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
3
Great
Clarendon
Oxford
and
of
Street,
University
furthers
the
Oxford
Oxford,
Press
is
University’s
education
by
a
OX2
Press
6DP,
department
objective
publishing
University
image:
Chaplin
in
of
United
of
the
University
excellence
worldwide.
Oxford
the
in
UK
and
p160
Kingdom
of
Oxford.
in
research,
is
a
certain
other
trade
Erin
mark
countries.
The
Oxford
University
Press
publisher
used
for
this
rights
of
the
authors
have
been
published
in
rights
a
SP1,
2021
reserved.
retrieval
No
system,
part
or
of
this
p22:
TZ0,
publication
transmitted,
in
any
may
form
be
or
reproduced,
by
any
means,
prior
permission
by
law,
by
in
writing
licence
or
of
Oxford
under
University
terms
agreed
Press,
with
or
the
as
the
rights
scope
of
organization.
the
Enquiries
above
should
be
concerning
at
address
sent
to
the
University
Press,
must
not
this
same
condition
British
Data
circulate
Library
on
to
and
thank
the
questions
the
International
from
past
corresponding
Baccalaureate
examinations.
past
paper
SP1,
TZ0,
The
for
their
questions
references,
are
sum-
the
this
work
any
acquirer.
Cataloguing
in
in
p8:
MATHL,
N16,
HP2,
MATSD,
TZ2,
Q6;
Sample
Q1;
Sample
student
student
answer,
Q9;
TZ0,
Sample
Q13;
Sample
student
answer,
p26:
student
answer,
M18,
answer,
p23–4:
MATHL,
N18,
HP1,
p17:
N17,
p14:
MATHL,
TZ2,
Q7;
MATSD,
HP2,
TZ0,
Sample
Q1;
student
p32:
N17,
MATHL,
SP1,
TZ0,
HP2,
Q2;
TZ0,
Sample
Q1;
Sample
student
student
answer,
p59:
answer,
M16,
p55:
MATME,
expressly
TZ2,
Q3;
Sample
student
student
answer,
p65:
answer,
M16,
p64:
M17,
MATSD,
SP1,
MATSD,
TZ2,
SP1,
Q13;
TZ2,
Q14;
Sample
student
reproduction
Rights
p83:
M19,
MATME,
SP2,
TZ2,
Q5;
Sample
student
answer,
p84–5:
Department,
MATME,
SP2,
TZ2,
Q6;
Sample
student
answer,
p106:
M17,
MATME,
above.
SP2,
You
David
(illustration),
appropriate
M18,
Oxford
SP1,
student
answer,
outside
like
adapt
answer,
M17,
MATME,
Sample
reprographics
p93
Harris
without
SP2,
permitted
Erin
stored
N18,
the
to
book,
student
answer,
in
(photos);
p152
here:
MATSD,
Sample
All
Curthoys
(photo),
asserted.
N16,
First
Judith
Harris
2021
answer,
moral
p69
David
(illustration).
would
permission
Sample
The
Harris
kind
marized
©
p150
It
scholarship,
registered
Shutterstock;
(photo),
any
other
form
and
you
must
impose
TZ2,
Sample
Sample
student
answer,
Publication
Q9;
student
answer,
p126–7:
M14,
answer,
p115:
M13,
MATME,
HP3,
SP1,
p138:
M17,
MATHL,
TZ0/DM,
p139:
N16,
MATSD,
p139:
M17,
MATSD,
SP1,
TZ1,
Q13;
p140:M17,
MATSD,
SP1,
TZ1,
Q9;
p107:
N17,
MATHL,
MATHL,
HP2,
TZ1,
TZ2Q9;
Q2;
SL
Sample
Practice
HP2,
Q5;
TZ0,
Sample
student
Q3;
student
answer,
questions,
Paper
1,
Q5,
Data
SP1,
TZ0,
Q11;
SL
Practice
questions,
Paper
1,
Q6,
Practice
questions,
Paper
1,
Q11,
available
SL
SL
Practice
questions,
Paper
1,
Q12
p141:
978-1-38-200728-3
10
9
8
7
6
5
4
3
2
1
M15,
MATSD,
M13,
MATME,
N17,MATHL,
Paper
used
in
the
production
of
this
book
is
a
natural,
recyclable
from
wood
grown
in
sustainable
manufacturing
process
conforms
MATSD,
to
the
environmental
regulations
country
Printed
in
of
the
TZ2,
SL
Q7;
Practice
SL
HP2,
TZ0,
Q5;
SP1,
TZ2,
Q12;
SL
questions,
Practice
Practice
Paper
questions,
questions,
1,
Paper
Paper
Q14,
1,
p141:
Q16
1,
p141–2:
SL
Practice
questions,
Paper
Q18,
2,
Q2,
p142–3:
p143:
MATSD,
SP2,
TZ0,
Q3;
SL
Practice
questions,
Paper
2,
Q3,
p143–4:
of
M17,
the
SP2,
Q9;
forests.
N17,
The
TZ1,
product
M17,
made
SP1,
MATSD,
SP2,
TZ1,
Q2;
HL
Practice
questions,
Paper
1,
Q6,
p145:
M09,
origin.
UK
by
Bell
and
Bain
Ltd,
MATHL,
HP3,
TZ0/DM,
Q1;
MATHL,
HP2,
TZ0,
Q8;
HL
Practice
HL
Practice
questions,
questions,
Paper
Paper
1,
MATME,
SP1,
TZ2,
Q7;
HL
Practice
questions,
Paper
1,
MATHL,
HP2,
TZ0,
Q7;
Sample
1
Q9,
Q10,
p145:
N13,
p145:
M16,
p146:
N18,
Glasgow.
TZ0,
Q5;
student
p170:
answer,
N16,
MATHL,
TZ2,
Sample
student
TZ0,
Sample
answer,
p195–6:
M17,
MATHL,
HP3,
MATME,
SP2,
SP1,
TZ2,
Sample
Paper
Q2,
Although
holders
we
the
p189:
Q3;
SP1,
SL
have
TZ1,
made
Q6;
Sample
p238:
MATHL,
will
TZ0/SE,
HP1,
effort
this
has
rectify
Q1;
answer,
Sample
M16,
Q1,
Q8;
answer,
M16,
MATHL,
TZ2,
Q3;
student
answer,
2,
p210:
M16,
Q2;
M18,
N17,
MATSD,
SP1,
TZ2,
Sample
Practice
HP2,
Sample
answer,
p218–19:
MATSD,
Q15;
student
questions,
Q11.
to
trace
and
not
been
possible
any
SP1,
TZ2,
student
Sample
TZ2,
HL
MATME,
p175:
SP1,
p224–5:
M19,
SP2,
Q2(a);
TZ2,
Sample
p178–9:
Paper
p226:
SP1,
answer,
student
MATME,
Q13,
N16,
MATME,
TZ2,
Sample
answer,
HP3,
every
M18,
student
Q2;
student
answer,
SP2,
p157:
MATME,
TZ2,
questions,
M19,
MATHL,
publication
publisher
MATME,
student
M16,
HP2,
p183:
Practice
M16,
Sample
student
answer,
M19,
answer,
N12,
Q6;
Sample
answer,
p158:
MATHL,
TZ0,
Q8;
Sample
p251:
before
notied,
Q1;
TZ0/SP,
p245:
2,
M19,
SP2,
MATSD,
Q15;
answer,
student
TZ0,
student
answer,
p169:
MATME,
HP3,
Q6;
student
student
errors
or
contact
in
all
all
omissions
copyright
cases.
at
the
If
earliest
opportunity.
Links
to
third
information
contained
in
party
only.
any
websites
Oxford
third
are
provided
disclaims
party
any
website
by
Oxford
in
responsibility
referenced
in
good
for
this
faith
the
and
for
materials
work.
/
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)
74
211
End-of-chapter practice questions
215
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
Applications
Level
of
and
Full
new
papers
syllabus
Level.
the
two
Oxford
University
companions
Mathematics:
with
Applications
HL
and
SL,
ISBN
978-0-19-842698-1.
exam
guide
for
the
and
Approaches
two
Oxford
There
new
is
which
a
sister
IB
Press
Analysis
and
978-0-19-842716-2
book
offers
support
to
their
examinations.
the
study
It
concepts,
strengthen
and
improve
The
book
exam
learn
course
is
tips
your
that
the
with
HL
and
common
annotated
informed
errors.
help
you
essential
to
IB
worked
example
by
past
All
best
may
be
or
succeed
a
questions
examination
to
check
papers
your
confidence
and
a
terms
practices
are
and
Interpretations
the
formula
booklet,
to
set
further
and
the
terms
and
your
own
in
the
examination,
you
will
broad
range
of
resources,
many
and
and
online.
The
authors
hope
that
this
you
through
this
critical
part
explain
making
your
preparation
for
the
of
IB-style
and
standard
more
level
students
must
and
two
The
take
of
your
to
each
internal
overall
studies.
(SL)
and
higher
level
complete
(SL)
or
and
as
DP
other
the
three
internal
(HL)
Papers
and
1
Paper
external
shown
in
assessment
papers
and
2
for
are
3
a
day
assessment
the
Mathematics
table
grade,
or
marks
below
from
2,
3,
4,
1
to
give
your
(lowest)
to
HL
weight
marks
weight
20
20%
20
20%
80
40%
110
30%
80
40%
110
30%
–
–
55
20%
5
AHL
for
HL
response,
syllabus
questions.
AHL
required
is
calculated
subjects
with
to
by
up
to
three
Theory
of
knowledge
carry
and
to
from
are
required
score
six
later.
questions.
Especially
Technology
two
marks
—
All
points
sat
questions.
problem-solving
additional
usually
required
Extended
for
their
T
opics
3
grades
(HL)
7(highest).
topics
Technology
combining
less
efficient.
assessment.
combined
your
2
diploma
exam
why
1,
IB
your
by
Including
¿nal
of
warn
Exploration
Extended-response
The
book
questions
boost
progress
Mathematical
Technology
Paper
are
Interpretations assessment
Description
Two
which
DP Mathematics: Applications and
1
Paper
to
skills
opportunities
skills,
Short-response
Paper
list,
notes.
need
of
SL
Assessment
Internal
notation
and
illustrated
which
complete
provide
monitor
the
missed.
knowledge
and
Applications
command
navigate
close
Practice
markschemes,
revise
examples
answers
examinations,
scored
textbooks,
and
of
external
marks
as
SL,
examinations.
topics
student
such
papers
glossary
All
against
intended
preparing
problem-solving
demonstrate
not
companions
students
approach
packed
is
978-0-19-842710-0.
will
your
book
the
stressful
material,
this
materials,
specimen
including
studies,
for
course
syllabus
will
This
guide,
Analysis
complements
Approaches,
and
your
papers,
available
ISBN
study
Mathematics:
use
Mathematics:
www.oxfordsecondary.
Prepared
To
University
at
examination
978-0-19-842704-9
Mathematics:
course,
any
replace
IB
and
online
and
and
past
Interpretations,
given
questions
Press
to
course
all
It
As
complements
are
to
com/ib-prepared-support.
Interpretations
Standard
solutions
out
and
present
satisfy
explains
your
the
how
Exploration
marking
criteria
to
in
select
a
topic
a
suitable
format
and
achieve
the
and
highestgrade.
Extended
essay
components.
The
Over view of the book structure
book
is
divided
into
sections
that
cover
3
for
assessment,
standard
level
(SL)
higher
with
of
level
(AHL)
material
in
HL,
will
five
set
and
a
The
Internal
questions
complete
set
of
at
the
end
practice
assessment
of
each
written
and
2
for
SL,
exclusively
and
for
Papers
this
book.
1,
2
These
give
you
an
opportunity
to
test
yourself
the
actual
exam
and
at
the
same
time
provide
practice
questions
for
every
topic.
chapter,
examination
section
1
practice
chapters
additional
a
Papers
IB-style
and
before
additional
contains
the
papers
internal
section
examination
and
The
final
outlines
papers.
the
The
answers
and
and
examination
solutions
papers
to
are
all
practice
given
online
questions
at
nature
www.oxfordsecondary.com/ib-prepared-support.
of
the
Mathematical
Exploration
that
you
will
have
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
and
the
follow
during
exam
Use
Get
ready
for
study
.
Have
plenty
of
water
enough
and
this
thinking
and
sleep,
eat
exam
sleep
reduce
physical
day
,
is
particularly
it
can
as
your
stress
Organize
your
exercises.
important
significantly
comfortable
place
temperature
possible
study
and
distractions.
computer
files
adequate
ventilation.
Keep
organized.
Papers
before
offline
Find
your
Plan
your
arrange
into
Bookmark
1
them
smaller,
2
2
that
are
are
and
you
can
use
your
1.5
hours
long,
HL
2
hours
long,
Paper
3
long.
are
sure
that
your
new
batteries.
required
for
calculator
is
Paper
to
consists
answered
underneath
and
useful
be
1
of
on
the
all
papers.
fully
Make
charged/has
short
the
response
exam
questions
paper
question.
online
2
consists
of
extended
by
Make
a
list
importance.
easily
of
your
Break
manageable
tasks
up
parts.
to
and
large
be
for
your
studying
time
involving
answered
in
response
you
can
complete
each
the
answer
reasoning
booklets
Create
and
task
sustained
provided.
tasks
3
consists
of
two
extended
response
make
problem-solving
sure
approach
material.
studies.
agenda
where
and
and
hour
Paper
an
your
a
all
papers
1
Calculators
your
lighting,
Eliminate
1
questions
3.
plan
the
Paper
and
to
A good
improve
environment.
with
reading
by
score.
2.
minutes
all
well,
is
night’s
time
questions
papers
positive
5
complete
effectively
.
SL
drink
be
to
itself.
GDC
1.
will
aim
your
identify
preparation
There
should
are
time.
some
you
before
questions,
also
to
be
the
answered
in
the
answer
booklets
provided.
deadline.
For
4.
Use
this
book
as
your
first
point
of
Papers
exam
Work
your
way
through
the
topics
identify
and
skills.
the
gaps
in
your
paper
extra
time
marks
on
the
topics
and
online
is
required.
resources
for
Check
more
your
Read
actively
.
Focus
on
information.
•
Make
sure
understanding
rather
using
your
example
the
Recite
own
and
answer.
key
words.
practice
Make
points
Try
to
for
and
every
before
future
worked
looking
7.
at
Optimize
carefully
,
reference.
terms.
as
6.
Get
ready
for
the
Practise
time
Learn
be
seen
by
allocated
to
each
part
a
good
should
that
for
be
you
the
display
guide
spent
have
exam:
to
on
all
the
examiner.
of
how
that
the
pens,
calculator
exam
paying
Keep
possible.
answering
exam-style
questions
a
many
part.
equipment
pencils,
ruler,
(GDC),
watch.
approach.
extra
your
Read
attention
answers
Double-check
as
to
short
all
all
questions
command
and
clear
numerical
values
units.
Label
axes
in
graphs
and
annotate
under
Use
the
exam
tips
from
this
book.
Use
constraint.
the
•
not
are
your
diagrams.
a
will
exams.
and
•
the
definitions
solve
problem
notes
on
than
graphic
memorizing.
written
textbook
required
5.
anything
where
minutes
improvement
3,
understanding
question
Spend
and
systematically
The
and
2
reference.
how
booklet
to
use
(allowed
the
in
Mathematics
all
exams)
correct
Ifyou
formula
notation
introduce
a
e.g.
put
variable
arrows
explain
on
vectors.
what
it
standsfor.
quickly
andefficiently
.
8.
•
Solve
you
as
can.
exam,
Your
but
papers
paper
many
at
you
the
will
questions
school
can
end
from
should
create
of
involve.
this
All
past
give
another
book.
papers
you
using
Know
questions
a
in
trial
not
each
papers
panic.
concentrate
realistic
these
the
what
all
as
Do
on
goals
goals.
improve
a
positive
things
and
Be
performance
to
Take
you
work
prepared
and
your
learn
future
attitude
can
and
improve.
systematically
to
reflect
from
your
on
Set
to
achieve
your
errors
in
order
results.
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
You
should
be
able
to
features:
Example
Note
Examples
offer
solutions
to
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
which
positive
given
will
show
in
always
the
be
are
taken
and
green
given.
typical
student
from
negative
and
An
red
past
responses
examination
feedback
pull-out
example
is
to
on
the
boxes.
shown
IB-style
papers).
student’s
The
correct
below.
S AMPLE STUDENT ANS WER
This feature assists in answering
par ticular questions, warns
Consider
against common errors and shows
=
c
+
id,
the
distinct
where
a,
b,
complex
c,
d ∈
how to maximize your score when
answering par ticular questions.
+
Find
the
real
part
terms
+
+
ib,
w
w
w
z
w
(b)
in
a
of
z
z
found
=
w
▼ Although
is
z
R
z
(a)
z
numbers
Find
the
value
of
the
real
part
+
w
of
correctly
z
w
when|z|=|w|
the
of
a
student
recall
the
and
does
b,
not
method
of
+
division
of
+
bi
(c
+
di)
complex
×
numbers.
The
approach
is
correct
a
to
+
a
both
and
the
c
a
c
+
c
+
i(b
a
c
+
i(b
a
+
a
c
+
i(b
+
d)
multiply
d)
numerator
denominator
c
i(b
+
c
i(b
d)
+
by
the
(a
c)
conjugate
+
i(b
of
a
2
+
2
+
d
2
a
in
the
student
on
part
part
2
+
b
2
=
c
+
d)
×
×
c
i(b
+
d
a
c
(a),
moves
the
(b)
a
i
d)
+
2
1
awarded
d)
i
×
1
Despite
difculties
i(b
1
2
b
a
to
+
d).
2
▲
d)
and
point
knowledge
C
is
for
and
(b
1
+
d)
×
understanding
he/she
a
c
(b
i
demonstrates
a
the
modulus
complex
paper
of
c
a
number.
Questions
Links provide a reference
d)
×
about
not
similar
to
past
IB
examinations
will
not
have
the
exam
icon.
to relevant material, within
Practice
questions
are
given
at
the
end
of
each
chapter.
The
questions
are
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.
The
paper
questions
at
a
level
Students
only
in
1
questions
are
of
set
at
level
further
of
1–3,
divided
into
group
at
2
a
groups.
level
of
Group
4–5
and
1
group
3
6–7.
following
the
a
are
higher
the
level
SL course
course.
need
These
to
be
aware
sections
will
that
be
some
sections
indicated
by
are
(HL).
vii
/
NUMBER
1
1 . 1
R E A L
the
meaning
of
decimal
places
that
exact
You should be able to:
signicant
gures
and
of
✔
approximate
decimal
nearest
✔
in
the
an
interval
value
A LG E B R A
N U M B E R S
You should know:
✔
AND
of
centred
a
measurement
on
its
rounded
a
number
places,
to
a
signicant
given
number
gures
or
to
of
the
unit
lies
value
✔
apply
the
laws
of
exponents
to
simplify
anexpression
✔
the
laws
of
exponents
✔
the
denition
✔
of
a
nd
the
upper
and
lower
bounds
of
a
logarithm
roundedvalue
✔
the
standard
scientic
form
of
notation,
a
number,
gives
represent
very
large
computer
and
GDC
and
a
also
known
convenient
very
small
way
as
✔
apply
the
formula
numbers
✔
apply
the
denition
an
✔
notation
for
is
not
appropriate
for
you
percentage
error
exponential
of
logarithms
expression
and
to
rearrange
solve
equations
standard
with
form
for
to
to
write
technology
in
anyassessment.
✔
calculate
form
✔
by
with
hand
write
down
GDC
in
numbers
written
and
technology
with
numbers
appropriate
given
by
in
a
standard
calculator
or
notation.
Approximation, errors and estimation
The
are
processes
fundamental
Artem
carries
in
a
mean
all
737
the
Note
a
to
possible
value
of
spreadsheet
went
large
8378
guesses
1581.083 969 465 6
exact
application
sweets
from
as
the
experiment
ranging
close
in
measuring,
an
of
of
counting,
out
number
as
of
to
and
1627.
took
Ar tem applied the idea of “ The
Artem
on
wisdom of the crowd”, in which
twosignificant
to
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
1600.
Hence
the
correct
to
to
figure
in
Similarly
,
estimate
two
and
two
significant
1581.083 969 465 6,
for
1627
the
exact
significant
figures,
he
which
rounded
values
Artem
2
were
examined
8,
and
rounded
down
to
0
get
the
same
when
NumbERs
the
is
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
m
0.04
2
m
to
rounds
Examine
to
place.
dp
Since
second
(ii)
0.041 792
0.042
m
m
to
rounds
2
to
figure.
(i)
67.0812°
to
(ii)
the
rounds
nearest
67.0812°
67.1°
to
to
(i)
8307.59
8308
(ii)
km
is
digit
to
7.
figure.
km
8307.59
to
rounds
the
km
nearest
rounds
km
to
0,
1,
round
place
third
it
is
7,
first
the
is
significant
the
Since
it
is
digit
to
8.
Examine
5,
figure
the
8,
to
2.
place.
units
significant
round
figure
first
4.
the
decimal
to
the
1.
decimal
round
the
to
round
fourth
it
to
the
significant
round
Since
Examine
to
the
it
third
(c)
the
Since
Examine
sf
is
decimal
significant
Examine
67°
degree
rounds
3
to
it
Since
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
<
v
E
2
+
A
2
Eape 1.1.2
The
base
Each
of
side
the
Cheops
measures
230
pyramid
is
m
nearest
to
the
a
square.
Regulations
metre.
football
pitch
measure
(a)
Find
the
upper
and
lower
bounds
for
require
for
within
that
the
dimensions
international
100
110
m
fixtures
long
and
of
a
must
64
75
m
wide.
the
A
length
of
one
side
of
the
base.
(c)
Calculate
the
smallest
value
of
where
F
is
the
F
2
area
(b)
Hence
find
area
m
the
interval
in
which
the
of
the
football
pitch
in
m
exact
2
A
of
the
baselies.
(d)
Hence
comment
football
the
pitches
on
the
that
fit
minimum
into
the
number
square
of
base
of
pyramid.
Solution
(a)
The
upper
and
the
bound
lower
is
230
bound
is
+
0.5
230
=
−
230.5
0.5
=
Add
m,
229.5
one
value
m
half
and
of
then
approximate
the
unit
subtract
to
the
one
approximate
half
of
the
unit
to
the
value.
2
(b)
Hence
if
the
2
then
229.5
area
of
the
base
2
m
2
≤
A
<
is
A
m
,
2
230.5
The
m
2
therefore
52 670.25
m
≤
A
The
smallest
value
of
<
is
F
(d)
The
is
area
more
an
of
the
than
base
six
international
53 130.25
of
times
the
m
≈
110
the
bound
of
the
area
is
the
upper
bound
of
in
length
the
lower
squared.
bound
of
the
area
of
the
base
and
6.38
75
the
Cheops
larger
football
×
side
Use
52 670.25
A
(c)
upper
2
largest
possible
area
of
pitch.
pyramid
area
than
pitch.
The aw f epnent and an ntrdctn t arth
Nte
When
you
were
younger
you
learned
that
3
×
2
is
a
way
to
represent
3
3
Don’t confuse 2
the
repeated
addition
2
+
2
the
repeated
multiplication
+
2.
Similarly
,
the
expression
2
call
2
represents
with 2 × 3
exponent.
3
is
In
You
say
“2
2
raised
×
to
×
the
2.
You
exponent
2
3
the
is
base
8”
or
and
“2
to
3
the
the
power
8”.
SL,
the
exponent
will
always
be
an
integer.
The exponent is known as the
Exponential
expressions
are
simplified
according
to
the
laws
of
power or index. The plural of index
exponents.
These
laws
are
not
in
the
formula
booklet
so
you
must
is indices. You can use these
know
them
before
your
exam.
terms in your assessments, but
you must understand the meaning
m
n
x
x
m
=
+
x
n
m
(x
n
)
mn
=
0
x
x
1
=
1,
x
m
=
x,
0
=
0
of exponent.
m
x
m
=
n
x
x
–
n
m
(xy)
m
=
x
m
y
1
–m
x
=
m
x
4
/
1 .1
RE Al
NumbERs
Cn pwer
The
base
powers
are
decimal
n
can
be
any
positive
commonly
equivalents
…
are
shown
in
and
this
However,
some
1
the
1
1
=
0.25
=
4
8
0.5
1
1
1
81
27
9
3
1
1
1
=
…
125
…
=
3
4
…
1
2
4
8
16
…
1
3
9
27
81
…
=
0.2
1
5
25
125
625
…
=
0.1
1
10
100
1000
10 000
…
1
0.001
=
1000
2
5
1
=
0.0001
10 000
0.04
25
1
1
n
10
1
1
n
625
0
2
…
5
used
1
=
0.125
1
n
following
commonly
2
1
16
of
the
table:
3
…
3
number.
encountered,
4
n
2
real
0.01
10
100
Eape 1.1.3
Use
the
laws
one
integer
of
exponents
to
write
each
expression
as
the
power
of
In calculations, it is best to use
and
hence
find
its
value.
exact values, for example a
3
(a)
3
×
fraction or a terminating decimal.
3
12
5
(b)
9
5
2
2
(c)
2
5
Solution
3
(a)
3
×
1
=
3
3
3
×
3
=
5
m
4
=
3
=
=
125
81
Apply
the
law
Apply
the
law
x
m
=
+
n
x
m
12
x
5
12
(b)
=
9
m
3
5
=
–
n
m
–
x
n
9
x
5
2
2
2
m
0
×
2
x
5
2
(c)
=
=
2
2
0
×
(
2)
Apply
5
the
laws
2
2
×
x
n
x
5
2
=
n
2
5
=
n
x
2
5
=
(2
×
5)
2
=
m
10
=
100
and
(xy)
m
m
=
x
y
2x
=
64
and
x
equation
like
2
2
You
learned
to
solve
equations
like
x
+
2
=
64,
=
64
=
64?
x
in
previous
The
years.
solution
to
But
this
how
do
equation
you
is
solve
written
an
as
x
=
log
64
and
you
read
it
2
as
“x
equals
“What
log
base
exponent
do
2
I
of
64”.
need
to
This
raise
just
means
2
by
in
=
log
you
order
to
ask
yourself,
get
64?”
6
The
answer
is
6.
The
expressions
6
64
and
2
=
64
are
just
2
rearrangements
of
each
other,
so
are
equivalent,
just
as
2 ×
32
=
64
64
and
=
32
are
equivalent.
2
The general definition of a logarithm is
x
a
= b ⇔ x = log
b, where a > 0, b > 0 and a ≠ 1
a
You say, “x equals log base a of b”
You
can
Two
The
of
calculate
the
most
number
applications
e
≈
in
logarithms
commonly
on
used
your
bases
2.718 281 828 459…
the
sciences.
This
GDC
is
is
for
an
why
for
any
positive
logarithms
irrational
log
x
is
are
10
number
referred
to
base.
and
with
as
e.
many
the
e
“natural
your
log”
of
x
and
this
is
written
ln
x
in
the
examination
and
on
GDC.
5
/
1
Eape 1.1.4
Solve
the
following
equations:
x
(a)
10
0.25z
3y
=
13.7
(b)
10
=
45
(c)
112.8e
=
745.03
(d)
3.1
log
v
=
−0.651
10
Solution
x
(a)
10
=
⇒
x
13.7
=
Apply
log
the
answer
13.7
to
definition
three
of
a
logarithm,
significant
use
your
GDC
and
give
the
figures.
10
x
=
1.14
(13.7)
log
1.13672056716
10
3y
(b)
10
=
45
⇒
3y
=
log
45
Use
your
GDC
only
at
the
final
step
in
order
to
be
efficient
and
10
log
accurate
45
with
your
working.
10
y
=
=
0.551
3
(45)
log
0.551070837925
10
3
0.25z
(c)
112.8e
=
745.03
Note
with
745.03
that
exponential
and
logarithmic
equations
can
also
be
solved
definition
of
logarithm,
graphs:
0.25z
⇒
e
=
112.8
1201.91
⎛
⇒
0.25z
=
ln
745.03
⎜
⎝
⎛
⇒
z
=
4 ln
⎞
⎟
112.8
745.03
⎜
⎝
y
(7.551
f2(x)=745.
=
7.55
⎠
f1(x)=112.8 
100
–5.91
(d)
3.1
log
v
=
, 745)
⎞
⎟
112.8
⎠
e
0.25 x
x
1
14.09
−0.651
Rearrange
the
0.651
solve
a
equation
then
apply
the
a
or
10
⇒
v
log
with
graph.
=
10
3.1
5.6
y
0.651
⇒
v
=
10
3.1
=
0.617
f4(x)
=
3.1 
log
(x)
10
0.5
x
8.43
–5.91
f3(x)=–0.651
(0.6166
, –0.651)
standard fr
k
A number
integer
is
Positive
expressed
written
values
negative
in
of
values
k
of
the
large
small.
the
form
standard
give
k
applications,
in
a
form,
numbers
give
positive
numbers
×
10
,
also
where
known
which
are
numbers
expressed
in
1
at
less
≤
a
as
10
and
scientific
least
than
standard
<
10,
1.
k
is
an
notation.
whereas
Most
form
are
many
10
often
either
in
very
k
or
very
number
has.
For
example,
The
Carolina
value
wants
of
to
a
tells
you
compare
how
the
distance
units
from
the
your
Earth
44
to
the
in
an
Sun
with
the
astronomy
number
book
14 959 800 000 000
cm.
that
She
2
the
to
find
out
distance
writes
this
which
from
number
the
in
is
larger.
Earth
to
standard
She
the
finds
Sun
form
is
by
6
/
1 .1
counting
the
number
of
decimal
places
(13)
from
the
right
until
RE Al
NumbERs
the
13
second
significant
digit
(4),
and
writes
1.495 98
GDC
which
gives
×
10
cm.
44
Carolina
types
2
into
her
her
1.759 218 604 441 6E13.
13
She
correctly
writes
this
as
1.759 218 604 441 6
13
with
“
×
power
10
of
10
×
10
by
replacing
“E13”
44
”.
Carolina
is
the
concludes
same
for
both
that
2
is
the
larger
number
since
the
numbers.
Eape 1.1.5
9
(a)
Neptune
Write
is
this
located
number
4.3514
as
a
×
10
km
from
(d)
Earth.
(i)
One Astronomical
The
decimal.
of
(b)
Nicole
reads
the
number
8.03E–5
from
V
oyager
141
Write
down
this
number
it
a
in
(AU)
spacecraft
from
the
is
149 598 000
reached
Earth
by
the
a
km.
distance
year1998.
her
Write
GDC.
AU
1
Unit
this
distance
in
kilometres
in
standard
standardform.
form
then
express
as
decimal
number.
(ii)
(c)
One
grain
of
pollen
weighs
0.000 002 5
Hence
justify
spacecraft
Write
this
weight
in
standard
the
estimate
that
the
V
oyager1
mg.
reached
approximately
5
times
as
form.
far
from
the
Earth
as
Neptune
by
1998.
Solution
9
(a)
To
find
4.3514
(b)
8.03E–5
×
10
you
move
the
decimal
point
9
places
to
the
right
and
write
4 351 400 000
–5
means
8.03
×
10
.
To
express
this
as
a
decimal,
you
move
the
decimal
point
5
places
to
the
left
–6
and
not
write
0.000 080 3.
expressed
in
Writing
standard
the
answer
as
80.3
×
10
would
not
be
awarded
any
marks
because
it
is
form.
–6
(c)
0.000 002 5
mg
can
be
written
as
2.5
×
10
mg
10
(d)
(i)
141
×
149 598 000
=
21 093 318 000.
10
2.109 331 8
×
10
10
2
(ii)
×
20
×
×
=
9
10
4
approximately
5
×
distance
be
written
as
2.109 331 8 ×
10
km.
10
=
9
4.3514
the
9
10
≈
So
5.
Hence
the
distance
reached
by
V
oyager
1
from
Earth
is
9
10
4
times
as
×
10
far
as
the
distance
from
Earth
to
Neptune.
Eape 1.1.6
You
are
given
Number
body
=
of
37.2
the
cells
following
in
the
information:
human
Average
length
trillion
nanometres
=
One
of
a
cell
is
10 000
–9
One
(a)
trillion
Write
down
average
(b)
Hence
line,
1 000 000 000 000
the
length
number
of
calculate
expressing
a
cell
the
of
in
in
the
standard
length
your
cells
if
answer
nanometre
human
(nm)
body
=
10
and
metre
the
form.
all
the
cells
in
km.
are
placed
in
a
straight
Solution
12
(a)
37.2
trillion
=
37.2
×
1 000 000 000 000
=
37.2
–9
10 000
nanometres
=
10 000
×
10
×
10
4
m
=
(10
13
=
3.72
×
10
–9
×
10
)
–5
m
=
1
×
10
m
12
37.2 × 10
13
(b)
The
length
of
all
the
cells
is
therefore
(3.72
×
10
is not written in
–5
)
×
(1
×
10
m)
standard form because 37.2 > 10.
8
=
3.72
×
10
m
You must write this number as
13
5
Since
1
km
=
1000
m,
the
distance
is
3.72
×
10
km
3.72 × 10
7
/
1
s AmPlE sTuDENT ANs WER
cos x + s
Let
p
n y
=
,
where
x
=
36º,
y
=
18º,
w
=
29
and
z
=
21.8
2
w
z
(a) Calculate the value of p. Write down your full
(b)
▼ Introducing
calculation
for
a
Write
separate
the
(i)
correct
reduce
more
efcient
accuracy
to
just
and
enter
it
expression
the
two
part
decimal
(a):
places
in
one
correct
to
three
significant
figures.
to
(ii)
the
k
Write
your
answer
part
(b)
in
cos
≤
a
<
10,
k
∈
correctly
writes
36°
down
10
,
where
+
sin
18°
cos
36°
+
sin
18°
full
calculator
correctly
of
gives
display
,
the
accuracy
21.8
0.039 062 5
28.621 670 11
then
number
to
the
(a)
0.039 062 5
(b)
(i)
required.
▼ Counting
the
is
source
common
=
2
29
a
×
ℤ
=
levels
form a
GDC.
▲ Student
the
the
calculation
1
on
to
to
is
(c)
entire
answer
display
.
denominator
(ii)
could
your
calculator
decimal
of
places
error.
0.04
(ii)
0.0391
Typing
3
this
expression
comparing
identied
into
with
the
the
0.0391
GDC
(c)
and
would
3.91
×
10
have
error.
1 . 2
S E Q U E N C E S
A N D
S E R I E S
Y hd knw:
Y hd e ae t:
✔
the
meaning
of
a
sequence
✔
identify
✔
the
meaning
of
a
series
✔
write
✔
the
notation
for
✔
apply
✔
what
sequences
and
sigma
notation
in
denes
geometric
an
arithmetic
sequence
and
if
and
that
sum
there
of
an
two
✔
equivalent
arithmetic
formulae
that
sum
✔
the
there
of
a
are
for
✔
and
of
the
formulae
to
arithmetic
sigma
calculate
including
or
geometric
notation
and
nancial
solve
problems
applications
approximation
for
is
not
an
when
exact
an
arithmetic
model
the
series
equivalent
geometric
meaning
annuity
two
understand
context,
use
apply
or
✔
is
a
sequence
are
sequence
formulae
sequence
✔
a
technology
nance
such
package
to
as
a
solve
graph,
spreadsheet
problems.
the
series
terms
depreciation,
ination,
amortization.
It’s
often
and
said
series
contexts
enable
and
A sequence
term.
that
mathematics
you
apply
is
an
to
formulae
used
the
represent
ordered
A commonly
is
list
science
number
to
solve
of
numbers,
notation
for
of
patterns.
patterns
seen
Each
the
term
in
first,
the
second,
sequence
third
is
in
real-life
problems.
a
each
of
sequence
which
is
u
,
1
meaning
Sequences
terms
and
calculated
so
with
on
a
until
u
,
2
is
called
u
,
...
3
the nth
,
a
u
n
term.
formula.
8
/
1.2
A series
is
formed
S
u
+
=
u
n
+
1
u
2
Sigma
+
by
…
adding
+
u
3
is
the
the
terms
sum
of
of
the
a
sequence;
first
n
terms
the
of
sEquENCE s
AND
sERiE s
series
a
sequence.
n
notation
gives
you
a
concise
way
to
write
this:
n
S
=
n
∑
i
This
can
be
read
as
“The
sum
of
all
=
u
i
1
the
terms u
,
where
i
takes
the
values
i
n
1,
2,
3,
and
so
on,
until
n.
Similarly
,
∑
i
=
u
=
u
i
+ u
k
+ u
k +1
+ ... + u
k +2
n
k
Eape 1.2.1
The
nth
term
of
a
sequence
is
found
with
the
formula
u
=
23
2n
n
(a)
Calculate
the
first
five
terms
of
the
sequence.
5
(b)
Hence
find
S
=
∑
5
i
=
u
i
1
75
(c)
Find
∑
i
=
u
i
73
Solution
(a)
u
=
23
2(1),
u
1
u
=
23
2(3),
u
3
u
=
23
2(2),
=
23
2(4),
Replace
n
with
the
values
1,
2,
3,
4
and
5
in
the
formula.
2
4
=
23
Find
the
answers
and
list
them.
2(5).
5
So
the
21,
19,
first
17,
five
15
terms
and
are
13.
5
(b)
∑
i
=
=
u
i
u
+
u
1
+
u
2
+
u
3
+
u
4
The
command
“Hence”
is
an
explicit
instruction
to
apply
the
1
previous
=
term
5
21
+
19
+
17
+
15
+
13
=
result.
85
75
(c)
∑
i
=
u
i
=
u
+
u
73
+
74
u
Replace
=
n
with
values
73,
74
and
75
in
the
formula
for
the
73
sequence
(23
2(73))
+
(23
and
add
S
S
75
(23
2(75))
=
these
terms
to
find
the
sum.
2(74))
(Finding
+
the
75
is
another
way
to
+
+
find
the
sum
required.)
72
−375
S
75
S
=
(u
72
+
u
75
=
u
+
75
+
u
74
u
74
73
+
+
u
u
72
71
…
+
u
2
+
u
)
1
(u
72
+
u
71
+
…
+
u
2
+
u
)
1
u
73
The previous example can be calculated on your GDC, however you must be
able to calculate expressions given in sigma notation by hand too.
1.1
1.2
sigma
notation
DEG
–375
75
(23–2.i)
Σ
i =73
9
/
1
An
arithmetic
any
two
sequence
consecutive
difference.
For
in
1.2.1
Example
is
terms
example,
is
one
2
is
the
in
constant.
common
because
u
than
the
terms
Informally
,
previous
of
the
you
term”.
sequence
can
is
called
=
−2
is
of
d
between
the common
the
always
sequence
true
for
this
n
“Each
common
term
of
the
difference
sequence
is
positive,
is
2
less
then
the
increase.
The
general
formula
first
term
and
common
this
sequence:
u
for
difference
difference
1
say
,
Ifthe
will
+
the
This
u
n
sequence.
which
the
nth
term
difference
of
d
an
can
arithmetic
be
found
sequence
by
with
investigating
1
the
terms
n
u
1
2
3
u
u
u
n
u
of
1
2
u
=
u
1
n
4
u
3
+
d
u
1
+
d
+
d
u
+
pattern
gives
+
d
u
+
2d
you
+
=
u
+
3d
3d
+
d
u
general
+
=
u
+
4d
=
u
1
formula
4d
+
…
d
1
1
the
…
6
1
1
The
2d
…
u
5
1
u
6
u
4
1
=
5
=
u
+
5d
1
u
n
+
(n
1)d.
1
These formulae are in
n
section 1.2 of the formula book .
The
formula
for
the
sum
of
n
terms
of
this
sequence
is
S
=
(u
n
+
u
1
).
n
2
By
substituting
the
formula
for
u
you
can
also
write
the
equivalent
n
n
formula
S
=
(2u
n
+
(n
1)d).
1
2
Eape 1.2.2
Adrian’s
swimming
ten
He
days.
on
day
one,
(a)
Find
(b)
Hence
total
tells
and
how
or
coach
Adrian
then
many
the
to
lengths
ten
planning
swim
increase
otherwise
over
is
the
50
how
training
lengths
number
Adrian
find
his
will
many
of
of
schedule
the
swimming
lengths
swim
on
lengths
over
day
by
5
pool
each
day
.
ten.
Adrian
will
swim
in
days.
Solution
(a)
The
number
swims
u
=
on
50
+
of
day
(10
lengths
ten
1)5
Adrian
is
=
Represent
first
95
lengths
term
the
50
difference
5
sequence
and
and
with
common
apply
the
10
formula
u
=
u
n
(b)
The
can
total
be
number
found
of
lengths
by
Since
total
the
+
(n
1)d.
1
question
number
of
asks
for
lengths
the
over
10
A common mistake for Example
S
=
(50
+
95)
=
725
lengths
the
ten
days,
for
the
sum
apply
the
formula
10
2
of
the
arithmetic
1.2.2 par t (a) is to answer
series.
The
other
formula
could
50 + (10)5 = 100.
be
applied
S
=
instead:
This is incorrect because on the
10
first day Adrian swims 50 + (0)5
(2(50)
+
(10
1)5)
=
725.
10
2
lengths, hence 50 + (10)5 would
But
be the 11th term of the sequence.
solution
the
form
is
used
more
in
the
direct.
10
/
1.2
The
formulae
problems
by
for
arithmetic
hand
and
with
sequences
and
series
can
be
used
to
sEquENCE s
AND
sERiE s
solve
technology
.
Eape 1.2.3
Adrian
for
to
plans
charity
know
a
training
through
some
schedule
online
facts
to
to
raise
sponsorships
publish
online.
money
and
Use
(a)
the
wants
in
Example
1.2.2
to
day
when
Adrian
swims
more
than
250lengths
the
(b)
information
first
the
number
of
days
needed
for
Adrian
to
have
find:
swum
more
than
5000
lengths
in
total.
Solution
(a)
50
+
5(n
1)
>
250
⇒
Use
u
=
u
n
5(n
1)
Hence
So,
on
more
>
n
200
>
41
day
42
than
⇒
(n
of
training,
250
1)
>
40,
+
(n
1)
d
to
write
the
inequality
1
needed.
Adrian
will
first
It
swim
is
a
good
Adrian
lengths.
255
idea
swims
lengths,
to
250
hence
check
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
create
inequality
.
a
graph
Notice
that
or
S
a
table
=
4950
to
and
36
S
=
5180.
The
final
answer
must
be
an
integer
37
and,
in
satisfy
A geometric
consecutive
For
sequence
terms
example,
the
is
is
one
in
constant.
common
which
You
ratio
call
the
ratio
this
r
between
quantity
of
the
sequence
you
can
say
,
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.
Informally
,
“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
previous
will
increase.
for
the
ratio
r
term”.
nth
term
can
be
of
a
If
r
>
1,
then
geometric
found
by
the
size
sequence
investigating
the
of
the
with
first
terms
of
1
this
sequence:
n
u
n
1
2
3
u
u
u
1
2
4
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
r
2.
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
=
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
<
r
problems
applying
the
<
r
1.
with
sequences
appropriate
by
first
interpreting
a
context
formulae.
Eape 1.2.5
A space
and
probe
falls
surface
of
the
pod
(a)
Find,
at
(b)
to
will
the
The
the
the
to
in
planet,
bounce
the
physical
bounce
with
112
If
the
a
probe
its
m
the
the
total
from
below.
On
upwards.
spacecraft
impact
with
Engineers
height
distance
a
after
estimate
each
travelled
the
by
bounce.
the
pod
impact.
probe
If
20
previous
the
fourth
units.
will
of
of
released
bounces
metre,
the
is
planet
65%
the
costs
pod
pod
condition
safety
probe
to
of
of
the
nearest
moment
the
protective
surface
Each
84,
a
probe
2
is
safety
number
measured
units
of
and
safety
in
the
units
safety
probe
falls
units.
begins
below
fail.
bounces
13
times,
predict
if
the
probe
will
fail.
Solution
(a)
There are other equivalent ways to
20
solve this problem.
1
The
2
height
after
1,
2,
3
that
the
and
3
20
0.65
pod
reaches
bounces
2
20
×
0.65,
×
,
Sketching
a
diagram
problem-solving
is
is
a
good
strategy
.
3
20
×
0.65
12
/
1.2
Therefore
the
answer
required
Use
is
the
the
sketch
context
to
and
sEquENCE s
AND
sERiE s
interpret
identify
the
2
20
+
2(20
×
0.65)
+
2(20
×
0.65
)
sequence
is
geometric
by
3
+
2(20
×
0.65
)
representing
“65%
of”
as
a
2
=
20
+
2(20
×
0.65
+
20
×
0.65
common
ratio
of
0.65
3
+
20
×
0.65
)
Apply
the
formula
for
the
sum
3
⎛
=
20
+
2
20 × 0.65(1
)⎞
0.65
⎜
of
⎟
⎝
1
three
terms
sequence,
1.1
three
RAD
1.2
_1–(0.65)
or
terms.
working
0.65 
a
geometric
just
add
Write
out
all
up
the
your
clearly
.
73.885
3
20 
of
⎠
0.65
+
20+2 
The GDC screen should correspond
1–0.65
to what you have written on your
2
20+2 
_20 
0.65+20 
(0.65)
3
+20 
(0.65)
+
exam paper. This helps you check
73.885
your working.
The
probe
travels
nearest
metre.
(b)
sequence
The
after
106,
each
The
units
74
of
m
the
safety
bounce
number
is
of
remaining
u
to
units
110,
108,
Since
the
number
units
decreases
amount
each
number
of
of
by
safety
the
bounce,
same
the
safety
safety
units
after
n
remaining
13
bounces
is
an
arithmetic
is
sequence.
=
u
110
+
(13
1)(
2)
=
86
13
Since
not
86
>
84,
predicted
the
to
probe
is
fail.
+
Arithmetic
sequences
can
be
modelled
by
linear
functions
with
domainZ
Arithmetic and geometric
sequences are linked to linear and
First
term
2,
common
First
term
10,
common
exponential models respectively in
difference
0.5
difference
–3
section 2.3 of this book .
nth
term
of
an
arithmetic
sequence
nth
7
15
6
10
5
5
m ret
m ret
4
htn
htn
3
2
term
of
an
arithmetic
sequence
0
–5
–10
1
–15
0
–20
0
2
4
6
8
10
12
n
n
Geometric
sequences
with
positive
common
ratio
can
be
modelled
by
+
exponential
First
term
nth
functions
40,
term
with
common
of
a
geometric
domain
ratio
Z
0.5
First
sequence
term
nth
45
3,
term
common
of
a
ratio
geometric
1.25
sequence
25
40
20
35
m ret
m ret
30
25
htn
htn
20
15
10
15
10
5
5
0
0
0
2
4
6
n
8
10
12
0
2
4
6
8
10
12
n
13
/
1
s AmPlE sTuDENT ANs WER
A comet
The
(a)
comet
Find
fifth
(b)
▼ The
student
uses
problem-solving
which
gives
the
a
list
method
correct
as
for
for
the
applying
fth
would
be
▼ The
term
more
the
maximum
writing
sum
of
an
and
marks
application
of
first
year
in
and
seen
is
seen
from
which
the
from
Earth
in
comet
Earth
the
was
every
year
seen
37
years.
1064.
from
Earth
been
seen
for
the
time.
up
how
to
the
many
year
times
the
comet
has
from
2014.
(a),
answer.
the
formula
the
sequence
1
1064
10
1397
19
1730
2
1101
11
1434
20
1767
3
1138
12
1431
21
1804
4
1175
13
1508
22
1841
5
1212
14
1545
23
1873
6
1249
15
1582
24
1905
7
1286
16
1619
25
1952
8
1323
17
1656
26
1989
9
1360
18
27
2026
successfully
list
down
Sun
efcient.
student
extends
of
the
Earth
a
the
was
Determine
(a)
However,
orbits
an
scores
for
(b),
after
incorrect
the
formula
arithmetic
for
the
1693
series.
x
However
it
consuming
method:
the
nth
of
and
full
a
very
and
lists
is
is
the
more
not
marks
time-
(b)
2014
=
formula
(2(1064)(x
1)37)
2
error-prone
applying
term
Use
for
is
for
efcient.
(a)
1212
(b)
26
recommended,
may
not
be
awarded
times
method.
Fnanca appcatn f arthetc and eetrc
eence
The
formulae
many
for
arithmetic
applications,
following
terms
Compound
including
and
in
and
financial
geometric
sequences
mathematics,
using
have
the
formulae:
interest:
(compounding)
sequences
the
period
amount
is
a
added
percentage
at
of
the
the
end
of
each
previous
interest
value
of
the
investment.
T
erm
Future
Symbol
value
FV
Meaning
The
of
value
specified
Present
value
Number
of
of
The
value
k
The
interest
yearly
,
periods
meaning
per
the
time
PV
compounding
term
of
investment
in
the
can
the
half-yearly
,
k
is
1,
2,
added
at
or
loan
now.
on
quarterly
4
loan
a
future.
investment
be
or
or
12
(compounded)
or
monthly
,
respectively
.
year
Nominal
annual
r
rate
of
interest
The
percentage
divided
equally
periods
per
For
of
years
of
n
The
per
is
total
year
Note
a
added
at
number
is
that
the
nominal
compounded
interest
Number
over
per
k
year
that
is
compounding
year.
example,
12%
interest
therefore
the
first
quarterly
the
of
annual
end
of
interest
means
each
rate
that
3%
quarter.
compounding
periods
kn.
value
of
n
is
zero.
14
/
1.2
The
formula
for
the
future
value
under
compound
interest
sEquENCE s
AND
sERiE s
is
kn
r
⎛
FV
=
PV
⎜
⎞
1 +
,
⎟
⎝
k
× 100
an
example
of
a
geometric
sequence
with
first
⎠
r
term
PV
and
common
ratio
1
+
k
Simple
is
interest:
the
amount
×
100
added
at
the
end
of
each
interest
period
constant
T
erm
Symbol
Capital
Meaning
C
The
of
term
amount
invested
(or
borrowed)
at
the
beginning.
Interest
rate
r
The
percentage
each
Number
of
n
The
interest
is
periods
n
interest
total
of
C
added
to
C
at
the
end
of
period.
duration
multiplied
by
of
the
the
investment
length
of
one
or
loan
interest
period.
Note
Interest
I
The
that
the
amount
first
value
added
to
of
the
n
is
zero.
investment
or
loan.
Crn
The
formula
for
the
future
value
A
under
simple
interest
is
A
=
C
+
,
100
an
example
of
an
arithmetic
sequence
with
first
term C
and
common
Cr
difference
100
Eape 1.2.6
Steven
invests
$4,350
simple
interest
at
Chimdi
invests
scheme
with
a
in
rate
the
a
of
same
nominal
savings
1.7%
per
amount
annual
scheme
in
paying
annum,
a
interest
compounded
and
money
for
7
different
investment
rate
Steven
1.15%
is
quarterly
.
years.
will
be
They
Steven
worth
both
predicts
more.
invest
that
their
his
Determine
if
correct.
Solution
Steven’s
4350
+
investment
4350
Chimdi’s
×
0.017
1.15
⎛
⎜
1 +
⎝
×
7
investment
7
4350
is
×
You
=
questions
is
$4867.65
leave
You
can
your
is
unless
correct
otherwise
to
two
decimal
places
in
financial
specified.
⎞
=
You
$4714.11
apply
the
formula
from
the
formula
booklet
or
use
your
GDC
⎠
app.
correct.
use
GDC
answer
worth
finance
Steven
your
4
⎟
4 × 100
worth
the
to
laws
solve
of
exponents,
logarithms
or
the
finance
app
on
problems.
Eape 1.2.7
Nicole
invests
interest
at
Calculate
a
€16,500
rate
how
of
in
2.1%
long
it
a
savings
per
will
scheme
annum
take
for
paying
compounded
Nicole’s
compound
monthly
.
money
to
double
invalue.
15
/
1
Solution
F inance
This
question
your
finance
can
be
solved
on
app.
Solver
Get to know your finance app
•
33.035881314315
N:
The
amount
€16,500
is
well in advance of the exam but
I(%):
2.1
entered
–16500.
money
as
–16
500
since
the
also make sure you know how
PV:
isbeing
given
to
the
to use algebra to solve financial
Pmt:
0.
investment
scheme.
problems.
FV:
33000.
•
PpY:
1
CpY:
12
The
number
year
PmtAt:
is
PpY
,
of
payments
which
is
per
1
END
•
F inance
Solver
info
stored
The
into
number
periods
per
of
compounding
year
is
CpY
,
Write down the formula with the
tvm.n, tvm.i
tvm.pv, tvm.pmt, ...
which
is
12
information substituted in, or the
data you enter in your finance app.
This helps to clarify your thoughts
and shows your method to the
For
your
working,
write
N
?,
2.1,
−16
=
FV
=
I
=
33
000,
PV
=
C/Y
=
down:
500,
12
examiner.
After
33
years
and
Nicole’s
money
doubled
in
one
will
month,
This
have
is
since
value.
for
in
the
33
the
correct
years
is
a
reduction
in
the
value
value
of
not
investment
when
compounded
Depreciation
is
interpretation,
an
asset
the
sufficient
to
double
interest
is
monthly
.
over
time.
Inflation is the fall in the purchasing value of money. For example, if the
inflation rate is 1.7% per year, prices increase by 1.7% by the end of the year.
Eape 1.2.8
Alexander
that
his
buys
car
Cornelia
will
buys
a
new
car
for
depreciate
a
vintage
by
car
$65,000.
He
20%
year.
for
per
$5500,
learns
which
(a)
Show
after
that
x
the
years
value
can
be
A(x)
of
written
Alexander ’s
car
as
will
x
A(x)
increase
in
that
car
her
value
will
Alexander ’s
by
17%
each
year
one
day
have
the
and
same
=
$65,000(0.8)
claims
value
(b)
as
car.
Hence
or
which
the
and
find
otherwise
two
this
cars
value
find
will
to
the
time
have
the
the
nearest
in
years
same
by
value
$100.
Solution
(a)
A(x)
is
the
future
value
of
Replace
the
information
given
in
the
formula
for
the
kn
Alexander ’s
⎛
car.
future
value,
20
⎛
A(x)
=
$65,000
⎜
⎝
1
FV
=
PV
⎜
1 +
⎝
x
So
r
⎞
⎟
k
× 100
⎠
⎞
⎟
100
⎠
The
value
is
depreciating
hence
r
=
−20%
x
which
simplifies
to
A(x)
=
$65,000(0.8)
16
/
1.2
(b)
Similarly
,
the
value
of
Cornelia’s
car
sEquENCE s
AND
sERiE s
is
x
B(x)
=
$5,500(1.17)
70000
y
x
Graph
both
functions
and
find
their
intersection
point.
20
f1(x)=65000
1–
100
x
17
f2(x)=5500
1+
100
10000
(6.497
–1
, 1.525 E+4)
x
1
The
10
graphs
Hence
value
intersect
after
6.50
$15,300
to
at
years,
the
(6.497,15 250).
both
nearest
cars
The
years
and
the
are
given
value
to
the
correct
level
to
of
three
significant
accuracy
figures
required.
have
$100.
s AmPlE sTuDENT ANs WER
Compound interest is linked to
Juan
buys
price
and
a
bicycle
in
a
sale.
He
gets
a
discount
of
30%
off
the
original
exponential models in section 2.3
pays
$560.
of this book .
(a)
T
o
Calculate
buy
the
annual
original
bicycle,
interest
thatthe
the
the
total
Juan
rate
of
amount
price
takes
a
of
the
loan
bicycle.
of
$560
75%, compounded
he
will
pay
will
be
for
6
months
monthly.
less
than
Juan
the
at
a
nominal
believes
original
price
of
bicycle.
(b)
Calculate
bicycle
the
and
difference
the
total
between
amount
the
Juan
original
will
price
of
the
▼ The student fails to apply
percentages,
pay
.
learning.
which
The
is
correct
prior
approach
is
560
FV
=
96
p
=
original
560
price
=
0.7
PV
=
560
560
+
N
=
6
30x
100
▼ The student does not apply the
K
=
12
%
=
75
=
560
+
formula
35.6
35.6
=
595.6
as
6
as
units
correctly
,
years.
x
▲ The
=
560
student
×
(
1
12
+
100
x
12
×
identies
reading
gaining
mark
trying
=
44042.464
noticed
time.
in
The
such
the
correct
approach
)
is
to
use
giving
a
12
×
0.5
nal
as
the
71
answer
of$805.68
a
−
595.6
=
to
43446.864 71
substitute
information
interest
formula,
for
months
the
x
compound
be
6
6
exponent,
correctly
Crucial
should
5minute
75
reading
the
▼ The
student
should
notice
that
values.
the
answer
context
of
is
the
unreasonable
in
the
question.
Ar tzatn and annte
Annuity:
A sum
of
money
P
(an
investment)
is
paid
to
a
bank
by
aninvestor.
Interest
from
is
the
regularly
bank
investment
is
to
added
the
used
to
investor
up.
P
and
over
Example:
payments
an
agreed
retirement
are
regularly
period
of
time
made
until
the
planning.
17
/
1
Amortization: A sum
by
a
of
money
L
(a
loan)
is
paid
to
a
borrower
bank.
In examinations the payments will
Interest
be made at the end of the period.
bank
is
regularly
over
buying
a
an
added
agreed
to
period
L
of
and
payments
time
until
the
are
loan
regularly
is
paid
made
off.
to
the
Example:
house.
Eape 1.2.9
Mats
an
is
takes
out
apartment.
3.75%
a
20
The
year
loan
nominal
compounded
of
£176,500
annual
monthly
.
to
interest
Mats
buy
(a)
Find
rate
make
repayment
that
Mats
makes
eachmonth.
his
(b)
repayments
the
Hence
find
the
the20
years
total
interest
Mats
pays
over
monthly
.
of
hisloan.
Solution
N
F inance
is
the
number
N:
480.
I(%)
3.75
Mats
makes
20
12
×
P/Y
PV:
=
monthly
payments
that
and
over
the
20
years,
so
is
equal
to
240
C/Y
repayments
176500.
calculated
Pmt:
are
both
monthly
12,
and
since
the
Mats
interest
makes
is
monthly
.
–710.458400674
PV
FV:
=
176 500
has
a
positive
sign
since
this
sum
is
0.
PpY:
12
cpY
12
added
to
Pmt
negative
is
makes,
PmtAt:
END
FV
Future Value, FV
of
(a)
The
repayment
(b)
The
total
paid
each
by
month
Mats
is
is
480
=
0
the
Mats’s
so
account
Edit
=
of
Solver
this
each
is
because
amount
at
the
this
is
is
start
the
of
the
loan.
payment
subtracted
from
Mats
his
month.
because
loan
account
when
the
loan
is
paid
off,
the
value
zero.
£710.46
×
£710.458..…
£341,020.03
Hence
the
total
£341,020.03
interest
£176,500
Mats
=
pays
is
£164,520.03
Eape 1.2.10
Yimo
Her
wishes
to
financial
each
are
Option
for
A:
a
B:
Find
(b)
Hence
period
the
in
a
savings
finds
of
25
equal
two
scheme
options
for
for
her
her
to
retirement.
choose
from,
years.
amounts
of
€350
every
month.
The
nominal
annual
interest
rate
is
4%
amounts
of
€450
every
month.
The
nominal
annual
interest
rate
is
2.1%
monthly
.
Deposit
compounded
(a)
advisor
Deposit
compounded
Option
invest
equal
monthly
.
value
state
of
each
which
investment
option
Yimo
after
should
25
years.
choose
and
justify
your
answer.
18
/
1.2
sEquENCE s
AND
sERiE s
Solution
Option
A:
F inance
N
is
25
×
the
12
Pmt
I(%)
4.
PV:
0.
Pmt:
is
Yimo
from
PpY:
12
cpY
12
300
negative
makes,
her
because
so
this
account
this
is
the
amount
is
subtracted
each
payment
month.
The
future
end
of
(FV)
is
positive
because
at
the
25
years
the
investment
moves
from
the
300.
I(%)
to
Yimo’s
account.
2.1
The
PV:
0.
Pmt:
interest
CpY
value
is
is
compounded
monthly
so
the
12.
–450.
12177347.15397276
FV:
PpY:
12
cpY
12
PmtAt:
END
Option A is
option
B
Option
slightly
total
worth
better
more,
of
are
Write
and
300
B:
option
compared
but
not
×
just
because
is
€350
far
300 ×
less
because
the
less.
than
In
€42,347.15
is
the
paid
sum
Write
in
of
money
correct
to
two
places.
full
sentences
to
answer
the
question
context.
A,
in
by
invested
meaning
interest:
under
pays
option
total
€135,000,
more
it
monthly
€105,000
€450 =
A earns
to
=
a
decimal
significantly
which
option
€179,945.34
€177,347.15
A is
deposits
Yimo
value
Solver
bank
that
=
B:
N:
in
is
END
F inance
a
which
179945.34158399
PmtAt:
(b)
payments,
–350.
FV:
(a)
of
300.
N:
Option
number
Solver
€74,945.34
option
B.
19
/
1
1 . 3
A L G E B R A
( A H L )
Y hd knw:
Y hd e ae t:
✔
the
laws
of
exponents
for
✔
the
laws
of
logarithms
rational
exponents
✔
simplify
that
laws
of
are
a
consequence
of
numerical
involve
when
exponents
an
innite
✔
geometric
series
rational
algebraic
expressions
exponents
the
apply
the
laws
numerical
✔
and
of
and
logarithms
algebraic
to
simply
expressions
converges.
✔
nd
the
sum
geometric
In
this
section
you
understanding
of
extend
and
exponents,
to
innity
of
a
convergent
series.
build
on
your
logarithms
and
knowledge
geometric
and
series.
Ratna epnent
The
rules
of
rational
exponents
m
m
n
n
x
Nte
x
=
enable
This calculation is more efficient
1
m
n
x)
= (
you
to
,
m,
find
n
∈
Z,
exact
n
≠
0
values
efficiently
and
rearrange
exponential
formulae.
1
than finding
2
2
= ... .
=
1
2
2
3
3
1
1
1
3
3
For
343
example,
343
=
=
=
=
2
343
(
=
=
2
2
3
( 7)
343 )
49
3
343
Eape 1.3.1
You
are
isV
=
given
that
the
volume
of
a
hemisphere
same
value
for
the
volume
as
for
the
surface
area
3
of
4S
,
where
S
is
the
total
surface
area
a
hemisphere.
of
243π
the
(b)
Find
S
(c)
Hence
in
terms
of
V
hemisphere.
find
the
radius
in
metres
of
the
2
(a)
If
S
=
15
m
,
find
V:
hemisphere
3
for
which
the
values
the
installation
of
S
and
V
2
ab
(i)
as
an
exact
value
in
the
form
are
where
c
equal.
π
+
a,
b,
c
∈ Z
The
budget
for
requires
that
the
2
(ii)
correct
to
three
significant
hemisphere
figures.
must
fit
in
a
square
of
area
100
m
and
3
contain
Teodora
airport
is
in
designing
the
shape
a
of
fountain
a
installation
at
less
(d)
hemisphere.
Determine
these
She
explores
and
wants
to
themes
know
of
if
harmony
it
is
in
possible
her
to
than
200
m
.
an
if
the
hemisphere
found
in
(c)
fits
requirements.
installation
have
the
Solution
3
3
4(15
(a)
(i)
V
3
)
4(5
3
)(3
3
)
=
=
4(5
2
)
=
3
)
=
5
243π
2(5
Apply
the
laws
of
exponents
GDC
to
find
laws
of
exponents
to
find
an
exact
value,
2
π
3
π
3
π
then
your
the
approximate
value.
2
(ii)
V
=
4.21
m
3
3
4S
4S
2
(b)
V
=
⇒
V
2
=
243π
⇒
243V
3
π
=
Apply
4S
1
2
243V
S
to
rearrange
theformula.
2
2
π
⎛
243V
π ⎞
3
243π
3
⇒
the
243π
=
⇒
4
S
=
=
⎜
⎝
3
3
V
⎟
4
⎠
4
20
/
1.3
A lg E b R A
(AHl)
2
243π
3
3
(c)
S
V
=
.
Hence,
if
V
=
Apply
S,
the
laws
of
exponents
to
solve
the
equation
4
V
=
S
2
1
243π
243π
3
3
V
V
=
⇒
=
V
⇒
volume
of
2
of
r
is
V
=
4
4
The
243π
3
3
V
hemisphere
2
3
πr
=
a
4
πr
3
=
⇒
4
3
729
terms
243π
3
hence
in
9
3
=
r
⇒
r
=
m
8
(d)
2
Since
4.5
<
5,
the
hemisphere
will
fit
You
in
find
exact
values
and
approximate
values
to
2
a
square
of
area
100
m
solve
.
the
problem
in
context.
243π
Since
=
191
to
three
significant
figures,
4
the
volume
of
requirement
the
for
hemisphere
volume
since
also
the
meets
the
volume
is
3
less
than
200
m
law f arth
The
laws
of
logarithms
are
m
For
example,
exponents.
if
a
Also,
equivalent
=
x
log
and
x
=
m
a
m
=
y,
and
then
log
a
logarithms,
as
forms
n
does
y
xy
=
of
+
the
laws
of
exponents.
n
a
follows
=
n
follow
x
+
log
from
from
the
the
laws
definition
of
of
a
log
xy
=
m
+
n.
log
xy
=
log
a
This
establishes
the
law
a
a
y.
The
other
laws
are
a
x
m
log
=
log
a
x
log
a
y
and
log
a
x
=
mlog
a
x.
All
three
laws
are
found
in
the
a
y
formula
booklet.
Eape 1.3.2
b
(a)
Write
3log
x
+
0.5
log
10
as
a
single
y
6
log
10
z
(b)
The
equation
ln
y
=
to
find
y
=
ax
can
be
written
in
the
form
10
logarithm.
m
ln
m
x
+
and
c.
c
Apply
in
the
terms
of
laws
a
and
of
logarithms
b
Solution
m
(a)
3log
x
+
0.5
log
10
y
3
=
6
log
10
log
0.5
x
+
log
10
z
Apply
log
x
log
x
y
log
mlog
x
a
z
x
10
0.5
from
left
to
right
6
y
=
a
Work
=
law
6
10
3
the
10
log
10
to
apply
the
laws
log
=
a
log
x
a
log
y
a
y
z
10
m
3
x
0.5
and
y
log
x
a
=
=
mlog
x
a
log
6
10
z
b
(b)
y
=
ax
b
⇒
ln
y
=
ln
ax
=
ln
a
+
ln
=
ln
a
+
b
b
Hence
m
=
b
and
c
=
ln
x
ln
x
a
21
/
1
s AmPlE sTuDENT ANs WER
1
⎛
Given
that
(p
⎜
10
+
2 q)
⎟
⎝
10
10
p
+
log
10
q),
p
>
0,
q
>
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)
=
pq
+
4q
of
2
logarithms,
setting
methodically
out
and
marks.
8
1
2
(p
2
+
4pq
)
=
pq
8
2
p
2
+
4pq
+
4q
2
p
+
4q
=
4pq
=
8pq
2
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|<
1,
r
n
decreases
in
size
as
n
increases
so
that
lim
n
r
=
can
be
→
0.
∞
u
1
A consequence
of
this
is
that
lim
n
→
=
S
S
=
.
This
fact
∞
n
∞
1
r
1
observed
in
making
a
link
between
the
rational
number
and
its
3
decimal
representation
The
latter
expression
first
term
0.3
and
0.3333….
can
be
common
=
0.3
written
ratio
as
0.03
an
0.003
infinite
0.3
n
+
+
0.0003
geometric
+
….
series
with
0.1:
∞
0.3
1
1
0.3(0.1)
n
+
=
=
1
1
=
0.1
0.9
3
Eape 1.3.3
A space
probe
then
60
data
every
probe
m
is
sets
east,
time
it
m
each
(a)
down
Write
fourth
on
a
south,
changes
represented
between
the
30
out
by
change
the
of
mission
then
15
in
m
direction.
the
origin
direction
position
of
the
which
west
The
and
it
travels
and
initial
the
so
on,
a
probe
as
north,
of
the
travelled
geometric
it
m
collecting
position
distances
follow
120
sequence.
changes
direction
for
time.
22
/
1.3
(b)
Represent
(c)
Hence
will
the
path
predict
converge
the
of
the
probe
coordinates
as
of
a
sum
the
of
A lg E b R A
(AHl)
vectors.
position
P
that
the
probe
to.
Solution
60
(a)
m
Representing
a
30
sketch
120
a
context
good
with
problem
m
solving
15
is
the
strategy
.
m
m
O
The
position
changes
time
(b)
is
the
direction
(45,
The
of
for
the
as
it
The
fourth
the
of
the
represented
⎞
⎞
by
probe
⎞
+
15
⎛
⎞
0
⎛
first
terms
⎜
⎟
⎜
⎟
⎜
⎝
0 ⎠
⎝
30 ⎠
⎝
0
⎟
⎜
⎠
⎝ 7.5 ⎠
⎟
common
ratios
horizontal
+
The
translations
y-coordinate
modelled
by
geometric
term
The probe gets closer and closer
the
to P. This is an application of the
vertical
can
be
identified
concept of a limit.
⎠
from
(c)
and
for
…
⎟
0
⎝
the
⎞
+
⎜
and
⎞
+
⎟
3.75
from
is
+
⎝ 120 ⎠
⎛
read
the
0
⎛
+
⎜
be
sum
60
⎛
can
diagram.
The
0
⎛
position
90)
path
infinite
probe
120
0.25,
an
hence
P
can
sum.
first
term
be
infinite
series
and
of
the
with
first
common
the
limit
ratio
of
the
120
y-coordinate
is
=
1
Similarly
,
the
(
96.
0.25)
x-coordinate
is
60
found
by
=
1
Hence
P
(
is
48.
0.25)
(96,
48)
s AmPlE sTuDENT ANs WER
Consider
term
of
a
geometric
sequence
with
a
of
4
and
a
fourth
–2.916
(a)
Find
the
common
(b)
Find
the
sum
to
ratio
of
infinity
this
of
sequence.
this
sequence.
23
/
1
a
(a)
▼ The
student
incorrectly
=
4
writes
2
down
the
terms
instead
This
sum
should
of
of
the
the
have
rst
four
fourth
been
a+
ar
+
a(1
+
r
+
r
4(1
+
r
+
r
3
ar
term.
+
ar
2
the
5
minute
Thecorrect
reading
approach
is
time.
to
4r
r
to
nd
r
)
=
−
2.916
=
3
+
r
)
solve
3
=
2.916
noticed
3
2.916
−
3
+ r
2
in
=
−
2
+
=
r
+
r
+
1
−
2.916
2.916
=
0.9.
4
3
r
2
+
r
+
r
3
r
+
r
+
1
+
r
+
1.729
=
−
0.729
2
=
0
3
using
GDC
araph
of
2
r
r
+
r
+
+
1.729
=
0
y
▼ The
the
no
student
correct
marks
equation
use
are
is
has
of
demonstrated
technology
,
awarded
but
because
the
(–1.28, 0)
incorrect.
x
▼ The
student
formula
well;
and
selects
sets
however
out
the
the
the
correct
working
formula
for
the
r
innite
the
sum
common
cannot
ratio
be
is
applied
less
than
=
−
1.28
if
u
−1.
1
(b)
S
=
∞
1
r
4
=
1
1 . 4
C O M P L E X
(−
1.2771)
=
1.756 62
=
1.76
N U M B E R S
Y hd knw:
( A H L )
Y hd e ae t:
2
✔
✔
that
dening
the
set
of
the
meaning
(Argand
✔
the
=
−1
complex
of
extends
the
real
numbers
to
✔
numbers
the
✔
the
real
complex
and
part,
plane
imaginary
argument
denitions
and
Euler
that
adding
numbers
✔
that
down
conjugate
the
of
✔
represent
a
✔
calculate
the
of
a
part,
conjugate,
complex
of
a
real
part,
given
complex
imaginary
complex
number
part
and
number
in
the
complex
plane
of
form
of
and
can
be
as
a
a
form,
complex
two
represented
as
two
complex
rotation
polar
form
and
complex
vector
calculate
form
✔
addition
numbers
a
given
complex
can
be
sums,
quotients
number
subtracting
multiplying
represented
Cartesian
a
conjugate,
modulus
and
argument
number
number
✔
✔
write
diagram)
terms
modulus
i
by
calculate
stretch.
✔
convert
and
powers
form
the
number
calculate
powers,
polar
apply
and
or
functions
with
of
numbers
different
hand
the
forms
and
quotients
in
by
and
of
a
technology
integer
powers
forms.
numbers
shift
Cartesian
technology
exponential
complex
phase
by
in
and
technology
complex
using
between
products
numbers
with
of
complex
in
✔
complex
hand
Cartesian
✔
of
differences,
to
nd
addition
different
phase
the
of
amplitude
sinusoidal
shift
angles.
24
/
1.4
Complex
real
numbers
provide
a
logical
and
useful
extension
to
the
set
C o m Pl E x
NumbERs
(AHl)
of
numbers.
b
2
For
example,
you
know
that
the
real
roots
of
ax
+
bx
+
c
=
0
are
x
=
–
2a
2
b
b
4 ac
2
,
x
x
1
=
±
whenever
b
4ac
≥
0
2
2a
2a
2
Hence
the
zeros
of
y
=
ax
+
bx
+
c
are
reflections
of
each
other
in
the
line
b
x
and
=
can
be
represented
as
the
points
2a
B
2
2
b
b
b
4 ac
+
A
,
2a
A
0
and
b
4 ac
,
B
2a
2a
0
as
shown.
2a
A complex number written in Car tesian form as z = x + yi where x, y ∈ R and
2
i
= −1 has real par t Re(z) = x and imaginary par t Im(z) = y.
The conjugate of z is z* = x − yi
Complex numbers are represented on the complex plane (Argand diagram)
on which the x-axis is called the real axis and the y-axis the imaginary axis.
2
When
2
b
−
4ac
<
0,
no
real
roots
exist.
However,
2
b
i
by
b
i
=
−1,
you
Im(z)
2
2
( 4 ac
writing
b
)
ac
b
b
)
x
can
write
the
roots
as
z
,
z
1
=
±
−
=
2a
laws
of
=
–
using
2
2a
the
±
−
2a
2a
2a
exponents.
2
b
4 ac
b
P
Hence
Re(z
)
=
and
−
Im(z
1
)
=
1
2a
2a
2
b
ac
Re(z)
b
2
Since
z
*
=
z
1
=
,
−
each
zeros
other
in
the
real
axis
and
can
If
z
z
=
x
=
w
then
+
λz
•
Addition
=
λx
+
yi
and
x
=
a
w
λyi
λ
for
and
∈
by
z
=
(x
(y
a)
Im(z
Hence
•
Multiplication
divide
bx
+
c
are
reflections
(x
w)
+
of
=
is
as
the
points
Q
a
+
y
shown.
bi,
=
b
(this
is
known
as
equating
real
and
ℝ
zw*
of
complex
like
Im(z)
+
defined
terms:
+
bi)
Im(w)
by
z
=
(xa
numbers,
+
yb)
+
(ya
+
b
and
+
w
Re(z
expanding
yb)
+
(ya
multiply
denominator
(xa
numbers
=
written
(x
+
a)
+
in
(y
Cartesian
+
b)i
and
b)i
yi)(a
the
=
and
+
w)
=
brackets
+
Re(z)
and
+
Re(w).
simplifying:
xb)i
top
and
bottom
by
the
simplify:
xb)i
=
2
w
represented
2a
adding
complex
conjugate
z
+
+
•
=
be
as
subtraction
defined
To
+
parts)
is
zw
=
and
form
w
ax
b
,
2a
imaginary
•
Q
2a
•
4 ac
b
and
2a
•
=
2
b
,
Given
y
2a
2
4 ac
b
P
of
2
2a
of
the
ww*
a
2
Eape 1.4.1
2
(a)
Find
the
using
roots
of
the
equation
x
10x
+
26
=
0
(c)
Let
your
answers
imaginary
technology
.
Find
z
+
parts
to
be
z
(a)
and
and
w
(b)
with
positive
respectively
.
w
2
(b)
Find
the
roots
of
the
equation
x
4x
+
7
=
0
byhand.
25
/
1
(d)
Represent
matter
in
z,
w
and
which
z
+
w
order
as
you
vectors
add
on
an
complex
Argand
diagram.
Comment
on
the
statement
“It
does
not
numbers.”
Solution
(a)
The
roots
are
(b)
The
roots
are
5
+
i
and
5
Use
i
the
polynomial
solver
on
your
GDC.
2
4 ±
(
4)
4(1)(7 )
{5.–i,5.+i}
2
cPoly
2(1)
4 ±
Roots(x
–10.
x+26.,x)
12
=
=
+
3i
and
w
2
and
3i
2
2(1)
(c)
Let
z
+
(d)
z
=
5
+
i
w
=
7
+
(1
=
2
3i
+
Apply
the
quadratic
formula.
3 )i
+
Im
Clearly
label
your
diagram.
4
z+w
3
The
confirms
commutative:
w
2
diagram
the
that
order
of
addition
the
of
complex
addition
does
numbers
not
is
matter.
1
z
Re
0
1
The
2
vector
adding
the
3
5
6
7
diagram
shows
complex
numbers
the
same
4
result
no
matter
that
the
gives
order.
s AmPlE sTuDENT ANs WER
Consider
where
a,
the
b,
c,
distinct
d
∈
complex
Find
the
real
part
+
the
z
+
value
of
the
real
part
+
is
z
the
in
method
of
numbers.
to
does
of
multiply
a
not
division
The
and
recall
of
correct
both
(a
c)
+
i(b
by
when
+
a
the
a
complex
approach
the
numerator
the
conjugate
+
c
a
c
a
difculties
moves
awarded
knowledge
and
a
on
in
of
a
to
point
+
di)
+
c
+
i(b
c
+
i(b
a
+
+
d)
d)
c
i(b
+
d)
c
i(b
+
c
+
i(b
d)
a
+
d)
of
2
2
b
part
(a),
2
part
for
2
+
b
2
=
c
i(b
1
2
+
d
×
d)
i
×
c
i(b
1
2
+
+
×
d)
i
+
d
a
c
(b)
the
understanding
demonstrates
modulus
(c
and
a
he/she
−
d).
student
is
bi
is
a
and
|z|=|w|
×
+
the
id,
b,
2
▲ Despite
+
w
a
denominator
c
w
terms
student
=
found
−
correctly
w
w
of
+
w
ib,
w
z
▼ Although
+
w
z
Find
a
of
z
(b)
=
R
z
(a)
numbers z
about
complex
the
number.
(b
1
+
d)
×
a
c
(b
i
d)
×
a
c
26
/
1.4
As
with
real
numbers
by
numbers,
you
hand
with
Assessment
and
tip:
make
can
carry
out
technology
sure
you
operations
as
know
with
C o m Pl E x
NumbERs
(AHl)
complex
appropriate.
where
to
find i
on
your
GDC.
Eape 1.4.2
6
Given
z
=
2
i,
z
1
=
0.7
+
and
13i
z
2
=
1
+
i,
(c)
find
(z
+
0.81z
2
)
giving
the
real
and
imaginary
1
3
partscorrect
to
the
nearest
unit.
2
(a)
z
+
z
1
3
z
+
z
1
3
(b)
z
2z
1
3
Solution
2
2
(a)
z
+
z
1
=
(2
i)
+
(1
+
i)
=
2
This
could
be
done
more
efficiently
by
finding
the
solution
3
2
=
(2
i)
+
1
+
2i
+
i
+
directly
i
An
(b)
z
1:=2–i
2.–i
z
3:=1+i
1.+i
z
from
alternative
+
3
z
1
1+z
3
3(3i)
z
+
z
is
9i
=
2z
1
1–2
method
=
=
i
i
z
z
GDC.
3
=
z
the
3i
(
3i)(3i)
9
3
3
z
1
3
=
z
i
from
GDC
3z
1
3
(c)
Use
z
1:=2–i
z
2:=0.7+
of
technology
is
by
far
the
most
efficient
method.
Use
of
2.–i
the
√13
∙
i
0.7+3.60555127546 
i
1213.5566219–1952.21449433
i
memory
facility
of
your
GDC
helps
you
check
your
answer.
6
(z
2+0.81z
1)
6
(z
+
0.81z
2
Since
)
=
1214
1952i
1
complex
numbers
can
be
represented
as
vectors,
we
can
find
their
Im
6
magnitude
and
direction.
5
2
For
a
complex
number
z
=
x
+
yi,
its
2
x
modulus |z|=
+
y
is
the
z
1
4
length
of
between
the
the
position
positive
vector
real
defined
axis
and
by z.
the
The
position
argument
vector
arg(z),
defined
the
by z
angle
where
3
⎛ y ⎞
−π
<
arg(z)
≤
π,
is
found
with
the
equation
arg(z)
=
arctan
⎜
⎟
2
⎝ x ⎠
1
For
example,
z
=
3
+
4i,
represented
in
Cartesian
form,
can
be
plotted
Re
on
an
Argand
diagram
by
the
coordinate
pair
(3,
4)
as
shown
0
on
0
1
2
3
4
5
6
theright.
⎛ 4 ⎞
2
|z|=
3
2
+
Im
=
4
5
and
arg(z)
=
arctan
⎜
⎟
=
0.927.
Hence
z
is
5
units
from
6
⎝ 3 ⎠
the
origin
in
the
direction
given
by
0.927
radians
measured
from
the
5
real
axis
as
shown
on
the
right.
z
1
4
Notice
that
the
Cartesian
coordinate
pair
(3,
4)
and
the
polar
coordinate
3
pair
(5,
0.927)
each
give
the
information
needed
to
locate z
on
the
2
Argand
diagram.
1
Therefore,
you
have
two
different
ways
to
represent
a
complex
0.927
rad
Re
number:
3
+
4i
and
5(cos(0.927)
+
sin(0.927)i)
are
equivalent.
0
0
1
2
3
4
5
6
27
/
1
The
generalised
form
of
this
result
is
given
by:
2
If z = x + yi, then z = r(cos(θ) + isin(θ)) where r =
x
2
+
y
and θ = arg(z).
iθ
You
write
which
is
this
as
known
z
rcis(θ)
=
as
the
for
short.
exponential
It
(or
can
the
be
shown
Euler
that
form)
of
rcis(θ)
a
=
re
,
complex
number.
Eape 1.4.3
(a)
Represent
z
=
(b)
Represent
w
−1
+
4i
in
exponential
form.
–0.871i
=
3.1e
in
cartesian
form.
Solution
2
(a)
r
=
(
2
1)
+
A sketch
second
=
4
Your
17
shows
that
z
is
in
the
GDC
angle(–1+4
confirm:
i)
1.81577498992
quadrant.
|–1+4∙
So,
can
arg(z)
=
π
−
arctan
(4)
≈
4.12310562562
i|
1.82
(–1+4∙
i)
Polar
1.81577498992∙
e
i
4.12310562562
Make sure you know how your GDC
3.1
cos(–0.871)
3.1
sin(–0.871)
1.99659187765
4
conver ts between Car tesian, polar
Q
–2.37141748203
and exponential forms of complex
–0.871∙
number.
13.1
1
i
2
e
Rect
1.99659187765–2.37141748203
i
1.82i
Hence
z
=
17 e
–0.871i
(b)
3.1e
Hence
Why
One
the
=
w
bother
reason
next
3.1(cos
=
2.00
with
is
to
(
0.871)
+
i sin
(
0.871))
2.37i
three
representations
visualize
and
of
complex
understand
numbers?
patterns,
as
seen
in
example.
Eape 1.4.4
π
i
n
6
(a)
Given
z
=
,
1.5e
find
1
(b)
z
=
(z
n
Represent
the
sequence
)
for
n
=
0,
,
z
,
2,
3,
4,
5
1
{z}
=
{z
,
0
z
1
z
2
,
3
z
,
4
z
}
on
an
Argand
5
diagram.
(c)
Describe
the
patterns
in
{z}
(i)
algebraically
(ii)
graphically
Solution
2
π
π
⎛
i
⎞
i
3
6
(a)
z
=
0
1,
z
1.5e
=
2
=
⎜
⎟
⎝
⎠
2.25e
Apply
i
i
=
3.375e
the
3
2
z
the
laws
of
exponents
to
2 π
π
,
z
3
=
5.0625e
Eulerform.
,
4
5π
i
6
z
=
7.593 75e
5
28
/
1.4
C o m Pl E x
NumbERs
(AHl)
(b)
Im
5
z
4
4
z
z
3
5
3
2
z
2
1
z
1
Re
z
0
0
–7
(c)
(i)
–6
–5
The
–4
–3
moduli
common
of
–2
{z}
ratio
–1
are
1.5.
0
a
1
2
3
geometric
The
sequence
arguments
are
an
with
first
arithmetic
term
1,
sequence
π
with
first
term
0
and
common
difference
6
π
(ii)
The
sequence
shows
a
power
is
stretch
factor
1.5
and
a
rotation
of
6
each
time
the
increased
by
1.
iθ
A consequence
laws
of
=
•
zw
the
exponents
ρ(cos(α)
w
of
can
isin(α))
+
r(cos(θ)
equivalence
be
used
to
show
+
that
rρ(cos(θ
α)
+
isin(θ
+
+
α))
Also,
patterns
iθ
n
z
=
(re
This
These
is
n
=
r
formulae
with
efficiently
is
that
+
isin(θ))
r(cos(θ)
=
the
and
in
the
(cos(θ
=
previous
α)
+
isin(θ
α))
ρ
example
can
be
generalized
as
n
e
=
known
calculate
seen
inθ
n
)
as
z
re
r
and
w
•
if
=
then:
z
=
isin(θ))
as
(cos(nθ)
r
De
give
Moivre’s
convenient
complex
without
isin(nθ)),
+
to
∈
Z.
theorem.
ways
numbers.
needing
n
to
represent,
For
example,
expand
brackets.
manipulate
be
and
powers
can
found
laws
exponents.
Eape 1.4.5
6
⎛
Bi
Given
z
=
⎞
z
–i
Ae
and
w
=
2e
,
find
Re
⎜
4
⎝
⎟
⎠
w
Solution
6
Bi
z
( Ae
6
6
)
A
6 Bi
Apply
6
e
=
4
i
w
(2 e
=
4
4
)
2
+
16
Write
⎛
Re
⎝
real
part
of
the
A
A
(6 B
cos(6B
⎟
⎠
w
the
6
=
4
down
6
⎞
z
⎜
of
4)i
e
4i
e
6
Hence
the
A
(6 B
=
+
polar
4)
form
of
+
4)i
,
e
16
16
6
A
(cos(6B
+
4)
+
isin(6B
+
4))
16
Despite
being
named
complex
numbers
example,
finding
functions
with
“imaginary”,
which
the
the
give
you
amplitude
same
period
there
useful
and
is
are
results
phase
useful
many
in
shift
in
of
applications
the
a
real
sum
electrical
of
world:
of
for
sinusoidal
engineering.
Eape 1.4.6
When referring to the adding of
Two
electricity
sources
V
and
V
1
are
V
=
5
sin
(3t
+
modelled
with
the
functions
sinusoidal functions of the form
2
π
π
)
and
V
1
=
2
sin
(3t
+
)
a sin(bt + c), the phase shift is c.
respectively
.
2
4
6
Elsewhere, the phase shift refers
(a)
Graph
V
,
1
(b)
Hence
(c)
By
V
and
V
=
2
V
+
1
compare
and
V
on
the
a
sin(bt
contrast
+
c)
axes.
2
to c in the form of the equation
the
features
(bt
applying
same
=
Im(ae
+
of
the
three
graphs.
always make it clear what is meant
c)i
),
a sin(b(x – c)). The question will
calculate
the
maximum
by the term in the context.
value
of
V
and
its
phase
shift.
29
/
1
Solution
Choose
(a)
a
viewing
window
in
radians
which
shows
RAD
1.1
the
details
of
all
V
+
three
graphs
clearly
.
y
V(x)=v1(x)+v2(x)
π
V2(x)=2 
sin
3
x+
6
2
x
0.52
π
V1(x)=5 
sin
3
x+
4
2π
(b)
The
period
of
both
V
and
V
1
is
and
the
2
3
period
of
V
of
V
looks
and
V
1
are
the
5
same.
and
2
of
V
is
and
V
1
phase
=
5
the
7.
The
phase
shifts
π
are
shift
of
and
sin
⎜
V
π
⎛
V
and
respectively
and
the
2
6
4
(c)
respectively
approximately
π
V
amplitudes
2
period
of
The
3t +
appears
⎞
4
π
⎛
+
⎟
⎝
between
2
sin
⎠
⎜
3t +
⎝
these
values.
⎞
Apply
⎟
6
V
=
V
1
⎠
then
2
(bt
a
⎛
π
⎛
⎝
=
⎞
⎞
3 t+
4
⎟
+
π
⎛
π
⎛
i
4
⎝
Im ⎜ 5e
+
)
Apply
Im(z
+
w)
=
Im(z)
Im(w)
π
π
i
3ti
⎠
Factorize
⎟
by
e
i
4
and
use
your
GDC
to
find
5e
6
+
2e
in
6
4
(5e
+ 2e
)
⎜
⎟
⎝
⎠
π
π

3ti
=
Im(e
=
6.95
0.711i
(6.95e
(3t
))
=
+
(3t
+
i

i
0.711)i
Im(6.95e
)
4
5
sin
form:
⎞
i
i
e
exponential
π
π
3 ti
0.711)
e
6
+2
e
0.710861208924
i
e
Hence
the
+
⎠
⎛
Im
c)i
⎞
⎞
2e
+
Im(ae
i
6
⎝
=
=
⎠
⎠
3 t+
⎠
c)
⎟
⎝
⎞
3 t+
⎝
+
i
6
Im ⎜ 2e
⎠
sin(bt
⎞
⎞
3 t+
⎝
Im ⎜ 5e
⎛
π
⎛
⎠
⎝
=
⎛
i
the
phase
maximum
shift
is
of
V
is
6.95
and
0.711
(bx + c)i
a sin(bx + c) = Im(ae
value
6.95115217254
iθ
) is a consequence of the formula r(cos(θ) + i sin(θ)) = re
which is in the formula book .
30
/
1.5
1 . 5
S Y S T E M S
O F
the
meaning
of
the
that
a
terms
zero,
root
✔
use
polynomial
or
with
most
of
linear
equation
technology
,
efcient
A polynomial
equations
can
be
however
function
has
the
=
a
x
N
and
✔
by
use
up
technology
solve
to
to
systems
three
solve
of
linear
variables
polynomial
equations.
hand
technology
is
the
form
n
+
a
n
∈
a
in
to
method.
f (x)
n
or
solved
n
where
technology
equations
system
E q u AT i o N s
Y hd e ae t:
andintercept
✔
oF
E Q U AT I O N S
Y hd knw:
✔
s Ys T E m s
each
a
∈
R
–1
x
n
2
+
…
–1
a
x
+
a
2
for
0
≤
i
≤
n.
x
+
a
1
Solving
0
polynomial
equations
These terms are easy to mix up.
i
by
to
hand
use
is
often
time-consuming
and
error-prone,
so
it
is
often
better
Make sure you know them so you
technology
.
can understand exam questions
and use the correct terminology in
The
zeros,
roots
and
intercepts
of
a
function
are
terms
that
describe
the
your internal assessment.
following
features:
The
zeros
of
The
solutions
The
point
function
of
where
y-intercept
The
a
(0,
point(s)
x-intercepts
the
the
of
are
equation
graph
of
solutions
f (x)
y
=
=
0
f (x)
are
of
the
its
equation
f (x)
=
0.
roots.
crosses
the
y-axis
is
the
x-axis
are
f (0)).
where
(β
,
0),
the
(β
1
zeros
f (x)
,
graph
0),
(β
2
,
of
0),
y
=
f (x)
crosses
…where
3
β
β
,
1
,
the
β
2
,
…
are
the
the
3
f (x).
Eape 1.5.1
(a)
Find
the
coordinates
of
all
the
axes
intercepts
of
(b)
Find
the
zeros
of
4
3
2
g(x)
=
2.1x
1.3x
9
h(x)
=
−0.3x
+
the
x
function
0.8x
+
2
Solution
(a)
The
y-intercept
is
The
x-intercepts
(0,
9)
The
but
are
(
1.78,
0)
and
(2.40,
y-intercept
the
can
be
x-intercepts
written
are
best
down
found
immediately
with
technology:
0)
2
PolyRoots(2.1
x
–1.3 
x–9.,x)
{–1.78368408134,2.40273170039}
2
(b)
The
zeros
of
the
function
are
1.30
and
3.27
Take
care
to
enter
the
coefficient
of
x
as
0.
DEG
1.1
Roots
of
a
Polynomial
4
a
X
+
4
a
=
4
a
=
3
a
=
2
a
=
1
a
0
=
.
.
.
+
a
X
1
+
a
=
0
0
–0.3
1.
0.
–0.8
2
OK
Cancel
31
/
1
You
the
should
also
use
examination.
Systems
substitution
or
reliable
efficient
and
technology
can
elimination,
to
be
solve
solved
however
systems
by
using
of
hand
equations
using
in
methods
technology
is
a
of
more
method.
Eape 1.5.2
Solve
the
system
of
equations:
Make sure you know how to
3α
−
4.01β
+
0.9z
=
2
enter the information to your GDC
α
+
β
=
z
in order to solve a polynomial
0.1α
=
7.1β
equation or to solve a system of
equations, and how to interpret
the output.
Solution
Entering
3
–4.01
+0.9
it
linSolve
15
+=z
0.1
,{
=7.1

=
0.519,
β
=
system
exactly
as
is
written
question
z
the
examination
2

0.007 30,
in
z}
you
{0.518645677344,0.007304868695,0.52595
α
the
z=2
=
0.526
to
can
make
check
you
it
easier
have
entered
everything
correctly
.
This
shows
this
gives
that
GDC
for
app
the
In the examination there will
solution
in
the
order
α,
β,
z
always be a unique solution to a
system of equations. So if your
GDC says otherwise, you will have
Write
the
order
correct
figures,
entered the system incorrectly.
answers
to
unless
in
three
told
the
correct
significant
otherwise.
s AmPlE sTuDENT ANs WER
▲ The
the
equations,
GDC.
it
is
student
This
fast
than
the
method
is
and
carefully
ready
the
best
solving
entry
much
and
by
up
method
accurate;
error-prone
of
for
sets
into
Boxes
the
4
student
exam
marks
with
have
is
are
on
sale
at
a
local
supermarket.
2
bananas,
3
kiwis
and
4
melons,
and
costs
$6.58.
better
Box
B
contains
5
bananas,
2
kiwis
Box
C
contains
5
bananas
and
and
8
melons,
and
costs
$12.32.
laborious
to
the
cost
of
each
type
of
4
kiwis
and
costs
$3.00.
fruit.
begin
knowledge
been
fruit
hand.
able
the
mixed
Box A contains
since
Find
▲ The
of
a
scored
that
quickly
.
Use the five-minute reading time
2B
+
3K
+
4M
=
6.58
5B
+
2K
+
8M
=
12.32
5B
+
4K
+
0M
=
3.00
Banana
:
$0.36
before the examination begins
Kiwi
:
$0.30
to look for oppor tunities like this
question in which use of the GDC is
Melon
:
$1.24
GDC
}
the most efficient method.
32
/
1.6
1 . 6
M AT R I C E S
A N D
M AT R I X
Y hd knw:
✔
the
terms
ofa
element,
matrix.
An
m
AND
m AT R i x
A lg E b R A
A L G E B R A
(AHl)
( A H L )
Y hd e ae t:
row,
×
m AT R i C E s
n
column
matrix
and
has
m
order
rows
✔
identify
and
matrix
the
in
meaning
the
context
of
each
of
a
element
of
a
problem
ncolumns
✔
✔
that
the
matrix
number
of
product
columns
in
AB
A
is
is
dened
equal
to
only
the
if
add,
the
andwith
rows
in
the
terms
multiply
matrices
by
hand
technology
nd
inverses
and
determinants
for
2
×
2
matrices
B
by
✔
and
number
✔
of
subtract
identity
matrix,
zero
of
a
matrix
and
and
for
n
×
n
matrices
with
technology
matrix,
✔
determinant
hand,
inverse
solve
systems
of
equations
using
an
matrix
inversematrix
✔
the
ways
in
which
matrix
algebra
is
different
✔
from
that
of
algebraic
expressions
and
nd
the
matrix
✔
the
denition
of
an
eigenvector
and
Some
a
of
give
the
practical
about
this
data
you
basic
a
sales
in
a
2
×
powerful
definitions
context.
total
For
one
3
way
of
sales
in
two
matrix
organize
matrices
example,
day
to
the
branches,
be
write
a
2
×
2
A
a
matrix
A
in
diagonalized
form
and
Street
of
the
matrix
information.
For
thiselement
is
is
just
example,
denoted
made
a
and
café
Mall,
it
nd
powers
of
A
data.
sense
of
collects
and
to
in
data
presents
water
212
145
⎝ 311
170
⎛
snack
76
⎞
=
Mall
that
manipulate
easily
⎜
see
of
S:
Street
S
can
and
management
coffee
We
eigenvectors
eigenvalues.
apply
Matrices
and
its
✔
associated
eigenvalues
formulae
a
in
S
⎟
convenient
the
Street
because
it
112 ⎠
table
branch
is
in
the
setting
76
out
snacks
first
row,
the
were
third
sold:
column.
1,3
2
×
3
is
down
this
the
a
order
sales
scalar
=
1.1
target
the
in
multiple
⎛
1.1S
of
212
145
matrix.
which
of
S
76
by
⎞
×
If
all
the
management
sales
rise
multiplying
by
10%,
each
wanted
they
element
1.1 × 212
1.1 × 145
1.1 × 76
⎝ 1.1 × 311
1.1 × 170
1.1 × 112 ⎠
⎛
⎝
to
could
by
write
calculate
1.1:
⎞
⎠
=
⎜
⎝ 311
⎟
170
⎜
112 ⎠
⎛
=
⎟
233.20
159.50
management
expresses
and
the
new
⎞
⎟
⎝ 342.10
The
83.6
⎜
the
187
123.20 ⎠
current
prices
in
2
in
€
of
their
products
P
1
proposed
prices
P
a
3
×
price
matrix
P
and
wishes
to
2
investigate
what
effect
these
changes
would
P
1
have
on
total
earnings:
Writing column and row headings
P
2
is not compulsory, but it may
coffee
⎛ 2.3
2.4 ⎞
help you plan your approach to a
⎟
⎜
P
=
2.1
water
⎜
⎜
snack
⎝ 3.4
2.1
⎟
problem in context.
⎟
3.1 ⎠
33
/
1
From
these
the
number
the
rows
process
of
of
matrices,
of
S
the
columns
and
matrix
the
145
SP
⎝ 311
170
76
S
⎞
of
P
to
the
can
be
the
matrix SP
number
of
combined
⎟
as
of
P,
follows
so
in
that
the
2.1
⎟
⎜
112 ⎠
⎟
⎝ 3.4
3.1 ⎠
212 × 2.3 + 145 × 2.1 + 76 × 3.4
212 × 2.4 + 145 × 2.1 + 76 × 3.1
⎜
⎞
⎟
⎝ 311 × 2.3 + 170 × 2.1 + 112 × 3.4
⎛
rows
because
⎟
⎜
2.1
⎛
equal
find
2.4 ⎞
⎜
=
is
can
multiplication:
=
⎜
of
columns
⎛ 2.3
212
⎛
management
1050.50
1048.90
311 × 2.4 + 170 × 2.1 + 112 × 3.1⎠
⎞
=
⎜
⎟
⎝ 1453.10
From
For
this,
1450.60 ⎠
the
example,
management
by
can
comparing
find
out
elements
about
SP
=
their
€1050
proposed
and
SP
1,1
to
see
that
virtually
the
total
earnings
unchanged
under
1234
⎛
for
the
the
proposed
741.50
443.60
661.50
394.80
PS
=
1098.30
⎝ 1684.90
order
answer:
this
fact,
only
not
compared
in
three
means
to
are
predicted
to
be
changes.
⎞
illustrates
that
in
general,
⎟
⎜
the
branch
⎟
⎜
changing
€1048.90
1,2
⎟
⎜
Finding
street
=
changes.
does
SP,
it
different
1020
of
multiplication
that
PS
605.60 ⎠
matrix
have
has
matrices
multiplication
different
little
of
elements
meaning
in
this
is
gives
a
different
not commutative.
and
different
context.
So
PS
order
≠
SP
In
when
is
true
ways!
The following results are true in general for matrix addition:
•
Matrix addition A + b is defined only if A and b have the same orders
•
Matrix addition is commutative: A + b = b + A
•
Matrix addition is associative: (A + b) + C = A + (b + C)
The following results are true in general for matrix multiplication:
•
Matrix multiplication Ab is defined only if the number of columns of A is
equal to the number of rows of b
•
Matrix multiplication is not commutative: in general, Ab ≠ bA
•
Matrix multiplication is distributive over addition: A(b + C) = Ab + AC and
(b + C)A = bA + C A
•
Matrix multiplication is associative: (Ab)C = A(bC)
You
can
use
technology
to
calculate
with
matrices.
Eape 1.6.1
⎛
(a)
Given
A
=
(3
4),
B
0
⎝
AB
+
Comment
(c)
Find:
(i)
(d)
⎟
1
AC
(b)
on
your
on
your
4 ⎠
C
1
1
⎞
=
⎛
,
⎜
and
D
⎝ 5
0⎠
(ii)
A(B
+
C)
(ii)
(BC)D
0
⎞
,
⎜
⎟
⎝ 0.5
1
=
find:
⎟
1⎠
(iii)
(B
+
C)A
results.
B(CD)
Comment
⎛
⎞
,
⎜
(i)
3
=
results.
34
/
1.6
m AT R i C E s
AND
m AT R i x
A lg E b R A
(AHl)
Solution
Enter
(a)
a
b+a 
c
a
(b+c)
[5.
–4.]
[5.
–4.]
[3
(b+c) 
a
“Error :
Dimension
the
–4]
matrices
in
your
GDC:
[3.
→a
–4.]
error ”
0
3
0.
3.
–1.
4.
→b
Hence
AB
However,
+
AC
(B
+
=
(5
C)A
4)
is
=
not
A(B
+
defined
C).
–1
4
because
1
1
0.5
0
1.
1.
0.5
0.
1.
0.
5.
1.
→c
the
number
which
is
of
not
columns
the
same
of
as
B
+
the
C
is
2,
number
1
0
→d
of
(b)
rows
This
of
A
which
illustrates
is
that
1.
5
matrix
1
multiplication
This
is
distributive
but
not
⎛
(c)
B(CD)
=
(BC)D
=
1.5
(d)
This
is
illustrates
order
of
zero,
•
The
m
that
0
to
know
identity
×
+
n
3
is
212
how
and
zero
A
=
A
+
0
0
⎝ 0
0
0⎠
⎜
76
⎜
⎝ 311
170
n
those
IB
=
×
in
BI
0
matrix
for
are
has
m
n
0
algebra
with
n
its
elements
matrices
A.
×
0
145
76
identity
equal
⎛
=
0⎠
⎟
⎝ 311
I
is
diagonal
170
has
all
which
matrices
its
are
matrix
of
0⎞
1
0
⎜
⎜
⎝ 0
0
and
×
⎜
⎟
⎜
be
B
6
3
⎝ 3
2
enable
Equation
and
4⎞
⎛ 1
clear
for
you
to
read
on
the
screen.
with
given,
X
the
concepts
to
zero
zero,
and
matrix
of
is
such
order
=
3
⎜
⎟
⎜
you
A
4⎞
×
3
6
3
⎛ 1
is
⎝ 3
to
2
=
all
⎜
⎞
⎟
0
equal
212
145
⎝ 311
170
⎛
⎜
0⎠
equal
to
76
⎞
+
1,
⎟
to
and
zero
is
112 ⎠
except
such
that
0
0⎞
1
0
⎟
⎜
⎟
⎜
⎝ 3
0
7
4⎞
⎛ 1
6
3
matrix
X
+
A
=
B
X
+
A
+
(−A)
For
example:
⎟
⎝ 0
2
.
⎟
1⎠
0
0⎞
0
1
0
⎝ 0
0
1⎠
⎟
⎜
⎟
1
⎟
2⎠
solve
0
elements
0
⎜
⎟
1
⎟
0⎠
7
0
B.
order
⎜
⎟
1
⎟
1⎠
A
7
⎜
⎟
⎜
0
⎛ 1
0
⎝ 0
112 ⎠
⎜
0
matrices,
The
⎞
⎜
matrix
n
212
⎛
=
⎜
0
your
essential.
all
×
⎞
⎟
main
all
0
out
⎜
⎛ 1
make
multiplication
⎛ 1
The
as
example:
⎜
identity
the
carry
all
0
⎝ 0
112 ⎠
n
for
For
⎛
⎞
⎟
The
GDC
1⎠
inverse
0
.
+
•
well
⎞
⎟
145
to
matrix
0
⎛
⎛
as
associative.
In
2×
time
⎞
⎟
4
that
save
calculations
0
⎜
⎝
can
commutative.
×
⎟
⎜
⎟
⎜
0⎠
⎟
⎟
equations
as
follows:
to
found.
Step
by
step
re-arrangement
the
equation
B
+
(−A)
Add
of
A
sides
of
to
both
the
equation
using
X
matrix
=
+
(A
+
(−A))
=
B
+
(−A)
Matrix
addition
is
algebra
associative
A
X
Solution
+
X
=
(0)
B
+
which
is
The
addition
Similarly
,
to
carry
to
out
of
A
solve
the
inverse
B
(
is
+
(
+
(
A)
gives
the
A)
zero
matrix
0
since
matrix
addition
A),
equal
(
A)
+
B
to
commutative.
carries
more
=
out
types
of
the
of
inverse
process
equations,
multiplying
by
a
you
of
adding
need
to
be
A.
able
matrix A.
35
/
1
⎛
For
example,
for
all
2
×
2
matrices
A
a
b
,
⎜
d
⎛
b
the
inverse
of
A
is:
⎟
⎝ c
1
⎞
=
d⎠
⎞
1
=
A
⎜
ad
ad
bc
is
⎟
a
.
bc ⎝
c
the
determinant
⎠
of
A
and
is
denoted
det
A
or
|A|
.
1
For
A
to
exist,
det
A
≠
0
hence
ad
≠
bc
1
I
and
A
enable
Equation
with
you
A
to
solve
XA
=
matrix
equations
as
follows:
B
Nte
and
be
–1
Do not confuse A
B
given,
X
to
found.
with the
1
Step
reciprocal of a number.
by
step
re-
1
XAA
=
Post
BA
multiply
both
sides
of
the
1
arrangement
the
equation
of
equation
using
1
X(AA
matrix
by
A
1
)
=
Matrix
BA
multiplication
is
algebra
associative
Nte
1
1
X(I)
=
AA
BA
gives
the
equal
A
identity
matrix
I
Because matrix multiplication
1
Solution
X
=
BA
1
,
which
is
not
B
to
since
is not commutative, the order of
multiplication matters. Hence if
matrix
multiplication
In
A
is
not
commutative.
1
you post/pre-multiply by a
fact,
B
is
the
solution
to
a
different
matrix on one side of an equation,
equation:
AX
=
B
which
can
be
derived
in
the
you must do the same to the
same
way
sides
by
other side.
as
above,
but
by
pre-multiplying
both
1
A
at
the
start
of
the
re-arrangement.
Eape 1.6.2
x
⎛
(a)
Solve
the
equation
y
⎞ ⎛
2
0.5
⎟ ⎜
⎝ z
t
the
0
1
⎞
equation
2
⎛
⎟
⎜
4
⎜
1
1
7
=
q
2
⎜
⎝
⎟
⎝ 3
0 ⎞ ⎛ p⎞
⎟ ⎜
0.7
⎜
3⎠
⎜
Solve
⎛
⎟
⎠ ⎝ 4
1
⎛
(b)
⎞
=
⎜
2
⎟
⎜
⎟ ⎜
⎟
⎜
⎠ ⎝ r ⎠
⎝
hand.
4⎞
⎟
using
11
⎟ ⎜
by
⎠
technology
.
⎟
⎟
0
⎠
Solution
(a)
2
0.5
⎛
=
2
(
3)
0.5
(4)
=
First,
−8
find
the
determinant
of
2
0.5
⎜
⎟
⎝ 4
4
3
2
0.5
⎞
3⎠
1
⎛
⎞
1 ⎛
3
0.5
4
2
⎞
=
⎜
⎝ 4
⎛
x
y
⎟
⎜
3⎠
8 ⎝
⎞
⎛
0
1
⎞
⎝ z
t
×
⎟
⎜
⎠
⎝ 3
1
⎛
=
⎜
so
⎟
⎟
⎜
8
3
⎛
⎞
⎟
⎝
2 ⎠
⎠
0.5
⎜
⎠
Post-multiply
⎞
4
⎝
2
⎠
⎛
of
1
⎛
1 ⎛
4
2
⎜
⎞
=
=
⎜
8 ⎝
⎟
17
1
2
2
⎜
,
y
=
16
z
=
4
2
by
the
inverse
⎞
⎟
3⎠
⎠
17
,
equation
⎟
8
1
1
the
5
Write
=
of
⎟
17
⎝
x
sides
⎟
4
⎜
so
0.5
⎝ 4
⎞
⎜
2.5 ⎠
both
⎟
out
the
solution
by
pre-multiplying
both
sides
by
the
5
and
t
inverse
=
8
matrix.
Use
technology
to
find
the
answer.
16
–1
1
⎛ p⎞
=
q
⎜
⎟
⎜
⎜
⎟
⎜
⎝ r ⎠
⎝
so
p
1
2
0 ⎞
⎛
⎟
⎜
0.7
4
2
1
1
⎜
⎟
⎜
(b)
⎛
=
−16.9,
7
q
1
2
0
–4
–16.9135802469
0.7
4
–2
11
6.45679012346
1
1
7
0
1.49382716049
4⎞
⎟
11
⎟
⎜
⎟
⎜
⎠
⎝
=
⎟
⎟
0
6.46,
⎠
r
=
1.49
36
/
1.6
m AT R i C E s
AND
m AT R i x
A lg E b R A
(AHl)
Eenvae and eenvectr
For
a
2
×
2
matrix
multiplying
x
is
(x
called
=
0
is
useful
The
x
an
by
A
and
eigenvector
of
procedure
Rearrange
•
Solve
solution
the
to
Ax
find
=
λx
the
to
1
vector
stretch
A
of
and
Ax
=
x
λ
x,
by
is
λx,
if
a
its
but
λ
and
Ax
=
λx
factor
of
then
the
λ.
this
associated
this
trivial
In
Identify
sets
a
particular
form
(A
−
polynomial
of
the
and
λI)x
|A
systems
x
and
solution
solution
procedure
do
is
not
eigenvectors
=
−
is:
0
λI|=
0
to
find
the
Ax
=
λ
x
and
Ax
=
λ
this
in
each
solution
set
to
form
the
2
is
the
from
x
2
x
1
to
case,
2
solution
eigenvectors
This
of
eigenvalue.
1
•
effect
λ
1
the
eigenvalues
the
characteristic
eigenvalues
Find
×
applications.)
•
•
2
to
a
is
a
just
always
in
A
applied
below.
You
will
not
be
able
to
use
technology
examination.
Eape 1.6.3
⎛
Find
the
eigenvalues
and
eigenvectors
of
the
matrix A
3
2
⎞
=
⎜
⎝
⎟
1
0⎠
Solution
⎛
⎜
3
2
⎞
⎛
x
⎞
⎜
⎟
1
⎝
⎜
0⎠
2
⎞
x
⎛
⎟
1
⎝
3
2
⎞
⎛
x
⎜
⎟
1
⎛
3
2
⎟
⎛ λ
⎞
1
⎛ λ
⎞
⎛
x
⎛
⎜
λ ⎠
λ
3
λ
⎟
definition
Ax
=
⎛
0
λx
x
⎛
⎛
⎞
0
Since
⎜
⎟
⎟
x
⎛
⎝ 0
⎝ λy⎠
λ ⎠
⎜
⎟
⎝ y⎠
matrix
multiplication
is
distributive.
Carry
out
the
matrix
subtraction.
⎟
⎝ 0⎠
⎛
⎞
only
unique
solution
x
⎞
is
=
⎜
⎟
0.
⎟
⎝ y⎠
⎝ 0⎠
⎝ y⎠
⎞
⎞
The
⎟
⎜
⎜
⎟
⇒
⎜
⎟
0
0⎞
=
⎝ 0⎠
⎝ y⎠
⎞
=
⎟
⎞
⇒
⎜
⎛ λ
⎛ λx⎞
⎞
⎝ y⎠
⎝ y⎠
⎞
x
⎜
=
⎟
1
⎜
⎝ y⎠
⎜
λ ⎠
⎞
⎜
Apply
⎛
=
⎟
⎝ 0
0⎠
the
⎞
⎟
×
⎜
2
⎜
λ ⎠
0⎞
×
⎝
x
⎛
⎟
⎝ 0
⎝ y⎠
λ
⎛ 3
0⎞
⎜
⎟
⎟
⎝
λ ⎠
⎝ 0
−
⎜
x
⎛
⇒
⎟
=
⎜
0⎠
0⎞
⎜
⎝ y⎠
×
⎝
⎛ λ
⎞
=
⎜
0⎠
Apply
⎟
⎝ y⎠
×
⎜
⎛
⎜
⎝ y⎠
3
⎛
⎞
λ
⎟
⇒
x
⎛
=
×
2
=
0
⇒
(3
−
λ)(−λ)
−
2(−1)
=
0
Normally
,
your
solution
in
an
exam
would
begin
at
this
line.
λ
1
For
other
solutions
to
the
system
to
exist
there
must
be
an
2
⇒
λ
3λ
+
2
=
1
or
0
⇒(λ
−
2)(λ
−
1)
3
=
0
⇒
λ
=
λ
=
infinity
2.
of
solutions,
so
λ
=
1
If
⎛
λ
=
3
1
2
⎞
⎛
⎟
1
⎝
x
⎞
⎛
=
⎜
0⎠
+
2y
=
or,
from
x
second
If
=
⎛
3
2
⇒
⎝
⎞
⎛
x
y
=
row,
⎞
⎛
=
⎜
0⎠
relationship
be
− x
−
x
+
2y
=
or,
from
not
have
a
unique
solution.
However,
the
=
found
between
fromthe
the
components
of
the
eigenvector
can
system.
of
y
⎟
⎝ y⎠
⇒
2y
=
−
x
multiplication
second
For
λ
1,
x
⎞
⇒
⎜
2x
the
=
x
2
⎟
⎝ y⎠
3x
⎛
does
then
⎟
1
system
⎟
⎝ y⎠
×
⎜
Each
⎞
⇒
⎜
multiplication
the
2
x
1
⎟
⎝ y⎠
3x
λ
0.
λ
then
×
⎜
2
that
row,
the
−
x
=
of
2y
solutions
are
of
the
You
describe
the
general
solution
for
each
system.
⎞
form
since
⎜
⎟
⎝
x⎠
x
=
y
37
/
1
For
λ
=
⎛
2,
2y
the
solutions
since
⎝
of
the
⎞
form
⎜
are
x
=
−2y
⎟
y
⎠
⎛
The
eigenvectors
are
therefore
1
⎜
⎟
⎝
1⎠
2
⎛
⎞
and
⎜
⎞
,
⎟
1
⎝
Choosing
particular
solutions
for
⎠
theeigenvectors.
associated
with
eigenvalues
λ
=
1
and
λ
=
Note
2
respectively
.
that
in
Example
7
⎛
could
choose
1.6.3,
you
⎜
⎟
⎝
7⎠
1
⎛
⎞
instead
of
3
2
7
7.
–1
0
–7
–7.
3
2
10
20.
–1
0
–5
–10.
⎞
⎜
⎟
⎝
1⎠
,
You can confirm your eigenvectors
⎛
are correct using technology in
10
2
⎛
⎞
or
instead
⎞
of
.
⎜
⎟
⎜
⎝
5⎠
⎝
⎟
1
⎠
the examination using matrix
You
can
confirm
that
these
with
technology
multiplication. But you will not
are
eigenvectors
be able to find eigenvalues nor
associated
with
1
and
2
respectively
.
eigenvectors with technology in
the examination.
The
•
procedure
Find
the
to
diagonalize
eigenvectors
x
A
and
is:
x
1
with
eigenvalues
Form
the
matrices
P
=
(x
x
1
and
1
⎛ λ
•
λ
2
)
and
D
λ
of
A
2
0⎞
1
=
⎜
2
⎟
λ
⎝ 0
⎠
2
1
•
Write
A
=
PDP
,
the
diagonalized
form
of
2
⎛ λ
2
Now
since
D
2
⎛ λ
0⎞
1
=
0⎞ ⎛ λ
1
⎛ λ
0⎞
1
λ
⎝ 0
=
⎟
⎜
⎟ ⎜
⎠
⎝ 0
λ
λ
⎠ ⎝ 0
2
2
0⎞
1
=
⎜
A.
⎟
⎜
⎠
⎝ 0
2
2
λ
⎟
,
it
can
be
shown
that
⎠
2
2
⎛ λ
0⎞
1
n
D
=
⎜
.
⎟
2
λ
⎝ 0
⎠
2
2
Also,
1
A
=
(PDP
2
1
)
=
(PDP
1
)(PDP
1
)
=
PD(P
n
Similarly
,
it
can
be
shown
that
A
n
=
PD
1
P)
1
DP
=
2
P(DD)P
=
PD
1
P
1
P
.
You
can
apply
this
formula
n
to
find
A
very
efficiently
.
Eape 1.6.4
⎛
Diagonalize
the
matrix
A
3
2
⎞
=
7
and
⎜
⎝
hence
find
A
⎟
1
0⎠
Solution
1
⎛
A
has
eigenvectors
⎞
⎜
⎟
⎝
1⎠
2
⎛
⎞
and
,
⎜
Form
⎟
1
⎝
P
with
(x
x
1
⎛ λ
associated
=
⎠
eigenvalues
D
1
)
and
2
0⎞
=
⎜
⎝ 0
⎟
λ
⎠
2
λ
=
1
and
λ
=
2
respectively
.
1
1
⎛
Therefore
A
2
1
⎞
⎛
⎟
⎜
⎠
⎝ 0
0
⎞
⎛
⎟
⎜
2⎠
⎝
1
2
1
⎝
1
⎛
A
1
2
1
1
1
⎛
⎟
⎜
⎠
⎝ 0
0
PD
1
P
1
⎞
⎛
⎟
⎜
2⎠
⎝
1
2
⎞
⇒
⎝
7
⎛
n
=
⎠
=
⎜
A
1
⎞
A
⎟
7
Hence
n
Apply
=
⎜
7
⎞
2
1
⎛
⎞
1
1
0
⎞
⎛
1
2
⎟
1
1
⎠
⎞
=
⎜
⎜
⎟
1
⎝
1
7
⎝ 0
⎠
2
⎟
⎜
⎠
⎝
⎟
1
1
⎠
8
⎛
=
1
⎜
7
1
⎝
2
255
⎛
=
2
⎞
⎛
⎟
1
2
1
1⎠
⎜
⎟
⎝
⎠
254
⎜
⎞
⎞
⎟
127
⎝
126 ⎠
7
Finding
A
with
diagonalization
takes
two
matrix
mulitplications
7
wheras
finding
A
by
hand
would
take
six.
38
/
PRACTICE
6.
SL
P R A CT I C E
Callum
1,
saving
money
to
buy
a
digital
camera.
Q U E ST I O N S
He
PA P E R
is
QUE STIONS
GROUP
1
has
n,
an
invests
annual
p
Euros
interest
(€)
in
rate
an
of
account
4.3%,
that
pays
compounded
monthly
.
1.
A cuboid
width
length
n +
4
and
height
After
+
2 n + 1,
where
a.
a.
Show
that
the
volume
V
2n
V
3
written
as
6
years
=
of
the
cuboid
can
Hence
or
which
V
2
+ 9n
otherwise,
+
Calculate
the
find
the
value
of
n
A bowl
is
in
the
correct
2
to
a.
Determine
shape
of
a
hemisphere.
Its
Hence,
as
6.3
cm,
correct
to
1
decimal
bought
it
six
the
the
find
volume
When
David
Find
the
upper
and
lower
bounds
of
fee
p.
Give
your
answer
places.
digital
camera
for
€2000
and
later
for
€900.
rate
which
the
value
of
camera
each
year.
In
this
question,
assume
that
the
distance
from
Earth
to
Mars
space
probe
is
252.65millionkm
and
that
the
hemisphere.
the
of
maximum
the
joins
is
at
the
percentage
error
in
a
straight
distance
to
Mars
line.
Find
the
in
metres
and
hemisphere.
a
£34.
travels
of
gym
in
January
2021,
The
fee
increases
by
your
answer
in
standard
form.
the
A space
monthly
a
years
express
3.
of
place.
a.
the
account.
radius
7
.
.
value
decimal
depreciated
measured
of
his
12 495
is
radius
in
for
.
2.
€2069.99
4n
sold
=
has
be
Callum
.
he
n ∈Z
2%
probe
leaves
Earth
at
a
speed
of
every
60 000
kilometres
per
hour.
twoyears.
.
a.
Calculate
how
February
2026.
much
the
monthly
fee
will
be
Find
the
speed
of
the
space
probe
in
metres
in
per
second,
and
express
your
answer
in
the
n
form a × 10
.
Find
the
total
amount
gym
membership
David
will
pay
for
he
stays
for
exactly
7
A sequence
of
80
cylindrical
concrete
Use
planned
in
the
garden
campus.
The
height
the
are
an
radii
in
of
of
a
each
new
is
10
sequence
m
and
your
a.
0.5
Find
m
the
and
common
first
column
difference
to
have
answers
taken
in
distance
with
GROUP
to
parts
a

and
to
find
the
seconds
for
the
probe
to
travel
to
account
which
Mars.
0.05
volume
2
first
8.
term
n ∈Z
university
column
arithmetic
< 10,
columns
the
is
a
years.
time
4.
1 ≤
his
c.
if
,
Pilar
deposits
pays
a
$17,000
in
a
bank
m.
nominal
interest
rate
of
7%,
compounded
greater
yearly
.
3
than
100
m
a.
.
Find
the
total
amount
of
concrete
needed
Find
how
much
interest
Pilar
Ximena
deposited
$20,000
in
account
pays
has
earned
after
for
6
years.
3
the
5.
Erin
80
columns,
measures
to
the
cuboid-shaped
the
nearest
height,
food
m
width
container
and
as
254
length
mm,
of
65
154
mm
Using
account.
Her
a
nominal
annual
interest
rate
of
mm
%,
compounded
monthly
.
After
five
years,
the
respectively
.
total
a.
bank
a
r
and
a
these
measurements,
find
the
amount
in
Ximena’s
account
is
$24,351.19
volume
.
Find
r,
correct
to
2
decimal
places.
3
of
the
container
in
mm
,
giving
your
answer
9.
correct
to
twosignificant
Eito
is
exploring
earthquakes.
.
Write
down
your
the
mathematics
answer
to
part
a
in
the
The
magnitude
R
of
be
modelled
by
the
R
formula
log
,
1 ≤
that
length
a
< 10,
the
are
n ∈Z
measurements
correct
to
the
of
height,
nearest
determine
to
if
hold
the
2.5
container
litres
of
will
where
x
is
a
reading
which
represents
from
a
seismograph
⎞
⎟
0.001
in
,
⎠
mm
width
the
size
of
the
movement
in
the
mm,
definitely
washing
x
⎜
⎝
Earth
able
=
10
a × 10
and
earthquake
⎛
can
Given
an
form
n
c.
of
figures.
caused
by
the
earthquake.
be
liquid.
39
/
1
a.
Find
the
x
which
.
magnitude
=
52 098
An
earthquake
the
size
correct
of
to
the
the
of
an
earthquake
a.
for
mm.
has
a
Find
the
The
distance
magnitude
movement
nearest
from
approximately
the
in
the
of
8.9.
Earth,
.
Find
x,
The
the
mm.
its
managers
value
Earth
149 600 000
to
the
Sun
Find
the
distance
from
of
original
the
kitchen
equipment
after
will
decide
to
the
when
the
equipment
value.
pass
re-fit
Find
before
falls
how
the
kitchen
below
many
kitchen
40%
of
complete
is
re-fitted.
is
km.
GROUP
a.
of
sevenyears.
years
10.
value
the
Earth
to
the
3
Sun
+
14.
Three
functions
g,
f
and
h
are
defined
for
as:
n ∈Z
incm.
n
n
k
g ( n) =
.
Write
your
answer
to
a
part
in
the
∑
form
( 5.7 + 3 k ) ,
f
( n) =
0.5 + 0.7
∑(
)
and
k =1
k =1
n
,
a × 10
1 ≤
a
< 10,
n ∈Z
n
k
h ( n) =
∑(
x
Madita
2
≤
x
draws
≤
p
a
with
graph
a
scale
of
1
the
cm
function
=
1
unit
f
on
( x) = 2
each
,
9.3 × 1.1
a.
axis.
Identify
which
arithmetic
c.
The
to
distance
f (p).
Find
from
p
the
correct
Earth
to
the
Sun
to
the
nearest
to
train
for
a
in
his
is
Magzhan
plans
then
to
is
going
.
centimetre.
run
4
increase
every
km
the
first
amount
marathon.
week
he
runs
of
by
c.
Hence
write
Find
how
many
12th
week
of
Using
km
Magzhan
will
run
in
his
notation,
for
the
write
total
down
run
after
Hence
this
ratio
of
Find
the
distance
Magzhan
20weeks
How
many
complete
week
in
training,
and
A trip
of
Bicycle
is
approximately
weeks
before
2015,
of
he
training
runs
the
Ebike
number
was
Assuming
in
of
1530.
that
geometric
hires
app
performance
customers
at
42
km
must
least
hire
long.
3
Magzhan
42
km
in
Predict
first
A new
The
term
arithmetic
and
common
sequence.
function
justify
your
represents
a
geometric
answer.
the
down
the
geometric
first
term
and
common
sequence.
least
g ( n)
value
for
all
of
n >
q
for
which
q
to
a
remote
jungle
region
is
planned.
and
80
cases
must
be
transported
on
in
In
the
2025
to
a
is
analysing
cars
cases,
or
The
vans.
and
2018,
the
number
the
hired
of
predict
nearest
by
figure
hires
the
whole
the
year
café
each
organizers
Each
van
car
can
of
the
carries
carry
7
4
trip
can
people
people
and
and
of
which
the
v
is
the
Ebike
was
number
of
cars
of
vans
hired,
two
equations
that
.
Use
technology
c.
Hence
to
hired,
write
down
represent
solve
the
and
this
a
c
the
system
of
context.
system.
1624.
number
follows
number
find
a
of
van
of
costs
the
cars
number
that
more
to
of
vans
should
hire
be
than
a
and
the
hired,
given
that
car.
number.
number
of
hires
Emilia
buy
an
is
comparing
apartment
loan
for
deals.
She
wishes
to
£250,500.
2000.
opens
€17,500
in
If
number
its
city
.
bicycles
sequence,
exceeds
value
roads.
10cases.
one
16.
invest
first
training?
sharing
business
13.
the
calculate
a.
.
answer.
distance.
A marathon
a
your
has
narrow
a.
the
which
write
68people
In
down
of
and
h ( n) >
an
15.
12.
justify
training.
sigma
expression
c.
and
an
week.
e.
.
Identify
series
km
d.
a.
series
represents
He
training,
0.5
function
equal
difference
11.
)
k =1
in
the
in
the
Galoisville.
kitchen
equipment
The
owners
by
A
for
years
20
requires
at
compounded
equipment.
decreases
Bank
5%
each
a
a
10%
deposit,
nominal
monthly
.
then
interest
a
rate
Repayments
loan
of
1.3%,
are
made
month.
eachyear.
40
/
PRACTICE
Bank
B
for
years
25
requires
at
compounded
a
a
15%
deposit,
nominal
monthly
.
then
interest
a
rate
Repayments
loan
subscriptions
of
1.1%,
are
made
.
each
stating
any
Hence,
predict
sells
assumptions
the
value
on
the
thirteenth
you
make.
of
and
y
the
value
day
,
of
x
month.
Assume
a.
Denise
QUE STIONS
Calculate
the
monthly
repayments
for
each
Denise
.
Hence
find
c.
Hence
describe
the
disadvantages
Bank
A in
total
the
of
repayment
advantages
Emilia
preference
to
for
on
each
of
her
first
32
working
days,
loan.
increases
she
sells
and
that
by
at
she
the
least
sells
number
1
y
but
no
of
subscriptions
more
subscriptions
than
on
3
her
each
first
day
day
.
and
choosing
the
each
that
loan.
deal
the
from
deal
c.
from
Bank
Determine
if
Denise
subscriptions
B.
in
will
total
in
sell
her
at
least
first
32
1000
working
days.
17
.
Teodora
of
a
is
planning
triangle
ABC.
To
2
c
formula
=
a
a
rock
find
garden
side
c
in
she
the
shape
uses
2.
the
2
+ b
Alex
has
If
a
=
10
m,
b
=
7
and
m
C
=
60°,
find
value
of
A:
If
T
eodora
approximates
7
to
m
Invest
interest
figures
and
uses
this
B:
find
the
value
instead
=
m
If
a
10
is
accurate
nominal
percentage
error
in
the
value
of
Chimdi
and
b
only
=
7
m
to
the
percentage
that
invests
pays
a
a.
Find
the
$9500
nominal
compounded
how
value
many
of
her
are
exact,
but
C
=
60°
nearest
degree,
find
the
error
the
in
value
of
c
a.
in
a
savings
pays
nominal
C:
To
the
annual
interest
rate
of
5%,
years
Chimdi
investment
must
wait
rate
$9500
in
a
interest
minimum
needed
in12
Invest
of
nearest
.
Option
B
yearly
savings
account
paying
rate
of
3.8%,
a
fund
per
€,
that
guarantees
a
year.
calculate
the
one
if
year
value
he
of
Alex’s
chooses:
.
Option
C
saving
account
Write
down
rate
of
r%,
for
nominal
Nicole
to
an
expression
for
the
value
V
of
which
investment
after
n
years
if
he
chooses:
compounded
annual
double
interest
her
.
Option
A
.
Option
B
.
Option
C
money
years.
is
told
grow
by
in
his
friend
Beth
value
faster
in
that
his
Option
B
investment
than
in
2
Denise
works
for
a
car
sharing
On
her
first
day
at
work,
she
sells
Her
the
subscriptions
number
of
amount,
x,
manager
each
asks
she
working
sells
27
subscriptions
her
to
day
,
and
41
on
the
sells
by
day
.
d.
on
the
Determine
y
x,
system
of
sells
on
for
the
the
number
sixth
day
claim
is
correct
and
justify
Find
the
number
of
investment
complete
would
be
years
worth
for
which
more
in
Options
A and
B
than
in
Option
C.
thirteenth
equations,
of
and
Beth’s
answer.
Find,
in
terms
to
the
subscriptions
for
the
nearest
interest
Alex
€,
the
will
difference
earn
after
20
in
the
years
by
of
choosing
and
if
your
total
a
are
sixth
e.
down
periods
the
workingday
.
Write
compounding
increase
both
working
the
frequent.
Alex’s
Denise
since
y
c.
subscriptions.
A,
subscription
more
scheme.
a.
compounded
doubles.
Option
same
in
after
A
will
1.
a
interest
€1100
Option
Alex
PA P E R
4%,
until
monthly
.
the
an
monthly
.
Alex’s
Find
paying
account
.
invests
.
account
quarterly
.
Nicole
a
savings
of
in
annual
investment
18.
a
c
payment
maximum
three
of
Option
c.
in
rate
Invest
compounded
7 ,
considers
three
a
significant
He
c
Option
.
invest.
the
annual
exact
to
options:
2 ab cos C
Option
a.
€21,000
option
A instead
of
Option
C.
Denise
number
of
41
/
1
6.
HL
P R A CT I C E
Simplify
these
algebraic
expressions:
Q U E ST I O N S
3
14
10
2
3
a.
PA P E R
1,
GROUP
9x
× 375 x
×
(2x
)
1
1
12
⎛
1.
a.
Find
the
eigenvalues
and
corresponding
729 x
.
⎜
eigenvectors
of
the
2
4
⎟
3
⎝
⎛
⎠
64 y
⎞
C =
matrix
3
⎜
1
⎟
⎝ 1.5
( 5y
1⎠
2
x
)
c.
5
2
.
Hence,
write
down
matrices
3
⎞
P
and
D
such
( 3y
that
4
x
)
1
C
=
PDP
n
c.
Find
a
general
expression
for C
in
terms
of
GROUP
2
7
.
ω
n
⎛
2.
The
magnitude
Richter
scale
⎛
is
R
of
an
earthquake
measured
by
the
on
Given
=
2 cis
1
the
expressions
formula
in
π
⎞
⎜
⎟
⎝
⎠
4
−π
⎛
ω
and
=
3cis
2
⎜
Euler
form
for
the
⎞
⎟
⎝
,
find
⎠
6
following:
⎞
I
R
=
log
⎜
where
⎟
I
is
the
intensity
of
a.
the
ω
ω
1
10
⎜
⎝
2
⎟
I
⎠
0
4
⎛ ω
earthquake
and
I
is
the
intensity
of
⎞
1
ground
0
.
movement
on
a
normal
⎜
⎟
⎝ ω
day
.
⎠
2
3
*
a.
The
San
1906
the
Francisco,
measured
8.25
earthquake
measured
how
5.9
many
Francisco
Dashur
California
on
near
on
Richter
Dashur,
the
times
the
Richter
more
earthquake
earthquake
Egypt
scale.
intense
was
scale,
in
the
in
of
(ω
c.
1
and
8.
1992
Two
voltage
The
a
Calculate
near
circuit
If
San
to
V
an
the
north-east
so
San
7.08
Francisco
times
more
earthquake
intense
of
1906.
the
are
connected
expression
V
=
A
=
in
2
total
voltage
is V
= V
+ V
V
and
=
.
2
3sin ( 20t + 13 ),
find
2
for
the
total
voltage
in
the
form
Asin ( 20t + B )
2
2
⎞
⎛
,
0
5
⎞
than
B =
⎟
⎝ 3
was
that
7 sin ( 20t + 5 )
=
⎜
2011
V
⎛
0
π
3
0 ⎠
⎞
coast
9.
in
and
1
comparison
Japan’s
V
1
earthquake.
earthquake
sources
1
⎛
.
)
,
⎜
0 ⎠
⎝ 2
C =
,
⎟
⎜
1⎠
⎝
and
⎟
the
Calculate
0
4
⎛
⎞
D =
the
measure
Richter
of
the
Japanese
earthquake
on
a.
Show
3 ⎠
scale.
that
written
3
⎝
a.
3.
⎟
⎜
the
the
number
0.151 515 151 515
can
Find
the
2 × 2
matrix
X
that
satisfies
the
be
equation
XA + XB
=
Solve
equation
C
3D.
as
.
the
BXA
=
D,
expressing
all
6
2
∑
2 n
0.15 × 10
⎛
answers
r =1
as
matrices
in
the
form
a
b
,
⎜
⎟
⎝ c
.
Show
that
the
number
0.51
can
be
written
⎞
q
d⎠
as
where
q ∈Q
and
a, b, c , d
∈R
are
all
exact
∞
r
∑
1
0.51 × ( 0.01)
values.
r =1
x
10.
c.
Hence
find
the
exact
value
of
Solve
the
a.
=
equation
4
series
has
common
ratio
1.5
(2
)
0.51
z
4.
A geometric
Find
the
values
of
x
for
which
the
sum
to
7i
z + 2i
infinity
5.
a.
Find
the
exact
values
of
the
roots
of
the
series
exists.
of
.
If
the
first
term
of
the
sequence
is
13,
2
3x
.
+
Find
x
7
the
=
0
exact
values
of
the
roots
of
and
the
sum
to
find
the
value
log
( a + 9)
infinity
of
of
the
series
is
16,
x
2
3x
+
x + 7
=
0
11.
log
a,
4
c.
V
erify
that
below
is
for
each
equation,
the
conjecture
terms
of
an
and
log
4
( a + 20 )
arithmetic
are
the
first
sequence.
true:
roots
of
ax
a.
Find
.
Hence,
the
value
of
a
b
2
The
three
4
+ bx + c
=
0
add
to
a
the
find
the
sum
of
the
first
20
terms
of
sequence.
42
/
PRACTICE
π
⎛
12.
Consider ω
=
cis
⎜
⎞
⎟
⎝
3
The
proposed
Given
that
⎜
w
in
Cartesian
1
⎞
3
⎛
=
⎟
⎝
and sin
⎠
π
⎜
⎝
2
=
⎟
3
are
represented
,
⎠
and
⎛
in
=
⎜
draw
modulus-
2π
⎛
cos
⎜
⎞
⎠
3
Hence,
2π
⎛
sin
and
⎟
⎝
form.
find
exact
values
Express
each
2
{1, ω , ω
4
a.
of
the
P:
p
1
2
3
4
2
1
1
0⎠
⎞
⎟
⎟
⎝
Find
the
matrix
RP
and
interpret
p
p
the
elements
,ω
}
in
the
the
points
matrix.
set
5
,ω
matrix
⎠
3
element
3
,ω
points
⎟
⎝
of
c.
the
for
⎞
⎜
the
⎜
lose
argument
in
p
win
form
to
express
form.
Cartesian
,
2
2
P
in
Find
p
3
⎞
2
.
and
p
1
awarded
a.
changes,
⎠
π
⎛
cos
QUE STIONS
The
modulus-argument
proposed
awarded
form.
changes,
have
and
1
sponsorship
2
,
to
payment
plans
s
and
1
s
respectively
,
based
on
the
number
of
points
in
T
,
2
d.
Hence,
with
reference
to
an
Argand
scored.
diagram
or
otherwise,
2
1 + ω
3
+ ω
4
+ ω
show
that
5
+ ω
+ ω
=
Matrix
0
E
shows
the
amount
£
that
1
would
T
like
to
win
under
T
and
2
sponsorship
plans
3
GROUP
s
3
and
s
1
2
s
s
1
13.
a.
Given
that
( ax + b ) = 2 + 2 log
log
3
( ax
b ),
3
T
⎛
1
the
laws
of
2
use
logarithms
to
find
p,
q
and
12 , 000
15, 600
10, 000
20, 000
⎞
in
r
E
=
⎜
T
⎟
2
terms
a
of
and
b
in
the
quadratic
equation
⎜
T
2
⎟
⎝ 8000
14, 000 ⎠
3
px
+ qx + c
=
0
The
.
Find
the
exact
values
of
r
and
s
if
the
sponsorship
awarded
function
f (x)
is
payments
for
each
point
piecewise
under
changes
p
and
p
1
continuous:
by
the
matrix
are
represented
2
S
sx
⎧
e
x
< 1
s
s
1
⎪
2
3
f
( x) =
3 + 2x
⎨
1 ≤
x
<
2
p
⎛
1
⎪
ln ( x
r )
x
≥
Find
the
2 x + 1
S
=
4
=
solution
of
the
equation
100 ⎠
.
Solve
the
equation
XS =
E
for
a
3 × 2
matrix
X
ln a
x
3
⎟
⎝ 40
p
⎞
2
exact
1
65
⎜
2
⎩
c.
70
.
Express
the
answer
in
the
form
,
and
interpret
its
elements
in
context.
ln b
where
a,
b ∈Q
16.
14.
a.
Given
z
=
2 +
and
i
ω
=
2
2 i,
find
Solve
the
1 + 2 log
y
=
log
10
in
Cartesian
equations
zw
z
and
simultaneous
10 x
10
form.
w
1 + log
x
=
log
10
.
Express
z
ω
and
in
Euler
form.
Hence,
find
10
(7
20 y )
zw
z
in
and
Euler
17
.
form.
Solve
the
simultaneous
equations
w
2
⎛
c.
By
comparing
the
two
forms
of
zw,
find
the
y
z
=
⎜
⎛
exact
value
cos
of
⎝
π
3
⎟
x
⎠
⎞
⎜
⎝
⎞
ln
⎟
12
⎠
2
15.
The
organizers
investigating
are
of
a
a
sports
proposed
league
change
ln
(x
ln
(
3
) + ln ( y
) =
10
are
to
how
points
xyz
) =
1
awarded.
18.
Three
teams
T
,
T
1
(win,
draw
or
and
have
T
2
their
four
represented
win
draw
in
sketch
below
shows
the
roads
connecting
results
3
lose)
The
the
matrix
villages
A,
B,
C
and
D
R:
lose
D
T
⎛
8
3
1
4
5
3
B
⎞
1
A
⎟
⎜
R
=
T
2
T
3
⎟
⎜
⎜
⎝
2
4
6
⎟
C
⎠
43
/
1
The
matrix
journeys
P
shows
between
the
the
number
four
of
c.
one-stage
villages:
more
to
A
⎛
A
B
C
D
0
1
1
0
=
from
B
1
0
1
d.
1
in
sector
If,
at
C
...
⎜
...
...
...
return
the
...
⎝
matrix
from
its
hours
initial
Zappy
is
position
first
and
is
D.
13
hours,
directly
to
thecomponents
⎟
...
...
...
Zappy
is
programmed
to
its
starting
position,
find
of
the
vector
that
represents
⎠
this
Complete
=
km
many
⎟
⎜
a.
t
50
how
⎟
⎜
D
⎞
than
after
⎟
⎜
P
Determine
journey
.
P
e.
A second
UA
V
named
Alpha
takes
off
at
the
2
.
Find
P
same
time
as
Zappy.
The
position
of
Alpha
after
2t
π
i
c.
Find
to
B
the
number
of
two-stage
and
identify
this
number
journeys
in
the
from
C
t
elements
hours
is
engineer
given
Nico
(
by a(t) =
states
that
3
1.51e
Alpha
)
.
and
Flight
Zappy
2
of
P
are
2
d.
Interpret
the
elements
of
in
meaning
2
2.
1.
An
Unmanned
Aerial
large
forested
area,
that
a
near
they
miss
will
be
after
within
V
ehicle
(UA
V)
is
other.
Flight
engineer
Determine
which
flight
6
hours
100
m
–
of
The
divided
into
BetaGamma
Zank
disagrees.
engineer
Information
is
correct.
Technology
store
surveying
has
a
of
P
each
PA P E R
danger
sectors
two
megastores,
one
in
Cairo,
Egypt
and
the
as
other
in
Toulouse,
three
types
France.
Each
megastore
has
shown.
salaries,
one
financial
Sector
B
employee
for
year
each
of
in
desk
earn
the
sales
room
C
Sector
the
work
in
$900
the
number
of
three
employee.
employees
after
t
hours
of
per
earn
week,
$700
office
earn
following
workers
who
different
For
the
work
on
per
those
week
$500
per
working
and
those
week.
the
UA
V
at
each
site
is
given
in
table:
IT
z
of
have
D
The
position
type
2020,
IT
who
The
who
Sector A
the
Sector
of
desk
Sales
room
Office
named
t
π
Cairo
52
15
19
Toulouse
34
9
12
i
Zappy
is
given
Distances
are
by
z(t) =
(
measured
3
1.5e
in
)
km.
a.
a.
Find
.
Hence,
the
values
z( 0) ,
of
z ( 1) ,
z(2),
z( 3) ,
z( 4)
Construct
represent
type
t
=
0,
1,
sketch
2,
3
the
and
positions
4
hours
on
of
Zappy
this
in
and
the
each
label
a
number
3 × 2
of
matrix
E
to
employees
of
each
megastore.
at
Argand
.
.
diagram:
What
would
the
matrix S
=
( 900
700
500 )
represent?
Im
.
What
c.
If
d.
Describe
e.
Find
SE
must
Re
are
=
( 66
the
the
800
what
total
pay
in
megastores
dimensions
x ),
the
calculate
value
amount
one
in
year
the
of
to
of
x
of
S?
the
financial
of
x
represents.
money
its
value
that BetaGamma
employees
year
of
in
both
2020.
44
/
PRACTICE
The
managers
year
after
of
2020,
BetaGamma
all
salaries
predict
will
that
increase
each
by
Scientists
1%
their
begin
initial
measuring
data
p
is
=
the
20,
y
0
and
one
of
each
type
of
employee
will
retire
f.
population
=
15
and
a
0
and
=
40
0
in
.
each
QUE STIONS
Find
the
number
year
later.
of
seals
in
each
group
one
megastore.
Predict
the
total
BetaGamma
amount
must
pay
in
of
money
one
year
that
to
in
both
megastores
in
of
answers
to
the
nearest
its
the
Explain
why
the
population
after
3
years
can
financial
be
year
your
integer.
c.
employees
Give
found
from:
2022.
3
⎛
⎞
7
7
6
6
0
⎟
⎜
PA P E R
3
⎛
⎞
20
⎟
⎜
⎜
0.5
⎜
1.
The
aim
of
this
question
is
to
explore
the
of
a
colony
of
seals
using
female
population
of
a
matrices.
colony
2
2
of
can
⎟
⎟
40
⎝
0
⎜
seals
⎟
15
⎟
⎜
⎠
⎟
3
⎝
The
0
long-term
⎜
population
0
3
⎠
be
Let
T
be
the
total
female
population
after
n
n
divided
into
three
groups:
pups,
young
seals
and
years.
adults.
are
Pups
those
seal
are
aged
over
2
seals
1-2
years
aged
years
0-1
and
years,
an
young
seals
seal
any
adult
is
d.
Use
an
extension
of
the
above
result
to
find:
old.
.
T
20
If,
in
year
n,
the
population
of
female
pups
is
,
p
n
.
T
30
female
young
seals
is
and
y
female
adults
is
a
n
,
n
It
then
be
the
population
found
of
females
in
year
n
+
1
is
given
that
for
n ≥
20
the
sequence
T
is
n
can
approximately
geometric.
e.
answer
from:
⎛
7
7
6
6
⎞
.
Use
your
to
part
d
to
find
the
T
for
0
⎛
⎟ ⎛
⎜
⎞
p
n+ 1
⎟
⎜
=
y
n+ 1
⎜
⎜
⎟
n+ 1
0
⎜
⎟
a
⎝
0.5
⎜
⎠
0
2
⎟
⎟
2
0
⎜
⎟
3
⎝
⎞
p
common
ratio
for
the
sequence
n
⎟
⎜
3
n ≥
20
n
⎟
⎜
y
n
⎜
⎜
⎟
a
n
⎝
.
Hence,
find
when
the
total
female
⎟
population
⎠
will
first
pass
50
000.
⎠
The
vector
q
gives
the
proportion
of
the
total
n
⎛
7
7
⎞
0
6
6
0
0
matrix
L
=
⎜
in
0.5
⎜
2
2
is
⎟
called
the
f.
vector
average
the
3
3
population.
change
of
the
groups
in
the
in
year
n
q
30
⎟
is
given
that
each
element
of
q
the
It
⎠
is
is
30
⎟
⎝
for
Find
Leslie
It
0
⎜
matrix
each
⎟
⎜
The
population
⎟
⎜
approximately
used
number
of
to
model
females
in
the
element
in
q,
equal
an
to
the
corresponding
eigenvector
for
L
a
.
.
Use
your
answer
to
part
f
and
the
relation
population.
Lq
a.
By
considering
.
the
average
young
.
the
the
entries
number
of
in
L,
write
young
down:
born
to
a
λ q to
find
λ
the
corresponding
eigenvalue.
.
Comment
on
your
answer.
female
probability
into
.
the
=
the
young
fraction
of
a
pup
seals
survives
and
moves
group
adults
that
die
each
year.
45
/
F U N CT I O N S
2
2 . 1
L I N E S
A N D
F U N C T I O N S
You should know:
✔
the
gradient
steepness
✔
that
if
and
parallel
two
their
of
lines
a
You should be able to:
straight
line
expresses
its
✔
direction
lines
are
gradients
have
context
equal
gradients
perpendicular
is
the
and
product
that
✔
of
the
denition
of
the
and
rearrange
of
a
gradient
interpret
any
straight
of
the
line
of
a
its
straight
in
a
meaning
three
into
line
any
forms
of
the
equation
other
–1
✔
✔
calculate
a
function
and
of
domain
nd
the
equation
of
a
straight
line
from
a
context
andrange
✔
✔
an
of
inverse
a
function
reverses
or
undoes
the
effect
nd
the
x-intercept
straight
line
from
and
its
the
y-intercept
of
a
equation
function
✔
represent
a
equation
as
function
as
a
table,
graph
or
1
✔
that
f
and
exists
(x)
denotes
only
if
the
f (x)
inverse
is
function
of
f (x),
appropriate
one-to-one
✔
use
technology
to
determine
the
range
of
a
1
✔
solving
✔
the
f (x)
=
a
is
equivalent
to
nding
f
function
(a)
from
its
graph
1
graph
graph
of
y
of
=
y
=
f
f (x)
1
(x)
in
is
the
found
line
y
=
by
reecting
✔
the
x
sketch
the
graph
of
apply
y
graph
y
=
f (x)
of
in
y
=
the
f
(x)
line
by
y
=
reecting
the
x
1
✔
the
domain
of
f
(x)
is
equal
to
the
range
of
f (x).
✔
=
equation
f (x)
to
technology
✔
apply
the
y
=
replacing
predict
if
f (x)
equation
using
by
the
an
value
x
value
of
y,
in
the
using
appropriate
by
to
replacing
predict
technology
or
the
the
a
y
value
value
inverse
of
in
x,
function
if
appropriate
1
✔
Steepness
of
any
is
a
apply
simple
modern
yet
building
and
interpret
important
with
stairs,
y
=
f
quantity
to
escalators
y
of
steepness.
The
formula
m
segment
[AB]
with
endpoints
A(x
,
1
y
When using the formula m =
2
ramps
must
finds
the
gradient
, y
of
1
y
)
1
and
B(x
,
2
y
).
2
1
to find the gradient of the line segment [AB]
x
1
, y
1
2
include
1
with end points A and B, it is not impor tant which point you call (x
which you call (x
design
y
2
x
or
The
x
2
line
context.
measure.
=
x
the
in
y
2
calculations
(x)
) and
1
). The gradient will be the same, provided you apply the
2
formula correctly.
46
/
2 .1
•
Any
vertial
line
segment
gradient
sine
the
[BL]
gradient
-
suh
denominator
as
of
[GM]-
the
has
no
formula
defined
is
M
has
m
zero.
=
3,
showing
that
[BL]
is
•
Any
whih
horizontal
has
(0, 5)
1
gradient
line
segment
m
-
1
2
1
1
8
4
4
as
[DK]
=
=
suh
sine
3
>
4
-
has
gradient
B
0
sine
the
numerator
(2, 6)
steeper
G
[CI]
of
the
formula
is
I
(1, 3)
Informally
,
[EJ]
and
[FQ]
hene
negative
“go
down”
from
left
to
right
(4, 1)
and
D
•
[EJ]
have
has
gradient
dereases
inreases
is
on
in
–1,
a
1
showing
:
1
ratio
[FQ]
has
that
with
gradient
m
the
amount
y
whih
the
amount
y
whih x
=
−5,
showing
that
it
is
the
F
Two
of
all
important
the
line
ways
to
(1, –5)
J
1
segments
ompare
gradients
m
and
m
are:
2
Q
If
m
=
m
1
•
If
then
the
lines
=
then
are
(5, –5)
shown.
1
•
(2, –2)
( −5 )
−
=
1.5
steepest
(9.19, 0)
[EJ].
7.5
•
K
(5, 0)
gradient.
E
y
(8, 2)
zero.
C
•
f unctions
3
=
2
than
and
(0, 7)
L
6
•
Lin E s
(1.5, –7.5)
parallel
2
m
×
m
1
−1
the
lines
are
perpendiular
2
Exle 2.1.1
The
diagram
B,
and
C
B
F
D
elow
are
shows
given.
(3, 5)
E
E
is
part
the
(6, 5)
C
of
a
plan
midpoint
for
of
a
setion
[BC]
and
(a)
(9, 5)
()
(2, 3.5)
F
of
is
ridge
the
the
[AD]
and
The
struture.
midpoint
Find
e
A
a
of
engineers
parallel,
oordinates
of
points A,
[AB].
oordinates
[DC]
The
of
G
and
of
H,
the
midpoints
of
respetively
.
predit
and
that
that
[FH]
[FE]
and
and
[EG]
[GH]
will
will
e
(1, 2)
H
perpendiular.
Determine
orret.
your
Justify
if
either
predition
is
reasoning.
G
D
(8, –2)
Solution
⎛
⎝
(a)
G
=
1 + 8
2 + (
2)
,
⎜
⎝
⎞
=
⎟
2
(4.5,
You
0)
apply
the
formula
for
the
oordinates
of
the
⎠
2
midpoint
of
a
line
segment
with
endpoints
(x
,
y
1
⎛
H
=
5 + (
9 + 8
,
⎜
⎝
2)
=
⎟
(8.5,
and
[FE]
has
0
say
that
1.5
=
2
1.5
gradient
[FE]
,
y
1.5
8.5
and
[GH]
whih
is
prior
learning:
+
x
1
⎝
y
x
2
+
1
2
y
2
,
⎜
⎞
⎟
2
⎠
and
[GH]
For
final
answers,
of
fration
the
numerator
and
denominator
8
a
should
e
integer
values.
3
=
4
),
2
3
=
4
=
4.5
to
3.5
gradient
6
has
(x
2
5
()
⎛
1.5)
⎠
2
2
)
1
⎞
so
it
is
true
8
are
parallel.
47
/
2
3.5
[FH]
has
=
8.5
5
has
5
=
6
4.5
line
segments
the
13
10
angle
this
etween
[FH]
and
[EG]
are
1.5
not
perpendiular
and
in
fat
alulation
them
proves
is
very
that
the
lose
to
seond
90°,
ut
predition
=
3
is
false.
Note
unneessary
So
look
and
6.5
0
gradient
The
4
=
2
[EG]
2
1.5
gradient
that
the
deimal
result
is
determined
approximations
of
without
the
perpendiular
4
eause
the
produt
of
their
frations.
gradients
The
exat
value
of
is
more
onvenient
13
40
10
4
is
×
=
≠
and
−1
easier
to
use
than
its
deimal
approximation
39
3
13
–0.307 692…
Geometri
properties
studied
Eulid
suh
as
parallel
and
perpendiular
lines
were
aee 
René
y
Desartes
in
300
made
bc.
the
But
it
was
powerful
in
link
17th
entury
etween
Frane
geometry
that
and
algera
It is easier and more accurate to
for
the
first
time.
This
means
that
you
an
apply
your
knowledge
use exact values when working
andunderstanding
of
algera
to
solve
prolems
in
geometry
–
and
with gradients.
with
graphs.
y
y
2
For
example,
the
formula
m
1
=
for
x
2
with
endpoints
(x
,
y
1
y
y
x
x
)
and
(x
1
,
2
the
gradient
of
the
line
segment
x
1
y
),
an
e
adapted
to
the
equation
2
1
m
aee 
=
,
whih
is
satisfied
y
)
y
all
points
(x,
y)
whih
lie
on
a
straight
1
line
through
(x
,
1
with
gradient
m.
1
When applying the point-gradient
This
an
e
written
as
y
−
y
form, you can choose any point
equation
(x
, y
1
=
m(x
x
1
of
a
straight
line.
),
the
point-gradient
form
of
the
1
When
this
is
rearranged
to
the
form
) on the line. The answer will
1
y
=
mx
+
c,
this
form
is
alled
the
gradient-intercept
form
eause
the
be the same.
oordinates
of
the
y-interept
an
e
written
down
as
(0,
c).
Exle 2.1.2
A solar
from
eletriity
where
installed
eah
at
unit
power
eletriity
points
with
A(2,
eah
station
supply
4)
and
supply
is
planned
ales
B(8,
run
1).
on
due
a
desert
north
A straight
salt
and
line
flat.
east
ale
The
from O
passing
entrane
as
to
shown.
through
A
the
Two
and
B
salt
flat
solar
is
is
at O(0,
panel
planned
units
to
0)
are
onnet
ale.
North
4
A
(2, 4)
Key:
salt
at
2
cable
B
(8, 1)
East
O
=
(0, 0)
–2
48
/
2 .1
(a)
Find
()
Hene
find
()
Hene
write
(d)
the
Two
gradient
the
one
solar
of
and
f unctions
[AB]
equation
down
further
exatly
of
Lin E s
the
the
straight
oordinates
panel
these
of
units
are
pointslies
of
line
the
passing
axis
planned
on
l
to
through
interepts
e
at
points
A
and
B
in
the
form
y
=
mx
+
c
of l.
points C
(0.9,
4.65)
and
D
(6.4,
1.8).
Show
that
l
Solution
4
(a)
[AB]
has
1
gradient
3
1
=
2
Write
=
8
6
the
gradient
in
its
simplest
form.
2
1
()
l
has
equation
y
1
=
(x
Apply
8),
the
point-gradient
form
sine
you
know
the
2
gradient
whih
an
e
simplified
and
the
oordinates
of
two
points
on
l
to
1
y
=
x
+
5
2
()
The
y-interept
of
l
is
(0,
5).
The
The
x-interept
question
asks
expliitly
for
the
oordinates.
1
is
found
y
solving
x
+
5
=
0,
whih
2
gives
x
=
10,
x-interept
so
of
l
the
are
oordinates
(10,
of
the
0)
1
(d)
For
C,
y
=
(0.9)
+
5
=
4.55
≠
4.65
For
a
point
to
e
the
y-oordinate
on
l,
oth
the
x-oordinate
and
2
hene
C
is
y
=
not
on
l.
(6.4)
+
must
satisfy
the
equation.
1
For
D,
5
=
1.8,
2
hene
A third
D
form
is
of
on
the
l
equation
of
a
straight
line
is
the general
form
aee 
ax
+
by
+
d
espeially
=
0.
Often,
if
a,
b
and
this
d
are
is
a
onvenient
integers.
For
way
to
example,
express
the
the
equation,
equation
for
l,
Your answer to the question “Find
1
whih
is
y
x
=
+
5,
an
e
written
as
2y
=
−x
+
10,
hene
l
is
point
is
on
l
if
the equation of the line l” should
2
x
+
2y
10
=
oordinates
0
in
general
satisfy
the
form.
Again,
a
only
never begin “l = ”. This is incorrect
its
use of notation because l is not a
equation.
variable. You write “l has equation”
For
D(6.4,
1.8),
6.4
+
2(1.8)
10
=
6.4
+
3.6
10
=
0,
onfirming
that
and then write the equation in one
D
is
on
l.
of the three forms.
ce   
The
onept
For
example,
etween
printed
the
in
of
a
funtion
Aaia
ollets
numer
Courier
Name
gives
of
the
vowels,
New
you
font
a
first
the
(size
powerful
way
of
students
numer
of
letters
36).
Aaia
Length
6
3
4.6
10
6
7.6
Barbora
7
3
5.4
Catarina
8
4
6.1
Nicole
6
3
4.6
Carolina
8
4
6.1
in
and
presents
V
owels
Anastasiia
ommuniate
names
Letters
Adrian
to
her
the
her
lass.
She
length
data
in
Name
relationships
to
this
wants
the
to
and
make
know
nearest
mm
if
of
preditions.
relationships
eah
name
exist
when
tale:
Letters
V
owels
Length
Jingyuan
8
3
6.1
Steven
6
2
4.6
Rea
3
2
2.2
Zank
4
1
3.0
Madita
6
3
4.6
Acacia
6
4
4.6
49
/
2
Aaia
the
sees
name,
numers
draws
from
the
of
letters
in
Number
tale
longer
mapping
differenes
the
the
and
the
name
in
numers
diagrams
the
that
to
greater
m.
of
The
the
of
relationship
vowels
explore
numer
is
not
further,
so
and
letters
etween
lear,
starts
so
to
in
the
she
pereive
the
relationships:
of
Number
letters
of
Number
vowels
of
Length
(cm)
letters
1dp
3
2.2
4
3.0
6
4.6
7
5.4
8
6.1
10
7.6
3
1
4
2
6
3
7
4
8
6
10
Aaia
1.3
1.4
letters
1.5
an
see
now
that
the
relationship
etween
the
numer
of
letters
DEG
and
the
length
is
one-to-one
eause
eah
different
numer
of
letters
Graph A
5.5
maps
to
exatly
one
length.
slewoV
4.0
It
was
Desartes
forming
a
who
oordinate
had
the
system,
idea
to
adapt
drawing
the
mapping
perpendiular
diagram
axes
and
y
plotting
2.5
1.0
3
4
5
6
7
8
9
points
to
graph.
Aaia
Graph
B
have
1.4
letters
1.5
Graph
the
uses
relative
her
sizes
GDC
to
of
eah
reate
variale
on
a
grid.
This
forms
a
graphs.
10
Letters
1.3
show
DEG
This
the
the
graph
same
means
different
B
is
of
a
function
x-oordinate.
that
points
the
Graph
variales
with
the
eause
are
same
A is
the
related,
no
two
graph
ut
x-oordinate.
different
of
there
The
a
relation.
are
set
points
of
at
least
two
x-oordinates
7.7
htgneL
5.5
(representing
the
D
8,
=
{3,
4,
6,
7,
set
numer
10}
word).
The
of
Aaia
researhes
of
the
of
letters
funtion
y-oordinates
in
L
{2.2,
the
word)
(where
3.0,
4.6,
L
is
the domain
represents
5.4,
6.1,
7.6}
length
is
of
the range
of
L
4.0
the
ontext.
She
finds
that
Courier
New
is
a
2.5
monospaced
3
4
5
6
7
8
9
font,
meaning
that
eah
harater
oupies
the
same
10
horizontal
spae.
Hene
L
is
diretly
proportional
to
n.
Aaia
applies
Letters
the
fat
that
funtion
1.3
1.4
*letters
1.5
L(n)
name
=
with
0.76n,
10
letters
has
length
7.6
L(n)
the
length
of
where
is
m
a
to
write
name
with
down
n
the
letters.
DEG
This
Graph
a
funtion
is
defined
4,
6,
for
all
real
numers,
however
the
domain
B
7.0
of
Lis
an
D
=
{3,
nowpredit
7,
the
8,
10}
in
length
the
ontext
using
L
as
a
of
Aaia’s
lass.
mathematical
Aaia
model
of
the
htgneL
5.5
ontext
f1(x):=0.76
she
is
exploring:
that
is,
she
uses
mathematics
to
represent
the
x
4.0
real-world
context.
2.5
3
4
5
6
7
8
9
10
A function f gives a unique value y = f (x) for each x in the domain of f.
Letters
This enables you to make predictions. x is the independent variable
and y the dependent variable.
aee 
The domain of a function will be the largest possible domain for which the
function is defined, unless otherwise stated.
50
/
2 .1
Lin E s
and
f unctions
Exle 2.1.3
Aaia
first
widens
names
short
as
to
one
her
a
researh
gloal
letter
of
the
ontext,
and
as
long
lengths
finding
as
20.
(ii)
of
names
She
ontext
with
G(n)
=
0.76n
whih
down
possile
as
the
length
of
the
longest
name.
models
()
this
Write
gives
Predit
the
length
of
the
name
the
Oluwapamilerinayo
length
in
m
Courier
of
New
a
name
with
font
(size
{1
n
n
letters
printed
in
()
36).
A silver
ompetition
winners’
The
domain
is
≤
≤
20|n
∈
N},
meaning
that
restrited
to
a
natural
(i)
Write
down
the
e
spaes
engraved.
for
The
length
the
spae
is
limited
to
10.8
m.
What
is
the
numer.
maximum
(a)
to
has
n
of
is
names
up
range
of
an
G.
numer
ontain
Courier
if
the
New
of
letters
engraving
font
(size
that
the
spae
mahine
uses
36)?
Solution
(a)
(i)
The
range
{0.76n|1
(ii)
L(20)
=
of
≤
n
15.2
G
≤
The
is
20,
n
∈
funtion
N}
You
m
in
()
The
is
()
name
L(17)
L(n)
=
=
has
17
12.92
10.8
has
range
letters,
hene
the
of
is
the
set
of
“outputs”
of
the
L
replae
the
G
the
equation
value
to
of
find
the
the
independent
value
of
the
variale
funtion.
length
m
14.2
solution
is
not
in
the
domain
of
L
ut
you
interpret
this
10.8
numer
n
=
≈
in
the
ontext
of
the
prolem.
14.2
0.76
The
maximum
ontain
With
every
is
numer
of
letters
the
spae
an
14.
operation
you
learn
in
mathematis,
it
is
extremely
useful
ne
to
know
how
addition
and
to
find
its
inverse.
sutration
are
You
learned
opposite
at
an
operations,
early
as
age
are
that
multipliation
Always use correct notation for
and
division.
When
you
make
an
error
on
a
omputer,
Ctrl+Z
will
the equation of a function.
“undo”
your
error.
Similarly
,
inverse
functions
reverse
or
undo
For
the
example, L(n) = 10.8 is correct
effet
of
a
funtion.
but L = 10.8 is not.
1
The inverse of the function f (x) is denoted f
(x). Do not confuse this with the
aee 
1
1
=
reciprocal of f (x), which is ( f (x))
.
f (x)
In internal assessment, use of
a convenient letter to denote
You
an
apply
Butimagine
nameof
L
to
you
length
find
were
7.6
the
length
asked
for
instead
a
to
given
find
numer
the
of
numer
a function can help your
letters.
of
letters
in
a
mathematical communication –
for example v(t) for velocity in
m.
terms of time, or C(n) for the cost
When
you
write
L(n)
=
7.6,
this
is
equivalent
to
of manufacturing n products.
7.6
0.76n
=
7.6
⇒
n
=
=
10
letters.
0.76
1
This
shows
that
the
expressions
L(10)
=
7.6
and
10
=
L
(7.6)
are
1.4
1.5
1.6
letters
RAD
1
equivalent,
where
L
is
the
inverse
funtion
of
y
L.
f4(x)=x
(7.6, 10.)
Therefore,
(10,
7.6)
is
on
the
graph
of
L
and
(7.6,
10)
would
e
on
the
x
1
graph
of
L
as
shown.
These
points
are
refletions
of
eah
other
in
the
f3(x)=
(10, 7.6)
0.76
line
y
=
x.
f1(x)=0.76
x
x
51
/
2
In
general,
of
L
if
(n,
L (n))
is
on
the
graph
of
L,
then
(L (n),
n)
is
on
the
graph
in
the
line
1
1
Hene,
y
=
the
graphs
of
L
and
L
are
refletions
of
eah
other
x.
Exle 2.1.4
Part
of
the
graph
of
f (x)
=
x
3
is
+ 2
shown
elow.
y
(a)
Use
the
graph
to
and
the
range
is
explain
why
the
domain
of
f
is
x
≥
3
12
f (x)
≥
2
11
1
10
()
Draw
the
()
Hene
or
graph
of
f
on
the
grid.
9
otherwise
determine
the
domain
and
the
8
1
range
of
f
7
6
5
4
3
2
1
x
0
0
1
2
3
4
5
6
7
8
9
10
11
12
Solution
(a)
f
is
only
end
defined
point
ontinues
window,
(3,
to
2)
for
on
x
the
inrease
hene
≥
the
3,
as
shown
graph
of
outside
range
is
f.
The
the
f (x)
y
Predit
the
graph
viewing
given
≥
hek
2.
than
The
()
what
the
funtion
window.
your
zero
You
an
predition.
for
f
to
ommand
have
term
like
hange
Logially
,
a
is
looks
real
outside
the
x
3
the
window
must
e
to
greater
value.
“Draw”,
so
points
must
e
y
12
plotted
aurately
and
joined
up
with
a
smooth
urve.
(5, 12)
11
10
9
1
f
(x)
8
(4, 7)
7
6
f(x) =
x
–
3
+
2
5
(3, 4)
(12, 5)
4
(7, 4)
3
(4, 3)
(2, 3)
2
(3, 2)
1
x
0
0
1
2
3
4
5
6
7
8
9
10
11
12
1
()
The
domain
of
f
is
x
≥
2
and
the
range
This
an
e
seen
from
the
graph
or
written
1
is
f
down
1
(x)
≥
3.
diretly
from
(a)
sine
the
domain
of
f
is
always
1
therange
domain
Note
that
if
f (x)
y
=
would
not
a
funtion,
x
f (x)
was
e
must
of
not
the
of
f,
the
range
of
f
is
always
the
f
one-to-one
graph
e
and
of
a
then
its
funtion.
refletion
For
the
in
the
inverse
line
of f (x)
to
e
one-to-one.
1
The domain of f
(x) is always equal to the range of f (x).
52
/
2.2
2 . 2
compositE
f unctions
C O M P O S I T E
F U N C T I O N S
F U N C T I O N S
( A H L )
Y hl kw:
✔
the
denition
✔
the
notation
of
invErsE
A N D
f unctions
(aHL)
I N V E R S E
Y hl be ble :
omposite
funtions
✔
nd
the
inverse
domain
(f
and
g)(x)
=
if
of
a
funtion,
restriting
the
neessary
f (g(x))
°
✔
1
✔
that
(f
f
Just
as
two
another
for
=
(f
f)(x)
or
example
applied
This
=
omposite
funtions
in
ontext.
x
°
more
numer,
funtions
apply
1
)(x)
°
y
funtions
adding,
together.
after
proess
“
”
°
with
the
the
“inside”
numers
an
an
e
defines
when
a
equation
new
(f
a
omined
omined
sutrating
Composite
another:
e
or
g(x)
is
funtion
g)(x)
=
many
make
multiplying
functions
=
to
in
express
found,
f (g(x)),
f (g(x)).
It
and
and
is
another
two
when
then
to
or
=
to
reate
funtion,
more
one
y
defines
helpful
ways
funtion
f (a)
the
is
is
found.
operation
understand
g
as
°
funtion
and
f
as
g is applied first in finding (f
the
“outside”
funtion.
g)(x)
°
Exle 2.2.1
(a)
For
the
funtions
f (x)
=
65x
and
g(x)
=
2.3x
315,
()
Identify
whih
of
the
funtions
c
p
or
p
°
determine
if
f
g
=
g
Erin
owns
sold
is
a
afé
models
f
°
Erin’s
profit
as
a
funtion
c
°
of
time.
°
in
whih
the
numer
of
offees
()
Find
the
smallest
numer
of
hours
h,
where
afé
needs
to
open
+
t
is
in
modelled
hours
The
profit
p(c)
=
the
afé
earned
2.3c
offees
her
y
315,
is
is
funtion
open
and0
modelled
where
c
is
c(t)
y
the
≤
=
t
the
65t,
≤
h
where
10.
∈
Z
order
,
for
to
whih
make
a
Erin’s
in
profit.
funtion
numer
of
sold.
Solution
(a)
(f
g)(x)
=
f (2.3x
315)
=
65(2.3x
315)
°
=
Apply
the
definition
( f
g)(x)
=
f (g(x))
arefully
,
°
149.5x
(g
20 475
f)(x)
=
g(65x)
starting
=
2.3(65x)
with
the
inside
funtion.
315
°
This
=
149.5x
is
Hene
f
g
≠
g
°
()
p
c
=
is
true
in
general.
Funtion
omposition
not
ommutative.
f
°
149.5t
315
models
Erin’s
profit
°
In
fat,
c
p
does
not
have
for
hours
an
appliation
in
this
°
eause
()
fat
315
the
149.5t
315
need
e
to
make
a
independent
variale
>
hours.
0
⇒
open
profit.
t
>
2.11
for
3
hours
Hene
h
=
in
is
time.
Erin
order
will
ontext.
Being
open
2
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
f
x
(The
restrition
have
the
The
same
final
step
x
≤
2
would
effet)
is
to
find
the
−1
equation
To
find
(a)
Write
of
the
f
inverse
of
a
funtion:
2
y
=
down
the
equation
y
=
(x
2)
for
y
x
=
(y
2)
+
3
+
3
f (x)
2
()
Transpose
()
Solve
x
aee 
2
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,
( f
f
−1
2
)(x)
=
f (
x
°
3
+
2)
=
(
x
3
+
2
2)
=
(
x
3 )
+
3
=
x
+
=
x
2
−1
−1
( f
f )(x)
=
f
3
3
2
2
((x
2)
+
3)
=
(x
2)
=
(x
2)
+
2
+
=
x
+ 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.
of
an
inverse
funtion.
the
syllaus,
the
onept
This
funtion
has
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
applies
notation
student
orretly
−1
()
Find
the
value
of
(
g
2)
the
funtions,
the
value
of
omposite
of
(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,
diret
✔
determine
✔
transfer
a
✔
reate
a
sketh
given,
laelling
and
✔
and
tehnology
nd
up
inversevariation
the
✔
the
formulae
the
stages
of
for
the
diret
and
modelling
for
inverse
asked
marks
you
to
“Draw”
the
inluding
nd
points
of
intersetion
key
features
of
a
graph
idea
the
and
of
from
of
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
given
funtion
apply
the
proess.
✔
When
to
graph
tehnology
✔
funtions
differenes
use
general
etween
and
graph
and
of
ui
sums
to
✔
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
stages
of
the
modelling
proess.
maximum
aee 
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
the
should
their
graph
of
the
a
funtion,
relationship.
points,
asymptotes
orret
loation.
the
same
take
are
orret
of
An
make
represent
Relevant
and
requirement
to
you
axes
features
of
auray
the
sketh
=
running
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,
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
that
you
graph
in
detail.
Show
the
an
see
the
10
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
with
domain
f (x)
of
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,
showing
point
and
+19
the
the
oordinates
horizontal
asymptote.
()
Hene
four
()
The
desrie
hours
after
heating
24ºC
ut
when
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)
1.3
*damped
1.4
DEG
30.69 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
the
c(t)
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
p(x)
=
r(x)
earrings
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
than
show
that
equal
at
are
=
Xtar
will
ost
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
y
making
and
of
earrings.
the
most
selling
profit
319
c(x)=3500+26.5 
x
pairs
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
of symmetry, use calculus or
c(x)
and
technology. In the context of
r(x)
provides
required.
the
the problem, choose the most
efficient method.
s ampLE studEnt ans WEr
0.5x
Let
(a)
f (x)
=
For
the
(i)
write
(ii)
find
(iii)
write
e
−
graph
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
∴
(iii)
y
=
x-intercept
is
(1.39,
sketh
shows
the
graph
on
main
0)
features
of
the
domain
given,
and
the
the
asymptoti
2
y
ehaviour
is
full
are
marks
skethed
well,
hene
awarded.
(b)
6
f(x)
5
4
3
2
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.
There
summarized
real-world
are
in
setions
several
the
tale
ontexts
examples
of
One
example
of
mathematis
appliation
models
you
of
shape
of
mathematial
must
One
sketches
of
know,
distinguishing
examples
application
feature
of
shape
Straight
line
Constant
Linear
f (x)
=
mx
+
Simple
c
eing
and
of
Equation
model
y
of
here.
Name
Name
examples
interest
gradient
y
y
2
m
1
=
x
x
2
1
Paraola
Axis
of
symmetry
2
Quadratic
f (x)
=
ax
+
bx
+
Projetile
c
motion
b
x
=
2a
Exponential
growth/
deay
x
f (x)
=
ka
+
c
Compound
Horizontal
asymptote
x
Exponential
f (x)
=
ka
+
interest,
c
y
=
c
depreiation
rx
f (x)
=
ke
+
c
In
siene,
for
Direct
As
n
f (x)
=
ax
,
x
inreases,
f (x)
+
n
∈
example
Z
Charles’
variation
inreases
law
In
n
f (x)
Inverse
=
siene,
for
ax
As
example
x
inreases,
f (x)
Boyle’s
variation
dereases
where
n
∈
Z,
n
<
0
law
Cui
V
olume
3
Cubic
f (x)
=
ax
of
a
ox
Can
have
a
loal
2
+
bx
+
cx
+
optimisation
d
maximum
prolems
=
asin(bx)
+
d
f (x)
=
aos(bx)
+
d
loal
wave
Periodi
Average
a
minumum.
Sine
f (x)
and
with
monthly
Sinusoidal
360
temperatures
period
=
b
Eah
model
represents
Eah
model
has
parameters
f (x)
the
=
+
c
has
the
that
respetively
.
so
that
m
the
of
and
For
etween
determine
features
parameters
parameters
make
relationship
parameters
determine
y-interept
these
to
mx
a
c
eah
model
the
the
For
y
there
variales.
and
example,
determine
model,
e
and
equation,
graph.
whih
an
the x
are
the
the
model
gradient
ways
determined
these
and
to
and
determine
then
applied
preditions.
60
/
2.3
Gr a p H s
and
Linear
To
determine
the
First
find
m
using
the
equation
x-
p(x)
the
two
points.
y
x
2
Replae
given:
1
=
x
(2)
modEL s
y
2
(1)
of
models
Information
parameters:
y
f E at u r E s
1
and
=
y-oordinates
mx
+
c
to
find
of
any
point
on
the
line
into
the
c.
f1(x)=p(x)
Or:
(1)
Replae
p(x)
=
the
mx
+
x-
and
y-oordinates
of
the
two
points
in
the
equation
c
(2.5, 3.45)
(2)
Hene
(3)
write
and
c
Use
your
down
GDC
to
a
system
solve
the
of
two
simultaneous
equations
in m
system.
x
(4, –3.0)
Or:
If
a
data
set
is
given,
use
the
linear
regression
tool
on
your
GDC.
Quadratic
To
determine
(1)
Replae
the
the
Information
parameters:
x-
and
models
y-oordinates
of
the
three
points
in
the
equation
given:
y
2
q(x)
(2)
=
Hene
a,
(3)
b
bx
write
and
Use
+
ax
+
c
down
a
system
of
three
simultaneous
equations
in
c
f2(x)=q(x)
your
GDC
to
solve
the
system.
(–1, 2.75)
If
one
of
of
two
the
parameters
simultaneous
(Note
that
of
y-interept
the
down
the
for
oth
is
given,
you
write
down
and
solve
a
system
equations.
the
are
parameter
linear
(0,
c
c)
x
and
and
so
the
if
quadrati
this
point
model,
is
given
the
you
oordinates
just
(4, –1)
write
diretly
.)
(3, –3.25)
Exponential
To
determine
the
Information
parameters:
x
For
the
given
models
at
least
To
find
the
(1)
Replae
e(x)
one
=
of
ka
the
x
+
c,
e(x)
=
ka
parameters:
the
y-oordinates
x-
and
given:
rx
+
c
or
e(x)
=
ke
+
c,
you
will
e
y
(–1, 11.2)
parameters.
remaining
models
of
the
point(s)
in
the
equation
fore(x)
f3(x)=e(x)
(2)
Hene
(3)
Solve
(Note
y
=
c,
write
y
that
down
hand
for
the
and
so
if
cdiretly
.
In
this
or
equation(s)
with
your
exponential
this
is
given
model,
the
remaining
parameter(s)
GDC.
model,
you
the
in
just
(2, 4.2)
the
write
horizontal
horizontal
down
the
asymptote
asymptote
is
parameter
of
e(x)
is
y
=
3.2)
x
61
/
2
Direct
variation
Information
given:
To
determine
the
parameters:
n
The
y
diret
variation
model
has
equation
d(x)
=
+
where
ax
n
∈
Z
.
n
This
(5, 25)
(1)
means
Replae
that
the
d(x)
x-
varies
and
diretly
as
y-oordinates
x
of
the
points
in
the
equation
ford(x)
(2)
Hene
(3)
Solve
write
down
equations
in
the
parameters a
and
n
f4(x)=d(x)
(Note
the
that
solveonly
if
equations
using
one
parameters
one
of
the
equation,
as
the
laws
shown
of
indies
is
given,
in
the
you
and/or
write
following
logarithms.
down
example
and
of
inversevariation.)
(2, 1.6)
aee 
x
n
In exam questions, y is propor tional to x
is equivalent to stating that y varies
n
directly as x
n
. Similarly, y is inversely propor tional to x
(n > 0) is equivalent
n
to stating that y varies inversely as x
Inverse
variation
Information
given:
To
determine
the
parameters:
n
y
The
inverse
This
means
variation
model
has
equation
i(x)
=
ax
where
n
∈
Z,
n
<
0.
n
i(x)
varies
inversely
as
x
(1, 9)
In
9
f(x)
this
example,
(1)
Replae
(2)
Hene
(3)
Solve
the
you
x-
write
are
and
down
given
that
n
y-oordinates
equations
in
=
of
the
2
the
point
in
remaining
the
equation
for
i(x)
parameter.
=
2
x
f(x)
=
the
equations.
9
x
The sinusoidal model has three features for which you must know
y
the terminology:
(360, 4)
Sinusoidal models are e, meaning that they have a
repeating pattern.
f7(x)=c(x)
⎛
(720, 1)
x
The graph of c(x) =
1.5os
x
⎜
⎝
⎞
⎟
2
+ 2.5 is shown.
⎠
The largest possible domain is R and the graph can be thought of as a
repetition of the red shape. The shor test length of such a shape that
can be used to make the whole graph is the e of the function.
y
(360, 4)
Hence, the period of c(x) is 720°
f10(x)=p(x)
The horizontal line in the middle of the graph is the l x
f7(x)=c(x)
The equation of the principal axis is y = d, where d is the average of the
(720, 1)
x
maximum and minimum values of the function.
4 + 1
Hence, the principal axis of c(x) is y = 2.5 because
= 2.5
2
62
/
2.3
Gr a p H s
and
f E at u r E s
of
modEL s
The distance from the principal axis to any maximum or minumum point of
y
the sinusoidal function is the le of the function.
(360, 4)
This can be found by subtracting the minimum value from the maximum
f10(x)=p(x)
value and dividing by 2
4
f7(x)=c(x)
1
Hence, the amplitude of c(x) is 1.5 because
= 1.5
(720, 1)
x
2
Sinusiodal
To
determine
The
method
s(x)
=
the
is
asin(bx)
parameters:
the
+
d
same
as
for
for
Information
finding
c(x)
=
models
the
aos(bx)
parameters
+
given:
y
of
d.
f6(x)=s(x)
(1)
Identify
the
maximum
and
minimum
values
of s
from
the
(45, 2)
(2)
information
given.
Identify
period
the
of
s
x
max
(3)
Find
d
+
min
=
,
whih
gives
you
the
equation
of
the
prinipal
2
axis
y
=
d
360
(4)
Find
b
=
period
max
(5)
min
|a|=
,
whih
gives
you
the
amplitude
of
s
2
(6)
The
sign
values
of
a
depends
a,
b
and
of
d
to
on
the
hek
orientation
your
model
of
the
on
graph.
your
So
use
your
(135, –4)
GDC.
Cubic
To
determine
the
parameters:
Information
3
For
a
least
(1)
ui
one
funtion u(x)
of
the
Replae
the
=
x-
Hene
(3)
Use
and
2
+
a,
bx
b,
+
c
cx
or
+
d,
you
will
always
e
given
at
+
d
y-oordinates
of
the
points
given
in
the
equation
f12(x)=u(x)
bx
write
given:
y
2
ax
(2)
ax
parameters
3
u(x)
=
models
+
cx
+
d
down
a
system
of
simultaneous
equations.
(7, 4.9)
your
GDC
to
solve
the
system.
(2, 3.9)
(0, 3.5)
(Note
that
for
the
ui
model,
the
y-interept
is
(0,
d)
and
so
if
(3, 1.7)
this
point
is
given
you
just
write
down
the
parameter d
diretly
.)
x
You
an
pratie
information
graphing
given.
given
the
The
these
for
eah
funtion
answers
methods
on
of
model
your
finding
aove.
GDC
and
the
parameters
Chek
your
omparing
with
answers
with
the
the
y
graphs
are:
2
p(x)
=
9.7
e(x)
=
3.2
i(x)
=
2.5x
q(x)
=
0.75x
d(x)
=
0.2x
s(x)
=
x
+
4(0.5
3x
1
3
)
9
3
sin
(2x)
1
2
x
3
u(x)
=
0.2x
2
1.8x
+
3x
+
3.5
63
/
2
s ampLE studEnt ans WEr
Jashanti
in
US
is
saving
Dollars
money
(USD),
an
to
uy
a
ar.
e
modelled
The
y
prie
the
of
the
ar,
equation
t
P
Jashanti’s
savings,
in
USD,
=
an
S
In
oth
equations,
for
the
(a)
Write
()
Use
is
the
time
e
=
modelled
400t
in
+
y
the
equation
2000
months
sine
Jashanti
started
saving
ar.
down
your
saved
Jashanti
months
extra
t
8500(0.95)
the
GDC
enough
does
not
amount
to
find
money
want
she
started
money
that
she
()
Calulate
how
y
=
8 500(0.95)
=
400t
money
how
to
to
after
of
long
uy
wait
it
saves
willtake
for
per
month.
Jashanti
to
have
thear.
too
saving.
Jashanti
long
She
and
wants
deides
to
ask
to
uy
her
the
ar
parents
two
for
the
needs.
muh
extra
money
Jashanti
needs.
t
1
y
+
2000
2
(8.64, 5456)
▲ The
the
GDC
is
used
intersetion
graphs
and
the
to
nd
etween
the
working
is
two
shown
learly
.
=
▲ The
working
is
learly
orret,
However,
gaining
full
months
2
written
P
and
8.64
=
(8500)(0.95)
=
7671.3
marks.
writing
2
8500(0.95
entering
more
)
it
–
400(2)
into
the
–
2000
GDC
then
would
e
S
=
400(2)
=
2800
+
2000
efient.
7671.3
2800
▲ The student orretly interprets
the
gradient
the
rate
at
of
the
whih
linear
funtion
Jashanti
as
=
4871.3
USD
saves.
(a)
400
USD
(b)
8.64
(c)
4871.3
months
USD
64
/
2.3
Gr a p H s
and
f E at u r E s
of
modEL s
s ampLE studEnt ans WEr
A population
population,
of
P,
mosquitoes
after
t
days
is
dereases
exponentially
.
modelled
y
The
size
of
the
t
P
(a)
Write
()
Find
()
Calulate
to
The
(d)
down
the
the
size
derease
population
Write
down
the
time
to
3200
exat
of
the
=
×
size
2
of
+
the
population
it
will
take
50,
where
initial
after
for
4
the
t
≥
0
population.
days.
size
of
the
population
60.
will
the
stailize
value
when
it
reahes
a
size
of k
of k
▼ The
student
misinterpreted
has
either
what
‘initial’
means,
0
(a)
3200
2
(b)
P
=
3200
P
=
250
×
+50
=
or
6450
has
inorretly
alulated
2
4
×
2
+
50
▲ The
lear
answer
gain
working
full
and
orret
marks.
5
(c)
3200
×
2
+
50
=
150
+
50
=
75
+
50
=
62.5
7
3200
×
2
3200
×
2
▼ Use
8
8
(d)
of
equation
a
P
GDC
=
60
orret
answer
▼ Use
of
to
solve
would
of
8.32
the
give
the
days.
days
––––
funtion
the
GDC
would
show
stale
population
ould
e
from
the
to
written
of
graph
the
50.
down
formula
for
the
long-term
Or,
this
diretly
P
the ellg e
The
modelling
proess
is
desried
y
this
yle
of
ontinuous
improvement:
Reject
Pose
a
real-
Develop
a
Test
Accept
the
Reect
on
and
Extend
world
problem
The
stages
you
in
(a)
oth
the
a
of
graph
its
Develop
the
data
as
an
in
this
1
and
real-world
you
a
e
want
for
to
The
suggest
appropriate
reasonale
paper
2,
problem:
model:
setion
proess
represented
then
may
model
modelling
paper
Pose
would
()
of
model
as
an
When
with
a
as
the
model
unpaked
your
given
a
internal
set
funtion?
into
of
that
help
assessment:
data,
What
skills
an
the
shape
preditions
make?
ontext
one
finding
for
e
well
or
of
the
your
the
more
representation.
domain
apply
of
prolem
the
You
models
then
parameters
and/or
from
apply
of
your
the
the
this
shape
setion
skills
model.
of
overed
Write
a
model.
65
/
2
()
T
est
the
well
does
fit
is
not
fit
is
good,
see
model:
the
good,
you
you
Reflect
makes
To
on
of
what
the
rejet
that
applied
to
apply
Extend:
You
are
other
e
the
and
(d).
If
made,
example
the
model:
appropriate?
your
related
may
fit
model
step
an
for
model
the
to
alulated
y
parameters,
graph
go
the
go
the
of
ak
fit
is
ak
altering
the
to
to
()
the
data?
step
good
how
().
ut
and
If
If
you
try
domain
the
an
to
or
the
make
the
model.
model
extent
have
your
ontinue
and
your
you
of
improvements,
parameters
(e)
graph
improvements
these
(d)
One
e
What
What
preditions
is
it
aout
the
preditions
reliale?
Can
ontext
an
e
your
that
made?
model
e
ontexts?
ale
to
adapt
your
model
to
represent
otherontexts.
The
following
two
examples
inlude
features
of
the
modelling
proess.
Exle 2.3.4
A new
drug
infetion.
patient’s
is
The
hours
nurse
to
monitors
loodstream
administered,
Time
administered
during
writing
down
in
Amount
(x)
drug
10
0.5
8.8
1
7.1
1.5
4.8
2
4.3
2.5
3.41
3
2.6
3.5
2.18
patient
the
the
the
in
amount
10
a
of
lini
the
to
after
the
following
data
set:
of
treat
drug
hours
μg
in
0
a
in
the
drug
4
1.8
4.5
1.3
5
1.2
5.5
0.94
6
0.74
6.5
0.6
7
0.49
8
0.3
9
0.21
10
0.11
an
was
n
The
nurse
wishes
to
explore
if
either
of
the
funtions f (x)
=
ax
x
where
data
n
∈
Plot
()
Hene
the
linear
<
0
and
g(x)
State
The
data
state
set
using
why
f (x)
model
with
a
reasons
why
an
may
e
deay
(d)
n
=
bc
are
appropriate
models
for
the
set.
(a)
()
Z,
model
nurse
knows
loodstream
Hene
make
write
in
this
falls
tehnology
.
and
negative
inverse
when
elow
a
are
3
oth
more
appropriate
than
a
gradient.
variation
appropriate
that
down
g(x)
the
μg,
model
amount
the
predition
to
model
drug
that
of
the
not
nurse
an
exponential
data.
drug
does
the
or
in
the
have
would
an
effet.
wish
to
ontext.
66
/
2.3
Gr a p H s
and
f E at u r E s
of
modEL s
Solution
(a)
1.1
1.2
DEG
1.3
Examine
the
key
and
onsider
features
of
the
shape
9
of
the
data.
6
gμ
3
0
0
1
2
3
4
5
6
7
8
graph
is
9
10
Hours
()
The
slope
onstant,
model
()
Both
that
a
is
of
hene
not
an
and
urves
in
asymptote
The
the
nurse
time
drug
You
may
given
well
will
e
the
=
model
models
a
on
gives
give
your
reasons
knowledge
understanding,
of
orret
are
ased
and
using
the
terminology
.
0.
want
whih
fall
in
time
amount
would
asked
values
You
feature
horizontal
y
at
the
in
the
oth
with
linear
have
inrease
derease
a
not
appropriate.
models
drug,
(d)
the
is
the
predit
level
of
μg
3
find
given
fits
the
elow
to
to
the
parameters
funtion,
and
of
then
a
model
omment
y
sustituting
ritally
on
how
data.
Exle 2.3.5
Use
the
The
nurse
of
f
and
same
data
uses
the
the
points
and
ontext
points
(2.5,
(3.5,
3.41)
Find
the
parameters
of
f
()
Find
the
parameters
of
g
()
Plot
of
f
and
g
model
is
(d)
Determine
(e)
Use
your
the
an
on
more
same
to
will
predit
e
(7,
and
elow
axes
as
to
the
when
for
the
2.3.4
0.74)
find
to
find
the
data,
model,
domain
3
(6,
0.49)
appropriate
appropriate
model
loodstream
the
Example
2.18)
and
(a)
eah
as
level
parameters
parameters
hene
justifying
your
the
identify
your
of g
whih
hoie.
model.
of
drug
in
the
μg
67
/
2
Solution
n
(a)
f (3.5)
=
a3.5
=
2.18,
Use
an
equation
solver
to
find
n
f (6)
=
a6
=
0.74
the
value
of
n
n
n
2.18
3.5
Hene
2.18
3.5
=
=
-2.00452130601
n
0.74
6
n
0.74
6
3.5
2.18
6
0.74
,n
nSolve
2.18
n
log
=
3.5
0.74
6
=
2.004 521 306 01
hene
2.18
a
=
=
26.85..,
=
26.9x
so
2.00452130601
3.5
2.00
f (x)
2.5
()
g(2.5)
=
bc
=
Replae
3.41
the
7
g(7)
=
bc
=
of
7
indies
funtion
3.41
0.49
=
c
Use
4.5
=
3.41
laws
and
the
inverse
of
c
to
find
c
the
unrounded
vale
of
c
in
3.41
the
=
the
0.49
4.5
b
apply
into
4.5
2.5
=
given
0.49
c
c
and
=
Hene
c
model,
values
0.49
c
⇒
the
0.649 777…
alulation
of
b
hene
3.41
=
10.019 414...
so
2.5
0.649777 …
x
g(x)
=
10.0
×
0.65
()
1.1
1.2
DEG
1.3
9
26.9
f(x):=
2.
x
gu rD
6
3
0
0
1
2
3
4
5
6
7
8
9
10
T ime
1.2
1.1
DEG
1.3
9
x
g(x):=10
(0.65)
gu rD
6
3
0
0
1
2
3
4
5
6
7
8
9
10
T ime
g
is
the
more
eause
it
appropriate
passes
points
than
f,
initial
point
Comment
more
goodness
through
inluding
(0,
model
the
shape
of
ritially
of
the
fit,
on
the
inluding
the
funtion.
10).
y
f
has
half
a
good
of
the
fit
for
data,
asymptote
is
the
ut
not
an
seond
the
vertial
appropriate
f6(x)=g(x)
feature
(d)
0
≤
of
x
a
≤
model
for
this
data.
10
f7(x)=3
(e)
The
level
elow
3
of
μg
drug
after
first
2.80
(2.797129658, 3)
falls
hours.
1
1
68
/
2.3
You
in
an
the
For
apply
example,
and
His
physis
Callum
Callum
admires
Callum
down
the
teaher
deides
modelling
the
modelling
proess
in
internal
assessment
as
well
and
f E at u r E s
of
modEL s
as
examination.
day
the
the
Gr a p H s
to
shape
tells
this
of
him
explore
if
fountain
the
the
his
on
his
way
to
ollege
every
water.
shape
is
physis
likely
to
teaher
is
e
paraoli,
orret
y
so
applying
proess.
arefully
the
noties
plaes
oordinates
parameters
of
a
axes
(0,0),
and
three
(1.20,3.66)
quadrati
points
and
as
shown:
(3.25,4.72)
in
and
writes
order
to
find
model.
2
The
the
quadrati
origin
Callum
of
to
model
e
on
his
sustitutes
simultaneous
is
f (x)
=
ax
model,
the
c
+
=
bx
+
c
ut
sine
Callum
plans
for
0.
oordinates
in
the
funtion
to
set
up
his
system
equations:
=
a(1.20)
f (3.25)
=
a(3.25)
then
into
GDC
his
enters
to
this
find
system
the
parameters:
2
f (1.20)
Callum
+
b(1.20)
=
3.66
+
b(3.25)
=
4.72
2
2
lin
Solve
a
(1.2)
+b
a
(3.25)
1.2=3.66
,{a
15
2
+b
b}
2
3.25=4.72
{–0.779362101313,3.98523452158}
Callum
a
≈
−
0.779
Callum
hand,
not
he
approximates
and
starts
his
taken
For
y
this
to
model
pleased
judges
b
the
3.99
fits
the
explore
if
Callum
arries
the
three
well
atually
is
on
his
what
model.
he
has
perfetly
.
funtion
model
is
found.
On
models
more
the
the
On
the
other
hand,
fountain
symmetri
than
and
This
a
third
gives
he
solves
He
finds
point
him
with
a
the
the
more
same
(5.01,
system
his
quadrati
f (x)
is
the
path
model
and
deides
to
appropriate.
proess
as
aove
with f (x)
=
ax
2
+
bx
+
cx
1.56).
of
three
equations
in
three
unknowns
whih
GDC.
3
that
he
eause
3
out
one
is.
rejets
model
graphs
points
the
quadrati
Callum
ui
then
ritially
how
water
reason,
parameters:
and
reflet
aout
that
a
≈
the
=
−0.0585x
2
0.519x
+
3.76x
69
/
2
Callum
reflets
on
what
he
has
found.
The
ui
funtion
fits
his
aee 
points
etter
perfetly
.
than
his
Also,
it
seems
quadrati
to
model
model
did.
the
path
Callum
taken
reflets
y
on
the
the
water
ontext
of
Real-world data can be “messy”
his
exploration
–
perhaps
one
the
water
has
lost
speed,
the
effet
of
and may not present itself as
gravity
on
extend
his
its
path
is
more
signifiant?
Callum
wonders
how
he
ould
“neat” nor “nice” – as seen in the
work
to
explore
larger
more
powerful
fountains
and
make
water fountain example.
preditions
Some
of
aout
ontexts
other
piees,
their
an
funtions.
eah
with
e
shapes.
modelled
These
its
only
piecewise
own
domain,
y
forming
functions
as
shown
a
funtion
from
an
have
two
the
next
example.
in
or
piees
more
Exle 2.3.6
In
a
large
and
of
ity
Vintair
vintage
there
whih
are
eah
eletroni
$50
“enh
the
devie
fee”
two
workshops,
offer
restoration
devies.
whih
is
a
AudioEx
AudioEx
fixed
just
e
put
on
the
workshop
a
restoration
muh
After
this,
restoration
enh
osts
$10
more
15
AudioEx
hours
of
would
work,
harge
find
than
for
Graph
V
and
A
on
the
same
set
of
axes
and
for
write
inspetion.
taking
Vintair.
a
()
to
For
how
servies
harges
harge
()
down
the
oordinates
of
any
intersetions
per
etween
the
two
graphs.
hour.
Vintair
(a)
Model
the
ost
of
a
restoration
funtion
A(t)
where
at
AudioEx
those
a
restoration
taking
reasonale
domain.
The
the
ost
of
a
t
A
is
hours
restoration
at
the
of
ost
in
work,
Vintair
is
$
of
and
laim
their
harges
are
always
less
than
with
of
AudioEx.
a
state
modelled
a
(d)
Explain
why
(e)
Erin
a
has
radio.
y
udget
Determine
maximum
funtion
Vintair ’s
times
of
$300
the
she
laim
to
is
false.
restore
differene
an
afford
a
rare
etween
at
eah
the
of
the
workshops.
⎧
V(t)
=
40 + 12t ,
0
≤ t
≤
8
t
>
8
⎨
9t + 64,
⎩
where
V
is
the
ost
in
$
for
t
hours
of
work.
Solution
(a)
A(t)
=
10t
+
50,
where
t
≥
0.
The
a
is
ost
for
onstant
rate
appopriate.
parameters
()
A(15)
=
$200,
restoration
()
1.2
203.11
1.3
V(15)
would
=
$199
ost
$1
so
more
*peice
1.4
the
15
at
hour
Find
AudioEx.
an
the
AudioEx
of
There
are
ost
$10
10
y
restoration
per
is
and
a
hour
fixed
so
a
ost
inreases
linear
of
$50
at
model
so
the
50.
replaing
the
time
in
eah
funtion.
RAD
Take
y
are
to
adjust
the
viewing
window
so
that
(14, 190)
you
40+12
V(x)=
an
see
the
graphs
learly
.
x,0≤x0≤8
5
9
x+64, 8<x
A(x)={50+10 
x
x≥0
(5, 100)
The
(5,
(d)
oordinates
100)
and
Vintair ’s
its
(14,
laim
servie
of
points
of
intersetion
are
190)
is
osts
the
false
more
eause
than
that
for
of
5
<
t
<
14,
AudioEx.
Interpret
aove
what
that
of
the
A
on
graphs
the
show:
interval
5
the
<
t
graph
<
of
V
is
14.
70
/
2.3
(e)
Find
y
40+12
V(x)=
the
differene
intersetions
x,0≤x0≤8
Gr a p H s
with
in
the
and
time
line
y
y
=
f E at u r E s
finding
of
the
modEL s
two
300.
5
9
x+64, 8<x
(25, 300)
You
b(x)=300
then
interpret
the
two
x-oordinates
26.2
and
(26.2, 300)
25
in
order
to
solve
the
prolem.
(14, 190)
A(x)={50+10 
x
x≥0
(5, 100)
x
Erin
at
You
an
afford
Vintair
might
a
than
at
maximum
AudioEx
onsider
the
of
on
1.2
a
hours
udget
quadrati
more
of
funtion
$300.
as
simple
and
familiar.
When you carry out the
However,
there
is
a
lot
of
knowledge
and
understanding
to
apply
modelling process you are involved
with
this
funtion,
muh
of
whih
is
a
onsequene
of
its
symmetry
.
in a “hands on” activity where you
The
quadrati
you
the
funtion
has
three
different
forms
and
they
eah
give
learn by actively “doing”. Reflect
following
information
aout
the
features
of
the
funtion.
on how this relates to the concepts
of evidence and interpretation you
Form
and
equation
Features
that
an
e
Features
that
an
e
learn in TOK .
(a
≠
0
in
eah
ase)
written
down
this
Standard
form
from
calculated
form
y-interept
from
this
form
(0,
c)
Axis
of
symmetry
2
f (x)
=
ax
+
bx
+
c
b
x
=
2a
Roots
the
of
f (x)
quadrati
(HL
Fatorized
f (x)
=
form
α)(x
a(x
x-interepts
β)
and
Axis
α)
(0,
of
with
formula
only)
α
x
0
symmetry
β)
(0,
=
+ β
=
2
Roots
of
f (x)
=
0
are
β,
as
y-interept
x
=
are
V
ertex
(ompleted
square)
α
and
the
x
=
zeros
V
ertex
of
(h,
aαβ)
(0,
f (x)
k)
Roots
form
of
f (x)
=
0
rearrangement
Axis
of
symmetry
2
f (x)
=
a(x
h)
2
+
k
y-interept
x
You
apply
relevant
out
the
in
information
the
y
ontext
of
a
given
to
=
you
prolem,
(0,
ah
+
k)
h
in
a
quadrati
and
you
use
in
large
your
funtion
GDC
to
if
it
is
arry
alulations.
Exle 2.3.7
A greenhouse
is
planned
for
a
park
a
y
V
ity
.
in
The
height
metres
is
of
the
modelled
roof
y
of
the
the
greenhouse
funtion
2
r(x)
=
0.06x
distane
+
from
7.02x
the
−
172.2
park
where
entrane
at
x
is
(0,
the
0)
horizontal
metres.
10
Entrance
Visitors
walk
from
the
entrane
to
point A
x
where
10
they
The
enter
the
vertex
V
greenhouse
is
the
and
highest
they
point
of
exit
the
at
point B.
greenhouse.
71
/
2
(a)
Find
the
()
Hene
oordinates
write
equation
()
Hene
of
find
down
the
the
of
a
axis
A
and
of
reasonale
of
height
B
domain
for
r(x)
and
find
V
the
symmetry
.
of
the
greenhouse.
20
(d)
The
arhitet
greenhouse
Calulate
skethes
suh
the
that
vertial
a
plan
the
for
a
horizontal
distane
distane
CD
from
=
the
20
walkway
inside
m
the
m.
walkway
to
the
ground.
Solution
(a)
A
=
(35,
0),
B
=
(82,
0)
Use
your
write
()
The
domain
for
The
equation
r(x)
is
35
the
axis
≤
t
≤
The
82.
GDC
the
to
find
oordinates
equation
of
the
the
as
zeros
was
axis
of
of
r(x)
required
symmtery
=
0
in
and
the
an
e
sure
to
question.
also
e
found
b
of
of
symmetry
is
y
replaing
the
values
of
a
and
b
in
the
formula
x
=
2a
35 + 82
x
=
=
58.5
2
()
The
is
height
r(58.5)
=
of
the
33.1
The
greenhouse
use
m
of
(d)
The
is
height
r(68.5)
=
of
the
27.1
ommand
the
solve
m
Note
that
in
previous
symmtery
Apply
walkway
the
term
the
the
to
fat
“Hene”
result.
find
that
You
the
the
is
an
use
expliit
the
maximum
paraola
is
instrution
equation
value
of
the
to
axis
diretly
.
symmetri
in
order
to
prolem.
ontext
of
this
prolem,
the
y-interept
(0,
172.2)
is
not
aee 
of
any
relevane.
information
It
is
not
always
the
ase
that
you
must
use
all
of
the
given.
Choose the most efficient
method for finding the solution
Many
to a quadratic equation in the
e
examination. Using your GDC is
patterns
faster and less error-prone than
sinusoidal
using the quadratic formula.
data
natural
modelled
phenomena
y
that
sinusoidal
define
funtion
as
well
for
eah
as
suh
make
monthly
funtions
periodi
enale
as
sine
ehaviour.
you
to
average
they
show
Knowing
ompare
and
temperatures
the
the
repeating
parameters
ontrast
an
sets
of
of
the
periodi
preditions.
Exle 2.3.8
The
average
24-hour
temperature
in
°C
month
in
Ushuaia,
Argentina
Jan
Feb
Mar
Apr
May
Jun
Jul
Aug
Sep
Oct
Nov
Dec
9.2
9.1
7.7
5.6
3.2
1.6
1.5
2.1
3.9
6.1
7.4
8.6
(a)
Represent
against
The
data
24-hour
()
(i)
the
twelve
months
an
e
months
using
Determine
the
in
January
to
Deemer
y
0,
1,
2,
…,
11
given
and
in
hene
this
plot
tale:
temperature
tehnology
.
modelled
temperature
from
is
y
°C
one
and
values
of
t
yle
is
the
the
of
a
time
sinusoidal
in
funtion T(t)
=
s
+
p
os(qt),
where
T
is
the
average
months.
parameters
s,
p
and
q
15
Plot
your
omment
The
average
model
on
on
fit.
the
24-hour
the
same
temperature
axes
for
as
the
points,
Stokholm
is
and
pmet_va
(ii)
0
represented
f1(x):=
y
this
graph
()
Compare
and
and
model:
ontrast
–10.35 
cos(30 
x)+6.85
–15
the
annual
24-hour
temperature
of
0
2
4
6
8
10
12
Month
Ushuaia
with
that
of
Stokholm.
72
/
2.3
Gr a p H s
and
f E at u r E s
of
modEL s
Solution
Choose
(a)
1.1
1.2
drug2
1.3
the
viewing
window
so
you
an
see
the
shape
of
DEG
the
data.
15
pmet_va
0
–15
0
Month
9.2
()
(i)
|p|
1.5
=
=
3.85
Determine
the
parameters
as
shown
to
the
parameters
orretly
aove.
Take
are
2
intepret
if
they
appear
in
a
9.2 + 1.5
s
=
=
5.35
2
different
order
prinipal
axis
to
what
you
expet:
for
example,
in T
the
360
q
=
=
30
has
equation
T
=
s
12
(ii)
1.1
1.2
drug2
1.3
f1(x):=3.85 
cos(30 
DEG
x)+5.35
15
pmet_va
0
–15
0
Month
T(t)
the
()
=
3.85os(30t)
graph
Compare
passes
and
+
5.35
fits
through
ontrast
the
all
data
the
means
data
for
give
Ushuaia
an
sinusiodal
funtion.
models
an
aount
of
the
similarities
and
differenes.
Differences
Stokholm
Both
sine
points.
Similarities
The
well
e
have
and
for
modelled
period
with
Stokholm
a
12months.
greater
The
a
has
greater
amplitude,
orientation
minimum
The
a
data
fit
in
of
and
a
of
greater
Stokholm
January
for
range
is
whereas
Ushuaia
is
temperatures,
value
the
on
opposite
Ushuaia
etter
the
than
has
that
shown
prinipal
of
a
for
as
y
its
axis.
Ushuaia’s:
it
has
maximum.
Stokholm.
aee 
The order of parameters in any exam question on modelling might not follow
the form you have studied. For example, the equation of a straight line can
be written y = mx + c or y = a + bx. Similarly, the parameters s, p and q in
s + pcos(qt) determine the principal axis, amplitude and period respectively.
aee 
Be aware of the dangers of extrapolation. For example, predicting average
temperatures 20 years into the future may not be reliable due to increases in
global temperatures due to climate change.
73
/
2
2 . 4
GRAPHS
AND
F E AT U R E S
Y hl kw:
✔
the
formulae
strethes
of
a
for
that
points
translations,
reetions
and
✔
apply
graph
vertial
the
the
saling
very
phase
large
makes
are
invariant
points
horizontal
logisti,
(with
logarithms
spread
a
that
equations
logarithmi,
✔
x-axis
and
under
names,
models
the
streth
invariant
✔
on
and
on
the
under
a
y-axis
are
✔
streth
features
pieewise
of
and
natural
✔
sinusoidal
or
(AHL)
small
it
easier
to
numers
represent
using
✔
oth
the
SL
and
to
AHL
any
of
setions
guide
apply
and
in
orret
the
interpret
linear,
your
range
omposite
transformations
order
quadrati,
sinusoidal
hoose
widely
in
transformations
and
ui,
exponential,
logisti
regression
tools
GDC
a
of
manageale
sale
for
data
with
a
wide
values
data
linearizing
data
the
on
very
interpret
funtions
power,
shift)
or
and
apply
✔
✔
M O D E LS
Y hl be ble :
of
✔
OF
on
data
gives
a
valuale
perspetive
plot
on
and
using
interpret
log-log
and
semi-log
graphs
tehnology
models.
✔
use
linearization
exponential
✔
apply
and
You
have
help
you
that
of
a
andan
the
seen
“mahine”
“output”
of
b
∈
What
(x,
f
y
point
+
gives
if
b
<
b).
a
0.
(mapping
onept
an
of
a
“input”
dependent
data
has
an
relationship
proess
with
diagrams
funtion.
(the
and
sophistiation
independent
variale
y).
relations)
Another
The
that
perspetive
is
variale x)
point
effet
(x,
y)
would
the
is
transformed
that
have
→
on
vertial
f (x)
funtion
the
Applying
In
modelling
→
the
x
The
power
if
(x,
y)
→
y
is
on
f
imagine
R.
with
(the
x
Now
the
a
determine
rigour.
representations
understand
graph
the
or
to
on
f (x)
graph
this
f
has
+
f (x)
of
=
the
has
of
to
to
f (x)
→
all
y
the
b
een
een
b
where
y
+
b
transformed
points
units
+
f?
b
graph
sin(x)
f (x)
graph
therefore
transformation
translation
diagram,
of
the
from
up
on
if
b
transformed
the
>
0,
to
graph
or
of
down
vertially:
74
/
2.4
Similarly
,
the
transformation
x
Eah
is
a
point
(x,
vertial
y)
of
of
Points
(x,
y).
x-axis
is
zero,
If
0
<
p
If
p
=
−1,
y
=
So
<
−f (x)
far
1,
on
eah
the
of
of
refletion
an
The
transformation
onsider
x
for
“output”
the
+
f (x
graph
a
y
f
a
of
has
e
represented
of
from
are
is
y
the
a
+
x-axis
they
loser
to
the
y
of
a)
an
in
e
x
has
must
a)
a)
een
units
to
=
f (x)
=
right
an
e
=
f (x)
y
if
a
the
p
of
modEL s
(aHL)
py
to
>
(x,
1,
py).
the
greater
distane
given
in
the
y
f.
a
This
vertial
than
from
that
the
This
>
0,
graph
the
same
same
+
and
represented
of
of
y.
y
as
of
x.
y:
order
means
(x
p.
value
value
→
in
of
x-axis.
a)
to
fator
The
the
given
“input”
y.
if
times
of
represented
the
p:
x-axis
that
of
transformed
the
f (qx)
e
p
sine
the
to
graph
is
of
f E at u r E s
are!
transformed
transforms
a
fator
invariant:
where
py)
f
transformed
and
y:
→
een
f (x
+
transformation
an
transformations
f (x
x
f ((x
of
is
f
→
a),
sine
translation
The
f
onsidered
You
on
(x,
is
also
py)
remain
point
of
graph
x-axis
graph
you
Hene
(x,
the
they
graph
the
the
a
f (x)
pf (x)
point
on
eah
×
→
streth
distane
p
Gr a p H s
a,
to
to
that
y
give
the
y).
This
the
left
if
→
y
point
is
a
a
the
(x,
y)
horizontal
<
0.
y:
x
→
f (qx)
q
⎛
Hene
the
point
(x,
y)
on
the
graph
of
f
has
een
transformed
⎞
x
, y
to
⎜
⎝
.
⎟
q
⎠
1
This
is
a
horizontal
streth
y
a
fator
of
q
75
/
2
⎛
Eah
point
(x,
y)
on
the
graph
of
f
has
een
transformed
⎞
x
, y
to
⎜
.
⎟
q
⎝
⎠
1
This
is
a
horizontal
streth
of
the
graph
of
f
y
a
fator
:
of
if
q
>
1,
q
⎛
eah
⎞
x
, y
point
⎜
⎝
is
q
times
0
<
q
<
1,
eah
, y
a
=
−1,
the
refletion
are
y-axis.
graph
of
the
of
1
is
further
from
the
y-axis
y
a
fator
of
of
.
⎟
q
⎝
q
the
⎞
x
point
⎜
If
to
⎠
⎛
If
loser
⎟
q
y
=
graph
q
⎠
f (x)
of
y
is
=
transformed
f (x)
in
the
to
that
y-axis.
of
y
Points
=
on
f (
x),
the
y-axis
invariant.
These
transformations
an
The graph of y = pf (q(x
e
summarized
into
one
statement.
a)) + b is a ver tical stretch with scale factor p,
a
1
, a horizontal translation by the vector
a horizontal stretch with scale factor
q
(
)
0
0
and a ver tical translation by the vector
(
)
of the graph of y = f (x).
b
This
statement
whih
effet
It
is
are
effet
important
tehnology
arry
not
vertial
horizontal
arried
you
is
and
out
the
formula
e
as
a
ooklet.
transformations
transformations
to
out,
in
aware
shown
of
in
the
the
onsideration
transformations
are
are
outside
inside
order
in
in
and
these
example.
omposite
the
f
that
parameters
those
whih
f.
whih
following
of
Note
orret
transformations
You
funtions
an
to
use
make
sure
order.
Exle 2.4.1
2
(a)
Given
f (x)
=
x
,
g(x)
=
3x
and
h(x)
=
x
+
()
2,
Hene
desrie
the
whih
transform
sequene
of
transformations
2
express
y
=
3x
+
2
as
a
omposition
the
graph
of
y
=
f (x)
to
the
2
graph
offuntions.
of
y
=
3x
+
2.
Solution
2
2
(a)
3x
+
2
()
Hene
=
h(3x
)
2
=
h(g(x
))
=
Apply
h(g(f (x)))
your
omposite
knowledge
and
understanding
of
funtions.
2
the
graph
sale
⎛
the
0
graph
of
fator
y
3
=
of
y
f (x)
=
y
followed
3x
a
+
2
is
otained
vertial
y
a
streth
vertial
from
of
translation
Write
the
a
detailed
omposition
aount
from
y
the
following
inside:
g
is
the
order
applied
of
first.
⎞
of
⎜
⎟
⎝ 2⎠
76
/
2.4
In
the
exam,
should
you
have
douts
regarding
the
orret
Gr a p H s
and
f E at u r E s
You
the
appliation
define
the
of
omposite
funtions
in
your
funtions
GDC
to
with
save
modEL s
(aHL)
order,
g(x):=3
verify
of
x
Done
h(x):=x+2
Done
tehnology
.
time,
in
a
alulator
2
page
On
or
your
your
dashed
Y
GDC,
to
=
editor,
use
for
olours
distinguish
example.
or
line
etween
f(x):=x
attriutes
graphs.
In
like
what
old,
solid
follows,
Done
and
the
“target”
2
ofy
=
Set
3x
up
+
the
2
is
shown
in
old,
for
example.
prolem:
This
sequene
eause
You
are
aiming
to
transform
it
is
does
inorret
not
This
transform
sequene
itdoes!
The
is
orret
vertial
eause
streth
of
the
2
the
graph
of
f (x)
=
to
x
the
graph
sale
fator
3
is
applied
efore
2
graph
of
f (x)
=
x
to
the
graph
of
2
of
y
=
3x
+
2
thetranslation.
2
y
=
3x
+
2.
1.1
1.1
1.1
y
y
y
2
f(x)=x
2
f8(x)=3
x
+2
f5(x)=f(x)
f2(x)=h(f(x))
2
f4(x)=g(h(f(x)))
f6(x)=g(f(x))
f(x)=x
f7(x)=h(g(f(x)))
x
This
analysis
whih
the
funtion
as
is
useful
in
helping
transformations
h(g(f (x)))
g(h(f (x))),
a
gives
take
the
onsequene
you
plae
orret
of
x
x
the
identify
does
not
sequene
fat
that
ases
where
matter.
and
this
The
is
omposition
the
in
omposite
not
of
order
the
same
funtions
is
2
not
ommutative,
as
seen
in
example
2.2.1.
This
is
why h(g(f (x)))
=
3x
+
2
2
whih
is
positions
not
the
of
and
However,
Given
as
a
s(x)
g
it
=
is
same
that
have
een
true
that
for
sin(x),
q(x)
=
omposition
means
h
funtion
you
of
get
2x
g(h(f (x)))
hanged,
any
three
and
funtions
the
as
4
orret
p(x)
the
=
3x
4x,
sin(2x)
answer
=
y
(p
y
6.
answers
funtions,
=
+
=
°
f
°
Beause
are
(g
4sin(2x)
s)
°
q(x)
applying
=
an
=
a
different.
h)
°
p
the
°
(f
e
(s
g)
°
°
h.
expressed
°
q)(x).
horizontal
This
streth
1
of
sale
to
fator
s(x)
=
sin(x)
followed
y
a
vertial
streth
of
sale
2
fator
4
to
the
result,
or
y
applying
a
vertial
streth
of
sale
fator
1
4
to
s(x)
=
sin(x)
followed
y
a
horizontal
streth
of
sale
fator
to
2
the
result.
aee 
Be aware that the order in which transformations are carried out can be
verified by applying composite functions and/or use of technology.
aee 
Read the exam question carefully. If a question asks for “the” sequence of
transformations – this means there is one correct answer, or “a” sequence of
transformations – this means there is more than one correct answer.
77
/
2
Exle 2.4.2
(a)
Desrie
the
sequene
of
transformations
required
⎛
the
graph
of
y
=
ln(x)
onto
the
graph
of
y
=
ln
()
translation
Determine
a
and
sequene
a
of
horizontal
graph
of
y
=
x
−
⎟
transform
5
y
a
⎠
3
streth.
transformations
required
2
the
to
⎞
⎜
⎝
vertial
x
to
transform
2
onto
the
graph
of
y
=
4(x
1)
Solution
x
⎛
(a)
ln
⎞
⎜
⎟
⎝
5
=
f (ln
(t(x))
−
5
and
t(x)
Give
where
1
f (x)
=
x
sequene
streth
3
full
aount
=
and
detailed
with
followed
is
sale
y
a
of
the
orret
notation
and
terminology
.
Another
orret
horizontal
fator
vertial
0
⎛
translation
a
using
x.
3
The
a
⎠
3
⎞
⎜
⎜
⎟
⎟
⎝
5⎠
2
()
4(x
1)
=
f (x)
=
4x,
s(x)
=
x
f (s(h(x))
h(x)
=
where
x
1
vertial
and
2
4
.
streth
followed
y
is
a
translation
of
1
⎜
1
sale
is
a
fator
horizontal
⎞
horizontal
translation
⎛
of
a
⎛
A sequene
answer
of
⎜
⎟
⎝ 0⎠
⎞
followed
⎟
⎝ 0⎠
y
a
vertial
fator
Note
that
of
sale
4.
there
involving
streth
an
different
e
equivalent
sequenes
of
transformations
transformations.
4
81
⎛
4
For
example,
sine
x
9.1
=
⎝
streth
Similarly
,
one
sine
vertial
an
e
⎞
x
⎜
16
vertial
3
2
9.1,
⎟
either
a
horizontal
or
a
⎠
applied.
ln(3x)
1
=
ln(x)
+
(ln(3)
1),
the
transformation
is
just
translation.
tye  el
In
addition
out
the
them
of
Equation
to
already
proess
preditions.
in
ut
SL
in
example
HL
of
application
it
overed
with
make
One
model
those
modelling
overed
Name
to
The
is
the
SL,
you
following
exponential
further
Name
and
in
of
used
sketches
examples
of
of
e
models,
model
to
shape
must
has
alulate
One
ale
as
to
well
arry
as
already
apply
een
half-life.
distinguishing
feature
shape
x
Exponential
f (x)
=
+
ka
c
Calulation
of
Exponential
growth/deay
half-life,
Horizontal
y
asymptote
=
c
x
f (x)
=
ka
+
c
for
example
rx
f (x)
=
ke
+
c
of
radioative
material.
Natural
logarithmic
f (x)
=
a
+
bln(x)
The
Rihter
magnitude
of
Logarithmi
an
V
ertial
x
asymptote
=
0
earthquake
78
/
2.4
Sinusoidal
f (x)
=
asin(b(x
c))
+
Sunrise
d
L
Logistic
f (x)
Gr a p H s
times
Inrease
Sine
in
and
f E at u r E s
wave
Logisti
of
modEL s
Phase
urve
Horizontal
(aHL)
shift c
asymptotes
=
kx
1 + Ce
L,
C,
k
>
height
of
a
y
=
L
and
y
=
0
0
plant.
Piecewise
f (x)
a
V
eloity/
(x),
x
1
=
≤
x
<
x
1
graph
time
of
Pieewise
Continuous
a
a
2
…
…
journey
(x),
x
n
(1)
Replae
of
(2)
two
Hene
aand
x
of
the
and
≤
x
≤
distint
x
n
the
write
parameters
y
=
a
(x
2
)
…
2
with
n
+
parts.
1
Natural
determine
)
2
{
a
To
(x
1
if
in
the
points
down
n(x)
equation
on
a
of
the
=
a
+
b lnx
n(x)
=
a
+
Information
b lnx
with
the
oordinates
of
models
given:
1.1
urve.
system
logarithmic
y
two
simultaneous
equations
in
b
f17(x)=n(x)
(3)
Use
your
GDC
to
solve
the
system.
(1, 3)
x
(3, –1.394)
Sinusoidal
To
determine
(1)
Carry
as
out
shown
the
the
parameters
steps
for
of
p(x)
finding
=
the
asin(b(x
c))
parameters
+
models
Information
d:
of s(x)
=
asin(bx)
+
d
given:
1.1
y
previously
.
(1.785, 2)
(2)
You
(t,
will
p(t))
need
in
to
order
know
to
find
the
oordinates
of
a
point
on
the
graph
c
x
Replae
p(t)
=
these
asin(b(t
oordinates
c))
+
in
the
equation
and
write
down
d
(1, –1)
(3)
Solve
this
equation
for
c
f19(x)=p(x)
Radian
measure
used
for
c
e
the
should
unit
of
the
referred
to
as
e
assumed
independent
unless
the
degree
symol
is
variale.
(0.2146, –4)
an
the
phase
shift.
79
/
2
Logistic
models
Information
L
given:
To
determine
the
parameters
of
g(x)
=
:
kx
1 + Ce
1.1
(1)
The
parameter
L
an
often
e
determined
from
the
ontext
of
the
y
prolem
given,
–
or
for
the
example
equation
if
the
arrying
apaity
of
the
horizontal
of
a
population
is
asymptote.
f15(x)=8.55
L
(2)
If
the
initial
ondition
g(0)
is
given,
solve
(2, 7.775)
=
g(0)
to
determine
C
1 + C
f18(x)=g(x)
(3)
Another
(0, 4.5)
point
on
the
urve
(a,
g(a))
is
required
to
determine
k.
L
Solve
=
g(a)
to
determine
k
ka
1 + Ce
This
x
is
not
Example
the
uniquely
together,
the
y
models
ould
designed
model
an
to
determine
the
parameters,
as
shown
in
models
Pieewise
You
way
2.4.4.
Piecewise
one
only
(piee)
as
seen
pratie
information
graphing
to
to
in
desried
desrie
another.
Example
these
given
your
e
a
“espoke”
where
insofar
ehaviour
requirement
is
that
as
they
hanges
the
piees
are
from
fit
2.4.6.
eah
funtion
ontext
A key
methods
for
as
on
of
finding
model
your
the
aove.
GDC
parameters
Chek
and
your
omparing
with
answers
with
the
graphsgiven.
8.55
The
answers
are:
n(x)
=
3
4lnx,
p(x)
=
3
sin(2(x
1))
1,
g(x)
=
1.1 x
1 + 0.9e
If you are given sufficient points, you can create a model using the linear,
quadratic, exponential, logarithmic, logistic or sinusoidal regression tool on
your GDC as appropriate.
At
HL,
you
setion
and
2.3
are
expeted
with
the
apply
sophistiation,
understanding
how
to
modelling
from
other
proess
the
for
modelling
example
topis.
an
e
The
y
proess
applying
following
applied
at
desried
in
knowledge
aount
shows
HL.
Wl l ellg
Steve
*world
1.4
pop
is
given
world
popuation
data
in
Geography
lass
and
he
DEG
represents
the
numer
years
of
data
on
after
a
set
1950
of
axes.
and
on
On
the
the
x-axis,
y-axis,
years
represents
population
is
the
measured
600
noitalupop
in
tens
(a)
of
Pose
millions.
a
real-world
problem:
What
will
the
world
population
e
600
in
the
year
2200?
0
–300
–200
–100
0
100
years
80
/
2.4
()
Develop
a
model:
Steve
examines
the
shape
of
the
graph
Gr a p H s
and
f E at u r E s
and
1.3
deides
that
neither
a
linear
funtion
nor
an
exponential
of
world
modEL s
pop
(aHL)
DEG
funtion
1200
are
appropriate
logisti
the
funtion.
an
only
parameters
This
eause
model
support
of
a
a
is
shape
resemles
appropriate
finite
logisti
the
arrying
model
with
for
the
apaity
,
a
part
of
ontext
so
regression
he
a
sine
our
noitalupop
planet
models,
estimates
tool.
600
0
()
T
est
the
illion
a
model:
in
2200,
population
Steve’s
and
of
a
model
predits
horizontal
almost
11
a
population
asymptote
illion,
greater
of
y
than
=
of
1095,
the
10.94
meaning
figures
for
–600
the
–400
–200
0
200
years
arrying
apaity
lass.
Aware
Steve
inspets
refinements
of
of
the
the
one
the
world
potential
fit
y
of
his
one,
he
for
has
error
funtion
going
ak
Knowledge
the
in
in
numers,
suh
and
to
model
f (x)
()
Trying
=
,
large
onsiders
after
Possible
L
For
enountered
Geography
the
following
eah:
refinement
to
improve
of
the
the
fit
model
of
the
it
ka
1 + Ce
an
e
shown
y
funtion
appliation
to
seqeuene
graph
that
hanging
have
of
y
transforms
these
=
and
the
data
set
through
a
of
graph
transforms.
logarithms
parameters
effets
on
the
graph
f (x):
•
L
-
vertial
•
C
-
horizontal
•
k
-
streth
horizontal
translation
streth.
L
For
the
of
f (x)
Estimating
=
,
it
an
e
the
point
in
the
shown
ka
1 + Ce
with
data
appliation
of
alulus
is
lnC
the
point
of
inflexion
the
rate
of
hange
greatest,
omparing
with
the
L
,
is
where
that
point
k
of
inflexion
of
the
model
2
and
adjusting
the
parameters
ifneessary
.
Changing
the
exploration
inlude
only
the
(reent)
data
will
parameters
points
only
(d)
from
e
of
hange
the
1600
Reflect
Given
on
the
model.
and
and
Steve
the
if
sample
the
model
Data
efore
apply
finds
of
modelling
more
modern
data
a
smaller
makes
the
representative
of
times.
an
Steve
world
population
has
least
at
Extend:
world’s
modify
his
that
fators
model
the
to
take
learns
explores
eah
a
how
for
that
is
the
the
of
his
population
in
2200.
ould
this
average
to
e
ut
world
ale
he
to
predit
knows
that
he
depth.
of
projeted
population
of
validity
,
some
of
refinement
ontriute
funtion
in
aount
aout
every
population
ountries,
human,
to
any
one
He
relate
the
degree
preditions.
targets
predition
with
iggest
and
of
With
how
learns
model
model:
wonders
explored
Steve
the
another
omplexity
population,
the
reliale
Exploring
estimates.
model,
(e)
most
to
China
to
in
fall,
order
aron
growth
and
and
to
of
India,
sets
make
footprint
and
aron
out
to
modified
of
a
emissions
other.
81
/
2
al  el  ble
You
have
already
Example2.3.4.
example
the
radioative
apply
seen
an
Half-life
example
of
the
required
is
time
exponential
amount
of
a
drug
in
the
element)
to
redue
to
half
exponential
models
to
for
a
deay
sustane
loodstream
of
alulate
its
in
initial
or
the
(for
amount
quantity
.
You
of
a
an
half-life.
Exle 2.4.3
x
Consider
This
the
g(x)
=
funtion
10.0
×
models
loodstream
x
0.65
the
hours
from
Example
amount
after
of
μg
10
drug
of
a
2.3.5.
administered.
1.2hours.
A in
drug
is
was
x
hours
The
after
It
is
found
amount
20
μg
of
of
that
drug
drug
B
the
B
in
was
half-life
the
is
loodstream
administered
is
qx
administered,
(a)
Find
the
model
x
≥
modelled
0.
half-life
of
drug
A predited
y
the
y
()
Find
the
()
Hene
h(x)
=
De
,
parameters
x
≥
D
0.
and
q.
g(x).
In
a
medial
to
another
trial,
μg
20
patient
at
of
the
drug
same
B
is
time
drug
admistered
as
drug A
predit
in
the
the
time
at
loodstream
oth
drug
A and
for
eah
drug
remains
whih
would
drug
at
B
this
the
e
and
level
equal
how
of
for
muh
of
time.
Solution
1
x
(a)
10.0
×
x
0.65
=
5
⇒
0.65
=
Write
down
and
solve
the
equation
or
use
an
2
⇒
x
=
log
0.5
=
1.61
equation
solver
to
find
the
value
of
t
0.65
The
half-life
is
1.61
hours
1.60904055074
t
(0.65)
=0.5,t
q(0)
()
h(0)
=
20
=
⇒
De
D
=
Interpret
20
the
initial
value
of
20.
1
q(1.2)
20e
=
10
⇒
1.2q
=
ln
2
⎛
ln
⎜
⎝
Hene
q
1
⎞
⎟
2
⎠
=
=
−0.578
1.2
()
After
4.72
same.
hours,
μg
1.31
the
level
of
eah
drug
will
e
remains.
Use
the
part
()
unrounded
to
avoid
value
0.577 622 650 466 63...
unneessary
aumulation
of
in
error.
y
8.81
–0.5776226505
f9(x)=20
x
e
(4.720433359, 1.308789999)
x
f8(x)=10
(0.65)
x
–0.63
13.02
Exle 2.4.4
The
marketing
department
of
a
shool
wants
to
(a)
Model
the
numer
of
views
V
after
t
hours
with
L
model
the
time
it
takes
for
its
soial
media
a
posts
logisti
model
V(t)
=
and
determine
kt
1 + Ce
to
e
viewed
y
the
shool
ommunity
.
The
shool
the
ommunity
onsists
alumni.
The
2hours
after
posting,
of
the
een
that
has
743
marketing
20memers
post
of
everyone
the
a
department
post
has
ommunity
seen
will
students,
at
y
171
some
parents
finds
een
and
memers.
time
look
that
viewed
after
at
7
()
y
hours
Assume
the
parameters
L,
C
and
k
and
post.
The
marketing
stop
have
department
monitoring
een
Predit
for
soial
viewed
how
y
media
95%
many
proedure
posts
is
to
when
they
of
the
ommunity
.
hours
the
post
will
emonitored.
82
/
2.4
Gr a p H s
and
f E at u r E s
of
modEL s
(aHL)
Solution
(a)
The
arrying
so
L
=
V(2)
=
apaity
is
Interpret
743
the
parameter
743.
ontext
in
order
to
identify
a
diretly
.
743
=
Set
20
up
and
solve
a
system
of
equations
using
2 k
1 + Ce
helaws
of
indies.
Use
the
unrounded
and
C
743
V(7)
=
=171
7 k
1 + Ce
Hene,
2k
20
+
7k
20Ce
=
743
=
171
+
171Ce
2 k
20Ce
=
723
5 k
⇒
=
171Ce
572
20
723
171
=
e
7 k
572
572
20
723
171
ln
⇒
k
=
=
0.476
5
723
⇒
C
=
=
93.7
2 ( 0.476 )
20e
()
95%
of
743
is
705.85
743
705.85
=
⇒
t
=
=
values
93.668 087 922 075
k
to
=
0.476 040 302 745 52
avoid
unneessary
15.7217
kt
1 + Ce
After
15.7
aumulation
hours
the
post
will
no
longer
1094.34
of
error.
y
743
f2(x)=
e
monitored.
-0.4760403027.x
1+93.66808792
f1(x)=705.85
e
(15.7217708, 705.85)
100
x
–6.42
10
53.58
s ampLE studEnt ans WEr
The
population
of
fish
in
a
lake
is
modelled
y
the
funtion
1000
f (t)
,
=
0
≤
t
≤
30,
where
t
is
measured
in
months.
0.2 t
1 + 24e
(a)
Find
the
population
()
Find
the
rate
at
of
whih
fish
the
when t
=
10.
population
of
fish
is
inreasing
▲ The
when
t
=
student
sustitution
the
()
Find
the
makes
the
orret
10.
value
of
t
for
whih
the
population
of
fish
answer
and
orretly
235.402...
as
interprets
235
in
is
thisontext.
inreasing
most
rapidly
.
1000
▲ The
(a)
f(10)
=
1
f(10)
=
24e
0.2(10)
1
+
reognizes
that
the
rate
of
hange
is
the
derivative.
235
(
dx
+
1000
d
(b)
student
0.2(10)
)
=
24e
′ (10)
=
0
▼ The
student
to
nd
the
of
f
ould
use
a
GDC
0
at
t=
numerial
10,
whih
derivative
is
36.0
sh
to
nd
the
permonth.
▼ Using
value
t
=
of
15.9
t
the
GDC
whih
months.
efient
maximizes
This
is
the
f ′
gives
most
method.
83
/
2
Exle 2.4.5
A Ferris
wheel
Passengers
and
Beth
different
The
pod
t
reserve
times.
is
takes
an
emark
vertial
where
in
amusement
a
tikets
Alex
15
time
for
a
departs
distane
the
viewing
of
in
pod
ride
at
Alex’s
minutes
minutes
and
park
the
at
has
S,
on
3
the
radius
m
from
Ferris
30
m.
the
ground.
wheel
ut
Alex
depart
at
10:00.
pod
from
the
ground
in
whole
metres
after
10:00.
One
pods
rotate
anti-lokwise.
is a(t),
revolution
of
the
Alex
(a)
Determine
the
()
Beth’s
departs
pod
parameters
c
of
the
minutes
model
after
a(t)
Alex.
=
p
Given
+
qos(rt),
that
4.5
t
≥
0
minutes
S
Beth
after
the
10:00,
model
()
Explain
(d)
Alex
Beth
b(t)
how
and
vertial
=
b
is
p
8.73
+
m
qos(r(t
represents
Beth
want
height.
aove
to
Find
at
c)),
the
film
the
t
fat
eah
what
ground,
≥
the
value
of c
in
c
that
Beth
other
time
find
for
they
departs
their
will
c
minutes
soial
media
photograph
after
when
eah
Alex.
they
will
first
e
at
the
same
other.
Solution
(a)
The
maximum
3
and
m
63
m
and
minimum
hene
p
=
33
heights
and
q
=
are
The
30.
In
value
this
of
q
ould
ontext,
e
when
t
either
=
0
the
positive
height
or
is
3
negative.
m,
so
q
2π
must
The
period
is
15
so
r
e
–30
=
15
()
b
(4.5)
=
8.73
=
33
30os
⎛
2π
⎞
( 4.5
⎜
⎝
=
45
The
os
the
c
fat
that
minutes,
her
(d)
domain
b
8.73
33
⎜
order
the
to
expression
avoid
the
in
your
aumulation
is
its
Alex
at
=
3.00
⎠
equation
has
the
error.
⎞
⎟
30
and
after
Beth
in
enter
een
very
represent
moving
start
for
of
journey
.
Beth
the
⎛
⎝
of
one
of
2π
()
alulator
only
c)
⎟
15
1
c
and
⎠
15
⇒
Rearrange
and
Alex
ground
at
will
oth
e
57.3
m
aove
10:09.
s ampLE studEnt ans WEr
At
an
amusement
rotates
the
at
a
park,
onstant
ground. A seat
a
Ferris
speed.
starts
at
The
the
wheel
with
ottom
ottom
of
of
diameter
the
the
wheel
111
is k
metres
metres
aove
wheel.
diagram
not
to
scale
111
k
ground
The
wheel
ompletes
one
revolution
in
16
minutes.
84
/
2.4
(a) After
After
t
8
minutes,
minutes,
the
the
height
π
⎛
y
h(t)
=
61.5
+
aos
Find
the
value
()
Find
when
third
time.
the
of
is
of
117
the
m
aove
seat
the
aove
ground.
ground
is
and
f E at u r E s
8
minutes
=
k
111
=
=
⎟
,
for
0
≤
t
≤
32
⎠
8
a.
seat
is
30
m
aove
the
ground
for
the
117m
above
(half
the
revolution)
student
h(t)
orretly
information
given
showing
applies
in
lear
the
reasoning.
6
6m
π
(b)
(aHL)
given
ontext,
117
modEL s
Find k
▲ The
(a)
of
⎞
t
⎜
⎝
()
seat
Gr a p H s
=
61.5
+
acos
▲ The
the
61.5
+
acos
replaes
orret
information
in
the
formula.
π
=
orretly
)
8
h(8)
student
t
(
t
(
)
8
π
117
=
61.5
+
acos
(8)
(
)
8
▼ The
value
of
a
an
e
found
π
55.5
=
acos
(8)
(
more
)
efiently
with
the
formula
8
(117 − 6)
for
55.5
=
acos
(π)
cos
π
amplitude:|a|=
2
use
55.5
=
55.5
=
a
=
a
(
+
(
55.5)cos
Pieewise
funtions
parts.
piees
ontinuously
ramp
at
a
GDC
to
of
enale
the
moves
) =
skate
park
has
orret
down.
The
hek.
you
one
three
equation
GDC
way
to
is
the
solve
is
written
most
the
equation:
30m
t
to
funtion
from
▼ The
efient
t
(
8
The
a
a
55.5
61.5
of
1)
π
(c)
and
1
=
model
must
piee
e
to
distint
phenomena
set
the
up
so
other.
piees
to
that
that
In
the
have
=
18.5
minutes.
distint
funtion
Example
2.4.6,
the
model.
Exle 2.4.6
Daniel
measures
reords
his
data
the
on
dimensions
this
of
the
ramp
at
a
skate
park
and
sketh:
1.2
(5.5, 0.51)
1.2
(0, 0)
3.6
3.6
Model
the
ross-setion
of
the
ramp
with
a
pieewise
funtion.
85
/
2
Solution
The
e
first
part
modelled
of
the
y
a
ramp
an
linear
The
urved
gets
loser
part
of
the
and
loser
in
smooth
ramp
to
the
1.2
funtion
with
horizontal
and
slope
a
way
so
3.6
that
y-interept
at
(0,
0),
skating
is
safer.
Therefore,
giving
Daniel
represents
the
urve
with
1
l(x)
x
=
an
exponential
deay
model.
3
The
seond
part
line
h(x)
The
urved
is
a
horizontal
Daniel
=
stores
the
parameters
in
1.2
his
part
is
GDC
memory
in
order
to
modelled
ensure
that
the
piees
h
and
c
x
y
c(x)
=
ka
join
at
x
=
4.8
4.8
c(4.8)
=
ka
c(5.5)
=
ka
=
1.2
=
0.51
5.5
1
0.51
⎛
Hene
a
=
⎜
and
⎟
⎝
The
1.2
0.7
⎞
1.2
model
=
4.8
⎠
an
⎧
k
a
e
written
as
1.1
0.286313010365
1
1
x
0
≤
x
≤
3.6
0.7
⎪
0.5
3
→a
⎪
1.2
r(x)
=
1.2
⎨
⎪
3.6 <
x
≤
4.8
4.8
x
≤
8.4
485.673096093
1.2
x
ka
<
→k
4.8
⎪
a
⎩
1.2
4.8
The
GDC
onfirms
and
ontinuity:
the
shape
k
a
1.1
y
4.1
1
aee 
x, 0≤x≤3.6
3
f3(x)=
1.2, 3.6<x≤4.8
In the modelling process, giving
x
k
1
distinct models distinct names
a
4.8<x≤8.4
x
is correct notation, and helps
–0.14
8.79
1
your communication.
In
some
the
ases,
you
parameters
ontinuous
For
of
use
the
another
domain
in
order
of
one
of
to
“join”
the
the
piees
piees
to
to
determine
make
a
urve.
example,
3
⎧
f ( x) =
x
⎨
0
≤
x
x
>
2
if
the
≤
2
3
1
(x
a)
⎩
1.1
is
a
ontinuous
funtion
two
piees
meet
3
at
one
point.
You
find
3
2.5
this
point
The
solution
This
2
in
y
value
solving
to
of
this
the
f (2)
=
3
2
equation
parameter
is
=
a
1
=
fores
=
1
(2
a)
for
the
parameter
a.
2.
the
two
piees
of f
to
meet,
as
shown
diagram.
1.5
The
old
line
is
1
3
⎧
0.5
f ( x) =
x
⎨
0
≤
x
x
>
2
≤
2
3
1
(x
2)
⎩
0
0.5
1
1.5
2
2.5
3
3.5
and
the
other
urves
are
those
for
different
values
of
the
parameter.
–0.5
86
/
2.4
Gr a p H s
and
f E at u r E s
of
modEL s
(aHL)
slg be  lezg 
The
appliation
useful
For
and
of
logarithms
revealing
example,
a
data
helps
perspetives,
set
of
you
represent
aiding
population
the
and
explore
modelling
density
for
197
data
in
proess.
ountries
1.1
measured
value
in
11 929
the
(Monao)
values
of
on
of
the
less
oth
and
data
than
axes
of
people
minimum
shows
1000
limit
per
a
the
value
very
people
per
square
high
1.93
has
of
km.
this
maximum
(Western
frequeny
square
usefulness
km
The
for
180
Sahara).
population
wide
range
ycneuqerF
A histogram
densities
numer
of
diagram.
120
60
Creating
a
new
y
is
data
set
in
whih
eah
population
density p
is
replaed
0
log
p
an
example
of
applying
logarithmi
saling
to
the
data.
This
10
0
histogram
For
of
the
example,
whih
when
saled
Armenia
saled
is
data
has
set
a
reveals
more
population
represented
y
aout
density
log
100
of
=
approximately
Armenia
logarithmi
densities.
For
in
densities
are
sale
one
the
the
an
example,
approximately
ountries
in
then
there
tenth
lass
modal
of
with
e
are
lass
used
8
of
2.
midpoint
This
1.0,
data
ompare
ountries
Armenia’s.
12000
1.1
this
to
9000
100,
with
is
32
set.
population
eause
meaning
24
population
that
there
their
densities
are
8
ycneuqerF
The
plaes
6000
popdensity
10
This
3000
thedata.
16
8
population
1
approximately
10
=
10.
There
is
only
one
ountry
0
(Monao)
with
Armenia.
This
population
density
approximately
100
times
that
0.5
of
1.0
1.5
2.0
2.5
3.0
3.5
4.0
logpopdensity
is
eause
Monao
is
in
the
lass
withmidpoint
4,
4
meaning
that
its
population
density
is
approximately
10
=
10 000.
On a logarithmic scale base a, with each increase of one unit on the logarithmic
scale, the value of the unscaled variable increases by a factor of a. This means
that you can use logarithmic scales to represent and explore the size of a
change in a variable in propor tion to itself.
You
an
model
a
data
set
y
appliation
of
graph
transforms,
algera
In chemistry, the acidity of a
to
determine
parameters,
regression
tools,
or
y
a
omination
of
these
solution (pH) is measured on a
methods.
Linearization
gives
you
another
view
of
a
data
set
whih
logarithmic scale base 10. This
is
valuale
eause
it
applies
logarithmi
well
giving
saling
to
make
patterns
in
means that a change in pH from
the
data
stand
out,
as
as
you
another
method
y
whih
to
3.7 to 4.7 represents an decrease in
determine
parameters.
acidity by a factor of 10.
For
example,
the
years
onsider
1920
to
the
task
of
modelling
China’s
population
over
1980:
Year
1920
1930
1940
1950
1960
1970
1980
Population
472
489
520.8
556.6
682.0
825.8
981.2
in
The
millions
graph
of
the
data,
where
years
are
the
numer
of
years
after
1900
1.1
is
shown.
950
The
data
an
e
modelled
y
a
pieewise
funtion
800
l( x) =
2.856 x +
409.6
20 ≤
x
<
50
160.0
50 ≤
x
≤
80
pop
⎧
⎨
14.18 x
⎩
650
2
The
r
values
for
eah
piee
are
0.977
to
data.
and
0.998
respetively
,
showing
500
f11(x):=l(x)
an
extremely
lose
fit
the
10
20
30
40
50
60
70
80
years
87
/
2
However,
population
anexponential
growth
funtion.
is
most
ommonly
Linearization
an
e
modelled
applied
to
y
explore
the
datafurther.
Lez  exel elh
The
an
steps
of
the
exponential
(1)
Apply
linearization
relationship
logarithmi
proess
are
to
determine
desried
saling
to
the
and
the
parameters
applied
as
of
follows:
dependent
1.1
variale
(This
6.8
The
is
y
an
popnI
an
graph
example
graph
However,
6.5
and
looks
this
is
exponential
it
of
very
against
a
semi-log
similar
evidene
model,
the
as
to
that
you
independent
variale
x.
graph)
the
the
will
one
data
on
the
very
previous
losely
page!
fit
see.
6.2
10
20
30
40
60
50
70
80
years
(2)
Apply
lny
E
=
linear
mx
+
regression
to
find
the
est-fit
line(s)
in
the
form
c
F
=LinRegM
The
T itle
Linear
lny
RegEqn
m*x+b
a
0.00557...
b
6.03647...
R...
model
=
is
(0.005 77..)x
+
(6.036..)
2
with
Use
r
value
your
given
y
0.982 87...
GDC
your
memory
to
store
the
values
to
the
full
auray
GDC.
2
r
0.98287...
(3)
Apply
mx
y
K
=
algera
+
c
c
e
=
e
to
find
the
exponential
model
mx
e
L
(4)
=ExpReg(
You
an
now
(6.036..)
y
Exponen...
T itle
=
e
write
the
exponential
model
(0.005 77..)x
e
whih
simplifies
to
x
y
RegEqn
a*b^x
a
418.415...
b
1.00559...
=
418.415(1.005 59)
This is exactly how your GDC works out exponential
2
regression. For this reason, r
2
r
value for the exponential
0.98287...
and linear regressions are equal.
(5)
*china
1.2
By
repeating
these
steps,
the
pieewise
exponential
model
DEG
e(x)
an
e
ompleted:
950
x
⎧ 418(1.006
⎪
e( x) =
800
)
20 ≤
x
<
50
50 ≤
x
≤
80
⎨
x
pop
⎪
⎩
217.7(1.019
)
650
2
has
r
values
showing
a
for
eah
etter
fit
piee
than
of
the
0.9829
linear
and
0.999
pieewise
respetively
,
model.
500
f11(x):=ex(x)
This
10
20
30
40
50
years
60
70
is
an
important
advantage
given
the
size
of
the
values
of
80
the
dependent
variale.
88
/
2.4
Refletion:
The
exponential
linearization
model
whih
proess
fits
the
showed
data
that
points
there
more
is
Gr a p H s
and
f E at u r E s
of
modEL s
(aHL)
an
losely
than
a
2
linear
The
model
sine
exponential
population
Note
and
model
that
the
model
1.9%
per
year.
hange
of
and
The
is
also
is
greater
more
With
the
rate
the
hange.
shows
model
14.18
modelling
of
value
e(x)
population,
population.
size
r
for
the
appropriate
exponential
to
the
model.
ontext
of
growth.
2.856million
of
eah
two
l(x)
has
two
per
year.
whereas
l(x)
expresses
growth
is
you
the
e(x)
an
more
growth
population
million
linearization
of
population
size
explore
proportional
of
hange
ontexts
appropriate
0.6%
increases:
expresses
the
rates:
than
where
modelling
Lez  we elh
The
a
steps
power
(1)
of
the
linearization
relationship
Graph
the
are
logarithm
as
of
proess
for
determining
the
parameters
of
follows.
the
aee 
dependent
variale y
against
x.
a
the
If it is not clear which type of
logarithm
of
the
independent
variale
(This
is
log-log
graph.)
linearization to apply, find the
(2)
Apply
linear
regression
to
find
the
est-fit
line(s)
in
the
best fit lines for the semi-log and
form
the log-log graphs and choose the
lny
=
m
lnx
+
c
best fit.
m
(3)
Apply
algera
to
find
the
power
model
y
=
ln(x)
+
c
e
c
=
e
m
ln(x)
c
e
=
e
m
x
Exle 2.4.7
Jonathan
is
phone
emit
to
distanes
interested
from
Distane
(m)
(D)
his
and
as
a
measures
sound
its
engineer.
loudness
at
He
the
uses
his
following
phone.
40
80
100
160
240
320
400
500
600
60
27
11
6
5
4
3
2.7
1
0.6
(y)
Use
linearization
Hene
or
a
to
determine
power
determine
signifiant
()
areer
20
exponential
()
sound
a
(x)
Loudness
(a)
a
in
the
data
is
est
modelled
y
an
of
your
model,
orret
to
five
inverse
square
law:
figures.
reads
inversely
proportional
is
the
funtion.
parameters
Jonathan
model
if
that
onsistent
loudness
to
with
the
this
follows
square
the
of
distane.
Desrie
if
it
is
his
laim.
Solution
(a)
Linearizing
that
the
the
log-log
data
shows
graph
DEG
1.1
4.5
has
y
=
–0.006390888673
x+3.18781930183
2
the
est
fit
eause
the
r
is
=
0.863333837653
3.0
greatest:
0.949
>
0.863
duolnl
Hene
the
data
is
est
1.5
modelled
Semi–log
y
a
power
graph
funtion.
0.0
0
100
200
300
400
500
600
dist
89
/
2
()
lny
=
(
1.201 05..)
(7.692…)
lnx
+
DEG
1.1
so
4.5
y
(7.692…)
y
=
e
(
=
–1.20105940944
=
0.948612196212
x+7.69289453816
1.201 05..)
x
3.0
duolnI
Hene
(
y
=
1.2011)
1.5
2192.7x
Log–log
()
Jonathan’s
modelled
data
y
a
is
power
3.0
relationship,
square
law
power
law.
However,
is
far
value
to
reflet
the
also
the
2,
on
olletion
and
so
inverse
3.5
4.0
4.5
5.0
5.5
6.0
6.5
Indist
a
Jonathan’s
from
of
is
graph
0.0
est
power
theoretial
he
his
may
need
data
methodology
.
aee 
In an examination question that examines your knowledge and
understanding of linearization, show all steps in your working to achieve
maximum marks instead of using your GDC.
90
/
PRACTICE
Design
SL
P R A CT I C E
A
has
a
deent
of
5
m,
as
QUE STIONS
shown
in
Q U E ST I O N S
thediagram.
PA P E R
1,
GROUP
1
y
1.
The
point
pointB
has
Write
has
oordinates
oordinates
down
of
1)
and
the
10
5).
the
oordinates
the
line
of
segment
M,
the
5
thgieH
midpoint
(6,
(–2,
)m (
.
A
AB.
x
0
10
The
point
C
has
oordinates
b.
Find
the
gradient
.
Find
the
equation
of
the
of
(0,
line
the
and
d
in
are
the
Horizontal
CM.
line
CM.
ax + by + d
form
30
40
50
60
8)
Write
Design
your
B
is
modelled
4 x + 50 y + 350
answer
20
=
0
where
a,
=
distance
y
the
(m)
equation
0
b
integers.
Determine
if
either
design
will
pass
the
safety
regulations.
2.
In
a
an
experiment,
surfae
the
distane
surfae.
in
varies
The
andelas
surfae
in
d
the
intensity
inversely
of
the
light
intensity
(d)
and
of
I
=
the
of
is
from
97
are
d
on
of
6.
the
from
given
when
d
L
the
=
e
aT
girder
girder
is
heated.
modelled
is
y
where
+ b,
Celsius °C.
that
=
A steel
an
measured
distane
You
light
square
soure
light
the
entimetres.
themeasurements
as
I
At
T
the
Its
length
linear
L
in
metres
funtion
is
the
temperature
in
231 °C,
the
length
steel
of
the
degrees
4.53metres.
32 m
.
Write
down
an
equation
that
shows
this
weremade.
information.
.
Find
an
equation
relating
I
and
d
At
b.
Hene
find
the
value
of
I
when
5x
3.
Consider
the
funtion
d
=
6.7 m
301 °C,
4.71
Show
that
the
.
6
2x
of
the
5
⎛
y-interepts
are
(2,
0)
and
⎜
0,
Determine
the
equation
of
x-
the
steel
girder
is
Write
down
3
the
an
piee
equation
of
that
shows
this
information.
and
⎞
respetively
.
⎟
⎝
b.
of
10
h( x ) =
oordinates
length
metres.
seond
.
the
b.
Hene
find
the
length
of
the
girder
at
276 °C
⎠
horizontal
GROUP
2
asymptote.
7
.
.
Explain
why
the
largest
possile
domain
of
The
numer
elerity
is
not
the
set
of
real
of
people
T
viewing
a
post
y
a
h
on
soial
media
is
modelled
y
the
numers.
t
.
Sketh
the
graph
of
funtion
M (t ) =
onstant
and
R × 2
,
0
≤ t
≤ 10,
where
R
is
a
h
t
is
the
time
in
hours
sine
the
post
1
4.
The
equation
of
line
L
is
y
2
=
(x
+
was
10 )
made.
Five
hours
after
the
post
was
made,
1
5
the
.
Write
down
the
gradient
of
numer
of
people
who
had
viewed
the
post
L
1
was
Point
A
lies
on
and
L
has
y-oordinate
99
328.
9
1
b.
Find
Line
L
the
is
x-oordinate
perpendiular
of
to
Find
b.
Interpret
.
Determine
the
value
of
what
R
R
A
L
2
the
.
and
passes
represents
in
this
ontext.
numer
people
through
1
y-interept
of
if
the
of
L
1
viewing
.
Determine
the
equation
of
L
.
Write
in
the
form
y
=
the
City
planners
are
a
ar
park.
exploring
Safety
two
designs,
A
State
a
of
the
ar
regulations
park
1500R
in
reason
state
that
must
not
exeed
why
it
would
not
e
andB,
the
domain
of
M
to
inlude
the
allreal
gradient
exeed
domaingiven.
reasonalefor
for
will
mx + c
.
5.
post
your
2
answer
the
numers.
0.09
91
/
2
8.
The
e
depreiation
modelled
y
of
the
David’s
ar
in
Euros
(€)
.
an
Apply
the
funtion
to
omplete
the
tale:
funtion
x
1
3
5.5
12
n
C ( n) =
13 300 × 0.86
+ 550,
n
≥
0
where
n
is
the
Depth
numer
of
years
after
David
ought
his
ar.
b.
.
Find
the
value
of
David’s
ar
after
6
Given
that
alulate
When
the
value
of
David’s
ar
falls
elow
an
exhange
it
for
a
€3500
disount
value
of
an
eletri
Calulate
after
how
The
many
omplete
the
two
e
ale
m
wide,
volume
of
the
water
it
holds
of
a
swimming
pool
is
modelled
funtions:
3
x
will
3
full.
perimeter
years
f
David
is
ar.
y
b.
pool
towards
11.
the
swimming
€2000,
when
he
the
years.
to
exhange
his
ar
2
( x) =
x
for
+
0.5x
+
5,
with
domain
2.5
≤
x
≤
4
8
4
x
thedisount.
and
g( x) =
1.4 ,
with
domain
3
≤
x
≤
4
81
David
e
predits
less
than
that
the
value
of
his
ar
will
never
€L
A top
as
a
view
shaded
measured
.
Write
down
the
greatest
possile
value
of
of
the
pool
region,
in
is
and
shown
all
in
the
distanes
diagram
are
metres.
L
2
2x
9.
Let
8
f ( x) =
+
d
3
2
x
+ 5
7
.
Sketh
the
graph
oordinates
of
of
all
f,
indiating
interepts
the
and
the
position
6
3
x
and
equation
of
the
horizontal
asymptote.
2
f (x)
=
–
x
+
0.5x
+
5,
–2.5
≤
x
≤
4
8
5
b.
P(t ) =
Let
f
population
after
(t ) ,
t
≥
0.
This
P
of
an
January
1,
2020.
is
island
used
in
to
model
thousands,
t
the
4
years
3
.
Predit
the
island’s
population
on
2
January1,
2023,
to
the
nearest
100.
1
.
Predit
the
long-term
trend
of
thepopulation.
x
–4
10.
The
depth
(in
metres)
of
a
swimming
pool
–3
–2
0
–1
1
2
3
4
5
when
–1
it
is
full
is
modelled
y
the
funtion
4
d:
x
g(x)
=
–
1.4,
–3
≤
x
≤
4
81
–2
⎧
1
0 ≤
x
<
3
3
≤
x
< 7
7
≤
x
≤ 12
⎪
1
⎪
d( x) =
1
x
⎨
4
–3
4
⎪
–4
2
⎪
⎩
where
of
the
x
is
the
distane
swimming
pool,
in
metres
from
as
shown
in
the
the
left
.
Find
the
range
b.
Find
the
greatest
of
f
end
diagram:
the
possile
x-oordinates
of
any
differene
two
points
etween
in
the
d
pool.
1
x
.
By
onsidering
the
graph
of
d( x) =
f
( x)
g ( x ),
–1
find
the
greatest
possile
differene
etween
–2
the
y-oordinates
of
any
two
points
in
the
–3
pool.
92
/
PRACTICE
GROUP
.
3
George’s
the
12.
The
volume
of
depth
of
the
set
water
is
off
at
from
least
the
4
m,
harour
and
he
if
wants
sail
in
the
afternoon.
Determine
the
earliest
varies
time
diretly
of
an
an
to
iosahedron
oat
QUE STIONS
with
the
length
of
at
whih
he
an
depart
the
harour
in
ue
hisoat.
its
edgex
14.
The
surfae
area
S
of
a
given
ue
an
e
2
An
iosahedral
with
edge
1.5
3
die
m
has
represented
y
V
V
the
funtion
S(V ) =
6
(
V
)
,
a
≥
0
where
is
the
volume
of
the
ue.
3
volume
of
7.36m
Thegraph
.
Find
the
equation
for
the
volume
V
of
of
S
is
shown
for
0
≤ V
≤
8
an
s
iosahedron
b.
An
in
terms
iosahedral
a
of
its
edge
greenhouse
planned
for
gardening
Find
volume
x
with
edge
5
m
is
show.
3
.
the
A jewel
in
the
volume
0.5 m
of
the
shape
greenhouse
in
m
of
an
iosahedron
orret
to
1
.
has
3
Calulate
edge
13.
The
y
length
depth
the
hours
The
of
in
upper
of
the
water
funtion
measured
of
the
,
after
lower
plae.
ounds
of
the
jewel.
in
h(t ) =
metres
and
deimal
a
r
harour
+
and
is
psin ( st° ) ,
t
modelled
where
represents
the
h
is
numer
midnight.
following
diagram
shows
the
graph
of
h
v
h
.
Write
down
the
value
of
S(0)
b.
Write
down
the
value
of
S(8)
6
(3, 5.2)
and
interpret
its
5
meaning
in
the
ontext
of
the
question.
4
The
range
of
S(V)
is
p
≤ S(V ) ≤
q
3
2
.
Write
down
.
Draw
the
the
values
graph
of
the
of
p
and
inverse
of
q
funtion
1
(V )
S
on
the
diagram
given.
1
(9, 1.2)
e.
In
the
ontext
of
the
question,
explain
the
t
1
0
2
4
6
8
10
12
14
meaning
16
15.
The
graph
minimum
has
at
a
(9,
maximum
at
(3,
5.2)
and
1.2)
In
an
top
of
Find
b.
Interpret
the
values
what
r
of
r,
p
and
a
tower.
(6) = 1
Its
horizontal
tower.
.
S
experiment,
of
its
of
a
footall
vertial
distane
Both
H
and
x
are
experiment
is
filmed
(7,
107)
(9,
y
the
is
kiked
height
x
from
H
is
the
measured
from
a
funtion
ase
in
the
of
the
metres.
The
s
represents
in
this
and
the
positions
(1,
143),
ontext.
and
63)
are
reorded.
H
is
modelled
2
funtion
H ( x) =
px
+ qx + r
93
/
2
h
17
.
A new
suuran
area
network
of
H(16,
representing
2),
roads.
of
a
ity
A health
a
is
planned
entre
position
is
16
with
planned
km
east
a
at
and
(1, 143)
2
km
The
(7, 107)
north
fire
named
the
(9, 63)
.
Find
station
Middle
the
roads
y-axis
.
and
Write
down
a
system
of
equations
in
p,
q
and
Find
the
determine
the
parameters
of
Find
the
Hene,
write
down
the
height
of
Determine
whih
the
the
oordinates
footall
is
at
its
of
the
the
isetor
and
of
Way
is
at
East
of
0).
20).
as
a
A road
line
L,
HF
L
P:
the
North
Avenue
is
the
x-axis.
whih
L
rosses
North
Way
.
at
East
F(2,
planned
through
points
Hene
find
the
plaza
whih
HF
rosses
North
Way
.
shortest
distane
from
the
to
the
fire
station
using
a
tower.
seletion
.
is
at
P(0,
H.
entral
b.
loated
points
and
plaza
r.
.
Hene,
is
equation
East
Avenue
.
entral
Way
pass
Avenue
x
the
perpendiular
Two
b.
of
point
maximum
of
Avenue,
at
the
East
roads
Way
Middle
and
Way
,
North
HF
height.
18.
You
are
given
the
24-hour
average
temperatures
1
.
Determine
the
16.
the
footall
oordinates
reahes
A roof
design
of
y
funtion
a
the
of
the
point
for
where
m
ground.
greenhouse
an
e
,
Beijing
=
1
where
represents
Month
modelled
m
=
0
represents
Feruary
(m)
and
so
January
,
on.
Temperature
°C
(T)
2
of
the
h
is
shown
measured
in
h( x ) =
5
for
≤
ax
x
+
≤
4x + c.
25.
Both
The
h
graph
and
x
0
–4.6
1
–1.8
2
4.7
3
13.6
4
20.0
5
24.4
6
26.0
7
24.7
8
19.8
9
12.6
10
3.9
11
–2.6
are
metres.
h
(0, 25)
(25, 0)
(–5, 0)
x
0
–5
5
.
Write
b.
Find
10
down
the
15
value
20
of
25
c
.
the
value
of
Plot
the
data
independent
.
Hene
find
the
using
tehnology
.
m
is
the
a
height
of
the
variale
and
T
the
dependent
roof.
variale.
The
roof
inthe
is
required
shape
of
ross-setional
a
to
fit
uoid
inside
with
a
uilding
b.
retangular
Assuming
this
model
data
Use
your
y
set
repeats
every
determining
the
year,
parameters
area.
in
.
the
data
answer
to

to
determine
if
the
the
model
T
=
kcos( pm°) + w,
where
k
<
0.
dome
.
Add
the
graph
of
T
to
the
data
plot
and
2
an
fit
in
a
retangular
area
of
1400
m
with
omment
length
32
on
how
well
your
model
fits.
m.
1
http:/
/www
.worldlimate.om/
gi-in
/
data.pl?ref=N39E116+1102+54511W
94
/
PRACTICE
PA P E R
1.
The
b.
2
management
Chumnet
ollet
susriers
after
to
of
the
data
their
soial
on
site
the
media
numer
every
month,
Use
exponential
thevalues
of
of
new
signifiant
figures.
w
weeks
.
Predit
week
.
New
susriers
1
320 000
2
300 000
3
280 000
4
265 000
5
250 000
and
of
B,
of
new
to
determine
orret
to
six
site
start-up.
w
A
regression
QUE STIONS
the
10,
numer
to
the
nearest
Predit
the
numer
of
susriers
of
susriers
in
1000.
weeks
when
the
numer
S
2.
new
The
graph
of
=
y
g(x)
shows
passes
first
part
of
falls
a
through
elow
funtion
the
point
50 000.
g.
The
(2,
graph
4)
y
5
Chumnet
management
graph
the
data
and
(2, 4)
then
4
model
the
models
as
data
with
linear
and
exponential
elow.
3
h
g
2
320 000
1
x
–3
–2
0
–1
1
–2
290 000
srebircsbus
–3
weN
–4
–5
260 000
.
Write
down
the
domain
of
b.
Write
down
the
value
g(2).
.
Draw
the
of
g
x
0
1
2
3
4
5
graph
of
y
=
1
0.5x
on
the
sameaxes.
Week
.
Linear
Exponential
model:
Hene,
g(x)
model:
=
find
1
the
0.5x
solutions
orret
to
to
the
the
equation
nearest
0.2
w
S
=
mw
+
S
c
=
e.
AB
For
g(x)
2
r
how
=
State,
with
model
of
new
is
have
values
exatly
of
k
two
would
the
solutions?
equation
Explain
2
=
0.994
r
=
0.998
your
.
k
many
reasons,
more
why
an
appropriate
answer
y
adding
a
sketh
to
the
graph.
exponential
for
the
numer
susriers.
95
/
2
findthe
HL
P R A CT I C E
eah
PA P E R
1.
.
b.
1,GROUP
Sketh
State
the
the
under
the
graph,
giving
reasons
for
answer.
1
funtion
largest
area
Q U E ST I O N S
f
( x ) = ln ( x + 2 )
possile
domain
b.
f
( x)
is
transformed
to
f
(x
2)
.
f
( x)
is
transformed
to
f
( x) + 3
.
f
( x)
is
transformed
to
2
4
of
f
( 0.5 x )
f
1
.
Find
( x)
f
4.
Sarah
akes
a
ake
and
takes
it
out
of
the
oven
to
1
2.
.
State
.
Given
the
that
g !
find
domain
f
and
range
( x ) = 3x + 2
of
f
g ( x) =
and
2 cos ( x )
ool.
She
measures
ake
over
time
Hene
write
.
Given
that
h ! g
down
h( x ) =
has
the
5 sin ( 4 x
the
period
ax + b,
same
1) + 3
find
a
of
f
and
amplitude
and
the
! g
b
C
(°C)
prinipal
these
range
Sarah
elieves
C
e
an
axis
t( x) =
as
same
2 cos ( x )
values
of
a
and
b,
the
f
( x) =
30
45
60
75
98
69
50
28
23
20
that
=
+
ae
18
a)
x + 2
the
value
determine
ake
is
taken
ake
a
funtion
of
the
form
t
is
the
time
in
minutes
after
from
the
oven
to
ool.
the
.
Show
b.
Find
ln ( C
that
the
18 ) =
equation
18 )
x
<
5
x
≥
5
.
Find
.
Sarah
the
on
kt + ln ( a )
of
the
regression
line
of
when
its
t
value
wants
of
a
in
order
to
ensure
that
f
is
of
k
and
of
a
graph
of
y
of
=
f
( x)
is
used
to
model
the
first
falls
the
ake
under
19 °C.
Predit
long
Sarah
will
have
to
wait
until
she
a
lassroom.
You
are
graph
on
an
interval
given
[p,
q]
the
ake.
designs
The
depth
of
water
D
in
a
hours
after
harour
is
measured
a
in
of
deorate
funtion.
5.
roof
to
a
deorates
sketh
the
4
how
ontinuous
the
y
where
temperature
for
of
⎨
Find
The
temperature
kt
2
(x
8
⎪
⎩
.
the
funtion
⎧
⎪
15
h ! t
of
Consider
data:
0
modelled
ln ( C
3.
following
the
as
the
For
the
of
so
C(t)
.
ollets
temperature
f
b.
s( x ) =
and
internal
1,
t(minutes)
that
the
on
metres
at
t
midnight.
whih
2
the
area
under
the
graph
is
36.2
m
.
t(hours)
D(m)
t(hours)
D(m)
0
6.5
6
3.4
1
4.6
7
4.7
2
3.4
8
6.5
3
2.3
9
7.6
4
2.1
10
8.1
5
2.6
11
8
y
10
9
y
=
f (x)
8
7
6
.
5
Use
sinusoidal
regression
and
write
model
your
in
to
the
model
the
data
form
4
f
(t ) =
a sin ( bt + c ) + d
3
The
2
depth
thesame
of
water
time
zone
in
a
different
is
modelled
harour
in
y
1
p
g (t ) =
q
3 cos ( 0.5t ) +
4.7
x
0
1
2
3
4
5
6
7
8
9
10
b.
Find
the
ofwater
The
arhitets
transform
the
graph
to
time
in
in
both
hours
for
harours
whih
is
the
aove
depth
3.7
m
model
eahmorning.
different
designs.
For
eah
transformation,
96
/
PRACTICE
10.
6.
Consider
the
funtions
f
( x ) = 0.2 x
3
A population
of
ateria
grows
The
dish.
numer
of
ateria
N
is
reorded
at
different
+ 7
times
Find
Petri
1
x
1
.
a
and
3
g ( x) =
in
QUE STIONS
as
follows:
1
( x)
g
and
( x)
f
t
(hours)
0
0.5
1
2
30
70
320
1070
1
b.
(
Find
! g )( x )
f
(
and
( x)
! g )
f
N
1
1
1
.
Hene,
.
Given
show
(
that
( x) =
! g )
f
1
(g
!
)( x )
f
.
that
h(x)
=
ln(x),
show
Use
logisti
regression
to
find
a
logisti
model
that
N
(t)
1
((
f
! g ) ! h
)( x ) = (
! ( g ! h)
f
)( x )
b.
Hene,
predit
population
GROUP
You
maximum
the
ateria.
possile
2
.
L
7
.
for
the
are
given
the
logisti
funtion
f
Find
the
time
taken
for
the
population
to
reah
( x) =
kx
1 +
L, k , C
where
>
Ce
90%
of
this
A similar
.
Explain
why
y
=
L
is
a
horizontal
asymptote
=
f
value.
experiment
takes
plae
in
a
larger
Petri
of
dish
y
maximum
0
at
the
same
time.
The
numer
of
ateria
is
( x)
1351
modelled
y N
(t ) =
2
2.5t
90e
1 +
b.
Show
that
f
.
Show
that
the
x-oordinate
of
are
has
no
turning
point.
.
Find
the
interval
in
whih
N
is
more
than
2
of
the
point
where
10%
lnC
the
values
f
hanging
fastest
is
x
greater
N
than
1
=
k
1
11.
Let
f
g ( x) =
and
( x) =
ln ( x + 2 ).
1
2
.
Hene,
where
find
the
the
y-oordinate
values
of
f
are
of
the
hanging
point
x
Find
fastest.
(
f
the
+ 1
two
! g )( x ) =
solutions
( g !
of
the
equation
)( x )
f
2
8.
The
graph
of
the
C ( n) =
funtion
n
is
2
transformed
to
the
graph
y
of
=
2n
12.
+ 1
A metal
girder
of
length
10
metres
is
eing
2
heated
.
Desrie
the
sequene
of
transformations
in
needed
that
transform
the
graph
of
C
to
the
an
oven
as
part
of
the
proess
T
graph
of
to
harden
the
girder.
The
temperature
y
2
of
b.
Find
the
image
applying
The
graph
of
the
point
(3,
9)
after
E(t ) =
T
of
the
funtion
v (t ) =
sin ( t )
the
oven
E
is
modelled
107 + 0.15t ,
the
hardening
the
girder,
in
where
proess
metres,
t
y
is
the
time
egan.
is
funtion
in
The
modelled
seonds
length
y
the
L
sine
of
funtion
is
5
L(t ) =
transformed
to
the
graph
of
y
=
2.1 sin(
10 + 1.2 × 10
t
3t)
3
Find
.
Desrie
a
sequene
transforms
the
of
graph
transformations
of
v
to
the
graph
S
of
L
as
a
funtion
of
E
that
y
3
GROUP
.
Find
the
image
of
the
⎛
π
⎝
3
point
after
,
2
⎠
13.
applying
3
3 ⎞
A small
usiness
produes
and
sells
x
units
S
ofa
produt.
The
usiness
reeives
revenueR,
4x
9.
Consider
the
n( x ) =
funtion
,
x
∈ R.
measured
2
x
The
graph
Newton’s
of
n
gives
a
shape
in
1000’s
the
oordinates
of
the
turning
points
of
the
equation
200
x,
of
the
of
n
+
Sketh
the
graph
where
R
≥
0
and
0
≤
x
≤
a
8
Sketh
the
graph
of y
=
R(x),
showing
the
asymptote.
oordinates
b.
y
n
.
and
modelled
1600
serpentine.
x
Find
Euros,
thefuntion
alled
R( x) =
.
of
+ 1
of
the
maximum
and
oth
axesinterepts.
.
Find
the
minimum
value
of
a
suh
that
b.
f
( x ) = n( x ) ,
x
≥
a
has
an
your
Hene,
find n
determine
the
largest
possile
value
of a
inverse.
The
Justify
Hene
usiness
explores
various
models
to
try
to
answer.
inrease
their
maximum
revenue.
1
.
( x),
stating
its
domain
and
range.
One
model
is
S( x ) =
1.1R ( 0.9 x )
97
/
2
.
Desrie
the
the
graph
transformations
of
R
to
that
of
that
transform
16.
The
graph
of
y
=
f (x)
is
shown
elow.
S
y
.
Add
a
sketh
axes
as
of
y
=
S(x)
on
the
same
set
of
(5, 9)
e.
the
on
the
effet
maximum
giving
your
whole
numer.
Chaya
is
to
of
the
revenue
answers
exploring
ommute
the
shool.
and
orret
time
First,
transformations
the
units
the
nearest
to
taken
she
for
yles
3
sold,
her
daily
km
at
an
(1, 3)
–1
average
speed
average,
for
of
10
the
remaining
has
an
v km
km
to
.
h
minutes
8
(8, 9)

part
Comment
on
14.
in
She
for
a
then
us
Chaya’s
waits,
whih
shool.
on
travels
The
us
–1
average
speed
of
20
km
h
(3, 1)
x
.
Show
that
ommute
Chaya’s
an
e
average
modelled
speed
y
the
for
her
funtion
330 v
C( v) =
⎧
a
17 v + 90
x
1 ≤
x
< b
b
≤
x
<
d
d
≤
x
<
8
⎪
2
1
b.
Find
The
equation
of
f
is
f
( v)
C
( x) =
(x
⎨
c)
⎪
e
⎩
.
Hene,
find
the
value
of
v
required
for
Chaya
–1
to
have
an
average
speed
of
15
km
.
for
h
Find
the
values
of
a,
b,
c,
d
and
e
herommute.
2.5
⎛
The
.
Chaya’s
the
friend
same
laims
journey
she
with
an
an
arry
average
km
h
stating
15.
In
.
Determine
your
eonomis,
O
is
related
•
the
if
this
laim
speed
is
of
true,
to
give
is
then
give
reasons.
the
quantity
of
output
per
b.
month
f
is
translated
y
the
vetor
⎜
⎟
1
the
graph
vertially
the
Find
graph
the
of
a
funtion
strethed
of
a
image
with
funtion
of
(3,
1)
h,
whose
sale
⎠
graph
fator
2
to
g
under
this
sequene
of
transformations.
to:
numer
of
⎝
–1
20
graph
out
⎞
of
hours
of
laour
per
month
.
N
Find
the
equation
of
g
and
17
.
•
the
in
numer
a
month
of
hours
of
physial
apital
used
K
Tony
arries
the
equation
O
=
AN
disharge
of
He
measures
the
is
olleted
from
a
down
the
experiment
voltage
V
voltage
to
investigate
through
at
following
a
various
resistor.
times
t
and
data:
c
K
where
A,
b,
c
∈R
t
Data
an
the
writes
b
y
out
fatory
in
three
(seonds)
5
10
20
30
40
50
60
30
27
14
10
5
4
2
separate
V
months.
(V
olts)
ct
Tony
Month
O
N
wants
to
know
whih
of
V
=
ae
or
K
q
(units/
(hours/
(hours/
month)
month)
month)
1
6546
75
37
2
9216
80
45
3
3939
30
41
V
=
pt
est
models
following
semi-log
and
the
finds
using
his
graph
equation
tehnology
.
data.
(See
of
of
lnV
the
GDC
Tony
plots
against
line
of
est
sreenshot
the
t
fit
on
the
nextpage.)
Find
b,
c,
and
A
98
/
PRACTICE
.
1.3
1.4
letters
1.5
Show
that
the
volume
of
the
QUE STIONS
ylinder
V
is
DEG
modelled
y
the
funtion
2
3.4
V ( x) =
2 πx
( 36
x
),
where
domain
of
OB
=
x
2.6
Vnl
b.
State
the
V
1.8
Xavier
is
an
artist.
He
defines
a
“perfet”
1.0
ylinder
volume
0
10
20
30
T ime
40
50
as
as
for
Write
down
the
.
(seconds)
produt-moment
of
the
line
of
est
fit
Find
the
shown
on
b.
Plot
a
the
surfae
same
value
for
its
area.
equation
insried
of
the
in
surfae
area
of
the
S
Hene
find
the
radius
and
height
of
the
perfet
the
ylinder
semi-log
its
has
orrelation
.
oeffiient
that
60
ylinder
.
one
that
exists
in
S
graph.
log-log
graph
of
ln V
against
ln t
using
PA P E R
2
2
x
tehnology
and
use
it
to
plot
the
missing
1
2
1.
points
1.3
on
1.4
this
Consider
the
f
funtion
( x) =
e
2π
diagram:
letters
1.5
.
Write
b.
Sketh
down
the
largest
possile
domain
of
f
DEG
of
the
the
graph
of
maximum
f,
showing
point,
the
the
oordinates
equation
of
the
3.4
asymptote
and
the
equation
of
the
axis
of
the
oordinates
of
of
are
2.6
Vnl
symmetry
.
1.8
.
Use
tehnology
to
find
the
1.0
point
where
Consider
1.4
1.8
2.2
2.6
3.0
3.4
3.8
the
the
values
f
g ( x) =
funtion
4.2
growing
(x
f
fastest.
15 )
lnt
.
Write
down
oeffiient
the
of
produt-moment
the
line
of
est
fit
orrelation
of
the
.
Find
the
oordinates
e.
Find
the
area
the
log-log
x-axis
of
enlosed
and
the
lines
the
y
x
maximum
the
=
graph
10
and
x
point
of
=
of g
g,
20
graph.
The
.
Hene,
determine
Tony’sdata,
e.
Hene,
whih
justifying
determine
the
model
your
est
sketh
h( x ) =
fits
shows
the
graph
of
y
=
g ( x)
and
pg ( qx )
answer.
parameters
of
Tony’smodel.
y
18.
A ylinder
6m.
In
is
the
insried
diagram,
in
O
a
is
sphere
the
S
of
entre
of
[BD]
is
the
radius
of
the
g(x)
y
=
pg(qx)
the
.
sphere,
=
radius
Determine
the
values
of
p
and
of
q
suh
that
ylinder.
the
axis
of
symmetry
y-oordinate
half
of
that
of
of
the
of
h
is
x
=
maximum
30
and
point
of
the
h
is
one
g
D
B
g.
Explain
h,
the
why
x-axis
the
and
area
the
enlosed
lines
x
=
y
20
the
and
x
graph
=
40
of
is
O
the
2.
same
Johannes
motion
and
the
your
Kepler
relationship
orit
as
is
answer
(1571–1630)
redited
etween
sun
for
and
its
with
the
part
e
studied
planetary
disovering
time
average
taken
for
distane
the
a
planet
from
the
to
sun.
99
/
2
Planet
d
(Average
from
sun,
distane
t
in
(Time
orit,
of
one
PA P E R
days)
In
millions
of
a
iology
57.9
toxin
V
enus
108.2
225
Earth
149.6
365
Mars
228.0
687
an
Jupiter
778.6
4333
Saturn
1427.4
10 760
Uranus
2838.2
30 686
Neptune
4497.8
60 191
Pluto
5913.5
90 739
she
against
against
graph.
ln
D
d
using
using
Hene
a
tehnology
,
tehnology
omment
on
then
on
why
a
the
investigates
population
variale
administers
ateria,
the
rate
at
whih
of
dangerous
ateria.
and
a
the
dose
t
as
of
the
toxin
variale
B
as
time
to
a
the
in
minutes
population
numer
of
after
of
ateria
1
100
t
kill
defines
remaining
Plot
Rea
88
Rea
.
projet,
km)
a
Merury
3
after
ateria.
t
Rea
minutes.
reords
The
the
initial
population
following
t
is
data:
B
1
plot
0
100
500
71
1 000
25
1 500
6
2 000
2
ln t
separate
saling
oth
.
Plot
the
values
of
B
against
t
on
your
GDC.
1
variales
data
b.
Use
in
etter
the
a
log-log
than
est-fit
an
graph
helps
unsaled
line
to
find
represent
graph
the
the
Use
regression
tools
to
model
the
data
with:
would.
.
a
ui
.
a
logisti
model
parameters
model.
n
aand
n
of
the
funtion
t
=
ad
Write
.
If
a
new
from
the
predit
to
planet
the
sun,
the
was
disovered
apply
time
nearest
it
your
would
day
,
8500
million
model
from
part
take
orit
the
stating
any
to
b
to
b.
.
Sketh
sun
you
Use
set
Use
your
your
model
dwarf
of
data
of
eah
model.
points
and
eah
model
on
axes.
graph
to
predit
the
time
planet
Ceres
to
orit
the
equations
to
omment
on
how
well
average
distane
from
the
model
the
ontext.
for
predited
that
the
toxin
would
eventually
sun.
eliminate
Ceres’
equations
make.
Rea
the
the
one
the
.
the
assumptions
.
that
down
km
sun
the
population
of
ateria.
is
.
Comment
on
how
well
eah
model
fits
413millionkm.
Rea’s
e.
Calulate
the
perentage
error
in
your
Rea
alulation,
given
that
the
atual
predition.
time
for
expands
experiment
orit
is
1682
days.
Comment
on
your
following
data
shows
the
time
required
of
Jupiter ’s
moons
to
omplete
set
y
repeating
her
and
ounting
the
numer
of
ateria
one
250
minutes.
for
She
eah
data
answer.
every
The
her
one
plots
her
data
and
uses
regression
to
fit
a
orit
112.41
logisti
ofJupiter.
model
B
as
=
2
the
moon
X
(Average
distane
Jupiter,
.
Y
from
km)
(Time
one
Europa
671 000
85.2
Ganymede
1 072 000
171.7
Callisto
1 883 000
400.5
adapted
model
to
model
this
in
data
on
the
next
page.
hours)
42.5
your
sreenshot
orit,
422 000
if
on
0.08928e
of
Io
Investigate
GDC
shown
0.00368t
1 +
part
b
an
e
set.
100
/
PRACTICE
.
*Model
1.4
Use
tehnology
QUE STIONS
to
plot
the
data
points
set
and
the
modified
of
axes.
of
Rea’s
DEG
expanded
on
the
data
same
set
Comment
model
on
how
the
90
modified
60
that
the
model
fits
population
with
Rea’s
should
expetation
initially
inrease.
pop
In
30
a
105
repliation
ateria
of
after
the
100
experiment,
Rea
reords
minutes.
0
e.
3.0
400
800
1200
1600
Calulate
the
perentage
error
etween
2000
time
thisvalue
modified
t
and
the
value
predited
y
the
model.
B
2
.
0
100
250
97
Find
the
derivative
of
the
model
600
B
=
3
0.011t
4 + e
500
0.0029t
+ e
71
g.
750
Hene,
show
that
the
time
at
whih
the
44
population
1000
25
1250
13
h.
Hene,
will
reah
demonstrate
its
maximum
that
the
is
t
=
parameters
95.9
in
the
p
general
model
y
whih
=
rt
1500
6
1750
3
2000
2
the
time
at
maximum
Rea
Other
researhers
population
of
tell
Rea
ateria
that
in
increases
general,
initially
the
while
reads
arries
the
whih
are
the
out
r
and
work
an
the
of
rate
of
population
reahes
its
s
another
idential
maximum
determine
st
+ e
q + e
researher
experiment,
inrease
of
B
who
laiming
is
the
that
same
3
the
.
toxin
is
Suggest
to
the
taking
an
effet.
size
improvement
shape
of
Rea’s
that
model
B
ould
for
it
e
to
made
etter
2
model
After
her
expanded
reading
widely
,
data
Rea
.
as
its
maximum
Determine
Rea’s
if
this
rate
of
laim
is
derease.
true
for
experiment.
set.
onstruts
a
modified
600
model
B
=
3
0.011t
4 + e
0.0029t
+ e
101
/
G EO M ET RY
3
3 . 1
T R I G O N O M ET RY
W O R K I N G
W I T H
T R I A N G L E S
You must know:
✔
the
of
✔
a
the
sine
rule,
triangle
AND
cosine
You should be able to:
rule
and
the
area
✔
construct
formula
denition
of
labelled
angles
of
elevation
and
✔
use
Pythagoras’
tangent
ratios
right-angled
how
to
use
from
written
statements
depression
✔
diagrams
theorem
to
nd
the
and
sine,
sides
cosine
and
and
angles
of
triangles
bearings.
✔
identify
relevant
right-angled
three-dimensional
unknown
lengths
objects
and
triangles
and
use
in
them
to
nd
angles.
Right-angled triangles
c
Pythagoras’
theorem
describes
the
relationship
between
the
lengths
of
a
the
sides
in
a
right-angled
triangle.
b
In
the
2
2
c
=
a
This
The formulae for the trigonometric
triangle
can
shown,
c
is
the
longest
side,
the
hypotenuse.
2
+ b
can
be
be
rearranged
so
that
the
lengths
of
one
of
the
shorter
sides
found.
ratios are not in the formula book ,
2
a
2
=
2
c
b
so they need to be memorized.
If
you
are
given,
trigonometric
Note
or
need
to
an
angle,
then
you
should
use
the
ratios.
opposite
sin θ
find,
=
adjacent
,
cos θ
opposite
=
hypoteneuse
,
tan θ
=
hypoteneuse
adjacent
These are often remembered
using the mnemonic device
SOH-CAH-TOA.
If
finding
angle
an
and
angle,
then
begin
take
the
by
finding
inverse
of
the
that
sine,
cosine
or
tangent
of
that
value.
4
In
this
example
tan θ
=
7
4
cm
1
⇒ θ
=
tan
θ
⎛
⎝
7
4 ⎞
⎜
⎟
7
≈
29.7°
⎠
cm
x
The inverse functions can also be
To
find
a
missing
side,
you
need
to
rearrange
the
formula.
30°
written as arcsinθ , arccosθ
7
cm
7
7
and arctanθ
Here
cos 30°
⇒
=
x
x
=
≈
8.08 cm
cos 30°
102
/
3 .1
WORK ING
WITH
T R I A NGl E s
Ange o eeation and depreion
The
angle
of
elevation
is
the
angle
you
look
up,
from
the
horizontal,
angle
to
see
an
object.
The
angle
of
depression
is
the
angle
you
look
of
down
elevation
through,
from
the
horizontal,
to
see
an
object.
horizontal
angle
Eampe 3.1.1
The
the
angle
point
cliff
is
to
15°.
and
the
a
of
A
boat
The
the
depression
on
foot
same
top
B
cliff
level
of
on
is
of
the
100
the
as
a
A
vertical
sea
m
cliff
the
from
below
100
high
C
is
boat.
m
at
Find,
C
correct
to
the
nearest
horizontal
distance
foot
cliff
of
the
of
depression
to
metre,
BC
the
from
B
the
the
boat.
Solution
Always
15°
clear
100
m
by
diagram
given
The
begin
drawing
and
add
a
the
information.
angle
of
depression
is
15°
outside
x
of
the
the
triangle,
parallel
but
because
horizontal
lines
it
100
is
equal
to
the
angle
of
elevation
tan 15° =
x
from
the
boat
to
the
top
of
thecliff.
100
x
=
=
373.205
tan 15°
The
The
distance
nearest
is
373
m,
to
two
sides
question
are
adjacent
(the
involved
the
in
opposite
the
and
the
the
opposite
is
given
metre.
and
the
found),
adjacent
so
the
needs
tan
to
be
formula
is
used.
Triange in 3 dimenion
When
able
considering
to
construct
calculate
angles
three
dimensional
right-angled
and
shapes
triangles
it
within
is
important
the
shape
in
to
be
order
to
lengths.
Eampe 3.1.2
A storage
2m,
container
length
wishes
to
4.6
m
the
mast
4
m
store
long.
The
rests
is
and
the
height
mast
base
at
made
of
point
in
2
m,
from
the
the
as
his
mast
shape
shown
boat
is
of
in
a
in
the
placed
at
cuboid.
the
It
has
diagram.
container.
point B
Georg
The
and
width
mast
the
top
is
of
D
Use the 5 minute reading time
to decide which right-angled
triangles need to be used.
D
4.6
2
m
m
A
2
B
4
m
m
C
103
/
3
Find:
(a)
AD,
the
storage
(b)
the
height
of
the
top
of
the
mast
above
the
floor
of
the
container
angle
the
mast
makes
with
the
floor.
Solution
2
2
AB
(a)
=
4
2
From
+ 2
=
the
diagram
to
find
AB
you
right-angled
Only
one
another
the
AB =
20
is
clear
to
use
length
needs
in
is
to
right-angled
The
Pythagoras
need
triangle
value
for
intermediate
Using
it
that
20
ABD
known
be
is
so
found,
triangle
AB
the
using
ABC
an
answer
so
it
is
best
triangle
kept
as
an
exact
number,
either
ABD
by
leaving
in
the
square
root
2
2
2
AD
=
AD =
(b)
Let
(
4.6
1.16
angle
20
≈
)
=
1.16
1.08m
ABD
symbol
or
use
full
by
your
As
all
can
⎛
1
⇒ θ
=
=
cos
decimal
sure
value
you
given
GDC.
the
sides
are
known
you
select
any
of
the
trig
ratios.
20 ⎞
⎜
⎟
⎝
4.6
=
making
θ
=
20
cos θ
the
by
⎠
4.6
13.540... ≈ 13.5°
soing non-right-anged triange
The
sine
and
The
sine
cosine
rules
can
be
used
for
any
triangle.
rule
The sine and cosine rule are in
a
b
sin A
c
=
section SL 3.2 in the formula books.
sin A
sin B
or
=
sin B
=
a
sin C
sin C
=
b
c
To be able to use the sine rule you must know either two sides and one of the
opposite angles, or two angles and one of the opposite sides.
The
cosine
rule
2
2
a
2
=
b
2
b
+ c
2bc cos A ,
which
can
be
rearranged
to
give
cos A
2
+
c
2
a
=
2 bc
The
letters
formula
length
a,
so
in
a
question
remember
and
are
that
similarly
for
unlikely
angle
B
and
A
is
to
be
the
always
same
as
opposite
those
the
in
side
the
with
C.
In the first form of the cosine rule, always place the side opposite the angle
being used on the left side of the equation.
The area o a triange
The formula for the area of a
There
are
two
formulae
for
the
area
of
a
triangle.
triangle is in section SL 3.2 of the
1
formula book .
The
first
is:
Area
×
=
base ×
height
2
When
the
height
of
the
triangle
is
not
known
the
following
formula
1
should
be
used:
Area
ab
=
sin C
2
104
/
3 .1
Notice
that
between
this
is
the
product
of
two
sides
and
the
sine
of
the
WORK ING
WITH
T R I A NGl E s
angle
them.
Eampe 3.1.3
The
following
diagram
shows
quadrilateral
ABCD
A
11
59°
B
82°
100°
Note that in the exam DB is read
6
D
as ‘the length of the line from D
to B’ which is what is being asked
C
for here. The name of the line
segment from D to B would be
AB
=
11
cm,
BC
=
6
cm,
!
B AD
=
59°,
!
ADB
=
100°,
!
CBD
and
=
82°
written[DB].
(a)
Find
DB
(b)
Find
DC
(c)
Find
the
area
of
triangle
DBC
Solution
DB
11
(a)
As
an
angle
and
the
length
of
=
sin 59°
sin 100°
the
opposite
side
are
given
the
11
DB =
sine
× sin 59°
rule
should
be
used.
sin 100°
9.574...
≈
In an exam, for your final
9.57
When
cm
deciding
which
form
of
answer, write down both a longer
the
sine
rule
to
use,
remember
expression from your calculator
it
is
easier
to
begin
with
the
as well as the answer rounded
unknown
quantity
on
top.
to three significant figures.
2
(b)
2
DC
2
= (9.574...)
9.574...
×
6
×
+ 6
cos
2
To
×
find
cosine
82°
side
=
DC
rule
we
as
length
need
there
to
is
opposite
use
no
the
the
known
angle
111.677...
given.
⇒ DC
=
=
111.677...
10.5677...
≈
10.6
cm
1
(c)
Area =
× 9.57...
We
× 6 × sin 82°
need
to
use
adjacent
to
the
the
two
sides
2
known
angle.
2
=
28.44...
≈
28.4
cm
Bearing
A bearing
gives
the
direction
of
one
point
from
another.
It
is
measured
N
in
degrees
mark
add
For
the
clockwise
north
line
from
at
the
each
north
point
line.
and
In
then
bearings
use
the
questions,
parallel
lines
always
rules
to
angles.
example:
180
If
the
bearing
Using
the
the
B
from
properties
anticlockwise
Hence
of
from
bearing
of
N
of
A
55°,
find
the
parallelograms,
at
A
is
B
is:
from
180
B
is
55
360
=
bearing
it
can
be
of
A
seen
from
that
55
=
125°
B
B
the
–
angle
N
125°
125
=
235°
55°
A
105
/
3
s AMPlE sTUDENT ANs WER
A ship
is
175
an
is
sailing
km
island
from
C
is
north
at
E.
north
of
A.
The
134°.
from
a
point
A
Point
D
is
60
bearing
of
E
from
This
is
shown
in
towards
km
the
north
A
is
point
of
055°.
following
C.
D.
Point
There
The
C
is
bearing
of
E
diagram.
N
D
60
134°
C
175
E
55°
▼ The
not
student’s
include
Student
angle
be
a
has
CEA
north
angle
is180
−
360
46
−
line
correctly
but
measured
The
diagram
the
from
−
=
78
46
=
at
A
E.
calculated
bearing
needs
(a)
Find
the
(b)
Find
CE
(c)
Find
DE
bearing
of
A
from
E
180
–
134
180
–
(99
to
north.
between
134
does
CE
so
and
north
bearing
is
236°
(a)
134°
A from
46°
E
=
+
=
46
46)
=
78
078°
55°
▲ Student has correctly identied
they
need
to
use
sine
rule
as
(b)
CE
175
they
a
have
a
side
and
an
opposite
b
=
angle
=
Sin
and
got
the
correct
CE
⇒
A
Sin
Sin
B
78˚
Sin
55˚
answer.
6
178.9
=
⇒
Sin
b
=
CE
b
=
(Sin
55°)(178.9)
55˚
146.6
=
146.6
km
C
▼ One
of
the
mark
cosine
awarded
rule
but
for
the
use
side
from
235
part
b
has
been
place,
and
the
put
in
wrong
the
wrong
triangle
has
(c)
DE
55°
been
used.
The
correct
and
the
triangle
to
2
C
use
is
CDE
correct
2
=
a
+
193
km.
▼ The
by
C
student
writing
out
has
the
lost
=
(235)
expected
that
calculated
line
of
when
time
complete
the
it
would
value
directly
2ab Cos C
+
(146.6)
2
55 225
+
21 491
=
76 716.6
−
−
−
68
(68 902)
2(239)(146.6) Cos 55°
−
(470)(146.6)(0.57)
(0.57) =
76 716.6
−
39 520.6
2
=
37 196
be
would
from
−
2
=
C
calculation
b
answer
2
is
146.6
2
the
be
C
=
C
=
37 196
Distance
can't
be
negative
second
192.9
working.
DE
=
192.9
km
106
/
3.2
3 . 2
S E C T O R S
O F
V O L U M E S
A
s E c TO R s
the
formulae
sector
in
a
for
circle
A
cIR clE;
A N D
S U R FA C E
the
formulae
points
in
for
length
when
of
an
arc
given
an
and
area
angle
in
of
a
✔
degrees
the
the
distance
three-dimensional
identify
relevant
between
for
volume
three-dimensional
pyramid,
space,
formulae
of
circle
a
for
with
the
and
solids
right-cone,
The
and
surface
including
sphere
and
circumference
radius
r
a
right-angled
lengths
objects
and
triangles
and
use
in
them
to
their
✔
nd
distances
area
of
✔
right-
nd
the
shapes
between
space
volume
when
when
and
the
points
given
surface
sides
in
are
three-
their
area
of
formed
coordinates
compound
by
polygons.
hemisphere.
C
and
area
A
are:
minor
=
2 πr
A
and
=
nd
angles
arc
r
2
C
ARE A s
two
dimensional
formulae
s U R fA c E
A R E A S
three-dimensional
midpoint
✔
AND
You houd be abe to:
unknown
✔
vOlUME s
C I R C L E ;
You mut know:
✔
Of
πr
sector
q
The
length
l
of
an
arc
and
the
area
A
of
a
sector
of
a
circle
with
an
O
angle
θ
at
the
centre
are
θ
l
given
θ
=
and
× 2 πr
A
the
following
formulae:
2
=
360
by
r
× πr
major
360
arc
s AMPlE sTUDENT ANs WER
These formulae are in section SL
This
out
diagram
of
four
shows
equal
a
metallic
sectors
of
a
pendant
larger
made
circle
3.4 in the formula book .
of
20°
radius
OB
smaller
Angle
=
9
circle
BOC
=
cm
of
and
four
radius
OA
equal
=
3
sectors
of
a
A
cm.
20°
O
▼ The
Find
the
area
of
the
student
his/her
would
method
struggle
method
20
2
Circle
is
not
making
pendant.
= πr
Inner
Sector =
being
clear.
to
The
work
examiner
out
the
used.
2
×πr
9 ×
360
▲ The
= π
9
=
student
answers
∴
Circle
=
9
π
=
7π
will
− 2π
be
exact
full
20
Sector
using
exact
to
ensure
correct
values,
number
to
the
3
sf.
make
from
nal
If
sure
your
answer
not
using
you
use
the
calculator.
2
×π
=
is
2π
× 4 ×
9
360
=
▼ The
18π
but
∴ Area =
7π
+
the
given
numerical
are
answer
correct
should
be
18π
given
=
answers
to
3
sf.
25π
or
▲ The
2
the
method
calculations
is
not
are
clear,
clearly
but
set
out.
78.54cm
107
/
3
voume and urae area
You
the
need
to
be
volumes
right
cones
the
areas
the
surface
able
of
find:
cuboids,
and
of
to
cylinders,
prisms,
right
pyramids,
spheres
parallelograms,
areas
of
triangles,
cylinders,
cones
trapezoids
and
spheres
and
circles
and
other
three-
These formulae are in two different
dimensional
shapes
whose
surface
is
made
up
of
polygons
or
circles.
sections of the formula book: Prior
Learning and SL3.1
Eampe 3.2.1
3
A cylinder
of
(a)
Write
The
cylinder
(b)
Use
(c)
radius
down
is
your
area
S
in
Find
the
an
to
r
cm
and
equation
be
height
in
terms
constructed
answer
to
part
terms
of
r.
values
of
(i) r
(ii)
(a)
h
h
cm
a
volume
h
for
of
r
and
has
a
minimum
so
it
to
find
such
has
an
that
the
surface
500
cm
volume.
surface
expression
the
of
for
area
area.
the
is
surface
a
minimum.
Solution
2
(a)
500
(b)
S =
=
πr
All
the
formulae
the
prior
needed
are
in
h
Always write out the formula with
learning
section
of
the
2
the values substituted in; there are
2 πrh + 2 πr
formula
no method marks for just writing
From
part
h
(a):
=
2
2
V
= πr
book.
500
Don't
forget
to
add
the
area
of
the
πr
h
two
Use
this
value
of
h
in
circles
at
the top
and
bottom
has
been
the
of the cylinder.
formula
for
S:
500
S =
2 πr
2
×
+
2 πr
2
πr
This question could have been
1000
set as par t of Topic 5: Calculus in
2
=
+ 2 πr
r
which case you might have been
asked to differentiate to find the
(c)
(i)
minimum. Because there is no
From
S
method specified here you should
r
is
≈
the
minimized
3.41
choose the easiest route, which is
h
=
graph
expression
to
can
one
be
variable
r,
a
drawn.
500
Y
=
2
to use the GDC.
the
reduced
when
cm
500
(ii)
Now
GDC
=1000/X+2*π*X
2
2
πr
π
≈
×
3.41
13.7
cm
Using your GDC to find key
Minimum
Y=348.73421
points on a curve is in section 2.3.
X=4.3012715
Finding maximum and minimum
Though
the
solution
is
giving
points using calculus is in
intermediate
answers
to
3
sf,
the
section 5.1
calculation
using
the
the
of
full
h
should
value
of
be
r
done
from
GDC.
coordinate ytem
Sometimes
two-
coordinates.
In
or
three-dimensional
these
cases,
2
d
=
(x
x
2
the
)
2
+ (y
1
endpoints
lengths
y
2
of
the
)
shapes
can
be
are
defined
calculated
by
using
a
set
the
of
formula
2
+ (z
1
line
z
2
)
1
where
(x
1
, y
, z
1
1
)
and
(x
, y
2
, z
2
)
are
2
segment.
108
/
3.2
Similarly
,
the
coordinates
of
the
midpoint
of
x
endpoints
(x
, y
1
, z
1
and
)
(x
, y
2
1
, z
2
)
are
2
the
+
line
y
x
1
s E c TO R s
+
segment
z
2
+
vOlUME s
AND
s U R fA c E
ARE A s
with
2
,
2
cIR clE;
z
1
,
(
A
Both formulae are in section
y
1
2
Of
2
)
2
SL 3.1 of the formula book .
Eampe 3.2.2
A right
the
square-based
origin
O
corner
A,
length
are
the
at
one
directly
pyramid
of
the
and
set
corners
opposite
metres,
is
the
O,
are
of
in
a
the
(15.2,
z-coordinate
coordinate
base.
15.2,
The
0)
system
with
coordinates
where
represents
the
the
of
units
height
the
of
above
base.
Remember that in exams [AB]
(a)
Find
the
coordinates
of
B,
the
midpoint
of
[OA]
should be read as the line segment
V
joining the points A and B
Let
the
vertex
The
pyramid
(b)
Write
(c)
Find
of
the
has
down
the
a
pyramid
height
the
length
of
be V.
10.4
coordinates
of
m
of
V
[OV]
A
C
is
a
corner
of
the
base
of
the
pyramid,
B
adjacent
C
are
to
(10.7,
O.
0,
The
coordinates
of
O
0)
C
2
(d)
Show
that
the
surface
area
of
the
pyramid
is
300
m
,
to
2
sf.
Solution
(a)
(7.6,
7.6,
0)
x
Using
(b)
(7.6,
7.6,
+
x
1
y
2
+
y
1
z
2
,
(
+
z
1
2
,
2
2
)
2
10.4)
A right
pyramid
directly
over
the
has
its
centre
vertex
of
the
base.
2
(c)
(7.6
2
0)
+ (7.6
0)
2
2
+ (10.4
0)
Using
(x
x
2
=
(d)
14.9559...
Let
the
OCV
≈
15.0
height
be
of
)
2
+ (y
1
triangle
Each
side
is
an
z
2
with
base
10.7
isosceles
2
15.0
+ (z
)
1
triangle
V
10.7
2
=
2
)
1
m
h
2
h
y
2
(
)
2
h
h
=
13.966...
≈
14.0
m
O
1
Area
× 10.7 ×
=
C
10.7
14.0
2
2
=
74.719...
≈
74.7
An
m
equivalent
find
Surface
area
4
=
of
pyramid
the
method
coordinates
of
is
to
the
An exam mark scheme would
include both the unrounded and
=
midpoint,
M,
the
of
of
[OC]
and
rounded answers. Often both need
2
×
74.7
298.87...
≈
300
m
length
distance
[VM]
formula.
using
the
to be seen for you to be awarded
the final mark .
109
/
3
3 . 3
V O R O N O I
D I A G R A M S
You mut know:
✔
properties
✔
terms
such
of
used
as:
You houd be abe to:
perpendicular
bisectors
in
V
oronoi
diagrams,
sites,
vertices,
edges,
✔
nd
the
of
edge
the
of
line
a
cell
is
the
segment
perpendicular
between
nearest
neighbour
when
given
a
segment
line
two
create
within
interpolation
a
cell
the
same
rainfall)
as
the
value
that
the
of
assigns
value
the
solution
to
the
“toxic
lies
at
the
vertex
new
site
nd
the
of
a
two
and
its
points,
or
bisector
the
equation
of
midpoint
cell
on
a
V
oronoi
diagram
when
a
is
added
equation
of
an
edge
of
a
cell,
identify
site
closest
to
a
given
point,
or
calculate
the
quantity
of
a
cell
site
waste
identify
where
when
it
is
appropriate
to
use
the
dump”
nearest
problem
perpendicular
all
✔
✔
a
new
area
(e.g.
either
a
sites
the
points
of
bisector
✔
✔
equation
cells
✔
✔
the
neighbor
interpolation
and
when
to
three
apply
the
toxic
waste
dump
problem.
cellsmeet.
Perpendiuar bietor
Look back on how to find the
gradient and equation of a straight
The
line from section 2.1
midpoint
perpendicular
of
[AB]
bisector
and
is
of
[AB]
is
the
perpendicular
line
to
which
passes
through
the
[AB].
1
If the lines
y = m
and
x + c
1
y = m
1
are perpendicular then m
x + c
2
2
= −
2
m
1
If asked to show two lines are perpendicular, show that m
×
m
1
=
−1
2
Eampe 3.3.1
(a)
Find,
in
the
bisector
(b)
V
erify
(c)
Find
form
y
of
the
line
that
the
point
the
distance
mx + c ,
=
joining
C(8,
the
A(1,
6)
equation
5)
lies
and
on
for
B(7,
this
the
perpendicular
7)
line.
CA
11
10
Solution
9
(a)
The
gradient
of
[AB]
Using
=
the
formula:
8
y
B
7
5
2
7
=
y
2
1
gradient
=
1
=
x
7
1
6
x
2
3
1
6
M
A
5
The
4
gradient
of
perpendicular
bisector
=
−3
3
The
line
will
pass
through
Using
the
formula
2
the
midpoint
of
[AB]
=
x
+
1
1
x
y
2
(
1 +
(
0
0
1
2
3
4
5
6
7
5
+
7
2
7
2
= ( 4,
y
2
)
2
)
,
+
1
,
2
6)
(which
is
section
in
in
the
the
Prior
Learning
formula
book)
In this diagram, M is the midpoint
Equation
is
Using
the
point-gradient
form
of
of [AB].
y
−
6
=
−3(x
4)
the
equation
of
a
straight
line.
1
The gradient of [AB] is
and
3
⇒
y
=
−3x
+
18
the perpendicular bisector has
gradient −3.
(b)
3
×
8
+
18
=
6
This
the
is
verified
value
for
x
by
substituting
into
the
equation.
110
/
3.3
2
(c)
CA
Using
2
=
(8
1)
+ (
=
49 + 121
=
170
6
the
DI A GR A M s
formula
5)
2
=
vORONOI
d
=
(x
x
1
13.038... ≈ 13.0
2
)
+ (y
2
(which
is
section
in
y
1
in
the
the
)
2
Prior
Learning
formula
book)
Every point on the perpendicular bisector of [AB] is an equal distance from
both A and B
A V
oronoi
diagram
All
diagram
is
points
The
divided
within
boundaries
meet
is
consists
called
a
of
a
into
cell
the
a
regions
are
of
set
are
the
of
called
closer
cells
vertex
of
to
points
cells,
that
called
called sites
and
each
containing
than
to
site
edges.
The
any
point
the
one
other
where
site.
site.
the
edges
cell.
The edges of the cells lie along the perpendicular bisectors of the sites.
In examinations, there will always be three cells meeting at a ver tex.
A ver tex is always an equal distance from the three adjacent sites.
The
toxic
diagram
waste
that
is
dump
problem
furthest
from
is
any
finding
of
the
the
point
on
the
V
oronoi
sites.
In an exam, the solution of the
Nearest
neighbour
interpolation
assigns
the
value
at
the
site
to
toxic waste dump problem will
all
points
in
the
cell.
For
example,
if
the
sites
are
weather
stations
always be at a ver tex of the
recording
temperature
for
a
forecasting
website
then
every
point
in
the
Voronoi diagram where three
cell
is
assumed
to
have
the
same
temperature
as
at
the
weather
station
cells meet.
in
that
cell.
Eampe 3.3.2
The
diagram
shows
a
map
12
of
an
island
with
transmission
D
towers
A(2,
6),
B(6,
2)
and
10
C
C(11,
9).
A,
B
to
show
A V
oronoi
and
C
diagram
hasbeen
the
area
for
8
constructed
which
is
A
6
closest
to
towers.
solid
each
This
lines
in
of
is
the
three
shown
the
with
the
4
diagram.
B
2
A fourth
tower
added
the
at
D
is
point
then
(6,
10).
2
The
perpendicular
between
D
and
(a)
Write
(b)
Complete
A
down
clearly
the
and
the
the
4
6
8
10
12
bisectors
between
equation
V
oronoi
added
D
of
and
the
diagram
C
are
shown
perpendicular
for
the
four
as
dashed
bisector
sites,
of
lines.
[DB].
indicating
cell.
In an exam you will be given a
The
strength
distance
of
of
the
signal
point
from
from
a
tower
the
at
tower.
a
point
is
proportional
to
the
copy of the Voronoi diagram
to complete.
111
/
3
(c)
A receiver
A fifth
tower
as
possible
It
given
is
(d)
Find
is
is
to
from
that
the
at
the
be
added
each
the
point
of
in
the
equation
coordinates
of
(10,
4).
order
four
of
From
boost
existing
the
the
to
which
the
tower
signal.
It
it
is
receive
to
be
the
strongest
placed
on
the
signal?
island
at
the
point
as
far
towers.
perpendicular
point
will
where
the
bisector
fifth
of
tower
[DC]
will
y
is
be
=
5x
33
built.
Solution
(a)
y
=
6
The
perpendicular
line
passing
bisector
through
the
will
be
midpoint
a
horizontal
of
[DB]
(b)
12
First
draw
the
final
perpendicular
bisector.
D
10
The
edges
of
the
new
cell
can
be
formed
by
C
tracing
around
the
perpendicular
bisectors
8
added
with
the
new
site,
moving
to
a
new
A
edgewhenever
6
edges
meet.
(10,
is
you
reach
a
vertex
where
three
4
B
2
2
4
(c)
Transmitter
(d)
Of
the
other
y
=
two
–
33
8
10
12
B
vertices,
vertices
5x
6
is
at
meets
y
the
the
=
one
point
furthest
is
from
where
the
the
line
This
=
x
=
5x
33 ⇒ 5 x
=
39
cell
is
which
an
contains
example
of
tower
the
toxic
B
waste
problem.
point
vertex
furthest
where
each
of
A,
B
from
B,
C
or
from
three
and
C
cells
to
P
the
other
meet.
will
be
sites
The
2.
will
distance
The
be
at
a
from
distance
7.8
5
The
the
6
39
=
in
question
dump
The
6
4)
coordinates
are
(7.8,
6)
is
where
the
D
to
Q
fifth
is
clearly
tower
more
should
be
than
2
so
this
built.
12
D
10
C
8
A
Q
6
P
4
B
2
2
4
6
8
10
12
14
Sometimes you will be asked to find the areas of the cells. These are likely
to be in the shape of triangles or trapezoids – see the formulae in the formula
book Prior Learning section.
112
/
3.4
3 . 4
T R I G O N O M E T R I C
You mut know:
✔
the
denition
✔
the
denitions
unit
of
a
of
F U N C T I O N S
radian
cosθ
✔
sinθ
and
in
terms
of
convert
✔
the
denition
identity:
+
θ
sin
=
vice
versa
put
your
GDC
tanθ
when
given
triangle
one
there
angle
might
✔
use
and
be
an
two
two
sides
possible
of
could
be
into
the
trigonometric
the
arc
given
a
radians
correct
mode
when
functions
correct
formula
for
nding
the
length
of
or
in
the
area
of
a
sector
when
the
angle
is
radians
triangles
✔
that
and
as
cos θ
✔
degrees
1
sin θ
of
( A H L )
2
θ
cos
between
and
using
2
Pythagorean
(AHl)
the
✔
the
f UNcTIONs
You houd be abe to:
circle
✔
TRIGONOME TRIc
solve
a
triangle
using
the
sine
rule
or
the
cosine
drawn.
rule
✔
in
the
ambiguous
solve
trigonometric
nite
interval.
case
equations
graphically
in
a
The unit ire
The
sine
function
can
be
defined
as:
sinθ
is
the
y-coordinate
The link between trigonometric
of
the
point
P
when
unit
circle
centred
it
has
rotated
at
the
origin.
through
an
angle
θ
around
functions and circular motion is
a
revisited in the section 3.6
P
An instance of the application
1
y
sin
q
of the use of these formulae can be
q
seen in Example 3.6.8 in the section
q
on motion with variable velocity.
Similarly
,
of
the
circle
The
the
point
cosine
P
centred
when
it
at
origin.
following
definitions
of
function
the
has
identities
sine
and
can
be
rotated
can
be
defined
through
derived
as:
an
cosθ
angle
directly
is
θ
from
the
x-coordinate
around
the
a
unit
unit
circle
cosine.
2
sin
sin θ
2
θ
+ cos
θ
=
1
and
tan θ
=
cos θ
The ambiguou ae o the ine rue
Another
is
sin θ
that
This
means
between
This
identity
0°
means
drawn
=
and
that
when
tell
which
can
(180°
there
'Ambiguous'
to
sin
that
are
180°
be
θ )
here
two
triangle
is
p
when
that
are
and
an
the
when
possible
there
sides
means
=
from
(π − θ )
two
sin θ
sometimes
given
sin
or
always
for
derived
0
radians.
p
<
1
possible
angle
without
in
circle
solutions
≤
two
unit
triangles
opposite
more
one
information
of
it
that
the
is
can
be
sides.
impossible
intended.
113
/
3
Eampe 3.4.1
A triangle
ABC
(a)
on
Show,
of
(b)
C
as
Find
C
the
is
the
to
be
constructed
same
and
C ′
area
of
diagram,
the
smaller
!
A
with
two
=
30°,
possible
AB
=
6
triangles
cm
that
and
BC
could
=
be
5
cm
formed,
labelling
the
two
positions
the
conditions
triangle.
Solution
(a)
B
The
two
given
6
cm
5
5
in
triangles
the
ABC
and
ABC ′
both
satisfy
question.
cm
cm
30
A
C
C
ˆ
(b)
Using
sin
the
!
( A CB)
sine
sin
rule
Of
the
the
30
two
possible
diagram
that
solutions
the
larger
sin( ACB) =
to
angle
0.6
itis
clear
from
isrequired.
=
6
5
sin 30
⇒ sin
!
( ACB) =
× 6
=
Since
0.6
sin θ
=
θ )
sin(180
5
⇒
!
AC ′ B =
36.869...
!
ACB =
and
!
ACB =
180
36.869... =
!
ABC
180
30
143.1...
To
find
the
necessary
=
143.1... =
!
AC ′ B
180
area
to
of
the
calculate
smaller
either
of
AC
the
or
two
the
triangles
angle
itwill
be
ABC
6.869...
1
Area
=
×
6
×
5
×
sin
(6.869...
Make
sure
you
avoid
errors
always
use
the
full
display
on
your
GDC
to
2
due
to
premature
rounding.
2
=
1.79
cm
Radian meaure
Radians
are
an
alternative
angle
measure
to
degrees.
Have your GDC set to radians, and
π
radians
are
equal
to
180°
change to degrees only when the
question uses degrees.
This
means
that
π
•
to
convert
from
degrees
to
radians
you
multiply
by
180
180
•
to
convert
from
radians
to
degrees
you
multiply
by
π
If
measuring
formulae
for
the
angle
the
length
1
These formulae are given in
l
=
rθ
and
A
=
of
a
of
sector
arc
(l)
in
radians
and
area
of
rather
than
sector
(A)
degrees,
are
the
simplified.
2
r
θ
2
section AHL 3.7 in the formula book .
minor
arc
r
sector
q
r
O
major
arc
114
/
3.4
TRIGONOME TRIc
f UNcTIONs
(AHl)
Eampe 3.4.2
The
diagram
below
shows
two
concentric
circles
with
centre O
and
radii
3
cm
and
6
cm.
π
The
points
P
and
Q
lie
on
the
larger
circle
and
!
POQ
= θ ,
where
0
< θ
<
2
P
O
Q
(a)
Show
(b)
Find
that
the
the
area
value
of
θ
of
the
that
shaded
region
maximizes
this
is
4.5(4 sin θ
–
θ)
area
of
area.
Solution
(a)
Area
of
triangle
POQ
=
1
Using
triangle
=
ab sin C
2
1
×
6
×
6
sin θ
×
18 sin θ
=
2
1
Area
of
sector
=
2
×
3
Working
×
θ
=
18 sin θ
=
4.5 ( 4 sin θ
=
4.5θ
in
radians
and
using
the
area
of
sector
formula.
2
Area
(b)
of
shaded
1.318 11...
≈
region
4.5θ
θ )
From
1.32
the
radian
GDC
mode.
–
you
Note
need
the
to
ensure
answer
your
could
also
and
finding
θ
=
be
is
in
found
using
1 ⎞
⎛
differentiation
GDC
arccos
⎝
▲ The
4
⎠
student
has
correctly
s AMPlE sTUDENT ANs WER
A rectangle
is
drawn
around
a
sector
of
spotted
the
is
to
equal
as
shown.
If
the
angle
of
the
the
of
the
radius,
rectangle
which
was
a
correctly
circle
length
sector
calculated.
is
2
1
radian
and
the
area
of
the
sector
is
7
cm
,
▼ The
find
the
dimensions
of
the
rectangle,
student
incorrectly
your
answers
to
the
nearest
the
radius
of
the
circle
be r, so
7
=
=
14
=
2
r
2
r
the
3.741657386
length
of
the
and
used
×
rounding
in
question,
leading
the
incorrect
rectangle
answer.
You
down
full
a
rounded
is
3.74
the
rest
of
to
the
the
should
wrong
always
write
as
as
1
the
So,
rounded
millimetre.
1
Let
has
giving
full
answer
answer
answer
and
from
well
always
the
subsequent
working.
▼ It
have
a
use
GDC
in
180
Angle
is
=
57.3
π
180
x
would
been
easier
57.2
=
=
61.4
to
stay
in
radians
and
calculate
x
2
π
−
1
.
as
height
=
2 ×
3.74 cos(61.4) =
In
this
case
the
calculator
2
3.59 cm
would
need
for
nal
the
to
be
in
radian
mode
calculation.
115
/
3
3 . 5
M AT R I X
T R A N S F O R M AT I O N S
You mut know:
✔
the
order
in
compound
You houd be abe to:
which
to
multiply
matrices
for
a
✔
nd
transformation
the
of
✔
area
✔
if
of
image
=
|detT|×
the
area
of
the
line
makes
an
angle
of α
of
line
with
the
the
matrix
formula
(1,
0)
for
book
and
(0,
a
or
given
by
transformation
considering
the
from
image
1)
object
✔
a
( A H L )
nd
the
image
of
a
given
point
under
a
x-axis
transformation
then
the
gradient
the
is
tan α
✔
nd
✔
use
the
point
iterative
complex
that
has
a
techniques
given
to
image
generate
a
shape.
linear tranormation
A linear
such
transformation
will
always
Examples
of
reflections
origin
the
as
a
y-axis
leave
linear
in
or
represented
point
passing
and
x-axis
the
be
(0,
0)
transformations
lines
centre
can
horizontal
2
rotations
the
and
a
×
2
matrix
and
as
unchanged.
are
through
by
origin,
vertical
about
the
origin,
enlargements
with
one-way stretches
the
with
invariant.
The general form of matrices
You
will
need
to
know
how
to
find
the
matrix
representing
representing each of these
a
transformation,
either
by
using
the
form
of
the
general
transformations are in section
matrix
given
in
the
formula
book
or
by
considering
the
effect
AHL 3.9 of the formula book .
of
the
transformation
on
two
points.
Eampe 3.5.1
(a)
Find
the
matrix
that
represents
(b)
Find
the
matrix
that
transforms
Make sure you know how to find
(4,
3)
to
(8,
a
reflection
the
point
in
(1,
the
2)
line
y
(2,
4)
to
=
2x
and
6)
the angle (α ) between a line and
the x-axis from its gradient (m)
Solution
using m = tan α
(a)
m = 2 ⇒
tan θ
⎛
=
⇒ θ
=
63.43...
sin
×
63.43...)
2
The
matrix
given
cos ( 2
×
63.43...)
sin
×
63.43...)
(2
⎝
(2
cos ( 2
×
63.43...)
0.6
0.8
⎞
0.6
⎠
reflection
formula
book
is
but
the
value
of
θ
needs
to
⎟
⎠
be
⎛
the
a
⎞
first
⎜
in
for
found.
=
0.8
⎝
(b)
Let
be
⎛
the
transformation
matrix
T
1
⎛
⎞
4
T
2
8
Because
we
points
matrix
be
⎞
a
are
given
two
equation
can
formed.
=
⎜
2
⎝
3
⎟
⎜
⎠
⎝
⎟
4
6
⎠
Note
that
we
need
to
1
⎛
⇒
T
2
8
⎜
⎝
⎛
T
⎞
⎛
⎟
⎜
⎠
⎝
1
4
⎞
post-multiply
both
sides
by
=
2
4
0
6
⎞
⎟
2
3
⎠
the
inverse.
The
final
answer
is
obtained
=
⎜
⎝
⎟
0
2
⎠
directly
from
the
GDC.
116
/
3.5
M AT R I x
T R A N s f O R M AT I O N s
(AHl)
The quickest way to find a matrix that represents a reflection in either axis,
the lines y = ±x, or a rotation by a multiple of 90° is to consider the effect on
the points (1, 0) and (0, 1) rather than use the formulae.
If (1, 0) goes to (a, b) and (0, 1) goes to (c, d) then the transformation matrix
⎛
is
a
c
b
d
⎞
⎜
. See Example 3.5.2
⎟
⎝
⎠
compound tranormation
If
a
point
with
represented
position
by
matrix
undergoes
a
image
after
From
of
x
this
vector
R,
the
transformation
we
the
can
transformation
two
see
R
x
undergoes
image
will
be
represented
the
followed
single
by
S
is
transformation
Rx.
If
by
transformations
that
the
is
this
matrix S,
S(Rx)
matrix
that
=
point
then
(SR)
then
Make sure you know the
the
order in which to multiply
x
matrices to represent a
represents
the
compound transformation.
SR
Eampe 3.5.2
(a)
Find
the
single
reflection
rotation
in
of
matrix
the
90°
line
y
that
=
represents
followed
x
a
by
(b)
a
clockwise.
Show
that
twice
will
performing
return
each
this
transformation
point
to
its
startingposition.
Solution
⎛
(a)
Reflection
in
y
=
x,
0
1
1
0
⎞
Either
⎜
⎛
of
90°
clockwise
is
0
1
⎜
1)
matrix
is
that
(1,
0)
goes
to
(0,
1)
and
0
1
⎜
1
⎝
0
0
1
to
(1,
0)
or
use
the
formula
in
the
formula
book
angle
45°
and
the
fact
that
y
=
x
makes
an
⎠
⎞
⎛
⎟
⎜
⎠
⎝
0
1
1
0
of
with
the
x-axis
to
find
the
matrix
for
reflection.
⎞
⎟
⎠
The
rotation
Note
⎛
goes
⎞
the
⎛
single
fact
⎟
1
⎝
The
the
⎠
(0,
Rotation
use
⎟
⎝
the
matrix
order
of
is
found
matrix
in
a
similar
multiplication
way
.
used
to
⎞
0
=
⎜
⎟
0
⎝
1
find
the
matrix
for
the
compound
transformation.
⎠
Squaring
the
matrix
gives
the
matrix
for
2
⎛
1
0
⎞
⎛
(b)
1
0
⎞
⎛
⎟
⎜
⎠
⎝
1
performing
⎞
0
the
transformation
twice.
=
⎜
⎝
0
1
⎟
⎜
⎠
⎝
0
⎛
1
1
0
0
1
⎟
0
1
⎠
⎞
=
⎜
⎟
⎝
Because
results
the
in
compound
the
identity
transformation
its
starting
⎠
twice
transformation
matrix
will
performing
return
each
the
point
to
position.
Aine tranormation
These
are
matrix
transformations
and
b
×
1
of
is
a
2
vector
transformation
x
undergoes
the
form
and
a
x′
linear
is
x′
=
the
Ax
+
b
image
where
of
x.
transformation
In
A
an
is
a
2
×
2
affine
followed
by
a
translation.
117
/
3
Eampe 3.5.3
⎛
Let
the
transformation
T
be
defined
by
x′
=
2
⎜
⎟
1
⎝
x′
is
the
image
of
x
under
the
transformation
(a)
Find
the
coordinates
of
the
image
(b)
Find
the
coordinates
of
the
point
position
when
transformed
by
of
(2,
that
⎛
⎞
1
3
x
+
⎜
⎟
4
⎝
⎠
⎞
1
,
where
⎠
T
2)
under T
remains
in
the
same
T
Solution
⎛
2
⎞
1
⎛
⎞
2
⎛
⎛
⎞
1
+
(a)
⎜
1
⎝
3
⎟
⎜
⎠
⎝
2
⎞
5
=
⎟
⎜
⎠
⎝
4
⎟
⎜
⎠
⎝
⎟
12
⎛
,
⎞
2
⎜
is
⎟
2
⎝
⎠
substituted
into
the
⎠
Be sure to note if the question
expression.
specifically asks for the
(5,
12)
coordinates of the points. If so,
⎛
these should be given, not the
2
⎛
⎞
1
⎞
a
⎛
⎛
⎞
1
+
(b)
⎜
1
⎝
3
⎟
⎜
⎠
⎝
b
⎞
a
=
⎟
⎜
⎠
⎝
4
⎟
⎜
⎠
⎝
⎟
b
⎠
position vector.
⎛
2a
+
b
⎞
1
⎛
⎜
⎞
a
=
⎟
⎜
⎜
a
+
3b
+
⎟
4
⎝
⎟
b
⎝
⎠
⎠
a
+
b
a
=
6,
So
=
1,
b
the
=
a
+
2b
=
−4
Form
equations
and
solve
simultaneously
.
−5
coordinates
invariant
two
point
of
are
the
(6,
5)
frata and e-imiar hape
Compound
on
an
shapes
object
image
is
and
then
can
then
made
be
formed
by
successively
up
of
repeating
on
the
transformed
a
transformation
subsequent
images
of
⎛
the
images.
original
⎞
x
⎛
transformation
can
be
written
in
the
form
⎜
⎟
⎜
⎛
A
is
a
2
×
2
matrix
and
b
⎞
x
is
⎛
a
n
⎜
is
⎟
y
n
⎝
the
position
vector
of
n
1
n
1
⎜
⎟
⎜
⎠
⎝
×
1
⎠
⎞
n
1
n
1
⎜
⎜
+
⎟
b
⎟
y
⎝
final
⎠
vector.
⎞
x
n
⎜
2
A
⎟
y
⎝
where
=
The
shape.
x
n
The
first
⎟
following
the
transformation.
⎟
y
⎠
Eampe 3.5.4
⎛
⎞
x
⎛
⎛
n
A transformation is
given
by the relation
⎜
⎟
1
1
⎝
⎛
⎟
y
n
an
expression
for
⎛
⎞
x
⎜
y
2
⎝
⎟
2
⎝
1
⎠
⎜
⎜
⎟
y
n
⎝
1
3
⎞
+
⎝
⎠
⎟
1
⎠
⎞
x
0
⎟
⎜
⎛
1
⎟
⎠
2
Find
n
⎜
⎜
⎜
⎞
x
⎞
=
in
the
form
A ⎜
⎟
⎜
⎠
⎝
⎟
+
b
⎟
y
0
⎠
Solution
⎛
⎛
⎞
x
⎛
1
⎜
⎜
⎟
1
⎞
1
⎝
Write
⎞
x
⎛
0
⎜
⎜
⎟
y
⎝
1
=
⎟
1
⎠
⎜
⎟
y
0
⎝
down
the
⎝
⎠
two
⎞
⎛
⎜
⎟
2
⎠
3
+
⎟
1
1
expressions for
⎜
⎜
⎝
⎛
⎛
⎞
x
⎛
2
⎜
⎜
⎟
⎟
y
2
⎝
1
⎜
⎝
⎟
⎟
2
1
⎠
⎜
1
1
1
⎞
1
⎞
1
⎜
=
⎜
⎝
2
1
⎜
⎠
⎜ ⎝
1
⎠
⎛
1
2
⎞
=
⎝
1
⎛
0
⎟
⎠
⎝
⎟
y
2
0
3
⎛
⎟
⎜
⎟
1
⎝
1
2
⎞
=
⎝
⎟
⎟
4
1
⎠
⎛
0
⎜
⎜
⎜
⎝
⎟
y
0
⎠
using the interative
⎟
2
⎠
3
then
substitute
⎞
⎝
⎟
1
the
first
expression
the
second.
into
⎠
⎠
3
⎞
+
⎝
7
⎟
⎜
⎠
⎝
⎠
⎞
x
⎠
+
⎜
⎠ ⎟
⎟
y
⎟
1
Simplify
5
⎝
right
hand
by
multiplying
out
⎞
+
⎜
the
⎠
side
⎛
⎛
and
⎟
1
⎞
x
formula,
⎞
+
⎠
⎛
⎞
+
⎜
⎜
0
⎞
x
⎟
4
⎞
⎟
y
⎝
⎜
⎜
⎜
⎝
⎛
0
⎟
⎜
⎠
⎞
x
⎟
y
2
⎟
1
⎟
2
⎝
⎛
⎝
⎜
⎟
⎞
⎜
⎜
⎠
⎛
⎛
3
+
⎟
y
⎝
⎛
⎛
⎛
1
⎜
⎠
⎛
⎞
x
⎞
1
=
⎞
x
⎠
the
matrices
the
two
and
adding
⎟
8
⎠
vectors.
118
/
3.5
M AT R I x
T R A N s f O R M AT I O N s
(AHl)
The area o an image
If
an
object
undergoes
a
transformation
represented
by
a
matrix T
then:
This formula is not given in the
The
area
of
the
image
=
|det
T|×
the
area
of
the
object.
formula book so it will need to
be memorized.
Eampe 3.5.5
1
It
is
given
that
sin 45° =
cos 45° =
2
(a)
Write
down
about
(0,
the
matrix
that
represents
a
45°
anti-clockwise
rotation
S
2
S
2
1
0)
S
3
S
0
A design
is
created
by
rotating
a
triangle
five
times
through
45°
–6
anti-clockwise
and
enlarging
it
by
a
factor
of
2
each
–4
–2
time.
2
–2
S
This
is
shown
in
the
diagram.
4
–4
(b)
Find
the
single
matrix
that
will
take
points
in
triangle S
to
points
in
0
–6
triangle
S
S
1
5
–8
(c)
V
erify
your
matrix
will
take
the
point
(2,
2)
onto
the
corresponding
–10
point
in
S
1
–12
(d)
Find
the
single
matrix
that
will
take
points
in
triangle S
to
points
in
0
triangle
S
5
(e)
Write
down
the
area
of
triangle
S
and
hence
find
the
area
of
–16
0
triangle
S
5
Solution
This
⎛
1
to
2
⎜
matrix
can
be
taken
from
the
formula
book.
It
is
acceptable
⎞
1
2
use
decimal
equivalents,
but
because
the
exact
forms
were
⎟
(a)
⎜
1
given
⎟
1
2
⎝
2
⎛
⎞
2
1
question
it
might
be
better
to
use
them.
⎛
2
⎜
⎟
⎟
⎝
⎠
1
⎜
1
⎟
⎜
⎞
1
⎛
2
occasion
the
order.
the
same
matrix
is
obtained
irrespective
of
1
1
⎞
⎝
⎟
1
1
⎠
⎟
2
⎝
1
this
=
⎜
⎜
On
⎟
2
⎟
⎞
1
⎜
0
⎜
⎛
the
⎠
⎛
(b)
in
⎟
⎜
2
⎛
⎞
⎠
⎞
0
From
the
diagram
it
is
clear
that
the
corresponding
point
=
(c)
⎜
⎝
1
⎛
1
1
⎟
⎜
⎠
⎝
2
⎟
⎜
⎠
⎝
⎟
4
⎠
is
(0,
4)
5
⎞
1
⎛
4
⎝
4
4
When
⎞
a
transformation
is
repeated
5
times
the
single
matrix
original
matrix
=
(d)
⎝
1
⎠
1
4
⎠
that
will
raised
to
represent
the
that
power
transformation
is
the
5.
1
(e)
Area
of
S
=
×
2
×
2
=
2
Whenever
an
area
is
asked
for
in
is
likely
that
a
question
on
matrix
0
2
⎛
4
⎝
4
4
det
=
Area
=
32
4
×
transformations
it
image=|det
the
the
formula
The
area
of
the
⎞
16 + 16
=
32
T|×
area
of
the
object
will
be
used.
⎠
2
=
64
119
/
3
3 . 6
V E C T O R S
You houd know:
✔
unit
vectors;
✔
components
base
of
a
You houd be abe to:
vectors
vector;
i,
j,
k
✔
column
use
position
✔
vector
vectors:
equation
OA
of
a
=
calculate
the
multiply
by
a
vector
line
in
two
and
geometric
sum
and
approaches
difference
of
two
to
vectors,
representation
!!!
"
✔
algebraic
and
three
✔
nd
dimensions
|v|
the
a
scalar,
from
nd
the
magnitude
of
a
components
resultant
as
the
sum
of
two
or
morevectors
!!!
"
✔
the
relative
✔
how
position
of
B
from
A
is
AB
✔
convert
the
parametric
to
model
velocity
✔
how
two
to
in
linear
two
model
and
motion
three
motion
with
equation
of
a
line
to
form
constant
dimensions
with
vector
variable
✔
velocity
nd
the
position
position
of
two
variable
velocity
of
an
object
objects
and
moving
the
with
relative
constant
or
in
dimensions
✔
✔
how
✔
f (t
to
model
projectile
and
circular
nd
a
velocity
vector
given
a
speed
motion
anddirection
–
a)
indicates
a
time
shift
of
a
✔
✔
the
denition
and
calculation
of
the
nd
times
of
two
the
denition
and
calculation
of
the
of
two
to
the
component
of
objects
are
other
the
angle
between
two
vectors
using
·
w
=
θ,
|v||w|cos
where
θ
is
the
angle
vectors
of
vector
a
acting
in
vector
b
is
(v
=|a|cos
two
ascertain
the
aib
direction
each
calculate
between
✔
two
vector
v
product
when
vectors
✔
✔
distances
scalar
closest
product
and
·
w
=
non-zero
whether
the
vectors
vectors
v
and
are
w,
and
perpendicular
0)
θ
b
✔
the
component
of
a
vector
a
use
|v
(and
perpendicular
to
vector
b,
in
the
plane
the
two
vectors,
×
w|
to
nd
the
area
of
a
parallelogram
hence
a
triangle)
formed
✔
a
by
×
acting
nd
the
component
of
a
vector
acting
in
a
given
b
=
is
a
sin θ
direction
b
or
perpendicular
to
that
direction.
Proper tie o etor
A vector
represents
and
direction.
will
meet
The
a
quantity
most
which
common
has
vector
both
magnitude
quantities
you
Other quantities such as force,
in
the
course
are
the
displacement
or
position
or electric field strength may be
of
an
object,
its
velocity
and
its
acceleration.
used in questions but you will not
require any knowledge of these
V
ectors
quantities beyond what is given in
the
can
base
be
written
vectors
i,
j
as
and
column
vectors
⎛
is
1
equivalent
to
2i
j
is
⎜
to
magnitude
of
a
vector
v
is
found
using
the
formula
the
v
2i
+
j
2k
⎟
2
⎠
length
2
is
equivalent
⎟
1
and
⎠
⎝
It
of
⎞
2
⎜
The
terms
⎞
2
⎝
in
k
the question.
⎛
or
=
v
of
v
2
+
v
1
is
written
as
⎛
v
⎜
v
1
v
where
v
=
⎝
The
magnitudes
2
AHL 3.10 of the formula book .
2
of
the
2
2
+ (
1)
two
=
5
and
2
vectors
2
+ 1
given
.
⎞
⎟
2
3
⎜
This formula is given in section
v
2
+
2
and
above
⎟
v
3
⎠
are:
2
+ (
2)
=
9
=
120
/
3.6
Any
vector
⎛
⎞
2
1
⎝
said
is
a
scalar
⎛
is
⎜
that
parallel
⎠
⎝
the
another
⎞
10
because
⎜
have
of
⎛
⎞
10
to
⎟
to
multiple
5
same
⎛
=
⎟
⎜
⎠
⎝
5
is
parallel
.
⎜
⎠
⎝
it.
⎞
2
5
⎟
to
v E c TO R s
Parallel
vectors
are
⎟
1
⎠
direction.
1
A unit
vector
has
a
magnitude
of
one,
v
hence
is
a
unit
vector
in
⎛
2
the
v
same
1
direction
⎛
v
⎛
2
and
⎝
5
1
⎜
unit
vectors
in
the
same
direction
⎜
⎟
1
⎜
⎟
2
⎝
The
⎠
sum
of
equivalent
two
to
or
the
more
vectors
combined
will
effect
of
give
the
the
single
individual
⎛
called
the
resultant
vector.
Let
the
vector
u
3
3
1
⎝
vectors.
This
is
vector
⎞
and
the
vector
v
⎠
⎞
⎟ .
⎜
which
⎟
4
⎝
⎛
vector
be
⎜
be
⎠
1
⎠
2
⎞
2
and
is
⎞
as
⎝
⎝
⎛
are
⎟
3
⎠
1
⎞
1
⎞
2
as
Then:
⎠
v
1
⎛
⎞
3
⎛
⎛
⎞
3
+
⎜
4
⎝
⎞
6
=
⎟
⎜
⎠
⎝
1
⎟
⎜
⎠
⎝
3
⎟
5
⎠
u
⎛
As
can
be
seen
in
the
diagram
⎟
5
⎝
⎛
⎞
3
⎜
⎟
4
⎝
⎛
followed
⎠
if
the
resultant
⎛
force
of
effect
4
is
equivalent
to
a
displacement
4
of
⎠
⎟
1
3
3
⎠
vectors
of
the
represent
two
forces
forces
would
acting
be
the
on
an
same
object
as
a
then
single
⎞
6
⎜
⎝
The
3
two
v
⎞
⎜
⎝
Similarly
,
the
by
⎞
6
⎜
+
u
⎟
5
⎠
position
vector
of
a
point
A
is
the
vector
from
the
origin
to
A.
!!!
"
It
is
written
as
OA
or
often
as
just
a
!!
"
The vector
AB = −a + b or b − a
A
a
+
b
a
B
b
O
Eampe 3.6.1
1
In
still
water,
⎛
direction
3
⎜
Find
While
boat
the
has
a
speed
of
2
ms
and
as
a
column
is
being
steered
in
the
⎞
⎟
4
⎝
(a)
a
.
⎠
velocity
travelling
of
with
the
the
boat
same
velocity
,
the
⎛
vector.
boat
0
enters
a
section
of
⎞
1
river
which
is
flowing
with
a
velocity
of
⎜
⎝
(b)
Find
the
whether
velocity
it
is
of
the
moving
boat
relative
upstream
or
to
⎟
1.5
ms
.
⎠
the
land,
and
state
downstream.
121
/
3
Solution
⎛
(a)
The
⎞
3
⎜
has
⎟
4
⎝
a
magnitude
velocity
is
in
⎛
as
⎜
must
⎠
2
+ (
4)
=
5
be
A unit
vector
in
⎛
direction
the
multiple
magnitude
same
is
The
⎟
4
⎝
a
1
⎛
of
required
⎜
5
the
vector
will
be
vector
in
the
.
The speed of an object is equal to
⎟
4
⎝
velocity
a
⎞
3
direction
⎠
⎜
The
have
2.
unit
⎟
4
⎝
but
⎞
3
⎛
therefore
it
⎠
2times
Note
of
⎞
3
as
⎜
the magnitude of its velocity vector.
so
⎟
4
⎝
2
same
⎞
3
⎠
direction
3
the
of
of
the
boat
⎠
is
therefore
2
⎛
⎞
3
⎛
⎞
1.2
=
⎜
5
⎛
4
⎝
⎟
⎜
⎠
⎝
⎛
⎞
1.2
⎟
1.6
(b)
⎠
⎞
0
⎛
+
⎜
1.6
⎝
The
⎞
1.2
⎜
⎠
⎝
1.5
⎟
⎜
is
relative
to
the
land
⎠
⎝
⎟
0.1
is
⎠
the
the
It
velocity
=
⎟
moving
velocity
velocity
of
of
the
the
boat
plus
river.
upstream.
The
0.1
moving
indicates
the
boat
is
upstream.
The aar produt
These formulae are in section
⎛
AHL 3.13 in the formula book .
v
⎞
⎛
⎟
⎜
The
scalar
(or
’dot’)
product
of
two
vectors
v
=
⎜
defined
as
v·w
=
v
w
1
Additionally
This
can
be
rearranged
cos θ
w
+
2
cos
w
1
as
v
2
=
v·w
v
vectors
+
1
+
to
v
1
v
3
+
2
v
=
⎟
w
2
⎜
⎟
⎜
⎠
⎝
⎟
⎟
w
3
⎠
3
where
a
w
⎟
w
3
give
w
2
v
and
2
⎝
is
1
v
⎜
⎞
w
1
⎜
θ
is
formula
the
for
angle
the
between
angle
v
and
between
w
two
w
3
3
=
The main use of the scalar product is
to find
the
angle
between
two vectors.
If two vectors are perpendicular then their scalar product is equal to 0.
Note
Eampe 3.6.2
!
AOB
The angle
!!
"
is the angle between
A triangle
OAB
has
vertices
at
(0,
0,
0),
(4,
8,
3)
and
(8,
6,
3)
!!
"
vectors OA and OB (α ) and not
between vectors
(a)
Find
the
lengths
(b)
Find
the
angle
(c)
Hence
(d)
Show
OA
and
OB
!!
"
!!
"
AO and OB ( β ).
find
that
the
that
!
AOB
area
of
the
!
angle O AB
triangle
=
OAB
90°.
A
Solution
2
OA
(a)
OA
=
4
=
89
2
+ 8
3
OA
OB
B
≈
equal
vector OA,
position
OB
2
=
8
=
109
the
magnitude
of
which
is
the
9.43
2
β
to
!!!
"
the
α
O
is
+ 3
+ 6
vector
of
A
2
+ 3
AO
≈ 10.4
122
/
3.6
4
×
8
+
8
×
6
+
3
×
3
This
!
cos( AOB) =
(b)
required
is
the
one
!!!
"
89
×
!!!
"
109
between
=
angle
v E c TO R s
vectors OA and OB
the
0.9036...
!
AO B
≈
25.4°
1
(c)
=
Area
×
89
×
109
Using
the
formula
for
area
2
andthe
lengths
of
sides
sin(25.363...)
alreadyfound.
=
21.0950... ≈
⎛
AO
(d)
⎞
4
!!!"
⎜
=
21.1
⎟
8
⎜
!!!"
This
,
⎟
3
⎝
is
the
angle
between
AO
!!!
"
⎠
AB
and
!!!
"
⎛
8
!!!
"
⎞
⎛
⎟
⎜
4
⎞
⎛
⎟
⎜
⎞
4
AB
⎜
AB =
⎜
⎟
3
⎝
⎜
⎝
⎠
2
⎜
⎟
3
!!!
"
calculated
using
!!!
"
!!!
"
AB = OB
OA
or
!!!
"
Using
key
fact
⎟
0
⎝
⎠
is
⎟
=
8
6
⎠
the
!!!"
!!!
"
AO ×
AB =
the
0
AB =
b
a
above
vectors
if
are
perpendicular
!!!"
!!!
"
AO ×
AB
Hence
=
4
!
O AB
=
×
4
+
8
×
2
+
3
×
0
=
0
90°
vetor produt
⎛
v
1
⎛
⎞
⎞
w
1
⎜
The
vector
(or
’cross’)
product
of
two
vectors
=
v
⎜
⎟
v
2
⎜
⎝
⎛
v
w
2
v
3
defined
as
w
3
⎜
v
1
⎟
3
⎝
⎟
⎠
3
⎟
⎟
v
w
1
⎝
v × w
⎠
w
w
1
⎜
Additionally
⎜
⎟
v
=
v × w
3
2
⎜
⎟
v
⎟
w
=
2
⎜
is
w
⎞
w
3
and
v
2
=
w
2
1
sin
,
⎠
It is very easy to make a mistake
where
θ
is
the
angle
between
v
and
when calculating a vector product.
w
Make sure you clearly set out
1
Using
the
formula
for
the
area
of
a
triangle
ab sin C
we
can
see
each step.
that
2
1
the
area
of
the
triangle
shown
sin θ
is
2
The formula book section 3.13
w
Hence
the
area
of
the
triangle
formed
by
the
vectors
θ
has the equivalent formula for the
v
1
v
and
w
is
|v
×
area of a parallelogram
w|
A =
×
2
where v and w form two adjacent
sides of the parallelogram.
The direction of the vector product of  and w is perpendicular to both  and w
Eampe 3.6.3
The
points
A
(2.1,
3.5,
4.2),
B
(3.2,
2.3,
1.4)
and
A gust
of
wind
strikes
the
sail
perpendicular
to
its
1
C(0.8,
(a)
3.9,
Find
1.6)
the
lie
area
at
the
three
of
the
sail.
corners
of
a
sail.
surface
(b)
with
Find
the
a
speed
velocity
of
of
12
the
ms
wind
as
a
column
vector.
!!!
"
Solution
Using
⎛
!!!
"
(a)
AB =
⎞
1.1
⎜
⎟
1.2
⎜
⎝
⎛
1.3
!!!
"
2.8
,
AC
=
⎜
the
formula
⎜
⎠
⎝
1.1
The
⎟
2.6
area
of
⎛
⎞
1.3
×
AC
=
⎜
⎜
⎝
⎜
⎟
×
1.2
2.8
⎞
⎛
⎟
4.24
⎜
⎠
⎝
2.6
⎜
=
0.4
⎟
the
sail
!!!
"
⎠
a
AB
is
the
area
of
the
triangle
!!!
"
and
AC
and
is
equal
to
⎞
!!!
"
AB
b
⎟
formedby
!!!
"
=
0.4
⎟
⎛
!!!
"
AB
⎞
⎟
!!!
"
1
6.5
⎟
⎜
⎠
⎝
AB
⎟
×
AC
2
1.12
⎠
123
/
3
1
Area
2
4.24
=
2
+ 6.5
2
+ (
As
the
wind
is
perpendicular
to
the
sail
it
is
perpendicular
to
1.12)
!!!
"
2
both
!!!
"
AB
and
AC
and
so
has
the
same
direction
as
!!!
"
!!!
"
AB ×
AC
1
=
× 7.841... ≈
3.92
2
⎛
4.24
⎞
⎛
13.0
⎞
⎜
19.9
⎟
12
1
⎜
⎟
(b)
6.5
3.92052...
⎜
≈
⎟
1.12
⎝
⎜
⎝
⎠
To
find
the
correct
magnitude
to
find
the
unit
vector
and
v
⎟
3.43
v
use
⎠
multiply
by
12.
component o etor in a gien diretion
The
components
along
For
each
of
example,
directions
⎛
of
base
if
coordinate
a
north
3
⎟
is
j
and
system
vertically
is
much
of
the
vector
is
acting
k
such
up
that
i,
j
and
respectively
k
represent
then
or
3i
+
j
+
0.8k
represents
a
body
the
moving
0.8
the
velocity
east
at
⎠
1
,
north
often
given
and
i,
how
⎟
⎝
It
vectors,
you
1
ms
1
1
ms
tell
⎞
⎜
3
vector
the
east,
⎜
vector
a
at
1
useful
1
ms
to
and
know
is
gaining
how
much
height
of
a
at
0.8
vector
ms
is
acting
in
a
direction.
Let  and w be two vectors and let the angle between them be θ.
i
The component of w acting in the direction of  is
cos θ
=
i
=
w
The component of w acting perpendicular to the direction of  is
θ
×
sin θ
v
×
=
=
Eampe 3.6.4
An
It
object
is
is
acted
Find
the
moving
on
by
a
force
component
(a)
in
the
(b)
perpenciular
in
direction
of
of
to
the
direction
of
2j
7i
the
+
of
5k
the
vector
3i
+
2j
+
4k.
Newtons.
force:
motion
the
motion.
2 j
5 k)
Solution
(3i
+
2 j
+
4 k)
i
(7 i
+
w
i
v
using
(a)
2
3
2
+
2
v
2
+
4
37
When answering a vector question,
=
≈ 6.87
N
29
you can use either component
form or column form to write
( 3i
+
2 j
4 k)
+
×
(7 i
2 j
+
5 k)
w
×
v
using
(b)
the vector.
2
3
( 18 i
13 j
2
+
2
2
+
v
4
20 k )
893
=
29
≈
5.55
N
29
124
/
3.6
v E c TO R s
The etor equation o a traight ine
The vector equation of a straight line can be written as r = a + λb where a is the
position vector of a point on the line and b is a vector in the direction of the line
and r is the position vector of the point on the line with parameter λ
⎛
⎜
In
three
dimensions
r
represents
the
P
⎞
x
⎟
y
vector
⎜
⎟
⎜
⎝
⎟
z
b
⎠
A
The
vector
equation
can
be
written
in
parametric
form
as
three
r
separate
⎛
x
⎜
r
equations.
⎞
⎛
⎟
=
⎟
⎜
⎟
z
⎝
⎜
=
y
⎜
⎛
⎞
2
⎟
+ λ
1
⎜
0
⎝
⎠
For
example
a
⎞
3
⎜
⎟
can
2
⎟
⎜
⎠
⎝
be
written
as
three
separate
equations:
⎟
4
⎠
O
x
=
2 + 3λ ,
y
=
2λ ,
1
z
=
4λ
The angle between two lines is the angle between the direction vectors of
the lines.
Eampe 3.6.5
A chair
lift
joins
an
upper
35,
40)
coordinates
(80,
coordinates
(110,
B
can
be
joining
(a)
65,
assumed
A
Find
and
the
joining
to
and
a
A
lower
0).
The
cable
lie
along
the
with
A pylon
station
joining
A
straight
B
with
and
line
P
top
of
the
the
cable
position
supports
pylon
at
of
a
the
makes
point
cable
on
contact
which
is
12
the
chair
with
m
the
lift.
line
higher
The
of
than
the
B
B
vector
A
station
and
equation
of
the
straight
(b)
Find
the
coordinates
of
P
(c)
Find
the
angle
by
with
the
horizontal.
line
made
the
line
of
the
chair
lift
B
Solution
!!!
"
(a)
The
direction
⎛
!!!
"
AB
=
110
⎜
of
⎞
⎛
⎟
⎜
⎟
⎜
⎠
⎝
the
80
65
⎛
⎟
⎜
=
0
40
Using
is
⎞
35
⎜
⎝
line
30
⎜
⎠
⎝
of
line
Any
point
40
r
80
=
⎝
z
=
⎛
⎟
⎜
35
⎜
(b)
⎞
is
=
110
⎜
12
⇒ λ
=
40
⇒
+
λ
⎜
⎠
⎝
⎝
be
used
so
⎛
⎟
⎜
0
+
λ
30
⎞
⎟
30
⎟
⎜
⎠
⎝
is
also
a
possible
answer.
⎟
40
⎠
⎟
40
40
=
⎠
12
x
=
80 + 0.7 × 30
=
101
y
=
35 + 0.7 × 30
=
56
P
are
of
can
⎟
0.7
Coordinates
line
30
⎟
40
the
⎞
65
⎞
30
on
⎠
⎜
⎜
a
⎟
r
⎛
AB = b
⎟
⎛
Equation
formula
⎞
30
⎟
the
(101,
56,
The
parametric
and
hence
the
equations
values
of
x
can
and
be
used
to
find
the
value
of
y.
12)
125
/
3
(c)
Direction
⎛
lift
of
the
The
chair
the
30
Angle
angle
between
horizontal
is
the
the
line
angle
of
the
chair
between
the
lift
and
direction
vector
and
the
horizontal.
⎟
0
⎝
⎠
between
= θ
horizontal
30
×
30
chair
lift
and
where
+
30
×
30
+
0
×
− 40
=
2
2
+
30
θ
below
⎟
⎜
cos θ
line
⎞
30
⎜
is
horizontal
2
30
=
0.7276...
≈
43.3
30
2
+
30
2
+
40
K inemati
The
position
will
be
In
v
this
is
lie
on
a
straight
equation
the
when
(displacement)
t
r
particle’s
=
line
gives
the
velocity
of
of
an
the
object
moving
form
=
object’s
and
a
is
r
a
position
the
+
with
constant
velocity
tv.
or
position
displacement
vector
of
the
at
time
t,
object
0
s AMPlE sTUDENT ANs WER
Distances
Ryan
and
ground.
The
this
Jack
position
r
=
5
⎜
of
Ryan's
⎛
⎟
⎜
+ t
⎟
takes
off
airplane
which
after
t
take
off
from
level
Ryan's.
seconds
after
it
takes
off
is
given
⎟
4
Find
the
speed
(b)
Find
the
height
of
airplanes
⎟
⎝
position
metres.
2
(a)
The
in
⎞
4
⎜
0
are
model
airplane
⎞
6
⎜
question
have
Jack's
⎛
by
in
⎠
of
Ryan's
of
Jack's
airplane.
Ryan's
airplane
airplane
s
after
seconds
two
after
it
seconds.
takes
off
is
given
by
If an object depar ts at time t = t
1
rather than t = 0
⎛
then the equation
r
=
39
⎜
)
⎛
⎟
⎜
+
44
⎜
becomes r = a + (t − t
⎞
s
⎞
4
⎟
6
⎟
⎜
⎠
⎝
that
the
⎟
and a is the
⎝
1
0
7
⎠
object’s position at t = t
1
(c)
▼ The
–4
Though
should
the
nal
be
in
brackets.
answer
Show
The
two
it
might
not
airplanes
be
awarded
it
collide
the
at
airplanes
the
point
(
are
perpendicular.
23,
20,
28).
followed
incorrect
long
after
Ryan's
airplane
takes
off
does
as
Jack's
technically
of
was
(d) How
correct
paths
airplane
take
off?
[5]
working!
2
▼ The
student
displacement
did
not
state
correctly
of
the
that
found
airplane
the
height
the
(a)
S
peed
(b)
t
=
=
is
8
⎛
m
=
⎜
⎝
direction
product
took
vectors.
equaling
the
The
0
is
to
show
=
36
=
6
⎛
⎞
5
⎟
6
0
+ 2
4
⎜
⎟
⎜
⎠
⎝
2
4
⎞
⎟
⎛
=
⎜
⎟
⎜
⎠
⎝
3
10
28
=
⎞
⎟
⎟
8
⎠
correct
⎛
scalar
sufcient
(c)
⎜
4
2
⎜
explanation
2
+ 4
2
⎜
student
2
+ 2
but
r
▲ The
−4
they
are
⎝
4
⎛
⎞
⎟
i
⎜
⎟
⎜
⎠
⎝
4
6
⎞
⎟
=
−
16 −
12
+
0
⎟
7
⎠
perpendicular.
126
/
3.6
(d)
If
Jack's
⎛
r
airplane
39
⎜
=
0
⎝
time
⎛
⎜
⎞
⎜
+ (t
⎜
⎠
⎝
⎜
28
Ryan's
5
⎜
=
− a)
⎟
⎛
⎟
20
⎝
⎛
⎟
when
23
a
⎞
44
⎜
is
6
⎟
⎜
⎠
⎝
20
=
⎛
⎟
⎜
t
⎟
⎜
0
⎠
⎝
6
+
after
Ryan's,
then:
⎞
4
⎟
6
▲ The
7
the
⎠
gets
to
(−23,
20,
28)
formula
⎛
For
Jack
⎜
⎜
⎟
2
to
move
the
start
time.
− 23 =
5
−
4t
→
t
=
7
▲ It
is
a
good
idea
to
check
the
⎟
4
other
⎠
⎞
39
⎜
=
⎛
⎟
+ (t −a )
44
⎜
⎟
28
⎝
used
⎞
4
⎛
⎟
20
correctly
coordinates
t
=
are
also
satised
the
student
7.
14
⎞
23
has
is
when
Check
student
⎟
plane
⎞
+
second
v E c TO R s
⎠
0
⎝
⎜
⎟
⎜
⎠
⎝
⎞
4
⎟
6
⎟
▼ Unfortunately
7
not
−23 =
−3
+ 4
(t
− a)
but
−23 =
−39 + 28 − 4a →
t
=
7
check
had,
the
might
16
a
=
=
did
⎠
their
small
have
answer
error
been
in
here.
the
If
last
they
line
spotted.
4
4
If
at
time
t
object
A
has
position
vector
r
and
object
B
r
then
The
relative
position
of
B
from
A
has
position
vector
!!!
"
A
is
the
AB =
vector
r
r
B
B
A
!!!
"
Their
distance
apart
is
AB
=
r
r
B
If
two
objects
collide
they
A
must
be
in
the
same
place
at
the
same
time.
Eampe 3.6.6
⎛
At
t
hours
after
13:00,
ship
A
has
position
r
21
⎛
B
has
position
r
⎞
7
=
⎛
2
6
1
⎝
relative
distances
in
(a)
the
two
ships
velocities
and
state
⎟
⎜
⎠
⎝
⎞
2
and
⎟
⎜
⎠
⎝
⎟
2
⎠
⎞
+ t
⎜
B
⎛
+ t
⎜
⎝
ship
⎞
=
A
to
a
port,
with
all
⎟
4
⎠
kilometres.
Note
Show
would
the
collide
time
this
if
they
would
maintained
their
occur.
As a velocity is defined by both
speed and direction, it is possible
Ship
B
changes
with
the
its
velocity
so
it
is
travelling
on
a
bearing
of
045°
for the velocity to change while the
same
speed.
speed remains constant.
(b)
Find
the
shortest
distance
between
the
two
ships.
Solution
(a)
The
ships
when
21
will
have
the
2t
7 + 2t
⇒ t
Substituting
⎛
r
=
r
A
of
2
3.5
equations
Having
found
and
see
if
The
speed
this
t,
an
also
alternative
gives
t
=
method
is
to
solve
4
gives
hence
collide
⎠
collision
of
16:30
the
velocity
of
B
is
of
B
is
the
magnitude
of
its
velocity
.
⎛
=
20
A bearing
of
045°
is
in
the
direction
of
the
1
⎝
20
velocity
of
B
⎛
1
⎞
is
⎛
=
⎜
2
⎝
1
1
⎞
vector
⎜
New
y
3.5
2
+
for
⎟
13
Magnitude
2
both
=
x-coordinate
⎞
,
⎜
⎝
(b)
into
=
B
Time
14
=
same
⎟
1
⎠
⎞
or
10
⎟
⎜
⎠
⎝
⎟
1
⎠
1
To
find
a
vector
with
the
correct
magnitude
use
v
v
⎛
3.162...
⎞
tofind
⎜
⎝
the
unit
vector
and
multiply
by
20
⎟
3.162...
⎠
127
/
3
!!!
"
d =
AB
=
r
r
B
A
⎞
⎛
⎛
⎞
7
+
⎜
t
⎟
⎜
⎜
⎠
⎟
10
⎝
⎞
14 + (2 +
⎟
⎜
7 + (
2 +
≈
⎛
⎟
⎜
⎠
⎝
14 + 5.16t
⎟ ⎟
2
⎠ ⎠
Simplify
⎞
expression
⎟
7 + 1.16t
⎝
+ (
minimum
as
possible
before
finding
an
for
the
magnitude.
last
stage
is
done
on
the
GDC.
2
14 + 5.16t )
Hence
much
A
⎠
This
(
=
as
r
B
2
d
r
⎟
⎜
⎠
⎞ ⎞
2
t
⎜
⎟
10 ) t
⎝
6
⎛
10 ) t
⎜
⎞
+
⎝ ⎝
⎠
⎛
21
⎜ ⎜
⎟
1
⎝
=
⎛ ⎛
10
=
7 + 1.16t )
value
of
d
≈ 13.8
km
Motion with ariabe eoity in t wo dimenion
Kinematics is also covered in
⎛
sections 5.4 and 5.5
If
an
object’s
position
x
is
a
function
of
t
with
x
=
x
⎜
⎞
⎟
then
its
velocity
y
⎝
⎛
Note
v
=
x
=
⎛
⎞
x
and
its
acceleration
is
⎟
⎜
a
!
! =
x
=
⎠
⎞
x
⎜
where
⎟
the
dot
notation
y
y
⎝
⎠
⎝
⎠
2
When x is in bold type then it
dx
signifies
differentiation
with
respect
to
time,
so
x
d
=
x
!
! =
x
and
2
represents a vector; when not bold
dt
dt
it represents a single value.
Eampe 3.6.7
A heavy
ball
is
thrown
from
a
height
of
2
m
at
an
(b)
Given
that
after
it
has
been
thrown,
1
angle
of
30°
to
the
horizontal
with
a
speed
of
3
1
ms
acceleration
the
(a)
Find
the
initial
velocity
the
component
vector,
horizontal
of
horizontal
the
ball
is
distance
9.81 j ms
from
the
,
find
point
of
with
projection
x
the
and
the
that
the
ball
hits
the
ground.
y
componentvertical.
Solution
(a)
(b)
3 cos 30i + 3 sin 30 j
=
2.598...i + 1.5 j
v
=
a dt
∫
=
c
i +
1
t
When
=
The
the
(
9.81t + c
2
components
velocity
can
be
found
from
a
diagram
of
vector.
) j
3
m/s
0
30°
v
=
2.598...i + 1.5 j
When
⇒ c
=
2.598...,
c
1
v
≈
r
=
=
2
2.60i + (
to
∫
=
(2.60t + c
)i +
2 j
t
=
⇒ c
=
0,
c
=
9.81
a
constant
term
both
of
the
components.
+ 1.5t + c
4
2
)
j
2
9.81
the
vertical
add
4
= ( 2.60t ) i +
When
to
0
3
r
need
2
t
(
3
When
r
=
you
9.81t + 1.5) j
9.81
v dt
integrating
1.5
(
object
2
t
+ 1.5t + 2
2
hits
the
displacement
=
)
j
ground,
0
hence
The
can
two
be
components
used
provide
two
equations
which
separately
.
2
t
+ 1.5t + 2
=
0
The
negative
value
of
t
is
disregarded
as
the
2
equation
⇒
t
=
is
only
valid
for
t
≥
0
0.809 5…
Horizontal
displacement
2.598... × 0.8095... ≈
2.10
m
is
Though
this
formula,
it
could
is
best
be
to
solved
solve
it
by
the
directly
quadratic
on
the
GDC.
128
/
3.6
The
motion
of
gravitational
of
motion
of
circular
a
body
force
with
a
is
which
called
variable
motion,
as
moves
only
projectile
velocity
.
illustrated
under
motion.
Another
in
the
the
It
is
special
example
action
a
of
special
case
v E c TO R s
is
case
that
below.
Even when the question gives vectors in terms of the base vectors, i, j and k
you can work entirely with column vectors.
Eampe 3.6.8
A particle
P
moves
so
that
its
displacement
⎛
point
O
at
time
t
seconds
is
given
by
r
from
3 cos 2 t
a
(c)
⎞
3 sin 2 t
P
is
always
a
constant
time
t.
Find
the
distance
(b)
and
Find
state
the
that
time
the
velocity
smallest
positive
value
from
velocity
is
in
the
direction
for
P
of
t
at
P
at
⎟
0
⎝
complete
which
⎞
1
⎜
to
the
of
distance.
taken
of
⎛
the
O
for
⎠
(d)
that
expression
⎟
⎝
V
erify
an
=
⎜
(a)
Find
⎠
one
fullcircle.
Solution
2
⎛
(a)
r
=
3
cos
2t
3
sin
2t
⎜
⎜
⎝
⎞
The
⎟
section
=
(3 cos 2t)
book
=
and
was
is
in
given
section
in
AHL
3.8
=
9 (cos
constant
will
The
and
formula
from
equal
complete
a
2t)
O
distance
particle
2
2t + sin
distance
the
book.
2
+ (3 sin 2t)
2
P
this
θ
⎠
2
(b)
of
+ cos
⎟
of
The
3.4
2
θ
sin
formula
is
to
=
is
being
moving
constant
in
a
means
that
the
circle.
3
therefore
3
units.
circle
when
2t
always
=
2π
⇒ t
≈
3.14
The
period
of
the
functions
3 cos 2t
and
3 sin 2t
2π
seconds
is
=
(see
π
sections
2.3
and
2.4)
2
⎛
(c)
v
6
sin 2 t
⎞
Each
=
⎜
⎝
6
cos 2 t
component
is
differentiated
rule.
⎛
This
occurs
when
6
cos 2t
=
0
and
6
sin 2t
is
If
moving
in
the
direction
1
⎜
⎝
⎞
⎟
0
then
the
⎛
vector
=
0 ⇒ 2t
2
If
was
of
to
be
1
⎜
⎞
.
⎟
0
a
negative
multiple
then
it
⎠
would
⎛
moving
in
the
direction
of
the
vector
positive
t
=
6 cos 2t
≈
=
0
could
also
be
solved
1
⎜
⎝
3π
6 sin 2t
it
4
4
For
multiple
⎝
or
=
positive
3π
π
t
a
or
=
2
Hence
be
3π
π
cos 2t
must
velocity
⎠
positive
6
the
⎠
chain
(d)
using
⎟
directly
be
⎞
⎟
0
⎠
from
2.36
4
a
GDC.
129
/
3
3 . 7
G R A P H S
You houd know:
✔
the
denitions
of:
adjacent
vertices,
a
simple
vertex,
You houd be abe to:
graphs,
vertices,
adjacent
graphs;
edges,
edges,
degree
complete
✔
represent
of
etc)
graphs,
directed
graphs;
and
subgraphs;
circuits,
out
trees,
cycles,
connected
degree
of
a
directed
adjacency
connected
calculate
✔
walks,
given
number
k-length
the
connected,
strongly
unweighted)
of
k-length
walks
(or
less
walks)
between
two
vertices
transition
undirected
matrix
or
for
directed
a
strongly-
graph
graphs
determine
th
that
the
construct
✔
✔
and
maps,
graph,
matrices,
and
(weighted
(circuits,
in
than
degree
graphs
structures
graphs;
✔
weighted
as
real-world
an
adjacency
matrix
A,
the
(i,
whether
an
Eulerian
trail
or
circuitexists
j)
k
entry
of
A
gives
the
i
j
number
of
k
length
walks
✔
connecting
and
nd
the
✔
that
given
a
transition
matrix
for
a
graph
a
minimum
matrix
algorithm
the
or
or
spanning
graphical
the
tree
using
method
graphical
for
method
either
Prim’s
for
Kruskal’s
th
(i,
j)
of
moving
entry
gives
the
gives
the
probability
✔
from
state
j
to
state
i
in
apply
the
Chinese
postman
algorithm
with
up
a
to
four
odd
vertices
randomwalk.
✔
✔
Eulerian
trails
✔
Hamiltonian
✔
Kruskal’s
convert
a
and
Prim’s
spanning
cycles
algorithms
for
✔
nding
to
a
of
least
use
classical
the
Chinese
for
determining
weighted
the
at
✔
least
the
problem
shortest
traversing
and
route
each
around
of
the
✔
a
use
the
Hamiltonian
weighted
the
salesman
cycle
complete
problem
of
least
an
problem
by
completion
of
a
table
neighbour
upper
algorithm
bound
for
the
for
travelling
problem
deleted
determining
edges
salesman
travelling
salesman
algorithm
once
the
problem
nearest
determining
trees
postman
graph
travelling
distances.
salesman
✔
practical
circuits
paths
and
minimum
and
to
a
vertex
lower
algorithm
bound
for
for
the
travelling
problem.
determine
weight
in
a
graph
Graph deinition
B
C
Consider
The
the
vertices
[AB],
[BC],
graph
are
[BD]
A,
G1.
B,
and
C
and
D
and
the
edges
are
[CD]
A
The
edges
[AB]
The
vertices
The
edge
from
D
one
edge
from
C
A
and
and
B
[BC]
are
are
adjacent
adjacent
because
because
they
they
are
share
a
vertex,
connected
by
an
B
edge.
D
Graph
to
to
itself
D
we
is
called
say
a
there
loop.
are
Because
multiple
there
is
more
than
edges
G1
A simple
graph
is
one
with
no
loops
or
multiple
edges.
130
/
3 .7
The
degree
of
a
vertex
is
the
is
1,
of
3,
of
number
of
edges
incident
to
it,
so
GR A P H s
the
Q
degree
of
A
B
is
C
is
3
and
of
D
is
5.
P
A directed
graph
onewayalong
by
an
is
the
one
in
edge.
which
The
it
is
possible
direction
of
to
travel
permissible
in
only
travel
is
indicated
arrow.
R
For
example
to
in
from
S
On
directed
Graph
G2,
it
is
possible
to
travel
from P
to
S
but
not
P
S
a
that
lead
leading
to
the
from
out-degree
graph,
vertex
the
is
the
in-degree
and
vertex.
the
of
a
vertex
out-degree
Hence
the
is
is
the
in-degree
of
the
number
of
Graph
G2
edges
number
of
edges
vertex P
is
1
and
the
2.
Adjaeny matrie
Note
th
In
an
adjacency
matrix
A,
the
(i,
j)
entry
(a
)
is
the
number
of
edges
ij
leading
from
vertex
i
to
vertex
This is the opposite to a transition
j.
matrix in which the a
entry gives
ij
The
adjacency
matrix
for
graphs
G1
and
G2
are
shown
the probability of moving from
below.
ver tex j to ver tex i
A
⎛
A
G1
=
B
C
D
0
1
0
0
1
0
1
1
0
1
0
2
P
⎞
⎜
A walk
is
⎝
a
1
route
2
2
from
S
0
1
0
1
1
0
0
1
⎜
0
1
0
0
0
0
0
0
⎟
S
⎝
⎟
⎟
⎜
⎠
⎞
⎟
⎜
R
⎟
0
R
⎜
Q
⎟
⎜
D
=
⎟
⎜
C
G2
⎟
⎜
B
⎛
P
Q
⎠
B
one
Note
vertex
to
another.
Its
length
is
the
C
number
of
edges
it
traverses.
In
The sum of a row or column of an
graph
G1
the
route
A
to
B
to
D
adjacency matrix gives the degree
is
a
walk
usually
of
length
written
as
2.
a
The
list
walk
of
is
A
of the ver tex represented by that
vertices,
row/column. For a directed graph
for
example
ABD
the sum of a row gives the out-
D
Graph
degree and the sum of a column
G1
gives the in-degree.
th
The (i, j)
n
entry of the adjacency matrix A
is the number of walks of length n
from i to j
Eampe 3.7
.1
(a)
Construct
an
adjacency
matrix
M
for
this
graph.
B
C
A
D
131
/
3
2
(b)
Find
(i)
(c)
Hence
than
The
connected
Georgina
three
(d)
or
or
graph
3
M
(ii)
otherwise
equal
to
3
represents
by
direct
wishes
to
M
find
from
the
A
to
airports
number
of
walks
of
length
less
D
A,
B,
C
and
D
and
shows
which
are
flights.
travel
between
airports
A
and
D
in
at
most
flights.
Explain
would
in
context
consider
is
why
the
fewer
number
than
the
of
possible
answer
to
routes
part
she
(c).
Solution
(a)
⎛
0
1
1
1
1
0
1
1
The
⎞
⎜
matrix
set
up
with
⎟
A
⎜
in
the
first
row/column,
⎟
⎜
1
1
0
0
1
1
0
0
⎟
⎜
then
B,
C
and
D
⎟
⎝
⎠
The
(b)
is
⎛
(i)
3
2
1
⎞
1
⎜
(ii)
⎛
4
⎜
5
4
⎜
5
5
5
5
⎞
5
5
⎟
2
2
⎟
5
the
powers
are
found
from
GDC
⎟
2
3
1
1
⎜
⎟
The
1
⎜
1
2
2
⎟
⎜
⎟
1
⎝
1
(c)
1 + 1 + 5
(d)
Some
2
=
2
5
⎝
⎠
number
5
2
2
⎠
number
of
1,
3,
2
the
7
and
top
three
of
leaving
going
back
the
walks
and
to
to
D
D
of
routes
is
the
⎟
⎜
routes
so
right
the
of
entries
corner
matrices
length
are
of
in
the
added.
involve
returning
directly
to
and
A
to
or
B
and
For
example,
trip
from
again
again.
A
ABAD.
to
would
B
be
and
The
back
unnecessary
.
Tranition matrie
Link to chapter 1 section 1.5
Matrices, and chapter 4 section
4.10 Transition matrices and
Markov chains. It is best to review
A transition
given
to
be
matrix
vertex
taken
from
(a
for
a
a
graph
starting
random
gives
vertex
the
probability
when
all
edges
of
are
reaching
equally
any
likely
walk).
these two sections before reviewing
th
The
(i,
j)
entry
is
the
number
of
edges
leading
from
vertex j
to
vertex
i.
this section.
A
B
C
D
1
1
1
⎞
3
2
2
⎟
1
1
⎟
2
2
0
0
Note
⎛
A
0
⎜
This is the opposite to an
⎜
adjacency matrix.
The
transition
matrix
for
graph
in
Example
3.7.1
is
1
0
B
⎜
3
⎟
⎜
1
C
1
⎜
3
3
1
1
3
3
⎜
⎝
⎟
⎟
⎜
D
⎟
0
0
⎟
⎠
In exams, all transition matrices
will be for strongly connected
graphs.
The
steady
matrix
time
state
gives
spent
at
the
vector
for
the
long-term
each
vertex
transition
proportion
on
a
random
of
walk.
B
Note
The
steady
state
will
always
exist
when
the
C
graph
is
connected
or
the
directed
graph
is
When the graph represents
strongly
connected
(which
get
any
to
means
you
can
A
connections between webpages,
from
vertex
any
other
vertex).
the steady state values can be
D
used to rank the webpages.
132
/
3 .7
GR A P H s
Eampe 3.7
.2
The
directed
webpages.
walk
graph
here
A search
through
the
shows
engine
graph
in
the
links
performs
which
it
a
is
between
(a)
Find
random
to
leave
a
site
by
any
of
the
matrix
for
this
random
equally
possible
Find
the
probability
of
the
search
engine
being
links.
at
site
E
begins
B
(c)
A
transition
walk.
(b)
likely
the
C
after
it
Find
the
each
of
following
search
at
site
long-term
the
three
links,
given
it
A.
probabilities
of
being
at
sites.
D
E
Solution
Note:
A
B
C
D
0
0.333
0
0
1
0.25
0
0.5
0
0
0.25
0.333
0
0.5
0
0.25
0.333
0.5
0
0
0.25
0
0
0.5
0
(a)
E
because
this
is
a
directed
graph
the
matrix
is
not
symmetrical.
⎛
A
⎟
⎜
T
=
C
⎜
⎜
D
⎝
the
there
first
are
column
four
there
edges
are
4
equal
probabilities
as
from A
⎟
⎟
⎟
⎜
E
In
⎟
⎜
B
⎞
⎠
3
3
(b)
From
(c)
Each
⎛
T
the
probability
column
0.222...
⎜
in
the
is
This
0.1875
steady
state
vector
is
is
the
Finding
entry
the
in
values
the
fifth
from
have
been
given
to
have
been
given
exactly
a
three
row
high
of
the
power
significant
first
of
column
T.
figures
The
but
of T
answers
could
⎞
⎟
as
fractions.
0.166...
⎜
⎜
⎜
⎟
0.222...
0.222...
⎜
⎝
⎟
⎟
0.166...
The
is
⎟
⎠
probability
0.222,
and
of
of
being
being
at
at
B
A,
and
C
and
E
is
D
0.167
There is no need to write out the whole matrix. You should state that you are
3
finding T
, so if you entered the values incorrectly or read the wrong value you
would get method or follow through marks.
Graph agorithm
A weighted
The
weights
travel
or
A cycle
graph
might
costs
is
repeating
a
one
in
which
represent
the
edges
distances
are
between
labelled
the
with
vertices
weights.
or
time
of
etc.
route
any
is
that
starts
and
ends
at
the
same
vertex
without
vertex.
133
/
3
A circuit
is
repeating
A tree
is
a
any
a
graph
with
There
are
the
already
the
edges
ways
least
and
ends
at
the
same
vertex
without
cycles.
tree
a
the
each
and
tree
possible
find
At
is
that
total
connects
edge
minimum
step,
which
add
does
the
vertices
in
the
weight.
spanning
the
not
all
edge
form
tree:
of
a
least
cycle.
weight
Repeat
that
until
is
all
connected.
algorithm:
edgeof
no
to
included
are
starts
minimum
algorithm:
not
a
with
spanning
two
Kruskal’s
that
edge.
graph
A minimum
Prim’s
route
Choose
weight
to
a
the
starting
tree
vertex.
already
At
formed
each
stage,
add
the
and
which
connects
to
newvertex.
Eampe 3.7
.3
Find
the
minimum
spanning
tree
for
the
graph
shown
B
using:
6
8
(a)
Kruskal’s
algorithm
5
4
6
C
(b)
Prim’s
algorithm
F
9
10
Write
down
the
order
in
which
the
edges
are
11
selected.
8
D
7
E
Solution
(a)
The
AF,
order
FB,
in
FC,
which
ED,
the
edges
are
selected
is
AE
Note
that
after
there
was
a
AB
AF
choice
couldnot
be
and
of
FB
two
selected
had
been
edges
selected
with
because
it
weight
would
6.
have
B
formed
Note
A
a
cycle.
that
when
connected
to
algorithm
the
end,
ED
the
was
rest
selected
of
the
tree.
edges
do
not
have
which
happens
it
In
was
not
Kruskal’s
to
be
connected
C
until
the
when
AE
is
added.
F
E
(b)
D
The
spanning
but,
beginning
FB,
FC,
AE,
tree
ED
is
with
the
same
vertex
A
as
the
for
part
order
is
(a)
In
Prim’s
AF,
adding
a
ED
could
the
tree.
algorithm,
new
not
vertex
be
because
to
added
the
you
tree
until
are
you
either
always
are
E
or
forming,
D
was
in
The question will often ask you to write down the order in which the edges are
selected – this is to ensure you have used the correct algorithm, so you might
not be awarded any marks if you don’t do this.
134
/
3 .7
The
graph
adjacency
in
Example
table
(or
3.7.3
matrix)
could
as
have
been
given
in
a
weighted
GR A P H s
B
6
follows:
8
5
4
6
C
A
B
C
D
E
F
10
8
4
F
A
6
9
10
11
8
B
6
8
5
D
C
8
11
7
6
E
Prim’s
algorithm
without
the
D
10
E
8
F
4
can
need
to
be
11
9
7
5
6
performed
draw
7
out
the
9
on
a
weighted
3
4
6
5
2
A
B
C
D
E
F
table
graph.
Cross
1
adjacency
off
above
6
row
column
Choose
the
column
(4)
Cross
a
2
A
off
and
the
write
a
1
entry
in
A
smallest
that
above
and
circle
row
that
it.
(F)
and
write
corresponding
column.
AF,
FB,
FC,
AE,
ED
Choose
any
the
smallest
entry
of
the
numbered
which
has
not
(5),
circle
(B)
and
Repeat
been
it
been
and
number
until
all
columns
crossed
cross
the
the
from
out
out
that
column
rows
row
3.
have
selected.
The chinee potman probem
A trail
is
a
A circuit
walk
does
a
trail
that
An
Eulerian
trail
is
An
Eulerian
circuit
same
Eulerian
A graph
Eulerian
at
for
which
which
trail
the
that
travels
point.
The
would
the
there
will
If
in
trail
is
an
pairs
every
that
edge.
finishes
traverses
Eulerian
vertex
idea
at
trail
has
at
the
same
all
the
edges
which
an
even
vertices
one
to
be
every
walk
of
of
the
vertex.
in
a
graph.
starts
and
ends
degree
will
contain
odd
degree
vertices
a
find
down
every
street.
even
no
degree
edges
of
vertices
routes
and
of
need
odd
odd
with
that
of
to
is
will
odd
least
at
the
an
have
degree
an
and
returns
minimum
to
its
starting
delivering
shortest
walk
weight
letters
will
be
an
repeated.
then
done
the
using
then
total
of
postman
the
be
degree
walk
and
a
then
degree
this
the
graph
is
repeated
the
of
to
name
have
and
edge
is
the
vertices
four
using
problem
behind
two
to
are
two
begins
postman
circuit
are
exactly
which
vertices
need
there
has
along
have
Eulerian
If
a
and
any
other.
Chinese
all
starts
repeat
circuit.
The
If
not
vertex.
A graph
ends
is
that
four ver tices with odd degree. 
route
the
these
In an exam there will be at most
between
route
need
of
to
these
two
leastweight.
be
connected
weight.
135
/
3
Eampe 3.7
.4
The
graph
mountain
shown
huts
represents
and
the
paths
weights
in
are
a
the
national
park.
distances
The
along
vertices
each
of
are
12
the
C
B
paths
in
kilometres.
A walker
would
like
to
trek
along
all
of
the
paths.
6
8
15
The
total
distance
of
all
the
paths
is
106
7
km.
11
A
(a)
Find
the
least
distance
he
would
have
to
walk
if
he
is
G
D
starting
10
and
finishing
at
the
same
mountain
hut,
and
list
the
paths
that
9
he
8
12
would
need
to
walk
along
F
twice.
8
The
walker
mountain
(b)
Write
he
receives
an
offer
to
be
dropped
and
picked
up
at
any
of
the
E
huts.
down
has
to
which
huts
he
should
choose
to
minimize
the
distance
walk.
Solution
(a)
The
odd
vertices
are
Whenever
there
B,
F,
D
and
are
and
DG
11
+
10
19
+
9
BD
and
GF
BG
and
DF
The
least
repeating
(b)
The
20
+
BF
choose
to
be
or
versa
and
=
=
19
distance
shortest
vice
three
are
four
vertices
of
odd
degree
combinations
to
consider.
of
be
shown.
The
G
weights
BF
there
is
for
all
them
must
21
28
=
39
106
+
21
=
127
km,
DG
distance
dropped
is
GF
at
B
so
he
and
should
picked
This
up
at
D
would
mean
he
only
has
to
walk
an
extra
8km.
The traeing aeman probem
A path
is
a
walk
that
does
A Hamiltonian
path
A Hamiltonian
cycle
same
The
a
is
path
a
pass
that
through
includes
Hamiltonian
any
all
path
vertex
the
that
more
vertices
starts
and
than
in
a
once.
graph.
ends
at
the
vertex.
A complete
other
is
not
graph
is
one
in
which
all
the
vertices
are
adjacent
to
all
vertices.
travelling
of
least
to
asthe
name
salesman
weight
is
looking
on
classical
that
for
the
the
a
problem
complete
travelling
vertices
shortest
to
find
weighted
salesman
might
way
is
to
graph.
to
towns
all
of
Hamiltonian
This
problem.
represent
travel
the
is
The
idea
and
the
often
a
cycle
referred
behind
salesman
towns
and
the
is
then
backhome.
In
many
real
life
or
‘practical’
problems
the
graph
connecting
the
If using a table of least distances,
vertices
is
not
complete.
In
this
case
the
distances
between
the
the solution to the practical
unconnected
towns
is
These
be
added
to
every
vertex
is
taken
as
the
shortest
distance
between
them.
problem is likely to involve
can
a
table
of
least
distances
or
shown
on
a
graph
passing through some ver tices
so
that
connected
and
the
solution
of
the
practical
more than once.
problem
is
the
same
as
the
solution
of
the
classical
problem.
136
/
3 .7
There
is
no
problem.
lower
the
algorithm
Instead,
bound
for
for
finding
algorithms
the
the
are
solution.
solution
used
The
to
closer
to
find
the
an
travelling
upper
together
these
GR A P H s
salesman
bound
and
are
better
the
a
solution.
The
weight
travelling
of
any
Hamiltonian
salesman
cycle
is
an
upper
bound
to
the
problem.
Eampe 3.7
.5
B
The
graph
shows
times
in
minutes
to
travel
between
towns A,
B,
C
and
D.
90
A salesman
wishes
to
visit
each
of
the
towns,
starting
and
finishing
at A
C
120
(a)
Complete
a
table
showing
the
shortest
travel
times
between
the
towns.
130
250
A
(b)
By
considering
takenby
the
the
cycle
salesman,
ABCDA,
stating
in
find
an
order
upper
the
bound
towns
he
for
the
passes
time
through
D
onthis
journey
.
Solution
(a)
Notice
A
B
C
D
A
0
120
210
250
B
120
0
90
130
C
210
90
0
220
D
250
130
220
0
the
table
not
the
is
minor
a
and
(b)
From
120
+
90
+
220
+
250
=
680
the
direct
shortest
route.
road
time
This
between
might
compared
with
be
C
and
because
the
roads
D
is
that
CB
BD
Refer
minutes
that
back
to
the
graph
to
see
the
actual
route
taken.
Passing
The
to
nearest
the
you
vertex.
The
a
is
route
to
vertex
the
C,
D,
and
until
B,
is
problem.
starting
is
A
often
In
then
all
algorithm
salesman
B,
algorithm
vertex
repeated
back
deleted
B,
salesman
starting
This
travelling
A,
neighbour
travelling
select
direct
through
the
used
nearest
move
vertices
vertex
often
is
to
to
the
have
an
upper
neighbour
nearest
been
then
used
find
bound
algorithm
adjacent
reached
and
the
taken.
to
find
a
lower
bound
to
the
problem.
Note
In
the
deleted
vertex
with
the
algorithm,
one
vertex
is
deleted
from
the
graph
Both the nearest neighbour
along
all
incident
edges.
The
minimum
spanning
tree
of
the
algorithm and the deleted ver tex
remaining
subgraph
is
then
found.
algorithm should be performed on
The
weight
of
the
minimum
spanning
tree
plus
the
weight
of
the
two
a complete graph or on a complete
edges
of
least
weight
that
connect
the
deleted
vertex
to
the
tree
give
a
table of least distances.
lower
bound.
137
/
3
s AMPlE sTUDENT ANs WER
The
weights
following
the
complete
C
D
E
F
A
–
4
9
8
14
6
B
4
–
1
14
9
3
C
9
1
–
5
12
2
D
8
14
5
–
11
12
E
14
9
12
11
–
7
F
6
3
2
12
7
–
Starting
at
an
upper
By
first
for
can
in
the
A,
use
bound
deleting
with
the
for
nearest
the
vertex
travelling
travelling
A,
Kruskal's
neighbour
use
the
table.
It
graph,
be
can
but
are
given
in
the
deleted
salesman
to
problem
find
for
problem
vertex
a
to
find
for
G.
algorithm
lower
bound
G.
neighbour
used
directly
from
be
helpful
to
errors
Nearest
Neighbor
.
draw
A
the
G
algorithm
salesman
algorithm
(a)
the
graph
table.
together
algorithm
edges
B
(b)
nearest
the
A
(a)
▼ The
of
could
be
made
could
be
lost.
4
B
AB(4)
in
doing
so
and
time
6
1
14
BC(1)
F
C
9
14
CF(2)
▲ The
the
student
algorithm
and
showed
correctly
to
all
the
the
rst
applied
few
steps,
so
points
FE(7)
was
EA(14)
awarded
some
marks.
E
D
▼ The student missed out edge ED
Hence,
▲ The
student
Kruskal’s
and,
correctly
though
upper
bound =
28
applied
not
told
to
(b)
do
so
the
in
the
order
selected,
marks
was
in
so
even
not
question,
which
they
wisely
the
were
though
edges
listed
were
awarded
the
nal
Bc(1),
CF(2),
CD(5),
FE(7)
four
answer
1
+
2
+
5
+
7
=
15
correct.
Lower
▼ The
two
15
+
student
edges
4
+
6
of
=
forgot
least
to
add
weight
bound
is
15
the
from
A:
25
138
/
PRACTICE
diagram
SL
P R A CT I C E
PA P E R
1.
1,
A logo
a
for
circle
The
a
and
sector
circle
Find
a.
GROUP
has
the
the
sports
a
of
a
Q U E ST I O N S
team
area
in
circle
radius
of
3
is
made
two
has
a
up
of
a
sector
colours,
as
shown.
centre
O
and
at
D
of
the
cm.
of:
triangular
section
B
b.
the
3
cm
a.
Calculate
on
horizontal
of
the
3
summits
adjacent
mountains
(108.3,
42.2,
are
4.5),
(102.8,
where
C
elevation
CD,
the
to
height
ground
D
of
is
250
in
from
48°
the
observation
Calculate
the
the
ground.
angle
of
depression
from
A
to
B
of
A triangular
postage
stamp,
ABC,
is
shown
in
the
two
2.9)
!
ABC
=
below,
26°
and
such
that
!
ACB =
AB
=
5
cm,
!
B AC
=
34°,
120°
distances
diagram
are
m
cm
39.1,
all
of
above
diagram
and
m
cm
6.
coordinates
angle
deck
O
B,
the
b.
3
point
C,
140º
The
250
sector.
From
2.
not
scale
1
triangle
the
to
A
QUE STIONS
kilometres.
C
to
not
scale
120º
Find
two
the
straight-line
distance
between
the
summits.
34º
26º
A
3.
A cake
of
a.
8
cm
Find
A piece
of
is
the
angle
in
the
and
the
of
a
shape
height
volume
the
cake
piece
is
of
at
35°
a
of
of
is
a
10
the
cylinder
of
a
5
The
Find
the
volume
of
circle
of
the
a.
Find
the
length
b.
Find
the
area
the
and
forms
an
GROUP
Let
A
be
B
be
a.
Find
A
Let
l
the
point
point
the
and
be
the
with
with
piece.
The
A,
coordinates
coordinates
midpoint
of
the
line
(
(2,
1,
8)
BC
of
the
postage
stamp.
2
B
V
oronoi
and
10).
segment
C,
unit
diagram
set
on
in
the
a
below
shows
rectangular
axes
three
area
represents
1
of
wells,
land.
km.
and
11
joining
B.
the
of
cake.
Each
4.
cm
cross-section
7
.
b.
B
radius
cake.
a
centre
with
cm.
removed.
sector
the
of
10
9
perpendicular
bisector
of
the
line
8
C
segment
b.
Find
A
is
joining
the
A
and
equation
of
B.
7
6
l.
5
5.
a
point
at
the
communications
top
of
tower
a
vertical
with
its
base
at
C.
The
4
A
tower
has
of
way
the
an
to
observation
the
top
of
deckD,
the
three-quarters
B
3
tower,A
2
1
0
1
2
3
4
5
6
7
8
9
10
139
/
3
a.
b.
Write
the
equations
forming
the
V
oronoi
diagram.
Find
the
three
area
of
of
each
of
the
boundaries
the
region
Find:
lines
within
the
the
distance
b.
the
area
from
covered
B
to
by
C
the
glacier.
containing:
10.
i.
a.
Jorge
wishes
to
bend
a
20
cm
length
of
wire
into
A
ii.
B
iii.
C
the
shape
of
a
the
sector
be
r
centre
A fourth
well
is
perpendicular
created
at
bisectors
the
point
between
D
D.
by
the
circular
sector.
cm
the
and
sector
Let
angle
the
radius
made
at
of
the
θ
be
The
and
each
r
of
A,
B
and
C
are
shown
as
dashed
lines
on
the
diagram.
θ
r
11
D
a.
10
Find
Jorge
9
an
expression
wants
the
possible
area.
b.
that
θ
for
sector
to
in
terms
contain
of
the
r.
largest
8
Show
the
area
of
the
sector
A
can
of
which
gives
be
given
C
7
by
the
expression:
6
2
A
=
10r
r
5
.
Hence
find
maximum
the
value
r
the
area.
B
A
3
11.
A type
of
candy
is
packaged
in
a
right
circular
3
2
cone
with
height
1
of
a
8
volume
of
100
cm
and
vertical
cm.
r
0
1
.
On
2
a
3
copy
of
4
the
5
6
7
diagram,
8
9
complete
10
the
8
V
oronoi
diagram
for
the
four
cm
wells.
l
8.
An
underground
tunnel
coordinates
(10.1,
coordinates
(8.5,
A relief
tunnel
surface
at
tunnel
is
C (9.0,
joining
A
3.2,
goes
0.8)
1.7,
0.9)
to
from
go
2.0,
0)
and
to
B.
to
a
the
All
from
a
point
point
point
B
on
with
with
the
midpoint
positions
A
of
are
a.
the
Find
the
radius,
r,
of
the
circular
base
of
thecone.
given
b.
Find
the
slant
.
Find
the
curved
height,
l,
of
the
cone.
inkilometres.
Find
9.
the
length
A geologist
is
approximate
The
120,
From
the
are
surface
C,
A(0,
230,
he
angle
the
trying
coordinates
glacier
C (
of
of
0,
140),
can
!
ACB
to
of
three
B(250,
where
both
out
area
of
the
cone.
a
the
triangular
corners
120,
75)
distances
A
surface
tunnel.
work
area
the
0),
see
as
relief
and
B
of
glacier.
the
and
are
and
in
metres.
measures
47°
140
/
PRACTICE
12.
A right
pyramid
ABCD,
with
The
AB
vertical
has
=
8
apex
cm,
height
of
V
BC
the
and
=
6
rectangular
cm
and
pyramid
V
A
=
base
13
14.
cm.
The
diagram
radius
r
Points
A
shows
a
circle
with
QUE STIONS
centre
O
and
circumference
of
the
cm.
is VM
V
diagram
to
circle
not
and
B
lie
!
AOB =
and
on
the
60°
scale
The
point
C
is
on
[OA]
such
that
!
BCO
=
90°
B
diagram
to
13
not
scale
cm
r
D
60º
A
M
A
O
C
C
8
cm
6
cm
B
a.
Calculate
VM
b.
Calculate
the
volume
of
the
pyramid.
a.
GROUP
13.
Find
an
expression,
in
terms
of
r,
for:
3
A garden
triangle,
consists
shown
of
as
a
lawn
ABC
in
in
the
the
shape
diagram,
of
i.
OC
ii.
the
The
area
a
and
area
of
the
triangle
OBC.
a
3
flower
bed
with
radius
the
a
length
formed
of
of
BC
4
by
m.
is
the
The
8m
arc
of
a
length
and
circle,
of
angle
AC
centre
is
!
ACB =
6
of
the
shaded
region
is
50
cm
shape
of
a
O
m,
b.
Find
the
value
of
r.
60°
15.
C
A large
sculpture
based
pyramid.
A
B,
and
have
is
in
Two
the
adjacent
coordinates
(0,
right
corners
0,
0)
of
and
square-
the
(8,
0,
base,
0)
60º
respectively
.
coordinates
a.
i.
Write
ii.
Find
The
(a,
vertex
a,
15).
down
the
of
All
the
the
distances
height
volume
pyramid, V,
of
of
the
are
the
has
inmetres.
pyramid.
pyramid.
O
b.
Write
The
4
4
the
.
decorative
of
the
garden
edging
to
go
wants
around
to
buy
the
flower
Find
the
length
of
the
side
Hence,
needs
find
to
the
length
of
value
a
light
of
a.
positioned
half
way
up
centred
Find
the
sides
face
which
horizontally
coordinates
of
has
the
of
pyramid
[AB]
on
the
are
as
that
its
base,
face.
light.
covered
in
panels.
bed.
Find
the
area
of
the
panels
used.
AB
16.
b.
is
steel
some
d.
a.
has
triangular
The
owner
pyramid
the
m
m
and
The
down
edging
the
owner
Barry
sea
is
at
level.
the
He
top
of
a
observes
cliff,
two
standing
yachts
in
80
the
m
above
sea.
buy
.
Seaview
Nauti
(S)
Buoy
is
at
(N)
an
is
angle
at
an
The
three-dimensional
and
the
two
yachts
at
S
of
depression
angle
of
diagram
and
of
25°
depression
shows
of
Barry
35°
(B)
N
141
/
3
X
lies
at
the
foot
of
the
cliff,
a.
!
SX N
and
=
State
the
level
of
pollution
that
would
that
the
be
70°
assigned
to
D
B
The
house
levels
An
a
are
owner
much
alternative
more
The
of
claims
than
method
is
will
be
pollution,
the
A new
,
for
to
cell
created
P
pollution
assigned
suggested
measure.
diagram
level
D
worse
accurate
V
oronoi
at
value.
give
for
the
around
the
point
D
D
D
will
then
be
found
using
6.5 a
the
formula:
+ 7.2 a
A
P
+ 3.2 a
B
C
=
D
a
where
a
is
the
area
of
the
new
cell
and
a
,
a
A
a
N
are
the
distance
accurate
to
three
between
the
significant
two
yachts,
Pollution
three
levels
in
a
city
environmental
environmental
the
regions
within
the
new
are
are
that
used
to
belong
to
stations
A,
B
and
C
figures.
recorded
stations
stations
of
respectively
.
The
17
.
areas
C
cell
Find
the
and
B
A,
B
at
and
shown
on
new
vertices
C.
V
oronoi
at
(4,
6),
diagram
(8,
6)
and
is
shown
(6,
below
with
0).
The
the
V
oronoi
10
diagram
with
coordinates
A(3,
4),
B(9,
4)
and
9
C(6,
7),
where
distances
are
given
in
kilometres.
8
C
7
10
6
9
D
5
8
B
A
C
4
7
3
6
2
5
B
A
1
4
3
0
2
3
4
5
2
1
b.
Find
the
value
of
a
.
Find
the
value
of
P
D
0
2
3
4
5
18.
On
a
particular
according
recorded
to
as
day
,
the
6.5
Air
at
the
levels
Quality
A,
7.2
at
of
B
and
placed
pollution
Health
Index
3.2
at
A cylindrical
with
are
on
a
water
container
flat
to
a
with
surface.
height
a
The
of
12
radius
container
cm,
as
taken
to
nearest
of
be
pollution
the
station
A house
is
same
to
as
that
situated
at
at
any
the
point
value
in
the
city
(6,
is
filled
recorded
at
not
scale
is
the
point.
the
point
D
with
12
coordinates
is
cm
C
to
level
8
shown.
diagram
The
of
cm
5)
8
cm
142
/
PRACTICE
a.
Find
the
volume
of
water
in
the
container.
(Note:
in
an
copyofthe
A heavy
ball
with
a
radius
of
2.9
cm
is
of
the
the
container.
water
As
a
increases
result,
to
h
the
cm,
as
in
A new
needs
thediagram.
and
diagram
Write
Give
an
to
be
equal
in
the
school
the
built
on
distance
the
island
from
should
the
a
form
the
line
each
owns
of
coordinates
straight
from
Abdallah
=
60
m,
of
of
a
be
of
schools
of
the
other
land
of
the
is
=
84
AB
where
=
new
near
the
40
m,
River
ABCD.
BC
=
The
115
m,
m,
shown
and
on
!
B AD =
the
90°.
This
diagram.
2
island
has
four
schools
which
are
shown
diagram
on
to
115
a
B
schools.
quadrilateral
are
point
C
An
A,
built.
distance
the
plot
sides
AD
information
1.
which
h
CD
PA P E R
is
down
new
lengths
of
the
C
Nile,
value
be
school
cm
2.
the
draw
scale
.
Find
to
a
diagram.)
school
to
the
b.
which
have
not
e.
h
on
would
height
shown
to
grid
you
dropped
V
oronoi
into
exam,
QUE STIONS
coordinate
grid
at
the
points
A(1,
1),
B(5,
not
scale
m
1),
B
C(5,
7)
and
D(7,
5).
Also
shown
is
the
V
oronoi
60
diagram
for
the
schools
A,
B
and
m
C.
40
m
8
A
D
84
C
m
7
a.
Show
that
BD
=
93
m,
correct
to
the
nearestmetre.
D
5
ˆ
b.
Calculate
.
Find
the
size
of
angle
BCD
4
the
area
of
ABCD
3
The
2
formula
estimate
A
the
that
the
area
of
ancient
a
Egyptians
quadrilateral
used
ABCD
to
is:
B
( AB + CD)( AD + BC)
1
Area
=
4
0
2
4
Abdallah
6
-1
of
d.
a.
Find:
his
uses
plot
of
this
formula
to
estimate
the
area
the
area.
land.
i.
Calculate
ii.
Find
the
Abdallah’s
percentage
estimate
error
for
in
Abdallah’s
be
modelled
estimate.
i.
the
midpoint
ii.
the
gradient
of
[BD]
3.
b.
Hence
show
of
that
the
equation
of
base
of
an
electric
iron
can
as
a
bisector
of
[BD]
is
pentagon
ABCDE,
where:
BCDE
rectangle
the
1
perpendicular
The
[BD]
y
=
x
−
+
is
a
with
sides
of
length
6
2
(x
.
Find
the
equation
of
the
+
3)cm
of
Complete
+
5)
cm
is
an
isosceles
triangle
with
AB
=
AE,
and
a
[CD]
height
d.
(x
perpendicular
ABE
bisector
and
the
V
oronoi
diagram
for
the
of
x
cm
schools
2
The
A,
B,
C
and
area
of
ABCDE
is
222
cm
D
143
/
3
diagram
to
B
not
scale
A
5
x
m
cm
A
30º
E
B
a.
Show
the
b.
(x
+
5)
that
other
Find
the
there
are
two
possible
end
of
the
pole.
area
of
the
smaller
of
positions
the
two
2
⎛
The
transformation
matrix
T
1
maps
D
a.
i.
Write
+
3)
an
equation
for
the
area
Show
onto
a.
the
value
of
b.
the
point
that
(b,
a
3).
and
Find:
the
value
of
b
using
that
the
the
information
equation
in
part
is
mapped
onto
(0,
–7)
of
3.
ii.
3)
1⎠
cm
down
ABCDE,
(4,
the
⎟
⎝ a
point
⎞
=
⎜
(x
pens.
cm
2.
C
for
An
aircraft
is
flying
such
that,
at
t
hours
after
given.
i
simplifies
11am,
its
airport
is
position
relative
given
the
by
to
its
destination
equation:
2
to
+
3x
19x
–
414
=
0
⎛
⎜
b.
Find
the
length
of
r
CD
226
Show
!
B AE
that
=
67.4°,
correct
decimal
Insulation
of
the
Find
of
the
x
is
the
is
the
wrapped
iron
length
point
element
Cto
in
around
the
⎜
⎠
⎝
⎟
c
⎠
directions
are
east
and
north
of
on
the
AB
iron
the
z
height
aircraft.
direction
gives
the
vertical
perimeter
of
the
ABCDE
the
perimeter
of
the
such
runs
that
in
a
BF
=
8
aircraft’s
velocity
remains
constant,
it
will
ABCDE
at
the
airport
at
11:30
a.
Find
the
values
of
a,
b.
Find
the
speed
am.
cm. Aheating
straight
line
b
and
c
from
of
the
aircraft.
F
4.
e.
y
and
land
F
and
respectively
,
If
d.
⎟
place.
tape
base
b
to
The
1
⎞
⎟
+ t
12
⎝
a
⎜
⎟
=
⎜
.
⎛
⎞
105
Calculate
the
length
of
Three
points
in
three-dimensional
space
have
CF
coordinates
A(1,
HL
P R A CT I C E
1,
GROUP
Find
A wall
5
m
long
makes
an
angle
of
30°
straight
section
of
4,
3)
and
C(1,
2,
4)
the
vector
!!!
"
river,
as
shown
in
ii.
AB
AC
with
b.
a
B(2,
!!!
"
1
i.
1.
5),
Q U E ST I O N S
a.
PA P E R
0,
Hence or otherwise,
find the area
of triangleABC.
the
⎛
diagram.
The
owner
of
the
field
on
the
one
5.
of
the
river
wants
to
make
a
triangular
pen
A particle
a
wall
AB
second
as
side.
one
For
of
the
the
sides
third
and
side,
the
he
river
will
has
use
at
t
=
⎛
a
2
3.5
m
tocut.
He
will
end
the
wall
of
long
attach
at
which
one
point
he
end
does
of
the
not
pole
given
by
⎞
!
! =
x
⎜
⎟
⎝
2 ⎠
the
particle
is
at
(0,
0)
and
has
⎞
the
find:
⎟
⎝ 1⎠
wish
to
0
,
⎜
pole
!
!
x
as
velocity
metal
acceleration
with
Given
the
6t
side
a.
an
expression
for
its
velocity
b.
an
expression
for
its
displacement
at
time
t
B
at
.
its
time
from
(0,
0)
t
distance
from
(0,
0)
at
t
=
2
144
/
PRACTICE
6.
Sameer
is
trying
to
design
a
road
system
QUE STIONS
to
diagram
to
connect
six
possible
are
a
roads
shown
town.
towns,
in
and
the
Each
A,
B,
the
D,
costs
graph.
edge
C,
E
of
Each
represents
and
F.
The
building
them
vertex
represents
a
and
road
not
scale
the
θ
weight
road.
of
each
Sameer
edge
needs
is
the
to
cost
design
of
the
building
lowest
that
cost
O
road
system
that
will
connect
the
six
towns.
a.
Show
that
the
shaded
area
can
be
expressed
as
B
5
50θ
A
50 sin θ
6
8
b.
Find
the
value
θ
of
for
which
the
shaded
area
is
D
equal
3
to
half
that
of
the
unshaded
area,
giving
4
C
your
answer
correct
to
four
significant
figures.
7
8
10.
Let
u
=
3i
+
j
+
k
and
v
=
m j
+ nk .
Given
that
v
is
7
4
a
unit
vector
values
E
Name
find
an
the
algorithm
lowest
that
cost
will
road
m
and
to
u,
find
the
possible
n
F
4
11.
a.
for
perpendicular
allow
Sameer
to
The
graph
shows
bus
between
10
the
costs
in
USD
to
travel
by
towns.
system.
E
b.
Find
the
lowest
cost
road
system
and
state
the
12
cost
the
of
building
it.
Clearly
show
the
steps
10
of
algorithm.
F
9
I
12
B
GROUP
2
6
8
7
10
7
.
In
triangle
ABC,
AB
=
5 cm,
BC
=
4
cm
8
and
9
J
π
!
B AC
A
D
G
=
7
9
11
a.
Use
the
cosine
rule
to
find
the
two
10
possible
10
C
values
for
AC.
H
b.
Hence,
find
the
area
of
the
larger
of
the
two
The
possible
total
exactly
8.
Triangle
T
has
coordinates
(2,
3),
(2,
5)
and
(6,
Find
the
area
of
once
Explain
M is
represented
by
the
2 sin θ
2 cos θ
cos θ
sin θ
the
routes
is
152USD.
why
the
graph
has
no
Euler
circuit.
would
like
to
travel
along
all
the
bus
for
the
least
possible
cost,
and
to
start
and
in
the
same
town.
⎟
⎠
b.
Show
that
the
area
of
the
image
of
T
under
Use
is
independent
of
the
diagram
20cm,
the
shows
centred
at
O,
circumference
measured
in
a
semi-circle
with
such
two
that
of
algorithm
to
find
least
possible
cost
and
state
the
bus
routes
need
to
be
repeated.
diameter
points
!
AOB =
postman
θ
which
The
Chinese
this
the
transformation
9.
of
⎞
finish
⎜
b.
each
matrix
routes
⎝
travelling
T.
A tourist
A transformation
⎛
for
4).
a.
a.
cost
triangles.
A
,
and
B
where
The
tourist
realizes
and
the
and
finishes
that
she
can
reduce
her
cost
on
θ
number
of
repeated
routes
if
she
starts
is
at
different
towns.
radians.
.
State
and
which
finish
would
towns
at,
save
and
on
she
how
bus
should
much
choose
money
to
start
she
travel.
145
/
3
12.
A security
museum.
guard
The
patrols
rooms
four
and
all
rooms
the
in
a
Let
are
shown
in
the
floor
=
SR
connecting
n
b.
doors
T
plan
Find
an
expression
i.
n
is
odd
ii.
n
is
even
for
the
matrix
T
,
when:
below.
A
B
x
⎛
⎞
n
⎛
D
1
⎞
n
Let
=
and
T
⎜
⎟
⎜
⎝ y
⎠
⎝ 0⎠
let
the
distance
of
(x
⎟
,
y
n
)
n
n
C
from
the
origin
be
d
n
While
patrolling,
the
security
guard
decides
he
N
is
the
smallest
value
of
n
such
that
d
>
20
n
will
leave
each
room
by
a
randomly
selected
door.
.
a.
Show
the
floor
plan
as
a
graph
with
as
the
rooms
and
the
edges
(x
,
y
)
N
N
a
boat
the
15.
vertices
Find
as
At
13:00,
is
sailing
directly
to
a
port
with
the
1
a
connecting
velocity
.
d.
Find
the
Find
Use
transition
the
steady
your
matrix
state
answer
to
for
vector
part

the
security
for
to
this
say
guard
spends
most
j
is
due
room
13.
Boat
the
speed
b.
Find
the
bearing
which
and
he
spends
on
that
situated
each
boat
board.
time
in
is
The
7km,
boat
least
time
in.
you
are
has
10
a
km
of
away
the
and
on
boat
B
it
which
both
transmitters
State
by
the
shaded
the
area
of
this
12
km
of
the
on
boat.
which
from
boat
Find
the
boat
reaches
Find
an
from
radio
boat
B,
boat
from
the
the
port.
remains
time,
to
the
transmitter
on
is
5
km.
The
can
be
region
boat
is
sailing.
Assume
that
the
when
the
constant.
nearest
minute,
port.
the
of
port
hours
for
in
the
displacement
terms
after
of
t,
where
t
of
is
the
the
13:00
boat
Two
students
are
standing
20
m
apart
and
are
region
detected
in
the
expression
number
transmitter
the
beanbags
towards
each
other.
is
A
throws
his
beanbag
towards
student
B
diagram.
from
Find
the
an
Student
represented
of
making.
marine
range
is
throwing
in
east
and
16.
A
due
room
d.
is
i is
matrix.
3
A
where
km h
north.
Find
.
assumption
GROUP
3.9j
a.
velocity
which
+
graph.
The
the
5.2i
doors.
and
b.
of
a
height
of
1.5
m
with
an
initial
velocity
of
region.
–1
and
5ms
After
⎛
0
at
release,
⎞
an
angle
the
of
30°
acceleration
of
the
beanbag
is
–2
ms
A
⎜
⎟
⎝
9.8 ⎠
a.
Find
the
column
b.
Find
at
it
Student
The
matrix
S
velocity
of
the
beanbag
as
a
represents
an
enlargement
vector.
the
time
after
14.
initial
B
displacement
t,
is
B
where
t
is
from
the
A
of
number
the
of
beanbag
seconds
released.
throws
her
beanbag
towards
student
A
scale
–1
at
factor
1.1,
centred
at
(0,
0),
and
the
matrix
t
=
a
reflection
in
the
line
y
=
initial
speed
of
the
beanbag
is
6ms
it
is
released
from
a
height
of
1
m.
x
.
a.
The
R
and
represents
0.
Given
that
the
beanbags
collide,
find
the
angle
Find:
of
i.
S
ii.
R
release
of
student
B’s
beanbag.
146
/
17
.
Using
suitable
a
units,
planet
from
star
by
equation
the
displacement
centred
at
(0,
0)
can
of
a
PA P E R
=
⎜
πt
⎛
3 cos
⎜
Sketch
particles
⎞
t
=
is
seconds
P
and
P
along
two
both
leave
from
a
point
A
2
different
straight
lines.
⎟
t
⎠ ⎠
2
seconds
after
leaving
A,
the
position
of
is
P
1
the
position
of
the
planet
at
3
⎞
2
⎛
⎞
times:
given
i.
in
time
question.
⎛
a.
metres
⎟ ⎟
⎝
⎝
in
1
⎟
⎜
⎜
are
⎟
⎠
2
this
Two
⎟
⎝
distances
⎞
⎞
⎜
⎜
r
πt
⎛
2 sin
and
2
Note:
in
⎛
QUE STIONS
modelled
1.
the
PRACTICE
by
r
=
⎜
⎟
1
+ t
⎜
2
⎟
⎜
⎟
⎜
⎟
⎝
2⎠
⎝
1⎠
0
ii.
t
=
1
a.
iii.
t
=
2
Five
Write
down
seconds
the
coordinates
after
leaving
A,
of
A.
is
P
at
point
B
1
i.
t
=
b.
3
Find:
!!!
"
b.
Find
an
expression
for
the
velocity
,
v,
of
AB
i.
the
!!!!
"
planet
at
time
t.
AB
ii.
.
Hence,
show
that
the
speed
of
the
planet
at
P
leaves
A
three
seconds
after
P
2
π
⎛
2
time
t
v
is
=
4 + 5 sin
2
πt
⎞
⎜
⎟
⎝
⎠
2
.
Two
seconds
1
after
it
leaves
A,
is
P
at
the
point
C,
where
2
4
⎛
⎞
!!!
"
d.
Hence,
values
write
of
t
at
down
the
which
first
the
two
speed
positive
is
AC
How
does
distance
your
of
the
answer
planet
to
part
from
the
⎜
d
relate
to
⎟
2
⎜
⎟
⎝
6⎠
maximum.
.
e.
=
Write
down
the
A graph
G
has
A
⎛
adjacency
matrix
star?
M
d.
A
B
C
D
E
F
0
1
0
0
0
1
1
0
1
0
1
0
Hence
write
down
⎜
0
1
0
1
0
0
0
⎜
0
1
0
1
0
E
0
1
0
1
0
in
⎞
Find
.
Hence
or
otherwise,
P
P
when
find
and
1
Find
⎜
1
0
t
=
the
Write
pass
.
List
all
0
and
0
of
1
0
walks
ending
how
through
at
a.
Use
⎟
to
Prim’s
find
tree
⎟
of
length
Find
the
for
algorithm,
the
length
of
the
graph
3
distance
beginning
the
at
minimum
represented
weighted
adjacency
the
in
order
all
walks
at
between
vertex
A,
spanning
by
the
table
below.
Write
down
which
the
edges
are
selected.
many
the
of
B
C
D
12
15
9
8
11
A.
of
these
walks
B
12
C
15
8
D
9
11
will
18
vertices.
length
3
starting
at
A
18
and
weighted
adjacency
table
shown
above
is
D.
probability
that
a
random
walk
table
of
beginning
at
A
will
finish
at
least
distances,
in
kilometres,
for
the
of
journeys
length
the
6
a
d.
3
⎟
The
ending
≥
G.
number
down
the
t
5
A
ii.
for
⎠
graph
starting
t,
2
A
i.
of
the
⎟
⎝
b.
terms
for
!
cos B AC
e.
1
⎜
the
P
equation
⎟
⎜
Draw
an
2
2.
a.
P
⎟
⎜
F
of
by:
⎟
⎜
D
for
⎟
⎜
B
C
given
vector
2
displacement
18.
velocity
the
between
four
hospitals
A,
B,
C
and
D.
D.
Rainer
needs
hospitals
his
and
to
deliver
uses
the
medicine
table
to
to
help
these
him
four
decide
route.
147
/
3
On
one
the
list.
occasion,
another
hospital
E
is
added
to
b.
Find:
⎛
x
⎞
1
i.
⎜
Rainer
knows
that
he
can
reach
E
only
by
first
⎟
⎝ y
⎠
1
going
to
either
A
or
B
and
the
distance
from
A
to
⎛
x
⎞
2
ii.
E
is
5
km
and
from
B
to
E
is
8
km.
⎜
⎟
⎝ y
⎠
2
b.
Copy
and
complete
the
table
of
least
distances
⎛
x
⎞
3
iii.
shown
below
so
it
includes
hospital
E
⎜
⎟
⎝ y
⎠
3
A
A
B
12
C
15
B
C
D
E
12
15
9
5
8
11
8
8
.
Plot
d.
From
these
9
11
E
5
8
18
V
erify
.
Beginning
at
algorithm
to
begin
A,
use
at
hospital
0.5,
set
of
show
coordinate
that
the
axes.
centre
of
1.5)
the
an
nearest
upper
this
point
is
invariant
under
the
transformation.
decides
he
point
(
wants
about
the
4,
2)
affine
transformation,
to
and
rotate
his
wishes
to
points
find
180°
the
A.
A,
that
will
do
this.
neighbour
.
find
a
diagram,
(
that
affine
George
to
is
on
18
D
needs
your
rotation
e.
Rainer
points
bound
for
Write
down
the
matrix
representing
a
rotation
Rainer ’s
of
180°
about
(0,
0)
journey
.
g.
d.
By
first
deleting
hospital
E,
use
the
By
considering
underA,
vertex
find
algorithm
a
lower
and
bound
your
for
answer
Rainer ’s
to
part
a
journey
.
x
⎛
n
⎜
⎝ y
Find
a
different
vertex
lower
bound
by
or
⎞
otherwise,
⎛
x
deleting
n
1
write
containing
down
the
smallest
⎜
⎠
⎝ y
n
1
order
the
least
length
of
to
to
do
his
calculations
the
route
an
expression
x
n
⎞
in
⎜
⎟
⎝ y
⎠
and
y
0
take.
V
erify
that
for
x
2
⎞
⎛
x
0
⎝ y
⎟
⎜
⎠
⎝ y
⎟
2
working
on
an
animation
that
⎞
=
A,
3
⎠
0
requires
⎛
rotate
points
about
a
centre
other
than
i.
Explain
geometrically
x
n
⎞
⎛
x
⎜
affine
has
been
told
transformations,
that
so
this
he
is
is
possible
whenever
using
n
is
⎜
⎠
⎝ y
these
can
be
⎟
⎠
0
even.
investigating
⎛
how
⎞
,
⎟
n
He
0
=
why
⎝ y
0).
terms
n
x
⎜
(0,
form
quickly
,
for
Rainer
⎛
is
more
⎛
h.
to
the
⎠
needs
0
needs
him
in
⎟
interval
of
George
A
+ b
⎟
George
1.
find
B
Hence,
PA P E R
point
⎞
M
n
In
.
invariant
to
=
e.
the
deleted
j.
constructed.
Find
an
expression
x
n
⎞
for
in
⎜
⎟
⎝ y
⎠
terms
of
x
0
n
and
y
,
when
n
is
odd.
0
a.
Describe
fully
the
⎛
by
the
transformation
0
1
⎞
1
0 ⎠
represented
k.
Find
an
⎜
⎝
⎟
for
the
point
An
⎛
x
n
affine
⎞
⎛
transformation,
0
1
x
⎞ ⎛
n
1
⎞
=
⎟
⎜
⎝ y
⎠
⎝
1
0
,
in
terms
of
θ,
a
and
b,
y
)
for
⎟
⎜
⎠
⎝ 2⎠
be
(4,
is
defined
θ
affine
transformation
anticlockwise
about
that
the
rotates
point
(a,
a
b).
as
⎞
⎟ ⎜
n
(x
1
0 ⎠ ⎝ y
n
Let
⎛
T,
+
⎜
expression
matrix
n
≥
1
⎟
1
0)
0
148
/
STAT I ST I C S
4
4 . 1
PROBABILIT Y
P R O B A B I L I T Y
Yu shud knw:
✔
the
meaning
of
the
likely
outcomes,
space
(U)
and
Yu shud e e :
terms
relative
trial,
outcome,
frequency
,
equally
✔
nd
the
✔
apply
that
the
probability
meaning
exclusive
of
the
sample
space
is
exactly
that
of
and
complementary
,
independent
probabilities
predictions,
mutually
the
formula
for
compound
enable
us
example
construct
V
enn,
or
of
a
table
to
the
choose
make
expected
events
to
sample
space
or
tree
diagrams,
outcomes
an
appropriate
represent
the
sample
recognize
when
diagram
with
which
to
space
number
events
can
be
represented
the
formula
for
a
conditional
that
sample
space
sampling
affects
the
tree
diagrams
–
the
quantifying
or
without
can
be
applied
replacement
recognize
with
problems
by
a
or
without
replacement
understanding
the
context
of
problem.
probability
.
mathematics
sciences
with
or
event
by
✔
V
enn,
In
complementary
occurrences
that
✔
for
probabilities
probability
✔
event
events
✔
of
an
1
✔
✔
of
event
✔
✔
probability
sample
nd
✔
AND
–
as
well
following
as
throughout
language
is
the
used
natural
when
and
the
human
representing
and
probabilities:
Experiment: a process by which you obtain an observation.
Trials: repeating an experiment a number of times.
Outcome: what may occur in a trial.
Event: an outcome or set of outcomes.
Sample space: the set of all possible outcomes of an experiment.
Probability
enters
qualifies
for
to
the
watch
relatives,
their
the
the
teams
played
your
seat
from
a
in
of
a
wonder
based
been
lives
are
on
playing
each
other,
your
charity
tickets
to
fund-raising
sold,
550
will
debating
who
will
team
watch
win
a
approach
are
knowledge,
you
stall.
you
Imagine
they
recently
,
media,
As
ways.
likely
prior
social
stadium
many
how
all
On
the
in
competition.
journalists
have
have
daily
you
opinions
probabilities
20 000
final
final,
sport
own
our
forming
read
has
the
You
that
won
game,
read
8
their
in
the
times.
the
the
win;
win.
emotions
you
on
to
your
team
stadium
friends,
They
and
all
have
how
own subjective
11
times
Before
buy
a
ticket
the
you
lottery
that
of
teams
find
ticket
the
prize.
149
/
4
Probability
imaginary
formulae
scenario
as
Relative
frequency
is
with
found
the
can
be
applied
the
information
as
experimental
in
this
follows:
(also
referred
to
probability)
formula:
Frequency
Relative
to
frequency
of
A
of
occurence
of
event
A
in n trials
=
n
Hence
the
relative
frequency
of
the
event
“your
team
wins”
after
8
11
trials
is
≈
0.727.
11
Having
found
your
team
held
many
because
this
has
a
good
years
the
Probability
relative
ago,
trials
(also
frequency
,
chance
to
critical
win.
you
But
reflection
should
all
take
referred
to
as
place
may
if
some
should
under
theoretical
reason
of
that
the
apply
11
to
identical
today
games
this
were
evidence
conditions.
probability)
is
found
with
theformula:
n( A )
Probability
of
A
=
P(A)
=
,
where
n(A)
is
the
number
of
outcomes
n(U )
that
make
A
happen
and
n(U)
is
the
number
of
outcomes
in
the
samplespace.
Hence
the
probability
of
the
event
n( A)
P(you
win
a
lottery
=
=
n(U )
This
are
probability
unlikely
winning
equally
as
to
even
you
are
carry
For
frequency
Increasing
is
you
not
three
you
in
to
an
the
number
of
prize”
is:
0.0275
a
hundred
have
ticket,
a
scale
since
and,
the
the
the
0
although
same
20000
you
chance
tickets
(impossible)
likelihoods
approximation
the
set
many
due
trials
exactly
from
theoretical
from
0.5,
lottery
of
are
all
lottery
.
on
experiment
exactly
a
the
as
in
do
compare
serves
a
20 000
buying
random
this
the
else
example,
at
out
prize,
represented
frequency
number
under
outcomes
enabling
probability
.
a
anyone
Probabilities
Relative
just
win
likely
(certain),
is
win
550
=
prize)
“you
to
{1,
2,
3,
results
10}
you
is
a
of
may
more
events.
0.5.
randomness
in
1
theoretical
probability
times,
the
to
of
to
choosing
However,
find
of
an
the
the
accurate
if
relative
experiment.
estimate.
Exmpe 4.1.1
Erin
rolls
a
biased
experiment
(a)
(b)
of
Complete
Erin
to
asks
verify
die
1115
her
repeating
oume
Fequeny
ree
ree
fequeny
fequeny
(ex)
 3 sf
table.
classmates
results
the
an
trials.
Erinʼs
two
in
1
223
2
236
236
0.212
1115
by
3
experiment:
190
190
0.170
1115
3
was
the
211
of
out
by
outcome
1200
trials
Maxine
in
4
163
5
140
0.146
carried
140
whereas
1115
Louis
reported
frequency
1000
trials.
estimates
rolling
These
photos
show
the
each
of
a
side
biased
has
a
die.
a
3
18.1%
List
for
in
relative
the
the
6
163
Totals:
1115 trials
in
three
probability
ascending
of
order
of
likelihood.
different
(c)
sides
of
a
Notice
different
Explain
how
Erin
a
comment
could
best
estimate
the
probability
of
rolling
how
3,
and
on
the
limitations
of
this
estimate.
shape.
150
/
4 .1
P r ob a bili t Y
Solution
(a)
oume
Fequeny
1
ree
ree
fequeny
fequeny
(ex)
 3 sf
223
223
0.200
1115
2
236
236
0.212
1115
3
190
190
0.170
Collaborating on probability
1115
4
163
163
experiments or creating probability
0.146
1115
5
simulations with technology
140
140
0.126
is a good way to increase your
1115
knowledge and understanding
6
163
163
0.146
about probability, and can lead you
1115
to find interesting topics to explore
Totals:
1115 trials
1
1.00
in your internal assessment.
(b)
The
estimates
of
the
probability
of
rolling
a
3
in
ascending
orderare:
assessmen p
Erin:
0.170
211
Maxine:
≈
0.176
=
0.181
Unless otherwise stated in the
1200
question, answers should be given
Louis:
18.1%
exactly or to three significant
(c)
Erin
190
can
+
put
211
all
the
+ 181
trials
+
1200
to
form
the
relative
frequency:
194
=
1115
together
+ 100
figures. Decimal representations
of a probability enable you to
≈
0.176
1105
easily judge its size, however a
This
would
rolling
An
event
symbols.
and
is
A
4”
to
be
the
because
be
A’.
event.
is
be
the
more
best
The
aware
For
same
of
table
that
by
A
is
the
as
“X
of
been
probability
carried
that
some
written
A
outcomes
or
is
or
used in probability calculations.
does
algebraic
not
examples.
are
It
can
discrete,
happen
is
describe
“X
is
less
less”
Written statement
Algebraic equivalent
Written
Algebraic equivalent
of event A
of A
statement of A'
of A'
X is prime
X is odd
X ∈
{2, 3, 5, 7, 11, ….}
X ∈
{1, 3, 5, 7, 9, ….}
fraction is exact and should be
of
out.
statements
the
3
the
statements
event
gives
different
if
have
written
below
example,
event
estimate
trials
represented
complement
denoted
same
than
3
can
The
important
the
a
likely
X is not prime
X ∉
{2, 3, 5, 7, 11, ….}
X ∉
{1, 3, 5, 7, 9, ….}
assessmen p
X is even
or X ∈
{2, 4, 6, 8, ….}
In the five-minute reading time
X is less than 4
X is at least 4
X < 4
X ≥ 4
given before you star t writing in
X is no more than 4
X is more than 4
X ≤ 4
X > 4
the exam, take time to read the
wording of events carefully to
Since
A
and
A’
cover
all
the
possibilities
in
the
sample
space,
it
is
a
rule
make sure you are clear exactly
that
P(A)
This
rule
+
P(A’)
=
1
what the event is asking for.
can
be
applied
to
solve
problems
efficiently
.
151
/
4
Exmpe 4.1.2
Callum
die
5,
numbers
with
6.
He
the
Find
the
as
faces
numbers
rolls
obtained
the
the
die
1,
2,
and
on
3,
a
5,
dodecahedral
8,
records
13,
his
4,
8,
16,
4,
number
5
C
8
probabilities:
8
(a)
P(C
(b)
P(2
(c)
P(C
is
<
prime)
C
is
≤
at
10)
least
4)
(d)
P(C
is
less
(e)
P(C
is
not
(f)
P(C
is
4
than
3
1
4)
prime)
or
more)
Solution
5
(a)
P(C
is
prime)
2,
=
3,
5
and
13
are
prime
and
12
5appears
8
(b)
P(2
<
C
≤
2
=
10)
twice.
=
First,
12
carefully
list
numbers
3
on
the
die
in
order:
1,2,3,4,4,5,5,6,8,8,13,16.
write
9
(c)
P(C
is
at
least
=
12
Your
5
P(C
is
not
prime)
list
helps
you
see
there
are
numbers
least
4.
This
is
on
the
die
that
are
at
7
1
is
answer.
4
9
(d)
the
3
=
4)
down
Then
=
12
the
complement
of
the
12
event
in
part
(b)
3
(e)
P(C
is
4
or
more)
=
P
(C
is
4
or
more)
is
the
same
4
event
assessmen p
as
P(C
is
at
least
4)
The expected number of
occurrences is similar to the mean
value of a data set. Hence it does
not need to be rounded to the
Use
is
of
one
about
the
of
formula
the
many
random
for
the
ways
expected
that
number
probability
of
occurrences
allows
us
to
make
of
an
event
predictions
events.
nearest or lowest integer value.
Expected
number
of
occurrences
of
A
in
n
trials
=
n
×
P(A)
This contrasts with using functions
to make a prediction: if N(5) = 14.7
predicts the number of elephants
Exmpe 4.1.3
in a game reserve 5 years after
opening, you would round this up
to 15 elephants.
Hannah
and
octahedral
it
95
times
Kotryna
die
with
and
are
exploring
sides
dice
numbered
counts
the
biased
six-sided
1,
occurrences
games.
2,
of
3,
4,
the
5,
Hannah
6,
7
event
and
“roll
has
8.
a
a
fair
She
rolls
factor
of360”.
Kotryna
rolls
a
outcomes
and
probabilities
following
table:
for
die
115
times.
Kotryna’s
die
The
are
possible
given
in
the
oume
3
6
8
11
15
21
Py
0.13
0.15
0.12
0.15
0.25
0.2
Kotryna
Predict
counts
which
number
of
the
of
occurrences
Hannah
occurrences,
or
of
the
Kotryna
justifying
event
should
your
“roll
a
expect
multiple
the
of
3”
greatest
answer.
152
/
4 .1
P r ob a bili t Y
Solution
The
probability
of
Find
Hannah’s
the
probabilities
of
each
7
event
since
is
all
of
event
the
and
multiply
by
the
number
8
of
numbers
of
360
on
her
except
expected
die
for
7.
number
are
Hence
of
trials
in
each
game.
factors
the
occurrences
7
for
Hannah
95
is
×
=
83.125
8
The
probability
event
0.2
=
is
0.13
+
of
Kotryna’s
0.15
+
0.25
The formula for expected
+
number of occurrences does
0.73
not appear in the formula book .
However, an equivalent form
Hence
the
expected
The
number
expected
numbers
of
appears in the formula book
of
occurrences
for
Kotryna
occurrences
is
are
very
close.
since
they
predict
in section 4.8; the mean of the
115
×
0.73
=
However,
83.95
binomial distribution. Therefore, you
the
Kotryna
should
expect
greatest
number
likely
average
of
the
can make a link between these two
occurrences,
of
number
the
conclusion
occurrences
par ts of the course and increase
isvalid.
since
83.95
>
83.125
your knowledge and understanding.
Py fmue nd dgms
Events
can
contexts.
tables
You
of
use
Once
a
them
raise
as
to
solve
has
overview
The
games
4,
6
and
serve
for
each
a
fair
in
for
that
probabilities
sample
as
space
useful
problems
been
shown
money
so
diagrams,
outcomes
context
Context
5,
combined,
V
enn
diagram,
to
be
following
charity
,
involve
three
rolling
octahedral
die
diagrams,
involving
Examples
you
dice
tree
of
choose
an
in
wider
diagrams
the
combined
4.1.5
and
games
fair
“B”
applied
and
sample
space.
events.
appropriate
examples.
4.1.4,
a
be
representations
understood,
the
can
X,
cubical
4.1.6: In
a
CAS
Y
and
Z
die
“A”
numbered
4,
6,
numbered
3,
5,
are
project
7,
designed.
8,
9,
1,
2,
3,
10
Exmpe 4.1.4
In
game
X,
the
10.
Then
the
number
Find
the
dice
player
A
and
chosen
randomly
B
are
shown
appears
probability
of
chooses
on
to
the
exactly
winning
game
a
number
player.
one
of
between
The
the
player
1
and
wins
if
dice.
X
Solution
The
U
events
relate
to
B
1
7
3
2
4
membership
8
of
a
set.
10
5
6
9
A V
enn
diagram
is
an
A
appropriate
U
The
probability
chosen
appears
that
the
only
number
on
die
A
is
From
you
is
2
by
.
The
probability
that
the
the
sample
the
can
representation.
V
enn
solve
writing
space.
diagram,
the
down
problem
the
number
10
probabilities
required.
4
chosen
appears
only
on
die
B
is
.
10
Hence
the
probability
of
winning
6
game
X
is
10
153
/
4
Exmpe 4.1.5
In
game
total
of
Y,
dice
the
A
and
numbers
B
on
are
the
rolled
two
and
dice.
T
is
The
defined
player
as
the
wins
if T
is
a
primenumber.
Find
the
probability
of
winning
game
Y
Solution
Two
1
2
3
4
5
6
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
7
8
9
10
11
12
13
8
9
10
11
12
13
14
9
10
11
12
13
14
15
10
11
12
13
14
15
16
random
combined
together
of
T,
to
an
adding
give
the
are
them
values
represent
the
space.
A sample
is
by
which
sample
quantities
space
diagram
appropriate
representation.
There
is
a
are
16
prime
outcomes
number
in
for
a
which
sample
T
space
Shading
of
48
equally
likely
prime
you
Hence
the
probability
16
game
Y
or
circling
the
outcomes.
of
numbers
count
the
helps
outcomes
winning
accurately
.
1
=
is
48
3
Exmpe 4.1.6
In
if
game
Z,
exactly
die
one
probability
of
A
of
is
rolled
the
two
winning
and
then
numbers
game
die
B
rolled
is
is
rolled.
a
The
factor
of
player
6.
Find
wins
the
Z
Solution
2
8
assessmen p
B
A
is
is
a
factor
of
6
One
event
occurs
after
a
4
another.
A tree
diagram
factor
Choosing to construct and apply
6
of
6
6
B
not
a
factor
of
6
is
an
appropriate
a probability diagram is often
8
representation.
the most efficient way to solve
2
a problem or find a probability.
A
8
B
You are encouraged to consider
applying a diagram to find a
2
not
6
factor
of
a
factor
of
6
Draw
the
events
in
order
6
B
probability before trying to apply
is
a
not
a
factor
of
6
from
left
to
right.
6
8
a formula.
154
/
4 .1
The
Z
probability
of
winning
Game
game
Z
is
won
if
the
number
rolled
on
die
A
is
a
P r ob a bili t Y
factor
of
is
6
4
and
the
number
rolled
on
die
B
is
not,
with
6
×
probability
6
4
6
×
6
2
2
+
×
8
6
8
7
=
8
or
if
the
number
rolled
on
die
A
is
not
a
factor
of
6
and
the
12
2
number
rolled
on
die
B
is
a
factor
of
6,
with
probability
2
×
6
The
Formulae
give
probabilities.
in
order
to
another
You
apply
way
must
them
to
represent
understand
the
final
combined
symbols
answer
events
used
in
is
found
and
the
by
adding
these
8
probabilities.
find
formulae
correctly
.
The event A ∪ B means A or B occurs. This use of “or ” includes the possibility
that h A and B occur. You say “A or B” or “A union B” for A ∪ B. You say “A and
B” or “A intersection B” for A ∩ B
The event A | B means A occurs given that B has occurred. You say, “A given B”
or “given that B has happened, A happens”.
These
symbols
are
applied
in
Example
4.1.7.
Exmpe 4.1.7
In
Jan’s
class
of
U
16
(a)
Find
the
following
probabilities:
B
students,
the
numbers
3
2
7
of
students
who
(i)
P(A)
(iv)
P(A
(ii)
P(B)
(v)
P(A
(iii) P(A
∩
B)
study
4
A
Art
are
or
Biology
,
or
represented
both,
in
B)
Hence
(i)
from
Jan’s
class
is
chosen
at
verify
that:
P(A
∪
B)
=
P(A)
P( A ∩
the
is
the
event
event
“the
“the
student
student
studies
studies
+
P(B)
P(A
∩
B)
random.
(ii)
is
B)
diagram.
A student
A
|
the
(b)
V
enn
∪
Art”
and
B)
P( A|B) =
B
P( B)
Biology”.
Solution
(a)
3
(i)
2
5
=
3
+
2
2
(ii)
+
P( A) =
+
+
7
+
There
4
is
a
total
of
5
Art
students
in
the
class
of
16.
16
7
9
P(B) =
=
3
+
2
+
7
+
4
16
2
(iii)
P( A ∩ B) =
2
students
study
both
Art
and
Biology
.
16
3
(iv)
+
2
7
12
=
3
+
2
+
2
(v)
+
P( A ∪ B) =
7
=
+
4
12
study
Art
or
Biology
.
This
includes
the
2
students
who
16
study
both.
Given
that
2
P( A|B) =
2
+
7
a
Biology
student
is
chosen,
the
sample
space
is
9,
9
of
whom
2
study
Art.
155
/
4
(b)
(i)
P(A)
5
+
P(B)
9
P(A∩B)
2
12
+
16
=
16
formula
Biology
=
16
The
=
so
subtracts
they
are
the
not
students
counted
who
twice
in
study
the
both
Art
and
total.
P( A ∪ B)
16
2
P( A∩ B)
2
16
(ii)
=
16
=
×
2
16
=
cancels
out.
The
sample
space
is
9,
of
whom
2
study Art.
9
P( B)
16
9
9
16
The
following
The
formula
A
B
or
will
formulae
P(A
∪
B)
=
are
true
P(A)
+
in
P(B)
general:
−
P(A
∩
B)
finds
the
probability
that
o c c u r.
P( A ∩ B)
The formula P( A | B) =
finds the probability that A occurs given than B
P(B)
has occurred. Another way to write the formula is P(A|B)P(B) = P(A ∩
B).
This is known as conditional probability.
How
A
two
and
time.
B
events
are
This
relate
mutually
means
of
other.
means
This
Sometimes
mean
the
each
same
P(A ∩
event
that
students
each
other
exclusive
that
occurrence
to
does
P(A |
think
thing.
B)
B)
the
if
=
=
they
0.
not
can
cannot
T
wo
affect
P(A),
easily
4.1.8
confused.
both
occur
events
are
in
way
and
any
at
the
the
P(A)P(B)
the
=
events
same
if
occurrence
exclusive and
illustrates
Two
independent
therefore
terms mutually
Example
be
the
of
P(A ∩
the
B).
independent
difference.
Exmpe 4.1.8
A survey
of
24
represented
numbers
Biology
of
students
by
A,
B
and
students
is
6.
A student
is
is
selected
C
made
find
out
how
respectively
.
The
results
in
each
at
random
to
set.
For
example,
from
the
many
the
are
students
study
presented
number
of
in
a
Art,
V
enn
students
who
class.
Biology
or
diagram
study
Chemistry
,
showing
both
Art
U
the
and
B
C
A
Determine
which
pairs
of
events
are
(i)
mutually
exclusive
(ii)
independent,
3
justifying
(a)
A
your
and
6
7
3
2
answers.
B
(b)
B
and
C
(c)
A
and
3
C
Solution
6
P( A ∩
B)
3
24
(a)
(i)
P( A|B) =
=
=
This
can
also
be
found
quickly
without
the
16
P( B)
8
6
24
formula
3
+
6
P( A) =
Since
from
6
+
7
+
set
B
in
the
3
=
P(A|B)
V
enn
8
=
P(A),
A
and
B
are
n(A
mutually
∩
B)
diagram.
Equivalently
,
independent.
(ii)
simplifying
3
24
Since
by
independence
demonstrated
=
6
≠
0,
A
and
B
are
by
showing
of
A
and
P(A)P(B)
=
B
can
P(A
be
∩
B).
not
exclusive.
156
/
4 .1
P r ob a bili t Y
3
P( B ∩ C )
3
24
(b)
(i)
P(B|C) =
=
=
Write
reasons
clearly
,
showing
how
you
5
P(C )
5
24
6
P(B)
+
7
+
3
24
P(B
|
probabilities
demonstrate
=
=
Since
replace
in
the
formulae
in
order
to
2
your
knowledge
and
understanding.
3
C)
≠
P(B),
B
and
C
are
not
independent.
(ii)
Since
P(B
∩
mutually
C)
=
3
≠
0,
A
and
B
are
not
exclusive.
0
(c)
P( A|C) =
(i)
=
0
Reflect
on
what
this
shows:
Since
A
and
C
are
5
24
3
+
mutually
6
P( A) =
3
if
one
they
occurs,
cannot
the
other
be
independent
cannot.
=
24
Since
because
exclusive,
8
P(A|C)
≠
P(A),
A
and
C
are
not
independent.
(ii)
Since
P(A
∩
C)
=
0,
A
and
C
are
mutually
exclusive.
Note
that
in
a
different
experiment
in
which
two
students
are
chosen,
2
9
9
the
probability
that
both
are
Art
students
is
(
=
)
if
the
9
student
can
be
chosen
twice
(with
replacement),
8
or
×
24
same
student
cannot
be
chosen
twice
(without
same
64
24
3
if
=
23
the
23
replacement).
You must note whether a sampling context involves replacement or not before
calculation of probabilities.
S aMPlE StUDENt aNS WEr
Events
A
P(A’
B)
(a)
∩
Find
and
=
B
are
independent
with
P(A
∩
B)
=
0.2
and
0.6
P(B)
▲ The
V
enn
(b)
Find
P(A
∪
diagram
recognizes
is
that
a
appropriate.
B)
(a)
P(B)
=
0.6
▼ The student fails to nd the
total
0.2
student
probability
for
B,
which
is
0.8
0.6
0.2
1
(b)
Applying P(A ∩ B) = P(A)
× P(B) gives P(A) =
▲ The
student
has
correct
formula
for
events,
gaining
a
applied
the
independent
follow
throughmark.
3
14
Applying P(A ∪ B) = P(A) + P(B) gives P(A ∪ B)
=
15
▼ The
formula
exclusive
which
is
answer
for
events
was
incorrect.
for
(b)
is
mutually
then
The
applied,
correct
0.85
157
/
4
S aMPlE StUDENt aNS WEr
In
3
a
class
own
of
21
students,
12
own
a
laptop,
10
own
a
tablet
and
neither.
The
following
and
“own
The
values
a
V
enn
diagram
shows
the
events
“own
a
laptop”
tablet”.
p,
q,
r
and
s
represent
Laptop
numbers
of
students.
Tablet
q
p
(a)
(i)
Write
(ii)
assessmen p
Find
(iii)
(b)
down
the
Write
the
value
down
A student
is
value
of
the
of
q
value
selected
p
at
of
r
and
random
of
s
from
the
class.
The command term “ Write down”
requires you to show little or no
(i)
working, whereas “Find” requires
Write
down
the
probability
that
this
student
owns
a
laptop.
you to show your working clearly
(ii)
Find
the
probability
that
this
student
owns
a
laptop
in order to access full marks.
or
(c)
Two
Let
(i)
tablet
but
not
students
are
randomly
L
a
be
the
Copy
F irst
event
and
student
a
both.
“student
complete
Second
11
selected
the
owns
a
following
from
the
class.
laptop”.
tree
diagram.
student
L
20
L
9
L'
20
L
9
L'
21
L'
(ii)
Write
down
student
owns
(a)
(i)
p
(ii)
▲ The
calculations
precise
and
are
=
a
the
owns
a
probability
laptop
that
given
the
that
second
the
first
laptop.
3
12
+
21
–
10
=
22
clear,
complete.
3
=
T
herefore
18,
q
=
so
4
must
be
counted
twice.
4
158
/
4 .1
(iii)
r
=
s
12
−
4
=
10
−
12
(b)
8
4
=
6
4
=
(i)
=
21
(c)
=
P r ob a bili t Y
0.571
7
▲ The
11
(i)
student
correctly
realizes
L
20
that
the
context
replacement”
12
is
and
“without
completes
the
L
21
tree
9
diagram
correctly
.
L'
20
12
9
L
20
L'
21
8
L'
20
12
(ii)
P(rst
student
owning
laptop)
=
=
P(x)
21
nd
P(2
student
P(
∴
P(y|X)
owning
y∩ x )
a
laptop)= P(y)
▼ Although
the
=
correct
the
student
answer,
found
considerable
P( x )
time
has
formula
P( first
AND
P( Only first
11
when
used
the
applying
answer
the
could
sec
ond owning)
=
12
been
have
been
found
from
the
what
should
immediately
owning)
tree
diagram.
be
This
realized
is
from
the
×
21
=
command
20
term
“Write
down”
12
21
×
11
×
20
×
11
=
20
159
/
4
4 . 2
P R O B A B I L I T Y
D I S T R I B U T I O N S
Yu shud nw:
✔
the
meaning
✔
probability
random
of
a
Yu shud e e :
discrete
distributions
random
are
variable
✔
generalizations
interpret
processes
probability
✔
a
function
distributions
or
a
can
be
the
sum
of
all
represented
random
apply
representation
✔
distribution
is
1
and
each
of
is
a
at
a
to
the
expected
predicts
its
value
average
✔
the
denition
of
✔
the
context
a
xed
✔
✔
of
the
binomial
the
normal
the
a
a
least
fair
of
after
zero
apply
✔
success
probability
those
determine
many
trials
continuous
of
can
properties
of
the
from
is
the
random
probability
of
an
event
fair
to
problems
games
context
if
the
binomial
appropriate
trials
be
and
a
modelled
binomial
✔
calculate
normal
✔
interpret,
by
notation
many
naturally
✔
sketch
distribution
and
write
of
the
probabilities
probabilities
with,
and
binomial
with
with
technology
technology
understand
distribution
the
and
the
distribution
represent
sets
normal
the
a
involving
calculate
normal
models
data
of
distributions
✔
distribution
distribution
discrete
of
variable
game
number
notation
probability
random
value
nite
probability
occurring
of
the
variables
calculate
distribution
✔
with
by
table
probabilities
events
discrete
including
✔
write
of
variable
✔
and
of
normal
the
distribution
normal
events
using
a
curve.
its
curve.
Just
as
patterns
sequences
can
be
find
For
numbers
functions,
generalized.
probabilities
example,
spinner
with
can
you
These
be
can
generalized,
find
patterns
generalizations
then
in
give
expressed
probability
you
as
events
convenient
that
ways
to
efficiently
.
consider
sides
an
experiment
numbered
2,
3
in
and
4
which
is
a
spun,
fair
three-sided
along
with
a
fair
2
2–
2
–
4
1
1
3
or
in
five-sided
spinner
with
sides
numbered
2,
1,
0,
1
and
2.
outcome
is
0
The
numbers
sample
space
2
T
are
added
to
give
1
0
1
2
0
1
2
3
4
3
1
2
3
4
5
4
2
3
4
5
6
a
discrete
values
The
and,
random
as
the
probabilities
t
total
T.
Each
shown
in
this
diagram.
2
is
a
variable
spinners
of
0
each
1
are
value
because
fair,
the
of
can
2
T
3
it
takes
value
be
4
of
only
T
discrete
varies
represented
5
integer
randomly
.
in
a
table:
6
1
2
3
3
3
2
1
15
15
15
15
15
15
15
P(T = t)
It
can
The
be
seen
table
possible
from
shows
the
table
exactly
that
how
this
the
probabilities
total
probability
all
is
add
to
a
total
distributed
to
of
1.
each
outcome.
160
/
4.2
The
table
can
be
generalized
=
where
and
represented
as
a
P r ob a bili t Y
DiStribUtioNS
function
Ne
f (t)
=
P(T
t),
⎧
You would not be expected to
t + 1
0
≤ t
≤
2
⎪
derive this function f (t) without
15
⎪
extra information being given. It
⎪ 1
f (t) =
t
⎨
=
t
3
∈!
is inser ted here to give you an
5
⎪
example of how a probability
⎪7
t
4
≤ t
≤ 6
distribution function can be found
⎪
15
⎩
from a probability distribution
like the one in the table on the
previous page.
This py dsun funn assigns to
The umu e dsun funn assigns to each
each value of the discrete random variable T the
value of the discrete random variable T the corresponding
corresponding probability of its occurrence and is often
probability of T being less than or equal to that value. It is
referred to as a “pdf ”.
often referred to as the “cdf ”.
Exmpe 4.2.1
Two
branches
their
of
a
customers’
café
chain
purchasing
are
investigating
Bytecafé
habits.
in
one
models
visit
probability
Alphacafé
finds
in
a
survey
that
the
number
with
purchased
by
a
customer
in
one
visit
a
number
discrete
distribution
of
items
random
function
purchased
variable
P(B
=
b)
=
B
with
f (b),
of
7
items
the
where
to
b
f (b) =
,
1
≤
b
≤
5,
b
∈

20
the
café
can
be
modelled
by
a
discrete
random
(a)
variable
A
with
the
following
Determine
selected
a
P(A = a)
in
which
café
that
a
randomly
distribution:
1
2
3
4
5
0.63
0.18
0.09
0.07
0.03
(b)
customer
is
more
least
three
purchases
Find
P(3
B
≤
<
5|B
>
in
1)
likely
one
and
to
make
at
visit.
interpret
the
result.
Solution
(a)
P(A
P(B
≥
≥
3)
3)
=
0.09
=
0.19
=
f (3)
A randomly
make
f (4)
3
+
20
to
0.07
at
+
Interpret
0.03
“at
expresses
+
4
=
+
+
2
=
20
selected
least
the
the
three”
and
probability
table
and
then
write
the
event
that
required.
the
function
to
find
the
probabilities.
9
+
20
Apply
f (5)
least
=
0.45
20
customer
three
is
purchases
more
in
likely
one
visit
at
Bytecafé.
P( 3
(b)
P(3
≤
f ( 3) +
B <
<
5)
P( B
>
1)
Apply
the
formula
for
conditional
probability
.
of
in
0.5
f (1)
Given
than
B
f ( 4)
=
1
≤
5|B > 1) =
that
one
items
are
a
customer
item,
it
is
just
purchased
in
Bytecafé
as
than
likely
buys
that
3
more
or
Interpret
the
meaning
the
event
context.
4
not.
161
/
4
Each
value
of
proportional
For
example,
customers
at
occurrences
a
to
a
random
its
variable
prediction
each
a
for
is
the
found
number
by
with
a
frequency
of
finding
items
the
purchased
expected
by
100
number
of
event:
1
Expected
occur
probability
.
Alphacafé
of
will
2
0.63 × 100 = 63
0.18 × 100 = 18
3
4
5
0.09 × 100 = 9
0.07 × 100 = 7
0.03 × 10 = 3
frequency
of of a items
purchased
Hence
63
to
2
buy
customers
items,
Finding
the
and
mean
are
so
of
63
×
l
expected
buy
only
one
18
frequency
×
2
+
9
×
3
+
table
7
×
4
+
gives
3
×
a
5
mean
63
It
process
predicts
purchase.
gives
the
the
likely
This
18
are
expected
+
18
+
9
expected
average
process
is
+
7
+
=
3
value
generalized
1.69
100
of
number
of
169
=
This
item,
on.
this
+
to
the
of
discrete
items
by
the
one
random
variable A.
customer
following
The expected value of a discrete random variable X is E(X) =
will
formula:
∑xP(X = x).
This is a measure of the central tendency (the mean) of the outcomes
of X in a number of trials.
This
formula
Alphacafé
when
confirms
the
expected
value
1
0.63
0.18
×
+
applied
2
×
this
of
+
3
the
answer
the
×
to
and
random
0.09
+
4
probability
×
gives
an
distribution
efficient
table
way
to
of
find
variable:
0.07
+
5
×
0.03
=
1.69
assessmen p
You can use your GDC to
1.1
efficiently manage probability
7–b
Done
f(b):=
distributions by using the GDC
20
memory and sigma notation as
shown.
f(3)+f(4)
1
1–f(1)
2
Here, the answer to
Example 4.2.1 (b) is found, as
5
well as the expected value of B
5
(b
f(b))
∑
2
b=1
The
in
concept
games
expected
of
of
the
expected
chance
gain
of
a
in
which
player
is
value
a
of
player
zero,
the
a
random
can
make
game
is
variable
a
gain
called
has
or
a
applications
loss.
If
the
fair.
Exmpe 4.2.2
Some
$k
to
of
$Z
students
play
.
The
design
students
according
z
P(Z = z)
a
to
the
game
of
chance
program
a
probability
for
a
CAS
spreadsheet
distribution
1
2
4
8
16
0.4
0.3
0.2
a
b
to
project
in
generate
shown
in
order
a
the
to
random
table
raise
money
number Z
for
and
charity
.
award
a
Players
cash
pay
prize
below.
162
/
4.2
(a)
Determine
should
(b)
Hence
The
charge
Find
the
the
price
estimate
$5
the
values
of
a
and
of
b
if
the
students
decide
that
the
probabilities
of
DiStribUtioNS
winning
$8
and
$16
equal.
find
students
they
(c)
be
the
P r ob a bili t Y
to
that
play
money
the
the
that
students
they
can
should
charge
persuade
350
to
play
people
in
the
game
their
for
school
the
game
to
community
be
to
fair.
play
the
game
if
game.
the
students
would
expect
to
raise
for
charity
by
charging
350
people
$5
to
play
game.
Solution
(a)
0.4
⇒
(b)
+
0.3
2a
The
+
+
0.2
0.9
=
+
1
expected
a
⇒
+
b
a
=
=
The
1
0.05
winnings
and
are
b
=
probabilities
×
0.4
=
$3
+
2
×
0.3
+
4
×
0.2
+
found
8
×
If
by
the
cost
0.05
+
16
×
ticket
should
cost
$3
for
to1.
to
the
game
play
then
is
the
the
same
expected
as
the
gain
is
expected
zero
and
0.05
thegame
The
add
0.05
winnings,
1
must
is
fair.
to
befair.
(c)
The
students
raise
350
×
5
would
350
×
expect
3
=
that
$700
they
for
would
charity
.
the nm dsun
Different
different
One
number
random
example
of
patterns
generalize
processes
a
discrete
are
to
different
modelled
probability
by
sequences.
different
distribution
is
Similarly
,
distributions.
the binomial
distribution
Context
of
the
binomial
is
success
distribution:
In
a
binomial
trial,
the
outcome
assessmen p
either
When
an
or
failure
experiment
and
involves
you
a
count
fixed
the
number
number
of
of
trials,
successes.
each
of
which
You are expected to be able to
has
the
same
probability
of
success,
and
the
trials
are
independent
of
read a context and determine
each
other,
the
binomial
distribution
enables
you
to
find
probabilities.
if the binomial distribution is
The
discrete
random
variable
is
the
number
of
successes.
appropriate. You could do this
For
example,
times
a
of
each
in
5
4
is
you
roll
rolled.
other.
The
a
The
cubical
number
random
die
of
5
times
trials
variable
is
is
the
and
5
count
and
they
number
of
the
are
4s
number
in the five-minute reading time
of
independent
before the examination begins.
rolled
1
trials,
and
the
probability
of
success
is
because
this
is
the
6
assessmen p
1
probability
of
rolling
a
4.
Hence,
5
are
and
the
parameters
of
this
6
binomialdistribution.
Define your random variable and
write its distribution to show your
1
Y
ou
write
“Let
F
be
the
number
of
4s
rolled
in
5
trials.
Then
F
~ B
(
5,
6
)
”
knowledge and understanding.
This is par t of your method and
The
latter
sentence
is
short
for
“F
is
distributed
binomially
with
you will be awarded method marks
1
parameters
5
.”
and
for correct writing.
6
163
/
4
Exmpe 4.2.3
For
these
reasons
three
and
contexts,
giving
a
determine
full
if
description
any
of
can
any
be
that
modelled
by
the
binomial
distribution,
identical
numbers
is
stating
your
do.
1
(a)
Alex
rolls
three
dice.
The
probability
of
rolling
three
.
Alex
rolls
the
three
36
diceuntil
ofrolls
(b)
(c)
has
torch.
this
first
a
box
She
are
drawn
After
5
three
labelled
having
rearranged.
11
identical
a
battery
A,
B,
from
The
C
and
at
numbers
random
D
in
pack
of
the
batteries,
random
whether
number
correctly
or
the
recorded
The
identical
times.
randomly
guessed
of
picks
experiment
Cards
rolls
then
stops.
The
random
variable A
is
the
number
made.
Barbara
her
he
a
or
cards
and
of
she
with
B
100
Carolina
not
which
it.
is
cards
experiment
is
repeated
is
A
does
is
7
does
which
to
work.
not
number
guess
correct,
label
don’t
She
the
tries
was
each
of
tests
variable
pack
and
five
25.
the
is
of
the
needs
replace
it
batteries
arranged
letter
card
The
She
is
on
put
random
in
in
working
the
that
box.
back
She
for
carries
out
work.
random
the
batteries
card,
and
variable X
A card
without
the
is
order.
pack
the
is
looking.
is
randomly
number
of
cards
times.
Solution
(a)
Since
not
(b)
the
number
follow
The
the
number
probability
not
(c)
constant
trials
does
not
X
does
follow
follow
of
is
the
the
number
the
trials
are
the
cards
being
a
but
are
binomial
is
Alex
replaced
at
of
or
roll
three
may
fail
Recognizing
that
replacement
is
identical
to
an
do
so
numbers
even
experiment
important
after
is
on
100
the
first
attempts.
without
information
about
thecontext.
distribution.
7
and
each
and
is
replaced.
distribution
fixed
independent
battery
not
could
attempt
the
working
binomial
trials
fixed,
distribution.
batteries
the
not
fixed,
choosing
since
B
trials
binomial
of
of
of
since
This
because
other
due
experiment
ensure
has
independent
been
carefully
designed
to
trials.
to
re-randomized.
1
X
~ B
(
7,
4
)
The probability distribution
When X ~ B(n, p), binomial probabilities are found with technology. The domain
function of X~B(n, p) is
of the binomial probability distribution function is 0 ≤ x ≤ n, x ∈ 
⎛
⎞
n
x
p
P(X = x) =
x
⎝
n
(1
x
p)
⎠
You
are
not
required
to
know
the
formula
for
the
binomial
pdf
in
the
This formula can be explored in the
examination.
You
should
always
find
binomial
probabilities
with
contexts of Pascal’s triangle and
technology
as
shown
pack
cards
in
the
next
example.
counting methods, but is not
needed for exams.
Exmpe 4.2.4
Cards
are
randomly
recorded
number
and
the
labelled
from
the
whether
of
A,
cards
B,
pack
or
not
with
experiment
(a)
Find
P(X
=
0)
(b)
Find
P(X
≥
6)
(c)
Find
the
(d)
Find
the
C
is
or
D
and
she
each
in
Carolina
was
label
repeated
probability
that
probability
a
25.
to
the
The
guess
card
is
random
which
is
the
letter
put
back
variable X
arranged
on
the
and
is
in
random
order.
card,
without
looking.
the
the
pack
is
number
randomly
of
cards
A card
is
After
drawn
having
rearranged.
guessed
The
correctly
times.
Carolina
that
100
tries
correct,
is
7
of
guesses
Carolina
fewer
guesses
than
more
five
than
cards
one
correctly
.
card
but
no
more
than
4
cards
correctly
.
164
/
4.2
P r ob a bili t Y
DiStribUtioNS
Solution
Use
1
X
~ B
(a)
(
7,
4
P(X
=
)
was
established
in
Example
technology
three
0)
=
and
write
your
answers
correct
to
4.2.3
significant
figures.
0.133
1.1
binomPdf(7,0.25,0)
0.133483886719
binomPdf(7,0.25,6)+binomPdf(7,0.25,7)
0.001342773438
(b)
P(X
≥
6)
=
P(X
0.001 34
≥
6)
=
1
equivalent
(c)
The
event
is
P(X
≤
0.987
4)
=
P(X
≤
4)
5)
way
P(X
<
The
cumulative
on
your
up
all
=
P(X
P(X
GDC
the
≤
≤
to
5)
(since
find
4)
the
since
X
the
is
effect
probabilities
in
is
discrete)
is
an
probability
.
probability
has
X
discrete.
function
of
this
adding
event.
1.1
4
(binomPdf(7,0.25,c))
∑
c=1
0.987121582031
binomCdf(7,0.25,4)
(d)
The
P(2
event
≤
X
≤
0.987121582031
is
4)
=
0.542
assessmen p
Make sure you know how to calculate binomial probabilities with your GDC
both with the pdf and the cdf. You first write the distribution, then the event
before using your GDC to find the probability.
If X ~ B(n, p) then E(X) = np and Var (X) = np(1 − p)
the nm dsun
There
are
because
type
of
many
the
examples
data
are
continuous
Context
of
distributed
the
not
of
counted
probability
normal
of
data
but
points
variables
with
is
data
from
mean
not
discrete
example
normal
data,
points
the
are
One
the
Measured
most
further
that
measured.
distribution
distribution:
symmetrically
,
Frequencies
random
this
distribution.
which
found
of
is
near
decrease
in
the
mean.
both
directions.
For
on
example,
the
the
following
heights
page.
of
The
1000
same
males
data
aged
is
17
then
is
shown
shown
in
a
in
a
dot
plot
histogram.
165
/
4
240
ycneuqerF
180
120
60
0
155
165
175
185
195
150
160
170
180
Height
190
200
Height
The data has the symmetric “bell”
The data is conveniently represented
shape characteristic of the normal
by a histogram. The shape of the
distribution.
histogram is modelled by the normal
curve. The y-values of the curve are
Most data points lie close to the
found from the normal pdf. Many
mean, and data points far from the
naturally occurring data sets are
mean are rare. You say that the data
modelled by the normal curve in
are distributed normally or that the
this way.
data follows a normal distribution.
When
and
we
write
variable:
of
one
randomly
down
the
who
value
is
distribution
“H
variance
49”.
we
These
is
a
17-year-old
H,
select
than
mean
follows
There
a
height
shorter
are
meaning
curve.
his
select
176
the
this
could
cm
and
normal
area
of
1
an
be
average
parameters
total
is
student
example
the
and
of
height
so
on.
variance
49.
distribution
determine
between
from
of
The
continuous
a
tall
mean
equation
normal
random
student,
write H
with
population
parameters
Y
ou
the
the
a
this
~
176
of
curve
of
or
this
N(176,
49),
and
the
and
normal
the x-axis.
2
The probability density
When X ~ N(μ, σ
), normal probabilities are found with technology. The domain
2
function (pdf) of X ~ N(μ, σ
) is
of the normal probability density function is x ∈ 
2
1
1
2
f ( x) =
e
σ
⎛
⎝
x
u
σ
⎞
⎠
. This formula
2π
You
are
not
required
to
know
the
formula
for
the
normal
pdf
in
the
can be explored in the contexts
examination
nor
the
concept
of
probability
density
.
You
should
always
of graph transformations and
find
normal
You
find
probabilities
with
technology
.
integration, but is not needed
for exams.
probabilities
distribution
function
curve
to
equal
the
of
events
(normal
probability
by
cdf).
you
using
This
the
normal
finds
the
cumulative
area
under
the
normal
require.
Because normally distributed data is continuous, P(X < a) = P(X ≤ a) is always
true. This contrasts with discrete probability distributions where “less than”
and “less than or equal” always apply to different events.
Exmpe 4.2.5
Ne
Given
that
follows
the
the
height
normal
H
of
a
randomly
distribution
with
selected
mean
176
17-year-old
and
male
variance
49,
The normal distribution notation
find
the
following
probabilities
and
represent
them
on
a
sketch.
gives you the variance, but your
(a)
P(150
≤
H
≤
200)
(b)
P(H
<
(c)
P(169
≤
H
≤
183)
(d)
P(162
176)
GDC always requires you to enter
the standard deviation.
<
H
≤
190)
166
/
4.2
P r ob a bili t Y
DiStribUtioNS
Solution
(a)
1.1
Ne
normCdf(150,220,176,7)
0.999898081101
Though the question gives the
P(150
≤
H
≤
200)
=
1
variance of the distribution, most
This
shows
normal
virtually
all
the
total
probability
of
1
under
the
calculators require you to input
the standard deviation. So in this
curve.
example, 7 was entered into the
GDC, not 49.
The x-axis is an asymptote to the
curve.
140
150
160
170
180
190
200
210
Ne
(b)
1.1
The lower bound of –9E999 stands
normCdf(–9. E999,176,176,7)
for –∞
0.500000000525
P(H
<
176)
=
0.5
The symmetric nature of the normal
curve means that the probability
the random variable is less than
the mean is always exactly 0.5 and
the mean is located at the axis of
symmetry of the curve. For every
normal distribution, the mean,
median and mode are all equal.
140
150
160
170
180
190
200
210
(c)
1.1
normCdf(169,183,176,7)
P(169
≤
H
≤
183)
=
0.682689480874
0.683
It is always true that approximately
68% of a normally distributed
data set lies within one standard
deviation of the mean.
140
150
160
170
180
190
200
210
167
/
4
(d)
1.1
normCdf(162,190,176,7)
P(162
It is always true that approximately
<
H
≤
190)
=
0.954499875972
0.954
95% of a normally distributed
data set lies within two standard
deviations of the mean. This reflects
the fact that outcomes far from the
mean are relatively rare. Similarly,
approximately 99.7% of data
always lies within three standard
deviations of the mean.
150
Note
that
160
the
170
values
68%,
180
95%
and
190
200
99.7%
shown
210
in
Example
4.2.5
assessmen p
give
you
within
a
valuable
2
and
1,
3
perspective
standard
on
how
deviations
of
probabilities
are
the
every
mean
for
distributed
normal
Be sure that you know how to
distribution.
However,
these
are
only
approximate
values
and
you
use the normal cdf on your GDC,
should
always
use
your
GDC
to
calculate
normal
probabilities
including how to enter parameters
since
using
these
approximate
values
would
introduce
unnecessary
and events.
rounding
In
errors.
problem
solving
applications
of
applications,
more
than
one
be
prepared
probability
for
questions
that
require
distribution.
Exmpe 4.2.6
The
time
standard
taken
for
David’s
deviation
(a)
Find
the
(b)
David
5
80minutes
five
on
at
commute
C
is
distributed
normally
with
mean
67
minutes
and
minutes.
probability
works
morning
that
days
least
David’s
per
week.
three
morning
commute
Find
probability
days,
the
stating
any
takes
that
assumptions
between
David’s
that
70
and
80
commute
you
minutes.
takes
between
70
and
make.
Solution
2
(a)
C
~
N(67,
5
Write
)
Then
P(70
≤
C
≤
80)
=
0
down
find
the
the
distribution
probability
and
using
the
event.
technology
.
.270
1.1
normCdf(70,80,67,5)
(b)
Assuming
each
independent
adverse
D
~
B(5,
the
–
five
ie
days
road
conditions
number
commute
80minutes
the
trials
weather
probability
,
David’s
of
takes
follows
0.270)
a
of
represent
conditions
do
not
days
D
between
binomial
or
affect
on
70
the
which
and
distribution.
Write
find
P(3
≤
C
≤
5)
=
0.269591842791
down
the
the
distribution
probability
using
and
the
event
then
technology
.
0.125
1.1
binormCdf(5,0.26959184279094,3,5)
0.125247959694
168
/
4.2
P r ob a bili t Y
DiStribUtioNS
S aMPlE StUDENt aNS WEr
Timmy
be
owns
modelled
and
daily
income
(b)
normal
needs
Calculate
shop
a
His
standard
Timmy
The
shop.
as
$820,
(a)
a
a
the
open
Calculate
Timmy
deviation
to
a
be
greater
with
$230.
than
that,
24
days
on
every
probability
makes
of
from
To
selling
a
mean
make
a
his
goods
daily
can
income
profit,
of
Timmy’s
$1000.
a
randomly
selected
day
,
profit.
for
the
income
distribution,
probability
makes
is
daily
a
profit
that,
on
month.
in
a
randomly
between
5
selected
and
10
days
P(T
>
1000)
month,
(inclusive).
▲ The student correctly
2
(a)
T~N(820,
230
)
interprets
as
P(T
>
1000)
=
an
the
application
mark
and
820
X~B(24
,
P(5
≤
x
for
its
≤
10)
=
P(x
≤
10)
−
P(x
≤
0.9927
−
binomCdf(24
,0.217
,5)
−
student
random
The
the
a
method
distribution
correctly
.
which
the
and
is
writes
not
0.5756
P(5
finds
probabilities
inverse
variable
develops
incorrectly
,
notation
≤
x
which
variable.
gains
parameters
inequality
0.417
cdf
and
writing
appropriate
normal
binomial
5)
GDC
The
problem
1000
▼ The
=
the
the
0.217)
binomCdf(24
,0.217
,10)
=
of
of
0.217
distribution,
(b)
context
normal
given
their
for
function
given
values
allows
probabilities,
you
using
of
to
the
find
your
≤
in
10)
gives
=
the
P(x
P(5
≤
examination.
≤
10)
P(x
x
≤
=
10)
≤
4),
0.613.
random
values
of
the
GDC.
Exmpe 4.2.7
A factory
and
fills
standard
bags
of
rice.
deviation
The
4.7
(a)
Find
the
interquartile
(b)
Find
the
mass
mass
of
a
bag
of
rice R
follows
the
normal
distribution
with
mean
500
g
g.
range
exceeded
by
of
R
90%
of
the
bags
of
rice.
Solution
2
(a)
R~N(500,
Let
the
Then
4.7
value
P(R
≤
q)
)
Write
of
the
=
0.25
lower
quartile
be
q.
Introduce
Write
1.1
invNorm(0.25,500,4.7)
496.829898178
q
=
496.82989..,
the
inter-quartile
range
your
GDC.
the
×
(500
496.82989..)
=
6.34
Let
the
value
required
be
a.
distribution.
variable
answer
to
help
solve
the
problem.
full
with
decimal
the
inverse
answer
in
normal
your
cdf
working
on
to
is
the
IQR
correct
to
three
significant
figures.
g
A sketch
(b)
the
event.
the
give
2
the
a
Find
Use
Since
down
Then
P(R
≤
a)
=
is
a
useful
tool
to
show
your
working
0.1
and
to
check
the
viability
of
the
answer:
1.1
0.1
invNorm(0.1,500,4.7)
a
=
494
493.976707637
g
a
The
answer
of
on
a
the
of
494
500
is
consistent
with
the
position
sketch.
169
/
4
S aMPlE StUDENt aNS WEr
The
weights
with
Anewborn
baby
weighs
than
less
Given
find
(b)
well-labelled
constructs
sketch
to
that
a
newborn
mean
is
w
babies
3.41
kg
considered
in
and
to
Australia
standard
have
a
low
are
normally
deviation
birth
0.57
weight
if
kg.
it
kg.
5.3%
of
newborn
babies
have
a
low
birth
weight,
w
A newborn
Find
student
of
distributed
(a)
▲ The
W
the
baby
has
probability
a
low
that
birth
the
weight.
baby
weighs
at
least
2.15
kg.
a
visualize
the
(a)
P
=
0.053
context.
W
=
2.488 631 27
0.053
▲ The
student
conditional
W
recognizes
probability
in
of
the
w
P(2.15
student
accuracy
in
<
X
<
3.41
2.49)
problem.
P
▲ The
2.49
the
(b)
context
=
uses
the
(2.15
<
X
<
2.49) =
0.039 465 67
=
full
calculation
to
0.039 5
avoid
0.03946567
accumulation
of
P
error.
=
0.053
4 . 3
P
=
0.744 635 283
P
=
0.745
T R A N S F O R M AT I O N S
O F
R A N D O M
VA R I A B L E S
Yu shud nw:
✔
✔
the
difference
of
a
of
at
the
random
least
concept
population
of
between
a
linear
and
a
unbiased
transformation
linear
independent
an
✔
combination
random
estimate
linear
variables
of
interpret
a
✔
parameter
the
a
linear
combination
variable(s)
nd
parameters
the
random
of
n
independent
is
normally
variable
a
useful
case
of
this
as
random
problem
or
and
apply
combinations
to
the
appropriate
of
and
a
a
of
linearly
a
linear
variables
transformed
combination
and
apply
of
these
normal
problem
solving
distributed
✔
✔
of
transformations
random
to
variables
( A H L )
context
independent
✔
C O M B I N AT I O N S
Yu shud e e :
variable
two
A N D
result
is
calculate
unbiased
estimates
of
population
that
parameters
with
technology
and
with
formulae
2
X
~
N( µ , σ
σ
⎛
2
) ⇒ X
~
⎞
µ,
N
⎝
n
✔
⎠
apply
the
central
probabilities
✔
the
statement
of
the
central
limit
limit
theorem
involving
a
linear
to
nd
combination
theorem.
of
independent
distributions
Y
ou
have
transformed
trigonometry
random
a
new
and
variables.
random
points
functions
and
variable
with
a
are
not
variables
in
the
respectively
.
random
different
whose
known.
functions
topics
T
ransforming
random
geometry
,
Y
ou
can
variable X
features
and
to
also
aX
transform
+
b
gives
parameters.
170
/
4.3
For
by
example,
the
red
translated
by
the
image
has
been
to
left
by
distribution
the
the
curve
the
green
under
in
curve,
is
units
whereas
a
probability
to
in
the
to
1.
two
define
horizontal
by
equal
the
transformed
10
represented
graph
below
t r a N S F o r M at i o N S
blue
a
distribution
different
In
of
oF
raNDoM
va r i a bl E S
(aHl)
it
is
represented
factor
each
c o M b i N at i o N S
represented
ways:
distribution
stretch
curve.
aND
5
defines
distribution,
the
the
area
X
–10
–8
–6
Random
–4
–2
2
Expectation
4
6
8
10
Variance
12
14
16
18
20
22
24
26
28
30
32
Comment
variable
X + b
E(X + b) = E(X) + b
The distribution is translated by b units
Var(X + b) = Var(X)
horizontally, hence the mean increases or
decreases by this value. The spread remains the
same because a translation does not change the
shape of the distribution.
2
aX
E(aX) = aE(X)
Var(aX) = a
All x-values are multiplied by a, so the mean is also.
Var(X)
Just as the range is a times larger, so too is the
standard deviation.
2
aX + b
E(aX + b) = aE(X) + b
Var(aX + b) = a
Var(X)
Apply the sequence of transforms X → aX → ax + b
to find the effect of the general transformation of
X to aX + b
Similarly
,
linear
combinations
following
expectation
and
of
two
random
variables
have
variance:
2
E(aX
±
bY)
=
aE(X)
±
the
bE(Y)
and
Var(aX
±
bY)
=
a
2
Var(X)
+
b
Var(Y)
assessmen p
For
the
You
latter
are
not
important
result,
the
expected
is
that
random
to
you
be
can
able
variables
to
apply
prove
must
these
be
independent.
results
–
what
is
It is impor tant to correctly
them.
distinguish between a transformed
random variable and a linear
These
results
can
be
applied
repeatedly
to
the
results
for
linear
combination.
combinations
E(aX
of
±
bY)
three
E(a
X
1
E(a
±
a
±
a
1
±
a
2
X
2
random
aE(X)
X
2
X
n
random
1
1
=
of
bE(Y)
variables,
X
3
±
±
)
=
E(a
3
…
variables:
can
±
a
X
n
)
applied
X
)
±
E(a
1
=
V
ar(aX
±
E(a
n
bY)
X
1
=
a
X
2
)
±
X
1
±
a
1
X
2
±
a
2
X
3
)
=
a
E(a
X
3
X
2
to
a
linear
combination
)
±
)
and
so
hence:
on,
3
…
±
E(a
2
X
n
)
n
2
V
ar(X)
+
b
V
ar(Y)
)
+
a
can
be
applied
2
Var(a
1
3
±
E(a
2
Var(a
)
2
1
2
Similarly
,
again
giving
1
2
be
X
1
giving
2
Var(a
2
1
again,
X
2
)
+
a
Var(a
3
2
X
3
)
and
soon,
3
hence:
Var(a
X
1
±
a
1
Again,
X
2
the
latter
independent
…
±
2
±
a
2
of
result
each
X
)
=
2
a
Var(a
n
n
1
is
only
true
X
1
)
+
a
1
Var(a
2
when
all
X
2
the
)
+
…
2
+
2
a
Var(a
n
random
variables
X
n
).
n
are
other.
Application:
Estimating
population
parameters
of
with
μ
a
parameters
population
are
with
sample
modelled
by
a
statistics:
random
if
the
variable X
2
E(X)
X
+
1
x
X
=
and
+ ! +
2
,
and
V
ar(X)
=
σ
,
you
calculate
the
sample
mean
X
n
by
=
taking
n
independent
samples
from
the
population.
n
171
/
4
Then
1
E( X )
= E
(
1
X
1
+
X
1
1
+ ! +
X
2
n
n
n
n
1
=
)
E(X
1
) +
E(X
) + ! +
1
1
E( X
2
n
)
=
( nµ )
=
µ
n
n
n
n
n
X
∑
i
Hence
E(X ) =
This
µ.
means
that
i
1
=
X
is
an
unbiased
estimator
of
μ
n
can
be
shown
that
the
unbiased
estimator
σ
of
n
2
2
It
s
is
s
1
n
n
1
n
2
s
where
2
=
n
is
the
variance
of
the
sample.
> 1,
Since
the
unbiased
n
n
estimate
of
the
population
variance
is
always
1
slightly
larger
than
the
samplevariance.
Exmpe 4.3.1
Y
are
independent
random
variables
with
E(Y )
i
=
10
and
V
ar(Y )
i
Find
the
expected
value
and
variance
of
each
of
the
following
(c)
(a)
P
=
=
2,
where
i
∈
{1,
2,
3,
4,
5,
6}.
i
R
=
random
3Y
6Y
variables:
2Y
1
Y
2
3
1
(b)
Q
=
Y
+
2Y
1
+
3Y
2
(d)
S
=
Y
3
+
Y
1
+
Y
2
+
Y
3
+
Y
4
+
Y
5
6
Solution
(a)
E(6Y
)
=
6E(Y
1
(b)
E(Y
+
2Y
1
+
+
Var(6Y
3Y
2Y
+
Var(3Y
Y
+
Var(Y
+
Var(Y
)
1
)
2
+
+
36Var(Y
)
+
2E(Y
=
Var(Y
)
3E(Y
)
=
)
+
Y
+
Y
3
4
)
2
3E(Y
4Var(Y
+
+
)
Y
)
+
)
+
72
=
60
9Var(Y
)
E(Y
)
)
=
+
Y
)
28
=
28
0
Var(Y
E(Y
)
+
)
3
E(Y
1
5
=
3
2
=
)
3
4Var(Y
6
Y
)
+
2
+
=
3
2E(Y
5
)
2
9Var(Y
4
Y
+
1
Y
Var(Y
+
1
3
+
)
2
1
=
3
Y
1
)
Y
Y
2
=
1
1
2
Y
E(Y
3
2Y
1
1
=
3Y
2
+
)
3
2Y
1
)
1
2
E(3Y
E(Y
60.
3
1
(d)
=
2
Var(Y
(c)
)
1
)
+
E(Y
2
)
+
E(Y
3
)
+
E(Y
4
)
+
E(Y
5
)
=
60
6
=
6
Var(Y
)
+
3
Var(Y
)
+
Var(Y
4
)
+
Var(Y
5
)
=
12
6
The example illustrates that whenever X, X
, X
1
, … X
2
are independent random
n
n
⎛
⎞
2
2
variables each with expected value μ and variance σ
2
the
mean
variables
is
example,
in
are
shown:
and
the
that
T
is
known,
the
the
time
commute,
and
T
of
it
be
would
diagram
time
taken
=
variance
W
+
modelled
below,
taken
riding
B,
by
is
by
to
also
the
a
a
two
1
)
home
normal
know
(B).
to
X
=
i
nσ
⎟
⎠
i =1
independent
its
The
to
total
normally
.
random
distribution.
normal
walk
For
distributions
her
bicycle
time
taken
The
diagram
(W)
for
the
shows
distribution.
2
B~N(25,
of
independent
distributed
2
W~N(5,
to
commuter
her
⎝
2
combination
useful
∑
⎜
σ
This must not be confused with Var(nX) = n
Now
, Var
3
)
2
T~N(30,
1
2
+
3
)
172
/
4.3
More
generally
,
independent
the
distribution
normal
t r a N S F o r M at i o N S
of
a
linear
distributions
is
normal.
aND
combination
c o M b i N at i o N S
oF
raNDoM
va r i a bl E S
(aHl)
of n
2
If
~ N( µ
X
i
σ
i
)
, i ∈ {1, 2, …, n},
i
2
X
then a
1
This
+ a
1
can
X
2
be
…
+
a
2
X
n
applied
to
μ
~N(a
n
1
solve
μ
+ a
1
2
…
+
µ
a
n
2
2
σ
a
n
1
2
+ a
1
2
σ
2
2
2
σ
+!a
2
)
n
n
problems.
Exmpe 4.3.2
Lisa
and
by
normal
to
a
write
(a)
On
Miguel
a
a
In
given
total
on
Lisa
taken
with
of
Lisa
takes
week,
time
one
day
,
report
one
psychologists.
distribution
report
Lisa’s
(b)
are
mean
his
and
more
more
is
Miguel
than
hours
taken
and
modelled
twice
30
L
as
long
reports
Lisa
to
write
one
Miguel’s
Miguel
write
a
deviation
by M~N(4.5,
as
and
by
standard
independently
two
than
time
5.7
clients
completes
is
The
to
report
2
on
hours.
a
client,
The
time
can
be
taken
modelled
for
Miguel
3.2)
report
each.
Find
the
probability
that
write.
completes
three.
Find
the
probability
that
the
hours.
Solution
(a)
P(L
>
2M)
E(L
=
2M)
P(L
=
5.7
2M
>
2
4.5
×
Interpret
0)
=
−3.3
the
context
and
to
a
write
Rearrange
the
event
linear
Determine
the
parameters
down
the
event.
combination.
2
Var(L
2M)
=
2
+
4
×
3.2
=
16.8
the
L
2M~N(
3.3,
and
use
technology
to
find
probability
.
16.8)
1.1
P(L
2M
>
0)
=
0.210
normCdf
0.9.E999,–3.3,√16.8
0.210375857927
(b)
Let
T
=
L
+
L
1
+
M
2
+
M
1
+
Write
M
2
not
E(T)
=
Var(T)
5.7
=
+
4
+
T~N(24.9,
5.7
4
+
+
4.5
3.2
+
+
4.5
3.2
down
the
random
variable.
Note
that
2L
+
3M
is
3
+
+
4.5
3.2
=
=
the
same
as
the
distribution
required.
24.9
17.6
17.6)
P(T
>
30)
=
0.112
Consider
a
sample
of
size
n
from
a
population
that
is
modelled
by
2
a
normal
random
variable
X
~
N( µ , σ
Each
).
of
the
n
data
points
~
N( µ , σ
i
2
in
the
sample
is
an
independent
random
variable X
)
and
i
X
+
X
1
X
+ ! X
2
n
=
is
a
linear
combination
of
these
is
that
random
variables
n
known
as
the
sample
A consequence
of
the
mean.
previous
key
point
you
can
write
down
2
σ
⎛
the
distribution
of
the
sample
mean
X
~
⎝
predictions,
as
shown
in
the
next
⎞
µ,
N
,
n
and
hence
make
⎠
example.
173
/
4
Exmpe 4.3.3
Lengths
of
3cm.
of
bamboo
A garden
with“average
Find
the
1
of
cm
centre
length
smallest
173
canes
cm
is
at
to
sell
the
normally
canes
in
with
a
mean
packages
of n
of
173
canes
cm
and
and
a
standard
wants
to
label
deviation
the
packages
cm”.
of
least
distributed
plans
173
value
are
n
required
so
that
the
probability
that
the
mean
length
of
the n
canes
is
within
0.9
Solution
B
=
the
length
A sample
of
of
a
randomly
size
n
is
chosen
cane.
taken.
Write
down
Solve
the
of
GDC
the
the
distributions
problem
input
with
to
and
the
technology
show
your
event.
and
provide
method.
The
a
sketch
range
of
2
B ~
3
⎛
2
N(173, 3
) ⇒ B
~
N
⎞
173,
⎝
the
n
functions
looking
P(172
<
B
<
174)
≥
is
[0,
1].
The
domain
can
be
found
from
⎠
at
the
table
of
values
of
the
function.
0.9
y
The
smallest
value
of
n
is
25
canes.
f2(x)=0.9
(24.35, 0.9)
3
f1(x)=NormCdf(172, 174, 173,
√x
0.1
x
–1.39
In
fact,
more
you
can
make
32.57
predictions
involving
the
sample
mean
in
many
contexts.
Central
limit
theorem:
When
sampling
from
any
population
of
discrete
2
or
continuous
data
with
mean
μ
and
σ
variance
,
the
distribution
,
if
is
of
2
σ
⎛
the
sample
mean
is
approximately
X
~
⎝
In
examinations,
can
make
which
n
>
30
predictions
distribution
will
about
models
be
the
n
considered
the
sample
⎞
µ,
N
n
sufficient.
mean
large
enough.
⎠
even
if
Therefore,
you
do
you
not
know
population.
Exmpe 4.3.4
All
of
residential
bedrooms
distribution
homes
has
is
in
mean
shown
a
city
3.30
are
and
surveyed
standard
in
a
census.
deviation
1.94
The
number
and
the
below:
x
1
A sample
of
probability
between
50
2
3
homes
is
that
3.5
the
and
4
taken
average
5
from
number
6
this
of
7
population.
bedrooms
in
8
Find
this
9
the
sample
is
5.5
174
/
4.3
t r a N S F o r M at i o N S
aND
c o M b i N at i o N S
oF
raNDoM
va r i a bl E S
(aHl)
Solution
Let
be
B
the
number
of
Write
average
down
applying
bedrooms.
the
the
distribution
central
limit
by
theorem.
2
⎛
B
Then
~
N
1.94
⎞
50
⎠
1.1
3.30,
⎝
1.94
normCdf
P(3.5
<
<
B
5.5
)
=
3.5,5.5,3.3,
0.233
50
0.233008078109
S aMPlE StUDENt aNS WEr
Adam
does
Thetime
the
crossword
taken
by
Adam,
in
X
the
local
minutes,
newspaper
to
complete
every
it
is
day
.
modelled
by
2
the
normal
(a)
Given
that
is
(b)
distribution
that,
he
to
the
the
find
five
).
chosen
crossword
the
value
that
randomly
the
of
day
,
in
the
less
probability
than
a
minutes
a
total
chosen
time
taken
crosswords
for
him
to
exceeds
minutes.
Beatrice
day
.
0.8,
5
randomly
probability
complete
120
a
completes
equal
Find
on
N(22,
also
The
does
time
the
taken
crossword
by
Beatrice,
in
Y
the
local
minutes,
newspaper
to
complete
every
the
2
crossword
(c)
Find
the
is
is
the
more
by
probability
time
the
modelled
taken
than
by
that,
normal
on
Beatrice
twice
crossword.
the
the
to
time
Assume
a
randomly
complete
taken
that
distribution
these
by
chosen
the
6
).
day
,
crossword
Adam
two
N(40,
to
times
complete
are
independent.
▲ The
student
provides
clear
2
Let
x
be
the
p(x
<
time
for
Adam
to
solve
puzzle
x
~
N(22,
5
)
and
a
(a)
a)
=
complete
clear
description
(b)
a
let
=
k
k=
26.2
be
x
+
E(k)
the
time
x
1
taken
+
x
2
=
var(k)
+
to
solve
x
+
3
5
aid
context,
part
puzzles
to
~
=
the
interpretation
scoring
full
marks
of
for
5
5E(x)
=
5var(x)
N(110,
>
random
x
4
120)
=
0.185 547
helps
and
0.186
(C)
setting
out
the
125)
random
p(k
the
(a).
▲ Carefully
k
of
including
0.8
variable
⇒
working,
let
y
N(40,
6
be
time
taken
for
Beatvice
to
solve
variable
the
score
and
student
full
the
solve
event
the
problem
marks.
puzzle
2
y
p(y
>
~
2x)
=
p(y
)
−
2x > 0)
▼ Although
let
m
=
y
−
E(m)
=
E(y)
full
−
2
E(x)
=
40
−
2(22)=
2
var(m)
m
~
p(m
=
=
the
student
gained
2x
var(y)
+
2
2
var(x)
=
(6
)
−
4
is
2
+
4(5
)=
136
marks
worth
for
this
pointing
question,
out
mathematics
random
written
with
capital
because
this
that
it
in
variables
letters
as
a
are
rule
N(−4, 136)
>
0)
=
0.365 8
from
other
helps
distinguish
them
variables.
0.366
175
/
4
4 . 4
P O I S S O N
D I S T R I B U T I O N
Yu shud nw:
✔
what
denes
distribution
✔
the
mean
of
the
is
an
the
( A H L )
Yu shud e e :
contexts
in
which
appropriate
Poisson
the
Poisson
✔
calculate
✔
select
Poisson
probabilities
with
technology
.
model
distribution
equals
the
between
distribution
the
as
Poisson,
appropriate
binomial
for
a
or
given
normal
context.
variance
✔
the
sum
of
two
distributions
independent
has
a
Poisson
Poisson
distribution.
Another
example
of
a
discrete
probability
distribution
is
the
Poissondistribution.
Context
of
situations
the
in
Poisson
which
distribution:
events
are
The
Poisson
independent
distribution
and
occur
at
a
The
random
models
uniform
Number of
buses arriving in
average
rate
number
of
in
an
interval
of
time
or
space.
variable
is
the
Frequency
occurrences.
15 minutes (x)
0
3
1
15
2
For
example,
a
frequency
of
20
that
bus
3
25
Hecollects
4
17
4:15pm
5
11
6
6
7
2
at
Note
his
arriving
1
data
hundred
3.11,
event
The
random
bus
stop
3
buses
is
in
per
a
at
each
On
the
or
X
~
the
the
nature
stop
you
and
the
is
15-minute
and
traffic
how
write
at
his
on
a
rate
bus
The
the
stop
of
a
this
in
a
a
sample
factors
buy
of
of
from
size
0
table.
to
one
8.
period.
at
as
people
distribution
to
Also,
the
is
buses
such
an
pm
buses
rate
the
tickets
is
minutes.
4
of
arrive
that
the
frequency
average
distribution
Poisson
from
complete.
that
amount
to
15
15-minute
buses
of
in
varies
assumption
the
every
with
reasons
number
was
uniform
them
Poisson
follows
from
run
David
intervals
findings
combination
takes
buses
arriving
The
congestion,
it
3
sample
number
period.
buses
maximum
buses
the
the
long
that
“X
of
his
the
its
Therefore,
15-minute
average
interval.
depends
of
until
that
about
presents
know
arriving
15-minute
be
hundred
not
X
minutes.
should
number
variable
advertises
five
sample
finds
assumption
model,
one
could
bus
independently
random
there
data
David
The
on
per
Monday
his
is
company
bus
stop
David
in
although
8
one
every
that
bus
arrive
the
embarking
from
the
driver.
appropriate
with
parameter
3”
Po(3).
If occurrences of an event are independent and occur at a uniform average
rate m in an interval of time or space, the number of occurrences X in the
interval follows a Poisson distribution with parameter m. You write X~Po(m).
Poisson probabilities are found with technology and the domain of the Poisson
probability distribution function is x ∈

176
/
4.4
For
X
~
Po(m),
average
The
m
mean
E(X)
times
for
in
=
the
longer
proportion,
as
You
required
are
not
examination.
technology
This
shorter
in
to
then
know
in
fact
that
is
that
the
can
if
the
event
expected
be
found
occurs
value
with
DiStribUtioN
(aHl)
on
of X.
direct
4.4.1
the
always
the
the
intervals
Example
should
shown
reflects
interval,
or
shown
You
as
m.
Poi SS oN
formula
find
following
for
the
Poisson
Poisson
pdf
probabilities
in
the
with
example.
Exmpe 4.4.1
The
a
number
Poisson
of
calls
received
distribution
following
with
in
an
mean
hour
C
by
7.3.
Find
one
hour.
an
the
IT
helpline
probabilities
follows
of
the
events:
(a)
Exactly
(b)
More
(c)
At
6
calls
than
least
4
15
are
but
calls
assumptions
received
no
more
are
made
you
in
than
in
a
8
calls
three
are
received
in
one
hour.
hour
interval,
stating
any
make.
Solution
(a)
C
~
Write
Po(7.3)
down
the
distribution
and
the
events.
P(C
(b)
P(4
=
6)
=
0.142
<
C
≤
8)
=
P(5
≤
C
≤
=
0.542
Find
the
probabilities
using
technology
.
8)
1.1
poissPdf(7.3,6)
0.141989080378
The probability distribution
poissCdf(7.3,5,8)
0.541884584152
function of X~Po(m) is:
x
m
(c)
Assuming
the
rate
Find
the
rate
by
direct
proportion.
P( X
=
x)
m
e
=
x!
continues
for
the
three
There
are
two
equivalent
ways
to
find
This formula can be explored in
hours,
and
writing
D
the
answer
with
technology
.
the context of functions, but is not
as
the
number
of
in
a
three-hour
D
~
Po(2
calls
1.1
P(D
≥
needed for exams.
period,
1–poissPdf(21.9,0,14)
0.950279643976
poissCdf(21.9,15,1000)
0.950279643976
1.9)
15)
=
assessmen p
0.950
The formulae for the mean and
As
the
value
X~Po(m).
2
and
for
10
For
are
mean
of
2.
m
increases,
example,
shown
This
is
bar
below,
because
so
does
charts
of
showing
for
the
spread
Poisson
a
greater
X~Po(m),
of
the
distribution
distributions
spread
Var(X)
=
for
with
mean
variance of the Poisson distribution
means
10
than
are in the formula book, section
AHL 4.17
m
P(X=x)
0.36
0.32
Po(2)
0.28
0.24
0.2
Po(10)
0.16
0.12
0.08
0.04
x
2
4
6
8
10
12
14
16
18
177
/
4
If
S~Po(λ)
S
+
T~Po(μ)
and
are
independent
Poisson
distributions
then
The expected value and variance
T~Po(λ
μ).
+
This
makes
sense
only
if
the
units
for
the
uniform
of a sum of independent random
average
rates
are
the
same.
It
would
buses
per
not
make
sense
to
add
the
rates
variables is covered generally in
1
bus
per
hour
and
5
day
to
find
6
buses
per
hour!
If
the
the
same
unit
rates
section 4.3 of this book .
are
not
direct
in
the
same
proportion,
unit
as
they
shown
should
in
the
be
changed
next
to
using
example.
Exmpe 4.4.2
Medical
staff
in
a
emergency
room
admissions
M
(a)
to
shift,
follows
the
For
a
the
maternity
(b)
Hence
(c)
The
find
find
the
work
a
the
>
during
ward
one
shifts
of
eight
distribution
distribution
follows
of
T,
the
shift,
a
hours.
with
stating
number
parameter
Poisson
total
The
of
with
admissions
assumptions
admissions E
admissions
distribution
number
any
1.7
of
you
per
to
hour.
The
mean
6.1
to
emergency
the
the
number
admissions
per
room
of
shift.
and
to
make.
24).
administrator
total
daily
Poisson
maternity
ward
P(T
hospital
days
hospital
admissions
works
is
at
a
least
five-day
25,
week.
stating
any
Find
the
probability
assumptions
you
that
on
exactly
three
of
these
make.
Solution
(a)
Assuming
for
is
the
1.7
×
that
the
duration
8
+
6.1
=
of
hourly
rate
a
the
19.7
shift,
hence
T
of
1.7
parameter
~
Convert
continues
of
using
T
the
hourly
direct
rate
to
the
rate
for
one
shift
proportion.
Po(19.7)
1.1
(b)
P(T
>
24)
=
0.141
1–poissCdf(19.7,0,24)
Find
(c)
Let
D
be
the
number
week
on
which
are
made.
the
probability
a
Then
of
total
T
~
days
of
B(5,
at
in
a
least
0.141)
Write
five-day
25
and
admissions
assuming
the
0.140558464438
probability
down
find
the
the
with
technology
.
distribution,
probability
with
the
event
technology
.
that
1.1
of
0.141
is
constant
for
all
five
binomPdf(5,0.14055846443808,3)
days
and
that
each
day
is
the
number
of
admissions
on
0.020511796032
the
other
P(D
=
3)
independent
of
the
admissions
on
days.
=
0.0205
S aMPlE StUDENt aNS WEr
2
A company
During
produces
rectangular
manufacturing
these
sheets
glass
of
sheets
glass
flaws
of
area
occur
at
5
m
.
the
2
rate
of
0.5
per
glass
(a)
Find
at
Glass
least
(b)
per
sheet
the
least
Find
.
It
is
assumed
a
probability
Poisson
that
a
that
the
number
of
flaws
distribution.
randomly
chosen
glass
sheet
contains
flaw.
with
flaw
the
m
follows
one
sheets
one
5
no
incur
flaws
a
loss
expected
earn
of
profit,
a
profit
of
$5.
Glass
sheets
with
at
$3.
P
dollars,
per
glass
sheet.
2
This
company
also
produces
larger
glass
sheets
of
area
20
m
.
2
The
rate
A larger
(c)
Find
of
occurrence
glass
the
sheet
is
of
flaws
chosen
probability
that
remains
at
it
at
0.5
per
5
m
random.
contains
no
flaws.
178
/
4.5
▼ The
DE S criPti vE
student
writes
S tat i S t i c S
the
inequality
2
X~Po(0.5)
where x
is
aws
per
5
m
correctly
,
(a)
p(x
≥
(b)
E(x
1)
=
1
− p(x
<
1)
=
of
0.0902
the
up
$5
(c)
event
the
0.9098
−
$3 ×
0.0902
=
0)
of
=
–
equivalent
P(X
the
answer
▲ Detailed
award
=
1
the
is
=
0)
GDC
form
would
better.
set
The
0.393
$4
.28
X~Po(2)
p(x
use
correct
×
but
of
working
follow
enables
through
the
marks.
0.135
The
correct
▲ Direct
applied
answer
is
proportion
correctly
,
probability
has
has
the
been
$1.85
been
correct
found
with
range
and
theGDC.
4 . 5
D E S C R I P T I V E
S TAT I S T I C S
Yu shud nw:
✔
the
difference
continuous
Yu shud e e :
between
discrete
and
✔
nd
data
median,
interquartile
frequency
✔
an
outlier
more
the
✔
than
mean,
central
dened
1.5
nearest
the
of
is
×
as
a
data
interquartile
item
which
range
(IQR)
✔
median
and
tendency
mode
are
all
from
values
the
interquartile
deviation
are
measures
✔
✔
intervals
the
variance
is
the
the
on
a
constant
world
make
form.
finding
is
of
and
how
increasing
this
This
an
or
mean
change
contains
sense
useful
the
by
a
a
cumulative
produce
and
✔
calculate
the
interpret
box
and
whisker
plots
table
closely
it
and
of
it
section
‘average’
fits
the
on
to
the
frequency
technology
calculate
the
mean,
quartiles
using
standard
deviation
and
technology
.
the
of
be
the
deviation
original
in
the
sorted
course
for
with
grouped
spread
midpoints
standard
the
a
the
needs
of
without
from
range
tables
information
value
decreasing
of
of
mean
standard
using
acting
lot
data
or
frequency
square
deviation
effect
mean
grouped
standard
The
it
the
estimated
the
of
by
from
are
from
curve
✔
measures
and
calculating
deviation
(IQR)
quartile
standard
when
range
percentiles,
is
✔
and
quartiles,
looks
data,
either
a
a
data.
form
and
at
of
but
summarized
three
ways
measure
straight
data,
line
of
or
a
into
this
how
to
is
a
done:
spread
out
continuously
function.
aeges
There
are
three
‘average’),
median
ways
which
and
to
you
measure
use
in
‘central
different
tendency’
situations.
(or
They
three
are
types
of
the mean,
mode
179
/
4
Men
To
find
the
number
of
mean,
all
values.
It
the
is
values
often
are
added
written
as
and
example,
the
mean
of
10,
12,
17,
divided
+
17
by
the
x .
10
For
then
24
x
is
12
+
+
24
=
=
15.75
4
∑ x
We
can
write
the
mean
as:
x
=
,
where
∑
x
represents
the
sum
or
n
total
of
the
data.
Ne
k
f
∑
x
i
k
i
i=1
The
formula
book
writes
the
formula
as
where
Rearranging this formula to give
n =
.
f
∑
i
n
i=1
the total in terms of the mean can
This
is
equivalent
to
the
above
and
is
used
when
the
data
is
given
in
a
be useful.
frequency
table.
If
the
data
is
given
individually
then
all f
are
equal
to
i
1
and
the
equation
above
is
formed.
Make sure you know how to find the mean (x ) on your GDC.
Exmpe 4.5.1
Eight
students
Aninth
Find
took
student
the
a
took
average
test.
the
score
Their
same
for
the
average
test
nine
a
score
day
later
was
and
14.5
points.
scored
19
points.
students.
Solution
Total
score
for
the
eight
∑ x
If
students
is
New
total
New
mean
14.5 × 8
=
=
116 + 19
x
x
∑
=
nx
n
116
=
then
=
135
135
=
=
15
9
Ne
For a large data set of size n use
Medn
the rule that the median is the
If
th
n +
value of the
(
the
data
being
considered
is
put
in
order
of
size
2:
3,
then
the
median
is
1
)
piece of data.
the
value
of
the
middle
number.
2
Consider
List
1:
the
2,
2,
ordered
3,
5,
6,
6,
data
7,
9,
9
List
3,
5,
5,
5,
7,
7,
8,
8,
9
assessmen p
In
list
1,
the
median
is
6.
Using
the
formula
the
(
Usually the median will be found
9
pieces
of
data
so
the
median
is
+
in
the
Note
box,
we
have
9
1
)
=
5th
piece
of
data.
2
using the statistical functions on
the GDC.
10
In
list
2
there
are
10
pieces
of
data
so
the
median
is
the
+
1
(
)
=
5.5th
2
piece
In
this
equal
The
assessmen p
In the exam, quar tiles will always
be obtained from the GDC, a
cumulative frequency curve or a
box and whisker plot.
of
(or
In
data.
case
to
we
the
mean
of
the
5th
and
6th
pieces
of
data,
which
is
6.
median
50%
take
of
is
the
addition
to
also
referred
way)
the
to
through
median,
as
an
the
the
50th
ordered
lower
percentile
list
of
quartile
the
(Q
)
as
it
lies
halfway
data.
is
the
value
that
1
is
25%
upper
of
the
way
quartile
through
(Q
)
is
the
data
(the
three-quarters
of
25th
the
percentile)
way
and
through
the
the
data
3
(the
If
75th
the
Q
1
=
percentile)
data
2.5,
in
Q
=
list
1
is
entered
into
a
GDC
it
will
give
the
values
8
3
180
/
4.5
DE S criPti vE
S tat i S t i c S
Mde
The
mode
common
is
the
number
number.
In
list
with
2
the
the
highest
mode
is
frequency
,
or
the
most
5.
Fequeny es
Exmpe 4.5.2
The
following
marks
out
of
8
were
pupils
in
the
achieved
by
a
class.
Find:
The formula for the mean is in
section 4.3 of the formula book:
(a)
the
number
of
class
(b)
the
mean
mark.
n
Mark
Frequency
4
5
5
6
6
8
7
4
8
2
x
f
∑
i
n
i
i =1
x
where
=
n =
∑
f
i
n
i =1
You met sigma notation in
section 1.2.
assessmen p
Solution
(a)
25
The
total
class
4
×
5
+
5
×
6
+
6
×
8
+
7
×
4
+
8
(b)
×
is
number
the
sum
of
of
students
the
in
the
In an exam you will usually be
frequencies.
calculating a mean using the
2
inbuilt function on your GDC.
When
adding
up
the
marks
in
this
25
A common error when doing this
frequency
table
it
is
much
easier
4 +
4
to
is forgetting to let the calculator
142
do
=
=
4 × 5
than
4 +
4 +
4 +
5.68
know there is a frequency list!
25
Guped d
When the data is grouped, it is not possible to know the exact values, so when
finding the mean it is assumed all data takes the value of the midpoint of
their group.
Exmpe 4.5.3
The
table
minutes
below
for
Time (t)
0 ≤
t <
10
74
shows
the
students
to
length
of
complete
time
a
in
(a)
Find
the
(b)
Write
(c)
Find
mean
of
this
data.
task.
down
the
modal
class.
Frequency
the
interval
in
which
the
median
lies.
4
10 ≤
t <
20
10
20 ≤
t <
30
22
30 ≤
t <
40
17
40 ≤
t <
50
14
50 ≤
t <
60
7
181
/
4
Solution
(a)
L1
L2
5
4
15
10
25
22
35
17
45
14
55
7
31.5
minutes
(b)
20
≤
t
<
30
(c)
30
≤
t
<
40
This
is
found
and
entering
The
modal
The
median
using
the
the
inbuilt
values
class
is
the
as
function
the
midpoints:
interval
with
74
is
the
time
of
+
the
5,
highest
GDC
15,
(1-variable
statistics)
25…
frequency
.
1
(
the
in
)
=
37.5th
person.
The
37.5th
person
is
2
in
the
the
be
interval
correct
30
≤
t
interval,
<
40.
but
The
is
GDC
unlikely
gives
to
be
the
the
median
actual
as
35,
median,
which
so
indicates
should
not
quoted.
Mesues f sped
assessmen p
The
For grouped data you will only be
asked for the interval in which the
range
value.
be
It
of
is
a
not
affected
set
of
data
usually
by
a
a
single
is
the
good
very
maximum
guide
large
or
to
value
the
very
minus
spread
small
of
the
the
value,
minimum
data
called
as
it
can
an outlier
median lies, not for its value.
The
interquartile
quartiles’
and
is
range
equal
(IQR)
to
Q
3
of
the
data
The
most
(σ ),
as
it
theone
The
uses
with
IQR,
σ
as
all
not
the
the
standard
shown
is
.
the
‘range
Q
.
the
range
by
measure
data.
highest
it
literally
takes
When
any
of
is
of
the
the
middle
50%
into
spread
Using
data
is
found
all
the
directly
from
the
is
more
data
from
Example
standard
different
deviation
account
is
outliers.
comparing
standard
deviation
the
This
between
1
affected
sophisticated
Unlikethe
assessmen p
and
is
sets
deviation
of
spread
data,
out.
points.
the
4.5.3,
GDC
the
and
is
usually
standard
X
The interquar tile range will usually
deviation
is
13.3
minutes.
be tested in a question on box
and whisker plots or cumulative
cnsn hnges n he d
frequency curves.
The
four
Their
values
mean
is
1,
2.25
2,
2,
4
and
are
their
shown
as
standard
blue
dots
deviation
in
is
the
diagram
1.09.
The
below.
mean
is
Ne
indicated
of
data
by
the
(shown
arrow
on
the
by
the
red
but
the
spread
diagram.
dots)
it
can
If
be
6
is
seen
added
that
to
the
all
four
mean
pieces
also
Sometimes a question will ask for
increases
by
6,
standard
deviation
of
the
data
is
unchanged
and
hence
the
the variance rather than for the
is
unchanged.
standard deviation. The variance is
equal to the square of the standard
2
deviation and has the symbol
(see the sample student answer
1
2
3
4
5
6
7
8
9
10
on page 183).
182
/
4.5
Similarly
,
if
This
though
time
deviation
1
2
all
is
3
the
data
the
5 × 1.09
4
5
is
multiplied
spread
=
6
has
by
5,
the
increased
mean
also
so
is
multiplied
the
new
15
16
by
DE S criPti vE
S tat i S t i c S
5.
standard
5.45
7
8
9
10
11
12
13
the
items
14
17
18
19
20
is
20.
S aMPlE StUDENt aNS WEr
A data
(a)
set
Find
The
has
items.
The
sum
of
is
800
and
the
mean
n
standard
multiplied
(b)
n
deviation
by
(i)
Write
(ii)
Find
of
this
data
set
is
3.
Each
value
in
the
set
is
10.
down
the
value
of
the
new
mean.
▲ The
the
value
of
the
new
the
800
has
correctly
used
for
the
mean.
student
has
calculated
formula
800
=
(a)
20 ⇒
=
n
(b)
student
variance.
n
n
=
40
20
(i)
new
mean
▲ The
increases
proportionally
to
increase
in
new
data
set
values
so
20 ×
10
=
200
3,
data
spread
∴
new
mean
=
200
is
mean
is
correctly
,
multiplied
the
by
as
when
xed
the
a
data
value,
so
mean.
2
(ii)
SD
=
stays
the
same;
variance
=
SD
2
3
=
▲ The
9
student
question
variance
=
9
not
Pesenn f d
of
a
and
set
and
the
Any
is
data:
a
the
plot
is
a
way
minimum,
of
displaying
lower
quartile,
five
summary
median,
upper
statistics
for
the
deviation.
▼ The
standard
deviation
should
also
becomes
then
30,
multiplied
and
the
the
variance
standard
be
that
by
10,
variance
so
is
900.
quartile
maximum.
outliers
highest
It
of
whisker
asking
noticed
the
it
A box
is
has
or
are
shown
lowest
useful
way
point
to
as
a
cross
which
compare
is
and
not
two
the
an
sets
whiskers
extend
only
to
the
outlier.
of
Ne
data.
An outlier is defined as any point
that is more than 1.5 × IQR above
the upper quar tile or below the
Exmpe 4.5.4
lower quar tile.
Julia
is
doing
hemisphere.
Africa
the
and
box
a
She
the
and
project
has
15
on
longest
recorded
longest
whisker
the
in
plots
the
South
rivers
lengths
in
of
America
the
the
and
southern
15
longest
they
are
rivers
shown
in
on
below.
South America
Africa
0
500
1000
1500
2000
2500
3000
3500
Length
4000
4500
5000
5500
6000
6500
7000
(km)
183
/
4
(a)
Use
(i)
the
an
plots
to
estimate
determine:
for
the
median
length
of
the
15
longest
rivers
in
Africa
(ii)
which
in
(iii)
the
the
and
685
two
the
700
their
752
Construct
a
or
South
continents
15th
collects
records
680
(b)
then
Africa
maximum
than
Julia
of
834
box
941
and
in
of
rivers
river
information
lengths
on
in
the
1004
whisker
in
the
Africa
South
15
kilometres
969
has
third
longest
river
combined
number
longest
America
1210
plot
the
1300
to
are
longer
America.
longest
in
that
rivers
table
1380
show
in
below.
1440
this
Australia
1472
1485
2508
data.
Solution
(a)
(i)
1750
(ii)
South
In
an
1800
is
Median
is
the
8th
longest
river
so
maximum
number
is
7
The
so
=
1004,
Q
=
752,
Q
1
=
1440
will
be
than
median
for
the
for
minimum
is
not
be
1700
to
accepted.
South
America
second
longest
the
minimum
These
Africa
for
could
values
for
can
is
South
included,
7th
outlier
but
all
the
be
than
America,
1st
to
greater
South
be
less
the
than
America.
obtained
from
3
the
Outlier
from
Africa.
the
Median
value
probably
maximum
longer
for
(b)
would
any
America
The
(iii)
exam,
above
1440
+
1.5(1440
752) =
list
or
from
the
GDC.
2472
Australia
It
400
600
800
1000
1200
1400
1600
1800
2000
2200
2400
2600
is
important
when
to
check
constructing
whisker
plot.
from
the
there
will
lower
box
be
In
outliers
a
box
and
this
case
it
and
no
for
whisker
outliers
is
clear
plot
below
that
the
quartile.
When an outlier is present, it is impor tant to decide whether or not it should
be included in the data. Sometimes it is an impor tant point and should
be included. Sometimes it might distor t the data and should be dealt with
separately, or might even be due to an error in calculation.
184
/
4.5
DE S criPti vE
S tat i S t i c S
Hsgms
A histogram
grouped
data.
represents
agood
all
the
way
have
the
different
is
similar
The
to
of
show
the
same
box
vertical
range
to
a
the
mean
and
axis
is
whisker
frequency
measurements
distribution
and
similar
of
plot,
and
as
data.
a
the
for
continuous
horizontal
continuous
The
standard
but
scale.
distributions
deviations
axis
but
It
is
below
very
distributions.
ycneuqerF
4
8
6
6
ycneuqerF
ycneuqerF
6
8
4
2
4
2
2
0
0
10
20
30
40
0
10
50
20
30
40
50
10
20
30
40
50
cumue fequeny u es
assessmen p
Cumulative
frequency
less
less
than
finding
(or
the
than
median
curves
or
and
show
equal
other
to)
a
the
number
given
percentiles
value.
for
of
values
They
grouped
are
which
are
useful
for
In an exam you will only ever
be asked to find the median
data.
for grouped data when using a
cumulative frequency curve.
When constructing a cumulative frequency curve, you should always plot the
point at the end of the interval, because it indicates how many of the items in
the data set are less than a given value.
Exmpe 4.5.5
The
scores
of
Mark (x)
0 ≤
students
Frequency
x ≤
10
10 <
x ≤
20
20 <
x ≤
30
30 <
x ≤
40
40 <
x ≤
50
(a)
60
Find
The
the
a
test
3
a
12
23
b
18
53
7
60
are
out
of
50
marks
are
given:
Cumulative frequency
3
values
marks
in
of
a
and
shown
b
on
the
cumulative
frequency
graph
here.
60
55
50
45
ycneuqerF
40
35
evitalumuC
30
25
20
15
10
5
0
5
10
15
20
25
30
35
40
45
50
Marks
185
/
4
(b)
From
the
graph,
estimate
(i)
the
median
(ii)
the
inter-quartile
(iii)
the
80
The
pass
the
value
of:
score
range
th
(c)
percentile.
mark
for
the
test
was
15.
Estimate
the
number
of
students
who
passed
the
test.
Solution
(a)
a
=
12
3
=
b
=
12 + 23
9
If
12
students
must
=
have
scored
been
9
less
who
than
scored
20
marks
between
there
10
and
20.
score
of
35
(b)
60
55
50
Q
3
45
ycneuqerF
40
35
Median
evitalumuC
30
25
20
Q
1
15
10
5
0
5
10
15
20
25
30
35
40
45
50
Marks
If
(i)
there
are
60
students
the
median
the
80
is
the
28
th
the
(ii)
34
22
(iii)
The
48
=
30
student.
12
th
in
the
th
student
scored
about
37
out
of
50
If
there
are
60
test.
students
percentile
is
the
80
th
score
of
the
× 60
=
48
student.
100
(c)
7
students
thetest.
the
scored
This
less
means
than
about
15
53
and
so
failed
students
passed
test
assessmen p
Clearly show the lines you are using on the cumulative frequency diagram.
Ne
There is no need to add one before dividing by two when finding the median
for a cumulative frequency curve, as the scale begins at 0
186
/
4.6
4 . 6
C O R R E L AT I O N
A N D
Yu shud nw:
✔
the
Pearson
(r)
to
model
a
linear
aND
r E Gr E SS ioN
R E G R E S S I O N
Yu shud e e :
product
coefcient
c o r r E l at i o N
moment
measures
the
correlation
goodness
of
✔
interpret
t
a
✔
and
b
the
in
calculate
a
r
meaning
linear
and
r
of
the
regression
using
parameters
y
=
ax
+
b
technology
s
✔
correlation
can
depending
on
be
strong,
moderate
or
weak
✔
how
close
it
is
to
a
straight
make
the
distinction
causation
and
positive
or
negative,
depending
on
✔
the
be
to
values
and
and
know
that
correlation
does
not
causation.
gradient
equation
used
for
✔
the
correlation
the
imply
signof
between
line;
of
the
predict
of
Spearman’s
x
regression
values
within
rank
of
the
y
line
from
range
correlation
of
of
y
on
x
values
data
coefcient
can
of
x,
given
r
is
s
used
can
to
be
measure
modelled
decreasing
✔
if
how
data
by
closely
an
two
variables
increasing
or
a
function
items
are
equal
when
calculating
r
,
s
the
ranks
should
be
averaged.
Se dgms
These
help
variables.
are
to
show
For
recorded
whether
example
and
the
the
there
heights
results
is
any
and
shown
on
relationship
weights
the
of
scatter
20
between
children
diagram
two
in
a
class
below.
62
60
58
56
Ne
54
The PMCC (r) can also be used to
52
)gk(
assess how well the data fits a
50
thgieW
linear model.
48
46
44
42
Ne
40
The line of best fit will always pass
38
through the point ( x , y )where
x
the mean of the x values and y
36
is
is
the mean of the y values.
130
There
is
132
an
drawing
you
to
their
134
a
136
138
140
142
144
Height
(cm)
approximately
line
find
height
the
of
best
fit
linear
weight
148
the
weight
follow
a
150
152
relationship
through
approximate
and
146
of
data.
a
similar
154
156
here
The
student
which
line
of
from
justifies
best
their
fit
allows
height
if
relationship.
187
/
4
62
60
58
56
54
52
)gk(
50
thgieW
48
46
44
42
40
38
36
130
132
134
136
138
140
142
Height
For
example,
with
If
a
height
set
they
a
a
of
have
there
One
the
a
is
this
150
linear
(Pearson)
of
lie
we
how
1
150
152
154
estimate
for
would
be
between
57and
close
say
close
to
a
line
with
correlation.
they
a
set
moment
and
148
an
relationship
product
between
146
model
positive
gradient
no
cm
points
strong
measure
values
of
data
negative
If
using
144
(cm)
have
we
of
a
say
data
If
a
the
strong
are
no
are
of
a
student
kg.
gradient
close
negative
is
points
correlation
58
positive
they
there
weight
to
a
we
line
say
with
correlation.
correlation.
to
a
straight
coefficient
(r)
line
which
is
takes
1.
10
10
10
9
9
9
8
8
8
7
7
7
6
6
6
5
5
5
4
4
4
3
3
3
2
2
2
1
1
1
0
0
0
1
2
3
4
5
1
r
=
1
2
r
3
≈
4
− 0.7
5
6
1
2
3
r
≈
4
5
6
0
A high value of the PMCC (r) does not necessarily imply causation.
A comparison of ice cream sales in a town on a given day and the number of
cases of sunburn on that day will probably show a good correlation, but it is
unlikely one is causing the other. Sometimes though a correlation can imply
a causation, for example noticing the significant correlation between the
number of smokers and the number of cases of lung cancer led doctors to
understanding that the increase in one was causing the increase in the other.
188
/
4.6
c o r r E l at i o N
aND
r E Gr E SS ioN
regessn
Ne
If
the
scatter
variables
is
diagram
indicates
appropriate,
eye’
you
can
This
line
is
use
the
then
one
that
a
rather
generated
linear
than
by
model
drawing
the
connecting
a
line
of
the
best
two
fit
‘by
For the straight line y
=
a
+
bx
we
call x the ndependen e
GDC.
and y the dependen e
called
the
least
squares
regression
line
for
y
on
x,
or
just
because it is given in terms of x.
the
regression
Using
the
this
range
There
are
line
for
equation,
of
the
two
y
we
data
on
x.
can
estimate
values
of
y
for
values
of
x
within
(interpolation).
limitations
to
using
the
regression
line
to
estimate
values.
Ne
1
It
would
outside
values
not
this
of
y
be
wise
range
to
of
beyond
assume
the
the
data,
given
this
so
it
line
remains
should
values
of
x
not
the
be
line
used
of
to
best
fit
estimate
(extrapolation).
To find a value of x from a value of
y you would use the regression
line for x on y, but this is not
2
The
regression
line
y
on
x
is
only
the
best
fit
line
for
estimating
required for assessment purposes.
values
of
y
from
estimating
a
values
given
of
x
value
from
of
x,
values
so
of
it
should
not
be
used
for
y.
S aMPlE StUDENt aNS WEr
Seven
adult
between
Their
The
men
their
waist
wanted
Body
sizes,
following
Mass
in
table
to
cm,
see
if
Index
were
shows
there
(BMI)
was
and
recorded
the
a
relationship
their
and
waist
their
BMI
size.
calculated.
results.
Waist (x cm)
58
63
75
82
93
98
105
BMI (y)
19
20
22
23
25
24
26
The
relationship
equation
(a)
(b)
y
=
ax
between
+
x
and
Find
the
value
(ii)
Find
the
correlation
the
can
be
modelled
by
the
regression
b
(i)
Use
y
regression
of
a
and
of
b
coefficient.
equation
to
estimate
the
BMI
of
an
▲ The
adult
work
man
whose
waist
size
is
95
student
clearly
(i)
a
=
0.141
(ii)
r
=
part
0.978
for
(b)
b
=
11.1
y
=
0.141x
y
=
0.141(95)
+
a
24
.495
y
=
24
.5
+
It
is
a
is
1
unit
In
the
is
save
the
x.
be
able
equation
Its
exam
correct
possible
the
value
with
to
gradient
of
by
from
the
GDC
for
substitution
less
on
most
regression
of
95
work
could
and
GDCs
equation
be
no
so
substituted
possibility
of
errors.
24
.5
important
regression
correct
b.
rounding
=
the
the
11.1
the
BMI
got
out
11.1
to
=
directly
and
part
▼ It
y
and
set
cm.
answers
(a)
has
of
y
=
the
units
interpret
the
value
of
a
and
b
in
the
ax + b
line
are
question
to
and
the
gives
number
above
the
the
of
value
increase
units
of
a
of
y
gives
in y
‘per ’
the
for
an
unit
of
increase
increase
of
x.
in
BMI
2
(measured
in
kg m
)
for
an
increase
2
measurement,
so
0.141
kg m
of
one
centimetre
in
the
waist
1
cm
189
/
4
b
is
the
value
of
y
when
involve
extrapolation
context
given.
be
the
BMI
In
for
a
the
x
is
and
so
exam
man
0.
with
Be
aware
that
this
value
might
question
a
waist
above
estimating
have
the
no
value
measurement
of
will
often
meaning
of
0
it
b
=
11.1
in
the
would
cm.
Spemn’s n en effen
Another
measure
coefficient.
14
variable
This
of
is
a
increases
correlation
measure
as
the
is
of
Spearman’s
the
degree
independent
to
rank
correlation
which
variable
the
dependent
increases.
12
For
the
data
in
for
this
data
is
the
diagram
on
the
left,
the
product
moment
correlation
10
8
y
only
not
seem
to
follow
the
value
of
y
moderate
a
straight
increases
(0.78)
and
misleading
line.
But
there
is
whenever
the
value
a
of
as
clear
x
the
data
does
relationship
does.
In
this
in
case
that
the
6
Spearman’s
rank
correlation
coefficient
(r
)
will
have
a
value
of
r
=
1.
If
s
s
4
the
value
of
y
always
decreases
as
the
value
of x
increases
then
r
=
1
s
1
Spearman’s
product
rank
correlation
coefficient
is
found
moment
correlation
coefficient
for
the
by
calculating
ranks
of
the
the
data.
0
1
4
6
8
x
assessmen p
For
example,
top
row
values
have
and
each
in
their
is
so
are
each
correlation coefficient is best in
otherwise
table
ranks
given
received.
You may be asked to decide which
the
For
given
have
the
below
in
the
rank
of
received
value
second
average
example,
a
the
the
3.5.
of
row.
the
two
5,
6
the
7s
and
they
would
the
7,
data
are
Whenever
ranks
Similarly
,
ranks
of
have
all
there
would
three
are
shown
had
5s,
are
a
the
equal
otherwise
ranks
which
given
in
3
and
would
rank
of
4
have
6.
which situation.
value
5
7
5
8
12
5
4
7
rank
6
3.5
6
2
1
6
8
3.5
Remember that the product
moment correlation coefficient is
used to find how well the data fits
Exmpe 4.6.1
to a sgh line.
An additional point is that the
Kathryn
PMCC can be easily distor ted by
so
an outlier to give a high correlation.
without
If one exists it should be excluded
below,
she
believes
organizes
that
a
knowing
along
people
test.
their
with
the
often
pay
too
Students
are
cost.
overall
costs
The
of
the
asked
much
to
for
rank
ranks
are
perfumes
six
perfumes
given
in
the
table
perfume.
or Spearman's correlation should
A
B
C
D
E
F
Rank
4
6
3
2
5
1
Cost ($)
18
18
120
74
68
35
be used.
(a)
Explain
why
moment
(b)
Find
the
it
would
correlation
Spearman’s
be
inappropriate
to
use
the
product
coefficient.
rank
correlation
coefficient
and
comment
on
result.
Solution
(a)
Because
table,
ranks
not
are
given
in
the
values.
It
to
would
say
that
looking
(b)
A
B
C
D
E
F
4
6
3
2
5
1
5.5
5.5
1
2
3
4
linear
The
r
=
also
be
justifiable
Kathryn
specifically
is
not
for
a
relationship.
cost
ranks
5
and
6
are
0.435
s
shared
There
is
only
correlation
for
a
A
and
B
moderate
between
preference
by
and
the
rankings
price.
190
/
4 .7
4 . 7
S TAT I S T I C A L
simple
and
random,
stratied
Yu shud e e :
convenience,
sampling
systematic,
quota
✔
methods
determine
the
terms:
null
and
alternative
whether
distributed
the
✔
t E S tS
T E S T S
Yu shud nw:
✔
S tat i S t i c a l
box
by
and
the
data
may
consideration
of
be
the
normally
symmetry
of
whiskers
hypothesis,
2
signicance
level,
how
the
p-value
and
critical
value
✔
carry
out
a
interpret
χ
the
test
using
technology
and
result
2
✔
for
to
nd
degrees
of
freedom
for
χ
a
test
independence
✔
carry
out
a
t-test
between
two
when
the
for
testing
means,
using
the
difference
technology
and
2
✔
how
to
nd
goodness
the
of
t
expected
test,
for
values
for
uniform,
a
χ
binomial
the
normal
all
data
is
given,
and
interpret
and
result.
distributions
2
✔
the
degrees
test
✔
the
is
always
the
the
is
means
Up
to
point
for
example:
the
However,
distribution
a
sample
population
the
a
goodness
one-tailed
had
a
t-test,
and
to
of
t
a
from
to
large
the
and,
to
statistics
out,
been
want
that
be
the
in
equal
pooled.
everyone
find
not
very
is
assumed
used
of
namely
distributed
are
data
have
try
often
of
for
marks
to
that
we
a
normally
we
the
data
population
take
between
variances
which
this
χ
a
1
assumptions
exams,
for
t-test
population
using
freedom
n
difference
two-tailed
✔
of
or
in
to
directly
find
the
the
describe
class.
infer,
entire
We
have
anything
populations,
not
about
been
a
wider
tested.
mean,
population.
population
to
and
In
standard
these
infer
deviation
cases
from
the
we
or
need
sample
the
to
Ne
the
statistics.
A sample that is unrepresentative
The
the
first
requirement
sample
is
in
order
representative
to
of
do
the
this
successfully
population
it
is
comes
to
ensure
that
is often referred to as sed
from.
Smpng
A random
an
equal
by
giving
sample
chance
of
number
Another
method
random
one
being
everyone
random
many
is
in
in
selected.
the
everyone
It
could
population
generator
is
which
to
select
systematic
in
be
the
found,
a
number
the
sample.
sampling.
population
This
and
saves
for
has
example,
then
using
having
to
a
generate
numbers.
In systematic sampling, a star ting number is selected randomly and a fixed
distance is chosen, normally the nearest integer to the size of the population
divided by the sample size. The sample begins with the first person chosen
and then each subsequent person is the fixed distance down the list from the
previous one. If you get to the end of the list you go back to the beginning of
the list.
191
/
4
For
example,
if
you
want
Ne
to
collect
a
sample
of
20
from
a
school
of
242,
242
you
would
=
calculate
12.1
and
so
would
choose
every
12th
person.
20
Because you star t at a random
Not
all
samples
will
be
random.
If
you
are
doing
a
survey
about
point, everyone has an equal
cafeteria
food
it
would
be
easy
to
ask
the
first
10
people
in
the
queue,
chance of being selected.
but
not
everyone
students
around
may
year
sampling
has
because
Suppose
sample
not
in
of
go
the
a
to
so
have
the
an
equal
cafeteria,
not
every
advantage
of
chance
of
lunchtimes
year
being
group
easy
to
is
being
might
asked.
do,
and
asked.
be
based
This
as
Some
type
such
is
of
called
sampling
a
12
matter
not
groups
convenience
Just
would
sample
school
could
but
is
random
there
have
when
if
are
10
equal
boys
does
does
it
mean
number
and
is
not
2
best
of
girls.
use
that
boys
For
it
is
and
some
stratified
representative.
girls. A random
surveys
this
might
sampling
In a stratified sample the population is divided into different groups and,
assessmen p
within each group, the sample is selected ndmy. In a propor tional
If you are asked in an exam to
stratified sample the propor tion of each group in the sample is the same
do either a stratified or quota
as the propor tion in the population as a whole.
sample, you should assume that
the size of the classes should
A common
alternative
to
stratified
sampling
is
quota
sampling.
be propor tional to those in the
overall population, even if this
is not explicitly mentioned in
In quota sampling, a population is divided into different groups and then
the question.
nenene smpng is used to select the number required in each group.
An
example
town
5
centre,
men
over
everyone
because
of
quota
in
which
30
is
years
equally
those
who
representative
of
sampling
the
old.
researcher
This
likely
to
choose
all
would
to
people
in
in
will
the
speak
survey
to
not
town
with
their
a
needs
sample
be
be
carried
speak
be
for
random,
centre,
the
to,
out
in
a
example,
as
not
and
might
be
biased
researcher
might
not
be
class.
Exmpe 4.7
.1
A large
with
to
the
100
they
(a)
service
it
randomly
are
firm
offers.
selected
the
firm
has
in
France,
find
the
in
the
few
the
form.
(b)
Write
of
the
There
survey
firm
clients,
how
decides
to
stratified
satisfied
send
by
the
a
its
clients
are
questionnaire
country
in
which
2050
clients
number
of
in
total
clients
and
from
of
these
France
65
that
are
based
should
be
to
is
ask
the
no
respond
the
next
type
of
to
100
the
clients
sampling
consideration
survey
of
that
used
the
so
the
they
in
the
country
firm
visit
instructs
to
complete
following
in
which
cases:
the
clients
based.
The
in
clients
down
are
(ii)
The
to
sample.
employees
(i)
wants
based.
Given
V
ery
its
accounting
numbers
part
surveyed
from
each
country
are
as
calculated
(a).
192
/
4 .7
S tat i S t i c a l
t E S tS
Solution
65
The
(a)
× 100
=
3.17
≈
formula
is
3
2050
number
in
the
class
× sample
number
(b)
(i)
convenience
(ii)
quota
in
Always
the
size
population
give
the
answer
on
sample
as
a
whole
number
Ss ess
There
the
many
population.
just
three;
same
to
are
to
mean,
see
if
a
tests
In
see
to
distribution
(a
the
if
if
can
be
Standard
two
see
sample
that
Level
normally
two
could
done
goodness
of
come
fit
MAI
course
distributed
categories
have
a
are
to
you
a
of
population
more
need
populations
independent
from
learn
to
have
each
with
about
know
the
other
a
and
given
test).
In any test, the null hypothesis (H
) is the initial belief that will be kept unless
0
there is strong evidence to reject it. It must be something that can be tested,
for example: the population means are the same or the two variables are
independent.
The alternative hypothesis (H
) is the one that will be accepted if there is
1
significant evidence to reject the null hypothesis.
Each
test
will
percentage,
the
null
the
values
have
for
a
example
hypothesis
in
hypothesis
the
is
significance
will
a
be
sample
true
is
less
5%
level,
significance
rejected
(or
more
than
normally
5%
if
the
level.
written
This
probability
extreme
values)
as
means
of
a
that
obtaining
when
the
null
(0.05)
Ne
the
es
The measure used to judge how far
2
The
χ
test
or
chi-squared
test
(pronounced
‘kai’
squared)
looks
at
the observed values ( f
)
o
how
far
a
sample
hypothesis.
If
it
is
is
from
too
far
what
would
away
then
be
the
predicted
null
under
hypothesis
is
the
hull
rejected.
are from the expected values
(f
) is often written as
e
n
2
⎛
2
χ
The values in the sample are referred to as the se ed values, and the values
( f
f
o
=
calc
∑⎜
⎝
)
⎞
e
⎟
f
but this is not
⎠
e
i =1
predicted under the null hypothesis are called the expeed ues.
required for the exam.
2
There
are
two
ways
to
decide
the
outcome
of
a
χ
test.
Ne
The critical value will depend on
2
The first is to consider the value of
χ
calc
given by your GDC and compare it with
the degees f feedm of the
2
a critical value
(χ
)
crit
for the significance level under consideration. If required
test. In the SL MAI course this is
the critical value for the test will always be given in the question.
the number of cells or groups,
2
If
χ
2
> 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
times
at
the
5%
the
die
is
and
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
when a die is rolled.
The
die
is
not
biased
The
die
is
biased
The
hypotheses
The
null
must
always
be
stated.
0
there are 6 possible outcomes
H
:
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 <
0.05
is
p-value with the significance level.
not
fair.
is
the
same.
Exmpe 4.7
.3
(a)
Given
and
(i)
Phil
a
X
follows
standard
P(8
<
X
believes
normally
of
size
6
in
of
2.4
the
cm.
Length, x (cm)
of
2.4,
P(X
length
(X)
with
mean
To
a
test
randomly
different
distribution
(ii)
< 10)
that
150
normal
deviation
distributed
deviation
length
a
of
this
as
mean
of
6.2
> 10)
pebbles
of
6.2
pebbles
shown
a
find:
from
cm
hypothesis
selected
classes
with
in
beach
a
standard
and
he
and
the
a
measures
records
table
are
the
their
below.
x ≤ 2.0
2 < x ≤ 4
4 < x ≤ 6
6 < x ≤ 8
8 < x ≤ 10
x > 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
at
the
(a)
hypotheses.
to
find
the
values
of a
and
b
2
is
a
χ
sufficient
degrees
of
test
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
<
X
< 10) =
0.1699... ≈
These
0.170
P
X
> 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
=
0.0569 >
there
is
reject
H
The
sample
size.
for
in
you
the
part
use
the
full
probability
(a)
0.05
insufficient
:
=
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
given
test
for
in
contingency
a
independence
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
down
(c)
Use
your
this
test.
the
number
of
degrees
of
freedom
for
this
test.
test. This can normally be read off
2
graphic
display
calculator
to
find
the
χ
the GDC but otherwise it can be
statistic
for
calculated for an n × m table as
(m − 1)(n − 1).
(d)
State
the
conclusion
for
this
test.
Give
a
reason
for
your
answer.
195
/
4
▲ Correct,
calculated
many
of
picked
is
3
With
could
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.
▼ It
sf.
might
incorrectly
GDCs
degrees
be
it
18.96
19.0
shown
a
2
loss
of
close
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
method
this
the
so
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
If
symmetrical
this
is
to
look
at
a
box
and
whisker
plot.
be quite strong evidence for a
it
is
fairly
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
valid
when
result
of
the
test
might
the
revision
not
be
Did not take
44
32
45
79
48
69
50
35
deciding
if
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
<
sufficient
This
is
evidence
at
the
who
did
level
that
the
population
mean
of
those
to
reject
the
revision
course
is
higher
is
a
than
for
10%
those
significance
saying
0.1
who
did
not.
Hence
it
one-tailed
test.
H
0
(b)
The
sample
there
one
might
group
was
not
randomly
be
other
factors
did
better
than
to
the
selected
explain
so
why
The
test
only
populations
what
other.
has
be
might
sure
it
eliminate
not
provided
having
cause
was
all
the
it.
If
the
the
revision
other
evidence
same
for
the
two
mean,
but
not
teacher
course
wanted
he
needs
to
to
factors.
assessmen p
In exams it is always assumed that the standard deviations of both
populations are equal. This means you must always choose the ped
option on your GDC.
196
/
4.8
4 . 8
S A M P L I N G
the
✔
how
a
✔
denition
to
obtain
validity
section
In
this
In
some
4.7
cases
just
and
validity
valid
scale,
directly
are
but
data
from
✔
select
relevant
✔
choose
✔
use
variables
relevant
and
from
many
appropriate
variables
data
to
analyse
valid
and
the
what
collection
to
assess
method
is
whether
a
data
reliable/valid.
methods
is
length
selecting
systematic,
how
data
required;
of
for
quota
a
sample
and
obtained
from
for
example
if
employees’
journey
to
a
stratified.
is
a
from
the
firm
work
sample.
wanted
then
they
question.
their
might
if
ask
which
the
the
choice
their
could
is
wants
employees
employees
method
firm
to
discover
how
satisfied
the
work.
multiple
questions
Whichever
reliability
correlation
validity
.
consider
difficult
from
Alternatively
,
the
with
they
for
convenience,
clear
that
provide
follow-up
criterion
shall
is
more
case
or
we
it
tests
considered
about
ask
employees
this
for
we
section
becomes
In
and
parallel
including
information
It
reliability
reliable
and
tests
population,
need
Yu shud e e :
sample
test-retest
In
of
(aHl)
( A H L )
Yu shud nw:
✔
S a M PliN G
used,
of
the
also
be
you
rank
questions
level
in
to
their
which
satisfaction
sample
satisfaction
do
not
could
could
be
ask
be
on
a
them
inferred.
interviewed
so
asked.
need
your
data
collection
to
be
both
reliable
A d es is one that measures
the quality that it claims to be
A valid
test
is
one
that
measures
the
quality
that
it
claims
to
be
measuring.
measuring.
A valid
on
will
calculus
measure
If
the
the
how
test
students’
hope
it
range
also
of
calculus
is
is
way
of
test
on
samples.
similar
should
You
be
T
est-retest.
test
will
know
The
validity
.
a
valid
student
how
In
test
the
of
the
the
be
the
might
results
do
each
not
it
to
but
be
to
test
is
test
to
predict
teacher
tests
content
regard
achieve,
might
teacher
words
by
with
they
similar
purpose
then
other
just
might
A reliable
by
grade,
test
well
the
A Mathematics
calculus.
used
represented
produce
is
in
would
for
of
a
the
test.
predicting
unlikely
their
time
it
valid,
wider
the
to
be
History
is
but
The
final
a
exam.
performed
all
valid
tests
A ee test is one that will
produce similar results when used
with similar respondents.
two
test
disadvantage
might
be
to
if
validity.
reliable.
should
One
going
measuring
A reliable
validity
know
probably
a
content
Mathematics
than
grade
have
students
criterion
qualities
test
the
IB
can
content
also
final
has
Mathematics
valid
have
well
result
test
is
is
tests
for
given
that
influenced
twice
the
by
reliability:
to
the
responses
the
same
in
responses
the
in
group.
second
the
first.
197
/
4
Parallel
a
forms.
similar
(parallel)
respondents
required
two
In
in
tests
both
Instead
have
one
not
making
are
of
cases,
correlation
is
the
given.
the
tests
the
respondents
This
test
and
has
the
before
it
is
the
same
advantage
but
difficult
more
to
test
that
work
ensure
the
is
the
difficulty
.
test
between
giving
done
two
equal
if
of
is
the
reliable
two
sets
then
of
there
should
be
a
high
results.
Exmpe 4.8.1
A school
school
introduces
to
do
ranked
on
rank
1
of
These
are
the
how
IB
shown
in
they
they
and
the
entrance
diploma.
well
means
ranks
an
the
did
test
Eight
for
pupils
performed
best
students’
in
students
the
final
in
are
the
coming
selected
entrance
into
and
test,
the
are
where
test.
points
total
for
the
IB
diploma
table.
Student
A
B
C
D
E
F
G
H
Entrance test
8
7
6
5
4
3
2
1
IB points total
28
30
32
30
35
29
36
42
By
finding
valid
You
for
may
an
appropriate
predicting
assume
evidence
for
a
a
a
the
statistic,
final
points
correlation
valid
state
whether
or
not
the
test
is
total.
coefficient
greater
than
0.7
will
be
test.
Solution
Ranks
for
points
total
The
are:
question
specify
A
B
C
D
E
F
G
H
8
5.5
4
5.5
3
7
2
1
PMCC
r
≈
0.719
does
whether
or
to
use
Spearman’s.
Because
ranks
and
are
you
not
are
given
looking
for
s
an
There
is
sufficient
evidence
that
association
than
entrance
test
is
a
valid
predictor
a
linear
points
relationship
of
Spearman’s
final
rather
the
should
be
used.
total.
198
/
4.9
4 . 9
N O N - L I N E A R
R E G R E S S I O N
Yu shud nw:
✔
the
meaning
regression
of
and
the
the
sum
of
terms
least
squares,
✔
nd
residual
square
r E Gr E SS ioN
(aHl)
( A H L )
Yu shud e e :
least
squares
technology
for
exponential,
✔
NoN - liN E a r
residuals
(SS
)
and
regression
linear,
power
curves
quadratic,
and
sine
using
cubic,
regression
the
res
2
coefcient
of
t
for
a
of
determination
(R
2
)
are
measures
✔
evaluate
R
using
✔
interpret
the
technology
model
goodness
of
t
of
a
curve
using
the
2
✔
that
many
model
by
and
itself,
is
different
In
section
factors
affect
the
the
coefcient
not
a
good
validity
of
way
of
a
value
determination,
to
decide
for
of
the
R
,
as
well
as
the
graph
and
the
context
model.
between
models.
4.6
you
considered
how
well
a
straight
line
the
fit
fitted
a
set
of
assessmen p
data
and,
squares
when
linear
appropriate,
how
to
find
best
line
using
least
regression.
In exams you will need to be
In
this
section
we
look
at
finding
the
best
fit
curves
for
a
set
of
able to fit linear, quadratic,
data.
cubic, exponential, power and
When
fitting
a
curve,
the
GDC
considers
the
residuals
which
are
the
sine functions.
distances
black
from
vertical
the
points
lines
from
to
the
the
proposed
points A,
B,
curve.
C,
D
They
and
E
in
are
the
shown
as
diagram
the
below
.
6
C
B
5
4
E
D
3
A
2
0
The
best
squares
fit
of
1
curve
the
2
is
3
4
described
residuals,
5
as
hence
the
the
6
one
7
8
with
name least
9
the
10
smallest
squares
11
sum
of
regression
the
curve
2
The
coefficient
of
determination
(R
)
is
a
measure
2
fits
a
set
giving
of
the
predicted
points.
R
=
proportion
from
the
of
how
well
a
curve
2
1
implies
of
a
variation
independent
perfect
in
the
fit.
R
can
dependent
be
thought
variable
of
that
as
can
be
variable.
Ne
When
deciding
consider
on
various
the
form
of
the
appropriate
curve
you
should
factors:
When considering a linear
2
1
The
context:
would
have
2
The
or
3
If
you
Is
maximum
hoping
to
other
function
expect
purpose:
all
the
it
or
Are
use
be
to
just
it
have
an
increasing
function, R
asymptote,
or
decreasing
or
to
fit
is equal to the square
of the product moment correlation
coefficient.
points?
trying
make
have
does
continuously
minimum
you
it
factors
to
periodic,
a
curve
as
closely
as
possible
predictions?
been
considered,
the
value
of
the
coefficient
This links with the section on the
of
determination
can
be
used
to
compare
how
well
appropriate
modeling process, section 2.3.
curves
fit
the
points.
199
/
4
Exmpe 4.9.1
An
oven
table
is
left
to
cool
and
its
temperature
(T °C)
t
minutes
after
it
has
been
turned
off
are
recorded
in
the
shown.
t minutes
T °C
0
5
10
15
20
25
100
78
58
41
30
22
bt
(a)
Find
the
least
squares
(b)
Give
the
value
of
the
exponential
coefficient
regression
of
line
determination
for
the
from
curve
your
in
the
GDC
form T
for
this
=
ae
data
and
comment
on
thevalue.
(c)
Give
one
reason
why
the
exponential
model
proposed
is
unlikely
to
be
a
valid
model
in
the
contextgiven.
Solution
0.0616 t
(a)
T
=
t
An
104e
equivalent
One
or
both
directly
To
of
from
convert
form
to
is
these
T
=
104
forms
the
GDC.
the
base
e
×
can
form
(0.9402...)
be
obtained
write
as
t
ln( 0.9402...)
T
=
104 ×
(e
0.0616 t
≈ 104e
)
2
(b)
0.998,
this
implies
a
very
good
fit
to
the
A high
data.
value
function
(c)
Because
imply
the
the
asymptote
is
temperature
eventually
reach
0˚
at T
inside
=
0
the
which
oven
is
for
the
R
does
best
not
model
necessarily
for
the
mean
the
data.
would
will
C.
assessmen p
Sometimes
it
may
be
necessary
to
decide
which
of
two
given
curves
If your GDC only gives one of these
are
the
best
fit.
In
this
case
you
would
calculate
the
sum
of
the
square
forms, make sure you know how
residuals
yourself,
and
the
curve
with
the
smallest
total
is
the
one
that
to conver t to the alternative form
fits
the
best.
in case this form is required in
the exam.
Exmpe 4.9.2
(a)
Find
(1,
The
the
5),
sum
(2,
of
least
7),
squares
(4,
square
7)
and
quadratic
(9,
11)
residuals
for
proposed
of
this
regression
curve
is
curve
for
the
following
four
points:
1.2
0.5
A second
(b)
By
model
is
considering
the
sum
of
the
the
form
square
y
=
3x
residuals
+ 2
for
the
new
model,
determine
which
curve
is
the
best
fit.
Solution
2
(a)
y
=
(b)
(5
0.009 61x
This
+ 0.585 x +
is
obtained
directly
from
the
GDC.
4.92
0.5
2
5)
2
+ (7
2
0 + 0.757
6.24...)
2
+
(7
8)
2
+
(11
The
values
2,
9
y
=
3x
+ 2
are
calculated
for
x
=
1,
4,
and
subtracted
from
the
y-coordinates
of
the
2
+ 1
+ 0
=
1.57
observed
The
of
11)
quadratic
curve
is
the
best
points
to
find
the
residuals.
fit.
200
/
4 . 10
4 . 1 0
coNFiDENcE
C O N F I D E N C E
H Y P O T H E S I S
iN t E r va l S
aND
I N T E R VA LS
T E S T I N G
Yu shud nw:
H Y P ot H E S i S
tE StiNG
(aHl)
A N D
( A H L )
Yu shud e e :
2
✔
that
to
1
in
be
χ
a
test
greater
from
the
all
than
expected
5
degrees
and
of
frequencies
you
should
freedom
for
need
✔
nd
subtract
normally
when
a
testing
condence
is
used
when
for
is
a
population
interval
when
σ
distributed
for
the
mean
of
a
population
test
for
a
population
mean
when
the
population
estimated
is
✔
intervals
every
✔
parameter
condence
σ
is
the
mean
normal
known
and
or
nding
normally
enough
for
distributed
the
central
or
the
limit
sample
theorem
size
to
large
apply
distribution
the
✔
perform
✔
test
a
paired
sample
t-test
t-distribution
unknown
for
proportion
using
the
binomialdistribution
✔
a
type
when
I
it
error
is
is
to
reject
the
null
hypothesis
✔
true
test
for
population
mean
using
the
Poissondistribution
✔
a
type
II
error
hypothesis
is
to
when
fail
it
is
to
reject
not
the
null
✔
true.
use
technology
that
the
to
test
population
the
hypothesis
product
ρ
correlationcoefcient
is
0
moment
for
bivariate
normaldistributions
✔
nd
the
type
and
following
variance,
the
I
type
errors
distributions:
Poisson
es esed
II
and
for
tests
normal
involving
with
known
binomial.
Ne
2
In
the
Higher
Level
course
there
are
two
refinements
for
the
χ
test
that
This is because when working
you
need
to
be
aware
2
of.
out the
χ
statistic you divide
by the expected values. If these
2
In both, the
χ
2
goodness of fit test and the
χ
are too small then they distor t
test for independence the
the distribution.
expected values need to be greater than five.
If
you
have
adjacent
an
expected
value
less
than
5
you
need
to
combine
two
groups.
assessmen p
Your calculator will often not
Exmpe 4.10.1
automatically display the
Yenni
was
conducting
favourite
genre
shown
the
in
of
film
a
survey
was
contingency
to
see
whether
independent
table
of
or
age.
not
Her
a
person’s
results
are
expected value matrix, so make
sure you always check it.
below.
Favourite genre
Age
Action
Comedy
Horror
Thriller
10–20
13
7
7
3
21–30
16
8
5
9
31–40
11
12
5
8
Over 40
8
7
6
7
Determine
null
whether
hypothesis
that
or
not
there
favourite
is
sufficient
genre
and
age
evidence
are
to
reject
the
independent.
201
/
4
Solution
The
expected
horror
two
is
4.88
number
so
we
of
over
must
40s
like
Notice
bottom
the
who
combine
the
that
it
does
not
observed
table
with
an
that
entry
there
less
is
a
than
cell
in
5.
rows.
You
it
Favourite genre
could
makes
Action
Comedy
Horror
Independent
10–20
13
7
7
3
21–30
16
8
5
9
Over 30
19
19
11
15
Thriller
Note
=
0.496
so
insufficient
evidence
to
films
the
any
hypothesis
favourite
genre
of
at
either
film
and
5
or
age
even
are
to
join
and
would
question
of
the
two
or
rows
Comedy
make
does
level.
rows
In
a
not
this
standard
columns.
as
or
case
values,
not
is
a
and
pairing!
particular
not
so
clear
natural
for
it
is
Horror
more
ask
it
Here
an
significant
appropriate
reject
conclusion
null
adjacent
Horror
significance
for
p-value
join
sense
whether
Age
the
matter
10%
can
be
drawn.
that
independent.
If estimating population parameters from the data you subtract 1 from the number
of degrees of freedom for each parameter estimated.
For
a
example,
normal
if
you
are
distribution
and
standard
best
choices
population
testing
you
deviation
would
mean
be
and
so
to
s
In
where
the
n
is
Example
This
of
is
case
as
and
of
the
one
the
an
not
know
work
sample
out
your
an
for
has
for
expected
as
the
data
estimate
the
mean
estimate
degrees
an
come
the
mean
values.
estimate
population
from
for
The
the
standard
the
as
number
of
freedom
would
be
n
1
2
=
n
3
cells.
illustrates
trickiest
can
to
or
1
the
number
4.10.2
the
trials
this
need
you
use
n
deviation.
will
whether
it
of
technique
is
easy
times
to
the
for
be
a
binomial
confused
trial
was
by
distribution.
the
number
repeated!
Exmpe 4.10.2
The
owner
quality
of
of
of
a
garden
seeds
germination,
from
in
germinate.
To
seeds
each
in
it
from
each
came
row.
If
from,
binomial
test
he
a
this
rate
Number germinating
Frequency
a
of
his
test
are
shown
30
14
number
with
appropriate
binomial
seeds
number
distribution
evidence
of
to
of
seeds
the
test
at
suggest
germinating
that
same
a
5%
that
per
would
mean
as
the
rate
of
expected
the
four
to
in
batch
follow
the
sample
found
in
level
germination
four.
for
part
to
a
table.
of
germinate
significance
rate
germinate
germinating
25
expected
follow
number
good
plants
of
20
the
is
independent
11
Find
there
results
many
4
(b)
if
The
the
is
a
than
owner
how
variable
often
fewer
centre
records
germination
expect
that
is
3
mean
an
and
there
a
2
the
Perform
garden
receiving
1
Find
(c)
seems
is
0
(a)
binomial
of
he
Though
it
the
batches
would
distribution.
batches
theory
100
suspects
company
.
some
of
the
centre
a
(a).
see
does
not
distribution.
202
/
4 . 10
coNFiDENcE
iN t E r va l S
aND
H Y P ot H E S i S
tE StiNG
(aHl)
Solution
(a)
x
=
(b)
For
This
2.16
a
binomial
distribution
x
=
is
Note:
np
obtained
the
n
directly
here
is
the
from
the
number
of
GDC.
seeds
in
eachtrial.
⇒ 2.16
=
4p
⇒
p
=
0.54
The
x
0
1
2
3
the
0.0447
P(X = x)
0.210
0.370
estimate
0.290
total
4.48
21.0
37.0
29.0
by
The
seeds
follow
a
binomial
The
seeds
do
follow
be
calculated
by
finding
of
seeds
that
germinate
and
8.50
probabilities
multiplied
:
also
400.
distribution
H
can
0.0850
The
(c)
p
number
dividing
Expected value
for
4
distribution.
calculated
function
by
case
are
In
this
of
parameter.
100
we
to
on
the
obtain
GDC
the
are
not
the
expected
using
testing
the
and
then
expected
for
a
binomial
values.
particular
value
0
H
:
not
a
binomial
1
distribution.
x
Because
0 or 1
2
3
4
25.5
37.0
29.0
8.5
two
Expected value
As
Number
of
degrees
of
freedom
=
4
1
1 =
=
evidence
0.0131<
0.05
to
the
reject
germination
follows
so
there
claim
a
is
that
the
value
need
of
p
to
has
be
values
is
less
than
5,
combined.
been
estimated
then
the
2
of
freedom
reduce
by
1.
sufficient
the
binomial
of
columns
degrees
p-value
one
rate
of
distribution.
the z-es nd he t-es
You
have
looked
We
at
first
already
met
in
detail.
consider
population
data
more
could
Consider
with
a
given
come
sample
of
normal
distribution
and
X
let
be
the
t-test
whether
a
have
the
a
size
with
n
a
section
single
mean,
from
in
set
of
rather
data
than
populations
taken
mean
distribution
of
from
of
the
μ
4.7;
at
could
testing
with
a
Higher
the
have
same
a
sample
is
from
two
sets
a
of
mean.
which
standard
it
come
whether
population
and
Level
follows
deviation
of
a
σ,
mean.
2
σ
⎛
From
section
4.3
we
know
that
X
~
⎜
⎟
n
⎝
Under
the
null
would
expect
hypothesis
that
the
⎞
µ,
N
⎠
Ne
population
mean
is
equal
to μ
we
2
σ
to
X
be
close
to
this
value.
the
null
hypothesis
Because the variance for
X
is
n
We
would
assumed
of
X
(for
therefore
value
being
example
hypothesis
fell
this
in
the
μ.
of
far
5%).
that
‘Too
away
From
the
critical
reject
far ’
by
the
is
chance
is
diagram
population
regions
chosen
mean
to
less
mean
than
below
is
if
we
equal
it
was
that
the
the
μ
far
from
at
reject
the
the
probability
significance
would
to
too
5%
level
the
would expect the sample mean to
be to μ
null
level
if
x
shown.
0.05
0.025
0.05
0.025
μ
μ
two-tailed
the larger the sample the closer we
test,
H
:
1
mean
≠
μ
one-tailed
test,
μ
H
:
1
mean
>
μ
one-tailed
test,
H
:
mean
<
μ
1
203
/
4
If
the
population
population
sample
is
normally
standard
mean
is
also
distributed
deviation
normally
in
the
and
we
question,
distributed
and
are
then
we
given
we
the
can
perform
assume
the
a z-test.
Even if we do not know that the
Exmpe 4.10.3
population is normally distributed, if
the sample size is large enough we
A sample
2
σ
⎛
can still assume that
X
~N
⎜
µ
distributed
⎟
n
⎝
of
size
10
is
taken
from
a
population
which
is
normally
⎞
with
a
standard
deviation
of
2.
at
the
⎠
by the central limit theorem, see
The
section 4.3.
the
hypotheses:
H
The
population
mean
μ
is
equal
The
population
mean
μ
is
greater
:
sample
mean
will
be
used
to
test,
to
5%
significance
level,
5
0
H
:
than
5
1
Ne
(a)
Find
(b)
It
the
critical
region
for
this
test.
Here you are given the population
is
found
that
x
=
6.5
standard deviation, so you use the
State
whether
Find
the
or
not
the
null
hypothesis
should
be
rejected
z-test.
(c)
p-value
and
verify
this
confirms
the
decision
we
will
in
part
(b).
Solution
(a)
We
require
that
under
H
This
is
a
one-tailed
test
and
reject
H
0
value
P(X
>
a) =
=
the
of
x
is
too
large.
0.05
To
a
if
0
find
a
we
use
the
inverse
normal
function
on
6.04
2
the
GDC
with
μ
=
5
σ
and
=
10
The
critical
region
is
X
> 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
<
0.05
so
reject
is
done
using
the
z-test
on
the
GDC
and
H
entering
0
If
the
population
deviation
of
the
is
is
normally
unknown
sample
statistics
mean
is
and
no
rather
distributed
needs
longer
to
be
than
but
the
data
and
a
below)
population
estimated,
normal
(see
then
t-test
the
standard
distribution
should
be
used.
There are several options for testing for a mean on the GDC so make sure you
Ne
can use all of them.
s
n
is denoted by s
1
on most
x
calculators and is used as the
1
z-test
or
t-test?
Use
the
z-test
when
the
population
standard
estimator for the population
deviation
is
known
and
the
t-test
when
it
has
to
be
estimated
standard deviation, σ . It should
from
the
data
s
(using
n
).
1
not be confused with the standard
2
Are
you
entering
statistics
(x
and
n
for
the
z-test
and
x,
n
and
deviation of the sample, s
n
s
on
which is denoted by
n
for
the
t-test)
or
are
you
entering
the
full
set
of
data
into
1
X
the
calculator?
most calculators.
3
Is
it
a
two-tailed
testing
for
the
hypothesis
or
or
a
mean
one-tailed
being
test?
smaller
And
than
if
that
one-tailed,
given
in
are
the
you
null
larger?
204
/
4 . 10
coNFiDENcE
iN t E r va l S
aND
H Y P ot H E S i S
tE StiNG
(aHl)
Exmpe 4.10.4
Ne
Anders
lives
near
shop
A,
where
they
sell
muffins
advertised
as
If prompted to include the degrees
weighing
100
g.
Anders
is
not
convinced
by
this
so
he
buys
six
of freedom for a t-test, it is n − 1.
muffins
and
Muffin
Weight (g)
(a)
records
their
weights
in
the
A
B
C
D
E
F
95
98
94
100
99
92
Perform
a
test
to
see
whether
there
table:
assessmen p
is
evidence
that
the
mean
Often a question will give you the
weight
of
the
muffins
(μ
)
is
less
than
100
g.
You
may
assume
A
sample standard deviation (s
).
n
the
weights
of
the
muffins
are
normally
distributed.
Don’t forget to conver t this to s
n
Another
them
shop
and
(B)
tests
to
opens
see
nearby
,
whether
so Anders
there
is
buys
evidence
10
muffins
that
the
before entering it into the GDC by
from
mean
n
weight
using s
n
=
s
1
n
n
of
muffins
sold
at
shop
B
(μ
)
is
greater
1
1
μ
than
B
A
(section 4.14 of the formula book).
The
10
muffins
deviation
(b)
of
3
Perform
have
a
mean
weight
of
98
g
and
a
standard
g.
the
test
and
state
whether
there
is
evidence
that
the
assessmen p
mean
weight
weight
in
of
shop
muffins
in
shop
B
is
greater
than
the
mean
A.
Tests will normally be done using
the inbuilt functions on the GDC and
You
may
assume
the
weights
of
muffins
from
shop
B
are
quoting p-values. Critical regions
normally
distributed
and
standard
deviations
of
both
are normally only used when
populations
are
equal.
finding type II errors.
Solution
(a)
H
:
µ
=
Always
100
give
:
µ
section 4.3. In the standard level
As
=
sufficient
:
µ
=
0.0177
<
the
evidence
to
:
µ
estimated
from
the
is
It
one-tailed
as
a
t-test.
you
s
µ
=
A
:
n
B
µ
data
this
is
a
test
are
is
testing
less
needs
to
than
be
whether
the
course the data will always be given
in full. In the higher level course
you might need to enter the correct
statistics instead.
100.
calculated
for
1
B
as
only
the
standard
assessmen p
µ
<
A
1
is
100
shop
H
deviation
reject
mean
0
standard
0.05
A
0
H
hypothesis.
< 100
p-value
(b)
and
A
1
H
null
Two sample t-tests were met in
alternative
H
the
A
0
B
deviation
of
the
sample
s
n
10
s
=
n
was
× 3
=
given.
It is easy to make an error in
10
1
9
entering data so make sure you
For
shop
A
write down any working you do
x
=
96.33...,
s
n
p-value
=
=
3.141...
so you can pick up method
1
0.162
>
Most
0.05
of
So
insufficient
evidence
GDCs
data
and
H
:
0
µ
=
A
not
allow
statistics
so
a
mix
the
marks, even if your final answer
is incorrect.
to
relevant
reject
do
data
for
shop
A needs
µ
B
to
be
collected
entered
option
into
on
first
the
the
and
then
2-sample
t-test
GDC.
205
/
4
Ped smpe es
It
is
so
easy
you
to
miss
may
paired
in
between
be
an
a
‘paired
test’
to
use
the
two-sample
way
,
you
need
tempted
obvious
the
sample
paired
as
to
two
sets
instead
of
data
t-test,
look
but
at
will
if
the
be
the
given
data
is
difference
values.
Exmpe 4.10.5
Five
runners
times
taken
results
are
at
to
a
club
run
shown
spend
400
m.
After
weeks
the
at
camp
an
overseas
they
again
training
time
camp.
themselves
Before
over
going,
400
m.
they
The
record
before
their
and
after
below.
Runner
Time before, T
two
(seconds)
1
2
3
4
5
55.2
60.1
56.8
53.5
59.3
55.0
57.8
57.0
53.4
58.8
1
Time after T
(seconds)
2
Perform
their
a
suitable
times.
You
test
may
at
the
5%
assume
significance
that
the
times
level
are
to
see
if
normally
there
is
evidence
that
the
camp
has
improved
distributed.
Solution
(a)
Let
the
mean
difference
in
times (T
T
1
be
)
µ
Note:
D
2
:
µ
=
0
>
0
:
µ
mean
difference
of
T
<
0
T
2
would
indicate
an
then
1
improvement.
D
1
The
:
the
D
0
H
was
D
H
H
µ
If
µ
1
null
always
Difference
p-value
0.2
=
evidence
that
hypothesis
for
a
paired
sample
test
will
D
the
running
2.3
0.131
to
>
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
the
r
≠
0
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
computer
affects
that
time
spent
playing
games
on
a
Make sure you know where to find
the
marks
obtained
in
class
assessments.
To
test
this test on your GDC. It is likely to
this,
he
selects
they
spend
8
students
and
asks
them
how
much
time
each
week
be in a menu for regression tests
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
is
and
are
see
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
ρ
=
0
reject
asking
at
for
is
correlation
insufficient
H
the
1
significance
whether
or
not
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
However,
construct
If
we
wish
to
test
H
µ
:
µ
=
a
a
result
being
Poisson
if
known
it
is
test
for
the
the
value
of
X
obtained
against
H
µ
µ
0
0
we
could
is
(μ).
mean
~
test
with
clear
Poisson
χ
the
distribution
obtained
or
to
in
our
:
>
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
10
m.
After
improvements
print
the
material
imperfections
To
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
the
conclusion
of
which
(c)
Find
the
p-value
the
gives
reduced.
significance
the
test.
to
level
the
same
for
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
:
<
24
To
find
the
Poisson
X
and
1.98253328235E
enter
the
cumulative
distribution
as
a
function
into
your
GDC
3.44000940596E
2
16
5.62622359137E
2
≤ 15) =
0.0344 ,
hence
read
off
from
the
table
the
last
value
which
2
15
P(X
(b)
region
F1
14
X
critical
1
the
critical
region
has
a
probability
It
more
of
less
than
0.05
is
≤ 15
19
≥ 15
reject
so
there
is
insufficient
evidence
to
H
0
(c)
P(X
≤ 19) =
0.18026... >
0.05
is
p-value
Hence
there
is
insufficient
evidence
to
reject
usual
rather
to
perform
than
finding
the
test
by
finding
the
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
<
Under
0.55
H
the
(assuming
Let
X
be
the
number
of
people
in
the
vote
for
the
≤ 11) =
0.183 >
is
(25,
0.55),
those
in
the
sample
were
randomly
and
independent.)
the
numbers
in
our
too
then
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
to
accept
In
fact
small
the
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
error
or
of
a
of
a
by
definition,
the
continuous.
For
binomial
distribution
they
a
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,
is
Example
probability
I
distribution
Poisson
(see
an
a
the
probability
only
the
of
for
the
calculating
critical
a
region,
type
II
error
given
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
sample
the
(a)
Find
(b)
Write
down
(c)
If
proportion
the
error
critical
is
and
the
random
region
for
H
explain
and
the
of
students
what
:
p
<
or
not
willing
in
0.2 ,
to
the
at
to
introduce
attend.
school
the
5%
She
who
a
homework
decides
would
significance
to
ask
attend
club,
a
the
and
sample
will
of
club.
level.
1
probability
of
be
students
0
Assume
whether
this
independent.
test.
a
in
type
the
would
I
error,
school
mean
to
who
in
3
sf
would
attend
is
15%,
find
the
probability
of
a
type
II
context.
Solution
(a)
Let
X
who
be
the
say
number
they
will
of
people
in
the
sample
a
is
of
attend.
the
X
≤
largest
a
is
entering
Assuming
is
H
true
X
~
B(50,
value
smaller
the
of
than
X
for
0.05.
cumulative
which
This
the
can
binomial
probability
be
found
function
by
into
a
0.2)
0
GDC
H
is
rejected
if
X
is
too
small,
so
the
as
a
function.
critical
0
X
Y
1
region
P(X
a
=
≤
a)
X
<
≤
a
where
0.05
0
1.4E–5
1
1.9E–4
2
0.0013
3
0.0057
4
0.0185
5
0.048
6
0.1034
7
0.1904
8
0.3073
9
0.4437
5
Hence
(b)
is
0.0480
the
critical
region
is
X
≤
5
A type
I
error
is
the
probability
of
rejecting
H
0
when
it
is
true
and
hence
is
the
P(X
≤
5)
=
0.0480
209
/
4
(c)
Given
in
the
X
~
B(50,
critical
0.15)
region
the
probability
of
being
is
Some
GDCs
will
being
in
critical
the
allow
the
direct
calculation
P(X
6|p
region,
≥
=
of
not
0.15)
P(X ≤ 5|p = 0.15) = 0.219 35
so
the
probaility
of
accepting
H
is
0
1
0.219 35
≈
This
would
with
the
was
not
0.781
be
the
probabilty
homework
20%
club
of
even
going
though
ahead
there
support.
S aMPlE StUDENt aNS WEr
A smartphone’s
fully
charged
working.
of
this
assuming
H
is
is
X
not
X
true
the
distribution
on
For
and
the
battery
lives
0.4
hours.
(a)
State
suitable
(b)
Find
the
be
on
used
that
number
before
the
smartphone
normally
region
battery
hours.
random
b
mean
hypotheses
critical
the
the
9.5
a
as
smartphone,
sample
are
defined
average,
each
the
is
claims
conducted
model.
measured
statistic
is
is,
life
can
A company
experiment
▼ The
battery
smartphone
of
battery
is
To
for
a
test
sample
the
of
calculated.
with
two-tailed
testing
b
at
the
of
a
this
battery
distributed
for
life
20
of
hours
a
stops
model
claim,
an
smartphones
life, b
hours,
It
be
can
standard
is
assumed
deviation
test.
5%
significance
level.
so,
It
0
is
then
found
that
this
model
of
smartphone
has
an
average
2
⎛
needed
is
X
~ N
0.4
⎞
20
⎠
9.5,
.
⎝
The
X
>
correct
9.68
answer
and
<
X
for
battery
part
(b)
is
(c)
life
Find
of
the
9.8
hours.
probability
of
making
(a)
H
:
μ
=
9.5
H
0
▲ Correct
a
Type
II
error.
9.32.
method
using
:
μ
≠
9.5
A
the
(b)
n =
20
0.025
inverse
normal
function
to
nd
0.025
the
2
boundaries
for
the
critical
region.
X
~
N(b,
0.4
X
~
N(9.5,
)
2
▼ The
student
has
given
critical
region
region.
and
The
)
the
z
acceptance
0.4
not
correct
the
answer
=
lnVNorm
(0.975,
9.5,
0.4)
=
10.28
= lnVNorm
(0.025,
9.5,
0.4)
=
8.716
1
is
z
2
[9.475,
9.525]
▲ Clearly
setting
out
the
correct
(c)
Critical
region
P(T
ype
=
II)
=
[8.72,
10.3]
P(Accepting
H
/H
0
expression
for
a
type
II
is
true)
1
error.
2
for
▼ Once
for
X
again
and
answer
not
for
the
X
part
is
distribution
used.
(c)
is
The
X~N(9.8,
P(8.72
<
×
<
10.3
0.4
=
)
0.89)
correct
0.0816.
▲ Apart
from
region
between
is
the
wrong
the
distribution
two
critical
the
method
is
correct.
The
acceptance
values.
210
/
4 . 11
4 . 1 1
TRANSITION
CHAINS
the
of
✔
initial
the
the
of
state
✔
the
denition
matrix
represents
✔
the
steady
represents
between
of
a
the
initial
state
✔
each
regular
the
probabilities
represent
the
transition
diagrams
matrix
You
have
as
the
of
an
studied
binomial
event
changes
state
A
the
is
consists
19
we
do
not
on
between
in
✔
calculate
steady
state
chain
probabilities
sequences
in
events.
below.
They
and
rain
of
independent
some
contexts
Consider
each
(R)
two
show
on
19
how
the
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
A
R
R
R
F
F
F
F
F
F
R
F
F
F
R
R
R
R
R
R
B
F
R
F
R
F
F
R
R
F
R
R
R
F
R
R
F
F
R
F
dependent
where
the
on
what
the
independent
trials
where
This
predict
is
an
weather
tomorrow’s
judgement.
not
weather,
This
is
a
the
use
the
was
model
state
example
of
of
a
F
today
.
probability
independent
we
simple
probability
weather
appropriate
with
with
matrices
transition
matrices
and
repeated
matrix
or
long-term
multiplication
by
solving
a
of
system
equations.
for
such
probability
weather
consecutive
4
trials
system
simulated
3
19
by
transition
events,
the
2
of
a
long-term
However,
(F)
in
state.
previous
fair
or
values
1
likely
.
predict
our
of
each
with
sequences
consists
tomorrow
in
contexts
depends
pattern
Sequence
being
distribution.
weather
equally
of
(aHl)
state
Markov
gives
transitions
interpret
of
probabilities
cHaiNS
MARK OV
✔
the
state
AND
Markov
Yu shud e e :
vector
transitioning
aND
(AHL)
system
transition
M at r i c E S
M AT R I C E S
Yu shud nw:
✔
traNSitioN
or
days.
R
Sequence B
of
F
or
R
are
a
weather
system;
events,
because
when
of
the
weather
Markov
today
we
in
chain.
A sequence of states in which the probability of the next state occurring
depends only on the current state is a Markov chain.
Markov
chains
matrices.
to
a
a
Both
future
presented
given
their
simulation
matrix
T
and
its
etats
erutuF
For
=
from
is
R
R
F
⎛
the
the
probabilities
state.
Sequence
probabilities
transition
and
of
A
given
in
the
was
in
transition
states
changing
generated
the
by
following
diagram:
state
0.7
F
0.37
0.7
⎞
F
R
⎜
0.63
⎟
⎝ 0.3
F
0.63 ⎠
0.37
0.3.
to
give
diagrams
0.3
R
example,
today
transition
current
using
Current
T
in
representations
state
computer
transition
are
the
probability
Another
is
way
to
it
is
say
fair
this
tomorrow
is
that
the
given
that
transition
it
is
raining
probability
0.3.
211
/
4
The probabilities in each column of a transition matrix t add to one. Each
column of a transition matrix shows the probabilities of transition from its
current state to each future state.
With
as
transition
shown
in
matrices,
Example
you
can
make
predictions
about
future
states
4.11.1
Exmpe 4.11.1
0.1
Electric
three
cars
are
zones:
rented
Central
by
(C),
ZEcar
Urban
in
(U)
a
large
and
city
Outer
for
commuter
(O).
In
one
journeys
rental
0.2
between
period,
0.5
C
customers
U
0.15
can
pick
up
a
car
from
a
zone
before
09:00
and
return
their
car
to
a
zone
after
0.75
17:00.
The
management
of
ZEcar
construct
this
transition
diagram
to
show
0.2
the
0.7
probabilities
of
a
car
transitioning
from
one
zone
to another
over
one
0.3
rental period.
O
(a)
Represent
columns
rows
the
transition
represent
represent
the
the
probabilities
location
location
of
of
the
an
in
a
transition
electric
car
at
the
matrix T
car
at
the
end
of
a
start
in
of
which
a
day
the
and
the
day
.
2
(b)
Find
T
0.1
2
.
Interpret
the
element
in
the
second
row
and
the
third
column
of T
Solution
Take
(a)
care
question
Current
enoz
T
=
C
erutuF
0.2
0.7 ⎞
rows
and
columns
as
the
requires.
0.5
0.2
⎟
0.15
⎜
⎟
⎜
O
the
O
⎜
U
represent
zone
U
⎛ 0.1
C
to
⎟
⎝ 0.75
0.3
0.1 ⎠
2
⎛ 0.565
2
(b)
T
0.33
0.18
0.34
0.225
⎜
=
Find
⎞
T
with
technology
.
⎟
0.24
The
⎜
⎟
⎜
⎟
⎝ 0.195
0.33
result
meaning
0.595⎠
can
of
be
the
confirmed
product
of
by
the
considering
row
and
the
column
shown:
The
element
that
a
0.225
represents
the
probability
⎛ 0.1
car
beginning
in
O
at
the
start
of
a
day
0.2
0.7
0.7 ⎞ ⎛
⎞
is
⎜
⎟ ⎜
⎟
0.2
found
in
U
at
the
end
of
the
next
day
.
⎜
⎟ ⎜
⎜
The
by
the
transition
⎟
0.1 ⎠ ⎝
of
the
probability
summing
O-C-U,
as
0.3
product
total
Just
⎟
⎟ ⎜
⎝ 0.75
the
O-U-U
probability
0.1
column
that
a
car
and
from
row
shown
transitions
probabilities
and
⎠
that
of
from
the
finds
O
to
the
U
transitions
O-O-U.
O
to
U
after
two
transitions
is
n
0.225,
all
the
This
the
same
process
transition
can
be
can
be
probabilities
applied
to
make
applied
between
long
term
n
times:
each
the
state
elements
after n
of
T
give
transitions.
predictions.
n
Each column of a transition matrix t
shows the probabilities of transition from
its current state to each future state after n transitions.
212
/
4 . 11
For
C,
example,
U
and
⎛
O
begin
respectively
.
with
putting
50,
50
and
information
100
you
cars
can
in
zones
write
down
With
this
state
probability
matrix
which
chosen
car
M at r i c E S
aND
Markov
cHaiNS
(aHl)
⎞
0.25
⎜
s
ZEcar
traNSitioN
⎟
=
,
0.25
0
⎜
the
initial
gives
the
⎟
0.5
⎝
initial
⎠
probabilities
zones.
Hence
you
of
finding
a
randomly
can
predict
the
distribution
of
the
in
each
cars
of
after
the
7
7
transitions
by
s
finding
=
T
s
7
where
0
7
⎛
7
0.2
0.7
0.15
0.5
0.2
0.75
0.3
0.1
⎜
s
T
0.1
=
0
Therefore,
by
⎛
⎟
⎜
trend
of
⎟
⎜
⎠
⎝
comparing
helping
migration
them
to
⎛
⎟
s
to
s
of
⎜
⎟
0.260
⎟
⎜
⎠
⎝
the
⎟
0.379
⎠
management
of
ZEcar
can
identify
0
their
manage
,
⎞
0.360
=
0.5
7
a
⎞
0.25
0.25
⎜
⎝
⎞
cars
and
from
plan
zone
their
O
to
zone
C
over
one
week,
business.
n
The elements of s
= t
s
n
predict the probabilities of being in each state,
0
n transitions after the initial state probability matrix s
0
n
Just
as
values
of
geometric
sequence
u
=
u
n
divergence
according
to
the
value
of
r,
1
exhibit
r
convergence
or
1
the
behavior
of
the
sequence
n
s
=
T
s
n
depends
on
T.
Markov
chains
are
classified
according
to
0
n
the
long
term
behavior
of
T
:
A regular
Markov
chain
with
transition
+
matrix
T
is
one
for
which
there
exists
for
n ∈Z
which
all
the
elements
n
in
T
are
outside
in
greater
the
than
syllabus
psychology
and
zero,
include
biology
⎛
For
example,
if
T
whereas
absorbing
0.4
0.75
and
of
Markov
periodic,
with
chains
applications
0.6
0.25
⎞
then
=
some
members
of
the
sequence
⎟
⎝
0.536 5
0.579 375
0.463 5
0.420 625
⎠
⎞
3
n
T
types
respectively
.
⎜
⎛
other
are
T
=
⎜
⎟
⎜
⎝
⎛
,
⎟
⎠
⎞
0.555 520 526 382 81
0.555 599 342 021 49
0.444 479 473 617 19
0.444 400 657 978 51
9
T
=
⎜
and
⎟
⎜
⎟
⎝
⎠
⎛
⎞
0.555 555 555 556
0.555 555 555 556
0.444 444 444 444
0.444 444 444 444
100
T
=
⎜
⎜
⎟
⎝
⎟
⎠
n
The
elements
of
T
converge
⎛
to
5
5
9
9
matrix
P
4
4
9
9
in
the
steady
state
⎟
⎟ .
⎜
=
found
⎞
⎜
probability
values
⎜
⎟
⎟
⎜
⎝
⎠
t(Ps
0
The
consequences
of
knowing
P
are:
s
=
T
can be written as
0
tu = u where u is the long-term
probability vector. Therefore u is the
n
•
) = Ps
s
n
s
becomes
0
=
Ps
n
in
the
long
term.
This
means
that
applying
0
eigenvector of t corresponding to the
P
predicts
the
long-term
probability
matrix
Ps
of
being
in
each
state.
0
eigenvalue of 1. See also section 1.6
n
•
Since
the
elements
of
T
converge
to
those
of
P,
s
=
T(Ps
n+1
This
is
because
Ps
is
the
long-term
probability
)
0
matrix,
so
=
Ps
.
0
carrying
out
0
tu = u can also be solved by
another
transition
does
not
make
a
change
to
the
state.
solving a system of linear equations.
•
The
solution
to
Tu
=
u
is
independent
of
the
values
of
s
0
See also section 1.5
213
/
4
These
consequences
are
applied
in
the
following
example.
Exmpe 4.11.2
Two
smartphone
subscriptions.
companies,
Each
year,
Alfa
25%
of
and
Alfa
Better,
operate
customers
in
switch
a
rural
to
area.
Better,
Customers
and
15%
of
sign
Better
up
for
yearly
customers
switch
toAlpha.
(a)
Write
down
a
transition
year,
assuming
(b)
Find
the
(c)
Hence
(d)
that
steady
find
the
matrix
there
state
are
T
no
additional
probability
long-term
representing
matrix
probability
P
the
losses
by
matrix
switches
or
gains
repeated
made
of
for
each
initial
of
the
smartphone
customers
subscribe
with
Alfa
(ii)
50%
of
the
smartphone
customers
subscribe
with
Better
equation
Tu
=
u,
where
u
is
the
long-term
two
companies
in
one
of
T
state:
80%
the
the
customers.
multiplication
(i)
Solve
between
probability
matrix,
by
solving
a
system
of
equations.
Solution
(a)
It
is
but
ynapmoc
T
=
Current
⎛
A
erutuF
⎝
essential
some
to
labelling
label
may
your
help
transition
you
plan
matrix,
your
work
company
A
B
0.75
0.15
0.25
0.85
⎜
B
not
and
be
consistent
and
row
with
the
meaning
of
the
column
headings.
⎞
⎟
⎠
100
⎛
0.75
0.15
⎞
⎛
(b)
⎜
0.25
⎝
⎛
(c)
0.375
0.375
0.625
0.625
⎞
=
0.85
0.375
=
⎟
⎜
⎠
⎝
0.375
⎞ ⎛
⎞
0.8
(i)
⎝
⎛
⎟ ⎜
0.625
0.625
0.375
0.375
⎠ ⎝
⎞ ⎛
0.2
⎛
⎟
⎜
⎠
⎝
⎞
0.5
(ii)
⎝
⎟ ⎜
0.625
0.75
0.15
0.625
⎞
(d)
⎜
0.375
your
GDC
high
powers
of
the
matrix
with
0.25q
+
until
you
can
identify
a
steady
state.
⎞
⎟
0.625
⎛
0.375
⎠
⎞
⎟
0.25
0.85
⎠
⎛
⎠ ⎝
⎞
q
⎜
⎜
0.5
⎟
1
q
⎝
⎟
⎜
⎠
⎝
⎛
=
⎟
0.625
⎞
q
⎜
⎟
⎜
⎠
⎝
⎟
1
⎠
q
⇒
0.75q + 0.15
0.15q
=
⎛
=
0.375 .
q)
=
1
–
q
⎠
Hence
give
the
same
answer.
0.375
⎞
u =
⎜
0.4
–
q ⇒
0.15
=
0.85(1
⎟
will
q
with
=
⎜
⎝
Experiment
⎠
=
⎜
⎛
P
⎟
⎝
⎟
0.625
⎠
214
/
PRACTICE
3.
Sl
PracticE
A survey
out
there.
1,
Gr o U P
Sheldon
is
more
The
in
through
comparing
Germany
the
following
two
data
the
cost
and
Italy
.
countries,
in
city
about
number
of
identical
T,
is
is
the
of
modelled
conducted
in
order
to
lives
beds
in
of
a
the
people
randomly
living
chosen
As
he
he
collects
t
the
P(T
Italy
1.80
1.50
2.00
1.70
1.70
3.00
1.80
1.90
4.
2.00
1.60
2.10
1.70
2.20
1.80
1.60
1.50
the
discrete
probability
represented
in
the
table.
travels
euros.
Germany
by
soft
distribution
drinks
large
1
home,
1.
a
QUE StioNS
find
Pa P E r
in
QUE STIONS
=
t)
1
2
3
4
5
6
7
0.47
3a
0.17
0.09
a
0.02
0.01
.
Find
the
value
.
Find
the
probability
home
has
more
than
be
thinks
has
that
believes
that
test
record
randomly
4
that
a
beds,
randomly
given
that
selected
it
has
2.
it
four
the
independent
and
a
exactly
A restaurant
Lisa
of
of
is
choices
choice
of
gender,
not.
selected
main
main
decide
course
is
course.
course
whereas
They
which
of
will
Michelle
to
do
chosen
a
by
60
customers.
1.80
The
.
Find
two
the
at
decides
the
hypothesis
Italy
means
for
the
costs
in
are
shown
than
5%
use
a
two-sample
significance
that
in
to
drinks
are
level
to
cheaper
State
table.
Female
Total
Beef
a
8
20
Fish
5
9
14
V
egetarian
8
b
15
Seafood
3
8
11
28
32
60
pooled
test
on
the
average
in
pasta
Germany
.
whether
tailed
the
Male
Total
.
in
the
countries.
Sheldon
t-test
sample
results
this
is
a
one-tailed
or
a
two-
.
test
Write
down
the
values
of
a
and
b.
Michelle
decide
to
perform
2
.
.
State
the
null
.
Find
the
p-value
.
Write
and
alternative
Lisa
hypotheses.
and
independence
for
the
the
the
conclusion
of
the
your
significance
test
for
level.
.
Write
down
the
null
and
alternative
test,
hypotheses
justifying
5%
χ
test.
.
down
at
a
for
the
test.
answer.
2
.
2.
Student
are
at
lockers
numbered
21
and
locker
is
in
a
with
ending
chosen
at
at
corridor
of
consecutive
53.
At
the
random.
Wiles
Find
of
the
The
starting
term,
critical
the
locker
number
49
or
higher
.
26
or
less
.
a
Use
not
a
A teacher
a
multiple
factor
of
the
test.
a
5%
significance
level
answer
to
part

to
write
down
of
the
test,
clearly
justifying
answer.
asks
many
the
students
minutes
they
in
a
spent
class
to
working
record
on
an
9
of
task,
designed
to
take
between
one
and
9
two
e.
for
conclusion
extended
d.
value
your
your
5.
of
for
is:
how
multiple
statistic
probability
the
.
χ
is7.81
a
.
that
the
Academy
integers
end
Find
hours.
The
results
are
shown
in
the
diagram.
120
215
/
4
10
.
Find
the
probability
that
a
randomly
selected
9
student
from
this
class
studies
both
Economics
8
7
ycneuqerF
and
Art.
6
5
.
Given
that
a
randomly
selected
student
4
3
studies
Economics,
find
the
probability
that
2
the
1
student
also
studies
Art.
0
60
65
70
75
T ime
80
85
90
.
Determine
studies
are
.
Find
the
.
Find
estimates
number
of
students
in
the
In
addition
the
whole
it
during
.
Write
for
6.
A fair
1,
the
class
total
and
3,
5
a
time
also
mean
taken
spent
spent
and
by
at
30
the
the
Find
the
fair
8
on
home
on
the
Find
or
.
a
the
numbered
cubical
is
die
thrown.
Find
or
a
the
size
and
probability
of
have
the
same
are
here:
shown
your
B,
at
random
reasoning.
C
and
D
are
the
same
range.
for
B
the
four
populations
C
D
deviation
task.
2,
3,
5
and
8
is
numbered
T
is
the
T
is
a
total
of
the
that
prime
Each
of
the
three
that
T
is
a
of
that
T
is
a
box
and
whisker
plots
below
number.
prime
Match
number
each
prime
to
of
and
frequency
A
number
one
box
the
four
whisker
populations.
plot
with
the
diagram.
B
C
4.
Each
of
the
three
over
the
results
in
the
the
distances
course
this
box
of
and
6
in
kilometres
weeks
whisker
and
that
four
showed
below
cumulative
a
5.1
frequency
frequency
is
3.3
.
Find
range
Match
one
to
one
with
of
the
the
correct
liquid
bottles
diagram.
plot.
7.3
interquartile
corresponds
populations.
A
B
C
13.2
for
the
lengths
of
the
runs
km.
the
value
of
a.
10.
Find
whether
shown
the
or
not
Carole
maximum
should
length
as
an
A machine
refills
your
in
reduce
order
a
class
18
study
of
30
students,
and
3
10
study
nor
to
waste.
The
volume
refilled
by
machine
is
normally
distributed
with
mean
students
ml
and
standard
deviation
3.7
ml.
Economics,
study
a
quality
control
procedure,
4%
of
the
bottles
neither
are
Economics
washing
answer.
In
Art
ml
outlier,
499.3
In
500
have
the
fullyjustifying
8.
student
on
correct
probability
recorded
ran
.
a
2
she
The
A,
histograms
10.
multiple
Carole
3.2
Justify
populations
diagrams
7
.
choosing
Frequency
.
Gr o U P
and
thrown.
probability
factor
Economics
task,
working
standard
the
when
corresponds
.
studies
students.
minutes
and
spent
die
four
A
mean
time
Art
independent.
The
.
.
events
standard
lesson.
and
numbers
the
time
tetrahedral
2,
two
to
for
the
down
thrown,
1,
of
their
the
the
class.
9.
deviation
if
(minutes)
rejected
for
containing
not
enough
washing
Art.
liquid.
Find
nearest
ml,
the
that
minimum
a
bottle
volume,
must
to
contain
the
in
order
tobeaccepted.
216
/
PRACTICE
11.
Natalia
the
wanted
mean
to
perform
distance
from
a
t-test
school
for
to
see
if
Gr o U P
students
was
greater
than
the
mean
Helen
school
for
the
students
in
grade
do
so,
she
50students
them
how
decided
from
far
to
collect
grades
they
9
lived
and
from
a
sample
10
very
tests.
The
and
to
of
out
her
ask
of
50
score
obtained
a
list
of
the
reaches
arranged
students
from
students
and
alphabetically
.
had
Grade
9
grade
10
to
stratified
decided
use
full
write
number
to
be
selected
115
Find
how
many
from
students
test
grade
.
9
2
Describe
how
she
she
grade
could
use
sampling
to
select
a
random
sample
each
year
carrying
values
from
shown
here.
out
grade
her
should
select
10
as
a
find
she
the
The
15
20
25
30
35
from
grade
Natalia
box
40
from
school
number
Explain
why
indicates
the
that
an
In
Ted’s
class,
the
and
=
name
45
and
t-test
plotted
of
shown
vowels
Frequency
Each
in
a
bag.
from
.
student’s
the
The
In
an
.
whisker
whisker
of
in
plot,
a
experiment
the
at
that
is
known
.
Find
.
If
not
be
15.
valid.
the
each
The
is
chosen
expected
least
The
3
is
vowels
in
this
her
score
in
the
tests
she
must
take
to
have
20.
scored
in
a
game
Y
is
discrete
probability
distribution:
5
6
7
8
0.2
p
q
0.1
0.3
E(Y)
=
6.2
values
point
find
this
a
of
p
scored
and
gains
the
price
fair
game.
is
played
of
of
10
a
q
a
cash
ticket
times.
prize
which
Find
of
would
the
that
at
least
7
USD
is
won
in
four
is
4
5
1
6
4
1
1
written
one
of
the
games.
grows
eight
adding
sets
of
plants
different
in
controlled
amounts
of
to
each
set.
The
data
in
the
table
shows
table.
3
carried
on
a
card
card
are
out
at
is
repeated
it
is
put
number
experiment
is
and
to
amount
each
of
plant
centimetres,
nutrient
and
of
the
each
put
(m)
in
grams
average
set
Mass
of
height
plants
(m)
given
(h),
after
in
six
weeks.
Height
(h)
chosen
A
0
5.2
B
5
9.2
once.
least
3
Find
the
C
10
14.4
vowels
D
15
18.5
E
20
21.2
F
25
21.0
G
30
21.0
H
35
21.6
in
in
of
25
back
times
his/her
times.
in
a
the
Each
bag.
student
name
is
time
Find
with
chosen.
is
repeated
twice,
but
Calculate
Spearman’s
the
card
probability
chosen
after
that
has
the
exactly
at
least
3
first
one
selection.
of
the
vowels.
rank
correlation
without
coefficient
names
for
each
2
there
vowels
replacing
the
aim,
4
that
game
Frances
.
.
her
name.
experiment
card
in
plot
1
experiment,
student’s
The
she
bag.
probability
the
name
until
her
the
Number
test
improve
50
might
number
is
of
points
the
y)
nutrients
student’s
every
expression
over
of
by
conditions,
12.
to
(km)
box
the
She
9.
exactly
.
in
succeeds
number
probability
Distance
test.
sampling
.
10
aims
scored
from:
make
5
she
takes
score
yUSD,
0
and
10
systematic
t-test,
50
group.
It
Before
of
find
P(Y
to
in
marks.
y
.
out
first
marks
she
modelled
.
score
students.
14.
.
her
down
mean
the
are
mean
had
.
Natalia
tests
the
the
nth
85
improve
school.
.
principal,
to
on
by
Assuming
She
keen
9.
8
To
is
distance
her
from
3
in
13.
grade10
QUE STIONS
two
and
comment
on
your
result.
Find
Frances
decides
modelthe
data
that
as
a
it
would
piecewise
be
better
to
function.
217
/
4
In
0
this
≤
m ≤
m >
.
model,
20
and
is
height
increases
approximately
linearly
constant
.
for
In
for
a
that
20
season
Brian
game
Find
the
h
form
.
the
line
=
of
best
fit
for
0
≤
m ≤
20 in
the
20,
the
meaning
of
Given
find
the
the
the
David
two
is
his
seat
the
probability
the
start
at
of
the
on:
at
least
two
occasions
.
exactly
two
occasions.
equation
is
h
=
function
value
of
is
of
experiment
in
the
tables
assumptions
that
you
make.
continuous
at
m =
20,
On
her
way
through
a
to
set
and
of
from
traffic
work,
Julia
passes
lights.
c.
designing
experiments
any
c
For
16.
in
find
a
18.
d.
not
games,
.
State
m >
is
20
am + b
Interpret
For
of
computer
on
his
A and
below,
of
games
laptop.
The
experiment
with
the
involving
of
outcomes
B
are
probability
a
period
times
of
each
lights(X).
60
days,
day
Her
she
records
she
has
to
stop
results
are
shown
the
at
in
number
traffic
the
table.
shown
Number
of
times
stopped
(X)
0
1
2
12
24
24
of
eachoutcome.
Frequency
Outcome
for
A
1
3
6
10
15
She
Probability
0.1
0.1
0.15
0.5
0.15
decides
that
X
is
the
outcome
of
experiment
A is
a
multiple
player
scores
a
point.
All
other
of
being
data
to
test
binomially
,
stopped
the
with
equal
hypothesis
the
to
0.5
Write
down
the
null
and
alternative
outcomes
hypotheses
score
this
of3,
.
the
use
distributed
probability
If
to
for
Julia’s
test.
zero.
.
Outcome
for
B
2
3
5
7
11
0.03
0.02
0.05
0.4
0.5
Find
the
times
Probability
expected
she
has
to
values
stop
if
for
the
the
null
number
of
hypothesis
istrue.
If
the
outcome
of
experiment
B
is
a
factor
of
10,
.
the
player
scores
a
point.
All
other
Perform
the
hypothesis
score
d.
is
the
number
of
points
scored
in
20
Y
is
the
number
of
points
scored
in
100
trials
of
Write
down
stating
any
the
distributions
assumptions
you
of
X
trials
and
of
of
.
why
the
Give
one
should
Hence
Y
show
are
Find
that
both
the
the
equal
be
rejected
at
and
variances
of
Y
values
of
X
and
8.
Pa P E r
2
Adam
17
.
of
Hence
X
Brian
him
find
10
distributions
of
is
for
a
be
the
hypothesis
null
true.
beekeeper
honey
six
of
who
collected
production
his
hives
is
in
his
shown
compare
that
always
to
He
seat
minutes
of
up
knows
is
gain
contrast
to
15
the
distribution
stadium
minutes
that
the
entry
to
distributed
and
the
Y
exactly
queue,
his
and
turns
team
off.
Find
standard
before
time
the
of
about
beehives.
in
this
The
table.
of
bees
(I)
Monthly
honey
taken,
stadium
normally
deviation
the
T,
game
for
and
with
2.5
the
probability
that
Brian
will
before
the
game
kicks
in
190
900
220
1100
250
1200
285
1500
305
1700
320
1800
grams
(P)
his
then
mean
minutes.
The
find
relationship
by
the
between
the
regression
variables
line
with
is
equation
his
P
seat
data
X
modelled
.
level.
Y
with
football
kicks
5%
QUE StioNS
monthly
data
the
why
production,
of
the
make.
expected
to
reason
not
Number
d.
null
B.
1.
of
state
A.
might
.
and
zero.
X
.
test
outcomes
=
aN
+
b
off.
218
/
PRACTICE
.
Write
.
Use
down
this
the
value
regression
of
line
a
and
to
of
b
estimate
production
hive.
that
production
30
low
has
270
honey
production
from
a
hive
2.
A health
The
has
200
hives
onthe
monthly
in
total.
He
collects
hives.
This
frequency
data
is
production
shown
in
the
of
become
regular
hives.
survey
heights
of
is
16
carried
to
out
in
18-year-old
a
large
city
.
students
are
data
measured.
honey
hives
probability
that
bees.
Adam
the
the
production
monthly
Calculate
QUE STIONS
all
The
heights
of
the
females
are
the
normally
distributed
standard
deviation
with
mean
163
cm
and
cumulative
10
cm.
The
heights
of
the
graph.
males
170
200
.
180
are
cm
normally
and
Find
the
male
is
a
distributed
standard
probability
taller
than
with
deviation
that
a
of
a
mean
11
cm.
randomly
175
cm.
the
females
of
selected
160
.
Given
that
15%
of
are
shorter
than
140
t
cm,
find
t
sevih
120
.
Find
fo
the
the
inter-quartile
range
of
the
heights
of
males.
rebmuN
100
The
students
surveyed
are
51%
female
and
49%
80
male.
If
the
person
is
female,
the
probability
they
60
are
taller
than
175
cm
is
0.115.
40
A student
20
d.
is
selected
at
Find
the
probability
than
175
cm.
random.
that
the
student
is
taller
0
0
600
800
1000
1200
1400
1600
1800
2000
e.
Monthly
production
of
Given
find
Adam’s
hives
production,
Type
are
as
of
labelled
defined
in
as
low,
this
hive
regular
or
Hl
Monthly
in
P
regular
1080
<
P
P
>
k
<
grams
1.
1080
of
low
the
probability
than
175
cm,
that
the
student
is
male.
QUE StioNS
≤
1,
Gr o U P
1
A sample
is
taken
from
normally
distributed
a
population
with
a
mean
which
equal
is
to
.
k
values
12.1
number
taller
(P)
The
high
the
is
honey
low
down
student
PracticE
Pa P E r
Write
the
high
table.
production,
.
that
honey
14.2
in
the
10.7
sample
9.8
15.6
are
given
10.6
11.8
in
the
17.2
table.
9.4
10.5
production
hives.
Adam
knows
that
128
of
his
hives
have
a
.
State
the
distribution
.
Find
the
95%
.
Given
of
the
sample
mean.
regular
confidence
interval
for
.
production.
2.
d.
that
Po ( 3.1)
A ~
and
B ~
Po ( 2.7 ) ,
find
Find:
E( A
.
the
value
of
the
number
and
Var( A
2 B).
k
.
.
2 B)
of
hives
that
have
a
Hence
explain
Poisson
high
why
A
2B
does
not
follow
a
distribution.
production.
3.
e.
Adam
bees
decides
in
each
suggests
that
a
low
that
low
to
increase
the
production
there
is
a
production
number
hive.
Research
probability
hive
of
of
becomes
0.75
a
regular
For
his
Science
collected
results
to
test
with
a
40
are
that
Internal
samples
from
the
mean
the
Assessment,
and
records
expected
difference
of
he
is
Antoine
how
results.
due
to
He
far
has
his
wants
random
error
0.
219
/
4
X
Let
.
be
State
the
the
why
sample
can
X
be
mean
of
assumed
Antoine’s
to
be
data.
normally
Gr o U P
7
.
distributed.
Ana
by
2
believes
her
that
favourite
distribution.
The
sample
mean
for
Antoine’s
data
is
–0.54
.
standard
Test,
not
at
deviation
the
5%
Antoine
mean
of
is
football
test
her
number
of
goals
State
assume
level,
his
whether
errors
have
during
a
38
shown
in
game
the
context
of
the
of
your
result
in
Given
Z
that
X
~
N ( 61, 25 )
~
follows
hypothesis,
a
Poisson
she
records
score
in
each
game
season
and
the
results
are
table.
of
goals
0
1
2
3
4
≥5
9
14
12
2
1
0
the
question.
Perform
4.
scored
a
Number
implication
goals
or
0.
one
team
they
Frequency
.
of
1.2
signifcance
can
number
and
the
the
To
the
Po ( 4.9 ) ,
are
~ B ( 7 , 0.35 )
Y
independent
of
and
random
an
goals
appropriate
scored
follows
test
a
to
see
Poisson
if
the
number
distribution.
variables,
8.
The
number
of
admissions
per
hour
on
Saturday
find:
evenings
.
E ( 2Y
.
Var ( 8
.
Var ( X
Poisson
+ 7 )
a
shift
to
distribution
of
4
Find
wishes
to
test
whether
or
not
a
coin
or
the
whether
it
is
more
likely
to
show
the
every
decides
mean
4.7.
Saturday
follows
Vicky
a
works
evening.
of
that
on
a
given
Saturday
to
flip
it
50
times
and
to
Heads
that
there
will
be
at
least
20
admissions
room
during
Vicky’s
shift,
Heads.
record
any
assumptions
that
you
make.
the
Vicky
number
probability
emergency
stating
He
with
room
is
to
fair,
hours
emergency
Z)
+ 2Y
evening
V
edant
hospital
2X )
.
5.
a
works
on
5
consecutive
Saturday
evening
appear.
shifts.
Let
p
be
the
probability
of
the
coin
showing
.
Find
the
most
likely
number
of
Saturday
Heads.
evening
.
Write
down
the
null
and
shifts
on
which
there
are
at
least
alternative
20
admissions
to
the
emergency
room
on
5
hypotheses.
consecutive
.
Find
the
critical
significance
region
for
a
test
at
the
assumptions
5%
Given
type
6.
Life
that
II
juice.
p
=
0.6
find
the
probability
of
a
The
produces
one-litre
bottles
company
wants
determine
theamount
of
In
office
in
a
vitamin
C
to
in
of
milligrams
orange
In
A,
week
mean
error.
drinks
that
evenings,
you
stating
any
make.
level.
9.
.
Saturday
of
the
number
follows
a
of
IT
Poisson
problems
that
distribution
occur
with
a
4.2
office
B,
the
number
Poisson
distribution
.
down
of
with
IT
a
problems
mean
of
follows
a
3.9
in
Write
the
distribution
of
the
total
thesebottles.
number
A random
sample
the
are
results
as
of
ten
bottles
is
analysed
offices,
and
firm
hoped
251
237
252
257
254
248
250
239
and
unbiased
variance
one-litre
estimates
of
stating
any
IT
problems
in
assumptions
the
you
two
make.
the
of
the
amount
of
population
vitamin
C
mean
in
introduces
will
help
new
reduce
software
the
which
it
is
problems.
252
In
Find
weekly
follows:
The
243
of
the
next
two
weeks
problems
reported.
.
the
there
are
only
12
IT
the
Test,
at
5%
significance
level,
whether
this
bottles.
is
sufficient
problems
evidence
has
been
that
the
number
of
IT
reduced.
220
/
PRACTICE
10.
A fabric
in
manufacturer
sheets
of
area
manufacturing
10
produces
square
process,
high
metres.
flaws
in
quality
During
the
silk
mean
the
fabric
of
the
rate
of
1.8
flaws
per
10
square
that
the
number
of
flaws
metres.
per
It
by
a
Poisson
sheet
Find
the
probability
Find
the
contains
at
than
that
a
randomly
Find
the
with
no
Silk
sheets
with
one
and
earn
profit
than
a
one
one
flaw
flaws
of
flaw
a
are
of
profit
sold
Any
loss
a
at
a
sheets
of
€350.
Find
the
expected
manufacturing
profit
70
research
survey
find
town
who
with
more
Gr o U P
the
the
would
proportion
company
would
research
and
asks
10
silk
gained
sheets,
Over
by
to
and
the
or
company
is
proportion
visit
a
new
is
greater
be
likely
company
people
not
results
Age
x
conducting
of
people
coffee
than
to
visits
from
they
0.1
build
the
each
are
would
shown
(x)
<
in
The
a
of
visit
in
the
Would
≤
x
<
30
4
30
≤
x
<
40
3
50
2
x
.
Name
x
≥
<
a
this
method
the
of
proportion
they
would
Test,
at
the
visit
1%
sample
of
70
A has
mean
corn
a
type
of
80
corn
plants
B.
period
deviation
bags
of
time,
it
of
the
flour
is
labelled
known
weight
of
1
that
the
kg.
the
bags
are
regularly
is
1
kg
checked
rather
to
than
ensure
less
that
than
1
the
kg.
then
bags
are
average
weight
measured.
taken
and
the
the
The
null
hypothesis
it.
town
five
the
centre
age
test
is
that
µ
=
hypothesis
is
that
µ
< 1000 g.
the
1000 g
and
the
alternative
ranges
coffee
table
the
mean
is
less
than
a
grams
then
the
null
shop
hypothesis
is
rejected
.
value
and
the
factory
is
fined.
below.
visit
Find
the
error
is
of
a
if
the
probability
of
a
type
I
0.05
owner
produce
sampling
a
the
of
the
of
the
bags
factory
with
an
has
set
average
.
Find
the
probability
of
.
Find
the
probability
that
least
one
a
the
machine
weight
type
the
II
of
980
to
g.
error.
machine
fails
at
shop.
significance
level,
the
proportion,
who
would
go
to
the
of
the
next
three
tests.
population.
sample
coffee
that
than
one
a
type
sample
produces
long
who
Residents
mobile
said
year,
in
an
phone
Pi
urban
10%
to
area
companies:
expects
to
Mu
retain
and
have
Pi,
76%
the
a
Mu
of
rest
choice
and
its
to
Fi.
of
three
Each
customers,
Fi.
the
p,
of
people
expects
coffee
shop
to
retain
70%
of
its
customers,
in
19%
to
Fi
and
the
rest
to
Pi.
Fi
expects
to
is
20%
of
its
customers
to
Pi
and
5%
to
Mu,
0.2
retaining
Give
with
15
lose
d.
that
checks,
losing
greater
cm.
these
Mu
hypothesis
town
a
3
50
Find
the
a
with
weight
losing
.
type
shop.
14.
.
mean
corn
8
20
20
≤
A has
70
3
bags
mean
a
The
40
than
standard
the
If
whether
255
fertilized
A factory
for
The
of
€100.
In
If
sample
is50g.
A market
to
a
discount
nearest€100
11.
that
with
probability
fertilized
13.
.
deviation
flaw.
earn
€150.
incur
standard
chosen
greater
sheets
a
distribution.
least
Silk
probability
fertilized
plants
sheet
and
is
.
.
cm
is
greater
modelled
250
cm.
plants
assumed
of
occur
.
at
12
height
QUE STIONS
criticism
of
the
question
asked
in
the
rest.
the
.
Construct
a
transition
matrix
to
show
the
survey
.
probabilities
12.
The
effects
growth
of
of
two
corn
types
are
of
being
fertilizer
compared
on
in
the
a
each
plants
fertilized
with
type
A have
of
253
10cm.
Corn
cm
and
a
standard
a
fertilized
deviation
with
type
transitioning
from
B
32%
of
the
residents
are
customers
of
mean
40%
are
customers
of
Mu
and
the
rest
have
the
are
all
of
with
plants
customer
study
.
Pi,
height
a
company
.
Currently
,
Corn
of
have
Fi.
Fi
predict
that
they
will
largest
a
share
of
the
market
after
three
years.
221
/
4
.
Determine
if
Fi’s
assumptions
claim
you
is
correct,
stating
any
Strawberries
make.
strawberries.
that
15.
.
Give
one
reason
why
you
might
the
a
quadratic
.
curve
a
cubic
model
the
points
shown
in
the
The
average
label
mass
on
of
in
containers
the
container
a
strawberry
of
n
states
in
the
is
12
g.
If
the
average
mass
is
above
curve
15
to
packaged
choose:
container
.
are
if
diagram.
g,
it
.
the
is
If
container
below
the
11
g
is
it
rejected
is
probability
as
rejected
of
a
too
as
heavy
,
being
package
of
n
too
and
light.
being
5
rejected
C
B
is
0.01,
find
n
(5, 4.3)
(3, 4.5)
4
17
.
A group
a
3
A
(1, 2)
scientists
conference.
or
D
2
of
rejected.
Each
is
assessing
poster
Concerns
are
is
posters
either
for
accepted
expressed
that
each
(6, 1.7)
decision
1
is
not
independent
of
the
previous
decisions.
To
address
these
concerns,
the
group
looked
at
0
1
2
3
4
5
6
7
the
Theory
suggests
that
the
model
or
y
will
either
250
into
be
most
25groups
accepted
2
y
=
3
0.4 x
+ 2.9 x
=
in
four
points
have
coordinates
4.5),
C(5,
4.3)
and
D(6,
each
5.
The
Find
the
curve
sum
and
of
state
the
is
of
shown
them
posters
in
the
table.
A(1,
of
2),
0
1
2
3
4
5
9
6
10
7
13
5
accepted
1.7)
square
which
divided
number
group
Frequency
.
and
+ 0.944 x
posters
B(3,
of
posters
2
0.155 x
Number
The
recent
residuals
model
you
for
each
would
.
Find
the
proportion,
p
,
of
posters
that
are
0
choose
your
on
the
basis
of
this
result.
Justify
accepted.
reasoning.
.
.
Find
for
the
least
squares
these
four
points.
cubic
regression
curve
State
the
state
what
value
from
your
this
indicates.
the
points
calculator
Given
that
18.
y,
horizontal
represent
the
a
distance,
x,
from
Izzy
of
an
object
projected
binomial
meaning
the
results
distribution,
of
the
result
above
and
in
the
come
state
the
context
of
measured
botanical
the
garden
heights
and
of
291
found
sunflowers
that
the
in
heights
height,
the
from
be
modelled
by
a
normal
distribution
initial
with
position
whether
question.
could
at
see
and
a
e.
a
likely
the
R
to
from
2
d.
Test
mean
189.5
cm
and
a
standard
deviation
ground
of15.3cm.
level,
two
state
one
models
squares
reason
might
be
why
either
preferred
of
to
the
the
first
least
.
model.
The
following
grew
310
sunflowers.
sunflowers
16.
Callum
grows
strawberries
and
raspberries
taller
farm.
are
normally
standard
The
.
and
Find
are
.
Find
that
the
g.
chosen
chosen
mass
mean
The
masses
of
that
the
that
Predict
than
195
the
cm,
garden
number
stating
of
any
that
you
make.
g
of
and
Callum’s
with
mean
Izzy
notices
measure
that
the
the
tape
she
heights
was
faulty
.
at
3
cm,
not
the
zero
.
What
are
the
correct
had
used
The
to
scale
started
mark.
g.
total
strawberries
the
strawberry
a
12
distributed
0.5
botanical
strawberries
with
deviation
probability
randomly
times
2.7
probability
randomly
Callum’s
normally
standard
the
of
distributed
deviation
raspberries
4g
masses
the
on
assumptions
his
season,
randomly
is
mass
more
mass
is
of
more
chosen
of
five
than
70g.
a
than
variance
the
.
four
of
the
values
of
the
mean
and
distribution
of
the
heights
to
part
of
sunflowers?
Hence
find
nearest
the
whole
true
answer

to
the
number.
raspberry
.
222
/
PRACTICE
Pa P E r
Large
2
the
1.
A manager
firm’s
is
cafeteria
employees
of
10
collecting
are
greater
fill
is
in
data
on
perceived,
a
survey
.
recorded,
where
how
and
Their
a
aircraft.
has
are
put
given
The
in
the
the
place
and
the
questionnaire
results
are
shown
Employee
of
mean
the
same
scores
score
with
standard
deviation
kg.
.
the
Find
a
employees
d.
are
The
survey
Second
survey
the
large
Find
the
a
B
C
D
E
7.2
4.1
6.1
5.4
3.9
7.3
5.2
6.3
6.7
4.2
moment
between
the
items
that
the
on
the
value
1.
In
Euclid
Perform
a
evidence
test
at
correlation
to
the
sets
of
data
all
and
the
necessary
that
the
there
that
is
there
results
of
significant
is
and
a
the
year,
kg
and
more
mass
than
item
of
of
a
three
hand
randomly
times
that
luggage.
Officers
at
sample
the
of
destination
three
hand
luggage.
total
mass
of
large
Find
the
airport
cases
and
probability
the
sample
exceeds
105
kg.
there
and
are
only
two
ResultsNow,
gym
who
franchises,
both
sell
membership
packages
starting
on
1.
Each
year,
ClearGym
retains
60%
of
its
second.
You
conditions
for
may
the
assume
use
of
and
the
rest
ResultsNow
the
rest
move
retains
move
to
to
70%
ResultsNow.
of
its
Each
customers
ClearGym.
first
Explain
why
this
Markov
chain.
context
can
be
modelled
as
a
that
this
test
.
are
25
and
.
survey
are
correlation
two
level
between
cases
obtained.
show
5%
is
the
of
city
members
.
of
3
January
comment
large
mean
that
chosen
random
two
annual
coefficient
hold
table.
A
product
case
Customs
ClearGym
.
7
probability
randomly
select
Pa P E r
First
the
the
changes
again.
in
of
in
out
indicates
survey
,
five
masses
distributed
chosen
results
carried
normally
satisfaction.
assessing
The
be
five
of
After
must
the
she
higher
cases
QUE STIONS
Sketch
a
transition
diagram
to
represent
this
satisfied.
context.
The
manager
claims
that
this
result
shows
her
.
survey
is
a
reliable
means
of
collecting
the
Hence,
matrix
.
Give
the
name
for
this
test
of
write
G
to
the
cafeteria
cafeteria
manager
has
claims
improved
and
label
a
transition
represent
the
probabilities
of
reliability
.
customers
The
down
data.
that
the
between
data
the
two
shows
changing
membership
between
the
companies.
two
At
the
start
of
2019,
ClearGym
had
17 530
surveys.
customers
d.
Carry
out
an
appropriate
test
of
the
claim
2.
The
cafeteria
masses
of
to
an
items
aircraft
Predict
the
number
of
hand
luggage
by
the
passengers
after
one
Find
the
year,
8956.
of
customers
each
gym
has
with
mean
9.4
kg
and
to
the
nearest
100
customers.
carried
are
eigenvalues
and
the
associated
normally
eigenvectors
distributed
had
manager.
e.
on
ResultsNow
of
d.
the
and
of
G
G
standard
1
deviation
.
Find
2.8
the
f.
Hence,
express
g.
Hence,
show
the
form G =
3
3(0.3)
PDP
kg.
probability
that
the
mass
of
a
that
randomly
n
⎡
chosen
in
item
of
hand
luggage
is
between
9
3 +
1
kg
n
4(0.3)
⎤
n
G
⎥
⎢
=
n
and
12
7
kg.
⎢
4
n
4(0.3)
⎥
4 + 3(0.3)
⎦
⎣
The
airline
sets
such
that
each
overhead
overhead
.
Find
eight
the
items
luggage
of
locker.
locker
the
hand
can
be
hand
The
hold
probability
will
luggage
too
dimensions
luggage
maximum
is
that
100
for
load
fit
into
stating
any
your
answer
result
to
part
for
part
g
to
verify
your
d
each
The
management
that
in
items
the
of
hand
of
ClearGym
are
concerned
assumptions
the
long
term
they
will
lose
customers
to
ResultsNow.
overhead
.
locker,
Use
kg.
eight
heavy
will
h.
that
you
Explain
if
the
management
of
ClearGym
are
make.
justified
in
having
this
concern.
223
/
CA LC U LU S
5
5 . 1
D I F F E R E N T I AT I O N
You should know:
✔
the
informal
You should be able to:
concept
of
a
limit
✔
nd
derivatives
n
f ( x) =
✔
the
derivative
of
a
function
f
gives
the
rate
n
ax
of
functions
of
the
form
1
+ bx
where
+ ...
all
exponents
are
of
integers
change
of
the
dependent
to
the
independent
to
the
gradient
variable
variable.
of
the
curve
notation
for
the
This
y
=
with
is
respect
equivalent
✔
nd
at
f ( x)
a
the
equations
given
point,
of
tangents
both
and
analytically
normals
and
using
technology
✔
forms
of
dy
for
rst
derivative,
dV
,
example:
✔
f ′( x),
dx
use
technology
and
✔
f ′(x)
>
0
for
increasing
f ′(x)
<
0
for
decreasing
functions
the
derivative
✔
local
of
maximum
where
f ′(x)
greatest
or
=
0,
nd
values
of
f ′(x)
given
f (x),
f (x)
and
but
least
=
✔
functions
ax
values
f ′(x)
the
=
ax
,
points
might
in
the
solutions
of
f ′(x)
=
0
solve
optimization
problems
in
context.
n−1
is
minimum
these
nd
and
n
✔
to
dt
not
∈ 
occur
be
given
n
the
domain.
The
of
of
a
function
at
that
derivative
of
a
the
The
of
derivative
change
of
the
function
at
a
point
or
the
gives
the
gradient
of
the
graph
point.
function
function.
The
gradient
larger
the
of
a
value
curve
of
the
gives
the rate
derivative,
the
Note
greater
the
rate
ofchange.
In the case of an equation like
A positive
value
for
the
variable
increasing
derivative
means
the
value
of
the
dependent
2
P
=
2t
+
3t −
7
the notation is
is
as
the
independent
variable
increases
and
a
dP
=
4t +
3
negative
value
means
it
is
decreasing.
can
written
dt
dy
The
derivative
be
as
or
f ′(x)
can
be
found
the
terms.
dx
The
derivative
=
ax
a
dy
n
y
of
n
⇒
=
polynomial
using
the
relation:
1
on
anx
each
of
dx
Note the special case of y = mx and y = c which have derivatives
dy
dy
=
m
=
and
dx
0 respectively.
dx
S AMPLE STUDENT ANS WER
3
Consider
the
function
(a)
Find
f ′( x)
(b)
There
are
graph
of
f
two
is
f ( x) =
points
11.
Find
at
x
2
3x
which
the
+
the
2x +
2
gradient
x-coordinates
of
of
the
these
points.
224
/
5 .1
3
(a)
f(x)
=
2
x
−
3x
1
+
2
f ’(x)
=
3x
f ’(x)
=
3x
11
3x
D i f f E R E N T i AT i o N
2x
+
2
1
−
6x
−
6x
+
2
▲ Gradient
f ′(x)
=
equal
to
11
means
11
2
+
2
2
(b)
=
−
6x
+
2
▼ This
could
be
solved
by
2
3x
−
6x +
2
=
11
−
6x
9
=
0
factorizing,
easiest
but
route
in
always
an
take
the
exam
2
3x
GDC
=
−
Poly
roots
▲ Clearly
{−1,
f(−1)
=
(−1)
=
−4
2
−
3(−1)
3
(2)
out.
3}
3
(1)
set
f(3)
=
3
=
8
+
2(−1)
+
2
2
−
3(3)
+
2(3)
+
2
▼ Unfortunately
just
asked
for
the
question
x-coordinates.
The
Answers:
student
probably
did
not
lose
any
2
(a)
f ’(x)
(b)
(−1,
=
3x
−
6x
(3,
8)
+
2
marks
because
x-coordinates,
−4),
they
but
did
time
the
was
lost
calculating
the
unnecessarily
in
y-coordinates.
Always
question
nd
read
the
carefully
.
Tanents and nrmals
The
tangent
curveat
to
that
a
curve
point
at
and
a
point
has
the
is
the
same
line
which
gradient
as
just
the
touches
the
curve.
t
en
ng
Ta
normal
to
the
perpendicular
to
curve
the
at
a
point
is
the
line
through
the
point,
tangent.
roN
lam
The
Example 5.1.1
2
x
Consider
the
curve
y
=
4
dy
(a)
Find
dx
Remember from section 3.3, that
2
x
(b)
Find
the
equation
of
the
tangent
to
the
graph
of
y
=
at
x
=
4
if a line has a gradient of m, then
4
(c)
Find
the
x-coordinate
of
the
point
at
which
the
normal
to
the
the line perpendicular to it has a
1
2
1
x
graph
of
y
has
=
gradient of
gradient
m
8
4
Solution
dy
2x
=
(a)
dx
x
2
x
=
4
1
can
2
be
written
4
1
1
× 2x
is
=
4
y
When
x
=
dx
is
y
=
x
x
or
2
2
=
2
The
equation
of
the
tangent
is
2
y
Equation
the
4
=
4 ,
so
4
derivative
(b)
2
x
as
=
mx + c
2x + c
The
of
first
m,
the
stage
which
tangent
gradient
of
is
is
to
the
and
the
is
find
the
gradient
equal
curve
at
to
x
=
value
of
the
4
225
/
5
16
When
x
4 ,
=
y
=
=
A point
on
the
find
value
line
is
needed
to
4
4
so
4
=
2
×
4
+
c
⇒
c
=
the
the
meets
=
2x
c,
so
we
can
−4
use
y
of
point
the
where
curve,
(4,
the
tangent
4)
4
Nte
An
alternative
y
y
−
=
m(x
−
method
x
1
is
to
use
)
1
All GDCs will be able to find the
(c)
numerical value of a derivative at
Gradient
of
tangent
=
If
8
the
gradient
of
a
tangent
is
x
a point, and some will also be able
=
8 ⇒
x
=
equal
16
to
m
then
the
gradient
2
of
1
to find the equation of a tangent
the
normal
is
equal
and
to
m
at point.
vice
versa.
optmzatn
Nte
Optimization is 'the action of
If
f ′(x)
>
0
then
is
decreasing.
the
function
f
is
increasing,
and
if
f ′(x)
<
0
the
function
making the best or most effective
When
f ′(x)
=
0
the
curve
will
have
a
local
maximum
or
minimum
point.
use of a situation or resource'.
This
fact
can
be
used
to
solve
optimization
problems.
In this topic it is finding the
maximum or minimum value of a
S AMPLE STUDENT ANS WER
function in a given context.
A potter
His
Nte
sells
monthly
1
P( x) =
In the HL course, f ′(x) =
x
vases
profit
3
x
per
in
month.
Australian
dollars
(AUD)
can
be
modelled
by
2
+ 7x
120,
x
≥
0
of
if
no
5
0 can
also occur at a horizontal point
(a)
Find
the
value
(b)
Find
P ′(x)
(c)
Hence
(a)
P(0)
P
vases
are
sold.
of inflexion.
▼ The
answer
is
correct
but
find
the
mark
line
by
with
is
not
in
danger
of
beginning
P ′(x)
losing
the
of
3
2
vases
that
will
maximize
the
profit.
the
1
candidate
number
a
=
−
( 0)
+
7( 0)
−
120
5
second
=
=
−120
1
(b)
P( x ) =
−
3
x
2
+
7 x
x
+
14 x
+
14 x
−
120
5
▲ The
occurs
maximum
when
of
P ′(x)
=
the
0.
It
curve
3
is
=
important
to
to
indicate
to
show
the
the
full
−
working
examiner
5
that
3
the
maximum
has
not
2
been
found
(c)
0 =
−
2
x
5
directly
from
the
GDC.
70
x
=
3
▼ The
be
a
whole
needs
the
number
to
be
context
of
vases
number,
so
rounded.
of
the
=
must
the
answer
Always
question
23.3
in
keep
mind!
226
/
5 .1
In
some
will
optimization
contain
two
information
variables.
about
(aconstraint)
questions,
a
In
second
which
will
the
this
function
case
the
relationship
allow
you
to
to
be
optimized
question
between
substitute
D i f f E R E N T i AT i o N
will
the
out
provide
two
one
variables
of
thevariables.
Example 5.1.2
Fred
in
makes
the
an
shape
open
of
a
metal
cuboid,
container
as
shown
here.
x
The
container
width
x
m
has
and
a
height
length
y
m.
x
m,
The
3
volume
is
36
m
x
y
Let
of
A(x)
the
be
the
outside
surface
container.
108
(a)
area
Show
A( x) =
that
2
+ 2x
x
(b)
Find
A′(x)
(c)
Hence
find
surface
the
height
of
the
container
which
minimizes
the
area.
Solution
2
(a)
A( x) =
2x
V
=
There
+ 3 xy
36
2
=
36
x
y
⇒
y
are
two
rectangular
square
sides,
as
sides
the
and
three
container
is
=
2
x
36
2
A(x)
=
‘open’.
2x
+
3x
×
The
equation
has
two
variables
so
we
2
x
2x
need
108
2
=
36
can
2
A( x) =
use
the
condition
that
the
volume
3
is
x
(b)
to
+
to
then
find
be
an
expression
substituted
into
for
y,
which
A(x).
1
+ 108 x
2x
m
First
write
using
negative
exponents.
If expressions are given as fractions,
2
A′( x) =
4x
108 x
they should be written with negative
exponents before differentiating.
108
(c)
4x
=
0
When
manipulating
an
expression
it
2
For example:
x
is
3
often
easiest
to
convert
a
negative
2
3
4x
=
108 ⇒
x
=
27
Rewrite
exponent
to
a
y =
as
fraction.
2
x
x
=
3
dy
2
y = 2x
Height
is
3
m.
Because
need
the
to
the
make
answer
Having
could
rather
question
it
from
written
then
than
clear
be
part
the
said
that
you
are
you
= −4 x
dx
using
(b).
first
solved
‘hence’
3
⇒
line,
using
this
the
equation
GDC,
analytically
.
Remember that if the domain is restricted then the maximum and minimum
values might occur at the end points rather than where the derivative is equal
to zero. In these cases you should always check by plotting the curve on
your GDC.
227
/
5
5 . 2
I N T E G R AT I O N
Yu shuld knw:
dy
✔
n
=
if
ax
Yu shuld be able t:
a
y
then
n+ 1
=
x
dx
+ c,
for
n ≠
−1,
which
✔
nd
its
can
be
written
ax
dx
n+ 1
=
derivative
x
of
a
function
when
given
or
rate
of
change
✔
nd
the
value
of
c
using
a
boundary
condition,
y
when
0
+ c
the
area
of
the
x-axis,
a
region
enclosed
by
a
curve
y
=
and
the
lines
x
=
a
and
x
=
b,
>
0,
can
be
found
from
calculating
use
trapezoidal
value
of
x
is
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
is
changing
on
is
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
=
=
19t
∫
19
Note
2t dt
1
∫
2
2 ×
t
that
+ c
19 dt
derivative
of
The
of
19t
=
because
19t
is
the
19.
2
2
=
19t
When
t
t
+ c
=
0,
value
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
the
at
t
t
=
0
curve
T
=
plot
or
it
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
to
find
for
the
>
(shown
b
a
area
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
with
,
≤ b
hall
can
b
y
be
equation
2
y
=
2.8 x
0.5 x
horizontal
0
≤
x
distance
from
theunits
are
measured
(a)
the
value
Find
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
and
the
4
trapezoids
to
find
an
approximate
value
for
the
area
between
the
curve
6
y
=
,
the
=
0.5
2 +
x-axis
lines
x
=
1
and
x
=
3
x
Solution
3
h
1
=
b
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) <
and
✔
n
of
curve
the
0
of
is
concave
curve
inexion
changes
(AHL)
Yu shuld be able t
x
✔
D i f f E R E N T i AT i o N
and
is
is
concave
a
the
down
point
✔
connect
✔
use
rule,
product
rule
and
rules
variables
with
related
rates
of
change
up
at
which
interpretation
second
between
the
of
the
this
local
minimum
derivative
maximum
to
distinguish
and
local
points.
context.
higher
level
you
need
to
be
aware
of
the
following
derivatives:
These are given in section 5.9
x
f (x)
sin x
cos x
tan x
n
e
ln x
x
, n ∈ 
and 5.11 of the formula book .
1
1
n
x
f ′(x)
sin x
cos x
nx
e
2
cos
1
x
x
The chan rule
The
chain
rule
is
Nte
used
to
differentiate
composite
functions.
For
example
This can be remembered by
dy
if
u
is
a
function
of
x,
and
y
=
g(u)
dy
then
=
g ′(u) × u′( x)
or
dx
dy
du
=
dx
regarding the terms as fractions
×
du
dx
and ‘cancelling’ du
The
way
this
differentiate
is
done
both
in
and
practice
then
is
to
multiply
separate
the
if
y
=
sin( x
functions,
them.
2
So
two
The chain rule formula is in
2
)
then
the
two
functions
are
differentiated
and
multiplied
u =
x
and
y
=
sin u.
section 5.9 of the formula book .
These
are
dy
together:
2
=
2 x × cos u =
2 x cos x
dx
Try
differentiating
the
answers.
the
functions
in
the
next
example
before
checking
Example 5.3.1
Differentiate
the
following
functions.
1
cos x
2
2
(a)
y
(b)
= (3 x + 1)
y
=
e
(c)
f ( x) =
sin
x
Make sure your GDC is in radian
mode when the question
involves calculus and
Solution
trigonometric functions.
1
2
(a)
u =
3 x + 1,
y
=
u
First
identify
the
two
functions.
1
du
dy
1
2
=
3,
=
dx
du
dy
1
Differentiate
u
1
dx
3 ×
one.
1
3
2
=
each
2
Find
2
u
=
the
product,
changing
all
(3 x + 1)
2
2
variables
back
u
the
to
x
u
(b)
u =
cos x ,
y
du
=
e
dy
=
dx
sin x ,
u
=
u
=
sin x × e
often
function
but
sometimes
not
shown
the
in
brackets,
brackets
are
e
du
dy
is
so
you
need
to
cos x
=
(sin x)e
consider
which
is
the
function
performed.
first
dx
231
/
5
2
(c)
f(x)
=
(sin x)
u
sin x,
2
=
f(u)
=
Rewriting
the
function
it
see
the
easier
to
makes
individual
u
It is wor th learning these common
functions
derivatives rather than using the
f'( x) =
that
composite
chain rule on each occasion.
make
up
the
practice
the
cos x × 2 u
=
function.
2 cos x sin x
After
plenty
of
middle
ax
f (x)
sin ax
cos ax
e
lines
are
often
omitted.
Think
ax
f ′(x)
a cos ax
a sin ax
ae
of
it
as
inside
‘the
derivative
function,
derivative
of
of
the
multiplied
the
outside
by
the
function’
Prduct and utent rule
dv
du
v
dy
These formulae are in
y
=
uv
dv
⇒
=
u
du
u
+
y
v
=
dy
u
dx
⇒
dx
=
2
dx
dx
v
dx
dx
v
section 5.9 of the formula book .
Again,
try
checking
differentiating
the
the
functions
in
the
next
example
before
answers.
Example 5.3.2
Differentiate
the
following
functions.
y
=
2t
x
2
(a)
x
ln x
f ( x) =
(b)
s
(c)
=
1
2 x
e
2
(2t + 1)
Solution
2
(a)
u =
The
x
,
v
=
of
dy
x
×
x
x
is
often
thought
as:
‘keep
the
first
the
same
and
+ ln x × 2 x
dx
=
rule
1
2
=
product
ln x
+
2x
ln
x
differentiate
the
second,
differentiate
the
first,
second
the
plus
keep
the
same.’
2 x
(b)
u =
x,
v
=
e
2 x
The
2 x
dy
derivative
of
e
2 x
is
2e
.
This
2 x
e
×
1
x
×
2e
=
can
2 x
dx
(e
be
worked
out
from
the
chain
2
)
rule
or
simply
learned.
2 x
e
(1
2 x)
1
=
2x
=
2 x
(e
2
2 x
)
If
the
question
just
says
e
‘differentiate’
simplify
,
will
there
though
make
any
is
the
no
need
to
simplification
subsequent
calculations
easier.
Because
derivative
1
(c)
2
u =
2t ,
v
= (2t + 1)
complex
1
dv
1
the
it
should
be
of
v
done
is
quite
first
1
2
=
2 ×
(2t
+
1)
=
using
the
chain
rule.
1
dt
2
(2t
1
(2t
+
1) 2
+
1) 2
1
×
2
2t
At
this
point
the
derivative
has
×
1
dy
(2t
+
1)
2
been
found
but
it
is
likely
that
=
2
dx
1
( ( 2 t + 1)
1
2( 2 t
+
2
some
)
simplification
is
required.
2t
Whenever
1) 2
(2t
+
you
have
fractions
1
1
(2t
1) 2
=
+
1) 2
within
×
fractions
remove
them
by
1
(2t
+
1)
(2t
2( 2 t
+
1)
2t
=
2t
+
+
1) 2
multiplying
top
denominators
2
and
bottom
within
the
by
the
fraction.
=
3
(2t
+
1) 2
3
(2t
+
1) 2
232
/
5.3
D i f f E R E N T i AT i o N
(AHL)
Related rates
If
the
rate
linked
of
change
quantity
can
of
one
quantity
sometimes
be
is
given,
the
calculated
rate
using
of
the
change
chain
of
a
Don’t forget the rate of change
rule.
is just another way of saying
the derivative.
Example 5.3.3
A helicopter
H
is
moving
vertically
upwards
with
a
speed
of
1
10ms
.
The
helicopter
situated
on
which
also
is
level
on
is
h
ground.
ground
m
directly
The
above
helicopter
level,
and
PQ
=
is
the
point
observed
40
Q,
which
from
is
point P,
m.
H
h
m
Nte
When tackling these questions,
P
write down the rate you want to
Q
40
m
find, then 'equals' and the rate you
ˆ
When
per
h
=
30 ,
show
that
the
rate
of
change
of
HPQ
is
0.16
radians
are given. Decide then which extra
second.
rate you need to ensure the two
sides are equal when cancelled,
as illustrated in the example.
Solution
You
dh
=
are
given
the
rate
of
change
10
dt
of
h
dθ
ˆ
Let
HPQ
θ
=
hence
we
need
If
dt
you
want
dθ
By
the
chain
what
what
you
are
you
given
×
dt
dt
then
dh
to
dh
and
down
dθ
dh
=
rule
write
the
find
chain
what
rule
you
can
be
need.
used
Here
you
40
dθ
h
=
40 tan θ
⇒
=
need
2
dθ
2
dθ
=
looking
for
an
equation
linking
and
h
θ
cos
10 ×
are
dh
2
θ
cos
so
θ
cos
=
dt
40
dθ
4
1
,
=
so
find
whichever
is
dh
dh
dθ
When
h
=
cos θ
30,
=
most
0.8
convenient.
2
dθ
⇒
Be
0.8
=
=
dt
careful
to
only
substitute
in
0.16
4
the
value
the
end.
for
the
variable
right
at
Secnd dervatve
Nte
The
derivative
of
a
function
gives
you
the
rate
of
change
of
the
The second derivative has the
dependent
variable
gradient
the
with
respect
to
the
independent
variable
(the
2
d
of
graph).
The
derivative
of
the
derivative
(the
second
y
notation
or
f ′′( x )
2
dx
derivative),
tells
The
gradient
The
curve
The
gradient
The
curve
is
of
you
the
curve
1
described
rate
is
as
of
change
increasing
concave
of
and
this
gradient.
f ′′( x) >
so
0.
Cur ve
up
1
–
At
a
point
is
of
curve
2
described
of
is
as
inflexion,
a
decreasing
concave
curve
and
so
f ′′ ( x ) <
+
0.
down
changes
0
its
concavity
from
concave
0
down
to
concave
up
or
vice
versa.
At
a
point
of
inflexion,
f ′′( x) =
0
+
–
Cur ve
2
233
/
5
Example 5.3.4
Jorge
is
cycling
5
y
2
=
up
a
hill
which
can
be
modelled
by
the
equation
3
(3 x
) ,
x
0
≤
x
≤
2.5 ,
where
x
and
y
are
the
horizontal
and
27
vertical
distances
(in
kilometres)
from
the
foot
of
the
hill.
y
0
2
x
dy
(a)
(i)
Find
dx
(ii)
Hence
and
show
find
that
the
the
summit
height
of
the
to
part
(b)
for
≤
of
hill
the
at
hill
this
occurs
when
x
=
2
point.
2
d
(b)
y
Find
2
dx
(c)
Use
the
the
hill
answer
is
steepest,
0
to
x
≤
find
the
values
of x
and
y
at
which
2
Solution
dy
(a)
5
(i)
dx
The
2
=
(6 x
3x
brackets
could
be
expanded
)
5
27
but
there
is
no
need
as
is
a
27
Nte
constant
f ′( x ) = 0 and
(ii)
f ′′( x ) > 0 at a
At
the
summit
of
Because
the
dy
minimum point
this
=
hill
needs
answer
f ′′( x ) < 0 at a
the
question
to
be
says
done
‘hence’
using
the
0
dx
f ′( x ) = 0 and
term.
5
from
the
previous
part.
2
(6 x
hence
3x
) =
0
Because
the
command
is
that’,
term
27
maximum point
‘show
a
GDC
cannot
2
⇒ 6x
3x
=
3 x(2
x) =
0
be
Hence
From
is
a
x
=
the
local
The
2
x
sketch,
is
2
x
0
working
=
given
2
should
stage
be
of
the
shown.
by
The
height
can
also
be
entering
the
equation
GDC,
the
command
found
into
by
your
) =
27
27
0.741
each
2
20
3
(3 × 2
≈
=
and
maximum.
height
5
or
used
km
as
term
is
‘find’.
2
d
y
5
=
(b)
10
(6
6 x) =
(1
x)
The
hill
is
steepest
when
the
2
27
dx
9
curve
(c)
The
hill
is
steepest
=
⇒
when
to
changes
concave
from
down
concave
(the
up
gradient
2
d
y
stops
0
x
=
increasing).
This
occurs
1
2
dx
at
10
When
x
=
1,
y
d
=
the
point
of
inflexion
when
2
y
=
27
0
2
dx
234
/
5.4
5 . 4
I N T E G R AT I O N
i N T E g R AT i o N
(AHL)
( A H L )
Yu shuld knw:
Yu shuld be able t:
n
✔
denite
and
indenite
integrals
of
x
where
✔
integrate
by
inspection,
or
substitution,
b
1
integrals
x
n
∈
,
including
n
=
−
1,
sin
x,
cos
x,
of
the
form
f (g(x))g′(x)dx
∫
e
and
2
cos
a
x
✔
✔
formulae
for
volumes
dx
or
of
nd
the
curve
b
=
of
the
region
enclosed
y-axes
in
by
a
and
the
x-
or
a
given
interval,
b
2
V
area
revolution
2
πy
V
=
∫
πx
dy
including
∫
a
cases
where
the
area
is
negative
a
2
dv
ds
✔
kinematics:
v
and
=
a
d
=
s
✔
dv
=
=
use
both
integration
and
differentiation
to
v
2
dt
✔
speed
is
the
✔
distance
dt
magnitude
of
nd
ds
dt
velocity
equations
for
displacement,
velocity
or
acceleration.
t
2
travelled
formula,
v(t)
∫
dt
t
1
Integration
main
is
the
families
of
reverse
of
functions
differentiation
can
be
deduced
and
so
from
the
the
integrals
of
the
Nte
corresponding
derivatives.
Remember to add the +
n
x
f (x)
x
, ( x ∈!
c term
1
1
x ≠ −1)
sin x
cos x
when integrating.
e
2
x
cos
x
n+1
x
x
∫
f ( x )dx
+ c
ln
x
+ c
cos x + c
sin x + c
tan x + c
e
+ c
n +1
These formulae are in
1
Because
f ( x) =
ln( x)
is
defined
only
for
x
>
0
and
is
f ( x) =
defined
sections 5.5 and 5.11 of the
for
x
formula books.
1
all
x
≠
0
the
integral
f ( x) =
of
is
given
as
ln
x
+ c ,
so
it
is
defined
over
x
the
same
domain.
interals  the rm ∫  (u(x))u´(x)dx
You
need
These
to
be
consist
able
of
a
to
find
integrals
composite
of
function
the
form
∫
multiplied
f
( u( x) ) u′( x) d x .
by
a
multiple
of
the
2
derivative
of
the
‘inside’
function,
for
example
∫
x sin( x
)dx ,
where
2
x
is
a
multiple
of
the
differentiating
by
can
and
be
spotted
integrating
them
derivative
the
chain
rule.
integrated
using
the
of
x
.
The
After
by
a
process
bit
of
inspection,
substitution
is
the
practice
but
reverse
the
initially
of
results
it
is
worth
technique.
Example 5.4.1
2
x
Find
(a)
∫
4 xe
x
2
dx
(b)
∫
sin x cos
x dx
(c)
dx
∫
2
x
+
3
Solution
2
(a)
Let
u =
x
du
the
substitution
needs
to
When
substituting,
be
spotted.
method
As
in
the
the
‘inside’
chain
rule
function
this
is
of
the
usually
composite
denoted
by u
du
=
dx
Using
2x
⇒ dx
=
all
x
terms
need
to
be
replaced
by
the
equivalent
2x
u
term.
and
This
includes
rearranging
as
if
the
it
dx.
were
This
a
is
substituted
by
differentiating
u
fraction.
235
/
5
2
x
u
4 xe
∫
dx
=
Replace
du
the
x
term
in
the
composite
with
u
and
dx
with
the
4 xe
∫
2x
expression
derived.
u
2e
∫
The
du
2
u
=
terms
x
2e
+
c
=
other
2e
+
of
term
u
in
which
x
will
can
be
then
cancel,
easily
leaving
an
expression
just
in
integrated.
c
2
(b)
sin x cos
∫
By
x dx
rewriting
the
integral
it
becomes
clearer
which
function
is u
2
=
∫
sin
x(cos x)
u =
cos x
Let
dx
Again
du
du
=
sin x
⇒ dx
the
substitution
for
dx
is
found
by
differentiating
u
andrearranging.
=
dx
sin x
2
∫
sin x(cos x)
dx
du
2
=
sin x(u)
∫
sin x
1
2
=
∫
u
3
du =
u
+ c
3
1
3
=
cos
x + c
3
x
(c)
dx
∫
It
x
+
often
easier
to
set
up
the
substitution
by
first
writing
a
quotient
signs
the
3
2
=
is
2
x( x
∫
as
a
product.
In
general,
1
+ 3)
dx
2
Let
u =
x
+ 3
du
du
=
2x
⇒ dx
=
dx
you
should
always
have
the
modulus
in
2x
2
solution,
2
∫
x( x
+ 3)
1
=
∫
du
xu
1
=
2x
2
this
case
x
because
+ 3
>
0
they
could
be
omitted.
2
1
∫
u
du
1
ln
in
dx
1
=
though
1
u
+ c
=
2
ln
x
+ 3
+ c
2
Integrals of the form f (ax) can be found using substitution, but the following
are so common the results should be learned and the integrals found directly:
1
∫
sinax d x = −
∫
a
1
ax
∫
1
cos ax + c
e
cos ax d x =
1
ax
dx =
e
1
+ c
∫
a
sinax + c
a
dx =
ax + b
ln ax + b
+ c
a
The dente nteral and areas
In
section
the
x-axis
5.2
it
and
was
stated
the
lines
the
HL
x
=
that
a
the
and
x
area
=
b
between
can
be
a
curve
found
by
y
=
f ( x),
calculating
b
f ( x) d x .
∫
For
course
you
also
need
to
be
able
to
calculate
this
a
value
directly
using
integration.
b
b
If
∫
f ( x )d x = F ( x ) + c then
∫
f ( x )d x = [ F ( x )]
this is equal to the area between a curve
x = a
= F (b)− F (a) and when
f ( x) > 0
a
a
and
y =
f ( x ), the x-axis and the lines
x = b
236
/
5.4
i N T E g R AT i o N
(AHL)
y
Example 5.4.2
1
1
Find
the
exact
area
between
the
curve
y
,
=
2x
lines
x
=
1
and
x
=
the
x-axis
and
0.8
the
1
5
0.6
0.4
Solution
The
is
question
asked
for
is
it
set
up
cannot
in
be
the
usual
way
calculated
but
because
directly
on
a
the exact
0.2
value
GDC.
0
x
1
2
3
4
5
5
1
Area
=
This
dx
∫
2x
1
integral
is
one
of
those
that
should
1
be
learned,
by
writing
but
it
can
also
be
calculated
5
1
⎡
⎤
=
x
5
1
1
⎣ 2
⎦
as
(2x
∫
1
1)
dx
and
using
the
1
u =
substitution
1
=
2x
1
1
ln 9
The
ln 1
2
values
are
substituted
and
subtracted.
2
1
=
ln 9
=
=
ln 3
Unless
told
otherwise
in
the
question,
any
2
of
If
the
curve
y
=
lies
f ( x)
these
entirely
answers
below
the
would
x-axis
be
acceptable.
between
the
values
y
b
x
=
a
and
x
=
b
then
the
definite
integral
∫
f ( x) d x
is
1
negative.
a
A
2
For
example,
considering
the
y
curve
=
1
x
x
–1
0
1
2
1
1
1
2
(1
∫
x
=
x
⎤
x
⎣
0
2
2
3
⎡
) dx
=
⎦
3
,
hence
the
area
of
A
is
B
,
–1
3
3
0
2
2
1
2
(1
∫
x
) dx
=
x
2
4
⎤
x
⎣
1
2
3
⎡
=
⎦
3
,
=
3
3
unlike
an
integral,
area
is
always
–2
3
1
4
positive,
hence
the
area
of
B
is
3
The
integral
between
1
=
0
and
2
x
x
=
2
will
0
∫
equivalent
(1
x
2
2
2
) dx +
be
to
2
2
(1
∫
x
–3
) dx
so
(1
∫
x
) dx
+
3
0
1
4
=
2
)
=
and
x
(
3
3
2
The
area
between
y
=
1
x
,
the
x-axis
2
of
A
plus
the
area
of
B,
x
=
0
=
2
is
the
area
4
+
hence,
and
When asked to evaluate areas in
=
3
2,
this
can
also
be
calculated
b
3
2
an exam, use
2
directly
as
∫
1
x
dx
=
∫
f ( x) d x
a
2
0
The
area
between
a
curve
f
and
the
y-axis
(shown
as
C
on
the
diagram)
is
(b )
1
given
by
the
formula
∫
f
f
(y) d y
y=f(x)
f(b)
( a)
Example 5.4.3
C
2 x
Find
the
area
between
the
curve
y
=
e
1 ,
the
y-axis
and
the
line
f(a)
y
=
4
a
b
Solution
A sketch
is
useful
to
identify
the
limits
4
of
integration.
see
the
The
to
lower
equation
make
x
the
From
limit
the
is
needs
sketch
we
can
0.
to
be
rearranged
subject.
237
/
5
1
2 x
y
=
e
1 ⇒
x
=
ln( y + 1)
The
integral
can
then
be
done
on
2
the
calculator.
In section 5.12 of the formula
4
1
ln( y + 1)
b
∫
0
book this is written as
x
∫
dx
≈
2.02
There
is
no
as
area
need
to
use
modulus
signs
2
dy ,
a
the
side
where a and b are values on the
of
always
y-axis.
the
is
entirely
axis,
use
but
them
it
on
is
when
the
a
positive
good
finding
habit
to
areas.
Vlumes  revlutn
The formulae for the volumes of solids of revolution formed by rotating a curve
around the axes x- or y-axis are:
y
y
=
f(x)
Around the x-axis:
b
2
Volume =
π y
∫
dx
a
x
These formulae are in
0
a
b
section 5.12 of the formula book .
y
Around the y-axis:
d
2
Volume =
π x
∫
dy
y
=
g(x)
c
d
x
c
x
0
S AMPLE STUDENT ANS WER
x
Let
▲ The
found
is
directly
good
easy
in
answer
to
to
so
the
with
and
if
GDC,
exact
they
f (x)
by
but
best
the
here,
4
−
be
.
The
diagram
but
it
values
if
shows
part
of
the
graph
of
f
are
y
needed
calculated
expanding
approach
nding
student
common
error
multiply
by
it
the
is
x
brackets,
the
one
directly
from
π
has
of
made
forgetting
(the
(a)
Find
(b)
The
but
correct
because
the
the
would
clearly
only
be
set
the
out
If
written
only
then
the
loss
all
graph
of
f
region
enclosed
by
the
graph
of
f,
the
x-axis
and
y-axis
is
rotated
360°
about
the
x-axis.
Find
the
working
of
the
solid
formed.
there
of
answer
the
the
answer
one
x
mark.
of
to
volume
been
x-intercept
the
the
3.425)
has
2e
GDC.
▼ The
is
=
been
f
could
exactly
the
have
parts.
▲ This
done
from
work
do
later
could
was
marks
in
(a)
0 =
this
4 − 2e
x
⇒ 2e
x
=
4 ⇒ e
=
2
⇒
x
= ln 2
≈
0.693
2
ln 2
x
part
would
have
been
lost.
(b)
π
∫
(4 − 2e
)
dx
=
1.09
0
238
/
5.4
i N T E g R AT i o N
(AHL)
K nematcs
V
elocity
(a)
is
(v)
the
is
rate
the
of
rate
of
change
change
of
of
displacement
velocity
,
(s),
and
acceleration
hence:
These formulae are in
2
ds
v
dv
=
and
a
d
=
s
=
section 5.13 of the formula book .
2
dt
dt
dt
Remember, displacement is the position of the object relative to a fixed point
which is not necessarily equal to the distance travelled by the object. Section 5.13
Nte
of the formula book gives the following equations for calculating the displacement
of an object and the distance travelled by the object between times t
t
By the chain rule,
and t
1
2
t
2
2
dv
displacement =
v(t)dt
∫
distance travelled =
v(t) dt
∫
t
a =
t
1
dt
1
ds
=
dv
×
dt
dv
= v
ds
ds
If boundary conditions are given, displacement can also be calculated by
which is useful if acceleration is
integrating velocity and finding an explicit equation for displacement at time t
given as a function of displacement.
Example 5.4.4
A particle
P
moves
in
a
straight
line.
Its
acceleration
at
time
t
is
2
given
(a)
a
by
Find
=
2t
the
7t + 5
values
of
for
t
t
≥
0
when
a
=
0
Kinematics in two dimensions is
(b)
Hence
or
velocity
otherwise,
of
P
is
find
all
possible
values
of
t
for
which
the
covered in section 3.6
decreasing.
1
When
t
=
0,
the
(c)
Find
an
(d)
Find
the
is
velocity
expression
total
of
for
distance
P
is
the
4
ms
velocity
travelled
of
by
P
P
at
time
when
its
t.
velocity
decreasing.
Solution
(a)
(2t
5)(t
1)
=
These
0
values
obtained
t
=
1,
could
directly
also
from
be
the
GDC.
2.5
6
4
2
Nte
0
0.5
1
1.5
2
2.5
3
Decreasing velocity is not always
3.5
the same as decreasing speed.
Think of acceleration as a ‘push’
(b)
1 < t
<
By
2.5
definition,
decreasing
velocity
(acceleration and force are linked
occurs
where
the
acceleration
by the equation F = ma). When a
is
negative.
par ticle is moving in the negative
2
(c)
v
=
∫
(2t
Remember
7 t + 5) dt
find
2
=
7
3
t
‘+
c’
integrating
term.
It
is
not
to
always
direction its velocity is negative,
so a negative acceleration
2
t
3
the
when
+ 5t + c
equal
to
the
initial
(a ‘push’ in the negative direction)
value.
2
will cause the velocity to become
When
t
=
0,
v
=
4 ⇒ c
=
4
more negative which will also
2
v
=
3
7
2
t
3
t
+ 5t +
result in an increase in speed.
4
2
Hence if a and v have the same
signs the speed is increasing and
(d)
distance
Using
=
the
distance
formula
from
if they have different signs the
the
formula
book.
2.5
2
∫
1
3
7
t
3
2
t
+ 5t
+
4
dt
≈
8.41
speed will decrease.
2
239
/
5
5 . 5
SOLUTIONS OF DIFFERENTIAL EQUATIONS (AHL)
Yu shuld knw:
✔
how
to
construct
✔
Euler ’s
Yu shuld be able t:
and
use
a
slope
eld
diagram
✔
set
a
method
solution
to
for
rst
nding
order
λ
the
the
x
solution
=
Ae
λ
t
equations
✔
of
p
✔
if
the
equations
distinct
eigenvalues
a
differential
for
positive
or
of
the
form
x
=
ax + by
y
=
cx + dy
use
Euler ’s
coupled
⎨
are:
✔
with
move
away
negative
or
complex
positive
from
with
use
real
the
solutions
move
complex,
the
solutions
real
imaginary
,
the
the
form
solutions
a
form
your
equations
future
paths
for
and
as
equilibrium
saddle
(one
qualitatively
real,
complex
identify
key
features
stable
populations
spiral
a
convert
second
order
differential
equations
2
circle
the
x
form
coupled
signs
inbuilt
points
dx
=
f
(
2
different
to
distinct,
points,
x,
, t
dt
dt
with
the
origin
ellipse
real
using
and
eigenvalues
trajectories
d
•
variable
GDC
diagrams/portraits
imaginary
sketch
of
or
single
part,
✔
•
solve
part,
and
•
to
origin
negative
towards
on
phase
such
all
method
differential
functions
✔
•
separation
⎩
eigenvalues
complex
solutions
by
variables
and
all
equation
coupled
analyse
•
from
2
✔
real,
equation
context
solve
t
+ Be
⎧
with
model/differential
2
p
1
differential
a
approximate
differential
1
✔
up
positive,
rst
order
)
into
differential
a
system
equations
of
using
one
dx
negative)
the
origin
is
a
saddle
point.
the
y
substitution
=
dt
A differential
equation
be
directly
,
is
one
containing
a
dy
integrated
for
derivative.
2 x
=
example,
⇒
y
=
the
equation
the
solution
the
variables.
is
can
given
in
terms
sometimes
be
of
the
found
dependent
by
the
∫
e
dx
=
can
2 x
e
+ c
2
variable
process
simplest
1
2 x
e
dx
If
The
then
of separating
Example 5.5.1
The
velocity
of
an
object
falling
in
a
viscous
liquid
is
proportional
The phrase ‘is propor tional to’
to
the
square
root
of
the
distance
h
that
it
has
fallen.
or ‘is inversely propor tional to’ is
1
equivalent to ‘varies directly as’
It
is
given
that
when
h
=
1
cm
the
velocity
of
the
object
is
5
cm
s
or ‘varies inversely as’. See
(a)
Find
an
(b)
Hence
equation
connecting
v
and
h
Section 2.3, page 62.
object
find
at
a
time
general
equation
for
the
distance
fallen
by
the
t
Solution
The
(a)
v
given
values
are
substituted
=
to
when
h
=
1,
hence
a
=
5
v
h
=
5
v
=
find
the
unknown
parameter.
5
240
/
5.5
1
V
elocity
dh
is
the
S oLUTioNS
derivative
of
DiffERENTiAL
E q U AT i o N S
(AHL)
of
1
2
(b)
=
5h
⇒
dh
∫
dt
=
∫
1
h
5 dt
displacement,
which
here
is
2
denoted
by
h
1
2
∫
h
dh
=
∫
5 dt
dh
can
be
treated
as
a
fraction
dt
1
and
2
2h
=
separated,
cross
multiplying
5t + c
is
then
used
to
put
the
same
2
5t
h
=
+
c
(
)
variables
on
the
same
sides.
This
2
1
gives
dh
=
The
5dt.
two
sides
1
2
h
are
then
Only
one
when
are
integrated.
‘+
both
c’
term
sides
is
needed
of
an
equation
equation
or
a
integrated.
A general
solution
will
still
constant
to
a
differential
have
of
general
an
equation
unknown
integration.
When the rate of growth of a population of size N is propor tional to N then the
growth is exponential.
dN
1
= kN ⇒
dt
∫
dN =
∫
N
⇒ ln N = kt + c
k dt
(the modulus signs are not needed
as N > 0)
kt +c
This can be rearranged to give N = e
c
= e
kt
e
c
kt
=
Ae
A = e
, where
. This is
such a common substitution that normally solutions would go straight from
kt
lnN = kt + c to N =
Ae
. As A is the value of N at time t = 0 it is often written as
kt
N = N
e
0
The
only
other
set
of
differential
equations
that
need
to
be
solved
Nte
without
the
using
numerical
methods
are
coupled
differential
equations
of
form:
These equations could also be
dx
=
ax + by
=
cx + dy
written as
dt
dy
dt
⎛
These
can
be
written
as
the
matrix
equation
⎞
x
⎜
⎟
y
⎝
⎛
When the eigenvalues, λ
and λ
a
b
solution is
=
c
d
⎞
=
⎜
⎝
⎟
⎠
⎛
x
⎜
⎞
y
=
cx + dy
⎟
y
⎝
⎠
This equation is given in
⎟
c
d
section 5.17 of the formula book .
⎠
⎛
λ
t
1
b
ax + by
are real and distinct then the
⎜
⎝
λ
a
=
⎞
, of
2
1
⎠
⎛
x
x
⎞
t
2
Ae
where x is the vector
+ Be
1
2
⎜
and p
⎟
1
y
⎝
and p
are
2
⎠
Finding eigenvalues and
the eigenvectors corresponding to the eigenvalues λ
and λ
1
2
eigenvectors was covered in
section 1.5, from page 37.
Coupled
equations
competing
of
this
form
are
often
used
to
model
populations.
241
/
5
Example 5.5.2
In
order
foxes.
to
The
control
early
population
foxes
were
(a)
Find
(b)
Given
of
population
growth
rabbits
of
the
of
rabbits
two
(measured
on
an
populations
in
100s),
y
is
island,
can
be
the
land
owners
modelled
the
population
x
=
1.5 x
y
=
x
of
by
the
foxes
introduced
equations
and
t
is
the
a
population
below,
time
in
where x
years
of
is
the
since
the
introduced.
general
that
values
a
of
equations
all
t,
for
parameters
find
the
long
the
in
population
the
term
general
ratio
of
of
rabbits
solution
foxes
0.5 y
to
and
are
foxes, t
positive
years
and
the
after
the
model
foxes
remains
are
introduced.
valid
for
high
rabbits.
Solution
⎛
(a)
λ
1.5
⎞
0.5
det
=
⎜
0
The
equation
can
be
written
as
⎟
λ
1
⎝
⎠
⎛
λ ( 1.5
λ ) + 0.5 =
⎞
x
⎛
0
⎜
⎟
⎞
0.5
⎜
y
⎝
1.5
=
⎟
1
⎝
⎠
0
⎠
⎛
⎞
x
⎜
⎟
y
⎝
⎠
2
λ
1.5 λ
+ 0.5
=
0
The
λ
=
0.5,
⎛
When
λ
=
1
0.5
⎞
0.5
⎜
⎟
1
⎝
0.5
⎠
⎛
⎞
x
⎜
⎛
step
is
to
find
the
eigenvalues.
⎟
To
⎜
y
⎝
⎞
0
=
⎠
0.5 y
=
0 ⇒
y
=
find
the
1.5
λ
eigenvectors
solve
either
⎟
0
⎝
⎠
⎛
x
first
1
⎞ ⎛
0.5
⎞
x
2x
⎟ ⎜
⎜
⎟
⎠ ⎝
0
⎜
y
λ
1
⎝
⎛
⎞
=
⎟
0
⎝
⎠
⎠
or
⎛
When
λ
=
0.5
⎞
0.5
1
⎜
⎟
1
⎝
x
y
=
0 ⇒
y
=
1
⎠
⎛
⎜
⎟
1
0
⎝
⎞
x
⎛
⎟
⎜
⎠
⎝
⎟
⎛
=
2
1.5
⎞
0.5
⎟
1
1
0
⎠
⎛
⎛
⎞
x
⎜
⎟
=
λ
1
⎜
⎠
⎝
⎞
⎛
⎝
⎠
2
⎟
⎜
⎠
⎝
form
of
the
solution
1
values
of
t
the
term
1
formula
can
also
terms
be
of
book
(section
written
to
5.17).
separately
,
The
two
giving
x
equations
and
y
in
t:
x
=
Ae
y
=
2 Ae
t
+ Be
t
+ Be
⎠
ratio
of
x
to
y
is
1
:
1
but
measured
in
100s.
x
is
the
number
of
0.5 t
> e
so
the
long
term
ratio
of
rabbits
foxes
from
⎟
1
The
t
directly
will
⎝
e
come
⎠
⎞
Be
⎜
as
can
⎞
⎟
1
the
0.5 t
dominate
⎠
⎠
+ Be
⎜
⎛
large
⎟
y
⎝
⎠
⎟
1
t
t
For
⎜
y
⎝
0.5 t
(b)
⎞
x
⎞
The
⎟
Ae
y
⎝
⎛
and
0.5 t
⎜
⎞
are
⎝
is
⎞
0
⎜
⎠
⎜
Solution
⎛
=
x
eigenvectors
⎛
⎞
y
⎝
⎛
Possible
x
rabbits
will
tend
to1:100
Phase darams / pr trats
Phase
y
as
t
diagrams
increases.
might
change
show
They
for
the
are
change
useful
different
in
for
values
the
sizes
of
indicating
of
initial
populations
how
the
of x
and
populations
populations.
242
/
5.5
S oLUTioNS
of
DiffERENTiAL
E q U AT i o N S
(AHL)
Example 5.5.3
(a)
Draw
(b)
Use
the
the
phase
phase
introduced
diagram
diagram
when
the
for
to
the
population
explain
rabbit
what
of
would
population
was
foxes
and
happen
rabbits
to
the
from
two
example
populations
5.5.2
if
5
foxes
were
200.
Solution
Only
(a)
y
=
the
When
2x
with
first
quadrant
drawing
real
a
distinct
corresponding
to
phase
is
needed
diagram
eigenvalues,
the
as
for
add
x,
a
>
0
system
the
eigenvectors.
y
two
lines
Because
the
=
eigenvalues
are
eigenvectors
Because
y
=
eigenvalue
the
the
trajectory
y-axis.
from
This
the
means
point
that
(2,
the
5)
ends
rabbits
becomes
extinct.
The
model
be
no
longer
valid,
but
it
of
this
line,
origin.
with
will
the
the
move
of
foxes
would
unless
they
have
is
then
an
clear
line
of
the
that
trajectories
eigenvectors
cannot
also
be
to
crossed
so
in
the
top
part
of
towards
the
direction
the
become
alternative
diagram
need
the
to
move
1
1
⎞
rather
⎟
1
⎝
food
⎛
source
parallel
⎠
⎜
extinct
towards
eigenvector
⎛
population
largest
will
the
then
trajectories
both
population
The
of
the
associated
the
along
⎟
1
⎝
at
is
from
direction
.
⎜
The
away
the
⎞
1
to
(b)
x
direction
⎛
is
positive,
⎠
⎞
than
rabbits.
⎜
⎟
1
⎝
⎠
y=2x
y=x
Note:
these
x
=
if
requested,
initial
2, y
=
5
values
and
t
=
the
particular
could
0
into
be
the
solution
found
by
general
for
substituting
solution.
imanary r cmplex eenvalues
If
the
as
eigenvalues
above,
the
= ( Ae
( a
p
a
solution
( a + bi ) t
x
for
time
the
can
be
bi ) t
+ Be
system
written
at
p
1
This
coupled
) =
e
can
split
the
form
a ± bi
then,
As
section
bit
p
+ Be
p
1
be
of
as
bit
( Ae
2
solution
are
into
)
2
two
parts.
in
1.4
bit
e
can
be
written
as
cos bt + i sin bt
and
represents
circular
motion,
In an exam you will not need to
2π
the
period
of
which
is
find the equation for any system
b
with complex or imaginary
If
a
=
0,
then
the
trajectories
form
circles
or
ellipses
centred
If
a
>
0,
then
as
t
increases
the
trajectories
spiral
away
If
a
<
0,
then
as
t
increases
the
trajectories
spiral
towards
on
(0,
(0,
0)
0)
eigenvalues but will need to
be able to draw and interpret
from
phase por traits.
(0,
0)
243
/
5
Example 5.5.4
Consider
x
=
y
=
the
system:
x + 3y
4x
Sketch
3y
the
trajectory
for
x
and
y
given
x
=
2,
y
=
0
at
time
t
=
0
Solution
The
⎛
λ
1
first
step
det
=
⎜
3
λ
the
eigenvalues.
0
λ )(
3
λ ) + 12
=
2
+ 2λ
2
+ 9
±
=
4
eigenvalues
mean
the
trajectory
is
a
spiral.
0
The
λ
find
⎠
Complex
λ
to
⎟
4
⎝
(1
is
⎞
3
actual
need
0
to
equation
know
the
is
not
sign
of
needed,
the
trajectory
spirals
inwards).
(2,
spirals
inwards.
real
W
e
but
part
know
to
draw
(here
the
it
the
is
trajectory
negative
trajectory
so
passes
you
the
through
36
=
=
1 ±
i
0)
and
W
e
still
need
to
find
whether
it
spirals
2
At
(2,
0)
x
=
2,
y
=
clockwise
or
at
anticlockwise.
This
is
done
by
calculating
the
values
8
x
and
y
The
fact
that
of
(2,
y
‘downwards’).
0)
is
negative
The
means
gradient
at
the
(2,
0)
spiral
can
be
is
clockwise
found
(motion
using
the
is
chain
(2, 0)
dy
rule
dy
=
dx
8
dx
=
÷
dt
=
4
2
dt
Numercal slutns
Most
differential
methods
need
to
equations
be
used
cannot
to
find
be
solved
qualitative
exactly
,
or
so
approximation
approximate
solutions.
Slpe elds
A differential
equation
A slope
is
field
indicating
them
a
will
diagram
by
a
give
that
small
the
gradient
shows
line
these
segment,
as
of
a
curve
gradients
in
the
at
at
any
many
diagram
value.
points,
below,
dy
which
is
the
slope
field
for
the
differential
equation
=
x
y
dx
For
example,
the
point
(2,
0)
has
a
gradient
of
2
−
0
=
2
Nte
5
From the equation we can see that
4
all maximum or minimum points
dy
= 0
will occur when
, which in
3
dx
this case is along the line y = x
2
1
Nte
–5
–4
–3
–2
–1
0
1
2
3
4
5
These diagrams are not phase
–1
por traits, as phase por traits show
trajectories over time. These
–2
diagrams simply show solutions to
–3
the differential equations.
–4
244
/
5.5
The
diagram
starting
indicates
values.
approximation
directions
of
the
of
It
is
for
the
solutions
the
graphs
possible
the
line
to
use
of
a
graph
segments.
which
pass
of
the
the
solutions
line
solution
diagram
A(2 ,
through
all
segments
particular
The
for
below
1)
and
S oLUTioNS
to
construct
an
by
following
the
3,
DiffERENTiAL
E q U AT i o N S
(AHL)
different
shows
B(
of
the
graphs
3)
5
4
B(–3, 3)
3
Nte
2
The accuracy of the values
obtained will depend on the size of
1
h and how far your approximation
is from the original point.
–5
–4
–3
0
3
4
5
–1
A(2, –1)
–2
–3
It
can
on
be
the
Both
seen
y
line
of
line
y
=
and
on
the
x
that
=
x
the
as
minimum
(or
the
graph
that
passes
solutions
x
y
=
1 ).
This
the
gradient
indicate
is
there
because
the
is
an
asymptote
gradient
of
dy
this
line
through B
lies
predicted.
particular
1
of
of
the
curve,
this
along
line
is
the
1
dy
,
is
also
1
dx
(
=
x
y
=
1
dx
)
Make sure you know how to use
Euler ’s methd
Euler ’s method using the in-built
Euler ’s
method
uses
the
same
idea
as
slope
fields.
It
assumes
that
the
functions on your GDC.
gradient
point
at
and
repeats
a
point
uses
the
remains
this
to
find
constant
an
for
a
small
approximation
distance
for
a
new
(h)
beyond
point.
It
the
then
process.
The formula for Euler ’s method is
The
formula
is
y
=
y
n+ 1
+
h ×
f (x
n
, y
n
);
x
n
=
n+ 1
x
+
h
where
h
is
a
constant
n
in section 5.16 of the formula book .
(step
length)
S AMPLE STUDENT ANS WER
▲ The
y
Consider
the
differential
=
equation
x +
y ,
(x +
y
≥
0)
dx
the
initial
Use
conditions
Euler ’s
using
a
method
step
length
y
to
of
=
1
when
find
h
=
an
0.1.
x
=
1.
Let
y
=
approximation
Give
your
c
when
for
answer
the
x
=
2.
value
correct
student
y
+ 0.1
n
x
+
to
4
of c
are
entering
into
GDC.
If
they
did
no
work
alone
would
marks
1.1 ,
1
=
=
2,
y
be
of
the
6
out
all
the
wasted
intermediate
one
or
it
is
two.
no
By
bad
thing
writing
to
this
here
further
the
student
marks
achieved
which
would
not
2.77011
10
have
when
should
though
answer
=
not
3
question.
1.14142
1
10
this
n
two
x
scored
for
this
y
n
y
other
the
dp.
write
=
have
available
stages,
x
the
they
writing
=
n+ 1
written
formula
▲ Time
y
has
satisfying
x
=
2,
y
=
been
given
otherwise.
2.77
▼ This
given
answer
to
the
accuracy
.
shown
for
this
been
has
not
requested
Without
above
the
question
the
been
degree
of
methods
implied
would
marks
not
have
awarded.
245
/
5
Slvn cupled derental euatns usn Euler ’s methd
Most
coupled
sonumerical
differential
methods
equations
need
to
be
cannot
used.
be
Consider
dx
system
of
differential
equations
solved
the
coupled
=
( x , y , t)
dy
=
f
( x , y , t),
f
1
2
dt
Euler ’s
x
=
method
x
n+ 1
+
h ×
f
n
uses
(x
1
the
, y
n
, t),
following
y
n
=
y
n+ 1
+
exactly
,
dt
formulae:
h ×
f
(x
n
2
, y
n
t
, t),
=
t
n+ 1
n
+
h,
where
h
is
n
These formulae are given in
a
constant
(step
length)
section 5.16 of the formula book .
Example 5.5.5
The
differential
rabbits
and
that
the
equations
foxes
in
used
example
for
5.5.2
modelling
is
the
modified
to
population
allow
for
of
the
fact
Make sure you know how to
as
populations
increase
the
resources
to
support
them
are
use Euler ’s method for coupled
restricted.
The
new
set
of
differential
equations
proposed
are
differential equations using the
inbuilt functions on your GDC.
2
x
=
1.5 x
y
=
2x
in
t
is
100s)
have
0.1x
0.2 xy
where
Nte
0.5 y
measured
and
two
y
is
the
non-zero
in
years,
number
points
of
x
of
is
the
foxes
number
(x,
y
≥
of
0).
rabbits
The
(measured
populations
equilibrium.
An equilibrium point is one at
(a)
dx
which both
dt
Find
the
numbers
of
rabbits
and
foxes
at
each
point
of
equilibrium.
dy
and
are equal
dt
A survey
finds
that
there
are
400
rabbits
and
6
foxes.
to zero.
(b)
(c)
Use
Euler ’s
find
the
method
number
a
rabbits
answers
to
State
equilibrium
the
the
of
with
nearest
step
and
whole
point
length
foxes
of
one
one-tenth
of
year
giving
later,
a
year
to
both
number.
the
populations
are
tending
towards.
Solution
(a)
y
=
2x
0.2 xy
=
0 ⇒
x
=
To
0
find
the
y
=
10
=
0
and
x
⇒
=
y
0,
=
x
=
0.5 y
=
0
simultaneously
.
0
0
Equilibrium
(0,
=
solve
dt
dt
If
points
dy
dx
or
equilibrium
0);
no
If
y
=
x
=
1.5x
point
foxes
or
is
rabbits.
10:
2
⇒
x
=
5
5,
x
−
=
10)
and
Hence,
=
0
10
Equilibrium
(5,
0.1x
points
(10,
10)
population
equilibrium
are
points
at
the
are
10
T
foxes
1000
(b)
x
and
either
500
or
=
x
2
( 1.5 x
y
=
6.91319756974
4.52156049834
7.18488387399
0.5
4.63670528999
7.43945823146
0.6
4.74461344509
7.67687705602
0.7
4.84734804162
7.89732346311
0.8
4.94461624434
8.10117156298
0.9
5.0367578048
8.28895112168
1.0
5.12413462759
8.46131389752
1.1
5.20712157004
8.61900259229
+ 0.1 ×
n
Always write down the formulae
sure of getting some marks, even
4340077248851
0.4
rabbits.
n+ 1
for Euler ’s method so you can be
0.3
0.5 y
n
0.1( x
n
y
n+1
+
0.1
n
)
n
×
(2x
n
)
0.2x
n
if you make an error entering them
y
)
n
into your GDC.
t
n+1
=
t
+
0.1
After
one
year
the
values
of
x
and
n
y
are
5.124…
and
8.461…
246
/
5.5
After
of
one
rabbits
number
(c)
year
1000
is
of
the
512
and
foxes
rabbits,
number
10
is
These
the
the
8.
need
to
be
population
S oLUTioNS
rounded
to
the
to
of
DiffERENTiAL
E q U AT i o N S
(AHL)
give
nearest
integer.
foxes
This
the
t
can
be
found
table
to
find
by
continuing
values
of
x
and
y
as
increases.
Secnd rder derental euatns
There
are
many
practical
processes
which
can
be
described
using
2
d
second
order
differential
equations
of
the
x
dx
=
form
f
(
2
x,
These
of
can
be
coupled
solved
first
by
order
dx
writing
the
second
order
, t
dt
dt
equation
as
)
a
system
equations.
dy
=
Let
y,
=
then
dt
f ( x , y , t)
dt
Example 5.5.6
The
motion
resistance
of
can
parachutist
be
falling
modelled
by
under
the
gravity
but
subject
to
air
equation:
2
2
d
x
In an exam, the question will
dx
=
9.8
0.004
2
(
)
always give the differential
dt
dt
equation, so knowledge from
where
x
is
the
distance
fallen,
t
seconds
after
he
leaves
the
plane.
outside the syllabus will not
The
initial
velocity
of
the
parachutist
can
be
taken
as
equal
to
be required.
zero.
2
d
(a)
State
x
dx
what
and
represent
in
the
context
of
the
question.
for
the
velocity
2
dt
dt
(b)
Use
the
the
differential
equation
to
find
the
limit
Nte
of
parachutist.
This limit is referred to as the
(c)
Use
Euler ’s
distance
method
fallen
(ii)
with
the
a
step
velocity
length
of
the
of
0.2
to
find
parachutist
(i)
after
the
termnal velcty
2seconds.
Solution
2
d
x
dx
is
(a)
the
acceleration
and
is
the
velocity
2
dt
dt
of
the
parachutist
2
dx
dx
(b)
9.8
0.004
(
9.8
1
=
)
0
⇒
=
dt
dt
=
49.5
m
s
The
limit
)
It
important
will
occur
when
the
acceleration
is
zero.
0.004
2
dx
dy
=
(c)
dx
y,
=
dt
9.8
0.004
dt
(
2
)
= 9.8
0.004 y
dt
2
x
=
x
n+ 1
+ 0.2 y
n
,
n
y
=
n+ 1
y
+ 0.2(9.8
n
0.004 y
is
been
t
=
t
n+1
show
the
formulae
that
have
used.
0.2
n
When
(i)
+
to
n
x
t
=
2
≈ 17.3
m
dx
1
(ii)
y
=
≈ 18.8
m
s
y
is
the
velocity
dt
247
/
5
6.
SL
P R A CT I C E
A straight
tunnel
has
a
uniform
cross-section
Q U E ST I O N S
in
the
shape
of
a
parabola
with
equation
2
PA P E R
1,
GROUP
1
y
4
1.
Consider
the
curve
y
=
2x
=
3.6 x
0.6 x
tunnel
in
metres
from
,
where
metres
at
a
y
is
the
height
horizontal
of
the
distance
of
x
7x
the
origin,
O.
dy
a.
Find
dx
The
b.
point
Find
P(
the
1,
9)
lies
gradient
on
of
the
the
curve.
tangent
at
point
P
2
y
c.
Find
the
equation
ax + by + c
=
0,
of
where
2
2.
a.
Find
the
value
this
a,
the
c
are
find
trapezoidal
an
rule
with
approximation
the
approximation
percentage
A curve
y
has
the
tunnel.
x
found
in
error
for
the
The
2
=
derivative
Find
3x
8x
the
width
height
of
the
b
part
dy
3.
maximum
O
dx
a.
find
the
trapezoids
1
for
1
Hence,
and
form
integers.
four
∫
c.
0.6x
x
2
to
the
–
dx
1
Use
and
in
3.6x
1
of
∫
b.
b
tangent
=
+
1.
cross-section
tunnel
is
50
m
of
long.
Find
dx
b.
the
equation
the
point
for
y,
given
that
it
passes
Find
the
volume
of
earth
removed
to
create
through
thetunnel.
4.
A curve
(2,
C
7)
is
defined
by
the
equation
GROUP
2
4
y
=
x
32 x + 8
1
7
.
The
equation
of
a
curve
is
y
=
3
x
2
3
point
P
dy
a.
2
4
x
6x
3
Find
dy
dx
a.
Find
dx
Let
P
be
the
minimum
point
on
the
curveC.
The
b.
Write
down
an
equation
satisfied
by
of
of
the
curve
at
a
is
30.
the
b.
x-coordinate
gradient
Find
the
coordinates
of
P
.
P
.
8.
The
average
cost
of
C
producing
x
light
bulbs
A
c.
Hence,
find
the
coordinates
of
P
dC
A
changes
5.
Mika
needs
to
estimate
the
area
of
a
fish
at
a
rate
2
=
of
6x
USD
per
unit.
design
dx
that
she
is
creating.
coordinate
she
plots
grid,
eight
She
places
marked
points
out
as
in
the
design
on
centimetres,
a
Given
and
4.5
the
USD,
average
cost
of
producing
12
units
is
find:
shown.
a.
the
b.
an
average
cost
of
producing
15
units
(3, 2)
(2, 1.8)
(4, 1.8)
(7, 1.8)
expression
for
the
total
cost
C
of
producing
2
(1, 1.4)
(5, 1.5)
x
units.
1
9.
A ramp
a
0
–1
consists
of
a
straight
Q
to
section
P
to
Q,
and
(6, 0.5)
(0, 0)
1
2
3
4
5
6
curved
section
R.
7
The
two
sections
can
be
modelled
by
the
–1
piecewise
function
–2
⎧
⎪
f
( x) =
ax + b
0
≤
x
<
4
4 ≤
x
≤
8
⎨
2
⎪
⎩
Use
the
value
trapezoidal
for
the
area
of
rule
the
to
find
an
0.2 x
2.4 x + 7.2
approximate
fish.
248
/
PRACTICE
d.
.
Explain
why
the
maximum
QUE STIONS
value
of
f
does
P
not
occur
values
.
12.
Write
A seller
to
of
when
found
down
ice
estimate
x
in
the
equal
to
maximum
uses
number
of
either
of
the
b
part
creams
the
is
the
value
of
following
sales
of
ice
f.
model
creams,
n,
0
4
given
8
the
1
The
two
branches
of
f(x)
need
to
be
continuous
expected
3
n =
temperature
T °C.
2
T
+ T
+
50
,
0
≤ T
≤
30
50
and
have
the
same
gradient
x
at
=
4
a.
Find
10.
Let
q
the
be
supply
For
a
values
the
at
a
of
a
and
number
price
particular
p
of
b
a
manufacturer
will
b.
Find
unit.
factory
,
the
when
units
per
Find
the
ice
supply
equation
c.
expected
the
an
temperature
expression
cream
Hence,
number
sales
find
for
when
the
is
of
ice
creams
sold
20 °C
the
the
rate
of
increase
temperature
temperature
at
which
is
the
is
in
T °C
the
rate
2
connecting
The
p
factory
and
q
is
receives
p
a
=
0.6q
of
payment
of
€240
perunit.
d.
State
of
a.
Find
the
supply
number
for
this
of
units
(q
0
that
)
they
increase
n
graph
of
the
ice
cream
significance
against
of
sales
this
point
greatest.
on
the
graph
T.
will
price.
GROUP
The
the
in
supply
function
is
shown
3
below.
3
13.
A function
f
is
given
by
f
( x) =
2x
+
2,
x
>
0
x
p
a.
Find
b.
Show
the
derivative
of
f.
240
graph
that
of
the
x-coordinate
of
the
point
on
given
by
the
f
1
is
S
equal
to
–4
is
2
14.
Consider
the
graph
of
the
parabola
R
2
( x) =
f
4x
x
q
A rectangle
ABCD
is
drawn
with
upper
vertices
q
0
B
b.
Find
the
curve,
area
the
of
the
q-axis
region
and
the
R,
line
bounded
q
=
by
the
and
as
C
on
shown
the
in
graph
the
of
f,
and
AD
on
the
x-axis,
diagram.
q
0
Economists
by
the
call
curve,
representing
c.
11.
Find
the
Consider
the
the
p-axis
price,
the
producer
the
of
region
and
the
bounded
horizontal
producer
surplus
S,
3
surplus
when
p
=
240
2
( x) =
a.
Find
b.
Find
the
values
c.
Give
the
range
9x
+ 12 x + 1,
0
≤
x
≤
0
of
of
x
such
values
that
of
x
O
A
a.
Let
D
3
f ′ ( x)
f ′ ( x) <
line
function
f
2x
area
f ′ ( x)
for
=
0
which
OA
=
x
.
Show
.
Hence
area,
that
AD
write
R,
of
=
4
down
the
2x
an
expression
rectangle
in
terms
for
of
the
x.
249
/
5
b.
Find
the
maximum
possible
area
of
y
the
rectangle.
dR
c.
.
Find
dx
Find
x
=
the
rate
Let
f
change
of
the
area
when
1
3
15.
of
thgieH
.
( x) =
2
ax
+ cx ,
+ bx
where
a,
b
and
c
are
real
numbers.
The
graph
of
f
passes
through
the
point
(2,
2)
x
0
5
a.
Show
The
that
graph
8a +
has
a
4b + 2 c
local
=
Find
two
c.
Find
the
other
1
16.
Let
f
( x) =
of
minimum
a,
b
at
(1,
.
Line
L
is
25
30
in
and
a,
b
the
and
c
c
normal
35
travelled
40
45
50
55
(m)
5)
to
the
the
point
P(a,
b)
the
maximum
a.
.
Find
the
values
.
State
the
maximum
graph
of
be
local
2
x
20
Distance
equations
values
15
2
Let
b.
10
coordinates
of
the
h
of
a
and
b
height
of
the
roller
3
coaster.
of
f
at
the
a.
Show
b.
Point
point
the
A
(3,
3)
equation
is
the
x-coordinate
of
L
is
x-intercept
of
x + 2y
of
L.
9
Find
=
b.
Find
h′ ( x )
c.
Find
the
for
≤
0
the
diagram,
a
the
shaded
region
R
is
the
x-axis,
the
of
the
roller
coaster
x
≤
55
Find
the
graph
of
f
and
the
angle
that
the
roller
coaster ’s
track
bounded
makes
by
gradient
A.
d.
In
steepest
the
line
with
the
horizontal
at
this
point.
L.
18.
The
diagram
There
is
a
shows
local
the
graph
minimum
at
of
the
a
function
point
f.
A
y
(3, 3)
R
x
A
A
c.
.
Write
.
Find
down
an
expression
for
the
area
of
R
2
17
.
The
the
height,
travelling
a
h,
area
of
a
of
car
distance
The
R
on
x
is
a
roller
given
coaster
by
the
after
a.
Find
x
x
is
given
x-coordinate
of
f ′( x) =
by
3x
5x
2
A.
coordinates
of
the
y-intercept
of
the
graph
of
f
+
5 x,
0
≤
x
≤
55,
where
all
are
distances
( 0, 1.5 )
5
b.
are
f
2
h( x ) =
500
the
of
equation
The
3
derivative
measured
in
Find
an
expression
for
f
( x)
metres.
The
c.
equation
Find
two
f
( x) =
possible
k
has
two
values
solutions.
for
k.
250
/
PRACTICE
PA P E R
1.
2
HL
A closed
cylindrical
can
P R A CT I C E
r
centimetres
height
h
centimetres
Q U E ST I O N S
r
with
PA P E R
radius
QUE STIONS
1,
GROUP
1
and
2
x
h
1.
Let
f
( x) =
x
e
3
has
a.
a
volume
Express
The
h
of
in
material
50
terms
for
.
cm
of
f ′ ( x)
a.
Find
b.
Hence
r.
the
base
and
the
and
top
of
the
or
otherwise,
find
the
range
of
values
of
can
x
for
which
f
is
an
increasing
function.
2
costs8¢
per
cm
material
for
the
curved
2
sides
costs
6¢
per
2.
.
cm
a.
Sketch
0
b.
Show
that
given
by
the
total
cost
of
the
material,
C,
≤
x
the
≤
graph
of
y
2( x
=
1)( x
3)
for
4
is
4
the
b.
equation
Find
2( x
∫
1)( x
3 ) dx
0
600
2
C
=
16πr
c.
Find
the
total
area
between
the
curve,
the
+
r
x-axis
and
the
lines
x
=
0
and
x
=
4
dC
c.
Find
3.
dr
d.
Given
that
there
is
a
minimum
value
for
The
rate
of
decay
of
a
radioactive
substance
dM
C,
is
directly
proportional
to
the
amount
of
dt
write
down
an
equation
it
must
satisfy
.
substance,
e.
Find
the
value
of
r
for
which
the
cost
of
the
is
=
Find
the
value
of
h
for
this
The
charity
committee
in
a
school
is
planning
doughnuts.
the
number
The
It
is
committee
calculates
Find
doughnuts,
n,
given
an
>
0
and
t
is
measured
in
that
when
t
=
0,
M
=
50 g
it
could
sell
expression
for
M
in
terms
of
k
and
t.
is
given
that
after
400
years,
50%
of
the
that
substance
of
k
to
It
sell
where
can.
a.
2.
kM
dt
minimum.
years.
.
remaining.
dM
Let
can
M,
at
has
decayed.
price
600
x
euros
is
given
by
the
function
,
n =
x
b.
≥ 1
Find
the
value
of
k.
2
x
a.
Write
down
would
an
receive
doughnuts
expression
from
for
x
sales
for
the
when
money
selling
they
4.
Garlic
parts
the
are
euros.
a
They
can
purchase
the
doughnuts
from
at
a
cost
of
0.75
euros
per
of
north
hoping
type
of
Write
down
an
expression
for
n
the
cost
the
profit,
P,
is
doughnuts
sales
of
the
Find
an
control
weevil
A team
the
which
is
of
spread
large
biologists
by
known
in
introducing
to
restrict
model
is
proposed
to
help
calculate
the
of
weevils
that
should
be
introduced
in
a
doughnuts.
found
from
by
the
subtracting
money
the
gained
cost
from
area:
of
x
=
2x
2y
y
=
2x
3y
the
doughnuts.
In
c.
America.
species
of
given
The
invasive
growth.
number
purchasing
an
doughnut.
A simple
b.
to
is
a
its
company
mustard
equation
for
the
profit
made
this
model,
x
euros
per
is
the
number
of
square
metres
when
affected
charging
x
and
y
is
the
number
of
weevils.
doughnut.
It
is
given
that
the
eigenvalues
and
dP
d.
Find
corresponding
dx
e.
Use
your
answer
to
part
d
to
find
the
price
2
2
⎛
⎞
are
the
charity
committee
should
sell
the
⎜
⎝ 2
doughnuts
in
order
to
maximize
their
find
the
number
of
doughnuts
1
⎞
⎛
and
–2,
⎜
⎟
matrix
2
⎞
1,
⎟
⎝ 2⎠
3⎠
they
⎜
⎟
⎝ 1⎠
Sketch
a
phase
portrait
for
this
system
with
need
x, y
to
the
profit,
a.
and
for
at
⎛
which
eigenvectors
≥
0
buy
.
251
/
5
2
b.
One
area
has
minimum
150
m
number
affected.
of
weevils
Find
that
the
GROUP
should
be
7
.
introduced
to
ensure
the
garlic
mustard
2
A particle
moves
along
a
straight
t
seconds
line
so
is
that
its
velocity
,
v,
after
is
given
by
removed.
t
v (t ) =
5.
Given
y
=
ln(2 x
1.4
for
2.7 ,
0
≤ t
≤
5
3)
a.
Find
when
b.
Find
the
acceleration
c.
Find
the
total
the
particle
is
at
rest.
dy
a.
Find
an
expression
for
dx
The
normal
point
(2,
0)
to
the
curve
intersects
y
the
=
ln(2 x
y-axis
at
3)
at
Find
the
value
of
the
distance
the
first
5
when
t
=
2
travelled
by
the
particle
seconds.
c.
2
dy
Consider
particle
c.
8.
6.
the
the
in
b.
of
differential
(y) =
0.2 ( y
0.5 )
+ 0.5
for
0
≤
y
≤ 1.
2
=
equation
f
Let
x
y
1.
Henriuses
as
(y)
f
a
model
to
create
a
barrel.
dx
Part
of
the
equation
slope
is
field
for
this
The
differential
the
shown.
region
line
y-axis.
y
enclosed
=
and
0
This
is
y
by
=
shown
the
is
1
in
graph
rotated
the
of
f,
360˚
diagram
the
y-axis,
about
where
the
the
y
units
are
in
metres.
1
2
f
(y)
=
–0.2(y
–
0.5)
+
0.5
x
0
a.
Find
The
the
barrel
water
rate
a.
.
Find
the
equation
of
the
curve
on
the
at
volume
is
in
filled
the
which
of
the
with
barrel
water
barrel.
water.
is
The
denoted
flows
into
volume
by
the
V
barrel
dV
locally
must
.
C
be
the
point
or
minimum
the
(0,
this
curve
solution
on
the
curve
slope
that
Euler ’s
find
the
method
value
of
y
with
on
this
a
passes
step
curve
be
t
=
by
3
0.6e
m
–1
min
,
where
t
is
began
to
the
be
time
filled.
in
It
minutes
is
given
since
the
at t
that
barrel
=
0,
of
t.
V
=
0
fielddiagram.
b.
Find
an
c.
Find
the
expression
for
V
in
terms
through
0)
Use
can
dt
that
b.
modelled
points
lie.
Sketch
Let
maximum
the
which
approximately
any
and
of
length
when
of
x
=
0.2
maximum
will
be
percentage
of
the
barrel
filled.
to
2
9.
The
size
of
a
population,
N,
can
be
modelled
by
pt
the
c.
On
the
slope
field
diagram,
sketch
the
equation
N
=
te
,
where
p
>
0
solution
dN
that
passes
through
(0,
0)
for
3
≤
x
≤
3
a.
Find
dt
b.
Hence,
find
the
value
of
t
and
=
0
corresponding
dN
value
of
N
for
which
dt
2
d
c.
N
Find
2
dt
d.
Hence,
partb
show
is
a
that
the
maximum
value
for
all
of
p
N
>
found
in
0
252
/
PRACTICE
10.
Consider
the
curve
y
tan ( 2 x
=
QUE STIONS
350
4)
14.
Consider
the
curve
f
,
(t ) =
t
>
0
t
1 +
6e
dy
a.
Find
a.
f ′ (t )
Find
dx
2
d
b.
y
The
function
represents
the
number
of
goats
on
Find
2
dx
an
Let
a
>
0
be
the
x-coordinate
of
the
point
of
b.
inflexion
whose
x-coordinate
is
closest
Write
Find
the
value
of
down:
to
theorigin.
c.
island.
a.
.
the
carrying
.
the
number
capacity
of
goats
of
at
the
t
=
island
0
2
11.
Consider
between
the
the
curve
curve,
y
=
.
x
the
The
x-axis
region
and
the
R
c.
is
line
x
=
Find
a
d.
R
the
rate
Find
The
the
piecewise
for
the
area
of
of
size
of
is
the
goats
at
the
is
rate
of
increase
greatest
this
and
in
state
time.
population
when
the
rate
greatest.
of
a
particle
is
given
by
and
in
the
function
2t
0
≤ t
<
t
3
3
⎨
6
⎪
⎩
expression
which
increase
acceleration
⎧
an
of
increase
a(t ) =
Find
at
population
a
a.
time
the
of
15.
the
≥
R.
–2
where
b.
Hence
or
otherwise,
find
an
expression
for
between
the
curve,
the
y-axis
and
the
=
is
The
velocity
of
a
particle
is
given
by
the
b.
m
s
that
the
velocity
of
the
particle,
v,
is
t
at
=
3
and
t
when
=
0, v
=
9
Find
an
expression
for
the
velocity
of
the
particle.
Find
an
expression
particle
at
Find
expression
an
particle
time
from
for
the
acceleration
of
t
its
for
initial
the
displacement
position
at
time
When
t
=
5,
find:
.
the
total
.
the
particle’s
the
of
the
t
distance
travelled
distance
by
from
the
its
particle
initial
position.
16.
GROUP
13.
a
4te
b.
a.
seconds
equation
2
=
given
continuous
t
v
in
a
a.
12.
measured
line
2
y
is
the
It
area
t
Water
is
pumped
into
a
tank
at
a
rate
of
3
Consider
the
system
of
coupled
W
= [ 4 sin ( 0.25t
At
the
2) +
4]
litres
per
hour.
differential
same
time,
water
is
removed
from
the
litres
per
equations
tank
x
=
3x
at
=
rate
= [3 sin ( 0.25t ) +
of
R
is
measured
4]
9y
hour,
y
a
4x
where
t
in
hours.
3y
Let
V
be
the
volume
of
water
in
the
tank.
dy
a.
Use
the
chain
rule
to
find
the
value
as
of
dV
dx
the
trajectories
a.

the
the
b.
Find
c.
Sketch
the
of
an
expression
for
dt
cross:
b.

Find
Given
the
tank
water,
find
initially
contains
100
litres
of
x-axis
y-axis
the
water
eigenvalues
the
point
trajectory
(2,
0),
for
of
a
this
system.
particle
indicating
clearly
that
the
c.
begins
in
an
the
Find
the
tank
during
expression
tank
at
minimum
a
time
for
volume
of
t.
amount
24-hour
the
of
water
in
the
period.
at
direction
motion.
253
/
5
d.
Find
of
17
.
the
difference
water
flowing
amount
of
24-hour
period.
Let
the
per
litre
bmg
between
into
water
per
the
tank
flowing
out
total
and
of
the
the
2.
amount
x
total
tank
The
over
=
graph
shows
3 cos 2 y +
4,
0
the
≤
y
relation
≤
π
y
a
π
concentration
and
the
the
of
a
chemical
concentration
of
A
a
be
a
mg
chemical
B
be
litre.
da
The
rate
of
reaction
of
a
chemical
A
and
is
dt
2
this
is
proportional
The
rate
to
ab
of
A
x
of
reaction
is
12
mg
per
litre
per
0
-7
second
when
the
concentration
of
A
is
4
mg
per
The
litre
and
the
concentration
of
B
is
2
mg
per
a
da
a.
Show
3
dt
solid
of
=
the
about
the
of
B
is
given
≤
6,
where
by
the
Calculate
the
0.2t
for
0
≤
t
moment
the
two
t
substances
is
Find
an
time
A
at
expression
time
of
the
solid
for
the
t,
where
0
≤
t
=
0.5
when
t
=
base
of
with
this
diameter
with
water
shape
is
made
Hence
write
14
at
a
rate
≤
6,
given
t
minutes
cm.
The
of
cm
after
solid
container
is
filled
2
–1
min
.
The
water
it
starts
being
filled
is
0
≤
h
≤
π,
and
the
volume
of
water
in
the
3
the
A
in
is
V cm
maximum
dV
b.
of
a
that
0
down
concentration
with
concentration
container
c.
formed.
mixed.
hcm,
a
form
from
depth
of
to
equation
the
are
volume
3
b.
y-axis
ab
concentration
1.2
360°
revolution.
A container
b
rotated
4
a.
The
is
2
=
that
curve
litre.
the
interval
0
≤
t
≤
.
Given
2
=
that
π ( 3 cos 2 h
+
4)
,
find
an
d h
6
d h
expression
18.
At
time
t,
a
particle
displacement,
x,
has
a
velocity
,
where
x
>
0,
v,
given
and
by
for
a
d t
d h
the
.
Find
the
value
π
of
h
when
=
d t
following
equation.
4
2
d
c.
.
h
Find
2
d t
2
0.1x
2
e
v
d
=
.
Find
the
values
of
h
for
=
d t
.
Find
an
0
2
x
a.
h
which
expression
for
the
acceleration
of
By
making
specific
reference
to
the
shape
of
d h
the
the
container,
at
interpret
the
values
of
h
d t
particle
in
terms
of
x
found
b.
Find
an
particle
expression
in
terms
of
for
t,
the
given
displacement
that
x
=
0
of
when
t
1.
=
0.
PA P E R
the
differential
In
this
to
model
following
system
of
a.
=
x
the
and
motion
subject
motion
By
be
of
an
modelled
!
! + 36 x
x
first
finding
the
eigenvectors
for
the
system,
construct a phase portrait for
the system
clearly
asymptotes.
object
use
a
variety
of
techniques
of
to
an
objectattached
to
a
externalforces.
object
attached
to
a
spring
the
equations
of
any
Given
find
x
=
an
3
x
=
equation
equation
hence
and
for
write
y
in
4
when
for
x
in
terms
down
the
t
=
0:
terms
of
=
0,
from
where
the
Use
t
x
the
is
differential
the
equation
displacement
equilibrium
=
differentiation
and
an
It
is
to
A sin 6t + B cos 6t
differential
of
by
position,
of
the
O
showing
x
c.
will
x + 2y
a.
b.
you
equations:
can
y
question
spring
coupled
The
=
3
2
Consider
x

part
the
1.
PA P E R
in
given
is
verify
that
a
solution
=
2
to
the
equation.
that
at
t
=
0,
x
and
x
=
0
t
long
term
ratio
of
x
:
b.
Find
c.
.
the
values
of
A
and
B
of
against
y
Sketch
the
graph
x
t
for
0
≤
t
≤
2
254
/
PRACTICE
.
Describe
relative
the
to
displacement
the
equilibrium
of
the
.
object
position.
Hence,
describe
displacement
equilibrium
The
motion
installation
to
reduce
of
of
its
described
a
object
the
=
is
refined
dampener
motion.
by
!
! + 5 x + 36 x
x
the
The
which
new
following
by
is
An
is
differential
so
equation
additional
that
the
Write
this
motion
a
pair
relative
to
the
O
force
now
acts
on
equation
the
object,
describing
its
is:
0
second-order
differential
of
first-order
coupled
2 cos ( t )
=
equation
.
as
object
position,
differential
!
! + 5 x + 36 x
x
d.
the
the
the
designed
motion
of
qualitatively
QUE STIONS
Write
this
second-order
differential
equation
differential
as
a
pair
of
first-order
coupled
differential
equations.
equations.
e.
.
Find
the
eigenvalues
for
the
system
in
.
Use
Euler ’s
method
with
a
step
length
of
0.1
of
x
and
x
=
displacement,
x,
partd
.
Sketch
phase
the
trajectory
portrait,
with
of
x
the
on
object
the
on
a
horizontal
y
on
the
vertical
axis,
given
that
an
when
=
=
0,
x
=
2
and
y
=
x
=
1,
given
that
at
for
t
=
the
0,
x
value
=
2
0
Sketch
the
motion
of
the
of
when
the
t
t
approximation
axis
h.
and
tofind
object
against
time
for
0
≤
t
≤
2
0
255
/
THE
AN
Th
intrnal
assssmnt
xploration.
ara
of
A S S E S S M E N T:
E X P LO R AT I O N
Appliations
an
INTERNAL
and
for
SL
and
Intrprtations
This
is
a
rport
Mathmatis.
in
Th
HL
Mathmatis:
your
th
mathmatial
a
is
whih
you
xploration
writing,
and
iliography
you
for
must
th
it
sours
sours
you
hav
and
provid
onsultd.
invstigat
is
a
Choc of topc
ompulsory
of
your
final
xploration
with
font
th
part
and
in
As
should
your
undrstand

ours
a
is
twn
suh
your
lass
what
and
as
th
12
and
12.
lassmats.
writ

in
is
worth
lngth
typd
Arial
should
you
whih
guid,
spaing
styl
xploration
anyon
th
grad.
doul-lin
siz
of
a
Th
This
to
your
th
20pags
in
al
of
20%
Choosing
long
rasonal
audin
mans
follow
for
that
whih
what
and
you
you
studnts
to
rush
xploration.
your
xploration
will

markd
using
th
th
namd
in
th
tal
low.
Th
in
ritria
A,
B,
C
and
D
is
way
marks
oth
SL
and
HL.
Stion
E,
‘us
xatly
of
th
th
diffrntly
gratr
twn
sophistiation
th
two
rquird
mathmatis’,
ourss
at

a
topi
dal
topi
that
of
this
daunting
of
raliz
xiting
frdom
hoi
is
a
thinking
tim
rally
is
wll
topi,
vry
to
that
many
It’stmpting
ut
it
is
arfully
invstd.
intrsts
in
xplor
on
hallng.
hoosing
that
opportunity
you
If
and
aout
you
that
you
to
important,
th
xploration
will

asir
and
njoyal
to
writ
aus
your
motivation
will
sam
gratr.
Choosing
th
right
topi
for
you
inrass
is
your
gradd
grat
Howvr,
to
of
an
ar

in
to
is
fiv
mor
awardd
a
pross
hoi
onsidr
ritria
hav
find
a
topi
want.
important
hoos
Your
your
hans
of
soring
tmpting
to
sarh
mor
marks.
rflt
HL.
It
a
Cto
Tt
Maxmum mak
A
Presentation
4
B
Mathematical communication
4
C
Personal engagement
3
D
Reection
3
E
Use of mathematics
6
is
topi,
ut
thoughts
tim
it
and
trying
to
is
ttr
onsult
onlin
to
look
your
undrstand
for
advi
insid
tahr
th
in
your
than
thoughts
own
to
of
hoosing
spnd
othrs.
To help you choose a successful topic:
•
Your teacher can show you successful topics from
previous years. By seeing the huge diversity of topics
that have been successful, you will be more aware of
what is possible, and your imagination will be engaged.
You should be familiar with the criteria before star ting
your exploration, and you should keep a copy of them
•
close to hand throughout the writing process.
Review and reflect on areas of the course that you
have enjoyed the most. Could your work in these
areas be extended or applied to other contexts that
Eah
mark
your
final
you
IB
gain
in
your
Mathmatis
xploration
total
adds
prntag,
1%
so
interest you?
to
doing
•
wll
in
your
xploration
will
hav
a
signifiant
Are there links between mathematics and your
fft
hobbies, interests or your plans for future study? In
on
your
final
grad.
You
an
ask
your
tahr
for
your life outside school or in CAS, there may be a topic
informal
fdak
an
giv
at
any
stag,
ut
your
tahr
“hiding in plain sight” just waiting for you to discover it.
only
writtn
advi
and
fdak
on
on
•
draft.
It
is
thrfor
a
good
ida
to
mak
your
Draw mind-maps star ting with a single area or an
draft
idea, and brainstorm how inquiry questions relate
as
omplt
as
you
an
so
that
you
an
gt
th
to this idea. You could even use an online random
maximum
nfit
from
this
fdak.
word generator to come up with ideas. Set yourself
Lik
all
intrnally
omponnts
adhr
to
in
th
and
th
IB
IB
xtrnally
Diploma
aadmi
assssd
programm,
honsty
poliy
the challenge of coming up with at least three good
you
must
ideas.
throughout
256
/
inTern Al
Trig
A ss e ss M e n T
function
•
Shape
Avoid topics that are not practical to carry out: before
Quadratics
star ting your exploration, you should check that
access to any data you will need can be carried out
in the time you have, leaving you enough time to
New
rides
v
old
write your exploration. Surveys need to be tested out
Waiting
time
before being given to a population, in order to find
T ime
Fun
of
day
fairs
any issues with the survey design.
Length
Compared
to
Cost
Adult
Day
In
th
aov
xampl,
you
an
s
of
ride
cost
aras
in
a
of
or
child
the
fun
week
fair
Th amt cta upackd
Cto A: Ptato
Achvmt
Dcpto
whr
a
mathmatis
qustion
that
you
is
usd
would
and
lik
this
to
ould
ask,
dvlop
suh
as
“Is
v
into
The exploration does not reach the standard
thr
0
a
onntion
of
a
tikt?”
twn
This
th
ould
lngth

a
titl
of
a
for
rid
an
and
th
described by the descriptors below.
ost
xploration.
The exploration has some coherence or
1
some organization.
Othr
xampls:
The exploration has some coherence and
2
•
A studnt
to
who
njoyd
study Aronautial
alulus
and
Enginring
who
at
wantd
shows some organization.
univrsity
The exploration is coherent and well
3
xplord
airraft
th
ross-stional
wings,
linking
aras
forward
to
and
his
shaps
of
organized.
futurstudis.
The exploration is coherent, well organized,
4
and concise.
•
A studnt
traking
whl
and
of
who
lovd
softwar
hr
to
skatoarding
tra
skatoard
modlld
th
data,
th
as
path
sh
usd
of
th
vido-
ak
prformd
making
a
link
to
triks
hr
hoy
.
This
th
ritrion
ohrn
doumnts,
•
A studnt
ut
knw
who
h
ofxploring
was
likd
th
struggling
ooking
rats
at
to
find
arrivd
whih
at
a
sizs
of
aks
ool
y
artils
that
you
that
ar
fl
ar
xploration.
or
ooks
that
and
Think
you
aout
hav
rad
wll
writtn,
and
prhaps
othrs
not.
In
a
wll-writtn
doumnt,
you
shaps
rainstorming

al
to
idntify
th
aim
y
rading
th
th
introdution.
word
your
organization
ida
should
and
of
th
topi
th
diffrnt
asssss
Th
mthodology
and
th
onlusion
“ooking”.
should
tals
to
oth
or
hlp
point

larly
diagrams
you
just
st
should
undrstand,
whn
out.
you

Rlvant
in
th
prhaps
nd
orrt
y
furthr
graphs,
pla
illustrating
xplanation.
a
You
To help you avoid choosing an unsuccessful topic:
nd
•
to
xplain
Avoid topics that are purely descriptive and historical
in
because such topics will limit your grade.
lmnts
your
lass
of
an
you
follow
writing
xploration.
•
what
Largr
ar
and
doing
so
that
undrstand.
add
to
th
data
sts
anyon
All
organization
or
matrial
ths
of
that
your
taks
Avoid topics whose conclusions are widely shared
up
too
muh
spa
in
th
ody
of
th
writing,
ut
online already because such topics can limit your
whih
may

usful
for
th
radr
to
rfr
to,
ar
grade and can lead to academic honesty issues with
st
put
in
an
appndix.
your work .
•
Avoid topics that are much more advanced than
the level of your course: this may sound like a
strategy to get a high grade, but this approach is not
recommended. Applying mathematics that you do not
fully understand can limit your grade, as well as use
up a dispropor tionate amount of your time.
•
Cohrn
rfrs
mntiond
aov
togthr.
skills
in
You
othr
and
writing
and
your
to
how
that
will
you
hav
sujts,
ssays.
radr
wll
th
hav
organizd
pratisd
for
your
xploration
that
it
with
no
whr
what
is
fl
is
fit
prsntation
xampl
A ohrnt
should
lmnts
in
planning
xploration
sur
whn
progrssing
“flows”
rading
logially
and
Avoid topics that are too easy: the level of
“jumps”
h/sh
is
lft
puzzld
as
to
mathematics must be commensurate with the level
going
on,
or
has
to
frquntly
rfr
ak
to
a
of your course or else your grade will be limited.
prvious
rsult.
Your teacher can advise you if your topic involves
mathematics that is too easy.
257
/
inTern Al
A ss e ss M e n T
This
ritrion
Mathmatis
onsistntly
.
asssss
how
wll
appropriatly
,
Mathmatis
you
ommuniat
rlvantly
an

and
ommuniatd
To get a good mark in this criterion you should:
•
with
notation,
your
writing.
symols
and
trminology
in
Write your introduction so that the reader will clearly
Howvr,
your
writing
an

understand your rationale and aims after reading it.
omplmntd
•
•
y
Make sure you describe and explain contexts which
Imagin
may be unfamiliar to the reader. For example, if you
triangls,
are exploring games of chance, you must supply

enough background information – for example,
first
the rules of a dice game – so that the reader can
300 bce,
understand what you are going to explore.
to
ommuniat
of
th
vry
rading
a
graphs
hard
to
any
mak
authors
typs
of
trigonomtry
or
mathmatis
writtn
othr
sns
txts
hav
typ
of
of.
wr
thought
You
txt
that
faturs
diagram.
In
fat,
hard
y
aout
yond
should
It
th
no
would
sin
pulishd
Mathmatis
word.
ommuniation.
th
Eulid
how
in
st
powr
ommuniat
Integrate your diagrams with your text by positioning
Mathmatis
appropriatly
y
ing
as
orrt
as
a
them conveniently and referring to features
profssional
mathmatiian
should

with
notation,
specifically. For example, use labels if appropriate to
symols,
graphs
t,
ut
also
you
should
think
aout
point out what you are referring to on a graph.
what
•
softwar
an
rat
graphs
or
diagrams
that
Swap explorations with a classmate before you
ar
spifi
to
your
xploration
and
how
thy
might
submit the draft. You each proof-read and tell each
nhan
your
ommuniation.
other if you lose touch with the development of
what you are presenting: do they get “stuck”? If so,
you may need to improve your coherence or your
organization.
To get a good mark in this criterion you should:
•
Be concise – this means you need to write enough
•
Communicate key terms clearly. For example, if
so that the reader understands well, but you do not
you are exploring an application of mathematics
repeat yourself. For example, copying and pasting
to revenue and cost functions in economics, state
content from early in your writing but just changing
the meaning of the independent and dependent
the parameters would be repetitive and would rule out
variables, their units, domain and range.
scoring maximum marks.
•
•
Always avoid word-processing or calculator notation
Consider including “signposts” in your writing to help
since it is not always the same notation used in
make links. For example, “In the previous section, I
Mathematics. Use an equation editor to communicate
showed that …”, “In this section I now aim to extend
x
correctly. For example, 5^x should be written as 5
this result to …” are examples of signposts. You may
.
4
Also, write 5x instead of 5*x. Write 5.67 × 10
instead
have used them in other subjects.
of 5.67E4.
•
Show in the final pages how you have achieved your
•
Always label scales and axes in each graph.
•
Consistently express your results to an appropriate
aim as you write your conclusions.
degree of accuracy.
Cto B: Mathmatca commucato
•
Invest time learning how mathematical software
can enable you to create powerful ways to
communicate your mathematics. For example,
Achvmt
Dcpto
v
three-dimensional diagrams are an appropriate
The exploration does not reach the standard
means of communicating the geometry and
described by the descriptors below.
trigonometry of solids or 3D vector equations of lines.
0
Traces of families of functions are an appropriate
The exploration contains some relevant
1
mathematical communication which is
way to communicate the effects of changing function
par tially appropriate.
parameters. These functionalities are features of
dynamic geometry software.
The exploration contains some relevant
2
appropriate mathematical communication.
The mathematical communication is
Not
3
that
it
is
not
ompulsory
for
you
to
us
relevant, appropriate and is mostly
multipl
forms
of
mathmatial
ommuniation.
consistent.
For
xampl,
if
you
ar
xploring
pattrns
in
The mathematical communication is
algrai
4
xpansions,
you
ould
ommuniat
your
relevant, appropriate and consistent
mathmatis
appropriatly
just
with
symols
and
throughout.
you
would
still

al
to
gain
maximum
marks.
258
/
inTern Al
A ss e ss M e n T
Cto C: Poa gagmt
•
Remember that you are writing a repor t of your
investigation. Make sure you repor t the decisions
Achvmt
Dcpto
you made in engaging with its development.
v
Maximum marks depend on you showing evidence
The exploration does not reach the standard
0
of your personal engagement. Only you can provide
described by the descriptors below.
this evidence by repor ting it in your writing.
There is evidence of some personal
1
engagement.
Not
that
xploring
mathmatis
mor
advand
not
rommndd
that
is
muh
There is evidence of signicant personal
2
engagement.
a
than
your
way
mathmatis
to
ours
dmonstrat
is
prsonal
There is evidence of outstanding personal
3
ngagmnt.
Th
ritria
ar
dsignd
so
that
you
engagement.
This
ritrion
ngag
If
you
rally
you
with
hav
asssss
th
hav
you
mad
xtnt
xploration
invstd
intrsts
th
a
tim
and
good
in
that
start
and
to
whih
mak
hoosing
you
find
towards
it
a
an
gain
at
lvl
a
maximum
marks
appropriat
to
y
th
applying
lvl
of
mathmatis
your
ours.
you
your
topi
own.
that
Cto D: rfcto
important,
soring
highly
intrst
has
Achvmt
Dcpto
in
this
ritrion.
alrady
n
This
is
ngagd,
aus
and
you
your
will

mor
al
v
to
The exploration does not reach the standard
0
dmonstrat
aout
is
how
ruial
is
that
th
tak
full
xploration
that
ngagmnt
you
you
within
show
your
ontrol
dvlops.
writtn
and
ownrship
Howvr,
vidn
xploration.
This
of
described by the descriptors below.
what
1
There is evidence of limited reection.
2
There is evidence of meaningful reection.
your
vidn
There is substantial evidence of critical
oms
in
many
diffrnt
forms,
som
of
whih
3
you
reection.
will
hav
nountrd
in
your
toolkit
lssons.
This
ritrion
analys
It
is
and
asy
to
asssss
valuat
nglt
how
wll
you
throughout
this
ritrion
rviw,
your
xploration.
whn
you
ar
so
To get a good mark in this criterion you should:
usy
•
Think independently and creatively.
•
Brainstorm inquiry questions both as par t of your
as
arrying
alulations,
solving
it
out
is
that
suh
a
mathmatial
analysis,
your
rlif
modlling
xploration
whn
prosss
you
or
prolm
rquirs.
finally
fl
suh
Somtims
you
hav
planning and as the exploration develops.
n
•
sussful
with
th
mathmatis
that
you
Write in your own voice, repor ting on how you
fl
your
work
is
don!
Or
may
our
arrs
as
progressed with your investigation, what you
mathmatiians
hav
n
onditiond
at
an
arly
wondered, what your ideas were, and how and why
ag
y
a
tndny
our
answrs
to
hk
with
our
tahr
that
you chose to engage with an inquiry question.
•
ar
Consider which of the following types of personal
onwith
engagement could be applied to your topic, and
littlrfltion.
a
happy
orrt,
fling
and
of
thn
just
suss,
moving
ut
with
apply them if appropriate:
In
o
creating mathematical models of a real-life context
o
designing and implementing a survey
o
applying technology in a resourceful and effective way
o
taking a risk and showing courage by exploring
your
tostat
xploration,
your
thyman,
rflt
you
ar
onlusions,
stat
any
xptd
ut
to
limitations
not
onsidr
thy
simply
what
hav
and
ritially
.
unfamiliar mathematics or contexts
o
linking to historical, local or global perspectives
o
showing how you address personal interest or
satisfy your curiosity
To get a good mark in this criterion you should:
•
Not leave your reflection until the end of your
exploration. Instead, reflect periodically throughout
the exploration when appropriate – for example each
o
creating a simulation or an experiment
o
including hypotheses or conjectures that did not
time you have established a result you can reflect
on it.
go quite as you expected and explaining why.
259
/
inTern Al
A ss e ss M e n T
Rquirmnts
•
for
HL
studnts:
Reflect on what your results mean in the real-world
context you are exploring. This is a frequent theme in
Achvmt
Mathematics: Applications and Interpretations which
v
Dcpto
you may have experienced in investigations, toolkit
The exploration does not reach the standard
0
lessons or exercises.
•
described by the descriptors below.
Some relevant mathematics is used. Limited
Reflect on the appropriate degree of accuracy given
1
understanding is demonstrated.
the context of your topic.
Some relevant mathematics is used. The
•
Think of reflection and personal engagement working
mathematics explored is par tially correct.
2
together. You may find that in reflecting on a result, a
Some knowledge and understanding
new line of inquiry suggests itself to you, which you
is demonstrated.
then engage with. If you feel that reflection drives the
Relevant mathematics commensurate
direction of your exploration, this is evidence of critical
with the level of the course is used.
reflection and you should repor t this in your writing.
3
•
The mathematics explored is correct.
Some knowledge and understanding
Aim to reflect critically: have you considered the
are demonstrated.
validity, implications and reliability of your results
and the appropriateness of the mathematics you
Relevant mathematics commensurate
applied to find them?
with the level of the course is used.
4
•
The mathematics explored is correct.
Do not hold back if you think some of your knowledge
Good knowledge and understanding
and understanding of TOK is relevant in writing a
are demonstrated.
critical reflection. After all, TOK is a course about
Relevant mathematics commensurate
critical thinking.
with the level of the course is used. The
mathematics explored is correct and
5
demonstrates sophistication or rigour.
Cto e: U of mathmatc
Thorough knowledge and understanding
This
ritrion
rlvant
asssss
mathmatis
ritrion
has
th
ritrion
only
th
most
for
to
in
what
your
marks
whih
xtnt
you
xploration.
availal
th
to
are demonstrated.
us
This
you.
rquirmnts
Relevant mathematics commensurate
This
for
is
with the level of the course is used. The
mathematics explored is precise and
th
6
awarding
of
marks
for
SL
is
not
xatly
th
sam
demonstrates sophistication and rigour.
as
Thorough knowledge and understanding
for
HL,
although
thr
ar
similaritis.
are demonstrated.
Rquirmnts
for
SL
studnts:
For
oth
SL and
HL,
th
mathmatis
you
us
Achvmt
Dcpto
must

rlvant,
maning
that
it
supports
th
v
dvlopmnt
of
your
xploration
towards
mting
its
The exploration does not reach the standard
0
aim.
Us
of
mathmatis
that
is
ovrly
ompliatd
described by the descriptors below.
(whr
1
simpl
mathmatis
would

nough
to
mt
Some relevant mathematics is used.
th
aims
of
th
xploration)
is
not
rlvant
–
and
this
Some relevant mathematics is used. Limited
will
2
limit
th
marks
you
ar
awardd
signifiantly
.
understanding is demonstrated.
For
oth
SL and
HL,
th
aim
is
for
you
to
Relevant mathematics commensurate with
dmonstrat
3
in
your
writing
that
you
hav
thorough
the level of the course is used. Limited
understanding is demonstrated.
knowldg
you
us.
It
and
is
undrstanding
not
nough
just
to
of
th
hav
mathmatis
a
orrt
Relevant mathematics commensurate with the
answr
to
dmonstrat
thorough
knowldg
and
level of the course is used. The mathematics
4
explored is partially correct. Some knowledge
and understanding are demonstrated.
undrstanding,
and
D
your
–
you
ut
must
knowldg
–
as
is
also
provid
and
th
th
as
with
vidn
ritria
C
to demonstrate
undrstanding.
Relevant mathematics commensurate with the
level of the course is used. The mathematics
Oviously
,
5
SL and
HL studnts
will
all
aim
for
explored is mostly correct. Good knowledge and
all
thir
mathmatis
to

orrt.
Th
dgr
to
understanding are demonstrated.
whih
your
mathmatis
is
orrt
is
proportional
to
Relevant mathematics commensurate with the
th
marks
you
an

awardd,
ut
you
must
also
level of the course is used. The mathematics
dmonstrat
6
your
knowldg
and
undrstanding
at
explored is correct. Thorough knowledge and
th
sam
tim.
understanding are demonstrated.
260
/
inTern Al
A ss e ss M e n T
Chckt
Hr
is
a
hklist
your
st
to
nsur
that
you
hav
don
To get a good mark in this criterion, SL and HL students
to
sumit
a
good
xploration.
should:
Communication
•
your exploration that the level of mathematics you
•
and
mathematical
presentation
Check with your teacher when you are planning
Hav
you
propose to use is not too simple nor too advanced,
faturs
and that it is relevant to your aims.
shool
Check with your teacher that your plans for using
Did
inludd
suh
or
you
as
titl
your
tahr ’s
start
a
with
ut
nam,
no
idntifying
andidat
numr,
nam?
an
introdution?
mathematics are achievable in the time and space
Hav
you
statd
a
lar
aim?
What
is
th
given. When using mathematics, it is best to focus on
thmof
your
xploration?
What
do
you
hop
quality rather than quantity.
itwill
•
do?
Don’t just use mathematics. Use it and then justify its
In
your
onlusion,
hav
you
answrd
or
use by showing why it is appropriate. For example, if
rspondd
to
th
aim
you
wrot
at
th
start
of
you use the formula for independent events, explain
your
xploration?
why the events are independent in the first place. If you
are applying the binomial distribution, explain why the
Do
context fits the requirements of the binomial distribution.
•
you
hoos
hav
this
a
lar
topi?
rational?
Why
is
this
Why
topi
did
of
you
intrst
toyou?
You are encouraged to use technology, but with
the same expectations for demonstrating your
Can
an
avrag
studnt
in
your
lass
rad
and
knowledge and understanding. For example, just
follow
your
xploration?
You
may
want
to
pr
finding the value of derivative at a point on a curve
dit
with
a
frind.
with technology does not necessarily demonstrate
Dos
th
ntir
papr
fous
on
th
aim
and
avoid
knowledge and understanding. You should explain
irrlvan?
what this value means in context.
Dos
For
HL,
your
SL.
thr
us
Th
of
dgr
to
th
addrss
your
is
to
xptations
whih
rigorous
highst
in
ritrion
your
and
pris
in
this
sophistiation
at
th
sur
first
ommnsurat
with
of
all
with
your
that
th
plad
E
than
mathmatis
grads
xploration,
making
mor
mathmatis
sophistiatd,
ass
ar
hlps
tahr ’s
your
us
Did
for
xptations
writing
inlud
titls
thm
all
Hav
you
at
at
or
that
it
involvs
a
Y
ou
that
gos
yond
stag
tals
th
and
plas
diagrams
and
not
attah
nd?
rrad
and
improvd
your
xploration?
of
y
Did
you
it
all
rfrns
and
aknowldg
Did
you
dirt
in
your
quots
iliography
appropriatly?
mathmatis
of
us
appropriat
mathmatial
HL
omplx
th
graphs,
gin
us
of
xptations
rprsntation?
(Not
omputr
SL
notation,
mathmatis
nily?
appropriat
languagand
mathmatis,
flow
to
guidan,
of
you
with
is
you
ritrion.
planning
on
th
of
suh
as
*,
^,
t.)
SL.
Did
you
dfin
Did
you
us
Did
you
think
ky
trms
appropriat
aout
th
whr
nssary?
thnology?
dgr
of
auray?
To get a good mark in this criterion, HL students should,
(For
your
topi,
how
many
dimal
plas
ar
in addition:
rlvant?)
•
Brainstorm “sophistication” when at the planning
stage. Consider looking at your topic from different
perspectives, using and understanding challenging
Did
to
you
your
nd
aim
with
and
a
onlusion
and
rlat
it
ak
rational?
concepts, identifying underlying structures to link
Personal
engagement
different areas of mathematics.
Did
•
you
addrss
why
you
think
your
topi
is
Review how you set out your mathematics. Is it
intrsting
or
why
it
appald
to
you?
expressed as rigorously as you would find in a
mathematics textbook? Are your logic and language
Did
you
us
th
prsonal
Did
you
ask
and
pronoun?
clear in your arguments and calculations?
•
Consider using logical connectives such as ∴,
⇔,
•
⇒,
(“Iwondr
if
answr
…”,
“What
prsonal
if
qustions
…”)?
to help communicate your logic.
Always use an appropriate level of accuracy.
261
/
inTern Al
Did
you
own
prsnt
way
ls’s
(as
A ss e ss M e n T
mathmatial
opposd
to
idas
opying
in
your
somon
Did
of
for
possil
your
limitations
and/or
topi?
thory)?
you
softwar
xplain
that
any
you
nw
had
to
mathmatis
or
Did
you
try
to
Did
you
rlat
add
th
“your
rsults
voi”
to
to
your
th
own
work?
mathmatis
what
mans
any
aras
approximat
whr
or
your
slightly
mathmatis
inaurat?
th
aftr
rsults
you
of
hav
you
diffrnt
mak
filds
links
twn
and/or
aras
your
of
topi
and
mathmatis?
lif?
Reflection
xplain

find
mploy?
Use
you
you
may
Did
Did
look
xtnsions
Did
Did
you
your
usd
of
mathematics
Did
you
us
Did
you
xplain
apply
mathmatis
familiar
from
unfamiliar
mathmatis
your
ours?
mathmatis,
to
a
nw
or
situation?
athniqu?
Did
Did
you
ask
invstigat
Did
you
qustions,
mak
mathmatial
onsidr
prsptivs
of
th
your
onjturs
and
you
world
you
(What
disuss
do
thy
th
mathmatial
situations,
if
this
modls
applid
to
for
your
ral-
topi?
idas?
historial
and
Did
you
apply
Did
you
look
prolm-solving
thniqus?
gloal
for
and
xplain
pattrns,
if
this
topi?
applid
Did
rat
impliations
man?
Why
ar
of
thy
your
to
your
topi?
rsults?
important?)
262
/
P R A CT I C E
E XA M
PA P E R S
Introduction to the exam-practice papers
At
this
point
you
will
Applications
and
Interpretation
skills
to
refine
practice
will
is
not
draw
a
line
A clean
This
be
will
prepare
All
given
All
Paper
you
the
1
do
the
that
you
answer
The
all
If
of
of
the
each
InPaper
to
you
of
you
are
by
it
so
put
will
these
HL
Papers
the
Diploma
not
start.
questions
are
to
you
run
that
be
out
you
at
the
just
then
and
1,
have
skills
2
and
from
the
picked
to
3,
the
up
the
of
some
test!
with
Programme
topics
In
course.
key
this
same
the
IB
Mathematics:
techniques
section
structure
Answers
to
you
as
and
will
the
these
find
external
papers
a
display
put
a
calculator
single
single
line
for
all
through
exams.
it.
line
though
the
Interpretation
formula
booklet is
If
it
is
incorrect
The
a
use
of
section
correction
that
is
wrong,
working.
required
for
all
papers.
start
on
to
this
a
can
a
Ensure
back
and
However,
on
then
have
done
this
to
answered
to
there
may
box
the
sure
at
you
not
write
question,
that
the
you
end
divide
of
during
this
rough
guide
not
waste
time:
it
the
2
is
at
of
each
time
to
be
the
a
difficulty
but
you
can
your
make
in
the
answer
sure
that
many
exam
if
should
significant
to
move
have
questions.
going
an
between
minutes
you
The
through
specially
the
what
wisely
prepared
in
answers
question.
how
gradient
paper
numerical
time
extended-response
complete
and
end
all
understand
your
a
examination
can
the
is
do
Paper
the
you
in
that
question
come
in
Make
instruction
question,
questions
space
You
stated
figures.”
attempted.
you
paper.
otherwise
followed
stuck
of
questions.
answered
of
the
“Unless
matter.
the
If
graphical
and
allocated
short-response
does
a
significant
be
that
2
questions
question
2
a
the
part
on
in
are
a
fresh
the
question
be
page.
question.
paper
inthe
question
that
you
thenadd
the
Paper
fresh
3
extra
question
is
it
the
next
time.
order
the
in
which
papers,
boxes
booklet.
well
should
onto
prepared
answer
the
with
under
Make
labelled
in
sure
the
of
in
incorrect,
some
of
the
each
working.
use
will
it
as
to
a
Obey
Make
not
basis
the
this
sure
go
the
in
to
booklets
instruction;
that
the
answer
your
answer
you
will
clearly
examiner
booklet.
working.
it
so
This
also
mark
if
you
can
Redraw
provided.
give
each
do
the
you
will
space
question
any
happen
There
in
diagram
to
on
particular
your
an
go
the
if
and
if
you
the
is
a
to
have
subparts.
question
there
answer
instruction
back
number
working
in
be
paper
diagram
booklet
and
work.
consists
page
start
no
to
become
transcribe
with
of
mistake,
time
have
marks
also
the
2,
to
you
contents
school.
three
you
At
end
Applications
instruction
sure
a
time
and
and
a
answering
make
on
now
the
booklet.
complete
HL
with
Additionally
1
bottom
reading
papers
mark
indicate
Paper
start
is
the
ruler
at
question.
at
make
your
for
number
a
at
that
questions
1
by
the
check
in
a
do
correct
questions
Paper
each
or
consists
easier
The
to
and
Put
It
Papers
Mathematics:
have
complete.
question.
and
strategy
The
SL
complete
you
minutes
will
are
approach.
pencil,
If
top
the
exactly
questions.
to
the
5
questions
take
pens,
provided
your
syllabus.
yourself
www.oxfordsecondary
.com/ib-prepared-support.
of
be
papers
figures
at
re-familiarized
papers:
will
allowed.
copy
will
There
at
you
require
fluid
be
that
available
You
exam
examination
assessment
are
your
have
two
compulsory
,
answer
paper
marks
may
problem-solving
questions.
Start
each
question
booklet.
there
Answers
extended-response,
will
must
be
be
be
the
advice
supported
given
for
a
“Full
by
correct
marks
working
method,
are
not
and/or
necessarily
explanations.
provided
that
this
is
given
Where
shown
for
an
by
a
correct
answer
written
answer
is
working.
263
/
PRACTICE
You
be
are
therefore
supported
these
you
as
part
should
by
of
E X AM
advised
suitable
your
have
PA P E R S
to
show
working.
answer.”
been
all
It
working.
For
example,
should
answering
all
of
Solutions
be
the
no
if
found
graphs
surprise
questions
to
you
are
you
from
used
to
when
have
graphical
find
a
(ii)
in
the
the
display
solution,
reading
done
Standard level
a
this
two
on
maximum
betweenthe
you
the
years
calculator
should
exam.
leading
possible
Earth
and
sketch
This
up
should
is
to
how
the
exam.
distance
Mars.
[1]
Practice paper 1
(b)
Time:
1
hour
30
Find
the
in100
min
distance
days
(take
travelled
a
year
as
by
the
equal
Earth
to
365days).
Technology
3
Maximum:
80
[Maximum
all
the
questions.
their
numerical
answers
must
be
given
exactly
to
three
significant
figures,
unless
in
the
Answers
or
should
marks
provided
of
This
hours
are
of
1
of
shown
resorts
were
ranking
and
sunshine
in
indicates
be
Where
may
this
is
supported
be
an
working
answer
awarded
shown
by
for
is
a
and/
per
ranked
the
day
on
mean
in
the
the
table
below,
where
popular
resort.
a
the
most
Resor t
A
B
C
D
E
Popularity rank
5
4
3
2
1
Hours of sunshine
5.5
4.6
8.2
7
.7
9.0
incorrect,
correct
method,
clearly
.
Calculate
1
holiday
question.
explanations.
some
5
otherwise
rank
stated
4]
or
resorts
correct
of
popularity
.
number
All
marks:
marks
A sample
Answer
[2]
required
[Maximum
marks:
Spearman’s
rank
correlation
4]
coefficient
for
r
this
data.
[4]
s
Tania
is
playing
a
game.
The
number
of
points
4
she
can
with
win
the
is
X,
where
following
X
is
a
random
probability
[Maximum
marks:
5]
Grain
from
silo
variable
distribution.
leaks
Tocalculate
x
P(X
(a)
(b)
2
=
x)
Find
Find
1
2
3
4
0.2
0.3
0.4
k
the
value
of
afarm
The
orbits
Sun
are
the
shown
of
the
Find
the
amount
of
measures
m
and
coneas
forms
grain
the
the
a
in
slant
cone.
the
cone,
height
diameter
of
of
the
1.0m.
volume
of
grain
1.2
m
in
the
cone.
[5]
[2]
marks:
of
base
as1.2
and
[2]
E(X).
[Maximum
worker
thecone
k.
the
a
5]
Earth
in
the
and
of
Mars
diagram
around
the
below.
Earth
1.0
5
Sun
[Maximum
The
Mars
rate
(profit),
of
P
The
orbits
of
Earth
and
Mars
can
be
considered
1000
with
228million
a
radius
150
kilometres,
million
kilometres
and
8]
change
of
an
kg)
is
the
in
amount
given
x
by
money
1000
gained
USD)
for
of
chemical
the
equation
X
per
a
day
factory
(measured
dP
2
=
circular
marks:
(measured
producing
in
m
0.6 x
− 3.6 x + 5,
0
≤
P
≤
3
dx
respectively
.
(a)
Given
that
the
factory
makes
a
loss
of
k
Giving
where
(a)
(i)
your
1 ≤
a
the
answers
< 10
and
in
k
minimum
betweenthe
the
∈ !,
form
500USD
find:
possible
Earth
a × 10
and
per
chemicalX,
distance
Mars
When
[2]
the
the
costs
greater
day
find
factory
of
is
when
the
produces
anexpression
close
producing
than
it
extra
to
more
profit
for
P
.
maximum
chemical
no
[5]
capacity
,
X
are
received.
264
/
PRACTICE
(b)
(i)
Find
the
factory
amount
should
of
chemical
produce
to
X
Carry
the
maximize
and
itsprofit.
out
state
the
test
clearly
at
the
the
5%
null
E X AM
significance
and
PA P E R S
level
alternative
hypotheses.
[2]
[5]
dP
(ii)
Write
down
the
value
of
at
8
this
[Maximum
marks:
7]
A souvenir
manufacturer
dx
point.
6
[Maximum
The
graph
[1]
marks:
shows
8]
the
depth
(d)
of
water
in
famous
statue
of
models
these
for
the
12-hour
period
between
different
is
sizes.
proportional
models
The
to
of
volume
the
cube
a
(V)
of
a
their
harbour
in
produces
high
height
(h).
tides.
The
statue
6
that
is
10
cm
tall
has
a
volume
of
3
225cm
sertem
5
(a)
Find
The
cost
root
of
an
expression
for
V
in
terms
of
h.
[3]
4
d
,htpeD
3
2
of
its
a
statue
volume.
is
proportional
The
10
cm
to
statue
the
square
costs
€12.
(6, 1.9)
1
(b)
Find
the
cost
of
a
20
cm
up
of
statue.
[4]
0
1
2
3
4
6
5
7
8
9
10
11
12
9
T ime,
t
[Maximum
high
low
t
=
tide
0
The
and
is
t
=
1.9
depth
m.
of
High
water
tide
is
6.3
occurs
m
at
and
at
of
times
spokes
edge
the
curve
t
the
is
is
d
sides
of
(i)
number
of
hours
the
value
umbrella,
and
is
formed
the
third
of
length
20
cm
20
triangles.
by
the
side,
at
the
cm.
[2]
b
[2]
(iii)
c
[1]
cannot
is
Find
enter
less
the
boat
[Maximum
order
to
hearing,
record
umbrella
are
identical
of:
(ii)
the
In
triangles
10
after
the
than
4
harbour
when
the
depth
m.
A view
7
the
the
the
made
a cos ( bt ) + c ,
=
a
water
(b)
of
of
is
tide.
Find
A boat
umbrella
Two
12.
where
∈!
ahigh
of
it
the
equation
a, b, c
(a)
tide,
6]
hours
An
At
marks:
a
percentage
cannot
marks:
test
of
enter
of
howmany
the
during
the
which
harbour.
the
umbrella
from
above
is
shown
in
diagram.
[3]
5]
whether
group
time
of
280
or
not
people
hours
each
loud
are
music
asked
week
they
affects
to
x
listen
x
to
music
ability
The
above
of
the
results
a
certain
groupis
are
volume.
also
shown
in
The
hearing
20
cm
tested.
the
table
below.
(a)
Find
the
the
value
of
the
length
marked
x
in
diagram.
[4]
Level of hearing
(1 indicates ‘good’ hearing)
(b)
1
2
Find
the
area
which
would
be
shielded
from
3
the
0
≤ t
<
2
52
54
22
2
≤ t
<
4
35
32
1
7
2
1
28
19
10
Time listening
rain
by
the
[Maximum
marks:
A farmer
picking
umbrella.
[2]
5]
(hours)
t
≥
4
are
is
placed
in
a
line,
pumpkins.
Twenty
each
from
1.5
m
pumpkins
the
next.
2
Use
is
the
an
loud
χ
test
for
association
music
and
independence
between
level
of
time
to
see
spent
hearing
if
there
listening
to
ability
.
1.5
m
265
/
PRACTICE
The
farmer
pumpkin
then
at
the
E X AM
parks
end
PA P E R S
his
truck
the
line.
of
next
to
N
a
x
Find
the
collect
distance
all
the
thetruck
to
itback
the
to
the
farmer
pumpkins,
collect
each
needs
given
one
he
and
to
walk
walks
then
to
from
N
carries
truck.
[5]
7
11
[Maximum
A total
of
funds,
A,
marks:
$18,000
km
6]
is
invested
in
three
different
H
B
and
fundApays
invested,
C.
out
fund
At
6%
B
the
end
interest
pays
out
of
on
4%
the
the
first
year,
money
interest
and
Find
paysout
8%
total
[Maximum
interest
paid
at
the
end
of
the
first
x.
[5]
marks:
6]
order
to
be
allowed
to
drive,
Sabine
must
take
year
a
is
of
interest.
In
The
value
fund
14
C
the
theory
test
and
then
a
practical
test.
She
will
$1210.
retake
The
amount
than
Let
the
the
funds
Form
invested
amount
amount
be
a,
b
three
in
fund
invested
of
and
money
in
A is
$4000
fund
more
she
B.
invested
any
test
that
passes
her
theory
probability
in
the
three
(a)
she
Complete
c
she
test
passes
the
fails.
The
(T)
the
is
probability
0.8
practical
probability
tree
and
the
test
(C)
is
0.7.
shown
below.
equations
in
a,
b
and
c
and
[1]
solve
C
0.7
them
to
find
the
amount
of
money
invested
T
ineachfund.
[6]
0.8
C
12
[Maximum
marks:
6]
2
Consider
the
equation
y
=
4.5
0.5 x
C
dy
(a)
(i)
Find
[1]
T
dx
(ii)
Hence,
to
The
the
find
curve
cross-section
of
the
at
a
equation
the
point
building
of
(1,
can
the
tangent
4).
be
[2]
C
modelled
(b)
Find
the
probability
Sabine
will
need
to
2
by
the
curve
y
=
4.5
0.5 x
,
3
≤
x
≤
3
retake
A builder
needs
to
do
some
work
on
the
roof.
to
reach
reach
the
the
highest
point,
her
ladder
least
on
the
roof
at
(1,
test.
[2]
Find
the
probability
she
takes
exactly
must
threetests
point
one
In
(c)
order
at
before
she
is
allowed
to
drive.
[3]
4).
Standard level
(1, 4)
Practice paper 2
Time:
1
hour
Technology
30
min
required
Maximum:
80
marks
Answer
questions.
x
0
–3
(b)
–2
Find
her
–1
the
1
minimum
2
length
3
required
All
for
ladder.
[3]
numerical
correct
stated
13
[Maximum
marks:
A boat
a
all
to
three
inthe
must
significant
be
given
figures,
exactly
unless
or
otherwise
question.
5]
Answers
sails
answers
distance
of
7
km
on
a
bearing
should
be
supported
by
working
and/or
of
explanations.
035°
a
from
a
harbour
distance
x
kilometres.
from
the
again
boat
and
is
sails
H.
140°,
It
When
the
straight
then
boat
back
sails
the
due
bearing
changes
to
H
west
of
for
H
direction
Where
be
an
answer
awarded
for
a
is
incorrect,
correct
some
method,
marks
may
provided
this
is
shownclearly
.
266
/
PRACTICE
1
[Maximum
marks:
(b)
16]
Find
the
amount
Sumitra’s
In
a
biology
experiment,
Pierre
is
feeding
of
plants
plants
The
are
with
nutrient
withnutrient
heights,
shown
h,
in
of
the
10
A and
a
second
set
table
fed
with
at
the
that
end
of
would
20
be
years
in
if
nutrient
took
no
money
out
of
the
account.
[2]
of
B.
plants
account
money
PA P E R S
one
she
set
of
E X AM
She
is
considering
she
would
She
plans
like
to
buying
buy
an
costs
apartment.
The
one
€130,000
A
to
pay
€20,000
as
a
deposit
and
the
rest
below.
in
5.8
6.5
6.2
7
.0
6.3
6.5
5.9
6.8
6.6
6.4
monthly
instalments,
month,
for
20
interest
per
years.
paid
The
at
bank
the
will
end
of
charge
each
her
6%
Height (cm)
(c)
(a)
For
these
data,
Find
the
median
the
lower
the
quartile
(b)
(i)
the
upper
Find
the
monthly
payment.
[3]
end
of
the
20
years
Sumitra
will
move
to
city
.
[1]
(d)
(iii)
Sumitra’s
monthly
.
[1]
another
(ii)
compounded
find:
At
(i)
year,
quartile.
Find
[1]
interquartile
range.
whether
or
buy
an
in
costs
it
is
better
apartment,
between
the
for
and
two
Sumitra
the
to
rent
difference
routes.
You
may
[1]
assume:
A piece
of
data,
x,
is
an
outlier
if
x
<
a
or
x
>
b
•
(ii)
(c)
(i)
(ii)
Find
the
Draw
State
a
values
box
one
of
plot
reason
a
and
b.
[3]
for
this
data.
[2]
why
the
box
her
€20,000
for
all
•
she
•
there
20
can
that
the
sample
of
from
a
are
heights
shown
of
in
the
the
8
table
fed
with
nutrient
3
below.
6.0
7
.2
6.9
7
.5
7
.0
7
.2
6.6
7
.1
[Maximum
The
B
Height (cm)
mass
are
thinks
plants
than
hypothesis
(d)
(i)
that
nutrient
nutrient
at
the
State
1%
the
Find
that
the
the
null
decides
and
two
to
taller
test
this
level.
You
test.
may
populations
with
no
Compare
or
significance
equal
p-value
level
theconclusion
2
[Maximum
Sumitra
marks:
rents
an
of
of
are
normally
standard
with
the
the
she
the
the
Find
are
A potato
is
than
220
g
15%
of
Find
test
and
state
test.
earning
other
as
from
a
with
interest
[7]
that
large
if
are
farm
of
are
190
g
and
a
g.
a
small,
small
a
mean
greater
as
as
She
in
the
pay
in
value
the
Find
the
if
its
randomly
than
220
medium
its
mass
mass
classed
is
as
chosen
g.
[2]
and
is
large.
greater
less
than ag.
small.
a.
to
Find
the
least
two
randomly
chosen
farm.
will
number
have
head
check
before
[3]
800
expected
restaurant's
potato
of
collects
(c)
(d)
increase
each
chef
their
randomly
probability
large
in
of
potatoes
category
.
randomly
size,
the
[4]
selects
replacing
selecting
that
the
each
another.
chef
selects
at
potatoes.
[4]
pays
[Maximum
rent,
find
rent
over
an
marks:
experiment,
13]
mould
is
being
grown
in
the
circular
dish.
The
area
of
the
dish
which
is
a
by
the
mould
is
recorded
each
day
.
[1]
€20,000
4%
35
mass
potatoes
from
in
her
bank
account
and
interest
per
year,
experience
has
shown
that
the
growth
of
it
the
is
and
the
5potatoes
[1]
period.
has
associated
inflation.
with
of
classed
potatoes
Past
Sumitra
a
classed
the
covered
20-year
no
probability
has
Potatoes
The
4
would
no
potatoes
(a)
a
amount
of
restaurant
13]
apartment.
no
€130,000
13]
deviation
In
Assuming
and
standard
A restaurant
assume
€400permonth.
(a)
for
costs
renting,
distributed
[3]
the
flat
other
normally
(b)
[2]
deviations.
(iii)
account
alternative
Pierre’s
p-value.
distributed
produces
significance
hypothesesfor
(ii)
B
A and
the
marks:
(X)
potato
Pierre
savings
normal
[1]
plants
the
plants
population.
The
in
plot
payments
couldcome
left
years
sell
buying
indicates
is
mould
can
be
modelled
by
the
equation
compounded
bt
A
monthly
at
the
end
of
each
=
ae
,
where
A
is
the
area
covered
and
t
the
month.
time
in
days
that
the
mould
has
been
growing.
267
/
PRACTICE
E X AM
PA P E R S
2
Initially
there
the
day
is
1.2
cm
covered
by
the
mould
and
Player
A
needs
to
pass
the
region
R
which
contains
player
B
than
ball
to
a
point
in
the
2
next
(a)
(i)
(when
Write
t
=
down
1)
there
the
is
value
3.4
of
cm
covered.
a.
Find
the
value
significant
of
b,
accurate
to
Using
the
your
figures.
area
when
t
of
=
answers
the
dish
to
(i)
Draw
a
dish
of
cm.
5
The
of
part
covered
(a),
(c)
of
sketch
show
the
of
the
diagram
and
on
it
region
R.
[3]
by
Find
the
area
of
R.
[2]
mould
6
[Maximum
marks:
11]
[1]
the
shape
growth
dish
Find
is
the
growth
of
a
circle
with
a
radius
nation’s
slows
covered
by
when
the
value
of
t
at
begins
to
slow.
80%
of
the
area
curve,
wealth
w(p),
shows
(w)
owned
the
by
percentage
the
poorest
mould.
which
the
rate
For
example,
in
the
diagram
shown,
the
percentages
are
given
their
of
p%
a
of
equivalents,
thepoint
poorest
in
this
time
the
area
of
A
as
indicates
in
which
decimal
that
the
of
60%
the
country
own
36%
of
the
[3]
wealth
After
players.
find
2.
in
opposing
thepopulation.
rate
the
is
the
to
[2]
A Lorenz
The
of
closer
three
(ii)
(b)
any
points
[1]
(d)
(ii)
to
those
the
dish
of
that
country
.
covered
1
bymould
can
A
up
be
modelled
by
the
equation
0.9
=
c
+
dt,
to
the
point
when
the
dish
is
0.8
completely
covered.
0.7
(d)
If
the
dish
is
completely
covered
when
0.6
=
6.0
find
the
values
of
c
and
d.
[5]
w
t
0.5
(e)
Write
down
the
rate
of
growth
of
the
0.4
A
2
mould(in
cm
per
day)
during
the
period
0.3
oflinear
growth.
[1]
0.2
5
[Maximum
marks:
0.1
14]
o
The
Red
team
are
playing
against
the
Blue
team
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
p
at
soccer.
the
Part
diagram
of
the
below,
playing
where
area
all
is
shown
on
measurements
are
(a)
in
Use
the
diagram
above
to
write
down
the
metres.
proportion
of
wealth
owned
by:
y
(i)
the
poorest
(ii)
the
richest
40%
[1]
8
D
7
10%
of
the
country
.
[2]
6
B
(b)
The
diagram
below
shows
the
wealth
5
distribution
in
two
countries.
State
which
4
C
ofthe
E
two
curves
(f
or
which
the
g)
represents
a
3
l
country
in
wealth
is
distributed
2
more
evenly
among
the
population.
1
Justifyyour
A
answer.
[2]
x
0
1
2
3
4
5
6
7
8
1
1
0.9
Player
A
currently
wants
to
and
are
has
the
ball
at
(4,
0)
and
0.8
pass
to
player
B
at
(4,
6).
Players
C,
D
0.7
E
all
on
the
opposing
team
at
the
points
0.6
with
coordinates
(2,
3),
(4,
7)
and
(7,
3).
w
0.5
The
line
l
is
the
perpendicular
bisector
of
B
and
E
0.4
(a)
Write
down
the
equation
of
l.
[2]
0.3
f
0.2
(b)
Show
that
the
equation
for
the
perpendicular
g
0.1
bisector
of
[BC]
is
2x
+
3y
=
19.5
[5]
o
(c)
Find
the
point
where
l
meets
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
p
2x
+
3y
=
19.5
[2]
268
/
PRACTICE
One
the
measure
Gini
of
inequality
within
a
country
is
The
results
are
shown
in
the
table
E X AM
PA P E R S
below.
coefficient.
Level of hearing
If
the
for
0
area
≤
given
In
p
≤
by
of
1
G
the
is
=
country
A,
1
B,
=
then
below
the
the
Gini
Lorenz
coefficient
curve
(1 indicates ‘good’ hearing)
(G)
1
2
3
is
2A
the
3
w
region
Lorenz
curve
has
the
+ 0.6 p
≤ t
<
2
52
54
22
2
≤ t
<
4
35
32
1
7
2
1
28
19
Time listening
equation
2
0.2 p
0
(hours)
0.2 p
t
(c)
(i)
Find
by
the
the
proportion
poorest
50%
of
of
wealth
the
≥
4
owned
population
2
Use
incountry
B.
Find
the
test
for
independence
to
see
if
there
[2]
is
(ii)
χ
the
Gini
coefficient
for
an
association
between
music
level
time
spent
listening
to
this
loud
country
.
and
of
hearing
ability
.
Carry
[4]
out
the
clearly
Higher level
3
test
the
at
the
null
[Maximum
5%
and
marks:
significance
alternative
level
and
state
hypotheses.
[5]
5]
Practice paper 1
Time:
2
hours
Technology
required
1.5
Maximum:
110
A farmer
Answer
all
the
is
numerical
correct
stated
to
answers
three
inthe
must
significant
be
given
figures,
exactly
unless
the
or
a
otherwise
next.
should
be
be
supported
by
working
collect
and/or
the
answer
awarded
The
the
placed
farmer
at
the
for
a
is
incorrect,
correct
some
method,
marks
distance
all
truck
back
an
pumpkins.
Twenty
in
a
line,
then
end
of
each
parks
the
his
1.5
m
truck
from
next
to
line.
question.
explanations.
Where
are
pumpkin
Find
Answers
picking
questions.
pumpkins
All
m
marks
to
the
to
the
the
farmer
pumpkins,
collect
each
needs
given
one
he
and
to
walk
walks
then
to
from
carries
it
truck.
[5]
may
provided
this
4
is
[Maximum
marks:
6]
shownclearly
.
1
[Maximum
marks:
Grain
from
5]
A total
of
funds,
A,
fund
leaks
a
silo
and
forms
a
calculate
the
amount
of
grain
in
a
height
of
the
Find
farm
of
the
base
the
worker
cone
of
the
volume
measures
as
1.2
cone
of
m
as
and
1.0
grain
the
in
A pays
pays
The
diameter
out
fund
out
is
m.
cone.
At
6%
the
end
interest
three
of
on
the
the
different
first
Let
money
B
pays
out
4%
interest
and
fund
8%.
interest
amount
the
the
funds
paid
at
the
end
of
the
first
invested
amount
amount
be
a,
b
in
fund
invested
of
and
money
in
A is
fund
$4000
year
more
B.
invested
in
the
three
c
m
Form
three
them
to
equations
find
the
in
a,
amount
b
of
and
c
and
money
solve
invested
ineachfund.
1.0
year,
$1210.
The
[5]
total
than
1.2
C.
in
slant
the
the
and
invested
the
C
cone,
B
is
cone.
invested,
T
o
$18,000
[6]
m
5
[Maximum
marks:
6]
2
2
[Maximum
marks:
Consider
5]
the
equation
y
=
4.5
0.5 x
dy
In
order
to
test
whether
or
not
loud
music
affects
(a)
(i)
Find
[1]
dx
hearing,
a
group
of
280
people
are
asked
to
(ii)
record
how
many
hours
each
week
they
Hence,
to
to
music
ability
of
above
the
a
certain
group
is
volume.
also
The
find
the
equation
of
the
tangent
listen
the
curve
at
the
point
(1,
4).
[2]
hearing
tested.
269
/
PRACTICE
The
E X AM
cross-section
of
a
PA P E R S
building
can
be
ones
modelled
that
she
fails.
Theprobability
she
passes
2
by
the
y
curve
=
4.5
0.5 x
,
3
≤
x
≤
3
her
theory
passes
A builder
order
to
reach
the
needs
reach
to
the
point
do
some
highest
(1,
work
point,
on
his
the
roof.
ladder
the
test
(T )
is
practical
0.8
test
and
(C)
theprobability
is
she
0.7
In
(a)
must
Show
this
information
on
a
probability
tree.
4).
(b)
[2]
Find
the
retake
probability
at
least
one
Sabine
of
will
the
tests.
she
takes
need
to
[2]
(1, 4)
(c)
Find
the
three
9
–2
is
–1
1
2
is
before
marks:
believed
mammal
0
–3
tests
[Maximum
It
x
probability
that
is
allowed
to
drive.
[3]
8]
the
linked
she
exactly
to
metabolic
its
mass
rate
(m)
by
(R)
the
of
a
power
law
3
d
R
(b)
Find
the
minimum
length
required
ladder.
[Maximum
marks:
A boat
a
sails
values
shown
distance
of
7
km
on
a
bearing
from
a
harbour
distance
x
km.
boat
is
140°,
H.
It
then
sails
due
west
for
the
When
boat
the
bearing
changes
of
H
from
the
direction
sails
ln R
for
four
mammals
straight
in
the
table
below.
back
to
Mouse
Cat
Sheep
Lion
–3.69
0.916
3.76
3.91
0.500
–0.386
–1.51
–1.47
again
(a)
and
and
of
ln R
a
ln m
5]
ln m
035°
of
[3]
are
6
cm
for
The
his
=
Find
the
product
moment
correlation
H
coefficient
for
these
data.
[1]
N
(b)
Find
the
data
in
least
the
squares
form
ln R
regression
=
a ln m + b,
line
for
this
a, b ∈!
[2]
x
(c)
Hence,
find
estimates
for
the
parameters
N
c
(d)
7
and
d.
[3]
Use
your
model
to
estimate
rate
of
elephant,
the
metabolic
km
Give
an
your
answer
significant
given
its
accurate
mass
to
is
3850
kg.
two
figures.
[1]
H
−1
The metabolic rate of an
Find
the
value
of
x.
(e)
7
[Maximum
marks:
Find
part
a
velocity
⎟
⎝
the
at
the
point
percentage
error
with
(a)
Find
(b)
Write
down
an
ofthe
drone
t
Find
the
(d)
Find
its
h
and
at
7:00am
10
[Maximum
2
marks:
⎛
coordinates
(4,
0,
the
matrix
A
speed
of
the
In
order
to
position
distance
be
after
for
7:00
[2]
the
(a)
(i)
−4
1
a + 2
⎟
Find
of
the
drone
from
(2,
2,
0)
am.
(ii)
[1]
when
at
t
=
2.
terms
of
Write
down
this
time.
drive,
of
the
matrix
in
a.
[2]
Sabine
is
no
the
values
of
a
for
which
inverse.
[2]
matrix
A
with
a
=
1
sequence
of
numbers
is
used
to
encode
written
as
a
[3]
matrix
to
inverse
[1]
7]
allowed
the
⎠
position
S
byfinding
the
product
⎛
must
The
encoded
message
2
×
3
AS.
2
−4
6
a
theory
fails
either
or
test
and
both
of
then
a
them,
practical
she
will
test.
retake
If
she
−10
8
−3
⎞
is
⎜
take
to
⎞
a − 3
⎜
drone.
expression
hours
marks:
answer
=
0).
a
[Maximum
your
7]
⎠
The
8
in
⎟
there
(c)
.
[1]
⎝
the
−1
h
(d).
Consider
is
ml g
−1
km
2
⎜
it
0.072
⎞
1
⎜
has
is
7]
⎛
A drone
elephant
[5]
⎝
⎟
⎠
the
(b)
Find
S.
[3]
270
/
PRACTICE
11
[Maximum
marks:
7]
given
that
vertices.
A rocket
Its
is
height
launched
in
each
kilometres
t
day
at
exactly
minutes
after
the
walk
Fully
can
justify
begin
E X AM
andendat
PA P E R S
any
of
the
youranswer.
[5]
12:00.
12:00
13
[Maximum
marks:
6]
t
is
given
by
the
equation
h
=
2
1,
t
≥
0
and
its
I
graph
is
shown
am
trying
to
get
fit,
so
I
decide
I
will
walk
to
below.
school
Iwill
10
the
more
drive
often.
the
If
next
probability
I
I
walk
day
walk
is
the
one
0.6.
day
,
If
next
I
the
drive
day
is
probability
one
day
,
0.8.
9
(a)
8
Write
down
the
transition
matrix
for
this
situation.
[2]
7
mk
(b)
If
I
drive
on
Tuesday
,
find
the
probability
6
h
that
I
also
drive
on
Friday
.
[2]
,thgieH
5
(c)
On
Monday
there
is
a
40%
chance
I
will
4
choose
to
walk.
Find
the
probability
I
will
3
walk
onFriday
.
[2]
2
14
[Maximum
marks:
8]
1
(a)
Find
the
matrix
that
represents:
0
1
2
T ime,
One
day
,
allows
it
the
to
3
t
4
(i)
a
reflection
(ii)
a
rotation
rocket
reach
Sketch
is
given
each
a
height
booster
in
half
engine
the
usual
about
that
the
new
graph
on
the
same
T
be
the
Write
down
rocket’s
launch
is
the
new
equation
for
compound
axes.
the
line
[2]
delayed
by
3
(i)
Sketch
the
new
graph
minutes.
on
the
60°
y
=
3x
[3]
anticlockwise
0).
[1]
y
that
represents
transformation
=
3x
(b)
Find
(c)
Describe
has
(b)
line
followed
about
(0,
of
by
a
a
the
reflection
rotation
of
in
60°
0).
the
height.
now
(0,
of
matrix
anticlockwise
The
the
time.
[2]
(ii)
in
minutes
Let
(a)
(i)
5
same
T.
the
[2]
the
single
same
transformation
effect
as
the
that
compound
axes.
transformation
represented
by
T.
[2]
[1]
15
(ii)
Write
down
rocket’s
the
new
height,
equation
stating
the
for
[Maximum
domain.
[2]
Part
of
the
dy
y
1
=
12
[Maximum
A graph
marks:
of
a
5]
network
paths
is
shown
6]
slope
is
field
shown
in
for
the
the
differential
diagram
equation
below.
2
dx
of
marks:
the
below.
3
The
weights
inminutes
Thesum
on
to
of
the
walk
all
the
graph
along
represent
each
weights
is
of
75
the
the
time
2
paths.
minutes.
1
B
C
–3
I
–2
–1
0
1
2
3
2
5
–1
4
6
D
6
A
Let
7
C
be
the
curve
that
satisfies
the
differential
5
equation
4
H
and
passes
through
(0,
0).
J
6
3
(a)
Sketch
(b)
Find
the
graph
of
C
on
the
diagram.
[1]
G
8
5
5
the
equation
for
C,
giving
your
answer
F
in
the
form
y
=
f (x).
[5]
9
E
16
Use
the
Chinese
theshortest
postman
time
required
algorithm
to
walk
to
all
find
of
thepaths,
[Maximum
Julie
is
points
marks:
playing
for
a
a
goal
9]
game
and
in
which
she
two
points
for
scores
a
three
basket.
271
/
PRACTICE
E X AM
PA P E R S
2
The
number
of
goals
and
baskets
she
scores
Initially
there
and
next
is
1.2
cm
covered
by
the
mould
2
follow
and
Poisson
4.5,
each
distributions
respectively
,
and
with
are
means
of
independent
3.2
of
the
t
=
1)
there
is
3.4
cm
other.
(i)
Write
down
the
value
of
a.
[1]
Find:
(ii)
(i)
the
expected
shescores
total
in
a
number
of
(ii)
the
variance
points
A season
(b)
she
consists
Explain
points
Any
player
20
points
(c)
Find
the
is
a
Julie
a
number
placed
on
a
the
[3]
season
score
in
of
that
can
a
areaof
The
dish
of
cm.
5
When
[1]
season
is
the
t
=
population
of
over
(c)
by
the
the
dish
accurate
to
three
[2]
to
part
(a),
covered
by
find
mould
2.
[1]
the
shape
80%
of
the
area
Find
is
the
the
growth
placed
on
After
[3]
of
a
circle
with
a
radius
this
t
begins
to
slow.
time,
A
up
+
dt,
the
be
to
at
area
of
point
is
covered
by
slows.
which
the
rate
of
[3]
the
modelled
the
dish
growth
of
can
c
of
the
value
bymould
=
rate
of
dish
by
the
when
covered
equation
the
dish
is
8]
a
unicellular
organism
logistic
function
covered.
is
(d)
k
modelled
answers
in
completely
The
b,
figures.
is
mould,
honour.
Julie
your
when
be
honour.
marks:
of
of
distribution.
roll
probability
a
Using
of
game.
number
in
normal
mean
of
[Maximum
mean
by
total
in
value
[2]
games.
whose
the
roll
40
the
scored
modelledby
scores
of
why
the
the
significant
game
of
Find
points
(b)
17
(when
covered.
(a)
(a)
day
P =
,
k
If
the
dish
is
completely
covered
when
> 0
t
1 + e
t
=
6.0,
find
the
values
of
c
and
the
rate
growth
d.
[5]
dP
(a)
Find
.
[2]
dt
(b)
Show
(e)
that
the
point
of
inflexion
on
Write
down
of
of
the
mould
2
P
in
cm
per
day
during
the
period
of
linear
k
occurswhen
P
=
.
growth.
[6]
[1]
2
2
[Maximum
marks:
14]
Higher level
The
at
Practice paper 2
2
hours
Technology
Maximum:
in
required
110
team
soccer.
the
Time:
Red
are
Part
diagram
of
playing
the
below
area
against
being
where
all
the
used
Blue
is
team
shown
measurements
on
are
metres.
y
8
marks
D
7
Answer
all
the
questions.
6
B
5
All
numerical
answers
must
be
given
exactly
or
4
correct
to
three
significant
figures,
unless
otherwise
C
E
3
stated
in
the
question.
l
2
Answers
should
be
supported
by
working
and/or
1
explanations.
A
x
0
1
Where
be
an
answer
awarded
shown
1
for
a
is
incorrect,
correct
some
method,
marks
may
provided
an
marks:
experiment,
circular
covered
Past
the
this
5
6
7
8
is
Player
A
wants
to
13]
and
are
dish.
by
the
The
can
area
mould
experience
mould
mould
has
be
is
is
of
being
the
grown
currently
has
the
ball
at
(4,
0)
and
E
dish
recorded
shown
that
modelled
by
the
the
which
each
=
ae
time
in
where
days
A
that
is
the
the
area
has
all
to
on
player
the
coordinates
B
at
(4,
opposing
(2,
3),
(4,
7)
6).
Players
team
and
at
(7,
the
C,
D
points
3).
is
The
line
l
is
the
perpendicular
(a)
Write
down
(b)
Show
that
bisector
of
B
and
E.
day
.
growth
the
equation
of
l.
[2]
of
the
equation
for
the
perpendicular
equation
covered
mould
pass
in
bt
A
4
1
with
a
3
clearly
.
[Maximum
In
2
and
been
t
bisector
of
[BC]
Find
point
is
2x
+
3y
=
19.5
[5]
the
growing.
(c)
the
where l
meets
2x
+
3y
=
19.5
[2]
272
/
PRACTICE
Player
A
needs
to
region
R
which
contains
than
to
any
of
pass
the
the
ball
those
opposing
to
a
point
points
in
closer
the
One
to
the
B
(i)
(ii)
Draw
a
show
the
Find
sketch
the
of
region
area
of
Gini
the
diagram
and
on
R.
R.
the
[Maximum
marks:
it
for
0
[3]
given
[2]
In
a
≤
p
curve,
wealth
within
a
country
is
≤
by
1
G
the
is
=
region
A,
1
share
then
below
the
the
Gini
Lorenz
curve
coefficient
(G)
is
2A
country
,
28%
of
the
the
poorest
income
50%
and
of
the
the
poorest
12]
w(p),
shows
the
percentage
of
ofthe
(w)
owned
by
the
poorest
p%
country
population.
For
example,
in
the
diagram
below,
(i)
Write
the
percentages
are
given
their
(ii)
in
down
the
Use
the
of
the
income.
coordinates
cubic
curve
of
four
passes
regression
through.
to
find
[2]
an
which
estimatefor
as
65%
of
points
the
share
a
(c)
nation’s
of
particular
80%
A Lorenz
inequality
coefficient.
area
country
3
of
PA P E R S
players.
If
(d)
measure
E X AM
the
Gini
coefficient
for
decimal
thiscountry
.
equivalents,
thepoint
poorest
in
wealth
60%
of
that
the
A
indicates
country
own
that
36%
[5]
the
of
the
4
[Maximum
marks:
13]
country
.
(a)
Find
the
eigenvalues
⎛
2
1
the
and
eigenvectors
for
⎞
2
matrix
[7]
⎜
⎟
1
⎝
0.9
5
⎠
0.8
(b)
Hence,
find
the
solution
to
the
system
of
0.7
coupled
0.6
that
at
t
differential
=
0,
x
equations
2
and
y
=
!
y
=
x + 5y
given
4.
w
=
below,
0.5
x
=
2x
2y
[6]
0.4
A
0.3
5
[Maximum
marks:
0.2
13]
2
The
y
graph
=
0.2 x
,
0
≤
x
≤ 10 ,
where
x
is
0.1
o
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
measured
in
centimetres,
the
to
form
y–axis
a
is
rotated
360°
about
bowl.
1
p
(a)
Use
the
diagram
proportion
of
above
wealth
to
write
owned
down
The
bowl
(a)
Find
the
poorest
the
richest
40%
10%
an
the
The
diagram
below
expression
bowl
in
two
of
the
country
.
the
[2]
shows
the
two
countries.
country
more
in
curves
(f
which
the
evenly
Justifyyour
or
among
liquid.
for
the
volume
it
is
filled
to
a
of
liquid
height
boiling
point
g)
State
represents
wealth
the
is
the
surroundings
of
the
liquid
evaporates
3
h.
[6]
in
cm
from
and
the
so
bowl.
is
close
the
to
liquid
The
rate
of
−1
s
is
equal
to
0.02A,
where
A
which
thearea
of
the
liquid
exposed
to
the
air.
a
(b)
distributed
Find
an
expression
evaporation
population.
answer.
of
wealth
is
ofthe
when
temperature
evaporation
distribution
with
[1]
gradually
(b)
filled
by:
The
(ii)
gradually
the
in
(i)
is
a
[2]
height
of
the
for
the
liquid
rate
of
when
it
is
filled
h.
[3]
3
The
bowl
is
to
filled
at
a
constant
rate
of
5
cm
−1
s
.
1
(c)
0.9
(i)
Find
the
value
ofevaporation
of
h
at
which
equals
the
the
rate
rate
at
0.8
whichthe
bowl
is
beingfilled.
[2]
0.7
0.6
(ii)
Find
the
volume
of
liquid
in
the
bowl
w
0.5
atthis
point.
[2]
0.4
6
[Maximum
marks:
11]
0.3
f
An
engineer
is
trying
to
create
an
efficient
0.2
g
network.
To
do
so,
he
needs
to
calculate
the
0.1
number
of
triangles
defined
as
a
it
contains.
A triangle
is
o
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
set
of
any
three
adjacent
vertices,
p
irrespective
of
order.
273
/
PRACTICE
(a)
E X AM
PA P E R S
Using
the
definition
down
the
number
given
above,
If
write
the
solar
panels
are
100%
efficient,
2
thethree
of
triangles
in
each
of
of
graphsshown.
[2]
power
equal
to
vector
G
G
1
per
the
the
amount
2
m
(watts
per
component
perpendicular
of
to
m
I
the
)
they
that
produce
lies
along
is
the
roof.
G
2
3
B
A
B
(e)
(i)
Write
down
tothe
roof.
a
vector
perpendicular
[1]
A
A
C
(ii)
Find
the
component
alongthe
C
B
D
D
the
engineer
graph
of
creates
his
an
adjacency
matrix
M
for
(f)
Use
network.
to
at
your
time
(i)
Write
down
the
adjacency
matrix
G
lies
perpendicular
to
the
calculate
the
[4]
to
parts
number
might
(c)
of
and
watts
expect
(e)
the
to
produce
20%
efficient.
at
time
M
T
forgraph
which
T.
answers
homeowner
(b)
I
C
roof
The
vector
of
.
ifher
solar
panels
are
[2]
[2]
2
8
(ii)
Hence,
length
find
the
three
number
from
A
to
of
A
walks
in
[Maximum
marks:
17]
of
graph
G
.
[2]
A factory
fills
bags
of
flour
1
The
which
are
all
labelled
2
as
(iii)
Explain
how
the
engineer
could
use
containing
weight
adjacency
matrix
to
find
the
triangles
in
a
simple
of
graph.
answer.
V
erify
method
your
by
using
M
the
number
of
g
triangles
in
G
the
flour
put
into
the
bags
is
normally
with
a
mean
of
kg
1
and
a
standard
of
50
g.
A bag
which
weighs
less
than
is
considered
to
be
underweight.
Let
the
of
flour
in
a
bag
be
X
the
(a)
graph
so
to
weight
find
set
[4]
950
(iv)
is
Fully
deviation
justifyyour
machine
number
distributed
of
kg.
an
.
Find
the
probability
that
a
bag
of
flour
[1]
2
chosen
7
[Maximum
marks:
A homeowner
amount
if
she
She
first
wants
electricity
covered
corners
the
of
17]
her
her
to
the
roof,
measure
she
roof
calculates
of
your
in
B (0,
12,
4.8),
solar
taking
as
of
C (5,
0,
6.6)
the
and
Find
the
!!!
"
AB ×
AC.
(b)
Find
(c)
Hence,
AB
vectors
!!!
"
generate
of
inspected
of
by
to
a
are
fined
€5000.
one
found
5
to
isinthe
shape
area
of
a
of
and
0,
4.8),
12,
6.6).
got
(b)
Using
part
homeowner
that
the
surface
sun
of
is
the
most
roof
⎛
of
the
vector
takes
given
[3]
measurements
when
it
shining
shines
in
and
⎜
your
answer
on
the
probability
(ii)
the
given
The
method
flour
the
the
is
of
direction
X,
was
(c)
less
(i)
at
that
written
time
is
to
factory
4
d.p.)
fails
from
an
loss
to
the
factory
over
10
[1]
so
the
that
weight
the
of
factory
the
bags
would
fail
the
than
a
mean
weight
of
the
5
bags,
g.
down
the
distribution
of
X.
[3]
Find
the
value
the
as
I
intensity
of
the
sun
at
time
a
given
that
the
of
the
factory
failing
the
=
k
8
⎜
has
not
changed.
[2]
T
⎞
owner
mean
of
weight
the
μ,
factory
of
flour,
decides
added
to
to
increase
the
bags
the
so
⎟
4
that
magnitude
they
are
less
likely
to
fail
the
inspection.
⎟
1
⎠
Let
the
of
T
⎝
Given
factory
⎠
⎜
(d)
the
checking
if
Write
(ii)
⎟
1
⎛
be
or
inspections.
changed
inspection
The
can
the
2
[4]
expected
inspection
is
(rounded
the
probability
It
if
⎟
⎝
occurs
and
4
⎜
This
taken
authority
.
finds
⎞
8
regularly
it
rectangle.
directly
[2]
inspection
of
The
is
measures
are
giving
places.
underweight,
find:
[3]
roof,
bags
and
bags
be
decimal
the
(a),
(i)
AC .
the
underweight,
of
[2]
the
in
weights
separate
find
is
four
flour
inspection
more
!!!
"
and
answer
weight
the
random
the
house
A (0,
The
In
origin
D (5,
!!!
"
(a)
to
panels.
the
measurements:
maximum
hope
coordinates
ground-levelcorners
thefollowing
could
the
at
of
I
is
900
watts
the
cost
Inspections
of
flour
take
be
place
€1
per
after
kilogram.
10
000
bags
have
2
per
m
,
find
the
value
of
k.
[2]
been
of:
filled.
Let
expected
plus
the
cost
the
random
payment
of
the
in
flour
variable
fines
for
per
10
C
be
the
total
inspection,
000
bags.
274
/
PRACTICE
The
owner
(d)
(i)
is
keen
Find,
μ
to
you
the
a
minimize
nearest
would
uses.In
of
to
nearest
in
g,
costs.
the
value
(a)
the
owner
calculations
use
the
(c)(ii)
rounded
to
(i)
Write
down
the
George
beats
(ii)
Use
binomial
value
Explain
[4]
he
uses
smaller
why
a
you
value
would
μ
for
valuefor
not
which
gives
George
a
C.
[1]
to
George
beats
against
the
wonders
decide
(b)
Higher level
their
is
stated
to
in
55
answers
must
significant
be
given
figures,
exactly
unless
or
a
paired
be
answer
for
a
may
each
are
that
not
way
take
into
race.
sample
assume
discusses
suggests
than
test
the
normally
is
to
see
less
if
the
than
differences
the
in
distributed.
[7]
supported
is
incorrect,
correct
marks:
question
by
working
(i)
is
findings
actually
their
with
the
position
test
average
State
and/or
should
position
one
assume
some
method,
marks
may
provided
this
George
is
which
Albert,
most
important
in
the
race
be
done
and
is
on
whether
significantly
less
reason
the
why
data
is
George
cannot
normally
collects
they
considers
which
tests
of
should
level
have
seasons.
three
different
was
two
be
and
and
35
competed
He
5.23
Albert’s
skiers
is
the
a
the
over
calculates
with
of
races
in
the
previous
his
mean
that
standard
deviation
of
performed
at
a
you
should
hypothesis,
position
was
6.21
with
deviation
of
4.56.
You
may
assume
both
populations
have
the
same
variance.
5%
clearly
and
mean
best.
state
State
the
why
the
sample
mean
can
now
the
be
alternative
from
methods
(ii)
significance
data
[1]
25]
that
and
that
another
astandard
three
his
his.
three
discover
null
does
Albert.
2.12,
All
in
it
optimal
distributed.
[Maximum
to
as
the
for
George’s
otherwise
clearly
.
This
best
is
time
position
1
this
question.
should
awarded
shown
is
[6]
race
explanations.
be
it
0.5
questions.
three
an
that
to
0.5.
mean
(c)
Where
probability
equal
hypothesis
test
marks
the
the
Answers
is
to
required
numerical
correct
Albert
George
perhaps
All
the
for
information
all
that
time
who
Answer
distribution
race
George
Maximum:
if
times
theirtimes
Technology
that
[1]
mean
You
hour
who
Perform
Practice paper 3
1
times
recommend
account
Time:
of
Albert.
thehypothesis
the
10g.
the
greaterthan
(ii)
number
PA P E R S
of
recommend
your
found
5
his
E X AM
regarded
as
following
a
normal
conclusion
distribution.
for
each
of
the
George
is
trying
test.
(iii)
significantly
To
test
this
to
better
he
work
out
skier
than
records
their
if
he
his
is
rival
results
Perform
recent
an
test
to
see
if
in
this
data
provides
that
George’s
significant
evidence
Albert.
their
average
position
is
less
eight
Albert’s.
[7]
races.
(d)
In
context
state
thedifferences
Race
appropriate
a
than
most
[1]
George
a
in
possible
the
reason
conclusions
for
of
Alber t
thetests.
Time
Position
[2]
Time
Position
(seconds)
(seconds)
1
5
90.2
7
91.3
2
10
95.6
15
97
.1
3
15
98.5
14
98.4
4
4
90.5
8
92.1
5
12
97
.6
15
99.0
6
13
92.5
4
89.2
7
9
97
.5
8
97
.4
8
7
95.0
8
95.1
275
/
PRACTICE
E X AM
PA P E R S
π
2
[Maximum
marks:
30]
Approximate
values
of
f (t)
at
intervals
of
are
4
The
aim
of
acomplex
the
question
sound
is
to
find
the
equation
of
taken
wave
below.
then
2
(a)
(i)
Expand
( cos t + i sin t )
.
z
=
The
the
graph
values
of
and
f
are
( t ) sin t
given
for
in
these
the
table
points
are
calculated.
[2]
It
Let
from
is
desired
to
use
all
these
values
to
find
an
cos t + i sin t
estimate
for
f (t).
2
z
can
be
written
in
r ( cos at + i sin at ) ,
the
a, r
form
∈!
f
t
(ii)
(iii)
Write
Use
down
your
the
values
answer
to
cos 2t
1
of
(a)(i)
a
and
and
r.
[2]
(a)(ii)
to
2
show
that
=
Write
t
4.65
3.29
1.19
1.19
−2.96
p
0
0
2.94
−2.08
−1.2
1
1.2
1
−4.65
3.29
0
0
2
Rearrange
the
expression
from
(a)(iii)
3π
2
tofind
an
expression
Hence
find
for
sin
t.
[2]
∫
4
π
2
(ii)
0
π
0
(i)
0
π
[1]
∫
(c)
( t ) sin t
[2]
sin 2t sin t dt .
down
f
4
2 sin
2 π
(b)
0
(t )
sin
t dt.
[2]
5π
(iii)
Use
your
answer
to
(c)(ii)
to
show
that
4
2 π
2
sin
t dt
= π .
[1]
∫
3π
0
It
in
is
thought
the
that
f
form
the
(t ) =
a
curve
sin t +
1
shown
a
sin 2t ,
2
can
a
1
,
be
a
written
∈!
2
7π
2
4
5
2π
4
(e)
Find
the
value
of
p
to
three
significant
figures.
[2]
3
(f)
(i)
Use
the
trapezoidal
intervals
2
to
find
an
rule
with
estimate
eight
for
2 π
f
( t ) sin t dt
∫
[3]
0
(ii)
1
Hence
write
down
an
estimate
for
a
.
[2]
1
1
2
(g)
It
is
given
sin
that
2t
(1
=
cos 4t )
2
0
p
p
p
Use
3p
2
this
result
to
show
that
2
2 π
f
–1
∫
( t ) sin ( 2t ) dt
= πa
.
[5]
2
0
The
table
below
shows
the
values
of
f
( t ) sin 2t
at
π
–2
intervals
of
for
0
≤ t
≤
2π
4
f (t) sin 2t
(h)
(i)
0
Use
4.65
the
0
2.96
0
trapezoidal
intervals
to
find
an
2.94
rule
0
with
estimate
4.65
0
eight
for
2 π
–5
f
∫
( t ) sin ( 2t ) dt
[2]
0
2 π
(d)
f
Show
∫
( t ) sin t dt
= π a
.
[2]
(ii)
Hence
write
down
an
expression
for
the
1
0
function
f (t).
[2]
276
/
INDEX
adjacency
of
edges,
graph
theory
130
adjacency
matrices,
graph
theory
131–2,
affine
transformations,
algebra
matrices
117–18
20–4
matrices
35–6
relationship
algorithms,
to
graph
geometry
theory
Cartesian
135
cdf
see
cells,
133–5,
137
complex
cumulative
V
oronoi
limit
central
tendency
rule,
Chinese
numbers
distribution
diagrams
central
chain
48
form,
theorem
of
function
111–12
174–5,
204
probability
differentiation
postman
25–8
distributions
231–2,
problem,
162
233
graph
theory
135–6
2
amortization
amplitudes,
angles
of
sinusoidal
elevation
annuities
also
depression
103
228
mean
and
sectors
107,
114,
transformation
areas
108,
104–5,
between
a
trapezoid
229,
rule
diagrams
images
119
freedom
115,
values
and
the
107,
areas
107,
trigonometric
axis
see
also
circuits,
columns
13
107,
of
of
theory
matrices
15
normal
207–8
random
v
command
variables
terms
normal
complement
completed
equations
67–8,
curves
2–3,
also
263–76
57–61
modelling
245
complete
78–81
167
179–81
mean;
median;
187,
distributions
probability
for
mode
189
relationship
approximations
whisker
plots
numbers
153
4
183–4
equations
integration
224–7,
228–30,
plane,
of
form
Argand
vectors
in
a
equations
243–4
25–7
direction
124
53–5
231–2
235
transformations
transformations
constant
71
28
given
compound
intervals
graphs,
changes
76–7
14–19
up/concave
connected
functions
136
diagrams
interest
231–4
151
quadratic
differential
compound
confidence
15
27
integration
order
theory
(Euler)
differentiation
208
form,
11–13,
15
25–7
functions
240–7
235–9
of
sequences
10–11,
24–30
form
form
sequences
probability
graph
complex
concave
differentiation
199
172–3
(vertex)
eigenvalues,
composite
224–55
differential
square
components
proportion
events,
exponential
163–4
population
(R2)
arithmetic
complex
complex
89,
of
Cartesian
105–6
88,
135
171–3
geometric
graphs,
polar
lines
113
v
papers
256–62
differential
relationship
33
distributions
practice
calculus
201–2
combinations
ratio,
asymptotes
134,
determination
common
and
195,
114
function
iv
,
of
rule
114–15
internal
bounds
5
distribution
107,
difference,
test
than
202
115
common
binomial
parameters
radians
graph
coefficient
10–11
terms
114,
iv–vi
best-fit
202–3
195–6
external
bearings
193–5,
114
x
25–7
interest
arcs
sectors
229–30
functions
command
see
206–7
greater
Poisson
123
assessment
averages
201–3
population
circumference
236–8
sequences
simple
115
109
114,
curve
integration
box
of
193–6,
coefficients
independence
2–4
triangles
linear
of
test
circles
surface
arithmetic
degrees
expected
integration
matrix
Argand
)
correlation
107–8
circles
area
and
63
(X
estimating
approximation
area
models
17
antiderivatives
see
chi-squared
18–19
down
matrices
curves
117–18
233
206
graph
in
of
data
theory
132
182–3
277
/
INDE X
contingency
continuous
tables,
random
convenience
systems
correlation
187–90
from
or
strength
188
distributions
195
165–70
rule
tests
rank
regions,
models
cumulative
of
dot
198
cumulative
coefficient
187,
188,
correlation
coefficient
190,
198
190
of
graph
elements
203
vectors
63,
equally
123–4
function
distributions
maximum
points
of
tangents
166–9
185–6
graph
and
229–30,
minimum
inflexion
81,
133,
forms
to
231,
validity
233,
2–3
Euler
225
descriptive
statistics
presentation
of
deleted
a
vertex
dependent
algorithm,
derivatives
50,
see
55,
René
48,
developing
models
differential
equations
Euler ’s
233–4
form
179–86
for
real-world
method
solutions
the
fields
differentiation
and
second
derivatives
models
probability
by
hand
31
II
in
vii
hypothesis
tests
209–10
8
form
of
complex
graph
solving
numbers
theory
differential
problems
149,
giving
interval
relationship
28
125
equations
245–7
151
preferred
4
2–3
over
decimal
approximations
48
iv–vi
terms
papers
v
263–76
and
strategies
vi
expected
frequency
,
probability
expected
number
of
occurrences
expected
value
random
of
experimental
probability
experiments,
definition
distributions
an
variable,
event,
162
probability
probability
frequency)
152
distributions
162
150
149
decay
models
exponential
equations,
exponential
(Euler)
exponential
functions,
exponential
models
exponential
regression
exponential
relationships,
82
graphical
form
of
61,
solutions
complex
geometric
60,
of
(relative
exponential
66,
6
numbers
sequences
28
13
78
88
linearization
of
data
88
powers
5
20–1
form
models
extrapolation
232
encountered
4–6
extending
226–7
6–8
for
real-world
beyond
data
73,
problems
189,
69,
81
190
233
factorized
233–4
theory
60,
solving
values
rational
rule
change
graph
answers
probability
standard
problems
of
over
82–4
3
circuits,
for
commonly
240–1
233
quotient
rounding
exponents
231–4
rates
discrete
in
to
identify
and
method
laws
related
variation
243–4
244–7
variables
231–2,
graphs,
eigenvalues
244–5
optimization
directed
81
242–3
224–7,
rule
product
65,
245–7
diagrams
separating
chain
problems
240–7
imaginary/complex
slope
231
241–2
numerical
trails
command
50
statistics
71
2–3
examinations
137
74
functions
Cartesian
coupled
to
preparation
descriptive
phase
theory
differentiation
also
type
GDC
practice
derivatives
also
Descartes,
graph
16
trigonometric
see
and
fractions
131
224
second
student
estimation
theory
error
approximation
179–86
graph
variables
depreciation
exact
197
183–6
vertex,
150
technology
(exponential)
Eulerian
reliability
I
estimation
234
136
and
240–4
48–9,
due
estimation
events
direct
three
37–8
33
outcomes
of
using
57
data
collection
55–6
111–12
equations
matrices
likely
type
236–8
points
226,
normals
theory
sketching
eigenvectors
advantage
Euler ’s
of
diagrams
and
sample
and
122–3
130
accumulation
165
curves
under
86
errors
(cdf)
52
area
70,
5
percentage
finding
52
vectors
versus
theory
of
equations,
69–70
frequency
160–2
54
differential
197
distributions
drawing
degree
functions
graphs,
(constant)
distributions
126–9
50–1,
products,
V
oronoi
curves
cycles,
function
probability
edges
correlation
distribution
normal
(scalar)
drawing
188
e
tests
t-test
60,
binomial
a
eigenvalues
validity
cubic
of
variables,
vectors
piecewise
104
products
random
displacement,
domain
206–7
moment
cross
discrete
108–9
test
Spearman’s
critical
normal
independence
coefficients
product
criterion
for
192
negative
chi-squared
cosine
test
restricting
two
positive
correlation
variables,
sampling
coordinate
data
chi-squared
131
62
distributions,
binomial
distribution
163–4
form,
quadratic
functions
financial
applications,
sequences
forms
of
equations
48–9,
forms
of
quadratic
functions
71
14–19
71
71
278
/
INDE X
fractals,
matrix
fractions
transformations
versus
frequency
functions
decimals,
tables,
histograms
118
probability
descriptive
statistics
hypotenuse
151
181,
hypothesis
185
type
46–101
finding
sum
graphs
55–90
inputs
lines
and
or
difference
outputs
of
two
functions
74–8
51
terminology
versus
50
relations
graphs
form
geometric
of
compound
functions
infinite
35
numbers
29
infinite
18–19
initial
goodness
of
inside
187
second
lines
derivatives
solving
110,
lines
tangents
and
curves
financial
finding
logarithmic
applications
sum
or
of
equations
sequences
difference
of
two
and
asymptotes
drawing
intercepts
of
intersection
functions
of
a
models
between
between
of
features
as
topic
use
reflections
functions
inverse
76
function
inverse
theory
on
GDC
30,
52,
56,
composite
74–8
functions
53
and
the
x
axis
229,
236–8
31,
238
57
14–19
58,
70,
73
iv
,
IQR
see
256–62
criteria
257–61
258
process
69–70
engagement
259
257–8
259–60
256–7
range
functions
260–1
(IQR)
between
two
51–2,
182,
183,
graphs
54,
186
64
55
35–6
of
variation
investments
74–8
window
functions
261–2
reflections
50–2
57
viewing
curve
mathematics
matrices
of
transformations
of
intersection
64
75,
a
revolution
choice
inter-quartile
57
graphs
of
213
235–9
assessment
reflection
57
55–90
two
view
chains
15
personal
13
55–6
31,
Markov
239
of
a
matrix,
“machine”
communication
58
89
representations
roots
of
11
22–4
functions,
228–30,
presentation
function
87,
sequences
sketching
functions
logarithmic
graphs
models
15–16,
75,
60,
76
62
18–19
inter-quartile
range
130–8
adjacency
matrices
algorithms
133–5,
Chinese
postman
features
130–1
minimum
131–2,
kinematics
135
integration
137
problem
vectors
135–6
see
spanning
transition
matrices
travelling
salesman
grouped
outside
checklist
17
59
location
versus
features
inverse
geometric
57–8,
determining
graph
and
modelling
arithmetic
series
outputs,
assessment
6
graphs
and
and
internal
and
sequences
probability
simple
225
solutions
exponential
74
powers
arithmetic
compound
to
55,
195–6
interest
244–5
46–8
normals
test
16
intercepts
equations
50,
exponents;
volumes
225
233
differential
straight
variables
kinematics
224–6
243–4
numbers
chi-squared
geometric
area
gradients
perpendicular
complex
also
integration
48
24,
equations
4
state
inputs
graphical
also
inflation
13
102–48
derivatives
209–10
imaginary
inequalities,
equations
errors
differential
see
22–4
of
102
201–10
eigenvalues,
geometry
form
II
triangle
imaginary
series,
gradient-intercept
angled
type
matrices
49
14–15,
185
193–6,
identity
geometric
fit
and
independence,
74–8
11–13
interest
exponential
I
independent
equations
sequences
tests
54
indices
general
right
function
see
of
of
identity
31
transformations
165–6,
58
46–52
notation
87,
trees
134–5,
also
Kruskal’s
137
239
126–9
velocity
algorithm,
graph
theory
134
132–3
problem
136–7
data
cumulative
frequency
descriptive
statistics
histograms
185
curves
185–6
laws
of
exponents
laws
of
logarithms
least
squares
curves
181–2
lines
4–6,
20
21–2
regression
199–200
189–90
limits
half-life,
exponential
Hamiltonian
paths
decay
and
78,
cycles,
infinite
82
graph
theory
136,
137
geometric
terminal
velocity
series
22,
23
247
279
/
INDE X
linear
combinations,
linear
equation
linear
functions,
linearization
systems
of
data
models
linear
regression
linear
relationships
linear
transformations
lines,
of
lists,
not
loans
local
88,
88,
187,
logarithmic
scales
logarithms
laws
for
graphs
87,
graph
sequences
lower
quartile
“machine”
view
differentiation
magnitude
of
and
minor
chains
227
5,
6,
130
180,
183,
185
functions
graph
of
scalar
a
and
rules
circle
eigenvectors
34,
of
maximum
and
minimum
points
data
points
of
182,
curves
183
57,
59,
226,
227
180
grouped
best
of
Poisson
of
sample
fit
207–8
173–4
spread,
and
lower
descriptive
bounds
statistics
of
approximations
whisker
cumulative
grouped
minimum
frequency
data
181–2,
spanning
class,
plots
grouped
curves
graph
data
181–2
modelling
process
65–73,
80–1
differential
expected
of
matrices
order
of
transformation
equations
244–7
values,
196,
statistical
tests
193
204
226–7
33
probability
sequence
76–7
149
182–4
chi-squared
sample
test
parallel
forms,
parallel
lines
4
reliability
47,
48,
123
theory
136
sequences
and
probability
product
error
105,
47,
48,
diagrams/portraits
phase
shift
meanings
coefficient
(r)
187,
188,
190
models
PMCC
product
70,
inflexion
110,
complex
62
225
functions
79,
of
81,
distribution
60,
242–3
80,
moment
form
models
29
sinusoidal
piecewise
form,
function
110–12
phase
of
8
correlation
sinusoidal
lines
point-gradient
198
3
perpendicular
see
69
121
series
moment
bisectors
of
data
distribution
perpendicular
derivatives
134–5,
103,
of
60,
85,
29,
correlation
231,
176–9,
79
87
equations
226,
30,
coefficient
48
233,
234
207–8
numbers
27
polynomials
185–6
theory
193–4
model/functions
105,
graph
see
test
206
quadratic
137
systems
of
224
equations
terminology
population
181
tests
order
polar
183–4
185–6
trees,
193
derivatives
Poisson
and
hypothesis
problems,
points
182
180
box
172–3
optimization
sum
187
upper
104–6
165–70
hypothesis
versus
other
181–2
distribution
measurements,
modal
2–3
data
tests
periodic/periods,
116–17
minimum
of
137
225
solutions,
percentage
36
and
estimates
theory
199–200
tests
Pearson
maximum
of
graph
two-tailed
pdf
34–5,
78
or
paths,
35
transformations
of
also
116–19
multiplication
median
alternative
patterns,
matrix
measures
null
33
addition
of
curves
parallelograms
35–6
211–14
matrix
239
203–5
to
parabolas,
37–8
131–3
multiples
126–9,
114
34
chains
6,
triangles
normals
paired
transformations
line
204
p-value,
functions
Markov
z-test
outliers
211–14
theory
inverse
196,
and
5,
combinations
t-test
outcomes,
74–8
120
arcs
63
78
33–8
function
angled
one-tailed
vectors
eigenvalues
mode
226,
89
of
80–90
188
distributions
observed
(Q1)
65–73,
130
algorithm,
regression
linear
14
87
theory
theory
(ln)
correlation
numerical
log-log
mean
neighbour
see
79
60–2,
(kinematics)
graph
nearest
5
models
problems
velocity
logarithm
normal
199–200
6
logarithms)
loops,
body
statistical
logistic
matrices
189,
21–2
(natural
parameters
natural
5–6
definitions
world
78–9
edges,
non-linear
minimum,
equations
Markov
190
18
or
real
89
170–1
recommended
a
also
non-right
89,
60,
of
negative
logarithmic
major
see
46–52
maximum
Loge
motions
88–9
89–90
187–8,
variables
best-fit
17,
types
61
61,
to
determining
13
116–17
functions
lines
sequences
multiple
linear
random
application
relationships
60,
models
171–3
87–90
relationships
matrices
variables
32
arithmetic
exponential
power
random
31
modelling
differential
31–2
65,
80–1,
equations
83,
241–3,
87–9
246–7
280
/
INDE X
population
posing
statistics,
real-world
samples
problems
position
vectors
positive
correlation
power
from
65,
regression
191–8
relationship
80
relative
121
linearization
of
data
4–5
see
also
representing
dependent
Markov
variables
chains
probability
rounding
sample
mean
scales
152
from
any
graph
distribution
theory
sinusoidal
134,
models
135
62
149
events
their
and
expected
complements
outcomes
experimental
formulae
replacement
effects
157,
distribution
probability
distributions
normal
probability
of
matrices
product
moment
p-value,
chi-squared
Pythagoras’
angled
roots
of
three
rotations,
matrix
rounding
2–3,
quadratic
quota
sampling
see
Pearson
radians
27,
79,
(pdf)
161,
164,
data
166
176–9
chains
213–14
coefficient
(PMCC)
187,
193–4
102
sampling
random
variables
linear
61,
scalar
multiples
walk,
188,
190
range
of
data
range
of
a
rates
of
scatter
diagrams
order
sectors
of
real
171–5
theory
132
182
52
rates
exponents
numbers
real-world
reflecting
the
sequences
8–19
(∑)
models
modelling
for
process
real-world
65–73,
problems
80–1
66,
81
inverse
matrix
functions
75,
76
transformations
regression
curve
regression
line
116–17
199–200
for
y
on
x
189–90
122–3
33
form)
6–8
equations
247
114–15
variables,
transformations
solving
applications
states
not
geometric
notation
levels,
figures
118
differential
equations
240–1
14–19
independent,
series
22–4
Markov
chains
211–14
statistical
tests
9
simple
interest
15
sine
rule
104
sine
rule
of
sine
waves
unit
theory
circles
130
113
60
also
sinusoidal
models
models/functions
V
oronoi
diagrams
graphs,
fields,
Spearman’s
standard
193
2–3
graph
differential
rank
equations
196,
z-test
204
62–3,
72,
78
55–6
244–5
coefficient
190,
198
182
distributions
t-test
60,
drawing
correlation
(σ)
29,
111–12
versus
deviation
normal
reflections
vectors
88
graphs,
slope
149
matrix
simple
sketching
2–8
173–4
8–9
infinite
sites,
20–1
problems,
on
graphs
sinusoidal
233
distributions
233–4
107,
shapes,
separating
see
224
probability
197–8
differential
circles
of
from
(standard
second
significance
change
related
117
187–8
derivatives
sigma
231
170–1
50,
116,
87
notation
significant
function
121
57
matrices
second
coefficient
191–2
differentiation
rational
correlation
of
numbers
financial
moment
114–15,
graph
vectors
33
products,
scaling
71
combinations
random
54
191–3
(dot)
sequences
transformations
200
152
collection
scalar
semi-log
192
random
82–4,
methods
71
113,
31,
transformations
matrices
scientific
test
product
function
199,
102–4
sampling
164
series
r
individual
159,
34
60,
of
definition
163–4
correlation
models
a
space,
59
forms
triangles
curve
of
sample
functions
graphs
sum
function
self-similar
quadratic
domain
regression
153–9
Markov
theorem
squares
predictions
150
165–70
distribution
product
153–9
87–9
vectors,
right
160–9
distribution
matrices,
164
191–2
mean,
function
distribution
Poisson
probability
sample
theoretical
diagrams
probability
binomial
159,
152–3
versus
and
the
of
157,
197–8
87
restricting
rows
151
88
data
least
a
87,
collection
sampling
residuals,
resultant
174–5
149–79
definitions
of
linearization
50
152
algorithm,
probability
graphs
211–14
results
axis,
using
data
probability
representative
exponents
81,
150
tests,
replacement,
89–90
predictions
principle
of
80,
correlation
representations
powers
Prim’s
see
61,
frequency
reliability
188
relationships,
tools
166–70
203
standard
form
of
quadratic
standard
form
(scientific
functions
notation)
71
6–8
281
/
INDE X
statistical
tests
chi-squared
paired
trials,
191–6
test
sample
triangles
193–6
test
definition
206
102–6,
functions
ratios
191–3
trigonometric
t-test
196,
203–5
trigonometry
z-test
203–4
t-test
149–214
descriptive
179–214
probability
149–79
steady
state
steepness
see
graph
transition
matrices
132
46
also
straight
vector,
type
type
and
unit
circles,
unit
vectors
and
of
linear
models
equations
48,
II
upper
60
equations
validity
stretches
of
75,
variables,
connected
areas
108,
76
graphs,
graph
theory
of
variable
132
lower
to
(Q3)
tests,
normal
vector
equations,
systems
equations
31–2
vector
products
vectors
of
outcomes,
distances,
tangents
to
terminal
velocity
terms,
testing
curves
sequences
models
test-retest
graph
theory
probability
153,
136
154,
156–8
and
series
graph
problems
137
185
real-world
reliability
problems
of
data
65,
66,
straight
of
two
121
complex
numbers
scalar
197
and
addition
properties
81
collection
197
independent
50,
55,
74
128–9
166–7
lines
125–6
vectors
123–4
120–9
kinematics
8–9
in
a
26,
27
given
direction
124
126–9
126–9
120–2
products
122–3
velocity
tests
data
see
collection
also
theoretical
three
dimensional
waste
trails,
tests
probability
volumes
toxic
197
statistical
geometry
dump
graph
matrices
random
order
problem
sequence
transition
diagrams,
transition
matrices
graph
theory
Markov
translation
transposing
x
and
y
trapezoid
rule,
travelling
salesman
trees,
graph
74–8
76–7
chains
area
under
theory
134–5,
of
inverse
curve
functions
of
quadratic
functions
71
154,
theory
158–9
revolution
graph
minimum
on
GDC
30,
points
52,
56,
58,
70,
73
238
110–12
theory
graphs,
and
131,
graph
population,
135
theory
modelling
133–8
80–1
54
zero
229–30
graph
153,
form
155
111–12
graphs
diagrams
weighted
problem,
probability
volumes
world
find
153,
130–1
diagrams
window,
108
walks,
74–6
to
square)
maximum
volumes
V
oronoi
211–12
211–14
graphs
diagrams,
theory
also
viewing
170–1
Markov
126–9
probability
(completed
graph
132–3
chains
of
diagrams,
see
graphs
247
128–9
121–4,
V
oronoi
of
240–1,
vertices
111
135
variables
of
239
vertex
116–19
transformations
integration
V
enn
238
theory
equations
vectors
shapes
103–4
108,
differential
variable
150
transformations
tree
183,
distributions
displacement
247
method,
209–10
113
salesman
kinematics
components
225
for
180,
dependent
191–2
tables
tests
4
data
sampling
least
functions
travelling
systematic
of
hypothesis
bounds
velocity
,
variance,
109
tables
204
125–6
192
of
errors,
196,
121
quartile
sampling
graphs
tests
trigonometric
solutions
49
stratified
surface
one-tailed
approximations
forms
strongly
102
203–5
or
I
113–15
102–6
two-tailed
lines
vector
196,
upper
gradients
109
trigonometric
sampling
statistics
149
136–7
matrices
zeros
of
a
z-test
204–5
35
function
31,
57
137
282
/
O X
F O
R
D
I B
P
R
E
P A
MATHEMATICS:
AND
R
E
D
APPLIC ATIONS
INTERPRETATION
Authors:
Peter Gray
Fully addressing the latest syllabus for Mathematics: applications
David Harris
and interpretation at Standard Level and Higher Level, this edition is
Also available
developed in cooperation with the IB and includes comprehensive
from O xford
coverage of both the First Teaching 2019 courses.
9780198426981
This is the ideal text for students studying for the IB examinations and
preparing International Assessments. The focused approach explains
the key points of the syllabus in clear, student-friendly terms. Each new
concept or technique is illustrated with exam-style examples.
Sample student answers highlight common errors and help explain
how exams are assessed. Practice questions of increasing diculty
cover all the topic areas, with sample exam papers helping
students to develop assessment condence.
9780198427049
O xford course books are developed with the IB. This means that
they are:
➜
The most comprehensive and accurate match to
IB specifications
➜
Written by exper t and experienced IB examiners and teachers
➜
Packed with accurate assessment suppor t, directly from the IB
➜
Truly aligned with the IB philosophy, challenging learners with
the latest materials.
What's on the cover?
Aerial view of a Thailand
cityscape at night
I B
D I P L O M A
P R O G R A M M E
FOR FIRST ASSESSMENT IN 2021
Suppor t material available at www.oxfordsecondary.com/ib-prepared-suppor t
How
to
get
in
contact:
web
www.oxfordsecondary.com/ib
email
schools.enquiries.uk@oup.com
tel
+44
(0)1536
452620
fax
+44
(0)1865
313472
ISBN
9
978-1-382-00728-3
781382
007283
/