IB Prepared IB Diploma Programme Mathematics: Applications and Interpretation [1 ed.] 1382007310, 9781382007313, 9781382007283

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O X

F O

R

D

I B

P

R

E

P A

R

E

D

MATHEMATICS:

APPLIC ATIONS

AND

INTERPRETATION

I B

D I P L O M A

P R O G R A M M E

Peter Gray

David Harris

/

O X

F O

R

D

I B

P

R

E

P A

R

E

D

MATHEMATICS:

APPLIC ATIONS

AND

INTERPRETATIONS

I B

D I P L O M A

P R O G R A M M E

Peter Gray

David Harris

/

Acknowledgements

Cover

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/

C ontents

Intr oduc tion

iv

4 Sta tis tic s and pr obability

4.1

Probability

149

4.2

Probability distributions

160

4.3

Transformations and combinations

1 Number and algebr a

1.1

Real numbers

2

1.2

Sequences and series

1.3

Algebra (AHL)

20

1.4

Complex numbers (AHL)

24

1.5

Systems of equations

31

1.6

Matrices and matrix algebra (AHL)

33

End-of-chapter practice questions

8

39

of random variables (AHL)

170

4.4

Poisson distribution (AHL)

176

4.5

Descriptive statistics

179

4.6

Correlation and regression

187

4.7

Statistical tests

191

4.8

Sampling (AHL)

197

4.9

Non-linear regression (AHL)

199

4.10

Condence intervals and

2 Func tions

hypothesis testing (AHL) 2.1

Lines and functions

2.2

Composite functions and

201

46

4.11

Transition matrices and

Markov chains (AHL) inverse functions (AHL)

53

2.3

Graphs and features of models

55

2.4

Graphs and features of models (AHL)

211

End-of-chapter practice questions

215

74

5 C alculus End-of-chapter practice questions

91

3 Geome tr y and trigonome tr y

3.1

Working with triangles

3.2

Sectors of a circle; volumes and

5.1

Dierentiation

224

5.2

Integration

228

5.3

Dierentiation (AHL)

231

5.4

Integration (AHL)

235

Solutions of dierential equations (AHL)

240

102

surface areas

107

5.5

3.3

Voronoi diagrams

110

End-of-chapter practice questions

3.4

Trigonometric functions (AHL)

113

3.5

Matrix transformations (AHL)

116

3.6

Vectors

120

3.7

Graphs

End-of-chapter practice questions

248

Internal assessment: an exploration

256

P r ac tice exam paper s

263

Index

277

130

139

Worked solutions to end-of-chapter practice questions and exam papers in this

book can be found on your suppor t website. Access the suppor t website here:

w w w.oxfordsecondary.com / ib-prepared-suppor t

iii

/

Introduction

This

book

(first

in

provides

examination

Mathematics:

for

both

Higher

full

in

coverage

2021)

IB

and

the

diploma

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of

and

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new

papers

syllabus

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the

two

Oxford

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companions

Mathematics:

with

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HL

and

SL,

ISBN

978-0-19-842698-1.

exam

guide

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and

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Oxford

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978-0-19-842716-2

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authors

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of

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iv

/

INTRODUCTION

Command terms

Command

term

terms

specifies

are

the

pre-defined

type

and

words

depth

of

and

the

phrases

response

used

in

expected

all

IB

from

Mathematics

you

in

a

questions.

particular

Each

command

question.

Command term

Denition

Calculate

Obtain a numerical answer, showing the relevant stages in the working.

Comment

Give a judgment based on a given statement or result of a calculation.

Compare

Give an account of the similarities between two (or more) items or situations, referring to both (all) of

them throughout.

Compare and contrast

Give an account of similarities and dierences between two (or more) items or situations, referring to

both (all) of them throughout.

Construct

Display information in a diagrammatic or logical form.

Contrast

Give an account of the dierences between two (or more) items or situations, referring to both (all) of

them throughout.

Deduce

Reach a conclusion from the information given.

Demonstrate

Make clear by reasoning or evidence, illustrating with examples or practical application.

Describe

Give a detailed account.

Determine

Obtain the only possible answer.

Dierentiate

Obtain the derivative of a function.

Distinguish

Make clear the dierences between two or more concepts or items.

Draw

Represent by means of a labelled, accurate diagram or graph, using a pencil. A ruler (straight edge)

should be used for straight lines. Diagrams should be drawn to scale. Graphs should have points

correctly plotted (if appropriate) and joined in a straight line or smooth curve.

Estimate

Obtain an approximate value.

Explain

Give a detailed account, including reasons or causes.

Find

Obtain an answer showing relevant stages in the working.

Hence

Use the preceding work to obtain the required result.

Hence or otherwise

It is suggested that the preceding work is used, but other methods could also receive credit.

Identify

Provide an answer from a number of possibilities.

Integrate

Obtain the integral of a function.

Interpret

Use knowledge and understanding to recognize trends and draw conclusions from given information.

Investigate

Observe, study, or make a detailed and systematic examination, in order to establish facts and reach

new conclusions.

Justify

Give valid reasons or evidence to suppor t an answer or conclusion.

Label

Add labels to a diagram.

List

Give a sequence of brief answers with no explanation.

Plot

Mark the position of points on a diagram.

Predict

Give an expected result.

Prove

Use a sequence of logical steps to obtain the required result in a formal way.

Show

Give the steps in a calculation or derivation.

Show that

Obtain the required result (possibly using information given) without the formality of proof.

“Show that” questions do not generally require the use of a calculator.

Sketch

Represent by means of a diagram or graph (labelled as appropriate). The sketch should give a general

idea of the required shape or relationship, and should include relevant features.

Solve

Obtain the answer(s) using algebraic and/or numerical and/or graphical methods.

State

Give a specic name, value or other brief answer without explanation or calculation.

Suggest

Propose a solution, hypothesis or other possible answer.

Verify

Provide evidence that validates the result.

Write down

Obtain the answer(s), usually by extracting information. Little or no calculation is required.

Working does not need to be shown.

v

/

INTRODUCTION

Preparation and exam strategies

In

addition

to

the

above

suggestions,

there

are

compulsory:

the

questions.

simple

rules

you

should

study

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vi

/

INTRODUCTION

Key features of the book

Each

chapter

checklists.

typically

Chapters

covers

contain

one

the

topic,

and

following

starts

with You

should

know

and

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should

be

able

to

features:

Example

Note

Examples

offer

solutions

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typical

questions

and

demonstrate

Provide quick hints and

common

problem-solving

techniques.

explanations to help you better

understand a concept.

Sample

student

Definitions to rules and concepts

questions

are given in a grey box like this one,

In

and explained in the text.

response

each

answers

(many

of

response,

answer

is

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show

in

always

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be

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typical

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red

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responses

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on

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IB-style

papers).

student’s

The

correct

below.

S AMPLE STUDENT ANS WER

This feature assists in answering

par ticular questions, warns

Consider

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=

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the

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chapter.

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another par t of this book or the headed

SL (standard

level)

and AHL (additional

higher

level).

Students

IB Mathematics: Applications and following

the

higher

level

course

should

attempt

all

the

questions.

Interpretations syllabus, that

relates to the text in question.

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paper

questions

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vii

/

NUMBER

1 1 . 1

R E A L

the

meaning

decimal

of

You should be able to:

signicant

gures

and

of

✔

approximate

places

decimal

nearest ✔

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value

A LG E B R A

N U M B E R S

You should know:

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centred

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a

number

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signicant

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number

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to

of

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nd

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upper

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standard

technology

given

by

a

calculator

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Approximation, errors and estimation

The

are

processes

fundamental

Artem

carries

number

of

ranging

from

mean

as

of

of

close

in

out

sweets

all

as

counting,

737

the

Note

a

an

in

to

possible

value

of

spreadsheet

went

application

experiment

a

large

8378

guesses

1581.083 969 465 6

exact

the

to

and

1627.

took

Ar tem applied the idea of “ The

Artem

on

wisdom of the crowd”, in which

twosignificant

to

measuring,

Artem

far

less

express

figures

Artem

reflected

time

his

by

to

asks

each

read

these

For

make

that

this

all

a

the

rounding

guess

the

estimate,

the

number

the

was

the

an

to

finding

calculated

that

example,

students

averaging

by

approximating

estimate

counting

estimate

applying

He

find

that

than

has

good

number.

amazed

he

students

a

and

mathematics.

which

provide

exact

was

in

131

sweets.

can

the

jar.

of

estimating

mean

close

131

to

as

the

numbers

on

sweets!

it

correct

to

rules:

estimates of a quantity made by

each member of a large group

Significant figures:

are collected and the average 1,2,3,4,5,6,7,8 and 9 are always significant.

found. The theory is that the 0 is significant if it is between two significant figures or at the end of the overestimates and underestimates number after the decimal point. balance out when the average is

Rounding to a given number of decimal places or significant figures: taken, leaving a good estimate of

the size of the quantity.

5,6,7,8,9 round up, whereas 0,1,2,3,4 round down

2

/

1 .1

To

express

third

to

his

estimate

significant

write

Hence

1600.

the

correct

to

to

figure

in

Similarly ,

estimate

two

and

two

figures,

1581.083 969 465 6,

for

1627

the

significant

significant

he

exact

which

rounded

values

Artem

2

were

is

down

the

8,

examined

and

to

0

same

get

when

NumbERs

the

rounded

to

RE Al

up

1600.

expressed

figures.

Eape 1.1.1

(a)

Write

(b)

(c)

0.041 792

Write

Write

67.0812°

8307.59

m

correct

correct

km

to:

to:

correct

to:

(i)

two

decimal

(ii)

two

significant

(i)

the

(ii)

three

(i)

the

(ii)

three

nearest

places

figures

degree

significant

nearest

figures

km

significant

figures

Solution

(a)

(i)

0.041 792

0.04

m

m

to

2

rounds

Examine

to

place.

dp

Since

second

(ii)

0.041 792

0.042

m

m

to

rounds

2

to

figure.

(i)

67.0812°

to

the

rounds

nearest

to

Since

degree

digit

(ii)

67.0812°

67.1°

to

rounds

3

to

(i)

8307.59

8308

km

to

7.

figure.

km

to

rounds

the

nearest

digit

(ii)

8307.59

km

rounds

to

1,

round

place

third

it

is

the

it

is

to

8.

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5,

is

the

8,

to

2.

place.

units

significant

round

figure

first

4.

the

decimal

to

the

1.

decimal

round

the

round

fourth

it

to

figure

round

the

to

the

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7,

first

significant

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km

0,

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to

the

is

third

(c)

the

it

Examine

sf

is

decimal

significant

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67°

it

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second

(b)

third

decimal

Examine

sf

the

the

fourth

place.

units

significant

Avoid premature rounding by

8310

km

to

3

sf

figure.

Since

it

is

7,

round

the

using the unrounded answers on

third

significant

figure

to

1.

your GDC. By doing this you avoid

accumulation of errors.

Valentina

that

his

sweets,

an

repeats

method

Artem’s

could

Valentina

estimate

of

asks

experiment

work.

154

With

a

she

different

students,

2303.980 519 480 5.

since

jar

averages

Artem

says

finds

hard

containing

their

his

it

guesses

estimate

is

to

imagine

2210

and

finds

better

since

Unless otherwise stated in the it

is

closer

to

the

exact

value:

the

difference

between

his

estimate

and

question, all final numerical the

exact

value

is

1627

−

1581.083 969 465 6

=

45.916 030 534 4

whereas

for

answers must be given exactly or Valentina

it

is

2269.435 064 935 1

−

2210

=

59.435 064 935 1.

correct to three significant figures.

But

of

in

the

fact,

Valentina’s

exact

value.

v

ε

You

find

is

less

the

when

expressed

percentage

error

ε

as

by

a

percentage

applying

the

v A

formula

error

E

=

×

100%

,

where

v

is

the

exact

value

and

v

E

is

the

A

v E

approximate

value

of

v

Artem:

Valentina:

45.916 030 534 4

59.435 064 935 1

× 2210

100%

=

2.69%

×

100%

=

2.82%

1627

3

/

1

Whenever a measurement is approximated to the nearest unit u as v

,

A

u

lies in an interval: v

the exact value v E

u

−

≤

A

v




n

200

>

(n

1)

>

40,

(n

1)

d

to

write

the

inequality

needed.

41

day

than

⇒

+

1

42

of

250

training,

Adrian

will

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It

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is

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lengths.

255

idea

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lengths,

to

250

hence

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your

lengths.

42

is

On

answer:

day

42

On

he

day

41,

swims

correct.

n

(b)

(2(50)

+

5(n

1))

>

5000

⇒

You

must

apply

this

formula

to

write

the

2

inequality n

>

because

n

is

not

known.

36.219 25...

Hence

after

37

swum

more

days

than

training

5000

Adrian

will

Use

have

your

solve

lengths.

the

GDC

to

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inequality .

a

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S

a

table

=

4950

to

and

36

S

=

5180.

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final

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37

and,

in

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consecutive

For

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example,

the

is

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in

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common

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ratio

of

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the

the

ratio

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r

between

quantity

sequence

4,

any

this

the

9,

13.5,

…

the

nearest

integer

does

not

inequality .

two

the common

6,

case,

is

ratio.

1.5

u n

+

1

because

=

1.5.

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you

can

say ,

“Each

term

of

the

sequence

u n

is

1.5

times

terms

The

of

larger

the

sequence

general

term

u

and

than

formula

common

the

will

for

previous

term”.

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r

>

1,

then

the

size

of

the

increase.

the

ratio

r

nth

can

term

be

of

a

geometric

found

by

sequence

investigating

the

with

first

terms

of

1

this

sequence:

n

1

u

2

u

n

3

u

1

4

u

2

5

u

3

u

4

u

=

u

1

n

(r)

u

1

(r)(r)

1

u

(r)(r)(r)

=

u

1

pattern

gives

you

the

general

formula

u

=

n

r

5

u

1

n

The

u

–

…

6

4

u

r

1

…

u

5

3

u

6

r

…

1

1

r

1

Eape 1.2.4

For

(i)

each

the

of

first

these

term

geometric

u

sequences,

(ii)

find:

the

common

ratio

r

(iii)

1

(a)

5,

10,

20,

40,

…

u 7

(b)

24,

12,

6,

3,

…

(c)

1.5,

3,

6,

12,

…

11

/

1

Solution

(a)

u n

The

(i)

u

=

common

ratio

is

equal

to

r

+

1

=

,

so

you

have

a

choice

of

u

5

n

1

how

to

find

r:

10

(ii)

r

=

=

2

5

20

40

10

,

or

are

equivalent

ways

to

find

r

=

2.

6

(iii)

u

=

5(2)

=

24

=

320

10

20

5

7

(b)

(i)

u

It

is

important

to

remember

that

the

7th

term

means

the

common

1

7

ratio 12

(ii)

r

=

is

raised

to

the

6th

power,

since

u

7

= 24

=

u

1

n

an

2

application

of

the

formula

u

=

u

n

–

–

1

6

r

=

u

1

r

.

This

is

1

1

r

1

6

⎛

(iii)

u

=

7

1

24 ⎜ ⎝

⎞ ⎟

2

=

0.375

⎠

(c)

(i)

u

=

1.5

1

3

=

(ii)

−2

1.5

6

(iii)

u

=

1.5(

2)

=

96

7

n

u

(r

1)

1

The

formula

for

the

sum

of

n

terms

of

this

sequence

is

S

=

.

n

r

If

r

By

is

greater

than

multiplying

1,

the

you

may

find

numerator

this

and

form

more

denominator

convenient

of

this

1

to

formula

use.

by

1,

n

u

(1

r

)

1

you

can

also

write

the

equivalent

form

S

=

,

which

is

more

n

1

convenient

You

can

and

then

to

use

solve

if

0




0,

find

p

in

terms

of

q.

10

2

1

(p

[

+

2q)

=

]

log

(pq)

10

2

2

1

log

(log

=

⎠

1

log

1

⎞

log

(p

[

+

2q)

=

]

log

(pq)

10

▲ The

student

correctly

applies

2

1

(p

[

+

2q)

=

]

1

pq

+

laws

the

working

gaining

2

(p

the

2q)

=

of

2

logarithms,

setting

methodically

out

and

marks.

pq

8

1

2

(p

2

+

4pq

+

4q

)

=

pq

8

2

p

2

+

4pq

+

4q

2

p

+

4q

=

8pq

2

=

4pq

2

2

4q

2

p

=

4q

p

= ▼ The

student

complicated

and

manipulations

2

−

4q

=

4q

spends

valuable

unsuccessful

because

the

time

on

algebraic

wrong

2

4pq

=

approach

p

has

factorization

2

p

been

chosen.

(which

is

Applying

prior

learning)

to

2

–

4pq

+

4q

quickly

leads

to

the

answer.

infnte eetrc ere

You

know

that

the

formula

for

the

sum

of

n

terms

of

a

geometric

series

n

u

(1

r

)

1

is

=

S n

1

r

n

When

|r|


0

⇒

open

profit.

variale

t

>

2.11

for

3

hours

Hene

h

=

hours.

in

is

time.

Erin

order

will

ontext.

Being

open

for

2

hours

would

mean

making

a

loss.

to

3.

53

/

2

To

find

For

the

inverse

example,

the

of

a

funtion,

graph

you

first

examine

its

graph.

of 1.6

1.7

letters

1.8

RAD

2

f (x)

=

(x

2)

+

3

shows

that

y 2

f7(x)=(x–2)

when

refleted

in

y

=

x

+3

the

f8(x)=x

graph

is

that

of

a

relation:

x

In

order

to

find

the

inverse 1.6

funtion,

the

domain

of

1.7

f

letters

1.8

RAD

y

f8(x)=x

muste

restricted

so

that

it

is 2

f7(x)={(x–2)

+3,x≥2

one-to-one.

For

example,

domain

see

the

of

f

if

to

you

x

graphs

≥

of

restrit

2,

you

two

the

now

funtions:

−1

f

and

(The

f

restrition

have

The

x

the

same

final

step

x

≤

2

would

effet)

is

to

find

the

−1

equation

To

find

of

the

f

inverse

of

a

funtion:

2

(a)

Write

y

=

down

the

equation

y

=

(x

2)

x

=

(y

2)

+

3

+

3

f (x)

2

()

Transpose

x

for

y

aee  2

()

Solve

the

equation

for

y

x

3

=

(y

2)

You can transpose x for y at the x

3

=

y

x

3

+

2

2

star t of the process or at the end

and still be awarded full marks.

=

y

−1

(d)

Hene

write

down

the

f

(x)

=

x

3

+

2

−1

equation

Note

that

if

f

(x)

the

=

y

restrition

x

≤

2

was

made,

you

would

take

the

negative

(f

f

−1

root

is

and

find

f

−1

(x)

=

−

x

+

3

2.

In

oth

ases,

the

identity

)(x)

=

x

°

true.

For

example,

−1

( f

f

2

)(x)

=

f (

x

°

3

+

2)

=

(

x

3

+

=

(

x

3 )

=

x

2

2)

2

−1

−1

( f

f )(x)

=

f

3

+

3

+

3

=

x

2

2

((x

2)

+

3)

=

(x

2)

=

(x

2)

+ 3

3

+

2

°

2

=

This

is

onsistent

Although

funtion

it

is

is

with

not

useful

the

x

+

2

definition

mentioned

here:

2

i(x)

=

in

x.

the

This

+

=

of

2

x

an

inverse

syllaus,

the

funtion

has

funtion.

onept

no

of

effet

the

at

all

−1

the

value

of

x.

Note

that

identity

on

−1

(f

f)(x)

=

i(x)

reflets

the

fat

that

f

must

°

reverse

or

“undo”

the

effet

of

f:

hene

the

omposition

has

no

effet.

54

/

2.3

Gr a p H s

and

f E at u r E s

of

modEL s

s ampLE studEnt ans WEr

T wo

funtions,

f

and

g,

are

x

2

f (x)

6

g(x)

(a)

Write

()

Find

down

the

defined

in

value

3

3

1

(g

tale.

6

2

5

2

value

of

following

1

7

the

the

9

of f (1)

f)(1) °

▲ Beause

the

student

orretly

−1

()

Find

the

value

of

(

g

2) applies

the

funtions,

notation

the

value

of

of

omposite

(g

f)(1)

an

°

(a)

f(1)

=

(b)

g(f(x))

e

3

=

g(f(1))

=

found

easily

g(x)

g(1)

=

the

tale.

5 ▲ Knowledge

(c)

from

=

−2

is

that

solving

equivalent

to

nding

2 1

g

(

2)

is

applied

orretly .

1

hence:

g

2 . 3

(

2)

=

1

G R A P H S

A N D

F E AT U R E S

Y hl kw:

✔

the

differene

“draw”

✔

the

minimum

funtions

horizontal

✔

the

etween

of

the

the

key

roots

features:

terms

✔

maximum

interepts,

of

equations

linear,quadrati,

✔

ommand

vertex,

tehnology

equations;

and

vertial

features

exponential,

use

✔

determine

✔

transfer

✔

reate

diret

and

given,

and

✔

and

nd

up

inversevariation

the

a

the

formulae

for

diret

and

for

inverse

a

the

stages

of

the

modelling

asked

to

“Draw”

the

funtions

inluding

points

of

intersetion

features

of

a

graph

of

laelling

idea

and

of

from

a

or

a

y

and

to

from

key

paper

information

features

to

give

a

shape

parameters

solving

sreen

graph

axes

the

a

of

a

system

model

of

y

setting

equations

sustitution

of

using

points

into

a

variation funtion

proess. ✔

When

nd

key

sketh

the

given

✔

to

graph

tehnology ✔

graph

differenes

tehnology

general

etween

to

and

of

ui

sums

✔

models

differene

use

zeros

asymptotes

names,

sinusoidal

the

their

values,

or

M O D E LS

Y hl be ble :

“sketh”

meaning

and

of

and

O F

graph

of

a

funtion,

to

gain

apply

the

stages

of

the

modelling

proess.

maximum

aee  marks

you

plotted

must

and

represent

joined

in

a

the

graph

straight

line

aurately ,

or

smooth

with

urve.

points

Axes

orretly

must

e

Using the scales given in the laelled

to

show

the

sales

and

independent

and

dependent

variales.

question is essential so that the Axes

and

straight

lines

should

e

drawn

with

a

ruler.

You

draw

the

graph fits on the graph paper. graph

on

graph

paper.

55

/

2

When

asked

general

idea

maximum

to

sketch

of

the

or

shape

minimum

e

approximately

in

e

given.

not

however

features

There

you

in

the

is

their

of

same

take

are

of

a

funtion,

relationship.

asymptotes

orret

the

orret

the

points,

the

should

graph

loation.

An

make

the

represent

Relevant

and

requirement

to

you

axes

features

of

auray

sketh

suh

interepts

indiation

of

the

sale

as

shows

as

should

must

in draw

relevant

plaes.

Exle 2.3.1

Two

paths

Path

A is

are

planned

modelled

y

in

a

the

forest.

equation

y

=

0,

running

east

from

the

0.1d

point

0

≤

d

≤

from

The

0)

8.

(0,

to

path

(8,

0).

Distanes

Path

are

B

is

modelled

measured

in

y

km

f (d)

and

d

=

−e

is

sin

the

(90d),

distane

east

0).

planners

point

(a)

(0,

where

of

the

the

garden

paths

want

ross

and

to

at

install

eah

a

water

point

fountain

where

the

at

eah

seond

turns.

Sketh

the

graphs

of

y

=

0

and

y

=

f (d)

for

0

≤

d

≤

8,

showing

all

aee  the

maximum

points,

minimum

points

and

zeros

of f

Use the domain and range given ()

Hene

find

the

numer

of

water

fountains

planned

and

the

in the question to determine the oordinates

of

the

water

fountain

that

is

furthest

from

path

A.

viewing window on your GDC.

Solution

Use

(a)

the

given

maximum

f(d)(km)

values.

domain

and

Adjust

for

minimum

the

the

d

maximum

–0.1d

f(d)=e

1

sin(90d)

and

(2.96, 0.742)

minimum

values

on

the

(6.96, 0.498)

aee  (0, 0)

(2, 0)

(6, 0)

(8, 0)

(4, 0)

d(km)

y-axis

so

graph

in

that

you

an

see

the

detail.

10

Show

the

relevant

points

in

You can display the graph with a

their

grid background in order to make

orret

plaes

and

inlude

(4.96, –0.608)

–1

points

(0.96, –0.907)

at

eah

end

of

the

transferring the graph from your

domain

if

they

are

relevant

to

GDC screen to paper easier.

the

question.

sale

()

Hene

the

furthest

numer

from

path

of

fountains

A has

on

planned

oordinates

the

is

(0.96,

Give

a

sense

of

axes.

9.

The

fountain

0.907).

56

/

2.3

There

are

other

tehnology

Feature

as

of

key

features

of

a

graph

the

graph

of

y

=

f (x)

How

V ertex,

to

Find

x-interept(s),

of

the

you

determine

and

f E at u r E s

of

modEL s

with

follows:

y-interept

roots

that

Gr a p H s

zeros

of

equation

maximum

f (x),

f (x)

and

f (0)

Solve

=

0

minimum

points

determine

f (x)

=

0

or

use

graphing

app.

Use

GDC’s

or

your

apply

your

your

GDC’s

graphing

knowledge

understanding

of

app

and

alulus.

a +

Asymptotes:

Funtions

of

the

form

f (x)

,

=

n

∈

Z

have

a

vertial

n

x

asymptote

domain

with

of

f (x)

is

The

horizontal

You

represent

part

of

1.1

the

equation

x

≠

x

is

asymptotes

of

0.

This

is

a

onsequene

of

the

fat

that

the

0.

asymptote

graph

=

the

y

=

with

0

eause

dashed

funtion

y

=

the

lines

range

eause

f (x)

they

is

y

are

≠

0.

not

f (x).

DEG

1.2

of

1.1

DEG

1.2

y

y

1 f1(x)= x=0

1

2

x

f1(x)= x x

x

y=0

y=0 x=0

A horizontal

also

known

asymptote

as

the

ehaviour

an

y

exploration

further

e

limit

shows

of

y

=

suggested

with

the

f (x)

y

long-term

as

the

your

x

ehaviour

inreases

shape

GDC

as

of

a

in

size.

graph,

shown

in

of

the

funtion,

Asymptoti

and

determined

Example

2.3.2

Exle 2.3.2

The

temperature

in

a

house

t

hours

after

an

eletrial

heating

unit

1.3t

turns

(a)

on

is

Sketh

of

the

modelled

the

graph

y

of

y-interept,

the

y

=

the

funtion

c(t)

for

0

≤

maximum

c(t)

t

=

≤

23te

4,

point

+19

showing

and

the

the

oordinates

horizontal

asymptote.

()

Hene

four

()

The

desrie

hours

after

heating

24ºC

when

ut

the

how

unit

the

the

temperature

heating

swithes

ontinues

heating

to

unit

unit

off

emit

turns

when

heat

swithes

in

as

house

hanges

in

the

on.

the

it

the

temperature

ools.

Find

the

reahes

value

of t

off.

57

/

2

Solution

You

(a)

of

–1.3t

c (ºC)

c(t)=23te

+

an

the

explore

funtion

the

ehaviour

outside

the

19

given

30

domain

with

a

tale

or

y

(0.769, 25.5)

widening

Both

(0, 19)

y

=

the

viewing

perspetives

window.

show

that

the

19

limiting

value

equation

y

=

of

is

the

19,

hene

asymptote

the

is

19.

4

t

1.2

(hours)

30.69

1.3

*damped

1.4

DEG

y

f4(x):=

X

f4( (0.769, 25.5)

()

At

t

=

19ºC

on.

0,

the

and

The

until

the

heater

temperature

0.769

reahing

25.5ºC

temperature

hours

a

it

23*xe^(-.

swithes

6.

19.0565...

7.

19.0179...

8.

19.0055...

9.

19.0017...

10.

19.0005...

inreases

have

maximum

efore

is

passed,

of

starts

to

x

derease,

loser

to

getting

gradually

–0.44

0.5

19.0005198757

3.56

19ºC.

Give

aee 

a

detailed

aount

of

the

() hanges

30.7

in

temperature,

using

y

–1.3·x

Don’t limit your view of the

f4(x)=23·

x·

e

your

+19

answer

to

(a).

function to only what your GDC

f9(x)=24

(0.337, 24)

first shows you: exploration of the

Use

behaviour of the function outside

your

GDC

intersetion

the viewing window can give you

After

valuable insights.

unit

0.337

hours,

swithes

the

of

to

y

find

=

c(t)

the

and

y

=

24.

heating

off.

aee 

When sketching or drawing asymptotic behaviour, take care to show the

shape correctly. The graph must be seen to get gradually closer to an

asymptote. It should not veer away from nor cross the asymptote.

One

sum

the

you

of

have

the

entered

funtions,

equations

again,

two

or

as

funtions

their

into

differene,

shown

in

your

GDC,

without

Example

you

it

an

graph

having

to

the

type

in

2.3.3

Exle 2.3.3

Xtar

Jewellery

edition

models

earrings

revenue

earned

with

y

the

the

ost

in

£

funtion

selling

the

x

of

making

c(x)

pairs

=

of

3500

x

+

pairs

of

26.5x,

earrings

with

limited

and

the

the

funtion

2

r(x)

(a)

=

77.6x

Find

the

order

()

By

0.08x

least

for

Xtar

numer

to

onsidering

find

the

make

the

numer

of

of

pairs

a

profit.

graph

of

earrings

of

earrings

p(x)

=

r(x)

that

will

that

c(x)

must

or

maximize

e

sold

in

otherwise,

Xtar ’s

profit.

58

/

2.3

Gr a p H s

and

f E at u r E s

of

modEL s

Solution

(a)

The

least

earrings

numer

Xtar

of

must

pairs

sell

is

In

of

order

revenue

79.

ost. ()

Sine

319.4

is

not

an

profit

for

319

and

e

The

=

e

profit,

greater

graphs

revenue

show

than

that

are

equal

Xtar

y

of

will

ost

at

=

78.02

y

4660.02

2

r(x)=77.6 

p(320)

the

found:

20000

p(319)

must

a

320 x

must

make

integer, and

the

to

=

x–0.08 

x

4600.00

make

making

and

the

most

selling

profit

319

c(x)=3500+26.5 

x

pairs

earrings.

aee 

(78.02, 5568)

(319.4, 4660)

x P(x)=r(x)–c(x) 0

1000

1000

In order to find the ver tex of

a quadratic function, you can

Sine

you

funtions

have

already ,

expression

the

p(x)

alulator

graph

entered

=

you

apply the formula for the axis

the

type

the

c(x)

and

r(x)

provides

required.

the

of symmetry, use calculus or

technology. In the context of

the problem, choose the most

efficient method.

s ampLE studEnt ans WEr

0.5x

Let

(a)

f (x)

For

the

=

e

−

graph

(i)

write

(ii)

find

(iii)

write

2,

of

down

the

for

4

≤

x

≤

4

f:

the

y-interept

x-interept

down

the

equation

of

the

horizontal

asymptote. ▲ The student expresses the

interepts

()

Sketh

the

graph

of

and

asymptote

and

(a)(i)

y

=

1

equation

of

the

f

the

y-intercept

is

(0,

to

the

with

level

the

of

orret

auray

notation

required.

1)

0.5x

(ii)

f(x)

=

0

=

e

2

f(x)

1.386

▲ The

∴

x-intercept

is

(1.39,

sketh

features

domain

(iii)

y

=

shows

the

main

0) of

the

graph

given,

and

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the

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2

y

ehaviour

is

full

are

marks

skethed

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hene

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6 f(x)

5

4

3

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1

(1.39, 0) x

–3

–2

–1

3

4

(0, –1)

–2

y=

–2

–3

59

/

2

tye  el

You

have

used

to

seen

in

previous

represent

models.

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summarized

real-world

are

in

setions

several

the

tale

ontexts

examples

of

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example

of

mathematis

appliation

models

you

of

shape

of

mathematial

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of

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line

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f (x)

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c

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y

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Name Name

examples

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gradient

y

y

2

m

1

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x

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of

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interest,

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c

depreiation rx

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siene,

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inreases law

In

n

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siene,

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x

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n

∈

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0) is equivalent

n

to stating that y varies inversely as x

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variation

Information

given:

To

determine

the

parameters:

n

y

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inverse

variation

model

has

equation

i(x)

=

ax

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n

∈

Z,

n




0,

graph

the

same

same

+

and

represented

of

of

of

x.

y.

y:

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means

(x

p.

value

value

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in

of

x-axis.

a)

to

fator

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the

given

“input”

y.

p:

x-axis

represented

the

f (x)

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of

f E at u r E s

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f (qx)

e

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py)

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y:

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transformation

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f (x

x

f ((x

of

is

f

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a),

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translation

(x,

f

onsidered

+

py)

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point

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x

graph

the

the

graph

also

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pf (x)

point

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eah

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streth

zero,

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=

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(x,

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a,

to

to

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y).

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give

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left

y

as

point

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if

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a

a

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times

0


χ calc

crit

then the observed values are too far from the expected values so

minus 1.

the null hypothesis is rejected.

193

/

4

The

assessmen p

most

common

method

though

is

to

consider

the p-value

for

the

test.

The p-ue for a test is the probability of obtaining the results in the sample

(or more extreme ones) when the null hypothesis is true. The null hypothesis

Remember the null hypothesis is

is rejected if the p-value is less than the significance level of the test.

2

rejected if

χ calc

is gee hn the

critical value (the observed values

Exmpe 4.7 .2

are too far from the expected)

or if the p-value is ess hn the

significance level (the probability

of the observed values occurring

A six-sided

die

in

below.

the

there

table

is

is

evidence

rolled

Test

that

60

at

times

the

the

die

and

5%

is

the

outcomes

significance

level

are

recorded

whether

or

not

biased.

by chance is even smaller than the Number

1

2

3

4

5

6

Frequency

8

9

7

10

6

20

significance level).

For

five

degrees

of

freedom

the

5%

critical

value

is

11.07

Ne

Solution

It is 5 degrees of freedom because :

H

The

die

is

not

biased

The

die

is

biased

The

The

when a die is rolled.

hypotheses

must

always

be

stated.

0

there are 6 possible outcomes

H

:

null

hypothesis

is

that

the

distribution

1

is

Expected

values

if

the

equal

The

die

to

the

observed

one

and

being

tested.

expected

values

are

2

60

is

fair

are

all

=

entered

10

into

a

GDC

and

a

χ

goodness

6

assessmen p

of

fit

test

is

performed.

2

χ

=

13

> 11.07

calc

You must always give a reason

for your conclusion by comparing

There

is

sufficient

evidence

The

alternative

The

p-value

The

conclusion

reasoning

would

be:

2

to

χ calc

reject

H

with the critical value or the

at

the

5%

level,

0

so

we

conclude

the

die

=

0.0234
10

13

25

37

43

26

6

Frequency

2

In

order

to

of

pebbles

perform

in

distribution.

each

His

Length, x (cm)

State

the

(c)

Use

(d)

Perform

the

results

he

classes

are

calculates

if

their

shown

in

the

expected

lengths

the

table

did

number

follow

this

below.

2 < x ≤ 4

4 < x ≤ 6

6 < x ≤ 8

8 < x ≤ 10

x > 10

6.0

21.0

43.1

46.0

a

b

null

your

of

test,

x ≤ 2.0

Expected value

(b)

χ

a

and

answer

alternative

to

part

(a)

hypotheses.

to

find

the

values

of a

and

b

2

is

a

χ

sufficient

degrees

of

test

at

the

evidence

freedom

5%

to

for

significance

reject

the

the

null

level

to

see

hypothesis.

if

there

State

the

test.

194

/

4 .7

(i)

P(8


10

=

0.05667... ≈

are

calculated

at

the 2

start (ii)

t E S tS

assessmen p

Solution

(a)

S tat i S t i c a l

of

the

question

but

will

0.0567 be

used

In a

χ

test all the expected

frequencies need to be greater

later.

than 5. In the Standard Level exam (b)

H

:

The

lengths

of

pebbles

can

0

this will always be the case.

2

be

modelled

by

a

N(6.2,

2.4

)

distribution.

H

:

They

cannot

be

modelled

by

1

this

(c)

distribution.

a

=

150 × 0.170 ≈

b

=

150 × 0.0567

The

25.5

expected

probability ≈

degrees

of

freedom

=

6

1 =

sure

values

5

found p-value

So

to

=

there

0.0569 >

is

reject

:

The

sample

size.

for

in

you

the

part

use

the

full

probability

(a)

0.05

insufficient

H

=

8.50 Make

(d)

×

value

evidence

lengths

of

0

pebbles

can

be

modelled

by

a

2

N(6.2,

the

2.4

)

distribution.

es f ndependene

Ne

2

A special

type

of

χ

test

is

to

test

whether

or

not

two

variables

are

Most GDCs will have a different 2

independent

of

each

other.

For

example,

you

might

want

to

test

function for the

whether

the

type

of

music

people

like

is

independent

of

their

χ

test for

age.

independence.

2

This

type

of

frequencies

test

are

is

called

the

normally

χ

test

given

in

a

for

independence

contingency

and

the

observed

table.

assessmen p

S aMPlE StUDENt aNS WEr

There are several ways to write

In

a

school,

preferred

obtained

students

drink.

are

The

in

grades

choices

organized

Grade 9

in

9

to

were

the

12

were

milk,

following

asked

juice

and

to

select

water.

their

The

data

table.

your conclusion to the test.

The following is an example of

acceptable phrasing:

Milk

Juice

Water

t

25

34

15

74

‘ There is sufficient evidence to

reject H

’ or ‘ There is insufficient

0

Grade 10

31

x

13

74

Grade 11

18

35

17

70

evidence to reject H

’.

0

You should avoid ‘Accept H

’ if the

0

Grade 12

t

9

36

26

71

83

135

71

289

result of the test is not significant.

2

χ

A

H

:

test

is

carried

out

at

the

5%

significance

the

preferred

drink

is

independent

the

preferred

drink

is

not

of

the

level

with

hypotheses:

grade

0

H

:

independent

of

the

grade

1

assessmen p 2

The

χ

critical

value

for

this

test

is

12.6

An exam question might ask you (a)

Write

down

the

value

of

x

for the degrees of freedom of the

(b)

Write

(c)

Use

down

the

number

of

degrees

of

freedom

for

this

test.

test. This can normally be read off

2

your

graphic

display

calculator

to

find

the

χ

the GDC but otherwise it can be statistic

for

calculated for an n × m table as this

test.

(m − 1)(n − 1).

(d)

State

the

conclusion

for

this

test.

Give

a

reason

for

your

answer.

195

/

4

▲ Correct,

calculated

many

▼ It

3

of

picked

sf.

is

An

will

actually

no

lead

to

been

rst

also

freedom

to

answer

imply

have

but

state

so

the

(a)

30

(b)

6

(c)

18.9

(d)

Reject

the

error

can

up.

With

could

might

incorrectly

GDCs

degrees

be

it

18.96

method

a

loss

this

the

so

2

was

that

performed

12.6.

T hey

are

not

independent.

the

t-test

is

used

to

see

if

two

samples

both

coming

from

populations

correctly .

the and

>

the -es

might

distributed

▲ Conclusion

18.9

marks.

The method

because

0

this

though

examiner

H

to

shown

of

close

19.0

same

normally

mean.

For

and

with

example,

the

if

same

you

standard

want

to

test

deviation

whether

could

have

students

in

reason

Germany

bothgiven.

you

score

could

scores

of

take

each

as

a

highly

in

random

Mathematics

sample

of

each

exams

and

as

students

compare

the

in

France,

average

sample.

Ne

The null hypothesis for a t-test is: H

: the means are equal. The alternative

0

It is not sufficient to say that if might be that the means are not equal (two-tailed test) , or it might be that one the average score of one sample is greater than or less than the other (a one-tailed test). Your GDC will ask you is greater than the other then to choose. that must be true for the whole

population, but if one is a long way The

t-test

should

be

used

only

if

the

background

population

is

normally

above the other then that could distributed.

One

way

to

check

this

is

to

look

at

a

box

and

whisker

plot.

be quite strong evidence for a If

it

is

fairly

symmetrical

and

the

inter-quartile

range

is

less

than

half

the

difference in the population means. range

then

it

is

likely

to

be

close

to

a

normal

distribution.

Exmpe 4.7 .4

A teacher

wants

to

helpful

offer

is

volunteer

course.

to

All

to

stay

the

see

to

if

his

a

revision

students.

behind

students

after

are

course

Half

school

then

his

to

given

he

plans

Perform

class

attend

a

(a)

test

to

the

see

course

scores

for

the

two

groups

are

recorded

in

the

table

there

at

is

the

any

improved

10%

significance

evidence

the

average

that

the

level

revision

scores.

may

assume

that

the

scores

come

from

and normal

listed

t-test

and You

the

if

a

distributions

with

the

same

standard

below. deviation.

Took revision course

23

45

78

82

56

90

65

72

(b)

State

why

the

result

of

the

test

might

not

be

Did not take 44

32

45

79

48

69

50

valid

35

when

deciding

if

the

revision

course

revision course

improves

scores.

Solution

(a)

H

:

The

revision

course

does

not

improve

the

An

alternative

general

statement

which

will

0

scores

in

the

always

test.

be

the

populations H

:

The

revision

course

does

improve

the

same

have

for

the

a

t-test

same

is:

The

two

mean.

score

1

p-value

There

is

=

0.0893


6.04

x

=

6.5

lies

in

the

critical

region,

so

H

is

rejected.

0

(b)

x

=

6.5

> 6.04

and

hence

we

reject

H

:

μ

=

5

0

This

(c)

p-value

=

0.008 85




0

:

µ

mean

difference

of

T




0.2

0.05

reject

camp

the

has

not

so

0.1

0.5

particular

insufficient

null

be

The

hypothesis

improved

that

the

value,

differences

same

way

as

mean

normally

are

for

difference

other

equal

to

a

0.

entered

any

is

into

the

GDC

in

the

t-test.

the

times.

cnfdene ne s assessmen p

Often

A p% confidence interval is often

to

instead

give

an

of

testing

interval

or

data

range

against

within

a

particular

which

the

mean

mean

is

it

is

more

likely

to

useful

lie.

thought of as ‘there is a p% chance

that the mean is in this interval’.

Though useful, this is not the

correct definition, and so should

not be quoted in an exam.

A p% confidence interval is such that if a sample was collected many times

then the population mean μ would fall within the confidence interval p% of

the time.

Confidence

If

the

intervals

population

distribution

degrees

of

is

are

normally

standard

used;

if

not,

obtained

deviation

then

the

is

directly

known

then

t-distribution

is

from

the

the

GDC.

normal

used,

with

(Z)

n

1

freedom.

tes f he en effen

It

is

possible

to

test

whether

or

not

two

categories

are

independent

by

2

using

the

χ

correlation

will

is

be

test.

a

done

linear

alternative

coefficient

important

for

An

when

to

between

values,

remember

is

the

rather

that

to

test

two

than

the

to

see

if

variables

the

is

frequencies,

correlation

product

equal

are

to

zero.

given

coefficient

is

moment

This

though

only

it

looking

relationship.

206

/

4 . 10

coNFiDENcE

iN t E r va l S

aND

H Y P ot H E S i S

tE StiNG

(aHl)

A necessary assumption for the test of the correlation coefficient is that both

variables are normally distributed.

assessmen p The

statistic

Inthe

same

imply

r

≠

0

the

the

What

way

in

this

that

can

say

evidence

variables

a

population

correlation

we

strong

used

are

is

is

sample

mean

is

the

that

if

it

not

ρ

the

is

is

sample

mean

coefficient

that

not

test

0,

for

not

we

the

being

not

equal

cannot

whole

sufficiently

also

correlation

to

0

that

from

to

0,

0,

does

just

population

far

equal

say

coefficient, r.

and

is

performed using the inbuilt

not0.

there

hence

In exams, this test will be

because

(ρ )

then

not

the

function on the GDC and the

is

data will always be given in

two

the question.

independent.

Exmpe 4.10.6

Ne

A teacher

is

convinced

that

time

spent

playing

games

on

a

Make sure you know where to find computer

affects

the

marks

obtained

in

class

assessments.

To

test

this test on your GDC. It is likely to this,

he

selects

8

students

and

asks

them

how

much

time

each

week

be in a menu for regression tests they

spend

gaming

and

compares

this

with

the

scores

obtained

in

or Linear regression tests. the

next

assessment.

The

responses

are

shown

in

the

table.

Student

A

B

C

D

E

F

G

H

Time on computer (hours)

12

0

15

8

9

4

0

8

Score in assessment

34

61

55

38

78

65

46

82

Perform

a

evidence

between

You

test

to

the

reject

time

may

at

10%

the

spent

assume

significance

hypothesis

on

that

a

that

computer

the

variables

level

to

there

and

are

see

is

no

score

in

if

there

is

any

correlation

the

normally

assessment.

distributed.

Solution

H

:

ρ

=

0

H

0

ρ

:

≠

Even

0

though

probably

p-value

=

0.836

>

0.1

there

teacher

is

looking

evidence

to

the

10%

a

negative

the

question

is

only

reject asking

at

for

is correlation

insufficient

H

the

1

significance

whether

or

not

ρ

=

0

level.

0

tes f he men f  Pssn dsun

2

It

is

possible

Poisson

it

the

use

it

being

is

not

distribution

population

Assume

X

is

if

but

a

it

see

if

particular

is

with

Poisson

to

due

a

to

a

mean.

the

different

then

it

is

sample

But

reject

H

Po(μ).

if

if

a

distribution

mean.

easy

to

come

from

significant

not

being

However,

construct

If

we

wish

to

test

H

µ

:

µ

=

a

if

it

test

the

value

of

X

obtained

against

H

0

0

we

could

is

a

result

is

Poisson

known

for

the

(μ).

mean

~

test

with

clear

Poisson

χ

the

distribution

obtained

or

to

in

our

:

µ

>

µ

test

is

very

then 0

1

unlikely

if

the

0

mean

were

µ 0

So

for

a

test

with

H

:

µ

>

µ

hypothesis

P(X

≥

a) ≤

is

rejected

0.05

if

µ

at

a

5%

significance

level,

the

null

0

1

=

if

X

≥

a

where

a

is

the

smallest

value

such

that

µ 0

In

the

diagram,

probabilities

There

are

critical

the

critical

coloured

two

region

blue

approaches

or

by

region

will

to

is

be

coloured

less

than

performing

calculating

P(X

≥

x)

blue.

The

sum

of

the

0.05

the

test,

where

x

either

is

the

by

finding

the

observedvalue.

a

207

/

4

Exmpe 4.10.7

Over

a

long

number

of

period

of

time

imperfections

it

in

is

known

printed

that

(a)

the

a

Poisson

distribution

with

the

critical

region

for

this

test.

material It

follows

Find

a

is

given

that

19

imperfections

are

found

in

mean thesample.

of

1.2

every

After

print

10

improvements

the

material

imperfections

To

m.

test

this

sample

of

it

is

occur

belief

at

length

on

the

felt

will

the

200

that

be

5%

m

is

machine

the

used

rate

at

(b)

State

(c)

Find

the

which

the

gives

reduced.

significance

conclusion

of

the

test.

to

level

p-value

the

same

for

the

result

test

as

in

and

part

confirm

this

(b)

a

checked.

Solution

(a)

Under

the

H

distribution

is

Po(24)

The

mean

is

1.2 × 20

=

24

0

H

:

µ

=

24

H

0

µ

:




0.05

It

is

more

p-value Hence

there

is

insufficient

evidence

to

reject

usual

rather

to

perform

than

the

finding

test

the

by

finding

critical

the

region.

H 0

tes f  ppun pp n usng he

nm dsun

Exmpe 4.10.8

assessmen p

A company

about

its

employed

popularity

by

in

a

a

political

town.

party

They

is

know

gathering

that

evidence

nationally

the

party

The tests for binomial and Poisson is

supported

by

55%

of

the

population

but

suspect

it

might

be

less

will always be one-tailed in in

this

town.

the exam.

They

say

to

select

they

a

randomly

support

believe

the

support

in

chosen

party .

the

Is

sample

this

town

is

of

25

sufficient

less

than

people

and

evidence

at

find

the

that

5%

11

level

55%?

Solution

H

:

p

=

0.55

H

0

:

p




is

(25,

0.55),

those

in

the

sample

were

randomly

and

independent.)

the

numbers

in

our

sample

supporting

the

party

0.05 were

There

be

party .

If P(X

would

sample chosen

who

distribution

0

1

insufficient

evidence

to

reject

H

at

the

too

to

accept

In

fact

small

the

then

this

alternative

would

be

strong

evidence

hypothesis.

0

5%

significance

level.

means

the

11

probability

is

not

in

the

is

greater

critical

than

0.05

which

region.

208

/

4 . 10

coNFiDENcE

iN t E r va l S

aND

H Y P ot H E S i S

tE StiNG

(aHl)

type i nd ype ii es

The are two possible errors when performing a hypothesis test.

Type I error: Rejecting H

when H

0

is true

0

Type II error: Failing to reject H

when H

0

If

we

let

region,

C

be

and

the

so

event

H

is

of

is not true.

0

our

rejected,

sample

we

will

statistic

write

falling

the

in

the

probability

critical

of

a

type

I

0

error

as

P(C|H

).

0

The

the

probaility

test

such

when

as

the

different

The

if

type

or

error

of

a

of

a

is

by

definition,

continuous.

the

For

distribution

a

they

significance

discrete

might

II

error

value

type

II

to

is

the

error

more

difficult.

parameter

as

P(C′|H

is

),

It

can

known.

with

H

1

alternative

The

1

of

distribution

be

slightly

be

calculated

We

will

standing

denote

for

this

1

value.

process

Find

level

4.10.9)

type

alternative

is,

binomial

Example

probability

I

distribution

Poisson

(see

an

a

the

probability

only

the

of

for

the

calculating

critical

a

region,

type

given

II

error

H

is

is

always

the

same:

true.

0

2

Find

the

probability

of

not

being

in

the

critical

region

given

H

is

1

true,

P(C′|H

),

or

1

P(C|H

1

)

1

Exmpe 4.10.9

The

not

principal

do

so

if

will

a

large

fewer

50students

She

of

to

test

than

find

the

school

20%

data

on

is

of

the

the

hypothesis

H

trying

to

students

proportion

:

p

decide

=

0.2

would

p

of

against

the

(a)

Find

(b)

Write

(c)

If

sample

the

critical

down

the

the

is

region

H

and

for

of

explain

and

the

of

students

what

:

p




0,

can

be

found

from

calculating

use

trapezoidal

value

of

y

when

x

is

0

technology

to

nd

the

area

of

a

region

enclosed

by

a

curve

y

=

f (x),

the

x-axis,

and

the

f ( x) dx

∫

lines

a

the

the

where

b

f (x)

example

f (x),

✔

✔

form

n+ 1

for

✔

general

as

a

n

∫

the

n+ 1

x

=

a

and

x

=

b,

where

f (x)

>

0

rule. ✔

nd

the

an

estimate

trapezoidal

width

when

for

the

rule,

given

value

with

either

of

an

intervals

a

table

of

area

of

using

equal

data

or

a

function.

Antdervatves

Nte An

antiderivative

rate

or

for

or

finding

integral

areas

is

under

useful

a

for

deriving

an

equation

from

a

curve.

This can be written as

a

n

∫

ax

dx

n +

= n +

dy

1

x

+ c

1

a

n

=

If

ax

y

then

n+ 1

=

x n +

dx

+ c,

n ≠

−1

1

Example 5.2.1 This formula can be found in

section SL 5.5 of the formula book .

Notice that it excludes the case

where n =

−1 as the formula in this

The

rate

after

which

heating

the

temperature

element

is

turned

T

on

is

is

changing

given

by

t

minutes

the

equation

dT

=

instance would involve dividing by 0.

a

at

19

2t ,

0

Find

the

≤ t

≤ 10

dt

(a)

When

t

=

0

rate

the

(b)

Find

an

(c)

Find

the

of

change

of

temperature

expression

for

maximum

the

is

when

t

=

4

5 °C

the

value

temperature

temperature

of

T

for

0

≤ t

at

time t

≤ 10

Solution

(a)

When

t

=

The

4

dT

In standard level exams, n will

rate

asking =

19

8

=

of

for

change

the

is

value

another

of

the

of

derivative.

11

dt

dT

always be an integer. Notice 11

way

˚C

per

that

the

units

of

are

˚C

per

dt

minute minute.

(b)

T

=

∫

19

1

=

19t

Note

2t dt

∫

2

2 ×

t

that

derivative

+ c

19 dt

of

19t

=

because

19t

is

the

19.

2

2

=

19t

When

t

t

+ c

=

0,

The

value

of

c

is

found

by

substituting

2

T

=

⇒

c

19

=

×

0

–

0

+

c

=

5

known

values

for

the

variables.

5

2

T

=

19t

t

+ 5

228

/

5.2

(c)

The

maximum

value

t

=

occurs

You

when

9.5

should

to

ensure

at

t

=

0

curve T

=

the

or

it

plot

t

is

=

the

curve

maximum

10.

to

find

your

does

Having

easier

on

not

plotted

the

i N T E g R AT i o N

GDC

occur

the

maximum

95.25 °C directly

from

the

curve

rather

than

dT

=

solving

0

dt

Areas bet ween a cur ve and the x-axs y

Integrals

The

x

=

can

be

notation

b

where

used

for

b

>

the

to

find

area

(shown

a

the

area

between

as

R

in

between

the

the

curve

y

diagram)

a

curve

=

f ( x),

is

given

and

the

the x-axis.

lines

by

the

x

=

a

and

definite

b

integral

written

as

f ( x) dx

∫

a

R

Example 5.2.2 x a

The

side

wall

modelled

by

in

a

a

concert

curve

hall

with

can

b

y

be

equation

2

y

=

2.8 x

0.5 x

horizontal

theunits

(a)

Find

,

0

≤

x

distance

are

the

≤ b

from

measured

value

where

of

the

in

x

is

the

point

O

and

metres.

b 0

x b

(b)

Write

the

down

area

of

an

the

integral

which

represents

wall.

At Higher Level you will need to 2

(c)

It

is

intended

to

repaint

the

wall.

If

one

can

of

paint

covers

4.5 m

know how to work out the value

of

wall,

find

the

minimum

number

of

cans

of

paint

needed.

of the integral you wrote down in

par t (b) using integration (anti-

Solution

(a)

b

=

derivatives) but for Standard

This

5.6

can

be

equation,

found

or

by

directly

factorizing

from

the

Level you will always use the

the

GDC

appropriate function on the

by

5.6 2

(b)

2.8 x

∫

0.5 x

dx

entering

the

equation

and

finding

calculator. You should though

the

0

zero

know how to use the notation

(x-intercept).

correctly. 2

(c)

Area

=

14.6

Number

of

m

Using

cans

The

the

area

integral

is

formula

obtained

for

directly

area.

from

the

GDC.

14.6

required

≈

3.24

4.5

To 4

cans

of

find

the

minimum

number

of

cans

you

paint need

to

round

up.

The trapezdal rule

y

Before

a

the

curve

arrival

and

trapezoids

the

and

particularly

of

GDCs,

x-axis

was

working

when

the

an

approximation

often

out

found

their

equation

by

areas.

defining

for

the

dividing

This

the

is

still

curve

area

the

area

useful

is

between

not

into

y

area

between

the

curve

y

=

f ( x),

the

lines

x

=

a

and

b

can

divided

into

n

trapezoids

each

with

height

h

x

f(x)

today ,

known.

y

The

=

=

b

with

b

>

a,

y 0

y 1

y 2

y 3

y 4

y –1

a

=

,

as

shown.

a

x

x 1

x 2

x 3

x 4

b

x

–1

n

229

/

5

The

area

of

the

trapezoids

can

be

found

using

the

trapezoidal

rule:

b

1

This formula is in section SL 5.8 f ( x) d x

∫

+

y

0

+ 2( y n

+

y

1

+ ... +

y

2

n

1

))

2

a

of the formula book

h( y

≈

The question may ask you if the trapezoidal rule gives an underestimate or an

overestimate of the actual area. This is usually clear from the diagram.

Example 5.2.3

Use

the

trapezoidal

rule

with

4

trapezoids

to

find

an

approximate

value

for

the

area

between

the

curve

6

y

=

,

2 +

the

x-axis

and

the

lines

x

=

1

and

x

=

3

x

Solution

3

h

1

=

b

=

0.5

Using

the

formula

h

4

n

You x

1

1.5

2

2.5

3

y

8

6

5

4.4

4

the

should

enter

required

individually .

1 Area

a

=

should × 0.5 ( (8 +

≈

4) +

values.

For

always

function

into

Alternatively

example,

write

when

down

the

your

they

x

=

1,

GDC

can

y

y-values

=

be

2

to

+

and

read

worked

6

=

obtain

8.

off

out

You

method

4.4) )

2(6 + 5 +

2

=

the

marks

in

case

you

make

an

error

in

your

calculations.

10.7

Example 5.2.4

A nature

reserve

trapezoidal

reserve.

rule

Each

is

bounded

and

unit

the

by

a

river

coordinates

represents

one

of

and

the

a

straight

points

fence,

shown

to

as

shown

find

the

in

the

diagram

approximate

below.

area

of

the

Use

the

nature

kilometre.

(2.5, 4.2) (1, 4)

(1.5, 3.5)

(2, 3.2)

(3, 2)

(3.5, 1.8)

(0.5, 1.8) Nature

reser ve

Fence

Solution

Width

of

trapezoids

=

The

0.5

width

difference

can

easily

between

be

the

found

from

x-values,

but

the

could

1

Area

≈

× 0.5 [ (0 + 0) + 2

10.25

+

4

+

3.5

+

3.2

+

4.2

+

2

+

1.8) ]

4

also

2

=

2(1.8

be

calculated

using

h

0

=

=

0.5

8

km

230

/

5.3

5 . 3

D I F F E R E N T I AT I O N

Yu shuld knw

the

derivatives

where

✔

if

if

that

At

the

0

point

concavity

in

sin

x,

cos

x,

tan

x,

e

n

,

ln

x,

x

✔

use

the

chain

quotient

f ′′( x) >

a

( A H L )

∈ 

f ′′( x)