0

5. Either use 2x + 5y - 8 = 0 or y = x +

a. Parallel line, 2x + 5y + c = 0Using (1, -2) 2 - 10 + c = 0

c = 8ˆ 2x + 5y + 8 = 0

b. perpendicular line 5x - 2y + d = 0 (swap coefficients, reverse the sign)Using (5, 6) 25 - 12 + d = 0

d = -13ˆ 5x - 2y - 13 = 0

6. Solving simultaneouslya. [3x + 4y = 10]× 5 Y [15x + 20y = 50]

[5x + 3y = 13]× 3 Y -[15x + 9y = 39] 11y = 11

y = 1Back substitute 3x + 4(1) = 10

3x = 10 - 4

x =

x = 2 ˆ (2, 1)Check 5(2) + 3(1) = 13 Tb. [0.1x - 0.2y - 0.7 = 0]× 40 Y [4x - 8y = 28]

[0.4x + 0.3y - 0.6 = 0]× 10 Y - [4x + 3y = 6] -11y = 22

y = ˆ y = -2

Back substitute 0.1x + 0.2(-2) - 0.7 = 00.1x + 0.4 - 0.7 = 0

0.1x = 0.3

x =

ˆ x = 3 and coordinate = (3, -2)Check 0.4(3) + 0.3(-2) - 0.6 = 0 T

7. A quick sketch can assist.

154 Answers

YEAR 12 MATHEMATICS


7. a. Midpoint of =

= (2, 3 )

Equation of median, m =

=

=

Gradient of median y + 1 = (x + 3)

10(y + 1) = 9(x + 3)ˆ 9x - 10y + 17 = 0

b. Altitude is a perpendicular line from (1, 5) to

Gradient of , m =

=

=

ˆ gradient of z line =

= -2Equation of z line y - 5 = -2(x - 1)

ˆ 2x + y - 7 = 0

c. perpendicular bisector cuts at the midpoint

Midpoint of = (-1, 2)

Gradient of =

=

ˆ gradient of perpendicular =

Equation of perpendicular bisector is y - 2 = (x + 1)

Using the midpoint 3(y - 2) = -2(x + 1)ˆ 2x + 3y - 4 = 0

8. Compare lengths of all three sides.If two sides equate thenªOPQ is isosceles

Length =

=

Length =

=

=

Length =

=

Since the length of = length

then ªOPQ is isosceles

Pages 61 - 66, COORDINATE GEOMETRY

PRACTICE TEST 1

QUESTION ONE

a. d =

=

=

d = 19.1 grid squaresˆ true distance = 19.1 × 20 m

= 382 m

b. gradient =

=

equation is (y - -2) = (x - -10)

14(y + 2) = 13(x + 10)ˆ 13x - 14y + 102 = 0

c. The fence line has the same gradient as the given

equation ˆ y = x + c

using (-2, 8) 8 = (-2) + c

8 = -1 + cˆ c = 9

ˆ fence equation is y = x + 9

QUESTION TWO

The drain line is a perpendicular bisector of the line whichjoins the two points.

Midpoint =

= (2, -3)

Gradient between the two points, m =

= -1ˆ the bisector gradient must be mz = 1ˆ equation is (y - -3) = 1(x - 2)

y + 3 = x - 2x - y - 5 = 0

or y = x - 5

QUESTION THREE

The median joins N(4, 11) to the midpoint of ST

midpoint =

= (-4, -1)

gradient =

=

=

equation y - 11 = (x - 4)

2(y - 11) = 3(x - 4)2y - 22 = 3x - 12

ˆ 3x - 2y + 10 = 0 is the equation of the median

Answers 155

YEAR 12 MATHEMATICS

QUESTION FOUR

Firstly, find where the altitude line intersects SP x + y + 12 = 0 x - y - 2 = 02x + 10 = 02x = -10 x = -5

Back substituting x + y + 12 = 0-5 + y + 12 = 0

y = -7ˆ the altitude meets line SP at point Q (-5, -7)ˆ the length of the altitude must be the distance

between Q(-5, -7) and T(2, 0)

d =

=

= 9.899 grid unitsgrid lines are 20m apart therefore multiply by 20= 198 m (3 sf)

QUESTION FIVE

Find a line through (6, 0) which is perpendicular to

y = x + 8, mz =

ˆ equation is y = x + c

at (6, 0), 0 = -8 + c ˆ c = 8

ˆ perpendicular line through (6, 0) is: y = x + 8

The point of intersection of these two lines appears to be(0, 8) since c = 8 in both line formulaeˆ the closest distance is between (0, 8) and (6, 0)

d =

=

= 10 units (× 20)= 200 m

Pages 67 - 72, COORDINATE GEOMETRY

- PRACTICE TEST 2

QUESTION ONE

a. L2 =

= (18, 21 )

b. m =

=

equation is (y - 15) = (x - 25)

8(y - 15) = -3(x - 25)ˆ 3x + 8y - 195 = 0

c. If parallel, then m = , ˆ equation is y = x + c

26 = (10) + c

26 - 4 = c c = 22

ˆ parallel line is y = x + 22

QUESTION TWO

The midpoint of T1 T3 =

= (6, 11)

Length from C to L3 =

=

= 13.04 m (2 dp)

QUESTION THREE

Altitude through T2 must be perpendicular to the otherside of the triangle T3H2

m of T3H2 =

=

ˆ Mz =

the equation through T2(25, 15) is

y - 15 = (x - 25)

2(y - 15)= -5(x - 25) 2y - 30 = -5x + 125

5x + 2y - 155 = 0

QUESTION FOUR

First find the gradients and the equations of each line

Gradient of T1H1 is =

ˆ Mz =

At (9, 0) equation of perpendicular

y - 0 = (x - 9)

4y = 27 - 3x3x + 4y = 27

Gradient of BH3 is = -1, ˆ Mz = 1

At (3, 15) equation of perpendicular: y - 15 = 1(x - 3) x - y = -12Therefore coordinates of D must be the intersection ofthe two legs. Solve these simultaneously.x - y = -12 6 4x - 4y = -483x + 4y = 27 6 3x + 4y = 27

7x =-21 x = -3

Back substitute y = -3 + 12 y = 9

ˆ coordinates of D are (-3, 9)

QUESTION FIVE

Midpoint of T3 B is = (5 , 25)

This point will be on a perpendicular line from T3T2

From Question 1b the equation is 3x + 8y - 195 = 0

ˆ gradient of T3T2 is =

ˆ mz =

156 Answers

YEAR 12 MATHEMATICS


equation of this line (y - 25) = (x - 5 )

3(y - 25) = 8(x - 5 )

or 8x - 3y + 31 = 0Lines meet at closest point to new light:solve 8x - 3y = -31 6 64x - 24y = -248

3x + 8y = 195 6 9x + 24y = 585 73x = 337

x = 4.616 (3 dp)Back substituting 8y = 195 - 3 × 4.616

y = 22.644ˆ coordinates are (4.616, 22.644)ˆ Distance between new light and existing cable

= 2.516 (3 dp)

ˆ 2.5 m is the closest distance

Pages 74 - 76, 2.5 SAMPLE STATISTICS

- Revision Summary

1. Modes are 36 & 40, Mean is = 76.23

2. a. Northern Southern 4 3 13 0 5 5 3 2 1 02 0 1 4 9

9 5 4 3 2 1 1 0 0 1 0 1 3 4 4 6 6 6 7 8 8 8 5 0 2 5 8 9b. Nth 5 (Low), 11 (LQ), 17 (Med), 24 (UQ), 34 (High)

Sth 2 (Low), 10.5* (LQ), 16 (Med), 19 (UQ), 30 (High) * should really be 10-11 accidents

c. Box and whisker plot for Labour Weekend nose totails.

Northern Direction

Southern Direction

0 10 20 30 40

d. Northern mean = 18.1 (18 accidents)Southern mean = 15.55 (15+ accidents)

3. a. 0 = 2.033 (3 dp) b. s = 1.217 (3 dp)

(Compare σ = 1.196)4. a. n = 25, the 13th price lies in the $70 - interval

b. 0 . $85.40c. Tarif Cum Freq

$50 0$60 1$70 4$80 13$90 17$100 19$110 21$120 24$130 25

5. Totals: 3x = 56, 3(x - 0 )2 = 72, 0 = 8, s = 3.4646. Totals 3f = n = 60, 3xf = 1053, 3f(x - 0 )2 = 832.85

0 = 17.55, s = 3.757

Page 77 - 78 SAMPLE STATISTICS, PRACTICE TEST 1

Note, this is a possible solution only.Sampling process:• Assign each member of the population (of 87 earthquakes)

a two digit number 00 - 86.• Using your calculator’s random function, start at, say, the

(8 + 7) = 15th number.• Choose the second and third digits of every three digit

random number.• Disregard numbers which are > 86.• Ignore repeated numbers (®). • Stop when you have a list of approx 30 quakes.

Number Date Time RMagnitude

17 19/02 00 04 6.5

06 23/01 20 10 6.3

46 14/06 17 10 6.8

76 14/11 21 38 7.0

51 20/06 04 03 4.7

96 - - -

02 10/01 23 48 5.5

57 25/07 15 43 5.0

18 22/02 02 25 6.4

36 03/05 07 21 4.9

41 19/05 01 54 6.9

58 05/08 14 14 5.2

51 ® - - -

75 8/11 07 54 5.1

46 ® - - -

64 26/09 01 55 7.5

84 12/12 21 47 6.6

34 19/04 21 11 5.5

91 - - -

93 - - -

44 06/06 07 41 5.7

93 - - -

95 - - -

93 - - -

59 13/08 04 58 4.8

63 24/09 19 24 5.6

55 23/07 07 34 6.0

Answers 157

YEAR 12 MATHEMATICS

Number Date Time RMagnitude

81 02/12 13 13 6.5

28 21/03 12 23 6.9

59 ® - - -

57 ® - - -

14 15/02 14 42 6.6

37 05/05 19 12 6.5

72 27/10 11 18 4.2

65 29/09 15 50 6.7

08 25/01 16 44 5.9

02 ® - - -

52 02/07 02 16 6.6

99 - - -

94 - - -

76 ® - - -

64 ® - - -

83 11/12 14 20 6.6

13 14/02 23 38 6.1

03 12/01 08 04 6.8

Though it is physically possible to test all the items in thispopulation of earthquakes, it is not essential, as a sample ofabout 30 is a useful representation of the population. Also,with the random function on the calculator (though it is nottruly random) we are generally satisfied that each quake inthe population had the same chance of selection in thesample. This simple random sampling method, though relianton the generated random numbers, has provided 30selections. If doubts rise over its authenticity, the exercisecould readily be repeated with another simple randomsample.

If this sample of 30 could be represented in a stem and leafplot these sample statistics of earthquake times would result.LQ = 07 21, UQ = 19 12, median = ½ way between 13 13 hours and 14 14 hours (13 43 ½)

The box and whisker plot of this sample would look like:

Box Plot of Earthquake Times During Day (sample n = 30)

From this sample it would appear that the times are skewedslightly to the later end of the day with a medium time of 13 : 43 : 30

The mean of the sample 0 . 12 26 and the standard deviationof the sample . 7hr 18 min. This would support the aboveresult (box plot) with both the mean and median appearing asp.m. times for this sample.

The distribution of the time in this sample appears to have areasonably even spread, i.e.

00 h - 09 h 12 quakes10 h - 19 h 12 quakes20 h - 23 h 6 quakes

The evenness of this distribution would suggest that quakescan occur at any time including the early hours of themorning and the latest hours of the night. Though randomsampling may be time consuming, we return to the theme,that ignoring the vagaries of calculator random programming,every earthquake in the population had the same chance ofselection for the sample.Pearson developed a measure of skewness of a distribution.

Skew =

=

= -0.528

Since the co-efficient of skewness generally lies between -3and +3, this suggests that the distribution of the sample timestend to negative skewness, so a few early hours are pullingthe mean away from the median.

Page 81 - 83 SAMPLE STATISTICS, PRACTICE TEST 2

Note, this is a possible solution only.The layout of the shellfish farm suggests that an ‘all in’sample may not provide the most representative 30 selectionsfrom the population. One possibility is to look at five possiblestrata.

Based on these proportions:

The top row should have × 30 .5 in the sample

The left group should have × 30 .7 in the sample

The middle group should have × 30 .7 in the sample

The right group should have × 30 .8 in the sample

The bottom end should have × 30 .3 in the sample

30 (Total)

Take a simple random sample of each strata / group in thepopulation. Note each site is already numbered e.g. Top end:

Assign the first two digits from the random numbergenerator of your calculator starting at (e.g. 4+5) 9th

number.Disregard any numbers > 45 or < 01Ignore repeated numbers (®).Stop when we have 5 for this strata.

158 Answers

YEAR 12 MATHEMATICS

Page 81 - 83 (cont)e.g. 62 37 68 78 14 34 94 68 82 86 99 89 63 49 34 02 29

ˆ 37, 14, 34, 02, 29 for the top endApply the same technique (using 3 digit numbers) to each ofthe other four strata, with sensibly applied numbering to suiteach case. The goal is to finish with a total sample size of atleast 30 shellfish sites

Example of possible strata selections:

Top End (001-045)

# Mass (kg)

37 500

14 191

34 245

02 413

29 165

Left Group (045 - 115)

77 508

59 359

49 345

107 175

47 185

109 399

88 366

Middle Group (116 - 171)

155 378

160 249

158 407

145 411

148 404

137 490

156 389

Right Group (172 - 227)

178 393

227 180

189 ® 520

180 407

174 373

196 458

188 139

200 240

Bottom End (228 - 255)

233 267

254 297

230 219

It is physically possible to have used the wholepopulation, but it is too time consuming, and a samplesuch as this has attempted to represent the wholepopulation. Also, with the areas of the shellfish farm splitinto strata, each sub-area is catered for and therandomness of the selection process, again is dependenton the generated random number function from thecalculator.

The sample provides these statistics when ordered.LQ = 240 kg, median = 369.5 kg, UQ = 407 kg

Visually this Box and Whisker Plot of the sample is:

The distribution is skewed with the median placed in theupper end of the interquartile range. The middle 50% ofthe sample has a mass between 240 kg and 407 kg. The mean of the sample, 0 = 335.7 kgThe sample standard deviation, s = 114.0 kgUsing Pearson’s skewness rule:

Skew =

=

= -0.889The data is negatively skewed with the smaller massespulling the mean away from the median. The masses ofthe shellfish lines are quite spread out. Both mean andmedian point towards the mid 300 kg mass, so if theowners of the farm were looking for a conservativeestimate of the mass of their crop, they should tendtowards the lower statistic of the two.

Another sample could produce a similar result, or not, butthe exercise could readily be repeated. It is important toconsider the sites of the areas and which are closer tosea, to shore, to nutrients, affected more by weather etc.Therefore the strata sampling would still be a sensibleoption for this farm.

Pages 88 - 91, PROBABILITY AND NORMAL DISTRIBUTION

Revision Summary

1. a. b. c. =

2. a. = b. = c. =

d. = e. 150 × = 25 times

3. Table of possibilities+ 1 2 3 4 5 61 2 3 4 5 6 72 3 4 5 6 7 83 4 5 6 7 8 94 5 6 7 8 9 105 6 7 8 9 10 116 7 8 9 10 11 12

a. = b. = c. =

d. Pr( 4 or 9) =

e. Pr (3 or 6 or 9 or 12) = =

Answers 159

YEAR 12 MATHEMATICS

Pages 88 - 91 cont

4. a.

b. P(WWW) = 0.7 × 0.8 × 0.75 = 0.42

c. P(any two W) = WWL or WLW or LWW= 0.14 + 0.105 + 0.18= 0.425

d. P(loses last two / wins first)= 0.2 × 0.25= 0.05

5. 0 = 4120 ml, s = 60 ml

a. Pr(4000 < V < 4180) = 13.5 + 34 + 34= 81.5% = 0.815

b. Pr(V < 4000) = 2.5%6. a. Pr(0 < z < 1.6) = 0.4452

b. Pr(-2.14 < z < 1.16)= Pr(0 < z < 2.14) + Pr( 0 < z < 1.16)= 0.4838 + 0.3770= 0.8608

c. 0 = 55, s = 8

i. Pr(x > 61)

= Pr ( z > )

= Pr( z > 0.75)= 0.5 - Pr ( 0 < z < 0.75)= 0.5 - 0.2734= 0.2266

ii. Pr(50 < x < 66)

= Pr ( < z < )

= Pr(-0.625 < z < 1.375)= Pr (0 < z < 0.625) + Pr(0 < z < 1.375)= 0.2340 + 0.4155= 0.6495

Pages 92 - 98, PROBABILITY & NORMAL DISTRIBUTION

PRACTICE TEST 1

1. a. One possible method and the suggested answers usingthis method. Use a 3 digit random number generator from acalculator. Assign the first digit in the random numberto bamboo. e.g. let a 1 in the first digit = bamboo present, (being1 out of 10 or 10%)ˆ 0, 2-9 in the first digit position = bamboo absent Assign the second and third digits (as a pair) for flaxe.g. let pairs 00 to -24 = flax present (f), (being 25 outof 100 or 25%) pairs -25 to -99 = flax absent

A table of this simulation based on the above is given.

RAN# RAN# RAN# RAN# RAN# RAN#

745 - 754 - 854 - 097 - 259 - 636 -

155 B 348 - 643 - 247 - 970 - 632 -

580 - 012 F 791 - 685 - 331 - 550 -

355 - 292 - 211 F 761 - 630 - 671 -

175 B 728 - 867 - 515 F 511 F 445 -

033 - 744 - 954 - 850 - 773 - 257 -

763 - 127 B 944 - 557 - 801 F 926 -

556 - 923 F 719 F 814 F 423 F 303 F

795 - 237 - 541 - 744 - 739 - 986 -

301 F 464 - 460 - 121 B,F 007 F 688 -

938 - 683 - 592 - 790 - 167 B

583 - 730 - 623 F 189 B 863 -

421 F 793 - 329 - 361 - 865 -

684 - 254 - 569 - 315 F 897 -

d. Letter given to 21 households

2. One possible approach (apart from recognisingindependence) is a tree diagram.

160 Answers

YEAR 12 MATHEMATICS


a. P(get letter) =

= 0.025 + 0.075 + 0.225= 0.325,

ˆ expect 200 x 0.325 = 65 lettersb. P(both) = 0.025

ˆ expect 0.025 x 200 = 5 would have bothc. From the simulation, 21 out of 80 would get a letter

ˆ expect 200 x = 52

ˆ 52 - 53 households are to receive a letter.d. From the simulation, 1 out of 80 had both.

ˆ expect 200 x = 2

ˆ 2 - 3 households with both3. Note: Example comment only. The simulated results have

lower values than the corresponding theoretical results.The random feature of a calculator is not truly random,but generated therefore a second sample of 80 numberscould have more than the theoretical values. Also there isno allowance in the theoretical case whether an operatorwho is monitoring these items, actually misses them in aparticular bag (or bags)(i.e. human error).

Section B

4. µ = 84 kg, σ = 8.5 kg

a. Pr(84 < x < 90) = Pr(0 < z < )

= Pr(0 < z < 0.706)= 0.2598

b. Pr(0 < z < ) + 0.5

Pr(0 < z < 1.176) + 0.5 = 0.3802 + 0.5= 0.9802

c. Pr(x < 100) = Pr(0 < z < ) + 0.5

= Pr(0 < z < 1.882) + 0.5= 0.4700 + 0.5= 0.97

5. a. i. Pr(100 < x < 105)

Pr( < z < ) = Pr(1.882 < z < 2.471)

= 0.4932 - 0.4700= 0.0232 = 2.32%

ii. Pr(x > 105) = Pr(z > )

= Pr(z > 2.471)= 0.5 - 0.4932= 0.0068= 0.68%

b. Pr(x > 104) = Pr(z > )

= Pr(z > 2.353)= 0.5 - 0.4907= 0.0093

Expect 850 x 0.0093 = 7.905 , i.e. 8 bags6.

Use inverse normali.e. Pr(0 < x < W) = 0.4543

ˆ Pr(0 < z < ) = 0.4543

ˆ = -1.688 (it is on the negative side)

or using x = σz + µ= -1.688 x 8.5 + 84= 69.65 kg= under 70 kg

Pages 99 - 104, PROBABILITY & NORMAL DISTRIBUTION

PRACTICE TEST 2

Possible Simulation1. 70% of the packs will have 1 wristband ˆ use random

numbers 000 - 69930% of the packs will have 2 wristbands ˆ use randomnumbers 700 - 999The packs with 2 wristbands could have 15 possibilities,i.e. Band 1 & Band 1 etc1 & 1 2 & 2 3 & 3 4 & 4 5 & 51 & 2 2 & 3 3 & 4 4 & 51 & 3 2 & 4 3 & 51 & 4 2 & 51 & 5Assume that the bands are randomly selected beforebeing packed. For the pairs with one wristband, assignthese random numbers (700 ÷ 5 = 140 possibilities)ˆ Band 1 000 - 139

Band 2 140 - 279Band 3 280 - 419Band 4 420 - 559Band 5 560 - 699

Answers 161

YEAR 12 MATHEMATICS

Pages 99 - 104 contFor the packs with two wristbands, assign these randomnumbers (300 ÷ 15 = 20 possible pairs).(1&1) 700-719 (2&2) 800-819 (3&3) 880-899(1&2) 720-739 (2&3) 820-839 (3&4) 900-919(1&3) 740-759 (2&4) 840-859 (3&5) 920-939(1&4) 760-779 (2&5) 860-879 (4&4) 940-959(1&5) 780-799 (4&5) 960-979

(5&5) 980-999 2. Here is a possible simulation exercise based on the above.

RAN# RAN# RAN# RAN# RAN# RAN#

406 3 347 3 991 5,5 814 2,2 737 1,2 969 4,5

307 3 065 1 507 4 679 5 637 5 665 5

301 3 938 3,5 596 5 715 1 459 4 578 4

297 3 836 2,3 246 2 662 5 227 2 223 2

135 1 112 1 597 5 740 1,3 021 1 494 4

3. Ten packets were needed in this simulation example toget all the wrist bands (see above).

4. a. After five bands (packets), based on this simulation

were collected so no student would have the fourdifferent bands.

b. After ten bands(packets), based on this simulation,

were collected every student would have a full set.5. The one band packs (5 possible choices) in 70% or (14%

each band) and the two band packs (15 possible choices)in 30% (or 2% each pair) provide a useful picture if wechoose one numbered band The diagram is of part of the tree diagrams.e.g. if 4 is the missing band, there is a 24% chance (from14 + 10) that the next packet will have a 4.

a. ˆ 25 students x 24% = 5.76 or almost 6 studentsb. If the next packet is a double band packet, then 10%

is possibleˆ 25 students x 10% = 2.5 or 2 - 3 students

Section B

6. µ = 315 g, σ = 6 g

Pr(315 < x < 323) = P( 0 < z < )

= P(0 < z < 1.333)= 0.4087

7. P(x < 323) = 0.5 + 0.4087= 0.9087

8. P(x < 310)

P(z < ) = P(z < -0.833)

= 0.5 - P(0 < z < 0.833)= 0.5 - 0.2975= 0.2025

ˆ 950 x 0.2025 . 192 packets9. P(310 < x < 321)

P( < z < ) = (-0.833 < z < 1)

= 0.2975 + P(0 < z < 1)= 0.2975 + 0.3413= 0.6388

ˆ 950 x 0.6388 . 607 packets10. P(321 < x < 324.5)

P( < z < ) = (1 < z < 1.5383)

= 0.4433 - 0.3413= 0.102

162 Answers

YEAR 12 MATHEMATICS


11. P(300 < x < 325)

P( < z < ) = P(-2.5 < z < 1.667)

= 0.4938 + 0.4522= 0.946

ˆ reject 1 - 0.946 = 0.054 or 5.4%12. P(0 < x < W) = 0.495

or P(0 < z < ) = 0.495

= - 2.576

or W = 6 x -2.576 + 315“Very light” packets have w = 299.5 g (1 dp)

13. P(0 < z < ) = 0.35

ˆ = -1.036

308 = µ - 1.038 x 6308 + 6.228 = µ µ = 314.228New mean = 314 (3 sf)

Pages 105 - 108, SEQUENCES - Revision Summary

1. T1: 1 + = 2, t2: 2 + = 2 , t3: 3 + = 3

T4: 4 + = 4

ˆ 2, 2 , 3 , 4

2. 24, 353. ...., ...., ...., 4, 7, 10 ˆ d = 3

a = 4 - 3 - 3 - 3= -5

4. a = 8, d = 6, t8 = a + (n - 1)d= 8 + 7 x 6= 50

S16 = [2a + (n - 1)d]

= [ 2 x 8 + 11 x 6]

= 8[16 + 66]= 656

5. T12 a + 11d = 41T1 a = 8

11d = 33ˆ d = 3

ˆ s13 = [ 2 x 8 + 12 x 3]

= [16 + 36]

= 3386. a = 7, d = 2

Sn = [ 2 x 7 + (n - 1) x 2]

352 = [ 14 + 2n - 2]

352 = [12 + 2n]

352 = 6n + n2

n2 + 6n - 352 = 0(n + 22)(n - 16) = 0ignore n = -27, ˆ n = 16 (terms)

7. 3, 9, 27, 81, 243; ˆ a = 3, r = 3ˆ t11 = a x r10

= 3 x 310

= 177 1478. T6 a x r5 = 160

T5 a x r4 = -80

by long division Y r = -2

ˆ if t6 = 160, then a x r5 = 160a x (-2)5 = 160

a =

a = -5ˆ Tn = (-5) x (-2)n - 1 is the general term for this GP

9. 3200 + 1600 + 800 + ...., ˆ a = 3200, r =

a. s12 =

=

= 6398

b. S4 =

=

= 6400

Answers 163

YEAR 12 MATHEMATICS

Pages 105 - 108 cont

10. a. 3k + 1 = 3 + 32 + 33 + 34 + 35 + 36

= 1092

b. (2n - n2)= 1 + 0 - 3 - 8 - 15 - 24 - 35 - 48

= -132

Pages 109 - 112, SEQUENCES - PRACTICE TEST 1

QUESTION ONE

a. a = 15, d = 4, Tn = a + (n - 1)dT16 = 15 + 15 × 4

= 75 min

= 1 hours

b. Sn = [ 2a + (n - 1)d]

S16 = [ 2 × 15 + 15 × 4]

= 8 [ 30 + 60 ]= 720 min (or 12 hours)

QUESTION TWO

Sn = where a = 24, r = 1.05

ˆ S20 = (i.e. 24 + 25.2 + 26.46 + ...)

= 793.6 minutes (or 13 hours, 13.6 minutes)

QUESTION THREE

T4 a x r3 = 49.13T2 a x r1 = 68.00

by division =

ˆ r2 = 0.7225 ˆ r = 0.85

since r = then T 2 = r × T1

and T1 =

=

= $80 (first monthly payment)

QUESTION FOUR

a or T1 = $495, 100% - 26% = 74% (or 0.74)

ˆ Tn = × 495

= $82.50ˆ Tn = a × rn - 1

= 82.50solving 495 × 0.74n - 1 = 82.50

0.74n - 1 =

(n - 1) log(0.74) = log ( )

n - 1 =

n - 1 = 5.95 n = 6.95 ( near the end of 7 years)

QUESTION FIVE

a = 96, r = (100 - 17.5)% = 82.5% or 0.825

ˆ S4 =

= $548.57 (2 dp)

QUESTION SIX

After 21 weeks, Sn = [ 2a + (n - 1) x d]

S21 = [2a + 20d]

= 21a + 210d= 11 235

In week 7, Tn = a + (n - 1)dT7 = a + 6d

= 595Solving simultaneously to find d, then a21a + 210d = 14 235 6 21a + 210d = 11235 a + 6d = 595 6 21a + 126d = 12495

84d = - 1260 d = -15

i.e. it reduces by 15 minutes each weekTo find a, T7: a + 6d = 595 minutes

a + 6 × -15 = 595first week a = 685 minutes

Pages 113 - 116, SEQUENCES - PRACTICE TEST 2

QUESTION ONE

a = 88, d = -2.75, n = 13ˆ Tn = a + (n - 1)d

T13 = 88 + 12 × -2.75= 55 cm (length of top rung)

QUESTION TWO

a = 6, d = 1.25, n = 31

Sn = [ 2a + (n - 1)d]

S31 = [ 2 × 6 + 30 × 1.25]

= [12 + 37.5]

= 767.25 km (Total distance travelled)

QUESTION THREE

a = 95 cm, r = 1.04, n = 10t10 = a × r9

= 95 × 1.049

= 135.2= 135 cm (nearest cm)

QUESTION FOUR

If growth in week 1 is T1 = 48 cm, r = (100 - 12.5%)= 87.5%= 0.875

As n 6 4, ˆ S 4 = 48 + 42 + 36.75 + 32.156 ...

=

= 384 cm (or 3.84 m)

164 Answers

YEAR 12 MATHEMATICS

Pages 113 - 116, cont

QUESTION FIVE

T1 = a = 201, T3 = a × r2

= 547.2

= (from )

r2 = 2.722r = 1.65 or 165%

ˆ Area of fourth ringA4 = 201 × 1.653

= 902.9 m2

QUESTION SIX

a = ?, n = 12, d = -150 mm, S12 = 25 500 mm

ˆ Sn = [2a + (n - 1)d]

25500 = [2a + 11 × -150]

25500 = 6[2a - 1650]25500 + 9900 = 12a

12a = 35400

a =

ˆ lowest piece = 2950mm or 2.95m

QUESTION SEVEN

First, compare some early figures for each payment plan.Plan A: Deposit $4950, M1 = 300, M2 = 320, M3 = 340...Plan B: Deposit $7995, M1 = 615, M2 = 605, M3 = 595 ...ˆ equate the two schedules by summing each then

combining them as an equation.

SA: 4950 + [ 2 × 300 + (n - 1)20]

= 4950 + [580 + 20n]

= 4950 + 290n + 10n2

SB: 7995 + [2 × 615 + (n - 1)(-10)]

= 7995 + (1240 - 10n)

= 7995 + 620n - 5n2

SA = SB

4950 + 290n + 10n2 = 7995 + 620n - 5n2

15n2 - 330n - 3045 = 015(n2 - 22n - 203) = 0( n - 29) (n + 7) = 0

ignore n = -7, ˆ n = 29 (monthly payments)Check

Plan A: 4950 + (2 × 300 + 28 × 20) = 21770

Plan B: 7995 + (2 × 615 + 28 × -10) = 21770

Pages 117 - 120, TRIGONOMETRY PROBLEMS

Revision Summary

1. a. x2 = 8.62 - 7.22

x =

x = 4.7 m (2 sf)

b. sin 62 =

ˆ x =

x = 126 cm (3 sf)

c. cos θ =

θ = cos-1

θ = 32.0E 1 dp

2. a. x2 = 1.142 + 1.062 - 2 × 1.14 × 1.06 × cos 61= 2.4232 - 2.4168 × cos 61

x =

x = 1.12 m (2 dp)

b. =

x =

x = 59.8 mm (3 sf)

3. a. cos θ =

θ = cos-1

θ = 72.9E

b. =

θ = sin-1

θ = 49.1E (1 dp)4. a. i. 0.52 rad ii. 2.88 rad iii. 4.15 rad

b. i. 60.7E ii. 75E iii. 337.5E5. a. l = 5.6 × 1.11

= 6.2 cm (1 dp)

b. l = 114 × 48E ×

= 95.5 mm (3 sf)

c. θ =

= 1.63 rad (2 dp)

6. a. A = × 64 × 26

= 830 cm2 (2 sf)

b. A = × 5.8 × 13.4 × sin 52

= 30.6 m2 (1 dp)

c. A = × 2452 × 0.72

= 21 600 mm2 (3 sf)d. A = A1 - A2

= × 34.52 × 95 × - × 34.52 × sin 95

= 394 cm2 (3 sf)

Answers 165

YEAR 12 MATHEMATICS

Pages 117 - 120, cont

7.

=

x =

x = 56 m (2 sf)Then

= sin 66

d = 56 × sin 66d = 51 m (2 sf)

The river is 51 m wide at this point.

Pages 121 - 124, PRACTICAL TRIGONMETRY PROBLEMS

PRACTICE TEST 1

PART A

Area of each ª

ªABX = ab sin x

= × 123 × 72 × sin 113E

= 4080 m2 (3 sf)

ªBCX = × 72 × 136.5 × sin 86E

= 4900 m2 (3 sf)

ªCDX = × 136.5 × 135 × sin 113E

= 8480 m2 (3 sf)

ªDAX = × 135 × 123 × sin 113E

= 6170 m2 (3 sf)TOTAL Area = 23,630 m2 (or 2.363 ha)

PART B

1. Diagonal AC p BXD = 48E + 113E = 161E

By Cosine RuleAC2 = 1232 + 136.52 - 2 × 123 × 136.5 × cos 161E

AC =

AC = 256 m (3 sf)

Diagonal BD; p BXD = 48E + 113E = 161E

By Cosine RuleBD2 = 1352 + 722 - 2 × 135 × 72 × cos 161E

BD =

BD = 204 m (3 sf)ˆ shortest distance is BD = 204 m

2. North West corner is pBAD = pBAX + pDAX

166 Answers

YEAR 12 MATHEMATICS

Pages 121 - 124 contBy Cosine RuleAB2 = 1232 + 722 - 2 × 123 × 72 × cos 113E

AB =

AB = 165 m (3 sf)

cos (pBAX) =

pBAX = Cos-1

= 23.7E ( 1 dp)AD2 = 1232 + 1352 - 2 × 123 × 135 × cos 48E

AD =

AD = 106 m (3 sf)

=

pDAX = sin-1 ( )

= 71.2E (1 dp)ˆ pBAD = 23.7 + 71.2

= 94.9E (1 dp)3.

Cos θ =

θ = cos-1

= 2.22 rad (2 dp)ˆ Area of segment

= Asector - Atriangle

= r2θ - r2 sin θ

= × 1.252 × 2.22 - × 1.252 × sin 2.22

= 1.11 m2 (2 dp)b. Volume = length × cross sectional area

= 256m × 1.11 m2

= 280 m3 (2 sf) of sediment

Pages 125 - 129, PRACTICAL TRIGONOMETRY PROBLEMS

- PRACTICE TEST 2

Part A

There are many possibilities, depending on the location ofQ. e.g. QR = 19.63 m, QL = 21.22 m, LR = 1.83 m

Part B

1. By the cosine rule,

cos(LQR) =

pLQR = cos-1

pLQR = 2.54E (2 dp)

2.

= 19.632 + 19.632 - 2 × 19.63 × 19.63 × cos 2.54E

=

= 0.87 m 3. a. Note - you could use triangle facts from several levels

with this situation. One solution is given below.

First find pWLR using the sine rule

=

pWLR = sin-1

= 82.2E 1 dp or 82E (2sf)then by co-interior angles on parallel linesα = 180 - 93

= 87Eand by the angle sum on a straight linepWLR + α + θ = 180E

θ = 180 - 87 - 82= 11E

ˆ bearing of S from L is 011E

b.

pWRL = 180 - 82 - 25 = 73E

By either cosine rule or sine rule find WL

=

WL =

WL = 4.14 m (from Wayne to post)Cos Rule would have been:WL2 = 4.292 + 1.832 - 2 × 4.29 × 1.83 × cos 73 etc

Answers 167

YEAR 12 MATHEMATICS


4. Area of the remaining (unpainted) area= Area of Whole Circle - Area of sector

Since p’s at a point add to 2π

pAOB = 2π - 2.17= 4.11 rad

ˆ Area = π × 4.572 - × 3.362 × 4.11

= 42.41 m2

5. Planned height of puck above C (TC) is 1.22 - 0.30 = 0.92 m

Since line PC is halfway between PL (4.9m) and PR(5.3m)

ˆ PC =

= 5.02 m

ˆ Angle of elevation pTPC =

= tan-1

= 10.4E

Pages 131 - 134, TRIGONOMETRIC EQUATIONS

Revision Summary

1. a. i. 30E × = 0.52 rad

ii. 1.99 radiii. 4.68 rad

b. i. 30E =

=

ii.

iii.

c. i. 0.92 × = 52.7E

ii. 150.7Eiii. 300E

2. a., b.

3. a. 2sin x = 1

sin x =

x = sin-1

ˆ x = 30E and x = 180-30

= 150Eˆ x = 30E or 150E

168 Answers

YEAR 12 MATHEMATICS


b. 5cos x = 1 + 2

cos x =

x = cos-1

x = 0.927 rad, 2π - 0.927ˆ x = 0.927 rad, 5.356 rad (3dp)

c. cos 2x = 0.85First revolution

2x = cos-1 (0.85)2x = 0.5548 or 2π - 0.5548

ˆ 2x = 0.5548 radians or 5.7284 radiansSecond revolution

2π + 0.5548, 4π - 0.5548= 6.8380 radians or 12.0116 radians

ˆ 2x = 0.5548, 5.7284, 6.830, 12.0116 x = 0.277, 2.864, 3.145, 6.006 radians (3 dp)

d. 2.5 sin 3x = -1.5

sin 3x =

3x = sin-1 (- 0.6)3x = -36.87

ˆ 3x = 180+36.87, 360-36.87, 540+36.87, 720-36.87900+36.87, 1080-36.87

x = 72.3E, 107.7E, 192.3E, 227.7E, 312.3E, 347.7E

e. sin( x - 45) = 0.4(x - 45) = sin-1 (0.4)(x - 45) = 23.6ECheck possibilities either side of [-270, 270]ˆ x - 45E = -360+23.6, -180-23.6, 23.6, 180-23.6

x - 45E = 336.4E, -203.6E, 23.6E, 156.4Ex = -291.4, -158.6E, 68.6E, 201.4E

(Note the value -291.4 is outside the domain)

4. y = 5sin or y = 5sin( x)

Max value is 5 5 sin( x) = 5

sin ( x) =1

x = sin-1 (1)

x =

ˆ x = πMin value is -5

5 sin( x) = -5

sin ( x) = -1

x = sin-1 (-1)

x =

ˆ x = 3π

Pages 135 - 139, TRIGONOMETRY EQUATIONS

PRACTICE TEST 1

QUESTION ONE

a. x = cos-1 (0.4)= 66.4E

andx = 360 - 66.4E

= 293.6Eˆ x = 66.4E, 293.6E(radian measures = 1.159, 5.121)

b. sin x = 3.09 - 4= -0.91

x = sin-1 (-0.91)= -65.5E

Domain is 0E # x # 360E and sin is negative in the thirdand fourth quadrants.ˆ solutions are x = 245.5E, 294.5E(radian measures = 4.285, 5.140)

c. tan x =

tan x = 1.6x = tan-1(1.6)x = 1.012x = π + 1.012

ˆ x = 1.012, 4.154(degree measures = 58.0E, 238.0E)

Answers 169

YEAR 12 MATHEMATICS


QUESTION TWO

cos 2x = 0.782x = cos-1(0.78)2x = 38.74E

Using a graph with 2 cycles

ˆ 2x = 38.74E, 360E-38.74E, 360E+38.74E, 720E-38.74Ex= 19.4E, 160.6E, 199.4E, 340.6E(In radians, x = 0.339, 2.803, 3.480, 5.945)

QUESTION THREE

a. 33 sin(45t) = 30

sin(45t)=

45t = sin-1 ( )

45t = 65.38E

t =

ˆ t = 1.453 seconds to reach 30 cm after starting at h = 0b. (45t)E 6 45 cycles in 360 seconds

ˆ = 8 seconds to return

ˆ 45 eight second cycles every 360s

= 1.453 s, 2.547 s to get to 30 cm

QUESTION FOUR

If y = 0.450.6 sin (512 π t) = 0.45

sin (512 π t) =

512 π t = sin-1 ( )

t =

= 0.00053 seconds (2 sf)

QUESTION FIVE

If depth, d = 3m then solve

2.3 cos ( ) + 3.8 = 3

2.3 cos ( t) = -0.8

cos ( t) =

t = cos-1 ( )

t = 1.926

t = 3.85 hˆ 3.85 hours (or 3h 51 min) each side of high tide has

enough water for safe passage in/out of harbour.Graphically, this would appear on (0, 2π)

Pages 140 - 143, TRIGONOMETRY EQUATIONS

PRACTICE TEST 2

QUESTION ONE

a. θ = cos-1 (0.4)θ = 66.4E

and θ = 360E - 66.4ˆ θ = 66.4E, 293.6E

b. 2.5 sin θ = 1.5 - 1

sin θ =

θ = sin-1 (0.2)θ = 11.5E (1 dp)

ˆ θ = 11.5E, 180E - 11.5Eˆ θ = 11.5E, 168.5E (1 dp)

c. tan θ = 1.1 - 2.7tan θ = - 1.6 θ = tan-1 (-1.6)ˆ θ = -1.012As this is outside the domain of [0, 2π]Draw a graph of tan θ on [0, 2π] to helpθ = -1.012 is matched when θ = π - 1.012, 2π - 1.012θ = 2.130, 5.271 (3 dp)

QUESTION TWO

a. θ - 30E = cos-1(-0.7)θ - 30E = 134.4E (second quadrant)or 360E - 134.4E (third quadrant)ˆ θ = 134.4E + 30E, 360E - 134.4E + 30Eˆ θ = 164.4E, 255.6EIt is useful to check with a substitution with one of yourown solutions.e.g. Evaluating cos(255.6E - 30E) = -0.69966

= -0.7 (1 dp)

170 Answers

YEAR 12 MATHEMATICS


b. 2θ = sin-1 (0.61)= 0.656 rad (3 dp)

ˆ 2θ = 0.656, π - 0.656ˆ 2θ = 0.656, 2.486ˆ 2θ = 0.328, 1.243

QUESTION THREE

38.6 + 1.6 sin ( t) = 40E

1.6 sin ( t) = 40E - 38.6E

sin ( t) =

t = sin-1 ( )

t =

t = 2.713 days (3 dp)ˆ during her third day.

QUESTION FOUR

First week

30 - 33 cos ( t) = 45

30 - 45 = 33 cos ( t)

= cos ( t)

cos-1 ( ) = t

ˆ t = 2.043

ˆ t = 2.043, 2π - 2.043

t = ,

t = 3.90, 8.10 (2 dp)ˆ from the end of the 4th week, and into the beginning

of the 8th week, weekly sales of snow boards are atleast 45 boards

Normal Distribution 171

YEAR 12 MATHEMATICS

Areas Under the Normal Curve

This table gives the area under the standard normal curve between 0 and z (the shaded area).

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.00 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359

0.10 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753

0.20 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141

0.30 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517

0.40 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879

0.50 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224

0.60 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549

0.70 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852

0.80 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133

0.90 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389

1.00 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621

1.10 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830

1.20 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015

1.30 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177

1.40 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319

1.50 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441

1.60 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545

1.70 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633

1.80 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706

1.90 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767

2.00 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817

2.10 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857

2.20 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890

2.30 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916

2.40 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936

2.50 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952

2.60 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964

2.70 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.4974

2.80 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.4981

2.90 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986

3.00 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990

3.10 0.4990 0.4991 0.4991 0.4991 0.4992 0.4992 0.4992 0.4992 0.4993 0.4993

3.20 0.4993 0.4993 0.4994 0.4994 0.4994 0.4994 0.4994 0.4995 0.4995 0.4995

3.30 0.4995 0.4995 0.4995 0.4996 0.4996 0.4996 0.4996 0.4996 0.4996 0.4997 3.40 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4998

3.50 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998

3.60 0.4998 0.4998 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999

3.70 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999

3.80 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999

3.90 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000

4.00 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000

172 Useful Formulae

YEAR 12 MATHEMATICS

FORMULAE SHEET

Pages for Extra Notes 173

YEAR 12 MATHEMATICS

174 Pages for Extra Notes

YEAR 12 MATHEMATICS

Pages for Extra Notes 175

YEAR 12 MATHEMATICS

176 Pages for Extra Notes

YEAR 12 MATHEMATICS

questions & answers ncea level 2 mathematics

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