15
Food packaging Optim ization Price vs profit
Pizza Packing theory
Pick and mix
Choosing the right
outlet
Food and drink
Diet
65
Food temperature
639
EXPLORATION
Research IA tip
Once you have chosen a topic, you will need to do some research. The purpose Try to avoid writing
a research report in
of this research is to help you determine how suitable your topic is.
Don't rely on the Internet for all your research—you should also make which you merely
explain a well-known
use of books and academic publications. result that can easily
Plan your time wisely—make sure that you are organized. be found online
Don't put it off—start your research in good time.
or in textbooks.
For Internet research: refine your topic so you know exactly what information
Such explorations
you are looking for, and use multiple-word searches. It is very easy to spend have little scope
hours on the Internet without finding any relevant information. for meaningful and
critical reflection and
Make sure that you keep a record of all the websites you use—this saves so
much time afterwards. You will need to cite them as sources, and to include it may be difficult
them in your bibliography.
Make sure that the sources are reliable—who wrote the article? Are they to demonstrate
personal
qualified? Is the information accurate? Check the information against engagement
another source.
IA tip
Research in your own language if you find this easier. Learning new
mathematics is not
These questions will help you to decide whether the topic you have chosen enough to reach
is suitable. the top levels in
Criterion C: Personal
What areas of mathematics are contained in the topic? engagement.
Which of these areas are contained in the syllabus that you are following?
Which of these areas are not in the syllabus that you are following but IA tip
are contained in the other 1B mathematics course? Popular topics
such as the Monty
Which of these areas are in none of the 1B mathematics courses? How Hall problem, the
Birthday paradox,
accessible is this mathematics to you? and so on are not
Would you be able to understand the mathematics and write an likely to score well
on all the criteria.
exploration in such a way that a peer is able to understand it all?
How can you demonstrate personal engagement in your topic?
Will you manage to complete an exploration on this topic and meet all the
top criterion descriptors within the recommended length of 12 to 20 pages
(double spaced and font size 12)?
Writing an outline
Once you think you have a workable topic, write a brief outline including:
Why you chose this topic.
How your topic relates to mathematics.
The mathematical areas in your topic, for example, algebra, geometry,
calculus, etc.
The key mathematical concepts covered in your topic, for example,
modelling data, areas of irregular shapes, analysing data, etc.
The mathematical skills you will use in the exploration, for example, integration
by parts, working with complex numbers, using polar coordinates, etc.
Any mathematics outside the syllabus that you need to learn.
Technology you could use to develop your exploration.
0 New key terms that you will need to define or explain.
How you are going to demonstrate personal engagement.
A list of any resources you have/ will use in the development of your
exploration. If this list includes websites you should include the URL and
the date when this was accessed.
Share this outline with your teacher and with your peers. They may ask
questions that lead you to improve your outline.
640
15
This template may help you write the outline for the exploration when IA tip
As you write
presenting a formal proposal to your teacher. your exploration,
remember to refer to
Mathematics exploration outline
the criteria below.
Topic:
Exploration title: IA tip
Remember that
Exploration aim: your peers should
Ex loration outline: be able to read and
Resources used: understand your
Personal en a ement:
work.
Writing your exploration
Now you should be ready to start writing your exploration in detail.
You could ask one of your classmates to read the exploration and give
you feedback before you submit the draft to your teacher. If your
exploration is related to another discipline, for example, Economics, it
would be better if the peer reading your exploration is someone who
does not study economics.
Mathematical exploration checklist
Work through this checklist to confirm that Have you used notation consistently
you have done everything that you can to through your exploration?
make your exploration successful. Have you defined key terms (mathematical
and subject specific) where necessary?
Does your exploration have a title? Have you used appropriate technology?
Have you given a rationale for your choice Have you used an appropriate degree of
accuracy for your topic/exploration?
of exploration? Have you shown interest in the topic?
Have you used original analysis for your
Have you ensured your exploration does not
include any identifying features—for example, exploration (eg simulation, modelling,
your name, candidate number, school name?
Does your exploration start with an surveys, experiments)?
introduction? Have you expressed the mathematical ideas
Have you clearly stated your aim? in your exploration in your own way (not
Does your exploration answer the stated aim?
Have you used double line spacing and just copy-and-pasted someone else's)?
Does your exploration have a conclusion that
12-point font? refers back to the introduction and the aim?
Do you discuss the implications and
Is your exploration 12—20 pages long? significance of your results?
Have you cut out anything that is irrelevant Do you state any possible limitations and/or
to your exploration?
Have you checked that you have not extensions?
repeated lots of calculations?
Do you critically reflect on the processes you
Have you checked that tables only contain
relevant information and are not too long? have used?
Is your exploration easy for a peer to read Have you explored mathematics that is
and understand? commensurate with the level of the course?
Have you checked that your results are correct?
Is your exploration logically organized?
Are all your graphs, tables and diagrams Have you clearly demonstrated
correctly labelled and titled?
understanding of why you have used the
Are any graphs, tables and diagrams placed
appropriately and not all attached at the end? mathematical processes you have used?
Have you used appropriate mathematical Have you acknowledged direct quotes
language and representation (not computer
appropriately?
notation, eg * etc.)?
Have you cited all references in a bibliography?
Do you have an appendix if one is needed?
641
Paper 1
Time allowed: 1 hour 30 minutes [l mark]
Answer all the questions. [l mark]
[2 marks]
All numerical answers must be given exactly or correct to three significant [l mark]
figures, unless otherwise stated in the question. [4 marks]
Answers should be supported by working and/or explanations. Where an [2 marks]
answer is incorrect, some marks may be awarded for a correct method, [2 marks]
provided this is shown clearly. [3 marks]
You are not allowed to use a calculator for this paper.
Section A (Short answers)
1 [Maximum mark: 51
Find the derivatives of the following functions.
a
1
b
x
c x sin x
d
2 [Maximum mark: 41
A single piece of wire has been bent to form the perimeter of a square,
with side length of 2 cm. The wire is straightened out and then bent
again to form an arc of a circle of radius 4 cm. The wire just forms the
arc, not the radii. Calculate the angle, in radians, that the arc subtends
at the centre of the circle.
3 [Maximum mark: 71
+Functions from R R are defined by
f (x) = 2x+l, g(x) = 3x— 2.
Find a
c
642
4 [Maximum mark: 81 [2 marks]
Let log x = p and log y = q. Find [2 marks]
[2 marks]
a log(xy) [2 marks]
[5 marks]
—x2
[4 marks]
b log [2 marks]
[5 marks]
C log N/¯x
[4 marks]
d log(l()()y).
[7 marks]
S [Maximum mark: 51
Find the equation of the straight line that goes through the point
(4, 3) and is perpendicular to the line y = 2x + 11. Give the answer
in the form y = mx + c.
6 [Maximum mark: 61
A piece of wood is 290 cm long. It is going to be cut into 10 pieces,
each of which is 2 cm longer than the piece cut off before it. All the
wood will be used up. Find
a the length of the first piece
b the length of the last piece.
? [Maximum mark: 51
41
Find dx giving the answer in the form In k, where ke Z.
2X+1
Section B (Long answers)
8 [Maximum mark: I II
a The probability distribution function for a discrete random
variable, X, is given by
x 123
a 0.2 0.3 0.1
i Find the value of a.
ii Find E(x).
Ab Toby takes part in a sports match and wins. random variable, Y,
represents Toby's winnings in euros. The probability distribution
function for Yis given by
-2 1 2 3
b c 0.1 0.2
If it is known that the game is a fair game, find the values of b and c.
643
PAPER 1
9 [Maximum mark: 131 [2 marks]
Let f (x) = x2 +6x +(8 + k). [5 marks]
[2 marks]
a Find the discriminant, D, of this quadratic, in terms of k. [2 marks]
[2 marks]
b Given that f (x) > 0 for all values of x, state, with reasons, the set
[l mark]
of values that can be taken by Di ii k.
c Write f (x) in the form f(x) = (x + p)2 +q, where pe Z and [4 marks]
[11 marks]
q is an expression involving k.
d Explain how the form obtained in c confirms the answer to b ii.
e If k = 4, write down the minimum point of the quadratic.
10 [Maximum mark: 161
A kmfarmer has a tractor in a field at point A, 2 due south of point B,
on a road which runs from west to east. The farmer wishes to get to
kmpoint C, which is 2 east of B, on the road. The tractor travels at
20 km/h on the road and 12 km/h in the field. The tractor will travel
kmin a straight line across the field, to join the road at point, P, x east of B.
a Sketch a diagram to represent this information.
b Show that the total time T for the tractor to travel from A to C,
is given by
2
12 20
c Use calculus to find the value of x that minimises the joumey time, T.
644
Paper 2
Time allowed: 1 hour 30 minutes [3 marks]
Answer all the questions. [4 marks]
All numerical answers must be given exactly or correct to three
[6 marks]
significant figures, unless otherwise stated in the question.
[2 marks]
Answers should be supported by working and/or explanations. Where [l mark]
an answer is incorrect, some marks may be awarded for a correct [l mark]
method, provided this is shown clearly. [2 marks]
You need a graphic display calculator for this paper. [3 marks]
Section A (Short answers)
1 [Maximum mark: 71
a i Factorize x 2 + 9x + 14.
ii Hence solve x2 +9x + 14 = 0.
b Solve x2 +4x + 2 = 0, giving your answers
i exactly
ii to 4 significant figures.
2 [Maximum mark: 61
The random variable Xis normally distributed, with mean equal to 8.
Given that P (X > 7) = 0.69146, find the value of the standard
deviation of X.
3 [Maximum mark: 91
Abi plants ten seeds. The probability that a seed grows into a plant
is always 0.8 (this is independent of what the other seeds do).
Let the random variable, X, be the number of seeds that grow
into plants.
Xa State the distribution that satisfies, including the parameters.
b Write down the mean of X.
C Calculate P (X = 7).
d Calculate P (X > 6).
X Xe Given that > 6 find the probability that = 7.
645
PAPER 2
4 [Maximum mark: 71 [1 mark]
[6 marks]
Maria sees and is frightened by a huge monster. The angle of elevation
from Maria's feet to the top of the monster's head is 400. Maria runs
m(on horizontal ground) 5 further away from the monster. The angle
of elevation from Maria's feet to the top of the monster's head is now 350.
a Sketch a diagram to represent this story.
b Find the height of the monster.
S [Maximum mark: 71
An object is moving in a straight line from the origin with velocity
v m/s given by v = 5sin (t 2 where t is the time in seconds from when the experiment
sta rted.
a Write down the initial velocity of the object. [1 mark]
[3 marks]
b Find the total distance that the object travels in the first 3 seconds. [3 marks]
c Find the displacement of the object from the origin, when t = 3.
6 [Maximum mark: 41 [4 marks]
6
+—Find the constant term in the expansion of x 2
x
Section B (Long answers)
? [Maximum mark: 161
Data is collected from ten football teams in the same league.
The amount that each team spent on players' wages, w, in millions
of pounds and the number of goals, g, that the team scored is recorded
in the following table.
Wages,w 200 185 190 150 160 120 50 100 40 10
32
Goals,g 92 8? 85 81 69 62 54 41
a For the amounts spent on wages calculate
i the mean ii the standard deviation. [3 marks]
[3 marks]
b For the goals scored calculate
[6 marks]
i the mean ii the standard deviation. [2 marks]
[2 marks]
c i Calculate the Pearson product moment correlation coefficient.
ii Describe the linear correlation, using two words.
iii Write down the line of best fit g on w.
d Another club in the same league spent €80 million on players'
wages. Estimate, to the nearest integer, the number of goals that
they scored.
e Yet another club in the same league scored 17 goals. Give two
comments on the validity of using the answer from c iii to
estimate the club's wage bill.
646
8 [Maximum mark: 1 31 [2 marks]
[2 marks]
A function is defined by f (x) — [4 marks]
[3 marks]
2x-10 [2 marks]
a i Write down the equation of the vertical asymptote. [4 marks]
ii Write down the equation of the horizontal asymptote.
[4 marks]
b i Write down the intercept on the y-axis. [3 marks]
ii Write down the intercept on the x-axis.
c i Find f'(x).
ii State what this indicates about the graph of f (x).
d Sketch the graph of f (x), showing the information found in
a, b and c.
e Write down the equation of the tangent to the curve at the point
where x = 2.
9 [Maximum mark: I II
Anna invests $200 in a bank that gives a compound interest rate of
5% per year.
a i Calculate how much money will she have after 3 complete
years. Give the answer to two decimal places.
ii Calculate how many complete years she would have to wait
until her money has more than doubled.
Anna wishes to buy a mountain bike which costs $350 but is
6%depreciating in value by per year.
b i Calculate how many complete years it will take for the bike
to cost less than $300.
ii Calculate how many complete years it will take until Anna
can afford the bike.
2%c If after two years the interest rate changes to only and the
depreciation rate changes to 4%, calculate how many complete
years (from when she first invested the $200) it will now take
until Anna can afford the bike.
Answers
Chapter 1 Exercise 1B Exercise ID
1 u = 69
Skills check ia -64, 256 2 11 ——40
3 7 = 6.97
ia 22
13
5 b
2a —4 116
33
12 6
53 6 c -1, 2, 8, 128, 32 768
3 a 128 5 6 u 12 = 36a
b c -81 d m, 3m + 5, 9m + 20,
27m + 65, 81m+ 200 Exercise IE
48
b9
2a
Exercise IA =b u
4u l, Ill 1
a -20, -23, -26
c 10 1 Eli = 105
b 49, 64, 81 2 = —4.86
du 5, u I = 14
c 30, 36, 42 = -71.74
125 125 125 u 3
d , or
f 4 60
2 1112 = —664
62.5, 31.25, 15.625 Exercise IC
s n=7;
6'7'8
the 7th term.
56 7
6 u = 52 seats
243 '729' 2187 4 12
"1=11
2+3- 4+5=2
Since they are held in 2050
2 a u = 10 x 511-1 , geometric (since n is a natural number),
the next year they will be
n held is 2054.
b un = —6n + 47, arithmetic b
6
1 rt•2
c = (-1)R+l —, geometric c
3
d u = un 1+u neither + I) = 2 6 + 12 = 20 8
n in 7 weeks
2n-1 5 5 Exercise IF
, neither 2 6 a u =2187
2n
5 b Not geometric
f u = —4 x 311, geometric c u -1.068 d u
3 100, 200, 300, . 4 68.3
1953 125
u = 100n, arithmetic f Not geometric
11 1 8
b un=6 — rt=l
rt=l
36 2
geometric g u
c 70, 77, 84 7 12
geometric rt=2 11 2 u = 536 870 912 cents or
30
11 $5 368709.12
n2ntn I or
648
3 Use an r value that is a factor Exercise 11 4 The series is I +2 + 3 +
of 64. For example, r = 2:
2, 4, 8, 16, 32 ia 3 s = 78
b
Exercise IG 42 But since there are two 12-
r = 1.5
2306 5 hour cycles in a 24-hour day:
2 = -56.3 (3 sf) 3 S = 78 x 156 chimes
2 24
1 9
2 5 The series is 5 +9 + 13 +
3
48 = 4752 line segments
4
Exercise IJ
or = 2460 i a Not converging as r = 1.5.
4 r = 0.5 3
b
6 u = 303 students s -0.199
10 4
a u = 536 870 912 grains d
30 —3
b n = 10th square n = 48
s =7056 14
8
48 -5
8, 16, 32, 64, 128 u 4 -5 2 1
ill u —5 4 -5 2
Exercise IH
As 1000 is a multiple of 2
This is geometric because you
are multiplying each previous 4, the largest multiple of I—r 55
22
1 4 less than 999 would be
11
height by — 996.
22
2 n = 249
s = 124 500 d Not converging as r
1
f 1
metres
512 3
—x—81 81
=s00
22
2 This is arithmetic because
f Not converging as r
you are adding more money
1 1
to your account every
month. 2 2
x = $123.57 2
635
3 This is arithmetic because 256 2 Any infinite geometric series
you are adding from year to where Irl < l, r $ O
2
year. 3 Any infinite geometric series
= 2400 seats where r < —l, r > I
n = 5.583
3 The series is I +2 +4 + 4 S 30
Finland did not gain Total distance = 48 ft.
independence in the year S = 63 family members
of the tiger. S 17 892 OOO gallons
4 This is geometric because
a rate implies you are
multiplying.
122%
649
ANSWERS
Exercise 1K A = 15043 348.839948... b 3 days = 72 hours = 6
875 348.84 Yen
ia 12-hour periods
d 12500 x 0.886
b i u = 127
28 3<365 = 5805.0510...
S = 1666
28 1+—A=300000 0.04 C = 5805 cellsmcL-l
365
c n = 39 730 c The limitation of the
general formula is that
23 1+—A=300000 0.04 white blood cell count
365 does not continue to
3 u = 16 decrease infinitely. Once
A = 324984.69581...
2 324984.70 - 300 000 the antibiotics killed the
4 infection, the patient's
24984.80 Mexican Pesos white blood cell count
3
A = 438 532.634627... would return to normal.
S Choose any r value
I 188 532.63 Swiss Francs 2 a This is an arithmetic
Example: sequence since the rate
2 r=O.000649855... decreases by —0.2% each
1 Fernando will pay an month.
annual simple interest rate
=if r of 3.38%. b
2 ii 23846 Columbian Pesos. where U represents the
. The series is 4 + 2 I 3 A $100 705.89 unemployment rate and t
6 = 1280 is the month starting with
4 P - $26410.18 January 2014.
By GDC, n 4.69116...
5 n 21.4 years c U-7.9-0.2(12-1)
A minimum of 5 rounds are
6 P 24 500 Brazilian Reals U-7.9-0.2(11)
required. U-5.70/0
8 a Since the geometric series Oliver:
d It is not realistic. There will
has only positive terms, A 425.78 GBP always be people you are
2 not capable of working,
3 Harry: or are switching jobs, or
looking for jobs.
b A 436.25 GBP
3 a This means that it
9a Harry earned more than
takes 1.23 years for the
6 Oliver. substance to decrease to
half of the original mass.
b i m=—ll m =4 8 Savings account:
b A = Ao — where A is the
c i Since the sum of an A = 20000(1 +0.012n)
amount remaining after
infinite series can GIC: At years, is the original
mass, and h is the half-life.
onlybe found when n = 6.03 years
7.2-1
1 Exercise 1M
| 123
2 1 1000 = 0.88
d 52
Exercise IL 12 500 2
A = 2400 12500 x 0.88, where C 5.040 650
1
1 = $900 represents the white blood
cell count and t is the time 2
A = 35200 every 12 hours.
1 = 3200 GBP 1.57985...
AZ 1.58 g
650
Exercise IN 12
n
+ 20X3 + 150X2 + 500X+ 625 LHS RHS
2
2 —b5 + 10114 — 40113 + — 80b + 32 1 3x-6 3
3 64x6 - 192x5 + 240+ - + x 3x—6
60x2- 12x+ 1
4 256x4 + 256x3y + 96x2y2 + n
16xy3 + YA
Since I for any xe R,
S x3 — 9x2y + 27xy2 — 27y3
6 243x5 + 1620xAy + 4320x3ya + nnn
5760x2y3 + 3840xy4+ 1024y5
o1 2
Exercise 10 nn
i a 451 068750x7 n 3
b 75 582720x7 LHS RHS
c -4320yA
d -19683 Exercise IP
2187 LHS RHS Chapter review
x7 -2(a- 28)
6(a-5) i This sequence is not
2 arithmetic since
+ 30
3 a -13 608x3 —2a + 56 18—6
b 11 340x5 -2(a-28)
LHS RHS ii This sequence is not
4 336 798x9
2 geometric since 18 —6
LHS RHS
b i This sequence is
(X— 3)2+5 X2 — 14 arithmetic since
-12 --14 = 2.
ii u = 2n- 18
iii
s = 72
6 c i This sequence is
-20. x2— 6x+14 geometric since
500 1
8 16.003 QED
1000 2
9 a x5 — 25x4 + 250x3 — 1250x2 3
+ 3125x- 3125 LHS =uii u rn-l
b 525+ 1 RHS 1
10 a (3 — 2x)4 = 16x4— 96x3 + 11 2
216x2-216x+81 iii u —7.81
(—2X+ = 16x4 — 96x3 + m +1 nt2+ m
216x2-216x+81 = 3937.5
1 1
d i This is a geometric
m+I m(m + l) 12 6
m 1 sequence since
63
b No, when the exponent is m(m + l)
ii
odd, the expansions will 1
not be the same. iii u = 48
(3x — 4y)3 s = 1533
QED
651
ANSWERS
e i This sequence is 13 1, 13, 78, 286, 715, 1287, 20 a x2-7X+12
arithmetic since 1716, 1716, 1287, 715,
286, 78, 13, 1 X2-X-6 X2-16 x
115 -110= 110 -
5. Each row in Pascal's
ii u = 5n + 100
iii u 7 = 135 triangle is symmetric. (x —3)(x +2) (x —4) (x +4)
iv S9 = 1125 x(x + 2)
2 = -600 b 1, 14, 91, 364, 1001, 2002,
3003, 3432, 3003, 2002, x2-7X+12
3 The first five terms are 1001, 364, 91, 14, 1
x
-4, 11 -19, 41, -79. 14 729x6 - 1458x5y+ 121502 -
4 540x3y3 + 135x2y4 — 18xy5 + y6 -7X+12
s r = 1.5 15 The coefficient is —870 912. x
x2 -7X+12
16
xx
6 An infinite sum can only i? a 10206x5 b -20412x6
x -7X+12 x2 -7X+12
be found for a converging 18
xx
geometric sequence.
QED
1
b
19
=-x2- 31x+51
2 16)
? n=20 =-x2- 31x+ 51
8 2x2 - 7x+ 3 - 3x2 -24x+ 48
=-x2- 31x+ 51
9 = 768
—x2— 31x+ 51 = —x2— 31X+ 51
9
QED
10 a 55, 51.15, 47.5695,
44.239635...
Exam-style questions
b It is a geometric sequence
47.5695 = 0.93 5 2
because 4 5 54 x 5x
4 13 -
51.15 21 a o 1
24
u 26.6 litres left in the tank 3 4 5 5
11
5 x5 x x
d 36.5 litres drained from the + 12
tank 45 (2 marks)
3 44 4
e S = 785 minutes or
13 hours and 5 minutes 5x 2 3 5x4 x5
11 a 5x
b u 10 = 81 cm 4 8 32 256 1024 (l mark)
c Eventually the spring will hit b Substituting x = O. I 800 100 5
the ground or the surface it is (l mark) 800 800 800
sitting on, so the length will
0.9755-1 - 705 141 (l mark)
become constant. Also, the 4 800 160
spring could break from too 8
much weight. 15 (l mark)
d 1500g or 1.5 kg
8 800
12 r = 4.50/0
652
22 a Using un = n = 28.76 (l mark)
26 = 49 5x2
(l mark) So Brad must wait (2 marks)
143 14d 29 years (l mark)
183 (l mark) 25 Require (3 x coefficient = 495
of term in x5)
Solving simultaneously 2
(l mark)
a = 108 (l mark) x coefficient of term in x
4) n2-n-110=O (l mark)
(2 marks)
5 88
43 (-2x)5 +1 x 44 (—2x)4 6
(1 mark)
2 54 2
x
b 100th term is a +99d (3 marks) 2? Require 2 (2 marks)
(l mark) 82 (l mark)
5 = 3 x (-114688) +1 x 286720 (l mark)
2
= 355.5 = -57344 (I mark) = 28 x (-2)6
= 28 x 64
(l mark) = 1792
23 a Money in the account
would be
3000 x 1.01 $3481)
(2 marks)
Therefore, interest gained is 1 44 41 4 2
1 2x 1
2x 41
28 a 0 2x
2 2x
3000 x 1.01510 -3000 = $481 4 1
(l mark) (-03 +
(2 marks)
4
b Total amount is
3000 x 1.01 + —r ——— x4 + 1 — 2x2 1 + —3 (l mark)
0152 + 16x 2x2 2
+1200 x 1.01510 = 27 - 27x + 9x2 -x3 (2 marks)
(l mark) 31 4 11 3
2x (27 - 27x + 9x2 -x3 4 — 2x2— 2
16x 2x2
= 3000 x 1.01511 +(1200 x 1.015) 3 9
2
1.01510-1 Therefore required term is 27><— (l mark)
1.015-1 (2 marks) 2 (l mark)
= 36
= $16 570 (l mark)
24 a 5500 x 1.02754 (2 marks) 29 120= (2 marks) 20
= $6130.41 (1 mark)
1-0.2 30 ar = 180 and ars =
b Consider
120=5—a 9
4 (2 marks)
5500 x 1.027* OOO Solving simultaneously
(2 marks)
a = 96 (l mark) (I mark)
12 ooo The 6th term is therefore 5 20 14
1.027* =
ar 81
5500
Using GDC: (l mark) 96 x 0.25 = 0.03072
(2 marks) 1 (l mark)
Therefore r =
3
653
ANSWERS
—180 180 n—l Chapter 2
So a = = 54() (l mark)
33 Skills check
a 540 540 (11—1)! 1
3 -810 ¯ (k—l)!(n—k)!
2 (3 marks)
(2 marks)
31 First part is geometric sum, x
a = l, r = 1.6, n = 16
(l mark)
(l mark)
Second part is arithmetic sum,
a=O, d=-12, n=16
(l mark) k!(n—k)!
(l mark)
Third part is I = 16 2 A(3, 0), B(-2, 4), E(-l, 1),
(l mark)
(l mark) R(2, -1)
3 a 17
Geometric sum: 3 3
d 2
1.616-1
4
= 3072.791
4 b —1 c 13 d -1
1.6-1
(l mark) 2
Sa
Arithmetic sum:
= x 0+ 15 x (-12)) n ooomono
2
= -1440 (l mark)
n=15 -1211+1) 34 1400 = 504+701-1)
.
= 129 (l mark) x
= 3072.791 - 1440 + 16 So the sum of the multiples omowooo
= 1648.8 of 7 is bY
(l mark) 1234 x
32 Required distance 129 c
= 20+ 2
= 122 808
6 (2 marks)
Sum of the integers from
500 to 1400 (indusive) is
100 (3 marks) 901
(l mark) 2
100
(l mark) = 855 950
1
(2 marks)
= 20+200
Therefore require
= 220 m
855 950-122808 733 142
(l mark)
•anon x
654
b
dY g -24
•azuxun x 3x2 — 6x
— 2) 15
fix + l) x g(x —2) x
= - -74x - 60
Exercise 2A 3 a Yes, it is a function. Every c
value of t will yield only
i a Functon one value of d. •mom
b Function
c Not a function as there are b d(t) = —75t+ 275 x
tickets with different costs
(adults vs child), so the c d(0) = 275 km 2a
same number of tickets
could a different total cost. d 0 < t < n, where n is the x
d Function amount of time it takes to
e Not a function b
f Function drive to Perth.
g Function
h Function 4 a Yes, it is a function. Every
i Not a function temperature in Celsius will
j Function only yield one temperature
k Function
I Function in Fahrenheit.
m Not a function b F(17) is asking what
temperature in OF is
2 Answers will vary
equivalent to 17 oC.
F(17) = 62.60F.
c F(C) = 100 is asking what
temperature in oc is
equivalent to 1000F.
37.7 37.80c
d F (o) = 320F
F(100) = 2120F
3 "All functions are relations, x
but not all relations are
functions. " g 176.6* 1770c
Exercise 2B S C(g) = log + 25
ia c
c 100) = 2250 C(14) = 165
d g = 7.5 gigs
d
3 Exercise 2C
1 x
2a
d6 ona6Bs x
e f(x- 2) =-3x2+ 12x- 12
655
ANSWERS j Not a function d
k Domain: x e R,] —00,00 [ or
d
(—00,00)
3a
Range: x
-o .5
1
x 2'
Domain: e R, I —00, 00 [ Or
(—00, 00) or [0,00) Domain: x e R,
Range: y 0, 00
2a Range: —l Sy 5, [—1, 5]
3 Answers will vary.
x a
5
-5
b x
10
Domain: x e R, 2.
7.5
Range: y e R,] b
5
(—00,co)
2.5 b
x0 -o x x
10
-2.5 10 zonoo
Exercise 2D Domain: x e l, | —00, 1 5
Not a function [ul or 1
Range: y e R, x
b Domain: {—5, —l, 3} 0, I —00,
Range: {4, 6, 14)
c Domain: {—12, —8, —5)
Range: {—8, 7)
d Not a function
e Domain: x e R ]
Range: y e R, I -00,
f Domain: x e R,]
O r 00)
Range: {4) 2. d
g Not a function
h Domain: x e R,] 25 x
O r ( —00, 00 2. 75
Range: 3 sy 5, [3, 51 x
Domain: x 2 0, Domain: x > l, II,
Range: y > —2, ] —2,
[O, 001 or [O,
Range: y O, I —00, O] or or
(—00, 01
656
4 Exercise 2F c f(g(x)) = I .06x + 25; this
b 16X2 + 36X— 6 represented only paying tax
6 fiii (h + 25X on the price of the fridge.
5 = —x + 3J¯x + 4
4 g(f(x)) = 1.06(x + 25); this
3 represents paying tax on
2
both the price of the fridge
1 and the delivery fee.
12345678910 iv
Exercise 2G
—x, fofof(-l) = 4686
5 2x+3, -1 < x '2
It is also possible to include —l vi g(h(9)) = 14
in the second interval rather
than the first and 2 in the third vii gof(2) = 22 =iii g(3) II
interval rather than the second.
fog(2) = —6
Exercise 2E
gof(2) + fog(2) = 22-6
= 16
a C(n) = 40+ 21n, where Rb x e orl —00, or b For f(x) :
C is the cost and n is the Domain: {2, 3, 5, 10}
number of hours. (—00, 00) or Range: {—4, l, 2, 6)
For g(x):
Rxeii orl —00, Domain: {—2, 3, 6, II)
Range: {-3, O, 2, 11)
b Domain: x 20, [O, or [O, 00) (—00, 00) or [O, 00)
Range: y 2 0, [0,
or [0, 00) iii x > O or [0,
iv x 2 0 or or [O, 00)
2a 2 Answers will vary. 2a = -2
4 fog(3)
3
2
5
b gof( —l) = —4
8
4 f(g(x)) = 3x2 - 24x + 42
b fof(9) =9
x 3 f(0) = o
.5
80 160 240 320 400 g(f(O)) =g(())
Femur Length (cm) x
g(f(l)) =—3
b Domain eg: 10 S fS 80 2 75
Range eg: 75.8 S h 228
oo—oxva
c cm
d f— 43.3 cm
0.025 f(g(—2)) =0
3 10000 1+ c Domain: x e R or | —00, d g(—0) = —4
12
b The equation satisfies the
vertical line test.
c Domain: t e 0. Range: y —6 or [ or 4 Answers will vary; g(x) contains
a point with an x-coordinate
Range: [10000m). [ —6, 00) of—l;f(x) must have a point
d Javier needs 27 years and with a y-coordinate of 2.
b g(x) = 1.06x
10 months to double his
money.
ANSWERS
Exercise 2H d g(h(x)) = x &c i ii
i a f(n) = x— 100 h(g(x)) = x
g(n) = 2.20n Since =g(f(x)) = x,
b f(n) represents that you these are inverses.
receive commission on
2 x-intercept 1 ozaouo
every new person who x
2
signs up after the first 100 anmuao
y-intercept: (O, 2)
people. iii g: x —x +6
Since you only need two
g(n) represents that you points to graph a line, you can
receive 2.20 GBP for each switch the coordinates to find
person (after the first 100) two point that the inverse
who sign up. 1
c i f(224) = 124
passes through: 0,— and
ii g(n) = 272.80 GBP
d i S(276) = 387.20 GBP 2
ii x = 152 people &3 a i ii
1
2
2 a f(x) = x— $25; this could .5 1
represent $25 off the price —y=x-6
of the TV.
2
g(x) = 1.10x; this could
represent a tax of 10%. x
y=2x-12
g-1 (x) = 2x— 12
X4 a i Domain: x > or [X,
b i You paid 699.99-25 or [X, 00)
= $674.99
—x RRange: y e or | —00,
4
ii After tax, the TV —iii f" I(x) or (—00, 00)
COSt I .10 x 674.99 Rxeii Domain: or | —00, c.o[
= 742.489 $742.49
c i P(x) $25) + RRange: y e or | —00,
49.99 Or ( —00, 00
ii P(525.99) $601.08 Riii Domain: x e or —00, c.o[
3 Answer will vary.
Exercise 21 x Range: y > 3 or 13, or
.5
(3,
f(g(x)) = —x— 2
iv Domain: x S I or [ —00, 11
Since these are 2
not inverses. or (—00, 11
=iii g-l (X) , x 20
Range: y 2 or [ —00, 2] or
2
(—00, 21
g(f(x)) = x = x, Note l: The domain restriction
is needed since the original b The domain of the function
Since f(g(x)) function g (x) = —2V¯x + 5 becomes the range of its
these are inverses. would have the same inverse, and the range of
the function becomes the
c f(g(x)) = x restriction. domain of its inverse.
Since f(g(x)) = = x, Note 2: The inverse, S Answers will vary. In order
can be simplified further if for the function to be a one-
these are inverses. to-one functi()n, the inverse
desired: must be a function.
1 5 25
4 24
658
6a Exercise 2J 4
b 2y=x+5 2x—4
2 2
2 4x — 8
—4
11+5
x
2
2x —4
2
-3 —4m
2 2x —4 + xm + nt2
d f(x) = II gives the same 3-3+x=x —4m — 8
answer as (l l). 2x — 4 + xm +
f(x) = n =b f(_f(x)) x
—4m — 8 X(2X — 4 +
f(x) = —2x —1
2x-1 —4m — 8 = 2.X2 — + .X2m + 1112
x=-2y-1 —4m — 8 = (2 + m)x2 — 4x + m2
2y=—x —l
x=x Chapter review c Yes
2 =xc f(f(x)) f Yes
i Yes b No i Yes
11
d No e Yes No
22
2 2 g Yes h No
11 j Yes
g of-I (X) 2 k Yes
22 m No
Rd fix) = n —x, n e is a self- 2 a Domain.
gor (x) = -3 inverse. Range.
4 b Domain:
3
4 Range:
-2 c Domain.
-3x2 -6x -3
4 5x +1 Range.
5 d Domain• .• ( —00, 00)
5X+1 or 001
gof-l = (x +2) 2(5x +1) RRange: y e or (—00, 00) or
4
g of-I (-5x -10) (5X+1) e Domain: —3 < x S 3 or
= - 30) -6(- (5x +1)
gor 4 (-3, 31 or | -3, 3]
X+2-10x-2
-3+6-3 Range: —3 < x —l or
5X+1
= (-3, -11 or | -3, -11
4 5X+1
Rf Domain: xe or (—00, 00)
gof-l(-l) = o Range: x 2—12.25 or
[12.25m) or [12.25,
659
ANSWERS Sa
g Domain: x 2 0 or [0, 00) or
Range: y s I or (—00, II or x
| -00, 11
h Domain: xe or (—00,00) or
Range: x 2 5 or [5, 00) or RDomain: x e or (—00, 00) 8 (fog) (x) = —x5 = (g(x) +
[5 Range: y 2 —2 or [ —2, 00) or 3 —8x6 =
3a b —2x2 =g(x) + 2
g(x) = —2x2 — 2
b f(-2) = -2 x
c g(-6) = 12
e 210) - = -16
f h(0) xf(-l) = 20 9 f(h-l = 264
10
2 3
h f(g(x)) = 4x2 —6 3
x
fog-I 24 -1
4
4a RDomain: x e or ( —00, 00) or
-00,
T 1400 1 2 3 Range: y 2—8.38 or [ —8.38, 00) Exam-style questions
1134 1020.6
1260 or [-8.38, [ —24 f (x) 26 (2 marks)
b This is a geometric sequence. =6 a Since _f(g(x)) =g(_f(x)) x, b
(2 marks)
c u = 1400 x 0.911—1 these are inverses. (2 marks)
d T(t) = 1400 x 0.9M -x -30 d 125< f (x) 250 (2 marks)
T(5) means what is the
8
temperature in the kiln
after 5 hours = 827 oc Since f(g(x)) $ x, these are
f T(t) = 700 oc means at not inverses.
what time did the internal
= 2x2 —x 4
temperature of the kiln
reach 700 oc. 32 12 a
t = 7.58 hours or 7 hours Since f(g(x)) these are = —8—2 = —10
and 35 mins not inverses.
(2 marks)
g Yes, a sum to infinity can d =g(_f(x)) x g (-2) = (-2)2-8 (-2) +15
Since f(g(x)) =
be found because this is = x,
a converging geometric these are inverses.
sequence. (2 marks)
h s = 14000
y=4x 2
i The kiln will reach the
—1 -l 4 (l mark)
minimum temperature 4 (l mark)
ii
equal to the ambient tempe-
2
rature of the room it is in.
b —4 x — or
660
d x2 —8x+15=27 (l mark) ff(10) = (l mark) f(x)
x2-8x-12=o = f(20) (l mark)
6
(l mark) 50 (l mark)
x = 10 or x = —2 c y=3x-10
x=3y-10
x
(2 marks)
13 a f (x) =128 —3 -15=177 (l mark)
2 3
(2 marks)
(2 marks)
(l mark) 18 a Range is f (x) 2
b f -399 3 ——,
(2 marks) Range is 5 < f-l (X) < 20 3
(2 marks) (I mark)
TO 1905 16 a NOT a function, since b x (3y+6) = I —2y (l mark)
3xy +6x = I — 2y (l mark)
(l mark) e.g. the value of x = 5 is
related to more than one
Range is —399 < f (x) < 1905 coordinate on the y-axis. (l mark)
(l mark)
(2 marks)
c Solving 128a —15 = 1162.6 b This is a function. Each 2
(l mark) value of x is related to only c Domain is, x * —
one value for y.
a=9.2 (l mark) 3'
(2 marks)
Domain is —3 x 3 c This is a function. Each (l mark)
Range is f (x) —2,
(2 marks) value of x is related to only
one value for y. (l mark)
Range is —l f (x) Sl
(2 marks) 19 a x2 = 2x—l (l mark)
(2 marks) d This is a function. Each
(x-1)2 -o (l mark)
b Domain is —l .5 < x 5 value of x is related to only
one value for y. (l mark)
(2 marks)
(2 marks) (l mark)
Range is —5 f (x) 4
i? a gf(x) = 2x2 —l (l mark)
(2 marks) = 2x2—l (l mark)
c Domain is O x 24 (x-1)2 -o (l mark)
(2 marks) (l mark) (l mark)
Range is O f (x) S 12 xy—x=k+y—l (l mark) 20 a C = 430 +14.5p (2 marks)
(2 marks) (l mark)
d Domain is —3 S x 3 So f is self-inverse
b Range is f (x) > l,
(2 marks)
(2 marks)
Range is O f (x ) < 9 b f (p) is a function since
there is only one value of C
(2 marks) which corresponds to each
value of P in the domain.
15 a Solving 3x —10 = 5 and
(l mark)
3x-10=50 (3 marks)
C = 430 +14.5p (I mark)
Domain is 5 < x < 20
(l mark) c-430
14.5 (l mark)
661
ANSWERS
1000-430 26 a x2 — 13
d = 39.3
(l mark) (2 marks)
14.5 (l mark)
1 Therefore k = 3
(l mark)
d (y-3)2 x- (2 marks)
She can therefore invite a
1 (l mark)
maximum of 39 people.
1
(l mark)
2Vi
c = $662 (l mark) (l mark)
—3 (l mark)
(l mark) 1 (l mark) c The domain of f-l (x) is
—66=241.375 (l mark) f gh(x) = x 24, (x e R) (l mark)
16 11
Katie will therefore need 17 The range of f-j (x) is
2X < 16
to charge a minimum of
24
$41.38 per guest. (l mark) (2 marks) (l mark)
(l mark) (x— 3)2 = x2 —6x+ 9 (l mark)
(l mark)
21 a h(x) 2 2, (h (x) € R) (l mark) = 2x2 — 18
(1 mark) (2 marks)
(2 marks)
b 2x2 18
(X) = 3X —6 (2 marks) Therefore g (x) = 2x2 + 12x
(l mark) (l mark)
x
hh (x) = (2 marks) (2 marks)
3 c Using GDC, or otherwise,
solving x = x (l mark)
(l mark) Chapter 3
x (l mark) Skills check
d (l mark) —3 9
3m(m —5) c
3 d
2
3
b (x + 6)(x—6)
(l mark)
d
e Because h(x) and It-I (x) 9X(X 2)
both intersect on the line
(l mark)
15 -15 h (4a - + 7b)
(2 marks) Exercise 3A
Therefore h(x) = x +2 (l mark) (2 marks) i a 1 2
b
2
5
(l mark) 25 a 3y+5 (l mark) 2 a 3 b -1 1
(l mark)
f (x) = x —15 4 2
4xy — 3x = By +5 (l mark) 3 Rate of change = —0.015
23 a The height of the candle
(1 mark) 4x-3 (l mark) decreases 0.015 cm each
So r (x) is self-inverse. additional second it burns.
(l mark) 4 a (342.93, 100) b 0.20
b rrrrrr (5) = rrrr (5) = rr (5) = 5
(2 marks)
662
Exercise 3B d Exercise 3G
a =—5 nt2 3 i2 b3
neither d4
e —6
5
b = —4 parallel 1 7
3
h
9
2 -1 2 a Domain: all real numbers
Range: all real numbers
3 8, 12
b For the first 40 hours Liam 2 b Domain: all real numbers
works in a week he is Range: all real numbers
paid $8 per hour. For each 3 a -3
hour over 40 hours that
Liam works, he earns $12 + 3) or 3a
per hour. c Change the equations to
Exercise 3C slope-gradient form: 12345
ia y=-3x-13 b No vertical line is a function
2 Exercise 3E as the y corresponding to
b (0,4)
i a x-6y-18=O the x-coordinate of the
3' b 2x+3y-12=O x-intercept is not unique (in
mc = O; (0,-2) fact, any y corresponds to it).
4a
2 2a b
b
3a b y=3x+7 57 3
b y=-10 22
4 a x=8 Exercise 3H
c x=9 3a
f-l(x) = —x + 5
Exercise 3D
6.x 18
c
b
b
Exercise 3F 3 a $615
12345 b (0.75, 2.5) b f-l(x) =—x1 —6.5, where x
10
(-3.58, -8.19) is the total cost of a t-shirt
c order in dollars and f-l (x)
d (1.18, 1.12)
is the number of t-shirts in
12345 b -5.05 the order
3 $1666.67 c 500 t-shirts
663
ANSWERS
Exercise 31 b Exercise 3K
i a The graph is reflected
—Fi a d 1 or F = 32d, where Axis: x = O vertex: (0, 4)
about the y-axis.
32 c b The graph is reflected
d = distance in cm and 123456 about the x-axis.
c The graph is compressed
F = force in newtons Axis: x = O vertex: (O, O)
horizontally with scale
b 11.5625 cm (exact) or d
1
cm11.6 to 3 s.f. Axis: x = 4 vertex: (4, —3)
factor
2 a y=o.16x+ 360 Vertical compression with 2
scale factor 1. g(x) = —x2
b The y-intercept represents d The graph is stretched
Frank's weekly salary of 4 vertically with scale
€360. The gradient shows b Reflection in x-axis,
that Frank's commission is factor 3.
vertical stretch with scale
16% of his sales. e The graph is translated to
factor 2; g(x) = —2.x2 the left by 6 units.
c €504
c Horizontal translation right f The graph is translated
3 a Plan A: 79.99 downwards by 3 units.
3, vertical translation up 2;
and Plan B: c(m) = 20m, g(x) = (x — + 2 2 a r(x) = 2f(x); s(x) = —f(x— 3)
mwhere c is the cost for d Horizontal translation left b r(x) =f(—x);
3, vertical stretch with
months at the gym scale factor vertical 3 a O sy<6
b Month 8 translation down 5; b
4a 3 (x +3)2-5
O s h s 40 2
p(h) =
12/1-160, 40 < h
b i €176 ii €404
5 a 1700
b The model predicts that -8-7 -6-5 1234 56 8
raising the price €20 will
result in 130 fewer printers
sold.
c 75 Euro
d d h(x) —4
e h(x) —4
Exercise 3L
q -6 + OOO
(461538,0) x-intercepts: (—2.81, 0),
(0.47 5, 0); y-in tercept:
( .333 0) (0, —4); vertex:
(-1.17,-8.08)
€75 2 x-intercepts: none;
Exercise 3J y-intercept: (O,
vertex: (0.726,-0.785)
Y
3 Domain:
Range: f(x) S 9.125
12 3
Axis: x = —3 vertex: (—3, O)
664
4 Domain: c a = 0.5, p = —2, q =—4;
Range: f(x) —30.752
a a=4, 16, —-20;
(9.92,0)
(1, 0), 0); (0, -20)
4. .00752
b a b-- , --14;
S Range: 0 S f(x) 8.1
0), (-1, O); (0, -14)
(081)
b c=ll; (4, 3);
(15.0) (0, 11)
6 Range: f(x) 2 —6 fix) = (x— 2)
C
(2.5)
Exercise 3M (-4.-8) (0,8) 5 i (3,-2)
3; (3,4) f(x) — 6x+ 7
1; (1,-5)
x=-6; (-6,-5)
(0, 5); x=4; (4,-11)
(-2,-3) h(x) = (f o g) (x)
(o, 3); x = -2; (-2.-2
(2, 0), (4, 0); 3; (3,-1) = (x2 —4x +4) —u +4—3
= x2 —6x+5
(-1,-16) (-2.4)
(-1, 16)
Exercise 3N d h(x) = (x— 5)
a
(-9, 0), (2, 0); (0, -18)
(0, 10)
665
ANSWERS
Exercise 30 3 -6±26; -2.54, -9.46 d 112; two distinct real roots
i a f(x) = x2—4x— 12
4 5±36; -0.196, 10.2 e 0; two equal roots
b f(x) = —
c f(x) = 4x2- 24x + 20 5 (0.5,4.5), (-1.5, 6.5) —199; no real roots
6 (1.17, 7.35), (-1.97, 1.07)
3x2 — 12x 6 7 2a 9 b k < 20
e f(x) = —0.5x2 — 1.5x + 5 8 (2.72, 7.64), (0.613, -0.0872) 4
f f(x) = —0.6x2 - 12x 9 x--0.802, 1.80
10 x--2.91, 0.915 25
2a
3 b p=12
b (1,4)
4
p = ±2Jä d 8
9
Exercise 3S 4a 33
5 4
2
d f(x) = —x2 + I()x— 26 2 3 ±JiF Exercise 3W
Exercise 3T
3 a (4, 80); The model rocket 2 i ax or x 2 —1
is predicted to reach a 4 6±445 3
maximum of 80 m, 4 s -1 ± 36
after it is launched. 6
2
b h(t) = —5(x— + 80 or
h(t) = —5t(t— 8) or 2 a x <-0.245 or 12.2
h(t) = — 5t2 + 40t where
-4±457 2 2
OStS8 b
3 4 1.26
mc 67.2 3
3
2 c -0.890
3 a k > 4 or k < —4
3 6
5
2
Exercise 3P 35x- 0.25x2 = 300 + 15x 1
1 b = 100 4
9
d -11, 11 b 0.25(x— k>—1
5 k < 0 or
x = 20, 60 4
f4 c 20 or 60
4 6 a p2-48 b
2 a 1,- 3 3 d 40
2
e 100 OOO Euros
3 77 Exercise 3X
c -2, d
mi 4 m, 12
4 3' 3 Exercise 3U
2 17.9m
51 1 5
3 b 0.211 s, 3.87 s
4' 3
mc 22.4
Exercise 30 c
4 3 Fare = 5.50 - 0.05x
ia b Number of riders
2 a -3 ± 36 = 800 + lox
3 3 c Revenue = (5.50 — 0.05x)
2 (800 + lox)
1 No real solution
d 10 or 20 decreases
3' 3 g 3±26
4 a Y = —(x— + 4 or
4 f 3, Y = —x(x — 4) or
2
f 3, 2 b If the centre of the object
3 23 is aligned with the centre
3a of the archway, it spans
2 (fog) (x) = 2±Vä
form x = 0.5 to x = 3.5.
b
2 Evaluating the function
= (2X+ 1)2-2 4 -2
-2 c r=l;s=2
= 4x2 + 4x— I Exercise 3V
i a —l l; no real roots
4
b 121,• two distinct real roots
b c —44; no real roots
3
Exercise 3R
i 4±N/Tö; 0.838, 7.16
-13.9
666
at x = 0.5 and x = 3.5 gives 8
1.75. Since 1.6 < 1.75, the
object will fit through the 5 9 f(x) = 2.x2 + 4x— 16
4
archway. 3 a -0.679, 3.68
b -4.92, 1.42
5 a A(x) —x) or (o,z)
a 18m b 26.6m
A(x) = 155x —x2 1 c 3.66s
g,o) d ostS3.66 e 1.72s
mb 77.5 by 77.5m
1Z3 5
c No; the touchline would 2a 1 12 A(4, 0); 13(0, 7); C(-4, 0)
not be longer than the 4 b y=-1.75x+7
c 2p cm by —l .75p+ 7 cm
mgoal line and 77.5 is
b y-—x-5 d Area = 2p(—l .75p + 7)
less than the minimum of
e 4 cm by 3.5 cm
m90 for the touchline.
f 14 cm2
d 90 < x S 120 (If the goal —2) or y = —x +1
line restrictions are also d Exam-style questions
taken into consideration 3
13 a 12y=-7x+168 (I mark)
the answer is 90 S x S b
110.) =y, +14 (l mark)
e 5850 m2 12
Chapter review b A (24,0) and B(o, 14)
(2 marks)
c Area
1 2
168 units
5 2
4
(2 marks)
4 a Vertical stretch with 14 a
-z scale factor 2, horizontal
b translation right 3
b Vertical compression with
scale factor vertical
2
translation up 5
c Reflection in the x-axis,
horizontal translation left 2,
vertical translation down I
3 d Vertical compression with (2 marks)
b (o, 5.9) and (-0.885, o)
(-5,0 scale factor —1
(2 marks)
3
o
e Reflection in the y-axis,
(2 marks)
vertical translation up 6
(l mark)
(3, O) and (—7, O); x = 2 (l mark)
b
c x=-2; (-2, 10) 15 a 2
_5 6a ii h =—3 2
-6
iii k=-7
b (-3,-7) c (2,-10)
7 a 11, -5 1
3'
d
(2 marks)
ANSWERS
b Using CJDC (l mark) Chapter 4
Therefore, the equation is
(2 marks) y=x2-5x-50 Skills check
16 a -18 b One coordinate is 5
c
2
(o, -50) (1 mark)
= 3(x2 -2x+1)-18 The other coordinate is 2
6
(l mark)
= 3x2 —6x —15 (l mark) (l mark)
b (1, -18) (l mark) 20 a f (x) = 2 [x 2 —2x —4]
2
(l mark) (l mark)
d f (x) GR, f (x) 2 —18 41 -6 -4 x
(l mark)
(2 marks) 246
-2
-10 (l mark) -4
-18-1
Ab horizontal translation -6
(2 marks) right I unit. (l mark)
(l mark)
-18x+8 A vertical stretch with scale Exercise 4A
factor 2. (l mark) 1 1 1
A vertical translation down ia b d -1
2
3 5
94
(2 marks)
h
5 10 units. (l mark) 57 8 11
(l mark) 3 22
1
4 21 a 'IWo equal roots
2x
1 b2 — 4ac = O (l mark) 2 a2 1
3
(l mark) 36 —4 (2k) (k) = O (l mark) b
2 1
x
b 8x2 + 6x — 5 —k = O d
4
No real solutions
3x
112 —4ac < O
(2 marks)
2
(l mark) b Equation of line of h
3
(l mark) symmetry is 3
b6 1
49
(l mark) 2a 4k 2k 3
(2 marks)
8 3 4
Therefore
7 11
2k
b
11 7
18 a -25+27
(2 marks) 3
(l mark) dx
2
Exercise 4B
(l mark) 4x2+6x+2=o
b Coordinates of the vertex
are (5, 2). (l mark) 2x2 + =
c Equation of symmetry is +1) = O (l mark)
(l mark)
x = ——1 or x = —l (2 marks)
19 a At (10,0), O = 102 +10b+c,
2
22
so c = —100 (2 marks) a A'(-6, 10) , B' (O, -16) , C' (1, 9)
b and (4 marks)
Line of symmetry is x = — ,
2
so b _ 5 (l mark) and
Solving simultaneously gives (4 marks)
=-100
668
2 a and c
20
7.5
15
10
7.5
2.5
100 200 300 400 500 600 700 800 900 1000
b 100
c (200, 5); string 5 cm long has
vibrations of frequency 200 Hz.
3 a4 64
c and d
100
-is -12B -to -7B
25 75
d The curves are in the 5 10 15 20 25 30 35 40 45 50 55
opposite quadrants. The
1.25 mins
negative reflects the
function in the x-axis. Exercise 4D
x=—l,y=()
Domain x e R, x * O
Range y e R, y * 0 Domain x e R, x
Range y e R, y # O
Exercise 4C
b 30s Domain xe R, x 5
Range ye R, y * O
Domain x e R, x 4
Range y e R, y # O
Domain x e R, x —5
Range y e R, y # O
x=—l,y=2
Domain x e R, x
Range y e R, y * 2
f
Domain x e R, x —l
Range y e R, y —2
Domain x e R, x 3
Range y e R, y 2
669
ANSWERS
f xe R, x ye R, y S a Asymptotes s
Domain x e R, x * 4 160
140
Range y e R, y —4 120
100
a xe R, x ye R, y
20
2 4 6 8 10 12 14 16 18 20
b 15 sessions
xe R, x ye R, y
xe R, x ye
c xe R, x ye R, y xe R, x ye R, y
xe R, x
xe R, x The linear functi()n is a line
of symmetry for the rational
670 function. The linear function
crosses the x-axis at the
same place as the vertical
asymptote.
Exercise 4E
ye R, y * -1 3 a 2: Translation of 2 units right Domain x e R, x * I
ye R, O Range y e R, y I
b 5: Reflection iny = O and a
Domain x e R, x * —l
translation of 2 units right Range y e R, y 2
c 1: Translation of 2 units
c
right and 2 units up
Domain x e R, x * —2
d 4: Translation of 2 units Range y e R, y 3
right and 2 units down x = 1.25, y=O.75
e 3: Translation of 2 units Domain x e R, 1.25
right and vertical stretch by Range ye R, y 0.75
a factor of 3
32 Domain x e R, x 2
Range y e R, y —3
29
28 C2 B; ii A; iii D; iv
-30 -20 -10 10 20 30 40 50 60 3
Domain x e R, x q
Range y e R, y I
5.56 c -272.220
20+10m c
35 Domain x e R, x # 5
25
15 Range y e R, y —
10
Domain x e R, x # 2
2468 10 12 14 16 Range y e R, y 3
c 4 months d IOAUD Domain x e R, x * 9
Range y e R, y 2
10 a
f
c (0.667. 0) and (0,1)
Domain x e R, x —2
12 Let f(x) Range
Domain xe R, x *—4
Range y e R, y
h
Domain x e R, x —3
Range y e R, y —3
b O, 14 x=l,y=2 c (-0.5,0)
13
13
2X+3 5 — 6x
9x-1 X+ll 12345
160 x = -2.5
140
120 Cha pter review
100
Domain x e R, x O
5 10 15 20 25 30 35 40 45 50 Range y e R, y * 0
Domain x e R, x —8
Range y e R, y * 0
b 20 c 50
671
ANSWERS 10 (-0.833,0)
b (0.5, O) and (O, l) c x = l,y=2 d (-0.5,0) 14 Consider the function
10
e -0.303, 3.30
11 x = 4 and y = 2
b Domain xe R, x
Range y e R, y * 2
c x = —l and y = 0.5
c 2.41
12 a
e Horizontal shift of 4 units
right and a vertical shift of
2 units up
c (2.67,0) and (0, 2.67) Exam-style questions
x--1.5, 1 16 a xe R' x —2 (l mark)
5
6 (4,0) b c 1.27
c
8a
(l mark)
c When x = O,
33 (goni)x = So one coordinate is (O, —5)
9 a The x-intercept is (l mark)
c (0, -0.833)
—When y = 0, x = 20
x = 2.5, y = 4 3.8
So the other coordinate is
c 2.375 20
(l mark)
i? a Domain is xe R, x —2
Range is f(x)e R, f (x) * 0
(2 marks)
672
b Domain is x e R, x —2 8=2a+b (1) (l mark) d
Range is f (x)e R, f (x) Substituting the second x
coordinate: 6
(2 marks)
aoBÄ0000
c Domain is xe R, x # O
(3 marks)
Range is R, f (x) * 4 38 38 23
(2 marks) -3=a+4b (2) 1
d Domain is x e R, x O (l mark)
Range R, f (x) $ O Solving (1) and (2)
(2 marks) simultaneously:
18 a (l mark) (l mark)
b (l mark)
(l mark)
Y
21 a 6 (l mark)
12
-57
(2 marks) x
x 18 (1 +0.82t) -1
(3 marks) c Solving 100 = (2 marks)
(l mark)
19 a y=10 (l mark) 3 0.034t Asymptotes are x = 3
3 (l mark)
b 2
282
= 24.8 months and y = 2 (2 marks)
11.36
(l mark)
d Horizontal asymptote at
= 434.12
0.034
(2 marks) Intersections with axes are
Therefore for t 2 0, 11
(2 marks) p 434 (l mark) at o,- and -—,O
-lox +23 34
(2 marks)
(l mark)
22 17-10x
20 a Vertical asymptote 2x-1
Chapter 5
occurs when c + 8x = O 12+5-10x
2x-1
(l mark)
(2 marks)
12+5(1-2x) Skills check
(l mark) 2x-1
3 12
(l mark) i a b
b 4 19
12
(l mark)
Substituting the first 2
coordinate: 1
2
5 10 b (l mark) 1 -2
(l mark)
(l mark) 2 b 3
c x 8
8 5
c
673
ANSWERS 3a -3 —3 2b
3x
3 at x =
b
2 so
c 12Ttx2 d 2x+2 125a2 15a
b2 1 5ac
e 2X + 3X¯2 f 6X2 — 2X + 6
(A solution correctly using the
x 1 -3 discriminant of a quadratic is
equally permissible)
4 a —x2 b 2
8
2
-2 -3
—x 3 +—x4
34
Exercise SE
1 b 13 15 Exercise 5G
a 10(2x+
4 Since 4 6x -1
2 b
25 1 -10 1 199 217
2 and 5, —
_
2 2' 24
6
Exercise SA 3 —1 41 Exercise 5F 29 2
4
i 10 23 1 d 12 x 2 — 2
x
Exercise 5B 1
2 y = 6 —4x
x2
1 419
—i Vertical asymptote at x = 2
6 27
1 3 y = —x +— and y 7 34 36
3
Horizontal asymptote aty = 5
4 -2 3
2 S Coordinates are
Exercise 5H
2 Vertical asymptotes at x = 1+6 7-146 i a 2x(3x- 1)
Horizontal asymptote aty = —l 3 27
3 Vertical asymptote at x = I I-Vi 7+146
Horizontal asymptote aty = —l
3 27
4 — 9x
4 Vertical asymptotes at x = ±N15
Horizontal asymptote aty = O 1 d 2x(5x — I ) (x2 —
Exercise SC -10 -9x
6 g(x) =
1 f'(x) = 7x6 2 f = 18x17 xg'(x) + ng(x) 2b
= x(—nx 't l)+nx-n
1 4 = —nx-n + nx-n = O -1
2 4 f(x) = —1 x-; 5
5
3 34 a f '(x) = 1 5a.x2 —4bx +4c Exercise 51
2 4 b
1 5ax2 — 4bx + 4c 0 16
Exercise SD ia b
i a 4x3 — x b 1 5x2 - 5 2 -3x2 -2x +3
c 24x3 — 6x
1 5a 1 5a d
f 24 2 4192 4c 3
2
2b
2
-1 -2 125a2 15a
1
2 b 3x5 The LHS is valid for all real
3
x and attains its minimum
—X—3 -1
c 2xQ— 2
2
Exercise 5J Exercise 5M iv Concave downward
x < I and concave
1 —15x2 2 upward for x > I
3x 2 b i (l, 0) min
ii No inflexion point
b 3 iii Increasing for x > I
decreasing for x < I
1-2x 4 2a2 = 8 a = ±2
iv Concave upward for
-2
d -4(2x3 -1)
431 Exercise 5N
2 Y = 24x- b 10,001
2 I-co 01
2 a No points of inflexion ci min
Exercise 5K ii Nowhere b fis concave up throughout ii (0, —2) = horizontal
its domain.
ia inflexion point
b c fis never concave down. -2
—,—2.59 = non-
3
x eii (0, (1,9 3 a (2, -38) horizontal inflexion point
-0.215)u (1.55, 4a 1 25 b 1 iii Increasing x > —l
3' 27 Decreasing x < —l
ii xe (-0.215,1.55) 3'
iv Concave upward
d
—-2
xeii (—1, l) 1
x < or x > 0; concave
2 a Increasing: nowhere 3
5 (0,0), (2,16) b downward x < O
Decreasing: Vx e R
c 3
b Increasing: x > 0
Decreasing: x < 0 Exercise 50
c Increasing: nowhere 6 a (1,0) b ia
7 a (-0.25, 0.992), (0, 1)
(note the function is 20
b -0.25; x > O 15
only valid here for x > l) 10
Decreasing: x e (1,00
c -0.25
8 a Coordinates of point of 12 4
inflexion are (2, O) -5
-10
1 b Function is concave up -15
when x < 0, x < 2 -20
d Increasing: x e 0,— -25
4 c Function is concave down -30
when O < x < 2 -35
(note the function is -40
9a -2 43
only valid here for x > O) b
-n on-horizontal
3 '27
1 b ( l, O) = horizontal inflexion 4
—4 310 3
Decreasing: x e 2
4' c = horizontal
3 ' 27 1
Exercise SL
-1
i a (0,—2) min b ( ) min inflexion x
56
d No inflexion points -6 -5 -4 -3 -2 123
c (O, 2) max; (4, —30) min 10 a i (-0.732, 3.39) max -2
(2.73,-17.4) min -3
4
ii (1,—7) = non-horizontal 3
2
3
3 c
inflexion point
1
4 (0, 21) iii Increasing: x < —0.732 or
x > 2.73 decreasing for -8 6 -2 24
-1
-0.732 2.73
675
ANSWERS x = 20 e Speeding up: t e (0, 1.5);
y = $10000 Slowing down: t e (l. 5, 3);
Exercise 5P
216 Chapter review
864
432
300
c- $164
Exercise SR 5x + x2
v(t) = 3t2 — 3 a(t) = 6t 123
b s(()) = Im, v(()) = —3ms-l, x=±3,y=6 b x = —
a(O) = O the particle is I unit
to the right of the origin and (19 - 33x)
is moving to the left.
dy
c The particle is moving away
from the origin at
9 m/ s-l and is accelerating
away from the origin at
12 ms-2
t=ls b 17m
mf d = 22
2a
c i v(O) = I ms-l
ii v(l) = 9ms-l
iii v(3) ——15 ms-l
d = 33m
3 t = l. 5s s(t) = 11.25
4 a 10m b 6.53 s
Exercise 50 c The diver hits the water 7 10
with a velocity of —8.06ms-l,
100 b P = 2x+ 100 and a constant vertical
acceleration of —2 ms-2.
. = 2005 (metres) Since both velocity and
(707.28 acceleration are negative,
the diver is speeding up
as he/she approaches the
water.
5 127.6m; 10.2s
t=o, 3, 11 9
3 27
b 10 (0, O) minimum
(—2, —4) maximum
ii 3<t<6
=c i t 1.5 =ii t 4.5 11 a b x = Oy
d 1.5, 4.5
676
negative or zero, whereas (Correct direction: I mark; 30
miles apart: I mark)
y = x has gradient l.
(l mark)
b Graph 2 (l mark) b d(t) =
y increases as x increases c (2 marks)
(l mark)
c Graph 3 (l mark)
Since the functions
represented by graphs
I and 2 are not defined
at infinity. (l mark)
12 a x = 0, x = b2 d Graph I (l mark)
—b2 as the function is t = 1.47 hours (? marks)
(? marks)
b X > increasing, decreasing. (l mark) d 15.4 miles
2 18 a i oct<2, 4.6StS5,
x < decreasing 8.5StSIO (3 marks) 20 a Letting x represent the
number of $10 increases
c If b > O concave upward, ii 2 < t S 4 and 5 < t S 7 above $320, then rental
income is
If b < 0 concave downward (2 marks)
(l mark) R(x) = (320 + 1 ox) (200 -5x)
13 a v(t) = 49 - 4.9t iii
I and
b h = 245m = =b f(t) t, g(t) 2, (l mark)
h(t) = — 3t+14,
—2 ms-l b
Maximize
c a = 3 -2 f(t) = —(2t —17) (5 marks)
s
—111 c
5 4
3
d Always speeding up since 2 (l mark)
acceleration is always 1 (l mark)
positive Which corresponds
kx — x3 b 27 to $360 rent (l mark)
b
Exam-style questions x = 180 (1 mark)
10
16 a f' (x) = 4x3 — 6x2 — 2x + 3 -1
-2
(l mark) -3 = $64 800
-4
b (2 marks)
(Up to two branches correct:
—4 (x2 +1) —(—4x).2x I mark; all branches correct: 21
2 marks; all branches correct b
+02 370 and
&and correct labels scale:
4x2 — 4 (2 marks) 438 (3 s.f.)
+ 02 (l mark) 3 marks) (2 marks)
c 19 a 112-9.8t
(2 marks)
c
(2 marks) 9 mi s/ho
d (l mark)
(l mark)
t = 11.4 s (3
(2 marks) d Either: double x-coordinate
Graph I (l mark) of maximum height
15 miles/ our
The gradient of the tangent
to the graph at any point is
ANSWERS
Or. solve Since 4 Between day I and day 3, the
34 number of people affected is
h(t) = 112t -4.9t2 = O 0—4
(l mark) increasing on average by 1410
(l mark) 23 a per day. Between day 4 and
(l mark) day 5, the number of people
Gives t = 22.8 s (3 s.f.) O—a affected is increasing on
(l mark) average by 2220 per day.
a 20 so a (l mark) (l mark)
(l mark)
b (l mark) —dlV
b = 900t —90t2 (2 marks)
dt
c After 5 days (reaches
¯c (l mark) 2250 cases) (2 marks)
00
x y = I is the equation of the d —d2=N900-180t
horizontal asymptote of dt2
the graph of f.
d (2 marks)
(Shape: I mark; domain: This tells you how the
I mark; maximum I mark) rate at which the disease
spreads varies. (l mark)
113m 2 25 a (l mark)
(2 marks) (2 marks)
(l mark)
ii
= —9.8 (l mark) xy—y=x+2 (l mark)
> O,for all x 2
which is constant. (l mark) (2 marks)
(l mark)
22 a i So the function has no b
maxima nor minima. (l mark) ddx [h(g-l
f'(2)g(2)-f(2)g'(2)
(l mark)
42 (l mark) e r (O) = — 24 m (l mark)
52 13 4
(2 marks) 3
16 4 3 (l mark) (x -1)2
(l mark) x-2 (l mark) (l mark)
(l mark)
(l mark) -2x2 +5x = o (l mark)
x = O or x = —5 (x -1)3
ii (g of)' 2
(l mark)
4 16 5 (l mark)
(l mark) (l mark)
(O, 2) and
33 2
b i False (l mark) (l mark)
as derivative changes sign. 24 a =1410
(l mark)
3-1
(l mark)
ii False (l mark) (l mark)
= 2220 (l mark)
678
26 a at (3, 3), tangent is (2 marks) - ((fxg)'
b
1
3
2+— = 3.5; The points (l, O), (2, O), (3, O), (2 marks)
(4, O), (5, O) do not lie on any
(l mark) (l mark)
3.5), (3, 3) (l mark) of these lines, so Peter does
not hit the targets. (l mark) 2(X+1) 4x
d i Let the plane be at the b
5
point a, 2+— when
—4
bullet is fired.
5
Differentiate: 3 (x -1)2
(l mark) 4x
Equation of tangent at
5
3
(3 marks: one for each
term correct)
(Shape: I mark; domain: (2 marks) Chapter 6
I mark)
This line passes through Skills check b 11
F(l, 5), G(2, 3.5), H(3, 3)
(l mark) (5,0), so solve 5 b I and 7
3
c Either: b6
Draw tangents at F, G and H
(2 marks)
(l mark)
a =1.62 (l mark) Exercise 6A
Bullet is fired when i a Discrete b Continuous
c Continuous d Discrete
plane is at (1.62, 3.85)
(l mark) 2 a Stratified b Systematic
c Simple random d Quota
3 a Stratified b Stratified
c Systematic
d Simple random e Quota
Conclude that none of the Exercise 6B
tangents intersects the x-axis x Continuous 6
at the target points (l mark) 11
3 b
OR: Time (mins) 4
y-3.85- -—i(x-1.62)
Calculate the equations of the 0<tS4 2
tangents to the curve: ii
8 < tg 12
at (l, 5), tangent is y (or alternative correct form)
(l mark) 12 16
at (2, 3.5), tangent is 16 20
3
4
679
ANSWERS
15 S a Hours Days 4 20 s 2.5
10
1 < 11<2 1 6 a r = 10, s = 13,
5 2 b t=18
Time 2<hS3 3
5 10 15 20 25 4 Exercise 6G 66 69 72
6
d Positive skew 8 59 62
2 a Continuous 6
58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73
b 6 b Negative skew 2 30.1 b 35 c 32.5 d 1.2
Time (mins) 8
20 5 Exercise 6C 3a
20 3
1 i8 c 13 runber of ctüjen in inernatonal in Portmany
30 < t S 40
d No mode 2 and 4
40 50
50 2 10 b 60 < y < 80
3a 3 b 30 < x s 35 3So SOO 1k 15k 2k 2.5k 3k 4k 45k 5k 55k 6k
10 Exercise 6D b 2000 + 1.5(1200) = 3800
8 6000 > 3800
6 3.65 b 12.805
4 c
2
Time c 3.35
10 20 30 40 50 60 70
2 20.5 b 43.8 c 2.30
3 2.6 o 200 400 600 800 1k 12k 14k 16k 1.8k 2k 2.2k 2.4k
d Positive skew 4 a 50 b8 c 7.08 d The outlier was removed
3 a Continuous
5 4.55 6 $21.89 as a single item of data
bf 60 b Right skewed distorted the analysis.
d4 4a
18 c2
16 20 30 40 50 60 70 80 90 100
14 8 220 b 40 < 60
12 b The morning exam
10 c 41.8 c This means that there is a
8
9 2 and 7 10 14 bigger difference between
6
4 11 the 25% and the 75% of
2
12 47.2 kg the scores
5a
Tine Exercise 6E
40 50 60 70 80 90 100
c Normal distribution 1 a 18 b 18.5 c4
d 3.5 e 6.5 f5
4a 29
Time (mins) 1 3 a $40468 b $200 c $400
0<1S10
2 Exercise 6F
20 < IS 30 3 8 7 c 12 b
30 < Is 40
40 < IS 50 10 d5 12 c Right skew
50 < Is 60 14
60 < Is 70 12 2a 6 b5 c 8.5 6 IA, 2C, 3B.
d 3.5 e 13
3 22.5 b 20.5
b Negative skew c 22.5 d 13 and 33.5
e More than 33.5
680
Exercise 6H b II min c 13.6 —8.2= 5.4 min d 15.6min 6 a Mean 58, standard
i a 18min
2 a 40 b 50 - 30 = 20 c 53 deviation 5
3a b Mean 480, variance 2500
Distance
7 40 b 60 c 90 d 50
CF
25 < d 50 75 < d 100 < d 125<d 8 800 b 65
100 125 s 150
25 S 50 75 c 75-55=20 d 100
220 236 240
o 32 134 e No. The 90th percentile is
more than 80 marks
b d 3, 2, 1.41 f 40
240 85.8, 34300, 185 9
120 2 a 3.2, 1.33 b 8.32, 3.33 10 20.125, 13.2
80
40 c 20.7, 13.3 b 27.5-10= 17.5
3 4.63, 2.67 b 4.24, 1.40 4, 4.125, 4, 1.36
4 62.37, 16.9 s 322, 91.7 12 10.9, 12.5, 6.92
6 a 24 months b We are assuming that the
b 20 to 30 hours number of items is equally
c 28.75 d 11.1 spread throughout the class
100 110 120 130 140 150 interval.
c 119 20 Exercise 6J 13 8
— 114 = 12 i a Mean = 4.4, median 14 a 80 b 50 c 6.25%
48 mode
4 90 d a = 10, 20
143 b Mean = 8.4, median
Pages 176 mode f 52.5, 22.5
196
100 < p s 200 200 c Adding 4 to each value m15 a = 12, n = 30
200 < p < 300 increases the mean,
300 < p s 400 b 21.1 c 32.3
median and mode by 4.
400 < ps 500 2 a Standard deviation = 3.02, 16 a Discrete b 2.73
500 < ps 600
variance = 9.11 c 1.34 d 23
600 < p s 700 b Standard deviation = 9.06,
700 < p s 800 Exam-style questions
variance = 82.0
b 200 1? a Discrete (l mark)
c The standard deviation is (I mark)
180 multiplied by 3 and the b Contin uous (l mark)
160 variance is multiplied by 9. c Contin uous
b 140 3 Mean = 21.2, median = 21, d Discrete (l mark)
standard deviation = 0.5. 18 a
40 As the mode is 5 there
20 4 The mean, median and
must be at least another 5.
standard deviation are doubled.
(l mark)
5 The variance is multiplied by 81
So we have l, 3, 5, 5, 6
100 200 300 400 500 600 700 800 Chapter review with another number to be
420 d 210 e 80 b 3 c 3.2 placed in order (I mark)
5 1B 2C 3A The median will be the
2 500 average of the 3rd and 4th
Exercise 61
i a 4.375, 4.48, 2.12 3 a 3 b 3 c 2.5 pieces of data. (l mark)
b 5.2, 10.4, 3.22 4 Mean 21.9, standard deviation For this to be 4.5 the missing
c -1.6, 8.24, 2.87 piece of data must be a 4.
1.1
5 a 32 min b 31 min Thus a = 5, b = 4 (2 marks)
681
ANSWERS
b 1+3+4 +5+5+6 d Median is approximately 23 a
6 the 100th piece of data x Frequency Cumulative
frequency
24 (2 marks) which lies in the interval 10
6 1 10
20 < h s 30. (l mark) 2 11
3 13 28
19 a An outlier is further than Will be 15 pieces of data 41
into this interval 4 15 56
1.5 times the IQR below the 71
lower quartile or above the Estimate is 5 15 83
6 12 93
upper quartile. (l mark) 20+1—5x10 = 23
10 99
b i Mode = 8 (l mark) 50 84 100
ii Median = 7 (l mark) (2 marks) 92
10 1
22 a Discrete (l mark)
iii Lower quartile b5 (l mark)
(l mark)
iv Upper quartile c i 4.79 (3 s.f.)
(l mark)
ii 1.26 (3 s.f.) (2 marks)
4d 5 ii iii 5.5
c IQR=6 1.5 x IQR=9 (l mark each) (4 marks for 6 correct,
3 marks for 4 or 5, 2 marks for
19-9=10 (l mark)
2 or 3, I mark for l)
19 is the (only) outlier
(l mark) o 1 23455.65 78 4 2 6b i
ii iii
(l mark) general shape (l mark each)
(l mark) median
20 a c 4.05 (3 s.f.)
10 (l mark) quartiles
(l mark)
Let the mass of the new (2.4140
f IQR=1.5 (3 marks)
student be s. (l mark) d No. It is bimodal at
11
(l mark) 5.5 + 2.25= 7.75 and 5. (2 marks)
(l mark) 4-2.25= 1.75 (l mark)
= 792 (l mark)
so 92 kg (l mark) So the 2 (unhappy)
b IQR = 10
candidates with grade I
are outliers. (l mark)
(l mark)
So the new student's mass of
92 is an outlier. (l mark)
21 a 200 (l mark) 24 80 < w s 90 (l mark)
b 35 (l mark) b
c Using mid-points 5, 15, 25... Mass w S 50 w < 60 w s 70 w S 80 w 90 w < 100 w s 120
as estimates for each interval
Cumu-
(l mark) 5 20 45 75 125 160 185 200
lative
i Estimate for mean is frequency
22.25 (2 marks) (2 marks) numbers
(l mark) labelling
ii Estimate for standard
deviation is 11.6 (3 s.f.)
(2 marks)
682
Chapter 7
200 6.5
190 Skills check 6.0
180
5.5
170 1296 b 64 c 343
5.0
160
150 2a 5 d3 4.5
140 4.0
y=4x-2130
3a u- 3.5
b 3.0
2.5
90 Exercise PA 2.0
SO 1.5
i a Strong, positive, linear
60 b Weak, negative, linear 20 30 40 50 60 70 80 90 100110120
so
40 c Strong, negative, linear Distance(km)
30
20 d Weak, positive, linear b Strong, positive, linear
10 correlation
e No correlation
0 to 20 30 40 50 60 70 SO 90
kg
120
2 i a Positive
(2 marks) scales (l mark) b Linear c Strong 1.4
points and curve
ii a Positive 1.2
i 85 73 97ii iii b Linear c Moderate
1.0
(l mark each) (l mark) lines iii a Positive c Weak
b Linear S 0.8
Student
25 iv a No correlation 0.6
20 0.4
15 b Non-linear
10 0.2
5 c Zero
0.0 01234567 8
o 123zt5678 grade v a Negative b Linear
c Strong weak
25 a i 7.5 ii 6.125 (2 marks)
b Strong, positive, linear
ii 6.9 (2 marks) vi a Negative b Non-linear correlation
c Sally's had the greater c Strong c The weight has increased.
d
26 a median. (l mark) Exercise 7B
i a 0.987 b Strong positive
3a
c As the floor area increases,
Rob's had the greater mean. Maximum dive time the house price increases.
(l mark) 55 2 -0.976
50 b Strong negative
not to scale 45
40 3 0.707
25 35 b Moderate positive
30
20 25 4 a 0.878 b Strong positive
20 20 c Yes
15
16 16 S a —0.449 b Weak negative
15 c Yes. The correlation between
10 4 GPA and game time is weak.
5 .4
6 0.970
1 2 3,156 7 b Strong positive
Grade c More practice can give a
(l mark) higher final grade.
scale (2 marks) 18 20 22 24 26 28 30 32 34
4iii (3 marks) Depth(m)
c The values of the median b Strong, negative, linear
and the mean are the same c As the maximum depth
due to the symmetry of the increases the time at that
depth decreases.
bar chart. (2 marks)
683
ANSWERS
Exercise 7C 2 5.75, 2.7825 a a = -8.46,b = 33.0
Note that answers may vary b b 3 c -0.952
d Strong, negative.
slightly for the line of best fit 6 8 0.911
by eye.
an pOiii
i a 48.5 b 29.3 b
80 x c An increase of one point
60 2 4 6 8 10 12 on the 1B diploma produces
40 a 2.19% increase in
Age(months)
an university grade
c 0.208x+ 1.589
20 d 68.7%
md 3.46 me 26.5
x
9a 6.416x+ 14.858
40 40 60 80
f No. Giraffes only grow b i 6.42; each pizza produced
to 6m. Extrapolation is raises costs by $6.42.
unreliable. ii 14.8; the cost to the
3 3 6 10 12 15 20 shop when no pizzas are
Distance(km) produced is $14.86.
Month! Rent (1000 rupees) 60 45 32 28 18 15 c $399.80
d i Unreliable as it is too far
a I I km b 33 000 rupees d y = -2.83x + 64.2
from the domain.
c 41500 rupees
70 ii 13 pizzas
f 5 km
60 10 a -0.984 b a--6.71, b=118
g No. Extrapolation is
@ 50 inaccurate. The rent is c V78000
negative.
40 Exercise 7E
2 i a y = 61.0 — 1.08x, 39 tickets.
30 b x = 52.1 -0.810, $24
e 2 x = 1.35 +0.159y, 9.3 min
3 x = 32.5 + 0.697y, 69.
20
4 149 b 19.7 km
10
x
5 10 15 20 25
Distance(m)
Exercise 7D 4 —0.988 Chapter review
b Y = -7.08x+ 202
1 A student who does not i a —l and I
c 149 km h-l
play sports does 35 hours of b A -0.6 B 0.9 C 0.5
d For every increase of one DO E -0.96
homework per week. second in the O to 9() time,
c Linear, strong negative
Each sport played decreases the the maximum speed of
horse of homework by 30 min. 2 a d f on the diagram
the car decreases by b 40 OC
2 A person with no criminal 7.08 kmh-l. c 1200 Dirhams
S a 0.889 b Y = 1.01x- 6.30
friends has I conviction. y=o.8x-20
c 69.6
Having one criminal friend 18
increases the number of 6 0.985 b Y = 1.74x- 48.0 16
conviction by 6. c 21.9 an
12
3 A new speaker costs $300. d For every cm that the cat
10
Each year its cost goes down grows in length, it grows
I .74cm in height. 8
by $40.
x
0
35 36 37 38 39 40 41 42 43 44 45
684
3 a a--2.65, 35.6 Exam-style questions b Strong, negative (2 marks)
b 22
c i i =4.625
c 2050 is too far outside of (2 marks) ii i =5.875
the domain. b The line of best fit goes
iii See above
through (i,y) (l mark) (5 marks)
= 58.5
y=108x+361 b 30 d See above
21.50C b 0.924 MI thru average (l mark)
c There is a strong positive (l mark) a 3.2 see above for lines
correlation: the hotter the
c Strong, positive (2 marks) drawn on (2 marks)
day, the more water sold.
d x on y (l mark) 100 = 70 m +c
d 18.3x- 227 e 131 140 = 100 m +c
f 36 oc is outside of the 12 i Perfect positive (l mark)
domain. Extrapolation is
ii Strong negative (l mark) 40=30 m m=—4 c 20
unreliable. (l mark)
iii Weak positive (l mark) 33
iv Weak negative (l mark)
11.8x+53750 (3 marks)
v Zero
b $136210 c i and ii b Positive (l mark)
a = 0.0874, 0.876 13 a r=O.979 (3 s.) (2 marks) c Line goes through (R,j)
b The car travels I km on b Strong, positive (2 marks) (l mark)
0.0874 litres of fuel c i y=1.23x-21.3 = —490 6 —2 = 126 —2
c 14.9 litres (2 marks) 3 33
kmd 5 is outside of the ii x=O.776y+20.8
domain and extrapolation is (2 marks) (2 marks)
unreliable. d Estimate is
3.90, 4 +6—2 -86 2
b An increase of I g of growth (I mark) —60
e 0.776x95+ 20.8 -95
hormone produces nearly 3 33
4 extra flowers. (l mark)
f It is extrapolation (l mark) (2 marks)
c A plant with no growth 14 a
16 400C (l mark)
hormone would have
b 700C (l mark)
2 flowers
10 c 1000C (l mark)
d An average of 8.84 flowers
9
e 2.56 g
f I kg is outside of the 8 d i (30,100)
domain and extrapolation is 7 (60,70)
unreliable. 6
a = 199 5 (0,40) (l mark)
4
3 T>80
2 130 — 80 t = 20
t = 50
b Every time I g is added the 1
mmspring expands 0.261 1234567 8 9 1011 x
13
c The spring was originally
(l mark scales 3 marks points (l mark)
mm199 long (2 marks) 6 points, I mark Interval is 20 t 50.
d 342 mm
3 points) (2 marks)
e 2kg is beyond the domain i? a
and extrapolation is unreliable
f 388g 14 15 16 16 18 18 19 19
10 a Y = -0.254x+ 33.7
2o 3 1 4 1 1 2 1 2
b 21 c -0.842
(3 marks) (2 marks for 5, I mark for 3)
d There is a strong, negative
correlation.
685
ANSWERS
b r = —0.0695(3s.f.) (2 marks) b No change r = 0.87 c P(multiple) = —2 = —1
c Very weak (negative) (l mark) 84
correlation so line of best fit (l mark) d
is meaningless. (l mark) iii The scatter diagram P (not a multiple of 4) I 3
has been stretched
25-year-old would be 4 4
extrapolation. (l mark) vertically. (l mark) 3
m 06 c i r = —0.87 (l mark) e P (less than 4) = —
18 i Gradient = —t— — 0.2 8
3 15
(l mark)
(2 marks) f P (mne
—3
1=06 (1 mark) 8
iii The scatter diagram
(l mark)
has been stretched
iv (l mark) horizontally and reflected 5
10
in the y-axis. (2 marks)
b = 0.6 (l mark)
vi Gradient p = 0.9-06 -0.1 iv Strong, negative
• (2 marks)
8-5 10
(2 marks) —3
vii P (Vowel) =
10
Chapter 8
1
(2 marks)
6 P (Even) = —
viii r (l mark) Skills check
(2 marks) 2
19 a i 0.849 (3 s.f.) 4 b4 4
35 —7
ii Strong, positive a c
(2 marks) 3 15 b P(Contains digit l) =
7 25
iii y = 0.937x+0.242 7
d 3
56 p(B) =
(2 marks) 3 0.625 b 0.7 7
b i 0.267 (3 s.f.) (2 marks) c 0.42 d 0.16 8 a x=o.2
20 a ii Weak, positive
e 15 f 0.0484 1
(2 marks)
iii The r value is too Exercise 8A 9 P(Sebastian wins) =
360
small for this to be —5 1
particularly meaningful Exercise 8B
i P(odd) =
(1 mark) i a i P(15 years old) = 0.18
10 2
No change r = 0.87 ii P(16 or older) = 0.22 +
—2 P(defective) = 30 = 1 0.27 + 0.13 = 0.62
(l mark) 150 5
b 1200 x 0.18 = 216
ii No change 1 5 (1 mark) —20 4
iii The scatter diagram has students
3 P(chorus) = 2 0.27
just been moved up by 5 7
and to the left by 4. b Probably not. The relative
4 a P(even) = —4 = 1
(l mark) frequencies are very
82
iv Strong, positive different.
(2 marks) —b P(multiple) = —2 = 1
c 450
84
686
3 a 10 of each a 6
b
1
Relative 0.1 0.1 0.15 0.138 0.138 0.15 0.138 0.0875
frequency
c
Relative 0.0925 0.1225 0.1375 0.125 0.14 0.145 0.1075 0.13 33%
frequency 30% 9
d There is a big difference between relative frequency of getting a b i P(News at 9 only) = 0.33
I and getting a 6. This suggests the dice is not fair. ii P(News at 6 only) = 0.24
iii P(Does not watch news)
Exercise 8C c P(Neither a card nor = 0.30
42
ia Exercise 8D
a present) =
20 9 1 50 25 42
i a P(Prime)=—
8 4a
5
—b P(neither) = 8 = 4 p
b P(Prime or multiple of 3)
38 19 c 3
5
2a AL
c P(Prime or multiple of 3)
s T,
1
A, Y, B,
2
19 21 16 S a {6)
b {2, 3, 4, 6, 8, 9, 10} 2 P(Have camera or female)
12+18+7 37
7 3, 3
b 63 members of the dub 3,
—40 {1, 2, 4, 5, 7, 8, 10) 26 13
f {1, 2, 3, 4, 7, 8, 9, 10)
c i P(Badminton) = b
63 9
P(T, R, 1, G, O, N, M, E, 26
—21 1 6a 6, 9, 12, 15)
b 2, 3, 5, 6, 10, 15) 2
P(Both) = 13
63 3
d
—7 1
iii P(Neither) =
63 9
iv 8 9 31 13 1
9 2
56 26 2
P(At least one sport) = 12 6 5 4 0.5 b 0.5
63 1
3a 10
46 23 8 11 13 14
P(Card or present) = 50 - 25
31 1 3
b P(Card but not a present) c b
63 Sa
50 25 15 5 4
4
62
6 0.1 b 0.9 3
5 c
a 27 37
64 b 64
64
ANSWERS
Exercise 8E P(E) = 0.4 S a 6x
i a No b Yes c No b P(F) = P(EnF) b drama comedy
d Yes e No f No g No i
2 Mutually exclusive ii P(EnF) * O 15
20
47 c 0.64
12 60
reality
c Yes, because the
27 b 0.63 c 0.57 0.3 b and D)
probability of A, B or C 9 a 0.27
winning is not equal to l.
Exercise 8H c No P(C) x P(D) P(Cand D)
Exercise 8F
i 12 d 0.6 e 75
b 0.43 b 0.316
27 23 catch
ii
27
1,2 1, 3 3 ao bO c 0.63 injected 30% not catch 80%
2, 3 2,4 10 11 not Injected 70% catch 90%
3, 3 3,4 13
4
0.3 not catch 10%
0.18 iii 0.46
—b i ii —iii iv c 0.036
4 16 i 0.09
b 0.343
Exercise 81
11 132 21
1105
1105 10 female eating carrots
91
10
91
23
19 27
10 15
63 11 11
12 11 11 Both female and eating carrots = 19
Exercise 8G 11 19
64
3 0.0375 70 46
c No P(F) x P(C) P(FandC)
25 125
4 a 0.2, 0.16 Chapter review
b Not independent 49 Exam-style questions
90
12 6561 15
(l mark)
30 b 2,4 or 6 — (l mark)
3 0.55
b 0.14
4 0.02 b 0.78 c Primes are 2,3,5 —
c 0.76 d (2 marks)
30 (2 marks)
(l mark)
d 4 or 5 —
e Impossible O
688