Mathematics Fundamentals PDF

DR RAMZAN RAZIM KHAN & DR DES HILL
MATHEMATICS
FUNDAMENTALS
1
Mathematics Fundamentals
1st edition
2017 Dr Ramzan Razim Khan & Dr Des Hill & bookboon.com
ISBN 978-87-403-1864-7
Peer review by Dr Miccal Matthews, The University of Western Australia
2
MATHEMATICS FUNDAMENTALS Contents
CONTENTS
1 Numbers and operations 5
2 Algebra 17
3 Simultaneous equations 25
4 Quadratics 31
5 Indices 41
6 Re-arranging formulae 53
7 Exponentials and logarithms 57
8 Equations involving logarithms and/or exponentials 65
9 Quadratic equations 77
10 Functions and graphs 87
11 Quadratic functions and Parabolae 103
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MATHEMATICS FUNDAMENTALS Contents
12 Calculus 111
Appendix 1: Answers to exercises 123
Appendix 2: Solutions to problem sets 141
Index 197
4
MATHEMATICS FUNDAMENTALS Numbers and operations
1. Numbers and operations
Types of Numbers
The natural numbers are the usual numbers we use for counting:
1, 2, 3, 4,
We also have the very special number zero
and the negative numbers
4, 3, 2, 1
which are used to indicate deficits (that is, debts) or to indicate

opposite directions. These numbers collectively form the integers
4, 3, 2, 1, 0, 1, 2, 3, 4,
The basic operations of arithmetic
Hopefully, the everyday operations +, , and are familiar:
(i) 3 + 4 = 7 (iii) 4 9 = 36
(ii) 11 6 = 5 (iv) 24 6 = 4
Arithmetic involving negative numbers
When negative numbers are involved we need to be careful, and

must use brackets if necessary.
Multiplication examples:
positive negative = negative
(i) 4 (7) = 28 (iii) (3) (6) = 18 negative positive = negative
negative negative = positive
(ii) (5) 6 = 30 (iv) 3 5 = 15 positive positive = positive
Other examples:
(i) 13 + (8) = 5 (iv) 16 13 = 29
(ii) 21 (3) = 24 (v) (12) 4 = 3
(iii) 16 + 13 = 3 12
(vi) =3
4
5
6 mathematics
MATHEMATICS fundamentals
FUNDAMENTALS Numbers and operations
BIDMAS. The order of operations
Consider the simple question:

What is 2 + 3 4?
Which operation (+ or ) do we perform first?
Mathematical expressions are not simply calculated from left to
right. They are performed in the following order:
B: Expressions within Brackets (. . .)

We will consider Indices in detail later
I: Indices (powers, square roots, etc)
DM: Divisions () and Multiplications ()
AS: Additions (+) and Subtractions ()
A couple of things to note:
Division and multiplication have the same precedence.
Addition and subtraction have the same precedence.
Operations with the same precedence are performed left to right.
Some BIDMAS examples
(i) 2 + 3 4 = 2 + 12 = 14
(ii) (4 + 6) (12 4) = 10 8 = 80
When brackets are involved we can
(iii) 4 + 7 (6 2) = 4 + 7 4 = 4 + 28 = 32 omit the multiplication sign. So
instead of 3 (4 + 8) 2 (10 6) we
(iv) 3 (4 + 8) 2 (10 6) = 3 12 2 4 = 36 8 = 28 could write 3(4 + 8) 2(10 6)
More examples
4233
= 233
= 63
=2
[2 3(4 6 2 (3))]
= [2 3(4 3 + 3)]
= [2 3 (1 + 3)]
= (2 3 4)
When brackets are nested we
= (2 12) usually use different sorts, e.g.
( ) [ ] { }
= (10)
= 10
6
MATHEMATICS FUNDAMENTALS
1. numbers and operations 7
Numbers and operations
Word problems
We are sometimes asked quesions in written or spoken conversa-

tion that we will be able to use mathematics to help us solve.
Our first task is to turn the word problem into a mathematics
problem.
Example: What is three times the difference between 12 and 4?
Answer:
3 (12 4)
= 38
= 24
Another example: Joy takes the product of the sum of two and
six, and the difference between nineteen and nine. She then result
the answer by four. What number does she arrive at?
Answer:
{(2 + 6) (19 9)} 4
= {8 10} 4
= 80 4
= 20
7
8 mathematics fundamentals
The number line
The integers sit nicely on a number line:

-5 -4 -3 -2 -1 0 1 2 3 4 5
Fractions
These are numbers that sit between the integers. For example, the
number one-half, which we write 12 , is exactly halfway between
0 and 1. 1

2

-5 -4 -3 -2 -1 0 1 2 3 4 5
Fractions are ratios of integers, that is,

p Note that p or q or both can be negative
q
where the numerator p and denominator q are integers. When the
numerator (top line) is smaller than the denominator (bottom line)
we call this a proper fraction.
The number four and two-fifths is written 4 25 and is between
4 and 5. To pinpoint this number, imagine splitting up the interval
between 4 and 5 into fifths (five equal pieces). Now, starting at 4,
move right along two of these fifths.

-5 -4 -3 -2 -1 0 1 2 3 4 5
We call 4 25 a mixed numeral, as it is the sum of an integer and a

proper fraction.
Converting mixed numerals to improper fractions
An improper fraction is one in which the numerator is larger than

the denominator.
Example: We wish to convert the mixed numeral 3 12 into an
improper fraction. We have three wholes and one half:
or three lots of two halves (making six) plus one half is

seven halves:
1 7
so 3 =
2 2
8
1. numbers and operations 9
Arithmetically, we write
1 32+1 6+1 7
3 = = =
2 2 2 2
Some examples
1 43+1 12 + 1 13
4 = = =
3 3 3 3
1 24+1 8+1 9
2 = = =
4 4 4 4
5 16+5 6+5 11
1 = = =
6 6 6 6
3 85+3 40 + 3 43
8 = = =
5 5 5 5
Note: Improper fractions turn out to be used more often in
maths, mainly because mixed numbers are confusing in algebra.
Multiplication of fractions
To multiply two fractions simply multiply their numerators and

denominators separately. For example,
1 5 15 5 Think of this as finding one-half of
= = five-sixths
2 6 26 12
Another example is
3 2 32 6
= =
5 7 57 35
Division of fractions
The notation 2 3 means the same as 23 , that is two-thirds. Now

think of 2 and 3 as fractions, that is
2 3
23 =
1 1
We can also use fraction multiplication to say that
2 2 1
=
3 1 3
Comparing these last two we have
2 3 2 1
=
1 1 1 3
That is, dividing by a fraction is equivalent to multiplying by its
reciprocal, that is, the fraction having been inverted. For example,
2 5 2 9 18
= =
7 9 7 5 35
9
10 mathematics
More examples
1 2 5 2 5 2 6 12
= = =
3 7 6 21 6 21 5 105
1 7 7
4 8 32 7 3 7 5 35
3
= 3
= = =
5 5
32 5 32 3 96
5 2 1 13 2 22 13 3 22 858
2 3 = = =
4 3 7 4 3 7 4 2 7 56
Factorization of numbers
This is the process of expressing a given number as a product of

two (or more) smaller numbers. For example,
28 = 4 7
108 = 2 54
= 269
64 = 2 32
= 2 2 16

= 222222 Later well write this as 64 = 26
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MATHEMATICS FUNDAMENTALS 1. numbers and operations
Numbers 11
and operations
Cancellation of numbers
The process of simplifying fractions by looking for factors which

are common to both the numerator and the denominator and then
eliminating (cancelling) them. For example,
18 2 9 9
= =
20 2 10 10
100 1
10 0 10 5
2 2
= = = =
250 1
25 0 25 55
5
24 4 6 6 32
2
= = = =
108 4 27 27 39
9
Adding and subtracting fractions
Consider the sum

1 2
+
3 5
We cant add these fractions yet because thirds are different to
fifths. However, we can make use of reverse cancellation to make
the denominators the same:
1 2 15 23
+ = +
3 5 35 53
5 6
= +
15 15
Now we just have a total of 5 + 6 = 11 fifteenths, that is
1 2 5+6
+ =
3 5 15
11
=
15
9
The same applies to subtraction. For example, 20 25 . Following
the above pattern the denominator would be 20 5 = 100, but we
can simplify the numbers a little as follows. Since 5 divides into 20
we could just do this instead:
9 2 9 24
=
20 5 20 5 4
9 8
=
20 20
98
=
20
1
=
20
Two examples
1 4 17 44 7 16 23
+ = + = + =
4 7 47 74 28 28 28
2 1 5 12 9 5 21 5 16
+ = + = = which if required can be simplified:
9 6 54 54 54 54 54 54 54
=
28 8
=
2 27 27
11
12 mathematics
Decimals
Decimals are a convenient and useful way of writing fractions with

denominators 10, 100, 1000, etc.
2 4
is written as 0.2 is written as 0.04
10 100
35 612
is written as 0.35 is written as 0.612
100 1000
The number 237.46 is shorthand for
4 6
2 100 + 3 10 + 7 1 + +
10 100
The decimal point separates the whole numbers from the fractions.
To the right of the decimal point, we read the names of the dig-
its individually. For example, 237.46 is read as two hundred and
thirty-seven point four six.
The places after the decimal point are called the decimal places.
We say that 25.617 has 3 decimal places.
Some fractions expressed as decimals
We can write
3 4
10 as 0.3 10 as 0.4
47 12 2
100 as 0.47 10 = 1+ 10 as 1.2
23 327 27
100 as 0.23 100 = 3+ 100 as 3.27
931 28
1000 as 0.931 1000 as 0.028
6 5
100 as 0.06 1000 as 0.005
Common fractions expressed as decimals

All fractions can be written as decimals. However, the decimal
sequence may go on forever.
Examples
1 1 1
= 0.1 = 0.01 = 0.001
10 100 1000
1 1 1
= 0.5 = 0.25 = 0.125
2 4 8
3 1 2
= 0.75 = 0.2 = 0.4
4 5 5
1 2 2
= 0.3333 = 0.6666 = 0.4
3 3 5
1 22 1 The indicate that the sequence goes
= 0.142857142857 = 3 + = 3.1428 on forever
7 7 7
12
Numbers 13
and operations
Fixing the number of decimal places

When working with decimals we usually limit ourselves to a
certain number of decimal places.
Examples To three decimal places
1 1 Note that we have rounded up the

= 0.333 = 0.143
3 7 2 to 3 because the 8 after the 2 in
1
7 = 0.14285714 is larger than 5 (out
of 10)
1 2
= 0.125 = 0.400 Note that we can pad out with zeros
8 5 to the left in the decimal for 25
Multiplying and dividing decimal numbers by 10

To multiply a decimal number by 10 simply move the decimal
point to the left, and to divide a decimal number by 10 move the
decimal point to the right
Examples
0.47 10 = 4.7 0.06 10 = 0.6
0.12 10 = 1.2
360
0.000234 10 = 0.00234
.
0.000234 100 = 0.0234 0.56 100 = 56.0
thinking
0.378 10 = 0.0378 0.04 10 = 0.004
2.571 10 = 0.2571 4.63 100 = 0.0463
360
thinking . 360
thinking .
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13
14 mathematics
Exercises 1
1.1 Evaluate the following
(i) 3(4 + 2 6) + 12 4 (ii) 6 + 2[25 3(2 + 5)]
1.2 What is the difference between the product of 6 and 7, and the
sum of 3 and 8?
1.3 Convert to improper fractions
1 3
( a) 1 (b) 2
4 8
1.4 Evaluate (that is, calculate)
3 1 1 2
( a) (b)
4 5 3 5
1.5 What fraction is two-thirds of four-fifths?
1.6 Evaluate
3 2 2 2+4
(i) 4 + 5 (ii) 3 51
1.7 Reduce to simplest form by using cancellation:
78
102
1.8 What is one-quarter of the sum of one-third and two-sevenths?
Problem set 1
1.1 Evaluate the following.
(a) 4 + 4 4 4 4
(b) 4 + 4 (4 4 4)
(c) 2 4 (8 1)
(d) 14 2 (20) (5)
(e) (4 12) [32 (4)]
(f) [4 (3 + 4) 21] [2 14 7 + 3]
(g) 3 8 [16 (4)]
(h) [3 (3 4) 19] [8 4 7 6]
(i) (8 4) (4 + 4) 8
(j) 4 + 4 4 (4 4)
14
Numbers 15
and operations
1.2 Compute the following, leaving your answer in the simplest

fraction form.
1 3 5 3 4 3
(a) + + (b) +
2 4 12 4 5 10
4 2 2 3 4
(c) +1 (d) 1 2 +1
24 3 3 5 5
4 3 2 2
(e) (f) 2 4
5 7 3 9
7 1 5 3 34
(g) 1 (h) 2
9 8 7 7 14
1.3 What is half of the difference between a third of 60 and

a quarter of 80?
1.4 What is the sum of half of 45 and a third of the product of two
thirds and 1 18 ?
1.5 James and his wife Sweet Li have a 12 year old daughter called
Lyn. James is 50 year old and Lyn is 12. Half of his age added
to five thirds of Lyns age gives the age of Sweet Li. How old is
Sweet Li?
1.6 Swee Khum put a third of her savings in the bank, a third in
bonds, a quarter of the remainder in stocks and the rest in fixed
deposit. If her total amount is $600,000 how much did she put in
fixed deposit?
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15
MATHEMATICS FUNDAMENTALS Algebra
2. Algebra
Algebra is a powerful tool for expressing fundamental relationships

and solving real-world problems.
An example: In the expression E = mc2 , the variables are the
energy (E) of an object and its mass (m), c2 means c c and c is a
constant representing the inconveniently large value of the speed of Such constants appearing in relation-
ships are called parameters
light in a vacuum, i.e. 29,792,458 metres per second.
The famous equation above summarises the relationship between An equation is something with an
equals sign in it
the variables E and m
We can also use algebra to express our rules for multiplying and
dividing fractions:
a c ac a c a d ad
= = =
b d bd b d b c bc
An example of algebra in action
A simplified version of the formula for the length of time it takes
a child to say are we nearly there yet? on a long car journey is
1+ A
T=
C2 This formula was proposed by Prof.
Dwight Barkley, Warwick University
where
T = time to say are we nearly there yet? (in hours),
A = number of activities provided and
C =number of children.
So, 2 children with 3 activities will take
1+3
T=
22
4
=
4
= 1 hour to get restless
Solving equations
In many situations we are told the numerical values of all variables

in an equation except one and asked to determine the value of that
variable. That is, we are asked to solve for the unknown variable.
Example. In the restlessness problem above we might know that
there were 2 children in the back and it took them an hour and a
half to complain. How many activities were they given? We have
1 1+ A 3 1+ A
1 = that is =
2 22 2 22
17
18 mathematics
FUNDAMENTALS Algebra
To answer questions like this we need to know how to re-arrange

equations.
Simple Equations
Here are some typical equations with an unknown variable:
x+4 = 7
3x
2x + 7 = +8
2
2 3
+4 =
x x
Our aim is to find out what value of the unknown variable x

makes the equation work. That is, we wish to solve for x.
Operations on equations
Lets look at
x+4 = 7
To get x on its own, we can remove the other operations (in this
case the + 4) by performing operations that undo them.
To maintain equality, operations must be applied to the entirety

of both sides. Operations include:
(1) Multiply (or divide) both sides of an equation by a non-zero

number
(2) Add (or subtract) the same number to (from) both sides of an
equation
Using these operations effectively is just a matter of technique

and practice.
Example:To solve
x+4 = 7
subtract 4 from both sides:
x + 44 = 7 4
x=3
The symbol means implies that, and the diagonal bars indicate
cancellation.
Its always a good idea to check that the value we have obtained
is in fact correct. Clearly 3 + 4 = 7 so we are correct.
Example: To solve 3x = 12 divide each side by 3,
3x 12
=
3 3
x=4
Check: 3 4 = 12
18
2. algebra 19
2. algebra 19
Example: To solve y 4 = 2y + 10 you need to get the two y

on the To
Example:
terms solve
same y 4 = 2y + 10 you need to get the two y
side,
terms on the same side,
y 4y = 2y + 10y
y 4 = 2y
y4 = 10(since
y ++10 y two ys minus one y is one y.)
4 = y + 10 (since two ys minus one y is one y.)
4 10 = y + 10 10
4 10
14 == yy + 10 10
14 = y
Check: 14 4 = 18 and 2 (14) + 10 = 18 hence
Check: 14 4 = 18 and 2 (14) + 10 = 18 hence
Example: This one presents a new problem: 3x + 2 = 2( x + 7).
Normally, This
Example: one presents
we perform a new problem:
the operations 3x + 2 =first
in the brackets 2( x but
+ 7).
Normally,
there is nothingwe weperform
can dothe operations
with x + 7. Wein need
the brackets
to expandfirstthe
but
there is nothing
brackets. We do we thiscan
withdothe
with + 7. We need to expand the
aidx of
brackets. We do this with the aid of
The Distributive Law: a(b + c) = ab + ac
The Distributive
where a, b and Law: a(b + c) =
c all represent ab + ac
terms.
where a, b and c all represent terms.
Basically, the term out the front, a is multiplied by (distributed
Basically,
amongst) theterm
each terminside
out the
thefront, a is multiplied
brackets, by (distributed
that is, b and c.
amongst) each term inside the brackets, that is, b and c.
Lets first test this law. Say a = 9, b = 4, c = 3. Then the
Lets first
left-hand sidetest this law.
(LHS) of theSay a = 9,isb = 4, c = 3. Then the
equation
left-hand side (LHS) of the equation is
9(4 + 3) = 9 7 = 63
9(4 + 3) = 9 7 = 63
and the right-hand side (RHS) of the equation is
and the right-hand side (RHS) of the equation is
9 4 + 9 3 = 36 + 27 = 63
9 4 + 9 3 = 36 + 27 = 63
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19
20 mathematics
Some examples
4(3 + x ) = 12 + 4x You need to be careful when negative
signs are involved
3(4 2x ) = 12 6x
and really careful when more negative
x (y 5z) = xy + 5xz signs are involved
Watch out when brackets are near other operations:
4x 4(3 + 5x ) = 4x 12 20x = 16x 12 The 4x part is not involved in the

expansion of the brackets
We can solve equations involving brackets by making use of the

Distributive Law.
Example:
3x + 2 = 2( x + 7)
3x + 2 = 2x + 14
expand the bracket
x + 2 = 14 then subtract 2x from both sides
x = 12 and finally, subtract 2 from both sides
and recall that the symbol means implies that.

Another example: To solve
2t 1 = 3 2 [2t 3 (1 t)]
first expand the brackets (from the inside out):
2t 1 = 3 2 [2t 3 + 3t]
2t 1 = 3 2 [5t 3]
2t 1 = 3 10t + 6
2t 1 = 9 10t
12t 1 = 9
12t = 10
and hence
10 which simplifies to t = 5
t= 6
12
Equations with fractions

Fractions create extra challenges and its a good idea to deal with
them early on. If we have an equation of the form
A C
=
B D
(where A, B, C, D are mathematical expressions) we can multiply
both sides by the denominators,
A C
=
B D
A C
BD
= B
D
B
D

AD = BC
20
This cross-multiplication is a short-
cut you may be familiar with
2. algebra 21
An example
2 3
=
x1 x2
2( x 2) = 3( x 1) IMPORTANT: When the denominators
are moved to the top line they must
2x 4 = 3x 3 be placed in brackets. We can then
expand the brackets if desired.
2x 1 = 3x
1 = x
Check: If x = 1 then the LHS (left-hand side) is
2 2 2 2
= = = = 1
x1 1 1 2 2
and similarly, the RHS (right-hand side) of the equation is 1
Example
x x
+ =5
2 3
We cant cross multiply here but we can still get rid of the fractions
by multiplying through by their denominators:
x x
23 + = 2 3 5
2 3
and expand the bracket:
x x
2 3 + 2 3 = 6 5
2 3
3x + 2x = 30
5x = 30
30
x=
5
6 6
Check: 2 + 3 = 3+2 = 5
x=6
Simplifying expressions
As you can see, when re-arranging equations we may need to
tidy up, or simplify the other side of an equation (or expression).
Example
6z [z + 4(z 5) 2(3 4z)] =
6z [z + 4z 20 6 + 8z] =
6z z 4z + 20 + 6 8z =
z + 26 8z =
7z + 26 =
21
22 mathematics
Mathematical modelling
We can use algebra to solve real-world problems. Given a problem

which is stated in words (a word problem) we can convert it into
an algebraic equation which we can hopefully solve to find the
solution to the real-world problem.
Example: A man is 12 times as old as his son. In 9 years time,
the man will be 3 times as old as his son. How old are each now?
Start by identifying the unknowns and assigning a letter to them. Let s
be the sons current age. We could call the fathers current age f
but we already know what it is in terms of s (ie. 12 s).
In 9 years time the sons age will be s + 9 and the fathers 12s + 9.
Now, we are told that
12s + 9 = 3 (s + 9)
12s + 9 = 3s + 27 9s = 18 s=2
Hence, the son is currently 2 years old, and his father 24. In 9 years
time, the son will be 11 and his father 33, which is correct!
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22
2. algebra 23
Algebra
Exercises 2
2.1 Solve the following equations
(a) 4y 3 = 3y + 7
(b) 9x + 8 = 8x 4
2.2 In the formula

5+t
x=
bt
if x = 2 and t = 1, then what is the value of b?

1 2
(a) 3x +1 = x 1
(b) 2y + 1 + 3(2y 4) = 2y 1
2.4 Sallys age is 12 years plus half her age. How old is she?
Problem set 2
2.1 Solve the following equations for the unknown.
x x
(a) x + 3 = 5 (m) =1
2 5
(b) 4y 5 = y + 10 y4 y
(n) + =4
(c) 8x 4 = 16 2 3
2z 2 3
(d) =4 (o) +4 =
z+2 y y
(e) 3 2x = 4 2y 1y
(p) =y
3 2
(f) 7x + 7 = 2( x + 1)
1y
1 (q) y +4 = 7
(g) x = 2 2
2 y
2y 3 6y + 7 (r) (3 y ) = 4 y
(h) = 3
4 3 d1 d
(i) t = 2 2[2t 3(1 t)] (s) =0
2 3
3 4 p
(j) (4a 3) = 2[ a (4a 3)] (t) = 8
2 3
x+3 2 d1 d
(k) = (s) =0
x 5 2 3
1 2 4 p
(l) = (t) = 8
p1 p2 3
23
24 mathematics fundamentals Algebra
1 1
2.2 If 2 cakes are served in slices of of a cake, how many slices
4 8
are served?
2.3 Lian earns $2,000 a fortnight and spends $1200 of it. What
proportion of his salary does he save?
2.4 Nadia buys a packet of 60 samosas for a tea party. She takes
out half for her family. She takes the remaining to the party. On
the way she meets a friend who takes a sixth of the samosas. A
little further on she sees a hungry cat and gives it a fifth of the
samosas she has. Then she suddenly feels hungry and eats a
quarter of the remaining samosas. How many arrive at the tea
party?
2
2.5 I have a naan recipe that calls for 4
cups of flour for 14 serv-
3
ings of naan. I want to make one serving only. How much flour
should I use?
2.6 Suppose Mike travels a certain distance on the first day and
twice the distance on the next day. If the total distance he trav-
elled is 60 km, how far does he travel on the first day?
2.7 Joy, Pam, Sandra and Lilin each make a donation to the Guide
Dogs Association. Sandra gives twice as much as Lilin, Pam
gives three times as much as Sandra and Joy gives four times as
much as Pam. If their total gift is $132, find the amount of Lilins
donation, and hence the amount donated by each.
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24
MATHEMATICS FUNDAMENTALS Simultaneous equations
3. Simultaneous equations
All of the equations that we have solved so far are called linear
equations in one unknown. Now we will look at situations where
we have two linear equations in two unknowns.
Example
David buys movie tickets for seven adults and one child for $92.
Grace buys tickets for ten adults and three children for $144. What
are the costs of the individual tickets?
Let a be the cost of an adult ticket, and c be the cost of a childs

ticket. Then the equations representing the information given are:
7a + c = 92 (1) Note: The (1) and (2) are labels

10a + 3c = 144 (2)
These are called simultaneous equations. (Many real-life problems

contain many equations in many unknowns and require computer
methods to solve them.)
A method of solving simultaneous equations
We manipulate the two equations to produce an equation in only

one unknown to solve.
Once we have solved this equation we substitute this value into
the remaining equation, and then solve for the second unknown.
Here is how the method performs the first step:
7a + c = 92 (1)
10a + 3c = 144 (2)
Start by multiplying both sides of equation (1) by 3 to produce

3c in both equations:
21a + 3c = 276 (3) Note that we have given this equation

a new label
10a + 3c = 144 (2)
Plan is to subtract equation (2) from equation (3) to get rid of c.

If we subtract equation (2) from equation (3) we get
3
21a + 3
c 10a c = 276 144
25
26 mathematics
FUNDAMENTALS Simultaneous equations
In tabular form we can write this as
21a + 3c = 276 (3)

() 10a + 3c = 144 (2) Note that the () is simply an indica-
tor of our intention
11a = 132
Since the terms are lined up we can perform the subtraction (or
addition) term by term.
We now have an easy equation to solve:
132 That is, the cost of an adult ticket is
11a = 132 a= = 12 $12
11
Because we eliminated a variable (in this case, c) this method is

usually called the elimination method.
We still need to determine how much a childs ticket costs.
All we have to do is substitute the value for a into any of the

labelled equations. For example, if we choose equation (1) we get
7a + c = 92
7 12 + c = 92
84 + c = 92
c = 92 84
That is, the cost of an childs ticket is
c=8 $8
We can check our calculations by substituting both values into

the other equation (equation (2)):
Note: The biggest source of errors is
10a + 3c = 10 12 + 3 8 = 120 + 24 = 144 getting subtractions wrong
Example
Solve the pair of simulataneous equations
3x 2y = 10 (1)
2x + 3y = 11 (2)
Notice that if we multiply equation (1) by 2 and equation (2) by
3 then each equation will have the term 6x and we will be able to
subtract one from the other. That is,
() 6x 4y = 20 (3)
6x + 9y = 33 (4)
Recall that the () is simply an indica-
13y = 13 tor of our intention
So y = 1, and substitution of this into say equation (1) gives
3x 2 1 = 10 3x 2 = 10 3x = 12 x=4
Finally, we should check our result by substituting both values into

equation (2):
2 4 + 3 1 = 8 + 3 = 11
26
MATHEMATICS FUNDAMENTALS 3. simultaneous equations
Simultaneous 27
equations
Another example
Solve the pair of simulataneous equations
4x 2y = 24 (1)
3x + 5y = 25 (2)
Notice that if we multiply equation (1) by 3 and equation (2) by 4

then each equation will have the term 12x and we will be able to
add the equations together. That is,
(+) 12x 6y = 72 (3) The (+ is simply an indicator of our

12x + 20y = 100 (4) intention
14y = 28
So y = 2, and substitution of this into say equation (1) gives

We should check our result by substi-
tuting both values into equation (2):
4x 2 (2) = 24 4x + 4 = 24 4x = 20 x=5 3 5 + 5 (2) = 15 10 =
25
The father and son problem revisited

A man is 12 times as old as his son. In 9 years time, he will be
only 3 times as old as his son. How old are each now?
Solution: Let s be the sons current age and m be the mans
current age. We are told that m = 12s and m + 9 = 3(s + 9)
We need to re-arrange these equations slightly:
m = 12s m 12s = 0 (1)
m + 9 = 3( s + 9) m + 9 = 3s + 27 m 3s = 18 (2)
If we subtract equation (1) from equation (2) we get
Hence the son is currently 2 and his
father is currently 24 years old
3s (12s) = 18 0 3s + 12s = 18 9s = 18 s = 2
Substituting this into the equation m = 12s gives m = 12 2 = 24
27
28 mathematics
FUNDAMENTALS Simultaneous equations
An example of modelling with simultaneous equations

Bruce is a building contractor. If he hires 8 bricklayers and 2
roofers, his daily payroll is $960, while 10 bricklayers and 5 roofers
require a daily payroll of $1500. What is the daily wage of a brick-
layer and the daily wage of a roofer?
Solution. Let B be a bricklayers daily wage and R a roofers

daily wage. We are told that
8B + 2R = 960 (1)
Note that we drop the $ signs when
10B + 5R = 1500 (2) doing algebra
We can keep the numbers smaller by replacing eq. (2) by a new

equation, eq. (3), which equals eq. (2) minus eq. (1).That is,
8B + 2R = 960 (1)
2B + 3R = 540 (3)
Multiply equation (3) by 4 to get
() 8B + 2R = 960 (1)
8B + 12R = 2160 (4)
Now it is simply a matter of subtracting equation (1) from equa-
tion (4) to get
1200 1
120 0
10R = 1200 R= = = 120
10 1 1
0
That is, a roofers daily wage is $120.
Substitution of R = 120 into, say equation (1) gives
8B + 2R = 960 8B + 2 120 = 960 8B + 240 = 960
and hence
We can verify these by substitution
720 8 90 into eq. (2) to get 10B + 5R = 1500, so
8B = 960 240 8B = 720 B= = = 90 10 90 + 5 120 = 1500 and hence
8 1
8 900 + 600 = 1500
That is, a bricklayers daily wage is $90.
28
MATHEMATICS FUNDAMENTALS 3. simultaneous equations
Simultaneous 29
equations
Exercises 3
3.1 Solve the pairs of simultaneous equations

2x + 3y = 10 (1)
( a)
x + 2y = 6 (2)
2x + 3y = 19 (1)
(b)
5x 6y = 20 (2)
2x + 4y = 4 (1)
(c)
5x 6y = 2 (2)
3.2 The sum of two numbers is 28 and their difference is 12.

What are the numbers?
Problem set 3
3.1 Solve the following sets of simultaneous equations.
(f) x + 3y = 3
(a) 2x + y = 8 2x + y = 8
3x y = 7
(g) 2x 3y = 5
(b) 2x + 3y = 19 5x + 2y = 16
4x y = 3
(h) x + y = 5
(c) x + 2y = 10 2x + y = 7
x y = 1
(i) 4x + y = 10
(d) 4x y = 0
2x + 3y = 10
2x y = 2
This one is too difficult for a Test
(e) 2x + y = 4 (j) 2x + 3y = 8 question
5x + 2y = 9 3x 2y = 7
3.2 Tickets for an ice-skating display are sold at $5 for adults and
$2 for children. If 101 tickets were sold altogether for a take of
$394, find the number of adults and children who attended.
3.3 An island contains foxes and rabbits. An ecologist counts both
species to study their interaction and how their populations
change over time. Last year she found the total number of foxes
and rabbits was 7,290 and that the fox population was one-eighth
of the rabbit population. How many of each species was present?
3.4 The Roman emperor Augustus was fond of gold and silver
sovereigns. Every gold sovereign was to weigh 50 grams and
every silver one to weigh 40 grams. One year Augustus sus-
pected that the jeweller was cheating him and delivering sub-
weight sovereigns. A consignment contained 30 gold and 20
silver sovereigns weighed 2, 250 grams, and a consignment of
15 gold and 25 silver sovereigns weighed 1, 550 grams? Was the
jeweller cheating Augustus?
29
MATHEMATICS FUNDAMENTALS Quadratics
4. Quadratics
In Chapter 2 we expanded single brackets. For example,
2(2x + 1) = 4x + 2
3( x 2) = 3x + 6
x (2y + 3z) = 2xy + 3xz
Double bracket expansion

To expand ( a + b)(c + d) we use the distributive law:
( a + b)(c + d) = ( a + b)c + ( a + b)d = ac + bc + ad + bd
Example
( x + 1)(2x + 3) = 2x2 + 3x + 2x + 3 = 2x2 + 5x + 3
Note that each term in the first set of brackets gets multiplied by
each term in the second set of brackets, or
FOIL - First (ac), Outside (ad), Inside (bc), Last (bd) As always,watch out for negative signs
Examples
(3x + 1)( x + 4) = 3x2 + 12x + x + 4 = 3x2 + 13x + 4

x2 means x times x
2 2
( x 4)(2x 3) = 2x 3x 8x + 12 = 2x 11x + 12
More than one variable

Expressions can contain any number of variables. For example,
In this expression xy means x times y
3x2 + 4xy + 3y z2
The expanding brackets idea is the same as for one variable.

Examples Note that we dont write expressions
like x3y. We always move the number
(i) x (3y + 2) = 3xy + 2x to the left of a term, and write 3xy
(ii) (2x + 4)(3y 5) = 6xy 10x + 12y 20
(iii) 3(2x + 1) 4(z 5) = 6x + 3 4z + 20 = 6x 4z + 23
31
Quadratic expressions
Double bracket expressions with one variable like this:
( x + 1)(3x + 2)
are an important type of expression called a quadratic.

When expanded out and terms collected they are always of the
form
Note that if a = 0 it isnt a quadratic
ax2 + bx + c (but b or c or both can be zero)
where a, b and c are numbers.

The numbers a, b and c are called the coefficients of the quadratic.
We say that a is the coefficient of x2 , b is the coefficient of x
and c is the constant coefficient.
An exercise
Two of the following are not quadratic expressions. Which two?
(a) x2 + 3x + 4 (f) 2x2 + 5x

(b) x2 12 (g) 3x 8
(c) 3x2 + 4x + 12 (h) 3 5x 2x2
(d) 4 x2 (i) x3 + 2x + 5
(e) x2 + 14 x + 3
5 (j) 27 x2 14 x + 3
5
Answer: ( g) and (i )
Factorization of quadratic expressions
Factorization is the reverse of expanding brackets. In other words,

we start with a quadratic expression, say:
x2 + 5x + 4
and return it to double bracket form (that is, two factors). In this
case the answer turns out to be:
( x + 1)( x + 4)
To see how this is done, well work with quadratic expressions

where a = 1 and note that
( x + A) ( x + B) = x2 + ( A + B) x + AB
So, to find A and B we can see that:
the constant coefficient is the product of A and B and
the coefficient of x is the sum of A and B.
32
MATHEMATICS FUNDAMENTALS 4. quadratics 33
Quadratics
Example: To factorize
x2 + 5x + 6 = ( x + A)( x + B)
we need to find numbers A and B such that
their product is 6
their sum is 5
To do this, look at the factorizations of 6:
23 or 16 or even (2) (3) or (1) (6)
The corresponding sums are
2+3 or 1+6 or even (2) + (3) or (1) + (6)
Only 2 and 3 sum to 5, so
x2 + 5x + 6 = ( x + 2)( x + 3)
Note that the order of brackets doesnt matter, we can write
x2 + 5x + 6 = ( x + 3)( x + 2)
33
34 mathematics
FUNDAMENTALS Quadratics
Example: To factorize
x2 7x + 12
We need the factorizations of 12. These are
1 12 26 34
and
(1) (12) (2) (6) (3) (4)
The corresponding sums are
13 8 7 13 8 7
We want the sum of the two numbers to be 7 so we must choose

3 and 4. That is,
x2 7x + 12 = ( x 3)( x 4)
or
x2 7x + 12 = ( x 4)( x 3)
Example. To factorize x2 3x 4, observe that the factorizations
of 4 are
1 (4) 2 (2) (1) 4
The corresponding sums are 3, 0 and 3 and we want the sum of

the two numbers to be 3 so we must choose 1 and 4. That is, Note: We can always check the answer
by expanding the brackets:
x2 3x 4 = ( x + 1)( x 4) ( x + 1)( x 4) = x2 4x + x 4
= x2 3x 4
Example. To factorize x2 22x + 21 we need to investigate the

factors of 21:
1 21 37 (1) (21) (3) (7)
The factors must add up to 22 so clearly we want 1 and 21 so
x2 22x + 21 = ( x 1)( x 21)
Special cases
Notice the following:
( x + 4)( x 4) = x2
+
4x 16 = x2 16
4x
and
( x 3)( x + 3) = x2 +

3x 9 = x2 9
3x
In general, we have the difference of two squares
( x + A)( x A) = x2 A2
Notice also that
( x + 3)2 = ( x + 3)( x + 3) = x2 + 6x + 9
34
Quadratics
and
( x 3)2 = ( x 3)( x 3) = x2 6x + 9
In general, we have the perfect squares
( x + A)2 = x2 + 2Ax + A2
and
( x A)2 = x2 2Ax + A2
A simple case
When b and c are both positive, we only have to consider factors
which are positive. For example, to factorize
x2 + 13x + 40
we only need to investigate the positive factorizations of 40:
1 40 2 20 4 10 58
The factors must add up to 13 so clearly we want 5 and 8 and so
x2 + 13x + 40 = ( x + 5)( x + 8)
Another example. To factorize x2 + 8x + 16 we investigate the

positive factorizations of 16:
1 16 28 44
The factors must add up to 8 so clearly we want 4 and 4 and so
x2 + 8x + 16 = ( x + 4)( x + 4)
Factorizing general quadratics
The most general form for a quadratic is
ax2 + bx + c a = 0
When a = 1 the factorizing becomes more difficult.

In general, when we see an expression of the form
ax2 + b + c
we want to factorize it by writing it in the form
( Ax + B)(Cx + D )
Note that
A and C multiply to produce the coefficient of x2 , i.e. a
B and D multiply to produce the constant coefficient, i.e. c
35
36 mathematics
Well then need to play around with factorizations of a and c to

get the right value of b.
Example
To factorize 7x2 + 15x + 2 we need to write it in the form
( Ax + B)(Cx + D )
The only factorisation of 7 is 1 7 so
7x2 + 15x + 2 = (7x + B)( x + D )
The numbers B and D must multiply to 2, so one of them must be

1 and the other must be 2. The question is, which one is which?
There are two possibilities:
(7x + 2)( x + 1) or (7x + 1)( x + 2)
Expand each of these to find out which:
(7x + 2)( x + 1) = 7x2 + 7x + 2x + 2 = 7x2 + 9x + 2 X
(7x + 1)( x + 2) = 7x2 + 14x + x + 2 = 7x2 + 15x + 2
36
Quadratics
A way to factorize when a = 1

We associate with a given quadratic ax2 + bx + c the associated
quadratic
x2 + bx + ac
and factorize it instead. If the answer is
x2 + bx + ac = ( x + A)( x + B)
then the factorization for the original quadratic is

B
ax2 + bx + c = ( ax + A)( x + )
a
Example. To factorize 6x2 + 25x + 4 instead factorize
x2 + 25x + 24 = ( x + 1)( x + 24)
(after a bit of work) and the required factorization is
6x2 + 25x + 4 = (6x + 1)( x + 4)
Example. To factorize 2x2 10x + 8 instead factorize
x2 10x + 16 = ( x 2)( x 8)
(after a bit of work) and the required factorization is
2x2 10x + 8 = (2x 2)( x 4)
Check:
(2x 2)( x 4) = 2x2 8x 2x + 8 = 2x2 10x + 8
Our first example To factorize 6x2 4x 10 instead factorize
x2 4x 60 = ( x 10)( x + 6)
(after quite a bit of work) and the required factorization is
6x2 4x 10 = (6x 10)( x + 1)
Check:
(6x 10)( x + 1) = 6x2 + 6x 10x 10 = 6x2 4x 10
Algebraic fractions
Algebraic fractions (or rational expressions ) are fractions which

involve algebraic variables. For example,
x+3 x x+y
x+1 x3 x2 y 4
1 x2 + 1 x 2 + y2 + z2
x+1 x2 4 1+z
Note that we can never divide by 0, so in for example
x
x3
we arent allowed to let x be 3, that is, we must have x = 3.
37
Simplifying rational expressions
Recall that we can cancel out common factors in numerical fractions

(if there are any): For example
56 87 7
= =
24 8 3 3
We can also cancel out common factors in algebraic fractions (if
there are any). To do this, we need to factorize the numerator and
denominator and see if there are any factors common to both.
Example
2 6
To simplify xx2 +xx12 factorise both quadratics:
x2 x 6 ( x 3)( x + 2)
=
x2 + x 12 ( x + 4)( x 3)
( x
3
)( x + 2)
=
( x
( x + 4) 3
)
x+2
=
x+4
In other words, the original expression produces the same number
as (the simpler) expression
x+2
x+4
whatever x value we choose to use (except those that would give 0
in a denominator).
For example, if x = 5 the original expression produces the value
52 5 6 25 5 6
=
52 + 5 12 25 + 5 12
14 72 7
= = =
18 9 2 9
The simplified expression also produces this value
5+2 7
=
5+4 9
Lets try another number. If x = 3 then the original expression
produces the value
32 3 6 0
= (huh?)
32 + 3 12 0
but the simplified expression produces the value
3+2 5
=
3+4 7
Whats gone wrong?
One of the few downsides of the number 0 is that it doesnt
always behave like other numbers. In particular,
Division by 0is not defined
When we have algebraic expressions in the denominator, values
of variables which make the denominator 0 are not allowed. So
x2 x 6 ( x
3
)( x + 2) x+2
= = x = 3, 4
x2 + x 12 ( x
( x + 4) 3
) x+4
38
MATHEMATICS FUNDAMENTALS 4. quadratics 39 Quadratics
Adding and subtracting algebraic fractions
The rules for adding and subtracting algebraic fractions are the
same as for numerical fractions. For example, to calculate
x+1 2x + 7
+
x3 x5
we still need a common denominator:
( x + 1)( x 5) (2x + 7)( x 3)
+
( x 3)( x 5) ( x 5)( x 3)
( x + 1)( x 5) + ( x 3)(2x + 7)
=
( x 3)( x 5)
x2 4x 5 + 2x2 + x 21
= (expand brackets)
x2 8x + 15
3x2 3x 26
= (group like terms)
x2 8x + 15
Multiplying and dividing algebraic fractions
The rules for multiplying and dividing algebraic fractions are the
same as for numerical fractions. For example, multiplication:
x+1 2x + 7 ( x + 1)(2x + 7)
=
x3 x5 ( x 3)( x 5)
and Recall that dividing by a fraction

2x + 1 4x + 5 2x + 1 x6 is equivalent to multiplying by its
=
x+3 x6 x+3 4x + 5 reciprocal
(2x + 1)( x 6) 2x2 11x 6

= =
( x + 3)(4x + 5) 4x2 + 17x + 15
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39
40 mathematics
Exercises 4
4.1 Expand brackets and collect terms in the following
(i) ( x + 2)( x 4) (iii) (2 t)(3t + 4)

(ii) ( x + 5)(3x + 4) (iv) y + 12 y + 14
4.2 Factorise the following quadratic expressions.
(i) x2 3x + 2 (iv) 3x2 + 8x + 4

(ii) x2 x 12
(iii) x2 + 14x + 40 (v) 4x2 5x 6
4.3 Simplify the following rational expressions.
x2 x 2 x2 1
(i ) (ii )
x2 + 2x + 1 x2 2x + 1
Problem set 4
4.1 Expand brackets and collect terms in the following
(a) ( x + 2)( x + 4) (i) x ( x 1)( x + 1)

(b) ( x 1)( x 2) (j) ( x 2y)( x + 2y)
(c) ( x 2)( x + 2) (k) (2x 3y)2
(d) ( x 1)(1 x ) (l) ( x2 1)( x2 + 1)
(e) (2x + 3)( x + 2) (m) ( x2 + x )( x2 x )
(f) ( x + 1)2 (n) ( x + 3y)( x 3y)
(g) (2x 1)2 (o) ( x + 1)( x + 2)( x + 3)
(h) ( x + 2)2 (p) ( x 1)( x + 1)( x + 2)
(a) x2 + 5x + 6 (d) x2 + x 6
(b) x2 5x + 6 (e) x2 2x 15
(c) x2 x 6 (f) x2 100
x2 9 x2 + 5x 14
(i ) (ii )
x+3 x2 + 10x + 21
40
MATHEMATICS FUNDAMENTALS Indices
5. Indices
Indices (sometimes called powers) are notation used to shorthand

expressions involving numbers multiplied by themselves a number
of times.
Some of the powers of 2 are
22 = 2 2 = 4
23 = 2 2 2 = 8
24 = 2 2 2 2 = 16
25 = 2 2 2 = 32

5 terms
For any positive integer (counting number) n,
2n = 2 2 2

n terms
Similarly,
3n = 3 3 3

n terms
For example,
34 = 3 3 3 3 = 81
In general For any number k and positive integer n,
kn = k k k

n terms
We say k to the nth power or k to the n. We can also take powers

of negative numbers, e.g.
(2)4 = (2) (2) (2) (2)

= 4 4
= 16
Note: 24 is read as (2)4 so 24 = (2 2 2 2) = 16.

We can alos take powers of fractions. For example,
3
1 1 1 1 1
= =
2 2 2 2 8
2
3 3 3 9
= =
7 7 7 49
41
42 mathematics
FUNDAMENTALS Indices
Properties of indicies (powers)
There are some shortcuts (or laws of indicies) we can develop.

Observe that
23 25 = 2 2 2 2 2 2 2 2

3 terms 5 terms
= 2 2 2 2
2 2 2 2
3 + 5 = 8 terms
8
=2
We say that 2 cubed times 2 to the 5th is 2 to the 8th .

In general, the rule is clearly
km kn = k k k k k k

m terms n terms
= k k k k k
m + n terms
That is,
k m k n = k m+n
Here are another two rules. Observe that
75 77777
3
=
7 777
7 7 7 7 7
= = 7 7 = 72
7 7 7
We say that 7 to the 5th divided by 7 cubed is 7 squared.

In general, the rule is
km
= k mn
kn
42
5. indices 43
Observe that
(3 5)3 = (3 5) (3 5) (3 5)
= 353535
= 3
33555

= 33 53

( jk)n = jn kn
A rule for fractions
3
2 2 2 2
=
7 7 7 7
222 23
= = 3
777 7
In general, the rule is n
j jn
= n
k k
A rule for powers of powers
(54 )3 = 54 54 54
= 5 5
5 5 5 5
5 5 5 5
5 5
4 terms 4 terms 4 terms
5 5 5 5 5 5 = 512
= 5 5 5 5 5 5
34=12 terms
( k m )n = k mn
The laws of indicies so far
We so far have five Laws of indicies. They are
km
k m k n = k n+m = k mn
kn
n
j jn
n
( jk) = j kn n
= n (km )n = kmn
k k
Some commonly occuring powers
22 = 4 23 = 8 24 = 16 25 = 32 26 = 64
and
32 = 9 33 = 27 34 = 81
Some commonly occuring squares (that is, to the power of 2)
22 = 4 32 = 9 42 = 16 52 = 25 62 = 36
72 = 49 82 = 64 92 = 81 102 = 100
The powers of 10
102 = 100 103 = 1, 000 104 = 10, 000
105 = 100, 000 106 = 1, 000, 000 etc.
43
The laws of indicies used in algebraic calculations
The laws can be very useful when working with calculations and
expressions. For example,
169 (24 )9 236

33
= 33
= 33 = 23633 = 23 = 2 2 2 = 8
2 2 2
y7
= y 73 = y 4
y3
( z 4 )2 = z 42 = z 8
(5x3 )2 = 52 ( x3 )2 = 52 x32 = 25x6
Simplifying algebraic expressions
(3y2 )(4y3 ) = 3y2 4y3 = 12y2 y3 = 12y2+3 = 12y5
12z5 43z5 z5
= = 4 = 4z52 = 4z3
3z2 3z2 z2
x5 x5 x5
= = = x 54 = x 1
x2 x2 x 2+2 x4
The power of 1
What does x1 represent? We can see the answer by cancelling
common factors:
x5 xxxxx
=
x4 xxxx
x x x x x
= =x
x x x x
So, x1 = x and we have another index law:
k1 = k
Examples:
51 = 5 (3)1 = 3
1 1
1 1 2 2
= =
2 2 7 7
The power of 0
Similarly, we can extend indices to powers other than counting
numbers. For example,
32 9
= =1
32 9
km
The index rule n = kmn gives us
k
32
= 322 = 30
32
Hence 30 = 1. In fact:
k0 = 1 for any value of k (except 0)
44
MATHEMATICS FUNDAMENTALS 5. indices 45
Indices
Examples:
0
0 2
5 =1 =1
7
0
0 987, 654, 321
140, 291, 090 = 1 =1
123, 456, 789
Negative powers
What about this situation:
32 3 3 11 1 1
4
= = = = 2
3 3 3 3 3 1 3 3 1 3 3 3
km
The index rule = kmn gives us
kn
32
= 324 = 32
34
1
Hence 32 = and in fact
32
1
k n = for any values of k, n
kn
Examples
1 1 1
51 = 103 = =
5 103 1000
1 1 1
(4)2 = = =
(4)2 (4) (4) 16
Note that
a 1 1 a b b
= 1 = 1 = 1 = This is another example of a reciprocal
b a b a a
b
45
46 mathematics
In all, we have nine Laws of Indices
km
k m k n = k n+m = k mn
kn
n
j jn
( jk)n = jn kn = n (km )n = kmn
k k
1
1 j k
k n = = k1 = k k0 = 1
kn k j
1
How do we interpret powers that are fractions, eg. 4 2 ? Using the
index law km kn = kn+m we can say
1 1 1 1
42 42 = 42+2
= 41
=4
1
That is, 4 2 is a number which, when multiplied by itself, gives 4.
1
Clearly, an answer is 2 because 2 2 = 4. We refer to 4 2 as the
square root of 4 and write

4
Note that (2) (2) = 4 as well so we will define the square root
(that is, power of 12 ) to represent only the positive answer. That is,
1
42 = 4 = 2
Some other examples
1
92 = 3 because 3 3 = 9

16 = 4 because 4 4 = 16
1
25 2 = 5 because 5 5 = 25

36 = 6 because 6 6 = 36
Note that

0=0 and 1=1
Unfortunately, most square roots arent whole numbers. For

example,

2 = 1.414 5 = 2.236
Mathematicians leave such answers in square-root form. Other

examples are

11 17 24 50
The index laws apply to fractional powers as well. For example,

1 1
4 2 42 2 1 1 1
= 1 = and = =
9 9 2 3 36 36 6
46
Indices
1
What about 8 3 ? Notice that
1 1 1 1 1 1
83 83 83 = 83+3+3
= 81
=8
1
That is, 8 3 is a number which, when multiplied by itself three
times, gives 8.
1
Clearly, an answer is 2 and 8 3 is an alternative way of express-
ing the cube root of 8:
1 3
8 3 = 8.
In general,
1
n
kn = k (the nth root of k).
Example
1
54 = 625
4
625 4 = 625 = 5 because That is, the 4th root of 625 is 5
Note that this time, (2) (2) (2) = 8 so there is only

1
one answer for 8 3 (that is, 2). In fact, we have
1
(8) 3 = 3 8 = 2
1
Note: On the other hand, (4) 2 = 4 does not exist. That is, there is no such number
Even and odd numbers
We refer to 2, 4, 6, 8, as the even numbers.

We refer to 1, 3, 5, 7, as the odd numbers.
So you cant take the square root of a
The situation is that if n is odd, then we can take the nth root of a negative number
negative number, but if n is an even, then we cannot.
Mathematicians had to invent a new type of number, called the

imaginary number i = 1 to handle this situation in practical
problems.
2
What if the numerator isnt 1. For example, what is 8 3 ?
The laws of Indices tell us. Recall that 8 = 23 . Now
2
2
3
8 3 = 23
3 2
= 2 3 using (km )n = kmn
= 22
=4
Examples
3 2 3
= 4 2 = 43 = 4 4 4 = 64
3 2
16 2 = 42

4 3 4
43 3
27 = 33 = 3
3
1 1 1
= 34 = = =
34 3333 81
47
48 mathematics fundamentals Indices
Surds
Surds

Recall that instead of 9 we can write 3, and 16 we can write 4.
Recall
Squarethat instead
roots, cube ofroots,9 etc.
we can
thatwrite
cant 3,
beand
written16 more
we can write for
simply, 4.
Square cuberoots, etc. that cant be written more simply, for
roots,
example 2 and 5 are called surds Thanks to fractional powers
example
and 2 andwe5have
index laws, are called surds
two very Thanks
useful to fractional
formulae powers
for simplifying
and index laws,
expressions we have
involving two very useful formulae for simplifying
surds:
expressions involving surds:

a a = a ab = ab
a a=a a b = ab
from
from k n k m = k n+m jn kn = ( jk)n
k n k m = k n+m jn kn = ( jk)n
Examples:
Examples:

77 = 7 35 = 15
7 7=7 3 5 = 15
Note: Similar expressions exist for other roots. For example
Note: Similar expressions exist for other roots. For example
3

3
3
( a )3 = a 3
a 3
b= 3
ab
3 3 3
( a) = a a b = ab
Simplifying surd expressions

Simplifying surd expressions
Expressions involving surds can be simplified using the usual laws
Expressions involving
of algebra, and surds
the rules just can be simplified
mentioned. Recallusing
that the usual laws
of algebra, and the rules just mentioned. Recall that
a(b + c) = ab + ac ( a + b)(c + d) = ac + ad + bc + bd
a(b + c) = ab + ac ( a + b)(c + d) = ac + ad + bc + bd
and
and
a a = a ab = ab
a a=a a b = ab
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48
Indices
Examples

(1 + 7)(1 7) = 1 7 + 7 7 = 1 7 = 6

( 3 + 7)( 3 + 7) = 3 + 3 7 + 7 3 + 7

= 3 + 21 + 21 + 7 = 10 + 2 21
We can sometimes simplify a surd itself. Especially if the number
under the root sign can be factorized. For example,

20 = 45 = 4 5 = 2 5

75 = 3 25 = 3 25 = 35 = 5 3
Exercises 5
5.1 Simplify
(i) 24 29
311
(ii) 35
(iii) (5 7)3
5.2 Remove the brackets in

2
(i) 73
(ii) (54 )7
5.3 Use the Laws of Indices to simplify the following

2
(i) x 3 x5 (iv) 3x 2 x 3 y 3
(ii) (5x3 )2 x6
5y2 z
(iii) (5 7)3 (v) (2y3 z5 )2
5.4 Simplify the following surds

(i) 48 (iii) (3 + 5)(2 5)

(ii) 3(5 + 2) (iv) (2 5 1)(2 5 + 1)
Problem set 5
5.1 Use index laws to simplify each of the following expressions

3
(a) x3 x4 x (f) 4x2
(b) x2 y3 x
(c) ( xy)2 (3x2 z)3
(g)
a2 b5 (2x )2
(d)
ab 5
2x4 y2 9x4
(e) (h)
6x2 y2 (3x )2
49
50 mathematics
5.2 Express the following numbers without indices
24 35
(a) (e)
42 272
512 (f) (2)4
(b)
255 (g) (3)3
(c) 41 (h) 06
(d) 102 (i) (6)0
(a) y2 y3 6a3 b2 8a3 b

(g)
(b) x 3 x3 a2 b 1 b
(c) ( a3 b4 )( a1 b) (2ab)3
(h)
(2a1 b)2
4g3 h5
(d) 2 1 0
12g5 h3 3x y
(i)
(e) (9x 1 y2 )2 2
1
3ab2 x6
3 2
4x
(f) (j) 3
c 12x
x
(a) 3x 2x 31 x xy yx
(c)
x y
22 (9 a ) b
(b) (d)
2x (32a )b
5.5 Evaluate each of the following expressions.
1 1 3
(a) 8 3 (f) 64 3 (k) 49 2
1 1 5
(b) 32 5 (g) 10000 4 (l) 64 6
1 1 1
(c) 81 2 (h) 144 2 (m) 64 6
1 2 3
(d) 125 3 (i) 32 5 (n) 81 4
1 3 2
(e) 81 4 (j) 36 2 (o) (27) 3
2 4
(a) 27 3 (i) 27 3
2
(b) 27 3 (j) 64 6
5
7
(c) 4 2
3
1 4 2
(d) 16 2 (k)
3 9
(e) 100 2
3 3
(f) 16 4 4 2
(l)
1 9
(g) 169 2
4 1
(h) 32 5 (m) 81 2
50
MATHEMATICS FUNDAMENTALS 5. indices 51 Indices
5.7 Simplify each of the following expressions
1 2 2 5
(a) 3x 2 4x 3 x 3 y 2
2 1
(e) 5
(b) 5x 5 6x 10 x 3 y 1
1 1 3
(c) (4x 2 )3 (9x 3 ) 2 3
x 7 y 5
4
1 5 2 (f)
(d) (64x2 ) 6 (32x 2 ) 5 x 27 y5
3
1
(a) 3x 2 2x 2
1
x 3 y5
(g)
(b) (8x3 y5 ) 23 x 2 y 7
1 1
(c) ( x 5 )3 5
(h) 7x 4 2x 2

3x 27 1
(i) (9x ) 2 (27x ) 3
1
(d)
x 12 1 1
1
(e) 3x 2 4x 3
2 (j) (9x ) 2 (27x ) 3
1 3
(f) ( x2 7) 7 (k) 3x 3 2x 4

(a) 12 (g) (1 2) 3

(b) 72 (h) (7 + 3 5) 3

(c) 150 (i) (6 2 5)(6 + 2 5)

(d) 3
96 (j) (3 + 7)2

(e) 2(3 2) (k) (4 3 1)(2 5 + 1)

(f) 5(2 3 + 2) (l) ( 5 3)(2 5 + 4 3)
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51
MATHEMATICS FUNDAMENTALS Re-arranging formulae
6. Re-arranging formulae
In Chapter 2, we saw how to solve equations with a single variable.

If there are several variables in the equation we can still isolate one
of them by the same methods.
Example. Solve 3P = 5v + 12 for v. We get
3P 12
3P 12 = 5v =v We call this process isolating v
5
H+4 3
Example. Solve = for H. Cross-multiply to get
t b+2
( H + 4)(b + 2) = 3t H (b + 2) + 4(b + 2) = 3t
H (b + 2) = 3t 4(b + 2)
and so
3t 4(b + 2)
H= We have isolated H
b+2
Equations involving indices
We can use index laws to isolate variables in more sophisticated

equations.
Example. To solve A = b5 for b, we need to undo the operation
raise to the power of 5.
This can be done by raising both sides to the power of 15 :
1
1
5
( A ) 5 = b5
1 1
A 5 = b 5 5 by index law (km )n = kmn
1
A = b1
5
1
A5 = b
More examples
1
Z = u3 u = Z3
1
K = p4 p = K4
5 2
h = w2 w = h5
1
R = g 7 g = R 7
53
54 mathematics
FUNDAMENTALS Re-arranging formulae
To solve c = ax2 + b for x, we proceed as follows:
c = ax2 + b c b = ax2
1
cb 2 cb 2 cb
=x = x or =x
a a a
4r3 c Note that we could also solve for b. We
To solve P = b2
for r, we proceed as follows: would get

4r3 c 4r3 c
P= Pb2 = 4r3 c b=
b2 P
13
Pb2 Pb2
= r3 =r
4c 4c
Exercises 6
6.1
1 2

(i ) Solve z3 p 4 = a4 w 3 for w (ii ) Solve y = x3 c for x
12s4 y3 1 1
(iii ) Solve m = for s (iv) Solve f 2 h 3 = vm 4 for h
7p2
Problem set 6
6.1 In each of the following scientific formulae, isolate the variable

indicated
(a) Temperature scales

5
Solve C= ( F 32) for F
9
(b) Volume of a sphere
4 3
Solve V= r for r
3
(c) Ideal gas law
Solve PV = nRT for T
(d) Newtons law of gravity

GMm
Solve F= for r
r2
(e) Brightness of a star
L
Solve b= for d
4d2
(f) Buckling of a column
1
2 E 3
Solve L = kd 3 for p
p
(g) Wavelength of gas particles

h2
Solve L= for m
2mkT
54
MATHEMATICS FUNDAMENTALS 6. re-arranging formulae
Re-arranging 55
formulae
6.2 In each of the following business formulae, isolate the variable

indicated
(a) Linear demand equation
Solve Q = a bP for P
(b) Compound interest
Solve A = P (1 + r ) n for r
(c) Money supply
Solve MV = PY for Y
(d) Relationship between marginal revenue and price elasticity

1
Solve R = P 1 for e
e
(e) Gravity model of trade
SD
Solve F =K for S
d2
55
MATHEMATICS FUNDAMENTALS Exponentials and logarithms
7. Exponentials and logarithms
Exponential equations
Question: 2 to which power gives us 8?. The answer is 3 because we

know that 2 2 2 = 23 = 8. We have just solved the equation
2y = 8
This is a new sort of equation because the variable to be isolated is

a power or exponent. . We call it an exponential equation.
Anything that grows (or shrinks) as a percentage of its size follows
an exponential equation and they arise in many areas of Science
(eg. population growth, radioactive decay) and Economics (eg.
compound interest, economic growth measurement).
Logarithms
To help in solving exponential equations, we can rewrite them as

logarithmic equations:
23 = 8 is the same as 3 = log2 8
The notation log2 8 is read logarithm to the base 2 of 8 and

poses the question what power of the base ( 2 ) gives us 8?.
In general, we say:
The notation logb x is read loga-
by = x is the same as y = logb x rithm to the base b of x and poses the
question what power of b gives us x?
Notice that the exponent (y) becomes an ordinary variable when
the equation is changed to logarithmic form. This will help us to
solve exponential equations.
Some examples of logarithms
log9 81 = 2 because 92 = 81
log3 27 = 3 because 33 = 27
log10 1000 = 3 because 103 = 1000
log2 32 = 5 because 25 = 32
log10 1, 000, 000 = 6 because 106 = 1, 000, 000
log3 81 = 4 because 34 = 81
57
58 mathematics
FUNDAMENTALS Exponentials and logarithms
A special case: logb b
log3 3 = 1 because 31 = 3
Clearly, this is true for any base, so logb b = 1
Another special case: logb 1
Clearly, this is also true for any base, so logb 1 = 0
Negative numbers
Negative bases create serious mathe-
We only consider positive bases for logarithms. We also dont matical complications
define logarithms with a base of 1.
Note that we cant take the log of a negative number. For ex-
ample, log2 (4) makes no sense because 2 raised to any power
will always give a positive number. However, the log of a positive
number may be negative. Recall the index law
1
k n =
kn
We have

1 1 1
log2 = 3 because 23 = =
8 23 8
Examples of applications of logarithms
Example: Binary Search in Computer Science is used to track down

specific values in an ordered list. Think of it this way:
I have chosen a number between 1 and 16. When you guess I will say
higher, lower or correct.
The maximum number of guesses required to identify the number

is given by
log2 16 = 4
In general, the maximum number of guesses is
log2 N
where N is the number of items in the list.

Example: pH in Chemistry measures the acidity or alkalinity of
an aqueous (water based) solution. It is related logarithmically to
the amount of activity of the hydrogen ion:
pH = log10 ( H + )
A pH of 7 is neutral (eg. pure water), lower values are acidic and

higher values are alkaline.
58
MATHEMATICS FUNDAMENTALS 7. exponentials and logarithms
Exponentials 59
and logarithms
Example: In many real-life situations (eg. newspaper articles,

balance sheets in accounting) numbers starting with 1 occur more
often than numbers starting with 2, etc.
Benfords Law gives the frequency of each starting digit:

d+1
P = log10
d
where P is the frequency and d is the starting digit (1, 2, 3 . . . , 9).

Around 30% of such numbers start with a 1, 18% with a 2, down
to just 5% starting with a 9.
There are many other applications, and in a wide diversity of
fields, for example,
Earthquakes (the Richter scale)
Signal processing (signal-noise ratio)
Psychology (Hicks law)
Physics (entropy)
Music (intervals)
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60 mathematics
FUNDAMENTALS Exponentials and logarithms
The Laws of Logarithms
Logarithms are obviously related to indices, and there are closely

related laws of logarithms. For all suitable values of x, y and b we
have
logb b = 1 (related to k1 = k)
logb 1 = 0 (related to k0 = 1)
logb ( xy) = logb x + logb y (related to km kn = kn+m )

x km
logb = logb x logb y (related to = k mn )
y kn
logb ( xr ) = r logb x (related to (km )n = kmn )
In any given investigation we are usually using the same base b

(for example, if we are discussing Binary Searches) and we can omit
it from the laws of logarithms and write
log b = 1
log 1 = 0
log( xy) = log x + log y

x
log = log x log y
y
log ( xr ) = r log x
Recently, we have used the laws of indices to simplify expressions

involving indices (powers).
Similarly, we can use the laws of logarithms to simplify expres-
sions involving logarithms.
Examples of simplifying equations involving logarithms
log2 4 + log2 16 = log2 (4 16) = log2 64
55
log3 55 log3 11 = log3 = log3 5
11
2 log7 6 = log7 62 = log7 36

log2 x2 + log2 x3 = log2 x2 x3 = log2 x5
1 1 3
3 log7 y 2 = log7 y3 2 = log7 y 2
60
Exponentials 61
and logarithms
We can use algebra and log laws to express logarithms as alge-

braic expressions based on simpler logarithms.
Example: Let log 2 = x and log 3 = y. Express each of the
following in terms of x and y:

9 1
log 12 log 18 log log
2 6
To do this we need to write the numbers in terms of 2 and 3.
Firstly, we have 12 = 3 4 = 3 22 so
log 12 = log(3 22 ) = log 3 + log 22 = log 3 + 2 log 2 = y + 2x
Similarly, we have 18 = 2 9 = 2 32 and so
log 18 = log(2 32 ) = log 2 + log 32
= log 2 + 2 log 3 = x + 2y
9 32
Also, 2 = 2 and so
2
9 3
log = log = log 32 log 2
2 2
= 2 log 3 log 2 = 2y x
1 1
Also, 6 = 32 and so

1 1
log = log = log 1 log(3 2)
6 32
= log 1 (log 3 + log 2) = log 1 log 3 log 2
= y x since log 1 = 0
Note that these are true no matter which base b we use!
Exercises 7
7.1 Convert the following exponential equations into logarithmic

form
(i ) 26 = 64 (ii ) 121 = 112
4
(iii ) 27 3 = 81 (iv) 12 = 144
7.2 Convert the following logarithmic equations into exponential

form
(i ) log5 125 = 3 (ii ) log4 64 = 3
(iii ) log7 x = 3 (iv) log4 8 = x
7.3 Use the laws of logarithms to simplify the following
(i) log2 9 + log2 3

(ii) log3 7 log3 28
(iii) log8 16 1
7.4 Let log4 3 = s and log4 5 = t. Express in terms of s and t:

5
(i ) log4 75 (ii ) log4
9
61
62 mathematics fundamentals Exponentials and logarithms
Problem set 7
7.1 Write the following logarithmic equations in exponential form.
(a) log10 100 = 2 (k) log2 1 = 0

(b) log10 1000 = 3 (l) log2 n = 1024
(c) log10 1 = 0 (m) log2 64 = 6
(d) log3 9 = 2 (n) log9 81 = 2

(e) log3 1 = 0 1 1
(o) log9 =
3 2
1
(f) log3 = 2 1
9 (p) log9 3 =
2
(g) logn 256 = 2
(q) logn 2 = 2
(h) logn 14 = 3 1
(r) log10 x =
(i) logn 62 = 4 2
(s) log5 100 = x
1
(j) log2 = 3 (t) logx 12 = 2
8
7.2 Write the following exponential equations in logarithmic form.
(a) 25 = 32 (g) 161 = 16 (m) 104 = 10000

1 1
(b) 23 = 8 (h) 81 2 = (n) 3x = 81
9
(c) 20 =1 1 (o) 25x = 5
(i) 32 =
9
(d) 82
= 64 1 (p) x5 = 7
1 (j) 33 =
(e) 81 = 27 2
8 (k) 35 = 243 (q) 3x = 81
1 1
(f) 8 3 = 2 (l) 25 4 = 5 (r) 203x+1 = 3
62
Exponentials 63
and logarithms
7.3 Simplify the following using logarithm laws.
(a) log 5 + log 2 (j) log5 20 + log5 4

2
(b) log 8 log 2 x
(k) log + log x 2 y3
y
(c) 3 log 2 2 log 3 1
(l) 2 log x 2 y2 log y2 x 1
(d) log 4 log 20
(e) log 12 log 3 log10 27
(m)
log10 9
(f) log 16 log 4
(n) log2 8 + log2 18 log2 3
(g) log2 12 log2 3
(o) log2 10 log2 5
(h) log10 25 + log10 4
(p) log xy + 12 log( xy)
(i) log10 25 log10 250 log( xy)
7.4 Let x = log 3 and y = log 4. Express the following in terms of x

and y.

(a) log 12 3
(e) log
(b) log 36 2

3 3
(c) log (f) log
4 8
(d) log 6 (g) log 144
7.5 Let log5 2 = x and log5 3 = y. Express the following in terms of

x and y.

(a) log5 6 3
(e) log5
(b) log5 12 2

3 3
(c) log5 (f) log5
10 8
(d) log5 60 (g) log5 30
63
EQUATIONS INVOLVING LOGARITHMS
MATHEMATICS FUNDAMENTALS AND/OR EXPONENTIALS
8. Equations involving logarithms and/or exponentials
In Chapter 2, we discussed how to solve equations which contain a

single variable, and in Chapter 6 we extended these ideas to cover
multiple variables and variables raised to powers.
Now we look at equations with variables as powers (these are
called exponential equations), e.g.
5(2x ) = 20
and equations involving logarithms, e.g.
log5 x log5 8 = 2
To solve these we need to be familiar with:
The laws of logarithms and the laws of indices/exponentials
Switching between exponential form and logarithm form:

y
b
= x logb x = y

exponential form logarithm form
Equations involving logarithms
To solve these we can collect log terms and then switch to the
exponential form.
A simple example: Solve log2 (3x ) = 5. Switch to exponential
form:
32
3x = 25 3x = 32 x=
3
A not so simple example: Solve log4 x log4 ( x 1) = 2. Use
the laws of logs to combine the terms on the left-hand side:

x
log4 =2
x1
Now switch to exponential form:

x x
= 42 = 16
x1 x1
Now cross-multiply to get
16
x = 16( x 1) x = 16x 16 16 = 15x x=
15
65
66 mathematics
FUNDAMENTALS AND/OR EXPONENTIALS
Equations involving exponentials

To solve, make the base on each side of the equation the same.
A simple example: Solve 4x = 16. Recall that 16 = 42 .Then
4 x = 42
and since the base is the same on both sides of the equation then
the indices must also be same. Hence x = 2.
A not so simple example: Solve 2x 23x1 = 8. Use the laws of
indices to combine the terms on the left-hand side:
2x+3x1 = 8
24x1 = 23 (make the bases the same)
3
4x 1 = 3 (equate the powers)
4x = 4 x=1
53x
A harder example: Solve = 10 Use the laws of indices to
52x1
simplify the left-hand side:
53x(2x1) = 10 5x+1 = 10
However, 10 cant be expressed as an easy power of 5 so switch to

logarithm form:
x + 1 = log5 10 x = log5 10 1
This is a mathematically exact answer but hard to visualise. We

know that log5 5 = 1 and log5 25 = 2 so log5 10 lies somewhere
on the numberline between 1 and 2. This means that the numerical
answer is between 0 and 1. A calculator tells us that it is
x 0.43 The symbol means is approxi-

mately equal to
Exponential equations These are equations of the form
y = a(b x )
for some value a and some positive base value b. We usually omit
the brackets and write y = ab x .
Example: The simplest example is
y = 2x
Exponentials grow very quickly. Lets look at y = 2x :

x 1 2 3 4 5 6 7 8 9 10 11 12
y 2 4 8 16 32 64 128 256 515 1,024 2,048 4,096
Exponentials with larger bases grow even faster. Example, y = 5x

x 1 2 3 4 5 6
y 5 25 125 625 3,125 15,725
66
8. equations involving logarithms and/or exponentials 67
AND/OR EXPONENTIALS
Exponential models
In general, exponentials describe situations where every one unit

increase in x changes y by a proportion of its current value. Appli-
cations of exponentials are many and varied:
animal populations: the more animals there are in a population,

the faster the population size increases.
viral infections: the more people who are infected, the greater
the spread of the infection will be.
compound interest: the more money you have in an account, the

greater the growth of your capital.
Since we are usually interested in quantities which change expo-

nentially over time, well replace x with t (for time) and y with a
suitable variable, e.g. P for population size at time t:
P = abt
These are all examples of exponential growth.
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67
68 mathematics
Initial population size

The initial population size is the size of the population when we
started our investigation. We call this time t = 0. We usually label
the initial population size as P0 . If we put t = 0 and P = P0 in the
equation P = abt we get
P0 = ab0 P0 = ab0 P0 = a1 P0 = a
In other words, a is just the initial population size. For this reason
we usually replace a with P0 and write
P = P0 bt
The bases b tells us how quickly the population grows. The

larger the value of b, the faster the population grows.
Example: Bacterial growth. If we start with N0 = 8 bacteria and
they double each generation, then after n generations well have
N = 8 (2n )
Example: Compound interest. If we invest A0 = $1, 000 at a 5%

interest rate compounded annually, then after t years well have

5 t
A = 1000 1 +
100
A more involved example: A bacterial colony numbered 3 mil-
lion at the start of a study (time t = 0), and after two weeks num-
bers 12 million. Assuming exponential growth, predict when the
bacteria population will exceed 150 million.
Using the units millions of bacteria for the colony size P and weeks
for time t, we have the exponential model
P = 3bt
We know that at t = 2 weeks we have 12 million bacteria, so
12 = 3b2
and from this we can find the growth rate of the bacteria:
12
12 = 3b2 = b2 b2 = 4 b=2 Note that b = 2 is also a mathemati-
3 cal solution but it makes no biological
sense, so we ignore it
Hence the growth model is

P = 3 2t
Note that we need the brackets here because without them it would
be 32 to the power of t. To find the time t when the population
reaches 150 million bacteria, we have to solve the exponential equa-
tion

150 = 3 2t
50 = 2t
log2 50 = t
68
MATHEMATICS FUNDAMENTALS 8. equations involving logarithms and/or exponentials
AND/OR 69
EXPONENTIALS
A calculator gives us the value t = 5.64 weeks, that is, about 5

weeks and 4 or 5 days, after the beginning of the study.
Exponential decay
We call exponential functions where the population gets smaller
over time exponential decay. For every one unit increase in t, the
population reduces by a proportion of the current value of P.
In such situations the base b is a fraction between 0 and 1. Appli-
cations of exponential decay models are also varied:
population decay (extinction): as the population gets smaller,

fewer animals die.
radioactive decay: the half life of a radioactive substance is the

time it takes the radioactivity to halve. It takes another half-life
to halve again (ie. reduce to one quarter of the original level).
depreciation of assets: a new car reduces in value rapidly just

after being bought but doesnt change much after a few years.
Example: A pigeon population initially has 1000 members, but

two years later it has declined to 250 members. Assuming the expo-
nential model P = P0 bt with t measured in years, after how many
more years will the pigeon population reach 100?
Since the initial population was 1000, then P0 = 1000 and so
P = 1000bt
After two years it has declined to P = 250, that is,

2 2 250 1 1 1 1
250 = 1000b b = = b= = =
1000 4 4 4 2
We wish to know when P = 100. That is,

t t
1 1 100 1 1
100 = 1000 = = t = log 1
2 2 1000 10 2 10
A calculator gives us the value t = 3.32 years, that is, about 3 years
and 4 months, after the beginning of the study.
A very special growth rate
To get around using logarithms with different bases for every
different growth rate, we tend to use an agreed common base (and
adjust the power to get different growth rates instead).
For many good mathematical reasons, this common base number
is 2.718281828459045 . . . It arises naturally in many varied branches
of Mathematics. This number is so widely used (and annoying to
write!) that it gets its own letter:
e = 2.718281828459045 . . .
where the letter e is in honour of Swiss mathematician Leonhard

Euler (1707-1783).
69
70 mathematics
The natural logarithm
We use logarithms with base e so much that they get their own
special notation:
loge is equivalent to ln
The l stands for logarithm and the n stands for natural

(since the base e occurs naturally in so many settings). An expres-
sion like
loge 250 ln 250
is equivalent to
loge 15 ln 15
Some examples of switching between exp and log forms:
ex = 4 x = ln 4
ln y = 7 y = e7
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AND/OR 71
EXPONENTIALS
The index and log laws for e and ln

The laws of indices and laws of logarithms still hold for e and ln:
es
es et = es+t = est
et
1
(es )t = est es = s
e
e0 = 1 e1 = e
s
ln(st) = ln s + ln t ln = ln s ln t
t
ln (sr ) = r ln s
ln 1 = 0 ln e = 1
Solving equations The method of solving exponential and loga-
rithmic equations is the same, except that now we take logarithms
with the base e (that is, natural logarithms).
Examples
0
(i ) e2x = 1 2x = ln 1 2x = 0 x= =0
2
(ii ) ln x + 2 ln 2 = ln 16 ln x + ln 22 = ln 16
ln x + ln 4 = ln 16 ln(4x ) = ln 16 4x = 16 x = 4
e3x
(iii ) = e2 e3x4 = e2 3x 4 = 2
e4
3x = 6 x = 2
Exponential models with base e
In this new format, our growth model is
P = P0 ekt
where k is used to set the required growth rate.

The bacterial colony example again: The model is now
P = 3ekt
After two weeks its size is 12 million so

ln 4
12 = 3e2k 4 = e2k ln 4 = 2k k=
2
In this case we can simplify this slightly:
A tip: dont replace k with this expres-
ln(22 ) 2 ln 2
k = = k = ln 2 sion in the growth model until it is
2 2 needed
To find the time t when the population reaches 150 million bacte-
ria we have to solve the exponential equation
150 = 3ekt
150
ekt = = 50
3
kt = ln 50
ln 50
t=
k
71
72 mathematics
and since k = ln 2 then

ln 50
t=
ln 2
Recall that our previous answer was
t = log2 50
A calculator says that they are numerically the same answer:
t = 5.64
Which model to use?
If we wish to describe exponential growth we use
P = P0 bt b>1
or
P = P0 ekt k>0
If we wish to describe exponential decay we use
P = P0 bt b fraction between 0 and 1
or Note that k is always positive. Its the
P = P0 ekt k>0 sign in the exponent that tells us if its
growth or decay
Another pigeon problem: A population of pigeons was mea-

sured to be 500 in 2012, but had reduced to 300 by 2015. Assuming
the population law P = P0 ekt , when will the population reach 100?
Let t be time in years, starting at t = 0 in 2012. Now P0 = 500
and so
P = 500ekt
Also, 2015 represents t = 3, so we have P = 300 and hence
300 3
300 = 500e3k e3k = e3k =
500 5
Now take logs of both sides (that is, convert to log form):
ln( 3 )
3k = ln 35 k = 35
but its best to leave it in mathematical
A calculator tells us that this is about k = 0.17 (that is, the popula- form
tion is decining at about 17% per year),
To find the time t when the population reaches 100 we solve
1
100 = 500ekt = ekt
5
and hence

1 1
kt = ln 5 t = ln 5 (k)
ln( 35 )
From before we have that k = 3 and hence
ln(1 ln( 3 ln( 15 )
5 ) 5 ) 3
t = 1 3 = 1
ln( 35 )
That is,
3 ln 15
t=
ln 35
A calculator tells us that this is about 9.5 years, so roughly the year
2021. It depends what month the measurements were done.
72
AND/OR 73
EXPONENTIALS
Exercises 8
1
(i ) log5 x + log5 8 = 2 (ii ) 22x+1 =
32
2
(iii ) 32x1 = 27
8.2 Simplify the following
16e5
(i ) ln 27 ln 3 + ln 4 (ii )
2e3
e5x+1
(i ) ln(2x ) + ln 6 = ln(4x ) (ii ) = e4
e3x2
8.4 A fish population numbered 20 at the start of a study (start

time t = 0) and after five years numbered 480. Assuming that
the population law P = P0 ekt predict when the population will
exceed 1, 000.
8.5 An industrial machine which cost $7000 in 2010 is valued at

$1000 in 2015. Assuming the decay law P = P0 ekt , when will
the machine be worth $700?
73
74 mathematics
Problem set 8
8.1 Solve the following exponential equations for the unknown.
(a) 2x = 64 1 (j) 3x+1 = 10

(f) 10x+1 =
100
(b) 42 x = 128 (k) 92x+1 = 50
3
(g) 3x = 81
(c) 9x = 27 (l) 7x1 = 4
x
(h) 4 2 1 = 128 x
(d) 5 x = 125 (m) 2 2 +1 = 64
(e) 82x = 32 (i) 4x = 7 (n) 52x1 = 20
8.2 Solve the following logarithmic equations for the unknown.
(a) log5 n = 4 (i) x = log4 72 log4 9

(b) log8 4 = n (j) x = log7 98 log7 2
1
(c) log2 n = (k) log5 x log5 7 = log5 6
3
1 (l) log3 63 log3 (7x ) = log3 2
(d) logn = 2
64 (m) log5 4 + log5 (2x 3) = 2
(e) log 1 n = 3
3
(n) log2 2 + log2 ( x + 2)
(f) log6 6 = n
log2 (3x 5) = 3
3
(g) logn 8 = (o) log2 4 + log2 ( x 1)
4
(h) log4 3 + log4 ( x + 2) = 2 log2 (3x 4) = 2
8.3 In each of the scientific formulae below, isolate the indicated

variable.
(a) pH. Using p as the symbol for pH and H for [ H + ]
p = log10 H find H
(b) Arrhenius equation

E
K = Ae RT find T
(c) Relationship of rate constants at two different temperatures

k E tT
loge = find K
K R tT
(d) Benfords Law

d+1
P = log10 find d
d
8.4 A population starts with 1000 individuals and doubles in size

every decade.
(a) Use the growth model N = N0 bt to find how long will it take
for the population to reach 500, 000.
(b) Repeat part (a) using the growth model N = N0 ekt .
74
AND/OR 75
EXPONENTIALS
8.5 An endangered species numbered 7, 200 at the start of the

year 2003 and at the start of 2014 numbered 800. Assuming the
population follows the decay law N = N0 ekt predict in what
year the population will fall below 120.
8.6 According to Wikipedia the number of articles on their site

grew exponentially between October 2002 and mid 2006. There
were 80, 000 articles in October 2002 and 166, 000 in October
2003. When does the model predict that Wikipedia reached
1, 500, 000 articles? Use the growth model N = N0 ekt .)
8.7 DDT is a pesticide that was very effective at controlling marlaria-

carrying mosquitos until its toxic effects on animals (including
cancer in humans) were established. Also, it remains active for
many years in the environment because it decays exponentially
with a half-life of 15 years. How long does it take for 100 grams
of DDT to decay down to 1 gram? Use model A = A0 ekt .
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75
MATHEMATICS FUNDAMENTALS Quadratic equations
9. Quadratic equations
In Chapter 4 we looked at quadratic expressions:
ax2 + bx + c with a = 0
For example: 2x2 + x 6 x2 3x + 4

Quadratic equations are equations of the kind
x2 4x + 3 = 0 2x2 + 7x 4 = 0
5x2 x 2 = 0 x2 20x + 40 = 0
In other words, a quadratic equation is a quadratic expression
equal to 0. The format ax2 + bx + c = 0 is called the standard form
of a quadratic equation.
Solving quadratic equations
If the left hand side can be factorized then the solution is easy:
x2 4x + 3 = 0
We can factorize the left-hand side:
( x 1)( x 3) = 0
We know that
0 anything = 0
so, if one of the factors above is 0, the entire left-hand side is 0 and
we have a solution to the equation. Hence
x1 = 0 or x3 = 0
and so
x=1 or x=3
As we can see, there can be more than one solution to a quadratic
equation. Well see later there can be zero, one or two solutions.
Some examples
Example. x2 x 6 = 0. Factorize to get
( x 3)( x + 2) = 0 x=3 or x = 2
Example. x2 9 = 0. Factorize to get
( x 3)( x + 3) = 0 x=3 or x = 3
77
78 mathematics
FUNDAMENTALS Quadratic equations
Example. t(t + 2) 1 = 3 t. Write this in standard form:

t ( t + 2) 1 = 3 t t2 + 2t 1 = 3 t t2 + 3t 4 = 0
Now factorize to get
(t + 4)(t 1) = 0 t = 4 or t=1
Example. x2 + 4x + 4 = 0. Factorize to get
x2 4x + 4 = 0 ( x 2)( x 2) = 0 x=2
We see that there is only one solution in this case.
Modelling using quadratic equations
Example: The product of two consecutive positive even numbers
is 48. Find the numbers.
As with earlier problems, we follow some general steps:
Step 1: Define an appropriate variable: Let x be the smaller of
the two. Then the other one is x + 2.
Step 2: Form a quadratic equation: The product of the two is
then x ( x + 2) and this must be 48 so
x ( x + 2) = 48 x2 + 2x 48 = 0
Step 3: Solve the quadratic equation:
x2 + 2x 48 = ( x + 8)( x 6) = 0 x = 8 or x=6
Step 4: Interpret the solution(s): The numbers must be positive
Check: 6 8 = 48
so x = 6 and hence the numbers are 6 and 6 + 2 = 8.
Example: A framer at a photo gallery wants to frame a print
with a border of uniform width all around the print. To make it pleas-
ing to the eye, the area of the border should equal the area of the
Its always a good idea to draw a
print. If the print measures 60 cm by 40 cm, how wide should the sketch
border be?
x
60 x
40
where x = is the width of the pathway. Note that the area of the
picture is
Apicture = 60 cm 40 cm = 2400 cm2
Observe is that the lengths of the sides of the frame are
40 + 2x cm and 60 + 2x cm
and hence the area of the frame is
Aframe = (40 + 2x )(60 + 2x ) cm2
78
MATHEMATICS FUNDAMENTALS Quadratic
9. quadratic equations 79 equations
We are told that Aframe = 2Apicture . That is,
(40 + 2x )(60 + 2x ) = 4800

2400 + 80x + 120x + 4x2 = 4800
4x2 + 200x 2400 = 0
x2 + 50x 600 = 0
( x + 60)( x 10) = 0
x = 60 or x = 10
The answer must be positive so x = 10 and hence the frame should

be 60 cm by 80 cm. Check: 60 80 = 4800 = twice
2400
Factorization of quadratics revisited
In many cases its very difficult to determine the factorization of
a quadratic expression. This is because it can involve square roots!
Some examples
(a) x2 7 = ( x A)( x B) where

A= 7 and B= 7
(b) x2 2x 1 = ( x A)( x B) where

A = 1+ 2 and B = 1 2
(c) x2 + 5x + 3 = ( x A)( x B) where

5 + 13 5 13
A= and B=
2 2
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79
The Quadratic Formula
There is a general formula that will give us the factorization. If
ax2 + bx + c = a( x A)( x B) a = 0
then

b + b2 4ac b b2 4ac
A= and B=
2a 2a
Since A and B have very similar structure, we write

b b2 4ac
A, B =
2a
(The symbol means plus or minus.) The plus expression
corresponds to A and the minus expression corresponds to B.
The values A and B are called the roots of the quadratic equa-
tion, even if they turn out not to involve square roots.
Lets check that the formula works on a simple example:
x2 3x + 2 = ( x 2)( x 1)
To use the quadratic formula, first identify the coefficients:
a=1 b = 3 c=2 The quadraticformula is

b b2 4ac
A, B =
Hence x2 3x + 2 = ( x A)( x B) where 2a

3 98 3 1 31 4 2
A, B = = = = , = 2, 1
2 2 2 2 2
Hence A = 2 and B = 1 are the roots of the quadratic equation, so
x2 3x + 2 = ( x 2)( x 1)
Example: Solve x2 2x 1 = 0. We have
a=1 b = 2 c = 1
First calculate
b2 4ac = (2)2 4 1 (1) = 4 + 4 = 8
and then

2 8 22 2 2 2 2
A, B = = = = 1 2
2 2 2 2

where weve used 8 = 4 2 = 4 2 = 2 2. That is, the roots
of the quadratic equation x2 2x 1 = 0 are

A = 1 + 2 and B = 1 2
Example: Solve 3x2 + 6x 4 = 0. We have
a=3 b=6 c = 4
80
MATHEMATICS FUNDAMENTALS 9. quadratic equations
Quadratic 81
equations
First calculate
b2 4ac = 62 4 3 (4) = 36 + 48 = 84
and then

6 84 6 2 21 21
A, B = = = 1
6 6 6 3

where weve used 84 = 4 21 = 4 21 = 2 21. That is, the
roots of the quadratic equation 3x2 + 6x 4 = 0 are

21 21
A = 1 + and B = 1
3 3
The number of solutions
This depends on the value of
b2 4ac
which is called the discriminant of the quadratic equation.
If the discriminant is positive, we have two solutions, one for the

positive component and one for the negative.
If the discriminant is zero, then the solutions are
b 0 b
A, B = A and B both =
2a 2a
We have one solution.
If the discriminant is negative, then we cant take its square root

and hence A and B are undefined. In other words, we have no
solutions to the equation.
The discriminant of the quadratic equation ax2 + bx + c = 0 is

given its own symbol:
= b2 4ac
Summary:
If > 0 then there are two distinct solutions
If = 0 then there is only one solution
If < 0 then there are no solutions
Example: The discriminant of the quadratic equation
2x2 + 5x + 4 = 0
is
= b2 4ac = 52 4 2 4 = 25 32 = 7
and hence there is no solution to the quadratic equation.
81
82 mathematics fundamentals Quadratic equations
An alternative form for quadratics
Instead of the standard form ax2 + bx + c, there is an alternative

form:
a ( x + )2 +
where the numbers and (the Greek letters alpha and beta) are
related to a, b and c. To convert from the alternative form to the
standard form is easy. We just expand the bracket and tidy up.
Example:
2( x + 3)2 + 5
= 2( x + 3)( x + 3) + 5
= 2( x2 + 6x + 9) + 5
= 2x2 + 12x + 23
Completing the square
Converting from the standard to the alternative form requires the

following process. It is called completing the square and the aim is
to re-arrange a quadratic in such a way as to achive our objective.
If the quadratic is of the form x2 + bx + c there are three steps:
(1) halve b. Call it .
(2) square and both add and subtract it from the quadratic
(3) write this quadratic in the form ( x + )2 +
If the quadratic is of the form ax2 + bx + c there is a preliminary
step of factoring out a:

b c
ax2 + bx + c = a x2 + x +
a a
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82
Quadratic 83
equations
Examples
(a) x2 + 6x + 14. Half of 6 is 3 and the square of this is 9 so write
x2 + 6x + 14 = ( x2 + 6x +9) + (149)
and observe that the first three terms are the expansion of ( x +
3)2 . So
x2 + 6x + 14 = ( x + 3)2 + 5
(b) x2 10x + 12. Half of 10 is 5 and the square of this is 25 so

write
x2 10x + 12 = ( x2 10x +25) + (1225)
and observe that the first three terms are the expansion of
( x 5)2 . So
x2 10x + 12 = ( x 5)2 13
Exercises 9
9.1 Solve the following quadratic equations
(i) (y 2)2 = 9y
(ii) 2( x 3)2 = 6 5x + x2
(iii) (2x + 2)2 ( x 5)2 = 2x ( x + 7)
Hint: Youll first have to write them in standard form
9.2 Solve the following quadratic equations
(i) x2 + 2x 4 = 0
(ii) 11x2 10x 1 = 0
(iii) ( x + 2)( x + 3) = x2 2x + 1
9.3 How many solutions do each of the following quadratic equa-

tions have?
(i) x2 4x + 2 = 0
(ii) x2 + 2x + 7 = 0
(iii) 4x2 + 4x + 1 = 0
9.4 Complete the square for the following quadratics
(i) x2 + 6x 4
(ii) x2 8x + 20
(iii) x2 + 18x 12
(iv) x2 10x
83
Problem set 9
9.1 Solve the following equations by factorising.
(a) x2 3x = 0 (f) ( x 2)( x + 1) = 4

(b) x2 5x + 4 = 0
15
(c) x2 5x 14 = 0 (g) x + =8
x
(d) x2 x 42 = 0
5
(e) 3x2 6x = 3 (h) x =
x4
9.2 Solve the following equations by using the quadratic formula.
3 2
(a) x2 5x + 4 = 0 (f) 2x + 4x + 1 = 0
(b) 3x2 + 7x + 2 = 0
(g) x2 + x 1 = 0
(c) x2 + 4x + 1 = 0
(h) 2x2 2x 3 = 0
(d) 2x2 + 3x 1 = 0
(e) x2 + 3x + 4 = 0 (i) 2 3x2 = 0
9.3 Determine the number of solutions for the following equations.
3 2
(a) x2 + 3x + 1 = 0 (e) 2x + 4x + 6 = 0
(b) 3x2 + 4x + 2 = 0
(f) x2 + 5x 1 = 0
(c) 2x2 + 3x + 5 = 0
(d) 2x2 + 4x + 2 = 0 (g) 4x2 5x + 4 = 0
9.4 Complete the square for the following quadratic expressions
(a) x2 + 2x + 3 (d) x2 6x
(b) x2 4x 8 (e) x2 + 12x + 36
(c) x2 + 10x + 7 (f) x2 20x 1
84
Quadratic 85
equations
9.5 The product of two consecutive odd numbers is 99. Find the
two numbers.

9.6 One positive number exceeds three times another positive num- Note that 25 + 12 68 = 29
ber by 5. The product of the numbers is 68. Find the numbers.
9.7 A group of zoologists was studying the effect on the body

weight of rats of varying the amount of yeast in their diet.
By changing the percentage P of yeast in the diet, the average
weight gain, G (in grams), over time was estimated to be
G = 200P2 + 200P + 20
What percentage of yeast would you expect to give an average

weight gain of 70 grams?
9.8 You wish to make a square-bottomed box with a height of
three cm and a volume of 75 cm3 . You will take a piece of square
cardboard, cut a three cm square from each corner and fold up
the sides. What sized piece of cardboard do you need? Hint: it might help to draw a picture.
9.9 A rectangular park has dimensions 40 metres by 50 metres. A

pathway is to be added all around the park which will increase
the total area to 3000 m2 . How wide is the pathway going to be? Hint: it might help to draw a picture.
exciting projects.
85
MATHEMATICS FUNDAMENTALS Functions and graphs
10. Functions and graphs
So far we have mostly examined ways to find specific values of

variables from equations. Now we will explore mathematical rela-
tionships between variables. For example,
The cost of a taxi ride is related to how far you travel.
The amount of interest earned on a deposit is related to how long you
leave it in the bank.
The distance travelled by a ball is related to how hard you throw it.
An example: a (simplified) taxi fare costs $2 per kilometre (km)
plus a $4 flagfall.
If you travel 5 km, the taxi ride costs $4 + $2 5 = $14
If you travel 10 km, the taxi ride costs $4 + $2 10 = $24
If you travel x km, the taxi ride costs $4 + 2x

If we define y to be the cost of the taxi ride, we can just write
y = 2x + 4
We say that y depends on x or y is a function of x:
y = f (x) where in this case f ( x ) = 2x + 4
For example, when x = 7, we have y = f (7) = 2 7 + 4 = 18. In
other words, a 7 km ride costs $18.
Using function notation f ( x ) to describe a variable like y is a
Note that f ( x ) does not stand for f
useful way to make clear that the value of y is determined by the multiplied by x
value another variable x via an algebraic expression.
Some examples:
y = f ( x ) = x2 + 1
f ( 2 ) = 22 + 1 = 4 + 1 = 5
f (0) = 02 + 1 = 0 + 1 = 1
f (1) = (1)2 + 1 = 1 + 1 = 2
y = g( x ) = 2x
g ( 1 ) = 21 = 2
g(2) = 22 = 4
1
g ( 12 ) = 2 2 = 2
87
Graphical representation of functions
Its useful to get a visual sense of how a function (y or f ( x )) changes

as x changes. Since two variables are involved, draw two number
lines at right angles (through the 0 point on each line):
y or f ( x )
4
3
2
1
4 3 2 1 1 2 3 4 x
1
2
3
4
These number lines are called the y-axis (vertical) and the
x-axis (horizontal).
To visually examine the behaviour of a function, say
y = 2x 3
start by finding out the value of y for different values of x.
x=2 then y = 2(2) 3 = 1
x=1 then y = 2(1) 3 = 1
x=3 then y = 2(3) 3 = 3
x=4 then y = 2(4) 3 = 5
x=0 then y = 2(0) 3 = 3
x = 1 then y = 2(1) 3 = 5
Now represent them on the picture by cross-referencing points

on the x-axis with their corresponding points on the y-axis:
y
4
4 3 2 1 1 2 3 4 x
1
88
MATHEMATICS FUNDAMENTALS 10. functions and graphs
Functions and89
graphs
If we tried x-values between the whole numbers, the plotted

points would follow the same pattern so we can join them up into a
smooth line:
y
4 3 2 1 1 2 3 4 x
3
360
4
We have now plotted a graph of the function y = 2x 3.

thinking .
360
thinking . 360
thinking .

89
90 mathematics
FUNDAMENTALS Functions and graphs
Co-ordinates
Rather than writing: when x = 1, y = 1 we use co-ordinates:
(1, 1)
These are like a map reference and the x-value is always given first.
To signify a point of interest we can give it a label:
y
4
3
2
1
4 3 2 1 1 2 3 4 x
1 P(1,1)
2
3
4
5
Example
To plot the graph of y = 4 x by proceeding at follows:
Calculate some co-ordinates, that is, find some points on the graph
Plot the points on a set of axes
Draw a continuous line which includes all of the points
Here is the graph of the function y = 4 x

y
6
5
4
3
2
1
3 2 1 1 2 3 4 5 x
1
2
3
Linear functions
Functions y = 2x 3 and y = 4 x are called linear functions

because their graphs are straight lines. Such functions are common
and easy to interpret. They have the general form
y = mx + c
where m and c are numbers.

If we understand how m and c relate to the straight line graph
they produce then we can draw the graph without plotting lots of
individual points. For example,
x=0 y = m0 + c y=c
90
MATHEMATICS FUNDAMENTALS Functions
10. functions and graphs
and graphs 91
Note that the co-ordinate (0, c) is on the y-axis. This is the place
where the line cuts the y-axis, and for this reason it is called the
y-intercept.
(0,c)
x
The y-intercepts of y = 4 x and y = 2x 3 are:
y y
6 4
y =4 x
y=2x 3
(0,4)
3 Note that we usually dont list all of

the numbers on the numberline. Just
4 4 x enough for us to get an idea of the
scale of things
3 3 x (0,-3)
4
3
The gradient
This is defined as the change in the y-values between two points on

the line divided by the change in their x-values.
Example: y = 2x 3
If x=2 then we have y=1
So y increases by 2 while x increases by 1, so the gradient is
change in y-value 2
= =2
change in x-value 1
The gradient is the same if you use different points:
If x = 1 then we have y = 5
So y increases by 10 while x increases by 5, so the gradient is
change in y-value 10
= =2
change in x-value 5
91
92 mathematics fundamentals Functions and graphs
Here is the graph of y = 2x 3 showing its y-intercept and

indicating its gradient.
y
4
3
2
1
gradient = 2
4 3 2 1 1 2 3 4 x
1
2
3 (0,-3)
4
5
Here is the graph of y = 4 x showing its y-intercept and

indicating its gradient.
y
6
5
(0,4) gradient = 1
4
3
2
1
3 2 1 1 2 3 4 5 x
1
2
3

92
Functions 93
and graphs
We observe that the gradient is the value of m in y = mx + c. A

useful way to understand the gradient is to think of it as the change
in y for every one unit increase in x. This is useful for describing
linear relationships in context.
Example: The taxi ride (from last time). The equation that de-
scribed the taxi fare was
y = 2x + 4
where
x = distance travelled (km)
y = cost($)
Recall that c is the y-intercept

The gradient of 2 represents the change in cost for every 1 km
travelled. In other words, $2 per km.
Physical meaning of the gradient
y
y = 2x + 1
6 y = x+1
5
4
y = 12 x + 1
3
2
1
2 1 1 2 3 4 5 x
1
2
3
4 y = -x + 1
y = -2x + 1
The larger the value of m, the steeper the line. A positive gradi-
ent means going uphill left-to-right, and a negative gradient means
going downhill left-to-right.
Horizontal lines
What is the equation of this horizontal line?
y
5
4
3
2
1
4 3 2 1 1 2 3 4 x
Hint: What is the gradient, and what is the y-intercept?

The gradient is zero, and the y-intercept is 3, so the equation is
simply y = 3. In general, the horizontal line passing through (0, c)
has equation
y=c
A special case is the x-axis. It has equation y = 0.
93
94 mathematics
Vertical lines
The only lines which dont fit the y = mx + c format are vertical:
y
5
4
3
2
1
4 3 2 1 1 2 3 4 x
The equation of this line is x = 2. In general, the vertical line

passing through ( a, 0) has equation
x=a
A special case is the y-axis. It has equation x = 0.

The x-intercept
This is where the line crosses the x-axis. That is, the place on the
line where y = 0.
Example: y = 2x + 6. We immediately observe that the gradi-
ent is m = 2 and the y-intercept is c = 6. The x-intercept is where
y = 0 so we solve
0 = 2x + 6 2x = 6 x=3
y
3 x
In general, the quick way to plot an accurate graph of y = mx + c

is to find both intercepts, mark them on the axes and draw the
straight line which contains them. We call this a sketch of the
graph.
Finding the equation of a line
We can work out the equation of a line given sufficient information.

Example: Find the equation of the line with gradient 2 which
contains the point (4, 7). We know that we are working with a line
y = mx + c and we are told that m = 2, so
y = 2x + c
94
Functions 95
and graphs
We also know that when x = 4 the equation gives y = 7. Substitute

these x and y values into the equation and solve for c:
7 = 2(4) + c
7 = 8+c
78 = c
c = 1
Hence, the equation of the straight line is y = 2x 1

Example: The equation of the line containing (2, 5) and (5, 11).
Method 1: Substituting the two co-ordinates into the equation of
a line y = mx + c gives us a pair of simultaneous equations to be
solved for m and c:
(5, 11) 11 = m5 + c (1)

(2, 5) 5 = m2 + c (2)
6 = 3m
6
So m= =2
3
Substituting m = 2 into say equation (2) gives
5 = 2(2) + c
5 = 4+c
c=1
Hence, the equation is y = 2x + 1.
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95
96 mathematics
Method 2: Use the definition of gradient to find m:

change in y value 11 5 6
m = = = = 2
change in x value 52 3
So the equation (so far) is y = 2x + c. To find c, substitute one of the
points, say (2, 5) into this equation and solve for c:
5 = 2(2) + c 5 = 4+c 54 = c c=1
Hence, the equation of the straight line is y = 2x + 1

y
(5,11)
(2,5)
Intersecting lines
We often wish to find the co-ordinates of the point where two
lines meet or intersect.
y
Example: Where do y = 4x 1 and y = 2x + 5 intersect?

The point of intersection has the same y-value on both lines, so
we can just equate the right-hand sides:
4x 1 = 2x + 5 6x = 6 x=1
Substituting back into either of the line equations gives
y = 4x 1 = 4(1) 1 = 4 1 = 3
Hence, the coordinates of the point of intersection are (1, 3).

Parallel lines Parallel lines have the same gradient. For example,
y = 2x + 1 and y = 2x 2
96
Functions 97
and graphs
x Note that parallel lines never intersect
-2
Perpendicular lines Perpendicular lines (that is, lines at right-

angles to each other) have gradients which multiply to give 1.For
example,

y = 2x 2 and y = 12 x + 1 2 1
2 = 1
If a given line has gradient m then any line perpendicular to it

must have gradient m1 .
Graphs of exponential and logarithm functions
Exponential growth and decay

When we discussed exponentials and logarithms we observed
that exponenial growth functions grow very quickly. For example,
the function y = 2x
x 1 2 3 4 5 6 7 8 9 10 11 12
y 2 4 8 16 32 64 128 256 515 1,024 2,048 4,096
The graph of 2x
To plot the graph of the function y = 2x well need to determine
some points on the graph, plot them and then draw a smooth line
between them. In all, we have:
x 4 3 2 1 0 1 2 3 4
1 1 1 1
y 16 8 4 2 1 2 4 8 16
and the graph is
97
98 mathematics
y
9 y = 2x
8
7
6
5
4
3
2
1
3 2 1 1 2 3 x
More graphs of exponentials
Here is a plot of the functions 2x , 3x , and between them e x .
y
10 y = ex
y = 3x
9 y = 2x
8
7
6
5
4
3 Note that the graphs never go below
the x-axes, and that all of the graphs
2 have y-intercept 1
1
3 2 1 1 2 3 x
These graphs of exponential growth functions give us a graphical
interpretation of things like population growth and compound
interest.
98
Functions 99
and graphs
The graph of 2 x
Here is a graph of 2 x , along with 2x .
y
y = 2 x 9 y = 2x
8
7
6
5
4
3 Note that the graphs are left-right-
symmetric
2
1
3 2 1 1 2 3 x
These graphs of exponential decay functions give us a graphical
interpretation of things like radioactive decay and depreciation.
The graph of log2 x
Here is a graph of log2 x, along with 2x .
y
y = 2x
8
7
6
5
4
3 y = log2 x
2
1
3 2 1 1 2 3 4 5 6 7 8 x
1
2 Note that there is a diagonal sym-
metry, and observe that the graph of
3 log2 x never crosses the y-axis
Exercises 10
10.1 Sketch the graphs of the following linear functions
(i) y = 4x 8
(ii) y = 9 3x
(iii) y = 2x + 1
10.2 Find the equation of the horizontal line which contains the
point (4, 5).
10.3 Find the equation of the line which is perpendicular to

y = 2x + 5 and has y-intercept 6.
10.4 Find the equation of the line containing (2, 4) and (3, 1).
99
Problem set 10
10.1 Find the indicated values of the following functions.
f ( x ) = 3x2 + 2x + 4 g( x ) = log2 x

h(u) = 11 u F (v) = 32v4
s3
G (t) = 5 3t H (s) =
s+1

(a) f (2) (e) h(7) 1
(i) G 4
(b) f (1)
(f) h(11) (j) G 52
(c) g(8) (g) F (3) (k) H (2)

(d) g 14 (h) F (2) (l) H (3)
10.2 Sketch the graphs of the following functions
(a) y = 3x + 6
(b) y = 4x 4
(c) y = 2t + 6
(d) z = 5x 4
10.3 Find the equations of the following lines.
(a) Gradient 2 and contains the point (0, 4)

(b) Gradient 3 and contains the point (1, 3)
(c) Contains the points (3, 4) and (6, 2)
(d) Contains the point (3, 1) and is parallel to the line y = 2x 1
(e) Perpendicular y = 4x 1 and contains the point (4, 2)
10.4 The line L1 has equation y = 3x + 1. Find the equations of the

following five lines.
(a) L2 which is parallel to L1 and contains the point (1, 4)

(b) L3 which is perpendicular to L1 and has x-intercept 3
(c) L4 perpendicular to L1 and intersects it at the point (2, 3)
(d) L5 which is horizontal and has the same y-intercept as L1
(e) L6 which contains (3, 7) and has the same y-intercept as L1
10.5 Sketch the following pairs of lines from the previous question
on the same axes.
(a) L1 and L3
(b) L2 and L6
(c) L4 and L5
10.6 Find the intersection point of each pair of lines in the previous
question.
100
Functions 101
and graphs
10.7 An application. This is an example of a problem in the area

of Operations Research (or Management Systems as it is called in
Business). Consider the lines
L1 : x + y = 4
L2 : x + 2y = 6
L3 : 2x + y = 7
(a) Sketch the graphs of all three on the same set of axes.
(b) Find the points of intersection between:
(i) L1 and L2
(ii) L1 and L3
(iii) L2 and the y-axis
(iv) L3 and the x-axis
(c) At each of those intersection points find the value of
z = 3x + 5y
(d) At which intersection point does z attain its greatest value?
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101
MATHEMATICS FUNDAMENTALS Quadratic functions and Parabolae
11. Quadratic functions and Parabolae
Recall that quadratics are expressions of the form
ax2 + bx + c
where a, b and c are given numbers.

Quadratic functions are functions of the form
y = f ( x ) = ax2 + bx + c
Here is the graph of y = x2 . We call this graph a parabola.
y
(3,9) 9 (3,9)
8
7
6
5
(2,4) 4 (2,4)
3
2
(1,1) 1 (1,1)
4 3 2 1 1 2 3 4 x
1
Quadratic functions
Different values of a, b and c produce different parabolic graphs.

We can obtain the graph of any function by choosing a series of
x-values and for each, calculating the corresponding y-value and
plotting that co-ordinate. Finally, draw a smooth curve which con-
tains all of the points.
Example: The quadratics function y = x2 2x 3. We have
x -3 -2 -1 0 1 2 3 4 5
y 12 5 0 -3 -4 -3 0 5 12
y
15
10
42 2 4 6x
5
103
104 mathematics
FUNDAMENTALS Quadratic functions and Parabolae
Obviously we would like to avoid plotting lots of points in order

to sketch a parabola. There are four main properties of parabolae:
the convexity
the yintercept
the x intercept(s) (if they exist)
the vertex (or turning point)

y
5
2 1 1 2 3 4 x
1
2
3
4
Each of these properties can be determined from the coefficients

a, b and c in
y = ax2 + bx + c
Convexity
y
5
4
3
2
1
21 1234
x
2
3
4
5
y = x2 2x 3
Concave up (like a smile)
y
4
3
2
1
1 1 2 3 4 5 6 x
2
3
4
5
6
y = x2 + 6x 5
Concave down (like a frown)
If the coefficient of x2 (that is, the value of a) is:

positive, the parabola is concave up (smile)
negative, the parabola is concave down (frown)
104
MATHEMATICS FUNDAMENTALS 11. quadratic functions
Quadratic and parabolae
functions 105
and Parabolae
The y-intercept
This is where the parabola passes through the yaxis. In other
words, the yintercept is the value of y when x = 0. So, for
y = ax2 + bx + c, the yintercept is:
y = a (0)2 + b (0) + c y=c
The co-ordinate of the yintercept is hence (0, c).
Example: The parabola y = 2x2 4x + 6

has y-intercept 6 and is concave up
Example: The parabola y = 5x2 3

has y-intercept 3 and is concave down
Example: The parabola y = x 4x2

has y-intercept 0 and is concave down
The x-intercept(s)
These are where the parabola passes through the x axis. In
other words, the x intercepts are the values of x when y = 0. So,
for y = ax2 + bx + c, we need to solve
ax2 + bx + c = 0
Recall that there may be 0, 1 or 2
That is, we have to find the roots of the quadratic equation. We do this solutions
either by factorising or using the quadratic formula.
Example: The parabola y = x2 2x 3. We must solve
x2 2x 3 = 0
Luckily, we can factorize:
( x 3)( x + 1) = 0 x = 3 or x = 1
Hence there are two x-intercepts and they are x = 3 and x = 1.
105
106 mathematics
The vertex (turning point)

The symmetry of parabolae tells us that the vertex has an x value
half way between the x intercepts.
Example: The parabola y = x2 2x 3. In this case, the x value
of the vertex is x = 1. To find the corresponding yvalue we subti-
tute this value into the function:
y = 12 2 ( 1 ) 3 = 4
The vertex is hence at (1, 4).

For the function y = x2 2x 3 we now have the four main
y
properties: 5
concave up 4
3
yintercept: (0, 3) 2
x intercepts: (1, 0) and (3, 0) 1
2 1 1 2 3 4 x
vertex: (1, 4) 1
2
and so we can draw the Here is a sketch of the parabola:
3
Note that we write in the coordinates of any points of interest
(for example, the vertex) which are not on one of the axes. 4
(1, 4)
A formula for the vertex x-coordinate
What if the x intercepts werent so easy to work with (or per-
haps didnt even exist)?
The quadratic formula provides the clue to a general method:
To solve ax2 + bx + c = a( x A)( x B) = 0 then

b b2 4ac
A, B =
2a
If we remove the term added on or taken off, we are left with an
x value halfway between the roots:
b
x=
2a
This is the x-coordinate of the vertex.
Example: The parabola y = x2 2x 3 has a = 1 and b = 2, so
b 2
x= = =1
2a 2
and substitution of this into y = x2 2x 3 gives us the
y-coordinate of the vertex: y = 12 2(1) 3 = 4.
An example A sketch of the graph of y = x2 6x + 8. We have:
The graph is concave up since a = 1 is positive.
The yintercept is c = 8.
To find the x intercepts we solve x2 6x + 8 = 0. We can do this

using the quadratic formula:

b b2 4ac 6 36 4 1 8
x= x=
2a 2
106
functions 107
and Parabolae
That is,

6 36 32 6 4 62 8 4
x= = = = , = 4, 2.
2 2 2 2 2
To find the vertex:

b 6
x= = =3 y = 32 6 3 + 8 = 9 18 + 8 = 1
2a 2
So the vertex is at (3, 1).
The sketch of y = x2 6x + 8 is:
y
10
9
8
7
6
5
4
3
2
1
1 1 2 3 x
1
2
(3, 1)
Another example
A sketch of the graph of y = x2 2x 2. We have that
The graph is concave down since a = 1 is negative.
The yintercept is c = 2.
To find the x intercepts we solve x2 2x 2 = 0. We can do

this using the quadratic formula:

b b2 4ac 2 4 4 (1) (2)
x= x=
2a 2 (1)

2 48 2 4
x= =
2 2
but we cant take the square root of a negative number and
hence there are no x-intercepts.
To find the vertex:

b 2
x= = = 1
2a 2
y = (1)2 2 (1) 2 = 1 + 2 2 = 1
So the vertex is at (1, 1).
107
108 mathematics
We dont have much to go on, just the vertex and the y-intercept.
Wed like to get a point symmetrically across from the y-intercept.
Its x-coordinate will be twice that of the vertex:
x = 2 y = (2)2 2(2) 2 = 4 + 4 2 = = 2
and from this we have another point on the graph: (2, 2).
The graph of y = x2 2x 2 is
y
1
3 2 1 1x
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108
functions 109
and Parabolae
Exercises 11
11.1 Sketch the graph of the quadratic function y = x2 2x 8

indicating
the convexity (concave up or down)

the yintercept
the vertex co-ordinates
whatever x intercepts exist
Problem set 11
11.1 Sketch the graphs of the following quadratic functions, in each

case indicating
the convexity (concave up or down)

the yintercept
the vertex co-ordinates
whatever x intercepts exist
(a) y = 2x2 + 8x + 6 (e) y = x2 4x + 12

(b) y = x2 2x 8 (f) y = x2 6x
(c) y = x2 4x + 4 (g) y = x2 4
(d) y = x2 + 4x 3 (h) y = x2 7x + 10
11.2 Complete the square for the following quadratic functions

and use the result to verify the co-ordinates of the vertex of the
parabola they repesent.
(a) y = x2 + 8x + 3 (b) y = x2 14x
109
MATHEMATICS FUNDAMENTALS Calculus
12. Calculus
Rate of change
In many situations we are interested in how quickly (or slowly)
something is changing. For example,
how quickly a car is accelerating
how quickly a virus is spreading
how quickly a company is growing
We refer to how quickly something of interest is changing as its

rate of change. In Mathematics we use functions to describe things
of interest. How can we calculate the rate of change of a function?
The basic Idea
Consider climbing a hill. Initially it may be quite steep, but as
we approach the top it starts to level off. The usually refer to the
steepness as its slope or gradient.
Its a measure of the rate of change of height. The simplest situa-
tion is a straight line, for example y = 2x + 1. Its slope/gradient is
m = 2 anywhere on it. y
What about a parabola? For example, y = 6x x2 . Its graph is: 10
9
What is its slope at
8
7
x = 0?
6
x = 1? 5
4
x = 3? 3
2
Notice that the slope/gradient depends on the value of x. How 1
do we find the gradient at any point x? Can we obtain a formula" 1 1 2 3 4 5 6 7x

1
or expression for the gradient of the graph in terms of x?
The simplest case

Lets look at the simplest case, y = x2 . We wish to know how y
10
steep it is, that is, what is its slope/gradient when say, x = 2. The 9
gradient we are after is the gradient of the line that just touches the 8
7
parabola at the point P(2, 4). This line is called the tangent to the
6
parabola at P. 5
4 P
3
2
1
1 1 2 3 4x
1
111
112 mathematics
FUNDAMENTALS Calculus
Chords
We approximate the tangent at the point to the graph of y = x2 at

the point x = 2 by a sequence of chords. A chord is a straight lines
joining two points on a curve.
y
25 Q
20
15
10
5
P
1 1 2 3 4 5x
Each chord joins our point of interest P to another point, Q on

the parabola. As Q moves in towards P, the chords gradient ap-
proaches that of the tangent. Lets take an arbitrary point Q on the
parabola. Its co-ordinates will be Q( a, a2 ) for some vlaue of a. Then
the gradient of PQ is
a2 4
a2
Graphically, we have
y
25 Q( a, a2 )
20
15
10
5
P(2, 4)
1 1 2 3 4 5 6x
112
MATHEMATICS FUNDAMENTALS 12. calculus 113 Calculus
Limits
2
What value does function aa24 approaches as a approaches 2?
Mathematicians refer to that value as the limit of the function as
a approaches 2, and write
a2 4
lim
a 2 a2
We cant substitute a = 2 in the expression because well get 00 . We

first factorise the numerator: a2 4 = ( a + 2)( a 2), so
a2 4 ( a + 2)
( a 2
)
lim = lim = lim ( a + 2) = 2 + 2 = 4
a 2 a2 a 2
(a 2
) a 2
So the gradient of the tangent line to the function y = x2 at x = 2

is m = 4. We can if needed calculate that the y-intercept is c = 4,
and hence the equation of the tangent line is y = 4x 4.
Examples
a2 11a + 30
( a 5
)( a 6)
lim = lim = lim ( a 6) = 5 6 = 1
a 5 a5 a 5 a 5 a 5
Our task is always to factorize the numerator. Observe that the

bottom line immediately tells us one of the factors of the top line.
Our job is to find the other.
Example
a2 + 3a + 2 ( a + 1)( a + 2)
lim = lim
a 3 a3 a 3 a3
but nothing cancels, and we will end up with 200 , but we cant di-
vide by zero! We say that the limit does not exist
113
114 mathematics
Tangents
Recall from last lecture our investigation of the tangent to the

parabola y = x2 at x = 2. Its gradient is that of the chord PQ as
a approaches 2. That is,
a2 4 ( a + 2)
( a 2
)
m = lim = lim = lim ( a + 2) = 2 + 2 = 4
a 2 a2 a 2
( a 2) a 2
We can do the same calculation for any point P( x, x2 ).
y
25 Q( a, a2 )
20
15
10
5
P( x, x2 )
1 1 2 3 4 5 6x
The gradient of the tangent line is that of the chord PQ in the

limit as a approaches x. That is,
a2 x ( a + x )
( a x
)
lim = lim = lim ( a + x ) = x + x = 2x
a x ax a x (a
x) a x
Example
The gradient of the tangent line at the point (3, 9) will be
m = 23 = 6
The derivative
We have an expression for the gradient of the tangent line to the

function f ( x ) = x2 . It is 2x. Note that this is another function. We
write it as
f ( x ) = 2x
and call this function the derivative of f ( x ) = x2 .
We could do the same sort of calculation for any function f ( x ).
For example, f ( x ) = x n or f ( x ) = e x or f ( x ) = log x.
114
MATHEMATICS FUNDAMENTALS 12. calculus 115
Calculus
If we are writing our function in the y = form then we write

dy
the derivative in the fractional form dx = , reminiscent of the
fraction in the limit calculation.
Example
dy
if y = x2 then = 2x
dx
The operation of finding the derivative is called differentiation.
Some derivatives
By performing a process similar to what we did previously, we
can show that for any number n,
if f ( x) = xn then f ( x ) = nx n1
Another way of writing this is

d
( x n ) = nx n1
dx
Examples
d 3
x = 3x2
dx
d 1 1 1 d 1
x2 = x 2 or equivalently x =
dx 2 dx 2 x

d 1 d 1 1
x = x 2 or equivalently = 2
dx dx x x
Two special ones are
d d 1
(x) = x = 1x0 = 1
dx dx
and
d d 0
(1) = x = 0x 1 = 0
dx dx
Derivatives of sums and multiples
If a function is the sum/difference of two others then its deriva-
tive is the sum/difference of the respective derivatives.
Examples
g( x ) = x2 + x3 g ( x ) = 2x + 3x2
h( x ) = x6 x4 h ( x ) = 6x5 4x3
If a function is a multiple of a another function then its deriva-
tive is a multiple of its derivative.
Examples
j( x ) = 4x2 j ( x ) = 4(2x ) = 8x
k ( x ) = 2x3 k ( x ) = 2(3x2 ) = 6x2

We can use both rules together, for example,
p( x ) = 4x3 7x 2 + 5 p ( x ) = 12x2 + 14x 3
115
116 mathematics
Note that for any constant, C we have
d d d
(C ) = ( C 1) = C (1) = C 0 = 0
dx dx dx
An application of derivatives
Derivatives can be used to determine the maximum (or mini-
mum) values of functions.
Example: What is the maximum value attained by the function
y = x2 + 6x 5
The solution to this problem is to realise that at its point of maxi-

mum value the tangent to the function will be horizontal:
y
4
2 2 4 6 x
That is, the gradient of the tangent line will be zero, and hence
the derivative of the function will be zero.
116
MATHEMATICS FUNDAMENTALS 12. calculus 117
Calculus
We can easily find the derivative of y = f ( x ) = x2 + 6x 5. Its
f ( x ) = 2x + 6
and this is zero where
2x + 6 = 0 2x = 6 x=3
Hence the maximum value of the function occurs when x = 3.

To find the actual value of the function we simply calculaute y(3).
That is,
ymax = 32 + 6(3) 5 = 9 + 18 5 = 9 5 = 4
Derivatives of exp and log functions

By performing a process similar to what we did for power func-
tions (like f ( x ) = x2 ) we can show that
if f (x) = ex then f (x) = ex
That is, e x is its own derivative.

We can also show that
1
if f ( x ) = ln x then f (x) = and x = 0
x
When constants (k or a or b) are involved we have the following:
d kx d kx
e = kekx and e = kekx
dx dx
d 1 d 1
ln( ax ) = and (logb x ) =
dx x dx ln(b) x
Antiderivatives
Not surprisingly, antiderivatives are the opposites of derivatives.

Since 3x2 is the derivative of x3 , then x3 is an antiderivative of 3x2 .
However, x3 plus any constant c is also an antiderivative because
d 3 d 3 d
x +c = ( x ) + (c) = 3x2 + 0 = 3x2
dx dx dx
We write this symbolically as follows:

3x2 dx = x3 + c
and say that the antiderivative of 3x2 is x3 + c where c can be any

constant. Similarly,

x3 c
x2 dx = +C where C = The symbol is actually a very tall,
3 3 very thin letter S (for sum)
Some antiderivatives
By considering the derivatives weve seen so far we can deduce
the following antiderivatives.

x n +1
x n dx = +C n = 1
n+1
117
118 mathematics

1
dx = ln x +C e x dx = e x +C
x
and more complicated things if constants like k and a are involved.
We can also calculate sums and multiples for antiderivatives. For
example,

5x3 + 7x 2 dx = 5 x3 dx + 7 x 2 dx

x4 x 1 5x4 7 Note the generic +C. It must always
=5 +7 +C = +C be there
4 1 4 x
Integrals
Another name for antiderivative is integral. We say for instance,

that the integral of 3x2 is x3 + C.
Actually, its called an indefinite integral. In general, we write

F(x) = f ( x ) dx
and say that F ( x ) is the integral of f ( x ). In addition, we call f ( x )

the integrand.
A definite integral is one which is to be evaluated between two
points. We write
b
b
f ( x ) dx = F(x) which means F (b) F ( a)
a
a
Example
5
5

2 3 Note that C will always cancel out
3x dx = x +C = 53 +
C 23 +
C = 125 8 = 117
2
2
Geometric meaning of the derivative

Recall that the derivative has the geometric interpretation of
being the gradient (slope) of a curve.
d
Example: Recall that dx ( x2 ) = 2x. Then when say, x = 2 we have
gradient = 2(2) = 4
y
8
4 P(2,4)
1 1 2 3 4 x
118
MATHEMATICS FUNDAMENTALS 12. calculusCalculus
119
The gradient of the tangent line to the parabola y = x2 is m = 2x,

where x is the x-value of the point of interest.
Geometric meaning of the integral
It turns out that integrals tell us the areas under curves. For
example, if the curve is the graph of y = x2 and the units are cm,
then what is the area of the shaded region below?
y
8
1 1 2 3 4 x
The answer is
3 3
x3 33 13 27 1 26
Area = x2 dx = = = = 8.67 cm2
1 3 1 3 3 3 3 3
Theres one thing we have to be careful of. If the area is below

the x-axis then the integral will be negative, and so the area must
be the negative of the integral because area must be positive.
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119
120 mathematics
Example: Line is the graph of y = 4 2x and the units are

metres. What is the area of the shaded region below?
y
1
1 1 2 3 4 5 6 x
1
The answer is
5 5
Area = (4 2x )dx = 4x x2 = 4(5) 52 + 4(2) 22
2 2
= (20 25) + (8 4) = (5) + 4 = 5 + 4 = 9 m2

which we can in fact check because the region is a triangle:
1 36 18
Area = base height = = = 9 m2
2 2 2
Exercises 12
12.1 Find if possible the following limits
a2 a 2 t2 5t+4
(i) lima2 a 2 (iii) limt6 t 6
s2 2s8 4z2 11z3
(ii) lims4 s 4 (iv) limz3 z 3
12.2 Differentiate the following functions
(i) f ( x ) = 2x3 + 4x2 + 3x + 5

(ii) g(t) = 2t(t2 + t + 1)
4
(iii) z = 3x4 + 1x + x + 4x
12.3 Integrate the following functions
(i) f ( x ) = 5x2 + 3x + 4
4
(ii) g( x ) = 3x4 + x12 + x
12.4 Find the area bounded by the graph of the function

y = 12 3x2 and the x-axis
120
12. calculus 121
Calculus
Problem set 12
12.1 Evaluate the following limits if they exist.
x2 16
( a) lim
x 4 x4
x2 + 2x 8
(b) lim
x 2 x2
x2 x 12
(c) lim
x 4 x1
x2 6x + 9
(d) lim
x 3 x3
4x2 16
(e) lim
x 2 x 2
x2 + 15x + 56
(f) lim
x 7 x+7
12.2 Differentiate each of the following functions.
( a) y = x3 + x2 + x + 6
(b) y = 3x4 x3 + x2 + x 2
1
(c) y = x2 + x + 3
x
12.3 Differentiate

1 2 1
y= x+ x + 2
x x
12.4 Perform the following integrations

( a) x2 + x + 6 dx

(b) 3x4 x3 2 dx

2
4
(c) x + x + 2 dx
x
12.5 Evaluate the following definite integrals.
1
( a) x3 + x2 dx
0
3
(b) 5x4 + 5x 2 dx
2
2
(c) x2 + 4x 3 dx
2
12.6 Find the AREA bounded by the graph of f ( x ) and the x-axis
between the points indicated for the functions below. [HINT:
Sketch the graph first!]
( a) f ( x ) = x2 + 1, x = 1 to x = 1
(b) f ( x ) = x2 x, x = 0 to x = 2
121
MATHEMATICS FUNDAMENTALS Appendix 1: Answers to exercises
Appendix 1: Answers to exercises
Exercises in Chapter 1
(i ) (ii )
In various places colours are used to
3(4 + 2 6) + 12 4 6 + 2[25 3(2 + 5)] focus attention
= 3 (4 + 12) + 3 = 6 + 2[25 3 7]
1.1
= 3 16 + 3 = 6 + 2[25 21]
= 48 + 3 = 6+24
= 51 = 6+8
= 14
1.2 Difference of the product of 6 and 7 and the sum of 3 and 8

is the difference of 42 and 11, that is, 42 11, which is 31. In
symbols we have
(6 7) (3 + 8) = 42 11 = 31
1.3 ( a) 1 14 = 41+1
4 = 4+4 1 = 5
4 (b) 2 38 = 28+3
8 = 168+3 = 19
8
1.4 3 1 31 3 1 2 1 5 15 5
( a) 4 5 = 45 = 20 (b) 3 5 = 3 2 = 32 = 6
1.5 Two-thirds of four-fifths is
2 4 24 8
= =
3 5 35 15
1.6
3 2 35 24 15 8 15 + 8 23
(ii ) + = + = + = =
4 5 45 54 20 20 20 20
2 2+4 2 6 24 63 8 18 8 18 10 10
(ii ) = = = = = which is the same as
3 51 3 4 34 43 12 12 12 12 12
1.7 In simplest form:

78 2 39 39 3 13 13
= = = =
102 2 51 51 3 17 17
1.8 One-quarter of the sum of one-third and two-sevenths is

1 1 2
+
4 3 7
1 13
=
4 21
13
=
84
123
124 mathematics
FUNDAMENTALS Appendix 1: Answers to exercises
2.1
(i ) 4y 3 = 3y + 7
4y 3 + 3 = 3y + 7 + 3 4y = 3y + 10
3y
4y 3y = 3y
+ 10
y = 10
Check:
4 10 3 = 40 3 = 37 and 3 10 + 7 = 30 + 7 = 37 hence
(ii ) 9x + 8 = 8x 4 9x = 8x 12 x = 12
Check:
9 (12) + 8 = 108 + 8 = 100

8 (12) 4 = 96 4 = 100
5+ t
2.2 In the formula x = bt , if x = 2 and t = 1, then
5+1 2 6
2= =
b1 1 b1
2(b 1) 6
2( b 1) = 6 =
2
2
b1 = 3 b=4
Check:
5+t 5+1 6
= = =2
bt 41 3
2.3 (i ) Cross-multiply:
1 2
= x 1 = 2(3x + 1)
3x + 1 x1
3
x 1 = 6x + 2 3 = 5x x=
5
(ii ) Expand the bracket:
2y + 1 + 3(2y 4) = 2y 1 2y + 1 + 6y 12 = 2y 1
4y 11 = 2y 1 2y = 10 y=5
2.4 Let a be Sallys age. If its 12 years plus half her age then
a
a = 12 + 2a = 24 + a a = 24
2
124
MATHEMATICS FUNDAMENTALS appendix 1: answers
Appendix to exercises
1: Answers 125
to exercises
3.1
2x + 3y = 10 (1)
( a)
x + 2y = 6 (2)
() 2x + 3y = 10 (1)
2x + 4y = 12 (3)
y = 2
Substituting this value into, say equation (2) gives
x + 2y = 6 x+4 = 6 x=2
Finally, check by substituting both values into equation (1):
2x + 3y = 10 4 + 6 = 10
2x + 3y = 19 (1)
(b)
5x 6y = 20 (2)
4x + 6y = 38 (3)
(+) 5x 6y = 20 (2)
9x = 18
Hence x = 2 and substituting this into, say equation (1) gives
2x + 3y = 19 4 + 3y = 19 3y = 15 y=5
5x 6y = 20 10 30 = 20
2x + 4y = 4 (1)
(c)
5x 6y = 2 (2)
multiply equation (1) by 5 and equation (2) by 2 to get
10x + 20y = 20 (3)

() 10x 12y = 4 (4)
32y = 16
and hence
16 1
61 1
y= = y=
32 1
62 2
Substituting this value into, say equation (1) gives
2x + 4y = 4 2x + 2 = 4 2x = 2 x=1
5x 6y = 2 53 = 2
125
126 mathematics
3.2 Let the numbers be m and n, with m larger than n. We are told
that
m + n = 28 (1)
m n = 12 (2)
Adding these equations together gives
40
m+m+n = 28 + 12
n 2m = 40 m=
2
= 20
Substitute this back into say equations (1) to get
20 + n = 28 n = 28 20 n=8
That is, the required numbers are 20 and 8. Check
20 + 8 = 28 and 20 8 = 12
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126
1: Answers 127
to exercises
4.1 Expanding brackets and collecting terms gives:
(i ) ( x + 2)( x 4) = x2 4x + 2x 8 = x2 2x 8
(ii ) ( x + 5)(3x + 4) = 3x2 4x 15x + 20 = 3x2 19x + 20
(iii ) (2 t)(3t + 4) = 6t + 8 3t2 4t = 3t2 + 2t + 8

1 1 1 1 1 1 3 1
(iv) y+ y+ = y2 + y + y + = y2 + y +
2 4 2 4 2 4 4 8
4.2 To factorize
(i ) x2 3x + 2. The factors of 2 are
12 (1) (2)
and the sum must be 3 so we have
x2 3x + 2 = ( x 1)( x 2)
(ii ) x2 x 12. The factors of 12 are
1 (12), 2 (6), 3 (4), 4 (3), 6 (2), 12 (1)
and the sum must be 1 so we have
x2 x 12 = ( x + 3)( x 4)
(iii ) x2 + 14x + 40. The positive factors of 40 are
1 40 2 20 4 10 58
and the sum must be 14 so
x2 + 14x + 40 = ( x + 4)( x + 10)
(iv) The original quadratic is 3x2 + 8x + 4, so the associated

quadratic is x2 + 8x + 12. After a little bit of work we find that
x2 + 8x + 12 = ( x + 2)( x + 6)
and hence the required factorization is

2 6
3x + 8x + 4 = (3x + 2) x +
3
That is,
3x2 + 8x + 4 = (3x + 2)( x + 2)
(v) The original quadratic is 4x2 5x 6, so the associated

quadratic is x2 5x 24. After a little bit of work we find that
x2 5x 24 = ( x + 4)( x 8)
127
128 mathematics fundamentals Appendix 1: Answers to exercises
and hence the required factorization is

8
4x2 x 5 = (4x + 4) x
4
That is,
4x2 5x 6 = (4x + 3)( x 2)
4.3 Factorizing and cancelling gives:
(i )
x2 x 2 +
( x 1)( x 2) x2
= =
2
x + 2x + 1 + 1)( x + 1)
( x x+1
(ii )
x2 + 5x 14 +
( x 7)( x 2) x2
= =
2
x + 10x + 21 + 7)( x + 3)
( x x+3
5.1
(i ) 24 29 = 24+9 = 213
311
(ii ) = 3115 = 36
35
(iii ) (5 7)3 = 53 73
5.2
2
7 72
(i ) = 2
3 3
(ii ) (54 )7 = 547 = 528
5.3 Using the Laws of Indices we have
(i ) x 3 x 5 = x 3+5 = x 2
( y 1 )7 y 17 y 7
(ii ) = =
y 3 y 3 y 3
= y7(3) = y7+3 = y4
(iii ) (5x3 )2 x6 = 52 ( x3 )2 x6 = 52 x 6 x6 = 52 x 6+6

1 1
= 52 x 0 = 52 1 = 52 = =
52 25
128
1: Answers 129
to exercises
2 2
(iv) 3x 2 x 3 = 32 x 2 x 3
1 4 3 1 43 x
= x x = x =
32 9 9
5y2 z 5y2 z 5y2 z

(v) 3 5 2
= 2 3 2 5 2 =
(2y z )
2 (y ) (z ) 4y6 z10
5 26 1(10) 5 8 1+10 5 8 11
= y z = y z = y z
4 4 4
5.4 Simplifying gives

(i ) 48 = 163 = 16 3 = 4 3

(ii ) 3(5 + 2) = 35 + 3 2 = 5 3+ 6
(iii )
(3 + 5)(2 5) = 6 3 5 + 2 5 5 5 = 6 + (2 3) 5 5

= (6 5) + (1) 5 = 1 5

(iv) 2
(2 5 1)(2 5 + 1) = 4 5 5 + 2
5 51
= 45 1 = 20 1 = 19
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129
130 mathematics
6.1 (i )
1
1
32
3 1 4 2 z3 p 4 2 z3 p 4
z p =a w4 3 = w3 =w
a4 a4
We usually write the isolated variable on the left-hand side:

1
32
z3 p 4
w=
a4
(ii )
y= x3 c y2 = x 3 c

y2 + c = x 3 x = 3 y2 + c
(iii )
12s4 y3
m= 7p2 m = 12s4 y3
7p2

7p2 m 7p2 m
= s4 s= 4
12y3 12y3
(iv)
1
1
3
2 1 1 vm 4
1 vm 4
f h = vm 3 4 h = 2
3 h=
f f2
Recall that the Laws of Logarithms are
logb b = 1 logb 1 = 0 logb ( xr ) = r logb x

x
logb ( xy) = logb x + logb y logb = logb x logb y
y
7.1
(i ) 26 = 64 so 6 = log2 64
(ii ) 121 = 112 so log11 121 = 2
4 4
(iii ) 27 3 =81 so = log27 81
3
1
(iv) 12 = 144 so log144 12 =
2
7.2
(i ) log5 125 = 3 so 125 = 53
(ii ) log4 64 = 3 so 64 = 43
(iii ) log7 x = 3 so x = 73
(iv) log4 8 = x so 8 = 4x
130
appendix 1: answers to exercises 131
MATHEMATICS FUNDAMENTALS Appendix 1: Answers to exercises
7.3
(i ) log2 9 + log2 3 = log2 27 since 9 3 = 27
(ii )
1 7 7
1 1
log3 7 log3 28 = log3 since = =
4 28 7
4 4
(iii ) In log8 16 1 write 1 as log8 8 to get Note that log takes precedence over
subtraction. If wed wanted the sub-
16
log8 16 1 = log8 16 log8 8 = log8 = log8 2 traction to be done first, wed write
8 log8 (16 1)
7.4 We have log4 3 = s and log4 5 = t. Now
(i ) log4 75 = log4 (3 25) = log4 (3 52 ) = log4 3 + log4 (52 )
= log4 3 + 2log4 5 = s + 2t

5 5
(ii ) log4 = log4 = log4 5 log4 (32 )
9 32
= log4 5 2 log4 3 = t 2s
8.1
(i ) log5 x + log5 8 = 2 log5 (8x ) = 2 8x = 52

25
8x = 25 x=
8
1
(ii ) 22x+1 = 22x+1 = 25 2x + 1 = 5
32
6
2x = 6 x= x = 3
2
2
(iii ) 32x1 = 27 32(2x1) = 33 34x2 = 33
5
4x 2 = 3 4x = 5 x=
4
8.2
27
(i ) ln 27 ln 3 + ln 4 = ln 3 + ln 4 = ln 9 + ln 4 = ln 36
16e5 16 53
(ii ) ( = e = 8e2
2e3 2
8.3
(i ) ln(2x ) + ln 6 = ln(4x + 2) ln(12x ) = ln(4x + 2)

2 1
12x = 4x + 2 8x = 2 x= =
8 4
e5x+1
(ii ) = e4 e5x+1(3x2) = e4
e3x2
1
5x + 1 3x + 2 = 4 2x + 3 = 4 2x = 1 x=
2
131
132 mathematics
8.4 The data gives immediately that P0 = 20 so P = 20ekt . We are

told that after 5 years there were 480 fish, so
480 ln 24
480 = 20e5k e5k = = 24 5k = ln 24 k =
20 5
This is about 0.64, that is, the growth rate is about 64%. To find
out when the population exceeds 1000 we solve
1000 100
P(t) = 1000 20ekt = 1000 ekt = = = 50
20 2
ln 50
kt = ln 50 t=
k
That is,
ln 50 ln 24 ln 50 5 5 ln 50
t = ln 50 k = = =
1 5 1 ln 24 ln 24
A calculator would tell us that t 6.15 years, that is, about 6
years and 2 months after the beginning of the study.
132
1: Answers 133
to exercises
8.5 The machine cost $7000 in 2010 and is valued at $1000 in 2015.
Assuming the decay law P = P0 ekt , when will the machine
be worth $700? Let t be time in years starting at 0 in 2010. Then
P0 = 7000, so
P = 7000ekt
Now, 2015 represents t = 5, so we have P = 1000 and hence
1000 1
1000 = 7000e5t e5k = e5k =
7000 7
Now take logs of both sides (that is, switch to log form):
ln( 17 )
1
5k = ln 7 k= 5
A calculator would tell us that this is about 0.39 (that is, decining
at 39% per year), but its best to leave it in mathematical form. To
find the time t when the machine will be worth $700 we solve
1
700 = 7000ekt = ekt
10
and hence

1 1
kt = ln 10 t = ln 10 (k)
ln( 17 )
From before we have that k = 5 and hence

1
1
ln( 10 ) ln( 17 ) ln 10
5
t = 1 5 = 1
ln( 17 )
That is,
1
5 ln 10
t=
1
ln 7
A calculator will tell us that this is about 6 years, so roughly the

year 2016.
133
134 mathematics
9.1 We need to write them in standard form and then factorize:
(i ) (y 2)2 = 9y y2 4y + 4 = 9y
y2 + 5y + 4 = 0
(y + 1)(y + 4) = 0 y = 1 or 4
(ii ) 2( x 3)2 = 6 5x + x2
2x2 12x + 18 = 6 5x + x2
x2 7x + 12 = 0 ( x 3)( x 4) x = 3 or 4
(iii ) (2x + 2)2 ( x 5)2 = 2x2 + 19
4x2 + 8x + 4 ( x2 10x + 25) = 2x2 + 19
x2 + 18x 40 = 0 ( x 2)( x + 20) x = 2 or 20
9.2 The quadratic formula is

2 b b2 4ac
ax + bx + c = a( x A)( x B) A, B =
2a
(i ) x2 + 2x 4 = 0. We have
a = 1, b = 2, c = 4 b2 4ac = 22 4 1 (4) = 4 + 16 = 20
and hence

2 20 2 + 2 5 2 2 5
A, B = = = = 1 5
2 2 2 2
(ii ) 11x2 10x 1 = 0. We have
a = 11, b = 10, c = 1 b2 4ac = 102 4 11 (1) = 144
and hence

10 144 10 12 22 2 1
A, B = = = , = 1,
22 22 22 22 11
(iii ) ( x + 2)( x + 3) = x2 2x + 1. We first need to put the

equation into standard form. We have
x2 + 5x + 6 = x2 2x + 1 2x2 + 7x + 5 = 0
We have
a = 2, b = 7, c = 5 b2 4ac = 49 4 2 5 = 49 40 = 9
and hence

7 9 7 3 7 + 3 7 3 10
A, B = = = , = 1,
4 4 4 4 4
134
1: Answers 135
to exercises
9.3 The discriminant is

= b2 4ac
(i ) x2 4x + 2 = 0. We have
= b2 4ac = (4)2 4 1 2 = 16 8 = 8 > 0 two solutions
(ii ) x2 + 2x + 7 = 0. We have
= b2 4ac = 22 4 1 7 = 4 28 = 24 < 0 no solutions
(iii ) 4x2 + 4x + 1 = 0. We have
= b2 4ac = 42 4 4 1 = 16 16 = 0 one solution
9.4 Completing the square gives
(i) x2 + 6x 4. Half of 6 is 3 and the square of this is 9 so write
x2 + 6x 4 = ( x2 + 6x +9) + (49) = ( x + 3)2 13
(ii) x2 8x + 20. Half of 8 is 4 and the square of this is 16 so

write
x2 8x + 20 = ( x2 8x +16) + (2016) = ( x 4)2 + 4
(iii) x2 + 18x 12. Half of 18 is 9 and the square of this is 81 so

write
x2 + 18x 12 = ( x2 + 18x +81) + (1281) = ( x + 9)2 93
(iv) x2 10x. Half of 10 is 5 and the square of this is 25 so

write
x2 10x = ( x2 10x +25) + (25) = ( x 5)2 25
10.1 We need to find the y-intercept and the x-intercept.

(i ) y = 4x 8. The y-intercept is c = 8 and to get the
x-intercept solve
0 = 4x 8 4x = 8 x=2
and hence the graph is
2 x
135
136 mathematics fundamentals Appendix 1: Answers to exercises
(ii ) y = 9 3x. The y-intercept is c = 9 and to get the

x-intercept solve
0 = 9 3x 3x = 9 x=3
y
9
3 x
(iii ) y = 2x + 1. The y-intercept is c = 1 and to get the

x-intercept solve
1
0 = 2x + 1 2x = 1 x=
2
12 x
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136
1: Answers 137
to exercises
10.2 Horizontal lines have equation y = c (that is, m = 0). Now, the
y-coordinate of the point (4, 5) is c = 5 so the equation of the
line must be
y = 5
10.3 We are told that the y-intercept is 6 and hence c = 6. Observe

that the given line (y = 2x + 5) has gradient 2. As the desired
line is perpendicular to this line then it must have gradient
1
m= The general equation of a line is
2 y = mx + c
and hence the equation of the desired line is
1
y = x+6
2
10.4 To find the equation of the line containing the points (2, 4)
and (3, 1) we can use either of Method 1 or Method 2.
Method 1: Substituting the two co-ordinates into the equation
of a line y = mx + c gives us a pair of simultaneous equations to
be solved for m and c:
(2, 4) 4 = m (2) + c (1)
(3, 1) 1 = m3 + c (2)
5 = 5m
5
So =1
m=
5
Substituting m = 1 into say equation (2) gives
1 = 3+c
c = 2
Hence the equation is

y = x2
Method 2: The points are (2, 4) and (3, 1). Use the defini-
tion of gradient to find m:
change in y value 1 (4) 1+4 5
m = = = = = 1
change in x value 3 (2) 3+2 5
So the equation (so far) is y = x + c. To find c, substitute one of
the points, say (3, 1) into this equation and solve for c:
1 = 3+c c = 2
Hence, the equation of the straight line is y = x 2

y
(3,1)
(-2,-4)
137
138 mathematics
11.1 To sketch the graph of y = x2 2x 8 we need
concave up
the yintercept is at (0, 8)
to find the x intercepts:
x2 2x 8 = 0
( x + 2)( x 4) = 0
So, the x intercepts are (2, 0) and (4, 0).

the vertex occurs halfway between the x intercepts, at x = 1
(or use
b (2)
x= = = 1)
2a 2(1)
The yvalue at x = 1 is then
y = (1)2 2 (1) 8
= 128
= 9
The vertex is (1, 9).
3 2 1 1 2 3 4 5 x
1
2
3
4
5
6
7
8
9
(1, 9)
10
12.1 In each case, we need to factorise the numerators
a2 a 2
( a 2
)( a + 1)
(i ) lim = lim = lim ( a + 1) = 3
a 2 a2 a 2 a 2 a 2
s2 2s 8
(s 4
)(s + 2)
(ii ) lim = lim = lim (s + 2) = 6
s 4 s4 s 4 s 4 s 4
t2 5t + 4 (t 1)(t 4)
(iii ) lim = lim does not exist
t 6 t6 t 6 t6
4z2 11z 3
( z 3)(4z + 1)
(iv) lim = lim = lim (4z + 1) = 13
z 3 z3 z 3 z 3 z 3
138
1: Answers 139
to exercises
12.2 Recall that

d d d
( x n ) = nx n1 (x) = 1 (1) = 0
dx dx dx
(i ) f ( x ) = 2x3 + 4x2 + 3x + 5. Then
f ( x ) = 2(3x2 ) + 4(2x1 ) + 3(1) + 5(0) = 6x2 + 8x + 3
(ii ) g(t) = 2t(t2 + t + 1) = 2t3 + 2t2 + 2t. Then
g (t) = 6t2 + 4t + 2
3x4 1
(iii ) z = 4 + x + x + 4x. We have
3 4 1
z= x + x 1 + x 2 + 4x
4
dz 3 1 1
= (4x3 ) x 2 + x 2 + 4
dx 4 2
1 1
= 3x3 2 + + 4
x 2 x
139
140 mathematics
12.3 Recall that Recall that

x n +1
x n dx = +C n = 1
n+1
(i ) If f ( x ) = 5x2 + 3x + 4 then

5x3 3x2
f ( x )dx = + + 4x +C
3 2
3x4 1
(ii ) If g( x ) = 4 + x2
+ x then we have
3 4 1
g( x ) = x + x 2 + x 2
4
and hence
3
3 x5 x 1 x2 3x5 1 2 3
g( x )dx = + + 3 +C = + x 2 +C
4 5 1 2
20 x 3
y
12.4 We have to integrate y = f ( x ) = 12 3x2 , but from where to
where? We need to know its x-intercepts. That is, 12
2 2 2
12 3x = 0 3x = 12 x =4 x = 2
9
We have
F(x) = f ( x )dx = 12x x3 + C
6
Hence the required area is
2
Area = F ( x ) = 12(2) 23 +
C 12(2) (2)3 +
C 3
2
= (24 8) (24 + 8) = 16 (16) = 16 + 16 = 32

4 2 2 4 x
140
MATHEMATICS FUNDAMENTALS Appendix 2: Solutions to problem sets
Appendix 2: Solutions to problem sets
Problem Set 1
1.1 Evaluate the following.
(a) 4+4444 (b) 4 + 4 (4 4 4)
= 4 + 16 4 4 = 4 + 4 (1 4)
= 4+44 = 4 + 4 (3)
= 4 = 4 12
= 8
(c) 2 4 (8 1) (d) 14 2 (20) (5)

= 247 = 7 (4)
= 87 = 11
= 1
(e) (4 12) [32 (4)] (f) [4 (3 + 4) 21] [2 14 7 + 3]

= (48) [36] = [4 7 21] [28 7 + 3]
= 84 = [28 21] [4 + 3]
= 77
= 1
(g) 3 8 [16 (4)] (h) [3 (3 4) 19] [8 4 7 6]

= 24 [16 + 4] = [(3) (1) 19] [2 7 6]
= 24 [12] = [3 19] [14 6]
= 24 + 12 = 16 8
= 36 = 2
(i) (8 4) (4 + 4) 8 (j) 4 + 4 4 (4 4)
488 4+440
1
= 8 but dividion by zero is not allowed
2
1 16
=
2 2
15
=
2
141
142 mathematics
FUNDAMENTALS Appendix 2: Solutions to problem sets
1.2 Compute the following, leaving your answer in the simplest

fraction form.
1 3 5 3 4 3
(a) + + (2 and 4 divide into 12) (b) + (4, 5 and 10 divide into 20)
2 4 12 4 5 10
16 33 5 35 44 32
= + + = +
2 6 4 3 12 4 5 5 4 10 2
6 9 5 15 16 + 6
= + + =
12 12 12 20
6+9+5 5
= =
12 20
20 5
1
= =
12 5
4
54 1
= =
34 4
5
=
3
4 2 2 3 4
(c) +1 (d) 1 2 +1
24 3 3 5 5
1 4 5 5 13 9
= + = +
64 3 3 5 5
1 52 25 39 + 27
= + =
6 32 15
1 + 10 13
= =
6 15
11
=
6
4 3 2 2
(e) (f) 2 4
5 7 3 9
43 8 38
= =
57 3 9
12 8 9
= =
35 3 38
4 2 33
=
1 3 2 19
43
=
1 19
12
=
19
7 1 5 3 34
(g) 1 (h) 2
9 8 7 7 14
7 9 7 17 14
= =
9
8 5 7 34
717
17
72
= =
185 7

17 2
49 2
= =
40 2
= 1
142
appendix 2: solutions to problem sets 143
1.3 What is half of the difference between a third of 60 and a quar-

ter of 80?

1 1 1 1
60 80 = (20 20) = 0
2 3 4 2
1.4 What is the sum of half of 45 and a third of the product of two
thirds and 1 18 ?
1 1 2 1
45 + 1
2 3 3 8
1 45 1 2 9
= +
2 1 3 3 8
45 1 91
= + =
2 4 4
1.5 James and his wife Sweet Li have a 12 year old daughter called
Lyn. James is 50 year old and Lyn is 12. Half of his age added
to five thirds of Lyns age gives the age of Sweet Li. How old is
Sweet Li?
1 5
50 + 12 = 25 + 20 = 45, so Sweet Li is 45
2 3
1.6 Swee Khum put a third of her savings in the bank, a third in
bonds, a quarter of the remainder in stocks and the rest in fixed
deposit. If her total amount is $600,000 how much did she put in
fixed deposit?
After Swee Khum put a third in the bank and a third in bonds, she is
left with one third of her savings, which is $200,000. She then puts a
3
quarter in stocks, so she is left with $200, 000 = $150, 000, so she
4
puts $150,000 in fixed deposit.
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143
144 mathematics
Problem Set 2
2.1 Solve the following equations for the unknown.
(a) x + 3 = 5
x+3 = 5
x+33 = 53
x=2
(b) 4y 5 = y + 10
4y 5 = y + 10
4y 5 + 5 = y + 10 + 5
4y = y + 15
4y y = y + 15 y
3y = 15
3y 15
=
3 3
y=5
(c) 8x 4 = 16
8x 4 = 16
8x 4 + 4 = 16 + 4
8x 20
=
8 8
20
x=
8
5
=
2
2z
(d) =4
z+2
First we get z out of the bottom line by multiplying both sides by
z + 2.
z + 2 2z
= 4( z + 2)
1 z+2
2z = 4z + 8
2z 2z = 4z + 8 2z
0 = 2z + 8
0 8 = 2z + 8 8
8 = 2z
4 = z
144
MATHEMATICS FUNDAMENTALS appendix 2: solutions
Appendix to problem
2: Solutions sets 145
to problem sets
(e) 3 2x = 4
3 2x = 4
3 + 3 2x = 4 3
2x = 1
2x 1
=
2 2
1
x=
2
(f) 7x + 7 = 2( x + 1)
7x + 7 = 2( x + 1)
7x + 7 = 2x + 2
7x + 7 7 2x = 2x + 2 7 2x
5x = 5
x = 1
1
(g) x =2
2
1
x =2
2
1
x=2
2
5
=
2
2y 3 6y + 7
(h) =
4 3
First multiply both sides by 3 and by 4 to remove the fractions.
2y 3 6y + 7
3 4 = 3 4
4 3

3 (2y 3) = 4 (6y + 7)
6y 9 = 24y + 28
6y 9 + 9 24y = 24y + 28 + 9 24y
18y = 37
37
y=
18
(i) t = 2 2[2t 3(1 t)]
First we simplify the bracket term.
t = 2 2[2t 3 + 3t]
= 2 2[5t 3]
= 2 10t + 6
= 8 10t
t + 10t = 8 10t + 10t
11t = 8
8
t=
11
145
146 mathematics fundamentals Appendix 2: Solutions to problem sets
3
(j) (4a 3) = 2[ a (4a 3)]
2
First we simplify the bracket term on the right-hand side, and them
multiply both sides by 2 to remove the denominator on the left-hand
side.
3
(4a 3) = 2[ a (4a 3)]
2
= 2[ a 4a + 3]
= 2[3 3a]
= 6 6a
3
2 (4a 3) = 2 (6 6a)
2

3(4a 3) = 12 12a
12a 9 = 12 12a
12a 9 + 9 + 12a = 12 12a + 9 + 12a
24a = 21
21
a=
24
7
=
8
x+3 2
(k) =
x 5
Cross-multiply to get
5( x + 3) = 2x
5x + 15 = 2x
3x = 15
x = 5.
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146
1 2
(l) =
p1 p2
Cross-multiply to get
p 2 = 2( p 1)
p 2 = 2p 2
p 2 + 2 p = 2p 2 + 2 p
0=p
x x
(m) =1
2 5
Multiply throught by 2 and by 5 to get
x x
=1
2 5
x x
2 5 2 5 = 2 5 1
2 5
5x 2x = 10
3x = 10
10
x=
3
y4 y
(n) + =4
2 3
Multiply through by 2 and by 3 to get
y4 y
2 3 + 2 3 = 2 3 4
2 3

3(y 4) + 2y = 24
3y 12 + 2y = 24
5y = 36
36
y=
5
2 3
(o) +4 =
y y
Multiply both sides by y to get

2 3
y +4 = y
y y

2
y + y 4 = 3
y
2 + 4y = 3
4y = 1
1
y=
4
2y 1y
(p) =y
3 2
147
2y 1y
2 3 2 3 = 23y
3 2

2(2 y) 3(1 y) = 6y
4 2y 3 + 3y = 6y
1 + y = 6y
1 = 5y
1
so y =
5
1y
(q) y +4 = 7
2
First transfer all the numbers to the right-hand side, and then multi-
ply both sides by 2.
1y
y =3
2
2y (1 y) = 6 (dont forget the brackets here!)
2y 1 + y = 6
3y = 7
7
y=
3
y
(r) (3 y ) = 4 y
3
Multiplying both sides by 3 gives
y 3(3 y ) = 3(4 y )
y 9 + 3y = 12 3y
4y = 21 3y
7y = 21
y=3
d1 d
(s) =0
2 3
d1 d
2 3 2 3 = 2 3 0
2 3
3(d 1) 2d = 0
3d 3 2d = 0
d=3
148
4 p
(t) = 8
3
Multiply both sides by 3 to get
4 p = 24
p = 28
p 28
=
1 1
p = 28
d1 d
(u) =0
2 3
d1 d
2 3 2 3 = 2 3 0
2 3
3(d 1) 2d = 0
3d 3 2d = 0
d=3
4 p
(v) = 8
3
Multiply both sides by 3 to get
4 p = 24
p = 28
p 28
=
1 1
p = 28
149
1 1
2.2 If 2 cakes are served in slices of of a cake, how many slices
4 8
are served?
1 1
2
4 8
9 8
=
4 1
9 4 2
=
4 1
= 18 slices
2.3 Lian earns $2,000 a fortnight and spends $1200 of that. What
proportion of his salary does he save?
Lian saves $2,000 - $1,200 = $800, which is
800
2000
8
=
20
2 2
= which is 5 of his salary.
5
2.4 Nadia buys a packet of 60 samosas for a tea party. She takes
out half for her family. She takes the remaining to the party. On
the way she meets a friend who takes a sixth of the samosas.
A little further on she sees a hungry cat and gives it a fifth of
the samosas she has. The she suddenly feels hungry and eats a
quarter of the remaining samosas. How many does she arrive
with at the tea party?
Nadia starts with half the samosas for her party, so she takes 12 60 =
30 samosas. Her friend takes 61 of these, so she is left with 56 30 = 25
samosas. She gives a cat 15 of these, so she is left with 45 25 = 20
samosas. She eats 14 of these, so she is left with 34 20 = 15 samosas.
She arrives at the party with 15 samosas.
2
cups of flour for 14 serv-
2.5 I have a naan recipe that calls for 4
3
ings of naan. I want to make one serving only. How much flour
should I use?
2
4 14
3
14 1
=
3 14
1
=
3
1
so I need 3 cup of flour.
150
MATHEMATICS FUNDAMENTALS Appendix
appendix 2: Solutions
2: solutions toto problem
problem sets 151
sets
2.6 Suppose Mike travels a certain distance on the first day and
twice the distance on the next day. If the total distance he trav-
elled is 60 km, how far does he travel on the first day?
Let x denote the distance travelled on the first day. The distance trav-
elled on the second day is 2x. Since the total distance travelled is 60km,
we must have
x + 2x = 60 3x = 60 x = 20
so Mike travels 20 km on the first day.
2.7 Joy, Pam, Sandra and Lilin each make a donation to the Guide
Dogs Association. Sandra gives twice as much as Lilin, Pam
gives three times as much as Sandra and Joy gives four times as
much as Pam. If their total gift is $132, find the amount of Lilins
donation, and hence the amount donated by each.
Let the amount given by Lilin be x. Then Sandra gives 2x, Pam gives
3 2x = 6x and Joy gives 4 6x = 24x. Total donation was $132, so
x + 2x + 6x + 24x = 132
33x = 132
132
x=
33

11
12
x= (via 11 times table)
3
11

12
x=
3
3 4
= =4
31

Thus Lilin gave $4, Sandra gave $8, Pam gave $24 and Joy gave $96.
The total of these amounts is $132 as required.
151
152 mathematics
Problem Set 3
Please note that these solutions present only one option for solving
these problems. You may have taken different steps. The important thing is
that you get the right answers at the end.
3.1 Solve the following sets of simultaneous equations.
(a)
2x + y = 8 (1)
(+) 3x y = 7 (2)
5x = 15
15
x = =3
5
Substitute this back into either of the equations, for example,

Eq. (1) to get We can check this solution by substi-
tuting these values of x and y into the
23+y = 8 LHS of the other equation (2):
6+y = 8 3 3 2 = 7 = RHS
y = 86 = 2 so the solution is correct: x = 3, y = 2.
(b)
2x + 3y = 19 (1)
4x y = 3 (2)
Multiply Eq. (1) by 2 to get
4x + 6y = 38 (3)
() 4x y = 3 (2)
7y = 35
35
y = =5
7
Note that 6y (y) = 7y.

Substitute this back into either of the original equations, for
example, Eq. (1) to get
2x + 3 5 = 19 We can check this solution by substi-

2x + 15 = 19 LHS of the other equation (2):
2x = 19 15 4 2 5 = 3 = RHS
2x = 4 x=2 so the solution is correct: x = 2, y = 5
152
Appendix to problem
to problem sets
(c)
x + 2y = 10 (1)
() x y = 1 (2)
3y = 9
We can check this solution by substi-
y = 3 tuting these values of x and y into the
LHS of the other equation (1):
Note that 2y (y) = 3y.
4 + 2 3 = 10 = RHS
example, Eq. (2) to get so the solution is correct: x = 4, y = 3
x3 = 1 x=4
(d)
4x y = 0 (1)
() 2x y = 2 (2)
2x = 2
x = 1 tuting these values of x and y into the
Note that y (y) = 0 and 0 (2) = 2.
Substitute this back into either of the original equations, for 2 1 4 = 2 = RHS
41y = 0 4=y
(e)
2x + y = 4 (1)
5x + 2y = 9 (2)
4x + 2y = 8 (3)
() 5x + 2y = 9 (2)
x = 1 We can check this solution by substi-
x = 1 LHS of the other equation (2):
Substitute this back into either of the original equations, for 5 1 + 2 2 = 9 = RHS
21+y = 4 y=2
(f)
x + 3y = 3 (1)
2x + y = 8 (2)
2x + 6y = 6 (3)
(+) 2x + y = 8 (2)
7y = 14
y = 2

x+32 = 3 x = 36 x = 3
153
2 (3) + 2 = 8 = RHS
so the solution is correct: x = 3, y =
MATHEMATICS FUNDAMENTALS Appendix 2:
2 Solutions to problem sets
(g)
2x 3y = 5 (1)
5x + 2y = 16 (2)
Do 5 Eq. (1) and do 2 Eq. (2) to get
10x 15y = 25 (3)

() 10x + 4y = 32 (4)
19y = 57
57
y =
19
1
93
y = = 3
1
91
2x 3(3) = 5 tuting these values of x and y into the
2x + 9 = 5
5(2) + 2(3) = 10 6 = 16 = RHS
2x = 5 9
so the solution is correct: x = 2, y =
4 3
x= = 2
2
(h)
x + y = 5 (1)
() 2x + y = 7 (2)
x = 2 We can check this solution by substi-
x = 2 LHS of the other equation (2):
Substitute this back into either of the original equations, for 2 2 + 3 = 7 = RHS
2+y = 5 y=3
154
Appendix to problem
to problem sets
(i)
4x + y = 10 (1)
2x + 3y = 10 (2)
Do 2 Eq. (2) to get
4x + y = 10 (1)
() 4x + 6y = 20 (3)
5y = 10
10
y =
5
= 2

4x + 2 = 10 tuting these values of x and y into the
4x = 10 2
2 2 + 3 2 = 10 = RHS
8
x= so the solution is correct: x = 2, y = 2
4
=2
(j) This one is too difficult for a Test question but if you like fraction
arithmetic ....
2x + 3y = 8 (1)
3x 2y = 7 (2)
Multiply Eq. (1) by 3 and Eq. (2) by 2 to get
6x + 9y = 24 (3)
() 6x 4y = 14 (4)
13y = 38
38
y =
13
example, Eq. (1) to get We can check this solution by substi-
38 LHS of the other equation (2):
2x + 3 =8
13
5

38 15 76
3 2 =
26x + 3 38 = 13 8 13 13 13 13
91
26x + 114 = 104 =
13
26x = 10
=

13 7
10
13 1
x= = 7 = RHS
26
25 so the solution is correct.
=
2
13
5
=
13
155
156 mathematics
3.2 Tickets for an ice-skating display are sold at $5 for adults and
$2 for children. If 101 tickets were sold altogether for a take of
$394, find the number of adults and children who attended.
Let a denote the number of adults and c the number of children.
We are told that 101 tickets were sold so we must have
a + c = 101 (1)
We are also told that
5a + 2c = 394 (2)
To solve this set of equations multiply Eq. (1) by 5 to get
5a + 5c = 505 (3)
5a + 2c = 394 (2)
Do Eq. (3) - Eq. (2) to get

111
5
5
a a + 5c 2c = 505 394 3c = 111 c=
3
It turns out that 111 = 3 37 so
111 37 3 37
c= = c= y = 37
3 1 3 1
a + c = 101 a + 37 = 101 a = 101 37 a = 64
That is, 64 adults and 37 children attended the display.
3.3 An island contains foxes and rabbits. An ecologist counts both

species to study their interaction and how their populations
change over time. Last year she found the total number of foxes
and rabbits was 7,290 and the foxes were only 1 eighth of the
rabbit population. How many of each species was present?
Let f be the number of foxes and r be the number of rabbits.
We are told that r + f = 7290. We are also told that the foxes
were only 1 eighth of the rabbit population, that is, the rabbit
population is eight times as large as the fox population. That
is r = 8 f which we can write as r 8 f = 0. Hence the two
simultaneous equations are
r + f = 7290 (1)
r 8f = 0 (2)
Do Eq. (1) - Eq. (2) to get

7290
r r + f (8 f ) = 7290 0 f + 8 f = 7290 9 f = 7290 f =
9
= 810
by using a calculator. The rabbit population is eight times this

much so
r = 8 810 = 6480
156
Appendix to problem
to problem sets
3.4 The ancient Roman emperor Augustus was fond of gold and
silver sovereigns. By royal decree every gold sovereign was to
weigh 50 grams and every silver one to weigh 40 grams. One
year a new jeweller was hired to prepare the sovereigns. Augus-
tus suspected that the jeweller was cheating him and delivering
sub-weight sovereigns. A consignment contained 30 gold and
20 silver sovereigns weighed 2, 250 grams, and a consignment of
15 gold and 25 silver sovereigns weighed 1, 550 grams? Was the
jeweller cheating Augustus?
Let g be the weight of a gold sovereign and s the weight of a
silver sovereign. Then
30g + 20s = 2250 (1)

15g + 25s = 1550 (2)
Do 2 Eq. (2) to get
30g + 20s = 2250 (1)

30g + 50s = 3100 (3)
Now do Eq. (3) - Eq. (1) to get
30g
30g
+ 50s 20s = 3100 2250 30s = 850
850 85
s= = = 28.333 grams approx.
30 3
but a silver sovereign should weigh 40 grams, so it looks like the
jeweller was cheating Augustus. However, we should first check
what the situation is with the gold soverigns. Substituting s = 85
3
into, say Eq. (1) gives
85
30g + 20 = 2250 3 30g + 20 85 = 3 2250
3
90g + 1700 = 6750
5050 505
90g = 5050 g= = = 56.111 grams approx.
90 9
so the gold soverigns are actually heavier than required!
157
Problem Set 4
4.1 Expand the following.
(a) ( x + 2)( x + 4) = x2 + 6x + 8 (i) x ( x 1)( x + 1) = x3 x

(b) ( x 1)( x 2) = x2 3x + 2 (j) ( x 2y)( x + 2y) = x2 4y2
(c) ( x 2)( x + 2) = x2 4 (k) (2x 3y)2 = 4x2 12xy + 9y2
(d) ( x 1)(1 x ) = x2 + 2x 1 (l) ( x2 1)( x2 + 1) = x4 1
(e) (2x + 3)( x + 2) = 2x2 + 7x + 6 (m) ( x2 + x )( x2 x ) = x4 x2
(f) ( x + 1)2 = x2 + 2x + 1 (n) ( x + 3y)( x 3y) = x2 9y2
(g) (2x 1)2 = 4x2 4x + 1 (o) ( x + 1)( x + 2)( x + 3) = x3 + 6x2 + 11x + 6
(h) ( x + 2)2 = x2 + 4x + 4 (p) ( x 1)( x + 1)( x + 2) = x3 + 2x2 x 2
(a) x2 + 5x + 6 = ( x + 2)( x + 3) (d) x2 + x 6 = ( x 2)( x + 3)

(b) x2 5x + 6 = ( x 2)( x 3) (e) x2 2x 15 = ( x 5)( x + 3)
(c) x2 x 6 = ( x + 2)( x 3) (f) x2 100 = ( x + 10)( x 10)

x2 9 ( x
( x 3) 3
+)
(a) =
x+3
x+ 3
= x3
x2 + 5x 14 +
( x 7
)( x 2)
(b) =
x2 + 10x + 21 (x+7
)( x + 3)
x2
=
x+3
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158
Appendix to problem
to problem sets
Problem Set 5
There are often several different (but equally efficient) ways to solve
index problems. Only one version is presented here so dont worry if you
get the same answer by a different method.

(a) x3 x4 x = x3+4+1 = x8
(b) x2 y3 x = x2+1 y3 = x3 y3
(c) ( xy)2 = x2 y2
a2 b5 a2 b5
(d) = = a21 b51 = ab4
ab a b
2x4 y2 1 2 x4 y2 1 1 x2
(e) 2 2
= 2 2 = x2 y0 = x2 or
6x y 3 2 x y 3 3 3
2 3 3 2 3 6
(f) 4x =4 x = 64x (note that 4 4 4 = 16 4 = 8 8)
(3x2 z)3 33 x6 z3 27 27 27x4 z3
(g) 2
= 2 2 = x62 z3 = x4 z3 or
(2x ) 2 x 4 4 4
4 5 4 5
9x 9x
(h) = = ( x42 )5 = ( x2 )5 = x10
(3x )2 9x2
5.2 Express the following numbers without indices

24 16 24 24
(a) = = 1 or by index laws = =1
42 16 ( 22 ) 2 24
512 512 512
(b) 5
= 2 5 = 10 = 51210 = 52 = 25
25 (5 ) 5
1
(c) 41 =
4
2 1 1
(d) 10 = 2 =
10 100
35 35 35 1
(e) 2
= 3 2 = 6 = 356 = 31 =
27 (3 ) 3 3
1 1
(f) (2)4 = =
(2)4 16
1 1
(g) (3)3 = =
(3)3 27
(h) 06 = 016 = 0001 000 = 10 but this is not defined.
(i) (6)0 = 1
159
160 mathematics

(a) y2 y3 = y2+(3) = y5
(b) x 3 x3 = x 3+3 = x0 = 1
(c) ( a3 b4 )( a1 b) = a3 b4 a1 b = a3+(1) b4+1 = a4 b3
4g3 h5 4 g 3 h 5 1 1
(d) = = g3(5) h53 = g2 h8
12g5 h3 12 g5 h3 3 3
1 1 2 4
(e) (9x 1 y2 )2 = (9)2 x 1(2) y2(2) = 2
x 2 y 4 = x y
(9) 81
1
3ab2 c
(f) = (recognising that a power of 1 represents reciprocal)
c 3ab2
6a3 b2 8a3 b 6a3 b2 b 3 6 a 3 b 2+1 3a3 b1 3 3
(g) 2 1 = 2 1 3 = 2+3 1+1 = 5 0
= a2 b1 or 2
a b b a b 8a b 8a b
4 4a b 4 4a b
(2ab)3 (2)3 a3 b3 8 3(2) 32
(h) = 2 12 2 = a b = 2a5 b
(2a1 b)2 2 a b 4
2 1 0
3x y
(i) = 1 (always!)
2
3 2 2
x6 4x 1 4 31 1 2 2 1 1 x7
(j) 3 = x 63 x = x3 x = x3 x4 = x7 or
x 12x 3 4 3 9 9 9
5.4 index laws to simplify each of the following expressions

(a) 3x 2x 31 x = 3x+1 x 2x = 3 2x (note that the is needed here)
22
(b) = 22 x
2x
xy yx
(c) = x y(y) y x = x2y y x
x y
(9 a ) b 9ab (32 ) ab 32ab

(d) 2a b
= 2ab = 2ab = 2ab = 1
(3 ) 3 3 3
1 1 1 2 2 2
(a) 8 3 = (23 ) 3 = 23 3 = 21 = 2 (i) 32 5 = (25 ) 5 = 25 5 = 22 = 4
1 1 5 15
(b) 32 = (25 ) = 2
5 5 = 21 = 2 3 3 3
(j) 36 2 = (62 ) 2 = 62 2 = 63 = 216
1 1 2 12
(c) 81 = (92 ) = 9
2 2 = 91 = 9 3 3 3 1
1 1 3 1 (k) 49 2 = (72 ) 2 = 72( 2 ) = 73 =
(d) 125 3 = (5 ) 3 = 5 3 = 51 = 5
3 343
1 1 1 5 5
(e) 81 4 = (34 ) 4 = 34 4 = 31 = 3 (l) 64 6 = (26 ) 6 = 25 = 32
1 1 1
(f) 64 3 = (43 ) 3 = 43 3 = 41 = 4 1 1 1
(m) 64 6 = (26 ) 6 = 21 =
14 14 1 2
(g) 10000 = (104 ) = 101 = 3 3
10 (n) 81 4 = (34 ) 4 = 33 = 27
1 1 1
(h) 144 2 = (122 ) 2 = 121 = 2 2
(o) (27) 3 = [(3)3 ] 3 = (3)2 = 9
12
160
Appendix to problem
to problem sets

2 2
(a) 27 3 = (33 ) 3 = 32 = 9
2 2 1
(b) 27 3 = (33 ) 3 = 32 =
9
7 7
(c) 4 2 = (22 ) 2 = 27 = 128
1 1
1
(d) 16 2 = (42 ) 2 = 41 =
4
3 3
(e) 100 2 = (102 ) 2 = 103 = 1000
3 3 1
(f) 16 4 = (24 ) 4 = 23 =
8
1 1 1
(g) 169 2 = (132 ) 2 = 131 =
13
4 4
(h) 32 5 = (25 ) 5 = 24 = 16
4 4
(i) 27 = (33 ) = 34 = 81
3 3
5 5 1
(j) 64 6 = (26 ) 6 = 25 =
32
3 3 3
2 2
2
4 2 2 8
(k) = = =
9 3 3 27
3 3 3
2 2
2
4 2 2 27
(l) = = =
9 3 3 8
1 1 1
(m) 81 2 = (92 ) 2 = 91 =
9
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161
162 mathematics

1 2 1 2 7
(a) 3x 2 4x 3 = 12x 2 + 3 = 12x 6
2 1 2 1 3
(b) 5x 5 6x 10 = 30x 5 +( 10 ) = 30x 10
3
1 1 3 64x 2 64x
(c) (4x 2 )3 (9x 3 ) 2 = 1
=
27x 2 27
1 5 2 1 1 1
(d) (64x2 ) 6 (32x 2 ) 5 = 64 6 x 3 5 2
(32x 2 ) 5
2
1 5 2 32 5 x 4x 2
= 1 1
(32x 2 ) 5 = 1
= 1
= 2x 3
64 x
6 3 2x 3 2x 3
2
3 52 5 5
x y x y y
3 2 7
(e) 5 = 2 = xy 2
x 3 y 1 x 3
3 4 3 2 5
x 7 y 5 x x77 x7
(f) 2 3 = 3 4
= 7
x 7 y 5 y5 y5 y5
1 1 1 1
(a) 3x 2 2x 2 = 6x 2 2 = 6x0 = 6
2 2 10 1 1
(b) (8x3 y5 ) 3 = 8 3 x 2 y 3 = 2 10
= 10
(23 ) x2 y 3 3 4x2 y 3

(c) ( x 5 )3 5 =x 53 5 = x15

3x 27 3x 93 3x3 3
3
(d) = = = 3x
x 12 x 43 x2 3
1 2 1 2 1
(e) 3x 2 4x 3 = 12x 2 + 3 = 12x 6

(f) ( x2 7 ) 7 = x27 = x14
x 3 y5 y5 y7 y12
(g) 2 7 = 2 3 = 5
x y x x x
1 1 1 1 1
(h) 7x 4 2x 2 = 14x 4 +( 2 ) = 14x 4
1 1 1 1 1 1 5
(i) (9x ) 2 (27x ) 3 = 3x 2 3x 3 = 9x 2 + 3 = 9x 6
1 1
1 1 3x 2 3x 2 1 1 1
(j) (9x ) 2 (27x ) 3 = 1
= 1
= x23 = x6
(27x ) 3 3x 3
1
1 3 3x 3 3 3 3
(k) 3x 3 2x 4 = 3 = 3 1
= 3+1
= 13
2x 2x x 2x 2x 12 4 4 3 4 3

(a) 12 = 4 3 = 4 3 = 2 3

(b) 72 = 36 2 = 36 2 = 6 2

(c) 150 = 25 6 = 25 6 = 5 6

(d) 3 96 = 3 8 12 = 3 8 3 12 = 2 3 12

(e) 2(3 2) = 3 2 2 2 = 3 2 2

(f) 5(2 3 + 2) = 2 15 + 10

(g) (1 2) 3 = 3 6

(h) (7 + 3 5) 3 = 7 3 + 3 15

(i) (6 2 5)(6 + 2 5) = 62 (2 5)2 = 36 22 5 = 16

(j) (3 + 7)2 = (3 + 7)(3 + 7) = 9 + 3 7 + 3 7 + 7 = 16 + 6 7

(k) (4 3 1)(2 5 + 1) = 8 15 + 4 3 2 5 1

(l) ( 5 3)(2 5 + 4 3) = 2 5 + 4 15 2 15 4 3 = 10 + 2 15 12 = 2 15 2
162
Appendix to problem
to problem sets
Problem Set 6
6.1 For each of the following formulae, isolate the indicated variable In each case, think of the operation
that has been applied to both sides.
These are listed for the first two parts
5 to get you started.
(a) Solve C = ( F 32) for F
9
9
C = F 32 [multiply both sides by 95 ]
5
9
32 + C = F [add 32 to both sides]
5
4 3
(b) Solve V = r for r
3
V 4 3
= r [divide both sides by ]
3
3V
= r3 [multiply both sides by 34 ]
4
1
3V 3 3V
[raise both sides to the power of 13 ]
3
r = or
4 4
(c) Solve PV = nRT for T

PV
= T
nR
GMm
(d) Solve F = for r
r2
r2 F = GMm
GMm
r2 =
F

GMm
r =
F
L
(e) Solve b = for d
4d2
4d2 b = L
L
d2 =
4b

L
d =
4b
163
1
2 E 3
(f) Solve L = kd3 for p
p
E
L3 = k 3 d2
p
L3 p = k 3 d2 E
k 3 d2 E
p = 3
L
h2
(g) Solve L = for m
2mkT
h2
L2 =
2mkT
h2
mL2 =
2kT
h2
m =
2kTL2
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We do not reinvent
light.
light every day.
Light is OSRAM
164
Appendix to problem
to problem sets
6.2 For each of the following business formulae, isolate the indicated
variable
(a) Solve Q = a bP for P

bP = aQ
aQ
P =
b
(b) Solve A = P (1 + r ) n for r

A
= (1 + r ) n
P
1
A n
= 1+r
P
1
A n
1 = r
P
(c) Solve MV = PY for Y

MV
= Y
P

1
(d) Solve R = P 1 for e
e
R 1
= 1
P e
1 R
= 1
e P
P R
=
P P
PR
=
P
P
e =
PR
SD
(e) Solve F = K for S
d2
d2 F = KSD
d2 F
= S
KD
165
166 mathematics
Problem Set 7
7.1 The required exponential forms are
(a) 102 = 100 1 (k) 20 = 1

1
(p) 9 2 = 3
(f) 32 =
9
(b) 103 = 1000 (g) n2 = 256 (l) 21024 = n (q) n2 = 2
(c) 100 = 1 (h) n3 = 14 (m) 26 = 64 1
(r) 10 2 = x
(d) 32 = 9 (i) n4 = 62 (n) 92 = 81 (s) 5x = 100
1 1 1
(e) 30 = 1 (j) 23 = (o) 9 2 = (t) x2 = 12
8 3
7.2 The required logarithmic forms are
(a) 5 = log2 32 (g) 1 = log16 16 1

(l) = log25 5
4
(b) 3 = log2 8 1 1
(h) = log81 (m) 4 = log10 10000
(c) 0 = log2 1 2 9
(n) x = log3 81
1
(d) 2 = log8 64 (i) 2 = log3 (o) x = log25 5
9
1 (p) 5 = logx 7
(e) 1 = log8 1
8 (j) 3 = log3
27 (q) x2 = log3 81
1
(f) = log8 2 (k) 5 = log3 243 (r) 3x + 1 = log20 3
3
7.3 (a) log 5 + log 2 = log(5 2) = log 10

8
(b) log 8 log 2 = log = log 4
2

8
(c) 3 log 2 2 log 3 = log 23 log 32 = log
9

4 1
(d) log 4 log 20 = log = log
20 5

12
(e) log 12 log 3 = log = log 4
3

16
(f) log 16 log 4 = log = log 4
4

12
(g) log2 12 log2 3 = log2 = log2 4 = 2
3
(h) log10 25 + log10 4 = log10 (25 4) = log10 100 = 2

25 1
(i) log10 25 log10 250 = log10 = log10 = 1
250 10
(j) log5 20 + log5 4 = log5 (20 4) = log5 80
2
x 1 1
(k) log + log( x 2 y3 ) = log x2 y 2 + log( x 2 y3 ) = log x2 y 2 x 2 y3
y
5

= log x0 y 2 = 52 log y
1 2
2
2 1 1
2

(l) 2 log x y log y x
2 = log x y
2 log y2 x 1 = log xy4 log y2 x 1

xy4
= log 1 2
= log x2 y2 = log ( xy)2 = 2 log( xy)
x y

log10 27 log10 33 3log 3 3
10
(m) = = =
log10 9 2
log10 (3 ) 2log103 2

8 18
(n) log2 8 + log2 18 log2 3 = log2 = log2 (8 6) = log2 48
3
166

10
(o) log2 10 log2 5 = log2 = log2 2 = 1
5
1

(p) log xy + 12 log( xy) log( xy) = log ( xy) 2 + 12 log( xy) log( xy)

= 12 log( xy) + 12 log( xy) log( xy) = 12 + 12 1 log( xy) = 0
7.4 In this question, we are given that x = log 3 and y = log 4.
(a) log 12 = log(3 4) = log 3 + log 4 = x + y

(b) log 36 = log 9 + log 4 = log 32 + log 4 = 2 log 3 + log 4 = 2x + y

3
(c) log = log 3 log 4 = x y
4
1
(d) log 6 = log 2 + log 3 = log 4 2 + log 3 = 12 log 4 + log 3 = 12 y + x
1
3
(e) log = log 3 log 2 = log 3 log 4 2 = log 3 12 log 4 = x 12 y
2
1
3
(f) log = log 3 log 8 = log 3 (log 4 + log 2) = log 3 log 4 + log 4 2
8

= log 3 log 4 + 12 log 4 = x y + 12 y = x 32 y

(g) log 144 = log 122 = 2 log 12 = 2(log 3 + log 4) = 2( x + y) = 2x + 2y
7.5 In the question, we are given x = log5 2 and y = log5 3.
(a) log5 6 = log5 3 + log5 2 = y + x

(b) log5 12 = log5 4 + log5 3 = log5 22 + log5 3
= 2 log5 2 + log5 3 = 2x + y

3
(c) log5 = log5 3 log5 10 = log5 3 (log5 2 + log5 5) = y ( x + 1) = y x 1
10
(d) log5 60 = log5 6 + log5 10 = log5 2 + log5 3 + log5 2 + log5 5 = 2x + y + 1

3
(e) log5 = log5 3 log5 2 = y x
2

3
(f) log5 = log5 3 log5 8 = log5 3 log5 23 = log5 3 3 log5 2 = y 3x
8
(g) log5 30 = log5 3 + log5 2 + log5 5 = x + y + 1
167
168 mathematics
Problem Set 8
8.1 (a) 2x = 64 1 92x+1 = 50

(f) 10x+1 =
100
2x = 26 2x + 1 = log9 50
10x+1 = 102
x=6 2x = log9 50 1
x + 1 = 2
log9 50 1
(b) 42 x = 128 x = 3 x= 0.39
2
16x = 128 (g) 3x = 81
3
(l) 7 x 1 = 4
x=8 x3 4
3 =3 x 1 = log7 4
x 3
(c) 9 = 27 x =4 x = log7 4 + 1 1.71
x 1
32 = 33
3 x
x = 4 or 4 3 1.59 (m) 2 2 +1 = 64
x
(h) 4 2 1 = 128 x
32x = 33 2 2 +1
= 26
2 x
(2 ) =2 2 1 7 x
2x = 3 +1 = 6
3 x 2
x= 2( 1) = 7 x
2 2 =5
2x 2
(d) 5 x = 125 2 = 7 x = 10
2
5 x = 53 x2 = 7 (n) 52x1 = 20
x = 3 x=9 2x 1 = log5 20
x
x = 3 (i) 4 = 7
2x = log5 20 + 1
x = log4 7 1.43 log5 20 + 1
(e) 82x = 32
x=
(j) 3x+1 = 10 2
(23 )2x = 25
x + 1 = log3 10 log5 4 + log5 5 + 1
6x = 5 =
2
5 x = log3 10 1 1.10 log5 4 + 2
x= (k) = 1.43
6 2
8.2 (a) log5 n = 4 (e) log 1 n = 3

3
n=5 4 3
1
=n
n = 625 3
(b) log8 4 = n 1
=n
27
8n = 4
23n = 22
(f) log6 6 = n
3n = 2
6n = 6
2
n= n=1
3
1
(c) log2 n =
3 3
1 3 (g) logn 8 =
n = 2 3 or 2 1.26 4
3
1 n4 = 8
(d) logn = 2
64 3
n 4 = 23
1 1 3
n 2 =
64 n4 = 23
1
n 2 = 2 1
n4 = 2
8
n 2 = 82 n = 24
n=8 = 16
168
Appendix to problem
to problem sets
(h) log4 3 + log4 ( x + 2) = 2 (l) log3 63 log3 (7x ) = log3 2

log4 [3( x + 2)] = 2 63
log3 log3 2 = 0
log4 (3x + 6) = 2 7x

9
3x + 6 = 42 log3 2 = 0
x
3x + 6 = 16
9
log3 =0
3x = 10 2x
10 9
x= = 30
3 2x
9
=1
2x
9 = 2x
9
(i) x = log4 72 log4 9 =x
2
72 (m) log5 4 + log5 (2x 3) = 2
x = log4
9
log5 [4(2x 3)] = 2
x = log4 8
log5 (8x 12) = 2
4x = 8
8x 12 = 52
(22 ) x = 23
8x = 25 + 12
2x = 3 37
3 x=
x= 8
2 (n) log2 2 + log2 ( x + 2) log2 (3x 5) = 3

2( x + 2)
log2 =3
3x 5

2x + 4
log2 =3
(j) x = log7 98 log7 2 3x 5

98 2x + 4
x = log7 = 23
2 3x 5
2x + 4
x = log7 49 =8
3x 5
7x = 49
2x + 4 = 8(3x 5)
7 x = 72
2x + 4 = 24x 40
x=2
44 = 22x
2=x
(o) log2 4 + log2 ( x 1) log2 (3x 4) = 2

2( x 1)
(k) log5 x log5 7 = log5 6 log2 =2
x 3x 4
log5 log5 6 = 0 4x 4
7 = 22
x 3x 4
log5 6 = 0 4x 4
7 =4
x 3x 4
log5 =0
42 4x 4 = 4(3x 4)
x
= 50 4x 4 = 12x 16
42
x 12 = 8x
=1
42 3
x = 42 =x
2
169
170 mathematics
8.3 (a) pH. Using p as the symbol for pH and H as the symbol for [ H + ]
p = log10 H p = log10 H H = 10 p
(b) Arrhenius equation

E K E K E
K = Ae RT =e RT loge =
A A RT

K E
RT loge =E T=
A K
R loge
A
(c) Relationship of rate constants at two different temperatures

k E tT
loge =
K R tT
k E t T
= e R ( tT )
K
K 1
= E t T
k e R tT )
(
k E t T E T t
K= E
t T
or ke R ( tT ) or even ke R ( tT )
(
e tT
R )
(d) Benfords Law

d+1 d+1
P = log10 = 10P d + 1 = 10P d
d d
1
1 = 10P d d 1 = d 10P 1 =d
10P 1
exciting projects.
170
Appendix to problem
to problem sets
8.4 A population starts with 1000 individuals and doubles in size every decade.
(a) Use the growth model N = N0 bt to find how long will it take for the population to reach 500, 000.
The initial population size is N0 = 1000 and doubling every decade means that measuring t in decades is a

good idea because we can then use b = 2. That is, N = 1000 2t . When N = 500, 000 we have

500, 000 = 1000 2t
500, 000
= 2t
1, 000
500

1000 t

1 =2

1000
500 = 2t
t = log2 500 8.97 decades = 89.7 years
(b) Repeat part (a) using the growth model N = N0 ekt .

Firstly, we need to get an expression for k. Since the population is doubling every decade, we know that there
will be 2000 individuals when t = 1 decades:
2000 = 1000ek(1) 2 = ek k = ln 2 0.69
When N = 500, 000 we have
500, 000 = 1000ekt so we can solve for t

kt
500 = e
kt = ln 500
t = ln 500 k
= ln 500 ln 2 from above
ln 500
= 8.97
ln 2
8.5 An endangered species numbered 7, 200 at the start of 2003 and at the start of 2014 numbered 800. Assuming
the population follows the decay law N = N0 ekt predict in what year the population will fall below 120.
If we use 2003 as our starting time (that is, t = 0) then 2014 is t = 11. This makes N0 = 7200 and N = 800
when t = 11:
N = 800 = 7200ek(11) and solve for k

800
= e11k
7200
1
800 11k

9 =e
800
1
= e11k
9
1
11k = ln = ln 91
9
= ln 9 using log laws
ln 9
k=
11
ln 9
= 0.20
11
171
172 mathematics
To find out when the population falls below 120 start with N = 120:
120 = 7200ekt so we can solve for t

120
= e11k
7200
1
120 11k

60 =e
120
1
= ekt
60
1
kt = ln
60
= ln 60 (see above)
t = ln 60 k
ln 9
= ln 60 from above
11
ln 60 11 11 ln 60
= = 20.50
1 ln 9 ln 9
That is, July 2023 is when the population will fall below 120.
8.6 According to Wikipedia (en.wikipedia.org/wiki/Wikipedia:Modelling_Wikipedia%27s_growth), the number
of articles on their site grew exponentially between October 2002 and mid 2006. There were 80,000 articles in
October 2002 and 166,000 in October 2003. When does the model predict that Wikipedia reached 1,500,000
articles? (Use the growth model N = N0 ekt .)
Firstly, we need to get an expression for k. It makes sense to use thousands of articles as the units of N (al-
though it doesnt matter if you use just articles instead) and set October 2002 as t = 0 years. This makes
N0 = 80 and there were 166 thousand articles at t = 1:

166 8
2 3 83
166 = 80ek(1) = e11k = e11k k = ln 0.73
80 4
2 0 40
When N = 1500 we have

1500
= e11k
80
2
75 0
= e11k
4
20

75
kt = ln
4

75
t = ln k
4

75 83
= ln ln from above
4 40
75
ln 4
= 83 4.02
ln 40
Hence Wikipedia would have reached 1,500,000 articles around October 2006. As it happens, growth of new
articles had started to slow down by then and this exponential model no longer worked.
172
MATHEMATICS FUNDAMENTALS appendix 2:Appendix
solutions
2: to problem to
Solutions setsproblem
173 sets
8.7 DDT is a pesticide that was very effective at controlling marlaria-carrying mosquitos until its toxic effects on
animals (including cancer in humans) were established. Furthermore, it remains active for many years in the
environment because it decays exponentially with a half-life of 15 years.
How long does it take for 100 grams of DDT to decay down to 1 gram? (Use the decay model A = A0 ekt .)
Firstly, we need to get an expression for k. Since the half-life is t = 15 years, we know that there will be 50
grams left at this time:

1 1 ln 2 ln 2
50 = 100ek(15) = e15k 15k = ln = ln 21 = ln 2 k = =
2 2 15 15
When A = 1 gram we have

1
= ekt
100
1
kt = ln
100

= ln 1001
= ln 100

360
t = ln 100 k
.
ln 2
= ln 100 from above
15
thinking
ln 100 15
=
1 ln 2
15 ln 100
= 99.66
ln 2
That is, t is between 99 and 100 years. Thats a long time for something to remain toxic!
360
thinking . 360
thinking .

173
174 mathematics
Problem Set 9
9.1 Solve the following equations by factorising.
(a) x2 3x = 0
(f) ( x 2)( x + 1) = 4
x ( x 3) = 0
x = 0 or x 3 = 0 Expand and convert to standard form:
x = 0 or x = 3 x2 x 2 = 4
x2 x 6 = 0
( x + 2)( x 3) = 0
(b) x2 5x + 4 = 0 x + 2 = 0 or x 3 = 0
( x 4)( x 1) = 0 x = 2 or x = 3
x 4 = 0 or x 1 = 0
x = 4 or x = 1
15
(g) x+ =8
(c) 2
x 5x 14 = 0 x
( x 7)( x + 2) = 0 Multiply through by x:
x 7 = 0 or x + 2 = 0 x2 + 15 = 8x
x = 7 or x = 2 x2 8x + 15 = 0
( x 5)( x 3) = 0
x 5 = 0 or x 3 = 0
(d) x2 x 42 = 0 x = 5 or x = 3
( x 7)( x + 6) = 0
x 7 = 0 or x + 6 = 0
x = 7 or x = 6
5
(h) x=
x4
(e) 3x2 6x = 3 Cross multiply:
2 x 5
3x 6x + 3 = 0 =
1 x4
Divide through by 3: x ( x 4) = 1 5
x2 2x + 1 = 0 x2 4x 5 = 0
2
( x 1) = 0 ( x 5)( x + 1) = 0
x1 = 0 x 5 = 0 or x + 1 = 0
x=1 x = 5 or x = 1
174
Appendix to problem
to problem sets
9.2 Solve the following equations by using the quadratic formula.

4 2 3
=
2
(a) x2 5x + 4 = 0, so a = 1, b = 5, c = 4: 2(2 3)
=
b b2 4ac 2
x= 1 2(2 3)
2a =
(5) (5)2 4 1 4 21

x=
21 = 2 3

5 25 16 x = 2 + 3 or x = 2 3
=
2
5 9
=
2
53 (d) 2x2 + 3x 1 = 0, so a = 2, b = 3, c = 1:
=
2
3 32 4 (2) (1)
5+3 53 x=
x= or x = 2 (2)
2 2
8 2 3 9 8
x = or x = =
2 2 4

x = 4 or x = 1 3 1
=
4
Note: we could have factorised this one:
3 1
=
( x 4)( x 1) = 0 4
3 + 1 3 1
x= or x =
4 4
1
x = or x = 1
2
(b) 3x2 + 7x + 2 = 0, so a = 3, b = 7, c = 2:

7 72 4 3 2 (e) x2 + 3x + 4 = 0, so a = 1, b = 3, c = 4:
x=
23
3 32 4 1 4
7 49 24 x=
= 21

6 3 9 16
7 25 =
= 2
6
7 5 3 7
= =
6 2
7 + 5 7 5 We cant take the square root of a negative
x= or x =
6 6 number so there are no solutions.
1
x = or x = 2
3
(c) x2 + 4x + 1 = 0, so a = 1, b = 4, c = 1:

4 42 4 1 1
x=
21

4 16 4
=
2

4 12
=
2

Recall that 12 = 2 3, so
175
(h) 2x2 2x 3 = 0, so a = 2, b = 2, c = 3:
3 2
(f) x + 4x + 1 = 0 (2) (2)2 4 2 (3)
2 x=
22
Multiply through by 2 to remove fraction:
2 4 + 24
=
3x2 + 8x + 2 = 0, so a = 3, b = 8, c = 2:
4
8 82 4 3 2 2 28
x= =
23 4
22 7
8 40 =
= 4
6
2(1 7)
=
Recall that 40 = 2 10, so 4
1 2(1 7)
=
8 2 10 =
6 4
2

2(4 10) 1 7
= =
6 2
1 2(4 10)

1+ 7 1 7
= x= or x =
63 2 2

4 10
=
3
4 + 10 4 10 (i) 2 3x2 = 0
x= or x =
3 3
No x term is the same as 0x:
3x2 + 0x + 2 = 0, so a = 3, b = 0, c = 2:

0 02 4 (3) 2
x=
2 (3)

24
(g) x2 + x 1 = 0, so a = 1, b = 1, c = 1: =
6

1 12 4 1 (1) 2 6
x= =
21 6

1 1 + 4 1 2 6
= =
2
63
1 5
= 6
2 =
3
1 + 5 1 5
x= or x = 6 6
2 2 x= or x =
3 3
176
Appendix to problem
to problem sets
9.3 Determine the number of solutions for the following equations.

All we need to do in each case is check the discriminant b2 4ac.
(a) x2 + 3x + 1 = 0 In this case we have a = 1, b = 3 and c = 1 so
b2 4ac = 32 4 1 1 = 9 4 = 5 > 0 two solutions
(b) 3x2 + 4x + 2 = 0
b2 4ac = 42 4 3 2 = 16 24 = 8 < 0 no solutions
(c) 2x2 + 3x + 5 = 0
b2 4ac = 32 4 2 5 = 9 40 = 31 < 0 no solutions
(d) 2x2 + 4x + 2 = 0
b2 4ac = 42 4 2 2 = 16 16 = 0 one solution
Note: you get the same conclusion if you divide through by 2 first: x2 + 2x + 1 = 0
b2 4ac = 22 4 1 1 = 4 4 = 0 one solution
3 2
(e) 2x + 4x + 6 = 0
3
b2 4ac = 42 4 6 = 16 36 = 24 < 0 no solutions
2
Note: you get the same conclusion if you multiply through by 2 first: 3x2 + 8x + 12 = 0
b2 4ac = 82 4 3 12 = 64 144 < 0 no solutions
(f) x2 + 5x 1 = 0
b2 4ac = 52 4 1 (1) = 25 + 4 = 29 > 0 two solutions
(g) 4x2 5x + 4 = 0
b2 4ac = (5)2 4 4 4 = 25 64 = 39 < 0 no solutions
177
178 mathematics
9.4 Complete the square for the following quadratic expressions
2 2
2 2
( a) x2 + 2x + 3 = x2 + 2x + +3
2 2
= x2 + 2x + 1 1 + 3
= ( x + 1)2 1 + 3
= ( x + 1)2 + 2
2 2
4 4
(b) x2 4x 8 = x2 4x + 8
2 2
= x2 4x + 4 4 8
= ( x 2)2 4 8
= ( x 2)2 12
2 2
10 10
(c) x2 + 10x + 7 = x2 + 10x + +7
2 2
= x2 + 10x + 25 25 + 7
= ( x + 5)2 25 + 7
= ( x + 5)2 18
2 2
2 2 6 6
(d) x 6x = x 6x +
2 2
= x2 6x + 9 9
= ( x 3)2 9
2 2
2 2 12 12
(e) x + 12x + 36 = x + 12x + + 36
2 2
= x2 + 12x + 36 36 + 36
= ( x + 6)2 36 + 36
= ( x + 6)2
(In other words, x2 + 12x + 36 is already a perfect square.)
2 2
20 20
( f ) x2 20x 1 = x2 20x + 1
2 2
= x2 20x + 100 100 1
= ( x 10)2 100 1
= ( x 10)2 101
178
Appendix to problem
to problem sets
9.5 The product of two consecutive odd numbers is 99. Find the two numbers.
Let n be the smaller number. The next odd number is then n + 2. Since the product of the two numbers is 99 we
have
n(n + 2) = 99 n2 + 2n 99 = 0
You might recognise that the factorisation is
(n 9)(n + 11) = 0
in which case n 9 = 0 or n + 11 = 0. Hence n = 9 or 11.

Answer: the numbers are either 9 and 11 or 11 and 9.
If you dont recognise the factorisation, the quadratic formula gives us that

b b2 4ac
n=
2a
2 22 4 1 (99)
=
21

2 400
=
2
2 20
=
2
2 + 20 2 20
x= or x =
2 2
x = 9 or x = 11
9.6 One positive number exceeds three times another positive number by 5. The product of the numbers is 68.

Find the numbers. Note that 25 + 12 68 = 29.
Let n be the smaller number. Then the larger one is 3n + 5. Since the product of the two numbers is 68 we have
n(3n + 5) = 68
3n2 + 5n = 68
3n2 + 5n 68 = 0
The quadratic formula then gives us that

b b2 4ac
n=
2a
5 52 4 3 (68)
=
23

5 25 + 12 68
=
6
5 29
=
6
5 + 29 5 29
x= or x =
6 6
34
= 4 or x =
6
Since we are told the number is positive, we have
Answer: the numbers are are 4 and 3 4 + 5 = 17.
179
9.7 A group of zoologists was studying the effect on the body weight of rats of varying the amount of yeast in their
diet. By changing the percentage P of yeast in the diet, the average weight gain, G (in grams), over time was
estimated to be G = 200P2 + 200P + 20. What percentage of yeast would you expect to give an average
weight gain of 70 grams?
We have to solve 200P2 + 200P + 20 = 70. That is
200P2 + 200P 50 = 0
Note that 200 = 4 50 so we can divide through by 50 to get
4P2 + 4P 1 = 0
a = 4, b = 4, c = 1

4 42 4 (4) (1)
P=
2 (4)

4 16 16
=
8
4 0
=
8
1
=
2
Answer: The percentage of yeast required is 12 %

Note: you can reduce the number of negative signs by dividing the quadratic equation through by 1 as well:
4P2 4P + 1 = 0
a = 4, b = 4, c = 1
This version leads to the same answer.

180
Appendix to problem
to problem sets
9.8 You wish to make a square-bottomed box with a height of three cm and a volume of 75 cm3 . You will take a
piece of square cardboard, cut a three cm square from each corner and fold up the sides. What sized piece of
cardboard do you need? Hint: it might help to draw a picture.
Let w = side length of of the piece of square cardboard:
3 w6 3
The base area of the box is (w 6)2 and its volume equals 75 cm3 , so we have
base height = (w 6)2 3 = 75

(w 6)2 = 25 w2 12w + 36 = 25
w2 12w + 11 = 0 (w 11)(w 1) = 0 (or use the quadratic formula)
w = 11 or 1
w = 1 isnt a plausible answer since we cant fold up 3 cm sides if the width is only 1 cm.
Answer: we need a square piece of cardboard of width 11 cm.
9.9 A rectangular park has dimensions 40 metres by 50 metres. A pathway is to be added all around the park
which will increase the total area to 3000 m2 . How wide is the pathway going to be? Hint: Draw a picture.
Let x = width of the pathway:
x
50 x
40
The side lengths of the total area (park and pathway) are 40 + 2x and 50 + 2x and the total area equals 75 cm3 ,
so we have
(40 + 2x )(50 + 2x ) = 3000

2000 + 180x + 4x2 = 3000 4x2 + 180x 1000 = 0
x2 + 45x 250 = 0
quad (dividing through by 4) (w 5)(w + 50) = 0 (or use the quadratic formula)
w = 5 or 50
w = 50 is negative so it isnt a plausible path width.
181
182 mathematics
Problem Set 10
10.1 Find the indicated values of the following functions.

f ( x ) = 3x2 + 2x + 4 g( x ) = log2 x h(u) = 11 u
s3
F (v) = 32v4 G (t) = 5 3t H (s) =
s+1

(a) f ( 2 ) = 3 ( 22 ) + 2 ( 2 ) + 4 (f) h(11) = 11 11 2 2
(j) G = 53
= 12 + 4 + 4 5 5
= 0
25 6
= 4 =0 = +
5 5
31
=
5
(b) f (1) = 3(1)2 + 2(1) + 4
(g) F (3) = 32(3)4
= 3 2 + 4
= 364 23
= 1 (k) H (2) =
= 32 2+1
=9 1
=
3
(c) g(8) = log2 8 1
=
=3 3
(h) F (2) = 32(2)4
= 344
1 1 3 3
(d) g = log2 = 30 (l) H (3) =
4 4 3 + 1

= log2 22 =1 6
=
2
= 2 =3

1 1
(i) G = 53
4 4
(e) h (7) = 11 7 20 3
=
= 4 4 4
17
=2 =
4
10.2 Sketch the graphs of the following functions
(a) y = 3x + 6 y
8
To find the x intercept: 6 (0,6)

5
1
(-2,0)
0 = 3x + 6 x
4 3 2 1 1 2 3 4
6 = 3x 1
2
2 = x
182
MATHEMATICS FUNDAMENTALS appendixAppendix
2: solutions to problem
2: Solutions sets 183sets
to problem
(b) y = 4x 4 y
8
To find the x intercept: 7
6 (0,6)
5
0 = 4x 4 4
4x = 4 3
2
x = 1 1 (3,0)
4 3 2 1
1
1 2 3 4 t
y
2
2
1
(-1,0)
x (d) z = 5x 4
4 3 2 1 1 2 3 4
1
2
To find the x intercept:
3
4 (0,-4) 0 = 5x 4
5 4 = 5x
6
4
7 =x
8 5
z
4
3
(c) y = 2t + 6 2
4
1 5, 0
To find the tintercept: x
4 3 2 1 1 2 3 4
1
2
0 = 2t + 6 3
4 (0,-4)
2t = 6
5
t=3 6
183
184 mathematics
10.3 Find the equations of the following lines.

(a) Gradient 2 and contains the point (0, 4)
Equation is y = 2x + c and point (0, 4) tells us that the y-intercept is 4 so the equation is y = 2x + 4.
(b) Gradient 3 and contains the point (1, 3)
The equation is y = 3x + c and the point (1, 3) is on the line so
3 = 3(1) + c
3 = 3+c 0=c
The equation is y = 3x.

(c) Contains the points (3, 4) and (6, 2)
Method 1 : The equation is y = mx + c and the points tell us that
(3, 4) 4 = 3m + c (1)
(6, 2) () 2 = 6m + c (2)
6 = 3m
2 = m
Substitute one of the points, say (3, 4), into y = 2x + c:
4 = 2(3) + c
4 = 6 + c 10 = c
The equation is y = 2x + 10.

Method 2 : The gradient of the line is
change in y value
m=
change in x value
2 4
=
63
6
= = 2
3
We now find c in the same way as for Method 1.

(d) Contains the point (3, 1) and is parallel to the line with equation y = 2x 1
The gradient is m = 2 and the point (3, 1) is on the line so
1 = 2(3) + c
1 = 6+c
5 = c
The equation is y = 2x 5.
(e) Perpendicular to the line y = 4x 1 and contains the point (4, 2)
The gradient is m = 14 and the point (4, 2) is on the line so
1
2 = (4) + c
4
2 = 1 + c
3=c
The equation is y = 14 x + 3.
184
Appendix to problem
to problem sets
10.4 The line L1 has equation y = 3x + 1. Find the equations of the following six lines.
(a) The line L2 which is parallel to L1 and contains the point (1, 4)
(b) The line L3 which is perpendicular to L1 and has x-intercept 3
(c) The line L4 which is perpendicular to L1 and intersects it at the point (2, 3)
(d) The line L5 which is horizontal and has the same y-intercept as L1
(e) The line L6 which contains the point (3, 7) and has the same y-intercept as L1
Note that the line L1 : y = 3x + 1 has gradient 3 and y-intercept 1
(a) L2 is parallel to L1 so y = 3x + c and it contains the point (1, 4) so
4 = 3(1) + c
4 = 3 + c
7=c
(b) L3 is perpendicular to L1 so y = 13 x + c and has x-intercept 3. Hence, (3, 0) is on L3 so
0 = 13 (3) + c
0 = 1 + c
c=1
The equation is y = 13 x + 1.
(c) L4 is perpendicular to L1 so y = 13 x + c and it intersects L1 at (2, 3) so the point (2, 3) must be on
L4 as well and hence
3 = 13 (2) + c
93 = 23 + c
c = 37
The equation is y = 13 x 73 .
(d) L5 is horizontal so it has the equation y = c. The line has the same y-intercept as L1 , that is c = 1 and
hence the equation of L5 is
y=1
(e) L6 has the same y-intercept as L1 so y = mx + 1 and it contains the point (3, 7) so
7 = 3m + 1
7 1 = 3m
6 = 3m
2=m
185
186 mathematics
10.5 Sketch the following pairs of lines from the previous question on the same axes.
(All x intercepts worked out via the same method as Question 2.)
(a) L1 : y = 3x + 1 and L3 : y = 31 x + 1 (c) L4 : y = 13 x 7

3 and L5 : y = 1
y y
4 6
3 4
2 2 (0,1)
1 (0,1)
( 13 , 0) (3,0) 12 10 8 6 4 2 2 4 x
x
2 (0, 73 )
4 3 2 1 1 2 3 4
1 (7, 0) 4
2 6
3
4
(b) L2 : y = 3x + 7 and L6 : y = 2x + 1
y
8
(0,7)
6
4
( 73 , 0)
2
(0,1)
8 6 4 2 2 4 6 8 x
2
4
( 12 , 0)
6
10
12
186
Appendix to problem
to problem sets
10.6 Find the intersection point of each pair of lines in the previous question.
(a) L1 : y = 3x + 1 and L3 : y = 13 x + 1 intersect where
1
3x + 1 = x + 1 9x + 3 = x + 3 8x = 0 x=0
3
Use, say, L1 to find the yvalue at the intersection: y = 3(0) + 1 = 1. Hence the intersection point is (0, 1).
(Check the graph to see that this must be the correct answer.)
(b) L2 : y = 3x + 7 and L6 : y = 2x + 1 intersect where
3x + 7 = 2x + 1 x = 6
Use, say, L6 to find the yvalue at the intersection:
y = 2(6) = 12 + 1 = 11
so the intersection point is (6, 11). (Check the graph to see that this answer is in the right area.)
(c) L4 : y = 13 x 7
3 and L5 : y = 1 intersect where
1 7
x =1 x 7 = 3 x = 10 x = 10
3 3
Since L5 is horizontal the intersection point is (10, 1). (Check the graph to see that this answer is in the
right area.)
10.7 Application: this is an example of a problem in the area of Operations Research (or Management Systems as
it is called in Business). Consider the lines
L1 : x + y = 4 L2 : x + 2y = 6 L3 : 2x + y = 7
(a) Sketch the graphs of all three functions on the same set of axes.
Re-arrange the equations into y = mx + c form:
1
L1 : y = x + 4 L2 : y = x + 3 L3 : y = 2x + 7
2
See below for the graphs.
(b) Find the points of intersection between:
(i) L1 and L2
1
x + 4 = x + 3 2x + 8 = x + 6 2=x
2
y = (2) + 4 = 2.
Hence, the intersection point is (2, 2).

(ii) L1 and L3
x + 4 = 2x + 7 x=3
y = (3) + 4 = 1.
Hence, the intersection point is (3, 1).

(iii) L2 and the yaxis This is just the yintercept of L2 : (0, 3).
187
188 mathematics
(iv) L3 and the x axis This is the x intercept of L3 :
7
0 = 2x + 7 2x = 7 x=
2
7
Hence, the intersection point is 2, 0 .
y
L3 8
L1 5
4
L2 (2, 2)
(0,3) 3
(3, 1)
2
1 1 2 3 4 5 x
1
7
2, 0
(c) At each of the intersection points in (b) find the value of z = 3x + 5y

(i) (2, 2): z = 3(2) + 5(2) = 16.
(ii) (3, 1): z = 3(3) + 5(1) = 14
(iii) (0, 3): z = 3(0) + 5(3) = 15

(iv) 72 , 0 : z = 3 72 + 5(0) = 21 1
2 or 10 2
(d) At which intersection point does z = 3x + 5y attain its greatest value?
At (2, 2) (greatest value 16).
Problem Set 11
11.1 Sketch the graphs of the following quadratic functions, in each case
indicating
the convexity (concave up or the vertex co-ordinates and

down),
the yintercept, whatever x intercepts exist.
(a) y = 2x2 + 8x + 6 trying to factorise first is an

concave up excellent shortcut.)
the vertex occurs halfway
the yintercept is at (0, 6)
between the x intercepts, at
x = 2 (or use
2x2 + 8x + 6 = 0 b (4)
x= = = 2)
2a 2(1)
x2 + 4x + 3 = 0
The yvalue at x = 2 is
( x + 1)( x + 3) = 0
then
So, the x intercepts are y = 2 (2)2 8 (2) + 6
(1, 0) and (3, 0). (You
= 8 16 + 6
can also use the quadratic
formula to find the roots but = 2
188
Appendix to problem
to problem sets
The vertex is (2, 2). (b) y = x2 2x 8

y concave up
7 the yintercept is at
6 (0, 8)
5 to find the x intercepts:
4
x2 2x 8 = 0
3
2
( x + 2)( x 4) = 0
1
So, the x intercepts are
4 3 2 1 1 2 3 4 x (2, 0) and (4, 0).
1 the vertex occurs halfway
2 between the x intercepts, at
(2, 2)
3 x = 1 (or use
b (2)
x= = = 1)
2a 2(1)
then
y = (1)2 2 (1) 8
= 128
= 9

y
32
11 1 2 3 4 5 x
2
3
4
5
6
7
8
9
10(1, 9)
189
190 mathematics
(c) y = x2 4x + 4 (d) y = x2 + 4x 3
concave up concave down
the yintercept is at (0, 4) the yintercept is at
to find the x intercepts: (0, 3)
x2 4x + 4 = 0
( x 2)( x 2) = 0 x2 + 4x 3 = 0
x2 4x + 3 = 0
So, the x intercept is (2, 0).
( x 1)( x 3) = 0
in this case the vertex oc-
curs halfway is also at the So, the x intercepts are
x intercept (or use (1, 0) and (3, 0).
b (4) the vertex occurs halfway
x= = = 2) between the x intercepts, at
2a 2(1)
x = 2 (or use
clearly 0 so the vertex is
b (4)
x= = = 2)
2a 2(1)
(2, 0).
Note: this is a feature of per- The yvalue at x = 2 is
fect squares. then
y
y = (2)2 + 4 (2) 3
9
= 4 + 8 3
8
7
=1
6 The vertex is (2, 1).
5 y
4 5
3 4
2 3
1 2
(2, 1)
x 1
2 1 1 2 3 4 5 6
1
2 1 1 2 3 4 5 6 x
1
2
3
4
5
190
Appendix to problem
to problem sets
(e) y = x2 4x + 12 (f) y = x2 6x
concave up concave up
the yintercept is at (0, 12) in this case c = 0 so the
the x intercepts look diffi- yintercept is at (0, 0).
cult so use the discriminant: we already know that one of
the x intercepts is (0, 0).
b2 4ac = (4)2 4(1)(12) Factorizing confirms this
= 16 48 and locates the other one:
<0
x2 6x = 0
So, there are no x ( x 6) = 0
x intercepts.
So, the other x intercept is
the x coordinate of the
vertex is =
(6, 0).
the vertex occurs halfway
b (4)
x= = =2 between the x intercepts, at
2a 2(1)
x = 3 (or use
b (6)
then x= = = 3)
2a 2(1)
y = (2)2 4 (2) + 12 The yvalue at x = 3 is
= 4 8 + 12 then
=8
y = (3)2 6 (3)
The vertex is (2, 8). = 9 18
y = 9
18
16
y
14
1
12
10 1 1 2 3 4 5 6 7 x
1
8
(2, 8) 2
6
3
4
4
2
5
8 6 4 2 2 4 6 8 x 6
2
7
8
9
(3, 9)
191
192 mathematics
(g) y = x2 4 (h) y = x2 7x + 10
concave up concave up
the yintercept is at the yintercept is at (0, 10)
(0, 4) to find the x intercepts:
x2 7x + 10 = 0
x2 4 = 0
( x 2)( x 5) = 0
( x 2)( x + 2) = 0
So, the x intercepts are So, the x intercepts are

(2, 0) and (2, 0). (2, 0) and (5, 0).
the vertex occurs halfway the vertex occurs halfway
between the x intercepts, at between the x intercepts, at
x = 0 (or use x = 72 (or use
b (0) b (7) 7
x= = = 0) x= = = )
2a 2(1) 2a 2(1) 2
then
then
y = (0)2 4 2
7 7
= 4 y= 7 + 10
2 2
(or realise that it is also the 49 49
= + 10
yintercept!). The vertex is 4 2
9
(0, 4). =
4
Note: this is a feature of the
difference of two squares. The vertex is 72 , 94 .
y y
4 12
3 10
2 8
1 6
4
432
11 1 2 3 4 x 2
2
3 864
22 2 4 6 8 x
4 4 7 9
5 6( 2 , 4 )
6 8
192
11.2 Complete the square for the following quadratic functions and use the result to verify the co-ordinates of the
vertex of the parabola they repesent.
( a) y = x2 + 8x + 3 (b) y = x2 14x
2 2 2 2
8 8 14 14
= x2 + 8x + +3 = x2 14x +
2 2 2 2
= x2 + 8x + 16 16 + 3 = x2 14x + 49 49
= ( x + 4)2 16 + 3 = ( x 7)2 49
y = ( x + 4)2 13
( x 7)2 cant be negative so the smallest value of y
(x + 4)2cant be negative so the smallest value of y occurs when
occurs when ( x 7)2 = 0
( x + 4)2 = 0
ie. x = 7. The yvalue at x = 7 is then 49, so
ie. x = 4. The yvalue at x = 4 is then 13,
the coordinates of the vertex are (7, 49).
so the coordinates of the vertex are (4, 13).
The x value of the vertex is also given by
The x value of the vertex is also given by
b (14)
b (8) x= = =7
x= = = 4 2a 2(1)
2a 2(1)
At x = 4: At x = 7:
y = (4)2 + 8(4) + 3 y = (7)2 14(7)

= 16 32 + 3 = 49 98
= 13 = 49
The answers are the same. The answers are the same.
Problem Set 12
12.1 Evaluate the following limits if they exist.
x2 16
( x 4
)( x + 4)
( a) lim = lim = lim ( x + 4) = 4 + 4 = 8
x 4 x4 x 4 x 4 x 4
x2 + 2x 8
( x 2
)( x + 4)
(b) lim = lim = lim ( x + 4) = 2 + 4 = 6
x 2 x2 x 2 x 2 x 2
x2 x 12 ( x 4)( x + 3) (4 4)(4 + 3) 07 0
(c) lim = lim = = = = 0
x 4 x1 x 4 x1 41 3 3
x2 6x + 9
( x 3
)( x 3)
(d) lim = lim = lim ( x 3) = 3 3 = 0
x 3 x3 x 3 x 3 x 3
4x2 16 4( x 2 4) 4(
x2
)( x + 2)
(e) lim = lim = lim = lim (4( x + 2)) = 4(2 + 2) = 4 4 = 16
x 2 x 2 x 2 x2 x 2
x 2 x 2
193
194 mathematics
x2 + 15x + 56 +
( x 7
)( x + 8)
(f) lim = lim = lim ( x + 8) = 7 + 8 = 1
x 7 x+7 x 7 x+ 7 x 7
12.2 Differentiate each of the following functions.
dy
( a) y = x3 + x2 + x + 6 = 3x2 + 2x + 1
dx
dy
(b) y = 3x4 x3 + x2 + x 2 = 12x3 3x2 + 2x + 1
dx
1 dy 1 1
(c) y = x2 + x+ 3 = 2x + 2
x dx 2 x x
12.3 To differentiate

1 1
y= x+ x2 +
x x2
we need to expand the brackets. We get

2
x x2 1 1 1 5 3
y = x x+ 2 + + 3 = x 2 x 2 + x 2 x 2 + x + x 3 = x 2 + x 2 + x + x 3
x x x
d
and because n
dx ( x ) = nx n1 we get
dy 5 3 3 5
= x 2 x 2 + 1 3x 4
dx 2 2
12.4 Perform the following integrations

x3 x2
( a) x2 + x + 6 dx = + + 6x + C
3 2
3x5 x4
(b) 3x4 x3 2 dx = 2x + C
5 4
3
4 x3 2x 2 4
(c) x2 + x + 2 dx = + +C
x 3 3 x
12.5 Evaluate the following definite integrals.

1 4 1 4 4
x x3 1 13 0 03 1 1 7
( a) x3 + x2 dx = + = + + = + =
0 4 3 0 4 3 4 3 4 3 12
3 3
5x2 5(32 ) 5(22 )
(b) 5x4 + 5x 2 dx = x5 + 2x = 35 + 2(3) 25 + 2(2)
2 2 2 2 2

45 20 45 20 25 418 + 25 443
= 243 + 6 32 + 4 = 237 + 28 = 209 + = =
2 2 2 2 2 2 2
2 2
x3 23 (2)3
(c) x2 + 4x 3 dx = + 2x2 3x = + 2(22 ) 3(2) + 2((2)2 ) 3(2)
2 3 2 3 3

8 8 8 8 16 16 36 20
= + 8 6 + 8 + 6 = + 2 + 14 = 12 = =
3 3 3 3 3 3 3
194
12.6 Find the AREA bounded by the graph of f(x) and the x-axis between
the points indicated for the functions below. The graphs are shown
below. Hint: Sketch the graph first.
y
2
2
( a) f ( x ) = x + 1, x = 1 to x = 1
1
The region is above the x-axis so
1 3 1 x
x 1 1
Area = x2 + 1 dx = +x
1 3 1

1 1
= +1 1
3 3
1 1 2 8
= +1+ +1 = +2 =
3 3 3 3
y
(b) f ( x ) = x2 x, x = 0 to x = 2 2
Some of the region is above the x-axis and

1
some below so
1 2
Area = x2 x dx + x2 x dx x
0 1 1 1 2
1 3 2
x3 x2 x x2
= +
3 2 0 3 2 1

1 1 8 4 1 1
= +0+
3 2 3 2 3 2
1 1 8 1 1 6
= + + 2 + = 1 = 1
3 2 3 3 2 3
195
MATHEMATICS FUNDAMENTALS Index
Index
x-intercept, 81 exponential models, 59 negative powers, 41

y-intercept, 78 number line, 7
factorization, 10, 28 numerator, 7
algebra, 15 fractions, 7
algebraic fractions, 32 function, 75 parabola, 89
antiderivative, 101 parallel lines, 83
area under a curve, 102 gradient, 78 perfect squares, 30
associated quadratic, 31 graph, 77 perpendicular lines, 83
proper fraction, 7, 8
BIDMAS, 6 horizontal axis, 76
horizontal lines, 80 quadratic, 28
cancellation, 10 quadratic equation, 67
chord, 96 improper fraction, 8 quadratic expressions, 67
co-ordinates, 77 indefinite integral, 101 quadratic formula, 69
coefficients, 28 integers, 5 quadratic functions, 89
completing the square, 72 integral, 101
convexity, 90 integrand, 101 rate of change, 95
intersecting lines, 83 rational expressions, 32
decimals, 11 RHS, 17
definite integral, 101 laws of Indices, 41
denominator, 7 laws of logarithms, 53 sigfig, 12
derivative, 98 LHS, 17 simultaneous equations, 23
difference of two squares, 30 limit, 97 sketch, 81
differentiation, 98 linear functions, 78 standard form, 67
discriminant, 70 logarithm, 51 straight line, 78
distributive law, 17 logarithmic equation, 51
tangent, 97
e, 61 mathematical modelling, 19
elimination method, 24 mixed numeral, 8 vertex, 91
exponent, 51 vertical axis, 76
exponential decay, 60 natural logarithm, 61 vertical lines, 81
exponential equation, 51 natural numbers, 5
exponential growth, 59 negative numbers, 5 word problems, 7
197

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DR RAMZAN RAZIM KHAN & DR DES HILL

7 Exponentials and logarithms 57

8 Equations involving logarithms and/or exponentials 65

10 Functions and graphs 87

11 Quadratic functions and Parabolae 103

Appendix 1: Answers to exercises 123

Appendix 2: Solutions to problem sets 141

1. Numbers and operations

We also have the very special number zero

and the negative numbers

which are used to indicate deficits (that is, debts) or to indicate

The basic operations of arithmetic

Hopefully, the everyday operations +, , and are familiar:

Arithmetic involving negative numbers

When negative numbers are involved we need to be careful, and

(i) 13 + (8) = 5 (iv) 16 13 = 29

(ii) 21 (3) = 24 (v) (12) 4 = 3

BIDMAS. The order of operations

Consider the simple question:

B: Expressions within Brackets (. . .)

Division and multiplication have the same precedence.

Addition and subtraction have the same precedence.

Operations with the same precedence are performed left to right.

Some BIDMAS examples

We are sometimes asked quesions in written or spoken conversa-

The number line

The integers sit nicely on a number line:

Fractions are ratios of integers, that is,

We call 4 25 a mixed numeral, as it is the sum of an integer and a

Converting mixed numerals to improper fractions

An improper fraction is one in which the numerator is larger than

or three lots of two halves (making six) plus one half is

MATHEMATICS FUNDAMENTALS Numbers and operations

To multiply two fractions simply multiply their numerators and

The notation 2 3 means the same as 23 , that is two-thirds. Now

This is the process of expressing a given number as a product of

The process of simplifying fractions by looking for factors which

Adding and subtracting fractions

Consider the sum

Decimals are a convenient and useful way of writing fractions with

Common fractions expressed as decimals

Fixing the number of decimal places

1 1 Note that we have rounded up the

Multiplying and dividing decimal numbers by 10

0.47 10 = 4.7 0.06 10 = 0.6

Deloitte & Touche LLP and affiliated entities.

Deloitte & Touche LLP and affiliated entities.

Discover the truth at www.deloitte.ca/careers

1.1 Evaluate the following

(i) 3(4 + 2 6) + 12 4 (ii) 6 + 2[25 3(2 + 5)]

1.3 Convert to improper fractions

1.7 Reduce to simplest form by using cancellation:

1.1 Evaluate the following.

1.2 Compute the following, leaving your answer in the simplest

1.3 What is half of the difference between a third of 60 and

We will turn your CV into

Algebra is a powerful tool for expressing fundamental relationships

In many situations we are told the numerical values of all variables

To answer questions like this we need to know how to re-arrange